Introduction to Electric Circuits 10th Edition [10 ed.] 019903141X, 9780199031412

Table of contents :
Cover
Introduction to Electric Circuits
Contents
From the Publisher
From the Preface to the First Edition (1959)
From the Authors of the Tenth Edition
PART I: The Basic Electric Circuit
1: Introduction
Key Terms
Learning Outcomes
1-1 Circuit Diagrams
1-2 The International System of Units
1-3 Calculators for Circuit Theory
1-4 Numerical Accuracy
1-5 Scientific Notation
1-6 SI Unit Prefixes
1-7 Conversion of Units
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
2: Current and Voltage
Key Terms
Learning Outcomes
2-1 The Nature of Charge
2-2 Free Electrons in Metals
2-3 Electric Current
2-4 The Coulomb
2-5 The Ampere
2-6 Potential Difference
2-7 The Volt
2-8 EMF, Potential Difference, and Voltage
2-9 Conventional Current and Electron Flow
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
3: Conductors, Insulators, and Semiconductors
Key Terms
Learning Outcomes
3-1 Conductors
3-2 Electrolytic Conduction
3-3 Insulators
3-4 Insulator Breakdown
3-5 Semiconductors
Summary
Review Questions
Integrate the Concepts
Practice Quiz
4: Cells, Batteries, and Other Voltage Sources
Key Terms
Learning Outcomes
4-1 Basic Terminology
4-2 Simple Primary Cell
4-3 Carbon-Zinc and Alkaline Cells
4-4 Other Commercial Primary Cells
4-5 Secondary Cells
4-6 Capacity of Cells and Batteries
4-7 Fuel Cells
4-8 Other Voltage Sources
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
5: Resistance and Ohm’s Law
Key Terms
Learning Outcomes
5-1 Ohm’s Law
5-2 The Nature of Resistance
5-3 Factors Governing Resistance
5-4 Resistivity
5-5 Circular Mils
5-6 American Wire Gauge
5-7 Effect of Temperature on Resistance
5-8 Temperature Coefficient of Resistance
5-9 Linear Resistors
5-10 Nonlinear Resistors
5-11 Resistor Colour Code
5-12 Variable Resistors
5-13 Voltage-Current Characteristics
5-14 Applying Ohm’s Law
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
6: Work and Power
Key Terms
Learning Outcomes
6-1 Energy and Work
6-2 Power
6-3 Efficiency
6-4 The Kilowatt Hour
6-5 Relationships Among Basic Electric Units
6-6 Heating Effect of Current
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
PART II: Resistance Networks
7: Series and Parallel Circuits
Key Terms
Learning Outcomes
7-1 Resistors in Series
7-2 Voltage Drops in Series Circuits
7-3 Double-Subscript Notation
7-4 Kirchhoff’s Voltage Law
7-5 Characteristics of Series Circuits
7-6 Internal Resistance
7-7 Cells in Series
7-8 Maximum Power Transfer
7-9 Resistors in Parallel
7-10 Kirchhoff’s Current Law
7-11 Conductance and Conductivity
7-12 Characteristics of Parallel Circuits
7-13 Cells in Parallel
7-14 Troubleshooting
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
8: Series-Parallel Circuits
Key Terms
Learning Outcomes
8-1 Series-Parallel Resistors
8-2 Equivalent-Circuit Method
8-3 Kirchhoff’s Laws Method
8-4 Voltage-Divider Principle
8-5 Voltage Dividers
8-6 Current-Divider Principle
8-7 Cells in Series-Parallel
8-8 Troubleshooting
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
9: Resistance Networks
Key Terms
Learning Outcomes
9-1 Network Equations from Kirchhoff’s Laws
9-2 Constant-Voltage Sources
9-3 Constant-Current Sources
9-4 Source Conversion
9-5 Kirchhoff’s Voltage-Law Equations: Loop Procedure
9-6 Networks with More Than One Voltage Source
9-7 Loop Equations in Multisource Networks
9-8 Mesh Analysis
9-9 Kirchhoff’s Current-Law Equations
9-10 Nodal Analysis
9-11 The Superposition Theorem
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
10: Equivalent-Circuit Theorems
Key Terms
Learning Outcomes
10-1 Thévenin’s Theorem
10-2 Norton’s Theorem
10-3 Dependent Sources
10-4 Delta-Wye Transformation
10-5 Troubleshooting
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
11: Electrical Measurement
Key Terms
Learning Outcomes
11-1 Moving-Coil Meters
11-2 The Ammeter
11-3 The Voltmeter
11-4 Voltmeter Loading Effect
11-5 Resistance Measurement
11-6 The Electrodynamometer Movement
11-7 Multimeters
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
PART III: Capacitance and Inductance
12: Capacitance
Key Terms
Learning Outcomes
12-1 Electric Fields
12-2 Dielectrics
12-3 Capacitance
12-4 Capacitors
12-5 Factors Governing Capacitance
12-6 Dielectric Constant
12-7 Capacitors in Parallel
12-8 Capacitors in Series
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
13: Capacitance in dc Circuits
Key Terms
Learning Outcomes
13-1 Charging a Capacitor
13-2 Rate of Change of Voltage
13-3 Time Constant
13-4 Graphical Solution for Capacitor Voltage
13-5 Discharging a Capacitor
13-6 Algebraic Solution for Capacitor Voltage
13-7 Transient Response
13-8 Energy Stored by a Capacitor
13-9 Characteristics of Capacitive DC Circuits
13-10 Troubleshooting
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
14: Magnetism
Key Terms
Learning Outcomes
14-1 Magnetic Fields
14-2 Magnetic Field around a Current-Carrying Conductor
14-3 Magnetic Flux
14-4 Magnetomotive Force
14-5 Reluctance
14-6 Permeance and Permeability
14-7 Magnetic Flux Density
14-8 Magnetic Field Strength
14-9 Diamagnetic, Paramagnetic, and Ferromagnetic Materials
14-10 Permanent Magnets
14-11 Magnetization Curves
14-12 Permeability from the BH Curve
14-13 Hysteresis
14-14 Eddy Current
14-15 Magnetic Shielding
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
15: Magnetic Circuits
Key Terms
Learning Outcomes
15-1 Practical Magnetic Circuits
15-2 Long Air-Core Coils
15-3 Toroidal Coils
15-4 Linear Magnetic Circuits
15-5 Nonlinear Magnetic Circuits
15-6 Leakage Flux
15-7 Series Magnetic Circuits
15-8 Air Gaps
15-9 Parallel Magnetic Circuits
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
16: Inductance
Key Terms
Learning Outcomes
16-1 Electromagnetic Induction
16-2 Faraday’s Law
16-3 Lenz’s Law
16-4 Self-Induction
16-5 Self-Inductance
16-6 Factors Governing Inductance
16-7 Inductors in Series
16-8 Inductors in Parallel
16-9 The DC Generator
16-10 Simple DC Generator
16-11 EMF Equation
16-12 The DC Motor
16-13 Speed and Torque of a DC Motor
16-14 Types of DC Motors
16-15 Speed Characteristics of DC Motors
16-16 Torque Characteristics of DC Motors
16-17 Permanent Magnet and Brushless DC Motors
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
17: Inductance in dc Circuits
Key Terms
Learning Outcomes
17-1 Current in an Ideal Inductor
17-2 Rise of Current in a Practical Inductor
17-3 Time Constant
17-4 Graphical Solution for Inductor Current
17-5 Algebraic Solution for Inductor Current
17-6 Energy Stored by an Inductor
17-7 Fall of Current in an Inductive Circuit
17-8 Algebraic Solution for Discharge Current
17-9 Transient Response
17-10 Characteristics of Inductive dc Circuits
17-11 Troubleshooting
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
PART IV: Alternating Current
18: Alternating Current
Key Terms
Learning Outcomes
18-1 A Simple Generator
18-2 The Nature of the Induced Voltage
18-3 The Sine Wave
18-4 Peak Value of a Sine Wave
18-5 Instantaneous Value of a Sine Wave
18-6 The Radian
18-7 Instantaneous Current in a Resistor
18-8 Instantaneous Power in a Resistor
18-9 Periodic Waves
18-10 Average Value of a Periodic Wave
18-11 rms Value of a Sine Wave
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
19: Reactance
Key Terms
Learning Outcomes
19-1 Instantaneous Current in an Ideal Inductor
19-2 Inductive Reactance
19-3 Factors Governing Inductive Reactance
19-4 Instantaneous Current in a Capacitor
19-5 Capacitive Reactance
19-6 Factors Governing Capacitive Reactance
19-7 Resistance, Inductive Reactance, and Capacitive Reactance
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
20: Phasors
Key Terms
Learning Outcomes
20-1 Addition of Sine Waves
20-2 Addition of Instantaneous Values
20-3 Representing a Sine Wave by a Phasor Diagram
20-4 Letter Symbols for Phasor Quantities
20-5 Phasor Addition by Geometrical Construction
20-6 Addition of Perpendicular Phasors
20-7 Expressing Phasors with Complex Numbers
20-8 Phasor Addition Using Rectangular Coordinates
20-9 Subtraction of Phasor Quantities
20-10 Multiplication and Division of Phasor Quantities
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
21: Impedance
Key Terms
Learning Outcomes
21-1 Resistance and Inductance in Series
21-2 Impedance
21-3 Practical Inductors
21-4 Resistance and Capacitance in Series
21-5 Resistance, Inductance, and Capacitance in Series
21-6 Resistance, Inductance, and Capacitance in Parallel
21-7 Conductance, Susceptance, and Admittance
21-8 Impedance and Admittance
21-9 Troubleshooting
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
22: Power in Alternating-Current Circuits
Key Terms
Learning Outcomes
22-1 Power in a Resistor
22-2 Power in an Ideal Inductor
22-3 Power in a Capacitor
22-4 Power in a Circuit Containing Resistance and Reactance
22-5 The Power Triangle
22-6 Power Factor
22-7 Power-Factor Correction
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
PART V: Impedance Networks
23: Series and Parallel Impedances
Key Terms
Learning Outcomes
23-1 Resistance and Impedance
23-2 Impedances in Series
23-3 Impedances in Parallel
23-4 Series-Parallel Impedances
23-5 Source Conversion
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
24: Impedance Networks
Key Terms
Learning Outcomes
24-1 Loop Equations
24-2 Mesh Equations
24-3 Superposition Theorem
24-4 Thévenin’s Theorem
24-5 Norton’s Theorem
24-6 Nodal Analysis
24-7 Delta-Wye Transformation
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
25: Resonance
Key Terms
Learning Outcomes
25-1 Effect of Varying Frequency in a Series rlc Circuit
25-2 Series Resonance
25-3 Quality Factor
25-4 Resonant Rise of Voltage
25-5 Selectivity
25-6 Ideal Parallel-Resonant Circuits
25-7 Practical Parallel-Resonant Circuits
25-8 Selectivity of Parallel-Resonant Circuits
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
26: Passive Filters
Key Terms
Learning Outcomes
26-1 Filters
26-2 Frequency Response Graphs
26-3 RC Low-Pass Filters
26-4 RL Low-Pass Filters
26-5 RC High-Pass Filters
26-6 RL High-Pass Filters
26-7 Band-Pass Filters
26-8 Band-Stop Filters
26-9 Practical Application of Filters
26-10 Troubleshooting
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
27: Transformers
Key Terms
Learning Outcomes
27-1 Transformer Action
27-2 Transformation Ratio
27-3 Impedance Transformation
27-4 Leakage Reactance
27-5 Open-Circuit and Short-Circuit Tests
27-6 Transformer Efficiency
27-7 Effect of Loading a Transformer
27-8 Autotransformers
27-9 Troubleshooting
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
28: Coupled Circuits
Key Terms
Learning Outcomes
28-1 Determining Coupling Network Parameters
28-2 Open-Circuit Impedance Parameters
28-3 Short-Circuit Admittance Parameters
28-4 Hybrid Parameters
28-5 Air-Core Transformers
28-6 Mutual Inductance
28-7 Coupled Impedance
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
29: Three-Phase Systems
Key Terms
Learning Outcomes
29-1 Advantages of Polyphase Systems
29-2 Generation of Three-Phase Voltages
29-3 Double-Subscript Notation
29-4 Four-Wire Wye-Connected System
29-5 Delta-Connected Systems
29-6 Wye-Delta System
29-7 Power in a Balanced Three-Phase System
29-8 Phase Sequence
29-9 Unbalanced Three-Wire Wye Loads
29-10 Power in an Unbalanced Three-Phase System
29-11 The AC Generator
29-12 Three-Phase Induction Motor
29-13 Three-Phase Synchronous Motor
29-14 Single-Phase Motors
29-15 The 30° Difference between Delta-Wye Configurations
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
30: Harmonics
Key Terms
Learning Outcomes
30-1 Nonsinusoidal Waves
30-2 Fourier Series
30-3 Addition of Harmonically Related Sine Waves
30-4 Generation of Harmonics
30-5 Harmonics in an Amplifier
30-6 Harmonics in an Iron-Core Transformer
30-7 rms Value of a Nonsinusoidal Wave
30-8 Square Waves and Sawtooth Waves
30-9 Nonsinusoidal Waves in Linear Impedance Networks
Summary
Problems
Review Questions
Integrate the Concepts
Practice Quiz
Appendices
1 Determinants
2 Calculus Derivations
2-1 Maxium Power-Transfer Theorem
2-2 Instantaneous Voltage in a cr Circuit
2-3 Energy Stored by a Capacitor
2-4 Instantaneous Current in an LR Circuit
2-5 Energy Stored by an Inductor
2-6 rms and Average Values of a Sine Wave
2-7 Inductive Reactance
2-8 Capacitive Reactance
2-9 General Transformer Equation
2-10 Maximum Transformer Efficiency
3 Multisim Schematic Capture and Simulation
Answers to Selected Problems
Glossary
Index

Citation preview

Introduction to

Electric Circuits

Tenth Edition

Introduction to

Electric Circuits Herbert W. Jackson Dale Temple Brian Kelly Karen Craigs Lauren Fuentes

Oxford University Press is a department of the University of Oxford. It furthers the University’s objective of excellence in research, scholarship, and education by publishing worldwide. Oxford is a registered trade mark of Oxford University Press in the UK and in certain other countries. Published in Canada by Oxford University Press 8 Sampson Mews, Suite 204, Don Mills, Ontario M3C 0H5 Canada www.oupcanada.com Copyright © Oxford University Press Canada 2019 The moral rights of the author have been asserted Database right Oxford University Press (maker) Eighth Edition published in 2008 Ninth Edition published in 2012 Updated Ninth Edition published in 2015 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by licence, or under terms agreed with the appropriate reprographics rights organization. Enquiries concerning reproduction outside the scope of the above should be sent to the Permissions Department at the address above or through the following url: www.oupcanada.com/permission/permission_request.php Every effort has been made to determine and contact copyright holders. In the case of any omissions, the publisher will be pleased to make suitable acknowledgement in future editions. Library and Archives Canada Cataloguing in Publication Title: Introduction to electric circuits / Herbert W. Jackson, Dale Temple, Brian Kelly, Karen Craigs, and Lauren Fuentes. Names: Jackson, Herbert W., author. | Temple, Dale, author. | Kelly, Brian, 1948– author. | Craigs, Karen, author. | Fuentes, Lauren, author. Description: Tenth edition. | Includes index. Identifiers: Canadiana (print) 20189066806 | Canadiana (ebook) 20190062096 | ISBN 9780199034703 (loose-leaf) | ISBN 9780199031412 (hardcover) | ISBN 9780199031474 (EPUB) Subjects: LCSH: Electric circuits—Textbooks. | LCGFT: Textbooks. Classification: LCC TK454 .J28 2019 | DDC 621.319/2—dc23 Cover image: TEK IMAGE/SCIENCE PHOTO LIBRARY/Getty Images Image credits for contents and chapter openers: © iStock.com/scorpion26; © iStock.com/konradlew; © iStock.com/Yukosourov; © iStock.com/imagestock, © iStock.com/Chepko; © iStock.com/oonal; © iStock.com/AlessandroZocc; iStock.com/vladm; Cover design: Laurie McGregor Interior design: Laurie McGregor Oxford University Press is committed to our environment. Wherever possible, our books are printed on paper which comes from responsible sources. Printed and bound in the United States of America 1 2 3 4 — 22 21 20 19

Contents From the Publisher xix

From the Preface to the First Edition (1959) xxvii From the Authors of the Tenth Edition xxix

PART I The Basic Electric Circuit

1 Introduction 1-1 1-2 1-3 1-4 1-5 1-6 1-7

2 2-1 2-2 2-3 2-4 2-5 2-6 2-7 2-8 2-9

Key Terms  3 Learning Outcomes  3 Circuit Diagrams  4 The International System of Units   4 Calculators for Circuit Theory  6 Numerical Accuracy  7 Scientific Notation  7 SI Unit Prefixes  10 Conversion of Units  12 Summary  15 Problems  15 Review Questions  17 Integrate the Concepts   18 Practice Quiz  18

Current and Voltage Key Terms  21 Learning Outcomes  21 The Nature of Charge  22 Free Electrons in Metals  23 Electric Current  24 The Coulomb  26 The Ampere  26 Potential Difference  28 The Volt  31 EMF, Potential Difference, and Voltage  32 Conventional Current and Electron Flow  33 Summary  35 Problems  35 Review Questions  37 Integrate the Concepts   38 Practice Quiz  38

vi

Contents

3

Conductors, Insulators, and Semiconductors

4

Cells, Batteries, and Other Voltage Sources

Key Terms  41 Learning Outcomes  41 3-1 Conductors  42 3-2 Electrolytic Conduction  43 3-3 Insulators  45 3-4 Insulator Breakdown  46 3-5 Semiconductors  47 Summary  48 Review Questions  48 Integrate the Concepts   49 Practice Quiz  49

4-1 4-2 4-3 4-4 4-5 4-6 4-7 4-8

5

Key Terms  51 Learning Outcomes  51 Basic Terminology  52 Simple Primary Cell  52 Carbon-Zinc and Alkaline Cells  53 Other Commercial Primary Cells  55 Secondary Cells  56 Capacity of Cells and Batteries  58 Fuel Cells  60 Other Voltage Sources  60 Summary  65 Problems  65 Review Questions  65 Integrate the Concepts   66 Practice Quiz  66

Resistance and Ohm’s Law

Key Terms  69 Learning Outcomes  69 5-1 Ohm’s Law  70 5-2 The Nature of Resistance  71 5-3 Factors Governing Resistance  72 5-4 Resistivity  73 5-5 Circular Mils  75 5-6 American Wire Gauge  77 5-7 Effect of Temperature on Resistance  79 5-8 Temperature Coefficient of Resistance  82 5-9 Linear Resistors  84

Contents

5-10 5-11 5-12 5-13 5-14

6

Nonlinear Resistors  86 Resistor Colour Code  88 Variable Resistors  91 Voltage-Current Characteristics  91 Applying Ohm’s Law  92 Summary  94 Problems  94 Review Questions  98 Integrate the Concepts   101 Practice Quiz  101

Work and Power

Key Terms  105 Learning Outcomes  105 6-1 Energy and Work  106 6-2 Power  107 6-3 Efficiency  110 6-4 The Kilowatt Hour  112 6-5 Relationships Among Basic Electric Units  113 6-6 Heating Effect of Current  113 Summary  116 Problems  116 Review Questions  120 Integrate the Concepts   121 Practice Quiz  121

PART II Resistance Networks

7 7-1 7-2 7-3 7-4 7-5 7-6 7-7 7-8 7-9 7-10 7-11

Series and Parallel Circuits Key Terms  125 Learning Outcomes  125 Resistors in Series  126 Voltage Drops in Series Circuits  128 Double-Subscript Notation  130 Kirchhoff’s Voltage Law  130 Characteristics of Series Circuits  131 Internal Resistance  133 Cells in Series  136 Maximum Power Transfer  137 Resistors in Parallel  139 Kirchhoff’s Current Law  141 Conductance and Conductivity  142

vii

viii

Contents

7-12 Characteristics of Parallel Circuits  145 7-13 Cells in Parallel  148 7-14 Troubleshooting  150 Summary  154 Problems  154 Review Questions  159 Integrate the Concepts   161 Practice Quiz  161

8

Series-Parallel Circuits

9

Resistance Networks

Key Terms  165 Learning Outcomes  165 8-1 Series-Parallel Resistors  166 8-2 Equivalent-Circuit Method  167 8-3 Kirchhoff’s Laws Method  171 8-4 Voltage-Divider Principle  173 8-5 Voltage Dividers  175 8-6 Current-Divider Principle  181 8-7 Cells in Series-Parallel  184 8-8 Troubleshooting  186 Summary  188 Problems  188 Review Questions  196 Integrate the Concepts   197 Practice Quiz  198

9-1 9-2 9-3 9-4 9-5 9-6 9-7 9-8 9-9 9-10 9-11

Key Terms  201 Learning Outcomes  201 Network Equations from Kirchhoff’s Laws  202 Constant-Voltage Sources  202 Constant-Current Sources  204 Source Conversion  206 Kirchhoff’s Voltage-Law Equations: Loop Procedure  208 Networks with More Than One Voltage Source  214 Loop Equations in Multisource Networks  216 Mesh Analysis  222 Kirchhoff’s Current-Law Equations  228 Nodal Analysis  231 The Superposition Theorem  237 Summary  242 Problems  242 Review Questions  252 Integrate the Concepts   254 Practice Quiz  255

Contents

10 Equivalent-Circuit Theorems Key Terms  259 Learning Outcomes  259 10-1 Thévenin’s Theorem  260 10-2 Norton’s Theorem  268 10-3 Dependent Sources  271 10-4 Delta-Wye Transformation  278 10-5 Troubleshooting  283 Summary  284 Problems  284 Review Questions  291 Integrate the Concepts   292 Practice Quiz  292

11

Electrical Measurement

Key Terms  295 Learning Outcomes  295 11-1 Moving-Coil Meters  296 11-2 The Ammeter  297 11-3 The Voltmeter  300 11-4 Voltmeter Loading Effect  302 11-5 Resistance Measurement  304 11-6 The Electrodynamometer Movement  311 11-7 Multimeters  312 Summary  315 Problems  315 Review Questions  318 Integrate the Concepts   319 Practice Quiz  320

PART III C  apacitance and Inductance

12 Capacitance

Key Terms  323 Learning Outcomes  323 12-1 Electric Fields  324 12-2 Dielectrics  327 12-3 Capacitance  328 12-4 Capacitors  330 12-5 Factors Governing Capacitance  333 12-6 Dielectric Constant  336 12-7 Capacitors in Parallel  338

ix

x

Contents

12-8

13

Capacitors in Series  338 Summary  342 Problems  342 Review Questions  345 Integrate the Concepts   346 Practice Quiz  346

Capacitance in DC Circuits

Key Terms  349 Learning Outcomes  349 13-1 Charging a Capacitor  350 13-2 Rate of Change of Voltage  352 13-3 Time Constant  354 13-4 Graphical Solution for Capacitor Voltage  356 13-5 Discharging a Capacitor  357 13-6 Algebraic Solution for Capacitor Voltage  362 13-7 Transient Response  366 13-8 Energy Stored by a Capacitor  370 13-9 Characteristics of Capacitive DC Circuits  372 13-10 Troubleshooting  375 Summary  376 Problems  376 Review Questions  382 Integrate the Concepts   385 Practice Quiz  385

14 Magnetism

Key Terms  389 Learning Outcomes  389 14-1 Magnetic Fields  390 14-2 Magnetic Field around a Current-Carrying Conductor  393 14-3 Magnetic Flux  396 14-4 Magnetomotive Force  397 14-5 Reluctance  398 14-6 Permeance and Permeability  399 14-7 Magnetic Flux Density  400 14-8 Magnetic Field Strength  401 14-9 Diamagnetic, Paramagnetic, and Ferromagnetic Materials  402 14-10 Permanent Magnets  404 14-11 Magnetization Curves  404 14-12 Permeability from the BH Curve  408 14-13 Hysteresis  410 14-14 Eddy Current  412

Contents

14-15 Magnetic Shielding  413 Summary  414 Problems  414 Review Questions  416 Integrate the Concepts   418 Practice Quiz  418

15 15-1 15-2 15-3 15-4 15-5 15-6 15-7 15-8 15-9

Magnetic Circuits Key Terms  421 Learning Outcomes  421 Practical Magnetic Circuits  422 Long Air-Core Coils  422 Toroidal Coils  425 Linear Magnetic Circuits  425 Nonlinear Magnetic Circuits  426 Leakage Flux  429 Series Magnetic Circuits  430 Air Gaps  433 Parallel Magnetic Circuits  435 Summary  438 Problems  438 Review Questions  442 Integrate the Concepts   443 Practice Quiz  443

16 Inductance

Key Terms  447 Learning Outcomes  447 16-1 Electromagnetic Induction  448 16-2 Faraday’s Law  450 16-3 Lenz’s Law  451 16-4 Self-Induction  453 16-5 Self-Inductance  454 16-6 Factors Governing Inductance  455 16-7 Inductors in Series  458 16-8 Inductors in Parallel  458 16-9 The DC Generator   459 16-10 Simple DC Generator   461 16-11 EMF Equation  463 16-12 The DC Motor   465 16-13 Speed and Torque of a DC Motor   467 16-14 Types of DC Motors   469 16-15 Speed Characteristics of DC Motors   471

xi

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Contents

16-16 Torque Characteristics of DC Motors   474 16-17 Permanent Magnet and Brushless DC Motors   476 Summary  477 Problems  477 Review Questions  479 Integrate the Concepts   482 Practice Quiz  482

17

Inductance in DC Circuits

Key Terms  485 Learning Outcomes  485 17-1 Current in an Ideal Inductor  486 17-2 Rise of Current in a Practical Inductor  487 17-3 Time Constant  490 17-4 Graphical Solution for Inductor Current  491 17-5 Algebraic Solution for Inductor Current  495 17-6 Energy Stored by an Inductor  499 17-7 Fall of Current in an Inductive Circuit  501 17-8 Algebraic Solution for Discharge Current  506 17-9 Transient Response  507 17-10 Characteristics of Inductive DC Circuits  509 17-11 Troubleshooting  510 Summary  512 Problems  512 Review Questions  516 Integrate the Concepts   517 Practice Quiz  518

PART IV Alternating Current

18 Alternating Current 18-1 18-2 18-3 18-4 18-5 18-6 18-7 18-8 18-9

Key Terms  523 Learning Outcomes  523 A Simple Generator  524 The Nature of the Induced Voltage  524 The Sine Wave  526 Peak Value of a Sine Wave  529 Instantaneous Value of a Sine Wave  529 The Radian  532 Instantaneous Current in a Resistor  533 Instantaneous Power in a Resistor  536 Periodic Waves  537

Contents

18-10 Average Value of a Periodic Wave  540 18-11 RMS Value of a Sine Wave  540 Summary  544 Problems  544 Review Questions  546 Integrate the Concepts   548 Practice Quiz  548

19 Reactance

Key Terms  553 Learning Outcomes  553 19-1 Instantaneous Current in an Ideal Inductor  554 19-2 Inductive Reactance  555 19-3 Factors Governing Inductive Reactance  556 19-4 Instantaneous Current in a Capacitor  558 19-5 Capacitive Reactance  559 19-6 Factors Governing Capacitive Reactance  560 19-7 Resistance, Inductive Reactance, and Capacitive Reactance  562 Summary  564 Problems  564 Review Questions  565 Integrate the Concepts   566 Practice Quiz  567

20 Phasors

Key Terms  571 Learning Outcomes  571 20-1 Addition of Sine Waves  572 20-2 Addition of Instantaneous Values  573 20-3 Representing a Sine Wave by a Phasor Diagram  575 20-4 Letter Symbols for Phasor Quantities  576 20-5 Phasor Addition by Geometrical Construction  576 20-6 Addition of Perpendicular Phasors  578 20-7 Expressing Phasors with Complex Numbers  581 20-8 Phasor Addition Using Rectangular Coordinates  585 20-9 Subtraction of Phasor Quantities  587 20-10 Multiplication and Division of Phasor Quantities  589 Summary  591 Problems  591 Review Questions  593 Integrate the Concepts   594 Practice Quiz  595

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xiv

Contents

21 Impedance

Key Terms  599 Learning Outcomes  599 21-1 Resistance and Inductance in Series  600 21-2 Impedance  601 21-3 Practical Inductors  604 21-4 Resistance and Capacitance in Series  607 21-5 Resistance, Inductance, and Capacitance in Series  608 21-6 Resistance, Inductance, and Capacitance in Parallel  610 21-7 Conductance, Susceptance, and Admittance  613 21-8 Impedance and Admittance  615 21-9 Troubleshooting  619 Summary  621 Problems  621 Review Questions  627 Integrate the Concepts   628 Practice Quiz  628

22 Power in Alternating-Current Circuits 22-1 22-2 22-3 22-4 22-5 22-6 22-7

Key Terms  633 Learning Outcomes  633 Power in a Resistor  634 Power in an Ideal Inductor  635 Power in a Capacitor  637 Power in a Circuit Containing Resistance and Reactance  639 The Power Triangle  641 Power Factor  645 Power-Factor Correction  648 Summary  657 Problems  657 Review Questions  660 Integrate the Concepts   662 Practice Quiz  662

PART V  Impedance Networks

23 Series and Parallel Impedances 23-1 23-2 23-3

Key Terms  667 Learning Outcomes  667 Resistance and Impedance  668 Impedances in Series  668 Impedances in Parallel  671

Contents

23-4 23-5

Series-Parallel Impedances  678 Source Conversion  682 Summary  684 Problems  684 Review Questions  688 Integrate the Concepts   689 Practice Quiz  690

24 Impedance Networks 24-1 24-2 24-3 24-4 24-5 24-6 24-7

Key Terms  693 Learning Outcomes  693 Loop Equations  694 Mesh Equations  700 Superposition Theorem  702 Thévenin’s Theorem  707 Norton’s Theorem  713 Nodal Analysis  716 Delta-Wye Transformation  724 Summary  729 Problems  729 Review Questions  735 Integrate the Concepts   736 Practice Quiz  736

25 Resonance

Key Terms  741 Learning Outcomes  741 25-1 Effect of Varying Frequency in a Series RLC Circuit  742 25-2 Series Resonance  745 25-3 Quality Factor  748 25-4 Resonant Rise of Voltage  749 25-5 Selectivity  751 25-6 Ideal Parallel-Resonant Circuits  753 25-7 Practical Parallel-Resonant Circuits  758 25-8 Selectivity of Parallel-Resonant Circuits  764 Summary  766 Problems  766 Review Questions  768 Integrate the Concepts   769 Practice Quiz  770

26 Passive Filters

Key Terms  775 Learning Outcomes  775

xv

xvi

Contents

26-1 26-2 26-3 26-4 26-5 26-6 26-7 26-8 26-9 26-10

Filters  776 Frequency Response Graphs   778 RC Low-Pass Filters   781 RL Low-Pass Filters   788 RC High-Pass Filters   791 RL High-Pass Filters   797 Band-Pass Filters  799 Band-Stop Filters  804 Practical Application of Filters   809 Troubleshooting  812 Summary  814 Problems  815 Review Questions  819 Integrate the Concepts   821 Practice Quiz  822

27 Transformers

Key Terms  825 Learning Outcomes  825 27-1 Transformer Action  826 27-2 Transformation Ratio  828 27-3 Impedance Transformation  831 27-4 Leakage Reactance  833 27-5 Open-Circuit and Short-Circuit Tests  835 27-6 Transformer Efficiency  837 27-7 Effect of Loading a Transformer  838 27-8 Autotransformers  841 27-9 Troubleshooting  843 Summary  846 Problems  846 Review Questions  849 Integrate the Concepts   851 Practice Quiz  851

28 Coupled Circuits 28-1 28-2 28-3 28-4 28-5

Key Terms  855 Learning Outcomes  855 Determining Coupling Network Parameters  856 Open-Circuit Impedance Parameters  857 Short-Circuit Admittance Parameters  864 Hybrid Parameters  867 Air-Core Transformers  874

Contents

28-6 28-7

Mutual Inductance  875 Coupled Impedance  878 Summary  882 Problems  882 Review Questions  885 Integrate the Concepts   886 Practice Quiz  886

29 Three-Phase Systems

Key Terms  891 Learning Outcomes  891 29-1 Advantages of Polyphase Systems  892 29-2 Generation of Three-Phase Voltages  895 29-3 Double-Subscript Notation  897 29-4 Four-Wire Wye-Connected System  899 29-5 Delta-Connected Systems  903 29-6 Wye-Delta System  909 29-7 Power in a Balanced Three-Phase System  913 29-8 Phase Sequence  915 29-9 Unbalanced Three-Wire Wye Loads  919 29-10 Power in an Unbalanced Three-Phase System  924 29-11 The AC Generator   927 29-12 Three-Phase Induction Motor   930 29-13 Three-Phase Synchronous Motor   932 29-14 Single-Phase Motors  934 29-15 The 30° Difference between Delta-Wye Configurations  935 Summary  937 Problems  938 Review Questions  940 Integrate the Concepts   943 Practice Quiz  943

30 Harmonics 30-1 30-2 30-3 30-4 30-5 30-6

Key Terms  947 Learning Outcomes  947 Nonsinusoidal Waves  948 Fourier Series  949 Addition of Harmonically Related Sine Waves  951 Generation of Harmonics  954 Harmonics in an Amplifier  956 Harmonics in an Iron-Core Transformer  958

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Contents

30-7 30-8 30-9

RMS Value of a Nonsinusoidal Wave  960 Square Waves and Sawtooth Waves  961 Nonsinusoidal Waves in Linear Impedance Networks  963 Summary  966 Problems  966 Review Questions  968 Integrate the Concepts   969 Practice Quiz  970

Appendices 1 2

Determinants 974 Calculus Derivations 977

2-1 2-2 2-3 2-4 2-5 2-6 2-7 2-8 2-9 2-10

Maxium Power-Transfer Theorem   977 Instantaneous Voltage in a CR Circuit   978 Energy Stored by a Capacitor   980 Instantaneous Current in an LR Circuit   980 Energy Stored by an Inductor   982 RMS and Average Values of a Sine Wave   982 Inductive Reactance   983 Capacitive Reactance   984 General Transformer Equation   985 Maximum Transformer Efficiency   985

3 Multisim Schematic Capture and Simulation   986 Answers to Selected Problems   988 Glossary  1000 Index  1013

From the Publisher

Since its first appearance in 1959, Introduction to Electric Circuits has been used as a core text by hundreds of thousands of college and university students enrolled in introductory circuit analysis courses. Through its many editions, this classic text helped shape the way the subject is taught, and it was acclaimed by instructors and students alike for its accessible writing style, its clear explanations of key concepts, and its comprehensive end-of-chapter problem sets.

In 2008, Oxford University Press proudly made this landmark work available once again, thoroughly updated to reflect the many advances made in the discipline. With this tenth edition, Oxford is proud to carry on the pioneering work of founding author Herbert W. Jackson by recognizing and addressing the changing needs of twenty-first-century students and instructors.

Highlights of the Tenth Edition New!

29

Expanded coverage of breadboards, band resistors, digital multimeters, nodal analysis, and three-phase systems. All-new sections on power in unbalanced three-phase systems and Delta-Wye configurations highlights a detailed expansion of key material, enhancing the book’s coverage of three-phase systems. Other exciting additions include expanded coverage of breadboards, colour codes for band resistors, digital multimeters, and nodal analysis.

s Three-Phase System

up to this point have rks we have considered with Most of the AC netwo e. However, systems of alternating voltag had only one source angles have significant phase nt differe with ic several AC voltages large amounts of electr uting distrib and ating advantages for gener industrial electric ts are also used for most l systems. energy. Polyphase circui contro al hanic types of electromec motors and for some

Chapter Outline

Chapter 29

29-15

s Three-Phase System

hase Systems 892 924 Advantages of Polyp 895 Three-Phase Voltages 29-2 Generation of ion 897 Notat ) 29-3 Double-Subscript ) − ( −0.5145 + j0.1392 899 m = ( 0.9714 + j0.2577 Connected Syste 29-4 Four-Wire WyeSystem 903 = 1.486 + j0.1185 A 29-5 Delta-Connected m 909 29-6 Wye-Delta Syste m 913 Therefore, IA = 1.5 A ced Three-Phase Syste 29-7 Power in a Balan IB = IBC − IAB nce 915 Similarly, 29-8 Phase Seque 919 1.005 A ∠+14.86º Three-Wire Wye Loads = 0.533 A∠+44.86º − 29-9 Unbalanced m 924 ) d Three-Phase Syste lance Unba ) − ( 0.9714 + j0.2577 an in r 29-10 Powe = ( 0.3778 + j0.3760 927 29-11 The AC Generator = −0.5936 + j0.1183 A Induction Motor 930 e -Phas Three 29-12 932 IB = 0.61 A Synchronous Motor and 29-13 Three-Phase Lamp B. Motors 934 935 Lamp A is brighter than 29-14 Single-Phase gurations, with a CBA phase sequence, en Delta-Wye Confi Hence betwe ence Differ 29-15 The 30°

29-1

jac31412_ch29_890-945.indd

10/31/18 09:05 PM to the original wyeline currents, we can return voltOnce we have solved for for the various phase use Ohm’s law to solve connected circuit and . = IBZB, and VCN = ICZC V , Z I BN = A A V ages: AN 29-60 to 29-65. and Review Questions 29-29 to 29-27 ms See Proble

890

B

Circuit Check

e ad from a 230-V 3-phas motor operates at full-lo efficient and CC 29-4. A 10-hp AC source. The motor is 80% 60-Hz delta-connected ate Calcul g. laggin its power factor is 0.88 (a) the line currents (b) the phase currents source powers a wye 4-wire wye-connected and ZC = CC 29-5. A 208-V 60-Hz 60º Ω, ZB = 100 ∠20º Ω, t. load that has ZA = 50 ∠ ts and the neutral curren curren line the ate Calcul . 40 ∠90º Ω

of rise to different types ing the phase split gives The method of obtain art motor uses a startmotors. The resistance-st g single-phase induction r wire than the runnin thinne and turns fewer ing winding made with nce and low reactance g winding has high resista winding. Then, the startin high reactance, giving g has low resistance and while the running windin to start the motor. with the starting the necessary phase split uses a capacitor in series g The capacitor-start motor tor and starting windin phase split. The capaci reaches winding to obtain the switch when the motor ugal centrif the by en are both switched out the phase shift betwe Since speed. g runnin approximately 75% of resistance-start motor, ts is greater than for the starting and running curren g torque and can be manudevelops more startin the the capacitor-start motor , the capacitor is left in In a capacitor-run motor sizes. larger in d facture This configuration has runs as a two-phase motor. torcircuit, and the motor than resistance and capaci ion operat r quiete the advantage of much pulsating torque. to be noisy due to the applistart motors, which tend motors used in small types of single-phase There are many other is essentially a DC series The universal motor, which or ances and power tools. can operate on either AC power tools. This motor many This in fans. used is and motor, small appliances -pole motor is used in on the stator, DC power. The shaded a portion of each pole around wire heavy of the motor has a loop the unshaded portion of to lag behind the flux in causing the magnetic flux to start the motor. Small the phase split necessary pole, and thus producing devices and clocks. motors are used for timing single-phase synchronous 29-76 and 29-77. See Review Questions

nce bet ween 29-15 The 30° Differe ons rati Delta-Wye Configu

onnected load as shown ire system with a wye-c es is Let’s consider a four-w respective load voltag phasor diagram of the in Figure 29-42(a). The b). 29-42( Figure N shown in A B C VC

120 V 120°

+ VA

VA

VB

−

− 120 V 0° − VC +

120 V 0°

+

alanced 29.1 0 Power in an Unb Phase System

Three-

in each arm will be differhase system, the power the In an unbalanced three-p will not be able to use all be the same, and we ent as the loads may not load. Instead, the total hase three-p ed with a balanc phase. factor √3 that we used each in power be the sum of the power in the load will

10/31/18 09:05 PM jac31412_ch29_890-945.indd

924

935

rations n Delta-Wye Configu The 30° Difference betwee

120 V −120°

VB

120 V 120°

120 V −120° (b)

(a)

ire system with Figure 29-42 Four-w

a wye-connected load

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From the Publisher

The best exercise sets available—over 2000 ­problems and review questions. From the very first ­edition of Jackson’s Circuits, instructors and students alike praised the ­book’s exercises for their variety and usefulness.

7-6

Internal Resistance

Complete

es in the circuit of Figure the table listing the voltag

8.0 Ω

A

lems have been added, and every e­ xercise has been checked for accuracy and relevance. Grounded in decades of classroom and lab experience, the problems give students the chance to test and refine their comprehension of the discipline’s key concepts.

133

A

Circuit Check CC 7-1.

New! For this edition, hundreds of new prob-

2.0 Ω

B

7-6.

“Circuit Check” problem sets, found throughout each chapter, give students the chance to test and refine their understanding of material they have just read.

C 5.0 Ω D

12 V

5.0 Ω 12 Ω E

F Figur e 7-6 VAF

VBF

VEF

VDF

VCF

VCE

VBE

VDE

VAC

VBC

ances in the circuit of

own resist CC 7-2. Find the unkn

VDC

VEC

Figure 7-7.

20 V R1

I R2

+ 100 V −

3.0 Ω

R3

50 V Figur e 7-7

a 100-Ω resistor will connected in series with ination is conCC 7-3. What resistance of 20 W when the comb dissipate heat at a rate e? sourc nected to a 120-V

7-6

766

Chapter 25

Internal Resistance

the e of a source is equal to that the terminal voltag depends on In Section 2-8 we noted t conditions. The EMF only under open-circui e and is independEMF of the source sourc the in action g y-convertin tial the nature of the energ n in Figure 7-8, the poten nt. However, as show ent of the circuit curre

Resonance

Summary

• The impedance of a series-resonant circuit is capacitive below the ant frequency and induc resontive above the resonant frequency. • In a series-resonant circuit at resonance, the impedance has its minim value, which equals the um resistance. • At the resonant frequ ency, the inductive reacta nce equals the capacitive reactance. • The quality factor of a resonant circuit deter mines the steepness of shoulders of the reson the ance curve. • In a series-resonant circuit with a high Q, inductance and capac voltages at resonance itance are much greater than the source voltage. • The half-power frequ encies define the band width of a resonant circui • Increasing Q increases t. the sensitivity and select ivity of a resonant circui • The impedance of a t. parallel-resonant circui t is inductive below the onant frequency and capac resitive above the resonant frequency. • In a parallel-resonan t circuit with a high Q, the tank current at reson is much greater than the ance source current. • In a parallel-resonan t circuit at resonance, the impedance has its imum value, which equal maxs the resistance. • The Q of a practical parallel-resonant circui t is equal to the Q of the • The internal resistance coil. of the source reduces the Q and the selectivity of a parallel-resonant circui t.

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Review exercises and end-of-chapter problem sets, now ranked for their level of difficulty, encourage students to reconsider chapter content and apply their newly learned skills to over 1500 problems.

B = beginner

I = intermediate

A = advanced

circuitS IM

walkthroug h

Problems

B

Section 25-2

25-1.

Series Resonance

Determine the resonant frequency of a 68-μF capacitor in series with a 22-μH coil that has a Q of 85. B 25-2. (a) What capac itance is needed to tune a 500-μH coil to series onance at 465 kHz? res(b) Use Multisim to verify the capacitance. B 25-3. What inductance in series with a 12-pF capac itor is resonant at 45 MHz? I 25-4. A variable capac itor with a range of 30 pF to 365 pF is conne in series with an induc cted tance. The lowest frequ ency to which the circuit can tune is 540 kHz. (a) Calculate the induc tance. (b) Find the highest frequ ency to which this circui t can be tuned. Section 25-3

A

25-5.

Quality Factor

A series RLC resonant circuit is connected to a supply voltage of 50 V at a frequency of 455 kHz. At resonance the maximum current measured is 100 mA. Determine the resistance, capacitance, and inductance if the quali ty factor of the circuit is 80.

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From the Publisher

Practice Quiz

Integrate the Concep

“Integrate the Concepts” exercises draw all the strands of the chapter together by presenting a scenario that requires students to make use of the various concepts that have been covered in order to solve a real-life problem.

385

ts

A 50-V source, a 20-kΩ resistor, a 1-μF capacitor, and an open switch form series circuit. Calculate a the following after the switch is closed: (a) the initial value of the current (b) the time constant (c) the capacitor volta ge 50 ms later (d) the time for the capac itor to fully charge (e) the energy stored by the capacitor after it is fully charged

Practice Quiz

1.

Practice quizzes at the end of each chapter help students test their retention of material and prepare for exams.

2.

Which of the following statements are true? (a) Uppercase letter symbols represent stead y-state values that are dependent on time. (b) Lowercase letter symbols represent instan taneous values that are dependent on time. (c) The time constant for charging a capacitor is directly proportional to the capacitance . (d) A negative rate of change of voltage indica tes that the voltage across the capacitor is decreasing. (e) Kirchhoff’s volta ge law can be applied to RC circuits to deter mine the voltage acros s the capacitor. What is the initial rate of change of potential difference across the capacitor in Figure 13-35 when the switch is closed ? (a) 2.4 V/s (b) 2.4 kV/s (c) 24 V/s (d) 0.24 V/s R1 2.0 GΩ V1 48 V

C1 0.010 μF

Figur e 13-35

3.

How long will it take the capacitor of Figur e 13-35 to charge to the battery voltage of 48 V? (a) 20 s (b) 20 ms (c) 100 ms (d) 100 s

jac31412_ch13_348-387.in

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10-5

ing 10-5 Troubleshoot

trace circuit theorems to help We can use equivalent-

Troubleshooting

283

circuit faults.

Example 10-9

the calculated value e circuit in Example 10-3, For the Wheatstone bridg if the measured value mine the probable cause for V3 is 33.33 V. Deter for V3 is (c) 66.67 V (b) 0 V (a) 100 V

Multisim and exercises. An ­introdu­ction to M ­ ultisim paves the way to a host of ­Multisim examples and exercises designed to a­ cclimatize s­ tudents to w ­ orking with the leading software for ­circuit ­simulation. All Multisim exercises and examples in the text—  ­denoted by the circuitSIM icon in the ­margin—are tied to the ancillary resource centre packaged with the book.

Solution V , so we deduce that e law gives E = V1 + 3 (a) Kirchhoff’s voltag (shorted) or I1 = 0 A. R , then either R1 = 0 Ω V1 = 0 V. Since V1 = I1 1 However, if R1 is open, open. is R or R 3 1 either For I1 to be 0 A, ude that R1 is not concl we V, 100 = Since V3 I3 = 0 A and V3 = 0 V. open. Remove these R is shorted or R3 is open. Therefore, either 1 with an ohmmeter. them check and t circui resistors from the V , so we deduce that e law gives E = V1 + 3 (b) Kirchhoff’s voltag R is shorted or ’s law, V3 = R3I3, so either 3 V1 = 100 V. From Ohm is open. However, if R3 R or R 3 1 either A, 0 I3 = 0 A. For I1 to be we conclude zero, not is V V = 0 V. Since 1 is open, I1 = 0 A and 1 or R3 is shorted. fore, either R1 is open that R3 is not open. There check them with an and t circui the from Remove these resistors

ohmmeter. V , so we deduce that e law gives E = V1 + 3 (c) Kirchhoff’s voltag and V3 are reversed, that the values of V1 V1 = 33.33 V. We note hanged. Checking the interc R have been suggesting that R1 and 3 the 150-Ω rewe discover that R1 is rs, resisto these on colour codes resistor. sistor and R3 the 300-Ω Multisim Solution ite. EX10-9 from the webs Download Multisim file with the given values as shown in Figure 10-5, The circuit is the same terfor the four resistors. 50-Ω resistor between s of R1 and R3. Insert a Interchange the value ect a ground to the ate a galvanometer. Conn minals A and B to simul ure the voltages meas to s meter atic. Insert bottom node of the schem uacross R1 and R3. ngs to verify the concl view the voltage readi Run the simulation, and ). sions of Example 10-9(c

circuitS IM walkthroug h

 xamples. Numerous worked examples ­throughout E each chapter show students how to perform the calculations that are essential to circuit analysis.

See Problem 10-42.

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From the Publisher

Chapter 22

650

Power in Alternating

Practical Circuits

-Current Circuits

ion Power-Factor Correct

raham Hughes Source: CP Photo/G

measure of how effect The power factor is a by a consumer convert ively devices operated l usefu to utility local electric current from the heat, light, or mechanpower output, such as practhe is ction corre or ical motion. Power-fact r factor of an inductive tice of raising the powe by inserting a capacitor load, such as a motor, , lower the power factor in parallel with it. The required to deliver a the greater the current y. given amount of energ ly bill commercial Electricity suppliers usual a power factor surand industrial customers of the customer’s charge if the power factor The power supplied to load drops below 90%. ured with a meter that these customers is meas e power and the reactmonitors both the usabl indicate the power can ive power, and hence factor each month.

Practical applications feature. “­Practical ­Circuits” boxes show students how circuit t­ heory is used in everyday situations.

Power meter

14.3 A A

70% lagging M power factor

120 V 60 HZ

308 μF

Figur e 22-13

Because

Effect of a series capa

citor on a lagging powe

r-factor load

are in series, the motor and capacitor ZT = 8.4 + j8.6 − j8.6 =

8.4 + j0 Ω = 8.4 Ω ∠0º

10/31/18 08:49 PM

300

Chapter 11

dd 650

jac31412_ch22_632-664.in

Electrical Measurem

ent

11-3 The Voltmeter

Moving-coil movement s can also be used in voltmeters. If a 1.0-m movement has a total resist A ance of 50 Ω, a voltage drop of V = IR = 0.0010 × 50 Ω = 50 mV appea A rs across the movement when it is reading full If we connect a 99.95-kΩ scale. resistor in series with the meter to bring the resistance up to 100 kΩ, total as in Figure 11-5, the voltag e we must apply to produce a 1.0-mA current through the meter is V = IR = 0.0010 A × 100 kΩ = 100 V 99.95 kΩ

Multiplier

Online icons—found throughout the chapters— direct students to a wide variety of interactive learning resources available on the mobilefriendly ancillary resource centre.

Figur e 11-5

1.0-mA movement (50 Ω)

V

Simple voltmeter

If we apply a 50-V poten tial difference to the voltm eter, the current through it is I=

V 50 V = = 0.50 mA R 100 kΩ

Thus, the meter will read half scale. With the 99.95 -kΩ multiplier resistor, the scale on the 1.0-mA movement correspond s to 0–100 V. To measure open-circu it voltage, a perfect voltm eter should draw no current. Since current must flow in a moving-co il movement to obtain reading, a 50-μA move a ment is much more accur ate than a 1.0-mA move for measurements in highment resistance circuits. The resistance of a standard 50-μA movement and its calibrating resistor is 1000 Ω. To obtain a full-scale range, the total 100-V resistance must be

100 V RT = = 2.0 MΩ 50 µA Therefore, the multiplier resistance must be 2000 kΩ − 1 kΩ = 1999 kΩ. Similarly, the total resist ance must be 10 MΩ for a 500-V scale and 20 kΩ for a 1.0-V scale. For any full-scale voltage with a 50-μA movement total resistance required , the is 20 kΩ for each volt of the full-scale readi Therefore, when used ng. as a voltmeter, a 50-μA movement has a voltm sensitivity of 20 kΩ/V eter . We can determine the total resistance of a movi coil voltmeter by multi ngplying the full-scale volta ge by the movement. For example, sensitivity of the when used as a voltm eter, a 1.0-mA movement requires a total resistance of 1000 Ω/V, so total resist ance for a 500-V scale is 500 V × 1000 Ω/V = 500 kΩ.

jac31412_ch11_294-320.in

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From the Publisher

Updated and revised instructor and student resources. A comprehensive package of ancillary resources includes a lab manual and an online library of learning materials for students, and an instructor's manual, a test generator, solutions manuals, and PowerPoint slides for instructors—all thoroughly updated and revised to reflect new and critical content in the tenth edition. For Students • Lab Manual. The perfect accompaniment for courses with a laboratory component, the revised lab manual is designed to give students hands-on experience through experiments carefully linked to chapter material. • Interactive Companion Site. This suite of online learning resources offers students easy, mobile-friendly access to online videos, quizzes, interactive animations, MultiSim example files, and other high-interest materials that expand on key elements from each chapter, allowing students to explore circuits in new and engaging ways. For Instructors • Instructor's Manual. The instructor's manual provides a variety of resources for instructors, including lesson plans, course outlines, suggestions for quizzes and assignments, and helpful hints for labs. • Expanded! Test Generator. The test generator—now expanded to include fifty percent more questions—offers an exhaustive store of ­multiple-choice, true/false, and short-answer questions in several formats, all suitable for tests and exams. • Solutions Manuals. The solutions manuals provide worked solutions to all end-of-chapter problems and questions and all exercises in the student lab manual. • PowerPoint Slides. These slides will enhance classroom instruction with lists of learning outcomes, chapter summaries, key equations and terms, illustrations, problems, and other content drawn from the text. Details on instructor’s supplements are available from your Oxford ­University Press sales representative or on our website: www.oupcanada.com/Jackson10e

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xxiv

From the Publisher

About the Authors When Herbert W. Jackson published the first edition of Introduction to Electric Circuits in 1959, he already had nearly two decades of teaching experience, first as a radar instructor for the RAF and RCAF during World War II—he was stationed in England during the Battle of Britain—and later at the veterans’ training institute in Toronto, Ontario, that would become the nucleus of what is today Ryerson University. J­ ackson headed the electronics and electrical engineering technology department at ­Ryerson for many years, during which he worked on early editions of this book. He later joined the Ontario Ministry of Education, where he oversaw the ­creation of that province’s community colleges in the 1960s and 1970s. So influential was he that some observers labelled him “the father of the ­Ontario college system.” Jackson continued to r­ evise Introduction to Electric Circuits even after retiring from teaching and public service. He and his wife, Elleda, had two sons, David and Robert. Herb Jackson died in 1999. Dale Temple taught electronics engineering technology at the College of the North Atlantic and has also served as coordinator of the electronics program there. Previous writing credits include work as the co-author of Canadian editions of Boylestad & Nashelsky’s Electronic Devices and Circuit Theory and Tocci’s Digital Systems: Principles and Applications. He lives in St. John’s, Newfoundland and Labrador. Brian Kelly for many years coordinated the introductory circuit analysis course at the College of the North Atlantic. Besides his extensive teaching experience, he is accomplished in numerous computer-aided design (CAD) software programs and helped prepare the circuit diagrams for previous editions of Introduction to Electric Circuits. Lauren Fuentes currently teaches in four different programs at Durham College in Oshawa, Ontario, and serves as the coordinator of the Electronics Engineering Technician and Technology Programs there. She is always interested in new challenges, and she has participated in several research projects at Durham and contributed to revisions of Jackson’s Introduction to Electrical Circuits textbook, among others. Karen Craigs is an electronics college professor in Toronto, Ontario. Besides developing courses and tinkering with electronics gadgets, she has gradually expanded her work over the years to include coaching teams for technology competitions, mentoring women in technology, and assisting with various electronics clubs and outreach programs.

From the Publisher

Acknowledgements Many individuals brought this new edition of Introduction to Electric Circuits to fruition. Cliff Newman, consulting acquisitions editor for Oxford, knew the book well from his many years as sales manager and editorial ­director at Prentice-Hall and felt it deserved a new lease on life. Elleda Jackson likewise believed her husband’s lifework should be made available to new generations of students, and she worked tirelessly to secure the contractual reversions necessary for Oxford to publish the eighth edition. Development editor Lauren Wing and copy editor Jess Shulman were instrumental in making suggestions for improvement and offering guidance through the many phases of production for the tenth edition. Various editing teams at Oxford University Press have contributed their outstanding hard work and dedication to this textbook, ensuring that it continues to be a relevant resource for students and educators. The publishers and authors would also like to thank the following reviewers, along with those who chose to remain anonymous, who provided valuable feedback at various stages in the development of this textbook: • Laura Curiel, Lakehead University • Terry Moschandreou, Fanshawe College • Mingbo Niu, Okanagan College • Gord Wilkie, Nova Scotia Community College

xxv

From the Preface to the First Edition (1959)

When a person first attempts to read a book in a foreign language, for example in French, armed only with a French–English dictionary, it is difficult for him to appreciate the content fully, because of the conscious effort required in translating the individual words. A prior study of the basic rules of the grammar of the language, along with sufficient vocabulary practice, would be a more logical approach to the task. Similarly, basic electric circuit theory constitutes the language by which the operation of electric and electronic apparatus is described. Therefore a study of the basic laws of this language along with sufficient practice to minimize the conscious effort required in its application are prerequisites to any serious study of electric and electronic equipment. To carry the analogy one step further, translation by searching through the dictionary for the desired word and substituting the equivalent English

xxviii

From the Preface to the First Edition (1959)

word will do little to assist in understanding the material being translated. Therefore the student is warned not to think of electric circuit theory and this text as a dictionary of formulas by which one may arrive at the correct numerical solution simply by selecting the right formula and inserting given data. In electric circuits, a formula is merely a convenient way of expressing a definition of the behaviour of the circuit. Since the task at hand is to understand circuit behaviour, the student should not apply a formula unless he is satisfied that he fully understands the definition it represents. In this respect electric circuit theory has an advantage over language study in that every definition can be readily and logically developed from a ­preceding step. . . . H.W. Jackson 1959

From the Authors of the Tenth Edition When Oxford University Press told us of their intention to revise and publish the tenth edition of Jackson’s Introduction to Electrical Circuits and that they wanted two female professors to join the team, we were honoured, delighted, and thrilled to be the first to start a new trend. This textbook has been a staple in many institutions for many years and we hope that our participation in this edition will provide many students, alumni, and professors a useful tool for teaching and reference. Karen Craigs Lauren Fuentes October 2018 Many thanks to my husband, Jeremy, and my family, John, Lynda, Theresa, and Mark, for their inspiration and love, always. —K.E.C Thanks to my husband, Andy, and my sons, Alberto and Kenny, for always supporting me in whatever new project I embark on. You give me strength as I continue to push new boundaries in everything I do. —L.F.

This tenth edition of Introduction to Electric Circuits is dedicated to the memory of Herbert W. Jackson.

PART

I

The Basic Electric Circuit

Electric circuit theory consists of a few fundamental laws and a series of equations stating relationships that apply to all electric and electronic circuits. Consequently, much of an introductory electric circuits course deals with the solution of numerical examples. However, developing an understanding of how circuits behave involves much more than entering numbers into a calculator or a computer program. To apply circuit principles to electrical and electronic systems, we also need to understand the physical basis for electric circuit theory. Part I develops these basic concepts in detail.

1 Introduction 2

Current and Voltage

3

Conductors, Insulators, and Semiconductors

4

Cells, Batteries, and Other Voltage Sources

5

Resistance and Ohm’s Law

6

Work and Power

Photo source:  © iStock.com/omada

1

Introduction Our standard of living depends on an abundant, convenient, and economical supply of energy and on the relative ease with which we can convert energy from one form to ­another. We need energy to heat, cool, and light our homes and to run appliances, vehicles, and industrial machinery. Electricity is an extremely useful form of energy. At the beginning of the nineteenth century, electricity was primarily a scientific curiosity. By the end of that century, the electric telegraph had revolutionized communications, and by the end of the twentieth century electricity had become a vital foundation of modern technology. To work effectively with electrical technology, we must first understand the basic principles that govern electric ­circuits. We can then develop techniques for analyzing the performance of practical electric and electronic circuits.

Chapter Outline 1-1

1-2

1-3

Circuit Diagrams  4

The International System of Units  5

Calculators for Circuit Theory  6

1-4

Numerical Accuracy  7

1-6

SI Unit Prefixes  10

1-5 1-7

Scientific Notation  7

Conversion of Units  12

Key Terms electricity  2 electric circuit  4 conductors  4 switch  4 source  4 load  4 circuit diagrams  4 schematic diagram  4 metric system  5 International System of Units (SI)  5

metre  5 kilogram  5 second  5 kelvin  5 ampere  5 significant digits  7 scientific notation  8 unit prefixes  10 engineering notation  10 dimensional analysis  12

Learning Outcomes At the conclusion of this chapter, you will be able to: • understand the schematic diagram of a simple electric circuit • identify the base units of the SI system of units • use the proper number of significant figures in a ­calculation

Photo sources:  © iStock.com/gwmullis

• use scientific notation to represent quantities • use engineering notation to represent quantities • use dimensional analysis to convert from one system of units to another

4

Chapter 1  Introduction

1-1  Circuit Diagrams The term electricity can mean either a flow of charged particles or the branch of physics that deals with such flows.

In a circuit, a load is any device that ­converts electric ­energy to some other form of energy.

Two energy conversions take place in the basic electric circuit illustrated in Figure 1-1. Chemical energy stored in the battery is converted into electric ­energy. Two copper electric conductors convey the electric energy to the lamp, which converts it into light and heat. A switch in this circuit allows us to start and stop the energy conversion at will by interrupting the transmission of electric energy from the energy source (the battery) to the load (the lamp).

Ba

tte

ry

+

−

Figure 1-1  Pictorial representation of a basic electric circuit

A diagram is usually the best and easiest way to show the interconnection of the components of an electric circuit. For more elaborate circuits, pictorial representations such as Figure 1-1 can be confusing and difficult to draw. To keep circuit diagrams simple and clear, we represent the various circuit elements by standard graphic symbols rather than by pictures. A circuit diagram consisting of lines and symbols is called a schematic ­diagram. Figure 1-2 shows a schematic diagram for the circuit of Figure 1-1. A table inside the back cover shows the standard schematic symbols for the most common circuit components.

+ −

Figure 1-2  Circuit diagram for a basic electric circuit

See Review Question 1-26 at the end of the chapter.

1-2  The International System of Units In 1883, the Scottish physicist William Thomson (Lord Kelvin) said, “when you can measure what you are speaking about, and express it in ­numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre

1-2   The International System of Units

and u ­ nsatisfactory kind.” To apply this philosophy to our study of electric ­circuits, we need a universally recognized system of measuring units. The English-speaking world in Lord Kelvin’s time still used an awkward set of units, including inches, feet, yards, rods, chains, fathoms, teaspoons, cups, pints, bushels, pounds, and several kinds of ounces. Most other countries used the metric system, a simple decimal system of units, which originated in France around 1800. As a result of a treaty signed in 1875 by representatives of 17 nations, the International Bureau of Weights and ­Measures (BIPM) was established to provide standards of measurements for use worldwide. In October 1965, the Institute of Electrical and Electronics Engineers (IEEE) adopted the International System of Units (SI) proposed at the eleventh General Conference on Weights and Measures (CGPM) held in France in 1960. SI unified and rationalized several earlier sets of metric units, allowing scientists and engineers in all fields to use units based on the same standards. Today, SI is structured around seven base units that can be used to derive another 22 named units in physics. In November 2018, after a unanimous vote of 60 nations at the twenty-sixth CGPM, the definition of four base units—the kilogram, the ampere, the kelvin, and the mole—will be redefined by setting exact numerical values for the constants that relate each of them to fundamental quantities of nature. Once the new definitions are adopted in May 2019, SI will be wholly derivable from universal natural phenomena and will finally resolve its objective after 220 years of use. The base unit of length or distance in the International System of Units is  the metre. Originally calculated as one ten-millionth (1 × 10−7) of the ­distance at sea level from the earth’s equator to the pole, the metre is now more accurately defined as the distance travelled by light through a vacuum in 1/299 792 458 of a second. Note that SI uses spaces instead of commas to separate groups of three digits in long numbers. The original metric unit of mass, the gram, was defined as the mass of one cubic centimetre of pure water at 4°C. The gram was too small a unit for many practical purposes. Hence the SI base unit for mass is 1000 grams—the kilogram. For over a century, the international standard for the kilogram was a platinum-iridium cylinder kept at the BIPM in Sèvres, France. As the last SI unit to be based on a manufactured object, the kilogram officially changes its definition in 2019 to be based on a fixed value of the Planck constant, h, which links the kilogram to the metre and second. The SI unit of time is the second. Originally defined as 1/86 400 of a mean solar day, we now define the second more accurately in terms of the decay of radioactive cesium in an atomic clock. SI still recognizes minutes and hours even though these units are not decimal multiples of a second. The SI unit for temperature is the kelvin (symbol K), which is equal to one Celsius degree. On the Celsius scale, the freezing point of water is 0°C (equivalent to 273.15 K) and the boiling point of water is 100°C (equivalent to 373.15 K). Absolute zero is 0 K, or −273.15°C. In 2019, the new definition will be based on a fixed value of the Boltzmann constant, k, which links the kelvin to the kilogram, metre, and second.

5

The abbreviation SI comes from the French name, Système Internationale ­d’Unités.

In the 1890s, the United States changed their most accurate objects for fundamental standards of length and mass from imperial system objects to metric system objects. Officials then used conversion tables to derive US customary measurements.

6

A degree symbol is not used with ­measurements in ­kelvins.

Chapter 1  Introduction

The SI base unit that extends the system of units to electrical measurement is the ampere. In order to relate electrical measurement to mechanical measurement, the ampere was defined in terms of the electromagnetic force between two current-carrying electric conductors. In 2019, the new definition of the ampere will be based on a fixed value of the elementary electric charge, e, which links the ampere to the second. We shall define the ampere in Section 2-5, with additional details described in Chapter 14. For equations, we need letter symbols to represent both the quantities being measured and the SI units of measurement. Table 1-1 shows the letter symbols for the SI base units used in this book. Note that italic (slanted) letters represent quantities while Roman (upright) letters represent units. TABLE 1-1  Some letter symbols Quantity length or distance

Quantity Symbol l or d

SI Base Unit

Unit Symbol

metre

m

mass

m

kilogram

kg

time

t

second

s

electric current

I

ampere

A

temperature

T

kelvin

K

Occasionally, we may encounter an abbreviation of a unit name. For ­example, electricians often shorten ampere to amp. Equations should always use the SI unit symbols.

Example 1-1 When the switch is closed in the basic electric circuit of Figure 1-2, the electric current through the lamp is one quarter of an ampere. Express this statement as an equation. Solution I = 0.25 A See Review Questions 1-27 and 1-28.

1-3  Calculators for Circuit Theory Numerical examples using SI units provide the best means of understanding the behaviour of electric and electronic circuits. Although the mathematics required can be kept reasonably simple, the calculations themselves can often be tedious if done by hand. Solutions of numerical examples in

1-5   Scientific Notation

7

this book assume the use of a suitable scientific or engineering calculator. For the more complex circuits, students could also use circuit simulation ­software. For this course, it is convenient to have a calculator with a key that gives answers in powers of 10 corresponding to SI unit prefixes. This key is often labelled Eng . Programmable features are not necessary, but the ­calculator must be able to evaluate exponentials and convert between rectangular and polar coordinates. Keys for these functions are often marked ex , →P , and →R . Some so-called “scientific” calculators do not have these functions.

1-4  Numerical Accuracy At first glance, it might seem that an electric current of 12 A is exactly the same as a current of 12.0 A. However, stating a current as 12 A means that the exact value is closer to 12 than it is to either 11 or 13, while a current of 12.0 amperes is closer to 12.0 than it is to 11.9 or to 12.1. Thus, 12.0 A is a more precise measurement than 12 A. We say that the figure 12 has two ­significant digits, whereas the figure 12.0 has three significant digits. A calculator treats any number we enter as an exact number. If we enter 34 ÷ 2.3, the calculator will show 14.782608. But, if the original measurements are each accurate to only two digits, the calculated result is also a­ ccurate to only two digits. We know only that the answer is closer to 15 than it is to 14, no matter how many digits are displayed by the calculator. When performing chain calculations with a calculator, we can leave the strings of digits contained in the calculator until the final answer. However, the final answer should not have more significant digits than the least accurate of the measurements used in the calculations. For most examples in this book, we do not need to record more than four significant digits in the intermediate steps. See Problem 1-1.

1-5  Scientific Notation Electrical measurements can involve very large quantities, such as a radio frequency of 455 000 Hz, and very small quantities, such as an inductance of 0.000 75 H. Scientific notation uses powers of 10 to eliminate long strings of zeros and to avoid confusion over whether the zeros at the end of a number are significant digits. If we shift the decimal point five places to the left in the frequency 455 000 Hz, we divide it by 100 000. We can restore the original value by then multiplying by 100 000 or 105. Hence,

455 000 Hz = 4.55 × 100 000 Hz = 4.55 × 105 Hz

The hertz (Hz) is the SI unit for frequency and is equal to one cycle per second. Chapter 16 describes inductance and the henry (H), the unit for inductance.

8

Chapter 1  Introduction

In the preceding calculation, we assumed that the trailing zeros are not significant digits. If they are significant, we include them after the decimal in order to show the precision of the measurement: f = 455 000 Hz = 4.550 00 × 105 Hz



If we shift the decimal point four places to the right in the inductance 0.000 75 H, we multiply it by 10 000. Consequently, to restore the original value, we must divide 7.5 by 10 000 or multiply by 10−4. Hence, 0.000 75 H =

7.5 H = 7.5 × 10 − 4 H 10 000

To express a quantity in scientific notation, we shift the decimal point until only one significant digit lies to the left of the decimal point, and then we multiply by the appropriate power of 10 to restore the quantity to its original value.

Example 1-2 Express 839 000 m in scientific notation, given that the trailing zeros are not significant. Solution To obtain one digit to the left of the decimal point, we must shift the decimal point five places to the left. To restore the quantity to its original value, we then multiply by 105:

839 000 m = 8.39 × 105 m

Example 1-3 Express 1/200 of a second in scientific notation. Solution The first step is to convert the fraction to decimal form. Here we treat the 1 in the numerator as an exact number: 1 s = 0.005 00 s 200 We then shift the decimal point three places to the right and multiply the result by 10−3:

0.005 00 s = 5.00 × 10−3 s

1-5   Scientific Notation

9

The following example illustrates the advantage of scientific notation.

Example 1-4 In Chapter 19, we shall derive this formula for inductive reactance: XL = 2πfL, where f is measured in hertz, L is in henries, and XL is in ohms ­(symbol Ω). Given f = 455 000 Hz and L = 0.000 750 H, calculate XL. Solution Substituting the known values into the formula gives

XL = 2 × 3.1416 × 455 000 × 0.000 750 Ω



If we enter the quantities in this form into a calculator, we may have trouble keeping track of the number of zeros. However, if we express f and L in scientific notation, we have XL = 2 × π × 4.55 × 105 × 7.50 × 10−4 Ω



Using the scientific notation feature of the calculator, we can enter this equation as follows: 2 ×

π

× 4.55

EE

5 × 7.5

EE

+/−

4 = 2144 = 2.14 × 103

Even in a longhand solution, scientific notation simplifies the compu­ tation. To multiply powers of 10, we simply add their indices algebraically. Hence, XL = 6.2832 × 4.55 × 7.50 × 10 (5+−4) = 2.14 × 103 Ω See Problems 1-2 to 1-7 and Review Questions 1-29 and 1-30.

Circuit Check CC 1-1. How many significant digits does each quantity have? (a) 435 m (b) 0.16 kg (c) 7.140 A CC 1-2. Perform the indicated operations, and express your answer in scientific notation with the correct number of significant digits. 651 × 0.00197 s 50 600 × 0.003 66 × 102 m (a) (b) 53.4 × 0.784 39 × 12 × 106 1 kg 1A (c) (d) 1 1 1 1 1 1 + + 6 + 5 + 720 180 500 18 × 10 39 × 10 56 × 103 CC 1-3. Simplify each expression to a power of 10: (a) (103)9 (b) (1/10−4)−2 (c) (10−2)−3 4 1/2 2 −4 1/2 (d) (10 ) (e) (10 /10 ) (f) (108)−1/2

stands for Enter Exponent. The π key retrieves the ­numerical value for π from the calculator’s permanent memory. The +/− key changes the sign. Some ­calculators have ­different labels for these keys. EE

A

10

Chapter 1  Introduction

1-6  SI Unit Prefixes Root units have no prefixes. All of the SI base units except the kilogram are root units.

For engineering applications, we generally use a variation of scientific ­notation in which a prefix added to the root unit indicates the appropriate power of 10. Table 1-2 lists the SI unit prefixes. This system is often called engineering notation. SI recommends using prefixes that give the numerical values between 0.1 and 1000. For electrical quantities, we usually shift the decimal point in multiples of three places. So, we seldom use the prefixes hecto, deka, deci, and centi when working with electric circuits. The other unit prefixes shown in lightface type in Table 1-2 are used with quantities much larger or much smaller than we shall encounter in this book. TABLE 1-2 SI unit prefixes Prefix exa peta tera giga mega kilo hecto deka deci centi milli micro

The symbol for micro is the Greek letter μ (mu).

nano pico femto atto

Multiplication Factor

Letter Symbol

1 000 000 000 000 000 000 = 10

18

E

1 000 000 000 000 = 1012

P

1 000 000 000 000 000 = 1015 1 000 000 000 = 109

T G

1 000 000 = 106

M

2

10 = 10

h

1

0.1 = 10−1

da

0.001 = 10−3

0.000 001 = 10−6

m

0.000 000 000 000 001 = 10−15

p

1 000 = 103 100 = 10

0.01 = 10

−2

0.000 000 001 = 10−9

0.000 000 000 001 = 10

−12

0.000 000 000 000 000 001 = 10

−18

k

d c μ

n f a

Example 1-5 Express 839 000 m in engineering notation. Assume that the trailing zeros are not significant digits. Solution Shifting the decimal point three places to the left gives 839, which we must multiply by 1000 or 103 to restore its proper magnitude. Hence,

839 000 m = 839 × 103 m = 839 km

The kilometre (km) is 1000 times as large as the root unit, the metre.

1-6   SI Unit Prefixes

Example 1-6 If 200 cycles occur in 1.0 s, how long does one cycle take? Solution The first step is to get the answer in decimal form.

1.0 s = 0.0050 s 200 If we shift the decimal point three places to the right to obtain the figure 5.0, we must multiply by 0.001 or 10−3 to restore the original magnitude. From Table 1-2, we see that we can express the answer in milliseconds:

5.0 × 10−3 s = 5.0 ms

When we make calculations involving quantities expressed in units with prefixes, we must remember that the prefixes represent powers of 10. These powers of 10 must be included in the computation for the answer to have the proper magnitude. We can demonstrate this point by using unit prefixes for the quantities in Example 1-4.

Example 1-7

Given XL = 2πf L, f = 455 kHz, and L = 750 μH, calculate XL.

Incorrect Solution If we forget the prefixes and just carry out the arithmetic we get 2 × π × 455 × 750 = 2.14 × 106

This answer appears to be much larger than the answer in Example 1-4. By omitting the prefixes, we have calculated the answer in milliohms (thousandths of an ohm). Correct Solution To make sure that the answer is in root units, we include the unit prefixes in the calculation by entering the corresponding exponents:

XL = 2 ×

π

× 455

EE

3 × 750

↑ kilo

EE

+/−

6 = 2144 Ω

↑ ↑ micro root unit

Since it is not good form in SI to leave a numerical value greater than 1000, we now make the conversion 2144 Ω = 2.144 × 103 Ω = 2.14 kΩ

Note that we round to three significant digits to match the precision of the given measurements.

11

12

Chapter 1  Introduction

Example 1-8

A calculator display reads 7.50 × 10−4 H. Express this answer in engineering notation. Solution Table 1-2 has no unit prefix for 10−4. Hence, we must start by simultaneously shifting the decimal point and changing the power of 10 (the exponent) until it is a positive or negative multiple of 3 and the numerical quantity (the significand) is between 0.1 and 1000. In this particular example, we can shift in either direction. We can lower the exponent from −4 to −6 and compensate by shifting the decimal point two places to the right, or we can raise the exponent from −4 to −3 and shift the decimal point of the significand one place to the left: 7.50 × 10−4 H = 750 × 10−6 H = 750 μH

or 7.50 × 10−4 H = 0.750 × 10−3 H = 0.750 mH

Engineering calculators have an Eng key that automatically converts a scientific notation answer to display an exponent that is a ­multiple of 3 with a significand between 0.1 and 1000. Some scientific calculators have EE↑ and EE↓ keys that adjust the exponent while shifting the decimal point one place at a time. In this example we can press the EE↓ key twice to lower the exponent from −4 to −6, or we can press the EE↑ key once to raise the exponent from −4 to −3. See Problems 1-8 to 1-18 and Review Question 1-31.

1-7  Conversion of Units When only a single conversion from a root unit to a prefixed unit is ­involved, as in the last step of Example 1-7, we can usually convert the ­calculator readout mentally to find an appropriate prefix. But some chain calculations may require several conversions before the final numerical answer is displayed. To determine the correct unit for the final answer, we can apply dimensional analysis. With this technique, we write the units in an equation along with the numerical quantities, and multiply by the required conversion factors. We cancel out the units that appear in both a numerator and a denominator, and then ­collect the remaining units to establish the proper units, or dimensions, of the answer.

1-7   Conversion of Units

13

Example 1-9 Convert 0.250 h to seconds. Solution We require two conversion factors—hours to minutes and minutes to ­seconds. We can write the first as 60 min = 1 h



We can rearrange this conversion factor by dividing both sides of the equation by 1 h: 60 min 1h = =1 1h 1h Thus a conversion factor expressed as a ratio is a dimensionless number equal to unity. We can, therefore, multiply any quantity by 60 min/1 h, or its reciprocal, without altering the value of the original quantity. Similarly, we can also multiply by the conversion ratio for minutes to seconds: 0.250 h = 0.250 h ×

60 min 60 s × = 900 s 1h 1 min

We can also use dimensional analysis to convert from one system of units to another. Since electrical engineering in North America has already adopted the International System of Units (SI), all calculations in this book are made in SI units. Still, we occasionally encounter quantities in the obsolescent foot-pound system of units. To make the necessary conversion into SI units for examples in this book, we need two basic conversion factors: 1 inch = 2.54 cm    and    1 pound = 0.4536 kg



The conversion from inches to centimetres is exact. The conversion from pounds to kilograms is an approximation, accurate to four digits.

Example 1-10 An electric conductor is 25 ft long. Express its length in SI units. Solution

25 ft = 25 ft ×

12 in 2.54 cm 1m × × = 7.6 m 1 ft 1 in 100 cm

See Problems 1-19 to 1-25 and Review Questions 1-32 to 1-36.

The “Weights and Measures” table at the back of the book provides a list of common metric and U.S. equivalents when converting a variety of measuring units.

14

Chapter 1  Introduction

Circuit Check

B

CC 1-4. Express the following as quantities having a metric prefix. (a) 3 × 10−6 m (b) 2 million kelvins (c) 5.6 × 102 V (d) 256 × 1012 Hz (e) 55 × 10−9 A CC 1-5. The speed of sound in dry air at 20°C is 343.2 m/s. Express this speed in kilometres per hour. CC 1-6. Derive the conversion factor between metric tonnes (1000 kg) and imperial short tons (2000 pounds).

15

Problems

Summary

• Schematic diagrams use lines and standard graphic symbols to represent electric circuits. • The International System of Units (SI) is a unified set of measurement standards used worldwide. The key base units for electrical work are the metre for distance, the kilogram for mass, the ­second for time, the kelvin for temperature, and the ampere for current. • The values of quantities may be expressed with scientific notation or with unit prefixes (engineering notation). • The value of a quantity expressed in one unit may be converted to ­another unit using dimensional analysis.

Problems B

B

B

B

Section 1-4  Numerical Accuracy 1-1.

Assuming that the numbers represent measured quantities, express the solutions to the following computations with the appropriate number of significant digits. (a) 1.47 + 120.6 + 9.581 = (b) 190 − 52.33 + 4.3 = (c) 567 × 0.0050 = (d) 2242 ÷ 37 = (e) 73.2 + 42 × 9.78 = 73.2 − 42 (f) = 9.78

Section 1-5  Scientific Notation 1-2.

Express each of the following numbers in scientific notation. (a) 455 (b) 73 000 (c) 10 000 (d) 865 (e) 0.000 21 (f) 0.963 1-3. Express each of the following numbers in conventional form. (a) 1.4 × 106 (b) 4.97 × 102 (c) 6.28 × 10 (d) 3.45 × 10−3 (e) 7.7 × 10−6 (f) 8.672 × 10−1 1-4. The distance between two towns is 25 km. Express this distance in metres, using scientific notation.

B = beginner

I = intermediate

A = advanced

16

Chapter 1  Introduction

B B B

B

B

I

1-5.

A coulomb is equivalent to the amount of charge on 6 241 510 000 000 000 000 electrons. Write this number in scientific notation with four significant digits. 1-6. The mass of an electron is 0.000 000 000 000 000 000 000 000 000 899 9 g. Express this mass in scientific notation. 1-7. The mass of the earth is about 5.98 million million million million kilograms. Express this mass in scientific notation.

Section 1-6  SI Unit Prefixes

Express each of the following quantities in engineering notation. (a) 68 000 Ω (b) 45 000 000 Hz (c) 1500 W (d) 0.0505 s (e) 0.0008 V (f) 0.000 000 000 39 F 1-9. Express the following quantities with root units, first in scientific ­notation and then in words. (a) 5.6 MΩ (b) 75 mm (c) 0.218 kW (d) 37 μV (e) 37 km (f) 37 kg 1-10. Perform the following calculations using a calculator, first in the form given, then in scientific notation, and finally in engineering ­notation. (a) 0.005 × 47 000 − 958 = (b) 0.005(47 000 − 958) = (c) 54 000 − 75 × 634 = (d) 0.00054 = 423 000 + 17 400 = (e) 560 − 0.04 1-8.

B B B B B B B

1-11. 1-12. 1-13. 1-14. 1-15. 1-16. 1-17.

B

1-18.

(f) √ 1252 + 4 × 5600 = Express 2250 g in kilograms. Express 850 mm in metres. Express 1/60 of a second in milliseconds. Express 0.000067 km in millimetres. Express 2.2 × 10−2 kg in grams. Express 0.044 m2 in square centimetres. Calculate the cross-sectional area, in square millimetres, of a conductor having a diameter of 3.0 cm. Express 4.08 × 107 cm3 in cubic metres.

Review Questions

B I I I I I I

Section 1-7  Conversion of Units 1-19. A steak weighs 8 ounces. Using dimensional analysis, determine the steak’s mass in grams. 1-20. Derive the conversion factor between miles and kilometres. 1-21. Derive the conversion factor between square feet and square metres. 1-22. Derive the conversion factor between ounces and grams. 1-23. Express a velocity of 50 miles per hour in metres per second. 1-24. Express a velocity of 50 ft/s in kilometres per hour. 1-25. The No. 14 AWG (American Wire Gauge) conductors used in house wiring have a diameter of 64.08 mils. (A mil is one thousandth of an  inch.) What is the cross-sectional area of this wire in square ­millimetres?

Review Questions

Section 1-1  Circuit Diagrams 1-26. A flashlight is one of the simplest complete electric circuits. It consists of a battery, electric conductors, a switch, and a small lamp. Using standard schematic symbols, draw a circuit diagram for a flashlight.

Section 1-2  The International System of Units 1-27. What do the letters SI stand for? 1-28. What is the advantage of a system of units that are all derived from a few base units?

Section 1-5  Scientific Notation 1-29. What is meant by scientific notation for expressing quantities? What is the advantage of expressing quantities in this manner? 1-30. How does the scientific notation for expressing quantities differ from using SI unit prefixes?

Section 1-6  SI Unit Prefixes 1-31. Before the adoption of SI units, wavelengths of light were often expressed in millimicrons. A micron is one millionth of a metre. What SI unit is equal to a millimicron?

Section 1-7  Conversion of Units 1-32. Why is it important to write down the proper units when recording a calculator solution to a numerical example? 1-33. What is the meaning of the term dimensional analysis? 1-34. How would you apply dimensional analysis when you are doing a chain calculation with a calculator?

17

18

Chapter 1  Introduction

1-35. One butcher shop advertises sirloin steak at $5.99 a pound; another sells sirloin steak at $12.59 a kilogram. Assuming equal quality, which is the better price? 1-36. How would you convert the fuel consumption for a vehicle from miles per gallon to litres per 100 km?

Integrate the Concepts

The hydroelectric power plant at Bay d’Espoir in Newfoundland and Labrador generates 604 million watts. (a)   Express this power using scientific notation. (b)   Express this power using engineering notation.

(c)  Convert 604 million watts to horsepower (hp) given that 1 hp ≈ 746 W. Give the answer to the appropriate number of significant digits.

Practice Quiz

1. Express each of the following numbers in scientific notation: (a)  8263 (b)  0.9 (c)   0.000 070 (d)  0.2 × 105 1.0 (e)   2 000 000 2. Express the following quantities using metric prefixes. (a)   50 thousand metres (b)  1.666 × 105 Hz (c)  4.70 × 10−7 A (d)   0.000 800 V 3. Perform the indicated operations: (a)  46.7 μA = ______________ mA = _____________ A (b)   0.000 652 A = _______________ mA = _____________ μA (c)  825 × 103 m =_______________ km = _____________ Mm (d)  403 μs = _______________ s = _____________ ms (e)   0.050 ng = _______________ μg = ______________ pg (f)   25 mH = _______________ H = _____________ μH (g)   31.4 kHz = ______________ MHz = _____________ Hz (h)   176 mV = _______________V = _____________ μV

Practice Quiz

4. Perform the indicated operations: (a)  633 μA + 2.61 mA = _________________mA (b)   0.477 mA + 630 μA = ________________ mA 1 (c)   = __________ Ω = __________ kΩ 1 1 1 + + 150 kΩ 220 kΩ 1.0 MΩ 24 V (d)   = ___________________ V = ____________________kV 2.5 × 10−4 100 V (e)   = ___________________ mV = _________________ μV 1.2 × 106 (f)   68.0 km × 145 μm = _____________ m2

19

2

Current and Voltage In the electric circuit of Figure 1-1, a battery converts chemical energy into electric energy and a lamp converts that electric energy into light and heat. Energy conversions are the primary function of any electric circuit. The concepts of current and voltage are essential for understanding how ­circuits convert energy.

Chapter Outline 2-1

2-2

The Nature of Charge  22

Free Electrons in Metals  23

2-3

Electric Current  24

2-5

The Ampere  26

2-4 2-6 2-7

2-8 2-9

The Coulomb  26

Potential Difference  28

The Volt  31

EMF, Potential Difference, and Voltage  32

Conventional Current and Electron Flow  33

Key Terms protons  22 neutrons  22 electrostatic force  22 Coulomb’s law  22 free electrons  23 electric current  24 coulomb  26 ampere  26

electromotive force (EMF)  28 generator  28 dynamo  28 potential difference  30 potential rise  30 potential fall (potential drop)  30 work  31

joule  31 volt  31 voltage  33 voltage drop  33 conventional current  34

Learning Outcomes At the conclusion of this chapter, you will be able to: • describe the interactions between like and unlike electric charges • describe the behaviour of free electrons in an electric ­conductor • calculate the quantity of electric charge associated with a group of electrons • perform calculations using the relationship between charge, current, and time

Photo sources:   © iStock.com/IanChrisGraham

• describe potential difference in terms of potential rise and potential drop • perform calculations using the relationship between ­voltage, work, and charge • differentiate between a source voltage and a voltage drop • differentiate between conventional current and electron flow

22

Chapter 2   Current and Voltage

2-1  The Nature of Charge The American ­scientist and ­statesman Benjamin Franklin ­introduced the terms positive charge and negative charge.

An atom consists of a nucleus with electrons orbiting around it. The n ­ ucleus contains protons bonded with neutrons. Protons have a positive charge, while neutrons have no charge. An electron is much lighter and smaller than a proton, and has a negative charge with exactly the same magnitude as the positive charge on a proton. All electrons have the same charge. In fact, all electrons are identical, as are all protons. Normally, the number of electrons orbiting the nucleus of an atom is equal to the number of protons in the nucleus, making the atom electrically neutral. However, when two dissimilar materials are rubbed together, some electrons from the surface of one material can transfer to the other. The material that gains the electrons has a net negative charge, while the material that loses electrons has a net positive charge. Early researchers used such transfers to demonstrate that there are two kinds of charge. ­Simple experiments like the one shown in Figure 2-1 reveal a key property of electric charges. Like charges repel, unlike charges attract.

Silk threads

+

+ (a) Like charges

Light balls

+

−

(b) Unlike charges

Figure 2-1  Interactions of (a) like charges and (b) unlike charges

Such experiments also show that the electrostatic force acts on the charges without contact between the charges. Therefore, the electrostatic force is a field force, similar to the gravitational force. However, the electrostatic force can either attract or repel, while the gravitational force always attracts. By 1785, a French physicist, Charles Augustin de Coulomb, had shown that the force between two electrically charged bodies is directly proportional to the product of the magnitudes of the two charges and inversely proportional to the square of the distance between them. This relationship is called Coulomb’s law. We can express this law with an equation:

2-2   Free Electrons in Metals

F= k



Q 1Q 2 d2

(2-1)

where F is the force between the two charges, Q1 is the charge on the first body, Q2 is the charge on the second body, d is the distance between the charged bodies, and k is Coulomb’s constant, about 8.99 × 109 N⋅m2/C2. See Review Question 2-27 at the end of the chapter.

2-2  Free Electrons in Metals In the atoms of some metals, such as copper and silver, the configuration of electrons is such that the repulsion between the outermost electron and the other electrons largely offsets the attraction between it and the positive nucleus. The atoms of such metals form a crystal lattice. Within this ­lattice, the outermost electron can wind up midway between its parent atom and a neighboring atom. Since the attractive forces balance at this point, the electron can leave its atom to orbit the adjacent n ­ ucleus. The thermal energy of the atoms also contributes to freeing the outermost electrons. Thus, the lattice consists of positive ions in fixed positions with a cloud of free electrons that can drift from ion to ion. The lattice is held together by the attraction between the positive metal ions and the common cloud of free electrons. For example, a copper atom has 29 electrons. The first 28 are tightly bound to the nucleus, but only a very weak attraction holds the 29th electron in orbit. The copper atoms form a cubical lattice with a metal ion in each of the eight corners of the cube and also in the centre of each of the six faces of the cube. Figure 2-2 shows a cross section of such a lattice structure with the outline of one crystal unit sketched in. There are ­roughly 8.5 × 1022 free electrons in a cubic centimetre of copper. +

+

+ +

+

+

+

+

+

+

+ + Copper ions

+

+ +

+

+ +

+ +

+

+

+

+

Figure 2-2  Lattice structure of copper

See Review Questions 2-28 to 2-31.

+ +

+ +

23

+

Free electrons

In the units for Coulomb’s constant, N stands for newtons (the SI unit for force) and C stands for coulomb (the SI unit for charge), which is described in Section 2-4.

A positive ion is an atom or group of atoms that has lost one or more electrons, while a negative ion is an atom or group of atoms that has gained one or more electrons.

24

Chapter 2   Current and Voltage

2-3  Electric Current The moving charged particles in electric current will create a magnetic field, which is described in Chapter 14-2.

An electric current is a flow of charged particles. In most circuits these charged particles are free electrons. Although there is a free electron for each copper ion, Figure 2-3 shows only one of these free electrons so that we may trace its motion through the lattice. In Figure 2-3(a), this free electron moves randomly from atom to atom. The random motion of all of the free electrons in the conductor averages out, so there is no net flow of electrons in any direction. Figure 2-3(b) shows the same conductor when connected in an electric circuit that ­conveys energy to the free electrons. This external source of energy superimposes a net electron drift to the right on the random motion of the free electrons. +

+ + +

+ +

+ +

+ +

+

+

+

(a) Random motion

+ + +

+ +

+ +

+

+

+ +

+

+

(b) Electron drift due to an external energy source

Figure 2-3  Motion of a free electron in a copper conductor

The battery in Figure 2-4 adds electrons to the left end of the conductor and, at the same time, removes the same number of electrons from the right end of the conductor. The battery uses chemical energy to maintain a ­surplus of electrons (a net negative charge) at its negative terminal, and

−

−

−

+ Free electrons − −

+

Copper conductor

−

+

−

+

−

+

− Terminal

Terminal +

Battery

Figure 2-4  Motion of free electrons due to energy from a battery

2-3   Electric Current

a deficiency of electrons (a net positive charge) at its positive terminal. The free electrons in the conductor are repelled by the negative charge at the negative terminal of the battery and are attracted by the positive charge at the positive terminal. Since these electrons are free to move from atom to atom in the lattice of copper ions, the result is a net electron drift from left to right in the conductor. To determine the net movement of free electrons in an electric conductor, we can consider an imaginary plane cutting across the conductor at right angles to its length. Figure 2-5(a) shows the conductor with no external source of energy applied to it. In any given time, the number of electrons crossing the plane in one direction equals the number crossing in the opposite direction. There is no net electron drift, and, hence, no electric current. In Figure 2-5(b), an external energy source causes more electrons to cross the imaginary plane from left to right than from right to left. Consequently, there is now a net electron drift to the right. This net drift of charge carriers is an electric current.

(a) No net electron drift

(b) Net electron drift to the right

Figure 2-5  Net movement of electrons in a conductor

Electric current is the net flow of charge carriers past a given point in an electric circuit in a given period of time. In an electric circuit, the current from the energy source to the load is equal to the current from the load back to the energy source. See Review Questions 2-32 to 2-37.

Circuit Check CC 2-1. How is the force between two electric charges affected by doubling the distance between them? CC 2-2. What is the net current produced by the random motion of 1 C of free electrons in a copper wire?

A

25

26

Chapter 2   Current and Voltage

2-4  The Coulomb Since each electron possesses the same elemental quantity of charge, we may express the net electron drift across the imaginary plane in Figure 2-5 in terms either of the number of electrons or of the total charge possessed by this number of electrons. A practical unit for expressing quantity of electric charge must represent the charge carried by many billions of ­electrons. The coulomb (symbol C) is the SI unit of quantity of electric charge. A coulomb is the quantity of electric charge carried by 6.24 × 1018 electrons. The letter symbol for quantity of electric charge is Q, while the ­letter e represents the charge on one electron.

The coulomb is named in honour of Charles Augustin de Coulomb (1736–1806).

Example 2-1

How much charge is carried by 2.40 × 1019 free electrons?

In 2019, the definition of one elementary charge, e, will be set to exactly 1.602 176 634 × 10−19 C. The total electric charge carried in one coulomb is the reciprocal of this value.

The letter symbol for current, I, comes from the French word ­intensité, which was used to denote the rate of current flow. As a unit of measurement, the letter C ­represents coulombs. As a letter symbol, C is used to represent the quantity ­capacitance, which is ­described in Chapter 12.

Solution

Q=

2.40 × 1019e = 3.85 C 6.24 × 1018e/C

See Problems 2-1 and 2-2 and Review Question 2-38.

2-5  The Ampere Electric current is measured in terms of the rate of charge flow. The SI unit of electric current, the ampere, is named after a French pioneer of electrical physics, André-Marie Ampère (1775–1836). The ampere (symbol A) is the SI unit of electric current. One ampere is equal to a flow of one coulomb of charge per second: 1 A = 1 C/s. The letter symbol for electric current is I. The relationship between current and charge is

I=

Q t

(2-2)

where I is current in amperes, Q is charge in coulombs, and t is time in ­seconds.

2-5   The Ampere

Example 2-2 Find the current in an electric heater when 75 C of charge pass through the heater in half a minute. Solution

I=

Q 75 C = = 2.5 C/s = 2.5 A t 30 s ↑ ↑ ↑ ↑ Step 1 Step 2 Step 3 Step 4

When using a calculator, we tend to simply enter the data to get a numerical answer in a single step. However, a systematic approach is often helpful, especially for more complex calculations. Here is the sequence of steps for solving Example 2-2. Step 1 Note that the problem states two pieces of information and asks for one. To express this information in equation form, write the symbol for the unknown quantity on the left of the equals sign and the symbols for the given data in their proper relationship on the right of the equals sign. Step 2 Substitute the given data into the equation, making sure that powers of 10 or unit prefixes are included to preserve the proper magnitude. Step 3 Perform the numerical computation. Step 4 The calculator displays the magnitude of the answer in root units. Express the answer in appropriate units, using unit prefixes if necessary.

Example 2-3 How long will it take 4.0 mC of electric charge to pass through a fuse if the current is 50 A? Solution Since I =

Q , t Q 4.0 mC t= = = 8.0 × 10 − 5 s = 80 × 10 − 6 s = 80 μs I 50 C/s

See Problems 2-3 to 2-11 and Review Question 2-39.

27

28

Chapter 2   Current and Voltage

2-6  Potential Difference In Section 2-3, we noted that no current flows in a conductor unless a device (such as a battery) imparts energy to the free electrons. We say that the battery is the source of an electron-moving force or electromotive force, usually abbreviated to EMF. Electromotive force is a property that distinguishes an energy source from the rest of a circuit. To help understand the relationship between energy and flow, consider the operation of the hydroelectric generating station shown in Figure 2-6. Water enters a tunnel at the base of the dam, flows down through a turbine, and then discharges into the river below. In falling 100 m, the water loses some of its gravitational potential energy while gaining kinetic energy. The turbine transfers some of this kinetic energy to the generator or dynamo, which converts most of it into electric energy.

g

Dam

Generator

100 m

Turbine g

Figure 2-6  Simplified cross section of a hydroelectric generating station

Since objects at the surface of the earth are 6400 km away from the earth’s centre of gravity, the difference in gravitational force acting on a cubic metre of water above and below the generating station is negligible. But there is an  appreciable difference in the potential energy of a cubic

2-6   Potential Difference

29

Source:  © iStock.com/IanChrisGraham

Seven Sisters Generating Station, the largest producer of electricity on the Winnipeg River, in Manitoba

metre of water above and below the station. The law of conservation of energy requires the difference in potential energy between a unit quantity of water above and below the generating station to be equal to the energy expended in raising the unit quantity of water the 100 m against the force of gravity. Figure 2-7 shows the electric circuit of Figure 2-4 with the battery turned on its side to parallel the water flow in Figure 2-6. Battery

−

−

−

+

−

− Conductor

Electrons − gain potential energy as they move away from + charge and toward − charge

+

−

− Electrons lose potential energy as they flow through conductor

Potential difference

+

+

+

+

Figure 2-7  Electric potential difference

30

Chapter 2   Current and Voltage

The free electrons in the conductor flow away from the negative ter­ minal of the battery and toward the positive terminal. To maintain the negative and positive charges at the two battery terminals, an equivalent number of electrons must move inside the battery from the ­positive ­terminal to the ­negative terminal. These electrons move away from the positive terminal and toward the negative terminal. So, the electrons inside the ­battery move against the electric forces acting on them, just as water moves against gravitational force when it is pumped uphill. The electrons acquire potential e­ nergy at the expense of the chemical energy stored in the battery. Consequently, an electron moving inside the battery has a greater potential energy when it arrives at the negative terminal than when it leaves the positive terminal. There is an electric potential ­difference ­between the negative and positive terminals of the battery, with electrons at the negative t­ erminal being at a higher potential than those at the positive terminal. Under the influence of gravity, the water in the hydroelectric generating station of Figure 2-6 always tends to fall to a lower potential energy level. Similarly, electrons at the negative terminal of an energy source tend to “fall” to a lower potential energy level. They can move to a lower potential via the external conductor connected between the battery terminals. When flowing from the negative terminal to the positive terminal through the ­external circuit, electrons lose as much potential energy as they gained in being moved inside the battery from the positive terminal to the negative terminal. This energy “lost” in the external circuit is converted into light, heat, or some other form of energy, depending on the nature of the circuit. In travelling the complete circuit around the closed loop of Figure 2-7, electrons experience a potential rise within the battery and a matching ­potential fall or potential drop in the circuit. The electric potential difference between any two points in a c­ ircuit is the rise or fall in potential energy involved in moving a unit quantity of charge from one point to the other. The letter symbol for potential difference is E or V. EMF is the energy per unit of charge that a battery or other energy source converts in the process of creating a potential difference between its terminals. Since a force has dimensions of mass times acceleration, EMF is not ­actually a force. Electromotive force is a traditional term that is gradually ­disappearing from common use. See Review Questions 2-40 to 2-42.

2-7   The Volt

31

2-7  The Volt Work is energy transferred to a body or system. For example, the force of gravity does work on a falling body. This work increases the kinetic energy of the body and decreases its potential energy. The letter symbol for work and energy is W. The joule (symbol J) is the SI unit of work and energy. One joule is equal to one newton metre: 1 J = 1 N⋅m = 1 kg⋅m2/s2. When a force moves a body, the work, W, done is equal to the magnitude, F, of the force times the distance, d, the body moves: W = Fd



(2-3)

We can express electric potential difference in terms of joules per coulomb, or volts. The volt (symbol V) is the SI unit of potential difference. The potential difference between two points is one volt if one coulomb of charge gains or loses one joule of energy when moving from one point to the other: 1 V = 1 J/C. This equation relates potential difference to charge and energy: E=



W Q

or

V=

W Q

(2-4)

where E (or V) is potential difference in volts, W is energy in joules, and Q is charge in coulombs. Note that EMF is measured in volts, while a force is measured in newtons.

Example 2-4 A power supply delivers 55 J when 50 C of electrons move from its negative terminal to its positive terminal. Find the potential difference between the terminals. Solution

E=

The joule is named in honour of the ­English physicist James Prescott Joule (1818–89).

55 J W = = 1.1 V Q 50 C

The volt is named in honour of the Italian physicist Alessandro Volta (1745–1827), who invented the first electric battery.

32

Chapter 2   Current and Voltage

Example 2-5 A current of 0.30 A flowing through the filament of a cathode-ray tube ­produces 9.45 J of heat in 5.0 s. What is the potential difference across the filament? Solution Since I = Q/t,

Q = It = 0.30 A × 5.0 s = 1.5 C V=



9.45 J W = = 6.3 V Q 1.5 C

See Problems 2-12 to 2-26 and Review Question 2-43.

2-8 EMF, Potential Difference, and Voltage At the beginning of Section 2-6, we used the term electromotive force to ­describe the property that an energy source must have to force charge ­carriers to flow in a circuit. EMF is the energy converted per unit quantity of electric charge moved from one terminal to the other inside the source. However, the EMF is equal to the potential rise between the terminals of the source when the circuit is open and no current flows (I = 0). When current flows in the circuit, inefficiencies in the energy-conversion process make the potential rise less than the internal EMF of the source. In the circuit shown in Figure 2-8, the left voltmeter measures the potential difference between the battery terminals. Even with the switch open, this potential difference is not the same as the EMF of the battery since the voltmeter itself draws some current from the battery.

Switch Battery

+

E

V

Lamp

− Voltmeters

Figure 2-8  Measuring source voltage and voltage drop in a basic electric circuit

For many circuit calculations, the energy source is assumed to be perfectly efficient, making its internal EMF and the potential difference between its terminals equal. Nonetheless, we should not use the term EMF to refer to the potential rise between the terminals of a generating device.

2-9   Conventional Current and Electron Flow

The term voltage is now commonly used as a simpler expression for p­ otential difference. We can maintain the distinction between rise and fall in potential by using the terms source voltage or applied voltage for potential rise between the terminals of a source and the term voltage drop for the ­potential fall across the circuit load. The letter symbol for source voltage or applied voltage is E. The letter symbol for voltage drop is V. The difference between a source voltage and a voltage drop is illustrated in Figure 2-8. When the switch is closed, both voltmeters show the same reading since the current through the lamp creates a voltage drop across the lamp equal to the source voltage (or applied voltage). When the switch is open, the voltmeter connected to the battery terminals still registers the source voltage, but the voltmeter connected to the lamp terminals reads zero. A voltage drop can appear across the lamp only when electrons are flowing through it. A voltage or potential difference must be measured between two points, from one point with respect to another, or across a circuit element. There is no such thing as a “voltage at a point.” On the other hand, we speak of current in or through a conductor or other component. The term amperage is sometimes used by electricians for the current normally carried by an electric device. Electrical codes often use ampacity to mean the maximum allowable current for a given size of conductor. See Review Questions 2-44 to 2-47.

Circuit Check

B

CC 2-3. How is the ampere related to the coulomb? CC 2-4. Consider a 4.0-V lithium-ion battery that can maintain a current of 1.0 A for 7.0 h. Assuming that the voltage of the battery remains constant while it discharges, determine the energy produced by the battery while supplying this current. CC 2-5. Calculate the work done when 15 C of charge moves between the terminals of a 12-V battery.

2-9 Conventional Current and Electron Flow The most common form of electric conductor is the metallic conductor ­described in Section 2-3. The charge carriers in metallic conductors are free electrons that flow from the negative terminal of the voltage source t­ oward the positive terminal. However, as we shall discover in Chapters 3 and 4, electric current can also result from the flow of positive and negative ions.

33

34

Chapter 2   Current and Voltage

For calculating the algebraic sum of the currents at a junction in a circuit, we need to define the direction of current flow in a way that we can apply to all types of charge carriers. Since most circuit conductors are metal, it would be logical to define current direction in terms of electron flow. ­However, experimenters established the convention for current direction long before the electron was discovered. Since Michael Faraday (1791–1867) showed that metals transfer from the positive terminal to the negative t­erminal in electrolytic cells, the direction of conventional current was taken to be from positive to negative. Positive charge carriers flow in the ­opposite direction to negative charge carriers such as free electrons and ­negative ions. Therefore, the laws and calculations for electric circuits were based on  current flowing from the positive terminal of the source, through the circuit, and back to the negative terminal. So, we need to keep in mind that conventional current is simply a mathematical convention for describing current and that electrons flow in the opposite direction. The circuit diagrams in this text show the direction of conventional c­ urrent. Some textbooks now define current as electron flow, and show currents leaving the negative terminal of a voltage source and flowing into its p ­ ositive terminal. However, the equations and calculations are the same as for conventional current. In fact, some authors of electrical and ­electronics textbooks produce two versions, one using conventional ­current and the other using electron flow. The main difference between the two versions is the direction of the current arrow in the schematic ­diagrams.

Problems

35

Summary

• Like charges repel each other and unlike charges attract each other. • The electrical properties of atoms depend on the configuration of negatively charged electrons that orbit positively charged nuclei. • The weak bonding of the outermost electron in the atoms of some ­metals produces free electrons that allow the metal to conduct current readily. • Electric current is produced when free electrons move through a conductor in response to the application of an electromotive force. • Electromotive force is an internal property of a voltage source, and it causes a potential difference to form across the source’s terminals. • An electric circuit is the configuration of a continuous path of conductors or devices that connect one terminal of a voltage source to its other terminal, creating a loop for current to flow. • Current (I) is the rate of charge flowing past one point in a circuit, and is measured in amperes (A). • Voltage (E or V) is the electric potential difference between two points in a circuit, and is measured in volts (V). E is a voltage source or applied voltage; V is a voltage drop. • Conventional current travels from positive to negative. Electron flow travels from negative to positive.

Problems B B

B B B B B B B

Section 2-4  The Coulomb 2-1. 2-2.

How many electrons must accumulate to produce a charge of 3.5 C? How much charge is carried by a trillion free electrons?

Section 2-5  The Ampere 2-3. It takes 7.5 s for 30 C of electric charge to pass a switch. Find the ­current through this switch. 2-4. In 1 min, 1 C of charge carriers arrive at the collector (output terminal) of a transistor. Find the current through the transistor. 2-5. If the net drift of electrons across the imaginary plane in Figure 2-5(b) is 1014 electrons per second, what is the current in amperes? 2-6. Express an electric current of 4.0 mC/min in appropriate SI units. 2-7. With the switch closed in Figure 2-8, what is the current through the lamp if 18 C passes through it in 1.0 min? 2-8. With the switch closed in Figure 2-8, how many coulombs of charge flow through each of the terminals of the battery in 15 ms if the current drain is a steady 0.50 A? 2-9. With the switch closed in Figure 2-8, how long will it take for 9.0 C of electric charge to flow through the switch if the current is 750 mA?

B = beginner

I = intermediate

A = advanced

36

Chapter 2   Current and Voltage

B B

B B B B B B B B B B B B B B B

2-10. How much charge has passed through a conductor if 20 A flows for 30 min? 2-11. How much current is flowing if 5.5 × 1020 electrons pass a point in half a minute?

Section 2-7  The Volt 2-12. If it takes 5 J of chemical energy to move 20 C of electric charge ­between the positive and negative terminals of a battery, find the ­potential difference between its terminals. 2-13. If a flow of 40 mC from one terminal of a lamp to the other releases 4.8 J of energy, determine the potential difference developed across the terminals. 2-14. What electric charge moves through a load that releases 54 J of energy and creates a potential drop of 120 V across its terminals? 2-15. If the voltage between the terminals of a battery is 1.5 V, how many electrons move from the positive terminal to the negative terminal during the conversion of 300 mJ of chemical energy? 2-16. How much energy is required to move 5 mC of electric charge through a potential rise of 60 V? 2-17. Determine the charge moving through a lamp that converts 2.1 J of electric energy into light and heat and creates a potential drop of 3.0 V between its terminals. 2-18. How much energy is required to maintain a 0.5-A current through a battery for half a minute if the potential difference between the battery terminals is 1.1 V? 2-19. How long does a 225-mA current have to flow in order to transfer 5.4  J of energy to a flashlight lamp with a 3.0-V drop between its ­terminals? 2-20. Find the applied voltage of a battery in a transistor radio if a steady current of 24 mA transfers 13 J of energy to the radio circuit in 1.0 min. 2-21. If a fully charged 12-V automobile battery can produce 3.5 × 106 J of electric energy, how long, in hours, can this battery maintain a current of 2 A? 2-22. Find the current through a load with a 120-V drop across its terminals when energy is delivered to the load at a rate of 18 J/s. 2-23. Assuming perfect efficiency, at what rate is chemical energy being converted in a battery with a source voltage of 3.0 V while a lamp connected to the battery draws a steady 120-mA current? 2-24. What is the voltage between the terminals of a 15-A electric heater that runs for 10 min and produces 1.98 MJ of heat? 2-25. How much current is flowing if 50 J of energy is transferred in 40 s at a potential of 30 V? 2-26. What is the potential difference when 1.25 J of energy is developed by a flow of 8.2 × 1017 electrons?

Review Questions

Review Questions

Section 2-1  The Nature of Charge 2-27. What effect would be observed if both balls in Figure 2-1 had a negative charge?

Section 2-2  Free Electrons in Metals 2-28. 2-29. 2-30. 2-31.

Distinguish among an atom, a positive ion, and a negative ion. Why is silver an excellent conductor of electricity? Define the term free electron. How are free electrons related to the ability of metals to act as ­electric conductors?

Section 2-3  Electric Current 2-32. What causes the random motion of free electrons shown in Figure 2-3(a)? 2-33. Why is the term drift appropriate for describing electric current in a metallic conductor? 2-34. Explain the forces acting on a free electron at the centre of the ­conductor shown in Figure 2-4. 2-35. Figure 2-5(b) shows the motion of electrons in current-carrying conductors. Why are some of the electrons moving in opposite ­ ­directions? 2-36. Describe the motion of free electrons in the conductors of Figure 2-8 when the switch is open. 2-37. What change in electron motion occurs in the circuit of Figure 2-8 when the switch is closed?

Section 2-4  The Coulomb 2-38. Describe the numerical relationship between the charge of a single electron and the unit of electric charge, the coulomb.

Section 2-5  The Ampere 2-39. What quantity could be measured in ampere seconds?

Section 2-6  Potential Difference 2-40. The water in Figure 2-6 loses potential energy as it flows through the turbine of the generating station. How is the energy replenished to maintain the operation of the station over the years? 2-41. In the hydroelectric generating station of Figure 2-6, gravity c­ reates a potential difference between the top of the intake pipe and the end of the discharge pipe below the turbine. What is the equivalent force that creates a potential difference in an electric ­circuit?

37

38

Chapter 2   Current and Voltage

2-42. Both gravitational force and electric force are measured in newtons. Explain why electromotive force cannot be measured in the same units.

Section 2-7  The Volt 2-43. Explain the relationship among the quantities in the equation V = W/Q.

Section 2-8  EMF, Potential Difference, and Voltage 2-44. Although charge carriers flow from the load back to the source in one of the conductors of an electric circuit, energy is not transferred from the load back to the source. Explain why not. 2-45. Explain what happens to the chemical reactions in the battery in ­Figure 2-8 when the switch is opened. 2-46. Why is the letter symbol E used in Example 2-4 and the letter symbol V in Example 2-5? 2-47. If we connect a voltmeter across the open switch terminals in the ­circuit of Figure 2-8, it will show the same reading as the voltmeter connected across the battery. Explain why.

Integrate the Concepts

A length of copper wire contains 1.5 × 1020 free electrons.

(a)   Find the total charge carried by these free electrons. (b)  Calculate the current when 1.5 × 1020 electrons flow through the wire in 5.0 s. (c)  Calculate the potential difference across the wire, given that it takes 48 J of energy to produce the current in part (b).

Practice Quiz

1. Which of the following statements is true? (a) The force between two electrically charged bodies is inversely proportional to the product of the magnitudes of the two charges and directly proportional to the square of the distance between them. (b) One ampere is equal to a flow of one coulomb per minute. (c) Voltage is a measure of the difference in electric potential between two points. 2. What amount of charge is carried by 10 mA of current in 0.5 s? 3. Determine the equivalent current for a flow of (a) 25 C in 1.0 s    (b) 30 C in 10 s

Practice Quiz

4. If current flowing through a conductor is 10 mA, how many electrons travel through it in 10 s? 5. How long will it take a 6.0-A fuse to safely transfer 4.0 × 1020 electrons between its two terminals? 6. What applied voltage will allow 17 C of charge to pass through a TV remote control while it consumes 51 J of energy? 7.

If the difference of potential between the two terminals of a device is 65 V, how much energy is used to move 10 μC from one terminal to the other?

8.

Determine the current flowing through a lamp if 10.8 kJ of energy is transferred in 1 min at a potential of 120 V.

39

3

Conductors, Insulators, and Semiconductors As we noted in Chapter 2, we can proceed with numerical analysis of electric circuits without reference to the physical process of electrical conduction. However, to develop a ­better understanding of the behaviour of circuit components, we shall pause in our numerical analysis to consider qualitatively the electrical properties of several important non-metals.

Chapter Outline 3-1

Conductors

3-3

Insulators

3-5

Semiconductors

3-2

3-4

42

Electrolytic Conduction 45

Insulator Breakdown

47

46

43

Key Terms printed circuit boards  42 breadboard 43 electrolytic cell  43 electrode 43

electrolyte 43 anode 43 cathode 44 electrolytic conduction  44

insulator 45 semiconductors 47

Learning Outcomes At the conclusion of this chapter, you will be able to: • differentiate between conductors and insulators • explain how molecular bonds and crystal ­structure d ­ etermine whether a material is a ­conductor, an i­nsulator, or a semiconductor

Photo sources:  © iStock.com/eldadcarin

• explain the process of electrolytic conduction • explain how a high potential difference can cause i­nsulator breakdown

42

Chapter 3   Conductors, Insulators, and Semiconductors

3-1 Conductors As discussed in Section 2-3, electric current is a net drift or flow of charged particles. It follows that the greater the number of charge carriers per unit volume, the better a material will conduct electricity. All metals can be used as electric conductors, although some metals have more free electrons per unit volume than others. The material most commonly used for conductors is copper. It is reasonably inexpensive, stable, and easy to shape. Copper has only 5% fewer free electrons per unit volume than silver, the most conductive element. Despite having only 60% as many free electrons per unit volume as copper, aluminum is used for some applications because it is much lighter and cheaper than either copper or silver. Some of the metal alloys used as heater elements have less than 1% of the number of free electrons per unit volume that copper has. The larger the diameter of a wire, the more current it can carry without overheating. For electronics, printed circuit boards have largely replaced wiring with individual conductors. A printed circuit board is a sheet of rigid insulating material coated with a film of copper. The circuit layout is printed onto the copper surface with a special ink, then any copper not covered by the ink is etched away with acid. Alternatively, the unwanted copper can be cut away by a computerized machine tool. The remaining copper is a n ­ etwork

Source:  © iStock.com/eldadcarin

A printed circuit board

3-2   Electrolytic Conduction

43

of traces, conducting paths that link the points where components are ­soldered to the circuit board. Students and electronic hobbyists have been experimenting with electric circuits for over 100 years. Since the 1970s, temporary circuit designs have typically been constructed on a solderless breadboard. Breadboarding allows a circuit prototype to be built relatively easily using actual throughhole components (which have long metal leads that are pushed through the holes of breadboards or printed circuit boards), and permits the user to observe and record electrical measurements around the circuit. Many issues can be addressed during the prototyping phase of a project, even before starting the design of a printed circuit board. Breadboards still appear to be popular among electronics learners but the through-hole component industry has rapidly declined in recent years. Manufacturers are instead replacing their stock with surface-mount technology (which uses short flat pins, rather than long metal leads, and is designed for use with printed circuit boards), offering improved features in smaller, less expensive packaging. In the near future, learners will experiment with different circuit designs exclusively through the use of powerful simulation tools, rather than breadboards.

The term breadboard emerged about 100 years ago, when enthusiasts built circuits on actual wooden boards, and the connections between the components were soldered to nails or thumbtacks pressed into the board.

Source:  Karen Craigs

Removing the protective backing of a breadboard reveals its hidden pattern of electrical connections.

See Review Questions 3-1 and 3-2 at the end of the chapter.

3-2  Electrolytic Conduction Free electrons are not the only charged particles that can act as charge ­carriers in an electric circuit. In solutions and in gases, current can result from a flow of ions. Figure 3-1 shows an electrolytic cell or electroplating bath, consisting of a silver electrode and an iron electrode immersed in a solution of silver ­nitrate (AgNO3). This current-carrying solution is an electrolyte. The electrode connected to the positive terminal of the battery is an anode, and the

Faraday originated the terms electrode, electrolyte, and ­electrolysis.

44

Chapter 3   Conductors, Insulators, and Semiconductors

electrode connected to the negative terminal of the battery is a cathode. Note that the symbol used for the battery can represent a cell or battery with any voltage. Battery +

Silver anode

Ag+

Electron flow

−

Electron flow

Cathode (steel spoon) NO3−

Silver nitrate solution Figure 3-1  Simple electrolytic cell

Svante Arrhenius was the first scientist to calculate a model of global warming, in 1896, due to the greenhouse effect. In 1903, he received the Nobel Prize for Chemistry, in recognition for his work with electrolytic dissociation.

From experiments with similar electrolytic cells in the 1830s, Michael Faraday reasoned that current must be flowing through the silver nitrate ­solution since he could measure an electric current in the copper conductors. Faraday also observed that, as current flowed through the cell, silver disappeared from the anode and the same amount of silver was deposited on the cathode. In 1887 the Swedish scientist Svante August Arrhenius (1859–1927) ­explained the nature of electrolytic conduction. Silver nitrate is an ionic compound consisting of a positive silver ion bonded to a negative nitrate ion. Arrhenius suggested that these ions tend to dissociate when the c­ ompound dissolves in water. The silver and nitrate ions in the electrolyte can then move independently. In the cell shown in Figure 3-1, the battery connected to the electrodes creates a potential difference between them. The positive silver ions are ­attracted to the negatively charged cathode. When a silver ion touches the cathode, an electron transfers from it to the ion and the resulting ­neutral atom is deposited on the cathode. At the same time, an equal number of negative nitrate ions are attracted to the positively charged anode. On reaching the anode, a nitrate ion gives up its surplus electron to the anode and takes an atom of silver from the anode into the electrolyte solution, thus maintaining the strength of the electrolyte. Since equal amounts of s­ ilver go into the solution at the anode and deposit onto the cathode, no chemical energy is used in the cell. The energy required to transport the ­silver from the anode to the cathode comes from the ­battery.

3-3  Insulators

Although the current in the copper conductors connecting the battery to the electrolytic cell is a unidirectional flow of free electrons, the ­current in the electrolyte results from positive and negative ions flowing simultaneously in opposite directions. However, Faraday had no way to observe these ions. All he could measure was the transfer of silver from the anode to the cathode. Since it was thought that the electric current carried the ­silver, the direction of electric current was assumed to be from the anode to the cathode. As noted in Section 2-9, positive to ­negative then became ­established as the conventional direction for electric ­current. See Review Questions 3-3 to 3-5.

3-3 Insulators A material that does not conduct electricity is an electric insulator. Dry air is an insulator since it consists mainly of oxygen molecules and nitrogen molecules, in which all the orbital electrons are firmly bound. Free ­electrons cannot readily drift from atom to atom in dry air the way they can in a metal conductor, and current will not leak from household circuits to the surrounding air. Even high-voltage transmission lines do not have to be ­insulated from the surrounding air, but they do need long glass or p ­ orcelain spacers to prevent current from leaking into the steel support towers. Some of the most effective insulators are polymers, compounds that have identical groups of atoms joined into long chains by covalent bonds in which electrons are shared between adjacent atoms. In some polymers, these covalent bonds hold the electrons so tightly that there is virtually no electron drift. Figure 3-2 shows the structure of polyethylene. Each carbon atom has six electrons, of which the inner two are tightly bonded to the carbon nucleus. Two of the outer electrons are shared with hydrogen atoms, ­ which also ­supply an electron each to form strong covalent bonds with the c­ arbon atom. The other two electrons are shared with the two adjacent ­carbon atoms, forming two more covalent bonds. This form of covalent bonding creates long, filament-like molecules that are excellent electric insulators. Many plastics and some types of synthetic rubber and varnish are polymers. Totally pure polyethylene would have no free electrons, making it an ideal insulator. However, mass-manufactured polymers usually contain some impurities with a few free electrons. For example, polystyrene used in electronic components typically has roughly 6 × 1010 free electrons per cubic centimetre. Although this number seems large at first glance, it is ­extremely small compared with the number of free electrons found in copper.

45

46

Chapter 3   Conductors, Insulators, and Semiconductors

H

H

H

C

C

C

H

H

H

Electron

H Hydrogen nucleus

C Carbon nucleus

Figure 3-2  Structure of a polyethylene molecule

For practical purposes, the leakage of the charge carriers through a layer of polystyrene is negligible. See Review Questions 3-6 and 3-7.

3-4  Insulator Breakdown Figure 3-3 shows a voltage source connected between the two conductors of a lamp cord. The potential difference gives rise to an electric force that acts on the electrons in the molecular bonds of the insulation between the conductors. If the potential difference is great enough, the force will produce free electrons by breaking some of the insulator’s molecular bonds, resulting in the rupture or breakdown of the insulator. The potential difference required to rupture a particular insulator will depend on the spacing between the two conductors and the strength of the bonds in the insulating material. Some insulating materials can withstand a much greater potential difference for a given thickness than others (see Table 12-1). + F

+ −

Electron in molecular bond

Parallel conductors Insulating sheath

− Figure 3-3  Electric stress on molecular bonds in an insulator

See Review Question 3-8.

3-5  Semiconductors

3-5 Semiconductors Column IV of the periodic table of elements includes carbon, silicon, and germanium. These elements each have four outer electrons that can form covalent bonds. These elements tend to form a crystal structure in which each atom shares an electron with each of four adjacent atoms. The resulting covalent bonds are not as strong as those in polymers such as polyethylene or polystyrene. At room temperature, crystals of carbon, silicon, and germanium have too few free electrons to be good conductors and too many to be good insulators. Materials that conduct more than insulators but less than conductors are called semiconductors. Crystals of highly purified silicon are the basis of most diodes and transistors (see Figure 3-4).

Si

Si

Si

Si

Si

Si

Si

Si

Si

Si

Si

Si

Si Silicon nucleus with 10 inner electrons Outer electron Figure 3-4  Covalent bonds in a crystal of pure silicon

See Review Question 3-9.

Circuit Check CC 3-1.

List an advantage and a drawback of aluminum as an electric conductor. CC 3-2. How does the flow of charge in electrolytic conduction differ from that in metals? CC 3-3. What factor determines whether a material is a conductor, a semiconductor, or an insulator?

A

47

48

Chapter 3   Conductors, Insulators, and Semiconductors

Summary

• Metals such as copper have a relatively large number of free electrons and therefore make excellent conductors in electric circuits. • If a voltage source is connected to the electrodes of an electrolytic cell, the movement of positive and negative ions through the electrolyte results in a current flow. • Insulators (such as polymers) have very few free electrons. • A high potential difference across an insulator can cause it to break down. • Semiconductors have more free electrons per unit volume than insulators have but fewer than conductors have.

Review Questions Section 3-1  Conductors 3-1. 3.2

Calculate the total charge on the free electrons in a cubic centimetre of copper. (Hint: Refer to Section 2-2.) Explain why printed circuit boards are not constructed with a rigid metal base.

Section 3-2  Electrolytic Conductors 3-3. Explain how the silver anode in the electrolytic cell shown in ­Figure 3-1 obtains the positive charge required to attract nitrate ions. 3-4. In a solution of an ionic compound, why is it important for the ions to dissociate before electric current is able to flow? 3-5. Describe the major difference between current flow in an electrolyte and in a metallic conductor.

Section 3-3  Insulators 3-6.

3.7.

Both copper and sodium chloride (common table salt) are crystalline solids. Explain why copper is a good conductor of electricity while sodium chloride is an insulator. Calculate the total charge on the free electrons in a cubic centimetre of polystyrene.

Section 3-4  Insulator Breakdown 3-8.

How do the properties of an insulator change when molecular bonds within the material are broken by a large potential difference?

Section 3-5  Semiconductors 3-9. Why are germanium and silicon neither good conductors nor good insulators?

Practice Quiz

Integrate the Concepts (a) Why is copper the most commonly used conductor in electric circuits? (b) Why does a current flow when a battery is connected to electrodes in an electrolyte? (c) What purpose do insulators serve in electric circuits?

Practice Quiz 1. Rank the following materials from most conductive to least conductive: aluminum, carbon, copper, glass, silver. 2. Why are polymers often excellent insulators? 3. What factors will affect the potential difference required to cause a breakdown in the insulating sheath shown in Figure 3-3? 4. What properties do the bonds in crystals of silicon and germanium have in common?

49

4

Cells, Batteries, and Other Voltage Sources A cell or battery is a chemical source of electric energy. We use these devices to power portable appliances such as laptop computers, cell phones, calculators, and TV remote controllers. Cells and batteries come in a wide range of sizes, types, and voltages, from miniature button cells for hearing aids and watches to large rechargeable batteries for trucks and ships.

Chapter Outline 4-1

Basic Terminology  52

4-3

Carbon-Zinc and Alkaline Cells  53

4-2

Simple Primary Cell  52

4-4

Other Commercial Primary Cells  55

4-6

Capacity of Cells and Batteries  58

4-5 4-7

4-8

Secondary Cells  56

Fuel Cells  60

Other Voltage Sources  60

Key Terms cell 52 battery 52 primary cell  52 secondary cell  52 cathode 52 anode 52 dry cells  53 carbon-zinc cell  53 polarization 53 local action  53 manganese-alkaline cells (alkaline cells)  53 zinc-chloride cell  55

heavy-duty cell  55 silver-oxide cell  55 zinc-air cell  55 lithium cell  55 lead-acid cell  57 nickel-cadmium (ni-cad) cell 57 nickel-metalhydride (ni-mh) cells 57 lithium-ion (li-ion) cell  57 rechargeable alkalinemanganese (RAM) cell  57 capacity  58

ampere-hour rating  58 fuel cell  60 hydroelectricity 61 steam turbine  62 gas turbines  62 wind turbines  62 photovoltaic (solar) cell  62 thermocouple 63 piezoelectricity 64 static electricity  64

Learning Outcomes At the conclusion of this chapter, you will be able to: • differentiate between primary and secondary cells • explain the basic construction and chemical action of a simple primary cell • describe the limitations of simple primary cells • describe the basic construction of commercial primary cells • describe the chemical actions, advantages, and ­disadvantages of various types of primary cells

Photo sources:  © iStock.com/hohl

• describe the basic construction of a leadacid secondary cell • list advantages and disadvantages of different types of secondary cells • compare the capacity of various types of cells • compare fuel cells to conventional batteries • discuss how various voltage sources convert mechanical energy, light, and heat into electric energy

52

Chapter 4   Cells, Batteries, and Other Voltage Sources

4-1  Basic Terminology Although not technically correct, most people, and even some manufacturers, refer to single cells as “batteries.”

A cell is a single unit for converting chemical energy into electric energy. The 1.5-V, AAA, AA, C, and D cells that we use every day are examples of cells. A battery is a set of interconnected cells used to provide higher voltage and/or current than is available from a single cell. For example, the 9-V batteries used in devices such as smoke detectors and the 12-V batteries in cars both contain six cells. In a primary cell the chemical reactions that produce electrical energy are irreversible, and the cell cannot be recharged. A secondary cell is rechargeable. A power supply connected to the secondary cell uses electric energy to restore the chemical energy in the cell. See Review Question 4-3 at the end of the chapter.

4-2  Simple Primary Cell A cathode is the ­terminal where ­conventional current leaves a device. ­Therefore, the cathode of a primary cell is its positive ­terminal while the cathode of an ­electrolytic cell or any other load is the ­negative terminal. Similarly, the anode of a voltage source is its negative terminal while the anode of a load is its positive ­terminal.

In its simplest form, a primary cell consists of two different metal electrodes immersed in an acidic liquid electrolyte. In solution, the acid molecules separate into positive and negative ions. For example, in the cell shown in Figure 4-1, sulfuric acid splits into two positive hydrogen ions and a negative sulfate ion: H2SO4 → 2H+ + SO42−



When the zinc electrode is immersed in the acid electrolyte, sulfate ions combine with the zinc to form zinc sulfate (ZnSO4), which precipitates into the electrolyte. The excess electrons from the sulfate ions remain on the zinc electrode, making it negative. Zn + SO42− → ZnSO4 + 2e



Load

Copper rod +

− Zinc rod

H2

+ +

− Dilute sulfuric acid

−

+ +

−

ZnSO4

−

Figure 4-1  A simple primary cell, the copper-zinc cell

4-3   Carbon-Zinc and Alkaline Cells

At the copper electrode, the hydrogen ions take electrons from the copper to form neutral hydrogen gas molecules (H2). This reaction leaves the copper electrode with a deficiency of electrons, making it positive.

2H+ + 2e → H2 ( gas )

These reactions slow as charge builds up on the electrodes, and cease when the negative charge on the zinc electrode is sufficient to repel sulfate ions, and the positive charge on the copper electrode repels hydrogen ions. At this equilibrium point, the potential difference between the electrodes is about 1.2 V. When the load is connected between the electrodes, electrons flow from the negative electrode of the cell through the external circuit to the positive terminal. This flow of electrons reduces the charge on the electrodes, so the chemical reactions start again. If the load stays connected, the reactions continue until the zinc cathode disintegrates or all of the sulfuric acid has become zinc sulfate. See Review Questions 4-4 and 4-5.

4-3  Carbon-Zinc and Alkaline Cells Dry cells use an electrolyte paste or a saturated absorbent spacer instead of a liquid electrolyte. The first dry cell was the carbon-zinc cell invented by the German chemist Carl Gassner (1855–1942) in 1886. Variations of this cell are still being used today. They are usually referred to as general-­ purpose or standard-duty cells. Negative ions from the electrolyte in the carbon-zinc cell react with the zinc electrode, making it negative, while hydrogen ions take electrons from the carbon rod electrode, making it positive. The carbon rod is surrounded by manganese dioxide to prevent polarization, hydrogen bubbles adhering to the electrode and insulating it from the electrolyte. The hydrogen gas produced at the electrode reacts with the oxygen from the depolarizer to form water. Impurities in the zinc electrode can cause local action, reactions that continue without any load connected to the cell. Local action limits the shelf life of carbon-zinc cells. These cells sometimes leak when nearly used up because the reactions in the cell weaken the zinc case. Table 4-1 summarizes a few characteristics of two common types of dry cells, the carbon-zinc cell and the manganese-alkaline cell, also called the alkaline cell. Compared to carbon-zinc cells, alkaline cells produce more energy, have a longer shelf life, and are much less likely to leak. See Figures 4-2 and 4-3 for a visual representation of their respective constructions.

53

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Chapter 4   Cells, Batteries, and Other Voltage Sources

TABLE 4-1  Comparison of common dry cells Cell Type

Nominal Voltage

Construction

Usage

Carbon-zinc cell (see Figure 4-2)

1.5 V

Anode: a zinc can (also the cell case) Cathode: carbon rod surrounded by a depolarizer Electrolyte: paste of ammonium chloride and zinc chloride (acidic)

• First used in flashlights, radios, and clocks. • Not as common today as the alkaline cell.

Manganese-alkaline cell, also called alkaline cell (see Figure 4-3)

1.5 V

Anode: zinc powder in a gel containing the electrolyte Cathode: manganese dioxide mixed with carbon powder inside a steel can (the cell case) Electrolyte: potassium hydroxide (alkaline)

• Most commonly used in portable devices, such as toys, flashlights, remote controls, music players, and clocks.

Carbon rod (positive electrode) Sealing compound Electrolyte paste (ammonium chloride and zinc chloride) Depolarizer (manganese dioxide, ammonium chloride, and graphite)

Zinc container (negative electrode) Figure 4-2  Commercial carbon-zinc cell

+

Steel can (current collector)

Carbon/manganese dioxide (cathode)

Outer jacket

Powdered zinc, potassium hydroxide electrolyte/anode

Brass pin (current collector) Separator

Insulator seal/vent − Figure 4-3  Manganese-alkaline cell

See Review Questions 4-6 to 4-8.

Steel base

4-4   Other Commercial Primary Cells

55

4-4  Other Commercial Primary Cells Alkaline and carbon-zinc cells are by far the most common types of cells in use today, but many other types have been developed for special applications. Table 4-2 summarizes the properties of a few other commercial types of cells. TABLE 4-2  Comparison of other commercial primary cells Cell Type

Nominal Voltage

Construction

Usage

Zinc-chloride cell, often called a heavy-duty cell

1.5 V

Similar to the carbon-zinc cell. Anode: a zinc can (also the cell case) Cathode: manganese dioxide mixed with carbon Electrolyte: zinc chloride dissolved in water

• Two to four times the operating life of carbon-zinc, depending on application and continuous use • Was commonly marketed for “heavy-duty” use in appliances until alkaline batteries surpassed it

Silver-oxide cell

1.5 V

Anode: zinc powder in a gel containing the electrolyte Cathode: silver oxide Electrolyte: either potassium hydroxide or sodium hydroxide

• Voltage remains almost constant over the life of the cell • Ideal for limited-space applications such as watches and calculators

Zinc-air cell

1.4 V

Similar to the alkaline cell. Anode: zinc powder in a gel containing the electrolyte Cathode: a thin layer of carbon with access to oxygen through venting holes Electrolyte: potassium hydroxide

• Lightweight with a high energy density • Shelf life of 3 years or more, if left sealed • Practical for use in hearing aids, digital cameras, cell phones, and larger batteries; can power navigation lights, such as railway signals • Not suitable for air-tight battery holders

Lithium cell

3.0 V

Anode: lithium foil layer pressed inside a ­stainless steel can Cathode: manganese dioxide mixed with carbon Electrolyte: lithium perchlorate dissolved in propylene carbonate

• Used in portable electronics and ­implanted medical devices • Greater capacity than similarly-sized alkaline cells • Voltage remains almost constant over the life of the cell • Shelf life of 10 years or more

Figure 4-4 shows how the voltages of various types of primary cells vary as the cells discharge. Figure 4-4 Discharge curves for common primary cells

3.5 Lithium

Terminal voltage (V)

3.0 2.5 2.0

Silver-oxide

1.5

Alkaline

1.0

Carbon-zinc & zinc-chloride

0.5 20

See Review Question 4-9.

40 60 Operating life (%)

80

100

56

Chapter 4   Cells, Batteries, and Other Voltage Sources

4-5  Secondary Cells Usually, secondary (rechargeable) cells are preferred for portable devices that get a lot of use, such as cell phones, laptop computers, and digital cameras. The most common secondary cell is still the lead-acid cell, invented in 1859 by the French physicist Gaston Planté (1834–89). This type of battery powers the starter motor in most cars. Once running, the car’s engine drives an alternator, which recharges the battery, reversing the chemical reactions that produced the electric energy to start the engine. Figure 4-5 shows the construction of a lead-acid secondary cell. Discharge

Charging supply A

V

Charge +

Load

−

Dilute sulfuric acid (H2SO4) Lead peroxide (PbO2) on lead-antimony grid

Spongy lead (Pb) on lead-antimony grid

Figure 4-5  Lead-acid secondary cell

Several types of rechargeable cells and batteries are made in the same sizes as primary cells. Some types work best when they are almost fully discharged before recharging (this is called “deep cycling”); if only partially discharged they exhibit what’s called a “memory effect,” preventing them from fully recharging. Others can be r­ echarged even if only partially discharged (“­shallow ­cycling”), which makes them the right choice for applications that call for shallow c­ ycling or “float” (­continuous) charging. Table 4-3 summarizes the properties of a few commercial types of secondary cells.

4-5   Secondary Cells

TABLE 4-3  Comparison of secondary cells Cell Type

Nominal Voltage

Construction

Usage

Lead-acid cell (see Figure 4-5)

2.1 V

Anode: a spongy form of pure lead on a grid of lead-antimony Cathode: lead peroxide on a grid of lead-antimony Electrolyte: dilute sulfuric acid

•  Common applications include car starter motors, emergency ­lighting, ­uninterruptible power supplies for ­computers, standby power systems for ships

Nickel-cadmium (ni-cad) cell

1.25 V

Anode: cadmium hydroxide Cathode: nickel hydroxide Electrolyte: potassium ­hydroxide in water

•  Self-discharges at a rate of roughly 20% a month •  Memory effect occurs if shallow cycled; unsuited for float charging •  Low cost and can be recharged up to 1000 times •  Commonly used in ­cordless power tools

Nickel-metal ­ hydride (ni-mh) cell

1.25 V

Anode: hydrogen storage alloy such as lanthanumnickel or zirconium-nickel Cathode: nickel hydroxide Electrolyte: potassium hydroxide

• Reduced memory effect; deep cycling not required • More energy density compared to ni-cads • Used in cell phones, digital cameras, laptop computers

Lithium-ion (li-ion) cell

3.6 V

Anode: Aluminum coated with a lithium compound such as lithium-cobalt ­dioxide, lithium-nickel ­dioxide, or  lithium-­manganese dioxide Cathode: copper coated with carbon Electrolyte: lithium salt such as lithium-phosphorous hexafluoride

• Low self-discharge rate and no memory effect •  Can be deep cycled, should not be float charged •  Twice the energy density compared to ni-cads •  More costly than ni-cad and ni-mh, thus typically used when size and weight must be minimized •  Used in cell phones, digital cameras, laptop computers

Rechargeable alkaline-­ manganese (RAM) cell

1.1 V

Anode: porous zinc gel designed to absorb hydrogen during the charging process Cathode: manganese dioxide Electrolyte: potassium hydroxide

•  Low self-discharge rate and no memory effect •  Short operating life, ­especially when deeply cycled •  Less expensive than ni-cad •  Suitable for low-cost consumer applications with shallow cycling

57

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Chapter 4   Cells, Batteries, and Other Voltage Sources

Various types of cells and batteries

Nickel Metal Hydride

Lithium Primary

Lithium Cylindrical

Lithium Rechargeable

Lead Acid

Source:  © Panasonic

Lithium Ion

Industrial Alkaline

Carbon Zinc

See Review Questions 4-10 and 4-11.

4-6  Capacity of Cells and Batteries The capacity of a cell is the total energy it can store. Capacity is determined by the size of the cell and the energy density of the chemicals in it. Battery capacity is usually expressed in terms of an ampere-hour rating, which is the product of the discharge current and number of hours to total discharge. This relationship can be expressed as an equation: Capacity = Life × Discharge current

Table 4-4 compares the ampere-hour ratings of several sizes of goodquality alkaline and ni-cad batteries. The capacity is reduced if the cell is discharged continuously at high current levels. Specifications for cells usually include a maximum discharge current. Exceeding this maximum can damage the cell.

4-6   Capacity of Cells and Batteries

Example 4-1 A wireless mouse is powered by AA cells. If the average current drawn from each cell is 50 mA under constant use, determine how long the cells will last if they are (a) alkaline (b) nickel-cadmium Solution (a) For alkaline cells, Life =



Capacity Discharge current

(b) For ni-cad cells,

Life =





=

2.0 A· h 2.0 A· h = = 40 h 50 mA 0.050 A

0.60 A· h 0.60 A· h = = 12 h 50 mA 0.050 A

See Problems 4-1 and 4-2 and Review Questions 4-12 and 4-13.

TABLE 4-4 Cell and battery capacities Battery Type

Capacity (A·h)

Typical Drain (mA)

Alkaline

Ni-cad 4.0

200

12.0

C

100

6.0

2.0

AA

 50

2.0

0.60

D

AAA

 10

1.0

0.30

9 Volt

 15

0.50

0.15

Circuit Check CC 4-1.

What is the difference between the cathode and anode of a ­device? CC 4-2. Why are alkaline cells often used in place of carbon-zinc cells? CC 4-3. Estimate how long a 9-V alkaline battery will last in a smoke detector with a continuous current drain of 50 μA.

A

59

60

Chapter 4   Cells, Batteries, and Other Voltage Sources

4-7  Fuel Cells A fuel cell produces a potential difference by combining hydrogen and oxygen to form water. This chemical reaction does not consume the ­electrodes. Instead, the chemical energy in the cell is replenished by feeding more ­hydrogen fuel into the cell (see Figure 4-6). Typical fuel cells produce less than 1 V at full-rated load, but many cells can be placed in series for higher voltage outputs, and they can be placed in parallel (or have their surface area expanded) to allow higher current to flow. Research and development of fuel-cell technology tends to fall within three main categories: stationary power generation, portable power generation, and power for transportation. Fuel cells are being developed as an energy source for electric vehicles, cell-phone towers, and electric grid substations, and as combined heat and power systems for buildings including homes, schools, hospitals, fire and police stations, wastewater treatment plants, and airports. The number of fuel-cell installations around the world is growing, including in the United States (where California has the most stationary fuel cells and Connecticut and Delaware have the largest installations [in the 15-30 MW range]) and South Korea (which boasts one of the world’s largest installations, at close to 60 MW). Canada is a global leader in hydrogen and fuel-cell research and development, exporting approximately 90% of this technology through the supply of parts, testing equipment, and expert services around the world.

−

Hydrogen (fuel) in

+

Electrolytic barrier

Water out

Oxygen (air) in

Catalytic electrodes

Figure 4-6  Simplified cross section of a fuel cell

See Review Question 4-14.

4-8  Other Voltage Sources All voltage sources create a potential difference by converting some form of energy into electrical energy.

4-8   Other Voltage Sources

Generators

Source:  © iStock.com/leezsnow

In 1831, Michael Faraday discovered that magnetism could produce electricity. Faraday built the first electric generator, a machine that converts mechanical energy into electrical energy. Large generators produce most of the electrical energy we use. Hydroelectricity harnesses the power of falling water to drive turbines connected to large generators. Figure 2-6 shows a cross s­ ection of a hydroelectric station. The largest hydroelectric stations in the world are installed in China, Brazil, Paraguay, and Venezuela, with generating capacities in the range of 10,000 to 22,000 MW.

A row of hydroelectric turbine generators at a generating station; the one in the foreground has been taken apart for maintenance.

61

The majority of the energy consumed in the world is used for heat and transportation. About 20% is used in the form of electricity.

Chapter 4   Cells, Batteries, and Other Voltage Sources

A steam turbine uses high-pressure steam to drive a generator. The heat to produce the steam may come from burning a fuel (primarily coal, oil, and natural gas) or from a nuclear reactor. Coal and natural gas are used in more than 50% of the world’s electricity generation. Gas turbines are essentially jet engines that use burning fuel to drive the turbine directly. Wind turbines use the power of wind to turn a ­propeller connected to a small generator. Denmark generates about 40% of its electricity with wind turbines, and the world’s largest offshore wind farms are found in the United Kingdom, Denmark, and Germany.

Photoelectricity A photovoltaic (solar) cell is a semiconductor device that produces a ­potential difference when light strikes its surface. These devices are made with two types of silicon crystal, as shown in Figure 4-7. The n-type s­ ilicon contains a small amount of an element such as antimony, arsenic, or phosphorus. When added to a silicon crystal lattice, each atom of these elements has a loosely bound outer electron that can move within the crystal. The p-type silicon contains a small amount of an element such as boron, gallium, or indium, which can loosely bind a free electron. At the junction of the two types of silicon, the p-type material binds the free electrons from the n‑type material, leaving almost no free charge carriers.

Source:  © iStock.com/hohl

62

Panels of Photovoltaic cells at a solar power farm

4-8   Other Voltage Sources

Light energy provides electron with sufficient energy to cross depletion layer Transparent layer of p-type silicon Depletion layer −

−

− −

n-type silicon

−

−

Load

− −

+ Figure 4-7  Cross section of a solar cell

Small solar cells power household devices such as calculators and garden lights. Larger arrays, consisting of hundreds of cells, provide electricity to pump water, run communications equipment, light homes, and run appliances. Solar cells also power the Hubble Space Telescope and some communications satellites.

Thermocouples

Source:  © Omega Engineering Inc. Reproduced with the permission of Omega Engineering Inc., Stamford, CT 06907 USA www.omega.com

A thermocouple is a junction of two dissimilar metals that develops a small potential difference proportional to the temperature at the junction. Although not practical for generating large quantities of energy, thermocouples can power small systems such as furnace controls. T ­ hermocouples are widely used to monitor pilot flames in gas appliances and to measure temperatures in engines, silos, and industrial ­machinery.

Thermocouples

63

64

Chapter 4   Cells, Batteries, and Other Voltage Sources

Piezoelectricity Certain crystals, such as quartz and Rochelle salt (potassium sodium tartrate), produce a potential difference when pressure is applied to them. This piezoelectric effect converts mechanical energy into electrical energy. Piezoelectricity has a variety of applications including microphones, buzzers, strain gauges, and barbecue lighters.

Friction Friction is another method of producing a potential difference by converting mechanical energy into electrical energy. When certain dissimilar materials are rubbed together, the friction between them can transfer electrons from one of the materials to the other, building up a charge of static electricity. Friction can generate very high voltages. One application is the Van de Graaff generator, in which a moving belt builds up a charge on a hollow metal sphere. Large Van de Graaff generators produce potential differences of millions of volts. Originally developed for particle accelerators, these generators are also used to produce radiation for medical treatments and tests of structural materials. See Review Questions 4-15 and 4-16.

Review Questions

65

Summary

• Cells produce a voltage as the result of chemical action. • A battery is a set of cells interconnected to provide a higher voltage and/ or current than is available from a single cell. • Secondary cells are rechargeable whereas primary cells are not. • A basic cell has two electrodes in contact with an electrolyte. • Polarization and local action can limit the output and shelf life of cells. • Common practical primary cells include carbon-zinc, manganese-­ alkaline, zinc-chloride, silver-oxide, zinc-air, and lithium cells. • Common secondary cells include lead-acid, nickel-cadmium, nickel-metal hydride, lithium-ion, and rechargeable alkaline-manganese cells. • The capacity of a cell is the total energy stored in the cell, usually measured in ampere-hours. • A fuel cell produces a voltage from the energy released as hydrogen and oxygen combine to form water. • Voltage sources convert chemical, mechanical, light, or heat energy to produce a potential difference.

Problems B B

Section 4-6  Capacity of Cells and Batteries 4-1. How long will an alkaline 9 V battery last in a device that draws 5 mA of current for an average of 15 min a day? 4-2. How long can a fully charged ni-cad AAA cell maintain a current of 10 mA?

Review Questions

Section 4-1  Basic Terminology 4-3.

Differentiate between primary and secondary cells.

Section 4-2  Simple Primary Cell 4-4. Describe the chemical reaction that takes place at the cathode of a simple zinc-copper primary cell. 4-5. At what point is the simple primary cell considered to be completely discharged?

Section 4-3  Carbon-Zinc and Alkaline Cells 4-6. What is polarization in a cell? How can this effect be prevented? 4-7. Compare the terminal voltage and the shelf life of the carbon-zinc cell and the alkaline cell. 4-8. Describe local action in primary cells, and explain why it is ­undesirable.

B = beginner

I = intermediate

A = advanced

66

Chapter 4   Cells, Batteries, and Other Voltage Sources

Section 4-4  Other Commercial Primary Cells 4-9.

Why do most hearing aids use zinc-air cells?

Section 4-5  Secondary Cells 4-10. Describe the various energy conversions that take place from the time a car’s battery is used to start the engine until the car’s alternator recharges the battery. 4-11. Why is the ni-mh cell a good choice for powering digital cameras?

Section 4-6  Capacity of Cells and Batteries 4-12. What is the capacity of a cell or battery? 4-13. How is the ampere-hour rating of a battery related to the total energy stored in the battery?

Section 4-7  Fuel Cells 4-14. Describe one advantage of the fuel cell over a conventional battery.

Section 4-8  Other Voltage Sources 4-15. How does a Van de Graaff generator produce a potential difference? 4-16. What type of voltage source could you use to measure pressure in a boiler? How does this sensor work?

Integrate the Concepts (a) What features do all cells have in common? (b) Choose a suitable type of cell for each application, and explain the reason for your choice: i) a flashlight ii) a cell phone iii) a memory chip that holds the settings required to start a ­computer (c) By what factor would an alkaline D cell outlast a ni-cad D cell powering the same portable device, assuming negligible self-discharge in the cells? (d) How might self-discharge affect your answer to part (c)?

Practice Quiz 1. What is the source of energy in a primary cell? 2. What is the term for two or more cells connected together? 3. Why can primary cells not be recharged? 4. List two advantages of lithium cells. 5. List three commonly used types of secondary cells.

Practice Quiz

6. How does continuous discharge at high current levels affect the capacity of a cell? 7. A wireless computer mouse is powered by AA ni-cad cells that have a capacity of 0.60 A·h. If the mouse draws an average current of 20 mA while in use, how long can it be used before the cells need to be ­recharged? 8. Which of the following characteristics apply for fuel cells?   (a) They combine hydrogen and oxygen to form water.  (b) They consume the electrodes.   (c) They are used for electric cars. (d) None of the above. 9. Which of the following statements are true?   (a) Every voltage source produces electric energy by converting some other type of energy.  (b) A photovoltaic cell is a semiconductor device that converts solar energy into electric energy.   (c) Thermocouples are often used to monitor temperatures. (d) Piezoelectric crystals convert heat into electric energy.

67

5

Resistance and Ohm’s Law Chapter 3 introduced the concepts of voltage and current. Now we shall explore how these two quantities are related.

Chapter Outline 5 -1

Ohm’s Law  70

5 -3

Factors Governing Resistance  72

5 -5

Circular Mils  75

5 -2

The Nature of Resistance  71

5 -4 Resistivity 73 5 -6

American Wire Gauge  77

5 -8

Temperature Coefficient of Resistance  82

5 -7

5 -9

Effect of Temperature on Resistance  79 Linear Resistors  84

5 -10 Nonlinear Resistors  86 5 -11

Resistor Colour Code  88

5 -12 Variable Resistors  91

5 -13 Voltage-Current Characteristics  91 5 -14 Applying Ohm’s Law  92

Key Terms Ohm’s law  70 resistance 70 ohm 70 resistor 72 resistivity 74 specific resistance 74 ohm metre  74 circular mil  75 mil  75 American Wire Gauge (AWG)  77

temperature coefficient of resistance 82 linear resistor  84 wire-wound resistor 84 precision resistor  85 carbon-composition resistor 85 nonlinear resistor  86 inrush current  86 thermistor 86 varistor 87

photoresistor 87 photoconductor 87 light dependent resistor (LDR) 87 nominal value  88 tolerance 88 reliability 88 potentiometer 91 rheostat 91 voltage-current characteristic 91 IR drop 93

Learning Outcomes At the conclusion of this chapter, you will be able to: • state Ohm’s law • use Ohm’s law to calculate the resistance of, voltage across, or current through an electric device • explain why conductors present an opposition to current • state the effect of length, cross-sectional area, and type of material on the resistance of a ­conductor • define resistivity • calculate the resistance of a material given its length, cross-sectional area, and resistivity Photo sources:  © iStock.com/David Cannings-Bushell

• calculate the resistance of a length of wire using the American Wire Gauge table • calculate the resistance of a material at different ­temperatures • describe the construction of different types of linear, ­nonlinear, and variable resistors • determine the resistance, tolerance, and reliability of a ­resistor using the resistor colour code • plot the voltage-current characteristic of a ­resistor

70

Chapter 5   Resistance and Ohm’s Law

5-1  Ohm’s Law In 1827 the German physicist and mathematician Georg Simon Ohm published a book describing his conclusions from careful measurements of currents in circuits like the one shown in Figure 5-l. He found the same current each time he closed the switch in a given circuit. Ohm also discovered that as long as the temperature of the conductor did not change, doubling the applied voltage doubled the current, and tripling the applied voltage tripled the current. Ohm found the same relationships when he varied the load by using conductors of different lengths and thicknesses. For a given circuit, the ratio of the applied voltage to the current is a constant, and E =k I



(5-1)

where E is the applied voltage, I is the current, and the value of k ­depends on the particular circuit. This relationship is known as Ohm’s law.

+ −

Voltage source

Load resistance

Figure 5-1  The basic circuit for Ohm’s resistance experiments

Ohm concluded that the E/I ratio for a given circuit is a property of that circuit. For a given applied voltage, the current decreases as the ratio increases. Therefore, this ratio indicates the opposition of the circuit to the flow of charge carriers. This property is termed resistance, and the unit of resistance, the ohm, is named for Georg Ohm. Resistance is the opposition of a circuit to current. The letter symbol for resistance is R. The ohm is the SI unit of electric resistance. The symbol for ohm is the Greek uppercase letter Ω (omega). One ohm equals one volt per ampere. We can substitute R for the constant k in Equation 5-1. If we want to ­determine the resistance of the load rather than the resistance of the whole

5-2   The Nature of Resistance

circuit, we must consider the voltage drop across the load, rather than the  voltage applied to the total circuit. The equation for Ohm’s law then becomes R=



V I

(5-2)

where R is the resistance in ohms, V is the voltage drop across the resistance in volts, and I is the current through the resistance in ­amperes.

Example 5-1 Find the resistance of a lamp if a current of 150 mA flows through the lamp when a voltage of 6.0 V is applied to its terminals. Solution

R=



V 6.0 V = = 40 Ω I 150 mA

Example 5-2 What value of resistance limits the current through the resistance to 20 μA when the voltage drop across the resistance is 480 mV? Solution

R=

V 480 mV = = 24 000 Ω = 24 kΩ I 20 μA

See Problems 5-1 to 5-8 and Review Questions 5-70 and 5-71 at the end of the chapter.

5-2  The Nature of Resistance As described in Chapter 2, a voltage applied to a conductor causes a net drift of free electrons along the length of the conductor. Repulsion from electrons entering the conductor from the circuit’s energy source accelerates free electrons along the conductor. Thus, energy from the source is transferred to the free electrons as kinetic energy. As the moving electrons collide with atoms in the conductor, some kinetic energy transfers from the electrons to the atoms. The transferred energy appears as heat since it increases the vibration of the atoms within the lattice structure of the c­ onductor.

71

72

Chapter 5   Resistance and Ohm’s Law

The collisions between the free electrons and the atoms reduce the speed at which the electrons drift in response to the applied voltage. The greater the rate of collisions in a material, the greater its resistance. Current flowing through a resistance always produces heat. Devices such as stove elements and incandescent lamps apply this heat. In many circuits, however, the heat produced is an unavoidable loss of energy from the system. Some applications require ventilation or other cooling to prevent the waste heat from damaging the circuit. A resistor is a component made to have a specific resistance. If we c­ onnect a resistor between a lamp and a voltage source, as shown in ­Fig­ure 5-2, the total resistance of the circuit increases and the current is correspondingly reduced. The rate at which the source transfers energy to the circuit is also reduced, and that energy is divided between the lamp and the resistor. Thus, adding the resistor to the circuit dims the lamp.

Resistor + −

Figure 5-2  Using a resistor to limit current

See Review Questions 5-72 to 5-74.

5-3  Factors Governing Resistance Suppose we have two pieces of wire that are identical except that one is twice the length of the other. For free electrons travelling the full length of these wires, the average interval between collisions with atoms in the wire is the same. So, electrons passing through the longer wire have twice as many collisions as electrons passing through the shorter wire. Consequently, the opposition of the longer wire to electric current is twice as great as that of the shorter wire. We can generalize this comparison: The resistance of a conductor is directly proportional to its length. Next we compare two pieces of wire that are identical except that one has twice the cross-sectional area of the other. The thicker wire has the same cross-sectional area as two pieces of the thinner wire joined together at both ends (connected in parallel). Each of the two smaller diameter wires will pass the same current when connected to a given voltage source. Therefore the thicker wire passes twice as much current as the thinner wire for a given applied voltage. Since R = V/I, the thicker

5-4  Resistivity

wire has half the ­resistance of the thinner wire. Again, we can generalize the comparison: The resistance of a conductor is inversely proportional to its crosssectional area. We noted in Section 3-1 that some materials possess more free electrons per unit volume than others. The intervals between collisions of a free electron with atoms in a material and the energy transferred by the collisions also depend on the molecular structure of the material. For example, a silver wire has a lower resistance than a copper wire with the same dimensions, and the copper wire has a lower resistance than an aluminum wire with the same dimensions. The resistance of a conductor is dependent on the composition of the conductor. Section 5-7 describes how temperature affects resistance. See Review Questions 5-75 and 5-76.

5-4 Resistivity The resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area. Therefore, we can calculate the resistance for any dimensions of a conductor if we know the resistance of a length of the material with a uniform cross-sectional area, assuming no change in temperature. R2 l2 A1 = × R1 l1 A2



(5-3)

 here R is the resistance of the conductor, l is the length, and A is the w cross-sectional area.

Example 5-3 A conductor 1.0 m long with a cross-sectional area of 1.0 mm2 has a resistance of 0.017 Ω. Find the resistance of 50 m of wire of the same material with a cross-sectional area of 0.25 mm2. Solution

R2 = 0.017 Ω ×

50 m 1 mm2 × = 3.4 Ω 1m 0.25 mm2

73

74

Resistivity is sometimes called specific resistance.

Chapter 5   Resistance and Ohm’s Law

Resistivity is a convenient quantity for calculating the resistance of a given conductor. The resistivity of a material is the resistance of a unit length of the material with unit cross-sectional area. The letter symbol for resistivity is the Greek letter ρ (rho). In SI, the resistivity of a material is the resistance between opposite faces of a cube of the material measuring 1 m along each side. We can apply ­dimensional analysis to determine the units for resistivity. ρ = resistance of unit area per unit length ohms × metre2 metre = ohm metres =



The ohm metre is the SI of resistivity. The unit symbol for ohm metre is Ω·m.

Since temperature affects resistance, values of resistivity are given for a specified temperature, usually 20°C. Table 5-1 lists the resistivities of the more common metallic conductor materials at 20°C. TABLE 5-1  Resistivity of some common conductors at 20°C Material

Constantan is an alloy of 50%–60% copper and 40%–50% nickel. Nichrome™ II is an alloy containing ­primarily nickel and chromium.

Resistivity (nΩ·m)

Silver

16.4

Copper (annealed)

17.2

Gold

24.4

Aluminum

28.3

Tungsten

55

Nickel

about 70

Iron

about 100

Constantan

about 490

Nichrome™ II

about 1100

If we set l1 = 1 m and A1 = 1 m2 in Equation 5-3, R1 becomes ρ. Rearranging the equation gives a formula for the resistance of any conductor:

R=ρ

l A

(5-4)

where R is the resistance of the conductor in ohms, l is the length of the ­conductor in metres, A is the cross-sectional area in square metres, and ρ is the resistivity of the conductor material in ohm metres.

5-5   Circular Mils

75

Example 5-4 Find the resistance at 20°C of 200 m of an aluminum conductor with a cross-sectional area of 4.0 mm2. Solution Using the value of ρ for aluminum from Table 5-1,

R=ρ×

l 200 m = 1.4 Ω = 2.83 × 10 − 8 Ω · m × A 4.0 × 10 − 6 m2

In this solution, converting square millimetres to square metres allows us the cancel out all the metre units.

Most conductors have a circular cross section. If we know the diameter of the conductor, we can calculate the cross-sectional area.

Example 5-5 Find the resistance at normal room temperature of 60 m of copper wire having a diameter of 0.64 mm. Solution

A = πr2 =

R=ρ×

π 2 π d = × 0.642 mm2 = 3.217 × 10 −7 m2 4 4

l 60 m = 1.72 × 10 −8 Ω · m × = 3.2 Ω A 3.217 × 10 −7 m2

See Problems 5-9 to 5-26 and Review Questions 5-77 and 5-78.

Note that CM stands for circular mil, while cm is the SI symbol for centimetre.

5-5  Circular Mils Many conductors are much less than an inch in diameter. To simplify area calculations for such conductors, the foot-pound system uses an area unit, called a circular mil, based on a mil, or one thousandth of an inch (see Figure 5-3). One circular mil (CM) is the area of a circle 1 mil in diameter.

0.001"

Figure 5-3  A circle with diameter of one mil

76

The mathematical ­formula for the area of a circle is πr2 = ( π/4 ) d2 . However, with circular mils the factor of π/4 is included in the units for the area.

Chapter 5   Resistance and Ohm’s Law

When using circular mils, the formula for the area of a circle becomes A = d2



(5-5)

 here A is the area of the circle in circular mils and d is the diameter in w mils.

Example 5-6 Find the cross-sectional area, in circular mils, of a wire 0.0280" in diameter. d  = 0.02800" = 28.0 mils A = d2 = 28.02 = 784 CM Solution

Resistivity can be defined in terms of a piece of wire 1 ft long with a crosssectional area of 1 CM, as shown in Figure 5-4. The volume of this wire is 1 circular-mil foot, and the units of ρ become CM-Ω/ft. Table 5-2 lists the ­resistivities for some metals. 1 ft 0.001" = 1 mil 1 Circular mil Figure 5-4  One circular-mil foot

TABLE 5-2  Resistivity of some metals at 20°C Material Silver

9.9

Copper

10.4

Gold

14.7

Aluminum Brass

Manganin is a ­copper alloy containing 13%–18% manganese and 1%–4% nickel.

Resistivity (CM-Ω/ft)

17.0 23.4

Tungsten

33

Nickel

42

Iron

60

Platinum

64

Manganin

290

Constantan

295

Mercury

576

Nichrome™ II

660

5-6   American Wire Gauge

77

Example 5-7 Calculate the resistance of 50.0 miles of copper wire 0.350″ in diameter. (1 mile = 5280 ft) d  = 0.350″ = 350 mils Solution

A = d2 = 122.5 × 103 CM 5280 ft l = 50.0 miles × = 2.54 × 105 ft 1 mile ρl 10.4 ( 2.54 × 105 ) = R= = 21.6 Ω A 122.5 × 103

See Problem 5-27.

Circuit Check

A

CC 5-1. What resistance passes a current of 21.28 mA when the applied voltage is 10.0 V? CC 5-2. What length of aluminium wire with a diameter of 0.25 mm has a resistance of 2.25 Ω at 20ºC? CC 5-3. An 6.5 m length of wire has a diameter of 0.8 mm and a ­resistance of 0.315 Ω at 20ºC. Of what material is this wire likely to be made?

5-6  American Wire Gauge In North America, most wire is produced in the standard American Wire Gauge (AWG) sizes listed in Table 5-3. The larger the gauge number, the smaller the wire size. Wires larger than 4/0 (sometimes called 0000) are usually specified in thousands of circular mils (MCM) instead of having a gauge number. The ratio of the diameter of each American Wire Gauge size and the next one is about 1.1229, and the ratio of their areas is about 1.26. The area ratio between wire sizes three AWG numbers apart is very close to 2. For example, #10 wire (10 380 CM) has about twice the area of #13 wire (5178 CM) and half the area of #7 wire (20 820 CM). Since resistance is inversely proportional to area, the resistance per unit length also differs by a factor of about two for wires three AWG numbers apart. For example, #1 wire (0.1239 Ω/1000′) has half the resistance of #4 wire (0.2485 Ω/1000′) and twice the resistance of #3/0 wire (0.0618 Ω/1000′).

The first M in MCM is the Roman n ­ umeral for 1000, not the ­metric prefix mega. MCM is sometimes written as kcmil, or kilo-circular mils.

78

Chapter 5   Resistance and Ohm’s Law

Example 5-8

What size of aluminum wire has a resistance of 1.62 Ω for a length of 2500′? Solution

Since R =

ρl , A

A=

ρl 17 × 2500 = = 26.2 MCM R 1.62

From Table 5-3, we see that this area matches AWG #6.

TABLE 5-3  American Wire Gauges AWG

The resistance of solid copper is also a function of the copper’s purity, or the percentage of copper content, as well as any treatment the copper wire receives after being drawn to its AWG size. These processes (such as annealing, bunching, stranding, tinning, or braiding) can each affect the overall wire resistance. The resistances listed in Table 5-3 for copper may be used as a general guideline.

Gauge

mils

circular mils 211 600

107.22

133 100

67.43

460

11.68

2/0

365

9.27

1

289

3

229

3/0

mm2

10.41

167 800

325

8.25

105 500

53.47

258

6.54

66 370

33.63

4

204

5.19

6

162

0 2

410

Resistance of Solid Copper at 20°C

Area

mm

4/0

7.35

83 690

85.03

42.41

5.83

52 640

26.67

182

4.62

33 100

16.77

7

144

3.67

20 820

10.55

9

114

13 090

6.63

5

8

4.12

41 740

26 250

128

3.26

102

2.59

10 380

 80.8

2.05

6 530

14

 64.1

1.63

4 107

16

 50.8

18

 40.3

10 11

12

13

15

The units Ω/1000 ft is sometimes listed as mΩ/ft.

Diameter

17

19

20 21

22

  90.7

 72.0

2.91 2.31

1.83

16 510

8 234 5 178

 57.1

1.45

 45.3

1.15

2 048

0.91

1 288

0.72

810

 35.9  32.0

 28.5  25.3

1.29 1.02

0.81

0.64

3 257

2 583

21.15

13.30 8.37

5.26 4.17

3.31

2.62

2.08

1.65 1.31

1.04

Ω/1000 ft

Ω/km

0.0618

0.202

0.0490 0.0779

0.0983 0.1239

0.1563 0.1970

0.2485 0.3133

0.3951

0.4982

0.405 0.511

0.644 0.812 1.02

1.29

1.63

0.9989

3.27

1.260

1.588

2.003 2.525

2.59

4.12

5.19

6.55

8.25

3.184

10.41

5.064

16.54

4.016

1 022

0.518

10.15

0.326

16.14

642

0.321

2.05

0.7921

0.824

0.411

0.255

0.6282

1 624

0.653

0.160

6.385

8.051

12.80

13.13

20.89

26.32 33.18

41.84

52.63

See Problems 5-28 to 5-33 and Review Questions 5-79 and 5-80.

5-7   Effect of Temperature on Resistance

5-7  Effect of Temperature on Resistance

Resistance

Checking the V/I ratio of conductors at various temperatures shows that the resistance of most conducting materials increases linearly with temperature except at very hot or very cold temperatures. Temperature has little effect on the resistance of some alloys, such as constantan. For a few materials, including carbon and other semiconductors, the resistance ­decreases as the temperature increases. Figure 5-5 shows the resistance of a conductor increasing with temperature. R1 is the resistance at temperature T1, and R2 is the resistance at ­tem­perature T. The line segment CF shows how the resistance varies for ­temperatures between T1 and T2. The steeper the slope of segment CF, the more the resistance increases with temperature.

F

R2 x

A Tx

C

R1

O

B T1

0°C

E D T2

Temperature Figure 5-5  Effect of temperature on resistance in the majority of materials

Extending CF to the temperature axis produce triangles ABC and ADF. Since these triangles are similar, DF AD AO + OD = = BC AB AO + OB

Substituting the quantities that the sides of the triangles represent gives

R2 x + T2 = R1 x + T1

(5-6)

where R1 is the resistance of the conductor at temperature T1, R2 is the ­resistance at temperature T2, and x is the difference between 0°C and the temperature at which the resistance would be zero if it continued to ­decrease linearly at very low temperatures. Table 5-4 lists values for x for some metals commonly used for conductors. The values for Nichrome™ II and constantan show that x is not the

79

80

Chapter 5   Resistance and Ohm’s Law

a­ ctual temperature at which the resistance becomes zero ohms. However, we can use x to calculate resistances within the range of temperatures where the resistance varies linearly with temperature. TABLE 5-4  Values of x for common conductors Conductor Material

x (°C)

Silver

243

Copper

234.5

Aluminum

236

Tungsten

202

Nickel

147

Iron

180

Nichrome™ II

6 250

Constantan (55% Cu, 45% Ni)

125 000

Example 5-9

A copper conductor has a resistance of 12.0 Ω at 20°C. Find the resistance at 100°C. Solution Since

R2 x + T2 = , R1 x + T1

R2 = R1

( 234.5 + 100 ) 12 × 334.5 x + T2 = 12.0 Ω × = 15.8 Ω = ( 234.5 + 20 ) x + T1 254.5

Note that the temperature units, °C, cancel out in this calculation.

Example 5-10 A precision resistor made of constantan wire has a resistance of 10 000 Ω at 20°C. What is its resistance when its temperature rises 20°C? Solution

R2 = R1

x + T2 125 000 + 40 = 10 000 × = 10 002 Ω x + T1 125 000 + 20

In addition to being able to calculate the effect that temperature has on r­ esistance, we can use the change in resistance of a certain material to ­calculate the temperature in locations where it is difficult to place a t­ hermometer.

5-7   Effect of Temperature on Resistance

Example 5-11 The copper winding of an electric motor that has been standing for several hours in a room at 20°C has a resistance of 0.20 Ω. When the motor has been in use for some hours, the resistance of the winding is found to be 0.22 Ω. Calculate the temperature rise in the winding. Solution



R2 x + T2 = R1 x + T1

0.22 234.5 + T2 = 0.20 234.5 + 20

T2 =

0.22 × ( 234.5 + 20 ) − 234.5 = 45.45ºC 0.20

temperature rise = T2 − T1 = 45.45oC − 20oC = 25oC

Substituting R = ρl/A from Equation 5-4 for R1 in Equation 5-6 gives R=ρ



l(x + T) A ( x + 20 )

(5-7)

With values for ρ and x, we can use Equation 5-7 to estimate the resistance of any metallic conductor at any temperature.

Example 5-12 Find the resistance at 40°C of 300 m of copper wire with a cross-­ sectional area of 1.50 mm2. Solution From Tables 5-1 and 5-4, ρ is 1.72 × 10–8 Ω · m and x is 234.5˚C.

R=ρ

l(x + T) A ( x + 20 )

= 1.72 × 10 − 8 Ω · m × = 3.44 ×

= 3.71 Ω

274.5 254.5

( 234.5ºC + 40ºC ) 300 m 2 × ( 234.5ºC + 20ºC ) 1.50 mm

See Review Questions 5-81 to 5-83.

81

82

Chapter 5   Resistance and Ohm’s Law

5-8 Temperature Coefficient of Resistance Now we shall derive an alternative method of showing the effect of temperature on the resistance of a conductor. In Figure 5-5, line segment CE is parallel to the temperature axis. Since resistivity is usually stated for a temperature of 20°C, we shall let T1 represent 20°C. Triangle ABC is similar to triangle CEF, so FE CE = BC AB

Substituting the quantities that the sides of the triangles represent gives ΔT FE = , where ΔT is the difference between T2 and 20°C R1 x + 20 ΔT Therefore, FE = R1 x + 20

(

Since CE and BD are parallel,

R2 = ED + FE = R1 + FE = R1 + R1

)

(

ΔT ΔT = R1 1 + x + 20 x + 20

) (

)

The quantity x +1 20 is the temperature coefficient of resistance at 20°C, the  proportion by which the resistance changes per degree of change in ­temperature from 20°C. We represent this coefficient by the Greek letter α (alpha), so R2 = R1 ( 1 + αΔT ) (5-8) where R2 is the resistance of a conductor at any specified temperature, R1 is its resistance at 20°C, α is the temperature coefficient of resistance at 20°C for the conductor, and ΔT is the difference between the specified temperature and 20°C.

Substituting R1 = ρl/A from Equation 5-4 gives a general equation for the resistance of a conductor:

R=ρ

l ( 1 + αΔT ) A

(5-9)

where R is the resistance of the conductor in ohms, ρ is the resistivity of the material in ohm metres at 20°C, l is the length in metres, A is the cross-­sectional area in square metres, α is the temperature coefficient of resistance of the material at 20°C, and ΔT is the difference between the temperature of the conductor and 20°C. Table 5-5 lists temperature coefficients for some materials commonly used for conductors and resistors.

5-8   Temperature Coefficient of Resistance

TABLE 5-5 Temperature coefficients of resistance at 20°C Conductor Material

α (K ) −1

Silver

0.003 8

Copper

0.003 93

Aluminum

0.004 3

Tungsten

0.004 5

Nickel

0.006

Iron

0.005 5

Nichrome™ II

0.000 16

Constantan (55% Cu, 45% Ni)

0.000 008

Carbon

−0.000 5

Example 5-12A What is the resistance of 300 m of copper wire with a cross-sectional area of 1.5 mm2 at 40°C? Solution From Table 5-1 and Table 5-5, ρ is 1.72 × 10−8 Ω · m and α is 0.003 93.

R=ρ

l ( 1 + αΔT ) A

= 1.72 × 10−8 Ω · m ×

= 3.44 × 1.079 = 3.71 Ω

300 m × ( 1 + 0.003 93 × 20 ) 1.5 × 10−6 m2

Example 5-13

A heating element made from Nichrome™ II has a resistance of 16.0 Ω at 1500°C. Find the resistance at normal room temperature. Solution In this example, R2 in Equation 5-8 is given and R1 is unknown. Hence,

R1 =

R2 16.0 16.0 = 12.9 Ω = = ( 1 + αΔT ) ( 1 + 0.000 16 × 1480 ) 1.237

If the required temperature is below 20°C, ΔT will be a negative quantity and the resulting resistance is less than the resistance at 20°C. See Problems 5-34 to 5-45 and Review Questions 5-84 and 5-85.

83

K−1 = 1/K , or “per ­Celsius degree”

84

Chapter 5   Resistance and Ohm’s Law

Circuit Check

CC 5-4. What length of iron wire 0.5″ in diameter has a resistance of 3.0 Ω? CC 5-5. An electric thermometer with a tungsten element has a ­resistance of 10 Ω at 20°C. It is immersed in a fluid and its ­resistance drops to 7.3 Ω. What is the temperature of the fluid? CC 5-6. In metres, what length of AWG #20 copper wire has a resistance of 2.5 Ω at 50ºC?

B

5-9  Linear Resistors For most conductors, a graph of current versus voltage is a straight line, ­indicating a constant resistance (see Figure 5-6). The smaller the resistance, the steeper the slope of the graph. A resistor that maintains a constant V/I ratio is a linear resistor. As current through a resistor increases, more heat is produced in the resistor, raising its temperature. This increase in temperature causes a slight increase in the resistance of most conductor materials. For the common conductor materials such as copper and aluminum, the change in resistance over the ranges of operating temperatures for most circuits is so small that these materials are usually considered to be linear resistors.

Current (mA)

20 5 kΩ

15 10

10 kΩ

5 0

0

25

50 Voltage (V)

75

100

Source:  © iStock.com/Gueholl

Figure 5-6  Current versus voltage for linear resistors

Wire-wound resistor encased in procelain

As the name suggests, wire-wound resistors have a metal wire wound on a hollow porcelain tube and sealed in position with a porcelain coating. These resistors are usually made with constantan or other alloys with a temperature coefficient of almost zero. The larger the size of a resistor, the more readily it can dissipate heat to the surrounding air, and hence the greater the power rating. (See Chapter 6-2 for more information about power ratings of resistors.) Inexpensive, mass-manufactured resistors often have resistances that vary by 10% or more from their nominal values. While such resistors

Resistance change (%)

Precision resistors

Carbon-composition resistor

Source:  Photo courtesy of Ohmite Manufacturing Co.

are adequate for many types of circuits, applications such as measuring ­instruments require resistors that are accurate to within 1% or less of the nominal resistance. Such precision resistors are often made by depositing a thin film of metal or carbon on a small ceramic cylinder, to which leads are ­attached. The film may be etched to adjust the resistance to the specified value before being coated with a layer of insulating material. Resistors used in electronic devices usually have resistances greater than a kilohm (a thousand ohms) and pass currents of only a few milliamperes. A carbon-composition resistor is commonly used when the current through the ­resistor produces less than 2 W of heat. The resistance element consists of finely ground carbon mixed with an insulating binder such as phenolic and pressed into a cylindrical shape with a wire lead embedded in each end. The resistance element is then sealed in a plastic jacket. The length and width of the cylinder, the proportion of carbon in the mixture, and the way the mixture is compressed determine the resistance. Carbon-composition resistors are much cheaper than wire-wound and film resistors, but the resistance of a carbon-composition resistor increases if the temperature varies appreciably from 20°C as shown in Figure 5-7. For moderate temperature variations, the resistance changes by only a few ­percent, so we can usually treat carbon-composition resistors as linear ­resistors.

85

Source:  © iStock.com/catetus

5-9   Linear Resistors

5 4 3 2 1 –50

–25

0 25 50 Temperature (°C)

75

100

Composition resistors are also made with a ceramic consisting of tin oxide and antimony bound with glass. Ceramic-composition resistors are particularly useful for circuits where the resistors must withstand voltage or ­energy surges. Integrated circuits (ICs) range from simple resistor networks to miroprocessors containing millions of microscopic resistors and transistors. ICs start as a wafer of highly purified silicon. Pure silicon is a poor conductor since it has few free charge carriers. The resistors are made by diffusing tiny regions of the silicon with precisely controlled amounts of elements that supply free electrons, making the regions more conductive.

Integrated circuits

Source:  © iStock.com/alex-mit

Figure 5-7 Resistance-temperature characteristic of carbon-composition resistors

Chapter 5   Resistance and Ohm’s Law

5-10  Nonlinear Resistors When an incandescent lamp is switched on, the temperature and the resistance of its tungsten filament increase dramatically. When white-hot, the filament of an ordinary 60-watt 120-V lamp has a resistance of 240 Ω, but its resistance at room temperature is about 18 Ω. An incandescent lamp is a nonlinear resistor. The inrush current at the instant the lamp is turned on is much greater than its normal operating current: Normal I =

V 120 V 120 V V = = 0.50 A    Inrush I = = = 6.6 A R 240 Ω R 18 Ω

Fortunately, the mass of the lamp filament is small enough that it gets white hot in less than a millisecond. Therefore, the current surge is brief, as shown in Figure 5-8. Nevertheless, switches used with incandescent lamps have to be designed to withstand the inrush current. To avoid large inrush currents, heating elements, such as those in stoves, are usually made from an alloy with a very small temperature coefficient.

7 6 Current (A)

86

5 4 3 2

Switch closes at t = 0

1 0

1 2 Time (milliseconds)

Figure 5-8  Inrush current of a 60-W incandescent lamp

A resistor with a large negative temperature coefficient, called a thermistor, can be used to limit inrush currents. Typically, such thermistors have a ­resistance of over 100 Ω at room temperature, but with a current of 1 A through them, their resistance drops to less than 1 Ω after 10 to 15 s. These thermistors contain semiconductive metal oxides with a ceramic binder. Heat produced by current through the resistor breaks covalent bonds in the metal oxides, creating enough free electrons to reduce the resistance to a fraction of its value at room temperature. Small thermistors are used to measure temperatures since a decrease in temperature of less than 20°C will more than double their resistance.

87

Source:  © Omega Engineering Inc. Reproduced with the permission of Omega Engineering Inc., Stamford, CT 06907 USA www.omega.com

5-10   Nonlinear Resistors

Varistors depend on the nonlinear resistance characteristic of zinc oxide or silicon carbide crystals, which are formed into wafers with a clay binder. Zinc oxide varistors (also called metal-oxide varistors or MOVs) are used to protect sensitive electronics from voltage surges. Silicon carbide varistors (commonly known by the trade name thyrite) can protect high-voltage systems. They are used as lightning arrestors on power transmission lines. Temperature has little effect on the resistance of a varistor. Instead a rapid increase in the number of charge carriers occurs when the potential difference across the varistor becomes greater than the threshold of the varistor. As Figure 5-9 indicates, the resulting decrease in resistance is such that the current through the varistor increases greatly without appreciably increasing the voltage drop across it. Thus, a varistor connected across the power input to a device can protect it from voltage surges.

Source: Hyper-Sense Technology Co., Ltd.

Thermistors

Varistors

Voltage surge supression Normal operating voltage

8 6 4

Source:  MARTYN F. CHILLMAID/SCIENCE PHOTO LIBRARY

Current (A)

10

2 0

30

60 90 Voltage (V)

120

150

Figure 5-9  Typical varistor characteristic

A photoresistor, photoconductor, or light dependent resistor (LDR) contains a thin zigzag strip of cadmium sulfide or cadmium selenide. Light falling on the strip breaks down valence bonds in the cadmium compound, creating additional charge carriers. The resistance can range from hundreds of kilohms in the dark to less than 100 Ω in bright daylight. Photoresistors are widely used in light meters, auto-exposure circuits in cameras, and controllers for outdoor lights.

Photoresistor

88

Chapter 5   Resistance and Ohm’s Law

Thermistors, varistors, and photoresistors all use semiconductors engineered to become more conductive under specific conditions. The schematic symbols for these nonlinear resistors are shown in Table 5-6. TABLE 5-6 Schematic symbols for nonlinear ­resistors General Thermistor

t°

or

Varistor

V

Photoresistor

See Review Questions 5-86 to 5-87.

5-11  Resistor Colour Code Many resistors are small enough that any printing on them would be hard to read. So, most resistors are marked with bands of colour, which have the further advantage of being visible no matter how the resistor is oriented (see Figure 5-10).

Some precision resistors use the first three bands for digits, the fourth band for the multiplier, and the fifth band for tolerance.

A single-band resistor only has one black colour band to indicate its value: zero ohms. It looks like a resistor but acts like a wire in the circuit.

First digit Second digit Multiplier

Reliability Tolerance

Figure 5-10  Colour bands on a resistor

A resistor has from three to six bands, but most common resistors have only four or five colour bands. The colours are read from the band nearest to an end of the resistor. On a 4-band resistor, the first three bands indicate two digits and a multiplier that give the nominal value of the resistor in ohms. The fourth band indicates the tolerance, which is the maximum percent deviation from the nominal value. A fifth band is sometimes added to show the reliability, which is the percentage of resistors with a resistance outside the specified tolerance after 1000 h of use, also known as the fail rate. Table 5-7 lists the meaning of the colour bands for 4-band resistors and 5-band reliability resistors. Table 5-8 lists the meaning of the colour bands for 5-band precision resistors.

5-11   Resistor Colour Code

89

TABLE 5-7 Resistor colour code for 4-band resistors and 5-band reliability resistors Band Colour Black

Brown Red

Orange Yellow Green

First First Digit 1

2

3

4

5

Second Second Digit

Third Multiplier

Fourth Tolerance

0

100

2

102

±1%

1

3

4

5

Blue

6

6

Grey

8

8

Violet White Gold

7

9

7

9

Silver

101

103

104

1%

0.1%

0.01%

0.001%

105

106 107

108 109

 0.1

 0.01

None

±2%

Fifth Fail Rate

±5%

±10%

±20%

TABLE 5-8 Resistor colour code for 5-band precision resistors First First Digit

Second Second Digit

Third Third Digit

Fourth Multiplier

Black

0

0

0

100

Red

2

2

2

102

Band Colour Brown Orange Yellow Green

1

3

4

5

1

3

4

5

1

3

4

5

Blue

6

6

6

Grey

8

8

8

Violet White Gold

Silver

None

7

9

7

9

7

9

101

103

104

Fifth Tolerance 1%

2%

105

0.5%

107

0.1%

106

 0.1

 0.01

0.25%

5%

10%

20%

Resistors are manufactured with standard nominal values such that any given resistance is within the tolerance range of at least one of the standard values. These standardized values for different tolerance ranges are also known as the E-series, and they can be valid for other types of components, such as capacitors, inductors, and Zener diodes. The lower the tolerance, the larger the number of standard values. Table 5-9 lists the two digits for the standard values for 5% and 10% tolerances. The standard values have these digits with any multiplier.

Six-band resistors have colour bands that indicate three digits, a multiplier, the tolerance, and a temperature coefficient. These highprecision resistors are used in applications where temperature is a critical factor.

90

The most common E-series are E6, E12, E24, E48, E92, and E192. For example, the E12 series contains 12 two-digit standard nominal values ranging from 10 to 82, with a 10% tolerance, as shown in Table 5-9.

Chapter 5   Resistance and Ohm’s Law

TABLE 5-9 Standard resistor digits

±5% Tolerance

±10% Tolerance

10

10

12

12

15

15

18

±5% Tolerance

±10% Tolerance

33

33

39

39

47

47

18

56

56

22

22

68

68

27

27

82

82

11

13

16

20 24 30

36 43 51

62 75 91

Example 5-14 Determine the specifications of a reliability resistor coded brown-black-red-gold-brown. Solution First digit: brown = 1 Second digit: black = 0 Multiplier: red = 102 Tolerance: gold = 5% Reliability: brown = 1% The nominal resistance is 10 × 102 = 1.0 kΩ. The tolerance is 5%, which means the actual resistance is between 950 Ω and 1050 Ω. The reliability of 1% means that one out of every 100 resistors will not lie within the tolerance range after 1000 h of operation at the resistor’s rated power.

Example 5-15 Interpret these resistor colour codes: (a) yellow-violet-orange-silver-red, on a precision resistor (b) yellow-violet-orange-silver-red, on a reliability resistor (c) green-brown-gold-gold Solution (a)  Resistance is 473 × 10−2 = 4.73 Ω ±2%. (b)  Resistance is 47 × 103 = 47 kΩ ±10% with a failure rate of 0.1%. (c)  Resistance is 51 × 0.1 = 5.1 Ω ±5% (reliability not specified). See Problems 5-46 to 5-50 and Review Questions 5-88 and 5-89.

5-13   Voltage-Current Characteristics

91

5-12  Variable Resistors

Source:  © Sergpet/Dreamstime/Getstock

Variable resistors have a moving contact, or wiper, that either rotates to run against a curved resistor (as shown in Figure 5-11) or slides along a straight resistor. The resistance element is usually carbon composition, cermet (a ceramic and metal mixture), or a wire coil. The resistance between the wiper and either end of the resistor depends on the position of the wiper, and varies from zero to the full value of the resistor.

Resistance element Wiper

Mechanism of a wire-wound variable resistor

A C

In Figure 5-12, the resistance between terminals A and B is constant. Since RAC + RCB = RAB, the resistance between terminals A and C increases as the resistance between terminals C and B decreases. When all three terminals of a variable resistor are used, it is called a ­potentiometer. Potentiometers provide a simple way of adjusting a v ­ oltage, and are commonly used as volume and tone controls in audio equipment. When only the wiper and one end of the resistor are connected, as in Figure 5-13, the variable resistor is called a rheostat. Rheostats can vary the current in a circuit, and are sometimes used for motor speed controls. See Review Question 5-90.

5-13  Voltage-Current Characteristics Although the resistance of a nonlinear resistor is not constant, the Ohm’s law relationship R = V/I still applies. It is often convenient to show variations in resistance by plotting a voltage-current characteristic. These graphs usually have voltage drop on the x-axis and current on the y-axis, as in Figure 5-14. A low resistance has a graph with a steep slope and a high resistance has a graph with a shallow slope. The two straight-line graphs in Figure 5-14 represent linear resistors. For a linear resistor, the resistance is constant, so we can find the resistance by  calculating the ratio of voltage to current at any point on the graph.

B Shaft

Figure 5-11 Crosssection of a variable resistor A C B

Movable arm (wiper)

Figure 5-12  Schematic of a potentiometer

C B

Figure 5-13  Schematic of a rheostat

Chapter 5   Resistance and Ohm’s Law

Low-linear resistance

Nonlinear resistance ΔI Current

92

ΔI High-linear resistance ΔI

ΔV Voltage Figure 5-14  Voltage-current characteristics of resistors

For a given ΔV, ΔI is much greater for a low resistance than it is for a high resistance. If the voltage-current characteristic is not linear, we can calculate a dynamic value of resistance from the ratio of the change in voltage drop to the associated change in current. Graphs of voltage-current characteristics are particularly useful for showing the behaviour of nonlinear components such as varistors. See Review Question 5-91.

5-14  Applying Ohm’s Law We can rearrange the equation for Ohm’s law (Equation 5-2) to get an expression for current, I = V/R. The total voltage drop V for a circuit connected between the terminals of a voltage source is equal to the applied voltage E of the source. Hence, the current is dependent on the applied voltage and the total resistance of the circuit:

I=

E R

(5-10)

where I is the current in the circuit (in amperes), E is the applied voltage (in volts), and R is the total resistance of the circuit (in ohms). Similarly, we can use Ohm’s law to find an expression for the voltage drop across a resistor:

V = IR

(5-11)

5-14   Applying Ohm’s Law

where I is the current through a resistor (in amperes), R is the resistance (in ohms), and V is the resulting voltage drop across the resistor (in volts). A voltage drop across a resistance is often referred to as an IR drop.

Example 5-16 Calculate the current through a 4.7 kΩ resistor connected across a 9.0-V source. Solution

I=

E 9.0 V = = 1.9 × 10−3 A = 1.9 mA R 4.7 kΩ

Example 5-17 Calculate the voltage drop produced by a 15-mA current through a 560-Ω resistor. Solution

V = IR = 15 mA × 560 Ω = 8.4 V

See Problems 5-51 to 5-69 and Review Question 5-92.

Circuit Check

C

CC 5-7. Determine the resistance and tolerance for resistors with the colour bands listed below. (a) brown-black-yellow-gold (b) red-red-black-silver (c) violet-brown-green-black-red, in a precision resistor CC 5-8. Find the minimum and the maximum acceptable resistances for the resistors in question CC 5-7. CC 5-9. How much current will flow if a 120-V source is connected to a 25-kΩ resistance?

93

94

Chapter 5   Resistance and Ohm’s Law

Summary

• Ohm’s law relates the resistance of an electric component to the voltage across it and the current through it. • When a voltage is applied to a conductor, it exhibits a resistance or ­opposition to electric current as free electrons collide with atoms in the conductor. • The resistance of a conductor depends on its length, its cross-sectional area, and its composition. • Resistivity is the resistance of a unit length of a conductor with unit cross-sectional area. • The American Wire Gauge specifies diameters for standard sizes of wire conductors. • The effect of temperature on the resistance of a material can be calculated using the temperature coefficient of resistance for the material. • Types of linear resistors include wire-wound, carbon-composition, ceramiccomposition, metal-film, and carbon-film. • Nonlinear resistors include the thermistor, varistor, and photoresistor. • Colour bands on resistors indicate their nominal resistance and tolerance. Additional bands can be used to indicate the failure rate and temperature coefficient. • Variable resistors can be used as either potentiometers or rheostats. • The voltage-current characteristic is a graph of the current through a ­device versus the voltage across it. B = beginner

I = intermediate

A = advanced

Problems B

Section 5-1  Ohm's Law 5-1.

B

5-2.

B

5-3.

B

5-4.

B

5-5.

B

5-6.

B B

5-7. 5-8.

Find the resistance of a ceiling fan that draws a 0.5-A current from a 120-V source. What resistance will limit the current in a circuit to 1.5 A when the applied voltage is 30 V? A 12.5-A current causes a 110-V voltage drop across a portable electric heater ­element. Find the resistance of this element. A current of 0.15 A causes a 1.0-V drop across a resistor. Find its ­resistance. A 15-μA current through a resistor causes a 45-mV drop across it. Find the resistance of the resistor. Calculate the resistance of an ammeter that reads 10 A with a 50-mV drop across it. What value of resistor has a 7-V drop for a current of 15 mA? What value of resistor draws 40 mA when connected to a 2.5-kV source?

Problems

Section 5-4  Resistivity I I

I I B B B B B I I I I I I

A

NOTE: Assume a temperature of 20°C for Problems 5-9 to 5-26. 5-9. A conductor with a cross-sectional area of 10 mm2 has a resistance of 1.72 Ω/km. Find the resistance of 500 m of wire of the same material with a cross-sectional area of 4 mm2. 5-10. A conductor with a cross-sectional area of 2.5 mm2 and a length of 50 m has a resistance of 1.42 Ω. What length of wire of the same ­material with a cross-sectional area of 1.5 mm2 has a resistance of 14.2 Ω? 5-11. A conductor with a diameter of 1.5 mm and a length of 70 m has a ­resistance of 0.65 Ω. Find the resistance of a conductor of the same material having a length of 40 m and a diameter of 0.8 mm. 5-12. A conductor 40 m in length and 1.2 mm in diameter has a resistance of 17.3 Ω. What diameter of wire of the same material has a resistance of 1.0 Ω/m? 5-13. Use Table 5-1 to calculate the resistance of a tungsten wire 250 m long and with a cross-sectional area of 5.0 mm2. 5-14. Find the resistance of a copper bar with a rectangular cross section 2 cm × 4 cm and a length of 1.5 m. 5-15. Calculate the resistance of 160 m of aluminum wire having a diameter of 1.6 mm. 5-16. Find the resistance of a tungsten filament 12 cm in length and 0.1 mm in diameter. 5-17. What length of silver wire 0.5 mm in diameter has a resistance of 0.04 Ω? 5-18. What diameter of copper wire has a resistance of 1 Ω/m? 5-19. Find the diameter of copper wire that will give a relay coil wound with 85 m of the wire a resistance of 30 Ω. 5-20. Constantan wire 0.4 mm in diameter is wound on a cylindrical core such that the mean diameter of each turn is 2.0 cm. How many turns of this wire are needed to make a 24-Ω resistor? 5-21. A carbon rod 0.50 cm in diameter and 10 cm in length has a resistance of 0.153 Ω. Find the resistivity of this carbon rod. 5-22. An electric conductor 40 m in length and 1.8 mm in diameter has a resistance of 17.3 Ω. What material is used for this conductor? 5-23. A stranded cable consists of eight copper conductors, each having a diameter of 1.0 mm. Twisting the wires to form the cable makes the length of each strand 3% longer than the length of the cable. Find the resistance of 1.0 km of this cable. 5-24. What diameter of solid copper conductor has the same resistance per kilometre as a stranded cable consisting of six aluminum ­conductors each having a diameter of 1.3 mm? Assume that the twisted aluminum strands are each 3% longer than the finished cable.

95

96

Chapter 5   Resistance and Ohm’s Law

I I

5-25. What diameter of copper wire will have a resistance of 10 Ω for a length of 100 m? 5-26. Calculate the resistivity of manganin (in nanoohm metres) given that 5.0 m of manganin wire 1.27 mm in diameter has a resistance of 1.7 Ω.

Sections 5-5 and 5-6  Circular Mils and AWG B

B B I I I I

B B I I I

I

NOTE: Assume a temperature of 20°C for Problems 5-27 to 5-33. 5-27. Determine the area in circular mils of wires with the following ­diameters: (a) 0.0063″ (b) 0.035″ (c) 0.565″ 5-28. Calculate the resistance of 2500 feet of aluminum wire, 0.150” in ­diameter. 5-29. What length of AWG #22 constantan wire is needed to make a 1.38 Ω resistor? 5-30. What is the diameter, in inches, of 1500 feet of copper wire if its ­resistance is 10 Ω? 5-31. A 1500-foot length of AWG #20 wire has a resistance of 50 Ω. ­Determine what material the wire is made of by calculating its ­resistivity. 5-32. What length of AWG #10 aluminum wire will have the same resistance as 1500″ of #12 copper wire? 5-33. What is the nearest AWG of Nichrome™ II wire that will have a resistance of 4.5 Ω for an 11′ length?

Sections 5-7  Temperature Effects on Resistance

5-34. A length of copper telephone line has a resistance of 24 Ω at 20°C. What does the resistance become on a hot summer day with a ­temperature of 36°C? 5-35. Find the resistance at −20°C of an aluminum conductor that has a ­resistance of 1.25 Ω at +30°C. 5-36. What is the resistance at 60°C of 25 m of copper wire having a diameter of 0.50 mm? 5-37. What length of Nichrome™ II wire with a diameter of 0.64 mm has a resistance of 48 Ω at 200°C? 5-38. If the resistance-temperature graph for brass is extended in a straight line until the resistance becomes zero, the corresponding temperature would be −480°C. Find the temperature coefficient of brass at 20°C. 5-39. The temperature coefficient of platinum at 20°C is 0.003 per Celsius degree. Find the temperature at which the resistance of platinum would become zero if the resistance-temperature graph were extended as a straight line.

Problems

B I I A

B A

B

B

B

B B

5-40. An incandescent lamp draws a 1.0-A current from a 110-V source to raise the temperature of its tungsten filament to 2800°C. Find the ­resistance of the filament at 20°C. 5-41. At what temperature does the resistance of a nickel rod increase to 105% of the rod’s resistance at 20°C? 5-42. A certain conductor has a resistance of 10.0 Ω at 20°C and 11.35 Ω at 50°C. From which of the materials listed in Table 5-5 is this conductor made? 5-43. An electric motor is set up 250 m from a 120-V source. The current drawn by the motor is 8.0 A. The temperature of the copper conductors feeding the motor is 40°C. What is the minimum diameter of wire that can be used without the ­voltage drop in the conductors exceeding 10% of the applied ­voltage? 5-44. The resistance of a length of aluminum wire is 10.5 Ω at 100ºC. What will its resistance be at −20 ºC? 5-45. A motor with copper windings was tested at an ambient temperature of 20ºC. The voltage across the motor was the 100 V, and the current through the motor at the start of the test was 5.0 A. D ­ uring the test the current dropped gradually to a steady-state value of 4.5 A. What was the temperature of the motor at the end of the test?

Section 5-11  Resistor Colour Code 5-46. Determine the value, tolerance, and reliability (if indicated) of resistors with the following colour bands: (a) orange-white-black-gold-brown, for reliability (b) red-red-red-red-red, for reliability (c) green-blue-red-silver-red, for precision (d) yellow-violet-yellow 5-47. State the colour code for each of the following resistors. (a) 10 kΩ ±5%, 1% reliability (b) 2.2 MΩ ±2%, 0.1% reliability (c) 0.47 Ω ±1%, reliability unspecified (d) 31.6 Ω ±2% (e) 825 Ω ±0.5% (f) 100 kΩ ±5% 5-48. The colour bands on a reliability resistor are yellow-orange-orangegold-red. (a) What is the minimum resistance for this resistor? (b) If 56 000 of these resistors are purchased, how many are likely to be out of tolerance after 1000 h of use? 5-49. What is the colour code for a 2% tolerance resistor with a value of 2.4 Ω? 5-50. A resistor is colour coded brown-green-orange-gold. What is the mini­mum value it can have?

97

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Chapter 5   Resistance and Ohm’s Law

B B B B B B B B B B B B B B B B B B B

Section 5-14  Applying Ohm's Law 5-51. Find the current through a 160-Ω resistor connected across a 240-V source. 5-52. What current through a 150-Ω resistor produces a voltage drop of 75 V across the resistor? 5-53. A 12-Ω heating element has a voltage drop of 95 V across it. Calculate the current flowing through this element. 5-54. Find the current that flows in a circuit with a total resistance of 17 Ω connected to a 117-V source. 5-55. Find the voltage that produces a 300-mA current when applied to a 15-Ω resistor. 5-56. Find the voltage drop across a 125-Ω resistor when a 0.2-A current flows through it. 5-57. What voltage drop is produced by a 160-mA current flowing through a 56-Ω resistor? 5-58. What applied voltage causes a current of 0.35 A to flow in a circuit with a total resistance of 124 Ω? 5-59. A fuse in the power supply of a transistor amplifier has a resistance of 0.02 Ω. Find the current that produces a 500-μV drop across the fuse. 5-60. What current will flow when a 50-μV source is connected to a 25-Ω resistor? 5-61. What current will flow through a 2.7-kΩ resistor connected across a 480-mV source? 5-62. Find the current that produces an 80-V drop across a 6.8-kΩ resistor. 5-63. What applied voltage causes a 200-µA current through a 22-kΩ ­resistor? 5-64. What IR drop is produced by a 5.0-mA current passing through a 15-kΩ load resistor? 5-65. What voltage drop appears across a 2.2-kΩ load resistor when the current through the resistor is 1.8 mA? 5-66. The contacts of a particular relay close when the current through the relay’s coil is at least 24 mA. The coil has a resistance of 3.7 kΩ. Find the minimum applied voltage for operating this relay. 5-67. How much voltage is necessary to send a current of 100 μA through a 15.0 kΩ resistance? 5-68. A 100 kΩ resistor has a current of 330 μA flowing through it. What is the voltage drop across this resistor? 5-69. An LED flashlight is powered by a 3.0 V battery. If the effective resistance of the LED is 120 Ω, how much current flow through it?

Review Questions Section 5-1  Ohm's Law

5-70. Why can Ohm’s law be described in terms of constant ­proportionality? 5-71. Express the ohm in terms of other SI units.

Review Questions

Section 5-2  The Nature of Resistance 5-72. Why does the V/I ratio of a circuit indicate its ability to oppose electric current rather than its ability to permit current? 5-73. Energy must be expended to force current to flow through a resistance. Where does this energy come from? Where does it go? 5-74. How is the resistance of a given conductor related to the number of free electrons in the conductor?

Section 5-3  Factors Governing Resistance 5-75. Why does shortening a conductor decrease its resistance? 5-76. Why is resistance inversely proportional to the square of the diameter of a conductor?

Section 5-4  Resistivity 5-77. Why does the term resistivity apply to the material of an electric conductor, rather than to a particular conductor? 5-78. Given an accurate resistance-measuring device, how would you go about determining the resistivity of a sample of an unknown alloy?

Section 5-6  American Wire Gauge 5-79. How are gauge number, diameter, and resistance related for conductors made in American Wire Gauge sizes? 5-80. North American electrical codes specify a limit of 15 A for current through AWG #14 house wiring. What is the reason for this limit? Is it possible for a current greater than 15 A to flow through #14 wire?

Section 5-7  Effect of Temperature on Resistance 5-81. A fuse contains a narrow strip of metal and is connected such that the current in the protected electric circuit flows through this strip of metal. A current greater than the rating of the fuse will melt the metal strip. Which has the greater resistance, a 10-A fuse or a 20-A fuse? Explain your reasoning. 5-82. Explain why the filament of an incandescent lamp is many times hotter than a Nichrome™ II wire-wound resistor that has the same resistance and carries the same current. 5-83. Before an electric motor was started, the resistance of its copper windings at room temperature was 50.0 Ω. After 30 min of operation, the resistance was 53.0 Ω. A half-hour later the resistance was 54.3 Ω, and after a further 40 min it was 54.7 Ω. After a total of 3 h, the ­resistance settled at a steady value of 55 Ω. Plot a graph of temperature against time for the motor. Explain the reasons for the shape of the graph.

99

100

Chapter 5   Resistance and Ohm’s Law

Section 5-8  Temperature Coefficient of Resistance 5-84. Use the data in Table 5-5 to draw a temperature-resistance graph for carbon, similar to Figure 5-5. Find the temperature x for carbon. 5-85. Find the temperature coefficient of copper at 0°C. Explain why the temperature coefficient at 0°C differs from the temperature coefficient at 20°C.

Section 5-10  Nonlinear Resistors 5-86. Draw a graph similar to Figure 5-5 for a resistor having a fairly pronounced negative temperature coefficient. Would such a resistor be useful as a current regulator? Explain. 5-87. Resistors made from resistance wire or metal film usually have a positive temperature coefficient of resistance while resistors made from semiconductor materials generally have a negative temperature coefficient. Account for this difference.

Section 5-11  Resistor Colour Code 5-88. What information does each of the five colour bands on a resistor ­convey? Provide an example for a precision resistor and a reliability resistor. 5-89. Alphanumeric markings are sometimes used for surface-mounted resistors. Look up this labelling system, and determine the resistances indicated by the following labels: (a) 220 (b) 2R2 (c) 102 (d) 4R7

Section 5-12  Variable Resistors 5-90. Differentiate between a potentiometer and a rheostat.

Section 5-13  Voltage-Current Characteristics 5-91. The voltage-current characteristic of a carbon-composition resistor is a straight line for all values of current (and voltage) within its ­normal operating limits. If the graph is extended beyond these limits, the voltage-current characteristic starts to curve slightly. In which direction will it bend? Why?

Section 5-14  Applying Ohm's Law 5-92. What variables determine the current in a circuit?

Practice Quiz

Integrate the Concepts

When a 120-V, 150-watt incandescent light bulb is switched on, the initial current through it is in the order of 12.5 A. After the bulb heats up, the current drops to about 1.25 A. How hot does the filament get?

Practice Quiz 1. Ohm’s law states that (a)   resistance equals voltage times current (b)   voltage equals current divided by resistance (c)   resistance equals voltage divided by current (d)   current equals resistance divided by voltage 2. When the voltage across a resistor is halved, the current through it is (a)  halved (b)  doubled (c)  quadrupled (d)  unchanged 3. The resistance that passes a current of 20 mA with an applied voltage of 1.0 V is (a)  5.0 Ω (b)  0.02 Ω (c)  20 Ω (d)  50 Ω

4. The resistance of a conductor is (a)   directly proportional to its length and its cross-sectional area (b)   inversely proportional to its length and its cross-sectional area (c)  directly proportional to its length and inversely proportional to its cross-sectional area (d)  inversely proportional to its length and directly proportional to its cross-sectional area 5. The resistance of an aluminum conductor at room temperature is 100 Ω. If the conductor is 20 m long, its radius is (a)   13.6 nm (b)  84.6 μm (c)   1.8 nm (d)  42.4 μm

6. The resistance of 125 miles of copper wire 0.025″ in diameter is approximately (a)  50 Ω (b)   10 kΩ (c)   11 kΩ (d)   270 kΩ

101

102

Chapter 5   Resistance and Ohm’s Law

  7. What parameter indicates how much effect temperature has on the resistance of a material?   8. What is the key property of a linear resistor?   9. Give two examples of nonlinear resistors. 10. A resistor colour-coded orange-blue-brown-silver has a resistance of (a)  360 Ω ±10% (b)  36 Ω ±10% (c)  3.60 Ω ±10% (d)  0.36 Ω ±10%

11. The colour bands for a resistor with a nominal value of 15 kΩ ±20% are (a)  brown-blue-orange-gold (b)  brown-green-orange (c)  brown-blue-orange (d)  brown-green-orange-gold 12. What is the difference between a rheostat and a potentiometer? 13. A resistance is (a)   directly proportional to voltage and directly proportional ­current (b)  directly proportional to voltage and inversely proportional ­current (c)  inversely proportional to voltage and directly proportional ­current (d)  inversely proportional to voltage and inversely proportional current 14. In the circuit of Figure 5-15, the value of the resistor is (a)  12.5 Ω (b)  1.25 Ω (c)  125 Ω (d)   none of the above

2.0 A + E 250 V −



Figure 5-15

R=?

to to to to

Practice Quiz

15. In the circuit of Figure 5-16, the current through the resistor is (a)   10 mA (b)   10 A (c)   100 mA (d)   none of the above

I=? + E 330 V –



Figure 5-16

R 33 kΩ

103

6

Work and Power Thus far we have discussed the relationships among v ­ oltage, current, and resistance in an electric circuit. Since voltage produces charge in motion (current) through resistors, the source must be supplying energy and the circuit is using that energy to do work. The next step is to calculate the energy and the power associated with the source and the resistors in the circuit.

Chapter Outline 6-1

Energy and Work  106

6-3

Efficiency  110

6-5

Relationships Among Basic Electric Units  113

6-2 Power 107 6-4 6-6

The Kilowatt Hour  112

Heating Effect of Current  113

Key Terms power 107 watt 107 efficiency  110

horsepower   111 kilowatt hours   111 British thermal unit (BTU) 114

Learning Outcomes At the conclusion of this chapter, you will be able to: • differentiate among work, energy, and power • perform calculations based on the equations relating power, voltage, current, and resistance • calculate the power rating for a resistor in a ­specific ­application

Photo sources:  iStock.com/Peter Englested Jonasen

• define efficiency in terms of energy and power • calculate the efficiency of a device • calculate the total energy consumed by an ­electrical ­system • calculate the heat produced by an electric ­current

106

Chapter 6   Work and Power

6-1  Energy and Work As explained in Section 2-7, work is energy transferred to a body or system. Since energy conversions involve such transfers, work is done whenever e­ nergy is converted from one form into another. For example, in the circuit shown in Figure 6-1, the generator does work on free electrons in its windings while converting mechanical energy into electric energy, and free electrons do work on atoms in the heater while converting the electric energy into heat. Mechanical energy input

Heat energy output

Electrical conductors

Electric heater (resistance)

Generator

Figure 6-1  Energy conversions in a simple electric circuit

The megagram is commonly called a tonne or metric ton (t): 1 Mg = 1 t = 1000 kg

Example 6-1 How much electric energy must be supplied to an electric motor to raise an elevator car having a mass of 4.0 t a distance of 40 m? Assume that 10% of the electric energy supplied to the motor is lost as heat. Solution The work, W, done by the motor to raise the elevator car is equal to the force acting on the car times the distance, d, it travels. The force equals the car’s mass, m, times the gravitational acceleration, g, which is about 9.8 m/s2. F = mg = 4.0 t × 9.8 m/s2 = 4.0 × 103 kg × 9.8 m/s2 = 39.2 kN W = Fd = 39.2 kN × 40 m = 1.568 MJ

The work done by the motor equals 90% of the electric energy supplied to the motor. Therefore,

Win =

1.568 MJ = 1.7 M J 0.90

See Problems 6-1 to 6-4 and Review Question 6-80 at the end of the chapter.

6-2  Power

107

6-2 Power If a more powerful motor were installed for the elevator in Example 6-1, it would transfer energy at a greater rate and raise the car more quickly. However, the work done would be the same as with the smaller motor. These comparisons illustrate the relationship between work and power. Power is the rate of doing work. The letter symbol for power is P. The watt (W) is the SI unit of power. One watt is equal to one joule per second: 1 W = 1 J/s.

The watt is named for the Scottish engineer and inventor James Watt (1736–1819), who developed the steam engine.

Equation 6-1 shows the relationship between power and work: P=



W t

(6-1)

where P is power in watts, W is work in joules, and t is time in seconds. When dealing with work and power, keep in mind that joules are a measure of energy while watts are a measure of the rate of transferring ­energy.

Example 6-2 At what rate must electric energy be supplied to the electric motor in Example 6-1 to raise the elevator 40 m in 4.0 min? Solution

P=

1.742 MJ Win = = 7.3 kW t 4.0 × 60 s

We can use the equations for potential difference and current to relate these quantities to power in a circuit. Since V =

W (Equation 2-4), Q

W = QV

108

Chapter 6   Work and Power

Similarly, I =

Q (Equation 2-2), so t t=



Q I

Substituting for W and t in Equation 6-1 gives So,

P=

QV W I = = QV × =V×I t Q/I Q P = VI

(6-2)

where P is power in watts, V is voltage drop in volts, and I is current in ­amperes. We can also substitute V = IR (Equation 5-11) into Equation 6-2 to get

P = IR × I

P = I2R



(6-3)

where P is power in watts, I is current through the resistance in amperes, and R is resistance in ohms. Similarly, substituting I = V/ R into Equation 6-2 gives P=



V2 R

(6-4)

where P is power in watts, V is voltage drop across the resistance in volts, and R is resistance in ohms.

Example 6-3 A lamp draws a current of 2.00 A when connected to a 120-V source. How much power is going to the lamp? Solution

P = EI = 120 V × 2.00 A = 240 W

6-2  Power

109

When designing circuits, we need to specify the power rating of a resistor as well as its resistance. Since a resistor converts all power it consumes into heat, the power rating is the rate at which the resistor can dissipate heat without damage. Heat dissipation depends on the surface area of the resistor and therefore on its physical size. Standard carbon-composition resistors are manufactured in sizes ranging from 1/8 W to 2 W, as shown in Figure 6-2.

2W

1W

1_ W 2

1 cm 1_ W 4

1_ W 8

Figure 6-2  Approximate sizes of carbon-composition resistors

Example 6-4 A 10-kΩ resistor is connected into a circuit where the current through it  is 50 mA. What is the minimum power rating required for this ­resistor? P = I 2R = (50 mA)2 × 10 kΩ = 25 W Solution

The resistor should be rated for at least 25 W.

If a resistor exceeds its power rating, it can overheat and fail, often causing irreparable damage to itself. It may also cause damage to the circuit board on which it is installed, and any nearby components.

110

Chapter 6   Work and Power

Example 6-5 What is the highest voltage that can be applied to a 3.3-kΩ, 2.0-W ­resistor without exceeding its heat-dissipating capability? Solution Since

P=

V2 R

V 2 = PR

V = √ PR

E = V = √ PR = √ 2.0 W × 3.3 kΩ = 81 V and

See Problems 6-5 to 6-28 and Review Questions 6-81 to 6-85.

Circuit Check

A

CC 6-1. A Miran CO2 surgical laser has an output power of 25 W. How much energy does its beam transfer in 1.2 s? CC 6-2. How much power is used by an electric toaster that has a resistance of 10 Ω and operates from a 120 V supply? CC 6-3. Determine which of the following resistors are likely to be damaged by overheating. Justify your answer. (a) 560 Ω, ½ W, with 75 V across it (b) 3 Ω, 20 W, with 4 A through it (c) ¼ W, with 0.25 mA through it and 40 V across it

6-3 Efficiency Almost all equipment converts some of its input energy into a form that does not produce useful work. This wasted energy is usually in the form of heat. For the motor in Example 6-1, some of this heat comes from friction and some of it from current heating the conductors in the motor. The wasted energy increases the cost of making the equipment as well as the cost of ­running it since the equipment has to be designed to dissipate the wasted energy safely. The efficiency of a device indicates how much of the input energy, Win, the device converts into useful work, Wout. Efficiency is the ratio of useful output energy to total input energy. The letter symbol for efficiency is the Greek letter η (eta).

η=

Wout Win

(6-5)

We usually express efficiency as a percentage. For example, the efficiency of the system in Example 6-1 is 90%.

6-3  Efficiency

111

Example 6-6 What is the efficiency of an electric hoist if its motor uses 60 KJ to raise a 300-kg mass through 18 m? Wout = 300 kg × 9.8 m/s2 × 18 m = 52.9 kJ Solution

η=



Since P =

52.9 kJ Wout = × 100% = 88% Win 60 kJ

W (Equation 6-1), t

W = Pt

Substituting for Wout and Win in Equation 6-5 gives η=



Wout Pout × t = Win Pin × t η=

So,

Pout Pin

(6-6)

For electric motors, Pout is mechanical power output, and Pin is electric power input. All types of power can be measured in watts. However in North America the mechanical power output of motors is often expressed in terms of horsepower (hp), a unit invented by James Watt: 1 hp = 550 foot-pounds/second ≈ 746 W



(6-7)

Example 6-7 Find the input power for an electric motor that has an output of 24 hp and an efficiency of 85%. Solution

Pout = 24 hp × Pin =

746 W = 17.9 kW 1 hp

17.9 kW Pout = = 21 kW η 0.85

See Problems 6-29 to 6-38 and Review Questions 6-86 to 6-89.

There are several ­different definitions for horsepower. Each gives a slightly ­different conversion factor between ­horsepower and watts.

112

Chapter 6   Work and Power

6-4  The Kilowatt Hour A joule is a relatively small quantity of energy. A typical home uses ­millions of joules of electrical energy each day. Since power × time = energy, we could use the product of any unit of power with any unit of time as a measure of energy. In fact, a joule is equal to a watt second. If we measure power in kilowatts and time in hours, we can use kilowatt hours (kW·h) as a unit for energy: W = Pt



where W is work or energy in kilowatt hours, P is power in kilowatts, and t is time in hours. Since 1 kW = 1000 W and 1 h = 3600 s, 1 kW·h = 1000 W × 3600 s = 3.6 × 106 J = 3.6 MJ



(6-8)

Many North American electric utilities measure their customers’ energy consumption in kilowatt hours. Most meters that record electric energy consumption are a form of motor with an aluminum disk that rotates at a speed directly proportional to both the applied voltage and the current. Since P = EI, the speed of the disk is proportional to the power used at any given moment, and the number of rotations indicates the energy consumed (power times time).

Example 6-8 At 7¢ per kilowatt hour, how much will it cost to leave a 60-W lamp burning for five days? Solution

W = Pt = 60 W × 24 h × 5 = 7200 Wh = 7.2 kW·h Cost = 7.2 kW · h ×

7¢ = 50.4¢ kW · h

See Problems 6-39 and 6-40 and Review Questions 6-90 and 6-91.

6-6   Heating Effect of Current

113

6-5 Relationships Among Basic Electric Units All of the quantities discussed in this chapter are interrelated. If we measure any three of these quantities for a given circuit, we can calculate values for all of the others. For example, given voltage, current, and elapsed time, we can determine resistance, work, and power. As shown for power in Section 6-2, straightforward algebraic manipulation of the defining equations gives formulas for each quantity in terms of various combinations of the others. Table 6-1 lists some of the more useful formulas. When dealing with the applied voltage for a circuit, we can substitute the symbol E for V in these formulas. See Problems 6-41 to 6-70. TABLE 6-1  Relationships among basic electric units Quantity

Defining Equations

Voltage

V = W/Q

Current

Resistance Power

Work and energy

Unit and Unit Symbol

Definition of Unit

volt, V

joules per coulomb

I = Q/t

ampere, A

coulombs per second

R = V/I

ohm, Ω

P = W/t

volts per ampere

watt, W

W = Fd

joules per second

joule, J

newton metres

Useful Derived Equation V = IR V = P/I V = √PR

transpose Equation 5-2 transpose Equation 6-2 transpose Equation 6-4

R = V 2/P R = P/I 2

transpose Equation 6-4 transpose Equation 6-3

I = V/R I = P/V I = √P/R P = VI P = I 2R P = V 2/R

W = Pt

6-6  Heating Effect of Current The heat produced by current flowing through a resistor is put to practical use in devices such as kettles, toasters, stoves, space heaters, and hot-water tanks. About 4.186 J of electrical energy produces the heat required to raise the temperature of 1 g of water by 1°C (or 1 K). The equation relating the quantity of heat to the change in temperature of a mass is

Q ≈ 4186 mCΔT

Derivation

(6-9)

where Q is the heat in joules, m is the mass in kilograms, C is the specific heat of the material, and ΔT is the change in temperature in kelvins. ­Specific heat is the ratio of the heat required to raise the temperature of a given mass of a substance by 1 K to that required to raise the temperature of the same mass of water by 1 K.

transpose Equation 5-2 transpose Equation 6-2 transpose Equation 6-3

Equation 6-2 Equation 6-3 Equation 6-4 transpose Equation 6-1

The relationship ­between electrical ­energy and heat was first measured by James Joule.

When analyzing problems with heat transfer, the letter symbol Q can represent the quantity of heat, measured in joules.

114

Chapter 6   Work and Power

Example 6-9 How long will it take to boil 4.3 kg of water starting from 20°C with a 120-V, 10-A heating element and an efficiency of heat transfer of 80%? Solution By definition, the specific heat of water is 1. So, the heat required is Q = 4186 mCΔT = 4186(4.3)(1)(100 − 20) = 1.44 MJ

Since heat transfer is only 80% efficient, the heating element must ­produce

Compared to most other substances, water has a very high specific heat. This means it is easy for water to store heat energy, which is an important factor in sustaining life on Earth.

1.44 MJ = 1.80 MJ 0.80



From Table 6-1, W = Pt = VIt. So,

t=

1.80 MJ W 1 min = = 25 min = 1500 s × ( 120 V ) ( 10 A ) VI 60 s

In the system of imperial units, the basic unit of heat is called the British thermal unit or BTU. One BTU is the amount of heat required to raise the temperature of one pound of water by 1°F. 1 BTU ≈ 1055 J. With foot-pound units, the equation relating the quantity of heat to the change in temperature of a mass becomes

Q = mCΔT

where Q is the heat in BTUs, m is the mass in pounds, C is the specific heat of the material, and ΔT is the change in temperature in degrees Fahrenheit. See Problems 6-71 to 6-79 and Review Questions 6-92 to 6-95.

6-6   Heating Effect of Current

Circuit Check

B

CC 6-4. A 220 V motor draws a current of 51.8 A when operating at full load. If the efficiency of the motor is 81.6%, determine its fullload output in horsepower. CC 6-5. Calculate the cost of electricity for a 30-day period for a home with the energy usage listed below. Assume that electricity costs 6.4¢/kW·h. Average Daily Use Purpose Power Duration

Lighting

Heating

Dryer

Stove

Water heater

Misc.

0.5 kW

10 kW

4 kW

5 kW

3 kW

5 kW

8h

6h

1h

3h

4h

5h

CC 6-6. A 12-Ω heater is connected to a 220-V supply. How much water can this heater raise from freezing to boiling in 20 minutes if the efficiency of heat transfer is 90%?

115

116

Chapter 6   Work and Power

Summary

• Work is done whenever one form of energy is converted into another. • Power is the rate of doing work. • Power is the product of the voltage across a device and the current through it. • A resistor is limited by its ability to dissipate heat quickly. • The efficiency of a device is the ratio of the useful output energy to the total input energy and is equal to the ratio of useful output power to the total input power. • The kilowatt hour is the most common unit of electrical energy monitored by utility companies. • The quantity of heat a substance can transfer depends upon its own mass, its specific heat, and the change in temperature. B = beginner

I = intermediate

A = advanced

Problems B B B B

B

Section 6-1  Energy and Work 6-1. How much energy does it take to hoist a 25-kg parcel 13 m? 6-2. How much energy is required to pump 1500 kg of water to the ­surface from a well 5.0 m deep? 6-3. What distance will a 100-kg pile-driver ram travel if it uses 14.7 kJ of energy? 6-4. Find the mass of an object that requires 81 kJ of energy to lift it 4.5 m.

Section 6-2  Power 6-5.

B

6-6.

B

6-7.

B

6-8.

B

6-9.

B

6-10.

B

6-11.

B

6-12.

I

6-13.

How much power is used by a toaster that draws a 4.5-A current from a 120-V source? How much power is used by a soldering iron that has a 110-V drop between its terminals while drawing a current of 1.2 A? If a 500-mA current through the heater of a guitar amplifier tube produces 3.0 W of heat, what is the voltage drop across the heater? At what rate is electric energy converted by the heater of a cathoderay tube if the heater draws 0.6 A from a 6.3-V power supply? What is the maximum current that a 1/2-W resistor can handle if a voltage drop of 80 V develops across it? At what rate is energy converted in a resistor that produces 6.0 kJ of heat in 10 min? At what rate does a 2.5-A current through a 91-Ω resistor produce heat? What power rating must a 12-kΩ resistor have to safely pass a current of 250 mA? What is the minimum common resistance value that will limit the current to 24 mA flowing through a 1/8-W resistor?

Problems

B B B B B B B B B B B A B I B

I B I B B I

6-14. At what rate does a 1.2-Ω motor-starting resistor convert electric energy while the current through it is 60 A? 6-15. What is the maximum current that a 18-kΩ, 10-W resistor can handle without overheating? 6-16. What is the maximum current that can flow through a 5-kW, 8-Ω resistor? 6-17. What power rating would you select for a 470-Ω resistor that has to develop a voltage across it of 3.3 V? 6-18. The armature winding of an electric motor has a resistance of 0.2 Ω. How much power is lost as heat when the voltage developed across the winding is 8 V? 6-19. What is the minimum value of resistance that can be connected across a 120-V source if the power drawn from the source is not to exceed 0.45 kW? 6-20. What power is released as heat from a 22-kΩ resistor that develops 400 V across its terminals? 6-21. Find the applied voltage for a 1.2-kW heating element that has a resistance of 34 Ω when hot. 6-22. Find the voltage drop across a 75-Ω 3.0-W decorative lamp. 6-23. How much energy does a 750-W electric heater use in 24 h? 6-24. How much heat is produced in 1 h in the winding of a motor that has an I2R loss of 360 W? 6-25. How much power is required to raise a 280-lb object 10 m in 10 s? 6-26. A voltmeter with a resistance of 180 kΩ draws 80 mW of power when connected across a voltage source. Find the potential difference of the source. 6-27. An electric heater produces heat at the rate of 45 kJ/min when the current through the heater is 9.4 A. Find the resistance of the heater. 6-28. A fuse element with a resistance of 0.02 Ω is designed to melt when the current through it produces more than 5 W of heat. What is the current rating of this fuse?

Section 6-3  Efficiency 6-29. Find the current drawn from a 240-V source by a 6.0-hp electric motor with an efficiency of 86%. 6-30. How much work can be done by an 80% efficient electric motor drawing 2 A from a 120-V source for 6 h? 6-31. Calculate the efficiency of an elevator that uses 2.8 × 106 J of electric energy to raise a 4.0-t mass 55 m. 6-32. What is the efficiency of a 1.8-hp electric motor when the electric power input is 2.2 kW? 6-33. How much energy does a pumping system with an overall efficiency of 75% use to raise 1250 kg of water 10.5 m? 6-34. Find the output in horsepower of the pumping system in Problem 6-33 if the system pumps the water at a rate of 1250 kg/h.

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Chapter 6   Work and Power

I B I

B

B

B

B B B B B B B B B

6-35. The output of a power supply for a transistor amplifier is 25 V at 2.4 A. The efficiency of the power supply is 80%. How much current does the power supply draw from a 120-V source? 6-36. A central air conditioner draws 12.0 A from a 230-V source. The efficiency of the air conditioner is 80%. How much work does the air conditioner do when it runs continuously for 1 h? 6-37. The following devices are simultaneously drawing energy from the 12.0-V storage battery in a car: • two 6-W tail lights • two parking lamps with a hot resistance of 12 Ω each • a radio drawing 5.0 A • the fan motor for the car's heater, which develops 0.05 hp at 75% ­efficiency How much power is the battery supplying? 6-38. If the overall efficiency of a radio transmitter is 48%, how much input energy does it need to produce an output of 50 kW from 7:00 A.M. to midnight?

Section 6-4  The Kilowatt Hour 6-39. How much money can be saved by operating a 12-Ω clothes dryer from a 240-V power line for 5 hours every week for a year during off-peak hours instead of on-peak hours? Assume that off-peak electric energy costs 7.7¢/kW·h, and on-peak electric energy costs 15.7¢/kW·h. 6-40. Find how much electric energy a household used in May given that: • The household pays 6¢/kW·h for electric energy. • The meter reading at the end of February was 73 067 kW·h. • The meter reading at the end of May was 74 267 kW·h. • The total bill for the energy used in March and April was $48.00.

Section 6-5  Relationships Among Basic Electric Units 6-41. Find the voltage drop produced by a 2.4-A current through a ­42‑Ω resistor. 6-42. At what rate is electric energy converted into heat in the resistor in Problem 6-41? 6-43. How long will it take the resistor in Problem 6-42 to convert 2.0 kW·h of energy? 6-44. What voltage must be applied to a 33-kΩ resistor to make it dissipate 180 mW of heat? 6-45. Calculate the power rating for a 72-Ω heater that passes a 6.0-A ­current. 6-46. How long will it take the heater in Problem 6-45 to consume 9400 J of electric energy? 6-47. How much charge passes through the heater in Problem 6-46? 6-48. Find the current that produces 550 μW of heat in a 1.2-kΩ resistor. 6-49. What resistance draws 12.5 μA from a 220-mV source?

Problems

B B B B I B B B B B B B B I B B B B B B B

B I B I

6-50. What current does a 40-W lamp draw from a 117-V source? 6-51. What resistance produces 50 W of heat when connected to a 110-V source? 6-52. Find the applied voltage that produces an 8-A current through a 650‑W heater. 6-53. Find the current that produces 400 W of heat in a 45-Ω resistor. 6-54. Find the applied voltage that makes a 2.7-kΩ resistor convert 0.15 kW⋅h in 8 h. 6-55. Calculate the efficiency of a motor with a 1-kW input and a 1-hp ­output. 6-56. Find the voltage drop created by a 6.8-mA current flowing through a 27-Ω resistor. 6-57. Find the applied voltage that delivers 50 μW to a 300-Ω resistance. 6-58. Find the resistance that has a voltage drop of 7.0 V when the current through the resistor is 18 mA. 6-59. What power rating should the resistor in Problem 6-58 have? 6-60. What current will flow when a 20-μV signal is applied to a 75-Ω load? 6-61. What power is fed to the load in Problem 6-60? 6-62. What current will a 4.5-kW load draw from a 240-V source? 6-63. If a 1.25-hp starter motor has 80% efficiency, what current does the motor draw from a 12-V battery? 6-64. How much work can a fully charged 1.5-V rechargeable battery do if its ­capacity is 2500 mA·h? 6-65. Find the resistance of a voltmeter that reads 120 V when a 40-μA current passes through it. 6-66. How long will it take for 2.8 C of electric charge to pass through the voltmeter in Problem 6-65? 6-67. What is the resistance of the copper bus bars feeding an aluminum refining cell if a 4000-A current through them causes a 620-mV drop across them? 6-68. How much electric energy is lost as heat in 8 h in the bus bars of Problem 6-67? 6-69. Find the resistance of an ammeter shunt that has a 5-mV drop across it when the current through it is 9.99 A. 6-70. Calculate the resistance of a 15-A fuse if heat must be developed at the rate of 4.3 W to melt it.

Section 6-6  Heating Effect of Current 6-71. How much energy is required to raise the temperature of 4.0 kg of water by 45°C? 6-72. How much heat must an electric stove element produce to increase the temperature of 1.0 kg of water from 20°C to 85°C if one-third of the heat is lost to the surroundings? 6-73. If heat is transferred to 1 kg of water at the rate of 1 kW, how long will it take to raise the temperature of the water by 45°C? 6-74. Find the power rating for a stove element that raises the temperature of 1.0 kg of water from 20°C to 80°C in 4 min, given that only 75% of input energy is transferred to the water.

119

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Chapter 6   Work and Power

A A A

A A

6-75. An electric kettle takes 15 min to raise the temperature 1.2 kg of water from 12°C to 95°C. The efficiency of this heating process is 89.4%. Find the heater current if the voltage is 120 V. 6-76. A heating element raises the temperature of 100 kg of water from 20°C to boiling point in 10 min. The efficiency of heat transfer is 85%. The current through the element is 90.6 A. Find its resistance. 6-77. A heating element operates at 240 V and has a resistance of 20 Ω. The element heats 15 kg of water for 15 min. The initial temperature of the water is 20ºC, and the heat transfer is 81% efficient. Find the final temperature of the water. 6-78. A 10-Ω heating element operating from a 550-V source heats 100 kg of water from 20°C to boiling point in 20 min. Calculate the efficiency of the heat transfer. 6-79. The 3.0-kW heater in a vat holding 10 kg of cooking oil takes 27 min to heat the oil to the desired operating temperature of 240ºC. Given that the oil has a specific heat of 0.45 and the efficiency of heat ­transfer is 86%, determine the original temperature of the oil.

Review Questions

Section 6-1  Energy and Work 6-80. What is the distinction between work and energy?

Section 6-2  Power

6-81. A 5-W and a 50-W 1000-Ω resistor are both made with NichromeTM II wire. Compare the conductor sizes, the overall dimensions, and the operating temperatures of these two resistors. 6-82. A 35-W soldering iron is used to solder miniature radio components, whereas a 150-W iron is needed to solder a lead to the radio chassis. Explain why the greater wattage is needed for soldering to the ­chassis. 6-83. SI defines the volt as the difference in electric potential between two points of a conductor carrying a constant current of 1 A when the power dissipated between these points is equal to 1 W. Show that this definition is consistent with the one used in Section 2-7. 6-84. Why is it common to find larger electric motors on passenger elevators than on freight elevators even though the freight elevators carry heavier loads? 6-85. What is wrong with the wording of the question: “How much power is consumed by a toaster drawing 3 A from a 110-V source?”

Section 6-3  Efficiency 6-86. Why would a 5-hp motor with a 90% efficiency be physically smaller than a 5-hp motor with a 60% efficiency?

Practice Quiz

6-87. Why is it possible to express efficiency in terms of the ratio of output power to input power even though efficiency is defined as the ratio of output energy to input energy? 6-88. What is wrong with the wording of the question: “How many joules are there in a horsepower?” 6-89. What is wrong with the wording of the question: “What is the efficiency of an electric motor that has a power input of 5 hp and a power output of 4 hp?”

Section 6-4  The Kilowatt Hour 6-90. Why is the kilowatt hour often used instead of the joule as a unit for electric energy? 6-91. What unit is equivalent to a watt second?

Section 6-5  Relationships Among Basic Electric Units 6-92. Why must electric kettles have thermostatic switches to open the ­circuit when the kettle boils dry? 6-93. Some small heaters contain an electric heating coil and a small fan. The heating element glows brighter when the fan is turned off. ­Explain why. 6-94. Does the heater coil in Question 6-93 draw more current when the fan is turned off? Explain. 6-95. (a) What is the SI unit for heat? (b) How is the BTU defined?

Integrate the Concepts Most building codes require every fire exit in a public building to have a lighted exit sign that remains on at all times. One of the most common types of exit signs was designed to use two 15-W incandescent light bulbs. These two bulbs can now be replaced by a single panel of light-emitting diodes (LEDs) that uses only 3 W. How long will it take a $20 LED panel to pay for itself in energy savings? Assume an energy cost of $0.075/kW⋅h.

Practice Quiz 1. Which of the following statements are true? (a)   Energy is the capacity to do work. (b)   Power is the amount of work done per second. (c)   Efficiency is the ratio of input power to output power. (d)  One BTU is the amount of heat required to raise the temperature of one pound of water by 1°C.

121

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Chapter 6   Work and Power

2. How much power is used by a hair dryer that draws 1.5 A when ­connected to a 120-V wall socket? (a)   80 W (b)   112 W (c)   125 W (d)   180 W 3.

The highest voltage that can be applied to a 1.5-kΩ, 10-W resistor without risk of damaging it is (a)   121 V (b)   12.2 V (c)   124 V (d)   122 V

4. What physical property affects the power rating of a resistor? 5. If an amplifier has an efficiency of 75% and an input power of 15 W, the output power is (a)   5 W (b)   11.25 W (c)   1.13 kW (d)   50 mW 6. At 10¢ per kilowatt hour, how much will it cost to leave a 100-W lamp on for one week? (a)  16.8¢ (b)  24¢ (c)  $1.68 (d)  $16.8 7. If 135 kJ of heat is needed to raise 5.1 kg of a substance from 18.0°C to 31.5°C, the specific heat of the substance is about (a)  0.20 (b)  0.50 (c)  2.0 (d)  26

PART Resistance Networks

II

Part II explains the principles and techniques we require for the analysis of any resistance network. We can apply the relationships among current, voltage, resistance, work, and power to all electric circuits, no matter how complex they may be. 7

Series and Parallel Circuits

8

Series-Parallel Circuits

9

Resistance Networks

10 Equivalent-Circuit Theorems 11

Electrical Measurement

Photo source:  © iStock.com/omada

7

Series and Parallel Circuits All the electric circuits in residential, automotive, and ­industrial applications are based on two fundamental types of circuits: series and parallel. Working with sophisticated circuits requires a good understanding of series and parallel connections.

Chapter Outline 7-1

Resistors in Series  126

7-3

Double-Subscript Notation  130

7-5

Characteristics of Series Circuits  131

7-2

7-4 7-6 7-7

7-8 7-9

Voltage Drops in Series Circuits  128

Kirchhoff’s Voltage Law  130 Internal Resistance  133

Cells in Series  136

Maximum Power Transfer  137 Resistors in Parallel  139

7-10 Kirchhoff’s Current Law  141 7-11

7-12

Conductance and Conductivity  142

Characteristics of Parallel Circuits  145

7-13 Cells in Parallel  148

7-14 Troubleshooting 150

Key Terms series circuit  126 equivalent circuit  127 active circuit element 129 passive circuit element 129 polarity 129 Kirchhoff’s voltage law 130

internal resistance  135 open-circuit voltage  135 voltage regulation  135 controlled source  135 maximum power transfer 138 overload 139 parallel circuit  139 branch 140

duality 141 equivalent resistance  141 Kirchhoff’s current law 141 conductance 142 siemens 142 conductivity 145 short circuit  152

Learning Outcomes At the conclusion of this chapter, you will be able to: • recognize series and parallel circuits • calculate the total resistance in a series circuit • label the polarity of voltages across voltage sources and resistors in a series circuit • identify the polarity of a voltage using doublesubscript notation • apply Kirchhoff’s voltage law to a series circuit • relate the ratio of resistances to the voltage drops across resistors in a series circuit • explain the effect of internal resistance on the terminal voltage of a practical voltage source • calculate the terminal voltage of cells in series and the r­ esulting circuit current Photo sources:  © iStock.com/powerofforever

• state the condition under which a voltage source delivers maximum power to a load • apply Kirchhoff’s current law to a parallel circuit • calculate the total conductance in a parallel circuit • calculate the equivalent resistance of two resistors in ­parallel • define conductivity • relate the ratio of conductances to the currents through two resistors in a parallel circuit • calculate the terminal voltage of cells in parallel and the resulting circuit current • troubleshoot series and parallel circuits

126

Chapter 7   Series and Parallel Circuits

7-1  Resistors in Series A series circuit can be identified by the connection between components or by the current through them. For example, in the circuit of Figure 7-1, R1 and R2 are connected in series because no other component or branch is connected to the junction of R1 and R2. None of the junctions in this circuit have a second branch, so all of the components are in series. Thus, a series circuit has only one path for current. Conventional current I R1 + −

E

R2 R3 Electron f low

Figure 7-1  Simple series circuit

Since resistors cannot store charge, the flow of electrons into R1 is equal to the flow of electrons out of R1, which in turn is equal to the flow of electrons into R2. Similarly, the flow of electrons through the source is equal to the flow of electrons through any other part of the series circuit. Georg Ohm was the first person to recognize this key property of series circuits. The current is the same in all parts of a simple series circuit. Conversely, two or more electric components are in series if a common current flows through them. Since the current in a series circuit is common to all components, it is not necessary to use a subscript with I to distinguish the current through the various components. In the circuit of Figure 7-1, I represents the current through R1, R2, and R3, as well as the current through the source and the connecting wires. Suppose that resistor Rl is made with 2 m of Nichrome™ wire, while ­resistor R2 contains 1 m and R3 contains 3 m of the same wire. An electron flowing through the circuit passes through a total of 6 m of Nichrome™ wire. Since the resistance of an electric conductor is directly proportional to its length, the total resistance of this circuit is the sum of the individual ­resistances. Generalizing,

7-1   Resistors in Series

The total resistance of a series circuit equals the sum of all the individual resistances in the circuit: RT = R1 + R2 + R3 + . . .



(7-1)

Once we know the total resistance of a series circuit, we can use Ohm’s law to find the common current: I = E/RT This current produces a total voltage drop VT, equal to the applied voltage E. Since the current through all the resistors is equal, the voltage drops across equal resistances are also equal.

Example 7-1 What current will flow in a series circuit consisting of a 45-V source, a 20-Ω, a 10-Ω, and a 30-Ω resistor? Solution

RT = R1 + R2 + R3 = 20 Ω + 10 Ω + 30 Ω = 60 Ω I=



E 45 V = = 750 mA RT 60 Ω

The current drawn from the source is exactly the same for a single 60-Ω resistor connected to its terminals as for the 20-Ω, 10-Ω, and 30-Ω resistors connected in series. Therefore, we can think of the resistances in series as being replaced by a single equivalent resistance Req. Figure 7-2 shows the equivalent circuit for the series circuit in Figure 7-1. When we analyze more elaborate circuits in later chapters, we simplify a circuit by replacing two or more components in series with a single equivalent component.

+

E

Req

−

Figure 7-2  Equivalent circuit of Figure 7-1

127

128

Chapter 7   Series and Parallel Circuits

Example 7-2 Three resistors are connected in series as shown in Figure 7-1. R1 and R3 are 10 kΩ and 22 kΩ, respectively, and the potential difference of the source is 16 V. What resistance for R2 gives a current of 340 μA? Solution

circuitSIM walkthrough

RT =

E 16 V = = 47 kΩ I 340μA

R2 = RT − ( R1 + R3 ) = 47 kΩ − ( 10 kΩ + 22 kΩ ) = 15 kΩ

Multisim Solution Download Multisim file EX7-2 from the website. The circuit is Figure 7-1 with an ammeter inserted and with the component values as shown below: E = 20 V   R1 = 8.2 kΩ  R3 = 12 kΩ

An ohmmeter indicates a total resistance of 35.2 kΩ. Calculate the value of R2. Double-click on R2 in the circuit to reveal the value of R2. Calculate the value of the current in the circuit. Run the simulation for the circuit, and check the value of the current as displayed on the ammeter.

See Problems 7-1 to 7-4 and Review Questions 7-48 and 7-49 at the end of the chapter.

7-2 Voltage Drops in Series Circuits Ammeter is short for ampere meter.

Figure 7-3 shows the simple series circuit of Figure 7-1 with a voltmeter and an ammeter added. Although the positions of R1 and R3 have changed in the circuit diagram, the connections between the components are the same as in Figure 7-1. The ammeter must be connected in series with the ­circuit in order to measure the current through it. Since the voltmeter is connected between points B and C, it measures the potential difference across R2. The ideal voltmeter draws negligible current and thus does affect the behaviour of the circuit. In the circuit of Figure 7-3, electrons experience a potential rise inside the voltage source as it uses energy to move them from the positive terminal to the negative terminal. As electrons flow around the external circuit, they

7-2   Voltage Drops in Series Circuits

Conventional current

A +

Electron flow V1

B +

+ +

V2

V −

E

−

R1 −

R2

− C + V3

R3

− −

A

+

D

Figure 7-3  Polarity of voltage drops in a series circuit

lose the potential energy they gained while moving through the source. Hence, the electrons at point D are at a higher potential than those at point C, and electrons at point B are at a higher potential than those at point A. Similarly, electrons at point B are at a lower potential than those at point C, and so on. If negative ions in the voltage source are free to move, they flow in the same direction as the electrons. Any positive charge carriers flow in the ­op­posite direction. Positive charge carriers inside a voltage source experience a ­potential rise as they move from the negative to the positive terminal. A voltage source is an active circuit element since it generates electric ­energy (at the expense of some other form of energy). Passive circuit elements consume electric energy. Electrons flow from the negative terminal to the positive terminal of a passive component. Thus, the polarity of the voltage drops indicates the direction of current flows in these components. Note that an ammeter is a passive circuit element. Electrons enter the ­ammeter through its negative terminal and leave through its positive terminal. In the circuit of Figure 7-3, electrons flow counterclockwise around the circuit. We can, therefore, mark the polarities of all voltage drops with − and + signs. The end of the resistor at which electrons enter is marked − and the end of the resistor from which electrons leave is marked +. In the circuit of Figure 7-3, the positive end of R3 is connected to the negative end of R2, and so on. So, point C is positive with respect to point D, and point B is even more positive with respect to point D. Similarly, point C is negative with respect to point A, and so on. For circuit diagrams, the nature of the actual charge carriers is not particularly important, but we often need to keep track of current direction.

129

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Chapter 7   Series and Parallel Circuits

The direction of both conventional current and electron flow are shown in Figure 7-3. The polarities of the voltage drops are exactly the same using ­either approach, but conventional current “flows” from the positive terminal of a passive circuit element to its negative terminal. The remainder of the text will show only the direction of conventional current unless the nature of the actual charge carriers is important. See Review Question 7-50.

7-3  Double-Subscript Notation In Figure 7-3, V1 represents the voltage drop across R1. The polarity of V1 is obvious since terminal A is positive with respect to terminal B. Double-subscript notation lets us indicate the polarity of voltage drops. For example, VAB represents the potential at point A with respect to point B. The second subscript normally designates the reference point. In Figure 7-3, VAB is a positive voltage, VBA is negative, and VAB  = −VBA. Double-subscript notation is essential for keeping track of voltage ­polarities and current directions in three-phase circuits such as those in Chapter 29. For most of the dc circuits in this book, single-subscript notation is adequate.

7-4  Kirchhoff’s Voltage Law Multiplying both sides of Equation 7-1 by I gives IRT = IR1 + IR2 + IR3 + . . . Since

E = IRT    and    V = IR,

E = V1 + V2 + V3 + . . .

(7-2)

In a series circuit, the sum of all the voltage drops across the individual resistances equals the applied voltage. The German physicist Gustav Robert Kirchhoff (1824–1887) discovered that a similar relationship applies to any complete electric circuit. Kirchhoff’s voltage law: In any complete electric circuit, the algebraic sum of applied voltages equals the algebraic sum of the voltage drops. Kirchhoff’s voltage law provides us with another method for solving ­ xample 7-1. E

7-5   Characteristics of Series Circuits

Example 7-1A What current will flow in a series circuit consisting of a 45-V source, a 20-Ω, a 10-Ω, and a 30-Ω resistor? E = IR1 + IR2 + IR3

Solution

45 = 20I + 10I + 30I = 60I 45 V I= = 750 mA 60 Ω



See Review Question 7-51.

7-5  Characteristics of Series Circuits Since the current is the same through all components of a series circuit, ­applying Ohm’s law to the voltage drops gives I=

V1 V2 V3 VT = = = R1 R2 R3 RT

By transposing variables in each pair of equal terms, we get V1 R1 = , V2 R2



V2 R2 = , V3 R3

and so on.

In a series circuit, the ratio of the voltage drops across two resistances equals the ratio of the resistances. We can use these ratios to save a step when solving for a specific para­ meter of a series circuit.

Example 7-3 A 20-kΩ resistor and a 15-kΩ resistor are connected in series to a 140-V source. Find the voltage drop across the 15-kΩ resistor. Solutions Often, drawing and labelling a schematic diagram, such as Figure 7-4, can help us see the relationships among the circuit parameters. R1 20 kΩ

+ 140 V −

15 kΩ

R2

Figure 7-4  Circuit diagram for Example 7-3

131

132

Chapter 7   Series and Parallel Circuits

Source:  © iStock.com/powerofforever

Long Method   Use the current to find the voltage drop. RT = R1 + R2 = 20 kΩ + 15 kΩ = 35 kΩ



I=



E 140 V = = 4.0 × 10−3 A = 4.0 mA RT 35 kΩ

V2 = IR2 = 4.0 mA × 15 kΩ = 60 V



Short Method   Use ratios to find the voltage drop. Christmas lights are connected in series: if one bulb fails, the whole string will go out.

Since



R2 15 kΩ V2 R2 = 60 V = ,   V2 = E = 140 V × RT 35 kΩ E RT

Since the same current flows through all components of a series circuit, changing any one component affects the current through all the components. Therefore, control components such as switches, fuses, and rheostats are connected in series with a load, as shown in Figure 7-5.

Switch + E Source −

Fuse

Rheostat Lamp

Figure 7-5  Control components in series with a load

The following characteristics can help us recognize a series circuit: • The current is the same in all parts of a series circuit. • The total resistance is the sum of all the individual resistances: RT = R1 + R2 + R3 + . . . . • The applied voltage is equal to the sum of all the individual voltage drops: E = V1 + V2 + V3 + . . . . • The ratios of voltage drops equal the ratios of resistances. • A change to any component of a series circuit affects the current through all components. See Problems 7-5 to 7-14 and Review Questions 7-52 to 7-54.

7-6   Internal Resistance

Circuit Check

A

CC 7-1. Complete the table listing the voltages in the circuit of Figure 7-6. 8.0 Ω

A

B

2.0 Ω

C

5.0 Ω 12 V

D 5.0 Ω 12 Ω

F

E

Figure 7-6  VAF

VBF

VCF

VDF

VEF

VBE

VCE

VDE

VAC

VBC

VDC

VEC

CC 7-2. Find the unknown resistances in the circuit of Figure 7-7. 20 V R1

I + 100 V −

R2

3.0 Ω

R3

50 V Figure 7-7 

CC 7-3. What resistance connected in series with a 100-Ω resistor will dissipate heat at a rate of 20 W when the combination is connected to a 120-V source?

7-6  Internal Resistance In Section 2-8 we noted that the terminal voltage of a source is equal to the EMF of the source only under open-circuit conditions. The EMF depends on the nature of the energy-converting action in the source and is independent of the circuit current. However, as shown in Figure 7-8, the potential

133

Chapter 7   Series and Parallel Circuits

Open-circuit voltage

Rint E

+ −

Constant potential difference

Terminal voltage depends on load current

Termin a

l volta

VT

ge

0

Vint E

Voltage

134

VT

Current

Figure 7-8  Equivalent circuit for a practical voltage source

difference between the two terminals of a practical voltage source decreases as the current drawn from the source increases. For example, a drop in terminal voltage causes a car’s headlights to dim when the starter motor is drawing a large current from the car’s battery. In some circuits, the change in terminal voltage of the source is small enough to ignore. However, in many circuits, particularly electronic applications, a decrease in the terminal voltage of the source can significantly ­affect the operation of the circuit. For such circuits we can treat the practical voltage source as consisting of an internal resistance connected in series with an ideal source that has a constant potential difference (see Figure 7-8). We can use this equivalent circuit to calculate the terminal voltage and ­energy losses of the actual source. When no current is drawn from the practical source, the voltage drop across the internal resistance is Vint = IRint = 0 since I = 0. Therefore, the opencircuit voltage of the practical source is equal to E, the constant ­potential difference generated by the ideal source. The following example illustrates how we can determine the internal resistance of a practical ­voltage source.

Example 7-4 A voltmeter that draws negligible current reads 6.0 V when connected to the terminals of a battery with no other load. When a 5-Ω resistor is added, as shown in Figure 7-9, the voltmeter reads 5.0 V. Find the internal resistance of the battery.

Rint

V 5.0 V Ideal source

RL = 5 Ω

+ E 6.0 V −

Figure 7-9  Circuit diagram for Example 7-4

7-6   Internal Resistance

Solution Since the open-circuit voltage is 6.0 V, the potential difference of the ideal source is 6.0 V. Applying Kirchhoff’s voltage law gives

E = Vint + VL Vint = E − VL = 6.0 V − 5.0 V = 1.0 V

From Ohm’s law, the current in the circuit with the 5-Ω resistor connected is VL 5.0 V I= = = 1A RL 5Ω Vint 1.0 V Rint = = = 1Ω I 1A Alternative Solution Vint Rint = RL VL 1.0 V Vint Rint = RL = 5Ω × = 1Ω VL 5.0 V The terminal voltage of the voltage source is equal to the total of the voltage drops in the rest of the circuit, VT. We can use Kirchhoff’s voltage law to write an equation for the terminal voltage:

VT = E − IRint

(7-3)

where VT is the terminal voltage of a practical voltage source, E is the open-circuit voltage of the source, I is the current drawn from the source, and Rint is its internal resistance. If we can increase the EMF, E, of the voltage source to match the increase in the voltage drop IRint, the terminal voltage VT will remain constant. This arrangement is one type of voltage regulation. Since the EMF of a battery depends on the chemical reaction in the cells, we cannot apply this type of voltage regulation to batteries. However, the operators of powergenerating stations can increase the voltage generated as the current demand increases. In electronic power supplies, the source voltage can be regulated by a feedback circuit that monitors the output terminal voltage. A voltage source with a regulation circuit is called a controlled source. Section 10-3 discusses controlled sources in more detail. For heavy-duty sources, such as large storage batteries or power generators, the terminal voltage under normal load is essentially the same as the open-circuit voltage. Some smaller voltage sources do have relatively high internal resistances. For the circuits in this text, we shall consider the internal resistance of the sources to be zero, unless otherwise stated. See Problems 7-15 to 7-18 and Review Question 7-55.

135

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Chapter 7   Series and Parallel Circuits

7-7  Cells in Series Most cells have a nominal voltage between 1.1 V and 1.5 V. The current rating of a cell depends on its size and type. Identical cells connected in series provide higher voltages than a single cell. The current rating will be the same as for each individual cell since the current in a series circuit is the same in all parts of the circuit. If cells with different capacities are connected in series, the lower capacity cells will be used up first. Most batteries for portable ­devices have cells connected in series, like those shown in Figure 7-10. S cells in series

Ecell

Rcell

Rcell

Ecell

Ecell

Ebat

+

Rcell

−

Figure 7-10  Equivalent circuit of a series-connected battery

If S is the number of cells connected in series, Ecell is the EMF of each cell, and Rcell is the internal resistance of each cell, then Ebat = SEcell

(7-4)

total battery internal resistance is    Rbat = SRcell

(7-5)

total battery EMF is  and

Example 7-5

How much current flows through an external resistance of 4.0 Ω connected to a battery consisting of ten cells connected in series, with each cell having an EMF of 1.5 V and an internal resistance of 0.20 Ω? Find the terminal voltage of the battery. Solution Ebat = SEcell = 10 × 1.5 V = 15 V

Rbat = SRcell = 10 × 0.20 Ω = 2.0 Ω

The total resistance in the circuit, RT = Rbat + RL = 2.0 Ω + 4.0 Ω = 6.0 Ω Applying Ohm’s law to the whole circuit gives

7-8   Maximum Power Transfer

I=

or

137

Ebat 15 V = 2.5 A = RT 6.0 Ω

VT = IRL = 2.5 A × 4.0 Ω = 10 V

VT = Ebat − IRbat = 15 V − ( 2.5 A × 2.0 Ω ) = 10 V

Figure 7-11 shows the equivalent circuit for the battery. IL

R bat

2.0 Ω RL = 4.0 Ω

VT ET

15 V

Figure 7-11  Equivalent circuit for Example 7-5

Multisim Solution Download Multisim file EX7-5 from the website. The battery consists of ten cells, each having an EMF of 1.5 V and an internal resistance of 0.2 Ω. These cells and an external resistance of 4.0 Ω are all connected in series. Run the simulation and check the values of VT and IL displayed on the meters. Note that these values match those calculated using Equations 7-4 and 7-5. See Problems 7-19 to 7-22.

7-8  Maximum Power Transfer To determine how the internal resistance of the voltage source and the ­resistance of the load affect the behaviour of a circuit, we examine how ­various parameters of a circuit change as the load resistance increases. The ­internal resistance and voltage of a voltage source are independent of the external circuit. For the circuit of Figure 7-12, these parameters remain constant at Rint = 6.0 Ω and E = 120 V, so we can calculate other parameters of the circuit using the following equations:

current:

I=

E 120 = RL + Rint RL + 6.0

circuitSIM walkthrough

Chapter 7   Series and Parallel Circuits

VL = IRL

load voltage:

PL = I2 RL

load power:

internal resistance power: Pint = I2Rint = 6.0 I2 η=

efficiency:

Rint

PL × 100% PL + Pint

6.0 Ω RL

VT

+ Constant 120 V voltage −

Figure 7-12  Circuit for observing the effects of changing RL

15

90

10

60

5

30

0

0

Load power

Load voltage and efficiency

Load current

0

8

4 6Ω

12 16 20 24 28 32 Load Resistance, RL (Ω)

36

40

600

100

450

75

300

50

150

25

0

Efficiency, η (%)

120

Load Voltage, VL (V)

20

Load Power, PL (W)

We can use a spreadsheet, a graphing calculator, or circuit simulation software to graph these parameters for values of RL from 0 Ω to 40 Ω, as shown in Figure 7-13. Load Current, IL (A)

138

0

Figure 7-13  Effects of varying load resistance

Although none of the graphs in Figure 7-13 are linear, we can see that as the load resistance increases, the current decreases while the load voltage and the efficiency increase. In fact, the graphs for the load voltage and the efficiency have exactly the same shape. The load power rises to a maximum when RL reaches 6.0 Ω and then tapers off gradually as RL increases further. So, the maximum power transfer occurs when the load resistance has the same value as the internal resistance of the source. From graphs like Figure 7-13, we find that for all circuits the condition for maximum power transfer to the load is

RL = Rint

(7-6)

7-9   Resistors in Parallel

139

Further, when the load resistance equals the internal resistance of the source, 1 I = Isc 2



(7-7)

where Isc is the short-circuit current (the current when RL = 0). VL =



1 E 2

(7-8)

η = 50%



(7-9)

The maximum power transfer does not coincide with maximum efficiency. When RL = Rint, the power lost as heat in the internal resistance of the source is equal to the power transferred to the load. A load resistance that is two to three times the internal resistance of the source results in appreciable reduction in wasted power with only a small reduction in power output. For a circuit where the output voltage is more critical than the power output (as in transistor voltage amplifiers), the load resistance should be roughly five times the internal resistance of the source. A load resistance much less than the internal resistance of the source reduces the power ­output and the efficiency while increasing the heat produced in the internal resistance of the source. Such overloads can ruin a voltage source. As shown in Figure 7-13, the slope of the load power graph is zero at maximum power. Appendix 2-1 derives Equation 7-6 using basic differential calculus. See Problems 7-23 to 7-29 and Review Questions 7-56 to 7-61.

7-9  Resistors in Parallel Figure 7-14 shows two different ways of drawing a circuit diagram for the same simple parallel circuit. Circuit diagrams usually show interconnecting conductors as either horizontal or vertical lines, as in Figure 7-14(a). However, to illustrate the nature of a parallel circuit, we redraw the circuit diagram by combining the junctions that are directly connected to each other, producing Figure 7-14(b). IT

+

E

IT I1

I2

I3

R1

R2

R3

−

+

E

A I1

I2

I3

R1

R2

R3

−

B

IT

IT (a)

(b)

Figure 7-14  S  imple parallel circuit: (a) Customary configuration, (b) Equivalent configuration

Using arrows to mark the direction of the current in each branch of a circuit makes it easier to see how these currents combine into the total current.

140

Chapter 7   Series and Parallel Circuits

We can identify a parallel circuit by connections among the components. In Figure 7-14, E, R1, R2, and R3 are all in parallel because they are all connected between the same two points, A and B. Since each of the resistors is connected directly across the voltage source, V1 = V2 = V3 = E. We can omit the subscripts since the voltage is common to all the components ­connected in parallel. Two or more electric components are in parallel if a common voltage appears across all of the components. For a series circuit, we determined the total resistance in order to find the current in the circuit. For a parallel circuit, we find the total current first, and use it to determine the resistance of the circuit. Given the applied ­voltage and the values of each resistance in the circuit of Figure 7-14, we can solve for the current in each branch by using Ohm’s law. If we think of the current in each branch in terms of electrons flowing through the branch, it is apparent that the current through the source must be the sum of the branch currents. In a simple parallel circuit, the total current is the sum of all the branch currents: IT = I1 + I2 + I3 + . . .



Example 7-6

(7-10)

For the circuit in Figure 7-14, assume that R1 is 40 Ω, R2 is 30 Ω, R3 is 20 Ω, and E is 120 V. What single resistance would draw the same current from the source? Solution





I1 =

V1 120 V = = 3.0 A R1 40 Ω

I3 =

V3 120 V = = 6.0 A R3 20 Ω

I2 =

V2 120 V = = 4.0 A R2 30 Ω

7-10   Kirchhoff’s Current Law



141

IT = I1 + I2 + I3 = 3.0 + 4.0 + 6.0 = 13.0 A

Req =

E 120 V = = 9.23 Ω IT 13.0 A

Multisim Solution

circuitSIM

Download Multisim file EX7-6 from the website. The circuit is Figure 7-14 with the component values as indicated below: E = 20 V   R1 = 80 Ω  R2 = 50 Ω  R3 = 40 Ω

walkthrough

Calculate IT, I1, I2, and I3.

Run the simulation and check the values of IT, I1, I2, and I3 as displayed on the ammeters. For a series circuit, the total resistance is greater than the resistance of any individual resistor. However, the resistance in a parallel circuit (e.g., the combination of resistors in Example 7-6) is less than the resistance of any ­individual resistor. Hence, when dealing with parallel circuits, we do not speak of total resistance. Instead, we speak of the equivalent resistance, Req, of two or more resistors in parallel. Parallel circuits have characteristics that, in many respects, are similar but opposite to those of series circuits. See Problems 7-30 to 7-34.

Circuit Check

B

CC 7-4. If a battery has an open-circuit terminal voltage of 6 V and an internal resistance of 1 Ω, what will the terminal voltage be when the load current is 250 mA? CC 7-5. Calculate the internal resistance of a source that has an efficiency of 80% when connected to a 10-Ω load.

7-10  Kirchhoff’s Current Law Kirchhoff extended the total-current principle of Equation 7-10 to apply to all electric circuits. Kirchhoff’s current law: At any junction in a circuit, the algebraic sum of the currents entering the junction equals the algebraic sum of the currents leaving the junction.

The “similarbut-­opposite” characteristic of series and parallel electric circuits is sometimes referred to as the principle of ­duality.

142

Chapter 7   Series and Parallel Circuits

Example 7-7 Find the current in the R2 branch of the circuit of Figure 7-15. IT = 12 A

Node X

I2

I1

+ 50 V −

R1

R2

10 Ω

Figure 7-15  Circuit diagram for Example 7-7

Solution

I1 =

V 50 V = = 5.0 A R1 10 Ω

The current flowing into junction X is IT and the currents flowing away from junction X are I1 and I2. Therefore,

IT = I1 + I2 I2 = IT − I1 = 12 A − 5.0 A = 7 A

See Problem 7-35.

7-11  Conductance and Conductivity Before the adoption of SI units, the common name for the unit of conductance was the mho (ohm spelled backward). The siemens was named in honor of the G ­ erman engineer Ernst Werner von Siemens (1816–1892).

As Examples 7-6 and 7-7 demonstrate, the total current is always greater than the current through any branch of a parallel circuit. Therefore, the equivalent resistance is always less than the smallest of the branch resistances. The more resistors we connect in parallel, the smaller the equivalent resistance becomes. In other words, the more resistors we connect in parallel, the more readily the circuit can pass current since there are more parallel branches for current to flow through. Conductance is a measure of the ability of an electric circuit to pass current. The letter symbol for conductance is G. The siemens (S) is the SI unit of conductance: 1 S = 11Ω . Conductance is the reciprocal of resistance:



G=

1 R

(7-11)

7-11   Conductance and Conductivity

where G is the conductance of a circuit in siemens and R is the resistance of the same circuit in ohms. Dividing both sides of Equation 7-10 by E (or V, since they are the same for simple parallel circuits), we get IT I1 I2 I3 . . . = + + + E V V V



Substituting Ohm’s law into Equation 7-11 gives G=

Therefore,

1 1 I = = R V/I V

GT = G1 + G2 + G3 + . . .



(7-12)

In parallel circuits, the total conductance is equal to the sum of the conductances of all the individual branches.

The equivalent resistance is simply Req = 1/GT . We can use conductances to find the equivalent resistance for a parallel circuit without calculating the total current. Try Example 7-6 using conductance:

Example 7-6A

For the circuit in Figure 7-14, assume that R1 is 40 Ω, R2 is 30 Ω, R3 is 20 Ω, and E is 120 V. What single resistance would draw the same current from the source? Solution

1 1 1 + + = 0.1083 S 40 Ω 30 Ω 20 Ω 1 1 = = = 9.23 Ω GT 0.1083 S

GT = G1 + G2 + G3 = Req

When only two resistors are connected in parallel, Equation 7-12 becomes GT = G1 + G2 =

1 1 R1 + R2 + = R1 R2 R1R2

143

144

Chapter 7   Series and Parallel Circuits

Req =

and

R1R2 R1 + R2

(7-13)

For two resistors in parallel, the equivalent resistance equals their product over their sum. We can rearrange Equation 7-13 to find the resistance R2 that we must connect in parallel with a given resistance R1 to obtain a desired equivalent resistance: R2 =



R1 − Req R1Req



(7-14)

Now we can use Ohm’s law and Kirchhoff’s current law to find the ­current through resistor R2: V1 = V2 = I1R1 = I2R2

I1R1 = (IT − I2)R1 I2 = IT



(

R1 R1 + R2

)



(7-15)

When N equal resistors are connected in parallel, we can collect the terms in Equation 7-12 and show that Req =



R N

(7-16)

where R is the resistance of each of the parallel resistors and N is the number of resistors.

Example 7-8 What is the equivalent resistance of a 1-kΩ and a 4-kΩ resistor in ­parallel? Solution

Req = =

R1R2 1 kΩ × 4 kΩ 4 000 000 = = 800 Ω = ( 1 kΩ + 4 kΩ ) R1 + R2 5000

7-12   Characteristics of Parallel Circuits

We defined resistivity as the resistance of a unit length and cross section of a material. Conductivity is defined in a similar way. The conductivity of a material is the conductance of a unit length and cross section of that material. The letter symbol for conductivity is the Greek letter σ (sigma). Conductivity is measured in siemens per metre. Since conductance is the reciprocal of resistance, conductivity is the reciprocal of resistivity: σ=



1 ρ

(7-17)

See Problems 7-36 to 7-38 and Review Questions 7-62 to 7-64.

7-12 Characteristics of Parallel Circuits

Since V = IR and R = 1/G, V = I/G. The voltage is the same across all components connected in parallel, so V=



IT I1 I2 I3 = = = = ... GT G1 G2 G3

(7-18)

By transposing variables in each pair of equal terms, we get I1 G1 R2 = = , I2 G2 R1

I2 G2 R3 = = ,    and so on. I3 G3 R2

In a parallel circuit, the ratio between any two branch currents equals the ratio of their conductances or the inverse of the ratio of their ­resistances.

Example 7-9 The total current drawn by a 12.5-kΩ resistor and a 50-kΩ resistor in parallel is 15 mA. Find the current through the 50-kΩ resistor. Solution I From the circuit diagram of Figure 7-16,

I1 + I2 = 15 mA

I1 R2 50 kΩ = = = 4.0 I2 R1 12.5 kΩ I1 = 4.0 I2

145

146

Chapter 7   Series and Parallel Circuits

Substituting for I1 in the first equation gives

4.0I2 + I2 = 15 mA 15 mA I2 = = 3.0 mA 5.0



There is more than one way to solve most circuit problems. Generally, it is best to select the method in which we can ­visualize what each step represents.

+ R1

E −

12.5 kΩ

R2

50 kΩ

Figure 7-16  Circuit diagram for Example 7-9

Solution II

Req =

R1R2 12.5 kΩ × 50 kΩ = = 10 kΩ ( 12.5 kΩ + 50 kΩ ) R1 + R2

V = ITReq = 15 mA × 10 kΩ = 150 V



I2 =



V 150 V = = 3.0 mA R2 50 kΩ

When the internal resistance of the source is negligible, altering the ­resistance of one branch of a parallel circuit does not affect the voltage across, or the current through, the other branches. Therefore, changes in one branch of a parallel circuit have a negligible effect on the other branches. In house wiring, lighting circuits are connected in parallel so that switching one circuit on or off does not affect the operation of the other circuits (see Figure 7-17). 15-A fuse

Switches

Power outlets

Voltage source

Lighting fixtures Figure 7-17  Parallel connection of loads in house wiring

Source:  © iStock.com/jokerproporduction

7-12   Characteristics of Parallel Circuits

Streetlights connected in parallel

The following characteristics can help us recognize parallel circuits: • The voltage is the same across all components. • The total conductance is the sum of all the individual branch conductances: GT = G1 + G2 + . . .. • The total current is the sum of all the individual branch currents: IT = I1 + I2 + I3 + . . . . • The ratio between branch currents is the same as the conductance ratio and the inverse of the resistance ratio. • Each branch is independent of any changes in the other branches, providing the voltage across the parallel circuit is constant. See Problems 7-39 to 7-45 and Review Questions 7-65 to 7-70.

147

148

Chapter 7   Series and Parallel Circuits

Circuit Check

C

CC 7-6. The switches S1, S2, and S3 shown in Figure 7-18 are considered closed in the configuration shown. Determine the conditions of each switch in order to turn on lamps L1, L2, and L3, respectively. S2 S1 S3 E

L1 L2

L3

Figure 7-18 

CC 7-7. Which resistors of the networks shown in Figure 7-19 are connected in series and which are connected in parallel? R2 R1

R1

R4

R3

V1

R2

R3

V1

(a)

R4

(b)

Figure 7-19

CC 7-8. A 33-Ω, a 56-Ω, and a 100-Ω resistor are connected in parallel. If all three resistors are rated at ¼ W, what is the maximum voltage that can be applied without exceeding the power rating of one of the resistors? Which resistor will burn out first if the voltage increases above this maximum?

7-13  Cells in Parallel A battery of identical cells connected in parallel (see Figure 7-20) has the same EMF as a single cell, but can deliver a greater maximum current. The total capacity of the battery is proportional to the number of cells. If the cells do not all have the same EMF, currents will flow between them since the cells

7-13   Cells in Parallel

with lower EMFs will act as loads for the cells with higher EMFs. Standard AA-size cells are sometimes connected in parallel to make a compact battery with enough current capacity to run motors in devices such as cordless electric razors.

E

Rcell

E

Rcell

E

Rcell

E

Rcell

+

Ebat

P cells in parallel

−

Figure 7-20  Equivalent circuit of a parallel connected battery

In Figure 7-20, P is the number of cells connected in parallel, Ecell is the EMF of each cell, and Rcell is the internal resistance of each cell. The battery EMF is Ebat = E



and the total internal resistance of the battery is Rbat =



Rcell P

Example 7-10

(7-19)

How much current flows through an external resistance of 4.00 Ω connected to a battery consisting of ten cells connected in parallel, with each cell having an EMF of 1.50 V and an internal resistance of 0.20 Ω? Find the ­terminal voltage of the battery. Solution



Ebat = 1.50 V Rbat =

0.20 Ω Rcell = = 0.020 Ω P 10

149

150

Chapter 7   Series and Parallel Circuits

I=



E 1.50 V = 0.373 A = RT 4.00 Ω + 0.020 Ω

VT = IRL = 0.373 A × 4.00 Ω = 1.49 V



Figure 7-21 shows an equivalent circuit for the battery. IL

R bat

20 mΩ VT

Ebat

RL = 4.0 Ω

1.5 V

Figure 7-21  Equivalent circuit for Example 7-10

circuitSIM walkthrough

Multisim Solution Download Multisim file EX7-10 from the website. The simulated circuit corresponds to the one shown in Figure 7-21, with a battery that consists of ten cells connected in parallel, each having an EMF of 1.5 V and an internal resistance of 0.20 Ω. Run the simulation and check the values of VT and IL displayed on the meters. Note that these values match those calculated using Equation 7-19. See Problems 7-46 and 7-47.

7-14 Troubleshooting The basic instruments used to measure circuit variables are the ammeter for current, the voltmeter for voltage, and the ohmmeter for resistance. Most analog meters have a pointer that moves along a graduated scale. Digital meters display a numerical value for the measurement. Chapter 11 discusses meters in more detail.

The Ammeter The ammeter is a two-terminal device that measures the current flowing through it. Therefore, the ammeter must be connected in series with the component. Since the current through components in series is the same, the current indicated by the ammeter is also the current flowing through the component.

7-14  Troubleshooting

151

The Voltmeter

The voltmeter is a two-terminal device that measures the voltage across it. The voltmeter must be connected in parallel with a component. Since the voltage across parallel components is the same, the voltage indicated by the voltmeter is also the voltage across the component.

The Ohmmeter

Source:  TREVOR CLIFFORD PHOTOGRAPHY/SCIENCE PHOTO LIBRARY

The ohmmeter is a two-terminal device that indicates the resistance of a resistor connected across its terminals. A circuit should be turned off before an ohmmeter is connected to it. An ohmmeter connected across a network of resistors in series or parallel measures the equivalent resistance of the network. To measure the resistance of an individual resistor in a network, we first disconnect the resistor from the network.

An ammeter is connected in series with a component while a voltmeter is connected in parallel (across it).

Troubleshooting Techniques

Malfunctions of a circuit can be caused by an open circuit, a short circuit, or a wrong value for a component. Troubleshooting is the process of locating and correcting such problems. An open circuit can occur if one of the components, such as a resistor, is burnt out (usually the result of an excessive current through it) or if a wire or circuit-board trace is not making a good connection. These faults ­prevent current from flowing in that part of the circuit. An ammeter will indicate zero current, suggesting an open circuit.

152

Chapter 7   Series and Parallel Circuits

A short circuit is a fault that creates a low-resistance connection between two points in a circuit. Short circuits are often caused by accidental contact between wires, component leads, or traces on a printed circuit board. Short circuits also sometimes occur within a component. Since V = IR, the voltage is zero when the resistance is zero. Thus, a voltmeter connected across a shorted component indicates zero voltage. If a wrong component is installed in a circuit, measurements of current and voltage can help to locate the error.

Example 7-11

The series circuit in Example 7-2 should have a current of 340 μA, as shown in Figure 7-22. However, the ammeter reads 640 μA instead. Determine the most likely fault. 340.4 +

DCμA

10 kΩ

−

R1

+ 16 V E −

R2

15 kΩ

22 kΩ R3

Figure 7-22  Measuring current in the circuit for Example 7-11

Solution Step 1 Use a voltmeter to check the applied voltage as shown in Figure 7-23. The voltmeter reads 16 V, verifying proper operation of the source. 16.00 +

DC V

10 kΩ

−

R1

+ 16 V E −

R2

15 kΩ

22 kΩ R3

Figure 7-23  Checking the applied voltage in the circuit for Example 7-11

Step 2 Check the voltage across R1 with the voltmeter. Ohm’s law tells us that VR1 should be 10 kΩ × 640 μA = 6.40 V. The voltmeter reads 6.40 V, so we conclude that there is no problem with R1.

7-14  Troubleshooting

Step 3 Check the voltage across R2. VR2 should be 15 kΩ × 640 μA = 9.60 V. The voltmeter reads 9.60 V, and we conclude that there is no problem with R2. Step 4 Check the voltage across R3. VR3 should be 22 kΩ × 640 μA = 14.1 V. The voltmeter reads 0 V. Since VR3 = IR3 and I is not zero, R3 must be zero. So, we conclude that R3 is shorted.

Example 7-12 Calculation of the total current for the circuit shown in Figure 7-24 gives IT = 1.8 A, but the ammeter in the circuit reads only 1.2 A. Determine the most probable cause. 1.20 +

+ 120 V E −

DC A

−

R1

150 Ω

R2

300 Ω

R3

200 Ω

Figure 7-24  Diagram for Example 7-12

Solution Step 1 Use a voltmeter to check the applied voltage. The voltmeter reads 120 V, indicating proper operation of the source. We conclude the problem must be in the resistor network. Step 2 For safety, turn off the power supply at this point before you start removing resistors from the circuit in the next few steps. Step 3 Remove R1 from the circuit and check its resistance using an ohmmeter. Its value is 150 Ω, so we conclude that there is no problem with R1.

Step 4 Remove R2 from the circuit and measure its resistance. Its value is 300 Ω, so there is no problem with R2. Step 5 Remove R3 from the circuit and measure its resistance. Its resistance cannot be measured even on the highest scale of the ohmmeter. Therefore, R3 is open and must be replaced. See Review Questions 7-71 to 7-75.

153

154

Chapter 7   Series and Parallel Circuits

Summary

• Two or more components are connected in series if a common current flows through them. • The total resistance in a series circuit is the sum of the resistances of all the resistors in the circuit. • Either plus and minus signs or double-subscript notation may be used to show the polarity of applied voltages and voltage drops in a circuit. • In a complete electric circuit, the sum of the voltage drops equals the sum of the applied voltages. • In a series circuit, the ratio of the resistances of two resistors equals the ratio of their voltage drops. • Any change to any component in a series circuit affects the current though all components. • The internal resistance of a practical voltage source causes the terminal voltage to drop when a load draws current from the source. • Cells connected in series provide a higher voltage than a single cell. • A practical voltage source delivers maximum power to a load when the resistance of the load equals the internal resistance of the source. • Two or more electric components are connected in parallel if a common voltage appears across them. • In a parallel circuit, the total current is the sum of all the branch c­ urrents. • The algebraic sum of the currents entering a junction equals the algebraic sum of the currents leaving the junction. • In a parallel circuit, the total conductance is the sum of the conductances of all the branches. • For two resistors in parallel, the equivalent resistance equals the product of their resistances divided by their sum. • In a parallel circuit, the ratio of any two branch currents equals the ratio of their conductances. • Cells connected in parallel can provide a greater current than a sin­gle cell. • The ammeter, voltmeter, and ohmmeter are basic instruments used for troubleshooting circuits. B = beginner

Problems

I = intermediate

A = advanced

Draw a fully labelled schematic diagram for each problem. B B B

Section 7-1  Resistors in Series 7-1.

Find the total resistance of a 10-Ω resistor, a 20-Ω resistor, and a 30-Ω resistor connected in series. 7-2. Find the total resistance of a series circuit consisting of R1 = 47 kΩ, R2 = 220 kΩ, and R3 = 3.3 kΩ. 7-3. A string of Christmas tree lights consists of 50 identical mini lamps connected in series to a 120-V source. Given that the hot resistance of each lamp is 8 Ω, determine the current.

Problems

B

B I I I

I

I

A

I I B

7-4. A 1.2-kΩ, a 2.0-kΩ, an 850-Ω, and a 500-Ω resistor are connected in series to a voltage source. Determine the applied voltage that will cause a current of 26.4 mA to flow in this circuit.

Section 7-5  Characteristics of Series Circuits 7-5. If the series circuit of Problem 7-1 is connected to a 240-V source, what is the voltage drop across each resistor? 7-6. What resistance must be connected in series with a heater rated at  200 mA with a hot resistance of 270 Ω in order to operate it safely from a 120-V source? What power rating must this resistor have? 7-7. A 5-V electric cooler draws a current of 2.5 A. What resistance must be connected in series with it in order to operate it in a car with a 12-V battery? What power rating should the resistor have? 7-8. An electric stove element is rated at 300 W when connected to a 110-V source. Assuming negligible change in resistance for any change in temperature, determine the total rate of energy conversion when two of these elements are connected in series. 7-9. Three resistors are connected in series to a 120-V generator. The first has a resistance of 200 Ω, the second passes a current of 320 mA, and the third has a voltage drop of 24 V. Calculate the resistance of the second and third resistors. 7-10. Three resistors are connected in series to a 120-V source. The voltage drop across R1 and R2 together is 57 V, and the voltage drop across R2 and R3 together is 78 V. The total resistance is 12 kΩ. Find the resistance of each resistor. 7-11. A 50-W 22-kΩ resistor, a 100-W 82-kΩ resistor, and a 150-W 36-kΩ resistor are connected in series. What is the maximum voltage that can be applied to this network without exceeding the power rating of any of the resistors? 7-12. A 110-V Christmas tree light set consists of twelve 3-W lamps in series. Find the hot resistance of each lamp. 7-13. What resistance must be connected in series with a 100-Ω resistor for the 100-Ω resistor to produce heat at the rate of 30 W when the combination is connected to a 120-V source? 7-14. Calculate the voltage drops across each resistor in Figure 7-25. 10 Ω

5.0 Ω

+ 150 V −

15 Ω 20 Ω



Figure 7-25

155

156

Chapter 7   Series and Parallel Circuits

I

I A

I

I

A

I A

A

Section 7-6  Internal Resistance 7-15. The high-voltage power supply for a television’s picture tube has an open-­circuit voltage of 18 kV and an internal resistance of 1.5 MΩ. (a) Find the voltage at the anode of the picture tube when the anode current is 800 μA. (b) What will the anode voltage be if the anode current is increased by 50%? (c) Find the short-circuit current of the power supply. 7-16. A certain generator has a terminal voltage of 110 V when a 5.5-Ω load is connected to its terminals. The terminal voltage becomes 105 V when the load is 3.5 Ω. Calculate the internal resistance of the generator. 7-17. A 20-Ω resistor dissipates heat at the rate of 10 W when connected to a certain generator. If a 30-Ω resistor is connected in series with the 20-Ω resistor, the heat dissipation of the 20-Ω resistor becomes 200 mW. Find the open-circuit voltage of the generator. 7-18. A primary D cell’s open-circuit voltage is measured to be 1.51 V. Touching the leads of an ammeter to the cell’s terminals for a moment gives a reading of 5.0 A. Find the terminal voltage of the cell when it is connected to a 2.5-Ω load.

Section 7-7  Cells in Series 7-19. A 12-Ω load is connected to the terminals of a battery consisting of 12 cells connected in series. Each cell has an EMF of 1.5 V and an ­internal resistance of 0.20 Ω. Calculate the load current and the ­terminal voltage. 7-20. A battery consisting of five identical cells in series produces a current of 1.5 A when the load resistance is 4.0 Ω. When the load resistance is 9.0 Ω, the load current is 0.75 A. Calculate: (a) the internal EMF of each cell (b) the internal resistance of each cell 7-21. A battery is made from six identical 1.5-V cells connected in series. When a 12-Ω load resistor is connected to this battery, the terminal voltage is 8 V. Find the internal resistance of each cell. 7-22. A marine emergency battery room contains an array of ten 24-V marine batteries, each having an internal resistance of 0.05 Ω, connected in series. At what load power level will this array deliver a load voltage of 220 V?

Section 7-8  Maximum Power Transfer

7-23. A storage battery has an open-circuit voltage of 4.5 V and an internal resistance of 0.05 Ω. (a) Find the terminal voltage of the battery when a 0.4-Ω load is connected to it. (b) What power will be dissipated by a 0.7-Ω load?

157

Problems

A

I

I

B

I

I

B B I I

(c) Find the efficiency of the system when a 0.35-Ω load is connected to the battery. (d) Find the maximum power that a load can draw from the b ­ attery. 7-24. A generator has an open-circuit voltage of 32 V and an internal resistance of 0.2 Ω. The generator is designed for a constant-duty power output of 300 W. (a) What resistance will draw energy from the generator at the rate of 300 W? (b) Find the efficiency of the generator with this load. 7-25. The magnetic cartridge on a turntable develops an open-circuit voltage of 100 mV and has an internal resistance of 1200 Ω. (a) Find the output voltage when the cartridge is connected to a 5.2‑kΩ load. (b) What value of load resistor must be used to obtain 75% ­efficiency? (c) What is the maximum power output of the cartridge? 7-26. A generator with an internal resistance of 1.0 Ω feeds a load at the end of a line, each wire of which has a resistance of 2.0 Ω. The generator is adjusted to produce 120 V across the load at full load. If the total heat dissipation of the two wires is not to exceed 10% of the total generated power, what is the maximum power that can be delivered to the load? 7-27. An audio amplifier has an internal resistance at its output terminals of 4 Ω. The speaker is connected by a cable, each wire of which has a resistance of 2 Ω. What condition must be met to obtain maximum power in the speaker? 7-28. A 30-V source with an internal resistance of 1.0 Ω is connected to a 7.0-Ω load. Calculate (a) power delivered to the load (b) efficiency of power transfer (c)  maximum power that this source can deliver (d) current flowing at maximum power point 7-29. Download Multisim file P7-29 from the website. Run the simulation, and adjust the value of RL until VT = E/2. Then compare the values of RL and Rint.

Section 7-9  Resistors in Parallel 7-30. Find the total current through a 10-Ω resistor, a 20-Ω resistor, and a 30-Ω resistor connected in parallel to a 120-V source. 7-31. Find the equivalent resistance for a parallel circuit consisting of R1 = 47 kΩ, R2 = 220 kΩ, and R3 = 3.3 kΩ. 7-32. The equivalent resistance of three resistors in parallel is 6 kΩ. If R1 is 10 kΩ and R2 is 20 kΩ, what is the resistance of R3? 7-33. What resistance connected in parallel with a 1-kΩ resistor will result in a total current of 50 mA drawn from a 10-V source?

circuitSIM walkthrough

158

Chapter 7   Series and Parallel Circuits

A

I

B I I

I I

A I I

B B

7-34. Two resistors connected in parallel draw a total current of 2 A from a 90-V supply. One of the resistors is three times the value of the other. Calculate the value of each resistor.

Section 7-10  Kirchhoff’s Current Law 7-35. A source connected to three resistors in parallel supplies a total current of 6.5 mA. One resistor is 10 kΩ and has a voltage drop of 4.5 V. The second resistor has a current of 750 µA. How much current is flowing through the third resistor?

Section 7-11  Conductance and Conductivity

7-36. A circuit element having a conductance of 150 μS is connected in parallel with a branch having a conductance of 750 mS. Find the equivalent resistance of this circuit. 7-37. What resistance must be placed in parallel with a 20-kΩ resistor to reduce the equivalent resistance to 15 kΩ? 7-38. Find the total conductance of a 10-kΩ, 22-kΩ, and 15-kΩ resistor connected in parallel. How much current will flow if a 20 V source is connected to this circuit?

Section 7-12  Characteristics of Parallel Circuits 7-39. Three lamps operating in parallel on a 110-V circuit are rated at 40 W, 60 W, and 100 W. What is the equivalent hot resistance of this load? 7-40. Three resistors in parallel pass a total current of 0.60 A. The first resistor has a resistance of 400 Ω, the second passes a current of 60 mA, and the third has a voltage drop across it of 150 V. Calculate the ­resistance of the second and third resistors. 7-41. If the three resistors in Problem 7-31 are each rated at 0.5 W, what is the maximum total current that the network can handle without overheating any resistor? 7-42. The total current passed by a 10-kΩ, a 15-kΩ, and a 20-kΩ resistor in parallel is 20 mA. What is the current through each branch? 7-43. Three resistors connected in parallel have an equivalent resistance of 2.5 kΩ. R1 has a resistance of 15 kΩ, R2 has a voltage drop of 25 V, and R3 produces heat at a rate of 25 mW. Determine R2 and R3. 7-44. A 500-Ω resistor and a 1-kΩ resistor are connected in parallel. How much resistance must be connected in parallel with this combination to draw a current of 100 mA from a 30 V source? 7-45. A 100-Ω resistor and a 50-Ω resistor are connected in parallel and  a  total current of 3A flows. What is the value of the source ­voltage?

159

Review Questions

B B

Section 7-13  Cells in Parallel 7-46. A 0.525-Ω resistor is connected to the terminals of a battery consisting of four 1.46-V cells connected in parallel. If the load current is 0.80 A, find the internal resistance of each cell. 7-47. Download Multisim file P7-47 from the website. Run the simulation, and measure IT with R1 = 150 Ω, R2 = 300 Ω, and R3 = 200 Ω. Then use a simulation to determine what happens to IT when R3 is open.

circuitSIM walkthrough

Review Questions

Section 7-1  Resistors in Series 7-48. Justify the statement that the current is the same in all parts of a simple series circuit. 7-49. Justify the statement that the total resistance of a series circuit is the sum of all the individual resistances.

Section 7-2  Voltage Drops in Series Circuits 7-50. In the electric circuit of Figure 7-26, one of the “black boxes” represents a voltage source and the other a resistor. The voltmeter and ammeter read correctly when their terminals are connected with the polarity markers as shown. Which “black box” is the source?

Section 7-4  Kirchhoff’s Voltage Law 7-51. Kirchhoff’s voltage law may be stated as “The voltage between any two points in a network is the same via any path between those two points.” Verify this statement for two ends of R2 in Figure 7-1, using the values given in Example 7-1.

Section 7-5  Characteristics of Series Circuits 7-52. Prove that the ratio between any two power dissipations in a series circuit is the same as the ratio between the two resistances. 7-53. Two 120-V lamps, rated at 100 W and 25 W, are connected in series to a 120-V source. Which one will glow more brightly? Explain. 7-54. Explain the visible effect when the 25-W lamp in Review Question 7-53 is (a) open-circuit (b) short-circuit

Section 7-6  Internal Resistance 7-55. How would you determine the internal resistance of a given flashlight battery?

1

−

V

+ A − 2 Figure 7-26

+

160

Chapter 7   Series and Parallel Circuits

Section 7-8  Maximum Power Transfer 7-56. Explain the shape of the graph of load power versus load resistance in Figure 7-13. 7-57. Express the condition for maximum power output in relation to short-circuit and open-circuit current. 7-58. Although there are two numerically correct answers to Problem 7-24(a), we select only one as the electrically correct answer. Why? 7-59. Generator B develops three times the open-circuit voltage of generator A but has three times as great an internal resistance. Each generator is developing the same power in its load. Which generator has the greater efficiency? 7-60. A generator is operated in such a manner that its terminal voltage is kept constant by increasing its generated potential difference as  the load current is increased. Is this system 100% efficient?­ Explain. 7-61. In electronic equipment, the load resistance is often made equal to the internal resistance. Why is the load resistance not set this way for a power-generating station?

Section 7-11  Conductance and Conductivity 7-62. What is the disadvantage of a series-circuit string of decorative lights as compared to a parallel-circuit string? 7-63. Why must the equivalent resistance of a parallel circuit always be less than that of the smallest of the branch resistances? 7-64. Show that the statement “the total current in a parallel circuit is the sum of all the branch currents” confirms Kirchhoff’s current law.

Section 7-12  Characteristics of Parallel Circuits 7-65. When working with parallel circuits, why is it preferable to think in terms of conductance rather than resistance? 7-66. If the two lamps in Question 7-53 are connected in parallel, explain the visible effect when the 25-W lamp is (a) open-circuit and (b) short-circuit. 7-67. An electric stove element has a resistance of 50 Ω with a centre-tap connection. Draw circuit diagrams to show three means of connecting it to a 110-V source to obtain three different rates of conversion of electric energy to heat. 7-68. Most homes have a lighting fixture that can be switched on or off from two different locations. This is accomplished with two single-pole, double-throw switches (which electricians call “threeway switches”). Draw a circuit diagram for a light controlled in this way. 7-69. Figure 7-17 is not a simple parallel circuit because the fuse is in series with the remainder of the circuit. Why must the fuse be located in the position shown?

Practice Quiz

7-70. Two 50-kΩ resistors are connected in series across a voltage source. If a voltmeter with a resistance of 100 kΩ is connected first across one resistor and then the other and finally across both, the sum of the first two readings does not equal the third reading. Is this third reading greater or less than the sum of the other two? ­Explain.

Section 7-14  Troubleshooting 7-71. How must an ammeter be connected into a circuit to measure c­ urrent through a component? 7-72. How must a voltmeter be connected into a circuit to measure voltage across a component? 7-73. Why is it often necessary to disconnect a resistor from a circuit b ­ efore measuring the resistance of the resistor? 7-74. List two probable causes for the ammeter in Figure 7-22 reading less than 340 μA. 7-75. List two probable causes for the ammeter in Figure 7-24 reading more than 1.8 A.

Integrate the Concepts One string of patio lights has 20 bulbs connected in series, with each bulb having a resistance of 30 Ω when hot. Another brand of patio lights has 20 bulbs connected in parallel with each bulb having a resistance of 12 kΩ when hot. Both strings run on 120 V. For each string, determine (a) how much current flows through each bulb (b) how much power each bulb consumes (c) how much power the whole string uses (d) what happens to the remaining bulbs when one burns out

Practice Quiz 1.

Four resistors are connected in series, and the current through the first resistor is 2 A. The current through the third resistor is (a)   0.67 A (b)   2 A (c)   6 A (d)   0.3 A

2.

When a third resistor is connected in series with two other resistors, the total resistance of the circuit (a)  decreases (b)   remains the same (c)  increases (d)  may either increase or decrease depending on the relative values of the resistors

161

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Chapter 7   Series and Parallel Circuits

3.

Kirchhoff’s voltage law states that (a)  the algebraic sum of applied voltages equals the algebraic sum of the voltage drops around a closed loop (b)  the algebraic sum of the resistances is equal to the sum of the voltages (c)  the algebraic sum of the individual currents around a closed loop is zero (d)  the voltages developed across each element in a series circuit are identical

4.

Identify the active and passive circuit elements in Figure 7-27. R1

R1

R2

Vs

R2

Vs

R3

R4 (a)

(b)

Figure 7-27 5.

Show the direction of conventional current through each component of the circuits in Figure 7-27.

6.

A 1.0-kΩ resistor, a 2.2-kΩ resistor, and a 10-kΩ resistor are in parallel. The equivalent resistance is approximately (a)   11 kΩ (b)   13 kΩ (c)  643 Ω (d)   1.7 MΩ

7.

If one resistor in a parallel circuit is removed, the total resistance (a)  decreases (b)  increases (c)   remains the same (d)  may either increase or decrease depending on the relative values of the resistors

8.

In a parallel circuit, the largest resistance has the (a)   most voltage (b)   least voltage (c)   most current (d)   least current

Practice Quiz

9.

Determine the current flowing in each branch of the circuit in ­Figure 7-28.

Vs 20 V



R1 510 Ω

R2 680 Ω

R3 1.0 kΩ

Figure 7-28

10. Conductance is a measure of the ability of an electric circuit to (a)   pass current (b)   limit current (c)   limit voltage (d)   pass voltage

163

8

Series-Parallel Circuits In Chapter 7, we considered only simple series and parallel circuits. ­Although electric networks are seldom that simple, they are often combinations of series and parallel ­circuits. This chapter describes techniques for analyzing such combinations.

Chapter Outline 8-1

Series-Parallel Resistors  166

8-3

Kirchhoff’s Laws Method  171

8-2

Equivalent-Circuit Method  167

8-4

Voltage-Divider Principle  173

8-6

Current-Divider Principle  181

8-5 8-7

Voltage Dividers  175

Cells in Series-Parallel  184

8-8 Troubleshooting 186

Key Terms series-parallel circuit  167 ladder network  168 voltage-divider principle  174

bleeder resistor  176 voltage regulation  176 chassis 177

ground 177 current-divider principle  181

Learning Outcomes At the conclusion of this chapter, you will be able to: • simplify a series-parallel circuit into a series or a parallel circuit • analyze a series-parallel circuit to calculate ­voltage, c ­ urrent, and power • apply Kirchhoff’s voltage and current laws to series-­parallel circuits • state the voltage-divider principle • calculate voltages across resistors in series using the ­voltage-divider principle • calculate an appropriate value for a bleeder resistor in a voltage divider

Photo sources:  © iStock.com/james steidl

• reference a voltage at any point in a circuit with respect to ground • state the current-divider principle • calculate currents through resistors in parallel using the current-divider principle • calculate the terminal voltage and total resistance of a s ­ eries-parallel combination of cells • troubleshoot series-parallel circuits

166

Chapter 8   Series-Parallel Circuits

8-1  Series-Parallel Resistors Considered as a whole, the circuit in Figure 8-1 is neither a series nor a ­parallel circuit. However, R2 and R3 are connected between the same two points in the circuit and must have the same voltage drop. Therefore these two resistors are in parallel, and we can calculate a single equivalent resistance for them. As measured from the terminals of the voltage source, the simplified circuit of Figure 8-2 is equivalent to the original circuit of Figure 8-1. In Figure 8-2, R1 is in series with the equivalent resistance of R2 and R3 in parallel. Therefore, we can solve the circuit of Figure 8-2 as a simple series circuit.

I1 I2

R1

+

I3 R2

E

R3

−

Figure 8-1  Simple series-parallel circuit

I1 I2 + I3

R1

+

Req =

E

−

R2R3 R2 + R3

Source:  © Sundeepgoel/GetStock/Dreamstime

Figure 8-2  Equivalent circuit for Figure 8-1

In the circuit shown in Figure 8-3(a), R2 and R3 have the same current through them and are therefore in series. We can replace them with an equivalent resistor, as shown in Figure 8-3(b). We can now solve the ­simplified circuit of Figure 8-3(b) as a simple parallel circuit.

R2

+ In this flashlight, a battery of cells connected in series provides power to two lamps connected in parallel.

E

R1

R3

−

+

R1

E

− (a)

(b)

Figure 8-3  (a) Series-parallel circuit; (b) Equivalent circuit

Req = R2 + R3

8-2   Equivalent-Circuit Method

We define a series-parallel circuit as one in which some portions of the circuit have the characteristics of simple series circuits while the other portions have the characteristics of simple parallel circuits. See Review Questions 8-40 to 8-42 at the end of the chapter.

8-2  Equivalent-Circuit Method We can solve series-parallel circuits by substituting the equivalent resistances for various portions of the circuit until the original circuit is reduced to either a simple series or a simple parallel circuit.

Example 8-1 Find the voltage drop, current, and power for each resistor in the circuit diagram of Figure 8-4. Solution Step 1 Draw a fully labelled schematic diagram for this particular circuit. I1

R1

R1 I2

12 Ω

+ 100 V −

R2

I3

10 Ω

R3

40 Ω

12 Ω

+ 100 V −

Veq Req

Figure 8-4  Circuit diagram for Example 8-1

Step 2 Since R2 and R3 are in parallel,

Req =

R2R3 10 Ω × 40 Ω = = 8.0 Ω ( 10 Ω + 40 Ω ) R2 + R3

The total resistance of the circuit is Step 3 From Ohm’s law,

RT = R1 + Req = 12 + 8.0 = 20 Ω IT =

E 100 V = = 5.0 A RT 20 Ω

Step 4 Since R1 is in series with the source,

I1 = IT = 5.0 A

8.0 Ω

167

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Chapter 8   Series-Parallel Circuits

Step 5 Applying Ohm’s law to R1, To verify our ­calcula­tions, we can check that the currents we calculated satisfy Kirchhoff’s current law for this circuit: I1 = I2 + I3.

Step 6 From Kirchhoff’s voltage law,

circuitSIM walkthrough

Veq = E − V1 = 100 V − 60 V = 40 V

Returning now to the original circuit,

V2 = V3 = Veq = 40 V

Step 7 Applying Ohm’s law to R2 and R3, V2 I2 = = R2 V3 I3 = = R3 Since P = VI,

Step 8 We can verify our ­calculations by ­checking that PT = P1 + P2 + P3.

V1 = I1R1 = 5.0 A × 12 Ω = 60 V

40 V = 4.0 A 10 Ω 40 V = 1.0 A 40 Ω

P1 = V1I1 = 60 V × 5.0 A = 0.30 kW

P2 = V2I2 = 40 V × 4.0 A = 0.16 kW P3 = V3I3 = 40 V × 1.0 A = 40 W

PT = VTIT = 100 V × 5.0 A = 0.50 kW Multisim Solution Download Multisim file EX8-1 from the website.

Connect a ground to the bottom node of the circuit. Insert current meters to measure I1, I2, and I3. Run the simulation, and note the currents displayed on the ammeters. Confirm that these values for I1, I2, and I3 match those calculated using equivalent resistances in the first solution for this example.

Example 8-2 Find the resistance at the input terminals of the ladder network in ­Fig­ure 8-5 with (a)   the output terminals open-circuit (b)  a 600-Ω load resistance connected to the output terminals

8-2   Equivalent-Circuit Method

100 Ω

200 Ω

A

Input

A

800 Ω

B

100 Ω

200 Ω

100 Ω

Output

800 Ω

B

100 Ω

Figure 8-5  Circuit diagram for Example 8-2

Solution (a)  Step 1 With the output terminals open-circuit, no current flows in the two right-hand 100-Ω resistors. Thus they are effectively out of the circuit, which is therefore equivalent to Figure 8-6(a). 100 Ω

200 Ω

A

Input

800 Ω

100 Ω

B

100 Ω

800 Ω

A

Input

200 Ω

800 Ω

100 Ω

(a)

1200 Ω

B (b)

100 Ω

Input

480 Ω

100 Ω (c)

Input

680 Ω

(d)

Figure 8-6  Equivalent circuits for Example 8-2(a)





Step 2 The two 200-Ω resistors and the right-hand 800-Ω resistor have the same current through them. Hence they have the characteristics of a ­series ­circuit: Req = 200 Ω + 800 Ω + 200 Ω = 1200 Ω

Substituting Req for the three resistors gives the equivalent circuit of Figure 8-6(b).

Step 3 Since the 800-Ω and 1200-Ω resistors in the equivalent circuit of Fig­ure  8-6(b) are connected between the same junction points, they are in parallel.

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Chapter 8   Series-Parallel Circuits

Req =



800 × 1200 = 480 Ω ( 800 + 1200 )

Step 4 Since the equivalent circuit in Figure 8-6(c) is a simple series circuit,



Rin = 100 Ω + 480 Ω + 100 Ω = 680 Ω



(b)  Step 1 A 600-Ω load resistor connected to the output terminals is in series with the two right-hand 100-Ω resistors. Hence, Req = 100 Ω + 600 Ω + 100 Ω = 800 Ω









Step 2 We can replace the two 800-Ω resistors in parallel in the equivalent circuit of Figure 8-7(a) with a single 400-Ω resistance, as in Figure 8-7(b). Step 3 The two 200-Ω resistors and the 400-Ω resistor of Figure 8-7(b) are in ­series, so we can replace them with a single 800-Ω equivalent resistance, as in ­Figure 8-7(c). Step 4 Once again, two 800-Ω resistors in parallel are equivalent to a single 400-Ω resistance, giving the equivalent circuit in Figure 8-7(d). Step 5 Finally, the total resistance of the series circuit in Figure 8-7(d) is

100 Ω

200 Ω

A

Input

Rin = 100 Ω + 400 Ω + 100 Ω = 600 Ω

800 Ω

100 Ω

B

100 Ω

A

800 Ω

200 Ω (a)

B

100 Ω

A

Input

200 Ω

800 Ω B

100 Ω

200 Ω

(b)

Input

100 Ω

800 Ω

100 Ω

800 Ω

A

800 Ω

Input

B (c)

Figure 8-7  Equivalent circuits for Example 8-2(b)

See Problems 8-1 to 8-13.

400 Ω

100 Ω (d)

400 Ω

8-3   Kirchhoff’s Laws Method

8-3  Kirchhoff’s Laws Method Kirchhoff’s laws provide us with a method for solving series-parallel ­circuits without reducing them to simple series or parallel circuits by substituting equivalent resistances.

Kirchoff’s Laws Solution to Example 8-1 Applying Kirchhoff’s current law to the circuit in Figure 8-4 gives I1 = I2 + I3

From Ohm’s law, I1 = V1/R1, I2 = V2/R2, and I3 = V3/R3. Therefore, V1 V2 V3 V1 V2 V3 = + = +    and    R1 R2 R3 12 Ω 10 Ω 40 Ω



Since R2 and R3 are in parallel, V2 = V3. Combining this equation with Kirchhoff’s voltage law gives V2 = V3 = E − V1 Substitution in the preceding equation gives

V1 100 V − V1 100 V − V1 = + 12 Ω 10 Ω 40 Ω



Multiplying by 120 gives

10V1 = 1200 − 12V1 + 300 − 3V1 25V1 = 1500

V1 = 60 V Therefore,

V2 = V3 = E − V1 = 100 − 60 = 40 V

We can now use Ohm’s law to determine the various currents: I1 =



I3 =

V1 60 V V2 40 V = = 5.0 A   I2 = = = 4.0 A R1 12 Ω R2 10 Ω V3 40 V = = 1.0 A R3 40 Ω

If we do not know all the resistances in a circuit, we cannot readily solve for the equivalent resistances to reduce the circuit to a simple series or parallel circuit. However, we can use Kirchhoff’s laws to calculate the unknown resistance.

171

172

Chapter 8   Series-Parallel Circuits

Example 8-3 A resistor passing a 20-mA current is in parallel with a 5.0-kΩ resistor. This combination is in series with another 5.0-kΩ resistor, and the whole network is connected to a 500-V source. (See Figure 8-8.) Find the resistance of the resistor that is passing the 20-mA current. I1

R1 5.0 kΩ

+ 500 V −

5.0 kΩ

I2

I3 = 20 mA

R2

R3

Figure 8-8  Circuit diagram for Example 8-3

Solution Applying Kirchhoff’s voltage law,

V1 + V2 = E Substituting V = IR,

I1R1 + I2R2 = E

Applying Kirchhoff’s current law,

I2 = I1 − 0.020 A Therefore,

R1I1 + R2(I1 − 0.020) = E

Substituting given values for R1, R2, and E,

5000I1 + 5000(I1 − 0.020) = 500

10 000I1 = 600



I1 = 60 mA Therefore, and

I2 = I1 − 0.020 A = 0.060 − 0.020 = 0.040 A = 40 mA V2 = I2R2 = 40 mA × 5 kΩ = 200 V

Since R2 and R3 are in parallel, V2 = V3 and

R3 =

V3 200 V = = 10 kΩ I3 20 mA

It is also possible to start the solution by applying Kirchhoff’s current law first and then substituting I = V/R into I1 = I2 + 0.020 A.

See Problems 8-14 to 8-17 and Review Questions 8-43 and 8-44.

8-4   Voltage-Divider Principle

Circuit Check

CC 8-1. Calculate the equivalent resistance of the network in Figure 8-9.

A

400 Ω

5.0 kΩ 500 Ω 15 kΩ 600 Ω

Figure 8-9

CC 8-2. Calculate the resistance measured from terminal T to ground in the network in Figure 8-10. T

10 kΩ

27 kΩ

2.2 kΩ

10 kΩ

2.2 kΩ

Figure 8-10

8-4  Voltage-Divider Principle For the series circuit of Figure 8-11, Kirchhoff’s voltage law states that E = V1 + V2 + V3. In other words, the total applied voltage is divided among the three resistors. We can connect a voltmeter across six possible combinations of the terminals A, B, C, and D, as shown in Figure 8-11. Thus, the ­series combination of R1, R2, and R3 acts as a voltage divider. A R1

200 V

20 kΩ B

+ 350 V −

R2

250 V 50 V

5 kΩ

150 V

C R3

100 V

10 kΩ

350 V

D

Figure 8-11  Voltage-divider principle

173

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Chapter 8   Series-Parallel Circuits

To solve for the six voltages in Figure 8-11, we could calculate RT, find I from Ohm’s law, and then calculate the voltage drop across each r­ esistance from V1 = IR1, V2 = IR2, and so on. However, as demonstrated in Example 7-3, we can calculate each voltage in a single step. Section 7-5 described a property of series circuits that is often called the voltage-­divider principle: In a series circuit, the ratio between any two voltage drops equals the ratio of the two resistances across which these voltage drops occur. Hence, for any resistor Rn in a voltage divider, Vn Rn = E RT



Vn = E

and

Rn RT

(8-1)

Example 8-4 What is the voltage between terminals B and D in the circuit of ­Figure 8-11? Solution

VBD = E



R2 + R3 R1 + R2 + R3

( 5.0 kΩ + 10.0 kΩ ) ( 20.0 kΩ + 5.0 kΩ + 10.0 kΩ ) 15.0 kΩ = 350 V × 35.0 kΩ = 150 V = 350 V ×



Using a variable resistor as a voltage divider provides a continuously variable terminal voltage, as shown in Figure 8-12. A E

RT

C Vout

B

D

Figure 8-12  Potentiometer circuit

See Review Questions 8-45 and 8-46.

8-5   Voltage Dividers

8-5  Voltage Dividers Voltage dividers are widely used in power supplies for electronic ­circuits so that a single voltage source can supply all the various voltages ­required by a piece of equipment, reducing the cost, size, and weight of the ­equipment. The series-dropping resistor of Figure 8-13 provides the simplest method of obtaining the required voltage drop across a particular circuit e­ lement. RS

+ 25 V −

Series-dropping resistor

15 V

RL

Figure 8-13  Series-dropping resistor

Example 8-5 A portion of an electronic circuit requires an operating voltage of 15 V and a current of 20 mA. If the supply terminal voltage is 25 V, what value of series-dropping resistor is required? Solution From Kirchhoff’s voltage law, the voltage drop across the series-­ dropping resistor must be VD = E − VL = 25 V − 15 V = 10 V ID = IL = 20 mA

Since the dropping resistor and the load are in series,

RS =

VD 10 V = = 500Ω ID 20 mA

To complete the design, we find the minimum power rating for RS: P = VI = 10 V × 20 mA = 0.20 W

A 0.5-W resistor would be a good choice since it will operate at a lower temperature than a 0.25-W resistor and thus be less likely to fail. The advantage of the simple series-dropping resistor is that the only drain on the power supply is the current through the load. But this circuit has the disadvantage that any change in load resistance changes the current

175

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Chapter 8   Series-Parallel Circuits

through the series-dropping resistor, which, in turn, changes the voltage drop across the resistor and the voltage supplied to the load. In circuits where the load is a transistor, RL can become very high. Under such circumstances, the voltage drop across a series-dropping resistor is ­almost zero and the voltage across the load is close to the full applied ­voltage, which may be high enough to damage the transistor. To prevent such damage, some voltage-divider designs include a bleeder resistor in parallel with the load, as in Figure 8-14. The bleeder resistor ensures there is always enough current through the series-dropping resistor to maintain an appreciable voltage drop across it. Hence, the load voltage cannot rise to the full applied voltage. RS Series-dropping resistor + Bleeder 25 V resistor −

IL = 20 mA

RB

RL

VL = 15 V

Figure 8-14  Voltage divider with bleeder resistor

In power supplies for electronic equipment, a bleeder current of 10–25% of the total current drawn from the source provides sufficient protection against excessive load voltage when the load resistance increases. The greater the bleeder current, the less variations in load current affect the load voltage. Thus, improved voltage regulation is achieved at the expense of extra current drain from the source and extra heat produced in the voltagedivider resistors. In designing voltage dividers for loads consisting of a ­single transistor, voltage regulation is more significant than a few extra ­milliamperes of bleeder current.

Example 8-6 Allowing a bleeder current of 50 mA, design a voltage divider to supply 15 V at 20 mA from a 25-V source. Calculate the open-circuit terminal voltage of the voltage divider. Solution Since the bleeder resistor is in parallel with the load, the voltage drop across RB is 15 V. Therefore,

RB =

VL 15 V = = 300 Ω IB 50 mA

8-5   Voltage Dividers

The power dissipated by the bleeder resistor is

PB = VLIB = 15 V × 50 mA = 0.75 W



The voltage drop across the series-dropping resistor is VD = E − VL = 25 V − 15 V = 10 V



ID = IB + IL = 50 mA + 20 mA = 70 mA

and

RS =



VD 10 V = 143 Ω = ID 70 mA

PS = VDID = 10 V × 70 mA = 0.70 W

We would use at least 1-W resistors for this voltage divider. Under open-circuit conditions, we can use the voltage-divider principle (Equation 8-1): VT = VB = 25 V ×



300 Ω = 16.9 V 143 Ω + 300 Ω

In circuits that obtain several different voltages from one power supply, it is convenient to have a common reference point for all voltage measurements. Often the circuit’s metal frame or chassis is connected to the circuit and used as the reference point. The chassis may also be connected to the earth, usually through the ground wire of the electrical system. The symbol at the bottom of Figure 8-15 indicates that the reference point is grounded. Hence, we can say that the voltage at point A is +250 V with respect to ground. The symbol in the middle of Figure 8-16 indicates the point where the chassis is electrically connected to the circuit. With respect to chassis, the voltage at point A in this circuit is +12 V and the voltage at point B is −12 V. +250 V 50 mA A R1

+150 V 25 mA B

+ 250 V E −

R2

+100 V 10 mA

IB = 10 mA

C

RB

Ground

D

Figure 8-15  Voltage divider with multiple output voltages

177

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Chapter 8   Series-Parallel Circuits

Example 8-7 Design a voltage divider for the specifications shown in Figure 8-15. Solution Step 1 Figure 8-15 uses a resistor to represent the load for each specified output. These resistors make it easier to trace the currents and draw in arrows showing the direction of the current in each branch of the circuit. None of the 50-mA current drawn from terminal A flows through the voltage-­divider resistors. Hence, this current does not enter into our ­calculations. Step 2 Starting with the bleeder resistor, calculate the current through each resistor of the voltage divider and mark the circuit diagram of­ Figure 8-15 a­ ccordingly. The only current through RB is the 10-mA bleeder current. Since R2 passes this bleeder current and the 10-mA current drawn by the 100-V load,

I2 = 10 mA + 10 mA = 20 mA

Similarly, R1 passes I2 plus the 25-mA current drawn by the 150-V load:

I1 = 20 mA + 25 mA = 45 mA

Step 3 Calculate the resistance and power rating for each of the three resistors of the voltage divider. and

RB =

VB 100 V = = 10 kΩ IB 10 mA

PB = VB IB = 100 V × 10 mA = 1.0 W

The voltage drop across R2 is the potential difference between terminals B and C: Hence, and

V2 = 150 V − 100 V = 50 V

R2 =

V2 50 V = = 2.5 kΩ I2 20 mA

P2 = V2I2 = 50 V × 20 mA = 1.0 W

For reliability, we would use 2-W resistors or even 5-W resistors for R2 and RB.

8-5   Voltage Dividers

V1 = 250 V − 150 V = 100 V

Similarly,

R1 =



V1 100 V = = 2.2 kΩ I1 45 mA

P1 = V1I1 = 100 V × 45 mA = 4.5 W For R1, we would use a 10-W resistor.

Transistor circuits can require positive or negative voltages relative to a common chassis connection, and many circuits require both. We can obtain these positive and negative voltages from one power supply by connecting a tap on a voltage divider to the chassis as shown in ­Figure 8-16. A IT = 0.50 A

+ R1

+ 24 V −

12 V

Load 1 400 mA

12 V

Load 2 200 mA

Chassis R2

B

−

Figure 8-16  Power-supply voltage divider for a transistor amplifier

Example 8-8 At full load, the power supply in Figure 8-16 has a terminal voltage of 24 V DC. Design a voltage divider to provide +12 V and −12 V with respect to  the chassis when the current drains are 400 mA at +12 V and 200 mA at −12 V. The total current drain on the power supply is 500 mA. Solution Since R1 is in parallel with load 1, V1 = 12 V. Applying Kirchhoff’s current law to junction A, I1 = 500 mA − 400 mA = 100 mA

R1 =

V1 12 V = = 120 Ω I1 100 mA

P1 = V1I1 = 12 V × 100 mA = 1.2 W

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Chapter 8   Series-Parallel Circuits

Similarly, V2 = 12 V, and I2 = 500 mA − 200 mA = 300 mA R2 =



12 V = 40 Ω 300 mA

P2 = 12 V × 300 mA = 3.6 W



See Problems 8-18 to 8-27 and Review Questions 8-47 to 8-51.

Circuit Check

B

CC 8-3. Determine values of R1 and R2 for the potentiometer in Figure 8-17 such that the current I is limited to 250 mA when the output voltage Vout is 50 V with no load connected.

+ Vin R 10 kV 1

−

I

+ R2 Vout

−

Figure 8-17

CC 8-4. Determine the voltage drop across each resistor shown in Figure 8-18. R1

R2

1.0 kΩ

2.2 kΩ R3 3.3 kΩ

V1 15 V R4 5.1 kΩ R5 10 kΩ

Figure 8-18

8-6   Current-Divider Principle

8-6  Current-Divider Principle In Example 8-6, the total current drawn from the voltage source divides between the load and the bleeder resistor. In circuits where current ­divides among parallel branches, we can apply the parallel circuit dual of the v ­ oltage-divider principle for series circuits. Section 7-12 described this ­current-divider principle: In a parallel circuit, the ratio between any two branch currents equals the ratio of the two conductances through which these ­currents flow. Since the voltage is common for resistors in parallel, V=



I1 I2 In IT = = = G1 G2 Gn GT In = IT

and

Gn GT

(7-18)

(8-2)

Substituting Gn = 1/Rn, and GT = 1/Req in Equation 8-2 gives In = IT



Req Req In    or    = Rn IT Rn

(8-3)

Hence, we can restate the current-divider principle: In a parallel circuit, the ratio between any two branch currents is the inverse of the ratio of the branch resistances. For two resistors in parallel,



Req =

R1R2 R1 + R2



I1 = IT

and

I1 = IT

Similarly,

I2 = IT



(7-13)

R1 R2 ( R1 R1 + R2 ) R2 R1 + R2 R1 R1 + R2

(8-4)

(8-5)

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182

Chapter 8   Series-Parallel Circuits

Example 8-9 Calculate the bleeder current in the voltage divider designed in Example 8-6 if the load resistance increases to 3.0 kΩ, as shown in ­Figure 8-19. RS 143 Ω

+ 25 V −

RB

300 Ω

RL

3000 Ω

Figure 8-19  Circuit diagram for Example 8-9

Solution The equivalent resistance of the bleeder resistor and the load resistance in parallel is Req =



300 Ω × 3 kΩ = 273 Ω ( 300 Ω + 3.0 kΩ )

RT = RS + Req = 143 + 273 = 416 Ω



IT =



E 25 V = = 60.1 mA RT 416 Ω

IB = IT



RL 3.0 kΩ = 60.1 mA × = 55 mA RB + RL 3.3 kΩ

Example 8-10

The ammeter in Figure 8-20 has an internal resistance of 50 Ω and reads full scale when the current through it is 1.0 mA. What shunt resistance must be connected in parallel with the meter so that it will read full scale for a circuit current of 1.0 A? RM = 50 Ω

A 1.0 mA

1.0 A

Shunt

Figure 8-20  Circuit diagram for Example 8-10

8-6   Current-Divider Principle

Solution From Kirchhoff’s current law, Ish = IT − IM = 1000 mA − 1.0 mA = 999 mA Applying the current-divider principle, Rsh IM = RM Ish



Rsh = RM ×



IM 1.0 = 50 mΩ = 50 Ω × Ish 999

In Example 8-2, we calculated the input resistance of a typical ladder network, but we did not compare the output voltage and current with the input voltage and current. Now we can use the current-divider principle to complete the circuit analysis without a series of tedious Ohm’s-law ­calculations.

Example 8-11 Find the output voltage for the ladder network in Figure 8-21 with (a)   the output terminals open-circuit (b)  a 600-Ω load connected to the output terminals Iin 100 Ω

+

A I2 200 Ω

204 mV

100 Ω

C

I1

I3

IL

800 Ω

800 Ω

RL

− 100 Ω

B

200 Ω

D

100 Ω

Figure 8-21  Circuit diagram for Example 8-11

Solution (a) From the solution of Example 8-2(a), we know that Rin = 680 Ω when the output terminals are open-circuit. Therefore,

Iin =

E 204 mV = = 300 μA Rin 680 Ω

This current splits into two parallel branches at junction A. Substi­ tuting into Equation 8-5 gives R1 800 Ω I2 = IT = 300 µA = 120 µA R1 + R2 800 Ω + 1200 Ω

183

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Chapter 8   Series-Parallel Circuits



Since the output terminals are open-circuit, IL = 0 and I3 = I2. Also, there will be zero voltage drop across the two right-hand 100-Ω resistors. Therefore,



VL = VCD = I2RCD = 120 μA × 800 Ω = 96.0 mV



I2 = 0.5 × 340 μA = 170 μA

(b)  From the solution of Example 8-2(b), Rin = 600 Ω when RL = 600 Ω. Then, E 204 mV Iin = = = 340 µA Rin 600 Ω This current splits into two branches at junction A. From Figure 8-7(c), we note that the two branches between junctions A and B have the same resistance. Therefore,

At junction C, I2 splits into two branches. Figure 8-7(a) shows that these branches have equal resistances, so

IL = 0.5 × 170 μA = 85 μA

VL = ILRL = 85 μA × 600 Ω = 51 mV See Problems 8-28 to 8-34 and Review Questions 8-52 to 8-54.

8-7  Cells in Series-Parallel A series-parallel battery connection, as shown in Figure 8-22, provides both greater voltage and greater capacity than the individual cells. The cells must be identical to avoid circulating currents within the battery. The total capacity of the battery is determined by the number of rows connected in parallel. S cells in series

Ecell Rcell Ecell Rcell Ecell Rcell Ecell Rcell Ecell Rcell Ecell Rcell

P rows in parallel

Ecell Rcell Ecell Rcell Ecell Rcell Ecell Rcell Ecell Rcell Ecell Rcell

+

Ebat

−

Figure 8-22  Equivalent circuit of a series-parallel connected battery

8-7   Cells in Series-Parallel

185

For a battery consisting of P parallel rows with S cells in series in each row, the EMF of the battery is the EMF of one series row, the same as for a ­series battery connection: Ebat = SEcell



The total internal resistance is the combination of P identical parallel rows of S resistors in series, with each resistor having a resistance of Rcell, so Rbat =



resistance of each row SRcell = number of rows P

(8-6)

Example 8-12 Twenty identical cells are connected in four parallel rows, each with five cells in series. Each cell has an internal resistance of 0.20 Ω. The battery delivers a current of 5.0 A to a 1.25-Ω resistor as shown in ­Figure 8-23. Calculate the EMF of each cell. Solution

Rbat =

SRcell 5 × 0.20 Ω = = 0.25 Ω P 4

Ecell =

Ebat 7.5 V = = 1.5 V S 5

Ebat = ILRT = 5.0 A × ( 0.25 Ω + 1.25 Ω ) = 7.5 V

Rbat

IL = 5.0 A

0.25 Ω RL = 1.25 Ω

Ebat

Figure 8-23  Equivalent circuit for Example 8-12

Multisim Solution Download Multisim file EX8-12 from the website. Connect a ground to the bottom node. Run the simulation, and check the value of IL displayed on the ammeter. Confirm that this value matches the one calculated using Equation 8-6. See Problems 8-35 to 8-39 and Review Question 8-55.

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Chapter 8   Series-Parallel Circuits

Circuit Check

C

CC 8-5. A 1.0-kΩ resistor, a 2.0-kΩ resistor, and a 4.0-kΩ resistor are ­connected in parallel and draw a total of 175 mA from a voltage source. Find the current through each resistor, and calculate the applied voltage. CC 8-6. A battery has three parallel rows, each with four cells ­connected in series. Given that each cell has an EMF of 1.5 V and an internal resistance of 0.2 Ω, find the current this battery will deliver to a 4-Ω load.

8-8 Troubleshooting We can use the voltage and current laws that apply to series-parallel networks to help diagnose faults in circuits.

Example 8-13 Determine the fault most likely to cause each of the following voltages in the voltage divider designed in Example 8-6: (a) VL = 21 V

(b)   VL = 0 V

(c)   VL = 25 V

Solution (a) Either the supply voltage is not 25 V or one of the resistors is shorted or open. Check the source voltage with a voltmeter. If the source voltage is 25 V, then we calculate the value of VL that results if any of the resistors are either shorted or open. RS shorted: RL shorted: RB shorted: RS open: RL open:

VL = 25 V VL = 0 V VL = 0 V VL = 0 V

VL =

RB 300 ( 25 V ) = 17 V E= RS + RB 300 + 143

RB open: The circuit is designed for RL =



VL =

VL 15 V = = 750 Ω IL 20 mA

RL 750 ( 25 V ) = 21 V E= RL + RS 750 + 143

The calculation for RB open yields the same result for VL as the measured value. Therefore, we conclude that RB is open.

8-8  Troubleshooting

187

(b) As noted in part (a), VL = 0 V if RS is open or either of RB or RL is shorted. Each resistor should be removed from the circuit and checked with an ohmmeter to determine which one is defective. (c) As noted in part (a), VL = 25 V if RS is shorted. However, VL will be 25 V if both RB and RL are open (leaving no complete path for current). It is more likely that we have one defective resistor, rather than two. Therefore, we first remove RS and check it with an ohmmeter. If it is not defective, then we check the other resistors. Multisim Solution Download Multisim file EX8-13(a) from the website. Run the simulation, and note the meter readings for VL and IL are 15 V and 20 mA, respectively. Replace RS with a short circuit. Run the simulation, and note that VL has become 25 V. Restore RS and replace RL with a short circuit. Run the simulation, and note that VL has become 0 V. Restore RL and replace RB with a short circuit. Run the simulation, and note that VL has become 0 V. Restore RB and replace RS with an open circuit. Run the simulation, and note that VL has become 0 V. Restore RS and replace RL with an open circuit. Run the simulation, and note that VL has become 175 V. Restore RL and replace RB with an open circuit. Run the simulation, and note that VL has become 21 V.

See Review Question 8-56.

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Chapter 8   Series-Parallel Circuits

Summary

• A series-parallel circuit can be simplified into an equivalent series or a­ parallel circuit. • Series-parallel circuits can be analyzed using Kirchhoff’s voltage and current laws. • Voltages across resistors in series can be calculated using the voltage-­ divider principle. • Bleeder resistors are used in voltage dividers to improve voltage ­regulation. • Voltages in a circuit are often referenced with respect to chassis or ground. • Currents through resistors in parallel may be calculated using the currentdivider principle. • Cells connected in series-parallel provide higher voltage and current to a load. • A fault in a resistive network is usually the result of an open or shorted resistor. B = beginner

I = intermediate

A = advanced

Problems I

Section 8-2  Equivalent-Circuit Method 8-1.

Calculate the equivalent resistance of each of the resistance networks in Figure 8-24. 60 Ω

400 Ω

40 Ω

30 Ω

20 Ω (a)



Figure 8-24

1.0 kΩ

600 Ω (b)

4.0 kΩ

189

Problems

I

8-2.

Calculate the resistance measured from terminal T to ground in each of the resistance networks of Figure 8-25. T

264 Ω T

240 Ω 600 Ω

900 Ω 300 Ω

10 kΩ

27 kΩ

10 kΩ

2.2 kΩ

2.2 kΩ (a)

(b)

Figure 8-25

I

8-3. How much current is drawn from the voltage source in each of the circuits in Figure 8-26? Short circuit 136 Ω

+ 50 V −

20 Ω

320 Ω

60 Ω

136 Ω

+ 50 V −

320 Ω

(a)

All resistors 2.2 kΩ

20 Ω

60 Ω

200 mV

(b)

(c)

Figure 8-26

A

8-4.

Determine the source current in each of the circuits in Figure 8-27. Short circuit

−48 V 16 Ω

12 Ω

18 Ω

20 Ω

+ 48 V −

8.0 Ω

16 Ω

32 Ω

12 Ω

20 Ω

18 Ω

8.0 Ω

4.0 kΩ

32 Ω

2.0 kΩ

+ 80 V −

6.0 kΩ

8.0 kΩ Open circuit (a) Figure 8-27

(b)

(c)

190

Chapter 8   Series-Parallel Circuits

I

100 Ω

200 Ω

800 Ω

I

8-5. By connecting two leads to various terminals of the resistance network in Figure 8-28, six different values of equivalent resistance can be obtained. Calculate all six. 8-6. Calculate RAB, RBC, and RCA in each of the networks in Figure 8-29. A

400 Ω

Figure 8-28

A 60 Ω

300 Ω

200 Ω

100 Ω

150 Ω

B

C

B

(a)

500 Ω (b)

C

Figure 8-29

I

8-7.

Determine the equivalent resistance of the circuit in Figure 8-30 (a) with the switch open (b) with the switch closed 40 Ω

200 Ω

500 Ω

300 Ω

400 Ω

+ 120 V −

Figure 8-30

A I

8-8. Find the voltage drop across the switch in the circuit of Figure 8-30 when the switch is open. 8-9. Find the input resistance of the ladder network in Figure 8-31 (a) with a 75-Ω load resistor connected as shown (b) with the load resistor disconnected 25 Ω

Input

Figure 8-31

50 Ω

100 Ω

50 Ω

100 Ω

25 Ω

100 Ω

Output

75 Ω

191

Problems

I I

A

8-10. What voltage applied to the input terminals of the ladder network in Figure 8-31 produces 200 μW of power in the 75-Ω load resistor? 8-11. For each set of values for the circuit of Figure 8-1, draw fully labelled diagrams of the equivalent circuits and calculate the voltage drop, current, and power for each resistor. (a) E = 40 V; R1 = 48 Ω; R2 = 40 Ω; R3 = 160 Ω (b) PT = 625 W; R1 = 8 Ω; R2 = 10 Ω; R3 = 40 Ω 8-12. For the circuit shown in Figure 8-32, calculate (a) the equivalent resistance (b) the current through the 18-Ω resistor (c) the voltage drop across the 8.0-Ω resistor (d) the power dissipated by the 20-Ω resistor 20 Ω 8Ω

11 Ω 30 Ω

3Ω

5Ω

19 Ω

18 Ω

+ 40 V −

12

9Ω

Ω

4Ω

I A

A

I

3Ω

10 Ω

Figure 8-32

8-13. Use Multisim to verify your answer to Problem 8-3(b).

Section 8-3  Kirchhoff’s Laws Method

8-14. For each set of values for the circuit of Figure 8-1, draw fully labelled diagrams of the equivalent circuits and calculate the voltage drop, current, and power for each resistor. (a) E = 320 V; R1 = 11.0 Ω; I2 = 1.6 A; V3 = 144 V (b) E = 100 V; PT = 75 W; R2 = 100 Ω; I3 = 0.50 A (c) E = 150 V; R1 = 5.4 Ω; R2 = 12 Ω; I3 = 2.0 A (d) E = 250 V; R1 + R2 = 20 kΩ; V2 = 120 V; I3 = 6.0 mA 8-15. A 12-V generator has an internal resistance of 50 mΩ. Two loads are connected in parallel to its terminals, one drawing a 12-A current and the other dissipating energy at the rate of 200 W. Find the terminal voltage of the generator with this load. 8-16. A digital voltmeter with an input resistance of 10 MΩ reads 4.5 V when measuring the automatic-gain-control (AGC) voltage of a radio receiver. When an analog voltmeter with an input resistance of 10 kΩ is also connected across the AGC voltage source, both meters read 0.40 V. Calculate the internal resistance of the AGC voltage source.

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Chapter 8   Series-Parallel Circuits

A

8-17. In the circuit of Figure 8-33, determine the resistances of R1 and R2 if (a) RL = 400 Ω and IL = 50 mA (b) RL = 200 Ω and IL = 75 mA R1

+ 48 V −

R2

RL

Figure 8-33

B I A A

Section 8-5  Voltage Dividers 8-18. What bleeder resistor is required to complete a voltage divider to give 250 V at 40 mA from a 500-V source if the series dropping resistor is 5.0 kΩ? 8-19. Design a voltage divider to supply 120 V at 75 mA from a 240-V source if the total drain on the source is to be 100 mA. 8-20. Design a voltage divider to deliver 100 V at 1.0 mA from a 240-V source so that a 100% increase in load current causes only a 5% reduction in load voltage. 8-21. Figure 8-34 shows the focus control for a cathode-ray tube. In the one extreme position, the focus-anode voltage is to be 800 V with a current of 10.0 μA. At the other extreme, the focus-anode voltage is to be 1400 V with a current of 14.0 μA. Calculate the resistance of R1 and of the focus control. Cathode ray tube

+ R1 2.0 kV Source

Focus control R2

1.0 MΩ

− Figure 8-34

I

circuitSIM walkthrough

8-22. (a) Design a voltage divider to supply 10 V at 10 mA and 25 V at 25  mA from a 40-V DC supply with a bleeder current (IB in ­Figure 8-35) of 10 mA. (b) Use Multisim to verify that your circuit design produces the required voltages and currents.

Problems

I1 R1 25 mA I2

+ 40 V E −

R2 10 mA

25 V

IB RB

10 V

Figure 8-35

I A

I

I I

B

8-23. Assuming that the load 1 in Figure 8-16 acts like a linear resistance, ­determine the voltage distribution (with respect to the chassis) in Example 8-8 if the load 2 becomes open-circuit. 8-24. A voltage divider consists of a 10-kΩ resistor with an adjustable tap. The tap is set so that it feeds 80 V at 5 mA to a load when the voltage divider is connected across a 200-V source. What is the resistance of the bleeder portion of the voltage divider? 8-25. A transistorized stereo amplifier requires the following DC supplies: −30 V at 2.0 A, −16 V at 250 mA, and +12 V at 2 A. All voltages are measured with respect to ground. Design a voltage divider to operate this amplifier from a single DC supply that ­develops a ter­minal voltage of 46 V with a total current drain of 2.5 A. 8-26. A voltage divider circuit is to supply a 100-mA load at 30 V. If the power supply is 50 V and the total current drain is to be 150 mA, design the circuit and sketch the circuit diagram. 8-27. Draw and design a voltage-divider circuit to provide the following load currents from a 50-V supply: 500 mA at 50 V   200 mA at 30 V   100 mA at 15 V Design for a bleeder current of 100 mA.

Section 8-6  Current-Divider Principle 8-28. Calculate the current through each resistor in the circuit of ­Figure 8-36(a).

193

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Chapter 8   Series-Parallel Circuits

I

8-29. What resistance must be connected across the terminals of the ­constant-current source shown in Figure 8-36(b) to reduce the terminal voltage to 75 V? IT = 15 mA + E −

8.0 kΩ

16 kΩ

40 mA

5.0 kΩ

(a)

(b)

Figure 8-36

I I

8-30. In the circuit in Figure 8-37, what resistance R will draw a current equal to 25% of the source current? 8-31. In the circuit in Figure 8-37, what resistance R makes the source current 160 mA?

50 Ω

+ 20 V −

50 Ω 100 Ω

R

Figure 8-37

A

circuitSIM walkthrough

8-32. (a) For the circuit of Figure 8-38, calculate the equivalent resistance for the resistor network, the current in the 5.2-Ω resistor, and the voltage across the 1.0-Ω resistor. (b) Use Multisim to verify your answers for part (a). 6.0 Ω

3.0 Ω

5.2 Ω 20 Ω

60 Ω

30 Ω

20 Ω

+ 30 V −

8.0 Ω

12 Ω 4.0 Ω

1.0 Ω

Figure 8-38

5.0 Ω

Problems

A

8-33.

For the circuit of Figure 8-39, calculate (a) the total resistance (b) the current in the 24-Ω resistor (c) the voltage across the 14-Ω resistor (d) the power to the 20-Ω resistor 3.0 Ω 5.0 Ω 6.0 Ω

20 Ω

27 Ω

10

24

Ω

Ω

+ 80 V −

15 Ω 18 Ω 22 Ω

8.0 Ω 6.0 Ω

14 Ω

Figure 8-39

A

8-34. For the circuit shown in Figure 8-40, calculate (a) the total resistance (b) the current in the 15-Ω resistor 12 Ω

6.

0 Ω

+ 50 V −

30 Ω

12

Ω

15 Ω 2.0 Ω

Figure 8-40

I

Section 8-7  Cells in Series-Parallel 8-35. A battery contains four parallel-connected rows of five identical cells in series. Each cell has an EMF of 800 mV and an internal resistance of 0.20 Ω. How much current will this battery deliver to a 6.0-Ω load?

195

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Chapter 8   Series-Parallel Circuits

I

A I

I

8-36. A battery is made up of four parallel-connected rows, each with six identical cells connected in series. Each cell has an internal r­ esistance of 2.0 Ω. The battery sends a current of 1.0 A through an 6.0-Ω load. Calculate the EMF of a cell. 8-37. The terminal voltage of a battery is 12.0 V when the load current is 1.6 A, and 10.0 V when the current is 2.4 A. Calculate the internal ­resistance of the battery. 8-38. The maximum sustained current for an alkaline C cell is 300 mA. Specify how to connect alkaline C cells to make a 12-V battery with a  capacity of 1.2 A. Given that the internal resistance of each cell is 0.150 Ω, find the ­actual terminal voltage of the battery when the ­current drain is 1.2 A. 8-39. A battery is made from 40 identical 1.5-V cells connected eight in series per row, five rows in parallel. When a 10-Ω load resistor is connected, the terminal voltage is 10 V. Find the internal resistance/cell.

Review Questions

Section 8-1  Series-Parallel Resistors 8-40. With respect to series-parallel resistance networks, what is meant by an equivalent resistance? 8-41. What characteristics must an equivalent resistance possess? 8-42. What circuit rules allow us to replace R2 and R3 in Figure 8-41 with an equivalent resistance? Rint

R2

+ E −

R1

R3

Figure 8-41

Section 8-3  Kirchhoff’s Laws Method 8-43. What is the advantage of solving series-parallel resistance networks by writing Kirchhoff’s law equations rather than using the equivalent-­ resistance method? 8-44. Write one Kirchhoff’s current law equation and two Kirchhoff’s voltage law equations for the network of Figure 8-41.

Section 8-4  Voltage-Divider Principle 8-45. What circuit condition must exist for the voltage-divider principle to be applied in network solutions?

Integrate the Concepts

8-46. Explain the operation of the volume-control circuit shown in ­Figure 8-42.

Input signal

Output to amplifier

Figure 8-42  Volume control

Section 8-5  Voltage Dividers 8-47. What is the purpose of a bleeder resistor in a practical voltage-­ divider circuit? 8-48. Calculate the load voltage in Examples 8-5 and 8-6 if the load current decreases to 10 mA. Account for the change in the load voltage. 8-49. Why does a decrease in the bleeder resistance of a given voltage d ­ ivider improve the voltage regulation of the output of the voltage divider? 8-50. In Chapter 2, we noted that voltage is measured between two points in a circuit. Yet circuit diagrams often have notations such as “Voltage at Point A is +12 V.” Explain. 8-51. A DC power supply for an amplifier has only two output terminals. Describe how to connect this power supply to obtain both positive and negative potentials with respect to the chassis of the amplifier.

Section 8-6  Current-Divider Principle 8-52. What circuit conditions must exist for the current-divider principle to be applied in network solutions? 8-53. The current-divider principle is sometimes called the parallel-circuit dual of the voltage-divider principle. Explain this duality relationship. 8-54. Explain why the ratio of the currents in two parallel branches is the inverse of the ratio of the resistances of the branches.

Section 8-7  Cells in Series-Parallel 8-55. How is the total internal resistance of a series-parallel combination of identical cells related to the internal resistance of the individual cells?

Section 8-8  Troubleshooting 8-56. What circuit conditions in Figure 8-14 would lead one to conclude that the bleeder resistor is open?

Integrate the Concepts Refer to the circuit of Figure 8-1 and its equivalent circuit in Figure 8-2. Set R1 = 48 Ω, R2 = 220 Ω, R3 = 330 Ω, and E = 450 V. Calculate Req, RT, I1, VR1, I2, and I3.

197

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Chapter 8   Series-Parallel Circuits

Practice Quiz 1.

The equivalent resistance of the circuit in Figure 8-43 is (a)  260 Ω (b)  473 Ω (c)  590 Ω (d)   143 Ω R2

R1

220 Ω

50 Ω

R3 220 Ω R4 100 Ω

V1

Figure 8-43

2.

The equivalent resistance of the ladder network in Figure 8-44 is (a)  763 Ω (b)  783 Ω (c)  842 Ω (d)   963 Ω 330 Ω

1.0 kΩ

100 Ω

100 Ω

1.0 kΩ

1.0 kΩ

100 Ω

Figure 8-44

3.

The total current in the circuit of Figure 8-45 is (a)   999 μA (b)   13.2 mA (c)   10.8 mA (d)   99.8 mA

Practice Quiz

R1 IT

IR3

100 Ω R2 10 kΩ

12 V

R3 1.0 kΩ

R4

100 Ω Figure 8-45

4.

The current through resistor R3 in Figure 8-45 is (a)   984 μA (b)   9.84 mA (c)   9.75 mA (d)   97.5 mA

5.

For the battery shown in Figure 8-46, the EMF of each cell is (a)   3.0 V (b)   3.5 V (c)   2.0 V (d)   2.5 V XMM1

−

V1

R1

V2

0.15 Ω

+

V4

R4

V3

0.15 Ω V5

0.15 Ω

R5

0.15 Ω RL

220 Ω

Figure 8-46

R2

R3

0.15 Ω V6

R6

0.15 Ω

199

9

Resistance Networks The techniques developed to this point work well for c ­ ircuits with resistors in series or in parallel. However, some circuits have components connected in patterns that are ­neither ­series nor parallel combinations. This chapter introduces methods of circuit analysis for such circuits.

Chapter Outline 9-1

9-2

9-3

9-4

Network Equations from Kirchhoff’s Laws  202

Constant-Voltage Sources  202

Constant-Current Sources  204

Source Conversion  206

9-5 Kirchhoff’s Voltage-Law Equations: Loop Procedure  208

9-6 Networks with More Than One Voltage Source  214

9-7 Loop Equations in Multisource 9-8 9-9

Networks 216

Mesh Analysis  222

Kirchhoff’s Current-Law Equations  228

9-10 Nodal Analysis  231 9-11

The Superposition Theorem  237

Key Terms source conversion  202 short-circuit current  204 loop equations  208 mesh equations  208

nodal equations  208 mesh analysis  222 mesh  222 planar 222

nodes 228 reference node  229 nodal analysis  231 superposition theorem  237

Learning Outcomes At the conclusion of this chapter, you will be able to: • calculate the internal resistance of constantvoltage and constant-current sources from their open-circuit voltages and short-circuit currents • convert a constant-voltage source to an ­equivalent ­constant-current source • convert a constant-current source to an ­equivalent ­constant-voltage source • analyze a circuit containing constant-voltage sources using the loop procedure based on Kirchhoff’s voltage law

Photo source:  © MARK SYKES/SCIENCE PHOTO LIBRARY

• analyze a planar circuit containing constantvoltage sources using mesh equations • analyze a circuit using Kirchhoff’s current law • analyze a circuit containing constant-current sources using nodal analysis • analyze a circuit containing more than one source using the superposition theorem

202

Chapter 9    Resistance Networks

9-1 Network Equations from Kirchhoff’s Laws The Wheatstone bridge in Figure 9-1 may appear at first glance to be a ­simple series-parallel circuit, but no two resistors in the circuit are actually connected in series or parallel. Hence, we cannot use the equivalent-­ resistance techniques of Chapter 8 to analyze a bridge circuit. However, there are two general methods we can use to solve such networks.

R1

R2 R5

+ E − R3

R4

Figure 9-1  Wheatstone bridge

One method uses various network theorems to simplify the original network so that we can apply equivalent-circuit techniques. The second method is based on Kirchhoff’s voltage and current laws, which apply to all circuits no matter how complex. To apply the Kirchhoff’s-laws method, we need to learn a few basic rules for writing an orderly set of simultaneous equations. This structured approach makes this method particularly suitable for circuit-analysis software. The disadvantage of Kirchhoff’s-laws techniques in an ­introductory course is that we can become more involved with plugging numbers into computer programs than with learning the principles of circuit behaviour. The equivalent-circuit and network-theorem methods let us analyze the same circuits with less emphasis on mathematical formats. An understanding of the principle of source conversion enables us to simplify the final formats for Kirchhoff’s-laws equations for various resistance networks. Hence, we shall discuss equivalent-circuit techniques in the next three sections of this chapter.

9-2  Constant-Voltage Sources The terminal voltage of most practical sources falls as the current drawn from the source increases. In Section 7-6 we explained this effect by assuming that a practical voltage source consists of a constant-voltage source in series with an internal resistance. In other words, the circuit of Figure 9-2(a) behaves exactly the same as the equivalent circuit of Figure 9-2(b).

9-2   Constant-Voltage Sources

RL

Battery

Rint

+

RL

E

− (a)

(b)

Figure 9-2  Equivalent circuit for a practical voltage source

In replacing a practical voltage source in a circuit diagram with an equivalent series constant-voltage source, we treat the actual voltage source as a “black box” that has only two terminals exposed. For example, consider a power supply that provides a DC voltage to operate a transistor amplifier. As shown in Figure 9-3(a), this power supply is essentially a “black box”— we can make measurements at the terminals but we cannot see how the ­terminals are interconnected inside the box. This black box outputs 16 V DC when connected to a 120-V 60-Hz power supply. Viewed from the load side, the power supply is simply a source of direct current. The details of how the black box converts its input power to direct current do not affect the operation of the load circuit.

Rint

A

+

0.20 Ω

Power supply

V

RL

+ 16 V −

RL

−

Terminal voltage (V)

(a)

(b)

16 14

0

10

20

30 40 50 Load current (A)

60

(c)

Figure 9-3  Equivalent circuit for a DC power supply

70

80

203

204

Chapter 9    Resistance Networks

We can measure the open-circuit terminal voltage of the power supply by replacing the load with a voltmeter that draws negligible current. When we connect progressively smaller values of RL , we find that the terminal voltage falls as the current drawn from the power supply increases, as shown by the graph in Figure 9-3(c). All that we can conclude from the data in Figure 9-3(c) is that the power supply behaves like a constant-voltage source with an internal resistance of

Rint =

ΔV 2.0 V = = 0.20 Ω ΔI 10 A

By extending the graph (as shown by the dashed line), we can find the short-circuit current, that is, the current when RL = 0 Ω and VL = 0 V. When RL = 0 Ω, the internal resistance is in series with just the voltage source, so

Rint =

open-circuit terminal voltage short-circuit terminal current

(9-1)

For the graph of Figure 9-3(c), the short-circuit current is 80 A. Hence, the internal resistance for the constant-voltage source of Figure 9-3(b) is

Rint =

Voc 16 V = = 0.20 Ω Isc 80 A

See Problems 9-1 and 9-2 and Review Questions 9-44 to 9-49 at the end of the chapter.

9-3  Constant-Current Sources We can use another approach to account for the properties of the blackbox power supply in Figure 9-3(a). Our measurements tell us that the sealed power supply has an open-circuit terminal voltage of 16 V and a short-­circuit terminal current of 80 A. Theoretically, the box could contain a ­special generator with a variable terminal voltage that always produces a current equal to the short-circuit terminal current. For this constant current to flow when the terminals of the power supply are open-circuit, the internal resistance must be in parallel with the constant-current generator, as shown in Figure 9-4(b). With no load connected, all of the current flows through the internal resistance, so

Rint =

open-circuit terminal voltage short-circuit terminal current



(9-1)

Note that the equation for internal resistance is the same for both c­onstant-voltage and constant-current sources. The arrow in the symbol for a constant-current source shows the direction of conventional current ­flowing through the source.

9-3   Constant-Current Sources

A

+ Power supply

V

RL

Rint 0.20 Ω

80 A

RL

− (a)

(b)

Figure 9-4  Constant-current source

Suppose that RL = 0.60 Ω in the circuits in Figures 9-3(b) and 9-4(b). With the constant-voltage source of Figure 9-3(b), the voltage-divider principle tells us that

VL =

and

RL 0.60 × 16 V = 12 V × Voc = Rint + RL 0.20 + 0.60 VL 12 V IL = = = 20 A RL 0.60 Ω

With the constant-current source of Figure 9-4(b), we switch to the ­current-divider principle: and

IL =

Rint 0.20 × Isc = × 80 A = 20 A Rint + RL 0.20 + 0.60 VL = ILRL = 20 A × 0.60 Ω = 12 V

For all values of load resistance, the constant-voltage source of Figure 9-3(b) and the constant-current source of Figure 9-4(b) are exact equivalents, as “seen” by RL. The constant-current equivalent source is the dual of the constant-­ voltage source. An American engineer, Edward L. Norton (1898–1983), devised the method for using constant-current equivalent sources for circuit analysis. In most practical electric circuits, we open the circuit (disconnect the load) to switch the circuit off. When we do so, both the current through the source and the current in the load become zero. The power dissipation in the internal resistance of the source is also zero under open-circuit c­ onditions. Hence, equivalent circuits with constant-voltage sources operate much like the actual circuits. If we disconnect RL in the current-source circuit of Figure 9-4(b), the current through Rint increases to the full constant-current value and the internal dissipation of the source becomes Voc × Isc = 1280 W. Theoretically, to switch off the circuit of Figure 9-4(b), we short-circuit RL so that the terminal voltage and the energy transfer to the load become zero. Constant-current equivalent sources are primarily just theoretical devices for network analysis. Although we will not encounter current sources in power-system networks, we do find devices that function as if they were current sources in low-power electronic circuits. For example, it is usually easier to analyze a transistor circuit if we show a current source in its equivalent circuit. Also, it is possible to use electronic feedback circuits to construct a DC power supply

205

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Chapter 9    Resistance Networks

that maintains a constant load current with a terminal voltage that varies as the load resistance varies. Such constant-current power supplies are less common than power supplies that maintain a constant terminal voltage. See Problem 9-3 and Review Questions 9-50 to 9-52.

9-4  Source Conversion For a given constant-voltage source, Rint in the equivalent constant-current source has the same value but appears in parallel with the ideal current source, as shown by the examples in Figure 9-5. Similarly, for a given constant-current source, Rint in the equivalent constant-voltage source has the same value but appears in series with the ideal voltage source. In order for us to replace a constant-voltage source in a network with its constant-current equivalent, there must be some resistance Rx in series with an ideal voltage source. Rx may be an internal resistance of the source, or it may be one of the network resistors in series with the voltage source. ­Similarly, there must be some form of resistance in parallel with an ideal current source before we can replace it with its constant-voltage equivalent. Applying Equation 9-1 to the constant-voltage source of Figure 9-5(a) shows that Ix for the equivalent constant-current source (Figure 9-5(b)) is the short-circuit current: Ix =



Eoc Rx

(9-2)

Similarly, given the constant-current source of Figure 9-5(c), Ex for the equivalent constant-voltage source of Figure 9-5(d) is the open-circuit voltage: Ex = IscRx



(9-3)

Rx 6.0 Ω

+

+

+ 120 V Eoc −

20 A

Ix

6.0 Ω

Rx

−

−

(a)

(b) Rx

−

− 40 A

− 32 V Ex +

0.80 Ω

Isc Rx

0.80 Ω

+

+ (c)

Figure 9-5  Source conversion

(d)

9-4   Source Conversion

Example 9-1 (a) Determine the equivalent constant-current source for a voltage source with an open-circuit voltage of 120 V and an internal resistance of 6.0 Ω. (b) Determine the equivalent constant-voltage source for a current source with a constant current of 40 A and an internal resistance of 0.80 Ω. (c) Check that the equivalent sources in part (a) produce the same results when connected to a 12-Ω load. Solution (a) The internal resistance for the constant-current source is the same as for the voltage source: Rint = 6.0 Ω Ix =



Eoc 120 V = = 20 A Rint 6.0 Ω

(b)  Rint for the constant-voltage equivalent source is the same as for the current source: Rint = 0.80 Ω

Ex = IscRint = 40 A × 0.80 Ω = 32 V

(c) When we connect a 12-Ω load to the voltage source of Figure 9-5(a), IL =



Ex 120 V = = 6.7 A Rx + RL 6.0 Ω + 12 Ω VL = 6.7 A × 12 Ω = 80 V



and



When we connect a 12-Ω load to the current source of Figure 9-5(b),



VL = Ix ×

and

Rx × RL 6.0 × 12 = 80 V = 20 A × Rx + RL 6.0 + 12 VL 80 V IL = = = 6.7 A RL 12 Ω

Once we are familiar with the concept, we can convert voltage sources to equivalent current sources (and vice versa) for network analysis ­purposes with just a bit of mental arithmetic. Analysis of transistor circuits often i­nvolves resistance networks containing both voltage and current sources. Such networks are usually easier to solve if we convert the sources so that they are either all voltage sources or all current sources. See Problems 9-4 to 9-6 and Review Questions 9-53 to 9-55.

207

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Chapter 9    Resistance Networks

Circuit Check

A

CC 9-1. Find the voltage across a 12-Ω load when it is connected to a 24-V source that has an internal resistance of 3.0 Ω. CC 9-2. Use an equivalent constant current source to verify the load voltage calculated in question CC 9-1.

9-5 Kirchhoff’s Voltage-Law Equations: Loop Procedure Kirchhoff’s laws let us analyze a circuit in several different ways, each with its own advantages and disadvantages. We can use Kirchhoff’s voltage law to write a set of either loop equations or mesh equations. With Kirchhoff’s current law, we can write a set of nodal equations. We shall consider the loop method first. In Section 7-4, we stated Kirchhoff’s voltage law: In any complete electric circuit, the algebraic sum of the source voltages must equal the algebraic sum of the voltage drops. This statement leads to equations in the form

E = V1 + V2 + V3 + . . .

(9-4)

Here is another way of stating Kirchhoff’s voltage law: In any closed loop, the algebraic sum of all voltage rises and voltage drops is zero. This statement leads to an equation in which we must carefully record the sign of voltage drops and voltage rises:

E − V1 − V2 − V3 = 0

(9-5)

V1 + V2 + V3 + . . . = E

(9-6)

To help set up our equations in a form that can be solved by determinants, we reverse Equation 9-4:

We start with an example that we can readily check with the equivalent resistance techniques of Chapter 8.

9-5   Kirchhoff’s Voltage-Law Equations: Loop Procedure

Example 9-2 Determine the magnitude and polarity of the potential difference between terminals A and B in the bridge network of Figure 9-6. 10 Ω Rint

R2

R1 20 Ω

+ 80 V −

A R3 30 Ω

15 Ω

B R4 10 Ω

Figure 9-6  Bridge network for Example 9-2

Solution Step 1 Draw a circuit diagram large enough to mark on tracing loops, voltage ­polarities, and any other data we need. We may change the location of components in the diagram so long as we do not change the electric ­connections (see Figure 9-7).

Rint

+

C

− 10 Ω

+ 80 V −

+

+ R 1 20 Ω − I1

A

15 Ω

− I2

B +

+ R3

R2

30 Ω

−

R4

10 Ω

−

D

Figure 9-7  Circuit diagram for Example 9-2

Step 2 Label the positive and negative terminals of the sources. We will draw current tracing loops on the circuit diagram, beginning at the positive terminal of the source and proceeding along a path through the network back to the negative terminal of the source. These tracing loops help us to write Kirchhoff’s voltage-law equations. We need enough tracing loops to include all components and to provide as many

209

210

Chapter 9    Resistance Networks

s­ eparate equations as there are unknowns. In this example, we require two tracing loops for the two unknown loop currents, I1 and I2. Note that reversing the polarity of a source in a network with more than one source alters all the current and voltage relationships in the network. Step 3 Mark the polarity of each voltage drop. In Section 7-2, we considered the polarities of voltage drops and voltage rises in relation to current direction, as shown in Figure 9-8. We can generalize this concept: In any passive component (such as a resistor) in an electric net­ work, the tracing direction is from positive to negative through that ­component.

+ E −

I

+ V −

Figure 9-8  Polarity of voltage drop and source voltage in relation to direction of current tracing loop

Once we have established the directions of tracing loops for I1 and I2, we can mark the polarity of the various voltage drops on the circuit diagram in Figure 9-7. Step 4 Write a Kirchhoff’s voltage-law equation for each loop. For the I1 loop,

VRint + VR1 + VR3 = E

Since I1 is the current through R1 and R3,

VR1 = I1R1     and    VR3 = I1R3

Current I1 leaves junction C through R1 and R3, and the current I2 leaves this junction through R2 and R4. Hence, from Kirchhoff’s current law, the current through Rint is I1 + I2, and

VRint = ( I1 + I2 ) Rint

Substituting for the voltages in the Kirchhoff’s voltage-law equation for the I1 tracing loop gives an equation in the standard form we shall use for the loop method: (I1 + I2)Rint + I1R1 + I1R3 = E

Writing the voltage drops as IR drops minimizes the number of ­ nknowns. u

9-5   Kirchhoff’s Voltage-Law Equations: Loop Procedure

Step 5 Substitute the known quantities into the equations. The equation for the I1 tracing loop becomes 10 Ω(I1 + I2) + 20 Ω(I1) + 30 Ω(I1) = 80 V

Collecting terms gives

Similarly, for the I2 tracing loop,

60I1 + 10I2 = 80

10 Ω ( I1 + I2 ) + 15 Ω ( I2 ) + 10 Ω ( I2 ) = 80 V

10I1 + 35I2 = 80

(1)

(2)

We then solve Equations 1 and 2 simultaneously. Elimination Method

6 × Equation 2: 60I1 + 210I2 = 480

subtract Equation 1: 60I1 + 10I2 = 80 200I2 = 400

I2 = 2.0 A



Substituting this value in Equation 1, we obtain I1 = 1.0 A

Determinant Method The determinant method provides a standard procedure that we can use to solve any array of simultaneous equations. Appendix 1 explains the matrix algebra for calculating determinants. 60I1 + 10I2 = 80







I1

I2

│8080 = │6010 │6010 = │6010

10I1 + 35I2 = 80

│ 2800 − 800 2000 = = = 1.0 A 2100 − 100 2000 10 35│ 10 35

│ 4800 − 800 4000 = = = 2.0 A 2000 2000 10 35│ 80 80

(1) (2)

211

212

Chapter 9    Resistance Networks

Simultaneous equations can also be solved using a programmable calculator. Select the program for solving simultaneous equations and input the coefficients. Run the program and the display will show the values of I1 and I2. Step 6 Find the voltage drops across the components of the network. In this example, we need to solve for just two voltages: and

circuitSIM walkthrough

VR3 = I1R3 = 1.0 A × 30 Ω = 30 V VR4 = I2R4 = 2.0 A × 10 Ω = 20 V

A voltage drop across a resistance is just another way of expressing a p­ otential difference between the two terminals of the resistance. Hence, terminal A in Figure 9-7 is 30 V more positive than junction D, and terminal B is 20 V more positive than junction D. Therefore, terminal A is 10 V more positive than terminal B: VAB = +10 V Multisim Solution Download Multisim file EX9-2 from the website.

The circuit is the one shown in Figure 9-7, except a voltmeter has been added to measure VAB.

Source:  © MARK SYKES/SCIENCE PHOTO LIBRARY

Run the simulation. Note that the voltmeter reading matches the value for VAB calculated using the loop procedure.

 Wheatstone bridge. A Invented by Samuel Hunter Christie in 1833, this circuit was improved and popularized by Charles Wheatstone in 1843.

Yet another approach to the potential difference between terminals A and B in Figure 9-6 is to use circuit simulation software that can capture schematics. Once the circuit is drawn on the computer screen, any current or voltage of interest can be displayed on the screen. Unfortunately, this method does not allow a student to learn anything about loop equations, determinants, or network theorems as they are buried within the computer program. However, circuit simulations are useful for predicting currents and voltages before a circuit is built. If we add a meter between terminals A and B of the bridge circuit of Figure 9-6, the network becomes a Wheatstone bridge, which can be used for high-precision resistance measurements. We now require one more tracing loop for our circuit analysis. In Figure 9-9 we have chosen a tracing loop for I3 passing through R1, RM (the resistance of the meter), and R4. We could choose a tracing loop passing through R2 , RM , and R3 instead. I3 would then pass in the opposite direction through RM, and have the opposite sign. If a tracing loop current has a negative sign, we simply treat the current as a negative quantity when we add the loop currents algebraically. Using the I3 loop shown in Figure 9-9, the Kirchhoff’s voltage-law equations for the three tracing loops are

9-5   Kirchhoff’s Voltage-Law Equations: Loop Procedure

R1(I1 + I3) + R3 I1 = E

R2I2 + R4(I2 + I3) = E

R1(I1 + I3) + RMI3 + R4(I2 + I3) = E

R1

+ E −

R2

I1

I2 R3

RM

I3

R4

Figure 9-9  Solving a Wheatstone bridge by loop procedure

Substituting the given values of resistance and source voltage, we can solve these three simultaneous equations for the three loop currents. We can then find the current through each component.

Total current:



Meter current:

I1 + I2 + I3



Current through R1: I1 + I3



Current through R3: I1



I3

Current through R2: I2

Current through R4: I2 + I3



Example 9-3

Find the current through a galvanometer with a resistance of 50 Ω when the meter is connected between terminals A and B of the bridge circuit of Figure 9-7. Solution For the I1 loop,

10 Ω(I1 + I2 + I3) + 20 Ω(I1 + I3) + 30 Ω(I1) = 80 V

213

214

Chapter 9    Resistance Networks

For the I2 loop,

10 Ω(I1 + I2 + I3) + 15 Ω(I2) + 10 Ω(I2 + I3) = 80 V

For the I3 loop,

10 Ω(I1 + I2 + I3) + 20 Ω(I1 + I3) + 50 Ω(I3) + 10 Ω(I2 + I3) = 80 V

Collecting terms gives

10I1 + 35I2 + 20I3 = 80



│ │

60I1 + 10I2 + 30I3 = 80 30I1 + 20I2 + 90I3 = 80

│ │

(1) (2) (3)

Since we are asked only to determine the meter current, we need only solve for I3. Using the determinant method,



I3 =



=



=

60 10 30

10 35 20

80 80 80

60 10 30

10 35 20

30 20 90

168 000 + 24 000 + 16 000 − 84 000 − 96 000 − 8000 189 000 + 6000 + 6000 − 31 500 − 24 000 − 9000

20 000 136 500 = 0.15 A

See Problems 9-7 to 9-9 and Review Questions 9-56 to 9-61.

9-6 Networks with More than One Voltage Source Most flashlights have two or more cells connected in series aiding to provide the lamp with a potential difference higher than that of a single cell. Since both voltage sources in Figure 9-10(a) make current flow in the same direction through any load, the total effective source voltage is the sum of the individual source voltages. In circuit diagrams, we can replace the individual series-connected voltage sources with a single equivalent source. If we load one cell backward in a flashlight, the lamp will glow dimly, if at all. With a series opposing connection, like the one shown in Figure 9-10(b), each source tries to make current flow in opposite directions through the load. The net terminal voltage for the equivalent source is the difference

9-6   Networks with More than One Voltage Source

­ etween the individual source voltages. We avoid series opposing con­ b nection of voltage sources in practical electric circuits other than battery chargers because the equivalent source voltage is actually reduced and the higher voltage source causes current to travel the wrong way through the lower voltage source when a load is connected. +

+

−

−

+

+ 6V

6V

+ 14 V −

− + 8V

−

− 2V

− − 8V

+

+ −

− (a)

+ (b)

+

Figure 9-10  Voltage sources: (a) Series aiding; (b) Series opposing

Figure 9-11(a) shows a circuit in which both a storage battery and a generator driven by an automobile engine provide current to a common load. The terminals of these two voltage sources (each with its own internal resistance) are connected in parallel. If the engine stops, the generator EMF must fall to zero and a heavy current—limited only by the two internal ­resistances in series—flows from the battery through the generator. In an electrical system for an automobile, the generator is automatically disconnected from the rest of the system when the engine is stopped. Parallel connection of voltage sources requires special precautions. Since a single power supply often feeds dozens of stages of an electronic system, we usually avoid cluttering circuit diagrams with long interconnecting leads by marking the DC supply voltages for individual stages as shown in the coupling circuit in Figure 9-11(b). Since both networks in ­Figure 9-11 include more than one voltage source, we cannot solve them by simple Ohm’s-law calculations. −15 V

2.2 kΩ Rgen

5.6 kΩ

Rbat Load

+

+ Egen

−

Ebat

6.8 kΩ

3.3 kΩ

RB

−

(a) Automobile electrical system

+10 V (b) Transistor coupling circuit

Figure 9-11  Examples of networks with more than one voltage source

215

216

Chapter 9    Resistance Networks

9-7 Loop Equations in Multisource Networks The procedure we developed in Section 9-5 for writing Kirchhoff’s voltagelaw equations for closed tracing loops applies equally well to multisource networks. As our first example, we shall use the automobile electrical system of Figure 9-11(a).

Example 9-4

An automobile generator with an internal resistance of 0.20 Ω develops an open-circuit voltage of 16.0 V. The storage battery has an internal resistance of 0.10 Ω and an open-circuit voltage of 12.8 V. Both sources are connected in parallel to a 1.0-Ω load. Determine the generator current, battery current, and load current. Solution I We can redraw the circuit diagram of Figure 9-11(a) as shown in Figure 9-12 to make it easier to draw the tracing loops. − 0.10 Ω +

+ 0.20 Ω − Rgen

Rbat

+ Egen

16.0 V

Igen

1.0 Ω

−

+ RL −

+ Ibat

Ebat

12.8 V

−

Figure 9-12  Circuit diagram for Solution I of Example 9-4

For the generator loop with current Igen, But from Ohm’s law,

VRgen + VL = Egen

VRgen = Igen Rgen = Igen × 0.20 Ω

Since Figure 9-12 shows both Igen and Ibat loops going in the same direction through the load, the load current is ­Igen + Ibat , and

VL = ILRL = 1.0 Ω × ( Igen + Ibat )

Hence, the Kirchhoff’s voltage-law equation for the generator loop becomes or

0.20Igen + 1.0 ( Igen + Ibat ) = Egen 1.2Igen + 1.0Ibat = 16.0

(1)

1.0Igen + 1.1Ibat = 12.8

(2)

Similarly, the Kirchhoff’s voltage-law equation for the battery loop becomes 0.10Ibat + 1.0(Igen + Ibat) = Ebat or

9-7   Loop Equations in Multisource Networks

Now we solve Equations 1 and 2 simultaneously. Elimination Method subtracting gives

5 × Equation 1: 6Igen + 5.0Ibat = 80

6 × Equation 2: 6Igen + 6.6Ibat = 76.8

− 1.6Ibat = 3.2

Ibat =−2.0 A



The negative value for Ibat is correct. This value tells us that the battery is not supplying current to the load, as we had supposed when setting up the direction for the tracing loop for Ibat. In this example, the storage battery is charging at a rate of 2.0 A. We do not need to change the direction of the Ibat tracing loop as long as we treat Ibat as a negative quantity. Substituting for Ibat in Equation 1 gives 1.2Igen + (1.0)(−2.0) = 16 1.2Igen − 2.0 = 16 Igen = 15 A

IL = Igen + Ibat = 15 + (−2.0) = 13 A Determinant Method

1.2Igen + 1.0Ibat = 16.0 1.0Igen + 1.1Ibat = 12.8





Igen

Ibat

│16.0 12.8 = │1.2 1.0

│1.2 1.0 = │1.2 1.0

│ 17.6 − 12.8 4.8 = = = 15 A 1.32 − 1.0 0.32 1.0 1.1│ 1.0 1.1

│ 15.36 − 16.0 −0.64 = = = −2.0 A 0.32 0.32 1.0 1.1│

16.0 12.8

IL = Igen + Ibat = ( 15 − 2 ) = 13 A

Solution II The tracing loops in Solution I are the normal choices from a circuit-­ operation point of view. However, all other choices lead to the same solution. To illustrate the loop technique further, we return to the circuit layout given in the original circuit diagram of Figure 9-11(a) and add tracing loops as shown in Figure 9-13.

217

218

Chapter 9    Resistance Networks

− Rgen

0.20 Ω +

Rbat

+ Igen

IL

+

RL

+ Egen

16.0 V

0.10 Ω

Ebat

−

12.8 V

1.0 Ω

−

−

Figure 9-13  Circuit diagram for Solution II of Example 9-4

Igen flows from the positive terminal of the generator through Rgen, Rbat, and the battery, back to the negative terminal of the generator. The IL tracing loop follows the same path as the Ibat loop in Solution I. But IL is not the same current as Ibat. The total load current is IL since the IL tracing loop is the only one that passes through RL. The battery current in this solution is the discharge current IL minus the charging current Igen. Note that the voltage drop across Rbat will be Rbat times the difference of the two loop currents. In writing loop equations, the direction of the tracing loop is always the positive current direction. Any current going against the tracing direction is negative. Hence, Kirchhoff’s voltage-law equation for the IL loop becomes 0.10 ( IL − Igen ) + 1.0IL = 12.8



−0.10Igen + 1.1IL = 12.8

or

(1)

In writing the equation for the Igen loop, we note that Igen enters the + terminal of the battery, just as it enters the + terminal for voltage drops across resistors. Thus, Ebat can appear as a positive term on the left-hand side of the equation. Alternatively, we can think of Ebat as a voltage source bucking the generator voltage. Then Ebat appears on the right-hand side of the equation as a negative quantity. The equation for the Igen loop becomes 0.20Igen + 0.10 ( Igen − IL ) + 12.8 = 16.0

or

│12.8 3.2 = │−0.1 0.3

0.30Igen − 0.10IL = 3.2

│ −1.28 − 3.52 −4.80 = = = 15 A +0.01 − 0.33 −0.32 1.1 −0.1│

(2)

Again we can use determinants to solve simultaneous Equations 1 and 2:



Igen

1.1 −0.1

As shown in Appendix 1, the letter D represents the matrix of all the ­coefficients of a system of equations. In this example,

9-7   Loop Equations in Multisource Networks

D=



IL =



│−0.1 0.3

│−0.1 0.3

1.1 0.1

│

12.8 3.2 D

│

=

219

−0.32 − 3.84 −4.16 = = 13 A −0.32 −0.32

Battery current = IL − Igen = 13 − 15 = −2 A

Again we find that the battery is charging rather than discharging. Multisim Solution Download Multisim file EX9-4 from the website. The circuit is the same as shown in Figure 9-12, except that ammeters have been added to measure the generator current, battery current, and load current. Run the simulation. Note that the ammeter readings match the c­ urrents calculated using loop equations.

Our next example illustrates how the Edison three-wire distribution ­system reduces power losses in transmission lines.

Example 9-5 (a) Calculate the total power loss in the three 1.0-Ω conductors in the three-wire distribution system shown in Figure 9-14. (b) Calculate the total power loss if the two loads are fed in parallel from a single 110-V source through a pair of 1.0-Ω conductors. R1 1.0 Ω

+ 110 V E1 −

I1

RL1

10 Ω

RL2

15 Ω

RN 1.0 Ω

+ 110 V E2 −

I2 R2 1.0 Ω

Figure 9-14  Three-wire distribution system

circuitSIM walkthrough

220

Chapter 9    Resistance Networks

Solution (a) The voltage drop across the resistance of the common or neutral conductor is 1.0 Ω times the difference of the two loop currents. As in the second solution for Example 9-4, current in the direction of a tracing loop direction is positive and any current flowing against the tracing ­direction is negative. Therefore, the Kirchhoff’s voltagelaw equation for the I1 loop becomes 1.0I1 + 10I1 + 1.0(I1 − I2) = E1

or   12I1 − I2 = 110

(1)

For the I2 loop,

1.0(I2 − I1) + 15I2 + 1.0I2 = E2

or   −I1 + 17I2 = 110



I1

│110 110 = │−112

I2 =

12 │−1

│ 1870 − ( −110 ) 1980 = = 9.75 A = 204 − ( +1 ) 203 −1 17│ −1 17

110 110 D

│

=

(2)

1320 − ( −110 ) 1430 = = 7.04 A 203 203

Power loss in line 1: P1 = I21R1 = 9.752 × 1.0 = 95 W Power loss in line 2: P2 = I22R2 = 7.042 × 1.0 = 50 W

Neutral current:   IN = I1 − I2 = 9.75 − 7.04 = 2.71 A Neutral power loss: PN = IN2RN = 2.712 × 1.0 = 7 W

Total power loss:  PT = P1 + P2 + PN = 95 + 50 + 7 = 152 W

(b) If the two loads are connected in parallel, the equivalent load resistance is Req =

10 × 15 = 6.0 Ω 10 + 15

Adding the resistance of two conductors, RT = 8.0 Ω. Hence, the circuit ­current becomes



I=

E 110 V = = 13.75 A RT 8.0 Ω

Power loss in each line = I2R = 13.752 × 1.0 = 0.19 kW Therefore, total power loss in the two conductors is PT = 2 × 0.19 kW = 0.38 kW

9-7   Loop Equations in Multisource Networks

The three-wire distribution system reduces transmission line losses by a­ rranging loop currents that largely cancel in the neutral lead. If we arrange for equal values of RL in Figure 9-14, the load is balanced and the neutral current is zero. Unbalanced load does cause some power loss in the neutral line, but not as much as operating both loads in parallel on a two-wire system.

Example 9-6 If RB is 1.0 kΩ in the coupling circuit shown in Figure 9-11(b), determine the magnitude and polarity of the voltage drop across RB. Solution The original circuit diagram does not show the actual voltage sources. We draw them on a working diagram (Figure 9-15) to enable us to draw complete tracing loops.

2.2 kΩ

5.6 kΩ 3.3 kΩ

− 15 V +

I1

6.8 kΩ

I3

I2

RB

1.0 kΩ

+ 10 V −

Figure 9-15  Circuit diagram for Example 9-6

We require three loops to include all components in this circuit. Several combinations of loops are possible. The loops drawn in Figure 9-15 are perhaps the most straightforward choices. Since volts = ­milliamperes  × kilohms, we can write the three Kirchhoff’s voltage-law ­equations without units as long as we remember that the currents are measured in milliamperes.

For the I1 loop, 6.8(I1 − I2) + 2.2I1 = 15 For the I2 loop, 3.3(I2 + I3) + 5.6I2 + 6.8(I2 − I1) = 10 For the I3 loop, 3.3(I3 + I2) + 1.0I3 = 10

Collecting like terms in each equation gives

9I1 − 6.8I2 = 15

−6.8I1 + 15.7I2 + 3.3I3 = 10 3.3I2 + 4.3I3 = 10

(1) (2) (3)

221

222

Chapter 9    Resistance Networks

│ │

│ │

We need to solve for just I3 since it is the only current through RB.



I3 = =



=

9 −6.8 0

9 −6.8 0

−6.8 15.7 3.3

−6.8 15.7 3.3

15 10 10 0 3.3 4.3

1413 + 0 − 336.6 − 0 − 297 − 462.4 607.59 + 0 + 0 − 0 − 98.01 − 198.83 317 310.75

= 1.02 mA

Hence, the voltage drop across RB is

VB = 1.02 mA × l.0 kΩ = 1.0 V

Since I3 is a positive quantity, the top end of RB is positive with respect to the chassis.

circuitSIM walkthrough

Multisim Solution Download Multisim file EX9-6 from the website. The circuit is the same as shown in Figure 9-15. Insert a voltmeter to measure the voltage across RB. Run the simulation. Note that the voltmeter reading matches the value calculated using loop equations.

See Problems 9-10 to 9-26 and Review Questions 9-62 to 9-65. Mesh equations were developed by James Clerk Maxwell (1831–79), the British physicist who also ­developed electromagnetic ­theory and predicted the existence of radio waves before they were discovered.

9-8  Mesh Analysis Now that we have a feel for writing equations for Kirchhoff’s voltage law, we can turn our attention to mesh analysis, a procedure that simplifies and speeds up writing the simultaneous equations for solving various resistance networks. The format for mesh equations is straightforward, but it cannot handle some of the networks that we can solve with the loop procedure. A mesh is a closed loop that does not enclose any circuit elements. Therefore, the circuit diagram cannot contain any conductors that cross without being joined at the point of crossing. In other words, the network diagram

9-8   Mesh Analysis

must be strictly two-dimensional or planar. The mesh format requires all sources to be voltage sources. We must convert any current sources to equivalent voltage sources before we use the mesh format. Instead of choosing convenient tracing loops, we draw a current loop for each mesh. All mesh currents must have the same direction, either clockwise or counterclockwise. The more common choice is clockwise. These rules provide a standard format for every mesh equation. In the loop procedure, we write equations for voltage drops around the current loops, and then rewrite the Kirchhoff’s voltage-law equations to collect the current terms. In the mesh format, we go directly to this second step. We can use the network shown in Figure 9-16 to illustrate the mesh format. The network is clearly planar and the four meshes have no internal electric components. 10 Ω

+ 20 V −

70 Ω

IA

20 Ω

30 Ω

80 Ω

40 Ω

60 Ω

50 Ω

IC

IB

+ 30 V −

ID 90 Ω

+ 40 V −

Figure 9-16  Mesh network for Example 9-7

For any particular mesh current, all other currents through common ­circuit elements are in the opposite direction. For example, IA and IB flow through the 30-Ω resistor in opposite directions. The mesh equation for mesh A has the form

RTIA − RAB IB − RACIC = ΣE

(9-7)

where RT is the total resistance around mesh A, RAB is the resistance that is common to mesh A and mesh B, RAC is the resistance that is common to mesh A and mesh C, and ΣE is the algebraic sum of the voltage sources around the mesh. Where IA passes inside a voltage source from its negative terminal to its  positive terminal, E is a positive. If the direction of IA through a voltage  source is from the positive terminal to the negative terminal, E is ­negative.

223

224

Chapter 9    Resistance Networks

Example 9-7 Find the magnitudes and polarities of the voltage drops across the 30-Ω, 50-Ω, and 60-Ω resistors in the network shown in Figure 9-16. Solution Using the mesh format, write a Kirchhoff’s voltage-law equation for each of the four meshes in Figure 9-16. In this example, the resistances are in ohms and the potential differences are in volts, so the currents will be in amperes. (10 + 30 + 50)IA − 30IB − 50IC = 20 (30 + 20 + 40 + 60)IB − 30IA − 60ID = 0 (80 + 70 + 50)IC − 50IA = −30 (90 + 60)ID − 60IB = 30 − 40 Collecting like terms in each equation gives 90IA − 30IB − 50IC



−30IA − 150IB



−50IA



− 60IB

+ 200IC

=

20

(1)

0

(2)

+ 150ID = −10

(3)

− 60ID =

= −30

(4)

Unfortunately, the calculations for solving fourth-order determinants are not as simple as those for second- and third-order determinants. We can get around this problem by eliminating one of the unknowns. From Equation 4, 60IB − 10 150

ID =

Substituting for ID in Equation 2 gives

−30IA + 150IB − (24IB – 4) = 0

We now have three simultaneous equations:



IA =

│ │

20 −4 −30

90 −30 −50

90IA − 30IB − 50IC =

−30IA − 126IB

−50IA

−30 126 0

−30 126 0

−50 0 200

│ │

−50 0 200

=

20

(1)

+ 200IC = −30

(5)

= −4

291 000 = 0.164 A 1 773 000

(3)

IB = IC =

│

│

ID =

90 −30 −50

90 −30 −50

20 −4 −30 D

−30 126 0 D

−50 0 200 20 −4 −30

│

│

9-8   Mesh Analysis

=

13 000 = 0.0073 A 1 773 000

=

−193 200 = −0.109 A 1 773 000

60 × 7.33 × 10−3 − 10 = −0.064 A 150

I30 = IA − IB = 0.164 A − 0.0073 A = 157 mA V30 = 30 Ω × 157 mA = 4.7 V  [positive at top] I50 = IA − IC = 0.164 A + 0.109 A = 273 mA V50 = 50 Ω × 273 mA = 14 V [positive on the right] I60 = IB − ID = 0.0073 A − 0.064 A = 71 mA V60 = 60 Ω × 71 mA = 4.3 V [positive on the right] Multisim Solution Download Multisim file EX9-7 from the website.

The circuit is the same as shown in Figure 9-16, except that voltmeters have been added to measure the voltage drops across the 30-Ω, 50-Ω, and 60-Ω resistors. Run the simulation. Note that the voltmeter readings match the values calculated using mesh analysis.

To demonstrate the application of mesh analysis, we can now repeat ­Examples 9-2 to 9-6 using the mesh equations.

Example 9-2A The circuit diagram of Figure 9-7 has two meshes. The I1 mesh is the same as the I1 loop. The second mesh is the closed loop through R3, R1, R2, and R4. The two mesh equations are

225

(10 + 20 + 30)I1 − (20 + 30)I2 = 80 V

(30 + 20 + 15 + 10)I2 − (20 + 30)I1 = 0 V IR3 = I1 − I2   and  IR4 = I2

The remainder of the solution is the same as in Example 9-2.

circuitSIM walkthrough

226

Chapter 9    Resistance Networks

Example 9-3A The circuit diagram of Figure 9-9 has three meshes. The I1 mesh is the same as the I1 loop. The second mesh is the closed loop through R1, R2, and RM. The third mesh consists of R3, RM, and R4. With the like terms collected, the three mesh equations become 60I1 − 20I2 − 30I3 = 80



−20I1 + 85I2 − 50I3 = 0



−30I1 − 50I2 + 90I3 = 0



We need to solve for both I2 and I3, since IM = I2 − I3.

(1) (2) (3)

Example 9-4A In the circuit diagram of Figure 9-12, we can write mesh-current equations by simply reversing the direction of Ibat. The second solution for Example 9-4 using the circuit configuration of Figure 9-13 already satisfies the rules for the mesh format and yields the same simultaneous equations when the terms are collected.

Example 9-5A The circuit of Figure 9-14 also satisfies the mesh format rules.

Example 9-6A Before we can write the mesh equations for the circuit shown in Figure 9-15, all mesh currents must be in the same direction. Reversing I3 so that all the mesh currents are counterclockwise gives

(2.2 kΩ + 6.8 kΩ) × I1 − 6.8 kΩ × I2 = 15 V

(3.3 kΩ + 5.6 kΩ + 6.8 kΩ) × I2 − 6.8 kΩ × I3 − 3.3 kΩ × I3 = 10 V

(1.0 kΩ + 3.3 kΩ) × I3 − 3.3 kΩ × I2 = −10 V

9-8   Mesh Analysis

Collecting terms gives

9.0 kΩ × I1 − 6.8 kΩ × I2 = 15 −6.8 kΩ × I1 + 15.7 kΩ × I2 − 3.3 kΩ × I3 = 10 −3.3 kΩ × I2 + 4.3 kΩ × I3 = −10

(1) (2) (3)

To solve for VB, we need to solve only for I3, which is −1.0 mA since we reversed the direction of I3. See Problems 9-27 to 9-31 and Review Questions 9-66 to 9-71.

Circuit Check

B

CC 9-3. Use loop equations to calculate the current in the 4.0-Ω resistor in circuit shown in Figure 9-17. 5.0 Ω Ix 4.0 Ω

+

10 V

2.0 Ω

−

− 2.0 Ω

10 V

1.0 Ω

+

Figure 9-17

CC 9-4. Use mesh analysis to calculate the unknown current, Ix, in the circuit shown in Figure 9-18. 5.0 Ω

6.0 Ω Ix

− 2.0 Ω

2V +

2.0 Ω

1.0 Ω

+ 10 V −

4.0 Ω

−

−

3V

+ 6V

10

+ 7.0 Ω

− + 4V

Figure 9-18

Ω

227

228

Chapter 9    Resistance Networks

9-9 Kirchhoff’s Current-Law Equations The loop-current and mesh-equation methods of solving resistance networks are based on Kirchhoff’s voltage law. With Kirchhoff’s current law we can write a different set of equations by considering currents entering and leaving junction points or nodes in a circuit. First we must identify the nodes in a given network. In the network shown in Figure 9-19, we do not consider the junction of the voltage source and R1 to be a node because the current does not branch at this point. The current law equation for this junction is simply I1 = I1, which is redundant. Similarly, the junction of R3 and R6 is not a node. As shown in Figure 9-19(b), we treat all connections to a common grounded conductor as a single node. Hence, the network in Figure 9-19 has only three nodes.

R1 I1

Node A I4 I

2

R2

Node B I5 I

3

Node A

R3 I6

+ E

R4

R5

R6

R1 +

Node B R3

E

−

R2

R4

R5

R6

− Reference node (a)

Reference node (b)

Figure 9-19  Identifying nodes in a network

The simplest statement of Kirchhoff’s current law is that the algebraic sum of currents at a node is zero. To use this statement to write equations, we have to remember which currents are positive and which are negative. For nodal circuit analysis, it is more convenient to state Kirchhoff’s current law in this form: The sum of the currents leaving a node in a circuit equals the sum of the currents entering the node. For the three nodes in the network of Figure 9-19, this statement leads to the following equations: Node A: Node B: Reference node:

I4 + I2 = I1 (1) I5 + I3 = I2 (2)

I1 = I4 + I5 + I3 (3)

9-9   Kirchhoff’s Current-Law Equations

Note that if we add Equations 1 and 2, we obtain Equation 3. Hence, Kirchhoff’s current-law equation for the third node is redundant. The number of Kirchhoff’s current-law equations we need to solve any network is ­always one less than the total number of nodes. We label the redundant node as the reference node. Choosing a reference node that is common to as many branch currents as possible usually gives simpler equations. The grounded conductor or chassis connection is a good choice for the reference node. The equations for nodes A and B in Figure 9-19 have five unknown currents. However, there are only two node voltages, VA and VB, with respect to the reference node in the network. To reduce the unknowns in the current-law equations to two, we use I = V/R to substitute V/R ratios for each current, so that we can solve for VA and VB. Once we know the node voltages, we can use Ohm’s law to find any other unknown in the network. In the network shown in Figure 9-19, V4 = VA and V5 = VB. For node A:

I4 =

VA R4

I1 =

V1 E − VA = R1 R1

I2 =

and

V2 VA − VB = R2 R2

Substituting these expressions in Equation 1 gives

VA VA − VB E − VA + = (4) R4 R2 R1

For node B:

I5 =



I3 =



and



I2 =

VB R5 VB R3 + R6

VA − VB R2

VB VB VA − VB + = (5) R5 R3 + R 6 R2

Given values for E and all the resistors, we can solve for VA and VB.

229

230

Chapter 9    Resistance Networks

Example 9-4B

An automobile generator with an internal resistance of 0.20 Ω develops an open-circuit voltage of 16.0 V. The storage battery has an internal resistance of 0.10 Ω and an open-circuit voltage of 12.8 V. Both sources are connected in parallel to a 1.0-Ω load. Determine the load current. Solution The Kirchhoff’s current-law equations for the two nodes in Figure 9-20 are the same except that all the signs are reversed. Hence, we need to write only one Kirchhoff’s current-law equation. IL = Igen + Ibat

0.20 Ω

0.10 Ω

Node

Rgen

+ 16.0 V −

Rbat

+ Egen

1.0 Ω

RL

Ebat

12.8 V

− Reference node

Figure 9-20  Circuit diagram for Example 9-4B

Note that we again assume that the battery is discharging. Applying Ohm’s law gives IL =

Egen − VL VL Ebat − VL ,   Igen = ,   and   Ibat = RL Rgen Rbat

Substituting these V/R ratios for the currents in the Kirchhoff’s currentlaw equation, we obtain VL 16.0 V − VL 12.8 V − VL = + 1.0 Ω 0.20 Ω 0.10 Ω

Multiplying the equation by 0.1 gives

0.10VL = 8.0 − 0.5VL + 12.8 − VL

and

VL =

IL =

20.9 = 13 V 1.6

VL 13 V = = 13 A RL 1.0 Ω

9-10   Nodal Analysis

Since we only required one nodal equation in the solution of Example 9-4B, we avoided the simultaneous equations of Example 9-4, which required two tracing loops. However, loop and mesh equations are free from algebraic fractions since we substitute a product—the IR drop—for each voltage drop. In the nodal equation of Example 9-4B, we substitute a V/ R ratio for each current term, leaving us with an equation in a more complex form. We can simplify such equations by using conductance instead of resistance. We then substitute I = VG for the currents. See Review Questions 9-72 to 9-74.

9-10  Nodal Analysis Nodal analysis is a circuit-analysis format that combines Kirchhoff’s currentlaw equations with the source conversions we developed in Section 9-4. Converting all voltage sources to equivalent constant-current sources allows us to standardize the way we write the Kirchhoff’s current-law equations. For nodal analysis, we consider source currents to flow into a node. If the arrows in a circuit diagram show that a source current actually flows out of a particular node, the current is a negative quantity for that node. Similarly, we consider all resistor currents to flow out of a node. We are thus restating Kirchhoff’s current law in the form: At any independent node, the algebraic sum of the resistor currents leaving the node equals the algebraic sum of the source currents ­entering the node. The Kirchhoff’s current-law equations then have the format

IR1 + IR2 + . . . = IS1 + IS2 + . . . (9-8)



The next step is to replace the resistor currents using I = VG. The voltage across each resistor equals the difference between the voltages, relative to the reference node, of the nodes at each end of the resistor. We always subtract the voltage of the adjacent node from the voltage of the node for which we are writing the equation. For node 1 in Figure 9-21, the current through R1 Node 1

IS1

Node 2

R3

R1

R2

IS2

Reference node

Figure 9-21  Circuit diagram for writing nodal equations

231

232

Chapter 9    Resistance Networks

equals V1G1. The voltage across R3 is V1 – V2. Hence the Kirchhoff’s currentlaw equation for node 1 becomes V1G1 + ( V1 − V2 ) G3 = IS1

Collecting the voltage terms gives

( G1 + G3 ) V1 − G3 V2 = IS1 (9-9)



Equation 9-9 illustrates the standard format for writing a nodal equation. We can use determinants to solve for the voltages in the same way that we solved for currents in loop and mesh equations. We can write Equation 9-9 directly by noting that the positive voltage term on the left-hand side of the equation is the unknown voltage for the node in question multiplied by the sum of the conductances connected to that node. From this positive term, we must subtract a term for the node voltage at every adjacent node connected by a resistor to the node in question. This term is the adjacent node voltage multiplied by the conductance between the two nodes. Using this procedure to write the nodal equation for node 2 gives ( G2 + G3 ) V2 − G3V1 = −IS2

Note the duality between this nodal equation and the mesh format equation Equation 9-7. Nodal analysis is particularly useful for networks where a common portion of the network is fed from several sources in parallel.

Example 9-4C

An automobile generator with an internal resistance of 0.20 Ω develops an open-circuit voltage of 16.0 V. The storage battery has an internal resistance of 0.10 Ω and an open-circuit voltage of 12.8 V. Both sources are connected in parallel to a 1.0-Ω load. Determine the load current. Solution Step 1 First we draw the equivalent circuit with the voltage sources converted to constant-current sources, as shown in Figure 9-22.

80 A

Igen 0.20 Ω

Rgen

1.0 Ω

RL

0.10 Ω

Figure 9-22  Circuit diagram for Example 9-4C

Rbat

Ibat

128 A

9-10   Nodal Analysis

The short-circuit current for the battery is Ibat = =

Ebat 12.8 V = = 128 A Rbat 0.10 Ω

Similarly, the short-circuit current of the generator is Igen =

Egen Rgen

=

16.0 V = 80 A 0.20 Ω

With a bit of mental arithmetic we can fill in the required data on the c­ ircuit diagram of Figure 9-22. Note that the generator and battery currents in this equivalent circuit are not the same as the real currents in the original circuit of Figure 9-20. Step 2 Next we find the conductance of each branch: Ggen =

1 1 1 1 = = = 5.0 S     Gbat = = 10 S Rgen 0.20 Ω Rbat 0.10 Ω GL =

and

1 = 1.0 S RL

Step 3 At first glance, Figure 9-22 appears to show six junction points, but there are only two because the same voltage appears across all parallel branches. For our nodal analysis, we can redraw the circuit in the form shown in F ­ igure 9-23. If we label the lower node as the reference node, the upper node is the only independent node. Consequently, we require only one Kirchhoff’s current-law equation for the circuit of ­Figure 9-23. Node

80 A

Igen 5.0 S

Ggen

1.0 S

GL

10 S

Gbat

Reference node

Figure 9-23  Equivalent circuit for Example 9-4C

Ibat

128 A

233

234

Chapter 9    Resistance Networks

At the single independent node, Equation 9-8 becomes I5 + I1 + I10 = Igen + Ibat

Or, we can go directly to the format of Equation 9-9, (G5 + G1 + G10)V = Igen + Ibat

V=



circuitSIM walkthrough

80 + 128 208 A = = 13 V 5.0 + 1.0 + 10 16 S

IL = VGL = 13 V × 1.0 S = 13 A

Multisim Solution Download Multisim file EX9-4C from the website.

The circuit is the same as shown in Figure 9-20. Insert an ammeter to measure the current through RL. Run the simulation. Note that the ammeter reading matches the value calculated using nodal analysis.

Nodal analysis allowed us to solve Example 9-4C with not much more than mental arithmetic since the circuit has only a single independent node. However, we do need to solve simultaneous equations when the network has more than one independent node.

Example 9-6B If RB is 1.0 kΩ in the coupling circuit shown in Figure 9-24, determine the magnitude and polarity of the voltage drop across RB.

2.2 kΩ

5.6 kΩ 3.3 kΩ

− 15 V +

RB

6.8 kΩ

+ 10 V −

Figure 9-24  Circuit diagram for Example 9-6B

Solution Step 1 In the circuit diagram of Figure 9-24, we can consider the 2.2-kΩ resistor to be the internal resistance of a 15-V constant-voltage

9-10   Nodal Analysis

235

source c­onnected across the 6.8-kΩ resistor. Similarly, we treat the 3.3-kΩ resistor as the internal resistance of a 10-V constant-voltage source connected across RB. Therefore, when we convert the voltage sources to constant-current sources, we obtain the equivalent circuit in Figure 9-25 with I1 =

15 V 10 V = 6.82 mA  and  I2 = = 3.03 mA 2.2 kΩ 3.3 kΩ

We must make sure that the directions of I1 and I2 are consistent with the polarities of the voltage sources they replace. Step 2 By converting each of the five resistances into its equivalent conductance, we obtain the values shown in Figure 9-25. V1 – V 2 5.6 kΩ

Node 1

Node 2

179 μS

− V1

−

I1

6820 μA

+

2.2 kΩ 454 μS

−

−

+

+

6.8 kΩ 147 μS

+ +

I2

+ 3030 μA

3.3 kΩ R bat 303 μS

−

−

+

1.0 kΩ 1000 μS

V2

−

Reference node

Figure 9-25  Equivalent circuit for Example 9-6B

Step 3 Choosing the bottom conductor (the chassis) as the reference node leaves two independent nodes in Figure 9-25. We need a nodal equation for each independent node. At node 1, all the currents are negative quantities since the source current actually leaves the node and resistor currents enter the node. Since volts × microsiemens = microamperes, the voltages will be in volts if we express the currents in microamperes and the conductances in microsiemens. For node 1,

(454 + 147 + 179)V1 − 179V2 = −6820

For node 2,  (179 + 303 + 1000)V2 − 179V1 = 3030 Collecting terms gives

780V1 − 179V2 = −6820

−179V1 + 1482V2 =

3030

(1) (2)

If we solve for V1, we find that it is negative, indicating that node 1 has a negative ­potential difference with ­respect to the chassis.

236

Chapter 9    Resistance Networks

780 │−179 = 780 │−179

│ 2 363 400 − 1 220 780 1 142 620 = = 1.02 V = 1 155 960 − 32 041 1 123 919 −179 1482│

Step 4 To find the voltage drop across RB, we need solve only for V2.

V2

−6820 3030

Hence, the voltage drop across RB is 1.02 V. Since V2 is positive, node 2 is positive with respect to the chassis.

circuitSIM

Multisim Solution Download Multisim file EX9-6B from the website.

walkthrough

The circuit is the same as shown in Figure 9-23. Insert a voltmeter to measure the voltage across RB. Run the simulation. Note that the voltmeter reading matches the value calculated using nodal analysis.

Although we had to solve simultaneous equations in Example 9-6B, they involved only two unknowns rather than three as in the loop and mesh ­solutions for this example. The Wheatstone bridge circuit of Figure 9-26(a) has four nodes, so we need three nodal equations to solve it. Since the loop or mesh procedure also needs three simultaneous equations, we would probably not use nodal analysis to solve the bridge circuit. However, using loop or mesh format to solve the circuit in Figure 9-26(b) requires four simultaneous equations while nodal analysis requires only one nodal equation to solve the whole network. Node

Node 1

Node 2

Node 3

RL

+ E

+

+ E1

−

Reference node

(a)

−

+ E2

−

+ E3

−

E4

−

Reference node (b)

Figure 9-26  (a) Wheatstone bridge; (b) Four parallel sources supplying a common load

See Problems 9-32 to 9-35 and Review Questions 9-75 and 9-76.

9-11   The Superposition Theorem

9-11  The Superposition Theorem The Kirchhoff’s-laws methods for analyzing resistance networks usually ­require solving a set of simultaneous linear equations. If there are no more than two tracing loops or independent nodes, the computation is not particularly onerous. But when we need three or more simultaneous equations, we start looking for alternative solutions. We can often avoid simultaneous equations in multi-source networks by applying a fundamental principle of linear networks known as the superposition theorem. In a network of resistors with multiple voltage sources, the current in any branch is the algebraic sum of the component currents that would be caused in that branch by each source acting independently with the others replaced by their respective internal resistances. This theorem is useful only for determining the current through or voltage drop across one branch of a network containing more than one source. However, in many cases, once we know the current through one branch of a network, we can solve the remainder of the network quite readily. In both the loop-current and nodal methods of solving resistance networks, the currents appearing in the simultaneous equations are not ­necessarily actual currents that we could measure at various points in the network. To obtain actual branch currents, we have to add the appropriate loop or mesh currents algebraically. Similarly, component currents calculated by the superposition theorem are only a means of calculating the ­actual current in a certain branch and do not represent currents that we can measure directly.

Example 9-4D

An automobile generator with an internal resistance of 0.20 Ω develops an open-circuit voltage of 16.0 V. The storage battery has an internal resistance of 0.10 Ω and an open-circuit voltage of 12.8 V. Both sources are connected in parallel to a 1.0-Ω load. Determine the load current. Solution Because there are two sources, there are two component currents through the load resistance. To apply the superposition theorem, we redraw the circuit of Figures 9-11(a) and 9-12 as the two equivalent circuits in Figure 9-27.

237

238

Chapter 9    Resistance Networks

IT

0.20 Ω

16.0 V −

I L1

Rgen

+ Egen

1.0 Ω

RL 0.10 Ω

Rbat

(a) 0.10 Ω IL2 0.20 Ω

Rgen 1.0 Ω

IT

Rbat

+

RL

Ebat

12.8 V

− (b)

Figure 9-27  Equivalent circuits for Example 9-4D

The first equivalent circuit of Figure 9-27(a) is a series-parallel network. Therefore, RT = Rgen +

RL × Rbat 1.0 × 0.10 = 0.20 + = 0.29 Ω RL + Rbat 1.0 + 0.10

Egen 16.0 V   IT = = = 55 A RT 0.29 Ω

From the current divider principle, the first component of the load current is ­IL1 = IT ×

Rbat 0.10 = 55 × = 5.0 A RL + Rbat 1.1

Similarly, for the circuit of Figure 9-27(b), RT = Rbat +



IT =

Rgen × RL

Rgen + RL

= 0.10 +

Ebat 12.8 V = 48 A = RT 0.267 Ω

0.20 × 1.0 = 0.267 Ω 0.20 + 1.0

The second component of the load current is IL2 = IT ×

0.20 = 8.0 A Rgen + RL 1.2 Since the component currents have the same direction in Figures 9-27(a) and (b), the actual load current is the sum of the component currents. Rgen

= 48 ×

IL = IL1 + IL2 = 5 + 8 = 13 A

9-11   The Superposition Theorem

Example 9-6C If RB is 1.0 kΩ in the coupling circuit in Figure 9-11(b), determine the magnitude and polarity of the voltage drop across RB. Solution Again there are two sources, so the superposition theorem yields two component currents. In the equivalent circuit of Figure 9-28(a), the 10-V source has been replaced by its zero-ohm internal resistance so that only the 15-V source contributes to the current through RB. Again we can solve the circuit as a series-parallel network. First we replace the 5.6-kΩ and 3.3-kΩ resistors and RB with their equivalent resistance, Req, then we find the total resistance for the equivalent circuit shown in Figure 9-28(b): Req = 5.6 kΩ + RT = 2.2 kΩ + IT =

2.2 kΩ

− 15 V 6.8 kΩ +

3.3 kΩ × 1 kΩ = 6.4 kΩ 3.3 kΩ + 1 kΩ

6.8 kΩ × 6.4 kΩ = 5.5 kΩ 6.8 kΩ + 6.4 kΩ

E 15 V = = 2.7 mA RT 5.49 kΩ

5.6 kΩ

2.2 kΩ

− Rbat 1.0 kΩ +

3.3 kΩ

IT

− 15 V 6.8 kΩ +

6.4 kΩ

IT

I B1

Req

Ieq (b)

(a) IB2

IB2

5.6 kΩ 3.3 kΩ 2.2 kΩ

6.8 kΩ

IT + 10 V −

3.3 kΩ 1.0 kΩ

Rbat

Req

(c)

Figure 9-28  Equivalent circuit for Example 9-6C

7.26 kΩ

IT + 10 V − (d)

1.0 kΩ

Rbat

239

240

Chapter 9    Resistance Networks

Now we use the current-divider principle twice to find the first component of the current through RB. In Figure 9-28(b), the current through Req is Ieq = 2.73 mA ×

6.8 kΩ = 1.41 mA 6.8 kΩ + 6.4 kΩ

IB1 = 1.41 mA ×

3.3 kΩ = 1.08 mA 3.3 kΩ + 1.0 kΩ

In the circuit of Figure 9-28(a),

Similarly, when the 10-V source operates alone, as in Figure 9-28(c), we start by replacing the 5.6-kΩ, 2.2-kΩ, and 6.8-kΩ resistors with an equivalent ­resistance. Req = 5.6 kΩ + RT = 3.3 kΩ + IT =

2.2 kΩ × 6.8 kΩ = 7.26 kΩ 2.2 kΩ + 6.8 kΩ

7.26 kΩ × 1.0 kΩ = 4.17 kΩ 7.26 kΩ + 1.0 kΩ

E 10 V = = 2.39 mA RT 4.17 kΩ

Applying the current-divider principle to the equivalent circuit in Figure 9-28(d) gives the second component of the current through RB: IB2 = 2.39 mA ×

7.26 kΩ = 2.10 mA 7.26 kΩ + 1 kΩ

In Figures 9-28(a) and (c) the two component currents flow in o­ pposite ­directions through RB. Hence, the real current through RB is and

IB = 2.10 mA – 1.08 mA = 1.02 mA

VB = 1.02 mA × 1.0 kΩ = 1.02 V

Because the component current IB2 is greater than IB1, the real current has the same direction through RB as IB2 in Figure 9-28(c). Thus, the top end of RB is 1.02 V positive with respect to the chassis.

See Problems 9-36 to 9-43 and Review Questions 9-77 to 9-81.

9-11   The Superposition Theorem

Circuit Check

C

CC 9-5. Use nodal analysis to calculate the voltage across the 4-Ω ­resistor in the circuit in Figure 9-29. 2.0 Ω

5A

1.0 Ω

1.0 Ω

5.0 Ω

2A

4.0 Ω

2.0 Ω

Figure 9-29

CC 9-6. Use the superposition theorem to calculate the unknown current in the circuit in Figure 9-30. Ix

+ 90 V − 36 Ω

Figure 9-30

12 Ω 2A

+ 60 V −

8.0 Ω

241

242

Chapter 9    Resistance Networks

Summary

• A constant-voltage source can be represented by a voltage source in series with a resistor. • A constant-current source can be represented by a current source in parallel with a resistor. • A constant-voltage source may be converted to an equivalent constantcurrent source and vice versa. • A circuit containing constant-voltage sources may be analyzed using the loop procedure based on Kirchhoff’s voltage-law equations. • A planar circuit containing constant-voltage sources may be analyzed using mesh equations. • A circuit containing constant-current sources can be analyzed using nodal analysis, which is based on Kirchhoff’s current law. • A circuit containing more than one source may be analyzed using the ­superposition theorem. B = beginner

Problems

I = intermediate

A = advanced

Solve each network problem with one method and use a different method to check the solution.

B B

B

B

Section 9-2  Constant-Voltage Sources 9-1.

As the current drawn from a DC power supply is increased from 600 mA to 800 mA, the terminal voltage decreases from 25 V to 23 V. Determine the equivalent constant-voltage source. 9-2. Find the voltage across a load resistance of 21 Ω connected to a constant-voltage source that has an open-circuit voltage of 12 V and a short-circuit current of 8 A.

Section 9-3  Constant-Current Sources

9-3. A 15-A constant current source has an internal resistance of 400 Ω. How much current will this source deliver to a 10-Ω load? Find the voltage across the load.

Section 9-4  Source Conversion

9-4. A voltage source has an open-circuit terminal voltage of 120 V and an internal resistance of 0.8 Ω. Determine the equivalent constantcurrent source.

243

Problems

B

9-5. Convert the voltage sources of Figure 9-31 into the equivalent ­constant-current sources. 180 Ω

+ 160 V −

68 kΩ

+

20 kΩ

E 25 mV

25 kΩ 100 kΩ

+ 100 V −

45 Ω

− 85 V + − (b)

(a)

(c)

Figure 9-31

B

9-6.

Replace the current sources in Figure 9-32 with equivalent constantvoltage sources. 33 kΩ

60 A

2.5 Ω

45 μA

(a)

22 kΩ

16 A

(b)

4.0 Ω

(c)

Figure 9-32

I

Section 9-5 Kirchhoff’s Voltage-Law Equations: Loop Procedure 9-7.

Determine the current through the 400-Ω resistor in Figure 9-33.

120 Ω

240 Ω

+ − 6.0 V

400 Ω 360 Ω

Figure 9-33

80 Ω

24 A

244

Chapter 9    Resistance Networks

I

9-8.

When the input voltage is 3.0 V, what is the current through a 250-Ω load connected to the output terminals of the bridged-T attenuator of Figure 9-34?

400 Ω 100 Ω Input

100 Ω 800 Ω

Output

Figure 9-34  Bridged-T network

A

9-9.

If 320 V DC is applied to the input terminals of the lattice network of Figure 9-35, determine the current through a 40-Ω resistor connected to the output terminals. (Hint: Redraw the circuit to eliminate the crossing wires.) 20 Ω 30 Ω

Output 30

Ω

Input

20 Ω

Figure 9-35  Lattice network

A

Section 9-7  Loop Equations in Multisource Networks 9-10. Determine the magnitude and polarity of the voltage drop across the 50-Ω resistor in Figure 9-36. 10 Ω

+ 20 V −

20 Ω

50 Ω

40 Ω

Figure 9-36

30 Ω

− 40 V +

Problems

A

245

9-11. Find the current drawn from each of the sources in Figure 9-37. 5.0

kΩ

20

kΩ

10 kΩ

+ 100 V −

+ 150 V −

+ 200 V −

Figure 9-37

I A I I I

I A

9-12. (a) Find the battery current when the open-circuit terminal voltage of the generator in Example 9-4 drops to 15.0 V. (b) Use Multisim to verify the battery current in part (a). 9-13. For the circuit of Example 9-4, find the open-circuit generator ­voltage that (a) makes the battery current zero (b) causes a 10-A charging current to flow into the battery 9-14. Find the neutral current in the circuit of Example 9-5 if a second 15-Ω load is connected in parallel with the existing 15-Ω load. 9-15. Calculate the total power loss in the three 1.0-Ω conductors in the three-wire distribution system shown in Figure 9-14 if the polarity of source E2 is reversed. 9-16. (a) Two batteries are connected in parallel to feed a 12-Ω load. Battery A has an open-circuit voltage of 6.3 V and an internal resistance of 1.5 Ω. Battery B has an open-circuit voltage of 6.0 V and an internal resistance of 2.1 Ω. Find the current drain from each battery. (b) Use Multisim to verify the current drains calculated in part (a). 9-17. If the 12-Ω load in Problem 9-16 is replaced by a battery charger with an open-circuit voltage of 7.2 V and an internal resistance of 0.20 Ω, find the charging current of each battery. 9-18. (a) Find the current through each voltage source in the circuit in ­Fig­ure 9-38. (b) Use Multisim to verify the currents calculated in part (a).

+ 120 V −

Figure 9-38

2.2 kΩ

walkthrough

circuitSIM walkthrough

circuitSIM walkthrough

3.3 kΩ 4.7 kΩ

circuitSIM

6.8 kΩ

− 90 V +

246

Chapter 9    Resistance Networks

B

circuitSIM

A

walkthrough

9-19. Determine the magnitude and direction of the current through the 2.2-kΩ resistor in the circuit of Figure 9-38. 9-20. (a) Determine the current through each of the five resistors in the circuit of Figure 9-39. (b) Use Multisim to verify the voltage calculated in part (a). 100 Ω

+ 50 V −

50 kΩ

300 Ω

Egen

A

90 V

50 kΩ

I

Figure 9-40

I

circuitSIM

I

walkthrough

−

+ 25 V −

Figure 9-39

V + Reference battery −

400 Ω

500 Ω

+ 80 V −

200 Ω

+

100 V

9-21. The 5.6-kΩ resistor in the circuit of Figure 9-11(b) is replaced with a variable resistor that is adjusted until the voltage across RB is exactly zero. What is the resistance of the variable resistor under these ­circumstances? 9-22. In the balance-detecting circuit of Figure 9-40, the voltmeter has a ­resistance of 100 kΩ. Find the voltmeter reading. 9-23. The current through the unknown resistance R in Figure 9-41 is 400 mA. Find this resistance. 9-24. (a) In Figure 9-42, a pulse generator circuit acts as a switch in series with a 1.0-kΩ resistor. Calculate the output voltage across a 42-kΩ resistor connected from the output terminal to ground first with the switch open and then with the switch closed. (b) Use Multisim to verify the voltage across the 42-kΩ resistor with the switch closed.

100 Ω

22 kΩ

R

Out

1.0 kΩ 4.7 kΩ

100 Ω

+

10 kΩ

−

200 V

+25 V Figure 9-41

−15 V

Figure 9-42

I

9-25. The circuit shown in Figure 9-42 is modified as shown in Figure 9-43. Calculate the voltage drop across a 42-kΩ resistor connected from the output terminal to ground (a) with the switch open (b) with the switch closed

Problems

22 kΩ

247

Out

1.0 kΩ 4.7 kΩ

+25 V

10 kΩ

−15 V

Figure 9-43

I

A

9-26. A trolleybus is halfway along a section of trolley wire that is fed at one end from a 600-V source and at the other end from a 596-V source. The internal resistances of the sources are negligible, but each trolley wire in the section has a resistance of 1.0 Ω. The trolleybus is the only one in the section at the moment, and draws a 50-A current from the trolley wires. Find the voltage drop across the trolleybus.

Section 9-8  Mesh Analysis 9-27. In the network shown in Figure 9-44, there is no electric connection where the leads of the 20-Ω and 40-Ω resistors cross. Redraw the network to show that it is planar. Determine the voltage drop across the 16-Ω resistor. 16 Ω

Ω

+ 12 V 40 − Ω

20

+ 12 V −

32 Ω

Figure 9-44

I

A

9-28. (a) Find the current drawn from each of the sources in the network shown in Figure 9-37 when a 25-kΩ resistor is connected across the top of the network (from the + terminal of the 100-V source to the + terminal of the 200-V source). (b) Use Multisim to verify the current calculated in part (a). 9-29. Use loop or mesh equations to determine the current drawn from each source in the network of Figure 9-54.

circuitSIM walkthrough

248

circuitSIM

Chapter 9    Resistance Networks

I

walkthrough

9-30. (a) For the circuit shown in Figure 9-45, write the mesh equations and solve to find Ix. (b) Use Multisim to verify the current Ix calculated in part (a). 2.0 Ω

9.0 Ω

4.0 Ω

5.0 Ω

1.0 Ω

50 V

Ix

20 V

15 Ω

30 V 20 V 3.0 Ω

Figure 9-45

I

9-31. For the circuit shown in Figure 9-46, find Ix by mesh analysis. 8.0 Ω

7.0 Ω

3.0 Ω

4.0 Ω

5.0 Ω

20 V

15 V

Ix

20 V

5V

5.0 Ω 1.0 Ω

7.0 Ω

35 V

4.0 Ω

6.0 Ω

Figure 9-46

B

Section 9-10  Nodal Analysis 9-32. Determine the voltage drop across each of the three resistors in the network shown in Figure 9-47.

Problems

10 Ω 3.0 A

5.0 A

30

20

Ω

Ω

6.0 A Figure 9-47

I

9-33. Determine the voltage drop across the 200-Ω resistor in the network shown in Figure 9-48. 300 Ω

20 mA

200 Ω

40 mA

400 Ω

Figure 9-48

I

9-34. Identify the nodes in Figure 9-49. Using the reference node shown, write the nodal equations. 10 A 2S

1S

5A

5A

4S

2S

2S

5A 1A 6S

4S 1S 3S

2A

1S

5S

Reference node

Figure 9-49

5A

3S

15 A

249

250

Chapter 9    Resistance Networks

I

9-35. For the circuit shown in Figure 9-50, find Ex by nodal analysis. 3S 5A

1S

Ex

4S

2S 5A

3S

4

5S

S

1S

15 A

5A

Figure 9-50

circuitSIM

I

walkthrough

Section 9-11  The Superposition Theorem 9-36. (a) In the network shown in Figure 9-51, find the voltage drops across the 2.2-kΩ resistor and the 10-kΩ resistor (b) Use Multisim to verify the voltage drops calculated in part (a). 4.0 mA

10 kΩ

+ 4.0 V

4.7 kΩ

2.2 kΩ

−

Figure 9-51

I

9-37. In the network shown in Figure 9-52, determine the current drawn from each of the voltage sources. 50 Ω

+ 200 V −

40 Ω

5.0 A

20 Ω

− 80 V +

Figure 9-52

I

9-38. Determine the voltage drop across the 30-Ω resistor in the network shown in Figure 9-53. 30 Ω

4.0 A

20 Ω

50 Ω

+ 100 V − 10 Ω

Figure 9-53

40 Ω

3.0 A

251

Problems

9-39. Use the superposition theorem to determine the voltage drop across the 30-kΩ resistor in Figure 9-54.

10

+ 150 V −

kΩ

kΩ

40 kΩ

25

15 30 kΩ

kΩ

+ 120 V −

20

kΩ

I

Figure 9-54

I A

9-40. Use the superposition theorem to determine the voltage drop across the 4.7-kΩ resistor in Figure 9-51. 9-41. Use the superposition theorem to find Ix in Figure 9-55. 15 Ω Ix

10 A

5.0 Ω 2.0 Ω

30 V 12 Ω

25 V

50 V

7.0 Ω

10 Ω

5.0 Ω

Figure 9-55

A

9-42. (a) Use the superposition theorem to find Ix in Figure 9-56. (b) Use Multisim to verify the current calculated in part (a). 40 V 8.0 Ω 50 V 10 Ω

2.0 A 6.0 Ω Ix

Figure 9-56

9.0 Ω

4.0 Ω

4.0 A

circuitSIM walkthrough

252

Chapter 9    Resistance Networks

A

9-43. Calculate Ix in the following circuit by superposition. Ix

8.0 Ω

+ 20 V −

2A

2.0 Ω

13.0 Ω

5.0 Ω

10.0 Ω

3.0 Ω

7.0 Ω

− + 32 V

Figure 9-57

Review Questions

Section 9-2  Constant-Voltage Sources 9-44. Define a constant-voltage source. 9-45. Why is the terminal voltage of a constant-voltage source not necessarily constant as the load resistance varies? 9-46. Describe a laboratory procedure for determining the internal resistance of an automobile storage battery. 9-47. What is the significance of the term short-circuit current? 9-48. How would you determine the short-circuit current of a car ­battery? 9-49. What is the significance of the term open-circuit voltage?

Section 9-3  Constant-Current Sources 9-50. Define a constant-current source. 9-51. Why is the current drawn from a constant-current source not necessarily constant as the load resistance varies? 9-52. Explain what happens to the terminal voltage of a constant-current source when a load resistor is connected to it.

Section 9-4  Source Conversion 9-53. Compare equivalent constant-voltage and constant-current sources in terms of their open-circuit voltage and short-circuit current. 9-54. Describe the two methods for determining the internal resistance of constant-voltage and constant-current sources. 9-55. Why can we not replace the source shown in Figure 9-1 with an equivalent constant-current source?

Review Questions

Section 9-5 Kirchhoff’s Voltage-Law Equations: Loop Procedure 9-56. Show that Equations 9-5 and 9-6 are algebraically the same and, therefore, represent two different ways of stating Kirchhoff’s voltage law. 9-57. Explain the procedure for determining the polarity of voltage drops around a current tracing loop, using Figure 9-7 as an example. 9-58. How does the loop procedure apply when determining the voltage drop across R in the circuit in Figure 9-41? 9-59. If, in solving Example 9-3, we had chosen the I3 tracing loop from the + terminal of the source through R2, RM, and R3 back to the − terminal of the source, the solution for I3 would have been −0.15 A. What is the significance of a negative current in solving loop equations? 9-60. Is it likely that I1 or I2 in Figure 9-9 would ever work out to be negative quantities? Explain. 9-61. If R1 = 40 Ω, R2 = 10 Ω, and R3 = R4 = 20 Ω in the Wheatstone bridge of Figure 9-9, what path would you select for the I3 tracing loop in order to avoid a negative value for I3?

Section 9-7  Loop Equations in Multisource Networks 9-62. How do you write loop equations for the circuit of Figure 9-11(b) when no complete loops appear in the circuit diagram? 9-63. Loop equations are based on Kirchhoff’s voltage law but the unknown quantities in the loop equations are currents. Explain the presence of current terms in voltage-law equations. 9-64. In the second solution to Example 9-4, how do we know that the battery current should be IL − Igen rather than Igen − IL? 9-65. In distributing loads on a three-wire residential service, why is it d ­ esirable to have the two values of RL in Figure 9-14 as nearly equal as possible?

Section 9-8  Mesh Analysis 9-66. Explain why the circuit in Figure 9-44 is actually planar even though this circuit diagram shows conductors crossing with no electrical connection. 9-67. Why are we more likely to encounter negative current terms with the mesh format for writing Kirchhoff’s voltage-law equations than with the loop procedure? 9-68. Explain why many of the currents calculated with either the loop or mesh formats cannot be measured in the actual network. 9-69. How do we determine whether the source voltage in a mesh equation is a positive or negative quantity? 9-70. How do we determine the polarity of the voltage drop across a resistor in a network where the actual current is made up of two different mesh currents? 9-71. In writing loop equations, we insert branch currents in the basic Kirchhoff’s voltage-law equation before we collect the loop-current

253

254

Chapter 9    Resistance Networks

terms. With the mesh format, we simply write the equations in terms of the mesh currents. What features of the mesh format permit us to take this shortcut?

Section 9-9  Kirchhoff’s Current-Law Equations 9-72. Nodal analysis is based on Kirchhoff’s current law, but the unknown quantities in the equations are voltages. Explain the presence of voltage terms in current-law equations. 9-73. When we write equations for node voltages, we require one less equation than the total number of nodes. Explain why. 9-74. What do we mean when we refer to “the voltage at node X”?

Section 9-10  Nodal Analysis 9-75. What factors would lead us to select nodal analysis in preference to loop or mesh analysis when solving a given network? 9-76. Under what circumstances are loop equations preferable to nodal analysis?

Section 9-11  The Superposition Theorem 9-77. What advantage does the superposition theorem have over ­Kirchhoff’s-law methods for analyzing networks? 9-78. What factors tend to limit the application of the superposition theorem in network analysis? 9-79. Are we likely to encounter negative values for component currents when applying the superposition theorem to a network problem? Explain. 9-80. Is it possible that we have to subtract component currents when we apply  the superposition theorem to a branch current in a network? ­Explain. 9-81. To use mesh equations, we convert any current sources to equivalent constant-voltage sources. To use nodal analysis, we convert any voltage sources to equivalent constant-current sources. Do we have to make such conversions when using the superposition theorem?

Integrate the Concepts

In Example 8-1 we used the equivalent-circuit method to determine the values of I1, I2 and I3 in the circuit shown in Figure 9-58. Solve for these ­currents using four other techniques. I1

+ 100 V −

R1 I2

12 Ω

Figure 9-58

R2

10 Ω

I3 R3

40 Ω

Practice Quiz

Practice Quiz 1.

When a load resistor is connected across a voltage divider output, the output voltage (a)  decreases (b)  increases (c)   remains the same (d)  will decrease only if the load resistance is less than the resistances in the voltage divider

2.

A practical voltage source can be represented by (a)   a constant-voltage source in parallel with an internal resistance (b)   a constant-voltage source in series with an internal resistance (c)   a constant-voltage source in parallel with the load resistor (d)   a constant-voltage source in series with the load resistor

3.

A constant-current source can be represented by a (a)   current source in parallel with its internal resistor (b)   current source in series with its internal resistor (c)   current source in parallel with the load resistor (d)   current source in series with the load resistor

4.

To convert a constant current source into a constant-voltage source you should use Ohm’s law to determine a constant-voltage source connected (a)   in series with the current source internal resistor (b)   in series with the load resistor (c)   in parallel with the current source internal resistor (d)   in parallel with the load resistor

5.

The equivalent constant-current source for the constant-voltage source of Figure 9-59 is (a)   a 1.8-A current source in series with a 0.15-Ω resistor (b)   an 80-A current source in parallel with a 0.15-Ω resistor (c)   an 80-A current source in parallel with a 100-Ω resistor (d)   a 1.8-A current source in parallel with a 100-Ω resistor R1 0.15 Ω V1 12 V

Figure 9-59

RL 100 Ω

255

256

Chapter 9    Resistance Networks

6.

The current in the circuit of Figure 9-60 is (a)   151 mA (b)   86 mA (c)   22 mA (d)   11 mA R1 100 Ω V1 12 V

V2 20 V

R2 50 Ω

V3 24 V R3 220 Ω Figure 9-60

7.

The current through resistor R2 in Figure 9-61 is (a)   0.75 A (b)   0.66 A (c)   0.50 A (d)   85 mA R1

R4

10 Ω

22 Ω R2 50 Ω

V1 20 V

R3

R5

33 Ω

10 Ω

R6 50 Ω Figure 9-61

V2 12 V

V3 15 V

Practice Quiz

8.

The voltage drop across resistor R5 in Figure 9-61 is (a)   37 V (b)   33 V (c)   25 V (d)   4.3 V

9.

Which method of circuit analysis uses simultaneous equations to ­ etermine the current flowing through each component? d (a)   Loop equations (b)   Mesh analysis (c)  Superposition (d)   (a) and (b)

10. When applying Kirchhoff’s current law to a parallel circuit, (a)   the maximum branch current corresponds to the smallest branch resistor (b)  the sum of currents entering a node is equal to the difference of all the branch currents (c)  the sum of the currents leaving a node equals the sum of the ­currents entering the node (d)   all of the above 11.

The current flowing through the resistor R3 in Figure 9-62 is (a)   667 mA (b)   674 mA (c)   818 mA (d)   1.0 A R1 100 Ω

10.0 V

Figure 9-62

R3 330 Ω

R2 220 Ω

2.00 A

R4 2.7 kΩ

257

10

Equivalent-Circuit Theorems With the Kirchhoff’s-laws methods for solving resistance networks, we can solve almost any network without having to tamper with the original configuration of the network. However, these methods can involve many tedious computations with possibilities for errors in either the equations or the calculations. This chapter introduces more methods for using equivalent circuits to simplify network calculations.

Chapter Outline 10-1

Thévenin’s Theorem  260

10-2 Norton’s Theorem 

10-3 Dependent Sources 

268

271

10-4 Delta-Wye Transformation  10-5 Troubleshooting 

283

278

Key Terms Thévenin’s theorem  261 Norton’s theorem  268 independent sources  271

dependent or controlled sources 271 delta-to-wye transformation 279

wye-to-delta transformation 280

Learning Outcomes At the conclusion of this chapter, you will be able to: • state Thévenin’s theorem • determine the Thévenin-equivalent source of a network consisting of resistors and one or more independent sources • state Norton’s theorem • determine the Norton-equivalent source of a network consisting of resistors and one or more independent sources • differentiate between independent and dependent sources • analyze circuits consisting of resistors, independent sources, and dependent sources

Photo sources:  © iStock.com/calvio

• determine the Thévenin-equivalent source of a network consisting of resistors, independent sources, and dependent sources • convert a resistive Y-network to its equivalent Δ-network • convert a resistive Δ-network to its equivalent Y-network • determine if a fault exists in a circuit by comparing ­measured and calculated voltages

260

Chapter 10   Equivalent-Circuit Theorems

10-1  Thévenin’s Theorem In Sections 9-2 and 9-3, we developed a technique for replacing any prac­ tical voltage source in a network with an equivalent constant-voltage or constant-current source. In this section, we shall expand this principle into one of the most useful theorems for simplifying circuit calculations. Suppose that the circuit of Figure 10-1(a) is sealed in a black box ­(represented by the thick line) with only the terminals A and B exposed. A high-resistance voltmeter connected across these terminals shows that the open-circuit output voltage of the circuit is 80 V. Similarly, connecting a very low resistance ammeter across terminals A and B shows that the short-circuit current is about 2.0 mA. Hence, the black box appears to have an internal resistance of RTh =

25 kΩ

20 kΩ

Eoc 80 V = 40 kΩ = Isc 2.0 mA RTh

A

A 40 kΩ

+ 100 V −

+ ETh 80 V −

100 kΩ B

(a)

B

(b)

Figure 10-1  (a) Original voltage source; (b) Thévenin-equivalent source

As far as we can tell without opening it, the box contains an 80-V source, as shown in Figure 10-1(b). Any load connected to terminals A and B of the equivalent circuit of Figure 10-1(b) draws exactly the same current with ­exactly the same voltage drop as the load would if it were connected to the network of Figure 10-1(a). In fact, all that we can determine from measurements at the output ­terminals of any network containing one or more voltage sources is that the network is equivalent to a simple constant-voltage source with a single ­internal resistance in series with it. The French engineer Léon Charles Thévenin (1857–1926) stated the principle of this equivalence as a theorem: Any two-terminal network of fixed resistances and voltage sources may be replaced by a single voltage source that has • an equivalent voltage equal to the open-circuit voltage at the ­terminals of the original network, and • an internal resistance equal to the resistance looking back into the network from the two terminals with all the voltage sources replaced by their internal resistances.

10-1   Thévenin’s Theorem

We can apply Thévenin’s theorem to any of the resistance networks we have solved so far by treating one branch of the network as a load and the remainder of the network as a two-terminal network containing one or more voltage sources. Having decided which branch of the original network to treat as a load, we remove it from the original network and place it in a Thévenin-equivalent circuit. Then, we apply Thévenin’s theorem to ­determine the rest of the equivalent circuit. Note that Thévenin uses an “ohmmeter” approach to determine the internal resistance “looking back into” the open-circuit terminals of the source network, rather than the short-circuit current method we used with our black-box technique. Both approaches lead to the same value for the internal resistance.

Example 10-1 What current does a 10-kΩ resistor draw when it is connected to a 100-V source through the T-network shown in Figure 10-1(a)? Solution Step 1 Connect the 10-kΩ resistor to a Thévenin-equivalent source consisting of a constant voltage source ETh with a series internal resistance RTh, as shown in Figure 10-2(b). According to Thévenin’s theorem, ETh equals the open-circuit voltage between terminals A and B of Figure 10-1(a). Since an open circuit draws no current, there is no voltage drop across the 20-kΩ resistor. Hence, the open-circuit terminal voltage is the same as the voltage drop across the 100-kΩ resistor. When terminal A is open-circuit, the 25-kΩ and the 100-kΩ resistors form a simple series circuit. Then, the voltage-divider principle provides us with the voltage drop across the 100-kΩ resistor. This voltage drop is the open-circuit voltage between terminals A and B, which is ETh in the Thévenin-equivalent circuit. ETh =

100 × 100 V = 80 V 25 + 100

A 20 kΩ

40 kΩ

+ 25 kΩ

RTh 10 kΩ

ETh 80 V −

100 kΩ B (a)

A

B (b)

Figure 10-2  Equivalent circuit for Step 2 of Example 10-1

261

262

Chapter 10   Equivalent-Circuit Theorems

Step 2 Replace the actual source in Figure 10-1(a) by its internal resistance, which is zero ohms. The original circuit of Figure 10-1(a) then becomes the series-parallel resistance network of Figure 10-2(a). The equivalent resistance for this network is RTh = 20 kΩ +

25 kΩ × 100 kΩ = 20 kΩ + 20 kΩ = 40 kΩ 25 kΩ + 100 kΩ

Step 3 Solve for the current through the 10-kΩ resistor in the Théveninequivalent circuit of Figure 10-2(b). I=

circuitSIM walkthrough

ETh 80 V = = 1.6 mA RT 40 kΩ + 10 kΩ

Multisim Solution We can use a circuit simulation to show that the circuits in Figure 10-1 are equivalent. Download Multisim file EX10-1(a) from the website. The circuit is the same as shown in Figure 10-1(a). Connect a 10-kΩ load resistor across terminals A and B. Then, add a meter to measure the current through this load resistor. Connect a ground to the bottom node of the schematic. Run the simulation, and record the meter reading for the current through the load. Repeat the simulation with a 50-kΩ load and with a 100-kΩ load, and record the load currents. Download Multisim file EX10-1(b) from the website. The circuit is the same as shown in Figure 10-1(b). Using the same procedure as for the first circuit, run simulations with a 10-kΩ load, a 50-kΩ load, and a 100-k Ω load. Record the load currents.

Comparing the load current values for the two circuits, we find that both circuits produce the same current for a given load. If we run simulations with very low and very high load resistances, we still find that the load currents are the same for both circuits. Therefore, the circuits are equivalent. To further demonstrate the usefulness of Thévenin’s theorem, we apply it to Example 8-3, which we solved originally using Kirchhoff’s-laws ­equations.

10-1   Thévenin’s Theorem

Example 10-2 A resistor passing a 20-mA current is in parallel with a 5.0-kΩ resistor. This combination is in series with another 5.0-kΩ resistor, and the whole network is connected to a 500-V source. Find the resistance of the resistor that is passing the 20-mA current. Solution Step 1 We select the unknown resistance passing the 20-mA current as the load, r­ emove it from the original circuit in Figure 10-3(a), and place it in the Thévenin-equivalent circuit of Figure 10-3(b). The voltage-divider principle then gives the open-circuit terminal voltage of the remaining circuit in ­Figure 10-3(a): ETh =

5.0 kΩ A

+ 500 V −

5.0 × 500 V = 250 V 5.0 + 5.0 RTh

IL = 20 mA

A IL = 20 mA

RL

5.0 kΩ

+ ETh 250 V −

RL

B

B

(a)

(b)

Figure 10-3  Circuit diagram for Step 1 of Example 10-2

Step 2 If we replace the original voltage source with its internal resistance of zero ohms, the circuit of Figure 10-3(a) becomes the simple parallel resistance network of Figure 10-4(a), and RTh =

5.0 kΩ × 5.0 kΩ = 2.5 kΩ 5.0 kΩ + 5.0 kΩ RTh

A 5.0 kΩ

+ ETh 250 V −

5.0 kΩ

IL = 20 mA

2.5 kΩ

RL

B (a)

A

B (b)

Figure 10-4  Equivalent circuit for Step 2 of Example 10-2

263

264

Chapter 10   Equivalent-Circuit Theorems

Step 3 We can now use Ohm’s law to solve the Thévenin-equivalent circuit of ­Fig­ure 10-4(b):

circuitSIM walkthrough

RT =

ETh 250 V = = 12.5 kΩ I 20 mA

RL = RT − RTh = 12.5 kΩ − 2.5 kΩ = 10 kΩ

Multisim Solution Download Multisim file EX10-2 from the website.

The circuit is the same as shown in Figure 10-3(a) with RL = 10 kΩ. Insert a meter to measure the current through RL. Connect a ground to the bottom node of the schematic. Run the simulation, and note that the load current is 20 mA. If we run the simulation with smaller load resistances, the load current increases. Similarly, if we use larger resistances, the load current decreases. Therefore, 10 kΩ is the only load resistance that passes a current of 20 mA.

Thévenin’s theorem allows us to solve a Wheatstone bridge network without having to solve simultaneous equations as required for the Kirchhoff’slaws solution in Chapter 9.

Example 10-3

In the Wheatstone bridge circuit of Figure 10-5, R1 = R4 = 300 Ω and R2  =  R3 = 150 Ω. Find the current through a 50-Ω galvanometer connected across terminals A and B when the source voltage is 100 V. Solution Step 1 Since we are asked to find the current through only one branch of the Wheatstone bridge, we can treat the meter as if it were connected to the Thévenin-equivalent circuit, as shown in Figure 10-5(b). Step 2 Applying the voltage-divider principle to Figure 10-5(a) gives and

V3 = 100 V × V4 = 100 V ×

150 = 33.33 V 300 + 150 300 = 66.67 V 300 + 150

10-1   Thévenin’s Theorem

R1

R2 A

+ E

RTh

265

B

B

+ ETh

− R4

R3

RM G

− A (a)

(b)

R1

R3

R2 RTh

R4

(c)

Figure 10-5  Solving Wheatstone bridge by Thévenin’s theorem

From Kirchhoff’s voltage law, ETh = VAB = V3 − V4 = 33.33 V − 66.67 V = −33.34 V Since VAB is negative, terminal A is negative with respect to terminal B. Step 3 Assuming that the voltage source in Figure 10-5(a) has negligible internal resistance, we short it out to determine RTh. Then, R1 is in parallel with R3, R2 is in parallel with R4, and the two pairs of parallel resistors are connected in series between terminals A and B. Therefore,

R1R3 R2R4 300 Ω × 150 Ω 150 Ω × 300 Ω + = + = 200 Ω R1 + R3 R2 + R4 300 Ω + 150 Ω 150 Ω + 300 Ω Step 4 Applying Ohm’s law to the Thévenin-equivalent circuit gives the meter current: ETh 33.34 V IM = = = 133 mA RTh + RM 250 Ω

RTh =

Multisim Solution Download Multisim file EX10-3 from the website.

The circuit is the Wheatstone bridge shown in Figure 10-5(a) with the given values for the four resistors. Insert a 50-Ω resistor between terminals A and B to simulate a galvano­ meter. Add a meter in series with the 50-Ω resistor to measure the current. Connect a ground to the bottom node of the schematic. Run the simulation. Note that the meter reading of 133 mA matches the current calculated using Thévenin’s theorem.

circuitSIM walkthrough

266

Chapter 10   Equivalent-Circuit Theorems

6.0 Ω + 6.0 V −

4.8 Ω

4.0 Ω + 8.0 V −

Figure 10-6 Network with two voltage sources

We can use the same method to determine the current through any other branch of the bridge circuit. Thévenin’s theorem also applies to networks with two or more voltage sources. The theorem’s usefulness for such ­networks depends on whether we can determine ETh without having to solve simultaneous equations. With only two sources, finding ETh is ­usually straightforward, as we can show with the network of Figure 10-6, which is similar to the automobile generator example in Chapter 9.

Example 10-4 Determine the current in the 4.8-Ω resistor in the network of Figure 10-6. Solution Step 1 Remove the 4.8-Ω resistor from the original circuit of Figure 10-6 and place it in the Thévenin-equivalent circuit of Figure 10-7(b). 1.2 V

− 6.0 Ω

+ 6.0 V −

0.80 V

+

− A

+

2.4 Ω

C

RTh

4.0 Ω

+ 8.0 V − D

B

A

+ ETh 7.2 V −

(a)

4.8 Ω

B (b)

Figure 10-7  Circuit diagram for Example 10-4

Step 2 The sum of the potential differences must be the same along both paths from point C to point D in Figure 10-7(a). One path is the 8.0-V source, and the other path consists of the 6.0-V source and the two resistors. Therefore, there must be a 2.0-V drop across the 6.0-Ω and 4.0-Ω resistors in series. From the voltage-divider principle, these voltage drops are 1.2 V and 0.8 V, respectively, as shown in Figure 10-7(a). Tracing either path from point A to point B shows that the open-circuit voltage ETh is 7.2 V. Step 3 Replace the sources by their internal resistances (zero ohms). The 6.0-Ω and 4.0-Ω resistors then appear in parallel between points A and B. Therefore, RTh =

6Ω × 4Ω = 2.4 Ω 6Ω + 4Ω

Step 4 Apply Ohm’s law to the Thévenin-equivalent circuit of Figure 10-7(b): IL =

ETh 7.2 V = = 1A RTh + RL 2.4 Ω + 4.8 Ω

10-1   Thévenin’s Theorem

267

Multisim Solution Download Multisim file EX10-4 from the website.

circuitSIM walkthrough

The circuit is the same as shown in Figure 10-6. Insert a meter to measure the current through the 4.8-Ω resistor. ­Connect a ground to the bottom node of the schematic. Run the simulation. Note that the meter reading of 1 A matches the current calculated using Thévenin’s theorem.

Practical Circuits Thévenin’s Theorem Applications I

A

Linear two-terminal circuit

B

+ V −

Load

+ V −

Load

(a)

Source:  © iStock.com/gwmullis

I

A

RTh

+

VTh

B

−

(b)

Figure 10-8

An electrical outlet

Often the load in a circuit varies while the other circuit elements are fixed. For example, various appliances may be connected to a household outlet, thus changing the load on that branch of the house wiring. When we are analyzing a circuit that has a varying load, Thévenin’s theorem can save us some tedious calculations

by letting us replace the unchanged portion of the circuit with a simple equivalent circuit, as shown in Figure 10-8. However, we can apply Thévenin’s theorem only to linear components (components with values that do not vary with voltage or current).

See Problems 10-1 to 10-18 and Review Questions 10-43 to 10-48.

268

Chapter 10   Equivalent-Circuit Theorems

10-2  Norton’s Theorem Figure 10-1 demonstrates how we can reduce a network containing one or more voltage sources into a simple equivalent constant-voltage source. In Chapter 9, we found that we can readily convert a constant-voltage source into an exactly equivalent constant-current source. Therefore, we should be  able to reduce any network containing one or more voltage sources to a  simple equivalent constant-current source, as shown in Figure 9-5. ­Norton’s theorem combines Thévenin’s theorem and source conversion into a single procedure: Any two-terminal network of fixed resistances and voltage sources may be replaced by a single constant-current source that has • a current equal to the current drawn by a short circuit across the terminals of the original network, and • a resistance in parallel equal to the resistance that would be ­measured from the two terminals if each voltage source were r­eplaced by its ­internal resistance.

Example 10-5 Determine the Norton-equivalent constant-current source for the network of Figure 10-9(a). 25 kΩ

+ 100 V −

20 kΩ

A

A IN

100 kΩ

2.0 mA

RN

40 kΩ B

B (b)

(a)

Figure 10-9  (a) Original voltage source; (b) Equivalent constant-current source

Solution I Step 1 Find the short-circuit current. When terminals A and B are shorted, the total resistance connected across the 100-V source is RT = 25 kΩ +

100 kΩ × 20 kΩ = 41.67 kΩ 100 kΩ + 20 kΩ

The current drawn from the 100-V source is

10-2   Norton’s Theorem

IT =

E 100 V = = 2.4 mA RT 41.67 kΩ

From the current-divider principle, IN = Isc =

100 kΩ × 2.4 mA = 2.0 mA 100 kΩ + 20 kΩ

Step 2 Replace the 100-V source with its internal resistance of zero ohms. Then the equivalent resistance connected between terminals A and B is RN = 20 kΩ +

25 kΩ × 100 kΩ = 40 kΩ 25 kΩ + 100 kΩ

IN and RN are the required values for the Norton-equivalent constantcurrent source of Figure 10-9(b). Solution II Step 1 We can avoid the current-divider calculations of Solution 1 by first finding the Thévenin-equivalent constant-voltage source as shown in Figure 10-10(a). ETh = Eoc =

40 kΩ

100 kΩ × 100 V = 80 V 100 kΩ + 25 kΩ A

A

RTh

+ 80 V ETh −

RL

IN 2.0 mA

RN

40 kΩ

V

RL

B

B (a)

(b)

Figure 10-10  (a) Thévenin-equivalent constant-voltage source; (b) Norton-equivalent constant-current source

Step 2 Find RTh with the same method we used to find RN in Solution 1. RTh = 40 kΩ Step 3 Convert the Thévenin-equivalent voltage source of Figure 10-10(a) into a constant-current source, obtaining the Norton-equivalent source of Fig­ure 10-10(b). IN = Isc =

ETh 80 V = = 2.0 mA RTh 40 kΩ

269

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Chapter 10   Equivalent-Circuit Theorems

Example 10-4A Determine the current in the 4.8-Ω resistor in the circuit shown in ­Figure 10-6. Solution Step 1 Remove the 4.8-Ω resistor from the original circuit and place it in the ­Norton-equivalent circuit of Figure 10-11(b). 6.0 Ω

4.0 Ω A

+ 6.0 V −

A

+ 8.0 V −

Isc

B

IN

3.0 A

RN

4.8 Ω

B (a)

(b)

Figure 10-11  Circuit diagram for Step 2 of Example 10-4A

Step 2 Replace the 4.8-Ω resistor in the original circuit by a short circuit, as shown in Figure 10-11(a). Drawing arrows for current directions shows that the short-circuit current is the sum of the currents resulting from connecting the 6-Ω resistor directly across the 6.0-V source and connecting the 4.0-Ω resistor directly across the 8.0-V source. Therefore, the short-circuit current is IN =

E1 E2 6V 8V + = + = 3A R1 R2 6Ω 4Ω A

6.0 Ω

A B

4.0 Ω

3.0 A

IN RN

2.4 Ω

4.8 Ω

B (a)

(b)

Figure 10-12  Circuit diagram for Step 3 of Example 10-4A

Step 3 Replace the sources with their negligible internal resistances. In the resulting circuit, shown in Figure 10-12(a), the 6.0-Ω and 4.0-Ω resistors are connected in parallel across terminals A and B. Hence, RN =

6Ω × 4Ω = 2.4 Ω 6Ω + 4Ω

10-3   Dependent Sources

Step 4 Apply the current-divider principle to the Norton-equivalent circuit of ­Figure 10-12(b): 2.4 Ω × 3A = 1A IL = 2.4 Ω + 4.8 Ω See Problems 10-19 to 10-26 and Review Questions 10-49 and 10-50.

10-3  Dependent Sources The EMF generated by the voltage sources we have encountered so far has been determined entirely by the energy-conversion process within the source. The open-circuit terminal voltage is, therefore, independent of what happens elsewhere in a network containing such constant-voltage or ­independent sources. The terminal voltage of a practical voltage source ­decreases as load current increases, but we can account for this effect by including an internal resistance in series with the ideal constant-voltage source in network ­diagrams. Constant-current sources are also independent sources. Transistors or other electronic devices can be connected such that they behave like voltage or current sources with an output that is dependent on or controlled by a voltage or current appearing in some other branch of the circuit. To distinguish dependent or controlled sources from independent sources in circuit diagrams, we use a diamond-shaped symbol with + and − signs for a dependent voltage source and a diamond with a current arrow for a dependent current source. As shown in Table 10-1, there are four types of dependent sources. TABLE 10-1 Dependent sources Type of Source Voltage-controlled voltage source   (VCVS) Current-controlled voltage source   (CCVS) Current-controlled current source   (CCCS) Voltage-controlled current source   (VCCS)

Magnitude V = kV1

Symbol

V = kI1 I = kI1

I = kV1

Figure 10-13 shows a simple network containing a current-controlled voltage source (CCVS) in series with a 120-V independent voltage source and its internal resistance. This dependent source is designed so that

271

272

Chapter 10   Equivalent-Circuit Theorems

Rint

+

− 5.0 Ω

+

+ 120 V −

I1 RL

5I1

− Figure 10-13  Network with a dependent source

its ­terminal voltage is V = 5 Ω × I1. Equations for dependent sources are commonly written without showing the units for the constant. For ­example, the equation for the CCVS in Figure 10-13 would be simplified to V = 5I1. The necessary units are understood to be included in the ­constant. As we increase RL in the circuit of Figure 10-13, I1 increases and so does the IR drop across Rint. The dependent source voltage also increases since its output is V = 5I1. Thus, the dependent source adds a voltage that ­exactly offsets the IR drop across Rint. The series combination of the two voltage sources acts as a voltage-regulated power supply. We think of the CCVS in Figure 10-13 as a source because the relationship between the polarity of the voltage between its terminals and the current through it is that of a source. However, dependent sources differ from real energy sources. If we disconnect RL in the circuit of Figure 10-13, I1 = 0 and the voltage across the CCVS disappears while the independent source maintains its voltage. Hence, we use the symbol V rather than E for dependent sources. Although Kirchhoff’s laws and network theorems hold true for networks containing dependent sources, we cannot always use these methods to solve such networks.

Example 10-6 Determine the load voltage VL in the network shown in Figure 10-14. 200 Ω

+

−

+ 5.0 V −

800 Ω −

+ V1

0.010V1

+ RL

500 Ω

−

Figure 10-14  Circuit diagram for Example 10-6

10-3   Dependent Sources

Solution I At first glance, it might seem that we could write two mesh equations to solve the network shown in Figure 10-14. But the rules for mesh equations require us to convert all current sources to equivalent voltage sources. We cannot convert the voltage-controlled current source (VCCS) in Figure 10-14 to an equivalent voltage-controlled voltage source (VCVS) because there is no resistor that we can use as Rint in parallel with the original current source. We can, however, apply Thévenin’s theorem to the left loop of the network without affecting V1. Figure 10-15 shows the resulting single-loop network. 0.010V1

+

− 160 Ω

+ 4.0 V −

+ + RL

V1

500 Ω

− − Figure 10-15  Equivalent circuit for Solution I for Example 10-6

The current in the equivalent circuit in Figure 10-15 is the output of the VCCS, I = 0.01V1 Applying Ohm’s law and substituting for I gives V1 = 4.0 V − 160I = 4.0 − 160(0.01V1) = 4.0 − 1.6V1 2.6V1 = 4.0 V1 = 1.54 V I = 0.01V1 = 15.4 mA VL = IRL = 15.4 mA × 500 Ω = 7.7 V

Solution II We cannot use the superposition theorem for this example since removing the voltage source will drop V1 to zero, making the output of the VCCS zero as well. But we can convert the independent voltage source to its equivalent constant-current source and use nodal analysis. 0.010V1

Node

+ + 25 mA

200 Ω

800 Ω

V1

RL

500 Ω

− −

Figure 10-16  Equivalent circuit for Solution II for Example 10-6

273

274

Chapter 10   Equivalent-Circuit Theorems

In the equivalent circuit of Figure 10-16, the single nodal equation is

1 1 + V1 = 25 mA − 0.01V1 200 Ω 800 Ω 25 mA = 1.54 V This equation reduces to V1 = 16.25 mS

(

)

VL = ILRL = 15.4 mA × 500 Ω = 7.7 V

Then

IL = 0.01V1 = 0.01 × 1.54 = 15.4 mA

and

In Figure 10-17, a current-controlled voltage source has replaced the voltage-controlled current source of Figure 10-14. The superposition theorem still does not work, and we cannot replace the left-hand part of the ­network with its Thévenin-equivalent source because we will lose the I1 current, which governs the voltage produced by the CCVS. However, we can use either mesh or nodal analysis techniques.

200 Ω

+

Node N

100I1

−

VB

I1

+ 5.0 V −

IA

+ 800 Ω −

IB

+ RL

500 Ω

−

Figure 10-17  Circuit diagram for Example 10-7

Example 10-7 Determine VL in the network shown in Figure 10-17. Solution I From Kirchhoff’s voltage law, (200 + 800)IA − 800IB = 5.0 (500 + 800)IB − 800IA = 100I1 = 100(IA − IB) Collecting terms gives

1000IA − 800IB = 5.0

900IA − 1400IB = 0

(1) (2)

10-3   Dependent Sources



IB

│1000 900 = │1000 900

│ −4500 = = 6.62 mA −680 000 −800 −1400│ 5.0 0

VL = IBRL = 6.62 mA × 500 Ω = 3.31 V

Solution II Figure 10-18 shows the Norton-equivalents for both voltage sources in the original network. Node N

+ 25 mA

IA 200 Ω

IB

+

800 Ω −

500 Ω −

RL 0.20I1

Figure 10-18  Equivalent circuit for Solution II for Example 10-7

At the independent current source, IA =

5.0 V = 25 mA 200 Ω

At the dependent voltage source, VB = 100I1 and

Since I1 =

IB =

VN VN , IB = 800 4000

VB 100I1 = = 0.200I1 500 500

For the single independent node, the nodal equation is

(

VN 1 1 1 VN = 25 mA − + + 200 800 500 4000

)

VN = 25 mA × 117.6 Ω = 2.94 V

Note that VN is not the same as VL in the original network of ­Figure 10-17. Substituting for VN in the equation for I1 gives

I1 =

2.94 V VN = = 3.676 mA 800 800 Ω

VB = 100I1 = 0.368 V

VL = VN + VB = 2.94 V + 0.37 V = 3.31 V

275

276

Chapter 10   Equivalent-Circuit Theorems

Multisim Solution Download Multisim file EX10-7 from the website.

circuitSIM walkthrough

The circuit is the same as shown in Figure 10-17. Insert a meter to measure the voltage across the 500-Ω load ­resistor. Run the simulation. Note that the meter reading of 3.31 V matches the voltage calculated using either Kirchhoff’s voltage law or Nortonequivalent sources. We can also use a simulation to check the output of the dependent voltage source. Place two meters in the circuit, one to measure the ­current, I1, and the other to measure the voltage, VB. Run the simulation. The meter readings confirm that VB equals 100I1.

We can include dependent sources in a Thévenin transformation as long as we are careful how we determine RTh. If the controlling variable for the dependent source is within the portion of the network included in the Thévenin-equivalent source, we cannot find RTh by simply calculating the equivalent resistance that would appear across the open-circuit terminals when the voltage sources are replaced by the internal resistances. ­Instead, we must derive the equivalent resistance from the short-circuit current of the network, as we did in Section 9-2:



RTh =

open-circuit terminal voltage (9-1) short-circuit terminal current

Example 10-8 Remove RL from the network in Figure 10-17, and determine the Thévenin-equivalent source that can replace the remainder of the ­network. Solution For the open-circuit condition shown in Figure 10-19(a), I1 =

5.0 V = 5.0 mA 200 Ω + 800 Ω

10-3   Dependent Sources

100I1

200 Ω

100I1

200 Ω

I1

+ 5.0 V −

I1

+ 800 Ω −

277

+ 5.0 V −

Voc = ETh

+ 800 Ω −

(a)

Isc

(b) 180 Ω RTh + 4.5 V ETh − (c)

Figure 10-19  Dependent source in a Thévenin transformation

The dependent source voltage is V = 100I1 = 0.50 V

Voc = ETh = 5.0 mA × 800 Ω + 0.50 V = 4.5 V

To find the short-circuit current, we can write two mesh equations for the circuit of Figure 10-19(b). (200 + 800)IA − 800IB = 5.0   and  800IB − 800IA = 100I1 = 100(IA − IB) Collecting terms gives

900IA − 900IB = 0





1000IA − 800IB = 5.0

Isc = IB

│1000 900 = │1000 900 RTh =

│ −4500 = 25 mA = −180 000 −800 −900│

(1) (2)

5 0

Eoc 4.5 V = = 180 Ω Isc 25 mA

Note that RTh is not the same as the resistance “seen” looking back into the open circuit terminals of Figure 10-19(a).

See Problems 10-27 to 10-35 and Review Questions 10-51 to 10-53.

We can check ETh and RTh by comparing the load voltage with the value found in ­Example 10-7:



VL =

500 × 4.5 V 180 + 500

= 3.31 V

278

Chapter 10   Equivalent-Circuit Theorems

Circuit Check

A

CC 10-1. Use Thévenin’s theorem to calculate the voltage across the 8.0-Ω resistor in the circuit shown in Figure 10-20. 120 Ω

+ 36 V −

40 Ω

60 Ω

12 Ω

20 Ω

18 Ω

8.0 Ω

Figure 10-20

CC 10-2. Use a Norton-equivalent source to calculate the voltage across the 40-Ω resistor in Figure 10-20. CC 10-3. Determine the voltage across the 15-Ω resistor in Figure 10-21. 1.5V1

V1

−

+ 25 V −

+

10 Ω 15 Ω

20 Ω

Figure 10-21

10-4  Delta-Wye Transformation Another useful type of circuit transformation applies to three-terminal ­resistance networks like the one in Figure 10-22. If we can state the conditions under which the wye (Y) circuit of Figure 10-22(a) is equivalent to the delta (Δ) circuit of Figure 10-22(b), we can substitute the wye connection for the delta, and vice versa. A

RA

RB

A RZ

RC

C

B (a)

B

RY RX

(b)

Figure 10-22  (a) Y-network; (b) Δ-network

C

10-4   Delta-Wye Transformation

For the circuits to be equivalent, the resistance between terminals A and B must be the same for both networks. For the wye circuit of Figure 10-22(a), the circuit between A and B is a simple series circuit, so RAB = RA + RB. The delta circuit of Figure 10-22(b) has two branches in parallel between terminals A and B, so RAB =



RA + RB =

Similarly,

RB + RC =

and

RC + RA =

RZ ( RX + R Y ) RZ + ( RX + RY ) RXRZ + RYRZ RX + RY + RZ

RXRY + RXRZ RX + RY + RZ

RYRZ + RXRY RX + RY + RZ

(1)

(2)

(3)

Subtracting Equation 2 from Equation 1 gives RA − RC =



RYRZ − RXRY RX + RY + RZ

(4)

Adding Equation 3 and Equation 4 gives 2RA =

RA =



Likewise, RB = RC =

and

2RYRZ RX + RY + RZ RYRZ RX + RY + RZ RXRZ RX + RY + RZ RXRY RX + RY + RZ

(10-1)

(10-2)

(10-3)

Equations 10-1 to 10-3 are the delta-to-wye transformation equations. Note that there is a simple pattern to the three equations: RA =

product of the two Δ sides connected to terminal A sum of all three Δ sides

279

280

Chapter 10   Equivalent-Circuit Theorems

We can solve Equations 1, 2, and 3 for the equivalent delta circuit for a given wye circuit. The wye-to-delta transformation is usually stated as



RX =



RY =



RZ =

RARB + RBRC + RCRA RA RARB + RBRC + RCRA RB RARB + RBRC + RCRA RC

(10-4)

(10-5)

(10-6)

Again the three equations have a pattern: RX =

sum of products of each pair of Y-arms opposite Y-arm

Inverting Equation 10-4 and substituting R = 1/G for each resistance gives 1 1 1 GA GA = GX = = RX 1 1 1 GA + GB + GC + + GAGB GBGC GCGA GAGBGC So,

GX =

Similarly,

GY =

and

GZ =

GBGC GA + GB + GC GAGC GA + GB + GC GAGB GA + GB + GC

(10-7)

(10-8)

(10-9)

Equations 10-7 to 10-9 are the duals of Equations 10-1 to 10-3. Inverting Equation 10-7 gives



RX = RBRC ( GA + GB + GC )

(10-10)

10-4   Delta-Wye Transformation

RY = RARC ( GA + GB + GC )

Similarly,

(10-11)

RZ = RARB ( GA + GB + GC )

and

(10-12)

The conductance forms of the wye-to-delta transformation equations are particularly useful for analyzing alternating-current circuits.

Example 10-3A

In the Wheatstone bridge circuit of Figure 10-23(a), R1 = R4 = 300 Ω and R2 = R3 = 150 Ω. Find the current through a 50-Ω galvanometer connected across terminals A and B when the source voltage is 100 V.

RA

R1

+

RM

R2

RC

R3

R4

E

G

E

RB

+ −

− R3

R4

(a)

(b)

Figure 10-23  Solving Wheatstone bridge by delta-wye transformation

Solution Step 1 Replace the delta connection of R1, R2, and RM with its equivalent wye ­circuit, as in Figure 10-23(b).





RA = RB =

RC =

300 Ω × 150 Ω = 90 Ω 300 Ω + 150 Ω + 50 Ω

300 Ω × 50 Ω = 30 Ω 500 Ω 50 Ω × 150 Ω = 15 Ω 500 Ω

281

282

Chapter 10   Equivalent-Circuit Theorems

Step 2 The load on the source is now a simple series-parallel circuit with an equivalent resistance of

Req = RA +

( R B + R3 ) × ( RC + R4 ) 180 Ω × 315 Ω = 204.55 Ω = 90 Ω + ( RB + R3 + RC + R4 ) 495 Ω

Therefore,

IT =

E 100 V = = 489 mA Req 204.55 Ω

Step 3 From the current-divider principle,

I4 = 489 mA ×

315 Ω = 311 mA 495 Ω

180 Ω = 178 mA 495 Ω Since R3 and R4 were not altered by the transformation, I3 and I4 are the same as for the original circuit. Therefore, in Figure 10-23(a), and

The right terminal of the meter is positive with respect to the left terminal.

I3 = 489 mA ×

V3 = I3R3 = 311 mA × 150 Ω = 46.65 V and

V4 = I4R4 = 178 mA × 300 Ω = 53.4 V

Step 4 From Kirchhoff’s voltage law, Therefore,

circuitSIM walkthrough

VM = V4 − V3 = 53.4 − 46.65 = 6.75 V IM =

6.75 V = 0.14 A 50 Ω

Multisim Solution We can use simulations to confirm that the two circuits shown in ­Figure 10-23 are indeed equivalent. If two circuits are equivalent, then the voltages across R3 in both circuits must be identical, and the voltages across R4 must also match. Download Multisim file EX10-3A(b) from the website. The circuit is same as shown in Figure 10-23(b). Insert multimeters to measure the voltages across R3 and R4. Connect a ground to the bottom node of the s­ chematic. Run the simulation, and record the voltage readings. Download Multisim file EX10-3A(a) from the website. The circuit is the same as shown in Figure 10-23(a). Insert a 50-Ω resistor between terminals A and B to simulate a galvanometer. Insert meters to measure the voltages across R3 and R4. Connect a ground to the bottom node of the schematic. Run the simulation and record the voltage readings. Comparing the two sets of voltage measurements confirms that the delta and wye circuits are equivalent. See Problems 10-36 to 10-41 and Review Questions 10-54 and 10-55.

10-5  Troubleshooting

283

10-5 Troubleshooting We can use equivalent-circuit theorems to help trace circuit faults.

Example 10-9 For the Wheatstone bridge circuit in Example 10-3, the calculated value for V3 is 33.33 V. Determine the probable cause if the measured value for V3 is (a) 100 V

(b) 0 V

(c) 66.67 V

Solution (a) Kirchhoff’s voltage law gives E = V1 + V3, so we deduce that V1 = 0 V. Since V1 = I1R1, then either R1 = 0 Ω (shorted) or I1 = 0 A. For I1 to be 0 A, either R1 or R3 is open. However, if R1 is open, I3 = 0 A and V3 = 0 V. Since V3 = 100 V, we conclude that R1 is not open. Therefore, either R1 is shorted or R3 is open. Remove these resistors from the circuit and check them with an ­ohmmeter. (b) Kirchhoff’s voltage law gives E = V1 + V3, so we deduce that V1 = 100 V. From Ohm’s law, V3 = R3I3, so either R3 is shorted or I3 = 0 A. For I1 to be 0 A, either R1 or R3 is open. However, if R3 is open, I1 = 0 A and V1 = 0 V. Since V1 is not zero, we conclude that R3 is not open. Therefore, either R1 is open or R3 is shorted. Remove these resistors from the circuit and check them with an ohmmeter. (c) Kirchhoff’s voltage law gives E = V1 + V3, so we deduce that V1 = 33.33 V. We note that the values of V1 and V3 are reversed, suggesting that R1 and R3 have been interchanged. Checking the colour codes on these resistors, we discover that R1 is the 150-Ω resistor and R3 the 300-Ω resistor. Multisim Solution Download Multisim file EX10-9 from the website. The circuit is the same as shown in Figure 10-5, with the given values for the four resistors. Interchange the values of R1 and R3. Insert a 50-Ω resistor between terminals A and B to simulate a galvanometer. Connect a ground to the bottom node of the schematic. Insert meters to measure the voltages across R1 and R3. Run the simulation, and view the voltage readings to verify the conclusions of Example 10-9(c).

See Problem 10-42.

circuitSIM walkthrough

284

Chapter 10   Equivalent-Circuit Theorems

Summary

• Any two-terminal network of fixed resistances and voltage sources can be replaced by a Thévenin constant-voltage source consisting of a single voltage source in series with a resistor. • Any two-terminal network of fixed resistances and voltage sources can be replaced by a Norton constant-current source consisting of a single current source in parallel with a resistor. • A dependent source has an output that depends on a voltage or current in some other branch of the circuit. • The Thévenin constant-voltage source may be determined for a network containing a dependent source provided the open-circuit v ­ oltage  and short-circuit current are used to determine the equivalent resistance. • A resistive Y-network can be converted to an equivalent Δ-network, and vice versa. • Comparing measured and calculated voltages enables us to identify a fault in a circuit. B = beginner

I = intermediate

A = advanced

Problems B

Section 10-1  Thévenin's Theorem 10-1.

Determine the Thévenin-equivalent constant-voltage source for the two-terminal T-network of Figure 10-24. 100 Ω

+ 200 V −

100 Ω

400 Ω

Figure 10-24 T-network

circuitSIM walkthrough

I

10-2.

(a) Determine the Thévenin-equivalent source for the π-network in Figure 10-25. (b) Use Multisim to verify the voltage of the Thévenin-equivalent source in part (a). 500 Ω

40 mA

I I

2.0 kΩ

Figure 10-25  π-network

2.0 kΩ

10-3. Use Thévenin’s theorem to determine the current through the 300-Ω resistor in the circuit of Figure 10-26. 10-4. Determine the current in the 100-Ω resistor in the circuit of Figure 10-26.

285

Problems

200 Ω

1.5 kΩ

100 Ω 300 Ω

500 Ω

− 200 V +

Figure 10-26

B B I I I A

10-5. Use Thévenin’s theorem to find the load resistance that draws 400 mA from the network shown in Figure 10-24. 10-6. (a) Use Thévenin’s theorem to determine the current through a 3.3-kΩ load connected to the network shown in Figure 10-25. (b) Use Multisim to verify the current calculated in part (a). 10-7. Find the maximum power that can be developed in a load connected to the network shown in Figure 10-24. 10-8. Determine the load resistance that can transfer the maximum power when connected to the network shown in Figure 10-25. 10-9. Solve for the current drawn from the 20-V source in Figure 10-27 by replacing the portion of the circuit to the left of the dashed line with its Thévenin-equivalent source. 10-10. Find the two load resistances that dissipate energy at the rate of 200 mW when connected to the network in Figure 10-25.

− 30 V +

circuitSIM walkthrough

15 Ω 50 Ω

+ 20 V −

50 Ω

circuitSIM walkthrough Figure 10-27

A

I

10-11. (a) Determine the Thévenin-equivalent source as “seen” by the 10-kΩ resistor in Figure 10-28. (b) Use Multisim to verify the voltage of the Thévenin-equivalent source in part (a). 10-12. Find the voltage drop across the 25-Ω resistor in Figure 10-29 by replacing the current source with an equivalent constant-voltage source. 100 Ω

+ 100 V −

+300 V

−150 V

22 kΩ

47 kΩ 10 kΩ

33 kΩ

25 Ω

5.0 A

Figure 10-28 Figure 10-29

68 kΩ

286

Chapter 10   Equivalent-Circuit Theorems

I A

circuitSIM walkthrough

10-13. Given E = 200 V, R1 = 8 kΩ, R2 = 2 kΩ, R3 = 2 kΩ, R4 = 14 kΩ, and RM = 25 kΩ for the Wheatstone bridge of Figure 10-23(a), use Thévenin’s theorem to determine the meter current. 10-14. (a) Use Thévenin’s theorem to determine the current through a 250-Ω load connected to the output terminals of the bridged-T attenuator of Figure 10-30 when the input voltage is 3.0 V. (b) Use Multisim to verify the current calculated in part (a). 400 Ω

100 Ω Input

100 Ω Output

800 Ω

Figure 10-30  Bridged-T network

A

10-15. Use Thévenin’s theorem to find the load resistance that draws 1.0 A from the output terminals of the lattice network in Figure 10-31 when 500 V is applied to its input terminals. 20 Ω 30 Ω

Output 30

Ω

Input

20 Ω

Figure 10-31  Lattice network

A

10-16. The switch in the pulse generator circuit of Figure 10-32 alternately closes for 1 ms and opens for 4 ms. Draw the open-circuit output-­voltage waveform (with respect to chassis) and determine its a­ mplitude. −24 V

5.6 kΩ 22 kΩ

500 Ω

+12 V

Figure 10-32

Out

27 kΩ

+24 V

287

Problems

A

10-17 Use Thévenin’s theorem to calculate the current through the 10-Ω load resistor in Figure 10-33. 7.0 Ω

8.0 Ω

5.0 Ω 10 Ω

4.0 Ω

RL

+ 50 V −

25 Ω 55 Ω

Figure 10-33

I

B B

B B I I

A

I

10-18. Use another method of analysis to verify your answer for Problem 10-17.

Section 10-2  Norton's Theorem 10-19. Determine the Norton-equivalent constant-current source for the two-terminal network of Figure 10-24. 10-20. (a) Determine the Norton-equivalent source for the network of ­Fig­ure 10-25. (b) Use Multisim to verify the current for the Norton-equivalent source in part (a). 10-21. Use Norton’s theorem to find the current that a 400-Ω load draws from the source shown in Figure 10-24. 10-22. (a) Use Norton’s theorem to determine the load resistance that draws a 950-μA current from the source shown in Figure 10-25. (b) Use Multisim to verify the load resistance calculated in part (a). 10-23. Replace each voltage source in Figure 10-27 with a Norton-equivalent source. Then use nodal analysis to determine the current through the centre 50-Ω resistor. 10-24. (a) Find the current through the 100-Ω resistor in Figure 10-29 by ­leaving the sources in their original form and using the super­ position theorem. (b) Use Multisim to verify the current calculated in part (a). 10-25. Use Norton’s theorem to calculate the current through the 10-Ω load resistor in Figure 10-33. 10-26. Compare the internal resistance of the Norton-equivalent source for the circuit in Figure 10-33 with the internal resistance of the Thévenin-equivalent source for the circuit.

Section 10-3  Dependent Sources 10-27. Figure 10-34 shows the equivalent output circuit for a typical transistor amplifier. Calculate the output voltage if the input current I1

circuitSIM walkthrough

circuitSIM walkthrough

circuitSIM walkthrough

288

Chapter 10   Equivalent-Circuit Theorems

is 20 μA, the forward current ratio hfe is 45, the output conductance hoe is 3.0 × 10−5 S, and the load resistance RL is 20 kΩ.

hoe

I = hfeI1

RL

Figure 10-34  Equivalent output circuit for a transistor amplifier

B I

10-28. Determine the Thévenin-equivalent source for the transistor circuit of Problem 10-27. 10-29. Determine the load voltage VL in the circuit of Figure 10-35. I1

2.2 kΩ

+ 40 V −

2.0I1

RL

6.8 kΩ

10 kΩ

Figure 10-35

circuitSIM walkthrough

I

10-30. (a) Determine V1 in the circuit of Figure 10-36. (b) Use Multisim to verify the voltage calculated in part (a). 2.0V1

2.2 kΩ

− 40 V +

RL

V1

10 kΩ

Figure 10-36

A

10-31. Find the Thévenin-equivalent voltage source for the network of ­Figure 10-37. 0.50V1

1.0 kΩ 2.0 kΩ

+ 36 V −

Figure 10-37

V1

289

Problems

A

10-32. (a) Find the Thévenin-equivalent voltage source for the network of ­Figure 10-38. (b) Use Multisim to verify the voltage calculated in part (a).

circuitSIM walkthrough

50 Ω 0.020V1

80 Ω

V1

− 20 V +

Figure 10-38

I

10-33. Determine the load voltage VL in the amplifier equivalent circuit of Figure 10-39. 200 Ω

+ 5.0 V −

V1 0.010V1

800 Ω

1000 Ω

RL

500 Ω

Figure 10-39

A

I

B

10-34. (a) Determine VL when a 2500-Ω feedback resistor is connected between the junction of the 200-Ω and 800-Ω resistors and the top of the VCCS in Figure 10-39. (b) Use Multisim to verify the load voltage calculated in part (a). 10-35. What value of RTh is “seen” looking back into the terminals of the dependent voltage source of Figure 10-40? (Hint: Connect a fixed voltage source to the terminals and calculate the terminal voltage and current.)

Section 10-4  Delta-Wye Transformation

10-36. Use a delta-to-wye transformation to determine the equivalent T-network for the π-network of Figure 10-41. 480 Ω

Input

1.2 kΩ

1.2 kΩ

Figure 10-41  π-network

Output

circuitSIM walkthrough 100 Ω

20I1

RTh I1

Figure 10-40

290

Chapter 10   Equivalent-Circuit Theorems

B

10-37. Use a wye-to-delta transformation to determine the equivalent π-network for the T-network of Figure 10-42. 3.3 kΩ

6.8 kΩ

Input

22 kΩ

Output

Figure 10-42 T-network

circuitSIM walkthrough

I

I

I

circuitSIM walkthrough

I

10-38. (a) Given E = 50 V, R1 = 5 Ω, R2 = 25 Ω, R3 = 10 Ω, R4 = 15 Ω, and RM = 50 Ω for the Wheatstone bridge of Figure 10-23(a), use a delta-to-wye transformation to determine the meter ­current. (b) Use Multisim to verify the meter current calculated in part (a). 10-39. Use a delta-to-wye transformation to determine the current through a 400-Ω load connected to the output terminals of the bridged-T ­ attenuator of Figure 10-30 when the input voltage is 5.0 V. 10-40. (a)  Use a delta-to-wye transformation to determine what voltage ­applied to the input terminals of the lattice network of Figure 10-31 will produce a 2.0-A current through a 20-Ω load connected to the output terminals. (b) Use Multisim to verify the input voltage calculated in part (a). 10-41. Use a delta-to-wye transformation to calculate the current delivered by the source in Figure 10-43. 20 Ω

+ 50 V −

25 Ω 15 Ω

20 Ω 30 Ω

5.0 Ω

Figure 10-43

A

Section 10-5  Troubleshooting 10-42. A voltmeter is used to measure the voltage across the 300-Ω resistor in Figure 10-26. Determine the most probable cause if the voltmeter reads (a) 0 V (b) −200 V (c) −64 V

Review Questions

Review Questions

Section 10-1  Thévenin's Theorem 10-43. What is a Thévenin-equivalent source? 10-44. What is the difference between the black-box method and the Thévenin’s-theorem method of determining the internal resistance of a Thévenin-equivalent source? 10-45. Show that the Thévenin-equivalent resistance found by using the short-circuit current is the same as that found by the procedure in the statement of Thévenin’s theorem. 10-46. Draw a Thévenin-equivalent circuit for the source shown in Figure 10-44 and express the equivalent voltage, ETh, and the ­equivalent internal resistance, RTh, in terms of the parameters of the original circuit. R1

+ E

−

R2 R3

Figure 10-44

10-47. Derive a single equation for IL when a load RL is connected to the ­circuit of Figure 10-44. 10-48. In a Thévenin-equivalent circuit, IL = ETh/(RTh + RL). Show that this expression is equal to the equation for IL in Question 10-47, thereby verifying Thévenin’s theorem.

Section 10-2  Norton's Theorem 10-49. What is a Norton-equivalent source? 10-50. Explain what happens to the terminal voltage of a Norton-­equivalent source when a load resistor is connected to its terminals.

Section 10-3  Dependent Sources 10-51. The constant for the current-controlled voltage source in Figure 10-13 is understood to include the unit ohm. What unit is included in the constant for (a) a voltage-controlled voltage source (b) a voltage-controlled current source 10-52. What are the restrictions on using the superposition theorem in a network containing dependent sources? 10-53. What are the restrictions on using Thévenin’s and Norton’s theorems in a network containing dependent sources?

Section 10-4  Delta-Wye Transformation 10-54. Derive the wye-to-delta transformation Equation 10-4.

291

292

Chapter 10   Equivalent-Circuit Theorems

10-55. Figure 10-23(b) shows a method for solving a Wheatstone bridge using a delta-to-wye transformation. Draw a diagram showing how we can solve the same bridge circuit using a wye-to-delta ­transformation.

Integrate the Concepts

In Example 8-1 of Chapter 8, we used the equivalent-circuit method to ­determine I1, I2, and I3 in the circuit shown in Figure 10-45. In Chapter 9 we used four other techniques to analyze this circuit. Solve for I3 using another two different techniques. I1

R1 I2

12 Ω

+ 100 V −

R2

I3

10 Ω

R3

40 Ω

Figure 10-45

Practice Quiz 1.

2.

A Thévenin-equivalent circuit consists of (a)   a constant-voltage source in series with a ­resistance (b)   a constant-current source in series with a ­resistance (c)   a constant-voltage source in parallel with a ­resistance (d)  a constant-current source in parallel with a ­resistance

single internal single internal single internal single internal

The Thévenin internal resistance for the network in Figure 10-46 is (a)   10 kΩ (b)   1.8 kΩ (c)   2.07 kΩ (d)   2.11 kΩ I2 10 mA

R3 1.2 kΩ

I1 15 mA

R1 1.0 kΩ

A

R2 10 kΩ

RL B

Figure 10-46

Practice Quiz

3.

The Thévenin-equivalent constant-voltage source for the network in Figure 10-46 is (a)   15 V (b)   4.9 V (c)   22.1 V (d)   12 V

4.

A Norton-equivalent circuit consists of (a)   a constant-voltage source in parallel with a single internal ­resistance (b)  a constant-current source in series with a single internal ­resistance (c)   a constant-voltage source in series with a single internal ­resistance (d)   a constant-current source in parallel with a single internal r­ esistance

5.

The Norton-equivalent internal resistance viewed between the terminals A and B of Figure 10-47 is (a)  13.64 Ω (b)  60.76 Ω (c)  395 Ω (d)   143.8 Ω R1 100 Ω A R2 150 Ω

V1 12 V

R3 150 Ω

B R4 220 Ω

22 V V2

Figure 10-47

6.

7.

The Norton’s constant-current source viewed between the terminals A and B of Figure 10-47 is (a)  −31.25 mA (b)  −25.32 mA (c)  +31.25 mA (d)   +25.32 mA

The output of a dependent source depends on some other variable in the circuit. Which of the following is a dependent source? (a)   voltage-controlled pressure source (b)   current-controlled light source (c)   voltage-controlled current source (d)   temperature-controlled voltage source

293

11

Electrical Measurement To understand electric circuit behaviour, we must be able to measure the circuit parameters. In this chapter, we consider how commonly used instruments measure current, voltage, and resistance.

Chapter Outline 11-1

11-2

Moving-Coil Meters  296

The Ammeter  297

11-3

The Voltmeter  300

11-5

Resistance Measurement  304

11-7

Multimeters 

11-4 11-6

Voltmeter Loading Effect  302

The Electrodynamometer Movement  312

311

Key Terms moving-coil movement  296 d’Arsonval movement  296 motor principle  296 diodes 297 sensitivity 297 resistance 297 shunt resistor  297

swamping or calibrating resistor 297 ammeter 299 voltmeters 300 multiplier resistor  300 voltmeter sensitivity  300 ohmmeter 305

galvanometer 309 electrodynamometer movement 311 wattmeter 311 volt-ohm-milliammeter (VOM) or multimeter  312 digital multimeter (DMM) 

312

Learning Outcomes At the conclusion of this chapter, you will be able to: • explain how a moving-coil movement measures current • calculate the calibrating and shunt resistors required for a multirange ammeter • calculate the multiplier resistors required for a ­multirange voltmeter • calculate the internal resistance of a movingcoil ­voltmeter • calculate the loading effect of a voltmeter

Photo sources: iStock.com/Zmiy

• design an ohmmeter using a moving-coil movement • use a Wheatstone bridge for accurate resistance ­measurements • explain the operation of the electrodynamometer ­movement • differentiate between analog and digital multimeters

296

Chapter 11   Electrical Measurement

11-1  Moving-Coil Meters The most common mechanism for meters that have a pointer that moves across a scale is the moving-coil or d’Arsonval movement. This movement can be used to measure current, voltage, and resistance. The operation of such analog meters is based on the motor principle, the fact that a magnetic field exerts a force on a current flowing perpendicular to it. Figure 11-1 shows the basic parts of a D’Arsonval movement. When current flows through the coil, it becomes an electromagnet with a north and a south pole. These poles are repelled by the permanent magnet, rotating the coil until the magnetic force is balanced by the spiral spring. As the coil rotates around the stationary iron core, the pointer moves to the right. The iron core increases the strength of the magnetic field in the air gap between the core and the permanent magnet. Often the spiral spring also provides an electric connection to the coil.

The French physicist Jacques-Arsène d’Arsonval (1851–1940) developed the moving-coil movement in 1885.

Scale

Volts

Pointer

Permanent magnet

Air gap

Spiral spring

N

Moving coil

Source:  ANDREW LAMBERT PHOTOGRAPHY/SCIENCE PHOTO LIBRARY

Figure 11-1  Moving-coil meter

A moving-coil mechanism

S

Soft-iron core

Pole face

11-2   The Ammeter

If the current in the coil increases, the electromagnetic force also increases, moving the pointer farther across the scale. The deflection of the pointer is directly proportional to the current, so the scale is linear. If the current in the coil is reversed, the direction of its magnetic field ­reverses and the pointer hits a stop at the left end of the scale. Since reverse deflection can damage some meters, it is important to ensure proper polarity on the meter connections. Moving-coil movements cannot measure alternating current directly since the alternating polarity makes the pointer vibrate back and forth. However, a meter can include diodes connected such that current always flows in the same direction through the coil. The pointer then shows the ­average magnitude of the alternating current. A moving-coil meter movement has two key characteristics: Sensitivity   Sensitivity is usually given in terms of the current that causes a full-scale deflection. This current is determined by the strength of the permanent magnet and the number of turns on the moving coil. Most moving-coil movements have a sensitivity of 1 mA or less. Resistance   The resistance of the movement comes from the resistance of the wire in the coil. Generally, a more sensitive meter has more turns on the moving coil and consequently a higher resistance. See Review Questions 11-23 and 11-24 at the end of the chapter.

11-2  The Ammeter

Commercial movements with strong magnets, delicate springs, and lowfriction pivots can have a sensitivity of 20 μA or less. Using a stronger spring and fewer turns with a thicker wire on the moving coil gives a ­sensitivity of several milliamperes. For meters up to 8 cm in diameter, the maximum wire size that can be conveniently used for the coil limits the current the meter can pass without damage to about 30 mA. To measure larger currents with these movements, we need to connect a shunt resistor in parallel. Sensitive 50-μA movements are commonly used for measurements in electronic circuits while more rugged 5.0-mA movements are used with electric power circuits. Usually a swamping or calibrating resistor is connected in series with the moving coil to bring the total resistance of the movement up to some convenient number, as shown in Figure 11-2. It is common practice to select a calibrating resistor such that the full-scale c­ urrent produces a 50-mV drop across the movement and its calibrating ­resistor. For a moving-coil movement with a sensitivity of 1.0 mA and a r­ esistance of 27 Ω, RT =

V 50 mV = = 50 Ω I 1.0 mA

297

A diode has a high ­resistance for current flowing through it in one direction and a low resistance for ­current flowing in the opposite direction.

298

Chapter 11   Electrical Measurement

Moving coil (27 Ω) Calibrating resistance (23 Ω)

Shunt

Figure 11-2  Moving-coil ammeter with shunt resistor

Therefore, the calibrating resistance is

R c = 50 − 27 = 23 Ω



The calibrating resistor is made from precision resistance wire having a small negative temperature coefficient to offset the positive temperature coefficient of the moving coil. Since accurate meters require precision resistors at key points in their circuits, we calculate the resistances here to four significant digits. To use the 1.0-mA movement to measure a current of up to 1 A, we arrange the meter circuit as shown in Figure 11-3, forming two parallel branches with 999 times as much current flowing through the shunt resistor as through the meter movement. Since the shunt and the movement are in parallel, the same voltage drop appears across both and their current ratio is inversely proportional to their resistance ratio. Therefore, Vsh = VM Rsh =

and

Ish × Rsh = IM × RM

1 50 × RM = = 50.05 mΩ 999 999

1.0-mA movingcoil movement and calibrating resistor (RT = 50 Ω)

V Shunt

50.05 mΩ

+

E

− Figure 11-3  Measuring current

Lamp

11-2   The Ammeter

If the current flowing through the lamp in Figure 11-3 is sufficient to make the movement read full scale, the current in the moving-coil branch is 1 mA. Therefore the current through the shunt resistor is 999 mA and the current through the lamp is 1 mA + 999 mA = 1 A. As long as this movement and shunt are used together, we can interpret the meter scale as being calibrated for 1 A full scale. If the lamp current is 0.6 A, this total current splits so that one part (0.6 mA) flows through the moving coil and 999 parts (0.5994 A) flows through the shunt. Therefore, we can interpret the 0.6 scale marking of the 1-mA movement as indicating 0.6 A. For currents up to about 25 A, the shunt is physically small enough to be placed inside the meter case. For larger currents, external shunt resistors are required. An ideal ammeter would have no internal resistance so that connecting the meter would not reduce the current being measured. For most circuits, connecting an ammeter has only a slight effect since we are adding just a ­little extra resistance to the circuit. For example, the total resistance of the meter and its shunt in Figure 11-3 is 0.05 Ω, while the hot resistance of the lamp is in the order of 200 Ω. Therefore, the change in circuit current due to adding the ammeter to the circuit is only 0.05/200, or 0.025%. In practice, the main problem with making current measurements is that we must place the  ammeter in series with a circuit in order to measure its ­current. ­Unless the meter is left in the circuit permanently, current measurements ­require opening the circuit to connect the meter and then restoring the original ­connections when we remove the meter. Because an ammeter has low r­esistance, we must guard against connecting an ammeter across a source. To make a test meter that can measure a wide range of currents, we arrange the meter circuit so that we can easily change the shunt resistance, as shown in Figure 11-4. The switch must be a make-before-break type so that the meter is not damaged by passing the entire current as we change the shunt.

V

Figure 11-4  Multirange shunt

See Problems 11-1 to 11-4 and Review Questions 11-25 to 11-29.

299

300

Chapter 11   Electrical Measurement

11-3  The Voltmeter Moving-coil movements can also be used in voltmeters. If a 1.0-mA ­move­ment has a total resistance of 50 Ω, a voltage drop of V = IR = 0.0010 A  × 50 Ω = 50 mV appears across the movement when it is reading full scale. If we connect a 99.95-kΩ resistor in series with the meter to bring the total ­resistance up to 100 kΩ, as in Figure 11-5, the voltage we must apply to produce a 1.0-mA current through the meter is V = IR = 0.0010 A × 100 kΩ = 100 V 99.95 kΩ Multiplier

1.0-mA movement (50 Ω)

V

Figure 11-5  Simple voltmeter

If we apply a 50-V potential difference to the voltmeter, the current through it is V 50 V I= = = 0.50 mA R 100 kΩ Thus, the meter will read half scale. With the 99.95-kΩ multiplier resistor, the scale on the 1.0-mA movement corresponds to 0–100 V. To measure open-circuit voltage, a perfect voltmeter should draw no current. Since current must flow in a moving-coil movement to obtain a reading, a 50-μA movement is much more accurate than a 1.0-mA movement for measurements in high-resistance circuits. The resistance of a standard 50-μA movement and its calibrating resistor is 1000 Ω. To obtain a 100-V full-scale range, the total resistance must be RT =

100 V = 2.0 MΩ 50 µA

Therefore, the multiplier resistance must be 2000 kΩ − 1 kΩ = 1999 kΩ. Similarly, the total resistance must be 10 MΩ for a 500-V scale and 20 kΩ for a 1.0-V scale. For any full-scale voltage with a 50-μA movement, the total resistance required is 20 kΩ for each volt of the full-scale reading. Therefore, when used as a voltmeter, a 50-μA movement has a voltmeter sensitivity of 20 kΩ/V. We can determine the total resistance of a movingcoil voltmeter by multiplying the full-scale voltage by the sensitivity of the movement. For example, when used as a voltmeter, a 1.0-mA movement requires a total resistance of 1000 Ω/V, so total resistance for a 500-V scale is 500 V × 1000 Ω/V = 500 kΩ.

11-3   The Voltmeter

301

Example 11-1 The meter in Figure 11-6 uses a 1.0-mA moving-coil movement with a resistance of 50 Ω, and has scales of 5 V, 50 V, 150 V, and 500 V. Calculate RA, RB, RC, and RD.

V

RA

RB

5V

−

RC

50 V

RD

150 V

500 V

Figure 11-6  Multirange voltmeter

Solution For the 5-V range, RT =

5.0 V = 5000 Ω and RA = 5000 Ω − 50 Ω = 4950 Ω 1.0 mA

Similarly, for the 50-V range,  RT = 50 kΩ and RB = 50 000 Ω − 5000 Ω = 45 kΩ for the 150-V range, RT = 150 kΩ and RC = 150 kΩ − 50 kΩ = 100 kΩ for the 500-V range, RT = 500 kΩ and RD = 500 kΩ − 150 kΩ = 350 kΩ Multisim Solution Download Multisim file EX11-1 from the website. The circuit is the same as shown in Figure 11-6 with the calculated values for RA, RB, RC, and RD. A 50-Ω resistor, Rm, represents the resistance of the moving-coil movement. Insert a meter to measure the current through the movement. Run the simulation, and check that the ammeter reading is 1 mA, the full-scale current for the meter movement. This reading confirms that the value of RA is correct. Move the positive lead of the voltage supply to the connection between RB and RC, and increase the voltage to 50 V. Run the simulation. Now a meter reading of 1 mA confirms that the value of RB is correct. Verify the calculations for RC and RD by running a simulation first with a lead from a 150-V supply connected between RC and RD and then with a lead from a 500-V supply connected to the right terminal of RD. The current readings should be 1 mA.

circuitSIM walkthrough

302

Chapter 11   Electrical Measurement

The selector switch and resistance network in a multirange meter

The voltmeter is a useful instrument for checking the operation of a circuit. With suitable clips on the test leads, we can connect a voltmeter across any part of a circuit without having to disconnect any of the components in the circuit. See Problems 11-5 to 11-8 and Review Questions 11-30 to 11-35.

Circuit Check

A

CC 11-1.

Determine the sensitivity of a voltmeter that uses a 50-µA meter movement. CC 11-2. Determine the total resistance for the 100-V scale range of a voltmeter with a 10-mA moving-coil movement. CC 11-3. Find the total resistance for the 30-V scale of a meter that has a sensitivity of 30 kΩ/V.

11-4  Voltmeter Loading Effect Since a voltmeter is connected in parallel with the portion of the circuit in which the voltage drop is measured, an ideal voltmeter should draw zero current and should, therefore, have an infinitely high internal resistance. However, the moving-coil type of voltmeter must draw some current in order to deflect the pointer. The smaller the current required for full-scale deflection, the more sensitive the voltmeter and the smaller the loading ­effect it has on the circuit being tested. When measuring voltage in circuits with fairly large currents, an extra 5 mA does not alter the circuit conditions appreciably. For such circuits, the more rugged and less sensitive voltmeters give adequate accuracy.

11-4   Voltmeter Loading Effect

Example 11-2 (a)   Find the voltage drop across the 200-kΩ resistor in Figure 11-7. (b)  What will a 20-kΩ/V meter set to a 150-V scale read when connected across the 200-kΩ resistor? (c)  What will a 1000-Ω/V meter set to a 150-V scale read when connected across the 200-kΩ resistor?

100 kΩ

+ 200 V − 200 kΩ

V

Figure 11-7  Effect of voltmeter sensitivity

Solution (a)   From the voltage-divider principle, V=

200 kΩ × 200 V = 133.3 V 300 kΩ

(b)  A 20-kΩ/V meter set to a 150-V range has a total resistance of 20 kΩ × 150 = 3.0 MΩ. This resistance is in parallel with the 200-kΩ resistor, thus ­reducing the resistance of the parallel combination to

R=

200 kΩ × 3000 kΩ = 187.5 kΩ 200 kΩ + 3000 kΩ

Therefore, V =

187.5 kΩ × 200 V = 130.4 V 287.5 kΩ

R=

200 kΩ × 150 kΩ = 85.7 kΩ 200 kΩ + 150 kΩ

(c)  A 1000-Ω/V meter set to a 150-V range has a total resistance of 1000 Ω × 150 = 150 kΩ. Therefore, the 200-kΩ resistor and the voltmeter in parallel have an equivalent resistance of and

V=

85.7 kΩ × 200 V = 92.3 V 185.7 kΩ

303

304

Chapter 11   Electrical Measurement

Multisim Solution Download Multisim file EX11-2 from the website. The circuit is the same as shown in Figure 11-7. Connect a ground to the bottom node of the schematic.

circuitSIM walkthrough

(a) Double-click on the voltmeter and view the value of the voltmeter resistance. Set it to a value of at least 1 GΩ. Run the simulation. The voltmeter reading of 133.3 V matches the value calculated using the voltage divider principle. (b) Set the simulated voltmeter resistance to 3.0 MΩ, the total resistance for a 150-V scale on an actual 20-kΩ/V meter. Run the simulation. The voltmeter now reads 130.4 V. (c) Set the voltmeter resistance to 150 kΩ, the total resistance for a 150-V scale on a 1000-Ω/V meter. Run the simulation. The voltmeter now reads 92.3 V.

However, in electronic circuits with small currents, the resistance of a ­voltmeter can alter circuit conditions considerably. Even the 20-kΩ/V meter in part (b) of Example 11-2 introduced an error in excess of 2%. Consequently, the simple moving-coil type of voltmeter is not suitable for voltage ­measurement in high-resistance circuits. However, such circuits can be measured accurately with an electronic voltmeter, like those described in Section 11-7. These meters have internal resistances much higher than those of moving-coil meters. See Problems 11-9 to 11-11 and Review Questions 11-36 and 11-37.

11-5  Resistance Measurement We can determine an unknown resistance in Figure 11-8 by applying Ohm’s law to the readings obtained from the voltmeter and ammeter. This method requires that the unknown resistance be connected into a special circuit with two separate meters. This method is useful for measuring very low resistances, such as those of motors, and for measuring the resistance of a component while it carries its normal operating current. Current-adjusting resistor

A + E

V

Rx

− Figure 11-8  Measuring resistance with a voltmeter and an ammeter

11-5   Resistance Measurement

With the voltmeter connected as shown in Figure 11-8, the ammeter indicates the sum of the currents through Rx and the voltmeter. Thus, the V/I ratio calculated from the meter readings equals the equivalent resistance of the two parallel branches. This ratio is less than the actual value of Rx unless the voltmeter current is negligible compared to the current through Rx. We can check for voltmeter loading by watching the ammeter reading as we disconnect the voltmeter. If there is any noticeable decrease in the current, we then connect the voltmeter on the source side of the ammeter. The a­ mmeter now reads only the current through Rx while the voltmeter now shows the sum of the voltage drops across the ammeter and Rx. However, if Rx is large enough to show a noticeable voltmeter loading effect, the voltage drop across the ammeter is insignificant compared to Vx and we can ­ignore the ammeter loading effect. For quick checks of circuit resistance, we can use an ohmmeter, a meter designed to measure resistance directly. The simple ohmmeter shown in ­Figure 11-9(a) consists of a 1.0-mA movement, a 4.5-V battery, and a resistance that passes a 1.0-mA current when we short-circuit the ohmmeter ­terminals. A portion of the total resistance is adjustable so that we can ­calibrate the meter to read exactly full scale when we connect the two test leads together. In this example, the total resistance, including the meter movement, is RT =

E 4.5 V = = 4.5 kΩ I 1.0 mA

As shown in Figure 11-9(b), the scale of an ohmmeter is nonlinear. The nonlinear scale makes low resistances easier to read accurately, but high ­resistances are crammed together at the left end of the scale. For circuit testing where we need a high-range ohmmeter, we could use a 50-μA 1-mA movement

V +

4.5 V

− 40.5 k 13.5 k 4.5 k

Unknown resistor

Zero adjust

0

∞

(a)

Figure 11-9  Simple ohmmeter

1.5 k

OHMS (b)

305

306

Chapter 11   Electrical Measurement

Example 11-3 Determine the resistance values that should be marked at full scale, centre scale, one-quarter of full scale, and one-tenth of full scale for the ohmmeter in Figure 11-9. Solution As we have already noted, the total internal resistance of the ohmmeter is adjusted so that the meter reads exactly full scale when the test leads are short-circuited. Therefore, the end mark on the scale represents 0 Ω. When the meter reads half scale, the current is 0.5 mA and the total resistance in the series loop is double the total ohmmeter resistance. Therefore, a centre-scale reading represents Rx = RM =

4.5 V = 4.5 kΩ 1.0 mA

For one-quarter of full scale, the current is 0.25 mA. Then, the total resistance in the loop is RT =

and

4.5 V = 18 kΩ 0.25 mA

Rx = 18 kΩ − 4.5 kΩ = 13.5 kΩ

Similarly, one-tenth of full scale represents

Rx =

4.5 V − 4.5 kΩ = 40.5 kΩ 0.10 mA

1.0-mA movement (50 Ω)

V 0.505-Ω shunt

movement rather than a 1.0-mA movement. With a 50-μA movement the total i­ nternal resistance of the ohmmeter in Figure 11-9(a) is 4.5 V

44.5 Ω Zero adjust

Figure 11-10  Low-range ohmmeter

RT =

E 4.5 V = = 90 kΩ I 50 µA

The centre-scale reading of the meter will also be 90 kΩ. These examples demonstrate that the resistance represented by a centre-scale reading is inversely proportional to the full-scale current for the meter. Therefore, we can convert the basic ohmmeter into a lowrange ohmmeter by placing a shunt across the moving coil, as shown in Figure 11-10.

11-5   Resistance Measurement

307

Example 11-4

Using a 1.0-mA movement with an internal resistance of 50 Ω and a 4.5-V battery, design an ohmmeter that reads 45 Ω at centre scale. Solution For a centre-scale reading, Rx = RM, so the total internal resistance of the ohmmeter is 45 Ω. Therefore, full-scale current is I=

E 4.5 V = = 0.10 A R 45 Ω

Since the meter movement passes 1.0 mA at full scale, the shunt current must be 99 mA. The shunt resistance is Rsh =

1 mA × 50 Ω = 0.505 Ω 99 mA

The equivalent resistance of the meter movement and shunt in ­parallel is Req =

50 Ω × 0.505 Ω = 0.500 Ω 50 Ω + 0.505 Ω

Therefore, the total resistance of the series resistor and the “ohms adjust” rheostat is RS = 45 Ω − 0.5 Ω = 44.5 Ω Multisim Solution Download Multisim file EX11-4 from the website. The circuit is the same as shown in Figure 11-10, with a 50-Ω resistor representing the resistance of the moving-coil movement. Place a short across the ohmmeter terminals. Insert a meter to measure the current flowing through the short. Connect a ground to the bottom left terminal of the circuit. Run the simulation, and check that the ammeter shows 100 mA. This reading confirms the calculation of the full-scale current for the ohmmeter. Replace the short across the ohmmeter terminals with a 45-Ω resistor. Run the simulation, and check that the ammeter shows 50 mA, which is the half-scale current for the ohmmeter. This reading verifies that the ohmmmeter design is correct.

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Chapter 11   Electrical Measurement

As with a voltmeter, we can readily connect an ohmmeter across a portion of a circuit. However, we must make sure that the circuit being measured is switched off. Current from sources in the circuit can cause an inaccurate reading or damage the ohmmeter. When we wish to measure a very high resistance, such as the leakage ­between the two conductors of a cable as in Figure 11-11, we use a voltmeter and a separate voltage source. This system has the added advantage of testing the resistance with the normal operating voltage applied.

V + 120 V −

Two-conductor cable

Figure 11-11  Using a voltmeter as a high-resistance ohmmeter

Example 11-5 The voltmeter in Figure 11-11 has a 20-kΩ/ V movement. The meter reads 120 V on its 150-V scale when the switch is closed, and 10 V when the switch is open. Calculate the leakage resistance of the cable ­insulation. Solution The resistance of the voltmeter is RM = 20 kΩ/ V × 150 V = 3.0 MΩ The voltage drop across the leakage resistance of the cable is Vleak = 120 V − 10 V = 110 V Since the resistance of the voltmeter and the leakage resistance of the cable form a simple series circuit, Rleak Vleak = RM VM

Rleak = 3.0 MΩ ×

110 V = 33 MΩ 10 V

11-5   Resistance Measurement

309

For precision resistance measurements, we can use a Wheatstone bridge as shown in Figure 11-12. If we adjust Ry such that there is no deflection of the galvanometer G when we close the switch, the voltage drops across Rx and RA must be exactly the same since a potential difference across the galvanometer would cause current to flow through the galvanometer. With no current through the meter movement, Ix = Iy. Hence,

Galvanometer is a term for a sensitive ammeter that can measure ­current flowing through it in either ­direction.

Vx = IxRx  and  Ix = Vx =

Therefore,

VA =

Similarly,

E Rx + Ry

ERx Rx + Ry

ERA RA + RB

Rx

RA G

+ E

−

RB

Ry

Figure 11-12  Measuring resistance with a Wheatstone bridge

Therefore, for perfect balance,

ERx ERA = Rx + Ry RA + RB

RxRA + RxRB = RxRA + RARy RxRB = RyRA and

Rx =

RARy RB

(11-1)

For the bridge to be balanced, the product of the resistances in one pair of opposite arms of the bridge must equal the product of the resistances in other pair of opposite arms. Since E does not appear in Equation 11-1, the magnitude of the source voltage used with a bridge circuit has no effect on the accuracy of the measurement. The source merely causes a deflection of

310

Chapter 11   Electrical Measurement

the galvanometer pointer if the bridge is not properly balanced. With precision resistors for RA, RB, and Ry, we can use Equation 11-1 to determine an accurate value for Rx.

Example 11-6 The Wheatstone bridge circuit of Figure 11-12 is balanced when RA = 1.0 Ω, RB = 50 Ω, and Ry = 17 Ω. Calculate Rx. Solution

Rx =

circuitSIM walkthrough

RARy RB

=

1.0 Ω × 17 Ω = 0.34 Ω 50 Ω

Multisim Solution Download Multisim file EX11-6 from the website. The circuit is a Wheatstone bridge as shown in Figure 11-12, with the given values for RA, RB, and Ry, and the calculated value for Rx. Set E = 10 V. Connect meters to measure the voltages across RA and Rx. Connect a ground to the bottom node of the circuit. Run the simulation, and note that voltage readings show that VA = Vx. These readings verify that the bridge is balanced.

See Problems 11-12 to 11-22 and Review Questions 11-38 to 11-44.

Circuit Check

B

CC 11-4. Find the voltmeter reading when the terminal voltage of a 12-V source with an internal resistance of 100 kΩ is checked with a 20-kΩ/V voltmeter set to its 50-V range. CC 11-5. Design a simple series ohmmeter that uses a 50-μA, 300-Ω movement, and a 1.5-V battery. Calibrate the meter for the quarter, half, and three-quarter scale positions. What deflection will show on the meter if an external resistance of 50 Ω is ­connected? CC 11-6. In a commercial Wheatstone bridge, the values of RA and RB are normally set as a variable ratio for determining low or high value resistances. Given that the RA/RB ratio is 0.01 and the bridge balances with an external resistance of 0.37 Ω, determine the value of the variable resistance.

11-6   The Electrodynamometer Movement

11-6 The Electrodynamometer Movement We can replace the permanent magnet in a meter movement with an electromagnet consisting of two stationary coils, as shown in Figure 11-13. Such electrodynamometer movements can measure alternating currents if the stationary and moving coils are connected in series. Whenever the current in the moving coil reverses direction, the magnetic field produced by the stationary coil also reverses. Therefore, the magnetic force moves the pointer clockwise, regardless of the current direction in the coils. Spring

Pointer

Moving coil Stationary coils

Figure 11-13  Electrodynamometer movement

The magnetic force acting on the moving coil is proportional to the product of the current in the moving coil and strength of the magnetic field from stationary coils. Since the magnetic field strength for a coil is proportional to the current through the coil, the reading of an electrodynamometer movement is proportional to the product of the stationary-coil and moving-coil currents. If the stationary coils are connected in series with a load as in Figure 11-14, the stationary-coil current is the same as the load current. If the moving coil in series with a multiplier resistor is then connected across the load, the moving-coil current is directly proportional to the load voltage. Therefore, the scale reading of the electrodynamometer movement is proportional to the product of the load current and load voltage. Since P = IV, the scale shows the power in the load. Thus, we can use an electrodynamometer as a wattmeter in both DC and AC circuits. ±

+ Load

E −

Multiplier ±

Figure 11-14  Electrodynamometer movement as a wattmeter

311

312

Chapter 11   Electrical Measurement

If we reverse the leads to either the current coil or the voltage coil when we connect a wattmeter into a circuit, the pointer is deflected counterclockwise instead of clockwise. Wattmeters usually identify one end of each coil with a ± mark, as shown in Figure 11-14. The marked terminal of the current coil (the stationary coil) should be connected to the voltage source and the marked terminal of the voltage coil (the moving coil) should be connected to the line containing the current coil. See Review Question 11-45.

11-7 Multimeters

Source:  © iStock.com/EricFerguson

The most common electrical test instrument is the volt-ohm-milliammeter (VOM), or multimeter. Analog multimeters have a single moving-coil movement with a multiposition switch for selecting various DC current ranges, DC and AC voltage ranges, and resistance ranges. The multimeter’s dial has a scale for each range. All these scales can be difficult to read accurately.

An analog multimeter

A digital instrument, on the other hand, displays data as discrete numbers. Since improved technology has reduced the size and cost of integrated circuits, digital multimeters (DMMs) have virtually replaced analog VOMs. DMMs have high input resistances (typically 20 MΩ), greater accuracy than moving-coil meters, and clear numerical readouts. DMMs are generally cheaper and more rugged than equivalent analog multimeters. In a­ ddition, many DMMs have extra features such as functions for testing diodes and transistors. The photographs show two types of digital multimeters.

Source: Photo © AEMC Instruments, a leader in electrical test and measurement instruments

11-7  Multimeters

Source:  Copyright © Tektronix. Reprinted with permission. All Rights Reserved.

Handlheld digital multimeter

Benchtop digital multimeter

Most electronic meters use a field-effect transistor (FET) to give them a very high input resistance, as shown in Figure 11-15. The voltage between the gate and source terminals of the FET controls the current from the source to the drain. Since the resistance between the gate and the source is extremely high, the FET draws only a tiny current from the circuit connected to the input terminals of the meter. Almost all of the current that goes to the measurement circuit comes from the meter’s battery.

Gate

−

FET

Drain Measurement circuit Source

20 MΩ + Figure 11-15  Input stage of an electronic meter

313

314

Chapter 11   Electrical Measurement

The measurement circuit inside a digital multimeter can vary between manufacturers but they typically follow similar logic principles. An example measurement process that converts an analog input into a digital output is shown in Figure 11-16. The red and black terminal leads of a DMM are attached to an analog input, such as a DC voltage. Many samples of the analog signal will be read by the circuitry in a very short amount of time (100 000 samples per second or more) but only a few of the samples will be saved into a buffer and averaged into a single, steady reading. This method of acquiring a steady reading can help to reduce fluctuations from noise and other small variations, thus improving the overall accuracy of the reading. The analog reading is sent to a comparator to determine if it is above or below the last reading. (If it is the first reading, it usually compares itself to half the maximum scale value on the DMM.) The output of the comparator drives either a high or a low signal into the Successive Approximation Register (SAR), which establishes the most significant digit of the reading. The digital output of the SAR then feeds back into the comparator through a digital-to-analog converter (DAC), where it is compared again so the next digit can be established, and so on. This iterative process continues until all the significant digits of the analog signal have been identified, perhaps requiring only a few microseconds to complete. After the analog input has transformed into a digital output, the DMM can then display the numerical value on the meter’s digital screen.

Analog input

Sample acquisition

Comparator High or low output

SAR

Digital output

DAC Figure 11-16  A DMM’s measurement circuit processes an analog input into a digital output

See Review Question 11-46.

315

Problems

Summary

• Since the deflection of the pointer in a moving-coil movement is proportional to the current through the coil, the movement can be used as the measuring device in an ammeter. • A voltmeter with a number of voltage ranges can be constructed from a moving-coil movement by placing multiplier resistors in series with it. • An analog voltmeter can reduce the voltage being measured. • An ohmmeter with a number of resistance ranges can be constructed using a moving-coil movement, a battery, shunt resistors, and series ­resistors. • The Wheatstone bridge can be used for accurate measurements of ­resistances. • The electrodynamometer movement is a practical means of measuring power in both DC and AC circuits. • Digital multimeters typically have higher input resistances and greater accuracy than moving-coil meters.

Problems B B B I

B B

B B

Section 11-2  The Ammeter 11-1. The full-scale voltage drop across a 5.0-mA movement is 75 mV. ­Calculate the shunt resistance required to use the movement to measure currents up to 1.0 A. 11-2. If the shunt resistance calculated in Problem 11-1 is used with a 1.0-mA movement having a total resistance of 50 Ω, what current would its full-scale reading represent? 11-3. Design an ammeter with a range of 50 mA from a 250-μA, 500-Ω movement. 11-4. A milliammeter has a range of 20 mA and a shunt resistor of 2.63 Ω. If the movement has a full-scale sensitivity of 1 mA, what is its r­ esistance?

B = beginner

I = intermediate

A = advanced

Section 11-3  The Voltmeter 11-5.

Calculate the multiplier resistors required for 15-V, 50-V, 150-V, and 500-V ranges in a voltmeter using a 50-μA, 1000-Ω meter movement. 11-6 (a) The resistance of the movement for a 200 Ω/V voltmeter is 10. Calculate the multiplier resistors required for 30-V, 150-V, and 1500‑V ranges. (b) Use Multisim to verify the resistor values calculated in part (a). 11-7. A voltmeter has a 2000-Ω, 75-μA movement and multiplier resistors of 38 kΩ, 160 kΩ, and 600 kΩ. What are the voltage ranges for this voltmeter? 11-8. Compare your answers to Problems 11-5 and 11-7. How does using a less sensitive meter movement affect the multiplier resistors required?

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Chapter 11   Electrical Measurement

I

Section 11-4  Voltmeter Loading Effect 11-9.

A voltmeter is constructed by using a 50-μA, 1000-Ω movement. It is used on the 20-V range to measure the voltage across R2 in the circuit shown in Figure 11-17. Find the percent error caused by the loading effect of the voltmeter. 100 kΩ R1

+

R2

10 V

250 kΩ

−

I

circuitSIM walkthrough

Figure 11-17

11-10. (a) Resistor R2 in the voltmeter shown in Figure 11-18 has burned out, and its original value is not known. R1 and R3 measure 122 kΩ and 375 kΩ, respectively. What value of resistance for R2 is required to repair the voltmeter? (b) Use Multisim to verify the resistance calculated in part (a).

V

−

I

circuitSIM walkthrough

circuitSIM walkthrough

B B B I

R1

R2

50 V

R3

100 V

250 V

Figure 11-18  Multirange voltmeter

11-11. Using the data from Problem 11-10, calculate the meter resistance, full-scale current, and ohms per volt rating for the voltmeter shown in Figure 11-18.

Section 11-5  Resistance Measurement 11-12. (a) What resistance corresponds to the centre-scale mark on a basic ohmmeter constructed with a 50-μA movement and a 1.5-V battery? (b) Use Multisim to verify the resistance calculated in part (a). 11-13. Find the battery voltage required for an ohmmeter with a 1.0-mA movement to use the same scale as the meter in Problem 11-12. 11-14. (a) What resistance must be connected to the terminals of the ohm­meter in Problem 11-12 for the pointer to read 20% of full-scale d ­ eflection? (b) Use Multisim to verify the resistance calculated in part (a). 11-15. Design an ohmmeter with 500 Ω at centre scale, using a 2.5-V battery and a 50-μA, 500-Ω movement.

Problems

I

B B B I B

11-16. The insulation resistance of a solenoid is checked by connecting one terminal of a 1.0-kV source to the solenoid and connecting a 20-kΩ/V voltmeter between the other terminal and the iron core of the solenoid. Calculate the insulation resistance given that the meter reads 360 V when set to 1500-V range. 11-17. What would the meter read in Problem 11-16 if the leakage resistance between the solenoid winding and the core dropped to 0.5 MΩ? 11-18. (a) The Wheatstone bridge in Figure 11-12 balances when RA is 5.0 kΩ, RB is 1000 Ω, and Ry is 42 kΩ. Calculate Rx. (b) Use Multisim to verify the resistance calculated in part (a). 11-19. Find the value of Ry that balances the Wheatstone bridge in Figure 11-12 when RA is 1200 Ω, RB is 1500 Ω, and Rx is 500 Ω. 11-20. The Wheatstone bridge in Figure 11-12 balances when VA is 2.4 V, IB is 0.24 A, and the total current drawn from the battery is 300 mA. Calculate Rx. 11-21. Figure 11-19 shows a variation of the Wheatstone bridge known as the slide-wire bridge. The slide wire is a length of uniform ­resistance wire 1.0 m long. The standard resistor, RS, is 37 Ω. Find Rx given that the bridge balances when the slider is 45 cm from the end connected to the standard resistor.

Rx

G

RS Slider

Resistance wire 1 metre

I

Figure 11-19  Slide-wire bridge

11-22. Figure 11-20 shows a variation of the Wheatstone bridge called the Varley loop, used for ­locating ground faults on long transmission lines. To trace the fault, the far ends of the two identical conductors are connected together. RA and RB have fixed values of 100 Ω and 1000 Ω, respectively. When the switch is in the line (L) position, the bridge balances with RC = 200 Ω. When the switch is thrown to the ground (G) position, the bridge balances with RC = 150 Ω. How far is the ground fault from the end of the cable at which the apparatus is located? Assume negligible resistance

317

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318

Chapter 11   Electrical Measurement

between the ground connection at the switch and the fault ground on the transmission line. RA

20 km

G

Transmission line RB



RC

L

Fault ground

G

Figure 11-20  Varley loop

Review Questions Section 11-1  Moving-Coil Meters 11-23. Describe how current passing through the coil in Figure 11-1 affects the pointer. 11-24. How would leaving out the soft-iron core affect the operation of a moving-coil movement in Figure 11-1?

Section 11-2  The Ammeter 11-25. What is the purpose of the calibrating resistor used with movingcoil movements? 11-26. What is the advantage of arranging that moving-coil movements have a 50-mV drop across their terminals when full-scale current flows through them? (Refer to Problem 11-2.) 11-27. Show that IM RS = RM IT − IM where RS is the resistance of an ammeter shunt, RM is the resistance of the moving-coil movement with its calibrating resistor, IM is the coil current required for full-scale deflection, and IT is the desired full-scale ammeter reading. 11-28. What undesired effect could result from connecting an external shunt to an ammeter as shown in Figure 11-21(a)?

(

)

Moving-coil movement

Shunt

E

Moving-coil movement

(a) Wrong

Shunt Load

Load

E

(b) Right

Figure 11-21  Connecting an external ammeter shunt

11-29. What is the effect of connecting an ammeter across a load?

319

Integrate the Concepts

Section 11-3  The Voltmeter 11-30. The basic moving-coil movement is a current-indicating device. How is it possible to calibrate it as a voltmeter? 11-31. What is the significance of voltmeter sensitivity? 11-32. Why is a 20-kΩ/V meter more suitable than a 1000-Ω/V meter for checking electronic circuitry? 11-33. Why is a 200-Ω/V meter satisfactory for use with electric machinery? 11-34. What would be the effect of connecting a voltmeter in series with a load? 11-35. What circuit information could be obtained from the voltmeter reading in Question 11-34?

Section 11-4  Voltmeter Loading Effect 11-36. Why is voltmeter loading more of a problem than ammeter resistance in practical measurements? 11-37. Why is the loading effect of a DMM considerably less than that of a 20-kΩ meter?

Section 11-5  Resistance Measurement 11-38. Why is a galvanometer used in the circuit of Figure 11-12 rather than a microammeter? 11-39. Draw a circuit diagram showing how to use a voltmeter and an ammeter to determine a resistance. Discuss the possible errors that might occur when measuring a 50-kΩ resistor with this method. 11-40. What precaution must be observed when checking the resistance of a circuit with an ohmmeter? 11-41. Lay out a scale for a simple ohmmeter that uses a 50-μA movement and a 1.5-V battery. 11-42. Lay out a scale for the low-resistance ohmmeter shown in Figure 11-22. 11-43. Would you select a 50-μA movement or a 5-mA movement for the Wheatstone bridge circuit of Figure 11-12? Explain your choice. 11-44. How does the accuracy of the galvanometer scale calibration affect the accuracy of the resistance measurements made with a Wheatstone bridge?

Section 11-6  The Electrodynamometer Movement 11-45. Why is it possible to use an electrodynamometer wattmeter in both DC and AC circuits?

Section 11-7  Multimeters 11-46. List the advantages of the DMM over an analog VOM. Describe a ­situation where an analog instrument might be more convenient.

Integrate the Concepts

Design a VOM that will accurately measure (a)   direct current up to 50 mA (b)  DC voltages up to 40 V (c)  100-Ω resistors

Ohms adjust 1.5 V

V 1.0-mA movement (50 Ω)

Figure 11-22  Low-resistance ohmmeter

320

Chapter 11   Electrical Measurement

Practice Quiz

1. What device can enable a moving-coil movement to measure alternating current? (a) transistor (b) resistor (c) diode (d) fuse 2. List two key characteristics of a moving-coil meter movement. 3. Why would an ideal ammeter have zero internal resistance? 4. Which of the following statements is true? (a) A galvanometer is a sensitive ammeter that can measure current flowing through it in only one direction. (b) An electrodynamometer is an instrument that can measure electric power in any DC and AC circuit.

PART

III

Capacitance and Inductance

In all the circuits we have considered so far, the passive components have only one basic property: resistance. Part III introduces the other two basic properties of circuits: capacitance and inductance. While resistance is opposition to the flow of current in a circuit, inductance is ­opposition to any change in the current. Similarly, capacitance is opposition to any change in the voltage between two points in a circuit. 12 Capacitance 13

Capacitance in DC Circuits

14 Magnetism 15

Magnetic Circuits

16 Inductance 17

Inductance in DC Circuits

Photo source:  © iStock.com/omada

12

Capacitance Suppose we have two metal plates that are close together but insulated from each other. If we connect one plate to each terminal of a voltage source, electrons are removed from one plate and deposited on the other. The first plate now has a net positive charge while the second has a net negative charge. If we remove the plates from the circuit and leave them isolated from each other, they will retain their net charge indefinitely. In this chapter we shall investigate how these plates—and other forms of capacitors—store charge.

Chapter Outline 12-1

Electric Fields  324

12-2 Dielectrics 327

12-3 Capacitance 328 12-4 Capacitors 330

12-5 Factors Governing Capacitance  333 12-6 Dielectric Constant  336 12-7

Capacitors in Parallel  338

12-8 Capacitors in Series  338

Key Terms electric field  324 strength (intensity) of an electric field  324 vector  324 electric lines of force (electric flux lines)  324 electric flux  325 electric flux density  325 voltage gradient  326 dielectric  327

polarized  327 dielectric strength  327 dielectric absorption  327 capacitor  328 capacitance  329 farad  329 electrolytic capacitor  331 working voltage  332 permittivity (absolute permittivity)  334

permittivity of free space  335 relative permittivity (dielectric constant)  336 ferroelectric dielectric  337 electrostatic induction  338 elastance  340

Learning Outcomes At the conclusion of this chapter, you will be able to: • describe the nature of an electric field • explain the relationships among the strength of an electric field, electric flux, electric flux density, and potential ­difference • explain how the electric charges in a dielectric behave in the presence of the electric field • describe the construction of various types of capacitors • define capacitance in terms of the charge on the plates of a capacitor and the voltage across it

Photo sources:  © sciencephotos/Alamy Stock Photo

• state the effect that the area of the plates, the spacing of the plates, and the type of dielectric have on the capacitance of a capacitor • define dielectric constant • calculate the capacitance of a capacitor given the dielectric and the area and spacing of the plates • calculate the total capacitance of capacitors in parallel • calculate the equivalent capacitance of capacitors in series

324

Chapter 12  Capacitance

12-1  Electric Fields In Section 2-1, we noted that Coulomb developed an equation to describe the force that any two electric charges exert on each other: F=k



Q1Q2 d2

(2-1)

where F is the magnitude of the force in newtons, Q1 and Q2 are the two charges in coulombs, d is the distance between the charges in metres, and k is Coulomb’s constant, 8.99 × 109 N · m2/C2. We can think of this force as resulting from an electric field. Some books use the script letter ℰ for ­electric field strength to avoid confusion with the letter symbol for EMF, E.

An electric field is that region in which an electric charge is acted upon by an electric force. The strength (or intensity) of an electric field is the force that the field exerts on a unit of charge. The letter symbol for electric field strength is the Greek letter E (capital epsilon). The units for electric field strength are newtons per coulomb, which are equivalent to volts per metre. Thus,

E=

F Q

(12-1)

where E is the strength of the electric field in newtons per coulomb, F is the force in newtons, and Q is the charge in coulombs.

Vectors are sometimes indicated by putting an arrow over the ­letter representing the variable. Thus, E → could be written as E. This alternative notation can prevent confusion in handwritten material.

A quantity that has direction as well as magnitude is called a vector. Such quantities are set in boldface, italic type. To refer to just the magnitude of a vector, we use the same letter symbol in lightface, italic type. For example, to fully specify a 10-N force directed upward, we write F = 10 N [up], but if we are concerned only with the magnitude of the force, we can write just F = 10 N. Electric field strength and many of the other quantities relating to forces and fields are vectors. Section 20-4 describes vectors and other types of variables in more detail. In diagrams, we represent an electric field with electric lines of force (or electric flux lines), which show the path that a tiny positive charge would follow because of the force exerted by the electric field. Figure 12-1 shows the lines of force around positive (a) and negative (b) point charges. The closer ­together the lines of force are, the greater the strength of the electric field. Note that the lines of force are directed away from the positive charge in (a), and directed toward the negative charge in (b). At every point, the electric field has both a magnitude and a direction. Figure 12-2 shows the electric fields around two parallel conductors (a) and between two concentric (or coaxial) conductors (b). These conductor

12-1   Electric Fields

+

−

(a)

(b)

325

Figure 12-1  Electric fields surrounding point electric charges

+

+

−

(a)

+

−

+

+ (b)

Figure 12-2  Electric field between (a) parallel conductors and (b) coaxial conductors

c­onfigurations have numerous applications in telecommunications, including radio, television, and computer networks. The electric field passing through a given surface is called electric flux, Ψ, and the flux per unit area is called the electric flux density, D. (Ψ is the Greek uppercase letter psi.) The greater the charge producing the electric field, the greater the flux. For example, doubling the charge in Figure 12-1(a) doubles the flux through a plane beside the charge. Similarly, increasing the charge by a factor of 10 increases the flux by the same factor. Thus, we can define electric flux and charge as equivalent:

ψ≡Q

(12-2)

If an electric field is uniform over a given area, A, then the flux density for that area is simply

D=

ψ Q = A A

(12-3)

In SI, the units for electric flux are coulombs and the units for flux density are coulombs per square metre. When two parallel conducting plates are connected to a voltage source, as shown in Figure 12-3, electrons flow into the plate connected to the negative terminal and an equal number of electrons flow out of the plate connected to  the positive terminal. On each plate, the force of repulsion between like charges causes the charge to be distributed evenly, except at the edges. As a result, the electric field between the parallel conducting plates is uniform. Therefore, a constant force is exerted on a charged particle as it moves from one plate to the other. The decrease in repulsive force from the charge on the plate that the particle leaves is balanced by the increase in attractive force from the charge on the plate that the particle is approaching.

One of Maxwell’s equations, also known as Gauss’s Law, describes how an electric field behaves around an electric charge: If an electric charge exists at a point in space, as in Figure 12-1, then the measure of the electric flux entering or exiting a surface surrounding that point is non-zero.

Because of the attraction between unlike charges, the charges are concentrated at the surface of each plate facing the other plate. Therefore, we speak of the charge on the plate rather than the charge in the plate.

326

Chapter 12  Capacitance

Uniform voltage gradient

+

−

+

−

+

−

+

−

Figure 12-3  Electric field between parallel plates

The work done on the charge when it moves from one plate to the other is W = Fd

(2-3)



W = QV

(2-4)

Therefore,

Fd = QV



This work can also be expressed in terms of the potential difference between the plates

Substituting for F from Equation 12-1 gives EQd = QV and

E=

V d

(12-4)

where E is the magnitude of the electric field strength in newtons per coulomb or volts per metre, V is the potential difference between the parallel plates in volts, and d is the distance between the plates in metres. This expression for electric field intensity leads to an interesting characteristic of electric fields. According to Kirchhoff’s voltage law, when we apply 400 V to the two parallel plates of Figure 12-3, there must be a 400-V potential difference between them. Since the electric lines of force begin at the positive plate and end at the negative plate, this potential difference appears across each line of force. Because the electric field between parallel plates is uniform, Equation 12-4 shows that the voltage gradient between the plates is also uniform. In other words, the potential difference is distributed evenly along the length of the electric line of force, somewhat like the way voltage is distributed along a linear potentiometer. Later in this chapter we shall see how the behaviour of charge on parallel conducting plates is a key to understanding capacitance. See Problems 12-1 to 12-6 and Review Questions 12-27 to 12-30.

12-2  Dielectrics

12-2 Dielectrics Figure 12-4 shows an electrical insulator, or dielectric, placed between the parallel charged plates. Although tightly bound to the molecules in the ­dielectric, the electrons orbiting each nucleus are attracted by the positive plate and repelled by the negative plate. As a result, the orbits of the electrons in each atom are displaced toward the positive plate and are no longer centred on the nucleus. Thus, the atoms in the dielectric become polarized. Nucleus

+ −

Electron orbit

+

−

+

−

+

−

+

−

+

−

+

−

Dielectric

Figure 12-4  Effect of an electric field on a dielectric (The displaced orbit is not drawn to scale.)

The extent of the polarization depends on the strength of the electric field. If we keep increasing the potential difference between the two parallel plates, we increase the electric field strength to the point where the ­resulting force tears the outer electrons free of their orbits, causing the ­dielectric to break down. The dielectric then becomes conductive and short-circuits the plates. The field strength required to break down a dielectric is called its d ­ ielectric strength. Some dielectrics are able to withstand a much stronger electric field than others. Table 12-1 lists typical values for the dielectric strength of some common dielectrics. The dielectric strength of a material sometimes depends on the manufacturing process and can vary considerably. After exposing a dielectric to a strong electric field between two parallel plates, we could remove the voltage source and temporarily short the plates together. We would expect to completely neutralize the charge on the plates. However, with some dielectrics there is still a small potential difference between the two plates, indicating that the electron orbits in the dielectric did not return fully to their original positions. The dielectric ­remains slightly polarized even though there is no electric field acting on it. This effect is called dielectric absorption.

327

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Chapter 12  Capacitance

TABLE 12-1  Dielectric strength Typical Dielectric Strength (kV/mm)

Material Air

3

Barium-strontium titanate

3

Porcelain

8

Transformer oil

16

Bakelite

16

Polystyrene

16

Paper

20

Mylar

24

Rubber

28

Teflon

60

Glass

120

Mica

200

See Problems 12-7 and 12-8 and Review Questions 12-31 to 12-34.

12-3 Capacitance The symbol at the right side of Figure 12-5 represents a capacitor that has two conducting plates separated by an insulator. Since the current is the same in all parts of a series circuit, no charge can build up on either plate of the capacitor until the switch is closed. When the plates have no charge, there is no electric field and no potential difference between them. When we close the switch, the galvanometer shows a surge of current with electrons flowing counterclockwise around the circuit. Since electrons cannot pass through the insulation between the plates of the capacitor, a negative charge builds up on the bottom plate and a positive charge builds up on the top plate. As time passes with the switch closed and these charges build up, the electric field between the plates gets stronger and the potential difference between them increases. According to Kirchhoff’s voltage law, the potential difference between the two plates cannot be greater than the potential difference of the source. The galvanometer reads zero Electrons flow counterclockwise as capacitor charges

+

E

V

−

+ −

G Figure 12-5  Charging a capacitor

12-3  Capacitance

329

as soon as enough charge has built up on the c­ apacitor to make its voltage drop equal to the applied voltage. If we suddenly double the applied voltage after the capacitor has finished charging, there will be another surge of current until the capacitor plates charge sufficiently to raise the potential difference between them to equal the increased applied voltage. There is a fixed ratio between the potential difference between any given pair of insulated conducting plates and the charge required to establish this potential difference:

Q = a constant V

(12-5)

If we now open the switch in Figure 12-5, there is no longer any conductive path between the two plates, and the surplus electrons on the bottom plate cannot flow to the top plate. Thus, the parallel plates store electric charge when a potential difference is applied between the plates. Since Q/V is a constant, the stored charge produces a potential difference between the plates. A capacitor is a component that can store electric charge. To determine just how much charge a capacitor can store when it is ­connected to a given voltage source, we must determine the factors that govern the value of the constant in Equation 12-5. This constant, which ­represents the ability of a capacitor to store an electric charge, is called its capacitance. The letter symbol for capacitance is C. The SI unit of capacitance is the farad, which is equal to one coulomb per volt. The unit symbol for farad is F. A circuit has a capacitance of 1 F when a charge of 1 C raises the potential difference by 1 V: 1 F = 1 C/V.

C=

Q V

(12-6)

where C is the capacitance in farads, Q is the charge in coulombs, and V is the potential difference in volts. One farad is a much greater capacitance than appears in most circuits. Consequently, capacitances are often expressed in microfarads, nanofarads, or even picofarads. However, when substituting for C in equations, we must remember that these equations were set up in terms of the basic unit of capacitance, the farad.

An older term for ­capacitance is capacity. However, the ability of a capacitor to store a charge at a certain potential difference is sometimes referred to as its capacity. In present-day usage, the term capacitance (which rhymes with resistance and inductance) has replaced the older form.

330

The farad is named in honour of Michael Faraday.

Chapter 12  Capacitance

Note that in defining the farad we spoke of the capacitance of a circuit. Capacitors are not the only type of component that has capacitance, although capacitors generally have much more capacitance for a given size. In fact, any pair of conductors that are insulated from each other have some capacitance. For example, an open-circuit transmission line, like the one in ­Figure 12-6, can have significant capacitance. + +

+

+

+

Transmission line

E

− −

−

−

−

G Figure 12-6  Capacitance between parallel conductors

We have already noted that a capacitor cannot charge instantly. It takes time, even if only a fraction of a second, for the charge to build up. Consequently, the full potential difference across a capacitor does not appear at the instant the switch is closed. This potential difference remains until there is a conducting path for electrons to flow from the negative plate to the positive plate. Since time is involved in any change in the p ­ otential difference between the plates of a capacitor, we can think of the capacitance as opposing any change in the potential difference across the capacitor. Capacitance is that property of a circuit that opposes any change in the voltage across the circuit. Note how this definition somewhat parallels the definition of resistance as that property of a circuit that opposes electric current through the circuit. See Problems 12-9 and 12-10 and Review Questions 12-35 to 12-38.

Circuit Check

A

CC 12-1. Calculate the force of attraction that is exerted on an electron located 0.03 mm from a point charge of 5 mC. CC 12-2. How much capacitance is present if a voltage of 50 V produces a charge of 11 µC? CC 12-3. How much voltage will produce a charge of 4 µC on a 50-nF capacitor?

12-4 Capacitors Parallel-plate capacitors can be constructed by supporting two metal plates with air between them, or by coating the two sides of a ceramic disk with metal, as in Figure 12-7(a). Some capacitors consist of interleaved parallel

12-4  Capacitors

331

Plastic film

Metal

Ceramic

Aluminum foil (a)

(b)

(c)

Figure 12-7  Construction of capacitors

Capacitors are sometimes referred to by the now obsolete term condenser.

Sources:  © sciencephotos/Alamy Stock Photo

Sources:  GIPHOTOSTOCK/SCIENCE PHOTO LIBRARY

plates with either air or mica dielectric, as in Figure 12-7(b). Others consist of two long strips of aluminum foil interleaved with strips of a dielectric and rolled up, as shown in Figure 12-7(c). Originally, waxed paper was used as the dielectric for such tubular capacitors. Modern tubular capacitors use thin films of a plastic such as polyethylene. Capacitors constructed by the methods shown in Figure 12-7 range in size from smaller than the head of a match to larger than a desktop computer. These types of capacitors usually have capacitances of less than 1 μF. An ingenious means of obtaining large capacitances in a compact package employs the electrolysis principle described in Chapter 3. Electrolytic ­capacitors have a tubular construction like that in Figure 12-7(c) but with thin gauze soaked in an aluminum hydroxide solution instead of the plastic dielectric. When first assembled, an electrolytic capacitor acts like a short circuit since the aluminum hydroxide electrolyte dissociates into positive and negative ions that act as charge carriers. However, passing a direct current through the electrolyte attracts the negative hydroxide ions to the

Capacitors: (top row, l–r) polypropylene plastic film, polyester film, ceramic, mica; (bottom row, l–r) silver mica, metallized polyester film, tantalum electrolytic, aluminum electrolytic

One set of the interleaved vanes can be rotated to adjust the capacitance of this ­air-dielectric tuning capacitor.

332

Chapter 12  Capacitance

Figure 12-8 Schematic symbols for capacitors

positive foil, where they react with the aluminum to form a very thin layer of insulating aluminum oxide. We now have a capacitor where one plate is the positive aluminum foil, the other is the electrolyte in contact with the negative foil, and the dielectric is the thin oxide layer. Electrolytic capacitors with capacitances from 1 to 5000 μF are readily available. Electrolytic capacitors must always be connected so that the polarity of the potential difference between their terminals is the same as the polarity of the forming voltage. Reversing the polarity of the operating voltage r­ everses the electrolytic action, which can then short out the capacitor by r­emoving the oxide film. The terminals of electrolytic capacitors have + and − signs or colour-coded wires to indicate the correct polarity. This ­polarity restriction greatly limits the use of electrolytic capacitors in alternating-­current circuits. However, two electrolytic capacitors connected in series opposing can be used as a starting capacitor for AC motors. The thinness of the oxide film enables electrolytic capacitors to have a  large capacitance compared to other types of capacitor of similar size. However, there is always some leakage of current through the dielectric in an electrolytic capacitor. The thin dielectric also makes the breakdown voltage of electrolytic capacitors relatively low. Capacitors are rated with a working voltage, which is the maximum DC voltage that can be applied continuously without breaking down the dielectric. In Figure 12-8 we can see that the two symbols most commonly used to show capacitors on circuit diagrams are both stylized representations of the plates in a capacitor. The version with parallel plates is an international standard, but North America generally uses the version with one curved plate. For polarized capacitors, the curved line indicates the negative plate. See Review Question 12-39.

Sources:  GIPHOTOSTOCK/SCIENCE PHOTO LIBRARY

Electrolytic capacitors

12-5   Factors Governing Capacitance

Practical Circuits Capacitors As you will see in later chapters, capacitors have a variety of applications in electronics and electrical engineering. Capacitors can store energy and pass alternating current while blocking direct current. Here are some common applications that make use of these properties.

Power Filtering Most power supplies for electronic devices use diodes to convert AC into a pulsating DC. Capacitors can smooth out these DC pulses by storing energy as the voltage peaks and then releasing it as the voltage drops.

Sources:  DAVID HAY JONES/SCIENCE PHOTO LIBRARY

Circuit Coupling Capacitors are often used to link two circuits, such as the stages of an amplifier. A coupling capacitor allows AC signals to pass from one circuit to the other but prevents DC from flowing between the two circuits. With this type of coupling, the two circuits can be powered by different DC voltages.

Bypassing When a capacitor is connected in parallel with a resistor or other component, AC will flow through the capacitor quite readily, effectively bypassing the other component. Thus, such bypass capacitors can provide a path for AC signals without significantly changing the DC voltage across the resistor or other components.

The blue and orange structure is a capacitor array at Fermilab, the US National Accelerator Laboratory at Batavia, Illinois.

12-5  Factors Governing Capacitance We can consider all the capacitors shown in Figure 12-7 to be parallel-plate capacitors. In Section 12-1, we noted that the strength of the electric field between two parallel plates depends on the quantity of charge per unit

333

334

Chapter 12  Capacitance

area on the surface of the plates. If we double the area of the plates, it takes twice as much charge to produce an electric field of a given strength. From Equation 12-4, V = Ed, so the enlarged plates hold twice as much charge at a given potential difference. Since C = Q/V (Equation 12-6), doubling the area of the plates doubles their capacitance. We can apply the same logic to any change in the area of the plates. The capacitance of parallel plates is directly proportional to their area. The positively charged plate in a capacitor attracts free electrons to the closer surface of the negatively charged plate. This attraction aids the applied voltage in transferring electrons to the negative plate. Similarly, the negatively charged plate helps to drive electrons from the positively charged plate into the positive terminal of the source. Moving the plates closer together increases these forces of attraction and repulsion, enabling the capacitor to store a greater charge for a given voltage source. The capacitance of parallel plates is inversely proportional to their spacing. Suppose we leave the switch closed after the charging current has ceased flowing in the circuit of Figure 12-5. If we then slide a sheet of glass or mica between the plates of the capacitor, the galvanometer shows an additional surge of charging current. Since the charge on the plates increased but the potential difference did not change, the capacitance must have ­increased. This simple experiment demonstrates that The capacitance of parallel plates depends on the type of dielectric between plates. To compare conductive materials, we defined the conductivity of a material as the conductance between parallel faces of a one-metre cube of the material (refer to the sections on resistivity [5-4] and conductivity [7-11] for details). We can define a similar quantity for comparing dielectric m ­ aterials. The permittivity (or absolute permittivity) of a material is the capacitance between opposite faces of a unit length and cross section of the material. The letter symbol for permittivity is the Greek letter ε (lowercase epsilon). The units for permittivity are farads per metre. Permittivity is the ability of a material to permit an electric field within it. Permittivity measures the extent to which polarization caused by an electric field reduces the strength of the field inside the material.

12-5   Factors Governing Capacitance

For parallel plates, C=ε



A d

(12-7)

where C is the capacitance in farads, ε is the permittivity in farads per metre, A is the area of each plate in square metres, and d is the distance between the plates in metres. Note the similarity between Equation 12-7 and the formula for the resistance of a conductor R=ρ

Just as

l A

(5-4)

Substituting for C = Q/V into Equation 12-7 and solving for ε gives ε=

Q Q d d × = × V A A V

For parallel plates, Equation 12-3 gives D = Q/A and inverting ­Equation 12-4 gives d/V = 1/E. Therefore, ε=



D E

(12-8)

where ε is the permittivity of a given dielectric in farads per metre, D is the electric flux density in coulombs per square metre, and E is the electric field strength in volts per metre. Parallel plates have capacitance even without a dielectric material between them. Applying a voltage to two parallel plates in a vacuum still produces an electric field between them. The ratio of electric flux density to electric field strength in a vacuum is a physical constant called the ­permittivity of free space, ε0



ε0 =

D = 8.85 × 10 − 12 F/m E

(12-9)

When we do place a dielectric between the plates of a capacitor, the electron orbits of atoms are forced off-centre, as shown in Figure 12-4. As a result, the negative charge of each atom of the dielectric is a little closer to the positive plate. This displacement has the same effect on the capacitance

335

336

Formulas for the ­capacitance of other configurations, such as parallel and ­concentric conductors, are given in electrical and electronics ­handbooks.

Chapter 12  Capacitance

as moving the parallel plates a little closer together. The permittivity of most solid and liquid dielectrics is considerably greater than that of free space, while the permittivity of air is very close to ε0. Therefore we can substitute ε0 into Equation 12-7 to find the capacitance of a parallel-plate air-dielectric capacitor

C = 8.85 × 10 −12

A d

(12-10)

where C is the capacitance in farads, A is the area of each plate in square metres, and d is the distance between the plates in metres. See Problems 12-11 and 12-12 and Review Questions 12-40 to 12-43.

12-6  Dielectric Constant Data tables for dielectrics commonly list the ratio between the permittivity of a given material and the permittivity of free space. This ratio is called the relative permittivity or dielectric constant of the material.

k or εr =

ε ε0

(12-11)

where k (or εr) is the dielectric constant of a material, ε is its permittivity, and ε0 is the permittivity of free space (8.85 × 10−12 F/m).

Therefore, the equation for the capacitance of any parallel-plate capacitor becomes

C=

k ε0 A d

(12-12)

where C is the capacitance in farads, k is the dielectric constant, A is the area of each plate in square metres, and d is the distance between the plates in metres. Like dielectric strength, the dielectric constant of a given type of material sometimes depends on the manufacturing process and can vary ­considerably. Table 12-2 gives typical values of dielectric constant for the dielectric materials listed in Table 12-1. Comparing Table 12-1 and 12-2, we note that there is no correlation between dielectric strength and dielectric constant. Many ­materials with a high dielectric constant cannot be used in a capacitor for high-voltage circuits because their dielectric strengths are too low. For example, distilled water has a dielectric constant of 80, but the dielectric strength of water is so low that it is not practical to use it as a capacitor d ­ ielectric.

12-6   Dielectric Constant

TABLE 12-2  Dielectric constants Typical Dielectric Constant

Material Air

1.0006

Barium-strontium titanate

7500

Porcelain

6

Transformer oil

4

Bakelite

7

Polystyrene

2.6

Paper

2.5

Mylar

3.0

Rubber

3

Teflon

2

Glass

6

Mica

5

Ceramics called ferroelectric dielectrics (such as barium-strontium t­ itanate) have molecules that are very easily polarized by an electric field. As a result, ferroelectric dielectrics have extremely high dielectric constants. The development of these ceramics made it possible to manufacture capacitors that are much more compact for a given capacitance and voltage rating. The term ferroelectric does not mean that these materials contain iron, but rather that they permit electric fields in a similar way to how iron permits magnetic fields. (Ferromagnetic materials are described in Chapter 14.) The dielectric constant of ferroelectric materials varies slightly with changes in the strength of the electric field.

Example 12-1 Use Tables 12-1 and 12-2 to calculate the capacitance and the breakdown voltage for a capacitor made by plating silver on each side of a barium-strontium titanate disk 1.0 cm in diameter and 0.20 mm thick. Solution From Table 12-2, the dielectric constant, k, is about 7500.

C=

kε0A 7500 × 8.85 × 10−12 F/m × π ( 5.0 mm ) 2 = d 0.20 mm = 2.6 × 10−8 F = 26 nF

From Table 12-1, the dielectric strength of barium-strontium titanate is about 3 kV/mm. Therefore, the breakdown voltage is approximately 3 kV/mm × 0.20 mm = 0.6 kV. See Problems 12-13 to 12-16 and Review Questions 12-44 and 12-45.

337

338

Chapter 12  Capacitance

12-7  Capacitors in Parallel + E

C1

+ −

C2

+ −

−

Figure 12-9 Capacitors in parallel

Connecting capacitors in parallel is like increasing the area of the plates of a single capacitor. Therefore, the total capacitance is greater than that of any individual one. Since both capacitors in Figure 12-9 are connected to the same battery, the total charge drawn from the battery is QT = Q1 + Q2



(12-13)

Since Figure 12-9 is a simple parallel circuit,

E = V1 = V2



QT Q2 Q1 + = E V1 V2



CT = C1 + C2

and Therefore,

The total capacitance of capacitors connected in parallel is the sum of all the individual capacitances. CT = C1 + C2 + C3 + . . .





(12-14)

Example 12-2

What single capacitance can be used to replace a 10-nF and a 0.05- μF ­capacitor connected in parallel? Solution CT = C1 + C2 = l0 nF + 50 nF = 60 nF See Problems 12-17 and 12-18.

12-8  Capacitors in Series When we connect capacitors in series, as shown in Figure 12-10, the plates connected directly to each other are charged by electrostatic ­induction. The positive charge on the top plate of C1 attracts free electrons from the top plate of C2 onto the bottom plate of C1. Similarly, the negative charge on the bottom plate of C2 repels electrons from the top plate of C2 onto the bottom plate of C1. The effect is much the same as increasing the spacing between

12-8   Capacitors in Series

C1

+ E

C2

−

+ − + −

Figure 12-10  Capacitors in series

the plates of a single capacitor. The capacitance of the s­ eries combination is less than that of either capacitor, but the combination can withstand a higher ­potential difference than either capacitor can by ­itself. Since the capacitors and the voltage source are a simple series circuit, E = V1 + V2 Since the current is the same in all parts of a simple series circuit, QT = Q1 = Q2



E V1 V2 = + QT Q1 Q2

1 1 1 = + Ceq C1 C2

or

Ceq =



1 C1C2 = 1 1 C1+ C2 + C1 C2

Therefore, The equivalent capacitance of series capacitors is Ceq =



1 1/C1 + 1/C2 + 1/C3 + . . .



(12-15)

Example 12-3 What single capacitance can be used to replace a 0.01-μF and a 0.05-μF ­capacitor connected in series? Solution Ceq =

( 10 × 10−9 ) × ( 50 × 10−9 ) C1C2 = = 8.3 × 10−9 F = 8.3 nF ( 10 + 50 ) × 10−9 C1 + C2

339

340

Chapter 12  Capacitance

Equation 12-15 has the same form as the equation for the equivalent r­esistance of resistors connected in parallel. To simplify calculations for parallel resistors, we considered conductance, which is the reciprocal of ­resistance. Similarly, for capacitors in series, we can use elastance, the reciprocal of capacitance.

The reciprocal of a farad is sometimes called a daraf ( farad spelled backwards). This term is not ­approved by SI.

Elastance is the opposition of material to the setting up of electric lines of force in an electric insulator or dielectric. The letter symbol for elastance is S. The units for elastance are reciprocal farads (F−1). S=



1 (12-16) C

where S is elastance in reciprocal farads and C is capacitance in farads. With elastance, Equation 12-15 for capacitors in series becomes

ST = S1 + S2 + S3 + . . . (12-17)



Since Q1 = Q2 in a series circuit and Q = CV,

Q1 = Q2 = C1V1 = C2V2 and

V1 C2 = (12-18) V2 C1

When capacitors are connected in series, the ratio between any two ­potential differences across the capacitors is the inverse of the ratio of their capacitances.

Example 12-4 Compare the following quantities for a 10-nF capacitor and a 40-nF capacitor connected in parallel to a 500-V source and the same capacitors connected in series to the source: (a) the total capacitance (b) the total charge transferred by the source (c) the charge on each capacitor and the potential difference across each capacitor Solution (a) Parallel: CT = C1 + C2 = 10 nF + 40 nF = 50 nF Series: Ceq =

C1C2 10 nF × 40 nF = = 8 nF C1 + C2 10 nF + 40 nF

12-8   Capacitors in Series

(b) Parallel: QT = CTVT = 50 nF × 500 V = 25 μC Series: QT = CeqVT = 8.0 nF × 500 V = 4.0 μC (c) Parallel: V1 = V2 = E = 500 V Q1 = C1V1 = 10 nF × 500 V = 5.0 μC Q2 = C2V2 = 40 nF × 500 V = 20 μC Series: Q1 = Q2 = QT = 4.0 μC

V1 =

V2 =

4.0 μC Q1 = = 400 V C1 10 nf 4.0 μC Q2 = = 100 V C2 40 nF

See Problems 12-19 to 12-26 and Review Questions 12-46 to 12-48.

Circuit Check

B

CC 12-4. What spacing between two 3 cm × 5 cm parallel plates will produce a capacitance of 47 pF if air is the only material between the plates? Find the maximum working voltage of this capacitor. CC 12-5. (a) Calculate the total capacitance of a 1.0-µF capacitor, a 250-nF capacitor, and a 0.50-µF capacitor connected in parallel. (b) Calculate the charge on each capacitor when the voltage across the combination is 50 V. CC 12-6. A 50-µF capacitor and a 100-µF capacitor are connected in series to a 150 V source. Calculate the total capacitance and the voltage drop across each capacitor.

341

342

Chapter 12  Capacitance

Summary

• An electric field is a region in which an electric charge is acted upon by an electric force. • The strength of an electric field at a certain point is the electric force ­acting on a charge at that point divided by the quantity of the charge. • The flux density of an electric field is the electric flux per unit area. • The electric field between parallel conductive plates is uniform except near the edges. • Dielectric strength indicates the electric field strength that a dielectric can withstand. • Capacitance opposes change in voltage and can store charge. • The maximum voltage that a capacitor can withstand is expressed as its working voltage. • Capacitance is the ratio of the charge on the plates of a capacitor to the voltage across it. • Capacitance is proportional to the area of the plates of a capacitor and inversely proportional to the spacing of the plates. • Capacitance is dependent on the type of dielectric in a capacitor. • Permittivity is the capacitance of a unit length and cross section of a ­material. • The total capacitance of capacitors connected in parallel is the sum of all the individual capacitances. • The equivalent capacitance of capacitors connected in series is the reciprocal of the sum of the reciprocals of the individual capacitances. • Elastance is the reciprocal of capacitance. B = beginner

I = intermediate

A = advanced

Problems B I I I B B

B B

Section 12-1  Electric Fields

Calculate the force of attraction that acts on a single electron located 0.5 cm from a point charge of +3 μC. 12-2. Two electric charges, one of which is four times as large as the other, exert a mutual force of repulsion of 24 N when they are 15 cm apart. Determine the magnitude of each charge. 12-3. Two parallel plates 1.0 cm apart are connected to a 250-V source. Find the force exerted on a free electron between the plates. 12-4. If 5 × 109 electrons are removed from one parallel plate and added to the other, what is the total electric flux between the plates? 12-5. Determine the electric flux density between two parallel plates measuring 25 cm by 50 cm and carrying an electric charge of 600 nC. 12-6. Calculate the electric field produced by a capacitor when 500 V is connected across plates with a spacing of 8 mm. 12-1.

Section 12-2  Dielectrics 12-7.

What is the maximum voltage that can be applied to a parallel plate capacitor if 0.4 mm of polystyrene is used as the dielectric? 12-8. Find the minimum thickness of rubber required to withstand an applied voltage of 21 kV.

Problems

B B

B B

B I

B I

B A

Section 12-3  Capacitance

12-9. If moving 3.0 × 1010 electrons from one parallel plate to the other produces a potential difference between the plates of 220 V, what is the capacitance of the plates? 12-10. How many electrons must be removed from one plate of a 270-pF capacitor and added to the other to raise the voltage between the plates to 420 V?

Section 12-5  Factors Governing Capacitance 12-11. A neutralizing capacitor in a radio transmitter consists of two ­aluminum disks, each 10 cm in diameter and 0.50 cm apart with air between them. What is its capacitance? 12-12. If the electric flux density between the parallel plates of the capacitor in Problem 12-11 is 1.0 μC/m2, what is the voltage between the plates?

Section 12-6  Dielectric Constant

12-13. Find the capacitance of a parallel-plate capacitor that consists of eleven 2 cm × 0.5 cm sheets of aluminum foil interleaved with ten sheets of mica 0.02 mm thick and connected as shown in Figure 12-7(b). 12-14. Two sheets of aluminum foil 2.5 cm wide and 1.0 m long and two sheets of 0.1-mm-thick polystyrene 3.0 cm wide and 1.0 m long are rolled as shown in Figure 12-7(c) to form a tubular capacitor. The polystyrene has a dielectric strength of 16 kV/mm and a dielectric constant of 3. ­Calculate the capacitance and voltage rating of this c­ apacitor. 12-15. Calculate the capacitance of a parallel plate capacitor with plates that are 6 cm × 3 cm, separated by 0.9 mm of polystyrene. 12-16. Two sheets of aluminum 4 cm × 10 m long are rolled up with two sheets of 0.5-mm mylar separating them. Calculate the capacitance and maximum working voltage.

Section 12-7  Capacitors in Parallel 12-17. Find the total capacitance and the charge stored by each capacitor when a 12-μF and a 68-μF capacitor are connected in parallel to a 250-V source. 12-18. Initially, the capacitors in Figure 12-11 are completely discharged. The switch is first moved to connect its left terminal. When the charging current ceases, the switch is moved to connect its right terminal. When the current ceases again, the switch is moved back to its left position, and the switching cycle is repeated. Sketch a graph of the voltage across the 10-μF capacitor as the switch is operated for five complete cycles. Label the exact voltage after each cycle of switch operation. Assume no leakage in either capacitor.

+ 300 V −



Figure 12-11

1 μF

10 μF

343

344

Chapter 12  Capacitance

I

I

B I

Section 12-8  Capacitors in Series 12-19. A 15-μF, a 22-μF, and a 47-μF capacitor are connected in series to a 240-V source. (a) Calculate the equivalent capacitance. (b) Calculate the total charge stored by this capacitor network. (c) Calculate the voltage across each capacitor. 12-20. After charging, the three capacitors in Problem 12-19 are disconnected from the source and from each other. They are then connected in parallel, with the three positive plates connected to each other. Find the voltage across each capacitor. 12-21. Two aluminum disks, each 6 cm in diameter, have a sheet of glass 0.3 cm thick separating them. Determine the capacitance. 12-22. Find the equivalent capacitance and the voltage across each of the capacitors in the circuit of Figure 12-12.

5 μF

+ 240 V −

I

8 μF 20 μF

Figure 12-12

12-23. Find the equivalent capacitance and the voltage across each of the capacitors in the circuit of Figure 12-13. 20 nF 30 V

I

I

Figure 12-13

330 pF

470 pF

12-24. Three capacitors have capacitances of 0.01 μF, 0.02 μF, and 0.05 μF, respectively, and a rated working voltage of 400 V for each capacitor. Find the highest voltage that can safely be applied to the series combination of the three capacitors. 12-25. Find the total elastance of the three capacitors of Problem 12-24 when connected (a) in series (b) in parallel

Review Questions

I

12-26. For the circuit shown in Figure 12-14, calculate the total capacitance and voltage across each capacitor. 20 μF

+ 100 V −



40 μF 10 μF

12 μF

Figure 12-14

Review Questions

Section 12-1  Electric Fields 12-27. Coulomb’s law of electrostatic force states that electric force is proportional to the product of two electric charges. Yet Figure 12-1 shows an electric field for a single isolated electric charge. Explain the purpose of sketches such as those in Figure 12-1. 12-28. Why does a potential difference exist between two electric conductors possessing unlike charges? 12-29. What is the significance of the arrows on the electric lines of force ­between the two parallel plates in Figure 12-3? 12-30. What characteristic of an electric field is referred to by the term electric field strength?

Section 12-2  Dielectrics 12-31. What is meant by a polarized atom? 12-32. What type of atoms can become polarized? 12-33. Why is the strength of a dielectric expressed in terms of the k ­ ilovolts per millimetre rather than the total potential difference? 12-34. Define dielectric absorption.

Section 12-3  Capacitance 12-35. Why must the ratio between the charge on a given pair of insulated conductors and the potential difference between them be a constant? 12-36. Define capacitance in terms of charge and voltage. 12-37. Why can we define capacitance as that property of an electric ­circuit that opposes any change in voltage across that circuit? 12-38. Define the farad in terms of flux density and strength of the electric field in the dielectric between the plates of a capacitor.

Section 12-4  Capacitors 12-39. Explain why a motor-starting capacitor consisting of two electrolytic capacitors connected in series opposing can be used safely in an AC circuit.

345

346

Chapter 12  Capacitance

Section 12-5  Factors Governing Capacitance 12-40. Why does the spacing between two parallel plates have an effect on their capacitance? 12-41. Why does the voltage applied to two parallel plates not affect their capacitance? 12-42. Why is it possible to have capacitance between two parallel plates when the space between them is evacuated? 12-43. Why does the presence of an insulating material between parallel plates increase their capacitance compared with that of the same plates in a vacuum?

Section 12-6  Dielectric Constant 12-44. Why is it possible to define the dielectric constant of a given insulating material as the ratio between the capacitance of a pair of parallel plates with that material as a dielectric and the same parallel plates with air as a dielectric? 12-45. Distinguish between the (absolute) permittivity and the relative permittivity of a given dielectric.

Section 12-8  Capacitors in Series 12-46. When two capacitors are connected in series, one plate of each has no conductive path to any voltage source. How is it possible for these capacitors to become charged? 12-47. Why must capacitors in series all store the same charge? 12-48. A 0.1-μF 400-V capacitor and a 0.05-μF 200-V capacitor are c­ onnected in series. What is the maximum voltage that can be safely applied to this combination?

Integrate the Concepts

Three capacitors, C1, C2, and C3, appear in a network such that C1 and C2 are in parallel and that combination is in series with C3. • C1 is a TeflonTM capacitor with square plates whose sides measure 1 cm and are 0.05 mm apart. • C2 is a ceramic-disc capacitor whose plates have a diameter of 1 cm and are 0.1 mm apart. • C3 is a mica capacitor with rectangular plates that measure 1 cm × 0.5 cm and that are 0.025 mm apart. Determine the total capacitance of the series-parallel combination.

Practice Quiz

1. Why is electric field strength described as a vector quantity? 2. What does the concentration of lines of electric force on a diagram represent?

Practice Quiz

3. Which of the following statements are true? (a) An electric field is directly proportional to the difference of potential between two points and inversely proportional to the distance between those two points. (b) A capacitor blocks alternating current and passes direct current. (c) The larger the capacitance, the more charge the capacitor can store. 4.

How are the permittivity and the dielectric constant of a material related?

5. The capacitance of parallel plates is (a) inversely proportional to their area and the distance between the plates (b) directly proportional to their area and the distance between the plates (c) directly proportional to their area and inversely proportional to the distance between the plates (d) inversely proportional to their area and directly proportional to the distance between the plates 6. The equivalent capacitance for the circuit of Figure 12-15 is (a) 46.87 nF (b) 5.87 nF (c) 30 nF (d) 72 nF

C1 10 nF

C2 22 nF

C3 40 nF

Figure 12-15

7. The equivalent capacitance for the circuit of Figure 12-16 is (a) 4.07 nF (b) 2.45 nF (c) 27 nF (d) 42 nF C1 10 nF

C2 10 nF

C3 22 nF

Figure 12-16

347

13

Capacitance in DC Circuits Now that we know something about capacitance as a basic property of electric circuits, we can examine the behaviour of practical electric circuits containing capacitance, either ­intentionally from capacitors, or as a byproduct of other components such as parallel traces on a printed circuit board.

Chapter Outline 13-1

Charging a Capacitor  350

13-2 Rate of Change of Voltage  352 13-3 Time Constant  354

13-4 Graphical Solution for Capacitor Voltage  356 13-5 Discharging a Capacitor  357

13-6 Algebraic Solution for Capacitor Voltage  362 13-7 Transient Response  366

13-8 Energy Stored by a Capacitor  370

13-9 Characteristics of Capacitive DC Circuits  372 13-10 Troubleshooting  375

Key Terms steady-state value  350 instantaneous value  351

time constant  354 transient response  368

transient phase  368 stray capacitance  374

Learning Outcomes At the conclusion of this chapter, you will be able to: • calculate the initial current through a capacitor when it is charging or discharging • calculate the initial rate of change of v ­ oltage across a ­capacitor when it is charging or ­discharging • calculate the time constant of a CR circuit • analyze a CR circuit by using universal ­exponential curves • calculate the capacitor voltage for any instant of time by using the charging and discharging equations

Photo sources:  Tomáš Hašlar/Alamy Stock Photo

• compute the transient response of the capacitor voltage in a CR circuit when the capacitor has an initial voltage across it • explain how energy can be stored by a capacitor • calculate the energy stored by a charged ­capacitor • describe applications of capacitors that use their energy-storage capabilities • describe the undesirable effects of stray capacitance on a circuit • troubleshoot a capacitor for defects

350

Chapter 13   Capacitance in DC Circuits

13-1  Charging a Capacitor We can think of charging a capacitor in terms of the quantity of charge transferred to its plates. However, Q = CV and C is a constant for a given capacitor, so the charge is directly proportional to the potential difference b ­ etween the plates of the capacitor. Since we can easily measure this ­potential difference with a voltmeter, it is more convenient to speak of the charging of a c­ apacitor in terms of the potential difference built up between its plates. At the instant we close the switch in the circuit of Figure 13-1, no e­ lectrons have had enough time to flow into the bottom plate and out of the top plate, so no potential difference has yet built up between the plates. Kirchhoff’s voltage law states that there must be a voltage drop in the c­ ircuit equal to the applied voltage. Since there is no potential difference across the capacitor, there must be an IR drop in the circuit. Although no r­esistance is shown, we shall assume that the resistance of the wiring and the internal resistance of the voltage source battery total 0.5 Ω. Therefore, at the instant we close the switch, the current is I0 =

+ 100 V E −

E 100 V = = 200 A R 0.5 Ω

v

1.0 μF

Figure 13-1  Charging a capacitor without a current-limiting resistor

A 200-A current charges the plates of the capacitor very rapidly, and the potential difference across the capacitor rises to 100 V in a few microseconds. When the potential difference across the capacitor equals the applied voltage, current ceases. Therefore, if we charge a capacitor by connecting it directly across a voltage source, there is a very short-duration pulse of very high current. Since such pulses can damage a source, we usually charge ­capacitors through a resistance that limits the initial current surge. When analyzing circuits that contain capacitors, we have to deal with voltages and currents that vary with time. It is conventional practice to use lowercase letters to represent quantities that vary with time. The resistance and capacitance of most circuits and components depend on physical ­factors (such as length, area, and material type) that are independent of time. Consequently, we represent resistance and capacitance with the ­uppercase letters R and C. In the preceding chapters, the source voltage and the resulting currents, voltage drops, and power were also independent of time. ­Accordingly, we represented all these steady-state values with ­uppercase letters. In a circuit like the one in Figure 13-2, the charging ­current and the

13-1   Charging a Capacitor

351

i

+ 100 V E −

vR vC

+ − +

2.0 MΩ 1.0 μF

−

Figure 13-2  Charging a capacitor through a current-limiting resistor

potential difference across the capacitor change from instant to instant and depend on how long the switch has been closed. Therefore, we represent the instantaneous values of the voltage and current with the lowercase letters i and v. Note that we represent time itself by lowercase t. Lowercase letter symbols represent instantaneous values that are dependent on time. Uppercase letter symbols represent steady-state values that are not dependent on time. As long as the switch in Figure 13-2 is closed, the sum of the potential difference across the capacitor and the IR drop across the resistor must equal the applied voltage. Therefore,

E = vC + iR

(13-1)

At the instant we close the switch, there is no potential difference across the capacitor. Hence, the initial current, I0, must make the IR drop in the ­circuit equal to the applied voltage:

I0 =

E R

(13-2)

The initial current in the circuit of Figure 13-2 is only 100 V/2.0 MΩ = 50 μA. This current transfers charge to the capacitor. Thus, a potential difference starts to build up across the capacitor. To satisfy Kirchhoff’s voltage law, the IR drop across the resistor must decrease. The instantaneous ­current decreases correspondingly, and the potential difference across the capacitor rises more slowly. Therefore, the rate of rise of the instantaneous voltage across the capacitor depends on the magnitude of the instantaneous current, which depends on the IR drop across the resistor, which, in turn, depends on the instantaneous voltage across the capacitor, which ­depends on how long the switch has been closed. Before we can sort out these interdependences, we must consider how we can express rate of change of voltage. See Review Question 13-27 at the end of the chapter.

A circuit like the one in Figure 13-2 is often referred to as a CR ­circuit since the only components other than the source and the switch are a ­resistor and a ­capacitor.

Chapter 13   Capacitance in DC Circuits

13-2  Rate of Change of Voltage

4 2 0

1 Time (seconds)

2

C it rcu

6

8 6

ΔV

Ci

rcu

Output Voltage (volts)

it A

10

Ci

8

Circ

10

uit B

Figure 13-3 shows the output potential difference for three circuits that are switched on at t = 0. In Figure 13-3(a), the potential difference increases as a linear function of time rather than immediately jumping to a steady-state value, as in a simple resistance network. We can read the rate of change of voltage directly from the straight-line graphs by noting the increase in ­potential difference during 1 s. For circuit A, the voltage rises from 0 to 5 V in the first second and from 5 V to 10 V during the next second. For both of these intervals, the rate of change of voltage is 5 V/s. Similarly, the rate of change of voltage in circuit B is 10 V/s. Note that the graph for circuit B is twice as steep as the graph for circuit A. The slope of a graph of voltage ­versus time represents the rate of change of the voltage.

Output Voltage (volts)

352

4 Δt

2 0

(a)

t1

1 2 Time (seconds)

3

(b)

Figure 13-3  Graphical determination of rate of change of voltage

In circuit C the output potential difference is a function of the square of the elapsed time. Since this graph is not a straight line, the rate of change of voltage is not the same for all time intervals. By drawing a tangent to the graph at time t1 in Figure 13-3(b), we can read the rate of change of voltage at that instant. Since the tangent is a straight line, we simply note the change in voltage, ΔV, over any convenient interval, Δt. Hence, the rate of change of voltage at time t1 is ΔV/Δt. To represent the change in voltage at any given instant, we use calculus notation. The symbol for rate of change of voltage with time is

dv dt

With differential calculus we find an equation for the rate of change of voltage at any instant: For circuit A,

v = 5t    and    

dv = 5 V/s dt

13-2   Rate of Change of Voltage

For circuit B, For circuit C,

v = 10t v = t2,

and

dv = 10 V/s dt

dv = 2t V/s dt

When we close the switch in the circuit of Figure 13-2, current flows i­nstantly at its initial value and then decreases as the voltage across the ­capacitor rises (see Figure 13-4). From Ohm’s law we know that the voltage drop across the resistance is always proportional to the current through it. Since the resistance is constant, the graph of the IR drop across the resistor has the same shape as the graph of the current. I0

Current (mA)

50 40 30

i

20 10 0

0

2

4 6 8 Time (s)

10

Figure 13-4  Instantaneous charging current

Since the initial value of capacitor voltage, v, is 0 V, the initial value of ­resistor voltage is E = I0R. Therefore

I0 =

E R

(13-3)

From Equation 12-6, we get v = q/C. The capacitance C does not change with time, so

dq dv 1 = × dt C dt

Since the current transfers charge to the capacitor, the rate of change of the charge on the capacitor is dq/dt = i . Therefore, dv/dt = i/C. When the switch is closed at t = 0,

dv I0 E/R = = dt C C

Therefore,

initial

E dv = dt CR

(13-4)

353

Chapter 13   Capacitance in DC Circuits

This initial rate of change of the capacitor voltage is represented by the slope of the tangent at t = 0 in Figure 13-5. Since the initial current, I0, in the CR circuit of Figure 13-2 is only 50 μA, it takes about 10 s for the voltage across the capacitor to rise to a value close to the applied voltage.

Capacitor Voltage, vC

354

τ

Slope =

100

dv E = dt CR

E

vC

80 60 40

63% of E

20 0

4 6 8 2 Time after switch is closed (seconds)

10

Figure 13-5  Rise of potential difference across a capacitor charging through a resistance

See Review Questions 13-28 to 13-30.

13-3  Time Constant To help analyze the charging and discharging of capacitors, we define the time constant of a CR circuit: The time constant of a CR circuit is the time it would take the potential difference across the capacitor to rise to the same value as the applied voltage if the potential difference increased at a constant rate equal to its initial rate of change. The letter symbol for time constant is the Greek lowercase letter τ (tau). As shown in Figure 13-5, if the rate of change of the capacitor voltage ­remains equal to the initial rate of change, the voltage across the capacitor when it is fully charged is V=E=τ×

Therefore,

E CR

τ = CR

(13-5)

where τ is the time constant of the CR circuit in seconds, C is the capacitance in farads, and R is the resistance in ohms. (Equation 13-5 also applies with the capacitance in microfarads and the resistance in megohms.) The time constant of a CR circuit is directly proportional to the capacitance. If we double the capacitance, twice as much charge must flow

13-3   Time Constant

355

into the capacitor in order to raise the potential difference across it to the ­applied voltage. But changing the capacitance does not affect the initial current, so the capacitor takes twice as long to charge. If we double the ­resistance instead, the initial current is cut in half and the capacitor again takes twice as long to charge to a given voltage. The applied voltage has no effect on the time constant. If we double the applied voltage, the capacitor must store twice the charge but the initial current is doubled. Therefore, it will take exactly the same time for the ­potential difference across the capacitor to reach the applied voltage. For every CR circuit, the relationship between the actual charging time and the time constant is the same. The instantaneous potential difference across a capacitor is always about 63.2% of the applied voltage one time constant after the switch is closed. For practical purposes, the potential ­differ­ence across the capacitor reaches a steady-state value equal to the ­applied voltage after a time interval equal to five time constants.

Example 13-1 (a)  Find the initial rate of change of potential difference across the capacitor in Figure 13-2. (b)  How long will it take the capacitor in Figure 13-2 to charge to a potential difference of 100 V? Solution (a) (b)

Initial

dv E 100 V = 50 V/s = = dt CR 1.0 μF × 2.0 MΩ

τ = CR = 1.0 μF × 2.0 MΩ = 2.0 s

Since the applied voltage is 100 V and V ≈ E after 5τ, the charging time is

t = 5τ = 5 × 2.0 s = 10 s

Multisim Solution (b) Download Multisim file EX13-1 from the website. The circuit is the same as shown in Figure 13-2. Connect a ground to the bottom node of the schematic. On the Menu bar, select Simulate/Analyses/Transient Analysis. Under Analysis Parameters, set the initial conditions to zero and the end time to 15 s. Under Output, add the voltage across the capacitor as the variable for analysis. At the bottom of the Transient Analysis dialog box, click Simulate. Use the graph to determine that the voltage reaches 100 V after about 10 s. See Problem 13-1 and Review Question 13-31.

circuitSIM walkthrough

Chapter 13   Capacitance in DC Circuits

13-4 Graphical Solution for Capacitor Voltage Because of the fixed relationship between the time constant and the charging time for a CR circuit, we can use an exponential curve to find the potential differences at any instant in any CR circuit. If we replot the graph of vC in Figure 13-5 with the horizontal (time) axis marked in time constants rather than seconds and with a vertical (voltage) axis marked in percentage of the applied voltage, we get the charging curve in Figure 13-6. 100 Potential Difference (percentage of applied voltage)

356

y = 1 − e−x

90 Charging

80 70 60 50 40 30

Discharging

20 10 0

y = e−x

1

2 3 Time (time constants)

4

5

Figure 13-6  Universal exponential curves of the charge and discharge of capacitors

Example 13-2 (a)  Find the potential difference across the capacitor in the circuit of ­Figure 13-2 3.0 s after the switch is closed. (b)  How long will it take the potential difference across the capacitor to rise from 0 V to 55 V? Solution

(a)  Since τ = CR = 1.0 μF × 2.0 MΩ = 2.0 s,

3.0 s =

3.0 s × τ = 1.5τ 2.0 s

13-5   Discharging a Capacitor

357

On the charging curve in Figure 13-6, y = 77% when t = 1.5τ, as shown in Figure 13-7(a). Therefore, vC = 77% of E = 0.77 × 100 V = 77 V 55 V × 100% = 55% of E (b) 55 V is 100 V On the charging curve in Figure 13-6, t = 0.80τ when y = 55 as shown in ­Figure 13-7(b).

100

Voltage (% of E)

Voltage (% of E)

Therefore, t = 0.80τ = 0.80 × 2.0 s = 1.6 s 77

0

1.5τ

100

55

0

0.80τ

Time (a)

Time (b)

Figure 13-7  Using the graph of Figure 13-6 to solve Example 13-2

Multisim Solution Download Multisim file EX13-2 from the website. The circuit is the same as shown in Figure 13-2. Check that the switch is closed. Connect a ground to the bottom node of the schematic. (a) On the Menu bar, select Simulate/Analyses/Transient Analysis. Under Analysis Parameters, set the initial conditions to zero and the end time to 10 s. Under Output, add the voltage across the capacitor as the variable for analysis. At the bottom of the Transient Analysis dialog box, click Simulate. Use the graph to determine that the capacitor voltage at t = 3 s is 77 V.

(b) The graph also shows that the capacitor voltage reaches 55 V at t = 1.6 s. See Problems 13-2 to 13-6 and Review Questions 13-32 and 13-33.

13-5  Discharging a Capacitor When the capacitor in Figure 13-8(a) is fully charged, the potential difference across the capacitor is equal to the applied voltage, 200 V. If we then turn the switch to position 2, the potential difference across the capacitor remains at 200 V for some considerable time since there is no c­ onducting

circuitSIM walkthrough

358

Chapter 13   Capacitance in DC Circuits

path between the plates except a bit of leakage through the dielectric. At the instant we turn the switch to position 3, the potential difference across the capacitor is still 200 V since it takes time for the surplus electrons to flow  from the bottom plate to the top plate of the capacitor. To satisfy Kirchhoff’s voltage law, the initial current has to be such that a 200-V IR drop develops across the very low resistance of the conductor connecting the two plates together. Therefore, the initial current is very large and the capacitor discharges in a fraction of a second. 100 kΩ

1

1 2

2

3

+ 200 V −

50 nF

+ −

(a)

3

+ 200 V −

100 kΩ 50 nF

+ −

(b)

Figure 13-8  Discharging a capacitor

To reduce the initial discharge current, we can add a 100-kΩ resistor as shown in Figure 13-8(b). The initial current is now limited to



I0 =

V0 200 V = = 2.00 mA R 100 kΩ

(13-6)

The reduced current takes much longer to discharge the capacitor fully. As the charge on the capacitor decreases, the potential difference across the capacitor must also decrease. Therefore, the initial current is the maximum value that the instantaneous discharge current can have. According to Kirchhoff’s voltage law, the IR drop must have the same magnitude as the potential difference of the capacitor. As this potential difference decreases, the current decreases and the capacitor discharges more slowly. This interdependence among the potential differences, the IR drop, the current, and the rate of decrease of the potential differences results in the exponential discharge curve shown in Figure 13-9(a). When the capacitor is discharging, the current through the 100-kΩ ­resistor in Figure 13-8(b) flows in the opposite direction to the current when the capacitor is charging. Consequently, the polarity of the IR drop across the resistance reverses when the capacitor is switched from charging to discharging. Therefore, the discharge current and the resistor voltage are shown as negative quantities in Figure 13-9(b).

Capacitor Voltage (% of E)

V0 100 80 60 40

vC

20 0

0

37% 1τ 2τ 3τ Time (a)

4τ

5τ

Current/Resistor Voltage (% of maximum)

13-5   Discharging a Capacitor

0

1τ

0

Time 2τ 3τ

4τ

5τ

20 i and vR

40 60 80 100

V I0 = R0 (b)

Figure 13-9  Instantaneous voltage and current as a capacitor discharges

With the same procedure we used for deriving Equation 13-4, we can show that the initial rate of change of the potential difference across the ­capacitor as it starts to discharge is dv V0 =− dt CR



(13-7)

where V0 is the initial voltage across the capacitor. The negative sign results from the discharge current flowing through the resistance in the direction opposite to that of the charging current. The ­negative rate of change of voltage indicates that the voltage across the capacitor is decreasing. We define the discharge time constant, τ, as the time it would take for the capacitor to completely discharge if it were to continue to discharge at the initial rate. Since the final voltage is 0 V, the total change in voltage, Δv, after the capacitor has discharged is −Vinit. If the capacitor continues to ­discharge at the initial rate, the time to discharge fully is Δt = τ and the ­discharge rate is dv/di = Δv/Δt. Equation 13-7 then gives

and

−V0 −V0 dv Δv = = = τ di Δt CR

τ = CR

(13-5)

Therefore, we can use the same formula for calculating the time constant for charging and for discharging. Again, we can use the exponential curve in Figure 13-6 to find the potential differences and currents at any instant in any CR circuit. After a time interval equal to one time constant, the ­potential difference of the capacitor drops to about 37% of its initial value—in other words, the capacitor loses about 63% of its initial potential

359

Chapter 13   Capacitance in DC Circuits

difference. For practical purposes, the charge on a capacitor is negligible after 5τ. For the CR circuit of Figure 13-8(b), τ = CR = 50 nF × 100 kΩ = 5.0 ms



The capacitor discharges fully in about 5 × 5.0 ms = 25 ms.

Example 13-3 When the capacitor in Figure 13-8(b) is fully charged, the switch is turned to position 3. (a) Find the initial rate of change of potential difference across the ­capacitor. (b) Find the potential difference across the capacitor after 7.0 ms. (c) How long does it take for the voltage drop across the 100-kΩ resistor to reach 110 V? Solution (a)

Initial

(b)

dv V0 200 V =− =− = −40 kV/s dt CR 50 nF × 100 kΩ τ = CR = 0.05 μF × 100 kΩ = 5.0 ms



7.0 ms is

7.0 ms = 1.4 τ 5.0 ms

On the discharging curve in Figure 13-6, y = 25% when t = 1.4τ, as shown in Figure 13-10(a). vC = 25% of V0 = 0.25 × 200 V = 50 V Capacitor Voltage (% of V 0 )



Capacitor Voltage (% of V 0 )

360

100

25 0

3τ

1.4τ

Time (a)

100 55

0

3τ

0.6τ

Time (b)

Figure 13-10  Using the graph of Figure 13-6 to solve Example 13-3

13-5   Discharging a Capacitor

361

(c) When the switch is in position 3, the resistor is connected directly across the capacitor. From Kirchhoff’s voltage law, vC + vR = 0    and    vR = −vC



110 V is

110 = 55% of Vinit 200

On the discharging curve in Figure 13-6, t = 0.6τ when y = 55%, as shown in Figure 13-10(b). t = 0.6τ = 0.6 × 5.0 ms = 3 ms

circuitSIM

Multisim Solution Download Multisim file EX13-3 from the website.

The circuit is as shown in Figure 13-8(b) with the switch in position 3. Connect a ground to the bottom node of the schematic. (a) Double-click on the capacitor in the circuit. Under the Values tab, set the initial condition to 200 V. On the Menu bar, select Simulate/Analyses/Transient Analysis. Under Analysis Parameters, set the initial conditions to Userdefined and the end time to 25 ms. Under Output, add the capacitor voltage as the variable for analysis. At the bottom of the Transient Analysis dialog box, click Simulate. Use the graph to determine that the capacitor voltage at t = 7.0 ms is 50 V. (b)  The graph also shows that the capacitor voltage reaches 110 V after about 3 ms. Since the resistor voltage and the capacitor voltage have the same magnitude, the resistor voltage also reaches 110 V when t = 3 ms.

See Problems 13-7 to 13-13 and Review Question 13-34.

Circuit Check

A

CC 13-1. A capacitor is charged through a 22-kΩ resistor and reaches full charge in 5.2 s. Determine the capacitance. CC 13-2. A 5.0-μF capacitor is connected in series with a 30-kΩ resistor and a 60-V source. What current will be flowing in this circuit 0.27 s after the capacitor starts charging?

walkthrough

362

Chapter 13   Capacitance in DC Circuits

13-6 Algebraic Solution for Capacitor Voltage When we require greater accuracy than we can read from the graphs of Figure 13-6, we can use the equations for the exponential curves. We start with the Kirchhoff’s voltage-law equation for the basic circuit of ­Figure 13-2. E = iR + vC



(13-1)

At every instant, the current is the rate of change of the charge on the ­capacitor:

i=



i=C

Since q = CvC ,

dq dt

dvC dt

(13-8)

Substituting for i in Equation 13-1, E = CR



dvC + vC dt

(13-9)

Equation 13-9 is a differential equation for an exponential function. ­ ppendix 2-2 shows how to solve this equation with basic integral calcuA lus. This solution gives an equation for the potential difference that builds up on an uncharged capacitor that begins charging at t = 0:

vC = E ( 1 − e−x )

(13-10)

where e = 2.7182818 . . . (the base of natural logarithms) and x = t/CR. Since

τ = CR,     x =

t τ

Thus, x represents elapsed time measured in time constants. Substituting for vC in Equation 13-1 gives E = iCR + E(1 − e−x)

0 = iCR − Ee−x

13-6   Algebraic Solution for Capacitor Voltage

iC =



E −x e R

363

(13-11)

When the switch in the circuit of Figure 13-8(b) is in position 3, there is no voltage source in the discharge loop. Hence, the applied voltage E is zero. To satisfy Kirchhoff’s voltage law, the sum of the IR drop across the resistor and the potential difference across the capacitor must be zero: 0 = iR + vC = CR



dvC + vC dt

As shown in Appendix 2-2, this equation is readily solved with integral calculus. The resulting equation for the voltage across a discharging ­capacitor is vC = V0e−x



(13-12)

where V0 is the initial voltage across the capacitor and x = t/CR. Equations 13-10 and 13-12 are the equations for the exponential curves in Figure 13-6.

Example 13-2A (a) Calculate the voltage across the capacitor in Figure 13-2 when the switch has been closed for 3.0 s. (b) How long will it take the voltage across the capacitor to rise from 0 V to 55 V? Solution (a) (b)

x=

t 3.0 s = = 1.5 CR 1.0 μF × 2.0 MΩ

vC = E ( 1 − e−x ) = 100 V × ( 1 − e−1.5 ) = 78 V 55 V = 100 V(1 − e−x)

e−x = 0.45 Hence,

−x = ln 0.45 = −0.7985 x = 0.80 =

t t = τ CR

t = 0.80 × 2.0 s = 1.6 s

The symbol ln is ­standard notation for a natural logarithm, based on powers of the mathematical ­constant e. We can calculate natural ­logarithms with the ln key on a scientific calculator.

364

Chapter 13   Capacitance in DC Circuits

Example 13-4 Starting from position 1, the switch in Figure 13-11 stays in each position for 1.0 ms. Calculate the voltage across the capacitor 3.0 ms after the switch first moves to position 2.

1 100 kΩ

100 kΩ

2 + 500 V −

10 nF

Figure 13-11  Circuit diagram for Example 13-4

Solution Step 1 When the switch first moves to position 2, the capacitor charges from 0 V toward 500 V. At any instant, vC = E(1 − e−x) = 500(1 − e−x)

At 1.0 ms,

x=

t 1 ms 1.0 ms = = = 1.0 CR 10 nF × 100 kΩ 1.0 ms

vC = 500(1 − e−1) = 500 × (1 − 0.3679) = 316 V

Step 2 As the capacitor discharges in position 1,

vC = V0e−x = 316 × e−x

After 1.0 ms,

x=

1.0 ms t = = 0.5 CR 10 nF × 200 kΩ

vC = 316 × e−0.5 = 316 × 0.6065 = 192 V

Step 3 When the switch returns to position 2, vC starts at 192 V. Now we need to calculate how long it would take the capacitor to charge from 0 V to 192 V. 192 = 500(1 − e−x)

13-6   Algebraic Solution for Capacitor Voltage

e− x =



500 − 192 = 0.616 500

We can use ­approximate values read from the ­exponential curves to check ­answers calculated by the ­algebraic method.

x = −ln 0.616 = 0.4845



x=

Since

t CR

t = 0.4845 × 1.0 ms = 0.4845 ms



One millisecond later, the capacitor will reach a voltage equivalent to having charged for a total time interval of 0.4845 + 1.0 = 1.4845 ms. Hence, 1.4845 ms t = = 1.4845 τ 1.0 ms

x=



vC = 500(1 − e−1.4845) = 500 × (1 – 0.2266) = 387 V



Example 13-5

The switch in Figure 13-12(a) is closed at t = 0 s and opened again at t = 1.0 s. Find the voltage across the capacitor at t = 2.0 s.

Solution The discharge circuit when the switch is open is quite straightforward. However, to calculate the charge time constant when the switch is closed, we need Thévenin’s theorem. Step 1 Considering the capacitor to be the load, we remove it from the original ­circuit of Figure 13-12(a) and place it in the Thévenin-equivalent circuit of Figure 13-12(b). We can calculate mentally that ETh = 400 V and RTh = 160 kΩ for the equivalent circuit of Figure 13-12(b). 200 kΩ

RTh

+ 500 V −

800 kΩ

(a)

2.0 μF

160 kΩ + E 400 V Th − (b)

Figure 13-12  (a) Capacitor-charging circuit and (b) its Théveninequivalent circuit

365

2.0 μF

366

Chapter 13   Capacitance in DC Circuits

Step 2 The charging time constant is and

τch = CRTh = 2.0 μF × 0.16 MΩ = 0.32 s 1.0 s =

1.0 = 3.125 τ 0.32

Since the switch was originally open, initial voltage across the capacitor is 0 V. We substitute into Equation 13-10 to find the voltage after 1.0 s of charging:

vC = E(1 − e−x) = 400(1 − e −3.125) = 400(1 − 0.044) = 382.4 V

Step 3 The discharging time constant is simply

τdis = CR = 2.0 μF × 800 kΩ = 1.6 s

1.0 = 0.625 τ 1.6 From Equation 13-12, at t = 2.0 s, and



circuitSIM walkthrough

1.0 s =

vC = 382.4 × e−0.625 = 382.4 × 0.535 = 205 V

Multisim Solution Download Multisim file EX13-5 from the website.

The circuit is the same as shown in Figure 13-12(a). Connect a ground to the bottom node of the schematic, and close the switch. On the Menu bar, select Simulate/Analyses/Transient Analysis. Under Analysis Parameters, set the initial conditions to zero and the end time to 1 s. Under Output, add the capacitor voltage as the variable for analysis. At the bottom of the Transient Analysis dialog box, click Simulate. Use the graph to determine that the capacitor voltage at t = 1 sec is about 382 V.

Open the switch. Double-click on the capacitor in the circuit. Under the Values tab, set the initial condition to 382 V. On the Menu bar, select Simulate/Analyses/Transient Analysis. Under Analysis Parameters, set the initial conditions to User-defined and the end time to 1 s. At the bottom of the Transient Analysis dialog box, click Simulate. Use the graph to determine that 1.0 s after the switch was closed the capacitor voltage is about 205 V.

See Problems 13-14 and 13-15.

13-7  Transient Response We can now consider another method for analyzing the response of CR networks to changes in applied voltage. If we charge a capacitor that is ­already partially charged, the voltage across it equals the initial voltage

13-7   Transient Response

367

plus an additional voltage that depends on the elapsed time. Using this ­approach to adapt Equation 13-10 gives vC = V0 + ( VF − V0 ) ( 1 − e−x )



(13-13)

where V0 is the initial capacitor voltage and VF is the steady-state voltage to which the capacitor can charge, and x is the charging time measured in time constants. Simplifying gives us a universal equation for instantaneous capacitor voltage in a DC CR network: vC = V0 + VF – V0 − VF e−x + V0 e−x vC = VF + ( V0 − VF ) e − x



(13-14)

where vC is the instantaneous voltage across the capacitor, V0 is the i­ nitial voltage across the capacitor, and VF is the final steady-state voltage.

When we close the switch at t = 0 in the purely resistive network of ­ igure 13-13(a), the output voltage of the two-terminal network instantly F drops from 500 V to its new steady-state value of 100 V, as shown in Figure 13-13(b). But in the CR network of Figure 13-13(c), the output voltage is the voltage across the capacitor, which takes five time constants to ­discharge from its initial voltage of 500 V to its final steady-state voltage of 100 V, as shown in Figure 13-13(d).

800 kΩ

Output Voltage, vout (V)

vout 200 kΩ

+ 500 V −

500 400 300 200 100 0

vout 800 kΩ

200 kΩ 2.0 μF

+ 500 V −

Output Voltage, vout (V)

(a)

0 1τ 2τ 3τ 4τ 5τ Time (b)

500 400 300

Transient

Steady state

200 100 0

(c)

Figure 13-13  Transient response of a DC CR network

0 1τ 2τ 3τ 4τ 5τ Time (d)

We can also derive Equation 13-14 by considering a charged capacitor discharging toward a given steady-state voltage.

368

Chapter 13   Capacitance in DC Circuits

Equation 13-14 represents the output voltage as consisting of two ­components: the steady-state response, VF, and the transient response (V0 − VF)e−x, which is shown in red in Figure 13-14. This transient response is negligible when t > 5τ.

Output Voltage (volts)

Transient response is the manner in which a voltage or current changes from one steady-state value to another. A transient phase is the time interval during which a network’s voltages and currents adjust from one steady state to another. 500

V0

400

Transient response

300

Steady state

200 100 0

VF 0

1τ

2τ 3τ Time

4τ

5τ

Figure 13-14 Separating Vout into steady-state and transient components

Example 13-6 If the switch in the two-terminal network of Figure 13-13(c) has been open for several minutes, what is the instantaneous output voltage 500 ms after the switch is closed? Solution From inspection of the network, V0 = 500 V and VF = 100 V. The discharge time constant is governed by the Thévenin equivalent resistance. x=

0.500 s = 1.5625 2.0 μF × 0.16 MΩ

vout = VF + ( V0 − VF ) e − x

= 100 + ( 500 − 100 ) e − 1.5625 = 100 + 400 × 0.2096

circuitSIM walkthrough

= 184 V

Multisim Solution Download Multisim file EX13-6 from the website. The circuit is the same as shown in Figure 13-13(c). Close the switch. Double-click on the capacitor in the circuit. Under the Values tab, set the initial condition to 500  V. On the Menu bar, select Simulate/­ Analyses/Transient Analysis. Under Analysis Parameters, set the

13-7   Transient Response

369

i­nitial conditions to User-defined and the end time to 500 ms. Under Output, add the capacitor voltage as the variable for analysis. At the bottom of the Transient Analysis dialog box, click Simulate. Use the graph to determine that 500 ms after the switch is closed the instantaneous output voltages is about 184 V. In the simple CR charging circuit of Figure 13-2, V0 = 0 and VF = 100 V for the capacitor. Since VR = E − vC, then V0 = 100 V and VF = 0 for the resistor. As long as we are careful to specify the correct V0 and VF, Equation 13-14 also ­applies to the instantaneous voltage drop across series charging resistors and shunt discharging resistors. The transient response approach lets us use a s­ ingle equation to determine the voltage across a partially charged capacitor.

Example 13-2B Given that the switch in Figure 13-2 has been closed for 3.0 s, find the ­voltage across (a) the capacitor (b) the series charging resistor Solution

x=

t 3.0 s = = 1.5 CR 1.0 μF × 2.0 MΩ

(a) For the capacitor, V0 = 0 and VF = 100 V.

VC = VF + (V0 − VF)e−x

= 100 V + (0 − 100 V)e−1.5 = 100 V − 100 V × 0.2231

= 77.7 V

(b) For the resistor, V0 = 100 V and VF = 0.

vR = VF + (V0 − VF)e−x

= 0 + (100 V − 0)e−1.5 = 100 V × 0.2231 = 22.3 V

See Problems 13-16 to 13-22 and Review Questions 13-35 to 13-39.

We can check our ­calculations by ­ver­ify­ing that E = vC + vR  = 77.7 V + 22.3 V = 100 V.

370

Chapter 13   Capacitance in DC Circuits

13-8  Energy Stored by a Capacitor The current that flows while a capacitor is charging transfers energy from the voltage source to the rest of the circuit. Some of this energy is converted into heat as the charging current flows through the series current-­limiting resistance in the circuit. This heat accounts for only some of the energy drawn from the source. The remainder is stored by the capacitor. Work is done on the charge that moves into a capacitor against the opposing potential difference building up between the plates. When we disconnect a charged capacitor from the source, the potential difference across the capacitor remains constant unless there is leakage of the charge through the dielectric. Since no current is required to maintain the potential difference between the plates of a charged capacitor, the ­capacitance stores electric energy. In fact, this energy is stored in the electric field between the plates of the capacitor. This electric field alters the orbits of electrons bound to the molecules of the dielectric, as shown in Figure 12-4. We might compare the electric energy stored by a capacitor to the mechanical energy stored in a cylinder of compressed air. Energy is used to force the air into the cylinder against the increasing pressure. This stored energy is ­released as kinetic energy when the compressed air is used to operate a paint sprayer or a pneumatic tool. Similarly, we can use the electric energy stored by a charged capacitor, as we shall discover in Section 13-9. At any instant, the rate at which a capacitor stores energy is power, p = vi. We can plot a graph of power versus time by calculating the ­product of the potential difference across the capacitance and the charging current at ­various times from 0 to 5τ. As shown in Figure 13-15, the power input to the capacitor peaks when the graphs of the potential difference and the charging current cross. At this instant, the instantaneous voltage has risen to half of its final value and the instantaneous charging current has dropped to half of its initial value. The peak power input occurs about 0.693 time constants after the capacitor begins charging.

iC vC Area = WC PC

0

1

2 3 4 Time (time constants)

5

Figure 13-15  Energy stored by a capacitor

13-8   Energy Stored by a Capacitor

After five time constants, the charging current is negligible and therefore the power input is also negligible. Thus, when a capacitor is fully charged, it has stored all the energy that it can at the particular voltage applied to it. This total energy is represented by the area under the power curve in Figure 13-15. Appendix 2-3 describes how to use integral calculus to calculate energy stored by a capacitor. However, we can simplify the calculation by considering a capacitor that charges at a constant rate equal to the initial rate of change of voltage. ­Suppose we connect the capacitor to a constant-current source with a terminal voltage that can vary while the current remains I = E/R, as shown in Figure 13-16(a). Since the voltage across the capacitor now rises at a ­constant rate from zero to its final value of V = E, as shown in Figure 13-16(b), the a­ verage voltage during this time interval must be 1⁄2V. Therefore, 1 W = Pt = Vav × I × t = V × It 2 Since It = q,

1 W = Vq 2

When the capacitor is fully charged, q = CV and

1 1 W = V × CV = CV 2 2 2



(13-15)

where W is the energy stored by the capacitance in joules, C is the ­capacitance in farads, and V is the potential difference across the capacitance in volts. The total energy stored by a fully charged capacitor is not affected by the rate at which it is charged.

Capacitor Voltage

I = E/R

Constant current source

(a)

V =E Vav

vC

0

t Time (b)

Figure 13-16  Determining the energy stored by a capacitor

371

Chapter 13   Capacitance in DC Circuits

Example 13-7 How much energy is stored in a 500-pF high-voltage filter capacitor when it is charged to 15.0 kV? Solution W=

1 1 CV2 = × 500 pF × ( 15 kV ) 2 = 5.63 × 10 − 2 J = 56 mJ 2 2

See Problems 13-23 to 13-26 and Review Questions 13-40 to 13-42.

Circuit Check

B

CC 13-3. A capacitor is connected to a 300-V source until it is fully charged. It is then discharged through a 2.2-MΩ resistor. After 20 min the capacitor voltage has fallen to 94 V. Calculate the ­capacitance of the capacitor. CC 13-4. Find the voltage across a 3000-µF filter capacitor when it is ­storing 750 mJ of energy.

13-9 Characteristics of Capacitive DC Circuits The ability of a capacitor to store electric charge and, as a consequence, to oppose a rapid change in potential difference between its terminals makes the capacitor a particularly useful component in DC circuits. Its ability to oppose any instant change in the voltage across it enables the capacitor to filter or smooth the output from a pulsating voltage source such as the rectifier in Figure 13-17(a). When the source voltage drops below the voltage across the capacitor, the capacitor maintains the load voltage by discharging some of its stored energy into the load. The c­ apacitor recharges the next time that the source voltage rises to its peak value, as shown in Figure 13-17(b).

+

Pulsating E source g −

DC

C

(a)

Load

Voltage

372

vC e

Time (b)

Figure 13-17  Filtering voltage fluctuations with a capacitor

13-9   Characteristics of Capacitive DC Circuits

This application of a capacitor is somewhat like the use of a reservoir tank with an air compressor for a paint sprayer. The tank stores enough compressed air to maintain a fairly constant pressure to the paint sprayer in spite of the fluctuations in the pressure developed by the piston action of the compressor. (The compressed-air-tank analogy provides a means of ­visualizing the charge and discharge action of capacitance, with the friction of the air hose representing resistance.) Energy storage in a capacitor is also useful when we wish to obtain a very high current for a short period of time without subjecting the source to a severe current surge. With a circuit patterned after the basic circuit of Figure 13-18, the capacitor charges slowly from the source and  then ­discharges rapidly when we connect the low-resistance load across it. The peak current drawn from the source is limited by the series resistor. This system is used in radar transmitters to generate micro­second pulses with a peak power of over a million watts. Capacitors can also produce short surges of very high current for spot-welding ­aluminum.

High resistance

+ E

C

Low-resistance load

− Figure 13-18 Obtaining short-duration, high-current pulses from a reservoir capacitor

The potential difference across the capacitor of a given CR circuit ­always rises along exactly the same exponential curve. Therefore, we can use a ­series CR circuit as the basis for a timing device. In the simple circuit of ­Figure 13-19, the neon lamp “strikes” (starts glowing) when the potential difference across it reaches 100 V. The lamp stops glowing when the ­voltage drops below 20 V. When a neon lamp is not struck, the gas is an ­insulator and the lamp acts like an open circuit. When the lamp strikes, the neon gas becomes ionized with the positive gas ions flowing in one ­direction and free electrons moving in the other direction. The resistance of the glowing lamp is about 100 Ω. It takes about 1.8 time constants for the potential difference across the capacitor in Figure 13-19 to rise to the striking voltage of the neon lamp. The low resistance of the ionized neon then quickly discharges the capacitor down to the extinguishing ­voltage of the lamp, whereupon the resistance of the neon lamp again becomes ­extremely high and the capacitor starts to recharge. The intervals b ­ etween the short flashes from the neon lamp, therefore, depend on the time ­constant CR.

373

374

Chapter 13   Capacitance in DC Circuits

R

+ 120 V

C

Neon lamp

− Figure 13-19 A CR timing circuit C

Leakage resistance

Figure 13-20  Equivalent circuit of a “leaky” capacitor

In some instances, capacitors can have undesirable characteristics. So far, we have assumed that there is negligible leakage of charge through the dielectric of capacitors. This assumption is reasonable for ceramic, mica, and polymer dielectrics. However, electrolytic capacitors often have enough leakage to affect the operation of a circuit. We can represent capacitor leakage in practical circuits by an appropriate high resistance in parallel with the capacitance, as shown in Figure 13-20. Capacitance in circuits is not limited to capacitors. As shown in ­Figure 12-6, there is some capacitance between any pair of electric conductors that are insulated from each other. In many circuit applications, this stray capacitance is small enough that its effect on the circuit is negligible. But in computers, where timing pulse voltages must rise and fall in a fraction of a nanosecond, stray capacitance can degrade circuit performance. In Figure 13-21, the stray capacitance of the conductors is directly in parallel with the input terminals of the transistor (even though this capacitance would not appear in the actual circuit diagram). Hence, the input signal to the transistor is the potential difference across this stray capacitance, which charges and discharges through the internal resistance of the pulse generator. If the time constant for the internal resistance and stray capacitance is more than a small fraction of the pulse duration, the capacitance will cause appreciable distortion by delaying the rise and fall of the voltage pulses. Internal resistance

Stray capacitance

Pulse generator

Pulse produced by pulse generator

Pulse appearing across stray capacitance

Figure 13-21  Effect of stray capacitance

See Review Question 13-43.

13-10  Troubleshooting

13-10 Troubleshooting Capacitors can fail in several different ways. • If a capacitor is subjected to a voltage higher than its working voltage, the dielectric often breaks down, allowing a high current to pass through the capacitor. Usually the capacitor burns out and the circuit b ­ ecomes an open circuit. • As capacitors age, the leakage current flowing through them tends to ­increase. Electrolytic capacitors in particular are prone to this problem because the electrolyte dries out somewhat over time. The increasing leakage current results in a lower leakage resistance for the capacitor. • A capacitor can short internally. A capacitor can be checked for faults with an ohmmeter or a capacitance meter. The capacitor should first be removed from the circuit and then fully discharged by shorting its leads together. To test the capacitor with an ohmmeter, connect it across the ohmmeter terminals. If the capacitor is shorted, the ohmmeter will give a steady reading very close to 0 Ω. If the capacitor is open, the meter will read infinity or over range. If the capacitor is not defective, it will be charged by the internal battery in an analog ohmmeter. When the meter is connected, the pointer immediately goes toward the zero ohms position and then moves toward the infinity position as the current decreases. Electrolytic capacitors usually have a resistance in the megohm range, while most other types should read a resistance of hundreds of megohms. If the capacitor has a significant leakage current, the pointer will stop at a fairly low value of resistance. Most digital ohmmeters cannot be used to measure the leakage resistance of ­capacitors. To test a capacitor with a capacitance meter, simply connect the capacitor leads to the terminals of the meter and read the capacitance from the dial or digital display of the meter. If no capacitance is displayed, the ­capacitor is shorted or open. See Review Questions 13-44 to 13-46.

375

376

Chapter 13   Capacitance in DC Circuits

Summary

• When a capacitor is charging or discharging, the initial value of current depends on the values of V0 and R. • When a capacitor is charging or discharging, the initial rate of change of capacitor voltage depends on the values of V0, C, and R. • The time constant for the charging and discharging of a series CR circuit is τ = CR. • The universal exponential curves provide a graphical method for analyzing a CR circuit when the capacitor is being charged or discharged. • The charge and discharge equations for a capacitor provide an algebraic method for analyzing a CR circuit. • The transient response of a CR circuit with an initial voltage across the capacitor can be determined with the universal equation for capacitor voltage. • A capacitor stores energy when it is charged. • Stray capacitance in circuits can have undesirable effects. B = beginner

I = intermediate

A = advanced

Problems B

Section 13-3  Time Constant 13-1. Find the following quantities when the switch in Figure 13-22 is first closed: (a) the initial current (b) the initial voltage across the capacitor (c) the initial rate of rise of voltage across the capacitor (d) the charging time constant (e) the time it will take the capacitor to charge

+ 250 V −

200 kΩ

20 μF

Figure 13-22

B

Section 13-4 Graphical Solution for Capacitor Voltage

13-2. (a) Given that vC = 0 until the switch in Figure 13-23 is closed, find: (i) the initial charging current (ii) the initial rate of change of voltage across the capacitor (iii) the charging time constant

Problems

(iv) the voltage across the capacitor at t = τ (v) the charge on the capacitor at t = τ (b) Use Multisim to determine the initial charging current and the capacitor voltage one time constant after the switch is closed.

B

I

I A

circuitSIM walkthrough

10 nF

+ 16 V −



377

33 kΩ

Figure 13-23

13-3. For the circuit of Figure 13-22, use the exponential curves of ­Figure 13-6 to find: (a) the voltage across the capacitor 3.0 s after the switch is closed (b) the charging current 4.0 s after the switch is closed (c) the time taken for the voltage across the capacitor to reach 100 V (d) the time taken for the voltage across the capacitor to rise from 50 V to 200 V 13-4. (a) For the circuit of Figure 13-23, use the exponential curves of ­Figure 13-6 to find: (i) how long it will take the capacitor to charge to 16 V (ii) how long it will take the capacitor to charge to 10 V (iii) the instantaneous voltage across the capacitor at t = 0.50 ms (iv) the instantaneous charging current at t = 0.25 ms (v) the time taken for the voltage across the capacitor to rise from 4 V to 12 V (b) Use Multisim to verify your calculations in part (a). 13-5. An unknown capacitor is charged by a 200-V source through a 1.5-MΩ resistor. Given that it takes 45 s for the voltage across the capacitor to reach 100 V, determine the capacitance of the ­capacitor. 13-6. For the circuit shown in Figure 13-24, determine the voltage across the capacitor 5.0 s after switch S1 is closed. t=0s

t=3s

S1

S2

10 kΩ

+ 50 V

−

Figure 13-24

+ 100 V −

100 μF

circuitSIM walkthrough

378

Chapter 13   Capacitance in DC Circuits

B

circuitSIM walkthrough

Section 13-5  Discharging a Capacitor 13-7.

B

13-8.

I

13-9.

A

13-10.

A

13-11.

I

13-12.

I

13-13.



A 33-μF capacitor has been charged to a voltage of 450 V. (a) How long will it take to discharge the capacitor through a 220‑kΩ resistor? (b) Find the maximum discharge current. The capacitor of Problem 13-7 is discharged by shorting its terminals with an electric conductor having a resistance of 0.4 Ω. (a) Find the peak discharge current. (b) How long will it take to discharge the capacitor? (c) Use Multisim to verify your calculations for parts (a) and (b). A 33-μF capacitor, which has been charged to a voltage of 450 V, is discharged through a 220-kΩ resistor. Using the universal exponential curves, determine: (a) how long it will take the voltage across the capacitor to drop to 75 V (b) the voltage across the capacitor 15 s after it starts to d ­ ischarge (c) the instantaneous discharge current 10 s after it starts to ­discharge (d) how long it will take the voltage across the capacitor to drop from 350 V to 50 V If the resistor in the circuit of Figure 13-19 is 5.0 MΩ and the capacitor has a capacitance of 8.0 μF, calculate the time interval between flashes of the neon lamp. The lamp strikes at 100 V, and its resistance then becomes 100 Ω. It extinguishes when the voltage across it drops to 20 V, and its resistance then becomes almost infinite. A capacitor is connected to a 100-V source until it is fully charged. It is then discharged through a 3.3-MΩ resistor. After 11 min the capacitor voltage has fallen to 40 V. Determine the probable value of the capacitor. A 10-kΩ resistor is connected in series with a 10-μF capacitor, a 100-V source, and a switch. Calculate: (a) the time constant (b) the voltage on the capacitor 50 ms after the switch is closed (c) the charging current 100 ms after the switch is closed In the circuit shown in Figure 13-25, S1 is closed at t = 0 s and S2 is moved to the lower contact at t = 15 s, disconnecting the voltage source from the capacitor. (a) How long it will take the capacitor to fully discharge after S2 is moved? (b) Calculate the voltage across the capacitor at t = 4.0 s.

Problems



(c) Calculate the voltage across the capacitor at t = 24 s. (d) Sketch a graph of the capacitor voltage from t = 0 s until the capacitor is fully discharged. t=0s

t = 15 s

S1

S2

+ 100 V −

A

A

10 kΩ

200 μF

10 kΩ

Figure 13-25



A

379

Section 13-6  Algebraic Solution for Capacitor Voltage 13-14. Repeat Problem 13-3 using algebraic expressions to solve for instantaneous values of voltage and current. 13-15. Repeat Problem 13-4 using algebraic equations to solve for instantaneous values of voltage and current.

Section 13-7  Transient Response

13-16. After a considerable period of-time in position 2, the switch in the circuit of Figure 13-26 is thrown to position 1 at t = 0. (a) Calculate iC and dvC/dt at t = 0. (b) How long must the switch remain in position 1 for vC to rise to 100 V? (c) The switch is thrown to position 2 at t = 5.0 min. Calculate iC and dvC/dt as the switch makes contact in position 2. (d) If the switch returns to position 1 at t = 5.5 min, calculate vC at t = 10 min. (e) Use Multisim to verify your calculations for parts (a) and (b). 5.0 MΩ

470 kΩ 1

+ 200 V −



Figure 13-26

2 50 μF

circuitSIM walkthrough

380

Chapter 13   Capacitance in DC Circuits

A

13-17. After a considerable period of time in position 1, the switch in the circuit of Figure 13-27 is thrown to position 2 for 2.0 ms and is then returned to position 1. (a) Calculate vC 2.0 ms after the switch moves to position 2. (b) Calculate vC 2.0 ms after the switch returns to position 1. (c) Draw a graph of the output voltage at terminal A with respect to ground for this 4-ms interval. +

48 V

− 2

68 kΩ

A

1

+ 12 V −

10 nF

22 kΩ

Figure 13-27

A

circuitSIM walkthrough

13-18. (a) Commencing from the open position, the switch in the circuit of Figure 13-28 closes for 1.0 ms and then opens. Calculate the voltage across the capacitor 1.0 ms after the switch reopens. (b) Use Multisim to verify the capacitor voltage calculated in part (a). 68 kΩ + 300 V −

A

470 kΩ

10 nF

Figure 13-28

13-19. Repeat Problem 13-18 by using the circuit in Figure 13-29. 220 kΩ + 15 kΩ 250 V

−



Figure 13-29

5.0 nF

Problems

A

381

13-20. For the circuit shown in Figure 13-30, S1 is closed at t = 0 s and S2 at t = 10 s. (a) Calculate the time for the capacitor to be fully charged. (b) Calculate the capacitor voltage at t = 4.0 s. 10 kΩ

t=0

2.0 kΩ

S1 + 50 V −

S2

t = 10 s

100 μF

Figure 13-30

A

(c) Calculate the capacitor voltage at t = 11 s. (d) Sketch the waveform of the capacitor voltage for the interval from 0 s to 15 s. (e) Use Multisim to verify your calculations for parts (a), (b), and (c). 13-21. The switch shown in Figure 13-31 is thrown to position 1 at t = 0 s. After 6.0 s, the switch is thrown to position 2. (a) Sketch the capacitor voltage waveform for the time period from 0 s to 50 s. (b) Calculate capacitor voltage at t = 4.0 s. (c) Calculate capacitor voltage at t = 20 s. (d) How long after the switch is closed to position 1 will the capacitor voltage reach 75 V? 5.0 kΩ

circuitSIM walkthrough

1 2 5.0 kΩ

+ 100 V −

25 kΩ 200 μF

Figure 13-31

A

13-22. (a) The switch in the network of Figure 13-32 has been in position 1 long enough for the circuit to reach a steady state. Find the output voltage with respect to ground (i) 2.0 s and (ii) 4.0 s after the switch is thrown to position 2. (b) Use Multisim to verify the output voltage calculated in part (a).

circuitSIM walkthrough

382

Chapter 13   Capacitance in DC Circuits

2.0 MΩ

5.0 μF

1 +

500 kΩ

2 −

200 V

Figure 13-32

B B B

B

Section 13-8  Energy Stored by a Capacitor 13-23. Calculate the energy stored by the capacitors in the circuits of ­Figures 13-22 and 13-23 when they are fully charged. 13-24. Calculate the energy stored by the capacitor in Problem 13-7 (a) before it starts to discharge (b) after it has been discharging for one time constant 13-25. A 100-μF capacitor, charged to a potential difference of 36 V, is discharged through a 15-kΩ resistor. How much energy is dissipated by the resistor as the capacitor is completely discharged? If the capacitor had been discharged through a 680-Ω resistor, how much energy would have been dissipated by the resistor? 13-26. In Problem 13-10, how much energy is transferred to the neon lamp with each flash?

Review Questions

Section 13-1  Charging a Capacitor 13-27. What factors determine the initial charging current of a capacitor?

Section 13-2  Rate of Change of Voltage 13-28. What factors determine the rate of change of potential difference ­between the plates of a capacitor? 13-29. When a discharged capacitor is connected to a voltage source, the current changes instantly from zero to a value of E/R. Why then can the voltage across the capacitor not change instantly? 13-30. Why must the rate of change of voltage across a capacitor diminish as the charge on its plates increases?

Section 13-3  Time Constant

13-31. Account for the exponential shape of the graph of vC in Figure 13-5.

Review Questions

Section 13-4  Graphical Solution for Capacitor Voltage 13-32. What is the advantage of using the time constant of a CR circuit as a unit of time in determining the instantaneous voltage? 13-33. Figure 13-6 shows a graph of the algebraic equation y = 1 − e−x. Why do we use this equation to solve for vC rather than the simple exponential equation y = e−x?

Section 13-5  Discharging a Capacitor

13-34. Why does the initial charge on a capacitor have no effect on the length of time required for the capacitor to discharge through a given resistance?

Section 13-7  Transient Response 13-35. In the circuit of Figure 13-28, we use the Thévenin-equivalent resistance to calculate the charge time constant, but not the discharge time constant. In the circuit of Figure 13-29, we use the Théveninequivalent resistance to calculate the discharge time constant, but not the charge time constant. Explain the difference. 13-36. Why are the charge and discharge time constants the same in the ­circuit of Figure 13-33? 13-37. Draw an accurately labelled graph of the instantaneous voltage across the capacitor in the circuit of Figure 13-33 plotted against time as the double-pole, double-throw switch is thrown to the opposite position. 500 kΩ

+ 120 V −



1.0 μF

Figure 13-33

13-38. When steady-state voltages are suddenly switched in a DC CR ­network, the transient current always starts from zero and returns to zero, having either a positive or negative direction, as shown by the graph of Figure 13-34. Does this mean that the current in ­resistors of the network must be zero except during the transient phase? ­Explain.

383

Chapter 13   Capacitance in DC Circuits

+

Current

384

0

Time

−



Figure 13-34

13-39. Draw superimposed graphs (in different colours) to show how a graph of vC is obtained by combining a sudden change in steadystate voltage with a transient having the form of either the positive or negative transient shown in Figure 13-34.

Section 13-8  Energy Stored by a Capacitor 13-40. Using the techniques we employed in producing Figure 13-15, plot a graph of the instantaneous power in a resistor when a charged ­capacitor is connected across it. 13-41. An air-dielectric capacitor is charged from a voltage source and then disconnected. The plates are then moved twice as far apart. Assuming no leakage of the charge, account for the difference in the energy stored by the capacitor before and after the plates are moved. 13-42. Instead of changing the spacing between the plates in Question 13-41, a dielectric with a dielectric constant of 2.0 is placed between  the plates after the capacitor is disconnected from the source. Account for any change in the energy stored by the ­capacitor.

Section 13-9  Characteristics of Capacitive DC Circuits 13-43. Briefly describe three applications of the capacitor.

Section 13-10  Troubleshooting 13-44. Name three problems that one may encounter with capacitors. 13-45. You use an analog ohmmeter to test a capacitor and determine that it is not defective. Describe the action of the pointer on the meter. 13-46. What would you conclude about a capacitor that is tested with an ohmmeter if (a) it indicates a leakage resistance of 10 kΩ (b) the pointer goes to the extreme right and stays there (c) the pointer does not move

Practice Quiz

Integrate the Concepts

A 50-V source, a 20-kΩ resistor, a 1-μF capacitor, and an open switch form a series circuit. Calculate the following after the switch is closed: (a) the initial value of the current (b) the time constant (c) the capacitor voltage 50 ms later (d) the time for the capacitor to fully charge (e) the energy stored by the capacitor after it is fully charged

Practice Quiz 1.

Which of the following statements are true? (a) Uppercase letter symbols represent steady-state values that are dependent on time. (b) Lowercase letter symbols represent instantaneous values that are dependent on time. (c) The time constant for charging a capacitor is directly proportional to the capacitance. (d) A negative rate of change of voltage indicates that the voltage across the capacitor is decreasing. (e) Kirchhoff’s voltage law can be applied to RC circuits to determine the voltage across the capacitor.

2.

What is the initial rate of change of potential difference across the capacitor in Figure 13-35 when the switch is closed? (a) 2.4 V/s (b) 2.4 kV/s (c) 24 V/s (d) 0.24 V/s R1 2.0 GΩ V1 48 V

C1 0.010 μF

Figure 13-35

3.

How long will it take the capacitor of Figure 13-35 to charge to the battery voltage of 48 V? (a) 20 s (b) 20 ms (c) 100 ms (d) 100 s

385

386

Chapter 13   Capacitance in DC Circuits

4.

How long after the switch is closed will the voltage across the capacitor of Figure 13-35 increase to 30 V? (a) 20 s (b) 19.8 s (c) 19.6 s (d) 19 s

5.

What is the rate of change of potential difference across the capacitor of Figure 13-36 when the switch is moved to position 2 with the capacitor fully charged? (a) 125 kV/s (b) −125 V/s (c) −125 kV/s (d) 125 V/s 1 R1 2

2.0 MΩ C1 1.0 μF

V1 250 V

Figure 13-36

6.

How long after the switch is moved to position 2 will the voltage across the capacitor of Figure 13-36 drop to 100 V? (a) 1.8 ms (b) 1.0 ms (c) 1.9 ms (d) 1.7 ms

7.

Which equation describes the voltage drop across the resistor during the discharge of the capacitor in a simple RC circuit? (a) VR = E + Ee RC −t

(b) VR = E 1 − e RC

( (

−t

(c) VR = E 1 − e RC t

(d) VR = Ee RC −t

) )

Practice Quiz

8.

The energy stored by a capacitor is (a) directly proportional to the potential difference across its­ terminals (b) inversely proportional to the potential difference across its terminals (c) directly proportional to the square of the potential difference across its terminals (d) inversely proportional to the square of the potential difference across its terminals

9.

How much energy is stored by a 100-μF capacitor when it is charged to 1000 V? (a) 50 mJ (b) 100 mJ (c) 5 J (d) 50 J

10. How much energy is stored by the capacitor of Figure 13-36 when it is fully charged? (a) 125 μJ (b) 31.3 mJ (c) 15.6 mJ (d) 62.5 mJ

387

14

Magnetism We now turn our attention to the third basic characteristic of electric ­circuits: inductance. Inductance depends on the relationship between magnetic fields and electric current just as capacitance depends on the relationship between electric fields and potential difference. An understanding of inductance requires an understanding of the nature of magnetism and its interactions with electricity.

Chapter Outline 14-1

Magnetic Fields  390

14-2 Magnetic Field around a Current-Carrying Conductor 393

14-3 Magnetic Flux  396

14-4 Magnetomotive Force  397 14-5 Reluctance 398

14-6 Permeance and Permeability  399 14-7 Magnetic Flux Density  400

14-8 Magnetic Field Strength  401

14-9 Diamagnetic, Paramagnetic, and Ferromagnetic Materials 402

14-10 Permanent Magnets  404

14-11 Magnetization Curves  404

14-12 Permeability from the BH Curve  408 14-13 Hysteresis  410

14-14 Eddy Current  412

14-15 Magnetic Shielding  413

Key Terms magnetic field  390 magnetic line of force (flux)  390 north pole  390 south pole  390 solenoid  394 electromagnet  394 magnetic flux  396 magnetic circuit  397 weber  397 magnetomotive force (MMF)  397 reluctance  398 reciprocal henry  398 permeance  399 permeability  399

relative permeability (μr)  399 magnetic flux density  400 tesla  400 magnetic field strength  401 nonmagnetic  402 diamagnetic  402 paramagnetic  402 ferromagnetic  402 magnetic domains  403 magnetized  403 saturated  404 retentivity  404 residual magnetism (remanence)  404

temporary magnet  404 permanent magnet  404 ferrites  404 magnetization curve (BH curve)  405 normal permeability  408 incremental permeability  409 differential permeability  409 initial permeability (μi)  409 average permeability  409 coercive force  410 hysteresis  411 hysteresis loop  411 eddy current  413

Learning Outcomes At the conclusion of this chapter, you will be able to: • state the properties of magnetic fields • use the right-hand rule to determine the direction of a magnetic field around a straight current-­carrying ­conductor or a solenoid • illustrate electromagnetic induction by means of a ­magnet, solenoid, and galvanometer • explain the relationship between magnetomotive force and the number of turns of wire in a solenoid and the current flowing through it • state the role of reluctance in a magnetic circuit • state the role of permeability in a magnetic circuit • state the relationship between flux density, magnetic field strength, and permeability

Photo sources:  © Tim@awe/Dreamstime/Get-stock

• differentiate among diamagnetic, paramagnetic, and ­ferromagnetic materials • differentiate between permanent and temporary magnets in terms of residual magnetism • determine the permeability of a magnetic material from its BH curve • explain the hysteresis characteristic of a magnetic ­material • explain how an eddy current is established in a solenoid • explain how electric devices may be protected from ­magnetic fields

390

Chapter 14  Magnetism

14-1  Magnetic Fields Two thousand years ago, Chinese mariners used a crude form of magnetic  compass for navigation. In ancient compasses and their modern counterparts, the pointer is a magnet mounted so that it is free to rotate around its centre. This pointer will deflect if another magnet, or any iron object, is brought close to it, demonstrating that magnetic force can act without physical contact. Like electric and gravitational force, magnetic force is a field force. The definition of a magnetic field is similar to that of an electric field. A magnetic field is that region in which a magnetic material is acted upon by a magnetic force.

Some theories of ­subatomic particles predict the existence of a magnetic ­monopole, a tiny form of magnet with only one pole. To date, there has been no ­direct experimental evidence of magnetic monopoles.

As shown in Figure 14-1, the strength and direction of a magnetic field can be represented by magnetic lines of force (or magnetic flux lines). Every magnet has two areas, called a north pole and a south pole, where the lines of force are concentrated. Like poles repel each other and oppo­ site poles attract. One pole of Earth’s magnetic field is in the Canadian arctic near 83° north latitude. The north pole of a magnet is the one that is ­attracted to this northern pole of Earth’s magnetic field.

N

S

Figure 14-1  The magnetic field around a bar magnet

Like electric lines of force, magnetic lines of force are imaginary. They are simply a way of representing a magnetic field. The strength and direction of magnetic fields have been measured by means ranging from ­observing patterns in iron filings or the deflections of compass needles to readings from sophisticated electronic sensors. The nature of magnetic fields is such that the lines of force must have the five characteristics ­described below. At any point in a magnetic field, the direction of a line of force is the direction of the magnetic force that the field would exert on a north pole at that point.

14-1   Magnetic Fields

If we bring one end of a bar magnet close to a compass needle, the compass needle will align itself with the magnetic lines of force. If we now bring the opposite pole of the bar magnet near the compass, the other end of its n ­ eedle will move toward the bar magnet. Hence, the direction of the ­magnetic field at one pole of the bar magnet must be opposite to that at the other pole. By convention, the positive direction of magnetic lines of force is defined so that the lines leave the north pole of a magnet and enter the south pole. Magnetic lines of force always form complete loops. Magnetic lines of force do not begin at the north pole of a magnet and end at the south pole. They continue between south pole and north pole ­inside the magnet to form complete closed loops. If we cut the magnet of Figure 14-1 in two and pull the two sections apart, the magnetic lines of force cross the gap and still form continuous loops. New north and south poles are formed on each side of the gap, as shown in Figure 14-2(a). No matter how many times we cut the magnet, each section has its own north pole and south pole.

N

S

N

S

(a)

N

S

S

N

(b)

Figure 14-2  Magnetic lines of force between unlike and like poles

391

Chapter 14  Magnetism

Magnetic lines of force tend to be as short as possible. When we separate the two magnets in Figure 14-2(a), we lengthen the magnetic lines of force, and the magnetic force opposes this motion. We can think of the magnetic lines of force as trying to pull the two parts back together in order shorten the loops. In this respect, the magnetic lines of force act somewhat like elastic bands. The tendency for magnetic lines of force to become as short as possible accounts for the attraction between north and south poles. Magnetic lines of force repel one another and cannot intersect. This characteristic accounts for the mutual repulsion of like poles and helps explain the pattern of the magnetic fields in Figure 14-1 and Figure 14-2. The magnetic lines of force diverge as they get farther and farther away from the poles. The magnetic lines of force are kept in equilibrium by each line tending to shrink its loop length and at the same time being p ­ revented from doing so by the repulsion of the next smaller loop. The magnetic field around a uniform magnet is symmetrical unless there are other magnetic materials in the field. The magnetic field is distributed so as to maximize the number of magnetic lines of force.

S

When a piece of soft iron enters the field of a magnet, the symmetrical pattern of the magnetic lines of force is altered as shown in Figure 14-3. Some of the lines of force detour to pass through the iron because it permits a greater concentration of lines of force than air does. As a result the number of lines of force increases. The magnetic lines of force that pass through the iron object will become shorter if it moves closer to the magnet. Therefore, the magnetic field attracts iron objects to the closer pole of a magnet.

N

392

Soft iron Bar magnet

N

S

Figure 14-3  Soft-iron bar in a magnetic field

Note that some characteristics of magnetic lines of force differ from those of electric lines of force. See Review Questions 14-27 to 14-30 at the end of the chapter.

14-2   Magnetic Field around a Current-Carrying Conductor

393

14-2 Magnetic Field around a Current-Carrying Conductor In 1819, the Danish physicist Hans Christian Oersted (1777–1851) discovered that a  current-carrying conductor deflected the needle of a nearby magnetic ­compass, as shown in Figure 14-4. Oersted thus demonstrated that a moving charge ­produces a magnetic field. The magnetic lines of force form concentric circles perpendicular to the axis of the conductor. As we move outward from the centre of the conductor, these circular lines of force are spaced farther and farther apart, and the strength of the magnetic field decreases. When the current is turned off, the lines of force collapse back into the conductor and the magnetic field disappears. Current-carrying conductor passing through sheet of cardboard

Compass needle

+

+

−

−

(a)

(b)

Figure 14-4  Detecting the magnetic field around a current-carrying conductor

Although the magnetic field around a straight current-carrying conductor has no north and south poles, the lines of force still have direction, as shown by the compasses in Figure 14-4(b). Reversing the direction of the current through the conductor causes all the compass needles to reverse their positions. We can determine the direction of the magnetic field around a straight current-carrying conductor by the right-hand rule illustrated in Figure 14-5. When the thumb of a right hand points in the direction of conventional current through a conductor, the fingers point in the direction of the magnetic lines of force when grasping the conductor. Right hand

Direction of flux

Conventional current

−

+

Figure 14-5  Right-hand rule for a straight conductor

Modern atomic theory suggests that all magnetic properties of matter stem from the motion of electrons within atoms.

394

Chapter 14  Magnetism

If we form the current-carrying conductor into a loop, as in Figure 14-6, the magnetic lines of force all pass through the centre of the loop in the same direction. For a given current, the total number of lines of force has not changed; therefore, we have strengthened the magnetic field by concentrating it into a smaller area.

−

+

Figure 14-6  Concentrating the magnetic field by forming the conductor into a loop

We can concentrate the magnetic field of the conductor even more by winding the conductor around a cylindrical core to form a solenoid, as in Figure 14-7. Because the current in the adjacent turns of the solenoid travels in the same direction around the coil, the solenoid acts like a single loop of a stranded conductor, with each strand carrying a current equal to the

S

− +

N

Figure 14-7  Magnetic field around a current-carrying solenoid

actual current through the solenoid. As shown in Figure 14-7, the magnetic field around the solenoid has a pattern similar to that around a bar magnet. This solenoid is a simple type of electromagnet. We can again apply the right-hand rule to find the direction of the magnetic lines of force around a solenoid, as illustrated in Figure 14-8. When the fingers of a right hand circling a solenoid point in the ­direction of conventional current, the thumb points in the direction of the magnetic lines of force through the centre of the solenoid.

Source:  © Tim@awe/Dreamstime/Get-stock

14-2   Magnetic Field around a Current-Carrying Conductor

Iron filings on a sheet of paper resting on top of a solenoid align with the lines of magnetic force around the solenoid.

Direction of flux

Right hand N

S

+

−

Conventional current

Figure 14-8  Right-hand rule for a solenoid

Because the solenoid acts as an electromagnet, we can treat the end of the solenoid where the direction of the magnetic lines of force is away from the coil (Figure 14-8) as its north pole. Therefore, the thumb in the righthand rule also indicates the north pole end of an electromagnet.

395

396

Chapter 14  Magnetism

A magnetic force exists between two parallel current-carrying conductors. Figure 14-9(a) shows a cross section of equal currents flowing in opposite directions through two parallel conductors. The cross on the conductor on the left represents the tail feathers of an arrow, indicating that conventional current is flowing into the page. Similarly, the dot on the conductor on the right represents the point of the arrow, indicating that conventional current is flowing out of the page. If we imagine a right hand with the thumb pointing down into the left conductor, we can verify the direction of the magnetic lines of force. Between the two conductors, the magnetic lines of force all have the same direction. Since they repel one another, there is a resultant magnetic force of repulsion between the two conductors.

×



×

(a)

×

(b)

Figure 14-9  Magnetic field around parallel current-carrying conductors

Starting in 2019, the definition of the ampere will be based on the fixed value of the elementary charge, e, rather than the magnetic force described in this section. While no longer ­considered the ­defining terms of the ampere, the magnetic force described in this section is still measurable and real.

In Figure 14-9(b), the conventional current in both conductors flows into the page. The magnetic lines of force between the conductors have opposite directions. The outer magnetic lines of force around each conductor have combined to form a shorter single loop that encompasses both conductors. The tendency of these lines of force to become even shorter results in a magnetic force of attraction between the two conductors. The magnitudes of these magnetic forces depend on the length and spacing of the two conductors and on the magnitude of the currents through them. In the International System of Units prior to 2019, the ampere was ­defined in terms of the magnetic force between two current-carrying electric conductors: An ampere is the constant current that produces a force of 2 × 10−7 N/m when flowing through two straight parallel conductors of ­infinite length and negligible cross section 1 m apart in a v ­ acuum. See Problems 14-1 to 14-4 and Review Question 14-31.

14-3  Magnetic Flux We have been using magnetic lines of force to represent magnetic flux in diagrams of magnetic fields. We can also think of magnetic flux as the average magnetic field passing through a perpendicular surface times the area of the surface.

14-4   Magnetomotive Force

397

The letter symbol for magnetic flux is the Greek letter Φ (phi). For current to flow in an electric circuit, the circuit must form a closed loop. Since magnetic lines of force always form closed loops, we can refer to the path we can trace around these loops as a magnetic circuit. Although there is no flow of magnetic particles comparable to the flow of electrons in an electric circuit, there are many similarities between the properties of magnetic circuits and those of electric circuits. Faraday connected a galvanometer across a solenoid, as shown in Figure 14-10. When he thrust a permanent magnet into the solenoid, he noticed a momentary deflection of the galvanometer’s pointer, indicating that the moving magnetic field induced a current in the solenoid. Faraday discovered that the induced voltage appeared only when he moved the magnet with respect to the electric circuit, and that the faster he moved the magnet, the greater the deflection of the meter pointer. Thus, the magnetic field links the magnetic circuit to the electric circuit and produces an EMF in the solenoid. Faraday’s discovery provided a means of specifying the basic unit of magnetic flux, the weber.

The SI unit of magnetic flux was named in honor of the German physicist Wilhelm ­Eduard Weber (1804–91), who ­established many of the fundamental theories of magnetism.

Solenoid

N

S

Motion

Bar magnet G Galvanometer

Figure 14-10  Faraday’s demonstration of electromagnetic induction

A change of one weber per second in the magnetic flux linking a single-turn coil induces an average EMF of one volt in the coil. The unit symbol for weber is Wb. A weber is equivalent to a volt second (1 Wb = 1 V·s). See Review Question 14-32.

14-4  Magnetomotive Force Just as an electric current cannot flow in an electric circuit until the circuit is connected to a voltage source, a magnetomotive force (MMF) is necessary to produce magnetic flux. Magnetomotive force in a magnetic circuit is the counterpart of electromotive force in an electric circuit. Like EMF, MMF is not actually a force that could be measured in newtons. Since magnetic flux in a solenoid appears only when current flows through the coil,

Although current ­standards avoid the use of script letters, the script letter 𝔉 is sometimes used to ­distinguish magnetomotive force from magnetic force, which also has the symbol Fm.

398

Chapter 14  Magnetism

­ agnetomotive force must be a direct result of the current. Therefore, the m unit of MMF is based on the current in a single-turn coil of wire. The letter symbol for magnetomotive force is Fm. The ampere is the SI unit of magnetomotive force. Since each turn of wire added to a solenoid increases the strength of the magnetic field in the solenoid by the same amount, the magnetomotive force is the product of the coil current and the number of turns in the coil. Fm = NI (14-1)

The ampere-turn is not an SI unit and is no longer used by the Institute of Electrical and Electronics ­Engineers (IEEE).

where Fm is the magnetomotive force in amperes (effective), N is the number of turns of wire in the coil, and I is the actual current through the coil in amperes. To avoid confusion between effective amperes of MMF and actual amperes of current, this book uses the ampere-turn (At) as the unit for MMF. For ­example, the MMF of a 10-turn coil carrying a current of 2.0 A is 20 At. See Problem 14-5.

14-5 Reluctance For a given magnetic circuit of nonmagnetic materials, the magnetic flux is directly proportional to the applied magnetomotive force. This constant is called the reluctance of the magnetic circuit. Reluctance in a magnetic circuit is the counterpart of resistance in an electric circuit. Some books use the script letter ℜ to ­distinguish reluctance from resistance, R.

Reluctance is the opposition of a magnetic circuit to the establishing of magnetic flux. The letter symbol for reluctance is Rm.

Rm =

Fm Φ

(14-2)

where Rm is reluctance, Fm is magnetomotive force in amperes, and Φ is the magnetic flux in webers.

Section 16-5 describes the henry, which is the SI unit for ­inductance.

Note how Equation 14-2 closely resembles the equation for Ohm’s law, R = V/I, with reluctance corresponding to resistance, magnetomotive force corresponding to voltage (or electromotive force), and magnetic flux corresponding to current. Reluctance can be expressed in ampere-turns per weber. The SI unit of ­reluctance is the reciprocal henry (symbol H−1), which is equal to one ­ampere per weber: 1 H−1 = 1 A/Wb. See Problems 14-6 to 14-8 and Review Question 14-33.

14-6   Permeance and Permeability

399

14-6  Permeance and Permeability When dealing with parallel electric circuits, we found it more convenient to think in terms of the conductances of the circuit elements rather than in terms of their resistances. Similarly, when dealing with magnetic circuits, we find it more convenient to think in terms of the ability of magnetic circuits to permit magnetic flux rather than in terms of their opposition to it. Permeance in a magnetic circuit is the counterpart of conductance in an electric circuit. Permeance is a measure of the ability of a magnetic circuit to permit magnetic flux. The letter symbol for permeance is Pm.

Pm =

1 Rm

(14-3)

We can express permeance in webers per ampere-turn. The SI unit of permeance is the henry. To compare the magnetic properties of various materials, we calculate the permeance per unit length and cross section, which is called ­permeability. Permeability indicates the ability of a material to permit magnetic flux, whereas permeance is a measure of the ability of a given magnetic circuit to permit magnetic flux. Permeability is comparable to permittivity, which is a measure of the ability of a material to permit electric flux. The letter symbol for permeability is the italic lowercase Greek ­letter μ (mu).

μ = Pm

l l = A RmA

(14-4)

where μ is the permeability in ampere-turns per metre, Pm is permeance in webers per ampere-turn, Rm is reluctance in ampere-turns per weber, l is the length of the magnetic circuit in metres, and A is the cross-sectional area in square metres. In SI, permeability is expressed in henrys per metre.

Note that the Greek letter μ (not italic) is also used for the prefix micro. Fortunately, this prefix is rarely needed in calculations for magnetic circuits. As with permittivity, it is often convenient to express permeability in terms of the ratio of the permeability of a material to the permeability of free space, μ0, which is a physical constant. This ratio is referred to as relative permeability, μr. Since it is a ratio, relative permeability has no units. μ μr = (14-5) μ0 where μ0 = 4π × 10−7 H/m. For all non-magnetic materials, μr = 1. For magnetic materials, μr ranges from 50 to 25 000.

Some books use the script letter 𝔓 to distinguish permeance from power, P.

400

Chapter 14  Magnetism

Example 14-1 A solenoid of 1000 turns is energized by a current of 250 mA. A fluxmeter measures a total flux of 1.0 mWb in this magnetic circuit. The core has an average path length of 25 cm and a cross-sectional area of 6.5 cm2. Calculate (a) the reluctance of the magnetic circuit (b) the permeability of the core (c) the relative permeability of the core Solution (a) Fm = NI = 1000 × 0.25 A = 250 At

Rm =

(b) μ =

Fm 250 At = 2.5 × 105 At/ Wb = Φ 1.0 × 10 − 3 Wb 1

RmA

(c) μr =

=

0.25 m = 1.5 × 10 − 3H/m ( 2.5 × 10 At/Wb ) ( 6.5 × 10 −4 m2 ) 5

μ 1.54 × 10 − 3 = = 1.2 × 103 μ0 4π × 10 − 7

See Problems 14-9 to 14-16 and Review Question 14-34.

Circuit Check

A

CC 14-1. Find the current that will develop a magnetomotive force of 180 At when flowing through a 1500-turn coil. CC 14-2. How much flux will be produced if a current of 4.0 A flows in a 1600-turn coil wound on a magnetic circuit having a reluctance of 5.8 MAt/Wb?

14-7  Magnetic Flux Density The tesla is named in honor of the inventor Nikola Tesla (1856–1943), who ­discovered key ­principles of alternating current and ­developed the first AC induction motor.

For calculations involving permeability we are not dealing with total flux in a magnetic circuit. Rather, we are interested in the magnetic flux density. Magnetic flux density is the magnetic flux per unit area. The letter symbol for flux density is B. The tesla is the SI unit of flux density. The unit symbol for tesla is T. One tesla equals one weber per square metre: 1 T = 1 Wb/m2

14-8   Magnetic Field Strength

B=



Φ A

401

(14-6)

where B is the magnitude of the magnetic flux density in teslas, Φ is the magnitude of the total magnetic flux in a magnetic circuit, and A is the cross-sectional area of the circuit in square metres. See Problem 14-17 and Review Question 14-35.

Circuit Check

B

CC 14-3. Determine the reluctance for the coil in Example 14-1 if the flux is 4 mWb. CC 14-4. Determine the reluctance of an iron rod 3.0 cm in diameter and 8.0 cm long. The relative permeability of the iron is 3500. CC 14-5. Calculate the flux density for the magnetic circuit in Example 14-1.

14-8  Magnetic Field Strength Since permeability is based on a unit cube of the material, we are not concerned with the magnetomotive force required for the full length of the magnetic circuit. Rather, we are interested in the magnetomotive force required to create a certain flux density in a unit length of the magnetic circuit. Magnetic field strength is magnetomotive force per unit length. The letter symbol for magnetic field strength is H. H=



Fm l

(14-7)

where H is the magnitude of the magnetic field strength in ampereturns per metre, Fm is the magnetomotive force in amperes, and l is the length of the magnetic circuit in metres. In SI, magnetic field strength is expressed in effective amperes per metre. Now that we have defined permeability, flux density, and magnetic field strength, we can derive a very useful relationship among these three magnetic quantities. Since Rm =

Fm , Φ

Substituting in Equation 14-4 gives

μ=

Pm =

Φ Fm

Φ l Φ l × = × Fm A A Fm

Magnetic field strength is also called magnetic field ­intensity, magnetizing force, and magnetomotive force gradient.

402

Chapter 14  Magnetism

But

Therefore,

Φ =B A

and μ=

Fm = H l B H

(14-8)

where μ is permeability in henrys per metre, B is the magnitude of the magnetic flux density in teslas, and H is the magnitude of the magnetic field strength in ampere-turns per metre (or amperes per metre). Comparing Equation 14-8 with Equation 12-8, we note that: • magnetic flux density in a magnetic circuit is the counterpart of electric flux density in a dielectric in an electric circuit • magnetic field strength in a magnetic circuit is the counterpart of electric field strength in an electric circuit • permeability in a magnetic circuit is the counterpart of permittivity in an electric circuit See Problem 14-18 and Review Question 14-36.

14-9 Diamagnetic, Paramagnetic, and Ferromagnetic Materials The way atoms in a material respond to a magnetic field determines the ­permeability of the material, which, in turn, governs the magnetic flux that can be established by a given magnetomotive force. However, the behaviour of materials in magnetic circuits is not as simple as the polarization of atoms of a dielectric shown in Figure 12-3. Most substances have no more effect on a magnetic circuit than free space does. These nonmagnetic materials have the same permeability as free space. A few materials have a permeability slightly less than that of free space. Such diamagnetic materials oppose magnetic flux slightly more than free space does. Silver, copper, and hydrogen are diamagnetic. A few materials have a permeability just slightly greater than that of free space. These paramagnetic materials are, therefore, slightly magnetic. Platinum, aluminum, and oxygen are paramagnetic. For most ­applications, we can treat diamagnetic and paramagnetic materials as if they were nonmagnetic. Some substances, including iron and its compounds (and, to a lesser extent, nickel and cobalt), have permeabilities many hundreds of times greater than that of free space. These substances are called ferromagnetic materials since iron is the significant element in this group. We can explain the magnetic properties of materials in terms of the ­magnetic fields produced by moving charges within the atoms in the materials. Every electron travelling in an orbit around the nucleus of an atom is somewhat

14-9   Diamagnetic, Paramagnetic, and Ferromagnetic Materials

like a tiny electric current flowing in a single-turn loop. Consequently, every orbital electron generates a tiny magnetic field (and magnetic flux) ­perpendicular to the plane of its orbit. Each electron also behaves as if it were spinning on an axis through its centre. Thus the charge on the electron circulates around the axis through the electron, generating an additional magnetic field. The ­positively charged nucleus can have a similar magnetic field. In most materials the orbital and spin magnetic fields all cancel out, so the atoms have no net magnetic field. For this reason, nonmagnetic materials have virtually the same permeability as a vacuum. However, an external magnetic field can shift the orbitals of the electrons, causing the atoms to have a slight net magnetic field that opposes the external field. Hence, the diamagnetism reduces the net strength of the magnetic field in a material. All materials are diamagnetic to some degree. The atoms in paramagnetic materials have a small net magnetic field. These atoms tend to behave individually like permanent magnets. When we expose paramagnetic materials to an external magnetic field, their atoms tend to align their magnetic fields with the external field, thus slightly increasing the total magnetic field and the resulting flux. In paramagnetic materials, this effect is stronger than the opposing diamagnetism. The atoms in ferromagnetic materials have appreciable net magnetic fields. As a result, these atoms tend to form magnetic domains, groups of atoms that have aligned their magnetic fields with each other. A domain usually contains from 1012 to 1018 atoms. Ferromagnetic materials are made up of crystals with the atoms arranged in a regular lattice. Initially, the magnetic fields of the domains within each crystal are aligned with the edges of the faces of the crystal lattice. Since half of the fields aligned with each of these axes are directed one way and the other half are directed the opposite way, the ferromagnetic material has no net magnetic field. When we place a ferromagnetic material inside a solenoid, as in Fig­ure 14-11, and gradually increase the current in the coil, the material ­becomes magnetized. At first the domains that have magnetic fields oriented close to the direction of  the magnetic field in the solenoid expand as adjacent atoms join them. As the solenoid current increases, the other domains shift to align with the lattice edges that are closest to being parallel to the magnetic field. This shift greatly increases the total field strength

Iron sample

−

+ Rheostat

Figure 14-11  Magnetizing an iron sample

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Chapter 14  Magnetism

in the domains, making the permeability appear very high. Then, as the ­magnetizing force increases even further, the orientation of the domains finally shifts to align with the magnetic flux rather than with the lattice edges of the crystals. If we increase the current to a point where all the domains are aligned with the magnetic field, any further increase in ­current can only increase the total flux by the same amount as if the iron were not present. The iron is now saturated and its permeability has dropped to that of any nonmagnetic material. The current that is needed to saturate a ­sample of  iron within the solenoid depends on the type and grade of iron used. See Review Questions 14-37 and 14-38.

14-10  Permanent Magnets When the current in the magnetizing solenoid of Figure 14-11 is switched off, the magnetic domains tend to return to their original orientations. In materials that have a low magnetic retentivity, such as soft iron, the ­residual magnetism or remanence is practically nil. Such materials are classified as temporary magnets. Residual magnetism in ferromagnetic materials is the magnetic counterpart of dielectric absorption (described in Section 12-2). In permanent magnets, a large proportion of the domains retain the orientation they were forced to take in the strong magnetic field produced by the energizing coil. The permanent magnets for meter movements, loudspeakers, and similar devices are often made of alnico, a very hard alloy containing iron, aluminum, nickel, and cobalt. Because of its hardness, an alnico magnet is usually cast in the desired shape and subjected to a very high magnetic flux density produced by an electromagnet while it is cooling. Ceramics containing iron oxides can have strong ferromagnetic properties. Among these ferrites are some that make excellent permanent magnets. Such ceramic magnets can be formed by grinding barium carbonate and ferric oxide into particles the size of a magnetic domain, pressing the powder into the desired shape, and firing it in an oxygen atmosphere. Unlike iron magnetic materials, ferrites are poor electric conductors. See Review Question 14-39.

14-11  Magnetization Curves The permeability of a sample of iron varies considerably as the current in the coil is increased. The permeability depends on what percentage of the magnetic domains has aligned with the magnetic field. This percentage, in

14-11   Magnetization Curves

turn, is dependent on the flux density. Even though we can draw a graph, as in Figure 14-12, showing how the permeability of a sample of iron varies with flux density, we are faced with the interdependence that becomes ­evident when we rearrange Equation 14-8:

B = μH

Permeability, µ

For a given magnetic field strength H, the flux density B depends on the permeability μ, which depends on the flux density B. However, we can disregard permeability and concern ourselves only with the manner in which the flux density of a given material varies with the magnetic field strength. Since B and H are both defined in terms of a cube with unit dimensions, the magnetization curve, or BH curve, (produced by plotting a graph of the manner in which B varies with H) is dependent only on the type of material and not on its dimensions.

Magnetic Flux Density, B

Figure 14-12  Variation of the permeability of cast steel with flux density

Magnetic Flux Density, B

The magnetization curve of Figure 14-13 can be divided into three ­sections. The magnetic flux density increases slowly at first. The lower knee of the BH curve corresponds to the conditions where the domains that are most nearly aligned with the applied magnetic field grow at the expense of neighbouring domains. In the steep portion of the BH curve, the domains are aligning with whichever axes of the crystals are closest to the direction of the applied magnetic field. When the magnetic field is great enough to i­ncrease the flux density beyond the upper knee of the BH curve, the magnetic field strength is great enough to force

Saturation Upper knee

Lower knee Magnetic Field Strength, H

Figure 14-13  Typical BH curve for cast steel

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Chapter 14  Magnetism

the domains to align themselves with the external magnetic field rather than with the axes of the crystals. F ­ inally, the BH curve becomes a straight line with a gradual slope when all the domains are aligned with the magnetic field. This slope shows the same increase in B for a given increase in H as for any nonmagnetic material. It is difficult to mark the exact point on the BH curve of Figure 14-13 where saturation occurs. Theoretical saturation occurs when all the domains are finally aligned with the external magnetic field. For some types of iron, this saturation requires considerable magnetic field strength. Practical saturation may be defined as the flux density beyond which it is impractical to magnetize a certain magnetic material. For some applications, this point is the upper knee in the BH curve where the steep, linear part of the slope ends. In high-quality audio transformers, magnetic field strength must be kept below this point, since any swing in H past this point causes distortion in the output signal. However, we can reduce the amount of iron needed for a power transformer if we design it so that the magnetic cores are magnetized to a point slightly beyond the upper knee of the BH curve. The gradual saturation of most types of iron, as shown in Figure 14-15, is due to the random orientation of the iron crystals. If iron can be manufactured so that all crystals have the same orientation, the flux density will rise very steeply with an increase in magnetic field strength. At a certain value of magnetic field strength, the iron will suddenly become saturated (as shown in Figure 14-14) because the same magnetic field strength is needed to force all of the domains to align themselves with the magnetic field. This type of iron was typically used for magnetic recording of audio and video signals as well as computer data and measurements in videotapes, floppy disks, scientific and medical research equipment, and hard drives. Magnetic Flux Density, B

406

Saturation

Magnetic Field Strength, H

Figure 14-14  BH curve for iron with all crystals having the same orientation

Figure 14-15 shows typical magnetization curves for a number of ferromagnetic materials used in practical applications. See Problems 14-19 and 14-20 and Review Questions 14-40 and 14-41.

14-11   Magnetization Curves

407

1.8 1.7

Sheet

1.6

steel el

t ste

Cas

1.5 1.4 1.3

Magnetic Flux Density, B (teslas)

1.2 1.1 1.0 0.9 0.8 0.7 n

t iro

Cas

0.6 0.5 0.4 0.3 0.2 0.1 0

0

400

800

1200

1600

2000 2400 2800 3200 3600 4000 4400 4800 Magnetic Field Strength, H (ampere-turns per metre)

Figure 14-15  Typical magnetization curves

5200

5600

6000

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Chapter 14  Magnetism

14-12  Permeability from the BH Curve As shown in Figure 14-12, the permeability of a ferromagnetic material is far from constant since it is quite dependent on flux density. However, we can determine permeability at a given magnetic flux density by applying the relationship μ = B/H (Equation 14-8) to the BH curve for a given type of magnetic material. The manner in which we apply this relationship depends on the conditions under which the magnetic circuit is to be operated. If the current in the energizing solenoid is to be kept constant, we can locate the appropriate value for magnetic field strength and the resulting value of flux density on the BH curve, as shown in Figure 14-16(a). The normal permeability for this flux density is simply μ=



B H

(14-8)

Magnetic Flux Density, B

Magnetic Flux Density, B

In some magnetic circuits, the magnetic field strength varies between two limits, as shown by ΔH in Figure 14-16(b). This situation occurs when a fluctuating direct current flows in the coil, as in many audio transformers.

μ= B H

ΔB μΔ = ΔB ΔH

ΔH Magnetic Field Strength, H (b) Incremental permeability

Magnetic Field Strength, H

dH

dB

X μd = dB dH

Magnetic Field Strength, H (c) Differential permeability

Magnetic Flux Density, B

(a) Normal permeability Magnetic Flux Density, B

408

Knee

μav = B H

Magnetic Field Strength, H (d) Average permeability

Figure 14-16  Determining permeability from a BH curve

14-12   Permeability from the BH Curve

In this case, we are interested in the permeability over a limited operating range. This incremental permeability is



μΔ =

ΔB ΔH

(14-9)

where μΔ is the incremental permeability, ΔB is the difference between the maximum and minimum magnitudes of magnetic flux density, and ΔH is the difference between the maximum and minimum magnitudes of magnetic field strength. If alternating current flows in the solenoid, the magnetic field strength is constantly changing. In this case, we are likely to be concerned with the permeability as the magnetic field strength swings through a certain value, rather than as it maintains that value. This particular value of permeability is called the differential permeability for a particular point on the BH curve. We determine differential permeability from the slope of the magnetization curve at the point in question. To calculate the slope of the BH curve at point X in Figure 14-16(c), we draw a tangent to the curve at point X and then determine dB/dH from the slope of the tangent. The differential permeability at point X is

μd =

dB dH

(14-10)

In electronic circuits where the current is in the order of microamperes, the magnetic field strength is so small that the B/H ratio is difficult to read from the magnetization curve. However, we can use a tangent to calculate the permeability for very small values of H and B. Thus, initial permeability (μi) is the differential permeability when the magnetic field strength is zero or very small. We can simplify the problem of solving magnetic circuits in which the magnetic field strength and, consequently, the permeability are continually changing by determining an average permeability. The average permeability is obtained by drawing a straight line through the origin of the BH graph and the upper knee of the curve as in Figure 14-16(d). Because this line is straight, we can pick any point on this line to read off the corresponding values of B and H for determining μav. As long as the flux density in a magnetic circuit is kept below saturation, this average permeability is reasonably accurate. See Problems 14-21 to 14-26 and Review Question 14-42.

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Chapter 14  Magnetism

Circuit Check

C

CC 14-6. For the magnetic circuit shown in Figure 14-17, the total flux ­produced is 2.0 mWb. Given that the coil has 1250 turns and the ring has a circular cross section with a diameter of 2.0 cm, calculate the magnetic field strength and the permeability.

2.5 A

+ E

R12.5 cm

−

Figure 14-17

CC 14-7. Use the magnetization curves in Figure 14-15 to determine the normal permeability of sheet steel at a flux density of 1.15 T. CC 14-8. Determine the average permeability of cast iron.

Magnetic Flux Density, B

14-13 Hysteresis

+ A

B

−

C

F

O E

+

Magnetic Field Strength, H

D

− Figure 14-18  Typical hysteresis loop

The BH curves of Figures 14-12 to 14-16 represent the manner in which the flux density of an unmagnetized sample of iron rises as the value of H is increased. However, they do not represent the manner in which the flux density drops as the magnetic field strength is decreased. When the magnetizing force has returned to zero, the residual magnetism of the iron will still produce an appreciable magnetic flux density. This is referred to as r­ etentivity. To get rid of this residual flux, it is necessary to pass some current through the solenoid in the opposite direction. The amount of ­reverse magnetic field strength or magnetizing force ­required to demagnetize a particular sample of material is called the coercive force. Figure 14-18 shows the complete cycle of magnetization of a piece of iron as we pass an alternating current through the solenoid. If we start with an unmagnetized sample and increase the magnetic field strength H from zero to the maximum positive value, the magnetic flux density B increases along the magnetization curve OA. As we reduce H to zero, the flux density decreases from saturation to a residual flux density at point B. To bring the flux density in the iron to zero, we must reverse the current through the coil and increase it until H reaches the value represented

14-13  Hysteresis

by OC. As we increase the reverse current, the iron saturates at point D (with the magnetic domains all oriented in the direction opposite to the orientation they had at point A). When we reduce the reverse current to zero, the flux density in the iron ­decreases only to the value at point E. We run current through the solenoid in the original direction to bring the flux density to zero at point F. If we again ­increase H to its maximum positive value, the iron again becomes saturated and the flux density returns to its value at point A. As shown in Figure 14-18, when the magnetic field strength is due to an alternating current, the flux density of ferromagnetic materials tends to lag behind the magnetic field strength that creates it. This lagging of the magnetization of the iron behind the magnetic field strength is called hysteresis. The graph that shows this lag is called a hysteresis loop. Since only ferromagnetic materials have residual magnetism, nonmagnetic materials show no hysteresis effect. Hysteresis occurs because some of the magnetic domains in a ferromagnetic material do not return to their original orientation. They have to be forced to do so by coercive force from a reversed magnetic field. Whenever motion is accomplished against an opposing force, an energy transfer must take place. This energy comes from the source of alternating voltage and is transferred to the molecules of the ferromagnetic material in the form of heat. The higher the frequency of the alternating current in the solenoid, the more rapidly the magnetic domains have to change their alignment and, therefore, the greater the hysteresis loss. The greater the retentivity of a particular type of iron, the greater the ­coercive force needed to demagnetize it. The greater the opposition by the magnetic domains to reorientation, the greater the heating of the ­ferromagnetic material. Therefore, hysteresis loss is also proportional to the retentivity of the iron. The iron used in the magnetic circuits of transformers for AC circuits should have as little retentivity as possible in order to minimize hysteresis losses. In graphs like Figure 14-18, the area within the hysteresis loop increases as the residual flux density increases. Therefore, the area within a hysteresis loop is a useful indication of hysteresis loss. Hysteresis can be used to advantage in some applications. Certain f­errites composed of oxides of magnesium, manganese, and iron have an almost rectangular hysteresis loop, as shown in Figure 14-19. As the magnetic field strength H is gradually increased in a positive direction, the flux density in  the core remains unchanged until the magnetic field strength reaches a ­critical level, at which the core suddenly magnetizes and ­saturates in the positive direction. A sudden demagnetization also occurs when the magnetic field strength is gradually increased in the negative direction. Consequently, these ferrite cores are always saturated in either the positive or negative ­direction. These two stable states of ferrite cores make them very useful for switching circuits.

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Chapter 14  Magnetism

Magnetic Flux Density, B

412

+

−

+ Magnetic Field Strength, H

− Figure 14-19  Rectangular hysteresis loop of a ferrite switching core

See Review Questions 14-43 and 14-44.

14-14  Eddy Current In Section 14-3, we found that moving a magnetic field in relation to an electric conductor induced a voltage in the conductor. The direction of the magnetic lines of force, the motion of the magnetic lines of force with respect to the conductor, and the flow of current are all at right angles to one another. If we place a metal core inside a solenoid, any motion or change in the magnetic field will induce a voltage in the metal. As we increase the current in the solenoid, more magnetic lines of force (or flux) must be set up. Crowding more lines of force into the core causes the lines of force already present to move outward along a radius from the axis of the solenoid. The direction of the lines of force themselves is parallel to the axis of the solenoid. When an alternating current flows in the ­solenoid, the lines of force move alternately away from the axis of the core and then back toward it. This changing magnetic field induces a voltage in the metal core of the solenoid. The resulting current must be perpendicular to the radius and the axis of the solenoid. Therefore, the current flows in a circle around the circumference of the metallic core, as shown in Figure 14-20. This circular current is an example of an eddy current. Eddy current can be produced in any conductive core material, including nonmagnetic ­materials. As the eddy current flows through the resistance of the core ­material, the resulting I2R power loss heats the core. Solenoid winding Eddy current flows around circumference of core Alternator Motion of magnetic lines of flux along radii of solenoid Solid metal core

Figure 14-20  Eddy current in a solid metal core in a solenoid (cross section)

14-15   Magnetic Shielding

Eddy-current losses can be greatly reduced by using a core made of thin laminations that are insulated from each other with a thin coat of varnish. As shown in Figure 14-21, the laminations are oriented so that the eddy current would have to flow from lamination to lamination through the varnish. Because of the high resistance of the varnish, the eddy current in a laminated core is very small and the resulting power loss is, in many cases, negligible. Coil Alternator

Laminated steel core

Figure 14-21  Cross section of solenoid with a laminated core

Eddy-current losses in alternating current machinery such as ironcore transformers and electric motors often reduce the efficiency of these devices. However, eddy currents may also be transformed into other types of energy, such as heat and electromagnetic forces. Coreless induction furnaces use eddy currents to melt steel and other ferrous metals while at the same time stirring the molten materials inside their crucibles. Some trains use eddy-current brakes, whose magnetic forces are proportional to the velocity of the train, since they typically produce a smoother stopping motion compared to air brakes. See Review Questions 14-45 and 14-46.

14-15  Magnetic Shielding Although there is no “insulator” for magnetic lines of force, it is possible to shield sensitive equipment from stray magnetic fields by surrounding it with a high-permeability material. As shown in Figure 14-22, most of the magnetic flux follows the lower reluctance path, greatly reducing the flux within the enclosure.

Shielded space

Stray magnetic field

Soft iron cylinder (not to scale)

Figure 14-22  Magnetic shielding

See Review Question 14-47.

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Summary

• A magnetic line of force (or flux) represents the path along which a theoretical isolated magnetic pole would move from one pole of a magnet to another. • Magnetic lines of force form complete loops extending from the north pole to the south pole of a magnet and then through the magnet back to the north pole. • Like poles of magnets repel one another. • A magnetic field exists around a current-carrying conductor. • The right-hand rule may be used to find the direction of the magnetic field around a current-carrying conductor or solenoid. • A moving magnetic field whose flux cuts across a solenoid induces a voltage across the solenoid. • Magnetomotive force is proportional to both the number of turns of wire and the current flowing through a solenoid. • Reluctance is the opposition of a magnetic circuit to the establishing of magnetic flux. • Permeability, a measure of the ability of a magnetic circuit to establish magnetic flux, is equal to the ratio of magnetic flux density to magnetic field strength. • A material is classified as diamagnetic, paramagnetic, or ferromagnetic depending on its permeability compared to that of free space. • Iron and steel alloys may be magnetized either temporarily or permanently depending on their retentivities. • The permeability of a material may be determined from its BH curve. • A magnetic material exhibits hysteresis as it is magnetized in one direction and then magnetized in the opposite direction. • An eddy current is established in a solenoid when it is energized by an alternating current. • Magnetic shielding may be used to protect electric devices from magnetic fields.

+ −

Figure 14-23 

B = beginner

Problems

I = intermediate

A = advanced

B B

GEN

B B

Figure 14-24 

B

Section 14-2  Magnetic Field around a Current-Carrying Conductor 14-1.

Sketch a diagram of the magnetic lines of force around the ­conductor in Figure 14-23 and indicate their direction. 14-2. What is the direction of conventional current through the ­conductor in Figure 14-24? 14-3. Sketch a diagram of the magnetic field around the coil in Figure 14-25 and indicate its direction. 14-4. What is the direction of conventional current through the coil of Fig­­­ure 14-26?

Section 14-4  Magnetomotive Force 14-5. A current of 5 A is flowing through the conductor in Figure 14-25. Find the resulting magnetomotive force.

415

Problems

B B B

B B B B I I B I

I

B

Section 14-5  Reluctance 14-6. Find the reluctance of a magnetic circuit in which a total flux of 2.0 × 10–3 Wb is set up by a 4.0-A current flowing in a solenoid ­consisting of 300 turns of wire. 14-7. Calculate the current that must pass through a 400-turn solenoid to produce a total flux of 1.5 mWb in a magnetic circuit whose reluctance is 1.2 × 106 At/Wb. 14-8. A 250-mA current through a solenoid produces a total flux of 6.0 × 10–4 Wb in a magnetic circuit whose reluctance is 2.5 × 105 At/Wb. How many turns of wire are in the solenoid?

Section 14-6  Permeance and Permeability 14-9. The core of a solenoid consists of a brass cylinder 7 cm in length and 1.5 cm in diameter. Find its reluctance. 14-10. The magnetic circuit in Problem 14-6 has a uniform cross-sectional area of 6.0 cm2 and an average path length of 20 cm. What is the permeability of the core material? 14-11. If the magnets in Figure 14-2(a) have a rectangular cross section of 1 × 2 cm and the spacing between them is 2.5 mm, find the reluctance of the air gap. 14-12. A piece of iron with a length of 8 cm and a rectangular cross section of 1.5 × 2.5 cm has a reluctance of 5.3 × 105 At/Wb. Find the permeability of the iron. 14-13. An iron rod is 50 cm long and has a diameter of 4.0 cm. A 250-turn coil is wound on the rod. When 2.0 A of current is supplied, a flux of 5 mWb is measured. Find the permeability of the iron. 14-14. Calculate the reluctance of an iron rod 100 cm long and 4.0 cm in ­diameter. The relative permeability of the iron is 3000. 14-15. Determine the flux produced if the solenoid in Example 14-1 has 2500 turns. 14-16. A flux of 1.5 mWb is produced when a current of 400 mA flows through a 300-turn coil. The magnetic core is a cylindrical rod of diameter 4.0 cm and length 25 cm. Calculate the (a) reluctance (b) permeability (c) relative permeability

Section 14-7  Magnetic Flux Density 14-17. If 150 turns of wire are wrapped around the iron bar in Problem 14-12 and a current of 1.6 A is supplied, what is the flux density in the iron?

Section 14-8  Magnetic Field Strength 14-18. A magnetic field strength of 2000 At/m produces a flux density of 1.0 T in a certain type of iron. What is its permeability at this flux density?

+ −

Figure 14-25 

S

N

GEN

Figure 14-26 

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Chapter 14  Magnetism

Refer to the typical magnetization curves of Figure 14-15 in solving the r­ emaining problems. B B

B B B B B B

Section 14-11  Magnetization Curves 14-19. What flux density is created in sheet cast by a magnetizing force of 1400 At/m? 14-20. What magnetic field strength is required in order to maintain a ­constant flux density of 1.4 T in sheet steel?

Section 14-12  Permeability from the BH Curve 14-21. Find the normal permeability of cast iron when the flux density ­remains at 0.625 T. 14-22 Find the average permeability of sheet steel. 14-23. Find the incremental permeability of cast iron between flux ­densities of 0.30 T and 0.35 T. 14-24. Determine the differential permeability of cast steel for a magnetic field strength of 1600 At/m. 14-25. Calculate the incremental permeability of cast steel between a ­magnetic field strength of 2000 and 3600 At/m. 14-26. Calculate the differential permeability of sheet steel at a flux ­density of 1.3 T.

Review Questions

Section 14-1  Magnetic Fields 14-27. In what way are gravity, electricity, and magnetism similar? 14-28. Sketch the magnetic field around a horseshoe-shaped perma­nent magnet. 14-29. With reference to the characteristics of magnetic lines of force, show why a nail is attracted to either pole of a horseshoe-shaped permanent magnet. 14-30. State six characteristics of magnetic lines of force.

Section 14-2  Magnetic Field around a Current-Carrying Conductor 14-31. Show with sketches how it is possible to determine the direction of the current in an electric conductor with the aid of a compass needle.

Section 14-3  Magnetic Flux 14-32. How is the magnitude of the weber established?

Section 14-5  Reluctance 14-33. Define the term reluctance.

Section 14-6  Permeance and Permeability 14-34. Define the term permeability.

Review Questions

Section 14-7  Magnetic Flux Density 14-35. Why is the flux density of a magnetic field greater in the centre of a current-carrying solenoid than around a straight wire carrying the same current?

Section 14-8  Magnetic Field Strength 14-36. What is the distinction between magnetomotive force and magnetizing force?

Section 14-9  Diamagnetic, Paramagnetic, and Ferromagnetic Materials 14-37. Compare the permeabilities of nonmagnetic, diamagnetic, and para­ magnetic materials. 14-38. Why can ferromagnetic materials become magnetized whereas para­ magnetic materials cannot?

Section 14-10  Permanent Magnets 14-39. Compare temporary and permanent magnets in terms of residual magnetism.

Section 14-11  Magnetization Curves 14-40. Account for the upper knee on the BH curve of ferromagnetic materials. 14-41. Sketch a BH curve for aluminum.

Section 14-12  Permeability from the BH Curve 14-42. What is the distinction between differential and incremental ­permeability?

Section 14-13  Hysteresis 14-43. Compare permanent and temporary magnets in terms of coercive force. 14-44. Define the term hysteresis.

Section 14-14  Eddy Current 14-45. Laminating the iron core of a transformer greatly reduces the eddy-current losses. What effect does the laminating process have on ­hysteresis losses? Explain. 14-46. Some magnetic circuits consist of powdered iron mixed with a ceramic binder. Compare such a core with a solid core of the same type of iron in terms of (a) permeability (b) hysteresis (c) eddy current

Section 14-15  Magnetic Shielding 14-47. How may a sensitive meter movement be protected from a nearby magnetic field?

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Integrate the Concepts Figure 14-27 shows a solenoid of 1000 turns energized by a current of 250 mA. A flux meter measures a total flux of 1.0 mWb through the solenoid. The core has an average path length of 25 cm and a cross-sectional area of 6.5 cm2. 250 mA

+ E

−

1000-turn solenoid

Figure 14-27 

(a) Draw the magnetic lines of force, given that the wire is wrapped around the solenoid so that the upper lead from the voltage source goes behind the solenoid. (b) Calculate the magnetomotive force. (c) Calculate the reluctance of the solenoid. (d) Calculate the permeance of the solenoid. (e) Calculate the permeability of the solenoid. (f) Calculate the flux density in the solenoid. (g) Calculate the magnetic field strength.

Practice Quiz 1. Which of the following statements are true? (a) Ferromagnetic materials cannot be magnetized. (b) Magnetic lines of force rarely form complete loops. (c) Magnetic lines of force repel each other and cannot intersect. (d) The right hand rule is used to determine the direction of the magnetic field around a current-carrying conductor. 2. The letter symbol for magnetic flux is (a) B (b) A (c) Φ (d) T 3. Magnetomotive force is (a) directly proportional to the number of turns of wire in a coil and directly proportional to the current through the coil (b) inversely proportional to the number of turns of wire in a coil and directly proportional to the current through the coil (c) directly proportional to the number of turns of wire in a coil and inversely proportional to the current through the coil (d) inversely proportional to the number of turns of wire in a coil and inversely proportional to the current through the coil

Practice Quiz

  4. Magnetomotive force can be measured in (a) milliampere-turns (b) ampere-turns (c) microampere-turns (d) all of the above  5. The magnetomotive force produced by a current of 250 mA flowing through in a 20-turn coil is (a) 200 mAt (b) 5.0 At (c) 12.5 mAt (d) 80 At   6. The opposition to magnetic flux is called (a) relative permeability (b) permeability (c) permeance (d) reluctance  7. The current that flows through a 600-turn solenoid that generates a total flux of 7.5 mWb in a magnetic circuit that has a reluctance 1.25 × 105 At/Wb is (a) 15.6 mA (b) 15.6 A (c) 1.56 A (d) 156 mA   8. What property of a magnetic circuit corresponds to resistance in an electric circuit?   9. Permeability is expressed in units of (a) henrys per turn (b) henrys per second (c) henrys per metre (d) henrys per weber 10. The tesla (T) is a unit of magnetic flux density that is equivalent to (a) 1 Wb/m2 (b) 1 Wb/cm2 (c) 1 Wb/cm (d) 1 Wb/m 11. The letter symbol for magnetic field strength is (a) Φ (b) H (c) B (d) T 12. Why should eddy currents in a core be minimized?

419

15

Magnetic Circuits Magnetic circuits are a crucial part of equipment ranging from speakers and relays to large industrial machines and power transmission systems. The principles of magnetic circuits described in this chapter will be of particular interest to students considering further study of motors and power systems.

Chapter Outline 15-1

Practical Magnetic Circuits  422

15-2 Long Air-Core Coils  422 15-3 Toroidal Coils  425

15-4 Linear Magnetic Circuits  425

15-5 Nonlinear Magnetic Circuits  426 15-6 Leakage Flux  429

15-7 Series Magnetic Circuits  430 15-8 Air Gaps  433

15-9 Parallel Magnetic Circuits  435

Key Terms magnetic circuit  422 long air-core coil  422 air-core coil  422 toroidal coil  425

linear magnetic circuits 425 nonlinear magnetic circuit  425

leakage flux  429 leakage factor  429 air gap  433 fringing 433

Learning Outcomes At the conclusion of this chapter, you will be able to: • analyze the magnetic field of a long aircore coil • determine magnetic field variables of a toroidal winding with a nonmagnetic core • use magnetization curves to analyze a nonlinear magnetic circuit

Photo sources:  © iStock.com/thiel_andrzej

• calculate magnetic field characteristics of a series magnetic circuit • calculate the effect of an air gap in a magnetic circuit • analyze the magnetic field of a parallel magnetic circuit

422

Chapter 15   Magnetic Circuits

15-1  Practical Magnetic Circuits For devices such as motors, generators, transformers, speakers, and relays, it is just as important to control the path and the strength of the magnetic fields as it is to control the current. With ferromagnetic materials, we can construct magnetic circuits (Figure 15-1) to control the path of the magnetic flux.

(a) Relay

(b) Meter movement

(c) Two-pole motor

(d) Loudspeaker

Figure 15-1  Common magnetic circuits

In most electric machines, the magnetomotive force is produced by current flowing through coils of wire. The shape and the location of the resulting magnetic field depends, in turn, on the reluctance of the magnetic circuit and the available magnetomotive force. We can determine the MMF required in much the same manner that we determined the applied voltage (resulting from an electromotive force) in a desired current in a circuit with a given resistance. See Review Question 15-16 at the end of the chapter.

15-2  Long Air-Core Coils A long air-core coil is a solenoid that has air or any other nonmagnetic material as a core (an air-core coil) whose core is at least 10 times as long as its diameter. We can calculate fairly accurately the relationship between the flux at the centre of this coil and the current through it.

15-2   Long Air-Core Coils

As shown in Figure 15-2, all the loops of the magnetic lines of force pass through the central part of the coil and through the space outside the coil. Since the permeability of the air outside the coil is the same as that of the nonmagnetic core, about half the lines of force do not travel the whole length of the core. Instead they leave and enter the core through the sides of the coil because magnetic lines repel one another and follow the shortest available path. Therefore, our calculations apply only to the centre of the coil.

+

−

Figure 15-2  Magnetic field around a long air-core coil

Figure 15-2 also shows that all the lines of force that are crowded into  the small cross section at the centre of the coil complete their loops  through a very large cross-sectional area outside the coil. Since the reluctance of a magnetic circuit is inversely proportional to its ­cross-sectional area, the r­ eluctance of the portion of the magnetic path within the coil is much greater than that of the return portion outside the coil. Therefore, For long air-core coils, the total reluctance is approximately equal to the reluctance of the nonmagnetic core of the coil itself. Transposing Equation 14-4 and substituting μ = μ0 gives

Rm =

l (15-1) μ0A

where Rm is the reluctance of the nonmagnetic core in ampere-turns per weber, l is the length of the coil in metres, μ0 is the permeability of the core material (nonmagnetic materials, 4π × 10−7 H/m), and A is the inside cross-sectional area of the coil in square metres.

423

424

Chapter 15   Magnetic Circuits

For air-core coils, μ = μ0, the permeability of free space. Equation 15-1 is the magnetic counterpart of the electric circuit equation R = l/σA, where R is the resistance of the conductor, l is the length of the conductor, σ is the conductivity of the conductor material, and A is the cross-sectional area of the conductor. Having calculated the reluctance of a magnetic circuit, we can readily determine the magnetomotive force for a required magnetic flux by transposing equation 14-2 to get Fm = ΦRm



(15-2)

Example 15-1 An air-core coil 20 cm in length with an inside diameter of 1.0 cm has 1000 turns. What current must flow in the coil to develop a total flux of 10−6 Wb at the centre of the coil? Solution Step 1 Solve for the reluctance of the core.

Rm =

=

l μ0A 4π × 10

−7

20 cm × π × ( 0.50 cm ) 2

= 2.03 × 109 At/Wb

Step 2 Solve for the magnetomotive force with Equation 15-2.

Fm = ΦRm = 10−6 Wb × 2.03 × 109 At/Wb = 2.03 × 103 At

Step 3 Solve for the current in the coil with Equation 14-1.

I=

Fm 2.03 × 103 At = = 2.0 A N 1000 t

See Problems 15-1 to 15-3 and Review Question 15-17.

15-4   Linear Magnetic Circuits

425

15-3  Toroidal Coils If we place an iron core in the coil of wire in Figure 15-2, the high permeability of the iron will reduce the reluctance of the core well below that of the return path outside the core. Now the reluctance of the return circuit outside the core is the major factor in determining the total reluctance of the magnetic circuit. The reluctance outside the coil is considerably more difficult to calculate than the reluctance of the core with its finite dimensions. We can make a magnetic circuit with a much lower reluctance by winding the coil completely around a ring-shaped core, as shown in Figure 15-3. Such toroidal coils confine the magnetic flux to a path completely within the coil.

Source:  © iStock.com/thiel_andrzej

Figure 15-3  A toroidal winding

A toroidal coil mounted on a printed circuit board.

See Review Question 15-18.

15-4  Linear Magnetic Circuits The calculations in Example 15-1 were straightforward because the reluctance of the magnetic circuit depended only on the physical properties of the circuit and was not affected by the applied magnetomotive force or the ­resulting magnetic flux. In such linear magnetic circuits the permeability, μ, is independent of the strength of the magnetic field. As shown in Figure 14-12, the permeability of ferromagnetic materials varies considerably with the magnetic flux density, except for a limited range at the hump of the graph. Hence, the reluctance of practical ferromagnetic circuits changes with variations in magnetomotive force and magnetic flux. Accordingly, ferromagnetic circuits are nonlinear magnetic circuits.

426

Chapter 15   Magnetic Circuits

Linear magnetic circuits are, therefore, limited to nonferromagnetic ­materials such as air, brass, aluminum, plastic, and wood.

Example 15-2 The brass core in the toroidal coil in Figure 15-3 has a circular cross section. The inside diameter of the core is 9.0 cm, and the outside diameter is 11.0 cm. If there are 2000 turns in the winding, what total flux will be produced by a 1.0-A current in the coil? Solution Step 1 From the dimensions given, it follows that the diameter of the cross section of the brass core is 2.0 cm, and A = π × (1.0 cm)2 = π × 10−4 m2

The effective length of the magnetic circuit is the average path length. In this example, the inside circumference is π × 9 cm and the outside circumference is π × 11 cm. Hence, the average path length is π × 10 cm. Step 2

Rm =

Step 3

l π × 10 cm = 7.958 × 108 At/Wb = μ0A 4π × 10 − 7 × π × 10 − 4 m2 Fm = NI = 2000 t × 1.0 A = 2000 At

Step 4

Φ=

Fm 2000 At = 2.5 × 10−6 Wb = Rm 7.958 × 108 At/Wb

See Problems 15-4 and 15-5.

15-5  Nonlinear Magnetic Circuits To work with nonlinear resistors, we require data such as voltage-current graphs for the resistors. To solve nonlinear magnetic circuits, we need comparable data on the properties of the ferromagnetic materials we wish to use in our magnetic circuit. This information is usually published in the form of detailed BH graphs like those in Figure 14-15.

15-5   Nonlinear Magnetic Circuits

Example 15-3 A toroidal coil has a circular cast-steel core with a square cross section. The inside diameter of the core is 12 cm and the outside diameter is 15 cm. If there are 2000 turns in the coil, what current must flow through it to develop a total flux of 3.0 × 10−4 Wb? Solution I Step 1 Determine the dimensions of the magnetic circuit:

Cross-sectional area:   A = (1.5 cm)2 = 2.25 cm2 = 2.25 × 10−4 m2 Average path length:   l = π × 13.5 cm = 42.4 cm = 0.424 m Step 2 Determine the magnetic flux density: B=



Φ 3.0 × 10 − 4 Wb = 1.33 T = A 2.25 × 10 − 4 m2

Step 3 Determine the permeability. From the magnetization curve for cast steel in Figure 14-15, a magnetic field strength of 1750 At/m is needed to produce a flux density of 1.33 T. μ0 =

Step 4 Rm = Step 5 Step 6

B 1.33 T = = 7.60 × 10 − 4 H/m H 1750 At/m

l 0.424 m = 2.48 × 106 At/Wb = −4 μ0A 7.60 × 10 × 2.25 × 10 − 4 m2

Fm = ΦRm = 3.0 × 10−4 Wb × 2.48 × 106 At/Wb = 744 At I=

Fm 744 At = = 0.37 A N 2000 t

Actually, the above procedure is a rather roundabout way of solving magnetic circuits if reference to BH curves is involved. Once we have determined the flux density, the magnetization curve tells us how much magnetic field strength we require for each unit length of the magnetic c­ ircuit. We can, therefore, solve for the MMF directly without calculating the permeability and reluctance.

427

428

Chapter 15   Magnetic Circuits

Solution II Step 1 Determine the dimensions of the magnetic circuit. As in Solution 1, A = 2.25 × 10−4 m2  and  l = 0.424 m

Step 2

B=



Φ 3.0 × 10 − 4 Wb = = 1.33 T A 2.25 × 10 − 4 m2

Step 3 From the BH curve for cast steel in Figure 14-15, H = 1750 At/m Step 4

Fm = H l = 1750 At/m × 0.424 m = 742 At Step 5

I=

Fm 742 At = = 0.37 A N 2000 t

In a simple magnetic circuit where all the magnetomotive force produces flux in a ferromagnetic core with a uniform cross section, solving for the flux is straightforward when we know the magnetomotive force.

Example 15-4 A current of 0.50 A is passing through a toroidal coil consisting of 960 turns wound on a cast-steel ring with a cross-sectional area of 12 cm2 and an a­ verage path length of 30 cm. Find the total flux in the core. Solution Fm = NI = 960 × 0.50 A = 480 At

H=

Fm 480 At = = 1600 At/m l 30 cm

On the BH curve for cast steel (Figure 14-15), B = 1.3 T when H = 1600 At/m. Therefore, Φ = BA = 1.3 T × 12 cm2 = 1.6 × 10−3 Wb See Problem 15-6.

15-6   Leakage Flux

Circuit Check

A

CC 15-1. An air-core coil is wound with 1750 turns and is 25 cm long by 3 cm in diameter. Calculate the flux produced when a current of 600 mA flows through the winding. CC 15-2. Calculate the flux produced by a coil wound on a sheet-steel core but otherwise identical to the coil in question CC 15-1.

15-6  Leakage Flux Toroidal windings are expensive because the wire has to be threaded through the centre of the core for each turn. Such windings are normally used only in circuits where the flux outside the core must be kept to a minimum. For other applications, the wire can be wound on a bobbin before the parts of ferromagnetic core are assembled to form a loop with one side passing through the bobbin, as shown in Figure 15-4. Iron core Leakage flux

Coil

Figure 15-4  Leakage flux in a magnetic circuit

Sections 7-9 to 7-12 described how the current through parallel resistors divides in proportion to the conductance of the parallel branches. Similarly, magnetic flux divides in proportion to the permeance of parallel paths. When the core loop is made of high-permeability iron, most of the magnetic flux flows through the core. However, some leakage flux flows through the air around the core. In the magnetic circuit of Figure 15-4, the total flux through the top section of the circuit is less than that through the centre of the coil ­because the leakage flux bypasses the top section. The ratio between the total developed flux and the useful flux is known as the leakage factor. Keeping a magnetic circuit compact helps to minimize the leakage flux. The cross-sectional area of the core should be large enough to ensure that the iron does not become saturated. If the iron does become saturated, its permeabilty drops drastically, greatly increasing the reluctance of the core. Air gaps in the magnetic circuit also greatly increase its total reluctance and hence the ­proportion of leakage flux. In well-designed magnetic circuits, the leakage flux

429

430

Chapter 15   Magnetic Circuits

is often only a very small percentage of the total flux. For the calculations in this chapter, we can ignore the leakage flux unless the leakage factor is given. See Review Question 15-19.

15-7  Series Magnetic Circuits In Figure 15-4, the straight section of the core and the U-shaped section are in series since the magnetic flux flows in a loop first through one section and then through the other. The total reluctance of a series magnetic circuit is similar to the total resistance of a series electric circuit. The total reluctance of a series magnetic circuit is the sum of all the individual reluctances. RmT = Rm1 + Rm2 + Rm3 + . . . (15-3)



Equation 15-3 has exactly the same form as the equation for the total ­resistance of a series electric circuit:

RT = R1 + R2 + R3 + . . . (7-1)

Reluctance of U-shaped cast-steel section Rmc Rms

Fm

Reluctance of I-shaped sheet-steel section

f RmT = Rmc + Rms

Figure 15-5  Schematic diagram of a series magnetic circuit 4.0 cm

15 cm

2.5 cm 10 cm

12.5 cm X

Cast steel Y

Average path

2.5 cm Laminated sheet steel Figure 15-6  Magnetic circuit for Example 15-5

15-7   Series Magnetic Circuits

There are no standard graphic symbols for drawing schematic diagrams of magnetic circuits. However, since electric and magnetic series circuits are similar, we could use Figure 15-5 to represent the magnetic circuit of Figure 15-6.

Example 15-5

Find the magnetomotive force that produces a total flux of 1.2 × 10−3 Wb in the magnetic circuit of Figure 15-6. The varnish on the laminations ­accounts for 10% of the thickness of the laminations. Solution I Step 1 Determine the reluctance, Rmc, of the U-shaped section. The dimensions in Figure 15-6 show that all three arms of the cast-steel ­section are 2.5 cm wide. Therefore, Ac = 2.5 cm × 4.0 cm = 10 cm2 = 1.0 × 10−3 m2

The average path length, lc, is equal to the average of the outside length of  the U-shaped section (12.5 cm + 15 cm + 12.5 cm) and the inside length (10 cm + 10 cm + 10 cm). lc =

Bc =



1 ( 40 cm + 30 cm ) = 35 cm = 0.35 m 2 Φ 1.2 × 10 − 3 Wb = = 1.2 T Ac 1.0 × 10 − 3 m2

On the BH curve for cast steel in Figure 14-15, Bc = 1.2 T when Hc = 1220 At/m. Bc 1.2 T   μc = = = 9.84 × 10 − 4 H/m Hc 1220 At/m

Rmc =

l 35 cm = = 3.557 × 105 At/Wb −4 μcAc 9.84 × 10 H/m × 10 − 3 m2

Step 2 Determine the reluctance, Rms, of the laminated section.

As = 2.5 cm × 4.0 cm × 90% = 9.0 cm2 = 9.0 × 10−4 m2



ls =

Bs =

1 ( 20 cm + 10 cm ) = 15 cm = 0.15 m 2 Φ 1.2 × 10 − 3 Wb = = 1.33 T As 9.0 × 10 − 4 m2

431

432

Chapter 15   Magnetic Circuits

On the BH curve for sheet steel, Bs = 1.33 T when Hs = 900 At/m Rms = Step 3

μs =

Bs 1.33 T = = 1.48 × 10 − 3 H/m Hs 900 At/m

ls 15 cm = = 1.126 × 105 At/Wb −3 μsAs 1.48 × 10 H/m × 9.0 × 10 − 4 m2

RmT = Rmc + Rms = (3.557 × 105) + (1.126 × 105) = 4.683 × 105 At/Wb Fm = ΦRmT = (1.2 × 10−3 Wb)(4.683 × 105 At/Wb) = 5.6 × 102 At

In many series magnetic circuits, the material and the cross-sectional area of the various sections are not the same, so we must calculate the flux density or the magnetomotive force for each section. The total voltage drop of a series electric circuit is the sum of all the individual voltage drops. Similarly, The total magnetomotive force of a series magnetic circuit is the sum of the MMFs for all of the sections of the circuit.

FmT = Fm1 + Fm2 + Fm3 + . . . (15-4)



Equation 15-4 has exactly the same form as the equation for the total voltage in a series electric circuit:

VT = V1 + V2 + V3 + . . . (7-2)



We can apply Equation 15-4 to solve Example 15-5 without calculating the permeability and reluctance of each section. Solution II As in Solution I, HC = 1220 At/m for the U-shaped section. Therefore,

Fmc = Hclc = 1220 At/m × 35 cm = 427 At

Similarly, for the laminated section, Hs = 900 At/m, and Fms = Hsls = 900 At/m × 15 cm = 135 At Therefore,

FmT = Fmc + Fms = 427 + 135 = 5.6 × 102 At

See Problems 15-7 and 15-8 and Review Question 15-20.

15-8   Air Gaps

15-8  Air Gaps In the magnetic circuits shown in Figure 15-1, each line of force crosses an air gap. In motors, meters, and speakers, such air gaps provide space for the electric conductors to move. Even in equipment with no moving parts, a small gap may be left between sections of the magnetic circuit. Such gaps increase the total reluctance of the circuit, which reduces the total magnetic flux and can prevent saturation of the iron in the circuit. Often, a sheet of a nonmagnetic material such as fibreboard or brass is inserted at a joint between sections to produce an “air” gap with an accurate and uniform width. In magnetic circuits like those in Figure 15-1, all the magnetic lines of force must pass through the air gaps. Thus, the gaps in each circuit are in series with the remainder of the circuit. However, since the air adjacent to a gap has the same permeability as the air in the gap, the magnetic lines of force tend to spread out as they cross the gap. This fringing makes the flux density in the air gap slightly less than that in the adjoining iron sections of the magnetic circuit (see Figure 15-7).

Iron polepiece

Iron polepiece

(a)

Iron field pole

Air gap Iron rotor

(b)

Figure 15-7  Fringing of magnetic lines of force in an air gap

If the distance across the air gap is small, we can make reasonably accurate calculations by assuming that the cross-sectional dimensions of the air gap are greater than the cross-sectional dimensions of the adjacent iron by an amount equal to the distance across the gap. With series magnetic circuits, it is much more difficult to use the applied magnetomotive force to calculate the total flux. Since the ferromagnetic

433

434

Chapter 15   Magnetic Circuits

Example 15-6

Find the magnetomotive force that produces a total flux of 1.2 ×  10−3 Wb in the magnetic circuit of Figure 15-6 with pieces of fibreboard 0.1 mm thick inserted at points X and Y. Assume that varnish accounts for 10% of the thickness of the laminations. Solution I Since the total flux is the same as in Example 15-5, the reluctance of the two steel sections is the same as we have already calculated, 3.557 × 105 At/Wb for the cast-steel section and 1.126 × 105 At/Wb for the laminated sheet-steel section. When we allow for fringing, the cross-sectional area of the gap is Ag = 2.51 cm × 4.01 cm = 10.07 cm2 = 1.007 × 10−3 m2 Rmg =

lg μgAg

=

4π × 10

−7

0.1 mm 4 −3 2 = 7.9 × 10 At/Wb H/m × 1.007 × 10 m

Since each magnetic line of force has to pass through both gaps, they are in series. Since the two gaps are identical, each has a reluctance of 7.9 × 104 At/Wb. RmT = (3.557 × 105) + (1.126 × 105) + 2(7.9 × 104) = 6.26 × 105 At/Wb

Fm = ΦRmT = (1.2 × 10−3 Wb)(6.26 × 105 At/Wb) = 7.5 × 102 At

Solution II The MMF required for the iron will be the same as in Example 15-5, 562 At. For each air gap,

Bg =

Hg =

Φ 1.2 × 10 − 3 Wb = = 1.19 T Ag 1.007 × 10 − 3 m2

Bg 1.19 T = 9.47 × 105 At/m = −7 μg 4π × 10 H/m

Fmg = Hg lg = 9.47 × 105 At/m × 0.1 mm = 94.7 At

FmT = 562 + ( 2 × 94.7 ) = 7.5 × 102 At

sections are nonlinear, we cannot easily determine how much of the total MMF applies to each section. Example 15-7 shows one method of tackling the problem in a series magnetic circuit. The simplest method for solving such problems is to make an educated guess at the answer, solve for the other parameters of the circuit, and check if these calculated parameters agree with the known values. If the known and the calculated values are not reasonably close, we adjust our estimate and repeat the process.

15-9   Parallel Magnetic Circuits

Example 15-7 A current of 0.50 A is passing through a coil consisting of 960 turns wound on a cast-steel core with a cross section 5.0 cm × 2.5 cm and an average path length of 30 cm. This core has a single air gap 0.25 mm wide. Find the total flux in the core. Solution The area of the air gap, with allowance for fringing, is: Ag = 5.025 cm × 2.525 cm = 12.7 cm2 FmT = NI = 960 t × 0.50 A = 480 At Assume that the MMF for the air gap is 210 At. Then

Hg =

Fmg lg

=

210 At = 8.4 × 105 At/m 0.25 mm

Bg = μgHg = 4π × 10 − 7 H/m × 8.4 × 105 At/m = 1.056 T Φ = BgAg = 1.056 T × 12.7 cm2 = 1.34 × 10 − 3 Wb

If this is so, the flux density in the iron must be

Bc =

ϕ 1.34 × 10 − 3 Wb = = 1.07 T Ac 12.5 cm2

On the BH curve for cast steel in Figure 14-15, Bc = 1.07 T when Hc = 885 At/m. Therefore,

Fmc = Hclc = 885 At/m × 30 cm = 266 At

and

FmT = 210 + 266 = 476 At

Since this value is just slightly less than the MMF calculated from known quantities, the total flux must be just slightly greater than 1.34 × 10−3 Wb. See Problems 15-9 to 15-12 and Review Questions 15-21 to 15-23.

15-9  Parallel Magnetic Circuits In the magnetic circuits of Figure 15-1(c) and 15-1(d), all of the magnetic flux passes through some portions of the magnetic circuit, but in other ­portions the flux divides between two symmetrical paths. Designers have found that such multiple paths can appreciably reduce the leakage flux. These paths are in parallel. When a magnetic circuit has two identical paths in parallel, the total flux splits equally between the two paths.

435

436

Chapter 15   Magnetic Circuits

Example 15-8 The core in Figure 15-8 is made of sheet-steel laminations stacked to a total thickness of 4 cm with varnish making up 5% of this thickness. The width of each outside leg of the core is 2.5 cm, and the width of the centre leg is 5.0 cm. The average path length is 25 cm. What is the maximum current that can pass through a 1000-turn coil on the centre leg of the core without the total flux in the core exceeding 2 × 10−3 Wb? Average path

CL

Figure 15-8  Magnetic circuit for Example 15-8

Solution I Since the core is symmetrical, we could divide the magnetic circuit down the centreline shown in Figure 15-8 and stack one section on top of the other without changing the total reluctance of the magnetic circuit. The cross-sectional area of the core then becomes A = 2 × 2.5 cm × 4.0 cm × 95% = 19 cm2

Therefore,    maximum B =

Φ 2 × 10 − 3 Wb = = 1.05 T A 19 cm2

On the BH curve for sheet steel (Figure 14-15), B = 1.05 T when H = 460 At/m. Therefore,

Fm = Hl = 460 At/m × 25 cm = 115 At

and              maximum I =

115 At Fm = = 115 mA N 1000 t

Solution II Each of the parallel halves of the magnetic circuit passes half of the total flux. The flux density in the iron is

B=

Φ 1 × 10 − 3 Wb = = 1.05 T A 9.5 cm2

15-9   Parallel Magnetic Circuits

As in Solution I,

Fm = Hl = 460 At/m × 25 cm = 115 At

and maximum I = 115 mA The same MMF in the other half of the magnetic circuit produces 1 × 10−3 Wb of flux, for a total of 2 × 10−3 Wb. See Problems 15-13 to 15-15 and Review Question 15-24.

Circuit Check

B

CC 15-3. A coil of 2000 turns is wound on the cast iron portion of the core shown in Figure 15-9. Calculate the current required to produce a flux of 3 mWb. 40 cm

5 cm Cast steel

5 mm 20 cm

Cast iron

10 cm

10 cm 10 cm

10 cm

Figure 15-9

CC 15-4. A coil is wound on the centre leg of the transformer core shown in Figure 15-10. How many turns are required if a current of 1.0 A through the coil is to produce a flux density of 0.31 T in the outside leg? 2.5 cm

2.5 cm

2.5 cm

Cast iron

Copper 15 cm 2.5 mm

2.5 cm 15 cm



Figure 15-10

15 cm

437

438

Chapter 15   Magnetic Circuits

Summary

• For long air-core coils, the total reluctance is approximately equal to the reluctance of the nonmagnetic core itself. • In a linear magnetic circuit, the permeability is constant. • In a nonlinear magnetic circuit, the permeability is dependent on the magnetic field strength. • Some leakage magnetic flux passes through the air in the area surrounding the ferromagnetic core of a coil. • The total reluctance of a series magnetic circuit is the sum of all the individual reluctances. • An air gap can substantially increase the total reluctance of a magnetic circuit. • The magnetic flux density in an air gap is somewhat less than that in ­adjacent ferromagnetic sections of a magnetic circuit. • The total magnetic flux divides among parallel paths in a magnetic ­circuit. B = beginner

I = intermediate

A = advanced

Problems I I

I

I

I

Section 15-2  Long Air-Core Coils 15-1. An air-core coil passes a 500-mA current. Its 1000 turns occupy a length of 12 cm on a 1.0-cm diameter form. What flux is developed at the centre of the coil? 15-2. An 10-cm-long coil of wire is wound on a hardwood dowel 5.0 mm in diameter. If there are 750 turns in the coil, what current must it pass to develop a flux of 6.0 × 10−7 Wb at the centre of the coil? 15-3. How many turns of wire must be wound on a 1.5-cm-diameter plastic rod to produce 4.0 × 10-7−7 Wb at the centre of the coil from a 450-mA current through the coil? The length of the coil is 15 cm.

Section 15-4  Linear Magnetic Circuits 15-4.

The core for a low-loss coil is made from a piece of flat plastic stock 8 mm thick by 25 mm wide by 20 cm long, bent into a circular loop with the ends joined together. The coil has 1000 turns of wire wound evenly around the circular form. What current must flow in the coil to develop a flux of 10−5 Wb in the core? 15-5. The toroidal coil shown in Figure 15-11 has 350 turns wound on a steel ring with an inside diameter of 18 cm and an outside diameter of 22 cm. Find the flux density in the steel when the current through the coil is 2.5 A.

439

Problems

2.5 A μr = 3000

+ E −

Figure 15-11

I

I

Section 15-5  Nonlinear Magnetic Circuits 15-6. If the core in Problem 15-4 were a cast-iron ring with the same dimensions, what current would produce a flux of 105 μWb?

Section 15-7  Series Magnetic Circuits 15-7. The horseshoe-shaped cast-steel electromagnet of Figure 15-12 is 2.3 cm in diameter and has an average path length of 25 cm. The ­dimensions of the sheet-steel bar are 12.0 cm × 4.0 cm × 1.0 cm. What current through the windings of the electromagnet develops a total flux of 0.5 mWb in the bar held by the magnet? Cast-steel electromagnet core

1000 turns

1000 turns

Sheet-steel bar Figure 15-12

I

A

A

15-8.

Find the total flux in the sheet-steel bar in Figure 15-12 if the current in the coils of the electromagnet is 0.5 A.

Section 15-8  Air Gaps 15-9.

The magnetic circuit of Figure 15-13 is made from a sheet-steel bar 15 cm long with a 12 mm × 20 mm rectangular cross section. The bar is bent into a ring, leaving an air gap 0.25 mm across. What current must flow through the 800-turn coil wound on this core in order to produce a total flux of 3 × 10−4 Wb in the air gap? 15-10. If a sheet of aluminum 1.5 mm thick is placed between the sheetsteel bar and the electromagnet of Figure 15-12, what current must flow in the coils to develop a total flux of 7.5 × 10−4 Wb in the sheetsteel bar if the leakage factor is now 1.15?

Air gap Sheet steel 800 turns

Figure 15-13

440

Chapter 15   Magnetic Circuits

A

15-11. In the magnetic circuit shown in Figure 15-14, the coil has 1000 turns. Find the current required to produce a total flux of 1.0 mWb.

Plastic 2.5 mm

Cast steel

5.0 cm

Plastic 2.5 mm

Cast iron

13.0 cm

2.5 cm

Sheet steel 2.5 cm

2.5 cm 10.0 cm

5.0 cm

Figure 15-14

A

15-12. When a 4.0-A current flows through a coil wound on the core shown in Figure 15-15, the total flux is 0.80 mWb. How many turns does this coil have?

5.0 cm

Cast iron Cast steel

Sheet steel

Aluminum 2.0 mm thick

15.0 cm

2.0 cm 2.0 cm 10.0 cm

8.0 cm

3.0 cm

Figure 15-15

A

Section 15-9  Parallel Magnetic Circuits 15-13. If the sheet-steel laminations of Figure 15-16 are stacked to a total thickness of 2.5 cm, how many turns of wire must be wound on the

Problems

Cast steel 2.5 cm in diameter 11.5 cm 2.0 cm 7.5 cm

2.0 cm

4.0 cm

1000 turns

2.0 cm

25 cm

2.5 cm Figure 15-17

Figure 15-16

A A

centre leg in order for a current of 100 mA to develop a total flux of 1.2 × 10−3 Wb in the centre leg? Varnish ­accounts for 5% of the thickness of each lamination. Assume no air gaps and negligible leakage flux. 15-14. For the magnetic circuit shown in Figure 15-17, calculate the current required to develop 0.80 mWb of flux in the centre leg. 15-15. For the magnetic circuit shown in Figure 15-18, calculate the current needed to develop a total flux of 1.0 mWb.

Plastic 1.0 mm thick

Sheet steel 2.5 cm

1000 turns 20 cm

Cast iron

Cast iron

2.5 cm 2.5 cm 5.0 cm Figure 15-18

15 cm

2.5 cm 5.0 cm

441

442

Chapter 15   Magnetic Circuits

Review Questions

Section 15-1  Practical Magnetic Circuits 15-16. State two methods of supplying a magnetomotive force to a magnetic circuit.

Section 15-2  Long Air-Core Coils 15-17. On what basis can we assume that the total reluctance of a long air-core coil is approximately equal to the reluctance of the nonmagnetic core of the coil itself?

Section 15-3  Toroidal Coils 15-18. Describe the effect of a toroidal winding on a magnetic circuit. What is the advantage of this type of winding?

Section 15-6  Leakage Flux 15-19. What causes leakage flux to occur in magnetic circuits? State two means of reducing leakage flux in magnetic circuit design.

Section 15-7  Series Magnetic Circuits 15-20. What characteristic of magnetic circuits allows us to calculate the total magnetomotive force needed in a circuit made up of several different sections?

Section 15-8  Air Gaps 15-21. Many practical magnetic circuits contain a gap in the form of a sheet of plastic or brass. Why can we treat such materials as air gaps? 15-22. In an all-iron magnetic circuit, the relationship between the applied magnetomotive force and the resulting total magnetic flux is quite nonlinear, as indicated by the graphs of Figure 14-15. Adding an air gap makes the relationship between applied MMF and total flux in the circuit more linear. Explain why. 15-23. What causes fringing of the magnetic lines of force in an air gap?

Section 15-9  Parallel Magnetic Circuits 15-24. If the two parallel paths for magnetic lines of force in the magnetic circuit of Figure 15-8 were not symmetrical, what complication would be introduced into our solution of a problem such as Example 15-8?

Practice Quiz

Integrate the Concepts Figure 15-19 shows a series magnetic circuit. Given that the current produces a total magnetic flux of 0.40 mWb, calculate: Cast steel I

2.0 cm

2000 turns

3.0 cm

Plastic 1.0 mm thick

Cast iron

2.0 cm

2.0 cm

6.0 cm

2.0 cm 3.0 cm

Figure 15-19

(a) (b) (c) (d) (e)

the magnetomotive force in the cast-iron section the magnetomotive force in the sheet-steel section the magnetomotive force in the plastic sections the total magnetomotive force the current through the coils

Practice Quiz 1. An example of a magnetic circuit is (a) a loudspeaker (b) a hard-drive read/write head (c) a relay (d) all of the above 2. The reluctance of a nonmagnetic core is (a)  directly proportional to the length of the coil and inversely ­proportional to the cross-sectional area of the coil (b) directly proportional to the length of the coil and to the crosssectional area of the coil (c) inversely proportional to the length of the coil and to the crosssectional area of the coil (d)  inversely proportional to the length of the coil and directly ­proportional to the cross-sectional area of the coil 3. The reluctance of a 400-turn air-core coil with a length of 50 cm and a radius of 5.0 cm is (a) 5.1 × 107 At/Wb (b) 1.3 × 107 At/Wb (c) 2.0 × 108 At/Wb (d) 5.1 × 108 At/Wb

443

444

Chapter 15   Magnetic Circuits

4

Which of the following statements are true? (a) The principal advantage of toroidal coils is that they are easy to make. (b) The ratio between the total developed flux and the useful flux is known as leakage factor. (c) The total reluctance of a series magnetic circuit is the sum of all individual reluctances. (d) The total magnetic flux divides between parallel paths in a magnetic circuit.

5.

The toroidal coil of Figure 15-20 has a brass core with an inside diameter of 12 cm and an outside diameter of 16 cm. The coil has 1000 turns. The total flux produced when a current of 1.5 A flows through the coil is (a) 54 μWb (b) 2.5 μWb (c) 5.4 μWb (d) 6.3 μWb

I I

Figure 15-20

6. If the core of toroidal coil in Question 5 were made of cast steel, the magnetic flux density necessary to develop a total flux of 880 μWb would be (a) 1.4 T (b) 600 mT (c) 400 mT (d) 700 mT 7. The current through the coil in Question 6 would be (a) 176 mA (b) 352 mA (c) 264 mA (d) 88 mA

Practice Quiz

  8. The total magnetomotive force of a series magnetic circuit is (a) zero (b) the product of all MMFs for all the sections of the circuit (c) the sum of all MMFs for all the sections of the circuit (d) the difference of all MMFs for all the sections of the circuit   9. When an air gap is inserted a magnetic circuit the total reluctance (a) becomes zero (b) does not change (c) decreases (d) increases 10. What is the advantage of symmetrical parallel paths in a magnetic circuit?

445

16

Inductance Chapter 12 described how capacitance in an electric circuit depends on electric fields. The third basic property of electric circuits, inductance, stems from the interaction of magnetic fields with electric circuits.

Chapter Outline 16-1

Electromagnetic Induction  448

16-2 Faraday’s Law  450 16-3 Lenz’s Law  451

16-4 Self-Induction  453

16-5 Self-Inductance  454

16-6 Factors Governing Inductance  455 16-7 Inductors in Series  458

16-8 Inductors in Parallel  458 16-9 The DC Generator  459

16-10 Simple DC Generator  461 16-11 EMF Equation  463

16-12 The DC Motor  465

16-13 Speed and Torque of a DC Motor  467 16-14 Types of DC Motors  469

16-15 Speed Characteristics of DC Motors  471

16-16 Torque Characteristics of DC Motors  474

16-17 Permanent Magnet and Brushless DC Motors  476

Key Terms

electromagnetic induction 448 induced voltage  448 primary winding  448 secondary winding  448 mutual induction  450 Faraday’s law  450 Lenz’s law  451 self-induction 453 counter EMF (CEMF) 453 self-inductance 454 inductance 454 henry 454 inductor 456

field system  460 armature 460 rotor 460 stator 460 yoke 460 shunt coil  460 series coil  460 commutator 460 brush 461 pulsating DC 463 wave winding  464 lap winding  464 torque 466 back or counter EMF 467

field excitation  469 shunt motor  470 load current  470 series motor  470 compound motor  470 cumulatively compounded  470 differentially compounded 470 universal motor  472 compound motor  472 permanent magnet motor 476 brushless motor  476

Learning Outcomes At the conclusion of this chapter, you will be able to: • demonstrate electromagnetic induction with a magnet, a conductor, and a galvanometer • explain electromagnetic induction in an iron core with primary and secondary windings • explain Faraday’s law relating induced voltage in an electric circuit to the rate of change of ­magnetic flux • use Lenz’s law to determine the polarity of an induced voltage • use Lenz’s law to determine the direction of current ­produced by an induced voltage • explain how self-induction occurs in a coil • express inductance in terms of induced voltage and the rate of change of current • express the inductance of a coil in terms of the number of turns in the coil and its reluctance • calculate the inductance of a coil given the number of turns and the permeability, Photo sources:  © iStock.com/albin

• •

• • • • • • • •

cross-sectional area, and length of the magnetic circuit calculate the total inductance of inductors in series when there is no mutual inductance calculate the equivalent inductance of inductors in parallel when there is no mutual inductance list the components of a DC machine graph the voltage produced by a rotating loop explain commutator action calculate generated voltage calculate the force on a current-carrying conductor calculate the speed and torque of a DC motor describe and draw circuits of different types of DC motors graph speed and torque characteristics

448

Chapter 16  Inductance

16-1  Electromagnetic Induction In Section 14-3, Figure 14-10 illustrates how Faraday discovered that thrusting a permanent magnet into a solenoid induces a current in the solenoid. Figure 16-1 shows another arrangement for demonstrating such ­electromagnetic induction. As the conductor is moved downward between the poles of the magnet, the galvanometer pointer swings one way from the zero mark at the centre of the meter’s scale. As the conductor is moved upward, the galvanometer pointer swings in the opposite direction. When the conductor is stationary, the pointer does not deflect.

Motion

N

S

G

Figure 16-1  Demonstrating electromagnetic induction

For current to flow through the galvanometer, there must be some voltage source in the circuit. Experiments show that this voltage appears only when the conductor is cutting across the magnetic lines of force (and the magnetic flux they represent). Moving the conductor parallel to the lines of force produces no deflection of the galvanometer pointer. In Figure 16-1, the conductor is moved while the magnetic field remains stationary, whereas the magnetic field in Figure 14-10 is moved while the electric conductor remains stationary. In both cases, an induced voltage appears in the conductor when the conductor and the magnetic field are moving perpendicular to each other. Figure 16-2 shows electromagnetic induction produced by a changing magnetic field without mechanical motion of either a magnet or a conductor. As the resistance of the rheostat decreases, the current in the primary winding (the winding connected to the source) increases. The magnetic flux in the iron core increases correspondingly. Since the secondary winding (the winding connected to the load) is wound on the same core as the primary winding, the magnetic flux through the secondary winding also increases. The pointer of the galvanometer deflects, indicating an induced voltage in

16-1   Electromagnetic Induction

Iron core

E

449

Magnetic field

+

G

−

Figure 16-2  Demonstrating mutual induction

Practical Circuits

Source:  © iStock.com/tbradford

Induction and Metal Detectors

Airport security

The operation of metal detectors is based upon the principles of electro­ magnetic induction. Figure 16-3 shows a metal-detecting security gate commonly used in airports. Current through the transmitting coils produces a magnetic field that induces a current in the receiving coils. When a metal object passes through the gate, the magnetic field induces eddy currents in the object. The resulting change in the magnetic flux through the receiving coil changes the induced current. Security gates can detect the location of a metallic object as small as a paper clip.

Transmitting coils

Receiving coils

Lines of magnetic force Figure 16-3  Metal-detecting gate

450

Chapter 16  Inductance

the secondary winding. Since most of the magnetic flux is ­confined to the iron core, it is more usual to think of the flux as linking the secondary winding rather than cutting across the turns of the secondary winding. When the resistance of the rheostat increases, the current in the ­primary winding and the magnetic flux linking the secondary winding both ­decrease. At the same time, the pointer of the galvanometer swings in the opposite direction, indicating that the induced voltage now has the ­opposite polarity. When the current in the primary winding is steady, there is no deflection of the galvanometer pointer. The generation of a voltage in a secondary winding by a changing current in a primary winding is an example of mutual induction. See Review Question 16-29 at the end of the chapter.

16-2  Faraday’s Law With apparatus like that shown in Figure 16-2, Faraday noted that the faster the flux in the core builds up or collapses, the greater the deflection of the galvanometer pointer. This relationship is now known as Faraday’s law: The voltage induced in an electric circuit is proportional to the rate of change of the magnetic flux linking the circuit. To express rate of change of flux at any given moment, we use the same calculus derivative notation as we used for the rate of change of voltage in Section 13-2. From the definition of the weber in Section 14-3, V=

dϕ dt

where V is the voltage in volts and dϕ/dt is the rate of change of magnetic flux in webers per second. Note the use of lowercase ϕ to indicate changing flux. In the secondary winding in Figure 16-2, the same changing flux links all the turns and thus induces identical voltages in each turn. Since these voltages are all in series, the total induced voltage is the number of turns times the voltage per turn:

eT = N

dϕ dt

(16-1)

where eT is the instantaneous value of the induced voltage between the terminals of the winding in volts, N is the number of turns in the winding, and dϕ/dt is the rate of change of magnetic flux in webers per second.

16-3   Lenz’s Law

Combining Equations 14-1 and 14-2 gives Φ=

Fm NI = Rm Rm

The number of turns in the primary winding is fixed. If the flux density stays below the saturation point of the iron core, the reluctance of the magnetic circuit is reasonably constant. Therefore, the rate of change of flux is ­directly proportional to the rate of change of current. When dealing with electromagnetic induction due to a changing current, we can restate Faraday’s law in this form: The magnitude of the induced voltage is directly proportional to the rate of change of current. See Review Question 16-30.

16-3  Lenz’s Law The Russian physicist Heinrich F. E. Lenz (1804–65) reasoned that electromagnetic induction must conform to the physics principle that reaction is opposite to action. We can use this statement of Lenz’s law to determine the polarity of an induced voltage: Any current resulting from an induced voltage opposes the change in the original magnetic flux. When we first close the switch in Figure 16-4(a), the magnetic flux through the primary winding increases from zero, inducing a voltage across the secondary winding. As a result, current flows through the closed circuit formed by the resistor and the secondary winding. According to Lenz’s law, the magnetic flux produced by this secondary current must o ­ ppose the increase in primary flux. Applying the right-hand rule to the primary winding, we find that the primary flux flows clockwise around the core. Rising primary flux

+

Collapsing primary flux

Opposing secondary flux

E

−

(a)

Sustaining secondary flux

+

−

+

+

−

E −

(b)

Figure 16-4  Determining the direction of an induced voltage

451

452

Chapter 16  Inductance

To oppose the increase in the primary flux, the flux produced by the secondary current must flow counterclockwise. Applying the right-hand rule to the secondary winding, we find the direction in which the secondary current flows through the resistor and can thus determine the polarity of the induced voltage. When we open the switch, the primary current stops flowing and the primary magnetic flux collapses, as shown in Figure 16-4(b). This r­ eduction in flux induces a voltage across the secondary winding. According to Lenz’s law, the secondary current now opposes the collapse of the primary flux. Thus, the secondary current now produces a clockwise flux. Applying the right-hand rule gives the polarity shown in Figure 16-4(b). The polarity of the induced voltage when the primary current is ­decreasing is, therefore, opposite to the polarity when the primary current is increasing. We can also use Lenz’s law to determine the direction of the induced voltage in Figure 16-1. Since the galvanometer completes the circuit, current flows through the conductor as it moves through the magnetic field. This current produces a magnetic field around the conductor. If we consider the two magnetic fields separately, the field of the stationary magnet is uniform between the poles of the magnet, as shown in Figure 16-5(a), while the conductor is surrounded by concentric circular lines of force, which are directed clockwise if the current is flowing out of the page, as shown in Figure 16-5(b). The two component fields combine to form the magnetic field shown in Figure 16-6. In Figure 16-5(b), the direction of the magnetic lines of force above the conductor is opposite to the direction of the lines between the poles of the permanent magnet field in Figure 16-5(a). Consequently, the flux from the conductor cancels some of the flux from the magnet and the resulting lines of force bend away from the conductor (see Figure 16-6). Below the ­conductor, the magnetic fields from the conductor and the magnet have the same direction. The resulting flux density is greater than that above the conductor, as shown by the spacing of the lines of force. Thus the induced current in the conductor bends some of the magnetic lines of force between the poles of the magnet to pass below the conductor.

N

S

Stationary permanent magnet field

Magnetic field around current-carrying conductor

(a)

(b)

Figure 16-5  Component magnetic fields for the demonstration of Figure 16-1

16-4  Self-Induction

Motion of conductor

453

Direction of magnetic force

N

S

Figure 16-6  Using Lenz’s law to determine the direction of the current produced by an induced voltage

Since magnetic lines of force tend to become as short as possible, the magnetic field exerts an upward force on the current-carrying conductor. In keeping with Lenz’s law, this force opposes the motion of the conductor since that motion produces a current that changes the original flux from the magnet. Consequently, when the conductor in Figure 16-6 moves downward through the stationary magnetic field, the induced voltage has a ­polarity that makes current flow out of the page. See Review Questions 16-31 to 16-33.

16-4 Self-Induction When we first close the switch in Figure 16-7, the increasing flux produced by the rising current in turn A induces a voltage into turn B. The same ­rising current in turn B induces a voltage into turn A. These voltages are in series, and according to Lenz’s law, both have a polarity such as to oppose the increase in current and magnetic flux. For a coil with N turns, Faraday’s law shows the total induced voltage is Ndϕ/dt, as in Equation 16-1. The generation of a voltage in a circuit by a changing current in the same circuit is called self-induction. As current flows from the voltage source in Figure 16-7, the voltage source provides energy to build up a magnetic field around the coil. The rising magnetic field induces an EMF in the turns of the coil. This EMF separates charge within the coil, causing a surplus of electrons at one end of the coil and a deficiency at the other. This process transfers energy from the magnetic field to the coil and produces a potential difference across the coil. Since the induced voltage opposes the change in current, the self-induced voltage is sometimes called a counter EMF or CEMF. When we open the switch in Figure 16-7, the magnetic field collapses and again induces an EMF in the coil. The polarity of the self-induced voltage is reversed so that it opposes the decrease in current and magnetic flux. See Review Question 16-34.

E

+

A B

− Figure 16-7  Self-induction

454

The henry is named in honour of the ­American physicist Joseph Henry (1797– 1878), who made an ­extensive study of electromagnetism and constructed the first electromagnetic motor.

The definition of the henry in terms of ­webers per ampere underlies the units for magnetic permeance (henrys or webers per effective ampere), magnetic reluctance (reciprocal henrys or effective amperes per weber), and magnetic permeability (henrys per metre) introduced in Section 14-6.

Chapter 16  Inductance

16-5 Self-Inductance Section 16-4 showed that the simple electric circuit of Figure 16-7 tends to oppose any change in current through it or in the magnetic flux linking it. This property is called its self-inductance Because it is usually apparent whether the electromagnetic induction in a circuit is self-induction or ­mutual induction, it is customary to refer to self-inductance as simply the inductance of the circuit. The letter symbol for inductance is L. The henry (symbol H) is the SI unit of inductance. We can define the unit of inductance either in terms of the change in flux ­resulting from a given change in current or in terms of the induced voltage resulting from a given change in current. The general definition of the henry is based on magnetic flux linkage: An electric circuit has an inductance of one henry when a change in current of one ampere produces a change in total linkage flux of one weber: 1 H = 1 Wb/A. The total linkage flux of a coil with N turns is N times as great as for a single turn carrying the same current. Therefore, L=N



dϕ di

(16-2)

where L is the inductance of a circuit in henrys, N is the number of turns linked by the magnetic flux, and dϕ/di is the change in flux for a given change in current in webers per ampere. Rearranging Equation 16-1 and Equation 16-2 gives N=

N=

and Therefore, and or

eL dΦ/dt L dϕ/di

eL L = dΦ/di dΦ/dt

L = eL ×

dt dΦ dt × = eL × di dΦ di L=

eL di/dt

(16-3)

16-6   Factors Governing Inductance

where L is the inductance of a circuit in henrys, eL is the induced voltage in volts, and di/dt is the rate of change of the current in amperes per second. Thus, we can also define the henry in terms of just electrical units: An electric circuit has an inductance of one henry when current changing at the rate of one ampere per second induces an EMF of one volt into the circuit. See Problems 16-1 to 16-5 and Review Question 16-35.

16-6  Factors Governing Inductance Even a straight wire has some inductance. As the current through a conductor increases, magnetic lines of force appear as tiny loops at the centre of the conductor and then expand outward. As they expand, they cut across the conductor, inducing a voltage into the conductor. However, the inductance of a straight wire is so small that we can ignore it, except at very high radio frequencies. The inductance of a coil can be quite large since it depends on the number of turns. Combining Equations 14-1 and 14-2 gives Φ= Dividing both sides by t gives

from which

Fm NI = Rm Rm

Φ N I = × t Rm t

dϕ N di = × dt Rm dt

Substituting for dϕ/dt in Equation 16-2, we get

eT =

N2 di × Rm dt

Now we substitute for the induced voltage in Equation 16-3:



L=

N2 Rm

(16-4)

455

Chapter 16  Inductance

This equation shows that the inductance of a coil is dependent on the ­ umber of turns in the coil and the reluctance of the magnetic circuit on n which the coil is wound. Inductors are circuit components constructed for the express purpose of adding inductance to a circuit. An iron core greatly reduces the reluctance of the magnetic circuit and correspondingly increases the inductance of a coil. Doubling the number of turns of wire in the coil of Figure 16-7 not only doubles the flux linking the coil for a given current (thus doubling the induced voltage per turn), but also doubles the number of turns that this flux links. Therefore, the total induced voltage (and consequently the inductance) is proportional to the square of the number of turns. Since Rm = L/μA (Equation 15-1), L=



N2μA l

(16-5)

where L is the inductance of the inductor in henrys, N is the number of turns in the coil, μ is the permeability of the magnetic circuit in henrys per metre, A is the cross-sectional area of the magnetic circuit in square metres, and l is the length of the magnetic circuit (or the length of the coil for air-core coils) in metres. Equation 16-5 is based on the assumption that all of the magnetic flux links all of the turns. For toroidal and iron-core coils, Equation 16-5 is reasonably accurate. However, where there is appreciable leakage flux, as in long air-core coils, we must use empirical formulas to calculate inductance. Electrical and radio handbooks list formulas for various ­ types of inductors.

Source:  Clive Streeter/Getty Images

456

Inductor with a ferite core

16-6   Factors Governing Inductance

Example 16-1 Find the inductance of a toroidal coil with 2000 turns of wire wound on a cast-steel ring, given that the ring has a cross-sectional area of 2.25 cm2, an average path length of 42.4 cm, and a permeability of 7.60 × 10−4 H/m. (See Example 15-3.) Solution L=

N2μA 20002 × 7.60 × 10 − 4 H/m × 2.25 × 10 − 4 m2 = 1.61 H = l 0.424 m

See Problems 16-6 to 16-13 and Review Questions 16-36 to 16-38.

Circuit Check

A

CC 16-1. A coil carrying a current of 8.0 A develops a voltage of 150 V when the current is switched over 0.5 s. Find the inductance of the coil. CC 16-2. How much voltage will be induced in a 6-H inductor carrying 4 A if the current is reversed over 20 ms? CC 16-3. The inductor shown in Figure 16-8 has a 600-turn coil wound on a sheet-steel core with a 5.0 cm × 5.0 cm cross section. (a) Given that the relative permeability of the steel is 4500, ­determine the inductance. (b) How many extra turns are required on the coil to raise the inductance by 25%? 25 cm

20 cm



Figure 16-8

457

458

Chapter 16  Inductance

16-7  Inductors in Series L1

+ E

L2

− Figure 16-9  Air-core inductors in series

The two inductors connected in series in Figure 16-9 have the same current flowing through them and experience the same rate of change of current. For the present, we shall assume that the two inductors are physically ­located so that the magnetic field of one cannot induce a voltage into the other. A changing current in the circuit of Figure 16-9 induces a voltage of e1 in L1 and e2 in L2. The total induced voltage is eT = e1 + e2 From Equation 16-3, LT =

Generalizing,

eT e1 + e2 e1 e2 = = + = L1 + L2 di/dt di/dt di/dt di/dt

When there is no mutual coupling between inductors in series, the total inductance is the sum of the individual inductances:

LT = L1 + L2 + L3 + . . .

(16-6)

Note that the equation for inductors in series is similar to the equation for the total resistance of resistors in series. See Problem 16-14 and Review Questions 16-39 and 16-40.

16-8  Inductors in Parallel The two inductors connected in parallel in Figure 16-10 have the same ­potential difference between their terminals. Since this voltage is produced by self-induction, the current in Ll must change at the rate of di1/dt, and the current in L2 must change at the rate of di2/dt. Therefore, the total current must change at the rate of (di1 + di2)/dt, and Leq =

eL di1 + di2 dt

Inverting both sides of this equation gives

di1 + di2 di1 di2 di1 di2 + l dt dt dt dt dt = = = + eL eL eL eL Leq l l l = + Leq L1 L2

16-9   The DC Generator

+ E

L1

i1

L2

i2

−

Figure 16-10  Iron-core inductors in parallel

Generalizing, The equivalent inductances of inductors in parallel is Leq =



1

1 1 1 + + + ... L1 L2 L3



(16-7)

For just two inductors in parallel, Equation 16-7 reduces to



Leq =

L1 × L2 L1 + L2

(16-8)

Note that Equations 16-6, 16-7, and 16-8 are valid only if there is no ­mutual induction between the individual inductors. Section 27-6 deals with ­combinations of inductors that have both self and mutual i­ nductions. See Problems 16-15 and 16-16 and Review Question 16-41.

Circuit Check

B

CC 16-4. What inductance in series with a 100-µH and a 0.25-mH inductor produces a total inductance of 0.5 mH? CC 16-5. Two inductors in parallel develop a voltage of 6.0 V when the current is changed at a rate of 32 A/s. Given that one of the inductors is 200 mH, determine the value of the other inductor.

16-9  The DC Generator A generator is a device that converts mechanical energy into electric energy. The DC generator is an application of Faraday’s law: a mechanical force (the prime mover) rotates coils in a stationary magnetic field, inducing a voltage in the coils.

459

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Chapter 16  Inductance

All motors and generators include two key components: • a field system of magnets or coils • an armature, which has conductors that cut the magnetic flux One of these key components is located on the rotor (the moving part of the machine) and the other is on the stator (the stationary part). For DC machines and small AC machines, the armature is on the rotor and the field system on the stator, as shown in Figure 16-11. For large AC machines, the field system is on the rotor and the armature is on the stator. Brush Rotor Frame (yoke) Commutator

Field winding Field pole core

Armature winding

Shaft

Field pole shoe

Figure 16-11  Construction of a DC generator

The frame or yoke is normally made of annealed steel. It provides the mechanical support for the poles and also serves as a return path for the magnetic flux created by the field coils. The poles are made of soft iron, which has good magnetic characteristics for developing a strong magnetic field. The poles have shaped pieces called shoes, which minimize the air gap between the poles and the rotating armature. The field poles are built of thin laminations insulated from each other by a layer of varnish. This construction is used in magnetic circuits to minimize the power loss through eddy currents. The field coils are wound on the poles and are either shunt coils or series coils. Shunt coils have a large number of turns of fine wire, and series coils have a small number of turns of heavy wire. The commutator consists of copper segments insulated from each other by plastic spacers. The armature coils ends are connected to the commutator segments. The commutator provides mechanical rectification for the voltage

16-10   Simple DC Generator

­ eveloped by the armature. The brushes are usually made with a mixture d of powdered carbon and graphite along with binders and other additives. The brushes are held against the commutator by a spring arrangement, and make an electrical connection to the moving armature. See Review Question 16-42.

16-10  Simple DC Generator In its simplest form a generator consists of a single loop of wire rotating in a magnetic field between two poles, as shown in Figure 16-12. Wire loop

S

N

Slip ring Brush V

Load

Figure 16-12  Single loop AC generator

Voltage is induced in the coil sides as they rotate through the magnetic flux. Since the sides of the coil move in opposite directions, the induced voltages in the two sides have opposite polarities, which add in series around the loop to give the total generated voltage. The induced voltage is determined by the rate of cutting the flux. Since the motion is rotational, the generated voltage will depend on the angle at which the flux is cut. In the vertical position, the conductors move parallel to the flux, and therefore no voltage is induced. As the loop rotates, the voltage increases and will be maximum when the coil is moving perpendicular to the flux, after 90º of rotation. Beyond this point the voltage decreases again and will reach zero after 180º of rotation when the conductors are again moving parallel to the flux. As the coil rotates farther, the side that was originally at the top and moving in the direction of the magnetic field is now at the bottom and moving in the opposite direction. Similarly, the side that was at the bottom is now at the top and also reverses its direction relative to the magnetic field. Thus, the polarity of the induced voltage

461

Chapter 16  Inductance

reverses as it begins to increase again. The cycle is completed as the coil rotates back to its initial position. If the speed of rotation is constant, the induced voltage has the shape of a sine wave, as shown in Figure 16-13. This alternating voltage is described in more detail in Section 18-2.

Voltage

462

Angle

90º 180º 270º 360º

Figure 16-13  Output voltage of a simple AC generator

To produce a DC output from the simple generator, the slip rings are replaced with a commutator, which in its simplest form consists of two copper segments insulated from each other, as shown in Figure 16-14.

S

N

Commutator V

Load

Figure 16-14 Simple DC generator

The commutator provides mechanical rectification by switching the connections to the armature just as the voltage induced in the coils reverses polarity. For the first 180º of rotation, brush 1 is connected to commutator segment A, and coil side A is positive (see Figure 16-15). Brush B is connected to commutator segment B and is negative. After 180º of rotation, sides A and B reverse polarity. At the same time the brushes move to different commutator segments. Brush 1 is now connected to coil side B and therefore remains positive, while brush 2 is now connected to coil side A and remains negative (see Figure 16-16).

16-11   EMF Equation

− A

+ 1 A

1

+ 2

B

A

B

B

2

− + −

+

−

1

2

Figure 16-15  Commutator connections at start of cycle − B

+ 1 B

1

+ A

A 2

B

A 2

− + −

+

−

1

2

Figure 16-16  Commutator connections after 180º of rotation

Voltage

Mechanical rectification is accomplished by physically switching the connections to the coil sides as the voltage polarity reverses. The resulting voltage is a rectified sine wave, referred to as pulsating DC (Figure 16-17).

90º

180º

270º

360º

Angle

Figure 16-17  Output voltage

See Review Questions 16-43 to 16-45.

16-11  EMF Equation When the armature of a generator is wound with more than one coil, the output voltage is the sum of the voltages in the individual coils. If the armature has two coils positioned at right angles to each other, the second

463

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Chapter 16  Inductance

coil is a quarter of a revolution behind the first coil, and the pulsating voltage in the second coil lags 90º behind the voltage in the first coil, as shown in Figure 16-18. When these two voltages are added together, the fluctuations in total voltage are much less than those in the individual coils. et

1

N

2

e1 e2

S

Figure 16-18  Output voltage from a two-coil armature

A practical generator has many coils wound in slots in the armature core. The output voltage is the sum of the coil voltages and has greatly reduced fluctuations. The commutator will have as many segments as there are coils on the armature. The output voltage can also be smoothed by using more than one pair of poles to produce the magnetic field. The voltage produced by a DC generator depends on the speed of rotation, the strength of the magnetic field, and the number of turns in each armature winding. The equation for generated voltage is given below: Eg =



ZΦN P × 60 A

(16-9)

where Z is the total number of armature conductors, Φ is the flux per pole in webers, N is the speed in revolutions per minute (r/min or RPM), P is the total number of poles in the field system, and A is the number of parallel paths through the armature. A depends on the way the armature coils are connected to the commutator. The two basic types of armature connections are wave winding, which results in two parallel paths, and lap winding, which results in as many paths as there are poles.

Example 16-2

An 8-pole, lap-wound armature has 96 slots with 10 conductors in each slot. The flux per pole is 20 mWb. Calculate the EMF generated when the armature is driven at 500 RPM. Solution Z = 96 slots × 10 conductors/slot = 960 conductors A = P = 8 parallel paths Eg =

ZΦN P 960 × 20 × 10 − 3 Wb × 500 RPM 8 × = × = 160 V 60 A 60 8

16-12   The DC Motor

465

Source:  © iStock.com/albin

Modern day applications for DC generators are limited to standalone battery charging units, welding generators, and some specialty manufacturing processes. Even in these applications, an AC generator with solid-state rectification is often used instead of a DC generator.

A field system with eight poles

See Problems 16-17 and 16-18 and Review Questions 16-46 and 16-47.

16-12  The DC Motor A motor is a device that converts electrical energy into mechanical energy. A DC motor has the same construction as the DC generator and in fact the two are interchangeable and require only an adjustment in brush position to reduce sparking. Motor action is based on the principle that a conductor in a magnetic field will experience a force when current flows through it (Figure 16-19). F I

N

S

Figure 16-19  Force on a current-carrying conductor

466

John Ambrose Fleming (1849–1945) was an English electrical engineer and physicist. In 1904, he invented the first vacuum tube that was able to detect radio signals. The vacuum tube was used in radios and radars for the next 50 years, until semiconductor electronics eventually replaced it.

Chapter 16  Inductance

The direction of force is given by Fleming’s left-hand (motor) rule. Extend the thumb, forefinger and middle finger of the left hand at right angles to each other. Orient the hand so that the forefinger is in the ­direction of the magnetic field and the middle finger in the direction of current flow. The thumb then indicates the direction of developed force. The magnitude of force created depends on three variables: • the strength of the magnetic field • the magnitude of the current • the length of the conductor in the magnetic field The magnitude of force, in newtons, is given by F = BlI 



(16-10)

where B is the flux density in teslas, l is the length of the conductor in metres, and I is the current in amperes.

Example 16-3 A rectangular coil has 100 turns and is in a magnetic field of flux density 0.800 T. Given that the length of the coil is 30.0 cm and the current flowing in the conductors is 15.0 A, calculate the force acting on each side of the coil. Solution

F = BlI = 0.800 T × (100 turns × 0.300 m) × 15.0 A = 360 N

A practical DC motor has many armature coils in slots on the rotor. The force on each conductor is offset from the centre of the rotor, so these forces each produce a torque on the rotor (Figure 16-20). The total torque turns the rotor.

N

S

x Current in Current out

Figure 16-20  Torque created in a DC armature

Source:  JEFF DALY, VISUALS UNLIMITED /SCIENCE PHOTO LIBRARY

16-13   Speed and Torque of a DC Motor

Multiple windings on a motor armature

A DC motor has a commutator that reverses the current direction in the armature coils as they pass from one field pole to the next, which has the opposite direction. As a result torque continues to turn the rotor in the same direction. See Problem 16-19 and Review Questions 16-48 and 16-49.

16-13  Speed and Torque of a DC Motor When a motor armature rotates, the conductors on the armature cut the flux provided by the field poles. Thus by Faraday’s law, an EMF will be induced in the armature conductors. This EMF opposes the applied voltage as per Lenz’s law and is therefore referred to as back or counter EMF (Eb). Since the back EMF is a generated EMF, it can be calculated using the generated EMF equation.

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Chapter 16  Inductance

Eg =



ZΦN P × 60 A

(16-11)

Solving this equation for N gives the speed equation for a DC motor: N=



60Eb A × ZΦ P

(16-12)

Example 16-4 A 440-V DC motor has four poles with a flux per pole of 20 mWb. The armature is wave wound with 44 slots and 12 conductors in each slot. The back EMF is 400 V. Calculate the motor speed. Solution

Z = 44 slots × 12 conductors/slot = 528 conductors A=2

N=

60Eb A 60 × 400 V 2 × = × = 1.1 × 103 RPM ZΦ P 528 × 0.02 Wb 4

Torque is the turning or twisting moment of a force about an axis, and is often used as a measure of the strength of the motor. The torque in a motor is created by the forces developed in the armature conductors, and consequently is proportional to the current flowing in the armature (IA) and strength of the magnetic field. The equation for torque in a motor is

T = KΦIA

(16-13)

K=

(16-14)

where T is the torque in newton metres (N⋅m), and K is the torque constant, which depends on the physical characteristics of the machine and is given by the following equation:

0.159Z P A

Combining for K in Equation 16-13 gives



T=

0.159Z P ΦIA A

(16-15)

16-14   Types of DC Motors

Example 16-5 Given that the full-load armature current for the motor in Example 16-4 is 60.0 A, calculate the full-load torque developed. Solution T=

0.159Z P 0.159 × 528 × 4 ΦIA = × 0.02 Wb × 60 A = 197 N . m A 2

As the motor rotates because of the torque developed in the armature, the motor shaft can do work on a mechanical load. The power produced by a motor depends on the torque and speed. In North America, the old imperial unit of power, the horsepower (hp), is still commonly used to rate the power output of electric motors: power ( in hp ) =



2πNT 33 000

(16-14)

where T is the torque in pound-feet (1 lb-ft = 1.356 N.m).

Example 16-6 Calculate the full-load horsepower output of the motor in Example 16-5. Solution

T = 197 N. m ×

power =

1 1b-ft = 145 1b-ft 1.356 N. m

2π × 1.1 × 103 RPM × 145 1b-ft 2πNT = 30 hp × 33 000 33 000

See Problems 16-20 to 16-22 and Review Questions 16-50 and 16-51.

16-14  Types of DC Motors motors are classified according to the method of field excitation. The field may be produced electromagnetically by current flow in the field coils or by permanent magnets. DC

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Chapter 16  Inductance

The shunt motor has field coils connected in parallel with the armature (Figure 16-21). Shunt coils have many turns of fine wire and require a relatively small current to produce a strong magnetic field. Ish

Il Ia Ra + E + −

Rsh Eb

− Figure 16-21  Shunt motor circuit diagram

The total current is called the load current and is the sum of the armature and field currents, IL = IA + Ish. Since the shunt field is connected in parallel with the source, Ish can be calculated from Ohm’s law, Ish = E/Rsh. The back EMF can be found using Kirchhoff’s voltage law, Eb = E − IARA. The series motor has field coils connected in series with the armature (Figure 16-22). Since the field is now developed by the large armature current, series coils require only a few turns of heavy wire.

Il

Ise Rse

Ia Ra

+ E + −

Eb

− Figure 16-22  Series motor circuit diagram

Since the same current flows through all components in a series circuit, IL = Ise = IA. The equation for back EMF now includes the voltage drop in the resistance of the series coil: Eb = E− IA (RA+ Rse). The compound motor has both shunt and series coils. The series coil may be connected to either aid or oppose the shunt field. With the aiding configuration, the motor is cumulatively compounded, and with the opposing configuration the motor is differentially compounded. The differentially compounded motor is rarely used because it tends to become unstable under heavy load and may even reverse its direction of rotation. The series coil

16-15   Speed Characteristics of DC Motors

may be connected in two positions, in series with the source (short shunt) or in series with the armature (long shunt), as shown in Figure 16-23. There is very little operational difference between these two series connections. Ish

Il Ish

Il Rse

Rse

Ia

Ia

Ra

E

+ E

Rsh +

−

Ra

+

Rsh

− +

Eb

−

Eb

−

(a)

(b)

Figure 16-23  Compound motor circuits: (a) Short shunt; (b) Long shunt

Example 16-7 A 440-V DC compound motor is connected long shunt and has shunt field, series field, and armature resistances of 110 Ω, 0.25 Ω, and 0.40 Ω, respectively. Calculate the back EMF when the motor operates with a load current of 60 A. Solution

Eb = E − IA ( Rse + RA )

IA = IL − Ish and Ish =

E 440 V = = 4.00 A  Rsh 110 Ω

IA = 60 A − 4.00 A = 56 A

Eb = 440 V − 56 A × ( 0.65 Ω ) = 440 V − 36.4 V = 404 V See Problems 16-23 to 16-24 and Review Questions 16-52 and 16-53.

16-15  Speed Characteristics of DC Motors The speed characteristic is the variation of the speed of a motor as a mechanical load is applied to the shaft. When a load is applied, the motor draws more current from its power supply and produces more torque.

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Chapter 16  Inductance

As shown in Section 16-13, the speed of a DC motor is given by 60Eb A N= × ZΦ P

For any given machine, Z, P, and A are constant, and therefore Eb N∝ Φ We can examine the speed characteristic for each type of DC motor by observing the variation of Eb and Φ as the load current increases. For the shunt motor, E Φ ∝ Ish = Rsh If E and Rsh are kept constant, the flux will remain constant. The back EMF is Eb = E − IARA, and will decrease as load current and therefore IA increase. Since RA is small, this decrease is small and the speed of the shunt motor remains relatively constant as a load is applied (Figure 16-24). For the series motor, the field is produced by the armature current, which is also the load current. Therefore, the flux will increase with increasing load, and the speed will decrease. The back EMF, Eb, will again decrease slightly as the load current increases, contributing to the reduction in speed. At light loads, the flux is small, and the motor speed is high. A series motor can never be operated at no load since the excessive speed would damage the motor. A series motor is usually coupled to its load by gears rather than a pulley system. Because a series motor has only a few turns on its field coil, the inductance is small and the flux produced stays synchronized with the armature flux even when alternating current flows. This motor will operate on AC and on DC, and is often referred to as the universal motor. The compound motor combines the characteristics of shunt and series motors. The series field will increase as a load is applied, and therefore the total flux will increase. The speed will drop more than in the shunt motor but not as much as in the series motor. Shunt Compound (cumulative) Speed, N

472

Series

Load Current (IL)

Figure 16-24  Speed characteristics of DC motors

16-15   Speed Characteristics of DC Motors

Example 16-8

A 220-V DC shunt motor has a shunt field resistance of 110 Ω and an armature resistance of 0.10 Ω. When unloaded, the motor draws a current of 7.0 A and rotates at 1250 RPM. A series winding of resistance 0.05 Ω is cumulatively connected long shunt. This winding increases the flux per pole by 20% when the motor is fully loaded and taking a current of 62 A. Calculate the full-load speed of this compound motor. Solution When the shunt motor is unloaded (Figure 16-25), Ish =



N∝

E 220 V = 2.00 A = Rsh 110 Ω

Eb Φ

IA = IL − Ish = 7.0 A − 2 A = 5 A

Eb1 = E − IARA = 220 V − ( 5 A × 0.10 Ω ) = 219.5 V



Ish = 2 A

Il = 7 A

Ia Ra = 0.10 Ω +

Rsh = 110 Ω

E + −

Eb

− Figure 16-25

When the compound motor is loaded (Figure 16-26),



IA = IL − Ish = 62 A − 2 A = 60 A

Eb2 = E − IA ( RA + Rse ) = 220 V − ( 60 A × 0.15 Ω ) = 211 V N1 ∝

Eb2╱Φ2 Eb2 Φ1 Eb1 Eb2 N → 2= × and N2 ∝ = Eb1 Φ2 Φ1 Φ2 N1 Eb1╱Φ1

Φ2 is increased by 20% because of the series field; therefore, Φ2 = 1.2Φ1 Φ1 1 = = 0.833 Φ2 1.2

N2 = 1250 RPM ×

211 V × 0.833 = 1001 219.5 V

RPM

473

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Chapter 16  Inductance

Ish = 2 A

Il = 7 A

Rse = 0.05 Ω Ia Ra = 0.10 Ω

+ E

Rsh = 110 Ω

− +

Eb

− Figure 16-26

See Problems 16-25 to 16-26 and Review Question 16-54.

Circuit Check

C

CC 16-6. How much flux per pole is required for a 12-pole lap-wound motor to rotate at 500 RPM if the back is 224 V? The motor has a total of 480 conductors. CC 16-7. A 550-V long-shunt compound motor has shunt field, series field, and armature resistances of 100 Ω, 0.24 Ω, and 0.32 Ω, respectively. What load current will produce a back of 500 V?

16-16  Torque Characteristics of DC Motors DC motors are used to drive a variety of mechanical loads, such as compres-

sors, fans, elevators, and electric vehicles. It is important to understand how the torque developed by the motor varies as the mechanical load is increased. From Equation 16-13, T ∝ ΦIA, so we can predict how torque varies as load is applied and IA and therefore IL increase. The shunt motor has a constant flux. Therefore, the torque produced will be directly proportional to the load current, and the torque graph is linear as load is applied (Figure 16-27). In the series motor, Φ ∝ IA, and consequently T ∝ I2A. The resulting graph is a parabola, and torque increases rapidly as load is applied. The cumulatively compounded motor has a combination of series and shunt characteristics with the series torque adding to the shunt torque as load is increased.

16-16   Torque Characteristics of DC Motors

The small torque developed at no load is necessary to overcome friction and other rotational losses, allowing the motor to turn. At starting, the back EMF is zero and a large current will flow. For a shunt motor: E − Eb E Eb = E − IARA ⟶ IA = ; therefore, at starting IA = RA RA Compound (cumulative)

Torque, T

Shunt

Series

No load torque

Load Current (IL)

Figure 16-27  Torque Characteristics of DC motors

Since RA is small, the starting current will be high. This current creates a large torque at starting which will help accelerate a heavy load. For this reason, the DC motor is well suited to traction and lifting applications.

Example 16-9 A 4.0-hp series motor when running at full load draws a current of 40 A and turns at 600 RPM. If the starting current is 70 A, calculate the fullload and starting torque. Solution

At full load, IA = 40 A Since hp =

T=

2πNT 33 000

hp × 33 000 4.0 hp × 33 000 = 35 lb- ft = 2πN 2π × 600 RPM

At starting, IA = 70A

For a series motor, T ∝ IA2 and Tst = Trun ×

I2Ast

I2Arun

I2Ast Tst = 2 Trun IArun 2 70 = 35 l b-ft × 2 = 107 lb-ft 40

See Review Problems 16-27 to 16-28 and Review Question 16-55.

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Chapter 16  Inductance

16-17 Permanent Magnet and Brushless DC Motors Many of the traditional DC motors with field windings have been replaced by permanent magnet motors. This design is typically used in fractionalhorsepower motors and even up to 3 hp sizes. Advances in magnetic materials enable the construction of motors with strong magnetic fields provided by permanent magnets. Such motors have a higher efficiency because they do not have any energy losses due to resistance in field windings. A brushless motor operates much in the same way as a traditional motor with brushes but uses electronic switching instead of brushes and a commutator. In a typical brushless motor, the field is on the rotor and is created by permanent magnets. Power is supplied directly to the armature, which is on the stator and therefore can be directly connected. A rotating magnetic field is produced by the armature as current flows and commutation is effected with electronic switches, often by using transistors. The rotor is pulled around in synchronism with this rotating magnetic field. Brushless motors have excellent reliability and long lives, and they can operate at high speeds. Brushless motors are used in computer fans, hard drives, and biomedical and aerospace applications—industries in which reliability is critical. See Review Question 16-56.

Problems

477

Summary

• Perpendicular motion of a conductor relative to a magnetic field p ­ roduces electromagnetic induction. • Changing current in a primary winding induces a voltage in a secondary winding on the same core. • Faraday’s law relates the induced voltage in an electric circuit to the rate of change of the magnetic flux. • The polarity of an induced voltage is always such that any resulting ­current opposes the change in the original magnetic flux. • Self-induction produces an induced voltage across a coil when the ­current through it is changing. • Inductance opposes any change in current in an electric circuit. • Inductance is proportional to the number of turns in a coil and the rate of change of flux with respect to current. • The inductance of a coil is proportional to the square of the number of turns, proportional to the permeability and cross-sectional area of the magnetic circuit, and inversely proportional to the reluctance and the length of the magnetic circuit. • The total inductance of inductors in series (when there is no mutual ­coupling) is the sum of the individual inductances. • The reciprocal of the equivalent inductance of inductors in parallel equals the sum of the reciprocals of the individual inductances, provided there is no mutual inductance. • All rotating electrical machines have a field system and an armature ­system. • A rotating coil in a magnetic field produces an alternating voltage. • A commutator can convert a generated voltage to DC. • Generated voltage depends on the strength of the field flux and the speed of rotation. • A DC motor has the same construction as a DC generator. • The force produced when a current flows through a conductor in a m ­ agnetic field depends on the strength of the field and the magnitude of the current. • The speed of a motor is directly proportional to back EMF and inversely proportional to magnetic flux. • The torque of a motor is directly proportional to the armature current and the magnetic flux. • DC motors are classified based on the configuration of the field s­ ystem.

Problems B B B B

Section 16-5  Self-Inductance 16-1.

Current changing at the rate of 250 mA/s in an inductor induces a voltage of 75 mV. Find the inductance. 16-2. A current rising at a uniform rate from 2.0 mA to 10 mA in 100 μs ­induces a voltage of 20 mV in a load. What is the inductance of the load? 16-3. Find the voltage induced into an 8-H inductor by a current that changes from 12 A to 6 A in 75 ms. 16-4. How long will it take the current in a 1.5-H inductor to rise from zero to 5 A if the voltage induced into the inductor is constant at 6 V?

B = beginner

I = intermediate

A = advanced

478

Chapter 16  Inductance

B

B

16-5.

Find the inductance of a coil if 250 V is induced across it by an 8-A current that reverses direction in 20 ms.

Section 16-6  Factors Governing Inductance 16-6.

I

16-7.

B B I

16-8. 16-9. 16-10.

I

16-11.

I

16-12.

I

16-13.



B

I B

B

B

B

Find the approximate inductance of a 15-turn air-core coil with an inside diameter of 3.0 cm and a length of 2.0 cm, assuming that the leakage flux is negligible. How many turns must an 80 μH inductor have if it has an air-core coil 3.0 cm long and 4.0 cm in diameter? Find the inductance of the inductor in Problem 15-9. Find the inductance of the coil in Problem 15-13. A 1-H inductor has 1400 turns. How many turns must be added to raise its inductance to 2 H? An 800-mH inductor has 600 turns. Where would you place a tap to obtain an inductance of 400 mH between one end and the tap? What is the inductance between the other end and the tap? An iron core has a relative permeability of 3000, a diameter of 4 cm, and a length of 15 cm. How many turns do you have to wind on this core to make a 10-H inductor? A 750-turn coil is wound on a magnetic core with a relative permeability of 2000, a length of 18 cm, and a cross-sectional area of 6.0 cm2. Calculate: (a) the inductance of the coil (b) the voltage induced when the current through the coil decreases steadily from 8 A to zero in 35 ms

Section 16-7  Inductors in Series 16-14. Three inductors with inductances of 10 mH, 20 mH, and 30 mH are positioned so that there is no mutual induction between them. Find the effective inductance when the three coils are connected in series.

Section 16-8  Inductors in Parallel

16-15. What inductance must be connected in parallel with an inductance of 40 μH to reduce the net inductance to 15 μH? 16-16. Find the equivalent inductance when the coils of Problem 16-14 are connected in parallel.

Section 16-11  EMF Equation 16-17. An 8-pole generator has a wave-wound armature with 52 slots, each containing 10 conductors. If the total flux produced by the field system is 0.32 Wb, calculate the EMF generated when the armature is driven at 600 RPM. 16-18. A 4-pole generator with a lap-wound armature has 120 slots with 4 conductors in each slot. If the flux per pole is 50 mWb and the generated EMF is to be 240 V, at what speed must this generator be driven?

Section 16-12  The DC Motor 16-19. What current will develop a force of 270 N on a 300-turn coil in a magnetic field of 1.8 T? The length of the coil in the field is 20 cm.

Review Questions

I B I

I

A

I A

B A

Section 16-13  Speed and Torque of a DC Motor 16-20. A 6-pole lap-wound DC motor has 24 coils on the armature, and each coil has 10 turns. When the flux per pole is 80 mWb, the motor turns at 1000 RPM. Calculate the back EMF generated in the motor armature. 16-21. Calculate the starting torque for the motor of Problem 16-20, given that the starting current is 55 A. 16-22. A 5.0-hp DC motor has a 4-pole wave-wound armature containing 16 coils with 20 turns each. The field system produces a flux per pole of 10 mWb. Given that the full-load speed is 450 RPM, calculate the full-load armature current.

Section 16-14  Types of DC Motors

16-23. A 220-V DC compound motor is connected short shunt and has shunt field, series field, and armature resistances of 55 Ω, 0.12 Ω, and 0.18 Ω, respectively. Calculate the back EMF when the motor has a load current of 72 A. 16-24. A 440-V DC compound motor is connected short shunt and has shunt field resistance of 75 Ω and series field resistance of 0.2 Ω. When the motor has a load current of 40 A, the back EMF is 420 V. Calculate the resistance of the armature.

Section 16-15  Speed Characteristics of DC Motors

16-25. Redo Example 16-8 with the series field connected differentially. 16-26. A 220-V DC 6-pole series motor has a field resistance of 0.15 Ω and an armature resistance of 0.20 Ω. The armature is lap-wound with a total of 248 conductors. When the motor delivers its full-load output, the load current is 52 A and the flux per pole is 35 mWb. When the motor is unloaded, the current drops to 7 A. Assuming the field core is unsaturated and has a linear magnetization curve, calculate the full-load and no-load speeds.

Section 16-16  Torque Characteristics of DC Motors 16-27. Calculate the full-load and no-load torque developed by the motor in Problem 16-26. 16-28. A 35-hp 440-V DC shunt motor has a field resistance of 80 Ω. When delivering its rated full-load horsepower, this motor rotates at 1150  RPM and draws a current of 75  A. When starting, the motor draws a current of 120 A. Calculate: (a)  the full-load and starting torque (b)  the full-load efficiency

Review Questions

Section 16-1  Electromagnetic Induction 16-29. What is meant by the term electromagnetic induction?

Section 16-2  Faraday’s Law 16-30. What factors govern the magnitude of the EMF induced into a c­ onductor?

479

480

Chapter 16  Inductance

Section 16-3  Lenz’s Law 16-31. What factors govern the polarity of the EMF induced into an electric conductor? 16-32. State the direction of conventional current in Figure 16-28. The arrow in the figure indicates motion.

S

N

G



Figure16-28

16-33. State the direction of conventional current in both the primary and secondary windings of Figure 16-29 when (a) the switch is first closed (b) the switch is opened

E

+ −



Figure16-29

Section 16-4  Self-Induction 16-34. What is meant by self-induction?

Section 16-5  Self-Inductance 16-35. How is the unit of inductance derived from other SI units?

Section 16-6  Factors Governing Inductance 16-36. Why is the self-inductance of a coil proportional to the square of the number of turns? 16-37. Why does the inductance of an iron-core coil depend on the current through it? 16-38. What effect does the addition of an air gap to the magnetic circuit of a coil have on its inductance?

Section 16-7  Inductors in Series 16-39. Why is it necessary to stipulate that there must be no mutual induction when one is deriving the equation for the total inductance of inductors in series? 16-40. Why is the total inductance of series inductors equal to the sum of the individual inductances?

Review Questions

Section 16-8  Inductors in Parallel 16-41. In what way is the equation for the equivalent inductance of inductors in parallel similar to the equation for the equivalent resistance of resistors in parallel? Compare these equations to the equation for the equivalent capacitance of capacitors in parallel.

Section 16-9  The DC Generator 16-42. Why is it important for the frame or yoke of a made from ferromagnetic material?

DC

generator to be

Section 16-10  Simple DC Generator 16-43. Explain why the voltage generated in a rotating coil is alternating. 16-44. Explain how a commutator converts the alternating voltage produced in the armature to DC voltage. 16-45. Why is the output voltage of a simple DC generator referred to a pulsating DC?

Section 16-11  EMF Equation 16-46. What can be done to smooth the output voltage of a DC generator? 16-47. For a particular generator, which factors determine the magnitude of generated voltage?

Section 16-12  The DC Motor 16-48. How does a DC motor differ in construction from a DC generator? 16-49. Why does a current-carrying conductor in a magnetic field experience a force?

Section 16-13  Speed and Torque of a DC Motor 16-50. What quantities determine the speed of a DC motor? 16-51. Why is torque an important characteristic of a DC motor?

Section 16-14  Types of DC Motors 16-52. Explain the difference between a short-shunt and a long-shunt compound motor. 16-53. Why is a differentially connected compound motor rarely used?

Section 16-15  Speed Characteristics of DC Motors 16-54. Which DC motor is best suited for driving a machine that uses magnetic tape for storage? Explain your answer.

Section 16-16  Torque Characteristics of DC Motors 16-55. Which DC motor is best suited for an electric golf cart? Explain your answer.

Section 16-17  Permanent Magnet and Brushless DC Motors 16-56. Describe how a brushless motor operates.

481

482

Chapter 16  Inductance

Integrate the Concepts

(a) Calculate the inductance of the air-core coil of Example 15-1. (b) Calculate the inductance of the toroidal coil of Example 15-2. (c) Calculate the total inductance of these two inductors connected in series (assume no mutual inductance). (d) Calculate the equivalent inductance of these two inductors connected in parallel (assume no mutual inductance).

Practice Quiz 1. Which of the following statements is true? (a)  An induced voltage will appear in the conductor if the conductor and the magnetic field are moving parallel to each other. (b)  The polarity of an induced voltage can be determined using Lenz’s law. 2. Faraday’s law states that (a)  the voltage induced in an electric circuit is inversely proportional to the rate of change of the magnetic flux linking the circuit (b)  the current induced in an electric circuit is directly proportional to the rate of change of the magnetic flux linking the circuit (c)  the voltage induced in an electric circuit is directly proportional to the rate of change of the magnetic flux linking the circuit (d) the current induced in an electric circuit is inversely proportional to the rate of change of the magnetic flux linking the circuit 3. Self-inductance is (a)  directly proportional to the number of turns linked by the magnetic flux and to the current flowing through the circuit (b)  inversely proportional to the number of turns linked by the magnetic flux and to the current flowing through the circuit (c)  directly proportional to the number of turns linked by the magnetic flux and inversely proportional to the current flowing through the circuit (d) inversely proportional to the number of turns linked by the magnetic flux and directly proportional to the current flowing through the circuit 4. If the permeability of the magnetic circuit of a coil decreases, the ­inductance (a)   decreases (b)   stays the same (c)   increases 5. If the length of the magnetic circuit of a coil decreases, the inductance (a)   decreases (b)   stays the same (c)   increases 6. If the number of turns in a coil is increased, the inductance (a)   decreases (b)   stays the same (c)   increases

Practice Quiz

7. The value of the equivalent inductor for the circuit of Figure 16-30 is: (a) 99 nH (b) 20 nH (c) 190 nH (d) 19 mH L1 0.0745 μH

L3 0.068 μH

L2 0.047 μH



Figure 16-30

  8.

The value of the equivalent inductor for the circuit of Figure 16-31 is (a) 0.48 mH (b) 12 mH (c) 0.91 mH (d) 1.2 mH

L1 1.0 mH



L2 10 mH

L3 1.0 mH

Figure 16-31

  9. A 250-turn coil with a length of 25 cm is in a magnetic field with a flux density of 2.0 T. If the force is 300 N, the current through the coil is (a)   4.8 mA (b)   24 mA (c)   1.67 A (d)  2.4 A 10.

The torque produced by a 2.5-hp DC motor operating at 1500 RPM is (a)   0.11 lb-ft (b)   8.8 lb-ft (c)   4.4 lb-ft (d)   18 lb-ft

483

17

Inductance in DC Circuits We now examine the behaviour of practical electric circuits when inductance is present. Often an inductor is used to give the circuit inductance for a specific purpose, as in the tuned and filtering circuits descibed in Chapter 25. However, some other components, such as wire-wound p ­ otentiometers, have some inductance in addition to their primary property.

Chapter Outline 17-1

17-2

Current in an Ideal Inductor  486

Rise of Current in a Practical Inductor  487

17-3 Time Constant  490

17-4 Graphical Solution for Inductor Current  491

17-5 Algebraic Solution for Inductor Current  495 17-6 Energy Stored by an Inductor  499 17-7

Fall of Current in an Inductive Circuit  501

17-8 Algebraic Solution for Discharge Current  506 17-9 Transient Response  507

17-10 Characteristics of Inductive DC Circuits  509 17-11 Troubleshooting 510

Key Terms LR circuit  490 discharge resistor  503

transient response  507 filter choke  510

Learning Outcomes At the conclusion of this chapter, you will be able to: • draw a graph of the rise of current in an ideal inductor • determine the time constant of a circuit ­containing ­resistance and inductance • use universal exponential curves to find ­instantaneous currents and voltages in a DC ­circuit containing ­resistance and inductance

Photo sources:  iStock.com/Marco Hegner

• calculate exact values of instantaneous currents and v ­ oltages in a DC circuit containing resistance and ­inductance • calculate the energy stored in an inductor • detect faults in an inductor with an ohmmeter or an LC meter

486

Chapter 17   Inductance in DC Circuits

17-1  Current in an Ideal Inductor An ideal inductor would have no resistance. However, all known materials have some resistance at room temperature. For the moment, we shall assume that the inductor in Figure 17-1 has negligible resistance.

+ 10 V −

E = VL

2.0 H

Figure 17-1  Voltage drop across an ideal inductor

According to Kirchhoff’s voltage law, a voltage drop exactly equal to the applied voltage appears across an external circuit connected to a voltage source. If this external circuit contains only resistance, current flowing through the resistance produces this voltage drop. When the circuit is switched on, the current instantly rises to a steady value of I = V/R = E/R. However, the voltage drop across an ideal inductor is produced by a changing current. Section 16-4 described how a changing current can produce a voltage across a coil by inducing an EMF in the coil. In Equation 16-3, eL represented the induced voltage. The lowercase e indicates a changing voltage resulting from an EMF. However, we do not think of the inductor in Figure 17-1 as a voltage source. In this circuit we treat the induced voltage as the ­voltage drop required by Kirchhoff’s voltage law. Thus, we can rewrite Equation 16-3 as

vL = L ×

di dt

(17-1)

Rearranging Equation 17-1 gives

di vL = dt L

Since the applied voltage is constant, the voltage across the inductor must also be constant. Therefore, we can represent the voltage drop with the uppercase letter V. The inductance of a coil depends on the number of turns and the reluctance of the magnetic circuit, so the inductance of a given air-core coil is constant. For such inductors, the ratio V/L is constant. Hence, the rate of change of current must also be constant, and di Δi V = = dt Δt L

where t is the number of seconds elapsed since the switch was closed.

17-2   Rise of Current in a Practical Inductor

Since the initial current is zero,

Δi i −0 i V = = = Δt t−0 t L



i=



V ×t L

For the circuit in Figure 17-1, i=



10 V × t = 5tA 2.0 H

The graph of current versus time is a straight line, as shown in Figure 17-2. The slope of this graph represents the rate of change of current. As shown above, this rate of change equals V/L. Increasing the inductance slows the rise of the current. This relationship shows how inductance opposes the change in current in a circuit.

Current (amperes)

10 8 6 4 2 0

1 Time (seconds)

2

Figure 17-2  Current in a DC circuit containing an ideal inductor

See Review Questions 17-27 to 17-29 at the end of the chapter.

17-2 Rise of Current in a Practical Inductor At room temperature, all practical inductors possess some resistance as well as inductance. Since there is only one path for current through an inductor, its inductance and resistance are effectively in series, as shown in Figure 17-3.

487

488

Chapter 17   Inductance in DC Circuits

vR + 10 V E −

vL

+ 1.0 Ω − + 2.0 H −

Figure 17-3  Equivalent circuit for a practical inductor

When we close the switch in the circuit of Figure 17-3, the voltage drop across the resistance of the coil plus the voltage drop across the inductance of the coil must equal the applied voltage in order to satisfy Kirchhoff’s voltage law: E = vR + vL Applying Ohm’s law and substituting the expression for vL from Equation 17-1 gives



E = iR + L

di dt

(17-2)

If we can solve for the instantaneous current in the circuit, we can then readily determine the voltage drop across the inductor. However, solving Equation 17-2 for i requires integral calculus. Therefore, we shall start by using our understanding of the current in an ideal inductor to determine the overall behaviour of a practical inductor in a circuit. The inductance of the practical inductor and the applied voltage in ­Figure 17-3 are the same as for the ideal inductor in Figure 17-1. Therefore, the current through the practical inductor cannot rise more rapidly than shown in the graph of Figure 17-2. Otherwise the induced voltage across the inductance would be greater than the applied voltage. Similarly, the current cannot rise to a value such that the IR drop across the resistance ­exceeds the applied voltage. The greatest value that the instantaneous ­current can have is such that vR = E. Then, vL is zero, so di/dt is also zero. Therefore, the current has reached a steady-state maximum value, which is

Im =

E R

(17-3)

The dashed black lines in Figure 17-4 show the two limits for the current.

17-2   Rise of Current in a Practical Inductor

Current (amperes)

Slope = 10

τ

di E = dt L

Im = E R i

8 6 4

63% of Im

2 0

2

4 6 Time (seconds)

8

10

Figure 17-4  Current in practical inductor

As shown in Section 17-1, the inductance limits the rate at which the c­ urrent can rise. At the instant we close the switch in Figure 17-3, the current through the resistance is zero and hence the voltage drop across it is also zero. Therefore, voltage across the inductor at this instant equals the applied voltage. Since di/dt = vL/L, the initial rate of change of current is the same as for the ideal inductor in Figure 17-1:

initial

di E = dt L

(17-4)

However, since the current is rising in order to induce a voltage in the i­ nductance, the voltage drop across the resistance must be rising. Because vR + vL = E, the induced voltage across the inductor decreases correspondingly. Consequently, the rate of change of current also decreases and the slope of the current graph curve in Figure 17-4 becomes more gradual. ­Although the current is now rising less rapidly, nevertheless it is still rising. As a result vL and, therefore, the rate of change of current decrease further, and the current graph becomes still more gradual. Eventually, the current reaches the steady-state maximum value given by Equation 17-3. When we close the switch in a DC circuit containing inductance and ­resistance in series, the current cannot instantly rise to E/R as it would if there were no inductance in the circuit. The effect of the inductance is to slow the rise of current, as shown by the red curve in Figure 17-4. Once the current reaches its maximum value, the effect of the inductance in the circuit disappears. Since vR = iR and R is a constant, the graph of the voltage drop across the resistance of the circuit must have the same shape as the graph of the current (see Figure 17-5). We can graph the instantaneous voltage across the inductance by subtracting the instantaneous IR drop from the constant applied voltage: vL = E − vR

489

Chapter 17   Inductance in DC Circuits

E 10 Instantaneous Voltage (volts)

490

τ vR

8 6 4 2 0

vL 2

4 6 Time (seconds)

8

10

Figure 17-5  Instantaneous voltage across the resistance and inductance

See Review Question 17-30.

17-3  Time Constant As with a capacitor, it is convenient to describe the current and voltage curves for an inductor in terms of a time constant related to the parameters of the circuit. The time constant, τ, of an LR circuit (a circuit containing inductance and resistance, but not capacitance) is the time it would take the current to rise to its steady-state value if the current increased at a constant rate equal to its initial rate of change. If the rate of change of current remains constant (as shown by the straight line in Figure 17-4), the steady-state current would be simply Im = τ × initial

di dt

But Im = E/R (Equation 17-3) and initial di/dt = E/L (Equation 17-4). Therefore,

and

Im =

E E =τ R L

τ

L R

(17-5)

where τ is the time constant in seconds, L is the inductance of the circuit in henrys, and R is the resistance of the circuit in ohms. Note that the time constant of an LR circuit, such as a practical inductor, is directly proportional to the inductance. Doubling the inductance reduces the initial rate of rise of current by half, and the current takes twice as long to reach the steady-state value. Since the time constant is inversely proportional to the resistance, doubling the resistance reduces the steady-state current by half. The current then takes half as long to

17-4   Graphical Solution for Inductor Current

491

reach the steady-state value. However, the applied voltage has no effect on the time constant. Doubling the applied voltage doubles both the ­initial rate of rise of current (E/L) and the steady-state current (E/R). Therefore, the current takes the same time to reach the steady-state value. The current in an LR circuit does not actually continue to rise at its initial rate. Instead, the current follows an exponential curve, as shown in Fig­ure  17-4. This exponential curve starts with a slope of E/L, and then levels off more and more slowly to finally merge with the horizontal line representing the steady-state current. For any values of L and R, the current reaches about 63.21% of the steady-state value one time constant after the circuit is switched on. Although theoretically the current never quite reaches the steady-state value, for practical purposes the current is E/R after five time constants have elapsed.

Example 17-1 (a) Find the steady-state current in the circuit of Figure 17-3. (b) How long does it take the current to reach this value after the switch is closed? Solution (a) (b)

Im =

τ=

E 10 V = = 10 A R 1.0 Ω

L 2.0 H = = 2.0 s R 1.0 Ω

Time to reach steady state = 5τ = 5 × 2.0 s = 10 s

To verify the calculation of the current in part (a) and the time in part (b), download Multisim file EX17-1 from the website and follow the instructions in the file.

See Problems 17-1 to 17-6 and Review Questions 17-31 to 17-34.

17-4 Graphical Solution for Inductor Current As with capacitors, we can use universal exponential graphs to make fairly accurate estimates of the current through an inductor at a given moment and the time when the current reaches a particular value. For inductors, the vertical axis of these graphs shows the percentage of the steady-state current and the horizontal axis shows time measured in time constants (see Figure 17-6).

circuitSIM walkthrough

Chapter 17   Inductance in DC Circuits

100 90 Current (% of maximum)

492

y = 1 − e−x

80 70 60 50 40 30 20 y = e−x

10 0

3 2 Time, x (time constants)

1

4

5

Figure 17-6  Universal exponential curves for inductive DC circuits

Example 17-2 (a) What is the instantaneous current in the circuit of Figure 17-3 1.0 s after the switch is closed? (b) How long does the current take to rise from 0 A to 5.0 A? Solution (a) We have already determined that the time constant of this circuit is



Therefore,

τ=

L 2.0 H = = 2.0 s R 1.0 Ω 1.0 s = 0.50τ

On the universal exponential graph, y = 39% when x = 0.50 as shown in Figure 17-7(a). Therefore the current at t = 1.0 s is i = 39% × (b)

E 10 V = 0.39 × = 3.9 A R 1.0 Ω

Im =

E 10 V = = 10 A R 1.0 Ω

5.0 A 5.0 A = = 50% Im 10 A

On the universal exponential graph, x = 0.70 when y = 50%, as shown in Figure 17-7(b). Therefore, t = 0.70τ = 0.70 × 2.0 s = 1.4 s

17-4   Graphical Solution for Inductor Current

100 Current (% of Im )

Current (% of Im )

100

493

50 39

0 0.5τ

50

0

0.7τ Time (b)

Time (a)

Figure 17-7  Using a universal exponential graph to solve Example 17-2

To verify the calculation of the current in part (a) and the time in part (b), download Multisim file EX17-2 from the website and follow the instructions in the file.

Example 17-3 In the circuit shown in Figure 17-8, switch S2 is closed 1.0 s after switch S1 is closed. What is the current 1.0 s after switch S2 is closed? S2 S1

4.0 Ω 1.0 Ω

+ 10 V −

2.0 Ω

Figure 17-8  Circuit diagram for Example 17-3

Solution Step 1 While S1 is closed and S2 is open, the total resistance in the circuit is RT = 4.0 Ω + 1.0 Ω = 5.0 Ω The time constant is

τ=

L 2.0 H = = 0.40 s R 5.0 Ω

circuitSIM walkthrough

Chapter 17   Inductance in DC Circuits

1.0 s = 2.5τ 0.40 s/T

Therefore,

From the curve of Figure 17-6, when t = 2.5τ, i = 92% of

E 10 V = 0.92 × = 1.84 A R 5.0 Ω

There are two ways of dealing with the 1-s interval after S2 is closed. Step 2 Consider the current curve for the circuit, which now has only 1.0 Ω of ­resistance. The steady-state current becomes Im =

E 10 V = = 10 A R 1.0 Ω

At the instant S2 is closed, i is already 1.84 A. Find the point on the graph corresponding to this current and mark off an additional 1.0 s, as shown in Figure 17-9(a). 1.84 A 1.84 = = 18.4% Im 10

From the curve of Figure 17-6, when i = 18.4% of Im, t = 0.2τ = 0.2

L 2.0 H = 0.2 × = 0.4 s R 1.0 Ω

Therefore, on the curve in Figure 17-9(a), the current at t = 1.4 s is the current 1.0 s after S2 is closed. In terms of time constants, 1.4 s =

1.4 τ = 0.70τ 2.0

Again, from Figure 17-6, when t = 0.7τ,

10 i 5 50%

1.84

0 0.2τ 0.7τ 1.0 s

Time (a)

Additional Current (amperes)

i = 50% of Im = 0.50 × 10 A = 5.0 A

Additional Current (amperes)

494

10 8.16 i

5 3.18

100% 39%

0

0.5τ 1.0 s

Time (b)

Figure 17-9  Graphical solution for Step 2 of Example 17-3

17-5   Algebraic Solution for Inductor Current

495

Alternative Step 2 Consider the additional rise in current after S2 is closed. At the moment S2 is closed, a current of 1.84 A is flowing in the i­nductance. As calculated in Step 2a, Im is now 10 A. Therefore, the maximum additional rise in current is 10 − 1.84 = 8.16 A. The time constant is now τ=

L 2.0 H = = 2.0 s R 1.0 Ω

Therefore, 1.0 s now corresponds to 0.5τ. During this time, the ­additional current rises to 39% of its maximum value, as shown in Figure 17-9(b). Therefore, the change in current is Δi = 39% of 8.16 A = 3.18 A

Therefore, the total instantaneous current 1.0 s after S2 is closed is i = 1.84 A + 3.18 A = 5.0 A To verify the calculation of the current, download Multisim file EX17-3 from the website and follow the instructions in the file.

circuitSIM walkthrough

See Problems 17-7 to 17-9 and Review Question 17-35.

17-5 Algebraic Solution for Inductor Current The graphic solution for an LR circuit is accurate enough for most purposes. If we need greater accuracy, we must return to Equation 17-2. Appendix 2-4 gives a calculus solution for this differential equation. The solution yields an exponential equation for the current: i=



E (1 − e − x ) R

(17-6)

where e is the base of natural logarithms (2.718 . . .) and x is the elapsed time measured in time constants (x = tR/L = t/τ).

Example 17-2A (a) What is the instantaneous current in the circuit of Figure 17-3 1.0 s after the switch is closed? (b) How long does it take for the current to rise from 0 A to 5.0 A?

Most scientific calculators have an ex key for calculating exponentials and a ln key for calculating natural logarithms.

496

Chapter 17   Inductance in DC Circuits

Solution

(a) x = t × i=



E 10 V ( 1 − e− x ) = × ( 1 − e− 0.50 ) R 1.0 Ω

= 10 A × ( 1 − 0.6065 ) = 3.93 A

(b)

R 1.0 Ω = 1.0 s × = 0.50 L 2.0 H

i=

5.0 A =

E ( 1 − e− x ) R

10 V × ( 1 − e− x ) 1.0 Ω

5 = 10 − 10e−x e−x = 0.5

−x = ln 0.5 = −0.693



t=x×



L 2.0 H = 0.693 × = 1.4 s R 1.0 Ω

Example 17-3A In the circuit shown in Figure 17-8, switch S2 is closed 1.0 s after switch S1 is closed. What is the current 1.0 s after switch S2 is closed? Solution Step 1 At t = 1.0 s, x=t× i=

R 5.0 Ω = 1.0 s × = 2.5 L 2.0 H

E 10 V ( 1 − e− x ) = × ( 1 − e− 2.5 ) = 2.0 × ( 1 − 0.0821 ) = 1.84 A R 5.0 Ω

Step 2 Using the method illustrated by Figure 17-9(a) gives

i=

1.84 =

E ( 1 − e−x ) R

10 V × ( 1 − e−x ) 1Ω

e − x = 0.816

17-5   Algebraic Solution for Inductor Current

x = 0.203 = t ×

R L −x = ln 0.816 = −0.203 2.0 H Therefore, t = 0.203 × = 0.406 s 1.0 Ω One second later, t = 1.406 s 1.0 Ω and x = 1.406 s × = 0.703 2.0 H E 10 V × ( 1 − e− 0.703 ) i = ( 1 − e−x ) = R 1.0 Ω = 10 A × ( 1 − 0.495 ) = 5.05 A from which

Alternative Step 2 Using the method illustrated by Figure 17-9(b), we can show that the maximum additional current is 10 − 1.84 = 8.16 A. Once S2 is closed, the time constant becomes τ = 2.0 s. For a time interval of 1.0 s, x=t×

R 1.0 Ω = 1.0 s × = 0.50 L 2.0 H

Δi = 8.16(1 − e−0.50 ) = 8.16(1 − 0.6065) = 3.21 A

The change in current during the first second after S2 is closed is Therefore, the total instantaneous current 1 s after S2 is closed is i = 1.84 A + 3.21 A = 5.05 A The LR network of Figure 17-10(a) includes a 4.0-Ω discharge resistor in parallel with a practical inductor. (Section 17-7 describes the function of this resistor.) To determine the time constant for the buildup of current in the inductor, we consider the 4.0-Ω resistor to be part of the source network as “seen” by the inductor. We can then use the Thévenin-equivalent circuit of Figure 17-10(b) to calculate the current through the inductor. 1.0 Ω

1.0 Ω + 10 V −

RTh

1.0 Ω

0.80 Ω

4.0 Ω 2.0 H

(a)

+ ETh 8 V −

2.0 H

(b)

Figure 17-10  (a) Inductor circuit with discharge resistor; (b) Thévenin-equivalent circuit

497

498

Chapter 17   Inductance in DC Circuits

Example 17-4 For the circuit of Figure 17-10(a), calculate the current in the inductor 1.0 s after the switch is first closed. Solution Step 1 With the inductor disconnected from the rest of the circuit, the opencircuit voltage across the 4.0-Ω resistor is and

ETh = 10 V ×

RTh =

4.0 Ω = 8.0 V 4.0 Ω + 1.0 Ω

4.0 Ω × 1.0 Ω = 0.80 Ω 4.0 Ω + 1.0 Ω

Hence, the time constant is Step 2 and

circuitSIM walkthrough



τ=

L 2.0 H = = 1.111 ( 1.0 + 0.80 ) Ω RT

x=

1.0 s t = = 0.90 τ 1.111 s

i=

ETh 8.0 V ( 1 − e−x ) = ( 1 − e − 0.90 ) RT 1.8 Ω

= 4.444 A ( 1 − 0.4066 ) = 2.6 A

To verify the calculation of the current, download Multisim file EX17-4 from the website and follow the instructions in the file.

See Problems 17-10 to 17-15.

Circuit Check 

A

CC 17-1. Calculate the resistance in series with a 240-mH inductor, given that the current reaches full value in 25 ms. CC 17-2. An inductor with 20-Ω resistance is connected to an 80-V source. After 15 ms, the current through the inductor is 1.8 A. Calculate the inductance.

17-6   Energy Stored by an Inductor

17-6  Energy Stored by an Inductor

Current

If we connect an ideal inductor to a voltage source having no internal ­resistance, the voltage across the inductance must remain equal to the applied voltage. Therefore, the current rises at a constant rate, as shown in Fig­ure  17-11(b). The source supplies electrical energy to the ideal inductor at the rate of p = Ei. Unlike resistance, inductance cannot convert this energy into heat or light. Instead, the energy is stored in the magnetic field as the rising current forces the magnetic lines of force to expand against their tendency to become as short as possible—somewhat as a rubber band stores energy when it is stretched.

+ E

Im i

Iav

− Time (b)

(a)

Figure 17-11  Determining the energy stored by an inductor

In resistance circuits where the current and voltage do not change with a change in time, the energy transferred from the source to the resistance is W = Pt = VIt. Although the voltage remains constant in the circuit of Figure 17-11(a), the current steadily increases as time elapses. However, since the rate of change of current is constant, the average value of the current, I, as it rises from zero to Im is 1⁄2 Im. Therefore, the energy stored by an inductor as the current rises from zero to Im is 1 W = V ×  Im × t 2 Since the rate of change of current is constant, Equation 17-1 becomes

V=L

di LIm = dt t

(17-1)

Substituting for V in the equation for W gives W=L×



Im 1 × Im × t t 2

1 W = LI2m 2

(17-7)

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Chapter 17   Inductance in DC Circuits

With a practical inductor, both the instantaneous voltage and current are changing and their rates of change are not constant. If we find the voltage across and the current through the inductance for given moment, we can use relationship p = vi to calculate the rate at which the inductance of the circuit stores energy at that moment. With a series of such calculations we can plot the graph in Figure 17-12.

vL iL Area = WL pL

0

1

2 3 4 Time (time constants)

5

Figure 17-12  Energy stored by a practical inductor

When the current in a practical inductor reaches its steady-state value of Im = E/R, the magnetic field ceases to expand. The voltage across the ­inductance has dropped to zero, so the power p = vi is also zero. Thus, the energy stored by the inductor increases only while the current is building up to its steady-state value. When the current remains constant, the energy stored in the magnetic field is also constant. Although no additional energy is stored by the inductance of the practical inductor, the resistance of the ­inductor dissipates energy at a steady rate of P = I2mR. Current must continue to flow to maintain the magnetic field. Thus, an inductor stores energy in a dynamic form, somewhat like the ­kinetic energy stored by the flywheel of an engine. By contrast, no current is required to maintain the energy stored by a capacitor (except to offset leakage through the dielectric). The electric field of a capacitor stores ­energy in a static form, somewhat like the energy stored in a tank of compressed air. Note the similarity between Figure 17-12 and Figure 13-15. The significant difference is that the voltage curve for one graph becomes the current curve for the other. Since p = vi, the power curve is the same in both figures. The rate of storage of energy, p, reaches its peak value at the same instant that iL rises to 1⁄2Im and vL drops to 1⁄2 E. The area under the power curve represents the energy stored by the inductance and is equal to the product of the average power and the elapsed time. Appendix 2-5 uses basic integral calculus to show that energy stored in the magnetic field of an inductor is

17-7   Fall of Current in an Inductive Circuit

1 w = Li2 2



(17-8)

where w is the stored energy in joules, L is the inductance in henrys, and i is the current in amperes.

Example 17-5 Find the maximum energy stored by an inductor with an inductance of 5.0 H and a resistance of 2.0 Ω when the inductor is connected to a 24-V source. Solution

Im =

E 24 V = = 12 A R 2.0 Ω

1 1 W =  LI2m =   × 5.0 H × (12 A)2 = 360 J 2 2 See Problems 17-16 and 17-17 and Review Question 17-36.

17-7 Fall of Current in an Inductive Circuit When the current through an inductor stops, the magnetic field collapses, releasing the stored energy. Since the inductance opposes any change in current, the current cannot cease instantly. Any change in the current takes time, even if only a few microseconds. At the instant we open the switch in Figure 17-13, the current in the ­inductor is the same as just before the switch was opened. The steady-state current before the switch is opened is I=

500 kΩ + 12 V −

E 12 V = = 6.0 A R 2.0 Ω

2.0 Ω

4.0 H

Figure 17-13  Interrupting current in an inductor

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Chapter 17   Inductance in DC Circuits

At the instant the switch opens, the current in the inductor is still 6.0 A. To demonstrate the effect of interrupting this current, we assume that the leakage resistance of the open switch is 500 kΩ. Since this leakage resistance is in series with the source and the inductor, the 6.0-A current flows through the leakage resistance, and the voltage drop between the switch contacts is V = IR = 6.0 A × 500 kΩ = 3.0 MV This extremely high voltage drop is matched by the voltage generated in the turns of the inductor by the rapidly collapsing magnetic field. Such a high voltage would cause arcing between the switch contacts and would probably damage the insulation of the inductor. Therefore, we must design DC circuits to prevent sudden interruptions of current in an inductor.

Example 17-6 Find the voltage across the switch contacts in Figure 17-14 at the instant that the switch is opened. 1.0 A

6.0 A

2.0 Ω + 12 V −

+ 12 Ω −

4.0 H

+ 12 V −

(a)

−

6.0 A

+

(b)

Figure 17-14  Circuit diagrams for Example 17-6

Solution With the switch closed, the steady-state current through the inductor is I=

12 V E = = 6.0 A R 2.0 Ω

At the same time, a 1.0-A current passes through the 12-Ω resistor, as shown in Figure 17-14(a). This current does not flow through the inductance of the circuit. When the switch is opened, the inductor momentarily acts as a generator as its magnetic field collapses. The load across this generator is 12 Ω and 2.0 Ω in series. Since the current through the inductor cannot change instantly, a current of 6.0 A flows through the 12-Ω resistor, as shown in Figure 17-14(b). The voltage drop across the 12-Ω resistor is then v = iR = 6.0 A × 12 Ω = 72 V

17-7   Fall of Current in an Inductive Circuit

503

The 72-V drop across the 12-Ω resistor and the 12-V potential difference of the battery are connected in series aiding across the switch. Therefore, the total voltage across the switch contacts at the instant we open the switch is v = 72 V + 12 V = 84 V To verify the calculation of the voltage, download Multisim file EX17-6 from the website and follow the detailed instructions in the file.

According to Kirchhoff’s voltage law, the initial induced voltage across the inductor must equal the voltage drop across the total resistance: Em = I0RT



(17-9)

where Em is the peak voltage induced by the collapsing magnetic field, I0 is the current through the inductor at the instant the switch is opened, and RT is the total resistance of the loop containing the inductor. In Example 17-6,

Em = 6.0 A × 14 Ω = 84 V

According to Lenz’s law, the induced voltage always opposes any change in current. When the current rises, the induced voltage cancels some of the applied voltage. But when the source is removed, the magnetic field around the inductor collapses. The polarity of the voltage induced by a collapsing field is opposite to that of the voltage generated by a rising field. This reversed polarity tends to maintain current flow in the original direction through the inductor. The 12-Ω resistor in the circuit of Figure 17-14 acts as a discharge resistor. As the Example 17-6 shows, this discharge resistor protects both the switch and the inductor from a high-voltage surge. Some such provision must be made in all direct-current circuits with appreciable inductance, such as DC motors and generators. Silicon carbide varistors are excellent for this purpose. The discharge resistor in Example 17-6 reduces the induced voltage, so the magnetic field must be collapsing at a much slower rate and, therefore, it takes longer to discharge the stored energy. Since the magnetic field is collapsing and the magnetic field strength depends on the current in the inductor, the current starts to decrease as soon as the switch is opened. We can define a discharge time constant, τ, as the time the current takes to drop to zero if it continues to fall at its initial rate of change. If the current does drop at a constant rate equal to its initial rate (as shown by the dashed line in Figure 17-15), then initial

di I0 = τ dt

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Chapter 17   Inductance in DC Circuits

Switch opened

100 80 Current (% of I0)

504

60

i

40 20 0

37% of I0 0

1

2 3 Time (time constants)

4

5

Figure 17-15  Fall of current in an inductive DC circuit

Since initial di/dt = Em/L (Equation 17-4) and Em = I0RT (Equation 17-9), τ=



L RT

(17-5)

Therefore, we determine the time constant of an LR circuit in exactly the same manner for rise and fall of current. The current through an inductor does not actually fall at a constant rate. Since the current is decreasing, the voltage drop across the resistance must also decrease. Kirchhoff’s voltage law requires the induced voltage to equal the voltage drop across the resistance, so the induced voltage must also decrease. As the induced voltage decreases, the magnetic field must collapse more slowly. Thus, the rate of change of current becomes progressively smaller. The graph of falling current in Figure 17-15 has the same ­exponential shape (but inverted) as the curve for rising current.

Example 17-7 The switch in the circuit of Figure 17-14 is closed for 3.0 s and then opened. Find the inductor current 0.25 s after the switch is opened. Solution Step 1 With the switch closed,

Charge τ =

L 4.0 H = = 2.0 s R 2.0 Ω

17-7   Fall of Current in an Inductive Circuit

505

3.0 s = 1.5τ 2.0 s/T

Therefore,

From the graph for current rise in Figure 17-6, iL = 78% of Im when t = 1.5τ. E 12 V = 4.7 A Hence, iL = 0.78 × = 0.78 × R 2.0 Ω Step 2 With the switch open, the time constant becomes Discharge τ =

L 4.0 H = 0.286 s = RT 12 Ω + 2.0 Ω

Expressing 0.25 s in time constants,

0.25 s = 0.875τ 0.286 s/T

From the graph for current decay in Figure 17-6, iL = 42% of I0 when t = 0.88τ. iL = 0.42 × 4.7 A = 2.0 A

Hence,

To verify the calculation of the current, download Multisim file EX17-7 from the website and follow the instructions in the file.

Figure 17-16 shows another way to prevent sparking at switch contacts in inductive circuits. The potential difference across the capacitor is zero while the switch is closed. Since the capacitor voltage cannot rise instantly when the switch opens, the switch contacts have time to move far enough apart to prevent arcing. Some of the energy released by the collapsing magnetic field of the inductor transfers to the capacitor as it charges. The capacitor then partially discharges, building up a smaller magnetic field around the inductor opposite to the original magnetic field. Several cycles of energy transfers take place until the stored energy is all dissipated in the

C + E

Inductive load

−

Figure 17-16  Suppressing a voltage surge with a capacitor

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Chapter 17   Inductance in DC Circuits

resistance of the load. Chapter 25 discusses the transient response of DC circuits containing both inductance and capacitance in more detail. See Problems 17-18 to 17-20 and Review Questions 17-37 to 17-41.

17-8 Algebraic Solution for Discharge Current When we write the Kirchhoff’s voltage-law equation for the circuit of Figure 17-14(b), the applied voltage in the circuit is zero. Therefore, 0 = iRT + L

di dt

As shown in Appendix 2-4, using calculus to solve for the instantaneous current gives i = I0e − x



(17-10)

where I0 is the initial current, e = 2.718 . . ., and x = tRT/L = t/τ.

Example 17-7A The switch in the circuit of Figure 17-14 is closed for 3.0 s and then opened. Find the inductor current 0.25 s after the switch is opened. Solution Step 1 With the switch closed,

x=t×

R 2.0 Ω = 3.0 s × = 1.5τ L 4.0 H

Therefore, E 12 V iL = ( 1 − e − x ) = × ( 1 − e−1.5 ) = 6.0 A × ( 1 − 0.223 ) = 4.66 A R 2.0 Ω Step 2 When the switch has been open for 0.25 s, and

x=t×

RT 14 Ω = 0.25 s × = 0.875 time constant L 4.0 H

iL = I0 × e − x = 4.66 A × e − 0.875 = 4.66 × 0.417 = 1.9 A

See Problems 17-21 and 17-22.

17-9   Transient Response

Circuit Check

B

CC 17-3. An inductor with an inductance of 40 H and a resistance of 2.5 kΩ is connected across a 120-V DC power supply that has negligible internal resistance. Calculate the maximum energy stored in the magnetic field of the inductor. CC 17-4. The switch in Figure 17-17 is set to position 1 for 1.0 s and then switched to position 2. Calculate: (a) the charge and discharge time constants (b) the current flowing after 1.5 s (c) the time the voltage across the 1.25 Ω resistor takes to fall to 2.5 V on discharge 1.2 H

0.80 H 1 2

+ 10 V −

0.50 Ω

1.25 Ω

Figure 17-17

17-9  Transient Response As we have seen throughout this chapter, the current in an LR network does not instantly change to a new steady state when we suddenly change the source voltage. The interval while the current reaches the final steadystate current is the transient response of the LR network. So far, we have used one exponential equation and curve for rising ­inductor currents and a different equation and curve for falling inductor currents. However, we can derive a single equation for any LR transient as long as we are careful in specifying the initial steady-state current (I0) and the final steady-state current (IF). For example, in Step 2b of Example 17-3A, I0 is 1.84 A (at t = 1.0 s) and IF is 10 A. The instantaneous ­current during the transient period is equal to the initial current plus an i­ ncreasing percentage of the additional current. In general,

iL = I0 + ( IF − I0 ) ( 1 − e − x )

(17-11)

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Chapter 17   Inductance in DC Circuits

Simplifying gives the universal equation for inductor current in a DC LR ­network: iL = I0 + IF − I0 − IFe−x + I0e−x iL = IF + ( I0 − IF ) e − x



(17-12)

where iL is the instantaneous inductor current, I0 is the initial inductor ­current, IF is the final steady-state inductor current (after at least five time constants), and x is the elapsed time measured in time constants. Equation 17-11 expresses instantaneous current as initial current plus a proportion of the difference between the initial and final currents. ­Equa­tion 17-12 expresses the same instantaneous current in terms of the final steady-state current and an exponential transient caused by the change in steady-state current.

Example 17-8 In the circuit of Figure 17-18(a), the switch is closed at t = 0 and opened again at t = 1.0 s. (a) Find the inductor current at t = 1.0 s. (b) Find the voltage across the switch contacts at t = 1.1 s. (c) Draw a graph of the voltage across the switch from t = 0 to t = 5 s.

2.0 H

1.0 Ω

+ 10.0 V −

4.0 Ω

Switch Voltage, vsw (volts)

508

20.6 V

20 16 12 8 4 0

1 2 3 4 Time, t (seconds)

(a)

(b)

Figure 17-18  Transient response in a DC LR network

Solution With the switch open, the steady-state current is IF =

and

τ=

10.0 V = 2.0 A 4.0 Ω + 1.0 Ω

L 2.0 H = = 0.40 s RT 5.0 Ω

5

17-10   Characteristics of Inductive DC Circuits

509

With the switch closed, the steady-state current is IF =

and

τ=

10.0 V = 10 A 1.0 Ω

2.0 H = 2.0 s 1.0 Ω

(a) When the switch closes at t = 0, the initial current through the coil is 2.0 A. At t = 1.0 s, i = IF + (I0 − IF)e−t/τ 

= 10 + (2.0 − 10)e−1.0/2.0 

= 10 − 8 × 0.6065 = 5.15 A

(b) At t = 1.1 s, the switch has been open for 0.1 s and i = IF + (I0 − IF)e−t/τ 

= 2.0 + (5.15 − 2.0)e−0.1/0.40 



= 2.0 + 3.15 × 0.7788 = 4.453 A

Since the switch is in parallel with the 4.0-Ω resistor, vsw = 4.453 A × 4.0 Ω = 18 V

(c) While the switch is open, the 2.0-A steady-state current through the 4.0-Ω resistor creates a steady voltage drop of 8.0 V. When the switch closes, the voltage across it instantly becomes zero. When the switch opens at t = 1.0 s, the 5.15-A current creates a voltage drop of 5.15 A × 4.0 Ω = 20.6 V across the 4.0-Ω resistor, creating the transient shown in Figure 17-18(b). Since 5τ = 2.0 s when the switch is open, the transient will have disappeared when t = 3 s. To verify the calculation of the current in part (a) and the voltage in part (b), download Multisim file EX17-8 from the website and follow the ­instructions in the file. See Problems 17-23 to 17-26 and Review Question 17-42.

17-10 Characteristics of Inductive DC Circuits The most important characteristic of inductance in a DC circuit is that the inductance develops an induced voltage that always tends to oppose any change in current. As a result, the current in an inductive circuit takes time to either rise or fall. The length of time needed for a given change in current depends on both the inductance and the resistance of the circuit.

In order to show the results more clearly, some of the answers in this example are given with more ­significant digits than are justified by the data.

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Chapter 17   Inductance in DC Circuits

When we close the switch in an inductive circuit, it takes time for the current to reach its steady-state level. This delay can be an advantage in that it reduces the surge in the current drawn from the source. However, an interruption in the current in an inductive circuit can induce a damaging high-voltage surge unless the circuit is designed so that current decays gradually. For this reason, switches for DC circuits are designed so that their contacts move apart as rapidly as possible to prevent arcing. There are applications for the high voltage produced by the collapsing magnetic field of an inductor. The ignition coil of a car stores energy gradually as current from the car battery builds up a magnetic field in the core of the coil. Then the current is suddenly interrupted by the breaker points, and the rapidly collapsing magnetic field induces a 10-kV ­voltage, which produces a spark between the electrodes at the end of a spark plug. Inductance can also be used to reduce, or filter, the variations in load current that occur when the load is fed from a DC source with a pulsating terminal voltage, such as a rectifier. As the source voltage in Figure 17-19 rises above its average value, the inductor stores additional energy in its expanding magnetic field, which opposes the increase in current. When the source voltage drops, the inductor opposes the corresponding drop in current and releases the stored energy. The inductor in such applications is often called a filter choke because it restricts the variations in load current and power.

Filter choke Pulsating DC source

+ e

Load

− Figure 17-19  Filter choke

17-11 Troubleshooting Three faults that commonly occur in inductors are   (i) open windings   (ii) shorted windings (iii) partially shorted windings Often these faults are the result of overheating, which can be caused by poor ventilation, a short within the inductor, or a fault in the power supply. It is good practice to check the power supply voltage when one replaces an inductor that has overheated.

17-11  Troubleshooting

Open Windings The heat from excessive current through a winding can cause the wire to burn out, or a break can occur where the wire had been nicked or scraped. An ohmmeter connected across the terminals of an open winding will show infinite resistance. Note that the inductor usually has to be disconected from the circuit in order to measure the resistance accurately.

Shorted Windings Occasionally the insulation on the winding may melt because of overheating, causing the turns to short to each other. The inductor then has a resistance close to zero ohms. Again, an ohmmeter can detect this fault once the inductor is removed from the circuit. The resistance of the inductor should be measured on the lowest scale since the normal resistance is often only a few ohms.

Partially Shorted Windings Sometimes overheating or physical damage causes just a few adjacent turns to short together. Such partial shorts reduce the resistance of the ­inductor somewhat, but the difference may be difficult to detect with an ohmmeter. A more accurate test can be done with an LC meter, which measures the inductance and capacitance of a component directly. See Review Question 17-43.

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Chapter 17   Inductance in DC Circuits

Summary

• The rate of change of current through an ideal inductor is constant. • The initial rate of change of current in a series DC circuit containing ­resistance and inductance is determined by the ratio of the applied voltage to the inductance. • The final value of current in a series DC circuit containing resistance and inductance is determined by the ratio of the applied voltage to the resistance. • The time constant of a circuit containing resistance and inductance is the ratio of inductance to the equivalent resistance. • For practical purposes, the current in a DC circuit containing resistance and inductance reaches its steady-state value after a time interval of five time constants. • Instantaneous currents and voltages in a DC circuit containing resistance and inductance may be determined by means of either an exponential graph or the corresponding equation. • The energy stored in an inductor is dependent on the inductance and the current flowing through it. • Inductor faults may be detected with an ohmmeter or an LC meter. B = beginner

I = intermediate

A = advanced

Problems B B B B B

I

Section 17-3  Time Constant

17-1. If a circuit has a resistance of 8.0 Ω and an inductance of 2 H, how long will the current take to reach its maximum value once the circuit is switched on? 17-2. In a solenoid with an inductance of 40 mH, it takes 25 ms for the current to reach its maximum value. Find the resistance of the solenoid. 17-3. What voltage must be applied to a 600-μH inductor if the initial rate of change of current must be 8.0 kA/s? 17-4. Calculate the steady-state current for a 100-mH inductor with a resistance of 8.0 Ω, given that the initial rate of rise of current is 600 A/s. 17-5. A choke with an inductance of 15 H and a resistance of 8 Ω is connected through a switch to a 36-V source. Find (a) the initial current when the switch is closed (b) initial rate of change of current when the switch is closed (c) the steady-state current (d) how long the current takes to reach its maximum value 17-6. An inductor with an inductance of 20 H and a resistance of 500 Ω is  connected in parallel with a 1000-Ω resistor. This combination is connected across a 120-V DC source with negligible internal resistance. (a) Calculate the initial current drawn from the source when the switch is closed. (b) Calculate the initial rate of change of the current through the ­inductor. (c) Calculate the steady-state current in the inductor. (d) Calculate how long the current takes to reach its steady-state value.

Problems

(b) Use Multisim to verify the initial source current in part (a), the steady-state inductor current in part (b), and the time in part (d).

A

I I

I I

I I

I

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Section 17-4  Graphical Solution for Inductor Current 17-7. (a) In the circuit of Problem 17-1, how long does the current take to reach 63% of its maximum value after the switch is closed? (b) How long does the current take to reach 50% of its maximum value? (c) Find the maximum value of the current if it is 2.6 A when the switch has been closed for 0.5 s. 17-8. (a) When will the current in the inductor of Problem 17-4 be 6 A? (b) Find the current through the inductor 10 ms after the circuit is switched on. 17-9. A 50-mH inductor with a resistance of 8 Ω is connected to a 12-V source. How long will the inductor current take to reach 500 mA?

Section 17-5  Algebraic Solution for Inductor Current 17-10. 17-11.

(a) When will the current through the choke in Problem 17-5 be 3 A? (b) Find the choke current 4 s after the switch is closed. (c) Use Multisim to verify your calculations in part (a) and part (b). A 20-Ω resistor is connected in series with the choke of Problem 17-5. This resistor is short-circuited 2.0 s after the switch is closed. Find the induced voltage across the choke 2.0 s after the 20-Ω resistor is short-circuited. 17-12. (a) A coil with a resistance of 20 Ω is connected to a 60-V DC source. After 25 ms the current is 1.2 A. Determine the inductance of the coil. (b) Use Multisim to verify the inductance calculated in part (a). 17-13. One type of time-delay relay is designed to operate 75 ms after the ­circuit is energized, with this delay equaling the time constant of the circuit. The supply voltage is 24 V and the relay trips when the current is 250 mA. Calculate the resistance and inductance of the relay. 17-14. For the circuit shown in Figure 17-20, find (a) an equation for the current as a function of time (b) the current flowing after 0.25 s (c) the time it takes for current to reach 3 A (d) the induced voltage after 0.25 s t=0s 10 Ω + 100 V −

I

513

1.0 H

Figure 17-20  17-15. Sketch a graph of current versus time for the circuit in Figure 17-20.

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Chapter 17   Inductance in DC Circuits

B

B

I

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A

A

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Section 17-6  Energy Stored by an Inductor 17-16. (a) How much energy is stored by the choke in Problem 17-5 when the current reaches its steady-state value? (b) At what rate is energy being drawn from the source when the current through the choke reaches its maximum value? 17-17. (a) How much energy is stored in the magnetic field of the inductor in Problem 17-6 when the current has reached its steady-state value? (b) At what rate is electrical energy being converted into heat when the current in the circuit has reached its maximum value?

Section 17-7  Fall of Current in an Inductive Circuit 17-18. The field coils of a DC motor connected across a 120-V source have an inductance of 8.0 H and a resistance of 40 Ω. (a) What is the maximum value for a discharge resistor that can be connected across the field coils to prevent the voltage ­between the coils from exceeding 200 V? (b) How long will this maximum resistance take to discharge the energy stored in the field coils? (c) How much energy is dissipated by the discharge resistor after the source is disconnected? (d) Use Multisim to verify the resistance calculated in part (a). 17-19. (a) Find the voltage across the switch contacts when the switch in the circuit of Problem 17-6 is opened. (b) How long does it take the coil to discharge its stored energy? (c) If the switch is opened 1.0 s after it was first closed, how long does the current in the coil take to drop 25 mA? 17-20. In the filter network of Figure 17-21, the 80-V source and the 40-mH inductor both have negligible internal resistance. (a) Find the output voltage 5 μs after the switch closes. (b) If the switch remains closed for at least five time constants, how long after the switch opens will the output voltage become 50 V? (c) Use Multisim to verify the voltage resistance calculated in part (a) and the time calculated in part (b). 40 mH + 80 V −

5.0 kΩ

5.0 kΩ

Figure 17-21  A

Section 17-8  Algebraic Solution for Discharge Current 17-21. The inductor in Figure 17-22 has negligible resistance. (a) Find the output voltage 5 μs after the switch closes.

Problems



515

(b) If the switch remains closed for at least five time constants, how long after the switch opens will the output voltage become +50 V with respect to ground?

5.0 kΩ 5.0 kΩ

40 mH

− 80 V +

Figure 17-22  A

17-22. After switch S1 in Figure 17-23 has been closed for 1.0 s, it is opened and S2 is closed simultaneously. (a) Calculate the time constant. (b) Write an equation for the current as a function of time. (c) Calculate the current flowing 30 ms after S1 opens. (d) Calculate the time the current takes to reach 3 A. (e) Calculate the induced voltage 30 ms after S1 opens. (f) Calculate the total energy stored in the magnetic field. (g) Sketch and label a graph of the current as a function of time. (h) Use Multisim to verify the calculation of the current in part (c), the time in part (d), and the induced voltage in part (e).

walkthrough

0.50 H

S1 + 100 V −

circuitSIM

S2

20 Ω

Figure 17-23 

A

A A

Section 17-9  Transient Response

17-23. (a) The switch in Figure 17-21 opens 10 μs after it closes. Find the output voltage with respect to ground 10 μs after the switch opens. (b) If the switch stays for 50 μs in each position alternately, draw a labelled graph of two complete cycles of the output voltage waveform. 17-24. (a) Repeat Problem 17-23 for the circuit of Figure 17-22. (b) Use Multisim to verify the output voltage calculated in part (a). 17-25. The switch in Figure 17-24 is closed for 2.0 s and then opened. Find the inductor current 1.0 s after the switch is opened.

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Chapter 17   Inductance in DC Circuits

1.0 Ω

5.0 H Discharge resistor

+ 20 V −

4.0 Ω 1.2 Ω

Figure 17-24 

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A

17-26. (a) Solve Problem 17-25 by replacing the source with an equivalent ­constant-current source. (b) Use Multisim to verify the inductor current calculated in part (a).

Review Questions

Section 17-1  Current in an Ideal Inductor 17-27. Why is the graph in Figure 17-2 a straight line? 17-28. What is the significance of lowercase letter symbols for variables? 17-29. Why are lowercase letter symbols used when one is representing the rise of current in inductive circuits?

Section 17-2  Rise of Current in a Practical Inductor 17-30. Why must the rise of current follow the exponential curve of ­Fig­ure 17-4?

Section 17-3  Time Constant 17-31. What effect does the resistance of an inductor have on the following? (a) the initial rate of change of current (b) the final steady-state current (c) the time the current takes to reach its steady-state value 17-32. What effect does the inductance of an inductor have on the following? (a) the initial rate of change of current (b) the final steady-state current (c) the time the current takes to reach its steady-state value 17-33. Justify this description of a time constant: “A time constant is the time it takes the current in an inductive circuit to rise to about 63% of its final steady-state value.” 17-34. Why does the voltage applied to an inductive circuit have no effect on the time the current takes to reach a steady-state value?

Section 17-4  Graphical Solution for Inductor Current 17-35. Draw a detailed graph of the actual inductor current in Example 17-3.

Section 17-6  Energy Stored by an Inductor 17-36. Why is there no further increase in the energy stored in an ­inductor when the current has reached its steady-state value? Where is the e­ nergy from the source going when a steady current flows through an inductor?

Integrate the Concepts

Section 17-7  Fall of Current in an Inductive Circuit. 17-37. What is the effect of suddenly disconnecting the voltage source in an inductive DC circuit? 17-38. How does a discharge resistor prevent insulation breakdown in an inductor? 17-39. In the instant after the circuit is switched off, why is the current through an inductor the same as the current the instant before the switch is opened? 17-40. Why are silicon carbide varistors particularly suitable for use as ­discharge resistors in LR circuits? 17-41. What effect does the resistance of the discharge loop of an inductive circuit have on the following? (a) the initial current when the switch is opened (b) the initial rate of fall of current (c) the time the current takes to fall to zero

Section 17-9  Transient Response 17-42. Derive Equation 17-12 by considering what happens in the circuit of Figure 17-8 when S2 is opened but S1 remains closed.

Section 17-11  Troubleshooting 17-43. Outline procedures for detecting three common faults in inductors.

Integrate the Concepts

In the inductive DC circuit shown in Figure 17-25, switch S1 is closed at time t = 0 while switch S2 stays open. After the inductor current reaches a steady state, S1 is opened at the same time as S2 is closed. Calculate: (a) the initial rate of change of current (b) the steady-state current before S2 is closed (c) the initial time constant (d) the current after 10 μs (e) the maximum energy stored in the inductor (f) the discharge time constant (g) the current 0.5τ after S2 is closed

S1

4.0 kΩ

2.0 kΩ S2 + 12 V −



Figure 17-25 

30 mH

517

518

Chapter 17   Inductance in DC Circuits

Practice Quiz 1. The resistance of an ideal inductor is (a)  zero (b)  low (c)  high (d)  infinite Complete the statements in questions 2–5. 2. The current in an inductor is _____________________ proportional to the voltage across the inductor. 3. At room temperature a practical inductor has a resistance that ­appears in _____________________ with the inductor. 4. The initial rate of change of current in an ideal inductor is _____________________ the initial rate of change of current in a practical inductor. 5. The time constant for an LR circuit is _____________________ proportional to inductance. 6. How long will the current take to reach 50% of the maximum value when the switch is closed in Figure 17-26? (a)   69.0 ns (b)   69.3 ns (c)   70.0 ns (d)   70.2 ns R 10.0 kΩ

12.0 V

L 1.00 mH

Figure 17-26 

7. What will the current be 110 ns after the switch in Figure 17-26 has been closed? (a)   1.0 mA (b)   1.2 mA (c)   0.95 mA (d)   0.8 mA

Practice Quiz

  8.

How long will the current take to reach the maximum value when the switch is closed in Figure 17-27? (a)  6.9 ms (b)   69 ms (c)  34 ms (d)   3.4 ms R

1.0 kΩ

L1 22 H

24 V

L2 10 H

Figure 17-27 

  9. What will the current in the circuit of Figure 17-27 be 7 ms after the switch has been closed? (a)   1.8 mA (b)   15 mA (c)   1.5 mA (d)   18 mA 10. The energy stored by an inductor is (a)   directly proportional to the resistance (b)   inversely proportional to the inductance (c)  directly proportional to the square of the maximum current flowing through it (d)  inversely proportional to the square of the maximum current flowing through it 11. When 15 A of current flows through an 8.0-H inductor, the energy stored in it is (a)   60 J (b)   900 J (c)   1.8 kJ (d)   120 J 12. Which of the following equations describes the current flowing through an inductor during discharge? (a)  iL = (b)  iL =

E E tR + e− L R R E E tR − e− L R R

519

520

Chapter 17   Inductance in DC Circuits

(c)  iL = (d)   iL = 13.

E tR eL R E −tR e L R

The switch in Figure 17-28 has been closed for 15 ms and then opened. The current through the resistor, R1, 2.0 ms after the switch has been opened is (a)   12 mA (b)   24 mA (c)   16 mA (d)   11 mA R1 1.0 kΩ

24 V

R2 1.0 kΩ

L1 22 H

L2 10 H

Figure 17-28 

14. List three types of faults that can occur in an inductor.

PART

IV

Alternating Current

In Part IV and Part V, we encounter voltage sources with EMFs that change continuously. Part IV describes alternating-current circuits and introduces mathematical techniques for analyzing the properties and behaviour of these circuits.

18

Alternating Current

19 Reactance 20 Phasors 21 Impedance 22

Power in Alternating-Current Circuits

Photo source:  © iStock.com/omada

18

Alternating Current In this chapter, we first examine how an alternating voltage is generated. We then investigate the characteristics of the sine wave, the most common form of alternating voltage.

Chapter Outline 18-1

A Simple Generator

524

18-2 The Nature of the Induced Voltage 18-3 The Sine Wave

526

18-4 Peak Value of a Sine Wave

529

18-5 Instantaneous Value of a Sine Wave 18-6 The Radian

532

18-7 Instantaneous Current in a Resistor

524

529 533

18-8 Instantaneous Power in a Resistor

536

18-10 Average Value of a Periodic Wave

540

18-9 Periodic Waves

537

18-11 RMS Value of a Sine Wave

540

Key Terms slip ring  524 brush 524 alternating voltage  525 alternator 525 cycle 526 period 526 phasor 526 phase angle  526 angular velocity  526 sine wave  528 amplitude 529 peak value  529

frequency 529 hertz 529 radian 532 periodic wave  537 peak-to-peak value  539 equivalent DC value  541 effective value  541 root-mean-square (RMS) value  541 form factor  543

Learning Outcomes At the conclusion of this chapter, you will be able to: • explain how an alternating voltage is produced across a loop of wire rotating within a magnetic field • graph a sine-wave voltage • calculate the instantaneous value of a sinewave voltage • convert frequency to angular velocity • calculate the instantaneous current through a resistor in a simple AC circuit

Photo sources:  © Lyroky/Alamy; Stock Photo

• calculate the instantaneous power in a resistor in an AC circuit • graph different types of periodic waves • calculate the average value of a half-cycle of a sine wave • calculate the RMS value of a sine wave

524

Chapter 18   Alternating Current

18-1  A Simple Generator Although batteries are convenient voltage sources for portable devices, generators produce almost all of the electric power used in homes and ­industry. These generators use electromagnetic induction to produce a ­potential difference. Chapter 16 described Faraday’s discovery that a voltage is induced in a conductor that moves perpendicular to a magnetic field. Figure 18-1 shows a simple arrangement that allows a loop of conductor to rotate continuously within a stationary magnetic field. Electrical connection to the rotating loop is maintained through a pair of slip rings and brushes.

Direction of rotation

N

Conductor formed into a loop Brush

Uniform magnetic field

Load S Slip ring

Figure 18-1 Simple AC generator

As the loop rotates, the two sides of the loop always cut across the magnetic lines of force in opposite directions. As a result, the two induced voltages are in series aiding and the sum of the two voltages appears across the  brushes. If we connect a load resistor to the brushes of this simple generator, the induced voltage will cause current to flow in the closed circuit c­ onsisting of the loop and the load. This current produces a magnetic field, which, according to Lenz’s law, opposes the rotation of the loop. To keep the loop rotating, this magnetic force has to be offset by a mechanical force. The mechanism that produces the mechanical force (a turbine, for ­example) transfers energy to the loop. Thus the generator converts ­mechanical energy into the electrical energy that flows to the load. When a conductor cuts across magnetic flux at the rate of 1 Wb/s, an EMF of 1 V is induced in the conductor. See Review Questions 18-32 to 18-34 at the end of the chapter.

18-2 The Nature of the Induced Voltage As the conductor loop in Figure 18-1 rotates through a complete revolution (360°), each side of the loop first cuts across the magnetic lines of force in one direction and then moves back across the same magnetic lines of force in the opposite direction. Consequently, the induced voltage appearing at

18-2   The Nature of the Induced Voltage

the brushes reverses its polarity with each half-revolution, and the generator develops an alternating voltage. A generator that produces alternating current is called an alternator. Figure 18-2(a) is a cross-sectional view showing 12 positions along the path of one side of the conductor in the generator of Figure 18-1. It is customary to consider the normal or positive direction of rotation as being counterclockwise, starting at the three o’clock position. In other words, the angle of rotation is measured counterclockwise from the positive x-axis of a grid with its origin at the centre of rotation. The 12 positions are spaced evenly at 30° intervals.

N 5

4 3 2

+

0 7

11 8 9 10

EMF

6

1

0

1 2 3 4 5 6 7 8 9 10 11

S − (a)

Position (b)

Figure 18-2  Nature of induced voltage

Faraday’s law states that the voltage induced in a rotating conductor is directly proportional to the rate at which it cuts across the magnetic lines of force. Thus, the induced voltage depends on the motion of the conductor in relation to the direction of the magnetic field. At position 0, the motion of the conductor is parallel to the magnetic lines of force. At that moment, the conductor does not cut any lines of force. Hence, the induced voltage is 0 V at the instant the conductor passes through point 0. At position 1, the conductor has started to cut across the lines of force, and some voltage is induced into the conductor. At position 2, the conductor is cutting across the magnetic lines of force at an even steeper angle. Thus, the rate of cutting and, therefore, the voltage induced into the conductor increases further. At position 3, the conductor is cutting across the magnetic lines of force at right angles, giving the greatest possible rate of cutting. Therefore, maximum voltage is induced into the conductor as it passes through point 3. From positions 3 to 6 in Figure 18-2, the rate of cutting across magnetic lines of force becomes progressively less, with the induced voltage reaching 0 V again just as the conductor passes through point 6, where it is again ­momentarily moving parallel to the lines of force. From positions 6 to 12, the motion of the conductor is symmetrical to its motion from position 0 to 6, but the conductor is now cutting across the magnetic lines of force in

525

526

Chapter 18   Alternating Current

the ­opposite direction. As a result, the polarity of the induced voltage is ­reversed during the second half of the revolution. The + sign on the graph of Figure 18-2(b) indicates the reference polarity of the terminal voltage. The – sign simply means that the polarity of the voltage is reversed with ­respect to the reference polarity. One complete rotation of the loop constitutes one cycle. The time taken for the loop to complete one cycle is the ­period of the waveform. See Review Question 18-35.

18-3  The Sine Wave

The international standard symbol for phase angle is the Greek letter ϕ (phi). Current North ­American standards permit the use of ­either ϕ or θ (the Greek letter theta) to represent the phase angle between current and voltage.

Figure 18-2(b) shows the magnitude and polarity of the induced voltage in a conductor as it rotates in a uniform magnetic field. To understand how circuits with an alternating applied voltage behave, we must be able to determine the value of the applied voltage at any instant. For such analysis, we can treat the alternating voltage as a phasor, a quantity that has a magnitude with an associated angle, much as a vector has a magnitude and a direction. As with vectors, this book uses boldface type to indicate a phasor and lightface type to indicate the magnitude of the phasor. In Figure 18-3, the phasor OX represents the position of the rotating conductor in the generator of Figure 18-1. This phasor pivots around the origin O. The end that is free to rotate is indicated by an arrowhead. By mathematical convention, phasor angles are measured counterclockwise from the reference axis, which is the positive x-axis. Thus, the position of a phasor is given by the phase angle ϕ through which it has rotated from the reference axis.

X

O

Rotation

ϕ

Reference axis

Figure 18-3  Phasor representation of the rotating conductor

Since OX pivots around O, the perpendicular phasor XY in Figure 18-4 represents the motion of the conductor at the moment it passes through the position represented by OX. If the angular velocity (the rate of rotation) of the conductor is constant, the speed of the conductor is also constant. Then the magnitude of the motion phasor XY is the same for any position of the conductor. The motion represented by XY can be resolved into a vertical component XZ and a horizontal component XW. The magnetic field is ­directed downward, so XZ is parallel to the magnetic lines of force and no induced voltage results from this component of the motion. Since XW cuts

18-3   The Sine Wave

Y

Z

ϕ

W

O

ϕ

X

ϕ

V Reference axis

Figure 18-4  Phasor diagram of the position and motion of a rotating conductor

across the magnetic lines of force at right angles, the induced voltage is directly proportional to the length of XW. When ϕ = 90°, XW = XY. Therefore, the ratio of XW to XY represents the ratio between the rate of cutting of the lines of force at angle ϕ and the maximum rate of cutting.

⦟

⦟

Since WX is parallel to OV, WXO = XOV = ϕ.

⦟ ⦟ ⦟ Since ⦟XWY = 90°, ⦟XYW = ϕ. Since ⦟OVX = 90° = ⦟YWX, all the corresponding angles of triangles XOV Since YXO = 90°, WXY = 90° − ϕ = OXV.

and XYW are equal. Therefore, these two triangles are similar and

XW XV = XY OX

Hence, the rate of cutting of magnetic lines of force at any instant is directly proportional to XV/OX. If OX remains constant, this ratio depends only on the phase angle ϕ. In the right triangle OVX, XV is the side opposite the angle ϕ and OX is the hypotenuse, so the ratio XV/OX is the sine of the phase angle ϕ. Therefore, The voltage induced into a conductor rotating in a uniform magnetic field is directly proportional to the sine of the phase angle. The geometrical construction in Figure 18-4 gives the relative magnitude of an induced voltage for phase angles less than 90°. We can use a similar procedure for phase angles greater than 90°. In Figure 18-5(a), the phasor OX has rotated through an angle of 120°. A perpendicular line drawn from point X to the reference axis forms a right triangle in which side XV is ­opposite an angle of 180° − ϕ, or 60°. Therefore,

sin 120° = sin (180° − 120°) = sin 60°

527

528

Chapter 18   Alternating Current

90°

90°

X

60° V O

180°

ϕ = 120°

0°

180°

ϕ = 210°

V O

0°

X

270° (a)

270° (b)

Figure 18-5  Determining the sine of angles greater than 90°

+1

sin ϕ

Most scientific ­calculators have a sin key that makes it easy to find the sine of any angle. Spread­ sheets also have ­built-in trigonometric ­functions.

Similarly, when ϕ = 210°, the right triangle has an angle opposite XV of 210° − 180°, or 30°, as shown in Figure 18-5(b). However, XV is below the reference axis, and sin 210° = −sin 30°. Referring to Figure 18-2, we see that XV now represents a negative value of induced voltage because the loop is moving in the opposite direction across the magnetic lines of force. Whenever XV is below the reference axis, it represents a negative quantity. Figure 18-6 shows how sin ϕ varies as ϕ increases. We can plot this sine curve by using a scientific calculator or a spreadsheet to calculate values of sin ϕ. An alternating voltage that varies in accordance with the sine curve is called a sine wave.

0

−1

90°

180° 270° 360°

Phase angle, ϕ

Figure 18-6  The sine curve

Figure 18-7 shows another method of producing a sine curve. This method is a projection of the geometric construction in Figure 18-5. Since the length of phasor OX is constant, the altitude VX of the tip of the phasor is proportional to the sine of angle ϕ. Therefore, VX is directly pro­portional to the induced voltage. When we project VX onto a graph such that point V1 is on the horizontal axis at a point corresponding to the value of ϕ, point X1 is at the point that represents sin ϕ. If we repeat this geometric projection at 5° intervals, we have enough points to draw a smooth sine curve.

18-5   Instantaneous Value of a Sine Wave

529

X1 X

O

ϕ

V

ϕ

V1

Figure 18-7  Drawing a sine curve by geometric projection

See Review Questions 18-36 and 18-37.

18-4  Peak Value of a Sine Wave

As shown in Figure 18-6, the greatest value of the induced voltage occurs when ϕ = 90°. When ϕ = 270°, the alternating voltage reaches the same maximum value, but the polarity is reversed. The magnitude of the alternating voltage at these two angles is the amplitude or peak value (Em) of the AC waveform. This peak value is measured in volts and does not have a polarity. The peak value of a given sine wave is a constant that indicates the maximum values of the sine wave. The peak value itself does not vary with time. In the simple AC generator, the peak voltage depends on the maximum rate at which the rotating loop cuts magnetic lines of force. This rate is determined by the strength of the magnetic field, the angular velocity of the loop, and the number of turns of wire in the loop. See Review Questions 18-38 and 18-39.

18-5 Instantaneous Value of a Sine Wave Figure 18-6 shows the manner in which the induced voltage changes from instant to instant as the loop rotates in the generator of Figure 18-1. During one complete revolution, the voltage varies through the whole range of possible values. As the loop rotates past 360°, the same waveform is generated again. The time the sine wave takes to go through one  complete cycle of instantaneous values is the period of the sine wave. The number of ­cycles completed in one second is the frequency of the sine wave. The letter symbol for frequency is f. The SI unit of frequency is the hertz (symbol Hz). A hertz is equal to one cycle per second. The induced voltage changes from instant to instant, so we represent its instantaneous value with the lowercase letter e. The magnitude of the ­alternating voltage at any instant is directly proportional to sin ϕ. Since the

The hertz is named in honour of the ­German physicist ­Heinrich ­Rudolf Hertz (1857−94), who ­discovered radio waves.

Chapter 18   Alternating Current

instantaneous voltage reaches its peak value at 90° and since sin 90° = 1, the instantaneous voltage at any instant is e = Em sin ϕ



(18-1)

In the simple two-pole generator of Figure 18-1, one complete revolution of the loop generates one complete cycle of the voltage sine wave. But the four-pole generator of Figure 18-8 generates one complete cycle of voltage as the loop moves through the magnetic flux under a north pole and then through the flux under an adjacent south pole. Thus, a 360° cycle of the voltage sine wave occurs while the loop rotates through only 180 mechanical degrees. The relationship between electrical degrees and mechanical degrees depends on the number of pairs of magnetic poles in the generator. Electronic circuits can generate sine waves of voltage without any mechanical rotation. To apply Equation 18-1, we must always express the phase angle ϕ in electrical degrees.

S

360 electrical degrees

N

Loop N

0° S

Figure 18-8  Electrical degrees in a four-pole generator

Source:  © Lyroky/Alamy Stock Photo

530

An automotive alternator

18-5   Instantaneous Value of a Sine Wave

Example 18-1 At what speed must the shaft of a six-pole alternator turn in order to produce a 60-Hz sine wave? Solution With three pairs of poles, one mechanical revolution generates three electrical cycles. Therefore, the shaft speed is

60 Hz = 20 r/s = 1200 r/min 3 cycles/r

Now we consider how to determine the angle ϕ in electrical degrees. Often, we know the frequency of the sine wave. Since frequency can be expressed in cycles per second, multiplying the frequency by the time elapsed since the instantaneous value passed through zero gives the angular position expressed in cycles. Multiplying this result by 360 then gives the phase angle in electrical degrees. Hence, Equation 18-1 becomes e = Em sin ( 360º × ft )



(18-2)

where e is the instantaneous source voltage, Em is the peak value of the voltage sine wave, f is the frequency of the sine wave in hertz, and t is the elapsed time in seconds.

Example 18-2 Find the instantaneous value of a 60-Hz sine wave 10 ms after the start of a cycle, given that the peak value is 150 V. Solution

e = Em sin ( 360º × ft )

= 150 sin ( 360º × 60 × 0.010 ) = 150 sin 216º

= −88 V

See Problems 18-1 to 18-6 and Review Questions 18-40 to 18-42.

531

532

Chapter 18   Alternating Current

18-6  The Radian The Babylonian system of dividing a circle into 360° is often convenient for measuring angles in geometry and trigonometry. However, this system is somewhat awkward for calculations such as finding the linear velocity of a rotating conductor from its angular velocity. In electrical engineering, the standard practice is to express the angular distance travelled by a rotating phasor in radians. The radian (symbol rad) is the SI unit for angles. A radian is equal to the angle formed by two radii of a circle when the arc between them has the same length as each radius. X

1 rad O

⦟

A

Figure 18-9  The radian

With most engineering calculators, angles can be entered in ­either degrees or ­radians.

In Figure 18-9, XOA is 1 rad when the length of arc AX equals the length of the phasor OX. Since the circumference of a circle is 2πr, there are 2π radians in one complete cycle. Hence, 90º =

and

π rad 2

180º = π rad

270º =

3π rad 2

360º = 2π rad

1 rad ≈ 57.3˚

(18-3)

Angular velocity can be expressed in radians per second. The letter symbol for angular velocity is the lowercase Greek letter ω (omega). The SI units for angular velocity are radians per second.

Since the phase angle is equal to angular velocity × elapsed time, we can rewrite Equation 18-2 as

e = Em sin ωt

(18-4)

Since there are 2π radians in a cycle,

ω = 2πf

(18-5)

18-7   Instantaneous Current in a Resistor

533

Substituting for ω in Equation 18-4 gives

e = Em sin 2πft



(18-6)

In Equation 18-4 and Equation 18-6, the angle is expressed in radians.

Example 18-3 Write the general equation for the instantaneous voltage of a 60-Hz generator with a peak voltage of 170 V. Solution

Since f = 60 Hz, Therefore,

2πf = 2 × 3.14 × 60 = 377 rad/s

e = Em sin 2πft = 170 sin 377 t

Since 60 Hz is the standard power-line frequency for North America, we will often use this conversion: 60 Hz = 377 rad/s



(18-7)

Example 18-2A Find the instantaneous value of a 60-Hz sine wave 10 ms after the start of a cycle, given that the peak value is 150 V. Solution

e = Em sin 2πft = 150 sin ( 2π × 60 × 0.010 rad ) = 150 sin 1.2π = 88 V See Problems 18-7 to 18-10 and Review Questions 18-43 and 18-44.

18-7 Instantaneous Current in a Resistor Now that we are acquainted with the sinusoidal nature of a basic AC voltage source, we can determine the current that a sine-wave voltage will ­produce in the simple circuit of Figure 18-10. At a particular instant, the voltage developed by the generator has a certain magnitude and polarity. If we consider only this one particular instant, the AC circuit of Figure 18-10 is just like a DC circuit. However, at the next instant, the magnitude of the ­alternating voltage changes, while the applied voltage in a DC circuit is ­constant. At any given instant, all the equations summarized in Table 6-1 apply to the AC circuit of Figure 18-10.

Since the 60-Hz frequency of the power grid is normally very precisely controlled, this book will treat 60 Hz as an exact ­measurement in most calculations.

Chapter 18   Alternating Current

e = Em sin ωt

R

Figure 18-10  Simple alternating-current circuit

Therefore, we can apply Ohm’s law to find the instantaneous current through the resistance in Figure 18-10: i=



e R

(18-8)

where i is the instantaneous current through the resistance, e is the instantaneous voltage applied to the resistance, and R is the resistance of the c­ ircuit.

Since resistance is determined by such physical factors as the type of material and its dimensions, the resistance of any circuit is a constant at a given temperature. Because R is a constant in Equation 18-8, the instantaneous current must stay exactly in step with the instantaneous voltage in order to satisfy Ohm’s law. If the instantaneous voltage is a sine wave, the instantaneous current must also be a sine wave, as shown in Figure 18-11. The current must reach its peak value at the same instant that the voltage reaches its maximum. The current and the voltage must become zero at the same instant, and the current must reverse its direction at the same instant that the voltage across the resistance reverses its polarity. Since the current through a resistor and the voltage across it have exactly the same phase angle at every instant, the current is said to be in phase with the voltage.

+ Em Instantaneous values

534

e = Em sin ωt

Im

i = Im sin ωt

0 π 2

π

3π 2

2π

Phase angle, ϕ − Figure 18-11  Instantaneous current through a resistor

18-7   Instantaneous Current in a Resistor

Substituting e = Em sin ωt in Equation 18-8 gives i=



Em sin ωt R

(18-9)

Since the peak value of the instantaneous current depends on the peak value of the applied voltage, Im =



Em R

i = Im sin ωt

and

(18-10) (18-11)

where i is the instantaneous current through a resistor and Im is the peak current. See Problems 18-11 to 18-14 and Review Questions 18-45 to 18-49.

Circuit Check

A

CC 18-1. A 100-Hz voltage sine wave has an instantaneous value of − 45 V at 18 ms from the start of a cycle. Find the peak value of this voltage. CC 18-2. Find the lowest frequency a 250-V sine wave could have if the instantaneous voltage is 75 V at 3 ms from the start of a cycle. CC 18-3. For the voltage wave in Figure 18-12, find (a) the frequency (b) the maximum value (c) the instantaneous voltage at t = 9 ms 30 V

3 ms

Figure 18-12

15 ms

535

Chapter 18   Alternating Current

18-8 Instantaneous Power in a Resistor We can apply Equations 6-2, 6-3, and 6-4 to the instantaneous values in an AC circuit. Therefore, p = vi = i2R =



v2 R

(18-12)

where p is the instantaneous power in a resistor in watts, v is the instantaneous voltage drop across the resistance in volts, i is the instantaneous current through the resistance in amperes, and R is the resistance of the circuit in ohms. Since R is a constant for a given circuit and since the instantaneous voltage and current in the basic circuit are both sine waves, the instantaneous power must be a sine-squared wave. Figure 18-13 shows a graph of the ­instantaneous power in a resistor. Note that the instantaneous power in a resistor is always positive because squaring a negative quantity results in a  positive quantity. We can interpret positive power as electric energy being converted into some other form of energy (such as heat in a resistor). Conversely, negative power represents some other form of energy being converted into electric energy. Since a resistor cannot generate electric energy, the instantaneous power in a resistor is never negative. As Figure 18-13 shows, the instantaneous power in the basic AC system pulsates, swinging from zero to maximum and back twice each cycle. The instantaneous power pulsates at twice the frequency of the voltage and current. The power that a current produces in a resistor does not depend on which way the current flows through it. Therefore, the pulse of energy transferred to the resistor during the first half of each cycle is the same as the energy pulse during the second half. This pulsating characteristic of instantaneous power can cause vibration in some types of small AC motors.

+ Instantaneous values

536

Pm Im

p = Pm sin2 ωt

0 π 2

π

3π 2

2π i = Im sin ωt

Phase angle, ϕ − Figure 18-13  Instantaneous power in a resistor

18-9   Periodic Waves

We can also show the pulsating nature of the power in an AC system by deriving an equation for instantaneous power. Substituting I = Im sin ωt in Equation 18-12, we get p = ( Im sin ωt ) 2 × R = I2mR × sin2 ωt



Since peak values are maximum instantaneous values, Equation 18-12 gives Pm = I2mR (18-13) Therefore,

p = Pm sin2 ωt

(18-14)

where p is the instantaneous power in a resistance and Pm is the peak power.

The relationship sin2 θ = 12(1 − cos 2θ) is a trigonometric identity, which means that the relationship applies for all values of θ. Therefore, 1 sin2 ωt = (1 − cos 2ωt) 2 and

1 1 p = Pm − Pm cos 2ωt 2 2

(18-15)

This equation tells us three things about the instantaneous power waveform that we can check by examining Figure 18-13: 1.  In any right-angled triangle, the cosine of one acute angle is also the sine of the other acute angle. Therefore, the general shape of the cosine curve and the sine curve are the same, and cos θ = sin (θ − 90°). Consequently, the fluctuations in instantaneous power are sinusoidal. 2. Since the angle in the cosine term is 2ωt, the frequency of the sinusoidal variation in power is twice the frequency of the instantaneous voltage and current. 3. Since the limits of the value of a cosine are +1 when ϕ = 0° and −1 when ϕ = 180°, the value of the expression 12(1 − cos 2ωt) can vary from 0 to +1. Therefore, the instantaneous power in a resistance is always positive. See Review Questions 18-50 to 18-53.

18-9  Periodic Waves Although the sine wave is by far the most important AC waveform, there are  many other types of periodic waves. In electric circuits, a periodic wave is any time-varying quantity, such as voltage, current, or power, that

537

Chapter 18   Alternating Current

Time

− 1 period (a) Sine wave

0

Time 1 period (b) Sine-squared wave

+

0

+

−

Time

Instantaneous value

0

Instantaneous value

+

+

0

− 1 period (c) Square wave

− 1 period (d) Pulse wave

+

+

0

Time

− 1 period (e) Sawtooth wave

Instantaneous value

Instantaneous value

Instantaneous value

c­ ontinually repeats exactly the same sequence of values with each cycle taking ­exactly the same time. Figure 18-14 shows how the instantaneous values of six common periodic waves vary as a function of time. The graphs of the sine wave and sine-squared wave are the same as the two  waves shown in Figure 18-13, but here the horizontal axis represents time rather than the phase angle. Since ϕ = ωt, the two curves have the same shape in both ­figures. As shown in Figure 18-14, the period of each waveform is the length of time it takes the instantaneous voltage or current to complete one cycle of values. Figure 18-14(b) shows four complete cycles of a periodic wave. All other graphs in Figure 18-14 show two complete ­cycles. The instantaneous values of a periodic wave change much too rapidly to be registered by a conventional electric meter. However, a cathode-ray tube or a liquid-crystal display (LCD) in an oscilloscope can quite readily respond to high-frequency instantaneous voltages. If the instantaneous voltage of a certain periodic wave, such as the sine wave of Figure 18-14(a), is applied to the vertical deflection circuit of an oscilloscope, the display will show a dot of light moving up and down exactly in step with the voltage. For most waves, this motion is still much too rapid for the eye to

Instantaneous value

538

0

Time

Time

− 1 period (f) Half-wave-rectified wave

Figure 18-14  Some common periodic waves

18-9   Periodic Waves

Source: © AEMC Instruments, a leader in electrical test and measurement instruments

follow, so what we see is a vertical line with a length proportional to the peak-to-peak value of the voltage of the periodic wave. In the sawtooth periodic wave of Figure 18-14(e), the instantaneous voltage is a linear function of elapsed time. Hence, if the horizontal deflection circuit of the oscilloscope generates a sawtooth wave having a period exactly twice that of the waveform fed to the vertical deflection circuit, the screen will display two cycles of that waveform. Multi-trace oscilloscopes can display two or more waveforms at the same time. See Review Question 18-54.

Source: © Tihis/Dreamstime/Getstock

A handheld multi-trace LCD oscilloscope

A dual-trace CRT oscilloscope

539

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Chapter 18   Alternating Current

18-10 Average Value of a Periodic Wave By definition, an alternating current or voltage is one in which the arithmetic average, or mean, of the instantaneous values over a cycle is zero. From an examination of Figure 18-14, it is fairly obvious that the sine, square, and sawtooth waveforms are alternating because the instantaneous values are symmetrical about the horizontal axis of the graph. The sine-squared wave, pulse wave, and half-wave-rectified wave are not alternating currents because their instantaneous values always have the same polarity. For complex periodic waves we can determine the mean of the i­ nstanta­neous values graphically by examining the area under a graph. Since the sine wave is symmetrical, each positive value during the first half-cycle has a matching negative value during the next half-cycle. Therefore, the mean value of a complete cycle of a sine wave is zero, and the mean value of the first half-cycle is equal in magnitude but opposite in polarity to the mean of the second half-cycle. Appendix 2-6 uses integral calculus to show that the mean value of the first half-cycle of a sine wave is 2/π, or about 0.637. Since i = Im sin ωt, the half-cycle average value of a sinusoidal current is

Iav =

2 Im ≈ 0.637Im π

(18-16)

Similarly, the half-cycle average value of a sinusoidal source voltage is

Eav = 0.637Em =

2 Em π

(18-17)

As we discovered in Section 18-9, the sine-squared wave of Figure 18-14(b) has a full-cycle average value of one-half the peak value of the sine-squared wave. The pulse waveform has an average value that depends on the ratio of the pulse duration to the period of the wave. The full-cycle a­ verage value of the half-wave-rectified wave of Figure 18-14(f) is the mean of 0.637Em for one half-cycle and zero for the next half-cycle, or 0.3183Em. See Problem 18-15 and Review Question 18-55.

18-11  RMS Value of a Sine Wave All electric circuits convert electric energy into some other form of energy, such as heat, light, or mechanical energy. For many of these energy conversions, it does not matter whether the energy source for the circuit produces direct current or alternating current. Therefore, we find equivalent steadystate values for alternating current and voltage that allow us to use the same relationships among voltage, current, resistance, power, work, and so

18-11   RMS Value of a Sine Wave

on, that we use for DC circuits. We can determine an equivalent DC value or effective value of an alternating current experimentally by finding the direct current that produces heat in a given resistance at the same rate as when the resistance is connected to a source of alternating voltage. We can also use algebraic analysis to find such equivalent values. To determine this equivalent DC value of an alternating current, we first find the average power in a resistive AC circuit. Since p = i2R (Equation 18-12), Pav = ( full-cycle average of i2 ) × R



In a DC circuit, P = I2R. Since the effective values of an alternating current are to represent DC equivalent values, we can use the same letter symbols for both, that is, uppercase italic letters without subscripts. Therefore, in an AC circuit, P = I2R where P is average power and I is the DC equivalent for the alternating ­current.

√ √

Solving for I gives I=

P = R

( full-cycle average of i2 ) × R R

= √ full-cycle average of i2

Thus I is equal to the square root of the mean of the squares of the instantaneous current over a full cycle. Root-mean-square (RMS) value, effective value, and equivalent DC value all mean the same thing. For the rest of the book, we use the term RMS. Figure 18-13 shows that the instantaneous power for a sine wave of alternating current swings alternately and symmetrically between zero and peak power. Therefore, the average power in a resistor through which a sine-wave alternating current is flowing is simply one-half the peak power. We can also determine this relationship from the general equation for instantaneous power: p = Pm sin2 ωt (Equation 18-14). As we have already shown, sin2 ωt = 12(1 − cos 2ωt). If we average a sine or cosine function over a complete cycle, the average must be zero, since for every positive value during the first half-cycle, there is a matching equivalent negative value during the second half-cycle. Hence, when averaged over a complete cycle, Equation 18-14 becomes 1 P = Pm (18-18) 2 For a sine-wave alternating current in an AC circuit,



P = I2R and Pm = I2mR

541

542

Chapter 18   Alternating Current

But Therefore,

1 1 P = Pm = I2mR 2 2 1 I2R = I2mR 2

I=



1 I2 = I2m 2

Im ≈ 0.707Im √2

(18-19)

The RMS (effective) value of a sine wave of current is 1/√2, or about 0.707, times the peak value. The RMS value of a sine-wave voltage should be such that the average power is the product of the RMS voltage across and RMS current through the resistance of the circuit, just as P = VI in the equivalent DC circuit. Since peak ­instantaneous power occurs at the instant when both the voltage across and the current through a resistance circuit reach their peak values, Pm = VmIm Substituting into Equation 18-18 gives P = VI = V



Therefore,

and

V=

Im 1 1 = Pm = Vm Im 2 2 √2 Vm ≈ 0.707 Vm √2

(18-20)

Vm = √ 2 V ≈ 1.414V

Since RMS values are used for the majority of AC circuit measurements and computations, all further AC currents and voltages are given as RMS values unless otherwise specified.

Example 18-4 Find the peak voltage of a 120-V 60-Hz electric service. Solution

Em = 1.414E = 1.414 × 120 V = 170 V

18-11   RMS Value of a Sine Wave

A term we may encounter occasionally when working with AC waveforms is form factor. Form factor is the ratio of the RMS value to the halfcycle average value of an AC wave. For a sine wave,

form factor =

I 0.707Im ≈ = 1.11 Iav 0.637Im

(18-21)

See Problems 18-16 to 18-31 and Review Questions 18-56 to 18-58.

Circuit Check

B

CC 18-4. A voltage wave with the equation e = 100 sin 500t V is applied to a 50-Ω resistor. Calculate the instantaneous current at t = 3.5 ms. CC 18-5. A voltage wave with the equation e = 163 sin 377t V is ­applied to a 100-Ω resistor. Calculate: (a) the RMS value of voltage (b) the RMS current in the resistor (c) the average power and the RMS power delivered

543

544

Chapter 18   Alternating Current

Summary

• A sine wave of alternating voltage is produced across a loop of wire as it rotates within a uniform magnetic field. • The peak value of the sine wave of alternating voltage produced by an AC  generator occurs when the loop of wire is parallel to the magnetic lines of force. • Zero voltage is induced across the loop of wire when it is perpendicular to the magnetic lines of force. • The instantaneous value of the voltage produced by an AC generator is proportional to the sine of the angle of rotation of the loop of wire. • The instantaneous current in an AC circuit obeys Ohm’s law. • The instantaneous power in a resistor in an AC circuit is a sine-squared wave. • The RMS value of a sine wave of current in the resistor in an AC circuit  equals the DC current that would produce the same amount of heat.

B = beginner

I = intermediate

A = advanced

Problems B B B I B B

B B I B

Section 18-5  Instantaneous Value of a Sine Wave 18-1.

If the peak voltage of a 50-Hz AC generator is 210 V, what is the instantaneous voltage after the loop has rotated through 25 degrees? 18-2. If the peak voltage of a 1.0-kHz audio oscillator is 20 V, what is the instantaneous voltage when ϕ = 210°? 18-3. An AM radio transmitter generates a sinusoidal carrier wave with a frequency of 900 kHz and feeds a peak voltage of 1.3 kV to the antenna. Determine the instantaneous voltage 0.7 μs after the start of a cycle. 18-4. The instantaneous voltage from a 400-Hz aircraft generator is 95 V when t = 1 ms. Find the instantaneous voltage at t = 2 ms. 18-5. Find the instantaneous voltage after the loop of the generator in Problem 18-1 has rotated for 10 ms. 18-6. Find the instantaneous voltage from the audio oscillator in Problem 18-2 when t = 600 μs.

Section 18-6  The Radian 18-7. Determine the instantaneous voltage when the loop of the generator in Problem 18-1 has rotated through 1.0 rad. 18-8. Find the instantaneous voltage from the audio oscillator in Problem 18-2 when ϕ = 5 rad. 18-9. Write the general voltage equation for a 400-Hz source with a peak voltage of 200 V. 18-10. An alternating voltage has an expression e = 75 sin 3000t V. Determine (a) the frequency of the sine wave (b) the peak value (c) the instantaneous value at t = 1.5 ms

Problems

I

I

I

B I

B B B I I I B I I B B

Section 18-7  Instantaneous Current in a Resistor

18-11. A 1-kHz sine wave current has an instantaneous value of 6 A when t = 0.4 ms. Determine (a) the peak value (b) the instantaneous value at t = 1.5 ms 18-12. If the instantaneous current drawn from a 50-Hz source is −263 mA when t = 0.013 s, what is the instantaneous current when t = 0.017 s?

Section 18-8  Instantaneous Power in a Resistor

18-13. A 50-Ω resistor is connected to a voltage source with an output of e = 75 sin 1400t V. Determine (a) the maximum power (b) the instantaneous power at t = 1 ms (c) the average power 18-14. What is the frequency of the power wave in Problem 18-13?

Section 18-10  Average Value of a Periodic Wave 18-15. The conductor in the simple generator of Figure 18-1 cuts across a total magnetic flux of 40 mWb during the first 180° of rotation. The frequency of the AC output is 60 Hz. Find the half-cycle average ­induced voltage.

Section 18-11  RMS Value of a Sine Wave 18-16. Calculate the peak voltage in the European 230-V 50-Hz system. 18-17. Determine the peak value of a microwave signal with an RMS value of 100 μV. 18-18. Find the RMS voltage drop across a 15-Ω resistor when the current flowing through it is 1.5 sin 377t. 18-19. Write the equation for the instantaneous voltage generated by a 117-V 400-Hz alternator. 18-20. If the symmetrical square wave of current shown in Figure 18-14(c) has a peak value of 40 mA, find the RMS value. 18-21. If the symmetrical sawtooth voltage waveform in Figure 18-14(e) has a peak voltage of 30 V, find the RMS value. 18-22. The average power of a 250-Ω soldering iron is 55 W. (a) Find the RMS current through the iron. (b) Find the peak voltage across the iron. 18-23. What reading does an AC ammeter show when connected in series with a 16-Ω load with a peak power of 288 W? 18-24. How long does a 10-Ω heater connected to a 120-V 60-Hz source take to convert 1.0 kW · h of electric energy into heat? 18-25. A fluorescent lamp has a resistance of 50 Ω. If it's connected to a 110-V AC outlet, determine its wattage rating. 18-26. Find the RMS value of the alternating voltage induced into the loop in Problem 18-15.

545

Chapter 18   Alternating Current

A

A

B B A

18-27. For a voltage wave with the equation v = 220 sin 377t, determine (a) the peak value (b) the RMS value (c) the frequency (d) the instantaneous value when t = 8 ms (e) how long the voltage takes to reach half its maximum value (f) the current the voltage produces in a 50-Ω resistor 18-28. For a 100-Hz voltage wave with an RMS voltage of 500 V, (a) write the equation of the voltage wave (b) find the instantaneous value of voltage at t = 6 ms (c) find the time when the voltage first reaches 300 V 18-29. The peak power in a resistor connected to a 500-V 60-Hz AC source is 1.0 kW. Calculate the resistance. 18-30. A 500-Ω resistor dissipates a peak power of 600 W when connected to an AC source. Find the RMS current. 18-31. The voltage waveform shown in Figure 18-15 has an RMS value of 150 V. Find the frequency of the voltage. Write the equation for this voltage. + Voltage

546

1.57 ms

Time

− Figure 18-15

Review Questions

Section 18-1  A Simple Generator 18-32. What conditions govern the generation of a voltage by electromagnetic induction? 18-33. Why does the angular velocity of the rotating loop in Figure 18-1 tend to decrease when a load resistor is connected to the brushes? 18-34. Why is it not possible to maintain a constant polarity of induced voltage in the loop in Figure 18-1?

Section 18-2  The Nature of the Induced Voltage 18-35. Why does the voltage across a loop of wire increase as it rotates from a position perpendicular to magnetic flux to a position parallel to the flux?

Section 18-3  The Sine Wave 18-36. Why do we describe the instantaneous voltage waveform from the simple generator of Figure 18-1 as a sine wave? 18-37. What is meant by the term angular distance?

Review Questions

Section 18-4  Peak Value of a Sine Wave 18-38. Why does doubling the number of turns in the rotating loop in a generator double the peak voltage? 18-39. Why is the peak value of an AC voltage never a negative quantity?

Section 18-5  Instantaneous Value of a Sine Wave 18-40. 18-41. 18-42.

How does the angular velocity of the loop in Figure 18-1 affect (a) the frequency of the voltage waveform (b) the peak value of the induced voltage Why is the answer to Example 18-2 a negative quantity? What is the significance of the term sine wave when describing the behaviour of an electric circuit?

Section 18-6  The Radian

18-43. With a diagram similar to Figure 18-5, show that sin 3π/4 = sin π/4. 18-44. Why are radian units preferred for the general equation for an alternating current?

Section 18-7  Instantaneous Current in a Resistor 18-45. What is the significance of a negative value of instantaneous ­current? 18-46. What effect does changing the peak value of an alternating current have on the waveform? 18-47. Why must the instantaneous current through a resistor have the same waveform as the instantaneous voltage across it? 18-48. What is meant by the statement that the current through a resistor is in phase with the voltage across it? 18-49. What factors govern the instantaneous value of the current through a resistor in an AC circuit?

Section 18-8  Instantaneous Power in a Resistor 18-50. Why can we substitute instantaneous AC values in the various equations we derived for DC circuits containing a voltage source and a load resistor? 18-51. What is the significance of a negative value of instantaneous power? 18-52. Why is the instantaneous power input to a resistor never negative? 18-53. Why does the instantaneous power in a resistor pulsate at twice the frequency of the applied voltage?

Section 18-9  Periodic Waves 18-54. What distinguishes a periodic wave from other types of waves?

Section 18-10  Average Value of a Periodic Wave 18-55. Explain why the mean value for a half-cycle of alternating current is different from the mean value for a complete cycle.

547

548

Chapter 18   Alternating Current

Section 18-11  RMS Value of a Sine Wave 18-56. Why is the average power in a resistor connected to a sine-wave source one-half the peak power? 18-57. Why is it necessary to consider average power when deriving an RMS value for alternating-current and voltage sine waves? 18-58. The equation for instantaneous voltage in an AC system is sometimes written in the form e = 2 E sin ωt. Suggest a reason for using this form.

Integrate the Concepts A 2-kHz sine wave generator with an RMS value of 123 V is connected to a 75-Ω resistor. (a) Determine the voltage at t = 0.1 ms. (b) Write an equation for the instantaneous current. (c) Calculate the peak power supplied to the resistor. (d) Write an equation for the instantaneous power supplied to the ­resistor. (e) Calculate the RMS values of the voltage and current.

Practice Quiz 1. Which of the following statements are true? (a) The alternating voltage is produced across a loop of wire when it rotates within an electric field (b) A phasor represents a time-varying quantity in terms of magnitude and direction. (c) The length of a phasor represents the direction. (d) One full cycle of a sine wave has 360 degrees. Complete the statements in questions 2–4. 2. The voltage induced into a conductor rotating in a uniform magnetic field is ______ to the sine of the phase angle. 3. Rotating a phasor counterclockwise from the reference axis ________________ the angle of the phasor. 4. The peak value of a sine wave of alternating voltage produced by an AC generator occurs when the loop of wire is moving ____________________ to the magnetic lines of force. 5. The peak value of the sine wave shown in Figure 18-16 is (a) 15 V (b) 7.5 V (c) 8.0 V (d) 16 V

Practice Quiz

8

Amplitude (V)

6 4 2 0 −2 −4 −6 −8

0

1

2

3

4 Time (ms)

5

6

7

8

Figure 18-16

  6. A full cycle of a sine wave has (a) π/2 radians (b) π radians (c) 2π radians (d) 4π radians   7. At what speed must the shaft of a four-pole alternator turn in order to produce a 500-Hz sine wave? (a) 750 RPM (b) 500 RPM (c) 125 RPM (d) 250 RPM   8. What is the instantaneous value of the sine wave of Figure 18-16 at t = 1 ms? (a) 1 V (b) −2 V (c) −1 V (d) −0.8 V   9. What is the instantaneous value of the sine wave of Figure 18-16 at 2.1 ms? (a) 4 V (b) 1 V (c) −1 V (d) −4 V 10. What is the frequency of the sine wave of Figure 18-16? (a) 1.1 KHz (b) 900 Hz (c) 500 Hz (d) 513 Hz

549

Chapter 18   Alternating Current

11. What is the frequency of the sine wave of Figure 18-17? (a) 323 Hz (b) 333 Hz (c) 160 Hz (d) 167 Hz 8 6 4 Amplitude (V)

550

2 0 −2 −4 −6 −8 0

0.5

1

1.5 2 Time (ms)

2.5

3

3.5

Figure 18-17

12. What is the instantaneous value of the waveform of Figure 18-17 at t= 2.2 ms? (a) 8 V (b) 7.5 V (c) 6.5 V (d) 7 V 13. What is the rms value of the sine wave of Figure 18-17? (a) 10 V (b) 6.9 V (c) 6.4 V (d) 9.5 V 14. What is the general equation for the instantaneous voltage of a 50-Hz generator with a peak voltage of 220 V? (a) 156 sin 314t V (b) 220 sin 314t V (c) 110 sin 314t V (d) 220 sin 377t V

Practice Quiz

15. What is the instantaneous current through the 20-Ω resistor in the circuit of Figure 18-18? (a) 11 sin 377t A (b) 1.1 sin 377t A (c) 11 sin 314t A (d) 1.1 sin 314t A

V 220 Vpk 60Hz 0°

+

R 20 Ω

−

Figure 18-18

16. What is the expression for the instantaneous power in the circuit of Figure 18-18? (a) 11 sin 377t W (b) 2.42 sin 377t kW (c) 11 sin2 377t W (d) 2.42 sin2 377t kW

551

19

Reactance In Chapter 18 we found that all the laws for DC circuits apply when an AC source is connected to a resistance. In this chapter we shall see that these laws also apply when an AC source is connected to an inductance or a capacitance. In an AC circuit, the current through a resistance and the voltage across it are in phase. However, this chapter will show that the AC current and voltage are not in phase for an ­inductance or capacitance.

Chapter Outline 19-1

Instantaneous Current in an Ideal Inductor  554

19-2 Inductive Reactance  555

19-3 Factors Governing Inductive Reactance  556 19-4 Instantaneous Current in a Capacitor  558 19-5 Capacitive Reactance  559

19-6 Factors Governing Capacitive Reactance  560

19-7 Resistance, Inductive Reactance, and Capacitive Reactance 562

Key Terms reactance 556 inductive reactance  556

displacement current  560 capacitive reactance  560

Learning Outcomes At the conclusion of this chapter, you will be able to: • predict the instantaneous current produced by a sine wave of voltage across an ideal inductor • calculate the inductive reactance of an ideal inductor given the AC voltage across the inductor and AC current through it • describe the effects of inductance and frequency on the inductive reactance of an ideal inductor • calculate the reactance of an ideal inductor

Photo sources:  © iStock.com/ Evgeny Rannev

• predict the instantaneous current produced by a sine wave of voltage across a capacitor • describe the effects of capacitance and frequency on the capacitive reactance of a capacitor • calculate the reactance of a capacitor • summarize the differences in the behaviour of resistance, inductance, and capacitance in AC circuits

554

Chapter 19  Reactance

19-1 Instantaneous Current in an Ideal Inductor L

A Figure 19-1 Inductance in an AC circuit

If there is no resistance in the circuit, any voltage drop that appears across the terminals of an inductor must be due to the voltage induced in the coil by a changing current through it. From Equation 17-1, the instantaneous voltage across the inductor in Figure 19-1 must be vL = L



Therefore,

L

di dt

di = Em sin ωt dt

di Em = sin ωt dt L

and

(19-1)

Since Kirchhoff’s voltage law requires the voltage across the inductor to be exactly equal to the applied voltage at every instant, we can represent the instantaneous voltage vL across the inductor by the blue sine curve in ­Figure 19-2. According to Faraday’s law of electromagnetic induction, the magnitude of this inductive voltage at any instant is directly proportional Maximum positive voltage Instantaneous Values

e = Em sin ωt

In the circuit of Figure 19-1, we assume that the inductor has negligible ­resistance. To satisfy Kirchhoff’s voltage law, at every instant the inductive voltage across the coil in Figure 19-1 must exactly equal the applied voltage. Hence, vL = e = Em sin ωt

Maximum negative rate of change of current Zero rate of change vL

+

iL

0 π 2

−

π

3π 2

2π

5π 2

ωt

Zero voltage Maximum positive rate of change of current

Maximum negative voltage

Figure 19-2  Instantaneous current in an inductor

19-2   Inductive Reactance

to the rate of change of current through the coil at that instant. When the maximum positive voltage appears across the inductor (at the point where ϕ = π/2 rad), the current must be changing at the greatest positive rate. Since ϕ = ω t, the slope of a graph of current versus phase angle is directly proportional to the rate of change of current with time. In Figure 19-2, the maximum rate of change is indicated by the steepest slope. One quarter-cycle later (when ϕ = π rad), the instantaneous voltage is momentarily zero, and the current must momentarily stop changing. ­Regardless of how great the current may be, as long as it is neither rising nor falling, there is no change in the magnetic flux linking the turns of the coil and, hence, no induced voltage. Therefore, the slope of the instantaneous current curve must be horizontal at this instant. Between π/2 rad and π rad, the voltage across the inductor has to decrease along a sine curve, and thus the rate of change of current has to decrease in the same manner. Another quarter-cycle later (when ϕ = 3π/2 rad), the i­ nstantaneous inductive voltage reaches its most negative value, and therefore the instantaneous current must be changing at its maximum negative rate. The instantaneous current through the inductor must have the sine-wave shape shown by the red curve in Figure 19-2 in order to induce the sinewave voltage that appears across the inductor. But the instantaneous current sine wave reaches its positive peak a quarter-cycle (π/2 radians) after the instantaneous voltage across the inductor reaches its maximum. The zero and minimum values of the current also occur a quarter-cycle after the corresponding points in the voltage wave. Therefore, For a sine-wave voltage drop to appear across an ideal inductor, the current through it must be a sine wave that lags behind the inductive voltage drop by π/2 radians. Therefore, the instantaneous current in the circuit of Figure 19-1 is

iL = Im sin ω t −

(

π 2

)



(19-2)

where the phase angle ωt − π/2 is measured in radians.

Appendix 2-7 shows how to derive this equation by using integral c­ alculus. See Review Questions 19-22 to 19-26 at the end of the chapter.

19-2  Inductive Reactance Since the current through the inductor in Figure 19-1 is a sine wave, it must have an RMS value. The ammeter reading shows this constant RMS value. For a given alternating-voltage source and a given inductance, the ratio V/I is constant.

555

556

Chapter 19  Reactance

This constant V/I ratio represents reactance, an opposition to the flow of alternating current. However, the current in an ideal inductor lags behind the voltage across it by π/2 radians, whereas the current in a resistor is in phase with the voltage drop. Furthermore, an inductor does not convert electric energy into heat as a resistor does. Thus, the reactance of an inductor differs from resistance. Inductive reactance is the opposition of inductance to alternating current. The letter symbol for inductive reactance is XL.

XL =

VL IL

(19-3)

Since inductive reactance is a V/I ratio like resistance, the ohm is the unit of inductive reactance: An AC circuit has an inductive reactance of one ohm when an alternating current of one ampere RMS through the inductance creates an inductive voltage drop of one volt RMS across the inductance. See Problems 19-1 and 19-2 and Review Question 19-27.

19-3 Factors Governing Inductive Reactance If the applied voltage in the circuit of Figure 19-1 remains sinusoidal, there are only three factors that can vary independently: the peak voltage of the sine wave of applied voltage, the frequency of the applied voltage, and the inductance. If we double the peak value of the applied voltage while keeping the frequency constant, the maximum rate of change of current must be doubled in order to develop the required inductive voltage drop. Since the frequency is unchanged, the current reaches its maximum value in the same length of time, but the slope of the current is twice as steep at every instant. Therefore, the current rises to twice the original peak value. Consequently, doubling the amplitude of the applied voltage also doubles the current. Since XL = VL/IL, there is no change in the inductive reactance. The same logic applies for any change in the applied voltage. The magnitude of the applied alternating voltage has no effect on the inductive reactance of an AC circuit. If we double the frequency of the applied voltage and leave the peak value the same, the instantaneous current in the inductance must have the same maximum rate of change and the same average rate of change over each quarter-cycle. Because the frequency is doubled, the period (the

19-3   Factors Governing Inductive Reactance

­ uration of a complete cycle) is half the original period. Therefore, the curd rent rises or falls for half as long in each part of the cycle, and the amplitude of the current sine wave is reduced by half. Since XL = VL/IL, the inductive ­reactance has been doubled. By considering other ratios of frequencies in the same way, we find that Inductive reactance is directly proportional to frequency. If we double the inductance but leave the source unchanged, Equa­ tion 19-1 shows that the maximum rate of change of current must be cut in half to develop the same voltage. Therefore, the amplitude of the current is reduced by half and the inductive reactance is doubled. Since the same reasoning applies for all other inductances, Inductive reactance is directly proportional to the inductance. We can derive a formula for inductive reactance by considering the a­ verage rate of change of current. The current rises from zero to Im in one quarter-cycle, and the duration of one cycle is 1/f. Therefore,

average

di = dt

1 4

Im = 4 f Im × ( 1/f )

From Equation 18-17, we know that Eav = 2Em /π. Substituting into Equation 19-1 gives 4fIm =

from which

Em × 2/π L

Em = 2πf L Im

XL = 2πf L



(19-4)

where XL is inductive reactance in ohms, f is frequency in hertz, and L is ­inductance in henrys. Since ω = 2πf (Equation 18-5),

XL = ωL

where ω is angular velocity in radians per second. Appendix 2-7 uses integral calculus to derive this equation. See Problems 19-3 to 19-10 and Review Question 19-28.

(19-5)

557

558

Chapter 19  Reactance

Circuit Check

A

CC 19-1. If a voltage of 30 V AC causes a current of 2 A to flow in an ideal inductor, how much inductive reactance is in the circuit? CC 19-2. What is the value of the inductor in question CC 19-1 if the frequency of the source is 60 Hz? CC 19-3. At what frequency will a 470-nH inductor have a reactance of 100 Ω?

19-4 Instantaneous Current in a Capacitor e = Em sin ωt

C

A Figure 19-3 Capacitance in an alternating-current circuit

If we connect a capacitor across a sine-wave voltage source, as in Figure  19‑3, Kirchhoff’s voltage law requires the voltage across the capacitor to be exactly the same as the applied voltage at every instant. The voltage across a capacitor can change only if the capacitor charges or discharges. Consequently, the capacitor in Figure 19-3 must charge and discharge in such a manner that the voltage across it is a sine wave equal to the applied voltage at every instant. Since q = Cv (Equation 12-6),

dq dt

By definition, i = dq/dt. Therefore,

=C

i=C

dv dt

dv dt

(19-6)

Since capacitance depends on such physical factors as the area of the plates and the dielectric constant of the material between the plates, the capacitance of a given circuit does not depend on the elapsed time. Since we can treat C in Equation 19-6 as a constant, this equation shows that the instantaneous current in Figure 19-3 is directly proportional to the rate at which the voltage across the capacitor is changing. The blue sine curve in Figure 19-4 represents the instantaneous voltage across the capacitor. This curve shows that the maximum voltage across the capacitor occurs π/2 radians after the maximum rate of change of voltage. At the exact moment when the voltage across the capacitor is greatest, the voltage is neither rising nor falling. Therefore, the instantaneous current must be zero at this instant. The maximum rate of change of voltage occurs when the voltage sine curve is steepest. At this instant the voltage is zero, indicating that the capacitor has just finished discharging its stored charge and is about to start building up an opposite charge. Therefore, the instantaneous current has its maximum positive value at the instant when

Instantaneous Values

19-5   Capacitive Reactance

Zero current

+

0 π 2

π

Maximum positive current

Maximum positive voltage

vC

3π 2

2π

5π 2

i C ωt

− Figure 19-4  Instantaneous current in a capacitor

the voltage across the capacitor changes from a negative polarity to a positive polarity. Similarly, the current reaches its maximum negative value just as the voltage changes from a positive to a negative polarity. The instantaneous current must have the sine-wave shape shown by the red curve in Figure 19-4 in order for the voltage across the inductor to match the applied voltage at every instant. The instantaneous current is at its maximum positive value at the instant that the voltage across the capa­ citance is just starting to increase from zero. When the voltage across the ­capacitance has reached its positive peak π/2 rad later, the instantaneous current has fallen back to zero. Therefore, For a sine-wave voltage to be developed across a capacitor, the current through it must be a sine wave that leads the instantaneous voltage by π/2 radians. Therefore, the instantaneous current in the circuit of Figure 19-3 is



ic = Im sin ωt +

(

π 2

)



(19-7)

where the phase angle ω t + π/2 is measured in radians. See Review Questions 19-29 and 19-30.

19-5  Capacitive Reactance Since the current through the capacitor in Figure 19-3 is a sine wave, it has an RMS value, which can be measured by an AC ammeter connected as shown in Figure 19-3. This current reading does not mean that current is flowing through the insulating dielectric of the capacitor. In an AC circuit, instan­taneous current flows in one direction for one half-cycle and then in the ­opposite direction for the next half-cycle. During each cycle,

559

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Chapter 19  Reactance

the ­capacitor builds up a positive charge, discharges, builds up a negative charge, and then  ­discharges again. The sine-wave alternating current in the circuit is a displacement current that transfers charge to and from the capacitance. Thus, capacitors can “pass” alternating current while blocking direct current. For a given alternating-voltage source and a given capacitance, both the sine-wave voltage across and the sine-wave current “through” the ­capacitor have fixed RMS values. Therefore, the ratio V/I is a constant. As with ­resistance and inductive reactance, this constant V/I ratio represents ­opposition to alternating current. Capacitive reactance is the opposition of alternating current. The letter symbol for capacitive reactance is XC. XC =



VC IC

capacitance

to

(19-8)

An AC circuit has a capacitive reactance of one ohm when alternating current of one ampere RMS creates an alternating voltage of one volt RMS across the capacitance of the circuit. See Problems 19-11 and 19-12.

19-6 Factors Governing Capacitive Reactance By varying the capacitance and the applied voltage in the circuit of  Figure 19-3, we can determine the factors that govern capacitive ­reactance. If we double the amplitude of the applied voltage while keeping the frequency constant, the capacitor has to store twice as much charge in the same time interval. Therefore, the peak charging current must be twice as great. Since XC = VC/IC, the capacitive reactance remains unchanged when both VC and IC are doubled. The same logic applies for any change in the applied voltage. The magnitude of the applied alternating voltage has no effect on the capacitive reactance of an AC circuit. If we double the frequency of the applied voltage and leave the peak value the same, the capacitor must store the same charge but in half the time. Therefore, the peak charging current must be twice as great. Consequently, IC is doubled. Since VC is unchanged and XC = VC/IC, the capacitive reactance is reduced by half. We can generalize this result:

19-6   Factors Governing Capacitive Reactance

Capacitive reactance is inversely proportional to frequency. Finally, we double the capacitance and leave the source unchanged. Since Q = CV, the capacitor must store twice as much charge at the peak of the voltage sine wave. Since the frequency is unchanged, the capacitor must build up the doubled charge in the same length of time. Therefore, the peak charging current must be twice as great. Again, IC is doubled and VC is unchanged, so the capacitive reactance is reduced by half. We can ­extend this reasoning to any capacitance. Capacitive reactance is inversely proportional to the capacitance. We can derive a formula for reactive inductance by considering the average rate of change of current, as we did for inductive reactance. First we rearrange Equation 19-6. dv i = dt C



(19-9)

We can substitute the average rate of voltage rise for dv/dt and an average value for i. During the first quarter of a cycle the voltage across the capacitor rises from zero to Em. The duration of a complete cycle is 1/f. Therefore,

average

dv Δv Em = = 1 1 = 4f Em dt Δt 4 × f

Equation 18-16 gives the half-cycle average for i. Since the first half of a sine wave is symmetrical, the half-cycle average is equal to the quarter-cycle ­average, Iav = 2Im/π. Substituting for both sides of Equation 19-9 gives 4f Em =



Im × 2/π C

Em 1 = Im 2πfC



and

XC =

1 2πf C

(19-10)

where XC is capacitive reactance in ohms, f is frequency in hertz, and C is capacitance in farads. Since ω = 2πf,

XC =

1 ωC

Appendix 2-8 gives a calculus derivation for this equation. See Problems 19-13 to 19-19 and Review Question 19-31.

(19-11)

561

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Chapter 19  Reactance

Circuit Check

B

CC 19-4. What is the capacitive reactance of a 470-pF capacitor at a frequency of 455 kHz? CC 19-5. What value of capacitance will cause a 25-mA current to flow when connected to a 22-V 5-kHz source?

19-7 Resistance, Inductive Reactance, and Capacitive Reactance Resistance, inductive reactance, and capacitive reactance are all forms of opposition to alternating current. Each results from one of the basic properties of an electric circuit, and each can be defined in terms of a V/I ratio. Although the instantaneous currents vary sinusoidally, the opposition to current in a given circuit at a given frequency is independent of the elapsed time. Therefore, we represent resistance, inductive reactance, and capacitive reactance with uppercase letters. There are important differences in the behaviour of resistance, inductance, and capacitance in AC circuits: • Resistance converts electric energy into heat, but inductance and capacitance alternately store energy during a charging quarter-cycle and return this energy to the circuit during the next (discharging) quarter-cycle. •  Resistance is independent of frequency, but inductive reactance is ­directly proportional to frequency and capacitive reactance is inversely proportional to frequency. • The current through a resistor is in phase with the voltage across it; the current through an ideal inductor lags behind the voltage across it by 90°, or π/2 radians; and the alternating current “through” a capacitor leads the voltage across it by 90°. These characteristics are summarized in Table 19-1.

TABLE 19-1 R, L, and C in AC circuits Resistance Inductive Reactance Capacitive Reactance

R=

XL =

XC =

VR IR

VL = ωL IL

VC 1 = IC ωC

IR is in phase with VR IL lags VL by 90º IC leads VC by 90º

19-7   Resistance, Inductive Reactance, and Capacitive Reactance

It is important to identify clearly the type of opposition when we discuss circuits. In addition to using a different letter symbol, we sometimes ­indicate the type of opposition after the units, as shown in the following examples. AC

Example 19-1 Find the resistance of a 660-W toaster operated from a 110-V 60-Hz source. Solution

R=



( 110 V ) 2 E2 = = 18.3 Ω(resistive) P 660 W

Example 19-2 What inductive reactance develops a 40.0-V induced voltage when the current through it has an RMS value of 80.0 mA? Solution

XL =

VL 40 V = = 0.50 kΩ ( inductive ) IL 80 mA

Example 19-3 Calculate the reactance of a 68-pF capacitor when it is used in a 4.5-MHz circuit. Solution

XC =

1 1 = = 0.52 kΩ ( capacitive ) 2πf C 2 × π × 4.5 MHz × 68 pF

Rather than consider DC and AC circuit analysis techniques separately, some textbooks treat steady-state direct current as a special zero-frequency case of alternating current. At zero frequency, ω = 0. Hence, XL = ωL = 0 and XC = 1/ω C = ∞. These values show that the steady-state DC response of inductance is a short circuit and the steady-state DC response of capacitance is an open circuit. See Problems 19-20 and 19-21 and Review Questions 19-32 to 19-34.

563

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Chapter 19  Reactance

Summary

• The current through an ideal inductor with a sine wave of voltage across it is a sine wave that lags behind the voltage by π/2 radians. • Inductive reactance is the ratio of the AC voltage across an ideal inductor to the AC current through it. • Inductive reactance is proportional to both frequency and inductance. • The current through a capacitor with a sine wave of voltage across it is a sine wave that leads the voltage by π/2 radians. • Capacitive reactance is the ratio of the AC voltage across a capacitor to the AC current through it. • Capacitive reactance is inversely proportional to both frequency and ­capacitance. • Resistance, inductance, and capacitance are the three basic properties that produce opposition to alternating currents. B = beginner

I = intermediate

A = advanced

Problems B B

B B B B B B B B

B

Section 19-2  Inductive Reactance 19-1.

Find the inductive reactance of an ideal inductor connected across a 220-V AC source when the current flowing through it is 5.0 A. 19-2. What value of inductive reactance limits the current through it to 24 mA when the voltage across it is 35 V RMS?

Section 19-3  Factors Governing Inductive Reactance 19-3. Calculate the 60-Hz reactance of a 4.0-H choke. 19-4. Find the reactance of a 250-mH choke at a frequency of 4.5 MHz. 19-5. At what frequency does a 0.5-H inductor have a reactance of 2.0 kΩ? 19-6. At what frequency will a 100-µH inductor have a reactive inductance of 1.5 kΩ? 19-7. What inductance has a reactance of 1400 Ω at 475 kHz? 19-8. What inductance draws a 160-mA RMS current when connected to a 110-V 25-Hz source? 19-9. The solenoid of a contactor for starting an industrial motor draws 2.0 A from a 220-V 60-Hz source. Assuming negligible resistance, find the inductance of the solenoid. 19-10. Find the current through a 30-mH inductor connected across a 50-V 5.0-kHz source.

Section 19-5  Capacitive Reactance 19-11. What capacitive reactance draws a 3.0-mA alternating current from a 50-V 400-Hz source?

Review Questions

B

B B B B B B B

I I

19-12. What is the current flowing through a capacitor that has a capacitive reactance of 250 Ω when connected to a 25-V RMS source?

Section 19-6  Factors Governing Capacitive Reactance 19-13. Find the reactance of a 0.05-μF coupling capacitor at 400 Hz. 19-14. Find the reactance of a 27-pF capacitor at a frequency of 88 MHz. 19-15. At what frequency does an 8.0-μF capacitor have a reactance of 160 Ω? 19-16. At what frequency does a 2.0-nF capacitor have a reactance of 80 Ω? 19-17. What capacitance has a reactance of 47 Ω at 500 Hz? 19-18. The input resistance of a transistor audio amplifier is 47 kΩ. What value of coupling capacitor should be used if its reactance is to equal the input resistance at 20 Hz? 19-19. A 9.7-kV 60-Hz transmission line has a capacitance connected across it drawing 50 A of current. What is the value of the capacitance?

Section 19-7 Resistance, Inductive Reactance, and Capacitive Reactance 19-20. Find the capacitor that has a reactance at 455 kHz equal to that of a 470-µF inductor. 19-21. At what frequency does the reactance of a 560-pF capacitor equal that of a 470-µH inductor?

Review Questions

Section 19-1  Instantaneous Current in an Ideal Inductor

19-22. Why must the current sine wave be exactly π/2 rad out of phase with the applied voltage waveform across an ideal inductor? 19-23. Referring to the definition of inductance in Chapter 16, explain why the current must lag behind the voltage by π/2 radians rather than lead it. 19-24. What is the advantage of expressing the phase difference between the voltage across and the current through an ideal inductor as π/2 rad rather than 90°? 19-25. Why is it convenient to label the horizontal axis of the graph in ­Figure 19-2 in radians? 19-26. Why does the ammeter in Figure 19-1 show a constant reading?

Section 19-2  Inductive Reactance 19-27. Why is it possible to express inductive reactance in ohms?

Section 19-3  Factors Governing Inductive Reactance

19-28. What is the meaning of the symbol ω in the formula XL = ω L?

565

566

Chapter 19  Reactance

Section 19-4  Instantaneous Current in a Capacitor

19-29. With reference to the definition of capacitance in Chapter 12, explain why the current leads the voltage across the capacitor by π/2 radians rather than lagging behind it. 19-30. Explain how the polarity of the voltage across a capacitor can change without any change in the direction of the current charging the capacitor.

Section 19-6  Factors Governing Capacitive Reactance 19-31. Explain why an increase in frequency increases the reactance of an inductor but decreases the reactance of a capacitor.

Section 19-7  Resistance, Inductive Reactance, and Capacitive Reactance 19-32. Why does the term inductive reactance not apply to DC circuits? 19-33. When we connect the terminals of a capacitor to an ohmmeter, the steady-state reading shows an infinitely high resistance. Nevertheless, the ammeter in Figure 19-3 indicates a constant current in the circuit. Explain. 19-34. A “black box” with a pair of terminals marked “1000 Ω ” contains a single electrical component. Suggest a laboratory procedure for ­determining whether the component is a resistor, a capacitor, or an inductor.

Integrate the Concepts An AC voltage source with a 20-V peak-to-peak value at a frequency of 200 Hz is connected across a capacitor with a value of 10 µF. (a) Determine the value of the capacitive reactance. (b) Calculate the peak value of the current. (c) Write an equation for the current. (d) What is the effect of tripling the frequency of the circuit? (e) How does reducing the capacitance to 1 µF affect the circuit? The same voltage source is connected across a 15-mH inductor instead of the capacitor. (f) Determine the inductive reactance of the circuit. (g) Calculate the peak value of the current. (h) Write an equation for the current. (i) What effect does doubling the frequency have on the circuit? (j) If the inductance in the circuit is doubled, what effect it will have on the circuit?

Practice Quiz

Practice Quiz 1. For the circuit of Figure 19-5, the instantaneous current of the 10-mH coil is given by (a) 11.7 sin 314t kA (b) 23.4 sin 314 kA (c) 11.7 sin(314t − 90°) kA (d) 23.4 sin(314t − 90°) kA

V 117 Vpk 50Hz 0°

+

L 10 mH

−

Figure 19-5

2. Inductive reactance is (a) directly proportional to the frequency (b) inversely proportional to the frequency (c) independent of the frequency (d) none of the above 3. Which of the following equations describes inductive reactance? 1 (a) XL = 2πL (b) XL = 2πL (c) XL = 2πf1 L (d) XL = 2πfL

4. If the inductor in Figure 19-5 were 30 mH, what would be the inductive reactance? (a) 3.77 Ω (b) 5.65 Ω (c) 9.42 Ω (d) 4.71 Ω 5. If the frequency is doubled, the inductive reactance of an ideal inductor will (a) decrease by half (b) stay the same (c) double (d) quadruple

567

568

Chapter 19  Reactance

6. The instantaneous current flowing through the 82-nF capacitor in Figure 19-6 is given by (a) −3.01 sin 314t mA (b) 3.01 sin 314t mA (c) 3.01 sin (314t – 90°) mA (d) 3.01 sin (314t + 90°) mA XMM1

+

V 117 Vpk 50Hz 0°

+

−

C 82 nF

−

Figure 19-6

7. Capacitive reactance is (a) directly proportional to the frequency (b) inversely proportional to the frequency (c) independent of the frequency (d) none of the above 8. Which of the following equations describes capacitive reactance? (a) Xc = 21f c (b) Xc = 2fC (c) Xc = 2π1f c (d) XL = 2πfC 9. The reactance of the 82-nF capacitor shown in Figure 19-6 is (a) 32.35 Ω (b) 38.32 kΩ (c) 38.32 Ω (d) 32.35 kΩ

10. If the frequency is doubled, the capacitive reactance of a capacitor will (a) decrease by half (b) stay the same (c) double (d) quadruple

Practice Quiz

11. Resistance is (a) directly proportional to the frequency (b) inversely proportional to the frequency (c) independent of the frequency (d) none of the above 12. Current flowing through a resistor is (a) 90° out of phase with the voltage across it (b) −90° out of phase with the voltage across it (c) in phase with the voltage across it (d) 180° out of phase with the voltage across it 13. Current flowing through a capacitor (a) leads the voltage across it by 90° (b) lags the voltage across it by 90° (c) is in phase with the voltage across it (d) is 180° out of phase with the voltage across it 14. Current flowing through an ideal inductor (a) leads the voltage across it by 90° (b) lags the voltage across it by 90° (c) is in phase with the voltage across it (d) is 180° out of phase with the voltage across it

569

20

Phasors

Our next step is to analyze AC circuits containing both resistance and reactance. Since this analysis involves computations with phasors, we need a type of algebra that can handle numbers in two dimensions. This chapter shows how we can use complex numbers to represent phasors and carry out circuit calculations involving phasor quantities. However, some aspects of the algebra of complex numbers are sig­nificantly different from the ordinary algebra we used to ­analyze DC circuits.

Chapter Outline 20-1 Addition of Sine Waves  572

20-2 Addition of Instantaneous Values  573

20-3 Representing a Sine Wave by a Phasor Diagram  575 20-4 Letter Symbols for Phasor Quantities  576

20-5 Phasor Addition by Geometrical Construction  576 20-6 Addition of Perpendicular Phasors  578

20-7 Expressing Phasors with Complex Numbers  581

20-8 Phasor Addition using Rectangular Coordinates  585 20-9 Subtraction of Phasor Quantities  587

20-10 Multiplication and Division of Phasor Quantities  589

Key Terms time domain  572 frequency domain  573 phasor diagram  575 scalar 576

polar coordinates  581 rectangular coordinates  581 real component  581 quadrature component  581

imaginary component  581 complex plane  581 complex number  582 complex conjugate  590

Learning Outcomes At the conclusion of this chapter, you will be able to: • graph the resultant of two sine waves having the same frequency • represent a sine wave by a phasor in polar coordinates • represent a sine wave by a phasor in rectangular coordinates • add phasors with the aid of a phasor diagram

Photo sources: iStock.com/wragg

• add phasors by adding their rectangular coordinates • subtract phasors with the aid of a phasor diagram • subtract phasors by subtracting their rectangular coordinates • multiply and divide phasors expressed in polar coordinates

Chapter 20  Phasors

20-1  Addition of Sine Waves

R

VR

Resistor Voltage

Figure 20-1 shows a simple AC series circuit containing resistance and ­inductance. The sine-wave voltage source causes a sine wave of current to flow in the circuit. Since all the components are connected in series, the current in the inductance and the current in the resistance must have the same magnitude at every instant. In fact, the sine-wave currents in the inductance and the resistance are one and the same, so we can represent both with the letter symbol i.

+ 0

Time

−

e L

VL

Inductor Voltage

572

+ 0

Time

− Figure 20-1  AC circuit with resistance and inductance in series

The sine wave of current through the resistance causes a sine-wave voltage drop across it that is exactly in phase with the current sine wave. As shown in Chapter 19, the instantaneous current through inductance lags ­behind the instantaneous voltage applied to it by π/2 radians. Consequently, the sine-wave voltage across the inductance and the sine-wave voltage across the resistance are π/2 radians out of phase, and the instantaneous voltage across the inductance reaches its positive peak π/2 radians earlier than the instantaneous voltage across the resistance. Kirchhoff’s voltage law states that the sum of the voltage drops in a series circuit must equal the applied voltage. In DC circuits where the source voltage is independent of time, we can find the sum of the voltage drops by simple arithmetic. Even when dealing with the charging of a capacitor in Chapter 13 and the rise of current in an inductor in Chapter 17, we were able to find the sum of the instantaneous voltage drops by simple arithmetic addition. Now we need to find the sum of two out-of-phase sine-wave voltages. As described in Chapter 18, we can represent a sine-wave voltage either by its instantaneous value, v = Vm sin ωt, or by its RMS value, V = 0.707 Vm. Since the instantaneous value varies continuously and depends on the elapsed time, it is said to belong to the time domain. In the time domain,

20-2   Addition of Instantaneous Values

we express the angular distance travelled by a sine wave in radians. We can use ordinary algebra for calculations with instantaneous values. Since the RMS value of a sine wave does not vary with the exact instant in time, we treat a sine-wave alternating voltage as a phasor quantity having a magnitude and a phase angle. Such quantities are said to belong to the frequency domain. Calculations with phasors require a specialized form of algebra, which is described later in the chapter. See Review Question 20-23 at the end of the chapter.

20-2 Addition of Instantaneous Values Since Kirchhoff’s voltage law applies to the circuit in Figure 20-1 at every ­instant, e = vT = vR + vL



(20-1)

Instantaneous Voltage

Therefore, we can add the instantaneous values of sine waves by simple arithmetic addition. Although the sum of the instantaneous voltages at one ­instant does not give us the overall picture of the addition of complete sine waves, we can repeat the calculation at intervals over a complete cycle to determine the nature of the resultant wave. To help with this time-consuming task, we can draw the sine curves for the two sets of instantaneous values on the same graph, as in Figure 20-2. Since angular velocity is constant for a given frequency, we can show the phase angle of the source, ωt, instead of time on the horizontal axis. vT

+ Em

vR

vL

π

3π 2

0 π 2

2π

ωt

− Figure 20-2  Addition of two out-of-phase sine waves

At t = 0, the instantaneous voltage across the resistance is zero; therefore, the instantaneous value of the total voltage is the same as the voltage across the inductance at that instant. As time elapses, the instantaneous voltage across the inductance is decreasing and the instantaneous voltage across the resistance is increasing. At approximately 1 rad, their sum reaches its peak value. At π/2 rad, the instantaneous voltage across the ­inductance is zero, and the instantaneous value of the total voltage equals

573

Chapter 20  Phasors

the ­instantaneous voltage across the resistance. Approximately 1 rad later, the instantaneous values of the voltages across the resistance and the inductance are equal in magnitude but opposite in polarity, so vT = 0. Careful examination of Figure 20-2 shows that the waveform for vT is a sine wave with the same period as vR and vL. In fact, any sum of sine waves with the same frequency has this property. The sum of sine waves of the same frequency is also a sine wave of the same frequency. Equation 20-1 can be expanded to

e = VmR sin ωt + VmL sin ( ωt + π/2 )



(20-2)

Since vR and vL do not reach their peak values at the same instant, the peak value of vT occurs at a time when the instantaneous values of the two component voltages are less than their peak values. Therefore,

Em ≠ VmR + VmL



E ≠ VR + VL

Since the RMS value of any sine wave equals the peak value divided by √ 2,

Therefore, arithmetic addition does not apply to peak and RMS values if the sine waves are out of phase, as in Figure 20-2. However, if the sine waves are exactly in phase, as in Figure 20-3, the resultant and component sine waves do reach their peak values at the same instant, and Em = Vm1 + Vm2    and    E = VT = V1 + V2

Thus, the RMS value of the sum of in-phase sine waves is simply the sum of the RMS values of the individual sine waves. + v1 e v2

Instantaneous Value

574

e v1 v2

0 π

2π

ωt

− Figure 20-3  Addition of instantaneous values of two in-phase sine waves

See Problems 20-1 to 20-3 and Review Questions 20-24 to 20-29.

20-3   Representing a Sine Wave by a Phasor Diagram

20-3 Representing a Sine Wave by a Phasor Diagram Once we accept that the addition of sine waves of the same frequency results in a sine wave of the same frequency, the tedious procedure required to produce Figures 20-2 and 20-3 is not warranted. The original information from which we constructed those graphs was simply their RMS values and the phase angle between them. RMS value and phase angle will also serve to identify the resultant. In Figure 18-3, the length of the motion phasor represents the distance from the conductor to its centre of rotation, and the position of the phasor indicates the angular distance the loop has travelled. If we draw phasors for the two sine-wave voltages in the circuit of Figure 20-1, the rotating phasor for the instantaneous voltage across the inductance will always stay ­exactly π/2 radians ahead of the rotating phasor for the instantaneous voltage across the resistance. Consequently, these voltage phasors do not move in relation to each other. Sine waves of the same frequency can be represented by “stationary” phasors with the angle between the phasors representing the phase angle between the sine waves. When drawing phasors for electrical quantities, we make the length of the phasors proportional to the RMS value of the sine wave they r­ epresent. These AC phasors belong to the frequency domain, where angles are ­usually expressed in degrees. In the simple series circuit of Figure 20-1, the current is common to all components. Therefore, in the phasor diagram for a series AC circuit, such as in Figure 20-4, we draw a phasor representing the RMS value of the ­current along the reference axis of the diagram (pointing toward three ­o’clock). Since the voltage across the resistance is in phase with the current through it, the phasor VR points in the same direction as the current phasor. Since the voltage across the inductance leads the current through it by 90°, we orient VL 90° counterclockwise from the reference axis. VL

VR

I

Reference phasor

Figure 20-4  Phasor diagram for the circuit of Figure 20-1

The phasor diagram of Figure 20-4 is much easier to draw than the instantaneous voltage graphs of Figure 20-2, and it shows the RMS values and phase angles at a glance. See Review Questions 20-30 to 20-32.

575

576

Chapter 20  Phasors

20-4 Letter Symbols for Phasor Quantities

A mathematical ­vector can have any number of ­dimensions.

Technical journals, ­including IEEE ­publications, usually represent a phasor quantity with a ­lightface italic letter (like a scalar), with the magnitude ­indicated by vertical bars, as in V = |V|∠ϕ .

In studying the behaviour of electric and magnetic circuits, we encounter three types of quantities. Scalars are quantities that have magnitude but no direction or angle. ­Resistance, inductance, and capacitance are scalars. Voltage and current in DC circuits are scalars, as are instantaneous values of voltage and current in AC circuits. Scalars are represented by lightface italic letter symbols (such as R, L, C, V, I, v, and i). Scalar quantities can be added by simple algebraic addition. Vectors are quantities that have both magnitude and direction. Electrostatic force, electric field intensity, and magnetic flux density are all vector quantities. This book uses boldface italic letter symbols (such as F, E, and B) to represent vector quantities, and the same letter symbols in lightface italic to represent magnitudes of vectors. For example, the magnitude of the vector F is the scalar quantity F. Most electric vector quantities have three dimensions. However, if all the vectors for a particular calculation lie in the same plane, we can treat them as being two-dimensional. Because vectors are multidimensional, most operations in vector algebra, such as addition and multiplication, differ substantially from the simple arithmetic operations for scalars. Phasors are quantities that have both magnitude and direction. Phasor notation is a convenient method for representing sine waves in AC circuits. For example, a sine-wave voltage can be represented as V = V∠ϕ, where V is the RMS magnitude and ϕ is the phase angle. We use bold Roman letters for phasors to distinguish them from vectors, shown by bold italic letters. As with vectors, the same letter in lightface italic indicates the magnitude of a phasor. Using phasor notation we can rewrite Equation 20-2 as

ET = VR + VL

(20-3)

See Review Question 20-33.

20-5 Phasor Addition by Geometrical Construction In a phasor diagram, the resultant of two phasor quantities is the diagonal from the origin to the opposite corner of the parallelogram with the two phasors as sides. The resultant can be constructed with ruler and compass, as shown in Figure 20-5. Using a radius equal to the length of E1, we trace an arc

20-5   Phasor Addition by Geometrical Construction

with the tip of the E2 phasor as the centre. Next, using a radius equal to the length of E2, we trace an arc with the tip of E1 as the centre. We then draw the resultant by joining the origin to the intersection of the two arcs.

E2

ET

ϕ

E1

Reference axis

Figure 20-5  Geometrical construction of a resultant phasor

Applying this technique to the series circuit of Figure 20-1 shows how the magnitudes and phase angles of E, I, VR, and VL are related. Figure 20-6 compares the sine-wave graph and the phasor diagram for this series circuit. Although we can determine RMS values and phase angles from the sinewave graphs in Figure 20-6(a), it is much easier to get this information from the phasor diagram in Figure 20-6(b).

vT

Instantaneous Voltage

+

vR

vL

π

3π 2

0 π 2

VL

2π

ωt

VT = E VR

I Reference phasor

− (a)

(b)

Figure 20-6  Sine-wave graph and phasor diagram for the circuit of Figure 20-1

If we draw the phasors carefully, we can measure the RMS voltages and the phase angle between the applied voltage E and the current I to ­two-figure accuracy. Although graphical solutions are not accurate enough for many of the AC circuit problems in later chapters, we can use approximate values from a phasor diagram to check the calculated answers. Phasor diagrams also help us to get the correct sign for the sine, cosine, or tangent of angles larger than 90°.

577

578

Chapter 20  Phasors

Example 20-1 Three AC voltage sources with the same frequency are connected in ­series. Find the total applied voltage given that the source voltages are E1 = 80 V ∠0º , E2 = 100 V ∠150º , and E3 = 40 V ∠45º



Solution Step 1 Construct a phasor for each voltage.

Step 2 Construct a resultant phasor for any pair of phasors. In Figure 20-7 it is convenient to start with E1 and E3. Step 3 Construct the resultant for the third source phasor and the resultant of Step 2. Step 4 Measure the length and phase angle of the final resultant. These measurements show that the total applied voltage is approximately 79 V ∠80°. ET

E2 E3

E1 + E3

E1

Reference axis

Figure 20-7  Phasor diagram for Example 20-1

See Problems 20-4 to 20-8.

20-6 Addition of Perpendicular Phasors In Section 20-2, we saw that the RMS values of sine waves that are in phase can be added arithmetically. Therefore, phasors that have the same phase angle can be simply added together. Similarly, when the angle between a pair of phasors is exactly 180°, the phasor sum is simply the difference of the two phasors. For two phasors with an angle of exactly 90° between them, we can use trigonometry to find the phasor sum.

20-6   Addition of Perpendicular Phasors

579

In the simple series circuit of Figure 20-1, the voltage drop across the r­ esistance is exactly in phase with the current and the voltage across the ­inductance leads the current by exactly 90°. Therefore, the two voltage ­phasors are exactly 90° out of phase. In the geometrical construction for their resultant in Figure 20-8(a), the parallelogram becomes a rectangle with side AC exactly equal to magnitude of the VL. Therefore, the triangle in Figure 20-8(b) shows exactly the same data as the conventional phasor diagram of Figure 20-8(a), which has all the phasors starting from the origin. Such triangular diagrams are useful for analyzing AC circuits that have both resistance and reactance.

B

C

C

e us ten o p VT Hy ϕ VR

VT

VL O

ϕ

VR

A

I

O

VL Opposite side

Adjacent side

(a)

A

(b)

Figure 20-8  Addition of phasors at right angles

Since triangle OAC is a right triangle, tan ϕ =



opposite side adjacent side

=

AC VL = OA VR

Given the RMS voltages across the inductance and the resistance, we can find the phase angle ϕ using trigonometric tables or a calculator with a tan–1 function. ϕ = tan−1



VL VR

(20-4)

where tan–1 means “the angle having a tangent of.” We now have a choice of several methods for finding the magnitude of the total voltage, ET. From trigonometry, sin ϕ = Therefore,

opposite side hypotenuse ET =

VL sin ϕ

=

AC OC (20-5)

Arctan is another term for tan–1.

580

Chapter 20  Phasors

cos ϕ =

Similarly,

Therefore,

adjacent side hypotenuse VT =

=

OA OC

VR cos ϕ

(20-6)

Another method uses the Pythagorean theorem, which states that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the other two sides. In Figure 20-8(b), OC2 = OA2 + AC2

VT = √ V2R + V2L



(20-7)

Note that it is customary to state the resistance component before the reactive component in Equation 20-7. Combining Equations 20-7 and 20-4 gives ET = √V2R + V2L tan−1



VL VR

(20-8)

Example 20-2 Given that the voltage drop across the resistance in the circuit of Figure 20-1 is 15 V RMS and the voltage across the inductance is 10 V RMS, find the magnitude of the total voltage and its phase angle with respect to the current. Solution or or

ϕ = tan− 1

ET =

ET =

VL 10 = tan− 1 = 33.7º VR 15

VR 15 = = 18 V cos ϕ 0.832 VL 10 = = 18 V sin ϕ 0.555

ET = √ V2R + V2L = √ 225 + 100 = 18 V

  ET = 18 V∠+ 33.7°

The positive angle shows that the voltage leads the current.

See Review Question 20-34.

20-7   Expressing Phasors with Complex Numbers

581

20-7 Expressing Phasors with Complex Numbers

Quadrature or imaginary axis

When we describe a phasor by stating its magnitude and phase angle with respect to the reference axis, we are using polar coordinates. These coordinates specify the position of the tip of the phasor in terms of a distance along a radius from the origin and an angle measured from the ­reference axis. Polar coordinates correspond to the measurements we o ­ btain with instruments such as voltmeters, ammeters, and ­wattmeters. By applying trigonometric relationships to the phasor diagram of ­Figure 20-8, we determined the polar coordinates of the total voltage from two perpendicular components, one of which lies along the reference axis. We can use the same method to find rectangular coordinates (or perpendicular components) for any phasor in the polar form E∠ϕ. As shown in Figure 20-9, the reference axis component (or real component) is the magnitude of the phasor multiplied by the cosine of its phase. The quadrature component (or imaginary component) is the magnitude of the phasor multiplied by the sine of its phase angle.

E

E

E sin ϕ ϕ

E cos ϕ

Reference or real axis

Figure 20-9  Determining the rectangular coordinates of a phasor

The rectangular coordinates of a phasor indicate distances along the axes of the complex plane. Figure 20-10 shows how angles in polar coordinates correspond to the four possible directions for coordinates in the complex plane. In the complex plane, multiplying a phasor by the j operator rotates it 90° counterclockwise. To rotate the phasor another 90°, we simply multiply once more by the j operator. Hence, multiplying a phasor by j2 changes its direction by 180°. In polar coordinates, reversing the direction of a phasor is the same as multiplying by –1. Hence j2 = –1. Since j = √−1 has no ­solution, numbers preceded by the +j or –j operator are called ­imaginary numbers. Multiplying a phasor by j three times rotates it a total of 270°. Since j3 = j2 × j and j2 = −1, the operator –j corresponds to a rotation of 270°.

Mathematicians use i (from imaginary) to represent √ −1. However, j is used in electrical calculations to avoid confusion with i, the symbol for instantaneous current.

582

Chapter 20  Phasors

+j

90° E ϕ 0° Reference axis

180°

Reference axis

−

270° or −90° (a)

+

−j (b)

Figure 20-10  Corresponding directions in (a) polar coordinates and (b) rectangular coordinates

A phasor can also be expressed in terms of a complex exponential: E = E cos ϕ + jE sin ϕ = Ee jϕ. Although this exponential form is useful for advanced AC theory, the rectangular-coordinate form is more convenient for our calculations.

The rectangular coordinates of a phasor consist of a real number (which can be positive or negative) and an imaginary number identified by either +j or –j. The real component is always listed before the imaginary component. The formula for converting a voltage phasor from polar coordinates to rectangular coordinates is E∠ϕ = +E cos ϕ + jE sin ϕ



(20-9)

This equation demonstrates how any sine wave quantity can be expressed in terms of a complex number that has real and imaginary components.

Example 20-3 Express 18 V ∠33.7º in rectangular coordinates. Solution Step 1 Sketch a phasor diagram to find the correct operators to use with the rectangular coordinates (see Figure 20-11). +j

18 V 18 sin 33.7° 33.7° 18 cos 33.7°

+

Figure 20-11  Phasor diagram for Example 20-3

20-7   Expressing Phasors with Complex Numbers

Step 2 Use trigonometry to find the components of E: E = +E cos ϕ + jE sin ϕ

  = +18 cos 33.7° + j18 sin 33.7°    = 15 + j10 V

Most scientific calculators have a function for converting between polar and rectangular coordinates. We enter the polar coordinates and then use the conversion function to display the rectangular coordinates without having to calculate cos ϕ and sin ϕ.

Example 20-4 State 4.0 A ∠240º in rectangular coordinates. Solution Step 1 Sketch a phasor diagram, as in Figure 20-12, to check the signs and approximate values of the coordinates. +j

240°

−

+

60° 4.0 A

−j Figure 20-12  Phasor diagram for Example 20-4

Step 2 Use either the trigonometric or the coordinate-conversion functions of a calculator. These functions automatically provide the correct signs for the rectangular coordinates:

I = −2.0 − j3.5 A

583

584

Chapter 20  Phasors

Example 20-5

Convert 48 ∠−45º to rectangular coordinates.

Solution A negative angle is simply one measured clockwise from the reference axis. From Figure 20-13, 48 ∠−45º = 48 cos (−45º ) + j48 sin (−45º ) = + 33.9 − j33.9

Again a calculator will handle the negative angle automatically. −45°

+

48

−j Figure 20-13  Phasor diagram for Example 20-5

Equation 20-8 indicates a procedure for converting from rectangular coordinates to polar coordinates. The magnitude of the polar quantity is simply the square root of the sum of the squares of the rectangular coordinates, leaving out the j operator. We find the phase angle by drawing a diagram of the perpendicular components. The angle between the phasor and its reference axis component is θ = tan−1



quadrature coordinate reference axis coordinate

We then use the diagram to determine the actual phase angle with respect to the reference axis.

Example 20-6

State −60 + j30 V in polar form.

Solution If we use the trigonometry function keys of a calculator, we must pay careful attention to the component diagram of Figure 20-14. Since the calculator cannot distinguish between 30/−60 and −30/60, it displays a value of −26.565° for angle θ. From Figure 20-14, it can be seen that

ϕ = 180° − 26.6° = +153.4°

20-8   Phasor Addition Using Rectangular Coordinates

V = √(− 60 ) 2 + 302 = 67.1 V

or

V=

30 30 = = 67.1 V sin 26.6º 0.995

V = 67 V∠+153.4º

Alternatively, we can use the coordinate-conversion function of a calculator to find the polar coordinates in a single step. +j

ϕ

+j30 −

θ −60

Reference + axis

Figure 20-14  Phasor diagram for Example 20-6

See Problems 20-9 and 20-10 and Review Questions 20-35 to 20-41.

20-8 Phasor Addition Using Rectangular Coordinates Although there are trigonometric procedures for adding phasors having ­angles between them of other than 0° or 90°, it is usually much easier to ­express the phasors in rectangular coordinates and then find the sum of the components. Since all the real components of the phasors are horizontal, we can simply add them algebraically. Similarly, the imaginary ­components are all vertical, so we can also add them algebraically. These two sums are the real and imaginary components of the resultant phasor. All that r­emains is to convert these rectangular coordinates into polar form. This procedure is the conventional method of phasor addition for AC circuit problems.

Example 20-7

Find the sum of E1 = 80 V ∠+60º and E2 = 80 V ∠−135º . Solution Step 1 Sketch a phasor diagram, as shown in Figure 20-15.

585

586

Chapter 20  Phasors

+j E1

ET 60°

45°

−

+ −135°

E2

−j

Figure 20-15  Phasor diagram for Example 20-7

Step 2 Convert each voltage into its rectangular coordinates: E1 = 80 cos 60°

+j80 sin 60°

= +40

+j69.28 V

E2 = 80 cos −135° +j80 sin −135° = −56.57 − j56.57 V

Step 3 Add the components: ET = −16.57 + j12.71 V Step 4 Convert the resultant into polar coordinates: 12.71 θ = tan−1 = −37.49º −16.57 From Figure 20-15,

  ϕ = 180º − 37.49º = +142.51º

  ET = √ 16.572 + 12.712 = 20.88 V

    ET = 21 V ∠+143°

As in the preceding examples, we can save several steps by using the ­coordinate-conversion function of a calculator. See Problems 20-11 to 20-16.

Circuit Check

A

CC 20-1. A voltage sine wave with a peak value of 300 V leads another voltage sine wave of 200-V peak value by 30º. Calculate the RMS resultant voltage of these voltage waves. CC 20-2. Three AC currents in parallel branches of a circuit are given by i1 = 2 sin 100t, i2 = 3 sin 100t +

(

π 3π , and i3 = 4 sin 100t − . 3 4

)

(

)

20-9   Subtraction of Phasor Quantities



(a) Construct a phasor diagram showing the currents. (b) Use the diagram to determine the RMS value of the total ­current. (c) Use phasor addition to calculate the RMS total current.

20-9 Subtraction of Phasor Quantities When two phasors are expressed in rectangular coordinates, one can be subtracted from the other by simple algebraic subtraction of first the real (or reference axis) components and then the imaginary (or quadrature) components. The resulting rectangular coordinates can then be converted into polar form.

Example 20-8

Subtract 4 − j5 from 3 + j6 and express the result in polar coordinates. Solution Step 1 Subtract the components:

3 + j6

4 − j5

−1  + j11

Subtracting gives

We can also subtract by reversing the + and − signs in the subtrahend and then performing an algebraic addition: 3 + j6

 −4 + j5

  −1 + j11

Step 2 Sketch a phasor diagram to help determine the phase angle of the polar form (see Figure 20-16). Then convert to polar coordinates: −1 + j11 = √ 12 + 112 ∠180º − tan − 1 ( 11/1 )



= 11.05∠+95.2°



+j11

ϕ

−1

Reference axis

Figure 20-16  Phasor diagram for Example 20-8

587

588

Chapter 20  Phasors

The second procedure in Step 1 of Example 20-8 suggests a method for  subtracting phasors on a phasor diagram. As shown in Figure 20-17, +4 − j5 and −4 + j5 represent phasors that are exactly equal in length but 180° apart in angular location. Therefore, we can use geometrical construction to subtract two phasor quantities by reversing the direction of the subtrahend phasor and then constructing the parallelogram for the adding phasors.

+j5

+4 −4

−j5 Figure 20-17  Reversing the signs of rectangular coordinates

Example 20-9

Use geometrical construction to subtract E1 = 40 V ∠30º from E2 = 50 V ∠135º. Solution From Figure 20-18, E2 − E1 is about 73 V ∠120°. 90°

E2 E1

E2 − E1 180°

0°

−E1

270°

Figure 20-18  Subtraction of phasors by geometrical construction

See Problems 20-17 and 20-18 and Review Question 20-42.

20-10   Multiplication and Division of Phasor Quantities

20-10 Multiplication and Division of Phasor Quantities Multiplication and division of phasor quantities in polar form is the simplest of all the complex algebra processes. To multiply quantities expressed in polar form, multiply their magnitudes and add their phase angles algebraically. E1 × E2 = E1 ∠ϕ1 × E2 ∠ϕ2 = E1E2 ∠ϕ1 + ϕ2



(20-10)

To divide phasor quantities expressed in polar form, divide their magnitudes and subtract their phase angles algebraically. E1 ÷ E2 = E1∠ϕ1 ÷ E2 ∠ϕ2 =



E1 ϕ1 − ϕ2 E2 ∠

(20-11)

Example 20-10

Calculate 80∠40º × 11∠−15º and 80 ∠40º ÷ 11∠−15º. Solution

80∠40º × 11∠−15º = 80 × 11∠40 + ( −15 ) = 880∠25° 80∠40º ÷ 11∠−15º =

80 40 − ( −15 ) = 7.27∠55° 11 ∠

Although the procedure for multiplying and dividing phasor quantities expressed in rectangular coordinates is straightforward, it does require skill in manipulating algebraic terms. It is often safer to do phasor multiplication and division by the simpler polar method. With a calculator that has coordinate-conversion functions, such conversions are the quickest way to do phasor multiplication and division. Example 20-11 demonstrates the procedure for multiplication and division in rectangular form.

589

590

Chapter 20  Phasors

Example 20-11

Calculate (8 + j6) × (2 − j4) and (8 + j6) ÷ (2 − j4). Solution 8 + j6 2 − j4 Multiply by 2

16 + j12

Multiply by −j4     − j32 − j2 24 Collect like terms

16    − j20 − j2 24

Since the operator j represents a rotation of 90° from the reference axis, j2 represents a rotation of 180° from the reference axis. Therefore, j2 = −1, and 16 − j20 − j2 24 = 16 − j20 + 24 = 40 − j20

The procedure for division is more involved. First, we set up the phasor quantities involved in fraction form. Then we need to eliminate any j term in the denominator. If we multiply both the numerator and denominator by the same quantity, the value of the fraction does not change. To eliminate the j term, we multiply the fraction by the complex conjugate of the denominator. The complex conjugate of a number has the same real component as the number but the opposite imaginary component.



( 8 + j6 ) ÷ ( 2 − j4 ) =

= =

8 + j6

2 − j4

( 8 + j6 ) × ( 2 + j4 ) ( 2 − j4 ) × ( 2 + j4 )

−8 + j44 20

= −0.4 + j2.2

Note that the division is straightforward once the j term is cleared from the denominator.

See Problems 20-19 to 20-22 and Review Question 20-43.

591

Problems

Summary

• The sum of two sine waves having the same frequency is also a sine wave of the same frequency. • The peak value of the sum of two sine waves having the same frequency is not the sum of their peak values unless the sine waves are in phase. • The polar coordinates of a phasor give its magnitude and phase angle. • The rectangular coordinates of a phasor give the values of its real and imaginary components. • Phasors may be added or subtracted on a phasor diagram. • Phasors may be added or subtracted when expressed in rectangular ­coordinates. • Phasors are simpler to multiply or divide when they are expressed in polar coordinates.

Problems

Solve Problems 20-1 to 20-3 by drawing sine curves on graph paper and adding ­instantaneous values to determine the resultants. I I I

I I

I I I

Section 20-2  Addition of Instantaneous Values 20-1. A 45-V sine wave leads a 125-V sine wave by 75°. Determine the RMS voltage value of the resultant wave. 20-2. A series circuit made up of a capacitor with a reactance of a 100 Ω and an inductor with a reactance of 75 Ω has a 3-A sine-wave current flowing through it. Find the RMS total voltage. 20-3. The currents in the two branches of a parallel circuit are i1 = 20 sin 314t and i2 = 50 sin (314t − π/4). Determine the total current.

Section 20-5  Phasor Addition by Geometrical Construction

20-4. Use phasor diagrams to solve Problems 20-1 and 20-2 by geometrical construction. 20-5. Three sources of alternating voltage each generate 120 V RMS, but E2 leads E1 by 120° and E3 lags behind E1 by 120°. Use a phasor diagram to determine the total voltage when all three sources are connected in series. 20-6. Use a phasor diagram to determine the total voltage in Problem 20-5 when the leads to source E1 are reversed. 20-7. Determine the sum of the following phasors in degrees using geometrical construction: 100∠+50º , 35∠+90º , 50∠+p/5 rad, 20∠−4p 20-8. An induction motor draws a 10-A current from a 120-V 60-Hz source. This current lags behind the applied voltage by 30°.

B = beginner

I = intermediate

A = advanced

592

Chapter 20  Phasors

A capacitor in parallel with the motor draws a 5-A current from the source. Draw a schematic diagram, and use a phasor diagram and geometrical construction to find the total current drawn from the source. B

B

Section 20-7  Expressing Phasors with Complex Numbers 20-9. Convert the following polar coordinates to rectangular: (a) 120∠−25º (b) 45∠180º (c) −35∠90º (d) 130∠3 rad (e) 7.5∠π/7 rad (f) 2∠−45º 20-10. Convert the following to polar coordinates: (a) 3 + j4 (b) −14 − j10 (c) 12.7 + j0 (d) −0.8 + j1.2 (e) 0 − j18 (f) 0.67 − j0.43 Give answers for the following problems in polar coordinates.

B B B B B B B B

B B

Section 20-8  Phasor Additon Using Rectangular Coordinates

20-11. Add 13 − j7 and 11 + j18. 20-12. Add 1.4 + j0.8, −0.9 + j2.1, and 0 − j3.9. 20-13. Add 170∠200º and 88∠−75º. 20-14. Add 1.8∠125º, 2.7∠−157º, and 1.3∠−66º. 20-15. Add 40 + j72 and 60∠100º. 20-16. Add 120 + j0, 120∠−120º , and −60 + j104.

Section 20-9  Subtraction of Phasor Quantities

20-17. Subtract 25 − j17 from 35 + j50 and express the result in polar coordinates. 20-18. Subtract −15 + j10 from 14 − j7 and express the result in polar ­coordinates.

Section 20-10  Multiplication and Division of Phasor Quantities 20-19. Calculate (a) 4.1∠−64º × 13∠13º (b) 4.1∠−64º ÷ 13∠13º 20-20. Calculate (a) (3 + j4) × (5 − j6) (b) (3 + j4) ÷ (5 − j6)

Review Questions

I I

20-21. Determine the equivalent impedance of a parallel circuit formed by Z 1 = 37 Ω ∠45º and Z2 = 45 Ω ∠−37º. 20-22. Calculate

( 47 + j13 ) ( −21 − j32 ) 51∠−111º + 19∠+70º

Review Questions

Section 20-1  Addition of Sine Waves 20-23. Why is a knowledge of complex algebra required in solving AC circuit problems?

Section 20-2  Addition of Instantaneous Values 20-24. RMS values in an AC circuit are defined as DC equivalent values. Why then can we not add RMS values by simple algebra as we can in a DC circuit? 20-25. Why is it possible to add instantaneous values in an AC circuit by simple algebra? 20-26. When writing an equation for the instantaneous value of the sum of two sine waves, how is the phase difference between them indicated? 20-27. Is it possible to add the instantaneous values of two sine waves if their frequencies are different? Explain. 20-28. What does the equation IT = I1 + I2 tell us about the current sine waves? 20-29. What is meant by simple algebraic addition?

Section 20-3  Representing a Sine Wave by a Phasor Diagram 20-30. What advantage does a phasor diagram have over a linear graph of instantaneous values plotted against time? 20-31. What information does a linear graph show that is not shown by a phasor diagram? 20-32. What condition is necessary for adding RMS quantities in an AC ­circuit?

Section 20-4  Letter Symbols for Phasor Quantities 20-33. What notation is used to distinguish scalar, vector, and phasor ­quantities?

Section 20-6  Addition of Perpendicular Phasors 20-34. When can we use the Pythagorean theorem to add RMS values?

Section 20-7  Expressing Phasors with Complex Numbers 20-35. 20-36. 20-37. 20-38.

Show with a phasor diagram that sin 210° = −sin 30°. Show with a phasor diagram that cos 135° = −cos 45°. What is meant by polar coordinates? What is meant by rectangular coordinates?

593

Chapter 20  Phasors

20-39. Describe how to convert the polar coordinates of a phasor into ­rectangular coordinates. 20-40. What is the significance of the operator j when one is dealing with phasor quantities? 20-41. What is the advantage of expressing phasor quantities in rectangular coordinates?

Section 20-9  Subtraction of Phasor Quantities 20-42. Explain the procedure for subtracting phasor quantities by geometrical construction on a phasor diagram.

Section 20-10  Multiplication and Division of Phasor Quantities 20-43. Outline two methods for finding the product of phasors that are ­expressed in rectangular coordinates.

Integrate the Concepts For the circuit of Figure 20-19: (a) Draw reasonably accurate graphs of e1 and e2. (b) Draw a reasonably accurate graph of the sum of e1 and e2. (c) Express e1 and e2 as phasors in polar coordinates. (d) Add e1 and e2 by using a phasor diagram. (e) Express e1 and e2 as complex quantities. (f) Use rectangular coordinates to determine the resultant of e1 + e2. (g) Determine e1 − e2. (h) Calculate e1 × e2. (i) Calculate e1/e2.

V1 100 V peak 60 Hz 30°

Figure 20-19 

+ −

−

V2 +

594

175 V peak 60 Hz 150°

230 Ω

Practice Quiz

Practice Quiz

1. The sum of two sine waves with the same frequency is (a) a sine wave that has double the frequency (b) a sine wave that has the same frequency (c) a sine wave that has half the frequency (d) not a sine wave 2. For the circuit of Figure 20-20, the voltage across the resistor can be expressed as (a) 153.5 sin (314t − 92.2°) (b) 180.9 sin (377t + 57.9°) (c) 180.9 sin (314t − 57.9°) (d) 153.3 sin (314t − 92.2°) V2 145 V peak 60 Hz 0° − +

+ −

V1 120 V peak 60 Hz 0°

2.2 kΩ

Figure 20-20 

3. The polar form for 10 + j8 is (a) 12.81∠38.66º (b) 12.81∠51.34º (c) 12.81∠−38.66º (d) 12.81∠−51.34º

4. The rectangular form for 25∠−60º is (a) 12.5 − j21.65 (b) −12.5 − j21.65 (c) 12.5 + j21.65 (d) −12.5 + j21.65 5. The sum of 330∠75º and 15∠−90º is (a) 364.5∠75.6º (b) 335.5∠74.3º (c) 335.5∠−74.3º (d) 364.5∠−75.6º

595

596

Chapter 20  Phasors

6. The sum of the phasors in Figure 20-21 is (a) 4.24∠45º (b) 3.16∠71.57º (c) 3.16∠18.43º (d) 4.24∠−45º +j 2 1 − −3

+

0 −2

1

−1

2

−1 −2 −3 −j Figure 20-21 

7. The difference between 75∠270º and 90∠−25º is (a) 89.55∠24.4º (b) 139.4∠−54.2º (c) 139.4∠125.8º (d) 89.55∠−155.6º 8. Subtracting −5 + j10 from 3 − j19 gives (a) −2 − j9 (b) 8 − j29 (c) −2 + j9 (d) −8 + j29 9. Subtracting −5 + j10 from 10∠45º gives (a) 12.42∠13.64º (b) 17.2∠83.08º (c) 12.42∠−13.64º (d) 17.2∠−83.08º 10. The product of 100∠80º and 50∠90º is (a) 5000∠170º (b) 5∠170º (c) 5000∠−170º (d) 5∠−170º

Practice Quiz

11. The product of 375 + j15 Ω and 525 − j50 Ω in polar coordinates is: (a) 0.72 Ω ∠7.7º (b) 197.924 Ω ∠3.2º (c) 197.924 Ω ∠−3.2º (d) 0.72 Ω + ∠−7.7º 12. Calculate Z1/Z2 for Z1 = 100∠80º and Z2 = 50∠90º. (a) −2∠170º (b) 2∠10º (c) 2∠− 10º (d) 2∠−170º

13. Determine Z1/Z2 for Z1 = 470 − j50 and Z2 = 325 + j45. (a) 1.4 − j0.35 (b) 1.44 − j0.046 (c) 150,500 + j37,400 (d) 1.4 + j0.046

14. Determine Z1 + Z2 × Z3/(Z2 + Z3) for Z1 = 50 − j25 Ω, Z2 = 10 + j60 Ω and Z3 = 220 − j45 Ω. (a) 99.83 Ω ∠−56.1º (b) 108.7 Ω ∠−46.5º (c) 74.9 Ω ∠−19.5º (d) 80.2 Ω ∠−21.1º

597

21

Impedance Circuit analysis for DC circuits containing resistance along with inductance or capacitance can be done with fairly straightforward algebra because the calculations involve only real numbers. In AC circuits containing inductance or capacitance, however, the calculations involve complex numbers (phasors). This chapter uses phasors to combine resistance and reactance into a single quantity that represents the total opposition to alternating current in a given circuit.

Chapter Outline 21-1

Resistance and Inductance in Series

21-2 Impedance

601

21-3 Practical Inductors

604

21-4 Resistance and Capacitance in Series

600

607

21-5 Resistance, Inductance, and Capacitance in Series

21-6 Resistance, Inductance, and Capacitance in Parallel 21-7

Conductance, Susceptance, and Admittance

21-8 Impedance and Admittance 21-9 Troubleshooting

619

615

613

608

610

Key Terms impedance 601 impedance diagram

602

total-current method admittance 614

611

susceptance 614 admittance diagram

615

Learning Outcomes At the conclusion of this chapter, you will be able to: • draw the phasor diagram for a series AC ­circuit containing resistance, inductance, and capacitance • calculate resistor, inductor, and capacitor voltages for a series AC circuit containing ­resistance, inductance, and capacitance • draw the impedance diagram for a series AC circuit ­containing resistance, inductance, and capacitance • calculate the total impedance of a series AC circuit • calculate the total impedance of a practical inductor • calculate the net reactive voltage of a series AC circuit

Photo sources:  © David J. Green/Alamy Stock Photo

• calculate the net equivalent reactance of a series AC circuit • draw the phasor diagram for a parallel AC circuit ­containing resistance, inductance, and capacitance • calculate resistor, inductor, and capacitor ­currents for a parallel AC circuit containing resistance, inductance, and capacitance • calculate the equivalent impedance of a parallel AC circuit • draw the admittance diagram for a parallel AC circuit • calculate the total admittance of a parallel AC circuit • detect faults in resistors, inductors, and capacitors in AC circuits

600

Chapter 21  Impedance

21-1 Resistance and Inductance in Series In Chapter 20 we noted that all rules and laws that apply to DC circuits also apply to AC circuits, although we keep in mind the sinusoidal nature of AC voltage sources. A key rule for series circuits is the one by which we define a series circuit: The same current must flow in all components in a series circuit. Because current is the only parameter common to all components in a ­series circuit, we often use the current as the reference in phasor diagrams for ­series circuits. The phase angles of other quantities are then calculated with respect to the common current, as shown in Figure 21-1.

VL VL

E

E

VR ϕ VR (a)

Reference I phasor

(b)

Figure 21-1 Phasor diagram of voltage and current relationships in a series circuit containing resistance and inductance

A second characteristic of series circuits is that the sum of the various voltage drops must equal the applied voltage. This voltage law applies to the simple algebraic sum of the instantaneous values in an AC circuit. The voltage law for series circuits also applies to RMS voltages provided we use phasor addition: In a series AC circuit, the phasor sum of the various voltage drops must equal the applied voltage. Since the voltage drop across the resistance must be in phase with the ­current, VR lies on the phasor reference axis in Figure 21-1. Similarly, the voltage across the inductance leads the current by 90°, so VL lies on the +j axis. Therefore,

E = VR + jVL



(21-1)

21-2  Impedance

From the resultant in Figure 21-1, we can see that the polar form of Equation 21-2 is E = √ VR2 + V2L tan − 1



VL VR

(21-2)

21-2 Impedance The current in the series circuit of Figure 21-1 must be a sine wave with a magnitude and phase angle with respect to the applied voltage such that the phasor sum of the voltage drop across the resistance and the voltage across the inductance equals the applied voltage. Therefore, for a given AC circuit, the magnitude of current is constant, and consequently the ratio ­between the applied voltage and the current is also constant. Just as VR/I represents the opposition of resistance to alternating current and VL/I represents the ­opposition of inductance to alternating current, the ratio VT/I or E/I represents the total opposition of the circuit to alternating current. In resistors, the voltage and current are in phase, and in inductors and capacitors the voltage and current are exactly 90° out of phase. But as shown in Figure 21-1, the angle ϕ between the applied voltage and the current is between 0° and +90° for a ­resistor and an inductor connected in series. Therefore, this total opposition is not equivalent to either a resistance or a reactance. The total opposition to current in an AC circuit is called impedance. The letter symbol for impedance is Z. The SI unit for impedance is the ohm. Z=



E I

(21-3)

where Z is the impedance of an AC circuit in ohms, E is the applied voltage in volts, and I is the current in amperes. Because the current is common to all components in a series circuit, we can divide every term in Equation 21-1 by I: Therefore,

E VR VL = +j I I I Z = R + jXL

(21-4)



We can also divide each term in Equation 21-2 by I:

Z = √ R2 + X2L tan−1

XL R

(21-5)

601

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Chapter 21  Impedance

We can derive similar equations for an AC circuit with resistance and ­capacitance in series. Therefore, The impedance of a series circuit is the phasor sum of the resistance and reactance. Like resistance and reactance, impedance is a constant V/I ratio, not a s­inusoidally varying quantity. But unlike resistance, inductive reactance, and capacitive reactance, the phase angle is not the same for all values of impedance. The equation Z = 5 kΩ tells us only the magnitude of the total opposition to electric current. To obtain the phase angle between the voltage and the current, we must recognize that impedance is a complex number that can be represented by a phasor. Hence, we can expand Equation 21-3: Z=



E∠ϕ = Z∠ϕ I∠0º

(21-6)

We could represent impedance with a standard common-origin phasor diagram as in Figure 21-2(a). However, to preserve a distinction between sinusoidal phasor quantities (such as voltage and current) and nonsinusoidal phasor quantities, we use a triangular phasor diagram for impedances, as shown in Figure 21-2(b). In such impedance diagrams, resistance is always drawn in the reference direction and inductive reactance in the +j direction. It follows that capacitive reactance is drawn in the −j direction. Z

+j

Z

Z

XL ϕ

Reference direction

R (a)

XL

ϕ

R (b)

Figure 21-2 Impedance diagram for a series circuit containing resistance and inductance

Example 21-1

A solenoid having an inductance of 0.50 H and a resistance of 100 Ω is connected to a 120-V 60-Hz source. Find the current. Solution Step 1 First, calculate the reactance of the solenoid:

XL = ωL = 377 × 0.50 = 188.5 Ω (inductive)

21-2  Impedance

603

Step 2 Draw and label a circuit diagram. Then draw an impedance diagram for the circuit, as shown in Figure 21-3. L = 0.50 H

XL

120 V 60 Hz

188.5 Ω Z

E R

XL = 188.5 Ω

100 Ω ϕ

R = 100 Ω (b)

(a)

Figure 21-3  Impedance diagram for Example 21-1

Step 3 Use the impedance diagram to find the rectangular coordinates of the ­impedance: Z = R + jXL = 100 + j188.5 Ω Step 4 Solve for the impedance in polar form.

ϕ = tan− 1

XL 188.5 = tan−1 = 62º R 100

Z = √ R2 + X2L = √ 1002 + 188.52 = 213 Ω

Z = 213 Ω ∠+62º Step 5 Solve for the magnitude of the current: E 120 V I= = = 0.563 A Z 213 Ω The impedance diagram in Figure 21-3 shows that the current lags behind the applied voltage by 62°. Alternatively, we can by rearrange Equation 21-6 and use phasor division to solve for both the magnitude and phase angle of the current. Since we are calculating this angle with respect to the applied voltage, the phase angle of the applied voltage is zero. I=

120 V∠0º E = = 0.563 A ∠−62° Z 213 Ω ∠62º

To verify the calculation of the magnitude and phase angle of the current, download Multisim file EX21-1 from the website and follow the instructions in the file. See Problems 21-1 to 21-7 and Review Questions 21-60 to 21-66 at the end of the chapter.

Although calculator shortcuts are possible for most of the ­examples in this ­chapter, full solutions are given to show the ­relationships between polar and rectangular coordinates. Students can practice using the coordinateconversion functions of a ­calculator to solve these ­examples and compare the results to the solutions given.

circuitSIM walkthrough

604

Chapter 21  Impedance

21-3  Practical Inductors

Source:  © David J. Green/Alamy

In practical circuits, the wire in an inductor’s winding has some resistance. Therefore, the practical inductor is not a pure reactance with a 90° angle. Hence the voltage drop across an inductor leads the current through it by something less than 90°. By expressing the impedance of an inductor in ­rectangular coordinates, we can determine the resistance and reactance components.

Example 21-2 The choke in the circuit of Figure 21-4(a) has an impedance of 80 Ω ∠80º. Find the total impedance when this choke is connected in series with a 75-Ω resistor. Z coil = 80 Ω 80° A choke wound on a ferrite core.

Vcoil

V

Zc

oil

ZT

E

75 Ω

Zcoil sin ϕ

ϕ

VR

R Zcoil cos ϕ

(a)

(b)

Figure 21-4  Practical inductor and resistor in series

Solution Draw an impedance diagram as shown in Figure 21-4(b). Then, express the impedance of the choke in rectangular coordinates: Zcoil = 80 cos 80° + j80 sin 80° = 14 + j79 Ω

Since the two resistance components are both along the + axis of the impedance diagram, they can be added by simple arithmetic: ZT = 75 + 14 + j79 Ω = 89 + j79 Ω

Changing back to polar form gives



ϕ = tan−1

79 = 41.6º 89

ZT = √ 892 + 792 = 119 Ω ZT = 119 Ω ∠+ 41.6°

21-3   Practical Inductors

We can also use trigonometric ratios to find the magnitude of the impedance: Z=



R X = cos ϕ sin ϕ

(21-7)

Example 21-3 A coil and a resistor each have an 80-V drop across them when connected in series to a 120-V 60-Hz source. The current in the circuit is 0.50 A. Find the resistance of the coil. Solution Step 1

ZT =



Zcoil =



120 V E = = 240 Ω I 0.50 A

R=



80 V Vcoil = = 160 Ω I 0.50 A 80 V VR = = 160 Ω I 0.50 A

Step 2 Since the coil has both resistance and inductance, the phase angle of its ­impedance must be somewhat less than +90°, as shown in Figure 21-5. Zcoil = 160 Ω ∠ϕ

In rectangular coordinates, the impedance of the coil is Zcoil = 160 cos ϕ + j160 sin ϕ

0.50 A A

E

Zc

ZT

oil

Vcoil = 80 V

VR = 80 V

ϕ

R Zcoil cos ϕ (a)

(b)

Figure 21-5  Impedance diagram for Example 21-3

Zcoil sin ϕ

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Chapter 21  Impedance

ZT = 160 + 160 cos ϕ + j160 sin ϕ

Step 3

Therefore, the magnitude of the total impedance is This example can also be solved by using the cosine rule, a trigonometric formula that relates the side lengths of any ­triangle to the cosine of one of its angles.

ZT = 240 = √ ( 160 + 160 cos ϕ ) 2 + ( 160 sin ϕ ) 2

Squaring both sides of the equation gives

57 600 = 25 600 + 51 200 cos ϕ + 25 600 cos2 ϕ + 25 600 sin2 ϕ cos ϕ =

32 000 − 25 600 ( cos2 ϕ + sin2 ϕ ) 5 − 4 ( cos2 ϕ + sin2 ϕ ) = 51 200 8

Step 4 Since sin2 ϕ + cos2 ϕ = 1,

cos ϕ =

Step 5 The resistance of the coil is

circuitSIM walkthrough

5−4 = 0.125 8

Rcoil = Zcoil cos ϕ = 160 × 0.125 = 20 Ω

To verify the calculation of the resistance of the coil, download Multisim file EX21-3 from the website, and follow the instructions in the file. See Problems 21-8 to 21-12 and Review Questions 21-66 to 21-68.

Circuit Check

A

CC 21-1. A 30-mH inductor is connected in series with a 50-Ω resistor and a 50-V 60-Hz source. (a) Determine the magnitude and phase angle of the current. (b) Determine the voltage drop across the inductor and the resistor. (c) Sketch the phasor diagram. CC 21-2. A series RL circuit is connected to a 100-V source. A current of 10 A flows in the circuit. Given that the voltage across the resistor is 50 V and the inductance is 30 mH, find the frequency of the voltage. CC 21-3. A 4.0-H inductor with a coil resistance of 200 Ω is connected in series with a 1.0-kΩ resistor and a 200-V 60-Hz source. Calculate the (a) phase angle of the inductor (b) phase angle of the circuit (c) current flowing in the circuit (d) voltage drop across the inductor

21-4   Resistance and Capacitance in Series

21-4 Resistance and Capacitance in Series In Figure 21-6(a), the current again is common to all components. Since the potential difference across the capacitor lags behind the current through it by 90°, the total-voltage phasor is in the fourth quadrant in the phasor diagram in Figure 21-6(b). For an AC circuit with resistance and capacitance in series, E = VR − jVC (21-8)



VC

VR

E

ϕ

Reference I phasor

VR

E

VC (a)

(b)

Figure 21-6  Series circuit containing resistance and capacitance

or

E = √ VR2 + VC2 −tan − 1 Z = R − jXC

and

or

VC VR

Z = √R2 + X2C −tan − 1

(21-9)

(21-10) XC R

(21-11)

Inductive reactance has a +j direction in impedance diagrams, and capacitive reactance always has a −j direction. Therefore, an inductive impedance has a positive angle between 0 and +90° and a capacitive impedance has a ­negative angle between 0 and −90°.

Example 21-4 What capacitance connected in series with a 500-Ω resistor limits the ­current drawn from a 48-mV 465-kHz source to 20 μA? Solution

Z=

E 48 mV = = 2.4 kΩ I 20 μA

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608

Chapter 21  Impedance

Applying the Pythagorean theorem to the impedance diagram in Figure 21-7 gives XC = √ Z2 − R2 = √ 24002 − 5002 = 2350 Ω XC =

Since

C=



1 2πf C

1 1 = = 0.15 nF 2πf XC 2 × π × 465 kHz × 2350 Ω

20 μA

R = 500 Ω

A

C

48 mV 465 kHz

Z = 2400 Ω

XC

500 Ω

(a)

(b)

Figure 21-7  Impedance diagram for Example 21-4

circuitSIM

To verify the calculation of the capacitance, download Multisim file EX21-4 from the website and follow the instructions in the file.

walkthrough

See Problems 21-13 to 21-22.

21-5 Resistance, Inductance, and Capacitance in Series In rectangular form the sum of the voltages in the circuit of Figure 21-8 is E = VR + jVL − jVC



(21-12)

VL VL E

VR VC

(a)

VX

E

ϕ

VR

I

VC (b)

Figure 21-8  A series circuit containing resistance, inductance, and capacitance

21-5   Resistance, Inductance, and Capacitance in Series

The voltage across the inductance leads the common current by 90°, and the voltage across the capacitance lags behind the current by 90°. Since these two voltages are exactly 180° out of phase, they can be added algebraically. Therefore, the imaginary terms in Equation 21-12 can be combined: E = VR + j ( VL − VC )



(21-13)

Unless the two reactive voltages are exactly equal, the circuit has an equivalent reactive voltage, VX, that is either inductive or capacitive depending on which of the original reactive voltages is greater. If VL is greater than VC, the circuit in Figure 21-8 appears to contain only inductance and resistance when measured at the terminals of the source. The overall circuit cannot be both inductive and capacitive at the same time. The current must either lead or lag behind the applied voltage. Converting Equation 21-13 into polar coordinates gives

E = √ VR2 + VX2 tan − 1

VX VR

where VX is the net reactive voltage, VL − VC.

(21-14)

Dividing Equation 21-13 and Equation 21-14 by the common current gives the impedance of a series circuit containing resistance, inductance, and capacitance: Z = R + jXL − jXC = R + j ( XL − XC )



Z = √ R2 + X2eq tan − 1

or

Xeq R

where Xeq is the net equivalent reactance, XL − XC.

(21-15)

(21-16)

Example 21-5 Find the total impedance at 20 kHz of a 1.5-mH inductor, a 100-Ω ­resistor, and an 80-nF capacitor connected in series. Solution Step 1 First find the reactances of the inductor and the capacitor:

XL = 2πf L = 2 × π × 20 kHz × 1.5 mH = 188.5 Ω (inductive) and XC =

1 1 = = 99.47 Ω (capacitive) 2πf C 2 × π × 20 kHz × 80 nF

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Chapter 21  Impedance

Step 2 Draw a schematic diagram and an impedance diagram, as in Figure 21-9. XL = 188.5 Ω

XL

188.5 Ω

XR

100 Ω

XC

99.5 Ω

Z

Xeq = XL − XC

ϕ R = 100 Ω XC = 99.5 Ω

(a)

(b)

Figure 21-9  Schematic and impedance diagrams for Example 21-5

Step 3 Find the impedance in rectangular coordinates:

Z = R + jXL − jXC = 100 + j188.5 − j99.5 = 100 + j89 Ω

Step 4 Convert the impedance to polar coordinates:

ϕ = tan − 1 Z=

Xeq R

= tan − 1

89 = +41.7º 100

R 100 Ω = = 134 Ω cos ϕ cos 41.7º

Z = 134 Ω ∠ + 42°

See Problems 21-23 to 21-32 and Review Questions 21-69 to 21-73.

21-6 Resistance, Inductance, and Capacitance in Parallel The characteristic of a parallel circuit is that the same voltage appears across all parallel branches. We use this common voltage as the reference phasor in phasor diagrams for any parallel AC circuits. Ohm’s law then gives the current through each branch of the circuit in Figure 21-10.

21-6   Resistance, Inductance, and Capacitance in Parallel

IR =

For the resistance,

E ∠0º E = ∠ 0º R ∠ 0º R

Therefore, IR is in phase with the applied voltage. IT

IC IC

IR

IT

IL

IX IL

ϕ

Reference E phasor

IR

(a)

(b)

Figure 21-10  A parallel circuit containing resistance, inductance, and capacitance

For the inductance,

IL =

E ∠0º E = −90º XL ∠+90º XL ∠

and IL lags behind the reference voltage by 90°. For the capacitance,

IC =

E ∠0º E = +90º XC ∠−90º XC ∠

and IC leads the reference voltage by 90°. The total current in the parallel circuit is the phasor sum of the branch ­currents:

IT = IR − jIL + jIC = IR + j ( IC − IL )

(21-17)

Converting Equation 21-17 to polar coordinates gives

IT = √ I2R + I2X tan− 1

where IX is the net reactive current IC − IL.

IX IR

(21-18)

We can apply Equation 21-6 to determine the equivalent impedance of the parallel circuit:

Zeq =

E IT

(21-19)

This method of solving for the equivalent impedance of a parallel circuit is called the total-current method. If the applied voltage is not known, we  can assume any convenient value in order to find the equivalent ­impedance.

611

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Chapter 21  Impedance

Example 21-6 Find the equivalent impedance at 20 kHz of a 1.5-mH inductance, a 100-Ω resistance, and an 80-nF capacitance all connected in parallel. Solution Step 1 Calculate the reactances:

XL = 2πf L = 2 × π × 20 kHz × 1.5 mH = 188.5 Ω (inductive) XC =



Note that dividing a quantity having the same direction as the reference phasor by a +j quantity creates a −j quantity, and dividing by a −j quantity creates a +j quantity.

1 1 =  = 99.5 Ω (capacitive) 2πf C 2 × π × 20 kHz × 0.08 mF

Step 2 Assume a convenient value of applied voltage, say E = 200 V ∠0º . Then

IR = IL =



and

IC =

E ∠ 0º 200 ∠ 0º = = 2.00 A ∠ 0º R 100 ∠ 0º

E ∠0º 200 ∠0º = = 1.06 A ∠−90º +jXL 188.5 ∠+90º E ∠0º 200 ∠0º = = 2.01 A ∠+90º −jXC 99.5 ∠−90º

To facilitate the phasor division, we expressed the direction associated with R, XL, and XC in polar form rather than with rectangular coordinate operators. Step 3 Draw a circuit diagram and a phasor diagram, as in Figure 21-11.

IC = 2.01 A IT

200 V 20 kHz

XC 99.5 Ω

R 100 Ω

XL 188.5 Ω

IX = 0.95 A IL = 1.06 A

(a)

ϕ IR = 2.0 A

(b)

Figure 21-11  Diagrams for Example 21-6

Step 4 Calculate the total current in rectangular coordinates.

IT = IR − jIL + jIC = 2.00 − j1.06 + j2.01 = 2.00 + j0.95A

E

21-7   Conductance, Susceptance, and Admittance

613

Step 5 Convert the total-current polar coordinates: ϕ = tan − 1 I=



IX 0.95 = tan− 1 = +25.4º IR 2.00

IR 2.00 A = = 2.214 A cos ϕ cos ( +25.4º )

IT = 2.21 A ∠+25.4°

Step 6 Calculate the equivalent impedance:

Zeq =

200 ∠0º E = = 91 Ω ∠−25° IT 2.21 ∠+25.4º

Note that the equivalent impedance is capacitive in this example because the capacitive branch draws a greater current from the source than does the inductive branch. When the same three components were connected in series in E ­ xample 21-5, the total impedance was inductive because there was a greater voltage across the inductance than across the capacitance. To verify the calculation of the magnitude and phase angle of the current in Step 5, download Multisim file EX21-6 from the website and follow the detailed instructions in the file.

See Problems 21-33 to 21-39 and Review Questions 21-74 and 21-75.

21-7 Conductance, Susceptance, and Admittance For parallel DC circuits, we found it convenient to think in terms of conductance, the reciprocal of resistance. Conductance is a measure of how easily a DC circuit passes current. We can use a similar approach for AC circuits. Since I = E/Z, we can rewrite Equation 21-17 as Dividing by E gives

E ∠ 0º E E E E = = −j +j Z Z R XL XC 1 1 1 1 = −j +j Z R XL XC

(21-20)

circuitSIM walkthrough

614

Chapter 21  Impedance

In this equation, we can replace 1/R with conductance, G. There are similar reciprocals for impedance and reactance. Admittance is the overall ability of an electric circuit to pass alternating current. The symbol for admittance is Y. Admittance is the reciprocal of impedance: Y = 1/Z Susceptance is the ability of inductance or capacitance to pass alternating current. The letter symbol for susceptance is B. Susceptance is the reciprocal of reactance: B = 1/X The SI unit for both admittance and susceptance is the siemens, the same as for conductance. When we divide 1∠0º by a phasor quantity with a +90° angle, the quotient has a −90° angle. Hence, the reciprocal of +jXL is −jBL. Similarly, the reciprocal of −jXC is +jBC. Thus, substituting admittance, conductance, and ­susceptance into Equation 21-20 gives Y = G − jBL + jBC = G + j ( BC − BL )



(21-21)

The polar coordinates of admittance are ϒ = √G2 + B2eq



tan− 1

Beq G



(21-22)

where Beq is the net equivalent susceptance, BC − BL.

Since BL = 1/XL ,

BL =

1 1 = (21-23) 2πfL ωL

Similarly,

BC = 2πfC = ωC

(21-24)

Example 21-6A Find the equivalent impedance at 20 kHz of a 1.5-mH inductance, a 100-Ω resistance, and an 80-nF capacitance all connected in parallel. Solution Step 1 Calculate the conductance and the susceptances: 1 1 G= = = 10.0 mS R 100 Ω

BC = 2πfC = 2 × π × 20 kHz × 80 nF = 10 mS ( capacitive ) BL =

1 1 = = 5.3 mS ( inductive ) 2πfL 2 × π × 20 kHz × 1.5 mH

21-8   Impedance and Admittance

Step 2 Draw an admittance diagram for the circuit, as shown in Figure 21-12. It is impossible to draw an impedance diagram for the circuit of Figure 21-11(a) because the equivalent impedance of a parallel circuit is smaller than the r­esistance, but the hypotenuse of a right triangle cannot be smaller than ­either of the other two sides.

BC = 10 mS

Y

Beq = BC − BL

ϕ G = 10 mS BL = 5.3 mS

Figure 21-12  Admittance diagram for Example 21-6A

Step 3 Using rectangular coordinates, calculate the admittance: Y = G + j(BC − BL) = 10 + j(10.0 − 5.3) mS = 10 + j4.7 mS Step 4 Convert the admittance to polar form: ϕ = tan−1



Y=



Beq G

= tan−1

4.7 = +25.2º 10

G 0.01 s = = 11.05 mS cos ϕ cos 25.2º Y = 11.05 mS ∠+25.2º

Step 5 Calculate the equivalent impedance: Z eq =

1 ∠0º 1 = = 91 Ω ∠−25° Y 11.05 mS ∠+25.2º

See Problems 21-40 to 21-51 and Review Questions 21-76 to 21-78.

21-8  Impedance and Admittance In the simple parallel circuit of Figure 21-10(a), each branch contained only one type of opposition to alternating current. Suppose, however, that one of the branches is a practical inductor, which has both resistance and i­ nductive

615

616

Chapter 21  Impedance

reactance. The current through this branch would now be ­inversely proportional to its impedance. The following example illustrates the procedure for finding the conductance and susceptance of a practical inductor.

Example 21-7 Find the conductance and susceptance of an inductor with a resistance of 30 Ω and an inductive reactance of 40 Ω (see Figure 21-13). 50 Ω 53.1° R

0.02 S −53.1°

30 Ω Z

XL

Y

G

BL

40 Ω

Figure 21-13  Finding the conductance and susceptance of an inductor

Solution Step 1 Find the impedance of the inductor:

Z = √ R2 + X2 tan−1 R = √ 302 + 402 tan− 1 X

40 = 50 Ω ∠+53.1º 30

Step 2 Calculate the admittance: Y=

1 1 = = 20 mS ∠−53.1º Z 50 Ω ∠+53.1º

Step 3 Split the impedance into its rectangular coordinates: Y = G − jB = Y cos ϕ − jY sin ϕ Therefore, and

G = Y cos ϕ = 20 cos 53.1° mS = 20 × 0.600 = 12 mS

BL = Y sin ϕ = 20 sin 53.1° = 20 × 0.800 = 16 mS

The rectangular coordinates of the impedance are the resistance and reactance of the equivalent series circuit, while the rectangular coordinates of the admittance are the conductance and susceptance of the equivalent parallel circuit. These “similar but opposite” characteristics of series and parallel AC circuits resemble those of series and parallel resistances in DC circuits (see Section 7-9). Table 21-1 compares the characteristics of series and parallel AC circuits.

21-8   Impedance and Admittance

617

TABLE 21-1 Series and parallel AC circuit characteristics Characteristic

Series Components

Parallel Components

IC

VL E Phasor diagram

VX

ϕ VR

VC

Sum of phasors Defining equation

Reference phasor

ϕ

IT

IR

IL

VT = V1 + V2 + V3 + . . . Z=

I

IX

Reference E phasor

IT = I1 + I2 + I3 + . . . Y=

VT I

IT V

XL

Z

Impedance diagram

Xeq

Not possible

ϕ

R

XC BC

Admittance diagram

Y

Not possible

Beq

ϕ

G

BL

Rectangular form

Z = R + jXeq

Y = G + jBeq

Resistive component

R (in ohms)

G (in siemens)

Reactive component

1 where XL = ωL and XC = ωc

where BC = ωC and BL =

Polar form

+j Xeq = +j ( XL − XC ) Z = Z ∠ϕ

Magnitude

Z = √ R2 + X2eq

Phase angle

ϕ = tan−1

Xeq R

See Problems 21-52 to 21-59 and Review Question 21-79.

+jBeq = +j ( BC − BL ) Y = Y ∠ϕ

Y = √ G2 − B2eq ϕ = tan−1

Beq G

1 ωL

618

Chapter 21  Impedance

Circuit Check

B

CC 21-4. Determine the total impedance of the network shown in Figure 21-14. L1 10 mH 2.0 V peak

R1 10 kΩ

+ 10 kHz

−

C1 10 μF

Figure 21-14

CC 21-5. For the circuit of Figure 21-15, determine the total admittance in polar coordinates.

+ 3 V peak 50 kHz − 0°

R 1 kΩ

C 0.68 μF

L 33 μH

Figure 21-15

CC 21-6. Determine the total impedance of the circuit shown in Figure 21-16. L2 10 mH C1 a

L1 10 mH

2.0 V peak

470 nF

+ 10 kHz

− b

C3 10 nF

Figure 21-16

470 nF C2

21-9  Troubleshooting

21-9 Troubleshooting The most common faults in circuits containing resistance, inductance, and capacitance are shorted and open components. The voltage-current relationships of resistors, inductors, and capacitors are a key to detecting these faults. These relationships may be stated as follows: • A current produces a voltage drop across a resistor. • A changing current (such as alternating current) produces a voltage across an inductor. • A changing voltage (such as alternating voltage) produces a current through a capacitor.

Testing a Series Circuit If the source voltage in the series AC circuit of Figure 21-9 is 500 V with a phase angle of 0°, the current is

I=

500 ∠0º E = 3.73 A ∠−41.7º = Z 134 ∠41.7º

Consequently, the resistor, inductor, and capacitor voltages are

VR = IR = 3.73 A × 100 Ω = 373 V

  VL = IXL = 3.73 A × 188.5 Ω = 703 V   VC = IXC = 3.73 A × 99.5 Ω = 371 V

If one of the components is open, an AC ammeter inserted into the circuit will show that no current is flowing. If the resistor is normal, VR = 0 V since I = 0 A. If the inductor is normal, VL = 0 V since the current is not changing. If the capacitor is normal, zero current means that VC is not changing, and therefore the AC voltage across the capacitor is also zero. When connected across the open component, an AC voltmeter will show a voltage drop equal to the source voltage (500 V). The reading for the other two components will be 0 V. However, if one of the components is fully shorted, an AC voltmeter connected across it will read 0 V, and the sum of voltage drops across the normal components will equal the source voltage. If one component is partially shorted, its voltage drop will be lower than calculated above, and the voltage drops across the other two components will be higher than the values calculated.

Testing a Parallel Circuit If the parallel AC circuit of Figure 21-11 is operating normally, the branch currents will be as calculated in Example 21-6. If the resistor, inductor, or

619

620

Chapter 21  Impedance

capacitor is open, an ammeter inserted in the branch with the open component will read zero current, while the currents in the other branches will be unchanged or will perhaps increase somewhat if the voltage source is not regulated. If a component is partially shorted, the current through it will be greater than normal. The currents through the other branches will either be unchanged or will decrease, depending on whether the increased load from the shorted component reduces the source voltage. If a component is fully shorted, the circuit will draw a very large current from the AC source, probably burning out a fuse or tripping a circuit breaker. We will then remove each component to test it, starting with any component that showed signs of overheating or other physical damage. We can use an ohmmeter to test the resistor and an LC meter to test the inductor and capacitor. See Review Questions 21-80 and 21-81.

Problems

621

Summary

• In a series AC circuit, the applied voltage equals the phasor sum of all the voltage drops. • Impedance is the total opposition to alternating current in an AC circuit. • The total impedance of a series AC circuit is the phasor sum of the resistance and the reactance. • A practical inductor consists of the resistance of the coil in series with its inductance. • In a series AC circuit, the net reactive voltage is the algebraic sum of the inductance and capacitance voltages. • In a series AC circuit, the net equivalent reactance is the algebraic sum of the inductive and capacitive reactances. • In a parallel AC circuit, the total current is the phasor sum of all the branch currents. • Admittance is the overall ability of an electric circuit to pass alternating current. • Susceptance is a measure of the ability of inductance or capacitance to pass alternating current. • Admittance is the reciprocal of impedance. • The rectangular coordinates of impedance are resistance and reactance. • The rectangular coordinates of admittance are conductance and susceptance. • The most common faults in AC circuits are open and shorted resistors, inductors, and capacitors.

Problems

Draw a circuit diagram for each problem. Also draw either a phasor diagram or an impedance diagram for each series circuit, and a phasor diagram or an admittance diagram for each parallel circuit.

I

Section 21-1  Resistance and Inductance in Series 21-1.

Calculate the magnitude of the unknown voltage in the AC circuits shown in Figure 21-17. 10 mV 10 V

E=?

VR = ?

16 mV 20 V

(a) Figure 21-17

(b)

B = beginner

I = intermediate

A = advanced

622

Chapter 21  Impedance

I

21-2.

B

21-3.

B

21-4.

B

21-5.

I

21-6.

I

21-7.

circuitSIM walkthrough

circuitSIM walkthrough

I I

A

circuitSIM walkthrough

I

A

circuitSIM walkthrough

(a) An inductor connected to a 120-V 60-Hz source has an IR drop of 72 V across the resistance of its winding. Find the magnitude and the angle of the voltage across its inductance with respect to the applied voltage. (b) Use Multisim to verify your calculations in part (a). The operating coil of a contactor is connected to a 240-V 60-Hz source and draws a 2-A current that lags behind the source voltage by 60°. Determine the impedance of the coil in: (a) polar coordinates (b) rectangular coordinates When a 400-Hz sine-wave voltage of 4.0 V ∠0º is applied to the voice coil of a loudspeaker, the resulting current is 0.50 A ∠−20º Express the impedance of the loudspeaker in both polar and rectangular ­coordinates. Determine the total impedance of a series circuit consisting of a 150-Ω resistor and a 10-mH inductor at a frequency of 750 Hz. Give your answer in polar coordinates. Two components in series have an impedance of 15 kΩ ∠45º at a frequency of 20 kHz. Determine the value of the two components. Determine the series equivalent circuit of a device that has an impedance of 5kΩ ∠+ 30º at a frequency of 200 Hz.

Section 21-3  Practical Inductors 21-8. (a) Find the resistance of a 4.0-H choke with a 750-Ω impedance at 25 Hz. Calculate the phase angle of the impedance. (b) Use Multisim to verify the resistance of the choke. 21-9. A certain choke draws a current of 1.2 A when connected to a 120-V 60-Hz source. When connected to a 12-V battery, it draws 0.5 A of current. Determine the resistance and inductance of the choke. 21-10. (a) Another choke draws a 5.0-A current when connected to a 117-V 25‑Hz source. When connected to a 120-V 60-Hz source, this choke draws a 3.0-A current. Determine the resistance and inductance of the choke. (b) Use Multisim to verify the resistance and inductance of the choke. 21-11. A GE 120-V 500-W lamp is operated on a 220-V 60-Hz system by connecting it in series with a choke with a resistance of 50 Ω. Determine the inductance the choke must have in order to provide the appropriate voltage to the lamp. 21-12. (a) A 15-W fluorescent lamp operates with a ballast inductance in series with it. When the total applied voltage is 120 V 60 Hz, the voltage across the lamp is 56 V RMS and the total voltage across the ballast inductor is 100 V RMS. Find the resistance and inductance of the ­ballast, given that the lamp has no reactance. (b) Use Multisim to verify the resistance and inductance of the ballast.

Problems

B

B

B

B B I I I B

I

I

623

Section 21-4  Resistance and Capacitance in Series 21-13. The voltages across a capacitor and a resistor connected in series in an AC circuit are measured as 60 V and 80 V RMS respectively. Find the magnitude and phase angle of the total voltage with respect to the common current. 21-14. (a) Find the magnitude and phase angle of the current that an impedance of 80 − j95 Ω draws from a 117-V AC source. (b) Use Multisim to verify the calculation of the magnitude and phase angle of the current. 21-15. The input circuit of a transceiver draws a current of 7 μA ∠17º from a 2.5-MHz signal generator when the source voltage is 350 μV ∠10º. Determine the impedance of the transceiver and express the result in polar coordinates. 21-16. Calculate the total impedance at 1.0 kHz of a 0.50-μF capacitor and a 500-Ω resistor in series. Give the answer in polar coordinates. 21-17. Find the impedance in polar coordinates of a 200-pF capacitor in ­series with a 1.2-kΩ resistor when the frequency of the applied voltage is 1.5-MHz. 21-18. At what frequency will a 20-nF capacitor and a 120-kΩ resistor in ­series have an impedance of 200 kΩ? 21-19. Determine the equivalent series circuit for a 4.5-MHz impedance of 500 Ω ∠−15º. 21-20. What value of capacitor must be connected in series with a 560-Ω ­resistor to limit the heat produced in the resistor to 5 W when the source voltage is 120 V at 60 Hz? 21-21. A 100-V 60-Hz source is connected in series with a 920-Ω resistor and a 47-uF capacitor. Determine: (a) the current of the circuit (b) the voltage across each component (c) the phase angle of the impedance 21-22. A heating element is rated 2.0 kW at 120 V. What capacitance could be connected in series with the element to power it from a 220-V 60-Hz supply?

Section 21-5 Resistance, Inductance, and Capacitance in Series

21-23. Determine the output voltage (magnitude and angle) of the components in the circuits shown in Figure 21-18. 4.5 V

80 mV

+

+ Vout = ?

10 V 0°

E

100 mV

−

− (a)

Figure 21-18

200 mV 0°

E

(b)

Vout = ?

circuitSIM walkthrough

624

Chapter 21  Impedance

I I

I B A

circuitSIM walkthrough

I

A

circuitSIM walkthrough

I A

I

21-24. Find the total 60-Hz impedance in polar form of a 10-μF capacitor in series with a 1.0-H choke having a resistance of 100 Ω. 21-25. For an RLC series circuit made up of a 820-Ω resistor, a 22-mH inductor and a 0.33-uF capacitor, determine the total impedance in polar coordinates when the frequency is (a) 250 Hz (b) 3 kHz (c) 15 kHz 21-26. At a frequency of 400 Hz, what value of capacitor must be connected in series with an impedance of 75 Ω ∠+60º to give a total impedance of 75 Ω ∠−60º? 21-27. Determine the total impedance of a series circuit formed by Z1 = 350 Ω ∠45º , Z2 = 720 Ω ∠−75º and Z3 = 500 + j120 Ω. 21-28. (a) A relay coil requires a 100-mA current through it in order to close the contacts of the relay. If it is operated from a dc source, the required voltage is 24 V. If the relay is operated from a 60-Hz source, the r­ equired voltage is 160 V. What capacitance in series with the relay coil will allow its operation from a 120-V 60-Hz supply? (Give two answers.) (b) Use Multisim to verify the capacitance. 21-29. At a frequency of 400 Hz, what value of capacitance must be connected in series with an impedance of 75 Ω ∠+60º to form (a) an inductive impedance of 50 Ω (b) a capacitive impedance of 50 Ω 21-30. A coil and capacitor are connected in series across a 400-Hz source. The voltage drops are 48 V across the coil, 52 V across the capacitor, and 6.0 V across the whole circuit. The current in the circuit is 100 mA. (a) Determine the inductance and resistance of the coil, and the capacitance of the capacitor. (b) Determine the total impedance in polar form. (c) Use Multisim to verify the resistance and inductance of the coil and the capacitance of the capacitor. 21-31. A 100-Ω resistor is connected in series with a 0.50-H inductor and a 220-V 60-Hz power supply. What size capacitor must added in series to give a current of 2.0 A? 21-32. An inductive circuit of 50 Ω resistance and 0.08 H inductance is connected in series with a capacitor across a 200-V 50-Hz supply. The current is 3.8 A leading. Find the value of the capacitor.

Section 21-6 Resistance, Inductance, and Capacitance in Parallel 21-33. Determine the magnitude and phase angle of the current with respect to the applied voltage in the ac circuits shown in Figure 21-19.

Problems

IT = ?

IT = 5.0 A

IL = ?

50 mA +

625

+ 750 mA

E −

E 3.75 A

− (a)

(b)

Figure 21-19

I

I I

I

B B

I

B B

21-34. (a) A given impedance draws a 5.0-A current from a 120-V 60-Hz source. The current through the impedance lags behind the voltage across it by 60°. Find the total current drawn from the source when a 100-mF capacitor is connected in parallel with the given impedance. (b) Use Multisim verify the total current drawn from the source. 21-35. Use the values of problem 21-27 to determine the equivalent total impedance of the circuit when all loads are connected in parallel. 21-36. A 75-W bulb is connected in parallel with a 68-μF capacitor and a 110-V 60-Hz ac source. (a) Determine the equivalent impedance in polar coordinates. (b) Determine the total current drawn from the source. 21-37. The bias circuit for one transistor in an audio amplifier has a 2.2-kΩ resistor and a 0.20-μF capacitor in parallel. Find the rectangular ­coordinates for the equivalent impedance of these components at 1.0 kHz. 21-38. A 200-Ω resistor, 0.50-H inductor, and 10-µF capacitor are connected in parallel to a 150-V 60-Hz source. Calculate the total current and the phase angle. 21-39. Find the equivalent impedance for the circuit in Problem 21-38.

circuitSIM walkthrough

Section 21-7  Conductance, Susceptance, and Admittance 21-40. (a) If the magnitudes of the applied voltages in the circuits shown in Figure 21-19 are 75 V RMS, calculate the admittance and impedance of each circuit in polar form. (b) Use Multisim to verify the admittance and impedance of each circuit. 21-41. Determine the total admittance and equivalent impedance in polar coordinates of a 20-mS conductance in parallel with a 75-mS inductive susceptance at a frequency of 50 Hz. 21-42. Calculate polar coordinates for the total admittance and equivalent impedance at 45 MHz of a 300-μS conductance in parallel with a 30-pF capacitor.

circuitSIM walkthrough

626

Chapter 21  Impedance

B B B I

B B B B I

I I I I I I I A

21-43. Calculate the 60-Hz admittance of a 0.50-H inductance in parallel with a 12-mS conductance. 21-44. Find the equivalent impedance of a device with an admittance of 0.050 − j0.015 S. Is this device inductive or capacitive? 21-45. What values of conductance and susceptance must be connected in parallel to form an admittance of 500 mS ∠−30º? 21-46. For an RLC parallel circuit made up of an 820-Ω resistor, a 22-mH inductor, and a 0.33-μF capacitor, determine the total admittance in polar form at frequencies of (a) 250 Hz (b) 3 kHz (c) 15 kHz 21-47. Calculate the polar coordinates for the equivalent impedances of the parallel components in Problem 21-46. 21-48. Find the polar coordinates for the total admittance of a circuit with 36 + j24 mS in parallel with 42 − j20 mS. 21-49. Find the equivalent impedance of an AC circuit consisting of two parallel branches with admittances of 64 μS ∠+15º and 48 μS ∠−45º. 21-50. Find the total admittance of a 1.0-kΩ resistor, 20-mH inductor, and 10-nF capacitor connected in parallel at a frequency of 10 kHz. 21-51. Find the equivalent impedance for the circuit in Problem 21-50 when the frequency is increased to 20 kHz.

Section 21-8  Impedance and Admittance 21-52. What values of resistance and capacitance in parallel have a 775-μS ∠55º admittance when the frequency applied is 15 kHz? 21-53. Determine the parallel equivalent circuit of a device that has an ­impedance of 2.4 kΩ ∠+30º at a frequency of 200 Hz. Compare this equivalent circuit with the answer to Problem 21-7. 21-54. Determine the parallel equivalent circuit at 4.5 MHz of an impedance of 150 Ω ∠−75º. Compare this equivalent circuit with the ­answer to Problem 21-19. 21-55. Determine the series components that have an equivalent admittance at 1 kHz of 100 + j55 mS. 21-56. Determine the series components that have an equivalent admittance at 25 MHz of 120 + j90 S. 21-57. What value of inductance must be connected in parallel with a 50-Ω resistor to form a total admittance at 60 Hz of 85 mS? 21-58. What value of resistance must be connected in parallel with an ­inductance of 1.5 H to produce a 400-Ω impedance at 60 Hz? 21-59. (a) Determine values for a set of parallel-connected components that has an admittance of 100 mS ∠−60º when connected to a 60-Hz source. (b) Determine values for a set of series-connected components that are equivalent to the parallel circuit in part (a).

Review Questions

Review Questions Section 21-2  Impedance

21-60. Why is the additional term impedance required for dealing with the opposition of a circuit to alternating current? 21-61. Why can we use the Pythagorean theorem to calculate the total impedance of a series circuit from its resistance and reactance? 21-62. Why is it necessary to state an angle as well as a magnitude when an impedance is given in polar form? 21-63. Why does a positive angle with an impedance always represent a circuit with inductive reactance? 21-64. How is the angle of an impedance related to the voltage across and the current through the impedance? 21-65. Why is it not necessary to state an angle when an impedance is given in rectangular coordinates?

Section 21-3  Practical Inductors

21-66. Why must the angle of the impedance of a practical inductor be less than +90°? 21-67. Explain why expressing the impedance of a practical inductor in ­rectangular coordinates gives the actual resistance and reactance of the inductor. 21-68. Describe how to determine the total impedance of a practical ­inductor and a resistor in series.

Section 21-5  Resistance, Inductance and Capacitance in Series 21-69. What is the significance of a total impedance having a 0° angle? 21-70. Under what circumstances does the total impedance decrease when a capacitor is added in series with a given impedance? 21-71. Under what circumstances is the magnitude of the RMS current drawn from a certain source unchanged when an inductance is connected in series with a given impedance? 21-72. Explain why an AC circuit containing both inductance and capacitance appears to the source as either an inductive circuit or a capacitive circuit. 21-73. What is meant by equivalent reactance?

Section 21-6  Resistance, Inductance, and Capacitance in Parallel 21-74. What is meant by the equivalent impedance of a parallel AC circuit? 21-75. Why is quadrant I the capacitive quadrant of a parallel-circuit phasor diagram, whereas quadrant IV is the capacitive quadrant of a series-circuit phasor diagram?

Section 21-7  Conductance, Susceptance, and Admittance 21-76. Why is it impossible to draw an impedance diagram for a parallel circuit?

627

628

Chapter 21  Impedance

21-77. What is the advantage of thinking in terms of admittance, conductance, and susceptance in dealing with parallel AC circuits? 21-78. Why does BL have a −j operator for phasor addition, whereas XL has a +j operator?

Section 21-8  Impedance and Admittance

21-79. Explain why Z = 20 + j40 Ω is not the equivalent impedance of Y = 50 − j25 mS.

Section 21-9  Troubleshooting 21-80. Describe a troubleshooting technique for detecting a shorted component in the circuit of Figure 21-20(a). IT

R

IR

I jXL

E

E

IL

IC jXL

R

−jXC

−jXC

(a)

(b)

Figure 21-20

21-81. Describe a troubleshooting technique for detecting an open component in the circuit of Figure 21-20(b).

Integrate the Concepts

In the circuits of Figure 21-20, the components are a 75-Ω resistor, a 9.1-mH inductor, and a 0.33-μF capacitor. The applied voltage is 240 V RMS with a phase angle of 10° and a frequency of 2.0 kHz. (a) Calculate the total impedance for the circuit in Figure 21-20(a). (b) Draw the impedance diagram for the circuit in Figure 21-20(a). (c) Determine the current in Figure 21-20(a). (d) Determine IR, IL , and IC for the circuit in Figure 21-20(b). (e) Draw the phasor diagram for the circuit in Figure 21-20(b). (f) Determine the total current drawn for the circuit of Figure 21-20(b).

Practice Quiz

1. Which of the following statements is true? (a) Impedance is the opposition to current in an AC circuit and has units of ohms. (b) Admittance is the ability of an electric circuit to pass AC current and has units of ohms.

Practice Quiz

2. The phase angle of the impedance of the circuit shown in Figure 21-21 is (a) 21.8° (b) −11.8° (c) −21.8° (d) 11.8° R1 3 kΩ XL 1.2 kΩ

+

25 VRMS 0° −

Figure 21-21

3. For what value of XL will the the phase angle of the impedance in Figure 21-21 become 57°? 4. The current flowing through the resistor in Figure 21-21 is (a) 7.74 mA ∠−21.8º (b) 7.74 mA ∠11.8º (c) 7.74 mA ∠−11.8º (d) 7.74 mA ∠21.8º 5. The voltage drop across the inductor of Figure 21-21 is (a) 9.29 V ∠−78.2º (b) 9.29 V ∠−68.2º (c) 9.29 V ∠78.2º (d) 9.29 V ∠68.2º 6. The total impedance of the circuit of Figure 21-22 is (a) 10.8 kΩ ∠−63.6º (b) 10.9 kΩ ∠63.8º (c) 10.9 kΩ ∠−63.8º (d) 10.8 kΩ ∠63.6º R1 2.2 kΩ L 10 VRMS 0°

+ −

Figure 21-22

R

Coil 10 kΩ 75°

629

630

Chapter 21  Impedance

7. The phase angle of the circuit shown in Figure 21-23 is (a) 84.3° (b) −84.3° (c) 5.7° (d) −5.7° R1 2.2 kΩ

+ X C 22 kΩ

+ 10 VRMS 45° −

Figure 21-23

8. If the value of XC in Figure 21-23 is changed to 2.2 kΩ, the phase angle of the circuit becomes 45°. True or false? 9. The total reactance of the circuit shown in Figure 21-24 is (a) j19.5 kΩ (b) j19.8 kΩ (c) −j19.5 kΩ (d) −j19.8 kΩ R1 2.2 kΩ 10 VRMS 45°

+

+

XC 22 kΩ

− XL 2.5 kΩ

Figure 21-24

10. The impedance of the circuit of Figure 21-24 is (a) 19.6 kΩ ∠−83.6º (b) 29.4 kΩ ∠−41.6º (c) 19.6 kΩ ∠83.6º (d) 29.4 kΩ ∠41.6º 11. The current flowing through the circuit of Figure 21-24 is (a) 340 μA ∠−38.6º (b) 510 μA ∠128.6º (c) 340 μA ∠128.6º (d) 510 μA ∠−30.6º

Practice Quiz

12. The voltage drop across the inductor of Figure 21-24 is (a) 1.27 V ∠38.6º (b) 11.2 V ∠38.6º (c) 11.2 V ∠−141.4º (d) 1.27 V ∠−141.4º 13. The impedance of the circuit of Figure 21-25 is (a) 98.97 Ω ∠+8.61º (b) 108.77 Ω ∠+7.83º (c) 98.97 Ω ∠−8.61º (d) 108.77 Ω ∠−7.83º R1 10 Ω 20 VRMS 60°

+

R2 100 Ω

XL 220 Ω

XC 330 Ω

−

Figure 21-25

14. The total current flowing through the circuit of Figure 21-25 is (a) 183.87 mA ∠−7.83º (b) 183.87 mA ∠−52.17º (c) 183.87 mA ∠7.83º (d) 183.87 mA ∠52.17º 15. The voltage across R2 in the circuit of Figure 21-25 is (a) 18.18 V ∠43.55º (b) 18.18 V ∠0.79º (c) 18.18 V ∠60.79º (d) 18.18 V ∠−16.45º

16. The current flowing through the inductor of Figure 21-25 is (a) 82.64 mA ∠−29.21º (b) 82.64 mA ∠−46.45º (c) 82.64 mA ∠29.21º (d) 82.64 mA ∠133.55º 17. The susceptance of the parallel portion in the circuit of Figure 21-25 is (a) 9.19 mS ∠7.83º (b) 10.11 mS ∠−8.62º (c) 10.11 mS ∠8.62º (d) 9.19 mS ∠−7.83º

631

22

Power in AlternatingCurrent Circuits

Most of the examples in this chapter deal with power in AC circuits at the standard North American power-line f­ requency of 60 Hz. Recall that power is the rate of doing work or converting energy from one form to another. Much of the terminology in this chapter has come from power-system applications.

Chapter Outline 22-1 Power in a Resistor  634

22-2 Power in an Ideal Inductor  635 22-3 Power in a Capacitor  637

22-4 Power in a Circuit Containing Resistance and Reactance 639

22-5 The Power Triangle  641 22-6 Power Factor  645

22-7 Power-Factor Correction  648

Key Terms real power  634 active power  634 reactive power  637 quadrature (or imaginary) power 637 volt ampere reactive  637

apparent power (S) 640 volt ampere  640 power triangle  642 phasor power  643 power factor  645 power-factor angle  645

lagging power factor  645 leading power factor  645 reactive factor  647 power-factor correction  651 power-factor improvement 651

Learning Outcomes At the conclusion of this chapter, you will be able to: • draw graphs of instantaneous voltage, ­current, and power for a resistor, ideal inductor, or capacitor ­connected to an AC circuit • calculate the real power in a circuit with an AC source • explain why the average power in an ideal inductor or capacitor in an AC circuit is zero • calculate the reactive power in a circuit ­containing ­resistance, inductive reactance, and capacitive reactance • draw graphs of instantaneous voltage, current, and power in an AC circuit containing resistance and i­nductive reactance • calculate the apparent power supplied by an AC source to a circuit containing resistance,

Photo sources:  CP Photo/Graham Hughes

•

•

• • •

inductive reactance, and capacitive reactance draw the power triangle for an AC circuit ­containing ­resistance, inductive reactance, and capacitive reactance calculate the power factor of an AC circuit ­containing ­resistance, inductive reactance, and capacitive reactance differentiate between lagging and leading power factors explain why increasing the power factor can improve ­circuit performance calculate the capacitance needed to increase the power factor of a circuit to a specified value

634

Chapter 22   Power in Alternating-Current Circuits

22-1  Power in a Resistor At any particular instant, all the power equations we derived for DC ­cir­cuits must apply to AC circuits. Therefore, p = vi



(22-1)

Since the voltage across a resistor and the current through it are in phase, v is always positive when i is positive and negative when i is negative. Therefore the instantaneous power in a resistor is never negative (see Figure 22-1). As shown in Section 18-8, the instantaneous power in a resistor in an AC circuit is a sine wave that has a mean magnitude of 1/2Pm and ­fluctuates at twice the frequency of the alternating current and voltage. Section 18-11 used this mean value of power in a resistor as the basis for the effective, or RMS, values of sine waves of voltage and current. It follows that the mean power in the resistance of an AC circuit is simply the product of the RMS value of the voltage drop across the resistance and the RMS value of the current through it: P = VRIR 

Instantaneous Value



(22-2)

p

+

0 π 2

π

i

3π 2 v

2π

ωt

−

Real power is also called true power because a resistance converts electric energy into heat and light that dissipates from the circuit, while the energy conversions in inductance and ­capacitance in AC ­circuits do not transfer any energy from the circuit.

Figure 22-1  Instantaneous power in a resistor

The average power input to the resistance of an AC circuit is referred to as real power or active power. The symbol for mean power in an AC ­circuit is P, the same as for power in a DC circuit. With RMS values, R = VR / IR for AC circuits, just as for DC circuits. T ­ herefore,

P = VRIR = IR2R =

VR2  R

(22-3)

where VR is the RMS voltage across the resistance and IR is the RMS ­current through only the resistance portion of the circuit.

22-2   Power in an Ideal Inductor

Example 22-1 Find the resistance that gives off 144 W of heat when the current through the resistance is 2.0 A RMS. Solution

R=



P 144 W = 36 Ω 2 = ( I 2.0 A ) 2

See Problems 22-1 to 22-4 and Review Questions 22-38 to 22-41 at the end of the chapter.

22-2  Power in an Ideal Inductor

Instantaneous Values

We can determine the nature of the instantaneous power in an ideal inductor by plotting the product of the instantaneous voltage and current versus time (see Figure 22-2). As explained in Section 19-2, the instantaneous current in an ideal inductor lags behind the voltage across it by π/ 2 radians. p

+ i 0 π 2

π

2π 3π 2 v

ωt

− Figure 22-2  Instantaneous power in an ideal inductor

When ωt = 0, the instantaneous voltage across the inductance of the circuit is zero, and the rate of change of current is also zero. Between 0 and π/2 rad, the instantaneous voltage is increasing. To generate this voltage, the rate of change of current is also increasing as the instantaneous current i­ncreases from its minimum value to zero. During this interval the instantaneous current is actually flowing against the applied voltage as the magnetic field around the inductor collapses, thus returning the energy stored in its magnetic field to the source. Since v is positive and i is negative, the instantaneous power, vi, is negative, indicating that energy is leaving the ­inductance. At π/2 rad, the current momentarily becomes zero, so the ­instantaneous power is also momentarily zero. Between π/2 and π rad, the instantaneous voltage across the inductor is positive but decreasing. The instantaneous current is positive with a positive rate of change that induces the positive voltage in the inductance. The product vi is now positive, and energy is stored in the magnetic field of the inductor. At π radians the instantaneous voltage and the instantaneous power momentarily become zero.

635

636

Chapter 22   Power in Alternating-Current Circuits

The instantaneous power again becomes negative between π and 3π/ 2 rad when the instantaneous current is decreasing, and the collapsing magnetic field returns energy to the system. Between 3π/ 2 and 2π rad, both v and i are negative, so vi is positive as energy is again transferred to the magnetic field of the inductor. As shown by the green curve in Figure 22-2, the instantaneous power in the inductance varies sinusoidally at twice the frequency of the instantaneous voltage and current, like the power in the resistance of an AC circuit. But the instantaneous power in the inductance fluctuates symmetrically between positive and negative values as the ideal inductor alternately stores and returns energy. Therefore, the average power in the inductance of an AC circuit is zero. From Equation 19-2, iL = Im sin (ωt − π/2). Substituting for v and i in Equation 22-1 gives p = v × i = Vm sin ωt × Im sin ωt −

(



Since sin(θ − 90°) = −cos θ and sin θ cos θ = 12 sin 2θ,

π 2

)

p = −Vmlm sin ωt cos ωt p=

and

1 VmIm sin 2ωt 2

(22-4)

Hence, the instantaneous power in an ideal inductor is a sine wave that has twice the frequency of the instantaneous voltage and current. At the beginning of each cycle the power is zero and then decreases to a negative value. Over a complete cycle, the average of sin 2ωt is zero, confirming that the average power in an ideal inductor in an AC circuit is zero. In Figure 22-3, the meters show the RMS values of the voltage across and the current through an ideal inductor. For the resistance of an AC circuit, the average power is VRIR. However, for the inductance, VLIL is the average amplitude of the power transferred to and from the magnetic field of the inductance.

A E

V

L

Figure 22-3 Measuring VL and IL

22-3   Power in a Capacitor

The product of the RMS voltage and current in an ideal inductor is called the reactive power, quadrature power, or imaginary power of the inductor. The symbol for reactive power is Q. Q = VLIL



(22-5)



To help distinguish reactive power from real power, the watt is not used as the unit for reactive power. The unit of reactive power is the volt ampere reactive (var). XL =

Since

VL (19-3) IL

Q = VLIL = I2LXL =



VL2   (22-6) XL

where Q is the reactive power in vars, VL is the RMS voltage across the inductance in volts, and IL is the RMS current through the inductance in ­amperes.

Example 22-2 Find the reactive power of an ideal 0.50-H inductor that draws 0.50 A from a 60-Hz source. Solution

XL = ωL = 377 × 0.50 H = 188.5 Ω (inductive) Q = I2LXL = ( 0.50 A ) 2 × 188.5 Ω = 47 var

See Problems 22-5 to 22-7 and Review Questions 22-42 and 22-43.

22-3  Power in a Capacitor

The instantaneous current in a capacitor leads the instantaneous voltage across it by π/2 rad. As with resistance and inductance, the instantaneous power for capacitance in an AC circuit is a sine wave with twice the frequency of the instantaneous voltage and current (see Figure 22-4). During the first quarter-cycle (from ωt = 0 to ωt = π/2), the potential difference across the capacitor is rising and the instantaneous power is positive. During this interval, the capacitor takes energy from the source and builds up a charge on its plates. During the next quarter cycle, the voltage across the capacitor decreases and the capacitor discharges the stored energy

637

Chapter 22   Power in Alternating-Current Circuits

Instantaneous Values

638

p

+ v

i

0 π 2

π

3π 2

2π

ωt

− Figure 22-4  Instantaneous power in a capacitor

back into the ­system. Therefore, the instantaneous power is now negative. ­Similarly, the power is positive during the third quarter of the cycle and negative during the last quarter. Because an ideal capacitor must discharge as much energy in the second and fourth quarter-cycles as it stores in the first and third quarter-cycles, the average power in the capacitance of an AC circuit is zero. Substituting for v and i in Equation 22-1 gives p=



1 VmIm sin 2ωt  2

(22-7)

As with inductance, the product of the RMS voltage across the capacitance and the RMS current in the capacitance is not the average power in the capacitance. The product of the RMS voltage and current in a capacitor is the ­reactive power of the capacitor. Q = VCIC = I2C XC =



VC2 XC

(22-8)

Example 22-3 Find the capacitance that has a reactive power of 50 var when connected across a 120-V 60-Hz source. Solution Since Q = VC2/XC,

XC =

Since XC = 1 /ωC, C=

( 120 V ) 2 VC2 = = 288 Ω ( capacitive ) Q 50 var

1 1 = = 9.2 × 10 − 6 F = 9.2 µF ωXC 377 rad/s × 288 Ω

See Problems 22-8 and 22-9 and Review Question 22-44.

22-4   Power in a Circuit Containing Resistance and Reactance

Circuit Check

CC 22-1. An AC current signal is given by the equation i = 250 sin 377t mA and it flows through a 75-Ω resistor. Determine the real power delivered to the resistor. CC 22-2. A 10-µF and a 33-µF capacitor are connected in series across a ­120-V 60-Hz source. Calculate the total reactive power ­delivered to the circuit.

A

22-4 Power in a Circuit Containing Resistance and Reactance If the series AC circuit in Figure 22-5 has equal resistance and inductive reactance, the instantaneous current lags behind the applied instantaneous voltage by π/4 radians (45°). As shown by the green curve in F ­ igure 22-6, the instantaneous power is negative from 0 to π/4 rad since the inductance is returning more energy to the system than the resistance is converting to heat. For the interval from π/4 rad to π rad, the instantaneous power is positive. At the peak of the instantaneous-power wave, both the resistance and the inductance are taking energy from the source. A R E

E

V L

VL 45° VR

(a)

I

(b)

Instantaneous Values

Figure 22-5  Ac circuit with resistance and inductance in series p

+ i 0 π 4

π 5π 4 v

2π

ωt

− Figure 22-6  Instantaneous power in an AC circuit with resistance and inductance

The instantaneous-power graph for a circuit containing both resistance and reactance is again a sine wave with twice the frequency of the voltage and current. In Figure 22-6, the instantaneous-power graph is neither a­ lways

639

640

Chapter 22   Power in Alternating-Current Circuits

positive (as it would be for just resistance) nor symmetrically positive and negative (as it would be for just reactance). The average power is p ­ ositive, and it corresponds to the real power in the resistance of the ­circuit. Over a complete cycle, the average of the reactive power is zero. At any given m ­ oment, p = VmIm sin ωt sin ( ωt + ϕ ) 



(22-9)

where ϕ is the phase angle of the current with respect to the applied voltage. Since sin (a + b) = sin a cos b + cos a sin b,

p = VmIm sin ωt(sin ωt cos ϕ + cos ωt sin ϕ)

= cos ϕ (VmIm sin2 ωt) + sin ϕ (VmIm sin ωt cos ωt)

The expressions in parentheses are the instantaneous real power in the ­resistance and the instantaneous reactive power in the reactance of the circuit, respectively. Therefore, p = pr cos ϕ + pX sin ϕ



 (22-10)

where pr is the instantaneous real power in the resistance of the circuit and pX is the instantaneous reactive power in the reactance of the circuit.  In the example above, the product of total RMS voltage and total RMS ­current represents neither real power nor reactive power. If we know only the voltmeter and ammeter readings in Figure 22-5, we cannot determine the real and reactive components of the power. The product of the total RMS voltage and the total RMS current in an AC circuit is called the apparent power of the circuit. The symbol for apparent power is S. S = VTIT 



(22-11)

At this point, we can talk only about the magnitude of the apparent power. Because apparent power is neither real power in watts nor reactive power in vars, apparent power is always expressed simply as the product of volts and amperes. The volt ampere (VA) is the unit of apparent power. Since



VT =Z IT S = VTIT = I2TZ =

(21-3) V2T Z

(22-12)

22-5   The Power Triangle

Example 22-4 A radio transmitter supplies a current of 8.0 A to an antenna system with an impedance of 500 Ω ∠+ 20º. Find the apparent power in this system. Solution

S = I 2 Z = (8.0 A)2 × 500 Ω = 32 × 103 VA = 32 kVA

See Problems 22-10 to 22-14 and Review Questions 22-45 and 22-46.

22-5  The Power Triangle The real power in the circuit of Figure 22-5 can be found from the product of VR and I, and the reactive power from the product of VL and I. Since VL leads VR by 90°, VT = √V2R + V2L



(21-2)

Multiplying each side of Equation 21-2 by the common current gives VTI = √( VRI ) 2 + ( VLI ) 2



Therefore,

S = √ P2 + Q2

(22-13)

Similarly, in the parallel circuit of Figure 22-7, the current in the capacitor branch leads the current in the resistor branch by exactly 90°. For this ­circuit, Equation 21-18 simplifies to IT = √ IR2 + IC2



Multiplying each side by the common voltage gives

Therefore,

VIT = √ ( VIR ) 2 + ( VIC ) 2 S = √ P2 + Q2 

(22-13)

641

642

Chapter 22   Power in Alternating-Current Circuits

IT E

R

C

IC

ϕ

IR

E

(b)

(a)

Figure 22-7  AC circuit with resistance and capacitance in parallel

Note that Equation 22-13 applies to both series and parallel circuits. Since power is the product of current (the common phasor of series c­ ircuits) and voltage (the common phasor of parallel circuits), the same equations for power apply to series and parallel circuits as well as to series-parallel circuits. Section 22-7 shows how these equations provide a simple method of solving series-parallel circuits without having to reduce them to equivalent series or parallel circuits. Although instantaneous power varies sinusoidally, it does so at twice the frequency of the voltage and current. Consequently, we cannot draw power phasors on the same phasor diagram with voltage and current. Therefore, it is customary to represent the Pythagorean relationship of Equation 22-13 with right triangles, as shown in Figure 22-8. In a power triangle, the horizontal side represents real power. The vertical side represents reactive power and forms a right angle at the right-hand end of the side representing real power. Since Equation 22-13 applies to either a series or a parallel circuit, we can develop a power triangle from either an impedance diagram or an admittance diagram. If we start with an impedance diagram, in which inductive reactance is drawn in the +j direction, the power triangle appears as in Figure 22-8(a). But if we start from an ­admittance diagram, in which inductive susceptance is drawn in the −j direction, the power triangle has the form shown in F ­ igure 22-8(b). To avoid confusion between inductive and capacitive r­eactive power in power triangle ­diagrams, we must choose one of these formats for all AC circuits. We will follow the modern North American convention of showing inductive ­reactive power in the +j direction. With this convention, the source current is the reference phasor when we are dealing with power in AC circuits.

A) r (V

t ren

pa

Ap

e pow

ϕ

Inductive reactive power (vars)

Real power (watts) ϕ Ap par ent pow er (VA )

Inductive reactive power (vars)

Real power (watts) (a)

(b)

Figure 22-8  Power triangle with (a) common current as reference axis and (b) voltage as reference axis

22-5   The Power Triangle

We can now determine the phase angle, ϕ, of the apparent power. With the power triangle drawn as shown in Figure 22-8(a), S has the same angle as the impedance of the circuit and as the conjugate of the admittance of the circuit (in other words, the angle of the admittance is −ϕ). When S is treated as a complex number, it is called the phasor power of the circuit. Thus, S = P + jQ 



(22-14)

We can now see where the terms real power for active power and quadrature (or imaginary) power for reactive power originated. Since inductive reactive power has a +j operator, capacitive reactive power has a −j operator. Since the voltage across the capacitance in a series circuit is 180° out of phase with the voltage across the inductance, the net reactive voltage is the difference between the two reactive voltages. Similarly, in a simple parallel circuit, the net reactive current is the difference between the capacitive and inductive branch currents. Since Q = VXIX, the net reactive power in both series and parallel AC circuits is the difference between the inductive reactive power and the capacitive reactive power. In Figure 22-2 the instantaneous power in an inductance is positive when the current is rising and building up a magnetic field around the i­ nductor. But Figure 22-4 shows that, when the current is rising, the capacitance is discharging its stored energy back into the system. Therefore, in an AC circuit containing both inductance and capacitance, the capacitance always returns energy to the circuit when the inductance takes energy from the circuit, and vice versa. Consequently, some energy is transferred back and forth between the inductance and the capacitance of the circuit, and the net reactive power to and from the source is the difference between the inductive reactive power and the capacitive reactive power.

Example 22-5 A circuit has two branches connected in parallel to a 120-V 60-Hz source. Branch 1 consists of a 75-Ω resistance and a 100-Ω inductive reactance in series, and branch 2 consists of a 200-Ω capacitive reactance. Determine the apparent power of the circuit. Solution I Step 1 Draw a circuit diagram as shown in Figure 22-9. Then calculate the current in branch 1: Z1 = R + jXL = 75 + j100 = 125 Ω ∠+53.1º



I1 =

E 120 V = = 0.960 A Z1 125 Ω

643

Chapter 22   Power in Alternating-Current Circuits

Source: Courtesy of Global Specialties

644

Inductive Q XL

ϕ

100 Ω

120 V 60 HZ

A wattmeter for measuring power in an AC circuit

XC R

200 Ω

75 Ω

S

92.2 var Net Q

P = 69.1 W 72 var

Capacitive Q

(a)

(b)

Figure 22-9  Diagrams for Example 22-5

Step 2 Draw a power triangle and calculate the net reactive power. Real power of the circuit is P = I21R = 0.960 2 × 75 = 69.1 W



Reactive power of the inductance is

Q = I21XL = 0.960 2 × 100 = 92.2 var



Reactive power of the capacitance is Q=



E2 1202 = 72.0 var = XC 200

Net reactive power of the circuit is

Q = 92.2 − 72.0 = 20.2 var (inductive) Step 3 Calculate the apparent power:

S = √P2 + Q2 = √69.12 + 20.22 = 72 VA

Solution II A somewhat similar procedure involves solving Steps 2 and 3 in terms of current rather than power. Step 1

Z1 = R + jXL = 75 + j100 = 125 Ω ∠+ 53.1° I1 =

120∠0° E = = 0.960 A ∠−53.1° Z1 125∠+53.1°

I1 = 0.58 − j0.77 A

22-6   Power Factor

Step 2 and Step 3

IC =

120 ∠0º E = = 0.60 A ∠+90° = 0 + j0.6 A −jXC 200 ∠−90°

IT = I1 + IC = ( 0.58 − j0.77 ) + ( 0 + j0.60 ) = 0.58 − j0.17 A IT = √0.582 + 0.172 = 0.60 A

S = ETIT = 120 × 0.60 = 72 VA

To verify the apparent power of the circuit, download Multisim file EX22-5 from the website and follow the instructions in the file. See Problems 22-15 and 22-16 and Review Questions 22-47 to 22-52.

22-6  Power Factor Since the hypotenuse of a right triangle must be longer than either of the other two sides, the apparent power that a generator must supply to a reactive load is always greater than the real power that the load can convert into some other form of energy. This relationship can be quite important since most industrial loads have appreciable inductive reactance. The power factor is the ratio between the real power and the apparent power of a load in an AC circuit. The power triangle of Figure 22-8(a) shows that the ratio of real power to apparent power is the cosine of the power-factor angle between real power and apparent power. Tracing the construction of the power triangle back through the impedance diagram to the phasor diagram for current and voltage, we find that the power-factor angle is the same as the phase angle, ϕ, between the voltage across and the current through the load. From the definition of power factor,

645

power factor =

P = cos ϕ S

(22-15)

We distinguish between inductive and capacitive loads by stating that inductive loads always have a lagging power factor because their currents lag behind their voltages, and that capacitive loads always have a leading power factor because their currents lead their voltages. Since the real power cannot be greater than the apparent power, the power factor cannot be greater than 1. It can be expressed as either a decimal fraction or a ­percentage.

circuitSIM walkthrough

646

Chapter 22   Power in Alternating-Current Circuits

Example 22-6 Find the real power and power factor of a load whose impedance is 60 Ω ∠+60° when connected to a 120-V 60-Hz source.

Solution I Step 1 Convert the impedance to rectangular coordinates to find an equivalent circuit consisting of resistance and reactance in series, as shown in Figure 22-10. Z = 60 cos 60° + j60 sin 60° Therefore,

120 V 60 HZ

R = 60 cos 60° = 60 × 0.500 = 30 Ω

60 Ω 60° Z

120 V 60 HZ

XL = 60 sin 60°

R = 60 cos 60°

Figure 22-10  Schematic diagram for Example 22-6

Step 2

I=

120 V E = = 2.0 A Z 60 Ω

P = I2R = 4.0 × 30 = 120 W

Step 3 S = I2Z = 4.0 × 60 = 240 VA

P 120 = = 0.50 lagging S 240 Although Solution I uses the definition of power factor, there is a more ­direct way to solve this example.

cos ϕ =

Solution II Step 1 As shown in Figure 22-11, Step 2

power factor = cos ϕ = cos 60° = 0.50 lagging S=

E2 1202 = = 240 VA Z 60

S=

240 VA

22-6   Power Factor

647

Q

60° P = 120 W

Figure 22-11  Power triangle for Example 22-6

Step 3 Since cos ϕ = P/S,

P = S cos ϕ = 240 × 0.50 = 0.12 kW

Multisim Solution To verify the real power and power factor of the load, download Multisim file EX22-6 from the website and follow the instructions in the file.

As shown in the second solution, we can express real power in terms of power factor and apparent power. Since S = EI, the real power in an AC ­circuit is P = EI cos ϕ 



(22-16)

Once we know the real power and apparent power of an AC load, we can determine reactive power: Q = √ S2 − P2 

(22-13)

Q = S√ 1 − cos2ϕ 

(22-17)

Substituting P = S cos ϕ gives

The ratio of reactive power to apparent power can be useful when we wish to solve directly for the reactive power of the load. The reactive factor is the ratio of the reactive power to the apparent power of an AC load.

circuitSIM walkthrough

648

Chapter 22   Power in Alternating-Current Circuits

From the power triangle in Figure 22-8(a) we can see that reactive factor =



Q = sin ϕ  S

(22-18)

See Problems 22-17 to 22-29 and Review Questions 22-53 to 22-56.

22-7  Power-Factor Correction The consequences of a low power factor in an industrial load are illustrated by Figure 22-12. Since the motor in Figure 22-12(a) has a 70% lagging power factor, the apparent power is S=



840 P = = 1.2 kVA cos ϕ 0.70

Therefore, the current drain on the source is I=



S 1.2 kVA = = 10 A V 120 V

10 A

7A

A

A

840 W M 70% lagging power factor

120 V 60 HZ

(a)

840 W M 100% power factor

120 V 60 HZ

(b)

Figure 22-12  Effect of load power factor

If we replace the original motor by one capable of developing a real power of 840 W at a 100% power factor, as in Figure 22-12(b), the apparent power is reduced by 840 VA and the current is only 7.0 A. The motor with a 70% power factor requires a current of 10 A to do the same work that the motor with a 100% power factor can do while drawing 7 A from the source. The size of the wires required for the windings of a generator and for the conductors connecting the load to the generator depends on the current the wires have to carry. The energy lost as heat in the wires depends on the square of the current. Therefore it is more efficient and more economical to feed power to a load with a 100% power-factor load than to a load with a feed of a 70% power factor. Electrical utility companies are very much concerned with maintaining a high overall power factor in their systems.

22-7   Power-Factor Correction

The additional current for the motor with a 70% lagging power alternately transfers energy from the system into the magnetic field of the inductance and then returns the stored energy back to the system. Although the average power of this process is zero, the reactive current component still has to flow through the windings and connecting conductors. Thus the reactive current increases the size of wire needed to deliver a given real power to the load. We can determine the reactive current components by expressing the current of each of the motors in rectangular coordinates. In the motor with a 70% lagging power factor, the current lags behind the applied voltage by cos−1 0.70 = 45.6°, and I = 10 cos 45.6° − j10 sin 45.6° = 7.0 − j 7.14 A

In a motor with a 100% power factor, the current is in phase with the ­applied voltage: I = 7.0 − j 0 A

The current in each motor has the same reference axis component. This component carries the real power to the load. The 70% power-factor load has an additional quadrature component that carries the reactive power. In an AC circuit containing both inductance and capacitance, the capacitance returns energy to the system while the inductance takes energy from the system, and vice versa. If a capacitor is connected to the motor with a 70% lagging power factor, the reactive power can flow back and forth ­between the capacitor and the inductance of the motor, thus reducing or eliminating the reactive component of the current that travels all the way from the source to the load and back. This arrangement would allow us to use lower power-factor loads yet obtain the advantages of a high system power factor. Consider the effect of connecting a 308-μF capacitor in series with the motor that has a 70% lagging power factor, as shown in Figure 22-13. The impedance of the motor is ZM = V/ I =

120 = 12 Ω 10

ϕ = cos−1 0.7 = 45.6°



ZM = 12 Ω ∠+45.6° = 12 cos 45.6°+ j12 sin 45.6° = 8.4 + j8.6 Ω XC =

1 1 = = 8.6 Ω ( capacitive ) ωC 377 × 308 µF

649

650

Chapter 22   Power in Alternating-Current Circuits

Practical Circuits Power-Factor Correction

Source:  CP Photo/Graham Hughes

The power factor is a measure of how effectively devices operated by a consumer convert electric current from the local utility to useful power output, such as heat, light, or mechanical motion. Power-factor correction is the practice of raising the power factor of an inductive load, such as a motor, by inserting a capacitor in parallel with it. The lower the power factor, the greater the current required to deliver a given amount of energy. Electricity suppliers usually bill commercial and industrial customers a power factor surcharge if the power factor of the customer’s load drops below 90%. The power supplied to these customers is measured with a meter that monitors both the usable power and the reactive power, and hence can indicate the power factor each month.

Power meter

14.3 A A

120 V 60 HZ

M

70% lagging power factor 308 μF

Figure 22-13  Effect of a series capacitor on a lagging power-factor load

Because the motor and capacitor are in series,

ZT = 8.4 + j8.6 − j8.6 = 8.4 + j0 Ω = 8.4 Ω ∠0º

22-7   Power-Factor Correction

Since the phase angle of the total impedance is now 0°, the current drawn from the generator is in phase with the source voltage. However, I=



V 120 V = 14.3 A = ZT 8.4 Ω

Therefore, connecting a capacitor in series with an inductive load reduces the net reactive power but increases the current drain on the source. Moreover, since the impedance of the motor itself is still 12 Ω, the voltage across the motor in Figure 22-13 becomes VM = IZM = 14.3 A × 12 Ω = 172 V, or 0.17 kV Although connecting the capacitor in series with the motor raises the system power factor to 100%, this arrangement aggravates the original problem by increasing the current drawn from the source and applying excessive voltage to the motor. To be practical, power-factor correction or improvement must reduce the apparent power drawn from the source without altering the voltage across the load. One of the characteristics of a parallel circuit is that a change in one branch does not affect the current in the other branches. In Figure 22-14, a capacitor is connected in parallel with the motor that has a 70% lagging power factor. Since the motor is still connected directly across the 120-V source, the motor current is still 10 A, lagging behind the applied voltage by 45.6°. Because of the lagging power factor, the motor has a reactive power of Q = VI sin ϕ = 120 × 10 × 0.714 = 860 var 7A

7.14 A

A

A A 10 A

120 V 60 HZ

840 W M 70% lagging power factor

Figure 22-14  Power-factor correction

158 μF

651

652

Chapter 22   Power in Alternating-Current Circuits

If the capacitor in parallel with the motor also has a reactive power of 860 var, all of the reactive power is transferred back and forth between the capacitor and the motor, and the net reactive power measured at the source is zero. Since the voltage across and the current through the motor are unchanged, the real power is still 840 W. Therefore the overall apparent power that the source must supply is 840 VA, and the current drawn from the generator is 7.0 A. Note that the motor current is still 10 A. As far as the source is concerned, the motor and capacitor in parallel are equivalent to a motor with a power factor of 100%. We can find the required capacitance by noting that the capacitor ­current must equal the reactive component of the load current or that the susceptance of the capacitor must equal the susceptance component of the load admittance. However, the simplest procedure is to work in terms of reactive power since the power equations apply to both series and parallel circuits. To obtain a power factor of 1 when a load has a lagging power factor, a capacitor with a reactive power equal to the reactive power of the load is connected in parallel with the load.

We can use a carefully drawn phasor diagram to verify that adding the capacitance does reduce the total ­current drawn from the source without affecting the load current.

Example 22-7 Find the capacitance that minimizes the source current when connected in parallel with a motor drawing 10 A at a 70% lagging power factor from a 120-V 60-Hz source. Solution Step 1 The reactive power of the motor is

Q = VI√ 1 − cos2 ϕ = 120 × 10√ 1 − 0.702 = 857 var (inductive)

Therefore, the capacitor must have a reactive power of 857 var when connected across a 120-V 60-Hz source. Step 2 Since Q = V 2/XC,

XC =

( 120 V ) 2 V2 = = 16.8 Ω ( capacitive ) Q 857 var

22-7   Power-Factor Correction



C=

1 1 = = 1.58 × 10 − 4 F = 158 µF ωXC 377 × 16.8 Ω

To verify the capacitance, download Multisim file EX22-7 from the website and follow the instructions in the file.

Example 22-8 A fluorescent lamp and its ballast inductance draw a 1.0-A current at a 50% lagging power factor from a 120-V 60-Hz source. Find the overall power factor when a 26.5-μF capacitor is connected across the fixture. Solution Step 1 Calculate S, P, and Q for the lamp and its ballast, and draw the power ­triangle for them: S = VI = 120 V × 1.0 A = 120 VA

P = S cos ϕ = 120 VA × 0.50 = 60 W

Q = √ S2 − P2 = √ 1202 − 602 = 104 var ( inductive )

Step 2 Calculate the reactance and reactive power for the capacitor: and

XC =

QC =

1 1 = = 100 Ω ( capacitive ) ωC 377 × 26.5 µF

V2 1202 = = 144 var ( capacitive ) XC 100

Step 3 From the power triangle of Figure 22-15,

Final Q = 144 − 104 = 40 var (capacitive)

Step 4

653

Final S = √ P2 + Q 2 = √ 60 2 + 40 2 = 72 VA

overall power factor = cos ϕ =

60 P = = 83% leading S 72

circuitSIM walkthrough

Chapter 22   Power in Alternating-Current Circuits

0V A

654

Q of capacitor 144 var

So

f la mp =

12

Q of ballast 104 var

P = 60 W Ov era ll S

Net Q 40 var capacitive

Figure 22-15  Power triangle for Example 22-8

circuitSIM walkthrough

To verify the overall power factor, download Multisim file EX22-8 from the website and follow the instructions in the file.

Example 22-9 Find the capacitance that raises the overall power factor to 91% lagging when connected in parallel with a load drawing 1.0 kW at a 70.7% lagging power factor from a 208-V 60-Hz source. Solution Step 1

Original S =

P 1000 W = = 1.41 kVA cos ϕ 0.707

Load Q = √ S2 − P2 = √ 1.412 − 1.02 = 1.0 kvar ( inductive )

Step 2

Final S =

P 1000 W = = 1.1 kVA cos ϕ 0.91

Overall Q = √S2 − P2 = √1.12 − 1.02 = 456 var ( inductive ) Step 3 From the power triangle of Figure 22-16,

QC = 1000 − 456 = 544 vars (capacitive) Therefore,



XC =

V2 2082 = = 79.5 Ω ( capacitive ) QC 544

22-7   Power-Factor Correction

C=

and

ri

O

al

n gi

1 1 = = 3.3 × 10 − 5 F = 33 µF ωXC 377 × 79.5 Ω Q of capacitor 544 var

S

al Fin

655

Q of load 1.0 kvar

S

Overall Q 456 var

P = 1.0 kW

Figure 22-16  Power triangle for Example 22-9

circuitSIM

To verify the capacitance, download Multisim file EX22-9 from the website, and follow the instructions in the file.

walkthrough

See Problems 22-30 to 22-35 and Review Questions 22-57 to 22-60.

Circuit Check

B

CC 22-3. For the network shown in Figure 22-17, determine (a) the total real power (b) the reactive power (c) the apparent power (d) the power factor (e) the capacitance that will make the power factor unity if connected in parallel with the network 4 – j2 Ω 10 A 0° 120 Hz

Figure 22-17

6 – j8 Ω

8 – j7 Ω

656

Chapter 22   Power in Alternating-Current Circuits

CC 22-4. For the circuit of Figure 22-18, find the (a) total real power (b) reactive power (c) apparent power (d) power factor (e) current in phasor form I

+

Load 1 0 var 100 W Load 2

E = 100 V 30°

1500 var 200 W (ind)

Load 3 700 var 200 W (cap)

− Figure 22-17

CC 22-5. The load on a 110-V 60-Hz supply is 5.0 kW (resistive), 10 kvar (inductive), and 5.0 kvar (capacitive). (a) Find the total apparent power. (b) Determine the power factor of the combined loads. (c) Find the current drawn from the supply. (d) Calculate the capacitance required to establish a unity power factor. (e) Find the current drawn from the supply when the power ­factor is unity.

Problems

657

Summary

• The average power in the resistance of an AC circuit is the real or active power of the circuit. • The average power in an ideal inductor or capacitor in an AC circuit is zero. • Reactive power in an ideal inductor or capacitor due to an AC source is the product of the RMS voltage and current. • The unit for reactive power is the volt ampere (reactive) or var. • Apparent power is the power supplied by an AC source to a circuit ­con­taining resistance and reactance. • The unit of apparent power is the volt ampere (VA). • A circuit containing resistance and reactance has both real and reactive power. • A power triangle shows the relationships among apparent power, real power, and reactive power in an AC circuit. • The power factor of a load in an AC circuit is the ratio of real power to ­apparent power. • The power factor is the cosine of the phase angle between the voltage across and the current through the load. • The reactive factor is the ratio of reactive power to apparent power. • The reactive factor is the sine of the phase angle between the voltage across and the current through the load. • A capacitor connected in parallel with an inductive load can increase the power factor and reduce the total current drawn from the AC source.

Problems B

B B B

I

B

Section 22-1  Power in a Resistor 22-1. A toaster draws a 6.0-A current from a 110-V 60-Hz source. (a) Calculate the real power of the toaster. (b) Calculate the peak value of the instantaneous-power input to the toaster. 22-2. The voltage drop across the heater of a certain cathode-ray tube is 6.3 V RMS when the current through it is 0.3 A. Find the average rate of conversion of electric energy into heat. 22-3. Determine the real power dissipated by a 120-Ω resistor when a sine-wave current of i = 2.7 sin 377t A flows through it. 22-4. What power rating must a 300-Ω dummy antenna resistor possess if the RMS voltage drop across it is 96 V?

Section 22-2  Power in an Ideal Inductor 22-5. An inductance of 3.0 mH passes a 1.0-kHz sine-wave current of 20 mA. (a) Find the average power input to the inductor. (b) Find the reactive power. (c) Find the peak rate at which the inductor stores energy. 22-6. A 20-mH inductor is connected to a 10-V 1-kHz source. Calculate the reactive power.

B = beginner

I = intermediate

A = advanced

658

Chapter 22   Power in Alternating-Current Circuits

I

I

B

I

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B B B

B

I I

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I

22-7. At what frequency will a 0.25-H inductor take a reactive power of 25 var from a 40-V source?

Section 22-3  Power in a Capacitor 22-8. A 68-μF capacitor is connected across a 230-V 50-Hz source. (a) Determine the reactive power of the capacitor. (b) If a 33-μF capacitor is connected in parallel with it, determine the total reactive power of the circuit. 22-9. An industrial capacitor is rated at 7.5 kvar when connected to a 208-V 60-Hz circuit. Find the capacitance.

Section 22-4 Power in a Circuit Containing Resistance and Reactance 22-10. (a) A solenoid having an inductance of 0.50 H and a resistance of 24 Ω is connected to a 120-V 60-Hz source. (a) Calculate the apparent power input to the solenoid. (b) Calculate the real power input to the solenoid. (c) Calculate the reactive power input to the solenoid. (d) Use Multisim to verify the apparent power in part (a) and the real power in part (b). 22-11. Find the average power input to the antenna in Example 22-4. 22-12. The reactive power in an impedance of 480 Ω ∠−60º is 300 mvar. At what rate is energy being dissipated by the impedance? 22-13. A coil with an internal resistance of 8 Ω and a 22-mH inductance is connected across a 120-V 60-Hz source. Find (a) the circuit current in polar coordinates (b) the real, reactive, and apparent power of the circuit 22-14. A 250-V 60-Hz power supply is connected to a circuit consisting of a 40-Ω resistance in series with a 0.80-H inductance and an 8.0-μF capacitance. Calculate (a) the current and phase angle (b) the apparent power, real power, and reactive power

Section 22-5  The Power Triangle 22-15. A series RLC circuit is formed by a 680-pF capacitor, a 470-mH inductor, and a 500-Ω resistor connected to a 300-V AC source. At what frequency will the net reactive power be zero? 22-16. (a) A series RC circuit has a current of 1.5 A, an applied voltage of 120 V at 60 Hz, and a power dissipation of 50 W. Find the resistance and capacitance. (b) Use Multisim to verify the resistance and capacitance.

Section 22-6  Power Factor 22-17. A relay coil with a power factor of 75% lagging draws 1 A from a 12-V 500-Hz source. Find the resistance and inductance of the relay coil.

Problems

A

I I

I

A

A

I

B A

A

I

22-18. (a) What series combination of resistance and inductance draws the same current from a 120-V 60-Hz source as a 500-W load with a 60% lagging power factor? (b) Use Multisim to verify the resistance and inductance. 22-19. What parallel combination of resistance and inductance draws the same current from a 120-V 60-Hz source as a 500-W load with a 60% lagging power factor? 22-20. (a) What current must a 110-V alternator supply to operate a 2.0-kW load with a 75% lagging power factor? (b) What real power could the alternator supply for the same magnitude of current in its windings to a load with a power factor of 1? 22-21. An apparent power input of 25 kVA must be provided to the “work coil” of a radio-frequency induction heater to generate heat in the steel work piece at the rate of 5 kW. Find the reactive factor of the load. 22-22. A crossover network used to feed the woofer (low-frequency driver) and the tweeter (high-frequency driver) in a speaker system has an 8.0-Ω resistor and a capacitor in parallel. The power input to these two components at 1.0 kHz is 500 mW with a 20% leading power factor. Find the capacitance. 22-23. An impedance coil having a 0.2 lagging power factor is connected in series with a 300-W lamp in order to supply the lamp with 120 V from a 208-V 60-Hz source. Find the voltage across the terminals of the impedance coil. 22-24. (a) An induction motor that draws 2.0 A from 120-V 60-Hz source at a 0.8 lagging power factor is connected in parallel with a 100-W lamp. Calculate the overall power factor. (b) Use Multisim to verify the overall power factor. 22-25. The ratio of the reactive power to the real power of a coil is known as its quality factor, Q. Determine the quality factor of the coil in Problem 22-15 at a frequency of 2.0 kHz. 22-26. (a) A series RLC circuit connected to a 500-V 60-Hz supply draws a current of 10 A. A power-factor meter indicates a leading power factor of 0.8. If the voltage across the capacitor is 800 V, find the value of each component. (b) Use Multisim to verify the resistance, inductance, and capacitance. 22-27. A parallel RLC circuit connected to a 203-V 50-Hz source draws 3 A of current. A power-factor meter reads 0.8 leading and the inductive reactive power is 400 var. Determine the resistance, capacitance, and inductance of the circuit. 22-28. An RL-series circuit draws 1.2 A from a 250-V AC source. A wattmeter in the circuit reads 135 W. What is the power factor and angle of lag?

659

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660

Chapter 22   Power in Alternating-Current Circuits

I

I

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I I

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A

A A I

A

22-29. An inductive circuit draws a current of 30.0 A at a power factor of 0.5 lagging when connected to a 5-kV 30-Hz supply. Find the resistance and inductance of the circuit.

Section 22-7  Power-Factor Correction 22-30. An induction motor draws 6.0 A at a 0.8 lagging power factor from a 208-V 60-Hz source. (a) What value of capacitance must be connected in parallel with the motor to raise the overall power factor to 1? (b) Calculate the magnitudes of the motor current, capacitor current, and source current with the capacitor in place. (c) Use Multisim to verify the calculation of the capacitance in part (a) and the currents in part (b). 22-31. What value of capacitance is necessary to produce an overall power factor of 0.96 lagging with the motor in Problem 22-30? 22-32. (a) Calculate the overall power factor when a 50-mF capacitor is connected in parallel with the motor in Problem 22-30. (b) Use Multisim to verify the overall power factor. 22-33. A synchronous motor capable of operating with a leading power factor draws 15 kW from a distribution transformer while driving an air compressor. The remainder of the load on the transformer is 80 kW at a 0.85 lagging power factor. (a) How much capacitive reactive power must the synchronous motor produce to raise the overall power factor to 0.96 lagging? (b) Find the reactive factor of the synchronous motor when it operates in this manner. 22-34. The power factor of a load connected to a 120-V 60-Hz source is raised from 0.707 lagging to 0.866 lagging by connecting a 53-μF capacitor across the load. Find the real power of the load. 22-35. The power factor of a load connected to a 120-V 60-Hz source is raised from 0.866 lagging to 0.966 leading by connecting a 110.5-μF capacitor in parallel with the load. Determine the RMS load current. 22-36. A 500-V 60-Hz induction motor has a power factor of 0.707 lagging. The motor draws a current of 10 A when fully loaded. Calculate the required capacitor to raise the power factor to 0.9 lagging. Find the new line current. 22-37. A lighting load operates from a 208-V 60-Hz supply and is rated 1.0 kW with a power factor of 0.9 lagging. A 40-µF capacitor is connected in parallel with this load. Calculate the total power factor and the percent change in line current.

Review Questions

Section 22-1  Power in a Resistor

22-38. Why can we state that p = vi in any AC circuit? 22-39. Why is the instantaneous power in resistance always a positive quantity?

Review Questions

22-40. Why can we say that the average power in the resistance in a sinewave AC system is one-half the peak power? 22-41. Why is the product VR × IR called the real power?

Section 22-2  Power in an Ideal Inductor

22-42. Describe how you could use an AC voltmeter and an AC ammeter to determine the peak instantaneous power in an ideal inductor in an AC circuit. 22-43. Why is the average power of an ideal inductor zero?

Section 22-3  Power in a Capacitor 22-44. What is the meaning of the term reactive power?

Section 22-4 Power in a Circuit Containing Resistance and Reactance 22-45. What is the meaning of the term apparent power? 22-46. Why is apparent power not expressed in watts?

Section 22-5  The Power Triangle 22-47. What is the apparent power of an ideal resistor? 22-48. What is the apparent power of an ideal capacitor? 22-49. Show that the net reactive power of an AC circuit is the difference between the inductive reactive power and the capacitive reactive power. 22-50. Why is the total real power of a network always the sum of the individual real powers regardless of how the components are connected? 22-51. Why is the apparent power of an AC circuit the square root of the sum of the squares of the real and reactive powers rather than the simple sum? 22-52. Why is the angle between the real power and apparent power in a power triangle the same as the angle between the total voltage and total current of a load?

Section 22-6  Power Factor 22-53. Why do we need to consider power factors when designing industrial wiring? 22-54. What is meant by a leading power factor? 22-55. Of what practical use is the reactive factor? 22-56. Derive an expression for reactive factor in terms of power factor.

Section 22-7  Power-Factor Correction 22-57. Why are AC generators and transformers rated in kilovolt amperes rather than kilowatts? 22-58. Why is connecting a capacitor in series with an inductive load not a satisfactory means of power-factor correction? 22-59. What is the purpose of power-factor correction? 22-60. Why does adding capacitance in parallel with an inductive load change the power factor?

661

662

Chapter 22   Power in Alternating-Current Circuits

Integrate the Concepts

A series RLC circuit consisting of a 50-Ω resistor, a 10-mH inductor, and a 22-μF capacitor are connected to a 12-V 500-Hz source. (a) Calculate the real power to the load when the source is connected across the resistor. (b) Calculate the reactive power to the load when the source is connected across the inductor. (c) Calculate the reactive power to the load when the source is connected across the capacitor. (d) Calculate the apparent power supplied by the source when it is connected across the resistor and inductor in series. (e) Draw the power triangle for the circuit of part (d). (f) Determine the power factor for the circuit of part (d). (g) Find the capacitance that minimizes the source current when connected in parallel with the load in part (d).

Practice Quiz

1. Which of the following statements are true? (a) The power dissipated in a resistor located in an AC circuit is true power. (b) The product of the RMS voltage and the current in an ideal inductor is known as reactive power and has units of volt amperes. (c) The reactive power in a capacitor has the same units as the reactive power in an ideal inductor. (d) Power factor is the ratio between real power and reactive power. (e) Power factors range from −1 to 1. 2. For the circuit shown in Figure 22-19, the power dissipated by the resistor is: (a) 8.41 W (b) 5.15 mW (c) 8.41 mW (d) 5.15 W R 1 kΩ

+

10 VRMS 0° −

Figure 22-19

C 3.3 kΩ

Practice Quiz

3. For the circuit of Figure 22-19, the reactive power of the capacitor is (a) 27.75 mvar (b) 17.0 var (c) 27.75 var (d) 17.0 mvar 4. The inductive reactive power of the circuit of Figure 22-20 is (a) 25 mvar leading (b) 50 mvar leading (c) 25 mvar lagging (d) 50 mvar lagging

R 500 Ω

+ 5 VRMS 45° −

L 1 kΩ

Figure 22-20

5. The apparent power of the circuit in Figure 22-21 is (a) 1.46 mVA (b) 145.7 mVA (c) 145.7 VA (d) 1.46 VA R 5 kΩ C 4.7 kΩ

+

100 VRMS 45° −

Figure 22-21

6. Complete this power triangle for the circuit of Figure 22-21. P = 1.06 W θ=? Q=? S = 1.46 VA Figure 22-22

7. The power factor of the circuit shown in Figure 22-21 is (a) tan +43.23° (b) sin +43.23° (c) cos−1 +43.23° (d) cos 43.23°

663

664

Chapter 22   Power in Alternating-Current Circuits

8. A lagging power factor indicates that the circuit has (a) only resistive components (b) a capacitor and a resistor in series (c) an inductor and a resistor in series (d) none of the above 9. The power factor of the circuit shown in Figure 22-23 is (a) 0.905 leading (b) 0.905 lagging (c) 0.425 leading (d) 0.425 lagging 1

5.0 VRMS 20°

+ −

R 10 Ω 2 L 20 Ω 3

Figure 22-23

10. To correct the lagging power factor of the circuit shown in Figure 22-23 to unity, we could connect a capacitor (a) in series with R and L (b) in parallel between terminals 1 and 2 (c) in parallel between terminals 2 and 3 (d) in parallel between terminals 1 and 3 11. The capacitive reactance needed to change the power factor of the circuit in Figure 22-23 from 0.447 lagging to 0.9 leading is (a) 20.3 Ω (b) 34.0 Ω (c) 51.4 Ω (d) 52.7 Ω

12. If the frequency of operation in the circuit of Figure 22-23 is 60 Hz, what capacitance will change the power factor of the circuit in Figure 22-23 from 0.447 lagging to 0.9 leading? (a) 78 μF (b) 52 μF (c) 131 μF (d) 50 μF

PART

V

Impedance Networks

Alternating currents in impedance networks obey all the basic laws of electric circuit behaviour that apply to direct-current circuits. However, steady-state AC quantities have phase angles as well as magnitudes, so calculations for AC circuits usually involve complex numbers. Chapter 23 and Chapter 24 review basic laws and circuit analysis techniques and adapt them to AC circuits. The remaining chapters deal with the circuit characteristics unique to AC systems. 23

Series and Parallel Impedances

24 Impedance Networks 25 Resonance 26

Passive Filters

27 Transformers 28 Coupled Circuits 29

Three-Phase Systems

30 Harmonics

Photo source:  © iStock.com/omada

23

Series and Parallel Impedances

Combinations of impedances in AC circuits behave much like combinations of resistances in DC circuits. Similarly, the currents through the impedances and the voltages across them behave much like currents through and voltages across resistances in DC circuits. Thus, the network equations developed for DC circuits are valid for AC circuits, except that phasors must be used to represent the impedances, ­currents, and voltages in AC circuits.

Chapter Outline 23-1 Resistance and Impedance  668 23-2 Impedances in Series  668

23-3 Impedances in Parallel  671

23-4 Series-Parallel Impedances  678 23-5 Source Conversion  682

Key Terms voltage-divider principle  670 current-divider principle  676

source conversion  683

Learning Outcomes At the conclusion of this chapter, you will be able to: • calculate the total impedance of impedances in series • calculate the equivalent impedance of impedances in parallel • calculate the total impedance of impedances in a series-parallel network

Photo sources:  © iStock.com/nsj-images

• convert an AC constant-voltage source to an equivalent AC constant-current source • convert an AC constant-current source to an equivalent AC constant-voltage source

668

Chapter 23   Series and Parallel Impedances

23-1  Resistance and Impedance Resistance and impedance both represent opposition to electric current. However, resistance opposes both direct and alternating current, while the reactance component of impedance opposes only changing current. Calculations for DC circuits can be done with scalar quantities and ordinary algebra. But impedance is a phasor quantity in AC circuits, and so ­calculations for impedance networks are based on phasor algebra.

Source:  © TOA Electronics Inc.

With phasor algebra, all the relationships for resistance networks also apply to impedance networks.

23-2  Impedances in Series At any particular instant, the circuit relationships in Figure 23-1(a) are ­exactly the same as in a DC circuit that has the voltages and currents that prevail at that instant. Consequently, Kirchhoff’s voltage law gives  hand-held analog A impedance meter

e = vT = vR + vL

(20-1)

ET = VR + VL

(20-3)

To extend Kirchhoff’s voltage law to the much more useful RMS values for AC circuit parameters, we must use phasor quantities. Then, the ­Kirchhoff’s voltage-law relationship for Figure 23-1(a) can be written as

Each term in this equation is a phasor, and the addition must be done with phasor algebra. Scalar Phasor

Scalar Phasor

v1 Z1 V1 vR

e E

vL

VR

e E

v2 Z2 V2

VL v3 Z3 V3

(a)

(b)

Figure 23-1  Kirchhoff’s voltage law applied to an AC circuit

We can apply the phasor form of Kirchhoff’s voltage law to the series circuit of Figure 23-1(b):

E = VT = V1 + V2 + V3

(23-1)

23-2   Impedances in Series

Dividing every term in Equation 23-1 by the common current in the ­series circuit gives VT V1 V2 V3 = + + I I I I



By definition, Z = V/I. Therefore,

ZT = Z1 + Z2 + Z3



(23-2)

The total impedance of impedances in series is their phasor sum. This relationship can be extended to any number of impedances in series:

Example 23-1

Find the total impedance when Z1 = 60 Ω ∠+60º and Z2 = 80 Ω ∠−45º are connected in series. Give the answer in polar form. Solution Step 1 Draw a circuit diagram and sketch an impedance diagram to indicate the approximate total impedance, as shown in Figure 23-2. Step 2 Convert the impedances to rectangular form, and add the components: Z1 = 60 cos 60° + j60 sin 60° = 30

+ j52

Z2 = 80 cos 45° − j80 sin 45° = 56.6 − j56.62 Adding gives

ZT = Z1 + Z2 = 86.6 −  j4.6 Ω

Z1 60 Ω +60°

60

Ω

Z1

60° ZT Z2 80 Ω −45°

−45° 80 Ω

Z2 (a) Figure 23-2  Impedances in series

(b)

Reference axis

669

670

Chapter 23   Series and Parallel Impedances

Step 3 Convert the total impedance into polar form:

ϕ = tan −1

Z=

−4.6 = −3º 86.6

86.6 = 86.7 Ω cos ( −3º )

Z = 87Ω∠−3°

Example 23-2

Z1 = 10 000 + j15 000 Ω, Z2 = 4700 + j0 Ω, and Z3 = 25 000 − j10 000 Ω are connected in series. Express the total impedance in rectangular coordinates. Solution Simply add the rectangular coordinates.

Z1 = 10 000 + j15 000 Z2 = 4700 + j0

Z3 = 25 000 − j10 000

ZT = 39 700 + j5000 Ω

It may appear from the simple solution of Example 23-2 that impedance should always be expressed in rectangular coordinates. However, as noted in Section 21-8, the rectangular coordinates of an impedance represent a r­ esis­tance and a reactance in series. Therefore, the addition process of ­Exam­ple 23-2 applies only to series impedances. We can now extend the voltage-divider principle of Section 8-4 to series impedances. With phasor quantities, Equation 8-1 becomes Vn = E



Zn ZT

(23-3)

Example 23-3 Find the voltage drop across Z1 in the circuit of Figure 23-2(a) when the ­applied voltage is 120 V ∠0º.

23-3   Impedances in Parallel

671

Solution I Step 1 Use Ohm’s law with ZT as determined in Example 23-1: Step 2

I=

120 V ∠0º E = = 1.38 A ∠+3º ZT 86.7Ω ∠−3º

V1 = IZ1 = 1.38 A ∠+3º × 60 Ω ∠+60º = 83 V∠+63°

Solution II Use the voltage-divider principle to solve for V1 without first finding I:

V1 = E

60 Ω ∠+60º Z1 = 120 V∠0º × ZT 86.7 Ω ∠−3º = 83 V ∠+ 63º

To verify the magnitude and phase angle of the voltage across Z1, download Multisim file EX23-3 from the website, and follow the instructions in the file. See Problems 23-1 to 23-5 and Review Questions 23-24 to 23-28 at the end of the chapter.

23-3  Impedances in Parallel As with parallel DC circuits, we can analyze parallel AC circuits by considering the circuit’s ability to conduct current. Two sets of parameters are proportional to a circuit’s ability to pass current: • the total of all branch currents • the total of all branch admittances

For parallel branches in an AC circuit, Kirchhoff’s current law gives

IT = I1 + I2 + I3 + . . .

(23-4)

Dividing every term in Equation 23-4 by the common voltage across parallel branches gives

IT I1 I2 I3 = + + + ... V V V V

Since admittance is defined as Y = 1/Z,

YT = Y1 + Y2 + Y3 + . . .

(23-5)

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Chapter 23   Series and Parallel Impedances

The total admittance of parallel branches is the phasor sum of the branch admittances. The equivalent impedance of a parallel AC circuit can be determined in two ways: by the total-current method (from Equation 23-4),

Zeq =

E IT

Zeq =

1 YT

(23-6)

and by the total-admittance method (from Equation 23-5),

(23-7)

For the special case of two branches in parallel, YT = Y1 + Y2 =



Inverting this equation gives

Zeq =

1 1 Z2 − Z1 + = Z1 Z2 Z1Z2 Z1Z2 Z1 + Z2

Example 23-4 Find the equivalent impedance Z2 = 80 Ω ∠−45º connected in parallel.

for

(23-8)

Z1 = 60 Ω ∠+60º

and

Solution I: Total-current method Step 1 Draw a schematic diagram and a current phasor diagram, as in F ­ igure 23-3.

0A

I2

3.

672

240 V 0°

E

Z1 60 Ω +60°

45°

Z2 80 Ω −45°

E

60°

IT

4.0 A I1 (a)

Figure 23-3  Total-current method for Example 23-4

(b)

23-3   Impedances in Parallel

Step 2 Assume a convenient value of applied voltage, such as E = 240 V ∠0º. Then determine the branch currents:

I1 =

I2 =

240∠0º E = = 4.0 A ∠−60º Z1 60 ∠+60º

240 ∠0º E = = 3.0 A ∠+45º Z2 80 ∠−45º

Step 3 Convert the branch currents to rectangular form and add the components: I1 = 4.0 cos 60° − j4.0 sin 60° = 2.0

− j3.464

I2 = 3.0 cos 45° + j3.0 sin 45° = 2.121 + j2.121 Adding gives

IT = 4.121 − j1.343 A

Step 4 Convert the total current to polar form:

−1.343 = −18º 4.121 4.121 IT = = 4.334 A cos ( −18º ) ϕ = tan−1

IT = 4.334 A ∠−18º

Step 5 Use the total current to calculate the equivalent impedance:

Zeq =

240 V ∠0º E = = 55 Ω ∠+18° IT 4.334 A ∠−18º

Solution II: Total-admittance method Step 1 Calculate the admittance of each branch (see Figure 23-4):

Y1 =

Y2 =

1 1 = = 0.0167 S ∠−60º Z1 60 Ω ∠+60º

1 1 = = 0.0125 S ∠+45º Z2 80 Ω ∠−45º

673

Chapter 23   Series and Parallel Impedances

S

Y2

0. 01 25

674

E

45°

Y2 0.0125 S +45°

Y1 0.0167 S −60°

Reference axis

60°

0.0

YT

167 S Y1 (a)

(b)

Figure 23-4  Total-admittance method for Example 23-4

Step 2 Calculate the total admittance:

Y1 = 0.0167 cos 60° − j0.0167 sin 60° = 0.008 33 − j0.014 43 S

Y2 = 0.0125 cos 45° + j0.0125 sin 45° = 0.008 84 + j0.008 84 S

YT = 0.017 17 − j0.005 59 S

Adding gives

Step 3 Convert the total admittance to polar form: ϕ = tan−1



YT =

−0.005 59 = −18º 0.017 17

0.017 17 = 0.018 05 S cos ( −18º )

YT = 0.018 S ∠−18º

Step 4 Use the total admittance to calculate the equivalent impedance:

Zeq =

1 1 = = 55 Ω ∠+18° YT 0.018 S ∠−18º

Solution III: Impedance formula for two parallel branches Step 1 Calculate the numerator for the formula:

Z1Z2 = 60 ∠+60º × 80 ∠−45º = 4800 ∠+15º

23-3   Impedances in Parallel

Step 2 Calculate the denominator: Z1 = 60 Ω ∠+60º = 30

+ j 51.96

Z2 = 80 Ω ∠−45º = 56.57 − j 56.57

Z1 + Z2 = 86.57 − j 4.61 Ω



Z1 + Z2 = 86.69 Ω ∠−3º

Adding gives from which

Step 3 Use phasor division to find the equivalent impedance: 4800 ∠+15º Z1Z2 = Z1 + Z2 86.69 ∠−3º



Zeq = 55 Ω ∠+ 18°



Although the total-current method may seem easier in Example 23-4, the total-admittance method is preferable for more elaborate AC networks.

Example 23-5

Find the equivalent resistance and reactance for Z1 = 10 000 + j15 000 Ω, Z2 = 4700 + j0 Ω, and Z3 = 25 000 − j10 000 Ω connected in parallel.

Solution Step 1 To calculate the admittance of each branch, first convert the impedances to polar form:

Z1 = 10 000 + j15 000 = 18.03 kΩ ∠+56.3º

Z2 = 4700 + j0

= 4.7 kΩ ∠0º

Z3 = 25 000 − j 10 000 = 26.93 kΩ ∠−21.8º

Step 2 Take the reciprocal of each impedance:

Y1 = 55.47 µS ∠−56.3º Y2 = 212.8 µS ∠0º

Y3 = 37.14 µS ∠+21.8º

675

676

Chapter 23   Series and Parallel Impedances

Step 3 Convert the branch admittances to rectangular form:

The rectangular ­coordinates of the ­admittance are not the reciprocals of the rectangular ­coordinates of the ­impedance.

Y1 =   30.77 − j46.15 μS



Y2 = 212.8  + j0

μS

Y3 =   34.48 + j13.79 μS



YT = 278.1   − j32.36 μS



Step 4 Convert the total admittance to polar form:

YT = 280 µS ∠−6.64º



Step 5 Take the reciprocal of the total admittance:

Zeq =

1 1 = = 3572 Ω ∠+6.64º YT 280 µS∠− 6.64º

Step 6 Convert the equivalent impedance to rectangular form to find the ­resistance and reactance components: Zeq = 3548 + j413Ω

Since the same voltage appears across parallel branches, we can establish the current-divider principle for AC circuits from the relationship

V = I1Z1 = I2Z2 I1 Z2 = I2 Z1

(23-9)

Therefore, the ratio of the currents in any two parallel branches of an AC ­circuit is the inverse of the ratio of the impedances of the two branches. However, it is more convenient to express the current-divider principle in terms of branch admittances. Since



V=

In IT = Yn YT

I n = IT

Yn YT

(23-10)

23-3   Impedances in Parallel

As with DC circuits, the current-divider equation (23-10) is the dual of the voltage-divider equation (23-3). When there are only two parallel branches, we can use Equation 23-8 to substitute for 1/YT in Equation 23-10: I1 = I T



Z2 Z1 + Z2

(23-11)

Example 23-6

What current does a 10-kΩ ∠− 45º load draw from the Norton-­equivalent source shown in Figure 23-5?

50 mA −30°

ZN 5.0 kΩ +30°

ZL 10 kΩ −45°

10

0

μS

YL 45°

Reference axis YT

30°

20

0μ

S YN

(a)

(b)

Figure 23-5  Diagrams for Example 23-6

Solution I: Using Ohm’s law Step 1 Step 2

Zeq = = =

Z1 × Z2 Z1 + Z2

5 kΩ ∠+30º × 10 kΩ ∠−45º ( 4330 + j2500 ) + ( 7071 − j7071 ) 50 × 106∠−15º ( 11 401 − j4571 )

50 × 106∠− 15º 12 283 ∠− 21.85º = 4.07 kΩ ∠+6.85º =

V = ITZeq = 50 mA ∠−30º × 4.07 kΩ ∠+6.85º = 203.5 V ∠−23.15º

677

678

Chapter 23   Series and Parallel Impedances

Step 3 IL =



203.5 V∠− 23.15º V = = 20 mA ∠+ 22° ZL 10k Ω ∠− 45º

Solution II: Using the current-divider principle Step 1 1 1 YL = = = 100 µS ∠+45º = 70.71 + j70.71 µS ZL 10 kΩ ∠−45º 1 1 YN = = = 200 µS ∠−30º = 173.2 − j100 µS ZN 5 kΩ ∠+30º YT = 243.91 − j29.29 µS YT = 245.7 µS ∠−6.85º

In polar form,

Step 2 From Equation 23-10,

circuitSIM walkthrough

I L = IT

100 µS ∠+45º YL = 20 mA ∠+22° = 50 mA ∠−30º × 245.7 µS ∠−6.85º YT

To verify the magnitude of the current through the load ZL , download ­Multisim file EX23-6 from the website, and follow the instructions in the file.

See Problems 23-6 to 23-9 and Review Questions 23-29 to 23-31.

23-4  Series-Parallel Impedances As long as we remember that we are working with phasor quantities, the procedure for solving series-parallel impedance combinations is essentially the same as for DC circuits.

Example 23-7 Calculate the total load in the circuit shown in Figure 23-6(a). 2.0 kΩ −60° Z1

Z1

Z2 5.0 kΩ +30°

Z3 10 kΩ −45°

(a) Figure 23-6  Circuit diagrams for Example 23-7

Zeq

(b)

23-4   Series-Parallel Impedances

Solution Step 1 Determine the equivalent impedance for Z2 and Z3 in parallel. Zeq =



5 kΩ ∠+30º × 10 kΩ ∠−45º Z2 × Z3 = ( Z2 + Z3 4330 + j2500 ) Ω + ( 7071 − j7071 ) Ω 50 × 106Ω ∠−15º 11 401 − j4571Ω 50 × 106Ω ∠−15º = 12 283Ω ∠−21.85º = 4.07 kΩ ∠+6.85º =



Step 2 Convert Z1 and Zeq to rectangular coordinates. For the simple series circuit of Figure 23-6(b), Z1 = 1000 − j1732 Ω



Zeq = 4041 + j485 Ω



ZT = 5041 − j1247 Ω = 5.2 kΩ ∠−14°



Example 23-8

Find the input impedance at 10 kHz for the π-network in Figure 23-7 when the output terminals are open-circuit. 4.7 kΩ

Input

1.0 nF

200 mH

1.0 nF

Output

Figure 23-7  π network for Example 23-8

Solution Step 1 Determine the phasor sum of the impedances of the resistor, inductor, and right-hand capacitor, all of which form a simple series circuit when no load is connected to the output terminals.

XL = 2 × π × 10 kHz × 200 mH = 12 566 Ω 1 XC = = 15 915 Ω 2 × π × 10 kHz × 1.0 nF Zs = R + j ( XL − XC ) = 4700 + j ( 12 566 − 15 915 ) Ω

679

680

Chapter 23   Series and Parallel Impedances

= 4700 − j3349 Ω

= 5771 Ω ∠−35.47º



Step 2 Use any of the methods for determining the equivalent impedance of two impedances in parallel. The total admittance method is shown here. BC = ωC = 2 × π × 10 kHz × 1.0 nF = 62.83 µS (capacitive)



Ys =



1 1 = = 173.3 µS ∠+35.47º Zs 5771 Ω ∠−35.47º

Y T = BC + Y s



= +j62.83 µS + ( 141.2 + j100.4 ) µS



= 141.2 + j163.23 µS



= 215.8 µS ∠+49.1º



Zin =



1 1 = = 4.6 kΩ∠−49° YT 215.8 µS ∠+49.1º

See Problems 23-10 to 23-20.

Practical Circuits Kirchhoff’s Current and Voltage Laws in Transmission Lines A transmission line is a system of conductors, such as wires, waveguides, or coaxial cables, designed for transmitting signals efficiently between two or more locations. As a signal travels down a pair of conductors, each new section of the line acts electrically as a small circuit element. In its simplest form, called the lossless model, the equivalent circuit of a transmission line consists of just inductors and capacitors. These elements are distributed uniformly down the length of the line, as shown in Figure 23-8. L C

L C

L C

L C

L C

Figure 23-8  Equivalent circuit of a lossless transmission line

This series-parallel combination of inductive and capacitive elements determines the AC impedance of the cable (Z 0), known as the characteristic impedance. At low frequencies the reactive inductance will be greater than the reactive capacitance of the cable. If the transmission line is being used at a particular frequency, the characteristic impedance of the line will be constant. When we terminate the cable with a load impedance, Z L, that equals the line impedance, maximum power is transferred to the load. We can then represent the signal source, the transmission line, and the load as a simple series circuit as shown in Figure 23-9. I + VS −

Figure 23-9

+ V0 − Z0 + ZL VL −

23-4   Series-Parallel Impedances

Circuit Check

681

A

CC 23-1. Two impedances are in series. If one of them has an impedance of 100 Ω ∠−75º , and the total impedance is 270 Ω ∠60º, what is the value of the second impedance? CC 23-2. For the circuit in Figure 23-11, determine the current flowing through the 25-Ω inductive reactance and the voltage across the 12-Ω resistor.

+

12 Ω

15 Ω

25 Ω 9Ω

+ 15 V 60°

+ 10 Ω + 10 Ω

−

Figure 23-11

Kirchhoff’s voltage law states that the algebraic sum of the voltage drops across each element in a circuit equals the voltage applied. Source:  © iStock.com/lucentius

Vs = V0 + VL

Applying Ohm’s law gives the current flowing through the system: I=

Vs V0 + VL = ZT Z0 + VL

If the transmission line has losses, we add resistance to the equivalent circuit for the transmission line, as shown in Figure 23-10. R

L

R G

C

L

R G

C

Figure 23-10 Equivalent circuit of a lossy transmission line

L G

C

In coaxial cable, a flexible dielectric holds the central copper conductor at a fixed distance from the shield conductor (usually copper braid or metal foil).

682

Chapter 23   Series and Parallel Impedances

CC 23-3. Calculate the current in the + j5 Ω inductor shown in Figure 23-12.

−j2 Ω 5Ω 80 V 25°

j9 Ω

+ 2Ω j3 Ω

−

−j4 Ω

j5 Ω Figure 23-12

23-5  Source Conversion Another useful equivalent-circuit technique is to convert a constant-­voltage source to an equivalent constant-current source, and vice versa. ­Although instantaneous voltages and currents vary continuously in AC ­circuits, the RMS values are the same for each cycle. Thus, we can speak of constantvoltage and constant-current sources in AC networks. For any “black-box” AC source, Equation 9-1 becomes



open-circuit RMS voltage Zint = short-circuit RMS current



(23-12)

As shown in Figure 23-13, the equivalent constant-current source for a ­constant-voltage AC source has the same internal impedance and a shortcircuit current of

Isc =

E Zint

(23-13)

Similarly, the equivalent constant-voltage source for a constantcurrent AC source has the same internal impedance and an opencircuit voltage of

Eoc = IZint =

I Yint

(23-14)

23-5   Source Conversion

Zint E

I

(a)

Zint

(b)

Figure 23-13  AC source conversion

In both cases, the phasor division gives an angle, as well as a magnitude, for the equivalent source. Source conversion provides a third solution for Example 23-6.

Example 23-6A

What current does a 10 kΩ ∠−45º load draw from the Norton-equivalent source shown in Figure 23-5? Solution Replace the constant-current source with the equivalent constantvoltage source, as shown in Figure 23-14. Ex = IZx = IZN = 50 mA ∠−30º × 5 kΩ ∠+30º = 250 V∠0º



ZT = Zx + ZL



= 5 kΩ ∠+30º + 10 kΩ ∠−45º



= ( 4330 + j2500 ) Ω + ( 7071 − j7071 ) Ω



= ( 11 401 − j4571 ) Ω



= 12 283 Ω ∠−21.85º



IL =



250 V∠0º Ex = = 20 mA ∠122° ZT 12 283 Ω ∠−21.85º 5.0 kΩ +30° ZN

50 mA −30°

ZN 5.0 kΩ +30°

ZL 10 kΩ −45°

(a)

250 V 0°

ZL 10 kΩ −45°

(b)

Figure 23-14  Source conversion for Example 23-6A

See Problems 23-21 to 23-23 and Review Questions 23-32 and 23-33.

683

684

Chapter 23   Series and Parallel Impedances

Summary

• The total impedance of impedances in series is their phasor sum. • The total admittance of admittances in parallel is their phasor sum. • The procedures for solving series-parallel resistance networks in DC ­circuits are valid for series-parallel impedance networks in AC circuits. • An AC constant-voltage source with an internal impedance can be converted to an equivalent AC constant-current source with the same internal impedance. • An AC constant-current source with an internal impedance can be converted to an equivalent AC constant-voltage source with the same internal impedance. B = beginner

Problems

I = intermediate

A = advanced

circuitSIM walkthrough

Draw schematic and phasor diagrams where applicable. B

Section 23-2  Impedances in Series 23-1.

B

23-2.

I

23-3.

I

23-4.

I

23-5.

Determine the total impedance of a series circuit when Z1 = 330 Ω ∠45º and Z2 = 220 Ω ∠−60º. Find the total impedance when Z1 = 5 kΩ ∠+30º , Z2 = 4.7 kΩ ∠−45º , and Z3 = 2.2kΩ ∠+75º are connected in series. A coil having a resistance of 100 Ω and an inductance of 0.50 H, a 60‑μF capacitor, and a 30-Ω resistor are all connected in series to a 208-V 60-Hz source. (a) Find the magnitude and phase of the current through the coil. (b) Find the magnitude and phase of the voltage across the coil. (a) What value of impedance must be connected in series with a 60-Ω ∠+45º impedance for the current drawn from a 120-V source to be 1.0 A lagging behind the source voltage by 30°? (b) Use Multisim to verify the impedance. Determine the output impedance (in rectangular form) for the network shown in Figure 23-15. 50 Ω

75 Ω

+

50 Ω

−

100 Ω Figure 23-15

Problems

B B I I

A

685

Section 23-3  Impedances in Parallel 23-6. Calculate the total admittance when the two components of Problem 23-1 are connected in parallel. 23-7. Find the equivalent impedance when the three impedances of Problem 23-2 are connected in parallel. 23-8. Find the overall power factor when the three components of Problem 23-3 are connected in parallel. 23-9. What impedance must be connected in parallel with a 10-kΩ ∠−60º impedance for the current from a 1.5-V 100-kHz source to be 150 µA ∠0º?

Section 23-4  Series-Parallel Impedances 23-10. (a) Determine the magnitude and phase of the two capacitor currents in the circuit of Figure 23-16. (b) Use Multisim to verify the magnitude and phase angle of the current through each capacitor.

circuitSIM walkthrough

−j200 Ω

120 V 0°

400 Ω

−j300 Ω

Figure 23-16

A

23-11. Determine the magnitude and phase of the voltage across the inductor in Figure 23-17.

j500 Ω + 100 mV 75°

−j300 Ω 1 kΩ

−

500 Ω Figure 23-17

A

23-12. (a) Determine the magnitude and phase angle of the current flowing through the output resistor in Figure 23-18. (b) Use Multisim to verify the magnitude and phase angle of the current through the output resistor.

circuitSIM walkthrough

686

Chapter 23   Series and Parallel Impedances

−j300 Ω

j500 Ω

Z1

Z3

Z2

250 V 15°

150 Ω −71°

Z4

500 Ω

316 Ω +75°

Figure 23-18

A

23-13. Find the magnitude and phase of the current drawn from the voltage source in Figure 23-19 (a) with the output terminals open-circuit (b) with the output terminals shorted 1.0 kΩ 5.0 kΩ

5.0 kΩ

3.0 kΩ 50 mV 0° 4.0 kΩ

Figure 23-19

I

circuitSIM walkthrough

A

23-14. (a) Find the magnitude and phase of the open-circuit output voltage in Figure 23-19. (b) Use Multisim to verify the magnitude and phase angle of the open-circuit output voltage. 23-15. Find the magnitude and phase of the voltage source required in ­Figure 23-20 for the total drain on the source to be 50 mA ∠+45º. 500 Ω f = 1.0 kHz

400 Ω

E

0.50 μF

100 mH

Figure 23-20

circuitSIM walkthrough

A

23-16. (a) Find the magnitude and phase of the open-circuit voltage between terminals A and B in the Hay bridge circuit shown in Figure 23-21. (b) Use Multisim to verify the magnitude and phase angle of the open-circuit output voltage between terminals A and B.

Problems

100 Ω 100 V 30°

687

3.3 kΩ

2.2 mH

f = 2.0 kHz

A

B 0.22 μF

330 Ω

A

A

50 Ω

Figure 23-21

23-17. If the source voltage in Figure 23-21 is 50 V ∠0º , determine the total source current with terminals A and B (a) open-circuit (b) short-circuit 23-18. For the circuit shown in Figure 23-22, (a) calculate the total impedance (b) calculate the total current (c) calculate the current in the 4.0-Ω resistor (d) calculate the power factor (e) Use Multisim to verify the magnitude and phase angle of the total current in part (b) and the power factor in part (d). j10 Ω

2.0 Ω

2.0 Ω

−j3.0 Ω

100 V 0° 4.0 Ω −j4.0 Ω

A

Figure 23-22

23-19. For the circuit shown in Figure 23-23, (a) calculate the source current (b) calculate the total real power and the total reactive power (c) calculate the power factor (d) sketch the phasor diagram 3 Ω −j3 Ω −j2 Ω j3 Ω 6Ω

8 Ω

2Ω

j6 Ω

j4 Ω

−j15 Ω 2 −j Ω

50 V 60°

j3 Ω

j9 Ω

Figure 23-23

−j9 Ω

12 Ω

−j8 Ω

circuitSIM walkthrough

688

Chapter 23   Series and Parallel Impedances

A

circuitSIM walkthrough

23-20. For the circuit shown in Figure 23-24, (a) calculate the total impedance (b) calculate the total apparent, active, and reactive powers (c) calculate the voltage across the −j11-Ω capacitance (d) Use Multisim to verify the apparent power and active power in part (b) and the magnitude and phase angle of the voltage across the –j11-Ω capacitance in part (c). j16 Ω

−j11 Ω

0Ω

j4

j40 Ω

24 Ω

100 V 30°

j4 Ω

−j1 Ω

−j8 Ω

j10 Ω j9 Ω

−j10 Ω 10 Ω

8Ω

3Ω

Figure 23-24

I I

circuitSIM walkthrough

B

Section 23-5  Source Conversion

23-21. Convert the 120-V source and −j200-Ω capacitor of Figure 23-16 into its equivalent current source and calculate the current through the −j300-Ω capacitor. 23-22. (a) Convert the 100-mV source and 500 Ω resistor of Figure 23-17 into an equivalent current source, and determine the current through the j500-Ω inductor. (b) Use Multisim to verify the calculation of the magnitude and phase angle of the current through the inductor. 23-23. Use source conversion to determine the voltage across the 25-Ω inductive reactance of the network shown in Figure 23-25. 9Ω

15 Ω

12 Ω 10 V 0°

25 Ω

+

15 V 60°

+

+

2 A 0°

10 Ω

+

+ 10 Ω Figure 23-25

Review Questions

Section 23-2  Impedances in Series 23-24. Write a statement for Kirchhoff’s voltage law as it applies to AC ­circuits.

689

Integrate the Concepts

23-25. Given the magnitudes of two impedances, why can we not c alculate the total impedance of these two impedances in ­ series? 23-26. Under what conditions could the magnitude of the total impedance of two impedances in series be less than that of either one by itself? 23-27. Under what circumstances is it more convenient to state impedances in rectangular coordinates? When is this form less convenient? 23-28. Write a statement for the voltage-divider principle as it applies to AC circuits.

Section 23-3  Impedances in Parallel 23-29. What is the advantage in thinking in terms of total admittance when dealing with parallel AC circuits? 23-30. Show how Equation 23-8 is derived. 23-31. Derive Equation 23-11 from Equation 23-10.

Section 23-5  Source Conversion 23-32. Describe how to determine the internal impedance of the constantcurrent source equivalent to an AC constant-voltage source. 23-33. Describe how to determine the open-circuit terminal voltage of the constant-voltage source equivalent to an AC constant-current source.

Integrate the Concepts

(a) Determine the total impedance, ZT, for the circuit in ­Figure 23-26(a). (b) Determine V1 for this circuit. (c)  Determine the equivalent impedance, Zeq, for the circuit in ­Figure  23‑26(b). (d) Determine I1 for this circuit. (e) Treating E and Z 1 in Figure 23-26(a) as a practical AC constantvoltage source, determine its equivalent AC constant-current source. I1

500 Ω +60° Z1

Z1 500 Ω +60°

V1

Z2 1 kΩ −49°

E 120 V −10°

(a) Figure 23-26

Z3 2.7 kΩ +30°

50 mA 45°

I Z2

1 kΩ −49°

(b)

Z3 2.7 kΩ +30°

690

Chapter 23   Series and Parallel Impedances

Practice Quiz 1.

Which of the following statements are true? (a) Kirchhoff’s voltage law can be applied to series AC circuits. (b) Kirchhoff’s current law can be applied to parallel AC circuits. (c) Source conversion can be used in DC and AC circuits to simplify the analysis. (d) An AC constant-current source with an internal impedance can be converted into an equivalent AC constant-voltage source with the same internal impedance connected in parallel.

2.

Increasing the frequency in a series AC circuit containing a resistor and an inductor causes total impedance of the circuit to (a) increase (b) decrease (c) stay the same (d) go to zero

3.

Increasing the frequency in a series AC circuit containing a resistor and a capacitor causes the total impedance of the circuit to (a) increase (b) decrease (c) stay the same (d) go to zero

4.

5.

6.

A series circuit consisting of Z1 = 25 + j75 Ω, Z2 = 330 − j55 Ω, and Z3 = 500 Ω ∠65º has a total impedance of (a) 712.97 Ω ∠−37.41º (b) 737.96 Ω ∠38.88º (c) 652.02 Ω ∠−29.71º (d) 712.97 Ω ∠37.41º

Increasing the frequency in a parallel AC circuit containing a resistor and an inductor causes the total impedance of the circuit to (a) increase (b) decrease (c) stay the same (d) go to zero Increasing the frequency in a parallel AC circuit containing a resistor and a capacitor causes the total impedance of the circuit to (a) decrease (b) increase (c) stay the same (d) go to zero

Practice Quiz

  7. A parallel circuit consisting of Z1 = 500 Ω ∠−15º and Z2 = 2.2 kΩ ∠+45º has a total admittance of (a) 2.26 mS ∠4.98º (b) 2.40 mS ∠−20.42º (c) 442.13 S ∠−4.98º (d) 2.40 mS ∠20.42º   8. The circuit of Figure 23-27 has a total impedance of (a) 199.35 Ω ∠48.63º (b) 282.96 Ω ∠−62.25º (c) 282.96 Ω ∠62.25º (d) 199.35 Ω ∠−48.63º R1

200 Ω L1 60 Ω + 20 V RMS 20° −

R2 25 Ω

R3 330 Ω

R4 100 Ω

Figure 23-27

  9. The total current in the circuit of Figure 23-27 is (a) 70.68 mA ∠82.25º (b) 100.33 mA ∠68.63º (c) 70.68 mA ∠−82.25º (d) 100.33 mA ∠−68.63º

10. The voltage across the 330-Ω resistor in Figure 23-27 is (a) 28.39 V ∠130.88º (b) 20 V ∠ 20º (c) 6.52 V ∠ 136.01º (d) 5.98 V ∠ 126.42º

691

24

Impedance Networks

As long as we use phasor algebra for the calculations, we can follow the same procedures in dealing with impedance networks in AC circuits as for resistance networks. The analysis of AC networks involves much more computation than the analysis of comparable DC circuits. However, we can reduce the computation required by choosing the network analysis technique that is the most straightforward for a given network.

Chapter Outline 24-1 Loop Equations  694

24-2 Mesh Equations  700

24-3 Superposition Theorem  702 24-4 Thévenin’s Theorem  707 24-5 Norton’s Theorem  713 24-6 Nodal Analysis  716

24-7 Delta-Wye Transformation  724

Key Terms superposition theorem  702 Thévenin’s theorem  707 Norton’s theorem  713

delta-to-wye transformation  724 wye-to-delta transformation  724

Learning Outcomes At the conclusion of this chapter, you will be able to: • use loop equations to analyze AC circuits • use mesh equations to analyze AC circuits • use the superposition theorem to analyze AC circuits

Photo sources:  © iStock.com/Daniel Brunner

• simplify an AC circuit by using Thévenin’s theorem or Norton’s theorem • use nodal equations to analyze AC circuits • perform delta-wye transformations on impedance networks in AC circuits

694

Chapter 24   Impedance Networks

24-1  Loop Equations For AC circuits, we restate Kirchhoff’s voltage law using phasor quantities: In any complete (closed-loop) AC circuit, the phasor sum of the voltage drops must equal the phasor sum of the source voltages. This law leads to equations in the form

V1 + V2 + V3 + . . . = E



(24-1)

As with DC circuits, we have a choice of two methods of writing ­ irchhoff’s voltage-law equations: loop equations and mesh equations. K The loop-equation procedure applies Kirchhoff’s voltage directly, whereas mesh equations provide a standardized format for setting up essentially the same equations. To write loop equations, we replace the voltage terms on the left side of Equation 24-1 with the IZ drop across the corresponding impedance. When the impedances are known, this procedure gives a set of simultaneous equations in which the loop currents are the unknown quantities. If we could omit Z4 in the network shown in Figure 24-1, we would have a simple T-network that we could solve by the series-parallel circuit techniques of Chapter 23. The addition of Z4 turns the network into a bridge ­circuit, which could have a different current through each of the five ­impedances. To avoid having to write five simultaneous equations, we draw a set of tracing loops that start from the upper terminal of the source and end at its lower terminal. We need three loops in order to include all the components in the bridged network. Once we solve the loop equations for the loop currents, we can use them to determine the current in each component. For example, with the loops shown in Figure 24-1, I2 + I3 = IL , the current through ZL. We use the sum of I2 and I3 since both tracing loops pass in the same direction through ZL. For DC circuits, we simply assumed that the loop currents followed the direction of conventional current from the positive terminal of the source, through the network, and back to the negative terminal of the source.

Z1

+ E

I1

Z4

Z2

Z3

−

Figure 24-1  Bridged-T network

I2

ZL I3

24-1   Loop Equations

However, in an AC circuit, the instantaneous current reverses its polarity twice each cycle. Nevertheless, we must adopt some method of marking the direction of an AC source since reversing the leads of an AC source will change the angle of the source voltage by 180°. In Figure 24-1, the lower lead is common to both the source and the load. In nodal analysis, we use this lead as the reference node. For AC circuits, we treat the voltage appearing at the upper terminal of the source as being positive with respect to the ­reference node (which is sometimes grounded). We can mark the direction for an AC source either with a + and a − sign beside the terminals or with an arrow. The + and − signs indicate that the phase angle for E is measured at the + terminal with respect to the − terminal and that the positive direction for tracing loops is from the + terminal of the source through the network to the − terminal. Obviously, the + and − signs on AC sources do not indicate instantaneous polarity. An arrow beside an AC source indicates that the phase angle for E is measured at the terminal beside the head of the arrow with respect to the terminal beside the tail of the arrow. The arrow also indicates the direction for currents in the tracing loops. To avoid possible confusion from using + and − signs in AC circuit diagrams, this book uses arrows to show directions for AC sources. There are several possible combinations of tracing loops for the network of Figure 24-1. Applying Kirchhoff’s voltage law to the loops shown gives these three loop equations: For the I1 loop, For the I2 loop, For the I3 loop, Collecting terms gives

Z1(I1 + I2) + Z3I1 = E

Z1(I1 + I2) + Z2I2 + ZL(I2 + I3) = E

(Z1 + Z3)I1 + 

Z4I3 + ZL(I2 + I3) = E

Z1I2 

Z1I1 + (Z1 + Z2 +  ZL)I2 + 

= E

ZLI3 = E

ZLI2 + (Z4 + ZL)I3 = E

Example 24-1 Find the current through the load ZL in the bridged-T network of Figure 24-1, given E = 20 V ∠0º , Z1 = Z2 = −j5.0 kΩ, Z3 = Z4 = 20 + j0 kΩ, and ZL = 5.0 + j5.0 kΩ. Solution Although we could use the loop equations we wrote for the tracing loops of Figure 24-1, we would have to solve for both I2 and I3 in order to determine IL. If we select the tracing loops shown in Figure 24-2, we need solve only for I3.

695

696

Chapter 24   Impedance Networks

20 kΩ −j5.0 kΩ

Z4

−j5.0 kΩ

Z1 20 V 0°

E

I1

20 kΩ

Z2 Z3

5.0 + j5.0 kΩ ZL

I3

I2

Figure 24-2  Circuit diagram for Example 24-1

Step 1 Write the three loop equations:

Z1I1 + Z3(I1 + I2) = E Z4(I2 + I3) + Z2I2 + Z3(I2 + I1) = E Z4(I3 + I2) + ZLI3 = E Collecting terms gives

(Z1 + Z3)I1 + 

Z3I2 

Z3I1 + (Z2 + Z3 + Z4)I2 + 

= E

Z4I3 = E

Z4I2 + (Z4 + ZL)I3 = E Step 2 If the impedances are expressed in kilohms and the applied voltage in volts, the currents will be in milliamperes. Since we need to add the impedances in the loop equations, we leave the impedances in rectangular form in this step. (20 – j5.0)I1 +   (20 + j0)I2 

= 20 + j0

  (20 + j0)I1 + (40 – j5.0)I2 +   (20 + j0)I3 = 20 + j0

(20 + j0)I2 + (25 + j5.0)I3 = 20 + j0

24-1   Loop Equations

Step 3 We now have to multiply the diagonals of the determinant matrices, so we write the impedances in the matrices in polar form:

I3 =







= = = =

│

│

20.61∠−14.04º 20∠0º 0

20.61∠−14.04º 20∠0º 0

20∠0º 40.31∠−7.13º 20∠0º

20∠0º 40.31∠−7.13º 20∠0º

│

20∠0º 20∠0º 20∠0º

0 20∠0º 25.5∠+11.31º

│

16 616∠−21.17º + 0 + 8000∠0º − 0 − 8244∠−14.04º − 8000∠0º 21 185∠−9.86º + 0 + 0 − 0 − 8244∠−14.04º − 10 196∠+11.31º ( 15 495 − j6000 ) − ( 7998 − j2000 ) ( 20 872 − j3628 ) − ( 7998 − j2000 ) − ( 9998 + j2000 )

7497 − j4000

2876 − j3628

8497∠−28.1º 4630∠−51.6º

  = 1.835 mA∠+23.5º

  IL = I3 = 1.8 mA∠+ 24º In the single-source network of Example 24-1, we started all tracing loops from the “positive” terminal. This keeps the number of negative terms in the loop equations to a minimum. If there is more than one voltage source in a network, at least one tracing loop starts and ends with each source, and the source direction determines the direction of the loop current.

Example 24-2 Two 60-Hz alternators connected in parallel feed a load with an impedance of 20 Ω ∠+15º. Alternator 1 has an open-circuit voltage of 120 V and an ­internal impedance of 10 Ω ∠+30º. Alternator 2 has an opencircuit voltage of 117 V and lags behind the voltage of alternator 1 by a constant 15°. The internal impedance of alternator 2 is 8.0 Ω ∠+45º. Find the load current.

697

698

Chapter 24   Impedance Networks

8.0 Ω 45° Z2

10 Ω +30° Z1 E1

20 Ω +15° ZL

I1

120 V 0°

I2

E2

117 V −15°

Figure 24-3  Circuit diagram for Example 24-2

Solution Step 1 Draw a loop starting from each source. Since the two loops shown in ­Figure 24-3 include all the circuit components, we can solve the circuit with just two loop equations: Z1I1 + ZL (I1 + I2) = E1

For the Il loop,

Z2I2  + ZL (I2 + I1) = E2

For the I2 loop, Collecting terms gives

 (Z1 + ZL)I1    + ZLI2 

= E1

ZLI1    + (Z2 + ZL)I2  = E2 I1 =

│(Z

Step 2

│E

E1

+ ZL ) ZL 2

1

ZL ( Z2 + ZL )

│

ZL ( Z2 + ZL )

│

and I2 =

│ ( Z Z+ Z )

│ ( Z Z+ Z ) 1

L

1

L

L

L

ZL ( Z2 + ZL )

Z1 + ZL = 10 Ω ∠+30º + 20 Ω ∠+15º

= ( 8.66 + j5 ) Ω + ( 19.32 + j5.176 ) Ω = 27.98 + j10.176Ω = 29.77 Ω ∠+19.99º

Z2 + ZL = 8.0 Ω ∠+45º + 20 Ω ∠+15º

= ( 5.657 + j5.657 ) Ω + ( 19.32 + j5.176 ) Ω = 24.98 + j10.833Ω = 27.23 Ω ∠+23.45º

│

E1 E2

│

24-1   Loop Equations



I1 =



= =



=



=



I2 = =



=



=

Step 3 From Figure 24-3,

│

20∠+15º 27.23∠+23.45º

3268∠+23.45º − 2340∠0º 810.6∠+43.44º − 400∠+30º

│

( 2998 + j1301 ) − ( 2340 + j0 ) ( 588.6 + j557.4 ) − ( 346.4 + j200 ) 658 + j1301

242.2 + j357.4 1458∠+63.17º 431.7∠+55.88º

│29.7720∠∠+19.99º +15º

│

120∠0º 117∠−15º

D 3483∠+4.99º − 2400∠+15º = D





│29.7720∠∠+19.99º +15º

20∠+15º 27.23∠+23.45º

= 3.377 A∠+7.29º





∠0º │117120∠−15º

( 3470 + j303 ) − ( 2318 + j621 )

1152 − j318

D

1195∠−15.43º 431.7∠+55.88º D

= 2.768 A∠−71.31º

IL = I 1 + I 2

= 3.377 A∠+7.29º + 2.768 A∠−71.31º

  = (3.35 + j0.429) A + (0.887 − j2.622) A   = 4.237 − j2.193 A    = 4.8 A∠− 27°

See Problems 24-1 to 24-27 and Review Question 24-37 at the end of the chapter.

699

700

Chapter 24   Impedance Networks

24-2  Mesh Equations The rules for writing mesh equations are the same for DC and AC circuits. However, in AC networks, the variables are phasor quantities. 1. Any current sources in the network must be converted to equivalent constant-voltage sources. 2. The circuit diagram must be laid out so that each component is part of a mesh, a closed loop with no components inside it. Mesh equations can be used only if such a layout is possible. 3. The mesh currents must be directed either all clockwise or all counterclockwise. (We shall choose clockwise.) In the equation for a given mesh, its current is a positive quantity while any adjacent mesh current that flows through a mutual component is a negative quantity. 4. If the direction of the mesh current is the same as the direction indicated for a source in the mesh, the source voltage is positive. If the directions are opposite, the source voltage is a negative quantity. Some meshes may not include a source. 5. The coefficient for the positive current in a given mesh is the sum of the impedances around the mesh. The coefficients for the negative adjacent mesh currents are the impedances of the mutual components. For ­example, the mesh equation for the IA mesh in Figure 24-4 is

( Z1 + Z3 ) IA − Z3I B − Z1I C = +E

(24-2)

These rules produce simultaneous equations with coefficients that can be entered directly into matrices for solving the equations.

Example 24-1A Find the current through the load ZL in the bridged-T network of ­Figure 24‑1, given E = 20 V∠0º, Z1 = Z2 = −j5.0 kΩ, Z3 = Z4 = 20 + j0 kΩ, and ZL = 5.0 + j5.0 kΩ, as shown in Figure 24-4. Solution Step 1 With the impedances in kilohms and the applied voltage in volts, the mesh equations for IA, IB, and IC are (in milliamperes), respectively, (20  − j5.0)IA − (20 + j0)IB − (0 − j5.0)IC = +20∠0º −(20 + j0)IA + (25 + j0)IB − (0 − j5.0)IC = 0 −(0 − j5.0)IA − (0 − j5.0)IB + (20 − j10)IC = 0

24-2   Mesh Equations

Step 2 As in the loop method, we now convert the coefficients to polar form and use matrices to solve the s­ imultaneous equations. To find IL , we need solve only for IB.

IB = =

= = =

│ │

+20.62∠−14.04º −20∠0º −5∠−90º +20.62∠−14.04º −20∠0º −5∠−90º

+20∠0º 0 0

−5∠−90º −5∠−90º +22.36∠−26.57º

−20∠0º +25∠0º −5∠−90º

−5∠−90º −5∠−90º 22.36∠−26.57º

│ │

0 + 500∠ −180º + 0 − 0 − 0 + 8944∠− 26.57º 11 524∠−40.61º− 500 ∠−180º − 500 ∠−180º − 625 ∠−180º − 515.4 ∠− 194.04º − 8944 ∠−206.57º ( − 500 + j0 ) + ( 8000 − j4000 ) ( 8750 − j7500 ) − ( −1625 + j0 ) − ( −500 + j125 ) − ( 8000 − j4000 )

7500 − j4000

2875 − j3625

8500∠−28.1º 4627∠−51.6º

IL = IB = 1.8 mA ∠ + 24° 20 kΩ Z4 IC

20 V 0°

E

−j5.0 kΩ

−j5.0 kΩ

Z1

Z2

IA

Z3

20 kΩ

IB

Figure 24-4  Circuit diagram for Example 24-1A

ZL 5 + j5.0 kΩ

701

702

Chapter 24   Impedance Networks

Example 24-2A Two 60-Hz alternators connected in parallel feed a load with an impedance of 20 Ω ∠+15º. Alternator 1 has an open-circuit voltage of 120 V and an ­internal impedance of 10 Ω ∠+30º. Alternator 2 has an opencircuit voltage of 117 V that lags behind the voltage of alternator 1 by a constant 15°. The internal impedance of alternator 2 is 8.0 Ω ∠+45º. Find the load current. Solution We can use the circuit diagram of Figure 24-3 for the mesh currents simply by reversing the direction of the I2 loop. Step 1 The mesh equations for I1 and I2, are, respectively,

│−EE

│

│Z −Z+ Z

│

(Z1 + ZL)I1 − ZLI2 = +E1  and  −ZLI1 + (Z2 + ZL)I2 = −E2



I1 =

│Z −Z+ Z 1 2

1

L

L

−ZL Z2 + ZL

−ZL Z2 + ZL

│

       I2 =

│Z −Z+ Z 1

1

L

L

L

L

+E1 −E2

−ZL Z2 + ZL

│

Step 2 Solving for I1 and I2 after substituting the known impedances and voltages gives

Step 3

I1 = 3.377 A ∠+7.29º  and  I2 = 2.768 A ∠+108.69º IL = = = = =

I1 − I2 3.377 A∠+7.29º − 2.768 A∠+108.69º ( 3.35 + j0.429 ) A − ( −0.887 + j2.622 ) A 4.237 − j2.193 A 4.8 A ∠−27°

See Problems 24-1 to 24-27.

24-3  Superposition Theorem In networks containing more than one source, we can avoid the simultaneous equations of the loop procedure by applying the superposition theorem. The theorem can be restated for AC circuits by introducing phasor quantities: The current in any branch of a network of impedances resulting from the simultaneous application of a number of sources distributed in any manner throughout the network is the phasor sum of the component currents that would be caused in that branch by each

24-3   Superposition Theorem

703

source acting independently in turn while the others were replaced in the network by their respective internal impedances. The superposition theorem can be used in networks containing current sources, as illustrated in the next example.

Example 24-2B Two 60-Hz alternators connected in parallel feed a load with an impedance of 20 Ω ∠+15º. Alternator 1 has an open-circuit voltage of 120 V and an ­internal impedance of 10 Ω ∠+30º. Alternator 2 has an opencircuit voltage of 117 V that lags behind the voltage of alternator 1 by a constant 15°. The internal impedance of Alternator 2 is 8.0 Ω ∠+45º . Find the load current. Solution Step 1 To find the current resulting from just source E1, we replace source E2 with its internal impedance, Z2, as shown in Figure 24-5(a). The total impedance of the resulting circuit is Z 2ZL ZT = Z1 + Z2 + ZL ( 5.657 + j5.657 ) × ( 19.319 + j5.176 ) = ( 8.66 + j5.0 ) + ( 5.657 + j5.657 ) + ( 19.319 + j5.176 )

= ( 8.66 + j5.0 ) +

= ( 8.66 + j5.0 ) + ( 4.72 + j3.5 ) = 13.38 + j8.5 Ω

= 15.85 Ω ∠+32.43º

I1 =

120 V ∠0º E1 = = 7.571 A ∠−32.43º ZT 15.85 Ω ∠+32.43º

10 Ω 30° Z1

E1

( 80.01 + j138.57 ) ( 24.98 + j10.83 )

120 V 0°

8.0 Ω 45° Z2

ZL 20 Ω 15°

10 Ω 30° Z1

8.0 Ω 45° Z2

20 Ω 15° ZL 117 V −15°

(a)

Figure 24-5  Circuit diagram for Example 24-2B

(b)

E2

To avoid repeatedly converting between rectangular and polar coordinates, we can use the procedure shown in ­Example 20-11 to do the phasor multiplication and division in rectangular coordinates.

704

Chapter 24   Impedance Networks

Applying the current-divider principle gives the first component of the load current: I1 = I1



Z2 Z2 + ZL

= 7.571 A ∠−32.43º ×



= 2.224 A ∠−10.87º



8 Ω ∠+45º 27.23 Ω ∠+23.44º

Step 2 We go back to the original circuit and replace source E1 with its internal ­impedance Z1, as shown in Figure 24-5(b). In the resulting circuit, ZT = Z2 +



Z1ZL Z1 + ZL

= ( 5.657 + j5.657 ) +



= ( 5.657 + j5.657 ) +



( 8.66 + j5.0 ) × ( 19.319 + j5.176 ) ( 8.66 + j5.0 ) + ( 19.319 + j5.176 )

( 141.42 + j141.42 ) ( 27.98 + j10.18 )

= ( 5.657 + j5.657 ) + ( 6.088 + j2.839 )



= 11.745 + j8.496 Ω



= 14.50 Ω ∠+35.88º

117 V∠−15º E2 = = 8.069 A∠−50.88º ZT 14.5 Ω ∠+35.88º The second component of the load current is

and

I2 =

IL 2 = I 2



Z1 Z1 + ZL

= 8.069 A∠−50.88º ×



= 2.710 A∠−40.87º



10 Ω ∠+30º 29.77 Ω ∠+19.99º

Step 3 The actual load current is the phasor sum of the two components:

IL = IL1 + IL2

= 2.224 A ∠−10.87º + 2.710 A ∠−40.87º

= ( 2.184 − j0.419 ) A + ( 2.049 − j1.773 ) A = ( 4.233 − j2.192 ) A = 4.77 A ∠− 27.4°

24-3   Superposition Theorem

Example 24-3 Calculate the reading of the AC milliammeter in Figure 24-6(a). Assume that the impedance of the meter is negligible. Z1

A

2.0 kΩ EA

20 V 0°

40 mA 30°

IB

Z2

j4.0 kΩ

(a) Z1 I1

2.0 kΩ EA

Z1

A

A I2

2.0 kΩ j4.0 kΩ

20 V 0°

Z2

IB

40 mA 30°

(b)

j4.0 kΩ

Z2

(c)

Figure 24-6  Circuit diagram for Example 24-3

Solution Step 1 To find the current resulting from just the voltage source EA, we replace the current source with an open circuit, as shown in Figure 24-6(b). Then IB ­becomes zero and the first component of the meter current is I1 =

20 V∠0º EA = = 4.472 mA ∠−63.43º ( 2 + j4 ) kΩ Z1 + Z2

Step 2 In the original circuit, we replace the voltage source with a short circuit, as shown in Figure 24-6(c). Applying the current-divider principle gives the second component of the meter current: I2 =



=



Z1 × IB Z1 + Z2

2 kΩ ∠0º × 40 mA ∠+30º 4.472 kΩ ∠+63.43º

= 17.89 mA ∠−33.43º

705

706

Chapter 24   Impedance Networks

Step 3 Since the two component currents have the same direction,

IM = I1 + I2 = (2 − j4) mA + (14.93 − j9.86) mA = 16.93 − j13.86 mA IM = √16.932 + 13.862 = 22 mA

circuitSIM

To verify the current through the milliammeter, download Multisim file EX24-3 from the website, and follow the instructions in the file.

walkthrough‑

Circuit Check

A

CC 24-1. Use mesh analysis to find the current in the 3-Ω resistor in Figure 24-7. j2 Ω

+

50 V 60°

−

50 V 60°

j2 Ω

+ −j4 Ω

2Ω

− + −

50 V 60° Ix

3Ω

+ 50 V 60°

j5 Ω

− 50 V 0°

− +



Figure 24-7

CC 24-2. Use superposition to calculate Ix in the circuit of Figure 24-8. 10 A –30° Ix j5

j8 –j2

1.0 Ω

100 V 30° 10 A 60° 5.0 Ω



Figure 24-8

–j3

–j2

24-4   Thévenin’s Theorem

CC 24-3. Use superposition to determine the voltage across the reactance XL2 in the circuit of Figure 24-9. XL1 10 Ω

XL2 25 Ω

R1 25 Ω

+X C2 10 Ω I V1

+

V2

+

10 V 0° +

+ XC3 10 Ω

2.0 A 0°

12 V 60°

XC1 25 Ω



Figure 24-9

See Problems 24-1 to 24-27 and Review Question 24-38.

24-4  Thévenin’s Theorem Since Thévenin’s theorem replaces a complex impedance network containing one or more sources with a single voltage source and its internal impedance, this theorem is one of the most useful circuit-analysis techniques. To apply Thévenin’s theorem to AC circuits, we state it in terms of impedances: Any two-terminal network of sources and fixed impedances may be replaced by a single voltage source that has • an equivalent voltage equal to the open-circuit voltage at the terminals of the original network, and • an internal impedance equal to the impedance looking back into the network from the two terminals with all sources replaced by their internal impedances. Thévenin’s theorem is particularly useful for analyzing amplifier c­ ircuits. We can represent the transistors (the active components) by Théveninequivalent sources. Then we apply Thévenin’s theorem to ­incorporate the coupling network into the signal source, as shown in ­Figure 24-10. 50 kΩ

50 kΩ

80 kΩ −30° ZTh

50 kΩ

Rint E

10 V 0° 400 Hz

−j150 kΩ

(a) Figure 24-10  Equivalent circuits for an amplifier

E Th

4.5 V −63.4°

(b)

707

708

Chapter 24   Impedance Networks

Example 24-4 Determine the Thévenin-equivalent source for the signal source and four-terminal network of Figure 24-10(a). Solution Step 1 With no load connected to the output terminals, the impedance across the original source is

Z = 50 kΩ + 50 kΩ − j50 kΩ = 111.8 kΩ ∠−26.56º

10 V∠0º E = = 89.44 µA∠+26.56º Z 111.8 kΩ ∠−26.56º Therefore, the open-circuit voltage is and

I=

ETh = VC = I X C = 89.44 µA∠+26.56º × 50 kΩ ∠−90º = 4.47 V∠−63.44º

Step 2 Next we find the impedance that would be measured at the output terminals if the voltage source were replaced by just its 50-kΩ internal resistance. Since all the given impedances are in kilohms, we can simplify the calculation by working in kilohms.

( 100 + j0 ) × ( 0 − j50 ) ( 100 + j0 ) + ( 0 − j50 ) −j5000 = ( 50 + j0 ) + ( 100 − j50 )

ZTh = ( 50 + j0 ) +

= ( 50 + j0 ) + ( 20 − j40 ) = ( 70 − j40 ) kΩ = 81 kΩ ∠−30°

By making it possible to remove the load from the source network, Thévenin’s theorem provides an alternative solution for Example 24-1.

Example 24-1B Find the current through the load ZL in the bridged-T ­ network of Figure 24-1, given E = 20 V∠0º, Z1 = Z2 = −j5.0 kΩ, Z3 = Z4 = 20 + j0 kΩ, and ZL = 5.0 + j5.0 kΩ.

24-4   Thévenin’s Theorem

20 kΩ −j5.0 kΩ

Z4

−j5.0 kΩ ZTh

Z1 20 V 0°

E

Z2

20 kΩ

Z3

ZL

ETh

(a)

(b)

Figure 24-11  Circuit diagrams for Example 24-1B

Solution Step 1 Removing the load from the original network gives the circuit in Fig­ure 24-11(a). We can see that this circuit is a series-parallel combination of components by redrawing the circuit as shown in Figure 24-12(a). We again express the impedances in kilohms throughout the calculations. ZT = Z3 +



Z1 ( Z 4 + Z2 ) Z1 + ( Z 4 + Z2 )

= ( 20 + j0 ) +



= 20 +



−j5.0 ( 20 − j5.0 ) −j5.0 + ( 20 − j5.0 )

−25 − j100 20 − j10

= 20 kΩ + ( 1 − j4.5 ) kΩ



ZT = ( 21 − j4.5 ) kΩ = 21.48 kΩ ∠−12.09º



IT =



20 V∠0º E = = 931.2 µA∠+12.09º ZT 21.48 kΩ ∠−12.09º A

Z2

Z4 E

Z1

A Z2

Z4

Z1

Z3

Z3 B (a)

B (b)

Figure 24-12  Determining Thévenin equivalent voltage and impedance

709

710

Chapter 24   Impedance Networks

From the current-divider principle, the current through Z4 and Z2 is

I2 = IT

= IT



Z1 Z1 + ( Z4 + Z2 ) −j5 kΩ

20 − j10 kΩ

= 931.2 µA∠+12.09º ×



= 208.2 µA∠−51.34º



5 kΩ ∠−90º 22.36 kΩ ∠−26.57º

Step 2 The voltage drops across Z2 and Z3 are

V2 = I 2 Z 2 = 208.2 µA∠− 51.34º × 5 kΩ ∠− 90º = 1.041 V∠−141.34º

  V3 = IT Z3 = 931.2 µA∠+ 12.09º × 20 kΩ ∠0º = 18.624 V∠+12.09º Therefore,

ETh = V2 + V3

= 1.041 V∠−141.34º + 18.624 V∠+12.09º

= ( −0.813 − j0.650 ) + ( 18.211 + j3.901 )



= ( 17.398 + j3.251 ) V



= 17.7 V∠+10.58º



Step 3 Figure 24-12(b) shows the series-parallel combination of impedances that is equal to ZTh. We find the equivalent of Z3 and Z1 in parallel, then calculate ZTh:



Z eq =

Z Th =

=

=

20 × ( −j5.0 ) −j100 Z3 × Z1 = = = ( 1.176 − j4.706 ) kΩ Z3 + Z1 20 − j5.0 20 − j5.0 Z 4 × ( Z 2 + Z eq )

Z 4 + ( Z 2 + Z eq )

20 × ( −j5.0 + 1.176 − j4.706 ) 20 + ( −j5.0 + 1.176 − j4.706 )

23.53 − j194.12

21.176 − j9.706

= ( 4.39 − j7.15 ) kΩ

24-4   Thévenin’s Theorem

Step 4 Returning to the Thévenin-equivalent circuit of Figure 24-11(b), we obtain I=



=



=



=



ETh ZTh + ZL

17.7 V∠+10.58º ( 4.39 − j7.15 ) kΩ + ( 5 + j5 ) kΩ 17.7 V∠+10.58º ( 9.39 − j2.15 ) kΩ

17.7 V∠+10.58º 9.64 kΩ ∠−12.92º

= 1.8 mA∠+24°



Thévenin’s theorem also allows us to avoid using simultaneous­ equations for the multisource network of Example 24-2.

Example 24-2C Two 60-Hz alternators connected in parallel feed a load with an impedance of 20 Ω ∠+15º. Alternator 1 has an open-circuit voltage of 120 V and an internal impedance of 10 Ω ∠+30º. Alternator 2 has an opencircuit voltage of 117 V that lags behind the voltage of alternator 1 by a constant 15°. The internal impedance of alternator 2 is 8.0 Ω ∠+45º. Find the load current. Solution 10 Ω 30° Z1

8.0 Ω 45° Z2 ZTh

A

A E1

120 V 0°

117 V −15°

E2

ZL 20 Ω 15°

ETh

B B (a)

Figure 24-13  Circuit diagrams for Example 24-2C

(b)

711

712

Chapter 24   Impedance Networks

Step 1 Removing ZL from the original circuit (Figure 24-3) leaves the simple series circuit shown in Figure 24-13(a). The Kirchhoff’s voltage law equation for this series circuit is (Z1 + Z2)I = E1 − E2

Therefore,

I= =



=



=



120 V∠0º − 117 V∠−15º 10 Ω ∠+30º + 8 Ω ∠+45º

( 120 + j0 ) − ( 113 − j30.28 ) ( 8.66 + j5 ) + ( 5.657 + j5.657 ) 7 + j30.28

14.317 + j10.657

31.078∠+76.98º 17.85∠+36.66º

= 1.741 A∠+40.32º



The voltage drop across Z1 is

V1 = I Z1 = 1.741 A ∠+40.32º × 10 Ω ∠+30º = 17.41 V ∠+70.32º

VAB = E1 − V1

= 120 V∠0º − 17.41 V∠+70.32º

= ( 120 + j0 ) V − ( 5.863 + j16.393 ) V

VTh = VAB = 114.137 − j16.393 V = 115.3 V∠−8.173º

Step 2 The Thévenin equivalent impedance across terminals A and B in Figure 24-13(a) is Z1 and Z2 in parallel:

ZTh =

=

Z1 × Z2 Z1 + Z2

80 Ω 2∠+75º 17.85 Ω ∠+36.66º

= 4.482 Ω ∠+38.34º = 3.515 + j2.780 Ω

24-5   Norton’s Theorem

Step 3 In the Thévenin-equivalent circuit of Figure 24-13(b),

IL =



=



=

=

ETh ZTh + ZL

115.3 V ∠−8.173º ( 3.515 + j2.780 ) + ( 19.319 + j5.176 ) Ω

115.3 V ∠−8.173º 22.834 + j7.956 Ω 115.3 V ∠−8.173º 24.18 Ω ∠+19.21º

= 4.8 A ∠− 27°

See Problems 24-28 to 24-30 and Review Questions 24-39 and 24-40.

24-5  Norton’s Theorem In Section 9-2, we used a “black-box” technique to find the open-circuit terminal voltage and internal resistance of a two-terminal source network. Starting with the same source network, Norton showed that the black box could contain an equivalent constant-current source. The same black-box parameters Eoc and Isc yield either a constant-voltage source with a voltage of Eoc and a series internal resistance of RTh = Eoc/Isc , or a constant-current source with a current of Isc and a parallel internal resistance of RN = Eoc /Isc. After we find the Thévenin equivalent constant-voltage source for a twoterminal network of voltage sources and fixed impedances, we can simplify further by converting it to the Norton equivalent constant-current source. Norton’s theorem, which combines these two steps into a single procedure, can be applied to AC circuits: Any two-terminal network of sources and fixed impedances may be replaced by a parallel combination of • a single constant-current source with a current equal to the ­current drawn by a short circuit across the terminals of the o ­riginal ­network, and •  an impedance equal to the impedance looking back into the ­network from the two terminals. Converting a constant-voltage source to an equivalent constant-current source is an application of Norton’s theorem. This theorem is particularly useful when a branch in a network is fed from several voltage sources in parallel. Norton’s theorem provides an even shorter solution for Example 24-2 than Thévenin’s theorem does.

713

714

Chapter 24   Impedance Networks

Example 24-2D Two 60-Hz alternators connected in parallel feed a load with an impedance of 20 Ω ∠+15º. Alternator 1 has an open-circuit voltage of 120 V and an ­internal impedance of 10 Ω ∠+30º. Alternator 2 has an opencircuit voltage of 117 V that lags behind the voltage of alternator 1 by a constant 15°. The internal impedance of alternator 2 is 8.0 Ω ∠+45º. Find the load current. Solution Step 1 Determine the short-circuit current of each alternator. I1 =



I2 =

120 V∠0º E1 = = 12 A ∠−30º Z1 10 Ω∠+30º

117 V∠−15º E2 = = 14.63 A∠−60º Z2 8.0 Ω ∠+45º

In Figure 24-14, the tracing directions for the constant-current sources match the tracing directions for the alternators in Figure 24-3. Hence, the sum of the two Norton currents divides among the three parallel impedances in the equivalent circuit. IT = I 1 + I 2



= 12 A∠−30º + 14.6 A∠−60º



= ( 10.39 + j6 ) A + ( 7.31 − j12.67 ) A



= ( 17.7 − j18.67 ) A

= 25.73 A∠−46.53º



12 A −30°

I1

Z1

ZL

Z2

I2

14.6 A −60°

Figure 24-14  Equivalent circuit for Examples 24-2D and 24-2E

24-5   Norton’s Theorem

Step 2

YT = Y1 + Y2 + Y3

= 0.1 S∠−30º + 0.05 S∠−15º + 0.125 S∠−45º

= ( 0.0866 − j0.05 ) S + ( 0.0483 − j0.0129 ) S + ( 0.0884 − j0.0884 ) S = 0.2233 − j0.1513 S = 0.2697 S∠−34.12º

From the current-divider principle (Equation 23-10), 0.05 S ∠−15º YL IL = IT = 25.73 A ∠−46.53º × = 4.8 A ∠−27° YT 0.2697 S ∠−34.12º Example 24-3 demonstrated that the superposition theorem applies to both voltage and current sources. However, when a network has both types of sources, it is usually simpler to analyze an equivalent circuit that has only voltage sources or only current sources. If we convert the constant-current source in Figure 24-15(a) into a constant-voltage source, we have to include Z2 and the milliammeter in the transformation. As a result, the meter current in the new equivalent circuit will not be the same as in the original circuit. But we can replace the constant-voltage source with its Norton-equivalent constant-current source, as shown in Figure 24-15(b). 2.0 kΩ

A IB 40 mA 30°

Z1 EA

20 V 0°

Z2

j4.0 kΩ

(a)

A IA Z1

2.0 kΩ

10 mA 0°

IB

40 mA 30°

Z2

j4.0 kΩ

(b) Figure 24-15  Circuit diagrams for Example 24-3A

715

716

Chapter 24   Impedance Networks

Example 24-3A Calculate the reading of the AC milliammeter in Figure 24-15(a). Assume that the impedance of the meter is negligible. Solution Step 1 The Norton-equivalent source on the left of Figure 24-15(b) has the same i­ nternal impedance Z1 as the voltage source and IA =

20 V∠0º EA = = 10 mA∠0º Z1 2.0 kΩ ∠0º

Step 2 The Kirchhoff’s current-law equation for the network of Figure 24-15(b) is

Step 3 The meter current is

(Y1 + Y2)V = IA + IB 10 mA∠0º + 40 mA∠+30º V= 500 µS∠0º + 250 µS∠−90º =

=

V= IM =

( 10 + j0 ) + ( 34.64 + j20 ) ( 500 + j0 ) + ( 0 − j250 )

44.64 + j20 mA 500 − j250 µS

48.92 mA = 87.5 V 559 µS

V 87.5 V = = 22 mA Z2 4.0 kΩ

See Problems 24-31 and 24-32.

24-6  Nodal Analysis As with other laws for DC networks, we can adapt Kirchhoff’s current law to AC circuits: The phasor sum of the currents leaving any node in an AC circuit equals the phasor sum of the currents entering that node.

24-6   Nodal Analysis

For nodal analysis, we mark the directions of all branch currents leaving and entering each independent node and then write a Kirchhoff’s ­current-law equation for the node. We then replace each impedance current with I = VY, the product of the voltage across the impedance and its admittance. For example, there is just one independent node in Figure 24-14, all currents leaving the upper node are impedance currents, and all currents entering the node are source currents. Therefore, the Kirchhoff’s currentlaw equation is IZ1 + IZL + IZ2 = I1 + I2

But few circuits are as straightforward as the one in Figure 24-14. Con­sequently, we generally use a nodal analysis format that is the dual of mesh equations. For this format, we restate Kirchhoff’s current law as ­follows: The phasor sum of the impedance currents leaving any node in an AC circuit equals the phasor sum of the source currents entering the node. Regardless of the actual directions of source currents, the nodal equations always have the form I Z1 + I Z2 + I Z3 + . . . = I S1 + I S2 + . . .



(24-3)

If an actual source current leaves the node, we enter it as a negative quantity in Equation 24-3. The signs for impedance currents are determined by these rules for the nodal analysis procedure: 1. All voltage sources must be converted to equivalent constant-current sources to provide the source-current terms for Equation 24-3. 2. A point that is common to as many branches as possible is used as the reference node. We number all the other independent nodes. 3. In the Kirchhoff’s current-law equations for the nodes, the voltage across an impedance connected between node n and the reference node is simply the node voltage Vn (with respect to the reference node). For example, the voltage across Z1 in Figure 24-16 is V1. Similarly, the voltage across an impedance connected between node n and an adjacent independent node A is Vn − VA , where VA is voltage at node A with respect to reference node. The current-law equations for the circuit in Figure 24-16 are Node 1: Node 2:

V1 Y1 + ( V1 − V2 ) Y3 = I S 1

V2 Y2 + ( V2 − V1 ) Y3 = −I S 2

(1) (2)

717

718

Chapter 24   Impedance Networks

Node 1 IS1

Node 2

Z3

Z1

V1

Z2

Reference

IS2

V2

Node

Figure 24-16  A circuit with three nodes

Collecting the terms gives

( Y1 + Y3 ) V1 −



Y3 V2 = I S 1

−Y3V1 + ( Y2 + Y3 ) V2 = −I S 2



(24-4) (24-5)

4. By noting the structure of Equations 24-4 and 24-5, we can write the nodal equations with coefficients that can be entered directly into matrices for solving the simultaneous equations. The left side of each nodal equation has one positive term and a varying number of negative terms. The variable in the positive term is the node voltage, Vn, and its coefficient is the phasor sum of all admittances connected to that node. Each negative term is an adjacent node voltage with the admittance ­between that adjacent node and node n as its coefficient. The right side of the equation is the phasor sum of all source currents entering node n.

Steps to follow when applying nodal analysis 1. Determine the number of nodes in the circuit. Select one node as your reference node, to which all voltages will be relative.  2. Assign a voltage polarity across each component where the voltage is unknown. Assign currents through node. The directions of the currents will be arbitrary. 3. Apply Kirchhoff’s current law to each node. 4. Express the current equations in terms of voltages using Ohm’s law, and solve the equations for the unknown node voltages. Consider the circuit in Figure 24-17. R1

+ VS1 −

R3

+ VS2

R2

−

Figure 24-17  Example circuit for applying nodal analysis

24-6   Nodal Analysis

1. We can see that the circuit has 4 nodes, as labelled in Figure 24-18. We’ll select Node B as our reference node. Node C

Node A R1

Node D

R3

+

+

VS1

VS2

R2

−

−

Node B Figure 24-18  Nodal analysis: identifying the nodes in the circuit

2. We assign the polarity of the voltage drops across each resistor and the current in each branch of the circuit. See Figure 24-19. Node A

Node C

R1

+

+ R3 −

− IR1

+ VS1 −

Node D

IR3

+ IR2

VS2

+ R2 −

−

Node B REFERENCE Figure 24-19  Nodal analysis: assigning voltage drops and currents relative to the reference node

3. We apply Kirchhoff’s current law to Node A. IR1 + IR2 + IR3 = 0

4. We express the currents of each branch in terms of circuit voltages using Ohm’s law.

719

720

Chapter 24   Impedance Networks

IR1 =

VR1 VS1−VA = R1 R1

IR3 =

VR3 VS2−VA = R3 R3

IR2 =

VR2 VA = R2 R2

We can now substitute the values into the formula from step 3 and solve for the voltage at Node A.

Example 24-2E Two 60-Hz alternators connected in parallel feed a load with an impedance of 20 Ω ∠+15º. Alternator 1 has an open-circuit voltage of 120 V and an internal impedance of 10 Ω ∠+30º. Alternator 2 has an opencircuit voltage of 117 V that lags behind the voltage of alternator 1 by a constant 15°. The internal impedance of alternator 2 is 8.0 Ω ∠+45º. Find the load current. Solution Step 1 Replacing each source by its equivalent constant-current source and maintaining the same direction for the currents produces the equivalent circuit in Figure 24-14. The short-circuit currents for the sources are I1 =



I2 =

and

120 V∠0º E1 = = 12 A∠−30º Z1 10 Ω ∠+30º

117 V∠−15º E2 = = 14.63 A∠−60º Z2 8.0 Ω ∠+45º

Step 2 The equivalent circuit has only two nodes. If we take the lower as the ­ref­erence node, the upper conductor is the only independent node, and (Y1 + YL + Y2)V = I1 + I2

Therefore

V=

12 A∠−30º + 14.63 A∠−60º 0.10 S∠−30º + 0.050 S∠−15º + 0.125 S∠−45º

24-6   Nodal Analysis





Step 3

= = =

( 10.39 − j6 ) + ( 7.31 − j12.67 ) ( 0.0866 − j0.050 ) + ( 0.0483 − j0.0129 ) + ( 0.0884 − j0.0884 ) 17.7 − j18.67 A

0.2233 − j0.1513 S 25.73 A∠−46.53º 0.2697 S∠−34.12º

= 95.4 V∠−12.41º IL =

95.4 V∠−12.41º V = = 4.8 A∠−27° ZL 20 Ω ∠+15º

There are simpler methods than nodal analysis for solving the network of Figure 24-16. For example, converting both current sources to equivalent constant-voltage sources gives a simple series circuit that can easily be solved by Ohm’s law. However, as a further example, we complete the ­solution by nodal analysis.

Example 24-5 Determine the magnitude of the current through Z3 in Figure 24-16, given that IS1 = 50 mA∠0º , IS2 = 100 mA∠+90º , Z1 = +j50 Ω, Z2 = − j25 Ω, and Z3 = 50 Ω (a resistor). Solution Step 1 Convert the impedances to admittances, and calculate the phasor sums for the coefficients in nodal equations (Equations 1 and 2).



Y1 = −j20 mS    Y2 = +j40 mS    Y3 = 20 mS

Y1 + Y3 = ( 20 − j20 ) mS = 28.3 mS∠−45º

Y2 + Y3 = ( 20 + j40 ) mS = 44.7 mS∠+63.4º

721

722

Chapter 24   Impedance Networks

Step 2 Next, substitute for the coefficients in the nodal equations. Since 1 mS   1 V = 1 mA, we can express the admittances in millisiemens provided we express the currents in milliamperes. For Node 1, For Node 2,

28.3∠−45º V1 − 20∠0º V2 = 50∠0º

44.7∠+63.4º V2 − 20∠0º V1 = −100∠+90º

│−10050∠∠0º+90º = ∠−45º │28.3 −20∠0º

Step 3 Solve for the node voltages:





Step 4 Step 5

V1

=

=

│ −20∠0º 44.7∠+63.4º│ −20∠0º 44.7∠+63.4º

2236∠+63.4º − 2000∠+90º 1265∠+18.4º − 400∠0º 1000∠0º 894∠+26.6º

= 1.12 V ∠−26.6º 28.3∠− 45º −20 ∠0º

50∠0º −100 ∠+ 90º V2 = D −2828∠+45º + 1000∠0º = D =

2236∠−116.6º 894∠+26.6º

= 2.5 V ∠−143.1º

VZ3 = V1 − V2

= (−2 − j1.5 ) − ( 1 − j0.5 )

= ( −3 − j1 ) V

= 3.16 V ∠−161.6º IZ3 =

VZ3 3.16 V = = 63 mA Z3 50 Ω

24-6   Nodal Analysis

Nodal analysis cannot be used for some circuits. For example, we cannot convert the voltage source in the bridged-T network of Example 24-1 to an equivalent constant-current source since the internal impedance of the given source is zero. See Problems 24-1 to 24-27 and Review Questions 24-41 and 24-42.

Circuit Check

B

CC 24-4. Find the Thévenin equivalent circuit for the circuit connected to the load resistor, RL , in Figure 24-20.

V1

L1

+

a

7.0 Ω

I1 1 A 45°

−

5.0 V 0°

+

C1 5.0 Ω

RL 10 Ω

R1

b

3.0 Ω Figure 24-20 

CC 24-5.  For the circuit of Figure 24-21, (a) determine the load impedance, ZL , for maximum power transfer (b) find the maximum power to this load

R1 100 Ω

R2 500 Ω

XC1 50 Ω

R3 330 Ω

a

+ I 10 A 15°

XL1 100 Ω

+

V1 50 V 45°

ZL

b



Figure 24-21

723

724

Chapter 24   Impedance Networks

CC 24-6. Find the voltage across the 2-Ω resistor in Figure 24-22 by nodal analysis. 5.0 Ω –j2.0 Ω

j2.0 Ω

10 A 30°



2.0 Ω

6 A 180°

Figure 24-22

24-7  Delta-Wye Transformation By using the same procedure as with the DC networks in Section 10-4, we can derive equations for converting an AC delta-network into an equivalent wye-network, and vice versa. The only difference is that the delta-to-wye and wye-to-delta transformation equations for AC networks have complex impedances instead of scalar resistances. Delta to wye:

ZA =



ZB =



ZC =

Wye to delta:

ZX =



ZY =

ZYZZ ZX + ZY + ZZ ZXZZ ZX + ZY + ZZ ZXZY ZX + ZY + ZZ ZAZB + ZBZC + ZCZA ZA ZAZB + ZBZC + ZCZA ZB

(24-6)

(24-7)

(24-8)

(24-9)

(24-10)

24-7   Delta-Wye Transformation

ZZ =



ZAZB + ZBZC + ZCZA ZC

(24-11)

For the wye-to-delta transformations, there is an alternative set of equations using admittances, similar to the DC equations with conductances. The ­alternative form is particularly useful for AC circuits because it requires fewer phasor calculations. ZX = ZB ZC ( YA + YB + YC )



ZY = ZAZC ( YA + YB + YC )



ZZ = ZAZB ( YA + YB + YC )



(24-12)

(24-13)

(24-14)

These transformation equations were developed originally for use with three-phase AC power circuits with the configurations shown in Figure 24-23(a) and (b). However, many four-terminal coupling ­networks in electronic circuits have a common input and output ­terminal. The T-network of Figure 24-23(c) is essentially the same as a wye-network. Similarly, the π-network of Figure 24-23(d) is essentially the same as a delta network. Therefore Equations 24-6 to 24-14 also apply to ­transformations between T- and π-­networks. A T-to-π transformation provides another solution for the bridged-T network of Example 24-1.

A

A

ZA

ZB

ZZ ZC

B

C

ZB

ZX

B

(a) Y-network

Input

ZY

(b) Δ-network

ZC ZA

(c) T-network

C

ZX Output

Input

ZZ

ZY

(d) π-network

Figure 24-23  Delta-wye transformations

Output

725

726

Chapter 24   Impedance Networks

Example 24-1C Find the current through the load ZL in the bridged-T network of Figure 24-1, given E = 20 V∠0º , Z1 = Z2 = –j5.0 kΩ, Z3 = Z4 = 20 +j0 kΩ, and ZL = 5.0 + j5.0 kΩ.

Solution If we replace the T-network consisting of Z1, Z2, and Z3 in Figure 24-24(a) with the equivalent π-network, we can solve the resulting circuit of Fig ure 24-24(b) as a series-parallel impedance network. Step 1 We label Z1, Z2, and Z3 as ZA, ZB, and ZC respectively in Figure 24-24(a). We then label the π-network of Figure 24-24(b) with subscripts that correspond exactly to the connections shown in Figure 24-23: ZX does not connect to the same terminal as ZA, ZY does not connect to the same terminal as ZB, and ZZ does not connect to the same terminal as ZC. We need not solve for ZY since it is connected directly across the source and thus does not affect the current through the load. Z4

E

Z4

(ZA)

(ZB)

Z1

Z2 ZL

Z3

(ZC)

ZZ E

ZY

(a)

ZX

ZL

(b)

Figure 24-24  Wye-to-delta transformation for Example 24-1C

Step 2 Next we calculate the common numerator for Equations 24-9 and 24-11, with the impedances expressed in kilohms. ZAZB + ZBZC + ZCZA = −j5.0 × ( −j5.0 ) + ( −j5.0 ) × 20 + 20 × ( −j5.0 )



Then,

= −25 + ( −j100 ) + ( −j100 )

= −25 − j200

= 201.56 kΩ ∠−97.12º

ZX =

ZAZB + ZBZC + ZCZA ZA

24-7   Delta-Wye Transformation

=



201.56 ∠−97.12º 5 ∠−90º

= 40.31 kΩ ∠−7.12º



ZZ =

and

=



ZAZB + ZBZC + ZCZA ZC 201.56 ∠−97.12º 20 ∠0º

= 10.08 kΩ∠−97.12º



Note that ZZ appears to have a negative resistance component. Therefore, we cannot construct an actual equivalent π-network with only passive circuit elements. Nevertheless, we can still use the theoretical transformation to find the current in the actual circuit. Step 3 Excluding ZY, the load on the source in Figure 24-24(b) is





ZT = =

=

=

Z4 × ZZ ZX × ZL + Z4 + ZZ ZX + ZL

20 ∠0º × 10.08 ∠−97.12º 40.31 ∠−7.12º × 7.07 ∠+45º + ( 20 + j0 ) + ( −1.25 − j10.0 ) ( 40 − j5 ) + ( 5 + j5 )

201.56 ∠−97.12º 284.99 ∠+37.88º + 18.75 − j10.0 45 + j0

201.56 ∠−97.12º 284.99 ∠+37.88º + 21.25 ∠−28.07º 45 ∠0º

= 9.49 ∠−69.05º + 6.33 ∠+37.88º

= ( 3.39 − j8.86 ) + ( 5.0 + j3.89 ) = 8.39 − j4.97 kΩ

= 9.75 kΩ ∠−30.64º

Step 4 The total current drawn from the source by Z4, ZZ, ZX, and ZL is

IT =

20 V ∠0º E = = 2.05 mA ∠+30.64º ZT 9.75 kΩ ∠−30.64º

727

728

Chapter 24   Impedance Networks

Step 5 The current-divider principle gives



IL = IT

ZX ZX + ZL

= 2.05 mA ∠+30.64º × = 1.8 mA ∠ + 24º

40.31 kΩ ∠−7.12º 45 kΩ ∠0º

See Problems 24-33 to 24-36 and Review Question 24-43.

Problems

729

Summary

• When loop equations are used to analyze AC circuits, phasor voltages must be entered into the voltage equations. • Mesh equations for AC circuits follow the same rules as for DC circuits, except that voltages appearing in equations are phasor quantities. • The superposition theorem may be used to analyze an AC circuit containing more than one independent source. • An AC circuit may be simplified for analysis purposes by means of Thévenin’s theorem or Norton’s theorem. • In nodal analysis of AC circuits, phasor currents must be entered into the current equations. • Delta-wye transformations for impedance networks in AC circuits use the same procedure as for DC circuits, except that the impedances in the equations are complex quantities.

Problems

Sections 24-1, 24-2, 24-3, and 24-6  Loop and Mesh Equations, Superposition Theorem, and Nodal Analysis I

For Problems 24-1 to 24-20, use one method to solve the problem and another to check the answer. 24-1. Find the source current drawn from each source in Figure 24-25. j5 Ω + V1 10 V 0°

2Ω

B = beginner

I = intermediate

A = advanced

−j3 Ω +

10 Ω

V2 5V 10°

Figure 24-25

I

24-2.

I

24-3.

(a) Determine the voltage across the 10-Ω resistor in Figure 24-25. (b) Use Multisim to verify the voltage across the 10-Ω resistor. Determine the value of the current flowing through the 100-Ω resistor in Figure 24-26. 100 Ω V1 15 V 0°

200 Ω V2 10 V 45°

+ −j50 Ω +

Figure 24-26

I

24-4.

Find the voltage drop across the capacitor in Figure 24-26.

circuitSIM walkthrough

730

Chapter 24   Impedance Networks

I

24-5.

Determine the currents of each branch in the circuit of Figure 24-27.

j1 kΩ

1.2 kΩ

1 kΩ

−j4.7 kΩ

j2.2 kΩ

+

+

50 V 60°

20 V 15°

j3.3 kΩ

Figure 24-27

circuitSIM walkthrough

I I

24-6.

(a) Determine the voltage across both resistors in Figure 24-27. (b) Use Multisim to verify the voltages. 24-7. Determine all three branch currents in Figure 24-28.

Z1 4.0 Ω 30°

Z2 4.0 Ω 30° ZL 10 Ω −45° 20 V 90°

20 V −90°

Figure 24-28

I A

24-8. Reverse the polarity of the 20 V∠+90º-source and determine all three branch currents in Figure 24-28. 24-9. Determine the currents in Z1 and Z4 in Figure 24-29. 50 Ω 45° Z1 20 Ω −30° Z2 E1

80 V 0°

20 Ω 30° Z3 Z4 40 Ω 0° E2

100 V −30°

Figure 24-29

A A

24-10. Determine the currents in Z2 and Z3 in Figure 24-29. 24-11. If the source voltage in Figure 24-30 is 10 V∠0º, calculate the magnitude of the source current.

Problems

731

E

2.0 + j3.0 Ω

2.0 − j3.0 Ω

Z1

Z2

Z3 4.0 − j3.0 Ω

Z4 5.0 + j5.0 Ω

Z5 4.0 + j3.0 Ω

Figure 24-30

A

I

24-12. (a) If the source voltage in Figure 24-30 is 25 V ∠−45º, find the magnitude of the voltage across Z4. (b) Use Multisim to verify the calculation of the magnitude of the voltage across Z4. 24-13. Calculate the magnitude of the current flowing through the capacitor in ­Figure 24-31. 200 Ω

200 Ω

circuitSIM walkthrough

−j1 kΩ +

2.0 A 0°

100 Ω

j500 Ω

25 V 75°

Figure 24-31

I

I

24-14. (a) Determine the magnitude of the voltage across the inductor in ­Fig­ure 24-31. (b) Use Multisim to verify the magnitude of the voltage across the capacitor. 24-15. Determine the magnitude of the voltage across each constant-­ current source in Figure 24-32.

circuitSIM walkthrough

40 mA 0°

5.0 kΩ 50 mA −45°

j2.0 kΩ

−j10 kΩ

2.0 kΩ Figure 24-32

I

24-16. (a) Find the magnitude of the current through the capacitor in Figure 24-32. (b) Use Multisim to verify the magnitude of the current through the capacitor.

circuitSIM walkthrough

732

Chapter 24   Impedance Networks

I

24-17. Determine the load current in Figure 24-33.

j10 Ω

−j50 Ω ZL 50 Ω 75°

50 V 0°

50 Ω

2.0 A 0°

120 V 30°

Figure 24-33

I A

24-18. Find the load voltage in Figure 24-33 if the direction of the 2-A∠0° current is reversed. 24-19. Determine the voltage drop across the 5.0-kΩ resistor in Figure 24-34. 10 kΩ

5.0 kΩ −j10 kΩ

200 V 0°

10 kΩ 150 V −30°

Figure 24-34

I

circuitSIM walkthrough

A

A

circuitSIM walkthrough

24-20. (a)  Determine the magnitude of the two capacitor currents in Figure 24-34 if the direction of the 200-V source is reversed. (b) Use Multisim to verify the magnitude of the two capacitor currents. 24-21. A 26.5-μF capacitor is connected in series with a 100-Ω resistor to a 120-V 60-Hz source. What resistance must be connected in parallel with the capacitor to make the current in the capacitor have a magnitude of 0.5 A? 24-22. (a) At what frequency will the output voltage of the network in ­ Fig­ure 24-35 be in phase with the input voltage? (b) Use Multisim to verify the frequency at which the output voltage will be in phase with the input voltage. 50 kΩ

Input

10 nF

100 kΩ

5.0 nF Output

Figure 24-35

A

24-23. If the two capacitors in the circuit of Figure 24-36 are identical, what must their capacitance be if the open-circuit output voltage of this network is 90° out of phase with a 60-Hz input voltage?

733

Problems

C1

Input

C2

100 kΩ

100 kΩ

Output

Figure 24-36

I

24-24. Find the current in the common neutral lead of the three-wire, ­single-phase system shown in Figure 24-37. 4.0 Ω 30° E1

Z1 20 Ω 15°

110 V 0° 6.0 Ω 20°

E2

Z2 24 Ω 30°

100 V 0° 4.0 Ω 30°

Figure 24-37

I A

24-25. The circuit in Figure 24-37 becomes a two-phase system if E2 =   110 V ∠90º. Find the current in Z1 under these circumstances. 24-26. (a) Figure 24-38 shows an equivalent circuit for a transistor. It includes a current-controlled voltage source with a magnitude equal to the mutual resistance rm of the transistor times the current through re. If the signal source has an open-circuit voltage, Eg, of 2.0 V RMS and an internal resistance, Rg, of 200 Ω, find the output voltage across a 2.0-kΩ load resistor. (b) Use Multisim to verify the magnitude of the output voltage.

circuitSIM walkthrough

Ie

rc

re

Rg

V = Ie rm

rb Eg re = 20 Ω r b = 200 Ω

rc = 2.0 MΩ rm = 2.0 MΩ

Figure 24-38  An equivalent circuit for a transistor

A

24-27. Figure 24-39 shows an equivalent output circuit of a transistor ­amplifier. To determine the output impedance, Zo, we short-circuit the input terminals and feed an AC signal, Eo, back into the output

RL

734

Chapter 24   Impedance Networks

terminals. From Ohm’s law, Zo = Eo / Io where Io is the current drawn from the signal source. Write an equation for the output impedance in terms of the circuit parameters. re Io I1

RL

h f I1

Eo

Figure 24-39  Output impedance of a common-collector transistor amplifier

A

Section 24-4  Thevenin's Theorem 24-28. Determine the Thévenin-equivalent impedance for the source shown in Figure 24-40 when E has a frequency of 1 kHz. 0.010 μF

E

48 V 0°

0.010 μF

33 kΩ E Out

Figure 24-40

A A

B A

I

24-29. Using Thévenin’s theorem, determine an equivalent source for the network of Figure 24-40 that will be applicable at any frequency. (Hint: Do not include the 33-kΩ resistor in the Thévenin transformation.) 24-30. Solve Example 24-2 by combining Thévenin’s theorem with the ­superposition theorem. That is, for each of the circuit diagrams in Figure 24-5, find the component current through ZL by Thévenin’s theorem rather than by series-parallel impedance calculations.

Section 24-5  Norton's Theorem 24-31. Find the Norton-equivalent source at 1 kHz for the source in Figure 24-40. 24-32. Determine the parameters for a Norton-equivalent source that will be applicable at any frequency for the network of Figure 24-40. Compare this source to the answer for Problem 24-31.

Section 24-7  Delta-Wye Transformation 24-33. Find the equivalent T-network for the circuit of Figure 24-41 when the input frequency is 400 Hz. 0.040 μF

Input

Figure 24-41

100 kΩ

500 kΩ

Output

Review Questions

I A

24-34. Determine the equivalent π-network for the transistor equivalent circuit of Figure 24-38. 24-35. Given that the ammeter in Figure 24-42 has negligible internal resistance, use a wye-delta transformation to find the ammeter reading.

E1

A E2

Z1 27 Ω 30°

110 V 0° 32 Ω

110 V 0°

Z2 41 Ω −45°

Figure 24-42

A

24-36. Find the source current in Problem 24-11 by using a wye-delta transformation.

Review Questions Section 24-1  Loop Equations 24-37. Why is it necessary to show a direction for each of the sources when writing loop equations for AC networks with more than one source?

Section 24-3  Superposition Theorem 24-38. What are the advantages and disadvantages of the superposition theorem as a general network theorem?

Section 24-4  Thévenin’s Theorem 24-39. Thévenin’s theorem is widely used in electronic circuit analysis. What is the main feature of this theorem that makes it so useful? 24-40. Thévenin’s theorem applies only for networks with impedances that are all linear and bilateral. (a) Extend the definition of a linear resistance to explain what a ­linear impedance is. (b) Research the term bilateral impedance and explain its significance for AC circuits.

Section 24-6  Nodal Analysis 24-41. Compare the merits of the six solutions for Example 24-2 and choose the method you think is most suitable. Explain your choice. 24-42. When the loop and nodal equations for an impedance network involve the same number of simultaneous equations, the choice of analysis technique then depends on whether we are solving for one or more branch voltages or one or more branch currents. Explain which technique you would choose.

735

736

Chapter 24   Impedance Networks

Section 24-7  Delta-Wye Transformation 24-43. Under what circumstances can we apply the equations that we developed for wye-delta transformations in three-terminal ­networks to four-terminal coupling networks?

Integrate the Concepts

For the circuit shown in Figure 24-43, solve for I using: (a) loop equations (b) mesh equations (c) the superposition theorem (d) Thévenin’s theorem (e) Norton’s theorem (f) nodal analysis I

j5.0 Ω

5.0 + j5.0 Ω 2.0 A 45°

4.0 + j5.0 Ω

2.0 + j2.0 Ω

5.0 −j3.0 Ω 10V 30°

Figure 24-43

Practice Quiz

1. Which of the following statements are true? (a) In any closed-loop in an AC circuit, the phasor difference of the voltage drops of all components should equal the phasor sum of the voltage source applied. (b) The superposition theorem is useful in the analysis of multiple sources in an AC circuit. (c) Thévenin’s theorem replaces the original circuit by an equivalent voltage source in series with an equivalent impedance. (d) Norton’s theorem replaces the original circuit by an equivalent current source in series with an equivalent impedance. (e) The phasor sum of the currents entering a node in a AC circuit equals the phasor difference of the currents exiting the same node. 2.

For the circuit in Figure 24-44 the loop equation that represents Loop 1 is (a) V1 = (R1 + jXL1) I1 + R2 (I1 − I2) (b) V1 = (R1 + jXL1) I1 + R2 (I1 − I2) (c) V1 = (R1 − jXL1) I1 − R2 (I2 − I1) (d) V1 = (R1 − jXL1) I1 + R2 (I1 − I2)

Practice Quiz

j5.0 Ω

R1

R3

10 Ω

10 Ω I1

+

V1

I2

Loop 1

15 V RMS 60°

Loop 2

C1 −j60 Ω

R2 20 Ω

Figure 24-44

3. For the circuit in Figure 24-44, the loop equation that represents Loop 2 is (a) 0 V = (R3 + jXC) I2 + R2 (I2 − I1) (b) 0 V = (R3 − jXC) I2 + R2 (I1 − I2) (c) 0 V = (R3 − jXC) I2 + R2 (I2 − I1) (d) 0 V = (R3 + jXC) I2 + R2 (I2 − I1)

4. Determine the value of the current I2 in the circuit of Figure 24-45. (a) 1.06 A ∠85.4º (b) 429 mA ∠74.4º (c) 330 mA ∠− 112.6º (d) 439 mA ∠127º R1

a

10 Ω

+ L1 20 Ω

V1

+

15 V RMS 60°

I1 R2 20 Ω

b

Figure 24-45

−

I2

−

V2 20 V RMS 0°

C1 60 Ω

737

738

Chapter 24   Impedance Networks

  5. The voltage drop across the capacitor C1 in Figure 24-45 is (a) 20 V ∠157.4º (b) 13 V ∠157.4º (c) 18 V ∠37º (d) 26 V ∠37º   6. The current flowing through resistor R2 in Figure 24-45 is (a) 330 mA ∠− 112.5º (b) 330 mA ∠67.47º (c) 330 mA ∠− 112.5º (d) 330 mA ∠67.35º   7. The superposition theorem is useful because (a) you can short all the sources (b) you can open all the sources (c) you can work with one source at a time (d) you need only Ohm’s law to solve the circuit  8. The Thévenin equivalent voltage source for the load connected ­between terminals a and b in Figure 24-45 is (a) 18.0 V ∠− 46.1º (b) 17.7 V ∠55.9º (c) 12.1 V ∠−137.3º (d) 17.7 V ∠55.9º   9. The Thévenin’s equivalent impedance for the load connected between terminals a and b in Figure 24-46 is (a) 24.89 Ω ∠−10.29 (b) 22.69 Ω ∠−26.24 (c) 80.52 Ω ∠−52.12 (d) 54.03 Ω ∠ + 65.67 10. The Norton equivalent current source between terminals a and b in Figure 24-46 is (a) 998.83 mA ∠−39.17 (b) 43.98 mA ∠+98.35 (c) 1.28 A ∠−65.21 (d) 1.25 A ∠+44.78

Practice Quiz

R2 75 Ω

I

5A 10°

R1 33 Ω

XL j50 Ω

a

XC −j75 Ω

b

Figure 24-46

11. The Norton equivalent impedance between terminals a and b in­ Figure 24-46 is (a) 22.69 Ω ∠−26.24 (b) 24.89 Ω ∠−10.29 (c) 54.03 Ω ∠+65.67 (d) 80.52 Ω ∠−52.12 12. In a delta-to-wye transformation, the value of ZC is given by (a) Zc = (b) Zc = (c) Zc =

ZYZZ ZX +ZY +ZZ ZXZZ ZX +ZY +ZZ

ZYZZ ZX + ZY +ZZ (d) none of the above

739

25

Resonance

In most of the AC network examples we have considered, the frequency of the source has remained constant. Most AC power systems are designed to keep the frequency as constant as possible. However, in electronics applications such as audio equipment, telecommunications, and computers, frequency is an important variable. Therefore, we must turn our attention to the behaviour of circuits in which the RMS voltage is constant but the frequency can vary.

Chapter Outline 25-1 Effect of Varying Frequency in a Series RLC Circuit  742 25-2 Series Resonance  745 25-3 Quality Factor  748

25-4 Resonant Rise of Voltage  749 25-5 Selectivity  751

25-6 Ideal Parallel-Resonant Circuits  753

25-7 Practical Parallel-Resonant Circuits  758

25-8 Selectivity of Parallel-Resonant Circuits  764

Key Terms resonant frequency 743 resonance  745 resonant circuit  745 tuning  746 varactor 746 quality factor (Q) 748

resonant rise of voltage  750 sensitivity  751 bandwidth  752 half-power frequencies  752 selectivity  753 antiresonance  755

tank current  757 tank circuit  757 resonant rise of tank current  757 damping  764 critically damped  764

Learning Outcomes At the conclusion of this chapter, you will be able to: • explain how frequency affects the impedance of a series-resonant circuit • draw graphs of impedance, current, phase angle, and capacitor voltage versus frequency in a series-­resonant circuit • calculate the resonant frequency of a resonant circuit • calculate the quality factor of a resonant circuit • describe how resistance and the L/C ratio affect the graph of current versus frequency for a series-resonant circuit • calculate capacitor, inductor, and resistor voltages at ­resonance in a series-resonant circuit

Photo sources:  © Antonio De Azevedo Negrão | Dreamstime.com

• calculate the bandwidth of a resonant circuit • describe how the quality factor, Q, affects the sensitivity and selectivity of a series-resonant circuit • explain the effect of frequency on the impedance and admittance of a parallelresonant circuit • draw graphs of impedance, total current, and output voltage versus frequency for a parallel-resonant circuit • describe the effect that the internal resistance of the source has on the selectivity of a parallel-resonant circuit

Chapter 25  Resonance

25-1 Effect of Varying Frequency in a Series RLC Circuit If the circuit connected to a variable-frequency source contains only resistance, the frequency of the source has no effect on the magnitude or phase of the current drawn from the source because I = E / R. But the reactance of an inductor is directly proportional to the frequency because XL = 2π f L. Similarly, the reactance of a capacitor decreases as the frequency increases since XC = 1/2πf C. In the circuit of Figure 25-1(a), resistance, inductance, and capacitance are all connected in series to a variable-frequency source. When the frequency is quite low, the inductive reactance is quite small and the capacitive reactance is quite large. Consequently, the net reactance is a large capacitive reactance, much greater in magnitude than the resistance, and the total impedance is practically equal to –jXC. As the frequency increases, the capacitive reactance decreases while the inductive reactance increases. We can show the effect of increasing frequency on the three ­passive components of the circuit by plotting graphs of their resistance or reactance as a function of frequency, as shown in Figure 251(b). Resistance is not ­frequency-dependent, so its graph is a horizontal line. Since XL is directly proportional to frequency, the graph of XL is a straight line that starts from the o ­ rigin and has a slope proportional to the inductance of the coil. To show that we subtract XC from XL when determining net reactance, we plot XC on the negative y-axis of the graph. Since XC is inversely proportional to the f­requency, the graph of XC is a ­reciprocal curve.

R E = 100 V f variable

(a)

100 Ω

L

0.10 H

C

0.10 μF

Resistance Capacitive Inductive reactance reactance

742

0

ive duct

In XL Frequency

reac

tanc

e

Resistance

XC tance ive reac Capacit

(b)

Figure 25-1 Series RLC circuit connected to a variable-frequency source

As the frequency increases, the net reactance, XL − XC, decreases until the frequency at which XL = XC. The net reactance at this frequency is zero, and therefore the total impedance of the circuit equals the resistance,  R.

25-1   Effect of Varying Frequency in a Series RLC Circuit

743

As the frequency increases further, the inductive reactance exceeds the capacitive reactance, and the total impedance becomes inductive. When the ­frequency is quite high, the total impedance is practically the same as the inductive reactance of the circuit. To examine the effect of frequency on the impedance of the series RLC circuit of Figure 25-1(a) in a more quantitative manner, we can calculate parameters of the circuit at various frequencies. Table 25-1 lists such parameters at selected frequencies from 159 Hz to 15.9 kHz.

TABLE 25-1  Effect of frequency on a series RLC circuit f (Hz)

R (Ω)

159

100

660 985

XL (Ω)

XC (Ω)

100

10k

100

415

2415

100

618

1618

1245

100

781

1281

1440

100

905

1105

1515

100

952

1052

1590

100

1000

1000

1675

100

1052

952

1760

100

1105

905

2040

100

1281

781

2575

100

1618

618

3840

100

2415

415

15 900

100

10 000

100

XL − XC (Ω)

−j9900 −j2000 −j1000 −j500 −j200 −j100 0

+j100

+j200

+j500

+j1000

+j2000

+j9900

Z = √R 2 + X 2 (Ω) 9900

ϕ = tan − 1 ( X/R )

2002 1005 509 223 141 100 141 223 509 1005 2002 9900

When the frequency of the source is 159 Hz, the capacitive reactance of the series circuit is 100 times as great as the inductive reactance, and the series RLC circuit behaves like a 100-Ω resistor in a series with a capacitance having a reactance of 9900 Ω. At a frequency of 15 900 Hz, XL greatly exceeds XC and the series RLC circuit behaves like a 100-Ω resistor in series with an inductance having a reactance of 9900 Ω. The frequency at which the inductive and capacitive reactances cancel out is known as the resonant ­frequency, fr, of the series RLC circuit. For the circuit of Figure 25-1(a), the resonant ­frequency is 1590 Hz. Figure 25-2 shows impedance diagrams for the series RLC circuit of Figure 25-1(a) at the resonant frequency of 1590 Hz and at two specific ­frequencies, one lower than the resonant frequency and the other higher. Figure 25-3 shows how the impedance of the series RLC circuit varies with the frequency of the source. The x-axis of this graph shows frequency with a logarithmic scale, where each unit on the axis corresponds to an increase by the same factor. The y-axis scales for impedance and current also have

−89.4°

I = E/Z (A)

−87.2°

0.0101

−63.4°

0.1965

−84.3°

0.499

−45°

0.448

+63.4°

0.707

+87.3°

0.995

−78.9°

0°

+45°

0.995

0.707 1.00

+78.7°

0.448

+89.4°

0.0499

+84.3°

0.1965

0.0101

Chapter 25  Resonance

XL = 905 Ω

XL = 1000 Ω

R Xeq = XL − XC = 200 Ω capacitive

Z

R Z

Xeq = 0 and Z = 100 Ω pure resistance

XL = 1105 Ω

Z R

Xeq = XL − XC = 200 Ω inductive

XC = 1105 Ω

XC = 1000 Ω

XC = 905 Ω

f = 1440 Hz

f = 1590 Hz = fr

f = 1760 Hz

(a)

(b)

(c)

Current (amperes)

Current Phase Angle

Figure 25-2  Impedance of a series RLC circuit when (a) f < fr , (b) f = fr , and (c) f > fr

+90° +45° 0°

−45° −90°

1.0

0.1

0.01 10 k Impedance (ohms)

744

1k

100

10 159

1590 Frequency (hertz)

15 900

Figure 25-3  Effect of frequency on impedance and current in a series RLC circuit

25-2   Series Resonance

logarithmic scales on which each unit represents a factor of 10. Note that using a logarithmic frequency scale produces impedance and current graphs that are symmetrical about the resonant frequency. If the source voltage remains constant at 100 V RMS as the frequency v ­ aries, the magnitude of the current reaches its maximum when the ­impedance reaches its minimum at 1590 Hz. At very low frequencies the impedance of the series RLC circuit of Figure 25-1(a) is almost completely capacitive reactance, so the angle of the impedance is almost −90°. Consequently, at these low frequencies, the current leads the applied voltage by almost 90°. At 1590 Hz, the net reactance becomes zero and the total impedance is simply the resistance of the circuit with a 0° angle. At very high frequencies, the impedance is almost completely inductive reactance, and the current lags behind the applied voltage by almost 90°. See Review Questions 25-24 and 25-25 at the end of the chapter.

25-2  Series Resonance When the frequency of the source is 1590 Hz in the series RLC circuit of Figure 25-1(a), • the inductive reactance and the capacitive reactance of the circuit are equal, • the impedance is at its minimum and is equal to the resistance of the ­circuit, • the current in the circuit is at its maximum and is equal to E/ R, and • the current through the circuit is in phase with the applied voltage. Each of these conditions shows that the circuit is in resonance. A ­circuit in which these conditions can occur is called a resonant circuit. Since XL = XC at resonance, 2π fr L =



and

fr =

1 2π fr C

1 2π√LC

(25-1)

where fr is the resonant frequency in hertz, L is the inductance in henrys, and C is the capacitance in farads. Equation 25-1 shows that we can bring an RLC circuit into resonance at a particular frequency by varying either the inductance or the capacitance of the circuit. Inductance can be varied by using a threaded shaft to move a ferromagnetic core in or out of a coil, thus changing the reluctance of the magnetic circuit. Mechanically variable inductance and capacitance

745

746

Chapter 25  Resonance

E

100 V

R

50 Ω

L

0.10 H

C

0.10 μF

Current (amperes)

are often used for preset tuning adjustments. Continuously variable tuning is usually done by rotating one set of plates of an air-dielectric capacitor in relation to a set of fixed plates. Electronic tuning uses varactors, solid-state devices with a capacitance that varies with voltage. Equation 25-1 also shows that the resistance of the series-resonant c­ ircuit has no effect on the resonant frequency. The resistance governs the minimum impedance of the circuit that occurs at resonance. As a result, the resistance governs the steepness of the “skirts,” or shoulders, of the resonance curve. In the series-resonant circuit of Figure 25-4(a), the inductance and capacitance are the same as in Figure 25-1(a). Therefore, the resonant 2.0

R = 50 Ω

1.0

R = 100 Ω

0.1 0.01 fr Frequency (log scale)

(a)

(b)

Source:  Johanson Giga-Trim® and Air Trimmer Capacitor

Figure 25-4  Effect of resistance on the shape of the resonance curve

Variable trimmer capacitors

Tuning circuit in a radio

25-2   Series Resonance

747

R E

100 V

100 Ω

L

0.20 H

C

50 nF

Current (amperes)

frequency is still 1590 Hz. However, the resistance has been reduced from 100 Ω to 50 Ω. By preparing a table similar to Table 25-1 for Figure 25-4(a), we can plot the resonance curve shown in Figure 25-4(b). Off resonance, the total impedance is not affected very much by the change in resistance. But at the resonant frequency, the impedance is only half that of the circuit of Figure 25-1(a), and thus the circuit current at resonance is twice as great. Note also from Equation 25-1 that as long as the product of L and C is constant, the resonant frequency of a series-resonant circuit remains constant. In Figure 25-5(a), the inductance is doubled and the capacitance is cut in half. Therefore, the LC product and the resonant frequency are unchanged. However, the L /C ratio has been increased by a factor of 4. As a result, at resonance, XL = XC = 2000 Ω as opposed to 1000 Ω, for the circuit of Figure 25-1(a). Similarly, at a frequency of 1515 Hz, XL − XC = 200 Ω instead of 100 Ω. Thus, an increase in the L /C ratio steepens the shoulders of the resonance curve.

1.0

L/C = 1 × 106 H/F L/C = 4 × 106 H/F

0.1 0.01 fr

Frequency (log scale) (a)

(b)

Figure 25-5  Effect of L/C ratio on the shape of the resonance curve

Example 25-1 A coil in a tuned circuit in a radio receiver has an inductance of 300 μH and a resistance of 15.0 Ω. What capacitance must be connected in series with the coil for the circuit to be resonant at 840 kHz? Solution

At resonance, XC = XL = 2πfL = 2 × π × 840 kHz × 300 µH = 1583 Ω 1 Since Xc = , 2πfC 1 1 C= = = 120 pF 2π f X C 2 × π × 840 kHz × 1583 Ω

To verify the capacitance, download Multisim file EX25-1 from the website, and follow the instructions in the file. See Problems 25-1 to 25-4 and Review Questions 25-26 to 25-33.

circuitSIM walkthrough

748

Chapter 25  Resonance

25-3  Quality Factor In an AC circuit containing both inductance and capacitance, the capacitance stores energy while the inductance returns energy stored in its magnetic field, and vice versa. Thus, AC circuits containing both inductance and capacitance exchange reactive power between the inductance and capacitance. Hence, the source needs to supply only the difference between the reactive power of the inductance and the reactive power of the capacitance. That is why the net reactance of a series circuit is the difference between the inductive and capacitive reactance, and why the reactive voltage of a series circuit is the difference between VL and VC. At resonance, the reactive power of the inductance and capacitance are equal and the source has to supply only the active power required by the resistance of the circuit. The letter Q can also stand for electric charge. Unfortunately, there are more electrical and magnetic quantities than letters in the alphabet!

The ratio of the reactive power of either the inductance or the ­capacitance at resonance to the active power of a resonant circuit is the quality factor, Q, of that circuit. Since the letter Q can represent either reactive power or quality factor, we use PQ to represent reactive power in any equations that also involve quality factor.

Q=

PQ P

(25-2)

Since the resistance in a resonant circuit is usually the resistance associated with the coil, we often refer to the quality factor, or Q, of a coil. Since PQ = I 2XL, P = I 2R, and the same current flows through all components in a series circuit,

Equation 25-3 is sometimes used to define Q.

and

Q=

Q=

I2XL I2R

XL ωL = R R

(25-3)

Equation 25-3 provides a simpler calculation of Q than Equation 25-2. Since ωL = 2πf L and fr = 1/2πp√LC,

Q=

√

1 L R C

(25-4)

The Q factor of a resonant circuit is, therefore, the single factor that takes into account the effect of both the resistance and the L/C ratio on the shape of a resonance curve. The higher the Q, the steeper the skirts of the resonance curve.

25-4   Resonant Rise of Voltage

Example 25-2 Find the quality factor of a 200-pF capacitor in series with a 50-μH inductance that has a resistance of 10.0 Ω. Solution I Solution II

Q=

√

√

1 L 1 50 µH 1 × 500 = 50 = = R C 10 200 pF 10

Since we usually want to know the resonant frequency of a circuit, it is often convenient to use fr to calculate Q.



fr =

1 1 = 1.592 MHz = 2 × π × √50 µH × 200 pF 2π√LC

XL = 2π f L = 2 × π × 1.592 MHz × 50 µH = 500 Ω Q=

XL 500 = = 50 R 10

See Problem 25-5 and Review Question 25-34.

Circuit Check

A

CC 25-1. A tuning capacitor has a range of 300 pf to 1500 pf. What size inductor must be placed in series with it so that the highest frequency that can be tuned is 200 kHz? What is the lowest ­frequency the circuit can tune? CC 25-2. A series-resonant circuit has a Q of 100 at the resonant ­frequency. The supply voltage is 10 V and resonant frequency is 1.0 MHz. If the current flowing at resonance is 100 mA, find the resistance, inductance, and capacitance.

25-4  Resonant Rise of Voltage Table 25-2 summarizes calculations of the component voltages in the series-resonant circuit of Figure 25-1(a). Plotting the data for VC produces the graph in Figure 25-6, which shows how the voltage across the capacitor varies with the frequency of the source.

749

750

Chapter 25  Resonance

TABLE 25-2 Voltages in the series RLC circuit of Figure 25-1 f (Hz)

VR = IR (V)

VC (volts)

159

1.01

VL = IXL (V) 1.01

VC = IXC (V) 101

1000

660

4.99

20.7

120

800

985

9.95

61.5

161

1245

600

1440

400

1515 1590

200 0

fr Frequency (log scale)

Figure 25-6  Voltage across the capacitance of the series-resonant circuit in Figure 25-1

19.65 44.8 70.7 100

153.5 406

252 495

675

747

1000

1000

1675

70.7

747

675

1760

44.8

495

406

2040

19.65

252

153.5

2575

9.95

161

61.5

3840

4.99

120

20.7

15 900

1.01

101

1.01

At frequencies well below the resonant frequency of the circuit, the voltage across the capacitor is almost the same as the applied voltage because the reactance of the capacitor accounts for most of the total impedance of the circuit. As the source frequency approaches the resonant frequency, the increasing current due to the decreasing total impedance causes the v ­ oltage across the capacitance to exceed the applied voltage. Because VL and VC are equal but 180° out of phase at resonance, VR is the same as the  applied ­voltage while VL and VC are both much greater than the applied voltage. As  we  increase the frequency to many times the resonant frequency, the impedance of the circuit is almost the same as the inductive reactance of the circuit and the voltage across the capacitance becomes very small. Since capacitive reactance decreases as frequency increases, the product IXC reaches a maximum at a frequency just slightly lower than the resonant frequency of the circuit (see Figure 25-6). This frequency difference is ­usually so small that we can ignore it. The frequency intervals in Table 25-2 are too large to show the exact frequency at which VC is highest. Similarly, the exact frequency at which VL is highest is slightly higher than the ­resonant frequency of the circuit. The increase in the voltage across the capacitance and the inductance at resonance is called resonant rise of ­voltage. EXC Since I = E /Z, VC = IXC = R At resonance, Z = R, and

VC =

EXC R

25-5  Selectivity

But XC /R = Q. Therefore, the resonant rise of voltage in a series-resonant circuit is VC = QE



(25-5)

where E is the source voltage at the resonant frequency. Because Q is usually much greater than 1, a voltage much greater than the source voltage appears across the capacitance and inductance of a series-resonant circuit at resonance. However, the voltage drops across the inductance and capacitance are equal and opposite at resonance, and the voltage drop across the resistance is equal to the applied voltage, so the phasor sum of the voltage drops does equal the applied voltage, as required by Kirchhoff’s voltage law. Resonant rise of voltage can be used in radio receivers as a means of increasing the input voltage of the signal to which the circuit is tuned. Figure 25-7(a) shows a typical antenna-input circuit of a radio receiver. Antenna current in the primary coil induces a voltage in the tuned secondary coil. The induced voltage acts in series with the coil and capacitor, as shown in Figure 25-7(b). Equation 25-5 shows that the voltage across the tuning capacitor at resonance is Q times the voltage induced in the coil by the antenna signal. The higher the Q of the tuned circuit, the greater its sensitivity. R Antenna

L To amplifier

(a)

To C amplifier E

(b)

Figure 25-7  A radio receiver input circuit: (a) actual circuit, (b) equivalent circuit

See Problems 25-6 and 25-7 and Review Question 25-35.

25-5 Selectivity Since the capacitance of the circuit of Figure 25-7 is variable, we can choose the frequency at which the resonant rise of voltage provides a voltage gain of Q. The shape of the resonance curve in Figure 25-6 is such that frequencies close to the resonant frequency receive almost as great an ­increase in voltage as the resonant frequency itself. Therefore, we can say that a ­resonant circuit selects a certain band of frequencies. This property has

751

Equation 25-5 is sometimes used to define Q.

many applications desirable in radio communications because any signal carrying information requires a bandwidth directly proportional to the rate at which information is transmitted. However, the skirts of the resonance curve are not steep enough that we can draw a definite boundary between the frequencies that are chosen and those that are rejected. Since the current is common to all components in a series circuit, we return to the graph of current versus frequency in a series-resonant circuit to define the bandwidth of a tuned circuit. As the frequency of the source decreases from the resonant frequency of a tuned circuit, the impedance of the circuit increases and the current decreases correspondingly. As shown in Figure 25-8(a), we consider the frequencies f1 and f2 at which the current is diminished to 1/√2 , or about 0.707, of its value at resonance to be the limits of the band of frequencies that the tuned circuit accepts. Then, the total bandwidth is Δf = f2 − f1



(25-6)

Since P = I 2R, the active power input to the tuned circuit at frequencies f1 and f2 is P=



Im 2 1 R = Pm 2 √2

( )

where Im is the current at resonance and Pm is the power at resonance. Consequently, the frequencies f1 and f2 on a resonance curve are referred to as the half-power frequencies.

Imax

1. 41

4R

0.707Imax

=

Δf

XL − XC = R

Z

The relationship ­between bandwidth and the rate of sending information is known as Hartley’s bandwidth law.

Chapter 25  Resonance

Current

752

45°

f1

fr

f2

R

Frequency (log scale) (a)

(b)

Figure 25-8  Bandwidth of a resonant circuit

For the current to drop to 0.707 of the current at resonance, the impedance of the tuned circuit must be √2, or about 1.414, times the impedance at resonance. At resonance, Z = R; hence, the circuit impedance at the ­half-power frequencies must be 1.414R. Applying trigonometry to the impedance diagram in Figure 25-8(b) shows that the net reactance at the half-power points equals the resistance of the tuned circuit. Since inductive reactance

25-6   Ideal Parallel-Resonant Circuits

753

is directly proportional to frequency and capacitive reactance is inversely proportional to frequency, XL at frequency f2 must have increased by 12R from its value at fr while XC decreased by 12R. Similarly, XL at frequency f2 must have decreased by 12R from its value at fr. Therefore, 2πf2L − 2πf1L = 2πL( f2 − f1) = R Δf =



R 2πL

(25-7)

Dividing both sides of Equation 25-7 by fr gives Δf



fr

=

and

R 1 R = = 2πf r L XL Q Δf =

fr  Q

(25-8)

Therefore, an increase in the quality factor of a resonant circuit not only increases the sensitivity by increasing the resonant rise of voltage, but also increases the selectivity by decreasing the bandwidth of the tuned circuit.

Example 25-3 (a) Determine the bandwidth of the resonant circuit of Example 25-2. (b) Find the voltage across the tuning capacitor at resonance when this ­circuit is connected to a 40-μV signal source. Solution (a) (b)

Δf =

1.592 MHz = 32 kHz Q 50 VC = QE = 50 × 40 µV = 2.0 mV fr

=

To verify the bandwidth in part (a) and the capacitor voltage in part (b), download Multisim file EX25-3 from the website, and follow the instructions in the file.

See Problems 25-8 to 25-16 and Review Question 25-36.

25-6  Ideal Parallel-Resonant Circuits The ideal parallel-resonant circuit of Figure 25-9 consists of an ideal inductor, a capacitor, and a high resistance, all connected in parallel to a voltage source. Since the current in the inductive branch lags behind the

circuitSIM walkthrough

754

Chapter 25  Resonance

E

C

Rp

L

Figure 25-9  Ideal parallel-resonant circuit

applied voltage by 90° and the current in the capacitive branch leads the applied voltage by 90°, IT = √IR2p + ( IC − IL ) 2



Because IL = E / XL and IC = E / XC, the current in the inductive branch decreases and the current in the capacitive branch increases as the f­ re­quency of the source increases. There is a particular frequency at which IL = IC. At this resonant frequency, the current drawn from the source is at its minimum, which is equal to the current in the resistance branch of the ideal parallel-resonant circuit. For the parallel RLC circuit of Figure 25-9, we can draw a series of admit­tance diagrams or a series of current phasor diagrams. As shown in Fig­ure 25-10(b), at the resonant frequency the currents in the inductance and capacitance branches cancel out and the total current is equal to the current in the Rp branch. At any frequency below the resonant frequency, XL is less than XC, making the total current greater than IR and inductive, as shown in Figure 25-10(a). Similarly, at any frequency above the resonant frequency,

IC IC

IC IR

IT = IR

E

IX IT lags E

IT leads E

IL

IX E

E

IR IL

IL

f < fr

f = fr

f > fr

(a)

(b)

(c)

Figure 25-10 Currents in a parallel RLC circuit when (a) f < fr , (b) f = fr , and (c) f > fr

25-6   Ideal Parallel-Resonant Circuits

Total Current

Impedance

the total current in the ideal parallel-resonant circuit is capacitive, as shown in Figure 25-10(c). In this respect, the behaviour of a parallel-resonant circuit is opposite to that of a series-resonant ­circuit. A graph of total current versus frequency for an ideal parallel-resonant circuit has the same shape as the corresponding graph for a series-resonant circuit, but it is inverted (compare Figures 25-3 and 25-11). For a constantsource voltage, total current is directly proportional to the total admittance of the circuit. Hence, the total-admittance graph for the parallel-resonant circuit has the same shape as the total-current curve of Figure 25-11. Since the equivalent impedance of the circuit is the reciprocal of the total ­admittance, the impedance curve for the ideal parallel-resonant circuit is inverted compared to the current curve. The symmetrical appearance of these resonance curves results from the logarithmic frequency scale, as in Figure 25-3. Because the resonance graphs for an ideal parallel-resonant circuit are the reciprocal curves of the corresponding series-resonant circuit, parallel resonance is sometimes called antiresonance.

fr Frequency (log scale) Figure 25-11  Parallel-resonance curves

Since the current in the inductance branch and the current in the capacitance branch are equal at resonance, it follows that the resonant frequency of the ideal parallel-resonant circuit is the same as for series resonance,

fr =

1 2π√LC

(25-1)

755

756

Chapter 25  Resonance

Practical Circuits

Source:  © Antonio De Azevedo Negrão | Dreamstime.com

Tank Circuits in Power Amplifiers Vcc

Cout +

Cin

Input

Q1

+

RL R

Audio power amplifier

Power amplifiers for audio or radio-frequency signals need to be efficient in order to minimize the cooling they require. To improve efficiency, a tank circuit formed by a capacitor and an i­nductor can be connected to a transistor as shown at the top of Figure 25-12. The ­components of the tank are selected so that the amplifier will amplify only signals that are close to the resonant frequency of the tank. The quality factor of the tank determines the steepness of the tank circuit ­frequency response. This arrangement is commonly used in radio transmitters. At high frequencies the tank circuit appears as a parallel RLC circuit connected to the load resistor (Figure 25-13). When tuned to the frequency of the input signal, the tank circuit acts

+C

L

−VBB Figure 25-12  Transistor amplifier with tank circuit

like a purely resistive circuit as the reactive capacitance and reactive inductance cancel each other. Then, all the voltage applied to the input of the amplifier appears at the output, multiplied by the gain of the amplifier.

+ Vin

RL

L

C

−

Figure 25-13 Equivalent RLC circuit

The total current in a parallel-resonant circuit has its minimum value at resonance. However, if the reactances of the inductive and capacitive branches are considerably less than the resistance Rp, the current in the inductance and capacitance is many times greater than the source current. Since the capacitance is charging when the magnetic field around the inductance is collapsing, and vice versa, we can consider this current as the medium that carries the energy back and forth between the capacitance and the

25-6   Ideal Parallel-Resonant Circuits

757

inductance. Therefore, at resonance, a considerable tank current flows in the tank circuit, the closed loop consisting of the inductance and capacitance branches, while only the resistance branch draws current from the source. As with series-resonant circuits, the quality factor of the parallel-­resonant circuit is defined as the ratio of the reactive power at resonance of either the inductance or the capacitance to the active power of the resistance: Q=



PQ P

=

Q=

E2/XL E2/Rp

Rp XL

(25-9)

Note that the ratio of resistance to reactance is inverted compared with Q for a series-resonant circuit in Equation 25-3. The resistance in parallelresonant circuits is usually 100 kΩ or so, whereas the resistance in seriesresonant circuits is usually only a few ohms. At resonance, the tank current in the ideal parallel-resonant circuit is the same as IL or IC. Since E = IRRp = ILXL, IL =



Rp

XL

IR

tank current = Q × source current 

and

(25-10)

We can think of Equation 25-10 as representing a resonant rise of tank current similar to the resonant rise of voltage in a series-resonant circuit. Figure 25-11 shows a resonant rise in impedance. When we use a parallelresonant circuit as the output load for a transistor amplifier with a com­ paratively high internal resistance, the resonant rise of the impedance, Zp, of the p ­ arallel-resonant circuit can produce a resonant rise in output v ­ oltage. Applying the voltage-divider principle in the equivalent circuit in Figure 25-14(a) gives Vout = E ×



High internal resistance

Vout

Zp ZT

=E×

I

Rint + Zp Zp

High Rint

Vout

E

(a)

(b)

Figure 25-14  Equivalent circuits for a parallel-resonant circuit at the output of an amplifier

Since the current graph of Figure 25-11 shows the total ­current drawn from the source, this graph shows a decrease in current at resonance rather than the ­resonant rise in tank current.

Chapter 25  Resonance

When the frequency of the signal source is considerably off resonance, the internal resistance of the source is much greater than the impedance of  the  parallel-resonant circuit. Consequently, most of the signal voltage ­generated by the constant-voltage source appears across the internal resistance of the source. But at the resonant frequency, the impedance of the parallel-resonant circuit is usually greater than the internal impedance of the signal source. Consequently, the output voltage is much greater at the resonant frequency than for frequencies off resonance. In equivalent circuits for transistor amplifiers, it is usually more convenient to convert the constant-voltage source to its equivalent constant-­ current source, as in Figure 25-14(b). From Ohm’s law, Vout = IZT



ZT consists of the internal resistance of the source in parallel with the impedance of the tuned circuit, Zp. Therefore, Vout is greatest at resonance since Zp is then at its maximum value. At resonance, the output voltage approaches the open-circuit voltage of the constant-voltage source, as shown in Figure 25-15. Off resonance, the output voltage is considerably less. Eoc Output Voltage

758

fr Frequency (log scale) Figure 25-15  Output voltage across a parallel-resonant circuit when the source has appreciable internal resistance

See Review Questions 25-37 to 25-39.

25-7 Practical Parallel-Resonant Circuits The losses in the dielectric of a practical capacitor are usually so small that they can be ignored. If we do need to take these losses into consideration, we can represent them as a resistance in parallel with the capacitance. Since the leakage resistance of capacitors is high compared to the resistance of practical inductors, we can omit such parallel resistance from the equivalent circuit of Figure 25-16. However, we must include the resistance of the coil as acting in series with its inductance. Thus, the equivalent circuit for

25-7   Practical Parallel-Resonant Circuits

a practical parallel-resonant circuit is a series-parallel circuit rather than a simple parallel circuit.

Rint

L C Rs

E

Figure 25-16  Practical parallel-resonant circuit

The resonant frequency of a practical parallel-resonant circuit can be defined by any of the following statements, all of which specify essentially the same conditions: The resonant frequency of a practical parallel-resonant circuit is the frequency at which • the reactive power of the inductance and capacitance are equal • the circuit has a power factor of 1 • the total impedance is completely resistive • the source current is in phase with the source voltage If the frequency of the voltage applied to the circuit of Figure 25-16 is such that XL = XC, the impedance of the inductive branch is slightly greater than the reactance of the capacitive branch because of the resistance in series with the inductive reactance. At a slightly lower frequency, there is enough current in the inductive branch to make the reactive power of the inductive branch equal to the reactive power of the capacitance. To find the actual resonant frequency of a practical parallel-resonant ­circuit, we convert the series network of the inductance L and resistance Rs in Figure 25-16 to an equivalent parallel network consisting of Rp and Xp as shown in Figure 25-17.

I

Rint

C

Rp =

Z2 Rs

Xp =

Z2 Xs

Figure 25-17  Parallel equivalent of a practical parallel-resonant circuit

The impedance of the practical inductor of Figure 25-16 can be expressed as Zind = Rs + jXs

759

760

Chapter 25  Resonance

The admittance of the practical inductor is Yind =



1 1 = Zind Rs + jXs

Multiplying by the conjugate of the denominator gives



Since



Yind =

R2s

( Rs − jXs ) jXs Rs − jXs 1 Rs = 2 − 2 = 2 2 2 Rs + jXs ( Rs − jXs ) Rs + Xs Rs + Xs Rs + X2s

jXs 1 Rs 1 Xs  , 2 = 2 2   and  − 2 2 = ( 2 2) = ( 2 + Xs Rs + Xs Rs + Xs j Rs + Xs j Rs + X2s ) Rs Xs Yind =

R2s

1 1 1 1 1 1 + 2 + ( 2 2) = 2 + 2 = R jX + Xs j Rs + Xs Z jZ p p Rs R Xs Xs s

At resonance, the reactances of the inductive and capacitive branches of the equivalent parallel circuit must be equal. Therefore,







1 R2s + ( 2πfr L ) 2 = 2πf r C 2πf r L

(25-11)

L = R2s + ( 2πf r L ) 2 C

fr2 =

fr =

L − CR2s 4π 2L2C

√

1 2 × 1 − CRs 2π√LC L

(25-12)

Examining the last factor in Equation 25-12 reveals several important properties of practical parallel-resonant circuits: • Since the factor is less than 1, the resonant frequency is somewhat lower than the resonant frequency of an ideal parallel-resonant circuit given by Equation 25-1. • The resonant frequency of a practical parallel-resonant circuit is dependent on the resistance of the circuit, unlike the resonant frequency of series-resonant circuits or ideal parallel-resonant circuits. • If the series resistance is large enough that CR2s/L > 1, there is no parallel-resonant frequency for that particular circuit.

25-7   Practical Parallel-Resonant Circuits

The resonant frequency of a practical parallel-resonant circuit is related to the quality factor of the circuit. For an ideal parallel-resonant circuit, Q=



Rp XL

(25-9)

For the equivalent circuit in Figure 25-17, Equation 25-9 gives Therefore,

Q=

Z2 Xs Xs × 2= Rs Z Rs

The quality factor of a practical parallel-resonant circuit equals the quality factor, Q, of the coil itself. Substituting 2π f r L/Q = Rs in Equation 25-11 gives

and

( 2πf r L/Q ) 2 + ( 2πfr L ) 2 1 + Q2 1 = = 2πf r L 2πf r C 2πf r L Q2 fr =

1 × 2π√LC

√1 +Q Q 2

2



(

) (25-13)

If the Q of the coil is greater than 10, the resonant frequency of the ­ arallel-resonant circuit is very close to that of the corresponding ideal p parallel-resonant circuit, and

f≈

1 2π√LC

At resonance, the impedance of a parallel-resonant circuit reaches its maximum value of

Zp = Rp =

R2s + X2s X2s = Rs + Rs Rs

There are several useful forms of this equation. Factoring out Rs and substituting Q = Xs /Rs gives

Zp = Rs ( 1 + Q2 )

(25-14)

Simply substituting for Xs /Rs gives

Zp = Rs + QXs



(25-15)

761

762

Chapter 25  Resonance

Since Rs is usually very small compared to QXs ,

Zp ≈ QXL

(25-16)

Zp ≈ Q2Rs

(25-17)

Substituting for XL from Equation 25-3 gives

Substituting for Q from Equation 25-3 gives

Zp ≈

X2L Rs

Zp ≈

L CRs

X2L ≈ XL × XC =

Since



(25-18)

ωL , ωC (25-19)

Again, if Q > 10, the approximations in the equations above are very close to the actual values.

Example 25-4 A coil in a tuned circuit in a radio receiver has an inductance of 50.0 μH and a resistance of 25.0 Ω. It forms a parallel-resonant circuit with a 200-pF capacitor. (a) At what frequency is the total impedance completely resistive? (b) What is the magnitude of the parallel impedance at this frequency? Solution (a) (b)

fr =

1 × 2π√LC

√1 − CRL

2 s

1 = × 2 × π × √50 µH × 200 pF

= 1.5915 MHz × 0.99875 = 1.59 MHz

√1 −

200 pF × ( 25 Ω ) 2 50 µH

XL = 2πfL = 2 × π × 1.59 MHz × 50.0 µH ≈ 500 Ω Q=

XL 500 Ω ≈ = 20 Rs 25.0 Ω

Zp ≈ QXL = 20 × 500 Ω = 10 kΩ

See Problems 25-17 to 25-23 and Review Questions 25-40 to 25-43.

25-7   Practical Parallel-Resonant Circuits

Circuit Check

B

CC 25-3. For the circuit shown in Figure 25-18, (a) determine the resonant frequency in radians per second and in hertz (b) find the Q of the circuit (c) calculate the values of V, IR, IL , and IC at resonance (d) determine the power dissipated at resonance (e) determine the bandwidth of the circuit (f) calculate the voltage at the half-power frequencies

10 μA

IR

IC

+

IL 10 mH

200 kΩ

VL

+ 10 nF

100 Ω

− Figure 25-18 

CC 25-4. A certain RLC parallel resonant circuit has the frequency response shown in Figure 25-19. The circuit has a quality factor of 20, and R = 200 Ω. Determine (a)  the frequency of resonance in hertz and in radians per second (b) the source current (c) the inductance (d) the capacitance 400 V 282.8 V

ω1 0.975 Mrad/s

Figure 25-19

ω2 10.25 Mrad/s

763

764

Chapter 25  Resonance

25-8 Selectivity of Parallel-Resonant Circuits

We usually express selectivity in terms of the bandwidth Δf between the half-power frequencies on either side of the resonant frequency. At these frequencies, the voltage across the equivalent parallel-resonant circuit in Figure 25-17 is 0.707 of the voltage at resonance. Since the equivalent circuit has a constant-current source, the equivalent impedance must be 0.707 of its value at resonance. An analysis similar to that for the bandwidth of series-resonant circuits in Section 25-5 shows that the susceptance of L and C must each change by an amount equal to 1/ Rp between the half-power frequencies. Since BC = ωC, 2πC ( f2 − f1 ) =



and

If Q > 10,

Δf =

1 Rp

fr 1 X = = fr × 2πCRp 2πfrCRp Rp



Δf ≈

fr Q

(25-20)

This equation is similar to Equation 25-8 for a series-resonant circuit. From Figure 25-14, we can see that the internal resistance of the signal source has a significant effect on the performance of a parallel-resonant ­circuit. The effective Q of a practical parallel-resonant circuit is seldom determined by Rp alone. In the equivalent circuit of Figure 25-17, the internal resistance of the source is directly in parallel with Rp. This parallel ­connection reduces the Q of the circuit and thus reduces selectivity by increasing Δf. Deliberately increasing the series resistance (or reducing the Q) of a tuned circuit is called damping. When Rs is large enough that resonance is impossible, the circuit is critically damped. See Review Question 25-44.

25-8   Selectivity of Parallel-Resonant Circuits

Example 25-5 Find the bandwidth when the parallel-resonant circuit of Example 25-4 is connected across a signal source having an internal resistance of 40 kΩ. Solution The additional parallel resistance does not enter into the equation for the resonant frequency. Hence, the reactance of either branch at resonance is still 500 Ω. The parallel resistance due to the coil is ( 500 Ω ) 2 X2 = = 10 kΩ Rs 25.0 Ω The equivalent parallel resistance is Rp ≈





Rint × Rp

40 kΩ × 10 kΩ = 8.0 kΩ Rint + Rp 40 kΩ + 10 kΩ Req 8.0 kΩ Q= = = 16 Xs 500 Ω

Req = Δf =

fr

Q

=

=

1.59 MHz = 99.3kHz 16

765

766

Chapter 25  Resonance

Summary

• The impedance of a series-resonant circuit is capacitive below the resonant frequency and inductive above the resonant frequency. • In a series-resonant circuit at resonance, the impedance has its minimum value, which equals the resistance. • At the resonant frequency, the inductive reactance equals the capacitive reactance. • The quality factor of a resonant circuit determines the steepness of the shoulders of the resonance curve. • In a series-resonant circuit with a high Q, inductance and capacitance voltages at resonance are much greater than the source voltage. • The half-power frequencies define the bandwidth of a resonant circuit. • Increasing Q increases the sensitivity and selectivity of a resonant c­ ircuit. • The impedance of a parallel-resonant circuit is inductive below the resonant frequency and capacitive above the resonant frequency. • In a parallel-resonant circuit with a high Q, the tank current at resonance is much greater than the source current. • In a parallel-resonant circuit at resonance, the impedance has its maximum value, which equals the resistance. • The Q of a practical parallel-resonant circuit is equal to the Q of the coil. • The internal resistance of the source reduces the Q and the selectivity of a parallel-resonant circuit. B = beginner

I = intermediate

A = advanced

circuitSIM walkthrough

Problems B B B I

A

Section 25-2  Series Resonance 25-1. Determine the resonant frequency of a 68-μF capacitor in series with a 22‑μH coil that has a Q of 85. 25-2. (a) What capacitance is needed to tune a 500-μH coil to series resonance at 465 kHz? (b) Use Multisim to verify the capacitance. 25-3. What inductance in series with a 12-pF capacitor is resonant at 45 MHz? 25-4. A variable capacitor with a range of 30 pF to 365 pF is connected in series with an inductance. The lowest frequency to which the circuit can tune is 540 kHz. (a) Calculate the inductance. (b) Find the highest frequency to which this circuit can be tuned.

Section 25-3  Quality Factor 25-5. A series RLC resonant circuit is connected to a supply voltage of 50 V at a frequency of 455 kHz. At resonance the maximum current measured is 100 mA. Determine the resistance, capacitance, and inductance if the quality factor of the circuit is 80.

Problems

B I

B I B A

I

A

A B A

767

Section 25-4  Resonant Rise of Voltage 25-6. Find the voltage that appears across the capacitor at resonance when the circuit of Problem 25-1 is connected to a 500-μV source. 25-7. Calculate the current that flows through the circuit of Problem 25-6 at resonance.

Section 25-5  Selectivity 25-8. Find the bandwidth of the circuit in Problem 25-1. 25-9. A tuned circuit consisting of 40-μH inductance and 100-pF capacitance in series has a bandwidth of 25 kHz. Calculate the quality ­factor of this circuit. 25-10. Determine the resistance of the coil in the tuned circuit of Problem 25-9. 25-11. The coil and capacitor of a tuned circuit have an L / C ratio of 1.0 × 105 H/F. The Q of the circuit is 80 and its bandwidth is 5.8 kHz. (a) Calculate the half-power frequencies. (b) Calculate the inductance and resistance of the coil. 25-12. A 470-μH inductor with a winding resistance of 16 Ω is connected in series with a 5600-pF capacitor. (a) Determine the resonant frequency. (b) Find the quality factor. (c) Find the bandwidth. (d) Determine the half-power frequencies. (e) Use Multisim to verify the resonant frequency in part (a), the bandwidth in part (c), and the half-power frequencies in part (d). 25-13. A series RLC circuit has a bandwidth of 500 Hz and a quality factor, Q, of 30. At resonance, the current flowing through the circuit is 100 mA when a supply voltage of 1 V is connected to it. Determine (a) the resistance, inductance, and capacitance (b) the half-power frequencies 25-14. A tuned series circuit connected to a 25-mV signal has a bandwidth of 10 kHz and a lower half-power frequency of 600 kHz. Determine the resistance, inductance, and capacitance of the circuit. 25-15. An AC series RLC circuit has R = 80 Ω, L = 0.20 mH, and C = 100 pF. Calculate the bandwidth at the resonant frequency. 25-16. A series-resonant circuit requires half-power frequencies of 1000 kHz and 1200 kHz. If the inductor has a resistance of 100 V, determine the values of inductance and capacitance.

circuitSIM walkthrough

768

Chapter 25  Resonance

B B B B I

I

I

Section 25-7  Practical Parallel-Resonant Circuits 25-17. Find the resonant frequency of a tuned circuit consisting of the coil and capacitor of Problem 25-1 connected in parallel. 25-18. Determine the impedance of the parallel-tuned circuit of Prob­ lem 25-17 at resonance. 25-19. Determine the tank current at resonance of the circuit in Problem 25-17 when connected to a 750-μV source. 25-20. Calculate the parallel-resonant frequency if loading the inductive branch of the tuned circuit of Problem 25-17 reduces Q to 15. 25-21. A capacitor and a coil with a Q of 90 form a parallel-resonant circuit tuned to 4.5 MHz. At resonance, the total impedance of the circuit is 60 Ω. Find (a) the inductance of the coil (b) the capacitance of the capacitor 25-22. A tuned circuit in a radio transmitter is formed by a 620-μF capacitor and a tank coil with an inductance of 98 nH. The voltage across the tank is 1500 V RMS. The overall quality factor, Q, of the circuit is 20. (a) Determine the current of the tank circuit. (b) How much power does this circuit draw from its source? 25-23. A 520-μF capacitor in parallel with a 240-μH inductor have a quality factor of 7 at resonance. Calculate (a) the resonant frequency (b) the total impedance at resonance

Review Questions

Section 25-1  Effect of Varying Frequency in a Series RLC Circuit 25-24. Describe the effect on the current in a series RLC circuit as the source frequency varies from well below the resonant frequency to well above it. 25-25. Describe the effect on the impedance of a series RLC circuit as the source frequency varies from well below resonant frequency to well above it.

Section 25-2  Series Resonance 25-26. Define the term resonance. 25-27. What property of an electric circuit makes resonance possible? 25-28. Derive an equation to determine the capacitance needed to tune a given inductance to series resonance at a given frequency. 25-29. Why does the resistance in a series-resonant circuit have no bearing on the resonant frequency? 25-30. How does the resistance in a series-resonant circuit affect its behaviour?

Integrate the Concepts

25-31. How does the L/C ratio of a resonant circuit affect its resonance curves? 25-32. Prepare a table similar to Table 25-1 for the circuit of Figure 25-4. 25-33. Prepare a table similar to Table 25-1 for the circuit of Figure 25-5.

Section 25-3  Quality Factor 25-34. How does the quality factor of a resonant circuit affect its behaviour?

Section 25-4  Resonant Rise of Voltage 25-35. What does resonant rise of voltage mean in connection with resonant circuits?

Section 25-5  Selectivity 25-36. What do the half-power frequencies indicate about a resonant ­circuit?

Section 25-6  Ideal Parallel-Resonant Circuits 25-37. How does a parallel-resonant circuit differ in circuit behaviour from a series-resonant circuit? 25-38. Why are parallel-resonant circuits capacitive at frequencies above the resonant frequency while series-resonant circuits are inductive at frequencies above the resonant frequency? 25-39. Explain what a tank current is.

Section 25-7  Practical Parallel-Resonant Circuits 25-40. Why does decreasing the resistance of the coil of a parallel-resonant circuit increase the total impedance at resonance? 25-41. Why does a change in the resistance of a coil in a parallel-resonant circuit affect the resonant frequency? 25-42. Does a resistor connected in parallel with the capacitor of a seriesresonant circuit affect the resonant frequency? Explain. 25-43. Why does the resonant frequency of a parallel-resonant circuit tend to shift as the loading on the circuit increases?

Section 25-8  Selectivity of Parallel-Resonant Circuits 25-44. What effect does the internal resistance of a source have on the quality factor, bandwidth, and selectivity of a parallel-resonant ­circuit?

Integrate the Concepts

(a) For a series circuit made up of a 910-μH coil with a winding resistance of 16 Ω, a capacitor of 130 μF, and a 30-V source, determine   (i) the resonant frequency (ii) the quality factor

769

770

Chapter 25  Resonance



(iii) the capacitor voltage at resonance (iv) the bandwidth

(b) The inductor of part (a) is connected in parallel with the 100-pF capacitor and a source with an internal resistance of 30 kΩ. Calculate   (i) the resonant frequency (ii) the quality factor (iii) the bandwidth

Practice Quiz

1. Which of the following statements are true? (a) At resonance, the ratio of either the inductive or the capacitive reactive power to the true power in the circuit equals the quality factor. (b) The higher the Q factor of a resonance circuit, the steeper the skirt of the resonance curve. (c) Bandwidth is directly proportional to the Q factor of a resonant circuit. (d) Selectivity indicates how well a resonant circuit responds to a ­certain frequency and discriminates against all others. 2. When the frequency of operation of a series RLC resonant circuit is below the resonance frequency, the circuit behaves as a (a) resistive circuit (b) inductive circuit (c) capacitive circuit (d) magnetic circuit 3. When the frequency of operation of a series RLC resonant circuit is above the resonance frequency, the circuit behaves as a (a) resistive circuit (b) inductive circuit (c) capacitive circuit (d) magnetic circuit 4. At resonance, a series RLC circuit behaves as a (a) resistive circuit (b) inductive circuit (c) capacitive circuit (d) magnetic circuit 5. Which of the following statements apply to a series RLC circuit at resonance? (a) XL = XC (b) The input impedance is at a maximum. (c) Current is at a minimum. (d) The current is in phase with the applied voltage.

Practice Quiz

  6. The resonant frequency of the circuit in Figure 25-20 is (a) 14.53 kHz (b) 14.79 kHz (c) 459.44 kHz (d) 2.18 kHz R1 100 Ω

+

E 100 V

−

L1 3 mH C1 0.04 nF

Figure 25-20 

  7. The capacitive reactance of the capacitor of Figure 25-20 is (a) 293.84 Ω (b) 27.38 Ω (c) 273.84 Ω (d) 29.38 Ω

  8. At resonance, the current flowing through the circuit of Figure 25-20 is (a) 0.1 mA (b) 1 mA (c) 0.1 A (d) 1 A   9. The quality factor, Q, of the circuit in Figure 25-20 is (a) 27.4 (b) 274 (c) 0.274 (d) 2.74 10. When the frequency increases from the resonant frequency, the voltage across the resistor in Figure 25-20 (a) becomes zero (b) drops (c) increases (d) stays the same 11. At resonance, the voltage across R1 in Figure 25-20 is (a) 10 V (b) 100 V (c) 1 V (d) 50 V

771

772

Chapter 25  Resonance

12. The bandwidth of the circuit in Figure 25-20 is (a) 53 kHz (b) 5.3 kHz (c) 530 Hz (d) 53 Hz 13. Which of the following statements apply to a parallel RLC circuit at resonance? (a) IL = IC (b) The input impedance is at a maximum. (c) The current is at a minimum. (d) The current is in phase with the applied voltage. 14. The resonant frequency for the circuit in Figure 25-21 is (a) 318 MHz (b) 159 MHz (c) 31.8 kHz (d) 15.9 kHz

I 1.0 A

R1 10 kΩ

L1 10 mH

C1 10 nF

Figure 25-21 

15.   At resonance, the current flowing through the circuit in Figure 25-21 is   (a) 100 μA   (b) 2.5 mA   (c) 5.0 mA   (d) 100 mA 16.  Which of the following statements apply to a practical parallel-resonant circuit at resonance?   (a) The total impedance is purely resistive.   (b) The circuit has a power factor of 1.   (c) The source voltage is in phase with the source current.   (d) XC = XL

Practice Quiz

17.  The resonant frequency for the practical parallel-resonant circuit in Figure 25-22 is   (a) 17.6 Hz   (b) 176 Hz   (c) 1.76 kHz   (d) 17.6 kHz

I 2.5 A

R1 10 kΩ

L1 100 mH

C1 82 nF

RS 16 Ω

Figure 25-22

18. At resonance, the impedance of the network circuit in Figure 25-22 is (a) 8.84 kΩ (b) 7.64 kΩ (c) 76.43 Ω (d) 884 Ω 19. The bandwidth of the parallel-resonant circuit in Figure 25-22 is (a) 318 Hz (b) 3.18 Hz (c) 3.18 kHz (d) 31.8 kHz

773

26

Passive Filters

There are situations, such as in communications systems, where we want a circuit to pass AC signals of frequencies below a critical frequency, but eliminate or attenuate signals of frequencies above that value. In other situations we may want a circuit to pass signals of frequencies above a critical frequency while attenuating those below that frequency. Yet again we may wish to pass signals of frequencies between two critical frequencies while attenuating all others, or, conversely, attenuate all signals of frequencies between two critical frequencies and pass all others. In this chapter we will use inductors and capacitors to form circuits that filter frequencies in these ways.

Chapter Outline 26-1 Filters 776

26-2 Frequency Response Graphs  778 26-3 RC Low-Pass Filters  781 26-4 RL Low-Pass Filters  788

26-5 RC High-Pass Filters  791 26-6 RL High-Pass Filters  797 26-7 Band-Pass Filters  799

26-8 Band-Stop Filters  804

26-9 Practical Applications of Filters  809 26-10 Troubleshooting  812

Key Terms filter  776 passive filter  776 active filter  776 low-pass filter  776 high-pass filter  776

band-pass filter  776 band-stop (band-reject) filter  776 passband  777 stopband  777

critical frequency  777 decibel  778 decade  780 Bode plot  785 centre frequency  802

Learning Outcomes At the conclusion of this chapter, you will be able to: • • • • •

explain the purpose of electric filters define the passband and stopband of a filter draw the electric circuits for passive filters determine the critical frequencies of filters draw the frequency response curves for passive filters • draw the phase plot for passive low-pass and high-pass filters

Photo sources:  © Broadcast Concepts Inc.

• calculate the centre frequency and bandwidth of band-pass filters • calculate the centre frequency and stopband of band-stop filters • troubleshoot filter circuits to determine defective components

776

Chapter 26   Passive Filters

26-1 Filters Suppose that the input voltage, vin, in Figure 26-1 is sinusoidal and that, depending on the frequency of vin, the circuit either passes vin (such that vout = vin) or blocks it (such that vout = 0 V). A circuit that acts in this way is known as a filter. vout

Filter

+ vin = E sin(2πf t) − (v)

RL

Rin

Figure 26-1  Electric circuit used as a filter

Filters are circuits designed to allow AC signals of a particular range of frequencies to pass through them while preventing signals of all other frequencies from doing so. These circuits may be constructed using passive components such as resistance and capacitance (RC filter), resistance and inductance (RL filter), or resistance, capacitance, and inductance (RLC filter). Such filters are called passive filters. Filter circuits that include amplifying devices, such as transistors or operational amplifiers, are called active filters. In this chapter we will investigate the behaviour of passive filters. Filters are classified according to the range of frequencies that they allow to pass through. Low-pass filters allow signals below a critical frequency to pass through while attenuating all others, as shown in Figure 26-2(a). Highpass filters allow signals above a critical frequency to pass through while attenuating all others, as shown in Figure 26-2(b). Band-pass filters allow signals between two critical frequencies to pass through while attenuating all others, as shown in Figure 26-2(c). Band-stop or band-reject filters attenuate signals between two critical frequencies while allowing all others to pass through, as shown in Figure 26-2(d). Vout

Vout E

E Passband

Stopband

Stopband fC

f

Passband fC

f (b)

(a) Vout

Vout E

E Stopband

Passband fC1

(c)

Stopband fC2

Passband f

Stopband fC1

Figure 26-2  Ideal frequency response curves for filters

(d)

Passband f C2

f

26-1  Filters

The passband of a filter is defined as the range of frequencies passed by the filter. The stopband is the range of frequencies attenuated by the filter. In a practical filter, the response curve undergoes a gradual change from the passband to the stopband, and vice versa. The critical frequency, fC, is defined as the frequency at which Vo = 0.707 E, where E is the maximum amplitude of vout in the passband. Typical frequency response curves for practical filters are shown in Figure 26-3. vout E Low-pass filter: 0.707 E

0

Passband

fC

f

Stopband (a)

vout E High-pass filter: 0.707 E

0

Stopband

fC

f

Passband (b)

vout E Band-pass filter: 0.707 E

0

Stopband

f1

f0 Passband (c)

f2

Stopband

f

vout E Band-stop filter: 0.707 E

0

f1

Passband

f0 Stopband (d)

f2

f Passband

Figure 26-3  Frequency response curves for practical filters

See Review Questions 26-29 to 26-33 at the end of the chapter.

777

778

Chapter 26   Passive Filters

26-2  Frequency Response Graphs Most frequency response graphs are drawn with the vertical axis representing the ratio of output voltage to input voltage (vout/vin ) , referred to as the voltage gain, Av vout Av = vin Commonly, this voltage gain is expressed as a logarithm: Av(dB) = 20 log10



vout vin

(26-1)

Av(dB) is referred to as the decibel, or dB, value of the voltage gain. Equation 26-1 is derived from the definition of the bel (symbol B), which is the common logarithm, or log base 10, of the ratio of two power levels. For example, the power gain, expressed in bels, is given by

AP(B) = log10



AP(dB) = 10 log10

Pout Pin

Since a decibel is one-tenth of a bel ( 1 dB = 0.1B ) , the power ratio expressed in decibels is Pout Pin

Assume that the circuit of Figure 26-1 supplies output power, Pout, to the resistive load, RL. The voltage across RL is vout, and the output power is

Pout =

v2out RL

Pin =

v2in Rin

The input resistance of the circuit is represented by Rin. The voltage across Rin is vin, and the input power is Thus,

Pout v2out v2in v2out Rin = ÷ = 2 × Pin RL Rin RL vin

If we can assume that then

RL = Rin Pout v2out vout = 2 = vin Pin vin

( )

2

26-2   Frequency Response Graphs

and AP(dB) = 10 log10



Pout vout = 10 log10 vin Pin

( )

2

= 20 log10

vout vin

Since the gain is now expressed as a ratio of voltages, we normally subscript the gain with a v: vout Av(dB) = 20 log10 vin

Example 26-1 Express the ratios of the following quantities in decibels: (a) Input power is 20 mW and output power is 8.0 W. (b) Input voltage is 20 mV and output voltage is 120 mV. (c) Input voltage is 0.50 V and output voltage is 0.40 V. (d) Input voltage is 15 V and output voltage is 0.45 V. Solution (a) (b) (c) (d)

AP(dB) = 10 log10

Pout 8.0 W 8.0 = 10 log10 = 10 log10 Pin 20 mW 0.020

Av(dB) = 20 log10

vout 120 mV = 20 log10 vin 20 mV

Av(dB) = 20 log10

vout 0.40 V = 20 log10 = 20 log10 ( 0.80 ) vin 0.50 V

= 10 log10 ( 400 ) = 10 ( 2.60 ) = 26 dB

= 20 log10 ( 6 ) = 20 ( 0.778 ) = 15.6 dB = 20 (−0.0969 ) = −1.9 dB

vout 0.45 V = 20 log10 ( 0.03 ) = 20 log10 vin 15 V = 20 (−1.52 ) = −30.5 dB

Av(dB) = 20 log10

Note that if the ratio is greater than 1 the dB value is positive, and if the ratio is less than 1 the dB value is negative.

On a frequency response graph, the horizontal axis represents frequency. This axis usually has a logarithmic scale to allow for easy plotting of the response curve over a wide range of frequencies. Semilog graph paper, shown in Figure 26-4, is used for this purpose.

779

780

Chapter 26   Passive Filters

1000 900 800 700 600

500

400

300

200

100 90 80 70 60

50

40

30

20

10 9 8 7 6

5

4

3

2

1

Figure 26-4  Three-decade semilog graph paper

A decade of frequency is a range of frequencies over which the frequency increases by a factor of 10. For example, the range from 100 Hz to 1000 Hz is a decade, as is the range from 30 Hz to 300 Hz. The frequency range of Figure 26-4 spans three decades: 1 Hz to 10 Hz, 10 Hz to 100 Hz, and 100 Hz to 1000 Hz. Note that the spacing of each decade is the same. For example, the spacing between 1 and 10 Hz is the same as the spacing between 10 and 100 Hz. Note also the unequal spacing within each decade. For example, the spacing between 1 and 2 Hz is greater than between 2 and 3 Hz. The spacing gets closer as we move to the higher frequencies in the decade. The position of a frequency within a decade is determined by the logarithm of that frequency. For instance, to determine where 2 Hz is to be located, we take the log of 2 ( log10 2 = 0.301 ) . We then place 2 Hz at 0.301 of the distance from 1 Hz to 10 Hz. Similarly, 5 is placed at the 0.699 point of the decade as log105 = 0.699. See Problems 26-1 to 26-4 and Review Questions 26-34 and 26-35.

26-3  RC Low-Pass Filters

26-3  RC Low-Pass Filters The circuit of Figure 26-5 is an RC low-pass filter. Note that vout is taken across the capacitor. +

+

R

vin

vout

C

−

−

Figure 26-5  RC low-pass filter

Voltage Gain versus Frequency We will go through a mathematical analysis of the circuit to determine the nature of the frequency response curve. The voltage-division rule can be used to determine an expression for the ratio of the amplitude of vout to vin: vout XC = vin √R2 + X2C where



XC =

1 2πfC

For the range of frequencies for which XC ≫ R,

vout XC XC XC = ≈ = =1 2 2 2 vin XC √R + XC √XC

Substituting 1/2πƒC for XC, we can say that for

1 ≫ R, 2πfC

vout ≈1 vin

Re-arranging the condition, we can say that for f ≪

1 , 2πRC

vout ≈ vin

Therefore at low frequencies, vin passes through the circuit. Now consider the range of frequencies for which XC = 1/2πƒC ≪ R. In this range, vout XC XC XC = ≈ ≈ 2 2 2 vin R √R √R + XC 1 vout XC ≈ ≪ R, vin 2πfC R Re-arranging the condition, we can say that Thus, for for f ≫

1 , 2πRC

vout XC ≈ vin R

781

782

Chapter 26   Passive Filters

Note that as frequency increases, XC decreases. Therefore,

for f ≫

vout ≈0 vin

1 , 2πRC

We have shown that at low frequencies the circuit passes vin ( vout ≈ vin ) , while at high frequencies the circuit attenuates vin ( vout ≈ 0 ) . Thus, the circuit functions as a low-pass filter.

Critical Frequency Let us derive an expression for the critical frequency. Since the amplitude of vout at low frequencies is approximately vin, the critical frequency, fC, is the frequency at which vout = 0.707 vout (max) ≈ 0.707 vin If f =

XC =

1 , 2πRC

1 =R 1 2π C 2πRC

(

)

vout XC R = = = 0.707 2 2 2 vin √R + XC √R + R2

Then,

Therefore, the critical frequency is fC =



1 2πRC

(26-2)

Gain Plot Let us determine the voltage gain, Av, at various frequencies, from which we can sketch the frequency response curve for the filter. From the voltagedivision rule we get vout XC = Av = 2 vin √R + X2C



We express f in terms of the critical frequency ( f = afC ) , and calculate the Av for different values of the coefficient a. Since XC =



1 ,   2πfC

XC =

Let f = 0.1fC. Then

Av =

1 1 = = 2πfC 2πafCC

√R2 + X2C XC

=

√R2 + ( R/0.1 ) 2 R

/0.1

1 R = a 1 2πa C 2πRC

(

=

)

√101R2 10R

=

10 = 1.00 √101

26-3  RC Low-Pass Filters

783

Table 26-1 summarizes calculations of Av for frequencies ranging from 0.1 fC to 100 fC. TABLE 26-1  Voltage gain and phase shift versus frequency of the low-pass RC and RL filters Frequency, f

Voltage Gain, Av

Voltage Gain, Av(dB)

0.1fC

0.995

   0.00

0.3fC

0.958

fC

0.707

3fC

0.316

−3.01

10fC

0.0995

30fC

0.0333

100fC

0.0100

−0.373

Phase Shift, ϕ (degrees) −5.71

−16.7

−45.0

−10.0

−71.6

−20.1

−84.3

−30.0

−88.1

−40.0

−89.4

These points are shown in Figure 26-6, with a smooth curve drawn through them to illustrate the typical frequency response curve of an RC low-pass filter. 10 9 8 7 6

5

4

3

2

1 9 8 7 6

5

4

3

2

1 9 8 7 6

5

4

3

2

1

Av 1

8.0

6.0

4.0

2.0

0.1fC

fC

10fC

Frequency

Figure 26-6  Typical frequency response of the RC and RL low-pass filters (AV vs f )

100fC

784

Chapter 26   Passive Filters

Example 26-2

For the circuit of Figure 26-5, C = 10.0 μ F . (a) Determine the resistance that will give a critical frequency of 300 Hz. (b) Draw and label the frequency response curve for the voltage gain, Av, using 3-decade semilog graph paper. Solution (a) R =

1 1 = = 53.1 Ω ( ) ( 2πfCC 2π 300 10.0 × 10−6 )

(b) When using 3-decade semilog graph paper, it is generally convenient to place the critical frequency in the centre decade. In Figure 26-7, the 3 in the centre decade represents 300 Hz. The centre decade then starts at 100 Hz and ends at 1000 Hz. The first decade starts at 10 Hz and the last decade ends at 10 kHz. In this manner we set the range of frequencies over which we will draw the frequency response curve. Set an appropriate scale for the voltage gain, and plot as many of the points given in the Table 26-1 as will fit on the graph. The response curve is plotted by drawing a smooth curve through these points, as shown in Figure 26-7. 0.1fC

0.3fC

Frequency (Hz) fC

3fC

10fC

30fC

Voltage Gain (Av)

10 9 8 7 6

5

4

3

2

10 9 8 7 6

5

4

3

2

10 9 8 7 6

5

4

3

2

0

⊥ 0.958

0.8 1.707 0.6

0.4 0.316 0.2 .0995 .0333 0

circuitSIM walkthrough

Figure 26-7  Frequency response curve for Example 26-2

To verify the calculation of R in part (a), download Multisim file Ex26-2 from the website, and follow the instructions in the file.

26-3  RC Low-Pass Filters

785

Figure 26-8 shows the frequency response curve for the data in Table 26-1 with the voltage gain expressed in decibels. A graph drawn on semilog graph paper with the vertical axis giving a variable in decibels and the horizontal axis representing frequency is referred to as a Bode plot, named after Hendrik Bode (1905–1982), an American engineer and scientist, who did extensive work in the development of electric filters. 10 9 8 7 6 5

4

3

2

1 9 8 7 6 5

4

3

2

1 9 8 7 6 5

4

3

2

1

Av (db) 0

−10

−20

−30

−40

0.1fC

10fC

fC

Frequency

Figure 26-8  Bode plot for RC and RL low-pass filters

Note in Figure 26-8 that above the critical frequency, the response curve slopes downward and closely resembles a straight line. The gain decreases 20 dB for each decade of increase in frequency. For example, as the frequency goes from 10fC to 100fC , the gain falls from −20 dB to −40 dB. The portion of the response curve above the critical frequency is referred to as the roll-off, and has a slope of −20 dB/dec.  

Example 26-3 Draw and label the Bode plot for the low-pass filter of Example 26-2. Solution Changing the vertical axis of Figure 26-7 to voltage gain in decibels will give us the Bode plot. Again we use the entries in Table 26-1, and set an appropriate scale for the voltage gain. The resulting Bode plot is shown in Figure 26-9.

100fC

786

Chapter 26   Passive Filters

0.3fC

3fC

fC

Frequency (Hz)

30 fC

Voltage Gain (dB)

4

3

2

1 9 8 7 6

5

4

3

2

1 9 8 7 6

5

4

3

2

1

10 K 9 8 7 6 5

0.1fC

0 3.01 –10

–20

–30

–40

Figure 26-9  Bode plot for Example 26-3

Phase Plot We have seen how the gain of the RC low-pass filter varies with frequency. The variation of the phase shift between vout and vin with frequency may also be important in certain situations, particularly in automatic control systems. The phase shift, ϕ, can be expressed as For f = 0.1fC, Then

ϕ = −tan−1

ϕ = −tan−1

XC =

R XC

R = 10R 0.1

R R = −tan−1 XC 10R

( )

= −tan−10.1 = −5.71º

Calculations of phase shift for frequencies ranging from 0.1fC to 100fC are summarized in Table 26-1. These points are plotted in Figure 26-10, with a smooth curve drawn through them to illustrate the typical phase plot of an RC low-pass filter.

26-3  RC Low-Pass Filters

787

10 9 8 7 6

5

4

3

2

1 9 8 7 6

5

4

3

2

1 9 8 7 6

5

4

3

2

1

0

−15 −30 −45 −60 −75

Phase shift (deg)

−90

0.1fC

10fC

fC

Figure 26-10  Phase plot of the RC and RL low-pass filters

Note from Figure 26-10 that the phase shift approaches 0° when f­ requency falls well below the critical frequency, approaches −90º as frequency rises well above the critical frequency, and is −45º at the critical frequency. Other names for the critical frequency include cutoff frequency, corner frequency, –3 dB frequency, and half-power frequency. The term −3 dB frequency comes from the fact that at the critical frequency, the voltage gain is approximately 3 dB below the maximum value in the passband, as indicated in Table 26-1 and in Figures 26-8 and 26-9. Similarly, the power delivered to a resistive load, RL, at the critical frequency is half the maximum value delivered in the passband. This maximum power is PL(max) =



VL2 (max) RL

At the critical frequency, fC

PL =

( 0.707VL(max) ) 2 RL

= 0.5

V2L (max) = 0.5PL(max) RL

See Problems 26-5 to 26-7 and Review Questions 26-36 to 26-41. 

100fC

788

Chapter 26   Passive Filters

26-4  RL Low-Pass Filters Source:  © Broadcast Concepts Inc.

The circuit of Figure 26-11 is an RL low-pass filter. Note that vout is taken across the resistor. L

+ vin

+ R

vout

−

−

Figure 26-11  RL low-pass filter Inductors in a low-pass filter

A mathematical analysis of the circuit will show that the variation with frequency of both the gain and phase shift is virtually identical to that of the RC low-pass filter described in Section 26-3. To determine the frequency response curve of voltage gain, we again employ the voltage division rule: where

vout R = Av = vin √R2 + X2L XL = 2πfL.

As the frequency approaches 0, and

Av ≈

R2 ≫ X2L

R R = =1 R √R2

As frequency becomes very high, and

Av ≈

X2L ≫ R2

R R = 2 X √XL L

Note that at high frequencies, XL becomes very large and Av approaches 0.

Critical Frequency, fC

At the frequency where R = XL = 2πfL,

Av =

√R + R2 R

2

= 0.707

Thus, this condition defines the critical frequency, and

fC =

R 1 = 2πL 2π ( L/R )

(26-3)

26-4  RL Low-Pass Filters

Gain and Phase Plots To determine the voltage gain for a number of values of frequency, we follow the procedure we used for the RC low-pass filter in Section 26-3: We express the frequency in terms of the critical frequency ( f = afC ) , and calculate the value of Av for various values of the coefficient a. Then, XL = 2πfL = 2πa



Av =

and

√R + X2L R

R L = aR 2πL

( )

2

Calculations of Av for frequencies ranging from 0.1fC to 100fC give the same values as for the RC low-pass filter, as listed in Table 26-1. Thus Figure 26-6 represents the frequency response curve of the RL low-pass filter, and its Bode plot is as shown in Figure 26-8. The phase shift, ϕ, can be expressed as

For example, with f = 0.1fC,

and

ϕ = −tan−1

ϕ = −tan−1

XL R

XL = 0.1R

XL 0.1R = −tan−1 R R

( )

= −tan−10.1 = −5.71º

Calculations of phase shift for values of frequency ranging from 0.1fC to 100fC result in the same values shown in Table 26-1. Thus Figure 26-10 represents the phase shift for both the RL and the RC low-pass filters.

Example 26-4 Consider the circuit of Figure 26-11 with the resistor having a value of 1.2 kΩ. (a) Determine the value of inductance required to obtain a critical frequency of 5.0 kHz. (b) Draw and label the Bode plot on 3-decade semilog graph paper. Solution (a) L =

R 1.2 kΩ = = 38 mH ( 2πfc 2π 5.0 kHz )

789

790

Chapter 26   Passive Filters



(b) Since we want 5 kHz in the centre decade, the frequency scale will start at 100 Hz and end at 100 kHz. Use the entries in Table 26-1, and set an appropriate scale for the voltage gain in decibels. The resulting Bode plot is shown in Figure 26-12. Frequency (Hz) 0.3fC

3fC

fC

10fC

Voltage Gain (dB)

10K 9 8 7 6

4

5K

3

2

15 K

1K 9 8 7 6

5K

4

3

2

1 9 8 7 6

5

4

3

2

1

1.5 K

0.1fC

0 3.01

−10

−20

−30

−40

Figure 26-12  Bode plot for Example 26-4

See Problems 26-8 to 26-10 and Review Questions 26-42 to 26-44.

Circuit Check

A

CC 26-1. Find the critical frequency for an RC low-pass filter constructed with a 220-Ω resistor and a 10-μF capacitor. CC 26-2. Find the critical frequency for an RL low-pass filter constructed with a 10-kΩ resistor and a 20-mH inductor.

26-5   RC High-Pass Filters

26-5  RC High-Pass Filters The circuit of Figure 26-13 is an RC high-pass filter. Note that vout is taken across the resistor. C

+

+

vin

vout

R

−

−

Figure 26-13  RC high-pass filter

As with the RC low-pass filter, we can use the voltage-division rule to determine an expression for the ratio of vout to vin: vout R = 2 vin √R + X2C 1 XC =  . 2πfC

where

For R ≫ XC ,

vout R R R = Av = ≈ = =1 2 2 2 vin R √R √R + XC

Hence,

for R ≫

vout ≈1 vin

1 , 2πfC

and for f ≫

Av ≈ 1

1 , 2πRC

Therefore, at high frequencies, vin passes through the filter. 1 , For R ≪ XC , or R ≪   2πfC Av =

Thus, for R ≪ and for f ≪

1 , 2πfC

1 , 2πRC

√R2 + X2C R

Av ≈

Av ≈

R XC

R XC

≈

R R = XC √X2C

Note that as frequency decreases, XC increases. Therefore, for f ≪ 1/2πRC, the ratio of vout to vin approaches zero, and Av ≈ 0. We have shown that at high frequencies the circuit passes vin, while at low frequencies the circuit attenuates vin. Thus, the circuit functions as a high-pass filter.

791

792

Chapter 26   Passive Filters

Critical Frequency

The amplitude of vout at high frequencies is approximately vin. ­Therefore, the critical frequency, fC, is the frequency at which vout = 0.707 vout (max) ≈ 0.707 vin. When f =

XC =

1 ,   2πRC

Av =

and

1 =R 1 2π C 2πRC

(

√R + R

2

X2C

=

)

√R + R2 R

2

= 0.707

Therefore, the expression for the critical frequency is the same as for the RC low-pass filter: fC =



1 2πRC

(26-2)

Gain Plot To determine the voltage gain at any frequency, we again follow the procedure used for the RC low-pass filter in Section 26-3. From the voltagedivision rule we get For f = afC ,

XC =

and for a = 0.1, Av =



=

√R2 + X2C R

√101R R

2

=

vout R = Av = 2 vin √R + X2C

1 1 = = 2πfC 2πafCC

=

1 R = a 1 2πa C 2πRC

(

)

√R2 + ( R/0.1 ) 2 R

√101 1

= 0.0995

Table 26-2 summarizes calculations of Av for frequencies ranging from 0.01fC to 10fC. These points are plotted in Figure 26-14, with a smooth curve drawn through them to illustrate the typical frequency response curve of an RC high-pass filter. Figure 26-15 shows the Bode plot for the data in Table 26-2.

26-5   RC High-Pass Filters

793

TABLE 26-2  Voltage gain and phase shift versus frequency of the high-pass RC and RL filters Frequency, f

Voltage Gain, Av

0.01fC

0.01

0.03fC

0.03

0.1fC

0.0995

0.3fC

0.287

fC

0.707

3fC

0.945

10fC

0.995

Voltage Gain, Av(dB) −40.0

Phase Shift, ϕ (degrees)

−30.5

89.4

−3.01

73.3

−20.0

88.3

−0.491

45.0

−10.8

84.3

−0.0435

18.4 5.71 10 9 8 7 6 5

4

3

2

1 9 8 7 6 5

4

3

2

1 9 8 7 6

5

4

3

2

1

Av ⊥

0.8

0.6

0.4

0.2

0

0.01fC

0.1fC

fC

Figure 26-14  Typical frequency response curve of the RC and RL high-pass filters (Av vs f )

Note that for frequencies below the critical frequency, this Bode plot slopes upward and closely resembles a straight line. The gain increases 20  dB for each decade of increase in frequency. For example, as the frequency goes from 0.01fc to 0.1fc, the gain rises from −40 dB to −20 dB. The portion of the response curve below the critical frequency is the roll-off, and has a slope of 20 dB/dec.

10fC

794

Chapter 26   Passive Filters

10 9 8 7 6 5

4

3

2

1 9 8 7 6

5

4

3

2

1 9 8 7 6

5

4

3

2

1

Av (dB)

0

−10

−20

−30

−40

0.1fC

0.01fC

10fC

fC

Figure 26-15  Bode plot for the RC and RL high-pass filters

Phase Plot The phase shift, ϕ, can be expressed as

For f = 0.1fC ,

and

ϕ = tan−1

ϕ = tan−1

XC =

XC R

R = 10R 0.1

XC 10R = tan−1 R R

( )

= tan−110 = 84.2º

Calculations of phase shift for frequencies ranging from 0.01fC to 10fC are summarized in Table 26-2. These points are plotted in Figure 26-16, with a smooth curve drawn through them to illustrate the typical phase plot of an RC high-pass filter. Note that the phase shift approaches 0° when frequency rises well above the critical frequency, approaches 90° when frequency falls well below the critical frequency, and is 45° at the critical frequency.

26-5   RC High-Pass Filters

795

Phase Shift (deg)

10 9 8 7 6

5

4

3

2

1 9 8 7 6

5

4

3

2

1 9 8 7 6 5

4

3

2

1

90 75 60 45 30 15

10

0.01fC

0.1fC

fC

10fC

Figure 26-16  Phase plot of the RC and RL high-pass filters

Example 26-5 Suppose that the circuit of Figure 26-13 contains a 15-kΩ resistor and a 2.2-nF capacitor. (a) Determine the critical frequency. (b) Draw and label the Bode plot on 3-decade semilog graph paper. Solution

(a) fC =

1 1 = = 4.8 kHz 3 2πRC 2π ( 15 × 10 )( 2.2 × 10−9 )

(b)  Since we want 4.8 kHz in the centre decade, the frequency scale will start at 100 Hz and end at 100 kHz. Use the entries in Table  26-2, and set an appropriate scale for the voltage gain in decibels. The resulting Bode plot is shown in Figure 26-17. A Bode plot of both gain and phase shift can be generated using the AC Analysis simulation in Multisim (see Figure 26-18).

circuitSIM walkthrough

796

Chapter 26   Passive Filters

0.03fC (145)

0.1fC (482)

0.3fC (1450)

Frequency (Hz) fC (4.82K)

3fC (4.5K)

10fC (48.2K)

Voltage Gain (dB)

10K 9 8 7 6

5

4

3

2

10K

9 8 7 6

5

4

3

2

1 9 8 7 6

5

4

3

2

1

0

–10

–20

–30

–40

Figure 26-17  Bode plot for Example 26-5

5

Magnitude

−5

−15

−25

−35 100

400

700 1k

4k 7k 10k Frequency (Hz)

40k

100k

Figure 26-18  Generating a Bode plot using Multisim

See Problems 26-11 to 26-13 and Review Questions 26-45 to 26-48.

26-6  RL High-Pass Filters

26-6  RL High-Pass Filters The circuit of Figure 26-19 is an RL high-pass filter. Note that vout is taken across the inductor. + Vin

R

+ L

Vout

−

−

Figure 26-19  RL high-pass filter

A mathematical analysis of the circuit will show that the variation of both the gain and phase shift with frequency is virtually identical to that of the RC high-pass filter described in Section 26-5. To determine the frequency response curve of the voltage gain, Av , we again employ the voltage division rule: vout XL = Av = 2 vin √R + X2L where XL = 2πfL As frequency approaches zero, XL2 ≪ R2, and XL XL Av ≈ = 2 R √R

Note that as frequency decreases, XL decreases and Av approaches zero. As frequency becomes very high, X2L ≫ R2, and

Av ≈

Critical Frequency

XL XL = =1 2 XL √XL

At the frequency where R = XL = 2πfL,

Av =

√X2L

+ X2L

XL

= 0.707

Since the voltage gain is 0.707, R = 2πfCL. Therefore, the expression for the critical frequency is the same as that for the RL low-pass filter:

fC =

Gain and Phase Plots

R 1 = 2πL 2π ( L/R )

(26-3)

To determine the voltage gain at any frequency, we follow the procedure we used for the RC low-pass filter of Section 26-3. For f = afC , R XL = 2πfL = 2πafCL = 2πa L = aR 2πL XL Av = and √R2 + X2L

( )

797

798

Chapter 26   Passive Filters

Table 26-2 summarizes calculations of Av for frequencies ranging from 0.01fC to 10fC. Thus Figure 26-14 represents the frequency response curve of the RL high-pass filter, and the Bode plot is as shown in Figure 26-15. The phase shift, ϕ, can be expressed as



ϕ = tan−1

For f = 0.1fC,

and

ϕ = tan−1

R XL

XL = 0.1R R R = tan−1 XL 0.1R

(

)

= tan−110 = 84.3º

Table 26-2 lists phase shifts for frequencies ranging from 0.01fC to 10fC , and Figure 26-16 shows the phase plot of the RL high-pass filter.

Circuit Check

B

CC 26-3. Find the critical frequency for an RC high-pass filter constructed with a 68-Ω resistor and a 1.0-μF capacitor. CC 26-4. Find the critical frequency for an RL high-pass filter constructed with a 1.2-kΩ resistor and a 5.0-mH inductor.

See Problems 26-14 to 26-16 and Review Questions 26-49 to 26-51.

Example 26-6 Suppose the circuit of Figure 26-19 uses a 33-mH inductor, and the criti­cal frequency is to be 20 kHz. (a) Determine the resistance required. (b) Draw and label the Bode plot on 3-decade semilog graph paper. Solution (a) R = 2πfCL = 2π ( 20 × 103 ) ( 33 × 10−3 ) = 4.1 kΩ (b) Since we want 20 kHz in the centre decade, the frequency scale will start at 1 kHz and end at 1 MHz. Use the entries in Table 26-2, and set an appropriate scale for the voltage gain in decibels. The resulting Bode plot is shown in Figure 26-20.

26-7   Band-Pass Filters

0.1fC

0.3fC

3fC

fC

10fC

Voltage Gain (dB)

−40

Figure 26-20  Bode plot for Example 26-6

26-7  Band-Pass Filters The band-pass filter exhibits the frequency response curve of Figure 26- 3(c). There are two different types of circuit arrangements commonly used to produce this characteristic.

Cascaded High-Pass and Low-Pass Filters The first circuit we will investigate is the combination of a high-pass filter with a low-pass filter as shown in Figure 26-21. Av

Av fC(HPF)

Vin

High-pass filter (HPF)

f

fC(LPF) Low-pass filter (LPF)

f

Vout

Figure 26-21  Band-pass filter formed from cascaded high- and low-pass filters

5

−30

1K 9 8 7 6K

−20

4

−10

Frequency (Hz) 3

2K

5

1K 9 8 7 6K

4

3

2K

5

1K 9 8 7 6K

4

3

2K

1K

0

799

Source: Alibaba.com

800

Chapter 26   Passive Filters

When fC (HPF) < fC (LPF) and Band-pass filter for cable television

• the frequency of vin is less than fC (HPF), then the high-pass filter attenuates vin and vout is approximately zero. • the frequency of vin is greater than fC (LPF), then the low-pass filter attenuates vin and vout is approximately zero. • the frequency of vin is greater than fC (HPF), but less than fC (LPF), then vin passes through both filters and vout is approximately equal to vin. Thus, the circuit acts as a band-pass filter if fC (HPF) < fC (LPF). The frequency response curve is shown in Figure 26-22. Av

1 0.707

fC1

fC2

f

Figure 26-22  Frequency response curve of cascaded high- and low-pass filters

Figure 26-23 shows the circuit for a band-pass filter using cascaded RC high-pass and low-pass filters. + vin

−

R2

C1 R1

C2

+ vout

−

Figure 26-23 Cascaded RC high-pass and low-pass filters

The critical frequencies of the high-pass and low-pass filters can be approximated using: 1 fC1 ≈ (26-4) 2πR1C1

fC2 ≈

1 2πR2C2

(26-5)

One problem with this circuit arrangement is the loading effect when the low-pass filter is connected to the high-pass filter. This loading is the reason why the expressions in Equations 26-4 and 26-5 are only appro­ximations. However, these approximations are reasonably accurate if R2 is much greater than R1, thus ensuring that the input impedance of the lowpass filter (the load on the high-pass filter) is much greater than the output impedance of the high-pass filter. We will set the condition that R2 ≥ 100R1. Note that vout is less than vin due to the voltage drops across C1 and R2. In other words, the voltage gain in the passband is less than 1. The bandwidth, BW, of the filter is the range of frequencies for which the voltage gain is 0.707 or greater:

BW ≈ fC1 − fC2

(26-6)

26-7   Band-Pass Filters

801

Example 26-7 A band-pass filter is to be constructed using cascaded RC high-pass and low-pass filters as shown in Figure 26-23. The critical frequencies are to be 1.0 kHz and 10 kHz. Given that C1 = 0.15 μF and C2 = 0.15 nF, (a) calculate the required values for R1 and R2 (b) draw and label a sketch of the frequency response curve for the filter Solution (a) Using Equation 26-4,

R1 =



Using Equation 26-5,



R2 =

1 1 = = 1.06 kΩ 3 2πfC1C1 2π ( 1.0 × 10 ) ( 0.15 × 10−6 ) 1 1 = = 106 kΩ 3 2πfC2C2 2π ( 10 × 10 ) ( 0.15 × 10−9 )

(b) The frequency response curve for the filter is shown in Figure 26-24. Av (mid) (0.707) Av(mid)

1 ( fC1)

f (kHz)

10 ( fC2)

Figure 26-24  Frequency response curve for Example 26-7

The critical frequency fC1 is known as the lower cutoff frequency, and fC2 is the higher cutoff frequency.

RLC Resonant Circuits

L

C Band-pass filters can be built using RLC resonant circuits, as shown in Vin R Figure 26-25. The frequency response curve for these filters is illustrated in Figure 26-25(c). −

+

+

+

Vout

Vin

−

−

(a) Av

R C

L

(b)

1

0.707

+

L

C

Vin

R

− (a)

+

+

Vout

Vin

−

−

R

+ L

C

Vout fC1 f0 fC2

− (b)

f (c)

Av Figure 26-25 Band-pass RLC resonant filters: (a) series RLC; (b) parallel RLC; (c) their frequency response curve 1 0.707

802

Chapter 26   Passive Filters

As discussed in Section 25-2, resonance for a series resonant circuit occurs at the frequency at which XL = XC. The total impedance at resonance is Z = R. The current in the circuit is maximum at resonance (see Figure 25-4). Since vout of the circuit in Figure 26-25(a) appears across the resistor, vout is maximum at resonance, and is equal to vin, making Av = 1. The resonant frequency, fr, is fr =



1 2π√LC



(25-1)

When a resonant circuit is used as a band-pass filter, the resonant frequency, fr, is often referred to as the centre frequency, fo. For the response curve of Figure 26-25(c), then fo =



1 2π√LC



(26-7)

The bandwidth, BW, of the series resonant filter is Δf, as shown in Figure 25-8, and Δf =

Hence,

and

fr Q



BW = fC2 − fC1 = Q=

where

Q=

or

(25-8) fo Q



XL R

1 R

√CL

(26-8)

(25-3)

(25-4)

As discussed in Sections 25-6 and 25-7, the impedance of a parallel resonant circuit is maximum at resonance (see Figure 25-11), such that the impedance of the tank circuit (the parallel combination of L and C) is much greater than R. Since vout of the circuit Figure 26-25(b) appears across the tank circuit, at resonance vout ≈ vin and AV ≈ 1. The resonant frequency is given by

fr =

1 2π√LC

√

1−

CR2s L

(25-12)

where RS is the resistance of the inductor coil.

√

Thus, the centre frequency, fo, for a parallel resonant band-pass filter is

fo =

1 2π√LC

1−

CR2s L

(26-9)

26-7   Band-Pass Filters

The bandwidth of the filter is approximated using Δf ≈



fr Q



BW = fC2 − fC1 ≈

or

(25-20) fo Q



(26-10)

where Q is the quality factor of the inductor.

Q=

2πfoL XL = Rs Rs

If Q > 10, the centre frequency may be approximated using fo ≈



1

2π√LC



(26-11)

(26-12)

Example 26-8 A series resonant band-pass filter is to be constructed with an inductance of 300 mH and a capacitance of 5.0 nF. (a) Calculate the centre frequency of the filter. (b) Calculate the resistance required to obtain a bandwidth of 200 Hz. (c) Draw and label a sketch of the frequency response curve for the filter. Solution (a) fo =

1 1 = = 4.11 kHz −3 2π√LC 2π√( 300 × 10 ) ( 5.0 × 10−9 )

(b) From Equation 26-8, Q=

From Equation 25-4,

√

fo BW

=

4110 Hz = 20.6 200 Hz

√

1 L 1 300 × 10−3 = = 376 Ω Q C 20.6 5.0 × 10−9 (c) The frequency response curve for the filter is shown in Figure 26-26. R=

Av

1 0.707

41 10

f (Hz)

BW = 200 Hz

Figure 26-26  Frequency response curve for Example 26-8

803

804

Chapter 26   Passive Filters

Example 26-9 A parallel resonant band-pass filter is to be constructed with a 10-nF capacitor and an inductor having an inductance of 100 mH and a coil resistance of 300 Ω. (a) Calculate the centre frequency of the filter. (b) Calculate the bandwidth of the filter. (c) Draw and label a sketch of the frequency response curve for the filter. Solution (a)

=

fo =

1 2π√LC

√

1−

2π√( 100 × 10−3 ) ( 10 × 10−9 ) 1

√

CR2s L

1−

( 10 × 10−9 ) ( 3002 ) 100 × 10−3

= 5010 Hz = 5.0 kHz

2πfoL 2π ( 5010 ) ( 100 × 10−3 ) = = 10.5 Rs 300 fo 5010 Hz BW ≈ ≈ ≈ 477 Hz Q 10.5 (c)  The frequency response curve for the filter is shown in Figure 26-27. (b) Q =

Av

1 0.707

5010 BW = 477 Hz

f (Hz)

Figure 26-27  Frequency response curve for Example 26-9

See Problems 26-17 to 26-22 and Review Questions 26-52 to 26-57.

26-8  Band-Stop Filters The band-stop filter exhibits the frequency response curve of Figure 26- 3(d). The stopband is that range of frequencies for which the output voltage is less than 0.707 of the maximum value in the passband. As with the b ­ and-pass filter, there are two different types of circuit arrangements c­ ommonly used to produce this characteristic.

26-8   Band-Stop Filters

The first circuit we will investigate is again a combination of a high-pass filter and a low-pass filter. However, instead of cascading them as we did to form the band-pass filter, we connect the two filters in a parallel arrangement as shown in the block diagram of Figure 26-28. HPF

Vo(HPF)

fC(HPF)

Vin

+ Vout

LPF fC(LPF)

+ Vo(LPF)

Figure 26-28  Band-stop filter formed by a parallel arrangement of highpass and low-pass filters

The circuit used to add the output voltages, vo(LPF) and vo(HPF), of the two parallel filters is called a summing circuit, and can be produced using an operational amplifier. Consider the operation of the complete filter when fc(LPF) < fc(HPF). Let f denote the frequency of vin.

• If f < fC(LPF), vin will pass through the low-pass filter, but not the highpass filter. So, vo(LPF) ≈ vin, vo(HPF) ≈ 0, vout ≈ vin, and Av ≈ 1. • If f > fC(HPF), vin will not pass through the low-pass filter, but will pass through the high-pass filter. So, vo(LPF) ≈ 0, vo(HPF) ≈ vin, vout ≈ vin, and Av ≈ 1. • If fC(LPF) < f < fC(HPF), vin will not pass through either filter. So, vo(LPF) ≈ 0, vo(HPF) ≈ 0, vout ≈ 0, and Av ≈ 0. Thus, the circuit behaves as a band-stop filter with fC1 = fC(LPF) and fC2 = fC(HPF). The frequency response curve for the overall circuit is shown in Figure 26-29.

Av 1 0.707

fC1

fC2

f

Figure 26-29  Frequency response curve of a band-stop filter formed by a parallel arrangement of high-pass and low-pass filters

The stopband is the range of frequencies between fC1 and fC2:

stopband = fC2 − fC1

805

806

Chapter 26   Passive Filters

Example 26-10 A band-stop filter is to be constructed using a parallel arrangement of RC high-pass and low-pass filters, as shown in Figure 26-28. For the low-pass filter, C1 = 0.022 μF and R1 = 10 kΩ. For the high-pass filter, C2 = 0.022 μF and R2 = 3.0 kΩ . (a) Calculate the critical frequencies of this filter. (b) Sketch the frequency response curve for the filter. Solution

(a) fC(LPF) =

fC(HPF) =

1 1 = = 723 Hz 3) ( ( 2πR1C1 2π 10 × 10 0.022 × 10−6 )

1 1 = = 2.4 kHz 3 2πR2C2 2π ( 3.0 × 10 ) ( 0.022 × 10−6 )

(b) The frequency response curve for the filter is shown in Figure 26-31. Av 1 0.707

723

f (Hz)

2410

Figure 26-30  Frequency response curve for Example 26-10

Band-stop filters can also be built using RLC resonant circuits similar to those used for the band-pass filters of Section 26-7. For a series resonant band-stop filter, vout is taken across the series combination of L and C, as shown in Figure 26-31(a). For a parallel resonant band-stop filter vout is across R, as shown in Figure 26-31(b). The frequency response curve for these filters is illustrated in Figure 26-31(c). Since the same circuit is used for both the band-pass and band-stop series resonant filters, the calculation of the centre frequency is the same:

fo =

1 2π√LC



(26-7)

The stopband is calculated using the equation for the bandwidth of the band-pass filter:

stopband = fC2 − fC1 =

fo Q



(26-13)

26-8   Band-Stop Filters

807

C R

+

+ L

+

Vout

Vin

DirecTV band-stop filter

+ L

Vin

R

Source: www.weaknees.com

Vout

C

−

−

−

−

(a)

(b)

Av Source: Dipol

1 0.707

fC1

f0

fC2 (c)

f

Figure 26-31 Band-stop RLC resonant filters: (a) series resonant; (b) parallel resonant; (c) frequency response curve

Q=

where

or

Q=

1 R

XL R

√CL

CDMA band-stop filter

(25-3)

(25-4)

CR 1− √ L 2π√LC

Similarly, calculations for the centre frequency of the band-pass and band-stop parallel resonant filters are the same:

fo =

1

2 s



(26-9)

The stopband is calculated using the equation for the bandwidth of the corresponding parallel resonant band-pass filter:

stopband = fC2 − fC1 ≈

fo Q



(26-14)

where Q is the quality factor of the inductor, given by



Q=

2πfoL XL = Rs Rs

If Q > 10, the centre frequency may be approximated using fo ≈

1 2π√LC



(26-11)

(26-12)

808

Chapter 26   Passive Filters

Example 26-11 A series resonant band-stop filter as shown in Figure 26-31(a) is to be constructed. The capacitor to be used is 0.10 µF. (a) Calculate the inductance required for a centre frequency of 10 kHz. (b) Calculate the resistance required for a stopband of 1.0 kHz. (c) Draw and label a sketch of the frequency response curve for the filter. Solution (a) From Equation 26-7,

L=

1 1 = 2.53 mH 2 = ( −6 ) [ ( ) C 2πfo 0.1 × 10 2π ( 10 × 103 ) ] 2

(b) From Equation 26-14,

Q≈

fo stopband

=

10 kHz = 10 1.0 kHz

mH = 16 Ω √CL = 101 √2.53 0.10 μF

From Equation 25-4,

R=

1 Q

(c) The frequency response curve for the filter is shown in Figure 26-32. Av 1 0.707

Stopband 1 kHz

10 kHz fo

f (Hz)

Figure 26-32  Frequency response curve for Example 26-11

Example 26-12 A parallel resonant band-stop filter as shown in Figure 26-31(b) is to be constructed with a 8.2-nF capacitor and an inductor having an inductance of 75 mH and a resistance of 200 Ω.

26-9   Practical Applications of Filters

(a) Calculate the centre frequency of the filter. (b) Calculate the stopband of the filter. (c) Sketch the frequency response curve for the filter. Solution (a) fo = =

√

1−

CR2s L

2π√( 75 × 10−3 ) ( 8.2 × 10−9 ) 1

(b) Q =

1 2π√LC

2πfoL Rs

stopband ≈

=

fo

Q

√

1−

( 8.2 × 10−9 ) ( 2002 ) = 6.4 kHz 75 × 10−3

2π ( 6400 ) ( 75 × 10−3 ) = 15.1 200 =

6400 Hz ≈ 424 Hz 15.1

(c) The frequency response curve for the filter is shown in Figure 26-33. Av 1

Stopband (424 Hz)

0.707

6400 Hz (fo)

f

Figure 26-33  Frequency response curve for Example 26-12

See Problems 26-23 to 26-25 and Review Questions 26-58 to 26-63.

26-9  Practical Applications of Filters

Analysis of the High- and Low-Frequency Limits of an Amplifier Most electronic amplifiers have low- and high-frequency limits. If the frequency of the input signal falls outside this range, the signal will be attenuated rather than amplified. For an audio amplifier, for instance, the low frequency limit may be in the neighborhood of 50 Hz and the high limit around 20 kHz. The schematic of an amplifier using a bipolar junction transistor is shown in Figure 26-34. Ci, CC, and CE are actual capacitors, and are vital to the operation of the amplifier. CP1, CP2, CP3, CP4, and CP5 are not actual

809

810

Chapter 26   Passive Filters

+ Vcc R1 Rs

CP3

Rc

Ci

CP4

B R2

+ Vs

CP1

Cc

CP2

CP5 RE

CE

RL

+ Vo

−

−

Figure 26-34  Bipolar junction transistor amplifier

capacitors, but rather parasitic capacitance effects caused by circuit wiring and junction capacitance between transistor terminals. The purpose of the circuit is to input a low-amplitude AC signal (VS) and produce a much larger amplitude output signal (VO) with the same waveform. However, the circuit will amplify the input AC signal only if its frequency falls between a lower limit (fL) and a higher limit (fH). If the frequency of the input signal is below fL or above fH, the signal will be attenuated rather than amplified. The values of fL and fH may be determined by analyzing the circuit in terms of RC low-pass and high-pass filters. To illustrate why the high-frequency limit exists, consider the AC equivalent circuit for the input section of the amplifier at high frequencies. The values of Ci, CC, and CE are such that at high frequencies their reactances are negligible (essentially short circuits). As shown in Figure 26-35, the equivalent input circuit then consists of • a voltage source, VS1, dependent on the value of VS • a resistance, Ri, dependent on the values of RS, R1, R2, and the input resistance as seen looking into the base (B) of the transistor • a capacitance, CiP dependent on the values of CP1, CP2, CP3, CP4, and CP5 This circuit is in fact a low-pass RC filter. Any signal with a frequency exceeding the critical frequency is attenuated such that Vi is effectively zero. Since Vi is the input signal to the amplifier, the output voltage, VO, is Ri

+ + Cip

VS1

Vi

− − Figure 26-35 High-frequency AC equivalent circuit of the input section of the amplifier of Figure 26-34

26-9   Practical Applications of Filters

also essentially zero. The high-frequency limit of the amplifier due to the input section, fHi, is then, 1 fHi = 2πRiC1P The RC circuit associated with the output section of the amplifier also produces a high-frequency limit, fHo, on the amplifier. The actual highfrequency limit on the amplifier would be close to the lower of these two values. To analyze the low-frequency limit, consider the AC equivalent circuit for the input section of the amplifier at low frequencies. The values of CP1, CP2, CP3, CP4, and CP5 are such that at low frequencies their reactances are very high (essentially open circuits). As shown in Figure 26-36, the equivalent input circuit then consists of

• a voltage source, VS2, dependent on the value of VS • a resistance, Ri, dependent on the values of RS, R1, R2, and the input ­resistance as seen looking into the base of the transistor • capacitance, Ci Ci

+ + Ri

VS2

Vi

− − Figure 26-36 Low-frequency AC equivalent circuit for the input section of the amplifier of Figure 26-34

This circuit is in fact a high-pass RC filter. Any signal with a frequency less than the critical frequency is attenuated such that Vi is essentially zero, thus making VO essentially zero. The low-frequency limit of the amplifier due to the input section is then,

fLi =

1 2πRiCi

The RC circuits associated with CC and CE also produce low-frequency limits, fLC and fLE, for the amplifier. The actual low-frequency limit on the amplifier would be close to the highest of these three values.

Audio System Crossover Networks As a general rule, a single speaker cannot reproduce good quality sound over the entire range of the human audio spectrum (say, 20 Hz to 20 kHz). To minimize distortion of the audio output, hi-fi systems use two or three

811

812

Chapter 26   Passive Filters

speakers, each reproducing sound over the portion of the audio spectrum for which it is designed. In 3-way crossover systems, a woofer is used to output the low frequencies, a tweeter the high frequencies, and a midrange speaker the frequencies in between. Figure 26-37 shows one scheme whereby a particular frequency range is selected and directed to each speaker.

High-pass filter

Band-pass filter

Audio input

Low-pass filter

Tweeter

Midrange speaker

Woofer

Figure 26-37  3-way parallel crossover network

The network is referred to as a parallel arrangement since each speaker with its filter is in parallel with the other two. The sound reproduced by the woofer consists only of the signals below the critical frequency of the lowpass filter (say, 400 Hz). The sound reproduced by the midrange speaker consists only of the signals within the passband of the band-pass filter (say, 400 Hz to 6 kHz). The sound reproduced by the tweeter consists only of the signals above the critical frequency of the high-pass filter (say, 6 kHz).

26-10 Troubleshooting If any of the components in a filter circuit is defective, the circuit will not operate as designed. The fault is most likely to be (i) a shorted resistor, inductor, or capacitor (ii) an open resistor, inductor, or capacitor (iii) an incorrect value of resistor, inductor, or capacitor in the circuit Depending on which component is defective and the type of defect, we may discover that (i) vout is zero for all frequencies of vin (ii) vout is approximately equal to vin for all frequencies (iii) the circuit operates as a filter, but not as designed (e.g., designed to be a band-pass filter but acts as a low-pass filter) (iv) the critical and centre frequencies differ from the designed values

26-10  Troubleshooting

Example 26-13 A band-pass filter is constructed using the series resonant circuit shown in Figure 26-38. L

C

+ Vin

−

+ R

Vout

−

Figure 26-38  Band-pass filter for Example 26-13

A two-channel oscilloscope is connected to the circuit such that we can view Vin on channel 1 and Vout on channel 2 as the frequency of Vin is varied. Determine the faulty component for each of the following conditions as observed on the oscilloscope: (a) Vout is approximately zero for all frequencies of Vin. (b)  Vout is approximately equal to Vin at low frequencies and approximately zero at high frequencies. (c)  Vout is approximately zero at low frequencies of Vin and approximately equal to Vin for high frequencies. Solution (a) If Vout is zero for all frequencies, a resistor may be shorted, or an inductor or capacitor could be open. Ohm’s Law ( v = iR ) says that if R = 0 Ω (shorted), then vR = Vout = 0 V . If either of the inductor or capacitor is open, then the current through R is zero, and again the voltage across R is zero. (b) The circuit is operating as a low-pass filter. Looking at Figure 26-38, imagine that the capacitor is shorted. Then, the circuit reduces to the circuit of Figure 26-11, an RL low-pass filter. The faulty component is most likely a shorted capacitor. (c) The circuit is operating as a high-pass filter. Looking at Figure 26-38, imagine that the inductor is shorted. Then the circuit reduces to the circuit of Figure 26-13, an RC high-pass filter. The faulty component is most likely a shorted inductor. See Problems 26-26 to 26-28 and Review Questions 26-64 and 26-65.

813

814

Chapter 26   Passive Filters

Summary

• Filters are circuits designed to allow AC signals of a particular range of frequencies to pass through them while attenuating signals of all other frequencies. • Passive filters may be constructed using resistors, capacitors and inductors. • Filters are classified as low-pass, high-pass, band-pass, and band-stop depending on the range of frequencies they pass or block. • For the purpose of analyzing the operation of a filter, the ratio of output voltage to input voltage is often expressed in decibels (dB). • The passive low-pass filter consists of (i) a resistor and capacitor in series with the output voltage across the capacitor or (ii) a resistor and inductor in series with the output voltage across the resistor. • The Bode plot is a graph of the voltage gain in decibels versus the frequency of the input signal plotted on semilog graph paper. • The phase shift of a low-pass filter approaches 0° at frequencies well below the critical frequency and −90° at frequencies well above the critical frequency. • The passive high-pass filter consists of (i) a resistor and capacitor in series with the output voltage across the resistor or (ii) a resistor and inductor in series with the output voltage across the inductor. • The critical frequency of the low-pass and high-pass filters is dependent on the resistance and capacitance in the RC filter and on the resistance and inductance in the RL filter. • The phase shift of a high-pass filter approaches 90° at frequencies well below the critical frequency and approaches 0° at frequencies well above the critical frequency. • A band-pass filter may be constructed by cascading a high-pass filter with a low-pass filter. • Band-stop filters may be constructed using a parallel arrangement of a high-pass and a low-pass filter. • For band-pass filters and band-stop filters constructed from high-pass and low-pass filters, the bandwidths and stopbands are determined by the critical frequencies of the high-pass and low-pass filters • The band-pass and band-stop filters may be constructed using the series or parallel resonant RLC circuit, the centre frequency being the resonant frequency of the circuit. • The passband or stopband of the resonant filter circuit is dependent on the centre frequency and the quality factor of the circuit. • A faulty component in a filter may be detected by observing the behaviour of the output voltage as the frequency of the input signal varies.

Problems

Problems B

B

B B

I

Section 26-2  Frequency Response Graphs 26-1.

Determine the ratio of output power to input power in decibels for the following: (a) input power of 5.0 W and output power of 40 W (b) input power of 1 W and output power of 25 W (c) input power of 50 mW and output power of 5 W 26-2. Determine the ratio of output voltage to input voltage in decibels for the following: (a) 10-V input voltage and 2.0-V output voltage (b) 100-mV input voltage and 7.5-V output voltage (c) 50-μV input voltage and 1.0-V output voltage 26-3. An amplifier has a 10-W output power when the input power is 350 mW. Determine the power gain in dB. 26-4. A 50-dB attenuator has an input power of 2 W. What is the value of the output power?

815

B = beginner

I = intermediate

A = advanced

Section 26-3  RC Low-Pass Filters 26-5.

For the RC low-pass filter shown in Figure 26-39, (a) determine the critical frequency (b) draw and label the Bode plot on 3-decade semilog graph paper (c) draw and label the phase plot on 3-decade semilog graph paper +

220 Ω

+ 10 nF

Vin

Vout

−

− Figure 26-39

B B

I

26-6. What new value of resistance would be required to change the critical frequency of the filter in Figure 26-39 to 1.0 kHz? 26-7. What new value of capacitance would be required to change the critical frequency of the filter in Figure 26-39 to 10 kHz?

Section 26-4  RL Low-Pass Filters 26-8.

For the RL low-pass filter shown in Figure 26-40, (a) determine the critical frequency (b) draw and label the Bode plot on 3-decade semilog graph paper (c) use Multisim to verify the critical frequency in part (a)

walkthrough

100 mH

+ Vin

− Figure 26-40

circuitSIM

+ 10 KΩ

Vout

−

816

Chapter 26   Passive Filters

B B

I

26-9. What new value of resistance would be required to change the critical frequency of the filter in Figure 26-40 to 5 kHz? 26-10. What new value of inductance would be required to change the critical frequency of the filter in Figure 26-40 to 30 kHz?

Section 26-5  RC High-Pass Filters 26-11. For the RC high-pass filter shown in Figure 26-41, (a) determine the critical frequency (b) draw and label the Bode plot on 3-decade semilog graph paper (c) draw and label the phase plot on 3-decade semilog graph paper 0.33 μF

+

+

Vin

1 kΩ

−

Vout

−

Figure 26-41

circuitSIM walkthrough

B B

I

26-12.   (a) What new value of resistance would be required to change the critical frequency of the filter in Figure 26-41 to 5.0 kHz? (b) Use Multisim to verify the new resistance in part (a). 26-13 What new value of capacitance would be required to change the critical frequency of the filter in Figure 26-41 to 20 kHz?

Section 26-6  RL High-Pass Filters 26-14. For the RL high-pass filter shown in Figure 26-42, (a) determine the critical frequency (b) draw and label the Bode plot on 3-decade semilog graph paper 2.2 kΩ

+ Vin

+ 330 mH

−

Vout

−

Figure 26-42

B B

I

26-15. What new value of resistance would be required to change the critical frequency of the filter in Figure 26-42 to 400 Hz? 26-16. What new value of inductance would be required to change the critical frequency of the filter in Figure 26-42 to 5.0 kHz?

Section 26-7  Band-Pass Filters 26-17. For the band-pass filter shown in Figure 26-43, (a) calculate the lower and upper cutoff frequencies (b) draw and label a free-hand sketch of the frequency response curve for the filter

Problems

R2

C1

+

68 nF

Vin

817

100 kΩ R1 3.3 kΩ

+ C2 82 PF

Vout

−

−

Figure 26-43

B I

26-18. Calculate the new values of capacitors required to change the lower cutoff frequency of Figure 26-43 to 1.0 kHz and the upper cutoff frequency to 10 kHz. 26-19. For the band-pass filter shown in Figure 26-44, (a) calculate the centre frequency (b) calculate the bandwidth (c) draw and label a free-hand sketch of the frequency response curve for the filter 25 mH

17 nF

+

+

Vin

Vout

100 Ω

−

−

Figure 26-44

B B B

26-20. (a) What new value of resistance would be required in Figure 26-44 to change the bandwidth to 200 Hz? (b) Use Multisim to verify the new resistance in part (a). 26-21. What new value of capacitance would be required in Figure 26-44 to change the centre frequency to 2.0 kHz? 26-22. For the band-pass filter shown in Figure 26-45, (a) calculate the centre frequency (b) calculate the bandwidth 100 Ω

+ Vin

+ 25 mH

15 nH

−

Vout

−

Figure 26-45

I

Section 26-8  Band-Stop Filters 26-23. A band-stop filter is to be constructed using a parallel arrangement of high-pass and low-pass filters. The high-pass filter consists of a 1.2-kΩ resistor and a 100-nF capacitor. The low-pass filter consists of a 2.7-kΩ resistor and a 220-nF capacitor.

circuitSIM walkthrough

818

Chapter 26   Passive Filters

I

circuitSIM walkthrough

(a) Calculate the lower and upper cutoff frequencies of the filter. (b) Draw and label a free-hand sketch of the frequency response curve for the filter. 26-24. For the stopband filter shown in Figure 26-46, (a) calculate the centre frequency (b) determine the stopband (c) draw and label a free-hand sketch of the frequency response curve for the filter (d) Use Multisim to verify the centre frequency in part (a) and the stopband in part (b). 10 Ω

+

+ 220 μH

Vin

Vout

0.5 μF

−

−

Figure 26-46

B

26-25. For the stopband filter shown in Figure 26-47, (a) calculate the centre frequency (b) determine the stopband 22 nF

+

50 mH

Vin

+ 100 Ω

Vout

−

−

Figure 26-47

B

B B

Section 26-10  Troubleshooting 26-26. The output voltage of the circuit of Figure 26-39 was monitored as the frequency of the input signal was varied. It was discovered that the amplitude of the output voltage was approximately the same as the input for all frequencies. Determine the faulty component. 26-27. When the circuit of Figure 26-40 was tested, it was discovered that the critical frequency was approximately 1.6 kHz. Determine a possible cause for the this value of critical frequency. 26-28. The output voltage of the circuit of Figure 26-45 was monitored as the frequency of the input signal was varied. Determine the faulty component if the circuit was found to operate as a (a) high-pass filter (b) low-pass filter

Review Questions

Review Questions B B B

B B

B B

B B B B B

B B B B

Section 26-1  Filters 26-29. What is the purpose of a filter? 26-30. Differentiate between passive filters and active filters. 26-31. Define the following classifications of filters: (a) low-pass filter (b) high-pass filter (c) band-pass filter (d) band-stop filter 26-32. How is the critical frequency determined in a practical filter? 26-33. Draw and label the frequency response curve for a practical (a) low-pass filter (b) high-pass filter (c) band-pass filter (d) band-stop filter

Section 26-2  Frequency Response Graphs 26-34. Show how the decibel can be used to express the ratio of (a) input and output power (b) input and output voltage 26-35. Describe the type of graph paper often used to plot frequency response curves of filters.

Section 26-3  RC Low-Pass Filters 26-36. The output voltage is across which component in the RC low-pass filter? 26-37. State the mathematical relationship that determines the critical frequency in an RC low-pass filter. 26-38. What is the value of the capacitive reactance in an RC low-pass filter at the critical frequency? 26-39. What is a Bode plot? 26-40. What value does the phase shift between input and output voltages approach in a low-pass filter when frequency is (a) well below the critical frequency? (b) well above the critical frequency? 26-41. List other common names for the critical frequency of a filter.

Section 26-4  RL Low-Pass Filters 26-42. The output voltage is across which component in the RL low-pass filter? 26-43. State the mathematical relationship that determines the critical frequency in an RL low-pass filter. 26-44. What is the value of the inductive reactance in an RL low-pass filter at the critical frequency?

819

820

Chapter 26   Passive Filters

B B B B

B B B

B B B B B B

B B B B B

Section 26-5  RC High-Pass Filters 26-45. The output voltage is across which component in the RC high-pass filter? 26-46. State the mathematical relationship that determines the critical frequency in an RC high-pass filter. 26-47. What is the value of the capacitive reactance in an RC high-pass filter at the critical frequency? 26-48. What value does the phase shift between input and output voltages approach in a high-pass filter when the frequency is (a) well below the critical frequency? (b) well above the critical frequency?

Section 26-6  RL High-Pass Filters 26-49. The output voltage is across which component in the RL high-pass filter? 26-50. State the mathematical relationship that determines the critical frequency in an RL high-pass filter. 26-51. What is the value of the inductive reactance in an RL high-pass filter at the critical frequency?

Section 26-7  Band-Pass Filters 26-52. Name two types of circuit arrangements commonly used to form band-pass filters. 26-53. What condition must be met for cascaded high-pass and low-pass filters to function as a band-pass filter? 26-54. Define the bandwidth of a filter. 26-55. The output voltage is across which component(s) in the (a) series resonant band-pass filter? (b) parallel resonant band-pass filter? 26-56. State the mathematical relationship that determines the centre frequency of a band-pass filter formed from a series resonant filter. 26-57. State the mathematical relationship that determines the bandwidth of a band-pass filter formed from a series or parallel resonant filter.

Section 26-8  Band-Stop Filters 26-58. Define the stopband of a band-stop filter. 26-59. Name two types of circuit arrangements commonly used to form band-stop filters. 26-60. What condition must be met for a parallel arrangement of highpass and low-pass filters to function as a band-stop filter? 26-61. The output voltage is across which component(s) in the (a) series resonant band-stop filter? (b) parallel resonant band-stop filter? 26-62. State the mathematical relationship that determines the centre frequency of a band-stop filter formed from a series resonant filter.

Integrate the Concepts

B

B B

26-63. State the mathematical relationship that determines the stopband of a band-stop filter formed from a series or parallel resonant filter.

Section 26-10  Troubleshooting 26-64. State three faults that can occur in filter circuits. 26-65. State four manners in which a filter may behave if it is not operating as designed.

Integrate the Concepts Refer to the block diagram of passive filter circuits in Figure 26-48. The amplitude of the output voltage of each filter will be either 0 V or the same as vin depending on the frequency of the input signal and the type of filter. Complete Table 26-3.

fo = 25 kHz

Vo3

BW = 5 kHz

Vin 10 Vp–p

fC = 10 kHz

Vo1

High-pass filter

fC = 50 kHz

Vo2

Low-pass filter

Series resonant band-pass filter

fo = 25 kHz

Vo4

Stopband = 5 kHz Parallel resonant band-stop filter

Figure 26-48  Block diagram for “Integrate the Concepts”

TABLE 26-3  Filter output voltages for Figure 26-43 Frequency (kHz)  2 15 27 40 70

Vo1 (volts, p-p)

Vo2 (volts, p-p)

Vo3 (volts, p-p)

Vo4 (volts, p-p)

821

822

Chapter 26   Passive Filters

Practice Quiz 1.

Match each description to the corresponding type of filter. (a) attenuates all signals above band-pass filter a specified frequency (b) attenuates all signals below high-pass filter a specified frequency (c) attenuates all signals between low-pass filter two specified frequencies (d) attenuates all signals outside a band-stop filter specified range

2.

For an RC low-pass filter consisting of a 10-kΩ resistor and a 3.6-nF capacitor, the critical frequency is (a) 44.2 kHz (b) 4.42 kHz (c) 4.42 Hz (d) 442 Hz

3.

In an RL low-pass filter with an inductance of 50 mH and a critical frequency of 10 kHz, the resistance is (a) 1 kΩ (b) 31.4 kΩ (c) 3.14 kΩ (d) 314 Ω

4.

In an RC high-pass filter with a resistance of 4.7 kΩ and a critical frequency of 20 kHz, the capacitance is (a) 1.69 nF (b) 1.69 μF (c) 5.32 μF (d) 5.32 nF

5.

In an RL high-pass filter with a resistance of 4.7 kΩ and an inductance of 50 mH, the critical frequency is (a) 14.96 MHz (b) 1.5 kHz (c) 47 kHz (d) 14.96 kHz

6.

To form a band-pass filter using a low-pass filter and a high-pass filter, they are arranged (a) in cascade (b) in series parallel (c) in parallel (d) in series (e) any of the above

Practice Quiz

7.

To form a band-stop filter using a low-pass filter and a high-pass filter, they are arranged (a) in cascade (b) in parallel c) in series d) in series parallel (e) any of the above

8.

Which combination can be used to make an RC high-pass filter with a critical frequency of 30 kHz? (a) 0.01 μF and 680 Ω (b) 10 nF and 220 Ω (c) 0.01 μF and 530 Ω (d) 10 nF and 100 Ω

9.

Which combination can be used to make an RL high-pass filter with a critical frequency of 45 kHz? (a) 25 mH and 10 kΩ (b) 10 mH and 2.83 kΩ (c) 50 mH and 15 kΩ (d) 2 H and 910 Ω

10. The critical frequencies of a band-pass filter are 20 Hz and 20 kHz. The bandwidth of this filter is (a) all the frequencies above 20 kHz (b) all the frequencies below 20 kHz (c) all frequencies between 20 Hz and 20 kHz (d) all the frequencies outside of 20 Hz and 20 kHz

11. The critical frequencies of a band-stop filter are 55 Hz and 65 Hz. The filter will pass (a) all frequencies outside of 55 Hz and 65 Hz (b) all frequencies below 55 Hz (c) all frequencies between 55 Hz and 65 Hz (d) all frequencies above 65 Hz 12. The bandwidth of a band-pass filter is (a) proportional to the centre frequency and the quality factor (b) inversely proportional to the centre frequency and the quality factor (c) inversely proportional to the centre frequency and proportional to the quality factor (d) proportional to the centre frequency and inversely proportional to the quality factor 13. The output voltage of a faulty RC low-pass filter is 0 V for all frequencies. Could this output result from the capacitor being open? 14. The output voltage of a faulty RL high-pass filter is the same as the input for all frequencies. Could this output result from the resistor being shorted?

823

27

Transformers

Section 16-1 introduced mutual induction, a process in which two windings are linked by magnetic flux. Because of this magnetic linkage, a changing current in a primary winding induces a voltage in a secondary winding. Since current in an AC circuit is continually changing, mutual induction is quite important in AC circuits.

Chapter Outline 27-1

Transformer Action  826

27-2 Transformation Ratio  828

27-3 Impedance Transformation  831 27-4 Leakage Reactance  833

27-5 Open-Circuit and Short-Circuit Tests  835 27-6 Transformer Efficiency  837

27-7 Effect of Loading a Transformer  838 27-8 Autotransformers  841 27-9 Troubleshooting  843

Key Terms transformer 826 general transformer equation 827 exciting current  828 magnetizing current  828 transformation ratio  829 step-up transformer  829

step-down transformer  829 reflected impedance  831 impedance matching  831 leakage flux  833 mutual flux  833 leakage reactance  834 copper loss  835

core loss  835 transformer efficiency  837 funicular phasor diagram  839 transformer voltage regulation  839 autotransformer  841

Learning Outcomes At the conclusion of this chapter, you will be able to: • calculate the magnetizing current in a transformer given its construction, dimensions, and flux density • describe the voltage induced in the primary winding of a transformer connected to an AC source • use the transformation ratio to calculate primary and ­secondary voltages and currents • compare the secondary voltage to the primary voltage for step-up and step-down transformers • calculate the reflected impedance given the load on the secondary winding • explain the primary leakage reactance in terms of leakage flux • determine transformer core loss from an open-circuit test Photo sources:  © iStock.com/wolv

• use a short-circuit test to determine equivalent resistance and reactance of a transformer • calculate the efficiency of a transformer for a given load • draw a funicular phasor diagram for a transformer under load • calculate the voltage regulation of a transformer • describe the construction and operation of an ­autotransformer • calculate current, voltage, and power for a given ­autotransformer under load • troubleshoot a transformer for open, shorted, or partially shorted windings

826

Chapter 27  Transformers

27-1  Transformer Action

Source:  Courtesy of Powervolt Incorporated

The operation of a transformer involves both self-induction and mutual induction. For the moment, we shall assume that the secondary terminals of the transformer in Figure 27-1 are left open-circuit. According to Kirchhoff’s voltage law, the voltage that appears across the primary coil must be exactly equal to the applied voltage at every instant. If the resistance of the primary coil is negligible, this voltage must be induced in the coil by a changing magnetic flux in the core. Since the applied voltage is a sine wave and the induced voltage is proportional to the rate of change of flux (Faraday’s law), the flux in the transformer core must be a sine wave lagging behind the a­ pplied voltage by 90°. ϕm

E

Primary winding

Secondary winding

ZL

Figure 27-1  Transformer action Transformer with multiple taps on both windings

Since each turn of the primary coil is linked by the same flux, Eav =



NΦm t

(27-1)

where Eav is the half-cycle average voltage induced in a coil (in volts), N is the number of turns in the coil, and t is the time (in seconds) the flux takes to rise from zero to its maximum value, Φm (in webers). The flux in the transformer is a sine wave, which rises from zero to Φm in one-quarter of a cycle. Therefore, Eav =

NΦm = 4 f NΦm 1/4 f

Since the half-cycle average value of a sine wave is 0.637 of the peak value, and the RMS value of a sine wave is 0.707 of the peak value, E = 1.11Eav

and

(27-2)



E = 4.44 f NΦm



(27-3)

where f is the frequency of the applied voltage in hertz, N is the number of turns in the coil, and Φm is the peak flux in webers.

27-1   Transformer Action

Equation 27-3 is the general transformer equation. Applying this equation to the primary winding of the simple transformer of Figure 27-1 gives Ep = 4.44 f NpΦm





(27-4)

To create this sine wave of flux in the core, an alternating current must flow in the primary winding. Combining Equations 14-1 and 14-2 gives Φp =



Therefore,

and

NpIp Fm = Rm Rm

Ipm = Φm

Ip =

Φm √2

Rm Np

×

Rm Np

(27-5)

To apply Equations 27-4 and 27-5 to a practical transformer, we must ensure that the magnetic flux in the core and the primary current are pure sine waves. Therefore the flux density, B, in the core must remain in the linear ­region of the magnetization curve for the iron in the core. The BH curve for sheet steel in Figure 14-15 shows that Bm should be less than 1.1 T. Once we decide on the maximum flux density and the cross-sectional area of the core, we can calculate the number of turns needed for the primary winding. To use Equation 27-5 to solve for the primary current, we must first determine the reluctance of the magnetic circuit of the transformer. Or we can use the magnetization curve of Figure 14-15 to find the magnetic field strength, H, that produces Bm. Combining Equations 14-1 and 14-7 gives NI = Hl

Therefore,

Im =

and

I=

Hml N

√2 N Hml



(27-6)

where I is the RMS primary current in amperes, Hm is the maximum magnetic field strength in amperes per metre, l is the length of the magnetic ­circuit in metres, and N is the number of turns.

827

See Appendix 2-9 for a calculus derivation of the general transformer equation.

828

Chapter 27  Transformers

For the moment, we will ignore losses in the core of the transformer. Even though the secondary of the transformer is an open circuit, there must be a sine wave of primary current in phase with the flux and lagging behind the applied voltage by 90°. This primary current is called the exciting current or magnetizing current of the transformer. In good transformers, this ­magnetizing current is usually less than 5% of the current that the primary winding carries when full load is connected to the secondary.

Example 27-1 The sheet-steel core of a transformer has a cross-sectional area of 5.0 cm2 and an average path length of 25 cm. If the peak flux density in the core is to be 1.0 T, (a) find the number of turns needed on a primary winding that is to be connected to a 120-V 60-Hz source (b) find the RMS current in the primary winding Solution (a) Φm = 5.0 cm2 × 1.0 T = 0.50 mWb

Np =

Ep

4.44 fΦm

=

120 V = 901 turns 4.44 × 60 Hz × 0.00050 Wb

(b) From the BH curve for sheet steel in Figure 14-15, H = 425 A/m when B = 1.0 T. Therefore, Ip =

Hml 425 At/m × 25 cm = = 83 mA √ 2 Np √ 2 × 901

See Problems 27-1 to 27-6 and Review Questions 27-32 to 27-40 at the end of the chapter.

27-2  Transformation Ratio In an ideal transformer, all the flux created by primary current links the secondary winding. Thus, the voltage induced in the secondary winding is

Es = 4.44 f NsΦm

(27-7)

Combining Equation 27-7 and Equation 27-4 gives



Ep Es

=

Np Ns

=a

(27-8)

27-2   Transformation Ratio

Source:  © iStock.com/wolv

Consequently, when all the primary flux links the secondary winding, the ratio of the primary voltage to the secondary voltage is the same as the ratio of the number of primary turns to the number of secondary turns. The ratio Ep /Es is the transformation ratio, a. Equation 27-8 shows that transformers can alter the voltage from a given AC source. A transformer that produces a secondary voltage higher than the source voltage is called a step-up transformer. Similarly, a step-down ­transformer produces a secondary voltage that is less than the primary voltage.

Substation step-down transformer

Example 27-2 If the transformer in Example 27-1 is to supply 6.3 V for the power ­supply of a calculator, how many turns must the secondary winding have? Solution

a=

Ns =

Ep Es

=

120 = 19 6.3

Np 901 = ≈ 48 turns a 19

When we connect a load to the secondary winding of the transformer circuit in Figure 27-1, the voltage induced in the secondary winding ­ causes current to flow in the secondary circuit. This current creates a ­magnetomotive force of NsIs ampere-turns. According to Lenz’s law, this

829

Chapter 27  Transformers

tends to reduce the magnetic flux in the core. But in a lossless transformer, the ­amplitude of the flux sine wave cannot change because it must always ­induce a voltage equal to the applied voltage into the primary winding. Therefore, when a load is connected to the secondary, additional primary current flows to develop a primary MMF component that exactly cancels the secondary MMF. Therefore, MMF

NpI′p = NsIs



(27-9)

where I′p is the additional, or load, component of the primary current. Rearranging Equation 27-9 gives Np Is = =a I′p Ns



(27-10)



If the load connected to the secondary terminals of the transformer is a  pure resistance, the secondary current is in phase with the secondary voltage. Since Ep and Es are both induced by the same changing flux, the additional primary current, I′p , is in phase with the applied voltage. Due to core loss, the magnetizing current, Imag, lags behind the applied voltage by slightly less than 90°. The total primary current, Ip, is the phasor sum of the load and magnetizing components, as shown in Figure 27-2. Ipʹ Imag

E Ip

Figure 27-2  Total primary current of a transformer with a resistance load

Source:  © iStock.com/Solidago

830

Step-down transformers supplying power to a farm in Alberta

See Problems 27-7 to 27-16 and Review Questions 27-41 to 27-43.

27-3   Impedance Transformation

831

27-3  Impedance Transformation When a load is connected to the secondary winding of the transformer in Figure 27-1, we can treat the source current as the phasor sum of two ­separate parallel branch currents. Hence, the right-hand circuit in Figure 27-3 is an equivalent circuit for the transformer circuit on the left. In  the equivalent circuit, the parallel branch carrying I′p has an ­impedance of Zp =



Ep I′p

This impedance is called the reflected impedance of the secondary ­circuit.

ZL

a2ZL

Open circuit

Ipʹ Figure 27-3  Reflected impedance

From Equation 27-8,

Ep =

Np Ns

Es = aEs

and from Equation 27-10,

Ip′ =

Ns Is Is = a Np

Therefore,

Zp =

Es N p Is Ns

( )

2

But Es /Is depends on the total impedance of the secondary circuit. In a good transformer, the impedance of the secondary winding will be small compared with the load impedance, and



Zp ≈

( ) Np Ns

ZL ≈ a2ZL

2

(27-11)

Many circuits use transformers for impedance matching. For example, a transformer can match a low-impedance load to the high-impedance output of an amplifier, thus enabling maximum power transfer to take place with reduced distortion.

In schematic diagrams, the primary winding, which is ­connected to the source, is usually shown as the left side of the transformer with the secondary winding on the right side.

832

Chapter 27  Transformers

Example 27-3 An audio transformer has 1200 turns in the primary winding, which is connected to the output of an amplifier. How many turns must be wound in the secondary winding to make a 4-Ω speaker appear as a 5-kΩ load to the amplifier? Solution

a2 =



Zp ZL

=

a = 35.36



Ns =



5000 = 1250 4

Np 1200 = = 34 turns a 35.36

Example 27-4 The audio output transformer of Figure 27-4 presents the correct load impedance to a transistor amplifier when either an 8-Ω load is connected to terminals A and B or a 16-Ω load is connected to terminals A and C. What load may be connected to terminals B and C alone to present the same reflected load? C B

Zp

16 Ω 8Ω

A Figure 27-4  Tapped audio output transformer

We can also solve this example by assuming convenient values for the number of primary turns and reflected impedance. Note that the impedance required is not simply the difference between the two loads.

Solution Let N1 represent the number of secondary turns between terminals A and B, and N2 the total number of turns between terminals A and C. 2 Np 2 Np 2 Np Then, Zp = 8 = 16 = Zx N1 N2 N2 − N1

√8 N1

=

( )

√ 16 N2

=

( )

√ Zx

(

)

N2 − N1

√ Zx = √ 16 − √ 8

Zx = (√ 16 − √ 8 ) 2 = 1.37 Ω

See Problems 27-17 to 27-22 and Review Questions 27-44 to 27-48.

27-4   Leakage Reactance

Circuit Check

A

CC 27-1. Determine the frequency required for a transformer that has a peak flux value of 20 mWb, with a 500-V primary with 300 turns. CC 27-2. Find the turns ratio for a transformer that has 300 turns in the primary winding and 150 turns in the secondary winding. CC 27-3. Determine the following quantities for the circuit shown in Figure 27-5: (a) secondary voltage (b) secondary current (c) load resistor (d) power in the load (e) reflected load Fuse

a=2

10 mA RL

+ 10 V

Figure 27-5

27-4  Leakage Reactance In Section 27-2, we assumed that all the flux created by the magnetizing current in the primary linked the secondary winding. However, even in the best transformers, some leakage flux follows a path outside the core, as shown in Figure 27-6. This portion of the primary flux does not link the secondary winding. Similarly, when load current flows in the secondary, some of the secondary flux does not link the primary winding. Flux that does link another winding is called the mutual flux. ϕm

E

Primary leakage flux

ϕleak

Secondary

Figure 27-6  Leakage flux

The effect of leakage flux is the same as if Φmutual/Φp of the primary turns were developing flux that linked the secondary winding and Φleak/Φp of the primary turns acted as a separate inductance in series with the primary winding. The reactance of these ineffective primary turns is called the

833

834

Chapter 27  Transformers

­primary leakage reactance. The equivalent circuit in Figure 27-7 shows the effect of the losses in a practical transformer. Rp and Xp represent the primary resistance and leakage reactance, respectively, while Rs and Xs represent the secondary resistance and leakage reactance. The parallel ­resistance RCL shows the effect of energy losses in the core. T is an ideal transformer, and Lmag is the self-inductance that determines the magnetizing current produced by the primary winding. Rp

Xp

Xs

Rs

T Lmag

RCL

ZL

Lmag

Figure 27-7  Equivalent circuit for a practical transformer

In most iron-core transformers, the primary and secondary leakage reactances represent approximately the same percentage of their respective windings. Therefore, the transformation ratio of the ideal transformer in the equivalent circuit of Figure 27-7 is still essentially Np / Ns. Therefore, the reflected impedance is Zref = a2 ( Rs + jXs + ZL )



(27-12)

We can simplify the equivalent circuit as shown in Figure 27-8(a). If the transformer is fully loaded, the magnetizing current in the branch of the equivalent circuit containing RCL and Lmag is quite small compared with the load current and it produces only a negligible voltage drop across Rp and Xp. Consequently, when we calculate the effects of loading a transformer, we can omit RCL and Lmag, reducing the equivalent circuit to the simple series circuit of Figure 27-8(b). In this series circuit, Re = Rp + a2Rs and Xe = Xp + a2Xs. Rp

Xp

a2Xs

RCL

Lmag

a2Rs

Re

Xe

a2ZL

a2ZL

Complete equivalent network for Figure 27-7

Simplified equivalent network

(a)

(b)

Figure 27-8  Equivalent circuits for an iron-core transformer

See Review Questions 27-49 and 27-50.

27-5   Open-Circuit and Short-Circuit Tests

27-5 Open-Circuit and Short-Circuit Tests We can determine the parameters for the equivalent circuits of Figure 27-8 from two simple tests on the actual transformer. If we leave the secondary terminals open-circuit, ZL in Figure 27-8(a) is infinitely large, and all the current flows through the branch with RCL and Lmag in parallel. If the transformer is designed with a 120-V primary winding, we probably use the 120-V winding as the primary winding for the test. However, if the transformer is designed to step a higher voltage down to 120 V, we use the 120-V winding as the primary to protect the measuring ­instruments. When testing a step-down transformer in this way, we must make sure that no one can touch the open-circuit high-voltage winding while power is connected to the low-voltage winding in the setup of Figure 27-9(a).

A

± W E (adjustable)

±

DANGER HIGH VOLTAGE

V

(a) A

± R

W ±

E

V

(b) Figure 27-9  Meter connections: (a) Open-circuit test; (b) Short-circuit test

For the open-circuit test, we connect a wattmeter, a voltmeter, and an ammeter as shown in Figure 27-9(a), and then feed the winding we are using as the primary with the voltage for which it was designed. Since the magnetizing current is only a very small fraction of the full-load primary current, we can assume that the I2R loss, or copper loss, due to magnetizing current alone in the primary winding is negligible. Therefore, the losses shown by the wattmeter are the core loss resulting from eddy currents and magnetic hysteresis in the iron core. The eddy currents produce a small counterflux in the core, just as if we had connected a resistor across the

835

836

Chapter 27  Transformers

secondary winding. The hysteresis loop of the iron causes the magnetizing current to lag behind the applied voltage by slightly less than 90°. The small in-phase component of the magnetizing current contributes to the ­active power reading on the wattmeter. From the voltmeter and ammeter readings, we can determine the total impedance of the branch consisting of Rp, Xp, RCL, and Lmag. An increase in load current slightly increases the voltage drop across Rp and Xp. This ­increased voltage, in turn, slightly reduces the current in the RCL and Lmag. Consequently, there is a slight reduction in magnetizing current as the transformer is fully loaded. However, for practical purposes, we can assume that the core loss remains constant as the load is increased from zero to full-rated load. For the short-circuit test, we short-circuit the low-voltage winding of the transformer, as shown in Figure 27-9(b). For a step-down transformer, the low-voltage winding is its normal secondary winding. The short circuit is equivalent to making ZL zero in Figure 27-8(b). Since Re + jXe is fairly small, we use rheostat R to adjust the voltage applied to the highvoltage winding so that the rated current flows in the windings. This applied v ­ oltage is less than 5% of the rated voltage for the high-voltage winding. Since reducing the applied voltage reduces the magnetizing current to less than 5% of its normal value, the core loss is now less than 0.25% of what they are ­during normal operation. Therefore, we can assume that the w ­ attmeter shows only the copper loss, from which we can determine Re.

Example 27-5 When the secondary winding of a small step-down transformer is short-circuited, an applied primary voltage of 6.0 V produces a primary current of 4.0 A and a wattmeter reading of 12 W. Determine Re and Xe. Solution



Re =

Ze =

P 12 W = 0.75 Ω 2 = ( I 4.0 A ) 2

E 6.0 V = = 1.5 Ω I 4.0 A

Xe = √ Z2e − R2e = √ 1.52 − 0.752 = 1.3 Ω ( inductive )

See Problems 27-23 and 27-24 and Review Questions 27-51 to 27-54.

27-6   Transformer Efficiency

27-6  Transformer Efficiency The core-loss data from the open-circuit test are needed for calculating the transformer efficiency. From Equation 6-6, η=



Pout PL = Pin PL + copper loss + core loss

(27-13)

The core loss remains constant and is equal to the wattmeter reading from the open-circuit test. But the copper loss is proportional to the square of current in the equivalent network of Figure 27-8(b). For practical p ­ urposes, the current, in turn, is proportional to the apparent power into the load. Hence, η=



SL cos ϕ + I2Re + PCL SL cos ϕ



(27-14)

Example 27-6 The transformer tested in Example 27-5 has a transformation ratio of 10 and develops a full-load secondary voltage of 12 V when the load current is 40 A at 0.866 lagging power factor. The open-circuit test indicates core loss of 5.0 W. Calculate the efficiency of the transformer with this load. Solution Step 1

SL cos ϕ = 12 V × 40 A × 0.866 = 416 W Step 2 In the equivalent network of Figure 27-8(b), I=

and Step 3

40 A Is = = 4.0 A a 10

copper loss = I2Re = (4.0 A)2 × 0.75 Ω = 12 W η=

416 W = 0.96 = 96% 416 W + 12 W + 5 W

837

838

Chapter 27  Transformers

Since transformers are usually operated at somewhat less than full load, they are designed so that maximum efficiency occurs at somewhat less than full load. Appendix 2-10 uses calculus to determine this condition for maximum efficiency: A transformer has maximum efficiency when the copper loss equals the core loss.

Example 27-7 What secondary current must be drawn from the transformer of Example 27-6 for maximum efficiency? Solution Step 1 Since I2Re = PCL for maximum efficiency,

I=

Step 2 The secondary current is

5.0 W = 2.58 A √RP = √0.75 Ω e

Is = aI = 10 × 2.58 A = 26 A

See Problems 27-25 and 27-26.

27-7  Effect of Loading a Transformer An important effect of the leakage reactance and resistance of the windings in Figure 27-7 is that the load voltage VL is not the same as the voltage induced in the secondary. To satisfy Kirchhoff’s voltage law,

Es = IsRs + jIsXs + VL

Similarly, the voltage induced in the primary by the mutual flux must be slightly less than the total applied voltage Vp. Therefore, as the current drawn by the load changes, the voltage across the load must also change, even though the applied voltage may remain constant. We can show the effect of load current on the load voltage by preparing phasor diagrams for the equivalent network of Figure 27-8(b). The current flowing in this equivalent circuit is the primary load current, which is 1/a times the actual load current. Likewise, the voltage across a2ZL is a times

27-7   Effect of Loading a Transformer

the actual load voltage. As we can see from Figure 27-8(b), the total applied primary voltage is the phasor sum of three voltages: Vp = aVL + IpRe + Ip Xe



(27-15)

Because the current is common to all components, we use it as a reference phasor in Figure 27-10. The phase angle of aVL with respect to the current is equal to the phase angle of the power factor of the load. The voltage IpRe equals ReIL /a and is in phase with the current. Similarly, IpXe equals XeIL/a and leads the current by 90°. Although we could add these phasor voltages by the customary geometrical construction, transformer phasor diagrams are less confusing if we add each phasor, with its proper direction and magnitude, to the tip of the preceding phasor, as shown in Figure 27-10. Such diagrams are called funicular phasor diagrams.

IpXe

Vp ϕ

aVL

ϕ

IpRe

Ip

Vp

aVL

IpXe

Ip (a)

IpRe (b)

Figure 27-10  Phasor diagrams for a transformer under load: (a) Lagging power-factor load; (b) Leading power-factor load

As mentioned in Section 27-4, the magnetizing current is small enough to ignore in loading calculations. Hence, Ip becomes zero when the transformer has no load. Since IpRe and IpXe also become zero, aVL then equals VP. But when the load has a lagging power factor, VL leads the current, and the load voltage drops appreciably under load. However, when the load has a leading power factor, the full-load voltage and no-load voltage are almost the same. The percentage change in secondary voltage from noload to full-load with respect to the full-load voltage is called the percentage transformer voltage regulation.



Voltage regulation =

VNL − VFL × 100% VFL

(27-16)

where VNL is the no-load secondary voltage, and VFL is the full-load ­secondary voltage.

839

840

Chapter 27  Transformers

Example 27-8 Calculate the percentage voltage regulation for the transformer of Example 27-6. aVL = 10 × 12 V = 120 V Solution

Since the power factor is 0.866 lagging, VL leads the current by cos−1 0.866, or 30°, as shown in Figure 27-11. IpRe =



IpXe =



40 A × 0.75 Ω = 3.0 V 10

40 A × 1.3 Ω = 5.2 V 10 5.2 V

2

.2

2

7

10

65

20

+

√

× 10

1 2=

V

120 sin 30°

1

30°

I 3.0 V

120 cos 30°

Figure 27-11  Phasor diagram for Example 27-8

Convert the load voltage into rectangular coordinates in order to carry out the phasor addition shown in Figure 27-11.

aVL = 120 cos 30º + j120 sin 30º = 104 + j60 V

Vp = ( 104 + 3 ) + j ( 60 + 5.2 ) = 107 + j65.2 V Vp = √ 1072 + 65.22 = 125.3 V

VNL =

Voltage regulation = =

Vp 125.3 V = = 12.53 V a 10

VNL − VFL × 100% VFL

12.53 − 12.0 × 100% 12.0

= 4.4%

See Problems 27-27 to 27-29 and Review Questions 27-55 to 27-57.

27-8  Autotransformers

Circuit Check

B

CC 27-4. A 100-kVA  60-Hz  66-kV/13.8-kV power distribution transformer was tested at full voltage and found to have a core loss of 4.3 kW and a copper loss of 4.5 kW. Calculate the transformer efficiency at (a) full-load with unity power factor (b) one-third load with power factor 0.9 lagging CC 27-5. A short-circuit test on a 25-kVA 550-V/120-V transformer found that 52 V applied to the primary produced the rated current in the short-circuited secondary, and the power taken was 1.2 kW. (a) Calculate the voltage regulation at full-load power with a power factor of 0.8 leading. (b) Sketch the phasor diagram.

27-8 Autotransformers In many applications, the primary and secondary circuits do not need to be electrically isolated from each other. For such applications, the ­autotransformer construction in Figure 27-12 can provide efficient v ­ oltage conversions at less cost. Instead of separate primary and secondary ­windings, the autotransformer has one continuous tapped winding.

120 V 60 Hz

Primary

200 turns

Secondary

100 turns Load Ip = 10 A

Ip = 5 A

IL = 15 A

Source: Courtesy of Superior Electric, Danaher Sensor & Controls Group

Figure 27-12  Autotransformer

Autotransformer with a moveable tap

841

842

Chapter 27  Transformers

Since the lower 100 turns in Figure 27-12 are connected to the load, they act as the secondary winding. The primary winding consists of only  the upper 200 turns. Hence, the primary voltage is not the same as the source voltage. Since the source voltage is connected across the ­complete winding, the RMS magnetic flux in the core is constant. The applied voltage divides according to the number of turns in each section of the winding, so the “primary” voltage is 80 V and the secondary voltage is 40 V. The load in Figure 27-12 draws 15 A at 40 V for an apparent power of 600 VA. If we assume negligible core and copper losses, the apparent power drawn from the source is also 600 VA, and I′p = 600/120 = 5.0 A. According to Kirchhoff’s current law, the secondary current must be 10 A with a direction as shown in Figure 27-12. This current also satisfies Equation 27-10. The direction of Is is opposite to the direction of I′p, and Np I′p =  NsIs. Therefore, the two components of the load current produce equal and opposite MMFs in the transformer core. As with a transformer with separate windings, the load current does not affect the magnetic flux in the core of an autotransformer, provided that the transformer losses are negligible. The apparent power delivered to the load by the secondary winding through transformer action is Ss = Vs × Is = 40 V × 10 A = 400 VA

The remaining 200 VA reaches the load by direct conduction because onethird of the load current flows through the primary. Therefore, an autotransformer used to reduce the voltage applied to a 600-VA load can be considerably smaller than a 600-VA transformer with separate primary and secondary windings. An autotransformer can also be connected to increase a load voltage. We can use a transformer with separate windings as an autotransformer by connecting them to act as a single continuous winding, as in Figure 27-13. 10 A

Secondary

11 A Load E

110 V 60 Hz 11 A

1.0 A

Primary 10 A

Figure 27-13  Separate-winding transformer connected as a step-up autotransformer

27-9  Troubleshooting

Example 27-9 A 10-A load requires a 120-V source for proper operation, but the available 60-Hz power outlet measures only 110 V. Show that a 120-V to 12-V two-winding filament transformer rated at 120 VA can provide the necessary voltage when connected as shown in Figure 27-13. Solution From the transformer specifications,

120 V = 10 Vs 12 V Since the primary winding is connected across the source, Vp = E. ­Therefore, a=



Vp

=

110 V = 11 V    and    VL = 110 V + 11 V = 121 V 10 Since IL flows through the secondary winding, Vs =

10 A Is = = 1.0 A a 10 Therefore, the current drawn from the source is

I′p =



I = 1.0 A + 10 A = 11 A

SL = 121 V × 10 A = 1210 VA



But the apparent power delivered to the load by transformer action is only Sp = 110 V × 1.0 A = 110 VA   or  Ss = 11 V × 10 A = 110 VA

See Problems 27-30 and 27-31 and Review Questions 27-60 and 27-61.

27-9 Troubleshooting Since transformers consist simply of windings on a core, they develop relatively few faults. The most common of these faults are open, shorted, or partially shorted windings.

Open Windings An open primary results in no primary current and consequently no ­induced voltage or current in the secondary (see Figure 27-14). An open secondary also results in no current in the secondary and no voltage across

843

844

Chapter 27  Transformers

the load. However, when the secondary is open, the magnetizing current still flows in the primary. Thus, by measuring the primary current, we can determine which of the two windings is open. Such faults can also be identified by ­disconnecting the transformer from the circuit and connecting an ohmmeter across each winding in turn (see Figure 27-15). If the winding is open, the meter will show infinite resistance.

E

V

Figure 27-14  Measuring the secondary voltage of a transformer

Ω

Figure 27-15  Testing the primary winding with an ohmmeter

Shorted Windings A shorted primary causes a very large primary current. Usually either the source or the transformer will burn out unless it is protected by a fuse or circuit breaker. If a short is suspected, disconnect the transformer from the source and test the primary with an ohmmeter. If there is a short, the meter will read 0 Ω. The meter scale selected has to be low enough to measure the DC resistance of the primary, which may be less than 1 Ω. A shorted secondary results in a very large secondary current but little or no load current since the short is connected across the load (see Figure 27-16). The primary current is also much greater than normal since the ­reflected ­impedance is very low, as given by Equation 27-11. A suspected short can be checked with an ohmmeter across the secondary winding with the source and load disconnected. If there is a short, the meter will read 0 Ω. Again the meter should be on a low scale. If part of the primary or secondary winding is shorted, the transformation ratio changes. For example, if the primary has 50 turns and the secondary has 100 turns, the transformation ratio, a, is 2, and the secondary voltage is twice the primary voltage. Suppose that 10 turns of the primary were

27-9  Troubleshooting

Short E

Load

Figure 27-16  Partially shorted primary or secondary

shorted. Now a = 100/40 = 2.5. The secondary voltage would be 2.5 times the primary voltage. A partially shorted winding could be d ­ etected by measuring the primary and secondary voltages with an AC voltmeter, and then calculating the transformation ratio. See Review Questions 27-54 and 27-55.

845

846

Chapter 27  Transformers

Summary

• The general transformer equation relates the voltage induced in the primary to the frequency of the source, the number of turns in the primary, and the maximum magnetic flux. • The magnetizing current depends on the magnetizing force, the length of the magnetic circuit, and the number of turns in the primary. • The transformation ratio relates the primary voltage to the secondary voltage and the primary current to the secondary current. • The impedance seen by the source connected to the primary equals the load times the square of the transformation ratio. • The effect of flux that leaks from the core of a transformer can be represented by leakage reactances in series with the windings. • An open-circuit test can determine the transformer core loss. • A short-circuit test can determine the equivalent resistance and reactance of the transformer. • The efficiency of a transformer depends on copper loss and core loss. • Funicular phasor diagrams are convenient for analyzing a transformer under load. • Percentage voltage regulation is a measure of the change in load voltage from no-load to full-load conditions. • An autotransformer has one continuous tapped winding that serves as the primary and secondary windings. • Open, shorted, or partially shorted windings can be detected with simple resistance, current, and voltage measurements. B = beginner

I = intermediate

A = advanced

Problems I

I

I I I I

Section 27-1  Transformer Action 27-1. A 1200/120-V 20-kVA 60-Hz transformer is designed to have an induced voltage of 5 V per turn. Determine (a) the number of turns (b) the full-load current 27-2. In the steel sheet core of a transformer with an average path length of 30 cm and a cross-sectional area of 4.0 cm2, which is connected to a 110-V 60-Hz source, must not exceed a flux density of 1.1 T. How many turns must the primary winding of the transformer have? 27-3. Assuming the magnetizing current to be a pure sine wave, calculate the RMS primary current in the transformer of Problem 27-2 when the secondary is open-circuit. (Use the BH curves of Figure 14-15.) 27-4. The primary winding of the transformer of Problem 27-2 is ­connected to a 110-V 25-Hz source. Find the RMS current. 27-5. What cross-sectional area must the core of the transformer in Problem 27-1 have if the flux density is not to exceed 1.1 T? 27-6. With the secondary winding open-circuit, the primary current of the ­trans­former in Problem 27-1 is measured as 300 mA. Find the

Problems

847

a­ p­proximate average path length of the magnetic circuit of the ­transformer. B B

B B

B B I B

B

I

B B

Section 27-2  Transformation Ratio 27-7. Determine the transformation ratio of a transformer having 200 turns on the primary winding and 30 turns on the secondary. 27-8. (a) Find the transformation ratio of a transformer that has an ­open-­circuit secondary voltage of 300 V RMS when the primary is connected to a 120-V 60-Hz source. (b) Use Multisim to verify the transformation ratio. 27-9. What RMS voltage must be applied to the primary of the ­transformer in Problem 27-7 to develop an open-circuit secondary voltage of 10.2 V? 27-10. (a) Find the load component of the primary current when a 2.0 kΩ resistor is connected across the secondary winding of the transformer in Problem 27-8. (b) Use Multisim to verify the load component of the primary current. 27-11. If the secondary of the transformer in Problem 27-8 has 1400 turns, how many turns does the primary have? 27-12. The primary of a certain transformer has 2.7 turns per volt of ­applied voltage. How many turns must a 5.0-V secondary winding have? 27-13. When a 0.80-Ω load is connected to the 24-V secondary of a ­step-down transformer, the primary current is measured as 8.0 A. ­Calculate the primary voltage of the transformer. 27-14. (a) How many turns would there be on a 7.5-V secondary winding on the transformer in Problem 27-2? (b) Use Multisim to verify the number of turns on the secondary winding of the transformer. 27-15. A 2300/230-V, 60-Hz transformer produces a flux of 20 mWb. Determine (a) the number of turns in the primary winding (b) the number of turns in the secondary winding 27-16. The no-load current of a 30-kVA 2000/220-V 60-Hz transformer is 5∠− 70º amps. The transformer is delivering full load power to a resistive load. (a) Calculate the total primary current. (b) Calculate the transformer phase angle. (c) Sketch the phasor diagram.

Section 27-3  Impedance Transformation 27-17. An audio output transformer has 3000 primary turns and 75 secondary turns. Find the reflected value of a 50-Ω load connected to its secondary. 27-18. The reflected impedance at the primary of a particular audio ­transformer is 7 kΩ at 1000 Hz when a 4-Ω load is connected to the secondary. What load should be connected to the secondary when this transformer is used with an amplifier requiring a 4500‑Ω load?

circuitSIM walkthrough

circuitSIM walkthrough

circuitSIM walkthrough

848

Chapter 27  Transformers

I

I

B

27-19. An audio output transformer has 1200 primary turns and two ­secondary windings of 50 turns each. If the circuit feeding the transformer must see a reflected load of 5 kΩ, calculate the load that must be connected to the secondary windings when they are ­connected (a) in series (b) in parallel 27-20. An impedance-matching transformer has a 6-kΩ primary winding and an 8-Ω secondary with a tap at 4-Ω with respect to a common secondary terminal. Find the reflected impedance when a 4-Ω load is mistakenly connected from the 4-Ω tap to the 8-Ω terminal. 27-21. What turns ratio for the transformer in Figure 27-17 will reflect 500 Ω into the primary circuit?

VS +

RL 1 kΩ

Figure 27-17

B

I I

27-22. The output of an amplifier is connected to an 8-Ω speaker through an impedance-matching transformer. The transformer has 1000 turns on the primary and 50 turns on the secondary winding. Determine the output impedance of the amplifier.

Section 27-5  Open-Circuit and Short-Circuit Tests

27-23. During a short-circuit test on a 2300/208-V 500-kVA 60-Hz distribution transformer, Vp = 95 V, IP = 218 A, and Pin = 8.2 kW. Calculate Re and Xe for this transformer [see Figure 27-8(b)]. 27-24. The open-circuit test on the transformer of Problem 27-23, with the high-voltage winding open and the low-voltage winding used as the primary, yields V = 208 V, I = 84 A, and P = 1.8 kW. Calculate the magnitude and phase angle of the exciting current when the transformer is used with the high-voltage winding as the primary. Draw phasor diagrams for the remaining problems.

I A

Section 27-6  Transformer Efficiency 27-25. Calculate the maximum efficiency and power output for the transformer of Problem 27-23 connected to a load with a 90% lagging power factor. 27-26. A 550/220-V 50-kVA 60-Hz transformer has Rp = 30 mΩ, Xp = 50 mΩ, Rs = 5.0 mΩ, and Xs = 10 mΩ. The power input to the transformer is 200 W when the secondary is open-circuit. Calculate the full-load efficiency when the load has a power factor of 1.

Review Questions

A A A

I

Section 27-7  Effect of Loading a Transformer 27-27. Calculate the percentage voltage regulation of the transformer of Problem 27-26 with a 50% lagging power-factor load. 27-28. Calculate the percentage voltage regulation for the transformer in Problem 27-23 with a 70.7% lagging power-factor load. 27-29. Calculate the percentage voltage regulation for the transformer in Problem 27-23 with an 86.6% leading power-factor load.

Section 27-8  Autotransformers

27-30. An autotransformer is connected as shown in Figure 27-18. Given that the load resistance is 15 Ω, calculate (a) the load current (b) the apparent power in the load (c) the source current (d) the current in each winding (e) the kVA rating of transformer

120 V 60 Hz

300 turns

Load

100 turns

Figure 27-18

I

27-31. A 220-V, 100-kVA load is to be supplied from a 1100-V source. (a) Sketch an autotransformer connection to provide this voltage. (b) Calculate the turns ratio and kVA rating of the autotransformer.

Review Questions

Section 27-1  Transformer Action 27-32. Why is mutual induction more significant in AC circuits than in DC circuits? 27-33. What is a transformer? 27-34. What is the purpose of an iron core in a power transformer? 27-35. What problems would be encountered in the use of iron cores in radio-frequency transformers? 27-36. What is the general transformer equation? What is its application? 27-37. What is exciting current? 27-38. What factors determine the exciting current? 27-39. The exciting current in a certain transformer lags behind the applied voltage by 87°. Account for this angle. 27-40. Account for the results obtained in Problem 27-4 and relate these ­results to a practical transformer.

849

850

Chapter 27  Transformers

Section 27-2  Transformation Ratio 27-41. What would be the effect of operating a 110-V 25-Hz transformer on a 120-V 60-Hz supply? 27-42. What is the relationship between the turns ratio of a transformer and its transformation ratio? 27-43. What is the relationship between the load current ratio and the turns ratio of a transformer?

Section 27-3  Impedance Transformation 27-44. By combining Equations 27-9 and 27-11, determine the ratio of power input to power output for a transformer. 27-45. What is meant by reflected impedance? 27-46. Why is the impedance ratio of a transformer equal to the square of the turns ratio? 27-47. What is the significance of impedance ratio in practical t­ ransformers? 27-48. Show the individual steps in reducing the initial equation in ­Example 27-4 to get an expression for Zx.

Section 27-4  Leakage Reactance 27-49. What is meant by mutual flux and leakage flux? 27-50. Define leakage reactance.

Section 27-5  Open-Circuit and Short-Circuit Tests 27-51. Why is the total core loss of a transformer essentially independent of load current? 27-52. Why can we assume that the wattmeter reading on an open-circuit transformer test is essentially only the core loss? 27-53. Why is the core loss on a short-circuit transformer test less than 1/400 of its normal value? 27-54. Explain why Re = Rp + a2Rs and Xe = Xp + a2Xs in Figure 27-8(b).

Section 27-7  Effect of Loading a Transformer

27-55. Draw a phasor diagram showing the relationship between primary terminal voltage and secondary terminal voltage for a load with a power factor of 1. 27-56. What is primarily responsible for the difference between the noload and full-load secondary voltages of a power transformer? 27-57. Why does a load with a leading power factor give better voltage regulation than a load with the corresponding lagging power factor?

Section 27-8  Autotransformers 27-58. Draw a circuit diagram for an autotransformer feeding a 208-V load from a 120-V source. Use calculations with a specific load to explain the operation of the autotransformer. 27-59. Using the load you chose in Question 27-58, show why an autotransformer can be appreciably smaller than the equivalent twowinding transformer.

Practice Quiz

Section 27-9  Troubleshooting 27-60. Describe the effects each of the following faults has on the operation of a transformer: (a) an open primary (b) an open secondary (c) a shorted primary (d) a shorted secondary (e) a partially shorted primary or secondary 27-61. How can each of the faults listed in Question 27-60 be detected?

Integrate the Concepts The sheet steel core of a transformer has a length of 15 cm and a ­cross‑ sectional area of 7 cm2. When the primary is connected to a 120‑V 60‑Hz supply, the secondary voltage is 40 V. The flux density is not to e­ xceed 1.2 T. (a) How many turns should the primary have? (b) Calculate the magnetizing current in the primary. (c) Calculate the transformation ratio. (d) How many turns should the secondary have? (e) Calculate the secondary and primary currents for a load of 50 Ω. (f) For the load of part (e), calculate the reflected impedance at the ­primary. (g) Under a short-circuit test, a 10-V source applied to the primary produces a primary current of 8.0 A and a power reading of 15 W. Calculate the equivalent resistance and reactance of the transformer. (h) The transformer develops a full-load secondary voltage of 25 V with a load current of 12 A at a lagging power factor of 0.900. An o ­ pen-circuit test shows a core loss of 7.0 W. Calculate the efficiency of the transformer with this load. (i) Calculate the percentage voltage regulation of the transformer for the load of part (h).

Practice Quiz 1. Which of the following statements are true? (a) Mutual inductance is defined as the inductance between two coils that are coupled by an electric field. (b) Transformers can be used for impedance matching. (c)  The reactance of ineffective primary turns is called primary ­leakage reactance. (d) A transformer has minimum efficiency when the copper loss equals the core loss of the transformer. (e) An autotransformer is a variation of a conventional transformer. (f) A step-up transformer can also function as a step-down transformer.

851

852

Chapter 27  Transformers

2. The operation of a transformer involves both (a) self-induction and permeability (b) mutual induction and permeability (c) self-induction and mutual induction (d) mutual induction and magnetic flux 3. Changing current in one magnetic circuit induces voltage in the other circuit because of (a) mutual inductance (b) self-inductance (c) reluctance (d) permeance 4. If the total flux in a cast iron core of a transformer connected to a 220-V 60-Hz source is 0.55 mWb, how many turns are on the primary winding? (a) 1000 turns (b) 1500 turns (c) 2000 turns (d) 2500 turns 5. In order to match a source with an internal impedance of 300 Ω to a 25-Ω load, we need to use a: (a) step-down transformer (b) step-up transformer (c) ideal transformer (d) autotransformer

6. If we increase the number of turns in the secondary winding of a transformer, the output voltage will (a) decrease (b) stay the same (c) increase (d) go to zero 7. If the RMS voltage across the load resistor in the circuit of Figure 27-19 is 275 V, the turns ratio of the transformer is (a) 0.4 (b) 0.5 (c) 2.5 (d) 4

+ −

110 V RMS 60 Hz 0°

Figure 27-19

a RL

50 Ω

Practice Quiz

8. The transformer of Figure 27-19 is (a) an air-core transformer (b) a unity transformer (c) a step-up transformer (d) a step-down transformer 9. The current in the primary winding of the transformer of Figure 27-19 is (a) 2.2 A (b) 5.5 A (c) 13.75 A (d) 34.78 A 10. If the primary winding of the transformer of Figure 27-19 has 80 turns and the secondary winding has 20 turns, the reflected resistance in the primary is (a) 3.13 Ω (b) 200 Ω (c) 800 Ω (d) 850 Ω 11. To perform an open-circuit test in a transformer, we (a) connect a wattmeter and a voltmeter at the primary winding (b) connect a wattmeter and an ammeter at the primary winding (c) connect an ammeter and a voltmeter at the primary winding (d) connect a wattmeter, an ammeter, and a voltmeter at the primary winding

12. To perform a short-circuit test in a transformer, we (a) connect a wattmeter and a voltmeter at the primary winding while leaving the secondary winding shorted (b) connect a wattmeter and a voltmeter at the primary winding while leaving the secondary winding open (c) connect a wattmeter, an ammeter, and a voltmeter at the primary winding while leaving the secondary winding open (d) connect a wattmeter, an ammeter, and a voltmeter at the primary winding while leaving the secondary winding shorted 13. In an autotransformer, the (a)  primary winding is electrically isolated from the secondary ­winding (b) primary winding is not electrically isolated from the secondary winding (c) primary winding is magnetically isolated from the secondary winding (d) primary winding is not magnetically isolated from the secondary winding

853

28

Coupled Circuits

Thévenin and Norton provided methods for simplifying twoterminal networks. These techniques are particularly useful when we need to determine output current, voltage, and power under varying load conditions. When both input and output conditions can vary, we have to deal with four terminals, one pair for the input and the other for the output. This chapter introduces techniques for analyzing such four-terminal networks.

Chapter Outline 28-1 Determining Coupling Network Parameters  856 28-2 Open-Circuit Impedance Parameters  857

28-3 Short-Circuit Admittance Parameters  864 28-4 Hybrid Parameters  867

28-5 Air-Core Transformers  874 28-6 Mutual Inductance  875

28-7 Coupled Impedance  878

Key Terms coupling network  856 port 857 two-port network  857 open-circuit impedance parameters (z-parameters) 857 z-parameter equivalent circuit 859

mutual impedance  860 coupled impedance  861 short-circuit admittance parameters (y-parameters) 864 y-parameter equivalent circuit 866

hybrid parameters (h-parameters) 868 h-parameter equivalent circuit 870 coefficient of coupling  874 bifilar winding  874 mutual inductance  875 mutual reactance  876

Learning Outcomes At the conclusion of this chapter, you will be able to: • write the equations for the open-circuit impedance p ­ arameters of a two-port network • explain the significance of the open-circuit impedance parameters of a two-port network • write the z-, y- and h-parameter equations for a two-port network • calculate the open-circuit impedance parameters of a two-port network • draw the z-, y- and h-parameter equivalent circuits of a two-port network • write the equations for the short-circuit admittance p ­ arameters of a two-port network • explain the significance of the short-circuit admittance parameters of a two-port network • calculate the short-circuit admittance parameters of a two-port network Photo sources:  Courtesy of Superior Electric, Danaher Sensor & Controls Group

• write the equations for the hybrid parameters of a two-port network • explain the significance of the hybrid parameters of a two-port network • calculate the hybrid parameters of a two-port network • illustrate how tight, loose, or minimum coupling ­between windings may be achieved • calculate the mutual inductance between two windings • calculate the coefficient of coupling between two windings • draw the z-parameter equivalent circuit of an air-core transformer • calculate the input impedance of an air-core transformer

856

Chapter 28   Coupled Circuits

28-1 Determining Coupling Network Parameters We have already found several ways of solving the impedance network of Figure 28-1(a). Since this network is not too elaborate, we can use a set of Kirchhoff’s voltage-law (loop) equations or Kirchhoff’s current-law (nodal) equations to solve for I1 and I2. Zp

Zs I1

E1

Zm

I2

E2

(a) Zp

Zs I1

E1

Zm

I2

ZL

(b) I1

1

E1

2

ZL

Coupling network 1

I2

2 (c)

Figure 28-1  Coupling networks: (a) between voltage sources; (b) between a source and a load; (c) of unknown internal composition

Thévenin’s theorem allows us to reduce the network to just two components: a two-terminal load and a two-terminal constant-voltage source. This equivalent circuit is particularly useful for analyzing electric power systems. Thévenin’s theorem gives us a method for determining the parameters of the Thévenin-equivalent source by laboratory measurements, even if we have no idea what the two-terminal “black box” actually contains. For electronic communications and control systems, we need still ­another way of viewing the circuit in Figure 28-1(a). Suppose we replace E2 by a load impedance, as in Figure 28-1(b). The circuit then becomes a T‑network, coupling the signal source E1 to the load ZL. Our experience with such networks tells us that I2 will be a negative quantity if we do not change its direction. However, the usual convention is to treat the input current I1 and the output current I2 of the generalized coupling network of Figure 28-1(c) as having a direction into the network at the upper or “live” terminals of the network.

28-2   Open-Circuit Impedance Parameters

If we knew that the network of Figure 28-1(c) actually had the form of  Figure 28-1(b), we could solve for I2 (and I1 if necessary) by simple ­series-parallel impedance calculations. But if the coupling network is sealed in a four-terminal “black box,” we need to define network parameters that we can measure at the terminals of the network. The simplest g ­ eneralized coupling network has one component or parameter that is common to both the input current I1 and the output current I2, one component through which only I1 flows, and one through which only I2 flows. We can therefore represent the coupling network, whatever its internal composition may be, by a simple T-network. To determine the parameters of an unknown network, we connect the network in the test circuit shown in Figure 28-2, using voltmeters with i­nternal resistances high enough that the voltmeter currents are negligible. To find out what we “see” when looking back into the output terminals, or port, we have to replace the actual load impedance with a voltage source, E2. I1

1

A E1

V1 V

Zp

Zs

2

Zm

1

I2

A V V2

E2

2

Figure 28-2  Determining open-circuit impedance parameters

See Review Questions 28-38 to 28-41 at the end of the chapter.

28-2 Open-Circuit Impedance Parameters To define the composition of a four-terminal, two-port network such as the one shown in Figure 28-1(c), we need four parameters. The test circuit of ­Figure 28-2 gives a set of parameters called the open-circuit impedance parameters (z-parameters) of the network. We start by opening the right-hand switch in Figure 28-2 so that I2 = 0. Leaving this switch open prevents the output circuit from affecting the measurement of input impedance. When we close the left-hand switch, the readings from the voltmeter measuring V1 and the ammeter measuring I1 give us a V/I ratio that represents only the open-circuit input impedance of the network. We now open the left-hand switch and close the right-hand switch. I1 ­becomes zero, making the voltage drop across Zp zero. However, there is a reading on the voltmeter measuring V1 as a part of I2 is coupled back to the

857

858

Chapter 28   Coupled Circuits

input side of the network. The ratio between the voltage, V1, at the input terminals (port 1) and the current, I2, flowing into the output terminals (port 2) is an impedance (in ohms) that indicates the amount of output-toinput (reverse-transfer) coupling in the network. The same technique also produces parameters for the output impedance and input-to-output coupling. Thus, there are four open-circuit impedance parameters: Open-circuit input impedance: z11 =



E1 ( with I2 = 0 ) I1

(28-1)

Open-circuit reverse-transfer impedance:

z12 =

V1 ( with I1 = 0 ) open-circuit  I2

(28-2)

z21 =

V2 ( with I2 = 0 ) open-circuit  I1

(28-3)

Open-circuit forward-transfer impedance:

Open-circuit output impedance:

Lowercase letter ­symbols are used for impedance ­parameters because they are equivalentcircuit parameters, rather than actual ­circuit elements.

z22 =

E2 ( with I1 = 0 )  I2

(28-4)

Having determined the four z-parameters, we now close both switches in Figure 28-2 so that the network actually does couple the output and input circuits. Consequently, both z11 and z12 are now present in the input circuit, and V1 consists of two voltages in series: the IZ drop due to I1 flowing in z11, plus a coupled voltage representing I2 flowing through the ­reverse-transfer impedance z12. Applying a Kirchhoff’s voltage-law equation to the input terminals (port 1) of the network gives

z11I1 + z12I2 = E1

(28-5)

z21I1 + z22I2 = E2

(28-6)

Similarly, for the output terminals (port 2),

The next step is to determine the equivalent circuit represented by these two Kirchhoff’s voltage-law equations. The sum of two voltages at the input terminals represents a series circuit. With the left-hand switch open and the right-hand switch closed in Figure 28-2, we discovered a voltage, V1, across the input port due to the output current I2. Hence, we can argue that we are looking into a Thévenin-equivalent voltage source that is ­dependent on

28-2   Open-Circuit Impedance Parameters

the current in some other part of the circuit. The open-circuit voltage of this current-controlled voltage source is z12I2, and z11 appears to be the internal impedance of the source. Looking into the output terminals, we see a similar Thévenin-equivalent dependent voltage source. Hence, Figure 28-3 is the z-parameter equivalent circuit for the open-circuit ­impedance parameters of a four-terminal coupling network.

1

I1

V1

Coupling network z22

z11 V = z12I2

I2

V = z21I1

1

2

V2

2

Figure 28-3  z-parameter equivalent circuit

We can now find the open-circuit impedance parameters for the T-network shown by dashed lines in Figure 28-2. With the left-hand switch closed and the right-hand switch open,

and

I1 =

E1 Zp + Zm

Zp + Zm =

E1 = z11 I1

(28-7)

At the same time, the voltmeter measuring V2 reads V2 = ZmI1 Hence,

Zm =

V2 = z21 I1

(28-8)

Similarly, with the left-hand switch in Figure 28-2 open and the right-hand switch closed,

and

Zs + Zm = Zm =

E2 = z22 I2

V1 = z12 I2

(28-9)

(28-10)

859

860

Chapter 28   Coupled Circuits

Rearranging Equations 28-7 to 28-10 gives the three T-network components in terms of the z-parameters: Zm = z12 = z21



Zp = z11 − z12





(28-11)



(28-12)

Zs = z22 − z21



(28-13)

If we know the z-parameters for a passive coupling network, we can readily find the equivalent T-network, or vice versa. For a T-network, the forward-transfer and reverse-transfer impedances must be equal. This equality is characteristic of any passive network containing only resistance and reactance elements. Consequently, we call Zm the mutual impedance of the coupling network. In active networks containing transistors, z12 and z21 must differ for amplification to take place. The z-parameter equivalent circuit of Figure 28-3 can represent both passive and active coupling networks.

Example 28-1 Check Equations 28-5 and 28-6 by writing Kirchhoff’s voltage-law equations for the T-network in Figure 28-2. Solution Therefore,

I1Zp + (I1 + I2)Zm = (Zp + Zm)I1 + ZmI2 = E1

( Z11−Z12 + Z12 ) I1 + Z12I2 = Z11I1 + Z12I2 = E1

I2Zs + (I1 + I2)Zm = (Zs + Zm)I2 + ZmI1 = E2 Therefore, ( Z22−Z21 + Z21 ) I2 + Z21I1 = Z21I1 + Z22I2 = E2

(28-5) (28-6)

Example 28-2 Write Equations 28-5 and 28-6 as they apply to the network in Figure 28-4. I1

1

E1 Zin

Zp

Zs

2

Zm

1 Figure 28-4  T-coupling network

I2

ZL

2

28-2   Open-Circuit Impedance Parameters

Solution In this example, ZL becomes the internal impedance of a source producing zero voltage. As Figure 28-4 shows, this internal impedance acts in series with Zs. There is no change in the equation for E1: z11I1 + z12I2 = E1



(28-5)

But the equation for the output voltage is

Therefore,

(ZL + Zs + Zm)I2 + ZmI1 = 0

z21I1 + ( z22 + ZL ) I2 = 0

(1)

The equations in Example 28-2 provide a means of developing a s­ impler e­ xpression for the total input impedance of the T-coupling ­network in ­Figure 28-4 (and of any generalized coupling network) than an equation for Zin in terms of series and parallel impedances. From Equation 1, I2 =



Substituting in Equation 28-5 gives z11I1 −



Hence,

Zin =

But from Equation 28-11, Therefore,

−z21I1 z22 + ZL

(28-14)

z12z21I1 = E1 z22 + ZL

E1 z12z21 = z11 − I1 z22 + ZL z12z21 = Z2m

Zin = z11 −

Z2m z22 + ZL

(28-15)

The last term in Equation 28-15 is the coupled impedance, an important parameter in coupled circuits:

coupled impedance =

Z2m z22 + ZL

(28-16)

861

862

Chapter 28   Coupled Circuits

Equation 28-15 represents an equivalent circuit consisting of two i­mpedances connected in series. The negative sign with the coupled impedance means that the coupled impedance reduces the input impedance of the n ­ etwork and increases the energy transferred to the load. Thus, the ­coupled-circuit technique gives us a series coupled impedance, in contrast to parallel r­ eflected impedance that we developed with the transformationratio t­ echnique for dealing with iron-core transformers. However, as far as the source is concerned, both techniques produce the same results. In Chapter 27, we saw that disconnecting the load made the reflected impedance in parallel with the primary impedance infinitely large so that Zin was simply the open-circuit input impedance of the transformer. With z-parameters, d ­ isconnecting the load makes ZL in Equation 28-15 infinite. Then the coupled impedance becomes zero, and Zin is again simply the open-circuit input impedance z11.

Example 28-3 Find the input impedance of the H-attenuator in Figure 28-5 when a 600-Ω resistive load is connected to the output terminals. 1

100 Ω

100 Ω

2

600 Ω 800 Ω

600 Ω

RL

E 100 Ω

100 Ω

1

2

Figure 28-5 H-attenuator

Solution The H-attenuator may be regarded as a balanced form of a T-network. The equivalent T-network has Zp = Zs = 200 Ω. Since the network is purely ­resistive, all the impedances have ϕ = 0°. Therefore,

We can use series-­ parallel resistance ­calculations to check this answer.

z11 = z22 = 200 Ω + 800 Ω = 1000 Ω Zm = 800 Ω

Zin = z11 −

Z2m 640 000 Ω = 1000 Ω − = 600 Ω z22 + ZL 1000 Ω + 600 Ω

28-2   Open-Circuit Impedance Parameters

863

Example 28-4 Calculate the power in a 75-Ω load connected to the circuit in Figure 28-6. 2.0 μF

E = 10 V f = 1.0 kHz

2.0 μF

6.0 mH

75 Ω

RL

Figure 28-6  Circuit for Example 28-4

Solution

z11 = z22 = −j = −j



1 + jωL ωC

1 + j ( 2 × π × 1.0 kHz × 6.0 mH ) 2 × π × 1.0 kHz × 2.0 µF

= −j79.58 + j37.70 Ω = −j41.88 Ω

Since j2 = −1,

Z2m = (+ j37.7 ) 2 = −1421

Substituting in Equation 28-16 gives   Zin = z11 −

= −j41.88 +

and Since

Z2m  z22 + ZL

Zin = 36.77 Ω

(28-15)

1421 75 − j41.88

= 14.44 − j33.82 Ω

Iin =

E 10 V = = 272 mA Zin 36.77 Ω −z21I1 I2 =  z22 + ZL 37.7 Ω I2 = × 272 mA = 119.4 mA 85.9 Ω

(28-14)

Since the load contains only resistance,

PL = I22RL = ( 119.4 mA ) 2 × 75 Ω = 1.1 W

See Problems 28-1 to 28-20 and Review Questions 28-42 to 28-52.

We do not need to ­calculate the angle of Zin to determine the magnitude of the load current.

864

Chapter 28   Coupled Circuits

Circuit Check

A

CC 28-1. Determine the equivalent T-circuit for the linear transformer shown in Figure 28-7. 30 mH A

C I1

I2

40 mH B



80 mH D

Figure 28-7

CC 28-2. Find the open-circuit z-parameters for the circuit shown in Figure 28-8. R1

R2

100 Ω

220 Ω R3 330 Ω



Figure 28-8

CC 28-3. For the T-network shown in Figure 28-9, calculate (a) the z-parameters (b) the voltage across the load resistance 5.0 mH

20 V 2 kHz



5.0 mH

1.2 μF

100 Ω

RL

Figure 28-9

28-3 Short-Circuit Admittance Parameters

We can represent the generalized coupling network by the π-network shown with dotted lines in Figure 28-10. If we apply Equations 28-1 to 28-4 carefully, we can derive equations relating the three π-network ­impedances to z-parameters for the network. However, it is simpler to work with admittances when we encounter a coupling network in the form of a π‑network, which is a dual for a T-network. Although the ­resulting shortcircuit admittance parameters (y-parameters) are not often used in circuit analysis, we derive these parameters of a ­generalized f­ our-terminal coupling network as an introduction to the hybrid p ­ arameters of Section 28-4.

28-3   Short-Circuit Admittance Parameters

I1

E1

1

A V1 V

2

Ym Yp

Ys

1

A

I2

V V2

E2

2

Figure 28-10  Determining the short-circuit admittance parameters

To prevent source E2 from affecting the measurement of the input admittance, we set the right-hand switch in Figure 28-10 so that the ammeter ­measuring I2 shorts the output terminals (port 2). Similarly, when we apply E2 to the output terminals, we set the left-hand switch to connect the ammeter measuring I1 directly across the input terminals (port 1), thus shorting them. With E1 applied to the input terminals, we can use the readings of the meters in the input circuit to find an I/V ratio that represents the input ­admittance ­of the network with the output terminals short-circuited. The reading on the ammeter measuring I2 indicates that a portion of the input current is coupled through to the output. Looking back into the short-­circuited output terminals, we see a Norton-equivalent source containing a ­dependent current source controlled by the voltage applied to the input terminals. To find the parameter that relates I2 to E1, we simply divide I2 by V1, giving a quantity in siemens. By reversing the position of the switches, we can find similar parameters for the output terminals. Thus, there are four short-circuit admittance parameters for any four-terminal, two-port network: Short-circuit input admittance:

y11 =

I1 E1

( with E2 = 0 )

(28-17)

Short-circuit reverse-transfer admittance:

y12 =

I1 E2

( with E1 = 0 )

(28-18)

Short-circuit forward-transfer admittance:

y21 =

I2 E1

Short-circuit output admittance:

y22 =

I2 E2

( with E2 = 0 )

(28-19)

( with E1 = 0 )

(28-20)

Since a Norton-equivalent current source has to be parallel in with the input admittance, the equivalent circuit has two paths for I1 to follow

865

866

Chapter 28   Coupled Circuits

­ etween the input terminals. Writing the Kirchhoff’s current-law equation b for I1 gives y11E1 + y12E2 = I1



(28-21)

Similarly, for the output terminals (port 2),

y21E1 + y22E2 = I2



(28-22)



Figure 28-11 shows the y-parameter equivalent circuit that satisfies Equations 28-21 and 28-22. Coupling network

I1

V1

y11

I = y12V2

I2

I = y21V1

y22

V2

Figure 28-11  y-parameter equivalent circuit

We can now find the y-parameters for the π-network shown with dashed lines in Figure 28-10. When the ammeter measuring I2 shorts the output terminals,   I1 = E1(Yp + Ym) Yp + Ym =

I1 = y11 E1

(28-23)

Similarly, Ys + Ym =

I2 = y22 E2

(28-24)

and

When the meter measuring I1 is connected directly across the input terminals, there can be no voltage drop across Yp and, hence, no current through Yp. Therefore, all the current through Ym must flow through the ammeter in the input circuit.As a result, I1 has the same magnitude as the current through Ym. But the current through Ym is caused by E2 and has the opposite direction to I1, as shown in Figure 28-10. Hence, and

I1 = −IYm = −YmE2

  Ym =

−I1 = −y12 E2

(28-25)

28-4   Hybrid Parameters

Ym =

Similarly,

−I2 = −y21 E1

(28-26)

Rearranging Equations 28-23 to 28-26 gives the equivalent π-network in terms of the y-parameters of the coupling network: Ym = −y12 = −y21



Yp = y11 + y12



Ys = y22 + y21



(28-27)





(28-28)



(28-29)

Example 28-5 Determine the short-circuit admittance parameters for the circuit in Figure 28-12. 1.4 mH E = 10 V f = 1.0 kHz

38 μF

38 μF

75 Ω

RL

Figure 28-12  Circuit for Example 28-5

Solution

Yp = Ys = +jωC = +j(2 × π × 1.0 kHz × 38 μF) = +j238.8 mS

   Ym = −j

1 1 = −j = −j113.7 mS ωL 2 × π × 1.0 kHz × 1.4 mH

The y-parameters are

    y11 = y22 = Yp + Ym = +j238.8 mS − j113.7 mS = +j0.13 S      y12 = y21 = −Ym = −(−j113.7 mS) = +j0.11 S

See Problems 28-21 to 28-26 and Review Questions 28-53 and 28-54.

28-4  Hybrid Parameters For analyzing circuits containing active devices such as transistors, it is more convenient to think of the input terminals of a four-terminal coupling network as a Thévenin-equivalent voltage source and the output terminals

867

868

Chapter 28   Coupled Circuits

as a Norton-equivalent current source. We then describe the coupling network in terms of four hybrid parameters (h-parameters). We determine these parameters using the same measurement techniques as for z-parameters and y‑parameters. To find the open-circuit voltage of the Thévenin-equivalent source at input terminals (port 1) in Figure 28-13(a), we feed V2 into the output terminals (port 2). In this circuit, we consider the Thévenin-equivalent source to be a voltage-controlled voltage source. The parameter that represents the fraction of the output voltage appearing at the input terminals is V1/V2, which is a ratio without units. This parameter is the open-circuit reversevoltage ratio, h12. Since we are treating the dependent source as a voltage-controlled voltage source, we short-circuit the output terminals while we measure the input voltage and current, as shown in Figure 28-13(b). The parameter h11 is V1/I1, which is expressed in ohms and represents the short-circuit input impedance of the network. Since h12V2 is a voltage source, the equivalent input circuit for the coupling network shows the dependent voltage source and input impedance in series, as in Figure 28-13(c). To determine the short-circuit current of the Norton-equivalent source at the output terminals (port 2) in Figure 28-14(a), we feed I1into the input 1 V1 V

I2

2 V2

Coupling network 1

2 (a)

I1 E1

1

A V V1

2

I2

Coupling network 1

2 (b)

1

h11

2 h12V2

1

2 (c)

Figure 28-13  Finding the Thévenin-equivalent input circuit of a four-terminal network: (a) Open-circuit reverse voltage; (b) Internal input impedance; (c) Network input parameters

28-4   Hybrid Parameters

I1

1

A

2

I2 A

Coupling network

E1 1

2 (a)

1

2 V2 V

Coupling network 1

A

I2 E2

2 (b)

1

h11

2

h12V2

h21I1

1

h22 2

(c) Figure 28-14  Finding the Norton-equivalent output circuit of a four-terminal network: (a) short-circuit forward current; (b) output admittance; (c) complete hybrid parameters

terminals and short-circuit the output terminals through the ammeter measuring I2. As long as the network impedances are linear (independent of voltage and current), I2 will be a constant fraction of the input current I1. The ratio I2/I1 is the short-circuit forward-current ratio, h21. The output impedance of a Norton-equivalent source is in parallel with the current source, so the fourth hybrid parameter is expressed as an admittance. Since we are treating this dependent source as a current-controlled current source, we leave the input terminals of the network open-circuit to make I1 zero while we measure I2 and V2. The parameter h22 is I2 /V2 , which is expressed in siemens and represents the open-circuit output a­ dmittance. These equations summarize the four hybrid parameters of a four-terminal coupling network: Short-circuit input impedance: h11 =



( with V2 = 0 )

(28-30)

( with I1 = 0 ) open-circuit

(28-31)

V1 I1

Open-circuit reverse-voltage ratio:

h12 =

V1 V2

869

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Chapter 28   Coupled Circuits

Short-circuit forward-current ratio: h21 =



I2 I1

( with V2 = 0 ) short-circuit

(28-32)

Open-circuit output admittance: h22 =



I2 V2

( with I1 = 0 )

(28-33)

Figure 28-14(c) shows the resulting h-parameter equivalent circuit. For the Thévenin-equivalent source for the network input, we can write a Kirchhoff’s voltage-law equation, as we did for z-parameters. For the Norton-equivalent source for the network output, we write a Kirchhoff’s current-law equation, as we did for y-parameters. The two unknowns in these equations are I1 and V2. h11I1 + h12V2 = E1



(28-34)

h21I1 + h22V2 = I2 (28-35)



The transistor amplifier equivalent circuit of Figure 28-15 is a typical ­example of hybrid parameters.

I1

1

h11

h12V2

E1

2

+ −

1

h21I1

h22

I2

RL

2

Figure 28-15  Hybrid parameters of a simple transistor amplifier

Example 28-6 Calculate the voltage gain when a 2.2-kΩ resistance load is connected across the output terminals of the transistor amplifier in Figure 28-15, given h11 = 50 Ω

h12 = 3.0 × 10−4

h21 = −0.95

h22 = 2.0 × 10 −5 S

28-4   Hybrid Parameters

Solution We define voltage gain as V2/E1. One approach is to derive a formula for voltage gain by rearranging Equations 28-34 and 28-35 and then substituting for the parameters. Here, we substitute the known values first and then solve for V2. Since there are no reactance elements in the equivalent circuit, we can solve Equations 28-34 and 28-35 using ordinary algebra. There are four variables in Equations 28-34 and 28-35. With only two equations, we must make two of the variables independent. We can pick some convenient value for E1, say 1.0 V RMS. From Ohm’s law, I2 =

−V2 RL

The negative sign results from the I2 direction for Equations 28-34 and 28-35 being established by a source in place of RL. Since h21 = I2/I1, it is also a negative quantity, as indicated in the given data. Substituting the known quantities into Equations 28-34 and 28-35 gives 50I1 + 3.0 × 10 − 4V2 = 1.0



−0.95I1 + 2.0 × 10−5V2 =



1.0 − 3.0 × 10−4V2 50 Substituting in Equation 2 gives From Equation 1,

and

−0.95

(

−V2  2.2 kΩ

(1) (2)

I1 =

1.0 − 3.0 × 10 − 4V2 −V2 + 2.0 × 10−5V2 = 50 2.2 kΩ V2 = 39.9 V

)

Therefore the voltage gain of the amplifier is V2 = 40 E1

Section 10-3 showed that we cannot use Thévenin’s theorem to find the equivalent internal resistance of a dependent source when the controlling ­element is included in the transformation. Therefore, when we want to ­determine input and output impedances of coupling networks, we must

871

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Chapter 28   Coupled Circuits

c­ alculate V/I. We determined h11 with a short circuit across the output terminals of the network. In the circuit of Figure 28-15, Zin differs slightly from h11 since the circuit has some reverse coupling.

Example 28-7 The source E1 in the equivalent circuit of Figure 28-15 has an internal resistance of 50 Ω. Determine the input and output resistances for the transistor amplifier circuit of Example 28-6. Solution For this example, we derive formulas and then substitute the given param­eters. In Example 28-6, we noted that

−V2 RL Substituting this expression for I2 in Equation 28-35 and collecting terms gives −h21ZLI1 V2 = 1 + h22ZL I2 =

We use V1 rather than E1 because V1 is not the same as E1 if the source has any internal resistance. The source resistance is not ­included in the input resistance of the ­network.

Substituting for V2 in Equation 28-34 and collecting terms gives and

V1 = h11I1 − Zin =

h12h21ZLI1 1 + h22ZL

V1 h12h21ZL  = h11 − I1 1 + h22ZL

(1)

Substituting the known parameters into Equation 1 gives



Rint = Zin = 50 −

−0.95 × 3 × 10−4 × 2.2 × 103 1 + 2 × 10−5 × 2.2 × 103

= 50 + 0.601  = 50.6 Ω

To obtain h22, we left the input terminals in Figure 28-14(b) opencircuit. To find the output impedance, we replace the voltage source E1 with its internal resistance Rint connected across the input terminals. The Thévenin-equivalent source in Figure 28-14(c) creates a current through h11 and Rint in the opposite direction to I1. Therefore, I1 =

−h12V2 h11 + Rint

28-4   Hybrid Parameters

Substituting for I1 in Equation 28-35, we obtain I2 =



−h12h21V2 + h22V2 h11 + Rint

= V2



h22 ( h11 + Rint ) − h12h21 h11 + Rint

V2 h11 + Rint = I2 h22 ( h11 + Rint ) − h12h21

from which

Dividing both numerator and denominator by h11 + Rint gives Zout =



h22

1  h12h21 − h11 + Rint

(2)

Substituting the known quantities into Equation 2, we obtain Zout =

2.0 × 10−5

1 = 44 kΩ −0.95 × 3.0 × 10−4 − 50 + 50

See Problems 28-27 to 28-30 and Review Questions 28-54 to 28-56.

Circuit Check

B

CC 28-4. In the network shown in Figure 28-16, Z1 = Z2 = +j2 kΩ and Z3 = −j4.7 kΩ. Find the input impedance Zin when ZL is a) 300 Ω b) 60 Ω ∠−50º Z3

Z1



Figure 28-16

Z2

ZL

CC 28-5. For the π-network shown in Figure 28-17, calculate a) the y-parameters b) Zin

873

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Chapter 28   Coupled Circuits

20 μF

20 V 2 kHz

0.8 mH

0.8 mH

100 Ω

RL

Figure 28-17 CC 28-6. Find the h-parameters for the circuit of Figure 28-17.

28-5  Air-Core Transformers We can apply coupled-circuit analysis to deal with magnetically coupled circuits used in radio circuits. The equivalent circuit for iron-core transformers that is based on the transformation ratio is valid only if the leakage flux is very small compared to the mutual flux. However, the mutual flux in loosely coupled air-core transformers is only a fraction of the total primary flux. This fraction is called the coefficient of coupling, k. By definition, k=



Φm Φp

(28-36)

The value of k can range from 1 when all the primary flux links the secondary, as in the ideal transformer of Figure 27-1, down to zero when the secondary winding is completely outside the magnetic field of the primary or is positioned at right angles to the primary flux, as in Figure 28-18(c). The physical arrangement of Figure 28-18(c) is used in radio circuits where magnetic coupling between tuned circuits is to be kept to a minimum and the physical dimensions are rather restricted. In order to obtain tight coupling with k close to 1 in radio-frequency coils where iron cores are not practical, the two strands of insulated wire are interwound on the coil to form a bifilar winding (Figure 28-18(a)). Magnetic coupling tuned of coils requires a very loose coupling with a coefficient of coupling of approximately 0.01, as with the two coils in Figure 28-18(b).

(a)

(b)

(c)

Figure 28-18 Variations in magnetic coupling: (a) tight coupling; (b) loose coupling; (c) minimum coupling

28-6   Mutual Inductance

(a)

(b) Radio-frequency transformers with ferrite cores: (a) The 2:1 turns ratio in this TV adapter matches 300-Ω twin-lead to a 75-Ω coaxial input. (b) The two windings on this AM antenna form a transformer that couples the tuned input circuit to the next stage of the table radio; the primary winding also serves as the inductance in the tuned circuit.

See Review Questions 28-58 to 28-60.

28-6  Mutual Inductance To develop an equivalent coupling network for an air-core transformer, we use a parameter that is common to both the primary and secondary circuits: the mutual inductance of the pair of coils. A pair of magnetically coupled coils has a mutual inductance of one henry when current changing at the rate of one ampere per second in one coil induces an average voltage of one volt in the other coil. The letter symbol for mutual inductance is M.

875

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Chapter 28   Coupled Circuits

Therefore, by definition, Es(av) =



MIp(m) t

(28-37)

where Es(av) is the average voltage induced in the secondary and Ip(m) is the maximum primary current. Since the instantaneous current must rise from zero to maximum in one quarter of a cycle of a sine wave, Es(av) =

But Thus, and

MIp(m) 1/4f

Eav =

= 4 fMIp(m)

2 Em π

2 E ( ) = 4 fMIp(m) π sm

Es = 2π fMIp

(18-17)

(28-38)

Rearranging Equation 28-38 gives Es = 2π f M = XM Ip



(28-39)

where XM is the mutual reactance of the magnetically coupled windings.

Example 28-8 A transformer with a primary inductance of 5.0 H is connected to a 120-V 60-Hz source. The open-circuit secondary voltage is 40 V. Neglecting losses, determine the mutual inductance between the two windings. Solution Es = ωMIp Ip =

But Therefore, and

M = Lp

Es =

Ep

ωLp M Ep Lp

Es 40 = 5.0 × = 1.7 H Ep 120

28-6   Mutual Inductance

Combining Equations 27-1 and 28-37

MIp(m) NΦm = t t



Φm = kΦp

But from Equation 28-36 Therefore,

MIp(m) = NskΦp

(1)

MIs(m) = NpkΦs

(2)

If we use the original secondary winding as the primary, and vice versa, it follows that

Solving Equations 1 and 2 for M and multiplying the resulting expressions gives M2 = k2



( )( ) NpΦp Ip(m)

NsΦs  Is(m)

(3)

From the definition of self-inductance and the definition of the weber, Eav =

LIm NΦ     and    Eav = t t

Therefore, Substituting in Equation 3 gives

L=

NΦ Im

M = k√ LpLs 



(28-40)

Example 28-9 If the secondary winding of the transformer of Example 28-8 has a self-­ inductance of 0.80 H, what is the coefficient of coupling between the ­windings? Solution

k=

√ LpLs M

=

√ 5.0 × 0.80 1.67

= 0.84

A method for determining mutual inductance experimentally is shown in Figure 28-19. We can measure the total inductance of the series-­ connected coils on an inductance bridge. First we connect the windings in series, using the right-hand rule to make sure current in both windings produces the magnetic flux that flows in the same ­direction around the

877

878

Chapter 28   Coupled Circuits

ϕp

ϕs

Figure 28-19  Determining mutual inductance

core. Then the four induced voltages are all in phase. These voltages are the self-induced voltage in the primary, the voltage mutually induced in the ­primary by current in the secondary, the voltage mutually induced in the secondary by current in the primary, and the self-induced voltage in the secondary. Therefore, and

E = I ( ωLp + ωM + ωM + ωLs )

LT = Lp + Ls + 2M

If we reverse the leads to the secondary, the mutually induced voltages in each coil are 180° out of phase with the self-induced voltages, and LT = Lp + Ls − 2M

Thus, the mutual inductance of the magnetically coupled coils in Figure 28-19 is one quarter of the difference between the total inductance with the coils connected series-aiding and with them connected series-opposing. We can extend Equation 16-6 for the total inductance of two ­inductances in series to include the magnetic coupling between them:

LT = Lp + Ls ± 2M

(28-41)

See Problems 28-31 to 28-34 and Review Questions 28-61 and 28-62.

28-7  Coupled Impedance A practical advantage that transformer coupling has over other forms of coupling networks is that for a given direction of E1 in Figure 28-20, we can reverse the phase of the output voltage either by reversing the direction of one of the windings or by simply reversing the leads to one of the windings. Where it is necessary to keep track of this phase relationship in a circuit diagram, we mark one end of each winding with a dot, as shown in Figure 28-20(a) and (b). These dots indicate that a changing mutual flux that induces an instantaneous voltage into the primary winding with a p ­ olarity such that the dotted end of the winding is positive with

28-7   Coupled Impedance

I1

I2

I1

E1

E1

ZL

ZL I2

(a) Ip

Ep

Rp

(b) Lp

Is(jωM)

Ls

+ −

+ −

Rs

Ip(jωM)

Is

ZL

(c)

Figure 28-20  Equivalent circuit of an air-core transformer

respect to the unmarked end also induces an instantaneous voltage into the secondary winding with a polarity such that the dotted end of the secondary winding is positive with respect to its unmarked end. Although reversing the secondary polarity reverses the current through the external circuit, I2 in Figure 28-20(b) still must flow into the dotted end of the secondary winding. Hence, the primary circuit is not affected by ­reversing the secondary leads. In selecting the direction for these current arrows, we must remember that the same mutual flux (which is produced by these currents) induces voltages into both primary and secondary windings. So if we draw I1 pointing into the dot end of the primary winding, we must also draw I2 pointing into the dot end of the secondary winding. When we write the Kirchhoff’s voltage-law equation for the primary loop, E1 equals the phasor sum of three voltages: an IR drop across the ­resistance of the primary circuit, a self-induced voltage due to I1 flowing through the primary winding, and a mutually induced voltage due to I2 flowing in the secondary circuit. Figure 28-20(c) shows one form of equivalent circuit for this transformer, in which the self-induced voltage is represented by an inductance and the mutually induced voltage by a current-controlled voltage source (CCVS). To satisfy Lenz’s law, the polarity of the CCVS must be such that it opposes changes in I1 due to the applied voltage E1. Hence, changing the polarity of the CCVS in the secondary circuit does not change the polarity of the voltage +jωMIs in the primary circuit. Consequently, ­Kirchhoff’s voltage-law equation for the primary loop is Ip ( Rp + jωLp ) + Is ( +jωM ) = Ep

or

IpZp + Is ( +jωM ) = Ep

(28-42)

where Zp is the open-circuit impedance of the primary circuit by itself, and ωM is the mutual reactance of the two windings.

879

880

Chapter 28   Coupled Circuits

Note that Lenz’s law gives us the same direction for I2 that we chose for  our generalized coupling network. (Compare Equation 28-42 with Equation 28-5.) So we write the loop equation for the secondary circuit of Figure 28-20(c) as Is ( ZL + Rs + jωLs ) + Ip ( +jωM ) = 0



Compare Equation 28-43 and Equation 28-14.

or

Is ( ZL + Zs ) + Ip ( +jωM ) = 0

(28-43)

where Zs is the open-circuit impedance of the secondary winding by itself. From Equation 28-43, Is =



−Ip ( +jωM ) Zs + ZL

(28-44)

Substituting in Equation 28-42 gives Since j2 = −1,

IpZp − Ip

( +jωM ) 2 = Ep Zs + ZL

(28-45)

( ωM ) 2 Zs + ZL

(28-46)

Ep = IpZp + Ip

Dividing by Ip gives Equation 28-47 is the equivalent of ­Equation 28-16.



Zin = Zp +

( ωM ) 2 Zs + ZL

(28-47)

where (ωM)2/(Zs + ZL) is the coupled impedance for transformer coupling. Note that ωM is an impedance with a 0º angle. We can check Equation 28-47 by considering the effect of loading the secondary winding of a pair of magnetically coupled coils. If the secondary is left open-circuit, ZL is infinitely large and, therefore, the coupled impedance is zero. Then, the total impedance simply equals the primary impedance. If we connect a resistance across the secondary winding, ZL has a 0° angle. Since Zs is largely the inductive reactance of the secondary winding, the total secondary-circuit impedance, Zs + ZL, is inductive. Consequently, the coupled impedance is capacitive since dividing (ωM)2 with its 0° angle by an impedance with positive angle gives an impedance with a negative angle. Because Zp is largely inductive, the capacitive impedance coupled into

28-7   Coupled Impedance

the primary circuit via the mutual inductance adds the coupled resistance component in series with the primary resistance component but r­ educes ­the total primary reactance. Thus, the total impedance decreases as the secondary is loaded with a resistance. The primary current increases correspondingly, transferring more energy to the secondary circuit. This ­result agrees with the analysis in Section 27-2, which showed that connecting a load to the secondary of a transformer causes additional current to flow in the primary.

Example 28-10 Assuming that the resistance of the windings is negligible, determine the total input impedance of the transformer in Examples 28-8 and 28-9 with (a) the secondary open-circuit (b) a 50-Ω resistance connected to the secondary Solution (a) (b)

Zin = Zp = ωLp = 377 × 5.0 = j1885 Ω = j1.9 kΩ Zin = Zp +

( ωM ) 2 Zs + ZL

= +j1885 + = +j1885 +

( 377 × 1.667 ) 2 +j ( 377 × 0.8 ) + 50 394 961 ∠0º 305.7 ∠80.59º

= +j1885 Ω + 1292 Ω ∠−80.59º

= (+j1885) Ω + (211.2 − j1275) Ω = 211.2 + j610 Ω = 0.65 kΩ ∠71°

See Problems 28-35 to 28-37 and Review Questions 28-63 to 28-65.

881

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Chapter 28   Coupled Circuits

Summary

• A two-port network may be represented by z-, y- or h-parameter equivalent circuits. • The open-circuit impedance parameters of a two-port network can be ­determined by representing the network by its equivalent T-network. • The short-circuit admittance parameters of a two-port network can be determined by representing the network by its equivalent π-network. • The coefficient of coupling is a measure of the magnetic coupling ­between two windings. • Mutual inductance between two coils relates the voltage induced in one coil to the changing current in the other. • An air-core transformer may be represented by a z-parameter equivalent circuit. • The input impedance of an air-core transformer is dependent on the coupled impedance. B = beginner

Problems

I = intermediate

Section 28-2  Open-Circuit Impedance Parameters

A = advanced

circuitSIM walkthrough

circuitSIM walkthrough

B B B I

B B B B B B B

In Problems 28-1 to 28-4, use the given data to calculate the open-circuit z-parameters for the T-network of Figure 28-21. 28-1. Zp = Rp = 330 Ω, Zs = Rs = 470 Ω, and Zm = Rm = 610 Ω. 28-2. Zp = Rp = 2.7 kΩ, Zs = +j3 kΩ, and Rm = 22 kΩ. 28-3. Zp = Zs = 20 − j300 Ω, and Zm = +j300 Ω. 28-4. Zp = Zs, which is a 50-μH inductance in series with a 200-pF capacitance; Zm is a 200-μH inductance, and f = 1.0 MHz. In Problems 28-5 to 28-8, find the equivalent T-network for a coupling network with the given z-parameters. 28-5. z11 = 50 Ω ∠0º , z12 = z21 = 10 Ω ∠0º , and z22 = 75 Ω ∠0º. 28-6. (a) z11  = 500 kΩ − j5.0 kΩ, and z12 = z21 = z22 = 500 kΩ + j0 Ω. (b) Download Multisim file 28-6A from the website and follow the instructions to verify the equivalent T-network. 28-7. z11 = z22 = 50 + j0 Ω, and z12 = z21 = −j1200 Ω. 28-8. z11 = 100 + j1500 Ω, z12 = z21 = 40 + j500 Ω, and z22 = 80 + j1200 Ω. 28-9.

Find the coupled impedance when a 25-Ω resistive load is connected to the output terminals of the network in Problem 28-5. 28-10. Find the coupled impedance when a capacitive reactance of 1240 Ω is connected to the output terminals of the network in Problem 28-8. 28-11. Determine Zin when a 680-Ω resistive load is connected to the output ­terminals of the T-network in Problem 28-1.

Problems

B I A

28-12. How much power will be delivered to the resistive load of Problem 28-11? 28-13. For the T-attenuator circuit of Problem 28-11, determine the ratio ­between the total power input to port 1 and the power into the load connected to port 2. 28-14. With the techniques used in deriving Equations 28-7 to 28-10, find the z-parameters for the transistor in Figure 24-35. (Note that the internal resistance of an ideal CCVS is zero.) In Problems 28-15 to 28-18, use the given data to calculate the open-circuit z-­parameters for the π-network of Figure 28-21. I1

1

2

Z3

I2

Rg Z1

E 1

Z2

ZL

2

Figure 28-21  π-coupling network

I I I I B I

28-15. Z1 = R1 = 3470 Ω, Z2 = R2 = 350 Ω, and Z3 = R3 = 440 Ω. 28-16. Z1 = 80 kΩ ∠80º , Z2 = 470 kΩ ∠0º, Z3 is a 20-nF capacitor, and f = 100 Hz. 28-17. Z1 = Z2 = −j3600 Ω, and Z3 = +j3600 Ω. 28-18. Z1 = Z2 = 150 μH, Z3 = 600 Ω, and f = 500 kHz.

28-19. Find Zin when a 300-Ω resistive load is connected to the output terminals of the π-network in Problem 28-15. 28-20. Find Zin when a load of 300 − j400 Ω is connected to the output terminals of the π-network in Problem 28-18.

Section 28-3  Short-Circuit Admittance Parameters

B B

B B I I

In Problems 28-21 and 28-22, use the given data to find the short-circuit y-­parameters for the π-network of Figure 28-12. 28-21. Yp = Gp = 3 mS, Ys = Gs = 9 mS, and Ym = Gm = 2 mS. 28-22. Yp = Ys = +j400 μS, and Ym = −j400 μS. In Problems 28-23 and 28-24, find the equivalent π-network for a coupling network with the given y-parameters. 28-23. y11 = 700 µS∠0º , y12 = y21 = 200 µS∠180º , and y22 = 500 mS∠180º. 28-24. y11 = y22 = 10 mS ∠−45º , and y12 = y21 = 10 mS ∠−135º. 28-25. Find the y-parameters for the network of Problem 28-18. 28-26. Find the z-parameters for the network of Problem 28-23.

883

884

Chapter 28   Coupled Circuits

Section 28-4  Hybrid Parameters

circuitSIM walkthrough

I A A

A

circuitSIM walkthrough

I

I

B I

I

A

28-27. Find the h-parameters for the network of Problem 28-1. 28-28. (a) Find the hybrid parameters for the transistor in Figure 24-35. (b)  Download Multisim file 28-28A from the website and follow the instructions to verify the hybrid parameters for the transistor. 28-29. For a common-base configuration, a certain transistor has the following parameters: h11 = 40 Ω   h12 = 4.0 × 10 −4   h21 = −0.96  h22 = 1.0 × 10 −5 S The transistor feeds a 4.7-kΩ resistive load and is fed from a 1.0-kHz signal source having an open-circuit voltage of 10 mV and an internal resistance of 250 Ω. Calculate the voltage gain (the ratio of output voltage to input voltage). 28-30. (a) For a common-emitter configuration, the transistor used in Problem 28-29 has the following parameters: h11 = 1000 Ω  h12 = 9.6 × 10−3  h21 = 24  h22 = 2.5 × 10−4 S Calculate the voltage gain for the same signal source and load as in Problem 28-29. (b) Use Multisim to verify the voltage gain.

Section 28-6  Mutual Inductance 28-31. An air-core transformer with adjustable coupling has a primary self-­ inductance of 500 mH and a secondary self-inductance of 250 mH. If a 100-μA current with f = 60 Hz is flowing in the primary, what ­coefficient of coupling is required to induce a secondary voltage of 55 mV? 28-32. Two coils wound on a common core have self-inductances of 1.6 H and 3.2 H respectively. When connected in series, their total inductance is 8 H. (a) Find the coefficient of coupling. (b) Find the total inductance when the leads of one of the coils are reversed. 28-33. Two coils wound on the same core have self-inductances of 5.3 H and 2.0 H. When the two coils are connected in series, the total inductance is 4.0 H. Determine the coefficient of coupling. 28-34. A transformer has its windings connected series aiding with a total inductance of 8.2 H. When the windings are connected series opposing, the total inductance is 5.8 H. Determine the mutual inductance.

Section 28-7  Coupled Impedance 28-35. An air-core transformer has a primary inductance of 210 mH, a secondary inductance of 100 mH, and a coefficient of coupling of 0.20. If the resistance of the windings can be ignored, calculate the coupled impedance at 5 kHz when a 5-kΩ resistor is connected across the secondary. 28-36. If the voltage applied to the primary winding of the transformer in Problem 28-35 is 20 mV RMS, find the voltage across the 10-kΩ resistor in the secondary circuit.

Review Questions

I

28-37. A 120-V/240-V 60-Hz transformer has a primary inductance of 4.5 H and a coefficient of coupling of 0.85. Determine the total input impedance when the secondary is connected to a load consisting of a 10-μF capacitor in series with a 1.5-kΩ resistor.

Review Questions Section 28-1  Determining Coupling Network Parameters 28-38. Why is I2 in Figure 28-1(b) a negative quantity? 28-39. What is the “black-box” technique of determining the equivalent ­circuit for a network? 28-40. Why does the equivalent circuit of a generalized coupling network require at least three components? 28-41. Give a practical example (with a circuit diagram) of a coupling network for which one of the components in its equivalent circuit has an impedance of 0 Ω.

Section 28-2  Open-Circuit Impedance Parameters

28-42. Why are the z12 and z21 parameters called transfer impedances? 28-43. With reference to Equations 28-5 and 28-6, discuss the merits of the directions chosen for I1 and I2. 28-44. What is the mutual impedance of a coupling network? 28-45. Under what circumstances does a coupling network have no mutual impedance? 28-46. Figure 24-35 shows the r-parameters for a transistor equivalent ­circuit. What is the reason for the dependent voltage source in this circuit? 28-47. Account for the negative sign in Equation 28-14. 28-48. Account for the negative sign in Equation 28-15. 28-49. At a given frequency, Zin in Figure 28-4 is the same if Zm is a pure ­capacitive reactance of X ohms or a pure inductive reactance of X ohms. Explain why. 28-50. Why is the H-attenuator network of Figure 28-5 called a “balanced” network, while the equivalent T-network is not? 28-51. Determine the open-circuit z-parameters for the π-network of Figure 28-21. 28-52. Use the z-parameters derived in Question 28-48 to find the equivalent T-network for Figure 28-21.

Section 28-3  Short-Circuit Admittance Parameters 28-53. Under what circumstances must we use the short-circuit parameters of a coupling network, rather than the open-circuit ­parameters? 28-54. Derive an equation for Yin when a load admittance replaces E2 in Figure 28-7.

885

886

Chapter 28   Coupled Circuits

Section 28-4  Hybrid Parameters 28-55. Why is the term hybrid used to describe the parameters in Fig­ure 28-15? 28-56. State the units in which each of the four hybrid parameters is ­expressed. 28-57. Is it possible to apply the Thévenin-equivalent-circuit technique used in developing the hybrid parameters to a passive coupling network? Explain with numerical examples.

Section 28-5  Air-Core Transformers 28-58. Define coefficient of coupling. 28-59. What is meant by loose coupling and tight coupling? 28-60. Express the coefficient of coupling in terms of mutual flux and leakage flux.

Section 28-6  Mutual Inductance 28-61. What effect has the coefficient of coupling on the mutual inductance of a pair of magnetically coupled coils? 28-62. What is mutual reactance?

Section 28-7  Coupled Impedance 28-63. How does the coupled impedance of loose-coupled coils differ from the reflected impedance of a transformer? 28-64. How is the coupled impedance related to the coefficient of coupling? 28-65. Why does an inductive secondary circuit appear as a capacitive coupled impedance?

Integrate the Concepts Given a two-port network, derive an equivalent circuit consisting of (a) a series input circuit and a series output circuit (b) a parallel input circuit and a parallel output circuit (c) a series input circuit and a parallel output circuit

Practice Quiz 1. Which of the following statements are true? (a)  Many complex passive and linear circuits can be modeled by a two-port network. (b)   Four parameters are required to define a two-port network. (c)  The short-circuit admittance parameters are commpnly used in circuit analysis.

Practice Quiz

2.

The open-circuit output impedance (z22) for the circuit of Figure 28-22 is (a)   676.83 Ω ∠−77.20º (b)   362.49 Ω ∠−65.56º (c)   676.83 Ω ∠77.20º (d)   362.49 Ω ∠65.56º Zp

ZS

−j330 Ω

j330 Ω

+ 5 V 10°

Zm 150 Ω

−

+ 20 V −15°

−

Figure 28-22

3. The open-circuit reverse transfer impedance (z12) for the circuit of Figure 28-22 is (a)   330 Ω ∠90º (b)   0 (c)   330 Ω ∠−90º (d)   150 Ω ∠0º 4.

The short-circuit input admittance (Y11) for the circuit of Figure 28-23 is (a)   7.32 mS∠24.42° (b)   4.29 mA∠−45° (c)   7.32 mS∠−24.42° (d)   4.29 mA∠45° Z2 150 Ω

+ 5 V 10°

Z1 –j330 Ω

Z3 j330 Ω

RL 500 Ω

−



Figure 28-23

5. The short-circuit forward-transfer admittance (y12) for the circuit of ­Figure 28-23 is (a)   3.03 mS ∠−90º (b)   6.67 mS ∠0º (c)   4.29 mS ∠−45º (d)   3.03 mS ∠90º

887

888

Chapter 28   Coupled Circuits

6. For the equivalent circuit of the common emitter transistor amplifier of Figure 28-24, the voltage gain across the output terminals of the amplifier is (a)   3.25 (b)   1.63 (c)   −3.26 (d)   −1.63 I1

+

I2

60 Ω

+ E1 4.0 ×10–4V2 2.0 V − peak-to-peak

–0.98 I1

4.0 × 10–5 S

RL 100 Ω

Figure 28-24

7.

The coeficient of coupling is a measure of the (a)   mutual inductance between two windings (b)   voltage induced in a winding (c)   magnetic coupling between two windings (d)   transformation ratio

8. Mutual inductance between two coils relates (a)   the voltage induced in the coils (b)  the changing current in one coil and the voltage induced in the other (c)   the changing current in both coils (d)  the voltage induced in one coil and the changing current in the other 9. Self inductance is (a)  directly proportional to the coeficient of coupling between the windings (b)  inversely proportional to the coeficient of coupling between the windings (c)  directly proportional to the square of the coefficient of coupling between the windings (d)   independent of the coeficient of coupling between the windings 10. Determine the coeficient of coupling between the windings of a transformer that has a primary inductance of 25 H, a self inductance of 25 H, and a mutual inductance of 3 H. (a)   0.254 (b)   0.504 (c)   0.425 (d)   0.405

Practice Quiz

11. The total inductance of two inductances connected in series is (a)  the algebraic sum of the primary inductance, the self inductance, and the mutual inductance (b)  the algebraic sum of the primary inductance, the self inductance, and double the value of the mutual inductance (c)  the algebraic sum of the primary inductance, the self inductance, and triple the value of the mutual inductance (d)   none of the above 12. Determine the total input impedance for the transformer of Figure 28-25, assuming that the resistance of the windings is negligible. (a)   18.85 Ω ∠90º (b)   13.19 Ω ∠90º (c)   9.42 Ω ∠90º (d)   16.85 Ω ∠90º 20 mH

120 V, 60 Hz + 25 mH



50 mH

+ 40 V −

Figure 28-25

13. Determine the total input impedance for the transformer of Figure 28-25 when a 500-Ω resistor is connected to the secondary. (a)   9.42 Ω ∠89.88º (b)   18.85 Ω ∠88.98º (c)   18.85 Ω ∠90º (d)   9.42 Ω ∠2.16º

889

29

Three-Phase Systems

Most of the AC networks we have considered up to this point have had only one source of alternating voltage. H ­ owever, systems with several AC voltages with different phase angles have significant advantages for generating and distributing large amounts of electric energy. Polyphase circuits are also used for most industrial electric motors and for some types of electromechanical control systems.

Chapter Outline 29-1 Advantages of Polyphase Systems  892

29-2 Generation of Three-Phase Voltages  895 29-3 Double-Subscript Notation  897

29-4 Four-Wire Wye-Connected System  899 29-5 Delta-Connected System  903 29-6 Wye-Delta System  909

29-7 Power in a Balanced Three-Phase System  913 29-8 Phase Sequence  915

29-9 Unbalanced Three-Wire Wye Loads  919

29-10 Power in an Unbalanced Three-Phase System  924 29-11 The AC Generator  927

29-12 Three-Phase Induction Motor  930

29-13 Three-Phase Synchronous Motor  932 29-14 Single-Phase Motors  934

29-15 The 30° Difference between Delta-Wye Configurations  935

Key Terms two-phase alternator  892 single-phase circuit  892 neutral 892 polyphase system  893 synchronous speed  895 three-phase (3ϕ) systems  896 balanced load  896 double-subscript notation 897 wye-connected three-phase source 899 star-connected source  900 phase voltage  900 wye-connected load  900

line current  901 unbalanced load  902 delta-connected source  904 delta-connected system  904 delta-connected load  905 phase current  905 line voltage  909 phase sequence (or phase rotation) 915 floating  919 three-phase alternators  928 distributed winding  929 induction motor  930 singly excited  930

synchronous speed  930 squirrel-cage rotor  931 wound rotor  931 slip or percent slip  932 pony motor  933 overexcitation 934 synchronous capacitors  934 starting winding  934 split-phase motor  934 resistance-start motor  935 capacitor-start motor  935 capacitor-run motor  935 universal motor  935 shaded-pole motor  935

Learning Outcomes At the conclusion of this chapter, you will be able to: • describe the construction of a two-phase alternator • draw a circuit consisting of a two-phase alternator c ­ onnected to a two-phase load • draw the waveform of the instantaneous power in a ­balanced two-phase system • illustrate how a rotating magnetic field is produced in a two-phase system • draw the schematic of an Edison three-wire single-phase system Photo sources:   © iStock.com/Digiphoto

• describe the construction of a simple threephase alternator • draw the waveforms for the voltages produced by a ­simple three-phase alternator • use double-subscript notation to specify a voltage rise or a voltage drop in a three-phase ­system • draw schematic and phasor diagrams of a four-wire wye-connected system • calculate the line currents in a four-wire wyeconnected system

892

Chapter 29   Three-Phase Systems

• draw the schematic and phasor diagrams of a delta-­connected system • calculate the line and phase currents in a delta-connected system • calculate the line voltages for a wyeconnected source • draw the schematic of a wye-delta system • calculate the magnitude of the line currents in a balanced wye-delta system • calculate power in a balanced three-phase system • discuss the effect of phase sequence on the ­operation of a three-phase system • calculate the line currents in a three-phase ­system for a given phase sequence

• calculate the line currents in a three-phase ­system with an unbalanced load • list the components of an AC generator • calculate the frequency of generated voltage • describe concentrated and distributed windings • describe excitation methods • explain the operation of an induction motor • calculate synchronous speed • describe the types of induction motor rotors • explain and calculate slip speed • describe the construction and operation of a synchronous motor • explain starting methods of a synchronous motor • describe various types of single-phase motors

29-1 Advantages of Polyphase Systems

N

A B

b a S

(a)

Instantaneous EMF

The two-phase alternator in Figure 29-1(a) has two identical loops mounted on the same rotor. Since both loops have the same number of turns and ­rotate at the same angular velocity, the voltages induced in them have the same magnitude and frequency. Loop A is mounted on the rotor 90° ahead (in the ­direction of rotation) of loop B. Consequently, the voltage in loop A always leads the voltage in loop B by 90°. A single-phase circuit has only one source of alternating current.

+

eA

eB

0 π 2

π

3π 2

2π

5π 2

ωt

−

(b)

Figure 29-1  Simple two-phase alternator

In Figure 29-2, identical loads are connected to the two windings of the two-phase alternator. It is customary to represent each alternator winding with a coil rather than using a generator symbol. In Figure 29-2, the coils are at right angles to each other to indicate that their voltages are 90° out of phase. Although each winding can be connected independently to its load as in Figure 29-2(a), two-phase circuits usually have a common or neutral lead as in Figure 29-2(b). This arrangement reduces the number of conductors needed from four to three. The neutral current is the phasor sum of the two

29-1   Advantages of Polyphase Systems

Line A

120 Ω EA

EA

120 V 90°

RA

Neutral EB

RB

120 Ω

120 V 0°

EB

(a)

Line B

(b)

Figure 29-2  Simple two-phase system

load currents. If the two loads are identical, the neutral current is only √2, or 1.414, times the current in each of the other conductors, as shown in the phasor diagram of Figure 29-3. With the two 120-Ω loads at 120 V, Line A and Line B each carry a current of 1 A and the neutral carries a c­ urrent of 1.414 A. The total of these three currents is 3.414 A. The equivalent single-phase system supplying the two 120-Ω loads in parallel has two conductors that each carry 2 A, making a total of 4 A. A polyphase system needs less copper than a single-phase system to supply a given power at a given voltage. EA

IN

IA 1

√2 45° 1

EB

IB

Figure 29-3  Phasor diagram for the neutral current in a simple two-phase system

Instantaneous Power

Figure 29-4 shows the instantaneous power to the two identical loads in Figure 29-2. The total instantaneous power supplied by the alternator at any instant is the sum of the instantaneous power to the two loads. Since the two voltages are 90° out of phase, the instantaneous power to one load is Pm

PT pA

0 π 2

π

pB 3π 2

2π

ωt

Figure 29-4  Instantaneous power in a balanced two-phase system

893

Chapter 29   Three-Phase Systems

A B

B A

Two-phase alternator

EA EB (a)

Instantaneous Current

894

ϕ

iA

0 π 4

π 2

3π 2

π

A B

2π

ωt

−

A B

iB

+

B

ϕ

A B

B

ϕ

A

A

A

(b)

(c)

(d)

B

Figure 29-5  Producing a rotating magnetic field in a two-phase system

greatest when the power to other load is zero. If we check carefully, we find that the sum of the two instantaneous powers is the same at every ­instant. A constant power is an important advantage for large machines since it a­ llows a steady conversion of mechanical energy into electric energy. The total instantaneous power of a polyphase source is constant if the load on each phase of it is identical. When the two-phase alternator is connected to a set of perpendicular coils as shown in Figure 29-5(a), each coil passes a sine wave of current, but the current in the B coils is 90° out of phase with the current in the A  coils. The magnetic flux produced by the two currents at any instant ­depends on the magnitude of the two instantaneous currents. At 0 rad, the current in the B coils is zero and the current in the A coils is at its maximum in the positive direction. The direction of the resulting magnetic field is ­indicated by the arrow in Figure 29-5(b). At π/4 rad, the coils all pass a current of 0.707Im. Therefore, the total flux is the phasor sum of two perpendicular components with equal magnitudes. This total flux has the same density as at 0 rad, but its direction has changed as shown in Figure 29-5(c). At π/2 rad, iA is zero and iB is at its positive maximum. The magnitude of

29-2   Generation of Three-Phase Voltages

895

the flux density is still the same, but the direction of the magnetic field, as shown in Figure 29-5(d), is now at right angles to its original direction at the start of the cycle. Calculating the magnetic flux phasors through the rest of the cycle shows that the coil arrangement of Figure 29-5(a) produces a rotating magnetic field with a constant magnitude. A polyphase source can develop a magnetic field that has a constant flux density and rotates at the frequency of the applied sine wave. A compass needle placed in the centre of the coils in Figure 29-5(a) will rotate with the magnetic field at a synchronous speed. Rotating magnetic fields greatly simplify AC motor construction. A single-phase system produces a magnetic field that does not rotate. Instead, this magnetic field has a varying magnitude and reverses its direction each 180°. See Problem 29-1 and Review Questions 29-33 to 29-39 at the end of the chapter.

29-2 Generation of Three-Phase Voltages The simple two-phase system that we used to show the advantages of polyphase electric power generation has been used to a limited extent in the distribution of electric energy. However, its main application in present-day equipment is in servomechanisms, such as the autopilot for an aircraft. If the aircraft deviates from its course, the navigation system p ­ roduces an error voltage that operates a servomotor connected to the rudder to bring the aircraft back on course. This error voltage is 90° out of phase from a r­ eference voltage. The direction of rotation of a two-phase motor depends on whether the sine-wave error voltage leads or lags behind the reference voltage. The two alternator coils in Figure 29-6 are 180° apart and have a common neutral lead connecting them to their loads. The neutral current is now the phasor sum of two equal currents that are 180° out of phase. With identical loads, the neutral current in this system is zero, and there is no I2R loss in the neutral conductor. Consequently, this system can use less copper than the two-phase system of Figure 29-2(b) to supply a given Line A

EA

RA

115 V

Neutral 230 V EB

RB

115 V

Line B

Figure 29-6  Edison three-wire single-phase system

The neutral ­conductor is often made the same size as the line conductors to ensure that the neutral will not be ­damaged should one of the loads fail or be ­disconnected.

A balanced load is one in which the ­impedance of each arm of the load is identical, both in magnitude and phase (power-factor) angle. A balanced load need not be resistive.

Chapter 29   Three-Phase Systems

power to a given load. However, the total voltage from Line A to Line B is greater than in the two-phase system. Since the instantaneous power in each coil in ­Figure 29-6 reaches its peak at the same instant, the total instantaneous power pulsates. These coils cannot produce a rotating magnetic field since the two currents are 180° out of phase. Therefore, this system is not a two-phase system. It is called the Edison three-wire single-phase system, and it is widely used for distributing electric energy to residences. Three-phase (3ϕ) systems are also widely used for electric power distribution. The alternator in Figure 29-7(a) has three coils mounted on the rotor at 120° intervals. This alternator produces three sine-wave voltages that are 120° out of phase with one another, as shown in Figure 29-7(b). Figure 29-8(a) shows one method of connecting these coils to a load. With a balanced load, the currents in each arm of the load are equal in magnitude and 120° out of phase with one another, as shown in the phasor diagram of Figure 29-8(b). The neutral current is the phasor sum of these three load currents. This neutral current is zero with a balanced load. As a result, a three-phase system can deliver a given power with less copper and less line loss than an equivalent two-phase system.

C

b

a

A B

eB

eA

N

c

S

Instantaneous EMF

896

eC

+ 2π 3

0 π 3

4π 3 5π 3

π

8π 3

2π 7π 3

3π

− (a)

(b)

Figure 29-7  Simple three-phase generator

IC

C EC EB

EA

RA

A IA

RC IA RB

Neutral B

IC

IB (a)

Figure 29-8  Simple three-phase system

IB (b)

ωt

29-3   Double-Subscript Notation

If the load resistors in Figure 29-8(a) are identical, pT =



e2A + e2B + e2C R

Careful inspection of Figure 29-7(b) shows that e2A + e2B + e2C is always constant. Therefore, the instantaneous power in a balanced three-phase system is a constant. We also find that the voltages generated by a threephase alternator can produce a rotating magnetic field of constant flux ­density, like a two-phase system. Consequently, the three-phase system has all the advantages of the two-phase system, but is more efficient. See Review Questions 29-40 to 29-46.

29-3 Double-Subscript Notation The three-phase circuit of Figure 29-8 represents an impedance network with three voltage sources acting simultaneously. Reversing the leads to one of the coils in an AC generator has the same effect on an AC circuit as reversing the leads to a battery in a DC circuit containing several DC voltage sources. So we cannot connect the six leads from the coils in Figure 29-8(a) in a random sequence, and then assume that we will obtain the current ­relationships shown by the phasor diagram of Figure 29-8(b). In Figure 29-8(a), we indicate how we connected the leads to each winding by drawing an arrow beside the symbol representing the alternator winding. We consider the tracing direction around the loop to be in the ­direction indicated by the arrow. However, the arrow system by itself does not show the connection of the leads to each coil when we write a Kirchhoff’s voltage-law equation for the loop. Neither does the letter ­ ­symbol EA give us any indication of which lead is which. This problem can be solved by using double-subscript notation. In a commonly used version, one end of each coil is indicated by a lowercase letter and the other by an uppercase letter, as in Figure 29-9(a). If a tracing C

EcC = VCc

b

c a

A

EaA = VAa

EbB = VBb

B (a)

(b)

Figure 29-9  Double-subscript notation for three-phase voltages

897

898

Chapter 29   Three-Phase Systems

loop enters coil A by the terminal marked with the lowercase letter and leaves by the terminal marked with the uppercase letter, we represent the induced voltage of that coil as EaA. In double-subscript notation, the first subscript indicates the end of a circuit component at which the tracing loop enters the component, and the second subscript indicates the end of the component from which the tracing loop leaves the component. When we use double-subscript notation to identify voltages, we must be careful to distinguish between a voltage rise inside the source and a voltage drop in the external circuit. In Figure 29-10, a load is connected directly across an alternator winding with terminals marked “a” and “A.” The ­direction of the induced voltage for this winding, as shown by the arrow, is from a to A inside the alternator. Therefore, we identify the voltage rise inside the alternator as EaA. To be consistent with the tracing loop, the voltage drop across the load VAa must reverse the order of the subscripts. From Kirchhoff’s voltage law, EaA = VAa



(29-1)

Hence, the phasors in Figure 29-9(b) can represent either the voltage rises inside the source or the voltage drops in the external loads. A

EaA

V

VAa

a Figure 29-10  Double-subscript notation for voltage rise and voltage drop

When we write loop equations based on Kirchhoff’s voltage law, we include both E and V in the same equations (as in Equation 29-1). So, proper designation of letter symbols and subscripts is very important. But when we write Kirchhoff’s current-law equations for nodal analysis, we use only the external voltage drop between the two nodes. Such equations require a further convention for subscripts: In the circuit external to the source, the second subscript with the voltage drop symbol V indicates the reference node. Thus, in Figure 29-10, terminal a is the reference node and the voltmeter is measuring the voltage at terminal A with respect to terminal a.

29-4   Four-Wire Wye-Connected System

In the following sections, when we are concerned with the connections of the alternator windings, we use the internal voltage rise symbol E. If we reverse the leads to coil A so that the tracing loop in a circuit enters the coil at end A and leaves at end a, we represent the source voltage with EAa, where EAa has the same RMS magnitude as EaA but their phase angles differ by 180°. Since EAa is 180° out of phase with EaA, EAa = −EaA



(29-2)

When we make calculations in the external circuit, we usually use the ­voltage drop symbol V with double subscripts. See Review Question 29-47.

29-4 Four-Wire Wye-Connected System If we take the tracing direction from the lowercase letter to the uppercase letter in each coil of the alternator in Figure 29-7, the three coil v ­ oltages are EaA = 120 V∠0º



EbB = 120 V∠−120º



Ec C = 120 V∠+120º



We can connect one end of each generator coil to a common neutral, as shown in Figure 29-11. To show that points a, b, and c are all connected to the neutral, we can replace these lowercase letters with the letter o. In circuits with a common neutral, the conventional tracing direction is from the generator to the load along the individual lines and back along the common neutral lead. Hence, the three source voltages become EoA = EaA, EoB = EbB, and EoC = EcC. As shown in Figure 29-11(b), we designate the equivalent ­voltage drops from line to neutral as VAN, VBN, and VCN. Connecting the ­alternator coils in this manner produces a wye-connected three-phase source ­(sometimes called C

C c

a

A

b

120° A N B

B (a)

EoC = VCN = 120 V 120°

EoA = VAN = 120 V 0°

120° 120°

EoB = VBN = 120 V −120° (b)

Figure 29-11  Wye-connected alternator windings

899

900

Like wye networks, wye-connected sources are named for their Y-shaped ­configuration.

Chapter 29   Three-Phase Systems

a star-connected source). The voltage measured from line to neutral with a wye-connected source is called the phase ­voltage of the source. We could also form a wye-connected 3ϕ source by connecting all the uppercase letter ends of the coils to the common neutral. However, we must avoid the situation shown in Figure 29-12. Coils A and C have their lowercase letter ends connected to the neutral, but the leads for coil B are reversed. Since the tracing direction is out along the line and back along the neutral, the voltage induced in coil B is EBb, which equals −EbB. If we connect such a source to a balanced load, the neutral current will have twice the magnitude of the current in each leg of the load. C B

EcC

EBb

b

C c

B a

A

A

EaA

N (a)

(b)

Figure 29-12  Incorrect wye connection

Figure 29-13 shows three impedances that form a four-wire ­wye-connected load. If we trace the wiring carefully, we note that ZA is connected directly across coil A of the alternator. The current in ZA, the current in line A, and the current in coil A must all be the same since the three components form a series loop. Because the convention for tracing direction defines the direction of each current as outward from the alternator along the individual lines and back along the neutral, we do not need double subscripts for currents in a four-wire wye system. For example, IA represents the current in ZA, line A, and the generator coil A. Hence, IA =



EoA VAN = ZA ZA

EoB VBN   IB = = ZB ZB and

 

IC =

EoC VCN = ZC ZC

IN = IA + IB + IC

(29-3A)

(29-3B)

(29-3C) (29-4)

29-4   Four-Wire Wye-Connected System

C

C

IC

A

IA

ZC o

N B

B

IN IB

ZA ZB

Figure 29-13  Four-wire wye-connected load

The three load impedances of the four-wire wye load can be independent single-phase loads. For example, ZA could be a bank of lamps while ZB is a heater, and ZC a single-phase motor. It is usually desirable to divide the total single-phase load among the three phases such that the line ­currents are as nearly equal in magnitude as possible. When ZA, ZB, and ZC are the windings of a 3ϕ motor, ZA = ZB = ZC and the load is ­balanced. A ­balanced load can be a combination of single-phase and threephase loads. For a four-wire balanced wye load, the three line currents are equal in ­magnitude, the neutral current is zero, and IL =



Vp Z

(29-5)

where IL is the magnitude of the current in each line, Vp is the magnitude of the phase voltage, and Z is the magnitude of the impedance between each line and the neutral.

Example 29-1

In the circuit of Figure 29-13, ZA = ZB = ZC = 30 Ω ∠( +30º, and the phase voltages are as shown in Figure 29-11(b). (a) Find the current in each line. (b) Verify that the neutral current is zero. (c) Draw a phasor diagram showing phase voltages and line currents. Solution (a)

IA =

IB =

IC =

120 V∠0º VAN = = 4.0 A ∠− 30º ZA 30 Ω ∠+30º

120 V∠−120º VBN = = 4.0 A∠− 150 º ZB 30 Ω ∠+30º 120 V∠+120º VCN = = 4.0 A ∠+ 90 º ZC 30 Ω ∠+30º

901

902

Chapter 29   Three-Phase Systems

(b)

IN = IA + IB + IC = 4.0 A∠−30º + 4.0 A∠−150º + 4.0 A∠+90º

  = ( 3.464 − j2.0 ) A + ( −3.464 − j2.0 ) A + ( 0 + j4.0 ) A = 0A



(c) Figure 29-14 shows the phasor diagram for the phase voltages and line currents. Note that each load current lags behind the voltage applied to that load by a power-factor angle of 30°. EoC = VCN IC ϕ ϕ

EoA = VAN

ϕ

IB

IA

EoB = VBN Figure 29-14  Phasor diagram for Example 29-1

In the example below, a four-wire wye-connected system has an ­unbalanced load. As a result, the line currents are not equal, and there is some neutral current.

Example 29-2 In the circuit of Figure 29-13, ZA is a 360-W lamp, ZB passes a current of 4.0 A at a 0.966 lagging power factor, and ZC = 60 Ω ∠−30º. Find the neutral current. Solution Because the lamp possesses only resistance, IA is in phase with VAN as shown in Figure 29-15. Therefore, PA 360 W = = 3.00 A VAN 120 V      IA = 3.00 A∠0º

IA =

29-5   Delta-Connected System

903

EoC = VCN

30° IC

15°

IB

IA

EoA = VAN

IN

EoB = VBN Figure 29-15  Phasor diagram for Example 29-2

Since ZB has a 0.966 lagging power factor, the power-factor angle is tan−1 0.966, or 15°. Therefore, IB lags behind VBN by 15° and

IB = 4.0 A∠−135º

IC =

120 V∠+120º VCN = = 2.0 A∠+150º ZC 60 Ω ∠−30º

IN = IA + IB + IC

= 3.0 A∠0º + 4.0 A∠−135º + 2.0 A∠+150º

= ( 3.0 + j0 ) A + ( −2.828 − j2.828 ) A + ( −1.732 + j1.0 ) A = −1.56 − j1.828 A = 2.4 A∠−131°

See Problems 29-2 to 29-7 and Review Questions 29-48 to 29-50.

29-5  Delta-Connected System The phasor diagram of Figure 29-9(b) suggests that the voltages generated in the three coils of a three-phase source can be connected such that their phasor sum is zero. To measure the sum of these voltages, we connect end b of coil B to end A of coil A and end c of coil C to end B of coil B, as shown in Figure 29-16(a). Finally, we connect a voltmeter ­between the remaining two terminals to read the total voltage. When we have to add several phasors diagrammatically, we can cut down on the geometric construction and obtain a less cluttered p ­ hasor diagram by using funicular phasors (as described in Section 27-7). Since the head of the phasor for EaA in Figure 29-16(a) represents the uppercase letter end

The three-phase source need not be a rotating machine. The secondary windings of three transformers fed from a threephase system can be considered a threephase source.

904

Chapter 29   Three-Phase Systems

a

EaA

A

V

60°

60°

b

C c

EcC

B (a)

EbB

60° (b)

Figure 29-16  Delta connection of a three-phase source

Like delta networks, delta-connected sources are named for their triangular configuration, which resembles the Greek letter delta, Δ.

of the coil, we can duplicate the circuit connection of Figure 29-16(a) by drawing the phasor EbB from the tip of the phasor EaA. Similarly, we construct the phasor for EcC from the tip of the phasor EbB. The tip of the phasor Ec C is then at the origin, completing an equilateral triangle. ­Therefore, the voltmeter in Figure 29-16(a) will read zero, and it is safe to connect terminal C to terminal a, making a delta-connected source. In practice, we would always check with a voltmeter before making this final connection. Reversing the leads to any one of the coils results in a voltmeter reading of twice the phase voltage, as shown in Figure 29-17. Connecting the delta with one pair of leads reversed would cause a very large current, limited only by the impedance of the three coils, to flow around the loop.

a

A

EaA

b

EbB

B 240 V −60°

V

120° C ECc

c

(a)

(b)

Figure 29-17  Incorrect delta connection

With a delta-connected source, we require only three wires to connect to a three-phase load, which can also be connected as a delta (see Figure  29-18). In a delta-connected system, each arm of the load is connected directly across one of the generator coils. We can label the three lines ­according to the uppercase letter of the phase voltages connected to each line. Thus, EaA becomes ECA, EbB becomes EAB, and EcC becomes EBC. The equivalent voltage drops become VAC, VBA, and VCB.

29-5   Delta-Connected System

905

C

Z BA

C

b c

ZAC

A

CB

A

Z

a

B B

Figure 29-18  Delta-connected load

In the delta-connected load, each arm of the load is connected from line to line. Consequently, we need to label each impedance with two subscripts to identify the two lines between which the impedance is connected. The order of the subscripts is determined by the tracing direction around each loop. For example, the tracing direction through generator coil B is from b to B. Therefore, the voltage generated in that coil is written as EAB. This loop then passes through the load impedance from line B to line A. Hence, the impedance is identified as ZBA. The impedance subscripts are in the same order as the subscripts for the voltage drop VBA. When working with delta-connected loads, it is convenient to use V rather than E since the voltages, currents, and impedances at the load then all have subscripts in the same order. The phase currents of the source are the currents flowing in each coil of the source. Similarly, the phase currents of the load are the currents flowing in each arm of the load. In each loop of a wye-connected system, the source phase current, load phase current, and line current are all one and the same. But in the delta system of Figure 29-18, each line carries current for two arms of the load. Moreover, the tracing direction for one of these currents is away from the source and the other is toward the source. Therefore, the line current to a delta load is the phasor difference between the two load phase currents flowing in that line. The load phase currents in Figure 29-18 are



IAC =

VAC ECA = ZAC ZAC

(29-6A)



ICB =

VCB EBC = ZCB ZCB

(29-6B)



IBA =

VBA EAB = ZBA ZBA

(29-6C)

If the angle between two currents is greater than 90°, their phasor difference is greater than either current.

In a wye-connected system, the phase voltage is lower than the line voltage, so motors or drives are able to run at lower speeds than if they were connected in a delta configuration.

906

Chapter 29   Three-Phase Systems

Even in a 3ϕ three-wire system, the conventional direction of the line current is outward along all three lines away from the source, as shown in Figure 29-18. Therefore,

(29-7A)

IC = ICB − IAC

(29-7C)

IB = IBA − ICB



We can verify Equation 29-8 by adding Equations 29-7A, 29‑7B, and 29-7C.

IA = IAC − IBA

(29-7B)

In a 3ϕ three-wire system, there is no neutral return path from the load back to the generator. Consequently, even with an unbalanced load, the phasor sum of the three line currents must be zero: IA + IB + IC = 0



(29-8)

This relationship is useful for checking calculations for 3ϕ three-wire AC circuits.

Example 29-3 Determine the three line currents when the impedances of the branches of the delta load in Figure 29-18 are ZAC = 60 Ω ∠0º, ZCB = 30 Ω ∠−30º, and ZBA = 30 Ω ∠+45º. Use the voltages given in Figure 29-11(b). Solution Step 1 Determine the phase currents.

IAC = ICB =

IBA =

120 V∠0º VAC = = 2.0 A∠0º ZAC 60 Ω ∠0º

120 V∠+120º VCB = = 4.0 A∠+150º ZCB 30 Ω ∠−30º

120 V∠−120º VBA = = 4.0 A∠−165º ZBA 30 Ω ∠+45º

Step 2 Determine the line currents (see Figure 29-19).

IA = IAC − IBA

= 2.0 A∠0º − 4.0 A∠−165º

     = ( 2.0 + j0 ) A − ( −3.864 − j1.035 ) A   = 5.864 + j1.035 A   = 6.0 A∠+10º

29-5   Delta-Connected System

VCB = EBC

IC

ICB IA IAC

IBA

VAC = ECA

IB VBA = EAB Figure 29-19  Phasor diagram for Example 29-3



IB = IBA − ICB

= 4.0 A∠−165º − 4.0 A∠+150º

  = ( −3.864 + j1.035 ) A − ( −3.464 + j2.0 ) A

= −0.40 − j3.035 A = 3.1 A∠−98º

  IC = ICB − IAC

= 4.0 A∠+150º − 2.0 A∠0º

= ( −3.464 + j2.0 ) A − ( −2.0 + j0 ) A = −5.464 + j2.0 A = 6.0 A∠+160º

To illustrate the relationship between line current and phase current in a delta system, we now use geometric construction to carry out the phasor subtraction for IB in Example 29-3. As noted in Section 20-9, on a phasor ­diagram, −ICB has the same length as ICB but the opposite direction. Since IB = IBA − ICB, we can construct IB by using the parallelogram method to add IBA and −ICB (see Figure 29-20). In a balanced delta load, the impedances of all three arms are equal in magnitude and angle. Since the phase voltages are equal and 120° out of phase, it follows that the phase currents are also equal and 120° out of phase, as shown in Figure 29-21. Therefore, ICB and −IAC are equal in magnitude and have a 60° angle between them.

907

To check the calculation, verify that the sum of the line ­currents is zero: IA + IB + IC = (5.864 + j1.035) + (−0.40 − j3.035) + (−5.464 + j2.0) = 0 + j0 A

908

Chapter 29   Three-Phase Systems

ICB

IBA −ICB IB Figure 29-20  Phasor subtraction of phase currents

IC Q

ICB

P

S 30° −IAC

120°

30° R

IAC

O

Figure 29-21  Line current and phase current for a balanced delta load

In Figure 29-21, we use the parallelogram OPQR to construct the linecurrent phasor. Since ICB and −IAC are equal in magnitude and have a 60° angle between them, OPR is an equilateral triangle that is bisected by the diagonal OQ, which bisects PR at right angles at point S. Therefore, OP = 2PS From the Pythagorean theorem,

OS = √OP2 − PS2 = PS√22 − 12 = √3 PS =

√3 OP 2

Because the diagonals of a parallelogram bisect each other, Hence,

OQ = 2OS = √3OP

The line current to a balanced delta load has a magnitude of √3 times the phase current in each arm of the load and is displaced 30° from the phase current. See Problems 29-8 to 29-13 and Review Questions 29-51 to 29-53.

29-6   Wye-Delta System

29-6  Wye-Delta System In the four-wire wye-connected three-phase system, the coils of the wye-­ connected source are connected from line to neutral. Hence, voltages ­measured from line to neutral are the phase voltages of the wye-connected source. However, it is customary to specify voltages for three-phase systems in terms of line voltages, which are measured between lines. In fact, unless otherwise specified, the voltages given for a three-phase system are line ­voltages. The phase voltage and line voltage of a delta-connected source are the same since the coils of a delta-connected source are connected from line to line. In the wye-connected source of Figure 29-22, each voltmeter reads a line voltage that depends on the voltages generated in two coils. In writing the subscripts for the three line voltages, we must decide whether to read clockwise or counterclockwise around the phasor diagram for the source. As long as we follow the same direction for all three line voltages, they will have the same relative angular positions. In Figure 29-22, the tracing ­arrows for the source are drawn in the direction that will give us the same subscripts for the line voltages in this wye-connected source as we had for the delta-connected source of Figure 29-18. C

C

V o

V

A V

B B

Figure 29-22  Line voltages of a wye-connected three-phase source

Following the tracing arrow from line A to line B through the ­wye-connected source in Figure 29-22 shows that

EAB = EAo + EoB = −EoA + EoB

Therefore, the line voltage in a wye-connected source is the phasor difference between two phase voltages. Since the angle between the phase voltages is greater than 90°, this phasor difference is greater than either phase voltage by itself. Using the voltages in Figure 29-11 gives

EAB = −120 V∠0º + 120 V∠−120º

= − ( 120 + j0 ) + ( −60 − j104 ) = −180 − j 104 V

= 208 V∠−150º

909

910

Chapter 29   Three-Phase Systems

We can also solve for EAB by geometric construction, as shown in Figure 29-23. First, we construct phasors for the given phase voltages EoA and EoB. Then we construct the phasor for −EoA by reversing the direction of the EoA phasor. EAB equals the phasor sum of −EoA and EoB. −EoA

30° 30°

EAB

EoA

120°

EoB

Figure 29-23  Determining the line voltages of a wye-connected source

If the coils of the wye-connected source are properly connected, the phase voltages are always equal in magnitude and 120° out of phase. We can use the same construction as for the line current to a balanced delta load to show that VL = √ 3 Vp ≈ 1.732 Vp



(29-9)

where VL is the line voltage and Vp is the phase voltage. For the phase voltages given with the wye-connected source of Figure 29-11, the three line voltages are

VBA = EAB = EAo + EoB = EoB − EoA = 208 V∠−150º VCB = EBC = EBo + EoC = EoC − EoB = 208 V∠+90º

VAC = ECA = ECo + EoA = EoA − EoC = 208 V∠−30º

(29-10A) (29-10B) (29-10C)

Figure 29-24(a) shows the phasors for all the line and phase voltages of a wye-connected three-phase source. The line voltages are always displaced 30° from the phase voltages since changes in the load do not affect the generated voltages. EBC ECA

EoC 30°

30° EAB

30° EoB

EoA

EBC

EoC

EoA

o

EoB EAB

ECA (a)

(b)

Figure 29-24  Phasor diagrams for a wye-connected three-phase source

29-6   Wye-Delta System

Because the opposite sides of a parallelogram are equal in length and parallel, a line joining the tips of the EoA and EoB phasors in Figure 29-24 would have the same magnitude and direction as EAB. Therefore, we can check the magnitude and direction of the line-voltage phasors by simply joining the tips of the phase-voltage phasors, as shown in Figure 29-24(b). With a wye-connected three-phase source, two different RMS voltages are available from the one system. For this reason, 3ϕ sources consisting of wye-connected transformer secondary windings rated at 120/208 volts are commonly used throughout North America. If we wish to apply 120 V, 3ϕ to the load, we connect ZA, ZB, and ZC as a four-wire wye load, as in ­Figure 29-13. But if we wish to operate the same load from a 208-V 3ϕ source, we can connect the three arms of the load in delta fashion, as shown in Figure 29-25. Since the various load impedances are connected from line to line, we use the same double-subscript identification as for the simple delta-connected system. C

C

Z AC A

ZCB

o

ZB

A

B

B

Figure 29-25  Wye-connected source feeding a delta-connected load

Example 29-4

A balanced load with an impedance of 52 Ω ∠+45º in each arm is deltaconnected to a 120/208-V 3ϕ wye-connected source, as in Figure 29-25. Find the magnitude of the current in each coil of the source. Solution Since each arm of the load is connected from line to line, the load voltage is the line voltage:

IBA = ICB = IAC =

Since the load is balanced,

208 V = 4.0 A 52 Ω

IL = 1.732Ip = 1.732 × 4.0 A = 6.9 A

As Figure 29-25 shows, the current in each winding of a wye-connected source is the same as the current in the line connected to that particular coil. Therefore, the current in each winding of the source is 6.9 A.

911

912

Chapter 29   Three-Phase Systems

When a delta load is connected to a wye-connected source, as in Figure 29-25, the phase voltages of the source are not used. So once we have determined the line voltages, we can disregard the phase voltages. Hence, the procedure for solving an unbalanced wye-delta circuit is exactly the same as that given in Example 29-3. See Problems 29-14 to 29-20 and Review Questions 29-54 and 29-55.

Circuit Check

CC 29-1. For the four-wire wye-connected load shown in Figure 29-26, Z1 = Z2 = Z3 = 75 Ω∠−45º, (a) determine the line currents (b) determine the phase voltages (c) verify that the load is balanced

A

B

120 V −120° + − −

+ 120 V 120°

− 120 V 0° +

IB

C N

IC

IN

Z1 A



Z2

Z3

IA

Figure 29-26

CC 29-2. For the delta-connected load shown in Figure 29-27, ZAC = 6.0 Ω∠38º, ZCB = 6.0 Ω∠−45º, and ZBA = 12 Ω∠45º. (a) Determine the value of the phase currents and the line ­currents. (b) How will the phase currents and the line currents be related if ZAC = ZCB = ZBA = 6 Ω∠38º? C

−

+

120 V 0°

+

120 V −120°

A

ZBA

IA B



ZCB

ZAC

–

+

120 V 120°

−

IC

IB

Figure 29-27

29-7   Power in a Balanced Three-Phase System

CC 29-3. A balanced load with an impedance of Z = 50 – j50 Ω in each arm is delta connected as shown in Figure 29-28. Find the ­magnitude of the current in each coil of the source.

C IC B

+ 120 V 120°

+ − −

120 V 0°

ZAB

− 120 V −120° +

ZBC

IB

ZCA

A IA



Figure 29-28

29-7 Power in a Balanced Three-Phase System In a balanced three-phase load, the power in each of the three arms is identical. Consequently, the total power is three times the power in each phase. Since the load phase voltage is the voltage across each arm of the load and the load phase current is the current through each arm of the load,

PT = 3VpIp cos ϕ

(29-11A)

Because the junction points on three-phase loads such as three-phase motors are not always accessible, it is more convenient to express the total power in terms of line voltage and line current, which can readily be measured. In the wye-connected system of Figure 29-13, VL = √ 3 Vp    and    IL = Ip

In the balanced delta-connected load of Figure 29-18, VL = Vp    and    IL = √ 3 Ip

In both cases, substitution in Equation 29-11A gives

PT = √ 3 VL IL cos ϕ (29-11B)

913

914

Chapter 29   Three-Phase Systems

Example 29-5 Find the total power input to the load in Example 29-4. Solution Since each impedance has a power-factor angle of +45°, cos ϕ = cos 45º = 0.707 lagging

and

PT = 1.732 × VLIL cos ϕ = 1.732 × 208 V × 6.93 A × 0.707 = 1.8 kW

Example 29-6 A 550-V 3ϕ motor delivers 15 hp to a mechanical load. The motor has an ­efficiency of 80% and a power factor of 0.90 lagging. Find the line current. Solution

PT =



From Equation 29-11,

IL =

hp × 746 W 15 × 746 W = = 14 kW η 0.80

PT 14 kW = = 16 A 1.732 VL cos ϕ 1.732 × 550 V × 0.90

Example 29-7 For the four-wire wye-connected load shown in Figure 29-26, determine the total power and the power factor of the load. Solution

We know that Z1 = Z2 = Z3 = 75 Ω ∠−45º. Step 1: Determine the line currents.

IA = IB =

IC =

120 V∠0º VA = = 1.6 A∠45º ZA 75 Ω ∠−45º

120 V∠120º VB = = 1.6 A∠165º ZB 75 Ω ∠−45º

120 V∠−120º VC = = 1.6 A∠75º ZC 75 Ω ∠−45º

29-8   Phase Sequence

Step 2: Determine the real power.

PA = PB = PC = VA × IA × cosθ zA

= 120 V × 1.6 A × cos ( −45º ) = 136.76 W

PT = PA + PB + PC = 3 × PA = 3 × 136.76 W = 407.29 W

Step 3: determine the reactive power

QA = QB = QC = VA × IA × sinθ zA = 120 V × 1.6 A × sin ( − 45º ) = −135.76 var ( capacitive )

QT = QA + QB + QC

= 3 × QA = 3 × ( −135.76 var ) = −407.29 var (cap)

Step 4: determine the apparent power

S = P ± jQ = 407.29 W − j407.29 var = 576 VA

Step 5: determine the power factor, PF

PF = cosθ Z = cos ( −45º ) = 0.707 ( leading ) or PF = 70.7%

See Problems 29-21 to 29-24 and Review Questions 29-56 and 29-57.

29-8  Phase Sequence In a three-phase service supplied by an electrical utility company, all three line voltages have the same magnitude and are displaced from one ­another by 120°. We can measure the magnitude of these voltages quite readily. However, the magnitudes do not tell us whether VBN leads VAN by 120° or lags behind it by 120°. The phase sequence (or phase rotation) of a three-phase system governs the direction of rotation of three-phase ­motors and the division of the current among the three lines feeding an unbalanced load. Figure 29-29(a) shows that VBN lags behind VAN by 120° and VCN lags behind VBN by 120°. Therefore, the instantaneous voltage to line A passes

915

916

Chapter 29   Three-Phase Systems

through its positive peak value first, then the instantaneous voltage to line B passes through its positive peak value, followed by the instantaneous voltage to line C. Thus, this system has an ABC phase sequence. If we reverse the leads to two of the generator coils, as in Figure 29-29(b), VBN now leads VAN by 120° and VCN leads VBN by 120°. Thus the phase sequence has been reversed and the system now has a CBA phase ­sequence. Once we have drawn a phasor diagram for the voltages in a 3ϕ ­system, we can easily read the phase sequence from the phasor diagram. Since the direction of rotation of a phasor is counterclockwise, the phase sequence is the order in which the voltage phasors would pass the ­reference axis if they rotate counterclockwise. Therefore, we can read the phase sequence from a phasor diagram by reading clockwise around the diagram.

A

Phase sequence is the order in which the phase voltages achieve their maximum (peak) value with respect to time.

VAN

B

VBN

N C

VCN

(a) ABC phase sequence

A

VCN

B

VBN

N C

VAN

(b) CBA phase sequence

Figure 29-29  Phase sequence of a three-phase source

When we are given the line voltage and phase sequence of a three-phase source, as in Figure 29-30(a), we can draw a phasor diagram by placing one phasor (usually VBA) along the reference axis. As we read clockwise around the phasor diagram of Figure 29-30(b), all the first subscript letters of the voltage phasors must follow the specified phase sequence. The second subscript letters must also match the phase sequence.

29-8   Phase Sequence

VAC 208 V 3ϕ source ABC phase sequence

A B

VBA

C VCB

(a)

(b)

Figure 29-30  Determining phase sequence from a phasor diagram

The following example shows the effect of phase sequence on the line currents to an unbalanced load.

Example 29-8

ZAB = 52 Ω ∠−30º, ZBC = 52 Ω ∠+45º, and ZCA = 104 Ω ∠0º are connected as a delta load to a 208-V 3ϕ source. Find the magnitude of the current in each line when the phase sequence of the source is (a) ABC and (b) CBA. Solution (a) Step 1 With an ABC phase sequence,

IAB =

208 V∠0º VAB = = 4.0 A∠+30º ZAB 52 Ω ∠−30º

ICA =

208 V∠+120º VCA = = 2.0 A ∠+120º ZCA 104 Ω ∠0º

IBC =



208 V∠−120º VBC = = 4.0 A∠−165º ZBC 52 Ω ∠+45º

Step 2 From the circuit diagram of Figure 29-31(a), IA = IAB − ICA

= 4.0 A ∠+30º − 2.0 A ∠+120º

= ( 3.464 + j2.0 ) A − ( −1.0 + j1.732 ) A = 4.464 + j0.268 A

Therefore, IA = 4.5 A

IB = IBC − IAB

= 4.0 A ∠−165º − 4.0 A ∠+30º

= ( −3.863 − j1.035 ) A − ( 3.464 + j2.0 ) A = −7.327 − j3.035 A

917

918

Chapter 29   Three-Phase Systems

Therefore,

IB = 7.9 A

IC = ICA − IBC

= 2.0 A ∠+120º − 4.0 A ∠−165º



= ( −1.0 + j1.732 ) − ( 3.863 − j1.035)



Therefore,

= 2.863 + j2.767 A

IC = 4.0 A

VCA

A

IC

ZAB From 3ϕ source

ICA

IA

ZCA

B ZBC

IAB

IBC IB

C VBC (a)

(b)

Figure 29-31  Phasor diagram for Example 29-7(a)

(b) Step 1 With a CBA phase sequence (as shown in Figure 29-32), IAB =



IBC =



ICA =

VBC IBC IB

208 V ∠0º VAB = = 4.0 A ∠+30º ZAB 52 Ω ∠−30º

208 V ∠+120º VBC = = 4.0 A ∠+75º ZBC 52 Ω ∠+45º

208 V ∠−120º VCA = = 2.0 A ∠−120º ZCA 104 Ω ∠0º

IA IAB

ICA

VAB

VCA IC Figure 29-32  Phasor diagram for Example 29-7(b)

VAB

29-9   Unbalanced Three-Wire Wye Loads

Step 2

Therefore, Therefore, Therefore,

IA = IAB − ICA

= 4.0 A ∠+30º − 2.0 A ∠−120º

= ( 3.464 + j2.0 ) A − ( −1.0 − j1.732 ) A = 4.464 + j3.732 A

IA = 5.8 A

IB = IBC − IAB

= 4.0 A ∠+75º − 4.0 A ∠+30º

= ( 1.035 + j3.863 ) A − ( 3.464 + j2.0 ) A = −2.429 + j1.863 A

IB = 3.1 A

IC = ICA − IBC

= 2.0 A ∠−120º − 4.0 A ∠+75º

= ( −1.0 − j1.732 ) A − ( 1.035 + j3.863 ) A = −2.035 − j5.595 A

IC = 6.0 A

See Problems 29-25 and 29-26 and Review Questions 29-58 and 29-59.

29-9 Unbalanced Three-Wire Wye Loads We have been able to solve unbalanced four-wire wye loads and delta loads with no particular difficulty because, in both cases, we know the phase voltage across each arm of the load. But with the unbalanced threewire wye load of Figure 29-33, we do not know what the phase voltages are because the wye junction point is floating. There are two fairly straightforward procedures for calculating the phase voltages using the techniques for solving AC networks.

919

920

Chapter 29   Three-Phase Systems

A B ZB

From 3ϕ 3-wire source

ZA Floating wye point

ZC C

Figure 29-33  Three-wire wye load

The first method uses Kirchhoff’s voltage-law equations for the wye load. Following the conventional tracing arrows through the wye load in Figure 29-33 gives three loop equations: IAZA − IBZB = EBA = VAB



IBZB − ICZC = ECB = VBC

ICZC − IAZA = EAC = VCA



(1) (2) (3)

However, these three equations are not independent, owing to the symmetrical relationship of the left-hand sides. So we need an additional equation. Applying Kirchhoff’s current law to the junction point of the load gives IA + IB + IC = 0



(4)

We can omit Equation 2, which does not contain IA, and rearrange Equations 1, 3, and 4: Z A IA − Z B IB



−ZAIA

│ │

IA +

│ │

= VAB

+ ZCIC = VCA

IB +

IC = 0

(1) (3) (4)

Using determinants to solve for IA gives

IA =

VAB VCA

ZA −ZA 1

−ZB 1

ZC 1

1

ZC 1

−ZB

=

0 + ( −0 ) + 0 − 0 − VAB ZC − ( −VCA ZB ) 0 + ( −ZB ZC ) + 0 − 0 − ZA ZC − ZA ZB

29-9   Unbalanced Three-Wire Wye Loads

Therefore,

IA =

Similarly,

IB =

and

IC =

VAB ZC − VCA ZB ZA ZB + ZB ZC + ZC ZA

VBC ZA − VAB ZC ZA ZB + ZB ZC + ZC ZA

VCA ZB − VBC ZA ZA ZB + ZB ZC + ZC ZA



(29-12A)



(29-12B)



(29-12C)

Example 29-9 The phase-sequence tester shown in Figure 29-34 is connected to a 208-V 3ϕ 60-Hz source with an ABC phase sequence. The choke has an ­inductance of 0.50 H with negligible resistance, and the lamps each have a resistance of 100 Ω. Assuming that the lamp resistances remain constant, determine the magnitude of the current in each lamp. ZA

A

ZB

Lamp A

B

VCA

Lamp B VAB ZC VBC

C (a)

(b)

Figure 29-34  Phase-sequence tester

Solution

ZC = +jωL = +j377 × 0.50 H = 188.5 Ω ∠+90º

From Equation 29-12A, IA = = =

VAB ZC − VCA ZB ZA ZB + ZB ZC + ZC ZA

( 208∠0º × 188.5∠+90º ) − ( 208∠+120º × 100∠0º ) ( 100∠0º × 100∠0º ) + ( 100∠0º ×188.5∠90º ) + ( 188.5∠90º × 100∠0º )

39 208∠+90º − 20 800∠120º 10 000∠0º + 18 850∠+90º + 18 850∠+90º

921

922

Chapter 29   Three-Phase Systems

= =

( 0 + j39 208 ) − ( −10 400 + j18 013 ) 10 000 + j37 700

10 400 + j21 195

10 000 + j37 700

IA =

Therefore, Similarly,



IB = = = =

23 609 = 0.61 A 39 003

( 208∠−120º × 100∠0º ) − ( 208∠0º × 188.5∠+90º ) 10 000 + j37 700

20 800∠−120º − 39 208∠+90º 10 000 + j37 700

(−10 400 − j18 013 ) − ( 0 + j39 208 ) 10 000 + j37 700

−10 400 − j57 221

10 000 + j37 700 58 158 = 1.5 A and IB = 39 003 Therefore, with an ABC phase sequence, Lamp B is brighter than Lamp A.

In Section 24-7, we applied the equations for transformations between delta and wye networks to AC circuits. We can use these equations to find an equivalent delta network for the original wye-connected load. Once we know the impedance of the delta arms, we can solve for the line currents in the usual way. For the circuit of Figure 29-35, Equations 24-9, 24-10, and 24-11 give

ZBC =



ZCA =



ZAB =

ZA ZB + ZB ZC + ZC ZA ZA ZA ZB + ZB ZC + ZC ZA ZB ZA ZB + ZB ZC + ZC ZA ZC

(29-13A)

(29-13B)

(29-13C)

29-9   Unbalanced Three-Wire Wye Loads

ZAB A

B

VBC

Z

BC

Z CA

VAB

VCA

C (a)

(b)

Figure 29-35  Wye-delta transformation of an unbalanced three-wire wye load

Example 29-10 Use a wye-delta transformation to determine the magnitude of the current in each lamp when the phase sequence of the source in Example 29-9 is changed to CBA. Solution Step 1 From Example 29-9,

ZAZB + ZBZC + ZCZA = 10 000 + j37 700 = 39 003∠+75.14º Therefore, Step 2

ZAB = ZBC =

ZCA =

IAB = IBC =

ICA =

39 003∠+75.14º = 206.9 Ω ∠−14.86º 188.5 ∠+90º

39 003∠+75.14º = 390 Ω ∠+75.14º 100 ∠0º

39 003∠+75.14º = 390 Ω ∠+75.14º 100 ∠0º

208 V∠0º VAB = = 1.005 A ∠+14.86º ZAB 206.9 Ω ∠−14.86º

208 V∠+120º VBC = = 0.533 A ∠+44.86º ZBC 390 Ω ∠+75.14º

208 V∠−120º VCA = = 0.533 A ∠+164.86º ZCA 390 Ω ∠+75.14º

Step 3 From Figure 29-35, IA = IAB − ICA

= 1.005 A∠+14.86º − 0.533 A ∠+164.86º

923

924

Chapter 29   Three-Phase Systems



    = ( 0.9714 + j0.2577 ) A − ( −0.5145 + j0.1392 ) A

Similarly,

IB = IBC − IAB



= 1.486 + j0.1185 A

Therefore, IA = 1.5 A

and

= 0.533 A∠+44.86º − 1.005 A ∠+14.86º

= ( 0.3778 + j0.3760 ) A − ( 0.9714 + j0.2577 ) A = −0.5936 + j0.1183 A

IB = 0.61 A

Hence, with a CBA phase sequence, Lamp A is brighter than Lamp B.

Once we have solved for line currents, we can return to the original wyeconnected circuit and use Ohm’s law to solve for the various phase voltages: VAN = IAZA, VBN = IBZB, and VCN = ICZC. See Problems 29-27 to 29-29 and Review Questions 29-60 to 29-65.

Circuit Check

B

CC 29-4. A 10-hp AC motor operates at full-load from a 230-V 3-phase 60-Hz delta-connected source. The motor is 80% efficient and its power factor is 0.88 lagging. Calculate (a) the line currents (b) the phase currents CC 29-5. A 208-V 60-Hz 4-wire wye-connected source powers a wye load that has ZA = 50 ∠60º Ω, ZB = 100 ∠20º Ω, and ZC = 40 ∠90º Ω . Calculate the line currents and the neutral ­current.

29-10  Power in an Unbalanced ThreePhase System In an unbalanced three-phase system, the power in each arm will be different as the loads may not all be the same, and we will not be able to use the factor √3 that we used with a balanced three-phase load. Instead, the total power in the load will be the sum of the power in each phase.

29-10   Power in an Unbalanced Three-Phase System

Example 29-11 For the four-wire wye-connected load shown in Figure 29-26, determine the total power and the power factor of the load, if Z1 = 5 Ω ∠0º , Z2 = 7.5 Ω ∠−45º, and Z3 = 8.3 Ω ∠33º Solution Step 1 Determine the line currents

IA = IB =

IC =

120 V∠0º VA = = 24 A∠0º ZA 5 Ω ∠0º

120 V∠120º VB = = 16 A∠165º ZB 7.5 Ω ∠−45º

120 V∠−120º VC = = 14.46 A∠−153º ZC 8.3 Ω ∠33º

Step 2 Determine the real power

PA = VA × IA × cosθ zA = 120 V × 24 A × cos ( 0º ) = 2.88 kW

PB = VB × IB × cosθ zB = 120 V × 16 A × cos ( −45º ) = 1.36 kW PC = VC × IC × cosθ zC = 120 V × 14.46 A × cos ( 33º ) = 1.46 kW PT = PA + PB + PC = 2.88 kW + 1.36 kW + 1.46 kW = 5.7 kW

Step 3 Determine the reactive power

QA = VA × IA × sinθ zA = 120 V × 24 A × sin ( 0º ) = 0 var

QB = VB × IB × sinθ zB = 120 V × 16 A × sin ( −45º ) = −1.36 kvar

QC = VC × IC × sinθ zC = 120 V × 14.46 A × sin ( −153º ) = −787.76 var QT = QA + QB + QC = 0 var + ( −1.36 kvar ) + ( −787.76 var ) = −2.15 kvar ( capacitive )

Step 4 Determine the apparent power

ST = PT ± jQT = 5.7 kW − j2.15 kvar = 6.32 kVA∠−25.59º

ST = 6.32 kVA Step 5 Determine the power factor, PF

PF = cosθ = cos ( −25.59º ) = 0.902 ( leading ) or PF = 90.2%

925

926

Chapter 29   Three-Phase Systems

Example 29-12 For the delta-connected load shown in Figure 29-27, determine the total power and the power factor of the load if ZAC = 20 Ω ∠0º, ZCB = 21 Ω ∠16º, and ZBA = 21 Ω ∠ − 13º. Solution Step 1 Determine the phase currents

IAC =

120 V∠0º VAC = = 6 A∠0º ZAC 20Ω ∠0º

IBA =

120 V∠−120º VBA = = 5.71 A∠−107º ZBA 21 Ω ∠−13º

ICB =

120 V∠120º VCB = = 5.71 A∠104º ZCB 21 Ω ∠16º

Step 2 Determine the real power

PAC = VAC × IAC × cosθ zA = 120 V × 6 A × cos ( 0º ) = 720 W

PBA = VBA × IBA × cosθ zB = 120 V × 5.71 A × cos ( −13º ) = 667.64 W PCB = VCB × ICB × cosθ zC = 120 V × 5.71 A × cos ( 16º ) = 658.66 W

PT = PAC + PBA + PCB = 720 W + 667.64 W + 658.66 W = 2.05 kW

Step 3 Determine the reactive power QAC = VAC × IAC × sinθ zA = 120 V × 6 A × sin ( 0º ) = 0 var

QBA = VBA × IBA × sinθ zB = 120 V × 5.71 A × sin ( −13º ) = −154.14 var QCB = VCB × ICB × sinθ zC = 120 V × 5.71 A × sin ( 16º ) = 188.87 var QT = QAC + QBA + QCB = 0 var + ( −154.14 var ) + ( 188.87 var ) = 34.73 var ( inductive )

Step 4 Determine the apparent power

ST = PT ± jQT = 2.05 kW + j34.73 var = 2.05 kVA∠0.97º ST = 2.05 kVA

Step 5 Determine the power factor, PF

PF = cosθ = cos ( 0.97º ) = 0.999 ( lagging ) or PF = 99.9%

29-11   The AC Generator

29-11  The AC Generator An AC generator, or alternator, has the same basic components as the DC generator, namely a field system and an armature system. Large AC ­generators have the armature mounted on the stator and the field on the rotor. As the field is driven past the armature conductors by the prime mover, a voltage is induced in the coils. Slow-speed generators usually have a salient (projecting) pole rotor, while high speed generators have a cylindrical (smooth) rotor. As described in Chapter 18, an AC generator produces a sinusoidal waveform, as shown in Figure 18-6. The voltage produced is dependent on the same factors as for the DC generator: speed of rotation, strength of the magnetic field, and the number of turns on the armature windings. In an AC generator the frequency of the generated voltage is as important as the magnitude of voltage. This frequency depends on the speed of rotation and the number of poles on the field system:

f=

PN 120

(29-14)

where f is the frequency in hertz, P is the number of poles, and N is the speed of rotation in revolutions per minute.

Source:  © Alvey & Towers Picture Library/Alamy Stock Photo

Slow-speed machines, which are generally driven by water turbines, have many poles to generate 60 Hz. High-speed turbo-alternators driven by steam turbines normally have only two or four poles.

Alternator in an automobile: part of the cover and stator have been removed to show the rotor.

927

Chapter 29   Three-Phase Systems

Example 29-13 A hydroelectric generating station has alternators with salient pole rotors containing 24 poles. What turbine speed is required to generate a 60-Hz current? Solution N=



f=

120f PN →N= 120 P

120 × 60 Hz = 300 RPM 24

Source:  © iStock.com/Digiphoto

Salient pole rotor assembly from a power plant alternator (detail)

Three-phase alternators are used to generate the electricity for largescale power systems. For a two-pole rotor, three coils are spaced 120º apart on the stator, as shown in Figure 29-36. A

B´

N

Rotor (field)

C´

Armature winding C

S

B

+ Instantaneous EMF

928

eA

eB

0

−

A´

Figure 29-36  Simplified three-phase, two-pole alternator

eC

29-11   The AC Generator

For large machines, placing the armature on the stator allows for much higher voltages to be generated since the load is connected directly rather than through slip rings. It would be very difficult to insulate the slip rings from the shaft at high voltages. The slip rings are connected to a DC supply at relatively low voltages to provide the electromagnetic field. This arrangement requires only two slip rings, while a rotating armature would require three. The simplified construction shown in Figure 29-36 would be ­somewhat inefficient because the armature windings for each phase is ­concentrated in a single pair of slots. As the rotor turns, the magnetic field is not cutting conductors for a significant portion of the rotation. A more efficient arrangement is the distributed winding with conductors evenly spaced around the circumference of the stator, as shown in Figure 29-37. Each phase occupies multiple slots, and the generated voltage for each phase is the vector sum of the voltages induced in the conductors in these slots.

A1

A2

A

3

A

4

N C4

C3

B

C1

1

C2 S

B2

B4

B3

Figure 29-37  Three-phase, two-pole distributed armature

Rotors in large alternators and in most small alternators are wound with field coils and fed DC current through slip rings. The generated voltage can be controlled by varying the field current (and hence the magnetic field strength). The DC current for the rotor may be produced by ­rectifying a portion of the armature output current. This method is referred to as self-excitation. Large alternators in power stations may use a separate DC generator called an exciter to provide the DC current for the field. The exciter can be mounted on the same shaft as the alternator. The exciter usually has a stationary field and a rotating armature, which allows a connection without brushes between the exciter and the rotor of the alternator. See Problem 29-30 and Review Questions 29-66 to 29-69.

929

930

Chapter 29   Three-Phase Systems

29-12  Three-Phase Induction Motor The induction motor is the most common motor for industrial applications. As the name implies, the magnetic field developed by the rotor is produced by magnetic induction instead of coils powered by a separate DC source. Since only AC power is required, an induction motor is referred to as a singly excited motor. The stator of the three-phase induction motor is identical to that of the three-phase alternator, and has a distributed winding wound for as many poles as the speed requires (Figure 29-38). Applying three-phase voltage to this armature produces a magnetic field that rotates around the stator at the synchronous speed. Rearranging equation 29-14 gives an expression for this speed: N=



120f P

(29-15)

N

3-Ph AC

S

Figure 29-38  Rotating field in a three-phase armature

Example 29-14 A three-phase induction motor has a six-pole stator and operates from a 50-Hz supply. Calculate the synchronous speed of this stator. Solution N= N=

120 f 120 × 50 Hz = 1000 RPM 6 P

29-12   Three-Phase Induction Motor

931

The rotor has a same laminated steel core with slots for the windings, like the rotor of an alternator. There are two types of rotor windings: squirrelcage and wound rotor (Figure 29-39). The squirrel-cage rotor is more ­common due to its simplicity and ruggedness. These rotors have copper or aluminum bars placed in the slots and welded to a heavy conducting ring at each end. The wound type has a conventional three-phase winding, normally wye connected with the three phases connected to slip rings. The slip rings are shorted under normal operation, but external resistance may be added to the rotor circuit to control the speed and to improve starting torque. Copper or aluminium rotor bar Sheet-steel laminated core

Three-phase rotor winding Slip ring

Shorting rings (a)

(b)

Figure 29-39  Induction motor rotors: (a) Squirrel-cage rotor; (b) Wound rotor

As the rotating stator field sweeps past the rotor, a voltage is induced in the rotor conductors, in accordance with Faraday’s law. Since the conductors are shorted, a current flows in them, producing a magnetic field and a torque that turns the rotor in the direction of the rotating stator field. The motor accelerates to a speed slightly less than synchronous speed. The

Source: GE

Source:  © Tektite/Dreamstime/Getstock

Single-phase induction motor featuring a squirrelcage rotor

Wound rotor

932

Chapter 29   Three-Phase Systems

motor can never reach synchronous speed since the force developed depends on the motion of the stator field relative to the rotor. This relative speed is referred to as the slip speed, NS. NS = N − NR



(29-17)

where N is the synchronous speed and NR is the rotor speed. Slip speed is normally expressed as a percentage of the synchronous speed, and is called simply slip or percent slip, S. S=



Ns N − NR × 100% = × 100% N N

(29-18)

Example 29-15 A 60-Hz three-phase squirrel-cage induction motor has a four-pole stator. When this motor is fully loaded, the slip is 5%. Calculate the full-load speed. Solution

N =



S=

120f P

=

120 × 60 Hz = 1800 RPM 4

NS → NS = S × N = 0.05 × 1800 = 90 RPM N

NR = N−NS = 1800−90 = 1710 RPM

The three-phase induction motor is the workhorse of industry. This motor is suitable for a variety of applications because of its simple, rugged design, good speed regulation, and high starting torque. See Problems 29-31 and 29-32 and Review Questions 29-70 to 29-72.

29-13  Three-Phase Synchronous Motor A synchronous motor is constructed just like an alternator and can be operated as an alternator. The rotor may be salient pole or cylindrical, and contains the field system, which is energized by direct current, as shown in Figure 29-40. As for the AC generator, the DC excitation is produced by a separate exciter or DC power supply.

29-13   Three-Phase Synchronous Motor

N

Slip ring N

+

3-Ph AC

–

DC

supply

S

S

Brush

Figure 29-40  Synchronous motor

When energized, the stator produces a magnetic field that rotates at the synchronous speed, as in the induction motor. If the rotor is now energized, it becomes magnetized with north and south poles. However, the rotor has significant inertia and cannot respond instantaneously to the torque produced by the interaction with the magnetic field from the stator. At one instant, an attractive force is produced by a south pole sweeping by a north pole, but before the rotor moves noticeably, the stator field has rotated far enough to reverse the force. Thus, the rotor will remain at rest, and the synchronous motor is not self-starting. To start a synchronous motor, it must be brought up to near synchronous speed by an external means. The DC excitation is then switched on, and the rotor poles “lock in” with the stator field causing the rotor to turn at exactly synchronous speed. The synchronous motor is a constant-speed machine: it can operate only at the synchronous speed. There are various methods of bringing a synchronous motor up to near synchronous speed. The exciter used for the DC field can be run as a DC motor coupled to the shaft, or a small induction motor, called a pony motor, can be coupled to the rotor shaft. The most common method is to place a squirrelcage winding in the field pole faces. The motor starts as an induction motor, and then locks into synchronism when the rotor winding is energized. At synchronous speed, there is no relative motion between rotor and stator field, so there will be no current flowing in the squirrel-cage starting winding. The synchronous motor is particularly useful for industrial machines that require constant speed, such as paper mills. A unique characteristic of the synchronous motor is that its operating power factor can be changed by adjusting the current supplied to the rotor. The field current can be increased until the motor operates at unity power factor, thus improving efficiency and avoiding the excess line current caused by the lagging power factor of the induction motor. If the field current is increased further, the synchronous motor has a leading power factor and can be used to

933

934

Chapter 29   Three-Phase Systems

improve the overall power factor of the electrical system and still deliver mechanical power. This mode of operation is called overexcitation. In large power generating stations, unloaded, overexcited synchronous motors are used to improve the station power factor. When used in this manner such motors are referred to as synchronous capacitors. See Review Questions 29-73 to 29-75.

29-14  Single-Phase Motors Three-phase motors are usually found in medium to large industries where three-phase power is readily available. In residences and smaller industries, single-phase motors are used in a variety of applications, such as washers, dryers, refrigerators, dishwashers, hair dryers, mixers, and blenders. Most single-phase motors are rated at less than 1 hp, but they are manufactured in sizes up to 10 hp. The majority of single-phase motors are induction motors with a squirrelcage rotor similar to the three-phase induction motor. The stator contains a single-phase winding in slots around the inside surface. When singlephase AC is applied to the stator winding, it produces a pulsating magnetic field that varies sinusoidally, rather than rotating like the field in the three-phase motor. As a result, no net torque is developed when the rotor is stationary. However, if the rotor is rotating, the motor will produce torque that continues the rotation. To make a single-phase motor self-starting, a starting winding is added to the stator in the same slots as the running winding. In order to start the motor, the current in the starting winding must be displaced electrically from the current in the running winding. This displacement produces a two-phase revolving field, which develops enough torque to start the rotor from rest. Once the rotor is rotating, a centrifugal switch opens and disconnects the starting winding (Figure 29-41). These types of motors are often referred to as split-phase motors. Centrifugal switch

Running winding E

Rotor

Starting winding

Figure 29-41  Schematic of a single-phase induction motor

29-15   The 30° Difference between Delta-Wye Configurations

The method of obtaining the phase split gives rise to different types of single-phase induction motors. The resistance-start motor uses a starting winding made with fewer turns and thinner wire than the running winding. Then, the starting winding has high resistance and low reactance while the running winding has low resistance and high reactance, giving the necessary phase split to start the motor. The capacitor-start motor uses a capacitor in series with the starting winding to obtain the phase split. The capacitor and starting winding are both switched out by the centrifugal switch when the motor reaches approximately 75% of running speed. Since the phase shift between starting and running currents is greater than for the resistance-start motor, the capacitor-start motor develops more starting torque and can be manufactured in larger sizes. In a capacitor-run motor, the capacitor is left in the circuit, and the motor runs as a two-phase motor. This configuration has the advantage of much quieter operation than resistance and capacitorstart motors, which tend to be noisy due to the pulsating torque. There are many other types of single-phase motors used in small appliances and power tools. The universal motor, which is essentially a DC series motor, is used in many power tools. This motor can operate on either AC or DC power. The shaded-pole motor is used in small appliances and fans. This motor has a loop of heavy wire around a portion of each pole on the stator, causing the magnetic flux to lag behind the flux in the unshaded portion of the pole, and thus producing the phase split necessary to start the motor. Small single-phase synchronous motors are used for timing devices and clocks. See Review Questions 29-76 and 29-77.

29-15  The 30° Difference between Delta-Wye Configurations Let’s consider a four-wire system with a wye-connected load as shown in Figure 29-42(a). The phasor diagram of the respective load voltages is shown in Figure 29-42(b). N A B C VC

120 V 120° + VA

VB

+

− 120 V 0° − − VC +

120 V −120°

120 V 120°

VA

120 V 0°

VB

120 V −120° (a)

(b)

Figure 29-42  Four-wire system with a wye-connected load

935

Chapter 29   Three-Phase Systems

A delta-connected load is added to the same four-wire system, as shown in Figure 29-43(a). N A B C

+ VA

− 120 V 0° − − V C +

− −

120 V 120°

120 V −120°

VCA

−

+

VAB

+

VB

+

936

+ VBC

(a)

Figure 29-43  Four-wire system with wye- and delta-connected loads

The phase voltages will be determined by calculating the difference between the voltages of two lines: VAB = VA − VB = 120 V∠0º − 120 V∠−120º = 208 V∠30º

VBC = VB − VC = 120 V∠−120º − 120 V∠120º = 208 V∠−90º VCA = VC − VA = 120 V∠120º − 120 V∠0º = 208 V∠150º

These phasor voltages will appear as shown in Figure 29-44(a). When they are re-drawn as shown in Figure 29-44(b), we can see that between the line voltages of a wye-connected load and the phase voltages of a deltaconnected load, we have a 30° difference. VCA

VC

208 V 150°

VCA VAB

VA

o

VBC

VC 120 V 120° 30°

VAB 208 V 30°

30° VA 120 V 0° VB 30°

120 V −120° VBC 208 V −90°

VB (a)

(b)

Figure 29-44  Phasor diagrams for a four-wire system with wye- and deltaconnected loads

Summary

Summary

• A two-phase alternator with two coils rotating 90° apart in a magnetic field produces two voltage sine waves, one of which leads the other by 90°. • The instantaneous power output of a two-phase alternator with identical loads in each phase is constant. • A set of perpendicular coils connected to a two-phase alternator produces a magnetic field with a constant flux density rotating at the frequency of the applied voltage. • The three-wire single-phase system is widely used for distributing electrical power for residential use. • A three-phase alternator has three coils rotating 120° apart in a magnetic field. • The neutral current in a three-phase system with a balanced load is zero. • The instantaneous power output of a three-phase alternator with identical loads in each phase is constant. • Double-subscript notation is used in three-phase systems to denote voltage rises and voltage drops. • In a four-wire wye-connected system, the three alternator coils and the neutral have a common connection point. • In a balanced four-wire wye-connected system, the magnitude of each line current is the magnitude of the phase voltage divided by the magnitude of the impedance from the line to the neutral. • In a delta-connected system, the alternator coils are connected in series. • In a balanced delta-connected system, the line current is √3 times the phase current. • A wye-delta system consists of a wye-connected source and a ­delta-connected load. • In a wye-delta system, each load impedance appears across a line voltage of the alternator. • In a balanced three-phase system, the total power is three times the power in each phase. • Phase sequence indicates the order in which the instantaneous phase voltages reach a given point in their cycles. • Phase sequence determines the direction of rotation of a three-phase motor, as well as the division of line currents for an unbalanced load. • In a three-phase system with an unbalanced three-wire wye load, the phase voltages are unequal because the wye junction of the load is floating. • An AC generator has the same basic components as a DC generator with slip rings instead of a commutator. • The frequency of an AC generator is as important as the voltage. • On large alternators the armature is stationary and the field rotates. • A distributed winding is more efficient than a concentrated winding. • The three-phase induction motor is the most common industrial motor. • In an induction motor, the rotor is energized by induction and does ­require connection to a power supply. • The two types of rotors for induction motors are squirrel cage and wound.

937

938

Chapter 29   Three-Phase Systems

• Slip speed is the difference between synchronous and rotor speed. • A synchronous motor is identical to the alternator and has constant speed. • An overexcited synchronous motor will act like a capacitor. • Most single-phase motors are split-phase induction motors. • Single-phase induction motors require a starting winding. • A variety of single-phase motors are used in industrial and residential applications. • In a four-wire system with wye- and delta-connected loads, the line voltages of the wye-connected load and the phase voltages of the deltaconnected load will always be 30° apart.

B = beginner

I = intermediate

A = advanced

Problems I

29-1. Draw a graph of the total instantaneous power if one of the resistors in Figure 29-2 is changed to 180 Ω.

Section 29-4  Four-Wire Wye-Connected System

B

29-2. A 60-Hz 3ϕ four-wire wye-connected source with EoA = 120 V∠0º, EoB = 120 V∠+120º, and EoC = 120 V∠−120º is connected to a balanced four-wire wye-connected load consisting of three 10-Ω ­resistors. (a) Draw a fully labelled schematic diagram showing tracing ­arrows. (b) Calculate the magnitude and phase of the current in each line. (c) Draw a phasor diagram of the phase voltages and currents. (d) Calculate the neutral current. (e) Use Multisim to verify the magnitudes of the line currents in part (b) and the neutral current in part (d). 29-3. Repeat Problem 29-2 with a balanced four-wire wye-connected load, each arm of which has a power of 500 W with a 90% lagging power factor. What will the neutral current be if line A becomes open-­circuit? 29-4. Repeat Problem 29-2 with a four-wire wye-connected load in which ZA = 30 Ω ∠+30º, ZB = 40 Ω ∠−30º and ZC = 50 Ω ∠−90º. 29-5. Repeat Problem 29-2 with an unbalanced four-wire wye-connected load in which ZA is a 50-Ω heater, ZC is a 5-µF capacitor, and ZB is a 0.10-H inductance with a resistance of 8 Ω. 29-6. A three-phase, four-wire, balanced wye-connected source has phase voltages of 200 V. The source supplies a wye-connected load consisting of a 50-Ω resistor in phase A, an impedance of 50 Ω ∠45º in phase B, and an impedance of 25 − j30 Ω in phase C. Draw the circuit diagram and calculate the line currents. 29-7. Calculate the neutral current for the circuit in Problem 29-6.

I

29-8.

I

circuitSIM walkthrough

Section 29-1  Advantages of Polyphase Systems

I I I I

Section 29-5  Delta-Connected System A delta-connected load consisting of three 11-Ω resistors is connected to a 60-Hz 3ϕ delta-connected source in which ECA = 550 V∠0º, EAB = 550 V∠+120º, and EBC = 550 V∠−120º. (a) Draw a fully labelled schematic diagram with tracing arrows. (b) Calculate the phase currents.

Problems

I I I I B A

I I I A I

B

B B B B

(c) Draw a phasor diagram of the phase voltages and currents. (d) Determine the line currents by geometric construction on a ­phasor diagram. (e) Use phasor algebra to calculate the line currents. (f) Use Multisim to verify the magnitudes of the phase currents in part (b) and the line currents in parts (d) and (e). 29-9. Repeat Problem 29-8 with a balanced delta load that has a total power of 30 kW and a 0.85 leading power factor. 29-10. Repeat Problem 29-8 with a delta load in which ZAC = 5 Ω ∠+45º, ZCB = 10 Ω ∠−36.9º, and ZBA = 11 Ω ∠0º. 29-11. Repeat Problem 29-8 with a delta-connected load in which ZCB = 15 kΩ, ZAC is a 25-H coil with an internal resistance of 10 Ω, and ZBA is 1-μF capacitor. 29-12. A 360-V, delta-connected, three-phase source powers a delta-­ connected load that has ZA = 150 ∠20º Ω, ZB = 100 ∠−45º Ω, and ZC = 200 ∠150º Ω. Calculate the load currents and line currents. 29-13. Sketch the phasor diagram for the circuit of Problem 29-12.

939

circuitSIM walkthrough

Section 29-6  Wye-Delta System 29-14. The load of Problem 29-8 is connected to the source of Problem 29-2. (a) Draw a fully labelled schematic diagram with tracing arrows. (b) Determine the line voltages by phasor construction. (c) Use phasor algebra to calculate the line voltages. (d) Calculate the phase currents. (e) Calculate the line currents. (f) Draw a complete phasor diagram for the circuit. (g) Use Multisim to verify the magnitudes of the line voltages in parts (b) and (c), the phase currents in part (d), and the line currents in part (e). 29-15. Repeat Problem 29-14 with EoA = 120 V∠0º, EoB = 120 V∠120º, and EoC = 120 V∠−120º. 29-16. (a) Repeat Problem 29-14 with the load of Problem 29-10. (b) Use Multisim to verify the magnitudes of the line voltages, the phase currents, and the line currents. 29-17. Repeat Problem 29-14 with the load of Problem 29-11. 29-18. (a) Repeat Problem 29-2 with the source of Problem 29-8. 29-19. A 208-V 60-Hz wye-connected source feeds a balanced deltaconnected load consisting of a 100-Ω resistor and a 75-μF capacitor in series on each arm. Calculate the load currents and the generator phase currents. 29-20 Draw a phasor diagram showing load voltages and currents for the circuit in Problem 29-19.

Section 29-7  Power in a Balanced Three-Phase System 29-21. 29-22. 29-23. 29-24.

Find the total power input to the load in Problem 29-2. Find the total power input to the load in Problem 29-8. Find the total power input to the load in Problem 29-4. Find the total power input to the load in Problem 29-10.

circuitSIM walkthrough

circuitSIM walkthrough

940

Chapter 29   Three-Phase Systems

B B A A A

B

B B

Section 29-8  Phase Sequence 29-25. Determine the phase sequence of the source in Problem 29-2. 29-26. Determine the phase sequence of the source in Problem 29-8.

Section 29-9  Unbalanced Three-Wire Wye Loads 29-27. Determine the current in each line in Problem 29-4 with the neutral lead open-circuit. 29-28. Determine the three line currents in Problem 29-27 if the phase sequence of the source is reversed. 29-29. The load of Problem 29-10 is connected to a 208-V 3ϕ source with a CBA phase sequence through three conductors, each having an impedance of 2 + j1 Ω. Determine the magnitude of the current in each line.

Section 29-11  The AC Generator

29-30 How many poles must a salient-pole alternator contain to generate 50-Hz voltage when turning at 500 RPM?

Section 29-12  Three-Phase Induction Motor 29-31 What is the synchronous speed of a 12 pole squirrel-cage induction motor when connected to a 208-V 60-Hz source? 29-32 The induction motor of Problem 29-31 is fully loaded and runs at 560 RPM. Calculate its full-load slip.

Review Questions

Section 29-1  Advantages of Polyphase Systems 29-33. What are the three main advantages of a polyphase power distribution system compared to a single-phase system? 29-34. Why is it desirable for the total instantaneous power input to a polyphase system to be constant? 29-35. Draw vectors for the magnetic flux in Figure 29-5 at 45° intervals for the rest of the cycle. 29-36. Why does the flux density of the rotating magnetic field in Figure 29-5 remain constant? 29-37. What is the meaning of the term synchronous speed? 29-38. How could we reverse the direction of rotation of the magnetic field in Figure 29-5? 29-39. Derive an equation to show the relationship between EA and the voltage between line A and line B in the two-phase system of Figure 29-2(b).

Section 29-2  Generation of Three-Phase Voltages 29-40. Why is the Edison three-wire system of Figure 29-6 unable to produce a rotating magnetic field? 29-41. What is the advantage of the Edison three-wire system for electric power distribution? 29-42. What is a balanced load in a polyphase system? 29-43. Explain how a balanced load can have a power factor of less than 1. 29-44. Draw a graph to show why the total instantaneous power input to a balanced three-phase load is constant.

Review Questions

29-45. Why are the currents in the three arms of a balanced three-phase load equal in magnitude and displaced 120° in phase from one another regardless of the power factor of the load? 29-46. Show (with diagrams) how a rotating magnetic field can be produced using a three-phase source.

Section 29-3  Double-Subscript Notation 29-47. Why is double-subscript notation used to represent voltages in a three-phase system?

Section 29-4  Four-Wire Wye-Connected System 29-48. Draw a phasor diagram for a wye-connected source in which the uppercase letter ends of the windings in Figure 29-11 are connected to the neutral. 29-49. Three transformers fed from a three-phase source have the following secondary voltages: EaA = 240 V∠0º , EbB = 240 V∠−120º , and EcC = 240 V∠+120º. Each secondary is centre-tapped, and these centre taps are connected to a common neutral, as shown in Figure 29-45. Draw a fully labelled phasor diagram for the resulting s­ tar-connected six-phase source. C

b

o

a

B

A

c

Figure 29-45  Star connection of three-phase transformer secondary windings

29-50. Describe how to use an AC voltmeter to test for an incorrect wye ­connection of a three-phase source.

Section 29-5  Delta-Connected System 29-51. What is meant by the terms phase current and line current in a threephase system? How do these terms apply to (a) a wye-connected load (b) a delta-connected load 29-52. What test should you perform when connecting the coils of a threephase source into a delta? 29-53. Use a phasor diagram to show the line currents to a balanced delta load are greater than the phase currents.

Section 29-6  Wye-Delta System 29-54. What is meant by the terms phase voltage and line voltage in a threephase system? How do these terms apply to (a) a wye-connected source (b) a delta-connected source

941

942

Chapter 29   Three-Phase Systems

29-55. Use a phasor diagram to show that the line voltage from a ­wye-connected source is √3 times the voltage generated in each winding and displaced 30° from the voltages in the windings.

Section 29-7  Power in a Balanced Three-Phase System

29-56. Explain why IL = √3 Ip in a delta-connected load only if the load is balanced load while VL always equals √3 Vp in a wye-connected source. 29-57. What is the advantage of expressing power input to a balanced three-phase load in terms of line voltage and line current rather than phase voltage and phase current?

Section 29-8  Phase Sequence 29-58. Explain how reversing the tracing arrows when writing the voltage subscripts for the wye-connected source of Figure 29-22 will affect the phase sequence of the source. 29-59. Why should we read clockwise around a phasor diagram when determining the phase sequence?

Section 29-9  Unbalanced Three-Wire Wye Loads 29-60. Why must the phasor sum of the three line currents in any threephase three-wire system be zero? 29-61. Explain the principle of operation of the phase-sequence tester of Figure 29-34(a). 29-62. Account for the changes in load phase voltages when the neutral lead becomes open-circuit in Problem 29-27. 29-63. What is meant by a floating wye point in a three-phase system? 29-64. Derive Equation 29-12B from the basic Kirchhoff’s law equations. 29-65. We can solve three-phase networks with the superposition theorem. The individual calculations are simpler, but there are many repetitive component-current calculations. However, we must use phase voltages for the source rather than line voltages. Using ­diagrams, ­explain this restriction.

Section 29-11  The AC Generator 29-66. Why do alternators driven by water turbines require many poles to generate the required frequency? 29-67. Why is a distributed armature winding used instead of a concentrated winding? 29-68. What advantages does a stationary winding have? 29-69. Describe the methods of field excitation used in a large alternator.

Section 29-12  Three-Phase Induction Motor 29-70. Describe the two types of rotors used in an induction motor. 29-71. How is torque produced in the rotor of an induction motor? 29-72. What is the slip speed of an induction motor?

Practice Quiz

Section 29-13  Three-Phase Synchronous Motor 29-73. Why is the synchronous motor not self-starting? 29-74. Describe the methods used to bring a synchronous motor up to near synchronous speed. 29-75. What unique feature does the synchronous motor have and how is it used?

Section 29-14  Single-Phase Motors 29-76. Explain how starting is accomplished in split-phase induction motors. 29-77. What advantage does a capacitor-run motor have over the other split-phase motors?

Integrate the Concepts

A three-phase alternator with windings that each produce 550 V is connected to a three-phase load with an impedance of 100 Ω and a lagging power factor of 0.95 in each arm. (a) Calculate the line currents if the alternator and load form a four-wire wye-connected system. (b)  Calculate the line currents if the alternator and load form a ­delta-connected system. (c) Calculate the magnitude of the line currents if the alternator and load form a wye-delta system. (d) Calculate the total power input to the system of part (a). (e) Calculate the line currents when the phase sequence for the system in part (b) is reversed.

Practice Quiz

1. Which of the following statements are true? (a) The total instantaneous power of a polyphase source will be constant if the load is balanced. (b) A balanced load must be purely resistive. (c) Wye-connected systems are named so for their Y-shaped configuration. (d) Delta-connected systems are named so for their triangular shape, which looks like the Greek letter delta (Δ). (e) In a balanced delta-connected system, the line current is √2 times

the phase current. (f) The phase sequence of a three-phase system governs the direction of rotation of three-phase motors, as well as the division of line currents for an unbalanced load.

2. To supply a given power at a specific voltage, a polyphase system ­requires ______________ copper wire than a single-phase system. (a) the same amount of (b) more (c) less (d) either more or less depending on the load

943

Chapter 29   Three-Phase Systems

3. The phase angle between the sine waves generated by a three-phase alternator is (a) 90° (b) 120° (c) 180° (d) 270° 4. What is the current in line AB for the three-phase system shown in Figure 29-46 if the balanced load equals 330 Ω ∠30º? (a) Zero (b) 364 mA∠90º (c) 364 mA∠−150º (d) 364 mA∠−30º C

b

IC

+

–

C

B

ZBC

ZCA

c –

+

944

– a

A

+ A

ZAB

IA B IB

Figure 29-46

5. What is the total power input to the load in Figure 29-46? (a) 21.84 W (b) 65.52 W (c) 37.83 W (d) 52.65 W 6. If the three-phase system shown in Figure 29-47 has a balanced load, the neutral current will be (a) IB − IC (b) IA − IC (c) IA − IB (d) Zero B IB

C EC

+ EB –

+

–

– EA

N

IC

IN

Z1

+ A

Figure 29-47

Z2

Z3

IA

Practice Quiz

7. What is the current in line C of Figure 29-47 if Z1 = Z2 = Z3 = 500 Ω ∠75º and the phase voltages are EA = 120 V ∠0º, EB = 120 V ∠120º and EC = 120 V ∠−120º? (a) Zero (b) 240 mA ∠−75º (c) 240 mA ∠95º (d) 240 mA ∠45º 8. A 208-V three-phase motor delivers 15 kW to a mechanical load. If the motor has an efficiency of 90% lagging and an impedance of 500 Ω ∠30º, what is the line current? (a) 80.19 A (b) 46.26 A (c) 83.34 A (d) 48.10 A 9. The total power of a balanced three-phase system is (a) three times the power in each phase and directly proportional to sin ϕ (b) three times the power in each phase and directly proportional to cos ϕ (c) one third the power in each phase and directly proportional to sin ϕ (d) one third the power in each phase and directly proportional to cos ϕ 10. In a three-wire unbalanced wye-load connection, the phase voltages are (a) equal (b) unequal (c) zero (d) either equal or unequal depending on the load 11. How fast must a 12-pole alternator rotate to generate 50-Hz voltage? (a) 5.0 RPM (b) 500 RPM (c) 600 RPM (d) 5000 RPM 12. What is the percent slip of a 12-pole induction motor if it operates at 1128 RPM when connected to a 60-Hz source? (a) 0.06% (b) 6.0% (c) 6.4% (d) 94%

945

30

Harmonics

Our analysis of electrical systems concludes with a look at circuits in which currents and voltages consist of multiple sine waves with different frequencies.

Chapter Outline 30-1 Nonsinusoidal Waves  948 30-2 Fourier Series  949

30-3 Addition of Harmonically Related Sine Waves  951

30-4 Generation of Harmonics  954

30-5 Harmonics in an Amplifier  956

30-6 Harmonics in an Iron-Core Transformer  958 30-7 RMS Value of a Nonsinusoidal Wave  960

30-8 Square Waves and Sawtooth Waves  961 30-9 Nonsinusoidal Waves in Linear Impedance Networks  963

Key Terms fundamental frequency 948 harmonic  948

sawtooth wave 950 pulsating DC waveform 950

half-wave-rectified sine wave  955 square wave  961

Learning Outcomes At the conclusion of this chapter, you will be able to: • recognize a harmonic in a nonsinusoidal waveform • write the general equation for the Fourier series of a periodic waveform • draw a complex wave given the equations of the fundamental and harmonics • determine the harmonic content of a halfwave-rectified sine wave • explain how even-order harmonics are produced in the output signal of an amplifier

Photo sources:  Cole-Parmer Instrument Company

• explain how odd-order harmonics are produced in an iron-core transformer • calculate the RMS value of an AC current or vol­tage containing harmonics • determine the harmonic content of square waves and sawtooth waves • determine voltage and current waveforms when a n ­ onsinusoidal wave is applied to a linear impedance ­network

Chapter 30  Harmonics

30-1  Nonsinusoidal Waves So far in our study of AC circuits, we have assumed that the sources produce sine waves. For the distribution of electric power, sine waves of ­voltage and current are desirable, and in designing AC machinery, we strive to obtain such waveforms. In practice, however, it is not always possible to produce an alternating source voltage that has a perfectly sinusoidal form. For example, in the simple a­ lternator of Chapter 18, a sine wave of induced voltage requires a uniform magnetic field. If the flux density is not uniform, the induced voltage is not a true sine wave. We can also generate special nonsinusoidal waveforms such as square waves, sawtooth waves, and short-duration pulses for radar, television, and computer systems. Cycle after cycle, the instantaneous value of these waves has exactly the same shape as the preceding cycle. Hence, these geometric waveforms are also periodic waves. In audio equipment, the waveforms are seldom sinusoidal. If we play a single note on a musical instrument and examine the waveform of the instantaneous voltage this note produces in a microphone, we see a complex nonsinusoidal alternating voltage, as shown by the blue waveform in Figure 30-1. Since this waveform repeats exactly the same sequence of instantaneous values cycle after cycle, it too is a periodic wave.

Instantaneous Value

948

+

0 Time

− Figure 30-1  A nonsinusoidal periodic waveform

Examining the complex wave of Figure 30-1 closely, we notice a component, somewhat sinusoidal in shape, that seems to go through five complete cycles during one cycle of the overall waveform. This pattern is more noticeable when we draw the sine wave that cuts across the complex wave halfway between the peaks of the minor excursions of the instantaneous voltage. It seems then that there are at least two distinct frequencies present in this nonsinusoidal waveform. One is the fundamental frequency at which the overall complex waveform repeats itself. This fundamental frequency is the lowest frequency present in the complex wave. The other ­predominant frequency appears to be exactly five times the fundamental frequency. This frequency is called the fifth harmonic of the fundamental ­frequency. Each harmonic must be an integral multiple of the fundamental frequency, as

30-2   Fourier Series

2.5 ms

Instantaneous Values

+ 0

0°

360°

Fundamental = 400 Hz

− + 0

0°

Second harmonic = 2 × 400 = 800 Hz

360°

− + 0

0°

Third harmonic = 3 × 400 = 1200 Hz

360°

− Figure 30-2  Harmonically related sine waves

shown in Figure 30-2. All waves that maintain the same complex waveform cycle after cycle contain harmonics. Every nonsinusoidal periodic wave is made up of a series of harmonically related pure sine waves. See Review Questions 30-15 and 30-16 at the end of the chapter.

30-2  Fourier Series To determine the behaviour of AC circuits in the presence of nonsinusoidal waveforms, we must find the fundamental frequency, the harmonics pre­ sent, their relative amplitudes, and their phase relationship with the funda­ mental sine wave. One method for determining the harmonic content of a nonsinusoidal wave is based on mathematical analysis developed by the French mathematician Jean-Baptiste Joseph Fourier in 1822. Fourier demonstrated that any periodic complex waveform is composed of a fixed term plus an infinite series of harmonically related cosine terms and an infinite series of harmonically related sine terms. If the instantaneous magnitude of the nonsinusoidal waveform is represented as a function of time by the mathematical expression f (t), the general equation for the Fourier series is f (t) = A0 + A1 cos ωt + A2 cos 2ωt + A3 cos 3ωt + . . .       + An cos nωt + . . . + B1 sin ωt + B2 sin 2ωt       + B3 sin 3ωt + . . . + Bn sin nωt + . . .

∞

= A0 + ∑ ( An cos nωt + Bn sin nωt )

(30-1)

n=1

where A0 is a steady-state term representing the average value or DC ­component of the complex wave.

949

Chapter 30  Harmonics

+

0 Time, t

A0 = 0

−

Instantaneous Value

The coefficient of each cosine or sine term can have any value. However, for most nonsinusoidal waveforms, the magnitudes of the coefficients diminish rapidly as the harmonic number increases. Often it is only necessary to include the first few sine and cosine terms of the infinite series to determine the shape of the complex wave. As in Figure 30-1, we can represent a waveform f (t) by a graph with the y‑axis representing instantaneous value and the x-axis representing time. Figure 30-3(a) shows the graph of a sawtooth wave in which the area between the x-axis and the positive portion of the wave graph equals the area between the x-axis and the negative portion of the wave. For this sawtooth waveform, the average value, A0, is zero. The waveform is an AC waveform since it has no DC component. The sawtooth waveform in Figure 30-3(b) has the same shape, but no negative instantaneous values. For this pulsating DC waveform, A0 equals half of the peak instantaneous DC value of the wave.

Instantaneous Value

950

+

A0

0

Time, t

A0 = 0.5 A

−

(a)

(b)

Figure 30-3  Average or DC component of a nonsinusoidal wave

A graph of cos ωt from −π/2 to +π/2 rad is symmetrical about the vertical axis at t = 0, as shown in Figure 30-4(a). The graph of sin ωt, on the other hand, is not symmetrical about this axis since sin ωt is positive from 0 to +π/2 rad, but negative from –π/2 to 0 rad. Consequently, if the graph of a nonsinusoidal wave is symmetric about the t = 0 axis, all the coefficients +

+

−π

π 2

−π 2

π

ωt

−π

−π 2

− cos ωt (a) Figure 30-4  Symmetry of cosine and sine waves

π 2

− sin ωt (b)

π

ωt

30-3   Addition of Harmonically Related Sine Waves

Instantaneous Value

π

π

2π ωt

−2π

Instantaneous Value

+

+

−2π

951

π 2

− t=0 Bn = 0 for all n

− t=0 An = 0 for all n

(a)

(b)

2π ωt

Figure 30-5  Effect of a phase shift on Fourier series for a square wave

of the sine terms in Equation 30-1 must be zero. Similarly, if the shape of the waveform is inverted on opposite sides of the vertical axis at t  = 0, like the sine wave in Figure 30-4(b), all the coefficients of the cosine terms must be zero. For example, the square wave in Figure 30-5(a) has no cosine components, but the same wave shifted by π/2 rad in Figure 30-5(b) has no sine components. The sawtooth wave in Figure 30-3(a) contains both sine and cosine components since it is not symmetric about any vertical axis. The Fourier coefficients for a periodic waveform can be determined using calculus. See Review Questions 30-17 and 30-18.

30-3 Addition of Harmonically Related Sine Waves A second method for determining the harmonic content of a nonsinusoidal wave is based on a reciprocity principle: If a nonsinusoidal wave can be duplicated by adding together harmonically related sine waves with certain magnitudes and phase ­relationships, then the complex wave contains this same series of harmonically related sine waves. We can see the result of adding harmonics to a fundamental sine wave by adding instantaneous values on a graph. Figure 30-6 shows a fundamental sine wave and a fifth harmonic with 25% of the amplitude of the fundamental. This harmonic is in phase with the fundamental because both sine waves start their cycles at the origin of the graph (that is, the i­ nstantaneous values of both waves are zero and increasing at t = 0). If the harmonic is out of phase with the fundamental by a half-cycle of the harmonic (π rad of the phase angle of the harmonic), we simply invert the harmonic waveform on the graph. The harmonic still has an instantaneous value of zero

Some music ­synthe­sizers use this principle to ­duplicate the sound of various ­musical ­instruments.

Chapter 30  Harmonics

eT = e1 + e5 + Instantaneous EMF

952

0

Fundamental Fifth harmonic Time, t

− Figure 30-6  Sum of a fundamental sine wave and a fifth harmonic

at the origin, but this is now decreasing while the instan­taneous value of the fundamental increases. If the harmonic lags behind the fundamental by a quarter-cycle (π/2 rad of the harmonic phase angle), we must draw the harmonic such that it reaches a negative peak at the ­instant that the fundamental starts its cycle. Adding the instantaneous values of the two sine waves in Figure 30-6 from instant to instant produces a waveform that duplicates the complex waveform in Figure 30-1. Since the fifth harmonic is an exact multiple of the fundamental frequency, the succeeding cycles have exactly the same shape as the cycle plotted in Figure 30-6. In Chapter 18, we described the shape of a sine-wave alternating voltage by writing the general equation for the sine-wave voltage: e = Em sin ωt

(18-4)



eT = e1 + e5 = Em ( sin ωt + 0.25 sin 5ωt )

(30-2)



eT = Em ( sin ωt − 0.25 sin 5ωt )

(30-3)



By substituting suitable values for Em and ω, we were able to determine the amplitude and frequency of the alternating voltage as well as the shape of its waveform. Similarly, we can write an equation to define the shape of a complex wave. Since we formed the complex wave in Figure 30-6 by adding the instantaneous values of two sine waves, the equation for the complex wave is the sum of the equations for the two sine-wave components: If the harmonic were π rad out of phase with the fundamental, the ­ olarity of all the instantaneous values of the harmonic sine wave would p reverse. The complex wave would then have a markedly different shape that is described by the equation The cosine is 1 at 0 rad and decreases to 0 at π/2 rad, whereas the sine is 0 at 0 rad and increases to 1 at π/2 rad. A cosine wave has the same shape as a sine wave but leads the sine wave by π/2 rad (since the cosine reaches

30-3   Addition of Harmonically Related Sine Waves

its maximum value π/2 rad ahead of the sine). Thus, cos θ = sin (θ + π/2). A cosine function can be a convenient way to express a π/2-rad phase difference between a harmonic and the fundamental when we write ­ equations for complex waves. If the negative peak of the fifth h ­ armonic in Figure 30-6 coincides with the start of the cycle of the fundamental, the equation describing the complex wave becomes eT = Em ( sin ωt − 0.25 cos 5ωt )



(30-4)

In practice, most nonsinusoidal waves do not consist of a fundamental and a single harmonic, as in Figure 30-6. Most complex waves consist of a series of harmonically related sine waves with the amplitude of each harmonic diminishing as the multiple of the fundamental frequency increases. However, we can get a fair approximation of the shape of the complex waveform by adding together the instantaneous values of the first few ­harmonics in the series. There are two series of harmonics that govern the general appearance of nonsinusoidal waves. Complex waves in which the positive and negative half-cycles are symmetrical, like the complex wave in Figure 30-1, are composed mainly of the odd-order harmonics (third, fifth, seventh, ninth, and so on). The even-order series of harmonics produces a symmetrical complex waveform if each even harmonic is either exactly in phase or π rad out of phase with the fundamental. When a complex wave is composed of an even-order series of harmonics, one or more of which is not exactly in phase or π rad out of phase with the fundamental, the positive and negative half-cycles of the nonsinusoidal wave can have quite ­different shapes.

Example 30-1

Determine the shape of the complex wave defined by e = 60 sin 2512t + 20 cos 7536t. Solution

Since ω = 2πf, For the fundamental, For the harmonic,

f=

f=

f=

ω 2π

2512 = 400 Hz 6.28

7536 = 1200 Hz ( third harmonic ) 6.28

Draw the fundamental sine wave on graph paper. Dividing each quarter-cycle into thirds (30° intervals) makes it easier to position

953

Chapter 30  Harmonics

the third ­harmonic properly. Draw the harmonic sine wave with 20/60, or 1⁄3, of the amplitude of the fundamental. Since the cosine has a positive coefficient, the harmonic has a positive peak at the start of the fundamental cycle. Add the instantaneous values of the two components’ waves every 10° or so, and plot the sums. Join the resulting points to show the waveform of the complex wave, as in Figure 30-7. eT = e1 + e3 + Instantaneous EMF

954

e3 = 20cos 7536t 0

60°

120°

180°

e1 = 60 sin 2512t

240°

300°

360° Phase angle of fundamental

− Figure 30-7  Complex waveform for Example 30-1

See Problems 30-1 to 30-4 and Review Questions 30-19 to 30-25.

Circuit Check

A

CC 30-1. (a) Write an expression for the total voltage when a 12-V 60-Hz source is connected in series with a 5.0-V 240-Hz source. The two sources are in phase at t = 0. (b) Determine whether the waveform of the total voltage is symmetrical. CC 30-2. Graph the waveform defined by e = 10 sin t + 6 sin 2t.

30-4  Generation of Harmonics In stating the AC network theorems in Chapter 24, we assumed that the impedance of every circuit component remained constant throughout ­ the  cycle of the sine-wave voltage and current waveforms. Impedances that do not vary with voltage or current are linear, like the linear resistors ­described in Section 5-9. However, the circuit of Figure 30-8(a) contains a diode, which is a nonlinear circuit element. An ideal diode has no resistance to current through it in one direction but infinite resistance to current in the opposite direction. As a result, the instantaneous current in the resistor in Figure 30-8(a) has a sine-wave shape during the positive half-cycle but ­remains zero during the negative half-cycle. Therefore, although a sine wave is applied to the circuit of Figure 30-8(a), the output waveform is as

e = Em sin ωt

R

(a)

Instantaneous current

30-4   Generation of Harmonics

+

0

−

ωt

2π

π

(b)

Figure 30-8  Complex waveform created by a half-wave rectifier

shown in Figure 30-8(b). This particular nonsinusoidal wave is called a half-wave-rectified sine wave. We should be able to duplicate the half-wave-rectified output by adding a series of harmonics to the input sine wave. Since the first half-cycle is not at all symmetric with the second half-cycle, the harmonics are the evenorder series. These harmonics must have positive peaks that coincide with the positive peak of the fundamental. Therefore all the harmonics must lag behind the fundamental by π/2 rad, and the harmonics are represented by negative cosine terms. In the complex waveform of Figure 30-8(b), all the instantaneous v ­ alues are positive quantities. Since the half-cycle average of a voltage sine wave is 2Vm/π, the average of a complete cycle of the half-wave-rectified ­waveform is Em/π, or about 0.318 of the peak instantaneous value of the applied voltage. Since this average is not zero, the waveform has a DC component in addition to the harmonically related sine waves. This DC component, which results from the nonlinear resistance of the diode, is a key property of a practical rectifier circuit. The DC component c­ orresponds to the first term of the Fourier series. The complete Fourier series for the ­half-wave-rectified sine wave developed by an ideal rectifier is



e=

Em π 2 2 1 + sin ωt − cos 2ωt − cos 4ωt π 2 3 15

(

2 2 cos 2nωt ···   − cos 6ωt − · · · − ( 2n ) 2 − 1 35

)

(30-5)

where n is a natural number (1, 2, 3, and so on). Equation 30-5 shows an important characteristic of nonlinear circuits: Whenever an AC waveform is applied to a nonlinear circuit element, it produces harmonics not contained in the input waveform. See Problem 30-5 and Review Question 30-26.

955

Chapter 30  Harmonics

30-5  Harmonics in an Amplifier In an ideal transistor amplifier, any change in the input signal voltage produces a directly proportional change in the output current. For example, increasing the input voltage by 1.0 V from its quiescent (no-signal) value causes the output current to increase by exactly the amount as it decreases when the input voltage decreases by 1.0 V. Thus, a graph of the dynamic transfer characteristic (output current versus input voltage) of an ideal ­amplifier is a straight line, rather than the curve shown in Figure 30-9. But many amplifiers do have slightly curved transfer characteristics, like the one in Figure 30-9. Because of the nonlinear characteristic, a 1.0-V increase in the input voltage in Figure 30-9 produces a 2.7-mA change in output current, whereas a 1.0-V decrease produces only a 2.3-mA change. Consequently the output voltage is not a pure sine wave. Since the positive and negative half-cycles of the output waveform are somewhat asymmetric, the transistor has added a series of evenorder ­harmonics that are neither in phase nor π rad out of phase with the ­fundamental. The most significant of these harmonics is the second, and we can approximate the output waveform by superimposing a fundamental with an amplitude of 2.5 mA and a second harmonic with an a­ mplitude of 0.2 mA on the quiescent DC component, I0, as shown in ­Figure 30-10. We can determine the amplitude of the second harmonic from the minimum and maximum of the output current. Since the positive peaks Output Current

956

Dynamic transfer characteristic

Input Voltage

7.7 mA

0

Output

imax

5.0 mA

I0

2.7 mA

imin Time

1.0 V 1.0 V Input

Figure 30-9  Nonlinear transfer characteristic of an amplifier

30-5   Harmonics in an Amplifier

imax 7.7 7.5

Output current

Fundamental

Output Current (mA)

Second harmonic 5.2 5.0 4.8

I0

2.7 2.5

0

imin

Time

Figure 30-10  Second harmonic in output of an amplifier

of the second harmonic coincide with the positive and negative peaks of the ­fundamental, the peak-to-peak amplitude of the fundamental equals the peak-to-peak amplitude of the output current. The RMS value of a sine wave is 1/√ 2 of the peak value, and the peak value is half of the peak-to-peak ­amplitude of the sine wave. Therefore, Ifund =



imax − imin 2√ 2



(30-6)

As Figure 30-10 shows, the second harmonic increases the difference between imax and I0, but decreases the difference between imin and I0. Therefore, peak-to-peak amplitude of second harmonic =  (imax − I0) − (I0 − imin) = imax + imin − 2I0

and

I2nd =

imax + imin − 2I0 2√ 2



where I2nd is the RMS value of the second harmonic current. See Problem 30-6 and Review Questions 30-27 and 30-28.

(30-7)

957

958

Chapter 30  Harmonics

30-6 Harmonics in an Iron-Core Transformer The transfer characteristic for a practical iron-core transformer is the hysteresis loop of the iron. As a result, the actual exciting current has a slightly different appearance from that shown in Figure 30-11.

In Chapter 27, we noted that when we connect the primary of a transformer across a sine-wave source, the flux in the core must also be a sine wave. If the permeability of the iron core were constant, the exciting current in the primary winding would also be a sine wave. However, because of saturation in the iron, the permeability decreases as the magnetic flux density ­approaches its maximum value. As the permeability decreases, a greater instantaneous primary current is required to maintain the total flux sine wave. We can use the BH curve of Figure 30-11 as an approximate transfer curve to determine the waveform of the exciting current. The peaks of the exciting-current waveform have a narrower shape and greater amplitude than those of a sine wave. The two half-cycles of the exciting-current waveform have the same shape, so this waveform must contain odd-order ­harmonics. Since the third harmonic is the predominant harmonic of this series, we can approximate the waveform of the exciting current in an ironcore transformer with negligible hysteresis by adding a third harmonic to the fundamental, as shown in Figure 30-12. If we reduce the amount of iron in the core of a given transformer, the core is closer to saturation (further towards the ends of the BH curve) when the magnetic flux reaches its positive and negative peaks. Consequently, the amplitude of the third harmonic increases appreciably. In a 60-Hz single-phase system, this third harmonic (180 Hz) flows in the lines ­ ­connecting the transformer to the source. This harmonic current can induce audible 180 Hz voltages into nearby telephone lines. B+

H−

Time

+

Pure sine wave of flux

−

Exciting current

Figure 30-11  Producing a sine wave of flux in an iron core having negligible hysteresis

30-6   Harmonics in an Iron-Core Transformer

Instantaneous Current

Exciting current + Fundamental Third harmonic 0

Time

− Figure 30-12 Exciting current of an iron-core transformer with negligible hysteresis

Source:  Cole-Parmer Instrument Company

Three-phase power distribution systems can avoid this problem. If the primary windings of a three-phase transformer bank are connected as a delta, the line current is the difference between the phase currents. With a balanced load, the fundamental frequency components of the phase ­currents are 2π/3 rad (120°) out of phase, and so the fundamental current in the line is √ 3 Ip. However, a 2π/3-rad phase shift at the fundamental frequency is the same as a 2π-rad phase shift for the third harmonic. Therefore, the third harmonic components of the­

This harmonic analyzer can measure and display the relative magnitudes of harmonics in voltage and current waveforms.

959

960

Chapter 30  Harmonics

transformer primary phase c­ urrents cancel out in each line current. The third-­ harmonic c­urrent necessary to operate the iron-core transformer properly flows around the delta, but not in the transmission lines. ­However, the phase difference between the fifth-harmonic components of the primary phase currents is five times 2π/3 rad, which is the same as –2π/3 rad. Thus, there will be a fifth-­harmonic component in the line current with a magnitude √ 3 times the fifth-harmonic amplitude of the phase current. See Problem 30-7 and Review Questions 30-29 to 30-33.

30-7 RMS Value of a Nonsinusoidal Wave In Chapter 18, we used average power to derive the RMS value of a sine wave of voltage or current. Since a periodic nonsinusoidal wave consists of a fundamental sine wave plus a series of harmonic sine waves, we can apply the principle of the superposition theorem, and treat the harmonically related sine waves as all acting independently to contribute to the total power. Therefore, PT =



E21 E22 E23 . . . + + + R R R

where E1 is the RMS voltage of the fundamental, E2 is the RMS voltage of the second harmonic, and so on. Since each of the terms is based on a sine wave of voltage, the definition for RMS value can be applied to the total power. Hence, the RMS value of a nonsinusoidal voltage wave is

E = √ PTR = √ E21 + E22 + E23 + . . .

(30-8)

Similarly the RMS value of a nonsinusoidal current wave is IT = √I21 + I22 + I23 + . . .



(30-9)

Example 30-2 A three-phase wye-connected alternator has a phase voltage of 122 V and a line voltage of 208 V. The harmonics other than the third and fifth are ­insignificant. Find the RMS value of the third harmonic voltage in each ­alternator coil.

30-8   Square Waves and Sawtooth Waves

Solution Since the fundamental phase voltages are 2π/3 rad out of phase, the line voltage contains a fundamental component that is √ 3 times the

fundamental component of the phase voltage. Since the third-harmonic phase voltages are 2π rad out of phase, there is no third harmonic in the line voltage. The fifth-harmonic phase voltages are out of phase by 600° or 10π/3 rad, which is equivalent to −2π/3 rad. Therefore, the line voltage contains a fifth-harmonic component that is √3 times the fifth-harmonic component of the phase voltage, and 208 V = 1.73√ E21 + E25  and  122 V = √ E21 + E23 + E25

Squaring both equations and solving for E3 gives E3 = √ 1222 − 1202 = 22 V

See Problem 30-8.

30-8 Square Waves and Sawtooth Waves In many applications we purposely distort an input sine wave to obtain a special wave shape. Consider, for example, the square wave of Figure 30-13(a). Although we can synthesize a required nonsinusoidal waveform by adding Input sine wave +

Voltage

Output square wave 0

−

Time

(a) Fundamental

+

Voltage

Third harmonic 0

Time Fifth harmonic

−

(b)

Figure 30-13 Production and synthesis of a square wave

961

Chapter 30  Harmonics

together the appropriate harmonics, in practice these waveforms are often produced by passing a sine wave through a nonlinear impedance network. For example, we can produce a square wave by passing a large-amplitude sine wave through a nonlinear network that limits the amplitude of the output voltage to a few volts. The square wave of Figure 30-13(a) has been produced by clipping the peaks of the input sine wave. A nonlinear impedance network that produces a square wave adds harmonics to the input sine wave. Since the square wave of Figure 30-13(a) is highly symmetric, it consists of a fundamental sine wave plus a series of odd-order harmonics. There is a simple relationship between the amplitude and the frequency of each harmonic: the amplitude of each harmonic is inversely proportional to its frequency. To produce the flat top and steep sides of the square wave, the harmonics have to be in phase with the fundamental. The Fourier series for the instantaneous value of a symmetrical square wave with no DC component is e = Em sin ωt +

(

1 1 1 sin 3ωt + sin 5ωt + sin 7ωt + . . . 3 5 7

)

(30-10)

Figure 30-13(b) shows the first three sine-wave components of a square wave.

Instantaneous Value

962

+

0

Time

− Figure 30-14  Sawtooth waveform

Another common nonsinusoidal wave is the sawtooth wave, shown in Figure 30-14. This waveform contains both odd-order and even-order ­harmonics. The Fourier series for the waveform of Figure 30-14 is e = Em sin ωt +

(

1 1 1 1 sin 2ωt + sin 3ωt + sin 4ωt + sin 5ωt + . . . 2 3 4 5

)

(30-11) See Problems 30-9 and 30-10 and Review Questions 30-34 and 30-35.

30-9   Nonsinusoidal Waves in Linear Impedance Networks

30-9 Nonsinusoidal Waves in Linear Impedance Networks We now consider how to analyze a circuit in which a nonsinusoidal wave is applied to a network of linear impedances. According to the superposition theorem, we can determine the net effect of several sources in a network on any branch of that network by determining the effect of each source acting independently with the other sources switched off. We can, therefore, think of the nonsinusoidal source of Figure 30-15(a) as being equivalent to harmonically related sine-wave generators connected in series, as in Figure 30-15(b). Applying the superposition theorem to this circuit, we first determine the current in the load resistor independently for each frequency. We can then use Equation 30-9 to calculate the total RMS current. If we wish to know the shape of the output waveform, we can graph the various harmonic components of the output and add their instantaneous values. Fundamental

Nonsinusoidal E waveform generator

Linear Z

Linear

Second harmonic

Z

RL Third harmonic

Network

RL

Network

Fourth harmonic (a)

(b)

Figure 30-15 Equivalent circuit for a linear impedance network with a nonsinusoidal source

Example 30-3 The nonsinusoidal waveform e = 45 sin 6280t − 15 sin 18 840t is applied to the circuit of Figure 30-16. (a) Determine the waveform of the input wave. (b) Determine the waveform of the output wave. (c) Determine the overall power factor of the circuit. 10 nF Nonsinusoidal waveform E generator



10 kΩ

Output

Figure 30-16  Schematic diagram for Example 30-3

963

Chapter 30  Harmonics

Solution (a) The fundamental has a frequency of



fr =

ω 6280 = = 1.0 kHz 2π 6.28

The harmonic has a frequency of fh =

18 840 = 3.0 kHz 6.28 Therefore, sin 18 840t represents a third harmonic. The equation for e ­indicates that the harmonic and the fundamental are 180° out of phase, and the amplitude of the third harmonic is one third of the amplitude of the fundamental. Therefore, the input voltage has the waveform shown in Figure 30-17. e Instantaneous Current

964

Fundamental

+

Third harmonic 0

Time

− Figure 30-17  Input-voltage waveform for Example 30-3

(b) Since vout = iR and the resistance is constant, the output voltage waveform has the same shape as the total instantaneous current waveform. At the fundamental frequency,

XC =

1 1 = = 15.92 kΩ ωC 6280 × 10 nF



Zf =



If =

10 kΩ ∠−57.87º = 18.8 kΩ ∠−57.87º cos (−57.87º )



ϕ = tan− 1

−15.92 kΩ = −57.87º 10 kΩ

Ef = 0.707Em = 0.707 × 45 V = 31.8 V

31.8 V∠0º Ef = = 1.69 mA ∠+57.87º Zf 18.8 kΩ ∠−57.87º

At the third harmonic, 1 XC = = 5.31 kΩ 18 840 × 10 nF ϕ = tan − 1

−5.31 kΩ = −27.97º 10 kΩ

30-9   Nonsinusoidal Waves in Linear Impedance Networks

10 kΩ ∠−27.97º = 11.32 kΩ ∠−27.97º cos (−27.97º )  E3 = 0.707 × 15 V = 10.6 V Z3 =





10.6 V∠+180º E3 = = 0.937 mA∠+207.97º Z3 11.32 kΩ ∠−27.97º Since the load resistance is constant, the output voltage waveform has the same shape as the output current waveform. In the output current, the ­amplitude of the harmonic is 55% of the amplitude of the fundamental since the reactance of the capacitor is less at the third-harmonic frequency than at the fundamental frequency. In the input wave, the 180° point of the third harmonic coincides with the 0° point of the fundamental at t = 0. But in the current waveform, the fundamental has been advanced 57.8° (of a fundamental cycle) and the third harmonic has been advanced only 27.9° (of a third harmonic cycle). Therefore, in a diagram of the output voltage waveform, the 57.8° point of the fundamental and the 207.9° point of the third harmonic coincide at t = 0 as shown in Figure 30-18. I3 =

Instantaneous Voltage

vout

+

Fundamental Third harmonic

0

Time

− Figure 30-18  Output-voltage waveform

(c) The total average power is

PT = I21 R + I23 R = [ ( 1.69 mA ) 2 + ( 0.937 mA ) 2 ] × 10 kΩ = 37.34 mW

The RMS input voltage is

ET = √E21 + E23 = √31.82 + 10.62 = 33.5 V

The RMS total current is

IT = √I21 + I23 = √ ( 1.69 mA ) 2 + ( 0.937 mA ) 2 = 1.93 mA S = ETIT = 33.5 V × 1.93 mA = 64.7 mVA

Therefore, the apparent power in the network is

The power factor is leading since the network is capacitive. By definition, power factor = 

37.34 mW P = = 57.7% leading S 64.7 mVA

See Problems 30-12 to 30-14 and Review Questions 30-36 and 30-37.

965

966

Chapter 30  Harmonics

Summary

• Any nonsinusoidal periodic wave consists of a series of harmonically ­related pure sine waves. • The harmonics of a fundamental frequency are integral multiples of that frequency. • The Fourier series expresses the harmonic content of a periodic waveform. • The coefficients of the sine terms in the Fourier series of a periodic waveform are zero if the wave is symmetric about a vertical axis at t = 0. • A fundamental combined with odd-order harmonics produces a symmetrical complex wave. • The shape of a complex wave can be determined from the equations of its harmonically related sine waves. • When an AC waveform is applied to a nonlinear circuit element, it produces harmonics not present in the input AC waveform. • Nonlinearity in the relationship between input and output signals of an amplifier results in even-order harmonics in the output signal. • Nonlinearity in the BH curve of an iron-core transformer results in oddorder harmonics in the exciting current. • The RMS value of an AC current or voltage containing harmonics is the square root of the sum of the squares of the RMS values of the fundamental and all the harmonics. • A square wave contains a fundamental and odd-order harmonics in phase with the fundamental, with the amplitude of each harmonic inversely proportional to its frequency. • The superposition theorem can be used to determine voltage and current waveforms when a nonsinusoidal voltage wave is applied to a linear ­impedance network. B = beginner

Problems

I = intermediate

The graphs for these problems can be produced by hand or with a spreadsheet or circuit-simulation software.

A = advanced

I I B

I

Section 30-3  Addition of Harmonically Related Sine Waves

30-1. Graph the complex wave described by Equation 30-3 if Em = 5 V and ω = 1 rad/sec. 30-2. Graph the complex wave described by Equation 30-4 if Em = 10 V and ω = 2 rad/sec. 30-3. A 10-A 1-kHz current source is connected in parallel with a 2-A 5-kHz source. (a) Write an expression for the total current in the circuit if it’s known that the sources are in phase at t = 0 (b) Is the total current symmetrical? 30-4. Sketch the shape of the complex wave defined by e = 50 sin 943t + 30 sin 4715t.

Problems

I

B

I

Section 30-4  Generation of Harmonics

30-5. Graph the complex wave described by Equation 30-5, using all terms up to and including the fifth harmonic if Em = 10 V and w = 1 rad/sec

Section 30-5  Harmonics in an Amplifier 30-6.

Section 30-6  Harmonics in an Iron-Core Transformer 30-7.

I A B

A

I

Sketch the nonsinusoidal wave defined by the equation

e = −35 sin 1256t −

(



A

When a sine wave is applied to a transistor amplifier, the collector current swings 8 mA above its quiescent value and 7 mA below. ­Determine the second-harmonic distortion in the output as a percentage of the fundamental current.

1 1 1 sin 3768t + sin 6280t − sin 8792t 9 25 49

)

Describe the shape of the resulting complex wave.

Section 30-7  RMS Value of a Nonsinusoidal Wave 30-8. A wye-connected three-phase source has a phase voltage of 110 V and a line voltage of 188 V. The induced voltage contains odd-order harmonics. Given that the amplitudes of the harmonics above the seventh are small enough to be ignored, find the RMS value of the third-harmonic in the phase voltage.

Section 30-8  Square Waves and Sawtooth Waves

30-9. Graph the complex wave that contains only the fundamental and the first two harmonic terms of Equation 30-10 if Em = 50 V and ω = 3 rad/sec. 30-10. Graph the nonsinusoidal wave defined by Equation 30-11, if Em = 75 V and the frequency is 60 Hz. Use only the first four harmonics. 30-11. Find the RMS voltage for the wave in Problem 30-10.

Section 30-9 Nonsinusoidal Waves in Linear Impedance Networks 30-12. (a) Find the RMS current in a series circuit consisting of a 50-mH ­inductance and a 50-Ω resistance when the applied voltage has the nonsinusoidal waveform of Problem 30-10. (b) Find the average power in this circuit. (c) Find the overall power factor of the circuit. 30-13. Determine the output voltage waveform if the positions of the ­capacitor and resistor in Example 30-3 are reversed.

967

968

Chapter 30  Harmonics

A

30-14. Determine the equation for the total current when a voltage ­described by the equation



e = 180 sin 377t − 60 sin 1131t + 36 sin 1885t

is applied to a circuit consisting of two parallel branches, one with an impedance of 90 Ω ∠−60º at the fundamental frequency, and the other with an impedance of 40 + j30 Ω at the fundamental ­frequency.

Review Questions

Section 30-1  Nonsinusoidal Waves 30-15. Suggest several reasons why it is desirable to use a sine-wave source for electric power distribution systems. 30-16. Define the terms fundamental and harmonic.

Section 30-2  Fourier Series 30-17. Under what conditions is the coefficient A0 in the Fourier series zero? 30-18. Is it necessary for a waveform to be a pulsating DC waveform for A0 in the Fourier series to be other than zero? Explain.

Section 30-3  Addition of Harmonically Related Sine Waves 30-19. Why is it possible to describe the shape of a nonsinusoidal wave by an equation such as Equation 30-2? 30-20. What is the significance of a negative value of t in Figure 30-5? 30-21. What is the significance of a – sign in front of a term in the equation for a nonsinusoidal wave? 30-22. What is the significance of a cosine term in the equation for a nonsinusoidal wave? 30-23. Describe the phase relationships among the fundamental and the harmonics shown in Figure 30-2. 30-24. Why is it usually possible to obtain a fairly good duplication of a nonsinusoidal wave by considering only the first five or six terms of the harmonic series? 30-25. Describe how odd-order and even-order of harmonics affect the shape of nonsinusoidal waves.

Section 30-4  Generation of Harmonics 30-26. Show that Equation 30-5 for a half-wave-rectified sine wave is consistent with the general form of the Fourier series in Equation 30-1.

Section 30-5  Harmonics in an Amplifier 30-27. Where do the harmonics come from when a transistor introduces harmonic distortion into a circuit? 30-28. A certain amplifier advances the phase of a 500-Hz sine wave by 2°. In order to pass a complex wave without changing its shape,

Integrate the Concepts

this amplifier must have a 4° phase advance at 1.0 kHz, a 6° phase ­advance at 1.5 kHz, and so on. Explain why.

Section 30-6  Harmonics in an Iron-Core Transformer 30-29. The waveforms shown for the transformer in Figure 30-11 are based on a sine-wave voltage source. Draw the waveforms that apply when the transformer is fed from a sine-wave current source. Which order of harmonics is introduced? 30-30. The BH-curve transfer characteristic in Figure 30-11 is based on only alternating current in the primary. Sketch the magnetic flux waveform when the transformer primary is connected in series with the collector of a transistor in which the collector current v ­ aries sinusoidally about its quiescent value. Which order of harmonics is introduced by this transformer? 30-31. Why does none of the third harmonic in the exciting current in the primary windings of a three-phase delta-connected transformer bank appear in the line current when the load is balanced? 30-32. Why does the amplitude of the seventh harmonic in the line current to a delta-connected transformer bank equal √ 3 times the amplitude of the seventh harmonic of primary current? 30-33. Suggest why the terminal voltage of an alternator contains oddorder harmonics rather than even-order harmonics.

Section 30-8  Square Waves and Sawtooth Waves 30-34. A symmetrical square wave of voltage has a peak value of 100 V. Find the RMS voltage. 30-35. What effect does reversing the sign in front of the even-order harmonic terms of Equation 30-11 have on the shape of the complex wave?

Section 30-9 Nonsinusoidal Waves in Linear Impedance Networks 30-36. Explain the significance of the superposition theorem in solving ­linear AC networks that have nonsinusoidal sources. 30-37. The output waveform in Example 30-3 appears across a resistor with a resistance that does not depend on frequency. Why then does the output waveform not have the same shape as the input waveform?

Integrate the Concepts

For the circuit of Figure 30-19: (a) Write the Fourier series expression for the total output voltage, eT, of the three generators in series. (b) Draw and label the graphs of V1, V2, V3, and VT. (c) Calculate the RMS value of eT. (d) Calculate the total average power delivered to the 33-Ω resistor.

969

Chapter 30  Harmonics

+

970

+ −

+ −

− V3 12 V 4 kHz 180°

V2 10 V 2 kHz −30°

R 33 Ω

V1 5V 1 kHz 0° L 5 mH

Figure 30-19

Practice Quiz

1. Which of the following statements are true? (a) A harmonic of a nonsinusoidal waveform is a multiple frequency of that frequency. (b) Fourier series represents mathematically the harmonic content of non-periodic waveforms. (c) Nonlinear devices generate harmonics of the input AC signal frequency. 2. If the fundamental frequency of a nonsinusoidal wave is 5 kHz, then its second and third harmonics are (a) 15 kHz and 20 kHz (b) 10 kHz and 15 kHz (c) 7.5 kHz and 15 kHz (d) 12.5 kHz and 25 kHz 3. The first coefficient of a Fourier series represents the (a) magnitude of the second harmonic of the complex wave (b) magnitude of the third harmonic of the complex wave (c) magnitude of the tenth harmonic of the complex wave (d) the DC component of the complex wave 4. The positive and negative half-cycles of a nonsinusoidal wave are always symmetrical when the wave is composed of a fundamantal and (a) even-order harmonics (b) odd-order harmonics

Practice Quiz

(c) both even-order and odd-order harmonics (d) just the first three harmonics 5. When a transistor adds a series of even-order harmonics that are neither in phase nor π-rad out of phase with the fundamental, the most important of those harmonics is the (a) second harmonic (b) third harmonic (c) fourth harmonic (d) eighth harmonic 6. We can approximate the waveform of the exciting current in an ­iron-core transformer with negligible hysteresis by adding the fundamental frequency and the (a) second harmonic (b) third harmonic (c) fourth harmonic (d) eighth harmonic 7. The RMS value of an AC current or voltage containing harmonics is given by (a) the sum of the squares of the RMS values of the fundamental and all the harmonics (b) the sum of the squares of the RMS values of all the harmonics (c) the square root of the sum of the squares of the RMS values of the fundamental and all the harmonics (d) the square of the sum of the square roots of the RMS values of the fundamental and all the harmonics 8. A Fourier series can represent (a) any combination of harmonics (b) asymmetical waveforms (c) sawtooth and square waveforms (d) all of the above 9. A sawtooth waveform is composed of (a) only odd-order harmonics (b) only even-order harmonics (c) odd and even-order harmonics (d) just the first five harmonics

10. For the nonsinusoidal wave e = 20 sin ωt + 15 sin 2ωt − 10 sin 3ωt, the RMS value is: (a) 6.71 V (b) 19.04 V (c) 26.93 V (d) 4.74 V

971

972

Chapter 30  Harmonics

11. The nonsinusoidal expression for the total current of Figure 30-20 is: (a) 5 + 50 sin 20t + 25 sin (40t − 70°) (b) 5 + 50 sin 20t + sin (40t + 0.54°) (c) 5 + 10 sin 20t + 0.5 sin (40t − 70°) (d) 5 + 50 sin 20t + 1.23 sin (40t + 35°) IT I1

I2

I1 = 5 + 30 sin 20t + 1.5 sin (40t + 20°) I2 = 20 sin 20t + 0.5 sin (40t − 90°)

Figure 30-20

Appendices Appendix 1

Determinants

Appendix 2

Calculus Derivations

Appendix 3

Multisim Schematic Capture and Simulation

Appendix 1

Determinants In solving simultaneous equations, we require as many equations as there are un­knowns. For the network examples in this text, we can limit our discussion to second­- and third-order simultaneous equations and write them in the general form shown below. Second-Order Equations a1x + b1y = k1 a2x + b2y = k2

Third-Order Equations (A-1) (A-2)



a1x + b1y + c1z = k1 a2x + b2y + c2z = k2 a3x + b3y + c3z = k3

(A-3) (A-4) (A-5)

In these equations, x, y, and z are the unknowns and a, b, c, and k are numerical coefficients or constants, which can be positive, negative, or zero. We can solve the second-order simultaneous equations above by the elimination method. Equation A-1 multiplied by b2: Equation A-2 multiplied by b1: Subtracting gives

Therefore,

x=

Similarly,

y=

a1b2x + b1b2y = k1b2 a2b1x + b1b2y = k2b1 x ( a1b2 − a2b1 ) = k1b2 − k2b1

k1b2 − k2b1 a1b2 − a2b1

(A-6)

a1k2 − a2k1 a1b2 − a2b1

(A-7)

Dy Dx and y = D D

(A-8)

We can use Equations A-6 and A-7 to solve any second-order simultaneous equations. The numerators and denominators in the solutions of simultaneous equations are called determinants. Since the same determinant appears in the denominators of both Equation A-6 and Equation A-7, we can rewrite these equations as



x=

Appendix 1  Determinants

These equations are an example of Cramer’s rule, a theorem that uses matrices and determinants to solve for any variable in a set of simultaneous equations. To calculate the determinants D, Dx, and Dy, we first write the simultaneous equations so that the sequence of the variables and constants matches that of Equations A-1 and A-2. Then, we write matrices of the coefficients and constants. Each matrix has as many columns and as many rows as there are unknowns. The matrix for the denominator determinant D simply has all the coefficients in the same arrangement as they appear in the equations. Column 1 Column 2 Substitution column a1x a2x

+ +

=

b1y

│kk x= │aa

=

b2y



│ b b│

2

b1 b2

1

1

1

│aa y= │aa

2

1

│ b b│

(A-1)

k2

(A-2)

(A-9)

2

k1 k2

2



k1

1

(A-10)

1

2

2

The matrix for the numerator Dx replaces a1 and a2, the coefficients of x, with the constants k1 and k2 from the right side of Equations A-1 and A-2. Similarly, the matrix for Dy replaces b1 and b2, the coefficients of y, with the same constants k1 and k2. To evaluate the determinant of a matrix, we multiply the numbers in the matrix as we move from left to right along each diagonal of the matrix. We then subtract the sum of the products from the upward diagonals from the sum of the products of the downward diagonals. The results of this procedure match the expressions in Equations A-6 and A-7. For example, applying this method to the denominator D gives

│aa

1 2

│=ab

b1 b2

1 2

− a2b1

To illustrate this procedure, we now use Cramer’s rule to solve the simultaneous equations in Step 5 of Example 9-2. 60I1 + 10I2 = 80 10I2 + 35I2 = 80

975

Cramer’s rule was developed by the Swiss ­mathematician Gabriel Cramer (1704–52).

976

Substituting into the original equations verifies the answers: 60 × 1 + 10 × 2 = 80 10 × 1 + 35 × 2 = 80

Appendix 1  Determinants





I1

I2

│8080 = │6010

│ 80 ( 35 ) − 80 ( 10 ) 2800 − 800 2000 = = = = 1A 60 ( 35 ) − 10 ( 10 ) 2100 − 100 2000 10 35│ 10 35

│6010 = │6010

│ 60 ( 80 ) − 10 ( 80 ) 4800 − 800 4000 = = 2 A = = 60 ( 35 ) − 10 ( 10 ) 2000 2000 10 35│ 80 80

Adding a third unknown greatly increases the computation required to solve the simultaneous equations. However, the procedure for evaluating third-order determinants is the same as for second-order determinants. Solving the three generalized Equations A-3, A-4, and A-5 for x, we obtain x=



k1b2c3 + b1c2k3 + c1k2b3 − k3b2c1 − b3c2k1 − c3k2b1 a1b2c3 + b1c2a3 + c1a2b3 − a3b2c1 − b3c2a1 − c3a2b1

The equations for y and z are similar, and have exactly the same denominator. The matrices for third-order simultaneous equations have three downward diagonals and three upward diagonals, with each diagonal corresponding to the product of three coefficients or constants. First, we set up the required matrices from Equations A-3, A-4, and A-5. Column 1   Column 2   Column 3   Substitution column + b1y

a1x

+ b2y

a2x

│ │

x=

│ │

+ b3y

a3x

k1 k2 k3

b1 b2 b3

c1 c2 c3

a1 a2 a3

b1 b2 b3

c1 c2 c3

+ c1z + c2z

│

+ c3z

       y =

a1 a2 a3

= k1 = k2

│

= k3 k1 k2 k3 D

c1 c2 c3

       z =

(A-3)

│

(A-4)

│

(A-5) a1 a2 a3

b1 b2 b3 D

k1 k2 k3

Again, we add the products for the downward diagonals and subtract the products for the upward diagonals. We can think of the 3 × 3 matrices as being printed around a vertical cylinder, so that there are three rows to each diagonal, no matter where we start. To simplify calculating the diagonal

A2-1   Maximum Power-Transfer Theorem

│

│

products, we can repeat the first two columns in sequence on the right-hand side of the original matrix. D=



a1 a2 a3

b1 b2 b3

c1 c2 c3

a1 a2 a3

b1 b2 b3

  = a1b2c3 + b1c2a3 + c1a2b3 − a3b2c1 − b3c2a1 − c3a2b1

Example Evaluate

│

Solution D=

4 5 6

3 −2 0

−2 4 1

│

D=

4 5 6

│

4 5 6

3 −2 0

3 −2 0

│

−2 4 1

= 4 ( −2 ) ( 1 ) +3 ( 4 ) ( 6 ) + ( −2 ) ( 5 ) ( 0 ) −6 ( −2 ) ( −2 ) −0 ( 4 ) ( 4 ) −1 ( 5 ) ( 3 ) = −8 + 72 + 0 − 24 − 0 − 15 = 25

Appendix 2

Calculus Derivations 2-1 Maximum Power-Transfer Theorem

In the circuit of Figure A-1, the load resistance, RL, is adjustable. To find the load resistance that produces the maximum power transfer from the source, we first find an equation that relates the power output to the source resistance and the variable load resistance: PL = I2RL

977

978

Appendix 2   Calculus Derivations

Since I = E / (Rint + RL),

PL =



E2RL ( Rint + RL ) 2

(A-11)

Rint 6.0 Ω + E

RL

−

Figure A-1  Circuit for determining maximum power transfer

Maximum power transfer occurs when the derivative of PL with respect to RL is zero: dPL E2 [ ( Rint + RL ) 2 − RL ( 2Rint + 2RL ) ] = = 0 ( Rint + RL ) 4 dRL



Since E ≠ 0 , the expression in square brackets must equal zero. Expanding and simplifying this expression then gives Rint + RL = 2RL

Therefore, maximum power output occurs when RL = Rint





(A-12)

2-2 Instantaneous Voltage in a CR Circuit When we throw the switch to position 1 in the circuit of Figure A-2, E = iR + vC



1

+ E

2

vR

− vC

Figure A-2  CR circuit

(A-13)

A2-2   Instantaneous Voltage in a CR Circuit

Since i = dq/dt and q = CvC,

i=C

dvC dt

Substituting for i in Equation A-13 gives E = CR



dvC + vC dt

( )

(A-14)

To solve this differential equation, we rearrange it in the form dvC E − vC = CR dt



dt dvC = CR E − vC



Multiplying both sides by −1 and then integrating gives dvC dt =− CR vC − E



ln ( vC − E )



]

vC 0

=

ln ( vC − E ) − ln ( 0 − E ) =

which becomes

ln

(

−t CR

−t + 0 RC

vC − E −t = −E CR

)

]

t 0

The exponential form of this equation is

vC − E = e−t/CR −E

Rearranging to solve for vC gives

vC = E − Ee − t/CR = E ( 1 − e − x )

(A-15)

where e is the base of natural logarithms (about 2.718) and x = t / CR.

If we throw the switch in the circuit of Figure A-2 to position 2 once the capacitor has charged, the applied voltage in the loop consisting of the ­capacitor and the resistor is zero, and

0 = vR + vC = CR

dvC + vC dt

979

980

Appendix 2   Calculus Derivations

Since vR = −vC, we can rewrite this equation as dvC −dt = vC CR

Integrating gives

ln vC



In



vC −t = V0 CR

( )

]

vC 0

=

−t CR

]

t 0

vC = e − t/CR V0

or

vC = V0e–x

where V0 is the initial voltage across the capacitor, and x = t / CR.

(A-16)

2-3  Energy Stored by a Capacitor To determine the energy stored by a capacitor, we calculate the area under the instantaneous power graph of Figure 13-15: Since p = vi and i = C

dv , dt

w=

∫ p dt

∫

(A-17)

0

w = Cv t

0



t

dv dt dt

= Cv dv

∫

v

0

and

w=

1 2 Cv 2

(A-18)

Hence, no matter how a capacitor charges, the energy it stores at any instant is proportional to the capacitance and to the square of the potential difference across it.

2-4 Instantaneous Current in an LR Circuit When we throw the switch to position 1 in the circuit of Figure A-3,

E = vR + vL = iR + L

( ) di dt



(A-19)

A2-4   Instantaneous Current in an LR Circuit

1

+ E

2

vR

− vL

Figure A-3

Rearranging this differential equation and integrating gives

( )] ( ) ( ) ( )



ln i −



ln i −

where x = tR / L.

i=

i

0

=−

R t L

i − E/R R =− t −E/R L

and



E R

t 0

E E R − ln 0 − = − t + 0 R R L

ln

Solving for i gives

]

di R = − dt i − E/R L

i − E/R = e − Rt/L −E/R

E E E − ( e−Rt/L ) = ( 1 − e−x ) R R R

(A-20)

If we throw the switch in the circuit of Figure A-3 to position 2 after the current has reached its steady-state value of I = E / R, the applied voltage in the loop consisting of the inductance and the resistor is zero, and 0 = vR + vL = iR + L

( )

di dt Rearranging this equation and then integrating gives



]

]

di dt = −R i L i −Rt t ln i = L 0 I0

ln

i −Rt = I0 L

()

i = I0 e−x 

or

i = e − Rt/L I0 (A-21)

where I0 is the initial current through the inductance, and x = tR / L.

981

982

Appendix 2   Calculus Derivations

2-5  Energy Stored by an Inductor To determine the energy stored in the magnetic field resulting from current through an inductor, we calculate the area under the instantaneous power graph of Figure 17-12: w=

di Since p = ivL = iL , dt

w=

w=

and

t

∫ p dt

(A-22)

0 i

∫ Li di 0

1 2 Li 2

(A-23)

Hence, the energy an inductor stores at any instant is proportional to its inductance and to the square of the instantaneous current through the ­inductor.a

2-6 RMS and Average Values of a Sine Wave To find the average heating effect of an alternating current, we integrate the instantaneous power over a complete cycle and divide by the duration of the cycle (2π rad):

Pav =

1 2π

∫

2π

0

p dϕ =

1 2π

∫

2π

2

i R dϕ

0

The DC current, I, that produces the same average power is Hence,

and

Pav = I 2R

R IR= 2π 2

I=

∫

2π

i2 dϕ

√2π1 ∫ 0

2π

i2 dϕ

(A-24)

0

Equation A-24 shows that the DC equivalent current for an alternating ­current is equal to the RMS value of the alternating current (the square root of the mean of the square of the instantaneous current). Equation A-24

A2-7   Inductive Reactance

gives the RMS value of an alternating current of any waveform. For a sine wave, i = Im sin ϕ and

I2 =

1 2π

∫

2π

I2m sin2 ϕ dϕ

From the trigonometric identity sin2 θ = 12(1 − cos 2θ),

I2 = =

I2m 4π

∫

[

2π

0

I2m 4π

0

]

( 1 − cos 2ϕ ) dϕ

ϕ−

1 sin 2ϕ 2

Since sin 0 and sin 4π are both zero,

2π 0



2πI2m I2m   I2 = = 4π 2 I=

and

√2 Im



(A-25)

The average value of one half-cycle of a symmetrical alternating-current waveform is

Iav =

1 π

∫

π

i dϕ (A-26)

0

For a sine wave Equation A-26 becomes

1   Iav = I sin ϕ dϕ π0 m =



∫

π

Im [ −cos ϕ ] π0 π

Iav =

and

2 I π m

(A-27)

2-7  Inductive Reactance When an ideal inductor is connected across a sine-wave voltage source, the instanta­neous voltage drop across the inductance must equal the instantaneous applied volt­age: Therefore,

L

di = Em sin ωt dt

di Em = sin ωt dt L

983

984

Appendix 2   Calculus Derivations

Indefinite integration of this equation gives =



i=

∫

Em sin ωt dt L

Em × ( −cos ωt + C ) ωL

Except during transients caused by opening or closing a switch in the circuit, the constant of integration, C, is zero. Since −cos θ = sin (θ − π/2), iL =



Em π sin ωt − ωL 2

(

)



(A-28)

The instantaneous current i in Equation A-28 has its maximum value when sin (ωt − π/2) = 1. Hence, Im =



Em ωL

Since IL and VL are both 1/√2 of their peak values, XL =



and

where ω = 2πf.

VL Em = = ωL IL Im XL = ωL

(A-29)

2-8  Capacitive Reactance The instantaneous current in a capacitor is

i=C

dv dt

(19-6)

Substituting Em sin ωt for the instantaneous voltage across the capacitor gives d ( Em sin ωt ) = ωCEm cos ωt i=C dt Since cos ϕ leads sin ϕ by 90° or π/2 rad,

iC = ωCEm sin ωt +

(

Therefore, i = Im when sin (ωt + π/2) = 1, and Im = ωCEm

π 2

)



(A-30)

A2-10   Maximum Transformer Efficiency

XC =

from which

1 ωC

(A-31)

2-9  General Transformer Equation An equation for Faraday’s law is

eT = N



dϕ dt

(16-1)

where eT is the instantaneous induced voltage between the terminals of the winding, N is the number of turns in the winding, and dϕ/dt is the rate of change of the magnetic flux. Since Equation 16-1 applies for the instantaneous voltage induced into a transformer winding connected to a sinusoidal voltage source, the flux in the core of the transformer must also be a sine wave: ϕ = Φm sin ωt

Hence,

dϕ = ωΦm cos ωt dt



e = ωNΦm cos ωt

and

(A-32)

The instantaneous voltage in Equation A-32 reaches its peak value when cos ωt = 1. Therefore, Em = 2πfNΦm



and

E=

√2

Em

≈ 4.44f NΦm

(27-3)

2-10  Maximum Transformer Efficiency The efficiency of a transformer is

η=

SL cos ϕ + I2Re + PCL SL cos ϕ

Substituting S = IV in Equation 27-14 gives

η=

IV cos ϕ + I2Re + PCL IV cos ϕ



(27-14)

985

986

Appendix 3   Multisim Schematic Capture and Simulation

If the current is the only variable, maximum efficiency occurs when the ­derivative of η with respect to I is zero.

( IV cos ϕ + I2Re + PCL ) V cos ϕ − IV cos ϕ ( V cos ϕ + 2IRe ) dη = ( IV cos ϕ + I2Re + PCL ) 2 dI

For this derivative to equal zero, the numerator must be zero. Therefore,

V cos ϕ ( I2Re + PCL − 2I2Re ) = 0

and

PCL = I2Re

(A-33)

Maximum efficiency occurs when core loss equals copper loss.

Appendix 3 circuitSIM walkthrough

Multisim Schematic Capture and Simulation Multisim is a software package designed to generate schematics and simulate electrical and electronic circuits. The user can assemble a circuit on the computer screen and then perform analyses to determine responses within the circuit. For this textbbook, simulation of circuit operation using Multisim has two objectives. First, it provides a check for the calculations in ­various examples throughout the chapters. Second, it is a means by which ­s tudents may determine answers to even-numbered problems in the chapters, answers being provided for most odd-numbered problems in the “Answers to Selected Problems” section at the end of the book. In virtually all of these problems, the schematic of the circuit is already drawn in the circuit window. Some modifications to the ­circuit, such as inserting a component or instrument, changing parameter values of components or reading measured values, are generally required of students. The documentation that accompanies the Multisim software will be particularly useful for students who are not already familiar with circuit simulation software. Students can access this documentation in the “Multisim User Manual,” in the “Multisim Help,” and in the tutorials in “Getting Started.”

A3-1   Multisim User Manual

3-1  Multisim User Manual The extensive manual leads the user through all aspects of the simulation software. The table of contents for Circuit Design Suite 11.1 encompasses nearly 20 pages, with the manual divided into 16 chapters. Most of the activities students will do with these Multisim problems are discussed in Chapters 1–3 and 8–10. • Chapter 1 explains the toolbars, pop-up menus and dialog boxes. • Chapters 2 and 3 cover the basics of placing components in the circuit window, setting component values, and wiring the circuit. • Chapter 8 details how simulation of circuit operation is performed. • Chapter 9 describes the various types of measuring instruments available. • Chapter 10 discusses the numerous analyses contained in ­Multisim (how they work, how to set parameters, and how to view results).

3-2  Multisim Help Multisim Help is available to the user on the Menu Bar. The Help button gives access to the entire Multisim User Manual, along with other items. Information on how to conduct an activity may be found using the Contents, Index, Search, or Favorites tabs.

3-3  Getting Started Chapter 2, entitled “Multisim Tutorial,” introduces the user to Multisim and most of the features students will need to know. The tutorials are designed for the novice user, with simple instructions and helpful diagrams. The user is led through such procedures as how to launch Multisim, how to place components in the circuit window, how to wire the circuit and measure responses with instruments, how to observe transients and how to understand how three-phase delta-connected and wye-connected voltage sources operate. For example: • Placing a resistor with a specified resistance in a circuit: The table of contents for the Multisim User Manual or Multisim Help shows that the user may consult Chapter 2, page 2-5. The table of contents for Getting Started also points the user to page 2-5 (Placing the Components). • Multimeters, oscilloscopes, wattmeters, and Bode plotters: The table of contents for the Multisim User Manual or Multisim Help directs the user to Chapter 9, page 9-7 for mutimeters, page 9-15 for ­oscilloscopes, page 9-13 for wattmeters, page 9-19 for Bode plotters.

987

Answers to Selected Problems Solutions to problems marked with the Multisim icon are on the textbook website.

Chapter 1

CC 1-1 (a)  3 (b)  2 (c)  4 CC 1-2 (a)  3.06 × 10−2 s (b)  3.8 × 10−1 m (c)  1 × 102 kg (d)  6 × 104 A CC 1-3 (a)  1027 (b)  106 10−8 (c)  (d)  102 (e)  103 (f)  10−4 CC 1-4 (a)  3 μm (b)  2 MK (c)  0.56 kV or 560 V (d)  256 THz (e) 55 nA CC 1-5 1236 km/h CC 1-6 1 tonne = 1.102 ton 1-1 (a) 131.7 (b) 142 (c) 2.8 (d) 61 (e) 480 (f) 3.2 1-3 (a)  1 400 000 (b) 497 (c) 62.8 (d)  0.00345 (e)  0.000 0077 (f) 0.8672 1-5 6.242 × 1018 electrons 1-7 5.98 × 1024 kg 1-9 (a)  5 600 000 ohms (b)  0.075 metres (c)  746 watts (d)  0.000 037 volts (e)  37 000 metres (f)  37 kilograms 1-11 2.25 kg 1-13 16.67 ms 1-15 22 g 1-17 707 mm2 1-19 226.8 g 1-21 1 sqft = 0.0929 m2 1. (a)  8.263 × 103 (b)  9 × 10−1 −5 (c)  7.0 × 10 (d)  2 × 104 −7 (e)  5.0 × 10 2. (a)  50 km (b)  166.6 kHz (c)  470 nA (d)  800 μV 3. (a)  0.0467 mA, 0.000 046 7 A (b)  0.652 mA, 652 μA (c)  825 km, 0.825 Mm (d)  0.000 403 s, 0.403 ms (e)  0.000 050 μg, 50 pg (f)  0.025 H, 25 000 μH (g)  0.0314 MHz, 3.14 × 104 Hz (h)  0.176 V, 1.76 × 105 µV 4. (a)  3.24 mA (b)  1.11 mA (c)  8.2 × 104 Ω, 82 kΩ (d)  9.6 × 104 V, 96 kV (e)  0.083 mA, 83.33 μA (f) 9.86 m2

Practice Quiz

Chapter 2 CC 2-1 reduced by a factor of 4 CC 2-2 zero CC 2-3 1 A = 1 C/s CC 2-4 101 kJ CC 2-5 180 J 2-1 2.18 × 1019 e 2-3 4A 2-5 16 μA 2-7 300 mA 2-9 12 s 2-11 2.94 A 2-13 120 V 2-15 1.25 × 1018 e 2-17 700 mC 2-19 8s 2-21 40.5 hr 2-23 360 mJ/s 2-25 41.7 mA

Practice Quiz 1. C 2. 5 mC 3. (a)  25 A (b)  3A 4. 6.24 × 1017 5. 10.7 s 6. 3.0 V 7. 650 μC 8. 1.5 A

Chapter 3 CC 3-1 Aluminum is lighter and cheaper than copper and silver, but has fewer free electrons per unit volume. CC 3-2 Charge is carried by positive and negative ions rather than by free electrons. CC 3-3 The key factor is the number of free electrons per unit volume, which is determined by the type and strength of the bonds among the atoms in the material.

Answers to Selected Problems

Practice Quiz 1. silver, copper, aluminum, carbon, glass 2.  Some polymers have covalent bonds that hold the outer electrons of the atoms so tightly that there are very few free electrons. 3. Spacing of the two conductors, and type of insulating material. 4.  Each atom in the crystals shares an ­electron with each of four adjacent atoms. These ­covalent bonds have moderate strength, so the crystals have fewer free electrons per unit ­volume than conductors, but more than i­ nsulators.

Chapter 4 CC 4-1 Conventional current enters the anode of a device and leaves the cathode. CC 4-2 Alkaline cells have the same nominal voltage as carbon-zinc cells, but produce more energy for a given size and are less likely to leak. CC 4-3 about 10 thousand hours (roughly 14 months) 4-1 400 days

Practice Quiz 1. chemical reactions 2. battery 3.  The chemical reactions in primary cells are not reversible. 4. greater capacity than most cells of the same size, almost constant voltage, long shelf life 5. The most common types are lead-acid, nickel-cadium, nickel-metal hydride, and lithium ion. 6. The capacity is reduced. 7. 30 h 8. A and C 9. A, B, and C

Chapter 5

470 Ω 3.9 m ρ = 24 nΩ-M material is gold 12 500 ft −40ºC 67 m (a)  100 kΩ ±5% (b) 22 Ω ±10% (c) 715 Ω ±2% CC 5-8 (a)  95 kΩ, 105 kΩ (b)  19.8 Ω, 24.2 Ω (c)  700.7 Ω, 729.3 Ω CC 5-9 4.8 mA 5-1 240 Ω 5-3 8.8 Ω 5-5 3 kΩ 5-7 466.7 Ω CC 5-1 CC 5-2 CC 5-3 CC 5-4 CC 5-5 CC 5-6 CC 5-7

989

5-9 2.15 Ω 5-11 1.31 Ω 5-13 2.75 Ω 5-15 2.252 Ω 5-17 47.9 cm 5-19 0.25 mm 5-21 3 × 10−5 Ω-m 5-23 2.82 Ω 5-25 0.60 mm 5-27 (a)  39.69 CM  (b)  1225 CM  (c)  319 225 CM 5-29 3.00 ft 5-31 Tungsten 5-33 AWG #18 5-35 1.015 Ω 5-37 13.65 m 5-39 −313.3ºC 5-41 28.3ºC 5-43 2.81 mm 5-47 (a) Brown–Black–Orange–Gold–Brown (b) Red–Red–Green–Red–Red (c) Yellow–Violet–Silver–Brown (d) Orange-Brown-Blue-Gold-Red (e) Grey-Red-Green-Black-Green (f) Brown-Black-Yellow-Gold 5-49 Red-Yellow-Gold-Red 5-51 1.5 A 5-53 7.917 A 5-55 4.5 V 5-57 8.96 V 5-59 25 mA 5-61 178 μA 5-63 4.4 V 5-65 3.96 V 5-67 1.5 V 5-69 25 mA

Practice Quiz  1.  2.  3.  4.  5.  6.  7.  8.  9. 10. 11. 12. 13. 14. 15.

C A D C D C temperature coefficient of resistance The ratio V/ I is constant. photoresistor, thermistor, varistor, tungsten lamp filament A B A rheostat is a variable resistor with only the wiper and one end connected to the circuit, while a potentiometer has all three terminals connected. B C A

990

Answers to Selected Problems

Chapter 6 CC 6-1 30 J CC 6-2 1440 W CC 6-3 (a)  damaged since the power is 10 W (b)  damaged since the power is 48 W (c)  undamaged since the power is 10 mW CC 6-4 12.5 hp CC 6-5 $230.40 CC 6-6 10.41 kg 6-1 3185 J 6-5 540 W 6-7 6V 6-9 6.25 mA 6-11 569 W 6-13 220 Ω 6-15 23.5 mA 6-17 1/8 W Resistor 6-19 32 Ω 6-21 202 V 6-23 64.8 MJ 6-25 1.245 kW 6-27 8.488 Ω 6-29 21.69 A 6-31 77.0% 6-33 171.5 kJ 6-35 625 mA 6-37 145.7 W 6-39 $99.84 6-41 100.8 V 6-43 8.267 hr 6-45 2.592 kW 6-47 21.76 C 6-49 17.6 Ω 6-51 242 Ω 6-53 2.981 A 6-55 74.6% 6-57 122.5 mV 6-59 1/4 W Resistor 6-61 5.333 pW 6-63 97.14 A 6-65 3.0 MΩ 6-67 155 μΩ 6-69 500.5 μΩ 6-71 847.7 kJ 6-73 3.14 min 6-75 4.32 A 6-77 64.6ºC

Practice Quiz 1. 2. 3. 4.

A, B D D surface area

5. B 6. C 7. C

Chapter 7 CC 7-1   VAF

VBF

12.0 V 9.0 V

VCF

VDF

VEF

VBE

8.3 V 6.4 V 4.5 V 4.5 V

VCE VDE VAC VBC VDC VEC 3.8 V 1.9 V 3.8 V 0.75 V −1.9 V −3.8 V

CC 7-2 2 Ω, 5 Ω CC 7-3 500 Ω or 20 Ω (Both answers work) CC 7-4 5.75 V CC 7-5 2.5 Ω CC 7-6 Bulb L1 will be on when S1 is closed and S2 is open (or vice versa). Bulb L2 will be on when S1 is closed, S2 is open, and S3 is closed. Bulb L3 will be on when S1 is open, S2 is closed, and S3 is open. CC 7-7 (a)  R2, R3, and R4 are connected in parallel, and this combination is in series with R1. (b) R2 and R3 are connected in parallel, and this combination is in series with R1 and R4. CC 7-8 2.9 V; the lowest resistance (33 Ω) will burn out first. 7-1 60 Ω 7-3 300 mA 7-5 40 V; 80 V; 120 V 7-7 2.8 Ω; 17.5 W 7-9 100 Ω; 75 Ω 7-11 4889 V 7-13 168.3 Ω 7-15 (a)  16.8 kV (b)  16.2 kV (c)  12 mA 7-17 12 V 7-19 1.25 A; 15 V 7-21 0.25 Ω 7-23 (a)  4 V (b)  25.2 W (c) 87.5% (d)  101 W 7-25 (a)  81.3 mV (b)  3.6 kΩ (c)  2.08 mW 7-27 8 Ω 7-31 3.041 kΩ 7-33 250 Ω 7-35 5.3 mA 7-37 60 kΩ 7-39 968 Ω 7-41 13.35 mA 7-43 56 kΩ, 36 kΩ

Practice Quiz 1. 2. 3. 4.

B C A The voltage sources are active circuit elements, and all the resistors are passive.

Answers to Selected Problems

5. 

(a) 

Practice Quiz

R1

Vs

  (b) 

1. 2. 3. 4. 5.

R2

Chapter 9

R1

Vs

R2

R3

R4

 6.  7.  8.  9. 10.

C B D I1 = 39.2 mA, I2 = 29.4 mA, I3 = 20 mA A

Chapter 8 CC 8-1 CC 8-2 CC 8-3 CC 8-4

A C C B C

1.4 kΩ 5.15 kΩ R1 = 39.8 kΩ , R2 = 200 Ω VR1 = 0.69 V, VR2 = 1.5 V, VR3 = 2.3 V, VR4 = 3.5 V, VR5 = 6.9 V CC 8-5 1.0-kΩ resistor, 100 mA; 2.0-kΩ resistor, 50 mA; 4.0-kΩ resistor, 25 mA; 100 V CC 8-6 1.4 A 8-1 (a) 36 Ω (b) 1800 Ω 8-3 (a)  250 mA (b)  981.5 mA (c) 55.94 μA 8-5 (a) 173.3 Ω (b) 93.3 Ω (c) 373.3 Ω (d) 293.3 Ω (e) 240 Ω (f) 360 Ω 8-7 (a) 361.4 Ω (b) 354.3 Ω 8-9 (a) 75 Ω (b) 77.38 Ω 8-15 10.44 V 8-17 320 Ω; 533.3 Ω 8-19 1.2 kΩ, 12 W; 4.8 kΩ, 3 W; 1.6 kΩ, 9 W 8-21 R1 = 817 MΩ; RF = 665 kΩ 8-23 +9 V; −15 V 8-25 1.6 Ω, 10 W; 28 Ω, 7 W; 64 Ω, 4 W; 24 Ω, 6 W 8-29 3.0 kΩ 8-31 250 Ω 8-33 (a) 20 Ω (b)  1.33 A (c)  18.7 V (d)  11.6 W 8-35 640 mA 8-37 2.5 Ω

CC 9-1 19.2 V CC 9-2 ISC = 8.0 A CC 9-3 −152mA CC 9-4 1.27 A CC 9-5 0.634 V CC 9-6 5.03 A 9-1 10 Ω; 31 V 9-3 14.6 A; 146 V 9-5 (a)  68 kΩ; 0.368 μA (b) 225 Ω; 1.09 A (c)  40 kΩ; 2 mA 9-7 −5.33 mA 9-9 1A 9-11 −5.714 mA; 2.143 mA; 3.571 mA 9-13 (a)  15.36 V (b)  18.56 V 9-15 324.75 W 9-17 473 mA; 481 mA 9-19 6.86 mA 9-21 20.78 kΩ 9-23 75 Ω 9-25 (a)  −3.50 V (b)  −7.80 V 9-27 1.78 V 9-29 I15 = −10.33 mA; I10 = −11.5 mA; I20 = 6 mA; I25 = 6 mA; I30 = 1.17 mA; I40 = 0 mA; I120 = 16.33 mA; I150 = −17.5 mA 9-31 0.729 A 9-33 6.67 V 9-35 −0.8 V 9-37 2.53 A; 3.84 A 9-39 −35 V 9-41 5.83 A

Practice Quiz  1.  2.  3.  4.  5.  6.  7.  8.  9. 10. 11.

A B A A B C D C D C C

991

992

Answers to Selected Problems

Chapter 10

CC 10-1 ETh = 12 V, RTh = 92 Ω, V8.0Ω = 960 mV CC 10-2 IN = 130 mA, V40Ω = 4.8 V CC 10-3 37.5 V 10-1 160 V; 180 Ω 10-3 320 mA 10-5 220 Ω 10-7 35.56 W 10-9 875 mA 10-11 237.1 V; 36 kΩ 10-13 4.76 mA 10-15 76 Ω 10-17 0.33 A 10-19 180 Ω; 0.889 A 10-21 275.9 mA 10-23 137.5 mA 10-25 0.33 A 10-27 11.25 V 10-29 33.86 V 10-31 18 V; 500 Ω 10-33 13.33 V 10-35 80 Ω 10-37 35.98 kΩ; 11.12 kΩ; 74.13 kΩ 10-39 8.75 mA 10-41 1.42 A

Practice Quiz 1. 2. 3. 4. 5. 6. 7.

A B C D B A C

Chapter 11 CC 11-1 CC 11-2 CC 11-3 CC 11-4 CC 11-5

20 kΩ/V 10 kΩ 900 kΩ 10.91 V Rint = 29.7 kΩ, R1/4 = 90 kΩ, R1/2 = 30 kΩ, R3/4 = 10 kΩ, I50Ω = 49.9 µA CC 11-6 3.7 mΩ 11-1 0.07538 Ω 11-3 2.513 Ω 11-5 299 kΩ; 700 kΩ; 2 MΩ; 7 MΩ 11-7 V1 = 3 V; V2 = 15 V; V3 = 60 V 11-9 15.2% 11-11 3 kΩ; 0.4 mA; 2500 Ω/V 11-13 30 V 11-15 Rsh = 5.05 Ω; Rs = 495 Ω 11-17 983.6 V

11-19 625 Ω 11-21 55.5 Ω

Practice Quiz 1. C 2. resistance and sensitivity 3. Any resistance in the ammeter reduces the c­ urrent being measured. 4. B

Chapter 12

8.00 × 10−3 N 220 nF 80 V 0.282 mm, 847 V (a)  1.75 µF Q1 µF = 50 µC, Q250 nF = 12.5 µC, (b)  Q0.5 µF = 25 µC CC 12-6 33.3 µF, V50µF = 100 V, V100µF = 50 V 12-1 2.70 × 10−10 N 12-3 4.01 × 10−15 N 12-5 4.80 μC/m2 12-7 6.4 kV 12-9 21.9 pF 12-11 13.9 pF 12-13 2.21 nF 12-15 46.0 pF 12-17 80 μF; 3 mC; 17 mC 12-19 (a)  7.50 μF  (b)  1.80 mC  (c)  120 V; 81.8 V; 38.3 V 12-21 50 pF 12-23 0.77 nF; 1.15 V; 28.85 V 12-25 (a)  17 × 107 reciprocal farads (b) 12.5 × 106 reciprocal farads CC 12-1 CC 12-2 CC 12-3 CC 12-4 CC 12-5

Practice Quiz 1. 2. 3. 4.

Field strength has both magnitude and direction. relative strength of an electric field A, C The dielectric constant of a material is the permittivity of the material divided by the permittivity of free space. 5. C 6. D 7. A

Chapter 13 CC 13-1 CC 13-2 CC 13-3 CC 13-4 13-1

47 μF 331 μA 470 μF 22.4 V (a)  1.25 mA (b)  0 V (d)  4 s (e)  20 s

(c)  62.5 V/s

Answers to Selected Problems

(a)  155 V (b) 460 μA (c)  2.0 s (d)  5.5 s 13-5 43.3 μF 13-7 (a)  36 s (b)  2.05 mA 13-9 (a)  13 s (b)  57 V (c)  0.52 mA (d)  14 s 13-11 220 μF 13-13 (a)  20 s (b)  86.5 V (c)  10.5 V 13-15 (a)  1.65 ms (b)  0.324 ms (c)  12.48 V (d)  0.2273 mA (e)  0.363 ms 13-17 (a)  46.1 V (b)  15.7 V 13-19 155.7 V 13-21 (b)  86.5 V (c)  9.2 V (d)  2.77 s 13-23 (a)  0.625 J (b) 1.28 μJ 13-25 64.8 mJ; W not affected by R 13-3

Practice Quiz  1. B, C, D, E  2. A  3. D  4. C  5. B  6. A  7. B  8. C  9. D 10. B

Chapter 14 CC 14-1 120 mA CC 14-2 1.1 mWb CC 14-3 6.25 × 104 At/Wb CC 14-4 2.57 × 104 At/Wb CC 14-5 1.54 T CC 14-6 3684 At/m, 1.73 × 10−3 H/m CC 14-7 1.9 × 10−3 H/m CC 14-8 3.5 × 10−4 H/m 14-5 20 At 14-7 4.5 A 14-9 3.15 × 108 At/Wb 14-11 9.95 × 106 At/Wb 14-13 3.98 × 10−3 H/m 14-15 2.5 mWb 14-17 1.21 T 14-19 1.25 T 14-21 1.95 × 10−4 H/m 14-23 2.5 × 10−4 H/m

Practice Quiz 1. 2. 3. 4.

C and D C A D

 5.  6.  7.  8.  9. 10. 11. 12.

B D C reluctance C A B Eddy currents cause power loss and heating in the core.

Chapter 15 CC 15-1 0.037 mWb CC 15-2 1.2 mWb CC 15-3 0.91 A CC 15-4 2172 turns 15-1 4.112 × 10−7 Wb 15-3 600 turns 15-5 5.25 T 15-7 200 mA 15-9 440 mA 15-11 6.66 A 15-13 1423 turns 15-15 3.166 A

Practice Quiz  1.  2.  3.  4.  5.  6.  7.  8.  9. 10.

993

D A A B, C, and D C D A C D The leakage flux is reduced.

Chapter 16 CC 16-1 9.38 H CC 16-2 2.4 kV CC 16-3 (a)  7.27 H (b)  71 turns CC 16-4 0.15 mH CC 16-5 3 H CC 16-6 0.56 Wb/pole CC 16-7 94.8 A 16-1 300 mH 16-3 640 V 16-5 313 mH 16-7 39 turns 16-9 17.07 H 16-11 69 mH 16-13 (a)  4.71 H (b)  1077 V 16-15 24 μH

994

Answers to Selected Problems

Practice Quiz  1.  2.  3.  4.  5.  6.  7.  8.  9. 10.

B C C A C C C A D B

Chapter 17

CC 17-1 48 Ω CC 17-2 502 mH CC 17-3 46.1 mJ CC 17-4 (a) 1.6 S, 0.46 S (b) 1.25 A 17-1 1.25 s 17-3 4.8 V 17-5 (a)  0 A (b)  2.4 A/s (c)  4.5 A (d)  9.38 s 17-7 (a)  250 ms (b)  173 ms 17-9 2.53 ms 17-11 8.934 V 17-13 60.7 Ω; 4.55 H 17-17 (a)  0.576 J (b)  43.2 W 17-19 (a)  360 V (b)  66.7 ms 17-21 (a)  −29.26 V (b) 3.76 μs 17-23 (a)  4.685 V 17-25 1.557 A

(c)  1.29 S

Practice Quiz

(c)  3.01 A

(c)  30.16 ms

Practice Quiz  1.  2.  3.  4.  5.  6.  7.  8.  9. 10. 11. 12. 13. 14.

CC 18-5 (a) 115.2 V (b)  1.15 A (c)  132.8 W 18-1 172.19 V 18-3 −947.66 V 18-5 0V 18-7 176.6 V 18-9 e = 200 sin 2513t volts 18-11 (a)  −6.31 A (b) −3.71 A 18-13 (a)  112.5 W (b) 70 mW (c) 45.65 W 18-15 9.6 V 18-17 141.44 μV 18-19 e = 165.5 sin 2513t volts 18-21 17.6 V 18-23 3A 18-25 242 W 18-27 (a)  220 V (b)  155.54 V (c)  60 Hz (d)  −129.31 V (e)  1.39 ms (f)  3.11 A 18-29 500 Ω 18-31 e = 212.2 sin 1000t volts

A directly series equal to directly B D C B C B D A open windings, partially shorted windings, shorted windings

Chapter 18 CC 18-1 47.32 V CC 18-2 16.2 Hz CC 18-3 (a) 33 Hz (b)  51.5 V (c)  e = 49.26 V CC 18-4 61 mA

 1.  2.  3.  4.  5.  6.  7.  8.  9. 10. 11. 12. 13. 14. 15. 16.

B, D directly proportional increases perpendicular B C D C A D C B B B A D

Chapter 19

CC 19-1 15 Ω CC 19-2 39.79 mH CC 19-3 33.86 MHz CC 19-4 7.44 Ω CC 19-5 36.17 nF 19-1 44 Ω 19-3 1508 Ω 19-5 636.6 Hz 19-7 469.1 μH 19-9 291.8 mH 19-11 16.67 k Ω 19-13 7.96 k Ω 19-15 124.3 Hz 19-17 6.8 μF 19-19 13.67 μF 19-21 310.23 kHz

Answers to Selected Problems

Practice Quiz  1.  2.  3.  4.  5.  6.  7.  8.  9. 10. 11. 12. 13. 14.

C A D C C D B C B A C C A B

Chapter 20 CC 20-1   342 V CC 20-2   (c) 0.5 20-1 143.4 V 20-3 iT = 65.68 sin (314t − 0.57 rad) A 20-5 0V 20-7 146.21 V∠+55º 20-9 (a) 108.8 − j50.7 (b)  −45 (c)  −j35 (d)  −128.7 + j18.3 (e) 6.8 + j3.3 (f) 1.41 − j1.41 20-11 26.4 ∠24.62º 20-13 198.12 ∠−133.74º 20-15 134.39 ∠77.28º 20-17 (b) 0.8256 ∠−16.39º 20-19 (a) 53.3 ∠−51º (b) 0.3154 ∠−77º 20-21 26.8 Ω ∠+8.9º

Practice Quiz  1.  2.  3.  4.  5.  6.  7.  8.  9. 10. 11. 12. 13. 14.

B C A A B B D B C A C C A D

Chapter 21

CC 21-1 (a)  975.4 mA ∠−12.75º (b) 11.03 V; 48.77 V CC 21-2 45.9 Hz

995

CC 21-3 (a) 82.5° (b)  51.5° (c)  0.104 A  (d)  157.9 V CC 21-4 10 000 + j626.41 Ω, or 10.02 kΩ ∠3.58º CC 21-5 117.25 mS∠89.51º CC 21-6 −j994.32 Ω, or 994.32 Ω∠−90º 21-1 (a)  22.36 V (b)  12.5 mV 21-3 120 Ω ∠60º; 60 + j103.9 Ω 21-5 157.3 Ω ∠17.44º 21-7 2.708 kΩ; 0.955 H 21-9 24 Ω; 257.51 mH 21-11 61.91 mH 21-13 100 V∠−36.87º 21-15 50 Ω ∠−7º 21-17 131.2 Ω ∠−23.85º 21-19 38.82 Ω; 244.1 pF 21-21 (a)  108.49 mA ∠3.51º (b)  99.81 V; 6.12 V (c)  −3.51° leading 21-23 (a)  199 mV ∠−5.74º 21-25 (a)  2.07 kΩ ∠−66.61º (b) 858.42 Ω∠+17.21º (c)  2.2 kΩ ∠68.08º 21-27 989.76 Ω ∠−19.35º 21-29 (a) 12.48 μF (b) 4.06 μF 21-31 18.6 μF or 11.3 μF 21-33 (a) 751.66 mA ∠+3.81º (b)  1.25 A ∠−90º 21-35 226.32 Ω ∠+14.85º 21-37 254.5 − j703.9 Ω 21-39 182.7 + j56.2 Ω 21-41 77.62 mS ∠−75.07º; 12.88 Ω ∠+75.07º; 21-43 13.12 mS ∠−23.85º 21-45 433 mS; 250 mS (inductive) 21-47 (a) 35.16 Ω ∠87.54º (b) 250.03 Ω ∠−72.25º (c) 32.63 Ω ∠−87.72º 21-49 10.3 kΩ ∠10.3º 21-51 148.8 Ω ∠81.4º 21-53 2.77 kΩ; 3.82 H 21-55 7.68 Ω; 37.71 μF 21-57 32.1 mH 21-59 (a) 20 Ω; 30.6 mH (b)  23.0 mH

Practice Quiz  1.  2.  3.  4.  5.  6.  7.  8.  9. 10. 11. 12. 13. 14.

A A 4.62 kΩ C C D B false C A B D B D

996

Answers to Selected Problems

15. C 16. A 17. B

10. F 11. A 12. C

Chapter 22

Chapter 23

              (e)  470 μF CC 22-4 (a)  500 W (b)  800 var (inductive) (c)  943.4 VA (d)  0.53 (e)  9.43 A ∠27.99º CC 22-5 (a)  7.07 kVA (b) 0.707 (c)  64.27 A ∠−45º (d)  1.1 mF (e)  45.45 A 22-1 (a)  660 W (b)  1320 W 22-3 437.27 W 22-5 (a)  0 W (b)  7.54 mvars (c)  7.54 mW 22-7 40.8 Hz 22-9 459.8 μF 22-11 30.07 kW 22-13 (a)  10.42 A (b)  868.61 W; 900.1 VAR (inductive), 1.25 kVA 22-15 8.9 kHz 22-17 (a) 9 Ω (b)  2.53 mH 22-19  28.8 Ω; 573 mH 22-21  98% 22-23   147.58 V 22-25  11.8 22-27  84.58 Ω; 327.93 mH; 59.12 μF 22-29  83.33 Ω; 765.73 mH 22-31  28.06 μF 22-33  (a)  21.87 kvars (capacitive)   (b)  82.5% leading 22-35   6.83 A 22-37   0.986 leading; 8.6%

Practice Quiz  1.  2.  3.  4.  5.  6.  7.  8.  9. 10.

A, B, C A B B A A A D B D

Chapter 24

CC 24-1 18.89∠+15.40º A

−j2 4

θ = 19.44° PTRUE = 78.46 W

−j

PA = 83.2 VA

CC 23-1 347.97 Ω∠71.7º CC 23-2 (a)  802.36 mA∠55.27º (b)  12 V∠150.02º CC 23-3 21.06 A∠−45.18º 23-1 346 Ω ∠7.11º 23-3 (a)  1.07 A ∠−47.98º (b)  228.5 V∠14.1º 23-5 70.71 ∠45º 23-7 1.7 kΩ ∠41.4º 23-9 10 kΩ ∠60º 23-11 22.06 mV∠92.10º 23-13 (a) 7.49 μA ∠−34.2º (b) 49.4 μA ∠82.5º 23-15 12.7 V ∠11.5º 23-17 (a)  130.68 mA ∠−2.57º (b)  171.51 mA ∠20.58º 23-19 (a)  5 A ∠6.87º (b)  150 W; 200 var (c)  0.60 lagging 23-21 0.229 A ∠106.9º 23-23 36.0 V∠−41.94º

50 V 60° 2.

0

j8

Ω

50 V 60° Ix

Practice Quiz  1.  2.  3.  4.  5.  6.  7.  8.  9.

A, C C A C C Q = 1 VAR C C B

50 V 60°

j4

PR = 27.69 var

CC 22-1 176.75 mA; 5.47 W CC 22-2 5.1∠41.7 var (capacitive) CC 22-3  (a)  78.46 W (b)  27.69 var (c)  83.2 VA (d)  0.943

50 V 60°

3Ω

j5

50 V 0° −j4

−j8

CC 24-2 16.0 A ∠− 4.0º CC 24-3 65.50 V∠−77.06º CC 24-4 VTh = 3.83 V ∠−112.5º, ZTh = −j5 Ω



Answers to Selected Problems

997

CC 24-5 (a) 330 Ω (b)  7.58 W 25-13 (a) 10 Ω; 3.18 mH; 35.40 nF CC 24-6 15.4 V ∠−124.3º (b)  15 250 Hz; 14 750 Hz 24-1 1.69 A ∠−59.11º; 1.52 A ∠120.45º 25-17 4.11 kHz 24-3 141.22 mA ∠+18.60º 25-19 3.11 μA 24-5 15.51 mA ∠−87.96º; 6.81 mA ∠−137.89º, 25-21 (a) 23.58 μH (b)  53.1 pF 12.29 mA ∠−62.86º 25-23 (a)  445.99 Hz (b) 4.71 Ω 24-7 0.935 A ∠−55.4º; 0.935 A ∠124.6º 1.87 A ∠124.6º Practice Quiz 24-9 1.01 A ∠52.5º; 1.434 A ∠−21.34º  1. B, D 10. B 24-11 2.847 A ∠8.28º  2. C 11. B 24-13 269.4 mA ∠−62.98º  3. B 12. A, B, C, D 24-15 153.3 V; 173.1 V  4. A 13. D 24-17 591.58 mA ∠−74.13º  5. A, D 14. A 24-19 20.1 V ∠35.7º  6. A 15. A, B, C, D 24-21 84.62 Ω  7. C 16. A 24-23 0.0265 μF  8. D 17. C 24-25 3.97 A ∠−8.63º  9. D 18. B re RL Zo = 24-27 re + ( 1 + hf ) RL Chapter 26 24-29 7.74 kΩ ∠−76.44º 24-31 3.01 mA ∠−90º CC 26-1 72 Hz 24-33 83.3 Ω ∠0.95º; 1.66 Ω ∠−89.1º; CC 26-2 80 kHz 8.29 Ω ∠−89.1º CC 26-3 2.3 kHz 24-35 1.63 A ∠−66.5º CC 26-4 38 kHz 26-1 (a)  9.03 dB (b)  13.98 dB (c)  20 dB Practice Quiz 26-3 14.56 dB  1. B, C 26-5 72.34 kHz  2. A 26-7 72.34 nF  3. C 26-9 314.16 kΩ  4. C 26-11 482.29 Hz  5. A 26-13 7.96 nF  6. D 26-15 829 Ω  7. C 26-17 19.41 kHz; 70.92 kHz  8. D 26-19 (a)  7.69 kHz (b)  636.31 Hz  9. C 26-21 253.30 nF 10. B 26-23 (a)  267.94 Hz; 1.33 kHz 11. D 26-25 (a)  4.79 kHz 12. C (b)  318.35 Hz 26-27 Since the critical frequency is off by a factor of Chapter 25 10, the most likely cause is a resistor 10 times too small or an inductor 10 times too large. CC 25-1 2.1 mH, 89.4 kHz CC 25-2 100 Ω, 1.59 mH, 15.9 pF Practice Quiz CC 25-3 (a)  15.8 kHz, 99.5 krad/s  1. A high-pass, B low-pass, C band-stop, (b)  9.95 D ­band-pass (c) VR = 95.2 mV,IL = IC = 0, IR = IT = 10µA  2. B (d)  952 nW (e)  1.59 kHz (f)  67 mV  3. C CC 25-4 (a)  159 kHz, 992 krad/sec (b)  2A  4. A (c)  4 mH (d)  2.51 nF  5. D 25-1 4.11 kHz  6. A 25-3 1.04 μH  7. B 25-5 500 Ω; 13.99 mH; 8.75 pF  8. C 25-7 26.7 μA  9. B 25-9 100.7 10. C 25-11 (a)  466.9 kHz (b) 3.95 Ω; 108.5 μH

998

11. 12. 13. 14.

Answers to Selected Problems

C D No Yes

Chapter 27 CC 27-1 18.77 Hz CC 27-2 2 CC 27-3 (a) 5 V (b)  20 mA (c)  250 Ω (d)  100 mW (e) 1 kΩ CC 27-4 91.9%, 86.2% CC 27-5 −0.62% 27-1 240 turns; 24 turns; 16.67 A; 166.67 A 27-3 113.1 mA 27-5 171 cm2 27-7 6.67 27-9 68.03 V 27-11 560 turns 27-13 90 V 27-15 43.2 turns 27-17 80 kΩ 27-19 (a) 34.72 Ω (b) 8.68 Ω 27-21 2.17 27-23 0.173 Ω; 0.40 Ω 27-25 211.5 kW; 98.3% 27-27 2.19% 27-29 −0.412% 27-31 (b) 4.0; 80 kVA

Practice Quiz  1.  2.  3.  4.  5.  6.  7.  8.  9. 10. 11. 12.

B, C, E, F C A B B A C C C D C B

Practice Quiz

Chapter 28 CC 28-1

L1

L2

10 mH

50 mH L3 30 mH

 

CC 28-2 z11 = 430 Ω, z12 = z21 = 330 Ω, z22 = 550 Ω CC 28-3 (a) z11 = −j3.5 Ω, z12 = z21 = −j66.3 Ω, z22 = −j3.5 Ω (b)  35.6∠−53.3º V CC 28-4 (a)  3.49 kΩ ∠87.30º (b)  3.51 kΩ ∠89.66º CC 28-5 (a)  y11 = j 6 Ω = y22, y12 = y21 = j4 Ω (b)  3.76∠84.1º Ω CC 28-6 h11 = j 1252 Ω, h12 = −18.9, h21 = 18.9, h22 = j0.29 S 28-1 z11 = 940 Ω; z12 = z21 = 610 Ω; z22 = 1080 Ω 28-3 z11 = 20 Ω; z12 = z21 = j300 Ω; z22 = 20 Ω 28-5 Zm = 10 ∠0º Ω; Zp = 40 ∠0º Ω; Zs = 65 ∠0º Ω 28-7 Zm = j10 kΩ; Zp = Zs = 33 − j30 kΩ 28-9 1.1 Ω ∠0º 28-11 729 Ω ∠0º 28-13 0.112 28-15 z11 = 643.5 Ω; z12 = z21 = 285.1 Ω; z22 = 321.2 Ω 28-17 z11 = z22 = 0 Ω; z12 = z21 = −j3600 Ω 28-19 512.7 Ω 28-21 y11 = 5 mS ∠0º; y12 = y21 = 9 mS ∠180º; y22 = 11 mS ∠0º 28-23 Ym = 500 μS ∠0º; Yp = 200 μS ∠0º; Ys = 700 μS ∠180º 28-25 y11 = y22 = 2.70 mS ∠−51.85º; y12 = y21 = 1.67 mS ∠180º 28-27 h11 = 595.46 Ω; h21 = −0.565; h12 = 0.565; h22 = 926 mS 28-29 14.78 28-31 4.13 28-33 −1.65 H 28-35 6.53 kΩ ∠88.96º 28-37 513.17 Ω ∠66.83º  1.  2.  3.  4.  5.  6.  7.  8.  9. 10. 11. 12. 13.

A, B B D C B B C D A D B C A

999

Answers to Selected Problems

CC 29-1 (a)  IA = 1.6 A ∠45º, IB = 1.6 A ∠165º ,   IC = 1.6 A ∠−75º (b)  VAN = 120 V∠0º, VBN = 120 V∠120º,    VCN = 120 V∠−120º (c)   IN = IA + IB + IC = ( 0.414 − j1.55 ) A + ( 1.13 + j1.13 ) A + ( −1.55 + j0.414 ) A = 0 CC 29-2 (a)  IAC = 20 A ∠−45º, IBA = 10 A∠−75º,   ICB = 20 A ∠165º (b)  IA = 11.44 A ∠−11.39º,   IB = 26.46 A ∠−34.11º,   IC = 39.20 A ∠153.50º (c)  I f the delta load is balanced, the line ­current will have a magnitude of √3 times the phase current in each arm of the load and will be displaced 30° from the phase ­current. CC 29-3 2.94 A CC 29-4 (a)  26.6 A (b)  15.4 A CC 29-5 IA = 2.4∠−60ºA, IB = 1.2∠−140ºA, IC = 3.0∠30º A, IN = 3.2 ∠−25.1º A 29-3 (b) 4.63 A ∠−25.84º; 4.63 A ∠94.16º; 4.63 A ∠−145.84º (d)  0 A 29-5 (b)  2.4 A ∠0º; 3.11 A ∠41.98º; 0.226 A ∠−30º (d)  5.29 A ∠21.84º 29-7 5.48 A ∠−178.1º 29-9 (b) 37.05 A ∠31.79º; 37.05 A ∠151.79º; 37.05 A ∠−88.21º (e) 21.39 A ∠31.79º; 21.39 A ∠151.79º; 21.39 A ∠−88.21º 29-11 (b) 185.44 mA ∠14.17º; 176.75 mA ∠−155.95º; 32.37 mA ∠124.63º 29-15 (c) 208 V ∠150º; 208 V ∠−90º; 208 V ∠30º (d) 56.56 A ∠31.79º; 56.56 A ∠151.79º; 56.56 A ∠−88.21º (e) 97.97 A ∠31.79º; 97.97 A ∠151.79º 97.97 A ∠−88.21º 29-17 (c) 208 V ∠30º; 208 V ∠150º; 208 V ∠−90º (d) 22 mA ∠−59.94º; 78.49 mA ∠120º; 13.87 mA ∠−90º (e) 100.49 mA ∠−59.99º; 90.77 mA ∠115.62º 12.17 mA ∠154.86º 29-19 1.96 A ∠−130.52º 29-21 4.32 kW 29-23 727.5 W 29-25 CBA 29-27 3.0 A ∠−1.71º; 4.58 A ∠154.4º; 2.36 A ∠−61.3º 29-29 30.24 A; 24.2 A; 17.9 A

Practice Quiz  1. A, C, D, F  2. C

 3.  4.  5.  6.  7.  8.  9. 10. 11. 12.

B D B D C D B B B B

Chapter 30

CC 30-1 (a)  VT = 12 V ( sin 377t + 0.417 sin 1510t ) (b)  asymmetrical CC 30-2 15

10 sin t + 6 sin 2t

10

10 sin t

5

6 sin 2t

e0 −5 −10



−15

0

1

2

3

4

5

t

6

7

8

9

10

7

8

9

10

6 4 5 sin t

2 Voltage

Chapter 29

0 −1.25 sin 5t

−2 −4 −6

0

1

2

3

4

5 Time

6

30-3 (a)  i = 14.14 sin 6283t + 2.83 sin 31416t A (b)  symmetrical 30-11 171.1 V 30-13 VCf = 26.9 V ∠−32.1º VC3 = 4.97 V ∠118º

Practice Quiz  1.  2.  3  4.  5.  6.  7.  8.  9. 10. 11.

A, C B D B A B C D C B B

Glossary

absolute permittivity  See permittivity. active circuit element  device that generates electric energy at the expense of some other form of energy

average permeability   the slope of the straight line drawn from the origin to the upper knee of the ­magnetization curve of a magnetic material

active filter  filter constructed with an amplifying device, such as a transistor or operational amplifier

back or counter EMF   voltage that is generated in the armature of a DC motor and opposes the applied v ­ oltage

active power  See real power.

balanced load   load connected with equal ­impedances in each arm

admittance  overall ability of an electric circuit to pass alternating current admittance diagram   phasor diagram of the conductances, inductive susceptances, capacitive susceptances, and resultant admittances in an AC circuit air-core coil   solenoid with air or any other ­nonmagnetic material as a core air gap   any nonmagnetic material through which magnetic flux passes in a magnetic circuit alkaline cell  See manganese-alkaline cell. alternating voltage   voltage that has a positive value for an interval of time followed by another interval of time during which it has a negative value alternator  generator that produces alternating c­ urrent American Wire Gauge (AWG)  the predominant ­standard for diameters of conductors in North ­America ammeter   instrument for measuring the current ­flowing in a branch of a circuit ampere (A)  SI base unit for electric current; 1 A = 1 C/s

ampere hour   a unit of charge, often used to indicate the capacity of a cell or battery amplitude   maximum absolute value of a periodic wave angular velocity  rate at which a body or vector ­rotates about a point or axis anode  the electrode from which negative charge leaves a device; the negative terminal of a primary cell antiresonance  resonance of a parallel-resonant c­ ircuit apparent power   power in volt amperes (VA) ­supplied by a source in an AC circuit armature  main conductors of rotating machinery; develops voltage in a generator and torque in a motor autotransformer   transformer with a tapped winding that serves as both the primary and secondary windings

band-pass filter   filter that allows signals between two critical frequencies to pass through while attenuating all others band-stop (band-reject) filter   filter that attenuates signals between two critical frequencies while allowing all others to pass through bandwidth  range of frequencies between the ­half-power frequencies of a resonant circuit BH curve  See magnetization curve.

bifilar winding   two strands of wire interwound on a coil to produce tight magnetic coupling between the windings (coefficient of coupling close to 1) bleeder resistor   resistor connected in parallel with the load on a voltage divider to improve voltage regulation Bode plot   graph drawn on semilog graph paper with the vertical axis giving a variable in decibels and the horizontal axis representing frequency branch   any portion of an electric circuit that contains one or more elements through which the same current is flowing breadboard   reusable, solderless prototyping board for electric circuits British thermal unit (BTU)   heat required to raise the temperature of one pound of water by 1°F brush   carbon rod used to provide an electrical ­connection from an external circuit to the rotating parts of motors and generators brush   connector that maintains contact with a ­moving surface, such as a commutator or slip ring; u ­ sually made with carbon and held in place with a spring brushless motor   motor using electronic switching instead of a commutator and brushes calibrating resistor   precision resistor connected in series with a moving-coil movement to set the total ­resistance to some specific value

Glossary

capacitance  the property of an electric circuit that opposes any change in the voltage across that circuit capacitive reactance   the opposition of capacitance to alternating current capacitor   component that can store electric charge capacitor-run   single-phase induction motor having two identical windings with a capacitor connected ­permanently in series with one winding capacitor-start motor  single-phase induction motor that uses a capacitor with the starting winding capacity  the total energy stored in a cell, often ­expressed as an ampere-hour rating carbon-composition resistor   a resistor constructed of a mixture of finely ground carbon and an insulating compound carbon-zinc cell   primary cell with a carbon rod ­cathode, a zinc anode, and an electrolyte consisting of a paste of ammonium chloride and zinc chloride

1001

complex conjugate   complex number that has the same rectangular coordinates as another complex ­number except that the imaginary component has the opposite sign complex number   a number possessing both real and imaginary components complex plane   a plane with one axis representing the real components and a perpendicular axis representing the imaginary components of complex numbers compound motor    motor with both series and shunt field coils conductance  ability of a circuit to pass current (the reciprocal of resistance) conductivity   conductance of a unit length and cross section of a particular material conductor  material that readily allows electric charge to flow through it controlled source  See dependent source.

cathode   the electrode through which negative charge enters a device; the positive terminal of a ­primary cell

conventional current   current defined as a flow of positive charge that leaves the positive ­terminal of a source and enters the negative terminal of the source

cell   electric component that produces a voltage as a result of chemical action

copper loss  I2R power loss in a transformer due to the resistance of the windings

cell capacity  total energy a cell can deliver to a load, often expressed as an ampere-hour rating

core loss   power loss in a coil due to eddy currents and hysteresis in the iron core

centre frequency ( fo)   resonant frequency of a ­resonant circuit used as a band-pass filter CEMF  See counter EMF. charge  the electrical property associated with p ­ rotons and electrons that produces a force between them chassis   metal frame housing an electrical or electronic circuit, which is often connected to a ­reference point in the circuit to serve as the ground choke  See filter choke. circuit diagram   an illustration of an electrical system consisting of lines to represent conductors and standard graphic symbols to represent electrical ­components circular mil   the area of a circle one mil in ­diameter coefficient of coupling  ratio of mutual magnetic flux to the total primary magnetic flux in a transformer coercive force   magnetic field intensity or magnetizing force required to demagnetize a particular sample of iron commutator   segmented connector arranged to ­mechanically rectify the voltage in a generator or to maintain constant torque direction in a motor

coulomb (C)  SI unit of electric charge, equal to the charge on 6.24 × 1018 protons or electrons; 1 C = 1 A⋅s Coulomb’s law   the relationship determining the force between two electric charges: The force is ­proportional to the magnitudes of each charge and inversely p ­ roportional to the square of the distance ­between them. counter EMF   self-induced voltage in a coil coupled impedance   impedance subtracted from the open-circuit input impedance to determine the total input impedance of a two-port network coupling network  See two-port network. Cramer’s rule   technique for using determinants to solve a set of simultaneous equations critical coupling   conditions that provide maximum energy transfer in a transformer critical frequency ( fC)   frequency at which a filter’s output voltage is 70.7% of the maximum output voltage critically damped   conditions that make parallel resonance of a tuned circuit impossible cumulatively compounded   connected such that the series flux aids the shunt flux

1002

Glossary

current  See electric current. current-divider principle   the principle that (i) the ratio between any two branch currents in a DC parallel network is the same as the ratio ­between the two ­conductances through which these currents flow, and (ii) the ratio between any two phasor branch currents in an AC parallel ­network is the same as the ratio ­between the two admittances through which these ­currents flow

differentially compounded   connected such that the series flux opposes the shunt flux differential permeability   slope at a point on the ­magnetization curve of a magnetic material digital multimeter (DMM)   multimeter that provides a digital display of the quantity being measured dimensional analysis  procedure for determining the units of an algebraic combination of quantities

cycle   portion of a periodic waveform that ­repeats ­itself

diode  two-terminal electronic device that permits ­current to flow through it in one direction only

damping   increasing the series resistance or ­reducing the quality factor, Q, of a tuned circuit

discharge resistor   resistor connected in parallel with an inductance to protect components from highvoltage surges

d’Arsonval movement  See moving-coil movement. decade  range of frequencies over which the frequency increases by a factor of 10 decibel (dB)  logarithmic unit used to represent change in power levels or sound intensity delta-connected system   a three-phase system in which a delta-connected source is connected to a ­delta-connected load such that each arm of the load is connected directly across one of the generator coils delta-connected load   three-phase load with each arm connected from line to line delta-connected (Δ-connected) source  threephase source with its windings connected in a ­series loop that resembles the Greek letter Δ

delta-to-wye transformation equations  equations for finding (i) a resistive Y-network equivalent to a ­resistive Δ-network or (ii) an impedance Y-network equivalent to an impedance Δ-network

displacement current   current that transfers charge to or from a capacitance distributed winding   armature winding that is ­distributed evenly in slots around the rotor double-subscript notation  method for ­identifying (i) polarities of voltages in DC circuits and (ii) the d ­ irection of tracing loops for voltages in threephase circuits dry cell   a cell in which the electrolyte is a paste or saturated absorbent spacer rather than a liquid duality   circuits that have similar but opposite ­electrical characteristics dynamo   machine that converts mechanical energy into electrical energy eddy current   a current that flows in a circle around the circumference of a conductor effective value  See root-mean-square (RMS) value.

dependent source   source with an output that ­depends on a voltage or current appearing in some other branch of the network

efficiency   ratio of useful output energy (or power) to total input energy (or power), usually expressed as a percentage

determinant  an array of the constants and co­ efficients contained in a set of simultaneous ­equations

elastance  opposition to electric flux in a material

diamagnetic  having a permeability slightly less than that of free space dielectric  electrically insulating material dielectric absorption  a small potential difference that remains across some dielectrics after the removal of an electric field dielectric constant  See relative permittivity. dielectric strength   electric field strength required to break down a dielectric

electric circuit  connection of electrical devices that provides a continuous path for electric current to flow electric current   rate of flow of charge in an electric circuit electric field   region in which an electric force acts upon an electric charge electric field strength   the force per unit charge at a given point electric flux   electric field passing through a given surface (defined as equivalent to electric charge)

Glossary

electric flux density   electric flux per unit area electric flux line  See electric line of force. electricity   energy carried by an electric current electric line of force   the path along which a massless electrically-charged particle would move in an electric field electrode   conductor through which electricity enters or leaves a cell or other device electrodynamometer movement  instrument that uses a fixed and a moveable coil to measure current electrolyte   electrically conductive solution that ­contains ions electrolytic capacitor  capacitor that has aluminum foil plates with an electrolyte between them, and a thin aluminum oxide coating for the dielectric electrolytic cell   cell that decomposes a compound in its electrolyte when current passes through it electrolytic conduction   conduction resulting from a compound separating into positive and negative ions in solution electromagnet   coil of wire wound around a core that becomes magnetized when current flows through the wire electromagnetic induction  (1) the process of ­inducing a voltage across a solenoid by moving a ­magnet through it (2) the generation of a voltage when an electric conductor cuts across magnetic lines of force electromotive force (EMF)   electric energy per unit charge produced by the conversion of some other form of energy; a type of potential difference electron  a subatomic particle with a negative charge electron flow  current defined as a flow of free ­electrons electrostatic force   the field force acting between two electric charges electrostatic induction  separation of charge on a body in an electric field EMF  See electromotive force.

engineering notation  use of SI unit prefixes to ­express quantities equivalent circuits   electric circuits that have the same current-voltage characteristics equivalent DC value  See root-mean-square (RMS). equivalent resistance   resistance equal to that of two or more resistors in parallel

1003

exciting current   primary current that induces a ­voltage in the primary winding of a transformer equal in magnitude to the applied voltage farad (F)  SI unit of capacitance: 1 F = 1 C/V

Faraday’s law   the principle that the voltage induced in a circuit is proportional to the rate of change of the magnetic flux linked with the circuit ferrite  ceramic material that contains ferric oxide and usually has ferromagnetic properties ferroelectric dielectric   dielectric with an extremely large dielectric constant ferromagnetic  having a permeability hundreds of times greater than that of free space field excitation   production of magnetic flux in a motor or generator field system   electromagnetic system that produces magnetic flux for rotating machinery filter   circuit designed to allow AC signals of a ­particular range of frequencies to pass through while preventing signals of all other frequencies from doing so filter choke   inductor used to restrict variations in load current and power floating  not directly connected to ground or to the chassis of a circuit form factor   the ratio of the RMS value to the ­half-cycle average value of an AC wave free electron   an electron that is not bound to an atom frequency  number of cycles a periodic waveform completes in one second frequency domain   mathematical model for ­representing a time-varying quantity in terms of frequencies fringing   the spreading apart of magnetic lines of force due to the force of repulsion between them as they pass through an air gap in a magnetic circuit fuel cell   cell that produces electric energy directly from a reaction with a fuel (often hydrogen) fundamental frequency   lowest frequency present in a periodic waveform funicular phasor diagram   diagram in which phasors to be added are drawn in a chain with the first phasor s­ tarting at the origin and each successive phasor starting from the tip of the preceding one

1004

Glossary

galvanometer   sensitive moving-coil meter in which the pointer is at the centre of its range when the current through the coil is zero gas turbine   turbine driven by hot gases from a ­burning fuel Gauss’s law   the relationship between the distribution of electric charge and the resulting electric field: the electric flux through a closed surface enclosing a volume is proportional to the total charge within that volume general transformer equation  equation relating the AC voltage induced in a transformer winding to the ­frequency of the applied voltage, the number of turns in the winding, and the magnetic flux generator  See dynamo.

ground  point that is directly connected to the earth; a point in a circuit chosen as a reference node from which voltages of all other nodes may be mea­sured or calculated half-power frequencies   frequencies at which the current in a series-resonant circuit or the voltage across a parallel-resonant circuit is equal to 1/ 2 of the ­magnitude at resonance half-wave-rectified sine wave   a sinusoidal waveform in which either the negative half or the positive half of the waveform is replaced with a 0-V line harmonic  any frequency that is an integral multiple of a fundamental frequency heavy-duty cell  zinc-chloride cell

henry  SI unit of inductance; 1 H = 1 Wb/A

hertz (Hz)  SI unit of frequency, equal to one cycle per second high-pass filter   filter that allows signals above a critical frequency to pass through while attenuating all others horsepower (hp)  imperial unit of mechanical power; 1 hp ≈ 746 W h-parameters  See hybrid parameters.

h-parameter equivalent circuit   two-port equiv­alent circuit consisting of an impedance in series with a dependent voltage source for the input network and an admittance in parallel with a dependent current source for the output network hybrid parameters   four parameters of a twoport ­network, derived by treating the input port as a Thévenin equivalent voltage source and the output port as a Norton equivalent current source hydroelectricity   electricity generated using water power hysteresis   lagging of the magnetic flux density of a ferromagnetic material behind an alternating magnetic field strength

hysteresis loop   graph that displays the hysteresis of a material imaginary component  imaginary-axis component of a phasor; magnitude of the phasor multiplied by the sine of its phase angle imaginary power  See reactive power.

impedance  total opposition to current in an AC ­circuit; ratio of the voltage across and the current through a component in an AC circuit impedance diagram   diagram of the phasors representing resistances, inductive reactances, capacitive ­reactances, and the resultant impedance in an AC ­circuit impedance matching   selecting a load impedance to achieve maximum power transfer from a source to the load incremental permeability   ratio of the change in magnetic flux density to the change in magnetic field strength between two points on the magnetization curve of a material independent source   voltage source with terminal voltage unaffected by the load, or current source with output current unaffected by the load induced voltage   the voltage produced by a changing magnetic flux inductance  See self-inductance.

induction motor  AC motor in which the rotor current is developed by induction from the stator field inductive reactance   opposition of inductance to ­alternating current inductor  circuit component designed to have inductance initial permeability   permeability of a magnetic ­material when the magnetic field strength is nearly zero inrush current   a large initial current supplied to a load when a source is turned on, but which quickly ­reduces to a much lower value instantaneous value   value of a quantity at a particular instant insulator   material that has relatively few free electrons and is thus a poor conductor of electricity intensity of an electric field  See electric field strength. internal resistance   the resistance of a practical voltage (current) source that causes the terminal voltage (current) to change as the load is changed International System of Units (SI)  international standard for metric units ionization   process in which an atom or group of atoms gains or loses one or more electrons

Glossary

IR drop   voltage drop across a resistor

joule (J)  SI unit of work or energy; 1 J = 1 W·s = 1 C·V kelvin (K)  SI unit of temperature; 0 K is absolute zero and 1 K = 1°C kilogram   the base unit of mass in the SI system of units

kilowatt-hour   unit of energy commonly used by ­utility companies; 1 kW·h = 3.6 MJ

Kirchhoff’s current law   principle that the algebraic sum of the currents entering any node in a circuit equals the algebraic sum of the currents leaving the node Kirchhoff’s voltage law   principle that the ­algebraic sum of the applied voltages in any complete circuit equals the algebraic sum of the voltage drops

1005

line current   current in any of the lines of a polyphase system line voltage   the voltage measured between lines in a polyphase source lithium cell   primary cell with a lithium anode, a ­cathode consisting of manganese dioxide mixed with carbon, and an electrolyte consisting of lithium ­perchlorate dissolved in propylene carbonate lithium-ion (li-ion) cell   a secondary cell with a ­positive electrode of copper coated with carbon, a negative electrode of aluminum coated with a lithium compound, and an electrolyte of lithium salt load  any device that receives electrical energy and converts it to another form load current   current flowing to an external load from a generator or drawn from a source by a motor

ladder network   circuit consisting of series-­parallel arrangements of components resembling a ladder

local action   electrochemical reactions caused by ­impurities in the anode of a cell

lagging power factor   power factor associated with an inductive load, in which the current lags behind the voltage

long air-core coil   solenoid with air or any other ­nonmagnetic material as its core and a length at least ten times greater than its diameter

lap winding   connection of armature coils that has as many parallel paths as the machine has poles

low-pass filter   filter that allows signals below a critical frequency to pass through while attenuating all others

lead-acid cell   secondary cell with a lead peroxide positive electrode, a lead negative electrode, and dilute sulfuric acid for the electrolyte leading power factor   the power factor associated with a capacitive load, in which the current leads the voltage leakage factor   ratio of the total magnetic flux to the useful flux in a magnetic circuit leakage flux   magnetic flux lines that flow outside the ferromagnetic material in a magnetic circuit leakage reactance   equivalent inductive reactance that would produce a flux equal to the leakage flux of the primary of a transformer when connected in series with the equivalent ideal transformer Lenz’s law   principle that the polarity of an induced voltage must be such that any current it produces opposes any change in the original magnetic flux light dependent resistor (LDR)  See photoresistor. linear impedance   impedance that is not affected by voltage or current linear magnetic circuit   magnetic circuit in which the permeability is independent of magnetic field ­intensity linear resistor   resistor with a resistance that is not ­affected by voltage or current

LR circuit   circuit containing inductance and ­resistance, but not capacitance magnetic circuit   path that traces the closed loops formed by magnetic lines of force magnetic domain   group of adjacent atoms in a ­ferromagnetic material that react collectively to an ­external magnetic field magnetic field   region in which a magnetic force acts upon a magnetic material magnetic field strength (or intensity)   ­magnetomotive force per unit length of a ­magnetic circuit magnetic flux   average magnetic field passing through a perpendicular surface times the area of the surface magnetic flux density   magnetic flux per unit ­cross-sectional area perpendicular to the magnetic field magnetic flux line  See magnetic line of force.

magnetic line of force   path along which a ­theoretical isolated magnetic pole would move from one pole of a magnet to another magnetization curve   a graph of magnetic flux ­density, B, versus magnetic field strength, H, for a ­ferromagnetic material

1006

Glossary

magnetized   having the magnetic fields of the ­molecules within a ferromagnetic material aligned with an external magnetic field magnetizing current  See exciting current.

magnetizing force  See magnetic field strength.

magnetomotive force (MMF)   work that would be required to carry an isolated magnetic pole of unit strength around a magnetic circuit magnetomotive force gradient  See magnetic field strength. manganese-alkaline cell   primary cell with a manganese dioxide positive electrode, a zinc negative electrode, and an electrolyte consisting of an alkaline solution of potassium hydroxide and zinc oxide maximum power transfer   the greatest amount of power that can be transferred from a source to a load (occurs when the internal resistance of the source in a DC circuit equals the load resistance) mesh  a closed loop that does not enclose any other circuit elements within a network mesh equations   circuit analysis equations based on Kirchhoff’s voltage law, which can be used to calculate branch currents metre  the SI base unit of length, representing the distance travelled by light in a vacuum during a time interval of 1/299, 792, 458 of a second metric system   decimal system of units devised by the French Academy in 1791, based on the metre, gram and second, from which the SI system of units is ­derived mho   common name for the unit of conductance ­before siemens was adopted by the IEEE mil  one-thousandth of an inch

motor principle   the principle that a magnetic field exerts a force on charge that moves perpendicular to the field moving-coil movement   meter movement that uses the interaction of a current-carrying coil and a ­permanent magnet to measure current multimeter  instrument that can measure current, voltage, and resistance multiplier resistor   resistor connected in series with a moving-coil movement in a voltmeter to increase the voltage that corresponds to a full-scale reading mutual flux   the magnetic flux that links the ­pri­mary winding of a transformer to the secondary; the total ­primary flux minus the primary leakage flux mutual impedance   impedance common to the input and output ports of a two-port network, equal to

both the open-circuit reverse-transfer impedance and the open-circuit forward-transfer impedance of a passive two-port network mutual inductance   the inductance common to a pair of coils, enabling a changing current in one coil to ­induce a voltage in the other mutual induction   generation of a voltage in a ­secondary winding by a changing current in the ­primary winding mutual reactance   ratio of voltage induced in a ­secondary winding to the current in the primary neutral   common point where the phases of a ­polyphase source are connected together neutron  neutral particle present in atomic n ­ uclei

nickel-cadmium (ni-cad) cell   secondary cell with a nickel-hydroxide positive electrode, a c­ admium-­hydroxide negative electrode, and a p ­ otassium-hydroxide electrolyte nickel-metal hydride (ni-mh) cell  secondary cell with a nickel/nickel hydroxide positive electrode, a ­hydrogen-storage alloy for the negative electrode, and potassium hydroxide for the e­ lectrolyte nodal equations  circuit analysis equations based on Kirchhoff’s current law, which can be used to ­calculate node voltages in a circuit node  point in a circuit at which two or more ­components are connected nominal value  rated value

nonlinear circuit element   circuit component with an impedance that changes with voltage or current nonlinear magnetic circuit   magnetic circuit with a permeability that depends on the magnetic field strength nonlinear resistor   resistor with resistance that changes with current or voltage nonmagnetic  having the same permeability as free space normal permeability   the ratio of magnetic flux ­density to magnetic field strength at a given point on the magnetization curve of a magnetic material north pole   the pole from which the magnetic lines of force exit a magnet; the pole that tends to point toward the north magnetic pole of Earth Norton’s theorem   the principle that (i) any two-­ terminal DC network of resistors and sources may be replaced by a single current source in parallel with a single resistor and (ii) any two-terminal AC network of impedances and sources may be replaced by a single current source in parallel with a single impedance

Glossary

ohm (Ω)  the SI unit for resistance, reactance, and ­impedance; 1 Ω = 1V/A ohmmeter   instrument that measures resistance ohm metre  the SI unit of resistivity

Ohm’s law   the principle that the ratio of applied ­voltage to current is a constant for a given conductor or linear resistor open-circuit forward-transfer impedance  ratio of output voltage to input current of a two-port network with the output port open-circuit open-circuit impedance parameters  four ­impedance parameters of a two-port network, two of which are determined with the input port opencircuit and the other two with the output port open-circuit open-circuit input impedance   ratio of input voltage to input current of a two-port network with the output port open-circuit open-circuit output admittance   ratio of output ­current to output voltage of a two-port network with the input port open-circuit open-circuit output impedance  ratio of output ­voltage to output current of a two-port network with the input port open-circuit open-circuit reverse-transfer impedance  ratio of input voltage to output current of a two-port network with the input port open-circuit open-circuit reverse-voltage ratio   ratio of input voltage to output voltage of a two-port network with the input port open-circuit open-circuit voltage   the terminal voltage of a practical voltage source with no current drawn from the source overexcitation   supplying excess field current to a synchronous motor so that it acts like a capacitor overload  a situation in which the load resistance is small compared to the internal resistance of a source, resulting in a considerable power dissipation in the source parallel circuit   an electric circuit consisting of branches connected to the same two points resulting in the same voltage across them

1007

resistance and inductance (RL filter), or resistance, capacitance, and inductance (RLC filter) peak-to-peak value   the magnitude of a waveform measured from the lower peak to the upper peak peak value   the maximum absolute value of an ­alternating waveform period  duration of one complete cycle of a periodic waveform periodic wave  time-varying function consisting of repeating cycles permanent magnet   magnet that maintains its ­magnetism in the absence of an external magnetic field permanent magnet motor   motor deriving its magnetic field from permanent magnets permeability  ability of a material to permit magnetic flux permeance  ability of a magnetic circuit to permit magnetic flux permittivity  ratio of the electric flux density in a material to the electric field strength permittivity of free space  ratio of the electric flux density in a vacuum to the electric field strength, about 8.85 × 10−12 F/m

phase angle   angular displacement between two sine waves having the same frequency; difference in the ­angles of the phasors representing the voltage across and the current through a load in an AC circuit phase current   current in any of the phases of a threephase source or in any of the arms of a three-phase load phase sequence (or phase rotation)  the order in which the line voltages lag behind each other in a three-phase system phase voltage   the voltage measured from line to neutral in a wye-connected three-phase source phasor  a complex number representing the ­magnitude and phase angle of a sine wave phasor diagram   phasors represented by arrows drawn on the complex plane

paramagnetic  having a permeability slightly greater than that of free space

phasor power   apparent power expressed as a ­complex number with the real component representing real power and the imaginary component representing reactive power

passband   range of frequencies passed by a filter

photoconductor  See photoresistor.

passive circuit element   device that consumes ­electric energy passive filter   filter constructed using passive components such as resistance and capacitance (RC filter),

photoresistor  a resistor with a resistance that ­changes with light intensity to which it is exposed photovoltaic cell   a semiconductor device that ­converts light energy into a potential difference

1008

Glossary

piezoelectricity   generation of potential difference in dielectric crystals subjected to pressure

primary cell  cell that cannot be recharged

planar  two-dimensional; a planar circuit diagram has no conductors crossing each other without an electrical connection

printed circuit board   a sheet of rigid insulating ­material on which conductive traces have been ­deposited, often by etching away portions of a coating of copper

polar coordinates   coordinates that express a ­complex number as a magnitude and a phase angle polarity   sign convention that indicates the direction of current that would result from a given potential ­difference polarization   formation of hydrogen bubbles on an electrode in a cell, insulating it from the electrolyte polarized  having charge on a material displaced such that one side becomes somewhat positive and the ­opposite side becomes somewhat negative polyphase system   circuit having two or more ­alternating voltages with the same frequency but ­different phase angles pony motor  small AC motor used to bring a ­synchronous motor up to speed port   any pair of terminals in a network

primary winding   input winding of a transformer

proton  subatomic particle with a positive charge

pulsating DC  DC voltage that varies from zero to maximum twice per revolution pulsating DC waveform   waveform with a magnitude that is either always positive or always ­negative quadrature component  See imaginary component. quadrature power  See reactive power.

quality factor (Q)  ratio of the reactive power of either the inductance or the capacitance to the ­active power of an RLC circuit radian (rad)   unit of angle equal to 1/2π of a ­circle; 1 rad ≈ 57.3° RC filter  See passive filter.

potential difference   difference in the potential ­energy per unit charge between two points

reactance   opposition of inductance or capacitance to alternating current

potential drop  See potential fall.

reactive factor   ratio of reactive power to apparent power

potential fall   decrease in potential energy per unit charge from one point to another in a circuit potential rise   increase in potential energy per unit charge from one point to another in a circuit potentiometer   variable resistor connected as a ­voltage divider power   rate of doing work or of converting energy from one form to another power factor   ratio of active power to apparent power in an AC circuit power-factor angle, ϕ   the angle between the sides representing the active power and the apparent power in the power triangle; the angle by which the load ­current lags the load voltage in a balanced three-phase system power-factor correction (or improvement)  addition of reactive power to an AC circuit (usually by means of a capacitor in parallel with the load) to increase the power

reactive power   power supplied to a reactance in an AC circuit real component   reference axis component of a phasor; magnitude of a phasor multiplied by the cosine of its phase angle real power   power supplied to a resistance in an AC circuit rechargeable alkaline-manganese (RAM) cell  ­ secondary cell with a manganese dioxide positive ­electrode, a porous zinc gel for the negative electrode, and potassium hydroxide for the electrolyte reciprocal henry  the SI unit of reluctance

rectangular coordinates   the real and imaginary components of a complex number reference node   node from which all other node ­voltages are specified for nodal analysis of a circuit

power triangle   right triangle with sides that ­represent the apparent power, real power, and reactive power in an AC circuit

reflected impedance   impedance seen by a source connected to the primary winding of a transformer; square of the transformation ratio times the load ­impedance on the secondary

precision resistor   resistor with a close tolerance (­generally 1% or better)

relative permeability   ratio of the permeability of a material to that of free space

Glossary

relative permittivity   ratio of the permittivity of a ­material to the permittivity of free space reliability  percentage of resistors that lie outside the tolerance range after 1000 hours of operation at their rated power reluctance  opposition to magnetic flux in a ­material remanence or residual magnetism  magnetic flux ­retained by a ferromagnetic material after the ­magnetizing source has been removed resistance   opposition of a material to the flow of electric charge resistance-start motor  single-phase induction motor that is started by a winding with high resistance and low reactance resistivity  resistance of a unit length and ­cross-sectional area of a material resistor   electrical component designed to a specific resistance resonance  state of an RLC circuit at its resonant ­frequency resonant circuit  AC circuit in which resonance can be achieved resonant frequency   frequency at which the ­inductive reactance of an RLC circuit equals the ­capacitive ­reactance

1009

saturated   the state of a magnetic material in which all the magnetic domains are aligned with an applied ­magnetic field sawtooth waveform   complex wave having a ­magnitude with a linear ramp followed by an ­instantaneous change scalar  quantity with magnitude but no direction or phase angle schematic diagram  See circuit diagram.

scientific notation   the use of base units and powers of 10 to represent numbers that are either much greater than or much less than 1 second   the SI base unit of time secondary cell   cell in which the chemical reactions are reversible so that it is rechargeable secondary winding  output winding of a ­transformer selectivity   ability of a resonant circuit to select the resonant frequency and reject all others self-inductance   opposition of a circuit to any change in current self-induction   generation of a voltage in a circuit by a changing current in the circuit semiconductor   element that is neither a good ­conductor of electricity nor a good insulator

resonant rise of tank current   increase in the tank current in a parallel-resonant circuit as the frequency of the applied voltage approaches the resonant ­frequency

sensitivity  (1) degree to which inductance (or ­capacitance) voltage at resonance exceeds the source voltage (2) the value of current that causes full-scale ­deflection of a D’Arsonval meter movement

resonant rise of voltage   increase in the inductance (or capacitance) voltage in a series-resonant circuit as the frequency of the applied voltage approaches the resonant frequency

series circuit   circuit consisting of one closed path with the same current flowing through all ­elements series coil   field coil connected in series with the ­armature

retentivity   extent to which magnetic domains in a ferromagnetic material remain aligned with the direction of an external magnetic field after the field is removed

series motor   motor with field coils connected in ­series with the armature

rheostat   variable resistor connected to provide an adjustable series resistance RL filter  See passive filter. RLC filter  See passive filter. root-mean-square (RMS) value   square root of the ­average of the squares of a set of values (for an ­alternating current, the RMS value equals the effective or DC equivalent value)

series-parallel circuit   circuit in which some portions of the circuit have the characteristics of simple ­series circuits while the other portions have the ­characteristics of simple parallel circuits shaded-pole motor   single-phase AC motor using copper rings around part of the stator poles to produce a phase shift for starting short circuit  a very low resistance connection ­between two points in a circuit

1010

Glossary

short-circuit admittance parameters  four admittance parameters of a two-port network, two of which are determined with the input port short-­ circuited and the other two with the output port short-­circuited

slip or percent slip   difference between synchronous speed and rotor speed

short-circuit current   current drawn from a network when the output terminals are s­ hort-circuited

solar cell  See photovoltaic cell.

short-circuit forward-current ratio  ratio of output current to input current of a two-port network with the output port short-circuited short-circuit forward-transfer admittance  ratio of output current to input voltage of a two-port network with the output port short-circuited short-circuit input admittance   ratio of input current to input voltage of a two-port network with the output port short-circuited short-circuit input impedance  ratio of input voltage to input current of a two-port network with the output port short-circuited short-circuit output admittance   ratio of output ­current to output voltage of a two-port network with the input port short-circuited short-circuit reverse-transfer admittance  ratio of input current to output voltage of a two-port network with the input port short-circuited shunt coil   field coil connected in parallel with the ­armature shunt motor   motor with field coils connected in ­parallel with the armature shunt resistor   resistor connected in parallel with an ammeter movement so that the full-scale ­reading ­corresponds to a current greater than the current ­capability of the movement

siemens (S)  SI unit of conductance, susceptance, and admittance; 1 S = 1 A/V significant digits  the number of digits appearing in a ­numerical value that express the accuracy of the ­number silver-oxide cell   primary cell with a silver-oxide ­positive electrode, an anode consisting of a gel of ­powdered zinc, and an electrolyte that is either ­potassium hydroxide or sodium hydroxide

slip ring   a conducting ring that makes an ­electrical connection to the rotating parts of motors and ­generators solenoid  cylindrical coil of wire that acts as a magnet when current flows through it source  device that produces electrical energy by ­converting some other form of energy source conversion   (1) the conversion of a voltage source in series with a resistor to a current source in parallel with the resistor, or vice versa, such that the two networks provide the same ­current and voltage to a load in a DC circuit; (2) the conversion of a phasor voltage source in series with an impedance to a phasor current source in parallel with the impedance, or vice versa, such that the two networks provide the same phasor current and voltage to a load in an AC circuit south pole   pole at which magnetic lines of force enter a magnet specific resistance  See resistivity. split-phase motor   single-phase induction motor that uses a winding with a phase shift for starting square wave   a periodic wave whose magnitude is one DC level for half the period and another DC level for the other half of the period squirrel cage rotor   rotor containing heavy ­conductors connected to shorting end rings star-connected source  See wye-connected ­(Y-connected) three-phase source. starting winding   winding added to a single phase motor to cause a phase shift for starting the motor static electricity   stationary electric charge, usually produced by friction between dissimilar materials steady-state value   value of a quantity after transient conditions in a circuit have passed steam turbine   turbine rotated by high-pressure steam step-down transformer   transformer with secondary voltage less than the primary voltage

sine wave   a periodic waveform with a magnitude proportional to the sine of the phase angle

step-up transformer   transformer with secondary voltage greater than the primary voltage

singly excited   method of developing rotor currents by induction from the stator using only one source

stopband   range of frequencies attenuated by a filter

single-phase circuit   circuit with only one source of alternating voltage

stray capacitance   capacitance introduced into a ­circuit due to pairs of electric conductors at different potentials running close to each other

Glossary

strength (intensity) of an electric field  the force per unit charge at a given point superposition theorem  (1) principle that in linear circuits a current (or voltage) due to a number of sources in the circuit equals the algebraic sum of the currents (or voltages) due to each source acting alone (2) a theorem that states that the current in any branch of a network of i­ mpedances resulting from the simultaneous application of a number of sources distributed in any manner throughout the network is the phasor sum of the component currents that would be caused in that branch by each source ­acting independently in turn while the ­others were replaced in the network by their respective internal ­impedances susceptance   ability of inductance or capacitance to pass alternating current

1011

network of sources and impedances may be replaced by a single voltage source in series with a single impedance three-phase alternator   alternator producing three voltages with 120º phase shifts between them three-phase (3 ϕ) system   circuit with a source that supplies three AC voltages of equal magnitude that are 120° out of phase with each other time constant   time it would take the potential difference across a capacitor or the current through an inductor to reach its final value if it rose or fell at the initial rate of change for the whole time interval time domain   mathematical model for analyzing quantities in terms of how they vary with time

swamping resistor  See calibrating resistor.

tolerance   maximum allowable deviation from the rated value of a component

switch  device that makes and breaks a connection in a circuit

torque   turning or twisting moment of a force around an axis

synchronous capacitor  overexcited synchronous motor, generally used to improve power factor in industrial plants and power systems

toroidal coil   coil wound on a ring-shaped core

synchronous speed   speed of the revolving magnetic field from the stator in an AC machine

total-current method   finding the total current in a parallel circuit by calculating the current in each branch and then adding these branch ­currents

synchronous speed   angular velocity of a rotating magnetic field in a polyphase circuit

transformation ratio   ratio of the number of turns in the primary winding of a transformer to the number of turns in the secondary

tank circuit   an inductance and a capacitance connected in parallel

transformer  device consisting of two or more coils with an AC current in one coil inducing an AC voltage in the other coil(s) by mutual induction

tank current   the current flowing around a tank ­circuit temperature coefficient of resistance  the proportion by which the resistance of a material changes per degree Celsius relative to the resistance at a reference temperature (often 20°C) temporary magnet   a magnet that does not ­maintain any significant magnetic flux in the absence of an ­external magnetic field tesla (T)  SI unit of flux density; 1 T = 1 Wb/m2

thermistor  resistor with a large negative temperature coefficient thermocouple   junction of two dissimilar ­metals that produces a small voltage dependent on ­temperature Thévenin’s theorem   the principle that (i) any twoterminal DC network of sources and resistors may be replaced by a single voltage source in series with a ­single resistor, and (ii) any two-terminal AC

transformer efficiency   ratio of output real power to input real power of a transformer, usually expressed as a percentage transformer voltage regulation   ratio of the change in secondary voltage from no load to the full-load voltage, usually expressed as a percentage transient phase  interval during which the currents and voltages in a network change from one steady state to another transient response   manner in which a voltage or current changes from one steady-state value to another; transient phase tuning  achieving resonance in an RLC circuit by varying the inductance or capacitance in the circuit two-phase alternator   alternator that produces two AC voltages 90° out of phase with each other two-port network   four-terminal network designed to connect a source to a load

1012

Glossary

unbalanced load   load in which the impedances in the arms are not equal

wave winding   connection of armature coils that has two parallel paths through the armature

unit prefix   a prefix used with a base unit to ­express a decimal multiple or sub-multiple of the base unit

wind turbine   turbine rotated by wind

universal motor   series motor that will operate with DC and AC power varactor   semiconductor device with a capacitance that varies with applied voltage varistor  component with a resistance that varies with applied voltage vector  quantity with both magnitude and ­direction volt (V)  SI unit of potential difference; 1 V = 1 J/C = 1 W/A voltage   change in potential energy per unit charge when moving from one point to another; potential ­difference voltage-current characteristic  graph of the current through a device versus the voltage across it voltage-divider principle   the principle that (i) the ratio between the voltages across any two r­ esistances in a DC series circuit equals the ratio of the resistances, and (ii) the ratio between the phasor voltages across any two impedances in an AC s­ eries circuit equals the ratio of the impedances voltage drop  See potential fall.

voltage gradient   distribution of voltage between two points; rate at which voltage varies with distance voltage regulation   ratio of the difference between ­no-load voltage and full-load voltage output of a voltage source to the full-load voltage, usually expressed as a percentage voltage rise  See potential rise.

weber (Wb)  SI unit of magnetic flux; 1 Wb = 1 V⋅s

wire-wound resistor   resistor constructed by winding a length of metal wire on an insulating form work   energy transferred from one body or system to another working voltage   maximum voltage that can be ­applied across a capacitor without risk of damaging it wound rotor   rotor containing multi-turn windings in slots wye-connected load   load connected to a threephase source such that each arm of the load is in ­series with a phase of the source wye-connected (or Y-connected) three-phase source   three-phase source with one terminal of each phase connected together wye-to-delta transformation equations  equa­ tions for finding (i) the equivalent resistive Δ-­network for a ­resistive Y-network in a DC circuit, and (ii) the ­equivalent impedance Δ-network for an impedance Y‑network in an AC circuit y-parameter equivalent circuit  equivalent cir­cuit of a two-port network consisting of an ­admit­tance in ­parallel with a dependent current source for both the input and output networks yoke   frame of rotating machines that also serves as a return path for magnetic flux y-parameters  See short-circuit admittance ­pa­rameters.

voltmeter  instrument for measuring voltage

zinc-air cell   primary cell with a zinc anode, an ­electrolyte consisting of an alkaline solution of ­potassium hydroxide and zinc oxide, and a thin carbon cathode electrode that catalyzes oxygen from the air for reaction with the zinc

voltmeter sensitivity  parameter indicating the load that a given voltmeter adds to a circuit, usually expressed as resistance per volt of full-scale reading

zinc-chloride cell   primary cell with a manganese ­dioxide/carbon cathode, a zinc anode, and an ­electrolyte consisting of zinc chloride in water

volt ampere (VA)  SI unit of apparent power volt ampere reactive (var)   unit of reactive power in an AC circuit

volt-ohm-milliammeter (VOM)  multimeter that can measure voltage, resistance, and small ­currents watt (W)  SI unit of power in a DC circuit and of real power in an AC circuit; 1 W = 1 J/s wattmeter   instrument used to measure the real power to a load

z-parameter equivalent circuit  equivalent circuit of a two-port network consisting of an impedance in ­series with a dependent voltage source for both the input and output networks z-parameters  See open-circuit impedance ­pa­rameters.

Index Page references in bold indicate glossary explanations. Page references followed by “t” indicate tables.

A

A (ampere), 6t; see also ampere abbreviations for unit names, 6 AC (alternating current), 522–51 direct current and, 540–3, 563 magnetic field strength and, 409, 411 measurement of, 297, 311–12 opposition to, 556, 562–3, 598–631 zero-frequency, 563 AC (alternating current) circuits impedance networks and, 692–739 impedances in, 666–91 mutual induction in, 824 power in, 632–64 reactance and, 552–69 AC (alternating current) machines, 460, 465 active circuit element, 129 active filters, 776 active (or real) power (P), 634 addition of harmonically related sine waves, 951–4 of instantaneous current, 573–4 perpendicular phasors and, 578–80 phasors and, 576–8 rectangular coordinates and, 585–7 of sine waves, 572–3 admittance (Y), 614–15 branch, 671–2, 676 coupling networks and, 864–7 impedance and, 615–18 open-circuit, 869, 870 admittance diagram, 615, 617t air-core coil, 422–4 air-core transformers, 874–5 air gaps, 433–5 algebraic solutions capacitor voltage, 362–6 discharge current, 506–7 inductor current, 495–8 phasors, 573, 576, 585–9, 668

alkaline cells, 53–4, 59t alkaline-manganese cells, 57 alnico, 404 alpha (temperature coefficient of resistance), 82–4 alternating current: see AC (alternating current) alternators, 525 AC, 927–9 three-phase, 928 two-phase, 892–5 wye-connected, 899–903 aluminum, 42, 73, 74 American Wire Gauge (AWG), 77–8t ammeter, 128, 150, 297–9 resistance and, 304–5 ampacity, 33 amperage, 33 Ampère, André-Marie, 26 ampere (A), 26–7 as base unit, 5 new definition of, 26, 396 ampere-hour rating, 58 ampere-turn (At), 398 amplifiers filters and, 809–12 harmonics and, 956–7 angles phase (ϕ), 526, 527–30, 617t power-factor, 645 of rotation, 525 SI symbol for, 532 angular velocity (ω), 526, 532–3 anode, 43–4, 52 antiresonance, 755 apparent power (S), 640 power factor and, 645 power triangle and, 643 applied voltage (V), 33 arctan, 579 armature, 460 Arrhenius, Svante August, 44

arrows: see type conventions atoms magnetic materials and, 402–4 attraction conducting plates and, 325 magnetic force of, 396 audio systems crossover networks, 811–12 autotransformers, 841–3 average values of periodic waves, 540 of sine waves, 826, 982–3

B

B (bel), 778 B (magnetic flux density), 400–1 B (susceptance), 614–15 balanced load, 896 balanced three-phase system power in, 913–15 band-pass filters, 776, 799–804 band-stop (or band-reject) filters, 776, 804–9 bandwidth, 751–2, 764 bandwidth law, 752 batteries, 50, 52, 50–67 capacity of, 58–9 electrolytic, 43–5 free electrons and, 24–5, 30 in parallel, 149–50 in series-parallel, 184–6 see also cells bel (B), 778 BH curve, 405–6, 958 bifilar winding, 874 bleeder resistor, 178 Bode plot, 785–6 Bode, Hendrik, 785 Boltzmann constant (k), 5 branch currents, 237 impedances and, 671–2, 676 branches, 140, 145–6, 147

1014

Index

breadboard, 43 bridge circuits, 202, 212–13 delta-to-wye transformation and, 281–2 Thévenin’s theorem and, 264–6 see also Wheatstone bridges British thermal unit (BTU), 114 brushes, 461, 524 brushless DC motors, 476 BTU (British thermal unit), 114 bypassing capacitors and, 333

C

C (coulomb), 23 calculators, 6–7 coordinate-conversion function and, 586, 589 engineering, 7 scientific, 495, 528 calculus derivations, 978–86 capacitive reactance, 984–5 capacitor energy storage, 980 general transformer equation, 985 inductive reactance, 983–4 inductor energy storage, 982 instantaneous current in LR circuit, 980–1 instantaneous voltage in CR circuit, 978–80 maximum power-transfer theorem, 977–8 maximum transformer efficiency, 985–6 RMS and average values of sine wave, 982–3 calibrating resistor, 297 Canada fuel cell research in, 60 capacitance (F), 321, 322–47, 329 DC circuits and, 348–87 energy and, 370–2 factors governing, 333–6 parallel circuits and, 610–13 series circuits and, 607–10 stray, 374 variable, 745–6 capacitance meter, 375 capacitive reactance, 984–5 capacitor-run motor, 935 capacitor-start motor, 935 capacitors, 328–33 applications of, 333 area of, 334

characteristics of, 372–4 charging, 350–7 construction of, 330–1 dielectric and, 334 discharging, 357–61 energy storage and, 370–2, 980 leakage and, 374, 375 in parallel, 338 parallel-plate, 335 power in, 637–9 rating of, 332 reactance and, 558–9, 561 in series, 338–41 spacing of, 334 symbols for, 332 synchronous, 934 troubleshooting, 375 undesirable characteristics of, 374 voltage and, 351–7, 359, 362–9 capacity, 329 cells/batteries and, 58–9 carbon, 47 carbon-composition resistors, 85, 107 cascaded high-pass and low-pass filters, 799–801 cathode, 44, 52 cells, 50–67, 50, 52 alkaline, 53–4, 59t capacity of, 58–9 copper-zinc, 52–4 dry, 53 electrolytic, 43–4 fuel, 60 general-purpose/standard duty, 53 heavy-duty, 55 lead-acid, 56, 57 lithium-ion (li-ion), 57 lithium, 55 manganese-alkaline, 53–4 nickel-cadmium (ni-cad), 57, 59t nickel-metal hydride (ni-mh), 57 in parallel, 148–50 photovoltaic, 62–3 primary, 52–5t rechargeable, 56 rechargeable alkaline-manganese (RAM), 57 secondary, 52, 56–8, 57t in series-parallel, 184–6 silver-oxide, 55 solar, 62–3 zinc-air, 55 zinc-chloride, 55 see also batteries centre frequency, 802–3, 807

ceramics, 404 charge, 22–3 capacitors and, 328–9, 350–61 conventional current and, 34 graphical solution for, 356–7 net negative/positive, 22 on one electron (e), 26 quantity of electric (Q), 26 chassis, 177, 179 circle formula for area of, 76 circuit boards printed, 42–3 circuit coupling, 333 circuit diagrams, 4 circuits single-phase, 892 transistor, 176, 179 see also electric circuits; specific circuits circular mils (CM), 75–7 CM (circular mils), 75–7 coefficient of coupling (k), 874 coercive force, 410–11 coils additional turns of, 398 air-core, 422–4 in DC generators, 460, 461–3 in DC motors, 466 inductance and, 455–6 motors and, 470–1, 472 series, 460 single-turn, 397–8 solenoid and, 394–5 toroidal, 425 colour code resistors and, 88–90, 89t commutator, 460–1, 462–3 complex numbers phasors and, 581–5, 582 complex plane, 581 composition resistors, 85 compound motor, 470–1, 472, 474 concentric (coaxial) conductors, 324–5 condenser, 331 conductance (G), 142–5, 613–15 conductivity (σ), 145 conductor loop, 524 conductors, 4, 24–5, 33, 42–5 concentric, 324–6 current-carrying, 393–6 induction and, 448, 450 insulators and semiconductors and, 40–9 magnetic fields and, 393–6

Index

parallel, 324–6, 330–1, 333–6 resistance of, 72–103 resistivity of, 74, 76t resistors and, 84–5 stranded, 394 temperature and, 79–84 conservation of energy law, 29 constantan, 74, 79–80t constant-current sources, 204–6, 206–7 equivalent, 205–6, 682–3 Norton’s theorem and, 268–71 constant-voltage sources, 202–4, 206–7 conversion of, 682–3 constants Boltzmann, 5 Coulomb’s, 23 dielectric, 336–7t Planck, 5 time, 354–5, 359, 362, 490–1, 504–5 controlled source, 135, 271–8 conventional current, 33–4 coordinates conversion of, 581–9 see also polar coordinates; rectangular coordinates copper, 23, 42–3 resistance of, 73, 74, 78 copper loss, 835–8 copper-zinc cell, 52–4 core loss, 830, 835–8 Coulomb, Charles Augustin de, 22, 26, 324 coulomb (C), 23, 26, 325 Coulomb’s constant (k), 23 Coulomb’s law, 23–4 counter (or back) EMF (CEMF), 453, 467–8, 470, 472, 475 coupled circuits, 854–89 coupled impedance, 861, 878–81 coupling coefficient of (k), 874 coupling capacitor, 333 coupling networks, 856–7 parameters of, 857–74 CR circuit, 351, 978–80 as timing device, 373 Cramer, Gabriel, 975 Cramer’s rule, 975 critical frequency, 777, 778–9, 782, 785 other names for, 787 RC filters and, 792 RL filters and, 788, 797 critically damped, 764 crystal structure, 23, 47 cumulatively compounded, 470–1, 474

current (I), 24–5 AC circuits and, 694–5, 702–3 base unit for, 6 charge and, 26 conventional, 33–4 discharge, 358 displacement, 560 eddy, 412–13 electron flow in, 34 exciting (magnetizing), 828 heat and, 113–15 induction and, 448, 450, 484–520 inrush, 86–8 instantaneous, 533–5 interrupting, 501–6 law, 141–2 line, 907–8, 913 magnetic fields and, 388 magnetizing, 828 Ohm’s law and, 92–3 other electric units and, 113t parallel circuits and, 140, 145–6, 147 phase, 905, 907–8 in phase with voltage, 534 primary, 827–8, 830 rate of change of, 487, 555, 556–7, 558 reversed, 410–11 rising, 487–90 series circuit and, 126, 127, 129–30 tank, 757 transformers and, 827–8, 830, 838–41 voltage and, 20–39, 70–103 see also instantaneous current; Kirchhoff’s current law; specific currents current-controlled current source (CCCS), 271t current-controlled voltage source (CCVS), 271t, 271–2, 274 current-divider principle, 181–4, 676–8 current simulations, 212–13 cycle, 526, 530, 538

D

d (distance), 5, 6t D (electric flux density), 325 damping, 764 daraf, 340 d’Arsonval, Jacques-Arsène, 296 d’Arsonval movement, 296–7 dB (decibel), 778–9 DC (direct current) alternating current and, 540–3, 563 measurement of, 311–12

1015

pulsating, 463 pulsating waveform, 950 as zero-frequency AC, 563 DC (direct current) circuits capacitance and, 348–87 characteristics of capacitive, 372–4 characteristics of inductive, 509–10 inductance and, 484–520 interrupted current and, 502–6 DC (direct current) generator, 459–63 DC (direct current) motors, 465–71 speed characteristics of, 467–9, 471–4 torque characteristics of, 474–5 types of, 469–71 decade, 780 decibel (dB), 778–9 “deep cycling,” 56 degrees mechanical v. electrical, 530 delta (�) circuits, 278 delta-connected load, 905 delta-connected sources, 904 delta-connected systems, 903–8, 904 line and phase current in, 907–8 delta-to-wye transformation, 278–82, 279, 281–2, 724–8, 922–3 delta-wye configurations difference between, 935–6 dependent sources, 271–8 determinants, 974–7 diamagnetic materials, 402–3 dielectric absorption, 327 dielectric constant (k), 336–7t dielectrics (electrical insulators), 327–8, 331–2, 335–6 dielectric strength, 327–8t, 336 differentially compounded, 470–1 digital multimeters (DMMs), 312–14 digital-to-analogue converter (DAC), 314 digits, significant, 7 dimensional analysis, 12–13 diodes, 297 direct current: see DC (direct current) discharge capacitors and, 357–61 discharge current, 506–7 discharge resistor, 503 discharge time constant, 504–5 displacement current, 560 distance (d) base unit for, 5, 6t distributed winding, 929, 930 division phasor qualities, 589–90, 703

1016

Index

double-subscript notation, 130, 897–9 dry cells, 53–4 duality, 141 dynamo, 28

E

e (charge on one electron), 26 E (electric field strength), 324 e (instantaneous value), 529–31 E (source voltage), 33 E-series, 88–9 eddy current, 412–13 Edison three-wire system, 219–21 effective value, 541 efficiency (η), 110–11 elastance (S), 340 electric circuits, 4 introduction to, 2–19 see also specific types electric fields, 324–6 dielectrics and, 327 magnitude and direction of, 324 electric field strength (E), 324, 335 electric flux (Ψ), 325 electric flux density (D), 325, 335 electric flux lines, 324 electric lines of force, 324 direction of, 466 electric units relationships among, 113 electricity, 2, 4 piezo-, 64 static, 64 electrodes, 43–4 primary cell and, 52–3 electrodynamometer movements, 311–12 electrolyte, 43–4 electrolytic capacitors, 331–2 electrolytic cell, 43–4 electromagnet, 394–5 electromagnetic induction, 448–50 see also induction electromotive force (EMF), 28, 30, 397 cells in parallel and, 149–50 counter or back, 453, 467–8, 470, 472, 475 equation, 463–5 series circuits and, 133, 135 voltage and, 32–3 electron flow, 33–4 electrons, 22 free, 23, 24–5, 30 electrostatic force, 22

electrostatic induction, 338 EMF: see electromagnetic force (EMF) energy storage capacitors and, 370–2, 980 dynamic/static, 500 inducers and, 982 inductors and, 499–501 energy law of conservation of, 29 other electric units and, 113t wasted, 110 work and, 31–2, 106 epsilon (electric field strength), 324–6 epsilon (permittivity), 334–6 equivalent circuits, 127 h-parameter, 870 inductance and, 459 resistance, 141–4 Thévenin’s theorem and, 260–7, 273, 707 y-parameter, 866 z-parameter, 859 equivalent-circuit methods, 202–8, 258–93 series-parallel circuits and, 167–70 troubleshooting and, 283 equivalent constant-current sources, 205–6, 682–3 equivalent DC value, 541 eta (efficiency), 110–11 excitation DC, 932, 933 exciter, 929, 932, 933 exciting (magnetizing) current, 828 exponentials calculators and, 495

F

F (capacitance), 329–30 F (farad), 329–30 f (frequency), 529 farad (F), 329–30 Faraday, Michael, 34, 44, 45, 330 magnetic field and, 61, 397, 448 Faraday’s law, 450–1 ferrites, 404 ferroelectric dielectrics, 337 ferromagnetic materials, 402–3 field-effect transistor (FET), 313 field excitation, 469 field system, 460 filter choke, 510 filters active, 776 applications of, 809–12

capacitive high-pass and low-pass, 799–801 inductance as, 510 low-pass, 776, 781–90, 799–801, 805–6 parallel, 805–6 passive, 774–823, 776 Fleming, John Ambrose, 466 “float,” 56 floating junction point, 919 form factor, 543 four-terminal networks, 854–89 four-wire wye connected system, 899–903 Fourier, Jean-Baptiste Joseph, 949 Fourier series, 949–51 Franklin, Benjamin, 22 free electrons, 23 electric current and, 24–5 potential difference and, 30 frequency (f), 529 amplifiers and, 809–12 centre, 802–3, 807 effect on RLC circuit and, 743t expression of, 531 filters and, 774–823 fundamental, 948–9 half-power, 752 impedance, 742–5 reactance and, 556–8, 561 resonant, 743–63 variable, 470, 742–5 voltage gain and, 781–2 see also critical frequency; resonant frequency frequency domain, 573 frequency response graphs, 778–80 friction, 64 fringing, 433 fuel cells, 60 funicular phasor diagrams, 839

G

G (conductance), 613–15 G (giga), 10t gain plots RC high-pass filters and, 792–4 RC low-pass filter and, 782–6 RL high-pass filters and, 797–9 RL low-pass filters and, 789–90 galvanometer, 309–10 induction and, 448, 450, 452 magnetic circuit and, 397 gas turbines, 62 Gassner, Carl, 53

Index

gauge, wire, 77–8 Gauss’s Law, 325 General Conference on Weights and Measures (CGPM), 5 general transformer equation, 826–7, 985 generators, 28, 61–2, 64 AC, 927–9 four-pole, 530 two-pole, 524, 530 geometric construction induced voltage and, 527–9 phasors and, 576–8, 579, 903, 907, 910 germanium, 47 giga (G), 10t gram, 5 graphical solutions capacitor voltage and, 356–7 discharge current and, 506–7 inductor current and, 491–5 rate of change of voltage and, 352–4 gravitational force and, 22 ground wire, 177

H

H (henry), 399, 454–5 H (magnetic field strength), 401–2 h- (hybrid) parameters, 864, 867–74 equivalent circuit, 870 half-power frequencies, 752 half-wave-rectified sine wave, 955 harmonics, 946–72, 948–9 amplifiers and, 956–7 even-order, 953, 956 generation of, 954–5 in iron-core transformer, 958–60 odd-order, 953 Hartley’s bandwidth law, 752 heat (Q) current and, 113–15 dissipation of, 107, 110 in inductors, 510–11 specific, 113 in wires, 42 see also temperature heavy-duty cell, 55 Henry, Joseph, 454 henry (H), 399, 454–5 reciprocal, 398 Hertz, Heinrich Rudolf, 529 hertz (Hz), 7, 529 high-pass filters, 776 cascaded, 799–801 parallel, 805–6

RC, 791–6 RL, 797–9 horsepower (hp), 111, 469 hours, 5 hp (horsepower), 111, 469 hydroelectricity, 61–2 hysteresis, 410–12, 411 transformers and, 835–6, 958 hysteresis loop, 411–12

I

I (electric current), 26 see also current; specific types imaginary (or quadrature) component, 581, 582, 585, 587 imaginary numbers (j), 581–2 imaginary power (Q), 637, 643 see also reactive power impedance (Z), 598–631 admittance and, 615–18 complex v. scalar, 724 coupled, 861, 878–81 coupling networks and, 857–64, 868 definition of, 601–3 frequency and, 742–5 linear, 954, 963–5 mutual, 860 nonlinear, 954–5, 956, 962 open-circuit, 857–74, 880 in phasors, 598` reflected, 831, 862 in series and parallel circuits, 666–91 series-parallel, 678–82 total-current method, 672–3 impedance diagram, 617t impedance matching, 831 impedance networks, 692–739 nonlinear, 954–5, 956, 962 impedance transformation, 831–3 independent sources, 271 induced voltage, 397, 448, 450–1 AC, 524–6, 527–8 DC generators and, 461–2 direction of, 452–3 phasors and, 527–8 polarity of, 451 inductance (L), 321, 388, 446–83, 454 DC circuits and, 484–520 equivalent, 459 factors governing, 455–6 as filter, 510 mutual (M), 875–8

1017

parallel circuits and, 610–13 self-, 454–5 series circuits and, 600–6, 608–10 total, 458 variable, 745–6 inductance and resistance circuits see LR circuits induction electromagnetic, 448–50 electrostatic, 338 mutual, 450, 824, 826 self-, 453 induction motor, 930–2 inductive reactance, 983–4 inductor current, 484–520 algebraic solution for, 495–8 graphical solution for, 491–5 troubleshooting, 510–11 universal equation for, 508 inductors, 456 energy stored by, 982 falling current in, 501–6 ideal, 486–7, 499–500 instantaneous current and, 554–5 in parallel, 458–9 power in, 635–8 practical, 487–90, 500–1, 604–6 rising current in, 487–90 in series, 458 inrush current, 86–8 instantaneous current (i), 572–3 addition of, 573–4 capacitor and, 558–9 ideal inductor and, 554–5 LR circuits and, 507–8, 980–1 resistors and, 533–5, 536–7 instantaneous power, 536–7, 893–4 instantaneous values (i and v), 351 sine wave, 529–31 instantaneous voltage in CR circuit, 978–80 universal equation for, 367 Institute of Electrical and Electronics Engineers (IEEE), 5 instruments basic measuring, 150–1 see also specific instruments insulators, 45–6, 327–8 breakdown of, 46 conductors and semiconductors and, 40–9 see also dielectrics integrated circuits (ICs), 85 internal resistance series circuits and, 133–5, 134, 137–9

1018

Index

International System of Units (SI): see SI units ions, 23, 43–4 IR drop, 93, 231, 350–1, 353, 358; see also voltage drop iron air gaps and, 433–5 hysteresis and, 410–11 induction and, 448–50 leakage flux and, 429 magnetic field and, 390, 392 permeability of, 402–5, 406 iron-core transformers, 827, 834, 835–6 harmonics and, 958–60

J

j (imaginary numbers), 581 Joule, James Prescott, 31, 113 J (joule), 31, 112, 107 joule (J), 31, 112 watts and, 107

K

k (dielectric constant), 336–7t K (kelvin), 5 k (kilo), 10t kcmil (kilo-circular mils), 77 kelvin (K), 5–6 kg (mass), 5, 6t kilo (k), 10t kilo-circular mils (kcmil), 77 kilogram (kg), 5, 6t kilowatt hour (kW·h), 112 Kirchhoff, Gustav Robert, 130 Kirchhoff’s current law, 141–2 network equations and, 202 in transmission lines, 680–1 Kirchhoff’s current-law equations, 208, 228–31 source conversions and, 231 Kirchhoff’s laws method series-parallel circuits and, 171–3 Kirchhoff’s voltage law, 130–1 inductors and, 486–7, 503, 506 network equations and, 202 reactance and, 552, 558 in transmission lines, 680–1 Kirchhoff’s voltage-law equations, 208–11, 213–14 mesh analysis and, 223 multisource networks and, 216–22 kW·h (kilowatt hour), 112

L

L (inductance), 446–83, 454 see also inductance l (length), 5, 6t ladder network, 169–70, 183 lagging power factor, 645 lead-acid cell, 56, 57 leading power factor, 645 leakage capacitor, 374, 375 leakage factor, 429–30 leakage flux, 429–30, 435, 456, 833 leakage reactance, 833–4 leakage resistance, 502 left-hand (motor) rule, 466 length (l), 5, 6t Lenz, Heinrich F.E., 451 Lenz’s law, 451–3 inductors and, 503 light dependent resistor (LDR), 87–8 line current in delta system, 907–8 power and, 913 line voltages, 909–12 power and, 913 linear impedance networks, 954, 963–5 nonsinusoidal waves in, 963–5 linear magnetic circuits, 425–6 linear resistors, 84–5 lithium cell, 55 lithium-ion (li-ion) cell, 57 load, 4 balanced, 896 Thévenin’s theorem and, 260–1 load current, 470 load resistance, 137–9 local action, 53 logarithms calculators and, 495 natural (nl), 363 long air-core coil, 422–4 loop equations, 208–11, 694–9, 898 multisource networks and, 216–22 loops closed, 208–11 hysteresis, 411–12 magnetic lines of force and, 391, 394 lossless model, 680 low-pass filters, 776 cascaded, 799–801 parallel, 805–6 RC, 781–7 RL, 788–90

LR (inductance and resistance) circuits, 490–8, 980–1 transient response and, 507–9

M

M (mega), 10t m (metre), 5, 6t m (milli), 10t M (mutual inductance), 875–8 magnetic circuits, 397, 420–46, 422 linear, 425–6 nonlinear, 425, 426–9 parallel, 435–7 practical, 422 series, 430–2 magnetic domains, 403–4 aligned, 406 magnetic field strength (H) (or intensity), 401–2, 405–6, 408–9, 410 permeability, flux density, and, 401–2 magnetic fields, 390–2, 524 concentration of, 394 current and, 296, 388 current-carrying conductor and, 393–6 electrodynamometer movements and, 311 magnetic flux (ϕ), 396–7, 422, 424, 451–2 permeance and, 399 reluctance and, 398 two-phase alternators and, 894–5 magnetic flux density (T), 400–1, 405, 407 permeability, magnetic field strength and, 401–2 residual, 410 transformers and, 827 magnetic flux lines, 390 see also magnetic lines of force magnetic flux linkage, 454–5, 456 magnetic lines of force, 390–2, 393 AC voltage and, 524–6 magnetic monopole, 390 magnetic poles generators and, 530 magnetic shielding, 413 magnetically coupled circuits, 874–8 magnetism, 388–419 residual, 404 magnetization, 403–4, 410–11 magnetization curves, 404–7, 405, 427 magnetizing current, 828

Index

magnetomotive force (MMF), 397–8, 422 air-core coils and, 422–4 nonlinear magnet circuits and, 426–9 series magnetic circuits and, 430–2 toroidal coils and, 425 magnets permanent, 404 magnitude, 617t reactance and, 556–8, 560 manganese-alkaline cells, 53–4 manganin, 76 mass (m), 5, 6t materials magnetic/nonmagnetic, 402–4 see also specific materials maximum power transfer, 137–9, 138 theorem, 977–8 maximum transformer efficiency, 985–6 Maxwell, James Clerk, 222 Maxwell’s equations, 325 mean power (P), 634 measurement electrical, 294–320 resistance, 304–10 mega(M), 10t megagram, 106 “memory effect,” 56 mesh, 222–3 mesh analysis, 222–7 mesh equations, 208, 694, 700–2 metal detectors, 449 metal-oxide varistors (MOVs), 87 metals as conductors, 42 free electrons in, 23 resistivity of, 76t see also specific metals metre, 5 metric system, 5 metric ton (t), 106 micro (μ), 10t microfarads, 329 mil, 75 milli (m), 10t minutes, 5 motor principal, 296 motors, 465, 471 AC, 460, 465 brushless DC, 476 capacitor-run, 935 capacitor-start, 935 compound, 470–1, 472, 474 DC, 465–7, 469–75

induction, 930–2 left-hand rule and, 466 permanent magnet, 476 pony, 933 resistance-start, 935 series, 470–1, 472, 474 shaded-pole, 935 shunt, 470, 472, 474, 475 single-phase, 934–5 singly excited, 930 split-phase, 934–5 synchronous, 932–4 three-phase, 930–4 three-phase synchronous 932–4 universal, 472, 935 moving-coil meters, 296–9 moving-coil movements, 300 mu (micro), 10t mu (permeability), 399–400 multimeters, 312–14 digital, 312–14 multiplication phasor qualities, 589–90, 703 multiplier resistor, 300 Multisim schematic capture and simulation, 986–7 multisource networks, 214–22 mutual flux, 833 mutual impedance, 860 mutual inductance (M), 875–8 mutual induction, 450 transformers and, 824, 826 mutual reactance, 876

N

n (nano), 10t N (newton), 23 nano (n), 10t nanofarads, 329 networks: see specific types network theorems, 202 neutral lead, 892–3 neutrons, 22–3 newton (N), 23 Nichrome™II, 74, 79–80t nickel-cadmium (ni-cad) cell, 57, 59t nickel-metal hydride (ni-mh) cell, 57 nodal analysis, 231–6 AC circuits and, 716–24 nodal equations, 208, 232 nodes, 228 analysis of, 231–6 more than one, 234–6 reference, 229, 718, 898

1019

nominal value resistors and, 88–90 nonlinear impedance networks, 954–5, 956, 962 nonlinear magnetic circuits, 425, 426–9 nonlinear resistors, 86–8 symbols for, 88 nonmagnetic materials, 402 nonsinusoidal waves, 948–54, 955 in linear impedance networks, 963–5 RMS value in, 960–2 north pole, 390, 391, 392 Norton, Edward L., 205 Norton’s theorem, 268–71, 713–16 notation double-subscript, 130, 897–9 engineering, 10 scientific, 7–9 single-subscript, 130 see also type conventions numbers complex, 581–5, 582 imaginary, 581–2 numerical accuracy, 7

O

Oersted, Hans Christian, 393 Ohm, Georg Simon, 70, 126 ohm (�), 70–1 ohm metre (�-m), 74 ohmmeter, 151, 261, 305–8 high-resistance, 308 low-range, 306 troubleshooting and, 375 Ohm’s law, 68–103, 70–3 application of, 92–3 omega (angular velocity), 526, 532–3 omega (ohm), 70–1 open-circuit forward-transfer impedance, 857 open-circuit impedance, 880 open-circuit impedance (z-) parameters, 857–64 open-circuit input impedance, 857 open-circuit output admittance, 869, 870 open-circuit output impedance, 857 open-circuit reverse-transfer impedance, 857 open-circuit reverse-voltage ratio, 869 open-circuit test, 835–6, 837 open-circuit voltage, 134 open windings, 510–11, 843–4 oscilloscope, 538–9 overexcitation, 934

1020

Index

overheating in inductors, 510–11 in wires, 42 overload, 139

P

P (peta), 10t p (pico), 10t P (power in DC circuit/active power in AC circuit), 107–10, 634 parallel circuits, 124–63, 139 capacitance in, 610–13 characteristics of, 145–8, 617t conductance and, 142–5 equations for, 617t impedances in, 671–8 inductance in, 610–13 magnetic, 435–7 resistance in, 610–13 v. series circuits, 141 testing, 619–20 troubleshooting, 151–2 parallel conductors, 324–6, 330–1, 333–6 parallel high-pass and low-pass filters, 805–6 parallel-plate capacitors permittivity and, 335 parallel reflected impedance, 862 parallel-resonant circuits, 753–63 paramagnetic materials, 402–3 passband, 777 passive circuit element, 129 passive filters, 774–823, 776 troubleshooting and, 812–13 peak-to-peak value, 539, 957 peak values, 529, 537 peak voltage reactance and, 556–8 percent slip (S), 932 period, 526, 538 periodic waves, 537–9, 948 average value of, 540 complex, 949 permanent magnet motors, 476 permanent magnets, 404 permeability (m), 399–400, 454 average, 409 BH curve and, 408–10 current and, 404–5 differential, 409 flux density, magnetic field strength and, 401–2 incremental, 409 initial, 409

materials and, 402–4 normal, 408 relative, 399 permeance, 399–400, 454 permittivity (ε), 334–6 absolute, 334 free space, 335 of relative, 336–7 peta (P), 10t phase angle (ϕ), 526, 527–30, 617t phase currents, 905, 907–8 phase plot RC high-pass filters and, 794–6 RC low-pass filters and, 786–7 RL high-pass filters and, 797–9 RL low-pass filters and, 789–90 phase sequence (phase rotation), 915–19 phase voltages, 900, 910–11, 912 in unbalanced systems, 919 in wye-connected systems, 924 phasor diagrams, 574, 576–8, 617t funicular, 839 series circuits and, 600–1 phasor power, 643 phasors, 526–9, 570–97 addition of, 576–8, 585–7, 617t, 669, 672 complex number expression of, 581–5 exponential expression of, 582 impedance and, 598 induced voltage in, 527–8 letter symbols for, 576 perpendicular, 578–80 “stationary,” 574 phi (magnetic flux), 396–7 phi (phase angle), 526, 527–30 photoconductor, 87–8 photoresistors, 87, 88 photovoltaic (solar) cells, 62–3 pi- (π-) network, 725, 866, 867 pico (p), 10t picofarads, 329 piezoelectricity, 64 planar network diagram, 222–3 Planck constant (h), 5 Planté, Gaston, 56 polar coordinates, 581, 584, 585–90, 617t rectangular coordinates and, 583 polarity dielectrics and, 327 electrolytic capacitors and, 332 moving-coil meters and, 297 polarization, 53 polyethylene, 45–6 polymers, 45–6

polyphase systems, 890–945, 893 advantages of, 892–5 polystyrene, 45–6 pony motor, 933 port, 857 potential difference, 28–30, 32–3 charging capacitors and, 350–1, 356 conducting plates and, 326 discharging capacitors and, 357–61 insulator breakdown and, 46 see also volt; voltage potential fall/drop, 30 see also voltage drop potential rise, 30, 32–3 potentiometer, 91 power (P), 107–10 AC circuits and, 632–64 capacitors and, 637–9 energy storage and, 370 ideal inductors and, 635–8 instantaneous, 536–7, 893–4 other electric units and, 113t phasor, 643 potential difference and current and, 107–8 pulsating, 536–7 resistance/reactance circuit, 639–41 resistors and, 634–5 in three-phase system, 913–15, 924–6 time and, 112 two-phase alternators and, 893–4 voltage drop and, 108 work and, 104–22 see also apparent power; reactive power; real (or active) power power amplifiers, 756 power factor, 645–8 power-factor angle, 645 power-factor correction (or improvement), 648–56, 651 electricity suppliers and, 650 power filtering capacitors and, 333 power rating, 109 power triangle, 641–5, 642 precision resistors, 85, 88 prefixes SI units, 10–12 primary cells, 52–5 commercial, 55 simple, 52–3 printed circuit boards, 42–3 protons, 22–3 prototype circuit, 43

Index

psi (electric flux), 325 pulsating DC, 463 waveform, 950 pulses capacitors and, 373, 374 Pythagorean theorem, 580

Q

Q (heat), 107, 100, 113–15 Q (quality factor), 748–9, 761 Q (quantity of electric charge), 26 Q (reactive, quadrature, or imaginary power), 637, 643 quadrature (or imaginary) component, 581, 584 quadrature power (Q), 637, 643 quality factor (Q), 748–9, 761

R

rad (radian), 532–3 radian (rad), 532–3 rate of change current and, 487, 555, 556–7, 558 reactance and, 555, 556–7, 558 rate of change of voltage, 371 algebraic solution for, 362–6 capacitors and, 351–4, 357–66 graphical solution for, 356–7 RC (resistance and capacitance) filters, 776 high-pass, 791–6 low-pass, 781–7 reactance, 552–69, 556 capacitive, 559–60, 560–3, 608, 984–5 factors governing, 556–8 inductive, 555–6, 562–3, 608, 983–4 leakage, 833–4 mutual, 876 reactive component, 617t reactive factor, 647–8 reactive power (Q), 637, 638 power triangle and, 641–2 quality factor and, 748 real (or active) power (P), 634 power factor and, 645 power triangle and, 641–2, 643 rechargeable (secondary) cells, 56–8 rechargeable alkaline-manganese (RAM) cell, 57 recharging, 56–7 reciprocal henry, 398 rectangular coordinates, 581, 582, 585–7, 617t polar coordinates and, 583

reference axis, 575, 584, 649 reference node, 229, 718, 898 reflected impedance, 831 relative permeability, 399 relative permittivity, 336–7 reliability resistors and, 88 reluctance, 398, 454 coils and, 423, 425 series magnetic circuits and, 430–2 remanence, 404 repulsion conducting plates and, 325 magnetic force of, 396 mutual, 392 residual magnetism, 404 resistance (R), 70–103, 562–3, 668 dynamic value of, 92 equivalent, 141–4 factors governing, 72–3 gauge and, 77–8 internal, 133–5, 137–9 leakage of, 50, 502 load, 137–9 measurement of, 297, 304–10 nature of, 71–2 Ohm’s law and, 68–103 other electric units and, 113t parallel circuits and, 610–13 series circuits and, 126–7, 600–10 specific, 74 temperature and, 79–84, 86 temperature coefficient of, 82–4 voltage drop and, 92–3 resistance and reactance containing circuit, 639–41 resistance networks, 200–57 resistance-start motor, 935 resistive component, 617t resistivity (ρ), 73–5, 76, 145 resistors, 72 calibrating, 297 bleeder, 178 carbon-composition, 85, 107 colour code for, 88–90, 89t composition, 85 discharge, 503 instantaneous current in, 533–5 instantaneous power in, 536–7 light dependent (LDR), 87–8 linear, 84–5 multiplier, 300 nonlinear, 86–8 parallel circuits and, 139–41 photo-, 87, 88

1021

power in, 634–5 power rating of, 109 precision, 85, 88 reliability of, 88 series-dropping, 175–6 series-parallel, 166–7 shunt, 297, 299 size of, 107 swamping, 297 symbols for, 88t temperature and, 82–3t tolerance of, 88–90 variable, 91 wire-wound, 84 resonance, 740–73, 745 resonant circuits RLC, 801–4, 806–9 resonant frequency parallel-resonant circuits and, 753–63 series-resonant circuits and, 743–53 voltage and, 749–51 resonant rise of tank current, 757 resonant rise of voltage, 749–51, 750 retentivity, 404, 410 rheostat, 91 rho (resistivity), 74 right-hand rule, 393–5, 451–2 RL (resistance and inductance) filters, 776 high-pass, 797–9 low-pass, 788–90 RLC (resistance, capacitance, and inductance) filters, 779 RLC circuits series, 742–7 RLC resonant circuits, 801–4, 806–9 RMS (root-mean-square) value, 540–3, 541 nonsinusoidal waves and, 960–1 reactance and, 555–6 sine wave, 982–3 roll-off, 785 root-mean-square (RMS) value: see RMS value rotation angle of, 525 rate of, 526 see also angular velocity rotors, 460, 466–7 cylindrical (smooth), 927 salient (projecting), 927 in single-phase motors, 934 squirrel-cage, 931, 934 in three-phase motors, 932 wound, 931

1022

S

Index

S (apparent power), 640 S (elastance), 340 s (second), 5, 6t S (siemens), 142 saturation, 404 theoretical/practical, 406 sawtooth waves, 950, 961–2 scalars, 576 schematic diagrams, 4 second (s), 5, 6t second-order determinants, 974–6 secondary cells, 52, 56–8 selectivity parallel-resonant circuits and, 764 series-resonant circuits and, 751–3 self-excitation, 929 self-inductance, 454–5 self-induction, 453 transformers and, 826 semiconductors, 47 conductors and insulators and, 40–9 resistors and, 88 semilog graph paper, 779–80 sensitivity, 751, 753 measuring instruments and, 297 series circuits, 124–63, 126 capacitance in, 607–10 characteristics of, 131–3, 600–1, 617t equations for, 617t impedances in, 668–71 inductance in, 600–6, 608–10 magnetic, 430–2 v. parallel circuits, 141 resistance in, 600–10 testing, 619 troubleshooting, 151–2 voltage drops in, 128–30 series coils, 460 series coupled impedance, 862 series-dropping resistor, 175–6 series motors, 470–1, 472, 474 series opposing connection, 214–15 series-parallel circuits, 164–99, 167 impedances in, 678–82 troubleshooting, 186–7 series-parallel resistors, 166–7 series-resonant circuits, 742–53, 745 band of frequencies and, 751–2 servomechanisms, 895 shaded-pole motor, 935 “shallow cycling,” 56 short circuit, 152 tests, 835–6

short-circuit admittance (y) parameters, 864–7 short-circuit current, 204 short-circuit forward current ratio, 870 short-circuit forward-transfer admittance, 865 short-circuit input admittance, 865 short-circuit input impedance, 868, 869 short-circuit output admittance, 865 short-circuit reverse-transfer admittance, 865 shorted windings, 510–11, 843–5 shunt coils, 460, 470 shunt motor, 470, 472, 474, 475 shunt resistor, 297, 299 SI units, 4–6, 10–12 new ampere definition and, 396 prefixes for, 10–12 see also specific units Siemens, Ernst Werner von, 142 siemens (S), 142 sigma (conductivity), 145 significant digits, 7 silicon carbine varistors, 87 silicon n-type and p-type, 62–3 as semiconductor, 47 silver, 23, 42 , 73, 74 silver nitrite, 44 silver-oxide cell, 55 simultaneous equations, 974–7 sine waves, 526–9, 528 addition of harmonically related, 951–4 addition of, 572–3 average values of, 982–3 half-wave-rectified, 955 instantaneous value of, 529–31 multiple, 946–72 peak value of, 529 reactance and, 555 RMS value of, 540–3, 982–3 single-phase circuit, 892 single-phase motors, 934–5 single-subscript notation, 130 single-turn coil, 397–8 singly excited motor, 930 slip (S), 932 slip rings, 524 solar cells, 62–3 solenoid, 394–5, 397 materials and, 403–4 source conversion, 202, 206–8, 682–3 Kirchhoff’s current law equations and, 231

sources controlled, 135, 271–8 conversions and, 202, 231, 682–3, 806–8 current-controlled current (CCCS), 271t current-controlled voltage (CCVS), 271t, 271–2, 274 delta-connected, 904 dependent, 271–8 energy, 4 independent, 271 multi-, 216–22 star-connected, 900 Thévenin-equivalent, 274–8 three-phase, 903, 906 variable-frequency, 742–5 voltage, 50–67 voltage-controlled, 271t, 273, 274 wye-connected, 899–90, 909–12 see also constant-current sources; constant-voltage sources source voltage (E), 33 South Korea fuel cell research in, 60 south pole, 390, 391, 392 specific heat, 113 specific resistance, 74 speed in DC motor, 467–9, 471–4 synchronous, 895, 930, 931–2, 933 split-phase motor, 934–5 square waves, 961–2 standard nominal values, 88–9 star-connected source, 900 starting winding, 934–5 static electricity, 64 stator, 460 steady-state current LR circuits and, 507–8 steady-state values, 350–1, 367–8 steam turbine, 62 step-down transformer, 829 step-up transformer, 829 stopband, 806 stray capacitance, 374 subtraction of phasor qualities, 587–8 Successive Approximation Register (SAR), 314 superposition theorem, 236–42, 702–7, 715 nonsinusoidal waves and, 960, 963 susceptance (B), 614–15 swamping resistor, 297

Index

switch, 4 symbols for capacitors, 332 for phasor quantities, 576 for unit names, 5–6 see also specific units symmetrical waveform, 953 synchronous capacitors, 934 synchronous motor, 932–4 synchronous speed, 895, 930, 931–2 in synchronous motor, 933

T

T (magnetic flux density), 400–1 see also magnetic flux density T (temperature), 5, 6t T (tesla), 400 t (time), 5, 6t T-network, 856, 860 tan, 579 tank circuits, 756, 757 tank current, 757 tapped winding, 841 tau (time constant), 354–5, 359, 362 temperature (T) base unit for, 5, 6t heat and, 113–14 measuring, 80, 86 resistance and, 72, 74, 79–82 see also heat temperature coefficient of resistance (α), 82–4 terminal voltage, 133, 135 Tesla, Nikola, 400 tesla (T), 400 thermistor, 86, 88 thermocouples, 63 theta (θ) (phase angle), 526 Thévenin, Léon Charles, 260 Thévenin-equivalent sources, 274–8 Thévenin’s theorem 260–7, 261, 273 AC circuits and, 707–13 applications of, 267 third-order determinants, 976–7 three-phase (3ϕ) systems, 890–945, 896 generation of voltages in, 895–7 power in, 913–15 three-phase alternators, 928 three-phase induction motor, 930–2 three-phase sources, 903, 906 three-phase synchronous motor, 932–4 three-wire systems, 906 thyrite, 87

time (t) base unit for, 5, 6t capacitors and, 350–7, 359, 362 time constant (τ) CR circuits and, 354–5, 359, 362 discharge, 504–5 inductors and, 490–1 time domain, 572–3 tolerance resistors and, 88–90 tonne, 106 toroidal coils, 425 torque, 466–7, 471 in DC motors, 467–9, 474–5 total-admittance method, 673–4 total-current method, 612 tracing direction, 897–90, 905, 909 transformation ratio (a), 828–30, 829 transformer efficiency, 837–8 maximum, 985–6 transformer equation general, 826–7, 985 transformers, 824–53, 826 coupling, 878 iron-core, 958–60 loading and, 838–41 primary current in, 827–8, 830 tests for, 835–6 troubleshooting, 843–5 transient phase, 368 transient response, 366–9, 368, 506, 507–9 transistor circuits, 176, 179 transmission lines, 680–1 trigonometry, 579–80, 605 troubleshooting capacitors, 375 equivalent-circuit theorems, 283 inductors, 510–11 passive filters, 812–13 resistance, inductance, and capacitance, 619–20 series and parallel circuits, 151–2 series-parallel circuits, 186–7 transformers, 843–5 true power, 634; see also real power tuning, 745 turbines, 62 two-port network, 857–74 two-terminal networks Norton’s theorem and, 268–71 Thévenin’s theorem and 260–1 type conventions arrows, 34, 204, 231, 271, 324, 695 boldface, 324, 526, 576, 574, 576

1023

for current direction, 34, 231, 271 double-subscript notation, 130, 897–9 for instantaneous values, 350–1 italic, 6, 397, 399, 541, 576 lightface, 10, 324, 526, 576 lowercase, 348–9, 486, 858, 897–8, 899 for phasors, 526, 576 for quantities, 6 Roman, 6, 576 for scalars, 576 script letters, 324, 397, 398, 399 single subscript, 130 forsteady-state values, 350–1 for units, 6 uppercase, 349, 541, 562 for vectors, 324, 576 see also notation

U

unbalanced load, 902 unbalanced three-phase system, 924–6 unbalanced three-phase wye loads, 919–24 United States fuel cell research in, 60 units, 5 conversion of, 12–13 see also SI units universal exponential graphs, 491–5 universal motor, 472, 935

V

V (volt), 31–2 V (voltage drop), 33 VA (volt ampere), 640 Van de Graaff generator, 64 var (volt ampere reactive), 637 varactors, 745 variable-frequency source, 742–5 variable resistors, 91 varistors, 87–8 vectors, 324, 576 velocity angular (ω), 526, 532–3 linear, 532 volt (V), 31–2 volt ampere (VA), 640 volt ampere reactive (VAR), 637 volt-ohm-milliammeter (VOM), 312–14

1024

Volta, Alessandro, 31 voltage, 32–3 AC, 552–69 alternating, 525 applied/source (E), 33 capacitors and, 351–69 current and, 20–39, 70–103 generated, 463 instantaneous, 367, 978–80 line, 909–12, 913 open-circuit, 134 other electric units and, 113t parallel circuits and, 140 phase with current, 534 resonant rise of, 749–51, 750 terminal, 133, 135 three-phase, 895–7 working, 332 see also induced voltage; phase voltages; potential difference voltage-controlled current source (VCCS), 271t, 273, 274 voltage-controlled voltage source (VCVS), 271t voltage-current characteristics, 91–2 voltage dividers, 173–80 principle of, 173–4, 670–1, 677 voltage drop (V), 33 resistance and, 91–3 series circuits and, 128–32 voltage gradient, 326 voltage law, 130–1 voltage regulation, 135, 178 voltage sources, 50–67 voltmeter, 32, 33, 128, 151, 300–2

Index

electronic, 304 resistance and, 304–5, 308 voltmeter loading effect, 302–4

W

W (watt), 107 W (work), 31 Watt, James, 107, 111 watt (W), 107 wattmeter, 311–12 wave winding, 464 Weber, Wilhelm Eduard, 397 Wb (weber), 397 weber (Wb), 397 Wheatstone bridge, 202, 212–13, 236 delta-to-wye transformation and, 281–2 galvanometer and, 309–10 Thévenin’s theorem and, 264–5 windings air-core transmitters and, 874 autotransformers and, 841–3 distributed, 929, 930 marking of, 878–9 partially shorted, 510–11, 845 in polyphase systems, 892–3 primary, 448, 450, 451–2 secondary, 448, 450, 452 starting, 934–5 transformers and, 826–31, 833, 876, 878–80 troubleshooting and, 510–11, 843–5 wave, 464 wind turbines, 62 wiper, 91

wire-wound resistors, 84 wire gauge of, 77–8 ground, 177 work (W), 31 energy and, 31–2, 106 other electric units and, 113t power and, 104–22 working voltage, 332 wye (Y) circuits, 278 wye-connected load, 900 wye-connected sources, 899–90, 909–12 wye-connected systems, 899–903 phase voltages in, 924 wye-delta systems, 909–12 wye-to-delta transformation, 280–1, 724–8, 922–3

Y

Y (admittance), 614–15; see also admittance yoke, 460 y-parameter equivalent circuit, 866 y-parameters, 864–7

Z

Z (impedance), 598–631, 610; see also impedance zinc-air cell, 55 zinc-chloride cell, 55 zinc oxide varistors, 87 z-parameter equivalent circuit, 859 z-parameters, 857–64