Circuit Theory [3 ed.] 9789339224752, 9339224752

Table of contents :
Title
CONTENTS
1 INTRODUCTION
2 BASIC CONCEPTS OF CIRCUITS
3 BASIC CONCEPTS OF AC CIRCUITS
4 MESH ANALYSIS
5 NODE ANALYSIS
6 NETWORK REDUCTION
7 THEOREMS IN CIRCUIT ANALYSIS
8 SERIES AND PARALLEL RESONANCES
9 COUPLED CIRCUITS
10 TRANSIENTS
11 ONE-PORT AND TWO-PORT NETWORKS
12 THREE-PHASE CIRCUITS
Question Paper
Appendix
Index

Citation preview

CIRCUIT THEORY Third Edition

ABOUT THE AUTHOR A Nagoor Kani is a multifaceted personality with efficient technical expertise and management skills. He obtained his BE degree in Electrical and Electronics Engineering from Thiagarajar College of Engineering, Madurai, and MS (Electronics and Control) through the Distance Learning program of BITS, Pilani. He is a life member of ISTE and IETE. Prof. Nagoor Kani started his career as a self-employed industrialist (1986–1989) and then changed over to teaching in 1989. He has worked as Lecturer in Dr MGR Engineering College (1989–1990) and as Assistant Professor in Satyabhama Engineering College (1990–1997). In 1993, he started a teaching centre for BE students, naming it the Institute of Electrical Engineering, which was renamed RBA Tutorials in 2005. A Nagoor Kani launched his own organization in 1997. The ventures currently run by him are •

RBA Engineering (involved in manufacturing of lab equipment, microprocessor trainer kits, and undertake electrical contracts and provide electrical consultancy);

•

RBA Innovations (involved in developing projects for engineering students and industries);

•

RBA Tutorials (conducting tutorial classes for engineering students and coaching for GATE, IES, IAS);

•

RBA Publications (publishing of engineering books); and

•

RBA Software (involved in web-design and maintenance).

His optimistic and innovative ideas have made the RBA Group a very successful venture. A Nagoor Kani is a well-known name in all major engineering colleges in India. He is an eminent writer and till now he has authored several engineering books (published by McGraw Hill Education and RBA Publications) which are very popular among engineering students. He has written books in the areas of Control Systems, Signals and Systems, Microcontrollers, Digital Signal Processing, Electric Circuits, Electrical Machines, and Power Systems.

CIRCUIT THEORY Third Edition

A Nagoor Kani Founder, RBA Educational Group Chennai

McGraw Hill Education (India) Private Limited NEW DELHI McGraw Hill Education Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited, P-24, Green Park Extension, New Delhi 110 016, Circuit Theory, 3e Copyright © 2016, by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. Print Edition: ISBN-13: 978-93-392-2433-2 ISBN-10: 93-392-2433-7 E-book Edition: ISBN-13: 978-93-392-2475-2 ISBN-10: 93-392-2475-2 Managing Director: Director—Product (Higher Education and Professional): Manager—Product Development: Specialist—Product Development: Head—Production (Higher Education and Professional): Senior Production Executive: Senior Graphic Designer—Cover: AGM—Product Management (Higher Education and Professional): Manager—Product Management: General Manager—Production: Manager—Production:

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Dedicated to Wife Mrs. C. Gnanaparanjothi, B.Sc., M.L., Elder Son N. Bharath Raj Younger Son N. Vikram Raj

CONTENTS PREFACE ...................................................................................................xv ACKNOWLEDGEMENTS ............................................................................... xvii LIST OF SYMBOLS AND ABBREVIATIONS ..................................................... xix CHAPTER 1 - INTRODUCTION ................................................................... 1.1 1.1

Basic Phenomena ...........................................................................................................1.1

1.2

Ideal Elements ................................................................................................................1.1

1.3

Electric Circuits ...............................................................................................................1.1

1.4

Units ...............................................................................................................................1.4

1.5

Definitions of Various Terms ............................................................................................1.5

1.6

Symbols Used for Average, RMS and Maximum Values ................................................1.7

1.7

Steady State Analysis and Transient Analysis.................................................................1.8

1.8

Assumptions in Circuit Theory.........................................................................................1.8

CHAPTER 2 - BASIC CONCEPTS OF CIRCUITS ............................................ 2.1 2.1

Basic Elements of Circuits ..............................................................................................2.1

2.2

Nodes, Branches and Closed Path .................................................................................2.2

2.3

Series, Parallel, Star and Delta Connections ..................................................................2.4 2.3.1

Series Connection ..............................................................................................2.4

2.3.2

Parallel Connection .............................................................................................2.4

2.3.3

Star-Delta Connection ........................................................................................2.6

2.4

Open Circuit and Short Circuit ........................................................................................2.7

2.5

Sign Conventions ............................................................................................................2.8

2.6

Ohm’s and Kirchhoff’s Laws ............................................................................................2.9

2.7

2.6.1

Ohm’s Law ..........................................................................................................2.9

2.6.2

Kirchhoff’s Current Law (KCL) ............................................................................2.9

2.6.3

Kirchhoff’s Voltage Law (KVL).............................................................................2.9

Voltage and Current Sources ....................................................................................... 2.10 2.7.1

Ideal and Practical Sources ............................................................................. 2.10

2.7.2

DC Source Transformation ............................................................................... 2.11

2.8

Power and Energy ....................................................................................................... 2.12

2.9

Resistance ................................................................................................................... 2.13

2.10

2.9.1

Resistance Connected to dc Source ............................................................... 2.13

2.9.2

Current Division in Parallel Connected Resistances........................................ 2.13

2.9.3

Voltage Division in Series Connected Resistances ......................................... 2.14

Network Topology ......................................................................................................... 2.14 2.10.1 Graph ............................................................................................................... 2.15

Contents

viii

2.10.2 Tree, Link and Cotree....................................................................................... 2.16 2.10.3 Network Variables ............................................................................................ 2.18 2.10.4 Solution of Network Variables .......................................................................... 2.18 2.11

Analysis of Series-Parallel Circuits .............................................................................. 2.19 2.11.1 Single Loop Circuit........................................................................................... 2.21 2.11.2 Single Node Pair Circuit ................................................................................... 2.21

2.12

Solved Problems .......................................................................................................... 2.22

2.13

Summary of Important Concepts ................................................................................. 2.31

2.14

Short-answer Questions ............................................................................................... 2.33

2.15

Exercises ...................................................................................................................... 2.38

CHAPTER 3 - BASIC CONCEPTS OF AC CIRCUITS ....................................... 3.1 3.1

Introduction .....................................................................................................................3.1

3.2

AC Voltage and Current Source ......................................................................................3.1 3.2.1

3.3

AC Source Transformation ..................................................................................3.3

Sinusoidal Voltage ...........................................................................................................3.3 3.3.1

Average Value .....................................................................................................3.6

3.3.2

RMS Value ..........................................................................................................3.6

3.3.3

Form Factor and Peak Factor .............................................................................3.7

3.4

Sinusoidal Current ...........................................................................................................3.7

3.5

Inductance .......................................................................................................................3.8

3.6

3.7

3.8

3.5.1

Voltage-Current Relation in an Inductance .........................................................3.9

3.5.2

Energy Stored in an Inductance ...................................................................... 3.10

Capacitance ................................................................................................................. 3.10 3.6.1

Voltage-Current Relation in a Capacitance ...................................................... 3.11

3.6.2

Energy Stored in a Capacitance ...................................................................... 3.11

Voltage-Current Relation of R, L and C in Various Domains ........................................ 3.12 3.7.1

Voltage-Current Relation of Resistance ........................................................... 3.12

3.7.2

Voltage-Current Relation of Inductance ........................................................... 3.13

3.7.3

Voltage-Current Relation of Capacitance......................................................... 3.14

Sinusoidal Voltage and Current in Frequency Domain ................................................. 3.15 3.8.1

Phase and Phase Difference ........................................................................... 3.15

3.8.2

Phasor Representation of Sinusoidal Quantities ............................................. 3.17

3.8.3

Phasor Diagram of a Circuit ............................................................................. 3.18

3.9

Power, Energy and Power Factor ................................................................................. 3.19

3.10

Resistance Connected to Sinusoidal Source ............................................................... 3.22

3.11

Inductance Connected to Sinusoidal Source ............................................................... 3.24

3.12

Capacitance Connected to Sinusoidal Source ............................................................. 3.26

3.13

Impedance ................................................................................................................... 3.29 3.13.1 Impedance Connected to Sinusoidal Source ................................................... 3.30

Contents 3.14

ix Conductance, Susceptance and Admittance ............................................................... 3.32 3.14.1 Conductance.................................................................................................... 3.32 3.14.2 Admittance ....................................................................................................... 3.32 3.14.3 Admittance Connected to Sinusoidal Source .................................................. 3.33

3.15

KVL, KCL and Ohm’s Law Applied to ac Circuits ......................................................... 3.35

3.16

Current and Voltage Division Rules for Impedances .................................................... 3.35 3.16.1 Current Division in Parallel Connected Impedances........................................ 3.35 3.16.2 Voltage Division in Series Connected Impedances ......................................... 3.35

3.17

Solved Problems .......................................................................................................... 3.36

3.18

Summary of Important Concepts ................................................................................. 3.58

3.19

Short-answer Questions ............................................................................................... 3.61

3.20

Exercises ...................................................................................................................... 3.65

CHAPTER 4 - MESH ANALYSIS ................................................................. 4.1 4.1

Introduction .....................................................................................................................4.1

4.2

Mesh Analysis of Resistive Circuits Excited by dc Sources ............................................4.1

4.3

Mesh Analysis of Circuits Excited by Both Voltage and Current Sources .................... 4.24 4.3.1

4.4

Supermesh Analysis ........................................................................................ 4.24

Mesh Analysis of Circuits Excited by ac Sources (Mesh Analysis of Reactive Circuits) ........................................................................... 4.30

4.5

Summary of Important Concepts ................................................................................. 4.36

4.6

Short-answer Questions ............................................................................................... 4.37

4.7

Exercises ...................................................................................................................... 4.39

CHAPTER 5 - NODE ANALYSIS ................................................................. 5.1 5.1

Introduction .....................................................................................................................5.1

5.2

Node Analysis of Resistive Circuits Excited by dc Sources ............................................5.1

5.3

Node Analysis of Circuits Excited by Both Voltage and Current Sources .................... 5.16 5.3.1

5.4

Supernode Analysis ......................................................................................... 5.17

Node Analysis of Circuits Excited by ac Sources ......................................................... 5.31 (Noda Analysis of Reactive Circuits)

5.5

Summary of Important Concepts ................................................................................. 5.35

5.6

Short-answer Questions ............................................................................................... 5.37

5.7

Exercises ...................................................................................................................... 5.39

CHAPTER 6 - NETWORK REDUCTION......................................................... 6.1 6.1

Introduction .....................................................................................................................6.1

6.2

Resistances in Series and Parallel ..................................................................................6.1 6.2.1

Equivalent of Series-connected Resistances .....................................................6.1

6.2.2

Voltage Division in Series-connected Resistances .............................................6.2

Contents

x

6.3

6.2.3

Equivalent of Parallel-connected Resistances ....................................................6.3

6.2.4

Current Division in Parallel-connected Resistances ...........................................6.4

Resistances in Star and Delta .........................................................................................6.5 6.3.1

Delta to Star Transformation ...............................................................................6.6

6.3.2

Star to Delta Transformation ...............................................................................6.6

6.4

Source Transformation ....................................................................................................6.7

6.5

Voltage Sources in Series and Parallel ...........................................................................6.9

6.6

6.7

6.8

6.9

6.10

6.5.1

Series Connection of Voltage Sources ...............................................................6.9

6.5.2

Parallel Connection of Voltage Sources ........................................................... 6.10

Current Sources in Series and Parallel ........................................................................ 6.11 6.6.1

Series Connection of Current Sources ............................................................ 6.11

6.6.2

Parallel Connection of Current Sources........................................................... 6.12

Inductances in Series and Parallel ............................................................................... 6.13 6.7.1

Equivalent of Series-connected Inductances ................................................... 6.13

6.7.2

Equivalent of Parallel-connected Inductances ................................................. 6.15

Capacitances in Series and Parallel............................................................................. 6.16 6.8.1

Equivalent of Series-connected Capacitances ................................................ 6.16

6.8.2

Equivalent of Parallel-connected Capacitances ............................................... 6.17

Impedances in Series and Parallel ............................................................................... 6.19 6.9.1

Equivalent of Series-connected Impedances .................................................. 6.19

6.9.2

Voltage Division in Series-connected Impedances .......................................... 6.20

6.9.3

Equivalent of Parallel-connected Impedances ................................................. 6.20

6.9.4

Current Division in Parallel-connected Impedances ........................................ 6.21

6.9.5

Reactances in Series and Parallel ................................................................... 6.22

Impedances in Star and Delta ...................................................................................... 6.23 6.10.1 Delta to Star Transformation ............................................................................ 6.23 6.10.2 Star to Delta Transformation ............................................................................ 6.23

6.11

Conductances in Series and Parallel ........................................................................... 6.24 6.11.1 Equivalent of Series-connected Conductances ............................................... 6.24 6.11.2 Equivalent of Parallel-connected Conductances.............................................. 6.25

6.12

Admittances in Series and Parallel............................................................................... 6.26 6.12.1 Equivalent of Series-connected Admittances .................................................. 6.26 6.12.2 Equivalent of Parallel-connected Admittances ................................................. 6.27 6.12.3 Susceptances in Series and Parallel ............................................................... 6.29

6.13

Generalized Concept of Reducing Series/Parallel-connected Parameters .................. 6.29

6.14

Solved Problems .......................................................................................................... 6.32

6.15

Summary of Important Concepts ................................................................................. 6.46

6.16

Short-answer Questions ............................................................................................... 6.48

6.17

Exercises ...................................................................................................................... 6.52

Contents

xi

CHAPTER 7 - THEOREMS IN CIRCUIT ANALYSIS......................................... 7.1 7.1

Introduction .....................................................................................................................7.1

7.2

Superposition Theorem ...................................................................................................7.1

7.3

Thevenin’s and Norton’s Theorems .............................................................................. 7.20

7.4

Maximum Power Transfer Theorem .............................................................................. 7.47

7.5

Reciprocity Theorem .................................................................................................... 7.69 7.5.1

Reciprocity Theorem Applied to Mesh Basis Circuit ........................................ 7.69

7.5.2

Reciprocity Theorem Applied to Node Basis Circuit ........................................ 7.70

7.6

Summary of Important Concepts ................................................................................. 7.80

7.7

Short-answer Questions ............................................................................................... 7.81

7.8

Exercises ...................................................................................................................... 7.87

CHAPTER 8 - SERIES AND PARALLEL RESONANCES ................................... 8.1 8.1

Introduction .....................................................................................................................8.1

8.2

Series Resonance ...........................................................................................................8.1

8.3

8.2.1

Resonance Frequency of Series RLC Circuit .....................................................8.1

8.2.2

Characteristics of Series RLC Circuit .................................................................8.2

8.2.3

Quality Factor of Series RLC Circuit ...................................................................8.3

8.2.4

Bandwidth of Series RLC Circuit ........................................................................8.6

8.2.5

Selectivity of Series RLC Circuit ...................................................................... 8.10

8.2.6

Voltage across R, L, and C at Resonance in Series RLC Circuit..................... 8.11

8.2.7

Solved Problems in Series Resonance............................................................ 8.12

Parallel Resonance ...................................................................................................... 8.16 8.3.1

Resonant Frequency of Parallel RLC Circuits.................................................. 8.17

8.3.2

Characteristics of Parallel RLC Circuit ............................................................. 8.25

8.3.3

Quality Factor of RLC Parallel Circuit .............................................................. 8.26

8.3.4

Bandwidth of RLC Parallel Circuit .................................................................... 8.29

8.3.5

Current through L and C at Resonance in Parallel RLC Circuit ....................... 8.35

8.3.6

Solved Problems in Parallel Resonance .......................................................... 8.36

8.4

Summary of Important Concepts ................................................................................. 8.43

8.5

Short-answer Questions ............................................................................................... 8.46

8.6

Exercises ...................................................................................................................... 8.52

CHAPTER 9 - COUPLED CIRCUITS ................................................................... 9.1 9.1

Introduction .....................................................................................................................9.1

9.2

Self-Inductance and Mutual Inductance ..........................................................................9.2 9.2.1

Self-Inductance ...................................................................................................9.2

9.2.2

Mutual Inductance ..............................................................................................9.2

9.2.3

Coefficient of Coupling........................................................................................9.4

Contents

xii 9.3

Analysis of Coupled Coils..............................................................................................9.5 9.3.1

Dot Convention for Coupled Coils.....................................................................9.6

9.3.2

Expression for Self-and Mutual Induced Emf in Various Domains....................9.9

9.3.3

Writing Mesh Equations for Coupled Coils .....................................................9.10

9.3.4

Electrical Equivalent of Magnetic Coupling

9.3.5

Writing Mesh Equations in Circuits with Electrical Connection

(Electrical Equivalent of a Transformer or Linear Transformer)........................9.11 and Magnetic Coupling................................................................................. ..9.14 9.4

Series and Parallel Connections of Coupled Coils.................................................... ..9.15

9.5

Tuned Coupled Circuits ............................................................................................ ..9.21 9.5.1

Single Tuned Coupled Circuits...................................................................... ..9.21

9.5.2

Double Tuned Coupled Circuits ......................................................................9.26

9.6

Solved Problems .........................................................................................................9.29

9.7

Summary of Important Concepts ................................................................................9.52

9.8

Short-answer Questions ..............................................................................................9.55

9.9

Exercises .....................................................................................................................9.58

CHAPTER 10 - TRANSIENTS.........................................................................10.1 10.1

Introduction..................................................................................................................10.1

10.2

Transient Response.....................................................................................................10.1 10.2.1 First and Second Order Circuits......................................................................10.2

10.3

Transient Analysis Using Laplace Transform...............................................................10.3 10.3.1 Some Standard Voltage Functions..................................................................10.3 10.3.2 s-Domain Representation of R, L, C Parameters............................................10.5 10.3.3 Solving Initial and Final Conditions Using Laplace Transform.........................10.9

10.4

Transient in RL Circuit ...............................................................................................10.10 10.4.1 Source-Free Response of RL circuit .............................................................10.10 10.4.2 Step Response of RL Circuit ........................................................................10.11 (Response of RL Circuit Excited by dc Supply) 10.4.3 RL Transient With Initial Current I0 ...............................................................10.16

10.5

Transient in RC Circuit...............................................................................................10.19 10.5.1 Source-Free Response of RC Circuit ...........................................................10.19 10.5.2 Step Response of RC Circuit........................................................................10.21 (Response of RC Circuit Excited by dc Supply) 10.5.3 RC Transient With Initial voltage V0 ............................................................. .10.25

10.6

Transient in RLC Circuit.............................................................................................10.29 10.6.1 Source-Free Response of RLC circuit ..........................................................10.29 10.6.2 Step Response of RLC Circuit ......................................................................10.29 (Response of RLC Circuit Excited by dc Supply) 10.6.3 s-Domain Current and Voltage Equation of RLC Circuit ..............................10.36

Contents

xiii 10.6.4 Initial Conditions in RLC Circuit ......................................................................10.37 10.6.5 Final Conditions in RLC Circuit ......................................................................10.39

10.7

Transient in Circuits Excited by ac Source ................................................................10.41 10.7.1 RL Circuit Excited by ac Source ....................................................................10.41 10.7.2 RC Circuit Excited by ac Source ................................................................... 10.43 10.7.3 RLC Circuit Excited by ac Source ................................................................. 10.45

10.8

Solved Problems in RL Transient................................................................................10.47

10.9

Solved Problems in RC Transient ...............................................................................10.62

10.10

Solved Problems in RLC Transient .............................................................................10.82

10.11

Summary of Important Concepts ...............................................................................10.92

10.12

Short-answer Questions .............................................................................................10.95

10.13

Exercises ...................................................................................................................10.98

CHAPTER 11 - ONE-PORT AND TWO-PORT NETWORKS.................................. 11.1 11.1

Introduction...................................................................................................................11.1

11.2

Parameters of a One-Port Network..............................................................................11.2

11.3

Parameters of a Two-Port Network...............................................................................11.3

11.4

Impedance Parameters (or Z-Parameters)...................................................................11.7

11.5

Admittance Parameters (or Y-Parameters)...................................................................11.8 .

11.6

Hybrid Parameters (or h-Parameters)..........................................................................11.9

11.7

Inverse Hybird Parameters (or g-Parameters) ...........................................................11.11

11.8

Transmission Parameters (or ABCD-Parameters) ......................................................11.12

11.9

Inverse Transmission Parameters (or A’B’C’D’-Parameters) .....................................11.13

11.10

Relationship Between Parameter Sets ......................................................................11.14

11.11

Properties of Two-Port Networks ...............................................................................11.20

11.12

Inter-Connection of Two-Port Networks .....................................................................11.20

11.13

Solved Problems .......................................................................................................11.22

11.14

Summary of Important Concepts ...............................................................................11.42

11.15

Short-answer Questions .............................................................................................11.43

11.16

Exercises ....................................................................................................................11.47

CHAPTER 12 - THREE PHASE CIRCUITS.........................................................12.1 12.1

Introduction ..................................................................................................................12.1

12.2

Three-Phase Sources ..................................................................................................12.1

12.3

Representation of Three-Phase EMFs ........................................................................12.2

12.4

Three-Phase Star-connected Source ...........................................................................12.4 12.4.1 Star-connected Source Three Wire-System ....................................................12.4 12.4.2 Star-connected Source Four Wire-System ......................................................12.7

12.5

Three-Phase Delta-connected Source .........................................................................12.8

12.6

Three-Phase Loads....................................................................................................12.13 12.6.1 Choice of Reference Phasor in Analysis of Three-Phase Circuits .................12.14

Contents

xiv 12.7

Analysis of Balanced Loads .........................................................................................12.15 12.7.1 Four-Wire Star-connected Balanced Load.......................................................12.15 12.7.2 Three-Wire Star-connected Balanced Load.....................................................12.17 12.7.3 Delta-connected Balanced Load......................................................................12.17 12.7.4 Power Consumed by Three Equal Impedance in Star and Delta.....................12.19

12.8

Analysis of Unbalanced loads......................................................................................12.20 12.8.1 Four-Wire Star-connected Unbalanced Load..................................................12.20 12.8.2 Three-Wire Star-connected Unbalanced Load................................................12.22 12.8.3 Neutral Shift in Star-connected Load.............................................................. 12.24 12.8.4 Delta-connected Unbalanced Load.................................................................12.24

12.9

Two Wattmeter Method of Power Measurement.......................................................... 12.26 12.9.1 Power Measurement in Balanced Load...........................................................12.27 12.9.2 Relation Between Power Factor and Wattmeter Readings..............................12.29

12.10

Solved Problems..........................................................................................................12.31

12.11

Summary of Important Concepts.................................................................................12.68

12.12

Short-answer Questions.............................................................................................. 12.72

12.13

Exercises..................................................................................................................... 12.76

APPENDIX APPENDIX APPENDIX APPENDIX APPENDIX APPENDIX APPENDIX APPENDIX APPENDIX APPENDIX APPENDIX APPENDIX

1 2 3 4 5 6 7 8 9 10 11 12

-

APPENDIX 13 APPENDIX APPENDIX APPENDIX APPENDIX APPENDIX

14 15 16 17 18

-

USING CALCULATOR IN COMPLEX MODE ......................................A.1 IMPORTANT MATHEMATICAL FORMULAE ......................................A.3 LAPLACE TRANSFORM .................................................................A.5 R,L,C PARAMETERS AND V-I RELATIONS IN VARIOUS DOMAINS .....A.8 CRAMER’S RULE ..........................................................................A.9 EQUIVALENT OF SERIES/PARALLEL CONNECTED PARAMETERS...... A.11 STAR-DELTA TRANSFORMATION ............................................... A.13 SUMMARY OF THEOREMS.......................................................... A.14 IMPORTANT EQUATIONS OF SERIES RESONANCE......................... A.15 PARALLEL RESONANT CIRCUITS.................................................. A.16 ELECTRICAL EQUIVALENT OF COUPLED COILS............................. A.17 EQUIVALENT OF SERIES AND PARALLEL CONNECTED COUPLED COILS........................................................................ A.18 INITIAL AND FINAL CONDITIONS IN RLC CIRCUITS EXCITED BY DC SUPPLY.......................................................................... A.19 SUMMARY OF PARAMETERS OF TWO-PORT NETWORK................. A.20 RELATIONSHIP BETWEEN PARAMETER SETS................................. A.21 CHOICE OF REFERENCE PHASOR.................................................. A.22 V-I EQUATION OF THREE PHASE LOAD......................................... A.23 TWO WATTMETER METHOD OF POWER MEASUREMENT............... A.28

ANNA UNIVERSITY QUESTION PAPERS.................................................................. Q.1 INDEX...................................................................................................................I.1

PREFACE The main objective of this book is to explore the basic concepts of Circuit Theory in a simple and easy-to-understand manner. This text on Circuit Theory has been crafted and designed to meet student’s requirements. Considering the highly mathematical nature of this subject, more emphasis has been given on the problem-solving methodology. Considerable effort has been made to elucidate mathematical derivations in a step-by-step manner. Exercise problems with varied difficulty levels are given in the text to help students get an intuitive grasp on the subject. This book with its lucid writing style and germane pedagogical features will prove to be a master text for engineering students and practitioners. Salient Features The salient features of this book on Circuit Theory are, - proof of important concepts and theorems are clearly highlighted by shaded boxes - wherever required, problems are solved in multiple methods - additional explanations for solutions and proofs are provided in separate boxes - different types of fonts are used for text, proof and solved problems for better clarity - keywords are highlighted by bold, italic fonts - easy, concise and accurate study material - extremely precise edition where concepts are reinforced by pedagogy - demonstration of multiple techniques in problem solving-additional explanations and proofs highlighted - ample figures and examples to enhance students’ understanding - practice through MCQ’s - pedagogy: Solved Numerical Examples: 221 Short-Answer Q/A:225 Figures: 1312 Numerical Problems:135 Review Questions (T/F):123 MCQs:160 Fill in the blanks:123

. . . . . . .

xvi

Preface

Organization

This text is designed for an undergraduate course in Circuit Theory for engineering students. This book is organized with 12 chapters. The fundamental concepts, steady state analysis and transient state analysis are presented in a very easy and elaborative manner. Throughout the book, carefully chosen examples are presented so that the reader will have a clear understanding of the concepts discussed. Chapter 1 starts with explanation of fundamental quantities involved in circuit theory, standard symbols and units used in circuit theory. The basic concepts of dc and ac circuits are presented separately in chapters 2 and 3.The mesh and node analysis of circuits are discussed separately in chapters 4 and 5, with special attention to dependent sources. The concepts of series, parallel and star-delta network reduction are discussed in Chapter 6. The analysis of circuits using theorems are presented in Chapter 7. The concepts of resonance are discussed in detail in Chapter 8. The analysis of coupled circuits are discussed in Chapter 9. The transient analysis of circuits are explained in Chapter 10 through Laplace transform. The concepts of one port and two port networks are discussed in Chapter 11. The analysis of three phase circuits and measurement of power in three phase circuits are presented in Chapter 12. The Laplace transform has been widely used in the analysis of Electric Circuits. Hence an appendix on Laplace transform is included in this book.All the calculations in this book are performed using calculator in complex mode. An appendix is also included to help the readers to practice calculations in complex mode of calculator. Since circuit theory is introduced as a course in the first year of engineering curriculum in most of the universities, this subject is considered tough by students entering engineering courses. Hence the author has taken special care in presenting the concepts in simple manner supported by carefully chosen solved problems. Online Learning Center The OLC of the book can be accessed at The author hopes that that the teaching and student community will welcome the book. The readers can feel free to convey their criticism and suggestions to [email protected] for further improvement of the book.

A Nagoor Kani Publisher’s Note McGraw Hill Education (India) invites suggestions and comments from you, all of which can be sent to (kindly mention the title and author name in the subject line). Piracy-related issues may also be reported.

ACKNOWLEDGEMENTS I express my heartfelt gratitude to my wife Ms C Gnanaparanjothi Nagoor Kani and my sons—N Bharath Raj alias Chandrakani Allaudeen and N Vikram Raj for the support, encouragement and cooperation they have extended to me throughout my career. I thank Ms T A Benazir, Manager, RBA Group, and all my office-staff for their cooperation in carrying out my day-to-day activities. It is my pleasure to acknowledge the contributions of our technical editors—Ms E R Suhasini, Ms C Mohana Priya and Ms F Shynimol for editing, proofreading and typesetting of the manuscript and preparing the layout of the book. I am also grateful to Ms Vibha Mahajan, Ms Koyel Ghosh, Ms Piyali Chatterjee, Mr Satinder Singh and Ms Jagriti Kundu of McGraw Hill Education (India) for their concern and care in publishing this work. Finally, a special note of appreciation is due to my sisters, brothers, relatives, friends, students and the entire teaching community for their overwhelming support and encouragement to my writing.

A Nagoor Kani

LIST OF SYMBOLS AND ABBREVIATIONS P

-

Active power

Y

-

Admittance

AC, ac

-

Alternating current

A

-

Ampere

ω

-

Angular frequency

ωr

-

Angular resonance frequency

S

-

Apparent Power

I ave

-

Average value of current

Vave

-

Average value of voltage

β

-

Bandwidth

B

-

Branch

C

-

Capacitance

XC

-

Capacitive reactance

BC

-

Capacitive susceptance

Q

-

Charge

k

-

Coefficient of coupling

j

-

Complex operator (j = - 1)

S

-

Complex Power

G

-

Conductance

C

-

Coulomb

kC

-

Critical coefficient of coupling

RC

-

Critical resistance

I

-

Current

i (0+)

-

Current at t = 0+

i (0−)

-

Current at t = 0–

i (∞)

-

Current at t = ∞

CC

-

Current Coil

I (jω), I

-

Current in frequency domain

I(s)

-

Current in Laplace domain

(t)

-

Current in time domain

ζ

-

Damping ratio

List of Symbols and Abbreviations

xx

E

-

DC source voltage

D

-

Determinant of matrix

DC, dc

-

Direct current

Y

-

Driving point admittance

Z

-

Driving point impedance

hB

-

Efficiency of battery

W

-

Energy

Req

-

Equivalent resistance

F

-

Farad

φ

-

Flux

Ψ

-

Flux linkage

kf

-

Form factor

p

-

Half period

H

-

Henry

Hz

-

Hertz

ωh

-

Higher cut-off angular frequency

fh

-

Higher cut-off frequency

j

-

Imaginary part

Z

-

Impedance

θ

-

Impedance angle

L

-

Inductance

XL

-

Inductive reactance

BL

-

Inductive susceptance

e, e(t)

-

Instantaneous value of ac source voltage

q

-

Instantaneous value of charge

-

Instantaneous value of current in time domain

C

-

Instantaneous value of current through capacitor

L

-

Instantaneous value of current through inductor

R

-

Instantaneous value of current through resistor

w

-

Instantaneous value of energy

p

-

Instantaneous value of power

-

Instantaneous value of voltage across capacitor

(t)

C

List of Symbols and Abbreviations

L

-

Instantaneous value of voltage across inductor

R

-

Instantaneous value of voltage across resistor

, (t)

-

Instantaneous value of voltage in time domain

J

-

Joule

K

-

Kelvin

kWh

-

kilowatt-hour

KCL

-

Kirchhoff’s Current Law

KVL

-

Kirchhoff’s Voltage Law

L

-

Laplace operator

L

-

Links

IL

-

Load Current

VL

-

Load Voltage

RL

-

Load Resistance

ωl

-

Lower cut-off angular frequency

fl

-

Lower cut-off frequency

Z

-

Magnitude of impedance

Y

-

Magnitude of admittance

Im

-

Maximum value of current

Vm

-

Maximum value of voltage

m

-

Mesh

M

-

Mho

M

-

Mutual inductance

ωn

-

Natural frequency

IN

-

Neutral current

N

-

Neutral point

N

-

Nodes

Ω

-

Ohm

Ω-m

-

Ohm-metre

O.C.

-

Open circuit

kp

-

Peak factor

φ

-

Phase difference between voltage and current

pf

-

Power factor

xxi

List of Symbols and Abbreviations

xxii

φ

-

Power factor angle

P

-

Power or Active power

PC

-

Pressure Coil

Q

-

Quality factor

Qr

-

Quality factor at resonance

rad/s

-

Radians/second

X

-

Reactance

Q

-

Reactive Power

R

-

Resistance

ρ

-

Resistivity

fr

-

Resonance frequency

s

-

Second

S.C.

-

Short circuit

S

-

Siemen

SDPT

-

Single Pole Double Throw

RS

-

Source Resistance

B

-

Susceptance

T

-

Tesla

t

-

Time

τ

-

Time constant

V

-

Volt

VAR

-

Volt-Ampere-Reactive

V

-

Voltage

(0 )

-

Voltage at t = 0+

(0−)

-

Voltage at t = 0−

(∞)

-

Voltage at t = ∞

V^ jωh, V

-

Voltage in frequency domain

V(s)

-

Voltage in Laplace domain

W

-

Watt

W-h

-

Watt-hour

W-s

-

Watt-second

Wb

-

Weber/Weber-turn

+

Chapter 1

INTRODUCTION 1.1

Basic Phenomena

The energy associated with flow of electrons is called electrical energy. The flow of electrons is called current. The current can flow from one point to another point of an element only if there is a potential difference between these two points. The potential difference is called voltage. When electric current is passed through a device or element, three phenomena have been observed. The three phenomena are, (i) opposition to flow of current, (ii) opposition to change in current or flux, and (iii) opposition to change in voltage or charge. The various effects of current like heating, arcing, induction, charging, etc., are due to the above phenomena. Therefore, three fundamental elements have been proposed which exhibit only one of the above phenomena when it is considered as an ideal element (of course, there is no ideal element in nature). These elements are Resistor, Inductor and Capacitor.

1.2

Ideal Elements

The ideal resistor offers only opposition to flow of current. The property of opposition to flow of current is called resistance and it is denoted by R. The ideal inductor offers only opposition to change in current (or flux). The property of opposition to change in current is called inductance and it is denoted by L. The ideal capacitor offers only opposition to change in voltage (or charge). The property of opposition to change in voltage is called capacitance and it is denoted by C.

1.3

Electric Circuits

The behaviour of a device to electric current can be best understood if it is modelled by using the fundamental elements R, L and C. For example, the incandescent lamp and water heater can be modelled as ideal resistance. The transformers and motors can be modelled using resistance and inductance. Practically, an electric circuit is a model of a device operated by electrical energy. The various concepts and methods used for analyzing the circuit is called circuit theory. A typical circuit consists of sources of electrical energy and ideal elements R, L and C. The practical energy sources are Batteries, Generators (or Alternators), Rectifiers, Transistors, Op-amps, etc. The various elements of electric circuits are shown in Figs. 1.1 and 1.2.

1. 2

Chapter 1 - Introduction Elements of Electric Circuits

Parameters or Loads

Energy Sources DC (Direct Current) Sources DC Voltage Sources

E Independent DC Voltage Source,

+-

Dependent DC Voltage Source Voltage Controlled DC Voltage Source,

mVx + -

RM Ix = Vx Current Controlled DC Voltage Source,

+ -

DC Current Sources I Independent DC Current Source, Dependent DC Current Source GM Vx = Ix Voltage Controlled DC Current Source, AI Ix Current Controlled DC Current Source, AC (Alternating Current) Sources AC Voltage Sources Independent AC Voltage Source,

o E+= EÐq - V

~

Dependent AC Voltage Source

mVx

Voltage Controlled AC Voltage Source, Current Controlled AC Voltage Source,

+ -

RM Ix = Vx + -

AC Current Sources I = IÐq o A

Independent AC Current Source,

~

Dependent AC Current Source GM Vx = Ix

Voltage Controlled AC Current Source, Current Controlled AC Current Source,

AI I x

Fig. 1.1 : Elements of electric circuits-energy source.

1. 3

Circuit Theory Elements of Electric Circuits

Energy Sources

Parameters or Loads Fundamental Parameters R

Resistance, Inductance

Self Inductance,

L

Mutual Inductance, M C Capacitance, Derived Parameters ± jX

Reactance,

Inductive Reactance,

+jXL = +j2pfL

Capacitive Reactance,

- jX C = - j

1 2pfC

Z = R ± jX

Impedance, Inverse Parameters G=

Conductance, m jB =

Susceptance,

1 R 1 ± jX

Inductive Susceptance,

- jB L = - j

1 2pfL

+jBC = +j2pfC Capacitive Susceptance, Y = G m jB

Admittance, Y=

1 1 = = G m jB Z R ± jX

Fig. 1.2 : Elements of electric circuits-parameters or loads.

1. 4

Chapter 1 - Introduction

The elements which generate or amplify energy are called active elements. Therefore, energy sources are active elements. The elements which dissipate or store energy are called passive elements. The resistance dissipates energy in the form of heat, inductance stores energy in magnetic field, and capacitance stores energy in electric field. Therefore, resistance, inductance and capacitance are passive elements. If there is no active element in a circuit then the circuit is called passive circuit or network. The sources can be classified into independent and dependent sources. Batteries, generators and rectifiers are independent sources, which can directly generate electrical energy. The transistors and op-amps are dependent sources whose output energy depends on another independent source. Practically, the sources of electrical energy are used to supply electrical energy to various devices like lamps, fans, motors, etc., which are called loads. The rate at which electrical energy is supplied is called power. The power in turn is the product of voltage and current. The circuit analysis rely on the concept of law of conservation of energy, which states that energy can neither be created, nor be destroyed, but can be converted from one form to other. Therefore, the total energy/power in a circuit will be zero.

1.4

Units

SI units are followed in this book. The SI units and their symbols for the various quantities encountered in circuit theory are presented in Table 1.1. In engineering application the large values are expressed with decimal multiples and small values are expressed with submultiples. The commonly used multiples and submultiples are listed in Table 1.2. TABLE 1.1 : UNITS AND SYMBOLS Quantity

Symbol for quantity

Unit

Unit symbol

Equivalent unit

Equivalent unit symbol

Charge

q, Q

Coulomb

C

Current

i, I

Ampere

A

Flux linkages

ψ

Weber-turn

Wb

-

-

Magnetic flux

φ

Weber

Wb

-

-

Energy

w, W

Joule

J

Newton-meter

N-m

Voltage

v, V

Volt

V

Joule/Coulomb

J/C

Power

p, P

Watt

W

Joule/second

J/s

Capacitance

C

Farad

F

Coulomb/Volt

C/V

Inductance

L, M

Henry

H

Weber/Ampere

Wb/A

Resistance

R

Ohm

Ω

Volt/Ampere

V/A

Conductance

G

Siemens

S

Ampere/Volt or mho

A/V or M

Coulomb/second

C/s

1. 5

Circuit Theory

TABLE 1.1: Continued... Quantity

Symbol for quantity

Unit

Unit symbol

Time

t

Second

s

Frequency

f

Hertz

Hz

Angular frequency

ω

Radians/second

rad/s

Magnetic flux density

-

Tesla

T

Temperature

-

Kelvin

o

Equivalent unit

Equivalent unit symbol

-

-

cycles/second

-

-

-

Weber/ meter square

Wb/m2

-

-

K

TABLE 1.2 : MULTIPLE AND SUBMULTIPLE USED FOR UNITS Multiplying factor

Prefix

Symbol

Multiplying factor

Symbol

1012

tera

T

10 −1

deci

d

109

giga

G

10 −2

centi

c

mega

M

10

−3

milli

m

kilo

k

10 − 6

micro

µ

hecto

h

10

−9

nano

n

deca

da

10 −12

10

6

103 10

2

101

pico

p

−15

femto

f

−18

atto

a

10 10

1.5

Prefix

Definitions of Various Terms

The definition of various terms that are associated with electrical energy like Energy, Power, Current, Voltage, etc., are presented in this Section. ENERGY : Energy is defined as capacity to do work or energy can be defined as stored work. The energy may exist in many forms such as Electrical, Mechanical, Thermal, Light, Chemical, etc. The energy is measured in joules, which is denoted by J (or the unit of energy is joules). In electrical engineering, one joule is defined as the energy required to transfer a power of one watt in one second to a load (or Energy = Power ´ Time). Therefore, 1 J = 1 W-s. In mechanical engineering, one joule is the energy required to move a mass of 1 kg through a distance of 1 m with an uniform acceleration of 1 m/s2.

1. 6

Chapter 1 - Introduction

Therefore, 1 J = 1 N - m = 1 kg - m2 - m s In thermal engineering, one joule is equal to a heat of 4.1855 (or 4.186) calories, and one calorie is the heat energy required to raise the temperature of 1 gram of water by 1o C. Therefore, 1 J = 4.1855 calories POWER :

Power is the rate at which work is done (or Power is rate of energy transfer). The unit of power is watt and denoted by W. One watt is defined as the power if the energy is transferred at the rate of one joule per second. An average value of power can be expressed as, Energy = W Time t A time varying power can be expressed as,

Power, P =

Instantaneous power, p = dw dt d q Also, p = dw = dw # = vi dt dq dt Hence, power is also given by the product of voltage and current.

.....(1.1)

.....(1.2) .....(1.3)

CHARGE : Charge is the characteristic property of the elementary particles of the matter. The elementary particles are electrons, protons and neutrons. There are basically, two types of charges in nature: positive charge and negative charge. The charge of an electron is called negative charge. The charge of a proton is called positive charge. Normally, a particle is neutral because it will have equal number of electrons and protons. The particle is called charged, if some electrons are either added or removed from it. If electrons are added then the particle is called negative charged. If electrons are removed then the particle is called positive charged.The unit used for measurement of charge is coulomb. One coulomb is defined as the charge which when placed in vacuum from an equal and similar charge at a distance of one metre repels it with a force of 9 × 10 9 N. The charge of an electron is 1.602 × 10 −19 C. Hence 1/(1.602 × 10 −19) = 6.24 × 10 18 electrons make up a charge of one coulomb. CURRENT : Current is defined as the rate of flow of electrons. It is measured in the unit of amperes. One ampere is the current flowing through a point if a charge of one coulomb crosses that point in one second. In SI units, one ampere is defined as that constant current in two infinite parallel conductors of negligible circular cross-section, one metre apart in vacuum, which produces a force between the conductors of 2 × 10 − 7 newton per metre length. A steady current can be expressed as, Charge Q Current, I = = Time t A time varying current can be expressed as, Instantaneous current, i =

dq dt

.....(1.4)

.....(1.5)

1. 7

Circuit Theory

where, Q = Charge flowing at a constant rate. t = Time. dq = Change in charge in a time of dt. dt = Time required to produce a change in charge dq. VOLTAGE : Every charge will have a potential energy. The difference in potential energy between the charges is called potential difference. In electrical terminology the potential difference is called voltage. The potential difference indicates the amount of work done to move a charge from one place to another. The voltage is expressed in the unit of volt. One volt is the potential difference between two points when one joule of energy is utilized in transfering one coulomb of charge from one point to the other. A steady voltage can be expressed as, Energy = W Charge Q A time varying voltage can be expressed as,

Voltage, V =

.....(1.6)

Instantaneous voltage, v = dw dq

.....(1.7)

Also, 1 V = 1 J = 1 J/s = 1 W 1C 1 C/s 1A ` Voltage, V = Power = P I Current

.....(1.8) .....(1.9)

One volt is also defined as the difference in electric potential between two points along a conductor carrying a constant current of one ampere when the power dissipated between the two points is one watt.

1.6

Symbols Used for Average, RMS and Maximum Values

The quantities like voltage, current, power and energy may be constant or varying with respect to time. For a time varying quantity we can define the value of the quantity as instantaneous, average, rms and maximum value. The symbols used for these values are listed in Table 1.3. TABLE 1.3 : Symbols of dc and ac Variables AC or Time varying Quantity

DC

Instantaneous value

Average value

Maximum value

rms value

Phasors or Vectors

Current

I

i

Iave

Im or Ip

I

I

Voltage

V

v

Vave

Vm or Vp

V

V

Power

P

p

P

Pm

-

S

Energy

W

w

W

Wm

-

-

1. 8

1.7

Chapter 1 - Introduction

Steady State Analysis and Transient Analysis

The circuit analysis can be classified into steady state analysis and transient analysis. The analysis of circuits during switching conditions is called transient analysis. During switching condition the current and voltage will change from one value to the other. In purely resistive circuits this may not be a problem because the resistance will allow sudden change in voltage and current. But in inductive circuits the current cannot change instantaneously and in capacitive circuits the voltage cannot change instantaneously. Hence when the circuit is switched from one state to the other, the voltage and current cannot attain a steady value instantaneously in inductive or capacitive circuits. Therefore during switching condition there will be a small period during which the current and voltage will change from an initial value to a final steady value. The time from the instant of switching to the attainment of steady value is called transient period. Physically, the transient can be realized in switching of tube lights, fans, motors, etc. In certain circuits the transient period is negligible and we may be interested only in steady value of the response. Therefore, the steady state analysis is sufficient. The analysis of circuits under steady state (i.e., by neglecting transient period) are called steady state analysis. The steady state analysis of circuits are discussed in this book in all chapters except Chapter 10. In certain circuits the transient period is critical and we may require the response of the circuit during transient period. Some practical examples where transient analysis is vital are starters, circuit breakers, relays, etc. The transient analysis of circuits are discussed in Chapter 10.

1.8

Assumptions in Circuit Theory In circuit analysis the elements of the circuit are assumed to be linear, bilateral and lumped elements.

In linear elements the voltage-current characteristics are linear and the circuit consisting of linear elements are called linear circuit or network. The resistor, inductor and capacitor are linear elements. Some elements exhibit non-linear characteristics. For example, diodes and transistors has non-linear voltage-current characteristics, capacitance of varactor diode is non-linear and inductance of an inductor with hystereris is non-linear. For analysis purpose the non-linear characteristics can be linearized over certain range of operation. In a bilateral element, the relationship between voltage and current will be same for two possible directions of current through the element. On the other hand, a unilateral element will have different voltage-current characteristics for the two possible directions of current through the element. The diode is an example of unilateral element. In practical devices like transmission lines, windings of motors, coils, etc., the parameters (R, L and C) are distributed in nature. But for analysis purpose we assume that the parameters are lumped (i.e., concentrated at one place). This approximation is valid only for low frequency operations and it is not valid in the microwave frequency range. All analysis in this book are based on the assumption that the elements are linear, bilateral and lumped elements.

Chapter 2

BASIC CONCEPTS OF CIRCUITS 2.1

Basic Elements of Circuits

Circuits and Networks An electric circuit consists of Resistors (R), Inductors (L), Capacitors (C), voltage sources and /or current sources connected in a particular combination. When the sources are removed from the circuit it is called network. R1

R1

+ E

~

L

R2

C

L

R2

C

E

Fig. a : Circuit.

Fig. b : Network.

Fig. 2.1 : Example of circuit and network.

DC Circuits The networks excited by dc sources are called dc circuits. In a dc source, the voltage and current do not change with time. Hence, the property of capacitance and inductance will not arise in the steady state analysis of dc circuits.This chapter deals with steady state analysis of dc circuits. Therefore in this chapter only resistive circuits are discussed. Active and Passive Elements The elements of the circuit can be classified into active elements and passive elements. The elements which can deliver energy are called active elements. The elements which consumes energy either by absorbing or storing are called passive elements. The active elements are voltage and current sources. The sources can be of different nature. The sources in which the current/voltage does not change with time are called direct current sources or in short dc sources. (But in dc sources the current/voltage change with load). The sources in which the current/voltage sinusoidally varies with time are called sinusoidal sources or alternating current sources or in short ac sources. The passive elements of a circuit are resistors, inductors and capacitors that will exhibit the property of resistance, inductance and capacitance respectively under ideal conditions. The resistance, inductance and capacitance are called fundamental parameters of the circuit. Practically, these parameters will be distributed in nature. For example, the resistance of a transmission line will exist throughout its length. But for circuit analysis the parameters are considered as lumped parameter.

2. 2

Chapter 2 - Basic Concepts of Circuits

The resistor absorbs energy (and the absorbed energy is converted to heat). The inductor and capacitor stores energy. When the power supply in the circuit is switched ON, the inductor and capacitor stores energy, and when the supply is switched OFF, the stored energy leaks away in the leakage path. (Hence, the inductors and capacitors cannot be used as storage devices). +

+ +

Is

E

E -

E = EÐq

Fig. a : Dc voltage source.

Vs

Is = Is Ðq

~

-

-

+ E

Fig. b : Dc current source.

Fig. c : Ac voltage source.

Is

Vs = RI or A vV

R

~

Fig. d : Ac current source.

L

C

Is = GV or A II

Fig. e : Dependent voltage source.

Fig. f : Dependent current source.

Fig. g : Resistance.

Fig. h : Inductance.

Fig. i : Capacitance.

Fig. 2.2 : Symbols of active and passive elements of circuits.

Independent and Dependent Sources The sources can be classified into independent and dependent sources. The electrical energy supplied by an independent source does not depend on another electrical source. The independent sources converts energy in some form to electrical energy. For example, a generator converts the mechanical energy to electrical energy, a battery converts the chemical energy to electrical energy, a solar cell converts light energy to electrical energy, a thermocouple converts heat energy to electrical energy, etc. The electrical energy supplied by a dependent source depends on another source of electrical energy. For example, the output signal (energy) of a transistor or op-amp depends on input signal (energy), where the input signal is another source of electrical energy. In the circuit sense, the voltage/current of an independent source does not depend on voltage/ current in any part of the circuit. But the voltage/current of a dependent source depends on the voltage/current in some part of the same circuit.

2.2

Nodes, Branches and Closed Path

A typical circuit consists of lumped parameters such as resistance, inductance, capacitance and sources of electrical energy like voltage and current sources connected through resistance-less wires. In a circuit, the meeting point of two or more elements is called a node. If more than two elements meet at a node then it is called principal node. The path between any two nodes is called a branch. A branch may have one or more elements connected in series. A closed path is a path which starts at a node and travels through some part of the circuit and arrives at the same node without crossing a node more than once.

Circuit Theory

2. 3

The nodes, branches and closed paths of a typical circuit are shown in Fig. 2.3. The nodes of the circuit are the meeting points of the elements denoted as A, B, C, D, E and F. The nodes A, B, C and D are principal nodes because these nodes are meeting points of more than two elements. +

E2

R3 E F

+

E2

A

R3 E

R2

C R2

A

F R4

B

R4

B

C

B

A

A

C

C R1

R1 R6

R5

E

R6

R5

E + E1 E

+ E1 E

D

D

Fig. a : Typical circuit.

D

D

Fig. b : Branches of the circuit in Fig. a.

E2

+

B

R3 E R2

R4

B

R1 R5

R4

R2 A

C

E

+ E1 E

R5

R1

R6

+ E1 E

R6

D +

E2

R3 E F

Fig. c : Nodes of the circuit in Fig. a. +

E2

R3

+

E

E2

R3 E

F R2

F R4

B

B

A

A

C

+

C

E2

R3 E F

R1 R6

R5

R2 R5

E

+

D

R4 C

+ E1 E D

B

A

E2

R3 E F

R2

B

R4

B

A

C

A

R1

C R1

R5

R5

E + E1 E

R6

R6

E + E1 E

D

D

D

Fig. d : Closed paths of circuit in Fig. a. Fig. 2.3 : A typical circuit and its branches, nodes and closed paths.

2. 4

Chapter 2 - Basic Concepts of Circuits

2.3

Series, Parallel, Star and Delta Connections

The various types of connections that we may encounter in electric circuits are series, parallel, star and delta connections. 2.3.1 Series Connection If two or more elements are connected such that the current through them are same, then the connection is called series connection. In a circuit if the current in a path is same then the elements in that path are said to be in series. R1

I

R2

R3

I

Fig. a : Resistances in series. V1 E

V3

V2

+

+

E

E

L3

L2

L

C

Fig. e : Resistance and inductance in series.

Fig. f : Resistance and capacitance in series.

Fig. g : Resistance, inductance and capacitance in series.

Fig. 2.4 : Examples of series connected elements. Ic

Ic

R2

B Id

R4

A C

A Ie

Ia

C If

Ia

R1

R6

R5

+

C

A

If

R1

L

R3

L

R3

~

C

L

R

I

+

Fig. d : Voltage sources in series.

Ib

C3

Fig. c : Capacitances in series. R

I

C2

C1

I

Fig. b : Inductances in series. R

I

I

L1

R6

+

E

~

E

E

E R7

D

C

D

D

R7

C

Fig. b : Series paths in the circuit of Fig. a.

Fig. a : Typical circuit.

Fig. 2.5 : A typical circuit and its series paths.

2.3.2 Parallel Connection If two or more elements are connected such that the voltage across them is same then the connection is called parallel connection. In a circuit if the voltage across two or more paths are same then, they are said to be in parallel. + V

+ R1

R2

R3

E

Fig. a : Resistances in parallel.

V

+

+ L1

L2

L3

E

Fig. b : Inductances in parallel.

V

C1

C2

C3

E

Fig. c : Capacitances in parallel.

V

R

L

E

Fig. d : R and L in parallel.

Circuit Theory

2. 5

+

+ C

R

V

R

V

C

L

I1

I2

I3

E

E

Fig. f : R, L and C in parallel. Fig. g : Current sources in parallel. Fig. e : R and C in parallel. Fig. 2.6 : Examples of parallel connected elements. + E

R1

+

R2 E

~ E

L

E

E

C

+

R2 R1

~

R1 R2

~ E

C

L

Fig. c : The voltage source, series combination of R1 and L and resistance R2 are in parallel.

Fig. a : The voltage source,series Fig. b : The voltage source,Resistance combination of R1 and L and series R1 and series combination of R2 combination of R2 and C are in parallel. and C are in parallel. Fig. 2.7 : Simple circuits with parallel branches. R6

G

L

R2

B

R3

A

C

R6

R3

B B +

L

G

C

R1

E

R5

R4

F +

R4 VAC

~

A A

C

E

D

C C

A

R2

R2

A

B

+

+ R1 R4

V AE

+

~

E

D

B

E

_

E

B

VAB

+

~

C _

C

R5 C

E

E

VBC

R4

R4

F

R3

B

_ B

+

E

D

R3

A

F

_

C

Fig. b : The path AGC is parallel Fig. c : The path BCD is parallel with the path ABC. with the path BED.

R1

E

B

R2

Fig. a : A typical circuit. A

_

+

E

R5

V BD

D E

E

Fig. e : The path AFEB is parallel Fig. f : The path BEDC is parallel Fig. d : The path ABE is with the resistance R3 . with the resistance R2 . parallelwith the path AFE. Fig. 2.8 : A typical circuit and its parallel paths. A

R1

R3

B

R5

C

R5

CC

D

D

+ + R2

E

R6

R4

VCE

R4

R6

E _ E

E

Fig. a : A typical circuit.

E

EE

E

Fig. b : R4 in parallel with series combination of R5 and R6.

2. 6

Chapter 2 - Basic Concepts of Circuits R3

B B

C

R1

A

+ +

VBE

R2

B B +

R4

VBE

E

R2

E _

_

E E

E

E E

Fig. d : The path EAB is in parallel with resistance R2 .

Fig. c : The path BCE is in parallel with resistance R2 .

Fig. 2.9 : A typical circuit and its parallel paths. 1

2.3.3 Star-Delta Connection

R1

N

R2 2

1

R1

N If three elements are connected R3 R2 R3 to meet at a node then the three 2 elements are said to be in star 3 3 3 connection. If three elements with a Fig. b : T-connection. node in between any two element are Fig. a : Star connection. 1 1 connected to form a closed path then 1 2 R2 R2 they are said to be in delta connection. R1 R1 R3 The star connection is also called 2 3 2 T-connection and delta connection is R3 3 3 3 also called P-connection. Fig. d : €-connection. Fig. c : Delta connection. Fig. 2.10 : Basic star and delta connections. R5

R1

R5

R3

B

R1

A

R3

B

A

C

B A

C

R3 C

+ R4

R2

E

R4

R2

E

D

D

Fig. a : A typical circuit.

D

Fig. b : Star connections in circuit of Fig. a.

R5

R3

B

C

R1 A

B

R3

R2

C

R4

D

Fig. c : Delta connections in circuit of Fig. a. Fig. 2.11 : A typical circuit and its star and delta connections.

Circuit Theory

2.4

2. 7

Open Circuit and Short Circuit

In a circuit if there is an open path or path of infinite resistance between two nodes then that path is called open circuit (O.C). Since current can flow only in closed paths, the current in the open circuit will be zero. 2‡

5‡ A

A

A Circuit N1

1‡ 10 V +E

O.C.

4‡

O.C.

O.C. 20 V +E

B B

B

3‡

1‡

Fig. 2.12 : Examples of open circuit (O.C.).

While applying KVL to closed paths the open circuit can be included as an element of infinite resistance in the path, because a voltage exists across the two open nodes of a circuit. In a circuit if there is a closed path of zero resistance between two nodes then it is called short circuit (S.C). Since the resistance of the short circuit is zero, the voltage across the short circuit is zero. 2‡

5‡

A

A

A Circuit N1

I

1‡ 10 V +E

S.C.

4‡

S.C.

S.C. 20 V +E

B 3‡

1‡

B

B

Fig. 2.13 : Examples of short circuit (S.C.).

In a circuit, if there are elements parallel to a short circuit then they will not carry any current, because the current will prefer the path of least resistance (or opposition) and so the entire current will flow through the short circuit. Hence, the elements parallel to a short-circuit need not be considered for analysis as shown in the example circuit of Fig. 2.14. 1W

10 V +-

2W

A

2W

4W

S.C.

1W

2W

1W

1W

Þ

2W

+

2W

10 V -

2W

3W

S.C.

B

B

5 V +-

A

2W

A

4W

S.C.

Þ

A

5 V +-

S.C.

6W B

Fig. 2.14 : Examples of short circuit.

6W

B

2. 8

2.5

Chapter 2 - Basic Concepts of Circuits

Sign Conventions

The elements of the circuit are two terminal elements. When a circuit is excited (i.e., power supply is switched ON) a voltage will be developed across the two terminals of the element such that one end is positive and the other end is negative, and a current will flow through the element. When an element delivers energy, the current will leave the element from positive terminal and when an element absorbs energy the current will enter at positive terminal. In a circuit, normally the sources will deliver energy and the passive elements resistance, inductance and capacitance will absorb energy. Therefore, in a voltage/current source, when it delivers energy the current will leave from positive terminal. In the parameters R, L and C the current will enter at the positive terminals when they absorb energy. I +

E E

I + V

I

R

E

I

I

+

+

V E

L

I

+ C

V

V

E

E

Fig. a :Voltage Fig. d : Inductance Fig. c : Fig. b : Current source absorbing Resistance source delivering energy. delivering energy. absorbing energy. energy.

Fig. e : Capacitance absorbing energy.

Fig. 2.15 : Sign conventions for sources when it delivers energy and parameters when they absorb energy. I

A chargeable battery is the best example E for understanding the concept of energy delivery V E +E I + and absorption by sources. When the battery is connected to load it delivers energy. When the Fig. a :Voltage source Fig. b : Current source battery is charged, it absorbs energy. When a absorbing energy. absorbing energy. source absorbs energy the current will enter the Fig. 2.16 : Sign conventions for sources when source at positive terminal as shown in Fig. 2.16. they absorb energy. The resistance always absorbs energy but the inductance and capacitance can deliver the stored energy temporarily. The inductance and capacitance stores energy when the supply is switched ON and when the supply is switched OFF the stored energy is discharged in the available paths or leakage paths. When the inductance and capacitance discharge energy, the current will leave from positive terminal as shown in Fig. 2.17. I

I

+

+

L

V

E

Fig. a : Inductance discharging energy.

C

V

E

Fig. b : Capacitance discharging energy.

Fig. 2.17 : Sign conventions for inductance and capacitance parameters when they discharge energy.

Circuit Theory

2.6

2. 9

Ohm’s and Kirchhoff’s Laws

The three fundamental laws that govern the electric circuit are Ohm’s law, Kirchhoff’s Current Law and Kirchhoff’s Voltage Law. 2.6.1 Ohm’s Law Ohm’s law states that the potential difference (or voltage) across any two ends of a conductor is directly proportional to the current flowing between the two ends provided the temperature of the conductor remains constant. The constant of proportionality is the resistance R of the conductor. \V a I

⇒

V = IR

..... (2.1)

From equation (2.1), we can say that, when a current I flows through a resistance R, then the voltage V, across the resistance is given by the product of current and resistance. 2.6.2 Kirchhoff’s Current Law (KCL) Kirchhoff’s Current Law states that the algebraic sum of currents at a node is zero. ∑I=0 I4

Hence, we can say that, the current cannot stay at a point. While applying Kirchhoff’s Current Law (KCL) to a node we have to assign polarity or sign (i.e., + or -) for the current entering and leaving that node. Let us assume that the currents entering the node are negative and currents leaving the node are positive. With reference to Fig. 2.18, we can say that, the currents I1 and I2 are entering the node and the currents I3 and I4 are leaving the node.Therefore, by Kirchhoff’s Current Law we can write,

R4 Node I1

I3

R3 R2 I2

Fig. 2.18 : Currents in a node.

-I1 - I2 + I3 + I4 = 0 ∴ I1 + I2 = I3 + I4

R1

..... (2.2)

From equation (2.2), we can say that, “the sum of currents entering a node is equal to sum of currents leaving that node”. This concept is easier to apply while solving problems using KCL. 2.6.3 Kirchhoff’s Voltage Law (KVL) Kirchhoff’s Voltage Law states that the algebraic sum of voltages in a closed path is zero. ∑V=0 A closed path may have voltage rises and voltage falls when it is traversed or traced in a particular direction.While applying KVL to a closed path we have to assign polarity or sign (i.e., + or -) to voltage fall and rise. Let us assume that the voltage rise as positive and voltage fall as negative. Consider the circuit shown in Fig. 2.19. Let us trace the circuit in the direction of current I. In the closed path ABCDEFGA, the voltage rise are E1 and E2 and voltage fall are IR1, IR2, IR3, IR4 and IR5.

R2

C +

R3

D +

E

+

E R1

IR4

IR1 +

R4

E F E E2

B + E1 E E A

E

IR3

IR2

I

I E

IR5 R5

+

+ I

G

Fig. 2.19 : A circuit with single closed path.

2. 10

Chapter 2 - Basic Concepts of Circuits

Therefore, by KVL we can write, E1 + E2 - IR1 - IR2 - IR3 - IR4 - IR5 = 0 ∴ E1 + E2 = IR1 + IR2 + IR3 + IR4 + IR5

..... (2.3)

From equation (2.3) we can say that, “the sum of voltage rise in a closed path is equal to sum of voltage fall in that closed path”. This concept is easier to apply while solving problems using KVL.

2.7

Voltage and Current Sources

The voltage and current are the two quantities that decide the energy supplied by the sources of electrical energy. Usually, the sources are operated by maintaining one of the two quantities as constant and by allowing the other quantity to vary depending on load. When the voltage is maintained constant and the current is allowed to vary then the source is called voltage source. When the current is maintained constant and the voltage is allowed to vary then the source is called current source. 2.7.1 Ideal and Practical Sources In ideal conditions the voltage across an ideal voltage source should be constant for whatever current delivered by the source. Similarly, the ideal current source should deliver a constant current for whatever voltage across its terminals. Is E I

+ +

V

Is

E E

E V

I

Fig. a : Characteristics of ideal voltage source. Fig. b : Characteristics of ideal current source. Fig. 2.20 : Characteristics of ideal sources.

In reality, the ideal conditions never exists (but for analysis purpose the sources can be considered ideal). In practical voltage source the voltage across the source decreases with increasing load current and the reduction in voltage is due to its internal resistance. In a practical current source, the current delivered by the source decreases with increasing load voltage and the reduction in current is due to its internal resistance. E

Is

V

I

Fig. a : Characteristics of practical voltage source.

Fig. b : Characteristics of practical current source.

Fig. 2.21 : Characteristics of practical sources.

Circuit Theory

Let, Es Is V I Rs

2. 11

= = = = =

Voltage across ideal source (or internal voltage of the source). Current delivered by ideal source (or current generated by the source). Voltage across the terminals of the source. Current delivered through the terminals of the source. Source resistance (or internal resistance).

The practical voltage source can be considered as a series combination of an ideal voltage source and a source resistance, Rs. The reduction in voltage across the terminals with increasing load current is due to the voltage drop in the source resistance. When the value of source resistance is zero the ideal condition is achieved in voltage sources. Hence, “the source resistance for an ideal voltage source is zero”. The practical current source can be considered as a parallel combination of an ideal current source and a source resistance, Rs. The reduction in current delivered by the source is due to the current drawn by the parallel source resistance. When the value of source resistance is infinite the ideal condition is achieved in current sources. Hence, “the source resistance for an ideal current source is infinite”.

+

IRs

V, E

I E +

Rs

E Vs I

E +E

}IRs

VV

V

sI

E

I

V = E E IRs

Fig. 2.22 : A practical dc voltage source. Is, I

I +

Ish Is

Is Vs V Rs

I Vs

V

}Ish V

E

V

I = Is E Ish

Fig. 2.23 : A practical dc current source.

2.7.2 DC Source Transformation The practical voltage source can be converted to an equivalent practical current source and vice-versa, with same terminal behaviour. In these conversions the current and voltage at the terminal of the equivalent source will be same as that of original source, so that the power delivered to a load connected at the terminals of original and equivalent source will be same. Rs +

A

A

+

IRs -

E +-

V

+

I

RL

Ish

Þ

Rs

Is

V

B

Fig. a : Voltage source.

I

RL

Is = E/Rs

B

Fig. b : Equivalent current source of the voltage source in Fig. a. Fig. 2.24 : Conversion of voltage source to current source.

2. 12

Chapter 2 - Basic Concepts of Circuits

A voltage source with series resistance can be converted to an equivalent current source with parallel resistance as shown in Fig. 2.24. Similarly, a current source with parallel resistance can be converted to an equivalent voltage source with series resistance as shown in Fig. 2.25. The proof for source conversions are presented in Chapter 6. Rs

A +

V Rs

Is

+ IRs

I

Rs

V

RL

Þ

E +-

B

-

A + I

V

RL

E = Is R s

B

Fig. a : Current source.

Fig. b : Equivalent voltage source of the current source in Fig. a. Fig. 2.25 : Conversion of current source to voltage source.

2.8

Power and Energy Power is the rate at which work is done or power is rate of energy transfer. Let, w = Instantaneous value of energy. q = Instantaneous value of charge. dq Now, Instantaneous power, p = dw = dw # dt dq dt dq d w We know that, = v and = i dq dt ` p = vi

Refer equations (1.5) and (1.7).

i.e., the power is the product of voltage and current. In circuits excited by dc sources the voltage and current are constant and so the power is constant. This constant power is called average power or power and it is denoted by P. \ In dc circuits, Power, P = VI The power is rate of work done and the energy is total work done. Hence the energy is given by the product of power and time. When time is expressed in second, the unit of energy is watt-second and when the time is expressed in hours, the unit of energy is watt-hour. \ Energy, E = P t in W-s or W-h The larger unit of electrical energy is kWh and commercially one kWh of electrical energy is called one unit. ` Energy, E =

Pt in kWh 1000 # 3600

Circuit Theory

2.9

2. 13

Resistance

(AU May’15, 2 Marks)

The Resistance is the property of element (or matter) which opposes the flow of current (or electrons). The current carrying element is called conductor. The resistance of a conductor (in the direction of current flow) is directly proportional to its length, l and inversely proportional to area of cross-section, a. ` Resistance, R α l a The proportionality constant is the resistivity, r of the material of the conductor. ρl ` R = a The unit of resistivity is ohm-metre(W-m). The resistivity of a material at a given temperature is constant. For example, the resistivity of copper is 1.72 ´ 10–8 W-m and that of aluminium is 2.69 ´ 10-8 W-m at 20o C. lumped resistance The resistance of a conductor is distributed R throughout the length of the conductor. But for analysis purpose the resistance is assumed to be concentrated at one place, which is called lumped resistance. For connecting the lumped resistance to other part of the resistance-less wire circuit, resistance-less wires are connected to its ends Fig. 2.26 : A lumped resistance with as shown in Fig. 2.26. (Normally, the term resistance resistance-less wires connected to its ends. in circuit theory refers only to lumped resistance). 2.9.1

Resistance Connected to dc Source

Consider a resistance, R connected to dc source of voltage, V volts as shown in Fig. 2.27. Since the resistance is connected across (or parallel to) the source, the voltage across the resistance I is also V volts. By Ohm’s law the current through the resistance is given by, I = V ⇒ V = IR R Power in the resistance, P = VI

+

..... (2.4) ..... (2.5)

2 P = VI = V # V = V and P = VI = IR ´ I = I2 R R R

2.9.2

2 or P = V R

R

V E

Using equation (2.4), the equation (2.5) can also be written as,

` Power, P = VI

+

V E

Fig. 2.27 : Resistance connected to dc source.

or P = I2 R

I I1

Current Division in Parallel Connected Resistances

I2 +

+

The equations (2.6) and (2.7) given below, can be used to determine the currents in parallel connected resistances shown in Fig. 2.28 in terms of total current drawn by the parallel combination and the values of individual resistances. Hence, these equations are called current division rule. The proof for current division rule is presented in Chapter 6.

+

V E

R1

V E

R2

V E

Fig. 2.28 : Resistances in parallel.

2. 14

Chapter 2 - Basic Concepts of Circuits I1 = I #

R2 R1 + R2

.....(2.6)

I2 = I #

R1 R1 + R2

.....(2.7)

The following equation will be helpful to remember the current division rule. In two parallel connected resistances, Total current drawn by Value of the # parallel combination other resistance Current through one of the resistance = Sum of the inidvidual resistances 2.9.3

Voltage Division in Series Connected Resistances

R2

R1

I +

+

E The equations (2.8) and (2.9) given below, can be used to V1 V2 determine the voltages across series connected resistances shown in Fig. 2.29 in terms of total voltage across the series combination and the values of individual resistances. Hence these equations +E are called voltage division rule. The proof for voltage division V rule is presented in Chapter-6. Fig. 2.29 : Resistances in series. R1 .....(2.8) V1 = V # R1 + R2

V2 = V #

E

R2 R1 + R2

.....(2.9)

The following equation will be helpful to remember the voltage division rule. In two series connected resistances, Total voltage across Value of the # series combination resistance Voltage across one of the resistance = Sum of the inidvidual resistances

2.10 Network Topology The topology is a branch of science which deals with the study of geometrical properties and special relations unaffected by continuous change of shape or size of figures. The concept of topology was first applied to networks by Kirchoff to study the relationship between the nodes and branches in a network. A circuit or network can be drawn in different shapes and sizes by maintaining the relationship between the nodes and branches as shown in Fig. 2.30. Therefore, “the network topology is the study of the properties of the network which are unaffected when we stretch, twist or distort the size and shape of the network”. A network consists of interconnections of various elements. The physical arrangement of the elements and the length of wires used for connecting the elements may give rise to different types of layout for the circuits.As long as the relationship between the nodes and branches are maintained the circuit response will be same.

Circuit Theory

2. 15 R3

3 2 L

L 1

3

+

R1

E E

R3

E

C

2

R1

R2

+E

1

0

C

C

R2

1 +

R3

E

R2

E

L

3

2

0

0 R1

Fig. 2.30 : Different shapes of a circuit.

2.10.1

Graph

The topological properties of networks are described by using the graph. A graph of a network consists of nodes and branches of the network. In a network the branches will have elements but in a graph the branches are drawn by lines. When arrows are placed on the branches of graph it is called oriented graph. The arrow indicates the direction of branch current and polarity of branch voltage.

Fig. a : Network.

Fig. b : Graph of network in Fig. a.

Fig. c : Oriented graph of network in Fig. a.

Fig. 2.31 : A typical network and its graph and oriented graph.

A sequence of branches traversed while going from one node to another node is called a path. A graph is said to be connected graph if there exists at least one path from each node of a graph to every other node of the graph. To draw the graph of a circuit first redraw the circuit by replacing the sources by their internal impedances. The ideal voltage sources are replaced by short circuits and ideal current sources are replaced by open circuits. Now, the circuit becomes a network consisting of R, L and C elements only. Then represent the nodes of the network as small circles and the elements connected between the nodes as lines. The series connected elements are considered as single branch. While drawing the graph of a network, the number of nodes and branches and the relationship between them has to be maintained. But the size and shape of graph and curvature of lines in the graph are not important. A typical circuit and its different graphs are shown in Fig. 2.32. In the graph, the nodes are represented by small circles and denoted by numerals 1, 2, 3 and 4. In the graph, the elements connected between the nodes are represented by lines. These lines are called branches and denoted by lower case letters a, b, c, d, e and f. This convention of denoting nodes by numerals and branches by lower case letters will be followed in this book.

2. 16

Chapter 2 - Basic Concepts of Circuits R6

R6

R3

R4

2

R3 3

1

R4

2

3

1

R1 R2

R1

R5

I

R2

R5

+

E E

4

4

Fig. b : The circuit of Fig. a after replacing sources by their internal impedance.

Fig. a : Typical circuit.

a

a

a

1

c

b 2

b

1

c

b 3

b

2

2

c

3

1 e

f

d

1

3

d 4

d e

3

e f

f

4

a 2 c

d

e 4

f

4

Fig. c : Various shapes of graphs for the circuit of Fig. a. Fig. 2.32 : A typical circuit and its different graphs.

2.10.2

Tree, Link and Cotree

When some of the branches in an original graph are removed the resultant graph is called subgraph. The tree is a subgraph which is obtained by removing some branches such that the subgraph includes all the nodes of the original graph, but does not have any closed paths. For a given graph, there may be more than one possible tree. Hence a tree can be defined as any connected open set of branches which includes all nodes of a given graph. A tree of a graph with N nodes has the following properties. l

The tree contains all the nodes of the graph.

l

The tree contains “N - 1” branches.

l

The tree does not have a closed path.

The branches removed to form a tree are called links or chords. By removing a link from a graph, one closed path can be eliminated. Alternatively, on adding a link to a tree one closed path is created. Hence by adding the links one by one to a tree all closed paths can be created. Therefore, the number of closed paths in a graph is equal to number of links.

Circuit Theory

2. 17

The cotree is complement of a tree. Hence every tree has a cotree. The links connected to the nodes of a graph form a cotree. The branches of a tree are called twigs and the branches of cotree are called links. A typical graph is shown in Fig. 2.33, and some possible trees of the graph and the cotree of each tree are shown in Table 2.1.

1

a

b

e

d

2

c

4

3

f For most of the trees the cotree will also be in the form of a tree. But for some possible tree, the cotree may have closed paths and cotree may not Fig. 2.33 : Graph. be connected (i.e., all the nodes are not connected in a cotree).

A definite relationship exists between number of nodes and branches in a tree. Any tree of the graph with B branches and N nodes will consists of “N - 1” branches and the remaining branches are links. Therefore for a graph with B branches and N nodes, the number of links or chords is given by, Link, L = B - (N - 1) = B - N + 1 TABLE 2.1 : THE TREES AND COTREES OF THE GRAPH IN FIG. 2.33 Tree

Cotree

Tree

Cotree

1

1

1 b

a

d

1 a

2 2

c

e

3

b

4

3 f

4

2

Twig: [a, d, e]

c 2 e

3

Link: [b, c, f]

d

4

f Link: [c, d, f]

Twig: [a, b, e]

1

1

1 a 2 2

d

e

3

1

b

c 4

2

d

Link: [a, b, f]

2

Twig: [b, c, d]

f

2

d

Twig: [b, c, f]

f

e

3

4

2

d

2

d

3

2

f Twig: [b, d, f]

4

Link: [a, c, e]

1 b e

4

2

3

a e

f Twig: [a, b, f]

e

3

1 c

4

c

4

3

1

b 3

1 a

Link: [a, d, e]

1

2

Link: [a, e, f]

b

a 4

4

1

1

3

e

3 f

c

b

a

4

3

1

2

a c

b

4

3 f

Twig: [c, d, e]

4

3

Link: [c, d, e]

Twig: [b, e, f]

4

2

d

c 3

Link: [a, c, d]

4

2. 18

2.10.3

Chapter 2 - Basic Concepts of Circuits

Network Variables

When a network is excited by connecting a source every branch will have a current flowing through it and so a voltage exist across the terminals of the branch. Hence a graph (or network) with B branches will have B number of branch currents and B number of branch voltages. These branch currents and voltages are called network variables. The branch currents are called current variables and branch voltages are called voltage variables of the network. An arrow is placed on the branch to indicate the direction of the branch current and polarity of branch voltage. The arrow placed on the branch is called reference or orientation. In a branch a single reference is used to represent both the direction of branch current and polarity of branch voltage. The current-voltage relation of a branch is obtained by Ohm’s law, by treating the branch as load. Hence the set of references for the branches of a graph are called load set reference. +

Vbr

Ibr

-

Þ

Vbr

Þ +

-

Ibr

Vbr = Branch voltage ; Ibr = Branch current

Fig. 2.34 : Orientation (or reference) of a branch.

The conventional direction of branch current and polarity of branch voltage are shown in Fig. 2.34. In a network, branch current directions can be assumed arbitrarily and the polarity of branch voltages can be fixed as per Ohm’s law, by treating the branches as loads. Alternatively, the polarity of branch voltages can be assumed arbitrarily and the direction of branch current can be fixed as per Ohm’s law, by treating the branch as load. 2.10.4

Solution of Network Variables

In a network or a circuit we may be interested in the voltage and current in the various branches which is normally referred to as response. In a network if all the branch currents are known then the voltages can be obtained by Ohm’s law. Alternatively, if the branch voltages are known then the currents can be obtained by Ohm’s law. Hence in order to determine the response on current basis first we have to solve B number of branch currents and to determine the response on voltage basis first we have to solve B number of branch voltages. For a unique solution of B number of variables, we have to form B number of equations involving the B variables and solve them. But in practice it can be shown that all the branch currents are not independent and so the independent current variables which are less than B, are sufficient to solve the currents. Similarly, all the branch voltages are not independent and so the independent voltage variables which are less than B, are sufficient to solve the voltages.

Circuit Theory

2. 19

Independent Current Variables In a network it can be proved that the branch currents of the links are independent current variables. When the links are removed in a network, all the closed paths are destroyed and so no current can flow in the network. The removal of a link is equal to making link current as zero. Therefore, when the link current are made zero, all the currents in the network become zero. Hence we can say that, the branch currents depends on link currents. Therefore, “the link currents are independent and branch currents are dependent”. In a network with N nodes and B branches we have “B – N + 1” links. Therefore in a network there will be B current variables in which “B - N + 1” are independent current variables and the remaining “N - 1” [i.e., B - (B - N + 1) = N - 1] currents are dependent current variables. In order to determine the response of a network on current basis, it is sufficient if we form “B – N + 1” equations involving independent current variables and solve them for a unique solution. Then the dependent current variables can be solved by using the independent current variables. Independent Voltage Variables In a network it can be proved that the branch voltages of the twigs (or tree branches) are independent voltage variables. In a graph when all the twigs are short circuited, then all the nodes will be short circuited. Eventually, the voltages of the nodes become zero. Also the short circuiting of nodes will lead to short-circuiting of all the branches and so all the branch voltages will become zero. Hence we can say that, the branch voltages depend on twig voltages. Therefore, “the twig voltages are independent and branch voltages are dependent”. In a network with N nodes and B branches we have “N - 1” twigs. Therefore in a network we have B voltage variables in which “N - 1” are independent voltage variables and the remaining “B - (N - 1)” voltages are dependent voltage variables. In order to determine the response of a network on voltage basis it is sufficient if we form “N - 1” equations involving independent voltage variables and solve them for a unique solution. Then the dependent voltage variable can be solved by using independent voltage variables.

2.11 Analysis of Series-Parallel Circuits A typical circuit consists of series-parallel connection of passive elements like resistance, inductance and capacitance and excited by voltage/current sources. The sources will circulate current through all the elements of the circuit. Due to current flow a voltage will exist across each element of the circuit. Basically, the circuit analysis involves the solution of currents and voltages in various elements of the circuit. The currents and voltages can be solved by using the three fundamental laws : Ohm’s law, Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL). Now, the questions are, What will be the direction of current and polarity of voltage of an element in a circuit and how to find them ? In practical cases we may come across circuits with single source or circuits with multiple sources. In circuits excited by single source, with little bit experience it is possible to predict the direction of currents because the current will leave from positive end of the source and flow through available paths and then return to the negative end of the source.

2. 20

Chapter 2 - Basic Concepts of Circuits

But in circuits with multiple sources it will be difficult to find the direction of current through various elements. A common procedure to determine the currents and voltages in various elements of a circuit is presented in the following section. Procedure for Analysis of Circuits Using Fundamental Laws

1. Mark the nodes of the given circuit as A, B, C, D, etc. We can mark all the nodes including the meeting point of two elements. 2. Determine the number of branches in the given circuit. Attach a current to each branch of the circuit and arbitrarily assume a direction for each branch current. Let, the branch currents be Ia, Ib, Ic, Id, etc. 3. Write Kirchhoff’s Current Law (KCL) equations at each principal node of the circuit. (Remember that a principal node is meeting point of three or more elements.) The KCL equation is obtained by equating the sum of currents leaving the node to sum of currents entering the node. i.e., by KCL, Sum of currents entering the node = Sum of currents leaving the node 4. Any circuit will have some independent currents and the remaining currents will depend on the independent currents. Hence using the KCL equations, try to minimize the number of unknown currents by expressing some branch currents in terms of other branch currents. (The ultimate aim is to choose some independent currents and to express other currents in terms of independent currents.) 5. Let, the number of independent currents in the given circuit be M. Now we have to identify or choose M number of closed paths in the given circuit. For each closed path write a Kirchhoff’s Voltage Law (KVL) equation. The KVL equation is formed by equating the sum of voltage fall in the closed path to sum of voltage rise in that closed path. i.e., by KVL, Sum of voltage fall in a closed path = Sum of voltage rise in a closed path 6. The M number of KVL equations can be solved by any technique to find a unique solution for M independent branch currents. 7. From the knowledge of independent branch currents, determine the dependent branch currents. 8. Once the branch currents are known it is easy to find the voltage across the elements by Ohm’s law. The voltage across the element is given by the product of resistance and current, i.e., by Ohm’s law, Voltage = Resistance ´ Current 9. If we are interested in calculating the power of an element then the power can be calculated from the knowledge of voltage and current in the element. In purely resistive circuits excited by dc sources, Power = (Current)2 ´ Resistance or

Power =

^Voltageh2 Resistance

or Power = Voltage # Current

Circuit Theory

2. 21

Some Important Basic Concepts

1. 2. 3. 4.

5.

When a source delivers energy, the current will leave from positive end of source and return to negative end of source. I In series connected elements same current will flow. + R V In parallel connected elements the voltage across them will be same. E When a current flows through a resistance, the polarity of voltage across the resistance will be such that the current entering point positive and Fig. 2.35. leaving point is negative as shown in Fig. 2.35. If the total voltage across two resistances R1 and R2 in series is V volts and V1 and V2 are the voltage across R1 and R2 respectively, then, V1 = V #

6.

and

V2 = V #

R2 R1 + R2

If the total current through two resistances R1 and R2 in parallel is I Amperes and I1 and I2 are the current through R1 and R2 respectively, then, I1 = I #

2.11.1

R1 R1 + R2

R2 R1 + R2

and

I2 = I #

R1 R1 + R2

Single Loop Circuit

A single loop circuit is one which has only one closed path. In a single loop circuit all the elements are connected in series and so current through all the elements will be same. A single loop circuit can be analyzed by using Kirchhoff’s Voltage Law (KVL) and Ohm’s law.

R1 + IR1

R2 E

+ IR2

+

E2 E

E

+ E1 E

I IR4

IR3

E + E + A single loop circuit is shown in Fig. 2.36. Let, I be the R3 R4 current through the circuit. By Ohm’s law, the voltage across a resistance is given by the product of resistance and current through Fig. 2.36 : Single loop circuit. the resistance.

Now, by using KVL we can write, IR1 + IR2 + E2 + IR3 + IR4 = E1 ` I =

⇒

I(R1 + R2 + R3 + R4) = E1 - E2

E1 − E2 R1 + R2 + R3 + R4

From the above equation, the current through the single loop circuit of Fig. 2.36, can be estimated. From the knowledge of current and resistance, the voltage across various elements can be estimated. From the knowledge of voltage and current, the power can be estimated. 2.11.2

Single Node Pair Circuit

A single node pair circuit is one which has only one independent node and a reference node. In a single node pair circuit all the elements are connected in parallel and so voltage across all the elements will be same. A single node pair circuit can be analyzed using Kirchhoff’s Current Law (KCL) and Ohm’s law.

2. 22

Chapter 2 - Basic Concepts of Circuits

A single node pair circuit is shown in Fig. 2.37. Let, V be the voltage of the independent node (node-1) with respect to reference node.The voltage of reference node is always zero. By Ohm’s law, the current through the resistance is given by the ratio of voltage and resistance.

Node-1

V

I1

V R1

V R2

V R3

R1

R2

R3

V R4

I2

R4

0V

Now, by using KCL we can write,

Reference node

Fig. 2.37 : Single node pair circuit.

V V V V 1 1 1 1 + + + I2 + = I1 ⇒ V c + + + m = I1 − I2 R1 R2 R3 R4 R1 R2 R3 R4 ` V =

I1 − I2 1 + 1 + 1 + 1 R1 R2 R3 R4

From the above equation the voltage of the independent node (node-1) can be estimated. This voltage is the voltage across all the elements in the single node pair circuit. From the knowledge of voltage and resistance, the current through various resistances can be estimated. From the knowledge of voltage and current, the power in various elements can be estimated.

2.12 Solved Problems I

EXAMPLE 2.1 2‡

9 V +E

SOLUTION The battery connected to resistive load can be represented by the circuit shown in Fig. 1. Let, I = Current delivered by the battery. Now, by Ohm’s law, I =

9 = 0.5 A 2 + 16

Power delivered to load, PL = I2 ´ 16 = 0.52 ´ 16 = 4 W Power loss in the battery, PLB = I2 ´ 2 = 0.52 ´ 2 = 0.5 W % Efficiency of battery, ηB = =

Load power # 100 Load power + Power loss PL 4 # 100 = # 100 = 88.9% PL + PLB 4 + 0.5

Battery

A 9 V Battery with internal resistance of 2 W is connected to a 16 W resistive load. Calculate, a) Power delivered to load, b) Power loss in the battery, c) Efficiency of the battery.

Fig. 1.

16 ‡

Circuit Theory

2. 23

EXAMPLE 2.2 An 8.4 A current generator with internal resistance of 200 W is connected to a 10 W resistive load. Calculate, a) Power delivered to load, b) Power loss in the current generator, c) Efficiency of the current generator.

SOLUTION The current generator connected to resistive load can be represented by the circuit shown in Fig. 1. Let, VL = Voltage across the load. Now, by Kirchhoff’s current law, VL V + L = 8.4 ⇒ VL c 1 + 1 m = 8.4 ⇒ 0.105 VL = 8.4 200 200 10 10 Current generator

8.4 = 80 V 0.105 V2 Power delivered to load, PL = L 10 ` VL =

VL VL

VL

200

8.4 A

2 = 80 = 640 W 10

Power loss in current generator, P LCG =

10

+

200 ‡

10 ‡

VL E

2 VL2 = 80 = 32 W 200 200

Fig. 1. % Efficiency of current generator, ηCG =

=

PL Load power # 100 = # 100 Load power + Power loss PL + PLCG 640 # 100 = 95.2% 640 + 32

EXAMPLE 2.3

IL = 160A IA

Two batteries A and B with internal emf EA and EB and with internal resistance RA and RB are properly connected in parallel to supply a current of 160 A to a load resistance RL. Given

+

IB

+ +

EA, RA +E VA

EB, RB +E

VB

VL

RL

E

that, EA = 120 V, RA = 0.15 W, RB = 0.1 W and IB = 60 A. Calculate a) EB and b) load power.

E

E

Fig. 1.

SOLUTION

IL = 160 A IA

IL

= Current supplied by the sources.

RAIA

IA, IB

RA = 0.15 ‡

+

+

E

E

VA, VB = Terminal voltage of sources.

+ IB = 60 A RB = 0.1 ‡ VA

RBIB

Let, EA, EB = Internal (or generated) voltage of sources.

+

+ VB

VL

RL

E

= Load current.

EA = 120 V +E

VL = Voltage across the load.

E

EB +E

E

Fig. 2.

The sources are connected parallel to the load as shown in Fig. 1. In Fig. 2, the sources are represented as ideal sources with source resistance connected in series with ideal source.

2. 24

Chapter 2 - Basic Concepts of Circuits By KCL, we can write, IL = IA + IB ∴ IA = IL - IB = 160 - 60 = 100 A By KVL, we can write, EA = RAIA + VA ∴ VA = EA - RAIA = 120 - 0.15 ´ 100 = 105 V Since the sources and load are in parallel, VA = VB = VL = 105 V By KVL, we can write, EB = RB IB + VB = 0.1 ´ 60 + 105 = 111 V Load power, PL = VL IL = 105 ´ 160 = 16800 W = 16800 kW = 16.8 kW 1000 VL IL = 105 = 0.65625 Ω 160

Also load resistance, RL =

RESULT EB = 111 V,

PL = 16.8 kW,

RL = 0.65625 W

EXAMPLE 2.4

4V

Determine the magnitude and direction of the current in the 2 V battery in the circuit shown in Fig. 1.

SOLUTION

E

+

2‡

+

2V E

3‡

Let us assume three branch currents Ia, Ib and Ic as shown in Fig. 2. The currents are assumed such that they leave from positive terminal of the sources. +

The nodes in the circuit are denoted as A, B, C, D and E.

: Ia

Currents entering node-A : Ib, Ic ∴ Ia = Ib + Ic

E

1.5 ‡

Fig. 1.

By KCL at node-A we get, Current leaving node-A

3V

..... (1)

Circuit Theory

2. 25 4V

With reference to Fig. 2, in the closed path ACBDA we get, Voltage fall

: 2Ia, 3Ib

Voltage rise

: 4 V, 2 V

2‡

+

E Ia

C

+

D

E

E

E

2Ia 3‡

2V +

A

E

Ib

∴ 2Ia + 3Ib = 4 + 2

3V

Put, Ia = Ib + Ic

+

⇒

∴ 2(Ib + Ic) + 3Ib = 6

5Ib + 2Ic = 6

..... (2)

+

B

3Ib 1.5 ‡

E

Ic

E

+ 1.5Ic

Fig. 2.

With reference to Fig. 2, in the closed path ADBEA we get, Voltage fall

: 2 V, 1.5Ic

Voltage rise

: 3Ib, 3 V

∴

2 + 1.5Ic = 3Ib + 3 -3Ib + 1.5Ic = 3 - 2 -3Ib + 1.5Ic = 1

Equation (2) ´ 1.5

⇒

Equation (3) ´ (-2)

⇒

On adding

..... (3) 7.5Ib + 3Ic =

9

6Ib - 3Ic = -2 13.5Ib

=

7

` Ib = 7 = 0.5185 A 13.5 Therefore, the current supplied by the 2 V battery is 0.5185 A in the direction B to A (Refer Fig. 2.). The currents supplied by other sources can be estimated as shown below: 1 + 3Ib = 1 + 3 # 0.5185 = 1.7037 A 1.5 1.5 From equation (1), Ia = Ib + Ic = 0.5185 + 1.7037 = 2.222 A

From equation (3), Ic =

EXAMPLE 2.5

10 ‡

In the circuit shown in Fig. 1, the current in 5 W resistor is 5 A. Calculate the power consumed by the 5 W resistor. Also determine the current through 10 W resistance and the supply voltage E.

20 ‡

5‡

30 ‡

SOLUTION

5A

Power consumed by 5 W resistor = (Current)2 ´ Resistance +

= 52 ´ 5 = 125 W

E

E

Fig. 1.

The resistances 20 W and 30 W in parallel in Fig. 1, can be replaced by single equivalent resistance as shown in Fig. 2. [Refer Chapter - 6 for calculating equivalent resistance]. Let, Is = Current supplied by source. I1 = Current through 10 W.

2. 26

Chapter 2 - Basic Concepts of Circuits I2 = Current through 12 W. V1 = Voltage across 5 W. V2 = Voltage across 10 W and 12 W in parallel. In Fig. 2, the current Is divides into I1 and I2 and flows through parallel resistances 10 W and 12 W. The

currents I1 and I2 can be calculated by current division rule.

I1

10 W

By current division rule,

+

12 12 I1 = Is # = 5 # = 2.7273 A 10 + 12 10 + 12 By Ohm’s law, V1 = 5 ´ 5 = 25 V

5W + V1

Is

+

Is = 5 A

V2 = 10 ´ I1 = 10 ´ 2.7273 = 27.273 V

V2

I2

+ E

By KVL, we can write,

V2

-

20 ´ 30 = 12 W 20 + 30

-

Fig. 2.

E = V1 + V2 = 25 + 27.273 = 52.273 V

RESULT Power consumed by 5 W resistor = 125 W Current through 10 W resistor = 2.7273 A Supply voltage, E = 52.273 V

EXAMPLE 2.6 8‡

In the circuit shown in Fig. 1, the voltage across 8 W resistor is 20 V. What is the current through 12 W resistor. Also calculate the supply voltage.

SOLUTION

+

E +E

20 V

E

18 ‡

12 ‡

Let, Is be the current supplied by the source. The Is divides into I1 and I2 and flows through parallel connected 18 W and 12 W resistances as shown in Fig. 2.

Fig. 1.

The current supplied by the source flows through 8 W resistance. Since the voltage across 8 W is known, the current Is can be calculated by Ohm’s law. By Ohm’s law, Is

Is = 20 = 2.5 A 8

By current division rule,

= 2.5 # Let,

18 = 1.5 A 18 + 12

V1 = Voltage across 8 W resistance. V2 = Voltage across parallel combination of 18 W and 12 W.

Given that, V1 = 20 V By Ohm’s law, V2 = 12 ´ I2 = 12 ´ 1.5 = 18 V

18 ‡

V2

Fig. 2.

+ 12 ‡ V2

E

E +E

+

18 18 + 12

I2 I1

E

I2 = Is #

8‡ E + V1 = 20 V

Circuit Theory

2. 27

With reference to Fig. 2, by KVL we can write, E = V1 + V2 = 20 + 18 = 38 V

RESULT Current through 12 W resistor = 1.5 A Supply voltage, E = 38 V

EXAMPLE 2.7 +

In the circuit of Fig. 1, show that the power supplied by the current source is double of that supplied by the voltage source when R = (10/3) W.

10 V

2A

R E

SOLUTION

Fig. 1.

Since the voltage source, current source and R are in parallel, the voltage across them will be same, as shown in Fig. 2. Let, I1 = Current supplied by voltage source. I1

I2 = Current supplied by current source. I

I

10 V +E

= Current through R.

R

10 V E

By Ohm’s law, I =

I2 = 2 A + 10 V E

+

Voltage across resistance = 10 = 3 A Resistance 10/3

2A

Fig. 2.

By KCL, we can write, I = I1 + I2 Given that, I2 = 2 A ∴ I1 = I - I2 = 3 - 2 = 1 A Now, Power supplied by 10 V source = 10 ´ I1 = 10 ´ 1 = 10 W Power supplied by 2 A source

= 10 ´ I2 = 10 ´ 2 = 20 W

From the above results it is clear that the power supplied by current source is 20 W, which is double the power supplied by voltage source.

EXAMPLE 2.8

C

10 A

Find the power dissipated in each resistor in the circuit of Fig. 1.

SOLUTION

2‡

1‡

The power dissipated in the resistors can be calculated from the knowledge of current through the resistors. Let us denote the current through the resistors as Ia, Ib and Ic as shown in Fig. 2. In the closed path ACBA using KVL, we can write, 2Ia + Ib = 2Ic ⇒

2‡

30 A

20 A B

A

Fig. 1.

I Ia + b = Ic 2

∴ Ic = Ia + 0.5Ib

..... (1)

2. 28

Chapter 2 - Basic Concepts of Circuits At node-A, by KCL, we get, Ia + Ic = 30

..... (2)

On substituting for Ic from equation (1) in equation (2), we get,

⇒

Ia + Ia + 0.5Ib = 30

2Ia + 0.5Ib = 30

..... (3)

At node-C, by KCL, we get, C

Equation (4) ´ 0.5

⇒

..... (4)

10 A Ib

2Ia + 0.5Ib = 30 0.5Ia - 0.5Ib =

2‡

5

+ 1‡ Ib

2Ia

E

⇒

= 10

+

Equation (3) ´ 1

Ia - Ib

E

Ib + 10 = Ia ⇒

On adding

2.5Ia

= 35

Ia 2‡

30 A

` Ia = 35 = 14 A 2.5

A

Ic

+

2Ic

20 A E

B

Fig. 2.

From equation (4) we get, Ib = Ia - 10 = 14 - 10 = 4 A From equation (1) we get, Ic = Ia + 0.5Ib = 14 + 0.5 ´ 4 = 16 A We know that, Power consumed by a resistor = ^Currenth 2 # Resistance

Power consumed by 2Ω resistor between nodes A and C = Ia2 # 2 = 142 # 2 = 392 W Power consumed by 2Ω resistor between nodes A and B = Ic2 # 2 = 162 # 2 = 512 W Power consumed by 1Ω resistor between nodes B and C = Ib2 # 1 = 4 2 # 1

EXAMPLE 2.9

= 16 W 2‡

1‡

In the circuit shown in Fig. 1, find a) the total current drawn from the battery, b) voltage across 2 W resistor and c) current passing through the 5 W resistor.

7‡

5‡

+ 10 V E

SOLUTION Fig. 1. Let, IT be the total current supplied by the source. This current flows through 1 W and 2 W in series and then it divides into I1 and I2 and flows through parallel combination of 7 W and 5 W as shown in Fig. 2. The parallel combination of 7 W and 5 W can be reduced to single equivalent resistance as shown in Fig. 3. (Refer Chapter 6 for calculating equivalent resistance.)

2‡

+

I1

7‡

IT

I2

5‡

10 V E

Fig. 2.

By Ohm’s law we can write, IT =

IT

1‡

10 = 1.6901 A 1 + 2 + 2.9167

2W

5´7 5+7 = 2.9167 W

IT

∴ Voltage across 2 W resistor = IT ´ 2 = 1.6901 ´ 2 = 3.3802 V

IT

1W

+ 10 V -

Fig. 3.

Circuit Theory

2. 29

By current division rule, Current through 5 Ω resistor, I 2 = I T #

7 5 + 7

= 1.6901 #

7 = 0.9859 A 5 + 7

RESULT a)

Total current supplied by the source, IT = 1.6901 A

b)

Voltage across 2 W resistor = 3.3802 V

c)

Current through 5 W resistor = 0.9859 A

EXAMPLE 2.10 2‡

Calculate the current in all the elements of the circuit shown in Fig. 1.

+ E 10 V

6‡

SOLUTION Since the circuit has only one source. The direction of current through

4‡

the elements can be found as shown in Fig. 2.

3‡

8‡

By applying KCL at node-A we get, Currents leaving node-A Current entering node-A

: :

Fig. 1.

Ib, Ic

A Ib

Ia

+ E 10 V

\ Ib + Ic = Ia \ Ic = Ia - Ib

Ic

Ia 2‡

6‡

B

D

..... (1)

4‡

From equations (1) we can say that, the current Ic can be expressed

3‡

8‡

in terms of Ia and Ib. Hence the circuit has only two independent currents, Ia

Fig. 2.

and Ib and they can be solved by writing two KVL equations in the closed paths

C

ABCA and ADCA. A Ib

With reference to Fig. 3, in the closed path ADCA we get, Voltage fall

: 2Ib, 3Ib, 4Ia

Voltage rise

: 10 V

Ia

2Ib +

+ 10 V E

E D

E

+

E

\ 2Ib + 3Ib + 4Ia = 10 4Ia + 5Ib = 10

4Ia

+ 3Ib

..... (2)

C

Fig. 3.

2. 30

Chapter 2 - Basic Concepts of Circuits With reference to Fig. 4, in the closed path ABCA we get, : 6(Ia - Ib), 8(Ia - Ib), 4Ia

Voltage rise

: 10 V

I c = I a E Ib + 6(Ia E Ib)

Ia

+ E 10 V

E

Voltage fall

A

\ 6(Ia - Ib) + 8(Ia - Ib) + 4Ia = 10

B E

14(Ia - Ib) + 4Ia = 10

E

14Ia - 14Ib + 4Ia = 10 18Ia - 14Ib = 10 Equation (2) ´ 14

⇒

56Ia + 70Ib = 140

Equation (3) ´ 5

⇒

90Ia - 70Ib =

On adding

146Ia

+

4Ia +

8(Ia E Ib)

C

..... (3)

Fig. 4.

50

= 190 ` Ia = 190 = 1.3014 A 146

10 − 4Ia 5 10 4 # 1.3014 = 0.9589 A − = 5

From equation (2), we get, Ib =

From equation (1), we get, Ic = Ia - Ib = 1.3014 – 0.9589 = 0.3425 A

RESULT The current through the elements (i.e., branch currents) are, Ia = 1.3014 A

;

Ib = 0.9589 A

;

Ic = 0.3425 A

EXAMPLE 2.11

A

Determine the current in all the resistors of the circuit shown in Fig 1.

(AU June’14, 8 Marks)

50 A

I1

I2

I3

2‡

1‡

5‡

SOLUTION Fig. 1.

Let, Voltage at node-A be VA Now, by Ohm’s law, VA 2

I1 =

; I2 =

VA 1

; I3 =

VA 5

By KCL, at node-A, I + I + I = 50

&

` 1.7 VA = 50

&

1

2

3

VA VA VA + + = 50 2 1 5

&

VA = 50 = 29.4118 V 1.7

0.5VA + VA + 0.2 VA = 50

B

Circuit Theory

2. 31 VA = 29.4118 = 14.7059 A 2 2 VA 29 4118 . I2 = = = 29.4118 A 1 1 V I3 = A = 29.4118 = 5.8823 A 5 5

` I1 =

RESULT The current in the resistors are, I1 = 14.7059 A

;

I2 = 29.4118 A

;

I3 = 5.8823 A

2.13 Summary of Important Concepts 1.

An electric circuit consist of parameters (R, L and C) and sources connected in a particular combination.

2.

A circuit without sources is called a network.

3.

The sources in which the current/voltage does not change with time are called direct current sources.

4.

The networks excited by dc sources are called dc circuits.

5.

The elements which can deliver energy are called active elements.

6.

The elements which consume energy either by absorbing or storing are called passive elements.

7.

The resistance, inductance and capacitance are called fundamental parameters.

8.

The voltage/current of an independent source does not depend on voltage/current in any part of the circuit.

9.

The voltage/current of a dependent source depends on the voltage/current in some part of the same circuit.

10.

Node is meeting point of two or more elements.

11.

Principal node is meeting point of more than two elements.

12.

The path between any two nodes is called a branch.

13.

A path that starts and ends at a same node after travelling through some part of a circuit is called closed path.

14.

The connection of two or more elements in which same current flows is called series connection.

15.

The connection of two or more elements such that same voltage exist across them is called parallel connection.

16.

When three elements meet at a node, then they are said to be in star or T-connection.

17.

When three elements connected to form a closed path with a node in between any two element then they are said to be in delta or P-connection.

18.

An open path or path of infinite resistance between two nodes is called open circuit.

2. 32

Chapter 2 - Basic Concepts of Circuits 19.

A closed path of zero resistance between two nodes is called short circuit.

20.

When current leave an element from positive terminal, it delivers energy.

21.

When current enter an element at positive terminal, it absorbs energy.

22.

Ohm’s law states that voltage across a conductor is directly proportional to current through it.

23.

Kirchhoff’s Current Law (KCL) states that the algebraic sum of currents in a node is zero. By KCL, the sum of currents entering a node = Sum of currents entering a node.

24.

Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of voltages in a closed path is zero. By KVL, the sum of voltage raise in a closed path = Sum of voltage fall in a closed path.

25.

In a source, when voltage is constant and current varies with load then it is called voltage source.

26.

In a source, when current is constant and voltage varies with load then it is called current source.

27.

In an ideal voltage source the source resistance is zero.

28.

In an ideal current source the source resistance is infinite.

29.

A voltage source with internal resistance Rs can be represented by an ideal voltage source in series with an external resistance of value Rs.

30.

A current source with internal resistance Rs can be represented by an ideal current source in parallel with an external resistance of value Rs.

31.

A voltage source E in series with resistance Rs can be converted to an equivalent current source Is (Is = E/Rs) parallel with resistance Rs.

32.

A current source Is in parallel with resistance Rs can be converted to an equivalent voltage source E (E = Is Rs ) in series with resistance Rs.

33.

When a source voltage depends on a voltage in some part of same circuit, then the source is called Voltage Controlled Voltage Source (VCVS).

34.

When a source voltage depends on a current in some part of same circuit, then the source is called Current Controlled Voltage Source (CCVS).

35.

When a source current depends on a voltage in some part of a same circuit, then the source is called Voltage Controlled Current Source (VCCS).

36.

When a source current depends on a current in some part of a same circuit, then the source is called Current Controlled Current Source (CCCS).

37.

The power is rate of work done and the energy is total work done.

38.

Commercially, one kilowatt-hour of electrical energy is called one unit.

39.

Resistance is the property of element by which it opposes the flow of current.

40.

The voltage-current relation in a resistance is governed by Ohm’s law. If V and I are voltage and current in a resistance R then, V = IR.

Circuit Theory 41.

2. 33

Current division rule: If I is the total current through two parallel connected resistances R1 and R2, then the currents I1 and I2 flowing through R1 and R2 are, I1 = I #

42.

R2 R1 + R2

and

I2 = I #

R1 R1 + R2

Voltage division rule: If V is the total voltage across two series connected resistances R1 and R2, then the voltages V1 and V2 across R1 and R2 are, V1 = V #

R1 R1 + R2

and

V2 = V #

R2 R1 + R2

43.

Network topology is the study of properties of the network which are unaffected when we stretch, twist or distort the size and shape of the network.

44.

A graph describes the topological properties of a network and consists of nodes and branches of a network.

45.

When arrows are placed on the branches of a graph it is called oriented graph.

46.

Tree is a subgraph obtained by removing some branches of a graph such that all nodes of a graph are included without a closed path.

47.

A tree has N nodes and N – 1 branches.

48.

The branches removed to form a tree are called links or chords and the branches of a tree are called twigs.

49.

In a graph with B branches and N nodes, the number of links, L = B – N + 1.

50.

The cotree is complement of a tree and it is obtained by connecting the links to the nodes of the graph.

51.

The branch currents and voltages are called network variables.

52.

The arrow placed on the branch is called reference or orientation.

53.

In a branch, a single orientation is used to represent current and voltage direction.

54.

When reference is placed on branch by treating it as load then the reference is called load set reference.

55.

In a graph, the link currents are independent current variables and twig currents are dependent current variables.

56.

In a graph, the twig voltages are independent voltage variables and link voltages are dependent voltage variables.

57.

A single loop circuit is one which has only one closed path.

58.

A single node pair circuit is one which has only one independent node and a reference node.

2.14 Short-answer Quesions Q2.1

What is the difference between a circuit and network? A network will not have any independent sources whereas, a circuit will have independent source. When independent sources are connected to a network it becomes a circuit.

2. 34 Q2.2

Chapter 2 - Basic Concepts of Circuits Define active and passive elements. The elements which can deliver energy are called active elements. The elements which consumes energy either by absorbing or storing are called passive elements.

Q2.3

List the active and passive elements of electric circuit. The active elements of electric circuits are voltage source and current source. (Both independent and dependent source.) The passive elements of electric circuits are Resistor, Inductor and Capacitor.

Q2.4

Define the dependent source of a circuit. If the electrical energy supplied by a source depends on the voltage or current in some other part of same circuit, then it is called dependent source.

Q2.5

What is node and principal node? In a circuit the meeting point of two or more elements is called a node. If more than two elements meet at a node then the meeting point is called principal node.

Q2.6

Define the branch of a circuit. The path between any two nodes in a circuit is called branch.

Q2.7

Define series and parallel connection. If two or more elements are connected such that the current through them are same, then the connection is called series connection. If two or more elements are connected such that the voltage across them is same then the connection is called parallel connection.

Q2.8

What is meant by open circuit and short circuit? A path of infinite resistance between any two nodes is called open circuit. The current through open circuit is zero. A path of zero resistance between any two nodes is called short circuit. The voltage across short circuit is zero.

Q2.9

Define Ohm’s law. Ohm’s law states that the potential difference (or voltage) across any two ends of a conductor is directly proportional to the current flowing between the two ends provided the temperature of the conductor remains constant.

Q2.10

Define Kirchhoff’s laws. 1. Kirchhoff’s current law states that the algebraic sum of currents in a node is zero. 2. Kirchhoff’s voltage law states that the algebraic sum of voltages in a closed path is zero.

Q2.11

Draw the characteristics of ideal and practical voltage source. E

E

I

Fig. Q2.11.1 : Characteristic of ideal voltage source.

I

Fig. Q2.11.2 : Characteristic of practical voltage source.

Circuit Theory Q2.12

2. 35

Draw the characteristics of ideal and practical current source. Is

Is

V

V

Fig. Q2.12.1 : Characteristic of ideal current source.

Q2.13

Fig. Q2.12.2 : Characteristic of practical current source.

A 10 A current source has a source resistance of 100 W. What will be the equivalent voltage source? Solution The current source can be converted to an equivalent voltage source as shown in Fig. Q2.13.1 below: A

A 100 W

10 A

Þ

100 W

+ - 10 ´ 100

= 1000 V B

B

Fig. Q2.13.1 : Current source to voltage source conversion.

Q2.14

Convert the voltage source shown in Fig. Q2.14.1 to current source. A

A

Solution

Rs = 5 ‡

Given that, E = 20 V, Rs = 5 W

E +E 20 V

Now, Is = E = 20 = 4 A Rs 5

Is B

Fig. Q2.14.1.

4A

Rs = 5 ‡

B

Fig. Q2.14.2.

The voltage source of Fig. Q2.14.1, can be represented by an equivalent current source of value 4 A with a source resistance of 5 W in parallel as shown in Fig. Q2.14.2.

Q2.15

A 2 kW, 220 V water heater is used to heat a water tank for 45 minutes. What will be the number of units of energy consumed? Solution Energy consumed = Power # Time = 2 kW # 45 Hours = 1.5 kWh = 1.5 Units 60

Q2.16

An electrical appliance consumes 1.2 kWh in 30 minutes at 120 V. What is the current (AU Dec’14, 2 Marks) drawn by the appliance? Solution Energy consumed in 30 minutes = 1.2 kWh Energy consumed in 1 hour (60 minutes) = 1.2 # 60 = 2.4 kWh 30 ∴ Power rating of the device, P = 2.4 kW = 2400 W ∴ Current, I = P = 2400 = 20 A V 120

2. 36 Q2.17

Chapter 2 - Basic Concepts of Circuits What will be the length of copper rod having a cross-section of 1 cm2 and a resistance of 1W ? Take resistivity of copper as 2 ´ 10 -8 W-m. Solution Given that, R = 1 Ω, a = 1 cm2 = 1 # 10- 4 m2, ρ = 2 # 10- 8 Ω − m We know that, Resistance, R =

ρl a

` Length, l = Ra ρ 1 # 1 # 10- 4 = 5000 m = 5 km = 2 # 10- 8

Q2.18

A

Determine the value of current I, in the circuit shown in Fig. Q2.18.

I2

I 2‡

Solution

+ 8V

2A +

E

The voltage at node-A = 8 V

2V E

Current through 2 Ω resistance, I 2 = 8 − 2 = 3 A 2

Fig. Q2.18.

By KCL, 2 + l = I 2 ⇒ I = I 2 − 2 = 3 − 2 = 1 A

Q2.19

Obtain the current in each branch of the network shown in Fig. Q2.19.1. using Kirchhoff’s Current Law. Solution

(AU May’15, 2 Marks)

5‡

2‡ I3

20 V + E

I1

10 ‡

I2

+ 8V E

Let VA be the voltage at node-A.

Fig. Q2.19.1.

By, KCL at node-A We get, I1 − I2 − I3 = 0

&

20 − VA VA − 8 VA − − = 0 5 2 10

V V V ` 20 − A − A + 8 − A = 0 5 5 2 2 10 8 − 0.8VA = 0

&

&

4 − 0.2 VA − 0.5VA + 4 − 0.1VA = 0

0.8VA = 8 5‡

V1 = 20V

` VA = 8 = 10 V 0.8

1

I2 =

VA − 8 = 10 − 8 = 1 A 2 2

I3 =

VA = 10 = 1 A 10 10

20 V + E

V2 = 8V 2

A I

20 − VA I1 = = 20 − 10 = 2 A 5 5

2‡

VA

1 a

20 E V A I2 a

5

V I3 a A 10

10 ‡

Fig. Q2.19.2.

VA E 8 2

+ E

8V

Circuit Theory 3‡

Find the voltage across AB in the circuit shown in Fig. Q2.20.1.

+E

5

V

7 V +E

4‡

With reference to Fig. Q2.20.2, by voltage division rule, Voltage across 4 Ω resistor, V2 = 7 # 4 = 4 V 4 + 3

B

Fig. Q2.20.1.

8 = 8V 8 + 2 3‡

On tracing from A to B, as shown in Fig. Q2.20.2.

Q2.21

Determine the currents I1, I2, I3 and I4 in the circuit shown in Fig. Q2.21.

V1

+E

E +

7 V +E

V2

I1

4‡

E

+

V 5

VAB = V2 + 5 + V3 = 4 + 5 + 8 = 17 V

+

10 V

A

8‡

E

Voltage across 8 Ω resistor, V3 = 10 #

2‡

8‡

+E

Solution

10 V

A

+

Q2.20

2. 37

V3 E

E

I2

V4

2‡

+

B

Solution

Fig. Q2.20.2.

By KCL at node-A

2A

I1 + 2 + 5 = 0

⇒ I1 = -7 A

A

B I1

At node-B I2 = 2 + 4 = 6 A

I2

4A

5A

At node-C 4 + I4 + 1 = 0

⇒

I4 = -5 A

I3

I4

D

At node-D

C 1A

I3 = 5 + 1 = 6 A

Q2.22

Fig. Q2.21.

Find the current I and voltage across 30 W resistor, in the circuit shown in Fig. Q2.22.1.

(AU June’14, 8 Marks)

Solution

2‡

8‡

+ E + 100 V

40 V

The given circuit is redrawn as shown in Fig. Q2.22.2.

E

I

30 ‡

Fig. Q2.22.1.

Now, by KVL, 40 + 8 I + 2 I + 30 I = 100

8‡ E

8 I + 2 I + 30 I = 100 - 40 40 I = 60 ⇒

I = 60 = 1.5 A 40

+ 40 V

+ 100 V E

I E + 30I E

Voltage across 30 W resistor = 30 I = 30 ´ 1.5 = 45 V

+ 8I

30 ‡

+

2I E 2‡

Fig. Q2.22.2.

2. 38 Q2.23

Chapter 2 - Basic Concepts of Circuits Determine the current through each resistor in the circuit shown in Fig. Q2.23. (AU June’14, 6 Marks) Solution

12 A + Vs

Since three equal resistances are connected in parallel, the current will divide equally in the three parallel paths.

E

` I1 = I2 = I3 = 12 = 4 A 3

4‡

4‡

4‡

I1

I2

I3

B

Fig. Q2.23.

` Vs = I1 # 4 = 4 # 4 = 16 V

Q2.24

What is a graph? A graph is topological description of a network and consists of nodes and branches.

Q2.25

Define tree, link and cotree. Tree: Tree is a subgraph and can be defined as any connected open set of branches which includes all nodes of a given graph. Link: The branches removed from the graph of a network to form a tree are called links. Cotree: The cotree is complement of a tree.

Q2.26

What are network variables? The branch currents and voltages are called network variables. The branch currents are called current variables and branch voltages are called voltage variables.

Q2.27

What is a single loop circuit? A single loop circuit is one which has only one closed path.

Q2.28

Define single node pair circuit. A single node pair circuit is one which has only one independent node and a reference node.

2.15 Exercises I. Fill in the Blanks With Appropriate Words 1.

The elements which consume energy either by absorbing / storing are called _________ elements.

2.

The sources in which the current / voltage do not change with time are called _________.

3.

The electrical energy supplied by _________source depends on another source of electrical energy.

4.

The path between any two nodes is called _________.

5.

In an electric circuit when elements are _________connected, the current will be same.

6.

In an electric circuit a path of infinite resistance is called _________.

7.

In an electric circuit a path of zero resistance is called _________.

8.

In an electric circuit the algebraic sum of _________ in a node is zero.

Circuit Theory 9.

2. 39

In an electric circuit the algebraic sum of _________ in a closed path is zero.

10. In an ideal _________ source the terminal voltage remains constant. 11. The _________ is given by the product of power and time.

ANSWERS 1. passive

5. series

9. voltages

2. dc sources

6. open circuit

10. voltage

3. dependent

7. short circuit

11. energy

4. branch

8. currents

II. State Whether the Following Statements are True/False 1.

The elements of electric circuits which can deliver energy are called active elements.

2.

The inductance and capacitance absorbs energy.

3.

The electrical energy supplied by an independent source depends on another electrical source.

4.

In an electric circuit the meeting point of two or more elements is called principal node.

5.

In parallel connected elements the voltage will be same.

6.

In an electric circuit, a voltage exists across open terminals.

7.

In an electric circuit, a current flows through short circuited terminals.

8.

In an electric circuit, when an element delivers energy the current will leave from negative terminal.

9.

In an electric circuit, when an element absorbs energy the current will enter at negative terminal.

10. An ideal voltage source can be converted to ideal current source. 11. In an ideal current source the terminal voltage remains constant. 12. The power is rate of work done and the energy is total work done.

ANSWERS 1. True

5. True

9. False

2. False

6. True

10. False

3. False

7. True

11. False

4. True

8. False

12. True

III. Choose the Right Answer for the Following Questions 1‡

1. The voltage VAB across the open circuit in the circuit shown in Fig. 1 is, a) 2 V b) 4 V

5‡ A +

20 V +E

4‡

E B

c) 8 V 5‡

d) 6 V

VAB

6‡

Fig. 1.

2. 40

Chapter 2 - Basic Concepts of Circuits 2. The current Isc through the short circuit in the circuit shown in Fig. 2 is, a) 4 A

3‡

12 V +E

4‡

2‡

Isc

5‡

b) 3 A c) 2 A

1‡

Fig. 2.

d) 1 A 3.

5‡

The voltage VL in the circuit shown in Fig. 3 is, a) 4 V b) 8 V

6‡

+ VL

2A

3A

4‡

E

c) 12 V d) 20 V 4.

Fig. 3.

The current Is delivered by the voltage source in the circuit shown in Fig. 4 is,

2‡

4‡ Is

a) 6 A 8‡

1‡

b) 5 A

+ E 20 V

c) 4 A

Fig. 4.

d) 2 A 5.

A

The node voltage VA in the circuit shown in Fig. 5 is,

VA

a) 8 V b) 6 V

5‡

10 A

2‡

2A

10 ‡

c) 10 V

Fig. 5.

d) 2 V 6.

The value of voltage sources E1 and E2 in the circuit shown in Fig. 6 are, a) 11 V, 16 V b) 16 V, 11 V

5‡

5‡

2A + E1 E

1A + E2

2‡ E

c) 5 V, 7 V

Fig. 6.

d) 7 V, 5 V 7. The current I1 and I2 in the circuit shown in Fig. 7 are, I1

2A

a) 7, 5 b) 3, 7 c) –5, –3 d) –7, –5

4A

5A

1A

Fig. 7.

I2

Circuit Theory

2. 41

8. The equivalent current source for the voltage source in the circuit shown in Fig. 8 is, 5‡

2‡

10 V + 2‡ E

1‡

3‡

Fig. 8. 7‡

6‡

2‡

a)

b) 10 A

10 A

9.

c) 2‡

d) 2‡

5A

10 ‡

2A

The equivalent voltage source for the current source in the circuit shown in Fig. 9 is, 5‡

5A 3‡

8‡

2‡

Fig. 9. a)

5‡

b) +

+ 3‡

5V

10.

+

6V E

3‡

d)

+

25 V

E

1.2 ‡

c)

15 V E

E

The equivalent current source for the dependent voltage source in the circuit shown in Fig. 10 with respect to terminals A -B is, 4‡

10 V +E

3Ix + E

A

Ix

5‡

2‡

Fig. 10. B

A

a)

1.5Ix

Ix

5‡

B

A

b)

Ix

5‡

2‡

B

A

c) 1.5Ix

2‡

d) Not possible Ix

5‡

B

1.5Ix

2‡

2. 42

Chapter 2 - Basic Concepts of Circuits 11. The equivalent voltage source for the dependent current source in the circuit shown in Fig. 11 with respect to terminals A-B is, A

a)

0.2Vx

B E +

2‡

0.1Vx A

b)

A

+ E

A

c)

12 V +E

2‡

0.4Vx

+ Vx E

4‡

6‡

B E +

2‡

0.4Vx

d)

Fig. 11.

Not possible 5A

12. In Fig. 12, the currents I1, I2 and I3 respectively are, a)

1.5 A, 1.5 A, 2 A

I1

b)

1 A, 1 A, 3 A

3‡

c)

1 A, 2 A, 3 A

d)

2 A, 2 A, 1 A

13.

B 2‡

7‡

B

4 V, 6 V, 10 V

b)

3.5 V, 6.5 V, 10 V

c)

2.5 V, 5 V, 12.5 V

3‡

6‡

Fig. 12.

In Fig. 13, the voltage V1, V2 and V3 respectively are, a)

I3

I2

+

V1

+

E

2‡

V2

+

E

4‡

V3

E

10 ‡

+E

20 V

d) 14.

Fig. 13.

2 V, 6 V, 12 V

The number of links and twigs in the graph shown in Fig. 14 respectively are,

15.

a)

5, 6

b)

5, 5

c)

6, 5

d)

11, 5

Fig. 14.

For the graph shown in Fig. 15, which of the following is not a proper tree? 1

2

4 3

5

Fig. 15. 1

a) 3

2

4

b) 5

3

1

2

4

c) 5

3

1

2

4

d) 5

3

1

2

4

5

Circuit Theory

2. 43

ANSWERS 1. c

6. b

11. c

2. b

7. d

12. d

3. d

8. c

13. c

4. a

9. d

14. a

5. c

10. b

15. c

IV. Unsolved Problems E2.1

Find the node voltages in the circuit shown in Fig. E2.1. R1 A

I2

R2 B I1

R1 12 V V1

+E

1

6V

V2

+E

2

3

R3

R2

R3

V3

2A

5A I3

+ E 3V

D

C 10 A

0

Fig. E2.2.

Fig. E2.1. E2.2

Find the branch currents I1, I2 and I3 in the circuit shown in Fig. E2.2.

E2.3

Find the value of the source voltage Vs in the circuit shown in Fig. E2.3.

E2.4

E2.5

E2.6

A 20 V source with internal resistance 0.2 W is connected in series with 30 V source with internal resistance 0.3 W to deliver a load current of 10 A to resistive load. Calculate, a) Load power, PL , b) Power delivered by each source to load.

5‡

3‡ 2A

1‡ 4‡ Vs

Fig. E2.3. Two current sources with internal resistance 50 W and 100 W are connected in parallel to supply a 4.8 kW load at 200 V. If the generated source current of the source with 50 W internal resistance is 12 A, what is the generated source current of the other source? What is the value of the emf, E of the battery in the circuit shown in Fig. E2.6. Say whether the battery is charging or discharging. 1‡

2‡

2‡

2‡ 1A

14 A + 24 V

1‡

+

+ E E

E

Fig. E2.6.

E2.7

2‡

+ E

3‡

E E

Fig. E2.7. What is the value of source voltage, E in the circuit shown in Fig E2.7.

1‡

2. 44

Chapter 2 - Basic Concepts of Circuits

E2.8

What is the value of load voltage, VL in the circuit shown in Fig. E2.8?Also, calculate the power delivered by the current source. 4‡

5‡ 4‡

+ 10 A

2‡

3‡

VL E

I

+ E

Fig. E2.8.

3‡

10 V 0.5 ‡

5‡

6‡

Fig. E2.9. E2.9

Determine the current, I delivered by the voltage source in the circuit shown in Fig. E2.9. Also calculate the power delivered by the voltage source.

E2.10 Determine the voltages V1 and V2 in the circuit shown in Fig. E2.10. 2A 2‡

V1

V2

4‡ 4‡ 4.5 A

4‡

4‡

1A

Fig. E2.10.

ANSWERS E2.1

V1 = 21 V, V2 = 9 V, V3 = 3 V

E2.2

I1 = 12 A, I2 = 7 A, I3 = 5 A

E2.3

Vs = 37 V

E2.4

PL = 450 W, PD20 = 180 W, PD30 = 270 W

E2.5

Is2 = 18 A

E2.6

E = 2 V, Charging

E2.7

E = 7V

E2.8

VL = 6 V, Ps = 610 W

E2.9

I = 2 A, Ps = 18 W

E2.10 V1 = 4 V, V2 = 3 V

Chapter 3

BASIC CONCEPTS OF AC CIRCUITS 3.1

Introduction

The sources in which the current/voltage sinusoidally varies with time are called sinusoidal sources. In sinusoidal sources the voltage/current will undergo cyclic changes and the number of cycles per second is called frequency. The time for one cycle is called time period. Since the nature of variation is identical in every period, the sinusoidal voltages and currents are also called periodic voltages and currents. In one period/cycle of sinusoidal quantity, the value of the quantity (i.e., voltage or current) will be positive for one half period/cycle and the value of the quantity will be negative for another half period/cycle. The nature of variations in positive half cycle is identical to that of negative half cycle but with opposite polarity. Since the sinusoidal voltage/current has alternate identical positive and negative half cycles, they are called alternating voltage/current. Therefore, the sinusoidal sources are called alternating current sources or in short ac sources. The circuits excited by sinusoidal sources are called ac circuits. In ac circuits the current and voltage varies with time and so all the three basic parameters resistance, inductance and capacitance exists in ac circuits. The basic concepts of resistance parameter are discussed in Chapter-2 and the basic concepts of inductance and capacitance are discussed in this Chapter.

3.2

AC Voltage and Current Source

The voltage and current are the two quantities that decides the energy supplied by the sources of electrical energy. Usually, the sources are operated by maintaining one of the two quantities as constant and by allowing the other quantity to vary depending on load. In ac sources, when the rms value of voltage is maintained constant and the rms value of current is allowed to vary then the source is called voltage source. When the rms value of current is maintained constant and the rms value of voltage is allowed to vary then the source is called current source. The voltage across an ideal voltage source should be constant for whatever current delivered by the source. Similarly, the ideal current source should deliver a constant current for whatever voltage across its terminals. + E = EÐq

~

-

Fig. a : Ac voltage source.

I s = Is Ðq

~

Fig. b : Ac current source.

Fig. 3.1 : Symbols for ac source.

3. 2

Chapter 3 - Basic Concepts of AC Circuits

In reality, the ideal conditions never exists (but for analysis purpose the sources can be considered ideal). In practical ac voltage source the voltage across the source decreases with increasing load current and the reduction in voltage is due to its internal impedance. In practical ac current source, the current delivered by the source decreases with increasing load voltage and the reduction in current is due to its internal impedance. Here, E, V = Magnitude of rms value of voltage. I, IS = Magnitude of rms value of current. E

Is

I

+ + E

Is

~

~

V

E

E

I

Fig. a : Characteristics of ideal voltage source.

V

Fig. b : Characteristics of ideal current source.

Fig. 3.2 : Characteristics of ideal sources. Is

E

I

V

Fig. a : Characteristics of practical Fig. b : Characteristics of practical voltage source. current source. Fig. 3.3 : Characteristics of practical sources.

Let, E = Voltage across ideal source (or internal voltage of the source). Is = Current delivered by ideal source (or current generated by the source). V = Voltage across the terminals of the source. I = Current delivered through the terminals of the source. Zs = Source impedance (or internal impedance).

The practical voltage source can be considered as a series combination of an ideal voltage source and a source impedance, Zs . The reduction in voltage across the terminals with increasing load current is due to the voltage drop in the source impedance. When the value of source impedance is zero the ideal condition is achieved in voltage sources. Hence the source impedance for an ideal voltage source is zero.

Circuit Theory

3. 3 + I Zs E

I

I

C

C

Ish

Zs

C

E

Is

V

~

~

Zs

V

E

E

E

I = Is E Ish

V = E E I Zs

Fig. 3.5 : Model of practical ac current source.

Fig. 3.4 : Model of practical ac voltage source.

The practical current source can be considered as a parallel combination of an ideal current source and a source impedance, Zs . The reduction in current delivered by the source is due to the current drawn by the parallel source impedance. When the value of source impedance is infinite the ideal condition is achieved in current sources. Hence the source impedance for an ideal current source is infinite. 3.2.1 AC Source Transformation The practical voltage source can be converted to an equivalent practical current source and vice-versa, with same terminal behaviour. In these conversions the current and voltage at the terminal of the equivalent source will be same as that of original source, so that the power delivered to a load connected at the terminals of original and equivalent source will be same. The voltage source with series impedance can be converted to an equivalent current source with parallel impedance. Similarly, the current source with parallel impedance can be converted to an equivalent voltage source with series impedance. I

I

A Zs

A

+

+

+

~

V

Þ

Þ

E

-

E = I s Zs

Is

~

Zs

V

-

B Is =

B

E Zs

Fig. 3.6 : Conversion of ac voltage source to current source and vice versa.

3.3

Sinusoidal Voltage

A sinusoidal voltage can be considered as a vector of length Vm rotating in space with an uniform angular speed ω rad/s as shown in Fig. 3.7. At any time instant the vector can be resolved into xcomp and ycomp. Now, the ycomp gives the value of sinusoidal voltage at any time instant. Therefore, the instantaneous value (i.e., the value at any particular time instant) of a sinusoidal voltage is given by, v = Vm sin ωt

y w ycomp

Vm

xcomp

Fig. 3.7.

x

3. 4

Chapter 3 - Basic Concepts of AC Circuits

Since the sinusoidal voltage is a rotating vector the value of the voltage repeats after an angular rotation of 2π radians or 360 o. The number of revolutions (or rotations) per second is called frequency and it is denoted by f. The unit of frequency is Hertz and denoted by Hz (or cycles per second). One rotation of the voltage vector is also called cycle, because the value of voltage repeats in every revolution. One revolution is equal to an angular motion of 2π radians. Hence the frequency can also be expressed in the units of radians per second (rad/s) which is denoted by ω and popularly called angular frequency. The relation between angular frequency (ω) and frequency (f ) is, ω = 2πf The time taken for one revolution or cycle is called time period (or simply period), and it is denoted by T. The unit of time period (T) is seconds. We know that, Frequency, f = Number of cycles per second. Hence, Time for one cycle = 1 f ` Time period, T = 1 seconds f The plot of the instantaneous value of the sinusoidal voltage with respect to ωt is called waveform. The instantaneous value of the sinusoidal voltage is computed for one cycle and the waveform is plotted in Fig. 3.8. when, ωt = 0 ; when, ωt = π ; 4 when, ωt = 2π = π ; 4 2 3 π when, ωt = ; 4 when, ωt = 4π = π ; 4

ν = Vm sin 0 = 0 ν = Vm sin π = Vm 4 2 π ν = Vm sin = Vm 2 3 ν = Vm sin π = Vm 4 2 ν = Vm sin π = 0

when, ωt = 5π ; 4 6 when, ωt = π = 3π ; 4 2 π 7 ; when, ωt = 4 when, ωt = 8π = 2π; 4

ν = Vm sin 5π = − Vm 4 2 3 ν = Vm sin π = − Vm 2 7 ν = Vm sin π = − Vm 4 2 ν = Vm sin 2π = 0

v Vm

0 p 2p 3p 4 p 5p 6p 7 p 8p 4 4 4 4 4 4 4 4 -Vm

wt

One cycle

Fig. 3.8 : Sinusoidal voltage waveform on an angular scale.

Since ω = 2πf = 2π/T, the instantaneous value of sinusoidal voltage can also be expressed as shown in equation (3.1). ..... (3.1) ν = Vm sin 2π t T

Circuit Theory

3. 5

Using the equation (3.1), the instantaneous value of voltage can be calculated for various time instants and the waveform is plotted as a function of t in Fig. 3.9. It can be observed that the waveform of Fig. 3.9 is same as that of Fig. 3.8 except the angular scale, which is replaced by time scale. when, t = 0 ; when, t = T ; 8 2 when, t = T = 8 3 when, t = T ; 8 4 when, t = T = 8

ν = Vm sin 0 = 0 ν = Vm sin 2π # T T; 2 π 4 ν = Vm sin T # ν = Vm sin 2π # T T; 2 π 2 ν = Vm sin T #

T = Vm 8 2 T = Vm 4 3T = Vm 8 2 T =0 2

when, t = 5T ; 8 6 when, t = T = 3T ; 8 4 T 7 when, t = ; 8 when, t = 8T = T; 8

ν = Vm sin 2π # 5T = − Vm T 8 2 π T 2 3 ν = Vm sin = − Vm # T 4 ν = Vm sin 2π # 7T = − Vm 7 8 2 π 2 ν = Vm sin #T=0 T

v Vm

0 T 2T 3T 4T 5T 6T 7T 8T 8 8 8 8 8 8 8 8 EVm

t

One cycle

Fig. 3.9 : Sinusoidal voltage waveform on a time scale.

The sinusoidal voltage described by equation (3.1), starts at ωt = 0 (i.e., the origin is at ωt = 0). At the origin, the value of the voltage is zero. But sometimes the voltage may be non-zero at the origin (i.e., at ωt = 0), such a sinusoidal voltage can be described by the equation (3.2). ..... (3.2) v = V m sin (ωt + φ) The equation (3.2), for sinusoidal voltage is the more generalized equation. v v = Vm sin wt

Vm

f=0 0

p

2p

3p

4p

wt

-Vm

v

v = Vm sin (wt + f)

Vm

f = Positive

0 |f|

p

3p

2p

4p

wt

-Vm

v

v = Vm sin (wt - f)

Vm

f = Negative |f|

0

p

2p

3p

4p

-Vm

Fig. 3.10 : Sinusoidal voltage waveforms.

wt

3. 6

Chapter 3 - Basic Concepts of AC Circuits

3.3.1 Average Value The average value of a time varying quantity is the average of the instantaneous value for a particular time period. Usually, for periodic waveforms the average is taken for one time period. In alternating quantities, the average value for one time period is zero because in one period, it has equal positive and negative values. Therefore, for alternating quantities the average is taken over half the period. The instantaneous value of the sinusoidal voltage is expressed by, v = Vmsinωt = Vmsin q, where, q = ωt. The total value over half period (π) is obtained by integrating the instantaneous value between limits 0 to π. Then the average is obtained by dividing this total value by half period (π). Let, Vave = Average value of sinusoidal voltage or alternating voltage Now, by definition of average value we can write, r

#

Vave = 1 π

r

#

ν dθ = 1 π

0

r

# sin θ dθ

Vm sin θ dθ = Vm π

0

0

= Vm 6 − cos θ @ 0r = Vm 6 − cos π + cos θ @ π π V V 2 m = m 61 + 1 @ = π π ` Vave = 2Vm π 3.3.2 RMS Value

.....(3.3)

(AU June’14, 2 Marks)

The rms value of a time varying quantity is the equivalent dc value of that quantity. (The rms value is also known as effective value.) For example, a 5 V dc is equivalent to 5 V rms value of ac. The rms stands for root-mean-square, which means that the value is obtained by taking the root of the mean of the squared function. Hence, to take the rms value, the function is squared and the mean (average) of the squared function is determined. Then take root of this mean value. For periodic waveforms the rms value is computed for one period. For alternating quantities, the rms value will be same if it is computed for half period or one period. The instantaneous value of the sinusoidal voltage is expressed by, v = Vmsinωt = Vmsin q, where, q = ωt. The total value of the squared function over half period (π) is obtained by integrating v 2 between limits 0 to π . The mean (average) is obtained by dividing the total value of squared function by half period (π). The rms value is obtained by taking square root of this mean value. Let, V = Rms value of sinusoidal voltage or alternating voltage. Now, by definition of rms value we can write, r

V =

#

1 π

r

ν2 dθ =

# (V sin θ) dθ

1 π

0

2

m

0 r

=

Vm2 π

r

# sin θ dθ 2

0

=

# ^1 −

Vm2 π 0

cos 2θh dθ 2

sin2 θ = 1 − cos 2θ 2

Circuit Theory

3. 7 r

Vm2 2π

=

# ^1 −

cos 2θh dθ =

Vm2 θ − sin 2θ r 2π 8 2 B0

0

=

2 m

V π − sin 2π − 0 + sin 0 = 2π 8 2 2 B

Vm2 6 π @ = Vm 2π 2

` V = Vm 2

.....(3.4)

3.3.3 Form Factor and Peak Factor The form factor is defined as the ratio of rms value and average value of a periodic waveform. The peak factor is defined as the ratio of peak value (or maximum value) and the rms value of a periodic waveform. ` Form factor, k f =

Rms value Average value

` Peak factor, k p = Maximum value Rms value

The form and peak factors are constants for a particular waveform. If the average value of a waveform is known then its rms value can be estimated using the form factor (or vice-versa). If the rms value of a particular waveform is known then its peak value can be estimated using the peak factor (or vice-versa).The form and peak factors for sinusoidal voltage can be estimated using equations (3.5) and (3.6) as shown below: Form factor, k f =

Vm 2 = π = 1.111 2Vm π 2 2

^For full sine waveh

..... (3.5) ..... (3.6)

Peak factor, k p =

Vm = Vm 2

2 = 1.414

^For full sine waveh

The form factor and peak factor of equations (3.5) and (3.6) are applicable for full sinusoidal waveform of voltage or current or any other quantity.

3.4

Sinusoidal Current

The discussions and analysis presented in Section 3.3 for sinusoidal voltage are applicable to sinusoidal currents. The equations derived for sinusoidal voltage in Section 3.3 are applicable for sinusoidal currents if we change the variable v by i. The instantaneous value of sinusoidal current is given by, i = I m sin ωt = Im sin q, where, q = ωt

The general equation for instantaneous value of sinusoidal current is given by, i = I m sin (ωt + φ)

3. 8

Chapter 3 - Basic Concepts of AC Circuits i Im

i = Im sin wt f=0 0

p

3p

2p

4p

wt

-Im

i Im

i = Im sin (wt + f) f = Positive

|f|

0

p

3p

2p

4p

wt

-Im

i

i = Im sin (wt - f)

Im

f = Negative

|f| 0

p

3p

2p

4p

wt

-Im

Fig. 3.11 : Sinusoidal current waveforms. r

Average value of sinusoidal current, Iave

= 1 π

#

r

0 r

Rms value of sinusoidal current, I =

#i

1 π

#I

i dθ = 1 π

m

sin θ dθ = 2I m π

0 r 2

dθ =

0

# (I

1 π 0

m

sin θ) 2 dθ = I m 2

Form factor, k f = I m / 2 = π = 1.111 2 I m /π 2 2 Peak factor, k p =

3.5

Im = 1m / 2

2 = 1.414

Inductance

The inductance is the property of element (or matter) by which it opposes the change in flux or current. The opposition is offered by the way of an induced emf opposing the current flow. Hence in an element with inductance property the current cannot change instantaneously, i.e., change in current will be delayed (whereas in purely resistive elements the current can change instantaneously). Note : The flux and current are inseparable in nature. Whenever flux exists in an element it is due to motion of electrons (i.e., current), or whenever current flows in an element the flux is created in the element. The flux in the straight current-carrying conductors is negligible, but the flux in the conductors in the form of a coil is appreciable. Hence, the property of inductance is predominant in coils and so commercial inductors are also made in the form of coils. Typically, a coil consists of a number of turns of copper or aluminium conductor wound on an iron core (Sometimes the coils may not have a core).

Circuit Theory

3. 9

The unit of inductance is Henry and denoted by H. The inductance of a coil is defined as the ratio of flux linkages (weber-turns) and current through the coil. The weber-turns refers to the product of flux, φ and number of turns, N of a coil. Hence, the inductance of a coil with N turns and carrying a current of I amperes is given by, Inductance, L =

Nφ I

..... (3.7)

In equation (3.7), if N = 1 turn, φ = 1 Weber, I = 1 Ampere, then L = 1 Henry. Therefore, “a coil is said to have an inductance of one Henry if a current of one ampere flowing through it produce a flux linkage of one weber-turn in it”. Practically, one Henry is a large value and so the smaller values mH (milli-Henry) and µH (micro-Henry) are practically used. Note : In this section the inductance refers to self-inductance. 3.5.1 Voltage-Current Relation in an Inductance We know that a current-carrying conductor will always have a flux associated with current. Faraday has observed that whenever the flux linkages of a conductor changes, an emf is induced in the conductor and he proposed this phenomena as laws of magnetic induction. Law I : Whenever the magnetic flux linked with a conductor changes an emf is always induced in it. Law II : The magnitude of induced emf is equal to rate of change of flux linkages. Consider an inductor with N turns carrying a current i as shown in Fig. 3.12. Let, φ be the instantaneous value of flux in the inductor. When the current, i is varied the flux, φ will also vary and so an emf is induced in the coil, in a direction opposing the current flow. (The direction of induced emf can also be found using Len’s law or Fleming’s right hand rule). i + Now, by Faraday’s Law we can write, L v dφ E & ν = N d c Li m ν = N Using equation (3.7) dt dt N N and L are constants ν = N # L di Fig. 3.12. N dt .....(3.8) ` ν = L di dt The equation (3.8) gives the voltage-current relation in an inductance. From equation (3.8), we can say that when, i = constant, then di = 0, and so v = 0. Hence for constant or direct current dt the inductance will behave as a short circuit (in steady state). Therefore, in steady state analysis of circuits excited by dc sources the inductances are considered as short circuit (or simply the inductances are neglected). On rearranging equation (3.8), we get ν dt = L di Integrating on both sides, we get 1 ν dt = i ⇒ ⇒ ν dt = L di ν dt = L di L

#

` i =

# 1 ν dt L #

#

#

#

.....(3.9)

3. 10

3.5.2

Chapter 3 - Basic Concepts of AC Circuits

Energy Stored in an Inductance

In an inductance the energy is stored as magnetic field. Let, dw be the energy stored in the inductance in the time interval dt. Now the energy, dw is given by the product of power, p and time, dt. ∴ dw = p dt = vi dt

Put, p = ni

= L di i dt dt = Li di

Put, ν = L di dt ..... (3.10)

Let the current, i raise from zero to a steady value of I (where, I is rms value in case of ac) to establish an energy of W. Now, the stored energy (or total work) is obtained by integrating the equation (3.10) between limits 0 to I. I

I

# Li di = L # i di

` W = 0

2 I 2 = L; i E = L I 2 0 2

0

..... (3.11) ` W = 1 LI2 2 The equation (3.11) can be used to compute the energy stored in an inductance when a steady current I flows through it.

3.6

Capacitance

The capacitance is the property of element (or matter) by which it opposes the change in charge or voltage. Since charge is a physical quantity it cannot change from one value to the other instantaneously. Whenever charge exists at a point then there should be some voltage at that point. Hence in an element with capacitance property the voltage cannot change instantaneously, i.e., change in voltage will be delayed, (whereas in purely resistive elements the voltage can change instantaneously). Note: The charge and voltage are inseparable in nature. Whenever charge exists in an element then there should be some voltage in it, or whenever voltage exist in an element there should be some charge stored in that element. The capacitance will exist between any two conductors separated by dielectric. The commercial capacitors will consist of two conducting plates separated by a dielectric. The unit of capacitance is farad and it is denoted by F. The capacitance of a capacitor is defined as the ratio of stored charge and the potential difference across its plates. The capacitance of a capacitor with a charge of Q coulombs and a potential difference of V volts across its plate is given by, Q Capacitance, C = ..... (3.12) V In equation (3.12), if Q = 1 coulomb and V = 1 volt then C = 1 farad. Therefore, “a capacitor is said to have a capacitance of one farad if a charge of one Coulomb establish a potential difference of one volt between its plates”. Practically, one Farad is a large value and so the smaller values µF (micro-Farad), nF (nano-Farad) and pF (pico-Farad) are used.

Circuit Theory

3. 11

3.6.1 Voltage-Current Relation in a Capacitance We know that, a potential difference exist between two charged conductors. The charge is directly proportional to voltage and the proportionality constant is capacitance C. Consider a capacitor with capacitance C and carrying a current of i as shown in Fig. 3.13. Let, q and v be the instantaneous value of charge and voltage in the capacitor respectively. Since charge is directly proportional to voltage we can write, i qαv + ∴ q = Cv v, q C E

On differentiating the above equation we get dq = C dν dt dt

dq = i ; Refer equation (1.5) dt

Fig. 3.13.

..... (3.13) i = C dν dt The equation (3.13) gives the voltage-current relation in a capacitance. From equation (3.13), we can say that when, v = constant, then dν = 0, and so i = 0. Hence, for constant or dc voltage dt the capacitance will behave as an open circuit (in steady state). Therefore, in steady state analysis of circuits excited by dc sources the capacitances are considered as open circuits (or simply the capacitances are neglected). `

On rearranging equation (3.13), we get i dt = C dv

On integrating both sides we get,

# i dt = # C dν ` ν = 1 C

⇒

# i dt = C # dν

# i dt

⇒

# i dt = Cν ..... (3.14)

3.6.2 Energy Stored in a Capacitance In a capacitance, the energy is stored as electric field. Let, dw be the energy stored in the capacitance in the time interval dt. Now the energy, dw is given by the product of power, p and time, dt. ∴ dw = p dt = vi dt

Put, p = ni

= νC dν dt dt = C v dv

Put, i = C dν dt

..... (3.15)

Let the voltage v rises from zero to a steady value of V(where V is rms value in case of ac) to establish an energy of W. Now, the stored energy (or total work) is obtained by integrating the equation (3.15) between limits 0 to V.

3. 12

Chapter 3 - Basic Concepts of AC Circuits V

#

` W =

V

# ν dν

Cν dν = C

0

V

2 2 = C ; ν E = CV 2 0 2

0

..... (3.16)

` W = 1 CV2 2

The equation (3.16), can be used to compute the energy stored in a capacitance when a steady voltage V exists across it.

3.7 Voltage-Current Relation of R, L and C in Various Domains The circuit variables like voltage, current, power and energy are functions of time, t. The time domain is the practical domain where we can physically realize any system or phenomena or activity. The voltage-current relations of the fundamental parameters in time domain are differential equations. The solutions of differential equations are tedious when compared to algebraic equations. Hence, it will be convenient if we transform the differential equation to algebraic equations. One such transform is Laplace transform. A brief discussion about Laplace transform is presented in Appendix 3. Let, i = i(t) = Current in time domain. v = v(t) = Voltage in time domain. L{v(t)} = V(s) = Voltage in s-domain or Laplace domain. L{i(t)} = I(s) = Current in s-domain or Laplace domain. 3.7.1 Voltage-Current Relation of Resistance Consider a resistance R connected to a source of voltage v(t) as shown in Fig. 3.14. i(t)

I(s)

+

+

I

+

v(t)

~

E

R

v(t)

R

E

Fig. a : Resistance in time domain.

+

V(s) E

R

Fig. b : Resistance in s-domain.

V E

Fig. c : Resistance in frequency domain.

Fig. 3.14 : Voltage-Current relation of resistance in various domains.

Let, i(t) = Current through the resistance. v(t) = Voltage across the resistance. By Ohm’s law, we can write v(t) = R i(t)

For simplicity

v = Ri

..... (3.17)

On taking Laplace transform of equation (3.17), we get ..... (3.18)

V(s) = R I(s) On substituting s = jω in equation (3.18), we get V^ jωh = R I^ jωh

For simplicity

V = RI

Circuit Theory

3. 13

In Summary, v = Ri

; Voltage-Current relation of resistance in time domain.

V(s) = R I(s)

; Voltage-Current relation of resistance in s-domain.

V = RI

; Voltage-Current relation of resistance in frequency domain.

3.7.2 Voltage-Current Relation of Inductance Consider an inductance L connected to a source of voltage v(t) as shown in Fig. 3.15. i(t)

I(s)

I

+

+

+

+

v(t)

~

-

v(t)

L

sL

-

V(s) -

jwL

V -

Fig. a : Inductance in time domain.

Fig. b : Inductance in Fig. c : Inductance in s-domain. frequency domain. Fig. 3.15 : Voltage-Current relation of inductance in various domains.

Let, i(t) = Current through the inductance. v(t) = Voltage across the inductance. By Faraday’s Law, we can write ν (t) = L d i (t) dt

For simplicity

ν = L di dt

.....(3.19)

On taking Laplace transform of equation (3.19), with zero initial conditions, we get V(s) = L sI(s) ∴ V(s) = sL I(s)

.....(3.20)

where, sL = Inductive reactance in s-domain On substituting s = jω in equation (3.20), we get V^ jωh = jωL I^ jωh

For simplicity

V = jωL I

where, ωL = XL = Inductive reactance In Summary, v = L di dt

; Voltage-Current relation of inductance in time domain.

V(s) = sL I(s)

; Voltage-Current relation of inductance in s-domain.

V = jωLI

; Voltage-Current relation of inductance in frequency domain.

3. 14

Chapter 3 - Basic Concepts of AC Circuits

3.7.3 Voltage-Current Relation of Capacitance Consider a capacitance C connected to a source of current i(t) as shown in Fig. 3.16. I(s)

i(t) i(t)

~

C

+ v(t) -

1 sC

I

+ V(s) -

1 jwC

+ V

-

Fig. a : Capacitance in Fig. b : Capacitance in Fig. c : Capacitance in frequency domain. s-domain. time domain. Fig. 3.16 : Voltage-Current relation of capacitance in various domains.

Let, i(t) = Current through the capacitance. v(t) = Voltage across the capacitance. dν (t) For simplicity i = C dν dt dt On integrating and rearranging equation (3.21), we get,

Now, i (t) = C

ν = 1 C

..... (3.21)

# i dt

On taking Laplace transform of equation (3.21) with zero initial conditions, we get, I (s) = CsV (s) ` V (s ) = 1 I ( s) sC

..... (3.22)

where, 1 = Capacitive reactance in s- domain sC On substituting s = jω in equation (3.22), we get, 1 I ( jω) jω C

V ( jω ) =

where,

For simplicity

V =

1 I = −j 1 I jω C ωC

1 = X C = Capacitive reactance ωC

In Summary, v = 1 C

# i dt

V (s) = 1 I (s) sC V = −j 1 I ωC

; Voltage-Current relation of capacitance in time domain. ; Voltage-Current relation of capacitance in s-domain. ; Voltage-Current relation of capacitance in frequency domain.

Circuit Theory

3. 15

TABLE 3.1 : R, L, C REPRESENTATION IN VARIOUS DOMAINS S.No.

Parameter

1.

Resistance, R

2.

Inductance, L

3.

Capacitance, C

Time-domain

I(s) + V(s) -

v -

i +

s-domain

R

I + V

R

i + v -

I(s) + V(s) -

L

I(s) + V(s) -

i + v -

1 sC

-

R I + V -

sL

C

3.8

Frequency domain

jwL

I +

V

-

1 jwC

Sinusoidal Voltage and Current in Frequency Domain The instantaneous value of sinusoidal voltage in time domain is represented as, v(t) = Vmsin(ωt ± φ) In frequency domain the rms value of sinusoidal voltage can be represented as, V ^ j ω h = V+ ! φ

For simplicity

V = V+ ! φ

where, V = Vm / 2 .....(3.23) V = Vm + ! φ = V+ ! φ 2 The instantaneous value of sinusoidal current in time domain is represented as, ∴ Vm sin ^ωt ! φh

⇒

i(t) = I m sin(ωt ± φ)

In frequency domain the rms value of sinusoidal current can be represented as, I ^ jω h = I + ! φ

For simplicity

I = I+ ! φ

where, I = I m / 2 . ∴ I m sin ^ωt ! φh

⇒

I = I m + ! φ = I+ ! φ 2

.....(3.24)

3.8.1 Phase and Phase Difference The sinusoidal quantities are rotating vectors and the y-component of the rotating vector gives the instantaneous value. The plot of the instantaneous value on an angular scale or time scale gives the waveform of the sinusoidal quantity. These waveforms are continuous waveform (of course with no beginning and end). For analysis purpose we assume a starting point (or origin) for the waveform. Normally, the origin of waveform is considered as time, t = 0. The position of the rotating vector at time t = 0, decides the phase (or phase angle) of the vector. Three different positions of a vector at time t = 0 and their waveforms are shown in Fig. 3.17. “The phase (or phase angle) of the vector is the angular position of the vector with respect to reference at time t = 0”. In Fig. 3.17(a), the vector lies on the reference line at time t = 0, hence the phase angle of the vector is zero.

3. 16

Chapter 3 - Basic Concepts of AC Circuits

In Fig. 3.17(b), the vector is lying at an angle φ ahead of reference at t = 0, hence the phase angle of the vector is φ degrees (or radians) leading. In Fig. 3.17(c), the vector is lying at an angle φ behind the reference at t = 0, hence the phase angle of the vector is φ degrees (or radians) lagging. y

y Vm

w x

y

w

Vm f

ref

x

ref

-f Vm

v

v

v

Vm

Vm

Vm

wt -Vm

f -Vm

v = Vmsin wt

x

ref

wt

wt f -Vm

v = Vmsin (wt+f)

t=0

t=0

w

v = Vmsin (wt - f)

t=0

Fig. a : The vector lies on Fig. b : The vector is lying ahead of Fig. c : The vector is lying behind their the reference line at time t = 0. reference (in the direction of rotation) reference (in the direction of rotation) by an angle f at time t = 0. by an angle f at time t = 0.

Fig. 3.17 : Different positions of a vector at t = 0.

In electrical engineering (or circuit theory) we are interested in phase difference between two sine waves, rather than the phase of a single sine wave. “The phase difference between two (or more) sine waves can be estimated only if the frequency of sine waves are same (or speed of rotation should be same), but the magnitudes need not be same”. Consider two voltage vectors with maximum values Vm1 and Vm2 rotating at same speed but having an angular spacing of φ radians between each other as shown in Fig. 3.18(a). The two vectors represent two sinusoidal voltages of same frequency, ω as shown in Fig. 3.18(b). With reference to Fig. 3.18, we can say that, the phase of voltage V m1 is φ1 and that of voltage Vm 2 is φ2. The phase difference between the two voltages is φ, where φ = φ2 − φ1. “The phase difference φ can be expressed either in degrees or in radians”. Also we can say that, the wave-2 is leading the wave-1 by an angle φ or the wave-1 is lagging the wave-2 by an angle φ. If “the phase difference between two sinusoids is zero then they are said to be in-phase”, because both the sinusoids have same phase. y

f

Vm2

Vm1 Vm2

w

wave-1 wave-2

Vm1 w f2

f

f1

f

f

wt

f

f2

f

f1

x

ref

Fig. a : Position of two vectors rotating at same speed at t = 0.

t=0 wt = 0

Fig. b : Waveform of the vectors shown in Fig. a. Fig. 3.18 : Two sinusoidal voltages with a phase difference of f.

Circuit Theory

3. 17

3.8.2 Phasor Representation of Sinusoidal Quantities The sinusoidal quantities are rotating vectors. When it is represented in a complex plane, the position of the vector at any time instant can be specified by two quantities: magnitude and phase angle. The magnitude of a rotating vector is constant but the phase angle varies with rotation. Hence, “the rotating vectors are functions of phase angle and so they are called phasors”. The sinusoidal quantity is given by the y-component (vertical component) of rotating vector and so the sinusoidal quantity is also called a phasor (or sinor or vector). The x-component (horizontal component) of a rotating vector gives the cosinusoidal wave. In general, both the sine wave and cosine wave are called phasors (or sinors or vectors). Throughout this book the terms phasor and vector are used synonymously to denote sinusoidal quantity. Since the sinusoidal quantity or phasor is a rotating vector we can take the position of the phasor at t = 0, for analysis purpose as shown in Fig. 3.19. Imaginary axis Complex plane a =Vmcos f b =Vmsin f w jb

a =Vmcos f b =Vmsin f

Imaginary axis

a

-f

Complex plane

m

w

-jb

Vm w

Vm

f a

Real axis

V

Real axis

Fig. a : A leading phasor.

a = Vm

a

Complex plane Real axis

Fig. b : A phasor with zero phase.

Imaginary axis

Fig. c : A lagging phasor.

Fig. 3.19 : Various position of a sinusoidal voltage phasor at t = 0.

From the theory of complex numbers the phasors (or vectors) shown in Fig. 3.19, can be represented as shown below: Vm +φ V m cos φ + j Vm sin φ Vm ejφ

; Polar form ; Rectangular form ; Eular form

In the above representation the phase angle φ can be positive, zero or negative. • When φ is positive, the voltage vector will be in the position shown in Fig. 3.19(a) at t = 0. • When φ is zero, the voltage vector will be in the position shown in Fig. 3.19(b) at t = 0. • When φ is negative, the voltage vector will be in the position shown in Fig. 3.19(c) at t = 0.

We know that the rms values (or effective values) are used for practical measurements/ applications. For example, the rated voltage 230 V of a ceiling fan at home refers to rms value of the sinusoidal voltage. The relation between rms value and maximum value for a sinusoidal voltage is V = Vm / 2 , where V is rms value and Vm is maximum value. From the relation, V = Vm / 2 , we can say that the rms value is the scaled down value of maximum value. Hence the rms value can also be represented as a phasor of same phase as that of maximum value but with reduced magnitude as shown in Fig. 3.20. But remember that the rotation of rms value of phasor will not produce the instantaneous value.

3. 18

Chapter 3 - Basic Concepts of AC Circuits Imaginary axis Complex plane a =V cos f b =V sin f

Imaginary axis

a =V cos f b =V sin f

a

-f

Complex plane

Real axis

V

jb -jb

V

V

f a

Real axis

Complex plane a

a=V

Real axis

Imaginary axis

Fig. a : A leading phasor. Fig. b : A phasor with zero phase. Fig. c : A lagging phasor. Fig. 3.20 : Phasor representation of rms values of sinusoidal voltage.

“In order to draw the rms vector take a snap-shot of rotating phasor at t = 0 and divide its magnitude by 2 ”. 3.8.3 Phasor Diagram of a Circuit When the excitation source in a circuit is sinusoidal, the voltage, current and power in various elements of the circuit will also be sinusoidal. Hence the various voltages and currents in a circuit can be represented by phasors. While drawing these phasors, one of the quantity (either a voltage or current) is chosen as reference and the phasors of the other quantities are drawn in relation to the chosen reference phasor. Such a diagram is called phasor diagram. Note : The phasor diagrams throughout this book are drawn only using rms phasors. I

+

VL

-

V is reference phasor.

R

+ V = VÐ0 o

VR

+

~

L

-

o

90

VL -

I lags V by an angle f .

V

-f

VR is in phase with I. I

V = VR + VL

VL leads I by an angle 90 o . VR

Fig. a : RL series circuit and its phasor diagram. VR

I

+

-

R

+ V = VÐ0 o

VR

I

+ VC

C

~

I leads V by an angle f.

V

o

90

-

-

V is reference phasor.

f

VR is in phase with I. V = VR + VC

VC lags I by an angle 90 o .

VC

Fig. b : RC series circuit and its phasor diagram. I

+ V = VÐ0o

~

V is reference phasor.

IC IL

IR

R

IR

IC

L

-

C

V

-f IL - I C

I L lags V by an angle 90 o .

I I = IR + IL + IC

I R is in phase with V.

IL

Fig. c : RLC parallel circuit and its phasor diagram. Fig. 3.21 : Phasor diagram of simple circuits.

I C leads V by an angle 90 o . I lags V by an angle f.

Circuit Theory

3.9

3. 19

Power, Energy and Power Factor Power is the rate at which work is done or power is rate of energy transfer. Let, w = Instantaneous value of energy. q = Instantaneous value of charge. dq Now, Instantaneous power, p = dw = dw # dt dq dt dq d w We know that, = ν and =i dq dt

Refer Chapter - 1 Equations (1.5) and (1.7)

∴ Instantaneous power, p = v i i.e., the power is the product of voltage and current. In circuits excited by dc sources the voltage and current are constant and so the power in constant. This constant power is called average power or power and it is denoted by P. ∴ In dc circuits, Power, P = V I In circuits excited by ac sources, the voltage and current are sinusoidal quantities which varies with time. When voltage and current are time varying quantities, the power is also a time varying quantity. For time varying quantities, the power is defined as average over a period of time. Since the average value of sinusoidal voltage and current are zero, we can take, the rms value of voltage and current. We know that, the rms values of voltage and current are complex and so the power is also complex. “The complex power is denoted by S and it is defined as the product of rms voltage and the conjugate of rms current”. ` Complex power, S = V I* where, I* = Conjugate of I Let, V = V+δ I = I+γ

then, I* = I+ − γ where, δ is phase of voltage and γ is phase of current. *

` S = V I = V+δ # I+ − γ = VI+ (δ − γ)

Let, δ − γ = φ where, φ = Phase difference between V and I ` S = VI+φ

Let,

S = S = VI

where, S = Apparent power and expressed in Volt-Ampere i.e., VA. (The larger units of S are kVA and MVA).

3. 20

Chapter 3 - Basic Concepts of AC Circuits

“The apparent power S is defined as the product of magnitude of rms voltage and rms current”. Since S is complex it can be expressed as a vector in complex plane as shown in Fig. 3.22. Imaginary axis Complex plane P = S cos f Q = S sin f S Q

Imaginary axis Complex plane

Imaginary axis Complex plane

P -f

S f

S

S

P

Real axis

Fig. a : Vector of S when f is positive.

P=S Q=0

P

Real axis

Real axis

Fig. b : Vector of S when f is zero.

Q

S

P = S cos f Q = -S sin f

Fig. c : Vector of S when f is negative.

Fig. 3.22 : Vector of complex power S for various values of f.

The real part of S is called active power or simply power. The imaginary part of S is called reactive power. The power is denoted by P and expressed in the units of watts, W. The reactive power is denoted by Q and expressed in the units of volt-ampere-reactive, VAR. With reference to Fig. 3.22, we can write, S = S cos φ + j S sin φ

Let, S = P + jQ ` P = S cos φ Q = S sin φ

We know that, S = S = VI ∴ P = VI cos φ in W ∴ Q = VI sin φ in VAR

In Fig. 3.22, the triangle formed by P, Q and S is also called power triangle. The larger units of power, P is kW or MW and larger units of reactive power, Q is kVAR or MVAR. ` P = VI cos φ in W = Q = VI sin φ in VAR =

VI cos φ VI cos φ in kW = in WM 3 10 106 VI sin φ VI sin φ in kVAR = in MVAR 103 106

In dc circuits, the V and I are constants and there is no phase difference between V and I. Hence φ = 0 and so, cos φ = 1 and sin φ = 0. Therefore, in dc circuits the complex power or apparent power is equal to active power and reactive power is zero.

Circuit Theory

3. 21

In ac circuits, the phase angle φ may be positive, zero or negative. (Remember that φ is phase difference between V and I. ) When φ is positive, • • • •

the current lags voltage. the circuit is inductive. the active power is positive. the reactive power is positive.

When φ is zero, • • • •

the current is in-phase with voltage. the circuit is resistive. the active power is positive. the reactive power is zero.

When φ is negative, • • • •

the current leads the voltage. the circuit is capacitive. the active power is positive. the reactive power is negative.

In summary, we can say that, the active power, P is always positive. The reactive power, Q is positive in inductive circuits and negative in capacitive circuits. In resistive circuits the active power is equal to apparent power. The power is rate of work done and the energy is total work done. Hence “the energy is given by the product of power and time”. When time is expressed in second, the unit of energy is watt-second and when the time is expressed in hours, the unit of energy is watt-hour. ∴ Energy, E = Pt in W-s or W-h The larger unit of energy is kWh and commercially one kWh of electrical energy is called one unit. “The ratio of active power and apparent power is defined as power factor”. The power factor is a measure of active power in the apparent power. ` Power factor =

Active power = P S Apparent power

From the power triangle of Fig. 3.22, we get, P = S cos φ S cos φ ` Power factor = P = = cos φ S S

Here, φ is the phase difference between voltage and current. Hence from the above equation we can say that, “the power factor is also defined as cosine of the phase difference between voltage and current”.

3. 22

Chapter 3 - Basic Concepts of AC Circuits

3.10 Resistance Connected to Sinusoidal Source

i

Consider a resistance R connected to ac source of voltage, v = Vm sin ωt as shown in Fig. 3.23. Since the resistance is connected across (or parallel to) the source, the instantaneous value of voltage across resistance is also v volts. \ Voltage, n = Vm sin ωt

.....(3.25)

+

+ v

v

R

~ E

E

Fig 3.23 : Resistance connected to ac source.

By Ohm’s law, the instantaneous current through the resistance is given by, Current, i = ν R = Vm sin ωt = Vm sin ωt R R

v = Vm sin ωt

= I m sin ωt

..... (3.26)

where, Vm = I m = Maximum value of current R From equations (3.25) and (3.26), we can say that, the voltage and current in a resistance are sinusoidal quantities of same frequency and are in-phase (i.e., the phase difference between voltage and current in a resistance is zero). The sinusoidal voltage and current can be expressed in polar form as shown below: Vm = Vm +0 o I m = I m +0 o

Since rms values are practically used (than the maximum values), the rms values of voltage and current are shown in circuit of Fig. 3.24. IÐ0o

+ VÐ0 o R -

+ VÐ0 o

~

-

Fig. a : Circuit showing rms value of voltage and current in resistance.

I

V

Fig. b : Phasor diagram of voltage and current in circuit of Fig. a.

Fig. 3.24 : Resistance connected to ac source.

Rms value of voltage across resistance, V = V+0o Rms value of current through resistance, I = I+0o The instantaneous value of power in the resistance is given by the product of the instantaneous value of voltage and current in the resistance. ∴ Instantaneous power, p = v × ip = V m sin ωt × Im sin ωt ∴ Instantaneous power, p = V m Im sin 2 ωt = Vm Im sin2q = Pm sin2q where,

q = ωt ; Pm = Vm Im

..... (3.27)

Circuit Theory

3. 23

From equation (3.27), we can say that the instantaneous power is squared sine wave and so the instantaneous power is always positive (because on squaring the negative cycle it becomes positive). v

Vm +

+

+ p

0

+ 3p

2p -

-

4p -

-

-

-

wt

i Im +

+

+ p

0

-

-

+ 3p

2p

4p

wt

4p

wt

p Pm = Vm Im

+

+

+

0

+

+

p

+

2p

+

3p

+

Fig. 3.25 : Waveform of voltage, current and power in a resistance.

Practically, power is measured as an average value which is obtained by taking average value of the equation (3.27), over one period of voltage or current (or the average can be taken for half period of voltage or current). r

#

` Power, P = 1 π

r

#V I

p dθ = 1 π

0

m m sin

2

θ dθ

0 r

2θ dθ # 1 − cos 2

= Vm I m π 0

r

=

Vm I m ^1 - cos 2θh dθ 2π 0

#

r = Vm I m 8θ − sin 2θ B 2π 2 0

= Vm I m 8π − sin 2π − 0 + sin 0 B e 2π 2 2 = Vm I m = Vm I m 2 2 2

sin 2π = 0 and sin 0 = 0

= VI rag

` P = VI e power

Alternatively, the expression for power can be obtained from complex power, Complex power, S = V I* = V+0 o # (I+0 o) * = V+0 o # I+0 o = VI+0 o

3. 24

Chapter 3 - Basic Concepts of AC Circuits

We know that, S = S = VI and +S = φ = 0 o ∴ Power, P = S cos φ = VI cos 0 o = VI Reactive Power, Q = S sin φ = VI sin 0 o = 0 “The resistance consumes only active power and the reactive power in a resistance is zero”.

3.11 Inductance Connected to Sinusoidal Source Consider an inductance L connected to ac source of voltage, v = Vm sin ωt as shown in Fig. 3.26. Since the inductance is connected across (or parallel to) the source, the instantaneous value of voltage across inductance is also v volts. \ Voltage, v = Vm sin ωt

..... (3.28)

The current i through the inductance is given by, i = 1 L = 1 L

+

v

~

E

L

i + v E

Fig. 3.26 : Inductance connected to ac source.

# ν dt # V sin ωt dt

= Vm L

m

# sin ωt dt

ν = Vm sin ωt

= Vm 8 − cos ωt B L ω

= Vm ^− cos ωth ωL = Vm ^− cos ωth XL = Vm sin ^ωt − 90 oh XL

sin (A − 90 o) = − cos A

= I m sin (ωt − 90 o)

.....(3.29)

where, ωL = X L = Inductive reactance Vm = I = Maximum value of current m XL

..... (3.30)

From equations (3.28) and (3.29), we can say that, the voltage and current in an inductance are sinusoidal quantities of same frequency, but have a phase difference of 90o. “The current in an inductance lags behind the voltage by 90o ” (or the voltage across the inductance leads the current in an inductance by 90o ). The sinusoidal voltage and current of equations (3.28) and (3.29) can be expressed in polar form as shown below: Vm = Vm∠0 o I m = I m∠− 90 o

Circuit Theory

3. 25

Since rms values are practically used (than the maximum values), the rms values of voltage and current are shown in circuit of Fig. 3.27. I = I Ð - 90 o

o

V

- 90 +

V = VÐ0 o

~

-

+ jwL

V = VÐ0 o -

I

Fig. a : Circuit showing rms value of voltage and current in inductance.

Fig. b : Phasor diagram of voltage and current in circuit of Fig. a. Fig. 3.27 : Inductance connected to ac source.

Rms value of voltage across inductance, V = V∠0o Rms value of current through inductance, I = I∠− 90 o From equation (3.30), we get, X L = Vm Im

Since, V = Vm

2 and I = I m

2 we can write,

XL = V I o Now, V = V+0 o = V +90 o = X L +90 o = jX L = jωL I I I+ − 90

..... (3.31)

From equation (3.31), we can say that, “the inductive reactance is given by the ratio of sinusoidal voltage to current in an inductance”. The instantaneous value of power in the inductance is given by the product of the instantaneous value of voltage and current in the inductance. ∴ Instantaneous power, p = v × i = Vm sin ωt × I m sin (ωt − 90 o )

sin(A − 90o ) = −cos A

= Vm I m sin ωt (−cos ωt) = −Vm Im sin ωt cos ωt ` p = − Vm I m sin 2ωt 2 = − Vm I m sin 2ωt = − Vm I m sin 2θ 2 2

..... (3.32)

where, q = ωt From equation (3.32), we can say that the instantaneous power is also a sinusoidal quantity whose frequency is double that of voltage or current.

3. 26

Chapter 3 - Basic Concepts of AC Circuits v

Practically, power is measured as an average value. With reference to Fig. 3.28, we can say that, for half-period of voltage or current wave, the power wave undergoes one full cycle and it is sinusoidal. Hence the average value of power over half-period or full-period of voltage will be zero.

Vm +

#

+ p -

+

2p

-

3p

-

wt

4p

-

i Im +

r

Power, P = 1 π

+

+

+

p

-

-

+

3p

2p -

wt

4p

-

p dθ

0

p

r

# − V2 I

= 1 π

m

m

sin 2θ dθ

0

Vm Im 2

+

# sin 2θ dθ

= − Vm I m 2π

+ p

r

+

-

-

+ 3p

2p

4p

wt

-

-

0 r = − Vm I m 8 − cos 2θ B 2π 2 0

Fig. 3.28 : Waveform of voltage, current and power in an inductance.

= − Vm I m 8 − cos 2π + cos 0 B 2π 2 2 V − 1 1 m Im = − + B = 0 2π 8 2 2

Alternatively, the expression for power can be obtained from complex power. Complex Power, S = V I* = V+0 o # ^I+ − 90 oh* = V∠0o × I∠ + 90o = VI∠90o We know that, S = S = VI and ∠S = φ = 90o ∴ Power, P = S cos φ = VI cos 90 o = 0 Reactive Power, Q = S sin φ = VI sin 90 o = VI The inductance consumes only reactive power and “the active power in the pure inductance is zero.” The reactive power of inductance is positive which means that it absorbs reactive power.

3.12 Capacitance Connected to Sinusoidal Source Consider a capacitance C connected to ac source of voltage, v = Vm sin ωt as shown in Fig. 3.29. Since the capacitance is connected across (or parallel to) the source the instantaneous value of voltage across capacitance is also v volts. ∴ Voltage, n = Vm sin ωt

.....(3.33)

i +

v

~

E

C

+ v E

Fig. 3.29 : Capacitance connected to ac source.

Circuit Theory

3. 27

The current i through the capacitance is given by, i = C dv dt d = C Vm sin ωt v = Vm sin ωt dt = C Vm d sin ωt dt = C Vm cos ωt ´ ω Vm cos ωt 1/ωC = Vm cos ωt XC = Vm sin ^ωt + 90 oh XC = Im sin (ωt + 90o) =

.....(3.34)

where, 1 = XC = Capacitive reactance. ωC Vm = I = Maximum value of current. m XC

.....(3.35)

From equations (3.33) and (3.34), we can say that, the voltage and current in a capacitance are sinusoidal quantities of same frequency, but have a phase difference of 90 o. “The current in a capacitance leads the voltage across it by 90o” (or the voltage across the capacitance lags the current in a capacitance by 90o ). The sinusoidal voltage and current of equations (3.33) and (3.34) can be expressed in polar form as shown below: Vm = Vm +0 o I m = I m + + 90 o

Since rms values are practically used (than the maximum values), the rms values of voltage and current are shown in circuit of Fig. 3.30. Rms value of voltage across capacitance, V = V∠0 o Rms value of current through capacitance, I = I∠+ 90o I = I Ð90 o

+ V = VÐ0 o

~

-

+ V = VÐ0 o -

I 1 jwC

90

Fig. a : Circuit showing rms value of voltage and current in capacitance.

o

V

Fig. b : Phasor diagram of voltage and current in circuit of Fig. a.

Fig. 3.30 : Capacitance connected to ac source.

3. 28

Chapter 3 - Basic Concepts of AC Circuits

From equation (3.35), we get, X C = Vm Im

Since, V = Vm

2 and I = I m

2 we can write,

XC = V I o Now, V = V +0 o = V + − 90 o = XC + − 90 o = − jXC = − j 1 = 1 I ωC jωC I I +90

..... (3.36)

From equation (3.36), we can say that, “the capacitive reactance is given by the ratio of sinusoidal voltage to current in a capacitance”. The instantaneous value of power in the capacitance is given by the product of the instantaneous value of voltage and current in the capacitance. Let, p = Instantaneous power in the capacitance ∴ p = v × i ∴Inst = V m sin ωt × I m sin(ωt + 90o )

sin(A + 90o) = cosA

∴Inst = Vm I m sin ωt cos ωt

sin 2A = 2 sinA cosA

= Vm I m 2 sin ωt cos ωt 2 = Vm I m sin 2ωt where,q = ωt 2

v Vm +

+

+ p

0

= Vm I m sin 2θ 2

4p

-

wt

-

+

+ p

0 -

+ -

3p

4p -

wt

-

+

+

2p

3p -

4p -

wt

Fig. 3.31 : Waveform of voltage, current and power in a capacitance.

# sin 2θ dθ 0

+ 2p

-

p

r

Vm I m sin 2θ dθ = Vm I m 2 2π

p

Vm Im 2

# p dθ r

+

+ 0

0

0

-

i

r

#

3p

Im

Practically, power is measured as an average value. With reference to Fig. 3.31, we can say that, for half period of voltage or current wave, the power wave undergoes one full cycle and it is sinusoidal. Hence, the average value of power over half period or full period of voltage will be zero.

= 1 π

-

..... (3.37)

From equation (3.37), we can say that, the instantaneous power is also a sinusoidal quantity whose frequency is double that of voltage or current.

Power, P = 1 π

-

+

2p

Circuit Theory

3. 29 r = Vm I m 8 − cos 2θ B 2π 2 0

= Vm I m 8 − cos 2π + cos 0 B 2π 2 2 = Vm I m 8 − 1 + 1 B = 0 2π 2 2

Alternatively, the expression for power can be obtained from complex power. Complex Power, S = V I* = V+0 o # ^I+90 oh* = V∠0 o × I∠−90o = VI∠−90o We know that, S = S = VI and ∠S = φ = −90 o ∴ Power, P = S cos φ = VI cos (−90 o ) = 0 Reactive power, Q = S sin φ = VI sin (−90 o ) = −VI The capacitance has only reactive power and “the active power in the pure capacitance is zero”. The reactive power of capacitance is negative which means that it delivers reactive power.

3.13 Impedance “The impedance is defined as the ratio of (sinusoidal) voltage and current”. The impedance is a frequency domain parameter but it is not a sinusoidal quantity. “The impedance is also defined as the total opposition offered to flow of (sinusoidal) current”. Hence the impedance is measured in the unit of Ohms, (similar to that of resistance). The impedance is a complex quantity and denoted by Z. The real part of impedance is resistance and the imaginary part of impedance is reactance. The unit of resistance, reactance and impedance are Ohm. There are two types of reactance: the Inductive reactance and Capacitive reactance. The inductive reactance is denoted by XL and equal to ω L. The capacitive reactance is denoted by XC and equal to 1/ωC. The inductive and capacitive reactances have exact opposite behaviours. Therefore, when expressed as a complex quantity the inductive reactance takes positive value and capacitive reactance takes negative value [Refer equations (3.31) and (3.36)]. ∴ Impedance, Z = R + jX

..... (3.38)

where, R = Resistance X = Reactance Also, Z = R + jX = R + jXL Z = R + jX = R − jXC

; when reactance is inductive. ; when reactance is capacitive.

Z = R + jX = R + j(XL – XC) ; when reactance is sum of inductance and capacitance.

3. 30

Chapter 3 - Basic Concepts of AC Circuits

The symbol used to represent the impedance is rectangle as shown in Fig. 3.32. The impedance is connected to other parts of circuits using resistance-less wires.

Z a R C jX

The magnitude of impedance is denoted by Z (i.e., without over-bar). The argument of impedance is called impedance angle and it is denoted by q.

Fig. 3.32 : Symbol for impedance.

In equation (3.38), the complex impedance is expressed in rectangular form. It can be expressed in polar form as shown below: Z = R + jX =

R2 + X2 + tan- 1 X = Z +θ R

where, Z = Z =

R2 + X2

and

` Magnitude of impedance, Z = Z =

..... (3.39)

θ = +Z = tan- 1 X R

R2 + X2

Impedance angle, θ = +Z = tan- 1 X R Since impedance is a complex quantity it can be represented as a point in a complex plane with polar coordinates Z and q as shown in Fig. 3.33. The line joining the origin and Z will be a vector of length Z and making an angle q with reference. Now, the vector Z can be resolved into horizontal and vertical components. The horizontal component is resistance R and vertical component is reactance X. The right-angled triangle formed by R, X and Z is called impedance triangle.

Imaginary axis

Z

Z

X

q

Real axis

R

Fig. 3.33 : Impedance triangle.

3.13.1 Impedance Connected to Sinusoidal Source Consider an impedance Z connected to ac source of voltage V volts rms value as shown in Fig. 3.34. Since the impedance is connected across (or parallel to) the source, the voltage across the impedance is also V volts. I

By Ohm’s law, the current through the impedance is given by, ..... (3.40) I = V Z Let, V be the reference phasor and q be the impedance angle. ` V = V+0 o

and

Z = Z+ ! θ

..... (3.41)

(+for inductive reactance and − for capacitive reactance) From equations (3.40) and (3.41) we can write, o I = V = V+0 = V + " θ = I+ " θ Z+ ! θ Z Z V where, I = Z

+ V

~

E

+ V E

Z

Fig. 3.34 : Impedance connected to ac source.

Circuit Theory

3. 31

The power factor angle is the phase difference between V and I. Let, φ be the phase difference between V and I. Now, φ = +V − +I = 0 − (" θ) = ! θ “It is interesting to observe that the power factor angle is same as that of impedance angle”. From the above analysis we can say that, the current through the impedance leads or lags the voltage by an angle φ. The impedance is a combination of resistances and reactances. Depending on the combination, the phase of current may be in-phase or leading or lagging the voltage. TABLE 3.2 : VARIOUS COMBINATIONS OF RESISTANCE AND REACTANCE Impedance, Z I C

V

E

Phase of current with respect to voltage

Phasor diagram of V and I

I in-phase with V

V

I

Z = R I C

V

E

I lags V

V

f

Z = R + j XL I C

V

E

I

I leads V

Z = R − j XC I C

V

V

V

E

Z = R + j XL − jXC and XL = X C (This condition is called resonance) I C

I

f

I in-phase with V V

I

E V

Z = R + j XL − jXC and XL > XC I C

V

I lags V

f I

E

Z = R + j XL − jXC and XL < XC

I

I leads V

f

V

3. 32

Chapter 3 - Basic Concepts of AC Circuits

3.14 Conductance, Susceptance and Admittance In certain circuits the inverse parameters will be useful for analysis. The inverse of resistance is conductance, the inverse of reactance is susceptance and the inverse of impedance is admittance. The MKS unit of conductance, susceptance and admittance is mho, (Ohm spelt in the reverse order). The SI unit of conductance, susceptance and admittance is Siemens, S. The circuit symbol used for an inverse parameter is same as that of original (or parent) parameter. The letters used to denote conductance, susceptance and admittance are G, B and Y respectively. Conductance, G = 1 ; R 3.14.1 Conductance

Susceptance, B = 1 ; X

Admittance, Y = 1

Z

The relationship between voltage and current in a conductance can be obtained from Ohm’s law. Consider a conductance carrying a current of I amperes as shown in Fig. 3.35. The conductance of the element is G and resistance is R = 1/G. By Ohm’s law, the voltage across the element is given by the product of current and resistance. I

+

` Voltage across conductance, V = IR = I G ` I = VG

or

V = I G

..... (3.42)

V _

Ga

1 R

Fig. 3.35 : Conductance.

The equation (3.42), gives the relation between voltage and current in a conductance. 3.14.2 Admittance The admittance is the inverse of impedance and so “the admittance is defined as the ratio of (sinusoidal) current and voltage”. The unit of admittance is mho or Siemens. The admittance is a complex quantity and denoted by Y. The real part of admittance is conductance and the imaginary part of admittance is susceptance. The unit of conductance, susceptance and admittance is mho or Siemens. There are two types of susceptance: Inductive susceptance and Capacitive susceptance. The inductive susceptance is denoted by B L and equal to 1/ωL. The capacitive susceptance is denoted by BC and equal to ωC. The inductive and capacitive susceptances have exact opposite behaviours. Therefore, when expressed as a complex quantity the inductive susceptance takes negative value and capacitive susceptance takes positive value [Refer equations (3.31) and (3.36)]. Admittance, Y = G + jB

..... (3.43)

where, G = Conductance B = Susceptance Also, Y = G + jB = G – jBL Y = G + jB = G + jBC

; when susceptance is inductive. ; when susceptance is capacitive.

Y = G + jB = G + j(–BL+ BC) ; when susceptance is sum of inductance and capacitance.

Circuit Theory

3. 33

The symbol used to represent the admittance is rectangle as shown in Fig. 3.36. The admittance is connected to other parts of circuits using resistance-less wires.The magnitude of admittance is denoted by Y (i.e., without overbar). The argument of admittance is called admittance angle and it is denoted by q.

Y a G C jB

Fig. 3.36 : Symbol for admittance.

In equation (3.43), the complex admittance is expressed in rectangular form. It can be expressed in polar form as shown below: Y = G + jB =

G2 + B2 + tan- 1 B = Y+θ G

where, Y = Y =

G2 + B2

and

` Magnitude of admittance, Y = Y =

θ = +Y = tan- 1 B G

G2 + B2

Admittance angle, θ = +Y = tan- 1 B G Since the admittance is a complex quantity it can be represented as a point in a complex plane with polar coordinates Y and q as shown in Fig. 3.37. The line joining the origin and Y will be a vector of length Y and making an angle q with reference. Now, the vector Y can be resolved into horizontal and vertical components. The horizontal component is conductance, G and the vertical component is susceptance, B. The right angled triangle formed by G, B and Y is called admittance triangle.

Imaginary axis

Y

Y

B

q

real axis

G

Fig. 3.37 : Admittance triangle.

3.14.3 Admittance Connected to Sinusoidal Source Consider an admittance Y connected to ac source of voltage V volts rms value as shown in Fig. 3.38. Let, Z = 1 Y or Y = 1 Z . Since the admittance is connected across (or parallel to) the source, the voltage across the admittance is also V volts. By Ohm’s law the current through the admittance is given by, I = V = V = VY Z 1 Y

.....(3.44)

and

Y = Y+ " θ

.....(3.45)

(− for inductive susceptance and + for capacitive susceptance). From equations (3.44) and (3.45) we can write, I = V Y = V+0 o Y+ " θ = VY+ " θ = I+ " θ

where, I = VY

+ V

Let, V be the reference vector and q be the admittance angle. ` V = V+0 o

I

~

E

+ V E

Y=

1 Z

Fig. 3.38 : Admittance connected to ac source.

3. 34

Chapter 3 - Basic Concepts of AC Circuits

The power factor angle is the phase difference between V and I. Let, φ be the phase difference between V and I. Now, φ = +V − +I = 0 o − ^" θh = ! θ “It is interesting to observe that the power factor angle is same as that of admittance angle”. From the above analysis we can say that, the current through the admittance leads or lags the voltage by an angle φ. The admittance is a combination of conductances and susceptances.Depending on the combination, the phase of current may be in-phase or leading or lagging the voltage. TABLE 3.3 : VARIOUS COMBINATIONS OF CONDUCTANCE AND SUSCEPTANCE Admittance, Y

I C

V

E

Phase of current with respect to voltage

Phasor diagram of V and I V

I

I in-phase with V

Y =G I C

V

E

V

I lags V

f

Y = G − j BL I C

V

E

I

I leads V

Y = G + j BC I C

V

E

I

f

I in-phase with V

Y = G − jB L + j BC and B L = BC (This condition is also called resonance) I C

V

E

V

E

Y = G − jB L + j BC and B L < BC

V

I

I lags V

V

f

Y = G − jB L + j BC and B L > BC I C

V

I

I leads V

I

f

V

Circuit Theory

3. 35

3.15 KVL, KCL and Ohm’s Law Applied to ac Circuits In ac circuits, the total opposition to flow of sinusoidal current is called impedance Z. The impedance is given by the ratio of sinusoidal voltage and current. ` Z = V I

or V = I Z

or I = V Z

The above equation is called Ohm’s law of ac circuits. From above equation we can say that, when a current I flows through an impedance Z, the voltage V across the impedance is given by the product of current and impedance, i.e., V = I Z. In ac circuits, the voltage and current will vary with time. Hence while applying KCL and KVL to ac circuits we have to consider the signs of voltage and current at a particular time instant. The sign conventions for ac circuits are applicable to particular time instant.

3.16 Current and Voltage Division Rules for Impedances 3.16.1 Current Division in Parallel Connected Impedances Consider two impedances Z1 and Z 2 in parallel and connected to an ac source of V volts as shown in Fig. 3.39. Let, I be the current supplied by the source and I1 and I 2 be the current through Z1 and Z 2 respectively. Since the impedances are parallel to the source, the voltage across them will be V volts. The equations (3.46) and (3.47) can be used to solve the currents in parallel connected impedances in terms of total current drawn by the parallel combination and the values of individual impedances. Hence these equations are called current division rule. I I1 = I #

Z2 Z1 + Z 2

I2

I1

..... (3.46)

+ V

~

Z2

Z1

E

I2 = I #

Z1 Z1 + Z 2

..... (3.47) Fig. 3.39 : Impedances in parallel.

The following equation will be helpful to remember the current division rule. In two parallel connected impedances, Total current drawn by Value of the # parallel combination other impedance Current through one of the impedance = Sum of the individual impedances Z1

3.16.2 Voltage Division in Series Connected Impedances +

V1

Z2

+

E

V2

E

Consider two impedances Z1 and Z 2 in series and connected I to an ac source of V volts as shown in Fig. 3.40. Let, I be the current supplied by the source and V1 and V2 be the voltages across Z1 and Z 2 respectively. Since the impedances are in series, E + V the current through them will be I amperes. Fig. 3.40 : Impedances in series.

~

3. 36

Chapter 3 - Basic Concepts of AC Circuits

The equations (3.48) and (3.49) can be used to solve the voltages in series connected impedances in terms of total voltage across the series combination and the values of individual impedances. Hence these equations are called voltage division rule. V1 = V #

Z1 Z1 + Z 2

..... (3.48)

V2 = V #

Z2 Z1 + Z 2

..... (3.49)

The following equation will be helpful to remember the voltage division rule. In two series connected impedances, Total voltage across Value of the # series combination impedance Voltage across one of the impedance = Sum of the individual impedances

3.17 Solved Problems

j2 W

EXAMPLE 3.1

V2

V1

-j5 W

Find the node voltages V1 and V2 in the circuit shown in Fig. 1. 0

5Ð90 A

2W

~

4W

~

0

10Ð0 A

0

Fig. 1.

SOLUTION

In the circuit of Fig. 1, the reference node is “0”. The voltage of the reference node is zero. To find the voltages V1 and V2 with respect to reference node, write two KCL equations at these nodes and solve the two equations for a unique solution. With reference to Fig. 2, at node-1, using KCL we get, Currents leaving node-1

V1 - V2 j2

: 5+90 , V1 , V1 − V2 , V1 − V2 2 j2 − j5 o

V2

Current entering node-1 : Nil

Node-1 5Ð900 A

V1

V2

` 5+90o + V1 + V1 − V2 + V1 − V2 = 0 2 j2 − j5

~

j5 + V1 + V1 − V2 − V1 + V2 = 0 2 j2 j2 j5 j5 ` c 1 + 1 − 1 m V1 + c− 1 + 1 m V2 = − j5 j2 2 j2 j5 j5 ^0.5 − j0.3h V1 + j0.3 V2 = − j5

0

V1 - V2 - j5

V1 2

0

Fig. 2. ..... (1)

Circuit Theory

3. 37

With reference to Fig. 3, at node-2, using KCL we get, Currents leaving node-2

:

Current entering node-2

:

10∠0 o A

` V2 − V1 + V2 + V2 − V1 = 10+0o j2 4 − j5 V2 − V1 + V2 − V2 + V1 = 10 j2 j2 4 j5 j5 c−

V1 - V2 j2

V2 − V1 , V2 , V2 − V1 j2 4 − j5 V1

Node-2 V2

V1

V2 - V1 - j5 V 2 4

~

0

10Ð0 A

1 + 1 V1 + 1 + 1 − 1 V2 = 10 m c m j2 j2 j5 4 j5

j0.3 V1 + ^0.25 − j0.3h V2 = 10

0

0

Fig. 3.

` ^0.25 − j0.3h V2 = 10 − j0.3 V1 ` V2 =

j0.3 10 V1 − 0.25 − j0.3 0.25 − j0.3

= 16.3934 + j19.6721 + ^0.5902 − j0.4918h V1 In order to solve V1 , let us substitute for V2 from equation (2) in equation (1). ` ^0.5 − j0.3h V1 + j0.3 616.3934 + j19.6721 + ^0.5902 − j0.4918h V1 @ = − j5 ^0.5 − j0.3h V1 − 5.9016 + j4.918 + ^0.1475 + j0.1771h V1 = − j5 ^0.5 − j0.3 + 0.1475 + j0.1771h V1 = − j5 + 5.9016 − j4.918 ^0.6475 − j0.1229h V1 = 5.9016 − j9.918

` V1 =

5.9016 − j9.918 = 11.6037 − j13.1149 V = 17.5113+ − 48.5 o V 0.6475 − j0.1229

From equation (2) we get, V2 = 16.3934 + j19.6721 + (0.5902 − j0.4918) V1 = 16.3934 + j19.6721 + (0.5902 − j0.4918) × (11.6037 − j13.1149) = 16.792 + j6.225 V = 17.9087Ð20.3o V

RESULT The node voltages are, o V1 = 17.5113Ð− 48.5 V o V2 = 17.9087Ð20.3 V

..... (2)

3. 38

Chapter 3 - Basic Concepts of AC Circuits

EXAMPLE 3.2 5 W -j2 W V1

In the circuit shown in Fig. 1, find the node voltages V1 and V2 and the current Is supplied by the voltage source.

V2

2W 0

E = 10Ð30 V

SOLUTION

j5 W

~

3W

Is

5W -j2 W

In the circuit of Fig. 1, the reference node is “0”. The voltage of the reference node is zero. To find the voltages V1 and V2 with respect to reference node, write two KCL equations at these nodes and solve the two equations for a unique solution.

0

Fig. 1.

Given that, E = 10+30 o V = 10 ^cos 30 o + j sin 30 oh = 8.6602 + j5 V

.....(1)

With reference to Fig. 2, using KCL we get, V1 E E 5 E j2

V1 − E , V1 − V2 , V1 j5 5 − j2 3

Currents leaving node-1

:

Current entering node-1

: Nil

V1 E V2 j5 V1

V2

E

Node-1

V1 3

` V1 − E + V1 − V2 + V1 = 0 5 − j2 j5 3 V1 E − + V1 − V2 + V1 = 0 5 − j2 5 − j2 j5 j5 3 c

1 E + 1 + 1 m V1 − V2 = 5 − j2 j5 3 j5 5 − j2

c

8.6602 + j5 1 + 1 + 1 m V1 − V2 = 5 − j2 j5 3 j5 5 − j2

0

Fig. 2.

Using equation (1)

( 0.5057 − j0.131) V1 + j0.2 V2 = 1.1483 + j1.4593

..... (2)

With reference to Fig. 3, at node-2, using KCL we get, V2 E V1 j5

V2 V2 − V1 , V2 , 5 2 − j2 j5

Currents leaving node-2

:

Current entering node-2

: Nil

V2 V1

V2 ` V2 − V1 + V2 + = 0 j5 5 2 − j2

Node-1

V2 2 E j2

V2 5

V2 − V1 + V2 + V2 = 0 j5 j5 5 2 − j2 0

1 + 1 + 1 V1 = 0 c m V2 − j5 5 2 − j2 j5

0

Fig. 3.

^0.45 + j0.05h V2 + j0.2 V1 = 0

` V2 =

− j0.2 V1 ⇒ 0.45 + j0.05

V2 = ^− 0.0488 − j0.439h V1

..... (3)

Circuit Theory

3. 39

In order to solve V1, let us substitute for V2 from equation (3) in equation (2). \ (0.5057 − j0.131) V1 + j0.2 (−0.0488 − j0.439) V1 = 1.1483 + j1.4593 [ 0.5057 − j0.131 + j0.2 (−0.0488 − j0.439 ) ] V1 = 1.1483 + j1.4593 [ 0.5935 − j0.1408 ] V1 = 1.1483 + j1.4593 ` V1 =

1.1483 + j1.4593 0.5935 − j0.1408

= 1.2795 + j2.7623 V

..... (4)

o

= 3.0442∠65.1 V In order to solve V2, let us substitute for V1 from equation (4) in equation (3). ` V2 = ( −0.0488 − j0.439 ) × ( 1.2795 + j2.7623 ) = 1.1502 − j0.6965 V = 1.3446∠−31.2 o V

..... (5)

With reference to Fig. 1, we can write, Current supplied by the voltage source, Is = E − V1 5 − j2 =

8.6602 + j5 − (1.2795 + j2.7623) 5 − j2

= 1.1182 + j0.8948 A = 1.4321∠38.7 o A

RESULT The node voltages are, o V1 = 3.0442∠65.1 V o V2 = 1.3446∠−31.2 V

Current delivered by the source, Is = 1.4321∠38.7 o A

EXAMPLE 3.3 5W

j2 W

Determine the current I 2 in the circuit shown in Fig. 1.

I2

+

SOLUTION

0

100Ð45 V

5W

2W -j4 W

-

Let, I T be the total current supplied by the source. This current I T divides into I1 and I 2 and flows through parallel impedances −j4 Ω and 2 + j2 Ω as shown in Fig. 2. The current I T is given by the ratio of source voltage and total impedance at the source terminals. IT

~

j2 W

Fig. 1.

j2 W

IT I2

+ o

100Ð45 V

~

-

I1

2W

Þ

-j4 W

+ 0

100Ð45 V

~

-

j2 W

Fig. 2.

ZT = 9 + j 2 W

3. 40

Chapter 3 - Basic Concepts of AC Circuits

The total impedance Z T at the source terminal is given by the parallel combination of −j4Ω and 2 + j2 Ω in series with 5 + j2 Ω. ^− j4h # ^2 + j2h ` Z T = ^5 + j2h + 6 − j4 | | , ^2 + j2h@ = ^5 + j2h + ; E − j4 + 2 + j2 = 5 + j2 + 4 + j0 = 9 + j2 Ω o 100 ^cos 45 o + j sin 45 oh Now, I T = 100+45 = = 9.1508 + j5.8232 A 9 + j2 9 + j2

By current division rule, I2 = IT #

^9.1508 + j5.8232h # ^− j4h − j4 = 2 − j2 − j4 + 2 + j2

= 14.974 − j3.3276 = 15.3393∠−12.5o A

EXAMPLE 3.4

5W

Determine the current IL in the circuit shown in Fig. 1.

IL

+ 0

10Ð0 V

SOLUTION

2W

-

~

~

-

j2 W

Fig. 1.

Let us mark the nodes of the circuit as A, B, C and D as shown o

j5 W

o

in Fig. 2. Let, I1 and I 2 be the current delivered by 10∠0 V and 5∠45 V sources respectively from their positive end as shown in Fig. 2.

5W A

By KCL at node-B, we can write, ..... (1)

0

10Ð0 V

I2

I1 B IL

I1

+

IL + I 2 = I1

5Ð450V

+

2W

I2

~

~

-

5Ð450 V

+

-j2 W

The currents I1 and I 2 can be solved by writing KVL equations

Fig. 2.

in the closed paths ABDA and BCDB.

D

C

j5 W

Consider the closed path ABDA shown in Fig. 3. By KVL, we can write, 5I1 + ^2 − j2h IL = 10+0 o

5W

A + I1

5^I L + I2h + ^2 − j2h I L = 10

+ 0

10Ð0 V

~

5 I1

B -

b 2 - j 2g I

+

e

I L = I1 - I 2

2W

L

-

5I2 + ^7 − j2h I L = 10

-j2 W -

` 5I2 = 10 − ^7 − j2h I L

Fig. 3.

Using equation (1)

7 − j2 ` I2 = 10 − IL 5 5 = 2 − ^1.4 − j0.4h I L

D

.....(2)

j

Circuit Theory

3. 41

Consider the closed path BCDB shown in Fig. 4. By KVL, we can write,

B

j5I 2 = 5+45 o + ^2 − j2h IL

+ I L -

b 2 - j 2g I

j5I 2 − ^2 − j2h IL = 5^cos 45 o + j sin 45 oh

j5 62 − ^1.4 − j0.4h@ I L − ^2 − j2h I L = 5^cos 45 o + j sin 45 oh

+ I2

-

j10 − ^2 + j7h I L − ^2 − j2h I L = 3.5355 + j3.5355

` IL =

+

-

D

^− 4 − j5h I L = 3.5355 + j3.5355 − j 10

5Ð450 V

~

L

C

j5 I 2

Fig. 4. Using equation (1)

3.5355 + j3.5355 − j 10 − 4 − j5

= 0.4434 + j1.0618 A = 1.1507 + 67.3c A

EXAMPLE 3.5 A series combination of 10 Ω resistance and 50 mH inductance is connected to a 220 V, 50 Hz supply. Estimate the current, active power, reactive power and apparent power. Also estimate the voltage across R and L and draw the phasor diagram. Z = R + j XL

SOLUTION Given that,

R = 10 Ω,

jXL = jwL

R

V = 220 V, f = 50 Hz

+

+

-

VL

VR

L = 50 mH

The RL series circuit excited by a sinusoidal source is shown in Fig. 1. Inductive reactance = jX L = jωL = j2πf L = j2π × 50 × 50 × 10−3 = j15.708 Ω Impedance, Z = R + jXL = 10 + j15.708 Ω = 18.621+57.5 o Ω Let the supply voltage be reference phasor. ` V = V+0 o = 220+0 o V

I

~

+ V = V Ð0 0

Fig. 1.

Let, I be the current through RL circuit. Now by Ohm’s law, 220+0 o Current, I = V = = 11.8146+ − 57.5 o A Z 18.621+57.5 o ∴ I = |I | = 11.8146 A Power factor angle, φ = +V − +I = 0 o − ^− 57.5 oh = 57.5 o ∴ Power factor = cosφ = cos 57.5o = 0.5373 lag

Since the current lags the voltage, the power factor is lag.

Apparent power, S = VI = 220 × 11.8146 = 2599.2 VA = 2599.2 kVA = 2.5992 kVA 1000 Active power, (or power)

P = VIcosφ = 220 × 11.8146 × cos 57.5o = 1396.6 W = 1396.6 kW = 1.3966 kW 1000

3. 42

Chapter 3 - Basic Concepts of AC Circuits Reactive power, Q = VI sinφ = 220 × 11.8146 × sin 57.5 o

VL

= 2192.2 VAR = 2192.2 kVAR = 2.1922 kVAR 1000 32.50

Voltage across resistance, VR = I # R = 11.8146∠− 57.5 o × 10

90 0

= 118.146∠−57.5 o V

V 0

E57.5

Voltage across inductance, VL = I # jXL = 11.8146∠− 57.5 o × j15.708 o

= 11.8146∠− 57.5 × 15.708∠90

I

o

VR

= 185.5837∠32.5 o V

Fig. 2 : Phasor diagram.

The phasor diagram of RL circuit with V as reference phasor is shown in Fig. 2.

EXAMPLE 3.6 A current of 50∠−30 o A is flowing through a circuit which consists of series connected elements, when excited by a source of 230∠45o V, 50 Hz. Determine the elements of the circuit and power. Also draw the phasor diagram.

SOLUTION Given that, V = 230∠45 o V and I = 50∠−30o A o ` Impedance, Z = V = 230+45 o = 4.6+75 o Ω = 1.1906 + j4.4433 Ω I 50+ − 30 Since the reactance is positive, the circuit is RL series circuit. (Also the current is lagging and so the circuit is inductive.)

Z

We know that, Z = R + jXL

jXL

R

∴ R = 1.1906 Ω

+

+

E

XL = 4.4433 Ω

E VL

VR

We know that, XL = 2πf L `

L = XL = 4.4433 = 0.0141 H = 14.1mH 2πf 2π # 50 *

I

The complex power, S = V I = 230+45 # ^50+ − 30 h o

o

= 230∠45 × 50∠30 = 11500∠75 Also, S = P + jQ

E

Fig. 1. o

= 2976.4 + j11108.1 VA VL

& P + jQ = 2976.4 + j11108.1

∴ Active power, P = 2976.4 W =

V

~

o *

o

+

2976.4 kW = 2.9764 kW 1000

V

Reactive power, Q = 11108.1 VAR = 11108.1 kVAR = 11.1081 kVAR 1000 Apparent power, S = |S | = 11500 VA = 11500 kVA = 11.5 kVA 1000 The RL series circuit is shown in Fig. 1. Let, VR and VL be the voltage across R and L.

600

Now, by Ohm’s law, o o VR = I # R = 50∠−30 × 1.1906 = 59.53∠−30 V o o o VL = I # jXL = 50∠−30 × j 4.4433 = 50∠−30 × 4.4433∠90

= 222.165∠60 o V The phasor diagram of the RL series circuit is shown in Fig. 2.

0

I

45 0 E30

Reference

VR

Fig. 2 : Phasor diagram.

Circuit Theory

3. 43

EXAMPLE 3.7 Consider a RL series circuit with an impedance angle of 50o at a frequency of 60 Hz. At what frequency the magnitude of the impedance will be twice the magnitude of the impedance at 60 Hz?

SOLUTION Let, R = Resistance of RL series circuit L = Inductance of RL series circuit Case i : f1 = 60 Hz, ω1 = 2πf1 Let, Z1 = Impedance of RL circuit at f1 Now, Z1 = R + jω1 L =

R2 + ω12 L2 + tan- 1

ω1 L = Z1 +θ1 R

R 2 + ω12 L2

where, Z1 = Z1 =

θ1 = +Z1 = tan- 1

.....(1)

ω1 L R

Given that, φ1 = 50o ω1 L = 50 o ⇒ R Case ii : Frequency = f2 , ω2 = 2πf2

ω1 L = tan 50 o ⇒ R

` tan- 1

ω1 L = 1.1918 R

.....(2)

Let, f2 = Frequency at which magnitude of impedance doubles Z 2 = Impedance of RL circuit at f2 Now, Z 2 = R + jω2 L =

R 2 + ω22 L2 + tan- 1

where, Z 2 = Z 2 =

ω2 L = Z 2 +θ2 R

R 2 + ω22 L2

θ2 = +Z 2 = tan- 1 To solve for f2 :

.....(3)

ω2 L R

Given that, Z2 = 2Z1 `

R 2 + ω22 L2 = 2 #

Using equations (1) and (3)

R 2 + ω12 L2

On squaring the above equation we get, R 2 + ω22 L2 = 4^R 2 + ω12 L2h On dividing by R2 we get, 2

1+

2

ω2 L R ω2 L R = ω1 L R ` f2 = `

Now,

2

ω L ω L ω22 L2 ω2 L2 = 4 e1 + 1 2 o ⇒ e 2 o = 4 =1 + e 1 o G − 1 2 R R R R =

ω L 4 =1 + e 1 o G − 1 = R

2.9464 1.1918

⇒

ω2 = 2.4722 ω1

4^1 + 1.1918 2h − 1 = 2.9464

⇒

2πf2 = 2.4722 2πf1

2.4722 # f1 = 2.4722 # 60 = 148.332 Hz

RESULT The frequency at which the magnitude of the impedance doubles = 148.332 Hz

Using equation (2)

3. 44

Chapter 3 - Basic Concepts of AC Circuits

Cross-Check Let, R = 2 Ω Here, ω1 L = 1.1918 R ` L = 1.1918 R = 1.1918 R = 1.1918 # 2 = 6.3227 # 10- 3 H = 6.3227 # 10- 3 H 2πf1 2π # 60 ω1 Z1 = Z2 =

R 2 + (ω1 L) 2 =

R 2 + (ω2 L) 2 =

R2 + ^2πf1 Lh2 = R2 + ^2πf2 Lh2 =

2 2 + (2π # 60 # 6.3227 # 10- 3) 2 = 3.1115 Ω 2 2 + (2π # 148.332 # 6.3227 # 10- 3) 2 = 6.2229 Ω

Z 2 = 6.2229 = 1.999968 = 2 Z1 3.1115

EXAMPLE 3.8 A resistance of 16 Ω is connected in parallel to an inductance of 20 mH and the parallel combination is connected to an ac supply of 230 V, 50 Hz. Determine the current through the elements and power delivered by the source. Draw the phasor diagram. I

Given that, R = 16 Ω,

230 Ð00V

SOLUTION L = 20 mH

V = 230 V, f = 50 Hz

IR

+

~

-

R = 16 W 1 W G= 16

IL

L = 20 mH - jBL = - j

1 wL

The parallel RL circuit is shown in Fig. 1.

Fig. 1. Conductance, G = 1 = 1 = 0.0625 M R 16 1 Inductive susceptance = − jBL = − j 1 = − j 1 = − j = − j0.1592 M ωL 2πfL 2π # 50 # 20 # 10- 3 Admittance, Y = G − jB L = 0.0625 − j0.1592 M = 0.171+ − 68.6 o M Let the supply voltage be reference phasor. ∴ V = V∠0 o = 230∠0o V Let, I be the current through RL circuit. Now by Ohm’s law, Current, I = V = V Y = 230+0 o # 0.171+ − 68.6 o Z = 39.33∠−68.6 o A ` I = I

= 39.33 A

Power factor angle, φ = +V − +I = 0 o − (− 68.6 o) = 68.6 o o

∴ Power factor = cos φ = cos68.6 = 0.3649 lag

Since the current lags the voltage, the power factor is lag.

Circuit Theory

3. 45 P = VI cos φ = 230 × 39.33 × 0.3649 = 3300.8 W = 3300.8 kW = 3.3008 kW 1000

Current through resistance, I R = V = V # G R = 230+0 o # 0.0625 = 14.375+0 o A

IR

V

0

0

E68.6

E90

Power,

Current through inductance, IL = V = V # ^− jBLh = 230+0 o # ^− j0.1592h jωL = 230+0 o # 0.1592+ − 90 o = 36.616+ − 90 o A The phasor diagram of RL parallel circuit with V as reference phasor is shown in Fig. 2.

I

IL

Fig.2:Phasor diagram.

Alternate method Inductive reactance = jX L = jωL = j 2πfL = j 2π × 50 × 20 × 10 Impedance, Z =

−3

= j 6.2832 Ω

R # jX L 16 # j6.2832 = = 2.1377 + j5.4437 = 5.8484+68.6 o Ω R + jX L 16 + j6.2832

Let, V be reference phasor,

` V = V+0 o = 230+0 o V

230+0 o Current, I = V = = 39.327+ − 68.6 o A ; Z 5.8484+68.6 o

` I = I

= 39.327 A

Power factor angle, φ = +V − +I = 0 o − (−68.6o ) = 68.6o ∴ Power factor = cos φ = cos 68.6 o = 0.3649 lag Power, P = VIcos φ = 230 × 39.327 × 0.3649 = 3300.6 W = 3.3006 kW o Current through resistance, IR = V = 230+0 = 14.375+0 o A R 16

Current through inductance, IL =

V = 230+0 o = 230+0 o = 36.606+ − 90 o A jωL j6.2832 6.2832+90 o

EXAMPLE 3.9 A series combination of 12 Ω resistance and 600 µF capacitance is connected to a 220 V, 50 Hz supply. Estimate the current, active power, reactive power and apparent power. Also estimate the voltage across R and C and draw the phasor diagram. Z = R - j XC

SOLUTION Given that,

V = 220 V,

f = 50 Hz

R = 12 Ω,

C = 600 µF

-j XC = -j

R +

+

VR

VC

The RC series circuit excited by a sinusoidal source is shown in Fig. 1. Capacitive reactance = − jX C = − j 1 = − j 1 ωC 2πfC = −j

1 = − j5.3052 Ω 2π # 50 # 600 # 10- 6

Impedance, Z = R – jX C = 12 – j 5.3052 Ω = 13.1204∠−23.9 o Ω

I

~

+ V = V Ð0 0 V

Fig. 1.

1 wC -

3. 46

Chapter 3 - Basic Concepts of AC Circuits Let the supply voltage be reference phasor. ` V = V+0 o = 220+0 o V Let, I be the current through RC circuit. Now by Ohm’s law, 220+0 o Current, I = V = 16.7678+23.9 o A Z 13.1204+ − 23.9 o ∴ I = |I | = 16.7678 A Power factor angle, φ = +V − +I = 0 o − 23.9 o = − 23.9 o Since the current leads the voltage, the power factor is lead.

∴ Power factor = cos φ = cos (−23.9 o ) = 0.9143 lead Apparent power, S = VI = 220 × 16.7678 = 3688.9 VA = 3688.9 kVA = 3.6889 kVA 1000 Active power,

P = VI cos φ = 220 × 16.7678 × cos (–23.9o)

(or power)

= 3372.6 W = 3372.6 kW = 3.3726 kW 1000 o

Reactive power, Q = VI sin φ = 220 × 16.7678 × sin(−23.9 )

Alternatively, Q = 1.4945 kVAR-capacitive

= − 1494.5 VAR = − 1494.5 kVAR = − 1.4945 kVAR 1000

VR

Voltage across resistance, VR = I # R = 16.7678∠23.9 o × 12

I 0

o

= 201.2136∠23.9 V

23.9

V

90 0 0

Voltage across capacitance, VC = I # ^− jXCh = 16.7678∠23.9o × (−j5.3052) = 16.7678∠23.9 o × 5.3052∠−90 o = 88.9565∠−66.1o V

E66.1

VC

Fig. 2 : Phasor diagram.

The phasor diagram of RC circuit with V as reference phasor is shown in Fig. 2

EXAMPLE 3.10 o

A current of 60∠25 A is flowing through a circuit consist of parallel connected elements, when excited by a source of 230∠−20o V, 50 Hz. Determine the elements of the circuit, active power and reactive power. Also calculate the current through the elements and draw the phasor diagram.

SOLUTION Given that, V = 230∠−20 o V

o

and I = 60∠25 A

o ` Admittance, Y = I = 60+25 o = 0.2609+45 o M V 230+ − 20

= 0.1845 + j0.1845 M Since the susceptance is positive the circuit is RC parallel circuit. (Also the current is leading and so the circuit is capacitive.)

Circuit Theory

3. 47

We know that, Y = G + jBC 1 ` G = 0.1845 M and R = 1 = = 5.4201 Ω G 0.1845 BC = 0.1845 M I

We know that, BC = ωC = 2πf C ` C =

IC

IR

BC = 0.1845 = 5.8728 # 10- 4 F 2πf 2π # 50 –3

= 0.5873 × 10

+ V

~

1 R

G=

-

jBC = jwC

F

= 0.5873 mF The complex power, S = V I

Fig. 1.

*

= 230∠−20 o × (60∠25 o )* = 230∠−20 o × 60∠−25 o = 13800∠−45 o VA Alternatively, Q = 9.7581 kVAR-capacitive

= 9758.1 − j 9758.1 VA Also, S = P + jQ ` Active power,

P = 9758.1W = 9758.1 kW = 9.7581 kW 1000

Reactive power, Q = − 9758.1VAR = − 9758.1 kVAR = − 9.7581 kVAR 1000 Apparent power, S = S

= 13800 VA IC

The RC parallel circuit is shown in Fig. 1. Let, IR and IC be the current through R and C.

I

Now by Ohm’s law, IR = V = V # G = 230+ − 20 o # 0.1845 = 42.435+ − 20 o A R

0

70 0

90

IC = V = V # jBC = 230+ − 20 o # j0.1845 − jXC = 230+ − 20 o # 0.1845+90 o = 42.435+70 o A

0

25 E20 0

Reference

IR

V

Fig 2 : Phasor diagram.

The phasor diagram of the RC parallel circuit is shown in Fig. 2.

EXAMPLE 3.11 A RLC series circuit consists of R = 75 Ω, L = 125 mH and C = 200 µF . The circuit is excited by a sinusoidal source of value 115 V, 60 Hz. Determine the voltage across various elements. Calculate the current and power. Draw the phasor diagram.

SOLUTION Given that, V = 115 V , f = 60 Hz , R = 75 Ω , L = 125 mH and C = 200 µF Inductive reactance = jX L = jωL = j2πfL = j2π × 60 × 125 × 10

−3

= j47.1239 Ω

1 Capacitive reactance = − jXC = − j 1 = − j 1 = −j = − j13.2629 Ω ωC 2πfC 2π # 60 # 200 # 10- 6 Total reactance = jX = jX L − jXC = j 47.1239 − j13.2629 = j33.861 Ω ∴ Impedance, Z = R + jX = 75 + j33.861 Ω = 82.2895 ∠24.3 o Ω

3. 48

Chapter 3 - Basic Concepts of AC Circuits The RLC series circuit is shown in Fig. 1. Let, I be the current through

Z = R + jX

the circuit and VR, VL and VC be the voltage across R, L and C respectively.

jX

Let, V be the reference phasor. ` V = V+0o V

+

-j XC = -j

jXL = jwL

R

+

VR

-

+

VL

1 wC

VC

Now by Ohm’s law,

` I = I

VX

+

115+0o I = V = = 1.3975+ − 24.3o A Z . 82 2895+24.3o

-

I

~

= 1.3975 A

+ V = V Ð0 0 V

VR = I # R = 1.3975+ − 24.3o # 75 = 104.8125+ − 24.3o V

Fig. 1.

o

VL = I # jXL = 1.3975+ − 24.3 # j47.1239 = 1.3975+ − 24.3o # 47.1239+90o = 65.8557+65.7o V VC = I # ^− jXCh = 1.3975+ − 24.3o # ^− j13.2629h = 1.3975+ − 24.3o # 13.2629+ − 90o = 18.5349+ − 114.3o V Apparent power,

S = VI = 115 × 1.3975 = 160.7125 VA

Alternatively, Q = 66.1355 VAR-inductive

Power factor angle, φ = ∠V − ∠I = 0o − (−24.3o ) = 24.3o Active power,

P = VI cos φ = 115 × 1.3975 × cos 24.3 o = 146.4739 W

Reactive power,

Q = VI sin φ = 115 × 1.3975 × sin 24.3 o

VL

= 66.1355 VAR VX

The phasor diagram of RLC series circuit is shown in Fig. 2. Here, VX = VL + VC

65.7

V

0

Reference

900

= IjXL + I^− jXCh

E24.3 0

E1

= I # j^XL − XCh

VC

= I # jX = 1.3975+ − 24.3o # j33.861 = 1.3975+ − 24.3o # 33.861+90o = 47.3207+65.7o V

14

.3 0 I VR

Fig. 2 : Phasor diagram of RLC series circuit.

EXAMPLE 3.12 An RLC parallel circuit consists of R = 50Ω, L = 150mH and C = 100µF. The circuit is excited by a current source of 5∠0 o A, 100 Hz. Calculate the voltage and current in the various elements. Determine the apparent, active and reactive power delivered by the source. Draw the phasor diagram.

SOLUTION Given that, I = 5+0o A R = 50 Ω

,

f = 100 Hz

,

L = 150 mH and C = 100 µF.

Let us analyze the parallel circuit in terms of admittance.

Circuit Theory

3. 49

Conductance, G = 1 = 1 = 0.02 M R 50 1 Inductive susceptance = − jBL = − j 1 = − j # = − j0.0106 M 2πfL 2π # 100 # 150 # 10- 3 Capacitive susceptance = j BC = j2πfC = j2π × 100 × 100 × 10

−6

= j 0.0628 M

Total susceptance = jB = jBC − jBL = j 0.0628 − j0.0106 = j0.0522 M Admittance, Y = G + jB = 0.02 + j 0.0522 = 0.0559∠69 o M The RLC parallel circuit excited by a current source is shown in Fig. 1. Let, V be the voltage across the source and parallel connected elements. Let, IR, IL and IC be the

0

I = 5Ð0 A

current through R, L and C respectively.

I B = I L+ I C

I

+

Now by Ohm’s law,

~

IL

IR

1 G= R

V

IC

jBC

-jBL

o

5+0 V = IZ = I = = 89.4454+ − 69o V Y 0.0559+69o

Y = G - jBL + jBC = G + jB

Fig. 1.

IR = V # G = 89.4454+ − 69o # 0.02 = 1.7889+ − 69o A IL = V # ^− jBLh = 89.4454+ − 69o # ^− j0.0106h = 89.4454+ − 69o # 0.0106+ − 90o = 0.9481+ − 159o A IC = V # jB C = 89.4454+ − 69o # j0.0628 = 89.4454+ − 69o # 0.0628+90o = 5.6172+21o A Let, IB = IL + IC = V # ^− jBLh + V # jBC = V # j^BC − BLh = V # jB = 89.4454∠−69 o × j 0.0522 = 89.4454∠−69 o × 0.0522∠90o = 4.669∠21o A

IC

) ) Complex power, S = V I = 89.4454+ − 69 o # ^5+0 oh o

= 89.4454∠−69 × 5 ∠0

IB

o

= 447.227∠−69 o VA

90o

= 160.2718 − j 417.5224 VA Apparent power, S = S = 447.227 VA

I

E690

IL IR

Also, S = P + jQ ∴ Active power,

0

E159

210

P = 160.2718 W

Reactive power, Q = −417.5224 VAR

V

Fig. 2 : Phasor diagram.

The phasor diagram of RLC parallel circuit with I as reference phasor is shown in Fig. 2.

EXAMPLE 3.13 A load absorbs 2.5 kW at a power factor of 0.707 lagging from a 230 V, 50 Hz source. A capacitor is connected in parallel to the load in order to improve the power factor to 0.9 lag. Determine the value of the capacitor.

3. 50

Chapter 3 - Basic Concepts of AC Circuits

SOLUTION Method - I Case i Given that, V = 230 V , P1 = 2.5 kW , cos φ1 = 0.707 lag , f = 50 Hz P Apparent power, S1 = = 2.5 = 3.5361 kVA cos φ1 0.707 Power factor angle, φ1 = cos–1 0.707 = 45o

Alternatively, Q1 = 2.5 kVAR-inductive

Reactive power, Q1 = S1 sinφ1 = 3.5361 × sin 45 o = 2.5 kVAR

Case ii Given that, V = 230 V , f = 50 Hz , cos φ2 = 0.9 lag The addition of capacitor to the load does not alter the active power but decrease the reactive power supplied by the source. Hence the active power remains same as that of 2.5 kW. ∴ P2 = 2.5 kW Apparent power, S 2 =

Alternatively, Q2 = 1.209 kVA-inductive

P2 = 2.5 = 2.7778 kVA cos φ2 0.9

Here, Q2 is inductive because the power factor is still lagging.

Power factor angle, φ2 = cos –1 0.9 = 25.8 o

Reactive power, Q2 = S2 sinφ2 = 2.7778 × sin 25.8 o = 1.209 kVAR Now, the reactive power supplied by the capacitor QC is given by, QC = Q2 − Q1 = 1.209 − 2.5 = −1.291 kVAR = −1291 VAR We know that, QC = V IC

⇒

QC = VIC ⇒

IC =

QC V

where, IC is the magnitude of current through capacitor. 2 2 V Capacitive reactance, XC = V = = V = 230 = 40.976 IC QC V QC 1291

Also, XC =

1 2πfC

` Capacitance, C =

1 1 = = 77.682 # 10- 6 F = 77.682 µF 2πfXC 2π # 50 # 40.976

Method - II Case i

+

Given that, V = 230 V , f = 50 Hz , P1 = 2.5 kW , cos φ1 = 0.707 lag Let, IL be the current through inductive load as shown in Fig. 1. We Know that, P1 = VILcos φ1

V = VÐ00

3

` IL = IL =

IL = IL Ð - f

P1 = 2.5 # 10 = 15.3742 A 230 # 0.707 V cos φ1

.....(1)

Fig. 1.

ZL

Circuit Theory

3. 51

Let the supply voltage V be reference phasor. ` V = V+0o = 230+0o V Since the power factor is lagging the current IL will lag the supply voltage V by an angle φ1, where φ1 = cos−1 0.707 ` IL = IL + − cos- 1 φ1

Using equation (1) -1

= 15.3742+ − cos

0.707 = 10.8696 − j10.8728 A

Case ii : When capacitor is added to inductive load I

The inductive load with capacitance in parallel is shown in Fig. 2. Here, the current through the load remains same as that of IL . Let, I be the current supplied by the source and IC be the current through the capacitor. The active power remains same after addition of capacitor.

+

C

ZL

-

We Know that, P2 = VI cos φ2 =

IL

V = 230Ð00V

∴ P2 = 2.5 kW

` I = I

IC

Fig. 2.

3 P2 = 2.5 # 10 = 12.0773 A V cos φ2 230 # 0.9

.....(2)

` I = I + − cos- 1 φ2 = 12.0773+ − cos- 1 0.9 = 10.8696 − j5.2644 A

Using equation (2)

By KCL we can write, I = I C + IL ` IC = I − IL = 10.8696 − j10.8728 − ^10.8696 − j5.2644h = − j5.6084 A Magnitude of capacitor current, IC = 5.6084 A Capacitive reactance, XC = V = 230 = 41.0092 Ω IC 5.6084 Also, XC = 1 2πfC ` Capacitance, C =

1 1 = = 77.619 # 10- 6 F = 77.619 µF 2πfX C 2π # 50 # 41.0092

Note : The slight difference in the capacitance value is due to approximation in calculations.

EXAMPLE 3.14 An inductive coil of power factor 0.8 lagging is connected in series with a 120 µF capacitor. When the series circuit is connected to a source of frequency 50 Hz, it was observed that the magnitude of voltage across the coil and capacitor are equal. Determine the parameters of the coil.

I

L

R

Z a R C jX L

SOLUTION The given circuit is RLC series circuit as shown in Fig. 1. Let, Z = Impedance of the coil I = Current through RLC series circuit

~

E + V, f = 50 Hz

Fig. 1.

C

3. 52

Chapter 3 - Basic Concepts of AC Circuits Given that, Z I = − jXC I

⇒ Z = XC R 2 + XL2 = Magnitude of impedance of the coil

where, Z =

` Magnitude of impedance, Z = XC =

1 1 = = 26.5258 Ω 2πfC 2π # 50 # 120 # 10- 6

Given that, power factor of the coil = 0.8 lag ∴ Power factor angle of the coil, φ = cos −1 0.8 = 36.9

o

Let us construct an impedance triangle for Z by using R and XL as two sides as shown in Fig. 2. Here, the impedance angle is same as power factor angle. With reference to Fig. 2, we can write, X cos φ = R and sin φ = L Z Z ∴ Resistance, R = Z cos φ = 26.5258 × cos 36.9 o = 21.2123 Ω Inductive reactance, X L = Z sin φ = 26.5258 × sin 36.9 o = 15.9266 Ω We know that, `

XL = ωL

Inductance, L =

XL XL = = 15.9266 = 0.0507 H = 50.7 mH ω 2πf 2π # 50

RESULT The parameters of the coil are R and L. Resistance of the coil, R = 21.2123 Ω Inductance of the coil, L = 50.7 mH

I1

~

I2

I3

Z3 = -j7 W

Z = 4 + j6W

Z1 = 12 W

SOLUTION

I4

230Ð00V = 230 V

Three impedances 12 Ω, 5 + j8 Ω and −j7 Ω are connected in parallel. This parallel combination is connected in series with an impedance of 4 + j6 Ω across a 230 V source. Determine the current through each impedance and the power.

Z2 = 5 + j8 W

EXAMPLE 3.15

The series-parallel connections of the impedances are shown Fig. 1. in Fig. 1. Let us name the impedances as Z1, Z 2, Z3 and Z 4 as shown in Fig. 1. Let, the current through the impedances be I1, I 2, I3 and I 4 as shown in Fig. 1. Let, supply voltage be reference phasor. ∴ Supply voltage = 230∠0 o V Let, Z eq be the equivalent impedance of the parallel combination of Z1, Z 2 and Z3 and the circuit can be modified as shown in Fig. 2. 1 1 Now, Zeq = = 1 + 1 + 1 1 + 1 + 1 12 5 + j8 − j7 Z1 Z2 Z3 = [12

−1

−1

+ (5 + j8)

+ (− j7)

−1 −1

]

= 6.2647 − j2.3785 Ω

Circuit Theory

3. 53

Let, Va and Vb be the voltage across Z 4 and Zeq as shown in Fig. 2. Z4 a 4 C j 6 ‡

I4

Now, by voltage division rule we can write,

+ Va

4 + j6 Va = 230 # 4 + j6 + 6.2647 − j2.3785

+ 230 V

= 121.8879 + j91.4379 V

+ Vb

~

E

E

Zeq a 6.2647

E

E j 2.3785 ‡

o

= 152.3731+36.9 V

Fig. 2.

By KVL we can write, Vb = 230 − Va = 230 − ^121.8879 + j91.4379h = 108.1121 − j91.4379 V = 141.5949 ∠−40.2 o V

Here, the voltage across Z 4 is Va and the voltage across Z1, Z2 and Z3 are Vb (because Z1, Z 2 and Z3 are in parallel). Now, the current through the impedances can be evaluated by using Ohm’s law, as shown below: o o I 4 = Va = 152.3731+36.9 = 152.3731+36.9o = 21.1304+ − 19.4 o A 4 + j6 Z4 7.2111+56.3 o I1 = Vb = 141.5949+ − 40.2 = 11.7996+ − 40.2 o A 12 Z1 o o I 2 = Vb = 141.5949+ − 40.2 = 141.5949+ − 40.2 = 15.009+ − 98.2 o A 5 + j8 Z2 9.434+58 o o o I3 = Vb = 141.5949+ − 40.2 = 141.5949+ −o40.2 = 20.2278+49.8 o A j 7 − Z3 7+ − 90

We know that, the complex power is given by the product of voltage and conjugate of current. Hence the complex power in each impedance can be obtained from the product of voltage and conjugate of current in the impedance. Let, S1, S 2, S3 and S 4 be the complex power of the impedances Z1, Z 2, Z3 and Z 4 respectively. Let, P1, P2, P3 and P4 be active power and Q1, Q2, Q3 and Q4 be the reactive power of the impedances. For impedance Z 4, * * S 4 = Va # I 4 = 152.3731+36.9 o # ^21.1304+ − 19.4 oh

= 152.3731∠36.9 o × 21.1304∠19.4o = 3219.7∠56.3o VA = 1786.4 + j2678.6 = P4 + jQ4 ∴ S4 = 3219.7 VA = 3.2197 kVA P4 = 1786.4 W = 1.7864 kW Q4 = 2678.6 VAR = 2.6786 kVAR For impedance Z1, * * S1 = Vb # I1 = 141.5949+ − 40.2 o # ^11.7996+ − 40.2 oh

= 141.5949∠−40.2 o × 11.7996∠40.2o = 1670.8∠0o VA = 1670.8 + j0 = P1 + jQ1 ∴ S1 = 1670.8 VA = 1.6708 kVA P1 = 1670.8 W = 1.6708 kW Q1 = 0

3. 54

Chapter 3 - Basic Concepts of AC Circuits For impedance Z 2, * * S 2 = Vb # I 2 = 141.5949+ − 40.2 o # ^15.009+ − 98.2 oh

= 141.5949∠−40.2 o × 15.009∠98.2o = 2125.2∠58 o VA = 1126.2 + j1802.3 VA = P2 + jQ2 ∴ S2 = 2125.2 VA = 2.1252 kVA P2 = 1126.2 W = 1.1262 kW Q2 = 1802.3 VA R = 1.8023 kVAR For impedance Z3, * * S3 = Vb # I3 = 141.5949+ − 40.2 o # ^20.2278+49.8 oh

= 141.5949∠−40.2 o × 20.2278∠−49.8o = 2864∠−90 o = 0 − j2864 = P3 + jQ3 ∴ S3 = 2864 VA = 2.864 kVA P3 = 0 Q3 = − 2864 VAR = − 2.864 kVAR

EXAMPLE 3.16 Two reactive circuits have an impedance of 20 Ω each. One of them has a power factor of 0.75 lagging and the other 0.65 leading. Find the voltage necessary to send a current of 12 A through the two circuits in series. Also determine the current drawn from 200 V supply if they are connected in parallel to the supply.

SOLUTION Case i : Impedances in series Let, Z1 and Z 2 be the impedances of the two circuits. The series combination of two circuits excited by a voltage source can be represented by the circuit shown in Fig. 1. Z

Now, the impedances Z1 and Z 2 can be expressed as shown below: Z1 = 20∠cos

−1

0.75

o

= 20∠41.4 Ω = 15.0022 + j13.2262 Ω Z 2 = 20∠− cos

−1

0.65

o

= 20∠−49.5 Ω = 12.989 − j15.2081 Ω

When the current is lagging the impedance will be inductive and the impedance angle is same as power factor angle. When the current is leading the impedance will be capacitive and the impedance angle is negative of power factor angle.

Let, Z be the total impedance of series combination. Now, Z = Z1 + Z 2 = 15.0022 + j13.2262 + 12.989 − j15.2081 = 27.9912 − j1.9819 = 28.0613∠−4.1 o Ω

Z2

Z1

I

+

~

E

V

Fig. 1.

Circuit Theory

3. 55

Magnitude of impedance, Z = | Z | = 28.0613 Ω Given that, the magnitude of current, I = |I | = 12 A ∴ Magnitude of supply voltage, V = IZ = 12 × 28.0613 = 336.7356 V If V is the reference phasor then, V = 336.7356+0 o V and I = 12+4.1 o A If I is the reference phasor then, V = 336.7356+ − 4.1 o V and I = 12+0 o A Case ii : When the impedances are in parallel The two circuits in parallel and excited by a 200 V source can be represented by the circuit shown in Fig. 2. Let, I1 and I 2 be the current through the impedances Z1 and Z 2 respectively. Let, I be the total current supplied by the source. Let, the supply voltage V be the reference phasor. o

I1

Z1

I2

Z2

o

` V = V+0 = 200+0 V o Now, I1 = V = 200+0 o = 10+ − 41.4 o A Z1 20+41.4

200+0 o I2 = V = = 10+49.5 o A Z2 20+ − 49.5 o I

By KCL we can write,

+

~

-

V = 200 Ð0 0 V

Fig. 2.

I = I1 + I 2 = 10+ − 41.4 o + 10+49.5 o = 7.5011 −j6.6131 + 6.4945 + j7.6041 = 13.9956 + j 0.991 A = 14.0306∠4.1o A

EXAMPLE 3.17 100 Ð00V

j2.5 W

In the circuit shown in Fig. 1, (a) determine the currents in all the branches,(b) Calculate the power and power factor of the source, (c) Show that power delivered by the source is equal to power consumed by the 2 Ω resistor.

+ 2W

~

-j1 W

-

SOLUTION

Fig. 1.

a) To find branch currents A

The circuit has three branches. Let us assume three branch

Ia

j2.5 W

B Ib

100 Ð00V

currents Ia, Ib and Ic as shown in Fig. 2. Let us denote the nodes

Ic

+

as A, B and C.

~

-j1 W

2W

-

By KCL at node-B, we can write, Ib + I c = I a

Fig. 2. C

` I c = I a − Ib

..... (1)

3. 56

Chapter 3 - Basic Concepts of AC Circuits With reference to Fig. 3, using KVL in the closed path A

j2.5Ia + 2Ib = 100+0

o

` j2.5Ia + 2Ib = 100

..... (2)

100 Ð00V

ABCA we can write,

Ia

Ib

+

+

2I b

~

-

With reference to Fig. 4, using KVL in the closed path BCB we can write,

B

B + j 2.5I a

-

Fig. 3.

C

Ib

Ic

+

+

2I b

Ej I c

E

C

E

Fig. 4.

− jIc = 2Ib ..... (3)

` 2Ib + jIc = 0

On substituting for Ic from equation (1) in equation (3), we get, 2Ib + j^Ia − Ibh = 0 ` jIa + ^2 − jh Ib = 0

..... (4)

Equation (2) × 1

⇒

j2.5Ia

Equation (4) × (−2.5)

⇒

−j2.5Ia − 2.5 ( 2 − j )Ib = 0

On adding ` Ib =

+ 2Ib

= 100

[ 2 − 2.5 (2 − j) ]Ib = 100

100 100 = = − 19.6721 − j16.3934 = 25.6073+ − 140.19 o − 3 + j2.5 2 − 2.5^2 − jh

From equation (2), we can write, 100 − 2 # ^− 19.6721 − j16.3934h 139.3442 + j32.7868 Ia = 100 − 2Ib = = j2.5 j2.5 j2.5 = 13.1147 − j55.7377 = 57.2598∠−76.76 o A From equation (1), we get, I c = I a − Ib = 13.1147 − j55.7377 − (−19.6721 − j16.3934) = 32.7868 − j39.3443 = 51.2147∠−50.19 o A

b) To find power and power factor of the source Let, E = Source voltage. Is = Current delivered by the source. φ = Phase difference between E and Is . Given that, E = 100+0 o V With reference to Fig. 2, we can write, Is = Ia = 57.2598+ − 76.76 o A

Circuit Theory Now,

3. 57

φ = +E − + I s = 0o − (−76.76o ) = 76.76o ` Power factor of the source = cos φ = cos ^76.76 oh = 0.229 lag Power delivered of the source = E # Is # cos φ = 100 × 57.2598 × 0.229 = 1311 W

..... (5)

c) To find power consumed by 2 Ω resistor Power consumed by 2 Ω resistor = Ib

2

# 2 2

= 25.6073 × 2 = 1311 W

..... (6)

From equations (5) and (6) we can say that, the power delivered by the source is equal to power consumed by the 2 Ω resistor. (Remember that the reactive elements does not consume power).

RESULT a) The branch currents are Ia = 57.2598+ − 76.76 o A

; Ib = 25.6073+ − 140.19 o A

; Ic = 51.2147+ − 50.19 o A

b) The power delivered by the source = 1311 W The power factor of the source = 0.229 lag c) The power consumed by 2 Ω resistor = 1311 W

(AU Dec’14, 16 Marks) Z3 = 6.25 + j1.25 ‡

+ 100 V 50 Hz

~

E

SOLUTION Given that, Z1 = 5 − jXc Ω ; Z 2 = 5 + j0 Ω ; V = 100 V

;

Z3 = 6.25 + j1.25 Ω

f = 50 Hz

With reference to Fig. 1, the equivalent impedance is obtained as shown below Zeq = Z3 +

Z1 # Z 2 Z1 + Z 2

= 6.25 + j1.25 +

(5 − jXC) # 5 5 − jXC + 5

= 6.25 + j1.25 +

25 − j5XC 10 + jXC # 10 − jXC 10 + jXC

Fig. 1.

Z1 = 5 E jXc ‡

Impedance Z1 and Z 2 are parallel and this combination is in series with an impedance Z3 , connected to a 100 V, 50 Hz ac supply. Z1 = 5 − jXc Ω , Z 2 = 5 + j0 Ω , Z3 = 6.25 + j1.25 Ω . Determine the value of capacitance such that the total current of the circuit will be in phase with the total voltage. Find the circuit current and power.

Z2 = 5 + j0‡

EXAMPLE 3.18

3. 58

Chapter 3 - Basic Concepts of AC Circuits 250 + j25XC − j50XC + 5XC2

= 6.25 + j1.25 +

10 2 + XC2 250 + 5XC2

= 6.25 + j1.25 +

10 2 + XC2

−j

25XC 10 2 + XC2

For the voltage and current to be in phase, the Zeq should be real and so imaginary part of the Zeq should be zero. ` j1.25 − j

25XC 10 2 + XC2

102 + XC2 = 20 XC

= 0

&

&

j1.25 =

j25XC 10 2 + XC2

&

102 + XC2 =

j25XC j1.25

XC2 − 20 XC + 100 = 0

2 ` XC = 20 ! 20 − 4 # 100 = 20 = 10 Ω 2 2

We know that, XC =

1 2πfC

&

C =

1 1 = = 3.1831 # 10- 4 F = 318.31 # 10- 6 F 2πfXC 2π # 50 # 10 = 318.31 µF

When XC = 10 W, Zeq = 6.25 +

250 + 5XC2 102 + XC2

2 = 6.25 + 250 +2 5 # 10 10 + 102

= 6.25 + 3.75 = 10 W ` Current, I = V = 100 = 10 A Zeq 10 Power, P = V # I = 100 # 10 = 1000 W = 1 kW (or Power, P = I2 × real part of Zeq = 102 × 10 = 1000 W)

3.18 Summary of Important Concepts 1.

The sources in which the current/voltage sinusoidally varies with time are called sinusoidal sources.

2.

The frequency is the number of cyclic changes that a sinewave will undergo in one second.

3.

The time for one cycle is called time period and it is also given by inverse of frequency.

4.

The angular rotation of a sinusoidal vector in one second is called angular frequency.

5.

The circuits excited by sinusoidal sources are called ac circuits.

6.

The relation between frequency, f and angular frequency, ω is given by, ω = 2πf.

7.

The plot of the instantaneous value of the sinusoidal voltage/current with respect to ωt or t is called waveform.

8.

In an ac source, when rms value of voltage is constant and the rms value of current varies then it is called ac voltage source.

Circuit Theory 9.

3. 59

In an ac source, when rms value of current is constant and the rms value of voltage varies then it is called ac current source.

10. In an ideal ac voltage source the source impedance is zero. 11. In an ideal ac current source the source impedance is infinite. 12. An ac voltage source with internal impedance Zs can be represented by an ideal ac voltage source in series with an external impedance of value Zs . 13. An ac current source with internal impedance Zs can be represented by an ideal ac current source in parallel with an external impedance of value Zs . 14. An ac voltage source E in series with impedance Zs can be converted to an equivalent current source Is aIs = E Zs k in parallel with impedance Zs . 15. An ac current source Is in parallel with impedance Zs can be converted to an equivalent voltage source E aE = I Zs k in series with impedance Zs . 16. The instantaneous value of a sinusoidal voltage(n) is expressed as, n = Vm sin(ωt ± φ). 17. The instantaneous value of a sinusoidal current(i) is expressed as, i = Im sin(ωt ± φ). 18. The average value of a time varying quantity is the average of the instantaneous value for a particular time period. 19. The rms value of a time varying quantity is the equivalent dc value of that quantity. 20. The form factor is defined as the ratio of rms value and average value of a periodic waveform. 21. The peak factor is defined as the ratio of peak value and the rms value of a periodic waveform. 22. The inductance is the property of element (or matter) by which it opposes the change in flux or current. 23. The inductance of a coil is defined as the ratio of flux linkages and current through the coil. 24. A coil is said to have an inductance of one Henry if a current of one ampere flowing through it produce a flux linkage of one weber-turn in it. 25. Faraday’s law says that, an emf is induced in a conductor when there is a change in flux linkage and the emf is equal to rate of change of flux linkages. 26. The n-i relation in an inductor is governed by Faraday’s law. Therefore, the voltage acorss an inductor is directly proportional to rate of change of current through it. 2 27. The energy, W stored in an inductance, L carrying a steady current, I is given by, W = LI . 2 28. The capacitance is the property of element (or matter) by which it opposes the change in charge or voltage. 29. The capacitance of capacitor is defined as the ratio of stored charge and potential difference across it.

3. 60

Chapter 3 - Basic Concepts of AC Circuits 30. A capacitor is said to have a capacitance of one Farad if a charge of one Coulomb establish a potential difference of one volt across it. 31. The current through a capacitor is directly proportional to rate of change of voltage across it. 32. The energy, W stored in a capacitance, C having a steady voltage, V across it is given by, 2 W = CV . 2 33. The phase (or phase angle) of a vector is the angular position of the vector with respect to reference at time t = 0. 34. The rotating vectors are functions of phase angle and so they are called phasors. 35. The diagram in which all the voltage and current phasors of a circuit are drawn with respect to a reference phasor is called phasor diagram. 36. The complex power, S is defined as the product of rms voltage V and conjugate of rms ) ) current I ai.e., S = V I k . 37. The magnitude of complex power, S is called apparent power, S. It is also given by the ) product of voltage and current di.e., S = S = V I = VI n . 38. The unit of apparent power is volt-ampere (VA). The higher units are kVA and MVA. 39. The complex power, S can be expressed as S = VI+ ! φ = VI cos φ ! jVI sin φ = P ! jQ, where, φ is the phase difference between V and I, P = VI cosφ and Q = VI sinφ. 40. The real part of complex power S is called active power, P(or simply power) and the unit of power is Watts (W). The higher units are kW and MW. 41. The imaginary part of complex power S is called reactive power, Q and the unit of reactive power is Volt-Ampere-Reactive (VAR). The higher units are kVAR and MVAR. 42. The power factor is defined as the ratio of (active) power and apparent power. It is also given by cosine of the phase difference between voltage and current. 43. When a resistance is excited by an ac source the current and voltage will be in-phase. 44. The resistance consumes only active power and the reactive power in a resistance is zero. 45. When an inductance is excited by an ac source the current lags the voltage by 90o. 46. The inductance consumes only reactive power and the active power in an inductance is zero. 47. When a capacitance is excited by an ac source the current leads the voltage by 90o. 48. The capacitance delivers only reactive power and the active power in a capacitance is zero. 49. The impedance of an element is defined as the ratio of sinusoidal voltage and current in that element. 50. The impedance is a complex quantity. The real part of impedance is resistance and the imaginary part is reactance.

Circuit Theory

3. 61

51. The admittance is inverse of impedance and it is defined as the ratio of sinusoidal current and voltage. 52. The admittance is a complex quantity. The real part of admittance is conductance and the imaginary part is susceptance.

3.19 Short-answer Questions Q3.1

A sinusoidal voltage is represented by the equation, 100 sin(503t + 30o) volts. What is the frequency and time period? Solution The general form of sinusoidal voltage is, v (t) = Vm sin ^ωt + φh

......(1)

Given that, v(t) = 100 sin (503t + 30o)

......(2)

On comparing equations (1) and (2), we get, ω = 503 rad/s Since, ω = 2πf Frequency, f = ω = 503 = 80 Hz 2π 2π Time period, T = 1 = 1 = 0.0125 second = 12.5 # 10- 3 second = 12.5 ms f 80

Q3.2

A sinusoidal current is given by the equation, i(t) = 7.072 sin 314t A. What is the rms and average value of the current? Solution The general form of sinusoidal current is, i(t) = Im sin (ω t + φ)

.....(1)

Given that, i(t) = 7.072 sin 314t

.....(2)

On comparing equations (1) and (2), we get, Im = 7.072 A ` Rms value of current, I =

Im = 7.072 = 5 A 2 2

Average value of current, Iave =

Q3.3

2Im = 2 # 5 = 4.502 A π π

Define form factor. The form factor is defined as the ratio of rms value and average value of a periodic waveform. ` Form factor, k f =

Rms value Average value

3. 62 Q3.4

Chapter 3 - Basic Concepts of AC Circuits Define peak factor. The peak factor is defined as the ratio of peak value (or maximum value) and the rms value of a periodic waveform. ` Peak factor, kp = Maximum value Rms value

Q3.5

What will be the inductance of a coil with 1000 turns while carrying a current of 2 A and producing a flux of 0.5 mWb? Solution Given that, N = 1000, φ = 0.5 mWb, I = 2 A Inductance, L =

Q3.6

-3 Nφ = 1000 # 0.5 # 10 = 0.25 H 2 I

A steady current of 3 A flows through an inductance of 0.2 H. What will be the energy stored in the inductance? Solution Given that, I = 3 A, L = 0.2 H Energy stored in inductance, W = 1 LI 2 = 1 # 0.2 # 3 2 = 0.9 Joules 2 2

Q3.7

A100 µF capacitance is charged to a steady voltage of 500V. What is the energy stored in the capacitance? Solution Given that, C = 100 µF , V = 500 V Energy stored in capacitance, W = 1 CV 2 = 1 # 100 # 10- 6 # 5002 = 12.5 Joules 2 2

Q3.8

When a sinusoidal voltage, v = 200 sin (377t + 30o) V is applied to a load, it draws a current of 10 (sin 377t + 60o) A. Determine the active and reactive power of the load. The rms current and voltage phasors in the polar form are, V = 200 +30 o V 2

;

I = 10 +60 o A 2

* * Complex power, S = V I = 200 +30 o # c 10 +60 o m = 200 +30 o # 10 + − 60 o 2 2 2 2

= 2000 +^30 o − 60 oh = 1000+ − 30 o = 866 − j500 VA 2 Since, S = P + jQ ; Active power,

P = 866 W

Reactive power, Q = −500 VAR

(or 500 VAR-Capacitive)

Circuit Theory Q3.9

3. 63

A load consisting of 3 Ω resistance and 4 Ω inductive reactance draws a current of 10 A when connected to a sinusoidal source. Determine the voltage and power in the load. Magnitude of impedance, Z =

R 2 + XL2 =

32 + 42 = 5 Ω

Voltage, V = I Z = 10 × 5 = 50 V 2

2

Power, P = I R = 10 × 3 = 300 W

Q3.10

When a sinusoidal voltage of 120 V is applied across a load, it draws a current of 8 A with a phase lead of 30 o. Determine the resistance, reactance and impedance of the load. Let, V be reference phasor. ` V = 120+0 o V

and

I = 8+ + 30 o A

o Impedance, Z = V = 120+0o = 15+ − 30 o = 12.99 − j7.5 Ω I 8+30

Since, Z = R + jX ; Resistance, R = 12.99 Ω Reactance, jX = –j7.5 W (or 7.5 W-Capacitive)

Q3.11 When a sinusoidal voltage of 100 V is applied across a load, it draws a current of 10 A with 30 o phase lag. Determine the conductance, susceptance and admittance of the load. Let, V be reference phasor. ` V = 100+0 o V and I = 10+ − 30 o A o Impedance, Y = I = 10+ − 30o = 0.1+ − 30 o = 0.0866 − j0.05M V 100+0

Since, Y = G + jB ; Conductance, G = 0.0866 M Susceptance, jB = j0.05 M

( or 0.05 M - inductive)

Q3.12 An inductive load consumes 1000 W power and draws 10 A current when connected to a 250 V, 25 Hz supply. Determine the resistance and inductance of the load. 2

We know that, P = I R ` Resistance, R = P2 = 1000 = 10 Ω I 10 2 ` Impedance, Z = V = 250 = 25 Ω I 10 We know that, Z =

R 2 + XL2

` Inductive reactance, XL = Inductance, L =

Z2 − R2 =

25 2 − 10 2 = 22.9129 Ω

XL = 22.9129 = 0.1459 H 2πf 2π # 25

XL = ωL = 2πfL

3. 64

Chapter 3 - Basic Concepts of AC Circuits

Q3.13 In an RC series circuit excited by sinusoidal source the voltage across resistance and capacitance are 60 V and 80 V respectively. What will be the supply voltage? VR

Let current, I through RC series circuit be the reference phasor. With reference to phasor diagram shown in Fig. Q3.13, we can write,

I

E53.1

E90

0

0

Supply voltage, V = VR + VC = 60+0 o + 80+ − 90 o = 60 – j80 = 100∠–53.1o V V

VC

Q3.14 In an RL series circuit with R = 20 Ω and XL = 30 Ω, for what value of R,

Fig. Q3.13.

the impedance of RL series combination will be doubled ? When R = 20 Ω and XL = 30 Ω, Z = R 2 + XL2 =

202 + 302 = 36.0555 Ω

Let, R 2 be the value of resistance when impedance is doubled. Now, R 2 =

(2Z) 2 − XL2 =

^2 # 36.0555h2 − 30 2 = 65.5744 Ω

Q3.15 A RC series circuit with R = 1.2 kΩ and C = 0.1 µF is excited by a sinusoidal source of 45 V and frequency 1kHz. Find the apparent power. XC =

Magnitude of impedance, Z =

^1.2 # 103h + c 2

R 2 + XC2 =

1 2πfC

2 1 = 1993.246 Ω -6 m 2π # 10 # 0.1 # 10 3

Magnitude of current, I = V Z 2 452 Apparent power, S = VI = V # V = V = = 1.0159 VA Z Z 1993.246

Q3.16 Determine the impedance of the RLC parallel circuit shown in Fig. Q3.16. R = 100 = 25 Ω 4 Z

;

1 1 + 1 + 1 R jXL − jXC

XL = 100 = 100 Ω 1 =

1 1 + 1 + 1 j100 − j25 25

;

XC = 100 = 25 Ω 4 = 16 − j12 Ω

4A

1A

4A

R

L

C

+ 100 V

~ E

Z = Z =

162 + 12 2 = 20 Ω

Fig. Q3.16.

Alternate method : Let total current be I. I = 4 + j1 − j4 = 4 − j3 A ; I = I = Now, Z = V = 100 = 20 Ω I 5

42 + 32 = 5 A

Circuit Theory Q3.17

3. 65

Determine the power factor of an RLC series circuit with R = 5 Ω, X L = 8 Ω and XC = 12 Ω. Impedance, Z = R + jXL − jXC = 5 + j8 − j12 = 5 − j4 Ω Power factor angle, φ = tan- 1 − 4 = − 38.7 o 5 Power factor = cos φ = cos (−38.7 o ) = 0.7804 lead

-1 If, Z = R + jX then, φ = tan X R

Since impedance angle is negative, pf is lead.

3.20 Exercises I. Fill in the Blanks with Appropriate Words 1.

The sources in which the current/voltage do not change with time are called ________.

2.

In sinusoidal voltage/current, the number of cycles per second is called ________ .

3.

The ratio of rms value and average value of a periodic waveform is called ________.

4.

The ratio of the peak value and the rms value of a periodic waveform is called ________.

5.

In a current-carrying conductor the induced emf is equal to rate of change of ________.

6.

In a capacitor the energy is stored as ________.

7.

In inductive circuit current ________ and in capacitive circuit current ________ the supply voltage.

8.

The ________ is defined as the ratio of active power and apparent power.

9.

In RLC circuit when total reactance is negative the current ________ the voltage.

10. In RLC parallel circuit the inductance and capacitance ________ are always in phase opposition.

ANSWERS 1. dc sources

2. frequency

3. form factor

4. peak factor

5. flux linkages

6. electric field

7. lags, leads

8. power factor

9. leads

10. current

II. State Whether the Following Statements are True/False 1.

The average value of sinusoidal voltage over one period is zero.

2.

The flux and current are inseparable in nature.

3.

The charge and voltage are inseparable in nature.

4.

In an inductance, energy is stored as electric field.

5.

In a capacitance, the energy is stored as magnetic field.

6.

In a resistance, the voltage and current are always in phase.

7.

In an inductance, the voltage and current are always in phase quadrature.

8.

The reactive power in a capacitive circuit is positive.

9.

In an RLC circuit, when the total susceptance is positive the current lags the voltage.

10. In an RLC series circuit the inductance and capacitance voltage are always in phase opposition.

3. 66

Chapter 3 - Basic Concepts of AC Circuits

ANSWERS 1. True

2. True

3. True

4. False

5. False

6. True

7. True

8. False

9. False

10. True

III. Choose the Right Answer for the Following Questions 1. The frequency of the sinusoidal voltage, v = 163 sin377t V is, a) 40 Hz

b) 50 Hz

c) 60 Hz

d) 70 Hz

2. The average and rms value of the sinusoidal current, i = 10 sin314t A are respectively, a) 3.183 A, 6.366 A

b) 6.366 A, 7.071 A

c) 15.708 A, 14.142 A

d) 7.854 A, 9.003 A

3. The form factor of the sinusoidal voltage, n = 230 sin314t V is, a) 1.111

b) 2.222

c) 0.901

d) 0.451

4. The peak factor of the sinusoidal voltage, n = 400 sin377t V is, a) 1.732

b) 0.866

c) 0.707

d) 1.414

5. The inductance of a coil with a flux linkage of 0.202 Wb-turn and carrying a current of 0.2 A is, a) 0.404 H

b) 0.101 H

c) 0.002 H

d) 0.402 H

6. The energy stored in a coil carrying a current of 20 A and having an inductance of 5 mH is, a) 2 Joule

b) 1 Joule

c) 0.5 Joule

d) 0.1 Joule

7. What is the charge in a 0.01 F capacitor when a voltage of 100 V exist in it? a) 10 Coulomb

b) 1 Coulomb

c) 0.1 Coulomb

d) 0.01 Coulomb

8. What is the energy in a 0.01 F capacitor when a voltage of 10 V exist in it? a) 2 Joule

b) 1 Joule

c) 0.5 Joule

d) 0.1 Joule

9. The phase difference between the voltages v1 = 230 sin(377t + 30 o) V and v2 = 230 sin(377t – 30 o) V is, a) 60o with v1 leading v2

b) 30o with v2 leading v1

c) 60o with v1 lagging v2

d) 30o with v2 lagging v1

10. When a load is connected to 230Ð10 o V, it draw a current of 10Ð-50 o A. What is the real and reactive power of the load? a) 1992 W, 1150 VAR

b) 1150 W, 1150 VAR

c) 1626 W, 2086 VAR

d) 1150 W, 1992 VAR

Circuit Theory

3. 67

11. The power factor of a load with active power 120 W and reactive power 100 VAR is, a) 0.64 lag

b) 0.64 lead

c) 0.768 lag

d) 0.768 lead

12. What is the value of the impedance drawing a current of 8Ð-37 o A, when connected to 220Ð10 o V supply? a) 27.5Ð47o W

b) 27.5Ð-47o W

c) 27.5Ð27o W

d) 27.5Ð-27o W

13. The resistance and reactance of the impedance 4Ð60 o W are respectively, a) 2 W, 3.46 W-capacitive

b) 2 W, 3.46 W-inductive

c) 3.46 W, 2 W-capacitive

d) 3.46 W, 2 W-inductive

14. What is the value of the admittance drawing a current of 20Ð-26o A when connected to 40Ð10 o V supply? a) 0.5+ − 36 o M

b) 0.5+36 o M

c) 2+ − 16 o M

d) 2+16 o M

15. The conductance and susceptance of the admittance 8+30 o M are respectively, a) 4 M, 6.93 M - inductive

b) 4 M, 6.93 M - capacitive

c) 6.93 M, 4 M - inductive

d) 6.93 M, 4 M - capacitive

ANSWERS 1. c 2. b 3. a

4. d 5. b 6. b

7. b 8. c 9. a

10. d 11. c 12. a

13. b 14. a 15. d

IV. Unsolved Problems E3.1 In the circuit shown in Fig. E3.1, determine the currents in all the branches. E3.2 Determine the current I 2 in the circuit shown in Fig. E3.2. E3.3 Determine the Voltage V 2 in the circuit shown in Fig. E3.3. j2.5 W

2W Ib

Ic

+

~

-

2W

-j1 W

20 Ð300V

100 Ð00V

Ia

4W

-j3 W I2

~

j2 W

1W

+ 0

12Ð0 A -j4 W

~

1W

2W

j2 W

-j5 W

V2

-

Fig. E3.1.

Fig. E3.2.

Fig. E3.3.

E3.4

A series combination of 5 Ω and 10 mH inductance is connected to a 115 V, 60 Hz supply. Estimate the voltage and current in the elements. Also calculate the active, reactive and apparent power.

E3.5

A parallel RL circuit connected to a 230 V, 50 Hz supply has active power of 2.5.kW and reactive power of 3.12 kVAR. Calculate the current through the elements, total current supplied by the source and the value of R and L.

3. 68

Chapter 3 - Basic Concepts of AC Circuits

E3.6

A current of 10∠30 o A is flowing through a circuit consist of series connected elements, when excited by a source of 200∠−30o V, 50 Hz. Determine the elements of the circuit, voltage across the elements and active, reactive and apparent power.

E3.7

A parallel combination of 10 Ω resistance and 400 µF capacitance is connected to a 160 V, 50 Hz supply. Estimate the current through the elements and total current drawn from the supply. Also calculate the apparent, active and reactive power.

E3.8

An RLC series circuit consists of R = 40 Ω, L = 70 mH and C = 450 µF. The circuit is excited by a sinusoidal source of value 100 V, 50 Hz. Determine the current and voltage in the elements. Also estimate the apparent, active and reactive power.

E3.9

A load absorbs 4 kW at a pf of 0.65 lagging from a 160 V, 50 Hz source. A capacitor is connected in parallel to the load to improve the pf to 0.8 lag. Determine the value of capacitor.

ANSWERS E3.1 Ia = 57.2598∠−76.8o A ; Ib = 25.5711∠−140.2o A ;

Ic = 51.2147∠−50.2o A

E3.2 I 2 = 6.2469∠171.3o A E3.3 V2 = 18.9734∠18.4o V E3.4 I = 18.3648∠−37 o A S = 2.112 kVA

; VR = 91.8242∠−37o V

;

; Q = 1.271 kVAR

; P = 1.6867 kW

E3.5 I = 17.3828∠−51.3 o A ; IR = 10.8696∠0o A R = 21.1599 Ω E3.6 R = 10 Ω S = 2 kVA

;

IL = 13.5652∠−90o A

; C = 183.78 µF

;

VR = 100∠30o V

; P = 1 kW

; Q = −1.7321 kVAR

; L = 54 mH

E3.7 I = 25.6955∠51.5 o A ; IR = 16∠0o A S = 4.1113 kVA

VL = 69.2336∠53o V

; P = 2.5593 kW

;

;

VC =173.205∠−60o V

Ic = 20.1062∠90o A

; Q = –3.2175 kVAR

E3.8 I = 2.3424∠−20.5 o A ; VR = 93.696∠−20.5o V ; VL = 51.512∠69.5 o V ;VC =16.5692∠−110.5o V S = 234.24 VA E3.9 C = 207.31 µF

; P = 219.4061 W

; Q = 82.0326 VAR

Chapter 4

MESH ANALYSIS 4.1

Introduction

Mesh analysis is a useful technique to solve the currents in various elements of a circuit. Mesh analysis is preferred, if the circuit is excited by voltage sources, and the current through various elements are unknown. Mesh analysis can also be extended to circuits excited by both voltage and current sources and to circuits excited by both independent and dependent sources. In a circuit, each branch will have a current through it. Hence, the number of currents in a circuit is equal to number of branches. In a circuit some of the currents will be independent and the remaining currents depend on independent currents. The number of independent currents in a circuit is given by number of links in the graph of the circuit. (Refer Chapter 2, Section 2.10.4.) In mesh analysis, the independent currents are solved by writing Kirchhoff’s Voltage Law (KVL) equations for various meshes in the circuit. If the graph of a circuit has B branches and N nodes then the number of links L is given by, L = B − N + 1. Hence L number of meshes are chosen in a given circuit. “Mesh is defined as a closed path which does not contain any other loops within it”. Let us denote the number of meshes by m. In a circuit the number of meshes, m is equal to links, L. For each mesh, an independent current is assigned called mesh current and for each mesh, an equation is formed by using Kirchhoff’s Voltage Law. The equation is formed by equating the sum of voltage rise to sum of voltage drop in a mesh. These m number of mesh equations are arranged in a matrix form and mesh currents are solved by Cramer’s rule. A simple procedure to form mesh basis matrix equation directly from circuit by inspection without forming KVL equations is also discussed in this chapter. Mesh analysis is applicable to planar circuits. “A circuit is said to be planar circuit if it can be drawn on a plane surface without crossovers”.

4.2

Mesh Analysis of Resistive Circuits Excited by dc Sources

A circuit with B branches will have B number of currents and in this some currents are independent and the remaining currents depend on independent currents. The number of independent currents m is given by, m = B − N + 1, where N is number of nodes. In order to solve the independent currents of a circuit we have to choose m meshes (or closed paths) in the circuit. For each mesh we have to attach a current called mesh current. The mesh currents are the independent currents of the circuit. Let, I1, I2, I3, ......,Im be mesh currents. For each mesh a KVL equation is formed by equating the sum of voltage rise to sum of voltage fall in the mesh. Since there are m meshes we can form m equations. In resistive circuits excited by dc sources the voltages and currents are real (i.e., they are not complex). For resistive circuits the m number of equations can be arranged in the matrix form as shown in equation (4.1), which is called mesh basis matrix equation. The formation of mesh basis matrix equation from the KVL equations is explained in some of the solved problems.

Chapter 4 - Mesh Analysis

4.2

The mesh basis matrix equation (4.1), can be written in the simplified form as shown in equation (4.2).

= = = =

..... (4.1)

↓

where, R I E m

↓

↓

Note : The bold faced letters represent matrices. R R11 R12 R13 g R1m V R I1 V R E11 V S W S W S W S R21 R22 R23 g R2m W S I2 W S E22 W S R31 R32 R33 g R3m W S I3 W = S E33 W S W S W S W h h h W Sh W S h S h W SR R R k R W SI W SE W mm T m1 m2 m3 X T mX T mm X RI=E Resistance matrix of order m × m Mesh current matrix of order m × 1 Source voltage matrix of order m × 1 Number of meshes

..... (4.2)

In equation (4.1), the elements of resistance matrix and source voltage matrix can be determined from the given circuit. Hence the unknowns are mesh currents, which has to be solved by any standard technique. Alternatively, the equation (4.1) can be formed directly from the circuit by inspection without writing KVL equations. A procedure to form mesh basis matrix equation by inspection is given below: Procedure to Form Mesh Basis Matrix Equation by Inspection Consider the mesh basis matrix equation shown below for a circuit with three meshes. Let, I1, I2, I3 be the mesh currents. R11 R12 R13 I1 E11 >R21 R22 R23 H >I2 H = >E22 H R31 R32 R33 I3 E33 ..... (4.3) The elements of equation (4.3) for circuits with independent sources are, R11 = Sum of resistances in mesh-1 R22 = Sum of resistances in mesh-2 R33 = Sum of resistances in mesh-3 R12 = R21 = Sum of resistances common between mesh-1 and mesh-2 R13 = R31 = Sum of resistances common between mesh-1 and mesh-3 R23 = R32 = Sum of resistances common between mesh-2 and mesh-3 E11 = Sum of voltage sources in mesh-1 E22 = Sum of voltage sources in mesh-2 E33 = Sum of voltage sources in mesh-3 The resistances R11, R22, R33 are called self-resistance of mesh-1, mesh-2, mesh-3 respectively. The resistances R12, R13, R21, R23, R31, R32 are called mutual-resistance between meshes.

4.3

Circuit Theory

The formation of the elements of resistance matrix and source voltage matrix are explained below: i)

The self-resistance R jj is given by sum of all the resistances in the jth mesh. The selfresistances will be always positive.

ii)

The mutual-resistance R jk is given by sum of all the resistances common between mesh-j and mesh-k. The common resistance R jk is positive if the mesh currents Ij and Ik flow in the same direction through the common resistance as shown in Fig. 4.1 and it is negative if the mesh currents Ij and Ik flow in the opposite direction through the common resistance as shown in Fig. 4.2. In circuit with only independent sources (reciprocal circuit), Rjk = Rkj. mesh-k

mesh-j Rjk

Ij

Ik

Fig. 4.1 : Example for positive Rjk .

Ij

Ik

mesh-j + E

Ij

Fig. 4.3 : Example for positive source voltage.

Rjk

Fig. 4.2 : Example for negative Rjk .

mesh-j + E

mesh-k

mesh-j

Ij

Fig. 4.4 : Example for negative source voltage.

iii) The source voltage matrix element E jj is given by sum of all the voltage sources in the j th mesh. A source voltage is positive if it is a rise in voltage in the direction of mesh current as shown in Fig. 4.3. A source voltage is negative if it is fall or drop in voltage in the direction of mesh current as shown in Fig. 4.4. Note : In circuit with both independent and dependent sources (nonreciprocal circuit) Rjk ! Rkj Solution of Mesh Currents In the mesh basis matrix equation [i.e., equation (4.1)], the unknowns are mesh currents I 1, I2 , I 3 ... Im. The mesh currents can be obtained by premultiplying the equation (4.1), by the inverse of resistance matrix. Consider the equation (4.2), RI=E On premultiplying both sides by R −1, we get, R−1 R I = R−1 E U I = R−1 E ∴ I = R−1 E

R−1 R = U = Unit matrix

..... (4.4)

UI = I

Chapter 4 - Mesh Analysis

4.4

The equation (4.4) will be the solution for mesh currents. The equation (4.4) can be solved by Cramer’s rule, by which the kth mesh current Ik is given by equation (4.5). ∆ ∆1k ∆ ∆ E11 + 2k E22 + 3k E33 + ...... + mk E mm = 1 ∆ ∆ ∆ ∆ ∆

Ik =

m

/

∆ jk E jj

j =1

..... (4.5)

where, Djk = Cofactor of Rjk. Ejj = Sum of voltage sources in mesh-j. D = Determinant of resistance matrix. Proof for Cramer’s Rule Consider the equation (4.4), for a circuit with three meshes. I = R

-1

E

I1 ⇒ > I2 H = I3

R11 R12 R13 - 1 >R21 R22 R23 H R31 R32 R33

E11

>E22 H

..... (4.6)

E33

We know that, R- 1 =

T Transpose of Rcof Adjoint of R Rcof = = Determinant of R Determinant of R ∆

where, ∆

= Determinant of R.

R cof = Cofactor matrix (matrix formed by cofactor of elements of R matrix). Let, ∆ 11 = Cofactor of R 11 ∆ 12 = Cofactor of R 12 and in general, ∆ jk = Cofactor of Rjk `

Rcof =

`

R- 1 =

>∆21

∆11 ∆12 ∆13 ∆22 ∆23 H ∆31 ∆32 ∆33

T Rcof

∆

=

Transpose

T Rcof =

>∆12

∆11 ∆21 ∆31 ∆22 ∆32 H ∆13 ∆23 ∆33

∆11 ∆21 ∆31 1 ∆12 ∆22 ∆32 H > ∆ ∆13 ∆23 ∆33

..... (4.7)

On substituting for R–1 from equation (4.7) in equation (4.6), we get, I1

>I2 H I3

=

∆11 ∆21 ∆31 1 ∆ ∆ ∆ > ∆ 12 22 32 H ∆13 ∆23 ∆33

E11

>E22 H E33

On multiplying the matrices on the right-hand side of the above equation and equating to the terms on the left-hand side we get, ∆ ∆ ∆ I1 = 11 E11 + 21 E22 + 31 E33 ∆ ∆ ∆ ∆ ∆ ∆ I2 = 12 E11 + 22 E22 + 32 E33 ∆ ∆ ∆ ∆13 ∆23 ∆33 I3 = E11 + E22 + E33 ∆ ∆ ∆ The above equations can be used to form a general equation for mesh current. In general, the k th mesh current of a circuit with m meshes is given by, Ik =

∆1k ∆ ∆ ∆ E11 + 2k E22 + 3k E33 + ...... + mk Emm = 1 ∆ ∆ ∆ ∆ ∆

m

/ j =1

∆ jk E jj

4.5

Circuit Theory

Short-cut Procedure for Cramer’s Rule A short-cut procedure exists for the Cramer’s rule which is shown below: Let us consider a circuit with three mesh. The mesh basis matrix equation for the three mesh circuit is, R11 R12 R13 I1 E11 >R21 R22 R23 H >I2 H = >E22 H R31 R32 R33 I3 E33 Let us define three determinants as shown below: E11 R12 R13 ∆1 = E 22 R 22 R 23 ; E33 R32 R33

R11 E11 R13 ∆2 = R 21 E 22 R 23 ; R31 E33 R33

R11 R12 E11 ∆3 = R 21 R 22 E 22 R31 R32 E33

Here, ∆ 1 = Determinant of resistance matrix after replacing the first column of resistance matrix by source voltage column matrix. ∆ 2 = Determinant of resistance matrix after replacing the second column of resistance matrix by source voltage column matrix. ∆ 3 = Determinant of resistance matrix after replacing the third column of resistance matrix by source voltage column matrix. Let, ∆ = Determinant of resistance matrix R11 R12 R13 ∆ = R21 R 22 R 23 R31 R32 R33 Now mesh currents I 1 , I 2 and I 3 are given by, I1 =

∆1 ∆

;

I2 =

∆2 ∆

;

I3 =

∆3 ∆

Cross-Check The equation for mesh currents obtained by short-cut procedure are same as equation (4.5), and they are verified as shown below: E11 R12 R13 Expanding along first column ∆ I1 = 1 = 1 E 22 R 22 R 23 ∆ ∆ E33 R32 R33 ∆31 ∆11 ∆21 1 = 6 E ∆ + E 22 ∆21 + E33 ∆31 @ = ∆ E11 + ∆ E 22 + ∆ E33 ∆ 11 11 I2 =

R11 E11 R13 ∆2 = 1 R 21 E 22 R 23 ∆ ∆ R31 E33 R33

Expanding along second column

∆32 ∆ ∆ = 1 6 E11 ∆12 + E 22 ∆22 + E33 ∆32 @ = ∆12 E11 + ∆22 E 22 + ∆ E33 ∆ I3 =

R11 R12 E11 ∆3 = 1 R 21 R 22 E 22 ∆ ∆ R31 R32 E33

Expanding along third column

= 1 6 E11 ∆13 + E 22 ∆23 + E33 ∆33 @ = ∆13 E11 + ∆23 E 22 + ∆33 E33 ∆ ∆ ∆ ∆

Chapter 4 - Mesh Analysis

4.6

Various Steps to Obtain the Solution of Mesh Currents and Branch Currents in a Circuit Step 1 :

Draw the graph of the circuit.

Step 2 :

Determine the branches B and nodes N. The number of mesh currents m is given by m = B − N + 1.

Step 3 :

Select m number of meshes of the circuit and attach a mesh current to each mesh.

Step 4 :

In the given circuit choose arbitrary direction for branch and mesh currents. Let us denote mesh currents by I1, I2, I3,....., and branch currents by Ia, Ib, Ic, Id, Ie,...... Write the relationship between mesh and branch currents. Preferably the directions of mesh currents are chosen in same orientation. For example, the direction of all the mesh currents can be chosen clockwise (alternatively, the direction of all the mesh currents can be chosen anticlockwise). When all the mesh currents are chosen in same orientation, all the mutual-resistances (Rjk) will be negative.

Step 5 : Form the mesh basis matrix equation by inspection and solve the mesh currents using Cramer’s rule. For a circuit with three meshes the mesh basis matrix equation and solution of mesh currents using Cramer’s rule are given below: R11 R12 R13 I1 E11 R R R I E = 21 22 23 H > 2 H > 22 H > R31 R32 R33 I3 E33

I1 =

E11 R12 R13 ∆1 = 1 E 22 R 22 R 23 ∆ ∆ E33 R32 R33

I2 =

R11 E11 R13 ∆2 = 1 R 21 E 22 R 23 ∆ ∆ R31 E33 R33

I3 =

R11 R12 E11 ∆3 = 1 R 21 R 22 E 22 ∆ ∆ R31 R32 E33

Step 6 : Solve the branch currents using the relationship between branch and mesh currents. Note : 1. After solving the branch currents if any of the current is found to be negative, then the actual direction is opposite to that of the assumed direction. If interested we can draw the circuit by indicating the actual direction of current. 2. If the directions of the current are already given in the circuit, then we have to solve for the given direction of the current.

4.7

Circuit Theory

EXAMPLE 4.1 Solve the currents in various branches of the circuit shown in Fig. 1, by mesh analysis.

SOLUTION The graph of the given circuit is shown in Fig. 2. It has 6 branches and 4 nodes. Hence, the number of meshes m in the circuit is, m = B − N + 1 = 6 − 4 + 1 = 3.

5‡ + 3‡ 50 V E 4‡

The circuit has 6 currents (corresponding to six branches) and in this 3 currents are independent (corresponding to three meshes).

50 V c

E

I2

+

Fig. 1.

+

b

I1

20 V

E

2‡

5‡ e

6‡

8‡

Let us assume three mesh currents I 1 , I 2 and I 3 as shown in Fig. 2. The directions of the current are chosen arbitrarily. The circuit with chosen mesh currents is shown in Fig. 3.

a

2‡

I1

I2

3‡

4‡

6‡

20 V + E

d I3

I3

8‡

f

Fig. 3.

Fig. 2.

Method I : Formation of mesh basis equation by applying KVL In this method, the mesh equations are formed by using Kirchhoff’s Voltage Law. The mesh equation for a mesh is formed by equating the sum of voltage fall to sum of voltage rise. The voltage rise and fall are determined by tracing the circuit in the direction of mesh current. 5I1

With reference to Fig. 4, the mesh equation for mesh-1 is formed as shown

+

below :

E + E

+

Voltage fall : 5 I 1 , 3I 1 , 4I 1

I1

50 V

E +

E

Voltage rise : 50, 3 I 2 , 4I 3

4I1 E +

∴ 5 I 1 + 3 I 1 + 4 I 1 = 50 + 3 I 2 + 4 I 3 12 I 1 − 3 I 2 − 4 I 3 = 50

3I1

4I3

+ E

I2 3I2

I3

Fig. 4.

..... (1)

+ 2I2 E

With reference to Fig. 5, the mesh equation for mesh-2 is formed as shown below : Voltage fall : Voltage rise :

+

+ E

3 I 2 , 2 I 2 , 6 I 2 , 20 3 I1

∴ 3 I 2 + 2 I 2 + 6 I 2 + 20 = 3 I 1 −3 I 1 + 11 I 2 = −20

4 I3 , 8 I 3

Voltage rise :

20, 4 I 1

−4 I 1 + 12 I 3 = 20

I2

E 4I1 +

20 V

4I3 E

..... (3)

+

I1

+

∴ 4 I 3 + 8 I 3 = 20 + 4 I 1

20 V

Fig. 5.

..... (2)

With reference to Fig. 6, the mesh equation for mesh-3 is formed as shown Voltage fall :

E

E

below :

6I2

I2

I1 3I1 3I2 E +

E

E

I3 8I3 +

Fig. 6.

+

Chapter 4 - Mesh Analysis

4.8

The equations (1), (2) and (3) are the mesh equations of the circuit shown in Fig. 3. The mesh equations are summarized here for convenience. 12 I 1 − 3 I 2 − 4 I 3 =

50

−3I 1 + 11 I 2 = − 20 −4 I 1 + 12 I 3 = 20 The mesh equations can be arranged in the matrix form as shown below and then solved by Cramer’s rule. 12 − 3 − 4 I1 50 11 0 H >I 2 H = >− 20 H 20 − 4 0 12 I3

>− 3

..... (4)

Method II : Formation of mesh basis matrix equation by inspection In this method, the mesh basis matrix equation is formed directly from the circuit shown in Fig. 3 by inspection. The circuit has three meshes. The general form of mesh basis matrix equation for three mesh circuit is shown in equation (5). R11 R12 R13 I1 E11 R 22 R 23 H >I 2 H = >E 22 H R31 R32 R33 I3 E33

>R21

..... (5)

The elements of resistance matrix and source voltage matrix are formed as shown below: R 11 = 5 + 3 + 4 = 12

R12 = R21 = −3

E 11 = 50

R2 2 = 3 + 2 + 6 = 11

R 13 = R31 = −4

E 22 = −20

R3 3 = 4 + 8 = 12

R 23 = R32 = 0

E 33 = 20

On substituting the above terms in equation (5), we get equation (6) and the solution of equation (6) will give the mesh currents. 12 − 3 − 4 I1 50 11 0 H >I 2 H = >− 20 H 20 − 4 0 12 I3

>− 3

..... (6)

Solution of mesh currents It is observed that the mesh basis matrix equation obtained in method I and II are same. In equation (6), the unknown are I 1, I2 and I3. In order to solve I1 , I 2 and I3, let us define four determinants ∆, ∆ 1 , ∆ 2 and ∆3 as shown below : 12 − 3 − 4 ∆ = − 3 11 0 ; − 4 0 12

50 − 3 − 4 ∆1 = − 20 11 0 ; 20 0 12

12 50 − 4 ∆2 = − 3 − 20 0 ; − 4 20 12

12 − 3 50 ∆3 = − 3 11 − 20 − 4 0 20

The determinants are evaluated by expanding along first row and the mesh currents are solved by Cramer’s rule. 12 − 3 − 4 ∆ = − 3 11 0 = 12 # 611 # 12 − 0 @ − (− 3) # 6 − 3 # 12 − 0 @ + (− 4) # 60 − (− 4) # 11 @ − 4 0 12 = 1584 − 108 − 176 = 1300 50 − 3 − 4 ∆1 = − 20 11 0 = 50 # 611 # 12 − 0 @ − (− 3) # 6 − 20 # 12 − 0 @ + (− 4) # 60 − 20 # 11 @ 20 0 12 = 6600 − 720 + 880 = 6760

4.9

Circuit Theory 12 50 − 4 ∆2 = − 3 − 20 0 = 12 # 6 − 20 # 12 − 0 @ − 50 # 6 − 3 # 12 − 0 @ − 4 20 12 + (− 4) # 6 − 3 # 20 − (− 4) # (− 20) @ = −2880 + 1800 + 560 = − 520

12 − 3 50 ∆3 = − 3 11 − 20 = 12 # 611 # 20 − 0 @ − ^− 3h # 6 − 3 # 20 − ^− 4h # ^− 20h@ + 50 # 60 − (− 4) # 11 @ − 4 0 20 = 2640 − 420 + 2200 = 4420

I1 = ∆1 = 6760 = 5.2 A ∆ 1300 I2 =

∆2 = − 520 = − 0.4 A ∆ 1300

I3 =

∆3 = 4420 = 3.4 A ∆ 1300

Here, the mesh current I 2 is negative. Hence, the actual direction of I 2 is opposite to that of assumed direction. Since there are six branches in the given circuit, we can assume six currents Ia , I b , I c , I d , I e and I f as shown in Fig. 7. The direction of branch currents are chosen such that they are all positive. The relation between mesh and branch currents can be obtained from Fig. 7 and the branch currents are evaluated as shown below: 2‡

5‡

Ia = I1 = 5.2 A Ib = I1 – I2 = 5.2 – (–0.4) = 5.6 A

+ I1

50 V

Ic = I1 – I3 = 5.2 – 3.4 = 1.8 A

Id

Ib

Ia

E

4‡

I2

3‡

6‡

20 V I e

Ic

I d = –I2 = – (–0.4) = 0.4 A

I3

I e = I3 – I2 = 3.4 –(–0.4) = 3.8 A

8‡

If

Fig. 7.

I f = I3 = 3.4 A

EXAMPLE 4.2 Determine the currents in various elements of the bridge circuit shown in Fig. 1, by using mesh analysis. 1‡

+

5

Ia

The circuit has 6 currents (corresponding to six branches) and in this 3 currents are independent (corresponding to three meshes).

10

1‡

Ic

1‡ 1‡

m = B – N + 1 = 6 – 4 + 1 = 3.

E

E Ib

It has 6 branches and 4 nodes. Hence the number of meshes m in the circuit is,

If

V

The graph of the given circuit is shown in Fig. 2.

Id

1‡

SOLUTION

Ie +

1‡

E 5V

Fig. 1.

V +

Chapter 4 - Mesh Analysis

4 . 10

Let us assume three mesh currents as shown in Fig. 2. The direction of the current are chosen arbitrarily. The circuit with chosen mesh currents is shown in Fig. 3. Id

E Ib

I3

I2 Ic

Ia

Ie

E

V

1‡ I2

I3

Ic

1‡ 1‡

Ia

If

1 + 0

Id

+

If

5V

Ib

1‡

1‡

Ie

I1 I1

+

1‡

Fig. 2.

E 5V

Fig. 3.

Method I : Formation of mesh basis equation by applying KVL In this method, the mesh equations are formed by using Kirchhoff’s Voltage Law. The mesh equation for a mesh is formed by equating the sum of voltage fall to sum of voltage rise. The voltage rise and fall are determined by tracing the circuit in the direction of mesh current. With reference to Fig. 4, the mesh equation for mesh-1 is formed as shown below: I2

Voltage fall : I1, I1, I1

I3

I3

+

E

+ 5V E

+

.....(1)

I1

E + I1

I1 E

3I1 − I2 − I3 = 5

E

+

I1

∴ I1 + I1 + I1 = I2 + I3 + 5

+

E

I

2

Voltage rise : I2, I3 , 5

Fig. 4. E

With reference to Fig. 5, the mesh equation for mesh-2 is formed as shown below: I2

+

Voltage fall : I2, I2, I2

+ E I2 I3

+

5

V E

Voltage rise : I1, I3 , 5

I2

+

E

∴ I2 + I2 + I2 = I1 + I3 + 5

I1 E

I1

−I1 + 3I2 − I3 = 5

I2 +

I3

E +

..... (2)

Fig. 5.

+

With reference to Fig. 6, the mesh equation for mesh-3 is formed as shown below: I3

I3, I3, I3

E

Voltage fall :

E

I2

∴ I3 + I3 + I3 = I1 + I2 + 10 −I1 − I2 + 3I3 = 10

I2

I3 I 3 + + E I3 E

E

..... (3)

10 V

+

+ E

Voltage rise : I1, I2 , 10

+

I1 I1

Fig. 6.

4 . 11

Circuit Theory

The equations (1), (2) and (3) are the mesh equations of the circuit shown in Fig. 3. The mesh equations are summarized here for convenience. 3I1 − I2 − I3 = 5 −I1 + 3I2 − I3 = 5 −I1 − I2 + 3I3 = 10 The mesh equations can be arranged in the matrix form as shown below and then solved by Cramer’s rule. 3 − 1 − 1 I1 5 3 − 1 H >I 2 H = > 5 H 10 − 1 − 1 3 I3

>− 1

..... (4)

Method II : Formation of mesh basis equation by inspection In this method, the mesh basis matrix equation is formed directly from the circuit shown in Fig.3 by inspection. The circuit has three meshes. The general form of mesh basis matrix equation for three mesh circuit is shown in equation (5). R11 R12 R13 I1 E11 R 22 R 23 H >I 2 H = >E 22 H R31 R32 R33 I3 E33

>R21

..... (5)

The elements of the resistance matrix and source voltage matrix are formed as shown below: R11 = 1 + 1 + 1 = 3

R12 = R21 = −1

E11 = 5

R22 = 1 + 1 + 1 = 3

R13 = R31 = −1

E22 = 5

R33 = 1 + 1 + 1 = 3

R23 = R32 = −1

E33 = 10

On substituting the above terms in equation (5), we get, 3 −1 −1 I1 5 >− 1 3 − 1 H >I2 H = > 5 H I3 10 −1 −1 3

..... (6)

Solution of mesh currents It is observed that the mesh basis matrix equation obtained in method I and II are same. In equation (6) the unknowns are I 1, I2 and I 3 . In order to solve I 1 , I 2 and I 3 , let us define four determinants ∆, ∆ 1 , ∆ 2 and ∆ 3 as shown below: 3 −1 −1 ∆ = −1 3 −1 ; −1 −1 3

∆1 =

5 −1 −1 5 3 −1 ; 10 − 1 3

3 5 −1 ∆2 = − 1 5 − 1 ; − 1 10 3

3 −1 5 ∆3 = − 1 3 5 − 1 − 1 10

The determinants are evaluated by expanding along first row and the mesh currents are solved by Cramer’s rule. 3 −1 −1 ∆ = − 1 3 − 1 = 3 # 63 # 3 − (− 1) # (− 1) @ − (− 1) # 6 − 1 # 3 − ^− 1h # (− 1) @ −1 −1 3 + (− 1) # 6 − 1 # ^− 1h − ^− 1h # 3 @ = 24 − 4 − 4 = 16 ∆1 =

5 −1 −1 5 3 − 1 = 5 # 63 # 3 − (− 1) # (− 1) @ − (− 1) # 65 # 3 − 10 # (− 1) @ + (− 1) # 65 # ^− 1h − 10 # 3 @ 10 − 1 3 = 40 + 25 + 35 = 100

Chapter 4 - Mesh Analysis

4 . 12

3 5 −1 ∆2 = − 1 5 − 1 = 3 # 65 # 3 − 10 # (− 1) @ − 5 # 6 − 1 # 3 − (− 1) # (− 1) @ + (− 1) # 6 − 1 # 10 − ^− 1h # 5 @ − 1 10 3 = 75 + 20 + 5 = 100 3 −1 5 ∆3 = − 1 3 5 = 3 # 63 # 10 − (− 1) # 5 @ − (− 1) # 6 − 1 # 10 − (− 1) # 5 @ + 5 # 6 − 1 # (− 1) − (− 1) # 3 @ − 1 − 1 10 = 105 − 5 + 20 = 120 I1 =

∆1 = 100 = 6.25 A 16 ∆

I2 =

∆2 = 100 = 6.25 A 16 ∆

I3 =

∆3 = 120 = 7.5 A 16 ∆

The relation between mesh and branch currents can be obtained from Fig. 3 and branch currents are evaluated as shown below: Ia = I1 = 6.25 A Ib = I2 = 6.25 A Ic = I1 − I2 = 6.25 − 6.25 = 0 Id = I3 = 7.5 A Ie = I1 − I3 = 6.25 − 7.5 = −1.25 A If = I2 − I3 = 6.25 − 7.5 = −1.25 A

EXAMPLE 4.3

A

(AU Dec’14, 16 Marks)

In the circuit shown in Fig.1, find (a) mesh currents in the circuit, (b) current supplied by the battery, (c) potential difference between terminals B and D.

2‡

+

6‡

10 V E D

B

SOLUTION

4‡

Since the given circuit has only one source, it is possible to predict the exact directions of the current.

3‡

8‡ C

The current will start from positive end of the supply and when it enters node-A, it will divide into two parts. These two currents will again meet at node-C and enter negative end of the supply through 4 Ω resistor.

Ib 2‡

The circuit has three branch currents and in this two are independent. Hence we can take two mesh currents. B

The actual directions of mesh and branch currents are shown in Fig. 2. Using the circuit shown in Fig. 2, the mesh basis matrix equation is formed as shown below: R11 R12 I1 E = G = G = = 11 G R 21 R 22 I 2 E 22

+ 10 V E I1

A Ic Ia

6‡

I2 4‡

3‡

8‡ C

..... (1)

Fig. 1.

Fig. 2.

D

4 . 13

Circuit Theory The elements of the resistance matrix and source voltage matrix are formed as shown below: R11 = 2 + 4 + 3 = 9

R12 = R21 = 4

E 11 = 10

R22 = 4 + 6 + 8 = 18

E 22 = 10

On substituting the above terms in equation (1), we get,

=

9 4 G 4 18

I1 = G I2

=

10 = G 10

..... (2)

In equation (2), the unknowns are I 1 and I2 . In order to solve I 1 and I 2 , let us define three determinants ∆, ∆ 1 and ∆ 2 as shown below: ∆ =

9 4 ; 4 18

∆1 =

10 4 ; 10 18

∆2 =

9 10 4 10

The determinants are evaluated by expanding along first row and the mesh currents are solved by Cramer’s rule. ∆ =

9 4 = 9 # 18 − 4 # 4 = 146 4 18

∆1 =

10 4 = 10 # 18 − 10 # 4 = 140 10 18

∆2 =

9 10 = 9 # 10 − 4 # 10 = 50 4 10

I1 =

∆1 = 140 = 0.9589 A 146 ∆

I2 =

∆2 = 50 = 0.3425 A 146 ∆

b) To find the battery current With reference to Fig. 2, the battery current is given by, Ia = I1 + I2 IB

∴ Battery current, Ia = I1 + I2 = 0.9589 + 0.3425 = 1.3014 A E

+

I

2

+

+

D

E

8‡

C

Fig. 3.

(AU May’15, 8 Marks)

Use branch currents in the network shown in Fig. 1 to find the current supplied by the 60 V source. Solve the circuit by the mesh current method.

I

1

3‡

∴ VBD = 3 × 0.9589 − 8 × 0.3425 = 0.1367 V

EXAMPLE 4.4

3

E

VBD

B 4‡

VBD = 3I1 − 8I2

6‡

10 V

The given circuit is redrawn as shown in Fig. 3. With reference to Fig. 3, using KVL , we can write, ⇒

I2

2‡

+

E 8

c) To find potential difference between the termianals ‘’B’’ and ‘’D’’

VBD + 8I2 = 3I1

A I1

7‡

60 V

+ E

I3

I2

I1 12 ‡

6‡

Fig. 1.

I4 12 ‡

Chapter 4 - Mesh Analysis

4 . 14 SOLUTION

7‡

The direction of the mesh currents are chosen to match the given branch currents as shown in Fig. 2. With reference to Fig.2, the mesh basis matrix equation is formed as shown below: R11 R12 R13 I1 R 22 R 23 H >I 2 H = R31 R32 R33 I3

>R21

I4 I1 60 V + E

I2

12 ‡

6‡

I3

12 ‡

E11

>E22 H E33

Fig. 2.

..... (1)

Now, I4 = I1 − I2 − I3 The elements of the resistance matrix and source voltage matrix are formed as shown below: R11 = 7 + 12 = 19

R12 = R21 = – 12

E11 = 60

R22 = 12 + 12 = 24

R13 = R31 = −12

E22 = 0

R33 = 6 + 12 = 18

R23 = R32 = +12

E33 = 0

On substituting the above terms in equation (1), we get, R VR V R V S 19 − 12 − 12 W SI1 W S60 W S W S− 12 W 24 12 I2 = S 0 W S WS W S W ..... (2) SS− 12 12 18 WW SSI3 WW SS 0 WW T XT X T X In equation (2), the unknowns are I1, I2 and I3. In order to solve I1, I2 and I3, let us define four determinants ∆, ∆1, ∆2 and ∆3 as shown below: 19 − 12 − 12 60 − 12 − 12 19 60 − 12 19 − 12 60 ∆ = − 12 24 12 ; ∆1 = 0 24 12 ; ∆2 = − 12 0 12 ; ∆3 = − 12 24 0 0 12 18 − 12 12 18 − 12 0 18 − 12 12 0 The determinants are evaluated by expanding along first row and the mesh currents are solved by Cramer’s rule. 19 − 12 3 = − 12 24 12 − 12

− 12 12 = 19 # 724 # 18 − 122 A − _− 12i # 7 − 12 # 18 − _− 12i # 12 A 18 + _− 12i # 7 − 12 # 12 − _− 12i # 24 A = 5472 − 864 − 1728 = 2880

31 =

60 − 12 − 12 0 24 12 = 60 # 724 # 18 − 122 A = 17280 0 12 18

19 3 2 = − 12 − 12

60 0 0

19 − 12 3 3 = − 12 24 12 − 12

− 12 12 = − 60 # 7 − 12 # 18 − _− 12 i # 12 A = 4320 18 60 0 = 60 # 7 − 12 # 12 − _− 12i # 24 A = 8640 0

4 . 15

Circuit Theory I1 =

31 = 17280 = 6 A 2880 3

I2 =

32 = 4320 = 1.5 A 2880 3

I3 =

33 = 8640 = 3 A 2880 3

I4 = I1 − I2 − I3 = 6 − 1.5 − 3 = 1.5 A

EXAMPLE 4.5

6‡

2‡

2‡

Solve the mesh currents shown in Fig. 1. + 25 V E

SOLUTION The mesh currents and their direction are given in the problem and so we need not assume the currents. Using the circuit shown in Fig. 1, the mesh basis matrix equation is formed as shown below: R11 R12 R13 I1 R 22 R 23 H >I 2 H = R31 R32 R33 I3

4‡ I1

5‡

I2

+ 10 V E

I3

Fig. 1.

E11

>R21

>E22 H E33

..... (1)

The elements of resistance matrix and source voltage matrix are formed as shown below: R11 = 2 + 4 = 6

R12 = R21 = −4

E11 = 25

R22 = 4 + 6 + 5 = 15

R13 = R31 = 0

E22 =

R33 = 5 + 2 = 7

R23 = R32 = −5

E33 = −10

0

On substituting the above terms in equation (1), we get, 6 −4 0 15 − 5 H 0 −5 7

>− 4

I1

>I2 H I3

=

>

25 0H − 10

..... (2)

In equation (2), the unknowns are I1, I2 and I3. In order to solve I1, I2 and I3, let us define four determinants ∆, ∆1, ∆2 and ∆3 as shown below: 6 −4 0 ∆ = − 4 15 − 5 ; 0 −5 7

∆1 =

25 − 4 0 0 15 − 5 ; − 10 − 5 7

6 25 0 ∆2 = − 4 0 −5 ; 0 − 10 7

6 − 4 25 ∆3 = − 4 15 0 0 − 5 − 10

The determinants are evaluated by expanding along first row and the mesh currents are solved by Cramer’s rule. 6 −4 0 ∆ = − 4 15 − 5 = 6 # 615 # 7 − (− 5) # (− 5) @ − (− 4) # 6 − 4 # 7 − 0 @ + 0 0 −5 7 = 480 − 112 = 368

∆1 =

25 − 4 0 0 15 − 5 = 25 # 615 # 7 − (− 5) # (− 5) @ − (− 4) # 60 − (− 10) # (− 5) @ + 0 − 10 − 5 7 = 2000 − 200 = 1800

Chapter 4 - Mesh Analysis

4 . 16

6 25 0 ∆2 = − 4 0 − 5 = 6 # 60 − (− 10) # (− 5) @ − 25 # 6 − 4 # 7 − 0 @ + 0 0 − 10 7 = −300 + 700 = 400 6 − 4 25 ∆3 = − 4 15 0 = 6 # 615 # (− 10) − 0 @ − (− 4) # 6 − 4 # (− 10) − 0 @ + 25 # 6 − 4 # (− 5) − 0 @ 0 − 5 − 10 = −900 + 160 + 500 = −240 I1 =

∆1 = 1800 = 4.8913 A 368 ∆

I2 =

∆2 = 400 = 1.0870 A 368 ∆

I3 =

∆3 = − 240 = − 0.6522 A 368 ∆

EXAMPLE 4.6

2‡

In the circuit shown in Fig. 1, find E such that I 2 = 0.

E

+ E

3‡

4‡

I1

SOLUTION In mesh analysis, when the solution of mesh current are obtained by

2‡

Cramer’s rule, the mesh current I 2 is given by, I2 =

I2

+ E

8.4 V

5‡ I3

∆2 ∆

1‡

..... (1)

Fig. 1.

In equation (1), if I2 = 0, then ∆2 = 0. Therefore, in order to find the value of E, we can form the mesh basis matrix equation. Then form the determinant ∆2 and equate the determinant to zero. Using Fig. 1, the mesh basis matrix equation is formed by inspection as shown below: R11 R12 R13 I1 R 22 R 23 H >I 2 H = R31 R32 R33 I3

>R21

E11

>E22 H E33

..... (2)

The elements of resistance matrix and source voltage matrix are formed as shown below: R11 = 2 + 4 + 2 = 8

R12 = R21 = −4

E11 = E

R22 = 4 + 3 + 5 = 12

R13 = R31 = −2

E22 = −8.4

R33 = 2 + 5 + 1 = 8

R23 = R32 = −5

E33 = 0

On substituting the above terms in equation (2), we get, 8 −4 −2 12 − 5 H −2 −5 8

>− 4

I1

>I2 H I3

E

=

>− 8.4 H 0

In equation (3) by Cramer’s rule, the unknown current I 2 is given by, I2 =

8 E −2 ∆2 , where ∆2 = − 4 − 8.4 − 5 ∆ 0 8 −2

..... (3)

4 . 17

Circuit Theory On expanding ∆ 2 we get,

8 E −2 ∆2 = − 4 − 8.4 − 5 = 8 # 6 − 8.4 # 8 − 0 @ − E # 6 − 4 # 8 − (− 2) # (− 5) @ + (− 2) # 60 − (− 2) # (− 8.4) @ 0 8 −2 = −537.6 + 42E + 33.6 = −504 + 42E On equating ∆2 = 0, we get, − 504 + 42E = 0 ∴ 42 E = 504 E = 504 = 12 V 42 12 ‡

EXAMPLE 4.7 Solve the current in 12 Ω resistor by mesh analysis.

4‡

SOLUTION

I1 + E

The mesh currents and their directions are given in the problem and so we need not assume the currents. Using the circuit shown in Fig. 1 the mesh basis matrix equation is formed as shown below: R11 R12 R13 I1 >R21 R22 R23 H >I2 H = R31 R32 R33 I3

40 V

+ E

5‡ 10 V

I2

I3

+ E

60 V

7‡

Fig. 1.

E11 >E22 H E33

..... (1)

The elements of resistance matrix and source voltage matrix are formed as shown below: R11 = 12 + 4 + 5 = 21

R12 = R21 = −4

E11 = 0

R22 = 4 + 7 = 11

R13 = R31 = −5

E22 = 40 − 10 = 30 V

R33 = 7 + 5 = 12

R23 = R32 = −7

E33 = 10 − 60 = −50 V

On substituting the above terms in equation (1), we get, 21 − 4 − 5 I1 11 − 7 H >I 2 H = − 5 − 7 12 I3

>− 4

>

0 30 H − 50

..... (2)

The current through 12Ω resistance is I1 . To solve the current I 1 by Cramer’s rule let us define the determinants ∆ and ∆1 as shown below: 21 − 4 − 5 ∆ = − 4 11 − 7 ; − 5 − 7 12

∆1 =

0 −4 −5 30 11 − 7 − 50 − 7 12

The determinants are evaluated by expanding along first row and the mesh current I1 is solved by Cramer’s rule. 21 − 4 − 5 ∆ = − 4 11 − 7 = 21 # 611 # 12 − (− 7) # (− 7) @ − (− 4) # 6 − 4 # 12 − (− 5) # (− 7) @ − 5 − 7 12 + (− 5) # 6 − 4 # (− 7) − (− 5) # 11 @ = 1743 − 332 − 415 = 996

Chapter 4 - Mesh Analysis

4 . 18 ∆1 =

0 −4 −5 30 11 − 7 = 0 − (− 4) # 630 # 12 − (− 50) # (− 7) @ + (− 5) # 630 # (− 7) − (− 50) # 11 @ − 50 − 7 12 = 40 − 1700 = −1660

I1 =

∆1 = − 1660 = − 1.6667 A 996 ∆

EXAMPLE 4.8

2‡

8‡

Solve the mesh currents in the circuit shown in Fig. 1.

4‡

SOLUTION

4‡ I1

With reference to Fig.1, the mesh basis matrix equation is formed as shown below: R11 R12 R13 I1 R 22 R 23 H >I 2 H = R31 R32 R33 I3

>R21

+ E

10 ‡

I3

I2 + E

10 V

3‡

1‡

5V

+ E

8V

20 V

+ E

Fig. 1.

E11

>E22 H

..... (1)

E33

Note : Here, the directions of the mesh currents are given in the problem itself . The elements of the resistance matrix and source voltage matrix are formed as shown below: R11 = 4 + 8 + 4 = 16

R12 = R21 = – 4

E11 = 10 − 5 = 5

R22 = 4 + 2 + 1 = 7

R13 = R31 = 0

E22 = 5 − 8 = −3

R33 = 1 + 10 + 3 = 14

R23 = R32 = 1

E33 = 20 − 8 = 12

On substituting the above terms in equation (1), we get, 16 − 4 0 7 1H 0 1 14

>− 4

I1

>I2 H I3

5

=

>− 3 H

..... (2)

12

In equation (2), the unknowns are I1, I2 and I3. In order to solve I1, I2 and I3, let us define four determinants ∆, ∆1, ∆2 and ∆3 as shown below: 16 − 4 0 ∆ = −4 7 1 ; 0 1 14

5 −4 0 ∆1 = − 3 7 1 ; 12 1 14

16 5 0 ∆2 = − 4 − 3 1 ; 0 12 14

16 − 4 5 ∆3 = − 4 7 − 3 0 1 12

The determinants are evaluated by expanding along first row and the mesh currents are solved by Cramer’s rule. 16 − 4 0 ∆ = − 4 7 1 = 16 # 67 # 14 − 1 # 1 @ − (− 4) # 6 − 4 # 14 − 0 @ + 0 0 1 14 = 1552 – 224 = 1328 5 −4 0 ∆1 = − 3 7 1 = 5 # 67 # 14 − 1 # 1 @ − (− 4) # 6 − 3 # 14 − 12 # 1 @ + 0 12 1 14 = 485 – 216 = 269

4 . 19

Circuit Theory 16 5 0 ∆2 = − 4 − 3 1 = 16 # 6 − 3 # 14 − 12 # 1 @ − 5 # 6 − 4 # 14 − 0 @ + 0 0 12 14 = –864 + 280 = – 584 16 − 4 5 ∆3 = − 4 7 − 3 = 16 # 67 # 12 − 1 # (− 3) @ − (− 4) # 6 − 4 # 12 − 0 @ + 5 # 6 − 4 # 1 − 0 @ 0 1 12 = 1392 – 192 –20 = 1180 I1 =

∆1 = 269 = 0.2026 A 1328 ∆

I2 =

∆2 = − 584 = − 0.4398 A 1328 ∆

I3 =

∆3 = 1180 = 0.8886 A 1328 ∆

EXAMPLE 4.9 5‡

Determine the power dissipation in the 4 Ω resistor of the circuit shownin Fig. 1.

2‡

6‡

+ 3‡

50 V

SOLUTION

4‡

E

The graph of the given circuit is shown in Fig. 2. It has 5 branches and 3 nodes. Hence the number of meshes m in the circuit is, m = B − N + 1 = 5 − 3 + 1 = 3.

+ 10 V E

Fig. 1.

The circuit has 5 currents (corresponding to five branches) and in this 3 currents are independent (corresponding to three meshes). Let us assume three mesh currents I1, I2 and I3 as shown in Fig. 2. The direction of the currents are chosen arbitrarily. The circuit with chosen mesh currents is shown in Fig. 3. Now, the current through 4 Ω resistor is (I2− I3) in the direction shown in Fig. 3. b d

I2

a

I3

I1

50 V

c

E

I1

I2 E I3 6 ‡ 4‡

3‡

Fig. 2. `

2‡

5‡ +

e

I2

I3

+ 10 V E

Fig. 3.

Power dissipated in 4Ω resistor = I 2 − I3

2

# 4

Using the circuit shown in Fig. 3, the mesh basis matrix equation is formed as shown below : R11 R12 R13 I1 R 22 R 23 H >I 2 H = R31 R32 R33 I3

>R21

E11

>E22 H

.....(1)

E33

The elements of resistance matrix and source voltage matrix are formed as shown below : R11 = 5 + 3 = 8

R12 = R21 = −3

E11 = 50

R22 = 3 + 2 + 4 = 9

R13 = R31 = 0

E22 =

R33 = 4 + 6 = 10

R23 = R32 = −4

E33 = −10

0

Chapter 4 - Mesh Analysis

4 . 20 On substituting the above terms in equation (1), we get, 8 − 3 0 I1 9 − 4 H >I 2 H = 0 − 4 10 I3

>− 3

>

50 0H − 10

..... (2)

Here, we have to solve the mesh currents I2 and I3. In order to solve I2 and I3, let us define three determinants ∆, ∆2 and ∆3 as shown below: 8 −3 0 ∆ = −3 9 −4 ; 0 − 4 10

8 50 0 ∆2 = − 3 0 −4 ; 0 − 10 10

8 − 3 50 ∆3 = − 3 9 0 0 − 4 − 10

The determinants are evaluated by expanding along first row and then the currents I2 and I3 are solved by Cramer’s rule. 8 −3 0 ∆ = − 3 9 − 4 = 8 # 69 # 10 − (− 4) # (− 4) @ − (− 3) # 6 − 3 # 10 − 0 @ + 0 0 − 4 10 = 592 − 90 = 502 8 50 0 ∆2 = − 3 0 − 4 = 8 # 60 − (− 10) # (− 4) @ − 50 # 6 − 3 # 10 − 0 @ + 0 0 − 10 10 = −320 + 1500 = 1180 8 − 3 50 ∆3 = − 3 9 0 = 8 # 69 # (− 10) − 0 @ − (− 3) # 6 − 3 # (− 10) − 0 @ + 50 # 6 − 3 # (− 4) − 0 @ 0 − 4 − 10 = −720 + 90 + 600 = −30 I2 =

∆2 = 1180 = 2.3506 A 502 ∆

I3 =

∆3 = − 30 = − 0.0598 A 502 ∆

`

Power dissipated in 4 Ω resistor = I 2 − I3

2

= 2.4104

# 4 = 2.3506 − (− 0.0598) 2

2

#4

# 4 = 2.41042 # 4 = 23.2401W

EXAMPLE 4.10 6‡

Determine the voltage E which causes the current I1 to be zero for the circuit shown in Fig. 1.

2‡ I2 20 V

SOLUTION

+ E

6‡

I1 1‡ 5‡

In mesh analysis, when the solution of mesh currents are obtained by Cramer’s rule, the mesh current I 1 is given by, I1 =

∆1 ∆

I3 E

4‡

+ E

Fig. 1. ..... (1)

In equation (1), if I1 = 0, then ∆1 = 0. In order to find the value of E, we can form the mesh basis matrix equation. Then form the determinant ∆1 and equate the determinant to zero.

4 . 21

Circuit Theory

Using the circuit shown in Fig. 1, the mesh basis matrix equation is formed by inspection as shown below: R11 R12 R13 I1 R 22 R 23 H >I 2 H = R31 R32 R33 I3

E11

>R21

>E22 H

..... (2)

E33

The elements of resistance matrix and source voltage matrix are formed as shown below: R11 = 6 + 2 + 5 = 13

R12 = R21 = −2

E11 = 20 − E

R22 = 2 + 6 + 1 = 9

R13 = R31 = −5

E22 = 0

R33 = 5 + 1 + 4 = 10

R23 = R32 = −1

E33 = E

On substituting the above terms in equation (2), we get, 13 − 2 − 5 I1 9 − 1 H >I 2 H = − 5 − 1 10 I3

>− 2

Now, ∆1 =

>

20 − E 0H E

20 − E − 2 − 5 0 9 −1 E − 1 10

..... (3)

On expanding ∆1 along column-1 we get, ∆1 = (20 − E) # 69 # 10 − (− 1) # (− 1) @ − 0 + E # 6 − 2 # (− 1) − 9 # (− 5) @ = (20 − E) × 89 + 47 E = 1780 − 89 E + 47 E = 1780 − 42 E On equating ∆1 to zero we get, 0 = 1780 − 42 E ` 42E = 1780

⇒

E = 1780 = 42.381V 42

EXAMPLE 4.11 10 ‡

5‡

In the circuit shown in Fig. 1, (a) Use mesh analysis to find out the power delivered to 4 Ω resistor, (b) To what voltage should the 80 V battery be changed

60 V

+ E

40 V

4‡ + E

80 V

+ E

so that no power is delivered to the 4 Ω resistor?

Fig. 1.

SOLUTION Let us assume two mesh currents I1 and I2 as shown in Fig. 2. The direction of the currents are chosen as clockwise. Now the current through 4 Ω resistor is I2. `

Power delivered to 4 Ω resistor = I 2

2

I2

a) To find the power delivered to 4 Ω resistor

I2

I1 60 V

# 4

4‡

5‡

10 ‡ + E

40 V

+ E

Fig. 2.

Using the circuit shown in Fig. 2, the mesh basis matrix equation is formed as shown below: R11 R12 I1 E = G = G = = 11 G R 21 R 22 I 2 E 22

..... (1)

The elements of resistance matrix and source voltage matrix are formed as shown below: R11 = 10 + 5 = 15 R22 = 5 + 4 = 9

R12 = R21 = −5

80 V

E11 = 60 − 40 = 20 E22 = 40 − 80 = −40

+ E

Chapter 4 - Mesh Analysis

4 . 22 On substituting the above terms in equation (1), we get, 15 − 5 I1 20 = G = G = = G − 5 9 I2 − 40 Let us solve I2 by Cramer’s rule. Now, I 2 =

∆2 ∆ where,

∆ = ∆2 =

` `

I2 =

15 − 5 = 15 # 9 − (− 5) # (− 5) = 110 −5 9 15 20 = 15 # (− 40) − (− 5) # 20 = − 500 − 5 − 40

∆2 = − 500 = − 4.5455 A ∆ 110 # 4 = − 4.5455 2 # 4 = 4.54552 # 4 = 82.6463 W

Power delivered to 4 Ω resistor = I 2

2

b) To find the change in voltage in 80 V source such that power delivered to 4 Ω resistor is zero Let us take the new value of 80 V source as E. Now in equation (1), the E22 is given by, E22 = 40 − E. The equation (1) for case (b) is given below:

=

15 − 5 I1 20 G = G = = G 40 − E − 5 9 I2

..... (2)

In equation (2), by Cramer’s rule, I2 is given by I2 = ∆2 / ∆. If power delivered to 4 Ω is zero, then I2 should be zero. For I2 to be zero, the determinant ∆2 should be zero. Now, ∆2 =

15 20 = 15 # (40 − E) − (− 5) # 20 − 5 40 − E = 600 – 15E + 100 = 700 – 15E

On equating ∆2 to zero we get, 0 = 700 − 15 E 15 E = 700 E = 700 = 46.6667 V 15 ∴ The value of 80 V should be reduced to 46.6667 V to make the power delivered to 4 Ω resistor as zero.

EXAMPLE 4.12

+

5‡

I

10 V

In the circuit shown in Fig. 1, find the current I by mesh method and the power supplied by each batteries to 1.25 Ω resistor.

15 ‡ 1.25 ‡

E

+ 20 V E

Fig. 1.

SOLUTION +

Let us assume two mesh currents as shown in Fig. 2. Now the current I

10 V E

5‡ I1

I

15 ‡ 1.25 ‡ I2

is given by sum of I1 and I2.

Fig. 2.

+ 20 V E

4 . 23

Circuit Theory Using the circuit shown in Fig. 1, the mesh basis matrix equation is formed as shown below: R11 R12 I1 E = G = G = = 11 G R 21 R 22 I 2 E 22

.....(1)

The elements of resistance matrix and source voltage matrix are formed as shown below: R11 = 5 + 1.25 = 6.25

R12 = R21 = 1.25

R22 = 15 + 1.25 = 16.25

E11 = 10 E22 = 20

On substituting the above terms in equation (1), we get,

=

6.25 1.25 G 1.25 16.25

I1 = G I2

=

10 = G 20

..... (2)

In equation (2), the unknowns are I1 and I2. In order to solve I1 and I2, let us define three determinants ∆, ∆1 and ∆2 as shown below and the mesh currents are solved by Cramer’s rule. ∆ =

6.25 1.25 = 6.25 # 16.25 − 1.25 # 1.25 = 100 1.25 16.25

∆1 =

10 1.25 = 10 # 16.25 − 20 # 1.25 = 137.5 20 16.25

∆2 =

6.25 10 = 6.25 # 20 − 1.25 # 10 = 112.5 1.25 20

I1 =

∆1 = 137.5 = 1.375 A 100 ∆

I2 =

∆2 = 112.5 = 1.125 A 100 ∆

∴ I = I1 + I2 = 1.375 + 1.125 = 2.5 A Let P10 and P20 be the power delivered by 10 V and 20 V sources. Now, P10 = 10 × I1 = 10 × 1.375 = 13.75 W P20 = 20 × I2 = 20 × 1.125 = 22.5 W Let P5 and P15 be the power consumed by 5 Ω and 15 Ω resistances respectively. 2

Now, P5 = I12 × 5 = 1.375 × 5 = 9.4531 W 2

P15 = I22 × 15 = 1.125 × 15 = 18.9844 W Let PL10 and PL20 be the power delivered to load (i.e., to 1.25 Ω resistor) by the 10 V and 20 V sources respectively. Now, PL10 = P10 − P5 = 13.75 − 9.4531 = 4.2969 W PL20 = P20 − P15 = 22.5 − 18.9844 = 3.5156 W

Cross-Check Power consumed by 1.25 Ω resistance, PL = I2 × 1.25 = 2.52 × 1.25 = 7.8125 W Also, PL = PL10 + PL20 = 4.2969 + 3.5156 = 7.8125 W

Chapter 4 - Mesh Analysis

4 . 24

4.3

Mesh Analysis of Circuits Excited by Both Voltage and Current Sources

The mesh analysis can be extended to circuits excited by both voltage and current sources. In such circuits if each current source has a parallel impedance then they can be converted to an equivalent voltage source with series impedance. After conversion the circuit will have only voltage sources and so the procedure for obtaining mesh basis matrix equation by inspection and its solution discussed in Section 4.2 and 4.4 can be directly applied to these circuits. In certain circuits excited by both voltage and current sources, the current source may not have a parallel resistance. In this situation the current source cannot be converted to voltage source. In this case the value of each current source is related to mesh currents and one of the mesh current can be expressed in terms of source current and other mesh currents. The remaining mesh currents can be solved by writing Kirchhoff’s Voltage Law equations. Alternatively, the mesh basis matrix equation can be formed directly by inspection, by taking the voltage of the current sources as unknown and relating the value of each current source to mesh currents. Here for each current source one mesh current is eliminated by expressing the mesh current in terms of the source current and other mesh currents. While forming the mesh basis matrix equation, the voltage of the current sources should be entered in the source matrix. Now in the matrix equation some mesh currents will be eliminated and equal number of unknown source voltages will be introduced. Thus, the number of unknowns will remain same as number of meshes m. On multiplying the mesh basis matrix equation, we get m equations which can be solved to give a unique solution for unknown currents. 4.3.1 Supermesh Analysis In circuits excited by both voltage and current sources, if a current source lie common to two meshes then the common current source can be removed for analysis purpose and the resultant two meshes can be considered as one single mesh called supermesh. In order to solve the two mesh currents of a supermesh, two equations are required. One of the equation is the KVL equation of the supermesh and the other equation is obtained by equating the source current to sum or difference of the mesh currents (depending on the direction of mesh currents). An example of formation of supermesh is shown in Fig. 4.5. Also, the example 4.14 is solved by using supermesh analysis technique. 5W

2W +

-

+ 5I2

2I1 +

10 V -

Mesh-1 I1 - 6I1 +

Mesh-2 + 3I2 -

4A

5W

2W +

-

-

+

2I1 3W

I2

6W

Fig. a : Two mesh circuit with mesh currents in same orientation.

Þ

+

10 V -

Supermesh

5I2 + 3I2 -

3W

Supermesh equations 2I1 + 5I2 + 3I2 + 6I1 = 10

- 6I1 + 6W

Fig. b : Supermesh of circuit shown in Fig. a and its equations.

I1 - I2 = 4

4 . 25

Circuit Theory 5W

2W +

5I2

2I1 +

10 V -

I1

4A

5W

2W +

-

-

+

5I2

2I1

_

I2 3I2 +

Þ

3W

+

-

-

_

+

10 V -

- 6I1 +

Supermesh

3W

3I2 +

I1 + I2 = 4

- 6I1 +

6W

Supermesh equations 2I1 + 6I1 = 5I2 + 3I2 + 10

6W

Fig. c : Two mesh circuit with mesh currents in opposite orientation.

Fig. d : Supermesh of circuit shown in Fig. c and its equations.

Fig. 4.5 : Examples of formation of supermesh. A

EXAMPLE 4.13

5‡

4‡

Find the voltage between A and B of the circuit shown in Fig. 1, by mesh analysis.

4‡ 1‡

4‡

10 A

E 10 V + B

+ 20 V E

Fig. 1.

SOLUTION

The graph of the given circuit is shown in Fig. 2. It has5 branches and 3 nodes. Hence, the number of meshes m in the circuit is, m = B − N + 1 = 5 − 3 + 1 = 3. The circuit has 5 currents (corresponding to five branches) and in this 3 currents are independent (corresponding to three meshes). Let us assume three mesh currents as shown in Figs. 2 and 3. A c 5‡

4‡ b a

I2

+

d I3

I1

10 A

4‡

E1

e E

Fig. 2.

I1

I2 B

4‡ E

1‡ E

20 V

I3 +

10 V +

Fig. 3.

The directions of mesh currents are chosen arbitrarily. Here, one of the mesh has 10 A current source which cannot be converted to voltage source, because the source does not have parallel impedance. Hence we can take this current as a known mesh current, but the voltage across the source E 1 is unknown. Therefore, the number of unknowns remain as three (i.e., unknowns are E1, I 2 and I3) and so we can write three mesh equations using KVL (corresponding to three meshes) and a unique solution is obtained by solving the three equations. The mesh equations can be obtained by two methods. Note : In solutions of simultaneous equations a unique solution can be obtained only if the number of unknowns are equal to number of equations.

Method I : Formation of mesh equations by applying KVL In this method the mesh equations are formed by using Kirchhoff’s Voltage Law. The mesh equation for a mesh is formed by equating the sum of voltage fall to sum of voltage rise. The voltage rise and fall are determined by tracing the circuit in the direction of mesh current.

Chapter 4 - Mesh Analysis

4 . 26

With reference to Fig. 4, the mesh equation for mesh-2 is formed as shown below: Voltage fall : 4 I 2 , 5I 2 , I 2

+ + E

Voltage rise : 4I1, I3, 10 V

4I1 I1

∴ 4 I 2 + 5I 2 + I 2 = 4I1 + I 3 + 10 − 4I1 + 10I 2 − I3 = 10 − 4 # 10 + 10I2 − I3 = 10

⇒

5I2

E

4I2 I2

E +

I1 = 10A

+ E I2 I3 I3 E + E 10 V +

Fig. 4.

10I2 − I3 = 50

..... (1)

With reference to Fig. 5, the mesh equation for mesh-3 is formed as shown below: Voltage fall : I3, 4I3, 10 V

+ E I2

Voltage rise : I2, 20 V ∴ I3 + 4I3 + 10 = I2 + 20 ⇒ − I2 + 5I3 = 10

I2

..... (2)

+

4I3

E

I3

E

E + E 10 V +

The mesh equations (1) and (2) are sufficient for solving I2 and I3.

20 V

I3 +

Fig. 5.

Method II : Formation of mesh equations by inspection In this method the mesh basis matrix equations is formed by inspection using the circuit shown in Fig. 3. R11 R12 R13 I1 R 22 R 23 H >I 2 H = R31 R32 R33 I3

>R21

E11

>E22 H

..... (3)

E33

R 11 = 4 + 4 = 8

R 12 = R 21 = − 4

E 11 = E 1

R 22 = 4 + 5 + 1 = 10

R 13 = R 31 = 0

E 22 = 10

R 33 = 1 + 4 = 5

R 23 = R 32 = − 1

E 33 = − 10 + 20 = 10

I1 = 10

On substituting the above terms in equation (3), we get, 8 − 4 0 10 10 − 1 H > I 2 H = 0 −1 5 I3

>− 4

E1

>10 H

..... (4)

10

On multiplying the matrices on left-hand side of equation (4), and equating to terms on right-hand side, we get the following equations. From row - 2,

−40 + 10I2 − I3 = 10

From row - 3,

−I2 + 5I3 = 10

⇒

10I2 − I3 = 50

..... (5) ..... (6)

Solution of mesh currents It is observed that the mesh equations obtained by both the method are same. From equation (5) we get, I3 = 10I2 – 50

..... (7)

4 . 27

Circuit Theory On substituting for I3 from equation (7) in equation (6), we get,

⇒

– I 2 + 5(10 I 2 – 50) = 10 `

49I 2 − 250 = 10

I2 = 260 = 5.3061 A 49 I3 = 10I2 − 50 = 10 × 5.3061 − 50 = 3.061 A

To find voltage across A and B Let us denote the meeting point of 4 Ω and 5 Ω as node-C and the meeting point of 1 Ω and 10 V source as node-D as shown in Fig. 6. There are two shortest path to find the voltage across A and B. They are closed path ABCA and ABDA. Let voltage across A and B be denoted as VAB. With reference to Fig. 6, in path-ABCA by KVL we can write,

C

+

5I2

A

E +

+ E

+ E 4I1

E +

I1

I3

I2

E + E D

AB

∴ VAB = 4I1 − 9I2 = 4 × 10 − 9 × 5.3061 = −7.7549 V

I2

4I2

V

VAB + 4I2 + 5I2 = 4I1

10 V

In path-ABDA by KVL we can write,

E

+

B

VAB + 10 + I3 = I2

Fig. 6.

∴ VAB = I2 − I3 − 10 = 5.3061 − 3.061 − 10 = −7.7549 V

EXAMPLE 4.14 In the circuit shown in Fig. 1, find the current supplied by the voltage source and the voltage across the current source by mesh analysis.

5‡

2‡

+ 10 V

SOLUTION

2‡ +

10 V _

10 A

4‡

4‡

E

Let us assume three mesh currents as shown in Fig. 2. The current delivered by the voltage source is I1. Let the voltage across the current source be E with current leaving point as positive. Also, the voltage across various elements of the circuit are shown in Fig. 2.

I1 +

3‡

3‡

5‡ +

E

Fig. 1.

E

+

2‡ +

E

2I1

5I2 3I3 _ + Mesh-3 + Mesh-1 + Mesh-2 4I1 4I2 4I3 4 ‡ E _ + _ 10 A _ I2 I1 I3

2I1

+ 10 V _

3‡

5‡ E

+

+ _ 4I1 4I2 _ + Mesh-1

Fig. 2.

E

+

5I2

E 3I3

Supermesh

+ 4I3 4 ‡ _

Fig. 3.

With reference to Fig. 2, the relation between mesh currents I2 and I3 is, I 3 – I 2 = 10

⇒

I 3 = 10 + I 2

..... (1)

Let us combine mesh-2 and mesh-3 and form a supermesh as shown in Fig. 3. The KVL equation for the supermesh is formed as shown below: 4 I2 + 5I2 + 3 I 3 + 4 I 3 = 4 I 1

⇒ 4 I1 = 9I2+ 7I3

Chapter 4 - Mesh Analysis

4 . 28 ∴ 4 I1 = 9I2+ 7(10 + I2)

Using equation (1)

∴ 4 I1 = 16I2 + 70

⇒ 16I2 = 4I1 – 70

` I 2 = 4 I1 − 70 16 16

⇒

..... (2)

I 2 = 0.25I1 − 4.375

The KVL equation for the mesh-1 is formed as shown below: 2I1 + 4I1 = 10 + 4I2 ∴ 6I1 = 10 + 4(0.25I1 – 4.375) ` 6I1 = 10 + I1 − 17.5

⇒

Using equation (2)

5I1 = − 7.5

I1 = − 7.5 − 1.5 A 5

⇒

∴ I2 = 0.25I1 – 4.375 = 0.25(– 1. 5) – 4.375 = –4.75 A ∴ I3 = 10 + I2 = 10 – 4.75 = 5.25 A With reference to Fig. 2 by KVL, E = 3I3 + 4I3 = 7I3 = 7 × 5.25 = 36.75 V ∴ Current supplied by the voltage source, I1 = –1.5 A Voltage across the current source, E = 36.75 V

(AU June’14, 8 Marks)

EXAMPLE 4.15 Find out the current in the each branch of the circuit shown in Fig 1.

3‡

1‡

SOLUTION + 5A

Let us assume four branch currents are Ia, Ib, Ic and Id as shown in Fig. 2. The current source of Fig.2 can be represented by an equivalent voltage source of value 50 V with a source resistance of 10 Ω in series as shown in Fig. 3. Let us assume two mesh currents I1 and I2 as shown in Fig. 3. Ia 3 W Ib 5A

10 W

Id

Fig. 1. 1W

10 W

+ 10 V

5W -

Þ

5 ´ 10 + = 50 V -

10 V

5‡ E

3W

Ic 1 W

10 ‡

+ 5W

I1

I2

10 V -

Fig. 3.

Fig. 2.

Using the circuit shown in Fig. 3, the mesh basis matrix equation is formed as shown below : R11 R12 I1 E = G = G = = 11 G R 21 R 22 I 2 E 22

..... (1)

The elements of the resistance matrix and source voltage matrix are formed as shown below: R11 = 10 + 3 + 5 = 18 R22 = 5 +1 = 6

R12 = R21 = − 5

E 11 = 50 E 22 = − 10

4 . 29

Circuit Theory On substituting the above terms in equation (1), we get,

>

18 −5

I1 50 −5 H > H = > H I2 6 − 10

.....(2)

In equation (2), the unknowns are I 1 and I2 . In order to solve I 1 and I 2 , let us define three determinants ∆, ∆ 1 and ∆ 2 as shown below: ∆ =

18 − 5 ; −5 6

∆1 =

50 − 5 ; − 10 6

∆2 =

18 50 − 5 − 10

The determinants are evaluated as shown below and the mesh currents are solved by Cramer’s rule. 3 = 18 −5 31 =

32 =

− 5 = 18 6 − _− 5 i2 = 108 − 25 = 83 # 6 50

− 10 18 −5

− 5 = 50 6 − − 10 − 5 = 300 − 50 = 250 _ # # i 6 50 = 18 _− 10 i − − 5 50 = − 180 + 250 = 70 _ # # i − 10

I1 =

31 = 250 = 3.012 A 83 3

I2 =

32 = 70 = 0.8434 A 83 3

The branch currents are, Ia = I1 = 3.012 A Ib = 5 - Ia = 5 − 3.012 A = 1.988 A Ic = I2 = 0.8434 A Id = I1 − I2 = 3.012 − 0.8434 = 2.1686 A

EXAMPLE 4.16

(AU June’14, 8 Marks)

Determine the current in each mesh of the circuit shown in Fig. 1.

+ E

3‡

10 A

10 V

2‡

1‡

SOLUTION Let, voltage across 10 A curret source be VS and I1 , I2 and I3 be mesh currets as shown in Fig. 2.Here I1 = 10 A. Using the circuit shown in Fig. 2, the mesh basis matrix equation is formed as shown below: R11 R12 R13 I1 >R21 R22 R23 H >I2 H = R31 R32 R33 I3

E11 E > 22 H E33

Fig. 1. + + 10 A

VS E

E

I1

3‡

..... (1)

I2

10 V

2 ‡ I3

Fig. 2.

The elements of resistance matrix and source voltage matrix are formed as shown below: R11 = 3

R12 = R21 = −3

E11 = VS

R22 = 3 + 2 = 15

R13 = R31 = 0

E22 = −10

R33 = 2 + 1 = 7

R23 = R32 = −2

E33 = 10

I1 = 10 A

1‡

Chapter 4 - Mesh Analysis

4 . 30 On substituting the above terms in equation (1), we get, R 3 −3 0 VW RS10 VW R Vs V S S W S− 3 5 − 2 W S I2 W = S− 10 W S W S W S W S 0 −2 3 W S I3 W T 10 X T X T X From row-2 we get,

.....(2)

−30 + 5I2 − 2I3 = −10 ∴ 5I2 − 2I3 = 20

.....(1)

From row-3 we get, − 2I2 + 3I3 = 10 Equation (1) × 3 Equation (2) × 2

.....(2)

⇒ 15I2 − 6I3 = 60 ⇒ − 4I2 + 6I3 = 20 Add

11I2

= 80

` I 2 = 80 = 7.2727 A 11 10 + 2 I 2 From equation (2), I3 = = 10 + 2 # 7.2727 = 8.1818 A 3 3 The mesh currents I1 , I2, and I3 are given by, I1 = 10 A I2 = 7.2727 A I3 = 8.1818 A

4.4

Mesh Analysis of Circuits Excited by ac Sources (Mesh Analysis of Reactive Circuits)

The reactive circuits consists of resistances, inductive and capacitive reactances. Therefore, the voltages and currents of reactive circuits will be complex (i.e., they have both real and imaginary components). In general the elements of the circuit are referred to as impedances. The general mesh basis matrix equation for reactive circuit is, ZI=E ..... (4.8) where, Z = Impedance matrix of order m × m I = Mesh current matrix of order m × 1 E = Source voltage matrix of order m × 1 m = Number of meshes The equation (4.8), can be expanded as shown in equation (4.9). R V R V R V S Z11 Z12 Z13 g Z1m W S I1 W S E11 W S Z 21 Z 22 Z 23 g Z 2m W S I 2 W S E 22 W S W S W S W S Z31 Z32 Z33 g Z3m W S I3 W = S E33 W S h S h W h h h W Sh W SS W S W SS W Z m1 Z m2 Z m3 g Z mm W SI m W E mm W T X T X T X

..... (4.9)

4 . 31

Circuit Theory

Note : The over bar is used to denote complex quantities. The formation of mesh basis matrix equation and the solution of mesh and branch currents are similar to that of resistive circuits except that the solution of currents involves complex arithmetic. Therefore, the kth mesh current of a reactive circuit with m-meshes is given by, Note : Refer equation (4.5). Ik = 1 ∆

m

/∆

jk E jj

..... (4.10)

j =1

where, ∆ jk = Cofactor of Z jk . th

E jj = Sum of voltage sources in j mesh.

∆ = Determinant of impedance matrix. Instead of using above equation for solution of mesh currents, the short-cut procedure for Cramer’s rule can be followed. Consider the mesh basis matrix equation for three mesh circuit consisting of reactive elements. Z11 Z12 Z13 Z 22 Z 23 H Z31 Z32 Z33

>Z21

>I2 H I1

I3

E11 = >E 22 H E33

Let us define four determinants as shown below: Z11 Z12 Z13 E11 Z12 Z13 Z11 E11 Z13 Z11 Z12 E11 ∆ = Z 21 Z 22 Z 23 ; ∆1 = E 22 Z 22 Z 23 ; ∆2 = Z 21 E 22 Z 23 ; ∆3 = Z 21 Z 22 E 22 Z31 Z32 Z33 E33 Z32 Z33 Z31 E33 Z33 Z31 Z32 E33

Here,

∆ = Determinant of impedance matrix. ∆1 = Determinant of impedance matrix after replacing the first column of impedance matrix by source voltage column matrix. ∆2 = Determinant of impedance matrix after replacing the second column of impedance matrix by source voltage column matrix. ∆3 = Determinant of impedance matrix after replacing the third column of impedance matrix by source voltage column matrix.

Now mesh currents I1, I2 and I3 are given by, ∆1 ∆ ∆2 I2 = ∆ I1 =

I3 =

∆3 ∆

Chapter 4 - Mesh Analysis

4 . 32

EXAMPLE 4.17

j5 W

5W

Solve the currents in various branches of the circuit shown in Fig. 1, by mesh analysis.

4W

+ 100Ð0oV -

4W -j2 W

2W

~

SOLUTION

2W

Fig. 1.

The graph of the given circuit is shown in Fig. 2. It has 5 branches and 3 nodes. Hence the number of meshes m in the circuit is, m = B − N + 1 = 5 − 3 + 1 = 3.

Ib Ia

The circuit has 5 currents (corresponding to five branches) and in this 3 currents are independent (corresponding to three meshes).

b

Ic

c

a

Z11 Z12 Z13 >Z21 Z22 Z23 H Z31 Z32 Z33

I1 E11 >I2 H = >E22 H I3 E33

Id

d I3

I1

Let us assume the mesh currents I1, I 2 and I3 and the branch currents Ia, Ib, Ic, Id and Ie as shown in Figs. 2 and 3. The directions of the currents are chosen arbitrarily. With reference to Fig. 3, the mesh basis matrix equation is formed as shown below:

I2

j5 W

5W

+ o 100Ð0 V

~

-

e

Fig. 2. Ib

Ia

Ie

Ie

Ic

4W

I2

..... (1)

2W

-j2 W

2W

I1

Id 4 W

I3

Fig. 3.

Z11 = 5 + j5 + 2 = 7 + j5

Z12 = Z 21 = − 2

o E11 = 100∠0 = 100

Z 22 = 2 + 4 − j2 = 6 − j2

Z13 = Z31 = 0

E 22 = 0

Z33 = −j2 + 4 + 2 = 6 − j2

Z 23 = Z32 = − (− j2) = j2

E33 = 0

On substituting the above terms in equation (1), we get, 7 + j5 − 2 0 j2 H − 2 6 − j2 0 j2 6 − j2

>

>I2 H I1

=

I3

100 0H 0

>

..... (2)

In equation (2), the unknowns are I1, I 2 and I3 . In order to solve I1, I 2 and I3 by Cramer’s rule, let us define four determinants ∆, ∆1, ∆2 and ∆3 as shown below: ∆ =

7 + j5 0 −2 j2 ; − 2 6 − j2 0 j2 6 − j2

∆1 =

100 0 −2 0 6 − j2 j2 0 j2 6 − j2

∆2 =

7 + j5 100 0 0 j2 ; −2 0 0 6 − j2

∆3 =

7 + j5 − 2 100 0 − 2 6 − j2 0 j2 0

The determinants are evaluated by expanding along first row and the mesh currents are solved by Cramer’s rule. 7 + j5 − 2 0 ∆ = j2 = (7 + j5) # 6(6 − j2) # (6 − j2) − j2 # j2 @ − 2 6 − j2 0 j2 6 − j2 − (− 2) # 6 − 2 # (6 − j2) − 0 @ + 0 = (7 + j5) × [36 − j24] + 2 × [−12 + j4] = 348 + j20 Note : All calculations are performed using the calculator in complex mode.

4 . 33

Circuit Theory ∆1 =

∆2 =

∆3 =

100 − 2 0 0 6 − j2 j2 = 100 # 6(6 − j2) 2 − (j2) 2 @ − (− 2) # 60 − 0 @ + 0 0 j2 6 − j2 = 100 # (36 − j24) = 3600 − j2400 7 + j5 100 0 0 j2 = (7 + j5) # 60 − 0 @ − 100 # 6 − 2 # (6 − j2) − 0 @ + 0 −2 0 0 6 − j2 = − 100 # 6 − 12 + j4 @ = 1200 − j400 7 + j5 − 2 100 0 = (7 + j5) # 60 − 0 @ − (− 2) # 60 − 0 @ + 100 # 6 − 2 # j2 − 0 @ − 2 6 − j2 0 j2 0 = 100 # (− j4) = − j400

I1 =

3600 - j2400 ∆1 = 9.9157 − j7.4664 = 12.412+ − 37 o A = ∆ 348 + j20

I2 =

1200 - j400 ∆2 = 3.3711 − j1.3432 = 3.629+ − 21.7 o A = ∆ 348 + j20

I3 =

− j400 ∆3 = − 0.0658 − j1.1456 = 1.147+ − 93.3 o A = 348 + j20 ∆

With reference to Fig. 3, the following relations between mesh and branch currents are obtained. Then the branch currents are evaluated using the mesh currents I1, I 2 and I3 . Ia = I1 = 12.412+ − 37 o A I b = I 2 = 3.629+ − 21.7 o A Ic = I1 − I 2 = 9.9157 − j7.4664 − (3.3711 − j1.3432) = 6.5446 − j6.1232 = 8.962+ − 43.1 o A I d = I 2 − I3 = 3.3711 − j1.3432 − (− 0.0658 − j1.1456) = 3.4369 − j0.1976 = 3.443+ − 3.3 o A I e = I3 = 1.147+ − 93.3 o A

EXAMPLE 4.18 +

In the circuit shown in Fig. 1, find the mesh currents. o

100Ð0 V

SOLUTION With reference to Fig. 1, the mesh basis matrix equation is formed by inspection as shown below: Z11 Z12 Z13 Z 22 Z 23 H Z31 Z32 Z33

>I2 H I1

+

~

50Ð90oV

-

I2

I3

Fig. 1.

>E22 H E11

=

I3

Z11 = 5 + j2 Z 22 = j2 + 4 − j12

I1

2W -j12 W

j2 W

~

-

>Z21

4W

5W

4 − j10

Z33 = −j12 + 2 = 2 − j12

..... (1)

E33

Z12 = Z 21 = − j2

o E11 = 100∠0 = 100

Z13 = Z31 = 0

E 22 = 0

Z23 = Z32 = − (− j12) = j12

o E33 = −(50∠90 ) = −j50

Chapter 4 - Mesh Analysis

4 . 34 On substituting the above terms in equation (1), we get, 5 + j2 0 − j2 j12 H − j2 4 − j10 0 j12 2 − j12

>

>I2 H I1

>

=

I3

100 0H − j50

..... (2)

To solve the unknowns (i.e., mesh currents) of equation (2) by Cramer’s rule, we can define four determinants ∆, ∆1, ∆2 and ∆3 as shown below: ∆ =

5 + j2 0 − j2 j12 − j2 4 − j10 0 j12 2 − j12

;

∆1 =

100 0 − j2 0 4 − j10 j12 j12 2 − j12 − j50

∆2 =

5 + j2 100 0 0 j12 − j2 0 − j50 2 − j12

;

∆3 =

5 + j2 − j2 100 0 − j2 4 − j10 0 j12 − j50

The determinants are evaluated by expanding along first row and the mesh currents are solved by Cramer’s rule. ∆ =

5 + j2 0 − j2 j12 = (5 + j2) # [(4 − j10) # (2 − j12) − j12 # j12] − (− j2) # 6 − j2 # (2 − j12) − 0 @ + 0 − j2 4 − j10 0 j12 2 − j12 = 296 – j276 + 8 − j48 = 304 − j324

100 0 − j2 ∆1 = 0 4 − j10 j12 = 100 # [(4 − j10) # (2 − j12) − (j12) 2] − (− j2) # 60 − (− j50) # j12 @ + 0 j12 2 − j12 − j50 = 3200 – j6800 − j1200 = 3200 − j8000

∆2 =

5 + j2 100 0 0 j12 − j2 0 − j50 2 − j12

= (5 + j2) # [0 − (− j50) # j12] − 100 # [− j2 # (2 − j12) − 0] + 0 = − 3000 − j1200 + 2400 + j400 = −600 − j800

∆3 =

5 + j2 − j2 100 0 = (5 + j2) # [(4 − j10) # (− j50)] − 0] − (− j2) # [− j2 # (− j50) − 0] − j2 4 − j10 0 j12 − j50 + 100 # 6 − j2 # j12 − 0 @ = –2100 – j2000 – j200 + 2400 = 300 − j2200 I1 =

3200 − j8000 ∆1 = 18.0595 − j7.0682 = 19.393∠−21.4 o A = ∆ 304 − j324

I2 =

− 600 − j800 ∆2 = 0.3891 − j2.2169 = 2.251∠−80 o A = ∆ 304 − j324

I3 =

300 − j2200 ∆3 = = 4.0731 − j2.8958 = 4.998∠−35.4 o A ∆ 304 − j324

4 . 35

Circuit Theory

EXAMPLE 4.19

3W

In the circuit shown in Fig. 1, Find E 2 such that the current in (1 + j1) Ω branch is zero.

1W

j1 W

6W

+

+

~

2W

j4 W

o

30Ð0 V

-

~

E2

-

SOLUTION

Fig. 1.

Let us assume three mesh currents I1, I 2 and I3 . as shown in Fig. 2. With reference to Fig. 2, the mesh basis matrix equation is formed by inspection as shown below:

~

-

Z11 Z12 Z13 >Z21 Z22 Z23 H Z31 Z32 Z33

I1 >I2 H = I3

E11 >E22 H E33

3W

1W

o

j4 W

j1 W

6W

+ 30Ð0 V

I2

I1

+

~

2W I3

E2

-

Fig. 2.

..... (1)

Z11 = 3 + j4

Z12 = Z 21 = − j4

o E11 = 30∠0 = 30

Z 22 = j4 + 1 + j1 + 2 = 3 + j5

Z13 = Z31 = 0

E 22 = 0

Z33 = 2 + 6 = 8

Z 23 = Z32 =

2

E33 = E 2

On substituting the above terms in equation (1), we get, 3 + j4 − j4 0 − j4 3 + j5 2 H 0 2 8

>

>I2 H I1

I3

30

=

> 0H

..... (2)

E2

It is given that the current through (1 + j1) Ω impedance is zero and so the mesh current I 2 is zero. When the mesh currents are solved by Cramer’s rule, I 2 is given by ∆2 / ∆. For I 2 to be zero, the determinant ∆2 should be zero. Therefore, the value of E 2 can be obtained by equating ∆2 to zero.

∆2 =

3 + j4 30 0 − j4 0 2 = (3 + j4) # 60 − 2 # E2 @ − 30 # 6 − j4 # 8 − 0 @ + 0 0 E2 8 = − E2 (6 + j8) + j960

Put ∆ 2 = 0, ∴ 0 = −E 2 ( 6 + j8 ) + j960 E 2 (6 + j8) = j960 `

E2 =

j960 = 76.8 + j57.6 = 96+36.9 o V 6 + j8

The value of voltage source, E 2 = 96∠36.9o V

Chapter 4 - Mesh Analysis

4 . 36

4.5 Summary of Important Concepts 1.

Mesh is defined as a closed path which does not contain any other loops within it.

2.

Mesh analysis is used to solve the independent current variables of a circuit.

3.

The number of current variables in a circuit is equal to number of branches.

4.

The number of independent currents in a circuit is given by number of links in the graph of a circuit.

5.

The number of links L in a circuit with B branches and N nodes is given by, L = B – N + 1.

6.

In mesh analysis the independent currents are solved by writing KVL equations for various meshes of a circuit.

7.

The mesh basis matrix equation for resistive circuit is, RR R R g R V RI V RE V 1m S 11 12 13 W S 1W S 11 W S R21 R22 R23 g R2m W S I2 W S E22 W S R31 R32 R33 g R3m W S I3 W = S E33 W S W S W S W h W Sh W h h h S h S h W SR R R g R W SI W SE W mm T m1 m2 m3 X T mX T mm X

8.

The mesh currents are solved by using Cramer’s rule.

9.

The kth mesh current Ik by Cramer’s rule is, Ik = 1 ∆

m

/∆

jk

E jj

j=1

where, m = Number of meshes in the circuit.

Djk = Cofactor of Rjk. Ejj = Sum of voltage sources in mesh-j. D = Determinant of resistance matrix. 10.

11. rule are,

For circuit with three meshes, the mesh currents by Cramer’s rule are, I1 =

∆ ∆11 ∆ E11 + 21 E22 + 31 E33 ∆ ∆ ∆

I2 =

∆ ∆12 ∆ E11 + 22 E22 + 32 E33 ∆ ∆ ∆

I3 =

∆13 ∆ ∆ E11 + 23 E22 + 33 E33 ∆ ∆ ∆

The mesh currents for a circuit with three meshes using short-cut procedure for Cramer’s ∆1 ∆ ∆ I2 = 2 ∆ ∆3 I3 = ∆ I1 =

4 . 37

Circuit Theory

R11 R12 R13 E11 R12 R13 R11 E11 R13 R11 R12 E11 where, ∆ = R 21 R 22 R 23 ; ∆1 = E 22 R 22 R 23 ; ∆2 = R 21 E 22 R 23 ; ∆3 = R 21 R 22 E 22 R31 R32 R33 E33 R32 R33 R31 E33 R33 R31 R32 E33

12.

4.6 Q4.1

When a current source lie common to two meshes then the common current source can be removed for analysis purpose and the resultant two meshes can be considered as one single mesh called supermesh.

Short-answer Questions What is mesh analysis? Mesh analysis is an useful technique to solve the independent current variables of a circuit.

Q4.2

When is mesh analysis preferred to solve the currents? The mesh analysis is preferred to solve the current variables when the circuit is excited by only voltage sources. Applying mesh analysis is straight forward and easier in circuits excited by only voltage sources. But the mesh analysis can also be extended to circuits excited by both voltage and current sources.

Q4.3

How is mesh analysis performed? In a circuit with B branches and N nodes the number of independent currents is given by, m = B – N + 1. Hence m number of meshes are selected in the given circuit and one mesh current is attached to each mesh. For each mesh a KVL equation is formed then the m number of mesh equations are solved by Cramer’s rule to get a unique solution for mesh currents.

Q4.4

How are the mesh currents solved using the mesh basis matrix equation? Consider the mesh basis matrix equation, RI=E On premultiplying the above equation both sides by R −1 we get, R−1 R I = R−1 E U I = R−1 E

R−1 R = U = Unit matrix

∴ I = R−1 E

UI = I

The above equation will be the solution for mesh currents and the kth mesh current is, Ik =

∆ ∆1k ∆ ∆ E11 + 2k E22 + 3k E33 + ...... + mk Emm = 1 ∆ ∆ ∆ ∆ ∆

m

/

∆ jk E jj

j =1

The above equation for mesh currents is also called Cramer’s rule.

Q4.5

What is supermesh? When a current source lie common to two meshes then the common current source can be removed for analysis purpose and the resultant two meshes can be considered as one single mesh called supermesh.

Chapter 4 - Mesh Analysis

4 . 38 Q4.6

What is the value of E in the circuit shown in Fig. Q4.6 if the value of I2 is zero? Solution

=

4 − 2 I1 10 G = G = = G − 2 5 I2 E

Here, I 2 =

3‡

2‡

The mesh basis matrix equation by inspection is,

10 V

+ E

E + E

2‡ I1

∆2 . Therefore, for I2 = 0 , ∆2 = 0 ∆

I2

Fig. Q4.6.

4 10 Now, ∆2 = = 4E + 20 −2 E ` 4E + 20 = 0

⇒

4E = − 20

⇒

E = − 20 = − 5 V 4 2‡

1‡

Q4.7

Find the value of E1 and E2 in the circuit shown in Fig. Q4.7. Solution The mesh basis matrix equation by inspection is,

>

4‡

+ E1 _

I1 = 3 A

E1 − E 2 5 − 4 I1 H H > H = > E2 − 4 9 I2

+

I2 = 1 A

3‡

E2 _

Fig. Q4.7.

On substituting for I1 = 3 A and I2 = 1 A in the above equation we get,

>

E1 − E2 5 −4 3 H > H = > H E2 −4 9 1

From row-2 we get, –4 ´ 3 + 9 ´ 1 = E2

⇒

E2 = –3 V

From row-1 we get,

⇒ 11 = E1 – E2

5 ´ 3 – 4 ´ 1 = E1 – E2

Q4.8

⇒

E1 = 11 + E2 = 11 – 3 = 8 V

Find the value of I2 and E2 in the circuit shown in Fig. Q4.8.

4‡

2‡

Solution The mesh basis matrix equation by inspection is, 9 7 − 5 I1 > H> H = > E H − 2 − 5 9 I2

⇒

+

I1 = 2 A

9 7 −5 2 > H> H = > E H − 5 9 I2 - 2

Fig. Q4.8.

From row-1 we get, 7 # 2 − 5I2 = 9

⇒

5I2 = 14 − 9

⇒

I2 = 14 − 9 = 1 A 5

From row-2 we get, –5 ´ 2 + 9I2 = −E2

⇒

−E2 = –10 + 9 ´ 1 = –1 V

⇒

+ E E 2

5‡

9V E

E2 = 1V

I2

4 . 39

Circuit Theory Q4.9

In the circuit shown in Fig Q4.9, find the power delivered to 1Ω resistor. Solution The mesh basis matrix equation is, 2 + 4 − 4 I1 16 − 5 > H> H = > H 5 − 10 − 4 4 + 1 I2

⇒

6 − 4 I1 11 > H> H = > H −5 − 4 5 I2

16 V

+ E

+ E

5V

I1

Now, ∆2 =

6 11 = 6 # a− 5 k − a− 4 k # 11 = 14 −4 −5

1‡

4‡

2‡

+ E

I2

Fig. Q4.9.

2 6 −4 = 6 # 5 − a − 4 k = 14 −4 5

∆ =

I2 =

∆2 = 14 = 1 ∆ 14

Power delivered to 1 Ω resistor = I22 ´ 1 = 1 ´ 1 = 1 W

Q4.10

Find the current I in the circuit shown in Fig. Q4.10. Solution

2‡

4‡

The mesh basis matrix equation is, I +

4 − 2 I1 10 > H > H = > H − 2 6 I2 0

10 V _

2 4 −2 Now, ∆ = = 4 # 6 − a − 2 k = 20 −2 6

∆1 =

10 − 2 = 10 # 6 − 0 = 60 0 6

∆2 =

4 10 = 0 − a− 2 k # 10 = 20 −2 0

I = I1 − I 2 =

2‡ I1

I2

Fig. Q4.10.

∆1 ∆2 ∆ − ∆2 − = 1 = 60 − 20 = 2 A 20 ∆ ∆ ∆

4.7 Exercises I. Fill in the Blanks with Appropriate Words 1.

Mesh analysis is used to solve ________ variables of a circuit.

2.

A circuit with B branches and N nodes will have ________ independent currents.

3.

In mesh basis matrix equation, the mesh currents are solved by using ________.

4.

The mesh equations are ________ equations of a circuit.

5.

The solution of mesh basis matrix equation RI = E will be in the form ________.

ANSWERS 1. Current

2. B – N + 1

3. Cramer’s rule

4. KVL

5. I = R–1E

10 V

Chapter 4 - Mesh Analysis

4 . 40

II. State Whether the Following Statements are True/False 1.

Mesh analysis can be used to solve the voltage variables from the knowledge of current variables.

2.

Mesh analysis can be applied only to circuits excited by voltage sources.

3.

The mesh currents are independent current variables of a circuit.

4.

Mesh analysis is applicable to non-planar circuits.

5.

The mesh equations are formed by using KCL.

ANSWERS 1. True

2. False

3.

True

4. False

5.

False

III. Choose the Right Answer for the Following Questions 1. Mesh analysis is based on, a) KCL

b) KVL

c) Ohm’s law

d) none of the above

2. A planar circuit has 6 branches and 4 nodes, then the number of meshes are, a) 6

b) 5

c) 4

d) 3

3. In a circuit with m meshes, the kth mesh current by Cramer’s rule is given by, k

a) I k = 1 ∆

/∆

c) I k = 1 ∆

/∆

jk E jj

j=1 m

jk E jj

j=1

j

b) I k = 1 ∆

/∆

d) I k = 1 ∆

/∆

jk E kk

k =1 m

jm E km

k =1

4. In mesh analysis, when all the mesh currents chosen in same orientation then the mutualresistances are, a) always negative c) positive or negative

b) always positive d) always zero 2A

5. The mesh currents I1, I2 and I3 in the circuit shown in Fig. 5 are respectively, a) 2 A, 4 A, 6 A

4A

I1

b) 2 A, 10 A, 6 A 6A

c) 2 A, 6 A, 0

I2

d) 2 A, 6 A, 4 A

I3

Fig. 5. 1‡

6. In the circuit shown in Fig. 6, the currents I1 and I2 are respectively,

1‡

a) 1 A, –1 A b) 2 A, 2 A c) 2 A, 0 d) 1 A, 0

5V

+ E

+ E

2‡ I1

I2

Fig. 6.

5V

4 . 41

Circuit Theory 1‡

1‡

7. In the circuit shown in Fig. 7, the value of E1 and I1 are respectively, a) 6 V, 3 A E1

b) 4 V, 2 A

+ E

I1

c) 7 V, 1 A

1‡

1‡ I2 = 1 A

d) 5 V, 3 A

Fig. 7.

a) 20 V

3‡

3‡

8. In the circuit shown in Fig. 8. What is the value of E1 for the current I1 to be zero? E1

+ E

b) 10 V

+ E

3‡ I1

10 V

I2

c) 5 V

Fig. 8.

d) 0 4‡

9. In the circuit shown in Fig. 9. What is the value of E1, for the power delivered to 1Ω resistor is 1 W? a) 10 V

E1

+ E

b) 9 V

2‡

3‡

1‡

I1

I2

c) 11 V

Fig. 9.

d) 7 V 1‡

10. In the circuit shown in Fig. 10, the value of I1 and I2 are respectively,

2‡

a) 2 A, 1 A 9V

b) 3 A, 2 A

d) 2 A, 5 A

Fig. 10.

ANSWERS 3. c 4. a

5. c 6. a

7. d 8. c

9. c 10. a

IV. Unsolved Problems E4.1

7‡ I1

c) 4 A, 1 A

1. b 2. d

+ E

Determine the mesh currents shown in the circuit of Fig. E4.1. 1‡

1‡

3‡

I2

3‡

I3

4‡

2‡

2‡

+ E 5V

+ E 10 V I1

Fig. E4.1. 5‡

5‡ I2

Chapter 4 - Mesh Analysis

4 . 42 E4.2

Determine the current through various branches of the circuit in Fig. E4.2 by mesh analysis. Take resistance of ammeter as 0.2 Ω. Ie

4‡

Id

Ib

3‡

A

Ic

12 V

5‡

Ia

1‡

+

1‡

+ E

1‡

+ E

E

1‡

1‡

9V

2‡

1‡

5‡

1‡

If

Fig. E4.3.

E

Fig. E4.2.

E4.3

In the circuit shown in Fig. E4.3, find the value of E by mesh analysis such that the current through the 5 Ω resistor is zero.

E4.4

By mesh analysis, determine the power delivered to 2 Ω and 1 Ω resistor in the circuit of Fig. E4.4.

E

+E

5V

2‡

10 V

+ E

VL 2‡

1‡ 1‡

5‡

2 ‡ 12 V +

+ E

+

3‡

5V

4‡

E

2‡

2‡

1‡

Fig. E4.4.

Fig. E4.5.

E4.5

In the circuit shown in Fig. E4.5, determine the voltage VL by mesh analysis.

E4.6

Determine the mesh currents of the circuit shown in Fig. E4.6.

50Ð0oV

~

IL

-j5 W

+ I1

-j2 W

2W 2W

-

1W

j3 W

4W

I3

j4 W

5Ð0o A

~

1W

j4 W

2W

-j3 W

I2

Fig. E4.6. E4.7

Fig. E4.7.

Determine the current I L in the circuit shown in Fig. E4.7 by mesh analysis.

j4 W

1W

~

2Ð30o A

4 . 43

Circuit Theory E4.8

Determine the active and reactive power delivered to the 2 + j4 Ω impedance in the circuit of Fig. E4.8. 2W

2 + j4 W

4W

1‡

2‡ -j5 W

-j3 W

8‡

5‡

5A 100Ð0oV

~

Fig. E4.9.

Fig. E4.8. E4.9

+ 10 V E

Determine the power delivered to the 8 Ω resistor in the circuit of Fig. E4.9 using mesh analysis.

E4.10 In the circuit shown in Fig. E4.10, form two supermeshes, and hence determine the current IL. 3‡

5‡

IL I1 6V

6‡

I4

+ E

2‡

2A I2

3A

5‡

10 V

+ E

6‡ 2‡

2A

I3

Fig. E4.11.

1‡

1‡

Fig. E4.10. E4.11 Determine the power delivered by each source to the 2 Ω resistor in the circuit of Fig. E4.11 by mesh analysis.

ANSWERS E4.1

I1 = −1.3043 A

;

I2 = 0.9938 A

;

I3 = 0.6366 A

E4.2

Ia = 1.8037 A

;

Ib = 0.9907 A

;

Ic = 0.813 A

Ie = 0.4766 A

;

If = 1.3271 A

E4.3

E = 12 V

E4.4

P2Ω = 0.0987 W

E4.5

VL = –6.25 V

E4.6

I1 = 16.1971∠31o A

E4.7

IL = 0.6114∠62.7 o A

E4.8

P = 108.9 W

E4.9

P = 16.0201 W

Id = 0.5141 A ;

P1Ω = 1.1857 W

;

;

;

; I 2 = 8.7841∠71.6o A

Q = 217.8 VAR

; I3 = 11.4531∠−104o A

Chapter 4 - Mesh Analysis

4 . 44 E4.10

I L = 1.0952 A 3‡

5‡ +

+

E

+

Supermesh-1

6V E

E4.11

+ _ Supermesh-2 + 2I2 2I3 5I4 _ + E

E I2 +

E I3 +

1‡

1‡

P 2X = 4.125 W 10V

E 3I4

5I1

; P2X = 11W 2A

5‡

Chapter 5

NODE ANALYSIS 5.1

Introduction

Node analysis is a useful technique to solve the voltage across various elements of a circuit. Node analysis is preferred when the circuit is excited by current sources and the voltage across various elements are unknown. Node analysis can also be extended to circuits excited by both voltage and current sources and to circuits excited by both independent and dependent sources. In a circuit each branch will have a voltage across it. Hence, the number of voltages in the circuit are equal to number of branches. In a circuit some of the voltages will be independent and the remaining voltages depend on independent voltages. The number of independent voltages in a circuit can be determined from the graph of the circuit. It is given by the branches of the tree (or twigs) of the graph. The voltages of the tree branches are same as node voltages. (Refer Chapter 2, Section 2.10.4.) In nodal analysis, the independent voltages are solved by writing Kirchhoff’s Current Law (KCL) equations for various nodes in the circuit. A tree of the graph with N nodes will have “N – 1” branches or twigs. Hence, the number of independent voltages, n = N − 1. The nodes of the circuit are same as nodes of the graph and so a circuit will also have N number of nodes. “A node is meeting point of two or more elements. When more than two elements meet at a point then the node is called principal node”. The voltage of a node can be expressed only with reference to another node. Hence one of the node is chosen as reference node and the node voltages are expressed with respect to reference node. For each node except the reference node, a voltage is assigned called node voltage. The voltage of the reference node is always zero. Using KCL, an equation is formed for each node by equating the sum of currents leaving the node to the sum of currents entering the node. These equations are arranged in the form of matrix and node voltages are solved by Cramer’s rule. A simple procedure to form node basis matrix equation directly from circuit by inspection without forming KCL equation is also discussed in this chapter.

5.2

Node Analysis of Resistive Circuits Excited by dc Sources

A circuit with N nodes and B branches will have “N − 1” independent voltages and “B – (N – 1)” dependent voltages which depend on independent voltages. Let us denote the number of independent voltages by n, where, n = N − 1. In order to solve the independent voltages of a circuit we have to identify the N nodes of the circuit and choose one of the node as reference node. For each node, except the reference, we have to attach a voltage called node voltage. The node voltages are the independent voltages of the circuit. Let V1, V2, V3, ....., Vn be the node voltages.

5. 2

Chapter 5 - Node Analysis

For each node except the reference node a KCL equation is formed by equating the sum of currents leaving the node to sum of currents entering the node. Since there are n independent nodes, we can form n equations. In resistive circuits excited by dc sources the voltages and currents are real (i.e., they are not complex). For resistive circuits the n equations can be arranged in the matrix form as shown in equation (5.1), which is called node basis matrix equation. The formation of node basis matrix equation from the KCL equations is explained in some of the solved problems. The node basis matrix equation (5.1), can be written as shown in equation (5.2).

..... (5.1)

↓

↓

Note : The bold faced letter represent matrices. RG11 G12 G g G V RV V RI V 13 1n S W S 1W S 11 W SG 21 G22 G23 g G2n W S V2 W SI22 W SG31 G32 G33 g G3n W S V3 W = S I33 W S W S W S W h h h W Sh W S h Sh W SG n1 G n2 G g G W S V W SI W n3 nn T X T nX T nn X ↓ GV=I

..... (5.2)

where, G = Conductance matrix of order n × n V = Node voltage matrix of order n × 1 I = Source current matrix of order n × 1 n = Number of nodes except reference node In equation (5.1), the elements of conductance matrix and source current matrix can be determined from the given circuit. Hence, the unknowns are node voltages, which has to be solved by any standard technique. Alternatively, the equation (5.1) can be formed directly from the circuit by inspection without writing KCL equations. A procedure to form node basis matrix equation by inspection is given below : Procedure to Form Node Basis Matrix Equation by Inspection Consider the node basis matrix equation shown below for a circuit with three nodes excluding the reference node. Let V1, V2, V3 be the node voltages. G11 G12 G13 V1 G22 G23 H > V2 H = G31 G32 G33 V3

>G21

>I22 H I11

I33

The elements of equation (5.3), for circuits with independent sources are, G11 = Sum of conductances connected to node-1 G22 = Sum of conductances connected to node-2 G33 = Sum of conductances connected to node-3

..... (5.3)

Circuit Theory

5. 3

G12 = G21 = Sum of conductances connected between node-1 and node-2 G13 = G31 = Sum of conductances connected between node-1 and node-3 G23 = G32 = Sum of conductances connected between node-2 and node-3 I11 = Sum of current sources connected to node-1 I22 = Sum of current sources connected to node-2 I33 = Sum of current sources connected to node-3 The conductances G11, G22, G33 are called self-conductance of node-1, node-2, node-3 respectively. The conductances G12, G13, G21, G23, G31, G32 are called mutual-conductance between nodes. The formation of the elements of conductance matrix and source current matrix are explained below: i) The self-conductance G jj is given by sum of all the conductances connected to the jth node. The self-conductances will be always positive. ii) The mutual-conductance G jk is given by negative of sum of all the conductances connected between node-j and node-k. In circuit with only independent sources (Reciprocal network), G jk = Gkj. iii) The source current matrix element I jj is given by sum of all the current sources connected to jth node. A current source is positive if it drives current towards a node as shown in Fig. 5.1, and it is negative if it drives current away from the node as shown in Fig. 5.2. Vj

Vj

Ij

Ij

Reference node

Fig. 5.1 : Example for positive current source.

Reference node

Fig. 5.2 : Example for negative current source.

Note : In circuit with both independent and dependent sources (nonreciprocal circuit) Gjk ! Gkj Solution of Node Voltages In the node basis matrix equation [i.e., equation (5.1)] the unknowns are node voltages V1,V2,V3 ... Vn. The node voltages can be obtained by premultiplying the equation (5.1), by the inverse of conductance matrix. Consider equation (5.2), GV=I

5. 4

Chapter 5 - Node Analysis

On premultiplying both sides by G−1, we get, G−1 G V = G−1 I U V = G−1 I

G−1G = U = Unit matrix

\ V = G−1 I

.....(5.4)

UV=V

The equation (5.4), will be the solution for node voltages. The equation (5.4), can be solved by Cramer’s rule, by which the kth node voltage Vk is given by equation (5.5). Vk =

∆l ∆l1k ∆l ∆l I11 + 2k I22 + 3k I33 + ...... + nk I nn = 1 ∆l ∆l ∆l ∆l ∆l

n

/

∆l jk I jj

..... (5.5)

j =1

where, ∆′jk = Cofactor of Gjk Ijj = Sum of current sources connected to node-j ∆′ = Determinant of conductance matrix Proof for Cramer’s Rule Consider the equation (5.4), for a circuit with three nodes excluding reference node. V1

G11 G12 G13

-1

⇒ >V2 H = >G21 G22 G23 H

V = G–1 I

V3

G31 G32 G33

I11 I > 22 H I33

.....(5.6)

We know that, G- 1 =

Transpose of Gcof GTcof Adjoint of G = = Determinant of G Determinant of G ∆l where, ∆’

= Determinant of G.

Gcof = Cofactor matrix (matrix formed by cofactor of elements of G matrix). Let, ∆’11 = Cofactor of G11 ∆’12 = Cofactor of G12 and in general, ∆’jk = Cofactor of Gjk `

Gcof =

` G- 1 =

∆l 11 ∆l 12 ∆l 13 ∆ > l 21 ∆l 22 ∆l 23 H ∆l 31 ∆l 32 ∆l 33

Transpose

T Gcof =

∆l 11 ∆l 21 ∆l 31 >∆l 12 ∆l 22 ∆l 32 H ∆l 13 ∆l 23 ∆l 33

∆l 11 ∆l 21 ∆l 31 T Gcof = 1 >∆l 12 ∆l 22 ∆l 32 H ∆l ∆l ∆l 13 ∆l 23 ∆l 33

On substituting for G–1 from equation (5.7) in equation (5.6), we get, V1

>V2 H V3

=

∆l 11 ∆l 21 ∆l 31 1 ∆l 12 ∆l 22 ∆l 32 H > ∆l ∆l 13 ∆l 23 ∆l 33

I11

>I22 H I33

.....(5.7)

Circuit Theory

5. 5

On multiplying the matrices on the right-hand side of the above equation and equating to the terms on the left-hand side we get, V1 =

∆l ∆l 11 ∆l I + 21 I22 + 31 I33 ∆l 11 ∆l ∆l

V2 =

∆l ∆l 12 ∆l I + 22 I22 + 32 I33 ∆l 11 ∆l ∆l

V3 =

∆l 13 ∆l ∆l I + 23 I22 + 33 I33 ∆l 11 ∆l ∆l

The above equations can be used to form a general equation for node voltage. In general, the k th node voltage of a circuit with n nodes excluding reference is given by, Vk =

∆l 1k ∆l ∆l ∆l I + 2k I22 + 3k I33 + ...... + nk Inn = 1 ∆l 11 ∆l ∆l ∆l ∆l

n

/

∆l

jk I jj

j=1

Short-cut Procedure for Cramer’s Rule A short-cut procedure exists for the Cramer’s rule which is shown below: Let us consider a circuit with three nodes excluding reference. The node basis matrix equation for this case is, G11 G12 G13 V1 G > 21 G22 G23 H >V2 H = G31 G32 G33 V3

I11 I > 22 H I33

Let us define three determinants as shown below : I11 G12 G13 l ∆ 1 = I22 G22 G23 I33 G32 G33 G11 I11 G13 ∆l2 = G21 I22 G23 G31 I33 G33 G11 G12 I11 ∆l3 = G21 G22 I22 G31 G32 I33

Here, ∆l1 = Determinant of conductance matrix after replacing the first column of conductance matrix by source current column matrix. ∆l2 = Determinant of conductance matrix after replacing the second column of conductance matrix by source current column matrix. ∆l3 = Determinant of conductance matrix after replacing the third column of conductance matrix by source current column matrix.

5. 6

Chapter 5 - Node Analysis

Let, ∆l = Determinant of conductance matrix. G11 G12 G13 ∆l = G21 G22 G23 G31 G32 G33

Now node voltages V1, V2 and V3 are given by, V1 = ∆l1 ∆l V2 = ∆l2 ∆l l V3 = ∆ 3 ∆l

Cross-Check The equation for node voltages obtained by short-cut procedure are same as equation (5.5), and they are verified as shown below: V1 =

I11 G12 G13 ∆l1 = 1 I 22 G 22 G 23 ∆l ∆l I33 G32 G33

Expanding along first column

= 1 6 I11 ∆l11 + I 22 ∆l21 + I33 ∆l31 @ ∆l =

V2 =

∆l ∆l11 ∆l I + 21 I 22 + 31 I33 ∆l 11 ∆l ∆l

G11 I11 G13 ∆l 2 = 1 G 21 I 22 G 23 ∆l ∆l G31 I33 G33

Expanding along second column

= 1 6 I11 ∆l12 + I 22 ∆l22 + I33 ∆l32 @ ∆l =

V3 =

∆l ∆l12 ∆l I + 22 I 22 + 32 I 33 ∆ 11 ∆l ∆l

G11 G12 I11 ∆l 3 = 1 G 21 G 22 I 22 ∆l ∆l G31 G32 I33 = 1 6 I11 ∆l13 + I 22 ∆l23 + I33 ∆l33 @ ∆l

=

∆l13 ∆l ∆l I + 23 I 22 + 33 I33 ∆l 11 ∆l ∆l

Expanding along third column

Circuit Theory

5. 7

Various Steps to Obtain the Solution of Node Voltages and Branch Voltages in a Circuit Step 1 : Draw the graph of the circuit. Step 2 : Determine the branches B and nodes N. The number of node voltages n is given by n = N – 1. Step 3 : Choose one of the nodes as reference. Let us denote the reference node as 0 (zero) and other nodes as 1, 2, 3, ....., n. Step 4 : Let us denote the node voltages by V1, V2, V3,....., and the branch voltages by Va, Vb, Vc, Vd, Ve,...... Write the relationship between node and branch voltages. Step 5 : Form the node basis matrix equation by inspection and solve the node voltages, using Cramer’s rule. For a circuit with three nodes excluding the reference, the node basis matrix equation and solution of node voltages using Cramer’s rule are given below: G11 G12 G13 V1 I11 G G G V I = 2 > 21 22 23 H > H > 22 H G31 G32 G33 V3 I33 V1 =

I11 G12 G13 ∆l1 = 1 I 22 G 22 G 23 ∆l ∆l I33 G32 G33

V2 =

G11 I11 G13 ∆l 2 = 1 G 21 I 22 G 23 ∆l ∆l G31 I33 G33

V3 =

G11 G12 I11 ∆l 3 = 1 G 21 G 22 I 22 ∆l ∆l G31 G32 I33

Step 6 : Solve the branch voltages using the relationship between branch and node voltages. Note : After solving the node voltages if any of the voltage is found to be negative, then that node has a potential lesser than reference node. V1

EXAMPLE 5.1

V2 1‡

Write and solve the node voltage equations for the circuit shown in Fig. 1. 2A

2A

2‡

1‡

SOLUTION 0

With reference to Fig. 2, the node equation for node-1 is formed as shown below: Currents leaving node-1 : V1 − V2 , V1 1 1 Current entering node-1

: 2A

Fig. 1. V1 V2 2A

` V1 − V2 + V1 = 2 1 1

2A

V1 E V2 V1 1 1

V1 − V2 + V1 = 2 2 V1 − V2 = 2

0

..... (1)

Fig. 2.

Reference node

5. 8

Chapter 5 - Node Analysis With reference to Fig. 3, the node equation for node-2 is formed as shown below: V2 E V1 1

Currents leaving node-2 : V2 − V1 , V2 , 2 A 1 2 Current entering node-2 : Nil

V2

V1

2A

` V2 − V1 + V2 + 2 = 0 1 2

V2 2

V2 − V1 + 0.5V2 = −2

2A

0

−V1 + 1.5 V2 = −2

..... (2)

Fig. 3.

The equations (1) and (2) are the node equations of the circuit, which are summarized here for convenience. 2 V1 − V2 = 2

..... (1)

−V1 + 1.5 V2 = −2

..... (2)

Equation (1) × 1

⇒

2 V1 − V2 = 2

Equation (2) × 2

⇒

− 2V1 + 3V2 = − 4

On adding ` V2 = − 2 = − 1V 2

2V2 = − 2

From equation (2), we get, V1 = 1.5V2 + 2 = 1.5 × (−1) + 2 = 0.5 V The node voltages are, V1 = 0.5 V V2 = −1 V

EXAMPLE 5.2 2A

Determine the voltages across various elements of the circuit shown in Fig. 1, by node method.

1 ‡ 4

1 ‡ 2

SOLUTION The graph of the given circuit is shown in Fig. 2. It has 7 branches and 4 nodes.

1 ‡ 2

1 ‡ 3

9A

Let us choose one of the node as reference as shown in Fig. 2. Let

1 ‡ 4

Fig. 1.

the voltages of other three nodes be V1,V2 and V3. The reference node is denoted by “0” to indicate that its voltage is zero volt. The circuit with chosen node voltages is shown in Fig. 3. g 2A V2

e

V1

V3

f

a

b

c

0

d

Reference node

Fig. 2.

V1 1 ‡ 2 1 ‡ 2

V3

V2

1 ‡ 4 1 ‡ 3

9A

0

1 ‡ 4

Reference node

Fig. 3.

Circuit Theory

5. 9

Method I : Formation of node basis matrix equation by applying KCL In this method, the node equations are formed by using Kirchhoff’s Current Law. The node equation for a node is formed by equating the sum of currents leaving that node to sum of currents entering that node. While writing the node equation for a node it is assumed that all the resistances connected to that node will draw current from that node. Hence, the current in the resistances will always leave the node. With reference to Fig. 4, the node equation for node-1 is formed as shown below: V1 V1 − V2 2A Currents leaving node-1 : , , 2A 1/2 1/2 Current entering node-1 :

Nil

V1 ` + V1 − V2 + 2 = 0 1/2 1/ 2

2A V2

V1

V3

V1 E V2 1/ 2

V1 1/ 2

2 V1 + 2 V1 − 2V2 + 2 = 0 4V1 − 2V2 = −2

Fig. 4.

0

..... (1)

With reference to Fig. 5, the node equation for node-2 is formed as shown below: Currents leaving node-2 :

V2 − V1 V2 − V3 V2 , , 1/2 1/4 1/3

Current entering node-2 :

9A

V2 E V1 1/ 2

V2

V1

V3

9A

V2 − V1 + V2 − V3 + V2 = 9 1/2 1/4 1/3

`

V2 E V3 1/ 4

V2 1/ 3

9A

2V2 − 2V1 + 4V2 − 4V3 + 3V2 = 9 − 2V1 + 9V2 − 4V3 = 9

0

0

..... (2)

Fig. 5.

With reference to Fig. 6, the node equation for node-3 is formed as shown below: Currents leaving node-3 :

2A

V3 − V2 V3 , 1/4 1/4

Current entering node-3 : 2 A

2A

V1

V2

V3 − V2 + V3 = 2 1/4 1/4

`

V3 1/ 4

1/ 4

4V3 − 4V2 + 4V3 = 2 −4V2 + 8V3 = 2

V3 V3 E V2

Fig. 6.

0

..... (3)

The equations (1), (2) and (3) are node equations of the circuit shown in Fig. 3. The node equations are summarized here for convenience. 4V1 − 2V2 = −2 −2V1 + 9V2 − 4V3 = 9 −4V2 + 8V3 = 2 The node equations can be arranged in the matrix form as shown below and then solved by Cramer’s rule. 4 − 2 0 V1 −2 9 − 4 H >V2 H = > 9 H 0 − 4 8 V3 2

>− 2

..... (4)

5. 10

Chapter 5 - Node Analysis

Method II : Formation of node basis matrix equation by inspection In this method, the node basis matrix equation is formed by inspection using the circuit shown in Fig. 3. The general node basis matrix equation for a circuit with three nodes excluding the reference is shown in equation (5). G11 G12 G13 V1 I11 >G21 G22 G23 H >V2 H = >I22 H ..... (5) G31 G32 G33 V3 I33 The elements of conductance matrix and source current matrix are formed as shown below: G11 = 2 + 2 = 4

G12 = G21 = –2

I11 = –2

G22 = 2 + 3 + 4 = 9

G13 = G31 = 0

I22 = 9

G33 = 4 + 4 = 8

G23 = G32 = – 4

I33 = 2

On substituting the above terms in equation (5), we get, 4 − 2 0 V1 −2 9 − 4 H >V2 H = > 9 H 0 − 4 8 V3 2

>− 2

..... (6)

Solution of node voltages It is observed that the node basis matrix equation obtained in method-I and II are same. In equation (6) the unknowns are V1, V 2 and V3 . In order to solve V1, V 2 and V3, let us define four determinants ∆l , ∆l1, ∆l2 and ∆l3 as shown below: 4 −2 0 4 −2 0 4 −2 −2 −2 −2 0 ∆l = − 2 9 − 4 ; ∆l1 = 9 9 − 4 ; ∆l2 = − 2 9 − 4 ; ∆l3 = − 2 9 9 0 −4 8 2 −4 8 0 2 8 0 −4 2 The determinants are evaluated by expanding along first row and the node voltages are solved by Cramer’s rule. 4 −2 0 = 4 # 69 # 8 − (− 4) 2 @ − (− 2) # 6 − 2 # 8 − 0 @ + 0 ∆l = − 2 9 − 4 = 224 − 32 = 192 0 −4 8 ∆l1 =

−2 −2 0 = − 2 # 69 # 8 − (− 4) 2 @ − (− 2) # 69 # 8 − 2 # (− 4) @ + 0 9 9 −4 = − 112 + 160 = 48 2 −4 8

4 −2 0 = 4 # 69 # 8 − 2 # (− 4) @ − (− 2) # 6 − 2 # 8 − 0 @ + 0 ∆l 2 = − 2 9 − 4 = 320 − 32 = 288 0 2 8 4 −2 −2 = 4 # 69 # 2 − (− 4) # 9 @ − (− 2) # 6 − 2 # 2 − 0 @ + (− 2) # 6 − 2 # (− 4) − 0 @ ∆l 3 = − 2 9 9 = 216 − 8 − 16 = 192 0 −4 2 V1 =

∆l1 = 48 = 0.25 V 192 ∆l

V2 =

∆l 2 = 288 = 1.5 V 192 ∆l

V3 =

∆l 3 = 192 = 1V 192 ∆l

Circuit Theory

5. 11

To solve branch voltages The given circuit has seven branches. Let us denote the branch voltages as Va, Vb, Vc, Vd, Ve, Vf and Vg as shown in Fig. 7. The sign of branch voltages are chosen such that they are all positive. The relation between branch and node voltages are obtained using the circuit shown in Fig. 7 and the branch voltages are solved as shown below: _

Va = V1 = 0.25 V Vb = V2 = 1.5 V Vc = V2 = 1.5 V

+

+ +

Va

Vd = V3 = 1 V

Vb

_

+

2A

_ Ve +

V1

Vg

V2

_ 9A

Ve = V2 − V1 = 1.5 − 0.25 = 1.25 V

Vf _ V3

+

+

Vc _

_

Vd

Fig. 7.

Vf = V2 − V3 = 1.5 − 1 = 0.5 V Vg = V3 − V1 = 1 − 0.25 = 0.75 V Note : The branch voltages are voltages across various elements in the circuit.

EXAMPLE 5.3

2‡

In the network shown in Fig. 1, find the current through the 2 Ω resistor, by node method.

1‡ 5A

4‡

4‡ 1‡

SOLUTION

4‡

The given circuit is redrawn as shown in Fig. 2. The graph of the circuit is shown in Fig. 3. It has 7 branches and 4 nodes. Let us choose one of the node as reference as shown in Fig. 3. Let the voltages of other three nodes be V1, V2 and V3. The reference node is denoted by “0”. V1

2‡

V2

V1

Fig. 1.

V2

f

1‡ 5A

4‡ 4‡

Reference node

a

4‡

1‡

b

V3

0

Reference node

Fig. 2.

0

d

c

g

e

V3

Fig. 3.

In the 2 Ω resistor, the current will flow from node-1 to node-2 if V1 > V2, and when V2 > V1, the current will flow from node-2 to node-1. Let the current through 2 Ω resistance be Ix. If V1 > V2 , then Ix =

V1 − V2 . 2

If V2 > V1,

V2 − V1 . 2

then Ix =

In both the cases it is enough if we solve the node voltages V1 and V2. The node basis matrix equation is formed by inspection using the circuit shown in Fig. 2. The general node basis matrix equation for a circuit with three nodes excluding the reference is shown in equation (1).

5. 12

Chapter 5 - Node Analysis G11 G12 G13 G 22 G 23 H G31 G32 G33

>G21

V1

>V2 H V3

I11

=

>I22 H

..... (1)

I33

The elements of conductance matrix and source current matrix are formed as shown below : G11 = 1 + 1 + 1 = 1.75 4 1 2

G12 = G 21 = − 1 = − 0.5 2

I11 = 5

G 22 = 1 + 1 + 1 = 1.75 2 4 1

G13 = G31 = − 1 = − 1 1

I 22 = 0

G33 = 1 + 1 + 1 = 1.5 4 1 4

G 23 = G32 = − 1 = − 0.25 4

I33 = 0

On substituting the above terms in equation (1), we get, 1.75

− 0.5 −1 1.75 − 0.25 H 1.5 − 1 − 0.25

>− 0.5

V1

>V2 H V3

5

=

>0 H 0

In order to solve the node voltages V1 and V2, let us define three determinants ∆’, ∆’1 and ∆’2 as shown below: 1.75 − 0.5 −1 ∆l = − 0.5 1.75 − 0.25 ; 1.5 − 1 − 0.25

5 − 0.5 −1 ∆l1 = 0 1.75 − 0.25 ; 0 − 0.25 1.5

1.75 5 −1 ∆l2 = − 0.5 0 − 0.25 1.5 −1 0

The determinants are evaluated by expanding along first row and the node voltages are solved by Cramer’s rule. 1.75 − 0.5 −1 = 1.75 # 61.75 # 1.5 − (− 0.25) 2 @ − (− 0.5) # 6 − 0.5 # 1.5 − (− 1) # (− 0.25) @ ∆l = − 0.5 1.75 − 0.25 1.5 − 1 − 0.25 + (− 1) # 6 − 0.5 # (− 0.25) − (− 1) # 1.75 @

= 4.4844 − 0.5 − 1.875 = 2.1094 5 − 0.5 − 1 = 5 61.75 1.5 − (− 0.25) 2 @ − 0 + 0 # # ∆l1 = 0 1.75 − 0.25 0 − 0.25 1.5 = 12.8125 1.75 5 − 1 = 0 − 5 # 6 − 0.5 # 1.5 − (− 1) # (− 0.25) @ + 0 ∆l2 = − 0.5 0 − 0.25 1.5 = 5 −1 0 ` V1 =

∆l1 = 12.8125 = 6.074 V 2.1094 ∆l

V2 =

∆l 2 5 = = 2.3703 V ∆l 2.1094

` Ix =

V1 − V2 = 6.074 − 2.3703 = 1.8519 A 2 2

Since V1 > V2, the direction of current I x is from node-1 to node-2.

Circuit Theory

5. 13 5A

EXAMPLE 5.4 4A

In the circuit shown in Fig. 1. Find the potential difference between A and D.

3A B

SOLUTION

C

A 3‡

2‡

The given circuit has four nodes. Let us choose the node-D as reference node and so the voltage of node-D is zero volt. All other node voltages are expressed

2‡

4‡

5‡

with respect to reference node. Let the voltages of node A, B and C be V1, V2 and V3 respectively. Now the voltage between A and D is V1.

Fig. 1.

D

The node basis matrix equation is formed by inspection using the circuit

5A

shown in Fig. 2. The general node basis matrix equation for a circuit with three 4A

nodes excluding the reference is shown in equation (1).

3A V2

G11 G12 G13 V1 I11 G 22 G 23 H >V2 H = >I 22 H G31 G32 G33 V3 I33

>G21

V3

V1

3‡

2‡

..... (1) 2‡

4‡

The elements of conductance matrix and source current matrix are formed as shown below:

Reference node 0

G11 = 1 + 2 G 22 = 1 + 2

1 = 0.75 4 1 + 1 = 1.33 2 3

0.75

0 − 0.5 1.33 − 0.33 H 0 − 0.33 0.53

V1

>V2 H V3

Fig. 2.

G12 = G 21 = − 1 = − 0.5 2

I11 = 5 − 4 = 1

G13 = G31 = 0

I 22 = 4 − 3 = 1

G 23 = G32 = − 1 = − 0.33 G33 = 1 + 1 = 0.53 3 3 5 On substituting the above terms in equation (1), we get,

>− 0.5

5‡

I33 = 3 − 5 = − 2

1

=

> 1H

..... (2)

−2

In order to solve the node voltage V1, let us define two determinants D’ and D’1 as shown below: 0.75 − 0.5 0 ∆l = − 0.5 1.33 − 0.33 ; 0 − 0.33 0.53

∆l1 =

1 − 0.5 0 1 1.33 − 0.33 − 2 − 0.33 0.53

The determinants are evaluated by expanding along first row and the node voltage V1 is solved by Cramer’s rule. 0.75 − 0.5 0 = 0.75 # 61.33 # 0.53 − (− 0.33) 2 @ − (− 0.5) # 6 − 0.5 # 0.53 − 0 @ + 0 ∆l = − 0.5 1.33 − 0.33 = 0.447 − 0.1325 = 0.3145 0 − 0.33 0.53 ∆l1 =

1 − 0.5 0 = 1 # 61.33 # 0.53 − (− 0.33) 2 @ − (− 0.5) # 61 # 0.53 − (− 2) # (− 0.33) @ + 0 1 1.33 − 0.33 = 0.596 − 0.065 = 0.531 − 2 − 0.33 0.53

` V1 =

∆l1 = 0.531 = 1.6884 V 0.3145 ∆l

Voltage between node A and D = VAD = V1 = 1.6884 V

5. 14

Chapter 5 - Node Analysis

EXAMPLE 5.5

2‡

In the circuit shown in Fig. 1, determine the power supplied by the current sources.

2‡

4‡

8A

10 A

SOLUTION 2‡

The given circuit has 4 nodes. Let us choose one of the node voltage as reference node, which is indicated by “0”. Let the voltages of other three nodes be V1, V2 and V3 as shown in Fig. 2.

2‡

V1

The power supplied by 8 A source in watts = V1 × 8

V2

2‡

The power supplied by 10 A source in watts = (V2 −V3) × 10 The node basis matrix equation is formed by inspection using the circuit shown in Fig. 2. The general node basis matrix equation for a circuit with three nodes excluding the reference is shown in equation (1). G11 G12 G13 V1 I11 >G21 G22 G23 H >V2 H = >I22 H G31 G32 G33 V3 I33

Fig. 1.

4‡ 10 A

8A

2‡

V3

0 Reference node

Fig. 2.

..... (1)

The elements of conductance matrix and source current matrix are formed as shown below: G11 = 1 + 1 = 1 2 2

G12 = G 21 = − 1 = − 0.5 2

I11 = 8

G 22 = 1 + 1 = 0.75 2 4

G13 = G31 = − 1 = − 0.5 2

I22 = 10

G33 = 1 + 1 = 1 2 2

G 23 = G32 = 0

I33 = − 10

On substituting the above terms in equation (1), we get, 1 − 0.5 − 0.5 0.75 0H 0 1 − 0.5

> − 0.5

V1

>V2 H

=

V3

>

8 10 H − 10

..... (2)

In equation (2), the unknowns are V1, V2 and V3. In order to solve V1, V2 and V3, let us define four determinants, ∆l , ∆l1, ∆l2 and ∆l3 as shown below: 1 − 0.5 − 0.5 ∆l = − 0.5 0.75 0 0 1 − 0.5 1 8 − 0.5 ∆l2 = − 0.5 10 0 1 − 0.5 − 10

∆l1 =

;

;

8 − 0.5 − 0.5 10 0.75 0 0 1 − 10

1 − 0.5 8 ∆l3 = − 0.5 0.75 10 0 − 10 − 0.5

The determinants are evaluated by expanding along first row and the node voltages are solved by Cramer’s rule. 1 − 0.5 − 0.5 ∆l = − 0.5 0.75 0 = 1 # 60.75 # 1 − 0 @ − (− 0.5) # 6 − 0.5 # 1 − 0 @ + (− 0.5) # 60 − (− 0.5) # 0.75 @ 0 1 − 0.5

= 0.75 − 0.25 − 0.1875 = 0.3125

Circuit Theory ∆l1 =

5. 15 8 − 0.5 − 0.5 10 0.75 0 = 8 # 60.75 # 1 − 0 @ − (− 0.5) # 610 # 1 − 0 @ + (− 0.5) # 60 − (− 10) # 0.75 @ 0 1 − 10

= 6 + 5 − 3.75 = 7.25 1 8 − 0.5 ∆l2 = − 0.5 10 0 = 1 # 610 # 1 − 0 @ − 8 # 6 − 0.5 # 1 − 0 @ + (− 0.5) # 6 − 0.5 # (− 10) − (− 0.5) # 10 @ 1 − 0.5 − 10

= 10 + 4 − 5 = 9 1 − 0.5 8 1 # 60.75 # (− 10) − 0 @ − (− 0.5) # 6 − 0.5 # (− 10) − (− 0.5) # 10 @ ∆l3 = − 0.5 0.75 10 = + 8 # 60 − (− 0.5) # 0.75 @ 0 − 10 − 0.5

= −7.5 + 5 + 3 = 0.5 V1 =

∆l1 = 7.25 = 23.2 V 0.3125 ∆l

V2 =

∆l 2 9 = = 28.8 V ∆l 0.3125

V3 =

∆l 3 = 0.5 = 1.6 V 0.3125 ∆l

Power supplied by 8 A current source = V1 × 8 = 23.2 × 8 = 185.6 W Power supplied by 10 A current source = (V2 − V3) × 10 = (28.8 − 1.6) × 10 = 272 W

EXAMPLE 5.6

1‡

Find the power in the 4 Ω resistor of the circuit shown in Fig. 1, by node method.

SOLUTION

1‡

1‡

10 A

To estimate the power in the 4 Ω resistor, first we have to determine the voltage across 4 Ω resistor. The given circuit has 4 nodes. Let us choose one of the node as reference node and it is indicated by “0”. Let the voltages of other three nodes be V1, V2 and V3 as shown in Fig. 2. Now the voltage across 4 Ω resistor is V2 Volts. V 22 ∴ Power in the 4 Ω resistor in watts = 4 . The node basis matrix equation is formed by inspection using the circuit shown in Fig. 2. The general node basis matrix equation for a circuit with three nodes excluding the reference is shown in equation (1).

20 A

4‡

Fig. 1. 1‡

V1

1‡

1‡

V2

10 A

4‡

0

V3 20 A

Reference node

Fig. 2. G11 G12 G13 G 22 G 23 H G31 G32 G33

>G21

V1

>V2 H V3

I11

=

>I22 H I33

.....(1)

5. 16

Chapter 5 - Node Analysis The elements of conductance matrix and source current matrix are formed as shown below: G11 = 1 + 1 = 2 1 1

G12 = G 21 = − 1 = − 1 1

I11 = 10

G 22 = 1 + 1 + 1 = 2.25 1 1 4

G13 = G31 = − 1 = − 1 1

I22 = 0

G33 = 1 + 1 = 2 1 1

G 23 = G32 = − 1 = − 1 1

I33 = 20

On substituting the above terms in equation (1), we get, 2

10 − 1 − 1 V1 2.25 − 1 H >V2 H = > 0 H 20 − 1 − 1 2 V3

>− 1

..... (2)

In order to solve the node voltage V2 , let us define two determinants D’ and D’2 as shown below: 2 −1 −1 ∆l = − 1 2.25 − 1 −1 −1 2

;

2 10 − 1 ∆l 2 = − 1 0 − 1 − 1 20 2

The determinants are evaluated by expanding along first row and the node voltage V2 is solved by Cramer’s rule. 2 − 1 −1 ∆l = − 1 2.25 − 1 = 2 # 6 2.25 # 2 − (− 1) 2 @ − (− 1) # 6 − 1 # 2 − (− 1) 2 @ + (− 1) # 6(− 1) 2 − (− 1) # 2.25 @ −1 − 1 2 = 7 − 3 − 3.25 = 0.75 2 10 − 1 ∆l2 = − 1 0 − 1 = 2 # 60 − 20 # (− 1) @ − 10 # 6 − 1 # 2 − (− 1) 2 @ + ^− 1h # 6 − 1 # 20 − 0 @ − 1 20 2 = 40 + 30 + 20 = 90

` V2 =

∆l 2 = 90 = 120 V ∆l 0.75

Power in the 4Ω resistor

5.3

2 V2 = 3600 W = 42 = 120 4

Node Analysis of Circuits Excited by Both Voltage and Current Sources

The node analysis can be extended to circuits excited by both voltage and current sources. In such circuits if each voltage source has a series impedance then they can be converted to an equivalent current source with parallel impedance. After conversion, the circuit will have only current sources and so the procedure for obtaining node basis matrix equation by inspection and its solution discussed in Sections 5.2 and 5.4 can be directly applied to these circuits. In certain circuits excited by both voltage and current sources, the voltage source may not have a series resistance. In this situation the voltage source cannot be converted to current source. In this case the value of each voltage source is related to node voltages and for each voltage source one of the node voltage can be expressed in terms of source voltage and other node voltages. The remaining node voltages can be solved by writing Kirchhoff’s Current Law equations.

Circuit Theory

5. 17

Alternatively, the node basis matrix equation can be formed directly by inspection by taking the current delivered by the voltage sources as unknown and relating the value of each voltage source to node voltages. Here for each voltage source one node voltage is eliminated by expressing the node voltage in terms of the source voltage and other node voltages. While forming the node basis matrix equation the current of the voltage sources should be entered in the source matrix. Now in the matrix equation some node voltages will be eliminated and equal number of unknown source currents will be introduced. Thus, the number of unknowns will remain same as n where n is number of nodes in the circuit except the reference node. On multiplying the node basis matrix equation we get n equations which can be solved to give a unique solution for unknowns. 5.3.1 Supernode Analysis In circuits excited by both voltage and current sources, if a voltage source is connected between two nodes then the voltage source can be short-circuited for analysis purpose and the shorted two nodes can be considered as one single node called supernode. In order to solve the two node voltages of a supernode, two equations are required. One of the equation is the KCL equation of the supernode and the other equation is obtained by equating the source voltage to difference of the node voltages. An example of formation of supernode is shown in Fig 5.3. Also, the example 5.15 is solved by using supernode analysis technique. 5V

V1

V1 2

10 A

V2 4

2W 4W 0

Supernode

V2

+-

V2 5

5W

V2 3

3W

Þ

V1 2

10 A

2W 4W

V2 5

0

Reference node

Fig. a : Circuit with two independent nodes.

V2 4

5W

V2 3

3W

Reference node

V1 V2 V2 V2 + + + = 10 2 4 5 3 V1 - V2 = 5

Fig. b : Supernode of the circuit shown in Fig. a and its equations. Fig. 5.3 : Example of formation of supernode.

EXAMPLE 5.7 In the circuit shown in Fig. 1, find the voltage across the 40 Ω resistor and the power supplied by 5 A source, by node analysis.

+

100 V E

SOLUTION +

E 60 V

The given circuit has 4 nodes. In this one of the node is chosen as reference. Let the voltages of other three nodes be V1, V2 and V3 as shown in Fig. 2. Here, the voltage sources does not have a resistance in series

5A

25 ‡

4A 20 ‡

and so they cannot be converted to current source. Let Is1 and Is2 be the currents supplied by 100 V and 60 V sources respectively.

Fig. 1.

40 ‡

5. 18

Chapter 5 - Node Analysis With reference to Fig. 2, we can write,

+

100 V E

V1 − V3 = 100 V and V2 − V1 = 60 V

Is1 E

From the above equations we can say that node voltages V2 and V3 can be expressed in terms of V1. Now the number of unknowns in the circuit are three and they are V1, Is1 and Is2. Therefore, we can write three

5A

V2

+

V1

V3 4A

60 V Is2

+ V1 E

40 ‡

20 ‡

25 ‡

node equations using KCL (corresponding to three nodes) and a unique solution for unknowns can be obtained by solving the three equations.

0

Reference node

Fig. 2.

The node equations can be obtained by two methods.

Method I : Formation of node equations by applying KCL In this method, the node equations are formed by using Kirchhoff’s Current Law. The node equation for a node is formed by equating the sum of currents leaving that node to sum of currents entering that node. While writing the node equation for a node it is assumed that all the resistances connected to that node will draw current from that node. Hence, the currents in the resistances will always leave the node. With reference to Fig. 3 the node equation for node-1 is formed as shown below: :

Is1

V1

E

Currents entering node-1 :

5A

V1 ` + Is2 = 5 + Is1 25

V2

Is2

60 V

5 A, I s1

0.04 V1 = 5 + I s1 − Is2

V3

+

5A

+

Currents leaving node-1

100 V E

V1 ,I 25 s2

V1 25

0

..... (1)

Fig. 3.

With reference to Fig. 4 the node equation for node-2 is formed as shown below: :

Currents entering node-2 : `

Is2

V2 20

V1

E

V2

V3

+

Current leaving node-2

4A

60 V

I s2, 4 A

V2 20

V2 = Is2 + 4 20 0

0.05 V2 = 4 + Is2

Fig. 4.

..... (2)

With reference to Fig. 5 the node equation for node-3 is formed as shown below:

Current entering node-3 :

V1

Nil

100 V E

Is1

+

V3 , I , 4A 40 s1

Currents leaving node-3 :

V2

V3 4A

`

V3 40

V3 + Is1 + 4 = 0 40

0.025 V3 = − 4 − I s1

..... (3)

Fig. 5.

0

Circuit Theory

5. 19

The equations (1), (2) and (3) are the node equations of the given circuit. On arranging the equations (1) to (3) in the matrix form we get, 0.04 0 0 0 0.05 0H 0 0 0.025

>

V1

>V2 H

=

V3

5 + Is1 − Is2 4 + Is2 H − 4 − Is1

>

.....(4)

Method II : Formation of node basis matrix equation by inspection In this method, the node basis matrix equation is formed by inspection using the circuit shown in Fig. 2. G11 G12 G13 G 22 G 23 H G31 G32 G33

>G21

V1

I11

>V2 H

=

V3

>I22 H I33

G11 = 1 = 0.04 25 G 22 = 1 = 0.05 20 G33 = 1 = 0.025 40

.....(5)

G12 = G 21 = 0

I11 = 5 + Is1 − Is2

G13 = G31 = 0

I22 = 4 + Is2

G 23 = G32 = 0

I33 = − 4 − Is1

On substituting the above terms in equation (5), we get, 0.04 0 0 0 0.05 0H 0 0 0.025

>

V1

>V2 H

=

V3

5 + Is1 − Is2 4 + Is2 H − 4 − Is1

>

.....(6)

Solution of node voltages It is observed that the node basis matrix equation obtained by both the methods are same. With reference to Fig. 2 the following relations can be obtained between node voltages. V2 − V1 = 60

V1 − V3 = 100

∴ V2 = 60 + V1

∴ V3 = V1 − 100

..... (7)

..... (8) Using equations (7) and (8), the equation (6) can be written as shown in equation (9). 0.04 0 0 0 0.05 0H 0 0 0.025

>

V1 + V1 H = V1 − 100

> 60

5 + Is1 − Is2 4 + Is2 H − 4 − Is1

>

.....(9)

On multiplying the matrices on the left-hand side of equation (9) and equating to the terms on the right-hand side we get the following three equations. 0.04 V1 = 5 + I s1 − I s2

.....(10)

0.05 (60 + V1) = 4 + I s2

.....(11)

0.025 (V1 − 100) = − 4 − I s1

.....(12)

On adding the above three equations, we get, 0.04 V1 + 0.05(60 + V1 ) + 0.025 (V1 − 100) = 5

5. 20

Chapter 5 - Node Analysis 0.04 V1 + 3 + 0.05V1 + 0.025V1 − 2.5 = 5 0.115V1 + 0.5 = 5 V1 = 5 − 0.5 = 39.13 V 0.115 ∴ V1 = 39.13 V From equation (7), V2 = 60 + V1 = 60 + 39.13 = 99.13 V From equation (8),

V3 = V1 − 100 = 39.13 −100 = − 60.87 V

With reference to Fig. 2, the voltage across the 40 Ω resistor is V3. ∴ Voltage across the 40 Ω resistor = V3 = − 60.87 V The negative voltage across the 40 Ω resistor indicates that the current through the 40 Ω is flowing towards the node. (Remember that while forming node equations it is assumed that the currents through resistances are leaving the node.) If we are interested in positive voltage across the 40 Ω resistor then the polarity of voltage across the 40 Ω resistor is assumed as shown in Fig. 6. V 3

E

Now, V40 = − V3 = −(−60.87) 40 ‡

V40

= 60.87 V With reference to Fig. 2, we can say that the voltage across 5 A current source is V1.

+

∴ Power supplied by 5 A source = V1 × 5 = 39.13 × 5 = 195.65 W

Fig. 6.

EXAMPLE 5.8 In the circuit shown in Fig. 1, find the value of E which will make the voltage across 10 Ω resistance as zero, by node analysis.

+

2

2A

A E

SOLUTION

5V

3‡

20 V E

E E

10 ‡

+

The given circuit has 5 nodes. In this one of the node is chosen

+

2‡

as reference. Let the voltages of other four nodes be V1, V2, V3 and V4 as shown in Fig. 2. Here, the voltage sources does not have series resistances

Fig. 1.

and so they cannot be converted to current source. Let us treat the currents supplied by the voltage sources as unknown quantities, and the values of

V4

voltage sources can be related to node voltages. Let I s1, I s4 and I s3 be

+

G12 G 22 G32 G 42

G13 G 23 G33 G 43

V G14 W G 24 W G34 WW G 44 W X

R V R V S V1 W SI11 W S V2 W SI 22 W S V W = SI W S 3 W S 33 W S V4 W SI 44 W T X T X

20 V E

.....(1)

V3 3‡

2‡

10‡

Is1

0

Is3

E E

+

V1 +

R S G11 SG 21 SG S 31 SG 41 T

A Is4 E 5 V V2

The node basis matrix equation of the given circuit is formed by inspection using the circuit shown in Fig. 2.

2

2A

the currents supplied by the 20 V, 5 V and E volt sources respectively.

Reference node

Fig. 2.

Circuit Theory

5. 21

G11 = 1 2 G 22 = 1 + 1 + 1 2 10 3 15 3 10 + + = = 28 = 14 30 30 15 G33 = 1 3 G 44 = 0

G12 = G 21 = − 1 2 G13 = G31 = 0 G14 = G 41 = 0 G 23 = G32 = − 1 3 G 24 = G 42 = 0 G34 = G 43 = 0

I11 = Is1 − 2 I 22 = − Is4 I33 = − Is3 + 2 I 44 = 2 + Is4 − 2 = Is4

V1 = 20 V2 = 0 (given) V4 − V2 = 5 ` V4 = 5 V3 = − E

On substituting the above terms in equation (1), we get, R 1 V R 0 0 W S 20 −1 S 2 S 2 W S S− 1 14 − 1 0 W S 0 S 2 15 3 W S S W S 1 1 0 W S− E S 0 −3 3 SS W S 0 0 0W S 5 0 T X T

V R V W S Is1 − 2 W W W S W S − Is4 W W = S W W W S W S− Is3 + 2 W WW W SS Is4 W X T X

.....(2)

From row-4 of equation (2), we get, I s4 = 0

..... (3)

From row-2 of equation (2), we get, c−

1 # 20 + 14 # 0 + − 1 # (− E) + (0 # 5) = − I m c m c m s4 2 15 3 − 10 + 0 + E + 0 = − Is4 3 `

E = − I + 10 s4 3

.....(4)

From equation (3), we know that, I s4 = 0, ` E = 0 + 10 3 E = 10 # 3 = 30 V

EXAMPLE 5.9 5‡

10 ‡ 2A

20 ‡

+

In the circuit shown in Fig. 1, determine the current delivered by 24 V source using node analysis.

The given circuit has only two principal nodes. Let us choose one of the principal node as reference and the voltage of other principal node as V1, as shown in Fig. 2. Let us take the voltage at the meeting point of 5 Ω and 24 V source as V2 and the voltage at the meeting point of 10 Ω and 36 V as V3. With reference to Fig. 2, we can say that,

+

24 V E

SOLUTION

E 36 V

Fig. 1. V1 10 ‡

5‡ V2

V2 = 24 V

and V3 = 36 V

+

+

Is1

20 ‡ V 3

2A

24 V E

E 36 V

0

Reference node

Fig. 2.

5. 22

Chapter 5 - Node Analysis With reference to Fig. 3, the node equation for node-1 can be written as shown below : Currents leaving node-1 : V1 − V2 , V1 − V3 , V1 5 10 20 Current entering node-1 : 2 A `

V −V V1 − V2 V + 1 3 + 1 = 2 5 10 20

V1

2A

V1 E V2 5

V V1 V2 V V − + 1 − 3 + 1 = 2 5 5 10 10 20

V2

V V ` c 1 + 1 + 1 m V1 = 2 + 2 + 3 5 10 5 10 20 Put,

V1 E V3 10

V1 20

2A

V3

0

0

Fig. 3.

V2 = 24 and V3 = 36 ` c 1 + 1 + 1 m V1 = 2 + 24 + 36 5 10 20 5 10 (0.2 + 0.1 + 0.05) V1 = 2 + 4.8 + 3.6 0.35V1 = 10.4 ` V1 = 10.4 = 29.7143 V 0.35

Let, I s1 be the current delivered by 24 V source as shown in Fig. 2. Is1 =

V2 − V1 = 24 − 29.7143 = − 1.1429 A 5 5

Since the current delvered by 24 V source is negative, we can say that it absorbs power in stead of delivering power.

EXAMPLE 5.10 2‡ 2‡

1‡

2A E

+

6V E

+

In the circuit shown in Fig. 1, solve the voltages across various elements by node method and hence determine the power in each element of the circuit.

SOLUTION

2V

Fig. 1.

The given circuit has 5 nodes and in this only two nodes are principal nodes. Let us choose one of the node as reference node, which is indicated by “0”. The voltage of reference node is zero volt. Let us choose three other nodes and assign node voltages V1, V2 and V3 as shown in Fig. 2. Let the current delivered by 2 V and 6 V sources be I s1 and I s2 respectively. With reference to Fig. 2, the following relation can be obtained for node voltages. V2 − V1 = 6 V

∴ V2 = 6 + V1 = 6 + 2 = 8 V In the circuit shown in Fig. 2, the voltage V 1 and V2 are known quantities, but the currents I s1 and I s2 are unknown quantities. Hence, the total number of unknowns are three (i.e., V 3, I s1 and I s2 ) and so three node equations can be formed and they can be solved to give a unique solution.

2‡

V3 2‡

1‡

Is2

E 6V

2A

0

V2 Is1

E 2V

+

;

+

V1 = 2 V

V1 Reference node

Fig. 2.

Circuit Theory

5. 23

Using Fig. 2, the node basis matrix equation is formed by inspection as shown below: G11 G12 G13 G 22 G 23 H G31 G32 G33

>G21

V1

>V2 H

I11

=

V3

>I22 H I33

.....(1)

G12 = G 21 = 0 V1 = 2 I11 = Is1 + 2 − Is2 1 1 G13 = G31 = 0 V2 − V1 = 6 I 22 = Is2 = = 1+2 3 − 1 = − 1 I33 = − 2 ` V2 = 6 + V1 = 6 + 2 = 8 G G = = 23 32 1+2 3 = 1 +1 = 1 +1 = 5 1+2 2 3 2 6

G11 = 0 G 22 G33

On substituting the above terms in equation (1), we get, R V R V R V 0 W S2 W SIs1 + 2 − Is2 W S0 0 S W S0 1 − 1 W S 8 W Is2 3W S W = S 3 W S S W S0 − 1 5 W SS V WW S W 3 S W −2 3 6 T X T X T X

..... (2)

The node equations are obtained by multiplying the matrices on the left-hand side and equating to the terms on right-hand side. From row-1 we get, From row-2 we get, From row-3 we get,

0 = I s1 + 2 − I s2

⇒

8−1 V = I ⇒ s2 3 3 3 − 8 + 5 V3 = − 2 3 6

I s1 = −2 + I s2

..... (3)

8 − V3 3

.....(4)

Is2 =

.....(5)

From equation (5), we can write, V3 = 6 # c − 2 + 8 m = 6 c − 6 + 8 m = 12 = 0.8 V 3 5 3 15 5 On substituting, V3 = 0.8 in equation (4), we get, Is2 =

8 − V3 = 8 − 0.8 = 2.4 A 3 3

On substituting, I s2 = 2.4 A in equation (3), we get, I s1 = −2 + I s2 = −2 + 2.4 = 0.4 A

voltage across 2 A source be E 2 as shown in Fig. 3. Now,

Va = 1 × I s2 = 1 × 2.4 = 2.4 V Vb = 2 × I s2 = 2 × 2.4 = 4.8 V Vc = 2 × Is1 = 2 × 0.4 = 0.8 V E 2 = V1 − V3 = 2 − 0.8 = 1.2 V

Va

1‡

+

Vb _ V3 +

2‡

Vc •

E 6V

Is1

_ E2

0

V2 Is2

2‡

E 2V

2A +

+

Let the voltage across the resistances be Va , Vb and Vc and the

+ _

+

In Fig. 2 it can be observed that the current through series combination of 1 Ω and 2 Ω is Is2 and the current through the 2 Ω resistance in series with 2 V source is Is1. Now the voltage across the resistances are given by the product of current and resistance.

V1 Reference node

Fig. 3.

5. 24

Chapter 5 - Node Analysis

Estimation of power in each element In dc circuits, the power in an element is given by the product of voltage and current in that element. The resistances always absorb power. The sources can either deliver power or absorb power. In a source if the current leaves at the positive end of the source then it delivers power. Power consumed by the 1 Ω resistor = Va × I s2 = 2.4 × 2.4 = 5.76 W Power consumed by the 2 Ω resistor = Vb × I s2 = 4.8 × 2.4 = 11.52 W Power consumed by the 2 Ω resistor = V × I = 0.8 × 0.4 = 0.32 W 1 c s1 in series with 2 V source Power delivered by 6 V source

= 6 × I s2 = 6 × 2.4 = 14.4 W

Power delivered by 2 V source

= 2 × I s1 = 2 × 0.4 = 0.8 W

Power delivered by 2 A source

= E 2 × 2 = 1.2 × 2 = 2.4 W

Note : It is observed that the sum of power delivered (14.4 + 0.8 + 2.4 = 17.6 W) is equal to sum of power consumed (5.76 + 11.52 + 0.32 = 17.6 W).

EXAMPLE 5.11

+

E 2V

Use nodal analysis to determine the values of voltages at various nodes

3

‡

2

‡

in the circuit shown in Fig. 1. 2

SOLUTION

‡

The given circuit has 4 nodes. In this one of the node is chosen as zero. Let the voltages of other three nodes be V1, V2 and V3 with respect to

2V E +

reference node as shown in Fig. 2. The voltage source in the circuit does not

Is3

have a series resistance and so it cannot be converted to current source. Let I s3

3

be the current supplied by the 2 V source. With reference to Fig. 2, we can write,

⇒

V3 = 2 + V1

‡

Fig. 1.

reference node and it is indicated by “0”. The voltage of the reference node is

V3 − V1 = 2 V

4

1A

‡

V1 2

‡

From the above equation we can say that the node voltage V3 can be expressed in terms of V1. Now, the number of unknowns in the circuit are three

2

V2

V3 1A

0

‡

4

‡

Reference node

Fig. 2.

and they are V1, V2 and I s3 . Therefore, we can write three node equations using KCL (corresponding to three nodes) and a unique solution for unknowns can be obtained by solving the three equations. The node basis matrix equation for the circuit shown in Fig. 2 is obtained by inspection as shown below: G11 G12 G13 G > 21 G22 G23 H G31 G32 G33

V1 V > 2H = V3

G11 = 3 + 2 = 5 G 22 = 3 + 2 = 5 G33 = 2 + 4 = 6

I11 I > 22 H I33

G12 = G 21 = − 3 G13 = G31 = 0 G 23 = G32 = − 2

.....(1) I11 = − Is3 V3 − V1 = 2 I 22 = 1 ` V3 = 2 + V1 I33 = Is3

Circuit Theory

5. 25

On substituting the above terms in equation (1), we get, 5 −3 0 5 − 2H 0 −2 6

>− 3

>

V1 V2 H = 2 + V1

>

− Is3 1H Is3

.....(2)

The node equations of the circuit are obtained by multiplying the matrices on the left-hand side of equation (2) and equating to the terms on the right-hand side. From row-1, we get, 5 V1 − 3V2 = − Is3

..... (3)

From row-2, we get, − 3 V1 + 5V2 − 2 × (2 + V1) = 1 ⇒ − 3 V1 + 5V2 − 4 − 2V1 = 1 ⇒ − 5 V1 + 5V2 = 5

..... (4)

From row-3, we get, − 2 V2 + 6 × (2 + V1) = Is3

⇒

⇒ 6 V1 − 2V2 = Is3 − 12

−2 V2 + 12 + 6V1 = Is3

..... (5)

On adding equations (3), (4) and (5), we get,

⇒

5 V1 − 3V2 − 5V1 + 5V2 + 6V1 − 2V2 = − Is3 + 5 + I s3 − 12

6 V1 = −7

` V1 = − 7 = − 1.1667 V 6 5 + 5 # (− 1.1667) 5 + 5 V1 = = − 0.1667 V 5 5 With reference to Fig. 2, we can write, From equation (4), we get,

V2 =

V3 − V1 = 2 ∴

V3 = 2 + V1 = 2 + ( −1.1667) = 0.8333 V

The node voltages are, V1 = −1.1667 V ;

V2 = −0.1667 V

and

V3 = 0.8333 V

EXAMPLE 5.12

(AU June’14, 8 Marks) 3‡

Use nodal analysis to determine the values of voltages at various nodes in the circuit shown in Fig. 1. 10 ‡

3‡

2‡

SOLUTION The given circuit has 5 nodes. In this one of the node is chosen as reference. Let the voltages of other four nodes be V1, V2, V3 and V4 as shown in Fig. 2. Here, V4 = 10 V. Let us convert the 10 V voltage source in series with 10 Ω resistance to equivalent current source as shown in Fig. 3.

+ 10 V

3W V1

2W

V1

V3

-

V2 2W

V3

V2

+ 10 V

3W

1‡

5A

Fig. 1.

3W

3W V4 10 W

S‡

E

5W

5A

0

Fig. 2.

1W

6W

Þ

10 = 1A 10

10W

5W

5A

0

Reference node

Fig. 3.

1W

6W

Reference node

6‡

5. 26

Chapter 5 - Node Analysis The node basis matrix equation is formed by inspection using the circuit shown in Fig. 3. G11 G12 G13 G 22 G 23 H G31 G32 G33

>G21

V1

>V2 H

I11

=

V3

>I22 H I33

.....(1)

With reference to Fig. 3, the elements of conductance matrix and source current matrix are obtained as shown below : G11 = 1 + 1 + 1 + 1 = 0.9967 10 5 3 3

G12 = G 21 = − c 1 + 1 m = − 0.6667 3 3

I11 = 1

G22 = 1 + 1 + 1 = 1.1667 3 3 2 G33 = 1 + 1 + 1 = 1.6667 2 1 6

G13 = G31 = 0

I22 = 5

G23 = G32 = − 1 = − 0.5 2

I33 = 0

On substituting the above terms in equation (1), we get, R V R V R V 0 W S V1 W S1 W − 0.6667 S 0.9667 S− 0.6667 W 1.1667 − 0.5 W S V2 W = SS5 WW S S W S 0 1.6667 W S V3 W S0 W − 0.5 T X T X T X To solve the node voltages V1 , V2 and V3, let us define shown below: 0.9667 − 0.6667 0 ∆l = − 0.6667 1.1667 ∆l1 = − 0.5 0 − 0.5 1.1667 0.9667 ∆l2 = − 0.6667 0

1 5 0

0 − 0.5 1.1667

four determinants ∆l , ∆l1, ∆l2 and ∆l3 as 1 − 0.6667 5 1.1667 0 − 0.5

0 − 0.5 1.1667

0.9667 − 0.6667 ∆l3 = − 0.6667 1.1667 0 − 0.5

1 5 0

The determinants are evaluated by expanding along first row and node voltages are solved by Cramer’s rule. 0.9667 − 0.6667 0 ∆l = − 0.6667 1.1667 − 0.5 = 0.9667 # 61.1667 # 1.6667 − ^− 0.5h2 @ + 0 − 0.5 1.6667 0.6667 # 6 − 0.6667 # 1.6667 − 0 @ = 0.8972 1 0 − 0.6667 2 ∆l1 = 5 1.1667 − 0.5 = 1 # 91.1667 # 1.6667 − _− 0.5 i C + 0.6667 # 75 # 1.6667 − 0 A 0 1.6667 − 0.5 = 7.2505 0.9667 1 0 ∆l2 = − 0.6667 5 − 0.5 = 0.9667 # 65 # 1.6667 − 0 @ − 1 # 6 − 0.6667 # 1.6667 − 0 @ 0 0 1.6667 = 9.1672 0.9667 ∆l3 = − 0.6667 0

− 0.6667 1.1667 − 0. 5

1 5 = 0.9667 # 60 + 0.5 # 5 @ + 0.6667 # 60 − 0 @ + 1 # 60.6667 # 0.5 − 0 @ 0 = 2.7501

Circuit Theory

5. 27

' V1 = 3 1 = 7.2505 = 8.0813 V 0.8972 3 ' V2 = 3 2' = 9.1672 = 10.2176 V 0.8972 3 ' V3 = 3 3' = 2.7501 = 3.0652 V 0.8972 3

The node voltages are, V1 = 8.0813 V ;

V2 = 10.2176 V ;

V3 = 3.0652 V

and

V4 = 10 V

(AU Dec’14, 12 Marks)

EXAMPLE 5.13 10 ‡

25 A

SOLUTION

1‡

20 A

4‡

10 ‡

2‡

+

Determine the node voltages and the currents across all the resistors of the circuit shown in Fig. 1, by node method.

E 20 V

Solution of node voltages

Fig. 1.

The given circuit has 4 nodes.In this one of the node is chosen as reference node, which is indicated

V1 10 ‡ V2

1‡

V3

by “0”. The voltage of reference node is zero volt. Let us Is

V2 and V3 as shown in Fig. 2. Let the current delivered

25 A

4‡

20 A

2‡

10 ‡

E 20 V

+

choose three other nodes and assign node voltages V1, by 20 V source be I s . With reference to Fig. 2, we get,

0

Fig. 2.

V3 = 20 V

Reference node

From the above equation we can say that the node voltage V3 is a known quantity. Now, the number of unknowns in the circuit are three and they are V1, V2 and I s . Therefore, we can write three node equations using KCL (corresponding to three nodes) and a unique solution for unknowns can be obtained by solving the three equations. The node basis matrix equation for the circuit shown in Fig. 2 is obtained by inspection as shown below: G11 G12 G13 G 22 G 23 H G31 G32 G33

>G21

V1

>V2 H

I11

=

V3

G11 = 1 + 1 = 0.35 4 10 G 22 = 1 + 1 + 1 = 1.6 10 2 1 G33 = 1 + 1 = 1.1 1 10

>I22 H

.....(1)

I33

G12 = G 21 = − 1 = − 0.1 10 G13 = G31 = 0 G 23 = G32 = − 1 = − 1 1

On substituting the above terms in equation (1), we get, R V 0.35 − 0.1 0 V1 S25 W >− 0.1 1.6 − 1 H > V2 H = S20 W SS I WW 0 − 1 1.1 20 s T X

V3 = 10 V

.....(2)

5. 28

Chapter 5 - Node Analysis

The node equations of the circuit are obtained by multiplying the matrices on the left-hand side of equation (2) and equating to the terms on the right-hand side. From row-1, we get, 0.35V1 − 0.1 V2 = 25

.....(3)

From row-2, we get,

⇒

− 0.1V1 + 1.6 V2 − 20 = 20

− 0.1 V1 + 1.6 V2 = 40

.....(4)

On Multiplying equation (3) by 16, we get, 5.6 V1 − 1.6 V2 = 400

.....(5)

On adding equations (4) and (5), we get, − 0.1 V1 + 1.6 V2 + 5.6 V1 − 1.6 V2 = 40 + 400

⇒

5.5 V1 = 440

` V1 = 440 = 80 V 5.5 From equation (2), we get, V2 =

40 + 0.1 V1 = 40 + 0.1 # 80 = 30 V 1.6 1.6

The node voltages are, V1 = 80 V

;

V2 = 30 V

and

V3 = 20 V

To solve branch voltages and currents The given circuit has five resistance branches. Let us denote the resistance branch voltages as Va, Vb, Vc, Vd and Ve and resistance branch currents as Ia, Ib, Ic, Id and Ie as shown in Fig. 3. The sign of branch voltages and currents are chosen such that they are all positive. The branch voltages and currents are solved as shown below: Va = V1 = 80 V Vb = V1− V2 = 80 − 30 = 50 V Vc = V2 = 30 V V1 Ib 10 ‡ V2 E + Vb

Ve = V3 = 20 V Ia =

Va = 80 = 20 A 4 4

Ib =

Vb = 50 = 5 A 10 10

Ic =

Vc = 30 = 15 A 2 2

Id =

Vd = 10 = 10 A 1 1

Ie =

Ve = 20 = 2 A 10 10

Id Ic

Ia + 25 A

Va E

1 ‡ V3 +

+ 4‡

20 A V

c

E

Fig. 3.

2‡

Vd

E

+ Ve E

Ie 10 ‡

+

Vd = V2 − V3 = 30 − 20 = 10 V

E 20 V

Circuit Theory

5. 29

EXAMPLE 5.14

(AU May’15, 16 Marks) 10 ‡

Use nodal analysis to solve the circuit shown in Fig.1.

2

1

SOLUTION

4‡

5‡

Let the node voltages be V1 and V 2. Let us convert the 25 V voltage source in series with 5 Ω resistance to equivalent current source as shown in Fig. 2.Similarly convert the 50 V voltage source in series with 2 Ω resistance to equivalent current source as shown in Fig. 2.The node basis matrix equation of the given circuit is formed by inspection using the circuit shown in Fig. 2.

+

2‡

25V

2‡

E

E + 50V Reference node

0

Fig. 1. 10 ‡

I11 G11 G12 V1 G = G = = G I 22 G21 G22 V2

V1

.....(1)

V2 25 a 5A 5

5‡

50 a 25A 2

2‡

With reference to Fig. 2, the elements of conductance matrix and source current matrix are obtained as shown below :

4‡

2‡

=

0

G11

= 1 + 1 + 1 = 0.8 2 5 10

G12 = G21

= − 1 = − 0.1 10

G22 = 1 + 1 + 1 = 0.85 10 4 2

I11

Reference node

= 5Fig. 2.

I22 = − 25

On substituting the above terms in equation (1), we get, R VR V R V S 0.8 − 0.1 W S V1 W S 5 W W S WS W = S W S WS W S S− 0.1 0.85 W S V2 W S− 25 W T XT X T X In order to solve the node voltages V1 and V2, let us define three determinants ∆’, ∆’1 and ∆’2 as shown below: 0.8 − 0.1 ; − 0.1 0.85

∆l =

∆l1 =

5 − 0.1 ; − 25 0.85

∆l 2 =

0.8 5 − 0.1 − 25

The determinants are evaluated by expanding along first row and the node voltages are solved by Cramer’s rule. 0.8

∆l =

− 0.1 5

∆l1 =

− 25

∆l 2 =

0.8

− 0.1 = 0.8 0.85 − ^− 0.1h2 = 0.67 # 0.85 − 0.1 = 5 0.85 − _− 25i _− 0.1i = 1.75 # # 0.85 5

− 0.1 − 25

= 0.8 # _− 25 i − _− 0.1 i # 5 = − 19.5

V1 =

∆l1 = 1.75 = 2.6119 V 0.67 ∆

V2 =

∆l 2 = − 19.5 = − 29.1045 V 0.67 ∆

5. 30

Chapter 5 - Node Analysis

EXAMPLE 5.15 1‡

5‡

Determine the node voltages and, hence, the power supplied by 5 A source into the circuit shown in Fig. 1, by supernode analysis technique.

5A

+E

6V 2‡

SOLUTION

4‡

Fig. 1.

Let us choose the reference node “0” and three node voltages V1, V2 and V3 as shown in Fig. 2. Now, the voltage across 5 A source is V1 and the power delivered by 5 A source is V1 ´ 5 Watts

V1

With reference to Fig. 2, the relation between node voltages V2 and V3 is, V2 − V3 = 6

⇒

V3 = V2 – 6

.....(1)

5A

Let us short-circuit node-2 and node-3 to form a supernode as shown in Fig. 3.

V1 5W

V3 - V1 1 V2 - V1 5 +

V2

V2 2

V1

V1

V3 4

6V

5W

1W

Þ

V3

V2 2

4W

2W

+ V1 V2 _

V3

+E

6V 2‡

4‡ Reference node

0

Fig. 2. 1W Supernode

V3 4

4W

2W

0

0

Is2

V1

V3 - V1 1 V2 - V1 5

1‡

5‡

0

0

Fig. 3 : Formation of supernode. The KCL equation of supernode is formed as shown below: V −V V V2 − V1 V + 3 1 + 2 + 3 = 0 5 1 2 4 0.2V2 − 0.2V1 + V3 − V1 + 0.5V2 + 0.25V3 = 0 – 1.2V1 + 0.7V2 + 1.25V3 = 0 – 1.2V1 + 0.7V2 + 1.25 (V2 – 6) = 0

Using equation (1)

– 1.2 V1 + 1.95V2 – 7.5 = 0 ` V2 =

7.5 + 1.2 V1 = 7.5 + 1.2 V1 = 3.8462 + 0.6154 V1 1.95 1.95 1.95

.....(2)

With reference to Fig. 4, the KCL equation of node-1 is formed as shown below: V −V V1 − V 2 + 1 3 = 5 5 1

V1 5‡

0.2V1 − 0.2V2 + V1 – V3 = 5 5A

1.2V1 – 0.2V2 – (V2 – 6) = 5 1.2V1 – 1.2V2 = 5 – 6

V2

Using equation (1)

1‡

V1 E V2 5

V1 E V3 1

Fig. 4.

V3

Circuit Theory

5. 31

1.2 V1 – 1.2(3.8462 + 0.6154V1) = –1

Using equation (2)

1.2 V1 – 4.6154 – 0.7385V1 = –1 0.4615V1 = 3.6154 ` V1 = 3.6154 = 7.834 V 0.4615 From equation (2), V2 = 3.8462 + 0.6154 V1 = 3.8462 + 0.6154 × 7.834 = 8.6672 V From equation (3), V3 = V2 – 6 = 8.6672 – 6 = 2.6672 V Power supplied by 5 A source = V1 × 5 = 7.834 × 5 = 39.17 W

5.4

Node Analysis of Circuits Excited by ac Sources (Nodal Analysis of Reactive Circuits)

The reactive circuits consist of resistances, inductive and capacitive reactances. Therefore, the voltages and currents of reactive circuits will be complex (i.e., they have both real and imaginary components). In general, the elements of the circuit are referred to as impedances. In node analysis the admittance (which is the inverse of impedance) is more convenient. The general node basis matrix equation for reactive circuit is, ..... (5.8)

YV=I

where, Y = Admittance matrix of order n × n V = Node voltage matrix of order n × 1 I = Source current matrix of order n × 1 n = Number of nodes except the reference node The equation (5.8), can be expanded as shown in equation (5.9). R V R V R V S Y11 Y12 Y13 g Y1n W S V1 W S I11 W S Y21 Y22 Y23 g Y2n W S V2 W SI 22 W S W S W S W S Y31 Y32 Y33 g Y3n W S V3 W = S I33 W S h Sh W h h h W Sh W SS W S W SS WW Yn1 Yn2 Yn3 g Ynn W S Vn W I nn T X T X T X Note : The over bar is used to denote complex quantities.

..... (5.9)

The formation of node basis matrix equation and the solution of node and branch voltages are similar to that of resistive circuits except that the solution of voltages involves complex arithmetic. Therefore, the k t h node voltage of a reactive circuit with “n” nodes excluding reference is given by, Vk = 1 ∆l

n

/ ∆l

jk

I jj

Note : Refer equation (5.5) ..... (5.10)

j =1

where, ∆l jk = Cofactor of Yjk I jj

= Sum of current sources connected to j th node

∆l = Determinant of admittance matrix

5. 32

Chapter 5 - Node Analysis

Instead of using the equation (5.10) for solution of node voltages, the short-cut procedure for Cramer’s rule can be followed. Consider the node basis matrix equation for a circuit with three nodes except the reference node. Y11 Y12 Y13 Y22 Y23 H Y31 Y32 Y33

>Y21

>V2 H V1

V3

I11 = >I 22 H I33

Let us define four determinants as shown below: Y11 Y12 Y13 ∆l = Y21 Y22 Y23 ; Y31 Y32 Y33

I11 Y12 Y13 ∆l1 = I 22 Y22 Y23 I33 Y32 Y33

Y11 I11 Y13 ∆l 2 = Y21 I 22 Y23 ; Y31 I33 Y33

Y11 Y12 I11 ∆l3 = Y21 Y22 I 22 Y31 Y32 I33

Here, ∆’ = Determinant of admittance matrix. ∆’1 = Determinant of admittance matrix after replacing the first column of admittance matrix by source current column matrix. ∆’2 = Determinant of admittance matrix after replacing the second column of admittance matrix by source current column matrix. ∆’3 = Determinant of admittance matrix after replacing the third column of admittance matrix by source current column matrix. Now, node voltages V1, V2 and V3 are given by, V1 =

∆l1 ∆l

V2 =

∆l 2 ∆l

V3 =

∆l 3 ∆l

EXAMPLE 5.16

Let us convert the 100∠0o V voltage source in series with 3 + j4 Ω impedance to current source IS in parallel with 3 + j4 Ω impedance.

~

4 - j3 W

10 W

~ Fig. 1.

10Ð45o A

SOLUTION

3 + j4 W

100Ð0oV

Determine the power consumed by the 10 Ω resistor in the circuit shown in Fig. 1, by nodal analysis.

3W

Circuit Theory

5. 33

100Ð0oV

3 + j4 W +

Þ

~

_

I s = 20Ð - 53.13o A

3 + j4 W

~

o IS = 100+0 = 100 = 12 − j16 A = 20+ − 53.13 o A 3 + j4 3 + j4

voltages of other two nodes be V1 and V2 with respect to reference node as shown in Fig. 2.

~

~ 0

The node basis matrix equation of the circuit of Fig. 2. is formed by inspection as shown below:

=

10 W

V2

10Ð45o A

which is denoted as “0”. The voltage of reference node is zero. Let the

3 + j4 W

has three nodes. Let us choose one of the node as reference node,

4 - j3 W

V1

Is = 20Ð-53.13o A

The modified circuit is shown in Fig. 2. The circuit of Fig. 2

3W

Reference node

Fig. 2.

Y11 Y12 V1 I11 G = G = = G Y21 Y22 V2 I 22

..... (1)

Y11 =

1 1 + + 1 = 0.38 − j0.04 3 + j4 4 − j3 10

Y22 =

1 + 1 = 0.493 + j0.12 4 − j3 3

Y12 = Y21 = − c 1 m = − 0.16 − j0.12 4 − j3 I11 = 20+ − 53.13 o = 20 cos (− 53.13) + j20 sin (− 53.13) = 12 − j16 I 22 = − 610+45 o @ = − (10 cos 45 o + j10 sin 45 o) = − 7.071 − j7.071 Note : All calculations are performed using the calculator in complex mode. On substituting the above terms in equation (1), we get,

=

0.38 − j0.04 − 0.16 − j0.12

12 − j16 − 0.16 − j0.12 V1 G = G = = G 0.493 + j0.12 − 7.071 − j7.071 V2

..... (2)

To determine the power consumed by the 10 Ω resistance, it is sufficient if we calculate V1 in equation (2). In order to solve V1, let us define two determinants ∆l and ∆l1 as shown below: ∆l =

0.38 − j0.04 − 0.16 − j0.12

− 0.16 − j0.12 ; 0.493 + j0.12

Now, the voltage V1 is given by, V1 = ∆l =

0.38 − j0.04 − 0.16 − j0.12

∆l1 =

12 − j16 − 7.071 − j7.071

− 0.16 − j0.12 0.493 + j0.12

∆l1 . ∆l

− 0.16 − j0.12 = 6(0.38 − j0.04) # (0.493 + j0.12) @ − [− 0.16 − j0.12] 2 0.493 + j0.12 = 0.1809 − j0.0125

5. 34

Chapter 5 - Node Analysis ∆l1 =

12 − j16 − 0.16 − j0.12 = 6(12 − j16) # (0.493 + j0.12) @ − 7.071 − j7.071 0.493 + j0.12 − 6(− 7.071 − j7.071) # (− 0.16 − j0.12) @ = 7.5532 − j8.4279

` V1 =

7.5532 − j8.4279 ∆l1 = = 44.759 − j43.496 = 62.412+ − 44.2 o V ∆l 0.1809 − j0.0125

Power consumed by the 10 Ω resistor =

V1 10

2

2 = 62.412 = 389.5 W 10

EXAMPLE 5.17

j3 W

Find the voltages across various elements in the circuit shown in Fig. 1, by node method.

SOLUTION

3W

~

6W

~

10Ð0o A

5Ð90o A

-j5 W

The graph of the given circuit is shown in Fig. 2. It has 6 branches and

Fig. 1.

3 nodes. Hence, the circuit will have 6 voltages corresponding to 6 branches. The branch voltages depends on the node voltages. In node analysis, the volt-

age of one of the node is chosen as reference and it is equal to zero volt. In the circuit of Fig. 3, the reference node is denoted by “0”. The voltages of other two nodes are denoted as V1 and V2 . j3 W

e c

b

0

5Ð90o A

a

V1

d

~

-j5 W

6W

3W

0

Reference node

Fig. 2.

V2

~

10Ð0o A

f

Reference node

Fig. 3.

The node basis matrix equation of the circuit shown in Fig. 3 is obtained by inspection as shown below:

=

..... (1)

Y11 Y12 V1 I11 G = G = = G Y21 Y22 V2 I 22

Y11 = 1 + 1 + 1 = 0.333 − j0.133 3 j3 − j5

I11 = 5+90 o = j5

Y22 = 1 + 1 + 1 = 0.167 − j0.133 j3 6 − j5

I 22 = 10+0 o = 10

Y12 = Y21 = − c 1 + 1 m = j0.133 j3 − j5 On substituting the above terms in equation (1), we get,

=

0.333 − j0.133 j0.133

j0.133 G 0.167 − j0.133

=

j5 V1 G = = G 10 V2

.....(2)

Circuit Theory

5. 35

To solve the node voltages by Cramer’s rule, let us define three determinants ∆l , ∆l1 and ∆l2 as shown below: j0.133 j5 j0.133 0.333 − j0.133 0.333 − j0.133 j5 ; ∆l1 = ; ∆l2 = ∆l = j0.133 0.167 − j0.133 10 0.167 − j0.133 j0.133 10 Now, the node voltages are given by, V1 = ∆l =

0.333 − j0.133 j0.133

j0.133 = 6(0.333 − j0.133) # (0.167 − j0.133) @ − [j0.133] 2 0.167 − j0.133

= 0.0556 − j0.0665 j0.133 = 6 j5 # (0.167 − j0.133) @ − 610 # j0.133 @ 0.167 − j0.133 = 0.665 − j0.495

j5 ∆l1 = 10

∆l 2 =

∆l1 ∆l 2 and V2 = . ∆ ∆

0.333 − j0.133 j5 = 6(0.333 − j0.133) # 10 @ − 6 j0.133 # j5 @ j0.133 10 = 3.995 − j1.33

` V1 = V2 =

0.665 − j0.495 ∆l1 = = 9.302 + j2.2227 = 9.564+13.4 o V ∆l 0.0556 − j0.0665 3.995 − j1.33 ∆l 2 = = 41.3339 + j25.5163 = 48.575+31.7 o V ∆l 0.0556 − j0.0665

To find branch voltages The branch voltages are denoted by Va, Vb, Vc, Vd, Ve and Vf as shown in Fig. 4. The polarites of branch voltages are chosen arbitrarily. The branch voltages depends on the node voltages. The relation between branch and node voltages are obtained with reference to Fig. 4 as shown below :

+

j3 W

+

~

-

-

V2

+

-j5 W

+ Va 3 W

Ve

+

+

6W

Vb

-

Vc -

Vd

~

-

10Ð0o A

V1

5Ð90o A

Va = Vb = V1 = 9.564+13.4 o V

Vf

-

o

Vc = Vd = V2 = 48.575+31.7 V Ve = Vf = V2 − V1

0

Reference node

Fig. 4.

= (41.3339 + j25.5163) − (9.302 + j2.2227) = 32.0319 + j23.2936 = 39.606∠36 o V

5.5

Summary of Important Concepts 1.

Node is defined as meeting point of two or more elements.

2.

When more than two elements meet at a point then the node is called principal node.

3.

Node analysis is used to solve the independent voltage variables of a circuit.

4.

The number of voltage variables in a circuit is equal to number of branches.

5.

The number of independent voltages in a circuit is given by number of twigs (or tree branches) in the graph of a cirucit.

6.

The number of twigs (or tree branches) n in a circuit with N nodes is given by, n = N – 1.

7.

In node analysis, the independent voltages are solved by writing KCL equations for various nodes of a circuit.

5. 36

Chapter 5 - Node Analysis 8.

In node analysis, one of the node is chosen as reference node and its voltage is considered as zero and all other node voltages are solved with respect to reference node.

9.

10.

The node basis matrix equation for resistive circuit is, RG G G g G V RV V RI V 1n S 11 12 13 W S 1W S 11 W SG 21 G 22 G 23 g G 2n W S V2 W S I 22 W SG31 G32 G33 g G3n W S V3 W = S I33 W S W S W S W h h h W Sh W S h Sh W SG G G g G W S V W SI W nn T n1 n2 n3 X T nX T nn X The node voltages are solved by using Cramer’s rule.

11.

The kth node voltage Vk by Cramer’s rule is, Vk = 1 ∆l

n

/ ∆l jk I jj j=1

where, n ∆l jk Ijj ∆l 12.

= = = =

Number of independent nodes in a circuit Cofactor of Gjk Sum of current sources connected to node-j Determinant of conductance matrix

For circuit with three nodes excluding the reference node, the node voltages by Cramer’s rule are, ∆l ∆l11 ∆l I + 21 I + 31 I ∆l 11 ∆l 22 ∆l 33 ∆l ∆l12 ∆l V2 = I + 22 I + 32 I ∆l 11 ∆l 22 ∆l 33 V1 =

V3 = 13.

∆l13 ∆l ∆l I + 23 I + 33 I ∆l 11 ∆l 22 ∆l 33

The node voltages of a circuit with three nodes excluding the reference using short-cut procedure for Cramer’s rule are, ∆l1 ∆l ∆l 2 V2 = ∆l ∆l 3 V3 = ∆l V1 =

G11 G12 G13 I11 G12 G13 l l where, ∆ = G 21 G 22 G 23 ; ∆ 1 = I 22 G 22 G 23 ; G31 G32 G33 I33 G32 G33 G11 I11 G13 G11 G12 I11 ∆l 2 = G 21 I 22 G 23 ; ∆l3 = G 21 G 22 I 22 G31 I33 G33 G31 G32 I33

14.

When a voltage source is connected between two nodes it can be short circuited for analysis purpose and the short circuited two nodes can be considered as one single node called supernode.

Circuit Theory

5.6 Q5.1

5. 37

Short-answer Questions What is node analysis? Node analysis is an useful technique to solve the independent voltage variables of a circuit.

Q5.2

When is node analysis preferred to solve the voltages? The node analysis is preferred to solve the voltage variables when the circuit is excited by only current sources. Applying node analysis is straight forward and easier in circuits excited by only current sources. But the node analysis can also be extended to circuits excited by both voltage and current sources.

Q5.3

How is node analysis performed? In a circuit with N nodes, one of the node is chosen as reference node and its voltage is considered as zero. The voltages of remaining N – 1 nodes are independent voltages of the circuit with respect to reference node. For each node (except the reference node) a KCL equation is formed and then the n (where, n = N – 1) number of node equations are solved by Cramer’s rule to get a unique solution for node voltages.

Q5.4

How are the node voltages solved using the node basis matrix equation? Consider the node basis matrix equation, GV=I On premultiplying the above equation both sides by G−1 we get, G−1 G V = G−1 I U V = G−1 I ∴ V = G−1 I

G−1 G = U = Unit matrix UV = V

The above equation will be the solution for node voltages and the kth node voltage is, Vk =

∆l ∆l1k ∆l ∆l I11 + 2k I22 + 3k I33 + ...... + nk Inn = 1 ∆l ∆l ∆l ∆l ∆l

n

/

∆l jk I jj

j =1

The above equation for node voltages is also called Cramer’s rule.

Q5.5

What is supernode? When a voltage source is connected between two nodes it can be short circuited for analysis purpose and the short circuited two nodes can be considered as one single node called supernode.

Q5.6

What is the value of Is2 in the circuit shown in Fig. Q5.6 if the value of V2 is zero? Solution

2

V1

‡

V2

The node basis matrix equation by inspection is, 4 5 − 2 V1 G G = G = = = − Is2 − 2 5 V2 Now, V2 = Here, ∆l 2 =

4A

∆l 2 . Therefore, for V2 = 0, ∆l 2 = 0 ∆ 5 4 = − 5Is2 + 8 − 2 − Is2

` − 5Is2 + 8 = 0 ⇒ 5Is2 = 8 ⇒ Is2 = 8 = 1.6 A 5

3

‡

3

‡

Is2

0 Reference node

Fig. Q5.6.

5. 38 Q5.7

Chapter 5 - Node Analysis Find the value of Is1 and Is2 in the circuit shown in Fig. Q5.7. Is2

Solution The node basis matrix equation by inspection is, Is1 − Is2 5 − 3 V1 G G = G = = = Is2 − 3 6 V2

3

‡

V1 = 3 V

On substituting for V1 = 3 V and V2 = 2 V in the above equation we get, Is1 − Is2 5 −3 3 G G = G = = = Is2 −3 6 2

V2 = 2 V 2

Is1

‡

2

‡

1

0

‡

Reference node

Fig. Q5.7.

From row-2 we get, –3 ´ 3 + 6 ´ 2 = Is2 ⇒ Is2 = 3 A From row-1 we get, 5 ´ 3 – 3 ´ 2 = Is1 – Is2 ⇒ Is1 = 9 + Is2 = 9 + 3 = 12 A

Q5.8

Find the value of V2 and Is2 in the circuit shown in Fig. Q5.8.

4

V1 = 4 V

Solution The node basis matrix equation by inspection is, 4 5 − 4 V1 G = G = = G = Is2 − 4 6 V2

1

4A

⇒

V2

‡

4 5 −4 4 G = G = = G = Is2 − 4 6 V2

2

0

From row-1 we get, 5 # 4 − 4V2 = 4 ⇒

‡

Is2

‡

Reference node

Fig. Q5.8.

4V2 = 16 ⇒ V2 = 16 = 4 V 4

From row-2 we get, – 4 ´ 4 + 6V2 = Is2 ⇒

Q5.9

Is2 = –16 + 6 ´ 4 = 8 A

In the circuit shown in Fig. Q5.9, find the power delivered to 2M conductance. Solution The node basis matrix equation by inspection is,

2A

1+3 4−2 4 − 3 V1 − 3 V1 2 G= G = = G⇒ = G= G = = G = 2+1 − 3 6 V2 − 3 3 + 1 + 2 V2 3 ∆l =

4 −3 = 4 # 6 − ^− 3h2 = 15 −3 6

∆l 2 = ` V2 =

3

‡

V2

V1 1

4A

4 2 = 4 # 3 − ^− 3h # 2 = 18 −3 3 ∆l 2 = 18 = 1.2 V ∆l 15

Power delivered to 2M conductance = V22 ´ 2 = 1.22 ´ 2 = 2.88 W

1

‡

2

‡

‡

1A 0

Reference node

Fig. Q5.9.

Circuit Theory Q5.10

5. 39

Find the voltage Vx in the circuit shown in Fig. Q5.10.

1

Solution

‡

+ Vx E V1

The node basis matrix equation is,

=

4

7 − 5 V1 9 G = G = = G − 5 10 V2 0

∆l =

2

‡

5

‡

9A

7 −5 = 7 # 10 − ^− 5h2 = 45 − 5 10

∆l1 =

9 −5 = 90 − 0 = 90 0 10

∆l 2 =

7 9 = 0 − ^− 5h # 9 = 45 −5 0

Vx = V1 − V2 =

V2

‡

0

Reference node

Fig. Q5.10.

∆l1 ∆l 2 ∆l1 − ∆l2 − = = 90 − 45 = 1V 45 ∆l ∆ ∆

5.7 Exercises I. Fill in the Blanks With Appropriate Words 1.

Node analysis is used to solve ________ variables of a circuit.

2.

A circuit with N nodes will have ________ independent voltages.

3.

In node basis matrix equation, the node voltages are solved by using ________.

4.

The node equations are ________ equations of a circuit.

5.

The solution of node basis matrix equation GV = I will be in the form ________.

ANSWERS 1. Voltage

2. N – 1

3. Cramer’s rule

4. KCL

5. V = G-1I

II. State Whether the Following Statements are True/False 1.

Node analysis can be used to solve the current variables from the knowledge of voltage variables.

2.

Node analysis can be applied only to circuits excited by current sources.

3.

The node voltages are independent voltage variables of a circuit.

4.

For node analysis any node can be chosen as reference node.

5.

The node equations are formed by using KVL.

ANSWERS 1. True

2. False

3. True

4. True

5. False

5. 40

Chapter 5 - Node Analysis

III. Choose the Right Answer for the Following Questions 1. Node analysis is based on, a) KCL

b) KVL

c) Ohm’s law

d) none of the above

2. A circuit has 8 branches and 5 nodes, then the number of independent voltages are, a) 7

b) 6

c) 5

d) 4

3. In a circuit with n independent voltages, the kth node voltage by Cramer’s rule is given by, k

a) Vk = 1 ∆l

/

c) Vk = 1 ∆l

/ ∆l

∆l jk I jj

j=1 n

jk

I jj

j=1

j

b) Vk = 1 ∆l

/ ∆l

d) Vk = 1 ∆l

/ ∆l

jk

I kk

jn

I kn

k =1 m

k =1

4. In node basis matrix equation the mutual-conductances are, a) always positive

b) always negative

c) positive or negative

d) always zero

5. The node voltages V1, V2 and V3 in the circuit shown in Fig. 5 are respectively. a) 6 V, 4 V, – 2 V

6V

b) 10 V, 4 V, 2 V

2V

V2

+E

V1

+E

V3

+ E 4V

c) 6 V, 2 V, –2 V 0

d) 10 V, 2 V, 2 V

Reference node

Fig. 5.

6. In the circuit shown in Fig. 6, the voltages V1 and V2 are respectively,

2

V1

‡

V2

a) 4 V, 2 V b) 3 V, 1 V

3A

1

‡

1

c) 3 V, 3 V

0

d) 2 V, 2 V

Fig. 6.

‡

3A

Reference node

7. The node voltages V1, V2 and V3 in the circuit shown in Fig. 7 are,

V1 6A

a) 1.5 V, 2 V, 1.67 V

V2

2‡

4‡ 4A

b) 24 V, 8 V, 15 V

Reference node

c) 2 V, 2 V, 2 V d) 10 V, 6 V, 8 V

3‡

5A

Fig. 7. V3

Circuit Theory

5. 41

8. If the node voltage V2 is zero, then the value of Is2 in the cirucit shown in Fig. 8 is? V1

1‡

1‡

V2

V3

a) –4 A 4A

b) 4 A

2‡

4‡

Is2

2‡

c) –2 A 0

Reference node

Fig. 8.

d) 2 A

9. In the circuit shown in Fig. 9 the power P1 and P2 delivered by current sources are, 1‡

V1

a) P1 = 4 W, P2 = 4 W b) P1 = 8 W, P2 = 4 W

4 A, P1

V2

2‡

1‡

c) P1 = 6 W, P2 = 2 W

2 A, P2

0

d) P1 = 8 W, P2 = 0

Reference node

Fig. 9. 10. In the circuit shown in Fig. 10, what is the value of Is1 for the power delivered to 1 W resistor is 25 W. V1

a) 35 A b) 17.5 A

V2

2‡

Is1

c) 8.75 A

2‡

d) 4.375 A

2‡

0

Reference node

Fig. 10.

ANSWERS 1. a

3. c

5. b

7. b

9.

2. d

4. b

6. c

8. a

10. b

d

IV. Unsolved Problems E5.1 Determine the branch voltages in the circuit shown in Fig. E5.1 by node method. V2

2‡ +

_ Vf

1‡

_

+

Ve

Va 8 ‡

_

_

5A

+

+ Vb

+

Fig. E5.1.

2‡ V3

V1 4A

Vc 2A

2‡

_ Vd

_

+

1‡

5‡

4‡

10 ‡

5‡

5‡

0

Fig. E5.2.

E5.2 Determine the node voltages in the circuit shown in Fig. E5.2.

1‡

5. 42

Chapter 5 - Node Analysis

E5.3 Determine the current I L in the circuit of Fig. E5.3 by node method. 0.1 ‡

IL

3A

5‡

0.4 ‡ 1‡

5A

0.5

0.2

4‡

2‡

‡

10 A

‡

4A

2A

1.6 ‡

2.5 ‡

Fig. E5.4.

0.25 ‡

Fig. E5.3.

E5.4 Determine the power supplied/absorbed by the current sources in the circuit shown in Fig. E5.4. E5.5 Determine the power absorbed by 10 Ω resistor in the circuit of Fig. E5.5. 10 ‡

o 5Ð0 A

2A

~

4A V1

1‡

5‡

3‡

V2

j4 W

3W

3‡

8W -j4 W

0

7‡

Fig. E5.5.

Fig. E5.6.

E5.6 Determine the node voltages V 1 and V 2 in the circuit shown in Fig. E5.6. E5.7 Determine the node voltages in the circuit shown in Fig. E5.7. j0.5 W

0.2 W

1+j2 W

V3

0.1 W

~

-j0.2 W j0.2 W -j0.1 W

~

10+j5 W

4+j8 W

2-j6 W

~

0

~

5Ð90o A

V2

10Ð0o A

0.4 W

3Ð90o A

6Ð0o A

V1

Fig. E5.8.

Fig. E5.7.

E5.8 In the circuit shown in Fig. E5.8, determine the active and reactive power in the impedance 1 + j2 Ω by node method. E5.9 In the circuit shown in Fig. E5.9, calculate the current I L by node method.

~

o

10+j5 W

A

-9 0

~

5Ð

5Ð0o A

IL

Fig. E5.9.

8-j4 W

~

10Ð30o A

2-j6 W

Circuit Theory

5. 43

E5.10 Determine the voltages across the resistors in the circuit shown in Fig. E5.10, by node analysis. +

+

10 + E

+

Vd

Vc _

5‡

+ E

+

15 V

Vb _

_

10 V

‡

+ E

4‡

_

2‡

Va

5V

+ Ve 1 ‡ _

Fig. E5.10.

E5.11 Determine the power delivered/absorbed by the sources in the circuit shown in Fig E5.11, by node analysis. 8‡

10 ‡ 15 V

4‡

+ E

4A

1A

2‡

5‡

1‡

Fig. E5.11.

E5.12 Determine the currents I 1, I 2 , I 3 a n d I 4 in the circuit shown in Fig. E5.12 by node analysis. I2

I1 2‡ 10 V

5‡

4‡

.

+ E

I3 5 A 8‡

I4 10 ‡

+ E

5V

Fig. E5.12.

ANSWERS Vb = 30.8772 V

;

Vc = 31.4035 V

Vd = − 0.5263 V ;

Ve = 7.0175 V

;

Vf = 6.4912 V

E5.2

V1 = − 3.3333 V ;

V2 = 0

;

V3 = 3.3333 V

E5.3

IL = − 6.199 A

E5.4

P5A = 30 W (Delivered) ;

E5.5

P10Ω = 40 W

E5.1

Va = 37.8947 V

;

P2A = − 2 W (Absorbed) ;

P3A = 15 W (Delivered)

v

5. 44

Chapter 5 - Node Analysis o

; V2 = 13.9754∠−153.4o V

V1 = 1.5132∠15.8 V

o

; V2 = 1.1186∠−34.2o V

E5.8

P = 45.5555 W

; Q = 91.111 VAR

E5.9

IL = 3.0403∠33 A

E5.10

Va = − 1.9512 V

E5.11

P4A = 80.5624 W (Delivered) ; P1A = –1.8748 W (Absorbed) ; P15V = 10.0485 W (Delivered)

E5.12

I1 = 5 A

E5.6

V1 = 7.8125∠53.1 V

E5.7

; V3 = 0.2925∠106.6o V

o

;

;

Vb = 8.0488 V

I2 = − 2.8572 A

;

;

I3 = 0 ;

Vc = 11.9512 V

I4 = 2.1429 A

;

Vd = − 1.9512V ; Ve = 3.0488V

Chapter 6

NETWORK REDUCTION 6.1

Introduction

(AU May’15, 2 Marks)

A typical network involves series, parallel, star and delta-connections of parameters like resistances, inductances and capacitances. Sometimes it may require to find the single equivalent value of the series/parallel/star-delta-connected parameters of the network. In such a case, the parameters of the network have to be reduced step by step, starting from a dead end. Basically, the network reduction will be attempted with respect to two terminals, and in any network reduction technique, the ratio of voltage and current should be same even after reducing the network. Reducing the series/parallel/star-delta-connected parameters to a single equivalent parameter, conversion of star-connected parameters to equivalent delta and vice versa are explained in this chapter. The network also involves series/parallel connection of sources for higher voltage/current requirement. The series and parallel connections of voltage sources and current sources and their reduction into a single equivalent are also discussed in this chapter.

6.2

Resistances in Series and Parallel

6.2.1 Equivalent of Series-connected Resistances R1 R2 Req = R1 + R2 Consider a circuit with + E E + series combination of two IR1 IR2 resistances R 1 and R 2 connected to dc source of voltage V as I shown in Fig. 6.1(a). Let the I E E + + current through the circuit be I. V V Now the voltage across R 1 and Fig. a : Resistances in series. Fig. b : Equivalent circuit of Fig. a. Fig. 6.1 : Resistances in series. R2 are IR 1 and IR 2 respectively.

By Kirchhoff’s Voltage Law, we can write, V = I R 1 + IR 2 = I (R 1 + R 2 ) Let,

V = I Req

..... (6.1)

where, Req = R 1 + R 2

From equation (6.1), we can say that, the series-connected resistances R 1 and R 2 can be replaced by an equivalent resistance, R eq given by the sum of individual resistances R 1 and R 2 .

6. 2

Chapter 6 - Network Reduction

This concept can be extended to any number of resistances in series. Therefore, we can say that the resistances in series can be replaced by an equivalent resistance whose value is given by sum of individual resistances. “When n number of identical resistances of value R are connected in series, then the equivalent resistance, R eq = nR”. R1

R2

Req = R1 + R2

Þ A R1

A

B R2

R3

B

Req = R1 + R2 + R3

Þ A

B

A R2

R1

R3

Rn

B

Req = R1 + R2 + R3 + .... + Rn

Þ B

A

B

A

Fig. 6.2 : Resistances in series and their equivalents.

6.2.2 Voltage Division in Series-connected Resistances

R1

I

+

R2

E

+

E

V2 V1 Consider two resistances R1 and R2 in series and connected to a dc source of V volts as shown in Fig. 6.3. Let I be the current supplied by the source and V1 and V2 be the voltages across R1 +E and R2 respectively. Since the resistances are in series, the current V Fig. 6.3 : Resistances in series. through them will be I amperes.

The equations (6.2) and (6.3) given below, can be used to determine the voltages in series-connected resistances in terms of total voltage across the series combination and the value of individual resistances. Hence, these equations are called voltage division rule. V1 = V #

R1 R1 + R2

..... (6.2)

V2 = V #

R2 R1 + R2

..... (6.3)

The following equation will be helpful to remember the voltage division rule. In two series-connected resistances, Total voltage across # Value of the resistance series combination Voltage across one of the resistance = Sum of the individual resistance

Circuit Theory

6. 3

Proof for Voltage Division Rule With reference to Fig. 6.3, by Ohm’s law, we can write, V1 = IR 1

..... (6.4)

V2 = IR2

..... (6.5)

By Kirchhoff’s Voltage Law, we get, V = V1 + V2

..... (6.6)

On substituting for V1 and V2 from equations (6.4) and (6.5) in equation (6.6), we get, V = IR 1 + IR 2 = I (R 1 + R 2 ) ` I =

..... (6.7)

V R1 + R2

On substituting for I from equation (6.7) in equation (6.4), we get, V1 =

R1 V # R1 = V # R1 + R2 R1 + R2

On substituting for I from equation (6.7) in equation (6.5), we get, V2 =

R2 V # R2 = V # R1 + R2 R1 + R2

6.2.3 Equivalent of Parallel-connected Resistances Consider a circuit with two resistances in parallel and connected to a dc source of voltage V as shown in Fig. 6.4(a). Let I be the current supplied by the source and I 1 and I2 be the current through R1 and R2 respectively. Since the resistances are parallel to the source, the voltage across them will be same. I V

..... (6.8)

I2 = V R2

..... (6.9)

V

R1

V

R2

E

Fig. a : Resistances in parallel. I

+

I = I 1 + I2 V

Using equations (6.8) and (6.9)

= Vc 1 + 1 m R1 R2 1 1 + 1 R1 R2

+ E

E

By Kirchhoff’s Current Law, we can write,

` V = I

+

+

I1 = V R1

= V + V R1 R2

I2

I1

By Ohm’s law, we can write,

+ E

V

E

Req Req a

R R 1 a 1 2 1 1 R1 C R2 C R1 R2

Fig. b : Equivalent circuit of Fig. a. Fig. 6.4 : Resistances in parallel.

6. 4

Chapter 6 - Network Reduction Let, V = I R eq

..... (6.10)

where, Req =

R1 R2 1 = 1 + 1 R1 + R2 R1 R2

..... (6.11)

Also, 1 = 1 + 1 R eq R1 R2

..... (6.12)

From equation (6.10), we can say that the parallel-connected resistances R 1 and R 2 can be replaced by an equivalent resistance given by equation (6.11). From equation (6.12), we can say that, the inverse of the equivalent resistance of parallel-connected resistances is equal to sum of the inverse of individual resistances. This concept can be extended to any number of resistances in parallel. Therefore, we can say that the resistances in parallel can be replaced by an equivalent resistance whose value is given by the inverse of sum of the inverse of individual resistances. “When n number of identical resistances of value R are connected in parallel, the equivalent resistance, R eq = R/n”. A

A R1

R2

Þ

B

Req =

B

A

R R 1 = 1 2 1 1 R1 + R2 + R1 R2

A R1

R2

R3

Þ

B

Req =

1 1 1 1 + + R1 R2 R3

B

A A R1 B

R2

R3

Rn

Req =

Þ

1 1 1 1 1 + + + ..... + R1 R2 R3 Rn

B

Fig. 6.5 : Resistances in parallel and their equivalents.

6.2.4 Current Division in Parallel-connected Resistances I

Consider two resistances R 1 and R2 in parallel and connected to a dc source of V volts as shown in Fig. 6.6. Let I be the current supplied by the source and I1 and I2 be the current through R1 and R 2 respectively. Since the resistances are parallel to the source, the voltage across them will be V volts.

I2

I1

+ V

+ E

V

E

+ R1

V

R2

E

Fig. 6.6 : Resistances in parallel.

Circuit Theory

6. 5

The equations (6.13) and (6.14) given below can be used to determine the currents in parallel-connected resistances in terms of total current drawn by the parallel combination and the values of individual resistances. Hence, these equations are called current division rule. I1 = I #

R2 R1 + R2

I2 = I #

R1 R1 + R2

..... (6.13) ..... (6.14)

The following equation will be helpful to remember the current division rule. In two parallel-connected resistances, Total current drawn by # Value of the other resistance parallel combination Current through one of the resistance = Sum of the individual resistance Proof for Current Division Rule With reference to Fig. 6.6, by Ohm’s law we can write, I1 = V R1

..... (6.15)

I2 = V R2

..... (6.16)

By Kirchhoff’s Current Law, we get, I = I1 + I2 R + R1 = V + V = Vc 1 + 1 m = Vc 2 m R1 R2 R1 R2 R1 R2 ` V = I#

R1 R2 R1 + R2

Using equations (6.15) and (6.16)

..... (6.17)

On substituting for V from equation (6.17) in equation (6.15), we get, I1 = I #

R1 R2 R2 # 1 = I # R1 + R2 R1 R1 + R2

On substituting for V from equation (6.17) in equation (6.16), we get, I2 = I #

6.3

R1 R2 R1 # 1 = I # R1 + R2 R2 R1 + R2

Resistances in Star and Delta

The star-connected resistances can be converted to an equivalent delta-connected resistances and vice versa. The conversion is valid if the ratio of voltage to current at any two terminals of the equivalent network is same as that in the original network. This means that the looking back resistance at any two terminals of the original network is same as that of equivalent network.

6. 6

Chapter 6 - Network Reduction

6.3.1 Delta to Star Transformation Consider three delta-connected resistances R 12, R23 and R31 as shown in Fig. 6.7(a). These resistances can be converted to an equivalent star-connected resistances R1, R2 and R3 of Fig. 6.7(b). The equations used to determine the star equivalent of delta-connected resistances are given below: R1 =

R12 R31 R12 + R23 + R31

R2 =

R12 R23 R12 + R23 + R31

R3 =

R23 R31 R12 + R23 + R31

From the above equations, we can say that when the three resistances in delta are equal with value R, then their equivalent star-connected resistances will consists of three equal resistances of value R/3. 1

1 R31

R1

R12

Þ R3

R2 2

2 R23 3

3

Fig. a : Delta-connected resistances.

Fig. b : Star-connected resistances.

Fig. 6.7 : Delta to star transformation.

The following equation will be helpful to remember the delta to star conversion equations. The product of resistances connected to the terminal in delta network Star equivalent resistance at one terminal = Sum of three resistances in delta network

6.3.2 Star to Delta Transformation Consider three star-connected resistances R 1, R2 and R3 as shown in Fig. 6.8(a). These resistances can be converted to an equivalent delta-connected resistances R12, R23 and R31 of Fig. 6.8(b). 1

1 R1 R3

R31

R12

Þ

R2

2

2 R23

3

3

Fig. a : Star connected resistances.

Fig. b : Delta connected resistances.

Fig. 6.8 : Star to delta transformation.

Circuit Theory

6. 7

The equations used to determine the delta equivalent of star-connected resistances are given below: R12 = R1 + R2 +

R1 R2 R3

R23 = R2 + R3 + R2 R3 R1 R31 = R3 + R1 + R3 R1 R2

From the above equations we can say that, when the three resistances in star are equal with value R, then their equivalent star-connected resistances will consist of three equal resistances of value 3R. The following equation will be helpful to remember the star to delta conversion equations. Sum of resistances Delta equivalent resistance between two terminals = connected to the # two terminals in star network

Product of the resistances connected to the two terminals in star network The third resistance in star network

6.4 Source Transformation “The practical voltage source can be converted to an equivalent practical current source and vice versa, with same terminal behaviour”. In these conversions, the current and voltage at the terminal of the equivalent source will be same as that of original source, so that the power delivered to a load connected at the terminals of original and equivalent source will be same. The voltage source with series resistance can be converted to an equivalent current source with parallel resistance as shown in Fig. 6.9. Similarly, the current source with parallel resistance can be converted to an equivalent voltage source with series resistance as shown in Fig. 6.10. Rs

+ E

+ -

IRs

A

A

-

+

I

V

RL

B

Fig. a : Voltage source.

Ish

Þ

Is

Is = E / Rs

Rs

+

I

V

RL

B

Fig. b : Equivalent current source of the voltage source in Fig. a. Fig. 6.9 : Conversion of voltage source to current source.

6. 8

Chapter 6 - Network Reduction A V Rs

Is

Rs

Rs

+ V

I

RL

Þ

B

E

+ -

A

-

+

I

E = Is Rs

V

RL

+

IRs

B

Fig. b : Equivalent voltage source of the current source in Fig. a. Fig. 6.10 : Conversion of current source to voltage source.

Fig. a : Current source.

Proof for Conversion of Voltage Source to Current Source Consider a voltage source with source resistance R s , delivering a current I to a load resistance R L as shown in Fig. 6.9(a). In Fig. 6.9(a), using KVL, we can write, E = IR s + V

..... (6.18)

On dividing the equation (6.18) throughout by R s, we get, E = I+ V Rs Rs Let, E = Is and V = Ish Rs Rs

..... (6.19) ..... (6.20)

From equations (6.19) and (6.20), we can write, Is = I + I s h

..... (6.21)

The equation (6.21) represent a current source with generated current, Is and delivering a load current, I to the load resistance, RL. The Ish is the current drawn by a parallel resistance of value Rs connected across the current source. Hence the equation (6.21), can be used to construct an equivalent current source as shown in Fig. 6.9(b). It can be observed that, current source of Fig. 6.9(b) is equivalent to voltage source of Fig. 6.9(a) with respect to terminals A-B.

Proof for Conversion of Current Source to Voltage Source Consider a current source with source resistance R s, delivering a current I to a load resistance, RL as shown in Fig. 6.10(a). In Fig. 6.10(a), using KCL, we can write, Is = V + I Rs

.....(6.22)

On multiplying equation (6.22) by R s, we get, IsRs = V + I Rs Let,

Is Rs = E

..... (6.23) ..... (6.24)

From equations (6.23) and (6.24), we can write, E = V + I Rs

..... (6.25)

The equation (6.25) represent a voltage source with generated voltage, E and delivering a load current, I to the load resistance, R L . The IR s is the voltage across a series resistance of value Rs. Hence the equation (6.25), can be used to construct an equivalent voltage source as shown in Fig. 6.10(b). It can be observed that the voltage source of Fig. 6.10(b) is equivalent to current source of Fig. 6.10(a) with respect to terminals A-B.

Circuit Theory

6.5

6. 9

Voltage Sources in Series and Parallel

A voltage source is designed to deliver energy at a constant voltage called rated voltage. The current delivered by the voltage source depends on the load and the current is limited by the power rating of the source. When the voltage requirement of a load is higher, then the voltage sources are connectedin series. When the current requirement of a load is higher, then the voltage sources are connectedin parallel. 6.5.1 Series Connection of Voltage Sources In series connection, voltage sources of different voltage rating may be connected in series. But in series connection, the current delivered by the voltage sources is same. Case i : Series connection of ideal voltage sources The series connection of ideal voltage sources is shown in Fig. 6.11(a). By applying KVL, to series-connected sources of Fig. 6.11(a) it is possible to determine the single equivalent source as shown in Fig. 6.11(b). I

I +

A

- +

E1

-+ -

E2

E3

+

-

En

Þ

B

+

A

-

B

Eeq

Eeq = E1 + E2 + E3 +.....+ En

Fig. a : Series connection.

Fig. b : Equivalent voltage source.

Fig. 6.11 : Series connection of ideal voltage sources.

Case ii : Series connection of voltage sources with internal resistance The series-connected voltage sources with internal resistance shown in Fig. 6.12(a) can be represented as ideal sources with a series resistance as shown in Fig. 6.12(b). The series-connected resistance of Fig. 6.12(c) can be represented by an equivalent resistance as shown in Fig. 6.12(d). The series-connected voltage sources of Fig. 6.12(c) can be represented by an equivalent source as shown in Fig. 6.12(d).

A

+

E

E1, R1

+

E

E2, R2

+

E

E3, R3

+

E

En, Rn

B

⇒

A

+

R1

E

E1

R2

+

E

E2

R3

+

E

E3

Rn

+

EB

En

Fig. b.

Fig. a : Series connection.

⇓ ⇐

A

Req

+

E Eeq B

Fig. d : Equivalent source.

A

R1

R2

R3

Rn

+

E + E + E

E1

E2

E3

+

E En B

Fig. c. Eeq = E1 + E2 + E3 + ..... + En Req = R1 + R2 + R3 + ..... + Rn

Fig. 6.12 : Series connection of voltage sources with internal resistance.

6. 10

Chapter 6 - Network Reduction

6.5.2 Parallel Connection of Voltage Sources Practically, voltage sources of identical voltage rating should be connected in parallel, but the current delivered by the parallel-connected sources may be different. If voltage sources with different voltage rating are connected in parallel, then current will circulate within the sources which produce excess heat and this in-turn may damage the source. Case i : Parallel connection of ideal voltage sources A

A

I1

I2

I3

In

Ieq

+

+

+

+

+

E

E

E

-

-

Þ

E

E

-

-

Ieq = I1 + I2 + I3 + .... + In

B

B

Fig. a : Parallel connection.

Fig. b : Equivalent source.

Fig. 6.13 : Parallel connection of ideal voltage sources.

The parallel-connected ideal voltage sources with same voltage rating shown in Fig. 6.13(a) can be converted to a single equivalent source as shown in Fig. 6.13(b), by using KCL. Note : The parallel connection of ideal voltage sources with different voltage rating is illegal. Case ii : Parallel connection of voltage sources with internal resistance Ieq

Ieq I1

+

+

+ E1, R1

I3

I2

E3, R3

E2, R2

E

E

A

In

+

+

Eeq

En, Rn

I1

Rn

+

E2

-

In

R3

+

+

E1

E

I3

R2

R1

Þ

E

E

I2

Eeq

+

En

E3

-

-

-

A

+

B

B

Fig. a : Parallel connection.

Fig. b.

ß A

E1 R1

R1

E2 R2

R2

E3 R3

En Rn

R3

Rn

B

Fig. c.

ß

Ieq =

E1 R1

R eq =

+

E2 R2

+

E3 R3

+ ...... +

1 1 1 1 1 + + + ...... + R1 R 2 R 3 Rn

A

A

En Rn

Eeq = Ieq ´ R eq

Req Ieq

Þ

Req

= R eq

+

Eeq B

Fig. d.

LM E1 + E2 + E3 + ..... + En OP Rn P MN R1 R2 R3 Q

B

Fig. e.

Fig. 6.14 : Parallel connection of voltage sources with internal resistance.

Circuit Theory

6. 11

The parallel-connected voltage sources with internal resistance shown in Fig. 6.14(a) can be represented as ideal sources with a series resistance as shown in Fig. 6.14(b). Using source transformation technique, the voltage sources can be converted to current sources as shown in Fig. 6.14(c). Now, the parallel-connected current sources can be combined as an equivalent single current source as shown in Fig. 6.14(d). Finally, again using source transformation technique, the current source of Fig. 6.14(d) can be converted to equivalent voltage source as shown in Fig. 6.14(e). Note : The conversion of parallel-connected voltage sources to a single equivalent voltage source can also be obtained by Millman’s Theorem.

6.6 Current Sources in Series and Parallel A current source is designed to deliver energy at a constant current called rated current. The voltage across the current source depends on the load and the voltage is limited by the power rating of the source. When the voltage requirement of a load is higher, then the current sources are connected in series. When the current requirement of a load is higher then current sources are connected in parallel. 6.6.1 Series Connection of Current Sources Practically, current sources of identical current rating should be connected in series, but the voltage across the series-connected sources may be different. If current sources with different current rating are connected in series, then sources with lesser current rating are forced to carry higher current which produce excess heat and this in-turn may damage the source. Case i : Series connection of ideal current sources The series-connected ideal current sources with identical current rating shown in Fig. 6.15(a) can be converted to a single equivalent source as shown in Fig. 6.15(b), by using KVL. I B

+

E1

I

I - +

E2

- +

I -

E3

+

I

I -

En

A

Þ

B

+

I -

Eeq

A

Eeq = E1 + E2 + E3 +.....+ En

Fig. a : Series connection.

Fig. b : Equivalent current source.

Fig. 6.15 : Series connection of ideal current sources.

Note : The series connection of ideal current sources with different current rating is illegal. Case ii : Series connection of current sources with internal resistance The series-connected current sources with internal resistance shown in Fig. 6.16(a) can be represented as ideal sources with a parallel resistance as shown in Fig. 6.16(b).

6. 12

Chapter 6 - Network Reduction

Using source transformation technique, the current sources can be converted to voltage sources as shown in Fig. 6.16(c). Now, the series-connected voltage sources can be combined as an equivalent single voltage source as shown in Fig. 6.16(d). Finally, again using source transformation technique the voltage source of Fig. 6.16(d) can be converted to equivalent current source as shown in Fig. 6.16(e). I2, R2

I1, R1

B

Fig. a.

In, Rn

I3, R3

A

ß

I1

I2

I3

In

R1

R2

R3

Rn

Fig. b.

ß

B

A

A

B

-

+

I1,R1

R1

Fig. c.

R2

-

+

I2,R2

Rn

I3,R3

R3

+

In, Rn

ß

B

A

Req

-

+

B

Þ

+

Ieq

A

Eeq

Fig. d.

Req

Ieq a

R eq a R1 C R 2 C R 3 C......C Rn Eeq a I1 R1 C I2 R 2 C I3 R 3 C......C In Rn

a

Fig. e.

Eeq R eq 1 R eq

I1 R1 C I2 R 2 C I3 R 3 C ...... C In Rn

Fig. 6.16 : Series connection of current sources with internal resistance.

6.6.2 Parallel Connection of Current Sources In parallel connection, current sources with different current rating may be connected in parallel. But in parallel connection the voltage across the sources is same. Case i : Parallel connection of ideal current source The parallel-connected ideal current sources shown in Fig. 6.17(a) can be converted to a single equivalent source as shown in Fig. 6.17(b), by using KCL. A

A

+ I1

I2

I3

In

E

-B

+

Þ

Ieq

E

-B Ieq = I1 + I2 + I3 +.....+ In

Fig. a : Parallel connection.

Fig. b : Equivalent of parallel connection.

Fig. 6.17: Parallel connection of ideal current sources.

Circuit Theory

6. 13

Case ii : Parallel connection of current sources with internal resistance The parallel-connected current sources with internal resistance shown in Fig. 6.18(a) can be represented as ideal sources with a parallel-resistance as shown in Fig. 6.18(b). The parallel-connected resistances of Fig. 6.18(c) can be represented by an equivalent resistance as shown in Fig. 6.18(d). The parallel-connected current sources of Fig. 6.18(c) can be represented by an equivalent current source as shown in Fig. 6.18(d). A I2 , R 2

I1, R1

I3, R3

In, Rn B

Fig. a.

ß A

R1

I1

R2

I2

In

R3

I3

Rn B

Fig. b.

ß A

I2

I1

I3

R1

In

R2

R3

Rn B

Fig. c.

ß A

Ieq

Req

Ieq = I1 + I2 + I3 + ....................... + In R eq =

B

1 1 1 1 1 + + + .......... + R1 R 2 R 3 Rn

Fig. d.

Fig. 6.18 : Parallel connection of current sources with internal resistance.

6.7

Inductances in Series and Parallel

6.7.1 Equivalent of Series-connected Inductances Consider a circuit with series combination of two inductances L 1 and L 2 connected to ac source of voltage v as shown in Fig. 6.19(a). Let, the current through the circuit be i and voltage across L 1 and L 2 be v 1 and v 2 respectively. L1

+

Leq = L1 + L2

L2

+

v1 E

i

+

v2

E

~E v

Fig. a : Inductance in series.

i

+

~E v

Fig. b : Equivalent circuit of Fig. a.

Fig. 6.19 : Inductance in series.

6. 14

Chapter 6 - Network Reduction

By Faraday’s Law, we can write, ν1 = L1 di and ν2 = L2 di dt dt

..... (6.26)

In Fig. 6.19(a), using Kirchhoff’s Voltage Law, we can write, v = v1 + v2 = L1 di + L2 di dt dt

Using equation (6.26)

= ` L1 + L2 j di dt Let,

ν = Leq di dt where, Leq = L1 + L2

..... (6.27)

From equation (6.27), we can say that, the series-connected inductances L 1and L2 can be replaced by an equivalent inductance, Leq given by the sum of individual inductances L 1 and L 2 . This concept can be extended to any number of inductances in series. Therefore, we can say that the inductances in series can be replaced by an equivalent inductance whose value is given by sum of individual inductances. “When n number of identical inductances of value L are connected in series, the equivalent inductance, L eq = nL”. L1

Leq = L1 + L2

L2

Þ A L1

B L2

A

B

Leq = L1 + L2 + L3

L3

Þ A L1

B L2

L3

A

B

Leq = L1 + L2 + L3 +....+ Ln

Ln

Þ A

B

A

B

Fig. 6.20 : Inductances in series and their equivalents.

Circuit Theory

6. 15

6.7.2 Equivalent of Parallel-connected Inductances Consider a circuit with two inductances in parallel and connected to an ac source of voltage v as shown in Fig. 6.21(a). Let, i be the current supplied by the source and i 1 and i2 be the current through L1 and L 2 , respectively. Since the inductances are parallel to the source, the voltage across them will be same. i

i i1

+

+ v

i2

~

v

L2

v

E

E

E

+

+ v

L1

~

E

+ v

E

Leq Leq a

L L 1 a 1 2 1 1 L1 C L 2 C L1 L 2

Fig. b : Equivalent circuit of Fig. a. Fig. a : Inductance in parallel. Fig. 6.21 : Inductance in parallel.

We know that, i1 = 1 L1

# ν dt

i2 = 1 L2

and

# ν dt

.....(6.28)

By Kirchhoff’s Current Law, we can write, i = i 1 + i2

= 1 L1

# ν dt

= e 1 + 1 o L1 L2

+ 1 L2

# ν dt

Using equation (6.28)

# ν dt

On differentiating the above equation, we get, di = 1 + 1 ν ⇒ eL dt L2 o 1 Let,

1 di 1 + 1 dt L1 L2

ν = Leq di dt where, Leq =

Also,

ν =

1 = 1 + 1 Leq L1 L2

.....(6.29)

L1 L2 1 = 1 + 1 L1 + L2 L1 L2

.....(6.30)

.....(6.31)

From equation (6.29), we can say that the parallel-connected inductances L 1 and L 2 can be replaced by an equivalent inductance given by equation (6.30). From equation (6.31), we can say that the inverse of the equivalent inductance of parallel-connected inductances is equal to sum of the inverse of individual inductances.

6. 16

Chapter 6 - Network Reduction A

A L1

L2

Leq =

Þ B

B A

L L 1 = 1 2 1 1 L1 + L 2 + L1 L 2

A L1

L2

L3

Leq =

Þ B

B A

1 1 1 1 + + L1 L 2 L 3

A L1

L2

L3

Ln

Leq =

Þ

B

1 1 1 1 1 + + +...+ L1 L 2 L 3 Ln

B

Fig. 6.22 : Inductances in parallel and their equivalents.

This concept can be extended to any number of inductances in parallel. Therefore, we can say that the inductances in parallel can be replaced by an equivalent inductance whose value is given by the inverse of sum of the inverse of individual inductances. “When n number of identical inductances of value L are connected in parallel, the equivalent inductance, Leq = L ". n

6.8

Capacitances in Series and Parallel

6.8.1 Equivalent of Series-connected Capacitances Consider a circuit with series combination of two capacitances C 1 and C2 connected to ac source of voltage, v as shown in Fig. 6.23(a). Let the current through the circuit be i and voltage across C1 and C2 be v1 and v2 respectively . Ceq

C2

C1

+v E

+

+v E 2

1

E Ceq a

i

i

~

+

+ v E

Fig. a : Capacitances in series.

C1 C2 1 a 1 1 C1 C C2 C C1 C2

~E v

Fig. b : Equivalent circuit of Fig. a.

Fig. 6.23 : Capacitances in series.

We know that, ν1 = 1 C1

# i dt

and

ν2 = 1 C2

# i dt

With reference to Fig. 6.23(a) using Kirchhoff’s Voltage Law, we can write, v = v 1 + v2

.....(6.32)

Circuit Theory

6. 17

# i dt

v = 1 C1

+ 1 C2

= e 1 + 1 o C1 C2 Let,

ν =

1 Ceq

Using equation (6.32)

# i dt

# i dt

where, C eq =

Also,

# i dt

.....(6.33)

C1 C2 1 = 1 + 1 C1 + C2 C1 C2

.....(6.34)

1 = 1 + 1 C eq C1 C2

.....(6.35)

From equation (6.33), we can say that the series-connected capacitances C 1 and C2 can be replaced by an equivalent capacitance given by equation (6.34). From equation (6.35), we can say that, the inverse of the equivalent capacitance of series-connected capacitances is equal to sum of the inverse of individual capacitances. This concept can be extended to any number of capacitances in series. Therefore, we can say that the capacitances in series can be replaced by an equivalent capacitance whose value is given by the inverse of the sum of the inverse of the individual capacitances. “When n number of identical capacitances of value C are connected in series, the equivalent capacitance, C eq = C/n”. C1

Ceq

C2

Ceq =

Þ B

A C1

C2

B

A

C3

Ceq

Þ B

A C1

C2

C3

Ceq =

B

A

1 1 1 1 + + C1 C2 C3

Ceq

Cn

Ceq =

Þ A

C C 1 = 1 2 1 1 C1 + C2 + C1 C2

B

A

B

1 1 1 1 1 + + +...+ C1 C2 C3 Cn

Fig. 6.24 : Capacitances in series and their equivalents.

6.8.2 Equivalent of Parallel-connected Capacitances Consider a circuit with two capacitances in parallel and connected to an ac source of voltage, v as shown in Fig. 6.25. Let, i be the current supplied by the source and i1 and i2 be the current through C1 and C2 respectively. Since the capacitances are parallel to the source the voltage across them will be same.

6. 18

Chapter 6 - Network Reduction i

i i1

+ v

i2

+

+

~ E

v

C1

+ C2

v

v

E

E

~ E

+ v

Ceq = C1 + C2

E

Fig. b : Equivalent circuit of Fig. a. Fig. a : Capacitances in parallel. Fig. 6.25 : Capacitances in parallel.

We know that, i1 = C1 dν dt

i2 = C2 dν dt

and

.....(6.36)

By Kirchhoff’s Current Law, we can write, i = i 1 + i2

Using equation (6.36)

= C1 dν + C2 dν dt dt = ` C1 + C2 j dν dt Let, i = Ceq dν dt

.....(6.37)

where, Ceq = C 1 + C2 From equation (6.37), we can say that, the parallel-connected capacitances C 1 and C2 can be replaced by an equivalent capacitance, Ceq given by the sum of individual capacitances C1 and C2 . This concept can be extended to any number of capacitances in parallel. Therefore, we can say that the capacitance in parallel can be replaced by an equivalent capacitance whose value is given by sum of individual capacitances. “When n number of identical capacitances of value C are connected in parallel, the equivalent capacitance, C eq = nC”. A

A C1

C2

B

B

A

A C1

C2

C3

Þ

Ceq = C1 + C2 + C3

B

B

A

A C1 B

Ceq = C1 + C2

Þ

C2

C3

Cn

Ceq = C1 + C2 + C3 +....+ Cn

Þ B

Fig. 6.26 : Capacitances in parallel and their equivalents.

Circuit Theory

6.9

6. 19

Impedances in Series and Parallel

6.9.1 Equivalent of Series-connected Impedances Consider a circuit with series combination of two impedances Z1 and Z2 connected to ac source of voltage V volts rms value as shown in Fig. 6.27. Let the current through the elements be I . Now, by Ohm’s law the voltage across Z1 and Z2 are I Z1 and I Z2 respectively. I+

I Z1

I Z2

+

E Z1

I

E

Z2

+

I Zeq

+

E

Zeq a Z1 C Z2

~E

+

V

~E V

Fig. a : Impedances in series. Fig. b : Equivalent circuit of Fig. a. Fig. 6.27 : Impedances in series and their equivalent.

By Kirchhoff’s Voltage Law, we can write, V = I Z1 + I Z2 = I ` Z1 + Z2 j Let, V = I Zeq

.....(6.38)

where,

Zeq = Z1 + Z2

From equation (6.38), we can say that, the series-connected impedances Z1 and Z2 can be replaced by an equivalent impedance Zeq given by the sum of individual impedances Z1 and Z2 . This concept can be extended to any number of impedances in series. Therefore, we can say that the impedances in series can be replaced by an equivalent impedance whose value is given by sum of individual impedances. “When n number of identical impedances of value Z are connected in series, then the equivalent impedance, Z eq = nZ". Z1

Z2

Zeq = Z1 + Z2

Þ B

A Z1

Z2

A

Z3

B

Zeq = Z1 + Z2 + Z3

Þ B

A Z1

Z2

Z3

A

B

Zeq = Z1 + Z2 + Z3 +.....+ Zn

Zn

Þ A

B

A

B

Fig. 6.28 : Impedances in series and their equivalents.

6. 20

Chapter 6 - Network Reduction

6.9.2 Voltage Division in Series-connected Impedances Consider two impedances Z1 and Z2 in series and connected to ac source of V volts as shown in Fig. 6.29. Let, I be the current supplied by the source and V1 and V2 be the voltages across Z1 and Z2 respectively. Since the impedances are in series, the current through them will be I amperes. The equations (6.39) and (6.40), can be used to solve the voltages in series-connected impedances in terms of total voltage across the series combination Z1 Z2 and the value of individual impedances. Hence, these equations E E + + are called voltage division rule. V1 V2 Z1 Z1 + Z2

.....(6.39)

V2 = V # Z2 Z1 + Z2

.....(6.40)

V1 = V #

I

~E

+

V

Fig. 6.29 : Impedances in series.

The following equation will be helpful to remember the voltage division rule. In two series-connected impedances, Total voltage across Value of the series combination # impedance Voltage across one of the impedance = Sum of the individual impedances

6.9.3 Equivalent of Parallel-connected Impedances Consider a circuit with two impedances in parallel and connected to an ac source of V volts rms value as shown in Fig. 6.30. Let, I be the current supplied by the source, I1 and I2 be the current through Z1 and Z2 respectively. Since the impedances are parallel to the I source the voltage across them will be same as that of source I1 I2 + + voltage. +

~

V

By Ohm’s law, we can write,

V

Z1

I1 = V and I2 = V Z1 Z2

.....(6.41)

Fig. a : Impedances in parallel.

By Kirchhoff’s Current Law, we can write, I = I1 + I2

I

Using equation (6.41)

= Ve 1 + 1 o Z1 Z2 V = I

f

+

+ V

`

1 1 + 1 Z1 Z2

Z2

E

E

= V + V Z1 Z2

V

E

~

V

E

Zeq a

1 1 Z1

p

Zeq

E

C

1 Z2

a

Z1 Z2 Z1 C Z2

Fig. b : Equivalent circuit of Fig. a. Fig. 6.30 : Impedances in parallel and their equivalent.

Circuit Theory Let,

6. 21

V = I Zeq

.....(6.42)

1 = Z1 Z2 1 + 1 Z1 + Z2 Z1 Z2

where, Zeq =

Also,

..... (6.43)

1 = 1 + 1 Zeq Z1 Z2

.....(6.44)

From equation (6.42), we can say that, the parallel impedance Z1 and Z2 can be replaced by an equivalent impedance given by equation (6.43). From equation (6.44), we can say that, the inverse of the equivalent impedance of parallel-connected impedances is equal to sum of the inverse of individual impedances. This concept can be extended to any number of impedances in parallel. Therefore, we can say that, the impedances in parallel can be replaced by an equivalent impedance whose value is given by the inverse of sum of the inverse of individual impedances. “When n number of identical impedances of value Z are connected in parallel, the equivalent impedance, Z eq = Z /n". A

A Z1

Z2

Þ

A

1 1

+

Z1

B

B

1

=

Z2

Z1 Z2 Z1 + Z2

A Z1

Z2

Z3

Þ

B

Zeq =

1

+

Z1

B

A

1 1

+

Z2

1 Z3

A Z1

B

Zeq =

Z2

Z3

Zn

Zeq =

Þ B

1 Z1

+

1

+

Z2

1 1 Z3

+...+

1 Zn

Fig. 6.31 : Impedances in parallel and their equivalents.

6.9.4 Current Division in Parallel-connected Impedances Consider two impedances Z1 and Z2 in parallel and connected to an ac source of V volts as shown in Fig. 6.32. Let I be the current supplied by the source and I1 and I2 be the current through Z1 and Z2 respectively. Since the impedances are parallel to the source, the voltage across them will be V volts.

6. 22

Chapter 6 - Network Reduction

The equations (6.45) and (6.46), can be used to solve the currents in parallel-connected impedances in terms of total current drawn by the parallel combination and the value of individual impedances. Hence these equations are called current division rule. I

I1 = I # Z 2 Z1 + Z 2

+

+ V

I2 = I #

~

V

+ Z1

V

E

E

Z1 Z1 + Z 2

I2

I1

.....(6.45)

Z2

E

.....(6.46)

Fig. 6.32 : Impedances in parallel.

The following equation will be helpful to remember the current division rule. In two parallel-connected impedances, Total current drawn by Value of the # parallel combination other impedance Current through one of the impedance = Sum of the individual impedances

6.9.5 Reactances in Series and Parallel The reactances in series and parallel combinations can be combined to give an equivalent reactance similar to that of impedance. In fact, the reactances are impedances with imaginary part alone. The equivalent reactances of series and parallel combinations of reactances are diagrammatically illustrated in Fig. 6.33.

c

jX2

jX1

j X eq = j X1 + X 2

h

Þ A

B

jX1

jX2

A

B

c

jXn

j X eq = j X1 + X 2 +......+ Xn

h

Þ B

A

A

A

A jX1

jX2

Þ

B

jX eq = j

F I GG 1 JJ = j X1 X2 GG 1 + 1 JJ X1 + X2 H X1 X2 K

j Xeq = j

F I GG J 1 GG 1 + 1 +...+ 1 JJJ H X1 X2 Xn K

B

A

A

jX1

B

B

jX2

jXn

Þ B

Fig. 6.33 : Series and parallel combinations of reactances and their equivalents.

Circuit Theory

6. 23

6.10 Impedances in Star and Delta The star-connected impedances can be converted to an equivalent delta-connected impedances and vice versa. The conversion is valid if the ratio of voltage to current at any two terminals of the equivalent network is same as that in the original network. This means that the looking back impedance at any two terminals of the original network is same as that of equivalent network. 6.10.1 Delta to Star Transformation Consider three delta-connected impedances Z12, Z23 and Z31 as shown in Fig. 6.34(a). These impedances can be converted to an equivalent star-connected impedances Z1, Z2 and Z3 of Fig. 6.34(b), using the equations given below: Z1 =

Z12 Z31 Z12 + Z23 + Z31

;

Z2 =

Z12 Z23 Z12 + Z23 + Z31

Z3 =

;

1

1 Z31

Z23 Z31 Z12 + Z23 + Z31

Z1

Z12

Þ

Z3

Z2

2

2 Z23

3

3

Fig. a : Delta-connected impedances.

Fig. b : Star-connected impedances.

Fig. 6.34 : Delta to star transformation.

6.10.2 Star to Delta Transformation Consider three star-connected impedances Z1, Z2 and Z3 as shown in Fig. 6.35(a). These impedances can be converted to an equivalent delta-connected impedances Z12, Z23 and Z31 of Fig. 6.35(b), using the equations given below: Z12 = Z1 + Z2 + Z1 Z2 Z3

;

Z23 = Z2 + Z3 + Z2 Z3 Z1

;

Z31 = Z3 + Z1 + Z3 Z1 Z2

1 1 Z1 Z3

Z31

Z12

Þ

Z2

2

2 Z23

3

3

Fig. b : Delta-connected impedances. Fig. a : Star-connected impedances. Fig. 6.35 : Star to delta transformation.

6. 24

Chapter 6 - Network Reduction

6.11 Conductances in Series and Parallel

+ I

6.11.1 Equivalent of Series-connected Conductances

V1

+

E

V2

E G2

G1

Consider a circuit with series combination of two conductances G 1 and G 2 connected to dc source of voltage, V volts +E V as shown in Fig. 6.36. Let, I be the current through the conductances Fig. a : Conductances in series. and V1 and V2 be the voltages across G1 and G 2 respectively. From Fig. 6.36, we can write,

+

I and V = I 2 G1 G2

..... (6.47)

By Kirchhoff’s Voltage Law, we can write, V = V1 + V2 =

Using equation (6.47)

I = V

f

1 1 + 1 G1 G2

Also,

1 1 C G1 G2

a

G1 G2 G1 C G2

Fig. b : Equivalent circuit of Fig. a.

p

Let, I = VGeq where,

V 1

Fig. 6.36 : Conductances in series and their equivalent.

= Ie 1 + 1 o G1 G2 `

+E

Geq a

I + I G1 G2

E

Geq

I

V1 =

V

..... (6.48)

Geq =

1 = 1 + 1 Geq G1 G2

G1 G2 1 = 1 + 1 G1 + G2 G1 G2

..... (6.49) .....(6.50)

From equation (6.48), we can say that the series-connected conductances G 1 and G 2 can be replaced by an equivalent conductance given by equation (6.49). From equation (6.50), we can say that, the inverse of the equivalent conductance of series-connected conductances is equal to sum of the inverse of individual conductances. This concept can be extended to any number of conductances in series. Therefore, we can say that the conductances in series can be replaced by an equivalent conductance whose value is given by the inverse of sum of the inverse of individual conductances. “When n number of identical conductances of value G are connected in series, the equivalent conductance, Geq = G/n”.

Circuit Theory

6. 25 G1

Geq

G2

Þ

Geq =

B

A G2

G1

A

G3

B Geq Geq =

Þ A

A

B Gn

G3

G2

G1

G1 G2 1 = 1 1 G 1 + G2 + G1 G2

B Geq Geq =

Þ A

1 1 1 1 + + G1 G2 G3

A

B

B

1 1 1 1 1 + + + ... + G1 G2 G3 Gn

Fig. 6.37 : Conductances in series and their equivalents.

6.11.2 Equivalent of Parallel-connected Conductances Consider a circuit with two conductances in parallel and connected to a source of voltage V volts as shown in Fig. 6.38. Let, I be the current supplied by the source and I1 and I 2 be the current through G 1 and G 2 respectively. Since the conductances are parallel to the source, the voltage across them will be same as that of source voltage. I

I I2

I1 V

+ E

+

+ V

G1

E

V

G2

E

V

+ E

+ V

Geq = G1 + G2

E

Fig. a : Conductances in parallel. Fig. b : Equivalent circuit of Fig. a. Fig. 6.38 : Conductances in parallel and their equivalent.

From Fig. 6.38, we can write, I 1 = VG1

and

I2 = VG2

..... (6.51)

By Kirchhoff’s Current Law, we can write, I = I 1 + I2 = VG 1 + VG 2

Using equation (6.51)

= V(G 1 + G 2 )

Let,

I = VG eq

..... (6.52)

where, G eq = G 1 + G 2 From equation (6.52), we can say that, the parallel-connected conductances G 1 and G2 can be replaced by an equivalent conductance Geq given by the sum of individual conductances G 1 and G 2 .

6. 26

Chapter 6 - Network Reduction

This concept can be extended to any number of conductances in parallel. Therefore, we can say that, the conductances in parallel can be replaced by an equivalent conductance whose value is given by sum of individual conductances. “When n number of identical conductance of value G are connected in parallel, the equivalent conductance, G eq = nG”. A

A G1

G2

Geq = G1 + G2

Þ

B

B

A

A G1

G2

G3

Geq = G1 + G2 + G3

Þ

B

B A

A G1

G2

G3

Geq = G1 + G2 + G3 +....+ Gn

Þ

Gn

B

B

Fig. 6.39 : Conductances in parallel and their equivalents.

6.12 Admittances in Series and Parallel 6.12.1 Equivalent of Series-connected Admittances Consider a circuit with series combination of two admittances Y1 and Y2 connected to ac source of voltage V volts rms value, as shown in Fig. 6.40. Let I be the current through the admittances and V1 and V2 be the voltages across Y1 and Y2 respectively. V V I +

From Fig. 6.40, we can write, V1 =

1

I and V2 = I Y1 Y2

2

E Y2

Y1

.....(6.53)

By Kirchhoff’s Voltage Law, we can write,

+

~E V

Fig. a : Admittances in series.

V = V1 + V2 =

+

E

I

I + I Y1 Y2

Using equation (6.53)

+

V

E Yeq

= Ie 1 + 1 o Y1 Y2 `

I = V

f

1 1 + 1 Y1 Y2

p

+

~E V

Fig. b : Equivalent circuit of Fig. a. Fig. 6.40 : Admittances in series and their equivalent.

Circuit Theory

Let,

6. 27 .....(6.54)

I = V Yeq

where,

1 = Y1 Y2 1 + 1 Y1 + Y2 Y1 Y2

Yeq =

..... (6.55)

1 = 1 + 1 Yeq Y1 Y2

Also,

..... (6.56)

From equation (6.54), we can say that, the series-connected admittances Y1 and Y2 can be replaced by an equivalent admittance given by equation (6.55). From equation (6.56), we can say that, the inverse of the equivalent admittance of series-connected admittances is equal to sum of the inverse of individual admittances. This concept can be extended to any number of admittances in series. Therefore, we can say that, the admittances in series can be replaced by an equivalent admittance whose value is given by the inverse of sum of inverse of individual admittances. “When n number of identical admittances of value Y are connected in series, the equivalent admittance, Y eq = Y /n ”. Y1

Yeq

Y2

Þ

Yeq =

B

A Y1

Y2

A

Yeq =

B Y3

Y2

1

=

Y1 Y2 Y1 + Y2

Y2

Yeq

Y3

Þ Y1

+

Y1

B

A

1 1

+

Y1

B

A

Yn

1

1 1

+

Y2

1 Y3

Yeq

Yeq =

Þ B

A

B

A

1

+

Y1

1

+

Y2

1 1 Y3

+ ... +

1 Yn

Fig. 6.41 : Series combination of admittances and their equivalents.

6.12.2 Equivalent of Parallel-connected Admittances Consider a circuit with two admittances in parallel and connected to ac source of voltage V volts rms value as shown in Fig. 6.42. Let I be the current supplied by the source and I1 and I2 be the current through Y1 and Y2 respectively. Since the admittances are parallel to the source, the voltage across them will be same as that of source voltage. I

I

+ V

~

E

+ V

E

I1 Y1

+

I2

+ Y2

V

E

Fig. a : Admittances in parallel.

V

~

E

+ V

Yeq a Y1 C Y2

E

Fig. b : Equivalent circuit of Fig. a.

Fig. 6.42 : Admittances in parallel and their equivalent.

6. 28

Chapter 6 - Network Reduction

From Fig. 6.42, we can write, I1 = V Y1 and I2 = V Y2

.....(6.57)

By Kirchhoff’s Current Law, we can write, I = I1 + I2 = V Y1 + V Y2

Using equation (6.57)

= V ` Y1 + Y2 j I = V Yeq

Let,

..... (6.58)

where, Yeq = Y1 + Y2 From equation (6.58), we can say that, the parallel-connected admittances Y1 and Y2 can be replaced by an equivalent admittance Yeq given by the sum of individual admittances Y1 and Y2 .

A

A Y1

Y2

Þ B

B A

A Y1

Y2

Y3

Yeq = Y1 + Y2 + Y3

Þ

B

B

A

A Y1

B

Yeq = Y1 + Y2

Y2

Y3

Yn

Þ

Yeq = Y1 + Y2 + Y3 + ... + Yn

B

Fig. 6.43 : Admittances in parallel and their equivalents.

This concept can be extended to any number of admittances in parallel. Therefore, we can say that, the admittances in parallel can be replaced by an equivalent admittance whose value is given by sum of individual admittance.

Circuit Theory

6. 29

6.12.3 Susceptances in Series and Parallel The susceptances in series and parallel combination can be combined to give an equivalent susceptance similar to that of admittance. In fact, the susceptances are admittances with imaginary part alone. The equivalent susceptances of series and parallel combinations of susceptances are diagrammatically illustrated in Fig. 6.44. jB1

jB2

jBeq jBeq = j

Þ A

B jB2

jB1

A

B jBeq

jBn

jBeq = j

Þ A

B

A A

B

1 1 1 1 + +...+ B1 B2 Bn

A jB1

jB2

jBeq = j(B1 + B2)

Þ

B

B

A

A jB1

B

B B 1 =j 1 2 1 1 B1 + B2 + B1 B2

jB2

jBn

jBeq = j(B1 + B2 +...+ Bn)

Þ B

Fig. 6.44 : Series and parallel combinations of susceptances and their equivalents.

6.13 Generalized Concept of Reducing Series/Parallel-connected Parameters In order to generalize the concept of reducing the series/parallel-connected parameters, they can be classified into two groups. Let, the parameters Resistance (R), Inductance (L), Reactance (X) and Impedance (Z) be group-1 parameters. Let, the parameters Conductance (G), Capacitance (C), Susceptance (B) and Admittance (Y) be group-2 parameters. The equivalent of series-connected group-1 parameters will be given by sum of individual parameters. The equivalent of parallel-connected group-1 parameters will be given by inverse of sum of individual inverses. The equivalent of series and parallel-connected group-1 parameters are summarized in Table 6.1. The equivalent of series-connected group-2 parameters will be given by inverse of sum of individual parameters. The equivalent of parallel-connected group-2 parameters will be given by sum of individual parameters. The equivalent of series and parallel-connected group-2 parameters are summarized in Table 6.2.

Z

Impedance

±jX

Reactance

L

Inductance

R

Z1

±jX1

L1

R1

Z2

±jX 2

L2

R2

Zn

±jXn

Ln

Rn

Þ

Þ

Þ

Þ

Zeq = Z1 + Z2 + ..... + Zn

Zeq

± jX eq = ± jX1 ± jX 2 ± ..... ± jX n

±jX eq

Leq = L1 + L2 + ..... + Ln

Leq

Req = R1 + R2 + ..... + Rn

Req

Z1

±jX1

L1

R1

Z2

±jX 2

L2

R2

Zn

±jXn

Ln

Rn

Þ

Þ

Þ

Þ

±jX eq

1 1 1 1 + + ..... + L1 L 2 Ln

Leq

1 1 1 1 + + ..... + R1 R2 Rn

Z1

1

+

Zeq

Z2

1

+ ..... +

1 Zn

1

1 1 1 1 + + ..... + ± jX1 ± jX 2 ± jXn

Zeq =

± jX eq =

L eq =

Req =

Req

TABLE 6.1 : SUMMARY OF EQUIVALENT OF SERIES/PARALLEL-CONNECTED GROUP-1 PARAMETERS Group-1 Parameter Series Connection of Parameters Parallel Connection of Parameters and their equivalent and their equivalent Resistance

6. 30 Chapter 6 - Network Reduction

Y

Admittance

±jB

Susceptance

C

Capacitance

G

Conductance

Group-2 Parameter

Y1

±jB1

C1

G1

Y2

±jB2

C2

G2

Yn

±jBn

Cn

Gn

Þ

Þ

Þ

Þ

1

Y1

1

Yeq

+

1 Y2

+ ..... +

1 Yn

1

1 1 1 + + ..... + ± jB1 ± jB2 ± jBn

1

1 1 1 + + ..... + C1 C2 Cn

Ceq

± jBeq =

Yeq =

1 1 1 1 + + ..... + G1 G2 Gn

±jBeq

Ceq =

Geq =

Geq

Series Connection of Parameters and their equivalent

Y1

±jB1

C1

G1

Y2

±jB2

C2

G2

Yn

±jBn

Cn

Gn

Þ

Þ

Þ

Þ

Yeq = Y1 + Y2 + ..... + Yn

Yeq

± jB eq = ± jB1 ± jB 2 ± ..... ± jBn

±jBeq

Ceq = C1 + C2 + ..... + Cn

Ceq

Geq = G1 + G2 + ..... + Gn

Geq

Parallel Connection of Parameters and their equivalent

TABLE 6.2 : SUMMARY OF EQUIVALENT OF SERIES/PARALLEL-CONNECTED GROUP-2 PARAMETERS

Circuit Theory 6. 31

6. 32

Chapter 6 - Network Reduction 18 W

82 W

6.14 Solved Problems

100 W 60 W

EXAMPLE 6.1

76 W

In the circuit shown in Fig. 1, find the total resistance across the

E

+ -

40 W

supply voltage.

Fig. 1. ß

SOLUTION 100 W

82 + 18 = 100 W

The step-by-step reduction of the given network are shown in Figs. 2 to 5.

Step-1 :

E

76 W

+ -

60 ´ 40 = 24 W 60 + 40

The series-connected 82 W and 18 W resistances in Fig. 1 are combined to

Fig. 2.

single equivalent resistance in Fig. 2. Also the parallel-connected 60 W and 40 W in

ß

Fig. 1 are combined to a single equivalent resistance in Fig. 2.

100 W

100 W

Step-2 : The series-connected 24 W and 76 W resistances in Fig. 2 are combined to a

+ -

E 24 + 76 = 100 W

single equivalent resistance in Fig. 3.

Fig. 3. ß

Step-3 : 100 W

The parallel-connected two 100 W resistances in Fig. 3 are combined to a

100 ´ 100 = 50 W 100 + 100

single equivalent resistance in Fig. 4. + -

E

Step-4 :

Fig. 4. ß

The series-connected 100 W and 50 W resistances in Fig. 4 are combined to a single equivalent resistance in Fig. 5.

100 + 50 = 150 ‡

RESULT

+ E

E

With reference to Fig. 5, we can say that,

Fig. 5. Total resistance across supply = 150 W

Circuit Theory

6. 33 2W

EXAMPLE 6.2 Find the total resistance as seen by the source in the circuit shown in Fig. 1.

A +

2W

V 4W

_ 4W B

Fig. 1.

SOLUTION

1W

3W

ß

The step-by-step reduction of the given circuit are shown in Figs. 2 to 5. A 2W

Step-1 :

3W + V 4W _

4W

The given circuit is redrawn as shown in Fig. 2.

2W

1W

Fig. 2. B

Step-2 :

ß A

The series-connected two 2 W resistances in Fig. 2 are combined to a single equivalent resistance in Fig. 3. Similarly, the 3

+ 2+2 = 4W

W and 1 W in series are converted to a single equivalent reisistance.

4W

_

3+1 = 4W

4W

V

Fig. 3. B

Step-3 :

ß A

The circuit of Fig. 3 is redrawn as shown in Fig. 4. + V

Step-4 :

Fig. 4.

4W

4W

_

B

4W

4W

ß

The parallel-connected four 4 W resistances in Fig. 4, are A

combined to a single equivalent resistance in Fig. 5.

+

RESULT

V

4 a 1‡ 4

•

With reference to Fig. 5, we can say that, B

Total resistance across supply = 1 W

EXAMPLE 6.3

Fig. 5.

(AU Dec’14, 4 Marks)

A 60 ‡

30 ‡

Find the equivalent resistance of the network shown in Fig. 1.

90 ‡

75 ‡

15 ‡

B

Fig. 1.

6. 34

Chapter 6 - Network Reduction

SOLUTION

1

A 30 W

60 W

R1 R2

The step-by-step reduction of the given network are shown in Figs. 2 to 6.

R3

2

3 90 W 75 W

15 W

Step-1 : B

Fig. 2. The delta-connected 30 W, 90 W and 60 W resistances in Fig. 2 are converted

ß

to equivalent star-connected resistances in Fig. 3. The resistances R1, R2 and R3 connected by dotted lines in Fig. 2 are the star equivalent resistances.

1

A

R1 = 10 W

R1 =

30 # 60 = 10 Ω 30 + 90 + 60

R2 =

30 # 90 = 15 Ω 30 + 90 + 60

R3 =

90 # 60 = 30 Ω 30 + 90 + 60

R3 = 30 W

R2 = 15 W 2

3

75 W

15 W

B

Fig. 3.

Step-2 :

ß A

The series-connected 15 W and 75 W resistances in Fig. 3 are combined

10 W

to a single equivalent resistance in Fig. 4. Similarly, the 30 W and 15 W in series are converted to a single equivalent reisistance. 15 + 75 = 90 W

Step-3 :

30 + 15 = 45 W

B

Fig. 4. The parallel-connected 90 W and 45 W resistances in Fig. 4 are combined

ß

to a single equivalent resistance in Fig. 5.

A 10 W

Step-4 : 90 ´ 45 90 + 45 = 30 W

The series-connected 10 W and 30 W resistances in Fig. 5, are combined to a single equivalent resistance in Fig. 6.

Fig. 5.

B

ß A

RESULT

10 + 30 = 40 ‡

With reference to Fig. 6, we can say that, Equivalent resistance across A-B, RAB = 40 W

B

Fig. 6.

Circuit Theory

6. 35 2W

EXAMPLE 6.4 Find the equivalent input resistance across terminals A and B of the bridged-T network shown in Fig. 1.

1W

3W

A 2W

SOLUTION

1W

B

Fig. 1.

The step-by-step reduction of the bridged-T network are shown

ß

in Figs. 2 to 5.

2W R1

Step-1 :

R3 R2

1W

The delta-connected resistances 1 W, 3 W and 2 W in Fig. 2 are converted A

to equivalent star-connected resistances in Fig. 3. The resistances R1, R2 and

2 # 1 = 2 = 1 Ω 2 + 1+ 3 6 3

R2 =

1# 3 = 3 = 1 Ω 2 + 1+ 3 6 2

R3 =

3 2W

R3 connected by dotted lines in Fig. 2 are star equivalent resistances. R1 =

3W

2

1

1W

B

Fig. 2. ß 1 R1 = W 3

2 # 3 = 6 = 1Ω 2 + 1+ 3 6

R3 = 1 W

A

R2 =

Step-2 :

1 W 2

1W

The series-connected two 1 W resistances in Fig. 3 are combined to a single equivalent resistance in Fig. 4. Similarly, the 1/2 W and 2 W in series are

2W B

W

converted to a single equivalent reisistance.

Fig. 3. ß

Step-3 :

1 W 3

The parallel-connected 5/2 W and 2 W resistances in Fig. 4 are converted

A

1 + 1 = 2W

to a single equivalent resistance in Fig. 5. 1 5 +2= W 2 2

Step-4 : B

Fig. 4.

The series-connected 1/3 W and 10/9 W resistances in Fig. 5, are

ß

combined to a single equivalent resistance in Fig. 6.

1 W 3 A

A

1 10 13 C a 3 9 9 a 1.4444 ‡

With reference to Fig. 6, we can say that, Equivalent resistance across A - B = 13 Ω = 1.4444Ω 9

B

5 ´2 2 = 5 +2 2 10 = W 9

Þ

RESULT

B

Fig. 6.

Fig. 5.

10 2 9 2

6. 36

Chapter 6 - Network Reduction

EXAMPLE 6.5

C

Find the equivalent resistance across terminals A-B in the network shown in Fig. 1. All the resistances are 3 W. F

SOLUTION The step-by-step reduction of the network are shown in Figs. 2 to 7. E

Step-1 :

D

The innermost delta-connected resistances in Fig. 2 are converted to equivalent star-connected resistance in Fig. 3. Since the delta-connected resistances are of equal value, the equivalent star-connected resistances will also have equal value which is one-third of delta-connected resistance.

B

A

Fig. 1. ß C

Step-2 : The series-connected 3 W and 1 W resistances in each branch of star-connection in Fig. 3 are combined to a single equivalent resistance of 3 W + 1 W = 4 W in each branch as shown in Fig. 4.

3W F 3W

3W 3W

3W

Step-3 : The star-connected resistances in Fig. 4 are converted to equivalent delta-connected resistance as shown in Fig. 5. Since the star-connected resistances are of equal value the delta-connected resistances will also have equal value which is three times the star-connected resistance.

E

3W

D

3W

3W B

A 3W

Fig. 2. ß C

C

C

1=

W 4´ A

3+

W

4 ´ 3 = 12 W

3+1=4 W

4W

12

3=

3=

12

Þ

4´

3W

F

3+

3W

Þ

3W

1=

4W

B

3W

3W

3W

D 3/3

=1

3/3 = 1 W

3W 3W

3W

W 3/3 = 3W

B

A A

B

3W

Fig. 5.

E 3W

1W

Fig. 3.

Fig. 4.

ß C 2.4 + 2.4 = 4.8 W 3 ´ 12 3 + 12 = 2.4 W

4.8 ´ 2.4 = 1.6 W 4.8 + 2.4

2.4 W

Þ

Þ 2.4 W

A

2.4 W

Fig. 6.

B

A

B

Fig. 7.

A

B

Fig. 8.

Step-4 : The network of Fig. 5 has three similar parallel connections of 12 W and 3 W resistances. Each parallel connection is converted to single equivalent resistance in Fig. 6.

Circuit Theory

6. 37

Step-5 : Since we need resistance across nodes A and B, we can eliminate node C, by combining the two series-connected resistances in the path ACB to a single equivalent resistance as shown in Fig. 7. Step-6 : The parallel-connected 4.8 W and 2.4 W resistances in Fig. 7 are converted to a single equivalent resistance in Fig. 8.

RESULT With reference to Fig. 8, we can say that, Resistance across A-B = 1.6 W

EXAMPLE 6.6

2‡ B

Determine the equivalent resistance at A-B in the network shown in Fig. 1.

2‡

2‡

SOLUTION

2‡

2‡

Method-I

2‡

2‡

Let us connect a voltage source of value V volts across A-B as shown in Fig. 2. Let I be the current delivered by the source. Let, RAB be the resistance across A-B. Now, RAB is given by, R AB = V I Due to the symmetry of network when a current enters a node it will divide equally in the outgoing path.

2‡

2‡

Fig. 1. I

T

2W

3

2W I 6 Q

A I

I 6

I

R

2W

3

B

I 3

I

2W

´2

+ I 6

_

2W O 2W W 2 I I 6 I 6 ´2 N 6

V

-

P I

2W 2W + I

3

M I 3

3

+

+

I 3

2W

3

I 6

2W

V = c I # 2m + c I # 2m + c I # 2m = c 2 + 2 + 2 m I 3 6 3 3 6 3

S I I

2W

With reference to Fig. 2, by KVL in the path AMNOSBA we get,

= c 4 + 2 + 4 m I = 10 I = 5 I 6 6 3

2‡ 2‡

A

Similarly, the currents entering a node from incoming branches will also be equal. The currents that will flow in the various path are shown in Fig. 2.

`

2‡

´2

I

V = 5 I 3

` R AB = V = 5 Ω I 3 Method-II

Fig. 2.

The given network can be redrawn as shown in Fig. 3, in which two of the resistive branches are considered as parallel combination of two 4 W resistances. 2W

2W

B1

B 2W

2W

2W

2W

2W

2W

2W

2W

4W

2W

Þ 2W

2W 2W

A

Fig. 1.

2W 2W

4W

2W

4W 2W

Fig. 3.

2W 2W

2W A

A1

A2

B2

2W

4W

B

6. 38

Chapter 6 - Network Reduction

The network of Fig. 3 can be considered as parallel combination two identical networks as shown in Figs. 4 and 5. Let, RA1B1 be the single equivalent resistance of the network shown in Fig. 4 and RA2B2 be the single equivalent resistance of the network shown in Fig. 5. Now, the equivalent resistance RAB at A-B of the original network is given by parallel combination of RA1B1 and RA2B2. Since, the network of Figs. 4 and 5 are identical, RA1B1 will be equal to RA2B2 and so R AB = R A1B1 2 or R AB = R A2B2 2 . Therefore, it is sufficient if we reduce the network of Fig. 4 into a single equivalent resistance. The step-by-step reduction of the network of Fig. 4 are shown in Figs. 6 to 10. 2‡ B1

2‡

B2

2‡ 2‡

2‡

4‡

4‡

4‡ 4‡

2‡

2‡

2‡

2‡

2‡

A2

A1

Fig. 5.

Fig. 4.

Step-1 :

1

The series-connected 4 W and 2 W resistances in Fig. 4 are combined to a single equivalent resistance in Fig. 6.

6W

Step-2 :

R3

The delta-connected resistances 2 W, 2 W and 6 W in Fig. 6 are converted to equivalent star-connected resistance in Fig. 7.

A1

2W R1 2W

B1

3 R2 2W

R1 = 2 # 6 = 12 = 6 Ω 2+2+6 10 5 2 2 4 # R2 = = = 2Ω 2+2+6 10 5 2 6 12 6 # Ω R3 = = = 2+2+6 10 5

6W

2

Fig. 6. ß 1

6 16 +2= W 5 5

6 W 5

6 W 5

Þ

B1

A1

A1

3

2W R1

6 W 5

B1 R3

2 32 +6= W 5 5

2 W 5

R2 6W

Fig. 8. 2

ß B1

A1 6 W 5

16 32 ´ 5 5 = 512 ´ 5 16 32 25 48 + 5 5 32 1 32 = ´ = W 5 3 15

Fig. 9.

Þ

B1

A1

6 32 18 C 32 C a 5 15 15 50 10 a a ‡ 15 3

Fig. 10.

Fig. 7.

Circuit Theory

6. 39

Step-3 : The series-connected 6 W and 2 W resistances in Fig. 7 are combined to a single equivalent resistance in 5 Fig. 8. Similarly, the series-connected 2 W and 6 W resistances are converted to a single equivalent. 5 Step-4 : The parallel-connected 16 W and 32 W in Fig. 8 are combined to a single equivalent resistance in Fig. 9. 5 5 Step-5 : The series-connected 6 W and 32 W resistances in Fig. 9 are combined to a single equivalent in Fig. 10. 5 15

RESULT With reference to Fig. 10, we get, R A1B1 = 10 Ω 3 Let, RAB be the equivalent resistance at A-B in the network of Fig. 1. Now, R AB =

R A1B1 = 10 # 1 = 5 Ω 2 3 2 3 4 kW

EXAMPLE 6.7 When a 6 V source is connected across A and B in the network shown in Fig. 1, find a) the total resistance between terminals A and B, b) the total current drawn from the source, c) the voltage across 3 kΩ resistance, and d) the current through 4.7 kΩ resistance.

3 kW

A 10 k W

5 kW

B

4.7 k W

Fig. 1.

SOLUTION

ß

a) To find the resistance between A and B

4 kW A

The step-by-step reduction of the given network to a single equivalent resistance are shown in Figs. 2 to 4. Step-1 :

3 kW

10 k W

B 5 ´ 4.7 = 2.4227 k W 5 + 4.7

Fig. 2.

The parallel-connected 5 kΩ and 4.7 kΩ resistances in Fig. 1 are converted to a single equivalent resistance in Fig. 2

ß

Step-2 : The series-connected 4 kΩ and 2.4227 kΩ resistances in Fig. 2 are converted to a single equivalent resistance in Fig. 3.

3 kW

10 k W

Step-3 :

B

The parallel-connected 10 k Ω, 3 k Ω and 6.4227 k Ω resistances in Fig. 3 are converted to a single equivalent in Fig. 4.

ß

Fig. 3.

A 1 1 1 1 C C 10 3 6.4227 a 1.6977 k‡

Let, RAB be the resistance across A and B. With reference to Fig. 4, we get, R AB = 1.6977 kΩ

RAB B

Fig. 4.

4 + 2.4227 = 6.4227 kW

A

6. 40

Chapter 6 - Network Reduction IT

b) To find the total current drawn from the source

A

The given network can be represented by a single equivalent resistance as

+

shown in Fig. 4. Let us connect a 6 V source across A and B as shown in Fig. 5. Let,

E

IT be the total current drawn from the source. Now, by Ohm’s law we get, Total current, I T =

RAB

6V

1.6977 k ‡

B

Fig. 5.

6 = 6 = 3.5342 # 10- 3 A = 3.5342 mA R AB 1.6977 # 103

c) To find the voltage across 3 kW resistance

4 k‡

In the network of Fig. 1, connect a 6 V source across A and

IT A

B as shown in Fig. 6. Now, by inspection we can say that, the 6 V source and 3 kW resistance are in parallel. Since, the voltages

+

+ 10 k ‡

are same in parallel connection, the voltage across 3 kW is 6 V.

6V •

3 k‡

6V •

+ 6V •

5 k‡ 4.7 k ‡

B

∴ Voltage across 3 kΩ resistance = 6 V

Fig. 6.

d) To find current through 4.7 kW resistance

The voltages and currents that will exist in various resistances when a 6 V source is connected across A and B of the network of Fig. 1 are shown in Fig. 7. For convenience the circuit of Fig. 7 is redrawn as shown in Fig. 8. I IT A 10 k W

+ 6V _

+ 6V _

+ V = 6V _

4 kW + V1

IT I

3 kW

5 kW 4.7 k W

I2 + V2 _

Þ

10 k W

+ 6V _

+ 6V _

3 kW

B

Fig. 7.

Fig. 8.

With reference to Fig. 8, by voltage division rule we get, V2 = V #

2.4227 = 6 # 2.4227 = 2.2633 V 4 + 2.4227 4 + 2.4227

With reference to Fig. 7, by Ohm’s law we get, I2 =

4 kW + V1

A

- V2 + I 1

B

V = V1 + V2 I = I1 + I2

I

V2 = 2.2633 3 = 0.4816 # 10- 3 = 0.4816 mA 4.7 # 103 4.7 # 10

\ Current through the 4.7 kΩ resistance = I2 = 0.4816 mA

RESULT Total resistance between A and B, R AB = 1.6977 kΩ Total current, I T = 3.5342 mA Voltage across the 3 kΩ resistance = 6 V Current through the 4.7 kΩ resistance = 0.4816 mA

+ V = 6V _

+ V2 _

5 ´ 4.7 5 + 4.7 = 2.4227 kW

Circuit Theory

6. 41 +

j0 .5

5

W

1.

EXAMPLE 6.8

+

.5 j1

0.

W

5

Find the equivalent impedance of the network shown in Fig. 1.

-j2 W

1W

3W

SOLUTION The step-by-step reduction of the given network to a single

j4 W

Fig. 1.

B

A

ß

equivalent impedance are shown in Figs. 2 to 5.

0.5 + j0.5 + 1.5 + j1.5 = 2 + j2 W

Step-1 : The series-connected 0.5 + j0.5 W and 1.5 + j1.5 W impedance in Fig.1 are converted to a single equivalent impedance in Fig. 2.

-j2 W

1W

3W

Step-2 :

Fig. 2. The parallel-connected 2 + j2 W impedance and –j2 W capacitive

j4 W A

B

ß

reactance are converted to a single equivalent impedance in Fig. 3.

b2 + j2g ´ b- j2g = 2 - j2 W 2 + j2 + b - j2g Step-3 : 1W

The series-connected 1 W and 3 W resistances and 2 – j2 W impedance in Fig. 3 are converted to a single equivalent impedance in Fig. 4.

Fig. 3.

3W j4 W A

B

ß 1 + 2 - j2 + 3 = 6 - j2 W

Step-4 : The parallel-connected 6 – j2 W impedance and j4 W inductive reactance are converted to a single equivalent impedance in Fig. 5.

j4 W

Fig. 4.

A

B

ß

RESULT

b6 - j2g ´ j4 = 2.4 + j3.2 W 6 - j2 + j4

With reference to Fig. 5, the equivalent impedance Z AB at terminals A-B is, Z AB = 2.4 + j3.2 W

Fig.5. A

B

EXAMPLE 6.9 Obtain the single delta-connected equivalent of the network shown in Fig. 1.

SOLUTION The given network has two star network between nodes 1, 2 and 3. We can convert the star network one by one to delta.

1

10 ‡

10 ‡

j10 ‡

j10 ‡

j5 ‡

5‡ 3

Fig. 1.

2

6. 42

Chapter 6 - Network Reduction

Consider the star-connection of 10 Ω, 10 Ω and 5 Ω shown in Fig. 2. The equivalent delta network is shown in Fig. 3. 1

10 W

10 W

R1

R2 5W

2

R12

1

2

40 W

Þ

R3

20 W 20 W R31

R23

3

3

Fig. 2.

Fig. 3.

The star-connected resistances are denoted by R 1 , R 2 and R 3 . The equivalent delta-connected resistances are denoted by R 12 , R 23 and R 31 and they are computed as shown below: R12 = R1 + R 2 +

R1 R 2 = 10 + 10 + 10 # 10 = 40 Ω R3 5

R 23 = R 2 + R3 +

R 2 R3 = 10 + 5 + 10 # 5 = 20 Ω R1 10

R31 = R3 + R1 +

R3 R1 = 5 + 10 + 5 # 10 = 20 Ω R2 10

Consider the star connection of j10 Ω, j10 Ω and j5 Ω shown in Fig. 4. The equivalent delta network is shown in Fig. 5. 1

j10 W

j10 W

X1

X2

j5 W

X3

2

j40 W

1

2

X12

j20 W j20 W

Þ X31

X23

3

3

Fig. 4.

Fig. 5.

The star-connected reactances are denoted by X1, X 2 and X3 . The equivalent delta-connected reactances are denoted by X12, X23 and X31 and they are computed as shown below: j10 # j10 X12 = X1 + X 2 + X1 X 2 = j10 + j10 + = j40 Ω j5 X3 j10 # j5 X 23 = X 2 + X3 + X 2 X3 = j10 + j5 + = j20 Ω j10 X1 j5 # j10 X31 = X3 + X1 + X3 X1 = j5 + j10 + = j20 Ω j10 X2 Using the delta equivalent shown in Figs. 3 and 5, the network of Fig. 1 can be transformed to the type shown in Fig. 6 and we can observe that R12 and X12 are in parallel in Fig. 6. Similarly, R23 and X 23 are in parallel and R31 and X31 are in parallel. The parallel combination of resistance and reactance can be combined to give a single equivalent impedance as shown below: Let, Z12 = Parallel combination of R 12 and X12 Z 23 = Parallel combination of R 23 and X 23 Z31 = Parallel combination of R31 and X31

Circuit Theory

6. 43

Z12 =

40 # j40 R12 X12 = = 20 + j20 Ω 40 + j40 R12 + X12

Z 23 =

20 # j20 R 23 X 23 = = 10 + j10 Ω 20 + j20 R 23 + X 23

Z31 =

20 # j20 R31 X31 = = 10 + j10 Ω 20 + j20 R31 + X31

The single delta equivalent of the network of Fig. 1 is shown in Fig. 7. X12 = j40 W

1

10 W

10 W j10 W

j10 W

Þ

2

1

2

Þ

R12 = 40 W

20 W

= R 31

=

X31 = j20 W

3

Fig. 1.

Z31 = 10 + j10 W

Z23 = 10 + j10 W

R

W

5W

23

20

j5 W

Z12 = 20 + j20 W

1

2

X23 = j20 W

3

3

Fig. 6.

Fig. 7. C

EXAMPLE 6.10

A C

C

Find the equivalent capacitance across terminals A-B in the network C

shown in Fig. 1.

C

C

SOLUTION

C

B C

Fig. 1.

Let us consider the capacitances as capacitive reactances as shown in

-jXC

Fig. 2 for convenience in applying reduction techniques.

A

The step-by-step reduction of capacitive reactances to a single equivalent are shown in Figs. 2 to 7.

-jXC

-jXC

-jXC

-jXC

-jXC

Step-1 : The delta-conected capacitive reactances in Fig. 2 are converted to star-connected capacitive reactances as shown in

Fig. 2.

-jXC

B -jXC

Fig. 3.Since the delta-connected reactances are of equal value,

ß

the equivalent star-connected reactances will also have equal value which is one third of delta-connected reactance.

X Ej C 3

A

Step-2 :

X Ej C 3

EjXC

The series-connected capacitive reactances – jXC/3, – jXC

X Ej C 3

and – jXC/3 in Fig. 3 are converted to a single equivalent reactance in Fig. 4. Similarly, the two – jXC/3 reactances in series are converted to a single equivalent.

X Ej C 3

Fig. 3.

B

X Ej C 3

X Ej C 3

EjXC

6. 44

Chapter 6 - Network Reduction

Step-3 : The parallel-connected capacitive reactances

X Ej C 3

A

–j2XC /3 and –j5XC /3 in Fig.4 are converted to a single equivalent capacitive reactance in Fig.5. E jXC

EjXC

Step-4 :

E jXC

2X C

a Ej

The series-connected capacitive reactances –jXC /3,

FG 1 C 1 IJ H 3 3K

a Ej

3

3

X Ej C 3

Fig. 4. X -j C 3

Step-5 :

F GG GH

-jXC

RESULT B

Let, CAB = Equivalent capacitance across terminals A-B

X -j C 3

Fig. 5.

XCAB = Equivalent capacitive reactance across terminals A-B With reference to Fig. 7, we get,

I JJ JK

2 5 ´ - jXC 3 3 2 5 + 3 3 10 3 = - jXC ´ 9 7 10X C = -j 21

A

The parallel-connected capacitive reactances –jXC and –j8XC /7 in Fig. 6 are converted to a single equivalent reactance in Fig. 7.

A E jXC

8XC = 15

.....(1) EjXC

a Ej

We know that, a Ej

XC =

5X C

B

–j10XC /21, and –jXC /3 in Fig.5 are converted to a single equivalent reactance in Fig. 6.

XCAB

FG 1 C 1C 1 IJ H 3 3K

1 2πfC

24XC 21 8X C 7

B

.....(2)

LM 1 C 10 C 1 OP N 3 21 3 Q

Fig. 6.

Using equation (2), the equation (1) can be written as, XCAB = 8 # 1 15 2πfC Let,

XCAB =

.....(3)

1 2πfC AB

LM MM MN

8 ´1 - jX C 7 8 +1 7

A

.....(4)

On equating equations (3) and (4), we get, 8 # 1 1 = 15 2πfC 2πfC AB

&

= - jX C ´

C AB = 15 C = 1.875 C 8

= -j

B

OP PP PQ

8 7 ´ 7 15

8X C 15

Fig. 7.

∴ Equivalent capacitance at A-B, CAB = 1.875 C farad.

EXAMPLE 6.11

3 + j2

‡

2 + j4

‡

A

Find the equivalent admittance of the network shown in Fig. 1. 4 + j4

‡

3 E j3

j2 + 2

j5

1 + j4

‡

+

‡

Fig. 1.

1

‡

‡ B

Circuit Theory

6. 45

SOLUTION

3 + j2

W

A

3 - j3

j2

W

+ 2

j5

1 + j4

b2 + j4g ´ b4 + j4g

+

W

W

Step-1 :

1

W

The step-by-step reduction of the given network to single equivalent admittance are shown in Figs. 2 to 5.

2 + j4 + 4 + j4 W = 1.44 + j2.08

B

Fig. 2. The series-connected 2 + j4 M and 4 + j4 M

ß 3 + j2

admittance in Fig. 2.

W

2 + j2 + 3 - j3 + 1 + j5 + 1.44 + j2.08 = 7.44 + j6.08

admittances in Fig. 1 are converted to a single equivalent A

W

1 + j4

W

Step-2 : The parallel-connected 2 + j2 M , 3 – j3 M , 1 + j5 M and 1.44 + j2.08 M admittances in Fig. 2, are converted to a single equivalent admittance in Fig. 3.

B

Fig. 3. ß

Step-3 :

W

The series-connected 3 + j2 M and 7.44 + j6.08 M are converted to a single equivalent admittance in Fig.4.

b3 + j2g ´ b7.44 + j6.08g

1 + j4

A

= 2 .1441 + j1.513

3 + j2 + 7.44 + j6.08 W

B

Step-4 :

Fig. 4.

The parallel-connected 1 + j4 M and 2.1441 + j1.513 M admittances are converted to a single equivalent admittance in Fig.5.

1 + j4 + 2.1441 + j1.513 = 3.1441 + j5.513

ß A

‡

RESULT With reference to Fig. 5, the equivalent admittance YAB at terminals A-B is,

B

Fig. 5.

YAB = 3.1441 + j5.513 M

EXAMPLE 6.12 j0.5

‡

Determine the equivalent susceptance for the circuit shown in Fig. 1. j0.2

‡

SOLUTION j0.2

‡

The step-by-step reduction of the given network are shown in Figs. 2 to 6. Let, jBAB be the equivalent susceptance across A-B. With reference to Fig. 6, we get, jB AB = j0.7 M

A

j0.4

‡

j0.4

‡

B j0.2

‡

Fig. 1.

6. 46

Chapter 6 - Network Reduction j0.5

W

j0.2

W

j0.2

A

j0.4

W

j0.4

j0.5 j0.2

W

j0.5

W

j0.2

Þ

W

W W

Þ A

B j0.2

W

j0.4

W

j0.4

W

W

Fig. 1.

j(0.2 + 0.2) W = j0.4

W B

A

j0.4

W

Fig. 2.

B

W j0.4 = j0.2 2

Fig. 3. ß

j0.5

W j0.5

B

W

Þ

j(0.2 + 0.5) = j0.7

A

Fig. 6.

W j0.4 = j0.2 2

B

Þ

A

Fig. 5.

A j0.4

W

W

j(0.2 + 0.2) W = j0.4

W

B

Fig. 4.

6.15 Summary of Important Concepts 1.

Resistances in series can be replaced by an equivalent resistance whose value is given by sum of individual resistances.

2.

When n number of identical resistances of value R are connected in series then they can be replaced by a single equivalent resistance of value nR.

3.

Voltage division rule: When a voltage V exist across series combination of two resistances R1 and R2, then the voltage V1 across R1 and V2 across R2 are given by, V1 = V #

R1 R1 + R2

;

V2 = V #

R2 R1 + R2 Similarly, voltages in two impedances Z1 and Z2 in series are, V1 = V #

Z1 Z1 + Z 2

;

V2 = V #

Z2 Z1 + Z 2

4.

Resistances in parallel can be replaced by an equivalent resistance whose value is given by the inverse of sum of the inverse of individual resistances.

5.

When n number of identical resistances of value R are connected in series then they can be replaced by a single equivalent resistance of value R/n.

6.

Current division rule : When a total current I flow through parallel combination of two resistances R1 and R2, then the current I1 through R1 and the current I2 through R2 are given by, I1 = I #

R2 R1 + R 2

;

I2 = I #

R1 R1 + R 2

Similarly, currents in two impedances Z1 and Z2 in parallel are, I1 = I #

Z2 Z1 + Z 2

;

I2 = I #

Z1 Z1 + Z 2

Circuit Theory 7.

6. 47

When three resistances R12, R23 and R31 are in delta-connection with respect to terminals 1, 2 and 3 then their equivalent star-connected resistances R1, R2 and R3 with respect to same terminals are given by, R1 =

R12 R31 R12 + R 23 + R31

;

R2 =

R12 R 23 R12 + R 23 + R31

;

R3 =

R 23 R31 R12 + R 23 + R31

Similarly, the star equivalent of delta-connected impedances are given by, Z1 =

Z12 Z31 Z12 + Z 23 + Z31

;

Z2 =

Z12 Z23 Z12 + Z 23 + Z31

;

Z3 =

Z 23 Z31 Z12 + Z 23 + Z31

8.

When three equal resistances of value R are in delta-connection then their equivalent star-connected resistances will consists of three equal resistances of value R/3.

9.

When three resistances R1, R2, and R3 are in star-connection with respect to terminals 1, 2 and 3 then their equivalent delta-connected resistances R12, R23 and R31 with respect to same terminals are given by, R12 = R1 + R 2 +

R1 R 2 R3

;

R 23 = R2 + R3 +

R2 R 3 R1

;

R31 = R3 + R1 +

R3 R1 R2

Similarly, the delta equivalent of star-connected impedances are given by, Z12 = Z1 + Z 2 + Z1 Z 2 Z3

;

Z 23 = Z 2 + Z3 + Z 2 Z3 Z1

;

Z31 = Z3 + Z1 + Z3 Z1 Z2

10.

When three equal resistances of value R are in star-connection then their equivalent deltaconnected resistances will consists of three equal reisistances of value 3R.

11.

A voltage source E with a resistance RS in series can be converted to a current source IS, (where IS = E/RS) with the resistance RS in parallel.

12.

A current source IS with a resistance RS in parallel can be converted to a voltage source E, (where E = IS RS) with the resistance RS in series.

13.

Sources are connected in series for higher voltage rating and connected in parallel for higher current rating.

14.

In group-1 parameters (resistance/inductance/impedance/reactance), the series combination of parameters can be replaced by an equivalent parameter whose value is given by sum of individual parameters.

15.

In group-1 parameters (resistance/inductance/impedance/reactance), the parallel combination of parameters can be replaced by an equivalent parameter whose value is given by the inverse of sum of the inverse of individual parameters.

16.

In group-2 parameters (resistance/inductance/impedence/reactance), the serious combination of parameter can be replaced by an equavalent parameter whose value is given by the inverse of sum of the inverse of individual parameters.

17.

In group-2 parameters (conductance/capacitance/admittance/susceptance), the parallel combination of parameters can be replaced by an equivalent parameter whose value is given by sum of individual parameters.

6. 48

Chapter 6 - Network Reduction

6.16 Short-answer Questions Q6.1

I2

I1

Determine the currents I1 and I2 in the circuit shown in Fig. Q6.1. Solution

10 A

12 ‡

8‡

By current division rule,

Q6.2

I1 = 10 #

12 = 6 A 8 + 12

I 2 = 10 #

8 = 4A 8 + 12

Fig. Q6.1.

Determine the voltages V1 and V2 in the circuit shown in Fig. Q6.2. Solution By voltage division rule,

Q6.3

V1 = 20 #

4 = 8V 4+6

V2 = 20 #

6 = 12 V 4+6

4‡

6‡

+ V E 1

+ V E 2

20 V

Fig. Q6.2.

Determine the resistance across A-B in the circuit shown in Fig. Q6.3.1. Solution

6‡

The given network can be redrawn as shown in Fig. Q6.3.2.

8‡

4‡

A B

A 2‡

6‡

4‡

6‡

8‡

2‡ 3‡

6‡

Fig. Q6.3.2.

3‡

Fig. Q6.3.1.

B

With reference to Fig. Q6.3.2, we get, R AB =

Q6.4

1 = 1.3548 Ω 1 + 1 + 1 +1 6 8+6 2+4 3

10 ‡

Determine the resistance across A-B in the circuit shown in Fig. Q6.4.1.

10 ‡

(AU Dec’14, 2 Marks)

5‡

B

Fig. Q6.4.1.

The given network can be redrawn as shown in Fig. Q6.4.2 A

10 ‡

5+5 = 10 ‡

10 ‡

A

Solution

10 W A

B

⇐

Fig. Q6.4.4.

5W

Fig. Q6.4.3.

With reference to Fig. Q6.4.4, we get, RAB = 10 = 5 W 2

10 ´ 10 = 5W 10+10

⇓ A

B

10 ‡

⇐

B

10 ‡ 5‡

Fig. Q6.4.2.

10 ‡

Circuit Theory 2‡ A

2

Determine the resistance across A-B in the circuit shown in Fig. Q6.5.1. (AU June’14, 2 Marks)

2‡

Q6.5

6. 49

‡

Solution The given network can be redrawn as shown in Fig. Q6.5.4

1.2 ‡

B 1‡

Fig. Q6.5.1.

⇓ A

2‡

2‡ B

2 ´ 3.2 2 + 3.2 = 12308 . W

⇐

B

‡

⇐

2

1.2308 + 1 = 2.2308 ‡

A

2W

A

2 + 1.2 = 3.2 ‡

B 1W

1‡

Fig. Q6.5.3.

Fig. Q6.5.2.

Fig. Q6.5.4. With reference to Fig. Q6.5.4, we get,

R AB = 2 # 2.2308 = 1.0546 Ω 2 + 2.2308

Q6.6

In the circuit shown in Fig. Q6.6, find the total resistance across A-B.

2‡ A

Solution 5‡

5‡

5‡

5‡

5‡

5‡

In Fig. Q6.6, the parallel combination of 6 numbers of 5 Ω resistances is equivalent to a single resistance of 5 W.

B

6

` R AB

Seven bulbs each rated at 75 W, 120 V are connected in parallel. Calculate the power and current consumed by them. Solution The parallel connection of 7 bulbs is equivalent to 7 resistances in parallel as shown in Fig Q6.7. Now, the total power is given by sum of power consumed by each bulb/resistance. I

` Total current, I = P = 525 = 4.375 A V 120

Fig. Q6.7.

75 W

75 W

75 W

75 W

75 W

+ E

75 W

We know that, in purely resistive loads, P = VI

75 W

\ Total power = 7 ´ 75 = 525 W V = 120 V

Q6.7

Fig. Q6.6.

= 2 + 5 = 2.8333 Ω 6

6. 50 Q6.8

Chapter 6 - Network Reduction Calculate the voltage across the terminals A-B for the circuit shown in Fig. Q6.8. 50 ‡

Solution

10 ‡

By voltage division rule,

A

+

VAB

20 + 4 = 28.5714 V = 100 # 50 + 10 + 4 + 20

100 V

4‡

_

20 ‡ B

A star-connected network consists of three resistances 3 W, 6 W and 10 W. Convert the star-connected network to equivalent delta-connected network. 1

1 31 = 18

Let R 1 , R 2 and R 3 be the resistances in star connection as shown in Fig. Q6.9.1 and R12 R23 and R31 be the resistances in equivalent delta connection as shown in Fig. Q6.9.2.

Q6.10

1

1

W

Solution

Fig. Q6.8.

R1 = 3 W

Þ R2 = 6 W 2

R3 = 10 W 3

R12 = 10.8 W

R

Q6.9

2

2

3

3

3

Fig. Q6.9.1.

R12 = R1 + R 2 +

R1 R 2 = 3 + 6 + 3 # 6 = 10.8 Ω R3 10

R 23 = R 2 + R3 +

R 2 R3 = 6 + 10 + 6 # 10 = 36 Ω R1 3

R31 = R3 + R1 +

R3 R1 = 10 + 3 + 10 # 3 = 18 Ω R2 6

2

R23 = 36 W

Fig. Q6.9.2.

A delta-connected network consists of three resistances 5 W, 6 W and 9 W . Convert the delta-connected network to equivalent star-connected network. 1

Solution Let R12, R23 and R31 be the resistances in delta connection as shown in Fig. Q6.10.1 and R1, R2 and R3 be the resistances in star connection as shown in Fig. Q6.10.2.

1

1

1 R1 = 2.25 W

R31 = 9 W

R12 = 5 W

Þ =

2

3

2

R23 = 6 W

R3 3

W 2.7

R2 = 1.5 W 2 2 3

3

Fig. Q6.10.1.

R1 =

R12 R31 = 5 # 9 = 2.25 Ω R12 + R 23 + R31 5+6+9

R2 =

R12 R 23 = 5 # 6 = 1.5 Ω R12 + R 23 + R31 5+6+9

R3 =

R 23 R31 = 6 # 9 = 2.7 Ω R12 + R 23 + R31 5+6+9

Fig. Q6.10.2.

Circuit Theory Q6.11

6. 51 0.1 H

What will be the equivalent inductance across A-B in the network shown in Fig. Q6.11.1.

A 0.5 H 0.2 H

0.4 H

Solution B

First, the parallel combination of 0.4 H, 0.5 H and

0.6 H

0.2 H has been reduced to a single equivalent as

Fig. Q6.11.1.

shown in Fig. Q6.11.2.

⇓ 0.1 H

In the network of Fig. Q6.11.2, all the inductances

A

1 1 1 1 C C 0.4 0.5 0.2 a 0.1053 H

are in series. Hence, the equivalent inductance LAB across A-B is given by, B

LAB = 0.1 + 0.1053 + 0.6 = 0.8053 H

Q6.12

Fig. Q6.11.2.

0.6 H

What will be the equivalent capacitance across A-B in the network shown in Fig. Q6.12.

A 2 mF

Solution

8 mF

10 mF

In the given network 2 µF and 8 µF capacitances are in parallel and the parallel combination is in series with 10 µF capacitance. Hence, the equivalent capacitance CAB across A-B is given by, C AB =

Q6.13

B

Fig. Q6.12.

^2 + 8h # 10 = 5 µF ^2 + 8h + 10

Determine the currents I 1 and I 2 in the circuit shown in Fig. Q6.13. 0

10Ð0 A

By current division rule,

Q6.14

I2

I1

Solution

I1 = 10+0 o #

1 − j5 = 8 − j14 A = 16.1245+ − 60.3 o A 2 + j4 + 1 − j5

I2 = 10+0 o #

2 + j4 = 2 + j14 A = 14.1421+81.9 o A 2 + j4 + 1 − j5

2 + j4 W

~

Determine the voltages V 1 and V 2 in the circuit shown in Fig. Q6.14.

Fig. Q6.13.

2 + j4 W

+

Solution

1 - j5 W

V1

1 - j5 W

-

+

+

~-

V2

By voltage division rule, V1 = 10+0 o # V2 = 10+0 o #

2 + j4 = 2 + j14 V = 14.1421+81.9 o V 2 + j4 + 1 − j5 1 − j5 = 8 − j14 V = 16.1245+ − 60.3 o V 2 + j4 + 1 − j5

0

10Ð0 V

Fig. Q6.14.

-

6. 52 Q6.15

Chapter 6 - Network Reduction The impedance of each branch of a delta-connected circuit is impedance of equivalent star-connected circuit?

3 Z . What will be the branch

Solution When three equal impedances are in delta, then equivalent star impedance of each branch will be 1/3 times of delta impedance. Given that, Zdelta =

3Z

` Zstar = 1 # Zdelta = 1 # 3 Z = 3 3

Z 3

A

3Z

A

3Z

3Z Z = 3 3

Þ

B

C

C

3Z

Z

3

3

B

Fig. Q6.15.2.

Fig. Q6.15.1. Q6.16

Z

Find the equivalent admittance at A-B in the network shown inFig Q6.16. 1 + j‡

Solution

A

Let, Z AB = Equivalent impedance at A - B Now, Z AB

2 + j2 ‡

(2 + j2) # (1 + j + 2 − j3) = 2 + j0.4 Ω = (2 + j2) + (1 + j + 2 − j3)

` Equivalent admittance at A - B , YAB =

2 E j3 ‡

B

Fig. Q6.16.

1 = 1 2 + j0.4 Z AB

= 0.4808 − j0.0962 M

Find the equivalent admittance and impedance at A-B in the network shown inFig Q6.17. Solution

1+j

Let, YAB = Equivalent admittance at A - B

‡

A

` Equivalent impedance at A - B , Z AB

1 = 1 = 1 + j0.6667 YAB

2 E j2

(1 + j) # (2 − j2 + 3 + j) = 1 + j0.6667 M (1 + j ) + (2 − j 2 + 3 + j )

‡

‡

Now, YAB =

3+j

Q6.17

B

Fig. Q6.17.

= 0.6923 − j0.4615 Ω

6.17 Exercises I. Fill in the Blanks With Appropriate Words 1.

When n number of resistances of value R are connected in series then the equivalent resistance is given by ___________.

2.

When n number of resistances of value R are connected in parallel then the equivalent resistance is given by ___________.

Circuit Theory

6. 53

3.

When three equal resistances of value R are in star-connection then its equivalent delta-connection will have three equal resistances of value ___________.

4.

When n number of capacitances of value C are connected in series then the equivalent capacitance is equal to ___________.

5.

The equivalent admittance of 3 identical parallel-connected admittances of value Y is equal to ________.

ANSWERS 1. nR

2. R n

3. 3R

4.

C n

5. 3 Y

II. State Whether the Following Statements are True / False 1.

When the star-connected resistances are converted to equivalent delta-connected resistances, the power consumed by the resistances remains same.

2.

Inductances connected in series can be replaced by an equivalent inductance whose value is given by sum of individual inductances.

3.

Capacitances connected in parallel can be replaced by an equivalent capacitance whose value is given by the inverse of the sum of the inverse of individual capacitances.

4.

Inductive susceptance is always negative and capacitive susceptance is always positive.

5.

When the voltage requirement of the load is higher, then the voltage sources should be connected in parallel.

ANSWERS 1. True

2. True

3. False

4. True

5. False

III. Choose the Right Answer for the Following Questions 1. When n number of identical resistances of value R are connected in parallel the equivalent value of the parallel combination is, a) nR

b) R n

c) 2R n

d) R 2n

2. The star equivalent of three identical resistances in delta of value R will be three identical resistances of value, a) R 3

b) 3R

2 c) R 3

d) 3R 2

3. In the network shown in Fig. 3, if the value of all the resistances are 1 W then what will be the equivalent resistance at A-B? a) 1 W B

A

b) 0.5 W c) 0.4 W d) 0.2 W

Fig. 3.

6. 54

Chapter 6 - Network Reduction

4. In the network shown in Fig. 4, if the value of all the resistances are 2 W then the equivalent resistance at AC and BD are respectively,

B

a) 0 W, 0.8 W b) 0.8 W, 0 W

A

C

c) 0 W, 0.4 W d) 0.8 W, 0.8 W

Fig. 4.

D

5. The equivalent value of two inductances 3 H and 6 H when connected in series and in parallel respectively are, a) 9 H, 3 H

b) 9 H, 2 H

c) 18 H, 2 H

d) 2 H, 3 H

6. In the network shown in Fig. 6, if the value of all the inductance is 2 H, then what is the equivalent inductance at A-B?

B

a) 4 H b) 3 H c) 2 H

A

Fig. 6.

d) 1 H

7. The equivalent value of two capacitances 6 mF and 12 mF when connected in series and in parallel respectively are, a) 6 mF, 18 mF

b) 18 mF, 4 mF

c) 4 mF, 18 mF

d) 72 mF, 18 mF

8. In the network shown in Fig. 8, if the value of all the capacitance is 2 mF, then what is the equivalent capacitance at A-B? A a) 4 mF b) 3 mF

D

C

c) 2 mF d) 1 mF

B

9. The equivalent reactance with respect to terminals A-B in the network shown in Fig. 9 is,

Fig. 8.

j10 ‡ j10 ‡

a) j10 W j5 ‡

b) j7.5 W c) j5 W d) j2.5 W

A

j5 ‡

B

j7.5 ‡ j5 ‡

Fig. 9.

Circuit Theory

6. 55

10. The equivalent value of two impedances 8 + j3 W and 2 – j3 W when connected in parallel is, a) 10 + j0 W

b) 6 + j6 W

c) 10 + j6 W

d) 2.5 - j1.8 W 3 + j4 ‡

11. In the network shown in Fig. 11, what is the equivalent impedance at A-B? a) 3 + j3 W

j2 ‡

Ej3 ‡ 3 + j3 ‡

A

b) 1 + j W

B j9 ‡

Ej2 ‡

c) 3 – j4 W 3 E j4 ‡

d) 1 + j2 W

Fig. 11.

12. In the network shown in Fig. 12, if the value of all the conductance are 3 M, then what is the value of equivalent conductance at A-B? B

a) 18 M b) 2 M c) 0.5 M A

d) 0.25 M

Fig. 12.

13. In the network shown in Fig. 13, what is the equivalent conductance at A-B? a) 0.9083 M

A 4

b) 15.5 M

6

‡

‡

2 3

4

‡

‡ 5

‡

c) 29 M

3

‡ 2

‡

‡

B

d) 20 M

Fig. 13.

14. In the network shown in Fig. 14, what is the equivalent susceptance at A-B?

Ej6

A

a) j0.5 M Ej6

b) j3 M

‡

j4

‡

‡

j4

c) j6 M

Ej6

‡

‡

B

d) j12 M Ej6

Fig. 14.

‡ 2 + j2

15. In the network shown in Fig. 15, what is the equivalent admittance at A-B?

2E

a) 0 + j4 M

2+

b) 4 + j4 M c) 4 + j0 M d) 4 - j4 M

‡

A

j2 ‡

Ej2

‡ j2

B 2 E j2

j2 ‡

Fig. 15.

‡

‡

6. 56

Chapter 6 - Network Reduction

ANSWERS 1. b

4. a

7. c

10. d

13. b

2. a

5. b

8. a

11. b

14. d

3. c

6. d

9. c

12. a

15 c

IV. Unsolved Problems E6.1

2‡

A

Find the equivalent resistance with respect to terminals A-B in the network shown in Fig. E6.1.

1.6 ‡

6‡

E6.2

Find the equivalent resistance across the source terminals A-B of the circuit shown in Fig. E6.2, and hence calculate the current delivered by the source.

4‡

B 1.2 ‡

Fig. E6.1. R 2‡ A

R

R

6‡

18 ‡

B

R R R

1 R

R + E

12 V

12 ‡

6‡

18 ‡

R R

A B

R

R

Fig. E6.3.

Fig. E6.2.

E6.3

Determine the equivalent resistance at A-B for the circuit shown in Fig. E6.3.

E6.4

Determine the equivalent resistance for the circuit shown in Fig. 6.4. 18 ‡ 10 ‡

+

16 ‡

18 ‡ 20 ‡

20 V

20 ‡

4‡

E

Fig. E6.4. In the network shown in Fig. 6.5, find the equivalent resistance between A and B. For the network shown in Fig. E6.6, find the equivalent resistance across A-B. C 4‡

8‡ 10 ‡

A

3‡

E6.5 E6.6

1‡

12 ‡

3‡

2‡ B A

Fig. E6.5.

12

9‡

6‡

‡

5‡ B

Fig. E6.6.

6‡

Circuit Theory

6. 57

E6.7

In Fig. 6.7, find the equivalent resistance across A-B.

E6.8

In circuit shown in Fig. 6.8, find the equivalent resistance across A-B. 2‡ 9‡

12 ‡ 6‡

10 ‡ 16 ‡

18 ‡

16 ‡

A

5‡

14 ‡

5‡

15 ‡

A

B

B

12 ‡

8‡

12 ‡

Fig. E6.7. 4‡

Fig. E6.8. E6.9

In Fig. 6.9, find the equivalent resistance RAB .

E6.10 Convert the circuit shown in Fig.E6.10 with multiple sources into a single equivalent current source at terminal A-B, with a single equivalent resistance in parallel. Also, calculate the voltage across the equivalent resistance. C

20

10 V

‡

4‡

A

3‡

16

‡

8‡

6‡

+ E

8‡ 2A

2‡ 18 ‡

A

B

5‡

E + 1V

8‡

3A

4‡

2‡

4‡

7‡

B

Fig. E6.10.

Fig. E6.9.

E6.11 Find the equivalent inductance across terminals A-B in the network shown in Fig. E6.11.

3H

j0.2 ‡

1.5 H

j0.6 ‡ j2.4 ‡

9H

j3 ‡

j0.6 ‡

B

A 6H

2H

j1 ‡

j2.3 ‡ A

j0.8 ‡

B

Fig. E6.11.

j0.2 ‡

Fig. E6.12.

E6.12 In Fig. 6.12, find the equivalent reactance across A-B.

j3 ‡

5A

6. 58

Chapter 6 - Network Reduction

E6.13 Find the equivalent capacitance across terminals A-B in the network shown in Fig. E6.13. Ej2 ‡

6 mF A 3 mF 3 mF

Ej12 ‡ Ej6 ‡

6 mF

3 mF

3 mF

Ej9 ‡ Ej4 ‡

A

6 mF

B

Ej2 ‡

Ej6 ‡

B Ej3.2 ‡

6 mF

Fig. E6.14.

Fig. E6.13.

Note : Consider 6 mF capacitor as reactance –jX and so reactance of 3mF capacitor is –j2X. Reduce the network by treating all the capacitors as reactances and finally convert reactance to capacitor.

E6.14 In Fig. 6.14, find the equivalent capacitive reactance across A-B. E6.15 Find the equivalent impedance across A-B in the circuit shown in Fig. 6.15. 2 + j2 W

5 - j4 W

A

3 + j6 W

4 + j2 W

5 - j3 W

B

+

~

o

-

Fig. E6.15.

24Ð0 V

ANSWERS E6.1

RAB = 5.6 Ω

E6.2

RAB = 3.75 Ω , I = 3.2 A

E6.3

R AB = 5 R Ω 6

E6.4

Req = 19 Ω

E6.5

RAB = 1.5686 Ω

E6.6

RAB = 1.9637 Ω

E6.7

RAB = 2.9084 Ω

E6.8

RAB = 6.4478 Ω

E6.9

RAB = 4.6663 Ω

E6.10

Ieq = 2.75 A, Req = 0.8 W, V = 2.2 V

E6.11

LAB = 2.875 H

E6.12

jXAB = j0.3522 Ω

E6.13

CAB = 9.6432 mF

E6.14

–jXAB = −j16.92 Ω

E6.15

Z AB = 5.3021 − j1.2792 Ω = 5.4542+ − 13.6 o Ω

Chapter 7

THEOREMS IN CIRCUIT ANALYSIS 7.1

Introduction

The theorems are useful tools for analyzing the circuits with lesser effort. They are derived from the fundamental laws and concepts. A given circuit can be analyzed by different methods, i.e., using KCL/KVL, using mesh/node method, using theorems, etc. For the analysis of a given circuit some methods will be easier and some other methods will be complicated. In most of the cases, the analysis of circuits using theorems will be much easier when compared to other methods. Remember that, the theorems do not always simplify the task of analysis, sometimes it may be more tough than other methods.

7.2

Superposition Theorem

The superposition theorem states that the response in a circuit with multiple sources is given by algebraic sum of responses due to individual sources acting alone. The superposition theorem is also referred to as principle of superposition. The Superposition theorem is an useful tool for the analysis of linear circuits with multiple sources. The linear circuit is a circuit composed entirely of independent sources, linear dependent sources and linear elements. A circuit element is said to be linear, if the voltage-current relationship is linear, i.e., n a i or n = ki, where k is a constant. The responses that can be determined by superposition theorem are listed below: i)

Current in resistance, inductance and capacitance.

ii) Voltage across resistance, inductance and capacitance. iii) Current delivered by independent voltage sources. iv) Voltage across independent current sources. v) Voltage and current of linear dependent sources. While calculating the response due to an individual source, all other sources are made inactive or replaced by zero value sources (Sometimes the bloodthirsty term killed is used). A zero value source is represented by its internal resistance (or impedance). “In an ideal voltage source, the internal resistance (or impedance) is zero, and in an ideal current source the internal resistance (or impedance) is infinite”.

7. 2

Chapter 7 - Theorems in Circuit Analysis

Therefore, while calculating the response due to one source all other ideal voltage sources are replaced by short circuit (or by their internal impedance) and all other ideal current sources are replaced by open circuit (or by their internal impedance). Procedure for Analysis Using Superposition Theorem 1.

When the internal impedances of the sources are specified, then represent them as an external impedance and so the sources become ideal sources. For a voltage source, the internal impedance is represented as an impedance in series with ideal voltage source. For a current source, the internal impedance is represented as an impedance in parallel with ideal current source.

2.

The response is either voltage or current in the elements. The response when all the sources are acting is called total response. If the polarity of total voltage response or directionof total current response are not specified in the problem, then assume a polarity for total voltage response and direction for total current response when all the sources are acting together.

3.

Determine the response due to each independent source by allowing one source to act at a time. While determining the response due to one source, replace all other independent ideal voltage source by short circuit (S.C.) and all other independent ideal current source by open circuit (O.C.).

4.

Denote the voltage response due to each source as Vl , Vll , Vlll ... and the current response as Il , Ill , Illl .... While determining the response due to each source, maintain the polarity of voltage response same as that of total response. Similarly, maintain the direction of current response same as that of total response.

5.

Determine the total response by taking the sum of individual responses.

Note : 1. Power cannot be directly determined from the superposition theorem. Hence, determine the power only using the total current and voltage response. 2. When all the independent sources are deactivated, there will not be any current or voltage in any part of the circuit. Hence, dependent sources will not contribute to the response when all the independent sources are deactivated (i.e., the response due to a dependent source acting alone will be zero). 1‡

EXAMPLE 7.1 1‡

Find the current through the 5 W resistor in the circuit shown in Fig. 1 using the superposition theorem.

10 A

1‡

5‡

Fig. 1.

20 A

Circuit Theory

7. 3

SOLUTION

1‡

Let IL be the current through the 5 W resistance when both the current sources are acting together as shown in Fig. 2. Let, IlL = Current through 5 W in the direction of IL when 10 A source alone is acting.

1‡

1‡ IL

10 A

IllL = Current through 5 W in the direction of IL when 20 A source alone is acting.

5‡

20 A

Fig. 2.

Now, by the superposition theorem, IL = IlL + IllL To find the response IlL when 10 A source is acting alone The 20 A current source is replaced by an open circuit as shown in Fig. 3. The circuit of Fig. 3 is redrawn as shown in Fig. 4. In Fig. 4, the 5 W resistance is in series with 10 A source and so the current through 5 W is also 10 A. ` IlL = 10 A 1W

1W

1W

1W

1W

5W

10 A

1W

Þ

I’L O.C

I¢L

5W

10 A

Fig. 3.

Fig. 4.

To find the response IllL when 20 A source is acting alone The 10 A current source is replaced by an open circuit as shown in Fig. 5. The circuit of Fig. 5 is redrawn as shown in Fig. 6. In Fig. 6, the 5 W resistance is in series with 20 A source and so the current through 5 W is also 20 A. ` IllL = 20 A 1W

1W

1W

1W

1W

I’’L O.C.

5W

Fig. 5.

1W I’’L

Þ 20 A

5W

Fig. 6.

To find the total response IL when both the sources are acting By the superposition theorem, IL = IlL + IllL ` IL = 10 + 20 = 30 A

20 A

7. 4

Chapter 7 - Theorems in Circuit Analysis

Cross-Check by Node Analysis

1‡

With reference to Fig. 7, the node basis matrix equation is, R V R V R V 1‡ V 1‡ S1 + 1 2 −1 − 1 W S V1 W V1 V3 S 10 W 1 W S W 1 S1 1 IL S W S 1 1 1 1 W S W 1 S W S − 1 1 + 1 + 5 − 1 W S V2 W = S 0 W 20 A 10 A 5‡ S 1 W S W S W 1 1 1 SS − − + W S 20 W 1 1 W S V3 W 1 1 T X T X T X 0 V R V R V R V IL a 2 S10 W S 2 − 1 − 1 W S V1 W 5 S − 1 2.2 − 1 W S V W = S 0 W S W S W S 2W Fig. 7. S20 W S − 1 − 1 2 W S V3 W T X T X T X 2 −1 −1 ∆l = − 1 2.2 − 1 = 2 # 9 2.2 # 2 − (− 1) 2 C − (− 1) # 9 − 1 # 2 − (− 1) 2 C + (− 1) # 9(− 1) 2 − (− 1) # 2.2 C −1 −1 2 = 6.8 − 3 − 3.2 = 0.6 2 10 − 1 ∆l 2 = − 1 0 − 1 = 2 # 90 − 20 # (− 1) C − 10 # 9 − 1 # 2 − (− 1) 2 C + (− 1) # 9 − 1 # 20 − 0 C − 1 20 2 = 40 + 30 + 20 = 90 V2 =

∆l 2 = 90 = 150 V ∆l 0.6

` IL =

V2 = 150 = 30 A 5 5

EXAMPLE 7.2 Using the superposition theorem, find the current through the 3 W resistance in the circuit shown in Fig. 1.

IL 10 V, + 1‡ E

+ 20 V, E 2‡

3‡

SOLUTION The internal resistance of the voltage sources are represented as a series resistance external to the source and the voltage sources are represented as ideal sources as shown in Fig. 2.

Fig. 1. 1‡

Let, IlL = Current through 3 W in the direction of IL when 10 V source alone is acting. IllL = Current through 3 W in the direction of IL when 20 V source alone is acting.

2‡ IL

10 V +E

3‡

+ 20 V E

Now, by the superposition theorem, IL = IlL + IllL

Fig. 2.

To find the response IlL when 10 V source is acting alone The 20 V source is replaced by short circuit as shown in Fig. 3. Let, Is1 be the current supplied by 10 V source. In Fig. 3, at node-A, the current Is1 divides between the parallel resistances 2 W and 3 W. The parallel-connected resistances 2 W and 3 W are combined to a single equivalent in Fig. 4.

Circuit Theory

7. 5 1W

1W

2W

A I’L

Is1

Is1

+

+ 3W

10 V

S.C.

Þ

2´3 2+3 = 1.2 W

10 V

-

-

Fig. 3.

Fig. 4.

With reference to Fig. 4, by Ohm’s law, Is1 =

10 = 4.5455 A 1 + 1.2

With reference to Fig. 3, by current division rule, IlL = Is1 #

2 = 4.5455 # 2 = 1.8182 A 2+3 2+3

To find the response IllL when 20 V source is acting alone The 10 V source is replaced by short circuit as shown in Fig. 5. Let, Is2 be the current supplied by 20 V source. In Fig. 5, at node-A, the current Is2 divides between the parallel resistances 1 W and 3 W. The parallel-connected resistances 1 W and 3 W are replaced by a single equivalent in Fig. 6. 1W

2W

2W

A I’’L

3W

S.C.

Is2

Is2 + 20 V -

Þ

Fig. 5. With reference to Fig. 6, by Ohm’s law, Is2 =

20 = 7.2727 A 2 + 0.75

With reference to Fig. 5, by current division rule, IllL = Is2 #

1 = 7.2727 # 1 = 1.8182 A 1+3 1+3

To find the response IL when both the sources are acting By the superposition theorem, IL = IlL + IllL = 1.8182 + 1.8182 = 3.6364A

+ 20 V -

1´3 1+ 3 = 0.75 W

Fig. 6.

7. 6

Chapter 7 - Theorems in Circuit Analysis

Cross-Check by Mesh Analysis With reference to Fig. 7, the mesh basis matrix equation is,

10 4 1 I1 H > H = > H 1 3 I2 − 10

>

∆ =

4 1 = 4 # 3 − 1 # 1 = 11 1 3

∆1 =

10 1 = 10 # 3 − (− 10) # 1 = 40 − 10 3

` I L = I1 =

2‡

1‡

I1 1+3 1 10 H > H = > H > 1 1 + 2 I2 10 − 20

IL +

+

I2

10 V

20 V

3‡ E

E

I1

IL = I 1

Fig. 7.

∆1 = 40 = 3.6364 A ∆ 11

(AU June’14, 8 Marks)

EXAMPLE 7.3

Using the superposition theorem, find the current through the 5W resistor in the circuit shown in Fig.1.

10 ‡

2A

+ 50 V

E

SOLUTION

2‡

I5 1‡

5‡

Let , I5 = Current through 5 W resistance shown in Fig. 1.

Fig. 1. Let, Il5 = Current through the 5 W resistance in the direction of I5 when 50 V source alone is acting. Ill

5

= Current through the 5 W resistance in the direction of I5 when 2 A source alone is acting.

Now, by the superposition theorem, I5 = Il5 + Ill5

10 ‡

+ 50 V

To find the response Il5 when 50 V source is acting alone The 2 A source is replaced by an open circuit as shown in Fig. 2. With reference to Fig. 2, by Ohm’s law, I'5 =

2‡ O.C

E

I’5 5‡

1‡

Fig. 2.

50 = 50 = 10 A 10 + 5 15 3

To find the response Ill when 2 A source is acting alone 5

The 50 V source is replaced by short circuit as shown in Fig. 3. In the circuit of Fig. 4, the 2 A source current divides between parallel resistances 10 W and 5 W.

Circuit Theory

7. 7

Therefore, by current division rule, I''5 = 2 #

10 = 20 = 4 A 10 + 5 15 3

10 W

2W

2W

10 W

2A

2A

Þ

S.C

5W

I’’5 1W

1W

I’’5

5W

Fig. 4.

Fig. 3.

To find the total current I5 when both the sources are acting By superposition theorem, I5 = I'5 + I''5 = 10 + 4 = 14 = 4.6667 A 3 3 3

EXAMPLE 7.4 3V

Using the superposition theorem, find the voltage VL and the power consumed by the 6 W resistor in the circuit shown in Fig. 1.

+E

1‡

2‡

SOLUTION Let, VlL = Voltage across the 6 W resistance with polarity same as that of VL when 3 V source alone is acting. VllL = Voltage across the 6 W resistance with polarity same as that of VL when 2 A source alone is acting.

4‡

+ 2 A VL _

6‡

Fig. 1.

Now, by the superposition theorem, VL = VlL + VllL Power consumed by the 6 W resistor, PL =

VL2 6

Note : Since power is proportional to square of voltage it is not a linear quantity and so the power cannot be determined directly by the superposition theorem. To find the response VlL when 3 V source is acting alone The 2 A current source is replaced by an open circuit as shown in Fig. 2. The circuit of Fig. 2 is redrawn as shown in Fig. 3. In Fig. 3, the voltage across series combination of 4 W and 6 W is 3 V. This 3 V divides into V1 and V2 and so by voltage division rule, we get, VlL = − V2 = − 3 # 3V

6 = − 1.8 V 6+4

3V

+-

1W

+-

1W

2W

Þ

+ 4W

O.C. V’L _

_ 3V

6W

Fig. 2.

2W

+

+ V1 _

+ V2 _ _ V’L +

4W

6W

Fig. 3.

7. 8

Chapter 7 - Theorems in Circuit Analysis

To find the response VllL when 2 A source is acting alone The 3 V source is replaced by short circuit as shown in Fig. 4. The circuit of Fig. 4 is redrawn as shown in Figs. 5 and 6. In the circuit of Fig. 6, the current 2 A divides between the parallel resistances 6 W and 4 W. Hence, by current division rule, I1 = 2 #

4 = 0.8 A 6+4

By Ohm’s law, VllL = I1 # 6 = 0.8 # 6 = 4.8 V I2

S.C.

I1 2A

1W

2W + 2 A V’’L 6 W _

4W

1W

Þ

2W

4W

Fig. 4.

+ V’’L _

6W

2A

Þ

4W 2A

Fig. 5.

1´ 2 + 1 + 2 V’’ L 2 _ = W 3

6W

Fig. 6.

To find VL and power in 6 W resistor By the superposition theorem, VL = VlL + VllL = − 1.8 + 4.8 = 3 V Power consumed by the 6 Ω resistor, PL =

2 VL2 = 3 = 1.5 W 6 6

Cross-Check Let,

PlL = PllL =

Let, PL, sup

(VlL) 2 (− 1.8) 2 = = 0.54 W 6 6 2 (VllL) 2 = 4.8 = 3.84 W 6 6 = PlL + PllL = 0.54 + 3.84 = 4.38 W

Here, PL, sup ¹ PL and so we can say that, the power calculated directly by the superposition theorem is not equal to actual power.

EXAMPLE 7.5 Find the voltage across the 2 W resistance in the circuit of Fig. 1 by principle of superposition.

E 3‡

SOLUTION Let VL be the voltage across the 2 W resistance when both the voltage sources are acting together as shown in Fig. 2. Let, VlL = Voltage across the 2 W resistance with polarity same as that of VL when 5 V source alone is acting. VllL = Voltage across the 2 W resistance with polarity same as that of VL when 10 V source alone is acting.

1‡

5V

+

1‡ + 10 V _

Fig. 1.

2‡

Circuit Theory

7. 9

Now, by the superposition theorem,

E

5V

+

VL = VlL + VllL

The 10 V source is replaced by short circuit as shown in Fig. 3. The circuit of Fig. 3 is redrawn as shown in Figs. 4 and 5.

-

+

-

1W

3W

Þ

+ S.C. V’L _

V2

-

1W +

2‡

Fig. 2.

+

3W _

1W

5V

+ + 10 V VL _ _

1‡

The parallel combinations of resistances are replaced by a single equivalent in Fig. 5. 5V

1‡

3‡

To find the response VlL when 5 V source is acting alone

_ V1

+

Þ

5V

_ V2 +

+

V1 = V’L _ +

2W

Fig. 3.

_ V2 +

V1 = V’L _ +

1W

2W

3 ´1 3 +1 = 0.75 W

1´ 2 1+ 2 = 0.6667 W

Fig. 5.

Fig. 4.

In Fig. 5, the source voltage 5 V divides between the series-connected resistances 0.6667 W and 0.75 W. Let, these voltages be V1 and V2. Since 1 W and 2 W are in parallel, the voltage across them will be same and so V1 = VlL . With reference to Fig. 5, by voltage division rule, VlL = V1 = 5 #

0.6667 = 2.353 V 0.6667 + 0.75

To find the response VllL when 10 V source is acting alone The 5 V source is replaced by short circuit as shown in Fig. 6. The circuit of Fig. 6 is redrawn as shown in Figs. 7 and 8. Figure 7 is same as that Fig. 4, except the 5 V source, which is 10 V in this case. Therefore, by voltage division rule, VllL = 10 #

0.6667 = 4.706 V 0.6667 + 0.75

S.C.

1W

3W

1W

3W + 1W

Þ

+ V’’L _

10 V -

10 V

Þ +

_

2W

Fig. 6.

1W

V’’L _

Fig. 7.

To find the response VL when both the sources are acting By the superposition theorem, VL = VlL + VllL = 2.353 + 4.706 = 7.059 V

10 V _

2W

3 ´1 3+1 =0.75 W

+

+

+ V’’L _

Fig. 8.

1´ 2 1+ 2 =0.6667 W

7. 10

Chapter 7 - Theorems in Circuit Analysis 8‡

EXAMPLE 7.6 Use the principle of superposition to find the current IL through the 8 W resistance in the circuit shown in Fig. 1.

SOLUTION The internal resistance of voltage source is represented as a series resistance external to the source and the internal resistance of current source is represented as a parallel resistance external to the source as shown in Fig. 2. Now, the sources can be treated as ideal sources. Let, IlL = Current through the 8 W resistance in the direction of IL when 10 V source alone is acting. IllL = Current through the 8 W resistance in the direction

IL 10 V, + 2‡ E

4‡

5 A, 2‡

Fig. 1. 2‡

8‡ IL

10 V +E

4‡

2‡

5A

of ILwhen 5 A source alone is acting. Now, by the superposition theorem,

Fig. 2.

IL = IlL + IllL To find the response IlL when 10 V source is acting alone

The 5 A source is replaced by an open circuit as shown in Fig. 3. Let, Is1 be the total current supplied by 10 V source. This current divides into I1 and I2 and flow through the two parallel paths in Fig. 3. 2W Is1

I1

8W

I2

2W

I’L

Is1 +

+ 10 V

4W

2W

O.C.

-

Þ

4 ´ (8 + 2) 4 + (8 + 2) = 2.8571W

10 V -

Fig. 3.

Fig. 4.

With reference to Fig. 4, Is1 =

10 = 2.0588 A 2 + 2.8571

With reference to Fig. 3, by current division rule,

IlL = I1 = Is1 #

4 = 2.0588 # 4 = 0.5882 A 4 + (8 + 2) 4 + 10

To find the response IllL when 5 A source is acting alone The 10 V source is replaced by short circuit as shown in Fig. 5. The parallel combination of 2 W and 4 W is replaced by a single equivalent in Fig. 6. With reference to Fig. 6, we can say that, the 5 A current divides between the parallel resistances 2 W and (8 + 1.3333) W. By current division rule,

IllL = − I 4 = − 5 #

2 = − 0.8824 A 2 + (8 + 1.3333)

Circuit Theory

7. 11 8W

2W

8W

I’’L 4W

S.C.

I4 I3

I’’L 2W

Þ

5A

4´2 4+2 = 1.3333 W

Fig. 5.

2W

5A

Fig. 6.

To find the response IL when both the sources are acting By the superposition theorem, IL = IlL + IllL = 0.5882 + (-0.8824 ) = –0.2942 A

EXAMPLE 7.7

(AU May’15, 8 Marks) 4‡

Compute the current I L in the circuit of Fig. 1 using the superposition theorem.

IL 47 ‡

SOLUTION

27 ‡

Let, IlL = Current through the 23 W resistance in the direction

20 A + E

of IL when 200 V source alone is acting. IllL = Current through the 23 W resistance in the direction

23 ‡

200 V

Fig. 1.

of ILwhen 20 A source alone is acting. Now, by the superposition theorem, IL = IlL + IllL To find the response IlL when 200 V source is acting alone The 20 A source is replaced by an open circuit as shown in Fig. 2. Let, Is1 be the total current supplied by 200 V source. This current divides equally between parallel-connected resistances 27 W and (4 + 23) W. I2

I1

4W I’L

Is1

Is1

I2

47 W 47 W O.C.

27 W

23 W

Þ

+ 200 V -

Is1 4W

200 V +-

With reference to Fig. 3, by Ohm’s law,

200 = 3.3058 A 47 + 13.5

With reference to Fig. 2, by current division rule,

I IlL = I1 = s1 = 3.3058 = 1.6529 A 2 2

23 W

Fig. 3.

47 W 27 = 13.5 W 2

Þ

27 W

Fig. 2.

Is1 =

I1 = I’L

200 V +-

Fig. 4.

7. 12

Chapter 7 - Theorems in Circuit Analysis

To find the response IllL when 20 A source is acting alone The 200 V source is replaced by short circuit as shown in Fig. 5. The parallel resistances 27 W and 47 W are replaced by a single equivalent in Fig. 6. In the circuit of Fig. 6, the 20 A source current divides between parallel resistances 23 and (4 + 17.1486) W. Therefore, by current division rule,

IllL = I3 = 20 #

(4 + 17.1486) = 9.5806 A 23 + (4 + 17.1486)

4W

4W I’’L

27 W

27 ´ 47 27 + 47 = 17.1486 W

Þ

23 W

20 A

I3

I4

47 W

S.C.

20 A

I’’L 23 W

Fig. 6.

Fig. 5. To find the total current IL when both the sources are acting By the superposition theorem, IL = IlL + IllL = 1.6529 + 9.5806 = 11.2335 A

EXAMPLE 7.8

2‡

Determine the current in the 5 W resistance in the circuit shown in Fig. 1, using the superposition theorem.

5‡ IL

+

E 4‡

25 V

SOLUTION

2‡

E

2‡

50 V +

Let, IlL = Current through the 5 W resistance in the direction of IL when 25 V source alone is acting.

Fig. 1.

IllL = Current through the 5 W resistance in the direction of ILwhen 50 V source alone is acting. By the superposition theorem, IL = IlL + IllL To find the response IlL when 25 V source is acting alone

The 50 V source is replaced by short circuit as shown in Fig. 2. The 2 W resistances in parallel are replaced by a single equivalent and then I’L is solved by mesh analysis. With reference to Fig. 3, the mesh basis matrix equation is,

>

2+4 −4 −4 4 + 5 + 1

∆ =

H > I1 H I

2

=

>

25 0

H ⇒

6 −4 = 6 # 10 − (− 4) 2 = 44 − 4 10

I 6 −4 H > I1 10 2

> −4

;

H

∆2 =

=

>

25 H 0

6 25 = 0 − (− 4) # 25 = 100 −4 0

Circuit Theory

7. 13 2W

2W

5W

2W

5W

I’L

I ’L

+ 4W

25 V

2W

S.C.

Þ

-

+ 25 V

4W

I1

2´2 = 1W 2+2

I2

-

Fig. 2.

Fig. 3.

∆ ` IlL = I 2 = 2 = 100 = 2.2727 A ∆ 44 To find the response IllL when 50 V source is acting alone The 25 V source is replaced by short circuit as shown in Fig. 4. The parallel-connected resistances 2 W and 4 W are replaced by a single equivalent as shown in Fig. 5 and then IllL is solved by mesh analysis. The mesh basis matrix equation is,

>

5 + 2 + 1.3333 − 2 −2 2+2

∆ =

8.3333 − 2 −2 4

2W

I

2

=

0 > 50 H ⇒

>

= 8.3333 # 4 − (− 2) 2 = 29.3332

8.3333 − 2 −2 4

;

H > I1 H I

> 50 H 0

=

2

0 −2 50 4

∆1 = 5W

2W

5W

= 0 − 50 # (− 2) = 100 2W

I’’L

I’’L S.C.

H > I1 H

-

4W

2W

50 V

Þ

+

4´2 4W 4+2 = 1.3333 W

2W I2

I1

50 V +

Fig. 5.

Fig. 4. ∆ 100 ` IllL = I1 = 1 = = 3.4091 A ∆ 29.3332 To find the total response IL when both the source are acting By the superposition theorem, IL = IlL + IllL = 2.2727 + 3.4091 = 5.6818

EXAMPLE 7.9

4‡

(AU Dec’14, 16 Marks)

Use the principle of superposition to find the current IL 32 V

through the 5 W resistance in the circuit shown in Fig. 1.

2‡

+E

SOLUTION Let, IlL = Current through 5 W in the direction of IL when 9 A source alone is acting.

IL 9A

5‡

10 ‡

4A

Fig. 1.

7. 14

Chapter 7 - Theorems in Circuit Analysis IllL = Current through 5 W in the direction of IL when 4 A source alone is acting. IlllL = Current through 5 W in the direction of IL when 32 V source alone is acting. Now, by the superposition theorem, IL = IlL + IllL + IlllL

To find the response IlL due to 9 A source The 32 V source is replaced by short circuit and 4 A source by an open circuit as shown in Fig. 2. The parallel combination of 2 W and 4 W is replaced by a single equivalent in Fig. 3. With reference to Fig. 3, we can say that, the 9 A current divides between the parallel resistances 5 W and 11.3333 W. By current division rule, I 'L = 9 #

11.3333 = 6.2449 A 5 + 11.3333 4W

S.C.

2W I’L

I’L 9A

5W

10 W

O.C.

Þ

9A

4 ´ 2 + 10 = 113333 . W 4+2

5W

Fig. 2.

Fig. 3.

To find the response IllL due to 4 A source The 32 V source is replaced by short circuit and 9 A source by an open circuit as shown in Fig. 4. The parallel combination of 2 W and 4 W is replaced by a single equivalent in Fig. 5. With reference to Fig. 5, we can say that, the 4 A current divides between the parallel resistances 10 W and (5+1.3333) W. By current division rule, I''L = 4 #

10 = 2.449 A 10 + (1.3333 + 5) 4W

S.C.

4 ´ 2 = 1..3333 W 4+2

2W

I’’L O.C.

I’’L

5W

10 W

Fig. 4.

4A

Þ 5W

10 W

Fig. 5.

4A

Circuit Theory

7. 15

To find the response IlllL due to 32 V source The current sources 9 A and 4A are replaced by open circuit as shown in Fig. 6. Let, us assume two mesh currents I1 and I2 as shown in Fig. 6. Now IllL = I 2 . The mesh basis matrix equation is,

>

4+2 +2 + 2 5 + 2 + 10

H > I1 H I

2

> 32 H 32

=

&

>2

6 2 17

H > I1 H I

2

> 32 H

4‡

32

=

I1

6 2 ∆ = = 6 # 17 − 2 # 2 = 98 2 17

32 V

2‡

+E

I’’’L

∆2 =

6 32 = 6 # 32 − 2 # 32 = 128 2 32

I''' L = I 2 =

O.C.

I2

5‡

10 ‡

O.C.

∆2 = 128 = 1.3061 A 98 ∆

Fig. 6.

To find the total response IL when all the sources are acting By the superposition theorem, IL = I'L + I'' L + I''' L = 6.2449 + 2.449 + 1.3061 = 10 A

EXAMPLE 7.10 In Fig. 1, find the component of Vx caused by each source acting

1.5 A

alone. What is the value of Vx when all the sources are acting together.

SOLUTION

20 ‡

Let, Vlx = Voltage across the 20 W resistance when 16 V source is acting alone. Vllx = Voltage across the 20 W resistance when 3 A source is acting alone. Vlllx = Voltage across the 20 W resistance when 10 V source is acting alone.

+

16 V +E

E+

_ Vx

10 V 3A

80 ‡

Fig. 1.

Vllllx = Voltage across the 20 W resistance when 1.5 A source is acting alone. The polarity of voltages Vlx , Vllx , Vlllx and Vllllx are chosen same as that of Vx. Now, by the superposition theorem, Vx = Vlx + Vllx + Vlllx + V llllx To find the response Vlx due to 16 V source The 10 V source is replaced by short circuit and the current sources are replaced by an open circuit as shown in Fig. 2. With reference to Fig. 3, by voltage division rule,

Vlx = 16 #

20 = 3.2 V 20 + 80

7. 16

Chapter 7 - Theorems in Circuit Analysis O.C.

20 W

20 W _

+

+

S.C.

V’x + 16 V -

80 W

O.C.

Þ

_ V’x

16 V +-

80 W

Fig. 2.

Fig. 3.

To find the response Vllx due to 3 A source The voltage sources are replaced by short circuit and 1.5 A source is replaced by an open circuit as shown in Fig. 4. With reference to Fig. 5, by current division rule,

I2 = 3 #

80 = 2.4 A 20 + 80

By Ohm’s law,

Vllx = − 20 # I2 = − 20 # 2.4 = − 48 V O.C.

20 W

20 W _

+

S.C.

V’’x

I1

V’’x 80 W

3A

S.C.

I2 _

+

Þ

80 W

3A

Fig. 4.

Fig. 5.

To find the response Vlllx due to 10 V source The 16 V source is replaced by the short circuit and the current sources are replaced by an open circuit as shown in Fig. 6. With reference to Fig. 7, by voltage division rule,

Vlllx = 10 #

20 = 2V 20 + 80 O.C.

20 W +

S.C.

20 W -+

_ V’’’x

+

10 V O.C.

Fig. 6.

80 W

_ V’’’x

-+

10 V

Þ

80 W

Fig. 7.

Circuit Theory

7. 17

To find the response Vllllx due to 1.5 A source The voltage sources are replaced by short circuit and 3 A source is replaced by an open circuit as shown in Fig. 8. With reference to Fig. 9, we can say that the current source is shorted and so no current will flow through the 20 W resistance. ` Vllllx = 0 1.5 A

1.5 A

I1 = 0

20 W +

_ V’’’’x

S.C.

+

S.C.

O.C.

80 W

I = 1.5 A 20 W _

S.C.

V’’’’x

I1 = 0

Þ 80 W

Fig. 8.

Fig. 9.

To find the response Vx due to all the sources By the superposition theorem, Vx = Vlx + Vllx + Vlllx + Vllllx = 3.2 + (–48) + 2 + 0 = –42.8 V

RESULT Component of Vx when 16 V source alone is acting, Vlx Component of Vx when 3 A source alone is acting, Vllx

= 3.2 V = – 48 V

Component of Vx when 10 V source alone is acting, Vlllx = 2 V Component of Vx when 1.5 A source alone is acting, Vllllx = 0 V The value of Vx when all the sources are acting,

Vx

= – 42.8 V

EXAMPLE 7.11 Using the superposition theorem, find the current through 2 + j2 W impedance of the circuit shown in Fig. 1.

10 W

SOLUTION

5W

20Ð0oV

-j5 W

2W

+

Let IL be the current through 2 + j2 W impedance branch as shown in Fig. 2.

~

_

j2 W

~

Let, I lL = Component of IL due to 20∠0 V source acting alone. o

I llL = Component of IL due to 10∠30o A source acting alone.

o

10Ð30 A

Fig. 1.

Now, by the superposition theorem,

10 W

5W

IL = I lL + I llL The 10Ð30o A current source is replaced by an open circuit as shown in Fig. 3.

+

20Ð0oV

To find the response IlL due to 20Ð0o V source

IL

~

_

j2 W

With reference to Fig. 4, by Ohm’s law, we get, I lL =

20+0 o = 20 = 1.6216 − j0.2703 A 10 + 2 + j2 12 + j2

-j5 W

2W

Fig. 2.

~

o

10Ð30 A

7. 18

Chapter 7 - Theorems in Circuit Analysis 10 W

IL¢

+ o

20Ð0 V

10 W

5W

2W

-j5 W

Þ

~

IL¢

+ o

20Ð0 V

_

2W

~

_ j2 W

j2 W

O.C.

Fig. 3.

Fig. 4.

To find the response I llL due to 10Ð30o A source

10 W

The 20Ð0o V voltage source is replaced by short circuit as shown in Fig. 5. With reference to Fig. 5, by current division rule we get, I llL = 10+30 o #

5W

IL¢¢

-j5 W

2W

S.C.

o 10 = 100+30 = 7.6975 + j2.8837A 10 + 2 + j2 12 + j2

j2 W

To find the response IL due to both the sources

10Ð30oA

~

Fig. 5.

By the superposition theorem,

IL = IlL + I llL = 1.6216 − j0.2703 + 7.6975 + j2.8837 = 9.3191 + j2.6134 A = 9.6786+15.7 o A

Cross-Check

5W

V

The 20∠0o V voltage in series with 10 W is converted to current source as shown in Fig. 6. With reference to Fig. 6, by KCL at node A we get,

V 2 + j2

V 10 20 = 2A 10

~

IL

-j5 W

2W

10 W j2 W

V + V = 2 + 10+30 o 10 2 + j2

o

10Ð30 A

~

Fig. 6.

1 ` Vf 1 + = 2 + 10+30 o ⇒ b 0.35 − j0.25 l V = 2 + 10+30 o 10 2 + j2 p o ` V = 2 + 10+30 = 13.4113 + j23.8652 0.35 − j0.25

Now, by Ohm’s law, IL =

V = 13.4113 + j23.8652 = 9.3191 + j2.6135 A 2 + j2 2 + j2 = 9.6786+15.7 o A

EXAMPLE 7.12 o

5Ð90 A

Using the superposition theorem, find the voltage V1 across the capacitance in the circuit shown in Fig. 1.

~

SOLUTION

j5 W

Let, Vl1 = Voltage across capacitance when 10Ð0o A source alone is acting.

Vll1 = Voltage across capacitance when 5Ð90o A source alone is acting.

Now, by the superposition theorem,

V1 = Vl1 + Vll1

+ o

10Ð0 A

~

-j4 W

V1

-

Fig. 1.

4W

Circuit Theory

7. 19

To find the response Vl1 due to 10Ð0o A source The 5Ð90o A source is replaced by an open circuit as shown in Fig. 2. The circuit of Fig. 2 is redrawn as shown in Fig. 3. With reference to Fig. 3, by current division rule,

I1 = 10 #

4 + j5 = 12.3529 + j9.4118 A − j4 + 4 + j5

Now, by Ohm’s law, we get, Vl1 = − j4 # I1 = − j4 # a12.3529 + j9.4118 k = 37.6472 – j49.4116 V O.C. I2

j5 W + + o

10Ð0 A

~

4W

-j4 W

V1

Þ

10 A

j5 W

-j4 W

V1¢

~

I1

-

4W

I1 + I2 = 10 A

Fig. 2.

Fig. 3.

To find the response Vll1 due to 5Ð90o A source The 10Ð0o A source is replaced by an open circuit as shown in Fig. 4. The circuit of Fig. 4 is redrawn as shown in Fig. 5. With reference to Fig. 5, by current division rule,

I 4 = j5 #

j5 = − 5.8824 + j1.4706 A j5 + (4 − j4)

Now, by Ohm’s law, we get, Vll1 = − (− j4 # I4) = j4 # a− 5.8824 + j1.4706 k

= − 5.8824 − j23.5296 V

o

5Ð90o A = j5 A

5Ð90 A

~

~ I3

j5 W j5 W

+ O.C.

-j4 W

V1

4W

-

Fig. 4.

Þ

I4

-j4 W +

4W

V1¢¢

Fig. 5. To find the total response V1 when both the sources are acting By the superposition theorem,

V1 = Vl1 + Vll1 = (37.6472 – j49.4116) + (–5.8824 – j23.5296) = 31.7648 – j72.9412 V = 79.5577∠– 66.5o V

7. 20

Chapter 7 - Theorems in Circuit Analysis

Cross-Check by Node Analysis

o

5Ð90 A = j5 A

With reference to Fig. 6, the node basis matrix equation is, R V R V R V S 1 + 1 − 1 W S V1 W S 10 − j5 W W S j5 − j4 W S j5 W S W S WS W= S 1 1 1 W S WS W S − + j 5 S W S j5 4 j5 W S V2 W T X T X T X

>

j0.05 j0.2 j0.2 0.25 − j0.2

V

=

2

>

10 − j5 j5

V1

j5 W V2

+ 10Ð0o A = 10 A

H

~

-j4 W

V1

4W

-

Fig. 6. 2

∆l =

j0.05 j0.2 = j0.05 # (0.25 − j0.2) − (j0.2) j0.2 0.25 − j0.2 = 0.05 + j0.0125

∆l1 =

10 − j5 j0.2 = (10 − j5) # (0.25 − j0.2) − (j5) # (j0.2) j5 0.25 − j0.2 = 2.5 − j3.25

V1 =

7.3

H > V1 H

~

2.5 − j3.25 ∆l1 = = 31.7647 − j72.9412 V = 79.5576+ − 66.5 o V ∆l 0.05 + j0.0125

Thevenin’s and Norton’s Theorems

Thevenin’s theorem will be useful to find the response of an element in a circuit by replacing the complicated part of the circuit by a simple equivalent voltage source. Similarly, Norton’s theorem will be useful to find the response of an element in a circuit by replacing the complicated part of the circuit by a simple equivalent current source. Consider a load impedance Z L connected to two terminals A and B of a circuit represented as a box in Fig. 7.1(a). Using Thevenin’s theorem, the circuit can be replaced by a voltage source in series with an impedance as shown in Fig. 7.1(b). Using Norton’s theorem, the circuit can be replaced by a current source in parallel with an impedance as shown in Fig. 7.1(c). Z th

A

+ Vth

A

ZL

~

-

Þ

Fig. b : Thevenin’s equivalent.

Circuit with sources and impedances

B

ZL

A B

Þ In

Fig. a : Original circuit.

~

Zn

ZL

Fig. c : Norton’s equivalent. B

Fig. 7.1 : Thevenin’s and Norton’s equivalent of a circuit.

These theorems can also be used to analyze a part of a circuit by replacing the complicated part of the circuit by simple equivalent circuit.

Circuit Theory

7. 21

Consider two parts of a circuit N1 and N2 connected through resistance-less wires as shown in Fig. 7.2(a). Now, one part of the circuit can be replaced by a simple equivalent circuit using Thevenin’s/Norton’s theorem for the analysis of another part of the circuit. Using Thevenin’s theorem, the circuit N1 is replaced by a voltage source in series with an impedance as shown in Fig. 7.2(b). Using Norton’s theorem, the circuit N1 is replaced by a current source in parallel with an impedance as shown in Fig. 7.2(c). Z th

A

+ Vth

Circuit N2

~

_

Þ A

B

Fig. b : Thevenin’s equivalent. Circuit N2

Circuit N1 B

Þ

A

Fig. a : Original circuit. In

Circuit N2

Zn

~

B

Fig. c : Norton’s equivalent. Fig. 7.2 : Thevenin’s and Norton’s equivalent of a circuit.

Thevenin’s Theorem Thevenin’s theorem states that a circuit with two terminals can be replaced by an equivalent circuit, consisting of a voltage source in series with a resistance (or impedance). The voltage source is called Thevenin’s voltage source and its value is given by the voltage across the two open terminals of the circuit. The series resistance (or impedance) is called Thevenin’s resistance (or impedance) and it is given by looking back resistance (or impedance), at the two open terminals of the network.The looking back resistance (or impedance) is the resistance (or impedance), measured at the two open terminals of the circuit after replacing all the independent sources by zero value sources. Rth A A Circuit with dc sources and resistance

A

Þ

Circuit with dc sources and resistances

Vth +-

B

A

+

Circuit with zero value sources and resistances

Vth _ B

B

Rth

B

Fig. a : Original circuit.

Fig. b : Thevenin’s Fig. c : To find Thevenin’s voltage. equivalent.

Fig. d : To find Thevenin’s resistance.

Fig. 7.3 : Thevenin’s equivalent of dc circuit.

7. 22

Chapter 7 - Theorems in Circuit Analysis A A Circuit with ac sources and impedances

A

Zth

Circuit with ac sources and impedances

+

Þ

Vth

~

-

A

+

Circuit with zero value sources and impedance

Vth

_

B Z th

B

B B

Fig. a : Original Circuit.

Fig. b :Thevenin’s equivalent.

Fig. d : To find Thevenin’s impedance.

Fig. c : To find Thevenin’s voltage.

Fig. 7.4 : Thevenin’s equivalent of ac circuit.

In order to calculate Thevenin’s resistance (or impedance), all the sources are replaced by zero value sources and the circuit is reduced to a single equivalent resistance (or impedance) with respect to two open terminals. The zero value sources are represented by their internal resistance (or impedance). For ideal voltage source, the internal resistance (or impedance) is zero and so it is replaced by short circuit. For ideal current source,the internal resistance (or impedance) is infinite and so it is replaced by open circuit.

Norton’s Theorem Norton’s theorem states that a circuit with two terminals can be replaced by an equivalent circuit, consisting of a current source in parallel with a resistance (or impedance). The current source is called Norton’s current source and its value is given by the current flowing when the two terminals of the circuit are shorted. The parallel resistance (or impedance), is called Norton’s resistance (or impedance), and it is given by looking back resistance (or impedance) at the two terminals of the circuit. The looking back resistance (or impedance), is the resistance (or impedance), measured at the two open terminals of the circuit after replacing all the independent sources by zero value sources. A A

A Circuit with dc sources and resistances

Þ

In

Circuit with dc sources and resistances

Rn

A Circuit with zero value sources and resistances

In B

B

Fig. b : Norton’s Fig. c : To find equivalent. Norton’s current. Fig. 7.5 : Norton’s equivalent of dc circuit.

Fig. d : To find Norton’s resistance.

B

Rn

B

Fig. a : Original circuit.

A A Circuit with ac sources and impedances

A

Þ

In

~

Circuit with ac sources and impedances

Zn

B

A In

B

Circuit with zero value sources and impedance B

Zn

B

Fig. a : Original Circuit.

Fig. b : Norton’s equivalent.

Fig. c : To find Norton’s current.

Fig. 7.6 : Norton’s equivalent of ac circuit.

Fig. d : To find Norton’s impedance.

Circuit Theory

7. 23

In order to calculate Norton’s resistance (or impedance), all the sources are replaced by zero value sources and the circuit is reduced to a single equivalent resistance (or impedance) with respect to two open terminals. The zero value sources are represented by their internal resistance (or impedance). For ideal voltage source, the internal resistance (or impedance) is zero and so it is replaced by short circuit. For ideal current source, the internal resistance (or impedance) is infinite and so it is replaced by open circuit. Relation Between Thevenin’s and Norton’s Equivalents Consider Thevenin’s equivalent of a given circuit as shown in Fig. 7.7. Rth A A Vth +-

Þ

Circuit N

B B

Fig. b : Thevenin’s equivalent.

Fig. a : Original circuit.

Fig. 7.7 : A circuit and its Thevenin’s equivalent.

Let us find Norton’s equivalent of the circuit N from its Thevenin’s equivalent. To find Norton’s current In the terminals A and B are short circuited as shown in Fig. 7.8(a). Now, In is the current flowing through the short circuit. By Ohm’s law we get, In = Vth / Rth. To find Norton’s resistance, the voltage source Vth is replaced by short circuit as shown in Fig. 7.8(b). With reference to Fig. 7.8(b) we can say that, the Norton’s resistance, Rn is same as that of Thevenin’s resistance, Rth. The Norton’s equivalent of the circuit N is shown in Fig. 7.8(c). Rth

Rth A

Vth +E

In a

A

Vth R th

B

Fig. a : To find Norton’s current.

S.C.

A

In a

B Rn = Rth

Fig. b : To find Norton’s resistance.

Vth R th

Rn = Rth

B

Fig. c : Norton’s equivalent.

Fig. 7.8 : Norton’s equivalent of circuit N.

Norton’s equivalent of a circuit can also be directly obtained from its Thevenin’s equivalent (or vice-versa) using source transformation technique as shown in Fig. 7.9. In fact, “the Thevenin’s equivalent is the voltage source model and the Norton’s equivalent is the current source model of a circuit”.

7. 24

Chapter 7 - Theorems in Circuit Analysis

From the above discussion it is evident that R th = R n (or Z th = Z n) and also the Thevenin’s resistance (or impedance) is given by the ratio of Thevenin’s voltage and Norton’s current. ` R th = R n =

Vth In

.....(7.1)

Z th = Z n = Vth In

.....(7.2)

The equations (7.1) and (7.2), can be used to determine the looking back resistance (or impedance) from the knowledge of open circuit voltage (Vth ) and short circuit current (In ). Rth = Rn A

A

Vth = InRn +

Þ

Þ

-

In =

Vth

Rn = Rth

R th

B

B

A

A

Zth = Zn

+

Þ

Vth = In Z n

Þ

~

In =

_

Vth Zth

~

Zn = Zth

B

B

Fig. 7.9 : Conversion of Thevenin’s equivalent to Norton’s equivalent (or vice-versa) using source transformation technique.

EXAMPLE 7.13

5A A

Determine Thevenin’s and Norton’s equivalents of the circuit shown in Fig. 1 with respect to terminals A and B. 12 A

SOLUTION

12 ‡

4‡

3A

To find Thevenin’s voltage Vth B

Thevenin’s voltage Vth is the voltage across terminals A and B

Fig. 1.

as shown in Fig. 2. The polarity of Vth is assumed such that terminal-A is positive and terminal-B is negative. 5A

5A A +

4W

(4 ´ 5) V 12 A

12 ‡

4‡

3A

Vth

12 A

12 W

I = 5A

_

Vth + _ 4´3 = 12 V

_ B

Fig. 2.

A +

+

Fig. 3.

_ B

Circuit Theory

7. 25

The 3 A current source in parallel with the 4 W resistance is converted to a voltage source in series with the 4 W resistance as shown in Fig. 3. Now, the 5 A source is in series with the 4 W resistance and 12 V source. By KVL we can write, Vth = (4 ´ 5) + 12 = 32 V To find Thevenin’s resistance Rth (and Norton’s resistance Rn) The current sources are replaced by an open circuit as shown in Fig. 4. With reference to Fig. 5, Thevenin’s resistance, Rth = 4 W \ Rth = Rn = 4 W O.C. A

O.C.

12 W

4W

A

O.C.

12 W

Þ

4W

B

B Rth

Fig. 4.

Rth

Fig. 5.

To find Norton’s current In The terminals A and B are shorted as shown in Fig. 6. Since the 4 W resistance is shorted no current can flow through it. By KCL at node-A we can say that, the current through the short circuit is 5 + 3 = 8 A. \ In = 8 A 5A

5A A

A 7A

I=0 12 A

12 W

4W

3A

In

Þ

12 W

12 A

3A

In = 5 + 3 = 8 A

7A (7 + 5) A

5A

B

Fig. 6.

Fig. 7.

Thevenin’s and Norton’s equivalent Rth = 4 ‡ A

Vth = 32 V +E

A

In = 8 A +E

B

Fig. 8 : Thevenin’s equivalent.

Rn = 4 ‡

B

Fig. 9 : Norton’s equivalent.

3A B 8A

7. 26

Chapter 7 - Theorems in Circuit Analysis

EXAMPLE 7.14

+

E + 12 V

8‡

20 V

Find the current through the 10 W resistance of the circuit shown in Fig. 1, using Thevenin’s theorem and confirm the result by mesh analysis.

E 2‡

10 ‡

5‡

SOLUTION Let us remove the 10 W resistance and mark the resulting open terminals as A and B as shown in Fig. 2.

Fig. 1.

Now, Thevenin’s voltage is the voltage measured across A and B and Thevenin’s resistance is the resistance measured between A and B. The polarity of Vth is assumed such that terminal-A is at a higher potential than terminal-B.

+

E + 12 V

8‡

20 V E 2‡

Vth

5‡

To find Thevenin’s voltage Vth

_ B

With reference to Fig. 3, by Ohm’s law, we get,

With reference to Fig. 3, by KVL, we can write,

⇒

8‡

Vth = 2 ´ 2.8571 + 12 = 17.7142 V

+

12 V E +

A +

No voltage +

20 V E

To find Thevenin’s resistance Rth

5‡

The voltage sources are replaced by short circuit as shown in Fig. 4. In Fig. 4, the 5 W and 2 W resistances are in parallel and the parallel combination is in series with the 8 W resistance. ` R th

Rth

Fig. 2.

I = 20 = 2.8571 A 5+2 Vth= 2I + 12

A +

2‡

2I E

Vth _

I

B

Fig. 3.

= b 5 | | , 2 l + 8 = 5 # 2 + 8 = 9.4286 Ω 5+2

S.C. A 8‡

Thevenin’s equivalent at A-B

S.C. 2‡

Rth= 9.4286 ‡ A 5‡

B

Vth=17.7142 V +E

Rth

Fig. 4. B

Fig. 5 : Thevenin’s equivalent. A

To find current through 10 W resistance

9.4286 ‡

IL

The 10 W resistance is connected to terminals A and B as shown in Fig. 6. Let, IL be the current through the 10 W resistance.

+ E 17.7142 V

10 ‡

With reference to Fig. 6, by Ohm’s law, we get, IL =

17.7142 = 0.9118 A 9.4286 + 10

B

Fig. 6.

Circuit Theory

7. 27

Cross-Check by Mesh Analysis Let us assume mesh currents as shown in Fig. 7. The mesh basis matrix equation is,

>

5+2 −2 − 2 2 + 8 + 10

H > I1 H 2 I

> 12 H 20

=

12 V E +

8‡ +

A IL

20 V

> −2

7 −2 20

H > I1 H 2 I

> 12 H

E

20

=

2‡

I1

7 −2 ∆ = = 7 # 20 − (− 2) 2 = 136 − 2 20 ∆2 =

10 ‡

5‡ I2 B

Fig. 7.

7 20 = 7 # 12 − (− 2 # 20) = 124 − 2 12

` I2 =

∆2 = 124 = 0.9118 A ∆ 136

EXAMPLE 7.15

10 ‡

20 ‡

Find Thevenin’s and Norton’s equivalents of the circuit shown in Fig. 1 with respect to terminals A and B. A 10 V +E

SOLUTION

+ E

50 V

B

To find Thevenin’s voltage Vth With reference to Fig. 2, in the closed path DACBD, by KVL we can write, 10I + 20I + 10 = 50

Fig. 1.

⇒ 30I = 50 – 10

` I = 50 − 10 = 1.3333 A 30

C

20 ‡

I _

With reference to Fig. 2, in the path DABD, we get,

10 ‡

A _

20I +

I

A+ 10 V +E

10I + Vth = 50 ⇒ Vth = 50 – 10 I

B

_

+ E

Vth

\ Vth = 50 – 10 ´ 1.3333 = 36.667 V I

To find Thevenin’s resistance Rth The voltage sources are replaced by short circuit as shown in Fig. 3. With reference to Fig. 4, R th = 20 | | , 10 = 20 # 10 = 6.6667 Ω 20 + 10 10 W

20 W

A

A S.C.

S.C.

Þ

20 W

10 W

B Rth

Rth B

Fig. 3.

D

10I +

Fig. 4.

B

Fig. 2.

I

50 V

7. 28

Chapter 7 - Theorems in Circuit Analysis

Thevenin’s and Norton’s equivalent Rth = 6.6667 ‡

A

A

In = 5.5 A

Vth = 36.667 V +E

Rn = 6.6667 ‡

B

B

Fig. 5 : Thevenin’s equivalent.

Fig. 6 : Norton’s equivalent.

Using source transformation technique, Norton’s equivalent is obtained from Thevenin’s equivalent shown in Fig. 5. In =

Vth = 36.667 = 5.5 A R th 6.6667

Rn = R th = 6.6667 Ω 10 ‡

20 ‡

Alternatively, Norton’s current can be directly determined, by shorting the terminals A and B and measuring the current through the short. A

With reference to the circuit shown in Fig. 7, we can write,

10 V +E

In

I1

+ E

I2

50 V

B

In = I1 + I 2 = 10 + 50 = 5.5 A 20 10

Fig. 7.

EXAMPLE 7.16

(AU May’15, 16 Marks)

Obtain the Thevenin and Norton equivalent circuits for the active network shown in Fig. 1.

3‡ A 3‡

6‡ E 10 +

20 V +E

V B

SOLUTION

Fig. 1.

To find Thevenin’s voltage Vth With reference to Fig. 2, by KVL we can write, 3I + 6I = 20 + 10

⇒

9I = 30 ⇒

I = 30 = 10 A 9 3

With reference to Fig. 2, we get, Vth + 10 = 6I

⇒

Vth = 6I - 10 = 6 # 10 − 10 = 10 V 3

3‡ C

E 3I C 20 V +E

3‡

6I E I

No voltage 6‡

A + Vth

E 10 +

V

Fig. 2.

_ B

Circuit Theory

7. 29 3‡

To find Thevenin’s resistance Rth

A

The voltage sources are replaced by short circuit as shown in Fig. 3.

6‡

3‡ S.C

R th = 3 # 6 + 3 = 5 Ω 3+6

S.C

Rth B

Fig. 3. Thevenin’s and Norton’s equivalent Rth = 5 ‡

A A In = 2A

Rn = 5‡

Vth = 10 V +E

B B

Fig. 5 : Norton’s equivalent.

Fig. 4 : Thevenin’s equivalent.

Using source transformation technique, Norton’s equivalent is obtained from Thevenin’s equivalent shown in Fig. 4. In =

Vth = 10 = 2 A R th 5

R n = R th = 5 Ω

EXAMPLE 7.17

IL

10 ‡

Using Thevenin’s theorem, find the current IL in the circuit shown in Fig. 1.

+ 10 V E

+ E 4V

3‡

SOLUTION

3‡

2‡

Let us remove the 10 W resistance and mark the resulting open terminals as A and B as shown in Fig. 2.

1‡

Fig. 1.

Now, we have to determine Thevenin’s equivalent of the circuit shown in Fig. 2, with respect to terminals A and

Rth A +

B. Let us assume Vth as shown in Fig. 2 with terminal-A as positive and terminal-B as negative.

B E

+ E 10 V

To find Thevenin’s voltage Vth

+ E 4V

3‡

In Fig. 3, by voltage division rule, we can write,

3‡

2‡

1‡

Va = 10 # 3 = 6 V 3+2 Vc = 4 #

Fig. 2.

3 = 3V 3+1

With reference to Fig. 3, using KVL in the path ABCA, we can write,

A + 10 V +E _

Vth + Vc = Va 2‡

\ Vth = Va – Vc = 6 – 3 = 3 V

I1

B E

Vth

Vc _

+ E 4V

I2

+

+ 3 ‡ Va _

3‡

_ Vd

Vb +

1‡

+ C

Fig. 3.

7. 30

Chapter 7 - Theorems in Circuit Analysis

To find Thevenin’s resistance Rth The voltage sources are replaced by short circuit as shown in Fig. 4. With reference to Fig. 5 we get, R th = b 3 | | , 2 l + b 3 | | , 1 l

= 3 # 2 + 3 # 1 = 1.95 Ω 3+2 3+1 Rth A

Rth

B B

A S.C.

S.C.

Þ

3W

3W 2W

3W

3W

1W 1W

2W

Fig. 4.

Fig. 5.

Thevenin’s equivalent at A-B Rth=1.95 ‡ A

Vth=3 V +E

B A

Fig. 6 : Thevenin’s equivalent.

1.95 ‡

To find IL

+ E

3V

IL 10 ‡

Connect the 10 W resistance between terminals A and B as shown in Fig. 7. With reference to Fig. 7, by Ohm’s law, IL =

B

Fig. 7.

3 = 0.251 A 1.95 + 10

(AU Dec’14, 16 Marks)

EXAMPLE 7.18 Using Thevenin’s theorem, find the current IL,

1‡

9V + E

50 V + E

IL 20 ‡

through the 20 W resistor shown in Fig. 1.

SOLUTION Let us remove the 20 W resistance and mark the resulting open terminals as A and B as shown in Fig. 2. Now, we have to determine Thevenin’s equivalent of the circuit shown in Fig. 2, with respect to terminals A and B. Let us assume Vth as shown in Fig. 2 with terminal-A as positive and terminal-B as negative.

20 V +E

10 ‡

5‡

Fig. 1.

+ 10 V E

2‡

Circuit Theory

7. 31 9V + E

1‡

9V + E

1‡

Rth 50 V + E

50 V + E

A

B

EV + th

+

+ 20 V +E

5‡

10 ‡

A

B

E Vth+

E

20 V +E

2‡

10 V

+ 10 V E

20 V 10 ‡ E

Fig. 2.

+ 5‡

E

10 V

2‡

Fig. 3.

To find Thevenin’s voltage Vth With reference to Fig. 3, using KVL, we can write, 20 + Vth = 50 + 10 ∴ Vth = 50 + 10 - 20 = 40 V To find Thevenin’s resistance Rth The voltage sources are replaced by short circuit as shown in Fig. 5. With reference to Fig. 6 we get, Rth = 0 W 1W

1W

S.C.

A S.C.

B

A Rth

S.C.

5W

10 W

A

B

S.C.

2W

Þ

Rth Rth

Þ Rth = 0 W B

Fig. 4.

Fig. 6.

Fig. 5.

Thevenin’s equivalent at A-B A

Rth = 0‡

+

Vth=40V E

B

Fig. 7 : Thevenin’s equivalent To find IL current through 20W

A

Connect the 20 W resistance between terminals A and B as shown in Fig. 8. With reference to Fig. 8, by Ohm’s law,

IL +

Vth=40V E

20 ‡

IL = 40 = 2 A 20

B

Fig. 8.

7. 32

Chapter 7 - Theorems in Circuit Analysis

Cross-Check With reference to Fig. 9, by KVL

9V

I1=1A 1 ‡

we get,

+

E

+ V E 1 IS1=1A

20 + V20 = 50 + 10

+

∴ V20 = 60 - 20 = 40 V +

` I20

V = 20 = 40 = 2 A 20 20

20 V E

B

A 20 ‡

50 V

E

20 =2A 10 +

E

I20=2A

+

V20=40V

+ 10 ‡

20 V E

10 V

10 V

2‡

10 V

E

E

IS1=1A

+

+ 5‡

10 =5A 2

IS2=8A

10 =2A 5

IS2=8A

E

I2=1A

By KVL ,

D

C

5A

Fig. 9.

V1 + 9 + V20 = 50 ∴ V1 = 50 - 9 - V20 = 50 - 9 - 40 = 1 V ` I1 =

V1 = 1 = 1A 1 1

By KCL, at node-A, IS1 + I 20 = I1 + 2

⇒

IS1 = I1 + 2 − I 20 = 1 + 2 − 2 = 1 A

⇒

IS2 = I 20 + 5 + 2 − I1 = 2 + 5 + 2 − 1 = 8 A

⇒

I 2 = 2 − IS1 = 2 − 1 = 1 A

⇒

I 2 = IS2 − 2 − 5 = 8 − 2 − 5 = 1 A

By KCL, at node-B, IS2 + I1 = I 20 + 5 + 2 By KCL, at node-C, I 2 + IS1 = 2 By KCL, at node-D, I 2 + 2 + 5 = IS2

EXAMPLE 7.19 Find Thevenin’s equivalent of the circuit shown in Fig. 1, with respect to terminals A and B.

1‡

3‡

A 2‡

SOLUTION 10 V

+ E

5A

5A E + 10 V

To find Thevenin’s voltage Vth Let us convert the 10V source in series with the 3 W resistance to an equivalent current source in parallel with 3 W resistance as shown in Fig. 2.

2‡

B

Fig. 1.

Also, the 5 A current source in parallel with the 1 W resistance is converted to an equivalent voltage source in series with the 1 W resistance as shown in Fig. 3, and the modified circuit is shown in Fig. 4.

Circuit Theory

7. 33 1W

5V

1W

3W

-+

5A

10 3 = 3.3333 A

2W

3.3333 A

Þ

+ 10 V -

3W

ß

A

3W

2W

5A

+ 10 V

1W -+

B

5 ´ 1 = 5V

Fig. 3 : Conversion of current source to voltage source.

Fig. 2 : Conversion of voltage source to current source.

Fig. 4 : Modified circuit. 5V

1W A

-+

5 + 3.3333 = 8.3333 A

The current sources in parallel in Fig. 4, can be combined as a single source as shown in Fig. 5. Also the resistances 3 W and 2 W are combined as a single resistance. The current source in Fig. 5 is converted to voltage source in Fig. 6.

2W

3´2 3+2 = 1.2 W

+ 10 V

B

With reference to Fig. 6, using KVL, we can write,

Fig. 5.

1.2I + I + 2I = 10 + 5 + 10 ⇒ 4.2I = 25

5V

` I = 25 = 5.9524 A 4.2

-+

_ 1.2 W

1.2I +

Also, Vth + 10 = 2I + -

\ Vth = 2I – 10

1W +

_ I

A +

+ 2I _

I

2W Vth + 10 V

8.3333 ´ 1.2 = 10 V

_

= 2 ´ 5.9524 – 10 = 1.9048 V

B

Fig. 6.

To find Thevenin’s resistance Rth

In the given circuit, the voltage sources are replaced by short circuit and the current sources are replaced by an open circuit as shown in Fig. 7. The Thevenin’s resistance, Rth is obtained by using network reduction technique as shown below: With reference to Fig. 11, we get, Rth = 1.0476 W 3W

1W

3W

1W A

A O.C. S.C.

2W

Þ

2W

O.C.

2W

2W

S.C. B Rth

B Rth

Fig. 7.

Fig. 8. ß 1W A

2.2 ´ 2 2.2 + 2 = 1.0476 W

1.2 + 1 = 2.2 W

2W

3´2 3+2 = 1.2 W

2W

B Rth

Þ B Rth

Fig. 11.

A

Þ

A

Fig. 10.

B Rth

Fig. 9.

7. 34

Chapter 7 - Theorems in Circuit Analysis

Thevenin’s equivalent at A-B Rth=1.0476 ‡ A

Vth=1.9048 V

+ E

B

Fig. 12 : Thevenin’s equivalent.

Alternate Method to Find Vth The voltage sources in Fig. 1, are converted to current sources as shown in Fig. 13. The node basis matrix equation is formed using the circuit of Fig. 13, as shown below. Now, Vth = V2 R V R V R S1 1 1 1 W S V1 W S 10 + 5 − 5 + + − S 3 2 1 W S 3 1 WS S WS W= S 1 1 1 S W S − + WS V 1 2 1 W S 2 W S 5−5 S T X T X T V1 1.8333 − 1 3.3333 > − 1 1.5 H > V2 H = > 0 H ∆l = ∆l2 =

V W W W W W X

1‡

V1

V2

5A

10 A 3

5A

3‡

10 2 a 5A

2‡

2‡

Vth

_ B

Fig. 13.

1.8333 − 1 = 1.8333 # 1.5 − (− 1) 2 = 1.74995 − 1 1.5 1.8333 3.3333 = 0 − (− 1) # 3.3333 = 3.3333 0 −1

` Vth = V2 =

∆l2 = 3.3333 = 1.9048 V 1.74995 ∆l 5A

EXAMPLE 7.20 Determine Thevenin’s and Norton’s equivalents at P-Q for the circuit shown in Fig. 1.

2A

SOLUTION

2V

4‡

Thevenin’s and Norton’s equivalents can be obtained by using source transformation techniques as shown below:

E+

Q

4‡ 5V

The 5 V source in series with the 2 W resistance is converted to current source as shown in Fig. 2. The 2 A source in parallel with the 4 W resistance is converted to voltage source as shown in Fig. 3. The modified circuit is shown in Fig 4.

2‡

+E

Fig. 1. 5A

-+

2W

5V

A +

2A

ß 5 = 2.5 A 2

4W

4W

ß 2 ´ 4 = 8V

2V

8V

-+

-+

2.5 A 4W 2W

-+

2W

Fig. 2.

Fig. 3.

4W P

Q

Fig. 4.

P

Circuit Theory

7. 35

The series-connected voltage sources in Fig. 4 are combined into a single source in Fig. 5. Also, the 4 W resistances in series are represented by a single equivalent resistance. 5A

5A -+

8W

10 V 2 + 8 = 10 V Q

P

Q 10 = 1.25 A 8

P

-+

1.25 A

ß

4 + 4 = 8W

8W

2.5 A 2.5 A 2W

8W

Fig. 5.

Fig. 6.

2W

Fig. 7. The 10 V voltage source in series with the 8 W resistance can be converted to current source as shown in Fig. 6. The modified circuit is shown in Fig. 7. In Fig. 7, three current sources are in parallel and they can be combined to give a single current source as shown in Fig. 8. Similarly, the resistances 8 W and 2 W in parallel are also represented by a single equivalent resistance in Fig. 8. 3.75 ´ 1.6 = 6 V 1.6 W

5 + 1.25 - 2.5 = 3.75 A

-+

Q

P 8´2 8+2

= 1.6 W

P

Q

Fig. 9.

Fig. 8. Here, Fig. 8 is Norton’s equivalent which can be transformed to Thevenin’s equivalent shown in Fig. 9, by source transformation technique. Thevenin’s and Norton’s equivalent Rth=1.6 ‡

P

Vth=6 V +E

P

In=3.75 A

Rn=1.6 ‡

Q

Fig. 10 : Thevenin’s equivalent at P-Q.

Q

Fig. 11 : Norton’s equivalent at P-Q.

EXAMPLE 7.21 Using Norton’s theorem, determine the current through an ammeter connected across A and B of the circuit shown in Fig. 1. Take resistance of the ammeter as 0.5 W.

1‡ A B

SOLUTION To find Norton’s current In The terminals A and B are shorted as shown in Fig. 2. The direction of Norton’s current is assumed such that it flows from terminal-A to terminal-B. The circuit of Fig. 2 is redrawn as shown in Fig. 3.

3‡

3‡

1‡

E + 12 V

Fig. 1.

7. 36

Chapter 7 - Theorems in Circuit Analysis

In the circuit of Fig. 3, the 3 W and 1 W resistances are in parallel and so they are represented as a single equivalent resistance as shown in Fig. 4.

1‡

In Fig. 4, two equal resistances are in series with 12 V source and so the source voltage 12 V divides equally between them. Since the voltage across the parallel resistances 1 W and 3 W is 6 V, the voltage across each resistance is also 6 V. By using Ohm’s law, the current through each resistance is calculated and marked in Fig. 3.

3‡ A In B

3‡

1‡

With reference to Fig. 3, at node A using KCL, we can write, In + 2 = 6

Fig. 2.

E + 12 V

∴ In = 6 – 2 = 4 A 6 = 6A 1

6 = 2A A 3

1W +

6 = 2A 3

+

6V

In = 4A

+

_ 6V

_ 6V

6 = 6A 1W 1

3W

+

B

3 ´1 = 0.75 W 3 +1

3 ´1 = 0.75 W 3 +1 _ + 12 = 6V 2

3W

_

+

_ 12 = 6V 2

_ 6V + 12 V

Fig. 4.

+ 12 V

Fig. 3. To find Norton’s resistance Rn The voltage source is replaced by short circuit as shown in Fig. 5. Norton’s resistance is determined by using network reduction techniques as shown below: A 1W

3W

A

0.75 W

Þ

B

Þ

Rn

Þ 0.75 W

1W 3W

1W

Rn

Rn B

S.C.

B

Fig. 5.

Fig. 6.

With reference to Fig. 8, we get, Norton’s resistance, Rn = 1.5 W Norton’s equivalent A

In=4 A

Rn=1.5 ‡

B

Fig. 9 : Norton’s equivalent at A-B.

0.75 + 0.75 = 1.5 W

A 1W

3W

A

3W

Rn B

1´ 3 = 0.75 W 1+ 3

Fig. 7.

Fig. 8.

Circuit Theory

7. 37

To find current through ammeter Connect the ammeter across terminals A and B as shown in Fig. 10. The ammeter can be represented by its internal resistance as shown in Fig. 11. In

A

4A

Þ

A

1.5 W

4A

I1

I2

1.5 W

0.5 W

B

Fig. 11.

Fig. 10.

Let, I2 be the current through ammeter. Now, by current division rule, Current through ammeter, I 2 = 4 #

1.5 = 3A 1.5 + 0.5

EXAMPLE 7.22

4‡

5‡

12 ‡

A

In the circuit shown in Fig. 1, determine the power delivered to the 15 W resistance using Norton’s theorem.

SOLUTION

12 ‡

120 V

Let us remove the 15 W resistance and determine Norton’s equivalent with respect to terminals A and B. Norton’s equivalent of the circuit of Fig. 2 is obtained by source transformation technique as shown below:

120 V

4W

5W

12 W

B

12 W

Þ

15 ‡

Fig. 1.

A

30 W

12 W

30 ‡

120 4 = 30 A

4W

5W

30 W

12 W

B

B

Fig. 2.

Fig. 3.

5W

12 W

12 W

A

5W

A

30 A

3W 4 ´ 12 = 3W 4 + 12

Þ

30 W

30 W + -

30 ´ 3 = 90 V

B

B

Fig. 4.

A

Fig. 5.

7. 38

Chapter 7 - Theorems in Circuit Analysis 5W

5W

A

3 + 12 = 15 W

90 V

90 15 = 6A

Þ

30 W + -

15 W

A

30 W

B

B

Fig. 6. 5W

6A

Fig. 7. 10 W

A

15 ´ 30 15 + 30 = 10 W

+ -

Þ

5W

A

6 ´ 10 = 60 V

B

B

Fig. 8. 10 + 5 = 15 W

Fig. 9.

A

A

60 15 = 4A

Þ

60 V +-

15 W

B

B

Fig. 10.

Fig. 11.

Here, Fig. 11 is Norton’s equivalent with respect to terminals A-B of the circuit of Fig. 2. Norton’s equivalent A

In=4 A

Rn=15 ‡

B

Fig. 12 : Norton’s equivalent.

A

To find current through 15 W resistance Connect the 15 W resistance to terminals A-B of Norton’s equivalent as shown in Fig. 13. Now, the current 4 A divides equally between parallel resistances. ` Current through 15 Ω resistance = 4 = 2 A 2 2

4A

2A

2A

15 ‡

15 ‡

B

Fig. 13. 2

Power through the 15 W resistance = Current ´ Resistance = 2 ´ 15 = 60 W

Circuit Theory

7. 39

EXAMPLE 7.23

R

I

_

+

In the network shown in Fig. 1, the resistance R is variable from zero to infinity. The current I through R can be expressed as I = a + bV where, V is the voltage across R with polarity as shown in Fig. 1, a and b are constants. Determine a and b.

V 2‡

2‡

2‡ 2‡

10 V

10 V

SOLUTION

10 V

Case i : Let R = 0 (zero)

Fig. 1.

When R = 0, the resistance can be represented as short circuit as shown in Fig. 2.

I=0

S.C. _

+

Given that, I = a + bV Since, R = 0, V is also equal to zero.

V=0

2‡

2‡

2‡ 2‡

\ a=I The voltage sources in the circuit of Fig. 2 are converted to current sources as shown in Fig. 3 and the modified circuit is shown in Fig. 4.

10 V

10 V

10 V

Fig. 2. I

A 2W

Þ

10 = 5A 2

2W 2‡

2‡

5A

5A

2‡

5A

2‡

10 V B

Fig. 3.

Fig. 4.

The current sources in parallel in Fig. 4 are combined to a single equivalent source in Fig. 5. Similarly, the resistances in parallel are combined as a single equivalent resistance. I

I + 1W

5A

1W

1W

10 A

-

I -

1W

I + I

Þ 5 ´ 1 = 5V

+ -

+ - 10 ´ 1 = 10 V

Fig. 6.

Fig. 5.

The current sources of Fig. 5, are converted to equivalent voltage sources as shown in Fig. 6. With reference to Fig. 6, by KVL, we can write, I=0 O.C.

B

5 + I + I = 10 5 = 2.5 A ∴ 2I = 10 –5 ⇒ 2I = 5 ⇒ I = 2

_ 2‡

When R = ¥, the resistance can be represented as open circuit as shown in Fig. 7.

2I1 +

10 V

E

_

2I1 2 ‡ _ I1

_

+ 2I2 _

V 2‡

+

A

Since, I = 2.5, a = 2.5 Case ii : Let R = ¥ (infinity)

+

2I2

F

D I2

10 V C

C

Fig. 7.

2‡

+

10 V

7. 40

Chapter 7 - Theorems in Circuit Analysis Given that, I = a + bV Since, R = ¥, I is equal to zero. b = −a V With reference to Fig. 7, we can write the following KVL equations. ∴ a + bV = 0

⇒

In the closed path ABCA, using KVL we can write, 2I1 + 2I1 = 10 I1 = 10 = 2.5 A 4 In the closed path DEFCD, we can write,

⇒

4I1 = 10

10 = 10 + 2I2 + 2I2 \ 4I2 = 10 – 10

⇒

4I2 = 0

⇒

I2 = 0

In the path BEDCB, V + 2I2 + 10 = 2I1 \ V = 2I1 – 2I2 – 10 = 2 ´ 2.5 – 2 ´ 0 – 10 = –5 V Since, V = − 5, b = − a = − 2.5 = 0.5 V −5

RESULT a = 2.5

;

b = 0.5

\ I = a + bV = 2.5 + 0.5 V B

EXAMPLE 7.24

500 ‡

2‡

Find the current through galvanometer shown in Fig. 1, using Thevenin’s theorem.

6‡

A

SOLUTION Let us remove the galvanometer and denote the resultant open terminals as P and Q. The source is represented as ideal source with its internal resistance (3.2 W) connected external to the source in series as shown in Fig. 1.

C G 8‡

4‡ D +E

Let us represent the circuit of Fig. 2, by Thevenin’s equivalent with respect to terminals P and Q. The polarity of Thevenin’s voltage is assumed as shown in Fig. 2, with terminal P as positive.

100 V, 3.2 ‡

Fig. 1. B

To find Thevenin’s voltage Vth The circuit of Fig. 2 is redrawn as shown in Fig. 3.

500 ‡

Since the 500 W resistance is left open, the potential at node-P will be same as that of node-B. Also the nodes Q and D are at same potential. Hence, the circuit

2‡

A

C + P Vth E QRth

of Fig. 3 is redrawn as shown in Fig. 4. Now, Vth is the voltage across B and D. Let, Is be the current supplied by the source and this current divides into I1 and I2 between two parallel paths as shown in Fig. 4. The series resistances and their parallel combination are represented as a single resistance in Fig. 5.

6‡

8‡

4‡

3.2 ‡ D +E

100 V

Fig. 2.

Circuit Theory

7. 41 2W

6W

B

I2

2W

6W

B +

500 W Vth

+ P Vth - Q

8W

Þ 4W

A

A

C

I1

_

8W D

100 V

3.2 W

Is

+-

100 V

3.2 W

+-

Fig. 3.

Fig. 4.

With reference to Fig. 5, by Ohm’s law, we can write, Is =

4W C

D

(2 + 6) ´ (8 + 4) 8 ´ 12 = = 4.8 W (2 + 6) + (8 + 4) 8 + 12

100 = 12.5 A 3.2 + 4.8

4.8 W Is

With reference to Fig. 4, by current division rule, we get, I1 = Is #

100 V

3.2 W

+-

Fig. 5.

(2 + 6) = 12.5 # 8 = 5A (8 + 4) + (2 + 6) 12 + 8

I2

` I 2 = Is − I1 = 12.5 − 5 = 7.5 A

+

2‡ 2I2

B E

+

With reference to Fig. 6, by KVL, we can write, Vth

2I2 + Vth = 8I1 \ Vth = 8I1 – 2I2

A

+ I1

= 8 ´ 5 – 2 ´ 7.5 = 25 V

8I1

E

E

8‡

D

Fig. 6. To find Thevenin’s resistance Rth The ideal voltage source of Fig. 2 is replaced by short circuit as shown in Fig. 7. The network of Fig. 7 is redrawn as shown in Fig. 8, and it is reduced to a single equivalent resistance with respect to P-Q. The step-by-step reduction of the network are shown in Figs. 9 to 14.

B 500 W

B

P

6W

500 W

2W

2W

A

6W

C P

Þ 8W

Q

Rth

A

C 3.2 W

4W 8W

4W

Rth

D 3.2 W

S.C.

Fig. 7.

D

Fig. 8.

Q

7. 42

Chapter 7 - Theorems in Circuit Analysis R12 = R1 + R 2 +

R1 R 2 = 2 + 3.2 + 2 # 3.2 = 6 Ω R3 8

R 23 = R 2 + R3 +

R 2 R3 = 3.2 + 8 + 3.2 # 8 = 24 Ω R1 2

R31 = R3 + R1 +

R3 R1 = 8 + 2 + 8 # 2 = 15 Ω R2 3.2 500 W

1

2W

R12

R1

500 W

1

P

P R12

6W

6W

6W

R2 R31

2

Þ

3.2 W 8W

R3

2

15 W

Rth

4W

R23

R31

24 W

Rth

4W R23

Q

3

Q

3

Fig. 9.

Fig. 10. 500 W

500 W

P

P 6´6 = 3W 6+6

Þ

15 W 24 ´ 4 = 3.4286 W 24 + 4

15 W

3 + 3.4286 W = 6.4286 W Rth

Rth Q

Q

Fig. 11.

Fig. 12.

500 W P

15 ´ 6.4286 15 + 6.4286 = 4.5 W

P

Þ

500 + 4.5 = 504.5 W

Rth

Rth

Q

Fig. 13. With reference to Fig. 14, Thevenin’s resistance, Rth = 504.5 W Thevenin’s equivalent at P-Q Rth=504.5 ‡ P

Vth=25 V

Q

Fig. 14.

+ E

Q

Fig. 15 : Thevenin’s equivalent at P-Q.

Circuit Theory

7. 43

To find current through galvanometer

504.5 W

504.5 W P

P IG

Connect the galvanometer across P-Q as shown in Fig. 16. The galvanometer has negligible resistance and so it can be represented as a short

IG 25 V +-

G

25 V +-

Þ

S.C.

circuit as shown in Fig. 17. Let, IG be the current through a Galvanometer. With reference to Fig. 17, by Ohm’s law, we get, IG = 25 = 0.0496 A = 49.6 # 10- 3 A = 49.6 mA 504.5

Q

Q

Fig. 16.

Fig. 17.

EXAMPLE 7.25 Determine the current through ZL in the circuit of Fig. 1, using Thevenin’s theorem.

Let us remove the load impedance and denote the resultant open terminals as A and B as shown in Fig. 2. Now, we have to determine Thevenin’s equivalent of the circuit shown in Fig. 2.

10Ð0o A

SOLUTION

IL

4W

10 W

2W

~

j6 W j8 W

+

~

-j2 W

o

10Ð0 V

-

ZL

Fig. 1.

To find Thevenin’s voltage Vth Let us convert the 10 V voltage source to current source. The modified circuit is shown in Fig. 3.

o

With reference to Fig. 3, using KCL, we can write,

10Ð0 A = 10 A

A

Vth + Vth = 10 + 10 10 + j8 4 + j6 4 + j6

4W

10 W

~

j6 W j8 W

+ 10Ð0o V =10 V

~

-

1 + 1 10 c m Vth = 10 + 10 + j8 4 + j6 4 + j6

Fig. 2.

^0.1379 − j0.1642h Vth = 10.7692 − j1.1538

Vth

B

+

10.7692 − j1.1538 0.1379 − j0.1642

= 36.4201 + j34.9992 V = 50.5111Ð43.9o V

10 A 4 C j6

~

4 C j6 Vth

~

j8 ‡

j6 ‡ _

To find Thevenin’s impedance Z th

B

Fig. 3.

The current source is represented by an open circuit and the voltage source is represented by a short circuit as shown in Fig. 4.

A 4‡ 10 ‡

With reference to Fig. 4,

j6 ‡

O.C.

Z th =

^10 + j8h # ^4 + j6h

10 + j8 + 4 + j6

A

Vth

4‡

10 ‡

10 C j8

10 A

` Vth =

Vth

A 4‡ 10 ‡ j6 ‡

O.C. j8 ‡

S.C.

= 3 + j3.5714 Ω

Zth

B

Fig. 4.

j8 ‡ S.C.

= 4.6642Ð50o W

Zth

B

Fig. 4.

7. 44

Chapter 7 - Theorems in Circuit Analysis

Thevenin’s equivalent at A-B Zth

A Vth = 50 .5111Ð43 .9oV

+ Vth +-

= 36.4201 + j34.9992V Z th = 4 .6642 Ð50o W

-

= 3 + j3.5714 W

B Zth

Fig. 5 : Thevenin’s equivalent at A-B.

A

To find current through ZL

IL

Connect the load impedance ZL across terminals A and B of Thevenin’s equivalent as shown in Fig. 6.

+ Vth

Now, by Ohm’s law,

2‡ ZL

~

_

Ej2 ‡

36.4201 + j34.9992 Vth = 3 + j3.5714 + 2 − j2 Z th + ZL = 8.6314 + j4.2872 A = 9.6375+26.4 o A

IL =

B

Fig. 6.

EXAMPLE 7.26 Find the current flowing in the 5 W resistance connected across terminals A and B of the circuit shown in Fig. 1, by Thevenin’s theorem.

+ -j10 W

j20 W

o

~

_

20Ð0 V

SOLUTION 4W

4W

Let us remove the 5 W resistance and denote the resultant open terminals as A and B as shown in Fig. 2. Now, the circuit of Fig. 2 should be replaced by Thevenin’s equivalent at terminals A and B. The polarity of Thevenin’s voltage is assumed as shown in Fig. 2 with terminal-A as positive.

5W A

B

Fig. 1.

To find Thevenin’s voltage Vth

+

With reference to Fig. 3, the voltage across the series combination of –j10 W and 4 W is 20Ð0o V. Hence, by voltage division rule,

_

4 4 − j10

B

Fig. 2. +

~

_

+ -j10 W

Va _

+

+

20Ð0o V

o

20Ð0 V

+

Vb _

_

Vth + Vd = Vb ` Vth = Vb − Vd = 2.7586 + j6.8966 − ^0.7692 − j3.8462h = 1.9894 + j10.7428 V = 10.9255+79.5 o V

_ Vth Zth

4 = 20 # 4 = 0.7692 − j3.8462 V 4 + j20 4 + j20

With reference to Fig. 3, by KVL, we can write,

4W

A

With reference to Fig. 3, the voltage across the series combination of j20 W and 4 W is 20Ð0o V. Hence, by voltage division rule, Vd = 20+0 o #

20Ð0 V

4W +

= 20 # 4 = 2.7586 + j6.8966 V 4 − j10

j20 W

o

~

_

Vb = 20+0 o #

-j10 W

_ Vd + 4W

4W +

+ 20Ð0o V V c _

_ Vth

Fig. 3.

j20 W

Circuit Theory

7. 45

To find Thevenin’s impedance Z th The voltage source 20Ð0o V in the circuit of Fig. 2 is replaced by short circuit as shown in Fig. 4, and is Z th determined by reducing the network of Fig. 4 to a single equivalent impedance across A and B as shown below: -j10 W

-j10 W

j20 W - j10 ´ 4 - j10 + 4

4W

A

B

Zth

4W

Þ

4W

4W

j20 ´ 4 j20 + 4 = 3.4483 - j1.3793 W = 3.8462 + j0.7692 W

j20 W

S.C.

Þ B

A

B

A

Fig. 5.

Fig. 4.

Fig. 6.

With reference to Fig. 6 we get,

Z th = 3.4483 − j1.3793 + 3.8462 + j0.7692 = 7.2945 − j0.6101 Ω = 7.32+ − 4.8o Ω Thevenin’s equivalent at A-B Zth

A

Vth = 1.9894 + j10.7428 V = 10.9255+79.5 o V Z th = 7.2945 − j0.6101 Ω = 7.32+ − 4.8 o Ω

+ Vth

~

_

B

Fig. 7 : Thevenin’s equivalent at A-B. To find current through 5 W resistance Connect the 5 W resistance across A and B of Thevenin’s equivalent as shown in Fig. 8. Let, IL be the current through 5 W resistance. Z th

A

Now, IL = Vth Z th + 5 1.9894 + j10.7428 = = 0.1182 + j0.8797 A 7.2945 − j0.6101 + 5 = 0.8876+82.3 o A

IL

+

5‡

Vth _

~

RESULT

B

Fig. 8.

o

Current through the 5 W resistance = 0.8876Ð82.3 A

EXAMPLE 7.27

SOLUTION Let us remove the 8 W resistance and denote the resultant open terminals as A and B as shown in Fig. 2. Now, the circuit of Fig. 2 should be replaced by Norton’s equivalent at terminals A and B.

20Ð30o A

Determine the voltage across terminals A and B in the circuit of Fig. 1 by Norton’s theorem.

8W

A

B

2W

4W

j5 W

-j12 W

~ Fig. 1.

7. 46

Chapter 7 - Theorems in Circuit Analysis

To find Norton’s current In Let us short circuit the terminals A and B in the circuit of Fig. 2 as shown in Fig. 3. The current flowing through the short circuit is Norton’s current. Let us assume the direction of Norton’s current as A to B.

2W

4W

j5 W

-j12 W

20Ð30o A

20Ð30o A

A S.C. B

B

A

~

In

2W

4W

~ j5 W

Fig. 2.

-j12 W

Fig. 3.

With reference to Fig. 3, by current division rule we can write, 2 + j5 = − 9.8632 + j6.26 A 2 + j5 + 4 − j12 = 11.6821+147.6 o A

Norton's current, In = 20+30 o #

To find Norton’s impedance Zn Let us replace the current source in Fig. 2, by an open circuit as shown in Fig. 4 and Zn is determined by reducing the network of Fig. 4 to a single equivalent impedance as shown below: Zn

2W

Zn

A

B

A

4W

B

2W

4W

j5 W

-j12 W

Þ

O.C. -j12 W

j5 W

Fig. 4.

Fig. 5.

With reference to Fig. 5, we can write, Norton's impedance, Zn = 2 + j5 − j12 + 4 = 6 − j7 Ω = 9.2195+ − 49.4 o Ω Norton’s equivalent at A-B

A

In

~

In = − 9.8632 + j6.26 A = 11.6821+147.6 o A Zn = 6 − j7 Ω = 9.2195+ − 49.4 o Ω

Zn

VL

B

Zn

To find voltage across 8 W resistance Let us connect the 8 W resistance across A and B of Norton’s equivalent as shown in Fig. 7. Let, VL be the voltage across the 8 W resistance. With reference to Fig. 7, by KCL we can write, VL + VL = I n Zn 8

⇒

VL c 1 + 1 m = In Zn 8

A VL 8

VL

Fig. 6 : Norton’s equivalent at A-B. In

~

Zn

Fig. 7.

+ VL _

8‡

B

Circuit Theory

7. 47

` VL =

− 9.8632 + j6.26 In In = = 1 1 +1 ^6 − j7h- 1 + 8- 1 Zn- + 8- 1 Zn 8

= – 31.3876 + j45.2219 V = 55.0473Ð124.8o V

RESULT Voltage across terminals A and B = 55.0473Ð124.8o V

7.4

Maximum Power Transfer Theorem

All practical sources have internal resistance or impedance. When the source delivers current to load, the current flows through internal impedance also and so a part of power is consumed by internal impedance of the source. Hence, when a load is connected to a source, the power generated by the source is shared between the internal impedance and the load. In certain applications, it is desirable to have a maximum power transfer from source to load. The maximum power transfer to load is possible only if the source and load has matched impedance. This situation arises in Electronics, Communication and Control circuits. For example, an antenna used in TV/Radio receives signal from atmosphere and the power level of the signal will be very low. This weak signal should be transferred to the input section of an amplifier to which it is connected. For good reception, the maximum power should be transferred from antenna to amplifier. This is possible only if the input impedance of amplifier is matched with antenna impedance. The sources may be dc or ac and the loads may be resistive or reactive. Hence, the matched impedance for maximum power will be different for different combinations of source and load. The following important six combinations of source and load are discussed in this book. Case i Case ii Case iii Case iv

: : : :

DC source with internal resistance connected to resistive load. AC source with internal resistance connected to resistive load. AC source with internal impedance connected to resistive load. AC source with internal impedance connected to a load with variable resistance and variable reactance. Case v : AC source with internal impedance connected to a load with variable resistance and fixed reactance. Case vi : AC source with internal impedance connected to a load with fixed resistance and variable reactance. Case i : DC source with internal resistance connected to resistive load

Theorem:

(AU Dec’14, 2 Marks)

“Maximum power is transferred from source to load, when the load resistance is equal to source resistance”. Consider a dc source of emf, E and internal resistance, Rs connected to a variable load resistance R. Now, the condition for maximum power transfer from source to load is, R = Rs

7. 48

Chapter 7 - Theorems in Circuit Analysis

The maximum power Pmax is, 2

Pmax = E 4R Proof: Consider a dc source of emf, E and internal resistance, Rs connected to a load resistance, R as shown in Fig. 7.10. Let, I be the current through the circuit. With reference to Fig. 7.10, by Ohm’s law we can write, E Rs + R

P = Power delivered to load.

Now, P = I2R

..... (7.4)

Using equation (7.3) in equation (7.4), we get, 2 E2 R P = c E m R = Rs + R R ^ s + Rh2

Rs I

R

LOAD

Let,

..... (7.3) SOURCE

I =

E +E

Fig. 7.10.

..... (7.5)

The condition for maximum power can be obtained by differentiating P with respect to R and equating (dP/dR) = 0 du = du # v − u # dv dv v2

On differentiating equation (7.5) with respect to R, we get, 2 2 2 dP = E # ^Rs + Rh − E R # 2^Rs + Rh dR ^Rs + Rh4

..... (7.6)

For (dP/dR) = 0, the numerator of equation (7.6), should be zero. \ E2(Rs + R)2 – 2E2R(Rs + R) = 0 ⇒ 2 E2 R(Rs + R) = E2 (Rs + R)2

.... (7.7)

On dividing equation (7.7) throughout by E2(Rs + R), we get, 2R = Rs + R ⇒ 2R – R = Rs

⇒

R = Rs

..... (7.8)

Equation (7.8) is the condition for maximum power transfer to load, which states that the maximum power is transferred from source to load when load resistance is equal to source resistance. On substituting R for Rs in equation (7.5), we can get an expression for maximum power. `

Maximum power, Pmax = P

Rs = R

=

E2 R ^R + Rh2

=

E2 R = E2 R = E2 4R (2R) 2 4R 2

..... (7.9)

Case ii : AC source with internal resistance connected to resistive load

Theorem: “Maximum power is transferred from source to load, when the load resistance is equal to source resistance.” Consider an ac source of emf, E and internal resistance, Rs connected to a variable load resistance, R. Now, the condition for maximum power transfer from source to load is, R = Rs

Circuit Theory

7. 49

E I = Rs + R

Rs

E

~

_

E E = Rs + R Rs + R Let, P = Power delivered to load. ` I = I =

Now, P = I 2 R =

R

I

+

LOAD

Consider an ac source of emf, E and internal resistance, Rs connected to a load resistance, R as shown in Fig. 7.11. Let, I be the current through the circuit. With reference to Fig. 7.11, by Ohm’s law we can write,

SOURCE

Proof:

Fig. 7.11.

E2 R ..... (7.10)

^Rs + Rh2

The equation (7.10) is same as that equation (7.5) and so the condition for maximum power transfer will be same as that of case (i). But here, I is rms value of current and E is rms value of source emf.

Case iii : AC source with internal impedance connected to resistive load

Theorem: “Maximum power is transferred from source to load, when the load resistance is equal to magnitude of source impedance.” Consider an ac source of emf, E and internal impedance, Zs ^ Zs = Rs + jXsh connected to a variable load resistance, R. Now, the condition for maximum power transfer from source to load is, R = Zs =

Rs2 + Xs2

Let, Z s = Rs + jXs R s2 + X s2

Fig. 7.12.

E E E = = Rs + jXs + R ^Rs + Rh + jXs Zs + R E ^Rs + Rh + jXs

=

~

_

.....(7.11)

With reference to Fig. 7.12, by Ohm’s law, we can write,

Magnitude of current, I = I =

R

I

+ E

` Magnitude of source impedance, Zs = Z s =

I =

Zs

LOAD

Consider an ac source of emf, E and internal impedance, Z s connected to a load resistance, R as shown in Fig. 7.12.

SOURCE

Proof:

..... (7.12) E ^Rs + Rh2 + X s2

..... (7.13)

Let, P = Power delivered to load. 2

Now, P = I R = I 2 R

..... (7.14)

Using equation (7.13) in equation (7.14), we can write, P =

E2 R R + ^ s Rh2 + X s2

..... (7.15)

7. 50

Chapter 7 - Theorems in Circuit Analysis The condition for maximum power can be obtained by differentiating P with respect to R and equating (dP/dR) = 0. On differentiating equation (7.15) with respect to R, we get,

du = du # v − u # dv dv v2

2 2 2 2 dP = E # 6^Rs + Rh + X s @ − E R # 2^Rs + Rh 2 2 dR 6^Rs + Rh + X s2 @

..... (7.16)

For dP = 0, the numerator of equation (7.16) should be zero. dR ` E 2 6^Rs + Rh2 + X s2 @ − E 2 R # 2^Rs + Rh

⇒

2E 2 R^Rs + Rh = E 2 6^Rs + Rh2 + X s2 @

On dividing the above equation throughout by E2 we get, 2R^Rs + Rh = ^Rs + Rh2 + X s2 2RRs + 2R 2 = R s2 + R 2 + 2RRs + X s2 2RRs + 2R 2 − R 2 − 2RRs = R s2 + X s2 R 2 = R s2 + X s2 ` R = Here,

R s2 + X s2

.....(7.17)

R s2 + X s2 = Zs = Magnitude of source impedance

The equation (7.17) is the condition for maximum power transfer. From equation (7.17) we can say that, the maximum power is transferred to load when load resistance is equal to magnitude of source impedance.

Case iv : AC source with internal impedance connected to a load with variable resistance and variable reactance

Theorem: ”Maximum power is transferred from source to load, when the load impedance is equal to complex conjugate of source impedance.” Consider an ac source of emf, E and internal impedance, Zs ^ Zs = Rs + jXsh connected to a load impedance Z, where Z = R + jX with R and X are individually variable. Now, the condition for maximum power transfer from source to load is, * Z = Zs ⇒

R + jX = ^Rs + jXsh*

⇒

R + jX = Rs − jXs

Proof : Consider an ac source of emf, E and internal impedance, Z s connected to a load impedance, Z as shown in Fig. 7.13. )

Now, I =

E E E = = Rs + jXs + R + jX ^Rs + Rh + j^Xs + X h Zs + Z

Magnitude of current, I = I = =

Zs

R

+ E

I

~

jX

_

LOAD Z = R + jX

I = Current through the circuit

SOURCE

Let, Z s = Rs + jXs ⇒ Z s = Rs − jXs

E ^Rs + Rh + j^Xs + X h

E ^Rs + Rh2 + ^Xs + X h2

Fig. 7.13. .....(7.18)

Circuit Theory

7. 51

Let, P = Power delivered to load, Now, P = I

2

R = I2 R

.....(7.19)

Note : In reactive loads, the power is consumed only by resistance and the active power in the reactance is zero. Using equation (7.18) in equation (7.19), we can write, P =

.....(7.20)

E2 R ^Rs + Rh2 + ^Xs + X h2

The condition for maximum power can be obtained by partially differentiating P with respect to X and then with respect to R and equating (∂P/∂X) = 0 and (∂P/∂R) = 0. On partially differentiating equation (7.20) with respect to X, we get, 2 2 2 2P = 0 # [^Rs + Rh + ^Xs + X h ] − E R # 2^Xs + X h 2 2X [^Rs + Rh + ^Xs + X h2] 2

=

du = du # v − u # dv dv v2

− 2E 2 R^Xs + X h [^Rs + Rh2 + ^Xs + X h2] 2

For 2P = 0 , the numerator of equation (7.21) should be zero. 2X ∴ -2E2 R (X s + X) = 0

.....(7.21)

..... (7.22)

In equation (7.22), E ≠ 0 and R ≠ 0, hence, ..... (7.23)

Xs + X = 0 ∴

X = -Xs

..... (7.24)

On partially differentiating equation (7.20) with respect to R, we get, 2 2 2 2 2P = E # [^Rs + Rh + ^Xs + X h ] − E R # 2^Rs + Rh 2R [^Rs + Rh2 + ^Xs + X h2] 2

..... (7.25)

For 2P = 0, the numerator of equation (7.25) should be zero. 2R ∴ E2 [(R s + R) 2 + (X s + X) 2 ] - 2E2 R(R s + R) = 0 2E2 R(R s + R) = E2 [(R s + R)2 + (X s + X) 2 ] On dividing throughout by E2 we get, 2 R (R s + R) = (R s + R)2 + (X s + X) 2

..... (7.26)

From equation (7.23) we know that Xs + X = 0, hence equation (7.26) can be written as, 2 R (Rs + R) = (R s + R)2 ⇒ 2 R = R s + R ⇒ 2 R-R = R s ⇒ R = R s

.....(7.27)

The equations (7.24) and (7.27) are the conditions for maximum power transfer. From these two equations, for maximum power transfer we can say that, R + jX = Rs - jXs. *

Here, Rs − jXs = Z s = Conjugate of source impedance. Hence, for maximum power transfer the load impedance should be equal to conjugate of source impedance. An interesting observation is that, when maximum power transfer condition is met, the circuit will behave as purely resistive circuit and so the circuit will be in resonance.

7. 52

Chapter 7 - Theorems in Circuit Analysis

Case v :AC source with internal impedance connected to a load with variable resistance and fixed reactance

Theorem: “Maximum power is transferred from source to load when load resistance is equal to absolute value of the rest of the impedence of the circuit”. Consider an ac source of emf, E and internal impedance, Zs ^ Zs = Rs + jXsh connected to a load impedance Z, where Z = R + jX with variable R and fixed X. Now, the condition for maximum power transfer from source to load is, R =

R2s + ^ Xs + Xh2

Proof : The statement of case (v) can be proved by proceeding similar to that of case (iv) and differentiating equation (7.20) with respect to R and equating (dP/dR) = 0. From equation (7.26) we get, 2 R (R s + R) = (Rs + R)2 + (X s + X) 2 2RRs + 2R 2 = R s2 + R 2 + 2RRs + ^Xs + X h 2 2RRs + 2R 2 − R 2 − 2RRs = R s2 + ^Xs + X h 2 ⇒ ` R =

R 2 = R s2 + ^Xs + X h 2

R s2 + ^Xs + X h 2

.....(7.28)

The equation (7.28) is the condition for maximum power transfer in case (v).

Case vi : AC source with internal impedance connected to a load with fixed resistance and variable reactance

Theorem: “Maximum power is transferred from source to load when load reactance is equal to conjugate of source reactance”. Consider an ac source of emf, E and internal impedance, Zs ^ Zs = Rs + jXsh connected to a load impedance Z, where Z = R + jX with fixed R and variable X. Now, the condition for maximum power transfer is, jX = –jXs Proof : The statement of case (vi) can be proved by proceeding similar to that of case (iv) and differentiating equation (7.20) with respect to X and equating (dP/dX) = 0. From equation (7.24) we get, X = - Xs The equation (7.29) is the condition for maximum power transfer in case (vi).

..... (7.29)

Circuit Theory

7. 53

TABLE - 7.1 Summary of Conditions for Maximum Power Transfer Case

Source emf

Source impedance

Load impedance

Variable element of

Condition for maximum

load impedance

power transfer

i

dc

Rs

R

R

R = Rs

ii

ac

Rs

R

R

R = Rs

iii

ac

Rs + jXs

R

R

iv

ac

Rs + jXs

R + jX

R, X

v

ac

Rs + jXs

R + jX

R

vi

ac

Rs + jXs

R + jX

X

R2s + X2s

R =

R + jX = Rs − jXs R2s + ^ Xs + Xh2

R =

jX = − jXs

Applying maximum power transfer theorem to circuits Generally it is desirable to have maximum power transfer to a particular element of a circuit. In this case remove the concerned element and create two open terminals. Then the circuit is represented by Thevenin’s equivalent with respect to open terminals. Now, Thevenin’s equivalent is considered as voltage source for load and apply maximum power transfer theorem. A A

A

A

Rs

R th = R s

Circuit with dc source and resistances

R

Circuit

Þ

Þ

R

Þ + _ Vth = E + _

B

B

+ _ E + _ B

B

Fig. a : Circuit with dc source and resistances. A Circuit with ac source and resistances and reactances

A

A

A Z th = Z s Z = R + jX

Þ

Circuit

Zs

Þ

Þ

Z = R + jX

+ B

B

Vth = E

+ E

~

_

~

_

B

B

Fig. b : Circuit with ac source and resistances and reactances. Fig. 7.14 : Applying maximum power transfer theorem to an element of a circuit.

Sometimes, it is desirable to have maximum power transfer to load by varying some parameter of a circuit. In this case, determine an expression for power, P delivered to load by relating the variable parameter to P. Let, Y be the variable parameter. Now differentiate P with respect to Y to get dP . Then form an equation by equating dP = 0 and solve the equation to get the condition dY dY for maximum power transfer. [Refer Examples 7.38 and 7.39.]

7. 54

Chapter 7 - Theorems in Circuit Analysis 15 ‡

EXAMPLE 7.28 In the circuit of Fig. 1, find the value of adjustable resistor R for maximum power transfer to R. Also, calculate the maximum power.

20 ‡

+ 100 V E

10 ‡

R

SOLUTION Let us remove the adjustable resistance R and denote the two open terminals by A and B, as shown in Fig. 2. Now, the circuit of Fig. 2, should be

Fig. 1. 20 ‡

15 ‡

A +

replaced by Thevenin’s equivalent. Let us assume the polarity of Vth as shown in Fig. 2. + 100 V E

To find Thevenin’s voltage Vth

Vth

10 ‡

Rth

In Fig. 3, the 20 Ω resistance is open and so no current will flow through it. Hence the voltage across the 20 Ω resistance is zero.

Fig. 2.

With reference to Fig. 3, by voltage division rule, we can write,

15 ‡

Vth

To find Thevenin’s resistance Rth

A no voltage +

+ Vth 10 ‡ E

100 V +E

The 100 V voltage source in the circuit of Fig. 2 is replaced by short circuit and resulting network is reduced to a single equivalent resistance as shown below:

S.C.

20 ‡

= 100 # 10 = 1000 = 40 V 15 + 10 25

20 W

15 W

20 W

A

10 W

Þ

Vth

E

Fig. 3.

A

B

10 ´ 15 = 6W 10 + 15

Rth

Rth

Fig. 5. B

B

Fig. 4.

E B

Rth = Rs = 26 ‡ A

Rth = 20 + 6 = 26 Ω To find R for maximum power and Pmax Thevenin’s equivalent of Fig. 2 is shown in Fig. 6. Now, Thevenin’s equivalent is the voltage generator for load resistance R.

Vth = E = 40 V

With reference to Fig. 5,

+ E

Rth B

∴ Vth = E

;

R th = Rs

Fig. 6.

Let us connect the adjustable resistance R across A and B of Thevenin’s equivalent as shown in Fig. 7.

A

E = 40 V

With reference to Fig. 7, by maximum power transfer theorem, for maximum power in R the value of R should be equal to Rs.

Rs = 26 ‡

+ E

I

R

∴ R = 26 Ω B

Fig. 7.

Circuit Theory

7. 55

2 We know that, maximum power, Pmax = E 4R 2 = 40 = 15.3846 W 4 # 26

Also, Pmax = I 2 R = c

2 40 m # 26 = 15.3846 W 26 + 26

RESULT The value of R for maximum power transfer = 26 Ω The maximum power in R, Pmax = 15.3846 W

EXAMPLE 7.29

10 ‡

15 ‡

In the circuit of Fig. 1, find the value of R for maximum power transfer. Also, calculate the maximum power. 12 V +E

SOLUTION Let us remove the resistance R and denote the two open terminals by A and B, as shown in Fig. 2. Now, the circuit of Fig. 2, should be

Fig. 1. 10 ‡

15 ‡

replaced by Thevenin’s equivalent. Let us assume the polarity of Vth as shown in Fig. 2. To find Thevenin’s voltage Vth

2A

R

+ A Vth _ B

12 V +E

2A

With reference to Fig. 3, by KVL, we can write,

Fig. 2.

Vth = 2 × 15 + 12 ∴ Vth = 42 V To find Thevenin’s resistance Rth Let us replace the voltage source by short circuit and current source by open circuit as shown in Fig. 4.

10 W 15 W 2 A _ + 2 ´ 15 + A Vth 12 V +_ B

2A

2A

10 W 15 W

Fig. 3.

15 W

10 W

A

A S.C.

O.C. B

Þ

Rth

Rth B

Fig. 4.

Fig. 5.

With reference to Fig. 5, R th = 15 Ω

To find R for maximum power and Pmax Thevenin’s equivalent of Fig. 2 is shown in Fig. 6. Now, the Thevenin’s equivalent is the voltage generator for load resistance R. ∴ Vth = E

;

R th = Rs

Vth = E = 42 V

Rth = Rs = 15 ‡ A

+ E

B

Fig. 6.

7. 56

Chapter 7 - Theorems in Circuit Analysis Rs = 15 ‡

Let us connect the resistance, R across A and B of Thevenin’s equivalent as shown in Fig. 7. power transfer to R the value of R should be equal to Rs.

E = 42 V

With reference to Fig. 7, by maximum power transfer theorem, for maximum

A

+ E

R

∴ R = 15 Ω 2 2 Maximum power, Pmax = E = 42 = 29.4 W 4R 4 # 15

Fig. 7.

B

EXAMPLE 7.30

2‡

Determine the value of resistance that may be connected across terminals A and B so that maximum power is transferred from the circuit to the resistance. Also, estimate the maximum power transferred to the resistance.

5V

4‡

A

E+

10 ‡

8‡

20 V

SOLUTION The circuit of Fig. 1 should be replaced by Thevenin’s equivalent as shown below:

B

Fig. 1.

To find Thevenin’s voltage Vth

2‡

5V

4‡

> >

2+8 −8 H − 8 8 + 4 + 10 10 − 8 H − 8 22

∆ =

I1 20 > H = > H I2 0

20 V

I1

8‡

I2

+ 10I2 10 ‡

Vth

E

I1 20 > H = > H I2 0

_ B

Fig. 2.

10 − 8 = 10 # 22 − (− 8) 2 = 156 − 8 22

∆2 =

A +

E+

With reference to Fig. 2, the mesh basis matrix equation is,

10 20 = 10 # 0 − (− 8) # 20 = 160 −8 0

` I2 =

∆2 = 160 = 1.0256 A ∆ 156

With reference to Fig. 2 by KVL, we can write, Vth = 5 + 10 I 2 = 5 + 10 × 1.0256 ∴ Vth = 15.256 V To find Thevenin’s resistance Rth Let us replace the voltage sources by short circuit and reduce the resulting network to a single equivalent resistance as shown below:

S.C.

8W

S.C.

4W

A

10 W

Þ Rth

Fig. 3.

B

2´8 2+8 = 1.6 W

A

A

10 W

Þ Rth

Fig. 4.

B

4 + 1.6 = 5.6 W

4W

2W

10 W Rth

Fig. 5.

B

Circuit Theory

7. 57 Rth = Rs = 3.5897 ‡ A

With reference to Fig. 5, Vth = E = 15.256 V

R th = 5.6 # 10 = 3.5897 Ω 5.6 + 10

To find the value of R for maximum power transfer and Pmax Thevenin’s equivalent of the given circuit is shown in Fig. 6. Now, the Thevenin’s equivalent is the voltage generator for the load that may be connected across A and B. ;

Fig. 6.

B

Rs = R th

A

Let us connect a load resistance R across A and B of Thevenin’s equivalent as shown in Fig. 7. Now, for maximum power transfer the value of R should be equal to Rs. ∴ R = 3.5897 Ω

Rs = 3.5897 ‡

E = 15.256 V

E = Vth

∴

+ E

+ E

R

B

2 Maximum power, Pmax = E 4R 2 = 15.256 = 16.2093 W 4 # 3.5897

Fig. 7.

EXAMPLE 7.31 4‡

Determine the value of R for maximum power

4‡

4‡

2‡

transfer to it and the maximum power.

SOLUTION

12 V +E

4‡

R

The given circuit can be reduced to a single voltage source with a resistance in series with respect to terminals of load resistance R, by source transformation technique as shown below: 4W

4W

+

2W

R

-

4W

4W

]

12 V

+

4W

Fig. 1.

2W

10 V

4W

-

Þ

+ E 10 V

2‡

12 4 = 3A

4W

2W R 2W

4W

Fig. 3.

Fig. 2.

ß

1W R + -

+ - 5 ´ 1 = 5V

3 ´ 2 = 6V

Fig. 5. ß

Þ

2W

4W

4W

4W

4W

3A

4 2 = 2W

R

Fig. 4.

2 2 = 1W

5A

10 2 = 5A

7. 58

Chapter 7 - Theorems in Circuit Analysis 4 + 2 = 6W

4 + 1 = 5W

Þ 6 V +-

6 = 1A 6

+ - 5V

R

6W

Fig. 6.

5 = 1A 5

5W

R

Fig. 7. ß

2.7273 W

Þ

R

5´6 5+6 = 2.7273 W

1A

Þ

2 ´ 2.7273 + E = 5.4546 V -

1A + 1A = 2 A

A Rs

R

6W

1A

5W

R

B

Fig. 10.

Fig. 9.

Fig. 8.

In Fig. 10, the given circuit has been reduced to the form of a voltage generator with respect to terminals of R. With reference to Fig. 10, The value of R for maximum power transfer = R s = 2.7273 Ω The maximum power, Pmax =

E 2 = 5.4546 2 = 2.7273 W 4#R 4 # 2.7273

EXAMPLE 7.32

10 ‡

Determine the value of R in the circuit of Fig. 1 for maximum power transfer to R from the rest of the circuit. 4‡

3‡

SOLUTION The given circuit should be replaced by Thevenin’s equivalent with respect to terminals of R as shown below:

2‡

+ 20 V E

R

Now, for maximum power transfer to R, the value of R should be equal to Rth. Hence, we have to find the value of Thevenin’s resistance Rth.

Fig. 1. 10 W

10 W Rth

Rth A

3W

3W

4W

Þ

A 2W

R B

Fig. 2.

+ 20 V -

A

4W A

Þ

Vth +-

Þ

Vth +-

R

+ 20 V -

2W

B

B

Fig. 3.

Fig. 4.

B

Fig. 5.

Circuit Theory

7. 59

To find Thevenin’s resistance R th Let us replace the voltage source by short circuit and remove the resistance R and denote the resulting open terminals as A and B as shown in Fig. 6. The network of Fig. 6 is reduced to a single equivalent resistance as shown below: 10 W 3W

3W

4W

A

4W

10 W

Þ

A 2W

Rth

2W

S.C. Rth B

B

Fig. 6.

Fig. 7. ß A

Þ

3 + 1.6667 = 4.6667 W

3W

A

4W

4W

3W

4W

Þ

A

10 W Rth

Rth

B

Fig. 10.

Rth

B 10 ´ 2 = 1.6667 W 10 + 2

2W B

Fig. 9.

Fig. 8.

With reference to Fig. 10, we can write, R th = 4.6667 # 4 = 2.1539 Ω 4.6667 + 4

RESULT The value of R for maximum power transfer = 2.1539 Ω

EXAMPLE 7.33

5‡

Determine the value of R in the circuit of Fig. 1 for maximum power transfer to R from the rest of the circuit.

10 ‡

+ 20 V

SOLUTION The given circuit should be replaced by Thevenin’s equivalent with respect to terminals of R as shown below:

12 A

5‡

E

5‡

R

4‡

Now, for maximum power transfer to R, the value of R should 8‡

be equal to Rth. Hence, we have to find the value of Thevenin’s resistance Rth.

Fig. 1.

7. 60

Chapter 7 - Theorems in Circuit Analysis 5W

5W 10 W

10 W

A

-

+

5W

12 A

A

-

+

20 V

20 V

5W

Þ

R

5W

12 A

4W

5W

4W 8W

8W B

B

Fig. 2.

Fig. 3. ß A

A Rth

Rth

Þ

Vth +-

R

Vth +-

B

B

Fig. 4.

Fig. 5. To find Thevenin’s resistance R th

Let us replace the voltage source by short circuit and current source by open circuit as shown in Fig. 6. Remove the resistance R and denote the resulting open terminals as A and B as shown in Fig. 6. Now, the network of Fig. 6 is reduced to a single equivalent resistance as shown below: 5W

10 W

A

A

A S.C. 5W 5W

5W

5 = 2.5 W 2

5W

O.C.

Þ

4W

Rth

Þ Rth

4W

8W

8W

8W

B

B

B

Fig. 6.

Rth

4W

Fig. 7.

Fig. 8.

With reference to Fig. 8,

RESULT

R th = 2.5 + 4 + 8 = 14.5 Ω

The value of R for maximum power transfer = 14.5 Ω 10 W

EXAMPLE 7.34 Determine the load impedance that can be connected across terminals A and B for maximum power transfer to load impedance. Also, calculate the maximum power transferred to load.

A

o

5Ð0 A

j2 W

4W

2W

j4 W

~ B

Fig. 1.

Circuit Theory

7. 61

SOLUTION Let us convert the given circuit to Thevenin’s equivalent with respect to terminals A and B by using source transformation technique as shown below: 2 + j2 W

10 W

10 W

A

5Ð0 A o

j2 W

4W

2W

j4 W

A Convert current source to voltage source

Þ

~

4W

+ o 5Ð0 ´ (2 + j2) = 5 ´ (2 + j2) = 10 + j10 V -

~ j4 W

B

B

Fig. 3.

Fig. 2.

Combine series impedance

ß

12 + j2 W A 4W 12 + j2 W

~

Þ

10 + j10 12 + j2

j4 W

= 0.9459 + j0.6757 A

A Convert voltage source to current source

4W

+ 10 + j10 V

~

-

j4 W

B

B

Fig. 4.

Fig. 5. ß

Combine parallel impedance 3.3425 + j2.2466 W A

A

0.9459 + j0.6757 A

(12 + j2) ´ (4 + j4) 12 + j2 + 4 + j4 = 3.3425 + j2.2466 W

~

Convert current source to voltage source

Þ

(0.9459 + j0.6757) ´ (3.3425 + j2.2466) = 1.6436 + j4.3836 V

Zth = Zs

+

~

-

Vth = E

B

B

Fig. 6.

Fig. 7. Zs

Now, the circuit of Fig. 7 is Thevenin’s equivalent of the given circuit with respect to terminals A and B. Thevenin’s source is the voltage source for load connected across terminals A and B and Thevenin’s impedance is the source impedance. `

E = 1.6436 + j4.3836 V Zs = 3.3425 + j2.2466 Ω

A

+ E

~

Let us connect a load impedance, Z across terminals A and B of Thevenin’s equivalent as shown in Fig. 8. Now, for maximum power transfer, *

I

Z

E

* Z = _Zs i = _3.3425 + j2.2466i Ω = 3.3425 − j2.2466 Ω = 4.0273+ − 33.9 o Ω

B

Fig. 8.

7. 62

Chapter 7 - Theorems in Circuit Analysis With reference to Fig. 8, we can write, I =

1.6436 + j4.3836 E = _3.3425 + j2.2466 i + _3.3425 − j2.2466i Zs + Z =

1.6436 + j4.3836 = 0.2459 + j0.6557 = 0.7003+69.4 o A 2 # 3.3425

Maximum power delivered to load, Pmax = I

2

# Real part of Z

= 0.70032 # 3.3425 = 1.6392 W

RESULT Load impedance across A and B for maximum power transfer = 3.3425 − j2.2466 Ω = 4.0273∠-33.9o W Maximum power transferred to load = 1.6392 W

EXAMPLE 7.35 In the circuit of Fig. 1, determine the value of R so that maximum power is transferred to it.

SOLUTION

100Ð45o V

2W

j2 W

4W

+ R

-j10 W

j4 W

~

-

Let us remove the resistance R and denote the resulting open terminals by A and B as shown in Fig. 2. The circuit of Fig. 2

Fig. 1.

should be converted to Thevenin’s equivalent with respect to

terminals A and B as shown in Fig. 3. Let us connect the load resistance R across A and B as shown in Fig. 4. Now, by maximum power transfer theorem, the value of R should be equal to magnitude of source impedance Zs , for maximum power transfer. Here, Zs = Z th . 2W

A

100Ð45o V

Zth = Zs

Zth

j2 W

4W

A

+

A

+

-j10 W

j4 W

~

Þ

-

~

+

Þ

Vth

-

~

R

Vth = E

-

B

B

Fig. 2.

B R = Zth = Zs

Fig. 3.

Fig. 4. To find Z th and the value of R for maximum power Let us replace the voltage source in the circuit of Fig. 2 by short circuit as shown in Fig. 5. The network of Fig. 5 is reduced to a single equivalent impedance as shown below: 2W

4W

4W

j2 W

j2 W

S.C.

j4 W

-j10 W

Þ Zth

2 ´ j4 2 + j4 = 1.6 + j0.8

-j10 W

B

Fig. 5.

A

A

A

Fig. 6.

Þ

(4 + j2) + (1.6 + j0.8) = 5.6 + j2.8

-j10 W

Z th

Zth

B

B

Fig. 7.

Circuit Theory

7. 63

With reference to Fig. 7, we can write, Z th =

_5.6 + j2.8 i # _− j10 i _5.6 + j2.8 i + _− j10 i

= 6.7308 − j1.3462 Ω = 6.8641+ − 11.3 o Ω Now, by maximum power transfer theorem, R = Z th = 6.8641 Ω

RESULT The value of R for maximum power transfer = 6.8641 Ω

EXAMPLE 7.36 Z = 2 + jX

4W

4W

value of the reactance jX for maximum power transfer to load.

+

50Ð0o V

resistance of 2 Ω and a variable reactance, jX. Determine the

+ j6 W

~

-j4 W

-

SOLUTION

~

-

50Ð90o V

In the circuit of Fig. 1, the load impedance, Z has a fixed

Let us remove the load impedance Z and denote the

Fig. 1.

resulting open terminals by A and B as shown in Fig. 2. The circuit of Fig. 2 should be reduced to Thevenin’s equivalent with respect to terminals A and B as shown in Fig. 3. Let us connect the load impedance Z across terminals A and B as shown in Fig. 4.

A

B

+

~

Zth = Rs + jX s

4W

+ j6 W

-

-j4 W

Rs + jXs A

~

-

50Ð90o V

50Ð0o V

4W

A

+

Þ

Vth

+

Þ

~

-

Vth

Z = 2 + jX

~

-

B

B

Fig. 2.

Fig. 3.

Fig. 4.

Now, by maximum power transfer theorem, for maximum power transfer to load, jX = − jX s.

To find Z th and the value of jX for maximum power Let us replace the voltage sources in the circuit of Fig. 2 by short circuit as shown in Fig. 5. The network of Fig. 5 is reduced to a single equivalent impedance as shown below:

7. 64

Chapter 7 - Theorems in Circuit Analysis Zth

4W

A

S.C.

Zth

4W

B

A

S.C.

-j4 W

j6 W

Þ

B

4 ´ j6 4 + j6 = 2.7692 + j1.8462 W

4 ´ (- j4) = 2 - j2 W 4 + (- j4)

Fig. 6.

Fig. 5. With reference to Fig. 6, we can write, Z th = _2.7692 + j1.8462 i + _2 − j2 i = 4.7692 − j0.1538 Ω Here, Z th = Rs + jXs = 4.7692 − j0.1538 ` jXs = − j0.1538 For maximum power transfer to load, jX = − jX s = − (− j0.1538) = +j0.1538 Ω ∴ jX = j0.1538 Ω

RESULT For maximum power transfer to load, the value of jX = j0.1583 Ω

In the circuit of Fig. 1, determine the impedance that can be connected across terminals A and B for maximum power transfer. Also, estimate the maximum power.

10Ð45o A

EXAMPLE 7.37 j2 W

4W

~

4W A

SOLUTION

j2 W

2W

Let us determine Thevenin’s equivalent of the given circuit with respect to terminals A and B as shown below:

B

Fig. 1.

To find Thevenin’s voltage Vth

In the given circuit, 4 Ω resistance is open and so no current will flow through it. Hence, there is no voltage across the 4 Ω resistance. The given circuit is redrawn as shown in Fig. 3. I2 I2

I1

10Ð45o A

~

I2

4W

2W

A + + No voltage Vth j2 W V

10Ð45o A

j2 W

~

th

-

+ Vth _

_

2W

Fig. 3.

A + Vth

_ B

I2

B

Fig. 2.

j2 W 4W

4W

j2 W

4W

Circuit Theory

7. 65

With reference to Fig. 3, by current division rule, o 4 40+45 o = 40+45 = 6 + j4 4 + _2 + j2 + j2 i 7.2111+33.7 o = 5.547+11.3 o A

I2 = 10+45 o #

Vth = I 2 # j2 = 5.547+11.3 o # j2 = 5.547+11.3 o # 2+90 o = 11.094+101.3 o V

To find Thevenin’s impedance Z th Let us replace the current source in the given circuit by an open circuit as shown in Fig. 4. The network of Fig. 4 is reduced to a single equivalent impedance as shown below:

O.C.

j2 W

4W

4W

4W

j2 W

4W

A

4W A

Þ

A

Þ j2 W

6 + j2 j2 W

2W

j2 W

2W

Z th Z th

Z th

B

B

B

Fig. 4.

Fig. 6.

Fig. 5.

With reference to Fig. 6, we can write, _6 + j2 i # j2 H + 4 = 70.4615 + j1.6923 A + 4 _6 + j2 i + j2 = 4.4615 + j1.6923 Ω

Z th = >

To find Z for maximum power and Pmax

Zth a Zs

A

Thevenin’s equivalent of the given circuit is shown in Fig. 7. Now, Thevenin’s equivalent is the voltage generator for the load that may be connected across A and B.

+ Vth a E

o

` E = Vth = 11.094+101.3 V

~

E

Zs = Z th = 4.4615 + j1.6923 Ω

B

Fig. 7.

Let us connect a load impedance Z across A and B of Thevenin’s equivalent as shown in Fig. 8. Now, for maximum power transfer to load, the load impedance Z should be equal to complex conjugate of source impedance Zs . *

Zs

*

` Z = Zs = _4.4615 + j1.6923 i = 4.4615 − j1.6923 Ω

A

With reference to Fig. 8, we can write,

+ o

I =

E 11.094+101.3 = 4.4615 + j1.6923 + 4.4615 − j1.6923 Zs + Z o = 11.094+101.3 = 1.2433+101.3 o A 2 # 4.4615

E

~

I

Z

E

B

Fig. 8.

7. 66

Chapter 7 - Theorems in Circuit Analysis 2

Maximum power in the load, Pmax = I # Real part of Z = 1.24332 # 4.4615 = 6.8966 W

RESULT The load impedance for maximum power transfer, Z = 4.4615 − j1.6923 Ω Maximum power transferred to load, Pmax = 6.8966 W

EXAMPLE 7.38

10Ð45 A

of R for maximum power transfer.

o

+

o

In the circuit of Fig. 1, the load impedance, Z has a fixed reactance of +j2 Ω and a variable resistance, R. Determine the value

20Ð30 V

10 W

~

4W

Z

-

~

SOLUTION Fig. 1.

The 20∠30o V voltage source in series with 10 Ω is converted

The circuit of Fig. 2 is redrawn as shown in Fig. 3. In this circuit the current sources in parallel

o

10Ð45 A

to an equivalent current source in Fig. 2. 20Ð30o 10

10 W

~

4W

Z

o

= 2Ð30 A

~

can be combined to give single equivalent current source and the 10 Ω and 4 Ω resistance in parallel

Fig. 2.

can be combined to give a single equivalent

ß

resistance as shown in Fig. 4.

o

10Ð45 A

source in Fig. 5.

o

resistance in the circuit of Fig. 4, is converted to voltage

2Ð30 A

The current source in parallel with 2.8571 Ω

~

10 W

~

4W

With reference to Fig. 5, we can write,

Fig. 3.

o I = 34.1229+42.5 2.8571 + R + j2

ß

2.8571W + o

~

I

Z = R + j2

Þ

11.9431Ð42.5 ´ 2.8571 = 34.1226Ð42.5o V

2Ð30o + 10Ð45o = 11.9431Ð42.5o A

-

Fig. 5. o ` I = I = 34.1226+42.5 = 2.8571 + R + j2

~

10 ´ 4 10 + 4 = 2.8571W

Fig. 4.

34.1226 = 2 _2.8571 + R i + 22

34.1226 2 _2.8571 + R i + 4

Z

Z

Circuit Theory

7. 67

Let, P = Power delivered to load. Now, P = I

2

# Real part of Z

34.1226 2 1164 R #R = 2 2 _2.8571 + R i + 4 _2.8571 + R i + 4

=

The condition for maximum power can be obtained by differentiating P with respect to R and equating (dP/dR) = 0.

9_2.8571 + Ri2 + 4 C # 1164 − 1164 R # 2 _2.8571 + Ri ` dP = 2 dR 9_2.8571 + Ri2 + 4 C

d (uv) = v du −2 u dv v

For dP = 0, the numerator of dP should be equal to zero. dR dR 2

∴ 1164[ (2.8571 + R) + 4] − 2 × 1164 R (2.8571 + R) = 0 2

2 × 1164 R (2.8571 + R) = 1164[ (2.8571 + R) + 4] On dividing either side by 1164, we get, 2

2 R (2.8571 + R) = (2.8571 + R) + 4 2

2

5.7142 R + 2R = 8.163 + 5.7142 R + R + 4 2

2

∴ 5.7142 R + 2R − 5.7142 R − R = 8.163 + 4 ∴ R2 = 12.163

⇒ R = 12.163 = 3.4875 Ω

Note : Here, the value of R for maximum power transfer is also given by, R =

2.85712 + 22 = 3.4875 Ω

RESULT The value of R for maximum power transfer = 3.4875 Ω

EXAMPLE 7.39 In the circuit of Fig. 1, the phase angle θ of the voltage source, 5∠θo V is continuously variable. Find the value of θ for maximum power transfer to 10 Ω resistance.

j8 W

-j8 W

+

+ o

5Ðq V

10 W

~

-

SOLUTION

5W

Let us assume two mesh currents as shown in Fig. 2.

o

10Ð0 V -

~

5W

Fig. 1.

Let, 5∠θo V = 5 cosθ + j5 sinθ V 10∠0 o V = 10 cos 0 o + j10 sin 0 o = 10 V With reference to Fig. 2, the mesh basis matrix equation is,

-j8 W

∆ =

15 − j8 10 10 15 + j8

j8 W +

+ o

5Ðq V

10 W

~

-

15 − j8 10 5 cos θ + j5 sin θ I1 H > H = > H > 10 15 + j8 10 I2

I

~

I1

I2

5W

5W

Fig. 2. = _15 − j8 i # _15 + j8 i − 10 # 10 = 15 2 + 8 2 − 100 = 189

-

o

10Ð0 V = 10V

7. 68

Chapter 7 - Theorems in Circuit Analysis ∆1 =

5 cos θ + j5 sin θ 10

∆2 =

15 − j8 10

Now, I1 =

∆1 ∆

;

10 = ^5 cos θ + j5 sin θh # ^15 + j8h − 10 # 10 15 + j8 = 75 cos θ + j40 cos θ + j75 sin θ − 40 sin θ − 100 = ^75 cos θ − 40 sin θ − 100h + j^40 cos θ + 75 sin θh

5 cos θ + j5 sin θ = _15 − j8 i # 10 − _5 cos θ + j5 sin θ i # 10 10 = 150 − j80 − 50 cos θ − j50 sin θ = _150 − 50 cos θ i + j _− 50 sin θ − 80 i I2 =

∆2 ∆

Let, I = current through 10 Ω resistance. With reference to Fig. 2, by KCL we can write, I = I1 + I 2 =

∆1 ∆ + 2 = 1 7 ∆1 + ∆2 A ∆ ∆ ∆

=

1 7_75 cos θ − 40 sin θ − 100 i + j _40 cos θ + 75 sin θ i + _150 − 50 cos θ i + j _− 50 sin θ − 80 iA 189

=

1 7_25 cos θ − 40 sin θ + 50 i + j _40 cos θ + 25 sin θ − 80 iA 189

= 25 7_cos θ − 1.6 sin θ + 2i + j _1.6 cos θ + sin θ − 3.2iA 189 ` I = I

= 25 189

2

2

_cos θ − 1.6 sin θ + 2 i + _1.6 cos θ + sin θ − 3.2 i

Let, P = Power delivered to 10 Ω resistance. 2 9_cos θ − 1.6 sin θ + 2i2 + _1.6 cos θ + sin θ − 3.2i2 C Now, P = I2 # 10 = 10 # 25 2 189 2 2 = 0.175 9_cos θ − 1.6 sin θ + 2i + _1.6 cos θ + sin θ − 3.2i C

The condition for maximum power transfer can be obtained by differentiating P with respect to θ and solving the equation obtained by equating (dP/dθ) = 0. ` dP = 0.175 72 _cos θ − 1.6 sin θ + 2i # _− sin θ − 1.6 cos θ i + 2 _1.6 cos θ + sin θ − 3.2 i # _− 1.6 sin θ + cos θ iA dθ = 0.175 # 2 [− cos θ sin θ − 1.6 cos2 θ + 1.6 sin2 θ + 2.56 cos θ sin θ − 2 sin θ − 3.2 cos θ − 2.56 cos θ sin θ + 1.6 cos 2 θ − 1.6 sin 2 θ + cos θ sin θ + 5.12 sin θ − 3.2 cos θ] = 0.35 73.12 sin θ − 6.4 cos θ A On equating dP = 0, we get, dθ 0.35 [3.12 sin θ – 6.4 cos θ] = 0 ∴ 3.12 sin θ – 6.4 cos θ = 0 ` 3.12 sin θ = 6.4 cos θ ⇒ ∴ θ = tan

–1

sin θ = 6.4 cos θ 3.12

(2.0513) = 64 o

RESULT The value of θ for maximum power transfer = 64 o

⇒ tan θ = 2.0513

Circuit Theory

7.5

7. 69

Reciprocity Theorem

(AU June’14, 2 Marks)

The reciprocity theorem states that, in a linear, bilateral, single source circuit, the ratio of excitation to response is constant when the position of excitation and response are interchanged. Here, the excitation is either a voltage source or current source and the response is either current or voltage in an element (R, L or C). “The reciprocity theorem will be satisfied only by circuits or networks which does not have dependent sources”. In circuits without dependent sources, the Z and Y matrices formed for mesh and node basis analysis will be symmetric, i.e., the element of matrices Z jk = Z kj and Yjk = Ykj . The networks which satisfy reciprocity theorem are called reciprocal networks. 7.5.1 Reciprocity Theorem Applied to Mesh Basis Circuit Consider a mesh basis circuit with single voltage source, E as shown in Fig. 7.15. Let I be the current in mesh-j when the source is in mesh-k as shown in Fig. 7.15(a). The reciprocity theorem implies that the same current I will be produced in mesh-k, if the source is shifted to mesh-j as shown in Fig. 7.15(b). It must be noted that currents in other parts of the circuit may not be same. R1 R2

1

R1 R3

2

Mesh-k

R5

1’

2 RL

RL

R4

R3

I

I E +E

R2

1

Mesh-j

R4 Mesh-k

2’

Fig. a : Source in mesh-k and response in mesh-j.

Mesh-j

R5

1’

+ E E

2’

Fig. b : Source in mesh-j and response in mesh-k.

Fig. 7.15 : Mesh basis circuit to demonstrate reciprocity theorem. Proof by mesh analysis : Consider a 2-port resistive network without sources shown in Fig. 7.16. In order to prove the reciprocity theorem we can connect a voltage source at port-1 and short circuit port-2 to observe the current response. Then the source can be shifted to port-2 and the same current response can be observed in short circuited port-1.

2-port resistive network

Port-1

Fig. 7.16.

Case i : Excitation in port-1 and response at port-2 Let us connect a voltage source E to port-1 and short circuit the terminals of port-2 as shown in Fig. 7.17. Let I be the current through short circuit, which is the response due to excitation, E. In the circuit of Fig. 7.17, the response I due to excitation E can be solved by mesh analysis. Let us consider two meshes mesh-k and mesh-j as shown in Fig. 7.17. Now the response, I = Ij.

Port-2

I Mesh-k +

E E

Ik

2-port resistive network

Mesh-j S.C. Ii Port-2

Port-1

Fig. 7.17.

In mesh analysis, the current in jth mesh Ij is given by (Refer equation 4.5 in Chapter-4), Ij =

∆1j ∆2j ∆3j ∆kj E + E + E + ..... + E + ..... ∆ 11 ∆ 22 ∆ 33 ∆ kk

.....(7.30)

7. 70

Chapter 7 - Theorems in Circuit Analysis Since the circuit of Fig. 7.17, has only one source in mesh-k, E11 = E22 = E33 = ..... = 0 and Ekk = E Hence the equation (7.30) can be written as, ∆kj ∆kj E = E ∆ kk ∆

Ij = `

The response, I = I j =

∆kj E ∆

..... (7.31)

From equation (7.31) the ratio of excitation to response is, E = ∆ I ∆kj

.....(7.32)

Case ii : Excitation in port-2 and response at port-1 Let us change the excitation source E to mesh-j as shown in Fig. 7.18, and estimate the current in mesh-k. [In mesh-k the voltage source is replaced by its internal impedance. Since the source is ideal the internal impedance is zero and so it is replaced by short circuit.] Now the response, I = Ik.

I Mesh-k S.C. IK

2-port resistive network

Mesh-j Ij

+ E E

Port-2

Port-1

Fig. 7.18.

In mesh analysis the current in kth mesh, I k is given by (Refer equation 4.5 in Chapter-4), Ik =

∆ jk ∆1k ∆ ∆ E + 2k E22 + 3k E33 + ..... + E + ..... ∆ 11 ∆ ∆ ∆ jj

.....(7.33)

Since the circuit of Fig. 7.18 has only one source in mesh-j, E 11 = E22 = E33 = ..... = 0 and Ej j = E Hence, the equation (7.33) can be written as, Ik =

∆ jk ∆ jk E = E ∆ jj ∆

` The response, I = Ik =

∆ jk E ∆

..... (7.34)

From equation (7.34), the ratio of excitation to response is, E = ∆ I ∆ jk

..... (7.35)

Conclusion When the 2-port network does not have dependent sources, ∆ kj = ∆jk and so the equations (7.32) and (7.35) are same. Hence, the reciprocity theorem is proved.

7.5.2 Reciprocity Theorem Applied to Node Basis Circuit Consider a node basis circuit with single current source, I s as shown in Fig. 7.19. Let, V be the voltage in node-j when the current source is connected between node-k and reference as shown in Fig. 7.19a . The reciprocity theorem implies that the same voltage V will exist in node-k if the source is shifted to node-j as shown in Fig. 7.19b. It must be noted that the voltages in other parts of the circuit may not be same.

Circuit Theory 1

7. 71 Node-k

Node-j R2

R1 Is

R4

1’

R5

R3 R6

Reference node

2

Node-k

1

+

+

V

V

_

_

2’

1’

Node-j R2

R1 R4

2

R3 Is

R6

R5

2’

Reference node

Fig. a : Source in node-k and response in node-j. Fig. b : Source in node-j and response in node-k. Fig. 7.19 : Node basis circuit to demonstrate reciprocity theorem. Proof by Node Analysis Consider a 2-port resistive network without sources shown in Fig. 7.20. In order to prove the reciprocity theorem we can connect a current source at port-1 and observe the open circuit voltage at port-2. Then the source can be shifted to port-2 and the same open circuit voltage can be observed in port-1.

2-port resistive network

Port-1

Port-2

Fig. 7.20.

Case i : Excitation in port-1 and response at port-2 Let us connect a current source, I s to port-1 as shown in Fig. 7.21. Let the voltage across the terminals of port-2 be V, which is the response due to excitation Is.

Node-j

Node-k

+ Is

In the circuit of Fig. 7.21, the response V due to excitation I s can be solved by node analysis. Let us consider two nodes, Node-k and Node-j as shown in Fig. 7.21. Now the response, V = Vj.

2-port resistive network

Vk E Port-1

+

+

Vj

V

E

E

Port-2

Fig. 7.21.

In node analysis, the voltage in jth node Vj is given by (Refer equation 5.5 in Chapter 5), Vj =

∆l kj ∆l 1j ∆l 2j ∆l 3j I + I + I + ..... + I + ..... ∆l 11 ∆l 22 ∆l 33 ∆l kk

..... (7.36)

Since the circuit of Fig. 7.21, has only one source in node-k, I11 = I22 = I33 = ..... = 0 and Ikk = Is Hence, the equation (7.36) can be written as, Vj =

∆l kj ∆l kj I = I ∆l kk ∆l s

Is = ∆l V ∆l kj

Node-j

Node-k

∆l kj ..... (7.37) ` The response, V = Vj = I ∆l s From equation (7.37), the ratio of excitation to response is, ..... (7.38)

Case ii : Excitation in port-2 and response at port-1

+ V E

+ Vk

2-port resistive network

E

+ Vj

Is

E Port-2

Port-1

Fig. 7.22.

Let us change the excitation source to node-j as shown in Fig. 7.22, and estimate the voltage at node-k. [In node-k the current source is replaced by its internal impedance. Since the source is ideal the internal impedance is infinity and so it is replaced by open circuit.] Now the response, V = Vk. In node analysis the voltage in node-k, Vk is given by (Refer equation 5.5 in Chapter 5), Vk =

∆l jk ∆l 1k ∆l ∆l I + 2k I22 + 3k I33 + ..... + I + ..... ∆l 11 ∆l ∆l ∆l jj

..... (7.39)

7. 72

Chapter 7 - Theorems in Circuit Analysis Since the circuit of Fig. 7.22, has only one source in node-j, I11 = I22 = I33 = ... = 0 and Ijj = Is Hence, the equation (7.39) can be written as, Vk =

∆l jk ∆l jk I = I ∆l jj ∆l s

∆l jk I ∆l s From equation (7.40), the ratio of excitation to response is, ` The response, V = Vk =

..... (7.40)

Is = ∆l V ∆l jk

..... (7.41)

Conclusion When the 2-port network does not have dependent sources, ∆l kj = ∆l are same. Hence, the reciprocity theorem is proved.

jk

and so the equations (7.38) and (7.41)

EXAMPLE 7.40

2‡

Prove the reciprocity theorem for the two port network of Fig. 1. 1‡

SOLUTION

4‡

1

2

Method-I : Proof by mesh Analysis Port-1

Let us connect a voltage source of E volts to port-1 and short circuit the port-2 as shown in Fig. 2. Let, I be the current through short circuit which is the response due to excitation E.

2’

1’

Fig. 1.

Let us assume three mesh currents I1, I2 and I3 as shown in Fig. 2. Now, the response I = I2. With reference to Fig. 2, the mesh basis matrix equation is, R V R V R V − 1 W SI1 W SE W S1 + 5 − 5 S −5 4 + 5 − 4 W SI 2 W = S 0 W S0 W S − 1 − 4 1 + 2 + 4 W SI W 3 T X T X T X R V R V R V 6 5 1 I E − − 1 S W S W S W S − 5 9 − 4 W SI 2 W = S 0 W S0 W S − 1 − 4 7 W SI W 3 T X T X T X 6 −5 −1 ∆ = −5 9 −4 −1 −4 7

6 E −1 ∆2 = − 5 0 − 4 −1 0 7

2‡ I3 1‡

E +E

I1

5‡

Fig. 2.

+ _− 1 i # 7_− 5 # _− 4 ii − _− 1 # 9 iA

= 0 − E # 7_− 5 # 7i − _− 1 # _− 4iiA + 0 = 39E

` The response, I = I2 =

∆2 = 39E ∆ 58

The ratio of excitation to response = E = 58 I 39

4‡ I

2 = 6 # 9_9 # 7 i − _− 4 i C − _− 5 i # 7_− 5 # 7 i − _− 1 # _− 4 iiA

= 282 − 195 − 29 = 58

Port-2

5‡

Case i : Excitation in port-1 and response at port-2

I2

S.C.

Circuit Theory

7. 73

Case ii : Excitation in port-2 and response at port-1

2‡

Let us interchange the position of source and response as shown

Ic 1‡

in Fig. 3. Let us assume mesh currents I a , I b and I c as shown in Fig. 3. Now the response, I = I b. With reference to Fig. 3, the mesh basis matrix equation is, R V R V R V − 4 W SIa W SE W S5 + 4 − 5 S −5 5 + 1 − 1 W SIb W = S 0 W S0 W S − 4 − 1 1 + 4 + 2 W SI W c T X T X T X R V R V R V 9 5 4 I E − − S W S W S aW S − 5 6 − 1 W SIb W = S 0 W S0 W S − 4 − 1 7 W SI W c T X T X T X

4‡

I

S.C.

5‡

Ib

+ E E

Ia

Fig. 3.

9 −5 −4 2 = 9 # 9_6 # 7 i − _− 1 i C − _− 5 i # 7_− 5 # 7i − _− 4 # _− 1iiA ∆ = −5 6 −1 + _− 4i # 7_− 5 # _− 1ii − _− 4 # 6iA −4 −1 7 = 369 − 195 − 116 = 58 9 E −4 = 0 − E # 7_− 5 # 7i − _− 4 # _− 1iiA + 0 ∆b = − 5 0 − 1 39E = −4 0 7 ` The response, I = Ib =

∆b = 39E ∆ 58

The ratio of excitation to response = E = 58 I 39

Conclusion It is observed that the ratio of excitation to response is same when the position of excitation and response are interchanged. Hence, the reciprocity theorem is proved. Method II : Proof by Node Analysis 2‡

Case i : Excitation in port-1 and response at port-2 Let us connect a current source of I s Amperes to port-1 and open circuit the port-2 as shown in Fig. 4. Let, V be the voltage across the open terminals

V1

1‡

V3 4 ‡

V2

+

of port-2, which is the response due to excitation Is . Let us assume three node voltages V 1, V2 and V3 as shown in Fig. 4. Now the response, V = V2. With reference to Fig. 4, the node basis matrix equation is, R V R V R V S Is W S1 + 1 − 1 W S V1 W −1 1 2 1 2 S W S W S W S −1 1 + 1 − 1 W SV2 W = S 0 W S W S 4W S W 2 4 2 S W 1 1 +1 +1W S W S −1 − S0 W S 4 1 4 5 W S V3 W 1 T X T X T X R V R V R V − 1 W S V1 W SIs W S 1.5 − 0.5 S − 0.5 0.75 − 0.25 W S V2 W = S 0 W S0 W S − 1 − 0.25 1.45 W S V W 3 T X T X T X

Is

5‡

V

E Reference node

Fig. 4.

7. 74

Chapter 7 - Theorems in Circuit Analysis 1.5 − 0.5 −1 ∆l = − 0.5 0.75 − 0.25 − 1 − 0.25 1.45

= 1.5 # 9_0.75 # 1.45 i − _− 0.25 i C − _− 0.5 i # 7_− 0.5 # 1.45 i − _− 1 # _− 0.25 iiA + _− 1 i # 7_− 0.5 # _− 0.25 ii − _− 1 # 0.75 iA 2

= 1.5375 − 0.4875 − 0.875 = 0.175 1.5 Is −1 = 0 − Is # 7_− 0.5 # 1.45 i − _− 1 # _− 0.25 iiA + 0 ∆l2 = − 0.5 0 − 0.25 0.975 Is − 1 0 1.45 = ` The response, V = V2 =

∆l 2 0.975 Is = ∆ 0.175

The ratio of excitation to response = Is = 0.175 = 0.1795 V 0.975 2‡

Case ii : Excitation in port-2 and response at port-1 Let us interchange the position of source and response as shown in Fig. 5. Let us assume node voltages Va, Vb and Vc as shown in Fig. 5. Now the response, V = Vb. With reference to Fig. 5, the node basis matrix equation is, R V R V R V S Is W S1 + 1 − 1 W S Va W −1 4W S W 2 S W S4 2 S −1 1 + 1 − 1 W S Vb W = S 0 W S W S 1W S W 2 1 2 S W 1 1 +1 +1W S W S −1 − S0 W S 1 1 4 5 W S Vc W 4 T X T X T X

+

Vb

V

1‡

Vc 4 ‡

Va

5‡

Is

E Reference node

Fig. 5.

R V R V R V SIs W S 0.75 − 0.5 − 0.25 W S Va W S − 0.5 1.5 − 1 W S Vb W = S 0 W S − 0.25 − 1 1.45 W S V W S0 W c T X T X T X ∆l =

∆lb =

0.75 − 0.5 − 0.25 2 = 0.75 # 9_1.5 # 1.45i − _− 1 i C − _− 0.5 i # 7_− 0.5 # 1.45 i − 0.5 1.5 −1 − _− 0.25 # _− 1 iiA + _− 0.25 i # 7_− 0.5 # _− 1 ii − _− 0.25 # 1.5 iA − 0.25 − 1 1.45 = 0.88125 − 0.4875 − 0.21875 = 0.175 0.75 Is − 0.25 = 0 − Is # 7_− 0.5 # 1.45 i − _− 0.25 # _− 1 iiA + 0 − 0.5 0 −1 − 0.25 0 1.45 = 0.975 Is

` The response, V = Vb =

∆lb 0.975 Is = ∆l 0.175

The ratio of excitation to response = Is = 0.175 = 0.1795 V 0.975

Conclusion It is observed that the ratio of excitation to response is same when the position of excitation and response are interchanged. Hence, the reciprocity theorem is proved.

Circuit Theory

7. 75 3‡

EXAMPLE 7.41 2‡

In the circuit of Fig. 1, calculate I x. Prove the reciprocity theorem by interchanging the position of 10 V source and Ix.

4‡

4‡

SOLUTION

10 V

+ E

Ix 5‡

Case i : To solve Ix in the given circuit

7‡

Let us assume three mesh currents as shown in Fig. 2. Now the

Fig. 1.

response, I x = I1 − I2

3‡

With reference to Fig. 2, the mesh basis matrix equation is, R V −5 −4W S2 + 4 + 5 S −5 5 + 4 + 7 −4W S −4 −4 3 + 4 + 4W T X R V R V R V 11 5 4 I 10 − − 1 S W S W S W S − 5 16 − 4 W SI 2 W = S 0 W S 0W S − 4 − 4 11 W SI W 3 T X T X T X

R V R V S10 W SI1 W SI 2 W = S 0 W SI W S 0W 3 T X T X

2‡ I3

10 V

Ix 5‡ I2

I1

..... (1)

Fig. 2.

11 − 5 − 4 = 11 # 7(16 # 11) − (− 4) 2 A − (− 5) # 7(− 5 # 11) − (− 4) 2 A ∆ = − 5 16 − 4 + (− 4) # 7(− 5 # (− 4)) − (− 4 # 16) A − 4 − 4 11 = 1760 − 355 − 336 = 1069

∆1 =

4‡

4‡ + E

7‡

10 − 5 − 4 = 10 # 7(16 # 11) − (− 4) 2 A − 0 + 0 0 16 − 4 0 − 4 11 = 1600

11 10 − 4 = 0 − 10 # 7(− 5 # 11) − (− 4) 2 A + 0 ∆2 = − 5 0 − 4 − 4 0 11 = 710 I1 =

∆1 = 1600 1069 ∆

;

I2 =

∆2 = 710 1069 ∆

` The respone, I x = I1 − I2 = 1600 − 710 = 1600 − 710 = 890 = 0.8326 A 1069 1069 1069 1069 Case ii : To prove the reciprocity theorem by interchanging the position of source and response Let us interchange the position of source and response as shown in Fig. 3. Let us assume mesh currents 3‡

I a , I b and Ic as shown in Fig. 3. Now, the response, I x = Ia. With reference to Fig. 3, the mesh basis matrix equation is, R V R V R V −5 − 4 W SIa W S 10 W S2 + 4 + 5 S −5 5 + 4 + 7 − 4 W SIb W = S − 10 W S 0W S −4 − 4 3 + 4 + 4 W SIc W T X T X T X R V R V R V 11 5 4 I 10 − − S W S W S aW S − 5 16 − 4 W SIb W = S − 10 W ..... (2) S 0W S − 4 − 4 11 W SI W c T X T X T X

2‡ Ic

Ix

4‡

4‡

+ 10 V E

Ia

5‡

Fig. 3.

Ib

7‡

7. 76

Chapter 7 - Theorems in Circuit Analysis On comparing equations (2) and (1), we can say that the ∆ is same in both the case. ∴ ∆ = 1069 10 − 5 − 4 = 10 # 7(16 # 11) − (− 4) 2 A − (− 5) # 7(− 10 # 11) − 0) A ∆a = − 10 16 − 4 − 4 # 7(− 10 # (− 4)) − 0 A 0 − 4 11 = 1600 − 550 − 160 = 890 Now the response, I x = Ia =

∆a = 890 = 0.8326 A 1069 ∆

Here, the response I x remains same after interchanging the position of source and response. Hence, the reciprocity theorem is proved.

EXAMPLE 7.42 2‡

In the circuit of Fig. 1, calculate Vx. Prove the reciprocity theorem by interchanging the position of 12 A source and V x.

4‡

SOLUTION Case i : To solve Vx in the given circuit

12 A

5‡

+ Vx 1 ‡ E

10 ‡

Let us assume three node voltages as shown in Fig. 2.

2‡

Fig. 1.

Now the response, Vx = V2. With reference to Fig. 2, the node basis matrix equation is, R S1 + 1 + 1 S 4 10 2 S −1 S 4 S −1 S 2 T

−1 4 1 +1 +1 4 1 5 −1 5

V −1 W 2W −1 W 5W 1 +1 +1W 5 2 2W X

R V R V R V S12 W S 0.85 − 0.25 − 0.5 W S V1 W S − 0.25 1.45 − 0.2 W S V2 W = S 0 W S − 0.5 − 0.2 1.2 W S V W S 0W 3 T X T X T X

R V R V S V1 W S 12 W S W W S S W = S W V 0 2 S W W S S W W S S V3 W S 0 W T X T X

2‡

4‡

V1

12 A

10 ‡

V2

5‡

+ Vx 1 ‡ E

V3

2‡

..... (1)

Fig. 2.

0.85 − 0.25 − 0.5 2 = 0.85 # 9_1.45 # 1.2i − _− 0.2i C − _− 0.25i # 7_− 0.25 # 1.2i − (− 0.5 # (− 0.2)) A ∆l = − 0.25 1.45 − 0.2 + _− 0.5i # 7_− 0.25 # _− 0.2ii − _− 0.5 # 1.45iA − 0.5 − 0.2 1.2 = 1.445 − 0.1 − 0.3875 = 0.9575

0.85 12 − 0.5 = 0 − 12 # 7_− 0.25 # 1.2i − (− 0.5 # (− 0.2)) A + 0 ∆l2 = − 0.25 0 − 0.2 − 0.5 0 1.2 = 4.8 ` The response, Vx = V2 =

∆l 2 4.8 = = 5.0131V 0.9575 ∆l

Circuit Theory

7. 77

Case ii : To prove the reciprocity theorem by interchanging the position of source and response Let us interchange the position of source and response as shown in Fig. 3. Let us assume node voltages as shown in Fig. 3. Now the response, Vx = Va. With reference to Fig. 3, the node basis matrix equation is, R V R V R V S 0 W S1 + 1 + 1 −1 − 1 W S Va W 2W S W 4 S W S 4 10 2 S −1 1 +1 + 1 − 1 W S Vb W = S 12 W S W S 5W S W 4 4 1 5 S W S 1 1 +1 +1W S W 1 − − Va S 0 W S 5 5 2 2 W S Vc W 2 T X T X T X R V R V R V + S 0W S 0.85 − 0.25 − 0.5 W S Va W Vx S − 0.25 1.45 − 0.2 W S Vb W = S12 W ..... (2) E S 0W S − 0.5 − 0.2 1.2 W S V W c T X T X T X On comparing equations (1) and (2), we can say that the ∆’ remains same.

2‡

4‡

10 ‡

5‡

Vb

Vc

2‡

1‡

12 A

Fig. 3.

∴ ∆’ = 0.9575 0 − 0.25 − 0.5 = 0 − (− 0.25) # 712 # 1.2 − 0 A + (− 0.5) # 712 # (− 0.2) − 0 A ∆la = 12 1.45 − 0.2 0 − 0.2 1.2 = 3.6 + 1.2 = 4.8 ∆l a 4.8 ` The response, Vx = Va = = = 5.0131V ∆l 0.9575 Here, the response Vx remains same after interchanging the position of source and response. Hence, the reciprocity theorem is proved.

EXAMPLE 7.43

8W

Ix o

50Ð30 V

In the circuit of Fig. 1, compute I x . Demonstrate the reciprocity theorem by interchanging the position of the source and I x .

8W

+ -j8 W

j4 W

~

3W

-

SOLUTION Case i : To solve I x in the given circuit

Fig. 1.

Let us assume three mesh currents I1, I 2 and I3 as shown in Fig. 2. Now, the response, I x = I3 .

o

o

8W

8W Ix

o

50Ð30 V

With reference to Fig. 2, the mesh basis matrix equation is, R V R V R V S50+30 o W 0 W SI1 W − j4 S8 + j4 S − j4 8 + j4 − j8 − (− j8) W SI 2 W = S 0 WW S S W S S 0 0 W − (− j8) 3 − j8 W I3 T X T X T X

+ -j8 W

j4 W

~

3W

-

I1

I2

I3

o

Here, 50∠30 = 50 cos30 + j50sin30 = 43.3013 + j25 V R V 0W S8 + j4 − j4 S − j4 8 − j4 j8 W S 0 j8 3 − j8 W T X

R V R V SI1 W S 43.3013 + j25 W SI 2 W = S 0W SI W S 0W 3 T X T X

Fig. 2. .....(1)

8 + j4 − j4 0 = (8 + j4) # 7(8 − j4) # (3 − j8) − (j8) 2 A − (− j4) # 7 − j4 # (3 − j8) − 0 A + 0 ∆ = j8 = 752 − j384 + 48 − j128 − j4 8 − j4 0 j8 3 − j8 = 800 − j512

7. 78

Chapter 7 - Theorems in Circuit Analysis

∆3 =

8 + j4 − j4 43.3013 + j25 = 0 − 0 + (43.3013 + j25) # 7 − j4 # j8 − 0 A 0 − j4 8 − j4 1385.6416 + j800 = 0 j8 0

` The response, I x = I3 =

1385.6416 + j800 ∆3 = ∆ 800 − j512 = 0.7747 + j1.4958 A = 1.6845+62.6 o A

Case ii : To demonstrate the reciprocity theorem by interchanging the position of source and response 8W

8W +

~

Now, the response, I x = Ia .

-j8 W

j4 W

With reference to Fig. 3, the mesh basis matrix equation is, R V 0W − j4 S8 + j4 S − j4 8 + j4 − j8 − (− j8) W S 0 − (− j8) 3 − j8 W T X R V R V 0 W SIa W − j4 S8 + j4 S − j4 8 − j4 j8 W SIb W S j8 3 − j8 W SIc W 0 T X T X

Ib

Ia

R V R V 0 W SIa W S SI b W = S 0 W S W S50+30 o W Ic T X T X R V 0W S 0W = S S 43.3013 + j25 W T X

Ic

-

o

Ix

50Ð30 V

Let us interchange the position of source and response as shown in Fig. 3. Let us assume mesh currents as shown in Fig. 3.

3W

Fig. 3.

..... (2)

On comparing equations (1) and (2), we can say that the value of ∆ remains same in both the cases. ∴ ∆ = 800 − j512

∆a =

0 0 43.3013 + j25

− j4 8 − j4 j8

` The response, I x = Ia =

0 = 0 − (− j4) # 70 − (43.3013 + j25) # j8 A + 0 j8 3 − j8 = 1385.6416 + j800 1385.6416 + j800 ∆a = ∆ 800 − j512 = 0.7747 + j1.4958 A = 1.6845+62.6 o A

It is observed that the response remains same after interchanging the position of source and response, which demonstrates the validity of reciprocity theorem. j5 W

2W

8W

Vx

~

4W

+

o

In the circuit of Fig. 1, compute Vx . Demonstrate the reciprocity theorem by interchanging the position of the source and response.

10Ð60 A

EXAMPLE 7.44

j4 W

SOLUTION Case i : To solve Vx in the given circuit

Fig. 1.

Let us assume two node voltages V1 and V2 as shown in Fig. 2. Now, the response, Vx = V2 .

j6 W

Circuit Theory

7. 79

With reference to Fig. 2, the node basis matrix equation is, R R V o S 10+60 S V1 W S W = S S S W 0 S S V2 W T T X

V W W W W X

0.1 − j0.4 j0.2

∆l =

∆l 2 =

0.1 − j0.4 j0.2

4W

+ 8W

Vx

~

j4 W

j0.2 V1 5 + j8.6603 H > H = > H 0.2019 − j0.3154 V2 0

0.1 − j0.4 j0.2

V2

2W

Here, 10∠60 o = 10cos 60 o + j10 sin60 o = 5 + j8.6603 A

>

j5 W

V1

o

V −1 W j5 W 1 +1 + 1 W j5 8 4 + j6 W X

10Ð60 A

R S 1 + 1 S 2 + j4 j5 S −1 S j5 T

j6 W

Fig. 2. ..... (1)

j0.2 = 7(0.1 − j0.4) # (0.2019 − j0.3154) A − (j0.2) 2 0.2019 − j0.3154 = − 0.06597 − j0.1123 5 + j8.6603 = 0 − 7 j0.2 # (5 + j8.6603) A 0 = 1.73206 − j

` The response, Vx = V2 =

1.73206 − j ∆l 2 = ∆l − 0.06597 − j0.1123 = − 0.1158 + j15.3555 V = 15.3559+90.4 o V

Case ii : To demonstrate the reciprocity theorem by interchanging the position of source and response Let us interchange the position of source and response as shown in Fig. 3. Let us assume node voltages Va and Vb as shown in Fig. 3.

j5 W

Va

4W

o

Vx

With reference to Fig. 3, the node basis matrix equation is, V −1 W j5 W 1 +1 + 1 W j5 8 4 + j6 W X

10Ð60 A

2W

Now, the response, Vx = Va

R S 1 + 1 S 2 + j4 j5 S −1 S j5 T

Vb

+

R R V 0 S S Va W S W = S S S W S 10+60o S Vb W T T X

V W W W W X

j4 W

~

8W j6 W

-

Fig. 3.

j0.2 Va 0.1 − j0.4 0 H > H = > H j0.2 0.2019 − j0.3154 Vb 5 + j8.6603 On comparing equations (1) and (2), we can say that the value of D’ remains same.

>

..... (2)

∴ D’ = –0.06597 – j0.1123 ∆l a =

0 j0.2 = 0 − 7(5 + j8.6603) # j0.2 A 5 + j8.6603 0.2019 − j0.3154 = 1.73206 − j

` The response, Vx = Va =

1.73206 − j ∆l a = ∆l − 0.06597 − j0.1123 = − 0.1158 + j15.3555 V = 15.3559+90.4 o V

It is observed that the response remains same after interchanging the position of source and response, which demonstrates the validity of the reciprocity theorem.

7. 80

7.6

Chapter 7 - Theorems in Circuit Analysis

Summary of Important Concepts 1.

Superposition theorem states that the response in a circuit with multiple sources is given by algebraic sum of responses due to individual sources acting alone.

2.

Superposition theorem is also referred to as principle of superposition.

3.

A circuit element is said to be linear, if the voltage-current relationship is linear.

4.

The principle of superposition is a combination of additivity property and homogeneity property.

5.

The property of additivity says that the response in a circuit due to a number of sources is given by the sum of the response due to individual sources acting alone.

6.

The property of homogeneity says that, if all the sources are mutiplied by a constant, then the respones is also multiplied by the same constant.

7.

While calculating the response due to one source, all other sources are made inactive or replaced by zero value sources.

8.

A zero value source is represented by its internal impedance.

9.

For an ideal voltage source the internal impedance is zero and for an ideal current source the internal impedance is infinite.

10. While calculating the response due to one source, all other ideal voltage sources are replaced by short circuit and all other ideal current sources are replaced by open circuit. 11. Thevenin’s theorem states that a circuit with two terminals can be replaced by an equivalent circuit consisting of a voltage source in series with a resitance(or impedance). 12. Thevenin’s voltage is given by the voltage across the two open terminals of the circuit. 13. Thevenin’s impedance is given by looking back impedance at the two open terminals of the network. 14. The looking back impedance is the impedance measured at the two open terminals of the circuit after replacing all the sources by zero value sources. 15. Norton’s theorem states that a circuit with two terminals can be replaced by an equivalent circuit consisting of a current source in parallel with a resistance (or impedance). 16. The Norton’s impedance is given by looking back impedance at the two open terminals of the network. 17. Thevenin’s equivalent is the voltage source model and the Norton’s equivalent is the current source model of a circuit. 18. The Thevenin’s and Norton’s impedance are same and it is given by the ratio of Thevenin’s voltage and Norton’s current. 19. The maximum power transfer to load is possible only if the source and load has matched impedance. 20. In dc source connected to resistive load, the maximum power transfer theorem states that maximum power is transferred from source to load, when the load resistance is equal to source resistance.

Circuit Theory

7. 81

21. In ac source connected to reactive load with resistance and reactance are independently variable the maximum power transfer theorem states that maximum power is transferred from source to load, when the load impedance is equal to complex conjugate of source impedance. 22. In general, the maximum power transfer theorem states that maximum power is transferred to a load impedance if the absolute value of the load impedance is equal to the absolute value of the looking back impedance of the circuit from the terminals of the load. 23. The reciprocity theorem states that in a linear, bilateral, single source circuit, the ratio of excitation to response is constant when the position of excitation and response are interchanged. 24. The networks which satisfy the reciprocity theorem are called reciprocal networks. 25. The reciprocity theorem will be satisfied only by circuits or networks which does not have dependent sources.

7.7

Short-answer Questions

Q7.1

State the superposition theorem. The superposition theorem states that the response in a linear circuit with multiple sources is given by algebraic sum of responses due to individual sources acting alone.

Q7.2

What are the properties of additivity and homogeneity? The property of additivity says that the response in a circuit due to a number of sources is given by sum of the response due to individual sources acting alone. The property of homogeneity says that if all the sources are multiplied by a constant, then the response is also multiplied by the same constant.

Q7.3

Find the current through the ammeter shown in Fig. Q7.3.1 by using the superposition theorem. Since the resistance of ammeter is not specified it can be represented by short circuit. The condition of the given circuit when each source is acting separately are shown in Figs. Q7.3.2 and Q7.3.3. 4W

10 V

S.C.

Þ

10 V

I’

+ -

Fig. Q7.3.2. 2.5 W

2.5 W

I’’ S.C.

I 10 V

+ E

A

Fig. Q7.3.1.

S.C.

4W

2.5 ‡

4W

2.5 W I’

+ -

4‡

S.C.

I’’ + 5V

Þ

Fig. Q7.3.3.

+ - 5V

E +

5V

7. 82

Chapter 7 - Theorems in Circuit Analysis With reference to Figs. Q7.3.2 and Q7.3.3, we can write, Response due to 10 V source, Il = 10 = 2.5 A 4 Response due to 5 V source, Ill = − 5 = − 2 A 2.5 Total response, I = Il + Ill = 2.5 + (− 2) = 0.5 A

Q7.4

4‡

2‡

+

Find the voltage VL in the circuit shown in Fig. Q7.4.1 by the principle of superposition.

24 V +E

4‡

4A

VL E

The condition of the circuit when each source is acting separately are shown in Figs. Q7.4.2 and Q7.4.3.

Fig. Q7.4.1. 4W + 12 V

4W

2W -

2A +

+ +

24 V -

2W

2A

4W

O.C.

V’L = 12 V

4W

S.C.

-

-

4A

V’’L = 4 ´ 2 = 8V

Fig. Q7.4.3.

Fig. Q7.4.2.

With reference to Figs. Q7.4.2 and Q7.4.3, we can write, VlL = 12 V

Q7.5

;

VllL = 8 V

;

` VL = VlL + VllL = 12 + 8 = 20 V

In the circuit shown in Fig. Q7.5.1, the power in resistance R is 9 W when V1 is acting alone and 4W when V2 is acting alone. What is the power in R when V1 and V2 are acting together ? Current through R when V1 is acting, Il = Current through R when V2 is acting, Ill =

9 = 3A 1 4 = 2A 1

2‡

2‡

V1 +E

R

+ V 2 E

1‡

Fig. Q7.5.1.

Total current when V1 and V2 are acting, I = I’ + I’’ = 3 + 2 = 5 A 2

Power in R when V1 and V2 are acting = I2 R = 5 × 1 = 25 W

Q7.6

P = I2 R

` I =

State Thevenin’s theorem.

P R

Thevenin’s theorem states that a circuit with two terminals can be replaced by an equivalent circuit consisting of a voltage source in series with a resistance (or impedance).

Q7.7

State Norton’s theorem.

A +

Norton’s theorem states that a circuit with two terminals can be replaced by an equivalent circuit consisting of a current source in parallel with a resistance (or impedance).

5 V +E

5‡ Vth

Q7.8

Find Thevenin’s voltage across terminals A and B in the circuit shown in Fig. Q7.8.1.

5A

+ E 10 V

Thevenin’s voltage, Vth = 5 + 10 = 15 V Note : Voltage across 5 Ω is 5 V .

E

Fig. Q7.8.1.

B

Circuit Theory Q7.9.

7. 83

(AU Dec’14, 2 Marks)

Find the value of In for the circuit shown in Fig. Q7.9.1

20 ‡

Let us remove the resistance RL and mark the resulting open terminals as A and B as shown in Fig. Q7.9.2.

+

The terminals A and B are shorted as shown in Fig. Q7.9.3. The 360 W resistance is short circuited and so no current will flow through it. Hence, the circuit of Fig. Q7.9.3 is redrawn as shown in Fig. Q7.9.4.

100 ‡

12 V E

In 360 ‡

RL

Fig. Q7.9.1. 20 ‡

20 ‡

+

+

100 ‡

20 ‡ 100 ‡

+

12 V

12 V A

E

A In

E

⇒

360 ‡

360 ‡

100 ‡

12 V E

⇒

A In

S.C.

S.C.

B

B

Fig. Q7.9.2.

B

Fig. Q7.9.3.

Fig. Q7.9.4.

With reference to Fig. Q7.9.4, by Ohm’s law, we can write, In =

Q7.10

12 = 0.1 A 20 + 100 A

Find Norton’s equivalent of the circuit shown in Fig. Q7.10.1. Norton’s current, In = 10 − 5 = 5 A

5‡

10 ‡

10 A

2‡

1‡

5A

Norton’s resistance, Rn =

B

1 = 0.5556 Ω 1 + 1 +1 +1 5 10 2 1

Fig. Q7.10.1.

A

5A

5.5556 ‡

B

Fig. Q7.10.2 : Norton’s equivalent.

5‡

10 ‡ A

Q7.11

Determine Thevenin’s equivalent of the circuit shown in Fig. Q7.11.1.

+ 20 ‡

200 V

Thevenin’s voltage is the voltage across 20 Ω resistance.

E

By voltage division rule, Thevenin's voltage, Vth = 200 #

B

20 = 160 V 20 + 5

Fig. Q7.11.1.

To find Thevenin’s resistance the 200 V source is replaced by short circuit as shown in Fig. Q7.11.2. With reference to Fig. Q7.11.2, we can write,

7. 84

Chapter 7 - Theorems in Circuit Analysis Thevenin's resistance, R th = 5 # 20 + 10 = 14 Ω 5 + 20

10 ‡

5‡

14 ‡ A

S.C.

A

160 V +E

20 ‡ Rth B

B

Fig. Q7.11.2.

A

In the circuit shown in Fig. Q7.12.1, using Thevenin’s theorem, determine the voltage across 90 W resistance after the switch is closed.

is open, the voltage across 90 Ω is 100 V. This is

~

100 V

70 ‡

90 ‡

E

0V E

10 +

~

B 90 ‡

~

100 V

90 ‡

+

Since the load is balanced when the switch

+

E

Q7.12

Fig. Q7.11.3 : Thevenin’s equivalent.

also Thevenin’s voltage at terminals A-B. To find Thevenin’s resistance, the voltage sources are replaced by short circuit. When the voltage sources are shorted the three 90 Ω resistances

Fig. Q7.12.1.

will be in parallel. ` Thevenin's resistance, R th = 90 = 30 Ω 3

30 ‡

A

The Thevenin’s equivalent at A-B is shown in Fig. Q7.12.2. With reference +

to Fig. Q7.12.2 by voltage division rule, 100 V +E

VL

70 ‡

E

Voltage across 70 Ω resistance, VL = 100 # 70 = 70 V 70 + 30

B

Fig. Q7.12.2. Q7.13

Find Thevenin’s equivalent of the circuit shown in Fig. Q.7.13.1. To find Thevenin’s voltage the current source is converted to voltage source as shown in Fig. Q7.13.2. By voltage division rule, V1 = 8 #

4 = 4V 4+4

4‡

2A A

+

8V E

4‡

3‡

By KVL, Vth = V1 + 6 = 4 + 6 = 10 V To find Thevenin’s resistance the voltage source is replaced by short circuit and current source is opened as shown in Fig. Q7.13.3. R th = 4 + 3 = 5 Ω 2

B

Fig. Q7.13.1.

Circuit Theory

7. 85

Thevenin’s equivalent is shown in Fig. Q7.13.4. 4W

3W

2 ´ 3 = 6V

4W

5W

O.C.

A

A

A

-+

+ + +

8V -

4W

V1 -

Vth

S.C.

3W

10 V +Rth

-

B

B

B

Fig. Q7.13.4 : Thevenin’s equivalent.

Fig. Q7.13.3.

Fig. Q7.13.2.

Q7.14

4W

Find V th at terminals A-B in the circuit shown in Fig. Q7.14.1.

4W A +

division rule, Vth = 60+0 o #

Q7.15

o

4 Ω and j3 Ω elements is 60∠0o V. Hence by voltage

60Ð0 V

In the given circuit the voltage across series combination of +

~

j4 W

j3 W

-j8 W

-

Vth

-

j3 = 21.6 + j28.8 V = 36+53.1 o V 4 + j3

B

Fig. Q7.14.1.

State maximum power transfer theorem.

(AU May’15, 2 Marks)

In purely resistive circuits, maximum power transfer theorem states that “Maximum power is transferred from source to load when the load resistance is equal to source resistance”. In general, the maximum power transfer theorem states that “Maximum power is transferred to a load impedance if the absolute value of the load impedance is equal to the absolute value of the looking back impedance of the circuit from the terminals of the load”.

Q7.16

The VI characteristics of a network is shown in Fig. Q7.16.1. Determine the maximum power that can be supplied by the network to a resistance connected across A-B.

I Network with linear resistors and independent sources

A + V E

I

20 V

V

E5 A

B

Fig. Q7.16.1.

When V = 0, I = −5 A

The condition V = 0 is equivalent to short circuiting terminals A−B and the current flowing through the short circuit is the Norton’s current. ∴ Norton’s current, I n = −I = − (−5) = 5 A When I = 0, V = 20 V The condition I = 0 is equivalent to open terminals A-B and the voltage across the open terminals is the Thevenin’s voltage. ∴ Thevenin’s voltage, Vth = 20 V Thevenin's resistance, R th =

Vth = 20 = 4 Ω In 5

7. 86

Chapter 7 - Theorems in Circuit Analysis The resistance, R to be connected for maximum power transfer across terminals A-B is R th. . 2 2 2 Maximum power transferred to R, Pmax = V th = V th = 20 = 25 W

‡

Determine the value of R in the circuit shown in Fig. Q7.17.1 for maximum power transfer.

4k

k

The value of R for maximum power transfer is given by the looking back resistance (or Thevenin’s resistance) from the terminals of R which is determined as shown below:

‡

3

Q7.17

4#4

4R th

+ E E

R

6k

‡

4R

W

k

3

Rth

Þ

S.C.

B

A

B

A

Þ

4

Fig. Q7.17.3.

Fig. Q7.17.4. 5‡

Determine the value of R in the circuit shown in Fig. Q7.18.1 for maximum power transfer. The value of R for maximum power transfer is given by the looking back resistance (or Thevenins’s resistance) from the terminals of R which is determined as shown below:

20 ‡

20 W

4W

Þ

5W

A

20 W

Rth

Fig. Q7.18.2. Q7.19

B

R

Fig. Q7.18.1.

A

O.C.

3A

E

Rth

Fig. Q7.18.3.

B

Find the value of R for maximum power transfer in the circuit shown in Fig. Q7.19. For maximum power transfer the value of R should be equal to absolute value of the looking back impedance from the terminals of R.

8‡

Em sin100t

S.C.

4W

4‡

+ 25 V

R = R th = 5 # 20 + 4 = 8 Ω 5 + 20 5W

4 kW

k

W

W

4

6kW

W

6k

6k

k

Fig. Q7.17.2.

Q7.18

4 kW

B

W

A

3 kW

W

W

3

4k

k

Fig. Q7.17.1.

4k

W

R = R th = 3 # 6 + 4 = 4 kΩ 3+6 2

4

‡

k

0.06 H

+ R

~

E

Fig. Q7.19.

Circuit Theory

7. 87

Here, Em sin ωt = Em sin 100t ; ∴ ω = 100 rad/s ` R =

Q7.20

82 + (ω # 0.06) 2 =

82 + (100 # 0.06) 2 = 10 Ω

State the reciprocity theorem. The reciprocity theorem states that in a linear, bilateral, single source circuit, the ratio of excitation to the response is constant, when the position of excitation and response are interchanged.

Q7.21

Two conditions of a passive, linear network are shown in Figs. Q7.21.1 and Q7.21.2. Using superposition and reciprocity theorem, find Ix. Ix

4A +

+

5V

1A

N

+ 10 V

N

10 V

E

E

E Port-1

Port-2

Port-1

Port-2

Fig. Q7.21.2.

Fig. Q7.21.1.

I’x

Let us replace the 10 V source in port-1 by short circuit as shown in Fig. Q7.21.3.

+ 10 V

N

S.C.

E

On comparing Figs. Q7.21.1 and Q7.21.3 by reciprocity theorem we can write, Ilx = 1A 10 V 5V

&

Fig. Q7.21.3.

Ilx = 1 # 10 = 2 A 5

Let us replace the 10 V source in port-2 by short circuit as shown in Fig. Q7.21.4. On comparing Figs. Q7.21.1 and Q7.21.4, by homogeneity property we can write,

Port-2

Port-1

I’’x + N

10 V

S.C.

E Port-1

Port-2

Imx = − 4 × 2 = −8 A

Fig. Q7.21.4.

By the principle of superposition, I x = Ilx + Illx = 2 + (− 8) = − 6 A

7.8

Here, source is multiplied by 2 and so response is also mulitpied by 2.

Exercises

I. Fill in the Blanks With Appropriate Words 1.

The principle of superposition is a combination of ________ and ________ property.

2.

While finding response due to one source, all other sources are replaced by their ________ .

3.

While finding response due to one source, all other ideal ________ sources are replaced by short circuit.

4.

While finding response due to one source, all other ideal current sources are replaced by ________.

5.

The ________ impedance is the looking back impedance from the open terminals of a network.

6.

The ________ equivalent is the voltage generator model of a network.

7.

The Norton’s equivalent is the ________ generator model of a network.

7. 88

Chapter 7 - Theorems in Circuit Analysis

8.

When load resistance and reactance are independently variable, maximum power transfer is achieved if load impedance is equal to ________ impedance.

9.

In purely resistive circuits ________ is transferred to load when load resistance is equal to source resistance.

10.

If a load impedance, R + jX with R alone variable is connected to a source with internal impedance Rs + jXs, then the condition for maximum power transfer is ________ .

11.

If a variable resistance R is connected to a source with impedance, Rs + jXs then the condition for maximum power transfer is ________ .

12.

The networks which satisfy reciprocity theorem are called ________ networks.

ANSWERS Rs2 + Xs2

1. additivity, homogeneity

6.

Thevenin’s

11. R =

2. internal impedances

7.

current

12. reciprocal

3. voltage

8.

conjugate of source

4. open circuit

9.

maximum power

5. Thevenin’s / Norton’s

10. R =

Rs2 + (Xs + X) 2

II. State Whether the Following Statements are True/False 1.

Superposition theorem can be extended to nonlinear circuits by piecewise linear approximation.

2.

Superposition theorem can be used to estimate the power directly.

3.

Superposition theorem is applicable to network containing linear dependent sources.

4.

Superposition theorem is applicable to a network containing time varying resistor.

5.

Thevenin’s equivalent can be determined only for linear part of the circuit.

6.

A circuit with a nonlinear resistance can be analyzed by using Thevenin’s theorem by replacing the rest of the circuit by Thevenin’s equivalent.

7.

The Thevenin’s and Norton’s impedance are same.

8.

In reactive circuits with variable reactance the maximum power transfer is achieved at resonance.

9.

The reactive circuit with fixed reactance will behave as purely resistive circuit when maximum power transfer condition is satisfied.

10. The reciprocity theorem will be satisfied by circuits with dependent sources. 11. The reciprocity theorem can be applied only to circuit with single source.

ANSWERS 1. True

4.

False

7. True

10. False

2. False

5.

True

8. True

11. True

3. True

6.

True

9. False-

Circuit Theory

7. 89

III. Choose the Right Answer for the Following Questions 1. The current through the ideal ammeter in the circuit shown in Fig. 1 is, 1.5 ‡

2‡

a) 1 A b) 1.5 A

+

E +

A

4V E

3V

c) 2 A d) 4 A

Fig. 1.

2. The voltage across the ideal voltmeter in the circuit shown in Fig. 2 is, a) 10 V b) 8 V

3A

V

2‡

5A

2‡

c) 6 V Fig. 2.

d) 4 V 3. The voltage VL in the circuit shown in Fig. 3 is,

2‡

a) 5 V b) 6 V

2‡

+ VL •

6 V +E

2A

2‡

c) 9 V Fig. 3.

d) 12 V 4. The current I2 in the circuit shown in Fig. 4 is,

2A

a) 8 A

1‡

1‡ I2

b) 7 A

1‡

3A

c) 4 A

5A

Fig. 4.

d) 2 A 5. The value of Vth and Rth in the circuit shown in Fig. 5 is, 15 ‡

6‡ A

a) 20 V, 30 W +

b) 5 V, 10 W

25 V •

10 ‡ Rth

c) 25 V, 4 W B

d) 10 V, 12 W

Fig. 5.

7. 90

Chapter 7 - Theorems in Circuit Analysis A

6. The value of Vth in the circuit shown in Fig. 6 is,

+ 5 V +E

a) 7 V

5‡ Vth

b) 5 V E +

3‡

c) 3 V

2V E

d) 2 V

B

Fig. 6. E 5V +

7. The value of Rth in the circuit shown in Fig. 7 is, a) 6 W

3‡

A

2A 6‡

b) 7 W

Rth

c) 13 W

B 4‡

d) 22 W

Fig. 7.

8. The value of RN in the circuit shown in Fig. 8 is,

10 ‡

3‡

A

a) 4 W b) 1.6 W

5‡

5A

2‡ Rth

c) 1.33 W

B

Fig. 8.

d) 1.2 W 9. The value of IN in the circuit shown in Fig. 9 is,

2A

a) 2 A

A 2‡

2‡

b) 3.5 A

IN

3A

c) 1 A B

d) 5 A

Fig. 9.

10. The value of IN in the cirucit shown in Fig. 10 is,

3‡

4‡ A

a) 1.5 A b) 3 A

1‡

3A

c) 0.5 A d) 1 A

IN

B

Fig. 10.

11. In the 2-terminal linear circuit shown in Fig. 11, the open circuit voltage measured across AB is 10 V and short circuit current through AB is 5 A. The value of resistance that can be connected across AB for maximum power transfer is, a) 10 W A Linear b) 5 W circuit B c) 50 W Fig. 11. d) 2 W

Circuit Theory

7. 91

12. The value of R for maximum power transfer in the circuit shown in Fig. 12 is, 0.04 H

2‡

a) 8 W b) 10 W

+

~

c) 8.25 W

•

em = sin 200t V

R

d) –j0.06 W

Fig. 12.

a) 2.4 + j1.2 W 3W

+

b) 2.4 – j1.2 W

~

_

R

Vm sin wt

c) 1.2 + j2.4 W

jX

j6 W

d) 1.2 – j2.4 W

IPQ

13. The value of impedance that can be connected across AB for maximum power transfer in the circuit shown in Fig. 13 is, j4 W 2W A

Z

Fig. 13.

B

14. The position of source and response in the circuit shown in Fig. 14a, are changed as shown in Fig. 14b. What is the value of IX ? a) 2 A

b) 4 A

c) 1 A R3

R2

R3

R2

d) 8 A

R1

R1

2A 10 V

+ E

R4

+ E

IX

R5

20 V

R4 R5

Fig. 14a.

Fig. 14b.

15. The position of source and response in the circuit shown in Fig. 15a, are changed as shown in Fig. 15b. What is the value of VX ? a) 6 V

b) 3 V

c) 2.4 V

R1

R3

R2

R4

R5

R3

R2 +

5A

d) 2 V R1

6V E

R6

Fig. 15a.

+ VX E

R5

R4

R6

2A

Fig. 15b.

ANSWERS 1. d

4. d

7. a

10. a

13. d

2. b

5. d

8. b

11. d

14. b

3. a

6. c

9. b

12. c

15. c

7. 92

III.

Chapter 7 - Theorems in Circuit Analysis

Unsolved Problems

E7.1

Using superposition theorem, determine the current I L in the circuit shown in Fig. E7.1.

E7.2

Determine the voltage VL in the circuit shown in Fig. E7.2 by using the superposition theorem.

E7.3

Calculate the current I y and voltage Vx in the circuit shown in Fig. E7.3 by using the superposition theorem. 3‡

5A

3‡

IL

+

2‡

VL

2‡

E +

4‡ 4A

20 ‡

10 ‡

5V

2A

6‡ 5‡

10 V

4‡

Fig. E7.2.

E 5‡

Fig. E7.3.

Determine the current I y and voltage V x in the circuit shown in Fig. E7.4 by using the superposition theorem. Iy

Vx

o

3Ð0 A

-

1W

~

1.5 W

2W

4W

j1 W

-j4 W

IL

5W

2W -j3 W

o

5Ð90 V

~

+ o

2Ð0 A

-

~

~

5W +

+

o

20Ð0 V

+

-j3 W

4W

o

+

o

10Ð-90 V

-

~

j3 W

j2 W

~

-

-

10Ð30 V

E7.4

x

1‡

Iy

+ V

E

Fig. E7.1.

+ 6V

2‡

1‡

10 V E

E +

E

+

+ 2A

4V

3‡

E

Fig. E7.5.

Fig. E7.4 E7.5

In the circuit shown in Fig. E7.5, determine I L by using the superposition theorem and hence estimate the active and reactive power in 2 - j3 W impedance.

E7.6

In the circuit shown in Fig. E7.6, determine I L using Thevenin’s theorem.

E7.7

In the circuit shown in Fig. E7.7, determine VL using Norton’s theorem. 4‡

3A

IL 2‡

5‡

2‡ +E

10 V

5V 4‡ 8‡

5‡

1‡

4‡

+

+E

+E

4‡

3V 10 ‡ 6‡

VL E

A

5‡

3‡

2A

+ E 5V

10 V +E

+E

B

12 V

Fig. E7.6. E7.8

Fig. E7.7.

Fig. E7.8.

Find Thevenin’s and Norton’s equivalent of the circuit shown in Fig. E7.8 with respect to terminals A and B.

Circuit Theory E7.9

7. 93

In the circuit shown in Fig. E7.9, determine the current I L by using Norton’s theorem. o

10Ð0 A

~ -j10 W -j10 W

+

2W

5W

~

IL

20 W

+

-j10 W

5W j4 W

+

VL

o

-

2W

4W

60Ð0 V

o

200Ð0 V

j20 W

j3 W

~

-

-

Fig. E7.9.

Fig. E7.10.

E7.10

In the circuit shown in Fig. E7.10, determine the voltage across 5 + j3 W impedance by using Thevenin’s theorem.

E7.11

Determine the value of R in the circuit shown in Fig. E7.11 for maximum power transfer. Also find the value of maximum power.

E7.12

Determine the value of Z L in the circuit shown in Fig. E7.12 for maximum power transfer.

E7.13

In the circuit of Fig. E7.13, calculate Vx . Prove reciprocity theorem by interchanging the position of 5 A source and Vx . j1 W

2W

4W

5A + V2Ðq2

2W R

1W

~

-

j5 W

2W

-j3 W

-j4 W

4W

+ 10 V -

+ 5W

3W

V1Ðq1 -

~

ZL

+ Vx 10 W -

8W

5W

3W

Fig. E7.11.

Fig. E7.13.

In the circuit of Fig. E7.14, calculate Ix . Prove the reciprocity theorem by interchanging the position of 10 V source and Ix . -j2 W

Ix

2W

j3 W

5W

4W

-j4 W

-+

5W

+ o

6W

~

+ -j5 W

-

Fig. E7.14.

4W

Fig. E7.15.

o

1W

3W

Ix

10Ð0 V

4W

10 V

5Ð0 A

E7.14

Fig. E7.12.

Vx

3W

j2 W

-

~

Fig. E7.16.

E7.15

In the circuit of Fig. E7.15, demonstrate the reciprocity theorem by interchanging the position of source and response, I x .

E7.16

Demonstrate reciprocity theorem in the circuit shown in Fig. E7.16 by interchanging the position of source and response.

7. 94

Chapter 7 - Theorems in Circuit Analysis

ANSWERS E7.1

IL = IlL(2 A) + IllL(4 A) + IlllL(10 V) = 2 + 4 + 0 = 6 A

E7.2

VL = VlL(5 V) + VllL(10 V) = 2 + (− 4) = − 2 V

E7.3

Vx = Vlx(4 V) + Vllx(6 V) + Vlllx(2 A) + Vllllx(5 A) = 1 + (− 1.5) + 4.5 + (− 6.25) = − 2.25 V I y = Ily(4 V) + Illy(6 V) + Illly(2 A) + Illlly(5 A) = 0.25 + (− 0.375) + (− 0.875) + (− 1.5625) = − 2.5625 A

E7.4

I y = I ly(5 V) + I lly(10 V) + I llly(2 A) = (− 0.1061 + j0.504) + (− 0.2122 + j1.008) + (0.5623 + j0.3289) = 0.244 + j1.8409 = 1.857+82.4 o A Vx = Vlx(5 V) + Vllx(10 V) + Vlllx(2 A) = (− 0.1061 + j0.504) + (− 0.2122 + j1.008) + (− 1.4377 + j0.3289) = − 1.756 + j1.8409 = 2.5441+133.6 o V

E7.5

IL = IlL(20 V) + IllL(10 V) + IlllL(3 A) = (3.0769 + j4.6154) + (− 0.1785 − j2.7678) + 0 = 2.8984 + j1.8476 A = 3.4372+32.5 o A ;

P = 23.6287 W

Q = −35.443 VAR

E7.6

Vth = 10 V

;

Rth = 1 Ω

;

IL = 1.6667 A

E7.7

In = 1.8 A

;

Rn = 5 Ω

;

VL = 6 V

E7.8

Vth = 13.25 V

E7.9

In = − 4 − j8 A = 8.9443+ − 116.6 o A ; Zn = 10 + j5 Ω = 11.1803+26.6 o Ω IL = − 2.0037 − j5.9988 A = 6.3246+ − 108.4 o A

E7.10

Vth = 44.6152 − j63.0768 V = 77.2606+ − 54.7 o V

;

Rth = Rn = 1.875 Ω

;

In = 7.0667 A

Z th = 3.8462 − j7.2308 Ω

= 8.1901+ − 62o Ω

VL = 45.9196 + j1.44 V

= 45.9422+1.8 o V

E7.11

R = 3.1429 Ω , Pmax = 0.6492 W

E7.12

ZL = 4.6432 + j3.443 Ω = 5.7804+36.6 o Ω

E7.13

Vx = 10.2273 V

E7.14

Ix = 0.1176 A

E7.15

I x = 0.3554 + j0.3928 A = 0.5297+47.9 o A

E7.16

Vx = 1.5529 + j2.1887 V = 2.6836+54.6 o V

Chapter 8

SERIES AND PARALLEL RESONANCES 8.1 Introduction In RLC circuits excited by sinusoidal sources, the inductive and capacitive reactances have opposite signs. Hence, when the reactances are varied there is a possibility that the inductive reactance may cancel the capacitive reactance and the circuit may behave as a purely resistive circuit. This condition of RLC circuit is called resonance. The resonance may be defined as a circuit condition at which the circuit behave as a purely resistive circuit. The inductive reactance, XL = wL = 2pfL, and so the inductive reactance can be varied by varying either frequency (f ) or inductance (L). The capacitive reactance, X C = 1 = 1 , and so the capacitive ractance can be varied ωC 2πfC by varying either frequency (f ) or capacitance (C). When the frequency of the sinusoidal source exciting the RLC circuit is varied, there is a possibility that “the inductive reactance is equal and opposite to that of capacitive reactance at a particular frequency”. Therefore, the total reactance is zero and the circuit will behave as purely resistive circuit. Now, the circuit will be in resonance and the frequency at which resonance occurs is called resonant frequency.

8.2 Series Resonance In a series RLC circuits, the resonance condition can be achieved by varying the frequency of exciting sinusoidal source. When the frequency is varied, at a particular frequency the inductive reactance will cancel the capacitive reactance and the circuit will behave as a resistive circuit. This condition of RLC series circuit is called series resonance. 8.2.1 Resonance Frequency of Series RLC Circuit Consider the RLC series circuit shown in Fig. 8.1 excited by a sinusoidal source of variable frequency. When the frequency of the source is varied by maintaining the voltage of the source as constant, the resonance occur at a particular frequency. The expressions for resonance frequency in the RLC series circuit of Fig. 8.1 are given below: Resonant angular frequency, ωr =

1 in rad/s LC

.....(8.1)

Z = R + jwL - j

1 wC

L

R

I

+

~

V, w

Fig. 8.1.

1 Resonant frequency, fr = ωr = in Hz 2π 2π LC

.....(8.2)

C

8. 2

Chapter 8 - Series And Parallel Resonance

Proof for resonance frequency in series RLC Consider the RLC series circuit shown in Fig 8.1. Let, Z = Impedance of RLC series circuit . Here, Z = R + jωL − j 1 = R + j aωL − 1 k ωC ωC At resonance frequency wr , the total reactance is zero. ωr L − 1 = 0 ωr C

` At ω = ωr, `

ωr L = 1 ωr C

`

ωr =

` fr =

ωr 2π

.....(8.3) Equating imaginary part of equation (8.3) to zero

&

.....(8.4)

ωr2 = 1 LC

1 LC

.....(8.5)

&

fr =

1 2π LC

w = 2pf

8.2.2 Characteristics of Series RLC Circuit Consider the RLC circuit shown in Fig. 8.1. Let, V = V+0 o V = Supply voltage. I = Current through the RLC series circuit. I r = Current at resonance. Z r = Impedance at resonance. The impedance at resonance is obtained by substituting w = wr in equation (8.3). ` Z r = R + jω r L − j 1 = R ωr C

` Impedance at resonance, Z r = R o Current at resonance, I r = V = V+0 = V +0 o = I r +0 o A R R Zr

where, I r = V = Magnitude of current at resonance. R

Using equation (8.4)

..... (8.6) .....(8.7) .....(8.8)

Let us examine the variation of impedance, Z of the RLC series circuit with frequency. At frequencies lower than resonant frequency, the capacitive reactance will be more than inductive reactance and so the total reactance will be capacitive. Since the capacitive reactance is inversely proportional to frequency, the capacitive reactance and hence the total reactance increases when the frequency is decreased from the resonant frequency. Therefore, the impedance of the series RLC circuit increases when the frequency is decreased from resonant value.

8. 3

Circuit Theory

At frequencies higher than resonant frequency, the inductive reactance will be more than capacitive reactance and so the total reactance will be inductive. Since the inductive reactance is directly proportional to frequency, the inductive reactance and hence the total reactance increases when the frequency is increased from the resonant frequency. Therefore, the impedance of the series RLC circuit increases when the frequency is increased from the resonant value. At resonant frequency the impedance of the RLC series circuit is equal to resistance and this value of the impedance is minimum. Since the impedance is minimum, the current is maximum at resonance. Also the current at resonance will be in-phase with supply voltage, V. Since the impedance increases for frequencies lesser or higher than resonant value, the current decreases when frequency is increased or decreased from the resonant value. X I

Z

XL

=w

L

XC Ir = 0

X

=

XL

w

wr =

R -X C = -

w wr

Fig. a : Current Vs Frequency.

wr

w

Fig. b : Impedance Vs Frequency.

1 wC

-X

Fig. 8.2 : Characteristics of series resonance.

Fig. c : Reactance Vs Frequency.

8.2.3 Quality Factor of Series RLC Circuit When a circuit consisting of resistor, inductor and capacitor is excited by a sinusoidal source, the resistor dissipates energy in the form of heat, the inductor stores energy in the magnetic field associated with it and the capacitor stores energy in the electric field associated with it. In the steady state (after the transient period), there is a possibility that the sum of energy stored in the inductor and capacitor being greater than the energy dissipated in the resistor. In a series RLC circuit, due to larger stored energy in the inductor and capacitor, the voltage across these devices will be greater than the supply voltage. In other words we can say that, there is a voltage magnification or amplification. The voltage magnification can be expressed by a factor called Quality factor (Q), which is defined as the ratio of maximum energy stored to the energy dissipated in one period. Maximum energy stored ` Quality factor, Q = 2π # .....(8.9) Energy dissipated in one period The term 2p is introduced to simplify the expression for quality factor.

8. 4

Chapter 8 - Series And Parallel Resonance

Quality factor at resonance, Q r = 2π # Here,

Maximum energy stored at resonance .....(8.10) Energy dissipated in one period at resonance

ωr L R

Qr =

At resonance, ωr =

.....(8.11) 1 LC

;

ω2r =

1 LC

and

L =

1 ω2r C

Therefore, the quality factor at resonance, Qr can also be expressed as shown below: Qr =

ωr L ω 1 = r # 12 = ωr CR R R ωr C

` Qr =

Qr =

ωr L = R

` Qr = 1 R

1 L = 1 R LC R

L C

1 ωr CR

.....(8.12)

L C

.....(8.13)

For frequencies less than resonant frequency the RLC series circuit behave as capacitive circuit. ` When ω # ωr ,

Q =

1 ωCR

.....(8.14)

For frequencies higher than resonant frequency the RLC series circuit behave as inductive circuit. ` When ω $ ωr ,

Q = ωL R

.....(8.15)

Note : In an RLC series circuit, when the inductor stores energy the capacitor discharges and vice-versa. At resonance, the sum of energy stored in inductor and capacitor is maximum. For w < wr , the energy stored in capacitor is maximum and for w > wr , the energy stored in inductor is maximum. Equation (8.15) can be used to calculate the Q-factor of a coil or RL series circuit and the equation (8.14) can be used to calculate the Q-factor of RC series circuit. Proof for quality factor at resonance, Qr in RLC series circuit Consider an RLC series circuit shown in Fig 8.3, excited by a sinusoidal voltage source of frequency, w. Let, I be the reference phasor. ` I = I +0 o A

R

L

C

i

+ v C

Let, i = Instantaneous value of current. \ i = Im sinwt

.....(8.16)

Let, wL = Instantaneous value of energy stored in inductor. wC = Instantaneous value of enrgy stored in capacitor. w = Total instantaneous energy stored in the RLC circuit. wr = Total intantaneous energy stored in the RLC circuit at resonance.

I

+

~

V, w

Fig. 8.3.

8. 5

Circuit Theory We know that, wL = 1 L i 2 2 1 = L _ Im sin ωt i2 2

Using equation (3.12) of Chapter - 3 Using equation (8.16)

= 1 L I m2 sin 2 ωt 2

.....(8.17)

We know that, wC = 1 C νc2 2 = 1 C< 1 2 C =

1 1 C< 2 C

` ωC =

Using equation (3.18) of Chapter - 3 2

# i dt F

1 C

# i dt

2

# Im sin ωt dt F

Using equation (8.16)

I 2 1 1 cos ωt 2 2 C < d− Im nF = m2 cos ωt 2 C ω 2ω C

Now, w = wL + wC =

` wr = w

νC =

I 2 1 L Im2 sin 2 ωt + m2 cos 2 ωt 2 2ω C

=

Im2 1 L sin 2 ωt + 2 cos 2 ωt F 2 < ω C

=

Im2 1 L sin 2 ωt + 2 cos 2 ωt F 2 < ω C

~ = ~r

=

.....(8.18) Using equations (8.17) and (8.18)

~ = ~r

I m2

L sin 2 ωr t + 12 cos 2 ωr t F 2< ωr C

=

Im2 LC 2 cos 2 ωr t F < L sin ωr t + 2 C

ωr =

=

Im2 L 7 2 sin ωr t + cos 2 ωr t A 2

sin 2 + cos 2 θ = 1

` ωr =

1 LC

&

1 = LC ωr2

Im2 L 2

.....(8.19)

From equation (8.19) we can say that, the energy stored in the RLC circuit at resonance is independent of time and it is constant. Therefore, the instantaneous energy is the maximum energy stored at resonance. Let, Wmr = Maximum energy stored at resonance ` Wmr = wr =

Im2 L 2

.....(8.20)

In the RLC series circuit, the enrgy is dissipated by the resistor. Let, WR = Energy dissipated in resistor in one period. WRr = Energy dissipated in resistor in one period at resonance. ` WR = Power # Time period = I2 R # T

I = RMS value of current

8. 6

Chapter 8 - Series And Parallel Resonance =d

Im 2 1 n R# f 2

=

I m2 2π R# ω 2

=

π Im2 R ω

I=

Im 2

ω = 2πf

T=

1 f

&

1 2π = f ω .....(8.21)

2

` WRr = WR

= ~ = ~r

` Qr = 2π # = 2π #

=

π Im R ωr

.....(8.22)

Wmr 1 = 2π # Wmr # WRr WRr Im2 L ωr # 2 π Im2 R

ωr L R

Using equations (8.20) and (8.22) .....(8.23)

8.2.4 Bandwidth of Series RLC Circuit From the current response (Refer Fig. 8.2(a)) of an RLC series circuit we can say that, the current is maximum at resonance and the current decreases when the frequency is decreased or increased from the resonant value. For practical applications we have to define a range of frequencies over which the current response is appreciable and this range of frequency is called bandwidth. Since the resistance being the load resistance in the practical circuits, the range of frequencies over which the response is appreciable can be decided based on power in the resistance. At resonance, the current is maximum and so power is maximum. For practical applications the frequency range in which the power is greater than or equal to 50% of the maximum power is chosen as useful range. It can be proved that when power is 50% of maximum value (or 1/2 times of maximum value), the current will be 1/ 2 times of maximum value. “The current response is maximum at resonance” and it decreases for increasing or decreasing frequency from resonance value. Therefore, when the frequency is decreased from resonant value we come across a frequency at which the power is 1/2 times that of maximum value (or current is 1/ 2 times that of maximum value), and this frequency is called lower cut-off frequency, wl. When frequency is increased from the resonant value we come across a frequency at which the power is 1/2 times that of maximum value (or current is 1/ 2 times that of maximum value), and this frequency is called higher cut-off frequency, wh. The two cut-off frequencies are also called half-power frequencies, and they lie on either side of the resonant frequency as shown in Fig. 8.4. It can be proved that, the resonant frequency is given by the geometric mean of the two half-power frequencies, i.e., ωr = ωl ωh Now, “bandwidth can be defined as the range of frequencies over which the power is greater than or equal to 1/2 times the maximum power”.

8. 7

Circuit Theory

Alternatively, “bandwidth can be defined as the range of frequencies over which the current is greater than or equal to 1/ 2 times the maximum current”.

I Ir 1 2

Ir = 0.707 Ir

The bandwidth is given by the difference between the cut-off frequencies and it can be denoted by β. The unit of bandwidth is rad/s or Hz.

b

wl wr wh

Fig. 8.4 : Current response of RLC series circuit.

The equations for cut-off frequencies and bandwidth are given below: Higher cut-off angular frequency, ωh = R + 2L

Lower cut- off angular frequency, ωl = − R + 2L

e

w

2

R + 1 in rad/s o 2L LC e

.....(8.24)

2

R + 1 in rad/s o 2L LC

.....(8.25)

Alternatively, ωh = ωr >

1 + 2Q r

ωl = ωr >−

1 + 2Q r

1+

1 4Q r2

1+

H in rad/s

1 4Q r2

.....(8.26)

H in rad/s

Higher cut- off frequency, fh = 1 > R + 2π 2 L Lower cut - off frequency, fl = 1 >− R + 2π 2L Bandwidth, β = R in rad/s L

.....(8.27)

e

R + 1 H in Hz o 2L LC

e

2

R + 1 H in Hz o 2L LC

.....(8.28)

2

.....(8.29)

.....(8.30)

Alternatively, Bandwidth, β = ωr in rad/s Qr Bandwidth in Hz =

β = R in Hz 2π 2πL

.....(8.31)

.....(8.32)

8. 8

Chapter 8 - Series And Parallel Resonance

Proof for cut-off frequencies and bandwidth

Z = R + jwL - j

Consider an RLC series circuit shown in Fig. 8.5, excited by a sinusoidal voltage source of frequency, w. R

1 wC

C

L

Let, V = V∠0 o = Supply voltage. I = Current through the RLC series circuit. Z = Impedance of RLC series circuit.

I

+

Here, Z = R + jωL − j 1 = R + j aωL − 1 k ωC ωC

~

-

V, w

Fig. 8.5.

2

R 2 + aωL − 1 k ωC

` Z = Z =

.....(8.33)

Here, I = V Z ` I = I =

` I =

V

V Z

= V Z

= Z

V 2 R + aωL − 1 k ωC

Using equation (8.33)

2

.....(8.34)

Let, P = Power in RLC circuit. Pr = Power in RLC circuit at resonance. Here, P = I 2 R =

V2 R 2 R 2 + aωL − 1 k ωC

Using equation (8.34)

.....(8.35)

At resonance, ωL − 1 = 0. On substituting this condition in equation (8.35), we get power at resonance, Pr. ωC 2 2 ` Pr = V 2R = V R R

.....(8.36) 2

2

Note : Alternatively, Pr = Ir2 R = a V k R = V R R

At half-power frequencies or cut-off frequencies the power will be equal to half the power at resonance. ` 1 # Pr = P 2 1 # V2 = 2 R

V2 R

2 R + aωL − 1 k ωC On cross multiplying the above equation, we get, 2

2 R 2 + aωL − 1 k = 2R 2 ωC

⇒

Using equations (8.35) and (8.36)

2 2 1 2 aωL − ωC k = 2R − R

2 ` aωL − 1 k = R 2 ωC

On taking square root of above equation, we get, ωL − 1 = ! R ωC

.....(8.37)

8. 9

Circuit Theory

Note : Equation (8.37) implies that the absolute value of total reactance at half-power frequencies is equal to resistance of the circuit. ` ωh L − 1 = R ; ωl L − 1 = − R ωh C ωl C On multiplying the equation (8.37) by w , we get, L ω2 − 1 = ! R ω LC L

ω2 " R ω − 1 = 0 L LC

⇒

` ω2 − R ω − 1 = 0 L LC

and ω2 + R ω − 1 = 0 L LC

The roots of quadratic ω2 − R ω − 1 = 0 are, L LC R! ω = L

The roots of quadratic ω2 + R ω − 1 = 0 are, L LC

R 2 4 a L k + LC 2

=

R ! 1 2L 2

=

R ! 2L

−R ! L ω =

2 4< 1 a R k + 1 F 4 L LC

R 2 4 a L k + LC 2

= − R ! 1 2L 2

R 2 1 a 2L k + LC

= − R ! 2L

2 4< 1 a R k + 1 F 4 L LC

R 2 1 a 2L k + LC

The cut-off frequencies are given by the positive roots of the two quadratic. ` Higher cut- off angular frequency, ωh =

R + 2L

R 2 1 a 2L k + LC in rad/s

2 Lower cut-off angular frequency, ωl = − R + a R k + 1 in rad/s 2L 2L LC Since, w = 2pf and f = ω , the cut-off frequency in Hz can be expressed as shown below: 2π

` Higher cut- o ff frequency, f h = Lower cut- off frequency, f l =

.....(8.38) .....(8.39)

1 1 + 2Qr = ωr2 >

1 + 1 2 H # ωr >− 1 + 2Qr 4Qr

1+ 1 2 + 1 H > 2Qr 4Qr

= ωr2 Q2 > Q3 Q3

Ir

Q2 Q1

Also, from the expression of bandwidth we can say that, w wr when quality factor is high, the bandwidth will be small and the Fig. 8.7 : Current response of circuit is highly selective. The current response of series RLC RLC series resonant circuit for various values of Q. resonant circuit for various values of Q are shown in Fig. 8.7. 8.2.6 Voltage Across R, L and C at Resonance in Series RLC Circuit Consider an RLC series circuit shown in Fig. 8.8, excited by a sinusoidal voltage source of V with variable frequency w. Let, the circuit resonate at a frequency wr. At resonance, I = I r = V R

Using equation (8.8)

.....(8.50)

Voltage across resistance at resonance, VRr = I r R = V # R = V .....(8.51) R

jwrL

R +

VRr

I = Ir

- +

VLr

- +

1 wrC -

-j VCr

~

+ V, w = w r

Fig. 8.8. Using equation (8.50)

8. 12

Chapter 8 - Series And Parallel Resonance

Voltage across inductance at resonance, VLr = I r # jωr L = V # jωr L R = jVQr

Using equation (8.50)

Qr =

ωr L 7equation (8.11) A R

= VQr∠90o V

......(8.52)

\ Magnitude of voltage across inductance at resonance, VLr = VQ r

......(8.53)

Voltage across capacitance at resonance, VCr = I r # e − j 1 o = V # e − j 1 o ωr C ωr C R = – jVQr

Qr =

Using equation (8.50)

1 7equation (8.12) A ωr CR

= VQr∠-90o V

.....(8.54)

\ Magnitude of voltage across capacitance at resonance, VCr = VQ r

.....(8.55)

From the above analysis we can say that the magnitude of voltage across inductor and capacitor is Qr times the supply voltage at resonance, but they are in-phase opposition. Therefore, the sum of inductance and capacitance voltage at resonance will be zero. (i.e., VLr + VCr = jVQ r − jVQ r = 0 ). At resonance the full supply voltage will be available across the resistance. Since the voltage across the inductance or capacitance is Q r times the supply voltage we can say that “the series RLC circuit amplifies the voltage”. 8.2.7 Solved Problems in Series Resonance EXAMPLE 8.1

12W

0.15H

22 mF

(AU June’14, 16 Marks)

For the RLC circuit shown in Fig. 1, determine the impedance at a) resonant frequency, b) 10 Hz below resonant frequency, and c) 10 Hz above resonant frequency. +

~

-

V

SOLUTION

Fig. 1.

Given that, R = 12 Ω ; L = 0.15 H Angular resonant frequency, ωr = Resonant frequency, fr =

and C = 22 µF = 22 × 10

1 = LC

–6

F

1 = 550.4819 rad/ sec 0.15 # 22 # 10- 6

ωr = 550.4819 = 87.6119 Hz 2π 2π

Let, Z be the impedance of an RLC series circuit. With reference to Fig. 1, we get, Z = R + jωL − j 1 = R + j eωL − 1 o ωC ωC

a) Impedance at resonance Let, Zr be the impedance at resonance frequency. At resonant frequency wr, the total reactance is zero, i.e., ωr L − 1 = 0 and the impedance is equal to resistance. ωr C ` Impedance of resonance, Zr = R = 12 Ω

8. 13

Circuit Theory b) Impedance at 10 Hz below resonant frequency

Let, Z1 be the impedance at 10 Hz below resonant frequency and w1 be the corresponding frequency. Now, w1 = 2p (fr − 10) = 2p × (87.6119 − 10) = 487.6499 rad/s ` Z1 = R + j eω1 L − 1 o ω1 C = 12 + j e 487.6499 # 0.15 −

1 o 487.6499 # 22 # 10- 6

= 12 − j20.0639 Ω = 23.3786∠−59.1o Ω

c) Impedance at 10 Hz above resonant frequency Let, Z 2 be the impedance at 10 Hz above resonant frequency and w2 be the corresponding frequency. Now, w2 = 2p (fr + 10) = 2p × (87.6119 + 10) = 613.3137 rad/s ` Z2 = R + j eω2 L − 1 o ω2 C = 12 + j e613.3137 # 0.15 −

1 o 613.3137 # 22 # 10- 6

= 12 + j17.884 Ω = 21.5369∠56.1o Ω

EXAMPLE 8.2

0.03 H 100 mF

5‡

For the circuit shown in Fig. 1, determine the frequency at which the circuit resonates. Also find the quality factor, voltage across inductance and voltage across capacitance at resonance. +

SOLUTION

E

Fig. 1.

Given that, R = 5 Ω ; L = 0.03 H ; C = 100 µF and Supply voltage, V = 20 V Angular frequency of resonance, ωr =

Resonant frequency, f r =

~

20 V

1 = LC

1 = 577.3503 rad/s 0.03 # 100 # 10- 6

ωr = 577.3503 = 91.8882 Hz 2π 2π

Quality factor at resonance, Qr =

ωrL = 577.3503 # 0.03 = 3.4641 R 5

Voltage across inductance at resonance, VLr = jQr V = j3.4641 # 20 = j69.282 V = 69.282∠90o V Voltage across capacitance at resonance, VCr = − jQr V = − j3.4641 # 20 = –j69.282 V = 69.282∠-90o V

8. 14

Chapter 8 - Series And Parallel Resonance Alternatively, VLr and VCr can be computed as shown below: Current at resonance, Ir = V = 20 = 4 A R 5 VLr = Ir # jω r L = 4 # j577.3503 # 0.03 = j69.282 V = 69.282+90 o V o 1 VCr = Ir # e− j 1 o = 4 # e− j o = − j69.282 V = 69.282+ − 90 V ω rC 577.3503 # 100 # 10- 6

EXAMPLE 8.3 A series RLC circuit has an impedance of 40 Ω at a frequency of 200 rad/s. When the circuit is made to resonate by connecting a 10 V source of variable frequency the current at resonance is 0.5 A and the quality factor at resonance is 10. Determine the circuit parameters.

SOLUTION Given that, supply voltage, V = 10 V We know that, Ir = V R We know that, Qr = 1 R ` C =

;

I r = 0.5 A ; Qr = 10

;

Z = 40 Ω

~ = 200 rad/s

` R = V = 10 = 20 Ω Ir 0.5 L ; C

` Q2r = 12 L R C

⇒ C = 12 12 L Qr R

1 # 1 # L = 2.5 # 10- 5 L 102 202

The magnitude of impedance, Z of the RLC series circuit is given by, Z =

2 R 2 + d ωL − 1 n ωC

` ωL − 1 = ωC

2 ⇒ Z 2 = R 2 + d ωL − 1 n ωC

⇒

2 Z 2 − R 2 = d ωL − 1 n ωC

Z2 − R2

On substituting w = 200 rad/s, Z = 40 Ω, R = 20 Ω and C = 2.5 × 10 –5L in the above equation we get, 200 L −

1 = 200 # 2.5 # 10- 5 L

402 − 202

⇒ 200 L − 200 = 34.641 L

On multiplying throughout by L/200 we get, L # 200L − L # 200 = L # 34.641 ⇒ 200 200 L 200

L2 − 1 = 0.1732 L

\ L2 – 0.1732 L – 1 = 0 ` L = 0.1732 !

0.17322 + 4 = 1.0903 H 2

` C = 2.5 # 10- 5 L = 2.5 # 10- 5 # 1.0903 = 2.72575 # 10- 5 F

RESULT

= 27.2575 # 10- 6 F = 27.2575µF

The circuit parameters R, L and C are, R = 20 Ω ; L = 1.0903 H

and C = 27.2575 µF

Taking positive value

8. 15

Circuit Theory EXAMPLE 8.4

An RLC series circuit consists of R = 16 Ω, L = 5 mH and C = 2 µF. Calculate the quality factor, bandwidth and half-power frequencies.

SOLUTION Angular resonant frequency, ωr =

1 = LC

1 = 10, 000 rad/s 5 # 10- 3 # 2 # 10- 6

10, 000 # 5 # 10- 3 Quality factor at resonance, Qr = ωr L = = 3.125 R 16 10, 000 Bandwidth, β = ωr = = 3200 rad/s Qr 3.125 Bandwidth in Hz =

β = 3200 = 509.2958 Hz 2π 2π

Lower angular cut − off frequency, ωr = − R + 2L

d

Higher angular cut − off frequency, ω h = R + 2L

d

Here,

R 2+ 1 n 2L LC

R 2+ 1 n 2L LC

R = 16 = 1600 2L 2 # 5 # 10- 3 1 = 1 = 108 LC 5 # 10- 3 # 2 # 10- 6 ` ωl = − 1600 + 16002 + 108 = 8527.1911 rad/s ωh = 1600 + 16002 + 108 = 11727.1911 rad/s

Lower cut − off frequency, fl =

ωl = 8527.1911 = 1357.1446 Hz 2π 2π

Higher cut − off frequency, fh =

ωh = 11727.1911 = 1866.4404 Hz 2π 2π

EXAMPLE 8.5 An RLC series circuit is to be designed to produce a magnification of 10 at 100 rad/s. The 100 V source connected to an RLC series circuit can supply a maximum current of 10 A. The half-power frequency impedance of the circuit should not be more than 14.14 Ω. Find the values of R, L and C.

SOLUTION Given that wr = 100 rad/s, Qr = 10, V = 100 V, Ir = 10 A. The current will be maximum only at resonance. Hence, 10 A current can be considered as current at resonant condition. Current at resonance, Ir = V R ` R = V = 100 = 10 Ω Ir 10

8. 16

Chapter 8 - Series And Parallel Resonance Quality factor at resonance, Qr = ` L =

ωr L R

Qr R = 10 # 10 = 1 H 100 ωr

Angular higher cut − off frequency, ωh = ωr < 1 + 2Qr = 100
In c1 − 100 m H 220

.125 MΩ = 2.0622 MΩ = 2062244.125 Ω = 2062244 106 Case ii : To find time for lamp to strike if R = 5 MΩ From equation (2), we get, Time, t = − RC ;In c1 −

vC (t) mE 220

6

Here, R = 5 M Ω = 5 × 10 Ω , C = 4 µF = 4 × 10 − 6 F , vC (t) = 100 V ` t = − 5 # 106 # 4 # 10- 6 ;In c1 − 100 mE 220 = 12.1227 seconds

RESULT 1. For lamp to strike at 5 seconds after switching, R = 2.0622 M Ω 2. When R = 5 M Ω, the time for lamp to strike after switching, t = 12.1227 seconds

EXAMPLE 10.17

t=0

1000 ‡

In the RC circuit of Fig. 1, when the switch is closed at t = 0, the current through the circuit is i (t) = 0.075 e−50t A. Find the value of Q0 and its polarity.

+ 50 V

SOLUTION

i(t) E

Given that, Initial charge = Q 0 Q ` Initial voltage, V0 = 0 C

Fig. 1.

Q0

20 mF

10. 72

Chapter 10 - Transients R

Let, L "i (t) , = I (s), L "50 , = 50 s

+

The s-domain equivalent of the given circuit is shown in Fig. 2. In Fig. 2, the polarity of Q 0 is assumed to be opposing the direction of current.

RI(s)

E

+ 50 s E

+ 1 I(s) sC E

1 sC

I(s) + E

V0 Q0 a s sC

With reference to Fig. 2, by KVL, we can write,

Fig. 2.

Q RI (s) + 1 I (s) + 0 = 50 sC sC s Q I (s) 1; Overdamped response. 2Lωn ζ2 − 1 C = Damping ratio. where, ζ = R 2 L 1 ωn = = Natural frequency of oscillation. LC

i (t) =

ωd = ωn 1 − ζ2 = Damped frequency of oscillation.

Circuit Theory 24.

10. 95

The ratio of resistance of the circuit and resistance for critical damping (or critical resistance) is called damping ratio, ζ. R ` Damping ratio, ζ = R = = R 2 RC L 2 C

25.

C L

The critical resistance, RC is the value of the resistance of the circuit to achieve critical damping and it is given by, R C = 2 L . C

10.12 Short-answer Questions Q10.1

What is transient? The state (or condition) of the circuit from the instant of switching to attainment of steady state is called transient state or simply transient.

Q10.2

Why does transient occur in electric circuits? The inductance will not allow sudden change in current and the capacitance will not allow sudden change in voltage. Hence, in inductive and capacitive circuits (or in general in reactive circuits) transient occurs during switching operation.

Q10.3

What are free and forced response?

(AU June’14, 2 Marks)

The response of a circuit due to stored energy alone is called free response and the response of a circuit due to an external source is called forced response.

Q10.4

What is a complementary function? The part of the response or solution which becomes zero as t tends to infinity is called complementary function. It is the transient part of the response or solution.

Q10.5

What is a particular solution? The part of the solution or response which attains a steady value as t tends to infinity is called particular solution. It is the steady state part of the solution.

Q10.6

Define time constant of an RL circuit.

(AU June’14, 2 Marks)

The time constant of an RL circuit is defined as the time taken by the current through the inductance to reach steady value if initial rate of rise is maintained.

Q10.7

What is the time constant of an RL circuit with R = 10 W and L = 20 mH ? -3 Time constant, τ = L = 20 # 10 = 2 # 10- 3 second = 2 ms R 10

Q10.8

What is the time constant of an RL circuit shown in Fig. Q10.8.1. Let us find an equivalent inductance at terminals A-B using Thevenin’s theorem as shown below:

0.3 H

E +E

0.9 H

Using Thevenin’s equivalent, the given network can be drawn as shown in Fig. Q10.8.5.

Fig. Q10.8.1.

10 ‡

10. 96

Chapter 10 - Transients Now, Time constant, τ = 0.3 H

E +-

0.3 H

A

12 W

0.9 H

Leq = 0.225 = 0.0225 second = 22.5 ms 10 R

S.C.

A

A

Leq

0.9 H

L eq =

Leq

Q10.9

0.3 ´ 0.9 0.3 + 0.9

Vth +-

10 W

= 0.225 H

B

B

B

Fig. Q10.8.3.

Fig. Q10.8.2.

Leq = 0.225 H

Fig. Q10.8.5.

Fig. Q10.8.4.

(AU June’14, 2 Marks)

Define time constant of an RC circuit.

The time constant of an RC circuit is defined as the time taken by the voltage across the capacitance to reach steady value if initial rate of rise is maintained.

Q10.10

What is the time constant of an RC circuit with R = 10 kW and C = 40 mF ? 3

Time constant, t = RC = 10 × 10 × 40 × 10

Q10.11

−6

= 0.4 second.

What is the time constant of an RC circuit shown in Fig. Q10.11.1. 10 ‡

Let us find an equivalent resistance at terminals A-B using Thevenin’s theorem as shown below: Using Thevenin’s equivalent, the given network can be drawn as shown in Fig. Q10.11.5.

E +E

40 ‡

0.5 F

Now, Time constant, t = R eq C = 8 × 0.5 = 4 seconds.

Fig. Q10.11.1. 10 W

E +-

10 W

A

40 W

0.5 F S.C.

10 ´ 40 10 + 40 = 8W

Req

40 W

Req =

Req B

Fig. Q10.11.2. Q10.12

B

Fig. Q10.11.3.

Req = 8 W

A

A

Vth +-

0.5 F

B

Fig. Q10.11.4.

Fig. Q10.11.5.

What is damping ratio? The ratio of resistance of the circuit and resistance for critical damping is called damping ratio.

Q10.13

What is critical damping? The critical damping is the condition of the circuit at which the oscillations in the response are just eliminated. This is possible by increasing the value of resistance in the circuit.

Q10.14

What is critical resistance? The critical resistance is the value of the resistance of the circuit to achieve critical damping.

Circuit Theory Q10.15

10. 97

Write the expression for critical resistance and damping ratio of an RLC series circuit. Critical resistance, RC = 2

L C

Damping ratio, ζ = R = R RC 2

Q10.16

C L

What is natural and damped frequency? The response of a circuit is completely oscillatory with a frequency, ωn in the absence of resistance and this frequency ωn, is called natural frequency. The response of underdamped circuit is oscillatory with a frequency of ωd and this oscillations are damped as t tends to infinity. The frequency of damped oscillatory response is called damped frequency.

Q10.17

Write the condition for underdamping and critical damping in the RLC series circuit. 2 The condition for underdamping is, c R m < 1 2L LC 2 The condition for critical damping is, c R m = 1 2L LC

Q10.18

An RL series circuit with R = 10 W is excited by a dc voltage source of 30 V by closing the switch at t = 0. Determine the current in the circuit at t = 2 t. t Current, i (t) = E ^1 − e- x h R

= 30 ^1 − e- 2h = 2.594 A 10

Q10.19

An RC series circuit is excited by a dc voltage source of 80 V by closing the switch at t = 0. Determine the voltage across the capacitor in a time of one time constant. t

Voltage across capacitor, vC (t) = E ^1 − e- x h = 80 ^1 − e- 1h = 50.5696 V

Q10.20 A 50 mF capacitor is discharged through a 100 kW resistor. If the capacitor is initially charged to 400 V, determine the initial energy. (AU Dec’14, 2 Marks) Initial energy = 1 CV02 = 1 # 50 # 10- 6 # 4002 2 4 = 4 Joules

Q10.21

An RLC series circuit with R = 5 W is excited by a dc source of 10 V by closing the switch at t = 0. Draw the initial and final condition of the circuit. 5W

O.C. _ + vL (0 + ) = 10V

5W

S.C. _ + vC (0 + ) = 0

+ 10 V

O.C. _ + vC(¥) = 10 V

+ +

-

S.C. _ + vL(¥) = 0

i(0 ) = 0

Fig. Q10.21.1 : Initial condition.

10 V -

i(¥) = 0

Fig. Q10.21.2 : Final condition.

10. 98

10.13

Chapter 10 - Transients

EXERCISES

I. Fill in the Blanks With Appropriate Words 1.

The time duration from the instant of switching till the attainment of steady state is called ________ .

2.

The current and voltage of circuit elements during transient period is called ________ .

3.

The response of a circuit due to ________ alone is called natural response.

4.

The complementary function is also called ________ .

5.

In _______ the current at t = 0− is equal to current at t = 0 .

6.

In circuits excited by dc source at steady state the _______ behave as a short circuit and ________ behave as an open circuit.

7.

In circuits excited by dc source when there is no stored energy at initial state ________ behave as a short circuit and ________ behave as an open circuit.

8.

The time constant of an RL circuit with R = 5 Ω and L = 0.2 H is ________ .

9.

The time constant of an RC circuit with R = 200 Ω and C = 10 µF is ________ .

+

10. The steady state current in an RC series circuit with R = 100 Ω excited by a dc source of 10 V is ________ . 11. The steady state voltage across the inductance in an RL circuit with R = 5 Ω, excited by a dc source of 20 V is ________ . 12. The ratio of resistance of the circuit and resistance for critical damping is called ________ . 13. The ________ is the condition for critical damping in the RLC series circuit. 14. The frequency of oscillatory response of a circuit with zero resistance is called ________ . 15. The resistance of the circuit at critical damping is called ________ .

ANSWERS 1.

transient period

6.

inductance, capacitance

11. zero

2.

transient response

7.

capacitance, inductance

12. damping ratio

3.

stored energy

8.

0.04 second

2 13. c R m = 1 2L LC

4.

natural response

9.

2 ms

14. natural frequency

5.

inductance

10. zero

15. critical resistance

II. State Whether the Following Statements are True/False 1.

Transients are due to energy storage elements.

2.

There is no transient in resistive circuits.

3.

The complementary function depends on the nature of the exciting source.

4.

The particular solution depends on the nature of the circuit.

Circuit Theory

10. 99 +

5.

In a capacitance the voltage at t = 0− is equal to the voltage at t = 0 .

6.

The steady state value does not depend on initial conditions.

7.

The transient in a circuit exist for a period of five time constant.

8.

The time constant of a circuit does not depend on R, L and C parameters.

9.

In circuits with energy storage elements, the response can attain steady state in a time of one time constant if initial rate of change is maintained.

10. An inductance with stored energy behave as a current source at t = 0. 11. A capacitance with stored energy behave as a voltage source at t = 0. 12. The oscillations in the response can be reduced by reducing the resistance of the circuit. 13. Damping ratio can be adjusted by varying the capacitance of the circuit. 14. The critical resistance does not depend on inductance and capacitance. 15. The natural frequency depends on damping ratio.

ANSWERS 1.

True

4. False

7. True

10. True

13. True

2.

True

5. True

8. False

11.

14. False

3.

False

6. True

9. True

12. False

True

15. False

III. Choose the Right Answer for the Following Questions 1. The time constant of the RL series circuit with R = 5 W and L = 0.1 H is, a) 0.05 ms

b) 20 ms

c) 0.5 ms

d) 2 ms

2. The current through the RL circuit excited by a 10 V dc source is given by i(t) = 2 (1 – e –10t) A. What is the value of R and L ? a) 5 W, 0.5 H

b) 2 W, 0.1 H

c) 5 W, 0.2 H

d) 20 W, 0.5 H

3. An RL series circuit with R = 10 W and L = 0.2 H is excited by a dc supply of 15 volts by closing the switch at t = 0. The voltage across the inductance is, a) 1.5 e–50 t V

b) 1.5 e–0.02 t V

c) 15 e–50 t V

d) 15 e–0.02 t V

4. The steady state voltage across the inductance in an RL circuit with R = 5W and excited by a dc source of 20 V is, a) 20 V

b) – 20 V

c) 4 V

d) 0 V

5. An RL circuit with R = 12 W and L = 0.2 H is excited by a dc source of 24 V by closing the switch at t = 0. The initial and final current in through the circuit are respectively, a) 2 A, 0 A

b) 0 A, 2 A

c) 0.50 A, 0 A

d) 0 A, 0.5 A

10. 100

Chapter 10 - Transients

6. The time constant of the RC series circuit with R = 200 W and C = 100 mF is, a) 0.05 ms

b) 5.0 ms

c) 2 ms

d) 20 ms

7. The current through the RC circuit excited by a 5 V dc source is given by, i(t) = 2.5 e –20 t A. What is the value of R and C ? a) 2 W, 0.025 F

b) 0.5 W, 20 F

c) 5 W, 0.2 F

d) 12.5 W, 0.1 F

8. An RC series circuit with R = 10kW and C = 1mF is excited by a dc supply of 20 V by closing the switch at t = 0. The voltage across the resistance is, a) 2 e–100 t mV

b) 20 e–0.01 t mV

c) 20 e–100 t V

d) 2 e–0.01 t V

9. The steady state current through the capacitance in an RC circuit with R = 100 W and excited by a dc source of 10 V is, a) 10 A

b) –10 A

c) 0.1 A

d) 0 A

10. An RC circuit with R = 150 W and C = 2 mF is excited by a dc source of 15 V by closing the switch at t = 0. The initial and final voltage across the capacitor are respectively, a) 10 V, 15 V

b) 15 V, 0 V

c) 0 V, 15 V

d) 0 V, 10 V

11. What is the time constant of the RC circuit shown in Fig. 11? 30 ‡

a) 4 ms b) 40 ms

20V +E

60 ‡

0.002 F

c) 0.25 ms Fig. 11.

d) 25 ms 12. What is the time constant of the RC circuit shown in Fig. 12? R

a) 3 RC b) 2 RC 9

C

2R

c) 9 RC 2

2C

Fig. 12.

d) 2 RC

13. What is the steady state current through the inductance in the circuit shown in Fig. 13? a) 0.5 A 60 ‡

+

b) 2 A

120 V

0.1 F E

12 mH

c) 3 A d) 4 A

Fig. 13.

Circuit Theory

10. 101

14. In series RLC circuit, what is the condition for critically damped response? R = 1 2L LC

a)

b)

R = L 2L C

c)

R = 1 2 a LC k 2L

2 d) a R k = 1 2L LC

15. An RLC series circuit with R = 10 W, L = 0.01 H and C = 1 mF is excited by a dc source of 16V by closing the switch at t = 0. The initial and final voltage across the capacitor are respectively, a) 0 V, 1.6 V

b) 0 V, 16 V

c) 1.6 V, 16 V

d) 1.6 V, 0 V

ANSWERS 1. b

5. b

9. d

13. b

2. a

6. d

10. c

14. d

3. c

7. a

11. b

15. b

4. d

8. c

12. d

IV. Unsolved Problems E10.1

A steady current of 12.5 A is established through an inductance of 0.4 H by connecting it to a current source. At time t = 0, the inductance is disconnected from the source and connected to a resistance of 32 W . Find an expression for the current through the resistance. Also draw the initial and final condition of the circuit.

E10.2

In the RL circuit shown in Fig. E10.2, the switch is closed at t = 0. Find the current through the di (t) d 2 i (t) circuit and voltage across inductance and resistance. Also determine at t = 0 +. and dt dt 2 t=0

20 ‡

t=0

8‡

1

2

+ 12 V

i(t)

0.2 H

E

Fig. E10.2.

10 V +E

+ 24 V E

0.8 H

Fig. E10.3.

E10.3

In the RL circuit shown in Fig. E10.3, the switch is closed at position-1 for a long time and then thrown to position-2 at t = 0. Determine the response i(t). Also draw the initial and final condition of the circuit.

E10.4

In the RL circuit of Fig. E10.4, the switch is closed to position-1 at t = 0. Then at t = 0.24 second the switch is moved to position-2. Determine the response i(t) and sketch the response. Also determine the time at which i(t) is zero.

10. 102

Chapter 10 - Transients 10 ‡

2‡

t=0

1 t=0

1

20 ‡

2

2 E +

10 V

i(t) 5V

1.2 H

+ E 12 V

+ e(t)

E +

0.7 H

~

E

e(t) = 20 sin(120t + 150) V

Fig. E10.4.

Fig. E10.5.

E10.5

The switch in the circuit of Fig. E10.5 is closed at position-1 for a long time. At time t = 0, the switch is moved to position-2. Find i(t) for t ≥ 0.

E10.6

A voltage of 12 V is established across a capacitor of 100 mF by connecting it to a voltage source. At time t = 0, the capacitor is disconnected from the source and connected to a resistance of 10 k W . Find an expression for current through the resistance. Also draw the initial and final condition of the circuit. In the RC circuit shown in Fig. E10.7, the switch is closed at t = 0. Determine the current and dv (t) d 2 vC (t) voltage in the resistance and capacitance. Also determine C at t = 0 +. and dt dt 2

E10.7

t=0

250 W

t=0

150 W

1

18 V

+ -

0.8 mF

2

5 V +-

+ vC(t)

+ 20 V -

Fig. E10.7.

160 mF

-

Fig. E10.8.

E10.8

In the RC circuit shown in Fig. E10.8, the switch is closed to position-1 for a long time. At time t = 0, the switch is moved to position-2. Determine and sketch the voltage across capacitance for t ≥ 0. Also draw the initial and final condition of the circuit.

E10.9

A capacitor of 100 mF has to be charged to a voltage of 160 V by connecting it to a dc source of 200 V. Determine the value of series resistance required to charge the capacitor in 0.1 second.

E10.10

In the RC circuit shown in Fig. E10.10, the capacitor has an initial charge of 300 mC. If the switch is closed at t = 0, determine the time at which the capacitor voltage is zero. Also estimate and sketch the current and voltage in the capacitor for t ≥ 0. t=0

+ 18 V

t=0

800 ‡

E 60 mF

E

Fig. E10.10.

300 mC +

240 ‡

+ e(t)

~

75 mF

E 0.45 mC +

E

Fig. E10.11.

Circuit Theory

10. 103

E10.11

The RC circuit shown in Fig. E10.11 is excited by a sinusoidal voltage source, e(t) = 60 sin (144t + φ) V by closing the switch at φ = 20o. Determine the current and voltage in the capacitor.

E10.12

An RLC series circuit is excited by a dc source of 80 V. If the initial current through the inductance is 2 A opposing the circuit current, determine the current through the circuit and voltage across the capacitor. Take R = 10 W, L = 0.1 H and C = 4 mF. Also draw the initial and final state of the circuits.

E10.13

An RLC series circuit with R = 40 W, L = 0.8 H and C = 200 mF is connected to a dc source of 100 V by closing the switch at t = 0. Determine the current and voltage in the inductance. Take initial charge on the capacitor as 4 mC opposing the capacitor voltage.

E10.14

In the RLC circuit shown in Fig. E10.14 find an expression for i(t) for t ≥ 0. Also calculate the di (t) d 2 i (t) value of at t = 0 +. and dt dt 2 t=0

32 ‡

50 mH

t=0

0.5 mF

+ e(t) i(t) E

~

i(t)

Fig. E10.15.

The RLC series circuit shown in Fig. E10.15 is excited by a sinusoidal source of value e(t) = 40 sin (415t + φ) V by closing the switch at φ = 0. Find an expression for the response i(t).

ANSWERS t

i (t) = 12.5 e- 80t A = 12.5 e- 0.0125 A i(0+) = 12.5 A

;

i(∞) = 0

32 W

12.5 A

32 W

i(0+) = 12.5 A

O.C.

t

i (t) = 1.5 ^1 − e- 40th A = 1.5 ^1 − e- 0.025 h A t

i(¥) = 0

Fig : Final condition.

Fig : Initial condition. E10.2

32.552 mF

E

Fig. E10.14.

E10.1

0.08 H

+

90 V

E10.15

147.2 ‡

t

vR (t) = 12 ^1 − e- 0.025 h V

;

vL (t) = 12 e- 0.025 V

di (0+) = 60 A/s dt

;

d 2 i (0+) = − 2400 A/s2 dt 2

10. 104 E10.3

Chapter 10 - Transients t

i (t) = 1.2 − 1.7 e- 0.04 A

i (0+) = − 0.5 A

;

;

i (3) = 1.2 A 20 W

20 W

24 V +-

24 V +-

0.5 A

+

i(0 ) = -0.5 A

Fig: Initial condition. E10.4

t

i (t) = − 1 + e- 0.12 A

;

i (t) = 0.6 − 1.4647 e-

i(¥) = 1.2 A

S.C.

Fig : Final condition.

for 0 # t # 0.24 sec

^t - 0.24h

0.06

A ; for

t $ 0.24 sec

t0 = 0.2935 second, (Time at which i(t) is zero) i(t) in A 0.6 0.4 0.2

t in sec

0 0.1

0.3

0.2

0.4 0.5

E0.2 t0

E0.4

t0 = 0.2935 sec

E0.6 E0.8 E0.8647

E1.0

− 2.857t

E10.5

i(t) = −2.272e

E10.6

i(t) = 1.2 e mA

−t

vC(t) = 12 e

−t

V

+ 0.238 sin (120t − 73.6o ) A

;

i(0 +) = 1.2 mA

;

i(∞) = 0

;

vC(0 +) = 12 V

;

vC(∞) = 0

+ vC (0 + ) =12 V +-

10 k W

i(0+) = 1.2 mA

vC(¥) = 0 V -

E10.7

i (t) = 0.12 e

t 0.12

A t

vC (t) = 18 ^1 − e- 0.12 h V d 2 vC (0+) = − 1250 V/s2 dt 2

10 k W

Fig : Final condition.

Fig : Initial condition. -

i(¥) = 0

-

t 0.12

;

vR (t) = 18 e

;

dvC (0+) = 150 V/s dt

V

Circuit Theory

10. 105 t

vC (t) = 20 − 15 e- 0.04 V

E10.8

vC(0+) = 5 V

;

vC(∞) = 20 V

i(0+) = 0.06 A

;

i(∞) = 0 250 W

250 W

vC(t) 20 V

5 V +-

+ - V0 = 5 V

+

i(0 ) = 0.06 A

20 V +-

+ vC(¥) = 20 V

i(¥) = 0 A

-

15 V 10 V 5V

t

0

Fig : Final condition.

Fig : Initial condition. E10.9

R = 1609 W

E10.10

vC (t) = 18 − 23 e- 0.048 V

Fig : Capacitor voltage.

t

t

i (t) = 28.75 e- 0.048 mA t0 = 0.0118 second , (Time at which vC(t) = 0) vC(t) in V

i(t) in mA

20 18

28.75mA

15 10 5

t

0 t0

t

0

Fig : Capacitor current. vC(t) = 10.278e

E10.11

i(t) = − 0.043e E10.12

i(t) = 900t e

− 55.556t

− 55.556t

− 50t

− 2e

vC(t) = 80 − 4500t e +

+ 0.233 sin (144t + 41.1 o ) A

− 50t

− 50t

A ; vR(t) = 9000t e

− 80e

− 50t

− 50t

– 20 e

− 50t

;

i(∞) = 0

;

vR(∞) = 0

; vL(∞) = 0

vL(0 +) = 100 V

2A

vR(0 ) = -20V

+

+ +

+

+ + S.C. vC(0 ) = 0V _

Fig : Initial condition circuit.

− 50t

;

vC(0 +) = 0

;

vC(∞) = 80 V

10 W _ + vR(¥) = 0V

-

vL(0 ) = 100V

i(0 ) = -2 A

V ; vL(t) = 100 e

– 4500t e

− 50t

V

vR(0 +) = −20 V

+

_

+ 21.593 sin (144t − 48.9 o ) V

;

10 W

80 V

Fig : Capacitor voltage.

i(0 ) = −2 A

+

t0 = 0.0118 sec

E5

S.C. _ + vL(¥) = 0V

+ 80 V

_

i(¥) = 0A

+ O.C. vC(¥) = 80 V _

Fig : Final condition circuit.

V

10. 106 E10.13

Chapter 10 - Transients i(t) = 2 e

− 25t

sin 75t A

vL(t) = 126.491 e E10.14

− 25t

i (t) = 3.603 ^e- 70.2t − e- 569.8th A di (0+) = 1800 A/s dt

E10.15

sin(75t + 108.4 o ) V

i(t) = – 0.159 e

− 240t

;

d2 i (0+) = − 1152038 A/s2 dt2

+ 0.0896 e

− 1600t

+ 0.261 sin (415t + 15.6o ) A

Chapter 11

ONE-PORT AND TWO-PORT NETWORKS 11.1 Introduction Generally, electrical and electronic devices will have input and output terminals for connecting source and load respectively. For an useful connection, a pair of terminals are required and such a pair of terminal is called a port. When the construction of the electrical/electronic devices are quite complicated to develop a mathematical model, an alternate approach is possible by modelling the devices based on voltage and current measurements at the input and output ports. In this approach, the devices are represented by an equivalent RLC network with one or two-ports. One of the popular model in electrical engineering is ABCD parameter model of a transmission line and another popular model in electronics is h-parameter model of a transistor. Practically, one-port and two-port networks are models of devices (operated by electrical energy) that are developed based on voltage and current measurements at the input and output terminals. Representation of One-Port Network A network with one pair of terminal (or two terminals) as shown in Fig. 11.1(a) is called oneport network, and it is represented by a rectangular box with two terminals as shown in Fig. 11.1(b). R1

R2

1 1 L

C

Port-1

R, L, C Network

1¢ 1¢

Fig. a : One-port network Fig. b : Representation of one-port network Fig. 11.1 : One-port network and its representation.

Representation of Two-Port Network A network with two pair of terminal (or four terminals) as shown in Fig. 11.2(a) is called twoport network and it is represented by a rectangular box with four terminals as shown in Fig. 11.2(b). R1

L

R2 2

1

1

C1 1¢

C2

Port-1 2¢

Fig. a : Two-port network.

2

R, L, C Network

1¢

Port-2 2¢

Fig. b : Representation of two-port network.

Fig. 11.2 : Two-port network and its representation.

11. 2

Chapter 11 - One-Port and Two-Port Networks

11.2 Parameters of a One-Port Network Source

1

The one-port networks are models of devices/RLC networks with one pair of terminals called port. The terminals of the port are marked as 1,1. When an excitation source (voltage/current source) is connected to the port as shown in Fig. 11.3, a voltage and current exist at the port terminals.

I +

V

V _

One port RLC Network

1¢

Fig. 11.3 : One-port network excited by a source.

For the one-port network, two network parameters can be defined and they are driving point impedance and driving point admittance. “The driving point impedance of a port is defined as the ratio of voltage to the current in that port and it is denoted as Z11”. ` Driving point impedance, Z11 = V I

.....(11.1)

“The driving point admittance of a port is defined as the ratio of current to voltage in that port and it is denoted as Y11”. ` Driving point admittance, Y11 = I V

.....(11.2)

Here, it can be infered that, Z11 = 1 or Y11 = 1 Y11 Z11 The driving point impedance and admittance are commonly referred to as immittance. Equations (11.1) and (11.2) are used to define the one-port network parameters when the network is purely resistive. When the one-port network has an inductance and a capacitance, then they are represented as a reactance either in frequency domain or in s-domain. In such cases, the one-port network parameters will be defined as shown below: One-Port Network Parameters in Frequency Domain Driving point impedance, Z11 = V I

.....(11.3)

.....(11.4) Driving point admittance, Y11 = I V Note : In frequency domain the inductance is specified as a reactance, jwL and capacitance is specified as a reactance, 1/jwC.

One-Port Network Parameters in s-Domain Driving point impedance, Z11 (s) =

V (s ) I (s )

.....(11.5)

Driving point admittance, Y11 (s) =

I (s) V (s )

.....(11.6)

Note : In s-domain the inductance is specified as a reactance, sL and capacitance is specified as a reactance, 1/sC.

Circuit Theory

11. 3

11.3 Parameters of a Two-Port Network The two-port networks are models of devices/RLC networks with two pair of terminals. A pair of terminals is called port and so four terminals form two-ports. In general, the two-port network is represented by a rectangular box with four terminals or two-ports as shown in Fig. 11.4. Note : The rectangular box represents any RLC network with or without dependent sources, or any electrically operated devices like transistor, FET, etc. The two-port network does not include any independent source. However the independent sources can be connected to the port terminals. The two-ports are called port-1 and port-2. The port-1 is also known as input port and its terminals are marked as 1,1′. The port-2 is also known as output port and its terminals are marked as 2,2′. The voltage across port-1 terminals and the current drawn by port-1 terminals are denoted as V1 and I1 respectively. The voltage across port-2 terminals and the current drawn by port-2 terminals are denoted as V2 and I2 respectively. 1 Port-1 (Input port) 1¢

I1

I2

+

+

V1 _

RLC network with or without dependent sources

V2 _

2 Port-2 (Output port) 2¢

Fig. 11.4 : Currents and voltages of a two-port network.

In the two-port network analysis, it is assumed that only the four variables V1, V2, I1 and I2 are available for measurement and all the internal variables are not accessable. Among the four variables, usually any two variables are chosen as independent variables and the other two are chosen as dependent variables. The dependent variables can be expressed as a linear combination of independent variables. For example, let X and Y be independent variables and U and V be dependent variables. Now the dependent variables can be expressed mathematically as shown in equations (11.7) and (11.8). U = k11 X + k12 Y

..... (11.7)

V = k21 X + k22 Y

...... (11.8)

where, k11, k12, k21 and k22 are constants. The constants k11, k12, k21 and k22 are known as parameters of two-port network. Equations (11.7) and (11.8) can be expressed in matrix form as shown below: k11 k12 X U H > H > H = > k21 k22 Y V

..... (11.9)

In a two-port network, four quantities (V1, V2, I1 and I2) are available for measurement, if we choose any two as independent variable and the other two as dependent variable, then we may have six possible combinations of independent and dependent variables as listed in Table 11.1. Each combination of independent and dependent variable will give rise to a set of parameters. The name of the parameters are chosen to indicate their dimensions (impendance or admittance) or lack of consistent dimensions (hybrid) or the principal application (transmission).

11. 4

Chapter 11 - One-Port and Two-Port Networks

In general, the two-port network parameters can be evaluated by making one of the independent variable as zero. In equations (11.7) and (11.8), if Y = 0, we get k11 = U and k21 = V X X In equations (11.7) and (11.8), if X = 0, we get k12 = U and k22 = V Y Y The above logic is used to evaluate the two-port network parameters listed in Table 11.1. TABLE 11.1 : SUMMARY OF PARAMETERS OF TWO-PORT NETWORK S.No Independent Dependent Parameters variables variables k11, k12, k21, k22 1.

2.

3.

Name of the parameter

I1,I2

V1,V2

Z11,Z12,Z21,Z22

V1,V2

I1, I2

Y11,Y12,Y21,Y22 Admittance parameter (Y-parameter or Short circuit parameter)

I1,V2

V 1, I 2

h11,h12,h21,h22

Impedance parameter (Z-parameter or Open circuit parameter)

Hybrid parameter (h-parameter)

Mathemetical equation V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2 V1 Z11 Z12 I1 H > H > H=> V2 Z21 Z22 I2 I1 = Y11 V1 + Y12 V2 I2 = Y21 V1 + Y22 V2 I1 Y11 Y12 V1 H > H > H=> I2 Y21 Y22 V2 V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 V1 h11 h12 I1 H > H > H=> I2 h21 h22 V2

I1 = g11 V1 + g12 I2 Inverse hybrid parameter(g-parameter) V2 = g21 V1 + g22 I2 I1 g11 g12 V1 H > H > H=> V2 g21 g22 I2

4.

V1,I2

I1, V2

g11,g12,g21,g22

5.

V2,I2

V 1, I 1

A, B, C, D

Transmission parameter V1 = AV2 − BI2 I1 = CV2 − DI2 (ABCD-parameter) V1 A B V2 H > H > H=> I1 C D − I2

6.

V1,I1

V 2, I 2

A′, B′, C′, D′

Inverse transmission parameter (A′B′C′D′-parameter)

V2 = Al V1 − Bl I1 I2 = Cl V1 − Dl I1 V2 Al Bl V1 H > H > H=> I2 Cl Dl − I1

Circuit Theory

11. 5

The mathematical equations listed in Table 11.1 are used to define the two-port network parameters when the network is purely resistive. When the two-port network has inductance and capacitance then they are represented as a reactance either in frequency domain or in s-domain. In such cases, the twoport network parmeters will be defined as shown in Tables 11.2 and 11.3. TABLE11.2:SUMMARY OF TWO-PORT NETWORK PARAMETERS IN FREQUENCY DOMAIN

S.No

Parameters

1.

Z11, Z12, Z21, Z22 Z = >

2.

Z11 Z12 Z21 Z22

>

h11 h12 h21 h22

g11 g12 g21 g22

H

H

A, B, C, D T = >

6.

Y21 Y22

g11, g12, g21, g22 g =

5.

Y11 Y12

h11, h12, h21, h22 h = >

4.

H

Y11, Y12, Y21, Y22 Y = >

3.

Name of the parameter

A B C D

H

Al , Bl , Cl , Dl Tl = >

Al Bl Cl Dl

H

H

Impedance parameter (Z-parameter or Open circuit parameter)

Admittance parameter (Y-parameter or Short circuit parameter)

Hybrid parameter (h-parameter)

Inverse hybrid parameter (g-parameter)

Mathemetical equation V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2

>

V1 V2

H=>

I1 H > H Z22 I2

Z11 Z12 Z21

I1 = Y11 V1 + Y12 V2 I2 = Y21 V1 + Y22 V2 I1 Y11 Y12 V1 H > H > H=> I2 Y21 Y22 V2 V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 V1 h11 h12 I1 H > H > H=> I2 h21 h22 V2 I1 = g11 V1 + g12 I2 V2 = g21 V1 + g22 I2

>

I1 V2

H=>

g11 g12 g21 g22

H> H V1 I2

Transmission parameter (ABCD-parameter)

V1 = A V2 − B I2

Inverse transmission parameter (A′B′C′D′-parameter)

V2 = Al V1 − Bl I1

I1 = C V2 − D I2 V1 A B V2 H > H > H=> I1 C D − I2

I2 = Cl V1 − Dl I1

>

V2 I2

H=>

Al Bl Cl Dl

H >

V1 − I1

H

11. 6

Chapter 11 - One-Port and Two-Port Networks

TABLE 11.3 : SUMMARY OF TWO-PORT NETWORK PARAMETERS IN s-DOMAIN S.No

1.

2.

3.

Parameters

Z11(s), Z12(s), Z21(s), Z22(s)

Name of the parameter Impedance parameter (Z-parameter or Open circuit parameter)

Y11(s), Y12(s), Y21(s), Y22(s) Admittance parameter (Y-parameter or Short circuit parameter)

h11(s), h12(s), h21(s), h22(s)

Hybrid parameter (h-parameter)

Mathemetical Equation V1 (s) = Z11 (s) I1 (s) + Z12 (s) I2 (s) V2 (s) = Z21 (s) I1 (s) + Z22 (s) I2 (s)

>

V1 (s) V2 (s)

H=>

Z11 (s) Z12 (s) Z21 (s) Z22 (s)

H >

I1 (s) I 2 ( s)

H

I1 (s) = Y11 (s) V1 (s) + Y12 (s) V2 (s) I2 (s) = Y21 (s) V1 (s) + Y22 (s) V2 (s)

>

I1 (s) I 2 ( s)

H=>

Y11 (s) Y12 (s) Y21 (s) Y22 (s)

H >

V1 (s) V2 (s)

H

V1 (s) = h11 (s) I1 (s) + h12 (s) V2 (s) I2 (s) = h21 (s) I1 (s) + h22 (s) V2 (s) V1 (s) h11 (s) h12 (s) I1 (s) H=> H > H > I 2 ( s) h21 (s) h22 (s) V2 (s)

4.

g11(s), g12(s), g21(s), g22(s)

Inverse hybrid parameter I1 (s) = g11 (s) V1 (s) + g12 (s) I2 (s) (g-parameter) V2 (s) = g21 (s) V1 (s) + g22 (s) I2 (s) I1 (s) g11 (s) g12 (s) V1 (s) H=> H > H > V2 (s) g21 (s) g22 (s) I2 (s)

5.

A(s), B(s), C(s), D(s)

Transmission parameter (ABCD-parameter)

V1 (s) = A (s) V2 (s) − B (s) I2 (s) I1 (s) = C (s) V2 (s) − D (s) I2 (s)

>

6.

A′(s), B′(s), C′(s), D′(s)

Inverse transmission parameter (A′B′C′D′-parameter)

V1 (s) I1 (s)

H=>

A (s ) B (s ) C (s ) D (s )

H >

V2 (s) − I 2 ( s)

H

V2 (s) = Al(s) V1 (s) − Bl(s) I1 (s) I2 (s) = Cl(s) V1 (s) − Dl(s) I1 (s)

>

V2 (s) I 2 ( s)

H=>

Al(s) Bl(s) Cl(s) Dl(s)

H >

V1 (s) − I1 (s)

H

Note : Refer Chapter 3, Section 3.7 for representation of R, L, C parameters in various domain.

Circuit Theory

11. 7

11.4 Impedance Parameters (or Z-Parameters)

(AU May’15, 16 Marks)

The equations defining Z-parameters are, V1 = Z11 I1 + Z12 I2

..... (11.10)

V2 = Z21 I1 + Z22 I2

..... (11.11)

The Z-parameters can be evaluated by conducting the following two tests. Test 1: Excite port-1 by a voltage source, while port-2 is open I2 = 0

I1 + V1

V1 _

+ 2-port network

V2 _

O.C.

Fig. 11.5 : Excite port-1 while port-2 is open.

With reference to Fig. 11.5, when port-2 is open, I2 = 0. Now we can measure the three port quantities V1, I1 and V2. From equation (11.10), when I2 = 0,

Z11 =

V1 in Ω I1

From equation (11.11), when I2 = 0,

Z21 =

V2 in Ω I1

Test 2: Excite port-2 by a voltage source, while port-1 is open I1 = 0

I2 +

+ O.C. V 1 _

2-port network

V2 _

V2

Fig. 11.6 : Excite port-2 while port-1 is open.

With reference to Fig. 11.6, when port-1 is open, I1 = 0. Now we can measure the three port quantities V2, I2 and V1. From equation (11.10), when I1 = 0, From equation (11.11), when I1 = 0,

V1 in Ω I2 V Z22 = 2 in Ω I2

Z12 =

Z-Parameter Model of a Two-Port Network From the above two tests the Z-parameters are evaluated and the units are found to be ohms. Hence, we can say that the equations (11.10) and (11.11), are KVL equations and so they can be used to construct a model of two-port network as shown in Fig. 11.7.

11. 8

Chapter 11 - One-Port and Two-Port Networks

+

Z22

Z11

I1

+Z I 11 1

_

Z22I2 Z12I2

V1

I2

_ + E

+ E

+

+

Z21I1

V2

_

_

Fig. 11.7 : Z-parameter model of two-port network.

Names of Z-Parameters Z11 = Input impedance when output port is open. Z12 = Reverse transfer impedance when input port is open. Z21 = Forward transfer impedance when output port is open. Z22 = Output impedance when input port is open.

11.5 Admittance Parameters ( or Y-Parameters)

(AU May’15, 16 Marks)

The equations defining Y-parameters are, I1 = Y11 V1 + Y12 V2

..... (11.12)

I2 = Y21 V1 + Y22 V2

..... (11.13)

The Y-parameters can be evaluated by conducting the following two tests. Test 1: Excite port-1 by a voltage source, while port-2 is shorted I2

I1 + V1

V1 _

+ 2-port network

V2 = 0 _

S.C.

Fig. 11.8 : Excite port-1 while port-2 is shorted.

With reference to Fig. 11.8, when port-2 is shorted, V2 = 0. Now we can measure the three port quantities V1, I1 and I2. From equation (11.12), when V2 = 0,

Y11 =

I1 in M V1

I2 in M V1 Test 2 : Excite port-2 by a voltage source, while port-1 is shorted

From equation (11.13), when V2 = 0,

S.C.

Y21 =

I1

I2

+

+

V1 = 0 _

2-port network

V2 _

V2

Fig. 11.9 : Excite port-2 while port-1 is shorted.

Circuit Theory

11. 9

With reference to Fig. 11.9, when port-1 is shorted, V1 = 0. Now we can measure the three port quantities V2, I2 and I1. I1 in M V2 I Y22 = 2 in M V2

From equation (11.12), when V1 = 0,

Y12 =

From equation (11.13), when V1 = 0, Y-Parameter Model of a Two-Port Network

From the above two tests, the Y-parameters are evaluated and the units are found to be mhos. Hence, we can say that the equations (11.12) and (11.13) are KCL equations and so they can be used to construct a model of two-port network as shown in Fig. 11.10. I1 +

Y11V1

V1

Y11

Y22V2 Y12V2

Y21V1

Y22

I2 + V2 _

_

Fig. 11.10 : Y-Parameter model of two-port network.

Names of Y-Parameters Y11 = Input admittance when output port is shorted. Y12 = Reverse transfer admittance when input port is shorted. Y21 = Forward transfer admittance when output port is shorted. Y22 = Output admittance when input port is shorted.

11.6 Hybrid Parameters ( or h-Parameters)

(AU May’15, 16 Marks)

The equations defining h-parameters are, V1 = h11 I1 + h12 V2

.....(11.14)

I2 = h21 I1 + h22 V2

.....(11.15)

The h-parameters can be evaluated by conducting the following two tests. Test 1: Excite port-1 by a voltage source, while port-2 is shorted I1

I2

+ V1

V1 _

+ 2-port network

V2 = 0 _

S.C.

Fig. 11.11 : Excite port-1 while port-2 is shorted.

11. 10

Chapter 11 - One-Port and Two-Port Networks

With reference to Fig. 11.11, when port-2 is shorted, V2 = 0. Now we can measure the three port quantities V1, I1 and I2. From equation (11.14), when V2 = 0,

h11 =

V1 in Ω I1

From equation (11.15), when V2 = 0,

h21 =

I2 ^ no dimensionh I1

Test 2: Excite port-2 by a voltage source, while port-1 is open With reference to Fig. 11.12, when port-1 is open, I1 = 0. Now we can measure the three port quantities V2, I2 and V1. I1 = 0

I2 +

+ 2-port network

O.C. V 1 _

V2 _

V2

Fig. 11.12 : Excite port-2 while port-1 is open.

From equation (11.14), when I1 = 0,

h12 =

V1 ^ no dimensionh V2

From equation (11.15), when I1 = 0,

h22 =

I2 in M V2

h-Parameter Model of Two-Port Network From the above two tests, the h-parameters are evaluated. The unit of h11 is ohm, the unit of h22 is mho, and h12and h21 are dimensionless constants. Hence, we can say that the equation (11.14), is KVL equation and equation (11.15), is KCL equation. Therefore, they can be used to construct a model of two-port network as shown in Fig. 11.13. I1 + V1

h11 + h I _ 11 1 h12V2

h22V2 + E

h21I1

h22

_

I2 + V2 _

Fig. 11.13 : h-Parameter model of two-port network.

Names of h-Parameters h11 = Input impedance when output port is shorted. h12 = Reverse voltage transfer ratio when input port is open. h21 = Forward current transfer ratio when output port is shorted. h22 = Output admittance when input port is open.

Circuit Theory

11. 11

11.7 Inverse Hybrid Parameters (or g-Parameters) The equations defining g-parameters are, I1 = g11 V1 + g12 I2

.....(11.16)

V2 = g21 V1 + g22 I2

.....(11.17)

The g-parameters can be evaluated by conducting the following two tests. Test 1: Excite port-1 by a voltage source, while port-2 is open I1

I2 = 0 +

+ V1

V1 _

2-port network

V2 _

O.C.

Fig. 11.14 : Excite port-1 while port-2 is open.

With reference to Fig. 11.14, when port-2 is open, I2 = 0. Now we can measure the three port quantities V1, I1 and V2. From equation (11.16), when I2 = 0,

g11 =

I1 in M V1

From equation (11.17), when I2 = 0,

g21 =

V2 ^ no dimensionh V1

Test 2: Excite port-2 by a voltage source, while port-1 is shorted I1

I2 +

+ S.C.

V1 = 0 _

2-port network

V2 _

V2

Fig. 11.15 : Excite port-2 while port-1 is shorted.

With reference to Fig. 11.15, when port-1 is shorted, V1 = 0. Now we can measure the three port quantities V2, I2 and I1. From equation (11.16), when V1 = 0,

g12 =

I1 ^ no dimensionh I2

From equation (11.17), when V1 = 0,

g22 =

V2 in Ω I2

g-Parameter Model of Two-Port Network From the above two tests g-parameters are evaluated. The unit of g11 is mho, the unit of g22 is ohm, and g12 and g21 are dimensionless constants. Hence, we can say that the equation (11.16) is KCL equation and equation (11.17) is KVL equation. Therefore, they can be used to construct a model of two-port network as shown in Fig. 11.16.

11. 12

Chapter 11 - One-Port and Two-Port Networks g22

I1 +

_

g11V1 g11

V1

g22I2 + E

g12I2

+

g21V1

I2 + V2

_

_

Fig. 11.16 : g-Parameter model of two-port network.

Names of g-Parameters g11 = Input admittance when output port is open. g12 = Reverse current transfer ratio when input port is shorted. g21 = Forward voltage transfer ratio when output port is open. g22 = Output impedance when input port is shorted.

11.8 Transmission Parameters (or ABCD-Parameters) The equations defining ABCD-parameters are, V1 = AV2 + B (−I2) = AV2 − BI2

.....(11.18)

I1= CV2 + D (−I2) = CV2− DI2

.....(11.19)

The ABCD-parameters can be evaluated by conducting the following two tests. Note: In transmission parameters the current I2 entering the port-2 is considered as negative. Test 1: Excite port-1 by a voltage source, while port-2 is open I2 = 0

I1 + V1

V1 _

+ 2-port network

V2 _

O.C.

Fig. 11.17 : Excite port-1 while port-2 is open.

With reference to Fig. 11.17, when port-2 is open, I2 = 0. Now we can measure the three port quantities V1, I1 and V2. From equation (11.18), when I2 = 0,

A =

From equation (11.19), when I2 = 0,

C=

V1 ^ no dimensionh V2

I1 in M V2

Test 2: Excite port-1 by a voltage source, while port-2 is shorted I1

I2 +

V1

V1 _

+ 2-port network

V2 = 0 _

S.C.

Fig. 11.18 : Excite port-1 while port-2 is shorted.

Circuit Theory

11. 13

With reference to Fig. 11.18, when port-2 is shorted, V2 = 0. Now we can measure the three port quantities V1, I1 and I2. From equation (11.18), when V2 = 0,

B =-

V1 in Ω I2

From equation (11.19), when V2 = 0,

D =-

I1 ^ no dimensionh I2

Names of ABCD-Parameters A = Reverse voltage transfer ratio when output port is open. B = Reverse transfer impedance when output port is shorted. C = Reverse transfer admittance when output port is open. D = Reverse current transfer ratio when output port is shorted.

11.9 Inverse Transmission Parameters (or A′B′C′D′-Parameters) The equations defining A’ B’ C’ D’ parameters are, V2 = A’ V1 + B’ (−I1) = A’ V1 − B’ I1

..... (11.20)

I2 = C’ V1 + D’ (−I1) = C’ V1 − D’ I1

..... (11.21)

The A’ B’ C’ D’-parameters can be evaluated by conducting the following two tests. Note:Ininverse transmission parameter the current I1 entering the port-1 is considered as negative. Test 1: Excite port-2 by a voltage source, while port-1 is open I1 = 0

I2 +

+

2-port network

V1 _

V2

V2 _

Fig. 11.19 : Excite port-2 while port-1 is open.

With reference to Fig. 11.19, when port-1 is open, I1 = 0. Now we can measure the three port quantities V2, I2 and V1. V From equation (11.20), when I1 = 0, Al = 2 ^ no dimensionh V1 I From equation (11.21), when I1 = 0, Cl = 2 in M V1 Test 2: Excite port-2 by a voltage source, while port-1 is shorted I1

I2 +

+ S.C.

V1 = 0 _

2-port network

V2 _

V2

Fig. 11.20 : Excite port-2 while port-1 is shorted.

11. 14

Chapter 11 - One-Port and Two-Port Networks

With reference to Fig. 11.20, when port-1 is shorted, V1 = 0. Now we can measure the three port quantities V2, I2 and I1. From equation (11.20), when V1 = 0,

V Bl = - 2 in Ω I1

From equation (11.21), when V1 = 0,

I Dl = - 2 ^ no dimensionh I1

Names of A’B’C’D’-Parameters A’ = Forward voltage transfer ratio when input port is open. B’ = Forward transfer impedance when input port is shorted. C’ = Forward transfer admittance when input port is open. D’ = Forward current transfer ratio when input port is shorted.

11.10 Relationship Between Parameter Sets All the six parameter sets discussed above have definite relationship between them.The complete relationship between the parameters are tabulated in Table 11.4. The general concept of converting one parameter set to other, involves rearrangement of the defining equations. The conversion of Z-parameters to other parameter sets are presented in this section. All other possible conversions are left as exercise to readers. Conversion of Z-Parameters to Y-Parameters The Y-parameters Y11,Y12,Y21 and Y22 can be expressed in terms of Z-parameters as shown below: Y11 =

Z22 ; DZ

` Y=>

Y12 = -

Z12 ; DZ

Y21 = -

H = D1 > Z -Z Y21 Y22 21 Y11 Y12

where, DZ =

Z22 - Z12 Z11

Z21 ; DZ

Y22 =

Z11 DZ

H

.... (11.22)

Z11 Z12

.....(11.23)

Z21 Z22

Proof: Let, P be a square matrix

Consider Z-parameter equations, V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2

⇒

V1 Z11 Z12 I1 = G = = G = G V2 Z21 Z22 I2

Let us premultiply the above equation by = ` =

Z11 Z12 - 1 G Z21 Z22

Here, =

Now, P -1 =

Transpose of cofactor matrix of P Determinant of P

Z11 Z12 - 1 G Z21 Z22

If, P is a square matrix of size 2 × 2, then its transpose of cofactor matrix is obtained by interchanging the elements of main diagonal and changing the sign of other two elements as V1 Z11 Z12 - 1 Z11 Z12 I1 = G = = G = G = G V2 Z21 Z22 Z21 Z22 I2 ..... (11.24) shown in the following example. 1 4 4 4 44 2 4 4 4 44 3 P11 P12 P22 − P12 UNIT MATRIX P == G ; P- 1 = 1 = G ∆P − P21 P11 P21 P22

Z11 Z12 - 1 Z22 − Z12 G = 1 = G ∆Z − Z21 Z11 Z21 Z22

..... (11.25)

where, ∆ p =

P11 P12 P21 P22

Circuit Theory where, ∆Z = Also, =

11. 15 Z11 Z12 = Z11 Z22 − Z12 Z21 Z21 Z22

Z11 Z12 - 1 G Z21 Z22

=

Z11 Z12 I1 I1 G = G = = G Z21 Z22 I2 I2

.....(11.26)

Using equations (11.25) and (11.26), the equation (11.24) can be written as shown below: 1 = Z22 − Z12 G =V1 G = = I1 G ∆Z − Z21 Z11 V2 I2 The above matrix equation can be expanded as following two equations. I1 =

Z22 Z V + c− 12 m V2 ∆Z 1 ∆Z

I2 = −

.....(11.27)

Z21 Z V1 + 11 V2 ∆Z ∆Z

.....(11.28)

The general form of Y-parameter equations are, I1 = Y11 V1 + Y12 V2 I2 = Y21 V1 + Y22 V2 On comparing the above Y-parameter equations with equations (11.27) and (11.28), we get, Y11 =

Z22 ; ∆Z

Y12 = −

Z12 Z ; Y21 = − 21 ∆Z ∆Z

;

Y22 =

Z11 ∆Z

Conversion of Z-Parameters to h-Parameters The h-parameters h11, h12, h21 and h22 can be expressed in terms of Z-parameters as shown below: Z12 Z ; h21 = - 21 ; Z22 Z22 R V S DZ Z12 W h11 h12 S Z22 Z22 W ` h= > H= S W h21 h22 S- Z21 1 W S Z22 Z22 W T X Z11 Z12 where, DZ = Z21 Z22 h11 =

Dz ; Z22

h12 =

h22 = 1 Z22

.... (11.29)

Proof: Consider Z-parameter equations, V1 = Z11 I1 + Z12 I2

.....(11.30)

V2 = Z21 I1 + Z22 I2

.....(11.31)

The equation (11.31), can be rearranged as shown below: I2 = −

Z21 I + 1 V2 Z22 1 Z22

.....(11.32)

11. 16

Chapter 11 - One-Port and Two-Port Networks On substituting for I2 from equation (11.32) in equation (11.30), we get, V1 = Z11 I1 + Z12 c−

Z21 I + 1 Vm Z22 1 Z22 2

` V1 =

Z11 Z22 − Z12 Z21 Z I1 + 12 V2 Z22 Z22

` V1 =

∆Z Z I + 12 V Z22 1 Z22 2

where, ∆Z =

⇒

V1 = Z11 I1 −

Z12 Z21 Z I + 12 V Z22 1 Z22 2

.....(11.33)

Z11 Z12 = Z11 Z22 − Z12 Z21 Z21 Z22

The general form of h-parameter equations are, V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 On comparing the above h-parameter equations with equations (11.33) and (11.32), we get, h11 =

∆Z ; Z22

h12 =

Z12 Z ; h21 = − 21 Z22 Z22

;

h22 =

1 Z22

Conversion of Z-Parameters to g-Parameters The g-parameters g11, g12, g21 and g22 can be expressed in terms of Z-parameters as shown below: Z12 Z ; g21 = 21 ; Z11 Z11 R V S 1 - Z12 W g11 g12 Z11 W S Z11 ` g= > H= S W DZ W g21 g22 S Z21 S Z11 Z11 W T X Z11 Z12 where, DZ = Z21 Z22 g11 = 1 ; Z11

g12 = -

g22 =

DZ Z11

.... (11.34)

Proof: Consider Z-parameter equations, V1 = Z11 I1 + Z12 I2

.....(11.35)

V2 = Z21 I1 + Z22 I2

.....(11.36)

The equation (11.35), can be rearranged as shown below: I1 =

1 V − Z12 I Z11 1 Z11 2

.....(11.37)

On substituting for I1 from equation (11.37) in equation (11.36), we get, Z V2 = Z21 c 1 V1 − 12 I2 m + Z22 I2 Z11 Z11 ` V2 =

Z21 Z Z − Z12 Z21 V + c 11 22 m I2 Z11 1 Z11

⇒

V2 =

Z21 Z Z V − 12 21 I2 + Z22 I2 Z11 1 Z11

Circuit Theory

11. 17

` V2 =

Z21 ∆ V + Z I Z11 1 Z11 2

where, ∆Z =

..... (11.38)

Z11 Z12 = Z11 Z22 − Z12 Z21 Z21 Z22

The general form of g-parameter equations are, I1 = g11 V1 + g12 I2 V2 = g21 V1 + g22 I2 On comparing the above g-parameter equations with equation (11.37) and (11.38), we get, g11 =

1 Z11

;

g12 = −

Z12 Z11

;

Z21 Z11

g21 =

;

g22 =

∆Z Z11

Conversion of Z-parameters to Transmission Parameters The transmission parameters A, B, C and D can be expressed in terms of Z-parameters as shown below: A =

Z11 ; Z21

B =

DZ ; Z21

C = 1 ; Z21

D =

Z22 Z21

R V S Z11 DZ W S Z21 Z21 W ` T = > H= S W C D S 1 Z22 W S Z21 Z21 W T X Z11 Z12 where, DZ = Z21 Z22 A B

.... (11.39)

Proof: Consider Z-parameter equations, V1 = Z11 I1 + Z12 I2

..... (11.40)

V2 = Z21 I1 + Z22 I2

..... (11.41)

The equation (11.41), can be rearranged as shown below: I1 =

..... (11.42)

1 V − Z22 I Z21 2 Z21 2

On substituting for I1 from equation (11.42) in equation (11.40), we get, Z V1 = Z11 c 1 V2 − 22 I2 m + Z12 I2 ⇒ Z21 Z21

V1 =

` V1 =

Z11 Z Z − Z12 Z21 V − c 11 22 m I2 Z21 2 Z21

` V1 =

∆ Z11 V − Z I2 Z21 2 Z21

where, ∆Z =

Z11 Z Z V − 11 22 I2 + Z12 I2 Z21 2 Z21

Z11 Z12 = Z11 Z22 − Z12 Z21 Z21 Z22

..... (11.43)

11. 18

Chapter 11 - One-Port and Two-Port Networks The general form of transmission parameter equations are, V1 = A V2 − B I2 I 1 = C V2 − D I2 On comparing the above transmission parameter equations with equations (11.43) and (11.42), we get, A =

Z11 Z21

;

B =

∆Z Z21

;

C =

1 Z21

;

D =

Z22 Z21

Conversion of Z-Parameters to Inverse Transmission Parameters The inverse transmission parameters A, B, C and D can be expressed in terms of Z-parameters as shown below : Al =

Z22 ; Z12

` Tl = =

Al Cl

where,

Z Dl = 11 Z12

D Bl = Z ; Cl = 1 ; Z12 Z12 R V S Z22 DZ W Bl S Z12 Z12 W G= S W Dl S 1 Z11 W S Z12 Z12 W T X Z11 Z12 DZ = Z21 Z22

.... (11.44)

Proof: Consider Z-parameter equations, V1 = Z11 I1 + Z12 I2

..... (11.45)

V2 = Z21 I1 + Z22 I2

..... (11.46)

The equation (11.45), can be rearranged as shown below: I2 =

1 V − Z11 I Z12 1 Z12 1

..... (11.47)

On substituting for I2 from equation (11.47) in equation (11.46), we get, Z V2 = Z21 I1 + Z22 c 1 V1 − 11 I1 m Z12 Z12

⇒

` V2 =

Z22 Z Z − Z12 Z21 V − c 11 22 m I1 Z12 1 Z12

` V2 =

∆ Z22 V− Z I Z12 1 Z12 1

V2 = Z21 I1 −

Z11 Z22 Z I + 22 V Z12 1 Z12 1

..... (11.48)

Z11 Z12 = Z11 Z22 − Z12 Z21 Z21 Z22 The general form of inverse transmission parameter equations are, where, ∆Z =

V2 = A’ V1 − B’ I1 I2 = C ’ V1 − D’ I1 On comparing the above inverse transmission parameter equations with equations (11.48) and (11.47), we get, Z A' = 22 Z12

;

∆Z B' = Z12

;

C' =

1 Z12

;

D' =

Z11 Z12

Circuit Theory

11. 19

TABLE 11.4 : SUMMARY OF RELATIONSHIP BETWEEN PARAMETER SETS [Z]

[Y]

[T]

[T]

[h]

[g]

Z11 Z12

Y22 Y − 12 ∆Y ∆Y Y21 Y11 − ∆Y ∆Y

A DT C C 1 D C C

Dl 1 Cl Cl DTl Al Cl Cl

∆h h12 h22 h22 1 h − 21 h22 h22

g 1 − 12 g11 g11 ∆g g21 g11 g11

D ∆ − T B B 1 A − B B

Al 1 − Bl Bl ∆ Dl − Tl Bl Bl

1 h − 12 h11 h11 ∆h h21 h11 h11

∆g g12 g22 g22 g 1 − 21 g22 g22

∆h h − 11 h21 h21 1 h − 22 − h21 h21

1 g22 g21 g21 g11 Dg g21 g21

[Z] Z21 Z22

[Y]

[T]

[T]

[h]

[g]

Z22 Z − 12 ∆Z ∆Z Z21 Z11 − ∆Z ∆Z

Y11 Y12 Y21 Y22 Y22 1 − Y21 Y21 ∆ Y − Y − 11 Y21 Y21

Z11 DZ Z21 Z21 1 Z22 Z21 Z21

−

Z22 DZ Z12 Z12 1 Z11 Z12 Z12

−

A

B

C

D

Y11 1 − Y12 Y12 ∆ Y − Y − 22 Y12 Y12

D B DT DT C A DT DT

∆Z Z12 Z22 Z22 1 Z − 21 Z22 Z22

1 Y − 12 Y11 Y11 ∆Y Y21 Y11 Y11

B ∆T D D 1 C − D D

1 Z − 12 Z11 Z11 ∆Z Z21 Z11 Z11

∆Y Y12 Y22 Y22 1 Y − 21 Y22 Y22

C ∆ − T A A 1 B A A

Dl DTl Cl DTl

Bl DTl Al DTl

−

1 h − 11 h12 h12 ∆h h22 h12 h12

Al Bl Cl Dl Bl Al ∆ − Tl Al

1 Al Cl Al

h11 h12 h21 h22 h22 h − 12 ∆h ∆h h21 h11 − ∆h ∆h

Cl 1 − Dl Dl ∆Tl Bl Dl Dl

∆g g − 22 g12 g12 g 1 − 11 g12 g12 −

g22 g12 ∆g ∆g g g − 21 11 ∆g ∆g g11 g12 g21 g22

Note : 1. In Table 11.4, the DZ , DY , Dh , Dg , DT and TT l represent the following determinants. ∆Z =

g g Z11 Z12 Y11 Y12 h11 h12 A B Al Bl ; ∆Y = ; ∆h = ; ∆g = 11 12 ; ∆T = ; ∆Tl = g 21 g 22 Z 21 Z 22 Y21 Y22 h 21 h 22 C D Cl Dl

2. The following relations holds good for parameters. -1

; >

-1

; >

>

Z11 Z12 Y11 Y12 H = > H Z 21 Z 22 Y21 Y22

>

Y11 Y12 Z11 Z12 H = > H Y21 Y22 Z 21 Z 22

-1

g g h11 h12 H = > 11 12 H g h 21 h 22 21 g 22

1

; >

1 A B Al Bl F H = < C D Cl Dl

g11 g12 h11 h12 A B Al Bl F = > H = > H ; < H g 21 g 22 h 21 h 22 C D Cl Dl

1

11. 20

Chapter 11 - One-Port and Two-Port Networks

11.11 Properties of Two-Port Networks The properties of a two-port network are passivity, reciprocity and symmetry. A two-port network is called passive network, if it has only passive elements like resistance, inductance and capacitance. If the two-port network have active elements like dependent source then it is called active network. If a two-port network satisfies the reciprocity theorem then it is called a reciprocal network. The reciprocity theorem says that, the ratio of excitation to response remains invariant when the position of source and response are interchanged. Hence, by the reciprocity theorem, we can say that in a reciprocal two-port network the ratio of excitation at port-1 and the response at port-2 will be same as the ratio of excitation at port-2 and the response at port-1. In a two-port network if the forward and reverse transfer impedances are same then the network will be a reciprocal network. Also, it can be observed that all the passive networks are reciprocal. A symmetrical two-port network is one which can be exactly divided into two identical sections. In a symmertrical two-port network, the looking back impedance from port-1 and port-2 will be same. Hence, in a symmetrical network the input and output ports can be exchanged. The conditions to be satisfied by two-port network for passivity, reciprocity and symmetry are listed in Table 11.5. TABLE 11.5 : CONDITIONS TO SATISFY PASSIVITY, RECIPROCITY AND SYMMETRY Parameter

Condition for passive and reciprocal network

Condition for electrical symmetry

Z

Z12 = Z21

Z11 = Z22

Y

Y12 = Y21

Y11 = Y22

ABCD

AD − BC = 1

A=D

A’ B’ C’ D’

A’D’ − B’C’ = 1

A’ = D’

h

h12 = −h21

Dh = 1

g

g12 = −g21

Dg = 1

11.12 Inter-Connection of Two-Port Networks Complicated two-port networks can be divided into two numbers of smaller two-port networks and they can be connected in series or parallel or cascade to represent the original network. When a complicated network is divided into smaller networks, it may be easier to determine certain parameters of smaller networks, and parameters of original network can be obtained either as sum or product of the parameters of smaller networks. When the sections of a two-port network can be connected in series to represent the original network then the Z-parameters of the original network can be obtained as sum of Z-parameters of individual network.

Circuit Theory

11. 21

When the sections of a two-port network can be connected in parallel to represent the original network then the Y-parameters of the original network can be obtained as sum of Y-parameters of individual network. When the sections of a two-port network can be connected in cascade to represent the original network then the ABCD-parameters of the original network can be obtained as product of ABCD-parameters of individual network. When the sections of a two-port network are connected, such that the input ports are in series and the output ports in parallel to represent the original network then the h-parameter of the original network can be obtained as sum of h-parameters of individual network. Two-Port Networks in Series When two two-port networks are connected in series as shown in Fig. 11.21, then overall Z-parameters are given by sum of individual Z-parameters as shown below: >

Zl11 Zl12 Zll11 Zll12 Z11 Z12 H + > H = > H Zl21 Zl22 Zll21 Zll22 Z 21 Z 22 I1

I2 + V1¢ _

+

Z11 ¢

Z12 ¢

Z ¢21

Z ¢22

I1

+ V2¢ _ V2

V1

_

Þ

I2

I1 + V1¢¢ _

Z11 ¢¢

Z12 ¢¢

Z ¢¢ 21

Z 22 ¢¢

I2

I1

I2

+ V1 _

Z11 Z12 Z 21

Z 22

+ V2 _

+ V2¢¢ _ I2

I1

Fig. 11.21 : Two-port networks in series.

Two-Port Networks in Parallel When two two-port networks are connected in parallel as shown in Fig. 11.22, then the overall Y-parameters are given by sum of individual Y-parameters as shown below: >

Yl11 Yl12 Yll11 Yll12 Y11 Y12 H + > H = > H l l ll ll Y 21 Y 22 Y 21 Y 22 Y21 Y22 I1

I¢2

I1¢

+ V1 _

Y11 ¢

Y12 ¢

Y21 ¢

Y22 ¢

I2 + V2 _

Þ I¢¢2

I1¢¢

+ V1 _

Y ¢¢ 11

Y ¢¢ 12

Y ¢¢21

Y ¢¢22

I2

I1 + V1 _

Y11

Y12

Y21

Y22

+ V2 _

Fig. 11.22 : Two-port networks in parallel.

+ V2 _

11. 22

Chapter 11 - One-Port and Two-Port Networks

Two-Port Networks in Cascade When two two-port networks are connected in cascade as shown in Fig. 11.23, then overall ABCD parameters are given by product of individual ABCD parameter matrices as shown below: >

A1 B1 A2 B2 A B H # > H = > H C1 D1 C2 D2 C D I¢2

I1 + V1 _

A1

B1

C1

D1

I2

I1¢

+ V2¢ _

+ V1¢ _

A2

B2

C2

D2

I2

I1 + V2 _

Þ

+ V1 _

A

B

C

D

+ V2 _

Fig. 11.23 : Two-port networks in cascade.

Input in Series and Output in Parallel When two two-port networks are connected such that the input in series and output in parallel as shown in Fig. 11.24, then overall h-parameters are given by sum of individual h-parameters as shown below:

>

hl11 hl12 hl21 hl22

H+>

hll21 hll22

H=>

h11 h12 h21 h22

h11 ¢

h12 ¢

h¢21

h¢22

H

I2

I¢2

I1 + V¢ _ 1

+

hll11 hll12

+ V _2

V1

Þ I1 + V ¢¢ _1

_

I2

I1

I1 I¢¢2 h¢¢11

h¢¢12

h¢¢21

h¢¢22

+ V1 _

h11

h12

h21

h22

+ V2 _

+ V _2

I1

Fig. 11.24 : Input in series and output in parallel. 2W

11.13 Solved Problems

2W

1

EXAMPLE 11.1

5W

4W

Determine the driving point impedance of the network shown in Fig. 1. 1¢

SOLUTION

Fig. 1.

The driving point impedance of the network shown in Fig. 1 is given by the looking back impedance from the port terminals, which can be obtained by step-by-step reduction of the given network to a single equivalent resistance as shown below: 2W

2W

2W

1

1

5W

1¢

4W

Þ

1

5W

1¢

Fig. 1.

Fig. 2.

4´2 4+2 8 4 = = W 6 3

Þ

5W 1¢

Fig. 3.

4 3 10 = W 3

2+

Circuit Theory

11. 23

With reference to Fig. 3, 10 3 = 10 5 + 3 50 50 3 = = = 2Ω 15 + 10 25 3 5 #

Driving point impedance, Z11

EXAMPLE 11.2

4W

1

Determine the driving point admittance of the network shown in Fig. 1. 6W

SOLUTION

3W 5W

1¢

Let us connect a source of voltage V1 to the port terminals as shown in Fig. 2. Let I1 be the current delivered by the source.

1W

2W

Fig. 1.

Now the driving point admittance is given by, Y11 =

I1 V1

4W 1

I1

The current variables in the circuit of Fig. 1 can be solved by mesh analysis. Let I1, I2 and I3 be the mesh currents. The mesh basis matrix equation is formed by inspection as shown below : R V R V R V -3 - 6 W SI1 W S V1 W S4 + 3 + 6 S W S W S W -3 3 + 5 + 1 - 1 W SI2 W = S 0 W S S W S W S W S0 W S -6 - 1 2 + 6 + 1 W SI3 W T X T X T X R V R V R V S V1 W S 13 - 3 - 6 W SI1 W S W S W S W S- 3 9 - 1 W SI2 W = S 0 W S W S W S W S0 W S- 6 - 1 9 W SI3 W T X T X T X The current I1 and hence Y11 can be solved as shown below :

V1

I1

3W

6W

1¢

I2 2W

I3

5W

1W

Fig. 2.

13 - 3 - 6 D = -3

9 -1

-6 -1

= 13 × [92 – (–1)2] – (–3) × [–3 × 9 – (–1) × (–6)] + (–6) × [(–3) × (–1) – (–6) × 9]

9 = 1040 – 99 – 342 = 599

V1 - 3 - 6 D1 = 0

9 - 1 = V1 # 69 # 9 - (- 1) # (- 1) @ = 80 V1

0 -1

9

D1 We know that, I1 = D 80V1 ` I1 = 599 From the above equation, we can write, I1 80 Driving point admittance, Y11 = = = 0.1336 M 599 V1

11. 24

Chapter 11 - One-Port and Two-Port Networks

EXAMPLE 11.3 Determine the driving point impedance and admittance of the network shown in Fig. 1.

2H

2H

1

1W

2W

SOLUTION 1¢

The s-domain representation of the given network is shown in Fig. 2. Now the driving point impedance is given by the looking back impedance from the terminals of the port, which can be obtained by step-by-step reduction of the network to a single equivalent impedance as shown below: 2s

2s

2s 1

1

1

1

Þ

2

Fig. 1.

2 ´ 2s Þ 2 + 2s 4s 2s = = 2 + 2s 1 + s

1

1¢

1¢

Fig. 2.

1

2s +

2s 2s + 2s2 + 2s = 1+ s 1+ s =

1¢

2s2 + 4s 1+ s

Fig. 4.

Fig. 3.

With reference to Fig. 4, we get, 2s2 + 4s 2s2 + 4s 1+ s 1+ s Driving point impedance, Z11 (s) = = 2s2 + 4s 1 + s + 2s2 + 4s 1+ 1+ s 1+ s 2^s2 + 2sh 2s2 + 4s s2 + 2s = = = 2 2 2 ^ h 2 s + 2.5s + 0.5 2s + 5s + 1 s + 2.5s + 0.5 1#

Driving point admittance, Y11 (s) =

1 s2 + 2.5s + 0.5 = Z11 ^s h s2 + 2s

EXAMPLE 11.4 Determine the driving point impedance and admittance of the network shown in Fig. 1.

j2W

-j3W

1

SOLUTION

2W

The driving point impedance of the network shown in Fig. 1 is given by the looking back impedance from the port terminals, which can be obtained by step-bystep reduction of the given network to a single equivalent impedance as shown below: j2W

-j3W

j2W

1

-j3W

1W

j1W

2W 4W

1¢

1¢

Fig. 1.

1

Þ

4W

j1W

j2W

1

2W

1W

4W

Þ

= 0.5 + j0.5 W

1¢

Fig. 1.

1 ´ j1 1 + j1

2W 4W

-j3 +0.5 + j0.5 = 0.5 - j2.5 W

1¢

Fig. 2.

Fig. 3. Þ j2W

1

1

b

j2 + 1.2 - j0.8 + 4 = 5.2 + j1.2 W

2 ´ 0 .5 - j 2 .5 2 + 0.5 - j 2 .5

4W

1¢

1¢

Fig. 5.

Fig. 4.

= 1.2 - j0.8 W

g

Circuit Theory

11. 25

With reference to Fig. 5, we get, Driving point impedance, Z11 = 5.2 + j1.2 Ω = 5.3367+13 o Ω Driving point admittance, Y11 =

1 5.2 + j1.2

= 0.1826 - j0.0421 M = 0.1874+ - 13 o M

EXAMPLE 11.5

1

2W

2W

2W 2

Determine the Z-parameters of the 2-port network shown in Fig. 1. 1W

SOLUTION

1W

1¢

The equations defining Z-parameters are,

2¢

Fig. 1. V1 = Z11 I1 + Z12 I2

.....(1)

V2 = Z21 I1 + Z22 I2

.....(2)

I1 1

To determine Z11 and Z21 Let us connect a source of voltage V1 to port-1 and leave

2W

2W

V1

Ia

Ib

1W

1W

Z21 =

V2 I1

+ V2

V2

O.C.

_ 1¢

2¢

Fig. 2.

When I2 = 0, from equations (1) and (2), we get, V1 I1

I2 = 0 2

+

_

port-2 as open circuit as shown in Fig. 2. Now, I2 = 0.

Z11 =

2W

The circuit shown in Fig. 2, can be reduced to a single equivalent resistance with respect to source terminals as shown below : I1

2W

2W Ia

V1

Ib

1W

I1

2W

1W

+ V2

Þ

V1

2W

Ib

1W

1W

_

Fig. 3.

I1

2W Ia

+ V2

Þ

1 ´ (2 + 1) 1 + (2 + 1) 3 11 = 2+ = W 4 4

V1

2+

_

Fig. 4.

Fig. 5.

With reference of Fig 5, Driving point impedance at port -1, Z11 =

V1 11 Ω = 2.75 Ω = 4 I1

With reference to Fig 4, we can say that the current I1 delivered by the source divides as Ia and Ib in the parallel paths. Hence, by current division rule, we get, I b = I1 #

I1 1 = 4 1 + ^2 + 1h

Now, by Ohm’s law, we get, V2 = 1 # I b =

I1 4

11. 26

Chapter 11 - One-Port and Two-Port Networks From the above equation, we get, Forward transfer impedance, Z21 =

V2 1 = = 0.25 Ω 4 I1

To determine Z22 and Z12 The given network can be divided into two identical sections and so it is a symmetrical network. ∴ Driving point impedance at port-2, Z22 = Z11 = 2.75 Ω The given network does not have any active element (dependent source) and so it is reciprocal. ∴ Reverse transfer impedance, Z12 = Z21 = 0.25 Ω Z-parameter matrix Z = >

Z11 Z12 Z21 Z22

H=>

2.75 0.25 0.25 2.75

H

EXAMPLE 11.6

0.5 F

Determine the Z-parameters of the network shown in Fig. 1.

2W

2W 2

1

SOLUTION 0.2 H

The s-domain equations defining Z-parameters are, 1¢

V1(s) = Z11(s) I1(s) + Z12(s) I2(s)

.....(1)

V2(s) = Z21(s) I1(s) + Z22(s) I2(s)

.....(2)

2¢

Fig. 1.

To determine Z11(s) and Z21(s) The s-domain equivalent of the given network is shown in Fig. 2. 1 2 = 0.5s s

2 s

I1(s)

2

2

2

1

V1(s)

0.2s

2

I2(s) = 0

2

1

+ V2(s)

0.2s

2 O.C.

_ 1¢

2¢

1¢

Fig. 2.

2¢

Fig. 3.

Let us connect a source of voltage V1(s) to port-1 and open circuit port-2 as shown in Fig. 3. Now, I2(s) = 0. 2 s

When I2(s) = 0, from equations (1) and (2), we get, Z11 ^s h = Z21 ^s h =

V1 ^s h I1 ^s h

I1(s)

2

Ib(s)

V2 ^s h I1 ^s h

In order to solve the above Z-parameters, let us choose two mesh currents as shown in Fig. 4. The mesh basis matrix equation is formed by inspection as shown below:

2

_

+ 2I (s) + b

V1(s)

0.2s Ia(s)

0.2s Ia(s) _

Fig. 4.

+ V2(s) _

Circuit Theory

11. 27

R S2 + 0.2s S -2 S T

V - 2 W I a ^s h V1 ^s h 2W > H=> H 2 + 2 + W I b ^s h 0 s X

R S2 + 0.2s S -2 S T

V - 2 W I a ^s h V1 ^s h 2 + 4s W > H=> H 0 s W I b ^s h X

2 + 0.2s D =

=

=

=

Da =

Db =

-2 2 + 4s m - ^- 2h2 2 + 4s = ^2 + 0.2sh # c s s

-2

4 + 8s + 0.4s + 0.8s2 - 4s 0.8s2 + 4.4s + 4 = s s 0.8 cs2 +

4.4 4 s+ m 0.8 0.8 s

0.8 b s2 + 5.5s + 5 l s V1 ^s h 0

2 4 cs + m -2 4 2 + 4s V1 ^s h = 2 + 4s = V1 ^s h # s s s 4^s + 0.5h V1 ^s h = s

2 + 0.2s V1 ^s h -2

0

= 2V1 ^s h

We know that, I1 (s) = Ia ^s h =

=

Da ^s h

= Da ^s h #

D

5^s + 0.5h 2

s + 5.5s + 5

4 (s + 0.5) 1 s V1 (s) # = s D 0.8^s2 + 5.5s + 5h

V1 ^s h

From the above equation we can write, Z11 ^s h =

= Z21 ^s h =

V1 ^s h

=

I1 ^s h

s2 + 5.5s + 5 5 (s + 0.5)

0.2 b s 2 + 5.5s + 5 l s + 0.5 V2 ^s h I1 ^s h

= 0.2s +

=

0.2sIa ^s h + 2Ib ^s h I a ^s h

2Ib ^s h = 0.2s + 2I (s) # 1 b I a ( s) I a ^s h

With reference to Fig. 4, by KVL, we get, V2(s) = 0.2 s Ia (s) + 2 Ia(s)

11. 28

Chapter 11 - One-Port and Two-Port Networks ` Z21 ^s h = 0.2s + 2 #

Tb T

#

T 1 = 0.2s + 2 # Tb # Ta Ta

= 0.2s + 2 # 2 V1 (s) #

= 0.2 s +

=

s 4 (s + 0.5) V1 (s)

2 s = 0.2 s + 0.1s + s s + 0.5 s + 0.5

0.2s^s + 5.5h 0.2s2 + 1.1s = s + 0.5 s + 0.5

To determine Z12(s) and Z22(s) The given network can be divided into two equal sections. Hence, it is symmetrical.

` Z22 ^s h = Z11 ^s h =

0.2s^s2 + 5.5s + 5h s + 0.5

The given network does not have any active elements like dependent source. Hence, it is reciprocal. ` Z12 ^s h = Z21 ^s h =

0.2 s^s + 5.5h s + 0.5

Z-parameter matrix R S Z11 ^s h Z ^s h = S SZ21 ^s h T

R 2 S 0.2s ^s + 5.5s + 5h V S Z12 ^s hW s + 0.5 W = S S W 0.2s ^s + 5.5h Z22 ^s h S X S s + 0.5 T

V 0.2s ^s + 5.5h W W s + 0.5 W 0.2 ^s2 + 5.5s + 5h WW W s + 0.5 X

EXAMPLE 11.7

2W

Determine the Z-parameters of the lattice network shown in Fig. 1.

2

1

j2 W

SOLUTION

V1 = Z11 I1 + Z12 I2

.....(1)

V2 = Z21 I1 + Z22 I2

.....(2)

j2

W

The frequency domain equations defining Z-parameters are,

2¢

1¢ 2W

Fig. 1.

To determine Z11 and Z 21 Let us connect a source of voltage V1 to port-1 and open circuit port-2 as shown in Fig 2. Now, I2 = 0. When I2 = 0, from equations (1) and (2), we get, Z11 = Z21 =

V1 I1 V2 I1

Circuit Theory

11. 29

I1

I2 = 0

2W

1

I1

2

+

1

+

j2

2W

W

V1

W

V2

O.C.

Þ

V _a 2

V1

+

j2

j2 W _

2¢

1¢

+ V2 _

V _b

I1

+ Vc _ 2¢

j2 W

+ Vd _

2W

Þ

2 + j2 2 = 1 + j1W

V1

1¢

2W

Fig. 2.

Fig. 3.

Fig. 4.

The circuit of Fig. 2 can be redrawn as shown in Fig. 3, and can be reduced as shown in Fig 4. With reference to Fig. 4, by Ohm’s law, we get, V1 = I1 ^1 + j1h

.....(3)

` Driving point impedance at port -1, Z11 =

V1 I1

= 1 + j1 Ω

With reference to Fig. 3, by voltage division rule, we get, Va = V1 #

2 2 + j2

Vc = V1 #

j2 2 + j2

With reference to Fig. 3, by KVL, we get, V2 + Va = Vc ` V2 = Vc - Va = V1 #

= V1 #

j2 2 + j2

- V1

2 2 + j2

j^2 + j2h j2 - 2 = V1 # = jV1 2 + j2 2 + j2

= j I1 ^1 + j1h

Using equation (3)

` Forward transfer impedance, Z 21 =

V2 I1

= j^1 + j1h = - 1 + j1 Ω

To determine Z12 and Z 22 The given lattice network is symmetrical and reciprocal. ` Z 22 = Z11 = 1 + j1 Ω Z12 = Z 21 = - 1 + j1 Ω Z-parameter matrix Z = >

Z11 Z12 Z21 Z22

H=>

1 + j1 - 1 + j1

- 1 + j1

1 + j1

H

Circuit Theory

11. 30

EXAMPLE 11.8

20 W 1

2

Determine the admittance parameters of the π-network shown in Fig. 1. 5W

SOLUTION

10 W

1¢

The equations defining Y-parameters are,

2¢

Fig. 1. I1 = Y11 V1 + Y12 V2

.....(1)

I2 = Y21 V1 + Y22 V2

.....(2)

To determine Y11 and Y21

I1

2

1

5W

V1

Let us connect a source of voltage V1 to port-1 and short circuit port-2 as shown in Fig. 2. Now, V2 = 0.

S.C. 2¢

1¢

Fig. 2. Þ

I1

I1 V1 I2 = V1

Y11 =

V1

Fig. 3. Þ I1

V1 1 V1 # ==4 5 20

⇒

20 ´ 5 = 4W 20 + 5

V1

With reference to Fig. 3, by current division rule, we can write, 5 1 = - I1 # 5 5 + 20

20 W I2

5W

The circuit of Fig. 2 can be reduced as shown in Fig. 4. With reference to Fig. 4, we get, V1 .....(3) I1 = 4 I1 1 ` Driving point admittance at port - 1, Y11 = M = 4 V1

I 2 = - I1 #

V2 = 0

10 W

When V2 = 0, from equations (1) and (2), we get,

Y21

I2

20 W

Using equation (3)

Fig. 4. I2 1 =V1 20

` Forward transfer admittance, Y21 =

I1

I2 1 M =20 V1

V1 = 0 S.C.

I2

20 W

5W

10 W

To determine Y12 and Y22

Fig. 5.

port-1 as shown in Fig 5. Now, V1= 0.

I1

Þ

Let us connect a source of voltage V2 to port-2 and short circuit

I2

20 W

When V1 = 0, from equations (1) and (2), we get,

Y22 =

I2 V2

Fig. 6. Þ

I1 V2

10 W

Y12 =

The circuit of Fig. 5 can be reduced as shown in Fig. 7. With reference to Fig. 7, we get, I2 =

V2 20/3

` Driving point admittance at port - 2,

.....(4) Y22 =

I2 3 M = 20 V2

V2

I2

20 ´ 10 20 + 10 20 = W 3

V2

Fig. 7.

V2

Circuit Theory

11. 31

With reference to Fig 6, by current division rule, we can write, I1 = - I 2 #

` I1 = -

10 1 = - I2 # 10 + 20 3

V2 I1 1 1 1 # = - V2 # ⇒ V = - 20 20 20/3 3 2

` Reverse transfer admittance, Y12 =

Using equation (4)

I1 1 M =20 V2

Note : It can be observed that, Y12 = Y21 and so the network is a reciprocal network. This is due to absence of any active elements like dependent source. Y - parameter matrix Y = >

R V S 1 - 1 W S 20 W 4 H = S 1 3 WW Y22 SS 20 20 W T X

Y11 Y12 Y21

EXAMPLE 11.9

1F

Determine the Y-parameters of the bridged-T network shown in Fig 1.

1W

1W 2

1

SOLUTION

1F

The s-domain equations defining Y-parameters are, I1(s) = Y11(s) V1(s) + Y12(s) V2(s)

.....(1)

I2(s) = Y21(s) V1(s) + Y22(s) V2(s)

.....(2)

1¢

2¢

Fig. 1.

To determine Y11(s) and Y21(s) The s-domain equivalent of the given network is shown in Fig. 2. 1 s

1 s

1

I1(s)

1 1 s

1¢

V1(s) 2¢

1

I3(s)

I2(s)

1

2

1

2

1

I1(s)

1 s

I2(s)

S.C. 2¢

1¢

Fig. 2.

V2(s) = 0

Fig. 3.

Let us connect a source of voltage, V1(s) to port-1 and short circuit port-2 as shown in Fig 3. Now, V2(s) = 0. When V2(s) = 0, from equations (1) and (2), we get, Y11 (s) =

I1 (s) V1 (s)

Y21 (s) =

I 2 ( s) V1 (s)

11. 32

Chapter 11 - One-Port and Two-Port Networks

The current I1(s) and I2(s) can be solved by mesh analysis. The mesh basis matrix equation for the circuit of Fig 3, is given below : V R V R V R 1 S V1 (s) W S1 + 1 - 1 W SI1 (s) W W S W S WS s s W S W S 1 WS 1 1+ 1 W S I2 (s) W = S 0 W S s s W S W S WS 1W S W S W S 1 1 2 + I ( s ) 0 W S W S sW S 3 X T X T X T 1 1 1+ -1 s s 1 1 D = 1+ 1 s s 1 1 2+ -1 s = c1 +

1 1 1 1 1 1 1 1 m ;c1 + m c2 + m - 1 E - ; c2 + m + 1 E - ; + c1 + mE s s s s s s s s

= c1 +

1 3 1 1 2 1 2 m 1 + + 2 H - >1 + + 2 H - ;1 + E s > s s s s s s

= 1+

3 1 1 3 1 1 2 1 2 + + + + - -1s s 2 s s 2 s3 s s 2 s3 s

=

1 2 s+2 + = s s2 s2 1 s 1 1+ s

V1 (s) 0

D1 =

1

0 = V1 (s) ;c1 + =

s2

D2 =

1 s 1 s

1 2+

1 s

1 1 3 1 mc2 + m - 1 E = V1 (s) f1 + + 2 p s s s s

s 2 + 3s + 1

1+

-1

V1 (s)

V1 (s)

-1

-1

0

1

0

1 2+ s

1 1 = - V1 (s) ; c2 + m + 1 E s s = - V1 (s) >1 + =-

^s + 1h2 s2

2 1 s 2 + 2s + 1 + 2 H = - V1 (s) > H s s s2

V1 (s)

Circuit Theory

11. 33

Now, I1 (s) =

D1 D

= D1 #

1 s 2 + 3s + 1 s2 s 2 + 3s + 1 V1 (s) # V1 (s) = = D s+2 s+2 s2

From the above equation, we can write, Driving point admittance, Y11 (s) =

I 2 (s) =

I1 (s) s 2 + 3s + 1 = s+2 V1 (s)

^s + 1h2 ^s + 1h2 D2 1 s2 V1 (s) # V1 (s) = D2 # == D D s+2 s+2 s2

From the above equation, we can write, 2

Forward transfer admittance, Y21 (s) =

^s + 1h I 2 (s) =s+2 V1 (s)

To determine Y12(s) and Y22(s) The given bridged-T network can be divided into two equal sections. Hence, the given network is symmetrical. ` Y22 (s) = Y11 (s) =

s 2 + 3s + 1 s+2

The given network does not have any active elements like dependent source. Hence, it is reciprocal. ` Y12 (s) = Y21 (s) = -

^s + 1h2 s+2

Y-parameter matrix

R 2 S s + 3s + 1 S s+2 Y11 (s) Y12 (s) Y (s) = > H = S S ^s + 1h2 Y21 (s) Y22 (s) SS s+2 T

V ^s + 1h2 W s + 2 WW s 2 + 3s + 1 W WW s+2 X

EXAMPLE 11.10

2 + j2 W

2 + j2 W

1

2

Determine the admittance parameters of the T-network shown in Fig. 1. -j2 W

SOLUTION The frequency domain equations defining Y-parameters are, I1 = Y11 V1 + Y12 V2

.....(1)

I2 = Y21 V1 + Y22 V2

.....(2)

1¢

2¢

Fig. 1.

To determine Y11 and Y12 Let us connect a source of V1 to port-1 and short circuit port-2 as shown in Fig. 2. Now, V2 = 0. When V2 = 0, from equations (1) and (2), we get, Y11 = Y21 =

I1 V1 I2 V1

11. 34

I1

Chapter 11 - One-Port and Two-Port Networks

2+j2 W

I1

I2

2+j2 W

1

I1

2+j2 W

2 V2 = 0

-j2 W

V1

S.C.

b 2 + j 2 g ´ b - j 2g Þ

V1

2 + j2 - j2

Þ

2 + j2 + 2 - j2

V1

= 4W

= 2 - j2 W

1¢

2¢

Fig. 2.

Fig. 3.

Fig. 4.

The circuit of Fig. 2, can be reduced to a single equivalent impedance with respect to source terminals as shown in Fig. 4. With reference to Fig. 4, by Ohm’s law, I1 =

V1 4

.....(3)

I1 1 = = 0.25 M 4 V1 With reference to Fig. 2, by current division rule, we get, Y11 =

`

I2 = - I1 # =-

- j2 2 + j2 - j2

V1 # ^- jh 4

Using equation (3)

` I2 = j0.25 V1 ` Y21 =

I2 V1

= j0.25 M

To determine Y12 and Y22 The given T-network is symmetrical and reciprocal. ` Y22 = Y11 = 0.25 M ` Y12 = Y21 = j0.25 M Y-parameter matrix Y = >

Y11 Y12 Y21 Y22

H = >

0.25 j0.25 j0.25

0.25

H

EXAMPLE 11.11 Determine the h-parameters of the two-port network shown in Fig. 1.

1

SOLUTION

10 W 2 20 W

The equations defining h-parameters are,

1¢

V1 = h11 I1 + h12 V2

.....(1)

I 2 = h 21 I1 + h 22 V2

.....(2)

2¢

Fig. 1.

Circuit Theory

11. 35

To determine h11 and h21 Let us connect a source of voltage V1 to port-1 and short circuit port-2 as shown in Fig. 2. Now, V2 = 0. When V2 = 0, from equations (1) and (2), we get, V1 h11 = I1 I2 h 21 = I1

I1

2

V1

V2 = 0

20 W

S.C. 2¢

1¢

In the circuit of Fig. 2, the 20Ω resistance is shorted and so it can be removed and the circuit can be drawn as shown in Fig. 3.

Fig. 2. Þ

Now, with reference to Fig. 3, by Ohm’s law, we can write, V1 = 10 I1. ` Input impedance, h11

I2

10 W

1

I1

V1 = = 10 Ω I1

I2

10 W

V1

Also in the circuit shown in Fig. 3, I2 = −I1. ` Forward current transfer ratio, h 21 =

I2 = -1 I1

Fig. 3.

To determine h12 and h22 Let us connect a source of voltage V2 to port-2 and open circuit port-1 as shown in Fig. 4. Now, I1 = 0. When I1 = 0, from equations (1) and (2), we get, V1 h12 = V2 I2 h 22 = V2 With reference to Fig. 4, by Ohm’s law, we can write, V2 I2 = 20 I2 1 ` Output admittance, h 22 = = = 0.05 M 20 V2

I2

I1 = 0 10 W

2

1 +

20 W

O.C. V1

V2

_ 1¢

2¢

Fig. 4.

Also in the circuit shown in Fig. 4, V1 = V2 .

V1 =1 V2 It can be observed that, h12 = – h21 and so the network is a reciprocal network. This is due to absence of any active elements like dependent source. ` Reverse voltage transfer ratio, h12 =

h-parameter matrix h11 h12 10 h = > H = > h 21 h 22 -1

1 0.05

H

EXAMPLE 11.12

1W

Determine the h-parameters of the bridged-T network shown in Fig. 1. 2W

SOLUTION

2W

1

The equations defining h-parameters are,

2

4W

V1 = h11 I1 + h12 V2

.....(1)

I2 = h21 I1 + h22 V2

.....(2)

1¢

2¢

Fig. 1.

11. 36

Chapter 11 - One-Port and Two-Port Networks

To determine h11 and h21 Let us connect a source of voltage V1 to port-1 and short circuit port-2 as shown in Fig. 2. Now, V2 = 0. When V2 = 0, from equations (1) and (2), we get, V1 I1 I2 h 21 = I1 The currents I1 and I2 can be solved by mesh analysis. The mesh basis matrix equation for the circuit of Fig 2, is given below: R V R V R V S V1 W S 6 4 - 2 W SI1 W S W S W S W S 4 6 2 W SI 2 W = S 0 W S W S W S W S0 W S- 2 2 5 W SI3 W T X T X T X 6 4 -2 h11 =

1W

2 = V1 # 630 - 4 @ = 26V1

0

2

5

V2 = 0

Fig. 2.

5

6

I2

2¢

2

0

2

1¢

-2

T1 =

4W

I1

2 = 6 # 630 - 4 @ - 4 # 620 + 4 @ - 2 # 68 + 12 @ = 20

4 -2

I2

2W

S.C. V1

6

V1

2W

1

4

T =

I3

I1

6 V1 - 2 4

0

2 = - V1 # 620 + 4 @ = - 24V1

-2

0

5

T2 =

` I1 = I2 =

D1 26V1 13 V1 = = D 10 20 D2 24V1 12 V1 ==D 10 20

V1 V1 10 Ω = = 13 13 I1 V1 10 12 V1 I2 12 10 Forward current transfer ratio, h 21 = = =13 I1 13 V1 10 To determine h12 and h22 Input impedance, h11 =

1W

Let us connect a source of voltage V2 to port-2 and open circuit port-1 as shown in Fig. 3. Now, I1 = 0. When I1 = 0, from equations (1) and (2), we get, V1 V2 I2 = V2

I1 = 0 1 + V1

h12 = h 22

_

Ib

2W

I2

2W

2

_ + 2Ib + 4Ia

4W

_

1¢

Ia

V2 2¢

Fig. 3.

Circuit Theory

11. 37

The above h-parameters can be solved by mesh analysis of the circuit of Fig. 3. Let the mesh currents be Ia and Ib. The mesh basis matrix equation for the circuit of Fig. 3, is given below:

>

6

-2

-2

5

H > H = > H Ia

V2

Ib

0

6 -2 T =

-2

5

V2 - 2 Ta =

= 6 # 5 - (- 2) 2 = 26

= 5V2

5

0

6 V2 Tb =

= 2V2 -2

0

` Ia =

Da 5V2 = D 26

Ib =

Db 2V2 = D 26

In the circuit of Fig. 3, by KVL, we can write, V1 = 2 Ib + 4 Ia = 2 #

2 5 24 V2 + 4 # V2 = V2 26 26 26

` Reverse voltage transfer ratio, h12 =

⇒

V1 24 12 = = V2 26 13

V1 12 = 13 V2

In the circuit of Fig. 3, I2 = Ia. ` I 2 = Ia =

5V2 26

⇒

I2 5 M = V2 26

` Output admittance, h 22 =

I2 5 M = 26 V2

h-parameter matrix h = > h 21

R S 10 S 13 H = S 12 h 22 SS 13 T

h11 h12

V 12 W 13 W 5 WW 26 W X

CONCLUSION 1. Here, h12 = −h21, hence the network is reciprocal. Also the network does not have any dependent source. h11 h12 2. D h =

h 21 h 22

10 12 10 5 12 12 10 2.5 12 12 25 144 13 13 # # # # = = + = + = + =1 13 26 13 13 13 13 13 13 169 169 5 12 13 26

Since, ∆h = 1, the network is symmetrical. Also the network can be divided into two identical sections.

11. 38

Chapter 11 - One-Port and Two-Port Networks

EXAMPLE 11.13 Determine the g-parameters of the network shown in Fig. 1.

1W

4W

1W

1

SOLUTION

2 2W

The equations defining g-parameters are, I1 = g11 V1 + g12 I2

.....(1)

V2 = g21 V1 + g22 I2

.....(2)

2W

1¢

2¢

Fig. 1.

To determine g11 and g21 Let us connect a source of voltage V1 to port-1 and open circuit port-2 as shown in Fig. 2. Now, I2 = 0. When I2 = 0, from equations (1) and (2), we get, g11 =

I1 V1

V2 V1 The circuit shown in Fig. 2 can be reduced to a single equivalent resistance with respect to source terminals as shown in Fig. 5. g21 =

I1

1W

4W

1W

1

V1

+ 2W

2W

I1

I2=0 2

V2

O.C.

Þ

V1

Ia

Ib

2W

2W

_ 1¢

I1

4W

1W

1W

+ V2 _

Þ

b

2´ 4+2

V1

2¢

Fig. 3.

Fig. 2.

Fig. 4.

With reference to Fig. 5,

`

g11

Þ

V1 = 2.5 I1

.....(3) I1

I1 1 = = = 0.4 M V1 2.5 V1

With reference to Fig. 3, by current division rule, we get, I b = I1 #

2 4+2+2

⇒ Vb = 0.25 I1

.....(4)

1+1.5 = 2.5 W

Fig. 5.

With reference to Fig. 3, by Ohm’s law, we get, V2 = 2 Ib ` V2 = 2 # 0.25 I1 Now, g21 =

V2 0.5 I1 = 2.5 I1 V1

∴ g21 = 0.2

⇒ V2 = 0.5 I1

.....(5)

g

2+4+2 = 1.5 W

Using equation (4)

Using equations (3) and (5)

Circuit Theory

11. 39

To determine g22 and g12 The given network does not have any active elements and so it is reciprocal. ∴ g12 = –g21 = – 0.2 The given network can be exactly divided into two sections and so it is symmetrical. ` Dg = 1 ⇒

` g22 =

g11 g12 = 1 g 21 g 22

⇒

g11 g 22 − g 21 g12 = 1

1 + 0.2 # ^- 0.2h 1 + g21 g12 = = 2.4 Ω g11 0.4

g-parameter matrix g = >

g11 g12 g21 g22

H=>

0.4 - 0.2 0.2

2.4

H

EXAMPLE 11.14

10 W

1

2

Determine the transmission parameters of the two-port network shown in Fig. 1.

4W

SOLUTION

4W

1¢

2¢

Fig. 1.

The equations defining transmission parameters are, V1 = A V2 – B I2

.....(1)

I1

I2 = 0

10 W

1

I1 = C V2 – D I2

+

.....(2) 4W

V1

4W

To determine A and C

V2 O.C. _

Let us connect a source of voltage V1 to port-1 and open circuit

1¢

2¢

Fig. 2.

port-2 as shown in Fig. 2. Now I2 = 0.

Þ

When I2 = 0, from equations (1) and (2), we get,

I1

V1 A = V2 I1 C = V2

V1

With reference to Fig. 2, by voltage division rule, we can write, 4 4 2 V2 = V1 # V1 = V1 = 7 4 + 10 14

2

4 ´ (10 + 4) 56 = 4 + (10 + 4) 18 28 = W 9

Fig. 3.

.....(3)

From the above equation, we get, A =

V1 7 = 2 V2

With reference to Fig. 3, by Ohm’s law, we can write, I1 =

V1 9 V1 = 28 28/9

` C =

I1 9V1 1 7 9 M # = I1 # = = 28 2V1 8 V2 V2

Using equations (3) and (4) .....(4)

11. 40

Chapter 11 - One-Port and Two-Port Networks

To determine B and D

I1

Let us connect a source of voltage V1 to port-1 and short circuit port-2 as shown in Fig. 2. Now, V2 = 0. When V2 = 0, from equations (1) and (2), we get, V1 I2

D =-

I1 I2

2 4W

V1

4W

S.C. V2 = 0 2¢

1¢

Fig. 4. Þ

B =-

I2

10 W

1

I1

In the circuit of Fig. 4, the 4Ω resistance is shorted. Hence it can be removed and the circuit is redrawn as shown in Fig. 5.

I2

10 W

V1

4W

With reference to Fig. 5, by current division rule, we get, I 2 = - I1

Fig. 5. .....(5) I1

From the above equation, we get, I1 7 = 2 I2 With reference to Fig. 6, by Ohm’s law, we get, D =-

V1 =

20 I1 7

` B ==-

Þ

4 2 = - I1 7 4 + 10

10 ´ 4 40 = 10 + 4 14 20 = W 7

V1

.....(6)

Fig. 6.

V1 1 = - V1 # I2 I2 20I1 7 # cm = 10 Ω 7 2I1

Using equations (5) and (6)

Transmission Parameter matrix R V S 7 10 W A B W S2 T = > H = S9 7 W C D W S S8 2 W T X

CONCLUSION

1. Here, A = D, hence the network is symmetrical. Also it can be observed that the network can be divided into two identical sections. 7 7 9 49 90 98 - 90 8 # - 10 # = = = =1 2 2 8 4 8 8 8 Here AD − BC = 1, hence the network is reciprocal. Also the network does not have any dependent source. 2. AD - BC =

EXAMPLE 11.15 Determine the inverse transmission parameters of the two-port network shown in Fig. 1.

SOLUTION

1

The equations defining inverse transmission parameters are,

4W

4W

2 4W

V2 = A′ V1 – B′ I1

.....(1)

I2 = C′ V1 – D′ I1

.....(2)

4W

8W 1¢

2¢

Fig. 1.

Circuit Theory

11. 41

To determine A′ and C′ I1=0 1 +

Let us connect a source of voltage V2 to port-2 and open circuit port-1 as shown in Fig. 2. Now, I1 = 0.

Al =

V2 V1

Cl =

I2 V1

I2 2

4W

V1

When I1 = 0, from equations (1) and (2), we get,

4W

4W

4W

8W

V2

_ 1¢

2¢

Fig. 2.

The circuit of Fig. 2, can be reduced to a single equivalent resistance with respect to source terminals as shown in Fig. 5. I2

4W

4W +

4W V1

I2

4W _

2 4W V2

8W

b 4 + 4g ´ 8

Þ

4+4+8 = 4W

_ 2¢

+ Vb _

Va

I2

+

4W

b 4 + 4g ´ 4 V2

Þ

4+4+4 32 8 = W 12 3

Fig. 4.

Fig. 3.

V2

=

Fig. 5.

With reference to Figs. 4 and 3, by voltage division rule, we can write, Vb = V2 #

V2 4 = 4+4 2

Vb V2 V2 4 1 # = = = 4+4 2 2 2 4 From the above equation, we get,

.....(3)

V1 = Vb #

V2 = 4 V1

Al =

With reference to Fig. 5, by Ohm’s law, we can write, I2 =

V2 3 V2 = 8 8 3

` Cl =

.....(4)

I2 1 3 4 V2 # = I2 # = = 1.5 M V1 V1 8 V2

To determine B′ and D′

Using equations (3) and (4)

The given network can be divided into two identical sections and so it is symmetrical. ∴ D′ = A′ = 4 The given network does not have any active elements like dependent source and so it is reciprocal. ` Al Dl - Bl Cl = 1 Al Dl - 1 4#4-1 = = 10 Ω 1.5 Cl Inverse Transmission Parameter ` Bl =

Tl = >

4 H = > 1.5 Cl Dl Al Bl

10 4

H

11. 42

Chapter 11 - One-Port and Two-Port Networks

11.14 Summary of Important Concepts 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

11. 12. 13.

14.

15.

16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

A pair of terminal of a network that can be used to connect a source/load is called a port. A network with one pair of terminals is called a one-port network. A network with two pair of terminals is called a two-port network. One-port and two-port networks are model of devices that are developed based on voltage and current measurements at the input and output terminals. The parameters of one-port network are driving point impedance and driving point admittance. The driving point impedance of a port is defined as the ratio of voltage to current in that port. The driving point admittance of a port is defined as the ratio of current to voltage in that port. The driving point impedance is inverse of driving point admittance and vice versa. The driving point impedance and admittance are commonly referred to as immittance. In a two-port network, four quantities (V1, V2, I1 and I2) are available for measurements. A two-port network parameter model is developed by choosing any two quantity as independent variables and the other two as dependent variables. The Z-parameter model of a two-port network is developed by choosing input and output currents as independent variables and input and output voltages as dependent variables. The Y-parameter model of a two-port network is developed by choosing input and output voltages as independent variables and input and output currents as dependent variables. The h-parameter model of a two-port network is developed by choosing input current and output voltage as independent variables and input voltage and output current as dependent variables. The g-parameter model of a two-port network is developed by choosing input voltage and output current as independent variables and input current and output voltage as dependent variables. The ABCD-parameter model of a two-port network is developed by choosing output voltage and current as independent variables and input voltage and current as dependent variables. The A′B′C′D′-parameter model of a two-port network is developed by choosing input voltage and current as independent variables and output voltage and current as dependent variables. A network with passive elements R, L and C is called passive network and a network with both passive and active elements (like dependent sources) is called active network. A reciprocal network is one that satisfies the reciprocity theorem. In a reciprocal two-port network, the forward and reverse transfer impedances will be same. Similarly, the forward and reverse transfer admittances will be same. A symmetrical two-port network is one which can be exactly divided into two identical sections. In a symmetrical two-port network, the input and output impedances will be same. Similarly, the input and output admittances will be same. When two number of two-port networks are connected in series then overall Z-parameters are given by sum of individual Z-parameters. When two number of two-port networks are connected in parallel then overall Y-parameters are given by sum of individual Y-parameters. When two number of two-port networks are connected in cascade then overall ABCD parameters are given by product of individual ABCD parameter matrices. When two number of two-port networks are connected such that the input in series and output in parallel then overall h-parameters are given by the sum of individual h-parameters.

Circuit Theory

11. 43

11.15 Short-answer Questions Q11.1

What are one-port and two-port networks? An RLC network with one pair of terminal (or two terminals) is called one-port network. An RLC network with two pair of terminal (or four terminals) is called two-port network.

Q11.2

What are the parameters of one-port network? The parameters of one-port network are driving point impedance and driving point admittance.

Q11.3

List the various parameter sets that are used to model a two-port network. The various parameter sets of two-port network are,

Q11.4

i)

Impedance parameters

iv)

Inverse hybrid parameters

ii)

Admittance parameters

v)

Transmission parameters

iii)

Hybrid parameters

vi)

Inverse transmission parameters.

List the h-parameters of a two-port network. The h-parameter of two-port network are, h11 = Input impedance when output port is shorted. h12 = Reverse voltage transfer ratio when input port is open. h21 = Forward current transfer ratio when output port is shorted. h22 = Output admittance when input port is open.

Q11.5

List the transmission parameters of a two-port network. The transmission parameters of two-port network are, A = Reverse voltage transfer ratio when the output port is open. B = Reverse transfer impedance when the output port is shorted. C = Reverse transfer admittance when the output port is open. D = Reverse current transfer ratio when the output port is shorted.

Q11.6

Determine the parameter z11 for the network shown in Fig. Q11.6.1. 1

1W

2W

2

Solution 3W

The parameter z11 is defined as, Z11 =

V1 I1

1¢

2¢

Fig. Q11.6.1.

I2 = 0

Hence connect a source of voltage V1 to port-1 and open circuit port-2 as shown in Fig Q11.6.2. I1

With reference to Fig. Q11.6.2, we get, V1 = ^1 + 3hI1 ` Z11 =

V1 = 4Ω I1

1

V1

1W

2W

3W

I2 = 0 2 O.C.

2¢

1¢

Fig. Q11.6.2.

11. 44 Q11.7

Chapter 11 - One-Port and Two-Port Networks In a two-port network on performing short circuit test, the following results were obtained. Test - 1 : I1 = 1 mA I2 = –0.5 mA V1 = 25 V

Test - 2 : I1 = –1 mA I2 = 10 mA V2 = 50 V

V2 = 0 V

V1 = 0 V

Determine the Y-parameters. Solution The equations defining Y-parameters are, I1 = Y11 V1 + Y12 V2

.....(1)

I2 = Y21 V1 +Y22 V2

.....(2)

In test-1, V2= 0V, on substituting V2 = 0 in equations (1) and (2), we get, Y11 =

I1 1 # 10- 3 = = 0.04 # 10- 3 = 40 # 10- 6 M = 40 µM V1 25

Y21 =

I2 - 0.5 # 10- 3 = = - 0.02 # 10- 3 = - 20 # 10- 6 M = - 20 µM V1 25

In test-2, V1 = 0V, on substituting V1 = 0 in equations (1) and (2), we get,

Q11.8

Y12 =

I1 - 1 # 10- 3 = = - 0.02 # 10- 3 = 20 # 10- 6 M = - 20 µM V2 50

Y22 =

I2 10 # 10- 3 = = 0.2 # 10- 3 = 200 # 10- 6 M = 200 µM V2 50

For the two-port network shown in Fig Q11.8.1, determine parameter, h21 . Solution By definition, h 21 =

I2 I1

` h 21 =

Q11.9

V2 = 0

I1 R R = - I1 =2R 2 R+R

1¢

2¢

Fig. Q11.8.1. I1

For the two-port network shown in Fig Q11.9.1, determine the transmission parameter, A.

I2

R

R

V2 = 0 S.C.

R

V1

I2 1 =2 I1

Fig. Q11.8.2.

1

Solution V1 By definition, A = V2

2

R

Hence, short circuit port-2 and connect a source of voltage V1 to port-1 as shown in Fig. Q11.8.2. With reference to Fig. Q11.8.2,by current division rule, I 2 = - I1 #

R

R

1

2R

2R

2

R 2 I2 = 0

1¢

2¢

Fig. Q11.9.1.

Circuit Theory

11. 45

Hence, open circuit port-2 and connect a source of voltage V1 to port-1 as shown in Fig. Q11.9.2. With reference to Fig. Q11.9.2, by voltage division rule,

V2 = V1 #

R 2 R 2R + 2

V1

Z12 = Z21 = 1Ω

Determine the Y-parameters. Solution Given that, Z = >

Z11 Z12 Z 21 Z 22

H = >

3 1

1 3

H

3 1 Let, Dz =

= 3#3-1#1 = 8 1 3

R V S 3 -1 W 3 -1 3 1 S 8 1 1 3 -1 8W Now, Y = Z- 1 = > H = > H= > H = S 1 3W D 8 z -1 3 1 1 -1 3 SW S 8 8W R V T X S 3 -1 W Y11 Y12 S 8 8W ` Y = > H = S 1 3W Y21 Y22 SW S 8 8W T X -1

Therefore, the Y-parameters are, 3 M 8

Y11 = Y22 =

Q11.11

;

Y12 = Y21 = -

1 M 8

The admittance parameter matrix of a two-port network is given as 0.9 0.2 Z = >

H

0.2 0.6 Determine the parameter, Z22 of the two-port network. Solution Z22 =

Y11 0.9 0.9 = = = 1.8 Ω ∆Y 0.9 0.2 0.9 # 0.6 - 0.22 0.2 0.6

Q11.12

The Y-parameters of a two-port network are Y11 = Y22 = 3 M Y12 = Y21 = 2 M . Determine the h-parameter, h22 of the network. Solution Y11 Y12 h 22 =

R 2

Fig. Q11.9.2.

The Z-parameters of two-port network are, Z11 = Z22 = 3Ω,

V2 _

I2 = 0 + V2 _

V1 ` A = = 5 V2

Q11.10

2R

2R +

R 1 V1 = V1 5 R ^4 + 1 h

=

I1

3 2

Y21 Y22 2 3 DY = = 3 Y11 Y11

=

3#3-2#2 5 M = 3 3

11. 46 Q11.13

Chapter 11 - One-Port and Two-Port Networks The Z-parameters of a two-port network are Z11 = Z22 = 4 Ω and Z12 = Z21 = 1 Ω . Determine the transmission parameter, B of the network. Solution Z11 Z12 B =

Q11.14

Z 21 Z 22 ∆z = Z 21 Z 21

4 1 =

1 4 1

-

4#4-1 = 15 Ω 1

The admittance matrix of a two-port network is, Y = >

H. 3 If two such identical network are connected in parallel then what will be the overall Y-parameters of the parallel combination. 2 -1

-1

Solution Let, Y0 = Admittance matrix of overall network 2 -1 2 -1 4 -2 ` Y0 = Y + Y = > H+> H = > H -1 3 -1 3 -2 6

Q11.15

The admittance matrix of a two-port network is, Y = >

2 -1

-1

3

H.

If two such identical network are connected in series then what will be the overall Z-parameters of the series combination. Solution The impedance matrix Z, is given by inverse of admittance matrix, Y. 1

2 -1 ` Z = Y- 1 = > H = -1 3

3 1 1 # > H 2 -1 1 2 -1

3

R S3 3 1 S5 1 1 = > H = > H = S1 5 6-1 1 2 1 2 S S5 T

V 1W 0.6 0.2 5W = > H 2 WW 0.2 0.4 5W X

3 1

Let, Z0 = Impedance matrix of overall network Now, Z0 = Z + Z = >

Q11.16

0.6 0.2 0.2 0.4

H+>

0.6 0.2 0.2 0.4

H= >

1.2 0.4 0.4 0.8

H

Two transmission lines are connected in cascade. Their individual ABCD parameters are,

>

A1 B1 C1 D1

H= >

1 10+30o 0

1

H ; >

A2 B2 C2 D2

H = >

1

0.25+ - 30o

0 1

H

Determine the overall ABCD parameters of the cascade combination of two transmission lines. Solution Let, A0, B0, C0, D0 be the ABCD parameters of the cascade combination.

Circuit Theory

11. 47

A0 B0 A1 B1 A2 B2 1 1 10+30 o Now, > H = > H#> H = > H# > C0 D0 C1 D1 C2 D2 0.25+ - 30 o 0 1 = >

1 + 2.5+0 o

10+30 o

0.25+ - 30 o

1

H = >

3.5

10+30 o

0.25+ - 30 o

1

0 1

H

H 5 KW

Find the frequency response V2 / V1 for the two-port circuit shown in Fig Q11.17.1.

+

(AU May’15, 2 Marks)

+ V1

Solution

1

-

jw ´ 1´ 10

-6

1250 W

+ +

W

+

Fig. Q11.17.1.

+ V2

Þ

V1

V2

Z

-

-

-

Fig. Q11.17.2.

Fig. Q11.17.3.

With reference to Fig.Q11.17.3, we get, 1 Z =

jw # 10- 6 1 jw # 10- 6

# 1250 = + 1250

1250 1 + 1250 # jw # 10- 6

=

1250 1 + j1.25 # 10- 3 w

By voltage division rule, 1250 1 + j1.25 # 10- 3 w V2 = V1 # = V1 # 1250 5000 + Z 5000 + 1 + j1.25 # 10- 3 w Z

= V1 #

1250 5000^1 + j1.25 # 10- 3 wh + 1250

` Frequency response,

V2 V1

=

1250 6250 + j6.25w

11.16 Exercises I.

V2

-

5000 W

5000 W

V1

1mF -

1250 W

Q11.17

Fill in the Blanks With Appropriate Words

1.

A network with two terminals is called ____________ network.

2.

A network with two pair of terminal is called _______________network.

3.

The ratio of voltage to current of a port is called _________________.

4.

The impedance parameters are also known as ___________________ parameters.

5.

The admittance parameters are also known as _________________ parameters.

6.

The parameter h21 gives the ________________ gain of the two-port network.

7.

The ABCD parameters are used for analysis of _______________.

8.

The inverse of Y11 gives ____________ of h-parameter.

11. 48 9.

Chapter 11 - One-Port and Two-Port Networks

The inverse of Z22 gives _____________ of h-parameter.

10. The negative of the inverse of Y21 gives the transmission parameter____________. 11. In a __________ two-port network Z12= Z21. 12. In a ___________ transmission line A = D. 13. When two-port networks are in series, then overall ____________ parameters are given by sum of ____________ parameters of individual sections. 14. In parallel connection of two-port networks, the overall ____________ parameters are given by sum of _____________ paramaters of individual sections. 15. If two 2-port networks are connected such that the inputs in series and output in parallel then the overall ____________ parameters are given by sum of ____________ parameters of individual sections.

ANSWERS

II.

1.

one-port

6.

current

11.

reciprocal

2.

two-port

7.

transmission lines

12.

symmetrical

3.

driving point impedance

8.

h11

13.

impedance, impedance

4.

open circuit

9.

h22

14.

admittance, admittance

5.

short circuit

10. B

15.

hybrid, hybrid

State Whether the Following Statements are True/False

1.

A two-port network consists of R,L,C parameters and sources.

2.

In two-port network analysis the assumed currents always flow away from the ports.

3.

In two-port network analysis, the network is modelled by using voltages and currents measured at port terminals.

4.

Using Z-parameters the two-port network is modelled as dependent voltage source.

5.

Using Y-parameters the two-port network is modelled as dependent current source.

6.

The units of all the h-parameters are same.

7.

The inverse of the parameter h12 gives the voltage gain of the two-port network.

8.

In transmission parameters, the current I2 entering the port-2 is considered as negative, because the input and output current flows in the same direction in transmission lines.

9.

The inverse of impedance parameter matrix gives admittance parameter matrix.

10. The transmission parameter C is same as the inverse of Z12 of impedance parameter. 11. For reciprocal network, the driving point impedances of both the ports should be same. 12. For electrical symmetry, the forward and reverse transfer admittances should be same. 13. A symmetrical network can be exactly divided into two identical sections. 14. All reciprocal networks are passive networks. 15. When two-port networks are in cascade, then overall transmission parameters are given by sum of individual transmission parameters.

ANSWERS 1. False

4. True

7. True

10. False

13. True

2. False

5. True

8. True

11. False

14. True

3. True

6. False

9. True

12. False

15. False

Circuit Theory

11. 49

III. Choose the Right Answer for the Following Questions 1. The driving point impedance of the network shown in Fig. 1 is, 1W

a) 0.5 Ω

1W

1

b) 1 Ω

2W

c) 2 Ω

1W

1W

2W

2W

1¢

Fig. 1.

d) 4 Ω 2. The driving point admittance of the network shown in Fig. 2 is,

W

W

c) 2 M

1 2

2

W

2

W

1

b) 4 M

2

W

W

4

1

W

a) 5 M

1¢

Fig. 2.

d) 1 M 3. The driving point impedance function of the network shown in Fig. 3 is,

1W

1

s+2 s2 + s + 1 c) 2 s + 1 s +s+1 a)

s s2 + s + 1 2 d) s + s + 1 s+1 b)

1F

1H

1¢

Fig. 3.

4. The driving point admittance of the network shown in Fig. 4 is,

1

a) 0.5 + j0.5 M

2W

2W

b) 0.5 − j0.5 M

j2 W

-j2 W

c) j0.5 M

1¢

Fig. 4.

d) 0.5 M

5. The input and output impedance of the 2-port network shown in Fig. 5 are respectively, a) 4 Ω, 3 Ω

4W

1

2

b) 4 Ω, 2 Ω c) 3 Ω, 2 Ω

8W

4W

1¢

2¢

d) 2 Ω, 1 Ω

Fig. 5.

6. The forward and reverse transfer impedances of the 2-port network shown in Fig. 6 are respectively, a) 3 Ω, 6 Ω

1

b) 6 Ω, 6 Ω c) 12 Ω, 6 Ω d) 12 Ω, 12 Ω

6W

6W

2

6W 1¢

2¢

Fig. 6.

11. 50

Chapter 11 - One-Port and Two-Port Networks

7. The individual Z-parameters of the two 2-port network in series are, Zl = >

2 0.5

0.5 4 H and Zll = > 2 1.5

1.5 H 4

The overall Z-parameters of the series combination is, a) >

2 1

b) >

1 H 2

−2 −1

−1 H −2

c) >

6 2

2 H 6

d) >

0.75 H 8

8 0.75

8. The input and output admittance of the 2-port network shown in Fig. 8 are respectively, a)

1 M, 1 M 10 10

b) 1 M, 1 M 6 6

10 W

10 W

1

2 10 W

10 W 10 W

c)

1 M, 1 M 10 6

d) 1 M, 1 M 5 10

1¢

2¢

Fig. 8.

9. The forward and reverse transfer admittance of the 2-port network shown in Fig. 9 are respectively, a) − 1 M, − 1 M 2 2

b) − 1 M, − 1 M 4 2

c) − 1 M, − 1 M 4 8

d) − 1 M, − 1 M 5 5

1

1W

2 1W

2W

2W

1¢

2¢

Fig. 9.

10. The individual Y-parameters of the two 2-port network in parallel are, Yl = >

0.5 − 1 H ; −1 2

Yll = >

4 − 0.7 H 3 − 0.7

The overall Y-parameters of the parallel combination is, a) >

− 3.5 − 0.3 H − 0.3 − 1

b) >

4.5 − 1.7 H 5 − 1.7

c) >

2 0.7

0.7 H 6

d) >

2.7 − 5.4

− 3.35 H 6.7

11. In a two-port symmetrical and reciprocal network, h11 = 4 Ω and h12 = 2. What is the value of h22 ? a) 4 M

b) 0.25 M

c) − 0.5 M

d) − 0.75 M

12. In a two-port symmetrical and reciprocal network, if h11 = 0.88 Ω and h22 = –0.5 M then the values of h12 and h21 are respectively, a) 1.2, –1.2

b) 0.25, –0.25

c) 1.76, –1.76

d) 0.44, –0.44

13. The h-parameter matrix of a two-port network is, h = >

1.25 1.5 H − 1.5 − 0.2

The g-parameter matrix of the network is, a) >

− 0.2 − 1.5 H 1.5 1.25

b) >

− 0.1 − 0.75 H 0.75 0.625

c) >

− 2.5 − 3 H 3 0.4

d) >

− 1.25 − 1.5 H 1.5 0.2

Circuit Theory

11. 51

14. The individual transmission parameters of the two-port networks in cascade are, T1 = >

2 0.75

4 H 2

1.5 2.5 T2 = > H 0.5 1.5

The overall transmission parameters of the cascade combination is, a) >

0.5 1.5 H 0.25 0.5

3.5 6.5 b) > H 1.25 3.5

c) >

3 10 H 0.375 3

d) >

5 2.125

11 H 4.875

15. The condition for reciprocal two-port network in inverse transmission parameters is, a) A′D′ – B′C′ = 0 ANSWERS 1. b 2. a 3. c

IV. E11.1

b) A′D′ – B′C′ = 1

c) A′B′ – C′D′ = 0

4. d 5. a

7. c 8. b

10. b 11. d

13. b 14. d

6. b

9. d

12. a

15. b

Unsolved Problems Determine the driving point impedance and admittance of the network shown in Fig. E11.1. 4W

2F

4W

1

10 W

2F

1

8W

1W

1¢

2W

1¢

Fig. E11.1.

E11.2 E11.3

d) A′B′ – C′D′ = 1

Fig. E11.2.

Determine the driving point impedance and admittance of the network shown in Fig. E11.2. Determine the impedance and admittance parameters of the T-network shown in Fig. E11.3. 1F 2

1

2W

2

2W

1

4W

2W

5W 1¢

2¢

1¢

2¢

1F

Fig. E11.3.

Fig. E11.4.

E11.4

Find the impedance parameters of the two-port network shown in Fig. E11.4.

E11.5

Determine the Y-parameters of the two-port network shown in Fig. E11.5. 1

1W

1W

1W

1W

2 1W

1¢

1F

2H

2 4W

2¢

Fig. E11.5.

E11.6

1

1¢

2¢

Fig. E11.6.

Determine the Y-parameters of the T-network shown in Fig. E11.6.

11. 52 E11.7

Chapter 11 - One-Port and Two-Port Networks Determine the h-parameters of the two-port network shown in Fig. E11.7. 4W 2

1

2W 2

1W

4W

4W

1W

1

2W

1¢

2¢

2¢

1¢ 4W

Fig. E11.7.

Fig. E11.8.

E11.8

Determine the g-parameters of the two-port network shown in Fig. E11.8.

E11.9

Determine the transmission parameters of the two-port network shown in Fig. E11.9. 3W 3W 1

3W

j2 W

2 -j4 W

3W 1¢

1¢

2¢

2¢

Fig. E11.10.

Fig. E11.9.

E11.10

j2 W

1

2

Determine the inverse transmission parameters of the two-port network shown in Fig. E11.10.

ANSWERS E11.1 Z11 = 4 Ω ; Y11 = 0.25 M

E11.3 Z = >

7 5

E11.5

5 9

H ;

R V S 9 - 5 W S 38 38 W Y = S 7 WW 5 SS 38 38 W T X

R V S 5 -1 W S 8 8W Y = S 5 WW 1 SS 8 8W T X

E11.7

R S 10 S 3 h = S S- 2 S 3 T

E11.9

R S5 S4 T = S S1 S4 T

V 2W 3W 5 WW 12 W X V 9W 4W 5 WW 4W X

E11.2

Z11 ^s h =

8s + 1 2

8s + 10s + 1

E11.4

R S 2s + 1 S 2s Z ^s h = S S 2s - 1 S 2s T

E11.6

R S 4s + 1 S 8s2 + 2s + 4 Y ^s h = S - 4s S SS 2 8s + 2s + 4 T

; Y11 ^s h =

8s2 + 10s + 1 8s + 1

V 2s - 1 W 2s W 2s + 1 WW 2s W X

E11.8

R V S 2 3 W S 5 5 W g = S W S- 3 17 W S 5 5 W T X

E11.10

Tl = >

0.5

j3

j0.25

0.5

- 4s 2 8s + 2s + 4 2s2 + 4s 8s2 + 2s + 4

H

V W W W W WW X

Chapter 12

THREE-PHASE CIRCUITS 12.1 Introduction In many countries the electrical energy is generated and distributed as 3-phase ac supply. For many reasons three-phase system is adopted for generation and distribution of electrical energy. The three-phase system are more economical than single-phase system. Also the three-phase motors are self starting and produce more uniform torque when compared to single-phase motors. The three-phase sources are three-phase alternators generating three emfs having equal magnitude but with a phase difference of 120o with respect to each other. Each generated emf is called a phase and so the three generated emfs are called three-phase. The three-phases are named as R-phase, Y-phase and B-phase in British convention and A-phase, B-phase and C-phase in American convention. The letters R,Y,B, have been chosen from the first letters, of three colours Red, Yellow, and Blue. In practical wiring, these coloured wires are used for wiring of three-phase circuits in order to provide clear visual distinction between different phase connections to loads. In a polyphase system, when the magnitude of emfs are equal and the phase difference between consecutive emfs are equal then the system is called balanced system and the emfs are called balanced emfs. The three-phase sources are designed to generate balanced emfs. Hence in the analysis of three-phase system the source emfs are always considered as balanced set of emfs.

12.2 Three-Phase Sources The three-phase sources can be represented by three independent sources. For operational convenience the three sources can be connected either in star or delta as shown in Fig. 12.1. The emfs generated by the sources vary sinusoidally and so instantaneous values of the emfs are represented by the following equations. eR = ER,m sin ωt

= Em sin ωt

E R, m = E Y, m = E B, m = E m

eY = EY,m sin (ωt − 120o ) = Em sin (ωt − 120o ) eB = EB,m sin (ωt − 240o ) = Em sin (ωt − 240o )

~

~

Y B

Fig. a : Star-connected source.

E

eY

eB

+

~

~

E

~

eY

eR

+

Y B

IPQ

+

Terminals to connect load

E

eB

eR

E E E

+

~

R

+

R +

IPQ

where, ER,m , Ey,m and EB,m are maximum values of generated emfs.

Terminals to connect load

Fig. b : Delta-connected source.

Fig. 12.1 : Three phase ac sources.

12. 2

Chapter 12 - Three Phase Circuits

In star connection the meeting point of the three sources is called neutral and there is no such neutral point in delta connection. The neutral point is denoted by N. The two types of threephase sources offer three terminals for connecting the load. The transmission lines or connecting wires from the source terminals to the load terminals are called lines, as shown in Fig. 12.2. R

R

phase

eR

~

N

~

+

+

E

load

E e Y

+

~

~

Y E

~

eY

eR

lines load

E

eB

eB

Lines

+

~

E

E

+ Phase

Y + B

B SOURCE

SOURCE

Fig. a : Star-connected source Fig. b : Delta-connected source with load. with load. Fig. 12.2 : Phase and lines in 3-phase system.

The voltage generated by each phase of three-phase source is called phase voltage and the voltage between the lines connecting the load is called line voltage. The current delivered by each phase of the three-phase source is called phase current and the current flowing through the line is called line current.

12.3 Representation of Three-Phase EMFs The three-phase emf can be represented by three rotating phasors (or vectors) having equal magnitude but maintaining a phase difference of 120o with respect to each other. The phasor will rotate at a constant angular speed of ω rad/s. The rotation of the phasors can be clockwise or anticlockwise. It is conventional practice to choose anticlockwise rotation for phasors (throughout this book anticlockwise rotation is followed for phasors). The reference point is chosen as ωt = 0. For the three-phase rotating phasors the order of reaching reference point is called phase sequence. In the three-phase rotating phasors, when the order of reaching reference is R-phase, Y-phase and B-phase, the phasors are said to have normal phase sequence or RYB sequence. When the order of reaching reference is R-phase, B-phase and Y-phase, the phasors are said to have reversed phase sequence or RBY sequence. The normal and reversed sequence three-phase emfs are shown in Fig. 12.3. RYB w

E Y, m

RBY w

E B, m

E R, m

E R, m

Reference

Reference

0

0

0 -12 0 0 -24

E B, m

0 -12 0 0 -24

E Y, m

Fig. a : Phasors rotating in RYB sequence. Fig. b : Phasors rotating in RBY sequence. Fig. 12.3 : Phasors representation of three phase emf.

Circuit Theory

12. 3

The instantaneous values of three emfs for RYB sequence with R-phase emf as reference are, eR = ER,m sin ωt

= Em sin ωt ER, m = EY, m = EB, m = Em

o

eY = EY,m sin (ωt − 120 ) = Em sin(ωt − 120o ) eB = EB,m sin (ωt − 240 o ) = Em sin(ωt − 240o )

The instantaneous values of three emfs for RBY sequence with R-phase emf as reference are, eR = ER,m sin ωt

= Em sin ωt o

eB = EB,m sin (ωt − 120 ) = Em sin (ωt − 120o) o

ER, m = EB, m = EY, m = Em

o

eY = EY,m sin (ωt − 240 ) = Em sin (ωt − 240 )

The rms phasors can be drawn by taking a “snapshot” of rotating phasors at ωt = 0 and reducing the length by 2 . But remember that the rms phasors are not rotating phasors. The threephase rms voltage phasors for RYB and RBY sequence are shown in Fig. 12.4. (Note that the phasor diagram in this book are drawn using rms phasors.) EY

EB

ER

ER

Reference

Reference

0

0

120

E

120

E

0

0

E24

E24

0

0

EB

EY

Fig. a : Rms phasors in RYB sequence.

Fig. b : Rms phasors in RBY sequence.

Fig. 12.4 : Phasors representation of three phase rms value of emf.

The rms values of three emfs for RYB sequence in the polar form with R-phase emf as reference are, E R = E R +0 o

= E+0 o

E Y = E Y + − 120 o = E+ − 120 o

ER = EY = EB = E

E = Em

2

E B = E B + − 240 o = E+ − 240 o

The rms values of three emfs for RBY sequence in the polar form with R-phase emf as reference are, E R = E R +0 o

= E+0 o

E B = E B + − 120 o = E+ − 120 o E Y = E Y + − 240 o = E+ − 240 o

ER = EB = EY = E E = Em

2

12. 4

Chapter 12 - Three Phase Circuits

12.4

Three-phase Star-connected Source

12.4.1

(AU May’15, 16 Marks)

Star-connected Source Three-Wire System

Let the phase sequence of the source emfs be RYB. eR, eY, eB

= Instantaneous values of source emfs.

E R, E Y, E B

= Rms values of source phase emfs.

ER, EY, EB

= Magnitude of rms values of source phase emfs.

VRY, VYB, VBR = Rms values of line voltages.

VRY, VYB, VBR

= Magnitude of rms values of line voltages.

I R, I Y, I B

= Rms values of phase and line currents.

IR

R

IR

Y

IY

+ B

IB

+

E

VRY

VBR

+

~

+

~

E

IB

E EY

~

+

EB

ER

N E

IY

E + V YB

E

IPQ

I R , I Y, IB = Magnitude of rms values of phase and line currents. The three-phase star-connected source with conventional polarity (or sign) of voltages and direction of currents for RYB sequence are shown in Fig. 12.5.

To load

Fig. 12.5 : Three phase star-connected source with conventional polarity of voltages and direction of currents for RYB sequence.

The instantaneous values of three emfs for RYB sequence with R-phase emf as reference are, eR = Em sin ωt eY = Em sin (ωt − 120o ) eB = Em sin (ωt − 240o )

Here, Em is the maximum value of rotating phasors. Hence the rms value E of the rotating phasor is E m . 2 The phase emfs in the polar form are, E R = E R +0 o

= E+0 o

E Y = E Y + − 120 o = E+ − 120 o E B = E B + − 240 o = E+ − 240 o ER = EY = EB = E = Magnitude of phase emf.

Also, E = E m

2.

Circuit Theory

12. 5

In a balanced star-connected supply the magnitude of line voltage is 3 times the magnitude of phase voltage and the line voltage leads the phase voltage by 30o. Therefore, the line voltages of star-connected source can be written as shown below: VRY =

3 E+30 o

= VRY +30 o

VYB =

3 E+ − 90 o = VYB + − 90 o

VBR =

3 E+ − 210 o = VBR + − 210 o = VL + − 210 o

where, VRY = VYB = VBR =

= VL +30 o = VL + − 90 o

3 E = VL = Magnitude of line voltage.

Proof for line voltages in star-connected source: Consider the star-connected source shown in Fig. 12.5. Now, the phase emf of the source are, E R = E+0 o E Y = E+ − 120 o E B = E+ − 240 o With reference to Fig. 12.6 in the path RNYR by using KVL, we can write,

R +

+

E Y + V RY = E R

~

` V RY = E R − E Y

ER VRY

E

o

o

N

= E∠0 − E∠−120

EY +

E

~

= E (1∠0 o − 1∠−120o ) = E(1.5 + j0.866)

E

Fig. 12.6.

= E^1.732+30 oh = E^ 3 +30 oh ` V RY =

Y

3 E+30 o

With reference to Fig. 12.7 in the path YNBY by using KVL, we can write, E B + V YB = E Y

EB

` V YB = E Y − E B

+ o

o

~

N E E

EY +

~

= E∠−120 − E∠−240

V YB

= E (1∠−120o − 1∠−240o ) = E(0 – j1.732) o

B

E

o

= E^1.732+ − 90 h = E^ 3 + − 90 h ` V YB =

Y +

Fig. 12.7.

3 E+ − 90 o

With reference to Fig. 12.8 in the path BNRB by using KVL, we can write,

R E

+

E R + V BR = E B ` V BR = E B − E R

EB o

o

= E∠−240 − E∠0 o

+

~

~

ER

NE E

VBR

o

= E (1∠−240 − 1∠0 ) = E( –1.5 + j0.866 )

` V BR =

= E^1.732+150 oh = E^ 3 + − 210 oh 3 E+ − 210 o

+ B

Fig. 12.8.

12. 6

Chapter 12 - Three Phase Circuits

From the above analysis, we can make following conclusions, 1. The magnitude of line voltage is

3 times the phase voltage.

2. The line voltage leads the phase voltage by 30o. It is conventional practice to assume one of the line voltages as reference. “When we take line voltage as reference we can say that the phase voltage lags the line voltage by 30 o”. By taking the line voltage VRY as reference, we can rewrite the line and phase voltages as shown below: Line voltages VRY = VRY +0 o

Phase voltages

= VL +0 o

E R = E R + − 30 o = E+ − 30 o

VYB = VYB + − 120 o = VL + − 120 o

E Y = E Y + − 150 o = E+ − 150 o

VBR = VBR + − 240 o = VL + − 240 o = VL +120 o

E B = E B + − 270 o = E+ − 270 o = E+90 o

Here, E R = E Y = E B = E VRY = VYB = VBR = VL =

3E

The rms phasors of the line and phase voltages of a star-connected source for RYB sequence are shown in Fig. 12.9. V BR

EB

120

o

Note : To remember suffix for line voltages of RYB sequence write as shown below:

o

90

VRY

Reference

R Y B R Y B

o o

E30

E150 o

E120

EY

Now the underlined letters will be the suffix for three consecutive line voltages. The suffix are RY, YB, BR.

ER

V YB

Fig. 12.9 : Phasor diagram of rms value of line and phase voltages of star-connected source for RYB sequence.

From Fig. 12.5, we can say that the current delivered by the source will also flow through the lines. Hence we can say that the phase current and line current are same in the star-connected system. The phase difference between the voltage and current depends on the load impedance. Also we can say that the source current is balanced if the load impedance is balanced. The source current will be unbalanced if the load impedance is unbalanced. Let φR, φY and φB be the phase difference between phase voltage and current. Now, the phase currents can be written as, I R = I R +^− 30 o ! φRh I Y = I Y +^− 150 o ! φYh I B = I B +^− 270 o ! φBh

“−” for lagging power factor “+” for leading power factor.

Circuit Theory

12. 7

For balanced system, IR = IY = IB = I φ R = φY = φB = φ

Line and Phase Voltages of RBY Sequence The line voltages of RBY sequence are VRB, VBY and VYR . By taking VRB as reference phasor, the line and phase voltages of RBY sequence can be expressed as shown below: Line voltages VRB = VRB +0 o

Phase voltages

= VL +0 o

E R = E R + − 30 o = E+ − 30 o

VBY = VBY + − 120 o = VL + − 120 o

E B = E B + − 150 o = E+ − 150 o

VYR = VYR + − 240 o = VL + − 240 o = VL +120 o

E Y = E Y + − 270 o = E+ − 270 o = E+90 o

Here, E R = E B = E Y = E VRB = VBY = VYR = VL =

3E

The rms phasors of the line and phase voltages of star-connected source for RBY sequence are shown in Fig. 12.10. V YR

EY

120

o

Note : To remember suffix for line voltages of RBY sequence write as shown below:

VRB

90

o

R B Y R B Y

Reference o

o

E30

E150 o

Now the underlined letters will be the suffix for three consecutive line voltages. The suffix are RB, BY and YR.

E120

EB

ER

VBY

Fig. 12.10 : Phasor diagram of rms value of line and phase voltages of star-connected source for RBY sequence.

12.4.2 Star-connected Source Four-wire System Since neutral point is available in star-connected system, it is possible to have four-wire supply system. In four-wire star-connected source, in addition to three terminals for phase there will be fourth terminal called neutral, as shown in Fig. 12.11. With reference to Fig. 12.11, by using KCL we can write, IR + IY + IB = I N

“In a balanced system the sum of three-phase current will be zero and so the neutral current is zero”. Therefore, in balanced system, I R + I Y + I B = 0 and so I N = 0

12. 8

Chapter 12 - Three Phase Circuits

“In unbalanced system the sum of three-phase currents will not be zero and so there will be a definite neutral current flowing through neutral wire and the neutral current is equal to sum of all the phase current”. Therefore, in unbalanced system, Neutral current, I N = I R + I Y + I B

IR

R

IR

Y

IY

+ B

IB

N

IN

+

E

VRY

VBR

+

~

IN

ER

N E

IB

~

EB

E EY

~

+

+

E

IY

E + V YB

E

IPQ

In a four-wire system, the relation between phase and line voltage and the relation between phase and line current remain same as that of 3-wire system.

To load

Fig. 12.11 : Three phase four wire star-connected source with conventional polarity of voltages and direction of currents for RYB sequence.

Points to remember 1. The voltages are always balanced in star-connected sources. 2. The currents may be balanced or unbalanced depending on load. 3. The phase and line currents are same in a star system. 4. The magnitude of line voltage is

3 times the magnitude of phase voltage.

5. The phase difference between the line voltage and phase voltage of source is 30o. The phase voltage of source lags line voltage by 30o. 6. In balanced four-wire system, the neutral current is zero.

12.5 Three-phase Delta-connected Source Let the phase sequence of source emfs be RYB. Let,

eR, eY, eB

= Instantaneous values of source emfs.

E R, E Y, E B

= Rms values of source phase emfs.

ER, EY, EB

= Magnitude of rms values of source phase emfs.

VRY, VYB, VBR

= Rms values of line voltages.

VRY, VYB, VBR

= Magnitude of rms values of line voltages.

Circuit Theory

12. 9

I RY, I YB, I BR

= Rms values of phase currents.

I RY , IYB, IBR

= Magnitude of rms values of phase currents.

I R, I Y, I B

= Rms values of line currents.

IR , I Y , I B

= Magnitude of rms values of line currents.

The three-phase delta-connected source with polarity (or sign) of voltages and direction of currents are shown in Fig. 12.12. The instantaneous values of three emfs for RYB sequence with R-phase emf as reference are, eR = Em sin ωt eY = Em sin (ωt − 120 o ) eB = Em sin (ωt − 240 o )

Here, Em is the maximum value of rotating phasors. Hence the rms value E of the rotating phasor is E m 2 .

~

~

IBR

E

B

ER

VRY

VBR

E

+

+

E

IRY

E

~

+

I YB

E +

Y

EY

IY

Y

V YB

B E

+

IB

IPQ

+

EB

IR

R

R

To load

Fig. 12.12 : Three phase delta-connected source with conventional polarity of voltages and direction of currents for RYB sequence.

The phase emfs in polar form are, E R = E R +0 o

= E+0 o

E Y = E Y + − 120 o = E+ − 120 o E B = E B + − 240 o = E+ − 240 o

Here, ER = EY = EB = E = Magnitude of phase emf. Also, E = E m

2.

With reference to Fig. 12.12, we can say that the phase voltage E R is same as line voltage VRY . Similarly, the phase voltage E Y is same as line voltage VYB and the phase voltage E B is same as line voltage VBR . Hence we can conclude that the phase and line voltages are same in delta-connected sources. Therefore, the line voltages are,

12. 10

Chapter 12 - Three Phase Circuits VRY = VRY +0 o

= VL +0 o

VYB = VYB + − 120 o = VL + − 120 o VYB = VYB + − 240 o = VL + − 240 o

Here, VRY = VYB = VBR = VL = Magnitude of line voltage. Also, VL = E. The rms phasors of the line voltage of delta-connected source are shown in Fig. 12.13. With reference to Fig. 12.12, we can say that the phase and line currents are not same. The line currents can be computed with the knowledge of phase currents by writing KCL equations at nodes R, Y and B.

VBR

VRY

The phase and line currents can be balanced or Reference 20 E1 unbalanced depending on the load impedance. If the 40 2 E load impedance is balanced then the currents are also balanced. If the load impedance is unbalanced then the currents will be unbalanced. Also the phase difference V YB between voltage and current depends on the nature of Fig. 12.13 : Phasor diagram of rms value of load impedance. line voltages of delta-connected source. Let φRY, φYB, and φBR be the phase difference between phase voltage and current. Now, the phase currents can be written as, o

o

I RY = I RY +^0 o ! φRYh I YB = I YB +^− 120 o ! φYBh

“−” for lagging power factor “+” for leading power factor.

I BR = I BR +^− 240 o ! φBRh

For balanced system, I RY = I YB = I BR = I φRY = φYB = φBR = φ

The relation between phase and line currents in a balanced delta system can be studied by considering the load as purely resistive, so that the phase voltage and current are in-phase. Therefore, φRY = φYB = φBR = φ = 0o. Now the balanced phase currents for RYB sequence can be written as, I RY = I RY +0 o

= I+ 0 o

I YB = I YB + − 120 o = I+ − 120 o I BR = I BR + − 240 o = I+ − 240 o

where, I RY = I YB = I BR = I = Magnitude of phase current.

Circuit Theory

12. 11

In a balanced delta-connected source, the magnitude of line current is 3 times the magnitude of source current and the line current lags the phase current by 30o. Therefore, for purely resistive load, the line currents can be expressed as shown below: IR =

3 I+ − 30 o

= I R + − 30 o = I L + − 30 o

IY =

3 I+ − 150 o = I Y + − 150 o = I L + − 150 o

IB =

3 I+ − 270 o = I B + − 150 o = I L + − 270 o

where, I R = I Y = I B = I L =

3I

Proof for line currents in delta-connected source: Consider the delta-connected source shown in Fig. 12.12. Now, the phase currents when the load is purely resistive are, I RY = I+0 o I YB = I+ − 120 o I BR = I+ − 240 o With reference to Fig. 12.14 in the node-R by using KCL, we can write, I R + I BR = I RY

= I^1.732+ − 30 oh = I^ 3 + − 30 oh

+E R

~

~

E

= I (1∠0o − 1∠− 240o ) = I(1.5 – j0.866)

+

EB

E

IRY

= I∠0 o − I∠− 240o

IBR

Fig. 12.14.

3 I+ − 30 o

` IR =

IR

R

` I R = I RY − I BR

With reference to Fig. 12.15 in the node-Y by using KCL, we can write, I Y + I RY = I YB ` I Y = I YB − I RY o

IRY

ER

E

= I∠−120 − I∠0

= I (1∠−120o − 1∠0o ) = I(–1.5 – j0.866) E

= I^1.732+ − 150 oh = I^ 3 + − 150 oh

+

~

o

~

EY

I YB

+

IY

Y

Fig. 12.15. ` IY =

3 I+ − 150 o

12. 12

Chapter 12 - Three Phase Circuits

EB

~

+

I B + I YB = I BR

E

With reference to Fig. 12.16 in the node-B by using KCL, we can write,

` I B = I BR − I YB IBR

= I ∠− 240o − I ∠− 120o

B

E

I YB

+

EY

= I (1 ∠− 240o − 1 ∠− 120o ) = I( 0 + j1.732) IB

= I^1.732+90 oh = I^ 3 + − 270 oh ` IB =

~

Fig. 12.16.

3 I+ − 270 o

From the above analysis we can make following conclusion for balanced loads. 1. The magnitude of line current is

3 times the phase current.

2. The line current lags the phase current by 30o . Line and Phase Voltages of RBY Sequence The line voltages of RBY sequence are VRB, VBY, and VYR . By taking VRB as reference phasor the line and phase voltages of RBY sequence can be expressed as shown below: Line voltages VRB = VRB +0 o

Phase voltages = VL +0 o

E R = E R +0 o

= E+0 o

VBY = VBY + − 120 o = VL + − 120 o

E B = E B + − 120 o = E+ − 120 o

VYR = VYR + − 240 o = VL + − 240 o

E Y = E Y + − 240 o = E+ − 240 o

Here, VRB = VBY = VYR = VL

V YR

ER = EB = EY = E = VL VRB

The rms phasors of the line voltages of delta-connected source for RBY sequence are shown in Fig. 12.17.

Points to remember

o

20 E1 o 40 E2

VBY

Fig. 12.17.

1. The voltages are always balanced in delta-connected sources. 2. The currents may be balanced or unbalanced depending on load. 3. The phase and line voltages are same in delta system. 4. The magnitude of line current is 3 times the magnitude of phase current. 5. The line current lags the phase current by 30o in balanced delta system.

Reference

Circuit Theory

12. 13

12.6 Three-Phase Loads The three-phase loads can be connected in star or delta and the load impedance may be balanced or unbalanced. In balanced load, the magnitude of load impedance of each phase will be equal and also the load impedance angle of each phase will be same. In unbalanced load, the load impedance of each phase may have different magnitude and/or different impedance angle. The three-phase loads can be classified as shown below. The balanced and unbalanced loads of star and delta system are shown in Fig. 12.18. Three phase load

Balanced load

Unbalanced load

Four wire star-connected load

Four wire star-connected load

Three wire star-connected load

Three wire star-connected load

Delta-connected load

Delta-connected load

R R R Z R = ZÐf Z R = ZÐf Z Y = ZÐf

N

Z RY = ZÐf

N

Z BR = ZÐf

Z Y = ZÐf

Y

Z B = ZÐf

Y

Z B = ZÐf

Y Z YB = ZÐf

B

B

B N

Fig. a : Four wire star-connected Fig. b : Three wire starbalanced load. connected balanced load.

Fig. c : Delta-connected balanced load.

R R R Z R = ZRÐfR Z R = ZRÐfR Z Y = Z Y Ðf Y

N N

Z RY = ZRY ÐfRY

Z BR = ZBRÐfBR

Z Y = Z Y Ðf Y

Y

Z B = ZBÐfB

Y

Z B = ZBÐfB

Y Z YB = Z YBÐf YB

B B

B

N

Fig. d : Four wire star-connected Fig. e : Three wire star Fig. f : Delta-connected unbalanced load. -connected unbalanced load. unbalanced load. Fig. 12.18 : Three phase loads.

12. 14

Chapter 12 - Three Phase Circuits

12.6.1 Choice of Reference Phasor in Analysis of Three-Phase Circuits In the analysis of three-phase circuits, it is conventional practice to choose one of the line voltage of the source as reference phasor. In RYB sequence, there are three line voltages and so we have three choice of reference phasor. In RBY sequence, there are three line voltages and so we have another three choice of reference phasor. The line voltages for six possible choice of reference phasor are listed in Table 12.1. TABLE 12.1 : CHOICE OF REFERENCE PHASOR Phase sequence

Reference phasor

RYB

VRY

Line voltages VRY = VL +0 o VYB = VL + − 120 o VBR = VL + − 240 o = VL +120 o

Phasor diagram VBR o

120

VRY o

E120 VYB VRY

RYB

VYB

VYB = VL +0 o VBR = VL + − 120 o VRY = VL + − 240 o = VL +120 o

o

120

V YB o

E120 VBR V YB

RYB

VBR

VBR = VL +0 o VRY = VL + − 120 o VYB = VL + − 240 o = VL +120 o

120

o

VBR o

E120 VRY V YR

RBY

VRB

VRB = VL +0 o VBY = VL + − 120 o VYR = VL + − 240 o = VL +120 o

o

120

VRB o

E120 VBY

RBY

VBY

VBY = VL +0 o VYR = VL + − 120 o VRB = VL + − 240 o = VL +120 o

VRB

120

o

VBY o

E120 V YR

VBY

RBY

VYR

VYR = VL +0 o VRB = VL + − 120 o VBY = VL + − 240 o = VL +120 o

o

120

V YR o

E120 VRB

Circuit Theory

12.7

12. 15

Analysis of Balanced Loads

12.7.1 Four-Wire Star-connected Balanced Load Let us assume a phase sequence of RYB. Let the reference phasor be VRY . The three-phase four-wire star-connected load with conventional polarity of voltages and direction of currents is shown in Fig. 12.19. R

R -

+

IR

+ VR

VRY

VBR

Z Y = ZÐf

+

Y

IY

-

VY

IR Z R = ZÐf IN

N

-

Z B = ZÐf VB +

Y IY

+

IB

B

V YB

B

-

+

IB

IN

N

Fig. 12.19 : Four wire star-connected balanced load with conventional polarity of voltages and direction of currents for RYB sequencce.

The line voltages of the supply/source for RYB sequence are, VRY = VL +0 o VYB = VL + − 120 o VBR = VL + − 240 o

where, VL= Magnitude of line voltage. Since the load impedance is balanced, the phase voltages of the load will be balanced. Since the load neutral is tied to source neutral the phase voltages of the load will be same as that of phase voltages of source. Hence we can say that, the magnitude of the phase voltage is 1/ 3 times the magnitude of line voltage and the phase voltage lag behind the line voltage by 30 o. Therefore, the phase voltages of the load are, VR =

VL + (0 o − 30 o) 3

VY =

VL + (− 120 o − 30 o) = V+ − 150 o 3

VB =

VL + (− 240 o − 30 o) = V+ − 270 o 3 where, V =

= V+ − 30 o

VL = Magnitude of phase voltgage. 3

12. 16

Chapter 12 - Three Phase Circuits

The phase currents are given by the ratio of phase voltage and phase impedance (Ohm’s law applied to ac circuit). Therefore the phase currents are, IR =

o VR = V+ − 30 = V + (− 30 o − φ) Z+φ Z ZR

= I+ (− 30 o − φ)

o I Y = VY = V+ − 150 = V + (− 150 o − φ) = I+ (− 150 o − φ) Z+φ Z ZY o I B = VB = V+ − 270 = V + (− 270 o − φ) = I+ (− 270 o − φ) Z+φ Z ZB

where, I = V Magnitude of phase currets. Z

Since the load is balanced, the neutral current will be zero. Neutral current, I N = 0.

`

In star system the line currents are same as that of phase currents. Therefore the line currents are, I R = I L + (− 30 o − φ) I Y = I L + (− 150 o − φ) I B = I L + (− 270 o − φ)

where, I L = I = Magnitude of line current. The power, P consumed by balanced three-phase star-connected load is given by, P = 3 V I cos φ

or

P =

3 VL I L cos φ

Proof for power consumed by balanced star-connected load: Let,

P = Power consumed by balanced three-phase load φ 1 = Phase difference between V R and I R φ 2 = Phase difference between V Y and I Y φ3 = Phase difference between V B and I B

Now, P =

Power consumed Power consumed Power consumed + + by R- phase load by Y- phase load by B- phase load

= V R I R cos φ1 + V Y I Y cos φ2 + V B I B cos φ3

V R = VY = V B = V I R = IY = I B = I φ1 = φ2 = φ3 = φ

= V I cos φ + V I cos φ + V I cos φ = 3V I cos φ = 3

VL IL cos φ 3

=

3 VL IL cos φ

..... (12.1) In balanced star system, VL and I = I L V = 3 ..... (12.2)

Circuit Theory

12. 17

12.7.2 Three-Wire Star-connected Balanced Load +

VRY

-

IR

+ VR

VBR

Z Y = ZÐf IY

-

Y

Y IY

+

+

VY

IR Z R = ZÐf

N

VB

Z B = ZÐf

+

The analysis of three-wire star-connected balanced load and four-wire star-connected balanced load are one and the same. Because, both the source and load neutrals will be at zero potential when the source and load are balanced. In balanced loads, the physical connection of source and load neutral by the neutral line has no significance. Hence for the analysis of threewire balanced star-connected loads, follow the steps given in Section 12.7.1.

R

R

-

Let us assume a phase sequence of RYB. Let the reference phasor be VRY . The three-phase three-wire star-connected load with conventional polarity of voltages and direction of currents is shown in Fig. 12.20.

IB

B

V YB

B

IB

+

Fig. 12.20 : Three wire star-connected balanced load with conventional polarity of voltages and direction of currents for RYB sequence.

12.7.3 Delta-connected Balanced Load Let us assume a phase sequence of RYB. Let the reference phasor be VRY . The three-phase deltaconnected load with conventional polarity of voltages and direction of currents is shown in Fig. 12.21. R IR

IRY

VRY

VBR +

Z RY = ZÐf

VBR

Z BR = ZÐf

+

VRY

-

-

+

-

R

I YB + V YB -

IY

-

Y

+

Y

IBR

B

Z YB = ZÐf

V YB

B

-

+

IB

Fig. 12.21 : Delta-connected balanced load with conventional polarity of voltages and direction of currents for RYB sequence.

The line voltages of the supply/source for RYB sequence are, VRY = VL +0 o VYB = VL + − 120 o VBR = VL + − 240 o

where, VL= Magnitude of line voltage. In delta-connected loads, the impedances are connected between two lines. Hence, the voltage across the impedance connected between two lines will be same as that of line voltage

12. 18

Chapter 12 - Three Phase Circuits

between those two lines. Therefore, the phase voltages will be same as that of line voltages of the source. The phase voltages are, VRY = V+0 o VYB = V+ − 120 o VBR = V+ − 240 o

where, V = VL = Magnitude of phase voltage. The phase currents are given by the ratio of phase voltage and phase impedance (Ohm’s law applied to ac circuit). Therefore the phase currents are, I RY =

o V RY = V+0 Z + φ Z RY

= V+−φ Z

= I+ − φ

o I YB = VYB = V+ − 120 = V + (− 120 o − φ) = I+ (− 120 o − φ) Z+φ Z Z YB o I BR = VBR = V+ − 240 = V + (− 240 o − φ) = I+ (− 240 o − φ) Z Z + φ Z BR

where, I = V = Magnitude of phase current. Z In balanced delta-connected loads, the relation between line and phase currents, will be same as that of balanced delta-connected source. Hence, we can say that, the magnitude of line current will be 3 times the phase current and line current will lag the phase current by 30 o. Therefore the line currents are, IR =

3 I + (− φ − 30 o)

IY =

3 I + (− 120 o − φ − 30 o) = I L + (− 150 o − φ)

IB =

3 I + (− 240 o − φ − 30 o) = I L + (− 270 o − φ)

where, IL =

= I L + (− 30 o − φ)

3 I = Magnitude of line current.

Alternatively, line currents can be computed from the following relation. I R = I RY − I BR I Y = I YB − I RY I B = I BR − I YB

The power, P consumed by balanced three-phase delta-connected load is given by, P = 3 V I cos φ

or

P =

3 VL I L cos φ

Circuit Theory

12. 19

Proof for power consumed by balanced delta-connected load: Let,

P = Power consumed by balanced three-phase load φ 1 = Phase difference between V RY and I RY φ 2 = Phase difference between V YB and I YB φ3 = Phase difference between V BR and I BR

Now, P =

Power consumed Power consumed Power consumed + + by R- phase load by Y- phase load by B- phase load

V RY = V YB = V BR = V I RY = I YB = I BR = I

= V RY I RY cos φ1 + V YB I YB cos φ2 + V BR I BR cos φ3 φ1 = φ2 = φ3 = φ = V I cos φ + V I cos φ + V I cos φ = 3V I cos φ = 3VL =

..... (12.3) In balanced delta system, I I = L and V = VL 3

IL cos φ 3

..... (12.4)

3 VL IL cos φ

Note : From equations (12.1) to (12.4), we can say that the expression (or equation) for calculating the power in balanced star and delta load is same. 12.7.4

Power Consumed by Three Equal Impedances in Star and Delta

For same load impedance and supply voltage the power consumed by delta-connected load will be three times the power consumed by star-connected load. Alternatively, power consumed by star-connected load will be one-third the power consumed by delta-connected load. Let, PD = Power consumed in delta connection PS = Power consumed in star connection Now,

PD = 3 PS

or

PS = 1 PD 3

..... (12.5)

Proof: Let us consider three equal impedances Z ∠φ connected in star. Let the supply voltage be VL volts. Now, Phase current in star, IS =

VL 3 = Z

Line current in star, IL, S = IS =

`

PS =

3 VL IL, S cos φ =

VL 3Z VL 3Z

3 VL

V L2 VL cos φ = cos φ Z 3Z

Let us reconnect the three equal impedances Z ∠φ in delta. Let the supply voltage be VL volts.

..... (12.6)

12. 20

Chapter 12 - Three Phase Circuits Now, phase voltage in delta, VD = VL Phase current in delta, ID =

VD V = L Z Z

Line current in delta, IL, D =

3 ID =

` PD =

3 VL IL, D cos φ =

3

3 VL 3

VL Z

V2 VL cos φ = 3 L cos φ Z Z

..... (12.7)

From equations (12.6) and (12.7) we can say that, PD = 3 PS or

12.8

PS = 1 PD 3

Analysis of Unbalanced Loads

In the analysis of unbalanced loads, the supply/source is always assumed to be balanced. Moreover, the line voltages of the source and load are same. Therefore, we can say that the line voltages are always balanced, for any type of load. 12.8.1 Four-wire Star-connected Unbalanced Load Let us assume a phase sequence of RYB. Let the reference phasor be VRY . The three-phase, four-wire unbalanced star-connected load with conventional polarity of voltages and direction of currents is shown in Fig. 12.22. R

R

+

-

IR

+ VR

VRY

VBR

IR

Z R = ZRÐfR

N

+

Y

IY

+ IY

VY

Z B = ZBÐfB VB

+

Y

-

Z Y = Z Y Ðf Y

IB

B

V YB

B

N

-

+

IN

IB

IN

Fig. 12.22 : Four wire star-connected unbalanced load with conventional polarity of voltages and direction of currents.

The line voltages of the supply/source for RYB sequence are, VRY = VL +0 o VYB = VL + − 120 o VBR = VL + − 240 o

where, VL = Magnitude of line voltage.

Circuit Theory

12. 21

In four-wire system, the load neutral is tied to the source neutral and so the phase voltages of source and load will be same. Since the source voltages are balanced, the phase voltages of the load should also be balanced. Here, “the significance of connecting the load and source neutral is that, it will not allow voltage unbalance even though the load is unbalanced”. Therefore, in a four-wire three-phase star-connected unbalanced loads the relation between line and phase voltages will be same as that of balanced loads. The magnitude of phase voltage will be 1/ 3 times the line voltage and the phase voltage lag behind the line voltage by 30 o. The phase voltages are, VL + (0 o − 30 o) = V+ − 30 o 3 V VY = L + (− 120 o − 30 o) = V+ − 150 o 3 V VB = L + (− 240 o − 30 o) = V+ − 270 o 3 VR =

where, V =

VL = Magnitude of phase voltage. 3

The phase currents are given by the ratio of phase voltage and phase impedance (Ohm’s law applied to ac circuit). Therefore, the phase currents are, o I R = VR = V+ − 30 Z + φ ZR R R

= V + ( − 30 o − φR) = I R + (− 30 o − φR) ZR

o I Y = VY = V+ − 150 = V + (− 150 o − φY) = I Y + (− 150 o − φY) Z Y +φY ZY ZY o I B = VB = V+ − 270 = V + (− 270 o − φB) = I B + (− 270 o − φB) Z B +φB ZB ZB

where, IR, IY and IB are magnitude of R-phase, Y-phase and B-phase currents respectively. Neutral current, I N = I R + I Y + IB

..... (12.8)

In star system the line currents are same as that of phase currents. Therefore the line currents are, I R = I R + (− 30 o − φR) I Y = I Y + (− 150 o − φY) I B = I B + (− 270 o − φB)

Let, P = Power consumed by three-phase load. Here, P =

Power consumed Power consumed Power consumed + + by R- phase load by Y- phase load by B - phase load

= VR I R cos φ1 + VY I Y cos φ2 + VB I B cos φ3 = VL IR cosφ1 + VLI Y cos φ2 + VLIB cosφ3

..... (12.9)

12. 22

Chapter 12 - Three Phase Circuits

where, VR = VY = VB = VL φ 1 = Phase difference between VR and I R φ 2 = Phase difference between VY and I Y φ 3 = Phase difference between VB and I B Also, φ 1 = φR

;

φ2 = φY

Note : The equation, P =

;

φ3 = φB

3 VL IL cosφ cannot be used to calculate power in unbalanced loads.

12.8.2 Three-wire Star-connected Unbalanced Load

(AU May’15, 16 Marks)

Let us assume a phase sequence of RYB. Let the reference phasor be VRY . R +

-

VRY

VBR

IR

+ VR

N

IY

-

Z R = ZRÐfR

Z Y = Z Y Ðf Y

Y

IR

+

+ Y IY

VY

Z B = ZBÐfB

VB

+

R

IB

B

V YB

B

+

IB

Fig. 12.23 : Three wire star-connected unbalanced load with conventional polarity of voltages and direction of currents for RYB sequence.

The line voltages of the supply/source for RYB sequence are, VRY = VL +0 o VYB = VL + − 120 o VBR = VL + − 240 o

where, VL = Magnitude of line voltage. In unbalanced star-connected load it will be easier to solve, the line currents by assuming two sources, across the lines whose values are equal to corresponding line voltages. Consider the circuit shown in Fig. 12.24 in which a voltage source of value VRY is connected across lines R and Y and a voltage source of value VYB is connected across lines Y and B. The circuit of Fig. 12.24, has two meshes, hence we can assume two mesh currents I1 and I2 as shown in Fig. 12.24. The mesh basis matrix equation is,

R

IR

+ VRY

~

ZY

E Y + V YB

~

E

B

ZR

I1

N

IY I2 IB

Fig. 12.24.

ZB

Circuit Theory

=

12. 23

ZR + ZY − ZY G − ZY ZB + ZY

I1 VRY G = G = = I2 VYB

ZR + ZY VRY − ZY − ZY Z R + Z Y VRY ; ∆1 = ; ∆2 = VYB Z B + Z Y − ZY ZB + ZY − Z Y VYB

Let, ∆ = Now, I1 =

∆1 ∆ and I2 = 2 ∆ ∆

From the mesh currents the line currents can be obtained as shown below: I R = I1 I Y = I2 − I1 I B = − I2

In a star-connected load, the phase currents are same as that of line currents. Therefore the phase and line currents in the polar form can be written as, I R = I R +γR I Y = I Y +γY I B = I B +γB

where, I R, I Y, and I B are magnitude of line and phase currents and γ , γ and γ are phase angle of line and phase currents with respect to R

Y

B

reference phasor. Now the phase voltages are given by the product of phase current and phase impedance (Ohm’s law applied to ac circuit.) Therefore the phase voltages are, VR = I R Z R = VR +δR VY = I Y Z Y = VY +δY VB = I B Z B = VB +δB

where, VR, VY and VB are magnitude of phase voltages and δR, δY and δB are phase angle of phase voltages with respect to reference phasor. Let, P = Power consumed by three-phase load. P =

Power consumed Power consumed Power consumed + + by R- phase load by Y- phase load by B- phase load

= VR I R cos φ1 + VY I Y cos φ2 + VB I B cos φ3 = VR I R cos φ1 + VY I Y cos φ2 + VB I B cos φ3

..... (12.10)

12. 24

Chapter 12 - Three Phase Circuits

where, φ 1 = Phase difference between VR and I R φ 2 = Phase difference between VY and I Y φ3 = Phase difference between VB and I B Here, φ 1 = δR − γR ; φ2 = δY − γY ;

φ3 = δB − γB.

Also, φ 1 = φR

φ3 = φB

; φ2 = φY

;

12.8.3 Neutral Shift in Star-connected Load In three-wire star-connected load, the load neutral is not connected to source neutral. Hence when the load is unbalanced, the load neutral will not be at zero potential. The voltage of load neutral with respect to source neutral is called neutral shift voltage (or neutral displacement voltage). The neutral shift voltage can be obtained by subtracting a phase voltage of load from the corresponding phase emf of the source. Consider the R-phase source and load as shown in Fig. 12.25. Let N’ be the source neutral and N be the load neutral.

+

+ ER

~

VR

N¢

VNN¢

-

Let, VNNl = Load neutral shift voltage with respect to source neutral.

N +

Fig. 12.25. o

In star-connected source, the phase emf will lag behind the line voltage by 30 and magnitude of phase emf will be 1/ 3 times the magnitude of line voltage. Hence E R will have a magnitude of VL / 3 and lag behind VRY by 30o ` ER =

VL + − 30 o 3

With reference to Fig. 12.25, using KVL, we can write, VNNl + VR = E R ..... (12.11)

` Neutral shift voltage, VNNl = E R − VR

12.8.4 Delta-connected Unbalanced Load Let us assume a phase sequence of RYB. Let the reference phasor be VRY . R

VBR

IR

IRY

VRY

VBR

Z RY = ZRY ÐfRY

-

VRY

-

-

+

+

R

Y

I YB + V YB -

IY

-

Y

+

Z BR = ZBRÐfBR

+ I BR

B

Z YB = Z YBÐf YB

V YB

B

+

IB

Fig. 12.26 : Three phase delta-connected unbalanced load with conventional polarity of voltages and direction of currents for RYB sequence.

Circuit Theory

12. 25

The line voltages of the supply/source for RYB sequence are, VRY = VL +0 o VYB = VL + − 120 o VBR = VL + − 240 o

where, VL = Magnitude of line voltage. In delta-connected loads, the impedances are connected between two lines. Hence the voltage across the impedance connected between two lines will be same as that of line voltage between those two lines. Therefore the phase voltages will be same as that of line voltages of the source. Since the line voltages are balanced, the phase voltages of the load are also balanced even though load impedances are unbalanced. Therefore the phase voltages are, VRY = V+0 o VYB = V+ − 120 o VBR = V+ − 240 o

where, V = VL = Magnitude of phase voltage. The phase currents are given by the ratio of phase voltage and phase impedance (Ohm’s law applied to ac circuit). Therefore, the phase currents are, I RY =

V RY = I RY +γRY Z RY

I YB =

V YB = I YB +γYB Z YB

I BR =

V BR = I BR +γBR Z BR

where, I RY , IYB and I BR are magnitude of phase currents and γ , γ and γ are phase angle of phase currents. RY YB BR By using KCL at nodes R, Y, and B the line currents can be calculated as shown below. The line currents are, I R = I RY − I BR I Y = I YB − I RY I B = I BR − I YB

Let, P = Power consumed by three-phase load. P =

Power consumed Power consumed Power consumed + + by R- phase load by Y- phase load by B- phase load

= VRY I RY cos φ1 + VYB I YB cos φ2 + VBR I BR cos φ3 = VL IRY cosφ1 + VLI YB cos φ2 + VLIBR cosφ3

..... (12.12)

12. 26

Chapter 12 - Three Phase Circuits where, VRY = VYB = VBR = VL

φ 1 = Phase difference between VRY and I RY φ 2 = Phase difference between VYB and I YB φ3 = Phase difference between VBR and I BR

12.9

Here, φ 1 = 0o − γRY

;

φ 2 = −120o − γYB

;

φ 3 = −240 o − γBR

Also, φ 1 = φRY

;

φ2 = φYB

;

φ3 = φBR

Two Wattmeter Method of Power Measurement (AU Dec’14, 16 Marks)

The power is generally measured by using wattmeters. Logically we may require one wattmeter for measuring power in one-phase and so we may require three wattmeters to measure power in three-phase. But it can be proved that “the power in any three-phase load(balanced/ unbalanced and star/delta) can be measured by using only two wattmeters”. A wattmeter will have a current coil (CC) and a pressure coil (PC). The pressure coil is also called voltage coil. The current coils of the two wattmeters employed for measurement are connected such that they carry any two line currents. The third line in which there is no current coil is called common line. The voltage coil of a wattmeter is connected between the line to which its current coil is connected and the common line. Similarly, the voltage coil of another wattmeter is connected between the line to which its current coil is connected and the common line. The possible connections of two wattmeters to measure the three-phase power are shown in Fig. 12.27. CC R

R

CC

3-phase load

Y

+ V YB

PC IB

+ P2

B

CC

IR

PC

VRB

E E

CC

+

PC

VBY

B

P1

IR

+ VRY

Y

P1

E

IY

3-phase load

PC P2

E

Fig. b : Wattmeters in lines R and Y.

Fig. a : Wattmeters in lines R and B. R E

E

V YR

Y

P1 CC

+

3-phase load

PC

VBR

B

IY

CC

+ P2

IB

PC

Fig. c : Wattmeters in lines Y and B. Fig. 12.27 : Possible connections of two wattmeters for measurement of 3-phase power.

Circuit Theory

12. 27

12.9.1 Power Measurement in Balanced Load Consider a balanced three-phase load (star or delta-connected). Let us connect wattmeters in the lines R and B and line Y be the common line for connecting the voltage coil as shown in Fig. 12.28. CC R

P1

IR

+ PC VRY

Y

3-phase balanced load

E E VBY

B

PC IB

+ P2

CC

Fig. 12.28 : Power measurement in balanced load.

Let, P1 = Power measured by wattmeter-1 P2 = Power measured by wattmeter-2 P = Power consumed by load φ = Load power factor angle Now, the load power and power factor angle in terms of wattmeter readings P1 and P2 are, Power, P = P1 + P2

..... (12.13)

Power factor angle, φ = tan- 1 c 3 P2 − P1 m P1 + P2

..... (12.14)

Power factor, cos φ = cos ; tan- 1 c 3 P2 − P1 mE P1 + P2

..... (12.15)

Proof for power measurement in three-phase load using two wattmeters : Consider the circuit shown in Fig. 12.28 for power measurement in three-phase load using two wattmeters. Now, the current through wattmeter-1 is I R and voltage across its pressure coil is V RY . Hence the power P1 measured by wattmeter-1 is, P1 = V RY # I R # cos θ1

..... (12.16)

where, θ 1 = Phase difference between V RY and I R . The current through wattmeter-2 is I B and voltage across its pressure coil is V BY . Hence the power P2 measured by wattmeter-2 is, P2 = V BY # I B # cos θ2 where, θ 2 = Phase difference between V BY and I B .

..... (12.17)

12. 28

Chapter 12 - Three Phase Circuits For balanced star and delta-connected load the line voltages and currents are, Line voltages

Line currents

V RY = VL +0 o

I R = IL + (− 30 o − φ)

V YB = VL + − 120 o

I Y = IL + (− 150 o − φ)

V BR = VL + − 240 o

I B = IL + (− 270 o − φ)

`

V RY = VL ;

I R = IL and

Refer section (12.7.1), (12.7.2) and (12.7.3)

I B = IL

V BY = − V YB = 1+180 o # V YB = 1+180 o # VL + − 120 o = VL + (180 o − 120 o)

Here,

` V BY = VL +60 o `

V BY = VL

Now, θ 1 = 0 o − (−30o − φ) = 30o + φ = φ + 30o θ 2 = 60 o − (−270o − φ) = 330o + φ = − 30o + φ = φ − 30o From the above discussions, the equations (12.16) and (12.17) for balanced load (star/delta-connected) can be written as, P1 = VL I L cos (φ+ 30o )

..... (12.18)

o

P2 = VL I L cos (φ− 30 )

..... (12.19)

Let us add the power measured by two wattmeters. ∴ P1 + P2 = VL I L cos (φ+ 30o ) + VL IL cos (φ− 30o ) = VL I L [cos (φ+30o ) + cos (φ− 30o ) ]

cos (C + D) + cos (C − D) = 2 cos C cos D cos 30 o =

3 2

= VL IL [2 cos φ cos30o ] = VL I L ; 2 cos φ # =

3E 2

3 VL I L cosφ

..... (12.20)

The equation (12.20) is same as that of equation for power in balanced load. Hence we can say that the sum of power measured by two wattmeters is equal to the power in three-phase load. Power factor From equations, (12.18) and (12.19) we can write, P1 − P2 = VL I L cos(φ + 30o ) − VL IL cos (φ − 30o) = VL I L [cos(φ + 30o ) − cos (φ − 30o )]

cos (C + D) − cos (C − D) = − 2 sin C sin D

= VL I L [ −2 sinφ sin30o] ` P1 − P2 = VL IL 8− 2 sin φ # 1 B 2 = −VL I L sinφ ∴ P2 − P1 = VL I L sinφ

..... (12.21)

Circuit Theory

12. 29

On dividing equation (12.21) by equation (12.20) we get, VL IL sin φ P −P = 2 1 P1 + P2 3 VL IL cos φ ` tan φ =

3

sin 30 o = 1 2

P2 − P1 P1 + P2

` Power factor angle, φ = tan- 1 c 3

P2 − P1 m P1 + P2

Power factor, cos φ = cos ;tan- 1 c 3

P2 − P1 mE P1 + P2

..... (12.22) ..... (12.23)

Equation (12.23) gives the power factor of balanced load in terms of wattmeter reading. One drawback in this method of power factor estimation is that we cannot determine whether the power factor is lagging or leading (But practically most of the loads are inductive in nature and so we can safely assume that the power factor is lag). 12.9.2 Relation Between Power Factor and Wattmeter Readings Case i : Wattmeter readings are equal Let, P1 = P2 = Px ` Power factor, cos φ = cos < tan- 1 c 3

P2 − P1 P −P 1 mF = cos < tan- c 3 Px Px mF P1 + P2 x+ x

= cos 7 tan- 1 0 A = cos 0 o = 1

Conclusion : When the wattmeter readings are equal, the power factor is unity. Case ii : One of the wattmeter readings is zero Let, P1 = 0 and P2 ! 0 Power factor, cos φ = cos < tan- 1 c 3

P2 − P1 P −0 1 mF = cos < tan- c 3 02 P mF P1 + P2 + 2

= cos 7 tan- 1 ^ 3 hA = cos 60 o = 0.5 Let, P2 = 0 and P1 ! 0 Power factor, cos φ = cos < tan- 1 c 3

P2 − P1 0−P 1 mF = cos < tan- 3 P 01 F P1 + P2 1+

= cos 7 tan- 1 ^− 3 hA = cos 7 − 60 o A = 0.5

Conclusion : When one of the wattmeter reading is zero, the power factor is 0.5.

12. 30

Chapter 12 - Three Phase Circuits

Case iii : One of the wattmeter readings is negative Let, P1 = −Px and P2 = +Py Power factor, cos φ = cos < tan- 1 c 3 = cos < tan- 1 c 3

Py + Px P2 − P1 1 mF = cos < tan- c 3 P P mF P1 + P2 − x+ y Py + Px mF Py − Px

Let, P1 = + Px and P2 = −Py ` Power factor, cos φ = cos < tan- 1 c 3 = cos < tan- 1 c 3 Let, c 3

− Py − Px P2 − P1 1 mF = cos < tan- c 3 P P mF P1 + P2 x− y Py + Px mF Py − Px

Py + Px m = A Py − Px

∴ cosφ = cos (tan−1A )

Here, (Py + Px ) > (Py − Px),

∴ A>

3 and tan −1A > 60o

tan 60 o =

3

o

Since, tan −1A is greater than 60 , cos (tan−1A) will be less than 0.5. Note : The value of cos 60o to cos 90o will lie in the range of 0.5 to 0.

∴ Power factor, cos φ < 0.5 Conclusion :

When one of the wattmeter readings is negative, the power factor will be less than 0.5.

Case iv : Both the wattmeter readings are positive Let, P1 = +Px and P2 = +Py Power factor, cos φ = cos < tan- 1 c 3

Py − Px P2 − P1 1 mF = cos < tan- c 3 P P mF P1 + P2 x+ y

= cos [tan−1B]

where, B =

Here, (Py − Px ) < (Px + Py ),

3

Py − Px Px + Py ∴B
0.5 Conclusion : When both the wattmeter readings are positive the power factor will be greater than 0.5.

Circuit Theory

12. 31

A Note on Power and Power Factor Estimation using Two Wattmeters It is possible to prove that the sum of power measured by two wattmeters is equal to power in three-phase, even if the load is unbalanced star/delta. The power factor is defined as ratio of active power by apparent power. In balanced loads, the ratio of real and apparent power of each phase is same, and so the power factor will be same for each phase. Also the three-phase load power factor will be same as power factor of each phase, (because the ratio of one phase active power by apparent power is same as the ratio of three-phase active power by apparent power). In unbalanced loads, the ratio of active and apparent power of each phase will be different and so the power factor of each phase will be different. Using two wattmeter readings we cannot determine the power factor of individual phase load. Therefore estimation of apparent power and hence the estimation of three-phase load power factor in unbalanced loads are not possible by two wattmeter method of power measurement. Note : VAR meters are available for measurement of reactive power in three-phase loads.

12.10 Solved Problems EXAMPLE 12.1 An unbalanced four-wire star-connected load has balanced supply voltage of 400V. The load impedances are ZR = 4 + j8 Ω, Z Y = 3 + j4 Ω and ZB = 15 + j10 Ω. Calculate the line currents, neutral current and the total power. Also draw the phasor diagram.

SOLUTION Let the phase sequence by RYB. Let the reference phasor be VRY . The star-connected load with polarity of voltages and direction of currents is shown in Fig. 1. R

R E

IR

+ VR

VRY

VBR

E Z Y a 3 C j4 ‡

Y

IY Y

E +

+

V YB

B

N

E

+

IB

IN

Fig. 1.

Z R a 4 C j8 ‡ IN

N

E E VY

Z B a 15 C j10 ‡ VB

+

+

B

12. 32

Chapter 12 - Three Phase Circuits The line voltages for RYB sequence, with VRY as reference phasor are, VRY = 400+0 o V VYB = 400+ − 120 o V VBR = 400+ − 240 o V = 400+120 o V

In a four-wire star-connected unbalanced load, the phase voltages are balanced because of neutral connection. Hence the phase voltages will have a magnitude of 1 3 times the line voltage and phase voltages lag 30 o with respect to line voltages. Therefore the phase voltages are, VR = 400 + (0 o − 30 o) = 230.9401+ − 30 o V 3 VY = 400 + (− 120 o − 30 o) = 230.9401+ − 150 o V 3 VB = 400 + (− 240 o − 30 o) = 230.9401+ − 270 o V = 230.9401+90 o V 3 In star system the line and phase currents are same. The phase currents are given by the ratio of phase voltage and phase impedance. Therefore the phase and line currents are, o o I R = VR = 230.9401+ − 30 = 230.9401+ − 30 = 25.8198+ − 93.4 o A o 4 j 8 + ZR 8.9443+63.4 o o I Y = VY = 230.9401+ − 150 = 230.9401+ −o 150 = 46.188+ − 203.1 o A 3 + j4 ZY 5+53.1 o o I B = VB = 230.9401+ − 270 = 230.9401+ − 270 = 12.8102+ − 303.7o A 15 + j10 ZB 18.0278+33.7 o = 12.8102+56.3o A

Neutral current, I N = I R + I Y + I B = 25.8198∠− 93.4 o + 46.188∠− 203.1 o + 12.8102∠− 303.7 o = −36.9083 + j3.0044 = 37.0304∠175.3 o A

–303.7o + 360o = 56.3o

VBR VB IB o

o

120

90

56.

IY

VRY

o

3

o

Reference o

E203.1

E30

o

50 E1

o

20 o E1 3.4 E9

VY

VR

IR V YB

Fig. 2 : Rms phasors of voltages and currents.

Circuit Theory

12. 33

Power, P =

Power consumed Power consumed Power consumed + + by R - phase load by Y- phase load by B- phase load

= VR IR cos φ1 + VY I Y cos φ2 + VB IB cos φ3 φ 1 = Phase difference between VR and IR = −30 o − (−93.4o ) = 63.4 o φ 2 = Phase difference between VY and I Y = −150 o − (−203.1o ) = 53.1 o φ 3 = Phase difference between VB and IB = −270 o − (−303.7o ) = 33.7 o Here, VR = VY = VB = V = 230.9401V ∴ P = V I Rcos φ1 + VIY cos φ2 + VIBcosφ3 = V [ I Rcosφ1 + IY cos φ2 + IBcosφ3] = 230.9401 × [25.8198 × cos63.4 o + 46.188 × cos 53.1 o + 12.8102 × cos 33.7 o ] = 11535.6 W = 11535.6 kW = 11.5356 kW 1000

EXAMPLE 12.2 A three-phase four-wire symmetrical 440 V, RBY system supply power to a star-connected load in which ZR = 10∠0 o Ω, Z Y = 10∠26.8o Ω and ZB = 10∠−26.8o Ω. Find the line currents, neutral current and the total power. Draw the phasor diagram.

SOLUTION The phase sequence is RBY. The line voltages for RBY sequence are VRB, VBY and VYR . Let VRB be the reference phasor. The star-connected load with polarity of voltages and direction of currents is shown in Fig. 1. IR

R

R

+ VR

VRB

V YR

-

Z R = 10 Ð 0 o W IN

N

o

Z Y = 10Ð 26.8 W

+ -

Y

IY Y

+

Z B = 10Ð - 26.8 o W

- VY

VB

+

+

B

VBY

B

-

+

IB

IN

N

Fig. 1. The line voltages for RBY sequence, with VRB as reference phasor are, VRB = 440+0 o V VBY = 440+ − 120 o V VYR = 440+ − 240 o V = 440+120 o V

12. 34

Chapter 12 - Three Phase Circuits

In a four-wire star-connected unbalanced load, the phase voltages are balanced because of neutral connection. Hence the phase voltages will have a magnitude of 1 3 times the line value and phase voltages lag 30 o with respect to line voltages. Therefore the phase voltages are, VR = 440 + (0 o − 30 o) = 254.0341+ − 30 o V 3 VB = 440 + (− 120 o − 30 o) = 254.0341+ − 150 o V 3 VY = 440 + (− 240 o − 30 o) = 254.0341+ − 270 o V = 254.0341+90 o V 3 In star system the line and phase currents are same. The phase currents are given by the ratio of phase voltage and phase impedance. Therefore the phase and line currents are, o IR = VR = 254.0341+o − 30 = 25.4034+ − 30 o A ZR 10+0 o IB = VB = 254.0341+ − 150 = 25.4034+ − 123.2 o A ZB 10+ − 26.8 o o I Y = VY = 254.0341+ −o270 = 25.4034+ − 296.8 o A = 25.4034+63.2 o A ZY 10+26.8 –296.8o + 360o = 63.2o Neutral current, IN = IR + IB + I Y

= 25.4034∠− 30 o + 25.4034∠− 123.2 o + 25.4034∠− 296.8 o = 19.5438 − j11.2836 o

= 22.5672∠− 30 A Power, P =

Power consumed Power consumed Power consumed + + by R-phase load by B-phase load by Y-phase load

= VR IR cos φ1 + VB IB cos φ2 + VY I Y cos φ3 φ1 = Phase difference between VR and IR = − 30 o − ^− 30 oh = 0 o φ2 = Phase difference between VB and IB = − 150 o − ^− 123.2 oh = − 26.8 o φ3 = Phase difference between VY and I Y = − 270 o − ^− 296.8 oh = 26.8 o V YR

VR = VB = VY = V = 254.0341V

VY IY

120

o

IR = IB = I Y = I = 25.4034 A ∴ P = V I cosφ1 + VIcosφ2 + VIcosφ3

VRB o

63.2

90 o

Here,

o

= V I [ cos φ1 + cosφ2 + cosφ3] = 254.0341 × 25.4034 × [cos 0 o + cos (−26.8 o ) o

+ cos ( 26.8 )] 17973 . 6 kW = 17.9736 kW = 17973.6 W = 1000

Reference

E30

o

E120

o

.2

VB IB VBY

E123 o E150

IR

VR

Fig. 2 : Rms phasors of voltages and currents.

Circuit Theory

12. 35

3.1 o

EXAMPLE 12.3

E36

E4

3o

.9 o

Reference

E5

In a four-wire three-phase system, two phases have currents of 10 A and 6 A of lagging power factor of 0.8 and 0.6 respectively, while the third phase is open circuited. Calculate the current in neutral and sketch the phasor diagram.

SOLUTION

IR

Let the given two phase currents be IR and I Y . Let the phase-B be open circuited and so the phase-B current IB is zero.

IY

∴ IR = 10 A at lagging pf of 0.8 = 10∠–cos − 10.8 = 10∠−36.9 o A IN

o I Y = 6 A at lagging pf of 0.6 = 6∠– cos − 10.6 = 6∠−53.1 A

Fig. 1 : Phasor diagram.

IB = 0 In 4-wire 3-phase system, the neutral current is given by sum of three-phase currents. ` Neutral current, IN = IR + I Y + IB = 10∠− 36.9 o + 6∠− 53.1 o + 0 = 11.5994 − j10.8023 = 15.8504 ∠− 43 o A

EXAMPLE 12.4

(AU Dec’14 & June’14, 8 Marks)

A balanced star-connected load of impedance 15 + j20 Ω per phase is connected to a three-phase, 440 V, 50Hz supply. Find the line currents and power absorbed by the load. Assume RYB sequence. Draw the phasor diagram. IR

The phase sequence is RYB. The line voltages for RYB sequence are VRY, VYB, and VBR . Let us choose VRY as reference phasor. The star-connected load with polarity of voltages and direction of currents is shown in Fig. 1. The line voltages for RYB sequence with VRY as reference phasor are,

E

VRY

VBR

R + VR

Y

IY

E

Z a 15 C j20 ‡

E Z a 15 C j20 ‡

Y

+

+

E VY

Z a 15 C j20 ‡ VB

B

V YB

VRY = 440+0 o V VYB = 440+ − 120 o V

+

+

R

E

SOLUTION

B

E

+

IB

Fig. 1.

VBR = 440+ − 240 o V = 440+120 o V Here,

VRY = VYB = VBR = VL = 440 V

In a three-phase balanced star-connected load, the magnitude of phase voltage will be 1 3 times the magnitude of line voltage. Also the phase voltage lag behind the line voltage by 30 o. Therefore the phase voltages are, VR = 440 + (0 o − 30 o) = 254.0341+ − 30 o V 3 VY = 440 + (− 120 o − 30 o) = 254.0341+ − 150 o V 3 VB = 440 + (− 240 o − 30 o) = 254.0341+ − 270 o V = 254.0341+90 o V 3

12. 36

Chapter 12 - Three Phase Circuits Here,

VR = VY = VB = V = 254.0341V

Given that, load impedance per phase, Z = 15 + j20 Ω = 25∠53.1 o Ω In star system, the line and phase currents are same. Therefore the line and phase currents are, o IR = VR = 254.0341+ −o 30 = 10.1614+ − 83.1 o A Z 25+53.1 o I Y = VY = 254.0341+ − o150 = 10.1614+ − 203.1 o A Z 25+53.1

VBR VB o

120

IB

o = VB = 254.0341+ − o270 = 10.1614+ − 323.1 o A Z 25+53.1 = 10.1614+36.9 o A

IB

IY

Here,

o

90

IR = I Y = IB = I = IL = 10.1614 A

36.9

φ = Impedance angle = 53.1o

Reference

E30 o

3.1

o

E120

50 E1

3 VLILcos φ VY

Here,

VRY

o

o

Power, P =

o

o

E203.1 E8

VR

IR o 3 × 440 × 10.1614 × cos 53.1

∴ P=

V YB

= 4649.7 W = 4649.7 kW = 4.6497 kW 1000

Fig. 2 : Rms phasors of voltages and currents.

EXAMPLE 12.5 The power consumed in a three-phase balanced star-connected load is 2 kW at a power factor of 0.8 lagging. The supply voltage is 400 V, 50 Hz. Calculate the resistance and reactance of each phase.

SOLUTION Given that, Power factor, cos φ = 0.8 lag Power, P = 2 kW = 2 × 1000 W = 2000 W Line voltage, VL = 400 V We know that, P =

3 VL I L cos φ

` Line current, I L =

P = 3 VL cos φ

2000 = 3.6084 A 3 # 400 # 0.8

In star-connected load, the line and phase current are same. ∴ Phase current, I = I L = 3.6084 A The magnitude of phase voltage in a star-connected balanced load will be 1 of line voltage. `

Phase voltage, V =

VL = 400 = 230.9401V 3 3

3 times the magnitude

Circuit Theory

12. 37

Now, Magnitude of impedance per phase, Z = V = 230.9401 = 64 Ω I 3.6084 We know that the impedance angle, θ is same as power factor angle φ. ∴ Impedance angle, θ = cos −1 0.8 = 36.9 o ` Impedance per phase, Z = Z+θ = 64+36.9 o Ω Let us express the impedance in rectangular form. `

Here, the power factor is lagging and so impedance angle is positive.

Z = 64+36.9 o Ω = 51.1798 + j38.4269 Ω

We know that, Z = R + jX ` Resistance of load, R = Real part of Z = 51.1798 Ω/phase Resistance of load, X = Imaginary part of Z = 38.4269 Ω/phase

EXAMPLE 12.6 For the circuit shown in Fig. 1, calculate the line current, the power and power factor when the supply voltage is 300 V, 50 Hz. The value of R, L and C in each phase are 10 Ω,1 H and 100 µF respectively.

SOLUTION

L

R C

R

L

C

Inductive reactance, X L = 2πfL = 2π × 50 × 1 = 314.1593 Ω 1 1 = = 31.831 Ω 2πfC 2π # 50 # 100 # 10- 6 Let, Z = Impedance per phase.

C L

Capacitive reactance, X C =

Fig. 1. Here, 1 = 1 + 1 + 1 = 1 − j 1 + j 1 R jXL R XL XC − jX C Z `

Z =

1 1 = 1 −j 1 1 − j 1 + j 1 + j 1 R XL XC 10 314.1593 31.831

=

1 1 = 0.1 − j0.0032 + j0.0314 0.1 + j0.0282

=

1 = 9.6246+ − 15.7 o Ω/phase 0.1039+15.7 o

∴ Magnitude of impedance, Z = 9.6246 Ω Impedance angle,

θ = −15.7 o

Given that, Line voltage, VL= 300 V `

Phase voltage, V =

VL = 300 = 173.2051V 3 3

Current per phase, I = V = 173.2051 = 17.9961 A Z 9.6246

R

12. 38

Chapter 12 - Three Phase Circuits In star system the line and phase currents are same. ∴ Line current, I L = I = 17.9961 A Power factor, cos φ = cosθ = cos(−15.7 o ) = 0.9627 lead Power, P = =

Here, the impedance angle is negative and so load is capacitive. Therefore power factor is lead.

3 V L I L cos φ 3 # 300 # 17.9961 # 0.9627 = 9002.3 W

= 9002.3 kW = 9.0023 kW 1000

EXAMPLE 12.7 Three equal inductors connected in star takes 5 kW at 0.7 power factor, when connected to a 400 V, 50 Hz, three-phase, three-wire supply. Calculate the line currents, (i) if one of the inductor is disconnected, (ii) if one of inductor is short circuited.

SOLUTION Given that, Power, P = 5 kW = 5 × 1000 W = 5000 W Power factor, cos φ = 0.7 Line voltage,

VL = 400 V

We know that,

P =

`

3 VL I L cos φ

Line current, IL =

P = 3 VL cos φ

5000 = 10.3098 A 3 # 400 # 0.7

In star system, line and phase currents are same. ∴ Phase current,

I = I L = 10.3098 A

In balanced star-connected load the magnitude of phase voltage is 1 `

Phase voltage, V =

3 times the line voltage.

VL = 400 = 230.9401V 3 3

Magnitude of impedance per phase, Z = V = 230.9401 = 22.4001 Ω/phase I 10.3098 We know that the impedance angle, θ is same as power factor angle, φ. ∴ Impedance angle, θ = cos −1 0.7 = 45.6 o Impedance per phase, Z = 22.4001∠45.6 o Ω/phase

Here impedance angle is positive because load is inductive.

Circuit Theory

12. 39

Case i : When one of the inductor is disconnected Let B-phase inductor be disconnected. With reference to Fig. 1, we can say that when one of the inductor is disconnected the load becomes two phase load. Let us assume RYB sequence. The line voltages of RYB sequence are VRY, VYB and VBR .

IR

R

R

+

Z V RY

Let VRY be reference phasor. Open

` VRY = 400+0 o V With reference to Fig. 1, we can say that the voltage across the load is VRY and total load impedance is 2 Z . Also IR = − I Y .

IR

Y

Z

Y

B

IB

B

By Ohm’s law, we can write,

IY

E

Fig. 1.

400+0 o = VRY = = 8.9285+ − 45.6 o A 2Z 2 # 22.4001+45.6 o `

I Y = − IR = − 1 # 8.9285+ − 45.6 o = 1+180 o # 8.9285+ − 45.6 o = 8.9285∠134.4 o A

Since, phase-B is open, IB = 0 In summary, the line currents when one of the inductor is disconnected are, IR = 8.9285+ − 45.6 o A

IR

R

I Y = 8.9285+134.4 o A

+

E

VRY

VBR

R

Z

IB = 0

Case ii : When one inductor is short circuited Let B-phase inductor be short circuited as shown in Fig. 2. Now the load can be treated as unbalanced load and analysed by mesh method. Let the phase sequence be RYB and VRY be the reference phasor. The line voltages are,

Short circuit Y

IY

E +

Z

B

Y

V YB

B

E

IB

+

Fig. 2.

VRY = 400+0 o V VYB = 400+ − 120 o V

IR

VBR = 400+ − 240 o V = 400+120 o V Let us connect two voltage sources VRY and VYB to unbalanced load as shown in Fig. 3. Let us assume two mesh currents I1 and I 2 as shown in Fig. 3. Now the line currents in terms of mesh currents are,

+

~

Z I1

VRY

E

IR = I1 IY

I Y = I 2 − I1 IB = − I 2

Z

+

~

V YB E IB

I2

Fig. 3.

12. 40

Chapter 12 - Three Phase Circuits The mesh basis matrix equation for the circuit of Fig. 3 is,

>

Z+Z −Z H Z −Z

>

2Z −Z H Z −Z

∆ =

I1 > H = I2 I1 > H = I2

> >

VRY H VYB

VRY H VYB

2Z − Z 2 2 2 = 2Z # Z − ^− Zh # ^− Zh = 2Z − Z = Z −Z Z

∆1 =

VRY − Z = VRY # Z − VYB # ^− Zh = Z ^VRY + VYBh VYB Z

∆2 =

2Z VRY = 2Z # VYB + Z # VRY = Z ^2VYB + VRYh − Z VYB

I1 =

o Z ^VRY + VYBh ∆1 = = VRY + VYB = 400 + 400+ − 120 2 o ∆ Z 22.4001+45.6 Z

= – 4.8021 – j17.1993 A

I2 =

o Z ^2 VYB + VRYh ∆2 = = 2 VYB + VRY = 2 # 400+ − 120 + o400 2 ∆ Z 22.4001+45.6 Z

= – 22.0982 – j21.6401 A Now the line currents are, IR = I1 = − 4.8021 − j17.1993 = 17.8571+ − 105.6 o A I Y = I 2 − I1 = − 22.0982 − j21.6401 − ^− 4.8021 − j17.1993h = −17.2961 − j4.4408 = 17.8571∠ –165.6oA IB = − I 2 = − ^− 22.0982 − j21.6401h = 30.9293+44.4 o A

EXAMPLE 12.8 Three similar resistors are connected in star across 400 V, 3-phase lines. The line current is 5 A. Calculate the value of each resistor. To what value should the line voltage be changed to obtain the same line current with the resistances delta-connected.

SOLUTION Case i : Star connection Line voltage in star, VL, S = 400 V ` Phase voltage, V = Given that, line current, I L = 5 A

VL = 400 = 230.9401V 3 3

Circuit Theory

12. 41

In star system line and phase currents are same. ∴ Phase current, I = I L = 5 A Resistance per phase, R = V = 230.9401 = 46.188 Ω I 5

Case ii : Delta connection The line current in delta connection should be maintained same as that in star connection. ∴ Line current, I L = 5 A In delta connection the phase current is 1 ` Phase current, I =

IL = 3

3 times the line current.

5 = 2.8868 A 3

The resistance per phase in delta connection is same as that in star connection. Resistance per phase, R = 46.188 Ω By Ohm’s law, Phase voltage, V = I R = 2.8868 × 46.188 = 133.3355 V In delta connection the line voltage is same as that of phase voltage. ∴ Line voltage in delta, VL, D = V = 133.3355 V

Conclusion The line voltage in star, VL, S = 400 V The line voltage in delta, VL, D = 133.3355 V Here, VL, S = 3 VL, D, (i.e., 3 × 133.3355 = 400 V ) Hence, we can say that when three equal impedances in star are reconnected to delta, then in order to maintain the same line current, the line voltage should be reduced to one-third. Alternatively when three equal impedances in delta are reconnected to star, then in order to maintain the same line current, the line voltage should be increased by three times.

EXAMPLE 12.9 A total 3-phase power of 100 kW is transmitted over transmission line of impedance 1 + j2 Ω/phase. The line voltage of the balanced 3-phase load is 11 kV. The load pf is 0.8 lagging. Find the line voltage and power factor at sending end.

SOLUTION tan f =

Given that, 3

Active power of load, P = 100 kW = 100 × 10 W

Q

S f

Power factor of load, cos φ = 0.8 lag ∴ φ = cos

–1

0.8

∴ tan φ = tan (cos –1 0.8) = 0.75 With reference to power triangle shown in Fig. 1,

P

Fig. 1 : Power triangle.

Q P

12. 42

Chapter 12 - Three Phase Circuits Reactive power of load, Q = P tanφ 3

= 100 × 10 3 × 0.75 = 75 × 10 VAR = 75 kVAR We know that, P =

3 VL IL cos φ 100 # 103 = 6.5608 A 3 # 11 # 103 # 0.8

P = 3 VL cos φ

` Line current, IL =

Let, Vd = Voltage drop in the line. Given that, line impedance, Zline = 1 + j2 Ω Since the current through the transmission line is IL, the line drop is given by, Vd = IL Zline

..... (1)

Let, Sline = Complex power consumed by line impedance. )

Now, Sline = Vd IL )

` Sline = IL Zline # IL

Using equation (1) ) = ^IL IL h Zline = IL2 Zline

2

)

2

= 6.5608 × (1 + j2 ) = 43 + j86 VA

IL IL = IL

= IL2

Let, Ps = Real power at sending end. Q s = Reactive power at sending end. Now, Ps = P + Real part of 6Sline @ 3

= 100 × 10 + 43 = 100043 W Qs = Q + Imaginary part of 6Sline @ 3

= 75 × 10 + 86 = 75086 VAR With reference to power triangle shown in Fig. 2, tan φs =

Qs Ps

` φs = tan- 1

Ss fs

Qs Ps

Ps

Power factor at sending end, cos φs = cos c tan

-1

Qs m Ps

= cos c tan- 1 75086 m = 0.7998 lag 100043 We know that, Ps =

3 VL, s IL cos φs

` Line voltage at sending end, VL, s = =

Ps 3 IL cos φs 100043 3 # 6.5608 # 0.7998

= 11007.5 V = 11007.5 kV 1000 = 11.0075 V

Fig. 2.

Qs

Circuit Theory

12. 43

EXAMPLE 12.10 A symmetrical 3-phase, 100 V, 3-wire supply feeds an unbalanced star-connected load with impedances of the load as ZR = 5+0 o Ω, Z Y = 2+90 o Ω and ZB = 4+ − 90 o Ω. Find the line currents, voltage across the impedances and the displacement neutral voltage. Also calculate the power consumed by the load. Sketch the phasor diagram.

SOLUTION Let the phase sequence be RYB. The line voltages for RYB sequence are VRY, VYB and VBR . The star-connected unbalanced load, connected to a balanced source is shown in Fig. 1. IR

R

R

-

+ +

+

ER N¢ -

+ EB

~

EY

VBR ZY

Y

+

VR

+ N

VNN¢

+

B

-

~

-

VRY

IY

Y

ZB

- VB

VY

+

B

+

~

ZR

V YB

-

+

IB

Fig. 1. The line voltages for RYB sequence with VRY as reference phasor are, VRY = 100+0 o V = 100 V VYB = 100+ − 120 o V VBR = 100+ − 240 o V = 100+120 o V Let us solve the line currents by mesh method. Consider two voltage sources, VRY and VYB connected across the load as shown in Fig. 2. Let I1 and I 2 be the mesh currents. Now the line currents in terms of mesh currents are, IR

IR = I1 I Y = I 2 − I1

+

~

IB = − I 2

VRY -

The mesh basis matrix equation for the circuit of Fig. 2 is,

=

` ∆ =

Z B = 4Ð - 90 o W

IY

ZR + Z Y VRY − Z Y I1 G = G = = G VYB − Z Y ZB + Z Y I 2

+

~

V YB

-

IB

ZR + Z Y − ZY 2 = ^ZR + Z Yh ^ZB + Z Yh − Z Y − Z Y ZB + Z Y

Z R = 5Ð0 o W = 5W

I1

Z Y = 2Ð90 o W = j2 W I2

Fig. 2. 2

2

= ZR ZB + ZR Z Y + Z Y ZB + Z Y − Z Y = ZR ZB + ZR Z Y + Z Y ZB = 5 # ^− j4h + 5 # j2 + j2 # ^− j4h = 8 − j10

= - j4 W

12. 44

Chapter 12 - Three Phase Circuits ∆1 =

VRY − ZY = VRY ^ZB + Z Yh + VYB Z Y VYB ZB + Z Y = VRY ZB + VRY Z Y + VYB Z Y = 100 # ^− j4h + 100 # j2 + 100+ − 120 o # j2 = 173.2051 – j300 ZR + Z Y VRY = ^ZR + Z Yh VYB + Z Y VRY − Z Y VYB

∆2 =

= ZR VYB + Z Y VYB + Z Y VRY = 5 × 100∠−120o + j 2 × 100∠−120o + j2 × 100 = −76.7949 − j333.0127 173.2051 − j300 ∆1 = = 26.7417 − j4.0729 A ∆ 8 − j10

` I1 =

I2 =

− 76.7949 − j333.0127 ∆2 = 16.5596 − j20.9271 A = 8 − j10 ∆

Now the line currents are, IR = I1 = 26.7417 − j4.0729 A = 27.0501+ − 8.7 o A I Y = I 2 − I1 = 16.5596 − j20.9271 − ^26.7417 − j4.0729h = − 10.1821 − j16.8542 A = 19.6911+ − 121.1 o A IB = − I 2 = − ^16.5596 − j20.9271h = − 16.5596 + j20.9271 A = 26.6864+128.4 o A The voltages across the impedance (i.e., phase voltages) are, VR = IR ZR = 27.0501+ − 8.7 o # 5+0 o = 135.2505+ − 8.7 o V VY = I Y Z Y = 19.6911+ − 121.1 o # 2+90 o = 39.3822+ − 31.1 o V VB = IB ZB = 26.6864+128.4 o # 4+ − 90 o = 106.7456+38.4 o V R

R +

+

~ N¢

VR

ER

-

VNN¢

Fig. 3. Let, VNNl = Neutral shift voltage. With reference to Fig. 3, using KVL we can write, VNNl + VR = ER ` VNNl = ER − VR

+

N

Circuit Theory

12. 45

Since the source is balanced, ER will have a magnitude of 1 o VRY by 30 .

3 times the line voltage and lag behind

` ER = 100 +^0 o − 30 oh = 57.735+ − 30 o V 3 ` Neutral shift voltage, VNNl = ER − VR = 57.735+ − 30 o − 135.2505+ − 8.7 o = – 83.6943 − j8.4094 V = 84.1157∠−174.3o V Power, P = VR IR cos φ1 + VY I Y cos φ2 + VB IB cos φ3 Here,

φ1 = Phase difference between VR and IR = − 8.7 − ^− 8.7h = 0 o φ2 = Phase difference between VY and I Y = − 31.1 o − ^− 121.1 oh = 90 o φ3 = Phase difference between VB and IB = 38.4 o − 128.4 o = − 90 o VBR

∴ P = V R I R cos φ1 + V Y I Y cos φ2 + V B I B cos φ3 IB

Here, φ2 = +90o ,

∴ cos φ2 = cos (+90 o ) = 0

128.3 o

and

o

φ3 = −90 ,

o

∴ cos φ3 = cos (−90 ) = 0

o

38.3

VRY

Reference

o

E8.7 o E31 o .2 121.2 E 5 IR o 1.7 o V E120

∴ P = V R I R cos φ1 = 135.2505 × 27.0501 × cos 0o = 3658.5 W = 3658.5 kW = 3.6585 kW 1000

VB

120 o

VR

Y

IY V YB

Fig. 4 : Rms phasor diagram of voltages and currents.

Note : Since Y-phase and B-phase loads are purely reactive loads the power consumed by them is zero.

EXAMPLE 12.11 A three-phase, 3-wire unbalanced load is star-connected. The phase voltages of two of the arms are, VR = 100+ − 10 o V and VY = 150+100 o V. Calculate the voltage between star point of the load and the supply neutral.

SOLUTION Let us assume RYB sequence. The line voltages for RYB sequence are VRY, VYB and VBR . The line voltage VRY can be calculated as shown below: With reference to Fig. 1, using KVL we can write, +

+

VRY + VY = VR

VR VRY

` VRY = VR − VY = 100∠−10 o − 150∠100 o

E

= 124.528 − j165.086 o

= 206.7864∠−53 V

E N VY

E

+

Fig. 1.

12. 46

Chapter 12 - Three Phase Circuits

Usually the source is balanced. Let us assume a star-connected balanced source. We know that the line voltage of source and load are same. Hence the R-phase emf ER of source will have a magnitude of 1 3 times the magnitude of VRY and phase lag by 30 o with respect to VRY . ` ER = 206.7864 +^− 53 o − 30 oh = 119.3882+ − 83 o V 3 R

The neutral shift voltage can be estimated from the knowledge of ER and VR . Consider the circuit shown in Fig. 2. By KVL we can write,

R +

~

VNNl + VR = ER

+ VR

ER

-

N¢

VNN¢

-

` Neutral shift voltage, VNNl = ER − VR

+

N

Fig. 2.

= 119.3882∠−83 o − 100∠−10 o = −83.931 − j101.1335 = 131.4245∠−129.7 o V

EXAMPLE 12.12

R

R

E

+

A three-phase balanced delta-connected load of 4 + j8 Ω is connected across a 400 V, 3-phase balanced supply. Determine the phase currents and line currents. Assume the phase sequence to be RYB. Also calculate the power drawn by the load. Sketch the phasor diagram.

SOLUTION

VRY

IRY VRY

VBR

VBR

Z

Z

Y

The phase sequence is RYB. The line voltages for RYB sequence are VRY, VYB and VBR . Let us choose VRY as reference phasor. The delta-connected load with polarity of voltages and direction of currents is shown in Fig. 1.

IR

IY

E +

I YB

Y

Z

IBR

V YB

V YB

B

E

IB

+

Fig. 1.

Z a 4 C j8 ‡

The line voltages of RYB sequence with VRY as reference phasor are, VRY = 400+0 o V VYB = 400+ − 120 o V VBR = 400+ − 240 o V = 400+120 o V Here,

In delta-connected load the phase and line voltages are same.

VRY = VYB = VBR = VL = V = 400 V

Now, the phase currents IRY, I YB and IBR are given by the ratio of phase voltage and impedance. Given that, impedance per phase, Z = 4 + j8 Ω = 8.9443+63.4 o Ω Now the phase currents are, 400+0 o IRY = VRY = = 44.7212+ − 63.4 o Z 8.9443+63.4 o o I YB = VYB = 400+ − 120 o = 44.7212+ − 183.4 o Z 8.9443+63.4 o IBR = VBR = 400+ − 240 o = 44.7212+ − 303.4 o = 44.7212+56.6 o A Z 8.9443+63.4

B

Circuit Theory Here,

12. 47

IRY = I YB = IBR = I = 44.7212 A

In a delta-connected balanced load, the line currents will have a magnitude of

3 times the magnitude

of phase current and lags the phase current by 30 o. Therefore the line currents are, IR = 44.7212 # 3 + ^− 63.4 o − 30 oh = 77.4594+ − 93.4 o A I Y = 44.7212 # 3 + ^− 183.4 o − 30 oh = 77.4594+ − 213.4o A = 77.4594+146.6o A IB = 44.7212 # 3 + ^− 303.4 o − 30 oh = 77.4594+ − 333.4 o A = 77.4594+26.6 o A VBR

Here,

IR = I Y = IB = IL = 77.4594 A IBR

3 VL I L cos φ

IY

14

6.6

Here, φ = Impedance angle = 63.4o

o

120

o

I YB o

V YB

Reference

o

IRY

= 24029.2 W = 24029.2 = 24.0292 kW 1000

VRY

E93.4 E6 3

E120 o

3 # 400 # 77.4594 # cos 63.4 o

IB

.4

E183.4

` P =

o

6 56. o 26.6 o

Power, P =

IR

Fig. 2 : Rms phasors of voltages and currents.

EXAMPLE 12.13 A delta-connected balanced 3-phase load is supplied from a 3-phase, 400 V supply. The line current is 20 A and the power taken by the load is 10,000 W. Find a) impedance in each branch, b) phase current, c) power factor and d) the power consumed if the same load is connected in star.

SOLUTION Case i : Delta connection Given that, Line voltage, VL = 400 V Line current, I L = 20 A Power, We know that, P =

P = 10,000 W 3 VL I L cos φ

` Power factor, cos φ =

P = 3 VL I L

10, 000 = 0.7217 3 # 400 # 20

In a delta-connected balanced load, the phase current is 1 is same as line voltage.

3 times the line current and phase voltage

12. 48

Chapter 12 - Three Phase Circuits ` Phase current, I =

IL = 20 = 11.547 A 3 3

Phase voltage, V = VL = 400 V Now, Impedance per phase, Z = V = 400 = 34.641 Ω I 11.547

Case ii : Star connection In star connection, the phase voltage is 1

3 times the line voltage.

VL = 400 = 230.9401V 3 3

` Phase voltage, V =

The impedance and power factor in star connection is same as that in delta connection. ` Phase current, I = V = 230.9401 = 6.6667 A Z 34.641 In star connection, the line current is same as phase current. ∴ Line current, I L = I = 6.6667 A Now, Power, P = =

3 VL I L cos φ 3 # 400 # 6.6667 # 0.7217 = 3333.4 W

Conclusion The power consumption in star connection is one-third of the power in delta connection. Also the line current in star is one-third of the line current is delta. This concept is utilized in star-delta starters of induction motors in order to reduce the starting current.

EXAMPLE 12.14 Three capacitors each of 100 µF are connected in delta to a 440 V, 3-phase, 50 Hz supply. What will be the capacitance of each of the three capacitors if the same three capacitors are connected in star across the same supply to draw the same line current.

SOLUTION Case i : When capacitors are connected in delta Let, C D = Capacitance per phase in delta. Given that, C D = 100 µF = 100 × 10

–6

F

` Capacitive reactance per X 1 1 4 C,D = 2πfC = D 2π # 50 # 100 # 10- 6 phase in delta connection = 31.831 Ω/phase Given that, line voltage, VL = 440 V In delta connection the phase voltage is same as line voltage. ∴ Phase voltage, V = VL = 440 V Now, phase current, I =

V = 440 = 13.823 A XC,D 31.831

In balanced delta-connected load the line current will be `

Line current, IL =

3I =

3 times the phase current.

3 # 13.823 = 23.9421 A

Circuit Theory

12. 49

Case ii : When capacitors are connected in star The line voltage and current in star connection should be maintained same as that in delta connection. ∴ Line voltage, VL = 440 V Line current,

IL = 23.9421 A

In star connection, the phase current is same as that of line current and phase voltage is 1 the line voltage. ∴

Phase current,

I = I L = 23.9421 A

Phase voltage, V = Now, capacitive reactance 4 per phase in star connection We know that, XC,S =

3 times

VL = 440 = 254.0341V 3 3

XC.S = V = 254.0341 = 10.6104 Ω/phase 23.9421 I

1 2πfCS

where, C S = Capacitance per phase in star connection `

1 1 = = 3 # 10- 4 F 2πfXC,S 2π # 50 # 10.6104

Capacitance, CS =

= 300 # 10- 6 F = 300 µF

Conclusion It is observed that, XC,S =

XC,D 3

or

XC,D = 3XC,S

When three equal impedances in delta are converted to star, in order to maintain the same line current for same line voltage, the star-connected impedance should have a value of one-third of the delta-connected impedance. Conversely when three equal impedances in star are converted to delta, in order to maintain the same line current for same line voltage, the delta-connected impedance should have a value of 3-times that of star-connected impedance. Also it is observed that, CD =

CS 3

or

CS = 3CD

Hence for a given line voltage and same line current when three equal elements are converted from star to delta or vice versa, the following relations will hold good. RS = 1 RD 3 LS = 1 LD 3

RD = 3RS

CS = 3 CD

CD = 1 CS 3

XS = 1 XD 3

XD = 3XS

ZS = 1 ZD 3

ZD = 3ZS

LD = 3LS

The suffix “S” stands for star-connected element per phase. The suffix “D” stands for delta-connected element per phase.

12. 50

Chapter 12 - Three Phase Circuits

EXAMPLE 12.15 A three-phase delta-connected load has ZRY = 100 + j50 Ω, Z YB = 20 − j75 Ω and ZBR = 70.7 + j70.7 Ω and it is connected to balanced 3-phase, 400 V supply. Determine the line currents, power consumed by the load. Sketch the phasor diagram. Assume RYB sequence and take VYB as reference phasor.

SOLUTION

R

R

+

The phase sequence is RYB. The line voltages for RYB sequence are VRY, VYB, and VBR . The threephase delta-connected load with conventional polarity of voltages and direction of currents is shown in Fig. 1.

E

IR

VRY

VBR = 400+ − 120 o VRY = 400+ − 240 Here,

+

Y

IBR

Z YB

I YB

+

Z BR

+

VYB = 400+0

IY

VBR Z RY

E

VBR

E

Y

E

VRY

The line voltages of RYB sequence with VYB as reference phasor are, o

IRY

+

B E

V YB

V YB

o

E

B

+

IB

Fig. 1.

VYB = VBR = VRY = VL = 400 V

In delta-connected load the phase voltage is same as line voltage. Now, the phase currents IRY, I YB, and IBR . are given by the ratio of respective phase voltage and impedance. Therefore the phase currents are, o IRY = VRY = 400+ − 240 = − 0.2144 + j3.5713 A 100 + j50 ZRY

= 3.5777+93.4 o A o I YB = VYB = 400+0 = 1.3278 + j4.9793 A 20 j 75 − Z YB

= 5.1533+75.1 o A o IBR = VBR = 400+ − 120 = − 3.8643 − j1.0354 A 70.7 + j70.7 ZBR

= 4.0006+ − 165 o A The line currents can be computed by writing KCL equations at nodes R, Y and B. R

With reference to Fig. 2, at node-R by KCL we can write,

IR IRY

IR + IBR = IRY ` IR = IRY − IBR = −0.2144 + j3.5713 − (−3.8643 − j1.0354 )

IBR

Fig. 2.

o

= 3.6499 + j4.6067 = 5.8774∠51.6 A With reference to Fig. 3, at node-Y by KCL we can write,

IRY

I Y + IRY = I YB ` I Y = I YB − IRY = 1.3278 + j4.9793 − (−0.2144 + j3.5713) = 1.5422 + j1.408 = 2.0883∠42.4o A

IY

Y

I YB

Fig. 3.

Circuit Theory

12. 51

With reference to Fig. 4, at node-B by KCL we can write, IB + I YB = IBR ` IB = IBR − I YB IBR

I YB

= −3.8643 − j1.0354 − (1.3278 + j4.9793)

Fig. 4.

= −5.1921 − j6.0147 = 7.9457∠−130.8 o A

B IB

In summary, the line currents are, IR = 5.8774+51.6 o A

I Y = 2.0883+42.4 o A

;

;

IB = 7.9457+ − 130.8 o A

VRY I YB

IR

E240

o

.1 51.6

o

IY

75 o

93.4 o

IRY

42.4

o

V YB

Reference

o

5 E16

o

0 E12

IB

E1

30

.8

o

IBR

VBR

Fig. 5 : Rms phasors of voltages and currents. Power, P = VRY IRY cos φ1 + VYB I YB cos φ2 + VBR IBR cos φ3 Here, VRY = VYB = VBR = VL = 400 V φ1 = Phase difference between VRY and IRY = − 240 o − 93.4 o = − 333.4 o φ2 = Phase difference between VYB and I YB = 0 o − 75.1 o = − 75.1 o φ3 = Phase difference between VBR and IBR = − 120 o − ^− 165 oh = 45 o ∴ P = V L I RY cos φ1 + VL I YB cos φ2 + VL I BR cos φ3 = VL [ I RY cos φ1 + IYB cos φ2 + IBR cos φ3] = 400 × [ 3.5777 × cos (–333.4o) + 5.1533 × cos (−75.1o) + 4.0006 × cos45 o ] = 2941.2 W = 2941.2 kW = 2.9412 kW 1000

EXAMPLE 12.16 A 300 V, 3-phase balanced source is connected to delta-connected load of impedance Z YR = 10+45 o Ω, ZBY = 8+0 o Ω and ZRB = 5+ − 45 o Ω. Determine the line currents and power consumed by the load. Assume RBY sequence. Sketch the phasor diagram.

12. 52

Chapter 12 - Three Phase Circuits

SOLUTION The phase sequence is RBY. The line voltages for RBY sequence are VRB, VBY and VYR . Let us take VBY as reference phasor. The three-phase delta-connected load with conventional polarity of voltages and direction of currents is shown in Fig. 1. R E

+

IR

IRB

+

E

R

V YR

VRB

Y

I YR

IY

+ E

VRB Z YR

Y

Z RB

Z BY

E

VBY

E

+

V YR

IBY

+

B

V BY

B

+

E

IB

Fig. 1. The line voltages for RBY sequence, with VBY as reference phasor are, VBY = 300+0 o VYR = 300+ − 120 o VRB = 300+ − 240 o Here,

VBY = VYR = VRB = VL = 300 V

In delta connection, the phase voltage is same as line voltage. Now, the phase currents IRB, IBY and I YR are given by the ratio of respective phase voltage and impedance. Therefore the phase currents are, o IRB = VRB = 300+ − 240 = 60+ − 195 o A = − 57.9555 + j15.5291 A o ZRB 5+ − 45 o IBY = VBY = 300+0 = 37.5+0 o A = 37.5 + j0 A o ZBY 8+0 o I YR = VYR = 300+ − 120 = 30+ − 165 o A = − 28.9778 − j7.7646 A o Z YR 10+45

The line currents can be computed by writing KCL equations at nodes R, Y and B. With reference to Fig. 2, at node-R by writing KCL equation we get, IR + I YR = IRB

R IR IRB

` IR = IRB − I YR = −57.9555 + j15.5291 − (−28.9778 − j7.7646 )

I YR

= − 28.9777 + j23.2937 A = 37.1793∠141.2 o A

Fig. 2.

Circuit Theory

12. 53

With reference to Fig. 3, at node-Y by writing KCL equation we get, I Y + IBY = I YR ` I Y = I YR − IBY IY

= − 28.9778 − j7.7646 − (37.5 + j0)

Y

I YR

IBY

Fig. 3.

= − 66.4778 − j7.7646 A = 66.9297∠− 173.3 o A With reference to Fig. 4, at node-B by writing KCL equation we get,

I RB

IB + IRB = IBY ` IB = IBY − IRB

IBY

= 37.5 + j0 − (− 57.9555 + j15.5291)

Fig. 4.

= 95.4555 − j15.5291 A = 96.7104 ∠− 9.2 o A

B

IB

In summary, the line currents are, VRB

IR = 37.1793+141.2 o A I Y = 66.9297+ − 173.3 o A IRB

o

2

1.

14

IR

IB = 96.7104+ − 9.2 o A

12 0

o

IBY

o

E195

Power, P = VRB IRB cos φ1

IY

I YR

o

65

E1

+ VBY IBY cos φ2 + VYR I YR cos φ3 Here,

VBY

Reference

E9.2 o

V YR

o

.3 E173 o 2 1 E 0

Fig. 5 : Rms phasors of voltages and currents.

VRB = VBY = VYR = VL = 300 V φ1 = Phase difference between VRB and IRB = − 240 o − (− 195 o) = − 45 o φ2 = Phase difference between VBY and IBY = 0 o − 0 o = 0 o φ3 = Phase difference between VYR and I YR = − 120 o − (− 165 o) = 45 o ∴ P = V L I RB cos φ1 + V L I BY cos φ2 + V L I YR cos φ3 = VL [ I RB cos φ1 + I BY cos φ2 + I YR cos φ3 ] = 300 × [ 60 × cos (− 45o ) + 37.5 × cos0o + 30 × cos 45 o ] = 30341.9 W = 30341.9 kW = 30.3419 kW 1000

IB

12. 54

Chapter 12 - Three Phase Circuits

EXAMPLE 12.17

(AU June’14, 8 Marks)

A symmetrical three-phase, three wire, 400V supply is connected to a delta-connected load. Impedance in each branch are Z RY = 10+300 Ω, Z YB = 10+450 Ω and Z BR = 2.5+600 Ω . Find its equivalent star-connected load.

SOLUTION The equivalent star load can be obtained by using delta to star transformation as shown below : R R ZR Z BR = 2.5Ð60 0 W

Z RY = 10Ð30 0 W

N ZY

Þ Y

ZB

Y Z YB = 10Ð45 0 W

B

B

Fig. 1 ZR =

ZY =

ZB =

Fig. 2

Z RY # Z BR 10+30 o # 2.5+60 o = = 0.7248 + j0.8646 Ω Z RY + Z YB + Z BR 10+30 o + 10+45 o + 2.5+60 o = 1.1282 + 50 o Ω Z YB # Z RY 10+45 o # 10+30 o = = 3.6955 + j2.5901 Ω Z RY + Z YB + Z BR 10+30 o + 10+45 o + 2.5+60 o = 4.5128 + 35 o Ω Z BR # Z YB 2.5+60 o # 10+45 o = = 0.4763 + j1.0227 Ω Z RY + Z YB + Z BR 10+30 o + 10+45 o + 2.5+60 o = 1.1282 + 65o Ω

EXAMPLE 12.18

(AU June’14, 8 Marks)

A symmetrical three-phase; three wire 440V supply is connected to a star-connected load. The impedance in each branch are Z R = 2 + j3 Ω, Z Y = 1 − j2 Ω and Z B = 3 + j4 Ω . Find its equivalent delta connected load.

SOLUTION The standard equations for converting star connected impedances to delta or vice-versa is based on the concept that the power drawn by the impedances in the star or delta connection is same. Hence, the equivalent delta load can be obtained by using star to delta transformation. R R Z R = 2 + j3W

N Z Y = 1 - j2 W

Z BR

Z RY

Þ Z B = 3 + j4 W

Y Y Z YB

B

B

Fig. 1

Fig. 2

Circuit Theory

12. 55

Z RY = Z R + Z Y +

ZR ZY _2 + j3 i # _1 − j2i = 2 + j3 + 1 − j2 + 3 + j4 ZB

= 2 + j3 + 1 − j2 + 0.8 − j1.4 = 3.8 − j0.4 Ω = 3.821 + − 60 Ω

Z YB = Z Y + Z B +

Z Y ZB _1 − j2 i # _3 + j4 i = 1 − j2 + 3 + j4 + 2 + j3 ZR

= 1 − j2 + 3 + j4 + 1.2308 − j 2.8462 = 5.2308 − j 0.8462 Ω = 5.2988+ − 9.20 Ω Z BR = Z B + Z R +

ZB ZR _3 + j4 i # _2 + j3 i = 3 + j4 + 2 + j3 + 1 − j2 ZY

= 3 + j4 + 2 + j3 − 8 + j = − 3 + j 8 Ω = 8.544+110.60 Ω The equivalent delta connected load impedances are, Z RY = 3.8 − j 0.4 Ω ;

Z YB = 5.2308 − j 0.8462 Ω ;

Z BR = − 3 + j 8 Ω

(AU June’14, 8 Marks)

EXAMPLE 12.19

A three phase, balanced delta−connected load of 4 + j8 W, is connected across a 400 V, 3 phase balanced supply. Determine the phase currents and line currents (Phase sequence is RYB).

SOLUTION R +

E

for RYB sequence are VRY, VYB and VBR . Let us take

IR

IRY

VRY as reference phasor. The three phase deltaVRY

connected load with conventional polarity of voltages

Z

VBR

E

and direction of currents is shown in Fig. 1. The line voltages for RYB sequence, with VRY as reference phasor are, VRY = VL +00

Y

= 400+00 V

VYB = VL + − 1200 = 400+ − 1200 V

+

Z VBR +

VRY

I YB + V YB E

IY

E

Y

IBR

B

Z

V YB

B

E

+

VBR = VL + − 2400 = 400+ − 2400 V Here,

C

R

E

The phase sequence is RYB. The line voltages

IB

Fig. 1

VRY = VYB = VBR = VL = 400 V

In delta connection the phase voltage is same as line voltage. Now, the phase currents IRY, I YB and IBR are given by the ratio of respective phase voltage and impedance. The load impedance per phase, Z = 4 + j8 Ω = 8.9443+63.40 Ω .

12. 56

Chapter 12 - Three Phase Circuits Therefore the phase currents are, 400+00 I RY = VRY = = 44.7212+ − 63.40 A Z 8.9443+63.40 0 I YB = VYB = 400+ − 120 0 = 44.7212+ − 183.40 A Z 8.9443+63.4 0 I BR = VBR = 400+ − 240 0 = 44.7212+ − 303.40 A Z 8.9443+63.4

The line currents can be computed by writing KCL equations at nodes R, Y and B. R

With reference to Fig. 2, at node-R by writing KCL equation we get,

IR IRY

I R = I RY − I BR

= 44.7212+ − 63.4 o − 44.7212+ − 303.4 o IBR

Fig. 2.

= 77.4594+ − 93.4 o A With reference to Fig. 3, at node-Y by writing KCL equation we get,

IRY

I Y = I YB − I RY

= 44.7212+ − 183.4 o − 44.7212+ − 63.4 o o

o

= 77.4594+146.6 = 77.4594+ − 213.4 A

IY

Y

I YB

Fig. 3.

With reference to Fig. 4, at node-B by writing KCL equation we get, I B = I BR − I YB

= 44.7212+ − 303.4 o − 44.7212+ − 183.4 o o

o

= 77.4594+26.6 = 77.4594+ − 333.4 A

IBR

I YB

Fig. 4.

B IB

(AU Dec’14, 8 Marks)

EXAMPLE 12.20

Three star-connected impedances Z1 = 20 + j37.7 Ω/phase are in parallel with three delta-connected impedance Z 2 = 20 + j37.7 Ω/phase . The line voltage is 398 V. Find the line current, power factor, power and reactive volt-ampere taken by the combination.

SOLUTION Given that, Z1 = 20 + j37.7 Ω/phase Z 2 = 20 + j37.7 Ω/phase Line voltage, VL = 398 V

Circuit Theory

12. 57

The star and delta-connected parallel loads, are shown in Fig.1 R

Z2

Z1

Z2 Z2

Z1

Y

Z1

B

Fig. 1. The delta-connected load can be converted to star and so that the total load will be a parallel combination of two star-connected loads as shown in Fig. 2. The two parallel star-connected loads shown in Fig. 2, are combined to give a single equivalent starconnected load as shown in Fig. 3 Let, Zeq = Equivalent impedance per phase in star connection. R R Z2

Z1

3

Zeq

Z2

Z2

3

Z1

3

Z1

Zeq

Y Y B

B

Fig. 2. ` Zeq

=

Fig. 3.

Z2 ^30 − j159.3h ^20 + j37.7h # 3 3 = = ^30 − j159.3h Z ^20 + j37.7h + Z1 # 2 3 3 Z1 #

^20 + j37.7h # ^10 − j53.1h

20 + j37.7 + 10 − j53.1

= 67.3653 + j11.7475 = 68.3819 + 9.9 o Ω

We know that, Line current, IL =

VL Zeq

3 =

398 3 = 3.3603 A 68.3819

Power factor, cos φ = cos 9.9 o = 0.9851 lag Power, P = 3 VL IL cos φ = 3 # 398 # 3.3603 # 0.9851 = 2281.9 W = 2.2819 kW Reactive power, Q = 3 VL IL sin φ = 3 # 398 # 3.3603 # sin 9.9 o = 398.3 VAR = 0.3983 kVAR

Zeq

12. 58

Chapter 12 - Three Phase Circuits

EXAMPLE 12.21 Determine the power and power factor of three-phase load if the two wattmeters used for power measurement read i) 1000 W each, both positive, ii) 1000 W each of opposite sign.

SOLUTION Case i : Wattmeter readings are equal and positive Let,

P1 = P2 = 1000 W

Power, P = P1 + P2 = 1000 + 1000 = 2000 W = 2 kW P2 − P1 mF P1 + P2

Power factor, cos φ = cos < tan- 1 c 3

= cos < tan- 1 c 3 # 1000 − 1000 mF = cos 7 tan- 1 0 A = cos 0 o = 1 1000 + 1000

Case ii : Wattmeter readings are equal but opposite in sign Let,

P1 = 1000 W, P2 = − 1000 W

Power, P = P1 + P2 = 1000 − 1000 = 0 W Power factor, cos φ = cos < tan- 1 c 3

P2 − P1 1 − 1000 − 1000 mF = cos < tan- c 3 # 1000 − 1000 mF P1 + P2

= cos < tan- 1 c 3 # − 2000 mF = cos 7 tan- 1 ^− 3hA = cos ^− 90 oh = 0 0

EXAMPLE 12.22 Each of two wattmeters connected to measure the power input to a 3-phase circuit read 10 kW on a balanced load, when the pf is unity. What does the instrument read when the pf falls to i) 0.866 lagging, and ii) 0.5 lagging, the total 3-phase power remaining unaltered.

SOLUTION When power factor is unity the power input to the load is 10 + 10 = 20 kW. This power remains same for any power factor. ∴ P1 + P2 = 10 + 10 = 20 kW

..... (1)

We know that, Power factor, cos φ = cos < tan- 1 c 3 `

cos- 1 (cos φ) = tan- 1 c 3

tan 7cos- 1 (cos φ) A =

`

P2 − P1 mF P1 + P2

P2 − P1 =

^P1 + P2h 3

3

P2 − P1 m P1 + P2

P2 − P1 P1 + P2

# tan 7cos- 1 (cos φ) A = 20 tan 7cos- 1 (cos φ) A 3

= 11.547 tan 7cos- 1 (cos φ) A

..... (2)

Circuit Theory

12. 59

Case i : Power factor, cos φ = 0.866 lag From equation (1), we get, P1 + P2 = 20 kW

..... (3)

From equation (2). we get, P2 − P1 = 11.547 tan [cos −1 (cos φ)] = 11.547 tan [cos −1 (0.866 ) ] = 11.547 tan 30 o = 6.6667 kW

..... (4)

On adding equations (3) and (4), we get, P1 + P2 = 20 P2 − P1 = 6.6667 2P 2 = 26.6667 ` P2 = 26.6667 = 13.3334 kW 2 From equation (3), we can write, P1 = 20 − P2 ∴ P1 = 20 − 13.3334 = 6.6666 kW The readings of wattmeter are, P1 = 6.6666 kW

and

P2 = 13.3334 kW

Case ii : Power factor, cos φ = 0.5 lag From equation (1), we get, P1 + P2 = 20 kW

..... (5)

From equation (2) we get, P2 − P1 = 11.547 tan [cos −1 (cos φ)] = 11.547 tan [cos −1 (0.5) ] = 11.547 tan 60 o = 20 kW On adding equations (5) and (6), we get, P1 + P2 = 20 P2 − P1 = 20 2P 2 = 40 ` P2 = 40 = 20 kW 2 From equation (5), we get, P1 = 20 − P2 ∴ P1 = 20 − 20 = 0 The readings of wattmeter are, P1 = 0

and

P2 = 20 kW

..... (6)

12. 60

Chapter 12 - Three Phase Circuits

EXAMPLE 12.23 A 500 V, 3-phase motor has an output of 3.73 kW and operates at a power factor of 0.85, with an efficiency of 90%. Calculate the reading on each of the two wattmeter connected to measure the input.

SOLUTION Given that, Power output = 3.73 kW Efficiency = 90% Power output = 3.73 = 3.73 = 4.1444 kW Efficiency 90 0.9 100 We know that, in two wattmeter method of power measurement, the power input is equal to sum of two wattmeter readings. Power input =

∴ P1 + P2 = 4.1444 kW

..... (1)

We know that, Power factor, cos φ = cos < tan- 1 c 3 `

cos- 1 (cos φ) = tan- 1 c 3

tan 7cos- 1 (cos φ) A =

`

P2 − P1 mF P1 + P2

P2 − P1 =

=

3

P2 − P1 m P1 + P2

P2 − P1 P1 + P2

P1 + P2 tan 7cos- 1 (cos φ) A 3 P1 + P2 tan 7cos- 1 0.85 A 3

= 4.1444 # tan 7cos- 1 0.85 A 3 = 1.4829 kW On adding equations (1) and (2) we get, P1 + P2 = 4.1444 P2 − P1 = 1.4829 2P2 = 5.6273 `

P2 = 5.6273 = 2.8137 kW 2

From equation (1) we get, P1 = 4.1444 − P2 ∴ P1 = 4.1444 − 2.8137 = 1.3307 kW Therefore the wattmeter readings are, P1 = 1.3307 kW

and

P2 = 2.8137 kW

cos φ = 0.85 Using equation (1)

..... (2)

Circuit Theory

12. 61

EXAMPLE 12.24 The input power to a three-phase motor was measured by two wattmeter method. The readings are 5.2 kW and 1.7 kW. The later reading have been obtained after reversing the current coil connections. The line voltage was 400 V. Calculate a) The total power, b) Power factor, and c) Line current.

SOLUTION Given that, P1 = 5.2 kW Here, P2 is negative, because the reading have been obtained after reversing the current coil connections.

P2 = −1.7 kW Power, P = P1 + P2 = 5.2 + (−1.7 ) = 3.5 kW Power factor, cos φ = cos < tan- 1 c 3

P2 − P1 mF P1 + P2

= cos < tan- 1 c 3 − 1.7 − 5.2 mF 5.2 + (− 1.7) = cos [tan −1 ( −3.4146 )] = cos (−73.7o ) = 0.2807 We know that, P =

3 VL I L cos φ

` Line current, I L =

=

P 3 VL cos φ 3.5 # 1000 3 # 400 # 0.2807

= 17.9972 A

EXAMPLE 12.25 Two wattmeters are used to measure power in a three-phase load. The wattmeter readings are 400 W and −35 W. Calculate a) Total active power, b) Power factor, c) Reactive power and d) Volt-amperes.

SOLUTION Given that, P1 = 400 W P2 = −35 W Total active power, P = P1 + P2 = 400 + (−35) = 365 W Power factor, cos φ = cos < tan- 1 c 3

P2 − P1 mF P1 + P2

= cos < tan- 1 c 3 # − 35 − 400 mF = 0.436 400 − 35 Here,

cos φ = 0.436 ∴

φ = cos −1 0.436 = 64.2 o

12. 62

Chapter 12 - Three Phase Circuits Consider the power triangle shown in Fig. 1. With reference to Fig. 1, we can write, tan φ = Q P ∴ Reactive power, Q = P tan φ

S Q f

= 365 ´ tan (64.2o ) = 755 VAR 2

Volt - amperes, S =

P +Q

2

2

365 + 755

=

2

= 838.6 VA

P

Fig. 1 : Power triangle.

EXAMPLE 12.26 Calculate the total power input and reading of the two wattmeters connected to measure power in a three-phase balanced load, if the reactive power input is 15 kVAR and the load power factor is 0.8. Also compute load kVA.

SOLUTION Given that, Reactive power, Q = 15 kVAR Power factor, cos φ = 0.8 ∴ Power factor angle, φ = cos −1 0.8 = 36.9 o Consider the power triangle shown in Fig. 1, with reference to Fig. 1, we can write, tan φ = Q P ` Active power, P = Q tan φ 15 = = 19.9781 kW tan 36.9 o Load kVA, S =

2

P +Q

2

=

2

19.9781 + 15

2

= 24.9825 kVA

S Q f P

Fig. 1 : Power triangle.

In two-wattmeter method of power measurement, we know that the sum of two wattmeter readings is equal to active power, P. ∴ P1 + P2 = P = 19.9781 kW

..... (1)

We know that, Power factor, cos φ = cos < tan- 1 c 3 cos- 1 (cos φ) = tan- 1 c 3 tan 7cos- 1 (cos φ) A = `

P2 − P1 =

3

P2 − P1 mF P1 + P2

P2 − P1 m P1 + P2

P2 − P1 P1 + P2

P1 + P2 tan 7cos- 1 (cos φ) A 3

= 19.9781 # tan 7cos- 1 (0.8) A = 8.6508 kW 3 On adding equations (1) and (2) we get, 2 P2 = 19.9781 + 8.6508 ` P2 = 19.9781 + 8.6508 = 14.3145 kW 2 From equation (1) we get, P1 = 19.9781 − P2 = 19.9781 − 14.3145 = 5.6636 kW

..... (2)

Circuit Theory

12. 63

EXAMPLE 12.27

P1

CC

R

PC

10 ‡

5‡

Ej2

Find the reading of wattmeter in the circuit shown in Fig. 1. Assume a symmetrical 360 V supply, with RYB sequence. Draw the phasor diagram.

j20 ‡

SOLUTION The phase sequence is RYB. The line voltages for RYB sequence are VRY, VYB and VBR . Let us take VRY as reference phasor. Now,

VRY = 360+0 o V

Y B

Fig. 1. o

VYB = 360+ − 120 V VBR = 360+ − 240 o V The polarity of voltages and direction of currents are shown in Fig. 2. CC R

+

P1

E

R IR

PC

0‡

+ VRY

+

Y

Ej25 ‡

VBR

E

10

+ j2

VBR

E

VRY

IRY

IBR

IY

E +

B Y

V YB

E

B

IB

+

Fig. 2. With reference to Fig. 2, IRY =

o VRY = 360+0 = 7.2 − j14.4 A = 16.0997+ − 63.4 o A 10 + j20 10 + j20

o IBR = VBR = 360+ − 240 = − 12.4708 − j7.2 A = 14.4+ − 150 o A − j25 − j25

By KCL at node-R, we get,

VBR

IR + IBR = IRY ` IR = IRY − IBR = 7.2 – j14.4 A – (– 12.4708 – j7.2)

120 o

VRY

= 19.6708 –j7.2

E20.1

o

E1 IBR

o

E120 o

50

= 20.9471Ж20.1o A The wattmeter reading, P1 = VYB IR # cos θ1 VYB = 360 V

Reference

o

3.

4

E6

IR

Here,

IRY

IR = 20.9471 A θ1 = − 120 o − ^− 20.1 oh = − 99.9 o

V YB

Fig. 3 : Rms phasors of voltages and currents.

∴ P1 = 360 ´ 20.9471 ´ cos (− 99.9 o ) = −1296.5 W

12. 64

Chapter 12 - Three Phase Circuits

EXAMPLE 12.28 A 440 V, 3-phase, three-wire system has a current of 10Ð −30o A in R-phase, 14Ð− 60o A in the Y-phase. a) Find the current in B-phase, b) Find the reading of two wattmeters with current coils in R and Y phase and voltage coils connected to B-phase. Phase sequence is RYB with VRY as reference phasor.

SOLUTION The line currents are IR, I Y and IB . Given that, IR = 10+ − 30 o A I Y = 14+ − 60 o A We know that in a three-phase three-wire system, the sum of line currents is zero. ` IR + I Y + IB = 0 ` IB = − ^IR + I Yh = − ^10+ − 30 o + 14+ − 60 oh = − 15.6603 + j17.1244 = 23.2054+132.4 o A The line voltages for RYB sequence with VRY as reference phasor are, VRY = 440+0 o V VYB = 440+ − 120 o V VBR = 440+ − 240 o V The connections of two wattmeters with polarity of voltages and direction of currents are shown in Fig. 1. With reference to Fig. 1, the reading of wattmeter, P1 is given by, R

Y

CC +

E

VRY

VBR

PC

VRB

CC

E +

E

IR

+

P2

IY

Three phase load

PC

V YB

B

P1

+

IB

E

Fig. 1. P1 = VRB IR # cos θ1 where, θ1 = phase difference between VRB and IR . VRB = − VBR = 1+180 o # VBR = 1+180 o # 440+ − 240 o = 440+ − 60 o V VRB = 440 V IR = 10 A

Circuit Theory

12. 65 o

o

θ 1 = − 60 − (−30 ) = −30

o

` P1 = 440 # 10 # cos (− 30 o) = 3810.5 W = 3810.5 kW = 3.8105 kW 1000 With reference to Fig. 1, the reading of wattmeter, P 2 is given by, P2 = VYB # I Y # cos θ2 where, θ2 = phase difference between VYB and I Y . Here, VYB = 440+ − 120 o V I Y = 14+ − 60 o A `

VYB = 440 V,

I Y = 14 A,

θ2 = − 120 o − (− 60 o) = − 60 o

` P2 = 440 # 14 # cos (− 60 o) = 3080 W = 3080 kW = 3.08 kW 1000

EXAMPLE 12.29 A delta-connected load consists of ZRY = 10 + j10 Ω, Z YB = 15 − j15 Ω, and ZBR = 20 + j10 Ω and it is connected to a 400 V, 3-phase supply of phase sequence RYB. Calculate the readings of wattmeter with current coil in line R and B. P CC

1

R

R

SOLUTION

+

+

IR

PC

+ VRY

VBR

VRY

VRY

VBR Z RY

Z BR

E

+

Y

E +

E E

V YB

VBY

B

VYB = 400+ − 120 o V VBR = 400+ − 240 o V

E

+

I YB

IY

PC

IBR

Z YB

B Y

o

VRY = 400+0 V

IRY

E

The phase sequence is RYB. The line voltages for RYB sequence are VRY, VYB and VBR . Let us take VRY as reference phasor. In delta connection the line and phase voltages are same. Hence the line and phase voltages are,

E

+ V YB

E

IB

+ P2

CC

Fig. 1.

The connections of wattmeters with polarity of voltages and direction of currents are shown in Fig. 1. With reference to Fig. 1, we can say that the current through the wattmeters are IR and IB . The voltage across the wattmeters are VRY and VBY . Hence the wattmeter readings P1 and P2 are, P1 = VRY # IR # cos θ1 P2 = VBY # IB # cos θ2 where, θ1 = Phase difference between VRY and IR . θ2 = Phase difference between VBY and IB .

12. 66

Chapter 12 - Three Phase Circuits Using Ohm’s law, the phase currents can be obtained as a ratio of phase voltage and impedance. Therefore phase currents are, o IRY = VRY = 400+0 = 20 − j20 A 10 + j10 ZRY o I YB = VYB = 400+ − 120 = 4.8803 − j18.2137 A 15 j 15 − Z YB o IBR = VBR = 400+ − 240 = − 1.0718 + j17.8564 A 20 + j10 ZBR

With reference to Fig. 2, by KCL, the line current IR is given by, IR = IRY − IBR

IR

R

IR + IBR = IRY

IRY

= 20 – j20 – (–1.0718 + j17.8564) IBR

= 21.0718 – j37.8564 A

B

Y

Fig. 2.

o

= 43.3258Ж60.9 A

R

With reference to Fig. 3, by KCL, the line current IB is given by, IB = IBR − I YB

IB + I YB = IBR IBR

= –1.0718 + j17.8564 – (4.8803 – j18.2137)

B

Y

= – 5.9521 + j36.0701 A o

= 36.5579Ð99.4 A The wattmeter reading P1 is given by, P1 = VRY # IR # cos θ1 Here, VRY = 400+0 o V and IR = 43.3258+ − 60.9 o A `

VRY = 400 V IR = 43.3258 A θ 1 = 0o − (−60.9 o ) = 60.9 o ` P1 = 400 # 43.3258 # cos 60.9 o = 8428.3 W = 8428.3 kW = 8.4283 kW 1000

The wattmeter reading P 2 is given by, P2 = VBY # IB # cos θ2

I YB

IB

Fig. 3.

Circuit Theory

12. 67

Here, VBY = − VYB = − 1 # VYB = 1+180 o # 400+ − 120 o = 400+60 o V and IB = 36.5579+99.4 o A VBY = 400 V

`

IB = 36.5579 θ 2 = 60 o − 99.4o = –39.4o ` P2 = 400 # 36.5579 # cos ^− 39.4 oh = 11299.8 W = 11299.8 kW = 11.2998 kW 1000 The readings of wattmeter are, P1 = 8.4283 kW P2 = 11.2998 kW

EXAMPLE 12.30 A balanced delta-connected load takes a line current of 15 A , when connected to a balanced 3-phase, 400 V system. A wattmeter with its current coil in one line and its pressure coil between the two remaining lines reads 2000 W. Describe the load impedance.

SOLUTION CC

Let the load impedance, Z = Z+φ Ω/phase

R

P

IR

R

+ PC

Given that, Line voltage, V L = 400 V

IRY

+ VRY

Line current, I L = 15 A

VRY

Z

Z

E

In delta connection the phase and line voltages are same, and the phase current is 1 3 times the line current.

B Y

V YB

\ Phase voltage, V = V L = 400 V B

Phase current, I =

Z

E +

Y

E

IL = 15 = 8.6603 A 3 3

Fig. 1.

Now, the magnitude of load impedance, Z is given by, Z = V = 400 = 46.1878 Ω/phase 8.6603 I Let the current coil of wattmeter be connected in line-R and its voltage coil be connected across line-Y and B as shown in Fig. 1. Now the reading of wattmeter, P is, P = VYB # IR # cos θ where, θ = phase difference between VYB and IR .

12. 68

Chapter 12 - Three Phase Circuits

Let the phase sequence of supply be RYB and VRY be the reference phasor. Now the line and phase voltages can be expressed as , VRY = VL +0 o = V+0 o VYB = VL + − 120 o = V+ − 120 o VBR = VL + − 240 o = V+ − 240 o Now the phase current IRY for an impedance of Z∠φ W/phase can be expressed as, o IRY = VRY = V+0 = V +^0 o − φh = I+ − φ Z+φ Z+φ Z

where, V = I = phase current Z In balanced delta-connected load, the line current IR will lag behind the phase current IRY by 30o. Also the magnitude of line current is 3 times the phase current. Hence the line current IR can be expressed as, IR = Now,

3 I+^− φ − 30 oh = IL +^− φ − 30 oh = 15+^− φ − 30 oh A

VYB = VL = 400 V IR = I L = 15 A IL = 15 A

θ 1 = − 120 o − (−φ −30 o ) = − 120 o + φ + 30o = φ − 90 o ` Power, P = 400 # 15 # cos ^φ − 90 oh = 400 # 15 # sin φ Given that,

cos ^φ − 90 oh = sin φ

P = 2000 W

` 2000 = 400 # 15 # sin φ ` sin φ =

2000 400 # 15

Now impedance angle, φ = sin- 1 c 2000 m = 19.5 o 400 # 15 `

12.11

Load impedance, Z = Z+φ = 46.1878+19.5 o Ω/phase

Summary of Important Concepts

1.

The three-phase sources are three-phase alternators generating three emfs having equal magnitude but with a phase difference of 120o with respect to each other.

2.

The three-phases are named R-phase, Y-phase and B-phase in British convention and A-phase, B-phase and C-phase in American convention.

3.

In a polyphase system, when the magnitude of emfs are equal and the phase difference between consecutive emfs are equal then the system is called balanced system and the emfs are called balanced emfs.

4.

Three-phase sources are always designed to generate balanced emfs.

Circuit Theory

12. 69

5.

For operational convenience and cost effective system the three-phase sources are operated in star/delta connection.

6.

In star connection, the meeting point of the three sources is called neutral and there is no such neutral point in delta connection.

7.

The transmission lines or connecting wires from the source terminals to the load terminals are called lines.

8.

The voltage generated by each phase of three-phase source is called phase voltage and the voltage between the lines connecting the load is called line voltage.

9.

The current delivered by each phase of the three-phase source is called phase current and the current flowing through the line is called line current.

10.

In three-phase rotating phasors, the order of reaching reference point is called phase sequence.

11.

In three-phase rotating phasors, when the order of reaching reference is R-phase, Y-phase and B-phase, the phasors are said to have normal phase sequence or RYB sequence. When the order of reaching reference is R-phase, B-phase and Y-phase, the phasors are said to have reversed phase sequence or RBY sequence.

12.

The maximum value phasors are rotating phasors and the rms phasors are non-rotating phasors.

13.

The rms phasors can be drawn by taking a snapshot of rotating phasors at ωt = 0 and reducing the length by 2 .

14.

The salient features of star-connected three/four-wire source are,

15.

g

The voltages are always balanced in star-connected sources.

g

The currents may be balanced or unbalanced depending on load.

g

The phase and line currents are same in star system.

g

The magnitude of line voltage is

g

The phase voltage of source lags line voltage by 30o.

g

In balanced four-wire system, the neutral current is zero.

3 times the magnitude of phase voltage.

The line and phase voltages of star-connected three/four-wire source for RYB sequence with VRY as reference phasor are, Line voltages

VRY = VRY +0 o

= VL +0 o

Phase voltages E R = E R + − 30 o

= E+ − 30 o

VYB = VYB + − 120 o = VL + − 120 o

E Y = E Y + − 150 o = E+ − 150 o

VBR = VBR + − 240 o = VL + − 240 o = VL +120o

E B = E B + − 270 o = E+ − 270 o = E+90 o

where, ER = EY = EB = E VRY = VYB = VBR = VL =

3E

12. 70

Chapter 12 - Three Phase Circuits 16.

The line and phase voltages of star-connected three/four-wire source for RBY sequence with VRB as reference phasor are, Line voltages

VRB = VRB +0

o

= VL +0

Phase voltages

o

E R = E R + − 30 o

= E+ − 30 o

VBY = VBY + − 120 o = VL + − 120 o

E B = E B + − 150 o = E+ − 150 o

VYR = VYR + − 240 o = VL + − 240 o = VL +120 o

E Y = E Y + − 270 o = E+ − 270 o = E+90 o

where, ER = EB = EY = E VRB = VBY = VYR = VL =

17.

3E

In a star-connected, four-wire system, Neutral current, I N = 0

- for balanced system

Neutral current, I N = I R + I Y + I B - for unbalanced system

18.

19.

The salient features of delta-connected source are, g

The voltages are always balanced in delta-connected sources.

g

The currents may be balanced or unbalanced depending on load.

g

The phase and line voltages are same in delta system.

g

The magnitude of line current is 3 times the magnitude of phase current.

g

The line current lags the phase current by 30o in balanced delta system.

The line and phase voltages of delta-connected source for RYB sequence with VRY as reference phasor are, Line voltages VRY = VRY +0

o

Phase voltages

= VL +0

o

E R = E R +0 o

= E +0 o

VYB = VYB + − 120o = VL + − 120o

E Y = E Y + − 120o = E+ − 120o

VBR = VBR + − 240o = VL + − 240o

E B = E B + − 240o = E+ − 240o

where, ER = EY = EB = E = VL VRY = VYB = VBR = VL = E

20.

The line and phase voltages of delta-connected source for RBY sequence with VRB as reference phasor are, Line voltages VRB = VRB +0

o

= VL +0

Phase voltages o

E R = E R +0o

= E+0o

VBY = VBY + − 120o = VL + − 120o

E B = E B + − 120o = E+ − 120o

VYR = VYR + − 240o = VL + − 240o

E Y = E Y + − 240o = E+ − 240o

where, ER = EB = EY = E = VL VRB = VBY = VYR = VL = E

Circuit Theory

12. 71

21.

In balanced load, the magnitude of load impedance of each phase will be equal and also the load impedance angle of each phase will be same.

22.

In unbalanced load, the load impedance of each phase may have different magnitude and/or different impedance angle.

23.

In analysis of three-phase circuits, it is conventional practice to choose one of the line voltage of the source as reference phasor. There are six choice for reference phasor. (Refer Table 12.1 for various choice of reference phasor).

24.

The apparent, active and reactive power in three-phase balanced star/delta-connected load are given by, Apparent power, S =

25.

3 VL I L

(Active) Power, P =

3 VL I L cos φ

Reactive power, Q =

3 VL I L sin φ

For same load impedance and supply voltage, the power consumed by delta-connected load will be three times the power consumed by star-connected load. Alternatively, power consumed by starconnected load will be one-third the power consumed by delta-connected load. Let, PD = Power consumed by delta-connected load. PY = Power consumed by star-connected load. Now, PD = 3PY

26.

(or)

PY = 1 PD 3

The power in unbalanced star-connected load is, Power, P = VR I R cos φ1 + VY I Y cos φ2 + VB I B cos φ3 where, VR = VY = VB = VL

φ1 = Phase difference between VR and I R . φ2 = Phase difference between VY and I Y . φ3 = Phase difference between VB and I B . 27.

In three-wire star-connected unbalanced load, the voltage of load neutral with respect to source neutral is called neutral shift voltage or neutral displacement voltage.

28.

The neutral shift voltage can be obtained by subtracting a phase voltage of load from the corresponding phase emf of the source. Let, E R = R-phase source emf. VR = R-phase load emf.

Now, Neutral shift voltage = E R − VR

12. 72

Chapter 12 - Three Phase Circuits 29.

The power in unbalanced delta-connected load is, Power, P = VRY I RY cos φ1 + VYB I YB cos φ2 + VBR I BR cos φ3 where, VRY = VYB = VBR = VL

φ1 = Phase difference between VRY and I RY . φ2 = Phase difference between VYB and I YB . φ3 = Phase difference between VBR and I BR . 30.

The power in any three-phase load (balanced/unbalanced and star/delta) can be measured by using only two wattmeters. The power is given by sum of two wattmeter readings.

31.

In two wattmeter method of power measurement, the power factor of balanced three-phase load in terms of two wattmeter readings P1 and P2 is, Power factor, cos φ = cos ; tan- 1 c 3 Also, Power factor angle, φ = tan- 1 c 3

32.

P2 − P1 m P1 + P2 E

P2 − P1 m P1 + P2

In two wattmeter method of power measurement, the following observations can be made regarding the power factor of balanced load. • • •

When wattmeter readings are equal, the power factor is unity. When one of the wattmeter reading is zero, the power factor is 0.5. When one of the wattmeter reading is negative, the power factor is less than 0.5.

•

When both the wattmeter readings are positive, the power factor is greater than 0.5.

12.12

Short-answer Questions

Q12.1

What is balanced voltage? In a polyphase system when the magnitude of phase voltages are equal and the phase difference between consecutive phasors are equal, then the voltages are called balanced voltages.

Q12.2

What is balanced impedance? When the impedances of all the phases of a three-phase load are equal, then the impedances are called balanced impedances.

Q12.3

What is phase sequence? In a set of rotating phasors, the order of reaching reference point is called phase sequence.

Q12.4

Write the relation between the line and phase value of voltage and current in a balanced starconnected source/load. In a star-connected system, the line current and phase current are same. In balanced starconnected system the magnitude of line voltage is the phase voltage by 30 o.

3 times the phase voltage and it leads

Circuit Theory Q12.5

12. 73

Write the relation between the line and phase value of voltage and current in a balanced deltaconnected source/load. In a delta-connected system the line voltage and phase voltage are same. In balanced delta-connected system the magnitude of line current is 3 times the phase current and it lags the phase current by 30o.

Q12.6

(AU May’15 & June’14, 2 Marks)

Distinguish between unbalanced source and unbalanced load.

In a three phase source, when the phase difference between any two phase is not equal to 1200 then sources are called unbalanced source, whereas in three phase load if the impedance angle of all the phases are not equal, then the load is called unbalanced load.

Q12.7

Write the equations for the phasor difference between the potentials of the delta connected networks. (AU Dec’14, 2 Marks) The phasor voltages in delta-connected network are VRY = VL + 0 o V VYB = VL + − 120 o V VBR = VL + − 240o V

Q12.8

Write the distortion power factor equation of the three phase circuits. Distortion power factor =

1 1 + THD2

(AU May’15, 2 Marks)

where, THD = Total Harmonic Distortion When excited by a voltage source, the current gets distorded and so THD is calculated for current as shown below : THD =

I22 + I32 + I24 + ..... + I2n I1

Where, I1 = RMS value of fundamental In = RMS value of nth harmonic. True power factor = Power factor of fundamental × Distortion power factor

Q12.9

A star-connected load has 6 + j8 W impedance per phase. Determine the line current if it is connected to 400 V, 3φ, 50 Hz supply. Impedance, Z =

6 2 + 8 2 = 10 Ω/phase

VL 3 Phase current, I = V = = Z Z

400 = 23.094 A 3 # 10

∴ Line current, IL = I = 23.094 A

Q12.10 A star-connected load each having a resistance of 20 W and an inductive reactance of 15 W are connected to a 400 V, 3-phase, and 50 Hz supply. What is the line current, power factor and power supplied? (AU Dec’14, 2 Marks) Phase current, I =

Vphase = Z

VL / 3 = R2 + XL2

400/ 3 = 9.2376 A 202 + 152

In star connection line current and phase current are same. ` Line current, IL = I = 9.2376 A

12. 74

Chapter 12 - Three Phase Circuits Power factor, cos φ = R = Z Power, P =

20 = 0.8 lag 20 2 + 15 2

3 VL IL cos φ =

3 # 400 # 9.2376 # 0.8

= 5120 W = 5.12 kW

Q12.11

A delta-connected load has 30 − j40 W impedance per phase. Determine the line current if it is connected to 415 V, 3 φ, 50 Hz supply. Impedance, Z =

30 2 + 40 2 = 50 Ω/phase

V Phase current, I = V = L = 400 = 8 A Z Z 50 ` Line current, IL =

Q12.12

3I =

3 # 8 = 13.8564 A

A star-connected balanced load draw a current of 35 A per phase when connected to a 440 V supply. Determine the apparent power. Apparent power, S =

3 VL IL =

3 # 440 # 35

= 26673.6 VA = 26.6736 kVA

Q12.13

A balanced delta-connected load of 4 – j6 W impedance is connected to 400 V 3-phase supply. What is the power and power factor of the load? Z =

4 2 + 6 2 = 7.2111 Ω/phase

I = V = 400 = 55.47 A 7.2111 Z Power factor = cos ctan- 1 − 6 m = 0.5547 lead 4 Power, P =

3 VL IL cos φ = 3VI cos φ = 3 # 400 # 55.47 # 0.5547

= 36923.1W = 36.9231 kW

Q12.14

In a three-wire system, the two line currents of a unbalanced load are I R = 10+ - 67o A and I B = 5+136 o A. Determine the line current I Y . We know that, IR + I Y + IB = 0 ` I Y = − ^IR + IBh = − ^10+ − 67 o + 5+136 oh

= –0.3106 + j5.7318 A = 5.7402∠93.1o A

Q12.15

What is neutral shift voltage? In three-wire star-connected load, the load neutral is not connected to source neutral. Therefore, when the load is unbalanced, the load neutral will not be at zero potential. The voltage of load neutral with respect to source neutral is called neutral shift voltage or neutral displacement voltage.

Q12.16

When a 3-phase star-connected unbalanced load is connected to a 400 V supply the R-phase voltage is 200 ∠16o V. Determine the neutral shift voltage. Let VRY be reference phasor. The R - phase source voltage, ER = 400 + − 30 o V 3

Circuit Theory

12. 75

With reference to Fig. Q12.16, +

+

Neutral shift voltage, VNN' = ER − VR

ER

= 400 + − 30 o − 200+16 o 3

~

VR

-

-

N¢

-

= 230.9401+ − 30 o − 200+16 o

VNN¢

N +

Fig. Q12.16.

= 7.7477 − j170.5975 = 170.7733+ − 87.4 o V

Q12.17

In a four-wire star-connected system, I R = 5+10o A, I Y = 7+85 o A, I B = 3+200 o A. What is the neutral current? Neutral current, IN = IR + I Y + IB = 5+10 o + 7+85 o + 3+200 o = 2.7151 + j6.8155 A = 7.3364+68.3 o A

Q12.18

Write the expression for the power measured by two wattmeters used in 3-phase balanced load, in terms of voltage, current and power factor. P1 = VL IL cos (φ + 30o) P2 = VL IL cos (φ − 30o) where, φ is the power factor angle.

Q12.19

Write the expression for power factor in two-wattmeter method of power measurement. Power factor, cos φ = cos < tan- 1 c 3

Q12.20

Q12.21

Q12.22

P2 − P1 mF P1 + P2

Write the relation between the power factor and wattmeter readings in two wattmeter method of power measurement. 1.

When the wattmeter readings are equal, the power factor is unity.

2.

When one of the wattmeter reading is zero, the power factor is 0.5.

3.

When one of the wattmeter reading is negative, the power factor will be less than 0.5.

4.

When both the wattmeter readings are positive, the power factor will be greater than 0.5.

If P1 and P2 are two wattmeter readings and φ is the power factor angle of a three-phase load, write the relation between wattmeter readings and power factor angle. 1.

When P1 = P2, then φ = 0o.

2.

When P1 = 0 and P2 ≠ 0 (or P1 ≠ 0 and P2 = 0), then φ = 60 o.

3.

When P1 is negative and P2 is positive (or P2 is negative and P1 is positive), then φ > 60o.

4.

When P1 and P2 are positive, then φ < 60o.

The readings of two wattmeter used for 3-phase power measurement are 5.2 kW and −1.6 kW. Determine the power and power factor of the load. Power, P = P1 + P2 = 5.2 + (−1.6) =3.6 kW P P Power factor, cos φ = cos < tan- 1 c 3 2 − 1 mF = cos < tan- 1 c 3 # − 1.6 − 5.2 mF = 0.2923 5.2 − 1.6 P1 + P2

12. 76 Q12.23

Chapter 12 - Three Phase Circuits In a star-connected 3-phase balanced load the total power measured by two wattmeter is 2400 W. What will be the power measured by two wattmeters if the load impedance is reconnected in delta. For same load impedance, the power consumed by delta-connected load will be three times the power consumed by star-connected load. Therefore the total power measured by wattmeters for delta-connected load will be 7200 W.

Q12.24

The line current of a delta-connected balanced load is 18A when connected to a 400 V supply. What will be the line current when the impedances are reconnected in star. When the balanced delta-connected impedances are reconnected in star, for same supply voltage the current drawn by star-connected load will be one-third the current drawn by delta-connected load. ` Line current in star connection = 18 = 6 A 3

12.13 Exercises I.

Fill in the Blanks with Appropriate Words

1.

In a balanced star-connected load the line voltage is ________ times the phase voltage.

2.

The line voltage ________ phase voltage by ________ in a balanced star-connected system.

3.

The ________ current is zero in a balanced four-wire system.

4.

In a balanced delta-connected load the phase current is ________ times the line current.

5.

The line current ________ phase current by ________ in a balanced delta system.

6.

In balanced three-phase load power consumed in delta connection is ________ times the power consumed in star connection.

7.

The voltage of load neutral with respect to source neutral is called ________ voltage.

8.

In balanced three-phase load for same line voltage the line current in star connection is ________ times the line current in delta.

9.

The ________ current is equal to sum of three line currents in four-wire system.

10.

In three-wire unbalanced star-connected load the sum of a ________ voltage of load and ________ voltage is equal to corresponding phase voltage of source.

11.

In a balanced three-phase load the total power consumed is equal to ________ times the power in one phase.

12.

In three-phase load ________ can be measured by using only ________ wattmeters.

13.

In two wattmeter method of power measurement, when pf is ________ , one wattmeter reading will be zero.

14.

In two wattmeter method of power measurement, the readings of two wattmeters are ________ when the pf is unity.

15.

In two wattmeter method of power measurement, one wattmeter reading is negative when pf angle is ________ than ________ .

Circuit Theory

12. 77

ANSWERS 1.

3

6.

three

11. three

2. leads, 30 o

7.

neutral shift

12. power, two

3. neutral

8.

1/3

13. 0.5

4. 1

9.

neutral

14. equal

3

5. lags, 30 o

10. phase, neutral shift

15. greater, 60o

II.

State Whether the Following Statements are True/False

1.

Rms phasors are rotating phasors.

2.

In balanced load the magnitude and argument of the impedances of all the phases will be equal.

3.

In three-phase system, the line voltages of source and load are same.

4.

In star-connected load the line current and phase current are not same.

5.

In delta-connected load the line voltage and phase voltage are same.

6.

In three-wire system the sum of three line currents is zero.

7.

In four-wire system the sum of three line currents and neutral current is zero.

8.

In unbalanced star-connected load the sum of three-phase voltage is zero.

9.

In unbalanced delta-connected load the sum of three-phase voltages is zero.

10.

In four-wire system the voltages are always balanced irrespective of balanced or unbalanced load.

11.

Two wattmeters are sufficient for measuring power in balanced and unbalanced loads.

12.

In a balanced load the power factor is same in all the phases.

13.

In unbalanced load the power factor is same in all the phases.

14.

The power factor of unbalanced load can be estimated from two wattmeter readings.

15.

In two wattmeter method of power measurement we cannot determine whether the power factor is lagging or leading.

ANSWERS 1. False

4. False

7. True

10. True

13. False

2. True

5. True

8. False

11. True

14. False

3. True

6. True

9. True

12. True

15. True

12. 78

Chapter 12 - Three Phase Circuits

III. Choose the Right Answer for the Following Questions 1. In a star-connected three-phase system, the relation between the line and phase values are, a) VL = V, I L = c) VL =

3I

b) VL =

V , I = I L 3

3 V, I L = I

d) VL = V, I L =

I 3

2. In a delta-connected three-phase system, the relation between the line and phase values are, a) VL = V, I L = c) VL =

3I

b) VL =

V , I = I L 3

3 V, I L = I

d) VL = V, I L =

I 3

3. Three identical impedances 3 + j6 W are connected in delta. What will be the equivalent star-connected impedances that draws same current when connected to same supply voltage? a) 3 + j6 W

b) 6 + j12 W

c) 1 + j2 W

d) 1.5 + j3 W

4. Three identical impedances 2 + j5 W are connected in star. What will be the equivalent delta-connected impedances that draws same current when connected to same supply voltage? a) 2 + j5 W

b) 1 + j2.5 W

c)

2 + j5 Ω 3

d) 6 + j15 W

5. A star-connected balanced load with impedance 4 + j5 W /phase is connected to a three-phase 200 V supply. What is the apparent power? a) 975.6 VA

b) 1281 VA

c) 6247 VA

d) 8200 VA

6. A balanced delta-connected load with impedance 5 + j2 W/phase is connected to three-phase 100 V supply. What is the power? a) 2586.3 W

b) 2986.3 W

c) 5172.5 W

d) 4479.6 W

7. In a three-phase unbalanced load the line currents and I R = 5+30 o A and I Y = 7+180 o A . What is the value of line current I B ? a) 11.6∠–167.6o A

b) 3.66∠–43.1o A

c) 3.66∠136.9o A

d) 11.6∠12.4o A

8. The line currents of four-wire star-connected load are I R = j20 A, I Y = 4 A and I B = − 2 − j3 A. What is the neutral current? a) 2 – j23 A

b) –6 + j17 A

c) 2 + j17 A

d) –2 – j17 A

9. A three-phase balanced inductive load draw a current of 10 A and consumes 6 kW when connected to 400 V supply. What is the power factor of the load? a) 0.866 lag

b) 0.5 lag

c) 0.866 lead

d) 0.5 lead

Circuit Theory

12. 79

10. A symmetrical delta-connected load draw a power of 8 kW when connected to 380 V supply. If power factor of the load is 0.866 lead, then the line current is, a) 10.5 A

b) 24.3 A

c) 8.1 A

d) 14 A

11. In two-wattmeter method of power measurement the expression for power factor angle is, a) tan 3

P2 − P1 P1 + P2

b) tan- 1 3

P2 − P1 P1 + P2

c) tan 3

P1 + P2 P1 − P2

d) tan- 1 3

P1 + P2 P1 − P2

12. In two-wattmeter method of power measurement what is the value of wattmeter readings when the load power is 8 kW with unity power factor? a) 8 kW, 0 kW

b) 4 kW, 4 kW

c) 12 kW, –4 kW

d) 16kW, –8kW

13. In two-wattmeter method of power measurement what is the value of wattmeter readings when the load power is 7 kW with 0.5 power factor? a) 7 kW, 0 kW

b) 3.5 kW, 3.5 kW

c) 9 kW, –2 kW

d) 14kW, –7kW

14. In two-wattmeter method, the readings of wattmeter for a balanced load are, P1 = 500 W, P2 = 400 W. What is the power factor of the load? a) 0.347

b) 0.949

c) 0.999

d) 0.982

15. In two-wattmeter method, the readings of wattmeter for a balanced load are, P1 = 800 W, P2 = 600 W. What is the reactive power of the load? a) 1200 VAR

b) 600 VAR

c) 346 VAR

ANSWERS 1. b

6. b

11. b

2. a

7. b

12. b

3. c

8. c

13. a

4. d

9. a

14. d

5. c

10. d

15. c

d) 692 VAR

12. 80

IV.

Chapter 12 - Three Phase Circuits

Unsolved Problems

E12.1

An unbalanced four-wire star-connected load is connected to a balanced supply of 380 V. Estimate the line currents, neutral current and the power consumed by the load. Also draw the phasor diagram. Take Z R = 2 + j4 Ω ; Z Y = 3 + j2 Ω ; Z B = 6 + j10 Ω

E12.2

A balanced star-connected load of 4 + j8 W per phase is connected to a three-phase, 400 V, 60 Hz supply. Find the line currents, active and reactive power of the load. Draw the phasor diagram.

E12.3

The power consumed by a three-phase balanced star-connected load is 2.25 kW at a power factor of 0.72 leading. The supply voltage is 360 V, 50 Hz. Describe the impedance per phase.

E12.4

An unbalanced three-wire star-connected load is connected to a symmetrical supply of 415 V. If the load impedances are Z R = 7+30 o Ω ; Z Y = 9+60 o Ω and Z B = 8+45 o Ω , calculate the line currents, phase voltages and displacement neutral voltage.

E12.5

Determine the values of three impedances Z R, Z Y and Z B connected in star to a 415 V, 3-phase supply, if the neutral shift voltage is 180 ∠ 75oV and the line current I Y = 12+ - 60 o A and I B = 16+90 o A.

E12.6

A delta-connected balanced three-phase load is supplied from a three-phase, 415 V supply. The line current is 19 A and the power consumed by the load is 12 kW. Find a) impedance per phase, b) current per phase, c) power factor and d) the power consumed if the same load is connected in star.

E12.7

A balanced delta load of 14 − j10 W per phase is connected to a 415 V, 3-phase supply. Determine the line and phase currents for RBY sequence, by taking VBY as reference phasor. Sketch the phasor diagram. Also calculate the power factor, active and reactive power of the load.

E12.8

An unbalanced delta load is connected to a 3-phase 415 V supply. Determine the phase and line currents for RYB sequence by taking VBR as reference phasor. The impedances of three arms are 70 + j45 W, 10 − j20 W and 30 + j10 W.

E12.9

The readings of two wattmeters connected to measure a 3-phase power is 8 kW each, when the power factor is unity. What will be the reading of wattmeters if the power factor falls to i) 0.7 lagging, ii) 0.5 lagging and iii) 0.4 lagging, the total 3-phase power remaining unaltered.

E12.10 A 415 V, 3-phase motor has an output of 5.595 kW and operates at a power factor of 0.82 with an efficiency of 85%. Calculate the readings of the two wattmeters connected to measure the input power.

Circuit Theory E12.11

12. 81

A three-phase load connected to a 415 V, 3-phase, 3-wire system draw a current of 12 ∠− 40o A in R-phase and 16 ∠− 200o A in the B-phase. Determine the readings of the two wattmeters connected to measure the power if current coil of one wattmeter is connected to R-phase and that of other to B-phase.

E12.12 The three impedances of star-connected load are 4 + j6 W, 8 − j10 W and 12 + j4 W . Calculate the readings of two wattmeters connected to measure the power consumed by the load if the current coils are connected to R and Y phase. Take supply voltage as 600 V, 50 Hz, RYB sequence.

ANSWERS E12.1

RYB sequence with VRY as reference phasor IR = 49.0582+ − 93.4 o A ; I Y = 60.8479+ − 183.7 o A ; IB = 18.8128+ − 329o A IN = 59.2178+ − 143.3 o A

E12.2

P = 18.0513 kW

RYB sequence with VRY as reference phasor IR = 25.8198+ − 93.4 o A ; I Y = 25.8198+ − 213.4 o A ; IB = 25.8198+26.6o A P = 8.0097 kW ; Q = 15.9951 kVAR

E12.3

Z = 41.4722+ − 43.9 o Ω/phase

E12.4

RYB sequence with VRY as reference phasor IR = 34.9248+ − 70.2 o A ; I Y = 30.8186+154.6 o A ; IB = 25.3461+50.9 o A VNN' = 43.3044+61.3 o V ; VR = 244.4736+ − 40.2 o V ; VY = 277.3674+ − 145.4 o V VB = 202.7688+95.9 o V

E12.5

ZR = 40.7744+75.6 o Ω ;

E12.6

a) Z = 37.8315+28.5 o Ω/phase ;

b) I = 10.9697 A

c) cos φ = 0.8787

d) Pstar = 4 kW

E12.7

Z Y = 32.3609+ − 70.9 o Ω ;

;

ZB = 5.0356+35.3 o Ω

IBR = 24.1213+35.5 o A ; I YR = 24.1213+ − 84.5 o A ; IRB = 24.1213+ − 204.5 o A IB = 41.7793+5.5 o A ; I Y = 41.7793+ − 114.5 o A ; IR = 41.7793+ − 234.5o A cos φ = 0.8141lag

E12.8

;

P = 24.4482 kW

;

Q = −17.4391 kVAR

IBR = 13.1235+ − 18.4 o A ; IRY = 4.987+ − 152.7 o A ; I YB = 18.5593+ − 176.6o A IB = 31.1246+ − 5.6 o A ; IR = 16.9856+173.7 o A ; I Y = 14.1418+175.2 o A

12. 82 E12.9

Chapter 12 - Three Phase Circuits i) P1 = 3.2879 kW

;

P2 = 12.7121 kW

iii) P1 = −2.583 kW ;

P2 = 18.583 kW

E12.10

P1 = 1.9648 kW

;

P2 = 4.6176 kW

E12.11

P1 = 3.8149 kW

;

P2 = −1.153 kW

E12.12

P1 = 15.1432 kW

;

P2 = 19.1062 kW

;

ii) P1 = 0 ; P2 = 16 kW

B.E/ B.Tech. DEGREE EXAMINATION, MAY/ JUNE 2014 Second Semester Electrical and Electronics Engineering EE 6201 – CIRCUIT THEORY (Regulation 2013) Time: 3 hours Maximum: 100 marks Answer all questions PART A – (10 × 2 = 20 marks) 1. Find the equivalent resistance of the circuit shown in Fig 1.

2 ‡

RT

2‡

2‡

1.2 ‡

1‡

Fig. 1

Chapter 6, SA - Q6.5 [ Page No - 6.49 ] 2. Define RMS voltage. Chapter 3, Section 3.3.2 [ Page No - 3.6 ] 3. What is reciprocity theorem? Chapter 7, Section 7.5 [ Page No - 7.69 ] 4. Why do you short circuit the voltage source and open the current source when you find Thevenin’s resistance of a network? Usually Thevenin’s resistance is obtained by network reduction technique. When this technique is applied the circuit should be converted to a network by deactivating or killing all the sources. An ideal voltage source is deactivated, when it is short circuited and ideal current source is deactivated, when it is open circuited. 5. Define quality factor in the resonant circuit. Chapter 8, SA - Q8.4 [ Page No - 8.47 ] 6. Determine the quality factor of a coil for the series resonant circuit consisting of R = 10 ohm, L = 0.1H, and C = 10 microfarad. Chapter 8, SA - Q8.6 [ Page No - 8.47 ] 7. Distinguish between natural and forced response. Chapter 10, SA - Q10.3 [ Page No - 10.95 ] 8. What is the time constant for RL and RC circuit? Chapter 10, SA - Q10.6 & Q10.9 [ Page No - 10.95 & 10.96 ]

Anna University Question Papers

Q. 2

9. Write the effect of power factor in energy consumption billing. Energy = Power × Time = VI cos f × Time = VI × Power factor × Time From the above relation it is clear that energy is directly proportional to power factor. Therefore, a high power factor will result in large energy consumption and higher value of billing. Note: The power factor is defined as ratio of active power and apparent power. The apparent power is the power supplied to consumer and active power is the power utilized by the consumer. If power factor is 0.8, then the consumer utilize only 80% of power and return back 20% to source. The power returned to the source increase the energy loss in transmission and so electricity boards insist for maintaining higher power factor at consumer end. In ideal case, the power factor should be unity so that the entire transmitted power is consumed which result is low transmission losses. 10. Distinguish between unbalanced source and unbalanced load. Chapter 12, SA - Q12.6 [ Page No - 12.73 ] PART B - (5 × 16 = 80 marks) 11. (a) (i) Find the current I and voltage across 30W of the circuit shown in Fig. 11(a)(i) (8) 2‡

8‡

+

E

40V

I E

C

100 V

30 ‡

Fig. 11(a) (i).

Chapter 2, SA - Q2.22 [ Page No - 2.37 ] (ii) Determine the current in all the resistors of the circuit shown in Fig.11(a) (ii) (8) A + 50 A E

I1

I2

2‡

1‡

I3 5‡

B

Fig. 11(a) (ii).

Chapter 2, Example 2.11 [ Page No - 2.30 ] (OR)

Q. 3

Anna University Question Papers

(b) (i) Determine the current through each resistor in the circuit shown in Fig 11.(b) (i) (6) 12 A

+ VS E

4‡

4‡

4‡

I1

I2

I3

Fig. 11(b) (i).

Chapter 2, SA - Q2.23 [ Page No - 2.38 ] (ii) When a dc voltage is applied to a capacitor, voltage across its terminals is found to build up in accordance with vc = 50(1 – e-100t). After 0.01 S the current flow is equal to 2 mA. (i) Find the value of capacitance in farad. (ii) How much energy stored in the electric field?

(10)

Chapter 10, Example 10.21 [ Page No - 10.77 ] 12. (a) (i) Determine the current in the 5W resistor in the network shown in the Fig.12(a)(i) (8) 10 ‡

2‡

2A

+ 50V

E 5‡

1‡

Fig. 12(a) (i).

Chapter 7, Example 7.3 [ Page No - 7.6 ] (ii) Find out the current in the each branch of the circuit shown in Fig.12(a)(ii) 3‡

1‡

+ 5A

10 ‡

10 V

5‡ E

Fig. 12(a) (ii).

Chapter 4, Example 4.15 [ Page No - 4.28 ] (OR)

(8)

Anna University Question Papers

Q. 4

(b) (i) Determine the current in each mesh of the circuit shown in Fig.12 (b) (i) C E

10 V 1‡

3‡

10 A

(8)

2‡

Fig. 12(b) (i).

Chapter 4, Example 4.16 [ Page No - 4.29 ] (ii) Determine the voltages at each node of the circuit shown in Fig.12 (b) (ii)

(8)

3‡ 10 ‡

3‡

2‡

+ 5‡

10 V

5A

E

1‡

6‡

Fig. 12(b) (ii).

Chapter 5, Example 5.12 [ Page No - 5.25 ] 13. (a) For the circuit shown in the Fig. 13 (a), determine the impedance at resonant frequency, 10 Hz above resonant frequency, and 10Hz below resonant frequency. (16) 10 mF

0.1H

10W

+

~

-

VS

Fig. 13(a).

Chapter 8, Example 8.1 [ Page No - 8.12 ] (OR) (b) Explain that how to derive Q factor of parallel resonance.

(16)

Chapter 8, Section 8.3.3 [ Page No - 8.26 ] 14. (a) A series RL circuit with R = 30W and L = 15H has a constant voltage V = 60v applied at t = 0 as shown in Fig. 14(a). Determine the current i,the voltage across resistor and the voltage across the inductor. (16) Chapter 10, Example 10.2 [ Page No - 10.48 ]

S

30 ‡

+ 60V

i E

Fig. 14(a).

15 H

Q. 5

Anna University Question Papers

(OR) (b) The circuit shown in the Fig. 14(b) consist of resistance, inductance and capacitance in series with 100V DC when the switch is closed at t = 0. Find the current transient (16) S

R 20 W

+ 100V

L

0.05 H

C

20mF

i(t) -

Fig. 14(b).

Chapter 10, Example 10.24 [ Page No - 10.82 ] Note : For the given values of R,L and C the response will be damped oscillatory same as that of Example 10.24 15. (a) (i) A symmetrical three-phase; three wire 440V supply to a star connected load. The impedance in each branch are ZR = 2 + j3W, ZY = 1 - j2W, and ZB = 3 + j4W.  Find its equivalent delta connected load. (8) Chapter 12, Example 12.18 [ Page No - 12.54 ] (ii) A three phase, balanced delta-connected load of 4 + j8W, is connected across a 400V, 3f balanced supply. Determine the phase currents and line currents. (Phase sequence is RYB). (8) Chapter 12, Example 12.19 [ Page No - 12.55 ] (OR) (b) (i) A symmetrical three-phase, three wire 400V, supply is connected to a delta-connected load. Impedance in each branch are ZRY = 10∠300 W, ZYB = 10∠450 W and ZBR = 2.5∠600 W. Find its equivalent star-connected load. (8) Chapter 12, Example 12.17 [ Page No - 12.54 ] (ii) A balanced star connected load having an impedance 15 + j20W per phase is connected to 3f, 440V, 50Hz. Find the line current and power absorbed by the load. (8) Chapter 12, Example 12.4 [ Page No - 12.35 ]

Anna University Question Papers

Q. 6

B.E./B.Tech. DEGREE EXAMINATION, NOVEMBER/DECEMBER 2014 Second Semester Electronics and Communication Engineering EE 6201 – CIRCUIT THEORY (Regulation 2013) Time: 3 hours

Maximum: 100 marks Answer all questions PART A - (10 × 2 = 20 Marks)

1. An electrical appliance consumes 1.2 kWh in 30 minutes at 120 V. What is the current drawn by the appliance? Chapter 2, SA - Q2.16 [ Page No - 2.35 ] 2. Calculate the equivalent resistance between the terminals ''a'' and ''b'', in Fig. 1. 10 ‡

10 ‡

a

10 ‡

5‡

b

Fig. 1

Chapter 6, SA - Q6.4 [ Page No - 6.48 ] 3. Calculate the value of IN for the circuit shown in Fig. 2. 20 ‡ +

100 ‡

12V E

IN 360 ‡

RL

Fig. 2

Chapter 7, SA - Q7.9 [ Page No - 7.83 ] 4. State maximum power transfer theorem for DC networks. Chapter 7, SA - Q7.15 [ Page No - 7.85 ] 5. Calculate the total inductance of the circuit, if the coefficient of coupling (k) between the two coils is 0.6, as shown in Fig.3.

47 mH

33 mH

k

Chapter 9, SA - Q9.12 [ Page No - 9.56 ] 6. Define quality factor of a series resonant circuit. Chapter 8, SA - Q8.4 [ Page No - 8.47 ]

Fig. 3

Q. 7

Anna University Question Papers

7. A coil of resistance 2.2 W and an inductance 0.01 H is connected in series with a capacitor across 220 V mains. Find the value of capacitance such that maximum current flows in the circuit at a frequency of 190 Hz. Also find the maximum current. Chapter 8, SA - Q8.22 [ Page No - 8.49 ] 8. A 50 mF capacitor is discharged through a 100 kW resistor. If the capacitor is initially charged to 400 V, determine the initial energy. Chapter 10, SA - Q10.20 [ Page No - 10.97 ] 9. Write the equations for the phasor difference between the potentials of the delta connected networks. Chapter 12, SA - Q12.7 [ Page No -12.73 ] 10. Three coils, each having a resistance of 20 W and an inductive reactance of 15 W are connected in star to a 400 V, 3-phase, and 50 Hz supply. Calculate (a) the line current, (b) power factor, and (c) power supplied Chapter 12, SA - Q12.10 [ Page No - 12.73 ] PART B – (5 × 16 = 80 marks) 11. (a) (i) Using node analysis, find the node voltages and the currents through all the resistors for the circuit shown in Fig. 4. (12) V1

25 A

10 ‡

4‡

1‡

V2

20 A

2‡

10 ‡

+ 20 V E

Fig. 4.

Chapter 5, Example 5.13 [ Page No - 5.27] (ii) Find the equivalent resistance between the terminals 'a' and 'b' for the network shown in Fig. 5. (4) a 4‡

2‡ 6‡

RT 3‡

3‡

b

Fig. 5.

Chapter 6, Example 6.3 [ Page No - 6.33]

Anna University Question Papers

Q. 8

(OR) (b) For the circuit shown in Fig. 6, find the (i) currents in different branches, (ii) current supplied by the battery, (iii) potential difference between terminals A and B. (16) 2‡

6‡

+ 10V E

A

B 4‡ 3‡

8‡

Fig. 6.

Chapter 4, Example 4.3 [ Page No - 4.12 ] 12. (a)

Find the current I, through the 20 W resistor shown in Fig. 7 using Thevenin's theorem. (16) 9V 1‡ +

20 ‡

50V + E

20 V

E

I

+ E

10 ‡

5‡

+ 10V E

2‡

Fig. 7.

Chapter 7, Example 7.18 [ Page No - 7.30 ] (OR) Find the current through 5W resistor using superposition theorem, in the circuit shown in Fig. 8. (16) 4‡

I1 32 V EE

+

(b)

9A

2‡

5‡

10 ‡ I2

Fig. 8.

Chapter 7, Example 7.9 [ Page No - 7.13]

4A

Q. 9

Anna University Question Papers

13. (a)

Impedance Z1 and Z2 are parallel and this combination is in series with an impedance Z3 ,connected to a 100 V, 50 Hz ac supply. Z1 = 5 − jX C Ω , Z2 = 5 + j0 Ω , Z3 = 6.25 + j1.25 Ω . Determine the value of capacitance such that the total current of the circuit will be in phase with the total voltage. Find the circuit current and power. (16) Chapter 3, Example 3.18 [ Page No - 3.57 ] (OR)

(b)

The switch in the circuit shown in Fig. 9, is moved from position 1 to 2 at t = 0. Find the expression for voltage across resistance and capacitor, energy in the capacitor for t > 0. (16) t=0

1 2

+

5 kW

100 V -

_ 50 V

i(t) + _ 1 mF

+

Fig. 9.

Chapter 10, Example 10.3 [ Page No - 10.64 ] 14. (a) (i) For a magnetically coupled circuit, derive the expression for mutual inductance(M) in terms of L1 and L2. (6) Chapter 9, Section 9.2.3 [ Page No - 9.5 ] 14. (a) (ii) For the coupled circuit shown in Fig. 10, find the value of V2 so that the current I1 = 0. (10) 2W

j2W

5W

+

~

10Ð900 -

I1

j8W

j2W

I2

+

V2 -

Fig. 10.

Chapter 9, Example 9.16 [ Page No - 9.48] (OR) 14. (b)

With neat illustration describe the parallel resonant circuit and the equivalent parallel network for a series RL combination. Also derive the unity power factor fp. (16) Chapter 8, Example 8.10 [ Page No - 8.39]

Anna University Question Papers

Q. 10

Note : 1. Explain any one parallel RLC circuit and derive its resonance frequency [Refer Section 8.3.1, Page No - 8.17] 2. The unity power factor frequency, fp is resonance frerquency of parallel RLC circuit. 15. (a)

Show that three phase power can be measured by two wattmeters. Draw the phasor diagrams. Derive an expression for power factor interms of wattmeter readings. (16) Chapter 12, Section 12.9 [ Page No - 12.26 ] (OR)

(b) (i) A 400 V(line-to-line) is applied to three star-connected identical impedances each consisting of a 4 W resistance in series with 3 W inductive reactance. Find (1) line current and (2) total power supplied. (8) Chapter 12, Example 12.4 [ Page No - 12.35 ] 15. (a) (ii) Three star-connected impedances Z1 = 20 + j37.7 Ω per phase are in parallel with three delta-connected impedance Z2 = 20 + j37.7 Ω per phase . The line voltage is 398 volts. Find the line current, power factor, power and reactive volt-ampere taken by the combination. (8) Chapter 12, Example 12.20 [ Page No - 12.56 ]

Q. 11

Anna University Question Papers

B.E/ B.Tech DEGREE EXAMINATION, APRIL/ MAY 2015 Second Semester Electronics and Communication Engineering EE 6201 – CIRCUIT THEORY (Regulation 2013) Time: 3 hours Maximum: 100 marks Answer all questions PART A – (10 × 2 = 20 marks) 1. Write briefly about resistance in a circuit. Chapter 2, Section 2.9 [ Page No - 2.13 ] 2. Obtain the current in each branch of the network shown below using Kirchhoff’s Current Law. 5‡

2‡

a E

C

I3

C

E

C

20 V + E

I1

10 ‡

+ E

I2

8V

E

b

Chapter 2, SA - Q2.19 [ Page No - 2.36 ] 3. State maximum power transfer theorem. Chapter 7, SA - Q7.15 [ Page No - 7.85 ] 4. Write briefly about network reduction technique. Chapter 6, Section 6.1 [ Page No - 6.1 ] 5. Define mutual inductance. Chapter 9, SA - Q9.4 [ Page No - 9.55 ] 6. Write the dot rule. Chapter 9, Section 9.3.1 [ Page No - 9.7 ] 7. Define the frequency response of series RLC circuit. Chapter 8, SA - Q8.3 [ Page No - 8.46 ] 8. Find the frequency response V2 /V1 for the two-port circuit shown below. 5 KW +

V1 -

1mF

1250 W

+

V2

-

Chapter 11, SA - Q11.17 [ Page No - 11.47 ] 9. Write the distortion power factor equation of the three phase circuits. Chapter 12, SA - Q12.8 [ Page No - 12.73 ]

Anna University Question Papers

Q. 12

10. Distinguish between unbalanced source and unbalanced load. Chapter 12, SA - Q12.6 [ Page No - 12.73 ] PART B – (10 × 16 = 80 marks) 11. (a) Use branch currents in the network shown below to find the current supplied by the 60 V source. Solve the circuit by the mesh current method (16) 7‡ I2

I3

I4

12 ‡

6‡

12 ‡

I1 60 V + E

Chapter 4, Example 4.4 [ Page No - 4.13 ] (OR) (b) Solve the network given below by the node voltage method.

(16)

10 ‡ 2

1 5‡ 4‡

2‡

E

E +

2‡

+

25V 50V ref

Chapter 5, Example 5.14 [ Page No - 5.29 ] 12. (a) (i) Compute the current in the 23 W resistor of the following figure shown below by applying the superposition principle. (8) 4‡

47 ‡

27 ‡

20 A

I23

+ E 200 V

Chapter 7, Example 7.7 [ Page No - 7.11 ] (ii) Derive the equation for transient response of RC and RL circuit for DC input. (8) Chapter 10, Section 10.4.2 & 10.5.2 [ Page No - 10.11 & 10.21 ]

Q. 13

Anna University Question Papers

(OR) (b) Obtain the Thevenin and Norton equivalent circuits for the active network shown below. (16) 3‡ a 6‡

3‡ + E

+ E

20 V

10 V b

Chapter 7, Example 7.16 [ Page No - 7.28 ] 13. (a) With neat illustration and necessary derivations, explain the linear transformer.

(16)

Chapter 9, Section 9.3.1, 9.3.3 & 9.3.4 [ Page No - 9.7 to 9.11 ] (OR) (b) Derive the mutual inductance and the coupling coefficient of the transformer with necessary illustration. (16) Chapter 9, Section 9.2.2 & 9.2.3 [ Page No - 9.2 & 9.4 ] 14. (a) Explain in detail with neat illustrations the High pass and Low pass networks and derive the network parameters. (16) Refer [ Page No - Q.14] (OR) (b) Explain the characterization of two port networks in terms of Z, Y and h parameters. (16) Chapter 11, Section 11.4, 11.5 & 11.6 [ Page No - 11.7 to 11.10 ] 15. (a) Discuss in detail the three phase 3-wire circuits with star connected balanced loads.

(16)

Chapter 12, Section 12.4 [ Page No - 12.4 ] (OR) (b) Explain in deatail the phasor diagram of the voltage of the voltage and currents of a three phase unbalanced circuits. (16) Chapter 12, Section 12.8.2 [ Page No - 12.22 ] Note: Explain theory from section 12.8.2 and draw phasor diagram shown in Fig.4. of Example 12.10.

Anna University Question Papers

Q. 14

14.(a) Explain in the detail with neat illustrations the High pass and Low pass network and derive the network parameters.

SOLUTION The low pass and high pass filters are best examples of two port network. A low pass filter will pass all low frequency signals less than a cut-off frequency. A high pass filter will pass all high frequency signals greater than a cut-off frequency. The low pass and high pass filters can be designed using LC network and popular configurations of such filters are constant-K, T-type and P-type filters shown in Figs.1 and 2. Here the product Z1 Z2 is a constant independent of frequency and so such filters are called constant-K filters. Z1

Z1

2

2

Z1

2

1

1

Z1

2

2Z 2

1¢

2¢

Fig.1: Constant-K T-type filter

2Z2

1¢

2¢

Fig.2: Constant-K P-type filter

Low pass Filter For a low pass filter, Z1 = jωL

and

` Z1 Z2 = jωL # Here, Let,

1 jωC

Z2 =

1 L = jωC C

L is a constant independent of frequency. C

L = K2 C L C

` K =

Here, K is also called design impedance or load impedance of the low pass filter. The cut-off frequency, fc of low pass filter is given by, 1

fc =

π LC

Z1

=

2

Z1

jwL 2

=

2

1

L/2 Z2 =

1¢

1 jwC

jwL 2 2

L/2

Z 1 = jwL 1

2Z 2 =

C

2

1 C 2

jw

2¢

Fig.3: Constant-K T-type low pass filter

1¢

L C 2

C 2

2Z 2 =

1 C 2

jw 2¢

Fig.4: Constant-K P-type low pass filter

Q. 15

Anna University Question Papers High pass Filter For a high pass filter, 1 jωC

Z1 =

Let,

Z2 = jωL

1 L # jωL = jωC C

` Z1 Z2 = Here,

and

L is a constant independent of frequency. C

L = K2 C L C

` K =

Here, K is also called design impedance or load impedance of the high pass filter. The cut-off frequency, fc of the high pass filter is given by, 1

fc =

4π LC

Z1

=

2

Z1

1 jw 2C

=

2

1 jw 2C

Z1 = 1

2

1

2C Z 2 = jwL

2

C

2C 2Z 2 = jw 2L

L

1¢

1 jwC

2L

2L

2Z 2 = jw 2L

1¢

2¢

2¢

Fig.6: Constant-K P-type high pass filter

Fig.5: Constant-K T-type high pass filter Z - Parameters of T-type filter

Consider the T-type filter shown in Fig. 7. The equations defining Z- parameters are,

.....(2)

V2 = Z21 I1 + Z22 I2

Let us connect a voltage source to port-1 and open circuit port-2 as shown in Fig.8. Now, I2 = 0 Therefore,when I2 = 0 , from equations (1) and (2) we get, V1 I1

1

.....(1)

V1 = Z11 I1 + Z12 I2

Z11 =

I1

;

Z21 =

V2 I1

Z1

2

2

I2

2 +

+ V1

1¢

Z1

Z2

V2

-

Fig.7: T-type filter

2¢

Anna University Question Papers

Q. 16 From Fig.9, by Ohm’s law,

I1

Z1 V1 = f + Z2 p I1 2 ` Z11 =

V1 I1

.....(3)

Z1

Z1

2

2

+ V1 -

Z1 = + Z2 2

Z2

2¢

Fig.8. Þ

Z2 Z1 + Z2 2

Z1 Z2 = f + Z2 p I1 # 2 Z1 + Z2 2 V1 ` Z11 = = Z2 I1

I1

Using equation (3)

The T-type filter is symmetrical. ` Z22 = Z11 =

Z1 + Z2 2

The T-type filter is reciprocal. ` Z12 = Z21 =

V2

1¢

From Fig.9, by voltage division rule, V2 = V1 #

I2

2 +

1

Z2

The Z- parameter matrix of constant-K, T-type filter is, R V R V W S Z11 Z12 W S Z1 + Z Z2 W 2 S W S2 W Z = S W= S Z1 W S W S Z Z + 2 2W SZ21 Z12 W SS W 2 T X T X The Z- parameter matrix of constant-K, T-type low pass filter is, R V R jωL V 1 1 S Z1 W S W + Z Z + 2 2 S2 W S 2 W jωC jωC S W Z = = S W ω j L Z1 1 1 S W S W + + Z2 W SS Z2 S W 2 jωC jωC W 2 T X T X The Z- parameter matrix of constant-K, T-type high pass filter is, R V R V S Z1 W W S 1 + jωL jωL Z2 W S 2 + Z2 W S jω2C W= S Z = S W 1 Z1 S W S jωL + jωL WW + Z2 W SS Z2 S W jω2C 2 T X T X

Z1 2

+

+ V1 -

Z2

V2

-

Fig.9.

O.C

Appendix-1 USING CALCULATOR IN COMPLEX MODE 1. Addition/Subtraction/Multiplication/Division of Complex Numbers Let, A1 = − 4 + j2 A2 = 3 + j5 Choose complex mode in claculator and enter the complex numbers as shown below: For addition

(–4 + 2 i) + (3 + 5 i)

For subtraction

(–4 + 2 i) – (3 + 5 i)

For multiplication (–4 + 2 i) × (3 + 5 i) For division

(–4 + 2 i) ÷ (3 + 5 i)

To perform the operation press = . To view the real and imaginary part of the result press SHIFT Re * I m . 2. Polar to Rectangular Conversion Let, A1 = 5+ − 30 o Method-1 : Choose complex mode in calculator and enter the complex number as shown below: (5+ − 30) SHIFT

a + bi

To perform the conversion press = . To view the real and imaginary part press SHIFT Re * I m . Method-2 : Choose complex mode in calculator and enter the complex number as shown below: 5 × cos 30 + 5i × sin 30 or 5 cos 30 + 5i sin 30 To perform the conversion press = . To view the real and imaginary part press SHIFT Re * I m . Method-3 : Choose normal computation mode in calculator and enter the complex number as shown below: SHIFT Re c (5, - 30) To perform the conversion press = . To view the real part press ALPHA E . To view the imaginary part press ALPHA F .

A. 2

Circuit Theory

3. Rectangular to Polar Conversion Let, A1 = 2 + j5 Method-1 : Choose complex mode in calculator and enter the complex number as shown below: (2 + 5 i) SHIFT

r+θ

To perform the conversion press = . To view the absolute value and argument press SHIFT Re * I m . Method-2 : Choose complex mode in calculator. i) To calculate the absolute value enter the complex number as shown below: SHIFT Abs (2 + 5 i) To view the absolute value press = . ii) To calculate the argument enter the complex number as shown below: SHIFT arg (2 + 5 i) To view the argument press = . Method-3 : Choose normal computation mode in calculator and enter the complex number as shown below: SHIFT Pol (2, 5) To perform the conversion press = . To view the absolute value press ALPHA E . To view the argument press ALPHA F . Note : The calculator treats the real and imaginary part as separate numbers, hence enclose the real and imaginary part of a complex number by parenthesis.

Appendix

A. 3

Appendix-2 IMPORTANT MATHEMATICAL FORMULAE Trigonometric Identities tan θ = sin θ , cos θ sec θ =

1 , cos θ

cot θ =

1 tan θ

cos (θ ! 90%) = " sin θ

1 sin θ

sin (θ ! 90%) = ! cos θ

cosec θ =

sin2 θ + cos2 θ = 1, 1 + tan2 θ = sec2 θ

tan (θ ! 90%) = − cot θ

1 + cot2 θ = cosec2 θ

cos (θ ! 180%) = − cos θ

sin (A ! B) = sin A cos B ! cos A sin B

sin (θ ! 180%) = − sin θ

cos (A ! B) = cos A cos B " sin A sin B

tan (θ ! 180%) = tan θ

2 sin A sin B = cos (A − B) − cos (A + B)

sin 2θ = 2 sin θ cos θ

2 sin A cos B = sin (A + B) + sin (A − B)

cos 2θ = cos2 θ − sin2 θ

2 cos A cos B = cos (A + B) + cos (A − B)

cos2 θ = 1 + cos 2θ 2

sin A + sin B = 2 sin A + B cos A − B 2 2

sin2 θ = 1 − cos 2θ 2

sin A − sin B = 2 cos A + B sin A − B 2 2

tan 2θ = 2 tan θ2 1 − tan θ

cos A + cos B = 2 cos A + B cos A − B 2 2

ji ji - ji - ji sin θ = e − e , cos θ = e + e 2j 2

cos A − cos B = − 2 sin A + B sin A − B 2 2

e! ji = cos θ ! j sin θ

Complex Variables A complex number, Z may be represented as, Z = x + jy = r+θ = re ji = r (cos θ + j sin θ) where, x = Re ^Zh = r cosθ; r= Z = j=

−1,

x2 + y2 ; 1 = − j; j

y = Im ^Zh = r sinθ θ = tan- 1 j2 = − 1

y x

A. 4

Circuit Theory The conjugate of the complex number, Z = x + jy , may be represented as, Z = x − jy = r+ − θ = re- ji = r (cos θ − j sin θ) )

The following relations hold good for a complex number, Z = x + jy . Z = x + jy =

re ji =

i

r ej2 =

r +θ 2

n

Z = (x + jy) n = r n e jni = r n +nθ , where n is an integer = (x + jy)1/n = r1/n e ji/n = r+ a θ + 2πk k , for k = 0, 1, 2, ........n − 1 n n ,n Z = ,n (re ji) = ,n r + ,n e ji = ,n r + j (θ + 2kπ) , where k is an integer. 1/n

Z

n

Demovier's theorem : ^e jih = e jni = cos nθ + j sin nθ Let, Z1 and Z2 be two complex numbers defined as, Z1 = x1 + jy1 = r1 +θ1 = r1 e ji1 Z2 = x2 + jy2 = r1 +θ2 = r2 e ji2 Now, Z1 = Z2 only if x1= x2 and y1= y2. Z1 ! Z2 = (x1 + x2) ! j (y1 + y2) Z1 Z2 = (x1 x2 − y1 y2) + j (x1 y2 + x2 y1)

or

Z1 Z2 = r1 r2 e j (i1 + i2) = r1 r2 + (θ1 + θ2)

Z1 = (x1 + jy1) # (x2 − jy2) = x1 x2 + y1 y2 + j x2 y1 − x1 y2 (x2 − jy2) Z2 (x2 + jy2) x22 + y22 x22 + y22 or Z1 = r1 e j (i1 - i2) = r1 + (θ1 − θ2) r2 r2 Z2 Derivatives and Integrals Let, U = U(x), V = V(x), and a = constant. d ^aUh = a dU dx dx d ^UVh = U dV + V dU dx dX dx d U = dx 9 V C

V dU − U dV dx dx V2

# a dx = ax + c # UV = U # V − # ; # V dU E

or

# U dV = U V − # V dU

Appendix

A. 5

Appendix-3 LAPLACE TRANSFORM

The Laplace transform is used to transform a time domain function to complex frequency domain called s-domain. In order to transform a time domain function f(t) to s-domain multiply the function by e-st and then integrate from 0 to ∞. The transformed function is represented as F(s). Here s = s + jw, and it is called complex frequency. This transformation was first proposed by Laplace (in the year 1780) and later adopted for circuit analysis for solving differential equations. Hence this transformation is called Laplace transform and the transformation is denoted by the script letter L. i.e., Symbolically the Laplace transform of f(t) is denoted as, F(s) = L [f(t)] Mathematically the Laplace transform of f(t) is defined as, t=3

F (s) =

# f (t) e

-st

dt

t=0

Definition of Laplace Transform Let f(t) be a function of t defined for all positive values of t, now the Laplace transform of f(t) denoted by L[f(t)] or F(s) is defined as, t=3

L{f(t)} = F(s) =

# f (t) e

-st

dt

t=0

Definition of Inverse Laplace Transform The s-domain function can be transformed to time domain by inverse Laplace transform. The inverse Laplace transform of F(s) is defined as, s = v + j~

1 L [F(s)] = f(t) = 2πj -1

# F (s) e

st

ds

s = v - j~

Here the path of integration is a straight line parallel to the jw-axis, such that all the poles of F(s) lie to the left of this line.

A. 6

Circuit Theory

Table - A3.1 : Laplace Transform Pairs

Sl.No.

f(t)

F(s)

1.

Unit impulse, δ(t)

1

2.

Unit step, u(t)

1/s

3.

t

1/s2

4.

t n - 1 ^n = 1, 2, 3 ...h (n - 1) !

1/sn

5.

e −at

1 s+a

6.

tn (n = 1, 2, 3 ... )

n! sn + 1

7.

t e −at

1 (s + a) 2

8.

1 t n - 1 e- at (n = 1, 2, 3 ...) (n − 1) !

1 (s + a) n

9.

tn e −at (n = 1, 2, 3 ... )

n! (s + a) n + 1

10.

sin wt

ω s2 + ω2

11.

cos wt

s s2 + ω2

12.

sinh wt

ω s2 + ω2

13.

cosh wt

s s2 − ω2

14.

e −at sin wt

ω ^s + ah2 + ω2

15.

e −at cos wt

s+a (s + a) 2 + ω2

Appendix

A. 7

Table - A3.2 : Properties of Laplace Transform Note : L{f(t)} = F(s); L{f1(t)} = F1(s); L{f2(t)} = F2(s) Property

Time domain signal

s-domain signal

Amplitude scaling

A f(t)

A F(s)

Linearity

a1f1(t) ± a2 f2(t)

a1 F1(s) ± a2 F2(s)

Time differentiation

d f (t) dt

s F(s) − f (0)

dn f (t) dt n

s n F (s) -

n

/

sn - K

d^K- 1h f (t)

K =1

dt K - 1

where n = 1, 2, 3 .....

Time integration

#

F (s) + s

f (t) dt

# ..... # f (t) (dt)

n

s

where n = 1, 2, 3 .....

+

t =0

s

/s

1 n - K +1

K =1

Frequency shifting

e! at f (t)

F (s " a)

Time shifting

f (t ! α)

e!as F (s)

Frequency differentiation

t f (t)

-

t n f (t)

# f (t) dt E n

F (s) n

;

;

# ..... # f (t) (dt)

dF (s) ds n

^- 1 h

where n = 1, 2, 3 .....

dn F (s) ds n

3

Frequency integration

1 f (t) t

# F (s) ds s

Time scaling

f(at)

1 s Fa k a a T

Periodicity

f(t + nT)

Initial value theorem Lt f (t) = f (0) t"0

Final value theorem

Lt f (t) = f (3)

t"3

1 1 - e- sT 0

# f (t) e

Lt s F (s)

s"3

Lt s F (s)

s"0

t =0

-sT

dt

K

E

t =0

A. 8

Circuit Theory

Appendix-4 R,L,C PARAMETERS AND V- I RELATIONS IN VARIOUS DOMAINS

S.No.

1.

2.

Parameter

Time domain

R v = iR

v

i +

Inductance, L

Inductance, L with initial current

v=L

_

I(s) +

L di ; i 0 = I0 dt

bg

sL I(s) LI0 _

I

-+

+

V

_

jwL

sL V(s)

+

V = I ´ jwL

_

+

v=

i

v

1 C

_

I(s) +

C 1 i dt C

i

v=

1 C

V(s) _

1 sC 1 V(s) = I(s) ´ sC

z

+ v _ V + 0 _

z

1 I(s) sC I(s) + _

i dt ; v(0) = V0

+

+ v _ _ V0 +

V0 s

I

+-

1/sC V(s)

V(s) = I(s) ´

_ V 1 + 0 sC s

1 V0 I(s) I(s) +sC _ -s+

C

z i dt ; v(0) = -V

_

V(s) = I(s) ´ sL + LI 0

C v=

sL V(s)

+

bg

i

Capacitance, C with initial voltage

+-

I0

di ; i 0 = -I 0 dt

v =L

Capacitance, C

sL I(s) LI0 _ I(s) +

_

v L

5.

R V a IR

V(s) = I(s) ´ sL - LI 0

i +

4.

_

sL V(s) = I(s) ´ sL

di dt

I0

3.

V

I(s) + V(s) _

_

v

i +

I +

R V(s) = I(s)R

L v=L

Frequency domain

I(s) + V(s) _

_

v

i +

Resistance, R

s-domain

0

+

1/sC V(s)

V(s) = I(s) ´

_ V 1 - 0 sC s

+

V _ 1 jwC

V=I´

1 jwC

Appendix

A. 9

Appendix-5 CRAMER’S RULE I. Cramer’s Rule for Mesh Basis Equation The mesh basis matrix equation for resistive circuit is, RR R R g R V R I V RE V 1m S 11 12 13 W S 1W S 11 W S R21 R22 R23 g R2m W S I2 W S E22 W S R31 R32 R33 g R3m W S I3 W = S E33 W S W S W S W h h h h W Sh W S h S h W SR m1 R m2 R m3 g R mm W SI m W SE mm W T X T X T X The kth mesh current Ik by Cramer’s rule is, Ik = 1 ∆

m

/∆

jk

E jj

j =1

where, m = Number of meshes in the circuit.

∆jk = Cofactor of Rjk. Ejj = Sum of voltage sources in mesh-j. ∆ = Determinant of resistance matrix. For circuit with three meshes, the mesh currents by Cramer’s rule are, I1 =

∆ ∆11 ∆ E11 + 21 E22 + 31 E33 ∆ ∆ ∆

I2 =

∆ ∆12 ∆ E11 + 22 E22 + 32 E33 ∆ ∆ ∆

I3 =

∆13 ∆ ∆ E11 + 23 E22 + 33 E33 ∆ ∆ ∆

The mesh currents for a circuit with three meshes using short-cut procedure for Cramer’s rule are, I1 =

∆1 ∆

I2 =

∆2 ∆

I3 =

∆3 ∆

R11 R12 R13 E11 R12 R13 R11 E11 R13 R11 R12 E11 where, ∆ = R 21 R 22 R 23 ; ∆1 = E 22 R 22 R 23 ; ∆2 = R 21 E 22 R 23 ; ∆3 = R 21 R 22 E 22 R31 R32 R33 E33 R32 R33 R31 E33 R33 R31 R32 E33

A. 10

Circuit Theory

II. Cramer’s Rule for Node Basis Equation The node basis matrix equation for resistive circuit is, RG G G g G V RV V 1n S 11 12 13 W S 1W SG 21 G 22 G 23 g G 2n W S V2 W SG31 G32 G33 g G3n W S V3 W = S W S W h h h W Sh W S h SG n1 G n2 G n3 g G nn W SVn W T X T X th The k node voltage Vk by Cramer’s rule is, Vk = 1 ∆l

RI V S 11 W SI 22 W SI33 W S W Sh W SI nn W T X

n

/ ∆l jk I jj j=1

where, n

= Number of independent nodes in a circuit

∆l jk = Cofactor of Gjk Ijj

= Sum of current sources connected to node-j

∆l

= Determinant of conductance matrix

For circuit with three nodes excluding the reference node, the node voltages by Cramer’s rule are, V1 =

∆l ∆l11 ∆l I + 21 I + 31 I ∆l 11 ∆l 22 ∆l 33

V2 =

∆l ∆l12 ∆l I + 22 I + 32 I ∆l 11 ∆l 22 ∆l 33

V3 =

∆l13 ∆l ∆l I + 23 I + 33 I ∆l 11 ∆l 22 ∆l 33

The node voltages of a circuit with three nodes excluding the reference using short-cut procedure for Cramer’s rule are, V1 =

∆l1 ∆l

V2 =

∆l 2 ∆l

V3 =

∆l 3 ∆l

G11 G12 G13 I11 G12 G13 l l where, ∆ = G 21 G 22 G 23 ; ∆ 1 = I 22 G 22 G 23 ; G31 G32 G33 I33 G32 G33 G11 I11 G13 G11 G12 I11 ∆l 2 = G 21 I 22 G 23 ; ∆l3 = G 21 G 22 I 22 G31 I33 G33 G31 G32 I33

Z

Impedance

± jX

Reactance

L

Inductance

R

Resistance

Group-1 Parameter

Z1

± jX1

L1

R1

Z

2

± jX2

L2

R2

Zn

± jXn

L3

Rn

Þ

Þ

Þ

Þ

Z eq = Z1 + Z2 + .....+ Zn

Z eq

± jXeq = ± jX1 ± jX2 ± ..... ± jXn

± jXeq

Leq = L1 + L2 +.....+ Ln

Leq

Req = R1 + R2 +.....+ Rn

Req

Series Connection of Parameters and their equivalent

Z1

± jX1

L1

R1

Z

2

± jX2

L2

R2

Summary of Equivalent of Series/Parallel-Connected Group-1 Parameters

Zn

± jXn

Ln

Rn

Þ

Þ

Þ

Þ

eq

=

Z

eq

eq

± jX

L eq =

R 1

2

n

1

2

n

=

=

Z1

1

+ Z

Z eq

2

1

+ .....+

1 Z

n

1

1 1 1 1 + + .....+ ± j X1 ± j X 2 ± j Xn

± jXeq

1 1 1 1 + + .....+ L L L

Leq

1 1 1 1 + + .....+ R R R

Req

Parallel Connection of Parameters and their equivalent

Appendix A. 11

Appendix-6

EQUIVALENT OF SERIES/PARALLEL CONNECTED PARAMETERS

Y1

Admittance

Y

± jB1

C1

G1

± jB

Susceptance

C

Capacitance

G

Conductance

Group-2 Parameter

Y 2

± jB2

C2

G2

Yn

± jBn

Cn

Gn

Þ Y

eq

=

eq

1

Geq

2

1

2

1

1

+ Y

2

1

Y eq

+ .....+

1 Y

n

1

1 1 1 + + .....+ ± jB1 ± jB 2 ± jBn

± jBeq

1

1 1 1 1 + + .....+ C C C

Ceq

n

n

1 1 1 + + .....+ G G G

Y1

=

=

=

eq

eq

C

G

± jB

Þ

Þ

Þ

Series Connection of Parameters and their equivalent

Summary of Equivalent of Series/Parallel-Connected Group-2 Parameters

Y1

± jB1

C1

G1

Y

2

± jB2

C2

G2

Yn

± jBn

Cn

Gn

Geq

Ceq

± jBeq

Þ

Y eq = Y 1 + Y 2 + .....+ Y n

Y eq

± jBeq = ± jB1 ± jB2 ± ..... ± jBn

Þ

Ceq = C1 + C2 +.....+ Cn

Þ

Geq = G1 + G2 +.....+ Gn

Þ

Parallel Connection of Parameters and their equivalent

A. 12 Circuit Theory

Appendix

A. 13

Appendix-7 STAR-DELTA TRANFORMATION 1

1 R1

R12

Þ

R3

R31

Þ

R2

2

2 R23

3

3

Fig. a : Star connected resistances.

Fig. b : Delta connected resistances.

Fig. A.7 : Star-delta transformation.

Star to Delta Transformation R1 R2 R3 R = R 2 + R3 + 2 R3 R1 R 3 R1 = R3 + R1 + R2

R12 = R1 + R2 + R23 R31

Product of the resistances connected to the two Sum of resistances terminals in star network connected to the Delta equivalent resistance between two terminals = # two terminals in The third resistance in star network star network

Delta to Star Transformation R1 =

R12 R31 R12 + R23 + R31

R2 =

R23 R12 R12 + R23 + R31

R3 =

R31 R23 R12 + R23 + R31

Product of resistances connected to the two terminals in delta network Star equivalent resistance at one terminals = Sum of three resistance s in delta network

A. 14

Circuit Theory

Appendix-8 SUMMARY OF THEOREMS

S.No

Theorem

Definition

1.

Superposition theorem

The superposition theorem states that the response in a circuit with multiple sources is given by algebraic sum of responses due to individual sources acting alone.

2.

Thevenin’s theorem

Thevenin’s theorem states that a circuit with two terminals can be replaced by an equivalent circuit, consisting of a voltage source in series with a resistance (or impedance).

3.

Norton’s theorem

Norton’s theorem states that a circuit with two terminals can be replaced by an equivalent circuit, consisting of a current source in parallel with a resistance (or impedance).

4.

Reciprocity theorem

The reciprocity theorem states that, in a linear, bilateral, single source circuit, the ratio of excitation to response is constant when the position of excitation and response are interchanged.

Nature of Source and Variable Element

Maximum Power Transfer Theorem

DC source with internal resistance connected Maximum power is transferred from source to a variable resistive load. to load, when the load resistance is equal to source resistance. AC source with internal resistance connected Maximum power is transferred from source to a variable resistive load. to load, when the load resistance is equal to source resistance. AC source with internal impedance connected Maximum power is transferred from source to a variable resistive load. to load, when the load resistance is equal to magnitude of source impedance. AC source with internal impedance connected Maximum power is transferred from source to a load with variable resistance and variable to load, when the load impedance is equal to reactance. complex conjugate of source impedance. AC source with internal impedance connected Maximum power is transferred from source to a load with variable resistance and fixed to load when load resistance is equal to absolute reactance. value of the rest of the impedence of the circuit. `

AC source with internal impedance connected Maximum power is transferred from source to a load with fixed resistance and variable to load when load reactance is equal to reactance. conjugate of source reactance.

Appendix

A. 15

Appendix-9 IMPORTANT EQUATIONS OF SERIES RESONANCE 1 in rad/s LC

1.

Angular resonant frequency, ωr =

2.

Resonant frequency, fr =

3.

Q-factor at resonance, Q r = ωr L ; R

When ω # ωr, Q =

Note : 4.

;

Qr = 1 R

L C

When ω $ ωr, Q = ωL R

β 2π

R 2 1 a 2L k + LC in rad/s

;

ωh = R + 2L

1 + 1 2 E in rad/s 4Q r

;

ωh = ωr ; 1 + 2Q r

ωl = ωr ; − 1 + 2Q r

fl =

ωl in Hz 2π

;

fh =

R 2 1 a 2L k + LC in rad/s

ωh in Hz 2π

Total reactance at half-power frequencies ωL − 1 = ! R ωC

7.

;

1 ωr CR

Half-power (or cut-off) frequencies ωl = − R + 2L

6.

1 ωCR

Qr =

ω Bandwidth, β = R in rad/s ; β = r in rad/s Qr L Bandwidth in Hz =

5.

1 in Hz 2π LC

;

ωh L − 1 = R ; ωh C

ωl L − 1 = − R ωl C

Selectivity Selectivity =

β ωr

;

Selectivity = 1 Qr

1 + 1 2 E in rad/s 4Q r

-

~

-

~

~

V, w

-

~

+

-

V, w

+

V, w

+

V, w

+

R

L

L

R1

L

R1

L

C

C

R2

C

C

R2

RLC parallel circuit

ωr =

ωr =

ωr =

wr =

1 LC

1 LC

1 LC

1 LC

L L − CR 22

CR12 L

L − CR12 L − CR 22

1−

wr in rad/s

fr =

fr =

fr =

fr =

1 2π LC

2π LC

1

2π LC

1

1 2π LC

L L − CR 22

CR12 L

L − CR12 L − CR 22

1−

fr in Hz

Resonant frequency

Rdynamic =

Rdynamic =

L R2 C

L R1 C

1 wr C

1 R2 R1 2 2 + 2 2 R1 + X Lr R 2 + XCr X Lr = wr L ; XCr =

Rdynamic =

Rdynamic = R

Rdynamic in Ω

Dynamic resistance

A. 16 Circuit Theory

Appendix-10

PARALLEL RESONANT CIRCUITS

Appendix

A. 17

Appendix-11 ELECTRICAL EQUIVALENT OF COUPLED COILS Electrical equivalent of Group-I coupled coils M Ij

Lj

M Lk

Ik

Ij

Lk - M

Lk

Note : In electrical equivalent the mesh currents orientation should be same as that of original circuit.

Þ

Þ

M

M Ij

Lj

Ik

Þ

Þ

Lj - M

Lj

M Ij

Ik

Lk

Lk

Lj

Ik

Electrical equivalent of Group-II coupled coils M Ij

Lj

M Lk

Ij

Ik

Lk + M

Lk

Note : In electrical equivalent the mesh currents orientation should be same as that of original circuit.

Þ

Þ

-M

M Ij

Lj

Ik

Þ

Þ

Lj + M

Lj

M Lk

Ik

Ij

Lj

Lk

Ik

A. 18

Circuit Theory

Appendix-12 EQUIVALENT OF SERIES AND PARALLEL CONNECTED COUPLED COILS

Connection

Circuit

Equivalent circuit

Series Aiding

M I

I

I

L1

Leq = L1 + L2 + 2M

L2

M

Series Opposing

I

I

I

L1

Leq = L1 + L2 E 2M

L2

I

I I1

Parallel Aiding

M

L1

I2

L2

L eq a

L 1 L 2 E M2 L1 C L 2 E 2 M

Parallel Opposing I

I I1

L1

I2

M

L2

L eq a

L 1 L 2 E M2 L1 C L 2 C 2 M

Appendix

A. 19

Appendix-13 INITIAL AND FINAL CONDITIONS IN RLC CIRCUITS EXCITED BY DC SUPPLY

Element

Initial condition t = 0+

Final condition t=∞

R

R

O.C.

S.C.

I0

S.C.

I0

S.C.

S.C.

O.C.

_

V0 +E

O.C.

+

V0 E+

O.C.

R

L I0 L I0 L

C

+

V0 C

_

V0 C

A. 20

Circuit Theory

Appendix-14 SUMMARY OF PARAMETERS OF TWO-PORT NETWORK

S.No Independent Dependent Parameters variables variables k11, k12, k21, k22 1.

I1,I2

V1,V2

Z11,Z12,Z21,Z22

Name of the parameter Impedance parameter (Z-parameter or Open circuit parameter)

Mathemetical equation V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2 V1 Z11 Z12 I1 H > H > H=> V2 Z21 Z22 I2

2.

V1,V2

I1, I2

Y11,Y12,Y21,Y22 Admittance parameter I1 = Y11 V1 + Y12 V2 (Y-parameter or Short I2 = Y21 V1 + Y22 V2 circuit parameter) I1 Y11 Y12 V1 H > H > H=> I2 Y21 Y22 V2

3.

I1,V2

V1, I2

h11,h12,h21,h22

Hybrid parameter (h-parameter)

V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 V1 h11 h12 I1 H > H > H=> I2 h21 h22 V2

4.

5.

V1,I2

V2,I2

I1, V2

V1, I1

g11,g12,g21,g22

A, B, C, D

Inverse hybrid parameter (g-parameter)

Transmission parameter(ABCDparameter)

I1 = g11 V1 + g12 I2 V2 = g21 V1 + g22 I2 I1 g11 g12 V1 H > H > H=> V2 g21 g22 I2 V1 = AV2 − BI2 I1 = CV2 − DI2 V1 A B V2 H > H > H=> I1 C D − I2

6.

V1,I1

V2, I2

A′, B′, C′, D′

Inverse transmission parameter (A′B′C′D′-parameter)

V2 = Al V1 − Bl I1 I2 = Cl V1 − Dl I1 V2 Al Bl V1 H > H > H=> I2 Cl Dl − I1

Appendix

A. 21

Appendix-15 RELATIONSHIP BETWEEN PARAMETER SETS

[Z]

[Y]

[T]

[T′]

[h]

[g]

[Z]

[Y]

[T]

Z11 Z12 Z21 Z22

Y22 Y − 12 ∆y ∆y Y21 Y11 − ∆y ∆y

A ∆T C C 1 D C C

Z22 Z12 − ∆z ∆z Z21 Z11 − ∆z ∆z

Y11 Y12 Y21 Y22

D − ∆T B B A −1 B B

Y22 − 1 Y21 Y21 ∆y Y − − 11 Y21 Y21

A B C D

Z11 ∆z Z21 Z21 1 Z22 Z21 Z21

−

Z22 ∆z Z12 Z12 1 Z11 Z12 Z12

−

Y11 − 1 Y12 Y12 ∆y Y − − 22 Y12 Y12

D ∆T C ∆T

[T′]

B ∆T A ∆T

∆z Z12 Z22 Z22 Z − 21 1 Z22 Z22

1 − Y12 Y11 Y11 ∆y Y21 Y11 Y11

B ∆T D D −1 C D D

1 − Z12 Z11 Z11 ∆z Z21 Z11 Z11

∆y Y12 Y22 Y22 Y − 21 1 Y22 Y22

C − ∆T A A 1 B A A

Dl Cl ∆Tl Cl

1 Cl Al Cl

Al − 1 Bl Bl ∆Tl Dl − Bl Bl Dl ∆Tl Cl ∆Tl

Bl ∆Tl Al ∆Tl

Al Bl Cl Dl

Bl Al ∆ − Tl Al

1 Al Cl Al

Cl − 1 Dl Dl ∆Tl Bl Dl Dl

[h]

[g]

∆h h12 h22 h22 h − 21 1 h22 h22

1 − g12 g11 g11 g21 ∆g g11 g11

1 − h12 h11 h11 h21 ∆h h11 h11

∆g g12 g22 g22 g − 21 1 g22 g22

h ∆h − 11 h21 h21 h − 22 − 1 h21 h21

1 g22 g21 g21 g11 ∆g g21 g21

−

1 − h11 h12 h12 h22 ∆h h12 h12

∆g g − 22 g12 g12 g 1 − 11 g12 g12 −

h11 h12 h21 h22

g22 g12 ∆g ∆g g21 g11 − ∆g ∆g

h22 h12 − ∆h ∆h h h11 − 21 ∆h ∆h

g11 g12 g21 g22

Note : 1. In the above table the ∆Z , ∆Y , ∆h , ∆g , ∆T and ∆T’ represent the following determinants. ∆z =

g g Z11 Z12 Y11 Y12 h11 h12 A B Al Bl ; ∆y = ; ∆h = ; ∆g = 11 12 ; ∆T = ; ∆Tl = g 21 g 22 Z 21 Z 22 Y21 Y22 h 21 h 22 C D Cl Dl 2. The following relations holds good for parameters. >

-1

; >

1

; >

Z11 Z12 Y11 Y12 H = > H Z 21 Z 22 Y21 Y22

Y11 Y12 Z11 Z12 H = > H > Y21 Y22 Z 21 Z 22

g11 g12 - 1 h11 h12 A B Al Bl - 1 ; > H = ; H = > H E g 21 g 22 h 21 h 22 C D Cl Dl

1 1 g11 g12 h11 h12 A BAl Bl ; = = H > H H > E ; g 21 g 22 h 21 h 22 C D Cl Dl

A. 22

Circuit Theory

Appendix-16 CHOICE OF REFERENCE PHASOR

Phase sequence RYB

Reference phasor VRY

Line voltages

VRY = VL +0 o VYB = VL + − 120 o VBR = VL + − 240 o = VL +120 o

Phasor diagram

VBR o

120

VRY o

E120 VYB

VRY

RYB

VYB

VYB = VL +0 o VBR = VL + − 120 o VRY = VL + − 240 o = VL +120 o

120

o

V YB o

E120 VBR

V YB

RYB

VBR

VBR = VL +0 o VRY = VL + − 120 o VYB = VL + − 240 o = VL +120 o

120

o

VBR o

E120 VRY

V YR

RBY

VRB

o

VRB = VL +0 VBY = VL + − 120 o VYR = VL + − 240 o = VL +120 o

120

o

VRB o

E120 VBY

VRB

RBY

VBY

VBY = VL +0 o VYR = VL + − 120 o VRB = VL + − 240 o = VL +120 o

120

o

VBY o

E120 V YR

VBY

RBY

VYR

o

o

VYR = VL +0 VRB = VL + − 120 o VBY = VL + − 240 o = VL +120 o

120

V YR o

E120 VRB

Appendix

A. 23

Appendix-17 V - I EQUATION OF THREE PHASE LOAD I. Three/Four wire star-connected balanced load R IR

+ VR

VBR

VRY

Z Y = ZÐf

Y IY

+

+

VY

+

-

IR

+

Z R = ZÐf

VR VRY

VBR

-

N

VB

Z Y = ZÐf

Z B = ZÐf IB

B

Y

IY

-

+

Y

IY

-

R

R IR

+

-

+

-

R

Y IY

+

V YB

VY

IR Z R = ZÐf IN

N

-

Z B = ZÐf VB + IB

B

V YB

-

+

B

IB

B

Fig. A.17.1 : Three wire star-connected balanced load with conventional polarity of voltages and direction of currents for RYB sequence.

-

IB

+

IN = 0

N

Fig. A.17.2 : Four wire star-connected balanced load with conventionalpolarity of voltages and direction of currents for RYB sequencce.

Line Voltages VRY = VL +0 o ;

VYB = VL + − 120 o ;

VBR = VL + − 240 o

VL= Magnitude of line voltage.

Phase Voltages VL + (0 o − 30 o) = V+ − 30 o 3 V VY = L + (− 120 o − 30 o) = V+ − 150 o 3 VL + (− 240 o − 30 o) = V+ − 270 o VB = 3 Phase Currents VR =

o VR = V+ − 30 = V + (− 30 o − φ) = I+ (− 30 o − φ) Z+φ Z ZR o V V 150 + − IY = Y = = V + (− 150 o − φ) = I+ (− 150 o − φ) Z+φ Z ZY o I B = VB = V+ − 270 = V + (− 270 o − φ) = I+ (− 270 o − φ) Z+φ Z ZB Line Currents

IR =

VL = Magnitude of 3 phase voltage.

V=

I = V = Magnitude of Z

phase current.

I L = I = Magnitude of line current.

I R = I L + (− 30 o − φ) ; I Y = I L + (− 150 o − φ) ; I B = I L + (− 270 o − φ) Power P = 3 V I cos φ

or

P =

3 VL I L cos φ

In star connection the line and phase currents are same.

A. 24

Circuit Theory

II. Delta-connected balanced load R -

IR

VRY

VBR +

Z RY = ZÐf

VBR

Z BR = ZÐf

+

VRY

IRY

-

+

-

R

I YB + V YB -

IY

-

Y

+

Y

IBR

B

Z YB = ZÐf

V YB

-

B

IB

+

Fig. A.17.3 : Delta connected balanced load with conventional polarity of voltages and direction of currents for RYB sequence.

Line Voltages VRY = VL +0 o ;

VYB = VL + − 120 o ;

Phase Voltages VRY = V+0 o ;

V = VL= Magnitude of phase voltage.

VYB = V+ − 120 o ;

VBR = V+ − 240 o In delta connection the line and phase voltages are same.

Phase Currents I RY =

VL= Magnitude of line voltage.

VBR = VL + − 240 o

o V RY = V+ 0 Z + φ Z RY

= V+−φ Z

= I+ − φ

o I YB = VYB = V+ − 120 = V + (− 120o − φ) = I+ (− 120o − φ) Z+φ Z Z YB o I BR = VBR = V+ − 240 = V + (− 240o − φ) = I+ (− 240 o − φ) Z+ φ Z Z BR

I = V = Magnitude of Z phase current.

Line Currents IR =

3 I + (− φ − 30 o)

= I L + (− 30 o − φ)

IY =

3 I + (− 120 o − φ − 30 o) = I L + (− 150 o − φ)

IB =

3 I + (− 240 o − φ − 30 o) = I L + (− 270 o − φ)

Power P = 3 V I cos φ

or

P =

3 VL I L cos φ

IL =

3 I = Magnitude of line current.

Appendix

A. 25

III. Three wire star-connected unbalanced load R +

-

VRY

VBR

IR

+ VR

Z R = ZRÐfR

+

+ Y IY

VY

Z B = ZBÐfB

VB

IB

V YB

-

+

~

Y +

B

IB

~

E

ZR

I1 ZY

E

V YB

B

IR

+ VRY

N

IY

-

R

Z Y = Z Y Ðf Y

Y

IR

+

R

N

ZB

IY I2 IB

B

Fig. A.17.4 : Three wire star-connected unbalanced load with conventional polarity of voltages and direction of currents for RYB sequence.

Fig. A.17.5 : Mesh analysis to solve line currents.

Line Voltages VRY = VL +0 o ;

VYB = VL + − 120 o ;

VBR = VL + − 240 o

VL= Magnitude of line voltage.

Line Currents I R = I1

= I R +γR The mesh currents I1 and I2 are solved by mesh analysis, and the line currents are estimated from mesh currents

I Y = I2 − I1 = I Y +γY I B = − I2

= I B +γB

Phase Currents I R = I R +γR I Y = I Y +γY I B = I B +γB

In star connected load the line and phase currents are same

I R, I Y, and I B are magnitude of line and phase currents and γ , γ and γ are phase angle R Y B of line and phase currents with respect to reference phasor

Phase Voltages VR, VY and VB are magnitude of phase voltages and δR, δY and δB are phase angle of phase voltages with respect to reference phasor

VR = I R Z R = VR +δR VY = I Y Z Y = VY +δY VB = I B Z B = VB +δB Power P =

Power consumed Power consumed Power consumed + + by R- phase load by Y- phase load by B- phase load

= VR I R cos φ1 + VY I Y cos φ2 + VB I B cos φ3 = VR I R cos φ1 + VY I Y cos φ2 + VB I B cos φ3 φ1 = Phase difference between VR and I R φ2 = Phase difference between VY and I Y φ3 =Phase difference between VB and I B

A. 26

Circuit Theory

V. Four wire star-connected unbalanced load R

R

+

-

IR

IR

+ VR

VRY

VBR

Z R = ZRÐfR

N

+

Y

IY

+ IY

-

Z B = ZBÐfB

-

Y

VB

VY

N

-

+

IB

B IN

V YB

B

+

Z Y = Z Y Ðf Y

IB

IN

Fig. A.17.6 : Four wire star-connected unbalanced load with conventional polarity of voltages and direction of currents. Line Voltages

VRY = VL +0 o ;

VYB = VL + − 120 o ;

VBR = VL + − 240 o

Phase Voltages V VR = L + (0 o − 30 o) = V+ − 30 o 3 V VY = L + (− 120 o − 30 o) = V+ − 150 o 3 VL + (− 240 o − 30 o) = V+ − 270 o VB = 3 Phase Currents

V=

VL = Magnitude of line voltage. VL = Magnitude of 3 phase current.

Since load neutral is connected to source neutral, phase voltages will be balanced even though load is unbalanced. IR, IY and IB are magnitude of R-phase, Y-phase and B-phase currents respectively.

o VR = V+ − 30 = V + (− 30 o − φR) = I R + (− 30 o − φR) Z ZR + φ ZR R R o I Y = VY = V+ − 150 = V + (− 150 o − φY) = I Y + (− 150 o − φY) Z Y +φY ZY ZY o I B = VB = V+ − 270 = V + (− 270 o − φB) = I B + (− 270 o − φB) In star connection the Z B +φB ZB ZB line and phase currents Line Currents

IR =

are same.

I R = I R + (− 30 o − φR) ; I Y = I Y + (− 150 o − φY) ; I B = I B + (− 270 o − φB) Power Power consumed Power consumed Power consumed P = + + by R- phase load by Y- phase load by B - phase load = VR I R cos φ1 + VY I Y cos φ2 + VB I B cos φ3 = VIR cosφ1 + VI Y cos φ2 + VIB cosφ3

VR = VY = VB = VL / 3 = V φ1 = Phase difference between VR and I R φ2 = Phase difference between VY and I Y φ3 =Phase difference between VB and I B

Appendix

A. 27

III. Delta-connected unbalanced load R

VBR

IRY

VRY

VBR

Z RY = ZRY ÐfRY

-

VRY

IR

-

-

+

+

R

Y

Y

+

+

I YB + V YB -

IY

-

Z BR = ZBRÐfBR

I BR

B

Z YB = Z YBÐf YB

V YB

-

+

B

IB

Fig. A.17.7 : Three phase delta-connected unbalanced load with conventional polarity of voltages and direction of currents for RYB sequence. Line Voltages

VRY = VL +0 o ; VYB = VL + − 120 o ; VBR = VL + − 240 o

VL = Magnitude of line voltage.

Phase Voltages VRY = V+0 o ;

VYB = V+ − 120 o ;

V = VL = Magnitude of

VBR = V+ − 240 o

phase voltage.

Phase Currents I RY = I YB

In delta connection the line and phase voltages are same.

V RY = I RY +γRY Z RY

I RY , IYB and I BR are magnitude of phase currents and γ , γ and γ are phase angle RY YB BR of phase currents.

V = YB = I YB +γYB Z YB

I BR =

V BR = I BR +γBR Z BR

Line Currents I R = I RY − I BR ;

I Y = I YB − I RY ;

I B = I BR − I YB

Power P =

Power consumed Power consumed Power consumed + + by R- phase load by Y- phase load by B- phase load

= VRY I RY cos φ1 + VYB I YB cos φ2 + VBR I BR cos φ3 = VL IRY cosφ1 + VLI YB cos φ2 + VLIBR cosφ3

VRY = VYB = VBR = VL = V φ = Phase difference between VRY and I RY φ2 = Phase difference between VYB and I YB φ3 = Phase difference between VBR and I BR

A. 28

Circuit Theory

Appendix-18 TWO WATTMETER METHOD OF POWER MEASUREMENT CC R

P1

IR

+ PC

VRY

-

Y

3-phase load

P2 = VBY IB VBY

j - ÐI j

BY

B

PC IB

+

B

e cos eÐV

P1 = VRY IR cos ÐVRY - Ð IR

P2

CC

Fig. a : Wattmeters in lines R and B. P1 R

CC

PC

VRB

CC Y

IY

+

3-phase load

e cos eÐV

j - ÐI j

e cos eÐV

j - ÐI j

P1 = VR B IR cos ÐVRB - Ð IR P2 = VYB I Y

PC P2

V YB

B

IR

+

YB

Y

-

-

Fig. b : Wattmeters in lines R and Y. R

-

-

V YR

Y

P1 CC

+

3-phase load

CC

+ P2

P1 = VYR I Y cos ÐVYR - Ð I Y P2 = VBR IB

PC

VBR

B

IY

BR

B

IB

PC

Fig. c : Wattmeters in lines Y and B. Fig. A.18 : Possible connections of two wattmeters for measurement of 3-phase power.

Reading of wattmeter

Power, P = P1 + P2 -1

Power factor angle, φ = tan

P −P c 3 P2 P1 m 1+ 2

Power factor, cos φ = cos ; tan- 1 c 3 P2 − P1 mE P1 + P2

pf

Equal

Unity

If one wattmeter is zero

0.5

If one wattmeter is negative

< 0.5

Both are positive

> 0.5

INDEX A ABCD parameters 11.12 A'B'C'D' parameters 11.13 AC circuit 3.1 AC source 2.1, 3.1 Active elements 1.4, 2.1 Active network 11.20 Active power 3.20 Additivity 7.80 Admittance 3.32 Admittance angle 3.33 Admittance triangle 3.33 Alternating - current 3.1 - voltage 3.1 Alternating current source 2.1 Ampere 1.6 Angular frequency 3.4 Anti-resonance 8.16 Apparent power 3.20 Average power 3.19 - value 3.6

B Balanced emf 12.1 - load 12.13 - system 12.1 Bandwidth 8.6 Bilateral element 1.8 Branch 2.2

C Calorie 1.6 Capacitance 1.1, 3.10 Capacitive reactance 3.14, 3.28 - in s-domain 3.14 Charge 1.6 Chord 2.16 Circuit theory 1.1 Closed path 2.2

Coefficient of coupling 9.4 Complementary function 10.2 Complex power 3.19 Conductance 3.32 Conductive coupling 9.1 Conductor 2.13 Connected graph 2.15 Co-sinusoidal voltage 10.4 Cotree 2.17 Coulomb 1.6 Coupled coils 9.1 Cramer's rule 4.3, 5.3 Critical coupling 9.25 Critical damping 10.30 Critical resistance 10.31 Current 1.1, 1.6 Current controlled current source 2.32 Current controlled voltage source 2.32 Current division rule 2.13, 3.35, 6.4, 6.25 Current magnification 8.24 Current source 2.10, 3.1 - ideal 2.10, 3.1 - practical 2.11, 3.2 Current variables 2.19 Cycle 3.4

D Damped oscillatory response 10.31,10.33 Damping ratio 10.32 DC circuit 2.1 DC source 2.1 Delta connection 2.6 Dependent source 2.2 Derived parameters 1.3 Direct current source 2.1 Dot rule 9.7 Double tuned coupled circuit 9.21 Double tuned coupled coil 9.21 Driving point admittance 11.2 Driving point impedance 11.2 Dynamic resistance 8.16

I. 2

E Effective value 3.6 Electric circuit 1.1 Electrical energy 1.1 Energy 1.5, 2.13, 3.21 Energy stored in - capacitance 3.11 - inductance 3.9 Exponential voltage 10.4

Index In-phase 3.16 Internal impedance 3.2 Internal resistance 2.10 Inverse parameter 1.3

J Joule 1.5

K

F

Kirchoff ’s current law 2.9 Kirchoff ’s voltage law 2.9

Farad 3.10 Form factor 3.7 Frequency 3.1, 3.21 Fundamental parameter 2.1

L

G G-parameters 11.11 Graph 2.15

H Half-power frequencies 8.6 Higher cut-off frequency 8.6, 8.29, 8.33 Homogeneity 7.1 H-parameters 11.9 Henry 3.9 Hertz 3.4

I

Lagging 3.16 Law of conservation of energy 1.4 Laws of magnetic induction 3.9 Leading 3.16 Lenz's law 9.7 Linear 7.1 - circuit 1.8 - element 1.8 - network 1.8 Line current 12.2 - voltage 12.2 Lines 12.2 Link 2.16 Load set reference 2.18 Looking back impedance 6.23 Looking back resistance 6.5, 7.21, 7.22 Lower cut-off frequency 8.6, 8.29, 8.33

Ideal - capacitor 1.1 - current source 2.10, 3.1 - inductor 1.1 - resistor 1.1 - voltage source 2.10, 3.1 Immittance 11.2 Impedance 3.29, 3.35 Impedance angle 3.30 Impedance triangle 3.30 Impulse voltage 10.3 Independent current variables 2.19 Independent source 2.2 Independent voltage variables 2.19 Inductance 1.1, 3.8 Inductive reactance 3.13, 3.24 - in s-domain 3.13

M Magnetic coupling 9.1 Mesh 4.1 - analysis 4.1 - current 4.1 Millman's theorem 6.11 Mutual-conductance 5.3 Mutual induced emf 9.1 Mutual-resistance 4.2

N Natural frequency of oscillation 10.32 Natural response 10.2 Negative charge 1.6

I. 3

Circuit Theory Network 2.1 Network topology 2.14 Network variables 2.18 Neutral 12.2 Neutral displacement voltage 12.24 - shift voltage 12.24 Node 2.2, 5.1 Node analysis 5.1 Node voltage 5.1 Normal phase sequence 12.2 Norton's current source 7.22 Norton's resistance 7.22 Norton's theorem 7.22

Potential difference 1.7 Power 1.4, 1.6, 2.12, 3.19 Power factor 3.21 Power factor angle 3.31 Power triangle 3.20 Practical - current source 2.11, 3.2, 3.3 - voltage source 2.11, 3.2 Primary winding 9.1 Principal node 2.2, 5.1 Principle of superposition 7.1 p-connection 2.6

Q O Ohm's law 2.9 - of ac circuits 3.35 One-port 11.1 Open circuit 2.7 Orientation 2.18 Oriented graph 2.15 Over damped response 10.35

P Parabolic voltage 10.4 Parallel aiding 9.17 Parallel connection 2.4 Parallel opposing 9.19 Parallel resonance 8.16 Parameters 1.3 Particular solution 10.2 Passive - circuit 1.4 - element 1.4, 2.1 - network 1.4,11.20 Path 2.15 Peak factor 3.7 Phase 3.15, 12.1 Phase angle 3.15 Phase current 12.2 - difference 3.16 - sequence 12.2 - voltage 12.2 Phasor 3.17 Phasor diagram 3.18 Planar circuits 4.1 Port 11.1 Positive charge 1.6

Quality factor 8.3, 8.26

R Ramp voltage 10.4 Rated current 6.11 Rated voltage 6.19 RBY sequence 12.2 Reactive power 3.20 Reciprocal network 11.20 Reciprocity theorem 7.69 Reference 2.18 Reference node 5.1 - phasor 12.14 Rejector circuit 8.29 Resistance 1.1, 2.13 Resistivity 2.13 Resonance 8.1 Resonant frequency 8.1, 8.16 Reversed phase sequence 12.2 Rms value 3.6 Rotating vector 3.17 RYB sequence 12.2

S Secondary winding 9.1 Selectivity 8.11 Selector circuit 8.29 Self-conductance 5.3 Self induced emf 9.1 Self-resistance 4.2 Series aiding 9.15 Series connection 2.4

I. 4

Circuit Theory Series opposing 9.16 Series resonance 8.1 Short circuit 2.7 Single loop circuit 2.21 Single node pair circuit 2.21 Single tuned coupled circuit 9.21 Single tuned coupled coil 9.21 Sinor 3.17 Sinusoidal current 3.7 Sinusoidal source 2.1, 3.1 Sinusoidal voltage 3.3,10.4 Source-free response 10.2 Source impedance 3.2, 3.3 Source resistance 2.11 Source transformation 2.11, 3.3 Stagger tuned circuit 9.21 Star connection 2.6 Steady state analysis 1.8 Step voltage 10.3 Subgraph 2.16 Supermesh 4.24 Supernode 5.17 Superposition theorem 7.1 Susceptance 3.32

T T-connection 2.6 Thevenin's resistance 7.21 Thevenin's theorem 7.21 Thevenin's voltage source 7.21 Three phase 12.1 Three phase alternators 12.1 Time constant 10.15,10.24 Time period 3.1, 3.4 Topology 2.14 Total response 7.2,10.1 Transient analysis 1.8,10.1 Transient period 1.8,10.1 Transient response 10.1 Transient state 10.1 Tree 2.17 Tuning circuit 8.10 Twig 2.17 Two-port 11.1

U Unbalanced load 12.13 Undamped response 10.32

Under damped response 10.33 Unilateral element 1.8 Unit 2.13, 3.21

V Vector 3.16 Volt 1.7 Voltage 1.1, 1.7 Voltage controlled current source 2.32 Voltage controlled voltage source 2.32 Voltage division rule 2.14, 3.36, 6.2, 6.20 Voltage magnification 8.3 Voltage source 2.10, 3.1 - ideal 2.10, 3.1 - practical 2.11, 3.1 Voltage variables 2.19

W Watt 1.6 Waveform 3.4 Weber-turns 3.9

X Y Y-parameters 11.8

Z Zero value source 7.1, 7.22, 7.23 Z-parameters 11.7