Basic Electrical Engineering Revised First Edition 0071328963, 9780071328968

Table of contents :
Title
Contents
1. INTRODUCTION
1.1 Charge
1.2 Voltage or Potential
1.3 Basic Circuit
1.4 Units and Symbols
Rules to Use SI Prefixes
Guidelines for Using SI Units
1.5 Basics of Experimentation
General Instructions and Precautions
2. OHM’S LAW
2.1 Ohm’s Law
2.2 Resistance
2.3 Duality
2.4 Conductors and Resistors
2.5 Variation of Resistance with Temperature
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
Experimental Exercise
3. NETWORK ANALYSIS
3.1 Network Components
3.2 Series and Parallel Combinations
3.3 Energy Sources
3.4 Combination of Sources
3.5 Types of Sources
3.6 Kirchhoff’s Laws
3.7 Loop-Current Analysis
3.8 Mesh Analysis
3.9 Node-Voltage Analysis
3.10 Nodal Analysis
3.11 Choice of Method of Analysis
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
Experimental Exercises
4. NETWORK THEOREMS
4.1 Introduction
4.2 Superposition Theorem
4.3 Thevenin’s Theorem
4.4 Norton’s Theorem
4.5 Maximum Power Transfer Theorem
4.6 Millman’s Theorem
4.7 Reciprocity Theorem
4.8 Tellegen’s Theorem
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
Experimental Exercises
5. ELECTROMAGNETISM
5.1 Introduction
5.2 Magnetic Field due to Electric Current
5.3 Force on Current-Carrying Conductor
5.4 Torque Experienced by a Coil
5.5 Force between Parallel Conductors
5.6 Electromagnetic Induction
5.7 Methods of Producing Induced EMF
5.8 Dynamically Induced EMF
5.9 Electromagnetic Induction and Lorentz Force
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
6. MAGNETIC CIRCUITS
6.1 Introduction
6.2 Magnetomotive Force (MMF)
6.3 Magnetic Circuit Theory
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
7. Self-And Mutual Inductances
7.1 Self-Inductance
7.2 Inductors
7.3 Mutual Inductance
7.4 Dot Convention
7.5 Coupled Coils in Series
7.6 Coupled Coils in Parallel
7.7 Energy Stored in Magnetic Field
7.8 Lifting Power of a Magnet
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
8. DC Transients
8.1 Introduction
8.2 The Simple RL Circuit
8.3 The Simple RC Circuit
8.4 DC-Excited Single-Capacitor RC and Single-Inductor RL Circuits
8.5 RC Timers
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
9. ALTERNATING VOLTAGE AND CURRENT
9.1 Introduction
9.2 Sinusoidal Functions—Terminology
9.3 Concept of Phasors
9.4 Algebraic Operations on Phasors
9.5 Mathematics of Complex Numbers
9.6 Power and Power Factor
9.7 Behaviour of R, L and C in AC Circuits
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
10. AC CIRCUITS
10.1 Series RL Circuit
10.2 Series RC Circuit
10.3 Complex Power
10.4 Parallel RL Circuit
10.5 Parallel RC Circuit
10.6 Series RLC Circuit
10.7 Parallel RLC Circuit
10.8 Network Theorems Applied to AC Circuits
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
Experimental Exercises
11. RESONANCE IN AC CIRCUITS
11.1 Introduction
11.2 Series Resonant Circuit
11.3 Different Aspects of Resonance
11.4 Resonance Curve
11.5 Parallel Resonant Circuit
11.6 Comparison between Series and Parallel Resonance
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
12.1 Introduction
12.2 Double-Subscript Notation
12.3 Concept of Three-Phase Voltages
12.4 Generation of Three-Phase Voltages
12.5 Three-Phase Loads
12.6 Star (Y)-Connected Three-Phase System
12.7 Delta (D)-Connected Three-Phase System
12.8 Voltages and Currents Relations in 3-f Systems
12.9 Power in Three-Phase System with a Balanced Load
12.10 Comparison between Two Three-Phase Systems
12.11 Measurement of Power
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
Experimental Exercise
13. TRANSFORMERS
13.1 Introduction
13.2 Principle of Operation
13.3 Ideal Transformer
13.4 Practical Transformer at No Load
13.5 Construction of Transformer
13.6 Transformer on Load
13.7 Practical Transformer on Load
13.8 Equivalent Circuit of a Transformer
13.9 Voltage Regulation of a Transformer
13.10 Efficiency of a Transformer
13.11 Autotransformers
13.12 Three-Phase Transformers
13.13 Some Special Transformers
13.14 Transformer Testing
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
Experimental Exercises
14. ALTERNATORS ANDSYNCHRONOUS MOTORS
14.1 Electromechanical Energy-Conversion Machines
14.2 Synchronous Machines
14.3 Construction of Alternators
14.4 Rotating Magnetic Flux Due to Three-Phase Currents
14.5 Armature Winding
14.6 EMF Equation
14.7 Armature Reaction
14.8 Equivalent Circuit of an Alternator
14.9 Power Delivered by an Alternator
14.10 Measurement of Synchronous Impedance
14.11 Synchronous Motors
14.12 Power Developed by a Synchronous Motor
14.13 Operation of a Synchronous Motor
14.14 Synchronous Condenser
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
15. INDUCTION MOTORS
15.1 Introduction
15.2 Principle of Working
15.3 Construction of Induction Motor
15.4 Rotor EMF, Current and Power Factor
15.5 Power Relations for an Induction Motor
15.6 Equivalent Circuit of an Induction Motor
15.7 Torque-Slip Characteristics
15.8 Starting of Induction Motors
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
Experimental Exercise
16. DC MACHINES
16.1 Importance
16.2 Construction of a DC Machine
16.3 Armature Current and Flux
16.4 Armature Winding
16.5 EMF Equation for a DC Generator
16.6 Types of DC Machines
16.7 Armature Reaction
16.8 Losses in a DC Machine
16.9 Efficiency of a DC Generator
16.10 Characteristics of DC Generators
16.11 DC Motors
16.12 Torque Developed by a DC Motor
16.13 Torque and Speed Characteristics of a DC Motor
16.14 Starting of DC Motors
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
Experimental Exercises
17. Fractional horse Power Motors
17.1 Introduction
17.2 Problem with Single-Phase Motor
17.3 Types of Single-Phase Motors
17.4 AC Series Motor
17.5 Universal Motor
17.6 Geneva Cam
17.7 Stepper Motors
17.8 Variable Reactance (VR) Stepper Motors
17.9 Permanent Magnet (PM) Stepper Motors
17.10 Hybrid Stepper Motors
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
18 Electrical Measuring Instruments
18.1 Classification of Instruments
18.2 Principle of Operation
18.3 Essentials of an Instrument
18.4 Moving Coil Instruments
18.5 Permanent Magnet Moving Coil (PMMC) Instruments
18.6 Dynamometer-Type Instruments
18.7 Moving-Iron Instruments
18.8 Ammeters and Voltmeters
18.9 Resistance Measurement
18.10 Meter Sensitivity (Ohms-per-volt Rating)
18.11 Multimeter
18.12 Measurement of Power
18.13 Measurement of Energy
Additional Solved Examples
Summary
Check Your Understanding
Review Questions
Multiple Choice Questions
Problems
19. ELECTRICAL INSTALLATION AND ILLUMINATION
19.1 Introduction
19.2 Distribution of Electrical Energy
19.3 Wires and Cables for Internal Wiring
19.4 Switches and Circuits
19.5 Electric Wiring Systems
19.6 Types of Wiring
19.7 Other Accessories
19.8 A Typical Control Circuit
19.9 Earthing of Installation
19.10 Safety Precautions in Handling Electrical Appliances
19.11 Testing of Electrical Installation
19.12 Incandescent or Filament Lamp
19.13 Fluorescent Tube
19.14 Compact Fluorescent Lamp (CFL)
19.15 Mercury Vapour Lamp
19.16 Sodium Vapour Lamp
19.17 Neon Lamp
Summary
Check Your Understanding
Review Questions
A Supplementary Exercises
B Supplementary Exercises
C Supplementary Exercises
D Supplementary Exercises
E Supplementary Exercises
Glossary
Index

Citation preview

Contents

Basic Electrical Engineering (Revised First Edition)

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ABOU T

TH E

AU T HO R

D C Kulshreshtha is working since July 2003 as Professor in Electronics and Communication Engineering, Jaypee University of Information Technology, Waknaghat, Distt. Solan (HP). He served for more than 25 years in Delhi College of Engineering, Delhi. From time to time, he has been a visiting faculty at Netaji Subhash Institute of Technolgy, New Delhi; Institute of Technolgy and Management, Gurgaon; Guru Tegh Bahadur College of Engineering, New Delhi; Maharaja Agrasen College of Engineering, Delhi; Galgotia College of Engineering and Technology, Greater Noida; and IEC College of Engineering and Technology, Greater Noida. He earned his BTech (Honours) in Electronics and Electrical Communication Engineering from IIT Kharagpur in 1967, and a masters degree in Advanced Electronics from Delhi College of Engineering, Delhi in 1980. For three years, he was at IIT Delhi, pursuing research in Digital Signal Processing. He secured 2nd rank in the All India Merit in Engineering Services Examination conducted by the UPSC, he was in-charge of Frequency Section, responsible for the management of electromagnetic spectrum on Station, Srinagar (J & K), and was responsible for installation of equipments and antenna system for the monitoring station. He has conducted or attended short-term refresher courses in Microelectronics at IIT Kharagpur; Measurement Techniques at University of Roorkee; Modern Process Control Techniques at DCE, Delhi; Digital Signal Processing at REC, Warangal; and Microwave Techniques and Systems at REC, Surathkal. He has authored/co-authored many excellent textbooks, namely, Electronic Devices, Applications and Integrated Circuits (1980), Basic Electronics and Linear Circuits (1984), Engineering Network Analysis (1989), Elements of Electronics and Instrumentation (1992), Electronic Devices and Circuits (2005), and Electronics Engineering (for UPTU) (2006).

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Basic Electrical Engineering (Revised First Edition)

D C KULSHRESHTHA Professor of Electronics and Communication Engineering Jaypee University of Information Technology Waknaghat, Distt. Solan (Himachal Pradesh)

Tata McGraw Hill Education Private Limited NEW DELHI 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

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Tata McGraw-Hill Published by the Tata McGraw Hill Education Private Limited, 7 West Patel Nagar, New Delhi 110 008. Basic Electrical Engineering (Revised First Addition) Copyright © 2012, 2009 by Tata McGraw Hill Education 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, Tata McGraw Hill Education Private Limited. ISBN (13 digit): 978-0-07-132896-8 ISBN (10 digit): 0-07-132896-3 Ajay Shukla Head—Higher Education Publishing and Marketing: Vibha Mahajan Publishing Manager—(SEM & Tech Ed.): Shalini Jha Asst. Sponsoring Editor: Tina Jajoriya Sr Copy Editor: Nimisha Kapoor Sr Production Manager: Satinder S Baveja Asst Manager—Production: Anjali Razdan Marketing Manager—Higher Education: Vijay Sarathi Sr Product Specialist—(SEM & Tech Ed.): John Mathews General Manager—Production: Rajender P Ghansela Asst General Manager—Production: B L Dogra Information contained in this work has been obtained by Tata McGraw Hill, from sources believed to be reliable. However, neither Tata McGraw Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither Tata McGraw Hill nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that Tata McGraw Hill and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Tej Composers, WZ 391, Madipur, New Delhi 110 063 and printed at Avon Printers, Plot No. 16, Main Loni Cover Printer: A. P. Offset

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Dedicated to

My Dear Wife Aruna

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FOREWORD

The present textbook on “Basic Electrical Engineering” is yet another contribution by Prof. D C Kulshreshtha. This textbook is based on teacher-taught-interaction pattern. The effort is praiseworthy. Having a long experience of teaching and authoring books, the author Writing a textbook is indeed a very challenging task. Besides possessing the expertise in the subject, one must have the talent of presenting the text in such a way that the student can understand it easily. The author has fully succeeded in achieving this goal. The most important aspect of this book is the perfect manuscript that has evolved after testing it in class room teaching for two years in Jaypee University and its constituent institutions. The appreciation tions of the experts during the review of this book at the pre-publishing stage helped in further enhancing the usefulness of the book. I highly admire the author for this contribution through an excellent textbook. I am sure that the teachers and students of various engineering colleges would approve this book as a basic textbook for teaching and learning of the subject.

Prof. D S Chauhan Vice Chancellor

PREFACE THE NEED FOR THIS BOOK Since the beginning of my career as a teacher, I have been teaching Basic Electrical Engineering—a core course usually taught in 1st or 2nd understanding the available textbooks, and therefore resorted to memorising the statements and formulae of the basic principles. It forced my teacher’s conscience to prepare a textbook that presented the important ideas in depth and left many of the details for future learning in further study. So, I authored this book emphasising important principles as the foundation of electrical engineering. The main goal of this book is to:

The book is a culmination of various stages such as drafting, testing and feedback on prepared material by students’ editorial board.

PREREQUISITES This book is meant to be used by students who have just passed 10+2, who lack the knowledge of higher mathematics. It uses derivatives at some places and integrals scantly. Simple algebra, vector algebra, complex-number algebra and matrices are extensively used with detailed explanations. Thus, it will serve the purpose of a textbook to the undergraduate students of all engineering disciplines and diploma students of Electrical Engineering and Electronics Engineering streams.

PEDAGOGY OF THE BOOK

examples, and many believe that only the equations are important. When the teacher probes their understanding of the in the class. The goals and needs of both the teacher and the students are met in the design and the pedagogy of this book. The principles and applications of the subject are clearly and concisely presented using a step-by-step approach.

STRUCTURE OF THE BOOK The book, spanning over 19 chapters, has been structured to cover all important topics required by the syllabus in a single volume. It begins with an introductory chapter, and ends with a chapter on electrical wiring and illumination needed by a practicing engineer. Chapters 2 to 4 explain the basic ideas, principles, circuit analysis techniques order transients in RC and RL dc circuits. Chapters 9 to 12 are on single-phase and three-phase ac circuits. Chapters 13 to 17 cover electrical machines. Lastly, Chapter 18 gives basic information of electrical measuring instruments.

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AIDS TO LEARNING This textbook addresses the viewpoints of both students and teachers in a number of ways: Objectives and Summary Each chapter begins with the learning objectives and ends with a summary giving Key Terms and Concepts, and Important Formulae. Causality In complicated situations, the cause-effect relationships are given. Understanding consists largely in knowing the causal connections between various factors in a problem so that the equations are written with a purpose. Students often have trouble solving problems because they memorise equations without clearly understanding what the variables mean. Major Equations Important major equations are boxed with a thick border, with a view to highlight them. This format, however, is sparingly used lest the students think that electrical engineering consists of a set of equations to be memorised. Key and Important Terms Key terms terms are italicised wherever they appear. Solved Examples/Problems The book has 528 examples/problems, solved step-by-step clearly bringing out the cause-effect relationships. Additional Solved Examples integrations of subject areas. Check Your Understanding Students can check their understanding of the principles studied in a chapter by answering 10 True/False (objective) questions given at the end of each chapter. Review Questions These questions allow students to review the key concepts and assess their understanding. MCQs by GRE, UPSC, NTPC, ONGC, Infosys, Accenture, etc. Practice Problems The book provides 840 numerical problems for the practice of the students in the category of in their attempts. Most of the problems have been taken from examination papers of different universities. Experimental Exercises Experimental Exercises aid students to perform experiments in the laboratory in a systematic way. Rich pool of pedagogy of more than 2000 solved problems and students’ practice problems needed by practicing engineers. Solved Examples/Problems: Multiple Choice Questions: Experimental Exercises:

528 244 16

Review Questions: Students’ Practice Problems: True/False Questions:

268 840 180

ONLINE LEARNING CENTRE This book is accompanied by a comprehensive website—http://www.mhhe.com/kulshreshtha/bee—designed to provide valuable resources for students, instructors and professionals. Students can access a sample chapter, 402 additional Multhe Publisher for the Solution Manual of numerical problems. Supplementary teaching materials include chapter-wise PowerPoint slides with animations for effective lecture presentations, and an exhaustive test bank.

SUPPLEMENTARY EXERCISES For a thorough understanding of electrical principles, one needs to work out an exhaustive number of numerical problems. Supplementary Exercises—in the category of Solved Problems and Students’ Practice Problems—is an additional

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feature of the book. These sets of exercises are meant to provide a standard methodology for sincerely learning the basic principles of Electrical Engineering. The questions and problems in this set of exercises are categorised into parts : I. Part A—DC Circuits : Assemblage of Ch. 2 : Ohm’s Law; Ch. 3 : Network Analysis; and Ch. 4 : Network Theorems. II. Part B—Electromagnetic Circuits : Assemblage of Ch. 5 : Electromagnetism; Ch. 6 : Magnetic Circuits; and Ch.7 : Self- and Mutual Inductances. III. Part C—AC Circuits : Assemblage of Ch. 9 : Alternating Voltage and Current; Ch. 10 : AC Circuits; Ch. 11 : Resonance in AC Circuits; and Ch. 12 : Three-Phase Circuits and Systems. IV. Part D—Electrical Machines : Assemblage of Ch. 13 : Transformers; Ch. 14 : Alternators and Synchronous Motors; Ch. 15 : Induction Motors; Ch. 16 : DC Machines; and Ch. 17 : Fractional Horse Power Motors. V. Part E—Miscellaneous : Assemblage of Ch. 8 : DC Transients; Ch. 18 : Electrical Measuring Instruments; and Ch. 19 : Electrical Installation and Illumination.

ACKNOWLEDGMENTS I gratefully acknowledge the assistance and guidance provided by my colleagues, Dr D S Chauhan, V C and Dr T S Lamba, Dean (Academic and Research) at JUIT, and Dr Krishna Kant at MNIT Allahabad. I highly appreciate the unstinted support given by Mr Hardeep Singh Rana, Computer Lab, and Mr Nitin Paliwal, Learning Resource Centre at JUIT, in the preparation of the manuscript on the computer. My sincere thanks are due to the Students’ Editorial Board for going through the manuscript to detect any slips, checking the solutions to the problems, and putting them into three categories—(i) Simple, (ii) Tricky, and (iii) Challenging. Avinash Kumar (051028) Pradeep Aluru (061148) Mohit Sain (071017) Uday Rangta (071342) Krishna Kumar (071143) Shubhrangshu Naval (071033) Monish Kaul (071250) Anurag Chaurasia (071337) Abhay Malhotra (071328) Mohak Narang (071094) Tanu Kundra (071028) Nitin Agarwal (071021) Sanya Bakshi (071582) Shrraddha (071044) Dhruv Garg (071006) Ravish Jain (051042) Prashant Navin Gupta (071070)

Pratik Golchha (061099) Daksh Arora (061041) Jasdeep Singh Bhatia (071043) Amit Singh (071343) Jayant Rajpurohit (061056) Manish Kumar (071074) Amar Pandey (071247) Utkarsh (071018) Mayank Agarwal (061072) Apoorva Ajmani (071088) Swati Sharma (071564) Swati Sharma (071222) Sweta Bansal (071119) Ankush Bansal (071219) Prateek Pandey (061098) Ketan Gulati (071275)

My special thanks are due to my favourite student Mr Gaurav Singla (081620) who has done the arduous and painstaking task of thoroughly checking and correcting the Solutions to Chapter-end Problems and Supplementary Exercises. book, which meets the requirements of two core courses of not only this university, but also many others.

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A note of acknowledgement is due to the following reviewers of the book for their valuable suggestions. Yogesh Pahariya Institute of Engineering and Science, IPS Academy, Indore, Madhya Pradesh Vinay Pathak Bhopal Institute of Technology (BIT), Bhopal, Madhya Pradesh A Mouleeshwaran Maharaja Engineering College, Coimbatore, Tamil Nadu R V S Lakshmi Kumar Vishaka Institute of Engineering and Technology, Vishakapatnam, Andhra Pradesh Leena G Jeevan Career Institute of Technology and Management, Faridabad, Haryana K P Singh M M M Engineering College, Gorakhpur, Uttar Pradesh A Lakshmi Devi Sri Venkateswara University College of Engineering, Tirupati, Andhra Pradesh K Lakshmi K S R College of Technology, Tiruchengode, Tamil Nadu M Rizwan Apostole Institute of Technology, Greater Noida, Uttar Pradesh K Manivannan Pondicherry Engineering College, Pondicherry S C Gupta Institute of Technology, BHU, Varanasi, Uttar Pradesh R Senthil Kumar Bannari Amman Institute of Technology, Sathyamangalam, Tamil Nadu K Siva Sankar Reddy K S R M College Of Engineering, Kadapa, Andhra Pradesh P Sujatha JNTU College of Engineering, Anantapur, Andhra Pradesh Bharti Dwivedi Institute of Engineering and Technology, Lucknow, Uttar Pradesh Despite the best efforts put in by me and my team, it is possible that some unintentional error might have eluded us. I shall acknowledge with gratitude if any of these is pointed out. Any suggestions from the readers for improvement in [email protected].

D C Kulshreshtha Publisher’s Note Tata McGraw Hill Education looks forward to receiving from teachers and students their valuable views, comments and suggestions for improvements, all of which may be sent to [email protected], mentioning the title and author’s name as the subject.

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CONTENTS About the Author Foreword Preface Walkthrough Scientific Calculator Standard Symbols Notations 1. INTRODUCTION 1.1 Charge 1 1.2 Voltage or Potential 4 1.3 Basic Circuit 6 1.4 Units and Symbols 7 Rules to Use SI Prefixes 8 Guidelines for Using SI Units 8 1.5 Basics of Experimentation 10 General Instructions and Precautions 10 2. OHM’S LAW 2.1 Ohm’s Law 14 2.2 Resistance 15 2.3 Duality 28 2.4 Conductors and Resistors 28 2.5 Variation of Resistance with Temperature Additional Solved Examples 32 Summary 39 Check Your Understanding 40 Review Questions 41 Multiple Choice Questions 41 Problems 42 Experimental Exercise 47 3. NETWORK ANALYSIS 3.1 Network Components 50 3.2 Series and Parallel Combinations 54 3.3 Energy Sources 57 3.4 Combination of Sources 62 3.5 Types of Sources 66 3.6 Kirchhoff’s Laws 67 3.7 Loop-Current Analysis 72 3.8 Mesh Analysis 75

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3.9 Node-Voltage Analysis 80 3.10 Nodal Analysis 84 3.11 Choice of Method of Analysis 85 Additional Solved Examples 85 Summary 93 Check Your Understanding 94 Review Questions 95 Multiple Choice Questions 95 Problems 96 Experimental Exercises 100 4. NETWORK THEOREMS 4.1 Introduction 104 4.2 Superposition Theorem 104 4.3 Thevenin’s Theorem 108 4.4 Norton’s Theorem 112 4.5 Maximum Power Transfer Theorem 113 4.6 Millman’s Theorem 116 4.7 Reciprocity Theorem 116 4.8 Tellegen’s Theorem 117 Additional Solved Examples 119 Summary 128 Check Your Understanding 129 Review Questions 130 Multiple Choice Questions 130 Problems 131 Experimental Exercises 134 5. ELECTROMAGNETISM 5.1 Introduction 141 5.2 Magnetic Field due to Electric Current 141 5.3 Force on Current-Carrying Conductor 144 5.4 Torque Experienced by a Coil 146 5.5 Force between Parallel Conductors 147 5.6 Electromagnetic Induction 148 5.7 Methods of Producing Induced EMF 150 5.8 Dynamically Induced EMF 150 5.9 Electromagnetic Induction and Lorentz Force 153 Additional Solved Examples 156 Summary 159 Check Your Understanding 160 Review Questions 161 Multiple Choice Questions 161 Problems 163 6. MAGNETIC CIRCUITS 6.1 Introduction 164 6.2 Magnetomotive Force (MMF) 164 6.3 Magnetic Circuit Theory 166 Additional Solved Examples 173 Summary 178 Check Your Understanding 179

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Contents Review Questions 180 Multiple Choice Questions Problems 180 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

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Self-Inductance 183 Inductors 186 Mutual Inductance 186 Dot Convention 189 Coupled Coils in Series 191 Coupled Coils in Parallel 193 Energy Stored in Magnetic Field Lifting Power of a Magnet 196 Additional Solved Examples 196 Summary 199 Check Your Understanding 200 Review Questions 201 Multiple Choice Questions 201 Problems 202

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Introduction 204 The Simple RL Circuit 205 The Simple RC Circuit 212 DC-Excited Single-Capacitor RC and Single-Inductor RL Circuits 216 RC Timers 219 Additional Solved Examples 220 Summary 228 Check Your Understanding 229 Review Questions 229 Multiple Choice Questions 230 Problems 231 9. ALTERNATING VOLTAGE AND CURRENT 9.1 Introduction 233 9.2 Sinusoidal Functions—Terminology 234 9.3 Concept of Phasors 242 9.4 Algebraic Operations on Phasors 244 9.5 Mathematics of Complex Numbers 246 9.6 Power and Power Factor 254 9.7 Behaviour of R, L and C in AC Circuits 256 Additional Solved Examples 262 Summary 266 Check Your Understanding 268 Review Questions 268 Multiple Choice Questions 269 Problems 269 10. AC CIRCUITS 10.1 Series RL Circuit 272 10.2 Series RC Circuit 275 10.3 Complex Power 276 10.4 Parallel RL Circuit 278 8.1 8.2 8.3 8.4 8.5

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Parallel RC Circuit 278 Series RLC Circuit 280 Parallel RLC Circuit 282 Network Theorems Applied to AC Circuits 283 Additional Solved Examples 286 Summary 294 Check Your Understanding 295 Review Questions 296 Multiple Choice Questions 297 Problems 298 Experimental Exercises 300 11. RESONANCE IN AC CIRCUITS 11.1 Introduction 310 11.2 Series Resonant Circuit 310 11.3 Different Aspects of Resonance 312 11.4 Resonance Curve 318 11.5 Parallel Resonant Circuit 320 11.6 Comparison between Series and Parallel Resonance Additional Solved Examples 326 Summary 329 Check Your Understanding 331 Review Questions 331 Multiple Choice Questions 332 Problems 333 10.5 10.6 10.7 10.8

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Introduction 335 Double-Subscript Notation 336 Concept of Three-Phase Voltages 337 Generation of Three-Phase Voltages 338 Three-Phase Loads 339 Star (Y)-Connected Three-Phase System 340 Delta (D)-Connected Three-Phase System 342 Voltages and Currents Relations in 3-f Systems 343 Power in Three-Phase System with a Balanced Load 350 Comparison between Two Three-Phase Systems 351 Measurement of Power 352 Additional Solved Examples 356 Summary 364 Check Your Understanding 365 Review Questions 365 Multiple Choice Questions 366 Problems 367 Experimental Exercise 369 13. TRANSFORMERS 13.1 Introduction 372 13.2 Principle of Operation 373 13.3 Ideal Transformer 374 13.4 Practical Transformer at No Load 379 13.5 Construction of Transformer 383 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9 12.10 12.11

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Contents Transformer on Load 384 Practical Transformer on Load 386 Equivalent Circuit of a Transformer 387 Voltage Regulation of a Transformer 390 Efficiency of a Transformer 394 Autotransformers 397 Three-Phase Transformers 400 Some Special Transformers 401 Transformer Testing 402 Additional Solved Examples 404 Summary 410 Check Your Understanding 411 Review Questions 412 Multiple Choice Questions 412 Problems 414 Experimental Exercises 417 14. ALTERNATORS ANDSYNCHRONOUS MOTORS 14.1 Electromechanical Energy-Conversion Machines 425 14.2 Synchronous Machines 431 14.3 Construction of Alternators 433 14.4 Rotating Magnetic Flux Due to Three-Phase Currents 436 14.5 Armature Winding 438 14.6 EMF Equation 440 14.7 Armature Reaction 442 14.8 Equivalent Circuit of an Alternator 444 14.9 Power Delivered by an Alternator 448 14.10 Measurement of Synchronous Impedance 449 14.11 Synchronous Motors 451 14.12 Power Developed by a Synchronous Motor 454 14.13 Operation of a Synchronous Motor 456 14.14 Synchronous Condenser 458 Additional Solved Examples 459 Summary 469 Check Your Understanding 470 Review Questions 471 Multiple Choice Questions 472 Problems 474 15. INDUCTION MOTORS 15.1 Introduction 476 15.2 Principle of Working 476 15.3 Construction of Induction Motor 479 15.4 Rotor EMF, Current and Power Factor 481 15.5 Power Relations for an Induction Motor 484 15.6 Equivalent Circuit of an Induction Motor 486 15.7 Torque-Slip Characteristics 491 15.8 Starting of Induction Motors 496 Additional Solved Examples 498 Summary 502 Check Your Understanding 503

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13.6 13.7 13.8 13.9 13.10 13.11 13.12 13.13 13.14

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Review Questions 504 Multiple Choice Questions 505 Problems 506 Experimental Exercise 508 16. DC MACHINES 16.1 Importance 513 16.2 Construction of a DC Machine 514 16.3 Armature Current and Flux 516 16.4 Armature Winding 516 16.5 EMF Equation for a DC Generator 519 16.6 Types of DC Machines 520 16.7 Armature Reaction 526 16.8 Losses in a DC Machine 527 16.9 Efficiency of a DC Generator 528 16.10 Characteristics of DC Generators 529 16.11 DC Motors 535 16.12 Torque Developed by a DC Motor 537 16.13 Torque and Speed Characteristics of a DC Motor 16.14 Starting of DC Motors 542 Additional Solved Examples 543 Summary 549 Check Your Understanding 550 Review Questions 551 Multiple Choice Questions 551 Problems 553 Experimental Exercises 555 17.1 17.2 17.3 17.4 17.5 17.6 17.7 17.8 17.9 17.10

18.1 18.2 18.3 18.4 18.5 18.6

513

538

Introduction 561 Problem with Single-Phase Motor 562 Types of Single-Phase Motors 566 AC Series Motor 571 Universal Motor 572 Geneva Cam 573 Stepper Motors 573 Variable Reactance (VR) Stepper Motors 574 Permanent Magnet (PM) Stepper Motors 579 Hybrid Stepper Motors 582 Summary 584 Check Your Understanding 585 Review Questions 585 Multiple Choice Questions 586 Problems 588 Classification of Instruments 589 Principle of Operation 590 Essentials of an Instrument 591 Moving Coil Instruments 596 Permanent Magnet Moving Coil (PMMC) Instruments 596 Dynamometer-Type Instruments 599

Contents 18.7 18.8 18.9 18.10 18.11 18.12 18.13

Moving-Iron Instruments 600 Ammeters and Voltmeters 602 Resistance Measurement 610 Meter Sensitivity (Ohms-per-volt Rating) Multimeter 613 Measurement of Power 613 Measurement of Energy 616 Additional Solved Examples 617 Summary 619 Check Your Understanding 620 Review Questions 621 Multiple Choice Questions 622 Problems 623

xvii

612

19. ELECTRICAL INSTALLATION AND ILLUMINATION 19.1 Introduction 624 19.2 Distribution of Electrical Energy 625 19.3 Wires and Cables for Internal Wiring 626 19.4 Switches and Circuits 627 19.5 Electric Wiring Systems 631 19.6 Types of Wiring 631 19.7 Other Accessories 634 19.8 A Typical Control Circuit 636 19.9 Earthing of Installation 637 19.10 Safety Precautions in Handling Electrical Appliances 638 19.11 Testing of Electrical Installation 639 19.12 Incandescent or Filament Lamp 641 19.13 Fluorescent Tube 643 19.14 Compact Fluorescent Lamp (CFL) 644 19.15 Mercury Vapour Lamp 645 19.16 Sodium Vapour Lamp 646 19.17 Neon Lamp 647 Summary 648 Check Your Understanding 648 Review Questions 649 A Supplementary Exercises B Supplementary Exercises C Supplementary Exercises D Supplementary Exercises E Supplementary Exercises Glossary Index

624

651 695 719 769 813 829 845

WALKTHROUGH OBJECTIVES

O B J E C T I V E S A er completing this Chapter, you will be able to :

Chapter objectives provide a concise statement of expected learning outcomes.

(EMEC) machines.

electromechanical system.

the stator in a synchronous machine.

ADDITIONAL SOLVED EXAMPLES EX AMP LE

5.9

A horizontal overhead line carries a current of 90 A in east-to-west direction. What is the magnitude and direction of the magnetic eld due to this current at a point 1.5 m below the line ?

(Xs)

(Zs).

Solution The magnitude of the magnetic eld is given by Eq. 5.2, as μ0 I 4 π × 10 − 7 × 90 = 12 ¥ 10 –6 T = 12 mT = 2π x 2 π × 1.5 By applying the right-hand thumb rule, we nd that the direction of the magnetic eld is from north-to-south. B=

EX AMP LE

5.1 0

What is the magnitude of the magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of uniform magnetic eld of 0.15 T ?

Solution Using Eq. 5.8, the force per unit length of the wire is given as Fu = EX AMP LE

F = I B sin q = 8 ¥ 0.15 ¥ sin 30° = 0.6 N m –1 l

5.1 1

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of uniform horizontal magnetic eld of magnitude 0.8 T. What is the magnitude of the torque experienced by the coil ?

S O LV E D E X A M P L E S Provided at appropriate locations, solved examples aid in learning the technique of applying concepts to practical problems.

Solution As shown in Fig. 5.17, the coil MNOP is kept such that the normal to its plane makes an angle of 30° with the uniform magnetic eld. The magnetic force F experienced by each of the sides MN and OP is given by Eq. 5.7, as F = IBln = 12 ¥ 0.8 ¥ 0.1 ¥ 20 = 19.2 N The normal distance between these two forces is x = l sin 30° = 0.1 ¥ 0.5 = 0.05 m Therefore, the torque experienced by the coil is t = Fx = 19.2 ¥ 0.05 = 0.96 Nm

IR

VRY

I1

I1 30° –I3

11°34¢ IB

138°26¢ 90° I2

71°34¢

30°

VYB

VBR

I3

I3 I2

11°34 IY

(a)

I2 (b)

(c)

FIGURES Figures are used exhaustively to illustrate concepts and methods described in the text.

Contents

xix

MULTIPLE CHOICE QUESTIONS Here are some incomplete statements. Four alternatives are provided below each. Tick the alternative that completes the statement correctly :

M U LT I P L E C H O I C E QUESTIONS Chapter-end MCQs help the students in clarifying concepts. Also, a useful tool to prepare for various competitive examinations.

. The speci c resistance r depends upon (a) the material, the area of cross-section and the length of the conductor (b) the area of cross-section and the length of the conductor only (c) the area of cross-section of the conductor only (d) the nature of the material of the conductor only 2. The resistance of a conductor increases when (a) its length increases (b) its area increases (c) both its length and area increase (d) its resistivity is kept constant 3. On increasing its temperature, the resistance of a conductor made of a metal (a) decreases (b) increases (c) remains constant (d) varies either way

5. The ‘ampere second’ could be the unit of (a) conductance (b) power (c) energy (d) charge 6. The polarity of voltage drop across a resistor is determined by (a) the value of the resistor (b) the value of current through the resistor (c) the direction of current through the resistor (d) the polarity of the source 7. If 110 V is applied across a 220-V, 100-W bulb, the power consumed by it will be (a) 100 W (b) 50 W (c) 25 W (d) 12.5 W 8. A resistance of 10 W is connected across a supply of 200 V. When another resistance of R ohms is connected in parallel with the above 10-W resistor, the current drawn from the supply doubles. The value of R is (a) 5 W (b) 10 W (c) 20 W (d) 40 W 9. Three resistances each of R ohms are connected in star. Its equivalent delta will comprise three

REVIEW QUESTIONS 1. What is the magnetic eld pattern due to a long current-carrying straight conductor ? 2. When a current-carrying conductor is placed in a magnetic eld, it experiences a force. How do you nd the magnitude and direction of this force ? 3. State and explain (a) Fleming’s left-hand rule, and (b) Fleming’s right-hand rule. 4. State how you will determine the nature of force between two parallel current-carrying conductors. 5. Sketch the magnetic eld around two adjacent parallel current-carrying conductors, when the currents owing through them are (a) in opposite directions, and (b) in the same direction. 6. Explain how the unit of current is de ned. 7. What is a solenoid ? Write the expression for the magnetic eld at a point (a) inside the solenoid, and (b) just at one of its ends. 8. As you move away from the middle of a solenoid, why does the magnetic eld decrease ? 9. What is the difference between a solenoid and a toroid ? 10. Suppose that you are sitting in a room with your back to the wall. Imagine that an electron beam travelling horizontally from the back wall to the

direction of the magnetic eld that exists in the room ? 11. What is meant by electromagnetic induction ? State and explain Faraday’s laws of electromagnetic induction. 12. State Lenz’s law. Show, by means of an example, that the Lenz’s law and Fleming’s right-hand rule give the same direction of induced emf in a circuit. 13. Show that Lenz’s law is a consequence of the principle of conservation of energy. 14. One end of a bar magnet is thrust into a coil. It is noted that the induced current in the coil is in clockwise direction as viewed from the front end. Is the end of the bar magnet its N-pole or S-pole ? 15. A metallic loop is placed in a nonuniform magnetic eld. Will an emf be induced in the loop ? 16. Two circular loops are placed coaxially but separated by a distance. A battery is suddenly connected to one of the loops establishing a current in it. Will there be a current induced in the other loop ? Do the loops attract or repel each other ? 17. The battery in the above question is suddenly disconnected. Is a current induced in the other

REVIEW QUESTIONS Review questions at the end of each chapter are meant to give good practice for answering theoretical questions in examinations.

PROBLEMS (A)

PROBLEMS WITH ANSWERS Problems given at the end of each chapter are divided, based on the difficulty level, into three categories—(A) Simple, (B) Tricky, and (C) Challenging with answers.

SIMPLE

P ROBLEM S

1. If the moving coil of an electric meter consists of 150 turns wound on a square former which has a length of 4 cm and the ux density in the air gap is 0.06 T, calculate the turning moment acting on the coil when it is carrying a current of 12 mA. [Ans. 172.8 ¥ 10 –6 Nm] 2. A moving coil instrument gives a full-scale de ection of 10 mA when the potential difference across its terminals is 100 mV. Calculate the series resistance to measure 1000 V on full scale. [Ans. 99 990 W]

(B)

TRIC KY

3. A moving coil instrument gives a full-scale de ection of 10 mA when a potential difference of 10 mV is applied across its terminals. Show how will you use the instrument to measure (i) currents up to 100 A, and (ii) voltage up to 500 V. [Ans. (i) 0.000 100 01 W; (ii) 49 999 W] 4. A moving coil ammeter has a resistance of 0.01 W and full scale de ection current of 0.25 A. How this meter can be made to read (i) voltage up to 250 V, and (ii) current up to 20 A ? [Ans. (i) 999.99 W; (ii) 1.2658 ¥ 10–4 W]

PR OBLEMS

5. The coil of a moving-iron instrument has a resistance of 500 W and an inductance of 1 H. An additional resistance of 2000 W is connected in series with the

instrument to make it a voltmeter. It reads 250 V when a dc voltage of 250 V is applied. What will it read when 250-V, 50-Hz is applied ? [Ans. 248 V]

EXPERIMENTAL EXERCISE 10.2 PARALLEL

A C

CIRCU IT

Objectives 1. To observe the variation of current I when the resistance in the parallel ac circuit is varied. 2. To draw the phasor diagram for the parallel ac circuit for four sets of impedances obtained by varying the resistance. 3. To calculate the circuit parameters (R, L and C) for the four sets of observations, assuming the ac supply frequency to be 50 Hz. 4. To compare the two values of power-factors—one obtained from the readings and the other obtained from the phasor diagram—for the four sets of observations.

Apparatus Single-phase ac supply; One Variac 0-250 V, 5 A; One choke coil with negligible resistance; One wattmeter, Three ammeters (MI type) 0-5 A; Three voltmeters (MI Type) 0-300 V; One rheostat 100 W, 5 A.

Circuit Diagram The circuit diagram is shown in Fig. 10.21.

E X P E R I M E N TA L EXERCISES Experimental exercies given at the end of relevant chapters help students to perform laboratory experiments in a systematic way.

CHE CK

YOU R

U N D E RS TA N D IN G

Before you proceed to the next Chapter, take this Test. Give yourself two marks for each correct answer and minus one for each wrong answer. If your score is 12 or more, go to the next Chapter; otherwise study this Chapter again. No.

Statement

True

False

Marks

1.

CHECK YOUR U N D E R S TA N D I N G

series resonant circuit

A set of 10 objective questions enable students gauge their mastery of chapter content. Z

SUMMARY TERMS

SUMMARY

AND

Self inductance of a coil arises when an emf is induced in itself by changing the current owing through it. Isolated inductances in series and parallel are added the same way as resistances. An inductor may be xed or variable, iron-cored or air-cored. When a magnetic eld is set up by an inductor, it stores energy. Mutual inductance is the property of two magnetically coupled coils because of which there is an induced emf in one coil due to change in current in the other coil. Coef cient of coupling (k) is the ratio of ux linkage between primary and secondary coils to the ux produced by primary current. The mutual voltage is present independently of and in addition to any voltage due to self-induction. Both L and M are measured in henrys (H). IMP ORTAN T

The chapter-end summary divided into: (1) Terms and Concepts and (2) Important Formulae can be used by students for a quick review during examinations.

C ONCEP TS

FOR MULAE

di . dt 1 2 Energy stored, W = LI . 2

Self-induced emf, e = L

EMF induced by mutual inductance, e2 = M

Coupling coef cient, k =

There is an important caution about the use of current divider. Many students apply this concept where it is not applicable. For example, consider the circuit of Fig. 2.12c. The circuit contains no resistances in series or in parallel. can be applied. However, many students take a quick look at resistances RA and RB and try to apply current division, writing an incorrect equation such as RB IA = s RA RB Remember, parallel resistances must be connected between the same pair of points (called nodes).

di1 . dt

N2 μ A L= . l d Φ 21 . M21 = N1 di 2 M ; 0 £ k £ 1. L1 L2

C O M M O N M I S TA K E S Wrong concepts or wrong solutions to problems are deliberately given at many places to emphasise the mistakes commonly committed by students.

*MNEMONICS

AID TO MEMORY/ MNEMONICS By giving examples drawn from day-today experience, aids to memory remove any confusion from the minds of the students.

As an aid to remember the sequence of colour codes given above, the student can memorise one of the following (all the capital letters stand for colours): (a) Bill Brown Realised Only Yesterday Good Boys Value Good Work (b) Bye Bye Rosie Off You Go Bristrol Via Great Western (c) B B Roy of Great Britain had a Very Good Wife.

SCIENTIFIC CALCULATOR The author strongly advises using an quires no programming. Such a calculator is capable of :

(such as CASIO fx-991ES), which re-

Directly solving quadratic and simultaneous equations (up to three variables), just by keying in the constants of the equation. Making calculations with complex numbers (even in mixed mode—polar and Cartesian), as is often required for the analysis of ac circuits errors. A word of caution about the use of a calculator is called for. While using a calculator, the student is tempted to show a false accuracy in the results obtained. Remember that the result of a calculation cannot produce a value of greater accuracy than the accuracy of the information used to formulate the calculation.

Note: In this book, we have used the calculator CASIO fx-991ES for solving equations and making other calculations. Therefore, often the intermediate steps of calculations have been omitted.

STANDARD SYMBOLS

Contents

xxiii

NOTATIONS A A, B, C … a, b, c, … B

Number of parallel paths Points on a curve, Nodes in a network Susceptance

Bm BW b E Eb Eg h f0 f1, f2 F Fu F G H i I, Irms or Ieff Ia I av/V av If Ifsd IL IL ILm

Bandwidth Step angle Emf Back emf Emf generated Resonant frequency Lower and upper cut-off frequencies Force Force per unit area MMF Conductance Instantaneous current RMS or effective value of current Armature current Average value of current/voltage Field winding current Current through an inductor Load current, Line current Peak value of current through an inductance

Im Im/Vm IN Iph IR IRm Isc Ise Ish IR IL IC kVA K Kf Kp kd kp Lsa Lso Lpa Lpo mr m0

Meter current Peak value of current/voltage Norton’s current Phase current Current through a resistance Peak value of current through a resistance Short-circuit current

Phasor current through resistance Phasor current through inductance Phasor current through capacitance Kilovolt-ampere Turns ratio, Transformation ratio Form factor Peak factor Distribution factor Pitch factor Two inductances in series aiding Two inductances in series opposing Two inductances in parallel aiding Two inductances in parallel opposing Relative permeability Permeability of free space or air

xxiv ns Ns pf P Pc Pd Pe Pg Ph Pi Pin P1 Po PR PS f f F Fm Q Re1 Re2 Req RL RL Rm Rp Rs Rse Rsh RTh R s S S T t

Contents Notations

Synchronous speed in rps Synchronous speed in rpm Power factor Number of poles Copper loss Power developed Eddy-current loss Air gap power Hysteresis loss Iron loss Input power Total loss Output power Rotor copper and iron losses Stator copper and iron losses Phase Angle Flux

Equivalent resistance referred to primary Equivalent resistance referred to secondary Equivalent resistance Resistance of an inductor Load resistance Meter resistance Resistance of parallel combination Resistance of series combination

Thevenin’s resistance Reluctance Slip Apparent power Complex power Number of turns Torque

tm tst qe qm V, Vrms or Veff VA VAR VR VC VC Vfsd VL VL VL VLm Voc Vph VR VR VRm VTh XC XL Xe1 Xe2 Xs YC YL Z Zeq ZTh Zs

Maximum torque Starting torque Electrical angle Mechanical angle RMS of effective value of voltage Volt ampere Volt ampere reactive Voltage regulation Voltage across a capacitance Phasor voltage across capacitance Voltage across an inductor Phasor voltage across inductance Load voltage, Line voltage Peak value of voltage across an inductance Open-circuit voltage Phase voltage Voltage across a resistance Phasor voltage across resistance Peak value of voltage across a resistance Thevenin’s voltage Reactance of a capacitor Reactance of an inductor Equivalent reactance referred to primary Equivalent reactance referred to secondary Synchronous reactance Phasor admittance of an capacitive branch Phasor admittance of an inductive branch Total number of conductors Equivalent impedance Thevenin’s impedance Synchronous impedance

INTRODUCTION

1

OBJECT IVES

All materials are made up of one or more elements. An element is a substance composed entirely of atoms of the same kind. An atom consists of a massive core (nucleus) carrying protons and neutrons. Electrons move around nucleus in orbits at distances that are large as compared to the size of the nucleus. Each electron has a mass of 9.11 ¥ 10 – 31 kg and a negative charge, – e, equal to – 1.602 ¥ 10 – 19 coulomb. Each proton carries a positive charge, e, equal in magnitude to that of an electron. A neutron carries no charge and its mass is almost same as that of a proton. Under normal conditions, an atom is supposed to be electrically neutral. The total positive charge on protons is equal to the total negative charge on electrons. If an atom loses electron(s), it becomes positively charged. On the other hand, if an atom acquires excess electrons, it becomes negatively charged. The charge is considered as the quantity of electricity. The unit of charge is coulomb (C).

The atoms in a metal (such as Cu, Al, etc.) are packed together to form a solid. In such a case, the loosely bound electron in the outermost orbit is free to move from one atom to the other throughout the solid. Such electrons are called free electrons. The path of a free electron deviates whenever it collides with a nucleus or with other free electrons. After the collision, the path may deviate in any direction. Thus, the free electrons move in a haphazard, zig-zag, or random way (from A to B, B to C, …) in the solid, as shown in Fig. 1.1a.

2 B B¢ D

X n1 electrons

n2 electrons

A

E¢ G G¢

B

F F¢

E

D

E

D¢

C¢ C Drift

A F

G C

Y (a) Random motion of electrons.

Battery (b) Drift of electrons.

When a free electron moves, it carries a negative charge with it. Hence, the free electrons in a metal work as charge carriers. In semiconductors (such as germanium and silicon), there are two kinds of charge carriers. First is the negatively charged electrons, and the second is the positively charged holes. In liquids, the molecules split into positive and negative ions. Both type of these ions work as charge carriers.

The random motion of an electron (see Fig. 1.1a) in a metal does not produce any net displacement. If you observe any imaginary plane XY, at any instant the number n1 of electrons crossing from left to right will be the same as the number n2 plane XY. Hence, . as shown in Fig. 1.1b. Negatively charged electrons experience a constant force towards the positive terminal. Instead of going straight from A to B, the electron takes a curved path and reaches B¢. Again, instead of straight path B¢C, it takes a curved path B¢C¢, and so on. This way, in addition to the random movement, the electrons drift towards positive terminal. This drift of charge carriers (here, electrons) results in an electric current, called drift current. The drift velocity (vd ). The magnitude of drift velocity is quite small, of the order of mm/s. The drift velocity of electrons must not be confused with the speed with which electrical signals travel along a wire. Remember, when you switch on an electric light, the current reaches the bulb in almost no time. Electrical signals travel along a wire almost at the speed of light (3 ¥ 108 m/s). This situation may be compared to the action of water takes quite a long time for a particular molecule of water to travel from the main water pipe in the street to the tap in the kitchen.

If one metre length of the conductor has n electrons, each having a charge – e, the total charge passing the plane XY in one second will be nevd. Since, electric current , the current I due to drift of free electrons is given as dq I= = – nevd (1.1) dt

Introduction

I

3

i

0

0

t (a) DC current.

t (b) Pulsating current.

The negative sign in the above equation indicates that the conventional direction of electric current is opposite to the motion of free electrons. The unit of current is ampere in one second. Although a current in a conductor has a magnitude in a certain direction, it is not a vector quantity. The

Types of Electric Current If the current

i

i

called unidirectional current (Fig. 1.2). If the magnitude of this current remains constant with 0 t time, it is called direct current or dc current (Fig. 1.2a). On the other hand, if the magnitude of the unidirectional current varies with time, it (a) Sinusoidal ac current. is called pulsating current (Fig. 1.2b). However, a current which keeps on reversing its direction continually is called alternating current or ac current (Fig. 1.3). Figure 1.3a shows a sinusoidal ac current, and Fig. 1.3b shows

0

t

(b) Square-wave ac current.

ac current.

N O T E The abbreviations ‘dc’ for ‘direct current’ and ‘ac’ for ‘alternating current’ have been taken as adjectives. Hence, it is perfectly alright to say, ‘ac current’ or ‘dc voltage’, etc.

We establish a graphical symbol for current by placing an arrow next to the conductor, as shown in Fig. 1.4. Thus, in Fig. 1.4a the direction of the arrow and the value ‘3 A’ indicate either that a or Fig. 1.4b, there are again two possibilities: either or a net same current. 3A (a)

–3 A The arrow is a fundamental part of the definition of (b)

I CURRENT i

tion of a current. Thus, to talk about the value of a current i without specifying the arrow is to discuss an

The current density in a conductor carrying current is the current per unit area of the cross section of the conductor. The area is in the direction normal to the current. The current density J is given as I J= (1.2) A 2 2 It is a vector quantity and its unit is ampere/metre (A/m ).

The absolute potential potential difference (pd) (also called voltage) between two points. If the energy required to move a charge of Q coulombs from point A to point B is W joules, the voltage V between A and B is given as W V= (1.3) Q The unit of voltage is volt (V), and is given as 1 volt = 1 joule/1 coulomb

While talking about potential difference or voltage, it is more appropriate to designate it as either voltage rise or voltage drop. In Fig. 1.5, the potential difference in all cases is 4 V. But in Fig. 1.5a there is a voltage rise of 4 V from B to A, and in Fig. 1.5b there is a voltage drop of 4 V from B to A. Double subscript notation is used to designate voltage rise or voltage drop. Thus, the notation VAB denotes the voltage of point A with respect to point B. It is simply the voltage rise from B to A, which is same thing as the voltage drop from A to B. For Fig. 1.5a, we can write, VAB = 4 V, as point A is 4 V above point B. The same thing can also be written as VBA = – 4 V, as point B is 4 V below point A. In general, we have VAB = –VBA (1.4) For Fig. 1.5c, the voltage of A with respect to B is VAB = – (– 4) V = 4 V. Therefore, this voltage representation is equivalent to that of Fig. 1.5a. B

A +

4V

–

(a) Voltage VAB = 4 V.

B

A –

4V

+

(b) Voltage VAB = –4 V.

B

A – –4 V

+

(c) Voltage VAB = –(– 4) V.

Introduction

5

Voltage or potential difference across a minus signs and then writing the value of the voltage. Thus, in Fig. 1.6a, placing the plus sign at terminal A indicates that the terminal A is v volts positive with respect to the terminal B. If v either that A is – 5 V positive with respect to B or that B is 5 V positive with respect to A.

. v

Fig. 1.6 In (a) and (b), terminal B is positive with respect to terminal A; in (c) and (d), terminal A is positive with respect to terminal B.

Electric Field in a Conductor b

E magnitude is given as V L where, V is the voltage across the terminals of the conductor and L volt/metre E=

Energy a b joule A

2A

B +

4V

–

A

2A

B +

(a) The element is absorbing energy.

Fig. 1.7

4V

–

(b) The element is delivering energy.

A

–2 A

B + –4 V

–

(c) Equivalent of Fig. (b).

Consider an element having a voltage v across it. A small charge Dq is moved through the element from the positive terminal to the negative terminal in time Dt. The energy DW absorbed by the element in this process is given as DW = vDq. Therefore, by dividing the two sides by Dt, we get W q ΔW Δq dW dq =v or Lt = v Lt or =v Δt → 0 Δt Δt → 0 Δt t t dt dt The left-hand side of the above equation is power. Thus, the power absorbed is given as p = vi (1.6) The unit of power is watt (W).

! While using above equation to calculate the power consumed by an element, the direction of current through the element and the reference polarity of the voltage across it must be in conformity. The passive sign convention (as in Fig. 1.7a). Thus, for the element in Fig. 1.7a, the power absorbed is p = vi = (4 V) (2 A) = 8 W The direction of current and voltage polarity for the element in Fig. 1.7b are not in conformity as per the passive sign convention. Here, the current of 2 A is leaving the positive terminal. This is equivalent to saying that a current of – 2 A is entering the positive terminal, as shown in Fig. 1.7c. These two representations are equivalent. We can now calculate the by the element in Fig. 1.7b, by using its equivalent in Fig. 1.7c, as follows : p = vi = (4 V) (–2 A) = –8 W The power absorbed is – 8 W. This is equivalent to saying that the by the element of Fig. 1.7b is 8 W. Horse power (hp) is also used as a unit of power [1 hp = 746 W].

If a charge Q t, and if the potential difference across the element is V, the energy expended or the work done is given as Energy = Work = Voltage ¥ Charge = V ¥ Q = V (It) = VIt = pt (1.7) When 1 watt of power is delivered to an electrical load for 1 second, the energy consumed is 1 watt-second (1 Ws) or 1 joule (J). Watt-second is a very small unit. Practical unit of electrical energy is kilowatt-hour (kWh). This is also called the commercial unit of electrical energy, as electric supply companies send the bills to the customers for kilowatt-hours of energy consumed. Heat is a form of energy. It is normally measured in kilocalories (kcal) [1 kcal = 4.2 kJ].

A basic electric circuit consists of a source, a pair of and a load, as shown in Fig. 1.8. Here, a battery is shown as the source and a resistance RL is shown as the load. The source forces a current I tor are needed.

Introduction

It is the voltage of the source (such as a battery) when nothing is connected to it. When connected in an electric circuit, it delivers energy (or power) to the other elements of the circuit. Note that the electromotive force (abbreviated as emf ) is not a force. It is a voltage and is measured in volts (V). It is called a force because it

The voltage across the terminals of a source is called its terminal voltage. In Fig. 1.8, V1 is the terminal resistance, r. Thus, the terminal voltage is slightly less than the emf of the source. From the source to the load, there will be some voltage drop due to the resistance of the conducting wires. The terminal voltage V2 across the load will be slightly less than the terminal voltage V1 across the source. Normally, the length of the connecting wires is small, and they offer very small resistance. The voltage drop across the wires is negligibly small. Hence, V1 = V2 for all purposes.

Electrical equipments (such as heater, motor, generator, etc.) cannot convert the entire input power into the required output power. The ratio of the output power to the input power is known as (h). It is usually expressed in percentage. For example, consider the A I C battery in the circuit of Fig. 1.8. The power generated due + + Outgoing conductor to the chemical reactions is the product of its emf E and r the current I. This can be considered as input power to the RL V1 V2 battery. The power delivered by the battery (i.e., its outE put power) is the product of its terminal voltage V1 and the Incoming conductor current I – – Output power V1 I Source B I D Load 100% 100% = h= Input power EI

The International System of Units (abbreviated as SI Units) is coherent, rational and comprehensive. It is now followed everywhere in the world—at least in engineering. It has seven base units as given in Table 1.1 and many derived units.

Physical quantity

Name of SI unit

Symbol

Length

metre

m

Time Electric current Thermodynamic temperature Amount of substance Luminous

second ampere kelvin* mole candela

*It should be written as kelvin only, and not degree kelvin or °K.

s A K mol cd

Symbol 1

10 2 10 3 10 106 109 1012 1015 1018

Factor –1

deca hecto kilo

da h k

10 –2 10 –3 10

giga* tera peta exa

G T P E

10 – 9 10 – 12 10 – 15 10 – 18

–6

P deci centi milli micro nano pico femto atto

Symbol d c m m n p f a

*Pronounced as jeega, as it is derived from the word ‘gigantic’. Note : Latin origin (except femto and atto, recently added, which have Danish origin). tion is micro (Greek letter m). hecto, and deca.

± 3n where n is an integer. However, the unit ‘centimetre (cm)’ owing to its convenient size and well-established usage, cannot be given up readily. mmF) or millinanofarad (mnF), use picofarad (pF). 2

instead of N/mm2, even though mathematically both forms are equivalent. 4. The rules for binding-in indices are not those of ordinary algebra. For example, cm2 means (cm)2 = (0.01)2 m2 = 0.0001 m2, and not c ¥ (m)2 = 0.01 m2.

Following are the rules and conventions regarding the use of SI units : 1. Full name of units, even when they are named after a person, are not written with a capital (or uppercase) initial letter, e.g., kelvin, newton, joule, watt, volt, ampere, etc. 2. The symbols for a unit, named after a person, has a capital initial letter, e.g., W for watt (after James Watt), J for joule (after James Prescott Joule) and Wb for weber (after Wilhem Eduard Weber). 3. The symbols for other units are not written with a capital letter, e.g., m for metre. 4. Units may be written out in full or using the agreed symbols, but no other abbreviation may be used. They are printed in full or abbreviated, in roman (upright) type, e.g., amp. is not a valid abbreviation for ampere.

Introduction

9

the preceding magnitude is measured, e.g., 50 kg, and not 50 kgs. 6. No full stops or hyphens or other punctuation marks should be used within or at the end of the symbol between the two symbols to avoid confusion. For example, while writing, say, metre second it should be abbreviated as m◊s to avoid confusion with ms, the symbol for millisecond.

(5 megawatt), 2 ns (2 nanosecond). 8. A space is left between a numeral and the symbol except in case of the permitted non-SI units for angular measurements, e.g., 57° 16 ¢ 44 ≤. 9. A space is left between the symbols for compound units, e.g., N m for newtons ¥ metres and kW h for kilowatt hour. This reduces the risk of confusion when an index notation instead of the solidus (/) is used. In the former notation, a velocity in metres per second is written as m s – 1 instead of m/s, but m s – 1 may mean ‘per millisecond’ This type of confusion will not occur if we follow the rule that the denominators of compound units are always expressed in the base units and not in their multiples or 10. When a compound unit is formed by dividing one unit by another, this may be indicated in one of the two forms as m/s or m s–1. In no case, should more than one solidus sign (/) on the same line be included in such a combination unless a parenthesis be inserted to avoid all ambiguity. In complicated cases, negative powers or parenthesis should be used. 11. Algebraic symbol representing ‘quantities’ are written in italics, while symbols for ‘units’ are written as upright characters, e.g., current I =3A energy E = 2.75 J terminal voltage V = 1.5 V 12. When expressing a quantity by a numerical value and certain unit, it has been found suitable in most applications to use units resulting in numerical value between 1.0 and 1000. To facilitate the reading of numerals, the digits may be separated into groups of three—counting from the decimal sign towards the left and the right. The groups should be separated by small space, but not by a comma or a point. In numerals of four digits, the space is usually not necessary. (It is recognised, however, that to drop the in use at present). A few examples are give below: (i) (ii) (iii) (iv)

Incorrect 40,000 or 40000 81234.765 764213.876 6 543.21

(i) (ii) (iii) (iv)

Correct 40 000 81 234.765 764 213.876 6543.21

Note (a) The recommended decimal sign is a full stop ( . ). The sign of multiplication of numbers is a cross ( ¥ ). (b) If the magnitude of a number is less than unity, the decimal sign should be preceded by a zero

10 13.

not

14. Full stop should not be used in a multi-word abbreviation. For example, write emf, pd, ac, dc, etc., and not e.m.f., p.d., a.c., d.c., etc. 15. The abbreviated forms ac and dc should be used only as adjectives. For example, dc motor, ac current. 16. A unit symbol should be used only after a numerical value of the quantity and not after the symbol of the quantity. For example, write 5 kg, 7.5 V, but m kilograms, I amperes. 17. A hyphen is inserted between the numerical value and the unit when the combination is used as an adjective. For example, a 240-V motor, a 2-W resistor, etc.

Doing experimental exercises in a laboratory helps in deeper understanding of physical concepts. At the end of each chapter, this book provides guidelines to perform suitable experiments. Conducting experiments serves the following purposes : (i) (ii) (iii) (iv) (v)

To be familiar with the basic components, measuring instruments and other equipments. To learn the techniques of measuring basic electrical and non-electrical quantities. To realise the limitations of accuracies of measuring instruments. To physically verify theorems/laws pertaining to a topic. To get training of technical report writing.

Before coming to the laboratory, a student should become familiar with the theoretical background and readings mechanically does not help in developing an understanding of the matter.

Since you are going to work with equipments and machines operating at high voltages, such as 220-V dc supply, 230-V, 50-Hz ac supply, or 440-V, 50-Hz, 3-phase ac supply, you should take utmost care to avoid electric shock hazards. Furthermore, enough care should be taken so as to get meaningful results, without damaging any equipment or instrument. Therefore, it is important to adhere to the following general instructions and precautions. (i) (ii) (iii) (iv)

Never work alone in the laboratory. Always put on shoes with rubber soles so as to provide insulation from ground. Don’t wear loose dress while working in the laboratory. Power should be switched off before changing any connection.

(vi) Keep away from moving parts.

connecting voltmeters or pressure coils of wattmeters as the current carried is small. Thicker wires are to be used where current carried is large.

Introduction

11

enable you to decide the range of all instruments to be used. (x) Select an instrument that gives the reading in its upper range of the scale. This provides better accuracy of the measurement. (xi) Switch on the supply only after the connections have been checked (preferably by another person). (xii) Don’t touch any live terminal while the supply is ON. (xiii) While applying electrical load, gradually increase it. Similarly, while removing the electrical load, gradually decrease it. (xiv) Never apply full voltage to a motor. While switching ON increase it.

Following power supply systems are normally available in the electrical engineering laboratories. It has two wires—phase/line wire and neutral wire (Fig. 1.9). Normally, it has four wires—three for phase/line and one for neutral (Fig. 1.10). L

R

230-V 50-Hz AC Supply

Y B N

N MCB

MCB

DC Supply positive and negative. Single phase ac supply is also available for lighting purpose. For low current requirements, the ac supply can be taken from the sockets. can be obtained by using a suitable rheostat (a variable resistance), as shown in Fig. 1.11. This arrangement has the disadvantage of incurring considerable power loss in the resistance. A C

230 V AC Supply B

Variable AC Supply

12 can be obtained by using an auto-transformer with a variable tap (also known as ‘variac’ or ‘dimmerstat’), as shown in Fig. 1.12. For 3-phase supply, we use a 3-phase variac (Fig. 1.13).

For measuring an electrical quantity, it is important to select a proper instrument with proper range. The following are the important instruments and equipments used in the laboratories. The basic principle of operation of these two instruments is the same. An ammeter is used for measuring amperes (or current). It has low resistance and must be connected in series with the circuit. A voltmeter is used for measuring voltage across two points of a circuit. It has high resistance and must be connected across (in parallel with) the two points. through them. Hence, their scale is uniform. However, these cannot be used for ac measurements. instruments is proportional to the square of the effective value (or rms value) of the current. Hence, their scale is non-uniform. A wattmeter is an instrument that measures the power (both dc and ac) going to an electrical load. It has two coils—called current coil and potential (or pressure) coil. The terminals of the current 1 and V2 1 are joined together to make a common terminal. The current coil is connected in series with the load and the pressure coil across the load, as shown in Fig. 1.14. Sometimes, the wattmeter can give negative reading. In such a case, the connections of either the current coil or the pressure coil are to be reversed and the reading is to be treated as negative. It is used for the measurement of rotating speed in rpm (revolutions per minute) of a machine. The tapered shaft of the tachometer is inserted into the tapered hole in the shaft of the machine. The reading of the tachometer is proportional to the speed of rotation of its shaft. A tachometer can be either analog type or digital type. R

Y N

B

Y

N

B

F

f

Variable 3-f supply

4-wire, 3-f supply

R

W M V1 Mains supply

L

I V2 Load

Introduction

13

It is a variable resistance made up of a closely wound wire of high resistivity (such as nickel-chromium-iron alloy) over a circular insulating (asbestos or porcelain) tube. These are available maximum current it can carry. Normally, it is 1000 W, 1.2 A and 100 W, 5 A. Rheostats are used as variable resistances and as potential dividers (as in Fig. 1.11). Commonly used loading devices are (i) lamp bank, and (ii) loading rheostats. A consists of a number of 230-V lamps (100 W, 60 W, 40 W, etc.) suitably connected and controlled by switches to provide different loads. A loading rheostat consists of a number of identical resistive elements, suitably connected in series, parallel and combinations thereof.

2

OHM’S LAW OB JE CT IV E S :

C

This is the most fundamental law in Electrical Engineering. It states that the potential difference between the , provided its temperature That is, V μ I or V = RI (2.1) The constant of proportionality R is called the resistance of the conductor. Its unit is ohm (W). The unit ohm volt is applied to the resistance. There is another way of stating Ohm’s law, I μ V or I = GV (2.2) The constant of proportionality G is called the conductance of the conductor. Comparison of Eq. 2.2 with Eq. 2.1 shows that the conductance is the reciprocal of the resistance, 1 G= (2.3) R The SI unit of conductance is siemens (S).

Ohm’s law is presented in graphical form in Fig. 2.1. The voltage is shown as independent variable (cause) and current as dependent variable (effect). The slope of the line is the reciprocal of resistance (1/R), and is called conductance. A conductor showing straight line v-i characteristics (as in Fig. 2.1) is said to have a linear resistance.

15 i i

A

1 Slope = R

+ v

R

0 v

– B

These two are special resistances. A short circuit (R i π 0) without any resulting voltage (v = 0). As shown in Fig. 2.2, the characteristic of a short circuit is a vertical line. An open circuit (R = ) permits voltage (v π 0) with no current (i = 0). Since power going into a resistance is p = vi, no power is required for the short circuit or the open circuit.

(

An ideal switch is also a special resistance. It can be changed from a short circuit to an open circuit to turn an electrical device ON or OFF. Ideal switches receive no electrical energy from the circuit. Different types of switches used in electric circuits are shown in Fig. 2.3. Figure 2.3a shows a switch in its open (OFF ) state. Figure 2.3b shows a throw switch which switches one input line between two output lines. Figure 2.3c shows a switch; the dashed line indicates mechanical coupling between the two components of the switch.

0)

(

)

(a)

(b)

(c)

16

Basic Electrical Engineering

through a conductor (Fig. 2.4) depends on how narL I I row is its cross section and how long is its length. In A other words, the electrical resistance of a conductor V is directly proportional to its length (L) and inversely proportional to its area of cross section (A), L L Rμ or R = r (2.4) A A Here, r is a constant of proportionality and is known as the resistivity of the material. The unit of resistivity is ohm metre (W m). EX A MP L E

2.1

A conducting wire has a resistance of 5 W. What is the resistance of another wire of the same material but having half the diameter and four times the length ?

Solution 2

Hence,

R2 ρ (L2 / A2 ) L2 A1 L2 π d12 /4 L2 ⎛ d1 ⎞ 4 ⎛ 2⎞2 × × ×⎜ ⎟ = × = 16 = = = = 2 R1 ρ (L1/ A1) L1 A2 L1 π d2 /4 L1 ⎝ d2 ⎠ 1 ⎝ 1⎠ R2 = 16R1 = 16 ¥ 5 = 80 W.

EX A MP L E

2.2

A copper wire has a resistance of 10 W. The wire is drawn so that its length increases three times. What is the resistance of the new wire ?

Solution If the length of the wire is made three times by drawing it, its area of cross section must decrease three times, is

R2 ρ (L2 / A2 ) L2 A1 3 3 = = × = × =9 ρ (L1 / A1) L1 A2 1 1 R1

Hence, the resistance of the second wire is R2 = 9R1 = 9 ¥ 10 =

J≡

I (V / R) V V (V / L ) E = = = = = A A RA ( ρL / A) A ρ ρ

where, E = V/L E = rJ This is often referred to as microscopic form of Ohm’s law.

(2.5)

Two or more resistances are said to be connected in series, if same (not them. Figure 2.5a shows three resistances connected in series with a voltage source of emf V. Figure 2.5b shows its equivalent circuit, in which the three resistances are replaced by a single resistance Rs.

17

I

R1

R2

R3

V1

V2

V3

Rs

I

V (a)

V (b)

The same current I the three individual voltages, V1, V2, and V3. That is,

V must be equal to the sum of

V = V1 + V2 + V3 = IR1 + IR2+ IR3 = I(R1+ R2+ R3) (2.6) From the equivalent circuit of Fig. 2.5b, we can write (2.7) V = IRs From Eqs. 2.6 and 2.7, it can be seen that Rs = R1 + R2 + R3 Thus, the equivalent resistance of a number of resistances connected in series is equal to the sum of indiIn general, for n resistances in series, we can write n

Rj

Rs = R1 + R2 + R3 + … Rn =

(2.8)

j =1

Obviously, if n identical resistances, each of resistance R, are connected in series, the equivalent resistance will simply be nR. That is, (2.9) Rsn = nR

Two or more resistances are said to be connected in parallel, if same (not merely equal) voltage exists across them. Figure 2.6a shows three resistances in parallel, connected across a voltage source of emf V. Figure 2.6b shows its equivalent circuit, in which the three resistances are replaced by a single resistance Rp. Same voltage V appears across all the three resistances. Hence, the three currents are given as V V V I1 = ; I2 = and I3 = R1 R2 R3 I

I

I1 R1

V

I (a)

I2 R2

I3 R3

Rp

V

I (b)

18

Basic Electrical Engineering

Since the total current I entering the combination divides into I1, I2 and I3, we have ⎛ 1 V V V 1 1⎞ + + =V⎜ + + ⎟ R1 R2 R3 ⎝ R1 R2 R3 ⎠ From the equivalent circuit of Fig. 2.6b, we can write V I= Rp Comparing Eqs. 2.10 and 2.11, we get 1 1 1 1 = + + Rp R1 R2 R3 I = I1 + I2 + I3 =

(2.10)

(2.11)

Thus, the reciprocal of equivalent resistance of a number of resistances connected in parallel is equal to the In general, for n resistances in parallel, we can write 1 1 1 1 1 = + + +… = Rp R1 R2 R3 Rn

n

1

∑R j =1

(2.12)

j

Obviously, if n identical resistances, each of resistance R, are connected in parallel, the equivalent resistance will simply be R/n. That is, Rp n =

R n

In case only two resistances are connected in parallel, the equivalent resistance is given as R + R2 1 1 1 = + = 1 Rp R1 R2 R1 R2 or

Rp =

R1 R2 R1 + R2

(2.13)

(2.14) (2.15)

For two resistances in parallel, using Eq. 2.15 is much more convenient than using Eq. 2.14.

The concept of voltage divider is very useful in analysing electric circuits. Consider the circuit of Fig. 2.7, in which two resistances R1 and R2 are connected in series with a voltage source The current I is given as V I= R1 + R 2 Therefore, the voltage V1 across resistance R1 is given as V1 = IR1 = or

V R R1 + R 2 1

V1 = V

R1 R1 + R 2

(2.16)

V2 = V

R2 R1 + R 2

(2.17)

Similarly, the voltage across R2 is

19

I I + R1

I

V1

I1

– +

V R2

I2

R1

+

R2 V

V2

–

–

Like voltage divider, the concept of current divider is also useful in analysing circuits. Consider the circuit in Fig. 2.8, in which two resistances R1 and R2 are connected in parallel. Same voltage V appears across both the resistances. The total current I entering the combination divides into I1 and I2, as shown. Therefore, we have I1 R1 V = I1 R1 and V = I2 R2 or I2 = R2 R ⎛ R + R2 ⎞ I = I1 + I2 = I1 + I1 1 = I1 ⎜ 1 ⎟ R2 ⎝ R2 ⎠ R2 or I1 = I (2.18) R1 + R2 Similarly, the current through R2 is given as R1 I2 = I (2.19) R1 + R2 the other EX A MP L E

2.3

Using voltage divider technique, determine the voltage across the four resistances in the circuit of Fig. 2.9a.

4W

8W

5A

5A 4W

(a)

4W

+ 12 V – + 12 V –

4W

4W

(b)

+ 16 V – + 8V –

8W

4W

20

Basic Electrical Engineering

Solution The voltage across the two parallel combinations divides independently in the two paths. The equivalent resistance seen by the current source is Rp = (4 + 4) || (8 + 4) =

8 × 12 = 4.8 W 8 + 12

Hence, the total voltage across the parallel combination is 5 ¥ 4.8 = 24 V. This voltage equally divides between the two Fig. 2.9b. EX A MP L E

2.4

Using the voltage divider and current divider techniques, determine the unknown currents through and voltages across the resistances in the circuit of Fig. 2.10. i1

2A

+ Vs –

R1 = 2 W

i3

+ v1 – R2 = 4 W

+ v3 –

+ v2 –

i2

R3 = 6 W

Solution

R2 and R3 can be replaced by their parallel equivalent Rp = 4 || 6 = 2.4 W (Fig. 2.11a). The 2-W resistance is now combined with 2.4 W to give 4.4 W (Fig. 2.11b). Clearly, the voltage across the current source is Vs = 2 ¥ 4.4 = 8.8 V. 2W

2A

+ Vs –

2W

2.4 W

2A

+ Vs –

(a)

4.4 W

2A

+ v1 – + + 8.8 V v4 – –

(b)

2.4 W

(c)

We now restore the original circuit. Figure 2.11c is the same as Fig. 2.11a, but now we know that a voltage of 8.8 V is applied across the series combination of the 2 W and 2.4 W. The voltages across these resistances have been marked as v1 and v4 source and not a voltage source which you are dividing.” True, but the voltage created by the current source divides in a series circuit. It does not matter whether we have a 2-A current source producing 8.8 V or an 8.8-V voltage source

v1 = 8.8 ¥

2 =4V 2 + 2.4

and v4 = 8.8 ¥

2.4 = 4.8 V 2 + 2.4

Voltage v4 is the same as the voltages v2 and v3 in the original circuit (Fig. 2.10). Obviously, current i1 = 2 A, and according to Ohm’s law, 4.8 V 4.8 V i2 = = 1.2 A and i3 = = 0.8 A 4 6

21

Vs and the resistance R I, and hence they are in series. They have the same voltage Vs across them, and hence they are in parallel. Consider the simple circuit of Fig. 2.12a

R1

I

IA R2

I

Vs

RA

R3

IB Is

R

R7

R5 R4

RB

RC

Vs

RD

RE

Vs R8 (a)

R6 (b)

(c)

combinations. Consider, for example, Fig. 2.12b. Which resistances are in series and which resistance are in parallel ? The circuit has only two resistances R1 and R8 in series. Note that these two resistances are physically placed parallel to each other, but electrically they are in series. The circuit has only two resistances R2 and R3 in parallel. No other combinations of resistances are in series or in parallel. A resistance (or any other element) need not be in series or in parallel with other resistance (or element). For example, the resistances R5 and R4 in Fig. 2.12b are neither in series nor in parallel with any other resistance. The resistances R5, R4 and R3 make what is called a delta-connection. In Fig. 2.12c, there are no resistances that are in series or in parallel with any other resistance. However, the resistances RB, RC and RD make what is called a star-connection. There is an important caution about the use of current divider. Many students apply this concept where it is not applicable. For example, consider the circuit of Fig. 2.12c. The circuit contains no resistances in series or in parallel. can be applied. However, many students take a quick look at resistances RA and RB and try to apply current division, writing an incorrect equation such as RB IA = s RA + RB Remember, parallel resistances must be connected between the same pair of points (called nodes).

There are two ways in which three resistances can be connected across three points of a network. One arrangement, shown in Fig. 2.13a, is called star or wye b, is called delta (D) or mesh connection In some cases, a network can be solved readily by means of star-delta R1, R2 and R3 in terms of RA, RB and RC. Similarly, RA, RB and RC in terms of R1, R2 and R3.

22

Basic Electrical Engineering A

A

A

RA

RA R2

R3

R2

R3

RC

RB B

RB

RC

B

B R1

R1 C

C (a) Star (Y) connection

C (b) Delta (D) connection

(c) Y and D superimposed

Figure 2.13c shows the two arrangements superimposed. Note that the resistance opposite to RA has been named as R1, that opposite to RB as R2 and that opposite to RC as R3. This has been deliberately done so that the star-delta transformation formulae become easy to remember.

Delta-to-Star Transformation If the two arrangements in Fig. 2.13 are to be equivalent, the resistance between any pair of terminals (AB, BC or CA) in the two circuits has to be the same, when the third line is left open. Keeping terminal A open, we equate the resistance between terminals B and C in the two circuits, to get R1 ( R2 + R3 ) RB + RC = (2.20) R1 + R2 + R3 Similarly, we can write R2 ( R1 + R3 ) RC + RA = R2 + R1 + R3 and

RA + RB =

R3 ( R1 + R2 ) R1 + R2 + R3

Adding above three equations and dividing by 2, we get R1 R2 + R2 R3 + R3 R1 RA + RB + RC = R1 + R2 + R3 Subtracting Eq. 2.20 from this equation gives RA = Similarly, we get

R2 R3 R1 + R2 + R3

(2.21)

(2.22)

R3 R1 (2.23) R1 + R2 + R3 R1 R2 and RC = (2.24) R1 + R2 + R3 Equations 2.22 to 2.24 are a set of equations which transforms a delta connection into its equivalent star connection. RB =

is equal to the product of the two delta resistances connected to the same terminal divided by the sum of the

N O T E For a symmetrical delta-connection, R1 = R2 = R3 = RD (say). Then, the equivalent resistance of the symmetrical star-connection is simply given as

R =

R RD RD = 3 RD + RD + RD

(2.25)

MN E M O N IC S

Remembering the above relation becomes easy if you recall that the same relation is obtained for the equivalent resistance of parallel combination of three resistances.

Star-to-Delta Transformation

To get the reverse transformation, we multiply Eq. 2.22 and Eq. 2.23

so as to get RA RB =

R1 R2 R32 ( R1 + R2 + R3 ) 2

(2.26)

RB RC =

R12 R2 R3 ( R1 + R2 + R3 ) 2

(2.27)

RC RA =

R1 R22 R3 ( R1 + R2 + R3 ) 2

(2.28)

Similarly, we can get

Adding Eqs. 2.26 to 2.28 gives RA RB + RB RC + RC RA = =

R1 R2 R32 + R12 R2 R3 + R1 R22 R3 ( R1 + R2 + R3 ) 2 R1 R2 R3 R2 R3 = R = RA R1 R1 + R2 + R3 R1 + R2 + R3 1

Therefore,

R1 = RB + RC +

Similarly,

R2 = RC + RA +

RB RC RA

RC RA RB RA RB R3 = RA + RB + RC

and

(2.29) (2.30) (2.31) (2.32)

the equivalent delta resistance connected between two terminals is equal to the sum of the two star resistances connected to those terminals plus the product of the

EX A MP L E

2.5

Calculate the effective resistance between points A and B for the combinations of resistances, given in Fig. 2.14.

24

Basic Electrical Engineering R

15 W A

R

R

3W 18 W

R

20 W

10

W

20 W

R

B

A

A

R R

B

R 20 W

R (b)

(a)

5W B

6W

(c)

Solution (a) The resistance Rp of the network within the dotted box is given by 1 1 1 1 1+ 2 +1 4 1 = + + = = = Rp 20 10 20 20 20 5

Therefore,

or

Rp = 5 W

RAB = 15 W + (20 W || 10 W || 20 W) = 15 W + 5 W = Alternatively, we can have much easier and quick solution. Just view the network within the dotted box as parallel combination of two resistances of 20 W each, which is then paralleled with a resistance of 10 W. That is, RAB = 15 W + (20 W || 20 W) || (10 W) = 15 W + (10 W) || (10 W) = 15 W + 5 W = (b) The resistance of the network within dotted box is 2 5 R+R= R 3 3 By symmetry, the resistance of the lower network will also be the same, i.e., 5 R2 = R 3 Thus, the entire network between points A and B is a parallel combination of three resistances, R1 (= 5R/3), R and R2 (= 5R/3). The combination of R1 and R2 is simply 5R/6. Thus, R1 = [(R + R) || R] + R =

RAB = (5R/6) || R =

(5 R /6) R 5 R2 /6 5 = = R (5 R /6) + R 11R /6 11

(c) The resistance of the network within dotted box is R1 = (3 W || 6 W) + 18 W = 2 W + 18 W = 20 W \ RAB = (20 W) || (20 W) + 5 W = 10 W + 5 W = 15 W EX A M P L E

2 .6

Determine the currents I1, I2 and I3

Solution Effective resistance connected across the battery is given as Reff = 2 + \

I =

1 + 2 = 2 + 6 + 2 = 10 W (1/12) + (1/ 20) + (1/ 30)

V 100 = = 10 A Reff 10

25 I1 12 W I 2W

2W

I2 20 W

I

A

I

I3 30 W

+

I1 RL

5W

R

10 W

120 V B

C

–

100 V

The voltage across the three resistances connected in parallel has to be the same. Therefore, applying Ohm’s law, we have 12 12 12 I1 = 20 I2 = 30 I3 fi I2 = I1 = 0.6I1 or I3 = I1 = 0.4I1 20 30 branches. Thus, 10 = I1 + I2 + I3 = I1 + 0.6 I1 + 0.4I1 = 2 I1 I2 = 0.6 I1 = 3 A

\

E XA MP LE

fi

I1 = 5 A

I3 = 0.4 I1 = 2 A

and

2.7

Calculate the supply current I in the network given in Fig. 2.16, if the 5-W resistance dissipates energy at the rate of 20 W.

Solution

Power dissipated by 5-W resistance is given as P = I 21RL

or

20 = I 21 ¥ 5

fi

I 21 =

20 =4 5

or I1 = 2 A

The concept of current divider gives I1 = I ¥ E XA MPLE

R R + RL

or

2=I¥

10 2 = I 10 + 5 3

fi

I=3A

2.8

Three 60-W, 120-V light bulbs are connected across a 120-V power line as shown in Fig. 2.17 Find (a) the voltage across each bulb, and (b) the total power dissipated in the three bulbs.

Solution (a) The resistance of each bulb is given as V 2 120 2 = = 240 W P 60 The combined resistance of bulbs B and C is RBC = 240/2 = 120 W. Therefore, using potential divider, the voltage across B (and also A) is given as 120 VB = VC = 120 ¥ = 40 V and VA = 120 – 40 = 80 V 240 + 120 R =

(b) P = PA + PB + PC =

(80)2 ( 40)2 ( 40)2 + + = 26.67 + 6.67 + 6.67 = 40 W 240 240 240

26

Basic Electrical Engineering

EX A MP L E

2.9

As the variable resistor R in Fig. 2.18 varies from a short circuit to an open circuit, it is desired that the equivalent resistance also vary between 30 W and 75 W. (a) Design R1 and R2 to accomplish this result. (b) Find R to give Req = (30 + 75)/2 W.

Solution (a) Minimum value of Req is obtained when R = 0 (i.e., a short circuit, because the parallel combination of R2 and R is reduced to zero). Maximum value of Req is obtained when R is an open circuit. Hence, R1 = 30 and R1 + R2 = 75 fi R1 = 30 W and R2 = 45 W (b) Required value of Req is (30 + 75)/2 W = 52.5 W. It means that the parallel combination of R2 and R should have a value, Rp = 52.5 – R1 = 52.5 – 30 = 22.5 W This value is exactly half of R2 = 45 W. Hence, the value of R should also be . R1

C

A R

Req

R2

R

2R R

4R 2R

8R 4R

16R 8R

...

to

B D

Fig. 2.18 EX A MP L E

Fig. 2.19

2.1 0

Calculate the equivalent resistance between terminals A and B in terms of the resistance network of Fig. 2.19

Solution terminals C and D has an equivalent resistance, which must be double of the original ladder. If the equivalent resistance across A-B is Rx, we can replace the remaining ladder across C-D by a single resistance 2R Thus, (R)(2 Rx ) Rx = R + (R || 2Rx) or Rx = R + or 2R 2x – 3RRx – R2 = 0 R + 2Rx Rx =

fi EXA MP LE

3 R ± 9 R2 + 8 R2 = 1.78R 4

(ignoring negative value)

2 . 1 1 a

Fig. 2.20

Solution

D) connection of three resistances R1, R2 and R3. Similarly, the RA, RB and RC. Note that in Fig. 2.20 we have marked the terminals according to our standard convention. Terminal A is opposite to resistance R1, terminal B is opposite to resistance R2, and so on. Therefore, for the required conversion, we use Eqs. 2.22 to 2.24, to get R2 R3 9×6 54 RA = = = =3W R1 + R2 + R3 3 + 9 + 6 18

27 RC

R1 = 3 W

RB

C

B

R3 = 6 W

B

R2 = 9 W

C

RA

A

A (a)

(b)

R1R3 3×6 18 = = = 1.0 W R1 + R2 + R3 3 + 6 + 9 18 R1R2 3×9 27 = = = 1.5 W RC = R1 + R2 + R3 3 + 6 + 9 18 RB =

EXA MP LE

2.12

Two coils connected in parallel across 100-V dc supply, take 10 A current from the supply. Power dissipated in one coil is 600 W. What is the resistance of each coil ?

Solution The effective resistance of the two coils in parallel is Reff =

100 = 10 W 10

600 =

100 2 R1

Power in one coil is given as fi

R1 =

10000 = 16.67 W 600

Since the two coils are connected in parallel, we should have 16.67 × R2 10 = fi R2 = 25 W 16.67 + R2 E XA MPLE

2.1 3

An electric boiler draws 12-A current at 115 V for a period of 6 hours. If electrical energy costs Rs. 2.50 per kW h, determine the cost of the boiler operation.

Solution Total energy consumed is W = 115 V ¥ 12 A ¥ 6 h = 8280 Wh = 8.28 kW h Therefore, the cost of the boiler operation is Cost = Energy ¥ Rate = 8.28 kW h ¥ 2.50 Rs/kW h = ` 20.70 EX A M PL E

2 .1 4

A toaster rated at 1000 W, 240 V is connected to a 220-V supply. Will the toaster be damaged ? Will its rating be affected ?

Solution The resistance, and the current rating of the toaster are R =

V2 240 2 P 1000 = = 57.6 W and Irating = Imax = = = 4.167 A V 240 P 1000

28

Basic Electrical Engineering

When the toaster is connected to 220-V supply, the current drawn is V 220 = = 3.82 A R 57.6 This current being less than the current rating, the toaster will not be damaged. The power consumed is P1 = 220 V ¥ 3.82 A = 840.4 W I =

Thus, the power rating is less than its original power rating.

Two circuits are said to be dual of each other, if the equations for the one have the same mathematical form as the current equations for the other. In dual circuits, the form of equations remain the same, if following Current Series Resistance Short circuit Closed switch

¤ Voltage ¤ Parallel ¤ Conductance ¤ Open circuit ¤ Open switch

Thus, the two forms of Ohm’s law, namely, V = RI and I = GV, are dual of each other. Similarly, the concept of voltage divider is dual of the concept of current divider. Note that Eqs. 2.18 and 2.19 can be rewritten, in terms of the two conductances G1 (= 1/R1) and G2 (= 1/R2), as G1 G2 I1 = I and I2 = I G1 + G2 G1 + G2 which have the same form as Eqs. 2.16 and 2.17. The principle of duality is very useful in electrical engineering.

A conductor is used to conduct current from one point to another in an electric circuit. Silver is the best conductor followed by copper and aluminum. Silver, being very costly, is rarely used for this purpose. Metallic conductors in the shape of bars, tubes, wires, or sheets are commonly used. Bus-bars are solid thick conductors to carry large currents (hundreds of amperes). Thin bus-bars are used in electronic circuits and computers to carry information signals. A wire may be single solid or may have a number of solid wires (called strands) twisted together to

them with a metallic braid. Such wires are called The property of a conductor to carry current is called its conductance, expressed in siemens (S). Ideally, a good conductor should have conductance (or, equivalently, its resistance must be zero).

Often, we deliberately introduce resistance at some point in an electrical or electronic circuit. The physical ‘resistor’. A resistor can be wire wound or carbon moulded.

29

Wire Wound Resistor A wire wound resistor consists of a resistance wire (made of constantan, manganin, or nichrome) wound on a heat-resisting (ceramic) tube. The ends of the wire are brought out to (such as rheostats and potentiometers). These resistors have resistances from a fraction of an ohm to thousands of ohm. The power rating varies from a fraction of a watt to several kilowatts.

Carbon Moulded Resistors

Electronic circuits need resistors of quite high value (say, of the order of megaohms). The cost of wire wound resistors of such high value is prohibitive. Since carbon has high resistivity, by powdering carbon and mixing it with a suitable binding material we can make resistors of high value. Such carbon moulded resistors are coated with some insulating material and baked to I IV resistance values range from a few ohms to tens of megaohms. The power rating is from 1/4 watts to a few watts.

II III

imprint their values on their surface. Therefore, it is usual practice to indicate the value by making four colour bands on its body (Fig. 2.21), using a standard colour code given in Table 2.1.

which the resistor was made by the manufacturer. In case the fourth band is not present, the tolerance is taken as ±20 %. EX A MP L E

2.1 5

as to satisfy the manufacturer’s tolerance.

Colour

Number

Black Brown Red Orange

0 1 2 3

Multiplier Tolerance (%) 1 101 102 103 4

* MNEMONICS

5 6

Blue Violet

6 7

10 107

White

9

109

Silver No colour

– –

As an aid to remember the sequence of colour codes given above, the student can memorise one of the following (all the capital letters

8

–1

10–2 –

(a

5 10 20

(b (c) B B R

Basic Electrical Engineering

Solution Using Table 2.1, we have 1st Band Yellow 4

2nd Band Violet 7

3rd Band Orange 103

4th Band Gold ± 5 % = 47 kW ± 5 %

47 103 5 W = 2.35 kW 100

Now 5 % of 47 kW =

Therefore, the resistance should be within the range 47 kW ± 2.35 kW or between 44.6 kW and 49.35 kW. EX A MP L E

2.1 6

to satisfy the manufacturer’s tolerance ?

Solution 1st Band Gray 8

2nd Band Blue 6

3rd Band Gold 10–1

5 % of 8.6 W =

8.6 100

4th Band Gold ±5 % = 86 ¥ 0.1 W ± 5 % = 8.6 W ± 5 % 5 = 0.43 W

The resistance should lie somewhere between the values (8.6 – 0.43) W and (8.6 + 0.43) W or 8.17 W and 9.03 W.

The resistance R of a material changes as its temperature changes. A rise in temperature increases the molecular movement within the material. As a result the drift of free electrons through the material is impeded. In other words the resistance of the material increases. Most conductors (pure metals) show this characteristic. For a moderate range of temperature, the change in resistance is usually proportional to the change in temperature. temperature, adopted as standard, is termed and is designated as a. Table 2.2 gives the value of a at 0 °C for different materials. In some materials such as semiconductors, electrolytes, etc., a rise in temperature makes more free electrons (or other charge-carriers) available to contribute towards the conduction of current. This increased availability of free electrons offsets the impeding effect of the enhanced molecular movements within the material. The net effect is a decrease in resistance of material with a rise in its temperature. Such materials are said to have . Assume that the resistance of a conductor at a standard temperature T0 (usually, T0 is taken as 0 °C) is R0 and at T1 is R1. Assuming that the variation of R with rise in temperature is linear, we can write R1 = R0 [1 + a0(T1 – T0)]

(2.33)

R2 = R0 [1 + a0(T2 – T0)]

(2.34)

where a0 T2, the resistance is

Ohm’s Law

Advance metal

*

Manganin

Constantan Copper

*

Platinum Platinum-iridium

0.003 7 0.00 1 2

Tantalum

0.003 3

31

*

International

German silver

0.000 36

Eliminating R0 from the above two equations, we get R1 R2 = 1 + a 0 (T1 - T0 ) 1 + a 0 (T2 - T0 ) If we put T0 = 0 °C, we get R1 R2 = 1 + a 0T1 1 + a 0T2

(2.35)

(2.36)

Obviously, similar relationship exists for resistivity r of the material. EXA M P L E

2. 1 7

The resistance of a transmission line is 126 W at 20 °C. Determine the resistance of the line at –35 °C. The temperature

Solution Given : R1 = 126 W, T1 = 20 °C, T2 = – 35 °C, a0 Using Eq. 2.36, we get

R2 R1 = 1 + a 0T2 1 + a 0T1 126 R2 = 1 + 0.00426 × 20 1 + 0.00426 ( −35) R2 =

0.8509 ¥ 126 = 98.8 W 1.0852

Basic Electrical Engineering EX A MP L E

2.1 8

The resistance of copper winding of a motor at room temperature of 20 °C is 3.42 W. After an extended operation of the motor at full load, the winding resistance increases to 4.22 W

Solution

R1 = 3.42 W, T1 = 20 °C, R2 = 4.22 W, a0 = 0.004 26

Using Eq. 2.36, R1 R2 = 1 + α 0 T1 1 + α 0 T2 3.42 4.22 = 1 + 0.004 26 × 20 1 + 0.004 26 T2

or or or

3.42(1 + 0.004 26T2) = 4.22(1 + 0.004 26 ¥ 20) T2 = 79.6 °C

\

The temperature rise = T2 – T1 = 79.6 – 20 = 59.6 °C

ADDITIONAL SOLVED EXAMPLES EX A MP L E

2.1 9

One metre long metallic wire is broken into two unequal parts P and Q. Part P of the wire is uniformly extended into another wire R. Length of R is twice the length of P and the resistance of R is equal to that of Q. Find (a) the ratio of the resistances of P and R, and (b) the ratio of the lengths of P and Q.

Solution Let the length of P be Then, the length of Q will be y = 1 – x. If k is the resistance per metre of the metallic wire, the resistances of the two parts are RP = kx and RQ = k(1 – x) When part P is extended to twice its length to make wire R, the area of cross section reduces to half (as the volume remains the same). Since the resistance of a wire is directly proportional to its length and inversely proportional to its area of cross section, the resistance of wire R will be 4 times that of P. That is, RR = 4RP = 4kx RR = RQ, or 4 kx = k(1 – x) fi x = 1/5. Therefore, y = 1 – x = 4/5. (a)

RP kx 1 = = = 1 : 4; RR 4 kx 4

EXA MP LE

(b)

lP x 1/ 5 1 = = = =1:4 4/5 4 lQ y

2.20

Reduce the networks, given in Fig. 2.22, to a single resistance between terminals A and B.

Solution (a) The resistance between terminals A and B is RAB = (5 || 6) + (4 || 3) =

5×6 4×3 + = 2.7 + 1.7 = 4.4 W 4+3 5+6

A

12 W

3W

10 W

6W

6W 3W

B

2W

6W

5W 4W

A

B (a)

(b)

(b) The resistance between terminals A and B is RAB = 2 + [{6 + (6 || 12)} || 10] + 3 = 2 + [{6 + 4} || 10] + 3 = 2 + [10 || 10] + 3 = 2 + 5 + 3 = 10 W EX A MP L E

2.2 1

Twelve identical wires, each of resistance 12 W, are arranged to form the edges of a cube as shown in Fig. 2.23a. A Calculate the potential difference between these two corners. 12 W 12 W 12 W

E

12 W

H

12 W

20 mA

12 W A

A

D

12 W

D, E B

12 W

H, C

12 W

F

12 W

G

12 W G

F C

B

12 W

20 mA

(a)

(b)

Solution Due to symmetry of the network, the current of 20 mA will equally divide into three branches AD, AE, and AB. Therefore, the points D, E, and C and F will be at the same potential. If the points at the same potential are of the given network is given in Fig. 2.23b. We have three sets of parallel combinations of equal resistances. The effective Rp1 = Rp3 = 12/3 = 4 W and that of the second set is Rp2 = 12/6 = 2 W. R

= Rp1 + Rp2 + Rp3 = 4 W + 2 W + 4 W = 10 W

V

= IR

= 20 mA ¥ 10 W = 0.2 V

Basic Electrical Engineering EX A MP L E

2.2 2

Figure 2.24a shows a network of 9 resistances. The number on each resistance represents its value in ohms. Find the resistance across the points E and F. D

D

D

6

6

A

A

6 A 16

16 4

10/3

4

2

O

F

7

E

10/3

B

C 2/3

2

7

F

E

1

10/3

16 (b)

16 (a)

2 O

6

C

B

16

16

6

C E

16

16

B 7

F

16 (c)

Solution The delta connection between the points A, B and (Fig. 2.24b), creating a new node O, RAO =

4×6 2×6 2×4 24 12 8 2 = = 2 W; RBO = = = 1 W and RCO = = = W 2+4+6 2 + 4 + 6 12 2 + 4 + 6 12 3 12

The network of Fig. 2.24a can now be drawn as in Fig. 2.24c. Reducing the series combination of two resistances into a single resistance gives us the network of Fig. 2.25 Now, converting the star into delta gives the network of Fig. 2.25b. Combining the parallel resistances, we get the network of Fig. 2.25c. The equivalent resistance between the points E and F is simply the parallel combination of the two branches, 8 × (8 + 32 / 3) 8 56 REF = = = 5.6 W 8 + (8 + 32 / 3) 80 D

D

8

G

16

16

16

16

I 4

8 16 (a)

F

8

32/3

32

16

O

E

D

16

E

H F

16 (b)

E

F 8 (c)

EX A MP L E

2.2 3

Determine the current drawn from the 5-V battery in the network shown in Fig. 2.26a. 5V

3W

I

5V

3W

G R1

2W

2W

B A

C 2W

3W

1W

4W

1W

2W 3W R

R2

4W

3

D (b)

(a)

Solution Current I drawn from the battery can be calculated easily, if we know the equivalent resistance Req of the network between the points A and (Fig. 2.26b). The star connection between points A, C and D can be converted into its equivalent delta connection (shown by dotted lines), using Eqs. 2.30 to 2.32, 2 × 3 + 3 × 2 + 2 × 2 16 = W 3 3 2 × 3 + 3 × 2 + 2 × 2 16 = =8W R2 = 2 2 2 × 3 + 3 × 2 + 2 × 2 16 = =8W R3 = 2 2 The resulting network is shown in Fig. 2.27a. Now, combining the parallel branches, we get the network of Fig. 2.27b. Combining the two series resistances, we get the network of Fig. 2.27c. We now combine the two parallel resistances to give an equivalent resistance of 16 32 × 32 3 9 16 32 = 15 W + 3 9 R1 =

I

5V

3W

G

I

5V

16/3 W

1W

8W

8W

3W

I

16/3 W C

A

G

4W

5V

G 16/3 W

C

A

B

A

8/3 W

8/9 W

32/9 W D (a)

D (b)

3W

(c)

Basic Electrical Engineering Thus, the network reduces to a simple circuit of Fig. 2.28 Finally, the two series resistances can be combined to get equivalent resistance of 77/15 W (Fig. 2.28b). The current I drawn from the battery is 5 75 I= = = 0.974 A 77 /15 77

5V

I

3W

G

77/15 W

32/15 W

G

A

C

A

5V

I

(b)

(a)

NO TE Hence, care should be taken during this process that no point of ultimate relevance is lost. EX A MP L E

2.2 4

and the pd between A and B. 5W

A

I

C

20 W

2.5 W

I1

5W

I2

D

15 V

10 W

2W A

B

B R

10 W

I3

C

Fig. 2.29

Fig. 2.30

Solution The equivalent resistance of the parallel combination is given by 1 1 1 1 + = + = 0.2 + 0.05 + 0.4 = 0.65 S RCD 5 20 2.5

fi RCD =

1 = 1.54 W 0.65

\ RAB = 1.54 + 10 = 11.54 W Total voltage V across A and B is given by P = \ \ EXA MPL E

I =

V2 R

fi

V=

PR =

V 75 = = 6.5 A R 11.54

VCD = V – 10I = 75 – 10 ¥ 6.5 = 10 V; I1 =

488 11.54 = 75 V

10 10 10 = 2 A; I2 = = 0.5 A; I3 = =4A 5 20 2.5

2.25

In the circuit shown in Fig. 2.30, voltage VAB = 5 V. Find the value of resistance

Solution Applying voltage divider concept, VAC = 15 ¥

10 = 10 V; 10 + 5

\

VBC = VBA + VAC = –VAB + VAC = –5 + 10 = 5 V

Again applying voltage divider concept to the second branch, R 15 R VBC = 15 ¥ ; or 5 = or 2+R 2+R EXA MP LE

10 + 5R = 15R fi

R=1W

2.2 6

In the circuit shown in Fig. 2.31a, determine (a) the current supplied by the 100-V source, and (b) the voltage across 6-W resistance. A 5W

B 5W

A IAB

I 5W

5W

3W

B IBE 5W

IAD

IBC

5W

3W E

E 100 V

10 W

10 W

I1 6W

D

C

I2

100 V

10 W

10 W

3W

I1 6W

D

I2 3W

C (b)

(a)

Solution (a) RBC = 10 || {5 + 3 + (6 || 3)} = 10 || {5 + 3 + 2} = 10 || 10 = 5 W RAD = 10 || (5 + 5) = 5 W; \ Rtotal = 5 + 5 = 10 W Therefore, the total current supplied by the 100-V source is 100 I= = 10 A 10 (b I1 through 6-W resistance. At A, the current of 10 A divides into two parts—IAD and IAB (see Fig. 2.31b). Note that the current IAB depends not only on RAB, but the total resistance on the right of point A (which is 10 W). Thus, the two currents will be equal. That is, IAD = 5 A and IAB = 5 A. Again at B, current IAB divides equally into IBC = 2.5 A and IBE = 2.5 A. The current IBE divides into I1 and I2. Thus, 3 2.5 2.5 I1 = 2.5 ¥ = A; \ V6W = I1 ¥ 6 = ¥6=5V 6+3 3 3 EXA MP LE

2.27

A resistance of R ohms is connected in series with a parallel combination of 8 ohms, 12 ohms and 24 ohms. The total power dissipated in the circuit is 80 W, when the applied voltage is 20 V. Calculate

Solution The equivalent resistance of the parallel combination is given by 1 1 1 1 3+ 2 +1 1 = + + = = ; \ Rp = 4 W Rp 8 12 24 24 4

Basic Electrical Engineering The total resistance of the circuit is Rtotal = R + 4 ohms. Power dissipated is given as P= EXA MPLE

V2 Rtotal

or

Rtotal =

V2 20 2 400 = = = 5 W; fi P 80 80

R = R total – Rp = 5 – 4 = 1 W

2.28

A combination of two resistances R1 and R2 connected in parallel across a 100-V supply takes 10 A current from the mains. Determine the resistance R2, if the power dissipated in R1 is 600 W.

Solution The power dissipated in R1 is given as

\ EXA MPL E

P =

V2 R1

I1 =

V 100 V 100 = = 6 A; I2 = I – I1 = 10 – 6 = 4 A; R2 = = = 25 W 4 R1 100 / 6 I2

fi

R1 =

V2 100 2 100 = = P 600 6

2.29

a) the quantity of oil needed per hour, and (b) the electrical energy generated

Solution Input power =

Output power 50 kW = = 142.9 kW 0.35 Efficiency

Heat energy per hour = 142.9 kW h = 142.9 ¥ 860 kcal

\

(a) Fuel needed per hour =

142.9 860 kcal = 9.8315 kg 12 500 kcal/ kg

(b) The power output of the generator is 50 kW. In one hour, the energy generated is 50 kW h, and the fuel needed is 9.833 kg. Thus, the energy generated per tonne (= 1000 kg) of fuel is = EXA MP L E

50 kW h ¥ 1000 = 5086 kW h 9.833

2 . 30

Two heaters A and B are connected in parallel across a supply voltage. Heaters A and B produce 500 kcal in 20 minutes and 1000 kcal in 10 minutes, respectively. The resistance of heater A is 10 ohms. (a) Calculate the resistance of heater B (b) If the two heaters are connected in series across the same supply voltage, how much heat will be produced in 5 minutes.

Solution (a) The amounts of heat produced by the two heaters are

and

HA =

V2 V 2 (20 60 s) V 2 (20 60 s) ¥ t or 500 ¥ 4.2 kJ = J= kJ RA 10 10 1000

(i)

HB =

V2 ¥ t or RB

(ii)

1000 ¥ 4.2 kJ =

V 2 (10 60 s) V 2 (10 60 s) kJ J= RB RB 1000

Dividing (i) by (ii), we get 0.5 = 2RB /10 = RB /5 fi

RB = 2.5 W

(b) The series resistance, R = 10 + 2.5 = 12.5 W and time t = 5 ¥ H =

V2 V 2 (5 60 s) ¥t= kJ R 12.5 1000

(iii)

Dividing (iii) by (i), we get H 500 EX A MP L E

4.2

=

10 1000 12.5 1000

5 60 kJ 20 60

or

H = (500 ¥ 4.2) ¥

1 kJ = 420 kJ = 100 kcal 5

2.3 1

8 lamps of 100 W each, 3 fans of 80 W each, 1 refrigerator of 1/2 hp, 1 heater of 1000 W. (a) Calculate the total current taken from the supply of 230 V. (b) Calculate the energy consumed in a day, if on an average only a quarter of the above load persists all the time.

Solution (a) The total load is given as Item 1. 2. 3. 4.

Load

8 lamps of 100 W each 3 fans of 80 W each 1 refrigerator of 1/2 hp 1 heater of 1000 W

8 ¥ 100 = 800 W 3 ¥ 80 = 240 W 1 ¥ 1/2 hp = (1/2) ¥ 746 W = 373 W 1 ¥ 1000 = 1000 W = 2413 W

Total load

P 2413 = ª 10.5 A V 230 (b) Energy consumed per day = 2413 W ¥ (1/4) ¥ 24 = 14478 Wh = 14.478 kW h \

Current taken from the supply, I =

SUMMARY TE RM S

O F

C O NC EPTS

Ohm’s law states that the potential difference between the two ends of a conductor is directly proportional to the resistance (measured in W) is called conductance (measured in S). short circuit (R i π 0) without any resulting voltage (v = 0). An open circuit (R = ) permits voltage (v π 0) with no current (i = 0). Resistance series parallel connected if the same potential difference is applied to each of them.

IMP O R TA N T Ohm’s Law

F O RM U LAE

V = RI or I = GV

40

Basic Electrical Engineering

L , where r is the resistivity of the material, measured in W m. A Rs = R1 + R2 + R3; if n equal resistances, each R ohms, are connected in series, their equivalent resistance is nR ohms. 1 1 1 1 = + + ; if n equal resistances, each R ohms, are connected in parallel, Rp R1 R2 R3

R=r

their equivalent resistance is R/n ohms. R1 R2 Voltage divider V1 = V ; Current divider I1 = I R1 + R2 R1 + R2 Delta-to-star

RA =

R2 R3 ; Star-to-delta R1 + R2 + R3

R1 = RB + RC +

RB RC RA

R1 = R0 [1 + a 0 (T1 – T0)]

CHECK YOUR UNDERSTANDING two marks for each correct answer and minus one for each wrong answer. If your score is 12 or more, go to the next chapter; otherwise study this chapter again.

6.

given as the ratio of the voltage across it to its conductance’. Potential difference can be expressed as voltage rise or voltage drop. a) of a substance depends on its temperature. A cassette player requires three 1.5-V batteries. If one battery is inserted backward, the net voltage developed will be 3 V. A battery charger supplies 5 A current into a 12.6-V car battery. In this process, the power supplied to the car battery is 63 W. A 1000-W resistor is rated at 5 W. Maximum current that can be

7.

Two circuit elements are said to be connected in series when same

8.

Two resistors are said to be in parallel if they share the same current. G are connected in parallel, their equivalent resistance will be 5/G. In a current divider consisting of three resistors, the resistances are in the

2.

4. 5.

10.

Your Score A N SW E RS

1. False 6. False

2. True 7. True

3. True 8. False

4. False 9. False

5. True 10. True

Ohm’s Law

41

REVIEW QUESTIONS 1. material. 2. Explain what is meant by ‘conventional current’. How does it differ from the current due to electron3. What is meant by the terms : (a) a short circuit, and (b) an open circuit. 4. Explain the meaning of and switches.

5. Describe the resistance parameter from three different points of view : circuit, energy and geometric. 6. How does the resistance of a conductor vary with 7. Why do we use colour bands to indicate the values 8. What do the third and fourth colour bands on a

MULTIPLE CHOICE QUESTIONS

. r depends upon (a) the material, the area of cross section and the length of the conductor (b) the area of cross section and the length of the conductor only (c) the area of cross section of the conductor only ( ) the nature of the material of the conductor only 2. The resistance of a conductor increases when (a) its length increases (b) its area increases (c) both its length and area increase ( ) 3. On increasing its temperature, the resistance of a conductor made of a metal (a) decreases (b) increases (c) remains constant ( ) varies either way 4. 1.

(a) increase in its resistance per degree celsius (b (c) increase in its resistance per ohm per degree celsius ( ) decrease in its resistance per degree celsius per ohm

5. The ‘ampere second’ could be the unit of (a) conductance (b) power (c) energy ( ) charge 6. The polarity of voltage drop across a resistor is determined by (a) the value of the resistor (b) the value of current through the resistor (c) the direction of current through the resistor ( ) the polarity of the source 7. If 110 V is applied across a 220-V, 100-W bulb, the power consumed by it will be (a) 100 W (b) 50 W (c) 25 W ( ) 12.5 W 8. A resistance of 10 W is connected across a supply of 200 V. When another resistance of R ohms is connected in parallel with the above 10-W resistor, the current drawn from the supply doubles. The value of R is (a) 5 W (b) 10 W (c) 20 W ( W 9. Three resistances each of R ohms are connected in star. Its equivalent delta will comprise three resistances each of value (a) 3R ohms (b) 2R ohms (c) R ) R 10. The current in a 5-W is 2 A. If this resistor is replaced by another of 10 W, the current in the branch will be (a) more than 2 A (b) less than 2 A (c) 2 A ( ) 1A

42

Basic Electrical Engineering

11. Three resistances each of R ohms are connected in delta. Its equivalent star will comprise three resistances each of value (a) 3R ohms (b) 2R ohms (c) R/3 ohms (d) R/2 ohms 12. A current of 10 A enters a parallel combination of two resistances of 2 W and 3 W. Then the currents through these resistances will be (a) 4 A, 6 A (b) 6 A, 4 A (c) 2 A, 8 A (d) 5 A, 5 A 13. Sixteen resistances each of 32 W are connected in parallel. The equivalent resistance of this

combination is (a) 2 W (b) 16 W (c) 32 W (d) 16 ¥ 32 W 14. Three resistors of 4 W, 6 W and 9 W are connected in parallel in a network. Maximum power will be consumed by (a) 4-W resistor (b) 6-W resistor (c) 9-W resistor (d) all resistors 15. The minimum number of resistors required to form a series-parallel combination is (a) one (b) two (c) three (d) four A N SW E R S

1. d 11. c

2. a 12. b

3. b 13. a

4. c 14. a

5. d 15. c

6. c

7. c

8. b

9. a

10. b

PROBLEMS ( A )

S IM P L E

PR OB LEM S

1. Two cubes of different materials measuring k metres and 2k metres on one side have equal resistances between any two faces. Find the ratio of their resistivities. [Ans. 2. Find the effective resistance between terminals A and B for the networks given in Fig. 2.32. [Ans. (a) 2 W; (b) 10 R/9; (c) 12 W; (d) 5 W; (e) 10 R/3]

3. The resistance of two coils is 25 W when connected in series, and 6 W when connected in parallel. Determine the individual resistances of the two coils. [Ans. 15 W , 10 W] 4. Determine the current Ix W resistor in the circuit of Fig. 2.33. [Ans. 4/3 A]

10. Determine the total resistance between points A and B in the network shown in Fig. 2.36. [Ans. 3 W] 12 W 6W 4W

8W

A

3W

5. When two resistances R1 and R2 are connected in parallel, they dissipate four times the power that they dissipate when they are connected in series with the same ideal source of emf. If R1 = 3 W R2. [Ans. 3 W] 6. If two electric bulbs, each designed to operate with a power of 500 W in 220-V line, are put in series in a 110-V line, what will be the power dissipated by each bulb ? [Ans. 31.25 W] 7. A resistance of 8 W is connected in series with a parallel combination of 12 W and 24 W. The whole circuit is connected across a 100-V supply. Find (a) the current drawn from the supply, (b) the voltage across 8-W resistance, and (c in 12-W and 24-W resistances. [Ans. (a) 6.25 A; (b) 50 V; (c) 4.17 A, 2.08 A] 8. Determine I, I1, I2, Vab, and Vbc in the network of Fig. 2.34. [Ans. 3 A, 1 A, 2 A, 6 V, 12 V] 2W

b I2

c

6W

18 V

9. Determine the voltage across and current through each resistor in the network of Fig. 2.35. [Ans. 10 A, 50 V, 30 V, 7.5 A, 2.5 A] 4W A

B

I1 I

5W

C I

I2

12 W

80 V

11.

1.9 W

a) the equivalent resistance between points A and B, (b) the current and power supplied by the battery. [Ans. (a) 8 W; (b) 6 A, 288 W] A 5W

48 V

2W

6W

3W

6W

9W

B

12. Determine the current in 15-W resistance in the circuit shown in Fig. 2.38. [Ans. 0.4 A]

I1 12 W a I

2.1 W

B C

44 ( B)

Basic Electrical Engineering

T R IC KY

PR OB LE MS

13. Calculate the current drawn from a 12-V supply with internal resistance 0.5 W ladder network, each resistance being 1 W, in Fig. 2.39. [Ans. 3.71 A] A

17. In the network shown in Fig. 2.43, calculate (a) the current in the other resistances, (b) the value of the unknown resistance and (c) the equivalent resistance between points A and B. [Ans. (a) 2 A, 1 A, 2 A; (b) 15 W; (c) 3 W] A

To infinity

(12 V, 0.5 W)

10 A 5A

V

I1

6W

B

I2 30 W

XW

I3 15 W

B

14. Determine the value of resistance R, if the power dissipated in 10-W resistance is 360 W in the circuit of Fig. 2.40. [Ans. 36 W]

18. Find the voltage V for the circuit shown in Fig. 2.44. [Ans. 310 V]

12 W 10 W

A

I

B 2A

18 W

50 W

20 W 100 W

50 W

V

R

I1 C

96 V

19. 15. The current in the 6-W resistor of the network shown in Fig. 2.41 is 2 A. Find the currents in all other resistors and the voltage across the network. [Ans. 3.5 A, 1.5 A, 2.5 A, 1 A, 46 V] 8W A

4W

8W

C

D 6W

B

a) R and Vs, and (b) the power delivered by the source Vs. [Ans. (a) 3 W and 8 V; (b) 16 W] A

B I1

I 1W

+ Vs –

+ 6V –

I2

C I3

R

6W

12 W

0.75 A 4W

20 W D

V

16. Find the equivalent resistance between points A and B in the network shown in Fig. 2.42. [Ans. 4 W]

20. In the circuit shown in Fig. 2.46, determine (a) the equivalent resistance between points A and B, (b) the total current, and (c) the power delivered to the 16-W resistance. [Ans. (a) 28 W ; (b) 3.5714 A; (c) 12.75 W]

24 W A

8W

8W

6W

4W

8W A B

100 V

B

24 W

8W

6W

24 W

4W

16 W

45 21.

a) the current in 15-W resistance, (b) the voltage across 18-W resistance, and (c) the power dissipated in 7-W resistance. [Ans. (a) 6.33 A; (b) 18 V; (c) 77.6 W] 9W 4W 18 W

2W

7W

15 W 125 V

22. If the total power dissipated in the network shown in Fig. 2.48 is 16 W, calculate the value of R and the total current. [Ans. 6 W; 2 A] 4W

2W

R

8W

parallel, each of 20 W. Calculate the value of resistance which should be shunted across the parallel combination so that the total current drawn by the circuit is 1.5 A with applied voltage of 20 V. [Ans. 5 ohms] 26. A resistance of R ohms is connected in series with a parallel circuit, consisting of 15-W and 10-W resistances. When a voltage of 20 V is applied across the circuit, a total power of 40 W is dissipated. Find R. [Ans. 4 ohms] 27. The circuit shown in Fig. 2.49 represents a batterycharger charging a battery. (a) Find the power into the battery being charged. (b) What is the time it would take to impart 1 kilocoulomb charge to the battery ? (c) What energy is given to the battery in this period of time ? [Ans. (a) 44.8 W; (b) 286 s; (c) 12.8 kJ] 28. In the circuit of Fig. 2.50, determine (a) the voltage vab, (b) the current i, and (c) the sum of powers into R1, R2 and the 2-V battery. [Ans. (a) 9 V; (b) –15 A; (c) 84 W] i = 3.5 A

I 14.8 V

12.8 V

8V

23. them are 0.6 A, 0.3 A, 0.2 A and 0.1 A, respectively. Find the value of each resistance. [Ans. 40 W, 80 W, 120 W, 240 W] 24. A 10-W resistance is in series with a parallel combination of 15-W resistance and 5-W resistance. If the current in 5-W resistance is 6 A, what total power is dissipated in the three resistances ? [Ans. 880 W] 25. A resistance of 10 W is connected in series with a combination of two resistances arranged in

Battery charger

R1

a

b –3 A

2V

7V

R2 i

12 A c

46 ( C)

Basic Electrical Engineering

C H A LL ENG I NG

P ROBLEMS

29. Determine the equivalent resistance at the terminals A-B of the networks given in Fig. 2.51. [Ans. (a) 4 W; (b) 36 W; (c) 18.75 W]

31. (a) Find the equivalent resistance of the parallel combination shown in Fig. 2.53. (b) If 10 A current enters the parallel combination at point a, W resistance ? (c in the 2-W resistance ? (d) What is the voltage Vab for this current ? [Ans. (a) 1.09 W; (b) 1.82 A; (c) –5.45 A; (d) 10.9 V] a 2W

4W

6W

b

32. (a) Two series resistances are to work as a voltage divider, with the smaller receiving 30 % of the total voltage. What are the resistors, given that their equivalent resistance is 200 W ? (b) If both resistors are 2-W resistors, what is the maximum total voltage the voltage divider can handle without exceeding this rating for either resistor ? [Ans. (a) 60 W and 140 W; (b) 23.9 V] 33. indicated unknowns using the concept of voltage and current dividers. [Ans. (a) 1.15 A; (b) 20 V, 0.16 A; (c) 0.8695 V; (d) –5.0 V] 30. A car radio designed to operate from a 6.3-V system uses 4.5 A of current, as shown in Fig. 2.52. (a) What resistance R should be placed in series with this radio if it is to be used in a 12.6-V system ? (b) What should be the power rating of this resistance ? [Ans. 1.4 W, Min. 28.35 W] 4.5 A

R +

12.6 V

6.3 V RADIO

34. An electric kettle is required to heat 0.6 kg of water from 10 °C to boiling point in 5 minutes, the supply 78 %. Calculate (a) the resistance of the heating element, and (b) the cost of the energy consumed at ` 2.30 per kW h. Assume 1 kcal = 4200 J. [Ans. (a) 54.53 W; (b) 20 p] 35. An electric kettle contains 1.2 kg of water at 20 °C. It takes 20 minutes to raise the temperature to 100 °C. Assuming the heat loss due to radiation, and by the kettle from the supply of 230 V. Assume

–

[Ans. 1.675 A]

47

36. The resistance of a coil embedded in a large transformer is 12 W at 25 °C. After the transformer has been in operation for several

hours, the resistance is found to be 13.4 W. If a = 393 ¥ 10–5 at 20 °C, what is the temperature of the core of the transformer ? [Ans. 53.26 °C]

EXPERIMENTAL EXERCISE 2.1 V-R

C H A RA CT E RIS TI C S

OF

LAMP S

Objectives

Apparatus Single-phase AC power supply 220 V; Variac 0-270 V, 15 A; Voltmeter (MI type) 0-250 V; Ammeter

0-1 A A 220 V AC Supply

V

MI 0-250 V Lamp

Variac (0-270 V, 15 A)

Resistance (W)

Circuit Diagram The circuit diagram is shown in Fig. 2.55. Carbon filament

Tungsten filament

Voltage (V)

48

Basic Electrical Engineering

Brief Theory voltage. Hence, the effect of voltage variation on resistance is similar to the effect of temperature variation on resistance. If R0 T0 (0 °C), its resistance R1 at a temperature T1 is given as R1 = R0 + a0(T1 – T0)R0 where a0 is called of the material at 0 °C. The resistance of tungsten increases with temperature (a0 positive), but that of carbon decreases with temperature (a0 negative).

Procedure

3. Increase the voltage up to 220 V in steps of 20 V, and after waiting for 2-3 minutes take readings of the voltmeter and ammeter. 5. Switch off the supply.

Observations No

V (in V)

I in (A)

R = V/I (in W)

V (in V)

I (in A)

R = V/I (in W)

1 2 3 4 5

Calculations For each set of readings of the voltmeter and ammeter, calculate the resistance of the lamp using the relation, R=

V I

Graphs Plot the graphs of resistance versus applied voltage for both the lamps, preferably on the same graph paper (as in Fig. 2.56).

Results 1. The variation of resistance with the applied voltage (and hence with the temperature) is shown in the graph, for both lamps.

Precautions 1. Before switching on the supply, the zero reading of the voltmeter and ammeter should be checked. taking the readings.

49

Viva Voce 1. Ans. : Copper, Aluminium, Silver, etc. 2. Ans. : 3.

4. 5. 6. 7.

Ans. : melt at such high temperature. Can you name some materials whose resistance is not much affected by rise in temperature ? Ans. : Manganin and constantan. That is why these materials are used in making rheostats. What is the effect of temperature on the resistance of insulating materials ? Ans. : Their resistance decreases with rise in temperature. Name a material that is bad conductor of electricity but good conductor of heat. Ans. : Mica. Can you think of some application of mica, in which this property is used ? Ans. :

NETWORK ANALYSIS

3

OB JE CT IV E S :

(KVL)

(KCL)

A network can be modelled in terms of (a) the interconnection of elements, components or devices, (b) a set of input signals, and (c) a set of output signals. The description of a network is written in terms of network variables. These network variables are ‘current through the components’ and ‘voltage across the components’. There are two kinds of components or devices in a network : (i) active, and (ii) passive. An active device supplies energy to the passive devices. The active device is also called a ‘source’, and the passive device is called a ‘load’. There are three basic passive devices or components : (i) Resistor (having resistance, R), measured in ohms (W), (ii) Capacitor (having capacitance, C), measured in farads (F), and (iii) Inductor (having inductance, L), measured in henrys (H). The energy received by a passive component is (i) Dissipated as heat, as in case of a resistance, or (ii) Stored in it, in the form of : (a (b

51 There are three different points of view you can look at a passive component : (i) We describe the component in terms of voltage-current relationship. This view point explains its behaviour when connected in a circuit. (ii)

We describe the component in terms of energy dissipated by it or stored in it.

(iii) We describe the component in terms of its geometrical dimensions and the properties of the material used.

R) The resistance element was discussed in detail in the last chapter. From circuit viewpoint, a resistance R is v = Ri or

R=

v i

(3.1)

From energy viewpoint, a resistance R is described by W = i2 Rt or

R=

W i 2t

From geometric viewpoint, a resistance R is described as L R =r A where, r is the resistivity of the material.

(3.2)

(3.3)

C) when a circuit is switched ON or OFF. At the time of switching ON, as there is no voltage across the capacitor* , it behaves as a short circuit. When the capacitor is fully charged, its voltage remains constant and the current i reduces to zero. It then behaves like an open circuit.

i) Circuit Viewpoint From circuit view point, a capacitance C

-

tion, i dv or C = (3.4) dv/dt dt It shows that the unit of capacitance is ‘ampere-second/volt’, also ‘coulomb/volt’. This unit is given a name farad (F). The above relationship has been represented graphically in Fig. 3.1a. Figure 3.1b shows circuit symbol of a capacitor. There are two important properties of a capacitor. Both these properties can be derived from Eq. 3.4 : i =C

(i) (ii) The voltage across a capacitor cannot change instantaneously. *

The physical device is called capacitor (or condenser), whereas its property is called capacitance. However, the two terms are often used interchangeably.

52

Basic Electrical Engineering

If the voltage is constant (dv = 0), Eq. 3.4 dictates that the current has to be zero (i = 0). If the voltage across the capacitor has to change dv, we must have dt nite which is a physical impossibility.

ii) It can be shown that the energy stored in a capacitor is W = ( 12 )Cv2 (3.5) Therefore, from energy point of view, a capacitor can be described in terms of energy stored as 2W C= 2 (3.6) v iii) Geometric Viewpoint Consider, for example, a parallel plate capacitor. If A is the area of each plate separated by a distance d and e is the permittivity of the material between the two plates, the capacitance of such a device is given as A C =e (3.7) d e = er e0, where er is the relative permittivity (a pure number) and e0 is the permittivity of free space (e0 = 8.85 ¥ 10 –12 F/m). Note that Eq. 3.7 gives the value of the capacitance C in terms of geometric

L) when there occurs a change in current through the circuit.

i) The behaviour of an inductance can be easily understood when we consider it from circuit point of view. The voltage v current i through the circuit, v di v = L dt or L = (3.8) di/dt The constant L is called inductance. Above equation shows that the unit of inductance is ‘volt-second/ ampere’ or ‘Vs/A’. This unit is given the name henry (H). The inductance has been represented graphically in Fig. 3.2a. Figure 3.2b shows circuit symbol of an inductance. The circuit element which exhibits the property given by Eq. 3.8 is called inductor. Ideally, an inductor has no resistance. But in practice, since an inductor is made by putting a conducting wire in a coil form, it has a little resistance. If the value of the inductance L is independent of current i, it is said to be a linear inductance.

53 There are two important properties of an inductor. Both these properties can be derived from Eq. 3.8 : (i) (ii) The current through an inductor cannot change instantaneously. If the current is constant (di = 0), Eq. 3.8 dictates that the voltage has to be zero (v = 0). If the current through di, we must have dt = 0), the voltage would have to An inductor has a property of opposing change in current. This property can be looked upon as the property of inertia ON or OFF. At the time of switching ON, the current i in the circuit is zero (though it tends to change very fast with time), it behaves as an open circuit short circuit.

ii)

It can be shown that the energy stored in an inductor is W = ( 12 )Li2 (3.9) Therefore, from energy point of view, an inductor can be described in terms of energy stored as 2W L= 2 (3.10) i iii) Geometric Viewpoint It can be shown that the inductance of a coil of N turns having an area of cross section A and a length l is given as N 2A L= (3.11) l where m is the permeability m = mr m0, where mr is the relative permeability (a pure number) and m0 is the permeability of free space (m0 = 4p ¥ 10–7 H/m). Note that Eq. 3.11 gives the value of the inductance L IMP O R TA N T

N O T E

Comparison of Eq. 3.4 with Eq. 3.8 shows that inductance (L) is the dual of capacitance (C). EX A MP L E

3.1

Find the value of capacitance parameter in each of the following cases : (i

having a total area of 0.113 m2.

(ii (iii

Solution (i) Using Eq. 3.7, we get C =e

A A 10 × 8.854 × 10 −12 × 0.113 = er e0 = = 0.1 mF d d 0.1 × 10 − 3

(ii) Using Eq. 3.6, we get C =

2W v

2

=

2

0.05

(100)2

= 10 mF

54

Basic Electrical Engineering

(iii) Using Eq. 3.4, we get C = EXA MP LE

i 5 × 10 −3 = = 5 mF d v/dt 100 / 0.1

3 .2

Find the inductance of the coil in each of the following : (i (ii) A current increases linearly from zero to 0.1 A in 0.2 s producing a voltage of 10 V. (iii) A current of 0.1 A increases at the rate of 0.5 A/s producing a power of 2.5 W.

Solution (i) Using Eq. 3.10, we get L =

2W i

2

=

2 0.2 = 10 H 0.2 0.2

(ii) Using Eq. 3.8, we get L =

v 10 = = 20 H di/dt 0.1/ 0.2

(iii) We know that instantaneous power is given as p = iv. Therefore, di

p = iL dt or

L =

p 2.5 = = 50 H i ( di /dt ) 0.1 0.5

The rules for combining inductances are the same as those for resistances. That is, the equivalent inductance of n inductances connected in series is given as n

Ls = L1 + L2 + L3 … + Ln =

Li

(3.12)

i =1

Same way, the equivalent inductance of n inductances connected in parallel is given by 1 1 1 1 1 = + + …+ = Lp L1 L2 L3 Ln

n

i =1

1 Li

(3.13)

Since the capacitance (C) is dual of inductance (L), the rules for combining capacitances will be dual of those for inductances. Hence, the equivalent capacitance of n capacitances connected in series is given by n

1 1 1 1 1 = + + …+ = Cs C1 C2 C3 Cn

i =1

1 Ci

(3.14)

Same way, the equivalent capacitance of n capacitances connected in parallel is given as n

Cp = C1 + C2 + C3 … + Cn =

Ci i =1

(3.15)

55 EX A MP L E

3.3

Three inductors are connected as shown in Fig. 3.3. Given that L1 = 2L2. Find L1 and L2 such that the equivalent inductance of the combination is 0.7 H. C4

C3

C2

C1

0.05 mF

0.20 mF

0.10 mF

0.05 mF

L1 0.5 H L2

220 V

Solution The equivalent inductance is given as Leq = 0.5 +

L1L2 L1 + L2

or

0.7 = 0.5 +

L1L2 L1 + L2

fi

(2 L2 ) L2 = 0.2 (2 L2 ) + L2

Hence, L2 = 0.3 H and L1 = 0.6 H. E X A M P LE

3.4

Determine the equivalent capacitance of the network shown in Fig. 3.4 and the voltage drop across each capacitor.

Solution The total capacitance (in mF) is given by 1 1 1 1 1 1 1 1 1 = + + + = = 55 + + + Cs C1 C2 C3 C4 0.05 0.10 0.20 0.05 \

Cs =

1 = 0.0182 mF 55

The charge transferred to each capacitor is Q = CsV = 0.0182 ¥ 10 −6

Q 4 × 10 = = 80 V C1 0.05 × 10 − 6 Q 4 × 10 − 6 V3 = = = 20 V C3 0.20 × 10 − 6

\

V1 =

EXA MP LE

–6

¥ 220 = 4 ¥ 10 –6 C Q 4 × 10 − 6 = = 40 V C2 0.10 × 10 − 6 Q 4 × 10 − 6 V4 = = = 80 V C4 0.05 × 10 − 6

V2 =

3.5

Two capacitors are charged as shown in Fig. 3.5. After the switch is closed, what voltage exists across each capacitor ?

Solution Since the voltage polarities across the two capacitors are same, on closing the switch the total charge available for redistribution is Q = 400 mC + 200 mC = 600 mC. Since, the two capacitors are connected in parallel, the equivalent capacitance, C = 2 mF + 10 mF = 12 mF. Hence the voltage across the parallel combination is given as V=

Q 600 × 10 − 6 = = 50 V C 12 × 10 − 6

2 mF 200 V 400 mC

+ –

+ C1

C2

–

10 mF 20 V 200 mC

56

Basic Electrical Engineering

EX A MP L E

3.6

A series combination of two capacitances C1 = 2 mF and C2 = 8 mF is connected across a dc supply of 300 V. Determine (a) the charge, (b) the voltage, and (c) the energy stored in each capacitor.

Solution (a) Charge : The equivalent capacitance of the series combination is given as CC 2×8 C = 1 2 = = 1.6 mF C1 + C2 2+8 charge on each capacitor is same. Hence, the charge on each capacitor is given as Q = CV = (1.6 mF) ¥ (300 V) = 480 mC (b) Voltage : The voltages across the two capacitors are given as Q 480 C Q 480 C V1 = = = 240 V and V2 = = = 60 V C1 2 F C2 8 F (c) Energy Stored : W1 = ( 12 )C1V12 = ( 12 )(2 ¥ 10 –6)(240)2 = 57.6 ¥ 10–3 J = 57.6 mJ W2 = ( 12 )C2V22 = ( 12 )(8 ¥ 10 –6)(60)2 = 14.4 ¥ 10–3 J = 14.4 mJ

EXA MP LE

3.7

In the network of Fig. 3.6 a, determine the value of capacitance C such that the equivalent capacitance between point A and B is 1 mF. C7

C1

C

C

C1

C7

1 mF

4 mF

4 mF

C2

12 mF C6

C8

8 mF

C4

C5

4 mF C9

4 mF

C3

6 mF

1 mF

2 mF

C2

8 mF

A

2 mF

A

B

B (b)

(a)

Solution Capacitances C3 and C4 are in parallel. Therefore, their equivalent capacitance C8 = 2 mF + 2 mF = 4 mF. Capacitances C5 and C6 are in series. Therefore, their equivalent capacitance C9 is (6 ¥ 12)/(6 + 12) = 4 mF. The given network reduces to that shown in Fig. 3.6b. Combining C2 and C8 gives C10 = 8/3 mF. Combining C7 and C9 gives C11 = 8 mF. Thus, the network reduces to that shown in Fig. 3.7a. Capacitances C1 and C11 are combined to give C12 = 8/9 mF (Fig. 3.7b). Capacitances C10 and C12 are combined to give C13 = 32/9 mF (Fig. 3.7c). Since the required equivalent capacitance between point A and B is 1 mF, we must have 1 1 9 = + 1 C 32

or

1 23 = C 32

or

C=

32 = 1.39 mF 23

57 C

C1

C

A

C12

1 mF C10

8/3 mF

A

B

8/9 mF C11

C13

C

A

32/9 mF

8 mF 8/3 mF B

B C10 (b)

(a)

(c)

tives (V = RI or I = GV), similarly an energy source could be taken as a voltage source or a current source. ideal voltage source and an ideal current source mean.

V) is independent of the output current (I). It means its terminal voltage is independent of the load resistance (RL) connected to it. For any value of RL, right from a, and its graphical characteristic is shown in Fig. 3.8b. Note that the source determines the voltage, but the current is determined by the load. Figure 3.9 shows the circuit symbols for different types of ideal voltage sources. I

I

+

Vs

Discharge

V

V

0 Vs

– V = Vs independent of I (a)

(b)

Charge

Basic Electrical Engineering

I

constant current, independent of its output voltage. Thus, the output current of such a source remains unchanged for RL varying

IS

I

+

IS

V

0

V

(i.e., an open circuit). The mathematical –

described by Fig. 3.10a, and its graphical I = IS characteristic is shown in Fig. 3.10b. Note independent of V that the source determines the current, (a) (b) but the voltage is determined by the load. Figure 3.11 shows the circuit symbols for different types of ideal current sources. Note that practically no such device as an ideal voltage source or an ideal current source exists. A source it is impossible. However, the concept of ideal sources helps us to describe the characteristics of practical sources.

Consider a load resistance RL connected to the terminals A-B of a practical voltage source such as a battery (Fig. 3.12a). What happens when we increase the load* on the source (by decreasing the value of the load resistance RL VL decreases. To account for this fact, a practical voltage source is modelled as (as shown in the dotted box in Fig. 3.12a). This resistance is named as ‘internal resistance’ or ‘source resistance’, RSV. The voltage of this ideal source is called the electromotive force (emf) of the practical voltage source. Note that the resistance RSV internal resistance is merely introduced to account for the non-ideality of the practical voltage source. The linear relationship between VL and IL is given as VL = VSV – RSV IL This characteristic line is plotted in Fig. 3.12b ideal voltage source. The open-circuit voltage (V ) and short-circuit current (ILSC) are given as V = VS (3.16) VS and ILSC = (3.17) RSV *

In electrical engineering, the term ‘load’ means the ‘load current (IL)’, and not the ‘load resistance (RL)’.

A practical current source is modelled as (as shown in the dotted box in Fig. 3.13a). This resistance RSI is called the internal resistance. (Note that a practical current source is dual of a practical voltage source.) The characteristic of a practical current source is described by the straight line V IL = IS – L RSI This characteristic line is plotted in Fig. 3.13b source. The open-circuit voltage (V ) and short-circuit current (ILSC) are given as V = RSI IS ILSC = IS

EX A MP L E

(3.18) (3.19)

3.8

A battery of emf 3 V and internal resistance of 1 W is used to supply power to a variable load resistance RL of range (a) 100 W to 1000 W; (b) 1 mW to 10 mW, as shown in Fig. 3.14. Determine in the two cases, the % change in VL and IL.

60

Basic Electrical Engineering

+ RS

1W VL

E

+

IL RL

100 W to 1000 W

RS

VL E

3V

IL

1W RL

1 mW to 10 mW

3V

–

– Load

Source

Load Source

(a) RL = 100 Ω to 1000 Ω.

(b) RL = 1 mW to 10 mW.

Solution (a) For the extreme values of RL, the terminal voltage (VL) and current (IL) are given as 3 = 0.0297 A; VL1 = E – IL1 Ri = 3 – 0.0297 ¥ 1 = 2.9703 V IL1 = (100 + 1) 3 IL2 = = 0.002997 A; VL2 = E – IL2 Ri = 3 – 0.002997 ¥ 1 = 2.9970 V (1000 + 1) 0.0297 0.002997 \ % change (a decrease) in IL = ¥ 100 % = 89.9 % 0.0297 2.9970 2.9703 ¥ 100 % = 0.89 % \ % change (an increase) in VL = 2.9703 3 = 2.997 A; VL1 = E – IL1 Ri = 3 – 2.997 ¥ 1 = 0.003 V (b) IL1 = (0.001 + 1) 3 = 2.970 A; VL2 = E – IL2 Ri = 3 – 2.970 ¥ 1 = 0.03 V IL2 = (0.01 + 1) 2.997 2.970 \ % change (a decrease) in IL = ¥ 100 % = 0.9 % 2.997 0.03 0.003 \ % change (an increase) in VL = ¥ 100 % = 900 % 0.003 C OMMEN TS a) the percentage change in IL is large (almost 90 %), but the percentage change in VL is quite small (less than 1 %). In the other instance, (b) the percentage change in IL is quite small (less than 1 %), but the percentage change in VL is large (900 %). In both cases, the load changes by the same order (10 times). It is more appropriate to view the source in case (a) as a practical voltage source, since its characteristic approaches to an ideal voltage source. Similarly, in case (b) the source is more appropriately treated as a practical current source. An energy source can be treated either as a voltage source or as a current source, depending upon which type suits better for the analysis of the network. Thus, the transformation of one type of source into another type is frequently needed while analysing networks.

A practical voltage source can be into a practical current source, and vice versa. Two sources would be equivalent if they produce identical value of IL and VL when they are connected to the same load RL, whatever be its value. The two equivalent sources should also provide the same open-circuit voltage and

61 the same short-circuit current. Equating Eq. 3.16 to Eq. 3.18, we get VS = RSI IS Equating Eq. 3.17 to Eq. 3.19, gives VS IS = R SV From the above two equations, it follows that RSV = RSI = RS (say) and VS = RS IS (3.20) where, RS would represent the internal resistance of either of the sources. As shown in Fig. 3.15, the two practical sources will be equivalent with respect to the load terminals. However, they are not equivalent internally. A RS RL

Voltage-source representation

VS A B

Source

RL

Load

B

A

IS

RS

RL

Current-source representation

B

EX A MP L E

3.9

In the circuit of Fig. 3.16a, replace the practical current source by its equivalent voltage source and check whether you get same IL and VL

Solution Using Eq. 3.20, the internal resistance and the voltage of the equivalent voltage source are given as RS = 2 W and VS = RS IS = 2 ¥ 3 = 6 V The equivalent voltage source is shown in Fig. 3.16b IL, VL and source power PS in the two cases. 2 = 1 A; VL = 1 ¥ 4 = 4 V and PS = I S2 R = 32 ¥ (2 || 4) = 12 W Case I : IL = 3 ¥ 2+4 6 4 V2 62 Case II : IL = = 1 A; VL = 6 ¥ =4V and PS = = =6W R 2+4 2+4 2+4 delivered by the ideal part of the two sources is different.

62

Basic Electrical Engineering

+

IS 3 A RS

2W

IL

VL

RL

+ RS

4W

VL VS

RL

4W

6V

–

– Load

Source

Load Source (b)

(a)

EX A MP L E

IL

2W

3.1 0

Find the currents through the two resistors in the circuit of Fig. 3.17. Then transform the current source and 2-W

2W 32 V

6W

Solution By current division rule, I2W = 16 ¥

6 = 12 A 6+2

fi

I6W = 16 – I2W = 16 – 12 = 4 A

Now, transforming the current source gives a voltage source of 16 ¥ 2 = 32 V in series with a 2-W resistor, as shown in Fig. 3.18. In this circuit, the current through both the resistors is the same and is given as 32 I2W = I6W = =4A 2+6 Note that the current in 6-W resistor is the same as for the original circuit, but the current in 2-W resistor is different. This result illustrates the fact that although a transformed source produces the same voltages and currents in the circuit exterior to the source, but the voltage and current inside the source usually change.

Two ideal voltage sources connected in series can be replaced by a single equivalent ideal voltage source by adding the voltages of the two sources, as shown in Fig. 3.19a. Its dual statement is that two ideal current sources connected in parallel can be replaced by a single equivalent ideal current source by adding the currents of the two sources, as shown in Fig. 3.19b.

63 A V1

A

∫

A

V1+V2

∫

I2

I1

A I1+I2

V2 B

B

B

(a) Ideal voltage sources.

B

(b) Ideal current sources.

Valid Two ideal voltage sources (of different values) cannot be connected in parallel. This connection contradual statement would be that two ideal current sources (of different values) cannot be connected in series. Thus, the combinations of ideal sources shown in Fig. 3.20 are invalid and such connections are not permitted. However, two ideal voltage sources of same value can be connected in parallel. In such a case, we must have V1 = V2 = V. Similarly, its dual is also true. Two ideal current sources of same value can be connected in series. In such a case, we must have I1 = I2 = I. A V1

V2 B

(a) Ideal voltage sources in parallel.

A

I1

I2

B

(b) Ideal current sources in series.

Invalid

practical voltage sources connected in series can be combined directly, as shown in Fig. 3.21a. The voltage and the internal resistance of the equivalent voltage source are given as V = V1 + V2 and r = r1 + r2 (3.21) Similarly, the dual is true for practical current sources. Two practical current sources connected in parallel can be combined directly, as shown in Fig. 3.21b. The current and the internal resistance of the equivalent current source are given as r1 r2 I = I1 + I2 and r = r1 || r2 = (3.22) r1 + r2 lent practical current sources and then combined directly, as explained above. Similarly, when two practical es and then combined directly, as explained above. The example given below explains the above process.

64

Basic Electrical Engineering A

A

A

A

r1

V1

r = r1+r2 ∫

I1

r2

r1

I2

∫

r2

I = I1+I2

r1r2 r = r +r 1 2

V = V1+V2

V2 B

B

B

(a) Practical voltage sources.

B

(b) Practical current sources.

Direct EX A MP L E

3.1 1

The network shown in Fig. 3.22a has four voltage sources connected across points A and B. Reduce this network to a single voltage source, by using source transformation. 6V

2W

3V

1W

2W

3A

A 2W

1W

A

1W 4A

2W

6A

1W 3A

8V

6V B (a)

B (b)

Solution First, all the voltage sources are transformed into their corresponding equivalent current sources, to give the network of Fig. 3.22b. The current sources in parallel are then directly combined to get the network of Fig. 3.23a. Next, the current sources are transformed into their corresponding equivalent voltage sources, to give the network of Fig. 3.23b. Finally, the two voltage sources in series are combined directly to give the network of Fig. 3.23c. Thus, the network of Fig. 3.22a is simply equivalent to a single voltage source of 32/3 V with internal resistance of 4/3 W.

65 2/3 W 4V

2/3 W A

A

4/3 W A

2/3 W 10 A

2/3 W

32/3 V

6A 20/3 V B

(a)

EX A MP L E

B (b)

B (c)

3.1 2

We take a benchmark example * of a circuit given in Fig. 3.24a. Using source transformation, we shall determine the voltage v across 3-W resistor

Solution First, we transform the 4-A current source in parallel with 1-W resistor into a voltage source of 4 V in series with 1-W resistor (Fig. 3.24b). The two voltage sources are combined to give a 10-V source in series with a 1-W resistor, as shown in Fig. 3.24c. We again transform this 10-V voltage source into a current source of 10 A in parallel with 1-W resistor, as shown in Fig. 3.24d. The two current sources, having opposite current-directions, are combined to give a 5-A current source (Fig. 3.24e). Transforming this current source into voltage source (Fig. 3.24 ), and then combining the two resistances we get a single voltage source of 5 V in series with a 3-W resistor (Fig. 3.24g). Finally, using voltage divider, v = 2.5 V. *

We call this our “benchmark example” because we will solve this problem by different methods presented in this Book.

66

(1) (2) (3) (4)

Basic Electrical Engineering

Voltage source Ideal source DC source Independent source

or Current source or or AC source or Dependent (controlled) source

We already know what is meant by a voltage source or a current source, and an ideal source or a practical source. A dc source (such as a battery) supplies dc power, whereas an ac source (such as the power supply in our homes) supplies ac power. The value of an independent source does not depend on the current through or the voltage across any dependent sources depends on some other current or voltage. These are also known as controlled sources. ics, while making the small-signal or ac equivalent of an electronic device (such as a bipolar junction transisinput terminals and a pair of output terminals. As shown in Fig. 3.25, there are four kinds of controlled (or dependent) sources :

a) Voltage Controlled Voltage Source

The proportionality constant Av is called voltage

ratio (having no units).

b) Current Controlled Current Source

The proportionality constant Ai is called current

ratio (having no units).

c) Current Controlled Voltage Source

resistance (with units of W). I1

The proportionality constant Rm is called mutual

I2

I2

I1

R2 V1

R1

Vs

V2

AvV1

(a) Voltage controlled voltage source (VCVS). I1

V1

R1

Ai I1

R2

V2

(b) Current controlled current source (CCCS).

I2

I2

I1

R2 V1

R1

Vs

Rm I1

V2

(c) Current controlled voltage source (CCVS).

V1

R1

GmV1

R2

V2

(d) Voltage controlled current source (VCCS).

67

d) Voltage Controlled Current Source

The proportionality constant Gm is called mutual

conductance (with units of S). EX A MP L E

3.1 3

For the circuit of Fig. 3.26, determine (a) the value of current I, (b) the power absorbed by the dependent source, and (c) the resistance ‘seen’ by the independent voltage source.

Solution (a

W resistor, gives V1 = 4I. Therefore, the value of dependent voltage source is 4.5V1 = 4.5(4I) = 18I. By applying KVL, we get 24 – 4I – 2I + 18I = 0 fi I = –2 A

4W

2W

I

V1 24 V

4.5V1

reference direction shown. (b) For the dependent source, the current and voltage reference do not P = – (4.5V1)(I) = – 4.5(4I)(I) = –18I 2 = –18(–2)2 = –72 W The negative sign indicates that the dependent source is supplying power instead of absorbing it. (c) The resistance “seen” by the source is equal to the ratio of the source voltage to the current going out of its positive terminal : 24 24 R= = = –12 W I 2 The negative sign of the resistance is a result of the action of the dependent source. It indicates that the remainder of the circuit supplies power to the independent source of 24 V. Actually, it is the dependent source alone that supplies this power, as well as to the two resistors.

The two laws given by (1824–1887) form the fundamental principles used in writing circuit equations. These laws relate to the topology (i.e., the way the circuit elements are connected) of the circuit. The laws do not depend on the nature of the elements of the circuit.

It states that branches meeting at a junction (also called a node), then

If there are k number of

k

I =0

(3.23)

j =1

KCL can be stated in another way. Note that this law is just a restatement of Since, charges cannot accumulate at a junction, the amount of charge entering it at an instant must be the same as the amount of charge leaving it.

Basic Electrical Engineering

It states that loop having k elements,

. For a closed k

V =0

(3.24)

j =1

This law is just a restatement of . The charges travelling around a closed loop transfer energy from one element to another, but do not receive energy themselves on the average. It means if you move a hypothetical test charge around a complete loop, the total energy exchanged would add up to zero.

Applying

mit error while applying KVL. The step is to mark the voltage polarity (+ and –) across each element in the closed loop. In a resistance, the polarity depends upon the assumed reference direction of current. As shown in Fig. 3.27a, the end into which the current enters is marked +. Note that, as shown in Figs. 3.27b and c, the polarity of the voltage (emf) across a battery does not depend upon the assumed direction of curinto the positive terminal of the battery (Fig. 3.27b), it acts as a load positive terminal, the battery acts as a source (Fig. 3.27c). I R

I

I

vR

E (b) A battery as a load.

E (c) A battery as a source.

(a) A resistance.

The next step is to go round the selected loop, and add up all the voltages with + or – signs. Any one of the following two rules can be followed : i) Rule 1 While travelling, if you meet a voltage rise, write the voltage with positive sign; if you meet a voltage drop, write the voltage with negative sign. ii) Rule 2 The above two rules give opposite signs to the voltages, which is immaterial as you are taking the algebraic sum. Whichever rule you follow, remain consistent throughout. In this book, we shall be following Rule 1. Applying this rule is easy, as it has a strong analogy with the physical height (altitude) of a place. When you travel up a slope, the height (above sea level) increases. When you travel down a slope, the height or altitude decreases. EX A MP L E

3.1 4

Find the voltage Vab across the open circuit in the circuit shown in Fig. 3.28.

d

a (+6) V

Vca = +4 Vx = ?

(–4) V

10 V

c

b

W resistance. Assuming a current I (in the clockwise direction) in the left loop, we write the KVL equation going clockwise around the loop starting from the bottom of 100-V battery, +100 – 40I – 60I = 0 fi I=1A We can now calculate the voltage V1 across the 60-W resistance, V1 = IR = 1 ¥ 60 = 60 V Note Vab by writing KVL equation around the rightmost loop. What loop ? There is indeed a loop even though no path exists for current. Thus, b and go counterclockwise, –10 + V1 + 0 ¥ 10 + 30 – Vab = 0 or –10 + 60 + 0 + 30 – Vab = 0 fi Vab = 80 V

Solution

EX A MP L E

3.1 5

The circuit of Fig. 3.29 has mixed reference-direction convention. The voltages are marked with both the +/– and the subscript notation. Determine the unknown voltage vx and vcd.

Solution loop starting from the bottom of the battery, + (10) – (+6) + (– 4) = 0 In this equation, the signs outside the parentheses come from the reference directions marked and the signs inside come *

a to c to b, – (– 4) + (+4) + vx = 0 fi vx = – 8 V Determining vcd d and walk towards point c and note down how high in voltage we have gone. Thus, vcd = – 6 + (+4) = – 2 V

There exists a systematic method of solving simultaneous equations. Let us consider, for example, following equations in three variables, x, y and , written in matrix form, ⎡a1 b1 c1 ⎤ ⎡ x ⎤ ⎡d1 ⎤ ⎢a ⎥⎢ ⎥ ⎢ ⎥ ⎢ 2 b2 c2 ⎥ ⎢ y ⎥ = ⎢d2 ⎥ ⎢⎣a3 b3 c3 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣d3 ⎥⎦ *

nonsense, like 1+3 = 5.

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Basic Electrical Engineering

Cramer’s rule gives the solution of the above equations as a

x =

;

y=

b

;

and

=

c

Here, the determinants are given as a1 b1 c1 D = a2 b2 c2 ; a3 b3 c3 d1 Da = d2 d3

b1 b2 b3

c1 a1 c2 ; Db = a2 c3 a3

d1 d2 d3

c1 c2 c3

and

a1 Dc = a2 a3

b1 b2 b3

d1 d2 d3

There are two ways of evaluating the value of a determinant. The , which is in common use, is to expand it in terms of its minors. To do this, we select any row or any column k, multiply each element in that row or column by its minor and by (–1) +k, and then add the products. The minor of the element appearing in row and column k is the determinant obtained when row and column k are removed; it is indicated by D . The second method, which is applicable to a third-order determinant (the one that we mostly come take the sum of the products of the numbers on the diagonals indicated by downward arrows, and subtract from this the sum of the products of the numbers on the diagonals indicated by upward arrows. For example, to understand these two methods, we evaluate the determinant 2 3 4 6 10 8 7

5 9

First Method D = (2) ¥

10 8 – (6) ¥ 5 9

3 4 3 4 + (7) ¥ 5 9 10 8

= (2) ¥ [10 ¥ 9 – (–5) ¥ 8] – (6) ¥ [(–3) ¥ 9 – (–5) ¥ 4] + (7) ¥ [(–3) ¥ 8 – (10) ¥ 4] = 2 ¥ (90 + 40) – (6) ¥ (–27 + 20) + 7 ¥ (–24 – 40) = 260 + 42 – 448 = –146.

Second Method 2 3 4 6 10 8 7 5 9

280 2 6 7 180

80 3 10 5 168

162

120

D = (180 – 168 – 120) – (280 – 80 –162) = (–108) – (38) = –146

71 EX A MP L E

3.1 6

Using KCL and KVL, determine the currents Ix and Iy in the network shown in Fig. 3.30. 10 W

50 V

I1

2W

50 V

I1 – Ix + Iy

I1 – Ix 5W

1

100 V

2

2W

3

3W

Iy

Ix

Solution Using KCL, the currents in other branches are marked as shown in Fig. 3.30. Writing KVL equations for the loops 1, 2 and 3, we get 100 – 10I1 – 5Ix = 0 5Ix + 50 – 2(I1 – Ix) + 2Iy = 0 –2Iy + 50 – 3(I1 – Ix + Iy) = 0 These equations can be written in matrix form,

fi fi fi

5Ix + 0Iy + 10I1 = 100 7Ix + 2Iy – 2I1 = –50 3Ix – 5Iy – 3I1 = –50

È5 0 10 ˘ È I x ˘ ⎡ 100 ⎤ Í ˙Í ˙ = ⎢ ⎥; 7 2 2 I Í ˙Í y˙ ⎢−50 ⎥ ÍÎ3 -5 -3˙˚ ÍÎ I1 ˙˚ ⎢⎣−50 ⎥⎦ Using calculator -991ES, we solve the above equations to get Ix = –3.87 A;

and

Iy = 0.51 A

The negative sign on Ix N O T E Using calculator (Casio simultaneous equations. EXA M PL E

3 . 17

Using Kirchhoff’s laws, determine R1, R2, I1, I2 and I3 in the circuit of Fig. 3.31.

Solution Applying KCL at nodes B and C, we get I1 + I2 = 20 and

I3 = 30 + I2

or

I1 + I2 + 0I3 = 20

(i)

or

0I1 – I2 + I3 = 30

(ii)

Applying KVL to the outer loop ABCDEFGHA, we get –0.1I1 + 0.3I2 + 0.2I3 – 120 + 110 = 0

or I1 – 3I2 –2I3 = –100

Using the calculator, we get the solution of the above equations as

(iii)

72

Basic Electrical Engineering A

I1

0.1 W

I2

B

0.3 W

R1

110 V

0.2 W

D

R2

120 V

20 A H

I3

C

30 A

G

I1 = 10 A; I2 = 10 A Now, applying KVL to the loop ABGHA, we get –0.1I1 – 20R1 + 110 = 0

F

and

E

I3 = 40 A

fi

R1 =

fi

R2 =

110

0.1I1 = 5.45 W 20

120

0.2 I3 = 3.74 W 30

Applying KVL to the loop CDEFC, we get 0.2I3 – 120 + 30R2 = 0

It is a general method of analysing a network. It can be applied to any network, howsoever complicated it may be. The variables are loop currents and the equations are based on KVL. The concept of loop currents is quite different from the branch currents. A branch current I1 is the Eq. (iii) in terms of the branch currents. However, a loop is a closed path and a loop current is a current In loop-current analysis, a loop current can be marked to go wherever we wish. The only limitations are resistance. resistance. The step-by-step procedure is given below.

Step 1 Recognise the independent loops* and label the loop currents. In the given circuit, we recognise two *

An independent loop does not pass through a current source. Furthermore, it is more convenient to apply loopcurrent method to a network that contains no current sources. Hence, in case a network contains current sources, convert all the current sources to their equivalent voltage sources before attempting to apply loop-current method.

73 independent loops and mark two loop currents I1 and I2 (Fig. 3.32b) in arbitrary directions. Step 2 Write KVL equations for each loop : S Voltage rise across an element in the loop = 0 Start from any point in the loop and going around the loop either clockwise or anticlockwise (irrespective of the loop-current direction) end at the same point. Step 3 Solve for the loop currents (using calculator in Equation Mode) and then compute the currents or voltages required. N O T E We write one KVL equation for each loop and we have one loop current for each loop. We always get the same number of equations as the unknowns.

We now write KVL equation for loop 1, starting from the top left corner and traverse (going around) the loop in clockwise direction : – (I1 – I2)(1) – I1(2) – 6 + 7 = 0 traversing direction is I1 – I2. Since (I1 – I2)(1) is a voltage drop, we write this term with a negative sign. a voltage drop across the 6-V source. The fourth and the last term is a voltage rise across 7-V source, hence written with positive sign. Same way, we write the second equation for loop 2. Let us arbitrary start from the bottom left corner and traverse the loop clockwise (against the direction of loop current I2) : +7 + (I2 – I1)(1) + I2(3) + I2(4) – 9 = 0 I2 – I1 and as we are traversing the loop from left-to-right, the voltage (I2 – I1)(1) is a voltage rise. Hence it has a positive sign. The third and fourth terms are both voltage rises, hence positive. The last terms is voltage drop across 9-V source, hence has negative sign. Simplifying these two equations, we get 3I1 – I2 = 1 and I1 – 8I2 = –2 These two equations can easily be solved to get I1 = 0.435 A and I2 = 0.304 A. We can now compute the voltI1 ¥ 2 = 0.435 ¥ 2 = 0.870 V, with positive (+) polarity at the top. The loop-current method is useful when we wish to determine only one current or voltage. resulting equations can then be solved for only that one loop current.

Have we Ignored

In loop-current method, we solve for the currents by writing only KVL equa-

into out of it also. For example, if we wrote KCL for node A in the circuit of Fig. 3.32b, each of the two loop currents, I1 and I2, contribute two equal and opposite terms, adding up to zero.

74

Basic Electrical Engineering 5W

5W

I1 8W

2A

10 V

(a)

I2

8W

(b)

5W

10 V

2A

10 V

5W

I1

2A

(c)

8W

10 V

I1

8W

2A

(d)

Counting Independent Loops The circuit of Fig. 3.33a appears to have two loops. But, these two loops are not independent. Suppose that we had marked the two loop currents I1 and I2 in the standard way, as shown in Fig. 3.33b I2 – I1 = 2 A The values of these two currents are constrained by the above relation. We identify independent loops by turning OFF all sources. By ‘turning OFF a voltage source’ means to ‘short circuit it’, and by ‘turning OFF a current source’ means to ‘open circuit it’. When we turn OFF both the sources in Fig. 3.33a, we are left with one loop containing two resistances. Thus, we have only one independent loop, requiring one unknown and one KVL equation. Suppose that we are interested to determine the current through 5-W resistance in Fig. 3.33a. We select the unknown loop current I1 passing through 5-W resistance (but not through the current source) and a known c. We now have only one unknown and so we need only one equation. This, we get by writing KVL around the loop of I1, 10 – 5I1 – 8(I1 + 2) = 0 fi I1 = – 0.462 A EX A MP L E

3.1 8

Using loop-current analysis, determine the current through 8-W resistance in the circuit of Fig. 3.33a.

Solution current I1 passing through 8-W the left loop, as shown in Fig. 3.33d.

W resistance. Therefore, we should select the unknown loop

75 Now, writing KVL equation around the loop of I1, we get 10 – 5(I1 – 2) – 8I1 = 0 fi I1 = 1.538 A Thus, the current through 8-W resistance is 1.538 A. EX A MP L E

3.1 9

Consider the benchmark example of Fig. 3.24a (redrawn in Fig. 3.34a), and solve it by using loop-current analysis. 2W

4A

3W

5A 1W

V

4A

1W

V

5A

4A

(a)

5A I 3W

2W

6V

(b)

6V

Solution We note that the given the voltage across 3-W resistance. So, we should select the unknown loop current I passing through 3-W resistance (but in Fig. 3.34b. Writing KVL equation around the loop of I, we get –2I – 3I + 6 – 1 ¥ (I + 5 – 4) = 0

fi

I=

5 A 6

Therefore, the unknown voltage v = 3I = 2.5 V.

In circuit terminology, a loop is any closed path. A mesh is a special loop, namely, the smallest loop one can have. In other words, a mesh is a loop that contains no other loops. In the fuller sense of loop-current analysis,

at least one loop current. In mesh analysis met. However, the application of mesh analysis is restricted only to planar networks. A network is said to be Fig. 3.35a appears to be nonplanar. But a closer look at it shows that it can be redrawn as in Fig. 3.35b, and hence it is a planar network. However, the network of Fig. 3.35c b), it often has the appearance of a multi-paned window. Each pane in the window may be considered to be a mesh. The mesh analysis not only tells us the minimum number of unknown currents to be taken, but it also ensures that the equations obtained by writing KVL equations for each mesh are independent.

76

Basic Electrical Engineering

(a)

(b)

(c)

To understand the beauty of mesh analysis, let us consider the circuit of Fig. 3.36a, redrawn in Fig. 3.36b. It has two meshes and the two mesh currents* I1 and I2 have been marked (both in clockwise direction). Let us write the KVL equations for these two meshes, 7 – 1 ¥ I1 – 2 ¥ (I1 – I2) – 6 = 0 and 6 – 2 ¥ (I2 – I1) – 3 ¥ I2 – 4 ¥ I2 – 9 = 0 or (1 + 2) I1 – 2I2 = 7 – 6 (3.25) and –2I1 + (2 + 3 + 4)I2 = 6 – 9 (3.26) 1W

1W

3W

3W

2W

2W 4W

7V

7V

6V

I1

I2 6V

9V

9V (a)

(b)

Let us write the above two equations in matrix form, −2 ⎤ ⎡ I1 ⎤ ⎡7 − 6⎤ ⎡(1 + 2) = ⎢ −2 ⎥ (2 + 3 + 4)⎦ ⎢⎣ I2 ⎥⎦ ⎢⎣6 − 9⎥⎦ ⎣ ⎡ 3 −2 ⎤ ⎡ I1 ⎤ ⎡ 1⎤ or = ⎢−2 9⎥⎦ ⎢⎣ I2 ⎥⎦ ⎢⎣−3⎥⎦ ⎣ or where

*

⎡ R11 R=⎢ ⎣− R21

4W

RI = E − R12 ⎤ ⎡ 3 −2 ⎤ = is a square matrix, and is called resistance matrix. R22 ⎥⎦ ⎢⎣−2 9⎥⎦

mesh current

(3.27)

(3.28)

77 ⎡ I1 ⎤ I = ⎢ ⎥ is a column matrix, and is called mesh current matrix. ⎣ I2 ⎦ ⎡ E1 ⎤ ⎡ 1⎤ E = ⎢ ⎥ = ⎢ ⎥ is a column matrix, and is called source matrix. ⎣ E2 ⎦ ⎣−3⎦

Resistance Matrix It is a square matrix. The elements on its principal diagonal (such as R11, R22, … etc.) are called self-resistances and have positive values. Thus, R11 is the self-resistance of the mesh 1, and it is equal to the sum of all the resistances in the mesh 1. The elements off the principal diagonal are called mutual resistances, and have zero or negative values. Thus, R12 is the mutual resistance between mesh 1 and mesh 2. It represents the total resistance common between these two meshes. If there is no resistance common to two meshes, their mutual resistance will be zero. The matrix is symmetrical about the principal diagonal (i.e., R12 = R21, … etc). Source Matrix It is a column matrix. Its each element is the algebraic sum of the voltage sources that forces the current in the same direction as the mesh current.

The beauty of mesh analysis lies in the fact that KVL equations in the matrix form (i.e., in the form of Eq. 3.27) inspection . Yes, it is possible, even for a multi-mesh network. The symmetry and sign pattern of the matrices minimises the chances of committing errors while writing the KVL equations. This method is, however, limited to independent voltage sources. Most of the networks that we deal with are of this category. The procedure is outlined below. 1. Make sure that the given network is planar. 2. Make sure that the network contains only independent voltage sources. If there is a practical current source, convert it into an equivalent practical voltage source. 3. Assuming that the network has m meshes, assign a mesh current in each mesh, I1, I2, I3, … Im, all in clockwise (or in counterclockwise) direction. 4. Write matrix equations directly by inspection of the network, keeping in mind the nature of the elements of resistance matrix and source matrix. 5. Solve the equations to determine the unknown mesh currents, either using Cramer’s rule, or using a Let us apply the above procedure to the network of Fig. 3.36a, and check whether we obtain the same equations as Eq. 3.27. The resistances of 1 W and 2 W are included in mesh 1. Hence, R11 = 1 + 2 = 3. The resistance common to mesh 1 and mesh 2 is 2 W. Therefore, R12 = R21 = 2. These are entered with negative sign in the resistance matrix. The self-resistance of the mesh 2 is R22 = source matrix represents the net source voltage that forces the current I1 around mesh 1. Since 7-V source aids current I1 and 6-V source opposes current I1, we have E1 = +7 – 6 = 1. In mesh 2, source of 6 V aids current I2 and source of 9 V opposes it. Hence, we have E2 be directly written by inspection of the network.

Basic Electrical Engineering

We cannot incorporate current sources into our normal mesh analysis. If a circuit has current sources, a modest extension of the standard procedure is needed. There are three possible methods.

First Method This reduces the number of meshes by 1 for each current source. Apply the standard procedure of mesh analysis to determine the assumed mesh currents. Go back to the original circuit, and get additional equations, one for each current source, by expressing the source currents in terms of assumed mesh currents. Solve Fig. 3.37a. The three mesh currents are marked. We convert a 13-A current source in parallel with a 5-W resistor into an equivalent 65-V source in series with a 5-W resistor, as shown in Fig. 3.37b. This reduces the number of meshes to two. We can write the mesh equations in the matrix form just by inspection, ⎡ 9 −5⎤ ⎡ I1 ⎤ ⎡10 ⎤ ⎢−5 11⎥ ⎢ I ⎥ = ⎢52 ⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦ Using Cramer’s rule, we get I1 = 5 A and I2 = 7 A We now go back to the original circuit of Fig. 3.37a I2 – I3 = 13 A fi I3 = I2 – 13 = 7 – 13 = –6 A 4W

75 V

6W

I1

I3

5W

4W

I2

13 A

13 V

75 V

6W

5W

I1

I2

13 V

65 V 1

3

2

1

(a)

2 (b)

4W

6W

I1

75 V 1

I3

5W 3

KVL

I2

13 A

13 V

2

(c)

We can assign unknown voltages to each current source, apply KVL around each mesh

Third Method A better approach is to use the so-called supermesh method. We create a supermesh from two meshes that have a current source as a common element; the current source is in the interior of the supermesh. Thus, the number of meshes is reduced by 1 for each current source present. If the current source lies on the perimeter of the circuit, then the single mesh in which it is found is ignored. The KVL is then applied to the meshes and supermeshes, using the assumed mesh currents. Let us apply this method to the circuit of Fig. 3.37a. ing this current source leads to a supermesh comprising the 5-W and 6-W resistors and the 13-V source (as shown by the dotted path in Fig. 3.37c). Going along the dotted arrow, the KVL equation for this supermesh is –5(I3 – I1) – 6I2 – 13 = 0 or 5I1 – 6I2 – 5I3 = 13 (3.29) The KVL equation for mesh 1 is 9I1 + 0I2 – 5I3 = 75 (3.30) We have only two equations for three unknowns. The third equation is obtained by applying KCL to either node of the current source, or simply, by noting that the current up through the current source in terms of the mesh currents is I2 – I3. This current must, of course, be equal to the 13 A of the source. Thus, we have 0I1 + I2 – I3 = 13 (3.31) In the matrix form, these three equations are ⎡ 5 −6 −5⎤ ⎡ I1 ⎤ ⎡13⎤ ⎢9 0 −5⎥ ⎢ I ⎥ ⎢75⎥ ⎢ ⎥ ⎢ 2⎥ = ⎢ ⎥ ⎢⎣0 1 −1⎥⎦ ⎢⎣ I3 ⎥⎦ ⎢⎣13⎥⎦ The solutions of these equations are the same as before : I1 = 5 A, I2 = 7 A and I3 = – 6 A. EX A MP L E

3.2 0

Apply mesh analysis to determine current drawn from the source in the network of Fig. 3.38. IS

60 V

7W

I1

12 W

I2

6W

I3

12 W

Solution There are three independent meshes, for which the three mesh currents I1, I2, and I3 are all marked with clockwise directions. The current IS drawn from the source is same as I1. To determine this current, we write the mesh equations in matrix form by inspection of the circuit, 0 ⎤ ⎡ I1 ⎤ ⎡60 ⎤ ⎡7 + 12 −12 ⎢ −12 12 + 6 − 6 ⎥⎥ ⎢⎢ I2 ⎥⎥ = ⎢⎢ 0 ⎥⎥ ⎢ ⎢⎣ 0 −6 6 + 12 ⎥⎦ ⎢⎣ I3 ⎥⎦ ⎢⎣ 0 ⎥⎦ Using calculator, we get IS = I1 = 6 A

or

0 ⎤ ⎡ I1 ⎤ ⎡60 ⎤ ⎡ 19 −12 ⎢−12 18 − 6⎥⎥ ⎢⎢ I2 ⎥⎥ = ⎢⎢ 0 ⎥⎥ ⎢ ⎢⎣ 0 − 6 18⎥⎦ ⎣⎢ I3 ⎥⎦ ⎢⎣ 0 ⎥⎦

Basic Electrical Engineering EX A MP L E

3.2 1

Apply mesh analysis to determine the three mesh-currents in the network of Fig. 3.39.

Solution The given network is a planar network having independent voltage sources. It has three meshes for which the mesh currents I1, I2, and I3 are marked all with clockwise directions. By inspection, the matrix equation is written as −4 0 ⎤ ⎡ I1 ⎤ È10 + 3 ⎡3 + 4 ˘ ⎢ − 4 4 + 5 + 6 −6 ⎥ ⎢I ⎥ = Í ˙ 3 15 + 30 2 ⎥⎢ ⎥ Í ⎢ ˙ ⎢⎣ 0 −6 6 + 7⎥⎦ ⎢⎣ I3 ⎥⎦ Î -30 + 31 ˚

or

0 ⎤ ⎡ I1 ⎤ È13˘ ⎡ 7 −4 ⎢− 4 15 − 6⎥ ⎢ I ⎥ = Í12˙ ⎥ ⎢ 2⎥ Í ˙ ⎢ ⎢⎣ 0 − 6 13⎥⎦ ⎢⎣ I3 ⎥⎦ Î 1˚

Using a calculator, we get I1 = 3 A, I2 = 2 A, and I3 = 1 A. N O T E The mesh analysis is quite similar to loop analysis. The principal difference is that the current paths selected in loop analysis are not necessarily meshes. Also, there is no convention regarding the direction of loop currents; they can be a mixture of clockwise or counterclockwise directions. For loop analysis, no current source needs to be transformed to a voltage source. But each current source should not applied to this loop because the current source voltage is not known. If the current through only one component is desired, the loops should be selected such that only one loop

This method is dual of the loop-current analysis. It is based on KCL and can be applied to any network. We This is done by expressing the currents in terms of node voltages. The procedure for node-voltage analysis consists of following simple steps. reference node (also called ground node or datum node wires is chosen as the reference node. Mark the reference node with r. 2. Count and label the independent nodes. Voltages on these nodes are referenced positive with respect to the ground node. 3. Label the independent nodes. 4. Write KCL in a special form, 1.

S Currents leaving the node in resistors = S Currents entering the node from current sources. The procedure is explained with the help of following example. EX A MP L E

3.2 2 W resistor in the circuit of Fig. 3.40a. a 3W

5A

2W

b

va

4W

6A

5A

vb

3W 2W

4W

6A

r (a)

(b)

Solution The circuit has three nodes. In this case, we chose the node at the bottom as reference node. We mark it as r, as shown in Fig. 3.40b. The remaining two nodes, labelled as a and b are the independent nodes. Let the voltage of these nodes be va and vb, respectively. We write KCL equations in terms of these node voltages. For node a, the KCL equation is va (0) v vb + a = + (5) (i) 2 3 The left side of the above equation represents the current leaving node a in the two resistors connected directly to node a. W resistor is merely the voltage va divided by the resistance between node a and reference node. We have written zero (0) for the voltage of the reference node as a reminder. The current from a to b through 3-W resistor is simply the voltage at a, minus the voltage at b, divided by the resistance between a and b. Similarly, we can write KCL equation for node b, vb va v (0) + b = – (+ 6) (ii) 4 3 i). This is so because we are now expressing the current referenced in the opposite direction. Note also that the current source term on the right side has a negative sign, because this current is leaving the node. The above two equations can be solved, using a calculator, to get va = 2.44 V and vb = –8.89 V (iii) We can now calculate the primary unknown, the voltage vab across the 3-W resistor, vab = va – vb = 2.44 – (–8.89) = 11.3 V

voltage sources, there are techniques to handle them. If one terminal of a voltage source with a series resistance is grounded (as in the circuit of Fig. 3.41), the KCL equation can be written in terms of this voltage. *

*

Basic Electrical Engineering

which uses the concept of constrained node or supernode. This method is especially suitable for the circuits

Supernode The two ends of a voltage source cannot make two independent nodes, since one nodevoltage can be determined in terms of the other node-voltage and the source voltage. Hence, we treat these end nodes and the voltage source together as a ‘supernode’. The supernode is usually indicated by the region enclosed by a dotted line. The KCL is then applied to both nodes at the same time. This is certainly possible. If the total current leaving one node is zero and the total current leaving the other node is zero, then the total current leaving the combination of two nodes is also zero. The procedure of using the concept of supernode is made clear in Example 3.24. An independent node is a node whose voltage cannot be derived from the voltage of another node. Independent nodes in a network can be counted by turning OFF all sources and number. EX A MP L E

3.2 3

Apply node-voltage analysis to the circuit of Fig. 3.41 to determine the current in 12-W resistor.

Solution The given network has two nodes. Node 2 is marked as reference node. Hence, there is only one independent node (node 1). We assign voltage V1 to this node. Writing KCL equation for node 1, we get V1 (0) V 60 V (0) I1 + I2 + I3 = 0 or + 1 + 1 = 0 fi V1 = 18 V 12 7 4 I1 in 12-W resistor is V 0 18 I1 = 1 = = 1.5 A 12 12 EXA MP LE

3. 2 4

Determine the current through 4-W resistor in the circuit of Fig. 3.42.

Solution Here, we have one node constrained to another independent node rather than to the reference node. The voltages at node a and b are unknown, but they are not independent. If we knew either of them, we could determine the

7W

1 I2

60 V

12 W

I I1 3 4W

2 (Ref. node)

va – vb = 6 or va – vb + 0vc = 6 (i) We can therefore treat the two constrained nodes a and b, as a supernode. Now, writing KCL equation for this supernode, we get va v vc + b = 2 or 0.33va + 0.25vb – 0.25vc = 2 (ii) 3 4 Applying KCL to node c gives vc v vb + c = –7 or 0va – 0.25vb + 0.45vc = –7 (iii) 5 4 Above three equations can be written in the matrix form as ⎡ 6⎤ −1 0 ⎤ ⎡ va ⎤ ⎡ 1 ⎢0.33 0.25 − 0.25⎥ ⎢v ⎥ = ⎢ 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ b⎥ ⎢⎣−7⎥⎦ ⎢⎣ 0 − 0.25 0.45 ⎥⎦ ⎢⎣ vc ⎥⎦ We solve the above equations using calculator to get vb = –8.8 V and vc = –20.4 V Finally, the current through 4-W resistor is v b vc 8.77 ( 20.42) = = 2.9 A 4 4 EXA MP LE

3.25

Consider the benchmark example of Fig. 3.24a (redrawn in Fig. 3.43a), and solve it by using node-voltage analysis.

Solution

OFF the sources to get the circuit of Fig. 3.43b. There are three nodes, two of which are independent. However, if we add the two series resistors to make a 5-W resistor we will have only one independent node (node a), and hence we will have to solve only one equation. The unknown voltage across 3-W resistor can then be determined by applying voltage divider rule. Taking node d as reference, we write KCL equation for node a as va (6) va (0) 6 1 + = +4 – 5 fi va = = 4.17 V 1 5 1.2 Using the voltage divider, the voltage across 3-W resistor is 3 v = 4.17 ¥ = 2.5 V 2+3

Basic Electrical Engineering

This method is dual of mesh analysis. It can be applied to a network that contains only independent current equivalent practical current sources. We can then proceed with nodal analysis, and write the KCL equations in the following matrix form, just by inspection of the network. ⎡ G11 − G12 ⎤ ⎡ V1 ⎤ ⎡ IS 1 ⎤ = or GV = IS (3.32) ⎢− G G22 ⎥⎦ ⎢⎣V2 ⎥⎦ ⎢⎣ IS 2 ⎥⎦ ⎣ 21 Here, the square matrix G is called conductance matrix, the column matrix V is called node-voltage matrix, and the column matrix IS is called node-current source matrix. 1, G11 = Self-conductance and is the sum of all conductances connected to node 1. G12 = G21 is the mutual-conductance between node 1 and node 2, and is sum of all conductances connected between node 1 and node 2. IS1 = Algebraic sum of source currents entering node 1. IS2 = Algebraic sum of source currents entering node 2. Note that all elements on the principal diagonal of matrix G are positive and all elements off-diagonal are negative (or zero), just as in the resistance matrix R occurring in mesh-analysis. The method of nodal analysis is made clear in the following example. EX A MP L E

3.2 6

Determine the current through 5-W resistor in the circuit of Fig. 3.44a. 5W

1

5W

2

0.2 S

1 V1

(a)

–2 A

–2 A 3 (b)

3A

1S

1W

3A

S

1W

0.5

2W

2W

3A

2 V2 –2 A

r (c)

Solution The given network is redrawn in Fig. 3.44b to emphasise the three nodes. Node 3 is taken as reference node r. The remaining two nodes are the independent nodes, for which we are to write nodal-analysis equations by inspection. in Fig. 3.44c. The conductances connected to node 1 are 0.5 S and 0.2 S. Hence, G11 = 0.5 + 0.2 = 0.7 S. Similarly, G22 G12 = G21 = 0.2 S. The current source of 3 A is entering node 1. Hence, IS1 = 3 A. Current leaving node 2 is –2 A. Hence, IS2 = – (–2) = 2 A. Thus, ⎡ 0.7 − 0.2 ⎤ ⎡ V1 ⎤ ⎡3⎤ = ⎢− 0.2 1.2 ⎥⎦ ⎢⎣V2 ⎥⎦ ⎢⎣2 ⎥⎦ ⎣

Using calculator, we get the solution of the above matrix equation as V1 = 5 V and V2 = 2.5 V Now, the current through 5-W resistor is given as V V 5 2.5 = 0.5 A I = 1 2 = 5 5

The primary consideration in choosing a suitable method of analysis is to answer the question, “How many equations must be solved ?” It is possible for a circuit to have more independent nodes than loops, or vice versa. In general, if a network has b branches and n nodes, it will have n – 1 independent nodes and b – (n – However, if the circuit is nonplanar, then there is no choice; only the nodal analysis can be applied.

easier. If the unknown is a voltage, nodal analysis might be the best, but if the unknown is a current, using If there is only one source and the circuit is not too complicated, the method of voltage and current dividers is favoured. EX A MP L E

3.2 7

Find the current I in the circuit shown in Fig. 3.45. 2W

Solution This circuit has two independent loops but only one independent node (if we combine the series resistors). Thus, we favour the nodal analysis. Choosing the node at the bottom as reference node, we designate the voltage at the top node as V. The single nodal equation is V 10 V−0 V 8 + + = 0 fi V = 6.91 V 2 1+ 3 6 6.91 \ I = = 1.73 A 1+ 3

6W

1W 10 V

8V 3W I=?

ADDITIONAL SOLVED EXAMPLES EX A MP L E

3.2 8

A series combination of two capacitances C1 and C2 is connected across a 200-V dc supply, and it is found that the potential difference across C1 is 120 V. This pd increases to 140 V, when a 3-mF capacitor is connected in parallel with C2. Determine the capacitances C1 and C2.

Solution Case I (Fig. 3.46a) : The charge on each capacitor is same. That is, C1V1 = C2V2 or 120C1 = 80C2

or

C1 = 2C2/3

(i)

Basic Electrical Engineering 3 mF C1

C2

C1

C2

120 V

80 V

140 V

60 V

200 V (a)

200 V (b)

Case II (Fig. 3.46b) : The capacitance of 3 mF in parallel with C2 gives an equivalent capacitance of (C2 + 3) mF. This is in series with C1. Again, the charge on these two must be the same, 140C1 = 60(C2 + 3)

(ii)

Solving (i) and (ii) gives C1 = 3.6 mF EXA MP LE

C2 = 5.4 mF

and

3.29

Calculate the current in each branch of the network of Fig. 3.47.

20 W

10 W

Ib

I5

2W

I3

I1

20 V

Ia

50 V

5W

I2

9V

10 V

1W

I4

Solution I1 =

20 10 10 = 2 A; I4 = = 2 A; I5 = = 5 A; 10 5 2

I2 =

20 − (10 + 9) 20 = 1 A; I3 = 1

50 10 = –2 A; 20

Ia = I1 + I2 + I3 = 2 + 1 + (–2) = 1 A; I2 + I3 + Ib = I4 + I5 EXA MP LE

3.30

Determine the currents in various resistances of the network of Fig. 3.48a.

fi

Ib = 2 + 5 – 1 – (–2) = 8 A

4V

1W

R1

4V

R3 3W

2V

R2

0.5 W

R4

1W

1W

0.5 W

I1 3W

2V

1W

I2

3V

3V

(a)

(b)

Solution There are two independent meshes. For writing mesh equations, we assign the two mesh currents I1 and I2 in the same (clockwise) direction, as shown in Fig. 3.48b. Writing mesh equations by inspection, ⎡1 + 0.5 + 3 −3 ⎤ ⎡ I1 ⎤ ⎡4 − 2 ⎤ = ⎢ −3 3 + 1⎥⎦ ⎢⎣ I2 ⎥⎦ ⎢⎣ 2 + 3⎥⎦ ⎣

or

⎡4.5 −3⎤ ⎡ I1 ⎤ ⎡2 ⎤ ⎢ −3 4 ⎥ ⎢ I ⎥ = ⎢5⎥ ⎣ ⎦ ⎣ 2⎦ ⎣ ⎦

Solving the above, we get I1 = 2.55 A, and I2 = 3.167 A. Therefore, Current through R1 = Current through R2 = I1 = 2.55 A Current through R3 = I1 – I2 = 2.55 – 3.167 = – 0.617 A Current through R4 = I2 = 3.167 A EXA M P L E

3 . 3 1

In the circuit shown in Fig. 3.49, the cells E1 and E2 have emf of 4 V and 8 V and internal resistance of 0.5 W and 1 W, respectively. Calculate the current in each resistor and potential difference across each cell. 1A

E1

E2

3W

R1

A

2A 4V

E1 4.5 W

E2

3W

E3 2A

R2 6W

3W

5W

4 mF

8V

1W 3W

1W

2W

4W

1A

Solution The 3-W resistance and 6-W resistance can be combined to give a 2-W resistance. Therefore, in the single loop obtained, the total resistance is Rt = 2 + 4.5 + 1 + 0.5 = 8 W, and the net emf driving the current anticlockwise is E = E2 – E1 = 8 – 4 = 4 V. Thus, the loop current, E 4 = = 0.5 A I = Rt 8

Basic Electrical Engineering This is the current through 4.5-W resistance. Using current divider, the current in 3-W resistance and 6-W resistance is, respectively, 3 6 I3 W = 0.5 ¥ = 0.333 A and I6 W = 0.5 ¥ = 0.167 A 6+3 6+3 The pd across cells E1 and E2 are and V2 = E2 – Ir2 = 8 – 0.5 ¥ 1 = 7.5 V V1 = E1 + Ir1 = 4 + 0.5 ¥ 0.5 = 4.25 V Note that the pd of cell E1 is more than its emf. This is so because this cell is working as a load, and is getting charged. The cell E2 is working as a source and is getting discharged. EX A MP L E

3 . 32

For the circuit shown in Fig. 3.50, R1 = 5 W, R2 = 9 W, E1 = 8 V, E2 = 6 V and E3 = 4 V. Find the potential of the point A.

Solution To determine the potential of the point A (with respect to the ground), we stand at ground, traverse a path to point A and see how high in potential we reach. The best alternative would be to go across E3, then across E2, and then E1, because all these voltages are known. Hence, VA = E3 + E2 – E1 = 4 + 6 – 8 = 2 V E XA MP LE

3 .3 3

A part of a circuit in steady state is shown in Fig. 3.51. Calculate the energy stored in the capacitor.

Solution get the current through the 5-W W resistance. The current through the 2-W resistance can be determined by applying KCL to the node at the bottom of the capacitor, as 2 – 1 = 1 A. We can now calculate the pd across the capacitor as V = 3 ¥ 5 + 3 ¥ 1 + 1 ¥ 2 = 20 V Therefore, the energy stored in the capacitor is 1 1 W = CV 2 = (4 mF) (20 V)2 = 800 mJ 2 2 E XA MP LE

3.34

In the circuit shown in Fig. 3.52a, E1 = 3 V, E2 = 2 V, E3 = 1 V, R = r1 = r2 = r3 = 1 W. (a) Find the potential difference between points A and B, and the current through each branch. (b) If r2 is short-circuited and the point A is connected E1, E2 and E3 and the resistance R.

A

R

C

r1

E1

r2

E2

E1

r1 I1 B

A

R

C

r2

E2

r1 I3 C

B

I5

E1 E2

B A

I2 r3

E3 (a)

r3

I6

I4

r3

E3

E3 (b)

R (c)

Solution (a) We assign the loop currents I1 and I2, as shown in Fig. 3.52b. Writing the two KVL equations for the two loops, we get E1 – I1 r1 – (I1 – I2)r2 – E2 = 0 fi 3 – I1 – I1 + I2 – 2 = 0 fi –2I1 + I2 + 1 = 0 E2 – (I2 – I1)r2 – I2 r3 – E3 = 0 fi 2 – I2 + I1 – I2 – 1 = 0 fi –2I2 + I1 + 1 = 0 Solving these two equations, we get I1 = I2 = 1 A. Thus, Current through r1 = I1 = 1 A; Current through r2 = I1 – I2 = 1 – 1 = 0 A; and Current through r3 = I2 = 1 A; Current through R = 0 A (no closed path) The pd across A and B = E2 + (I1 – I2) ¥ r2 + 0 ¥ R = E2 + 0 ¥ r2 + 0 ¥ R = 2 V (b r2 and connecting point A to point B, the circuit changes to that shown in Fig. 3.52c. It has two nodes C and B. The voltage VCB = E2 = 2 V, irrespective of the currents in other branches. If node B is taken as reference (i.e., grounded), the voltage of node C is VC = VCB = 2 V. The currents in other branches can easily I3 =

VC

E1 r1

=

2

3 1

= –1 A; I4 =

VC

E3 r3

=

2 1 V 2 = 1 A; I6 = C = = 2 A 1 R 1

The current in branch containing E2 can be found by applying KCL at node C, I5 = I3 + I4 + I6 = (–1) + 1 + 2 = 2 A EXA M PL E

3 .3 5

Determine the voltage drop Vab across the open circuit in Fig. 3.53. Also, state which resistors have no effect on the result obtained. 10 W

5W

V1

V2

a 4W

6A

8A

11 W

18 W

Vab 13 W

9W

b 15 V

Solution

W resistance and 9-W resistance, as there is no closed path for the current to

10-W resistance is 6 ¥ 10 = 60 V, irrespective of the value of 4-W resistance. Similarly, the pd across 5-W resistance is 5 ¥ 8 = 40 V, irrespective of the value of 11-W resistance. The pd across the parallel combination of 15-V source and 18-W resistance remains 15 V, irrespective of the value of the 18-W resistance. Thus, 4-W, 11-W, 9-W, 18-W and 13-W resistances will have no effect on the voltage drop Vab. Note that we can conclude from above that a resistance connected in series with an ideal current source can be ignored and is redundant to the circuit. Similarly, a resistance connected in parallel with an ideal voltage source can be ignored and is redundant to the circuit. Vab = –15 + 5 ¥ 8 – 6 ¥ 10 = –15 + 40 – 60 = –35 V

Basic Electrical Engineering EXA M P L E

3. 36

Find V1 in the circuit of Fig. 3.54. 16 W

I1

20 kW

I1 10I2

24 V

4V1

0.5I1

4W

V1

36 V

I2

5 kW

V

W resistance, from bottom to top. Therefore, we have V1 = – (0.5I1) ¥ 4 = –2I1 Writing KVL equation for the input loop, and using the above result, we get 24 – 16I1 – 4V1 = 0 or 24 – 16I1 – 4(–2I1) = 0 fi I1 = 3 A \ V1 = –2I1 = –2 ¥ 3 = – 6 V

Solution In the output loop, the current 0.5I1

EXA MP L E

3. 37

Determine the voltage V in the circuit shown in Fig. 3.55.

Solution Applying KCL at the upper node, I1 – I2 + 10I2 = 0 fi I1 = –9I2. Now, writing KVL equation for the left loop, 36 – (20 kW) ¥ I1 = 0 or 36 – (20 kW) ¥ (–9I2) = 0 fi I2 = – 0.2 mA Since a current of 10I2 W resistance, we have \ V = –10I2 ¥ (5 kW) = –10 ¥ (–0.2 mA) ¥ (5 kW) = 10 V EXA MP LE

3 . 38

Find the currents i1 and i2 in the network of Fig. 3.56. 1W

2W 12 W 1W

4V

i1

I i2

3V V2

9A

5W

3i2

Solution Writing loop equation for the output loop gives 3i2 – (i2 – i1) ¥ 1 – i2 ¥ 2 – 3 = 0 Writing loop equation for the input loop, we get 4 – i1 ¥ 1 – (i1 – i2) ¥ 1 – 3i2 = 0

fi

i1 = 3 A

fi i1 + i2 = 2

fi i2 = –1 A

10 W

V1

3I

EX A MP L E

3.3 9

Determine the voltage V1 and V2 in the circuit of Fig. 3.57.

Solution The network has two nodes. Taking the bottom node as reference, we can write nodal voltage equation for the top node, in terms of its voltage V1, V1 V V + 3I = 4I + 1 ; where I = 1 . 10 10 5 V1 V1 9V1 + = fi V1 = 10 V 9 =4¥ 5 10 10 9 =I+

\ Now, for the left loop,

V2 – 12 ¥ 9 = V1 E XA MPLE

fi

V2 = V1 + 108 = 10 + 108 = 118 V

3.40

Find current I in the circuit of Fig. 3.58, by applying nodal-voltage analysis. 1A 6W

4A

3W

1

2

6W

2W

1

I 18 V

2

I

6W

6W

12 V

2A

2W

8W

7A

Solution Let the two nodal-voltages be V1 and V2. Writing KCL equations at the two nodes, V1 − 18 V1 V1 − V2 = 1 fi 2V1 – V2 = 12 + + 6 6 3 V2 − 12 V2 V2 − V1 + 1 = 0 fi 4V2 – 2V1 = 6 + + 6 6 3 Solving these two equations, we get V1 = 9 V and V2 = 6 V. V V 9 6 \ I = 1 2 = =1A 3 3 EXA MP LE

(i) (ii)

3.41

Find current I in the circuit of Fig. 3.59, by node analysis.

Solution Let V1 and V2 be the voltages of the independent nodes 1 and 2. Writing the node equations by inspection, we get

⎡1 1 ⎢2 + 2 ⎢ 1 ⎢ − ⎣ 2

1 ⎤ 2 ⎥ ⎡ V1 ⎤ ⎡4 + 2 ⎤ 1 1 ⎥ ⎢V ⎥ = ⎢⎣7 − 4 ⎥⎦ + ⎥ ⎣ 2⎦ 2 8⎦ −

or

− 0.5 ⎤ ⎡ V1 ⎤ ⎡6⎤ ⎡ 1 ⎢− 0.5 0.625⎥ ⎢V ⎥ = ⎢3⎥ ⎣ ⎦ ⎣ 2⎦ ⎣ ⎦

Solving the above, we get V1 = 14 V and V2 = 16 V. V V 14 16 \ I = 1 2 = = –1 A 2 2

Basic Electrical Engineering EX A MP L E

3.4 2

Determine the current through 2-S resistor in the circuit of Fig. 3.60a.

Solution Apparently, the circuit has 7 nodes. But a careful inspection indicates that it has only 4 nodes, as shown in Fig. 3.60b. The bottom node is made reference node r, remaining three are assigned voltages V1, V2 and V3. The node analysis equation in matrix form are written directly as −3 −4 ⎤ ⎡ V1 ⎤ ⎡( −3) + ( −8)⎤ ⎡( 4 + 3) ⎥ ⎢V ⎥ = ⎢ − ( −3) ⎥ ⎢ −3 (3 + 2 + 1) −2 ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢ ⎢⎣ − 4 −2 ( 4 + 2 + 5)⎥⎦ ⎢⎣V3 ⎥⎦ ⎢⎣ − ( −25) ⎥⎦

or

⎡ 7 −3 − 4 ⎤ ⎡ V1 ⎤ ⎡−11⎤ ⎢ −3 6 −2 ⎥ ⎢V ⎥ = ⎢ 3⎥ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢ ⎢⎣− 4 −2 11⎥⎦ ⎢⎣V3 ⎥⎦ ⎢⎣ 25⎥⎦

Solving the above for V2 and V3 only, we get V2 = 2 V and V3 = 3 V. Therefore, the current I through 2-S resistor is I = G(V2 – V3) = 2(2 – 3) = –2 A 4S –3 A –3 A V2

V1 3S

3S

2S

4S –8 A

5S

–25 A

–25 A

1S

r (b)

(a)

EX A MP L E

5S

1S

–8 A

V3

2S

3.4 3

Find the node voltages in the circuit shown in Fig. 3.61a. 65 A 13 V

5S

V1 1 75 A

6S

4S

V2

V1

V2 2

V3

2

1 13 A

(a)

75 A

4S

5S

6S

13 A

(b)

Solution First Method source of 65 A and a parallel resistor of 5 S, as shown in the circuit of Fig. 3.61b. Now, we can write the nodal equations in matrix form for the two nodes just by inspection, ⎡ 9 −5⎤ ⎡ V1 ⎤ ⎡10 ⎤ ⎢−5 11⎥ ⎢V ⎥ = ⎢52 ⎥ ⎣ ⎦ ⎣ 2⎦ ⎣ ⎦

Solving these equations, we get V1 = 5 V Now, from the original circuit shown in Fig. 3.61a Thus, V3 = V2 – 13 = 7 – 13 = –6 V.

and

V2 = 7 V V3 more negative than V2 by 13 V.

Second Method Another approach is to use the concept of supernode. The voltage source is enclosed in a region by a dotted line, as shown in Fig. 3.61a. The KCL is then applied to this closed surface : 6V2 + 5(V3 – V1) = –13 (i) The KCL equation for node 1 is 9V1 – 5V3 = 75 (ii) Since, there are three unknowns, we need another independent equation. This is obtained from the voltage drop across the voltage source, V2 – V3 = 13 (iii) The solutions are, of course, the same : V1 = 5 V, V2 = 7 V and V3 = – 6 V In general, for the supernode approach, the KCL equations must be augmented with KVL equations, the number of

SUMMARY TE RM S

A N D

C ON CE PT S

R, L and C. There are three different points of view you can look at a passive component : (i) Circuit view point, (ii) Energy view point, and (iii) Geometric view point. capacitor, if the voltage across it remains constant. The voltage across a capacitor cannot change instantaneously. inductor, if the current through it remains constant. The current through an inductor cannot change instantaneously. ON a circuit, a capacitor behaves as a short circuit. Under steady state, it behaves as an open circuit. ON a circuit, an inductor behaves as an open circuit. Under steady state, it behaves as a short circuit. C) is dual of inductance (L). Series combination and parallel combination of inductances follow the same rule as for resistances. Series combination and parallel combination of capacitances follow the inverse rule as for resistances. ideal voltage source has terminal voltage (V) which is independent of the output current (I). ideal current source delivers a constant current (I), independent of its output voltage (V). VS = RS IS. i) either a voltage source or a current source, (ii) either an ideal source or a practical source, (iii) either a dc source or an ac source, and (iv) either an independent source or a dependent source. i) VCVS, (ii) CCCS, (iii) CCVS, and (iv) VCCS. Kirchhoff’s Current Law (KCL) states that Kirchhoff’s Voltage Law (KVL) states that . loop-current analysis is a general method and can be applied to any network. The variables are currents and the equations are based on KVL. mesh analysis can be applied to only (i.e., a network which can be drawn on a sheet of paper without crossing lines).

94

Basic Electrical Engineering

node-voltage analysis

–

IM P O R TA N T

F OR MU LAE

resistance R W v R R i i 2t capacitance C i 2W C C dv/dt v2

R

r

L A

C

e

A d

L

inductance v L L di /dt

2W i

transforming VS RS IS

L

2

RSV

R

N2A l

RS

CHECK YOUR UNDERSTANDING two one

minus

12

S. No.

Statement V

V

n

True

False

V

n

Your Score

Marks

Network Analysis

95

A N SW E R S

1. 6.

2. 7.

3. 8.

4. 9.

REVIEW QUESTIONS 1.

7. 8.

2.

9. a

ON

b 10.

3. a

ON

b

11.

4.

12.

5.

13. 14.

6. 15.

MULTIPLE CHOICE QUESTIONS Here are some incomplete statements. Four alternatives are provided below each. Tick the alternative that completes the statement correctly

4. a b c d

1. a b c d

5. a b c d

2. 2

a C V c Q 2/2C 3.

2

b C V/2 d CV 6.

a b c d

a b c d

5. 10.

Basic Electrical Engineering 7. When a number of capacitors are connected in series, the total capacitance is (a) greater than any of the capacitors (b) greater than the largest capacitor (c) smaller than the smallest capacitor (d) None of the above 8. If ordinate represents voltage and abscissa the current, the characteristics of an ideal current source would be represented by (a) a vertical line (b) a horizontal line (c) a diagonal line at 45° to either axis extending (d) a diagonal line at 45° to either axis extending from the fourth to the second quadrant 9. Kirchhoff’s current law states that (a) the sum of currents in a series circuit is zero (b) the sum of currents in a parallel circuit is zero (c) the sum of currents in a series-parallel circuit is zero (d) the sum of currents entering a node is equal

to the sum of currents leaving the node 10. Kirchhoff’s voltage law states that (a) the sum of emfs in a closed loop is zero (b) the sum of voltage drops and emfs in a closed loop is zero (c) the total voltage drop in a series circuit is (d) the sum of potential differences across each element in a circuit is zero. 11. The number of equations required to analyse a given network by loop-current method is equal to (a) the number of independent loops (b) the number of nodes (c) the number of branches (d) None of these 12. The number of equations required to analyse a given network by nodal analysis is equal to (a) the number of independent loops (b) the number of nodes (c) the number of branches (d) one less than the number of nodes A N SW E RS

1. a 11. a

2. c 12. d

3. b

4. c

5. d

6. d

7. c

8. a

9. d

10. b

PROBLEMS ( A )

S IM P L E

PRO BL EMS

1. If a 12-V car battery has a 0.04-W internal resistance. What is the battery terminal voltage when the battery delivers 40 A ? [Ans. 10.4 V] 2. If a 12-V car battery has a 0.1-W internal resistance.

into the positive terminal ? [Ans. 12.4 V] 3. Find the voltage Vab across the open circuit in the circuit shown in Fig. 3.62. [Ans. 80 V] 4. For the circuit shown in Fig. 3.63, Find V1, Vad, Vbc and Vac + Vce. [Ans. 9 V, –3 V, 4 V, 2 V] d

e

b

5V

3V

–2 V a

12 V V1

c f

5. For the circuit shown in Fig. 3.64, determine the unknown currents using KCL. [Ans. I5 = 2 A, I3 = 1.5 A]

6. For the circuit shown in Fig. 3.65, determine the current in the 10-W resistor for R = 6 W, using nodevoltage analysis. [Ans. 0 A] 10 W

I1 = 4.5 A Is = 7 A

Single node 12 V

I4 = 0.5 A

I5 = ?

I2 = 3 A

I3 = ?

0.2 W

A

2A

R

7. Determine the voltages at nodes B and C in the network shown in Fig. 3.66. [Ans. VB = VC = 114 V] B

30 A

0.3 W

0.1 W

C

D

20 A

120 V

116 V

G

( B)

TRI C KY

P RO B LE MS

8. Using KCL, determine i and iab in the circuit shown in Fig. 3.67. [Ans. –15 A, 6 A] 9. v2, v3, i, R1 and R3, in the circuit of Fig. 3.68. [Ans. i = –0.6 A, v2 = –2 V, v3 = 15 V, R1 = 1.79 W, R3 = 3 W] 10. Apply mesh analysis and determine the loopcurrents I1, I2 and I3 in the circuit shown in Fig. 3.69. [Ans. I1 = 4.5 A, I2 = –2.5 A, I3 = 0.5 A]

11. in the circuit shown in Fig. 3.70. [Ans. I1 = – 0.25 A, I2 = – 4.75 A, I3 = – 4 A] 12. Find the loop currents in the network shown in Fig. 3.71, by using the loop-current method. [Ans. I1 = 4.9 A, I2 = 1.36 A, I3 = 3 A] 13. Determine the current through the 3-W resistor in the circuit shown in Fig. 3.72, by using mesh analysis. [Ans. I3W = –1.35 A] R3

–18 A

5A

5.6 A i

v3

3A

R2 a

b

R1

25 V

i 12 V 12 A

v2 v1 = 10 V

Basic Electrical Engineering 20 W

I1

200 V

2W

10 W

I3

I2

100 V

I1

12 V

3A

I3

1W

2W

I3

I2

1W

10 W

10 W

20 W

4A

2W 5V

3W

10 V 3W

2W

9V

I1

I2

12 V

14. Determine the current through the 1-W resistor in the circuit shown in Fig. 3.73, by using mesh analysis. [Ans. 1.39 A] 15. Transform the voltage source within the dotted box in the circuit shown in Fig. 3.74 into its equivalent current source and then apply nodal analysis to determine voltage at node b. [Ans. 7.28 V] 16. Using source transformation, reduce the network shown in Fig. 3.75 to a single loop circuit and then determine current through the 20-W resistor. [Ans. 1.375 A] 17. Find the current downward in the 50-W resistor in vb using node-voltage analysis. [Ans. 0.0448 A] 18. For the circuit shown in Fig. 3.77, solve for Va and Vb using node-voltage analysis, and then determine 2W

2W

I2

I3

12 W

12 V

1W

6V

10 V

5W

4W

19.

20.

21.

22.

the voltage across the three resistors. [Ans. Var = – 6.557 V, Vbr = + 4.393 V, Vab = –10.95 V] Using nodal analysis, determine the voltages of nodes 1 and 2, for the circuit shown in Fig. 3.78. [Ans. 19.43 V, 18.28 V] By applying nodal analysis to the circuit of Fig. 3.79, determine the currents I1 and I2. [Ans. 4.75 A, 4.25 A] Find the current I by using node-voltage analysis for the circuit given in Fig. 3.80. [Ans. –2.255 A] Using node-voltage analysis, determine the current I in the 15-W resistor of the network shown in Fig. 3.81. [Ans. 3.5 A]

a

6W

b

3W 10 V

5W

1A

I1 24 V 4W

r

10 W

10 W

5A

20 W

1.5 A

100 W

a

a

5V

150 W

b

50 W

0.1 A

10 V

b 10 V

r

2A 4W

1 a

2

b 7W

15 W

V1

10 A

10 W

2W

V2

2A

8W

r

1W 4A 1

2 I1

9A

( C )

1

15 W

80 W

2W

2A

PROBLEMS

2

voltage across the load. [Ans. (a) 3 A and 1 A; (b) 4 A; (c) 4 V] 24. For the circuit shown in Fig. 3.82, determine currents I1 to I5 by using node-voltage analysis. [Ans. I1 = 1.25 A, I2 = 3.75 A, I3 = –2.5 A, I4 = 5 A, I5 = –7.5 A]

10 W

2W

I 400 V

1W

3A

23. Two batteries having emfs of 10 V and 7 V, and internal resistances of 2 W and 3 W respectively, are connected in parallel across a load of resistance 1 W. Calculate (a) the current supplied by each battery, (b) the current through the load, and (c) the 20 W

2

I 3W

C HA L L EN G IN G

2W

I2

6W

3W

6V

1

1W

1

I1 90 W

200 V

10 V

I2

2

I3 2W

2W I5

I4

2W

25 V

100

Basic Electrical Engineering

EXPERIMENTAL EXERCISE 3.1 K IR C HH O F F ’S

C U RR ENT

LAW

Objective To verify Kirchhoff’s Current Law (KCL). Apparatus DC power supply 220 V; Three ammeters (MC type) 0-5 A; Three rheostats 100 W, 5 A. Circuit Diagram The circuit diagram is shown in Fig. 3.83.

Brief Theory Kirchhoff’s Current Law states that at any instant the algebraic sum of currents meeting at a junction n branches meeting at a junction, KCL can be stated as n

Ij = 0 j =1

Procedure 1. 2. 3. 4. 5. 6. 7.

Make connections as given in Fig. 3.83. Set all the three rheostats to their maximum value. Switch ON the supply. Note the readings of the ammeters A1, A2 and A3, giving the values of the currents I1, I2 and I3 in the circuit. Change the settings of rheostats so as to get different readings in the ammeters. Note the readings. Switch OFF the supply.

Observations S. No. 1. 2. 3. 4. 5.

I1 (in A)

I2 (in A)

I3 (in A)

I1 + I2 (in A)

101 Calculations For each set of settings of the rheostats, add the readings in columns 2 and 3, and record it in column 5 of the table. Results I3 in column 4 is nearly the same as that of (I1 + I2) in column

1. 5. 2.

Precautions 1. Before switching on the supply, the zero reading of the ammeters should be checked. 2. The terminals of the rheostats should be connected properly. 3. While setting the rheostats, care should be taken that the current I3 as recorded by the ammeter A3 does not exceed 5 A.

Viva Voce 1. In an ac circuit consisting of resistances, inductances and capacitances, the three currents meeting at a junction were found to have the values I3 = 4.5 A, I1 = 2.8 A and I2 = 3.2 A. Here, I1 + I2 = 6 A, whereas I3 = 4.5 A. Therefore, will you say that KCL is not applicable ? Ans. : No. KCL is applicable to both the dc as well as ac circuits. However, in ac circuits, we consider phasor sum rather than algebraic sum. 2. What do you mean by algebraic sum ? Ans. : junction is assigned a ‘positive’ sign, then the current leaving the junction must be assigned a ‘negative’ sign. 3. Is KCL based on the principle of conservation of energy ? Ans. : No. It is based on the principle of conservation of charge. Since the accumulation of electric charge at a junction is not possible, the amount of charge entering a junction must be the same as the amount of charge leaving the junction. 4. What do you mean by the term ‘node’ in reference to an electric circuit ? Ans. : A node is a point in a circuit where more than two elements are joined together.

EXPERIMENTAL EXERCISE 3.2 K IR C HH O F F ’ S

V O LTA G E

LAW

Objectives To verify Kirchhoff’s Voltage Law (KVL). Apparatus

Two rheostats 100 W, 5 A.

Circuit Diagram The circuit diagram is shown in Fig. 3.84. Brief Theory Kirchhoff’s Voltage Law states that at any instant the algebraic sum of voltages around a loop or a closed circuit is zero. This statement simply tells us that if we start form a point in an electric circuit and go around a loop so as to come back to the same point, the net potential rise (or potential drop) is zero. For a loop having k elements, the KVL can be stated as k

Vj = 0 j =1

where V represents the voltage drop (or voltage rise) of th element in the loop.

102

Basic Electrical Engineering 0-10 A A

Rheostats 100 W, 5 A

I MC V2

220 V DC Supply

MC V3 0-300 V

0-300 V 0-300 V V1 MC

Switch

Kirchhoff’s voltage law can also be stated like this : In any closed circuit or a loop, the algebraic sum of the products of current and resistance in each of the conductors is equal to the algebraic sum of the emfs.

Procedure 1. 2. 3. 4. 5.

Make connections as given in Fig. 3.84. Set both the rheostats to their maximum value. Switch ON the supply. Note the readings of the ammeter A, and the three voltmeters V1, V2 and V3. Change the settings of the rheostats so as to get different readings in the voltmeters V1 and V2. Note the readings.

6. 7. Switch OFF the supply.

Observations S. No.

I (in A)

V1 (in V)

V2 (in V)

V3 (in V)

V1 + V2 (in V)

1. 2. 3. 4. 5.

Calculations For each set of settings of the rheostats, add the voltages V1 and V2 recorded in columns 3 and 4, and record the same in column 6 of the table. Results 1.

V3 in column 5 is nearly the same as that of (V1 + V2) in column

6. 2. Kirchhoff’s Voltage L

Precautions 1. Before switching ON the supply, the zero reading of the ammeter and voltmeters should be checked. 2. The terminals of the rheostats should be connected properly. 3. While setting the rheostats, care should be taken that the current I as recorded by the ammeter A does not exceed 5 A, the current rating of the rheostats.

103 Viva Voce 1. While applying in a loop ? Ans. : No. In ac circuits, we consider phasor sum rather than algebraic sum. 2. What do you mean by the term ‘loop’ ? Ans. : A loop is a closed path in an electrical circuit. That is, if we start from one point, we should be able to come back to the same point by traversing the loop. 3. Is a loop different from a mesh ? Ans. : Yes. A mesh is a special loop that has no loop within it. Thus, all meshes are loops, but all loops are not meshes. 4. In using KVL, is it necessary that a voltage drop is written with a negative sign ? Ans. : No. It depends upon the sign convention chosen. We may take voltage drop as positive and voltage rise as negative, or vice versa. Both ways, we get the same result.

NETWORK THEOREMS

4

OB JE CT IV ES :

Simple circuits can be solved by using Ohm’s law, Kirchhoff’s laws, voltage divider, current divider, series and parallel combination of sources and resistors, etc. Special techniques, known as network theorems and network reduction methods, have been developed which drastically reduce the labour of solving a more complicated network. The network theorems provide simple conclusions and good insight into the problems. Some of these have universal applications, whereas some are limited to the networks containing linear* elements only.

This theorem states that the response in a linear circuit at any point due to multiple sources can be calculated by summing the effects of each source considered separately, all other sources being made inoperative (or turned OFF). The theorem is applicable only to a linear network (containing independent and/or dependent sources).

When a voltage source is made inoperative or turned OFF, no voltage drop exists across its terminals but a current can still flow through it. Hence, it acts like a short circuit (Fig. 4.1a). *

A linear element obeys Ohm’s law, e.g., resistance, inductance, etc. Semiconductor diodes and transistors are not linear elements.

105

Similarly, when a current source is made inoperative or turned OFF, no current flows through it but a voltage can appear across its terminals. Hence, it acts like an open circuit (Fig. 4.1b). By making a source inoperative or turned OFF means that the voltage source is replaced by a shortcircuit and the current source is replaced by an open circuit. EXA M P L E

4. 1 I in the circuit shown in Fig. 4.2a.

0.1 W

0.3 W

0.1 W

80 mV

0.5 A

0.3 W

0.1 W

0.3 W

80 mV

0.5 A I2

I1

I

(c)

(b)

(a)

Solution

I1 due to the 0.5-A current source, and turn OFF the 80-mV voltage source by shorting it (Fig.4.2b). Applying current divider, we get I1 = – 0.5 ¥

0.3 0.1 + 0.3

= –0.375 A

Next, consider the voltage source, and turn OFF the current source by opening it (Fig.4.2c). Ohm’s law gives I2 =

80 × 10 − 3 = 0.2A 0.1 + 0.3

By the principle of superposition, the total current is given as I = I1 + I2 = –0.375 + 0.2 = – 0.175 A Note that in redrawing the circuit for each source, we are always careful to mark the response current in the original direction and also assign a suitable subscript to indicate that we are not working with the original variables. This prevents the possibility of committing errors when we add the individual currents.

106

Basic Electrical Engineering

EXA M P L E

4. 2 Ix in the network given in Fig. 4.3. 10 V –+ 120 A

50 W

150 W

40 A

Ix

Solution (Fig. 4.4a). Current I1 is I1 =

10 50 + 150

= 0.05 A

10 V –+ 50 W

150 W I1

50 W

150 W

40 A

120 A

50 W I3

I2

(a)

150 W

(b)

(c)

Next consider the 40-A current source alone (Fig. 4.4b). Applying current divider, the current I2 is given as 150 I2 = 40 ¥ = 30 A 50 + 150 Lastly consider the 120-A current source alone (Fig. 4.4c). The current I3 is 50 I3 = –120 ¥ = –30 A 50 + 150 Applying the principle of superposition, we get Ix = I1 + I2 + I3 = 0.05 + 30 – 30 = 0.05 A EXA MP L E

4 .3

Consider our benchmark example (Fig. 3.24a) discussed in Example 3.12, wherein we had calculated the voltage across 3-W resistor as 2.5 V, by using source transformation. The same problem was solved in Example 3.19 by using loop-current analysis v across 3-W resistor by applying the principle of superposition. The circuit is redrawn in Fig. 4.5a for convenience.

Solution

4-A current source, while turning OFF the remaining two sources. The turned-OFF 5-A current source is replaced by an open circuit and the turned-OFF 6-V voltage source is replaced by a short circuit, as shown in Fig. 4.5b

107

2W 5A 4A

1W

2W + v

3W

4A

1W

i 3W

v4

3W

+ v6

–

6V (a)

(b)

2W 5A 1W

2W + v5

3W

1W

–

–

6V (c)

(d)

i =4¥

1

=

2

A 1 + (2 + 3) 3 Thus, voltage v4 across 3-W resistor due to 4-A current source is v4 = i ¥ R = ( 23 A) ¥ (3 W) = 2.0 V The polarity of this voltage is the same as the original polarity markings for v summation will appear with a + sign. Next consider 5-A current source, with 4-A and 6-V sources turned OFF (Fig. 4.5c). From the perspective of the 5-A source, the 2-W and 3-W resistances are in series and together in parallel with the 1-W resistor. Using current divider, the voltage v5 across 3-W resistor due to 5-A current source is given as

1 È ˘ A ¥ ( 3 W) = –2.5 V v5 = Í- 5 ¥ 1 + (2 + 3) ˙˚ Î Note that the actual polarity of this voltage v5 is opposite to the polarity marked on v. Hence, the minus sign. We now consider 6-V voltage source, with 4-A and 5-A sources turned OFF (Fig.4.5d). The three resistances are connected in series. The resulting voltage v6 is calculated by voltage divider, as 3 v6 = 6 ¥ = 3.0 V 1+ 2 +3 The polarity of this voltage is same as that of the original voltage v. Hence, the sign of v6 summation. Using principle of superposition, we now obtain the total voltage v across 3-W resistor, as v = + v4 + v5 + v6 = + 2.0 – 2.5 + 3.0 = +2.5 V

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EX A MP L E

4.4 Is to reduce the voltage across 4-W resistor to zero.

For the circuit shown in Fig. 4.6a

2W

2W 4W

10 V

Is

2W 4W

10 V

4W

I1

6W

6W (a)

I2

Is

6W (b)

(c)

Solution To solve this problem, we apply the principle of superposition. The current I1 (from top to bottom) in 4-W resistor due to 10-V source in Fig. 4.6b is 10 5 I1 = = A 2+4+6 6 The current I2 (from top to bottom) in 4-W resistor due to current source Is in Fig. 4.6c is I2 = –Is ¥

2+6

2

Is 3 The voltage across 4-W resistor can be zero, only if the current through this resistor is zero. That is, I1 + I2 = 0

or

2+6+4 5

=–

⎛ 2 ⎞ + − Is = 0 6 ⎝ 3 ⎠

fi Is =

5

3

6

2

= 1.25 A

This theorem was first proposed by a French telegraph engineer M.L. Thevenin in 1883. Often, we need to find the response (current, voltage or power) in a single load resistance in a network. Thevenin’s theorem enables us to do this without solving the entire network. It is specially very helpful and timesaving when we wish to find the response for different values of the load resistance. Thevenin’s Theorem states that it is possible to simplify any linear circuit containing independent and dependent voltage and current sources, no matter how complex, to an equivalent circuit with just a single voltage source and a series resistance, between any two points of the circuit.

Procedure The procedure to apply Thevenin’s theorem to a network will be explained in steps, by taking an example. Consider the circuit shown in Fig. 4.7a. Suppose that we are interested to find the current in resistor R2. We proceed as follows. 1. Designate the resistor R2 as ‘load’ (Fig. 4.7b). 2. Pull out the load resistor and enclose the remaining network within a dotted box (Fig. 4.7c). 3. Temporarily remove the load resistor R2, leaving the terminals A and B open (Fig. 4.7d).

109

4. Find the open-circuit voltage across the terminals AB as follows : 28 − 7 21 = = 4.2 A; VAB = 7 + 4.2 ¥ 1 = 11.2 V I= 4 +1 5 5. This is called Thevenin voltage, VTh = VAB = 11.2 V. 6. Turn OFF all the sources in the circuit (Fig. 4.7e) and find the resistance between terminals A and B. This is the Thevenin resistance, RTh. Thus, 1× 4 = 0.8 W RTh = 1 W || 4 W = 1+ 4 7. The circuit within the dotted box (Fig. 4.7c) is replaced by the Thevenin’s equivalent, consisting of a voltage source VTh in series with a resistor RTh, as shown in Fig. 4.7f. 8. The load resistor R2 is again connected to Thevenin’s equivalent (Fig. 4.7f) forming a single-loop circuit, and the current I2 through this resistor is easily calculated as VTh 11.2 = =4A (4.1) I2 = RTh + R2 0.8 + 2

The equivalent circuit replaces the circuit within the box only for the effects external to the box. We can no longer ask questions about the circuit in box. For example, if we are interested in the current in R1 (= 4 W) or the total power consumed by the resistors R1 and R3 in the box of Fig. 4.7c, the equivalent circuit of Fig. 4.7f is just useless.

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The example of the circuit considered above was simple. With a little more effort, we could have computed this result directly from the original circuit of Fig. 4.7a. You may wonder why solve it by this roundabout method. It is because we get the following benefits from the concept of Thevenin’s theorem. 1. The theorem enables us to derive the same simple equation as Eq. 4.1, no matter how complicated the original circuit is. There could have been hundreds of sources and thousands of resistors in the original circuit. We still would have reduced the circuit to two parameters, Thevenin’s voltage VTh and Thevenin’s resistance RTh. 2. Thevenin’s equivalent circuit leads to one of the most useful ideas of electrical engineering, namely, the idea of ‘output impedance’ of a circuit. 3. We gain the freedom to ask many additional questions such as : What value of R2 makes the voltage across it 20 V ? What value of R2 draws the maximum power from the circuit ? The answer to the first question is that no value of R2 will give 20 V, as VTh is only 11.2 V. The second question leads us to an interesting and important result, named as maximum power transfer theorem, which we shall soon see. EX A MP L E

4.5

By using Thevenin’s equivalent of the circuit within the dotted box, determine the voltage across the load resistor RL in Fig. 4.8a. RTh

a 10 W 50 V

10 W 20 W

1.5 A

10 V

+ RL = 5 W

VL

VL

VTh

–

–

b

b

(a) 10 W

a

+

RL

(b) RTh

10 W

20 W

RTh

a +

10 W VTh

VL

27.5 V

RL = 5 W

– b (c)

(d)

Solution We replace the circuit in the box with a simpler circuit shown in Fig. 4.8b. VTh is the open-circuit voltage between terminals a and b with RL W resistor and then the open-circuit voltage. With 1.5-A current source turned OFF, the current (downward) through 20-W resistor due to the two voltage sources is

111

I1 =

50 − 10 10 + 10 + 20

= 1.0 A

The current (downward) due to the 1.5-A current source acting alone is 10 = 0.375 A 10 + (10 + 20 ) Adding these two currents, we get the current with all sources ON as I = 1.375 A, and the voltage across 20-W resistor, which is same as the open-circuit voltage or Thevenin’s voltage, as VTh = 1.375 ¥ 20 = 27 .5 V with positive (+) at the top. The polarity is important because we require that the two circuits in the box in Fig. 4.8a and b be fully equivalent as seen by the load. To compute RTh, we turn OFF all three sources within the box, as shown in Fig. 4.8c, and get RTh = 20 W || (10 W + 10 W) = 10 W Thus, as far as RL is concerned, the circuit of Fig. 4.8a reduces to the equivalent circuit of Fig. 4.8d. Using voltage divider, we can now easily determine the voltage across RL, 5 VL = 27. 5 ¥ = 9. 17 V 5 + 10 I2 = 1.5 ¥

EXA MP LE

4 . 6

Let us again consider our benchmark example* (Fig. 3.24a) to determine voltage across 3-W resistor by applying Thevenin’s theorem.

Solution We treat the 3-W resistor as load and enclose the remaining circuit within a box (Fig. 4.9a). Thevenin voltage VTh is the open-circuit voltage between terminals a and b (with RL removed). This can be calculated by source transformation. The 4-A current source in parallel with 1-W resistance is transformed into a 4-V voltage source in series a 2W

2W 4A

1W

RL 3W

5A 6V

2W

1W 5A 4V 6V

10 A

b

b

(b)

(c) RTh

a

a 2W

2W 1W

1W

b (d)

5A

1W

b

(a)

5A

a

a

3W RTh

VTh

RL 3W

5V

b (e)

a

b (f)

* This problem was solved in Example 3.12 by using source transformation, in Example 3.19 by using loop-current analysis, and again in Example 4.3 by applying the principle of superposition.

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with 1-W resistance (Fig. 4.9b). Combining 4-V and 6-V voltage sources and then converting it into current source, we get the circuit of Fig. 4.9c. Combining the two current-sources, we get the circuit of Fig. 4.9d. From this, VTh = 5 ¥ 1 = 5 V, with the + at the top. To compute RTh, we turn OFF all the sources in the circuit within box and get the circuit of Fig. 4.9e. Thus, RTh = 3 W. Thevenin’s equivalent circuit is shown in Fig. 4.9f. Using voltage divider, the voltage across the load is given as VL = 5 ¥

3 3+3

= 2.5 V

Note that we got the same result in Example 4.3.

An American engineer named E.L. Norton came up with an equivalent circuit similar to Thevenin’s equivalent circuit. He stated that a two-terminal linear network containing independent voltage and current sources may be replaced by an equivalent current source IN in parallel with a resistance RN. The procedure for determining the Norton’s equivalent circuit is as follows. 1. Short circuit the two terminals of the network and determine the current through this short circuit. This is Norton’s current IN. 2. Turn OFF all the sources in the network and determine the resistance between the two terminals. This is Norton’s resistance RN. However, this is the same resistance RTh that we found in Thevenin’s equivalent. Therefore, we may say, RN = RTh = Req (say) (4.2) 3. Norton’s equivalent circuit is current source IN in parallel with resistance RN, as shown in Fig. 4.10a. a

a RTh

IN

RN

VTh

b (a) Norton’s equivalent circuit

b (b) Thevenin’s equivalent circuit

If two circuits are equivalent to the same circuit, they must be equivalent to each other. Thus, the two circuits of Fig. 4.10 are equivalent. The open-circuit voltage between terminals a and b in the circuit of Fig. 4.10a is obviously INRN. This must be same as VTh of Fig. 4.10b. That is, (4.3) VTh = IN RN This equation is very useful. It can be applied in the laboratory to find the output impedance of a source. In practice, it may not be possible to get inside a circuit and turn OFF internal sources. But we can measure the open-circuit output voltage and the current that would flow when the output terminals

are shorted. From these values, the output resistance can be computed, Voc VTh Ro = = Isc IN EX A MP L E

(4.4)

4.7

Obtain Norton’s equivalent circuit with respect to the terminals AB for the circuit of Fig. 4.11a and then determine the W if it were connected across terminals AB.

5W

A

10 W

5W

10 V

5V

10 V

A

I1

IN

B

10 W I2

5V

B

(a)

(b) Norton’s equivalent

5W

A

10 W

IL IN

2.5 A

RN

10 W 3

RL

5W

B

(c)

(d)

Solution When terminals AB are shorted (Fig.4.11b), the current IN is given as IN = I1 + I 2 =

10

+

5

= 2.5 A 5 10 Turning OFF the sources (Fig.4.11c), the two resistances of 5 W and 10 W appear in parallel and the equivalent resistance is given as 5 × 10 10 RN = = W 5 + 10 3 Hence, the Norton’s equivalent of the given circuit is as shown within dotted box in Fig. 4.11d. Using current divider, the current through 5-W resistance connected across AB is given as IL = 2.5 ¥

(10 / 3) (10 / 3) + 5

=1A

This theorem states that maximum power is absorbed from a network when the load resistance is equal to the output resistance of the network as seen from the terminals of the load.

Proof of the Theorem As per Thevenin’s theorem, a two-terminal network can be converted into a single voltage source VTh in series with resistance RTh (Fig.4.11). Let us find such a value of RL that

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Basic Electrical Engineering

consumes maximum power from the network. The power consumed by the load resistance is given as 2 VTh RL VTh ⎤ 2 P = I2RL = ⎢⎡ ¥ R = L ⎥ ( RTh + RL ) 2 ⎣ RTh + RL ⎦ Power P becomes maximum when differential of P with respect to RL reduces to zero. That is, 2 2 ( RTh + RL )2 ¥ VTh - VTh RL ¥ 2( RTh + RL ) dP = dRL ( RTh + RL )4

fi

RL = RTh

(4.5)

(4.6)

Impedance Match When the load resistance is made equal to the output impedance of the circuit, we say the impedance match has been done. Importance of Maximum Power Transfer Often we deal with small amount of power in electronics. We wish to make full use of the power that is available. Obtaining the maximum power out of a circuit is then important. For example, we connect a ‘rabbit ears’ antenna to the TV set to receive power from radio waves originating at a transmitter miles away. The antenna does not collect much power. So, the TV receiver is designed to make maximum use of the power provided by the antenna. Optimisation In practice, we always have some constraints. When a best possible design is made under the constraints, we say the design is the optimum. Thus, the TV receiver input is optimised when its input impedance is matched to the output impedance of the antenna because this gives maximum power to the receiver. Available Power The maximum power that can be extracted from a circuit is called the available power, Pavl, and is given by Pavl = P (when RL = RTh ) = I2RL 2 2 2 VTh ⎤ ⎡ VTh ⎤ ⎡ VTh ¥ R = ¥ R = (4.7) =⎢ L Th ⎥ ⎢R + R ⎥ 4 RTh Th ⎦ ⎣ RTh + RL ⎦ ⎣ Th Note that a circuit or a source with low output impedance can supply large amount of power to an external load.

EX A MP L E

4.8

The open-circuit voltage of a standard car-battery is 12.6 V, and the short-circuit current is approximately 300 A. What is the available power from the battery ?

Solution From Eq. 4.4, we determine the output impedance of the battery, Ro =

Voc 12.6 = = 0.042 W Isc 300

Therefore, the available power from Eq. 4.7 is Pavl =

2 (12.6)2 VTh V2 = oc = = 945 W 4 RTh 4 Ro 4 0.042

115 EXA M P L E

4. 9

A stereo sound system is operated from a battery, which is made from 8 dry cells connected in series. Each cell has an emf of 1.5 V and an internal resistance of 0.75 W. How much is the available power from this battery ?

Solution The open-circuit voltage of the battery is Voc = 8 ¥ 1.5 = 12 V, and output impedance of the battery is Ro = 8 ¥ 0.75 = 6 W. Therefore, the available power is Pavl =

2 (12)2 Voc = =6W 4 Ro 4 6

Comment Compare the available power in the above two examples. Though, the two sources have almost the same emf, but the car-battery can supply much larger power. EXA M P L E

4. 1 0

A sound system is designed to give 25 W into an 8-W speaker. (a) What would be the voltage provided by this b) What power will be delivered to the load, if (i) the load is a short circuit, (ii) the load is an open circuit ? (c) If different speakers of load resistance 2 W, 4 W, 6 W, 8 W, 16 W and 32 W are used in the system, calculate the power delivered to the speaker in each case, and draw a curve showing the variation of the output power Po with load resistance RL.

Solution (a) We assume that the speaker is represented by 8-W impedances are matched (Ro = 8 W Thevenin’s equivalent as shown in Fig. 4.12a. Using Eq. 4.7, Thevenin’s voltage is given by 2 VTh fi VTh = 28.28 V 4 8 The voltage Vo across the speaker will be half of this value, as the voltage VTh divides equally between Ro and RL. Thus, Vo = 14.14 V.

25 =

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Basic Electrical Engineering

(b) (i) If the load is a short circuit (RL = 0), the output voltage Vo would be zero, and hence the output power, Po = Vo I L = 0 ¥ I L = 0 (ii) If the load is an open circuit (RL IL would be zero, and hence the output power, Po = Vo IL = Vo ¥ 0 = 0 (c) Using Eq. 4.5, we calculate the output power Po for different values of RL as given below. Speaker resistance, RL (in W)

0

2

4

6

8

16

32

Output power, Po (in W)

0

16

22.22

24.49

25

22.22

16

0

The graph showing the variation of the output power Po with load resistance RL is shown in Fig. 4.12b. Note 8 W.

This theorem is helpful in finding a single equivalent voltage source for a number of voltage sources connected in parallel. It states that a number of parallel voltage sources V1, V2, V3 …, Vn with internal resistances R1, R2, R3…, Rn, respectively (as shown in Fig. 4.13a), can be replaced by a single voltage source V in series with equivalent resistance R (as shown in Fig. 4.13b), as follows: ±V1G1 ± V2G2 ± V3G3 º ± VnGn G1 + G2 + G3 º + Gn 1 1 = R= G G1 + G2 + G3 … + Gn V=

and

(4.8) (4.9)

This theorem states that in a linear bilateral network, if a voltage source V in a branch A produces a current I in any other branch B, then the same voltage source V acting in the branch B would produce the same current I in branch A. In other words, it simply states that V and I are interchangeable. The ratio V/I is known as the transfer resistance. The theorem is illustrated in the following example.

117 EX A MP L E

4.1 1

In the network of Fig. 4.14a transfer the voltage source to branch B (as in Fig. 4.14b established ? Also, determine the transfer resistance from branch A to branch B. 2W

3W

2W

36 V

I

3W

36 V

I¢

12 W

12 W 1W

4W Branch A

Branch B (a)

1W

4W Branch A

Branch B (b)

Solution For the network of Fig. 4.14a, the equivalent resistance for the voltage source is Req = 2 + [12 || (3 + 1)] + 4 = 2 + 3 + 4 = 9 W The current supplied by the voltage source = 36/9 = 4 A. Using current divider, the current I in branch B is given as 12 I =4¥ =3A 12 + 4 Now, for the network of Fig. 4.14b, the equivalent resistance for the voltage source is Req = 3 + [12 || (2 + 4)] +1 = 3 + 4 + 1 = 8 W The current supplied by the voltage source = 36/8 = 4.5 A. Using current divider, the current I ¢ in branch A is given as 12 I ¢ = 4.5 ¥ =3A 12 + 6 I and I ¢ have the same value 3 A. Thus, the reciprocity theorem is established. The transfer resistance is given as Rtr =

V 36 = = 12 W I 3

It was given by D.H. Bernard Tellegen in 1952. It simply states that the sum of instantaneous power delivered to each branch of a network is zero, b

ÂV I

k k

=0

(4.10)

k =1

The only requirement (or limitation) is that the voltages Vk satisfy KVL and that the currents Ik satisfy KCL. The power of the theorem lies in the fact that the voltages Vk and the currents Ik are arbitrary except for the Kirchhoff constraints. The theorem is illustrated in Example 4.12.

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EX A MP L E

4.1 2

Consider the network shown in Fig. 4.15. It has six branches. Arbitrary reference directions have been selected for all branch currents, and corresponding branch voltage is indicated with current entering the plus terminal. Select a set of branch voltages that satisfy KVL. Select two different sets of branch currents, each satisfying KCL. Show that both the sets of branch currents along with the set of branch voltages satisfy Eq. 4.10. R6 + V2 – B R2 + –

V1

– V6 + C I2

I1

I6

+ V3 – I3

R3 A

+

V4 – D R4

I4 R5

+ V5 –

I5

Solution Let us arbitrarily choose V1 = 4 V and V2 = 2 V. Applying KVL around the loop ABCA, we see that we must have V3 = 2 V. Similarly, around loop ACDA, if we choose V4 = 3 V, then as per KVL we are required to let V5 = –1 V. Now, around the loop BCDB, the values we have selected for V2 and V4 require that V6 = –5 V. This set of branch voltages is listed in Table 4.1. Next, we apply KCL successively to nodes B, C and D. At node B, we arbitrarily let I1 = 2 A, and I2 = 2 A. This selection requires that I6 = 4 A. At node C, we arbitrarily select I3 = 4 A, and then it is required that I4 = –2 A. At node D, I4 and I6 have already been selected, so that as per KCL we must let I5 = –6 A. This set of branch currents is also listed in Table 4.1.

Branches Items Vk (in V) Ik (in A) Vk Ik (in W) Ik¢ (in A) Vk Ik¢ (in W)

1

2

3

4

5

6

4 2 8 –1 –4

2 2 4 3 6

2 4 8 2 4

3 –2 –6 1 3

–1 –6 6 –1 1

–5 4 –20 2 –10

For each branch, we multiply Vk and Ik , and put the result in the third row of Table 4.1. Adding all the entries in this row, we get 6

ÂV I

k k

= 8 + 4 + 8 – 6 + 6 – 20 = 0

k =1

currents I k¢, satisfying KCL at each node, as indicated in 4th VkI k¢ and put the products in 5th row. Summing up all the entries in the 5th row, we get

119 6

 V I ¢ = – 4 + 6 + 4 +3 + 1 – 10 = 0 k k

k =1

Proof of Tellegen’s Theorem : We will again consider the circuit shown in Fig. 4.15 and note that the things that we have done for this network will lead to the same results in general. We will now make use of node-to-datum voltages and currents with double subscript to indicate direction. For example, for the branch 2, we will write V2I2 = (VB – VC)IBC Summing a similar product for each of the six branches, we get 6

ÂV I

k k

= VBIAB + (VB – VC)IBC + VCI CA + (VC – VD)ICD + VDIDA+ (VD – VB) IDB

k =1

This equation is next rearranged by factoring with respect to the node-to-datum voltages, 6

ÂV I

k k

= VB(IAB + IBC – IDB) + VC(– IBC + ICA + ICD) + VD(– ICD + IDA + IDB)

k =1

= VB(KCL at node B) + VC(KCL at node C) + VD(KCL at node D) Each product term in above equation vanishes because each KCL summation is equal to zero. Although, we have shown the validity of Tellegen’s theorem for a specific example, the procedure used is identical to that used for a general network. Note that the theorem depends only on the two Kirchhoff’s laws. Therefore, it is applicable to a very general class of networks composed of elements that are linear or nonlinear, passive or active, time-invariant or time-varying. This generality is one of the reasons that makes the Tellegen’s theorem a powerful tool.

ADDITIONAL SOLVED EXAMPLES EX A MP L E

4.1 3

Using Thevenin’s theorem, calculate the current I3 in the circuit of Fig. 3.47.

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Basic Electrical Engineering

W resistance. Hence, we treat this resistance as the load resistance and remove it (Fig. 4.16a). To determine, VTh, we write KVL equation for the loop bacdeb,

Solution The current I3

VTh + 50 – 20 + 10 = 0 fi V Th = –40 V To determine RTh, we short circuit all the voltage sources ( Fig. 4.16b). Its equivalent circuit is given in Fig. 4.16c. Obviously, RTh = 0 W. Hence, VTh − 40 I3 = = = –2 A 20 + 0 RL + RTh EXA MP LE

4.14

Using Thevenin’s Theorem, determine the current I2 in the circuit of Fig. 3.47. W resistance. Hence, we treat this resistance as the load resistance and remove it (Fig. 4.17a). To determine, VTh, we write KVL equation for the loop bacdefb, VTh – 20 +10 + 9 = 0 fi VTh = 1 V

Solution The current I2

a+ V – Th b a 9V f

c

b

c

f

20 W

e

d

20 W

e

d

(a)

2W

2W

10 V

5W

20 V

b

f

c

20 W

10 W

50 V

a

b

5W

b

10 W

a

a

e

d

(b)

(c)

To determine RTh, we short circuit all the voltage sources (Fig. 4.17b). Its equivalent circuit is given in Fig. 4.17c. Clearly, RTh = 0 W. Hence, VTh 1 I2 = = =1A RL + RTh 1+ 0 EXA MP LE

4.15

Determine the current through the resistance R = 0.5 W in the circuit shown in Fig. 4.18, using superposition theorem.

Solution turning OFF 20-V source. The circuit reduces to that shown in Fig. 4.19a. The 2-W and 2-W resistances come in parallel and the circuit reduces to that shown in Fig. 4.19b. The circuit is further c. The current I is given as 15 I= = 9.375 A 1 + 0.6

1W

15 V

A

1W

R = 0.5 W

B

2W

2W

20 V

121

From Fig. 4.19b, we get the current I1 through R due to 15-V source as 1 1 I1 = I ¥ = 9.375 ¥ = 3.75 A (from A to B) 1 + (0.5 + 1) 2.5 Now, consider 20-V source alone (by turning OFF 15-V source). The circuit reduces to that shown in Fig. 4.20a. It can be reduced to that shown in Fig. 4.20b, and then to that shown in Fig. 4.20c. From this circuit, we get 20 I¢ = = 7.5 A (2 / 3) + 2 1W

1W

A

0.5 W

B

2W A

2W

I¢1 0.5 W

B

B

I¢1 0.5 W

20 V

2W

(a)

2W

I¢ 2W

20 V

(b)

2/3 W

(c)

From Fig. 4.20b, we get the current I 1¢ through R due to 20-V source as 2 2 I 1¢ = I ¢ ¥ = 7.5 ¥ = 5 A (from B to A) 2 + (0.5 + 0.5) 3 Hence, by superposition theorem, the net current through R (from A to B) is Inet = I1 – I 1¢ = 3.75 – 5 = – 1.25A (from A to B) = 1.25 A (from B to A) E X A M P L E

4. 16

Find the voltage across the 5-W resistance in the network shown in Fig.4.21, using Thevenin’s theorem. 2W

4W

A

1W 5W

20 V 10 V B

12 V

20 V

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Basic Electrical Engineering

W resistance (Fig.4.22a Using node-voltage analysis, for node C we can write

Solution

Voc across AB.

Voc − 20 Voc + 10 Voc − 12 + + = 10 2 1 4

fi

Voc = 1.714 V

Thus, VTh = Voc = 1.714 V. 2W

2W

4W

C

A Voc

1W 20 V

A

RTh

4W

C

A

I 12 V

1W

5W

VTh

B

10 V

B

B (a)

(b)

(c)

Next, to determine RTh we turn OFF the sources (Fig. 4.22b), RTh = 2 W ||1 W || 4 W = 0.571 W The Thevenin’s equivalent circuit is shown in Fig. 4.22c. Therefore, using voltage divider, the voltage across 5-W resistance is 5 5 V AB = VTh ¥ = 1.714 ¥ = 1.54 V 5 + 0.571 5.571 EXA MP LE

4.17 I1

Draw Norton’s equivalent circuit across terminals AB in the circuit through a 2-W resistance when connected across the terminals AB.

20 V

10 W

5W

A

10 W

Solution We short circuit the terminals AB, and then determine the current Isc through this short (Fig. 4.24a). Due to the short circuit, no W resistance connected across the short. Hence, this resistance can be removed from the circuit (Fig. 4.24b). We W resistance and 5-W resistance are in parallel, and both are in parallel with 20-V source. Hence, the current through 10-W resistance is 20 Isc = =2A 10

B

Thus, IN = Isc = 2 A. RN, we turn OFF the voltage source (Fig.4.24c). The circuit reduces to that shown in Fig. 4.24d. Thus, RN = 10 W || 10 W = 5 W Norton’s equivalent is drawn in Fig. 4.24e, with IN = 2 A and RN = 5 W. The 2-W resistance is now connected across AB, and the current I through this resistance is found by using current divider, RN 5 I = IN ¥ =2¥ = 1.43 A RN + 2 5+2

10 W

I1

20 V

5W

10 W

10 W

I1

A

ISC

5W

20 V

B

(a) 10 W

10 W

A

5W

ISC

10 W

B

(b)

A

A

RN

B

(c)

A

A I

10 W

RN

IN

RN

B

B

(d)

EX A MP L E

2W

RN

IN

B (f)

(e)

4. 18 3W

Using Norton’s theorem, determine the power consumed by 2-W resistor RL in the network shown in Fig. 4.25.

2W

12 V

Solution First step is to remove the load resistor RL and

4A

x

5W

RL = 2 W

determine Norton’s equivalent across terminals xy. To determine IN, we short circuit terminals xy (Fig. 4.26a Isc

y

through 5-W resistance; and hence this resistance can be removed from the circuit (Fig. 4.26b). We now apply node-voltage analysis, to determine the voltage at node 1, 3W

12 V

2W

5W

4A

2W

12 V

Isc

x

IN

RN

RL = 2 W

RN

y (d)

Isc

x

IN

y (c)

x

y

(b)

x

5W

2W

1

4A

y

(a) 3W

3W

x

y (e)

124

Basic Electrical Engineering V1 − 12 V1 =4 + 3 2

fi

V1 = 9.6 V

\

IN = ISC

V1 9.6 = = 4.8 A 2 2

To determine RN, we turn OFF the sources (Fig. 4.26c), RN = (3 + 2) || 5 = 2.5 W Norton’s equivalent is shown in Fig. 4.26d. We now connect the load RL across xy, and get current I through this load using current divider, RN 2.5 I = IN ¥ = 4.8 ¥ = 2.67 A RN + RL 2.5 + 2 P = I2RL = (2.67)2 ¥ 2 = 14.257 W

\ EXA M PL E

4 . 19

Again consider the circuit shown in Fig. 4.8a solved in Example 4.5, and determine the voltage across the load resistance RL = 5 W, using Norton’s theorem.

Solution terminals a-b (Fig. 4.27a IN (Fig. 4.27b), combine the two current sources of 5 A and 1.5 A, and then transform the resulting current source of 6.5 A back to voltage source (Fig. 4.27c). The short circuit effectively removes the 20-W resistor from the circuit, so that 65 − 10 = 2.75 A 10 + 10 Norton’s resistance RN is same as Thevenin’s resistance RTh determined in Example 4.5 as 10 W. Hence, RN = 10 W. Norton’s equivalent with RL connected across a-b is shown in Fig. 4.27d, from which we get RN 10 I L = IN ¥ = 2.75 ¥ = 1.833 A RN + RL 10 + 5 IN =

VL = ILRL = 1.833 ¥ 5 = 9.17 V a 10 W

10 W 50 V

a 10 W

20 W

1.5 A

IN

5A

10 W

b

10 V

10 V

(a)

IN

b

(b) a

a +

10 W

10 W

20 W

1.5 A

20 W

65 V

IN

2.75 A

10 W

– b

b

10 V (c)

VL

(d)

RL = 5 W

125 EX A MP L E

4.2 0

A voltmeter is connected across the 500-kW resistor in the circuit shown in Fig. 4.28a. Determine the reading of the voltmeter, if (a) it is assumed to be ideal, (b) it has an input resistance of 10 MW. 308 kW

800 kW

+ 10 V

500 kW

+ V VM –

10 MW (Voltmeter)

V¢

3.85 V

–

Voltmeter (a)

(b)

Solution load the circuit at all. The reading of the

(a voltmeter would be

500 kΩ = 3.85 V 500 kΩ + 800 kΩ (b) If the voltmeter has its own input resistance, it would load the circuit and would change the voltage to be measured. V = (10 V) ¥

Thevenin’s equivalent across the terminal where we are going to connect the voltmeter. Thevenin’s voltage is the open circuit voltage, that is, VTh = 3.85 V, and Thevenin’s resistance is RTh = 800 kW || 500 kW = 308 kW Thevenin’s equivalent is drawn in Fig. 4.28b, from which the voltmeter reading is given as 10 MΩ V ¢ = (3.85 V) ¥ = 3.85 ¥ 0.97 = 3.73 V 10 MΩ + 308 kΩ The loaded voltage, which is what the voltmeter will indicate, proves to be 3 % lower than the unloaded voltage of 3.85 V, which is what we are seeking to measure with the voltmeter. Note that, in practice, we can correct for loading error if we know the output resistance of the circuit and the input resistance of the voltmeter. EXA MP LE

4.21

For the circuit shown in Fig. 4.29a, determine the following : (a) Replace the circuit in the dotted box by Thevenin’s equivalent circuit. 18 kW

18 kW

a

a +

5 mA

12 kW

RL

6 kW

5 mA

12 kW

6 kW

Voc = VTh –

40 V (a)

b

40 V (b)

b

126

Basic Electrical Engineering

(b) Find Vab for RL = 3 kW. (c) What value of RL receives maximum power from the circuit ? (d) What value of RL makes the current in the 6-kW resistor to be 0.1 mA ?

Solution (a) First, we remove RL Voc = VTh across ab (Fig. 4.29b). We transform the 5-mA current source to voltage source (Fig. 4.30a). From the single-loop circuit, using voltage divider, we get 6 kΩ VTh = Voc = (60 – 40) ¥ = 3.33 V 6 kΩ + (12 + 18) kΩ To determine the Thevenin’s resistance, we turn OFF all sources (Fig. 4.30b) in the given circuit, and get RTh = (6 kW) || (12 kW + 18 kW) = 5 kW The Thevenin’s equivalent circuit is shown in Fig. 4.30c. 12 kW

18 kW

18 kW

a

5 kW

a

+ 60 V

6 kW

40 V

a

RTh

VTh

12 kW

6 kW

VTh

RTh

3.33 V

– b

b

(a)

b

(b)

(c)

(b) Connecting RL = 3 kW across ab, voltage vab is obtained by using voltage divider as RL 3 Vab = VTh ¥ = 3.33 ¥ = 1.25 V RL + RTh 3+5 (c) According to maximum power transfer theorem, the required value of RL = RTh = 5 kW. 12 kW

18 kW

30 kW

a

a

0.1 mA 60 V

RL

6 kW

+ 6 kW Vab

20 V

IL RL

40 V – b (a)

b (b)

(d) We use the equivalent circuit of Fig. 4.30a, and connect load resistance RL (Fig. 4.31a shown in Fig. 4.31b. Since the current through 6-kW resistor is required to be 0.1 mA, the voltage Vab = IR = (0.1 mA) ¥ (6 kW) = 0.6 V V30 k 19.4 V \ V30kW = 20 – 0.6 = 19.4 V and I30kW = = = 0.647 mA 30 k 30 k V 0.6 V = 1.1 kW \ Applying KCL: IL = 0.647 –0.1 = 0.547 mA fi RL = ab = IL 0.547 mA

127 EX A MP L E

4.2 2

For the circuit shown in Fig. 4.32a, determine the following : (a) Find the value of R to receive maximum power from the circuit. (b) For the value of R in part (a

Solution (a) To receive maximum power, the value of resistance R should be equal to the output resistance of the remaining circuit, which is the same as the Thevenin’s resistance. We remove R, turn OFF all the sources (Fig. 4.32b), and then determine RTh, RTh = Rab = 2W || 6W = 1.5 W Thus, when R = RTh = 1.5 W, it would receive maximum power. (b R = 1.5 W, let us mark the two mesh currents, I1 and I2, as shown in Fig. 4.32c, and write the mesh equations by inspection, in matrix form, ⎡ 8 − 6 ⎤ ⎡ I1 ⎤ È4˘ fi I1 = 3.25 A ⎢ − 6 7.5 ⎥ ⎢ I ⎥ = Í8 ˙ ⎣ ⎦ ⎣ 2⎦ Î ˚ Therefore, the power supplied from the 12-V source is P = VI1 = 12 ¥ 3.25 = 39 W E XA MP LE

4 . 23

For the circuit shown in Fig. 4.33a

Is such that the current in the 120-W resistor is zero.

Solution I1 through 120-W resistor due to 12-V source working alone (Fig. 4.33b), and then current I2 due to current source Is working alone (Fig. 4.33c), 12 80 I1 = = 0.06 A and I2 = Is ¥ = 0.4Is 80 + 120 80 + 120

128

Basic Electrical Engineering

Since the net current through 120-W resistor is required to be zero, hence we must have 0.06 I1 + I2 = 0 or 0.06 A + 0.4Is = 0 fi Is = = –0.15 A 0.4 EXA MP LE

4.24

Obtain the Thevenin’s equivalent of the circuit shown in Fig. 4.34a.

Solution The circuit contains no independent source. So, the current I in 12-W resistor will be zero, and hence the dependent voltage source 8I will also be zero. Obviously, Thevenin’s voltage VTh = 0 V. To determine RTh, we apply a known voltage source (say, 12 V) across terminals ab (Fig. 4.34b Is supplied to the circuit. 12 V I = =1A fi Voltage of the dependent source = 8I = 8 V 12 Applying KCL to node a, 12 12 12 − 8 12 V Is = = 4 A; \ RTh = s = =3W + + Is 12 6 4 4

SUMMARY TE RM S

A N D

C ON CE PT S

Most circuit problems can be solved by applying Kirchhoff’s Laws to produce simultaneous equations; the Superposition theorem states that the response (current or voltage) at any point in a linear network due to multiple sources can be determined by summing the effects of each source considered separately, while all other sources being turned OFF. Thevenin’s theorem states that it is possible to simplify any linear circuit containing independent and dependent , voltage and current sources, no matter how complex, to an equivalent circuit with just a single voltage source and a series resistance connected to a load. output impedance’ of a circuit. Norton’s theorem states that a two terminal linear network containing independent and dependent, voltage and current sources can be replaced by an equivalent current source IN in parallel with a resistance RN. Maximum power transfer theorem states that maximum power is absorbed from a network when the load resistance is equal to the output resistance of the network as seen from the terminals of the load. impedance match has been done. Millman’s theorem helps us in converting a number of practical parallel voltage sources into a single practical voltage source.

129 Reciprocity theorem states that in a linear bilateral network, if a voltage source V in a branch A produces a current I in any other branch B, then the same voltage source V acting in the branch B would produce the same current I in branch A Tellegen’s theorem states that the sum of instantaneous power delivered to each branch of a network is zero. I MP O RTA N T

FO R MU L AE

VTh = INRN

and RTh = RN V V output resistance can be computed by Ro = oc = Th Isc IN RL = RTh Pmax = b

Tellegen’s theorem :

ÂV I

k k

2 VTh 4RTh

=0

k =1

CHECK YOUR UNDERSTANDING Before you proceed to the next chapter, take this test. Give yourself two marks for each correct answer and minus one for each wrong answer. If your score is 12 or more, go to the next chapter; otherwise study this chapter again. S. No. 1.

3. 4.

6.

7. 8.

9.

10.

Statement

True

False

The superposition theorem can be applied to any network, howsoever complex, containing any number of linear and nonlinear elements. linear. By ‘turning OFF’ a current source means ‘open-circuiting’ its terminals. For a given network, the value of Thevenin’s resistance RTh is same as that of Norton’s resistance RN. source is 100 %. Thevenin’s theorem is especially useful in such applications as determining the load for an electronic circuit which will result in maximum average power delivery to the load. Thevenin’s and Norton’s theorems are valid for a load containing any kind of elements, linear or nonlinear, time-varying or time-invariant. Reciprocity theorem means that the interchange of an ideal voltage source in one branch and an ideal ammeter in another branch will not change the reading of the ammeter. When the resistance of a voltmeter connected across two terminals of a circuit has the same value as the Thevenin’s resistance, it gives the correct reading. It is possible to directly derive Norton’s equivalent from the Thevenin’s equivalent of a circuit. Your Score

Marks

Basic Electrical Engineering A N SW E RS

1. False 6. True

2. True 7. True

3. True 8. True

4. True 9. False

5. False 10. True

REVIEW QUESTIONS 1. State superposition theorem and illustrate its use to solve a network. 2. Explain why superposition theorem is not applicable to a network containing nonlinear elements. 3. State and explain Thevenin’s theorem, by means of a simple example.

4. What are the advantages of Thevenin’s theorem ? 5. How will you convert Thevenin’s equivalent circuit into Norton’s equivalent circuit and vice-versa ? 6. State and explain the importance of the maximum power transfer theorem. 7. State Tellegen’s theorem. Give some of the implications of this theorem.

MULTIPLE CHOICE QUESTIONS Here are some incomplete statements. Four alternatives are provided below each. Tick the alternative that completes the statement correctly . 1. Thevenin’s theorem is based on the idea of (a) an equivalent current source (b) an equivalent source of emf (c) an equivalent power source (d) an equivalent resistance 2. Norton’s theorem is based on the idea of (a) an equivalent current source (b) an equivalent source of emf (c) an equivalent power source (d) an equivalent resistance 3. In a circuit, maximum power is transferred to the load when (a) the total circuit resistance is equal to the load resistance (b) the internal resistance of the source of emf is (c) the total series resistance is equal to the equivalent resistance of the parallel combination of resistances (d) the internal resistance of the source is equal to the load resistance 4. The superposition theorem is used when a circuit contains

5.

6.

7.

8.

(a) a single voltage source (b) a single current source (c) a number of sources (d) passive elements only When a circuit has a large number of branches and a large number of unknown currents, the circuit can easily be solved by applying the (a) Kirchhoff’s voltage law (b) Kirchhoff’s current law (c) reciprocity theorem (d) loop-current method The superposition theorem is essentially based on the concept of (a) duality (b) reciprocity (c) linearity (d) nonlinearity While Thevenizing a circuit between two terminals, VTh equals to (a) the short-circuit terminal voltage (b) the open-circuit terminal voltage (c) the emf of the battery nearest to the terminals (d) the net voltage available in the circuit While calculating Thevenin’s resistance RTh, all the current sources in the circuit are (a) treated in parallel with other voltage sources (b) converted into equivalent voltage sources (c) replaced by open circuits (d) replaced by short circuits

A N SW E RS

1. b

2. a

3. d

4. c

5. d

6. c

7. b

8. c

PROBLEMS ( A )

S I M P L E

PR O BL EMS shown, using the principle of superposition. [Ans. 0.0455 A] 7. Find the current I in 10-W resistor in the circuit shown in Fig. 4.41, using Thevenin’s theorem. [Ans. –2 A] 8. Find Thevenin’s equivalent at terminals ab of the network shown in Fig. 4.42. [Ans. 100 V, 4 W] 9. across the resistance R in the circuit shown in Fig. 4.43. [Ans. 12.24 V] 10. Solve the above problem, by applying Norton’s theorem. [Ans. 12.24 V] 11. Find the current in RL in the circuit of Fig. 4.44 by using Thevenin’s theorem, and then verify your results by using loop-current method. [Ans. 0.625 A] 12. Solve Prob. 2, by using Thevenin’s theorem. [Ans. – 0.5 A]

1. Determine the current Ix through the 10-W resistor in Fig. 4.35, using superposition theorem. [Ans. – 1 A] 2. Determine the current through R = 3 W resistor (from A to B) in the circuit shown in Fig. 4.36, using superposition theorem. [Ans. – 0.5 A] 3. Determine the current through 4-W resistor (from A to B) in the circuit shown in Fig. 4.37, using superposition theorem. [Ans. 0.516 A] 4. Find the current I through 4-W resistor in the circuit shown in Fig. 4.38, using superposition theorem. [Ans. 2.916 A] 5. Find the current I through 20-W resistor in the circuit shown in Fig. 4.39, using superposition theorem. [Ans. 0.714 A] 6. For the circuit shown in Fig. 4.40, determine the current I in 50-W resistor with the reference direction Ix

10 W A

50 V + –

30 W

5A

2W

20 W

R=3W

1W

2A 20 W

2W

2V

12 W

1W

A

4W

3W

4V

B

1W

A

2W

4W

15 V

B

2W

B

5A

2W

4V

Basic Electrical Engineering 13. Solve Prob. 3, by using Thevenin’s theorem. [Ans. 0.516 A]

14. Find the current in the 3-W resistor of the circuit shown in Fig. 4.45, using Thevenin’s theorem. [Ans. 5.143 A]

150 W

100 W

10 W

A

B

3W 5V

50 W

0.1 A

6W

10 V

15 W

4A

9V

I

RL = 2 W

a 5W

30 V

10 W

3W

2W

R = 10 W

b 2W

2W

10 V

2W 10 V

A

2W C 12 W

3W 1A

–

+ –

3W

T R IC KY

1W

1V

36 V D

( B)

2W

+

6W

6A

B 2W

PR OB LE MS

15. In the circuit shown in Fig. 4.46, determine the current through 1-W resistor connected across AB, using superposition theorem. Verify the result using Thevenin’s theorem. [Ans. 406.25 mA] 16. Find Thevenin’s equivalent circuit for a dc power supply that has a 30-V terminal voltage when delivering 400 mA and a 27-V terminal voltage when delivering 600 mA. [Ans. 36 V, 15 W]

17. Determine the current I through the 3-W resistor in the circuit of Fig. 4.47, by using Thevenin’s theorem. [Ans. 3 A] 18. Determine the current through 6-W resistor connected across AB in the circuit of Fig. 4.48, by using Thevenin’s theorem. [Ans. 1 A] 19. In the circuit shown in Fig. 4.49, determine (a) the value of R so that the load of 20 W draws maximum

A 2W

C

1W

6W

A

4W 3W

I + –

10 V

2W

6W

6V

3W

10 A

15 V B

B

current in part (a) ?

power, and (b) the value of the maximum power drawn by the load. [Ans. (a) 30 W, (b) 180 W] 20. Using Thevenin’s theorem, calculate the current W resistor connected between terminals A and B of the circuit shown in Fig. 4.50. [Ans. 10 mA] 21. A Wheatstone bridge is shown in Fig. 4.51. The galvanometer connected across BD has a resistance of 20 W. Using Thevenin’s theorem, compute the current through this galvanometer. [Ans. 5.14 mA] 22. Refer to the circuit shown in Fig. 4.52. (a) Find the current in 3-W resistor. (b) What resistance, replacing the 3-W resistor, would draw one-half the

[Ans. (a) 0.667 A; (b) 8.4 W] 23. of V to make the current in the 5-W resistor to be zero. [Ans. 30 V] 24. A circuit, as shown in Fig.4.54a, has a variable load and ideal meters to monitor load voltage and current. The table in Fig. 4.54b shows partial results of a series of tests. Fill in the blanks in the table with the missing data. [Ans. 0 V, 0.0333 A, 12.85 V, 0 A] 25. Find the current in the 9-W resistor of the circuit shown in Fig. 4.55, using Thevenin’s theorem. [Ans. 2 A] B

30 W

A

A B

R 180 V

60 W

20 W (Load)

20 W

10 W

100 W

0.5 V

60 W

C

50 W 5W

4W D

5V 10 V

4W

6V

3W

2W

5W

4W

1A

10 W 20 V

V

Basic Electrical Engineering

I

RL

0.15 A

0

V

A V

RL

6Ix Ix

300 W

10 V

4W

6W

9W

20 V (a)

( C )

+–

(b)

C HA L L E NGI NG

PROBLEMS

26. Develop a Thevenin’s equivalent circuit for the part of the circuit shown in the box in Fig. 4.56. Use this I, as shown. [Ans. – 0.0625 A] 27. A student is testing a circuit containing batteries and resistors. The output voltage is 6.26 V when measured with a good (assumed ideal) voltmeter, but 6.05 V when a 600-W resistor is connected across the output terminals. What current would result if the output terminals were short-circuited ? [Ans. 0.301 A]

6W I 1A

2W

4W

8W

EXPERIMENTAL EXERCISE 4.1 S U PE RP O S IT ION

T H EOREM

Objectives To verify superposition theorem. Apparatus Two DC power supplies 12 V and 6 V, Three ammeters (MC type) 0-2 A, Three rheostats 100 W, 5 A. Circuit Diagram The circuit diagram is shown in Fig. 4.57.

Brief Theory The theorem of superposition states that in a linear network containing several sources (including dependent sources) the response (i.e., the current or voltage) at a point in the network equals the sum of the responses of each source working alone with all other sources made inoperative.

Procedure 1. 2. 3. 4.

Make connections as given in Fig. 4.57, keeping the rheostats at their maximum value. Set the three rheostats such that the readings in the three ammeters are within their range. Note the readings of the three ammeters. Without changing the settings of rheostats, disconnect the 12-V source and short circuit the point A to point E. Note the readings of the three ammeters. 5. Next, replace the 12-V source at its place. Disconnect the 6-V source and short circuit the point C to F. Again, note the readings of the three ammeters. 6. Repeat steps 2 to 5 for four different settings of the rheostats. 7. Disconnect the circuit.

Observations and Calculations With both sources S. No. A1

A2

A3

With 6-V source (A) A1 A2 A3

With 12-V source (B) A1 A2 A3

A1

A+B A2 A3

1. 2. 3. 4. 5.

Calculations Add algebraically the readings of the corresponding ammeters in step 4 (i.e., column A) and in step 5 (i.e., column B), and put the sum in the last column (A + B). Results 1. The sums recorded as A + B in the last column is found almost the same as the readings of the three ammeters 2.

Precautions 1. Before connecting the dc supply, the zero reading of the ammeters should be checked. 2. The terminals of the rheostats should be connected properly. 3. While setting the rheostats, care should be taken that the currents recorded by the ammeters do not exceed 5 A, the current rating of the rheostats. 4.

Viva Voce 1. Under what conditions is the superposition theorem applicable ? Ans. : The elements of the network should be linear and bilateral. 2. What do you mean by the term ‘linear’ ? Ans. : It means that the V-I characteristic of the element is a straight line. 3. Q. : What do you mean by ‘bilateral’ ? Ans. : An element which behaves in exactly same manner if connected in the reverse order is said to be bilateral.

Basic Electrical Engineering 4. Q. : Can the superposition theorem be applied to determine ‘power’ in an element of a circuit ? Ans. : determine the current or the voltage using superposition theorem and then calculate the power

EXPERIMENTAL EXERCISE 4.2 TH E VE N I N ’ S

TH E OR EM

Objectives To verify Thevenin’s theorem. Apparatus 220-V dc variable supply, One variac (0-270 V) 15 A, One ammeter (MC type) 0-5 A, One voltmeter (MC Type) 0-250 V, Four rheostats 100 W, 5 A.

Circuit Diagram The circuit diagrams are shown in Fig. 4.58.

Brief Theory Sometimes, our object of analysing an electrical circuit is to determine the response (i.e., the current, or the voltage, or the power consumed) in a single resistance. In such a case, this resistance can be treated as a load resistance. The remainder of the network can be replaced by a simple equivalent circuit containing a voltage source in series with a resistance, using Thevenin’s theorem. Thevenin’s theorem specially becomes very useful and timefollows: 1. Any two terminals AB of a network consisting of linear passive and active elements can be replaced by a simple equivalent circuit containing a voltage source VTh in series with a resistance RTh. 2. The Thevenin’s equivalent voltage source VTh is equal to the open-circuit voltage between the terminals AB (i.e., under the condition when no external resistance is connected across AB) caused by the active elements of the network.

3. The Thevenin’s equivalent resistance RTh is the equivalent resistance looking back into the network across the terminals AB, when all its sources are made inactive. According to Thevenin’s theorem, the actual current IL through a load resistance RL connected across the terminals AB in the original network is given as VTh IL = (i) RL + RTh

Procedure 1. Make connections as given in Fig. 4.58a, keeping the rheostats at their maximum value. Keep the switch S closed and switch on the supply. Set the four rheostats such that the reading in the ammeter is within its range. 2. Note the readings I2 and V2 of the ammeter and voltmeter, respectively. The actual load current is same as I2 (i.e., ILa = I2). The load resistance RL is given as RL =

V2 I2

(ii)

3. Open the switch S, and again take the reading V3 of the voltmeter. This is the open-circuit voltage between terminals A-B and gives the Thevenin’s equivalent voltage VTh. 4. Without changing the settings of rheostats, connect the circuit as shown in Fig. 4.58b. Note the readings I4 and V4 of the ammeter and voltmeter, respectively. Thevenin’s equivalent resistance RTh is given as V RTh = 4 (iii) I4 5. Calculate the current through the load, using Thevenin’s theorem, as VTh ILc = (iv) RL + RTh 6. For another setting of the rheostats, repeat Steps 2 to 5. 7. Disconnect the circuit.

Observations and Calculations Step 2 S. NO.

V2

l2 (= lLa)

Step 3 RL Eq. (ii)

V3 (= VTh)

Step 4 V4

l4

RTh Eq. (iii)

ILc Eq. (iv)

1. 2.

Results 1. From the Table, it is observed that the actual value of load current ILa in column 2 (under Step 2) tallies with the calculated value of the load current ILc in the last column. 2.

Precautions 1. Before connecting the dc supply, the zero readings of the ammeter and voltmeter should be checked. 2. The terminals of the rheostats should be connected properly. 3. While setting the rheostats, care should be taken that the currents recorded by the ammeters do not exceed 5 A, the current rating of the rheostats.

Basic Electrical Engineering

Viva Voce 1. Ans. : resistance measured between terminals AB is then the Thevenin’s equivalent resistance. 2. What do you mean by making the sources inactive ? Ans. : It means that sources are replaced by their internal resistances. The voltage source is short-circuited and the current source is open-circuited. 3. Do you know another network theorem that serves the same purpose as Thevenin’s theorem ? Ans. : Yes, it is called Norton’s theorem. 4. What is the difference between the two ? Ans. : equivalent current source.

EXPERIMENTAL EXERCISE 4.3 MAX I MU M

PO W ER

TRAN SFER

TH EO RE M

Objectives To verify maximum power transfer theorem. Apparatus 12-V supply, One ammeter (MC type) 0-2 A, Two voltmeters (MC Type) 0-15 V, Two rheostats 100 W, 5 A.

Circuit Diagram The circuit diagram is shown in Fig. 4.59.

Brief Theory The maximum power transfer theorem states that the power transferred from the source to the load is maximum when the load resistance is equal to the source resistance, i.e., RL = RS

Procedure 1. 2. 3. 4.

Connect the circuit as shown in Fig. 4.59. Keep the setting of the rheostat RL at its maximum position. Adjust rheostat RS such that the reading of the ammeter is around the middle of its scale. Note the readings IL and VS of the ammeter and the voltmeter across the rheostat RS, respectively. Calculate the value of RS, as V (i) RS = S IL

5. Reduce the resistance of the rheostat RL in steps and for each setting note the readings of IL and VL of the ammeter and the voltmeter across the rheostat RL, respectively. Calculate the value of RL, as V (ii) RL = L IL 6. Also, for each setting of the rheostat RL calculate the power consumed by the load, as P = VLIL 7. For another different setting of the rheostat RS, repeat steps 4 to 6. 8. Draw the graphs for the two values of RS.

(iii)

Observations and Calculations (a) For RS1 = S. No.

VL

IL

Eq. (i) RL Eq. (ii)

PL Eq. (iii)

VL

(b) For RS2 = Eq. (i) RL IL Eq. (ii)

PL Eq. (iii)

1. 2. 3. 4. 5.

Graph The graphs for the two values of RS is drawn in Fig. 4.60. Results 1. From the graphs given in Fig. 4.60, it is observed that as the value of RL is increased the power P increases, becomes maximum, and then decreases. 2. Also, it is seen that for lower value of RS the maximum power is greater. 3. The value of RL for which the power becomes maximum is almost same as that of corresponding RS. 4.

Precautions 1. Before connecting the dc supply, the zero readings of the ammeter and voltmeters should be checked. 2. The terminals of the rheostats should be connected properly. 3. While setting the rheostats, care should be taken that the currents recorded by the ammeters do not exceed 2 A, the current rating of the rheostats.

Viva Voce 1. Can you give some reason why the value of RL for which maximum power transfer takes place is a little greater than that of resistance RS ? Ans. : The dc supply itself has some internal resistance. Therefore, the actual resistance of our source is slightly more than the resistance of the rheostat RS. 2. From a source of open-circuit voltage 10 V and of resistance 5 W, how much maximum power can be drawn ? Ans. : For drawing maximum power, the condition is that RL = RS. Thus, E ⎞2 10 2 E2 ⎛ =5W Pmax = I 2L RL = ⎜ ⎟ R L = 4R = 4 5 ⎝ RS + RL ⎠ S

140

Basic Electrical Engineering

3. Is the maximum power transfer theorem applicable to ac circuits ? Ans. : Yes, but for ac circuits the condition for maximum power transfer is that the load impedance should be equal to the complex conjugate of the source impedance. 4. Is it always possible to operate a circuit at maximum power transfer conditions ? Ans. : No. For example, the domestic and industrial power supplies have very low source resistance so as to provide good voltage regulation. If the load resistance is also made so low, undesirably high currents would 5. Then, where do you make use of this theorem ? Ans. : This theorem is utilised in electronic communication circuits, where the signal power is quite low. In such cases, the load is properly matched with the source so that maximum signal power is transferred to the load.

5

ELECTROMAGNETISM OB JE CT IV ES

(a) (

( )

(c)

( )

)

For several centuries, electric and magnetic forces were thought to be completely different, with nothing in common. It is now realised that these forces are very closely related to each other. These forces are as basic as the gravitational force, pervading all matter, and determining their properties. In a rough sense, electricity and magnetism give rise to each other. In 1820, Hans Christian Oersted, a

the consequences of this mutual relation is the electromagnetic wave, in the form of radio and light waves.

a shows a straight

the right-hand thumb rule. It says, “Stretch the thumb of your right hand along the current, the curl (natural

142 In some situations, the conductor is placed perpendicular to the paper. The conductor is represented by a small circle and the direction of current is then shown by putting a dot (∑) or a cross (¥). As a convention*, (∑ b) and (¥) represents a current entering the paper c b c

right-hand thumb rule a

b,

a it is North Pole.

South P

(a) Single loop (Clockwise current).

b,

(b) Single loop (Anticlockwise current).

-

way of remembering the convention is to imagine an arrow pointing in the direction of the vector. If the vector points out of the paper, you see the head of the arrow, namely, the ∑ (dot). If the vector points into the paper, you see the tail of the arrow, namely, the ¥ (cross).

Electromagnetism

143

B and is measured in tesla (T) after the great Yugoslav inventor

Magnetic Field due to a Long Straight Wire

x from a long

straight wire carrying current I is given as B =

m0 I 2px

where, m 0

p ¥ 10

Tm/A.

Magnetic Field due to a Circular Loop

r,

carrying current I is given as B=

m0 I 2r

Magnetic Field due to a Solenoid

a shows a solenoid material so that the adjacent turns are electrically insulated from each other. Generally, the length of the

n turns per unit length and carrying current I is given as B = m 0 nI Bend = EX A MP L E

m0 nI 2

5.1

inside the solenoid ?

Solution

B = m0 nI, where n is the number of turns per metre. Therefore, the

required current is I= EX A MP L E

B 20 ¥ 10- 3 = = 8.0 A m0n 4p ¥ 10-7 ¥ (20 ¥ 102)

5.2

a b

c

Solution B = m0 nI. Note that the radius of the solenoid does not enter in this equation. Therefore, a, the two ends of these letters point in the direction of current in the coil.

144 to get n, simply multiply the number of turns per layer by the number of layers and divide the product by the length of the solenoid, 350 4 n = = 2800 m 0.50 –2 \ B = m0 nI p ¥ 10 ¥ 2800 ¥ 2.1 ¥ 10 T (b m nI 2.1 × 10 − 2 –2 B = 0 = = 1.05 ¥ 10 T 2 2 (c negligible,

r P I (a) A solenoid.

I

(b) A toroid.

Magnetic Field due to a Toroid If a solenoid is bent in a circular shape and the ends are joined, we get a toroid b. Alternatively, one can start with a nonconducting ring and wind a conducting wire on it. If r is the radius of the toroid, and it has a total number of n the toroid at any point (such as P) due to current I is given by B=

m0 nI 2pr

B, a magnetic material such as iron is introduced inside the cylindrical or toroidal region enclosed by the conducting windings.

a shows a small conductor of length dl

B. If it carries a

current I dF = I dl ¥ B q E, given as F = qE. We notice that the force is proportional to I, dl and B, and is perpendicular to both dl and B. a, the angle between the length vector dl B dF = IdlB For a conductor of length l, carrying current I force on the conductor is

B, the F = IBl

Electromagnetism

145

If the current-carrying conductor is placed at an angle q c), its effective length is l sin q F = IBl sin q B, the angle q = 0, and the force on the conductor too reduces to zero.

b. The

a b B

c F P

(a) Uniform magnetic field.

(b) Field due to current.

(c) Resultant magnetic field.

146

is zero. c on the conductor.

a and breadth b, carrying a steady current I. Suppose that it B q with B

a

F2 = IBb The conductor

ON

F2

q

q) = IBb cos q

ON

ON

are of same

forces.

F1 F1 = I Ba. In case the coil has n turns of wire, the force becomes F1 = IBan Note they do not act along the same line. The perpendicular distance between these forces can be found from the b, as x = b sin q t = F1x = (IBan) ¥ (bsin q) = BInab sin q = BInA sin q where, A = ab made radial by introducing a core and suitably shaping the magnetic poles. This way, the angle q for any orientation of the coil. In such case, the torque on the coil becomes t = BInA

147

Electromagnetism

It is important to note that the torque on a coil depends, besides other things, on its area and not on its shape. It means that a coil having a circular shape or any other irregular shape, but the same area as A, will

Consider two parallel conductors carrying currents I1 and I2, separated by a distance r

a.

I1 I 2 r The force is found to be attractive when the direction of currents in the two conductors is the same and Fμ

b c I1

I2 P

Q

I

I

I

I

r Field due to P

P

F

Attractive forces

Repulsive forces

(b)

(c)

Top view of conductor Q

(a)

a) produces B= B

m0 I1 2pr BI2

148

F = BI2 =

m0 I1 I 2 2pr

Unit of Current The SI unit of current is based on 1 ampere each, and the separation between the two conductors is 1 m, the force will be μ0 m II F= 0 1 2 = = 2 ¥ 10 N 2π 2pr Thus, one ampere of current is 2 ¥ 10 EX A MP L E

. 5.3

the force between them and specify its nature.

m0 I1I 2 4p ¥ 10- 7 ¥ 80 ¥ 30 = = 2.4 ¥ 10 – 4 N m –1 2p ¥ 2 2pr Since the two currents are in the same direction, the force will be attractive.

Solution F =

EX A MP L E

5.4

Two straight wires A and B B near its centre.

Solution B, m0 I1I 2 4p ¥ 10- 7 ¥ 4 ¥ 6 F= = 2p ¥ 0.03 2pr B is Fnet = F ¥ l

¥ 10

¥

Since the currents are in opposite directions, the force Fnet is away from it.

¥ 10

¥ 10

Nm

= 24 mN

. It means that its direction is normal to wire A

Faraday observed that when a bar magnet is suddenly brought close to a coil, the galvanometer G showed a b motion. c. One circuit, which we shall call the

circuit, consists of a battery B, a coil P

Electromagnetism N

S

N

0

S

0 G

G

(a) Deflection when the magnet is moved closer.

(b) Deflection when the magnet is moved away.

i B K

149

P

i

0

S

B G

P

K

(c) Deflection when the key is closed .

0

S

G

(d) Deflection when the key is opened.

circuit, consists of a coil S, and a galvanometer G.

d short time, roughly the time during which the current in the primary circuit is changing.

changed. circuit. Stated mathematically, e=

F 2 - F1 Dt

or

e=

dF dt

If a coil has N turns, emf is induced in each turn. The total emf induced in the coil then becomes dF e= N dt Later, Lenz described the direction or the sense of the induced emf. (

) the the induced current in the loop

a, the N-pole of the bar magnet is being pushed towards the coil. The induced emf and hence the current through the coil must be such that the right-

150

R. 2

R

e = -N

dF dt

There are three requirements to producing an induced emf in a conductor, namely, (i) the presence of a conductor, (ii iii the conductor. Following are three methods by which this can be achieved. (1 (2) By using a stationary conductor or a moving permanent magnet (or an electromagnet). As we shall see a, fed by a

(3 generators and motors.

statically induced emf motional emf or dynamically induced emf.

Consider a closed loop of wire of area A placed in a magnetic field. If the field is uniform (constant in space but not necessarily with time) and perpendicular to the surface of the a F (measured in webers) passing through the loop is given as F = BA. However, if the normal to the plane q with the magnetic b), the effective area of the loop reduces to Acos q the loop becomes

A

A

B

B q

(a)

(b)

F = BA cos q

(i) (ii) (iii)

A cos q

, , and (i.e.,

q).

Electromagnetism

151

(i) Induced emf by Changing B

a and b. Change in B through the coil is accomplished by relative motion of the loop and the magnet. For this, it does not matter whether the magnet moves or the coil moves or both move. When the magnet and the coil come closer, B increases and an induced emf is produced.

(ii) Induced emf by Changing A

B

a). Let l of a velocity perpendicular in the conductor is given as

b. It slides over U-shaped conducting rails situated in the magnetic x (from position X to position Y) in time t, its average linear v = x/t, and the change in area of the coil is clearly lx. The induced emf e= -

Blx dF ⎛ x⎞ = = –Bl = – Blv ⎝t⎠ t dt Magnetic field

N B b X

x a

l

B

Conductor

a X Y

Y

q

l b

x

S

Direction of motion of conductor (a)

(b)

(c)

In case the conductor moves at an angle q velocity v

c), the component of its v sin q. The induced emf then becomes e = –Blv sin q motional

emf.

(iii) Induced emf by Changing q B

w a q b t from the instant when the coil is perpendicular to

q = 0 at t = 0, then at any instant, F = BA cos q = BAcos wt e= -

d ( BA cos w t ) dF = = BAw sin wt dt dt

If the coil has N turns, the net induced emf is given as e

N

dF = NBAw sin wt = Em sin wt dt

152 N M, N M B 1 2

P

B

q = wt

O

O, P

(a)

(b)

or a generator. It is important to note that the induced emf in a coil depends, besides other things, on its area and not on its shape. It means that the induced emf in a coil having a circular shape or any other irregular shape, but the same area as A

Although the direction of motional emf could be venient to use , as illus-

in mutually perpendicular directions. Now, adjust -

IMP O R TA N T

Motion

Motion Field

d ce du rent n I ur c

O

N

S

Field

ed uc nt d In urre c

P

N O T E

the thumb represent the same quantities. It may therefore be helpful to associate F Field F C C thuMb with Motion of the conductor Often, students get confused which rule to apply where. In electrical engineering, we come across two types of situations. One may be called generator action, and the other motor action. In generator action, the induced emf action, the motion of a conductor is the result when a current is passed through the conductor placed in a magnetic You can easily remove the confusion by noting that it is your right hand that (usually) can be associated with the generator action of the right hand. applies to the motor action.

most of the

Electromagnetism

A charge q moving with a velocity v given by

B

153

Lorentz force,

F = q(v ¥ B) dragged in with velocity v a. Free electrons present in the rod also move with velocity v B and v. According to right-hand screw rule (for determining the direction of cross product of two vectors), the force F on a positive b Fe downward from Q to P. Because of this force, the electrons accumulate at the end P, providing it negative polarity. The other end Q deprived of electrons, becomes positively charged. F +Q+ +

B

+

B v

v

– – –P– Fe (a)

(b)

a force on the electrons in a direction opposite to the Lorentz force. When enough charges accumulate, this force cancels the Lorentz force and the free electrons stop drifting. If E accumulation of charges, the electrostatic force on an electron is eE (e is the charge on an electron). From evB. Thus, we must have eE = evB or E = vB l of the rod PQ. Hence, the electric potential difference between P and Q is VQP = l ¥ E = l ¥ vB = Blv laws.

B

154 emfs, being equal in magnitude and opposite in polarity while going round the loop, cancel each

B

v Q

When the loop is partly inside and partly out-

R

F P

Hence, an emf is induced. It is only the conductor PQ of length l where the motional emf Blv is induced. If R is the resistance of the loop, the current in the loop due to this emf is I = Blv/R. As the loop is pulled away (as shown in

S

I 2R loss in the loop. Thus, the power supplied in pulling the loop against the force F is P = I 2R =

(Bl v) 2 R2

¥R=

B 2l 2 v 2 R

Since, power = force ¥ velocity, the force is given as F= EX A MP L E

P v

=

B 2l 2 v R

5.5

(a (b (c

Solution The magnitude of the induced emf is given as e = Blv sin q. (a) Since q = 0, the induced emf is e = Blv sin q = Blv ¥ 0 = 0 V (b) Here, q e = Blv sin q ¥ 0.2 ¥ ¥ 1 = 0.5 V (c) Here, q e = Blv sin q = ¥ 0.2 ¥ ¥ 0.25 V EX A MP L E

5.6

¥ 10 1100 1000 3600

Solution The speed of the aeroplane is v Hence, the induced emf is e = Blv sin q EX A MP L E

¥ 10

¥

¥

¥ 1 = 0.604 V

5.7

A conducting rod of 1 m length is placed with its one end at the centre and the other end at the circumference of a

Electromagnetism

155

B = 1.0 Wb/m2, parallel to the

Solution q is the angle between the rod and the radius OP at time t, the area of the arc formed by the rod and the radius OP is A = pR 2 ¥ q = 2p

1 R2q 2

where R is the radius of the circle. Hence, the induced emf is d

(

1 R 2q

2 d (BA) dF dA e = = =B¥ =B¥ dt dt dt dt

=

1 2

¥ 1.0 ¥ (12) ¥

Metallic ring

)=

1 BR2 dq 2 dt

¥ 2p = 157 V

B = 0.5 T

B = 0.5 T

10 cm

R

v = 1 cm/s

w

v = 1 cm/s

q P

O

3 cm

EX A MP L E

5.8

(a) How much voltage is developed across the cut, if the loop moves in a direction normal to the shorter side a) ? How long the induced voltage last ? (b) How much voltage is developed across the cut, if the loop moves in a direction normal to the longer side b) ? How long the induced voltage last ? (c) Is a force required to pull the loop if it has a cut ? (d) Find the force needed to pull the loop if it has no cut, and has a resistance of 1 mW.

Solution (a) The induced emf is e = Blv

¥

¥ 0.01 = 0.15 mV

t = l/v = (10 cm)/(1 cm/s) = 10 s (b) The induced emf, e = Blv ¥ 0.1 ¥ 0.01 = 0.5 mV Now, the time for which the induced voltage last is given as t = l/v 3s (c I 2R loss or heat produced. If we neglect friction, no force is required to pull the coil.

156

(d) For Fig. For Fig.

(0.5)2 × (0.03)2 × 0.01 B 2l 2 v = = 2.25 mN 1 × 10 − 3 R (0.5)2 × (0.1)2 × 0.01 B 2l 2 v b F1 = = = 25 mN 1 × 10 − 3 R a F1 =

ADDITIONAL SOLVED EXAMPLES EX A MP L E

5.9

Solution B=

4p ¥ 10- 7 ¥ 90 m0 I = = 12 ¥ 10 2p ¥ 1.5 2px

T = 12 mT north-to-south.

EX A MP L E

5.1 0

Solution Fu = EX A MP L E

F = IB sin q = 8 ¥ l

0.6 N m –1

¥

5.1 1

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and

Solution F F = IBln = 12 ¥ 0.8 ¥ 0.1 ¥ The normal distance between these two forces is x=l ¥ t = Fx EX A MP L E

¥

0.96 Nm

5.1 2

side of the loop is 2.0 cm away from the conductor. What is the net force on the loop ?

157

Electromagnetism

10 cm D A

FCD

N

I2

25 cm

FAB

I1 C

P

W

E

B S

x

2 cm

Solution The current I1 in AB and the current I2 in straight conductor are in the same direction. Hence, the force FAB is

,

4 π × 10 − 7 × 15 × 25 m0 I1I 2 ¥l= ¥ ¥ 10 2 π × 0.02 2pr The current I1 in CD and the current I2 in straight conductor are in the opposite directions. Hence, the force FCD is , and its magnitude is 4 π × 10 − 7 × 15 × 25 m II FCD = Fu ¥ l = 0 1 2 ¥ l = ¥ ¥ 10 2 π × (0.02 + 0.10) 2pr Therefore, the net force on the loop is Fnet = FAB FCD 0.78125 mN FAB = Fu ¥ l =

EX A MP L E

5.1 3

A long vertical wire carrying a current of 10 A in the upward direction is placed in a region where the horizontal

Solution west to the wire. If x B= EX A MP L E

m0 I = 2.0 ¥ 10 2px

or

x=

m0 I 2p ¥ 2.0 ¥ 10- 3

=

4p ¥ 10- 7 ¥ 10 2p ¥ 2.0 ¥ 10- 3

= 10

m = 1 mm

5.1 4

or repulsive ?

Solution Since the

l >> r, we can say that for the

(4p ¥ 10- 7) ¥ 300 ¥ 300 m0 I1I 2 = = 1.2 N m –1 2p ¥ 0.015 2pr Since the currents in the two wires are in the opposite direction, the force between them will be repulsive.

F=

158 EX A MP L E

5.1 5

the counter torque that must be applied to prevent the coil from turning.

Solution The counter torque required to prevent the coil from moving must be equal (and opposite) to the torque t = BInA sin q = 1.0 ¥ EX A MP L E

¥ [p ¥ (8 ¥ 10 )2] ¥

¥

3.1 Nm

5.1 6 2

Solution Using Em = NBAw EX A MP L E

¥

¥ 10 ) ¥

¥

1.5 V

5.1 7 B

instant when the radius is 2 cm.

Solution Let the radius of the loop be r at a time t. F = BA = B ¥ (pr 2

e= EX A MP L E

d ( r 2) dF dr = Bp = Bp ¥ 2r = 0.02 ¥ p ¥ 2 ¥ 0.02 ¥ (1 ¥ 10 ) = 2.5 mV dt dt dt

5.1 8

A semicircular conducting loop ACDA of radius r a velocity w loop is R current and time, for two time-periods of rotation.

B directed into

w i

A

q C

O q D

B i (a)

(b)

(c)

Electromagnetism

159

Solution When the loop rotates through an angle q (q < p A(t) =

\ The induced emf, e =

Bw r dF = dt 2

The current through the loop, i =

q pr 2 wtr 2 wr 2 qr 2 = = = t p 2 2 2 2

F = BA =

2

Bw r 2 t 2

Bw r 2 e = 2R R b), the direction of the induced

c), and is given by pr 2 w r 2 A(t) = t 2 2 Hence, the induced current will have the same magnitude but

SUMMARY TE RM S

A N D

C ON CE PT S right-hand thumb rule says, “If you stretch the thumb of your right

For a current-carrying coil, the right-hand thumb rule A dot inside a circle represents a current coming out of the plane of the paper, whereas a cross inside a circle means a current entering the plane of paper. current and inversely proportional to its distance. current and inversely proportional to its distance. B) is measured in tesla (T), which is also equal to Wb/m2. F C thuMb

with

Field or F Current Motion of the conductor

right-hand generator action). motor action). One ampere parallel conductors situated in vacuum and separated by one metre, produces on each of the conductors a force of 2 ¥ 10 newton per metre length.

160 IMP O R TA N T

F O RM U LAE B=

At the centre of a circular loop, B =

m0 I . 2px

m0 I . 2r

m nI a) at the centre, B = m0 nI b) at either end, B = 0 . 2 m nI Inside the toroid, B = 0 . 2pr Force on a straight conductor, F = IBl sin q. m II F = BI2 = 0 1 2 . 2pr dF e N . dt

F = BA cos q. Induced emf in a conductor, e = –Blv sin q. Induced emf in a coil, e = NBAw sin wt.

CHECK YOUR UNDERSTANDING two

minus

one for each wrong answer. If your score is 12

1.

The right-hand thumb rule gives the direction of force on

current-carrying conductor is directly proportional to the current and inversely proportional to the square of its distance from the conductor. 2

.

is inversely proportional to the square of its radius.

7. 8.

10.

When two parallel conductors carry current in the same direction, the mutual force between them is repulsive. Statically produced emf is directly proportional to the rate of change

The emf induced in the secondary winding of a transformer is statically induced emf. Your Score

Electromagnetism

161

A N SW E RS

1. False 6. True

2. False 7. False

3. True 8. True

4. False 9. False

5. False 10. True

REVIEW QUESTIONS 1. current-carrying straight conductor ? 2. When a current-carrying conductor is placed in a

room ? 11. What is meant by electromagnetic induction ? State induction.

3.

a

12.

(b 4. State how you will determine the nature of force between two parallel current-carrying conductors. 5. parallel current-carrying conductors, when the a) in opposite directions, and (b) in the same direction. 6. 7. a) inside the solenoid, and (b) just at one of its ends. 8. As you move away from the middle of a solenoid, 9. What is the difference between a solenoid and a toroid ? 10. Suppose that you are sitting in a room with your

rule give the same direction of induced emf in a circuit. 13. principle of conservation of energy. 14. One end of a bar magnet is thrust into a coil. It is noted that the induced current in the coil is in Is the end of the bar magnet its N-pole or S-pole ? 15. A metallic loop is placed in a non-uniform magnetic 16. separated by a distance. A battery is suddenly connected to one of the loops establishing a current in it. Will there be a current induced in the other loop ? Do the loops attract or repel each other ? 17. The battery in the above question is suddenly disconnected. Is a current induced in the other loop ? If yes, when does it start and when does it end ? Do the loops attract or repel each other ?

MULTIPLE CHOICE QUESTIONS (c) newton/metre

2

(d) tesla/metre

3. . 1. An electric charge in uniform motion produces (a (b (c (d 2. B is (a) weber (b) weber/metre2

carrying conductor is directly proportional to (a) the resistance of the conductor (b) the distance from the conductor (c (d 4. Two long parallel wires carrying sinusoidally varying currents in the opposite directions (a) attract each other (b) repel each other

162 parallel to one of the sides of the loop and is in the plane of the loop. If a steady current I is established

(c) do not affect each other (d) get rotated to be perpendicular to each other 5. A wire is placed parallel to the lines of force in

(a (b) move away from the wire (c) move towards the wire (d) remain stationary

Then (a (b (c

I

i

(d 6. paper with N on the left side. A conductor is placed

conductor out of the paper, it tends to move (a) upward (b) downward (c) to the right (d) to the left 7.

r

B with distance r from a long straight wire carrying a current ? i B

B

9. loop being r metre. It carries a current i ampere. r

r

(a)

part is (a) zero

(b)

B

(b

m 2i (c) 0 ◊ (p + 1) 4p r

B

m0 2i (p ◊ 4p r

(d)

10. A length of wire carries a steady current. It is bent same length is now bent more sharply to give a r

at the centre of this loop developed due to the same current is (a (b (c (d) unaltered

r

(c)

(d)

8. A rectangular loop carrying a current i is situated near a long straight wire such that the wire is

A N SW E RS

1. c

2. b

3. d

4. b

5. b

6. a

7. c

8. c

9. d

10. a

Electromagnetism

163

PROBLEMS ( A )

S I M P L E

P RO B LEMS

1. A circular coil of wire consisting of 100 turns, each

the coil ?

3. A long straight wire in the horizontal plane carries the magnitude and direction of B east of wire. [Ans. mT, vertically up]

[Ans. 4.

2. B at a point 20 cm [Ans. mT]

from the wire ?

is the magnetic force on the wire ?

( B)

TRI C KY

P RO B LE MS

5.

7.

solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre. [Ans. 6. A straight horizontal conducting rod of length

rod through the wires. (a should be set up normal to the conductor so as to b) What will be the total tension in the wires if the direction of as before ? (Ignore the mass of the wires.) [Ans. (a) Horizontal magnetic b

( C)

C H A LL EN GI NG

in the wire.

mT parallel to the current. Find the magnitude away from the wire. [Ans. mT] 8. A rectangular loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform to the loop. What is the voltage developed across the cut if the velocity of the loop is 1 cm/s in a direction normal to (a) the longer side, (b) the shorter side of the loop ? For how long does the induced voltage last in each case ? [Ans. (a (b

PROBLEM S

9. A small piece of metal wire is dragged across the

10.

[Ans. 81 mN]

[Ans. B perpendicular to the plane of a square frame made of

copper wire. The wire has a diameter of 2 mm and with time at a steady rate dB/dt = 0.02 T/s. Find the current induced in the frame. The resistivity of copper is 1.7 ¥ 10 W m. [Ans.

MAGNETIC CIRCUITS

6

OB JE CT IV E S : (

)

(H)

(m)

(R )

(G).

closed loops magnetic circuit.

permeability

magnetomotive force mmf

ampere turns F.

H I shown in Fig. 6.1. The mmf is the total

N

l IN

165

the mmf per metre length of the magnetic circuit the force

N turns I A m2

H H =

F IN = l l

l metres

H N

I

the coil.

m B in the permeability of free space

H. m0 B = m0 H m 0 is 4p ¥ 10 –7

B = m0H

B B = mH m is called the and is denoted by the symbol m m m0

m = m0 m

m

Paramagnetic materials ferromagnetic materials

m m m

m can be as high as 100 000. m m

density B characteristics

H magnetisation .

B (T)

m = 1. diamagnetic materials is a bit m

2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0

Silicon steel Low-carbon mild steel

Cast iron Mumetal Ferrite

2000

4000

6000

8000

10000 H (A/m)

166

Basic Electrical Engineering

Reluctance (R) and Permeance (G) We know that the electrical resistance of a conductor of length l and area of cross section A is given as l 1 l R = or R = A A where, r is the resistivity and its reciprocal s (= 1/r) is the conductivity of the material. Just as the current I in an electric circuit is limited by the presence of resistance of the electric circuit, the F in a magnetic circuit is limited by the presence of the reluctance of the magnetic circuit. Thus, the by the symbol R . Thus, by analogy, the reluctance of a specimen of magnetic material is given as 1 l 1 l R = = A m r m0 A

(6.3)

The reciprocal of reluctance is known as permeance (G). Thus, permeance in magnetic circuits is analog of conductance (G) in electric circuit. Note

of turns carrying a dc current. Consider a simple magnetic circuit of Fig. 6.1. The coil wound around the toroid has N turns and carries a current of I amperes. The toroid has a cross-sectional area of A metre2 and a mean circumference of l metres. Then, the value of mmf, F = NI H is set up throughout the length l H, we should have F = Hl (6.4) B F =B¥A

(6.5)

Dividing Eq. 6.5 by 6.4, we get F or

=

F=

B A A A BA = = m = m r m0 Hl H l l l

F l /( m r m0 A)

(6.6)

Similar equation in electrical circuits gives the electric current I as the ratio of emf E to the resistance R. That is, E (6.7) R Comparison of Eq. 6.6 with 6.7 suggests that the quantity l/(m r m 0 A) should represent some sort of resistance in magnetic circuits. This quantity is called reluctance Eq. 6.3. I=

167

differences too : 1. 2. 3.

S. No.

Electric Circuits Quantity

Magnetic Circuits Quantity Units

Units

E

F F

I W

R 1 l R= ⋅ σ A s G

R 1 l R = ◊ m r m0 A P m P

J

G B

E H 8.

E Ohm’s law : I = R

EX A MP L E

F F= R

6.1

a c

Solution a H = b c

200 4 NI = = 1333 A/m l 0.6

B = mH = m m0 H = 1 ¥ 4p ¥ 10 –7 ¥ F = BA ¥ 10 –6 ¥ ¥ 10 –6

= 1675 mT 0.8375 mWb

b

168

Basic Electrical Engineering

EX A MP L E

6.2

.

Solution

B=

\

F = Hl

EX A MP L E

¥

0.015

=

200 × 10 − 4 B 0.75 H= = 4 π × 10 − 7 0 ¥ 10 = 1492 At A

6.3

a

b

m

Solution a

0.4 1 l = = 1.675 ¥ 106 A/Wb 380 × 4 π × 10 − 7 × 500 × 10 − 6 m r m0 A F = FR = ¥ 10 –6 ¥ ¥ 106 F 1340 I= = = 6.7 At 200 N R =

b \

a. Let l1 and l m1 and m

A1 and A R1=

l1 m1 A1

R =

R = R1+ R =

l m A

l1 l + 2 m1 A1 m 2 A2

I l2

B

l1

l1 A (a) Composite magnetic circuit.

(b) Steel ring with an air gap.

l2

169

F=

mmf of coil F = = total reluctance R

NI l1 l + 2 m1 A1 m 2 A2 b

A

R =

l1 and l

ˆ 1 Ê l1 ˆ 1 Ê l1 l1 l + l2 ˜ = + l2 ˜ + 2 = Á Á m0 A Ë ( m1/m0 ) m0 A Ë m r ¯ ¯ m1 A m0 A m m1 m0 R l

a I leak called a as fringing

b

I L d a

d

b

L F

L I

L I

C

N

S

F

F (a)

L

(b)

170

Basic Electrical Engineering

b as follows : Total flux through exciting winding l = Useful flux

a

I.

F L = FM + F N P P I

IM

IL

L C

M

N

L

IN M

N

L E

Q

Q (b) Electric circuit equivalent to the given magnetic circuit.

(a) Magnetic circuit to illustrate Kirchhoff’s laws.

In a closed magnetic circuit the algebraic sum of HL HM and lM

lL is the length HN and lN

HL lL + HM lM HL lL + HN lN 0 = HM lM – HN lN

and

a b.

171 EX A MP L E

6.4 a

b

Solution

B

N

For ferromagnetic core : lc

Ac b,

B

¥ 10 –6 m . Hc

\

Fc = Hc lc

For air gap : lg

¥ 10 –6 m . B 0.9 Hg = = = 717 000 At 4 π × 10 − 7 0 F g = Hg lg = 717 000 ¥ 0.001 = 717 At

\

\

¥

Ag

F = Fc + F g = 180 + 717 = 897 At F 897 I = = = 0.224 A 4000 N

M

1. B

I.

172

Basic Electrical Engineering

2.

Flux, F B = F/A

F = BA Flux density, B

¥ 0.9

B = m0mrH

H = B/m0mr Field strength, H

H = F/l

F = Hl MMF, F

3.

F = NI

I = F/N Current, I

method. 4. m =

0.9 B = m0 H 4 π × 10 − 7 × 820

0.22 lc = = −7 4 π × 10 × 873 × 50 × 10 − 6 m0 m r A lg 0.001 Rg = = = −7 m0 A 4 π × 10 × 50 × 10 − 6 Rc =

and

5.

The F F1 plot F

Fm F Fm

eddy current loss of thin sheets called *

F Fm

173

F F1

No air gap 0.2 mm gap

0

1.0 mm gap

0.5

F/Fm

1.0

is

ADDITIONAL SOLVED EXAMPLES EX A MP L E

6.5

600 mm R 2 mm

lA I

A

lB

E

40 mm 50 mm

lC C

B

174

Basic Electrical Engineering

Solution Cross-sectional area of the core, Ac = width ¥ depth = 40 ¥ 50 = 2000 mm2 For air gap : lg = 2.0 mm = 2 ¥ 10 Bg =

= 2 ¥ 10 –3 m2 m; Ag = 2500 mm2 = 2.5 ¥ 10 –3 m2; F = 2.5 ¥ 10 –3 Wb.

2.5 × 10 − 3

= 1 T; Hg =

Bg

\ Fc \

−3

=

1

= 796 000 At/m 2.5 × 10 4 π × 10 − 7 0 –3 \ mmf, F g = Hglg = 796 000 ¥ 2 ¥ 10 = 1592 At For the core : Since 8 % is taken up by the insulation between the laminations, the effective area of cross section (through Ag

=

–3

Ac = 2 ¥ 10 –3 ¥ 0.92 = 1.84 ¥ 10 –3 m2 ¥ leakage factor = 2.5 ¥ 10 –3 ¥ 1.2 = 3.0 ¥ 10 –3 Wb 3.0 × 10 − 3 Bc = c = = 1.63 T Ac 1.84 × 10 − 3

mmf, F c = Hclc = 4000 ¥ 0.6 = 2400 At Total mmf needed, F = F g + Fc = 1592 + 2400 = 3992 At F 3992 I= = =5A 800 N

\

EXA M P L E

6. 6

A toroid of uniform cross-sectional area 0.001 m2 consists of three materials as shown in Fig. 6.10. Material A is silicon steel with length lA = 0.3 m, material B is low-carbon mild steel with length lB = 0.2 m, and material C is cast iron with length lC a mWb, (b) the current in the coil, and (c required, use the data given in Fig. 6.2.

Solution (a is,

600 × 10 − 6 = 0.6 T A 0.001 For this value of B, the corresponding value of H is found for the three sections of the toroid from Fig. 6.2, as HA = 20 At/m, HB = 700 At/m and HC = 2500 At/m BA = BB = BC = B =

=

600 mWb is F = F A + F B + FC = HA lA + HB lB + HC lC = 20 ¥ 0.3 + 700 ¥ 0.2 + 2500 ¥ 0.1 = 6 At + 140 At + 250 At = 396 At (b) To produce above mmf, the current required in the coil is F 396 I = = = 3.96 A 100 N B (c) Since B = mH = m r m0 H, we have m r = . Therefore, the relative permeabilities of the three materials are given H 0 as B 0.6 0.6 B m rA = = = 23 885; m rB = = = 682 m 0 HB 4 π × 10 − 7 × 20 4 π × 10 − 7 × 700 0 HA B 0.6 and m rC = = = 191 4 π × 10 − 7 × 2500 0 HC

175 reluctances FA 6 At F 140 At RA = = = 10 000 At/Wb; R = B = = 233 333.3 At/Wb 600 Wb 600 Wb F 250 At RC = C = = 416 666.6 At/Wb 600 Wb

and EX A MP L E

6.7

m

Solution

B=

For the air gap : B

Hg =

=

0

A

=

600 × 10 − 6 4 × 10 − 4

1.5 4 π × 10 − 7

= 1.194 ¥ 106

F g = Hg lg = 1.194 ¥ 106 ¥ 1 ¥ 10

For the core

H

\

Fc ¥ F = Fc + F g F 2394 I = = = 7.98 A N 300

\

Ac = 4 cm2

I

lg = 1 mm

300

600 l2

l3

I l1

I lc = 40 cm

EX A MP L E

6.8 l1

of l1

l =l

l and l

of 100 m

Solution For air gap \

= 1194 At

Hg =

Bg =

Bg 0

=

0.16 4 π × 10 − 7

A

=

¥ 10

100 × 10 − 6 6.25 × 10 − 4

= 0.16 T

F g = Hg lg

¥ 10 ¥

l1

2 mm

176

Basic Electrical Engineering

For path l1 : B1 \

H1 =

0.16 B1 = 800 × 4 π × 10 − 7 m r m0

F 1 = H1 l1

¥

¥ 10

The two paths l and l l1 two paths. −6 100 Wb 50 × 10 Wb For path l : F = = m B = 2 = = 0.167 T 2 2 A2 3 × 10 − 4 m 2 0.167 B2 \ H = = F = H l = 166 ¥ 18 ¥ 10 800 × 4 π × 10 − 7 m r m0 l1 and l NI = F g + F 1 + F F 299.462 I = = = 0.499 A 600 N EX A MP L E

6.9 a I

Solution

b shows B

R 1 and R F

F =

A2 A1

l1 l F1 = 2 F 0 A1 0 A2 l1 0.02 m 2 1.0 mm ¥ F1 = ¥ l2 0.01 m 2 2.0 mm

F = F1 + F

R1 =

3.0 mWb

1 × 10 − 3 l1 = = 4 π × 10 − 7 × 0.01 m0 A1

R =

R = R 1|| R =

C

N

79577 2

F

2 mm

B

1 mm

A

F

+ –

2 × 10 − 3 l2 = 4 π × 10 − 7 × 0.02 m0 A2

F1 R1

Cast iron

F2 R2

21 cm

I (a)

(b)

Cast steel

0.2 mm

R 1F 1 = R F fi

F1 and F

F

0.2 mm

and C

177

F = FR EX A MP L E

NI = FR

fi

I=

R 3.0 × 10 − 3 × 39 788.5 = = 238.7 mA N 500

6.1 0

Solution

B=

A

=

0.8 × 10 − 3 10 × 10 − 4

= 0.8 T

For total air gap Fg = Hglg =

B

¥ lg =

0

0.8 4p ¥ 10- 7

¥

¥

¥ 10

For cast iron F i = Hili =

0.8 p ¥ 0.21 B pD ¥ = ¥ -7 2 166 ¥ 4p ¥ 10 m r m0 2

Fs = Hsls =

0.8 p ¥ 0.21 B pD ¥ = ¥ -7 2 800 ¥ 4p ¥ 10 m r m0 2

For cast steel

F = Fg + Fi + Fs EX A MP L E

1782.2 At

6.1 1 a

8 cm

B H F

A

B

F F2

Rg

G 34 cm

15 cm

1 mm

R2

R + F

E

D (a) Magnetic circuit.

C

– (b) Its electrical equivalent circuit.

F1

R1

178

Basic Electrical Engineering

Solution

b

F is applied F in it. R R R 1 and R

R1

Rg F

F1 = F = F B= B

A

=

10 − 3 8 × 10 − 4

H ¥

F DG = Hl

B

F g = Hglg =

¥ lg =

0

1.25 4 π × 10 − 7

¥ 0.001 = 994.7 At

F F1 = F B1 =

1

A1

=

B

0.5 × 10 − 3

=1T

5 × 10 − 4 H ¥

F 1 = H1l1

F = F DG + F g + F 1

1137.7 At

SUMMARY TE RM S

A N D

C ON CE PT S

Magnetomotive force mmf F B H m

B m m

R .

. The

H

179 IMP O R TA N T

F O RM U LAE

F = NI H= m = m m0

R =

NI F = l l

B = mH m0 = 4p ¥ 10–7 mmf (F ) F . Reluctance (R ) F R =

1 l 1 l = A m r m0 A

CHECK YOUR UNDERSTANDING two one

12

S. No.

Statement

True

False

Marks

which the magnetic domains can be aligned.

.

F=

mmf of coil (F ) total reluctance (R )

Your Score A N SW E RS

1. 6. False

2. False 7. False

3. 8.

4. False 9.

5. False 10. False

180

Basic Electrical Engineering

REVIEW QUESTIONS 1.

2.

5.

a b

3. 6. 4.

MULTIPLE CHOICE QUESTIONS b c d

completes the statement correctly.

4.

1. a c

¥ ¥

to a b c d

b d

2. a b

5. W

c d

a c

b d

a c

b d

6. 3. is a

8

intensity A N SW E RS

1. a

2. d

3. c

4. a

5. a

6. c

PROBLEMS ( A )

S IM P L E

PRO BL EMS

1. coil is 4 cm and the mean length of the magnetic

[Ans.

mWb]

Magnetic Circuits 2. Repeat Prob. 1, if the core is made of cast iron. Fig. 6.2 may be used. [Ans. 2000 A/m, 0.57 T, 228 mWb] 3.

2

long with a cross-sectional area of 25 cm , if an [Ans. 3.77 T] 4. A cast iron ring with a mean diameter of 10 cm and 2 a cross-sectional area of 3 cm has a radial gap of

( B )

T R IC KY

181

0.15 cm. Calculate the required current through the coil of 1000 turns wound uniformly over the ring to relative permeability of cast iron as 238. [Ans. 1.345 A] 5. An iron ring with a mean diameter of 50 cm has the relative permeability of the iron is 400 when a Ans. 0.0637 T]

PR OB LE MS

6. A ring-shaped core is made of a ferromagnetic material having relative permeability of 1000. The core has two sections. The thicker section has a mean length of 10 cm and area of cross section of 6 cm2. The thinner section has a mean length of 20 cm and area of cross section of 4 cm2

2

in air gap. [Ans. 2605.72 At] 9. A circular ring of mean length 4p cm and of crosssectional area 10 cm2 has an air gap 0.4p mm long. The relative permeability of iron is 1000. The ring is wound with a coil of 2000 turns. Determine the

thicker section. The current through the coil needed 0.5 A. Compute the number of turns of the coil. [Ans. 955] 7. An iron ring with a mean length of magnetic path of 20 cm and of small cross-sectional area has an air gap 1 mm long. A current of 1 A in a coil of 440 turns wound uniformly over the ring produces p ¥ 10 –3 T. Neglecting leakage and fringing, determine the relative permeability of iron. [Ans. 19.9] 8. A ring-shaped electromagnet has an air gap 6 mm long and 20 cm2 in area. The mean length of the iron core is 50 cm and its cross-sectional area is 10 cm2. Assuming the permeability of iron to be 1800, determine the ampere turns required to set up ( C)

C H A LL ENG I NG

[Ans. 3.67 mWb] 10. An iron ring with a mean length of magnetic path 25 cm and a small cross-sectional area has an air

p mT, determine the relative permeability of iron. Neglect the leakage and fringing effects. [Ans. 312.5] 11. A ring-shaped electromagnet has an air gap 6.5 mm in length and 22 cm2 in area. The mean length of the iron-core is 52 cm and its cross-sectional area is 11 cm2. Assuming the relative permeability of iron to be 1800, calculate the ampere turns required 2 in the air gap. [Ans. 3097.77 At]

P ROBLEMS

12. and across the ring, an iron bar of dimensions 18 cm ¥ 3 cm ¥ air gap, as shown in Fig. 6.16. Neglecting required to be applied to one-half of the

in the other half of the ring. The B-H characteristics of the two materials are given in the following table. [Ans. 694.703 At] Quantity

For cast steel

B (T) 1.0 1.1 1.2 H (At/m) 900 1020 1220

For iron 1.2 590

1.4 1.45 1200 1650

182

Basic Electrical Engineering F1

A I

A

C

I

l2

B l3

N 12 cm

A1 = 1 cm2

A2 = 2 cm2 0.2 mm

0.2 mm

I

F2

l1

D

18 cm

F3

A3 = 1 cm2

0.25 mm

13.

a m = b m of 1 cm

[Ans. a

b

SELF- AND MUTUAL INDUCTANCES

7

OB JE CT IV ES :

We have seen in Chapter 5 that a statically induced emf can be self-induced too. Whenever the current

self-inductance of current, eμ L is called the

di dt

or e = L or simply

di dt

(7.1) *

.

An inductor can be linear or nonlinear. For a linear inductor, the magnitude of inductance is independent air-cored inductor is linear. When iron (or any other ferromagnetic material) is used as core, the inductor becomes nonlinear.

Energy Stored in an Inductor

I in time , an emf e is induced in the coil. During this process, the total energy W absorbed by the inductor is given *

184 as t

t

W=

Ú

ei dt =

Ú

Ê di ˆ ÁË L ˜¯ dt

I

=

Ú

Li di =

Li 2 2

I

= 0

1 2 LI 2

(7.2)

Inductance from Geometrical Viewpoint e= N

dF di = L dt dt

L =N

or

L= N

dF di

(7.3)

(7.4)

I

F=

NI R

L=

N 2m A l

and

R=

l mA

(7.5)

properties of the core material.

Determination of L

L can be determined by using any one of the three

relations given in Eqs. 7.1, 7.4 and 7.5. EX A MP L E

7.1

in the coil.

Solution Using Eq. 7.1, the emf induced in the coil is e= L EX A MP L E

10 4 di =4¥ = 240 V dt 0.1

7.2

Solution Using L=N

e= L

di dt

¥

I

¥

10

0.01 = 0.15 H 10 ( 10) = 300 V 0.01

Self- and Mutual Inductances EX A MP L E

185

7.3

Solution Using

EX A MP L E

L = N

dF di

e = L

di =4¥ dt

¥

[0.4 ( 0.4)] mWb = 4 mH [10 ( 10)] A 10

¥

( 10) =8V 0.01

7.4

Solution Cross-sectional area of the solenoid is given as A = pr 2 = p

EX A MP L E

)2 = 1.77 ¥

¥

m2

L =

(900)2 × 4 π × 10 − 7 × (1.77 × 10 − 4 ) N 2 m0 A = = 0.6 mH 0.30 l

W =

1 1 2 ¥ LI = 2 2

¥

) ¥ 52 = 7.5 mJ

7.5

Calculate ( ) the permeability of iron, (b) the relative permeability of iron, (c) the inductance of the coil, and ( ) the

Solution A = pr 2 = p ¥ (1 ¥

)2

¥

m2

(

\

H =

3000 0.5 NI = l 0.2

m =

B 1.2 –4 = = 1.6 ¥ 10 Tm/A H 7500

(b mr =

= 0

1.6 ¥ 10 - 4 4p ¥ 10 - 7

= 127

(c L =

3000 × 1.2 × 31416 . × 10 − 4 NBA NF = = = 2.26 H I 0.5 I

186 ( e = N EX A MP L E

dF dt

¥

1.2 × 31416 . × 10 − 4 (1 − 0.1) = 101. 78 V 0.01

7.6 W

Solution V =

+ L

di =1¥ dt

¥

¥

10000 =4V 1

) are available in different types ranging from large high current iron-cored m

Iron core

Air core

General

Tapped

(a) Fixed

Slider

Movable core

(b) Variable

) conductive, (b) electrostatic, and (c and circuit b Figure 7.2c

, resistor

12

C.

187

Self- and Mutual Inductances

r1

Vs

R12 1

r1

r2

r2

Vs

2

1

(a) Conductively coupled circuits

2

r2

F21

N1 F11

V1

C

F12

r1

i1

N2 F22

1

(b) Electrostatically coupled circuits

i2

V2 2

(c) Magnetically coupled circuits

mutually induced emf e2 in the second coil is dependent on the rate of change of current di e2 μ di1 or e2 = M 1 dt dt M is called L), the unit of mutual inductance (M

1

or simply

.

Note

transformer*.

c. Current F12,

F1 coil

F1 = F11 + F12 F22 Let N1 and N2 be the number of turns in coils F1 F12 induced in coil is given as dF12 e2 = N 2 dt

F21 and

establishes a total F11 F2 produced by the current 2 F2 = F22 + F21. 1 is time-varying, the 1

(7.7) to coil

M12

dF12 di1 = N2 dt dt

or M12 = N 2

dF12 di1 to coil

M21 = N1

*

dF21 di2

We shall study more about transformers in Chapter 13.

is given by

as

188

M12 = N 2 M21 = N1

and

12

i1 21

(7.11)

i2

Furthermore, because of the reciprocity of coupling M12 = M21 = M

k)

N12 m A N 2mA and L2 = 2 l l due to the current 1 is given as mmf N i mA Ni F1 = = 11 = 11 l /mA R l is only a part of F1 F12 = kF1 inductance from coil to coil k F1N2 kN1 N 2 m A M12 = N 2 12 = = i1 l i1 to coil k F2 N1 kN 2 N1m A 21 M21 = N 21 = = i2 l i2 Multiplying Eq. 7.14 and 7.15, putting M12 = M21 = M L1 =

M2 = k 2 \

k=

(7.12)

(7.13) £k£ (7.14)

(7.15)

Ê N 2 m A ˆ Ê N 22 m A ˆ N1 N 2 m A N 2 N1m A = k2 Á 1 = k 2L1L2 ¥ Ë l ˜¯ ÁË l ˜¯ l l M

L1 L2

k the product of inductances of coil and coil the value of k is unity, and the coils are said to be

M to the square root of

k said to be magnetically EX A MP L E 2

.

7.7

is placed centrally inside the solenoid. Assuming k = 1, calculate ( ) the mutual inductance, and (b) the emf

Self- and Mutual Inductances

Solution ( ) Using Eq. 7.14, the mutual inductance is given as 1 × 2000 × 500 × 4 π × 10 − 7 × 30 × 10 − 4 kN1 N 2 m A M = = = 5.38 mH 0.70 l (b e2 = M di1 ¥ ¥ 1.4 V dt EX A MP L E

7.8

Calculate L1, L2, k and M.

Solution Using Eq. 7.4, the self-inductance of the second coil is given as 2

L2 = N2 k=

21

=

2

Using Eq. 7.12, L1 = L2 ¥

N12 N22

EX A MP L E

I2 0.5 × 10 − 3

0.8 × 10 − 3 = 0.226 H 6

= 0.625 0.8 × 10 − 3 (600)2 ¥ = 0.028 H (1700)2 ¥

M = k L1 L2

¥

0.226 = 0.05 H

0.028

7.9

Solution 1

L1 = N1 and

I1

L2 = N 2

2

I2

and

M = N2 k =

0.25 × 10 − 3 = 0.06 H 5

¥

0.15 × 10 − 3 = 0.024 H 5

¥

F12 = kF1 \

¥

12

¥

I1

M L1L2

¥

=

0.15 × 10 5

¥

Wb

−3

= 0.024 H

0.024 = 0.6325 0.06 0.024

We have as yet considered only a mutual voltage induced across an

189

190 in the other coil. di1 di +M 2 dt dt di2 di v2 = L2 +M 1 dt dt v1 = L1 ,

(7.17)

the assumed voltage reference.

i2

the mutual voltage) depends not only on the current directions and

i1 1

is 1

1,

is also 2 2 1

and

2

produce

current 2 as the self-induced voltage in that coil.

dotted dotted , the current 1 enters the dotted terminal of L1, then the voltage v2 is sensed positively at the dotted terminal of L2, and v2 = 1 of voltage v2 convenient to represent v2 Fig. 7.4b; then obviously, v2 M( 1 ). undotted undotted

Self- and Mutual Inductances

191

c and is equivalent to Fig. 7.4 , and Fig. 7.4b is equivalent to Fig. 7.4c.

Figure 7.5 same,

1

=

2

= v1 = L1

di di +M dt dt

and

v2 = L2

di di +M dt dt

di ˆ Ê di di ˆ di Ê di v = v1 + v2 = Á L1 + M ˜ + Á L2 + M ˜ = (L1 + L2 + 2M) Ë dt ¯ Ë ¯ dt dt dt dt Lsa is the equivalent inductance of the di v = Lsa dt

Lsa = L1 + L2 + 2M

(7.21)

192

b

v1 = L1

di di -M dt dt

and

v2 = L2

di di -M dt dt

di ˆ Ê di di ˆ Ê di v = v1 + v2 = Á L1 - M ˜ + Á L2 - M ˜ = (L1 + L2 Ë dt dt ¯ Ë dt dt ¯ Lso is the equivalent inductance of

v = Lso

di dt

M)

di dt

(7.22)

(7.23)

Lso = L1 + L2

M

(7.24)

Note to its other end (compare Figs. 7.5 and b any circuit involving coupled coils.

M M inductance Lsa of series aiding combination and the equivalent inductance Lso of series opposing combination, Lsa Lso = 4M or

M=

Lsa

Lso

(7.25)

4

From the values of L1, L2 and M EX A MP L E

7.1 0

Solution M =

L sa

L so 4

=

1.4

0.6 4

0.2 mH

Self- and Mutual Inductances

L1 + L2 + 2 ¥

Lsa = L1 + L2 + 2M L1 = L2 k = EX A MP L E

193

fi L1 + L2

0.2 mH = 0.4 (0.5mH) (0.5mH)

M = L1L2

7.1 1

Solution L1 + L2 + 2M L1 + L2 M

() ( )

) from ( M = 0.25 H

4M Adding Eq. ( ) and ( 2(L1 + L2 k = Using the algebraic relation, (L1 L1

L1 + L2 M L1L2

or

L2)2 = (L1 + L2)2 L2 =

)

L1 L2 =

M2 k

2

=

(0.25)2 (0.6)2

)

L1L2, and using Eqs. ( ) and (

2

( L1 + L2 ) - 4 L1L2 =

(1.3)2 − 4 × 0.1736

)

) and ( L1 = 1.149 H

and

L2 = 0.151 H

) and b voltages across both have to be the same, i.e., v1 = v2 = v.

194

v = L1

L1 =

1

+

2

L1

and

v = L2

di2 di +M 1 dt dt

di1 di di di + M 2 = L2 2 + M 1 dt dt dt dt 2 by ( 1) in the above equation to get

di1 d ( +M dt dt

(L1 + L2

or

di1 di +M 2 dt dt

M)

fi 1

1)

= L2

d ( dt

1)

+M

di1 dt

di1 di = (L2 M) dt dt L2 - M di di1 = dt L1 + L2 - 2M dt

by (

(7.27)

2

L1 - M di di2 = L1 + L2 - 2M dt dt

v = L1

L1 - M di1 di Ê L2 - M ˆ di Ê ˆ di +MÁ + M 2 = L1 Á Ë L1 + L2 - 2M ˜¯ dt Ë L1 + L2 - 2M ˜¯ dt dt dt

Ê L L - M 2 ˆ di = Á 1 2 ˜ Ë L1 + L2 - 2M ¯ dt

Lpa is the equivalent inductance of the v = Lpa

Lpa =

di dt

L1L2 - M 2 L1 + L2 - 2M

(7.31)

b

b

M

L1 L2 - (- M ) L L - M2 = 1 2 L1 + L2 + 2M L1 + L2 - 2 (- M ) 2

Lpo =

(7.32)

Self- and Mutual Inductances EX A MP L E

195

7.1 2

) the mutual induction assists the self-induction, and (b) the mutual induction opposes the self-induction.

Solution ¥

M = k L1L2

8

6

( ) From Eq. 7.31,

Lpa =

8 × 6 − 32 L1L2 - M = = 4.875 H L1 + L2 - 2 M 8+6−2×3

(b) From Eq. 7.32,

L po =

8 × 6 − 32 L1L2 - M 2 = = 1.95 H L1 + L2 + 2M 8+6+2×3

2

value I

L W=

1 2 LI 2

(7.33)

) part of the coil.

collapses and the energy stored is used in generating the induced emf or current. and putting L=

W=

m m AN 2 N 2mA = 0 r l l

1 2 1 1 N 2I 2 LI = m mr A = m mr 2 2 2 l

N 2I 2 l2 NI

=

2 1 Ê NI ˆ m mr ( ) Á ˜ Ë l ¯ 2

2 1 1 Ê NI ˆ m m r ( ) Á ˜ = m m r H2 (Volume) Ë l ¯ 2 2 is given as 1 Wu = m m r H2 2 B = m mr H 1 Wu = 1 m mr H 2 = BH 2 2

H),

W=

(7.34)

(7.35)

196

Wu =

or

B2 2m0 m r

3

A be their area of cross section, and F be tive force through a small distance

Ê B2 ˆ = Á ˜( Ë 2m0 ¯

)

magnet), F=

B 2A 2m0

Fu =

B2 2m0

And, the force per unit area, 2

(7.37)

ADDITIONAL SOLVED EXAMPLES EX A MP L E

7.1 3

An air-cored coil is required to have a length of 2.5 cm and an average cross-sectional area of 2 cm2. Find the number m

Solution L =

N = EX A MP L E

N 2m 0 A l

Ll = m A

400 × 10 − 6 × 2.5 × 10 − 2 4 π × 10 − 7 × 2 × 10 − 4

7.1 4

L1, L2, M and N2.

= 200

Self- and Mutual Inductances

197

Solution 250 × 4 × 10 − 3 = 0.333 H i1 3 v2 = 1 3 M¥ fi M = 0.07 H 3 × 10 − 3 M = k L1L2 , the self-inductance of the second coil is given as 1

L1 = N1

L2 =

M2

=

(0.07)2

=

(0.75)2 0.333 k L1 2 L μ N , the number of turns in the second coil is given as 2

L2 L1

N2 = N1 EX A MP L E

= 0.0262 H

0.0262 = 70 0.333

¥

7.1 5 2 2

2

Solution L1 = N1 L2 = N2

and

\ Mean value of self-inductance, L =

i1 2

i2

=

N1B1 A 1000 × 1.0 × 20 × 10 − 4 = = 0.5 H 4 i1

=

N 2 B2 A 1000 × 1.4 × 20 × 10 − 4 = = 0.31 H 9 i2

L1 + L2 0.5 + 0.31 = = 0.405 H 2 2

e = L EX A MP L E

1

di dt

¥

9 4 = 40.5 V 0.05

7.1 6

125 cm2 each coil, (b

) the self-inductance of c

Solution (

N12 m0 m r A (100)2 × 4 π × 10 − 7 × 2000 × 125 × 10 − 4 = = 157.1 mH 2 l N 2 m m A (150)2 × 4 π × 10 − 7 × 2000 × 125 × 10 − 4 L2 = 2 0 r = = 353.4 mH 2 l L1 =

and (b

M = k L1L2 = 1 ¥

157.1 353.4 = 235.6 mH

198 (c e2 = M EX A MP L E

di1 dt

¥

¥

5 0 = 58.9 V 0.02

7.1 7 1

2 and

3

4

2

3

1

4

2

4

1

3

Solution Let L

) and ( \ EX A MP L E

L + L + 2M = 4 or L+M =2 L+L M L M 2 + 0.4 2 0.4 L= = 1.2 H and M= = 0.8 H 2 2 0.8 M k = = = 0.667 or 66.7 % 1.2 1.2 L1L2

() )

7.1 8

) series aiding, (b) series opposing, (c) parallel aiding, and ( ) parallel opposing combinations.

Solution M = k L1L2

¥

200

800 ¥ ¥

( ) (b)

Lsa = L1 + L2 + 2M Lso = L1 + L2 M

(c)

Lpa =

200 × 800 − (200)2 L1L2 - M 2 = = 200 mH L1 + L2 - 2M 200 + 800 − 2 × 200

( )

Lpo =

200 × 800 − (200)2 L1L2 - M 2 = = 85.71 mH L1 + L2 + 2M 200 + 800 + 2 × 200

EX A MP L E

1400 mH 600 mH

7.1 9 2

Solution F =

60 2

¥

Self- and Mutual Inductances

F =

B = \

H =

B 2A 2m0

2 m 0F = A

2 × 4 π × 10 − 7 × 294 6 × 10 − 4

111 . B = 4 π × 10 − 7 × 800 m0 m r ¥

I =

Ampere turns 496.8 = = 0.4968 A Total turns 2 500

SUMMARY TE RM S

A N D

C ON CE PT S

Self-inductance of a coil

Mutual inductance one coil due to change in current in the other coil. (k by primary current. Both L and M IMP O R TA N T

F O RM U LAE e= L

Energy stored, W =

di . dt

1 2 LI . 2

EMF induced by mutual inductance, e2 = M N 2m A . l dF M21 = N1 21 . di 2 L=

k=

M L1L2

£ k £ 1.

di1 . dt

2

199

200 v1 = L1

di1 di +M 2. dt dt Lsa = L1 + L2 + 2M

(

Lso = L1 + L2

(

2

( ) Parallel aiding, Lpa =

( ) Parallel opposing, Lpo =

L1L2 - M 2 L1 + L2 + 2M

B2 . 2m0 m r

Wu = Fu =

L1L2 - M L1 + L2 - 2M

M

B2 . 2m0

CHECK YOUR UNDERSTANDING two one

12

1. as L = (2 ¥

I 2.

as capacitances. proportional to the permeability of the core material. in one coil due to a change in current in the other coil.

to any voltage due to self-induction. L = L 1 + L 2.

Lpa =

L1L2 + M 2 . L1 + L2 + 2M

Your Score A N SW E RS

1. False 6.

2. 7. False

3. 8.

4. False 9. False

5. False 10. False

Self- and Mutual Inductances

201

REVIEW QUESTIONS 1.

M L1 and L2 is

6. given by M = k L1L2

2. calculating the inductance.

L1 and L2 are placed

7.

3. 4.

k

them is M is measured.

combination.

5.

8.

coupling.

MULTIPLE CHOICE QUESTIONS coils are connected in series, the net inductance ( (c

. 1.

iple of statically induced emf is utilised in ( ) transformer ( ) motor (c) generator ( ) battery

7.

2.

be ( (c

( ) the coil resistance ( (c ( ) All of the above

8.

3. ( ) henry (c) Wb/A

( ) Vs/A ( ) All of the above

( ) 32 A/s (c) 2 A/s

( ( ) 12 A/s

( (c

4. 9.

is reduced to half and that of the other is doubled,

5.

( ) decrease (c) remains the same 6.

( ) increase ( ) Any of the above

( (c 10. coupled coils depends on ( ) the number of turns of the coils ( ) the cross-sectional area of their common core (c) the permeability of the core ( ) All of the above

202 11. inductance is L1 opposing self-inductance is L2 inductance M is given by ( ) L1 + L2 (b) L1 L2 (c L1 + L2) ( L 1 L 2)

is ( (c

b ) 1 L1 = L2 =

14. k turns ratio N1 N2 must be ( ) 1 (c

12.

(b) 2 ) 4

15.

b

( ) 8 2 (c

be ( (c

13.

b

A N SW E RS

1. 11.

2. c 12.

3. 13.

4. c 14.

5. 15.

6. b

8. b

7.

9. b

10.

PROBLEMS ( A )

S I M P L E

PR O BL EMS

1. Calculate the resistance and inductance of an aircored solenoid of mean diameter 1 cm and length

5. ) the mutual inductance of b) the emf induced in the second

1.72 ¥ Wm. [Ans. 2.75 W m 2. Calculate the inductance of an air-cored toroid having a mean diameter of 25 cm and circular

b

Ans. (

m

6.

2

m

Ans. 3. A toroid has a core of square cross-sectional area

m

2

Ans. 7.

[Ans. series combination.

4.

[Ans.

8. calculate ( ) the average emf induced in the second coil, (b

2 2

[Ans. [Ans. ( ) 2.25 V; (b

.

Self- and Mutual Inductances ( B )

TRI C KY

203

PR O BL EM S coil open, a current of 5 A in second coil produces L1, L2 and M. Ans.

9. An iron-core solenoid has a mean length of 2

14. Determine the self-induced emf in the solenoid, if the current through the coil changes from 2 A to Ans. 2 10. in

of the search coil is 25 m Ans.

m

15. magnetise in the same direction, and then in the

Ans. 11. Ans. 12.

Ans. 16. m the combination if (

Ans. 13.

Ans. ( ) ( C )

C HA L L EN G IN G

PROBLEMS 18.

17. established. When this current is reduced to zero in 2 ms, the voltage induced in another coil lying in

the self-inductance of each coil and their mutual changes from 6 A to and the number of turns in the second coil. Ans.

Ans.

DC TRANSIENTS

8

OB JE CT IV E S :

We have seen that both the inductance and capacitance are energy-storing elements. When a network contain-

state’. Now, if we switch off the source, or switch over the network to another source, the circuit starts attaining steady-state condition is called transient time. which contain only resistors and inductors or only resistors and capacitors.

the circuit, known as transient response. circuit analysis. Our interest lies mainly in the solutions themselves and their meaning and interpretation. answers for new circuits by just plain thinking. An engineer must always remember that these mathematical techniques are only tools with which meaningful and informative answers can be obtained; they do not constitute engineering in themselves. We shall begin our study of transient analysis by considering the simple RL circuit.

205

RL Consider the simple series RL circuit of Fig. 8.1a i(t), and we shall assume the value of i(t) = I0 at t = 0. You may wonder how there can be a current without any b, which has a special switch S, called make-before-break I0 the inductance L

I0 is not changing with time, there is no induced emf in I0 =

V0 R0

(8.1)

(t) = I0 at t = 0 For the circuit of Fig. 8.1a vR + vL = 0

or Ri + L

di =0 dt

or

di R + i =0 dt L

(8.2)

i(t value I0 at t

R di = - dt i L i (t )

Ú

I0

or

1 di = i

t

Ú 0

Ê Rˆ ÁË - L ˜¯ dt

or

ln i

i I0

(8.3)

=

R t L

t

or 0

i(t) = I0 e–Rt/L

ln i – ln I0 =

R (t – 0) L (8.4)

a. As the time passes, the current I0 to zero. At any instant, the power being dissipated in the resistor, 2

pR = i 2R = I0 Re –2Rt/L

206

Basic Electrical Engineering

∞

∞

WR =

∫ pR dt = ∫ e 2 I0 R

0

− 2 Rt / L

0

L ˆ - 2 Rt / L 2 Ê dt = I0 R Á e Ë 2R ˜¯

0

=

1 2 LI0 2

For the series RL assumed value I0 i(t)/I0 versus t, as shown in Fig. 8.2a. Since the function we are plotting is e –(R/L)t, the curve will not change if R/L for every series RL circuit having the same R/L or L/R ratio.

Suppose that we double the ratio L/R in a series RL e – (R/L)t

t is also doubled, then

that with larger L/R ratio, the current takes longer to decay to any fraction of its original value. We might L/R. t = 0 to time constant’ of the circuit as the time that would be required for the current to drop to zero if it continued to drop at its initial rate. t (tau).

207

R - Rt/L R d = (i/I 0 ) t = 0 = - e t =0 L L dt a), we have 1 1 R L tan q = fi = or t = L R Rt/L must be dimensionless, the ratio L/R time constant t, the response of the series RL circuit may be written simply as i(t) = I0 e–t/t

(8.5)

(8.6)

(8.7)

An important meaning of time constant t is obtained by determining the value of i(t)/I0 at t = t. We have i( ) = e–1 = 0.368 or i(t) = 0.368I0 I0 t can also be determined graphically from this fact, as indicated by Fig. 8.2b.

i(2t) = 0.368i(t) = 0.368 ¥ 0.368I0 = 0.1354I0 Suppose that we are asked, “How long does it take for the current to decay to zero this question, let us calculate the value of current i(t i(3t) = 0.0498I0, i(4t) = 0.0183I0, i(5t) = 0.0067I0, … We can now say that it takes about for the current to decay to zero. At the end of this time interval, the current is less than one percent of its original value.

L/R L allows greater energy storage for the same initial L/R by reducing R time is required to dissipate the stored energy. EX A MP L E

8.1

a t = 0. + (a) Determine the current i(0 ) = I0. (b) Find vR across 20-W resistor at the instant just after the switch is opened. (c) Find vL across the inductor immediately after the switch is opened.

Solution (a equivalent resistance faced by the 24-V source is R eq = 20 W + (20 W || 10 W) = 26.67 W

208

Basic Electrical Engineering

V 24 = = 0.9 A Req 26.67

I=

IL = 0.9 ¥

20 = 0.6 A 20 + 10

i(0+ ) = I0 = 0.6 A (b I0 20-W resistor is given as vR = (–I0)R = –0.6 ¥ 20 = – 12 V (c b) so as to oppose the decay in e = L

EX A MP L E

di dt

t =0

= L ¥ ⎛ I0 × R ⎞ = I0 R = 0.6 ¥ (20 + 10) = 18 V ⎝ L⎠

8.2

a are R = 0.8 W and L i = 20 A at t a) i(0), (b) the power being absorbed by the inductor at t = 1 s, and (c) the time at which the energy stored in the inductor is 100 J.

Solution t =

L 1.6 = R 0.8 i(t) = I0 e – t/t = I0 e –t/2

(i)

t = –1 s. Hence,

(a)

20 = 12.13 A 1.6487 e Since I0 is the value of current at t = 0, we have i(0) = I0 = 12.13 A. (b t = 1 s, 20 = I0 e–(–1)/2 or

I0 =

20

1/ 2

=

i(1) = I0 e –1/2 = 12.13e –0.5 = 7.36 A p(1) = i 2R = (7.36)2 ¥ 0.8 = 43.33 W absorbed by the inductor is – 43.33 W.

DC Transients

209

i(t

(c

W = i

1 L{i(t)}2 2 –t/2 e fi

1 ¥ 2 e

i(t)}2

t/2

fi i(t

t = 0.1632 s

Growth of Current in Series RL Circuit a

R

V

L

t i

t

– *

)

I = V/R. +

i

t

I = V/R.

A series RL circuit connected to a voltage source. t vL + vR = V

L

di + iR = V dt

i(t V L di +i = R dt R

L/R)

( L/R) ln( I 0 - i ) =t+k -1 k – (L/R I = k – (L/R I – i) = t – (L/R

\

⎡I − i⎤ ln ⎢ 0 ⎥ = ⎣ I0 ⎦ *

Rt L

di +i=I dt

L/R

( L/R) di = dt ( I 0 - i)

I – i) = t + k t

I

i L/R

I –i

I ]=t

⎡ I0 − i ⎤ –Rt/L ⎢ ⎥ =e I ⎣ 0 ⎦

i –

t t

i +)

210

Basic Electrical Engineering

fi

e–Rt/L) e–t/t

i =I i(t) = I time constant

t = L/R b

a t=t

i(t)/I i (t ) I0

i(t)/I

e

i(t

i(2t) = i(t

I – i(t

I

I

I

i(t i t

I ,i t

I

t

I i t I, for the current

it takes about

Rate of Growth of Current di dt

t =0

⎛ I ⎞ = − − 0 e − t /τ ⎝ τ ⎠

t =0

I0

=

V R V = R L L

=

L EXA M P L E

8. 3 W

a

b) Once

Solution

t = L/R I =

(a i =I

e

V 140 = R 10

–t/t

t

e

) = 3.479 A

(b i(t) = I e–t/t

e–t

t 1.4 EXA M P L E

e–t fi t = 0.7834 s

8. 4 W

a a

t

–

, (b

t

+

t , (c

tÆ

(d

t

211

DC Transients

Fig. 8.5 Solution t

(a

–

i –) = I

\ t

(b

+

V1 20 = = 0.25 A R 80 b

i +) = I

\

= 0.25 A

tÆ

(c

I (d

=

+ t (t = L/R

=

V1 + V2 20 + 40 = = 0.75 A R 80 I

W i

I

+ (I

–I

e ¥

e–t/t e ) = 0.682 A –2

General RL Circuit RL

RL

e –t/t

EXA M P L E

8. 5 a

(e) i2

t

a) iL

), (b) i2

), (c) iL +), (d) iL

¥

)

212

Basic Electrical Engineering

Solution W and 5 W in series. Hence, 20 iL(0 ) = I0 = = 0.667 A 25 + 5 20 (b R3. Hence, i2(0 –) = = 0.667 A. 30 (c) Since the current in an inductor cannot change instantaneously, (a

+ iL(0 ) = iL(0 ) = I0 = 0.667 A (d) At t = 0, the switch is opened so that the 20-V source is disconnected. Resistances R1 and R2 being in series, make a net resistance R12 = 20 + 40 = 60 W. Similarly, resistances R4 and R5 are in series and make a net resistance R45 = 25 + 5 = 30 W b. equivalent resistance R eq that is connected across the inductor terminals. 60 × 30 R eq = R 45 + (R12 || R3) = 30 + = 50 W 60 + 30 L 2 t= = = 40 ms Req 50

iL(t) = I0 e –t/t iL(20 ms) = I0 e–t/t = 0.667e– (20 ms)/(40 ms) = 0.667e–0.5 = 0.405 A (e

iL i2 (20 ms) = –iL(20 ms) ¥

R12 60 = – 0.405 ¥ = – 0.27 A R12 + R3 60 + 30

RC RC circuit has a greater practical importance than the RL between using a capacitor and using an inductor in designing an electronic circuit, he chooses an RC over an RL lower cost, smaller size, and lighter weight.

Consider a capacitor with an initial stored energy connected across a resistor through a switch, as shown in Fig. 8.7a V0 be closed at t stored energy keeps dissipating in the resistor. During this process, the voltage v across the capacitor keeps on decreasing.

iC + iR = 0 or

C

dv v =0 + dt R

dv v =0 + dt CR i is replaced by v, and L/R by CR. RC circuit, or

(8.11)

v = V0e–t/RC = V0e–t/t (8.12) where, t = RC is the time constant of the RC b. At t v = V0, which is correct initial condition. As t t, the slope of the curve at t = 0 is given as –V0/t.

RC R or C provide larger time constants and slower dissipation of the stored energy. A larger resistance will dissipate a smaller power with a given voltage across it, thus requiring a greater time to convert the stored energy. A larger capacitor stores a larger energy with a given voltage across it, again requiring a greater time to dissipate this initial energy.

Consider the circuit of Fig. 8.8a, in which a series combination of R and C is connected through a switch S across a voltage source of emf V0 v across it is zero. Now, let the switch be closed at t = 0. Since the voltage across a capacitor cannot change

214

Basic Electrical Engineering

instantaneously, we must have v(0+) = v(0–) = 0

(8.13)

entirely decided by V0 and R i(0+) = I0 =

V0 R

(8.14)

v across the capacitor goes on increasing. As v increases, the voltage vR (= V0 – v) decreases which results in a decrease in current i process continues, till the capacitor is fully charged so that the voltage v becomes the same as the source voltage V0 i v(t Æ ) = V0 and i(t Æ ) = 0 v and current i versus time t is shown in Fig. 8.8b.

Response of the Circuit vR + v = V0 or iR + v = V0 (8.15) Note that the current in the capacitor is the same as in R. Hence, the current i is related to the voltage v across the capacitor by the relation, i =C

dv dt

Substituting this value of i RC

dv dt

+ v = V0

dv RC

(V0

v)

= dt

(8.16)

215

RC ln (V0 - v) = t + k or –RC ln(V0 – v) = t + k -1 k, we apply initial condition, at t = 0, we must have v = 0, –RC lnV0 = k \ –RC ln(V0 – v) = t – RC lnV0 or –RC [ln (V0 – v) – lnV0] = t t ⎡V − v ⎤ ⎡ V0 − v ⎤ – t/RC or ln ⎢ 0 ⎥ = RC or ⎢ V ⎥ = e V ⎣ 0 ⎦ ⎣ 0 ⎦ fi v = V0 (1 – e–t/RC) v = V0 (1 – e–t/t) where, t = RC is the time constant of the circuit. i(t) = I0e–t/t where I0 Note that, as shown in Fig. 8.8b, in one time constant t and the current drops to 0.368 times the initial value I0. EX A MP L E

(8.17)

(8.18)

(8.19) V0

8.6

a, the single-pole double-throw switch S has been in position b for a long time so that the 5-mF capacitor is fully discharged. Now, at t = 0, the switch is thrown to position a. Determine (a) v(0+), (b) i(0+), (c) time constant t, (d) v and i at t = 15 ms.

Solution (a) Since the voltage across a capacitor cannot change instantaneously, we have v(0+) = v(0 ) = 0 V V0 3V = = 2 mA R 1.5 k (c t = RC = (1.5 kW)(5 mF) = 7.5 ms –t/t – (15 ms)/(7.5 ms) ) = 3(1 – 0.1353) = 2.594 V (d) At t = 15 ms : v = V0 (1 – e ) = 3(1 – e (b) i(0+) = I0 =

–t/t

i = I0 e

= (2 mA)e– (15 ms)/(7.5 ms) = 2 ¥ 0.135 = 0.27 mA

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Basic Electrical Engineering

EX A MP L E

8.7

b, the single-pole double-throw switch S has been in position a for a long time so that the 5-mF capacitor is fully charged. Now, at t = 0, the switch is thrown to position b. Determine (a) v(0+), (b) i(0+), (c) time constant t, (d) v and i at t = 1.2 ms.

Solution (a) Since the voltage across a capacitor cannot change instantaneously, we have +

–

v(0 ) = v(0 ) = V0 = 3 V +

(b) At t = 0 , the capacitor behaves as a voltage source of emf V0. Hence, V0 3V i(0+) = –I0 = =– = –30 mA R 100 (c t = RC = (100 W)(5 mF) = 0.5 ms (d) At t = 1.2 ms : v = V0 e–t/t = 3e – (1.2 ms)/(0.5 ms) = 3 ¥ 0.0907 = 0.2721 V i = –I0 e –t/t = – (30 mA)e – (1.2 ms)/(0.5 ms) = –30 ¥ 0.0907 mA = – 2.721 mA

RC RL RL or RC circuits with a single energy-storing element (C or L), all voltages and currents vary exponentially t = 0 s, are shown in Fig. 8.10. Here, the variable x represent either voltage v or current i a 8.10b Note that the curve in Fig. 8.10a can represent any of the following: 1. Decaying of current through an inductor. 2. Decrease of voltage across an inductor when current through it is decaying. 3. Decrease of voltage across an inductor when current through it is growing.

217 4. Decrease of voltage across a capacitor during its discharging. 5. Decrease of current through a capacitor during its discharging. 6. Decrease of current through a capacitor during its charging. Similarly, the curve in Fig. 8.10b can represent any of the following:

(t) xf in Fig. 8.10a and the initial value xi in Fig. 8.10b history of the energy-storing element, respectively. that they never actually reach them. However, after For a single-capacitor RC circuit, the time constant t = CR ; and for a single-inductor RL circuit, t = L/R . Here, R xi is the immediately after switching, and xf is the , then the variation of x with time t and depends upon time constant of the circuit and the difference (xf ~ x i ). For the curve of Fig. 8.10a, the x(t

x(t ) = xf + ( xi - xf )e - t / t

(8.20)

xf = 0. b x(t x(t) = x i + (xf – xi)(1 – e– t /t ) = x i + (xf – x i)(1 – e–t /t ) + xf – xf = xf + (x i – xf) – (x i – xf)(1 – e – t /t ) = x f + (x i – xf ){1 – (1 – e– t /t )} = xf + (xi – xf )e– t /t

initial t we mean the steady-state condition of the circuit, its state after a long period of time. would reach steady state.

Determining Initial Values fundamental principle is that the stored electrical energy in a capacitor and the stored magnetic energy in an inductor must be continuous. From continuity of energy, we conclude that the voltage across a capacitor

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Basic Electrical Engineering

inductor-current) before the switch is thrown and carry this value over to the moment immediately after the switch is thrown. capacitor acts as a battery only zero volt (i.e., a short circuit). Similarly, an energised inductor is modelled as a current source, as shown in zero ampere (i.e., an open circuit). + –

vC (0 )

–

vC (0 )

C

iL(0–)

+

iL(0–)

L

– t = 0–

t = 0+

t = 0–

t = 0+

Determining Final Values time derivatives must eventually vanish. As a result, the current through a capacitor and the voltage across an inductor must approach zero, as suggested in vC Æ constant fi iC = C

d vC Æ0 dt

as t Æ

iL Æ constant fi

diL Æ0 dt

as t Æ

vL = L

shown in Figs. 8.13 and 8.14, respectively. + vC (t)

vC ( )

C t

–

EX A MP L E

+ –

iL(t) iL( )

L t=

8.8 a, the switch is opened at t = 0 s. Determine the current i(t) for t > 0 s.

219 Solution of the capacitor is R

= (1 kW + 10 kW) || (1 kW) = 917 W t = R C = (917 W) ¥ (10 mF) = 9.17 ms b

switch, the voltage across the capacitor is determined by voltage division, –

vC (0 ) = (30 V) ¥

1 kW = 15 V 1 kW + 1 kW

At the instant of opening the switch, the capacitor acts as a battery of 15 V, as shown in Fig. 8.15c +

30 – i(0 )(1 kW) – 15 = 0

i(t)

+

fi i(0 ) = 15 mA

i( ) is easily derived as 30 V i( ) = = 2.5 mA (1 + 1 + 10) kW

i(t) = i( ) + [i(0+) – i( )]e–t/t = 2.5 + (15 – 2.5)e–t/9.17 ms mA

RC An RC timer, the circuit that measures time. A simple timer consists of a series combination of a switch, a capacitor, a resistor, and a dc voltage source. At the beginning of the time interval to be measured, the switch is closed to start the charging of the capacitor. At the end of a measure of the time interval. A voltmeter connected across the capacitor can have a scale calibrated in time to give direct reading of the time interval elapsed. As can be seen from Fig. 8.10b, the capacitor voltage changes almost linearly, for times much less than constant or less.

RC

RL

RC and RL circuits, it is possible to make RL practically, RC timers are considered to be a better choice due to the following reasons : negligible. techniques.

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Basic Electrical Engineering

ADDITIONAL SOLVED EXAMPLES EX A MP L E

8.9

(a) the current at t = 1 s, (b of current.

W and its time constant is 1.8 s. At t = 0, a 10-V source is connected to it. Determine c) the initial rate of growth

Solution Since t = L/R, L = tR = 1.8 ¥ 3 = 5.4 H. Since the current cannot change abruptly in an inductor, we have +

i(0 ) = i(0

–

i( ) = I0 =

V0 10 V = = 3.33 A R 3 +

i(t) = i( ) + [i(0 ) – i( )]e

–t/t

= 3.33 + [0 – 3.33]e

–t/1.8

= 3.33(1 – e

–t/1.8

) mA

t = 1 s is given as i(1 s) = I0 (1 – e–t/t) = (3.33 mA)(1 – e–1/1.8) = 1.419 A

(a (b

I0 = I0 (1 – e –t/t) 2

fi

e –t/1.8 = 0.5

fi t = 1.25 s

(c di dt

EX A MP L E

= t =0

V0 10 = = 1.85 A/s L 5.4

8.1 0

When a dc voltage is applied to a series RL circuit, after a lapse of 1 second the current reaches a value 0.741 times dc voltage source is removed and the terminals of the circuit are shorted. What would be the value of the current after 1 s of this sudden change ?

Solution

I0

t = 1 s, we have i = I0 (1 – e –1/t) or 0.741I0 = I0 (1 – e –1/t) During the decay of current, we have i(t) = I0 e–t/t t = 1 s, we have i1 = I0 e –1/t = 0.259I0 0.259 times the steady-state value. EX A MP L E

fi e –1/t = 0.259

8.1 1

At t = 0, suddenly a dc voltage source of 120 V is applied to a series RL circuit having R = 20 W and L = 8 H. Determine (a) the current in the circuit at t = 0.6 s, and (b) the time at which the voltage drops across R and L are same. L 8 = = 0.4 s R 20 V 120 I0 = 0 = =6A R 20 (a) At t = 0.6 s, the current is given as

Solution

t=

i(0.6 s) = I0 (1 – e –t/t) = 6(1 – e–0.6/0.4) = 4.66 A

221 R, vR = iR = I0 (1 – e –t/t)R LI di LI R L, vL = L = 0 e –t/t = 0 e–t/t = I0 e –t/tR

(b

dt

vR = vL or I0 (1 – e –t/t)R = I0 e–t/t R fi t = 0.693t = 0.693 ¥ 0.4 = 0.277 s

\ EX A MP L E

L

e–t/t = 0.5

8.1 2

At t = 0, suddenly a dc supply of 30 V is applied to a series RL circuit having R = 12 W and L = 18 H. Determine (a) the time constant of the circuit, (b) the initial rate of change of current, (c) the current at t = 3 s, (d) the energy t = 3 s, and (e) the energy lost as heat till t = 3 s.

Solution (a

t=

L 18 = = 1.5 s R 12

(b di dt

t = 3 s is given as

(c

= t =0

V0 30 = = 1.67 A/s L 18

V0 30 (1 – e –t/t) = (1 – e –3/1.5) = 2.16 A R 12 t = 3 s is given as 1 2 1 W1 = Li1 = ¥ 18 ¥ (2.16)2 ª 42 J 2 2 i1 = I0 (1 – e –t/t) =

(d

t

(e energy lost,

2

Wlost = W0 – W1 = EX A MP L E

1 2 1 ⎛ 30 ⎞ LI 0 – 42 = ¥ 18 ¥ – 42 = 56.25 – 42 = 14.25 J ⎝ 12 ⎠ 2 2

8.1 3

a has been in the condition shown for a long time. At t = 0, the switch is opened. Determine (a) i(0+), (b) v(0+), (c) vL(0+), (d) i and v at t = 20 ms, and (e) i and v at t = 50 ms.

Solution (a

a is given in Fig. 8.16b. Since the current in an inductance cannot change instantaneously, we must have i(0+) = i(0– ). When the switch had been closed for long, a constant current I0

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Basic Electrical Engineering

other parallel branches of 200 W and 17 W + – i(0 ) = i(0 ) = I0 = 20 mA W v(0+) = iR = (20 mA) ¥ (17 W) = 0.340 V (c) After the switch is opened, the current I0 (= 20 mA) in the inductor starts decaying through the 200-W resistance, so that i(t) = I0 e –t/t, with L 5 mH time constant, t = = = 25 ms 200 W R LI (5 mH) (20 mA) di ⎛ 1⎞ + \ vL(0 ) = - L = –LI0 − = 0 = =4V ⎝ τ⎠ dt t = 0 25 s

(b

(d) At t = 20 ms, the value of current i is given as i(t) = I0 e–t/t = (20 mA)e –20 ms/25 ms = 20 ¥ 0.449 = 8.98 mA W resistance, the voltage v(t) = v(0+) = 0.340 V (e) At t = 50 ms, the value of current i is given as i(t) = I0 e –t/t = (20 mA)e –50 ms/25 ms = 20 ¥ 0.1353 = 2.706 mA v(t) = v(0+) = 0.340 V

and EX A MP L E

8.1 4 a, the switch is closed at t = 0. Find these currents at t = 5 ms : (a) iL, (b) ix, and (c) iy. 800 W

800 W 40 W

120 mA

t=0

iL

ix 200 W

40 W

0.8 H

iy

200 W

120 mA iL

iy

(a)

(b)

Solution For t iL(0– ) = IL0 = (120 mA) ¥

200 = 100 mA 200 + 40

(a) When switch is closed (Fig. 8.17b), the current through the inductor cannot change instantaneously. Hence, iL(0+) = iL(0– ) = 100 mA. For t equivalent resistance given as R eq = 40 + (800 || 200) = 200 W t = L/R eq t = 5 ms, the current in the inductance branch is given as iL(at t = 5 ms) = IL0 e –t/t = (100 mA)e– (5 ms)/(4 ms) = 28.7 mA

W resistance branch. W is due to the current iL (acting as a source current). Hence, by current division, 800 i (at t = 5 ms) = –iL = – (28.7 mA) ¥ 0.8 = –22.9 mA 800 + 200 iL and current source of 120 mA. Hence, the principle

(b

(c of superposition gives

iy (at t = 5 ms) = iy1 + iy2 = –iL ¥

200 + (120 mA) 200 + 800

= –(28.7 mA) ¥ 0.2 + (120 mA) = 114.26 mA EXAMP LE

8.15

After being closed for a very long time, the switch in the circuit of Fig. 8.18a is opened at t = 0. Find (a) iL(0+) and w L(0+), (b) iL and v10 at t = 1 s.

Solution (a) When the switch had remained closed, in the steady-state condition, the inductance behaved as a short circuit and the current IL0 superposition, we get 12 V IL0 = IL1 + IL2 = +2A=3A+2A=5A 4 Since the current in inductance cannot change instantaneously, we have iL(0+) = iL(0– ) = I = 5 A 1 2 1 LI L 0 = ¥ 4 ¥ 52 = 50 J 2 2 (b) After the switch is opened (Fig. 8.18b), the time constant for the decaying current, L 4 = = 0.4 s t = R 10 At t = 1 s, the current in the inductance is given as w L (0+) =

–t/t

–1/0.4

= 5e = 0.41 A iL (t) = IL0 e At t = 1 s, the voltage across 10-W resistance is given as v10 (t) = –iL (t)R = –0.41 ¥ 10 = – 4.1 V EXA M P L E

8 . 1 6

Determine the energy stored in the inductor of the circuit shown in Fig. 8.19 after 20 ms of throwing the switch.

224

Basic Electrical Engineering

Solution

IL0 = 5 mA. After the switch is thrown, the current starts decaying,

with time constant, L 5 mH = = 25 ms R 200 At t = 20 ms, the current in the inductance is given as t =

iL (t) = I L0 e–t/t = (5 mA)e– (20 ms/25 ms) = 2.25 mA Hence, the energy stored in the inductor, 1 1 2 w(t) = Li L2 = ¥ (5 mH) ¥ (2.25 mA) = 12.65 nJ 2 2

EX A MP L E

8.1 7

is closed at t = 0. Determine (a) the current in the circuit at t = 0.6 s, and (b) the time at which the voltage drops across R and L are the same.

Solution

V 120 = =6A R 20 L 8 t= = = 0.4 s R 20 t > 0, the current in the circuit is given as i(t) = iL = I0 (1 – e –t/t) = 6(1 – e –t/0.4) A I0 =

t = 0.6 s is given as

(a

–t/0.4 ) = 6(1 – e –0.6/0.4) = 4.66 A i(0.6 s) = 6(1 – e (b R at any time is given as vR(t) = iR = 6(1 – e –t/0.4)20 = 120(1 – e –t/0.4) Whereas the voltage across L at any time is given as 1 –t/0.4 di d [6 (1 e t / 0.4 )] vL(t) = L =8 =8¥6¥ e = 120e –t/0.4 dt 0.4 dt Applying the given condition, we get 120(1 – e –t/0.4) = 120e –t/0.4 or t = 0.277 s

EXA MP L E

8. 18

Determine i at t = 2, 0 , 0+, 2, and 4 ms, in the circuit of Fig. 8.21a.

225

Solution current IL0 can be determined by current division, IL0 = (10 mA) ¥

80 = 8 mA 80 + 20

t i (–2 ms) = iL(0 – ) = IL0 = 8 mA iL(0+) = iL(0– ) = IL0 = 8 mA t > 0, the equivalent circuit is shown in Fig. 8.21b. At t = 0+, by current division, we have 30 30 = (8 mA) = 4.8 mA 30 + 20 30 + 20 iL through the inductor starts decaying through the equivalent resistance of the parallel combination of 20-W +

+

i (0 ) = iL(0 )

and 30-W, R eq = (20 || 30) = 12 W

and

t=

So that, for t > 0, the current iL is given by iL(t) = IL0 e –t/t = (8 mA)e –t/(2.08 ms)

L 25 mH = = 2.08 ms R 12

iL(2 ms) = (8 mA)e–(2 ms)/(2.08 ms) = 3.05 mA iL(4 ms) = (8 mA)e –(4 ms)/(2.08 ms) = 1.17 mA

and

30 = 1.83 mA 30 + 20 30 i (4 ms) = (1.17 mA) = 0.702 mA 30 + 20 i (2 ms) = (3.05 mA)

and EX A MP L E

8.1 9

Determine v(0+) and i(0+) for the circuit of Fig. 8.22.

Solution

t < 0), the capacitor has been fully charged to the source v(0– ) = 1.5 V t = 0+ ), the voltage across the capacitor remains the same, as it cannot change instantaneously. Hence, v(0+) = V0 = v(0– ) = 1.5 V Now, the capacitor starts discharging through the 5-mW resistance. Hence, we have

226

Basic Electrical Engineering +

i

) =I =

V0 1.5 V = R 5m

300 A

Fig. 8.22 EX A MP L E v(t

Fig. 8.23

8.2 0 i(t

+

t

Solution t v –) = V

¥

(3 + 2 ) (3 + 2 ) + 1

–

t

, the current i V 6 i –) = = = 1 mA R (1 + 3 + 2) kΩ v +) = V = v – ) = 5 V W

W

W W

R +

t

V 5V i +) = I = C = Req 10 k +

t

t = RC

W) ¥

m

, the current i = 0.5 mA

i v(t) = V e –t/t

v

i(t) = I e –t/t

t v

e

v

e

i

= 1.834 V

e

= 0.1839 mA

t = 0.677 V

i

e

= 0.0677 mA

DC Transients EXA M P L E

8. 2 1

v +

i +

a

Fig. 8.24 Solution V I

¥

v –) = V = V t

+

1000 1000 + (200 + 800) =I

¥

W

¥

W

b v +) = v – ) = V = 4 V c. 4V 4V i C +) = = (200 W || 800 W) 160 i i +

¥

800 = 20 mA 800 + 200

,

t 4V i +) = 200

EXA M P L E

+

, the current i

20 mA

8. 2 2 a

t

t

Fig. 8.25

a) vC, (b) vR, (c) v

W

227

228

Basic Electrical Engineering

Solution voltage division,

20 = 10 V 20 + 4 Since the voltage across the capacitor cannot change instantaneously, we have vC (0– ) = vR (0– ) = 12 ¥ +

–

vC (0 ) = vC (0 ) = V0 = 10 V + At t = 0 , the equivalent circuit of the given circuit is given in Fig. 8.25b. equivalent resistance, –3 R eq = 5 + 20 = 25 W with t = RC = 25 ¥ (50 ¥ 10 ) = 1.25 s + –t/t For t > 0 , vC decays with time as vC (t) = V0 e t = 1 s, we have (a) vC (1 s) = V0 e–t/t = 10e –1/1.25 = 4.49 V (b vR(1 s) = vC ¥

20 20 = 4.49 ¥ = 3.59 V 20 + 5 20 + 5 (c) From Fig. 8.16a, the voltage across the open switch is given as vSW (1 s) = 12 – 3.59 = 8.41 V

SUMMARY TE RM S

A N D

C ON CE PT S

time constant (t time-constant RL circuit is given as t = L/R. Charging of a capacitor is the process of increasing the charge held in a capacitor. Discharging of a capacitor is the process of reducing the charge held in a capacitor. RC circuit is given as t = RC. i (in RL circuit) and the voltage vC (in an RC circuit) to reach its energised inductor acts as a current source. A special case is an unenergised inductor, which acts as an open circuit. inductor acts as a short circuit. energised capacitor acts as a voltage source. A special case is an unenergised capacitor, which acts as a short circuit. capacitor acts as an open circuit. IMP O R TA N T

F O RM U LAE

Current growth in an RL circuit, i(t) = I0 (1 – e –Rt/L), where I0 = V0/R. Current decay in an RL circuit, i(t) = I0 e–t/t. Charging of a capacitor, v(t) = V0 (1 – e–t/t) and i(t) = I0 e –t/t, where I0 = V0 /R. Discharging of a capacitor, v(t) = V0 e–t/t and i(t) = I0 e –t/t, where I0 = V0/R. RC and RL circuits, v(t) = v( ) + [v(0+) – v( )]e–t/t V and i(t) = i( ) + [i(0+) – i( )]e–t/t A.

229

CHECK YOUR UNDERSTANDING two marks for each correct answer and minus one

12

S. No.

Statement

True

1.

When a series RC circuit is connected to a dc voltage, the circuit

2.

When a resistance R is connected to a dc voltage V0, the current instantaneously rises to the value V0/R. When a series RL circuit is switched to a dc voltage V0, the current at the instant of switching is zero. When the value of L is in henrys and of R is in ohms in a series RL circuit, its time constant is R/L seconds. When a series RL circuit is switched on to a dc voltage, the initial rate of change of current in the circuit is minimum at t = 0. When a series RC circuit is connected to a dc voltage, the charging

3. 4. 5. 6.

False

Marks

When a series RL circuit carrying current I0 is suddenly open-circuited, the current at t = 0+ is zero. When a series RC circuit is switched on to a dc voltage, the current is zero at t = 0+. During discharging of a capacitor through a resistor, the voltages across R and C are equal. RL circuit cannot change instantaneously.

7. 8. 9.

Your Score A N SW E RS

1. False 6.

2. 7. False

4. False 9.

3. 8. False

5. False 10.

REVIEW QUESTIONS (d) Does the supply voltage affect the time taken

1. With reference to the rise of current in a series RL circuit, answer the following :

RL circuit and

(a) What prevents the current from rising to its

2.

(b

3. With reference to the energy stored in the inductor, RL circuit increases (a) by increasing the value of inductance, and (b) by decreasing the value of resistance.

current ? (c) What are the factors that decide how fast the

230

Basic Electrical Engineering RL

4.

(c

(a (d (b 7. (c i

ii) the

RC

a b

a

5.

RC

8. b

RC

6.

V

9.

t 10.

(a t

+

(b

MULTIPLE CHOICE QUESTIONS b d

(a (c RC

4. RL

1.

E L

(a (c

(b (d

E

RL

2.

R L V , the

V b) 0 R

V0 L R (c) L

(a)

V, b (d) V /C

(a (c) V /R RC

5. V

b (d) V /C

(a (c) V /R 6. RL t

V

d (a) t (c) t

3.

L/R

b) t = L/R (d) t = A N SW E RS

1. c

2. a

3. b

4. c

5. a

6. d

PROBLEMS ( A )

S I M P L E

P RO B LEMS

RL a) i(2t)/i(0), (b) i(4t)/i(2t), (c) t/t if i(t) = 0.1i(t), and (d) t/t if i(t) – i(t) = 0.1i(0). [Ans. (a) 0.1353; (b) 0.1353; (c) 3.30; (d) 0.759] 2. An inductor coil has an inductance of 15 H and a resistance of 10 W supply of 20 V. Calculate (a) the time constant of the circuit, (b) the initial rate of change of current, (c) the current after 2 seconds, (d) the rate of change of current after 2 seconds, and (e) the energy stored 1.

3.

4.

5.

6.

( B)

[Ans. (a) 1.5 s; (b) 1.33 A/s; (c) 1.47 A; (d) 0.351 A/s; (e) 16.2 J] A circuit having a 100-mH inductance in series with a 500-W resistance is connected to a 50-V dc source. Determine the instantaneous values of circuit currents during the period from 0 to 1 ms at an interval of 0.2 ms. [Ans. 0, 63.2 mA, 86.5 mA, 95 mA, 98.2 mA, 99.3 mA] Determine the current in a series RL circuit having R = 3 W and L = 12 H, at t = 6 s after a dc supply of 120 V is connected to it. [Ans. 31.07 A] A series RC circuit has a resistance of 90 kW and a capacitance of 25 m 160-V battery at t = 0. Calculate (a) the time constant of the circuit, (b) i(t) at t = 0+, 5 s, and . [Ans. (a) 2.25 s; (b) 1.778 mA, 0.193 mA, 0] A circuit having a 120-mF capacitor and a 10-kW resistance in series is connected across a 230-V dc supply. Determine (a) the initial rate of rise of voltage across the capacitor, (b) the initial charging

TRI C KY

7.

8.

9.

10.

11.

current, (c and (d) the ultimate energy stored in the capacitor. [Ans. (a) 191.67 V/s; (b) 23 mA; (c) 0.0276 C; (d) 3.174 J] An 8-mF capacitor is connected in series with a 0.5-MW resistor across a 200-V dc supply. Calculate (a) the time constant, (b) the initial charging current, (c) the time taken for the potential difference across the capacitor to grow to 160 V, and (d) the current and pd across the capacitor, 4 s after it is connected to the supply. [Ans. (a) 4 s; (b) 400 mA; (c) 6.44 s; (d) 147 mA, 126.4 V] A 20-mF capacitor is charged at a constant current of 5 m capacitor and the corresponding charge on it. [Ans. 150 V, 3 mC] A 2-mF capacitor is connected in series with a 1-kW resistor. Calculate the values of charging current and capacitor voltage at t = 0, 2, 4, 6, and 8 ms from the instant the circuit is switched on to a 10-V dc supply. [Ans. 10, 3.68, 1.35, 0.5, 0.18 mA; 0, 6.32, 8.65, 9.5, 9.82 V] A 1-mF capacitor has been charged to 100 V. Calculate the value of the resistance that must be connected across this capacitor for discharging it so that the initial discharge current is limited to 1 mA. [Ans. 100 kW] A 16-mF capacitor is charged to 30 V. How long will it take for its voltage to fall from 8 V to 5 V, if it is discharged through a 1.5-kW resistor ? [Ans. 11.28 ms]

P RO B LE MS

12. A capacitor of 10 mF is connected to a dc supply through a resistance of 1.1 MW. Calculate the time charge. [Ans. 25.33 s] 13. A 3.5-mF capacitor in series with a resistance R is connected through a switch across a 230-V

dc supply. A voltmeter is connected across the capacitor. What should be the value of R so that the voltmeter reading is 165 V, 5.65 s after the switch is closed ? [Ans. 1.281 MW] 14. A 2-mF capacitor and a resistance R are connected in series across a 150-V dc supply. A neon lamp

Basic Electrical Engineering which strikes (becomes on) at 90 V is connected across the capacitor. (a) What should be the value of R so that the lamp strikes 5 seconds after the switch is closed ? (b R = 1 MW, how much time will the lamp take to strike ? [Ans. (a) 2.73 MW; (b) 1.83 s] 15. A 10-mF capacitor charged to 40 V is discharged through a 0.2-MW resistance. Find (a) the time constant of the circuit, (b) the time taken by the

( C)

C H A LL ENG I NG

and (c) the total energy dissipated in the resistor during the transient period, and (d) the total energy initially stored in the capacitor. [Ans. (a) 2 s; (b) 1.4 s; (c) 8 mJ; (d) 8 mJ] 16. A 12-mF capacitor is allowed to discharge through its own leakage resistance. Using an electronic voltmeter, it is observed that the voltage across the capacitor falls from 120 V to 100 V in 300 s. Calculate the leakage resistance of the capacitor. [Ans. 137 MW]

P ROBLEMS

17. For the circuit of Fig. 8.26, determine the steadystate current when the switch S is open. Calculate the value of current at t = 0.8 s after the switch is closed. [Ans. 1 A, 1.465 A]

18. to position a when t = 0 and then moved to position b when t = 20 ms. Determine the voltage across the capacitor when t = 30 ms. [Ans. 1.14 V]

ALTERNATING VOLTAGE AND CURRENT

9

OB JE CT IV E S

(a)

( )

Today, the vast majority of electrical power is generated, distributed, and consumed in the form of ac power. The term ‘ac’ means ‘alternating current’. Such a current reverses its direction periodically. In ac power, the voltages and currents vary with time sinusoidally. Such a variation is shown graphically in Fig. 9.1, and is mathematically represented as i = Im sin wt (9.1) You may wonder why we should select only the sinusoidal variation. There are many technical and economical reasons for doing this. (i) A sinusoidal wave, even after repeated differentiation or integration, remains a sinusoidal of the same frequency. Furthermore, the sum or difference of a number of sinusoids of same frequency but of different amplitudes and phase angles is a sinusoid of the same frequency. It is because of this fact that when a sinusoidal voltage is applied to a circuit containing linear passive elements, all currents and voltages in the circuit are also sinusoidal. (ii is capable of doing work. Majority of electric motors used in commercial and industrial applications work on this principle. (iii) Use of sinusoidal voltages underlines the operation of transformers which enable bulk power transmission at high voltages over a long distance.

234 i

i

Im

Im

i 0 q

90° p/2

p

180° 270° 3p/2

360° 2p wt

i 0

t

3T/4 T/4

T/2

T t

–Im

–Im One cycle (a) Current i versus angle w t.

One period (b) Current i versus time t.

or a transformer, the voltage across the element is also sinusoidal. This is not true for any other waveform. If an electric company generates a sinusoidal waveform of voltage, the waveforms for all its customers are sinusoids too.

sinusoidal waveform, shown in Fig. 9.1.

(1) Cycle The values of sine wave repeat after every 2p radians. One complete set of positive and negative values of the function (which goes on repeating) is called a cycle. As seen in Fig. 9.1, in one cycle the value of current i increases from zero to a maximum value, decreases through zero to a negative maximum value, and then again increases towards zero. (2) Maximum (or Peak Value) It is the maximum value (denoted as Im), positive or negative, of the quantity. It is also sometimes called the amplitude of the sinusoid. (3)

It is the value of the quantity at any instant.

(4) Time Period (or Periodic Time) It is the duration of time required for the quantity to complete one cycle. It is denoted as T.

(5)

It is the number of cycles that occur in one second. It is denoted as f. The unit of frequency is hertz (Hz) which is same as cycles/second. Obviously, frequency is the reciprocal of time period, 1 f = (9.2) T The frequency of ac supply in India is 50 Hz, which corresponds to time period, T = 1/50 = 0.02 s or 20 ms. In USA, the frequency of ac supply is 60 Hz. AC supply of 400 Hz is used in airborne and some naval applications*. *

Because motors and transformers are smaller and lighter at higher frequency.

Alternating Voltage and Current

235

(6)

The values of a sine function repeat after every 2p radians. Angular frequency, denoted as w, is equal to the number of radians covered in one second. Its unit is rad/s. Since one cycle covers 2p radians and there are f cycles in one second, the angular frequency is given as w = 2pf or w =

(7)

2p T

(9.3)

It is one-half of the cycle, when it includes either all positive or all negative values.

(8) Phase It is the fraction of the time-period or cycle that has elapsed since it last passed from the chosen zero position or origin. The phase at time t from the chosen origin is given by t/T.

(9)

It is the equivalent of phase expressed in radians or degrees. It is denoted as q. Thus, phase angle, q = 2pt/T. As can be seen in Fig. 9.1a, the maximum value of sinusoidal current occurs at a phase angle of p/2 radians or 90°.

In earlier classes, we were introduced to sines and cosines through the study of a right-angle triangle. The sine of the angle (between the base and hypotenuse) equals to the ratio of the perpendicular to the hypotenuse; the cosine is the ratio of the base to the hypotenuse. There is another way to learn about these functions. You can get a sinusoidal waveform for an ac voltage by imagining a crank (a bar or a rod) AB, as shown in Fig. 9.2a. Let this crank rotate counterclockwise about point A with a uniform angular speed w. It describes an angle q = wt in time t, starting from the instant when the crank was horizontal. The vertical projection of the crank is its length times the sine of the angle q. As the crank rotates, the vertical projection generates a sine wave function of time (see Fig. 9.2b). If the length of the crank is Vm, the sinusoidal waveform generated can be expressed as v = Vm sin wt

Mathematically, this waveform can be described as a sine function or a cosine function, but we simply call it a sinusoid, or sinusoidal waveform. The term ‘sinusoid’ merely denotes the type of variation. We have adopted the sine function as the standard mathematical form for a sinusoidal waveform.

236

If the rotating crank had an initial (i.e., at time t = 0) counterclockwise displacement of f degrees, the resulting waveform would be as shown in Fig. 9.3a. Then the initial instantaneous voltage (at time t = 0) becomes V0 = Vm sin f. This shift of the waveform on the time axis or angle axis is called phase shift and the angle f is called phase angle. The phase is related to the time origin when we use a mathematical description. However, in an ac circuit, what matters are the phase differences of the various sinusoidal voltages and currents. We can write a general expression for a sinusoidal voltage by incorporating the phase shift, as v = Vm sin (wt + f) (9.4) If f is a positive number, the waveform is shifted towards left, and f is known as the angle of lead as shown in Fig. 9.3a. On the other hand, if f is a negative number, the waveform is shifted towards right, and f is known as the angle of lag as shown in Fig. 9.3b. In Fig. 9.3c, the angle of lead is 90°, and Eq. 9.4 becomes v = Vm sin (wt + 90°) = Vm cos wt The waveform in Fig. 9.3c is also seen to be a cosine wave. In Fig. 9.3d, the angle of lead is 180°, and Eq. 9.4 becomes v = Vm sin (wt + 180°) = –Vm sin wt Also, as it should be, the waveform in Fig. 9.3d is negative of sine wave.

Alternating Voltage and Current

237

Note that the resulting position of the waveform is the same whether the phase shift is 180° (p rad) to the left, or 180° to the right. It means that a phase lead of 180° is the same as the phase lag of 180°. This is so because the sinusoids are generated by a rotating crank. This concept can be extended further. A phase lead of 90° (as in Fig. 9.3c) and a phase lag of 270° (= 360° – 90°) is one and the same thing. N O T E There are some traditional inconsistencies in electrical engineering which one must tolerate. The mathematical unit for wt is radians and hence the correct unit for phase should also be radians. However, it has become normal practice in electrical engineering to express wt in radians and the angle f in degrees. EX A MP L E

9.1

A sinusoidal voltage is given by v = 20sin wt volts. (a) At what angle will the instantaneous value of voltage be 10 V ? (b) What is the maximum value for the voltage and at what angles ?

Solution (a) The angle can be determined by using the given equation, –1 10 = 20sin wt fi q = wt = sin (10/20) = 30°

(b) The maximum value is Vm = 20 V. This occurs twice in one cycle when sin wt = ±1, that is at the instants when wt = 90° or 270°. E X A M P L E

9 .2

The equation for an ac voltage is given as v = 0.04 sin(2000t + 60°) V. Determine the frequency, the angular frequency, and the instantaneous voltage when t = 160 ms. What is the time represented by a 60° phase angle ?

Solution Comparing the given equation with the general equation given in Eq. 9.3, we get 2000 w w = 2pf = 2000 rad/s and f= = = 318.3 Hz 2p 2p The instantaneous value of the voltage at t = 160 ms is v = 0.04 sin (2000 ¥ 160 ¥ 10

–6

rad + 60°) V = 0.04 sin (0.32 rad + 60°) V

180° ⎞ = 0.04 sin ⎛ 0.32 × + 60° V = 0.04 sin (18.3° + 60°) V = 0.0392 V = 39.2 mV ⎝ ⎠ π 1 1 Time period, T = = = 3.14 ms f 318.3 Thus, full-cycle of 360° corresponds to 3.14 ms. Therefore, the angle 60° corresponds to 60 t= ¥ 3.14 ms = 0.52 ms 360 EXA MP L E

9 . 3

A sinusoidal voltage is 20 V peak-to-peak, has a time of 5 ms between consecutive peak and trough, and at t = 0 is –3.6 V and decreasing. Find the equation for the instantaneous value of the voltage, and the value of the voltage at t = 12 ms.

238

Solution Figure 9.4 shows the sinusoid. The maximum value or peak value is half of the peak-to-peak value. The period is the time between the two consecutive peaks. Thus, 1 1 1 = = = 100 Hz 10 ms 10 × 10 − 3 T The angular frequency is w = 2pf = 2p ¥ 100 = 628.3 rad/s. Therefore, the sinusoid voltage is of the form Vm = 20/2 = 10 V; T = 2 ¥ 5 ms = 10 ms;

\ f=

v = 10 sin(628.3t + f) Now, we are given that at t = 0, v = –3.6 V. Putting these values, we get –3.6 = 10sin (f) or sin f = – 0.36

fi f = –158.9°

or 338.9°

As can be seen from Fig. 9.4, the given sinusoid is shifted to right by angle f, which is less than 180°. Also, we know that shifting rightward means there is a lag of angle f. Therefore, the equation of the given sinusoidal voltage is v = 10 sin (628.3t – 158.9°) V The value of voltage at 12 ms would be

Ê Ë

v = 10 sin Á 628.3 ¥ 0.012 ¥

180∞ ˆ - 158.9∞˜ V ¯ p

= 10 sin (432° – 158.9°) V = 10 sin(273.1°) V = – 9.985 V E X A MP L E

9. 4

An alternating current of frequency 60 Hz has a maximum value of 12 A. (a) Write down the equation for its instantaneous value. (b) Find the value of current after 1/360 second. (c) Find the time taken to reach 9.6 A for the

Solution (a) The angular frequency, w = 2pf = 2p ¥ 60 = 377 rad/s. Therefore, the equation for the instantaneous value is given as i = 12sin 377t A (i) (b) The value of current at t = 1/360 second is 1 180° ⎞ ⎛ i = 12 sin 377 × × A = 10.4 A ⎝ 360 π ⎠ (c) Putting i = 9.6 A in Eq. (i), we have 180° 180° ⎞ 9.6 = 12 sin ⎛ 377t × fi 377t ¥ = 53.13° ⎝ ⎠ π π 5313 . π = 2.46 ms \ t = 377 × 180

Alternating Voltage and Current

239

An average value a continuous variation of the value of a quantity with time t (or angle q), repeated after each cycle. The area under the waveform is found by integration and full cycle is normally taken as 2p radians or T seconds. Thus, the average value Vav of the instantaneous voltage v, taken over full cycle is given as 2π

Vav

Area under full cycle = = Length of one cycle

Vav =

or

1 T

∫ v dθ 0

2π

=

1 2π

2π

∫

v dθ =

0

1 2p

2p

Ú v d (w t )

(9.5)

0

T

(9.6)

v dt 0

Consider the full cycle of a sinusoidal wave shown in Fig. 9.1. It has same area of the positive and negative loops. Hence, the algebraic sum of the two areas is zero. Accordingly, the average value of a sine wave over a full cycle is zero. This average value does not convey any useful information about the waveform. Therefore, average value of a symmetric periodic waveform, it is understood to be determined for only half-cycle, either positive or negative. Thus, the average value of a sinusoidal current i, depicted in Fig. 9.5a, is determined as follows. Iav

Area under half cycle 1 = = Length of half cylcle p =

p

Ú 0

1 i d (w t ) = π

π

∫I

m sin

wt d(wt)

0

Im I 2I π [ − cos ω t ]0 = m [–cos p + cos 0°] = m = 0.637Im p π

(9.7)

b) is same as above. For this waveform, one cycle is from 0 to p, as it repeats itself after p. Figure 9.5c shows the current at the p, and remains zero from p to 2p. Therefore, the average value of this current is Iav

1 = 2p I = m 2π

2p

p 1 ÈÍ i d (w t ) = I m sin w t d (w t ) + 2p Í 0 Î0 I π [ − cos ω t ]0 = m = 0.318Im

Ú

Ú

i

i

i

Im

Im

Im

0

p

2p

3p wt

(a) Sinusoidal ac current.

0

p

2p

3p

wt

(b) Full-wave rectifier output.

0

2p

Ú p

˘ 0 d (w t ) ˙ ˙ ˚ (9.8)

p

2p

3p

wt

(c) Half-wave rectifier output.

240

i = Im sin wt

Let an ac sinusoidal current I which gives R (Fig. 9.6b). This would be the

R (Fig. 9.6a

effective value of the ac sinusoidal current. i

I

R

R

(a)

(b)

i Im 0

p 2

3p 2

2p

p

wt

(c)

p

a 2

Im R 0

( )

p 2

Pav p

(a) (d)

3p 2

2p

(d)

wt

( )

Figure 9.6c shows one cycle of current i. Since, power p = i2R, we can get the variation of instantaneous power with wt, as shown in Fig. 9.6d. Note that two cycles of variation of p are accommodated from 0 to 2p. It means that the power p varies at double the frequency, but it remains positive all the time. It has an average value, Pav. Let us determine the average power, Pav. The instantaneous value of the heating power p of current i in resistor R is given as 2 p = i2R = (I m2 sin2 wt)R = I m R¥

2 2 (1 - cos 2w t ) I m R I m R = cos 2wt 2 2 2

(9.9)

term in the above expression is constant, (i.e., independent of t). The second term is a cosine function varying with time at a frequency of 2w. The average value of this term is zero. Hence, the average value of power p is

Alternating Voltage and Current

241

I m2 R 2 The power in resistor R due to dc current I is I 2R. Hence, we put Pav =

I m2 I2 R or I 2 = m 2 2 Taking square root, we get the effective value of ac sinusoidal current as I Ieff = m = 0.707Im (9.10) 2 Value g the effective value of ac current i instantaneous power p absorbed in resistor R, as p = i2R. We then found the average of this power over one cycle, as follows. T T È1 T ˘ 1 2 1 Í Pav = = = i R dt p dt i 2 dt ˙ R T T ÍT ˙ 0 0 Î 0 ˚ 2 This power, we equated to I eff R. Thus, we found that I 2R =

2 I eff

1 = T

Ú

Ú

fi Ieff =

1 T

T

Ú i dt 2

0

Ú

T

Ú i dt = 2

0

1 2p

2p

Ú i d (w t ) 2

(9.11)

0

The quantity under the square root is the average or mean of the squared function i 2. We therefore conclude that the effective value of an ac wave is the square root of the mean of the squared function. Or, simply Ieff is the root mean square (rms) value, which is denoted by symbol Irms. When we say that the domestic ac power supply is 220 V, it means it is a sinusoidal ac voltage of rms or effective value of 220 V. Its peak value, according to Eq. 9.10, is 2Veff = 2 ¥ 220 = 311 V Since it is the effective value of voltage and current which is used in everyday usage, the symbols V and I (without any subscript) are taken to mean the effective values. The subscript ‘eff’ or ‘rms’ is used only when there is a possibility of some confusion. It is interesting to note that the rms value is always greater than the average value, except for a rectangular wave. For a rectangular wave, since the heating effect remains constant with time, both the rms and average values are same. b and c. For (Fig. 9.5b), one cycle is of only p. Hence, the rms value is given as Vm =

Ieff =

=

1 p

p

Ú 0

I m2 2p

2

i d (w t ) =

1 p

p

Ú

I m2

2

sin w t d (w t ) =

0

I m2 2p

p

Ú (1 - cos 2w t ) d (w t ) 0

p

I sin 2w t ˘ È = m ÍÎw t ˙ 2 ˚0 2

It is not surprising that this is the same as the rms value of full sinusoidal ac wave of Fig. 9.5a. The two waveforms are the same except that the value of i from p to 2p Squaring positive or negative values of the same magnitude gives the same result.

242 (Fig. 9.5c), the rms value is given as

For Ieff =

1 2p

2p

Ú i d (w t ) = 2

0

1 2p

p

Ú 0

I m2

1 sin w t d (w t ) + 2p 2

p

=

I m2 (1 - cos 2w t ) d (w t ) + 0 = 4p

Ú 0

I m2 4p

2p

Ú 0 d (w t ) p

p

Im Im sin 2w t ˘ È = = Íw t ˙ 2 4 2 ˚0 Î

The ratio of the effective value to the average value is known as the form factor of a waveform of any shape (sinusoidal or nonsinusoidal). Thus, V (9.12) Form factor, Kf = rms Vav The peak factor or crest factor or amplitude factor (or maximum) value to its rms value. Thus, Kp =

Peak factor,

Vm Vrms

(9.13)

Let us calculate these two factors for a sinusoidal voltage waveform,

And

Kf =

Vrms 0.707Vm V / 2 = m = = 1.11 Vav 2Vm /p 0.637Vm

Kp =

Vm Vm = = Vrms Vm / 2

2 = 1.414

Importance of the Form Factor and Peak Factor except when calculating hysteresis loss for different waveforms. An ac magnetising current with higher form factor produces higher hysteresis loss per cycle. The peak factor of an ac voltage assumes importance while testing the dielectric insulation, since the dielectric stress produced depends on the maximum or peak value of the voltage and not on the rms value.

We have seen in Fig. 9.2 how a rotating crank produces a sinusoidal waveform. We call this rotating crank a phasor. Thus, a phasor is just a bar that produces the time-function sinusoid when rotated counterclockwise at proper angular frequency (w) and projected on the vertical axis. Figure 9.5a shows a voltage phasor of amplitude Vm drawn in the reference direction (positive real axis), and a current phasor of amplitude Im drawn at a negative angle q with the reference direction. When both these phasors rotate counterclockwise, time-function voltage and current sinusoids are produced as shown in Fig. 9.7b. These time functions can be represented mathematically as v(t) = Vm sin wt and i(t) = Im sin (wt – q)

Alternating Voltage and Current

243

v w O

P Vm q Im

Vm

i I m

0

wt q

Complex plane (a) Rotating voltage and current phasors.

(b) Waveforms produced.

Phasor Diagram It is a diagram containing the phasors of inter-related sinusoidal voltages and currents, with their phase differences indicated. Thus, Fig. 9.7a is a phasor diagram showing the amplitudes and phasor relationship of voltage and current. This diagram and the time-function waveforms shown in Fig. 9.7b convey the same information about the voltage and current. Obviously, drawing the phasor diagram needs much less effort than drawing the time-varying waveforms. Therefore, electrical engineers prefer to represent inter-related sinusoidal quantities by phasors rather than by time waveforms. Suppose an observer is standing at point P in Fig. 9.7a. The phasors are rotating counterclockwise at the speed w radians per second or f revolutions per second. In every revolution, the phasor Vm passes the observer before the phasor Im does. Thus, the phasor Im lags (or follows) the phasor Vm by q degree. The angle q can also be seen as time lag measured in units of time. Same conclusion can be drawn from Fig. 9.7b. Compared to the voltage sine wave, the current sine wave is shifted to the right, or later in time. Therefore, it is correct to describe Im sin (wt – q) as lagging Vm sin wt by an angle q, or Vm sin wt as leading Im sin (wt – q) by an angle q. The leading or lagging relationship between the two sinusoids should be recognisable both graphically and mathematically. Note that the phase of two sine waves can be compared only if 1. Both have the same frequency. 2. Both are written with positive amplitude. 3. Both are written as sine functions, or as cosine functions.

The sine and cosine are essentially the same functions, but with a 90° phase difference. Thus, sin wt = cos(wt – 90°). We can add to or subtract from 360° the argument of any sinusoidal function, without changing the value of the function. Suppose that we are to compare the phase of v1 = Vm1 cos (50t + 10°) with that of v2 = Vm2 sin (50t – 30°). v1 as a sine function. v1 = Vm1 cos (50t + 10°) = Vm1 sin[(50t + 10°) + 90°] = Vm1 sin (50t + 100°) We can now say that v1 leads v2 by f = 100° – (–30°) = 130°. It is also correct to say that v1 lags v2 by 230°, since v1 may be written as v1 = Vm1 sin (50t – 260°). Useful Formulae – sin wt = sin (wt ± 180°); – cos wt = cos (wt ± 180°) ± sin wt = cos (wt ± 90°); ± cos wt = sin (wt ± 90°)

244 EX A MP L E

9.5

Determine the phase difference of the sinusoidal current i1 = 4 sin (100pt + 30°) A with respect to the sinusoidal current i2 = 6 sin (100pt) A in terms of time and draw the phasor diagram to represent the two phasors.

Solution Comparing the expressions of the given sinusoids with the standard form, we get w = 2pf = 100p

fi

f = 50 Hz

fi T=

1 1 = = 0.02 s = 20 ms f 50

Thus, the time for full revolution (360°) for either of the phasors is 20 ms. The time taken to cover 30° is 30 Im1 = 4 A t = (20 ms) ¥ = 1.67 ms 360 The phasor i1 leads the phasor i2 by 30° (or 1.67 ms). Taking i2 as the reference phasor, the phasor i1 can be drawn 30° leading i2. The phasor diagram is drawn in Fig. 9.8, in terms of the maximum values of the two sinusoidal currents.

30° Im2 = 6 A

While analysing an ac circuit, often we need to perform algebraic operations on sinusoids or phasors. For example, suppose that the currents, i1 = Im1 sin wt and i2 = Im2 sin (wt + q2) (9.14) i3 parallel combination ? Do we simply add Im1 and Im2 to get Im3, the peak value of i3? No, in ac circuits we i1 and i2 in following three ways : 1. By using the plots of waveforms. 2. By using trigonometrical identities. 3. By using the concept of phasors.

(1) sinusoids and then make point by point summation of the two sine waves, as shown in Fig. 9.9b that the addition of two sinusoids is also a sinusoid of the same frequency. The amplitude Im3 and the phase q3 can be read from the plot of the resulting sinusoid. The current i3 can then be expressed as i3 = Im3 sin (wt + q3) (9.15) This method, though straight forward, is quite laborious and time consuming. More so, when you add more than two sinusoids.

(2)

This proves to be a better way than the graphical procedure depicted in Fig. 9.9b. We proceed as follows. i1 = i1 + i2 = Im1 sin wt + Im2 sin (wt + q2) = Im1 sin wt + Im2 [sin wt cos q2 + cos wt sin q2] = Im1 sin wt + (Im2 cos q2) sin wt + (Im2 sin q2)cos wt = [Im1 + (Im2 cos q2)] sin wt + (Im2 sin q2)cos wt (9.16)

245

Alternating Voltage and Current Reference line

i3 = Im3 sin(wt + q3) B

Im2

C3 C2 C1

Im3 q2

O

i2 = Im2 sin(wt + q2)

Im2 Im3

D

i1 = Im1 sin wt

Im1

q3

Im1 A

wt

q3 q2

(a) Addition using phasors.

(b) Addition using plots of waveforms.

For the convenience of writing the expression, let us put a = Im1 + (Im2 cos q2) and b = Im2 sin q2 Therefore, Eq. 9.16 can be written as i3 = a sin wt + b cos wt We now use a standard technique to express i3 as a single sinusoid of the form of Eq. 9.15, a b ⎛ ⎞ i3 = a 2 + b 2 sin ω t + cos ω t 2 2 ⎜ a2 + b2 ⎟ a +b ⎝ ⎠ a

Putting

2

a +b

2

= cos q3

b

and

2

a + b2

= sin q3

(9.17) (9.18)

(9.19)

(9.20)

Equation 9.19 can now be written as i3 =

a 2 + b 2 (sin wt cos q3 + cos wt sin q3) =

a 2 + b 2 sin (wt + q3)

Comparing this equation with Eq. 9.15, and from Eqs. 9.20, we get b (9.21) a The values of a and b can be computed from Eqs. 9.17 using the known quantities, Im1, Im2 and q2. This procedure, though does not require us to plot the sinusoid curves, is too cumbersome to be used in routine calculations. Im3 =

a2 + b2

and

q3 = tan –1

(3) Addition Using Concept of Phasors two methods. The sinusoidal currents i1 and i2 can be considered to be produced by the phasors OA and OB

246 of amplitude Im1 and Im2 rotating counterclockwise in circles C1 and C2, as shown in Fig. 9.9a. Let us freeze these rotating phasors at t = 0, and treat them as vectors. By applying the law of parallelogram or law of triangles of vectors, we can determine the sum of these two vectors as OD at an angle q3 with vector OA. We circle C3 produces the resulting sinusoidal current i3 of amplitude Im3 and phase q3. Thus, we see that the laws of addition (or subtraction) of phasors are exactly the same as those for vectors*.

Important Points about Addition of Phasors Whenever we talk of ac sinusoidal quantities, we peak values Im1 and Im2 of the ac currents i1 and i2, and adding them in a phasor diagram to get peak value Im3 and then getting the rms value of i3. Instead, the phasor diagram can directly be drawn in terms of the rms values I1 and I2, so as to directly get the rms value I3 of the resultant current. This is done by representing the phasors by complex numbers. Many of us are not very familiar with complex numbers. So, let us refresh our knowledge of these complex numbers, since we are going to use them extensively in ac circuit.

Most of us were introduced to complex numbers through the study of quadratic equations. We discovered that the solution to certain equations, like x 2 + 4 = 0 required the introduction of a new type of number : x=± 4 = ± i2 These new numbers are called imaginary and the symbol i is chosen (by mathematicians) to identify these numbers. The use of the word ‘imaginary’ to identify such numbers is rather unfortunate. The terms ‘real’ and ‘imaginary’ are merely technical terms, just like ‘positive’ and ‘negative’. The imaginary numbers are in fact as real as the real numbers, as would be clear from Example 9.6, given below. The ‘imaginary numbers’ need to be distinguished from the ‘real numbers’ because different rules must be applied in the mathematical operations involving them. The solutions of other quadratic equations are combination of the real and imaginary numbers, such as 2 ± i 2 . These combinations are called complex numbers. Complex numbers are appropriately named because the rules for manipulating them are more complicated than for real numbers.

Complex Plane All numbers can be represented as points in a complex plane, such as shown in Fig. 9.10. The horizontal axis represents real numbers, and the vertical axis represents imaginary numbers. Note that we are writing j instead of i to indicate imaginary numbers. This is customary among electrical engineers. It avoids confusion between imaginary numbers and currents, which have traditionally been symbolised by i. In Fig. 9.10, we have shown three complex numbers, z1, z2 and z3, each having real and imaginary components. In general, a complex number** z (represented by point A in Fig. 9.10) can be expressed as z = a + jb (9.22) *

**

This is the reason why some people refer to a phasor diagram as a vector diagram. In fact, phasors are rotating vectors. The calculations with phasors are done at a particular instant. One of the phasors is taken as reference and all others are expressed with respect to this reference. We have used bold letters for the symbols of complex numbers (also of vectors and phasors). While writing with a pen, it is not possible to make bold letters. So, we put a bar or an arrow over the symbol.

Alternating Voltage and Current

247

This complex number represents phasor OA. Given the real part a and imaginary part b, we can determine the amplitude r and phase angle q from the right angle triangle OAB as follows : b (9.23) a The phasor OA can then be expressed as OA = r–q. This representation is called polar form. Conversely, given r and q, we can express the phasor OA in rectangular or Cartesian form a and b, as follows : a = r cos q and b = r sin q (9.24) r =

a2 + b2

and

q = tan –1

Since e jq = cos q + jsin q, we can express the phasor OA (or the complex number) in the following forms : z = a + jb = r (cos q + jsin q) = re jq = r–q

(rectangular or Cartesian notation) (trigonometric notation) (exponential notation) (polar notation)

Addition, Subtraction and Multiplication For these operations, just use ordinary algebra plus two more rules : (1) keep real and imaginary parts separate, and (2) treat j 2 as –1. For example, whenever we add complex numbers, we add the real parts and the imaginary parts separately : z1 + z2 = (3 + j4) + (–7 – j3) = (3 – 7) + j(4 – 3) = – 4 + j1 and similarly for subtraction. Thus, complex numbers are added and subtracted like vectors in a plane. This is one of the few properties common between complex numbers and vectors. Similarly, multiplication of z1 and z2 is z1 z2 = (3 + j4) (–7 – j3) = 3(–7) + j4(–j3) + 3(–j3) + j4(–7) = –21 – j 212 – j9 – j28 = –21 + 12 – j9 – j28 = (–21 + 12) – j(9 + 28) = –9 – j37

248

Division and Conjugation Division requires a trick to get the results in standard form : z1 3 + j4 3 + j4 − 7 + j3 = = × z2 − 7 − j3 − 7 − j3 − 7 + j3 (3) ( − 7) + ( j 4) ( j 3) + (3) ( j 3) + ( j 4) ( − 7) − 21 − 12 + j 9 − j 28 = = 49 + 9 ( − 7) 2 − ( j 3) 2 33 j19 33 19 = = j 58 58 58 in the denominator. To force the denominator to be real, we multiply numerator and denominator by the complex conjugate of z2. The complex conjugate of a complex number z is the same number except that the sign of the imaginary part is changed, and it is represented by the symbol z*. This forces the denominator to at a glance. The process of forcing the denominator to be real and positive is called rationalisation of the quotient. Note that the polar form is merely a compact notation. When we write r–q, we actually mean re jq. No mathematical operation can be performed on the notation r–q. The equivalent exponential form re jq, being a legitimate mathematical function, can be subjected to any mathematical operation. Thus, the multiplication of two complex numbers is performed as follows : (z1)(z2) = (r1e jq1)(r2e jq2) = r1 r2 e j(q1 +q2) Although the above proof rests on exponential form, the results are more easily expressed in polar form : (z1)(z2) = (r1–q1)(r2–q2) = r1 r2–(q1 + q2) (9.25) This means that to multiply two complex numbers, just multiply their magnitudes and add their angles. In a similar way, we can show that dividing two complex numbers requires dividing the magnitudes and subtracting the angles : z1 r r –q = 1 1 = 1 –(q1 – q2) z2 r2 r2 –q 2

(9.26)

Rotation of Phasor by 90° Consider a phasor represented by complex number A = 3 + j4 = 5–53.1°. If this phasor is multiplied by j, we get another phasor B = jA = j3 + j 24 = – 4 + j3 = 5–143.1° = 5–(90° + 53.1°) Thus, a phasor, when multiplied by j, rotates counterclockwise (in positive direction) by 90°. If multiplied by – j, it rotates clockwise (in negative direction) by 90°. EX A MP L E

9.6

An ac current, denoted by a phasor in complex plane as I = 4 + j Determine the power consumed by the resistor.

Solution

I in the polar form, I = Ir + jIi = 4 + j3 =

4 2 + 32 –tan –1 (3/4) = 5–36.87° A

P = I 2R = 52 ¥ 10 = 250 W

W.

Alternating Voltage and Current

249

It is interesting to determine the power due to the two components of the current. The power due to the ‘real’ component, Pr = I r2R = 42 ¥ 10 = 160 W The power due to the ‘imaginary’ component, 2 2 Pi = I i R = 3 ¥ 10 = 90 W The sum of these two powers, Pt = Pr + Pi = 160 + 90 = 250 W This is seen to be the same as the actual power. Thus, we conclude that the imaginary component of current gives as much actual power as does the real component.

Let us now see how we use the concept of complex numbers to add two phasors I1 and I2, given as and I2 = I2–q° (9.27) I1 = I1–0° Here, I1 and I2 are the rms values of the two sinusoidal currents. If Im1 and Im2 are the peak values of these sinusoids, respectively, we should have Im1 = 2 I1 and Im2 = 2 I2 . The instantaneous values of the two ac currents, if needed, can be written as i1 = Im1 sin wt and i2 = Im2 sin (wt + q°) (9.28) We draw the phasors I1 and I2 on the complex plane, as shown Fig. 9.11. Phasor I1 is drawn as an arrow OA (of length I1) along the real axis, as its phase angle is 0°. Hence, we can write I1 = I1 + j0 (9.29) Phasor I2 is drawn as an arrow OB (of length I2), at an angle q2 with the real axis, as its phase angle is q2. Its components OM and ON along the two axes, are given as OM = OB cos q2 = I2 cos q2 and ON = OB sin q2 = I2 sin q2 Hence, phasor I2 expressed as a complex number is I2 = I2 cos q2 + jI2 sin q2 (9.30) The addition of phasors I1 and I2 is now obtained by adding the complex numbers of Eqs. 9.29 and 9.30, I3 = I1 + I2 = (I1 + j0) + (I2 cos q2 + jI2 sin q2) = (I1 + I2 cos q2) = jI2 sin q2 In Fig. 9.11, the real component of I3 is simply the sum of the real components of the two phasors, i.e., OC = OA + OM. Similarly, for the imaginary component. The magnitude OD of phasor I3 is given as I3 =

( Real part) 2 + ( Imaginary part) 2 = ( I1 + I2 cos θ 2 ) 2 + ( I2 sin θ 2 ) 2

j-axis B N

O

I2

q2 q3 I1 A

D I3

M

C

Real axis

(9.31)

250 And the phase angle q3 is given as q3 = tan –1 EX A MP L E

Imaginary part I2 sin θ 2 = tan–1 Real part I1 + I2 cos θ 2

(9.32)

9.7

A sinusoidal current of 10–0° A is added to another sinusoidal current 20–60° A. Find the resultant current.

Solution I1 = 10–0° A = (10 + j0) A and I2 = 20–60° A = (20cos 60° + j20 sin 60°) A = (10 + j10 3 ) A We can now get the summation of the two currents as I = I1 + I2 = (10 + j0) + (10 + j10 3 ) = 20 + j10 3 = 26.46–40.9° A EX A MP L E

9.8

Find the expression for the sum of the currents i1 = 10 2 sin wt A and i2 = 20 2 sin (wt + 60°) A Also, determine the rms value of the sum of these two currents.

Solution values of the two currents as complex numbers, we get Im1 = 10 2–0° A and Im2 = 20 2 –60° A Using fx-991-ES, the summation of these two currents is given as Im = Im1 + Im2 = (10 2 ∠ 0° + 20 2 ∠ 60° ) A = 37.42–40.9°A Therefore, the expression for the sum of the two currents can be written as i = 37.42 sin (wt + 40.9°) A The rms value of the sum is given as I 37.42 I= m = = 26.46 A 2 2 EX A MP L E

9.9

i1 = 5sin wt,

i2 = 5sin (wt + 30°)

and

i3 = 5sin (wt – 120°)

Find the rms value of the resultant current (expressed in polar form) that leaves the junction.

Solution Since we are to determine the rms value of the resultant current in polar form, we work with rms values. First, we express the currents as complex numbers in polar form, 5 5 I1 = –0° A; I2 = –30° A 2 2

and

I3 =

5 ––120° A 2

Using fx-991ES calculator, the addition of the three currents is given as 5 5 5 I = I1 + I2 + I 3 = –0° + –30° + ––120° = 5––15° A 2 2 2

Alternating Voltage and Current

251

Alternatively, we could work with peak values, expressed as complex numbers, add them up and then get the rms value of the resultant. Thus, Im1 = 5–0° A; Im2 = 5–30° A and Im3 = 5––120° A Using fx-991ES calculator, the addition of these peak values is obtained as Im = Im1 + Im2 + Im3 = 5–0° + 5–30° + 5––120° = 7.07––15° A Therefore, the rms value of the resultant current is given as I 7.07–-15∞ = 5––15°A I= m = 2 2

The ac power all over the world is supplied in the form of sinusoidal voltage. However, often we come across waveform and of sinusoidal waveform are the same. Following example will make the process clear. EX A MP L E

9.1 0

Determine the average and rms value of the resultant current in a wire carrying simultaneously a dc current of 10 A and a sinusoidal current of peak value 10 A.

Solution The resultant current at any instant is given as i = 10 + 10 sin q A i (A)

The waveform of this resultant current is shown in Fig. 9.12. For this type of unsymmetric waveform, the average is found over one cycle (and not over half-cycle as in a sinusoidal waveform). As can be seen from Fig. 9.12, the current goes as much positive as negative around the value of 10 A. Hence, its average value is 10 A.

20 10 0

current.

p

0

2p

3p

4p q

First Method It is based on using the formula for determining the rms value,

1 2p

Irms =

100 2p

=

50 p

=

Second Method

Ú

2p

1 2p

i 2 dq =

0

Ú

2p 0

Ú

2p 0

(10 + 10 sin q ) 2 dq

(1 + 1sin q ) 2 d q =

50 p

Ú

2p

2p 0

q sin 2q ˘ È Íq - 2 cos q + 2 - 4 ˙ = Î ˚0

(1 + sin 2 q + 2 sin q ) d q 50 (2 π + π ) = π

150 = 12.247 A

It is based on using the very basis of the meaning of effective (or rms) value. If two (or more than R, the net heating power is the sum of heating powers of the

individual currents, 2

2

2

2

2

I R = I 1 R + I 2 R + I 3 R + I 4 R + …;

2

2

2

2

2

or I = I 1 + I 2 + I 3 + I 4 + …

252 the rms values are added on square basis. Since the rms value of the dc current is 10 A and of the sinusoidal ac current is 10 / 2 A , the rms value of the resultant current i is easily determined as I=

2 ⎛ 10 ⎞ I12 + I 22 = (10) + ⎜ ⎟ ⎝ 2⎠

2

=

100 + 50 =

150 = 12.247 A

N O T E

standard formula. EX A MP L E

9.1 1

One full cycle of an alternating voltage waveform is given in Fig. 9.13. Determine its average value.

9.11. Solution total period. The area between 0 and 1 ms = 10 ¥ (1 ¥ 10 –3) = 10 ¥ 10 –3 The area between 1 and 3 ms = –5 ¥ (2 ¥ 10 –3) = –10 ¥ 10 –3 The area between 3 and 4 ms = 20 ¥ (1 ¥ 10 –3) = 20 ¥ 10 –3 The area between 4 and 5 ms = 0 ¥ (1 ¥ 10 –3) = 0 The area between 5 and 8 ms = 5 ¥ (3 ¥ 10 –3) = 15 ¥ 10 –3 \ The total area between 0 and 8 ms = (10 – 10 + 20 + 0 + 15) ¥ 10 –3 = 35 ¥ 10 –3 Therefore, the average value is Total area 35 × 10 − 3 Vav = = = 4.375 V Total period 8 × 10 − 3 EX A MP L E

9.1 2

Determine the effective value of the voltage wave form shown in Fig. 9.14a.

Alternating Voltage and Current

253

Solution The given waveform is periodic with a period of 20 ms. The squared values of the voltages are plotted for two cycles in Fig. 9.14b. The area under the squared curve from 0 and 10 ms = 400 ¥ (10 ¥ 10–3) = 4000 ¥ 10 –3 The area under the squared curve from 10 and 20 ms –3 –3 = 100 ¥ (10 ¥ 10 ) = 1000 ¥ 10 The area under the squared curve from 0 and 20 ms = (4000 + 1000) ¥ 10 –3 = 5000 ¥ 10 –3 Since the length of the full cycle is 20 ¥ 10–3 s, the average value of the squared waveform is 5000 × 10 − 3

= 250 20 × 10 − 3 The square root of this quantity is the rms value of the waveform of Fig. 9.14a, Vrms = EXA MP LE

250 = 15.81 V

9 .1 3

Determine the rms value, the average value and the form factor for the current waveform shown in Fig. 9.15a.

Solution The given waveform is periodic with a period of 3 seconds. Hence, 10 2 × 2 + 0 2 × 1 = 8.16 A 3 10 × 2 + 0 × 1 = 6.67 A Iav = 3 I 8.16 = 1.22 Form factor, k f = rms = Iav 6.67 Irms =

\ EXA MP LE

9.14

Determine the average value, the rms value, the form factor and the peak factor for the saw-tooth voltage waveform of peak value Vm = 10 V and time period T = 5 ms, shown in Fig. 9.15b.

Solution Area under the curve in one cycle (1/ 2) Vm = T Duration of one cycle 10 =5V = 2

Vav =

T

=

Vm 2

(9.33)

254

y = mx + c or

v=

2 ⎛ Vm ⎞ 2 ⎛V ⎞ 2 t+0 fi v = m t ⎝ T ⎠ ⎝ T ⎠

Thus, the rms value is given as Vrms =

=

1 T

T

Ú

v 2 dt =

0

1 T

T

Ú 0

2

Ê Vm ˆ 2 ÁË T ˜¯ t dt =

T

Vm2 ⎡ t 3 ⎤ Vm ⎢ ⎥ = 3 T3 ⎣ 3 ⎦0

(9.34)

10 = 5.77 V 3

\

Form factor, kf =

Vrms 5.77 = = 1.154 Vav 5

and

Peak factor, kp =

Vm 10 = = 1.733 Vrms 5.77

Consider a general ac circuit of Fig. 9.16a. An ac voltage source is delivering power to a load. Let us assume that the current i lags the voltage v by an angle q. We can then write v = Vm sin wt and i = Im sin (wt – q) For one complete cycle, the waveforms for v and i are drawn in Fig. 9.16b by thin and thick lines, respectively. Let V = Vm / 2 be the effective value of the voltage across the load, and I = I m / 2 be the effective value of the current through the load. Apparently, it seems that the power going to the load should be equal to VI. In fact, the product VI is given the name ‘apparent power’. But, is the actual power consumed by the load same as VI ? As we shall see, this need not be true always. At any instant, the power p consumed by the load is given as the product vi. If you multiply the value of v and the value of i from instant to instant, you get the waveform of the instantaneous power p, as shown in Fig. 9.16b by a dashed line. At points O, A, B, C and D, either the voltage or the current has zero value.

Alternating Voltage and Current

255

Hence, at these points the power is zero. During the period form O to A, the voltage v has positive values but the current i has negative values. Therefore, the power (p = vi) has negative values. However, during the period from A to B, both the voltage v and current i have positive values. Hence the power p has positive values. From B to C, voltage is negative and current is positive; hence the power is again negative. From C to D, both the voltage and current are negative; hence their product (= p) is again positive. It is seen that for some duration, the power p has positive values and for some it has negative values. A positive power means that power p cycle of voltage (or current), there are two cycles of variations of power.

The average power going into the load is the average of the waveform of instantaneous power p, and is shown as Pin in Fig. 9.16b p = vi = [Vm sin wt] [Im sin (wt – q)] = Vm Im sin wt sin (wt – q) Vm I m [cos{wt – (wt – q)} – cos{wt + (wt – q)}] 2 V Im = m [cos q – cos (2wt – q)] = VI [cos q – cos (2wt – q)] 2 2 = VIcos q – VIcos (2wt – q) (9.35) The second term in the above expression represents a sinusoidal waveform of angular frequency 2w; its VI cos q is constant with time t. It therefore represents the average value of power Pav (= Pin) delivered to the load. Thus, the average power or actual power or real power consumed by the load in an ac circuit is given as =

P = VI cos

The product VI is known as apparent power, and the real power is VI cos q. The power factor the factor by which the apparent power is to be multiplied to get the real power. Thus, power factor (pf ) = cos q

(9.36)

(9.37)

where, angle q is the phase angle. If the current i lags the voltage v, as is usually the case, the pf is called lagging pf and is assigned a positive sign. On the other hand, if the current i lead the voltage v, the pf is called leading pf and is assigned a negative sign. The magnitude of power factor varies from 0 to 1. It can also be expressed in percentage. Thus, a pf of 0.8 can be expressed as pf of 80 %. In case the phase angle is zero, the circuit is said to have unity pf. The real power is then same as the apparent power. EX A MP L E

9.1 5

In an ac circuit, the instantaneous voltage and current are given as v = 55sin wt V

and

i = 6.1 sin (wt – p/5) A

Determine the average power, the apparent power, the instantaneous power when wt (in radians) equals 0.3, and the power factor in percentage.

256 Solution Here, the phase angle, q = p/5 The rms value of the voltage, V =

Vm 55 = = 38.89 V 2 2

The rms value of the current, I =

Im 6.1 = = 4.31 A 2 2

Therefore, using Eq. 9.36, the average power is given as Pav = VIcos q = 38.89 ¥ 4.31 ¥ cos p/5 = 167.62 ¥ 0.809 = 135.6 W The apparent power is Pa = VI = 38.89 ¥ 4.31 = 167.62 VA Note that the apparent power is expressed in volt amperes (VA) and not in watts (W), since it is not a power in reality. The instantaneous power at wt = 0.3 is given by Eq. 9.35, as p = VIcos q – VIcos (2wt – q) = 135.6 – 167.62 ¥ cos (2 ¥ 0.3 – p/5) = –31.95 W The power factor is given as pf = cos q = cos p/5 = 0.809 = 80.9 %

R L

C

We shall now study the steady-state response of the basic circuit parameters (R, L and C) to a sustained sinusoidal function. The ‘steady state’ condition implies that the sinusoidal function had been recurring for a long time in the past, is recurring now, and will continue to recur in future too for a reasonably long time. We shall establish the phase angle relationships between the current through and the voltage across each of the circuit parameter. is connected in a circuit.

Consider the circuit of Fig. 9.17a. A resistance R is connected across the terminals of an ac voltage source A. Suppose that the ac voltage is a sine wave, v = Vm sin wt, as shown in Fig. 9.17b. At any instant B, if the value of the voltage is v, the current in R is given as i = v/R. Since the current remains proportional to the voltage all the time, the waveform of the current is also a sine wave. The two waveforms are in phase with each other. This fact is illustrated in the phasor diagram of Fig. 9.17c.

t.

Alternating Voltage and Current

For a source voltage v = Vm sin wt, the current in the resistive circuit is given as v Vm sin t Vm i= = = sin wt = Im sin wt R R R

257

(9.38)

Power Waveform In a purely resistive circuit, the current and the voltage are in phase. The instantaneous power p, given by the product vi

Vm Im, as shown P = Pav =

V I Vm I m = m m = VI 2 2 2

Power and Power Factor For a purely resistive circuit, the current is in phase with voltage; the phase angle q = 0° (see the phasor diagram of Fig. 9.17c). Therefore, the real power is and

Pr = VIcos q = VIcos 0° = VI = Apparent power pf = cos q = cos 0° = 1

Consider the circuit of Fig. 9.19a. A sinusoidal ac voltage source A is connected across a coil of inductance L v and the resulting current i. For our convenience, we start with the assumption that the current is given as i = Im sin wt (9.39) and then we determine what should be the applied voltage v. inductance, the induced emf is given as e = -L di = -L d (Im sin wt) = –w LIm cos wt = wLIm sin (wt – p/2) (9.40) dt dt This shows that the induced emf e lags the current i by p/2, as we have shown in the waveform plot of Fig. 9.19b as well as in phasor diagram of Fig. 9.19c. The induced emf e opposes the applied voltage v. Hence, v = –e = wLIm cos wt = wLIm sin (wt + p/2) (9.41)

258 On comparing this with Eq. 9.39, we conclude that the applied voltage v leads the current i by p/2, as shown in Fig. 9.19b. This fact is also shown in phasor diagram of Fig. 9.19c. In terms of phasors, if I = I–0° = I + j0, then the voltage phasor can be written as V = V–90° = 0 + jV

N O T E We are at full liberty to take any phasor as reference, while drawing a phasor diagram. However, in electrical engineering, it is a normal practice to take the voltage as reference. Hence, the phasor diagram of Fig. 9.19c for a purely inductive circuit is redrawn as shown in Fig. 9.19d. It can now be seen that the current lags applied voltage by p/2.

Inductive Reactance A look at Eq. 9.41 shows that the peak or maximum value Vm of v is given as Vm = wLIm fi

Vm = wL Im

Since, the rms values of v and i are V = Vm / 2 and I = I m / 2 , respectively, the ratio of rms voltage to rms current for a purely inductive circuit is given as Vm / 2 Vm V = = = wL (9.42) I Im Im / 2 This ratio V/I for purely inductive circuit is called inductive reactance, and is represented by XL. Like resistance, the inductive reactance is expressed in ohms (W). Hence, XL = wL = 2p f L (9.43)

Alternating Voltage and Current

259

For a given ac voltage V, the current I in a purely inductive circuit can be determined, using Ohm’s law, V V V I= = = (9.44) XL w L 2p f L As can be seen from Eqs. 9.43 and 9.44, the inductive reactance is proportional to the frequency and the current produced by a given voltage is inversely proportional to the frequency. This variation is shown in Fig. 9.22a. Taking the current phasor as reference (as in phasor diagram of Fig. 9.19c), the voltage phasor across an inductor is given as V = jX LI, which is along +j axis. Hence, we associate +j with an inductive reactance.

Power and Power Factor In a purely inductive circuit, the current lags the applied voltage by p/2, as shown by the voltage and current waveforms (drawn for one complete cycle) in Fig. 9.20. The power vi from instant to instant. At wt = 0°, 90°, 180°, 270°, 360°, either the voltage or the current has zero value. Hence, at these instants, the power too is zero. Between 0° and 90°, the voltage is positive but the current is negative; and between 180° and 270°, the voltage is negative but the current is positive. Hence, during these intervals the power is negative. Between the 90°–180° and 270°–360°, both are either positive or negative. Hence, during these intervals the power is positive. over a complete cycle is zero. That is, the power consumed by the circuit is zero. In fact, in a purely inductive source.

and

P = VIcos q = VIcos 90° = 0 pf = cos q = cos 90° = 0 (lagging)

Consider the circuit of Fig. 9.21a. A sinusoidal ac voltage source having v = Vm sin wt is connected across a capacitance C. Using the basic relationship between the applied voltage v and the resulting current i for a

260 capacitor, we get dv d = C (Vm sin wt) = CVmw cos wt = wCVm cos wt dt dt = wCVm sin (wt + p/2) This shows that the resulting current i leads the applied voltage v, as illustrated in Fig. 9.19b. i =C

(9.45)

The phasor diagram for the circuit is shown in Fig. 9.21c. Here, we have taken the voltage phasor as reference; that is V = V–0° = V + j0. The resulting current phasor will then be I = I–90° = 0 + jI. From Eq. 9.45, it follows that the maximum value Im of the current is wCVm. Hence, if V and I are the rms values, the ratio Vm / 2 Vm V 1 = = = = 1 I Im C Im / 2 2pf C This ratio V/I for purely capacitive circuit is called capacitive reactance, and is represented by XC. Like inductive reactance, the capacitive reactance is expressed in ohms (W). Hence, 1 XC = = 1 (9.46) C 2pfC For a given voltage V, the resulting current in a purely capacitive circuit is given as V I= = 2pfCV (9.47) XC Equations 9.46 and 9.47 show that for purely capacitive circuit, the reactance varies inversely as the frequency and the current varies directly as the frequency, as shown in Fig. 9.22b.

Alternating Voltage and Current

261

Taking the current phasor as reference, we have I = I–0° (as in phasor diagram of Fig. 9.21d). The voltage phasor across the capacitor is then given as V = V––90° = – jV = – jX C I, which is along –j axis. It is for this reason that we associate –j with a capacitive reactance.

Capacitive reactance

Current

Current

Inductive reactance

O

Frequency (a) Purely inductive circuit.

Frequency

O

(b) Purely capacitive circuit.

Power and Power Factor Like an inductive circuit, the current and voltage are in quadrature* in

a purely capacitive circuit. However, in a capacitive circuit, the current leads the applied voltage by p/2, as shown by the voltage and current waveforms (drawn for one complete cycle) in Fig. 9.23. The power vi from instant to instant. Between 0°-90° and 180°-270°, same sign. Hence, during these intervals the power is positive. Between 90°-180° and 270°-360°, the voltage and the current are of opposite sign. Hence, during these intervals the power is negative.

alternately from source to the load and then from load to the source. The average power consumed by the circuit is zero. P = VI cos q = VIcos 90° = 0 pf = cos q = cos 90° = 0 (leading)

and

*

Two phasors are said to be in quadrature when the phase angle between them is 90°.

262

R L

C

All the three parameters are passive in nature. That is, none of these can act as a source of energy in a circuit. However, there exist marked differences in their properties and their behaviour in a circuit. The resistance is a dissipative parameter. It consumes electrical power and dissipates it as heat. But, inductance and capacitance do not consume any power. They temporarily store energy in one half-cycle and give away the same in next half cycle. The comparison among the three parameters is listed in Table 9.1.

Property Current Frequency dependency Power Phase difference Reactance

Resistance

Inductance

Capacitance

V/R Independent

V/XL XL μ f

V/XC XC μ (1/f )

VI = I 2R = V 2/R 0° R

Zero 90° lagging jX L = jwL = j2pf L

Zero 90° leading – jXC = 1/jwC = 1/j2pf C

ADDITIONAL SOLVED EXAMPLES EX A MP L E

9.1 6

Find the average and rms values of the two waveforms given in Fig. 9.24.

Solution (a 0 to T/2 and then divide this area by T T/4 and dividing it by T/4, since the two right-angle triangles in the positive half-cycle are mirror-image of each other. Thus, the problem reduces to that of Example 9.14. Hence, the average value is given by Eq. 9.6 as V 10 Vav = m = =5V 2 2 cycle. Since the square of a negative value is a positive quantity, the area under the squared curves would be same for the four time durations : (i) from 0 to T/4, (ii) from T/4 to T/2, (iii) from T/2 to 3T/4, and (iv) from 3T/4 to T. quarter cycle (from 0 to T/4). Thus, the problem reduces to that of Example 9.14, and the rms value is given by

Alternating Voltage and Current

263

Eq. 9.34 as

Vm 10 = = 5.77 V 3 3 (b) This is an unsymmetrical waveform. Hence, its average has to be found over one cycle. Thus, Area under the curve in one cycle (1/ 2) × Vm × (T / 2) + 0 Vm Vav = = = 4 Duration of one cycle T 10 = = 2.5 V 4 The rms value is found as follows. Vrms =

T

Vrms =

1 v 2 dt = T

Ú 0

=

EXAMP LE

1 T

T /2 T ˘ 1 È Í v 2 dt + 0 dt ˙ = T Í ˙ T /2 Î 0 ˚

Ú

T/ È 2 Ï 3 ¸ 4˘ Í 32Vm Ì t ˝ ˙ = ÍÎ T 2 Ó 3 ˛0 ˙˚

Ú

32Vm2 T 3 / 64 = ◊ 3 T3

1 T

È T /4 ˘ Í2 v 2 dt + 0˙ = Í ˙ Î 0 ˚

Ú

1 T

È T /4 Ê V ˆ 2 ˘ 2 ˙ m Í2 t dt ÁË T / 4 ˜¯ Í ˙ Î 0 ˚

Ú

Vm2 V = m = 4.08 V 6 6

9.17

An alternating current of frequency of 60 Hz has a maximum value of 12 A. (a) Write down the equation for its instantaneous value. (b) Find the value of the current after 1/360 second. (c) Find the time taken to reach 9.6 A for

Solution (a) The angular frequency, w = 2pf = 2 ¥ 3.141 ¥ 60 = 377 rad/s. Therefore, the instantaneous value is given as i = 12 sin 377t A 377 1 377 × 1 × 180° (b) i = 12sin = 12 sin = 10.39 A 360 360 × π –1 9.6 = 0.927 rad (c) 9.6 = 12sin 377t fi 377t = sin 12 0.927 = 0.00246 s = 2.46 ms \ t = 377 EX A MP L E

9 .1 8

An alternating current has a maximum value of 10 units. At time t = 0, the current has a value of 5 units, and is increasing in the positive direction. Find the expression of the current in cosine form.

Solution The current is 5 units at t = 0 and is increasing in positive direction, hence the waveform would be as shown in Fig. 9.25. This is just a sine wave with a phase lead of f, having equation, i = 10 sin (wt + f). The phase angle f can be determined from the given condition, 5 = 10sin (w ¥ 0 + f) fi f = sin–1 (5/10) = p/6 \ i = 10sin (wt + p/6) = 10cos{(p/2) – (wt + p/6)} = 10cos{– (wt – p/3)} = 10cos (wt – p/3)

264 i (A) 20÷2

i 10 5

p 0

10÷2

3p/4

p/2

2p

wt

p/2 t

0

p

3p/4 2p wt

f f

EX A MP L E

9.1 9

An alternating current varying sinusoidally with a frequency of 50 Hz has a rms value of 20 A. At what time, measured from negative maximum value, instantaneous current will be 10 2 A?

Solution The peak value of the current is Im = 20 2 A . At negative maximum value, we put t = 0. Hence, the given

current waveform should be as shown in Fig. 9.26. This is just a sine wave with a phase lag of f = p/2. Hence, the equation of the current should be i = 20 2 sin (100 pt – p/2) A. Therefore, the required time is given by 10 2 = 20 2 sin (100pt – p/2) fi \

t =

EX A M P L E

100 pt – p/2 = sin–1 (1/2) = p/6

π/6 + π/2 = 0.00667 s = 6.67 ms 100 π

9 . 20

Determine the average and rms values of a current given by i = 10 + 5cos 314t A.

Solution The given current is seen to be the combination of a dc current of 10 A and a sinusoidal current of peak value 5 A. The average value of the sinusoidal current being zero, the average of the overall current would be the same as the dc current. That is, Iav = 10 A. The rms value can easily be found by adding the rms values of the two components on square basis, 2 ⎛ 5 ⎞ Irms = (10) + ⎜ ⎟ ⎝ 2⎠

EXA MP LE

2

=

112.5 = 10.6 A

9.21

Obtain the sum of the three voltages, v1 = 147.3cos(wt + 98.1°) V,

v2 = 294.6cos (wt – 45°) V

and

v3 = 88.4 sin (wt + 135°) V

Solution We plot the above phasors in complex plane, in terms of their peak values. First, we write the voltages in terms of sine functions. Since, sin (90° + q) = cos q, we can write

and

v1 = 147.3sin (90° + wt + 98.1°) V = 147.3sin (wt + 188.1°) V v2 = 294.6 sin (90° + wt – 45°) V = 294.6 sin (wt + 45°) V v3 = 88.4 sin (wt + 135°) V

The plot of these three phasors (in terms of peak values) is shown in Fig. 9.27.

Alternating Voltage and Current

265

Vm2 = 294.6 V j-axis

Vm3 = 88.4 V 135° 188.1°

45° O

Real

Vm1 = 147.3 V

imaginary parts separately, the addition is found as Vm1 = ( − 145.8 − j 20.8) V Vm 2 = (208.3 + j 208.3) V Vm3 = ( − 62.5 + j 62.5) V Vm = ( 0 + j 250.0) V = 250 ∠ 90° Therefore, the expression for the sum of the three voltages can be written as v = 250 sin (wt + 90°) V E X A MP L E

9. 22

(a) What reactance will be offered (i) by an inductor of 0.2 H, (ii) by a capacitance of 10 mF, to an ac voltage source of 10 V, 100 Hz ? (b) What, if the frequency of the source is changed to 140 Hz ?

Solution (a) (i) XL = 2pfL = 2p ¥ 100 ¥ 0.2 = 125.66 W 1 1 (ii) XC = = = 159.15 W 2π × × 100 10 × 10 − 6 2pf C (b) (i) XL = 2pf L = 2p ¥ 140 ¥ 0.2 = 175.9 W 1 1 = = 113.7 W (ii) XC = 2p f C 2π × 140 × 10 × 10 − 6 EXA M PL E

9 . 23

Study the ac circuits given in Fig. 9.28. Draw the phasor diagram and determine the values of the unknown quantity in each case.

Solution (a) The two currents I1 and I2 are in phase with the applied voltage (Fig. 9.29a). Hence, the unknown current is Ia = I1 + I2 = 10–0° + 10–0° = 20–0° A

266

(b) The current I2 is in phase with applied voltage, but the current I1 lags the voltage by 90° (Fig. 9.29b). Hence, the unknown current is Ib =

I12 + I22 =

10 2 + 10 2 = 10 2 A fi

Ib = 10 2 –– 45° A

(c) Both currents I1 and I2 lag the applied voltage by 90° (Fig. 9.29c). Hence, the unknown current is Ic = I1 + I2 = 10 + 10 = 20 A fi Ic = 20––90° A (d) The inductive reactance is taken as j25 W. Hence, the unknown current is given as 250 + j 0 250 ∠ 0° Id = = = 10–– 90° A 25 ∠ 90° j 25 Hence, the current Id is 10 A, and lags the applied voltage by 90° (Fig. 9.29d). (e) The capacitive reactance is taken as –j25 W. Hence, the unknown voltage is given as Ve = (5 + j0) (–j25) = (5–0°) (25––90°) = 125––90° V Hence, the voltage Ve is 125 V, and lags the current by 90° (Fig. 9.29e).

SU MMARY TE RM S

A N D

C ON CE PT S

alternating system, the voltages and currents vary in a repetitive manner. A cycle of variation is the sequence of change before the repetition commences. sinusoidal variation. sum or difference of a number of sinusoids is a sinusoid.

Alternating Voltage and Current

267

period. frequency is the number of cycles completed in one second. Vm = 311 V, T = 20 ms). phasor is a line drawn to represent a sinusoidal alternating quantity. It is drawn to scale and its angle relative to the horizontal represents its phase shift in time. same way as for vectors. complex numbers. Phasor diagrams can be used to represent rms quantities in which case they are frozen in time. j, rotates counterclockwise (in positive direction) by 90°; when multiplied by –j, it rotates clockwise (in negative direction) by 90°. average value of a symmetric periodic waveform is determined for half-cycle only. effective value of an ac current is the equivalent dc current that produces the same amount of heat. It is given by root of mean of squares of the values over one cycle. Hence, it is also called rms value. VI is called apparent power (measured in volt amperes); the actual or real power is given by P = VIcos q, where cos q is called power factor. Thus, power factor is the factor by which the apparent power is to be multiplied to get the real power.

reactance of an inductor linearly increases with frequency. reactance of a capacitor inversely decreases with frequency. IMP O R TA N T f=

F O RM U LAE

1 T i = I m sin wt, where w = 2pf or w =

2p T

Iav = 2I m , p

and Irms =

Im . 2

Iav = 2I m , p

and Irms =

Im . 2

and Irms =

Im . 2

Iav =

Im

,

Kf =

Vrms = 1.11 Vav

(for sinusoidal waveform)

Kp =

Vm = 1.414 Vrms

(for sinusoidal waveform)

268 C HEC K

YOUR

UNDERSTAND ING

Before you proceed to the next chapter, take this test. Give yourself two marks for each correct answer and minus one for each wrong answer. If your score is 12 or more, go to the next chapter; otherwise study this chapter again. S. No. 1. 2.

4. 5. 6.

8. 9. 10.

Statement

True

False

Marks

A sinusoidal current i = 2 sin wt W. The power consumed by the resistor is 40 W. In an ac circuit, the current and voltage are given as i = 10 sin wt and v = 20sin (wt + 60°). It means the current lags voltage by 60°. given by Irms = Im / 2 . When a phasor is multiplied by –j, it rotates counterclockwise by 90°. For an ac circuit, I = 2 A, V = 10 V and P = 10 W. It shows that the power factor is 0.5. A complex number does not always represent a sinusoidal function. does not contribute to the real power. If a constant-voltage ac source is connected across an inductor, the resulting current goes on increasing as the frequency of the source increases. In a purely capacitive circuit, the voltage lags the current by p/2. The capacitive reactance is directly proportional to the frequency of the ac source. Your Score A N SW E RS

1. False 6. True

2. True 7. False

3. True 8. False

RE VI EW

4. False 9. True

5. True 10. False

Q UESTI ONS

1. What are the requirements for a waveform to be 2. How is the frequency related to the period of an ac waveform ? 3. What is a phasor and how is the concept of phasor helpful in solving ac circuit ? 4. How is the phase angle related to time in a sinusoidal voltage waveform ? 5. What is meant by the average value of an ac voltage waveform ? 6. waveform ? Why is it also called rms value ?

7. A square wave has equal positive and negative peak values. What are its average and effective values ? 8. What advantages do you get by using complex algebra in solving ac problems ? 9. Why are the multiplication and division of complex numbers preferably done in polar form ? 10. What is the difference among a vector, a complex number and a phasor ? 11. What is the difference between apparent power and real power in an ac circuit ? 12. Describe the phasor relationship between current and voltage for (a) an inductor, and (b) a capacitor.

Alternating Voltage and Current M U LT I P L E

CH O ICE

QUE STION S

Here are some incomplete statements. Four alternatives are provided below each. Tick the alternative that completes the statement correctly . 1. A sinusoidal voltage is expressed as v = 20 sin (314.16t + p/3) V Its frequency and phase angle, respectively, are (a) 314.16 Hz, 60° (b) 60 Hz, 60° (c) 50 Hz, 60° (d) 50 Hz, –60° 2. A sinusoidal voltage v1 leads another sinusoidal voltage v2 by 180°. Then, (a) voltage v2 leads voltage v1 by 180° (b) both voltage have their zero values at the same time (c) both voltages have their peak values at the same time (d) All of the above 3. The rms value of an ac sinusoidal current is 10 A. Its peak value is (a) 7.07 A (b) 14.14 A (c) 10 A (d) 28.28 A 4. If A = 10–45° and B = 5–15°, then the value of A/B will be (a) 50–60° (b) 2–60° (c) 2––30° (d) 2–30° 5. In an ac circuit, the active power and apparent

269

6.

7.

8.

9.

10.

power are equal in magnitude. Then the power factor of the circuit is (a) 1 (b) 0.8 (c) 0.6 (d) 0 When a phasor is multiplied by –j, it gets rotated through ___ in the counterclockwise direction (a) 90° (b) 180° (c) 270° (d) None of the above For a purely inductive circuit, the current leads the voltage by (a) 90° (b) 180° (c) 270° (d) 360° In a purely capacitive circuit, the voltage lags the current by (a) 90° (b) 180° (c) 270° (d) 360° The power consumed by a pure inductance connected to an ac source is (a) zero (b) very low (c) very high (d If a 10-W resistance is connected to an ac supply v = 100 sin (314t + 37°) V, the power dissipated by the resistance is (a) 10 000 W (b) 1000 W (c) 500 W (d) 250 W A N SW E RS

1. c

2. d

3. b

4. d

5. a

6. c

7. c

8. a

9. a

10. c

PROBLEMS ( A )

S I M P L E

P RO B LEMS

1. The maximum value of a sinusoidal alternating current of frequency 50 Hz is 25 A. Write the equation for instantaneous value of the alternating current. Determine its value at 3 ms and 14 ms. [Ans. i = 25sin100pt A, 20.225 A, –23.78 A] 2. Calculate the rms value of a triangular wave in which voltage rises uniformly from zero to Vm, and completes the cycle by falling instantly back to zero. Find also its form factor and peak factor. [Ans. 0.577Vm, 1.1554, 1.733]

3. An ac voltage is mathematically expressed as v = 141.42 sin (157.08t + p/2) volts. Find its (a) effective value, (b) frequency, and (c) periodic time. [Ans. (a) 100 V; (b) 25 Hz; (c) 40 ms] 4. The equation of an alternating current is given as i = 62.35sin 323t amperes. Determine its maximum value, frequency, rms value, average value and form factor. [Ans. 62.35 A, 51.4 Hz, 44.1 A, 39.7 A, 1.11]

270 5. Find the rms value of the current given by i = 10 + 5cos (628t + 30°). [Ans. 10.6 A] 6. Determine the effective value of the voltage waveform represented by v = 200sin wt + 100 sin 2wt + 50 sin 3wt V [Ans. 162.02 V] 7. Find the rms value of a resultant current in a wire, which carries simultaneously a direct current of 5 A and a sinusoidal alternating current with a peak value of 5 A. [Ans. 6.1237 A] 8. The waveform of a nonsinusoidal voltage has a form factor of 1.15 and a peak factor of 1.4. The peak value of the voltage is 322 V. Find (a) the rms

( B)

T R IC KY

value, and (b) the average value of this voltage. [Ans. (a) 230 V, (b) 200 V] 9. A 10-W resistor is connected across 200-V, 50-Hz ac supply. Find the peak value, average value and the rms value of the current and the power dissipated by the resistor. [Ans. 28.28 A, 18 A, 20 A, 4 kW] 10. Two phasors A and B are given as A = 3 + j1, and B = 4 + j3. Calculate the values of (a) A + B; (b) A – B; (c) AB; (d) A/B. Express the results in both polar and rectangular coordinates. [Ans. (a) 7 + j4 = 8.06–29.7°; (b) –1 – j2 = 2.24––116.6°; (c) 15.8–55.3° = 8.99 + j12.99; (d) 0.632––18.44° = 0.6 – j0.02]

PR OB LE MS

11. A 60-Hz sinusoidal current has an instantaneous value of 7.07 A at t = 0, and rms value of 10 2 A. Assuming the current wave to enter positive half at t = 0, determine (a) the expression for current, (b) the magnitude of current at t = 0.0125 s and at t = 0.025 s after t = 0. [Ans. (a) i = 20sin (120pt + 20.7°) A; (b) –18.71 A, –7.07 A] 12. Determine the average value and effective value of the three voltage-waveforms shown in Fig. 9.30. [Ans. (a) 0.7Vm, 0.7746 Vm; (b) 0.54Vm, 0.584Vm; (c) 0.5433Vm, 0.674Vm] 13. A 200-mH inductance is connected across 230-V, 50-Hz ac supply. Find the inductive reactance and the current in the circuit. [Ans. 62.832 W, 3.66––90° A] 14. A 40-mF capacitor is connected to a 230-V, 50-Hz ac supply. Find the capacitive reactance and the

current in the circuit. [Ans. 79.57 W, 2.89–90° A] 15. A voltage v = 141 sin (314t + 60°) V is applied separately to (a) a 20-W resistor, (b) a 0.1-H inductor, and (c) a 100-mF capacitor. Write, in each case, the expression for instantaneous current and power factor. [Ans. (a) i = 7.05sin (314t + 60°) A, 4.985 A, 497 W, 1; (b) i = 4.49 sin (314t – 30°) A, 3.175 A, 0 W, 0; (c) i = 4.43sin (314t + 150°) A, 3.13 A, 0 W, 0] 16. The instantaneous values of two emfs are e1 = 30sin wt V e2 = 20sin (wt – p/4) V. Derive the expression for the instantaneous value of (a) e1 + e2, and (b) e1 – e2. [Ans. (a) e = 46.35sin (wt – 17.76°) V; (b) e = 21.24 sin (wt + 41.72°) V]

Alternating Voltage and Current ( C)

C H A LL ENG I NG

271

P ROBLEMS

17. There are three conducting wires connected to a two wires are i1 = 10sin314t A i2 = 15sin (314t – 45°) A. What is the current leaving the junction in the third

wire ? What is its value at t = 0 ? [Ans. i3 = 23.17sin (314t – 27.23°) A, –10.6 A] 18. Following impedances are connected in parallel. Find the equivalent impedance of this combination. Z1 = (8 + j6) W; Z2 = (8 – j6) W; Z3 = (8.66 + j5) W. [Ans. (3.89 + j0.79) W]

10

AC CIRCUITS OB JE CT IV E S

RL Figure 10.1a shows a series RL circuit. We know that a practical inductor possesses inductance and resistance effectively in series. Therefore, the following analysis of R and L in series is equivalent to the analysis of a circuit containing a practical inductor. I) for the given applied ac voltage V. Note that the reference polarity of V is shown by means of an arrow. Let VR and VL be the voltage drops across resistance R and inductance L, respectively. Unlike dc circuits, here we cannot obtain the total voltage V just by adding the magnitudes of the voltages across the resistor and inductor. That is, to write V = VR + VL

is wrong.

This does not mean that KVL is not applicable here. KVL is a basic law and is applicable to all circuits whether dc or ac. In an ac circuit, the voltages and currents are phasors. So, the two voltages must be added by treating them as phasors, V = VR + V L

We shall now explain the procedure of drawing the phasor diagram for an ac circuit. As an example, we take the series RL circuit of Fig. 10.1a. 1 Mark the source voltage V, showing its polarity, either by an arrow or by using + and signs. Mark the source current I showing its direction by an arrow. As a convention, the current I must leave the positive terminal of the source. 2

Mark ‘the voltage across’ and ‘the current through’ each individual component of the circuit, following

273

marked VR and IR for the resistance R, and VL and IL for the inductance L. 3

Draw the phasor diagrams for individual components. (i) For resistance, R : The current is in phase with the voltage. Draw the voltage phasor VR along x direction). Draw the current phasor IR also along the b). ii) For inductance, L : The current lags the voltage by 90°. Draw the voltage phasor VL along the reference direction. Draw the current phasor IL 90° lagging, as shown in Fig. 10.1c.

4 To get complete phasor diagram, superimpose all the individual phasor diagrams, by recognising the common phasor among them. For this, you may have to rotate some phasor diagrams. I = IR = IL. In Fig. 10.1b, the current phasor IR = I is already along the reference direction. But, in Fig. 10.1c, the current phasor IL = I is not along the reference direction. To align the two current phasors, we rotate the phasor diagram of Fig. 10.1c d). We can now superimpose Fig. 10.1d and Fig. 10.1b, to get the complete phasor diagram of the circuit, as shown in Fig. 10.1e. 5

VR and VL, by drawing the parallelogram OABC

The resultant of this addition is given by the diagonal OB. The phasor OB must be equal to supply voltage V, as per KVL. From Fig. 10.1e, it becomes clear the current phasor I lags the supply voltage V by an angle BOA = q.

274

Basic Electrical Engineering

Step 6 Once the phasor diagram is drawn, we can take the help of complex algebra to make calculations. Imagine that the phasor diagram is drawn in the complex plane. That is, mark the reference direction (+x-axis) as the positive real axis and the y-axis as the imaginary axis. We can then write I = I–0°; VR = IR and VL = jI XL = I jXL = I jwL Therefore, Eq. 10.1 can be rewritten as V = VR + VL = IR + IjwL = I(R + jwL) (10.2)

Complex Impedance In general, for an ac circuit, the ratio of the voltage phasor to the current phasor is a complex quantity, called complex impedance (represented by symbol Z). Its real part is called resistance and its imaginary part is called reactance. Thus, Complex impedance = (resistance) + j(reactance) or Z = R + jX (10.3) For the series RL circuit, from Eq. 10.2, we have V Z = = R + jwL = Z–q (10.4) I Z =

where,

R 2 + (w L ) 2

and

q = tan–1

wL R

(10.5)

From the phasor diagram of Fig. 10.1e, we can separate the voltage triangle OAB. If each side of this triangle is divided by I, the result is the impedance triangle, shown in Fig. 10.1f. Note

Voltage Phasor as Reference resulting current in a circuit. This can be done by taking the voltage as reference, and then writing the current as V V – 0∞ V I = = = –-q (10.6) Z Z –q Z This shows that the current lags the applied voltage by an angle q, given by Eq. 10.5. For a given ac voltage v = Vm sin wt volts, the equation of the resulting current is Vm i= sin{wt – tan–1(wL)/R} amperes (10.7) 2 2 R + (w L ) EXA M P L E

1 0.1

For the series RL circuit shown in Fig. 10.2a, (a) (b) (c) (d) (e)

Calculate the rms value of the steady-state current and the relative phase angle. Write the expression for the instantaneous current. Find the average power dissipated in the circuit. Determine the power factor. Draw the phasor diagram.

Solution (a) Since the phase angle of the applied voltage is given as 0°. In the complex form, the applied voltage can be written as

AC Circuits

275

A series RL circuit. 141 Vm – 0∞ = 100–0° = 100 + j0 volts – 0∞ = 2 2 The impedance, Z = R + jwL = 3 + j100p ¥ 0.0127 = 3 + j4 = 5–53.1° ohms V 100– 0∞ \ Current, I = = = 20––53.1° A Z 5–53.1∞ Thus, the rms value of the steady-state current is 20 A, and the phase angle is 53.1° lagging. (b) The expression for the instantaneous current can be written as

V = V–0° =

i = 20 2 sin (100pt – 53.1°) A = 28.28 sin (100pt – 53.1°) A (c) Average power, P = VI cos q = 100 ¥ 20 ¥ cos 53.1° = 1200 W Or, P = I 2R = (20)23 = 1200 W (d) pf = cos q = cos 53.1° = 0.6 lagging. Alternatively, Average power 1200 P pf = = = = 0.6 lagging Apparent power VI 100 20 (e) Taking the current as reference, the phasor diagram is drawn in Fig. 10.2b, where I = 20 A; VR = IR = 20 ¥ 3 = 60 V; VL = IXL = 20 ¥ 4 = 80 V and V = 100 V The same phasor diagram is redrawn in Fig. 10.2c, by rotating it clockwise by an angle 53.1°, so that the applied voltage becomes the reference phasor.

RC For the series RC circuit of Fig. 10.3a, we can apply KVL and get Ê Ë

V = VR + VC = IR – jIXC = I(R – jXC) = I Á R - j

1 ˆ ˜ = IZ wC ¯

(10.8)

where Z is the complex impedance for the series RC circuit. Since the voltage across a capacitor lags the current by 90°, we have put in the above equation, VC = –jIXC = I(–jXC) Also, note that –j is associated with XC (whereas +j is associated with XL). Since the current is common to both the resistor and the capacitor, we take current phasor I as reference. For the resistance, the voltage VR is drawn in phase with the current I; but for the capacitance, the voltage VC is drawn lagging the current I by 90°. The complete phasor diagram for series RC circuit is drawn in Fig. 10.3b. Here, the supply voltage V is the phasor sum of VR and VC. From this phasor

276

Basic Electrical Engineering

c. Note that a capacitive circuit has an impedance triangle in fourth quadrant. For an impedance triangle to be in either the second or third quadrant, a circuit must have a negative resistance. This may occur if a circuit contains dependent sources.

We have seen that, if the terminal voltage is V = V–q and the current is I = I–f in an ac circuit, the average power absorbed by it is given as P = VI q–f By using Euler’s formula, the above equation can be written as P = VI Re[e j q–f) Ve jq Ie – jf Ve is simply the voltage phasor. The second term Ie – jf is seen to be the complex conjugate * I of the current phasor I = Ie jf. Hence, Eq. 10.10 can be written as P = Re [VI* complex power S) as jq

S = P + jQ = VI* = VIe j q – f) It is seen that the magnitude VI of S is the apparent power q – f) of S is the pf angle. The real part P of S is the average real or actual) power absorbed by the circuit. The imaginary part Q of S is given the name reactive power Q = VI q–f It is obvious that the dimensions of Q should be the same as that of P. However, to avoid confusion, the units of Q are taken as Name 1. Apparent power 2. Average power 3. Reactive power EX A MP L E

10. 2

Symbol S P Q

Value VI VI VI

q–f q–f

Units

277 a e) the reactive power.

b

c

d) the

Solution

a b. The voltage phasor VR is in phase with the current I. The voltage phasor VC lags the current I by 90°. The rated current of the lamp is P 750 I= = VR 100

a) From D

b, we can determine the voltage across the capacitor, VC =

c) d) e)

VR2 =

(230)2 (100)2

VC 207 1 = W fi I 7.5 2pf C 1 \ C = ¥ 10 F = 115 mF 2p ¥ 50 ¥ 27.6 V 100 Phase angle, f = cos –1 R = cos –1 = 64°12¢ V 230 Power factor, pf = cos f 0.435 leading Apparent power = VI ¥ 1725 VA Reactive power = VI sin f ¥ ¥ ¢ = 1553 VAR XC =

Now,

b)

V2

EX A MP L E

10. 3

W and a capacitor of reactance W. Find the impedance, power factor, supply voltage, voltage across resistor, voltage across capacitor, apparent power, active power and reactive power.

Solution Taking current as the reference phasor, I = 0.9–0° A. Impedance, Power factor, Supply voltage, Voltage across resistor, Voltage across capacitor, Apparent power, Actual power, Reactive power,

Z j 277.3–– 64.4° W pf = cos q 0.432 leading V = IZ – – 249.6––64.4° V VR = IR –0°) ¥ 108–0° V VC = IXC – ––90°) = 225––90° V Papp = VI ¥ 0.9 = 224.6 VA Pa = VI cos q ¥ 0.9 ¥ 97.06 W Pr = VI sin q ¥ 0.9 ¥ 202.59 VAR

278

Basic Electrical Engineering

RL a. Through this circuit, we shall introduce the concept of complex admittance represented by Y. The current IR through resistor R and the current IL through inductor L are given as V V V and IL = = IR = jw L jXL R Using KCL, the total current I is given as V V 1 ˆ Ê1 = VÁ + = VY + R jw L Ë R jw L ˜¯ where Y is the complex admittance of the parallel RL circuit. After rationalisation, we can write 1 1 1 1 Y = + = - j R jw L R wL In general, the real part of complex admittance Y is called conductance G) and the imaginary B). Thus, we can write part is called susceptance

I = IR + IL =

Y = G + jB

j

For parallel RL circuit, G = 1/R and B = –1/wL 1 ˆ Ê1 I = IR + IL = VY = V G + jB) = V Á - j ˜ ËR wL¯ This shows that in a parallel RL circuit, the current I lags the voltage V by the admittance angle, given by -R -1/w L q = tan -1 = tan -1 w L 1/ R IR

I

V

IR

IL

R

L

O

V

q

I IL (a) The circuit.

(b) Phasor diagram.

Phasor Diagram Sometimes it is hard to know where to start drawing the phasor diagram. But, the rule is simple start with the quantity that is common to the components of the circuit. Here, we are dealing with a parallel circuit, therefore the voltage V is the common quantity. Hence, we start the phasor diagram by taking the voltage phasor as reference. The current IR is in phase with V and current IL lags voltage V by 90° b). The resultant current phasor I is then found by adding phasors IR and IL.

RC For the parallel RC C are given as

a, the current IR through resistor R and the current IC through capacitor IR =

V R

and

IC =

V V = = V jwC) - jXC - j (1/w C )

279 Using KCL, the total current I is given as V I = IR + I C = + V jwC ) = V ÊÁ 1 + jw C ˆ˜ = VY R ËR ¯ Therefore, for this circuit, the complex admittance is given as Ê1 ˆ G + jB) = Á + jw C ˜ ËR ¯ Thus, the conductance G = 1/R and susceptance, B = wC. Equation 10.19 shows that the current phasor I leads the voltage phasor V by the admittance angle, given by B wC q = tan–1 = tan -1 = tan–1 wCR 1/ R G

Y

I IR R

V

IC

IC

I

C q O

(a) The circuit.

V

IR (b) Phasor diagram.

Phasor Diagram Taking voltage V as reference, the current IR is in phase with voltage V, but the current b). The resultant current phasor I is then found by adding phasors

IC leads voltage V IR and IC. EX A MP L E

10. 4 j

An ac sinusoidal voltage, V

I

j10) A. Find the

reactive power.

Solution Given: V =

j

IMP O R TA N T

N O T E

–

I

j

–

While converting the rectangular form of a phasor into its polar form, you should be careful. For example, for the current phasor I, the angle is calculated as 10 q = tan–1 wrong value !) 4 I j10, then also you would have got the same result for the angle.) The correct way Then, decide the actual angle depending upon the quadrant in which the phasor lies. The given current phasor I j10 Hence, the actual angle should be q

Impedance,

Z=

200–36.87∞ V = 10.77–111.8∞ I

–

4.83 – j17.93) W

280

Basic Electrical Engineering

Since the imaginary part of the impedance is negative, the circuit is capacitive. Power factor, pf 0.26 leading Active power, Pa = VI cos q ¥ ¥ 560 W Reactive power, Pr = VI sin q ¥ ¥ 2080.8 VAR EX A MP L E

Z

10. 5

j10) W. Determine the values of the two elements.

Solution The nature of the impedance indicates that the circuit is inductive. j10) W

Z = R + jXL \

R = 10 W

EX A MP L E

Z

XL = 10 W fi

and

L=

10 XL = pf 2p ¥ 50

31.8 mH

10. 6

j10) W. Determine the values of the two elements.

Solution The nature of the impedance indicates that the circuit is capacitive. Since the circuit has two elements Y \

G + jB) =

1 1 1 = = Z 10 - j10 14.14–- 45∞

fi R=

G

1 1 = = 20 W G 0.05

and

–

j

B

fi C=

B

=

0.05 = 159 mF 2pf

RLC a shows a series RLC circuit. Writing KVL equation, we get applied voltage as V = VR + VL + VC = IR + I jXL ) + I jXC) = I[R + j XL – XC)] = IZ where, Z is the complex impedance of the given circuit. We can write 1 ˆ Ê Z = R + j XL – XC) = R + j Á w L ˜ Ë wC ¯ 2

or

Z=

1 ˆ w L - (1/w C ) Ê – tan -1 R2 + Áw L ˜ wC ¯ R Ë 2

w L - (1/w C ) 1 ˆ Ê and q = tan–1 R2 + Áw L ˜¯ Ë R wC Usually, we take the applied voltage as the reference phasor, i.e., V = V–0°. In such case the resulting current I is given as V V –0∞ w L - (1/w C ) I = I–f = = –- tan -1 2 2 Z R R + (w L - 1/w C )

or

Z = |Z| =

281 In a series RLC depends upon the relative values of the terms wL and 1/wC. There can be following three possibilities : (1) w 1/w The phase angle f the voltage. The circuit behaves as an inductive circuit. (2) w 1/w The phase angle f the voltage. The circuit behaves as a capacitive circuit. (3) w = 1/w The phase angle f = 0. The current is in phase with voltage. The circuit behaves as a purely resistive circuit. This is a special case, and is called resonance. We shall talk about this phenomenon a little later.

Phasor Diagram Since it is a series circuit, we take current I as the reference phasor. The voltage VR across resistor R will be in phase with the current I. The voltage VL across inductor L leads the current I by 90°. The voltage VC across capacitor C lags the current I by 90°. For wL > 1/wC, the phasor diagram is as b. For the case, wL < 1/wC c. Since the phasors VL and VC VL + VC. We then add to it the phasor VR EX A MP L E

10. 7 a

a

e diagram for the circuit.

Solution XL = wL a) The impedance,

b

f

pf L = 100p ¥

W; XC = 1/wC

Z = R + j XL – XC 2

2

j -1

c d) the voltage g) the average power. Also, draw the phasor

p ¥ 100 ¥ 10 (12 + j15.3) W

= 12 + 15.3 – tan (15.3 /12) = 19.4–51.9° W V 100– 0∞ b) The current, I = = = 5.15–– 51.9° A Z 19.4–51.9∞ c) The phase angle, f = –51.9°

W

282

Basic Electrical Engineering

d) The voltage, e) The power factor, f) The apparent power, g) The average power,

VR = I R ¥ 61.8 V; VL = I XL VC = I XC ¥ 163.8 V pf 0.617 lagging Papp = VI = 100 ¥ 515 VA Pavg = VI 317.77 W

¥

242.6 V

b.

RLC Figure 10.9a shows a parallel RLC circuit. Writing KCL equation, we get total current I supplied by the applied voltage as V V V I = IR + IL + I C = = VG + V jYL) + V jYC) + + R jX L - jX C or I = V[G + j YC – YL)] = VY where, Y is the complex admittance of the given circuit. Obviously, 1 1 1 C 1 = and YC = = = wC G = ; YL = wL XL R XC 1 Note that +j is associated with YC XL) and –j with YL. The complex admittance of the circuit can be written as Y = G + j YC – YL) = G 2 + (YC - YL )2 – tan -1 YC - YL G Take the applied voltage as the reference phasor, i.e., V = V–0°, the resulting current I can be written as Y - YL I = I–f = VY V–0°) ( G 2 + (YC - YL ) 2 ) – tan -1 C G IC

I

V

IR IR

IL

IC

R

L

C

V

IL + IC

I

IL (a) The circuit.

(b) Phasor diagram.

283

However, the condition required to get maximum power transferred from a source to a load is a bit different in ac circuits, as shown below.

Consider that a load ZL = RL + jXL is connected across the terminals AB of Thevenin’s equivalent of an ac circuit, as shown in Fig. 10.10a. The ac voltage source VTh is the Thevenin’s equivalent voltage source and ZTh = RTh + jXTh is Thevenin’s equivalent impedance.

Let the voltage across and the current through the load be given as VL = VL–q V and IL = IL–f A with q > f Therefore, the load impedance is given as ZL

RL + jXL) =

VL VL –q VL = = –q – f = ZL–qL I L –f I L IL

The load impedance triangle is given in Fig. 10.10b. The power absorbed by the load is given as P = VLIL cos qL Now, from the circuit of Fig. 10.10a, the load current phasor is given as IL =

VTh VTh VTh = = ZTh + Z L ( RTh + jX Th ) + ( RL + jX L ) ( RTh + RL ) + j ( X Th + X L )

Therefore, the magnitude of load current, IL =

VTh ( RTh + RL ) 2 + ( X Th + X L ) 2

Hence, the magnitude of the load voltage is given as VTh VL = ILZL = ◊ RL2 + X L2 2 2 ( RTh + RL ) + ( X Th + X L )

284

Basic Electrical Engineering

From the load impedance triangle of Fig. 10.10b, we get RL R cos qL = L = ZL RL2 + X L2 P = VLIL cos qL =

or

P=

VTh 2

( RTh + RL ) + ( X Th + X L )

2

◊ RL2 + X L2 ◊

VTh 2

( RTh RL ) + ( X Th + X L )

2

◊

RL RL2

+ X L2

2 VTh RL

( RTh + RL ) 2 + ( X Th + X L ) 2 XTh + XL) absorbs no power. Its presence in the expression reduces power. XTh + XL XL = – XTh 2 VTh RL

P=

( RTh + RL ) 2

RL = RTh condition for maximum power transfer in ac circuits as RL = RTh and XL = –XTh fi

RL + jXL = RTh – jXTh

or Z L = Z*Th

That is, to get maximum power transferred, the load impedance must be equal to complex conjugate of the Thevenin’s impedance. Purely Resistive Load

What would be the condition for maximum power transfer, if the load is purely RL = RTh. But it is not correct. Let us derive

P=

2 VTh RL

( RTh + RL ) 2 + ( X Th ) 2

=

2 VTh RL 2 RTh

+

2 X Th

+ RL2 + 2 RTh RL

Power P becomes maximum when the differential of P with respect to RL 2 2 V 2 ( R 2 + X Th + RL2 + 2 RTh RL ) - VTh RL (2 RL + 2 RTh ) dP = Th Th =0 2 2 dRL ( RTh + X Th + RL2 + 2 RTh RL ) 2

285 2 2 2 VTh ( RTh + X Th + RL2 + 2 RTh RL - 2 RL2 - 2 RTh RL )

or

2 2 ( RTh + X Th + RL2 + 2 RTh RL ) 2

fi

2 2 RTh + X Th

RL =

EX A MP L E

10. 8 a) Find the current i t

c nected in place of the capacitor of

1 6

b) Find the voltage across the capacitor by applying

mF, so as to consume maximum power from the remaining circuit.

Solution

a) The equivalent resistance faced by the voltage source, ( j )(1 - j 2) j + 1 - j2

Zeq \

=0

I =

–

W

Vs 28.28–0∞ V = Zeq 2.5–36.87∞ kW

–

fi

Im

Therefore, the expression for i t) can be written as i = 16 sin (3000t – 36.87°) mA b) Thevenin’s equivalent voltage is given as j1 1.5 + j1 Thevenin’s equivalent impedance is given as VTh

ZTh = 1 +

–0° ¥

(1.5) ¥ ( j1) (1.5 + j1)

–

j

W

286

Basic Electrical Engineering

Now, applying voltage division, the voltage across the capacitor is given as VC = VTh c) We have ZTh

–

¥

- j2 = 16.0–8.2° V 1.46 + j 0.69 - j 2

W. The load impedance that consumes maximum power is given as

j

ZL = \

- j2 Z Th - j 2

Z*Th

j

RL = 1.46 kW;

and XC

W W

1 1 = = 483.1 nF w X C 3000 ¥ 0.69 ¥ 103

fi C=

ADDITIONAL SOLVED EXAMPLES EX A MP L E

10. 9 W. Determine the value of R for which

A circuit consists of a resistance R

Solution The power factor, cos q sin q = 1 − cos2 θ = 1 (0.8)2 From the impedance triangle, we know that X 3 X tan q = C fi = C R 4 R EX A MP L E

fi or

tan q =

3 60 = 4 R

sin cos fi

=

0.6 3 = 0.8 4

R = 80 W

10. 1 0

a

b

c) the power

drawn by the coil.

Solution a) The iron choke has both resistance R and inductance L. When connected to a dc supply, only its resistance is effective in limiting the current; when connected to an ac supply, its impedance becomes effective in limiting the current. Hence, 20 V 65V R = =5W and Z= W fi XL = Z 2 R2 W 4A 5A X 12 \ L = L = 38.2 mH 2 p ¥ 50 pf R 5 b) The power factor, pf = cos q = = = 0.38 (lagging) Z 13 c) The power, P = VI cos q ¥ ¥ 123.5 W EX A MP L E

10. 1 1

287 Solution On changing the frequency of the ac supply, only the reactance part of the choke changes. Thus, V 240 V = 60 A I1 V 240 V Z = = 40 A I2

At f At f i

fi

p¥

R

W fi R

L)

i)

p ¥ 100L)

ii)

ii), we get p) L

p) L X1

fi

R =

EX A MP L E

W

Z1 =

fi pf1L Z12

L= p¥ X12

=

20

= 8.22 mH

(200p ) 2 - (100p ) 2 ¥ ¥ 10

4

2

2.58

2

W

= 3.05 W

10. 1 2

W choke coil. Sketch a neat phasor diagram, indicating the current and all voltages.

Solution A choke coil has some resistance, say r, and some inductance, say L. Though the resistance is distributed throughout the coil winding, we can show it as lumped parameter in its equivalent circuit, as given in Fig. 10.10a. Because of the presence of this resistance, the voltage across the choke coil, VCh, leads the current I by an angle less than q b. As per KVL, the phasor voltages VR and VCh must add up to the supply voltage V. That is, V = VR + VCh. Hence, ( 440)2 − (200)2 − (300)2 V = VR + V Ch VRVCh cos q fi cos q = 2 × 200 × 300

Now, I =

VR 200 = R 100

EX A MP L E

VCh ¥ I ¥ cos q

P

¥

¥

318 W

10. 1 3 W

W

a

b) the power

c) the power factor of the whole circuit

Solution \

X1 Z1

pf L1 = 100p ¥ j W

–

Z W W

and and

X Z

pf L = 100p ¥ j W

–

W W

288

Basic Electrical Engineering

\

¥

a) V1 = IZ1 b) P1 = I R1 c) pf = cos q

¥

EX A MP L E

W

j

Z = Z1 + Z

–

W

fi

I=

ª 174 V and V = IZ ª 109 W and P = I R1 0.469 (lagging)

¥

V 230 = Z 85.4 ª 76 V ¥ ª 182 W

10. 1 4

current and power will be taken from the same source, if the two coils joined in series are connected to it ?

Solution 100 W; R1 = 8 100 For coil B : Z = = 10 W; R1 = 10 For the two coils joined in series, we have R = R1 + R For coil A : Z1 =

\

Z =

2

R +X

2

¥

P =I R EX A MP L E

P I2 P I2

= =

120

W

82 500

W

10 2 W

fi X1 =

Z12

R12

fi X =

Z22

R22

W W W

X = X1 + X

and

V 100 W; I = = = 4.52 A Z 22.115 140.46 W

10. 1 5 RA

LB b) the inductance LA

W and inductance a) the resistance RB of

c) the voltage drop across each coil.

Solution a) S = fi

I =

P 2 + Q2 =

32 + 2 2

S 3.6055 103 = V 240

The real power is given as ¥ 10 8.3 W

P = I RA + RB fi RB RA b) The reactive power is given as Q = I XA + XB ¥ 10 But, XB pfLB p¥ ¥ \

LA =

W;

RA + RB)

XA + XB) \ XA

fi

fi

W

RA + RB

W

XA + XB W

XA = 4.155 = 0.0132 H 2p f 2p ¥ 50

c) ZA = RA2 + XA2 = \ VA = IZA

52 + 4.1552 ¥

W; ZB = 97.63 V

and

RB2 + X B2 = VB = IZB

8.32 + 4.712 ¥

W 143.336 V

289 EXA M P L E

10 . 16

a b c W resistance, in series with the electric bulb so as to drop the excess voltage. Calculate the value of the component used, the total power consumed and the power factor in each case. Giving reasons, state which alternative is the best. P 100 = V 120 V 120 fi R= R = = 144 W; pf = 1 I 0.833 ¥ 200 W 207.84 1 fi XC = W fi C= = 12.76 mF 100p X C 0.833

Solution Rated current of the electric bulb, I = a) VR = V – Vb Total power consumed, Pt = VI b) VC =

240 2 120 2

120 = 0.5 (leading); P = VI cos f ¥ 240 2 c) Vr = I r ¥ VL = 240 − (120 + 8.33)2 XL 243.46 \ L = = = 0.775 H 100 100 128.33 pf = cos f = = 0.535 (lagging); P ¥ 240 b

¥

pf = cos f =

100 W fi

¥

XL =

202.8 0.833

W

107 W

a) is the worst, b) is better than c), as it involves less wastage of power and, furthermore, a capacitor is much cheaper than a choke coil.

EXA M P L E

10 . 17

W resistance is connected in series with a parallel combination of a capacitance C a. This circuit is connected to a voltage source of angular frequency w = 1000 rad/s. Find the value of capacitance C

Solution The inductive reactance, XL = wL = 1000 ¥ W. The impedance of the circuit is given as Z = R ± jX, because the net reactance X of the parallel combination of L and C can be either inductive or capacitive. b. The voltage VR is in phase with current

290

Basic Electrical Engineering

I. If X is inductive, the voltage phasor VX1 leads the current phasor I by 90°, and the source voltage V1 leads the current I X is capacitive, the voltage phasor VX lags the current phasor I by 90°, and the source voltage V lags the current I X fi X=R W R Case I X inductive) : The net reactance X is given as 1 1 1 1 1 1 1 1 1 1 1 1 + = fi = or = = = jX jX jX X XL XC XC XL X 15 20 60 L C 1 1 fi XC W fi C= = = 16.67 mF 1000 60 w XC Case II X capacitive) : The net reactance X is given as 1 1 1 + = jXL - jX C - jX

fi

XC =

EX A MP L E

60 7

1 1 = X XL

fi

fi C=

1 XC

or

1 1 1 1 1 7 + = + = = XC XL X 15 20 60

7 1 = = 116.67 mF w XC 1000 60

10. 1 8

j W and Z j W, are connected in parallel. If the potential difference across j0) a b) the power consumed by each branch and the total power, and c) the pf of each branch and the overall pf. Two impedances, Z1

Solution Z1

j

W

–

W; Z

j

W

–

W; V

j

–0° V

a) The currents are given as I1 =

V 230– 0∞ V = = 11.98––51.3° A Z1 19.2–51.3∞ W

I =

V 230– 0∞ V = = 25.73–26.56° A Z2 8.94–-26.56∞ W j

I = I1 + I b)

P = VI cos q

¥

¥

j

j j j

7011.42 W

P1 = I1R1 ¥ 1722.24 W; P = I R ¥ 5296.26 W R 12 R 8 c) pf1 = 1 = = 0.625 (lag); pf = 2 = = 0.895 (lead); pf Z1 19.2 Z2 8.94 EX A MP L E

30.56–4.03° A

0.998 (lead)

10. 1 9 W a

b) the

power dissipated in the coil. Also, draw the phasor diagram.

Solution

a

current I is given as I=

b. The magnitude of the VR 130 = R 50

A

291

a) From the parallelogram OACB, the angle q is given by ¥ ¥ VL = VC sin q Vr = VC cos q

\ and \

L =

and

r =

b) P = IVr EX A MP L E

¥

XL

=

¥ cos q ¥ ¥

fi

cos q

q

VL /I V 179.46 = L = = 0.22 H w I 2.6 × 2 π × 50

Vr 13.84 = = 5.32 W I 2.6 b.

36 W

10. 2 0 t V and i v-i relationship.

The terminal voltage and current for a series circuit are v

Solution The simplest RC

a

the given expressions of v and i, we have V 141.4 V= m = = 100 V 2 2

and

b. From I=

Im 7.07 = 2 2

fi

R + XC =

From the circuit, we have V = IZ

¥

R2 + XC2

t

100 2 52

i)

292

Basic Electrical Engineering

From D

b, VR 5R IR = = V 100 V

Putting this value of R

fi

R = 16 W

ii)

i), we get

XC

W

XC

EX A MP L E

10. 2 1

b

i) the voltage V

fi C=

1 1 = = 41.67 mF w XC 2000 12

a i) the current I ii) the voltage V1 by using voltage division ii) the voltage V1, by applying Thevenin’s theorem.

Solution a

i) The impedance Z of the two parallel branches on the right, Z =

( j 20)(15 - j 30) ( j 20) + (15 - j 30)

j

–

W

Thus, the total impedance across the voltage source, –

Z = 10 + Z \

I =

W

100– 20∞ = 2.32––28.6° A 43.1– 48.6∞

ii) Using voltage divider, the voltage V across the two parallel branches, V =V \ b

Z2 10 + Z 2 –

V1

–

Ê 37.2–60.3∞ ˆ ÁË 10 + 37.2–60.3∞ ˜¯

–

Ê - j 30 ˆ ÁË 15 - j 30 ˜¯ = 77.3–5° V

i

j j voltage divider, VTh = 100– ZTh =

¥

15 - j 30 10 + 15 - j 30

10 ¥ (15 - j 30) (10 + 15 - j 30)

–

– W

293

\

V = VTh

ZL ZTh + Z L

–

j 20 = 86.42–31.64° V 8.59–-13.24∞ + j 20

¥

ii) Now, we treat –j ¥

VTh = 100– ZTh \

V1 = VTh

EX A MP L E

j 20 10 + j 20

–

10 ¥ j 20 (10 + j 20)

–

ZL ZTh + Z L

–

W ¥

- j 30 = 77.29–5.07° V - j 30 + 23.35–9.87∞

10. 2 2 a

IR

b) the Norton’s theorem.

Solution (- j 5)(4) – W. - j5 + 4 Applying current division, we get

a) Z = j

–

I

¥

2 2 + 2.65– 23.3∞

–

Again applying current division, we get –

IR

¥

- j5 = 6.85––7° A 4 - j5

W

b we get

–

IN

and \

ZN =

¥

2 = 11.09– 2 + j3

(- j 5) ¥ (2 + j 3) (- j 5 + 2 + j 3)

I R = IN

ZN ZN + R

– –

W ¥

6.37–11.31∞ = 6.84––6.99° A 6.37–11.31∞+ 4

294

Basic Electrical Engineering

EX A MP L E

10. 2 3 a

V

b) the Thevenin’s

theorem.

Solution a) Z =

(40)(20 + j 50) 40 + (20 + j 50)

–

W

Therefore, using voltage divider concept, the voltage across Z is given as

V

Ê ˆ 27.58– 28.39∞ ÁË - j 20 + 27.58– 28.39∞ ˜¯

–

V

Ê 20 ˆ ÁË 20 + j 50 ˜¯ = 81.22–6.03° V

–

W

–

W resistor as load and temporarily remove it from the W resistor. Hence, using voltage divider,

b

VTh = Voc = V

W

40 40 - j 20

–

–

Thevenin’s impedance, ZTh = j

j

j

(- j 20)(40) (- j 20 + 40)

j

W

From the Thevenin’s equivalent circuit, using voltage divider, we get V

W

= VTh

ZL ZTh + Z L

–

¥

20 = 81.23–6.03° V (8 + j 34) + 20

SUMMARY TE RM S

A N D

C ON CE PT S

If a circuit contains both resistance and inductance, the current lags the voltage by an angle less than 90° but greater than 0°. If a circuit contains both resistance and capacitance, the current leads the voltage by an angle less than 90° but greater than 0°.

295 If a circuit contains resistance, inductance and capacitance, the current may lead, lag or be in phase with the voltage depending on the relative values of inductive and capacitive reactances. The quantity Z is called complex impedance, whose real part is called resistance R) and imaginary part is called reactance X), either inductive or capacitive. The inverse of impedance is called admittance Y), whose real part is called conductance G) and the imaginary part is called susceptance B), either inductive or capacitive. We associate +j with XL and –j with XC. IMP O R TA N T

FO R MU LAE

In a series RL circuit,

Z = R + jwL

In a series RC circuit,

Z =R–j

The complex power, The apparent power, The active or real power, The reactive power, For a series RLC circuit,

S S P Q

wC)

I=

and

(V∠ 0° ) I= = Z

V = Z

V 2

R + (w L) V

2

–- tan -1 (w L / R )

2

R + (1/ ω C )

2

∠ tan −1 (1/ ω C R)

= P + jQ = VI* = VIe j q – f) = VI = VI q–f = VI q–f

Z = R + j XL – XC) = R + j wL – 1/wC) = I = I–f =

(V– 0∞ ) = Z

and

V– 0∞ 2

R + (w L - 1/ w C )

2

R 2 + (w L - 1/ w C ) 2 – tan -1

– - tan -1

w L - (1/ w C ) R

w L - (1/ w C ) R

For a parallel RLC circuit, I = I–f = VY where,

G =

V–0°) ( G 2 + (YC − YL )2 ) ∠ tan −1

1 1 1 ; YL = = wL XL R

and YC =

YC − YL G

1 C = = wC XC 1

CHECK YOUR UNDERSTANDING Before you proceed to the next chapter, take this test. Give yourself two marks for each correct answer and minus one for each wrong answer. If your score is 12 or more, go to the next chapter; otherwise study this chapter again. S. No. 1.

Statement All phasors can be represented by complex numbers. Hence, all complex numbers, such as impedance and admittance, represent phasors. RL circuit, the current always lags the voltage. RL circuit is called conductance, and is given as G = 1/R. RC circuit can be written as Y R) – jwC.

True

False

Marks

296

Basic Electrical Engineering

RC circuit, if the current phasor is represented as I–0°, then the voltage can be represented as V––q, where angle q has positive values. the voltage across it leads the current through it by an angle less than 90°. phasor, because the voltage across each of the components is the same. as Y j resistance in series with an inductance.

10.

as Z j W resistance in parallel with a capacitance. In a series RC circuit, the current lags the voltage by an angle Your Score A N SW E RS

1. False 6. True

2. True 7. True

3. False 8. False

4. False 9. False

5. True 10. True

REVIEW QUESTIONS 1. In a series RL circuit, when can you neglect the inductive reactance as compared to the resistance ? 2. What is the impedance of a parallel RL circuit ? How does the series resistance in inductive branch affect this impedance ? 3. How is that in a purely inductive circuit, although P 4. How and why does the frequency of the ac supply affect the magnitude of current in an ac circuit ? 5. Show the wave shape of the current and voltage for a series RL circuit. 6. a) a parallel RC b) a series RC circuit. 7. a) a parallel RL b) a series RL circuit. 8. For a series ac circuit having resistive and reactive components, a) How do you determine the active power con

sumed ? Give two equations for calculating the active power. b c d) How is the VAR calculated ? 9. What do you understand by the term ‘power factor’ ? leading ? When is it lagging ? 10. What is an impedance triangle ? Draw the impedance triangle for a series RL circuit and series RC circuit. 11. Derive expressions for calculating the impedance, admittance, conductance and susceptance of a series RL circuit. 12. In a series RLC circuit, the current is seen to be lagging the voltage. What does this mean ? What is the nature of the circuit ?

AC Circuits

297

MULTIPLE CHOICE QUESTIONS Here are some incomplete statements. Four alternatives are provided below each. Tick the alternative that completes the statement correctly . 1. In a series RL circuit, the phase difference between the applied voltage and the current increases if (a) XL is increased (b) XL is decreased (c) R is increased (d) supply frequency is decreased 2. The power factor of a series RL ac circuit is given by (a) XL/R (b) R/XL (c) R/Z (d) Z/R 3. For greater accuracy, the value of phase angle q in an ac circuit should be determined from (a) cos q (b) sin q (c) tan q (d) sec q 4. A low power factor of an ac circuit means that (a) it causes less voltage drop in the line (b) it draws more active power (c) it draws less line current (d) it draws more reactive power 5. The impedance of a circuit is 15.5 ––30° W. It means that the circuit is (a) capacitive (b) inductive (c) purely resistive (d) None of the above 6. If the active and apparent power of an ac circuit are equal in magnitude, its power factor is (a) 0.5 (b) 0.707 (c) 0.8 (d) 1 7. For the ac circuit given in Fig. 10.20, the power factor is (a) 0.4 lagging (b) 0.6 lagging (c) 0.75 lagging (d) 0.8 lagging 8. The power consumed in the circuit of Fig. 10.20 is (a) 4000 W (b) 2400 W (c) 2000 W (d) 1200 W 6W

8W

RL

XL

200 V

Fig. 10.20

9. The reactive power drawn in the circuit of Fig. 10.20 is (a) 4000 VAR (b) 3200 VAR (c) 2000 VAR (d) 1200 VAR 10. The apparent power drawn in the circuit of Fig. 10.20 is (a) 4000 VA (b) 3200 VA (c) 2000 VA (d) 1200 VA 11. In a series RLC circuit, the inductive reactance is 10 W and the capacitive reactance is 15 W. The total reactance in the circuit is (a) 25 W (b) 18.03 W (c) 5 W (d) 1.5 W 12. A series RL circuit has a resistance of 6 W and a reactance of 8 W. The impedance of the circuit is (a) 10 W (b) 14 W (c) 2 W (d) 13.43 W 13. The resistance and the reactance in a series RC circuit are 7.5 W each. In this circuit, (a) the voltage leads the current by 45° (b) the current leads the voltage by 45° (c) the voltage leads the current by 60° (d) the current leads the voltage by 15° 14. In an RC circuit, the resistance and reactance are 1 W and 10 W, respectively. In this circuit, (a) the voltage leads the current by 84.3° (b) the current leads the voltage by 5.7° (c) the voltage leads the current by 5.7° (d) the current leads the voltage by 84.3° 15. In a series RLC circuit, R = 5 W, XL = 10 W and XC = 15 W. If this circuit is connected to a voltage source v = 100 sin (314t + 30°) V, the rms value of the current will be (a) 14.14 A (b) 10 A (c) 5 A (d) 3.33 A 16. The impedance of an RL circuit is 25 W at a frequency of 50 Hz. At a frequency of 60 Hz, its impedance will be (a) greater than 25 W (b) exactly 25 W (c) less than 25 W (d) 0 W 17. The impedance of an RC circuit is 20 W at a frequency of 50 Hz. At a frequency of 60 Hz, its impedance will be (a) greater than 20 W (b) exactly 20 W (c) less than 20 W (d) 0 W

298

Basic Electrical Engineering a c

18. The maximum and minimum values of power factor in an ac circuit can be

b) +1 and –1 d) +10 and –10 A N SW E RS

1. a 11. c

2. c 12. a

3. c 13. b

4. d 14. d

5. a 15. b

6. d 16. a

7. b 17. c

8. b 18. b

9. b

10. a

PROBLEMS ( A )

S I M P L E

PR O BL EMS

1. A series circuit having a resistance of 10 W and an i t A. Obtain the total voltage across the series circuit, and the angle by which the current lags the voltage. [Ans. 2. W and W in series. A voltage, v t) V is applied to the circuit. a) Find the complex impedance and draw the impedance triangle. b) Determine the rms and instantaneous values of the current. c) Calculate the power delivered to the circuit. d) Find the equation for the voltage appearing across the capacitor. e) Find the value of the capacitance. [Ans. a) 10– W; b) 10 A, i t c d) v t e mF] 3. A series RC circuit, having R W and C mF, a

b

c) the current d) the power factor of the

circuit. [Ans. a c 4. A series RL circuit, having R – ( B)

T R IC KY

W b

W;

d W and L

a) the current drawn by the circuit, and b) the power factor of the circuit. [Ans. a – b 5. Two impedances Z1 j W Z j W supply. Determine the current, active power, apparent power and power factor of the circuit. [Ans. 6. Given for an ac circuit v tV i t a b c) d) reactive power. [Ans. a b c d 7. A coil has a resistance of 10 W and draws a current a) the inductance of the coil, b c) the reactive power. [Ans. a b c 8. W resistor is connected in parallel to a a impedance of the circuit ? [Ans. a –

b) the input b

–

W]

PRO BLE MS

9. An adjustable resistance R in series with a capaci m a) the value of the resistance so that the voltage across

b) the [Ans. a

c) the pf. W b

c

AC Circuits 10. When a two-element series circuit is connected to an ac source, v = 200 2 sin (314t + 20°) V, the resulting current is found to be i = 10 2 cos (314t – 25°) A. Determine the values of the elements of the circuit. [Ans. R = 14.14 W; C = 225.2 mF] 11. A parallel RL circuit, having R = 50 W and XL = 40 W, is connected across 100-V ac source. Calculate (a) the current drawn from the source, (b) the apparent power, (c) the real power, and (d) the reactive power. [Ans. (a) 3.2 A; (b) 320 VA; (c) 200 W; (d) 249.7 VAR] 12. A 46-mH inductive coil has a resistance of 10 W. (a) How much current will it draw if connected

( C )

C HA L L EN G IN G

across a 100-V, 60-Hz source ? (b) What is the power factor of the coil ? [Ans. (a) 5––60° A; (b) 0.5 lagging] 13. A capacitor is connected across the coil of Prob. 12 to make the power factor of overall circuit unity. Determine the value of the capacitance. [Ans. 153 mF] 14. A two-element series circuit draws 600 W of power and 10 A of current at 100 V and 500/p Hz. Specify the values of these elements. [Ans. R = 6 W and L = 8 mH or C = 125 mF] 15. A parallel RL circuit has R = 4 W and XL = 3 W. Obtain its series equivalent such that the series circuit draws the same current and power at a given voltage. [Ans. R = 1.44 W, XL = 1.92 W]

PROBLEMS

16. A 100-V, 60-W lamp is to be operated from a 220-V, 50-Hz mains. What (a) pure resistance, and (b) pure inductance, placed in series with the lamp, will enable it to be used without being over-run ? Which method would be more economical ? [Ans. (a) 200 W, (b) 1.039 H; Second method] 17. A choke coil, when connected to a 230-V, 50-Hz supply, takes a current of 15 A. On decreasing the frequency of the supply to 40 Hz, the current increases to 17.2 A. Determine the resistance and inductance of the coil. [Ans. R = 8.872 W; L = 39.74 mH] 18. A capacitor is used in series with a tungstenits rated illumination when connected to a 220-

V, 50-Hz supply. Calculate (a) the value of the capacitor, and (b) the power factor of the current drawn from the supply. [Ans. (a) 81.2 mF; (b) 0.455 (leading)] 19. It is desired to run a bank of ten 100-W, 100-V lamps in parallel from a 230-V, 50-Hz supply by inserting a choke coil in series with the bank of lamps. If the choke coil has a pf and inductance. [Ans. 3.76 W, 18.4 W, 0.058 H] 20. A non-inductive resistor is connected in series with an iron-cored coil and a capacitor. The circuit is connected to a single-phase ac supply. When voltages across the elements are as indicated in the Fig. 10.21. Find the applied voltage and the power loss in the coil. [Ans. 34.17 V, 1.9 W] I1

10 W

0.12 H A

I I2

40 mF

20 W B +

Fig. 10.21

299

200 V, 50 Hz

Fig. 10.22

–

300

Basic Electrical Engineering

21. Two branches A and B are connected in parallel a) the currents in each branch, c) the power factor. [Ans. a – – b – c 22. Three circuit elements R W, XL W, and XC = 10 W a) Determine the admittance of each element and hence obtain b) If this circuit is connected

R and L. [Ans. R = 10 W, L

b

current in each branch and the total input current. [Ans. a) YR j0) S, YL j YC j0.10) S, Y j b) IR –0° A, IL –– 90° A, IC = 1.0–90° A, I – 23. The resistance R and the inductance L of a coil are to be determined experimentally. The available W resistor. W resistor is connected in series with the coil

24. When a capacitor of unknown value is connected in series with the given circuit and the combination put across the same voltage source, the magnitude of the input current is found to increase. Determine the values of the original circuit elements. [Ans. R W, L 25. the value of C out of phase with the input voltage at frequency w Ans. mF] 26. C such that the input voltage V and the input current I are in the same phase. [Ans. mF]

j12 W I

C

20 W

6W I

V

15 mH

C

V = 100–0° V

EXPERIMENTAL EXERCISE 10.1 S E RIE S

R L

C I RC U IT

Objectives 1. To observe the variation of current I when the impedance of a series RL circuit is varied. To draw the locus diagram for a series RL circuit by varying the resistance.

Apparatus Circuit Diagram

W

301

Brief Theory For the series RL

a, the quantities V, I, VR and VL represent the rms values of the supply voltage, the current, the voltage across the resistance and the voltage across the inductor, respectively. According to the Kirchhoff’s voltage law, the phasor sum of voltages VR and VL must be equal to the phasor V. That is, V = VR + V L i) b, the current I lags the supply voltage V by an angle f1 for the inductive load. The voltage VR I, but the voltage VL phasor AP) leads the current by 90°. The phasor OP is then the phasor sum of voltages VR and VL, and hence must represent the supply voltage V.

Keeping the supply voltage constant, if we increase the resistance R, the voltage VR phasor OB) and the voltage VL i). Whatever be the value of resistance R, the phasor VL is always perpendicular to phasor VR. Hence we must have V = VR + VL This shows that the locus of the tip of the phasor OA is a semicircle of the diameter equal to supply voltage V. For a given value of the resistance R, the phase angle can be calculated by V ⎛V ⎞ cos f = R or f = cos –1 R ⎝ V⎠ V

Procedure 1. Switch on the power supply. Adjust the variac so that the voltmeter V

ii)

iii)

302

Basic Electrical Engineering

Adjust the rheostat R W). Note the readings of the ammeter A and voltmeters VR and VL. Go on increasing the value of resistance R in steps, and for each step note the above readings. iii). Draw the phasor diagram taking the phasor V I lagging at an angle F. Mark the phasor VR along phasor I, and join point A to point P. 9. Draw a semicircle taking OA as diameter. 10. Switch OFF the supply.

Observations and Calculations Supply voltage, V = …… V S. No. 1.

VR

VL

I

f

Results 1. The current I goes on decreasing as the resistance R increases. From the phasor diagram drawn, it is seen that the locus of point A for different set of readings falls on the semicircle.

Precautions 1. Before switching ON The terminals of the rheostat should be connected properly. the current rating of the rheostat. The supply voltage V should remain constant throughout the experiment. If required, the variac may be adjusted in between.

Viva Voce 1. What would be the power factor of the series RL Ans. : As power factor, pf = cos f = R/Z R = 0. 2. What is the power factor of circuit having pure inductance ? Ans. : For such a circuit, f = 90°, hence pf 3. What is the frequency of your ac supply ? Ans. : 4. Why are the ac quantities expressed in rms values and not in average values ? Ans. : quantity we cannot take average value. 5. What is meant by rms value ? Ans. : It means the root of mean of squares energy transfer point of view. 6. Ans. : Vm = 2 ¥ ª

2

303 7. Ans. : We do not associate any algebraic sign with amplitude. 8. What does the form factor of an ac wave indicate ? Ans. : It indicates how far the ac wave departs from a sinusoidal wave. The form factor of a sinusoidal wave is is more than 1.11, the wave is said to be peaky. 9. If a series RL circuit has R W and XL W Ans. : No. As Z = R + jXL; Z =

2

R +

X L2

=

W? 2

2

4 +3

W.

EXPERIMENTAL EXERCISE 10.2 PAR A L L E L

AC

C IR C U IT

Objectives 1. To observe the variation of current I when the resistance in the parallel ac circuit is varied. To draw the phasor diagram for the parallel ac circuit for four sets of impedances obtained by varying the resistance. R, L and C) for the four sets of observations, assuming the ac supply

Apparatus

W

Circuit Diagram

Brief Theory

R in series with an inductance L, and the other branch has only a capacitance C. Various voltage and current relationships are VR = I1R, VL = I1 XL, VC = I XC, I = I1 + I and V = VC = VR + VL i)

304

Basic Electrical Engineering

Keeping the supply voltage constant, if we increase the resistance R, the voltage VR phasor OA) and the voltage VL i). Whatever be the value of resistance R, the phasor VL is always perpendicular to phasor VR. Hence we must have V = VR + VL ii) This shows that the locus of the tip of the phasor OA is a semicircle of diameter equal to supply voltage V. The current I leads the voltage V by 90°.

Procedure 1. ON the power supply. With the help of the variac, increase the supply voltage so that some observable readings are obtained in all the meters. Note the readings of the wattmeter and all the three ammeters. R, L and C). Repeat the experiment with four more values of the applied voltage, keeping the setting of the rheostat different every time. OFF the supply.

Observations S. No. 1

V

I

I1

I

I

VR

VL

VC

P

Calculations 1. For each set of observations, calculate the values of the circuit parameters, R, L and C as follows : V V V 1 XL R = R ; XL = L fi L = ; XC = C fi C = 2p ¥ 50 ¥ X C 2p ¥ 50 I1 I1 I2 Z=

V I

305

pf =

P VI

Measure the phase angle f and calculate the value of cos f. pf), one obtained from measurement and the other obtained form the phasor diagram. S. No. 1

R

L

C

Z

pf

pf from phasor diagram)

Results 1. For different sets of observations, the calculated values of R, Z and pf are different; but the values of L and C are found to be almost same. pf are almost same for each set of observations.

Precautions

A1 the current rating of the rheostat.

Viva Voce 1. While solving a parallel ac circuit, we normally work with admittances rather than impedances. Why ? Ans. : For a parallel circuit, the equivalent admittance is the sum of admittances of individual branches. 2. What is the sign of inductive reactance ? Ans. : It is conventionally taken as positive. 3. Is the sign of inductive susceptance also positive ? Ans. : No. If the sign of inductive reactance is taken positive, the sign of its susceptance will be negative. 4. What is the sign of capacitive reactance ? Ans. : It is negative.

EXPERIMENTAL EXERCISE 10.3 TH R EE - A M M ET ER

MET HOD

Objectives Apparatus Circuit Diagram

W a.

306

Basic Electrical Engineering

Brief Theory The power consumed by the inductive load is given as P = VI cos f = VI cos f

i)

voltmeter and an ammeter. The power is normally measured by a wattmeter. However, it is possible to measure power in an ac circuit by using three ammeters. b shows the phasor diagram for the ac circuit. The current I through the rheostat is in phase with the applied voltage V. The current I through the inductive load lags the voltage V by an angle f. The total current I1 is the phasor sum of currents I and I . Therefore, we can write I1 = I + I \

power factor, pf = cos f =

I 12

i),

- 2I 2 I 3

I cos f = ii), we get

I I cos f

I 22

I 32

ii) iii)

P V

P V fi P I1 – I – I ) iv) I V iii iv), we can calculate the power factor and the power consumed by the load by taking the readings of the three ammeters and the voltmeter. I1 = I + I

I

Procedure a.

1.

ON the power supply. With the help of the variac, change the supply voltage so that some observable readings are obtained in all the meters. Note the readings of the three ammeters and the voltmeter.

Switch OFF the supply.

307 Observations and Calculations S. No. 1

V in V)

I1 in A)

I in A)

I in A)

P in W)

pf

Results 1. For all sets of observations, the value of the power factor remains the same. On changing the power supply voltage, the power consumed by the inductive load also changes.

Precautions 1. Before switching ON The terminals of the rheostat should be connected properly. the current rating of the rheostat.

Viva Voce 1. Ans. : The apparent power. 2. In an ac circuit, can these two powers ever have the same value ? Ans. : Yes. It is possible when the load is purely resistive. 3. Is the load in your experiment purely inductive or simply inductive ? Ans. : It is simply inductive load. 4. What is the basic difference between these two types of load ? Ans. : An inductive load includes a resistance in addition to an inductance. On the other hand, a purely inductive load has only inductance. In practice, it is not possible to have a purely inductive load, as an inductance has some resistance of its own. 5. wattmeter ? Ans. : method.

EXPERIMENTAL EXERCISE 10.4 T H R E E -V O LT M E T E R

M ETHOD

Objectives Apparatus Circuit Diagram Brief Theory

W a.

b shows the phasor diagram for the ac circuit. The current I is in phase with the voltage V across the rheostat. The voltage V across the inductive load leads the current I by an angle f. The total voltage V1 is the phasor sum of the voltages V and V . Therefore, we can write

308

Basic Electrical Engineering

V1 = V + V \

Power factor, pf = cos f =

V V cos f V12

- V 22

i)

- V32

ii)

2V2V3 As the power consumed by the inductive load is given as P = IV cos f, we have P V cos f = I i), we get

P I fi P V1 – V – V ) iii) V I ii iii), we can calculate the power factor and power consumed by the load by taking the readings of the three voltmeters and the ammeter. V1 = V + V

V

Procedure a.

1.

ON the power supply. With the help of the variac, change the supply voltage so that some observable readings are obtained in all the meters. Note the readings of the three voltmeters and the ammeter.

Switch OFF the supply.

Observations and Calculations S. No. 1

I in A)

V1 in V)

V in V)

V in V)

P in W)

pf

309 Results 1. For all sets of observations, the value of the power factor remains the same. On changing the power supply voltage, the power consumed by the inductive load also changes.

Precautions 1. Before switching ON The terminals of the rheostat should be connected properly. the current rating of the rheostat.

Viva Voce 1. Ans. : Instead of using three separate voltmeters, we can use a single voltmeter with probes to measure the three voltages. 2. Why have you taken the current as reference phasor in your phasor diagram ? Ans. : Because in the series circuit, the current remains the same for all the elements.

RESONANCE IN AC CIRCUITS

11

OB JE CT IV E S

1 2 0

Resonance is the condition that exists in ac circuits under steady state when the input current is in phase with the input voltage. When in resonance, the ac circuit is purely resistive and draws power at unity power factor. The condition of resonance can be obtained by connecting an inductor and a capacitor in series (or in parallel) across an ac voltage source of variable frequency. At resonance, the impedance of the circuit is minimum (or maximum). Hence, for a given applied voltage the current becomes maximum (or minimum).

Consider the circuit shown in Fig. 11.1a. An inductor and a capacitor are connected in series across an ac voltage source of voltage V and frequency f. An inductor (say, an iron-cored coil) has a small resistance R and an inductance L. A capacitor usually has no losses, and hence it is represented as a pure capacitance C with no resistance. The general expression for the total impedance of this series RLC circuit is Z = R + j(XL – XC) (11.1) If the reactance term (XL – XC) is zero, the impedance Z of the circuit is R alone, and the condition of resonance VL (= IXL) and VC (= IXC) must be equal, and hence the phasor diagram is as shown in Fig. 11.1b. The voltage VL cancels the voltage VC. The applied voltage V is then same as the voltage VR. If the resistance is small compared to the inductive and capacitive reactances, the

311 voltages VL and VC are many times the supply voltage V, as shown in the phasor diagram in Fig. 11.1b.

We shall represent the condition of resonance by putting subscript ‘0’, e.g., f0, Z0, XC0, etc. Thus, the resonance in a series RLC circuit requires that 1 XL0 – XC0 = 0 or XL0 = XC0 or w 0 L = (11.2) w 0C Solving the above for w 0, we get the frequency of resonance as 1 1 w0 = or f 0 = (11.3) LC 2p LC When resonance occurs, the impedance of the circuit assumes a minimum value given as Z 0 = R + j0 = R, and the current has a maximum value given by V V I0 = = (11.4) R Z0

Let the values of R, L, C and V in the circuit of Fig. 11.1a frequency f of the ac source is varied from 0 to . Since +j is associated with XL and –j with XC, we plot the values of XL on the positive imaginary axis and of XC on the negative imaginary axis. Since XL = 2pfL, XL varies directly with the frequency f. The plot of XL versus f is a straight line starting from the origin, as shown in Fig. 11.2a. Since XC = 1/2pfC, XC varies inversely as frequency f. The plot of XC versus f is a rectangular hyperbola. At a frequency f0, the values of XL and XC are seen to be equal. Hence the net reactance of the circuit, X = (XL – XC) is zero at this frequency. At f0, the circuit behaves as purely resistive. Below f0, X has negative values (i.e., the circuit is capacitive). Above f0, X has positive values (i.e., the circuit is inductive). The impedance of a series RLC circuit is given as X − XC Z = R 2 + ( X L − XC ) 2 ∠ tan −1 L (11.5) R At f = 0, the inductance behaves as a short circuit (XL = 0) and the capacitance behaves as an open circuit (XC = ). The impedance Z I = 0. As the frequency is increased, XL increases but XC decreases. The circuit remains capacitive in nature for f < f0. At f = f0, XL equals XC, the impedance has minimum value and the current has maximum value. As the frequency is further increased, XL becomes

312

Basic Electrical Engineering

Effects of variation of frequency in a series RLC circuit. greater than XC, the impedance increases and consequently the current goes on decreasing. It approaches zero as f b. I=

V = Z

V – 0∞ Ê X - XC ˆ R 2 + ( X L - X C ) 2 – tan -1 Á L ˜¯ Ë R RLC

=

Ê X - XC ˆ –- tan -1 Á L ˜¯ Ë R R + ( X L - XC ) V

2

2

Ê X - XC ˆ -1 Ê X C - X L ˆ q = - tan -1 Á L ˜¯ ˜¯ = tan ÁË Ë R R c.

q

Importance of Resonant Circuit RLC resonant circuit is also called an acceptor circuit selectivity *

Variation of Voltage across C and L Consider a series RLC f

V I I=

*

tuning

V 2

R + ( X L - XC )

2

=

V 2

R + (w L - (1/w C )) 2

.

313 Therefore, the voltage across capacitor, VC = IXC = fi

VC2 =

V 2

R + (w L - (1/w C )) V2

2

◊

1 wC

=

V2

w 2 C 2 {R 2 + (w L - (1/w C )) 2 } R 2w 2 C 2 + (w 2 LC - 1) 2 VC is maximum, we should have dVC =0 dw

or

(11.7)

dVC2 =0 dw

È 0 - {2w R 2C 2 + 2(w 2 LC - 1) 2w LC } ˘ V2◊Í ˙ =0 {R 2w 2 C 2 + (w 2LC - 1) 2 }2 ÍÎ ˙˚ or 2wR 2C 2 + 2(w2LC – 1)2wLC = 0 2 1 (2L – CR2) = 1 – R or w2 = LC 2 L2 C 2L2 Thus, the frequency at which the voltage VC is maximum is given as or

R2 1 - 2 LC 2L

1 2p Now, the voltage across inductance is given as

fC =

V

◊ wL R 2 + (w L - (1/w C )) 2 V 2w 4 L2 C 2 V 2w 2L2 VL2 = 2 = R + (w L - (1/w C )) 2 w 2C 2 R 2 + (w 2 LC - 1) 2 VL is maximum, we should have VL = IXL =

fi

or or

dVL =0 dw

or

2w 2LC – w 2C 2R – 2 = 0

or

w2 =

(11.8)

(11.9)

dVL2 =0 dw

w 2(2LC – C 2R) = 2 1

LC - (R 2C 2/ 2 ) Thus, the frequency at which the voltage VL is maximum is given as 1 fL = 2 p LC - ( R 2 C 2/ 2)

(11.10)

The Plots of VC and VL versus Frequency The variations of VC with frequency (Eq. 11.7) and of VL with frequency (Eq. 11.9) are plotted in Fig. 11.3a. At f = 0, the capacitor behaves as an open circuit, and hence the current in the circuit is zero. The entire source voltage V appears across the capacitor. With increase in frequency, the voltage VC also increases, reaches a maximum value at fC (given by Eq. 11.8) and then falls again. As the frequency f VC approaches zero, as the capacitor tends to behave as a short circuit. At f = 0, the inductor behaves as a short circuit, and hence the voltage VL is zero. As the frequency increases, the voltage VL rises, reaches a maximum value at fL (given by Eq. 11.10) and then falls again. As

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the frequency f VL approaches the source voltage V, as the inductor tends to behave as an open circuit. The point of intersection of these two curves gives the resonant frequency f0. At this frequency, VC = VL, and the circuit becomes purely resistive. If the resistance R in a series RLC circuit is small compared to XL0 (or XC0), the frequencies fC and fL come close together and coincide at f0, as shown by the curves in Fig. 11.3b. Also, under this condition, at f = f0 the voltages across C and L are equal and each is greater than the source voltage V.

At resonance, the circuit impedance is a minimum being equal to the resistance R. The resonant current I0 is dependent on the resistance, as given by Eq. 11.4. This current is considerably higher than that at other frequencies. The power dissipation (I 02 R) in the circuit is maximum, and also the peak rates of energy storage in the two reactive components (L and C) become equal and maximum. As the circuit operates at unity power factor at resonance, it has no reactive power. The energy stored by the reactive components is a constant and it oscillates between electric and magnetic modes of storage. The predominance of this energy oscillation over the energy associated with the resistance is important to electronics and communication engineers. The quality of a resonant circuit to accept current (and power) at the resonant frequency to the exclusion of other frequencies is measured by a factor termed quality factor (Q factor), described below.

quality factor of a resonant circuit as follows : 2p(Maximum energy stored in L or C per cycle) Q = Energy dissipated per cycle

(11.11)

315 Q in various forms for a resonant series RLC circuit. Thus, if Im is the peak value of the current at resonance and T0 is the time period at the resonant frequency, we get Q= Replacing Im by its rms value I0 (Im =

2p[(1/ 2) L( I m ) 2 ] I 02 RT0

2 I 0 ) and T0 = 1/f0 = 2p/w 0, we get

Q =

2 p[(1/ 2) L( 2 I 0 ) 2 ] I 02 R (2p /w 0 )

=

w0L R

Since at resonance, we have from Eq. 11.2, w 0L = 1/w 0C 1 Q = w 0CR Another form for Q can be obtained by putting w 0 = 1/ LC in the above expression,

(11.12)

(11.13)

1 LC 1 L = = (11.14) CR w 0CR R C A capacitor usually has no losses; hence it is represented as a pure capacitance C with no resistance. However, in practice an inductor (say, an iron-cored coil) always has a small resistance R in addition to an inductance L. The quality factor Q of a series inductor-capacitor circuit, therefore, is the same as the quality factor Q of the coil used. In fact, Q Q¸ the better is the coil. Coils having Q less than 10 are described as low-Q coils. The coils having Q equal to or greater than 10 are described as high-Q coils. Coils having Q as high as 200-300 are used in electronic circuits. Q =

Let us determine the voltage drops across each element of the series RLC circuit at resonance. V Voltage drop across resistance, VR = I0 R = R = V R X w L V Voltage drop across inductance, VL = I0 X L0 = X L0 = V L0 = V 0 = VQ R R R 1 XC0 V Voltage drop across capacitance, VC = I0 XC0 = XC0 = V =V = VQ w 0CR R R V appears across R. More surprising is the fact that the voltage VL (and voltage VC) is Q times the supply voltage. For example, if the supply voltage is 230 V and the Q of the coil is 10, at resonance the voltage VR will also be 230 V, but the voltages across the coil and the capacitor will each be 2300 V ! Therefore, extreme care must be taken while working on series ac circuits that may become resonant. The multiplication of voltage by the Q of the coil in a series resonant circuit is often called Q gain in electronics. Since, the voltage increases Q times in a series resonant circuit, it is also known as a voltage resonant circuit. EX A MP L E

11. 1

A series RLC circuit has R = 12 W, L = 0.15 H and C = 100 mF. It is connected to an ac source of voltage 100 V, whose frequency can be varied. Determine (a) the frequency of the source at which the current supplied by it is

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Basic Electrical Engineering

maximum, (b) the value of this current, (c) the frequency at which the voltage across the capacitor is maximum, (d) the frequency at which the voltage across the inductor is maximum, (e) the inductive reactance at resonant frequency, (f) the capacitive reactance at resonant frequency, (g) the quality factor of the circuit, and (h) the voltages across each element, under resonant condition.

Solution (a) The current becomes maximum at the resonant frequency of the circuit, 1 1 = = 41.09 Hz f0 = 2p LC 2p ¥ 0.15 ¥ 100 ¥ 10 - 6 (b) The maximum current supplied by the source is V 100 = = 8.3 A I0 = R 12 (c) The frequency at which the voltage across the capacitor is maximum, fC =

1 2p

1 R2 1 - 2= LC 2 L 2p

1 0.15 ¥ 100 ¥ 10- 6

-

(12) 2 2 (0.15) 2

= 40.09 Hz

(d) The frequency at which the voltage across the inductor is maximum, 1 1 fL = = = 42.11 Hz -6 2 2 2p 0.15 ¥ 100 ¥ 10 - {122 ¥ (100 ¥ 10 - 6) 2 / 2} 2p LC - ( R C / 2) (e) The inductive reactance, XL0 = 2p f0 L = 2p ¥ 42.11 ¥ 0.15 = 39.68 Hz (f) The capacitive reactance, XC0 = 1/2pf0C = 1/2p ¥ 42.11 ¥ 100 ¥ 10–6 = 37.79 Hz X 37.79 w L (g) The Q of the circuit, Q = 0 = L0 = = 3.14 12 R R (h) The voltage drops across the elements are VR = V = 100 V; VL = VC = QV = 3.14 ¥ 100 = 314 V NOT E The curves drawn in Fig. 11.3b correspond to this series RLC circuit. EXA MP LE

1 1. 2

A series combination of a resistance of 4 W, an inductance of 0.5 H and a variable capacitance is connected across a 100-V, 50-Hz supply. Calculate (a) the capacitance to give resonance, (b) the voltage across the inductance and the capacitance, and (c) the Q factor of the circuit.

Solution (a) For resonance, we should have fi

XL0 = XC0 or 2pf0 L = 1/2pf0C 1 1 = = 20.3 mF C = 2 (2p ¥ 50) 2 ¥ 0.5 (2pf 0) L

V 100 = = 25 A R 4 \ The voltage across L = I0 XL = 25 ¥ (2p ¥ 50 ¥ 0.5) = 3927 V and the voltage across C = voltage across L = 3927 V

(b) At resonance, I0 =

317 (c) From Eq. 11.12, Q= E XA MP LE

2pf 0L w 0L 2p ¥ 50 ¥ 0.5 = = = 39.27 4 R R

11.3

The voltage applied to a series resonant circuit is 0.85 V. The Q of the coil used is 50 and the value of the capacitor is 320 pF. The circuit is required to resonate at a frequency of 175 kHz. Find the value of the inductance, the circuit current and the voltage across the capacitor under resonance. Draw the phasor diagram.

Solution Since f0 =

1 , the required inductance is given as 2p LC 1 1

L=

= = 2.58 mH 4p 2 f 02C 4p 2 ¥ (175 ¥ 103 ) 2 ¥ 320 ¥ 10 -12 The reactance of the coil (and also of the capacitor) at resonance is given as 3 –3 XL0 = 2pf0 L = 2p ¥ 175 ¥ 10 ¥ 2.58 ¥ 10 = 2836.86 W

The resistance of the coil is given as XL0 2836.86 = = 56.74 W 50 Q V 0.85 = = 14.98 mA \ I = R 56.74 The voltage across the capacitor (and also across the inductance part of the coil) is R =

VC = QV = 50 ¥ 0.85 = 42.5 V Using the above numerical values, we can draw the phasor diagram, as shown in Fig. 11.4.

EX A MP L E

11. 4

A coil with inductance 1.0 mH and resistance 2.0 W is connected in series with a capacitor and a 120-V, 5-kHz supply. a) the current at the resonance frequency, and (b resonance frequency.

Solution The required value of the capacitance is given as C=

1 4p 2 f 02 L

=

1 4p ¥ (5 ¥ 103 ) 2 ¥ 1 ¥ 10 - 3 2

= 1.01 mF

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Basic Electrical Engineering

V 120 = = 60 A R 2 2 2 –3 2 (b) The maximum instantaneous energy, Um = (1/2)LI m = LI rms = 1 ¥ 10 ¥ (60) = 3.6 J (a) The current, I0 =

EXA MP LE

11.5

For the circuit shown in Fig. 11.5, R1, R2 and R3 are 0.51 W, 1.3 W and 0.24 W respectively; C1 and C2 are 25 mF and 62 mF respectively; and L1 and L2 are 32 mH and 15 mH respectively. Determine (a) the resonance frequency f0, (b) the quality factor of the overall circuit Q, (c) the quality factor of coil-1, Q1, and (d) the quality factor of coil-2, Q2. R3

L1, R1

C1

C2

L2, R2

Solution Req = R1 + R2 + R3 = 0.51 + 1.3 + 0.24 = 2.05 W Leq = L1 + L2 = (32 + 15) mH = 47 mH; Ceq = C1C2

C1 + C2

(a) \

f0 =

(b) Q =

1 Req

=

62 × 25 μF = 17.8 mF 62 + 25

1 1 = = 174 Hz 3 2p LeqCeq 2p 47 ¥ 10 ¥ 17.8 ¥ 10 - 6

Leq Ceq

=

1 2.05

47 × 10 − 3 17.8 × 10 − 6

= 25

-3 (c) Q1 = w 0 L1 = 2p ¥ 174 ¥ 32 ¥ 10 = 68.6 0.51 R1 -3 = 12.6 (d) Q2 = w 0 L2 = 2p ¥ 174 ¥ 15 ¥ 10 1.3 R2

The current versus frequency curve for a resonant circuit is called its resonance curve. For a series resonant circuit, the resonance curve is shown in Fig. 11.2b. It has been redrawn in Fig. 11.6 for two values of resistances R1 and R2 (R1 < R2). The range of frequencies within which the current does not drop below 0.707(= 1/ 2 ) times the maximum value is called the passband or bandwidth. Thus, for R = R1, the bandwidth is BW = f2 – f1 (11.15) The frequencies f1 and f2 are often called lower and upper cutoff frequencies. At these frequencies, the current reduces to I 0 / 2 . Since the power dissipated at f0 is given as P0 = I 02 R, the power dissipated at the cutoff frequencies is ( I 0 / 2 ) 2 R. = P0/2. That is, the power at f1 or f2 is half the maximum power P0. Hence, these frequencies are also called half-power frequencies.

As we can see from Fig. 11.6, the lower the value of R, the sharper is the resonance curve. The curve sharply falls to zero on the two sides of resonance frequency. The response of the circuit at the desired frequency

319 I I0 (1) For R = R1 0.707I0

R1 < R2

I0¢ 0.707I 0¢

(2) For R = R2

O

f¢1

f1

f0

f 2¢

f2

f

(i.e., f0) is high, but it is quite low for the nearby other frequencies. In other words, it has a good capability of selecting the desired frequency and of rejecting the nearby undesired frequencies. We say that such a resonant circuit has high selectivity. For larger values of R, not only the peak value of current falls, but even the response curve becomes less sharp (see the curve for R2 in Fig. 11.6). Its bandwidth increases. In other words, its selectivity reduces. We use such resonance circuits with high quality factor to tune in the desired broadcasting station in the radio receiver.

f0 f1

f2

At f1 and f2 the current drops to 1/ 2 of its resonant value I0. It means that at these frequencies, the impedance must be equal to 2 times the resonant value R. That is, 1 R 2 + ( X L − XC ) 2 = 2R fi XL – XC = ±R or w L = ±R (11.16) wC For XL < XC, which corresponds to f1 or w 1, Eq. 11.16 yields 2

R 1 Ê Rˆ 4 + + 2L 2 ÁË L ˜¯ LC Similarly, for XL > XC, which corresponds to f2 or w 2, Eq. 11.16 gives

w1 = -

w2 =

R 1 + 2L 2

(11.17)

2

4 Ê Rˆ ÁË L ˜¯ + LC

Multiplying Eqs. 11.17 and 11.18, we get 2 2 1 ÈÊ R ˆ 4 ˘ 1 Ê Rˆ 1 ˙- Á ˜ = w 1w 2 = Í Á ˜ + = (w 0) 2 4 ÍÎ Ë L ¯ LC ˙˚ 4 Ë L ¯ LC Hence, f0 = f1 f 2

(11.18)

or

w0 =

w1w 2 (11.19)

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Basic Electrical Engineering

BW From Eqs. 11.17 and 11.8, we get È R 1 BW = f2 – f1 = Í + ÍÎ 4pL 4p BW =

2 ˘ È 4 ˙ Í R 1 Ê Rˆ - + + ÁË L ˜¯ LC ˙˚ ÍÎ 4p L 4p

2 ˘ 4 ˙ Ê Rˆ + ÁË L ˜¯ LC ˙˚

R 2pL

(11.20)

BW

Q

f0

From Eq. 11, we have R/L = w 0/Q = (2pf0)/Q. Substituting this into Eq. 11.20, we get BW =

f 1 Ê 2pf 0 ˆ 1 Ê Rˆ R = = = 0 Á ˜ Á ˜ 2p Ë L ¯ 2p Ë Q ¯ Q 2p L

(11.21)

As indicated in Fig. 11.6, the resonance frequency is not centrally located with respect to the two half-power frequencies, especially when the BW is large (i.e., when the Q is small). Actually, as can be seen from Eq. 11.19, f0 is the geometric mean of f1 and f2; and the geometric mean is always less than the arithmetic mean. However, if Q > 10, the resonant frequency f0 half-power frequencies f1 and f2 (as for the curve (1) in Fig. 11.6), and we can then write f1 ª f0 -

BW 2

and

f2 ª f0 +

BW 2

(11.22)

For low-Q coils, we cannot use the above approximations. The exact relations are given in Eqs 11.17 and 11.18. EX A MP L E

11. 6

A series ac circuit has a resonance frequency of 150 kHz and a bandwidth of 75 kHz. Determine its half-power frequencies.

Solution From Eq. 11.21, Q = 150/75 = 2 (which is less than 10). Hence, we cannot use the approximate relations given in Eq. 11.15. Using the exact relations given in Eqs. 11.17 and 11.18 and working in kHz, we have 75 = f2 – f1 Eliminating f2 between the two equations, we get 2

f 1 + 75f1 – 22 500 = 0

and fi

150 =

f1 = 117.1 kHz

f 2 f1

or

–192.1 kHz

Ignoring the negative value, we have f1 = 117.1 kHz. Hence, f2 = 75 + f1 = 192.1 kHz.

In a parallel resonant circuit, the ac source is connected across a parallel combination of an inductor and a capacitor, as shown in Fig. 11.7a. In general, both the inductor and the capacitor have some losses. In a circuit, these losses are accounted for by inserting equivalent series resistances R1 and R2, respectively. By varying the frequency of the source, resonant condition may reach when the reactive (or wattless) component of line current I reduces to zero.

321

Figure 11.7b shows the phasor diagram of the circuit under resonance condition. The branch current I1 through the inductor lags the supply voltage V by an angle q1. The branch current I2 through the capacitor leads the voltage V by an angle q2. Under resonance condition, the reactive components of these two currents are equal in magnitude (but opposite in phase). That is, I1 sin q 1 = I2 sin q 2 (1/w 0 C ) V (w 0 L) V or = ¥ ¥ R12 + (w 0 L) 2 R12 + (w 0 L) 2 R22 + (1/w 0 C ) 2 R22 + (1/w 0 C ) 2 w C (w 0 L) (1/w 0 C ) w0 L or = 2 or = 2 20 2 2 2 2 2 2 2 R1 + w 0 L R2 + (1/w 0 C ) R1 + (w 0 L) R2 w 0 C + 1 L(R22w 02C 2 + 1) = C(R 12 + w 02L2)

fi or

w0 =

1

CR12 - L

LC

CR22 - L

=

f0 =

or

w 02LC(R22C – L) = CR 21 – L

or

1

R12 - ( L/C )

LC

R22 - ( L/C ) R12 - ( L/C )

1 2p

LC

R22 - ( L/C )

(11.23)

It is possible to get a capacitor having negligible losses. It means that the resistance R2 in series with capacitor C can be ignored. However, in practice an inductor does have some losses. The circuit of Fig. 11.8a therefore represents a practical parallel resonant circuit. By putting R2 = 0 and R1 = R in Eq. 11.23, we get the resonance frequency of this circuit as f0 =

1 2p

LC

R 2 - ( L/C ) 1 = - ( L/C ) 2p LC

1 - R 2C/L

(11.24)

Note that if R2C/L > 1 fi R2 > (L/C), f0 as given by the above equation is imaginary, and therefore resonance cannot occur.

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Basic Electrical Engineering

Ideal Parallel Resonant Circuit It would have been ideal if the inductor too were lossless, though it is not physically possible. In such a case, we could ignore resistance R in Fig. 11.8a, so that the circuit reduces to that shown in Fig. 11.8c. The resonance frequency of this circuit can be obtained from Eq. 11.24, as 1 (11.25) f0 = 2p LC Note that this equation is same as Eq. 11.3 for series resonance circuit.

In the practical circuit of Fig. 11.8a, at resonance we have V R w L and q1 = cos–1 = sin–1 0 For branch 1 : Z1 = R 2 + (w 0 L) 2 ; I1 = Z1 Z1 Z1 V V For branch 2 : Z2 = 1/w 0C; I2 = = = Vw 0C and q2 = 90° Z2 1/w 0 C Equating the reactive components of the two currents, we get V w0 L L ◊ I1 sin q1 = I2 sin 90° or = Vw 0C fi Z1 = Z1 Z1 C Since the reactive components of the two currents cancel each other, the line current is VR V V R V VR I = I1 cos q1 + I2 cos q2 = ⋅ +0= 2 = = = L/C L/CR Z1 Z1 Z0 Z1 Thus, the effective or equivalent or dynamic impedance of the parallel resonance circuit is given as Z0 =

L CR

(11.26)

parallel resonant circuit of Fig. 11.8a, the admittances of the two branches are given as R wL 1 1 - j 2 Y1 = = 2 = G – jB L and Y2 = = +jwC = + jBC 2 2 R + jw L (1/ jw C ) R + (w L ) R + (w L ) Thus, the conductance and susceptance of the inductive branch, respectively, are R wL G= 2 and BL = 2 2 R + (w L ) 2 R + (w L)

323 The conductance of the capacitive branch is zero and its susceptance is BC = w C. Total admittance of the circuit is given as wL R - j 2 + jw C = G + j(BC – BL) Y = Y1 + Y2 = 2 2 R + (w L ) R + (w L) 2 From the above expressions, it can be seen that as frequency increases, the value of conductance G decreases. The plot of G versus frequency is shown in Fig. 11.9a by a dashed curve. The inductive susceptance BL is considered negative, since –j is associated with it. For low frequencies, wL < R and (wL)2 R and (wL)2 >> R2, hence the magnitude of BL is inversely proportional to the frequency. Consequently, the plot of BL versus f is a rectangular hyperbola in the high-frequency region. The overall plot of BL versus f

The capacitive susceptance is considered positive. Its magnitude is seen to vary directly with frequency. Thus, as shown in Fig. 11.9a, the plot of BC versus f is a straight line. By algebraic addition of BL and BC, the plot of the net susceptance B versus f can be determined (as shown by dotted curve). Since the net admittance of the circuit is given as Y = G 2 + B 2 , the plot of Y versus f is as shown by solid curve (V-shaped) in the

Current versus Frequency Since the impedance is inverse of admittance (Z = 1/Y), the Z verses f plot Y versus f plot, as shown in Fig. 11.9b. At resonance frequency f0, the impedance has maximum value Z0, given by Eq. 11.26. The plot of current I versus f resonance frequency, the line current is seen to have minimum value given by V V VCR I0 = = = (11.27) Z0 L/CR L If the resistance R is zero (as in an ideal tank circuit of Fig. 11.8c), the line current I0 at resonance would be zero.

Bandwidth As shown in Fig. 11.9b, the half-power frequencies f1 and f2 where the impedance reduces to 1/ 2 (or 0.707) times the impedance Z0 at resonance. The range from f1 to f2 is called bandwidth (BW). To examine how the nature of the circuit (inductive or capacitive) varies with

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Basic Electrical Engineering

frequency, let us write the expression for the impedance, ( B - BL ) 1 G 1 -j 2 C Z= = = 2 2 G + j ( BC - BL ) Y G + ( BC − BL ) G + ( BC - BL ) 2 At the resonance frequency f0, (BC – BL) = 0, hence the circuit is purely resistive; the power factor is unity. For frequencies below f0, BC < BL (see Fig. 11.9a), the imaginary part of Z is positive, and hence the circuit is predominantly inductive. For frequencies above f0, BC > BL, the imaginary part of Z is negative, and hence the circuit is predominantly capacitive. At frequencies f1 and f2, the impedance Z = Z 0 / 2 , which is possible only when R = X, so that Z = R ± jX = 2R – ± 45° W. The impedance angle is 45°; the power factor is 1/ 2 (or 0.707).

It can be seen from the phasor diagram of Fig. 11.8b that both the branch-currents I1 and I2 are much larger than the line current I. This shows that in a parallel circuit, the current taken from the ac source is greatly Q-factor. At resonance, the reactive components of the two branch currents are equal and opposite in phase. That is, I2 sin 90° = I1 sin q 1 or I2 = I1 sin q1 Current through the capacitor I I sin 1 w L \ Q = = 2 = 1 = tan q1 = 0 Line current I I1 cos 1 R S O M E

IM P OR TANT

POINT S

Following are some of the important points about parallel circuit at resonance : The reactive (or wattless) component of line current is zero; hence the circuit power factor is unity. The circuit is purely resistive; the impedance has maximum value given as Z0 = R0 = L/CR. The line current is minimum, given as I0 = V/(L/CR). Since the circuit rejects the current at resonance (i.e., it has minimum value), the parallel resonant circuit is also called rejector circuit or anti-resonant circuit. 5. Since the circulating current between the two branches is many times the line current, the parallel resonant circuit is also called current resonant circuit. 6. The circuit is also called a tank circuit.

1. 2. 3. 4.

The resonance in a series and in a parallel circuit differs in its characteristics and properties. These differences are summarised in Table 11.1. EX A MP L E

11. 7

An inductor coil having a resistance of 20 W and inductance of 200 mH is connected in parallel with a variable capacitor. This combination is connected in series with a resistance of 8 kW. The circuit is then connected across a 230-V, 1-MHz ac source. Determine (a) the value of capacitance to cause resonance, (b) the Q-factor of the circuit, (c) the dynamic impedance of the parallel resonant circuit, and (d) the line current.

325

S.No.

Property

Series circuit

1. 2.

Impedance at resonance Current at resonance

3.

Resonance frequency

5.

Nature of the circuit : (i) Below f0 (ii) Above f0

Parallel circuit

Z0 = R, the minimum I0 = V/R, the maximum 1 , f0 = 2p LC R) (i) Capacitive (ii) Inductive

Z0 = L/CR, the maximum I0 = V/(L/CR), the minimum 1 f0 = 1 - R 2 C/L , 2p LC Current (affected by R) (i) Inductive (ii) Capacitive

Solution The circuit diagram is shown in Fig. 11.10. 6 –6 XL = 2pf L = 2p ¥ 1 ¥ 10 ¥ 200 ¥ 10 = 1256.6 W

(a) Since R 10,

2

2

11. 13

A series RLC circuit has a resistance of 1 kW and half-power frequencies of 20 kHz and 100 kHz. Determine (a) the bandwidth, (b) the resonance frequency, (c) the inductance, and (d) the capacitance.

Solution (a) The bandwidth,

BW = f2 – f1 = 100 – 20 = 80 kHz

(b) The resonance frequency, f0 =

f1 f 2 =

20 100 = 44.72 kHz

328

(c) Q =

Basic Electrical Engineering 44.72 f0 = = 0.56; but Q = BW 80

(d) The capacitance, C = E XA MP LE

1 (2pf 0 ) 2 L

=

2pf 0 L R

fi

L=

0.56 × 103 QR = = 2 mH 2p f 0 2 π × 44.72 × 103

1 (2p ¥ 44.72 ¥ 103 ) 2 ¥ 2 ¥ 10 - 3

= 6.33 nF

11.14

A series RLC circuit having R = 5 W operates from a 20-V source. Determine the power at half-power frequencies.

Solution I0 = \

V V 20 2 2 = = = 4 A; P0 = I0 R = (4) ¥ 5 = 80 W Z0 R 5 Phalf-power = (1/2)P0 = (1/2) ¥ 80 = 40 W

E XA MP LE

1 1 . 15

A 240-V, 100-Hz ac source is connected to a series RLC circuit consisting of a coil and a variable capacitor. The coil has a resistance of 55 mW and an inductance of 7 mH. The capacitor is varied so as to achieve resonance. Determine (a) the value of the capacitance, (b) the circuit quality factor, and (c) the half-power frequencies.

Solution (a) Since, f0 =

1 , the value of the capacitance is given as 2p LC 1 1

= 361.86 mF (2p ¥ 100) 2 ¥ 7 ¥ 10 - 3 2p ¥ 100 ¥ 7 ¥ 10 - 3 2pf 0 L (b) The quality factor, Q = = = 79.97 55 ¥ 10 - 3 R f 100 (c) The bandwidth, BW = 0 = = 1.251 Hz Q 79.93 Therefore, the half-power frequencies are 1.251 f1 = f0 – BW = 100 – = 99.3747 Hz and f2 = f0 + BW = 100.6255 Hz 2 2 2 C =

EXA MP LE

(2pf 0)2 L

=

11.16

A coil of resistance and inductance 20 W and 0.2 H, respectively, is connected in parallel with a 100-mF capacitor. Determine the frequency at which the circuit behaves as a non-inductive resistance. Find the value of this resistance.

Solution The circuit behaves as a pure resistance at resonance frequency, given by f0 =

1 2p LC

1 - R 2C / L =

1

2 π 0.2 × 100 × 10 − 6 0.2 L The effective resistance, Z0 = R0 = = = 100 W CR 100 × 10 − 6 × 20 EXA MP LE

1−

(20)2 × 100 × 10 − 6 = 31.83 Hz 0.2

1 1 . 17

The medium-wave band in a radio receiver spreads from 570 kHz to 1560 kHz. The radio is tuned to the desired

329 amplitude-modulated carrier frequency by using a series resonant circuit consisting of a coil and a variable capacitor. (a) If the coil has an inductance of 20-m (b) If the Q circuit. (c) Find the Q of the circuit at the upper tuning frequency.

Solution (a) Since the tuning frequency (or resonance frequency) is given as f0 =

1 2p LC

, we have C =

1 (2pf 0) 2 L

Therefore, at the lower tuning frequency, the value of the capacitor should be C1 =

1 1 = = 3.9 nF 2 (2 p f 01) L (2p ¥ 570 ¥ 103) 2 ¥ 20 ¥ 10 - 6

The value of the capacitor at the upper tuning frequency should be C2 =

1 2

(2pf 02 ) L

=

1 (2p ¥ 1560 ¥ 103 ) 2 ¥ 20 ¥ 10 - 6

= 0.52 nF

Thus, the range of the tuning capacitor is from 0.52 nF to 3.9 nF.

2pf 01L w 01L = , the resistance of the coil is given as R R 2pf 01L 2p ¥ (570 ¥ 103) ¥ (20 ¥ 10 - 6) R = = = 1.433 W 50 Q1

(b) Since Q1 =

Bandwidth, BW =

f01 570 103 = = 11.4 kHz Q1 50

(c) At the upper tuning frequency, Q2 =

2p ¥ (1560 ¥ 103) ¥ (20 ¥ 10 - 6) 2pf 02 L w 02L = = = 136.8 1.433 R R

SU MMAR Y TERMS

A ND

C ONC E PT S

Resonance is the condition that exists in ac circuits under steady state when the input current is in phase with the input voltage. Resonant frequency (f0) is the frequency at which resonance occurs. RLC circuit, the impedance is minimum and purely resistive. The power factor is unity. RLC circuit, when f < f0, the circuit is capacitive and the power factor is leading; when f > f0, the circuit is inductive and the power factor is lagging. Bandwidth (BW) or passband is the range of frequencies within which the current does not drop below 0.707 (=1/ 2 ) times the maximum value (I0). f1 and f2 at which the current reduces to 0.707 times the maximum value (or, equivalently, at which the power reduce to half the maximum value) are called lower and upper cut-off frequencies or half-power frequencies.

330

Basic Electrical Engineering

Q factor, the narrower is the BW and the greater is the selectivity of a tuned circuit. RLC Q times. Hence it is called voltage resonant circuit. anti-resonant or tank circuit) has a coil and a capacitor in parallel. It is said to resonate when the reactive (or wattless) component of line current I reduces to zero. capacitance in other branch. Hence, it is called a rejector circuit. f < f0, the circuit is inductive and pf is lagging; when f > f0, the circuit is capacitive and pf is leading. parallel resonant circuit is also called current resonant circuit. tank circuit. IMP O R TA N T

F O RM U LAE

1 LC

w0 =

f0 =

or

1 . 2p LC

Z0 = R.

f0 =

VC is maximum,

fC =

VL is maximum,

fL =

1 2p

R2 1 - 2 . LC 2 L

1 2p LC - ( R 2C 2 /2)

.

f1 f2 . 1 L w0L = . R C R f R BW = f2 – f1 = 0 . Also, BW = . Q 2pL Q =

f1 ª f0 – BW

Q coils (Q > 10),

2

(a) General circuit :

f0 =

(b) Practical circuit :

f0 =

(c) Ideal circuit :

f0 =

and f2 ª f0 +

1 2p LC

R12 - ( L/C )

1

1 - R 2C / L .

2p LC 1 . 2p LC

R22 - ( L/C )

.

BW . 2

331 Z0 =

L . CR

Q =

w 0L . R

CH E CK

YOU R

UNDERSTANDIN G

Before you proceed to the next chapter, take this test. Give yourself two marks for each correct answer and minus one for each wrong answer. If your score is 12 or more, go to the next chapter; otherwise study this chapter again. S. No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Statement

True

False

Marks

The ratio of bandwidth to the resonance frequency of a resonant circuit is called its quality factor. In a series resonant circuit, the lower the resistance in the circuit, the steeper is its current response. When a capacitor is connected in parallel to an inductive circuit, the phase angle increases and the power factor decreases. In a practical resonant circuit, the value of the resistance affects the resonant frequency. The impedance of both the series and parallel resonant circuit increase with increase in frequency. In a series resonant circuit, the impedance for the frequencies above resonant frequency is inductive. When the frequency is much greater than the resonant frequency of a series resonant circuit, the angle of impedance Z approaches 0°. For a series resonant circuit, the resonance curve is a plot of frequency versus voltage. At half-power points of a resonance curve of a series resonant circuit, the value of current is 1/ 2 times the current at resonance. A parallel ac circuit draws maximum current when in resonance. Your Score A N SW E RS

1. False 6. True

2. True 7. False

3. False 8. False

RE VI EW

4. True 9. True

5. False 10. False

Q UESTIO N S

1. electrical circuits. What is the practical application of this phenomenon ? 2. In a series RLC circuit, the resonance condition can be achieved in three ways. Explain them.

3. An ac voltage is connected to a circuit containing R, L and C in series. The supply frequency is gradually varied. Explain the observations. 4. For a series RLC circuit, discuss the variation of the following quantities with frequency of the ac source : XL, XC, XL – XC, Z and I.

332

Basic Electrical Engineering expressions for the resonance frequency, the quality factor and the bandwidth for a practical parallel resonant circuit. 8. Show that a practical parallel resonant circuit behaves as purely resistive with R0 = L/CR, at resonance. 9. Show that a parallel resonant circuit is inductive when f < f0, and capacitive when f > f0.

5. RLC circuit. How is the bandwidth affected by Q of the circuit ? 6. At resonance in a series RLC circuit, the voltage across the capacitor may become many times the supply voltage. Explain how. 7. How is a practical resonant circuit different from a theoretical parallel resonant circuit ? Derive the

M U LT I P L E

C HOI CE

Here are some incomplete statements. Four alternatives are provided below each. Tick the alternative that completes the statement correctly. 1. In a series RLC circuit, if the frequency of the source is below the resonance frequency, then (a) XC = XL (b) XC > XL (c) XC < XL (d) None of the above 2. (a) voltage (b) current (c) both voltage and current (d) None of the above 3. (a) voltage (b) current (c) both voltage and current (d) None of the above 4. In a parallel ac circuit, if the supply frequency is greater than the resonance frequency, the circuit is (a) resistive (b) inductive (c) capacitive (d) None of the above 5. A parallel resonant circuit draws a current of 2 mA at resonance. If the Q of the circuit is 100, the current drawn by the capacitor is (a) 0.02 mA (b) 1 mA (c) 2 mA (d) 200 mA 6. If a parallel resonant circuit is shunted by a resistance, then (a) the circuit impedance is decreased (b) the Q of the circuit is increased (c) the gain of the circuit is increased (d) None of the above 7. A series RLC circuit has unity power factor if operated at a frequency of (a) 1/LC Hz (b) 1/pLC Hz (c) 1/2pLC Hz (d) 1/ 2 LC Hz

QUE STIONS

8. In a series RLC circuit at resonance, (a) its impedance is maximum (b) its admittance is maximum (c) its impedance is purely reactive (d) its current is minimum 9. A series RLC resonant circuit implies that it has (a) zero pf and maximum current (b) zero pf and minimum current (c) unity pf and minimum current (d) unity pf and maximum current 10. A series RLC circuit, consisting of R = 10 W, XL = XC = 20 W, is connected across 100-V, 50-Hz supply. The magnitude and the phase angle (with reference to the supply voltage) of the voltage across the inductance are, respectively, (a) 100 V, 90° (b) 100 V, – 90° (c) 200 V, – 90° (d) 200 V, 90° 11. A resonant circuit with a high-Q coil has (a) large bandwidth (b) high losses (c) low losses (d 12. If the Q factor of a resonant circuit is high, then (a) its power factor is high (b) its impedance is high (c) its bandwidth is large (d) None of these 13. In a series resonant circuit, if the value of inductance L is increased, (a) its resonance frequency decreases (b) its bandwidth decreases (c) its Q increases (d) All of the above 14. In a series resonant circuit, if the value of capacitance C is increased, (a) its bandwidth increases (b) its resonance frequency decreases

Resonance in AC Circuits (c) its Q decreases (d) Both (b) and (c) 15. In case of a series resonant circuit, a change in supply voltage will change (a) the resonance frequency (b) the bandwidth of the circuit (c) the current drawn (d) the Q of the circuit 16. For a series RLC circuit, at half-power frequencies, we must have (a) XC = XL (b) XC = 2XL (c) R = 0 (d) XC – XL = ± R 17. For a series RLC circuit, the resonance frequency f0 is related to the half-power frequencies f1 and f2 as (a) f0 = f1 f 2 (b) f0 = (f1 + f2)/2

333

(c) f0 = f2 – f1 (d) f0 = f1 + f2 18. The power factor of a series RLC circuit at its halfpower frequencies is (a) unity (b) leading (c) lagging (d) leading or lag ging 19. In a parallel resonant circuit, the current through the capacitor is (a) equal to the line current (b) greater than the line current (c) less than the line current (d) zero 20. A parallel resonant circuit can be employed (a) to amplify voltage (b) to reject a small band of frequencies (c) as a high impedance (d) Both (b) and (c) A N SW E RS

1. b 11. c

2. a 12. d

3. b 13. d

4. c 14. d

5. d 15. c

6. a 16. d

7. d 17. a

8. b 18. d

9. d 19. b

10. d 20. d

PROBLEMS ( A )

S IM P L E

PRO BL EMS

1. A 120-V, 20-Hz source is connected to a series circuit consisting of 5.0-W capacitive reactance, a 1.6-W resistor, and a coil with resistance and inductive reactance of 3.0-W and 1.2-W, respectively. Determine (a) the input impedance, and (b) the circuit current. [Ans. (a) 5.97––39.56° W; (b) 20.1–39.56° A] 2. For Prob. 1, calculate (a) the voltage across the coil and (b) the resonance frequency. [Ans. (a) 64.98–61.36° V; (b) 40.92 Hz] 3. A series RLC circuit consists of R = 1 W, L = 140 mH and C = 100 mF. Determine the frequency at which resonance occurs. If the applied voltage is 220 V at 50 Hz, determine the current and voltage drops across R, L and C. [Ans. 42.5 Hz, 18 A, 18 V, 791.64 V, 572.94 V] 4. A coil with resistance and inductance of 20 W and 3.0 mH, respectively, is connected in series with a

capacitor and a 12-V, 5.0-kHz source. Determine (a) the value of capacitance that will cause the system to be in resonance, and (b) the circuit current at the resonance frequency. [Ans. (a) 338 nF; (b) 0.6 A] 5. What is the maximum stored energy in the capacitor of Prob. 4 ? [Ans. 1.08 mJ] 6. A 24-mF capacitor is connected in series with a coil whose inductance is 5.0 mH. Determine (a) the resonance frequency, (b) the resistance of the coil if a 40-V source operating at the resonance frequency causes a current of 3.6 mA, and (c) the Q of the coil. [Ans. (a) 459.5 Hz; (b) 11.1 kW; (c) 1.3 ¥ 10 –3] 7. A coil having a resistance of 2 W is connected in series with a 50-mF capacitor. The circuit resonates at 100 Hz. (a) Calculate the inductance of the coil. (b) If the circuit is connected across a 100-V, 100-Hz

334

Basic Electrical Engineering

ac source, determine the power dissipated in the coil. (c) Calculate the voltage across the capacitor and the coil. [Ans. (a) 50.66 mH; (b) 5 kW; (c) 1591.5 V, 1594.6 V] 8. A coil having resistance and inductance of 5.0 W and 32 mH, respectively, is connected in series with a 796-pF capacitor. Determine (a) the resonance ( B)

T R IC KY

PR OB LE MS

10. A generator supplies a variable frequency voltage of constant amplitude of 100 V (rms) to a series RLC circuit, having R = 5 W, L = 4 mH and C = 0.1 mF. The frequency of the generator is varied until a maximum current is obtained. Determine the maximum current, the frequency at which it occurs, and the resulting voltage across the inductance and the capacitance. [Ans. 20 A, 7.958 kHz, 4 kV] 11. A series RLC circuit, having a variable inductor, is connected to a sinusoidal source 200 2 sin100pt V. By varying the inductance, the maximum current obtained is 0.314 A, and under this condition the voltage across the capacitor is 300 V. Find the values of the circuit elements. [Ans. R = 637 W, C = 3.332 mF, L = 3.04 H] 12. A coil with resistance and inductance of 40 W and ( C )

frequency of the circuit, (b) the quality factor, and (c) the bandwidth. [Ans. (a) 31.53 kHz; (b) 1268; (c) 24.9 Hz] 9. The circuit of Prob. 8 is connected to a 120-V source and operates at resonant frequency. Calculate (a) the input current, and (b) the voltage across the capacitor. [Ans. (a) 24 A; (b) 152 kV]

C HA L L EN G IN G

50 mH, respectively, is connected in series with a 450-pF capacitor and a generator. Determine (a) the resonance frequency, and (b) the circuit impedance at the resonance frequency. [Ans. (a) 33.553 kHz; (b) 40 W] 13. A 60-V source having an internal resistance of 10 W is connected to the circuit of Prob. 12. Determine (a) the circuit current, and (b) the voltage across the capacitor. [Ans. (a) 1.2 A; (b) 12.65 kV] 14. The bandwidth of a series RLC circuit is 100 Hz and the resonance frequency is 1000 Hz. If the circuit resistance is 10 W, determine L and C. [Ans. 15.92 mH, 1.59 mF] 15. Find the half-power frequencies of a series RLC circuit having Q = 60 and f0 = 12 kHz. [Ans. 11900 Hz, 12100 Hz]

PROBLEMS

16. A series RLC circuit consists of a 100-W resistance, and inductance of 0.318 H and a capacitance of unknown value. When this circuit is energised by a 230–0° V, 50-Hz sinusoidal ac supply, the current was found to be 2.3–0° A. Find (a) the value of the capacitance, (b) the voltage across the inductance, and (c) the total power consumed. [Ans. 31.86 mF, 230–90° V, 529 W] 17. A series RLC circuit has a Q of 5.1 at its resonance frequency of 100 kHz. (a) Assuming that the power dissipation of the circuit is 100 W when drawing a

current of 0.80 A, determine the circuit parameters. (b) What is the bandwidth of the circuit ? (c) Determine the half-power frequencies of the circuit ? [Ans. (a) R = 156 W, L = 1.26 mH, C = 2.01 nF; (b) 19.6 kHz; (c) 90.7 kHz, 110.3 kHz] 18. A practical parallel resonant circuit consists of a 65-pF capacitor in parallel with a coil whose inductance and resistance are 56 mH and 60 W, respectively. Determine (a) its resonant frequency, and (b) the quality factor at resonance frequency. [Ans. (a) 2.63 MHz; (b) 15.4]

THREE-PHASE CIRCUITS AND SYSTEMS

12

OB JE CT IV E S :

4

3

3

3 3

3

two-phase system three-phase system* ac supply.

*

f

(

)

336 The use of polyphase systems

f

f

1.

2. 3.

4. 5. 6.

V or V

V V a

equivalent b

V =V +V

V = V + V = Vac + V = V + V + V Iab a

a

c

a

b c

+ + –

Vab

b

Icd ?

Icd ?

–

d b (a)

d (b)

(c)

Three-phase Circuits and Systems

337

write V = V + V

I by the direct path

I c I c

a

ea e

ec

b E

=E

=E

=E

w c

the phase order abc

phase sequence

b onto the

phase rotation

338 b

ea e

ec

Ea + E + Ec Ea ea = E sin wt e =E wt wt

ec = E

wt

E

t ea

e =E

E

t

e =E

E

t ea + e + ec

E

E

a RYB

R ¢

Y ¢

¢

B

w ¢ ¢

¢ ¢

¢ ¢

b or c

¢

b shows

¢

¢

Magnetic field R

R B¢

Y¢

Y¢

120°

w

N 120°

S Y

B

B¢

w

w

B

R¢

R¢

(a)

(b)

Y

Three-phase Circuits and Systems

339

w

clockwise

w b

a

Note ¢

b ¢

¢

e

e

e

two

i

D

ii

1 Start

R eR

Finish

R¢

L1 2 3

Start

Y eY

Finish

Y¢

L2

Loads

4 5

Start

B eB

Finish

B¢

L3 6

a b

Z Z

Z

340

Basic Electrical Engineering

Three-phase loads. In a three-phase load, if all the three impedances are equal (both in resistive as well as in reactive parts), the load is said to be a balanced load. An example of such a balanced three-phase load is a three-phase motor, which has three identical windings. To such a balanced load, if a balanced three-phase supply is applied, the currents will also be balanced. Conversely, if it is carrying balanced currents, the voltages across the circuit will also be balanced.

¢, Y¢ and B¢ at N, as shown in Fig. 12.6a. This way, the three conductors 2, 4 and 6 of Fig. 12.4 are replaced by a single conductor NM of Fig. 12.6a. a. Let i , iY and iB be the instantaneous values of the three phase currents. Then the current in the common wire MN is (i + iY + iB), having positive direction from M to N.

Star-connected four-wire three-phase voltage system.

Three-phase Circuits and Systems

341

four-wire star-connected b star or neutral point.

I i = I sin wt The instantaneous currents in L i =I

L wt

i + i + i = I {sin wt balanced load

i =I wt

wt

wt I ¥ *

zero at every three-wire star-connected

Note

f R local neutral point

*

neutral

342

Basic Electrical Engineering

If an unbalanced star-connected load is connected to a 3-wire 3-f system (shown in Fig. 12.8), the currents in the three phases of the load would be different. However, as per KCL, the sum of the three line currents must still be zero. This can happen only if the voltage drops across different load branches are different. In fact, the voltages across the three loads get adjusted such that the sum of the line currents becomes zero. This results in a situation known as neutral shift or . The potential of the neutral point does not remain zero but depends on the load impedances. Under such a situation, it is quite possible that one phase voltage becomes much larger than the other phase voltages. The equipment connected across this phase may get damaged due to overvoltage. On the other hand, the undesirable. In view of this, unbalanced star-connected loads are not normally used on a three-wire threephase system. : distribution of electric power to domestic, industrial and other consumers. The domestic and other singlephase loads are connected between a phase and neutral. Three-phase loads and other loads needing higher voltages are connected between the lines. In such cases, the currents in the three lines may not be equal; their phasor sum would return through the neutral conductor.

D ¢ and Y thereby replacing conductors 2 and 3 by a single conductor. Similarly, we join Y¢ and B together replacing conductors 4 and 5 by another single conductor. The result is shown in Fig. 12.9a. ¢ two points is zero at every instant. It will then be ensured that no circulating current is set up when these two conductors are connected together. The instantaneous value of the total emf from B¢ e + e Y + e B = Em[sinq + sin(q – 120°) + sin (q – 240°)] = Em[sinq + {sinq ◊ cos 120° – cosq ◊ sin 120°} + {sinq ◊ cos 240° – cosq ◊ sin 240°}] = Em[sinq – 0.5sinq – 0.866 cosq – 0.5sinq + 0.866cosq] =0 ¢ together, as shown in Fig. 12.9b, without creating any circulating current in the circuit. The three line conductors are joined to the junctions thus formed.

Three-phase Circuits and Systems 1 Start

R

R R

eR Finish

eR

R¢

R¢ R

Y B¢ Start

Y

Y eY

Finish

343

eY

Y¢

B

Y¢

R

eB eR eY Y¢

R¢ Y

Y

B Start

B

Finish

B¢

eB

eB 6

(a)

Fig. 12.9

B

B B¢ (b)

(c)

Connecting the three phase windings of Fig. 12.4 to make delta (D)-connected three-phase system.

The circuit derived in Fig. 12.9b is conventionally drawn as in Fig. 12.9c. In appearance, it looks like Greek letter delta (D). Hence, this connection is given this name. Note that in Fig. 12.9c

i) line voltages, and (ii) phase voltages. Similarly, i) line currents, and (ii) phase currents. We shall now determine the relations connected system.

ERN, EYN and EBN.

V

E. phase voltages,

line voltages. In Fig. 12.10, VRN phase R. Thus, VRN, VYN and VBN

344 R

IR

Y

IY

R ERN

IR IB

IY

Y

EYN

VRN

EBN

B

B

Three-phase load VYN

IB VBN

N

IN

Line conductors

Generator

V V

V

a a

b

a V V V V N V V

N to Y =V =V V a

=V

+V

=V

=V

+V R to Y

R to

–V =V V V to get –V

V

Analytical Analysis write

V V

= Vph–

|V

| = |V

V

= Vph–

| = |V

| = Vph V

= Vph–

= V – V = Vph– Vph– Vph – Vph = Vph – Vph ( - 1/2 - j 3/2) = Vph (3/2 + j 3/2) V V

= Vph

(9/4 + 3/4) =

3Vph

Vph– j

Three-phase Circuits and Systems

345

VRN is given as

VRY

⎛ 3 /2⎞ –1 ⎛ 1 ⎞ q = tan –1 ⎜ ⎟ = tan ⎜ ⎟ = 30° ⎝ 3/ 2 ⎠ ⎝ 3⎠

(12.7)

Hence the phasor VRY 3Vph –30∞

VRY =

(12.8)

Similarly, we can get VYB =

3Vph –- 90∞

and

VBR =

3Vph –150∞

(12.9)

VL VL =

(12.10)

3Vph

Geometrical Analysis

a, it is evident that the line voltages

VRN and –VYN is 60°, we have VRY = 2(VRN cos 30°) which is same as Eq. 12.10.

or VL = 2Vph ( 3/2) =

3Vph

R R and N (12.11)

I L = Iph IB

VBR

–VYN

VBN

30° 120°

VRY

IB¢B

120°

VRN

30°

120°

VYN

120°

IY

IY¢Y

IR¢R

–IB¢B

VYB (a) Star-connected system.

(b) Delta-connected system.

Phasor diagrams.

Let IR¢R, IY¢Y, IB¢B

IR

346

Basic Electrical Engineering

A star-connected generator supplying power to a three-phase load. b. |IR¢R | = |IY¢Y | = |IB¢B | = Iph (say) IR¢R = Iph–0°; IY¢Y = Iph––120°; IB¢B = Iph––240° = Iph –120° IR¢R phase current IB¢B IR = IR¢R – IB¢B b. From the symmetrical

IR¢R and –IB¢B 120°. Also IR = 2(Iph cos 30°) =

3Iph

are related as IL =

3I ph

(12.12)

VRY is same as the phase voltage VRR¢ delta-connected system, we have VL = Vph

(12.13)

(i line voltages and line currents. Thus, when we say that a 3-phase, 11-kV circuit is carrying a current VL = 11 kV and IL D connection. (ii Iph = Vph/Z Iph.

Z

Vph and

Three-phase Circuits and Systems EXA M P L E

347

12 .1 W resistance

A 400-V, 3-f and 24-W (a) Y-connected, and (b) D-connected.

Solution Each impedance, Z = R + jX = (32 + j24) W. \

Z = (a) Y connection \

Vph =

R2 + X 2 =

VL 400 = V 3 3

IL = Iph =

32 2 + 24 2 = 40 W

fi Iph =

EXA M P L E

IL =

=

Z

400 / 3 10 = A 40 3

10 = 5.78 A 3

(b) For D connection Vph= VL = 400 V fi \

Vph

3I ph =

Iph =

Vph Z

=

400 = 10 A 40

3 ¥ 10 = 17.32 A

12 .2

a) the current in each line, and (b) the current in the neutral conductor.

Circuit diagram for Example 12.2. Solution (a Vph =

and

VL 415 = = 240 V 3 3

IR = 10 ¥ 1000/240 = 41.67 A IY = 8 ¥ 1000/240 = 33.33 A IB ¥ 1000/240 = 20.83 A a.

348 IR = 41.7 A

IV = 14.59 A

30°

IN

30°

IB = 20.8 A IH = 10.83 A

IY = 33.3 A (a) Phasor diagram of line currents.

(b) Horizontal and vertical components of current IN.

b I =I IV = I – I

I I b I =

EX A MP L E

(10.83)2 + (14.59)2 = 18.2 A

12. 3 W W

m

a

b IR

R

R1 100 W Y

IY

B

IB

C3 I1

R2 20 W

I2

30 mF

I3 L2

191 mH

12.3 Solution a Z =R

W

f

Three-phase Circuits and Systems R22 + X22 = (20)2 + (2 πf L2)2 = (20)2 + (2 π × 50 × 0.191)2 60 f = tan 20 Z pfC p¥ ¥ ¥ W f V V V V I 415 415 I = = = 4.15 A V Z1 100 415 415 I = = = 6.56 A V f Z2 63.25 415 415 I = = = 3.91 A V Z3 106.1 Z =

The phase current I I

b

349 W

a

R is I =I –I a I

I

I I =

( 4.15)2 + (3.91)2 + 2 × 4.15 × 3.91 × cos 30° =

60.53 = 7.79 A

350

Basic Electrical Engineering

The line current IY and I1 (reversed) is

I2

I2 – I1 (see Fig. 12.16b q Y = f2

\

( 4.15)2 + (6.56)2 + 2 × 4.15 × 6.56 × cos 11°34′ = 10.66 A

IY =

Similarly, the line current IB and I2 (reversed) is given as

I2

I3, as shown in Fig. 12.16c

I3

qB \

IB =

(6.56) 2 + (3.91) 2 + 2 ¥ 6.56 ¥ 3.91 ¥ cos138.44° = 4.47 A

Vph and the current is Iph. The average active power P1 = Vph Iph cos f where f consumed is P = 3P1 = 3Vph Iph cos f

(12.14)

practice to mention line voltage and line current in a three-phase system. Hence, let us determine the VL and IL. For a star-connected system, we have VL =

3Vph and IL = Iph. Hence,

P = 3 (VL / 3 ) I L cos f = For a delta-connected system, we have VL = Vph and IL =

3VLILcos f 3 Iph . Hence,

P = 3VL ( I L / 3 ) cosf =

3VLIL cos f

any balanced load (connected in either Y or D), the total power is given as P=

3VL IL cosf f

VL and IL.

EXA M P L E

12 .4

A 400-V, 3-f W inductive reactance. Determine (i) the line current, (ii the three impedances are (a) star-connected, and (b) delta-connected.

W resistance and iii) the total power in kW, when

Three-phase Circuits and Systems

Solution a For star-connected load VL = \

IL = Iph

3Vph VL 400 = 3 3

Vph =

20 2 + 152

Zph =

W

i Vph

IL = Iph =

=

Z ph

231 = 9.24 A 25

ii cos f =

R ph Z ph

=

20 = 0.8 (lagging) 25

iii 3 VLIL cos f =

P= b For delta-connected load VL = Vph \

IL =

3 ¥

¥

¥

5.12 kW

¥

15.36 kW

3I ph

VL = Vph

i

Iph = \

Vph Zph

IL =

ii iii

=

400 25

3I ph =

3 ¥

27.71 A

pf = 0.8 (lagging) P=

f

f

f

3VL IL cos f = 3 ¥

¥

f f

351

352

S. No.

Star-Connected System VL =

3 Vph

Delta-Connected System

IL = Iph

VL = Vph

f

IL =

3 Iph

f

i

ii a ¥

iii

3-f

353

Three-phase Circuits and Systems

Suppose that the three loads L1, L2 and L3 are connected in star, as shown in Fig. 12.18b. The current coils 1 and W2

Proof Let vRN, vYN and vBN iR, iY and iB line (and phase) currents. \

Total instantaneous power = iRvRN + iY vYN + iB vBN. 1 is iR, and the potential across its potential coil is vRN – vYN,

we have W1, p1 = iR(vRN – vYN) W2, p2 = iB(vBN – vYN) 1 and W2 is p1 + p2 = iR(vRN – vYN) + iB(vBN – vYN) = iRvRN + iBvBN – (iR + iB)vYN

N is zero, i.e., iR + iY + iB = 0 fi (iR + iB) = –iY p1 + p2 = iRvRN + iBvBN – (iR + iB)vYN = iRvRN + iB vBN + iY vYN = Total instantaneous power

\

the sum of the two wattmeter readings gives the total power under all conditions.

f (lagging), connected to a 3-wire, 3-f system, as shown in Fig. 12.19a W1 and W2 are connected in the line conductors R and Y, respectively. Their potential coils are connected Let IR, IY and IB

Since VRB = VRN VYN *

Here, point N

VRN, VYN and VBN

*.

Since the

IR = IY = IB = IL (say) and VRN = VYN = VBN = Vph (say) f b. VBN, and VYB = VYN VBN, we can determine the line voltages VRB and VYB b that the line voltage VRB lags the phase voltage VRN VYB leads VRB and the line current IR f). Similarly,

354

V W =V W =V

f

I f f

I I W1 W2

=

f f

VL IL VL IL

cos(30° − φ ) cos(30° + φ )

cos(30° − φ ) − cos(30° + φ ) 2 sin 30° sin φ W1 - W2 = = W1 + W2 cos(30° − φ ) + cos(30° + φ ) 2 cos 30° cos φ or

tanf =

f

È W - W2 ˘ 3Í 1 ˙ Î W1 + W2 ˚

f

f W

W + W = VL IL = VL IL

f

f

VL IL f

VL IL

3

VL IL f

f 3 VL ILcos f

P=W +W This is an

W f

Three-phase Circuits and Systems

IM PO RTAN T

1. 2.

f f

3. 4.

f f

POINTS

W W W

EX A MP L E

12. 5

Solution P=W +W

4.5 kW

È W - W2 ˘ 3Í 1 ˙ = Î W 1 + W2 ˚ 1 f = tan 3

1 ⎡ 3 − 1.5 ⎤ 3⎢ ⎥= 3 ⎣ 3 + 1.5 ⎦

tan f = \

pf = cos f

EX A MP L E

3 / 2 = 0.866

12. 6

a

b

c

Solution a b

P=W +W

3.5 kW

tan f = \

È W - W2 ˘ 3Í 1 ˙ = Î W1 + W2 ˚

⎡ 5.2 − ( − 1.7) ⎤ 3⎢ ⎥ ⎣ 5.2 + ( − 1.7) ⎦ ¢

f = tan

¢ = 0.281

pf = cos f c 3 ¥

¥ IL ¥

fi

IL = 17.3 A

355

356

ADDITIONAL SOLVED EXAMPLES EX A MP L E

12. 7 W

W

Solution

–

j

Z

W

Vph = VL / 3 = 220/ 3 = Vph – = Vph – = Vph –

V V V =V =V =V

V V V

j j

=

I = 3VLIL cos f =

127–0° V 127––120° V 127–+120° V

j j j

I

EX A MP L E

j

j

–V –V –V

I =

P=

V

j j

220–30° V 220–– 90° V 220–+150° V

j

j

j

VRN 127– 0∞ V = = 12.7––53.13° A 10– 53.13∞ W Z VYN 127–-120∞ V = = 12.7––173.13° A 10–53.13∞ W Z

127–+ 120∞ V VRN = = 12.7–66.87° A Z 10–53.13∞ W

3 ¥

¥

¥

2904 W

12. 8

Solution

Z Vph = VL V

–

j

W

V

= 220–0° V

V

= 220––120° V

V

I

=

220– 0∞ V = 22––53.13° A 10–53.13∞ W

I

=

220–-120∞ V = 22––173.13° A 10–53.13∞W

I

=

220–120∞ V = 22–66.87° A 10–53.13∞ W

= 220–+120° V

j j j

Three-phase Circuits and Systems

357

The line currents are

Power consumed, P = EX A M P L E

IR = IRY – IBR = (13.2 – j17.6) – (8.65 + j20.2) = 38.1–– 83.13° A IY = IYB – IRY = (–21.85 – j2.63) – (13.2 – j17.6) = 38.1–156.87° A IB = IBR – IYB = (8.65 + j20.2) – (–21.85 – j2.63) = 38.1–36.81° A 3VLIL cos f = 3 ¥ 220 ¥ 38.1 ¥ cos 53.13° = 8710.85 W

12 .9

Three identical impedances of 12–30° W are connected in delta, and another set of three identical impedances of 5–45° W are connected in star. Both these sets are connected across a 3-f, three-wire, 400-V supply. Find the line current and the total power supplied.

Solution The delta-connected impedances can be converted into its equivalent star connection. Each equivalent impedance is given as ZA =

Z2 Z3 Z1 + Z2 + Z3

ZY =

or

Z 12 ∠ 30° = = 4–30° W 3 3

Thus, each phase has two impedances in parallel. The equivalent load impedance in each phase is (5– 45∞) (4– 30∞) Z eq = = 2.24–36.65° W (5– 45∞) + (4– 30∞) Thus, we have a single star-connected balanced three-phase load each of impedance Zeq. V 400 The line to neutral voltage, Vph = L = = 231 V 3 3 The line current, IL = Iph = Power, P =

3VLIL cos f =

EX A M P L E

Vph Z eq

=

231 = 103.1 A 2.24

3 ¥ 400 ¥ 103.1 ¥ cos 36.65° = 57308 W ª 57.3 kW

12 .10

A balanced three-phase, star-connected load of 150 kW takes a leading line current of 100 A from a 1100-V, 50-Hz, three-phase supply. Determine the constants of the load per phase.

Solution For the star-connected load, the total power is P = 3I L2 R. Hence the resistance of the load is given as R=

P 3I L2

=

150 1000 3 (100)2

=5W

Vph

V / 3 = L , the load impedance is Z Z 1100/ 3 V / 3 = Z= L = 6.35 W IL 100 Since the load takes a leading current, the load must be capacitive. Hence, we have Since the line current is given as IL = Iph =

Z = \

C =

R2 + XC2 1 2p fX C

XC = Z 2 R2 = 6.352 1 = = 812.84 mF 2p ¥ 50 ¥ 3.916 or

52 = 3.916 W

358

Basic Electrical Engineering

EXA M P L E

12 .1 1 f

f, 400-V supply. The line current is 20 A and the total power b c) total power

a

Vph = VL = 400 V; Iph = I L / 3 = 20/ 3

Solution

Z ph =

(a

I ph

P = 3 VLI L

f=

(b

Vph

400 = 34.63 W 11.55 10 000 = 0.7217 3 400 20 =

Vph = VL/ 3 = 400/ 3 = 231 V; Iph = IL

(c \

IL = Iph =

Total power consumed, P = EXA M P L E

Vph Z ph

=

231 = 6.67 A 34.64

3 VLIL cos f =

3 ¥ 400 ¥ 6.67 ¥ 0.7217 = 3.34 kW

12 .1 2 R1 = 100 W, R2 = 200 W

L3 a) the total

power in the system, and (b) the total volt ampere reactive.

An unbalanced delta-connected load fed from a three-phase supply. Solution X3 = 2pf L = 2p ¥

¥ 0.3 = 94.2 W. V = 100–0° V; V = 100––120° V and Vca = 100––240° V The phase impedances are Z = (100 + j0) W = 100–0° W; Z = (0 + j94.2) W = 94.2–90° W Zca = (200 + j0) W = 200–0° W

I

=

I ca =

Vab 100 ∠ 0° = = 1–0° A; Zab 100 ∠ 0°

I

=

Vca 100 ∠ − 240° = = 0.5––240° A Zca 200 ∠ 0°

and

Vbc 100 ∠− 120° = = 1.06––210° A Z bc 94.2 ∠ 90°

Three-phase Circuits and Systems

a Active power P

=

\ b Volt-ampere reactive

2 Vab 100 2 = 100 R1 P=P +P

P

Pca =

+ Pca

150 W ¥ ¥ 106 VAR

= V I sin f \ EX A MP L E

2 Vca 100 2 = 200 R2

12. 1 3

Solution

Zph =

210 103

P = 3VLI L

pf = cos f =

Vph Iph

=

3

VL / 3 IL

=

11 . 103 160 1.1 ¥ 103/ 3

W

160

R ¥

R = Zph cos f \

C =

EX A MP L E

2.734 W

C W

¥

XC = Zph sin f

1 1 = = 1106 mF 2p ¥ 50 ¥ 2.878 2pfX C

12. 1 4

a b

c

Solution a 13.856 103 P = 3 400 25 3VLI L Vph 400/ 3 V / 3 Zph = = L = W Iph IL 25 Rph = Zph cos f ¥ 7.392 W Xph = Zph sin f

pf = cos f =

\ \

L =

X ph 2p f

=

¥

5.544 = 17.6 mH 2p ¥ 50

b Q

VphIph sin f

¥ (400/ 3 ) ¥

¥

10.392 kVAR

c S

VphIph

¥ (400/ 3 ) ¥

17.321 kVA

W

359

360 EX A MP L E

12. 1 5 W

Solution Z

W j

j

Z = – jX \

Zph = Z ||Z =

– pfC

W W

j

Vph Zph

=

VL / 3 Z ph

pf = cos f P =

a

3 ¥

W

–

W

415/ 3

= 3.39 W 70.5 0.7046 (leading) =

3 ¥

3 VLIL cos f = 3 VLIL =

EX A MP L E

–

(100– 0∞) ¥ (99.5–- 90∞) (100– 0∞) + (99.5–- 90∞)

IL = Iph =

m

¥

¥

¥

1.717 kW

2.437 kVA

12. 1 6

b

c

d

Solution a

P=W +W

b

f = tan \

IL =

d

Po

Solution \

È W - W2 ˘ 3Í 1 ˙ = tan Î W1 + W 2 ˚

pf = cos f

c

EX A MP L E

340 kW ⎡ 255 − 85 ⎤ 3⎢ ⎥ = tan ⎣ 255 + 85 ⎦

0.756 (lagging) 340 103

P = 3VL cosf

3

1.6 103 0.756

¥

= 162.28 A

¥

292.4 kW

12. 1 7

pf = cos f f

P =W +W f f =

È W - W2 ˘ 3Í 1 ˙ Î W1 + W 2 ˚

W = 17.9125 kW

fi

W –W =

1 W +W 3

W = 7.0875 kW

f=

1 ¥ 3

¥

Three-phase Circuits and Systems EX A MP L E

12. 1 8 –

Z

W Z

–

W

Z

–

a W

Solution

W

V –

V

–

V

I =

200– 0∞ VRY = 10 – 60∞ Z1

I =

VBR 200–+120∞ = 10– 60∞ Z3 I =I –I I =I –I

–

V

–

I =

200–-120∞ VYB = 10– 0∞ Z2

–

–

– –

– –

– –

active or real S ∫ P + jQ = V

I*

–

–

–

j

W =0W S ∫ P + jQ = V

I* = –V –

I*

–

–

j

W = 8000 W P=W +W EX A MP L E

8000 W

12. 1 9 b

Solution

W

361

362 – –

I =I –I I =I –I active or real I* = –V –

S ∫ P + jQ = V

– –

I*

– –

–

–

–

–

j

W = 5999.8 W S ∫ P + jQ = V

*

I =V –

W = 2000 W P=W +W

I

*

–

–

j 7999.8 W

N O T E

EX A MP L E

12. 2 0 W f

W

W RYB R

A

R C

B D

W

I2

I1 60 W

56 W

45 W Y B

–

Solution Let V I

–

V

=I =I +I = –

VRY V 440 ∠ 0° 440 ∠ (120° − 180°) + RB = + 60 + j 45 − j 56 75 ∠ 43° 56 ∠ − 90° ª – V

P=I

V

¥

–

V

=V ¥

– ¥

–2439.8 W

363

Three-phase Circuits and Systems EX A MP L E

12. 2 1

f j

W

f a

b

c

d

f Solution

V

V fi

a For Motor Delta-Connected pf = cos f d IL = I = P \ \

IL = I

3 ¥

3 VLcos f d I –I

–

I 123 Iph = I = L ––fd = – 3 3

fd

fd ¥

123––66.87° A 71––36.87° A

–

364

Basic Electrical Engineering

(b) For Load (Star-Connected

Z = (3 + j4) W – W 440 Vph = VRN = – ––30° V 3 Vph 254 ∠− 30° IL = IRs = Iph = = = 50.8––83.13° A Z ph . ° 5 ∠ 5313

(c) Total Line Current IL = IR = IRd + IRs = (123– (d) Total Power Consumed P1 =

––83.13°) = 172.3––71.6° A 3 VLIRd cos f d =

3 ¥ 440 ¥ 123 ¥ 0.8 = 74 991 W

P2 = 3 VLIRs cos f s = 3 ¥ 440 ¥ ¥ Total power consumed, P = P1 + P2 = 74 991 + 23 229 = 98 220 W = 98.22 kW

SUMMARY TE R MS

A N D

C O NC EPT S

Three-phase systems motor drives. Red, Yellow and Blue, although some systems use letters a, b and c star (Y) or delta (D) phase values (represented as Vph and Iph termed the line values (represented as VL and IL).

not. IMP O RTA N T

F O R MU LAE VL =

3Vph and IL = Iph.

VL = Vph and IL = any balanced load, P =

3 Iph .

3 VLIL cos f, where f is the phase angle in one phase.

W1 = VLIL cos (30° – f) and W2 = VLILcos (30° + f) P = W1 + W2 W - W2 ˘ f = 3 ÈÍ 1 ˙ Î W1 + W 2 ˚

Three-phase Circuits and Systems

365

CHECK YOUR UNDERSTANDING two one

12

S. No.

Statement

True

False

Marks

f f

total power P P

Your Score A N SW E RS

1. 6. True

2. True 7.

4. True 9. True

3. 8.

5. 10. True

REVIEW QUESTIONS 1.

5.

2.

a b

3.

6. a

4.

b

366

Basic Electrical Engineering 9.

7.

connected to a three-phase supply ? 10.

8. is not normally used in a three-wire, three-phase system.

MULTIPLE CHOICE QUESTIONS Here are some incomplete statements. Four alternatives are provided below each. Tick the alternative that completes the statement correctly.

(a b) 30 A (c d) 0 5. Three unequal impedances are connected in star in a three-wire, 3-f

V

1. represents (a

(a

(b

(b (c)

(c (d and a 2. In a three-phase, star-connected system, the three phase voltages are 100–0° V, 100––120° V and 100––240° V. Then the three line voltages must (a) 173.2–0° V, 173.2––120° V and 173.2––240° V (b) 173.2––30° V, 173.2– 173.2––270° V (c) 173.2–30° V, 173.2––90° V and 173.2– (d – ––120° V and ––240° V 3. f, star-connected load, the three phase currents are 10––30° A, 10– 10– (a) 17.32––30° A, 17.32– 17.32–90° A (b) 17.32–0° A, 17.32––120° A and 17.32–120° A (c) 10–0° A, 10––120° A and 10–120° A (d) 10––30° A, 10– ––270° A 4.

phase currents 3 phase currents (d) zero 6. Three unequal impedances are connected in delta in a three-wire, 3-f system. Then (a unequal (b (c (d 7. The phase sequence R B Y denotes that (a 120° (b 120° (c 120° (d 120° 8.

(a) unity (c) 0.707 leading

(b) 0.707 lagging (d) zero

(a) unity (c

(b) 0.8 d) zero

9.

367

Three-phase Circuits and Systems a b c d

10.

A N SW E R S

1. b

2. c

3. d

4. d

5. d

6. d

7. b

8. a

9. c

10. b

PROBLEMS ( A )

S IM P L E

PR OB LEM S a

1.

b i

W resistance

ii

iii a

b [Ans. a

–

[Ans. a i iii

b

2.

ii b i

ii

W

– iii

7. a

b [Ans. a

b

[Ans.

W

3. a

8.

f

b

f

c

a b

[Ans. a

b

c [Ans. a

4. j

W

W

b

9. a

b [Ans. a

–

a

b b

5.

[Ans. a a c [Ans. a

b b

pf

c

6. W

b

10.

W

[Ans.

368 f

11.

12.

Ans. Ans. ( B)

TRI C KY

PR O BL EM S

13. W

[Ans. a b

W per

j

18.

[Ans. a b

–

[Ans.

14.

] 19.

a b Ans. a

f

b

15.

a

W

[Ans.

b –

Ans. a

W Z –

20.

W b

–

Z

16.

W

Z

–

W

W

f ABC –

[Ans. W m

[Ans.

–

W

– –

21. f

17. Ans. ( C )

C HA L L E NGI NG

22.

Z Z

j

PROBLEMS W are

j

23.

f

W

j f

Ans.

f

W

Three-phase Circuits and Systems a

369

25.

b

f

c

f

d b

[Ans. a –

–

– –

c

a b

d

c

24. a [Ans. a 26.

f

b

W

[Ans. a

W

[Ans.

b

EXPERIMENTAL EXERCISE 12.1 TW O -WAT TM ETER

Objectives Apparatus Circuit Diagram

Brief Theory

MET HOD

b

W c

370 W

W P=W +W

tan f =

È W - W2 ˘ 3Í 1 ˙ Î W1 + W 2 ˚

or

f = tan

i È W - W2 ˘ 3Í 1 ˙ Î W1 + W 2 ˚

ii

pf = cos f VL

IL P=

1. 2. 3. 4. 5. 6. 7.

iii

3 VLIL cos f

iv

ON

8. 9.

Observations S. No.

IL in

VL in

P in

P in

Calculations S. No.

P=W +W in Eq. i

Phase angle Eq. ii

Power factor Eq. iii

P in Eq. iv

Results 1. 2. 3.

i ii

iii iv

Three-phase Circuits and Systems

371

Precautions 1. 2. 3.

Viva Voce single

1. two Ans. : two

three

2. i

Ans. :

ii

3. Ans. : Q=

3 W –W

4. ii

Ans. :

W =W

5. ii

Ans. : W = –W 6. Ans. :

P

f=

ii

3 fif

7. R Y B

Ans. : 8. Ans. : 9.

C

Ans. :

L L1 A

L2 B

C C

A B

C

TRANSFORMERS OB JE CT IV E S :

*

Applications

*

13

Transformers

373

practice is to generate ac voltage at about 11 kV, then step up by means of a transformer to higher voltages of 132 kV, 220 kV and 400 kV for the transmission lines. This conversion aids the transmission of huge electrical power at low cost. High-voltage lines carry low currents, and hence the cost of lines and the power loss are tremendously reduced. At distribution points, other transformers are used to step the voltage down to practically needs almost no maintenance and supervision. A transformer also electrically isolates the end user from contact with the supply voltage. Apart from the above, small-size transformers are used in communication circuits, radio and TV circuits, telephone circuits, instrumentation and control systems. Audio transformers are used to couple stages of match devices such as microphones and record player cartridges to the input impedance of wires.

A transformer operates on the principle of mutual induction between two coils. Figure 13.1a shows the general construction of a transformer. The vertical portions of the steel core are termed limbs, and the top and bottom portions are called yokes N1 and N2 turns, are wound on the limbs. the core. The coil P is connected to the supply and is therefore called primary; coil S is connected to the load and is termed the secondary. Basically, two principles are involved in the operation of a transformer. Firstly, an electric current produces

is transferred from one circuit to the other. Core Core F

I1 V1

E1

N1

P

I1

I2 S

Primary

N2

E2

Load

E1

I2 N1

N2

E1

Secondary (a) Construction.

(b) Circuit symbol.

Figure 13.1b shows the circuit symbol of a transformer. The thick line denotes the iron core. By having different ratios N1/N2 of the two windings, power at lower or at higher voltage can be obtained. When N2 > N1, the transformer is called a step-up transformer; and when N2 < N1, the transformer is called a step-down transformer.

374

Basic Electrical Engineering

V

a

F = Fm sin wt = Fm sin pft Fm

f N

e = -N

d dF = -N (Fm sin w t dt dt

w t = wNFm sin (wt p

NwFm Em = wNFm

given as E=

2pfNFm w NFm Em = = 2 2 2

E is fN Fm

E = 4.44 f NFm

or

emf equation of transformer F

*

EX A MP L E

13. 1 a b

Solution (a Fm =

6400 E = 4.44 f N1 4.44 50 480

60 mWb

(b E

fN Fm

ideal transformer *

skin

¥

¥

¥

266.4 V

375

b

Conditions for Ideal Transformer m

(i (ii (iii (iv

IR

ZL V

a

V E E is the same as voltage V E =V E mutually induced emf in

V V E counter emf or back emf

E

F

I1 V1

E1

N1

V1 = –E1

I2 N2

E2

Load ZL

V2

F

E1 E2 = V2 (a) The circuit.

(b) The phasor diagram.

F

b

E and E F

V

E

V

F

F

ratio of secondary voltage to the primary voltage is known as transformation ratio or turns-ratio. It is K. Let N and N E and E E

fN Fm

376

Basic Electrical Engineering

fN Fm

E

and transformation ratio or

K=

V2 E2 N2 = = V1 E1 N1 the voltage K *

K K

(i (ii

N >N N V V V larger

I2 N1 1 = = I1 N2 K

or

I I

E >E

V

E

I¢

I I¢

I I

I

Note

I

EX A MP L E

13. 9 –E1 = V1 = 440 V f1 = 42.49° 36.87°

I¢1 = 30 A

I1 = 33.9 A

f0 = 78.46°

Solution

I0 = 5 A

f

f V2 110 1 = = V1 440 4 I¢ – –

K= I¢ =K¥I ¥ I = I¢ + I – = 33.9– 42.49° A

f0 = 36.87°

I2 = 120 A

f

0.737 (lagging)

F

E2 = 110 V

(Phasors not drawn to scale.)

386

Basic Electrical Engineering

ideality conditions (iii

iv

I R loss or copper R

loss R

a F

FL

FL

Fu I1

I2 FL

V1

FL

1

2

+

X1

X2

V1

V2

–

– Ideal transformer (b) Its effect accounted for.

(a) Its definition.

F EL

X and X Note

+

FL FL b EL = I X and EL = I X primary and secondary leakage reactances

EL X and X

387 FL and FL F

V = I R + jI X E = I (R + jX E = I R + jI X + V = I (R + jX

and

E V

V resistance R and leakage reactance X I V

E X

E I

E is resistance R and leakage reactance

I

I I I and Im Im

I

I¢ R

I

X

388

the following points in mind : F. 2. The induced emfs E1 and E2 F by 90°. 3. The values of emfs E1 and E2 are proportional to the number of turns on the primary and secondary windings. 4. The magnitude and phase of the current I2 is decided by the load. 5. The resistive voltage drops are always in phase with the respective current phasor. 6. The inductive voltage drops lead the respective current phasor by 90°. 7. The secondary induced emf E2 is obtained by vector sum of the terminal voltage V2 and the impedance drop I2 Z2. Hence, V2 must be drawn such that the phasor sum of V2 and I2 Z2 is E2. 8. The primary balancing current I1¢ and the secondary current I2 are in inverse proportion to the number of turns on the primary and secondary windings. 9. The primary current I1 is the vector sum of the no-load current I0 and the primary balancing current I1¢. 10. The primary voltage V1 is obtained by adding vectorially the impedance drop I1 Z1 to the negative of E1. 11. The phase angle f1 between V1 and I1 is the power-factor angle of the transformer. Fig. 13.16. V1

I1R1 –E1 f f1 0 V2 I2R2

V1

I1X1 I¢1 I0

I2 f2 = 0

I1R1 –E1 f0 f1

I1

F

I2 I2R2

E2 (a) Resistive load.

I¢1 I0

f2

E1 I2X2

I1X1

I2X2

I1

I¢1

I1X1 V1 I1R1 I1 –E1 f1 f0 I0

F f2

V2

E1 E2

(b) Inductive load.

F

I2

E1 V2 E2 I2R2 I X 2 2 (c) Capacitive load.

I0 of a transformer is only about 3-5 percent of the full-load primary current, not R0-X0 in Fig. 13.15 is shifted to the left of impedance R1-X1. This results in a circuit shown in Fig. 13.17a.

Transformers

Using the impedance transformation, we can now transform the impedances from the secondary side to the primary side and remove the ideal transformer from the circuit, as shown in Fig. 13.17b. This can further Fig. 13.17c. Here, the total resistance and total leakage reactance as referred to primary are given as Rel = R1 R2/K 2 Xel = X1 X2/K 2

circuit R0-X0 Fig. 13.18a. Alternatively, we could transfer the impedances from primary side to the secondary side, so as to get the b. Here, the total resistance and total leakage reactance as referred to secondary are given as Re2 = K2R1 + R2 and Xe2 = K2X1 + X2

390

Basic Electrical Engineering

EX A MP L E

13. 1 0

R W this transformer (a c e g

W and X

X b d f

Solution

I =

VA 50000 = V1 4400

VA 50000 = V2 220 V2 220 1 K= = = V1 4400 20 ]= I =

(a (b (c (d

Rel = R R =K Xel = X X =K

(e Zel = (f (g

Z

=

+ (R /K R +R + (X /K X +X Re21 +

X e21

¥ ]= ¥ =

Re22 + X e22 =

( 7.05)2 + (11.2)2 = (0.0176)2 + (0.028)2 = 0.0331 ¥

I R +I R I Rel I R

¥ ¥

909 W

909.8 W ¥

909 W

R W

W and

391 V V

and V

(i

V

regulation down = regulation down =

(ii

regulation up = regulation up =

inherent regulation.

V2( 0 )

V2

V2( 0 ) V2( 0 )

V2( 0 )

V2

V2( 0 ) V2

V2 V2( 0 ) V2 V2

¥

¥ ’ means ‘

V b

V ª E and E = V = E = KE = KV

K

OA AB BC AC OC

V IR IX IZ V KV

V

V I2Ze2

C

V2(0) = KV1 O

f

A f2 = f

V2

E F

f B

I2

I2Re2

D I2Xe2

G

392

Basic Electrical Engineering

IZ V

leading power factor f

=I R

f + I X sin f

=I R

f

f

IR

I X sin f

f ± I X sin f

percent regulation I2 Re 2 cos φ ± I2 Xe 2 sin φ ¥ V2( 0 ) = Vr f ± V sin f I2 Re 2 Vr ¥ V2( 0 ) I2 Xe 2 V ¥ V2( 0 )

and

or

OC

OC = OF + FC = FC

ª OC] 2

\

FG =

FC (DC DF ) = 2OC 2OC

= (I R

2

=

(DC BE ) 2OC

2

=

f + I X sin f

( I 2 X e2 cos f - I 2Re 2 sin f ) 2 2V2(0)

( I 2 X e2 cos f - I 2 Re2 sin f ) 2 2V2(0)

393

IR

f

I X sin f

I R

f ± I X sin f

( I2 Re 2 cos φ ± I 2 Xe 2 sin φ ) ¥ V2( 0 ) 1 = (Vr f ± V sin f (V 2V2(0)

( I 2 X e 2 cos f + I 2Re 2 sin f ) 2 2V2(0) ( I 2 X e 2 cos f I 2Re 2 sin f ) 2 2V2(0) ( I 2 X e 2 cos f I 2Re2 sin f ) 2 ¥ 2V2(0) ¥ V2(0)

f

Vr sin f

only if the condition for zero regulation is given as R f I X sin f f = e2 Xe 2 f IR f

load has leading power factor. I R I X sin f

f and d (I R df

fi

f + I X sin f tan f =

or EX A MP L E

I R sin f + I X

f

Xe 2 R e2

13. 1 1 R

R

W

W a

Solution

R

W R

W X

b W 250 K= 6600

c

W and

394

Basic Electrical Engineering I =

\

R

and (a

f

40000 250

=K R +R

¥

X

¥

=K X

W W

f

I 2Re 2 cos f + I 2 X e 2 sin f ¥ V2(0)

\ = f

(b

160 × 0.0343 × 1 + 0 ¥ 250

f

f=

2.195 %

1 − cos2 φ

I2Re2 cos f + I 2 X e2 sin f ¥ V2(0)

\ = f

(c

160 × 0.0343 × 0.8 + 160 × 0.0502 × 0.6 ¥ 250 f f

3.68 %

I 2Re2 cos f - I 2 X e2 sin f ¥ V2(0)

\ =

h=

160 × 0.0343 × 0.8 − 160 × 0.0502 × 0.6 ¥ 250

Power output Po Power output = = Power output + Power losses Po + P1 Power input

h

Major Losses or

2

P =I R +I R =I R =I R The copper losses are variable Po = VI

f = VI ¥ pf

fi

VI =

Po pf

0.172 %

395

2

Ê VA ˆ P = Á P Ë VA FL ˜¯

or Fm in a normal transformer the core losses

Pi = Ph + Pe constant at all loads

h=

Po Po V2I 2 cos f2 = = Po + P1 Po + Pc + Pi V2I 2 cos f2 + I 22Re2 + Pi

(i) *

(ii)

(iii)

I I V2 cos f2 h= V2 cos f2 + I 2Re 2 + Pi /I 2

d (V dI

f +I R

+ Pi /I I R = Pi or

or

Pi I22 P = Pi

R

I *

. case of a distribution transformer. The primary of a distribution transformer remains energised all the time. But the load on the secondary is intermittent and variable during the day. It means that the core losses occur therefore, are designed to have minimum core losses. This gives them better hall-day =

EXA M P L E

Output energy (in kW h) in a cycle of 24 hours Total input energy (in kW h))

13 . 12 a b c

d

Solution a E2 = 4.44f N2Fm fi and

250 E2 = 4.44 50 0.06 4.44 f Fm E 5000 N1 = 1 N2 = ¥ 19 = 380 turns E2 250 N2 =

19 turns

b Output power, Po = 0.5 ¥ kVA ¥ Copper losses, Pc

\

2

¥ 150 ¥ 1 = 75 kW 2

¥

¥ 1800 W = 0.45 kW

Iron losses, Pi = 1500 W = 1.5 kW 75 Po h= ¥ 100 = ¥ 100 = 97.47% Po + Pc + Pi 75 + 0.45 + 1.5

c

\ d

Output power, Po kVA ¥ ¥ 0.8 = 120 kW Copper losses, Pc = 1800 W = 1.8 kW Iron losses, Pi = 1500 W = 1.5 kW Po 120 h= ¥ 100 = ¥ 100 = 97.3% Po + Pc + Pi 120 + 1.8 + 1.5 x be the fraction of full-load kVA Pc = Pi

or

x2 ¥ 1800 = 1500 fi x =

1500 /1800 = 0.913

Therefore, the load kVA kVA

kVA ¥ x = 150 ¥ 0.913 = 137 kVA

397 EX A MP L E

13. 1 3

(i (ii (iii

Solution (a ¥ Po 80 kVA = = pf 1

Ê kVA ˆ P = Á Ë kVAFL ˜¯ Pi Pl = P + Pi ¥

\

2

P

=

2 ⎛ 80 ⎞ ¥ ⎝ 200 ⎠

(b ¥ Po 160 kVA = = pf 0.8 P =P Pi P = P + Pi ¥

\

\

FL

(c P Pi Pl = P + Pi ¥

\

Wo Wl \

h

=

Wo ¥ Wo + Wl

a

1760 ¥ 1760 + 65.46

96.41 %

Ii

I1

+

I2

+ V1 Vi

Io

Ii +

+ N1

N2

–

+

+

V2 –

V1 Vo

Vi

–

–

(a) Special connection of a transformer.

Io

– N2

–

N1

+ V2

+ Vo

–

–

(b) Redrawn in the standard way.

secondary drawn in the usual position. Figure 13.20b

the new secondary. Also, the primary and secondary are not electrically isolated from each other. Obviously, the voltage V2 = Vo. From Fig. 13.20b, it is obvious that N1 N1 + N 2 Vi = V1 + V2 = V2 +V2 = Vo N2 N2 N2 Vi or Vo = N1 + N 2 Hence, the new turns-ratio becomes N2 N1 + N2 potential divider

EXA M P L E

13 . 14

A single-phase, 12-kVA, 120-V/120-V transformer is connected as an autotransformer to make a 240-V/120-V transformer. What is the apparent power rating of the autotransformer ?

100 A +

+ 120 V

Solution Figure 13.21 shows the transformer connection with rated

200 A – voltage and current. The current rating on both primary and secondary 240 V + + 100 A windings is 12 kVA 120 V 120 V I1 = I2 = = 100 A 120 V – – – In autotransformer mode, the input apparent power is 240 ¥ 100 = 24 kVA, and the output apparent power is 120 ¥ 200 = 24 kVA. Thus, the apparent Fig. 13.21 power capacity of the 12-kVA transformer is doubled by the autotransformer connection. In effect, half the apparent power is transformed and half is conducted directly to the secondary side.

399

N a

V

E is N E /E = N /N = K

emf E K=

N2 V2 = N1 V1 I

I

E

I

common winding

series winding

variacs (variable autotransformers

I1

I2 X

+

X I1

Y

V1 N1

I2

I1 +

N2 – Z

+

I2 Y

N2

+

(I2 – I1) V2 – I2

–

– I1

(a) A step-down autotransformer.

Z

I2

(b) A step-up autotransformer.

(a)

a N V I

I V I = V I + V (I V I V (I

I

Load

N1

V1 (I1 – I2)

Load

V2

I

proportional to the current carried by it, and its length is proportional to the number of turns. Therefore, Weight of copper in a winding μ NI = kNI For a two-winding transformer: Weight of copper in primary = kN1I1 Weight of copper in secondary = kN2I2 Total weight of copper = k N1I1 + N2I2 For an autotransformer a The portion XY of the winding has N1 – N2 turns and carries current I1. The portion YZ of the winding has N2 turns and carries current I2 – I1. Therefore, Weight of copper in portion XY = k N1 – N2 I1 Weight of copper in portion YZ = kN2 I2 – I1 Total weight of copper = k N1 – N2 I1 + kN2 I2 – I1

k N1 – 2N2 I1 + N2I2]

Therefore, the ratio of copper-weights for the two cases is ⎡ ⎛ N 2 ⎞ ⎤ ⎛ I1 ⎞ + ⎛ N 2 ⎞ ⎟⎥ ⎜ ⎟ ⎜ ⎟ ⎢1 − 2 ⎜ ⎝ N1 ⎠ ⎦ ⎝ I2 ⎠ ⎝ N1 ⎠ k[( N1 - 2N 2 ) I1 + N2 I2 ] [1 - 2K ] K + K = ⎣ = =1–K K+K k ( N1I1 + N 2 I 2 ) ⎛ I1 ⎞ + ⎛ N 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ I2 ⎠ ⎝ N1 ⎠ Evidently, the saving is large if K is close to unity. A unity transformation ratio means that no copper is needed at all for the autotransformer. The winding can be removed all together. The volt-amperes are conductively transformed directly to the load ! The use of autotransformer has following disadvantages : Y and Z, full primary high voltage appears across the load. 3. The short-circuit current is larger than that in two-winding transformer. Autotransformers are used in following areas : 1. Boosting or buckling of supply voltage by a small amount. 3. Continuously varying ac supply as in variacs.

Modern large transformers are usually of three-phase core type, schematically shown in Fig. 13.23. Three similar limbs are connected by top and bottom yokes. Each limb has primary and secondary windings arranged

401

EX A MP L E

13. 1 5

a (b

Solution (a V V \

V

3300 VL1 = 3 3 72 ¥ 840 = V = 163.3 V =

(b V

=V

V \

A current transformer (

V

)

=V

¥

72 840

¥

3

¥

3 = 490 V

and protective relaying, where they facilitate the safe measurement of large currents. The current transformer isolates the measurement and control circuitry form the high voltages typically present on the circuit being measured.

voltage transformer (VT), also referred to as potential transformer protection in high-voltage circuits. It is designed to present negligible load to the supply being measured and to have a precise voltage ratio to accurately step down high voltages so that metering and protective relay

leakage transformer, also called a inductance than the power transformers. The leakage inductance is increased by providing loose coupling between the primary and secondary windings, or sometimes by magnetic bypass or shunt in its core. The output and input currents are low enough to prevent thermal overload under all load conditions—even if the secondary is shorted.

installations.

resonant transformer is a kind of leakage transformer. It uses the leakage inductance of its secondary application of the resonant transformer is to couple between stages of a super heterodyne receiver, where the

compared with the full-load output of the transformer.

transformer is connected as shown in Fig. 13.24.

variac left open. The ratio of the voltmeter readings, V2 /V1, gives the transformation ratio of the transformer. The reading of ammeter A, Io, gives the no-load current I0,

403 I R loss I R Wo Pi = Wo

I =I

I =

Wo V1

Im =

I 02

I w2

R =

V1 Iw

X =

V1 Im

low

of ammeter A I W P

Rel = EX A MP L E

Wsc I sc2

Zel =

Vsc Isc

X =

Z e21

Re21

13. 1 6

(i (ii a

b

Solution (a given as I =

Im =

Wo 120 W = = 0.6 A 200 V V1 I 02

I w2 =

(1.3)2 (0.6)2 = 1.15 A

(b R =

V1 200 V = = 333 W 0.6 A Iw

and

X =

V1 200 V = = 174 W 115 . A Im

404

Basic Electrical Engineering

K= IFL =

\

and

Wsc

R

=

X

=

R

=K R

X

=K X

2 Isc

=

200 W

12 kVA 400 V

W

(30 A )2

Z e21 Re21 =

V2 200 V 1 = = V1 400 V 2

Z

and

=

W

(0.733)2 (0.222)2

(2) 2 =(1) ¥ 2

2 = 1 ¥

Vsc 22 V = Isc 30 A

0.055 W 0.175 W

ADDITIONAL SOLVED EXAMPLES EX A MP L E

13 .17

a

b

c

Solution (a K= \

N2 40 = N1 500

E = KE ª KV

¥

240 V

(b I =

kVA 25 kVA = = 8.33 A 3 kV V

and

I =

I1 8.33 A = = 104.125 A K 0.08

(c Fm = EX A MP L E

13 .18

E1 3000 = = 0.027 Wb 4.44 f N1 4.44 50 500

W

Transformers

Solution Fm =

230 E1 = 4.44 50 4.44 f N1

50

= 0.02072 Wb

0.02072 m 2 2 = = 0.02072 m = 207.2 cm Bm 1 Due to the insulation of laminations from each other, the gross area is about 10 % greater than the active area. Thus, \

Active core area, A =

2

Gross area = 207.2 ¥ 1.1 = 227.92 cm a= EXA M P L E

227.92 = 15.09 ª 15 cm

13 . 19

A single-phase transformer, having a core of cross-sectional area 150 cm2 1.1 Wb/m2 from a 50-Hz supply. If the secondary winding has 66 turns, determine the output in kVA when connected to a load of 4-W

Solution Fm = BmA = 1.1 ¥ 0.015 = 0.0165 Wb E2 = 4.44Fm f N2 = 4.44 ¥ 0.0165 ¥ 50 ¥ 66 = 242 V = V2 V2 242 = = 60.5 A 4 ZL 242 60.5 V I Output kVA = 2 2 = = 14.6 kVA 1000 1000 I2 =

\ EXA M P L E

13 . 20

A 11-kV/400-V distribution transformer takes a no-load primary current of 1 A at a power factor of 0.24 lagging. Find a b c

Solution Iw = I0 cos f0 = 1.0 ¥ 0.24 = 0.24 A

a b

Im =

I02

Iw2 =

(1)2 (0.24)2 = 0.971 A

Pi = V1I0 cos f0 = 11000 ¥ 1.0 ¥ 0.24 = 2640 W

c EXA M P L E

13 . 21

N2/N1 are 2.5 W and 6 W, whereas the secondary winding resistance and reactance are 0.25 W and 1 W, respectively. Its magnetising current and core-loss current are 51.5 mA and 20.6 mA, respectively. While in operation, the output voltage for a load of 25–30° W is found to be 50 V. Determine the supply voltage, the current drawn from the supply and the power factor. j W = 6.5–67.38° W; Z2 V2 as the reference phasor, i.e., V2 = 50–0° V.

Solution Given : Z1

j

W = 1.03–75.96° W

406

Basic Electrical Engineering V2 50– 0∞ V = – ZL 25–30∞ W E =V +I Z j

\

I =

⎛N ⎞ E =E ⎜ 1⎟ ⎝ N2 ⎠

–

⎛N ⎞ I ⎜ 2⎟ ⎝ N1 ⎠

I¢

– ¥

–

¥ b Im

\ \

Im – I = I¢ + Im + I – = 1.044556–148.17° A V E +I Z – = 108.601–183.647° V V and I pf =

\

EX A MP L E W

–

–

–

–

E

– E

I

– –

E –

I – –

–

0.814 lagging

13. 2 2

W

W

W a

b

Solution

c

I

K R

= R + R /K

W

and

X

= X + X /K

W

\

Z

=

Re21 + X e21 =

W

(2.36)2 + (6.14)2

(a 24 Vsc = Ze1 6.578 I = I¢ I = I ¢/K 14.592 A I R ¥ 31.4 W 31.4 P f = = = 0.3586 VI 24 3.648 I =

I (b (c EX A MP L E

13. 2 3

OC test SC test a

b

Transformers

Solution a

V Ze1 =

= 10.5 A, P

= 360 W. Then,

Psc 360 W Vsc 86 V = = 8.19 W; Re1 = 2 = = 3.265 W Isc (10.5 A )2 Isc 10.5 A (8.19)2 (3.265)2 = 7.51 W

Z e21 Re21 =

Xe1 =

VA 20 000 = = 9.09 A V1 2200 f = 0.8, and sin f = sin [cos–1 0.8] = 0.6

The full-load primary current, I1=

% Regulation =

= 86 V, I

I1 ( Re1 cos φ + Xe1 sin φ ) ¥ 100 V1

9.09 (3.265 × 0.8 + 7.51 × 0.6) ¥ 100 = 2.94 % 2200 The short-circuit test has been conducted for a short-circuit primary current of 10.5 A, whereas full-load primary current is only 9.09 A. Therefore, the full-load copper loss is given as =

2

Full-load copper loss =

⎛ 9.09 ⎞ ¥ 360 = 269.58 W. ⎝ 10.5 ⎠

The open-circuit test gives the core loss. Hence, Pi = 148 W. The full-load output power, Po = VA ¥ pf = 20 000 ¥ 0.8 = 16 000 W Po 16 000 \ h= ¥ 100 = ¥ 100 = 97.45 % Po + Pc + Pi 16 000 + 269.8 + 148 b pf = cos fsc = EXA M P L E

Re1 3.265 = = 0.399 (lagging) Z e1 8.19

13 . 24

Solution At a power factor of 0.8, the full-load output power, kVA ¥ pf

Po

Pin =

Po

¥ 0.8 = 160 kW 160 kW = = 163.26 kW 0.98

Pc + Pi = Pin – Po

i

we have 2

i

ii

⎛ 3⎞ P = Pi ⎝ 4⎠ c Pc = 2.09 kW

ii and

Pi = 1.17 kW

2 2 ⎛ 1⎞ ⎛ 1⎞ P + Pi = ¥ 2.09 + 1.17 = 1.69 kW ⎝ 2⎠ c ⎝ 2⎠ Po (160 / 2) hhalf-load = ¥ 100 = = 97.93 % Po + P1 (160 / 2) + 1.69

Total losses, P1 = \ EXA M P L E

1 3.2 5

A single-phase, 150-kVA, 5000-V/250-V, 50-Hz transformer has the full-load copper losses of 1.8 kW and core losses a b full rated kVA c kVA, with unity power factor, and d kVA

Solution a

\ b

E = 4.44fNFm, the number of turns on the secondary winding, E2 250 N2 = = = 19 turns 4.44 f Fm 4.44 50 0.06 E 5000 N1 = N2 1 =19 ¥ = 380 turns E2 250 kVA, the current is also full load. Therefore,

\ c

Power output Po = V2I2 ¥ cos f ¥ 0.8 = 120 kW Copper losses, Pc = 1.8 kW and the iron losses, Pi = 1.5 kW Po 120 h= ¥ 100 = ¥ 100 = 97.32 % Po + Pc + Pi 120 + 1.8 + 1.5 kVA, the current is half the full-load current. Thus,

Power output, Po = 0.5 ¥ V2I2 ¥ cos f = 0.5 ¥ ¥ 1 = 75 kW Copper losses, Pc I2 2Re2 = 0.25 I 22 Re2 = 0.25 ¥ Iron losses, Pi Po 75 \ h= ¥ 100 = ¥ 100 = 97.47 % Po + Pc + Pi 75 + 0.45 + 1.5 d x be the fraction of the full-load kVA x2

Pi x ¥ 1.8 = 1.5 2

or kVA EXA M P L E

fi

x = 0.913 ¥ 0.913 = 137 kVA

1 3.2 6

A single-phase, 50-kVA, 2400-V/240-V, 50-Hz transformer is used to step down the voltage of a distribution system. a

409 (b (c

Solution (a I I = ZL = K=

(b

50 kVA 240 V

V2 240 V = = 1.152 W I2 208.33 A

V2 240 = V1 2400 Z

= 115.2 W

= ZL/K

(c I ¢ = KI EX A MP L E

¥

20.833 A

13. 2 7

43.48 A X 43.48 A

230 V

Y 47.83 A 2530 V

a b

kVA

4.35 A

2300 V

c

d 43.48 A

Solution 10 000 2300 10 000 230 I

(a I (b

kVA

2300 47.83 = 110 kVA 1000 VI V (I

(c

(d

K=

N2 V 2300 = 2 = N1 V1 2530

¥ ¥

I 90.9%

100 kVA 10 kVA

Z

47.83 A

Load

410

Basic Electrical Engineering

SUMMARY TE RM S

A N D

C ON CE PT S primary and secondary.

mutual induction ideal

i

ii

iii

(iv exciting current or no-load current I I

i

Im

R X

IR iron loss

Losses

copper loss

R and R X and X I primary balancing current or load component of primary current I ¢ autotransformer IMP O R TA N T

F O RM U LAE Fm f N V2 E N K= = 2 = 2 V1 E1 N1 E

I =

I2 N 1 = 1 = and Z I1 N2 K

= Z /K

2 Im + Iw2

n Ph = KhB m fV

Pe = KeBm f t V R = R + (R /K R regulation down = regulation up =

=K R +R

X and

X

= X + (X /K =K X +X

V2( 0 ) V2 V2( 0 )

V2( 0 ) V2 V2 IR

f±I X R f = e2 Xe2

sin f

pf pf

ii

411 Xe 2 Re2 Power output Po h= = Power output + Power loss Po + P1 f=

Pi = Wo R

=

Wsc 2 Isc

W I = Io I = o Im = I02 Iw2 V1 Vsc Z = X = Z e21 Re21 Isc

V R = 1 Iw

P = Pi V X = 1 Im

CHECK YOUR UNDERSTANDING two one

minus

12

Fm

Im

P P

Your Score A N SW E R S

1. False 6. False

2. 7.

3. 8.

4. False 9. False

5. 10. False

412

Basic Electrical Engineering

REVIEW QUESTIONS 1.

14.

2. 3.

15. (a b 16. a

4.

c

(b 5. 17.

6. 7.

18. in a transformer only if the load has a leading

8.

19.

9.

20.

10.

21.

11. 22. (a

b

c 23.

12. 24. 13. 25.

MULTIPLE CHOICE QUESTIONS are provided below each. Tick the alternative that completes the statement correctly. 1. laminated sheets so as to

(a (b (c losses (d

Transformers 2. A close magnetic coupling between primary and secondary windings results in a b c d 3. The number of turns in the primary winding of a transformer depends on a b c d 4. a b c d 5. The emf induced in the windings of a transformer a b c p/2 d p/2 6. A single-phase, 150-kVA, 1100-V/400-V transformer has 100 turns on the secondary winding. The number of turns on its primary winding will be a b c d 7. A single-phase, 5-kVA, 200-V/100-V transformer delivers 50 A at the rated voltage. The input current will be a b c d 8. In a single-phase, 10-kVA, 230-V/1000-V transformer, the no-load current will be about a b c d 9. A single-phase transformer has a turns-ratio of 4 : 1. If the secondary winding has a resistance of 1 ohm, this resistance as referred to the primary will be a W b W c W d W 10. A transformer is supplying a unity power-factor load. The power factor at the primary terminals will be

413

a b c d 11. When the secondary winding of a transformer is short-circuited, the power factor of the input is a b c d 12. Under no-load condition, the power factor of a transformer is a b c d 13. If the full-load copper loss of a transformer is 100 W, its copper loss at half load will be a b c d 14. If the full-load core loss of a transformer is 100 W, its core loss at half load will be a W b W c W d W 15. A single-phase transformer is supplying power to a load at a terminal voltage of 11 kV. When the load is disconnected, the terminal voltage becomes 11.5 kV. The voltage regulation of this transformer for this load is a b c d 16. when a minimum b c d loss 17. A distribution transformer should be selected on the basis of its a b c d

414 18. In a transformer, the iron losses do not vary with load current because a b c d 19. It is economical to use an autotransformer when the turns-ratio is

a b c d 20. An auto-transformer steps down voltage V1 to V2. If an open circuit develops in the common winding, the voltage across the load may become a V1 b V1 – V2 c V1 + V2 d V2 A N SW E R S

1. b 11. d

2. c 12. c

3. a 13. d

4. d 14. b

5. c 15. c

6. d 16. b

7. b 17. a

8. b 18. a

9. a 19. a

10. a 20. a

PROBLEMS ( A )

S IM P L E

PR OB LEM S

1. A single-phase, 250-kVA, 11-kV/415-V, 50-Hz transformer has 80 turns on the secondary. Calculate a b c [Ans. a b c 2. The primary winding of a 50-Hz transformer has 480 turns and is fed from a 6400-V supply. a b winding has 20 turns [Ans. a b 3. A single-phase, 50-Hz transformer has 80 turns on the primary winding and 400 turns on the secondary winding. The net cross-sectional area of the core is 200 cm2. If the primary winding is a b [Ans. a b 4. A 10-kVA, single-phase transformer has its primary connected to a 2000-V supply. It has 60 turns on the secondary winding and the voltage across it is found to be 240 V. Assuming the transformer a

b secondary currents. [Ans. a b 5. A 200-kVA, 3300-V/240-V, 50-Hz, single-phase transformer has 80 turns on the secondary winding. a b c primary turns. [Ans. a b c 6. A single-phase transformer with 10 : 1 turns-ratio and rated at 50 kVA, 2400-V/240-V, 50 Hz is used to step down the voltage of a distribution system. at 240 V. Find the value of the load impedance of

[Ans. 1.152 W, 0.047 Wb] 7. The primary of a single-phase transformer takes 1 A at power factor of 0.4 when connected to a 240-V, 50-Hz supply and the secondary is on open circuit. The number of turns on the primary is twice that on the secondary. A load taking 50 A at a lagging power factor of 0.8 is now connected

Transformers across the secondary. What is now the value of the Ans. 25.9 A] 8. A single-phase transformer has 100 turns on the primary winding and 400 turns on the secondary winding. The net cross-sectional area of the core is 2 250 cm . If the primary winding is connected to a a b [Ans. a b 9. The no-load current of a single-phase transformer is 5.0 A at 0.3 power factor when supplied from a 240-V, 50-Hz source. The number of turns on the a b c the magnetising current. [Ans. a b c 10. A transformer on no load takes 1.5 A at a power factor of 0.2 lagging when its primary is connected to a 50-Hz, 230-V supply. Its transformation ratio N2/N1 the secondary is supplying a current of 40 A at a drops in the windings. [Ans. 14.5 A] 11. The no-load current of a 50-Hz, 230-V transformer is 4.5 A at a power factor of 0.25 lagging. The number of turns on the primary winding is 250. a b c core. [Ans. a b c ( B )

T R IC KY

415

12. A 100-kVA transformer has 400 turns on the primary and 80 turns on the secondary winding. The primary and secondary resistances are 0.3 W and 0.1 W, respectively. The primary and secondary leakage reactances are 1.1 W and 0.035 W, respecto the primary side. [Ans. 3.426 W] 13. A single-phase, 100-kVA, 1100-V/220-V transformer has following parameters : R1 = 0.1 W, X1 = 0.3 W, R2 = 0.004 W and X2 = 0.012 W. Detera actance as referred to the high voltage winding, and b as referred to the low voltage winding. [Ans. a W, 0.6 W b W, 0.024 W] 14. In a 50-kVA, 11-kV/400-V, single-phase transformer, the iron and copper losses are 500 W and 600 W, respectively under rated conditions. Calcua b c the iron and copper losses for this load. [Ans. a b c 15. A 18-kVA, 600-V/300-V, 50-Hz, single-phase transformer gave the following test-results :

a

b

referred to the low voltage side. [Ans. a b

W, 0.2621 W]

PRO BLE MS

16. A single-phase, 50-Hz transformer has 30 turns on primary and 350 turns on secondary. The net cross-sectional area of the core is 250 cm2. If the primary winding is connected to a 230-V, 50-Hz a b c [Ans. a b c 17. A single-phase, 50-Hz, 100-kVA, 2400-V/240-V transformer has no-load current of 0.64 A and core

the two components of the no-load current. If this transformer supplies a load current of 40 A at 0.8 lagging pf on its low voltage side, determine the primary current and its power factor. Ignore the resistance and leakage reactance drops. [Ans. 0.5697 A, 0.2917 A; 4.584 A, 0.762 lagging] 18. An ideal 50-Hz, core-type transformer has 100 primary-turns and 200 secondary-turns. The primary a

416 b reference to the secondary voltage vector when secondary delivers a current of 8 A at a lagging power factor of 0.8. [Ans. 8.26 ¥ 10–3 m2 a b –143.1° A] 19. A 500-kVA, 11000-V/400-V, 50-Hz, singlephase transformer has 100 turns on the secondary a b values of the primary and secondary current, and c [Ans. a b c 20. A single-phase, step-up transformer has a turnsof 7.5 mWb at 50 Hz. When working on no load, it takes 0.24 kVA at a power factor of 0.26 lagging from the supply. If it supplies a load of 1.2 kVA a b c [Ans. a b c 21. The primary and secondary windings of a 30-kVA, 6000-V/230-V, single-phase transformer have

( C )

C HA L L E NGI NG

resistances of 10 W and 0.0016 W, respectively. The total reactance of the transformer as referred to the primary side is 23 W. Calculate the percentage regulation of the transformer when supplying fullload current at a power factor of 0.8 lagging. [Ans. 1.85 %] 22. A step-down transformer with turns-ratio 8 : 1, has the resistances of the primary and secondary windings as 0.85 W and 0.012 W, respectively, and the leakage reactances as 4.8 W and 0.07 W, respectively. Determine the voltage to be applied to the primary to obtain a current of 150 A in the secondary when the secondary terminals are short-circuited. Ignore the magnetising current. [Ans. 176.6 V] 23. A single-phase, 50-Hz step-down transformer has a turns ratio 6 : 1. The primary and secondary winding resistances are 0.90 W and 0.03 W, respectively, and leakage reactances are 5 W and 0.13 W, a to the high voltage side to get a current of 100 A in the low voltage winding on short circuit, and b [Ans. a

b

PROBLEMS

24. A single-phase transformer has Z1 j W and Z2 j W. The input voltage is 6600 V and turns-ratio is 10.6 : 1. The secondary feeds a load which draws 300 A at 0.8 power factor I0, determine the secondary voltage and the output in kW. [Ans. 25. A single-phase, 10-kVA, 4000-V/400-V transformer has following parameters : R1 = 13 W, X1 = 20 W, R2 = 0.15 W, X2 = 0.25 W, R0 = 12000 W and X0 = 6000 W a sistance and leakage reactance as referred to b c input current when the secondary supplies a load current of 25 A at a power of 0.8 lagging. [Ans. a W, 45 W b ––63.5° A; c ––42.9° A] 26. A single-phase, 200-V/2000-V transformer is fed

resistance and leakage reactance as referred to the low voltage side are 0.16 W and 0.7 W, respectively. The resistance representing core losses is 400 W and the magnetising reactance is 231 W. A load impedance of Z j W is connected a b c [Ans. a – b c 27. A single-phase, 100-kVA, 2000-V/200-V, 50-Hz transformer has impedance drop of 10 % and a b power factor is the regulation zero? [Ans. a b 28. The primary and secondary windings of a 500-kVA, 11-kV/415-V, single-phase transformer have resistances of 0.42 W and 0.0019 W, respectively. Its core losses are 2.9 kW. Assuming

417 a b Ans. (a

on (a

b

b

29. b

Ans. (a 31. Ans. 30. Ans.

EXPERIMENTAL EXERCISE 13.1 L OA D

T E S T

ON

A

SIN G LE-PHASE

TRANSFORMER

Objectives 1. 2. 3. 4.

Apparatus

Circuit Diagrams Brief Theory alternately relative polarities

(i

and S

a and S

a a

and S V =V

V

and S b

the voltage V = V + V (ii

(iii

Pin

K=

N2 E = 2 N1 E1

V2 V1

h=

Po Po = Pin Po + P1

Po

418

Basic Electrical Engineering

3

Pl

a

419

h

V2

max

V2(0) V2(FL)

O

O

I2 (a) Efficiency versus load-current.

I2(FL)

I2

(b) Output voltage versus load-current.

Procedure (i) 1. 2. ON 3. 4. If V < V 5. If V > V 6. 7. 8. 9. OFF

a

and S and S V

V

V /V

(ii) 1. 2. 3. 4. 5. 6. 7.

b ON

V Pi I

ON

and ammeter A 8. 9. 10.

OFF

(i)

If V < V

= = = and S

V >V

and S

(ii) S. No.

V1

V2

Turns-Ratio = V2 /V1

1 2 3 (iii) S. No. 1. 2. 3. 4. 5. 6.

W

V2

I2

Output = V2 I 2

= V2 I 2/W

a ( ) V V \

= =

% Regulation =

V2( 0 ) V2( FL ) V2( FL )

¥ 100% =

1. The polarities of the primary and secondary windings have been marked. 2. 3. ________ amperes.

1. Before switching on the supply, the zero reading of the wattmeter, voltmeters and ammeters should be checked. 2. Meters of proper range should be selected. 3.

Viva Voce 1. Can you name some applications of transformers ? Ans. : To step-up generated voltage before transmission and to step-down at consumer end. Current and voltage circuits, e.g., radio, TV, etc. 2. Can you name the material normally used for making the core of transformers ? Ans. : 3. What is the purpose of adding silicon to steel ? Ans. : 4. Any harm if you add more than 4 % silicon ? Ans. : By adding more silicon, the resistance increases further. But, the steel become very brittle.

421 5. Ans.: 6. Ans. : 7. Ans. : 8. Ans. : 9. Ans. : 10.

Ans. :

EXPERIMENTAL EXERCISE 13.2 O P E N - C I R C U I T A N D S H O R T- C I R C U I T T E S T S O N A T R A N S F O R M E R

Objectives I

1. 2.

Im R and X R

X

Z of the

Apparatus

Circuit Diagrams Brief Theory a

(i) Wo

Vo

Io Pi = Wo

I

Im R and X I =

Wo Vo

Im =

I02

Iw2

R =

Vo Iw

X =

Vo Im

(ii) W

422

Basic Electrical Engineering

Fig. 13.30 (Vsc) and the ammeter (Isc) are noted. Since the applied voltage is low, the iron losses are negligibly small compared to the copper loss. Therefore, the wattmeter indicates the full-load copper loss, i.e., Copper loss, Pc = Wsc The equivalent resistance Req, the equivalent reactance Xeq and the equivalent impedance Zeq, as referred to the winding on which the measurements are made, can be calculated as follows: W V Re1 = 2sc ; Ze1 = sc ; and Xe1 = Z e21 Re21 Isc Isc

(i) 1. 2. 3. 4. 5. 6.

Connect the circuit as shown in Fig. 13.30a. Put the variac at low voltage output. Switch ON the ac supply. Adjust the variac to the rated voltage of the transformer. Record the ammeter, wattmeter and voltmeter readings. Switch OFF the supply.

(ii) 1. Connect the circuit as shown in Fig. 13.30b.

423 2. 3. 4. 5.

OFF

Observations (i) Io = Wo = Vo = (ii) I = W = V =

(i) I =

Wo Vo

R = (ii) R

=

X

=

I02

Im =

Vo Iw Wsc

W

X =

Vo Im

Z

=

W

2 Isc

Z e21 Re21 =

Iw2 =

( )2 ( )2 W

Vsc Isc

W W

( )2 ( )2 =

Results 1. R =

X = Im

2. 3.

I

Precautions 1. 2. 3.

Viva Voce 1. Ans. :

2.

W

W

R

=

W

X I

=

W

Z

=

W

424 Ans. : in turn depends upon the voltage applied. In open-circuit test, full rated voltage is applied to the transformer. Hence, the iron losses occur at full rated value. Furthermore, as the secondary winding is open, the secondary current is zero. Hence there is no copper loss in the secondary winding. However, in the primary winding, there the wattmeter reading gives the iron losses. 3. What does the reading of wattmeter indicate in the short-circuit test ? Justify your answer. Ans. : It indicates the full-load copper losses of the transformer. In short-circuit test, full rated current is made of low voltage the magnetisation level of the core is very low. Hence the iron losses are negligibly small. Thus, the wattmeter indicates only the full-rated copper losses of the transformer. 4. Ans. : of the insulation. 5. How do the iron losses vary with the load on the transformer ? Ans. : Iron losses do not depend on the load. These losses depend only on the magnetisation level of the core, which in turn depend upon the voltage applied. 6. How do the copper losses vary with the load on the transformer ? Ans. : Copper losses or I2R 7. What is the phasor relationship between Iw, Im and I0 ? 2 . The iron-loss component I is in phase Ans. : The current I0 is the phasor sum of Iw and Im. I0 = Iw2 + Im w with the applied voltage. The magnetising component Im lags behind the applied voltage by 90°. 8. What is the magnitude of no-load current as compared to the full-load current ? Ans. : The no-load current is about 3–5 % of the full-load current. 9. Of what order is the power factor of a transformer under no-load condition ? Ans. : It is about 0.2 lagging.

ALTERNATORS AND SYNCHRONOUS MOTORS

14

OB J EC T I VE S : (EMEC)

(Xs)

14.1

(Zs)

ELECTROMECHANICAL ENERGY-CONVERSION MACHINES

In the previous chapters, we saw how electrical power is transmitted by means of three-phase circuits and how transformers are used to raise and to lower the system voltages. Next, we should consider to what purpose the electrical power is used at the consumer’s end.

426 A large amount of electrical power is used to provide heat during winters, either for heating a room or for heating water in a geyser. During summers, we use electrical power to provide cooling, as in air conditioners or refrigerators. Some heat conversion is used to provide light as in the common light bulb. But a lot of electrical power is used to drive machines in industry and homes. Useful gadgets like fans, tape recorders, mixies, etc., all run with the help of a motor electrical power mains and convert it into mechanical energy which is used to turn the blades of the fan or mixie. Reverse conversion takes place in a machine called generator. You have a generator (also, called dynamo) in your car. While the car is running, the dynamo converts mechanical energy (drawn from the engine) into electrical energy which keeps on charging the battery. Motors and generators are electromechanical energy-conversion (EMEC) devices or machines *. The process of electromechanical energy conversion is essentially reversible (except for a small amount which is lost as heat energy). In both the motor and the generator, it is the that couples the electrical between the two systems. The energy to be transferred from one system to the other is temporarily stored in d then released in the other system.

Fig. 14.1 The interchange of energy between the electrical system and mechanical system takes place through the induced emf (e) and current (i electromagnetic torque (te) and angular speed (w), respectively. When the electrical system is characterised by direct current, the machines are called dc motors and dc generators. Similarly, if the electrical system is characterised by alternating current, the machines are called ac motors and ac generators. Basically, the ac machines are not different from the dc machines. They differ merely in the constructional details. Only for our convenience, we discuss different type of machines in different chapters.

*

Not all machines deal with large amount of energy. Some devices operate at very low power levels and are termed transducers (such as a microphone or a loudspeaker), which deal with ‘signals’ to operate electronic circuits.

Alternators and Synchronous Motors

Essentially, an electrical rotary m rotor.

427

stator e i

ei

is the generator action. te w

tew

motor action.

tew into electrical power ei. ei = t ew

Power Considerations for a Generator *

tm w RL

e is i

Fig. 14.2 t e. This is the reaction torque t te t

tm tm

tm tm = t e + t

w t mw = tew + t w *

alternators.

428 tmw supplied by the prime mover, only the power tew is converted by the generator into electrical power ei. The remaining power tf w is wasted as frictional loss. Now, let us see what happens at the output side. If r is the resistance of the coil, the current i in the circuit is given as e i= fi e = ir + iRL r + RL Multiplying both sides by i, we get ei = i2r + i 2RL or

ei = i 2r + iV

(14.3)

where, V = iRL ei generated by the generator, only the power Vi is delivered to the electrical load RL and the remaining power i2r is in Fig. 14.3.

Fig. 14.3

Power Considerations for a Motor The same rotary machine can work as a motor. An electrical source of voltage V is connected to the terminals AB of the armature coil (Fig. 14.4). Current i te is developed. This causes the coil (and hence the rotor) to rotate at a speed w. As a result, the mechanical load (say, a water pump) connected to the shaft of the motor also rotates at a speed w.

Fig. 14.4

Alternators and Synchronous Motors

429

As the conductors of the c e is induced. This is the reaction emf and it opposes the applied voltage V. The difference of the applied voltage V and the induced emf e is responsible for causing the current i V e i= fi V = e + ir (14.4) r where, r is the resistance of the coil. Note that in a motor the direction of the induced emf e is opposite to that of the current i; whereas in a generator, the emf e and the current i are in the same direction. Multiplying both i, we get Vi = ei + i2r

(14.5)

Th Vi supplied by the electrical source, only the power ei is made available for conversion to mechanical power, the remaining power i2r is wasted as copper loss in the coil. Now, next let us see what happens at the output side. The motor converts electrical power ei into mechanical power tew tew developed by the motor, a portion tf w is lost in overcoming the friction; the remaining part tmw is supplied to the mechanical load. That is, t ew = tf w + tmw

(14.6)

Thu

Fig. 14.5

How a Generator Differs from a Motor With reference to Figs 14.2 and 14.4, we can bring out the essential differences between a generator and a motor as follows : 1. The directions of induced emf e and current i are same in a generator but opposite in a motor*. t e and rotation w of shaft are opposite in a generator but same in a motor. te t f are same in a generator but opposite in a motor. *

In fact, we can conclude that a circuit having e and i in the same direction acts as a source of electrical energy. On the other hand, a circuit with e and i in the opposite direction acts as a consumer of electrical energy.

430

How Mechanical Force is Developed in a Machine A mechanical energy system must have a mechanical force associated with a displacement of its point of action. There are two ways in which a mechanical force can be developed in an electromagnetic system : (i) by alignment, and (ii) by interaction.

attracted towards one another. The force of alignment acts in a direction that will increase the magnetic energy stored in the arrangement. In this case, the force will try to bring the poles together thereby decreasing the

N

N

Force

Force

Force (lateral) Poles

S (a) Force of attraction.

S (b) Lateral force of alignment.

Fig. 14.6 In Fig. 14.6b, the two poles are not situated opposite to one another. In addition to the force of attraction, there is a lateral force that tries to move the poles laterally. If the poles move laterally, the cross-sectional magnetic energy. The force of alignment can give rise to linear motion, as in an electromagnetic relay shown in Fig. 14.7a. When the coil is energised, the armature is pulled in the direction indicated. Such relays are used to operate switches in many gadgets. The force of alignment can also be used to produce a rotary motion, as in the reluctance motor shown in Fig. 14.7b. The lateral force of alignment on the ends of the rotating piece, termed the rotor Figure 14.7c shows a simple machine which uses the force of interaction to give rise to rotary motion. about the axis of rotation.

Type of Rotary Machines rotary machines. Essentially, an electrical stator, and the moving part, called the rotor. The machine has a magnetic system (usually excited by a current in winding*) on one part and an armature

*

A winding is a series connection of coils.

Alternators and Synchronous Motors

431

Fig. 14.7 exists three important categories of doubly-excited rotating machines as follows :

1. Synchronous Machines 2. Asynchronous Machines 3. Commutator Machines

synchronous machine can be used as a generator or a motor. As a motor, it has highly specialised properties and serves a narrow range of applications. However, the synchronous generator (also called alternator) is the workhorse of the electric power industry for generating ac electric power.

General Characteristics of the Synchronous Machine Figure 14.8 shows the external connections of a synchronous machine. It has two electrical connections : a rotating shaft. As we shall see later, the machine has the following characteristics : by the dc exciting current.

432

enter directly into the energy-conversion process. 3. The armature circuit is placed on the stator and carries three-phase currents. mechanical system exchanges real power only. For generator action, the mechanical input power, minus the losses, becomes three-phase output power. For motor action, the real power from the electrical input minus the losses becomes mechanical output power.

, and extremely useful property of the synchronous machines.

Fig. 14.8

Synchronous Speed In a generator, if the armature windings on the stator are designed for two poles, the induced emf passes through one complete cycle in one revolution of the rotor. One revolution of rotor represents 360 mechanical degrees, and one cycle of emf represents 360 electrical degrees. Thus, in a two-pole machine, the mechanical and electrical angles are identical. Now, suppose that the machine has 4 poles. Then one cycle of emf would machine two cycles of emf is generated when the rotor completes one revolution. It means that if the machine has P poles, the number of cycles of emf in one revolution will be P/2. Therefore, in general the electrical angle qe and the mechanical angle qm in a machine are related as ⎛ P⎞ qe = ⎝ ⎠ qm 2 If the rotor has a speed of ns

(14.7) f of the induced emf would be

P f = ◊ns 2 ns must remain constant. Therefore, a synchronous generator must run at a constant speed known as synchronous speed. Alternatively, the synchronous speed, Ns (expressed in

Alternators and Synchronous Motors

433

revolutions per minute*) is given as Ns =

2 120f ¥ f ¥ 60 = P P

(14.8)

120 50 120f = = 3000 rpm P 2 Note that, as a machine cannot have less than 2 poles, the maximum speed a synchronous machine can have is 3000 rpm. Ns =

EXA M P L E

1 4.1

be needed on the generator, if it still runs at the same speed ?

Solution The speed of rotation is given as Ns =

P=

120 60 120 f = = 1200 rpm P 6 120 20 120 f = =2 N 1200

on the rotor and the armature winding is placed on the stator, as opposed to the system in a dc machine (Chapter 16).

Advantages of Placing Armature on Stator

large rotating torque. Furthermore, the connections from the rotating armature winding are made through a slip-ring-and-brush to collect thousands of amperes of current at kilovolts of voltage from the rotating slip-rings. This arrangement is impractical.

*

The speed of rotation of a machine is normally expressed in revolutions per minute (rpm).

434

Stator

Fig. 14.9

Fig. 14.10

Rotor exciter*.

i

1.

* **

ii

**

Alternators and Synchronous Motors

435

h needs only two poles. For such high-speed alternators*, cylindrical type of rotor construction (Fig. 14.11a) is preferred. The centrifugal force on a high-speed rotor is enormous. For instance, a mass of 1 kg on the outside of a rotor of 1 m diameter, rotating at 3000 rpm, has a centrifugal force (= mv 2/r = mrw 2 on it. To withstand such a force, the rotor is usually made of solid steel forging with longitudinal slots cut as shown in Fig. 14.11a actual rotor there are more slots and more conductors per slot. The winding is in the form of insulated copper a represent the end connections. distribution of rotor mmf. To reduce the centrifugal force on the conductors, the diameter of the cylindrical rotor is made small and axial length is made long. Because of the smooth surface, this rotor has less windage loss compared to the salient-pole-type rotor.

2.

In a hydroelectric power station, the alternators are driven by

water

their rotors. It is more convenient to build a rotor having large number of poles in projected or salient form (as shown in Fig. 14.11b for 4 poles). To accommodate such large number of poles, the rotor-diameter is made larger N and S. Unlike the cylindrical-type rotor, the mmf in salient type is concentrated by the coils wound around the a sine wave and thus improve the waveform of the generated emf. Stator core (slots omitted)

Phase R Phase Y N

S

S

N

Phase B (a) Cylindrical type.

Fig. 14.11 *

Also called turbo-alternators or turbo-generators.

(b) Salient type.

436

Figure 14.12a shows a cylindrical magnetic structure with one winding excited by a single-phase current. The current is entering into the bottom conductors and coming out of the top conductors. If the current i is dc, Bm = Ki. However, it decreases sinusoidally as we move away in the air gap from this direction, and reduces heavy concentration of conductors in the centre of the coil, tapered to few conductors at the edge of the qm, given as B = Bm cos qm. If the current is alternating (i = Im cos w t air gap is function of time t as well as angle qm, B(t, qm) = Bm cos wt cos qm

(14.9) –iR

+iB

+iY By

qm BR

S

N

BB

i –iY

–iB

+iR (a)

(b)

Fig. 14.12 Next, we consider Fig. 14.12b which shows a two-pole magnetic structure wound with three coils separated by 120° in space. These coils are supplied three-phase (RYB) currents. Note that the coil R corresponds to the coil of Fig. 14.12a. The current iR enters into the bottom conductors and returns from the top. The current iY iB sinusoidally and share slots between their most dense regions. As shown in Fig. 14.12b density from coil R lies in the horizontal plane, and the maxima from coils Y and B are displaced 120° and 240° in space, respectively.

Analysis of Three-Phase System phase currents supplied to the three-phase structure of Fig. 14.12b can be represented as iR(t) = Im cos (wt) iY(t) = Im cos (wt – 120°)

Alternators and Synchronous Motors

437

iB(t) = Im cos (wt – 240°) = Im cos (wt +120°)

and

BR(t, qm) = Bm cos (wt) cos (qm)

(14.10)

where Bm coils Y and B are respectively given as BY(t, qm) = Bm cos (w t – 120°) cos (qm –120°) BB(t, qm) = Bm cos (wt + 120°) cos (qm + 120°)

(14.11) (14.12)

B(t, qm) = BR(t, qm) + BY(t, qm) + BB(t, qm) = Bm cos (wt) cos (qm) + Bm cos (wt –120°) cos (qm –120°) + Bm cos (wt + 120°) cos (qm + 120°) Using trigonometric identities, we obtain B(t, qm) =

1 1 Bm{cos (wt – qm) + cos (wt + qm)} + Bm{cos (wt – qm) + cos (wt + qm – 240°)} 2 2 1 + Bm{cos (wt – qm) + cos (wt + qm + 240°)} 2

⎤ 1 ⎡3 = Bm ⎢ cos(ω t − θ m ) + {cos(ω t + θ m ) + cos(ω t + θ m − 240°) + cos(ω t + θ m + 240°)}⎥ 2 ⎣2 ⎦ The three bracketed terms add to zero at all times because they form a balanced three-phase set (considering 3 Bm cos (wt – qm) = Br cos (wt – qm) (14.13) 2 where, Br = (3/2)Bm is the magnitude of the rotating flux density with all coils operating together. B(t, qm) =

properties : 1. The magnitude is 50 % greater than what each coil gives individually. qmax = wt, qm = 0°.

and if t qm tp = q m/w. For a value of t

f, namely the duration of one cycle, the electrical angle is q = wt = (2p f )(1/f ) = 2p electrical radians

cycle. th –q m for RYB

+q m

438

rotor*. Let us consider why synchronous machines are constructed with stationary armature winding and rotating poles. Suppose, we have a 20-MVA, 11-kV, three-phase synchronous machine. Then, assuming unity power factor, the line current IL 20 ¥ 106 = 3 ¥ 11 ¥ 103 ¥ IL fi IL = 1050 A Hence, if the machines were made with stationary poles and rotating three-phase windings, three slip-rings brush-gear would be subjected to a working voltage of 11/ 3 = 6.35 kV. Further, it is usual to connect the armature windings in star and to suitably ground the star-point for which a fourth slip-ring would be By having stationary arm 1. It is easier to provide insulation to armature winding for high voltages, as the insulation of the stationary winding is not subjected to mechanical stress due to centrifugal forces and also more space is available for providing insulation on the stator. the need of slip-rings.

is light in weight, and therefore can run with high speed.

Pitch Factor (kp) In a full-pitch coil AB (as in Fig. 14.13a), the emfs induced in the two coil-sides Ea and Eb are in phase and the resultant emf of the coil is simply (as shown in Fig. 14.13b),

E)

Er = Ea + Eb = E + E = 2E However, due to some advantages, the coils are usually wound a little short-pitched (as in Fig. 14.13c). In such a short-pitch coil, the emfs induced in the two coil-sides Ea and Eb E) are out of phase by an angle b. The two emfs must then be added vectorially to get the resultant emf Er of the coil (as shown in Fig. 14.13d), Er = OQ = 2 ¥ OS = 2 ¥ OP ¥ cos (b/2) = 2E cos (b/2)

(14.14)

The factor by which the emf per coil is reduced because of the pitch being less than full-pitch is known as pitch factor (or coil span factor) kp. Thus, kp =

Phasor sum of the coil-side emfs 2E cos (b / 2) = = cos (b/2) Arithmetic sum of the coil-side emfs 2E

(14.15)

This pitch factor has been determined for the fundamental component of induced emf. However, because of

*

As we shall see in Chapter 16, the reverse is the case in dc machines. The armature winding is placed on rotor and

Alternators and Synchronous Motors

180 °

(E l

Q

tric ec

A

439

al)

Er = 2E

N B S

S Ea = E

N

O (b) Phasor sum of coil-side emfs

(a) Full-pitch coil in a 4-pole machine

Eb = E Q Er b

A N S

Eb = E

S

b

B

P Ea = E

S

N b/2 O (d) Phasor sum of coil-side emfs

(c) Short-pitch coil in a 4-pole machine

Fig. 14.13 induced emf. For the nth harmonic, the pitch factor is given by kpn = cos (nb/2)

(14.16)

The concept of short-pitch coil helps to design a winding in which the induced emf can be made free from a particular harmonic. Suppose that we wish to remove 6th harmonic altogether. Then, for 6th harmonic, the pitch factor must be made zero, kp6 ∫ cos (6b/2) = 0 fi

6b/2 = 90º or b = 30º

connections is reduced.

Distribution Factor (kd) The coils comprising a phase of the winding are distributed in two or more slots per pole. Compared to a concentrated winding, a distributed winding results in an induced emf of waveform nearer to sinusoidal, lesser armature reaction and better heat dissipation. However, in a distributed winding, the emf in the adjacent coils will be slightly out of phase with respect to one another, and their resultant will be less than their algebraic sum. Figure 14.14 shows the component emfs of coils (E1, E2 and E3) and the resultant emf (Er) due to a phase group for a winding having 3 slots per pole per phase. Each component emf is displaced from the component emf of the adjacent slot by the slot angle a

440 electrical degrees. The distribution factor (or breadth factor) kd for q

or

kd =

AD 2 AF 2 R sin (qα / 2) Phasor sum of component emfs = = = q BC q 2 BG q × 2 R sin (α / 2) Arithmetic sum of component emfs

kd =

sin (qa / 2) q sin (a / 2)

(14.17)

If q (the number of slots per pole per phase) is very large, the angle a becomes very small so that sin (a/2) ª a kd =

sin (qa / 2) qa / 2

(14.18)

The total angle qa (expressed in electrical radians) is called the phase spread.

A E1 = E 2 = E 3 = E

R E1 qa/2

B

a O

a

a /2

a

R

F

G E2

Phase spread = qa

C a

Er

E3 D

Fig. 14.14

We know that if a conductor of length l with a relative speed v

B e = Blv

Let d be the diameter of the armature (on the stator) and ns be the rotational speed of the rotor in revolutions per second. Then, the relative speed v = pdns. If F P is the total number of poles,

B=

Flux per pole Number of poles FP = Cylindrical area of armature surface p dl

Alternators and Synchronous Motors

441

Thus, the average value of the induced emf in a conductor is e=

FP ◊ l ◊ pdns = FPns = F ¥ 2f p dl

To get the rms value, we must multiply the average value by the form factor (which is 1.11 for a sinusoidal waveform). Moreover, if the winding consists of T concentrated turns (i.e., 2N conductors), the rms value of the net induced emf in a concentrated winding is given as Ec = (2fF) ¥ 1.11 ¥ 2 N = 4.44fFT

(14.19)

In practice, the coils are short pitched and the winding is distributed. Hence, the rms value of the induced emf is reduced by the pitch factor kp and distribution factor kd, to give E = (2fF) ¥ 1.11 ¥ 2T ¥ kp ¥ kd = 4.44fFTkpkd EX A MP L E

(14.20)

14. 2

Determine the distribution factor for a machine having 9 slots per pole for the following cases : (a) a three-phase winding with 120° phase group, and (b) a three-phase winding with 60° phase group.

Solution The slot angle, a =

180 = 20° 9

(a) Since one phase occupies 120°, the number of slots in one phase group, q = \

kd =

sin (6 × 20° / 2) sin (qa / 2) = = 0.831 6 × sin (20° / 2) q sin (a / 2)

(b) Since one phase occupies 60°, the number of slots in one phase group, q = \

EXA MP LE

kd =

sin (3 × 20° / 2) sin (qa / 2) = = 0.960 3 × sin (20° / 2) q sin (a / 2)

120 = 6. 20

60 = 3. 20

14.3

A three-phase, 50-Hz, 20-pole, salient-pole alternator with star-connected stator winding has 180 slots on the stator. a) the speed, (b) the generated emf per phase, and (c) the line emf.

Solution Total number of armature conductors, Z = 180 ¥ 8 = 1440. Therefore, the number of turns per phase, T = (a) The speed, Ns =

1440/2 = 240. 3

120 50 120 f = = 300 rpm P 20

(b) Since the coils are full-pitch, the pitch factor, kp = 1. Next we calculate the value of distribution factor, kd. 180 No. of slots per pole = =9 20 Electrical angle per pole 180 = = 20° \ Slot angle, a = Number of slots per pole 9 Number of slots per pole per phase, q =

9 =3 3

442 sin (3 × 20° / 2) sin (qa / 2) = = 0.960 3 × sin (20° / 2) q sin (a / 2) Therefore, the rms value of the generated emf per phase, \

kd =

E = 4.44fFTkpkd = 4.44 ¥

¥

¥ 240 ¥ 1 ¥ 0.960 = 1278.7 V

(c) Since, the stator winding is star-connected, the line emf, EL =

3 ¥ 1278.7 = 2214.8 V

two slots per pole per phase. Suppose its rotor is excited by dc current and is rotated clockwise at synchronous speed by a prime mover. An emf is induced in the stator winding and if the generator is on no load the stator a. This R

R

Y1 N

B1

S Y

B

Y

B R1 (a) Due to rotor currents alone.

(b) Due to stator currents alone.

R

R

Y1

B1

N S

N

S B

Y

B

Y

R1 (c) Resultant flux for case (i).

Fig. 14.15

(d) Resultant flux for case (ii) and case (iii).

Alternators and Synchronous Motors

443

Now, suppose the generator is connected to an electrical load resulting in stator-currents. The magnetic a, the induced emf in phase RR1 is maximum and is towards the paper in conductor R and coming out of the paper in R1 stator currents) separately and then superimposing them. We shall consider three cases of the electrical load: (i) when the current and the generated emf are in phase, (ii) when the current lags the generated emf by 90°, and (iii) when the current leads the generated emf by 90°.

Case (i b at the instant when the rotor (unexcited) is in the a. c.

c.

Case (ii R reaches its maximum value, the poles of the rotor will have moved forward by 90° to the position as shown d b. It means

d

Case (iii a d. It means d.

Variation in Terminal Voltage in Fig. 14.16a emf E when the generator is on no load. Let us see how the terminal voltage changes with increase in load current. For unity power-factor load (i.e., resistive load), the fall in terminal voltage is comparatively small. the demagnetisation effect of armature reaction. If the load is purely inductive (pf = 0), the demagnetisation effect of armature reaction becomes much more predominant, and hence the fall in terminal voltage is also much more rapid. With capacitive load (power factor leading), the effect of armature reaction is to increase the magnetic magnetising effect of the armature reaction, the terminal voltage increases with an increase in the load current.

444

Fig. 14.16

Voltage Regulation V when supplying its rated current at a

unaltered. As shown in Fig. 14.16b, if the power factor is 0.8 lagging, the terminal voltage increases to a value E1 when the load is fully removed. However, if the power factor is unity, the terminal voltage increases to a value E2 (i.e., by a lesser amount) on removing the load. This change in terminal voltage between full load and no load, expressed as a per-unit value or a percentage of the full-load voltage, is termed as voltage regulation. Thus, Per-unit voltage regulation = and

Percentage voltage regulation =

E

V

(14.21)

V E

V V

¥ 100

(14.22)

where, V is the terminal voltage at full load and E is the terminal voltage when the load is removed.

machine can be conceptually divided into two components—the excitation emf E (in the stator winding). On the

reactance of the called synchronous reactance (Xs) of the winding. The net voltage V generated in the stator can be regarded as the difference between the excitation emf E and the voltage drop due to the synchronous reactance. The excitation emf E V is in phase with the E and V is the rotor power angle, dR

Alternators and Synchronous Motors

445

e rotor and stator. It means that the synchronous machine will have, in addition to the synchronous reactance, a leakage reactance (Xl usually Xl 3 fi f > 60° or cos f Hence, if one of the wattmeters reads negative, the power factor has to be less than 0.5. 7. Ans. : 8.

pull-out torque’ of an induction motor ? Ans. : The maximum torque the motor can develop is also called pull-out torque.

9. Ans. : If the load on an induction motor is increased beyond the maximum torque it can develop, the motor slows down, the slip increases and the operation becomes unstable. Ultimately, the machine stops. It pulls out of the motion. 10. plugging of an induction motor ? Ans. : In case of a running motor, if suddenly two of the three-phase supply connections to the stator are interquickly. This amounts to braking or plugging of the machine. 11. Ans. : If one of the lines of the three-phase supply gets open due to some fault, the three-phase motor continues drawn by the two remaining lines increases and the motor gets overheated, and it may even get damaged. 12. Do you know any other machine whose output-speed characteristic is similar to that of a three-phase induction motor ? Ans. : Yes, dc shunt motor. 13. In what way, an induction motor is similar to a transformer ? Ans. : (i) Both work on the principle of electromagnetic induction. (ii) The stator of the induction motor is analogous to the primary of the transformer and the rotor to the secondary of the transformer. (iii) Both can be represented by similar equivalent circuits. (iv) Both machines can be tested through similar tests. The no-load test on an induction motor is similar to the open-circuit test on a transformer. 14. In what way, an induction motor differs from a transformer ? Ans. : (i) The windings of a transformer are cylindrical, whereas those of an induction motor are distributed. (ii) The induction motor has an air gap in its magnetic path and hence the magnetising current of an induction motor is much higher than that of a transformer. (iii lower than that of a transformer. (iv) An induction motor converts electrical energy into mechanical energy, whereas a transformer transforms electrical energy from one voltage level to another.

DC MACHINES OB JE CT IV E S

alternator

16

514

Basic Electrical Engineering

Portable devices powered by batteries require dc motors, such as portable tape players, walkman, windowlifters, etc. Also, the dc motor is readily controlled in speed and torque and hence is useful for control systems. Examples are robots, elevators, machine tools, rolling mills, etc.

Figure 16.1 shows the basic structure of a dc machine (a motor or a generator). The machine has following important parts.

Figure 16.1 shows the magnetic structure of a four-pole dc machine. Its main components are described below.

Slot

Field coil Yoke Pole shoe

N S

S

qm

Pole core Armature core

Shaft

(i)

N

It is the outermost cylindrical part which serves two purposes. First, it acts as a supporting frame

forged steel. Usually, small machines have cast-iron yokes.

(ii) Poles

The machine has salient poles. The pole cores cross section of the pole core is rectangular. By attaching a pole shoe, the end of the pole is made to have a cylindrical surface. The cross-sectional area of the pole shoe is considerably larger than that of the pole core of cast steel, or forged steel. Each pole carries a permanent magnets.

(or exciting coil). Small machines generally use

(iii) Field Coils coils are identical and are connected in series such that on excitation by a dc source, alternate N and S poles

F. Therefore, the cross section of the yoke should be selected accordingly.

Flux

DC Machines

F

N

0°

90° S

N 180°

270°

qm

360°

S

S

(i) Armature

(ii) Commutator

Brushes

Ia +

V

N

S +

–

Vf If

–

515

516

(iii) Brushes

reactance voltage

I is the V

Ia

V.

*

a ¢ b ¢ *

¢

¢

DC Machines

N1

517

S1 1

1¢

2

4 1¢ 5

11 10 9 S2

3

8

7

6 N2 1

(a) Arrangement of double-layer winding.

2

3

1¢

2¢

(b) Arrangement of overlap of end connections.

¢

¢ ª

(1) Lap Winding

(2) Wave Winding

3¢

Total number of slots Total number of poles

a

b

P Q (c) Ends of a coil.

a

518

Basic Electrical Engineering

Thus, if a machine has P poles, the number of parallel paths in armature winding is A=P A=2

and

(for lap winding) (for wave winding)

Hence, it may be said that, in general, lap windings are used for low-voltage, heavy-current machines, and wave windings are used for high-voltage, low-current machines.

EXA M P L E

16 . 1

that the average emf generated in each conductor is 2.1 V, and each conductor is capable of carrying a full-load current of 200 A. Calculate the terminal voltage on no load, the output current on full load and the total power generated on full load, when the armature is (a) lap-wound, and (b) wave-wound.

Solution (a) With the armature lap-wound, the number of parallel paths, A = P = 8. Therefore, the number of conductors per path is Z 480 = = 60 A 8 Therefore, the terminal voltage on no load, ⎛ Z⎞ E = e ¥ ⎝ ⎠ = 2.1 ¥ 60 = 126 V A

519 The output current on full load, IL = Full-load current per conductor ¥ No. of parallel paths = 200 ¥ 8 = 1600 A The total power generated on full load, Po = IL ¥ E = 1600 ¥ 126 = 201 600 W = 201.6 kW (b) With the armature wave-wound, the number of parallel paths, A = 2. Therefore, the terminal voltage on no load, 480 ⎛ Z⎞ = 504 V E = e ¥ ⎝ ⎠ = 2.1 ¥ A 2 The output current on full load, IL = 200 ¥ 2 = 400 A The total power generated on full load, Po = IL ¥ E = 400 ¥ 504 = 201 600 W = 201.6 kW Note that the total power generated by a given machine is the same whether the armature is lap-wound or wavewound.

Let there be P number of poles and let F Z be the total number of conductors and let A be the number of parallel paths on the armature winding. Let the rotational speed of the rotor be N rpm. Consider one revolution of the rotor. As the rotor makes N revolutions in one minute, it makes N/60 revolutions in one second. In other words, the speed of rotation is N/60 rps. Therefore, the time (in seconds) taken in making one revolution is 1 60 Dt = = N / 60

N

P

As F ductor on the armature is PF

-

DF = PF Therefore, the induced emf per conductor is given by Faraday’s law as e=

DF Dt

=

PF NPF = 60/N 60

(16.1)

The conductors are connected to make coils, and the coils are connected to form parallel paths. The brushes collect the emf from all these identical parallel paths. The net emf E generated in the machine is the same as the total emf in one parallel path. As the armature has Z number of conductors, and there are A number of parallel paths, the number of conductors per parallel path is Z/A. Therefore, using Eq. 16.1 we can write the expression for the net emf E generated in the dc machine as ⎛ Z⎞

NPF Ê Z ˆ

E = e⎝ ⎠ = Á ˜ A 60 Ë A ¯ or

E=

FZ NP 60 A

Remember, the number of parallel paths, A = P (for lap winding) and A = 2 (for wave winding)

(16.2)

520 EX A MP L E

16. 2

Solution

Z A

EX A MP L E

0.02

FZ NP 60 A

E

¥ P 780 1200 60 4

4

312 V

16. 3

Solution E

FZ NP 60 A

KN

N2 E1 N1

600 ¥ 500

where K E2 E1 EX A MP L E

N2 N1

E

216 V

16. 4 a b

Solution (a E

KN

where K E2 E1

N2 N1

N

E2 N1 E1

250 ¥ 220

852 rpm

b E

K¢FN

2N 2

2

1 N1

1

E2 E1

where K¢ E2 E1

¥

42 %

N1 N2

250 220

750 600

DC Machines

521

a -

I b

Let E E I E

F E

E

522 ID

R

A IM ED

If

D

B

I

EB

Ra

I

E

E

I

generator

E

E

E

E

I I

motor I *

E V

Ra

Ia

As Generator V

Ia

E

E V

As Motor

E IaRa

Ia

V

E E V

E + IaRa

permanent magnet generators separately excited generators self-excited generators *

V

back emf

DC Machines

523

Ia Ra E Ise Rse Ish Rsh IL V RL

(1) Permanent Magnet Generators Separately Excited Generators

Self-Excited Generators (a) a ( ) (c)

-

b -

Short-Shunt c Long-Shunt

and

524

Basic Electrical Engineering

Self-excited dc generators.

525 EXA M P L E

16 . 5 W and that

of the armature winding is 0.02 W. Calculate the emf induced in the armature.

Solution V 440 = =4A Rsh 110 \ Armature current, Ia = IL + Ish = 496 + 4 = 500 A Therefore, the generated emf is Ish =

Eg = V + Ia Ra = 440 + (500 ¥ 0.02) = 450 V EXA M P L E

16 . 6

W and 50 W, respectively. It supplies power to 100 lamps, each of 60 W, 200 V. Calculate the total armature current, the current per path and the generated emf. Allow a brush drop of 1 V at each brush. P 60 = = 0.3 A V 200 Since all the lamps are connected in parallel, the total load current is

Solution

The current taken by each lamp, I1 =

IL = 100 ¥ I1 = 100 ¥ 0.3 = 30 A V 200 Ish = = =4A Rsh 50 \ Armature current, Ia = Ish + IL = 30 + 4 = 34 A For lap winding, the number of parallel paths, A = P = 4. Thus, I 34 The current per path, Ic = a = = 8.5 A A 4 The generated emf, Eg = V + Ia Ra + brush-drop = 200 + 34 ¥ 0.2 + 2 ¥ 1 = 208.8 V EX A MP L E

16 .7

A short-shunt compound-wound dc generator supplies a load current of 100 A at 250 V. The generator has following winding resistances : W;

armature = 0.1 W

W

Find the emf generated, if the brush drop is 1 V per brush.

Solution

\

Refer to Fig. 16.9c

Ise = IL = 100 A

Vse = IseRse = 100 ¥ 0.1 = 10 V Vsh = V + Vse = 250 + 10 = 260 V V 260 Ish = sh = =2A Rsh 130 The armature current, Ia = IL + Ish = 100 + 2 = 102 A The generated emf, Eg = V + Vse + IaRa + brush-drop = 250 + 10 + 102 ¥ 0.1 + 2 ¥ 1 = 272.2 V

526

a

A

A

N

A

C

S

B (a) Flux due to field current alone.

B (b) Flux due to armature current alone.

D

B

q (c) Resultant flux due to field and armature currents.

b

in F

c

a

and b

the demagnetisation effect c

q *

and the component. *

q

527 when the machine is working as a generator and if the ‘short-circuit’ or ‘excessive-overload’ condition occurs, the demagnetising component may even reverse the polarity of the main poles. RE M EDY The adverse effect of armature reaction can be neutralised by shifting the brushes to the magnetic neutral plane and by increasing the air gap at pole tips.

Various losses occurring in a dc machine are as follows.

(i Armature Copper Loss It is given as I a2Ra. This loss amounts to about 30 to 40 % of the full-load losses. 2 It is given as I sh2Rsh for shunt-wound machine and as I se Rse for series-wound machine. This loss amounts to about 20 to 30 % of the full-load losses. For shunt-wound machine, it remains practically constant; but for series-wound machine, it increases with the load.

(ii

(iii

This loss occurs due to the resistance of the brush contact with the commutator. This is usually included in armature copper loss.

Since the current in the armature winding is alternating at a frequency f This loss amounts to about 20 to 30 % of the full-load losses. There can be two types of magnetic (or iron) losses :

(i Hysteresis Loss

= B1.6 max f

(ii Eddy-current Loss = B 2max f 2

There are two types of mechanical losses.

(i Air Friction (or Windage) Loss It occurs due to rotation of the armature. (ii Bearing Friction Loss Mechanical losses are about 10 to 20 % of the full-load losses. Mechanical losses taken together are also called stray losses.

528

hm

he =

Total watts generated in armature Mechanical power supplied at the input EI hp ¥ 746 Total watts available to the load Total watts generated

VI EI

Total watts available to the load Mechanical power supplied VI hp ¥ 746

, hc

hm ¥ he

hc

P Pc

P

I a Ra Output Output + Total losses

h

VI VI + I 2Ra + Pc 1 P ˆ Ê IR 1+ Á a + c ˜ Ë V VI ¯

d dI

Ï Ê IRa Pc ˆ ¸ + ˜˝ Ì1 + ÁË V VI ¯ ˛ Ó I Ra

I EX A MP L E

VI VI + I a2Ra + Pc

Ia ª I

Ra P - c2 V VI

Pc

Pc /Ra

I

Pc /Ra

16. 8 W

b

VI VI + ( Pv + Pc )

W c

a

DC Machines

529

Solution a IL

30 kW 200 V

Ish

200 V 50

Ia E

b c

h

EX A MP L E

IL + Ish ¥

V + IaRa

207.7 V

¥

¥ 1985.8 W 30 000 30 000 + (1000 + 1985.8)

I shRsh + I a Ra Output Output + Losses

90.95 %

16. 9 W

Solution

V ¥ IL ¥ Po 40.95 kW 0.90 Pin P V 210 V Rsh 52.5 IL + Ish

P

\

Pin

\ Ish \

Ia

I sh Rsh

¥

\ I a Ra Ra

I a Ra

Armature copper loss

3000

Ia2

1992

0.0757 W

Pc

\

Ia

Pc Ra

\

IL

Ia Ish

1550 0.0757

143.1 A 139.1 A

Open-Circuit, Magnetisation, or No-Load Characteristic E

I

Load (or External) Characteristic IL

Internal Characteristic

performance characteristic

V and voltage regulation curve E Ia

530

Basic Electrical Engineering

Open-Circuit Characteristic (OCC) saturation curve N E open-circuit characteristic OCC

I

. E

E

a

Fig. 16.11 b E residual magnetism

I I

Rh E

E

linearly

I

I

The Field Resistance Line

b E

DC Machines

531

E E.

E /I

RF

RF

R

Rh

Building Up of Voltage I

Critical Field Resistance E a

E

less than C E1

Eg

E2

A

Eg

A

N1

B

N2

O

If

O

If

Critical Speed

N b

532 N

N N

N N N

N is the

V

Separately Excited Generator

IL

T

a

N

I

E V

IL

b V

and IL

Shunt Generator b

a

DC Machines

indicates the unstable

b IL

Ish

V

A

+

V –

Strong field

E0

A

Weak field

Load

A1 0

IL

(a) The circuit arrangement.

(b) The load characteristic.

Series Generator

a

b

I

IL

IL

E V

IL booster

I1 +

If

A

b

V

A

a V

c

Load d

–

A1

0

(a) The circuit arrangement.

Compound-Wound G b

b c and

IL (b) The load characteristic.

533

Terminal voltage, V

534

C E0

A B

A : Ideal B : Level-compounded C : Over-compounded D : Under-compounded E : Differential-compounded

D E Load current, IL

Case I B V

level-compounded Case II

V over-compounded

Case III

V under-compounded.

Case IV differential-compounded

EX A MP L E

reverseV

16. 1 0

¥ ¥

Solution

Nse

3000 100

30

DC Machines

535

E E

FZNP 60 A

V and hence it is treated as

back emf

E V IL

E Ia V IL

and

I

E + IaRa Ia + I

A Z and P

E

written as E k

where

kNF ZP 60 A

E V

kNF + Ia Ra fi N

V - I a Ra kF

536 I aR a

V Nª

V kF

Nμ

V V

F N N0

Nf Nf

EX A MP L E

¥

16. 1 1 W

W

Solution

Ish

\ EX A MP L E

N the

V Rsh Ia E

250 250 IL Ish V IaRa

¥

246 V

16. 1 2 W

Solution E

N EX A MP L E

Solution

¥

V IaRa 60AE b F ZP

60 2 426 ª 626 rpm 0.023 888 4

460 kF

kF

16. 1 3

F

N¢

V¢ kF ¢

V¢ k(0.7F )

V¢ 0.7kF

200 ª 559 rpm 0.7 0.511

DC Machines

537

Ia VIa

E Ia + I a Ra

VIa

I a Ra E Ia td E Ia Pm 2pt d N 60

N

2pt d N watts 60

E Ia FZNP ¥ Ia 60 A

td

E

FZ Ê P ˆ Á ˜ Ia 2p Ë A ¯

Z P and A t d μ Ia ¥ F

E Ia

Note

t sh

EX A MP L E

td

16. 1 4 W a

Solution a

E

V Ia Ra E

FZNP 60 A

td

FZ Ê P ˆ Á ˜ Ia 2p Ë A ¯

¥ N

60AE b F ZP

60 6 458 ª 636 rpm 0.05 864 6

b 0.05 ¥ 864 Ê 6 ˆ ¥Á ˜ ¥ Ë 6¯ 2p

ª 756 Nm

b

538

Basic Electrical Engineering

EXA M P L E

16 . 15

A dc generator runs at 900 rpm when a torque of 2 kNm is applied by a prime mover. If the core, friction and windage losses in the machine are 8 kW, calculate the power generated in the armature winding.

Solution

The power required to drive the generator,

2 π × 2000 × 900 2ptN = = 188 495 W = 188.5 kW 60 60 Therefore, the power generated in the armature, Pd = Pin – Plosses = 188.5 – 8 = 180.5 kW Pin =

When no load is connected to the shaft of a dc motor, it develops only that much torque which overcomes the rotational (frictional) losses and the iron losses. How does the motor react to the application of a shaft load ? To answer this question, we require to know the performance characteristics of the motor.

The speed characteristic of a motor represents the variation of speed with the input current. Its shape can be easily derived from the expression of Eq. 16.13, namely N=

V - Ia Ra kF

Shunt Motor F, therefore, remains constant. Since the drop Ia Ra at full load rarely exceeds 5 % of V, the speed N is almost constant. Its speed characteristic may be represented by curve A in Fig. 16.18a. Thus, constant speed motor. In actual practice, the drop in speed with F partially compensates the drop due to IaRa.

(2) Series Motor in series with the armature. If Rse represents the resistance of this winding, the back emf is given as Eb = V – Ia(Ra + Rse) N=

V - Ia ( Ra + Rse) kF

F Ia and then less rapidly due to the magnetic saturation. Hence, the speed is roughly inversely proportional to the current, as indicated by the curve B in Fig. 16.18a. Thus, variable speed motor. Note motor should never be used when there is a risk of the load becoming very low. For instance, the load should never be belt-connected, as it has the risk of breaking or slipping. The load to a dc series motor is either directly connected or geared to the shaft.

539

(3) Compound Motor a

stabilised series motors

Fig. 16.18

Torque Characteristics of DC Motors t t μI ¥F

t =kI ¥F

k

(1) Shunt Motor

F

t μI

b

(2) Series Motor

F

t μ I2 b.

(3) b. b

540 EX A MP L E

16 .16 W

Solution ¥

E

V Ia Ra + Rse

E

FZNP 60 A

kF N

k¥

¥

E

k ¥

V Ia Ra + Rse k¥F ¥N

N EX A MP L E

222.5 k × Φ2

222.5 ª 938 rpm 14.83 0.016

16 .17 W

a b c

Solution a

E Ia

b

V

Eb Ra

250 200 0.2

250 A

V

Eb

250 250 0.2

0

Eb

1250 A

E Ia

c

Ra

250 0 0.2

V

E Ia

Ra

W

DC Machines E

Ia EX A MP L E

V

250

Eb

( 250) 0.2

Ra

16. 1 8 W

a

2500 A

W

b

Solution a E \ b

2πτ d N 60

EX A MP L E

¥

V Ia Ra + Rse

N

60AE b F ZP

60 6 458 0.05 864 6

td

60Eb I a 2pN

60 × 458 × 110 2 × π × 636

636 rpm

756 Nm

E Ia

16. 1 9 W

Solution

N E E2 E1

N2 N1

¥

V IaRa E2 210.1

450 700

fi

E

R E

R EX A MP L E

V Ia Ra + R ¥

R

3.411 W

16. 2 0 W

Solution E

V Ia Ra + Rse

¥

W

541

542

E F

¥

V Ia Ra + Rse

F E

\

E2 E1

N2F2 N1F1 E2 N1 E1 ¥ 0.6

N

kFN N 2 ¥ 0.6F1 N1F1

0.6 N2 N1

224 1000 ª 1713 rpm 218 0.6

starter

no-volt release and OFF

ON

ON

No-Volt Release

ON

OFF ON

DC Machines ting

Star

Resistance

3

5 6

1

OFF

ON

4

2 Brass arc

No-volt release

Soft iron keeper

Starter arm Spiral spring

Overload release A

A1

Switch

+ DC supply –

Shunt field

OFF

Overload Release

OFF

ADDITIONAL SOLVED EXAMPLES EX A MP L E

16. 2 1 W

543

544 Solution

¥

Z A

P 0.08

FZNP 60 A

E

¥

Rw

R R1 0.39 4 4 E IaRa

Ra V EX A MP L E

520 1000 60 4 W Rw 1.56 4 4

4

W

W ¥

687.98 V

16. 2 2 W

Solution

IL

V RL Ish

\

Ia E

W 250 12.5 V Rsh

¥

V + IaRa FZNP 60 A 60AEg

E F EX A MP L E

250 250 21 A 255.04 V

60 2 255.04 778 500 8

ZNP

9.83 mWb

16. 2 3 W

Solution

Ia

E Ia E

E

500 1000 500

Po V

¥

V + IaRa Po 250 1000 V 500

¥

V + IaRa FZNP 60 A

N

KE

K

W

DC Machines

N1

EX A MP L E

K ( E1 - E2) KE1

N2 N1

515

507.5 515

1.456 %

16. 2 4 W a

b

Solution a For IL Vt

¥

ILRt

¥

V Vd

Vt V

8V

b For IL Vt

Vd EX A MP L E

¥

W

Solution

W

Ish

\

Ia V + Ia Ra + Rse

EX A MP L E

V Rsh IL + Ish

500 250 52 A 506.16 V

16. 2 6

a b

Solution a

2.5 V

16. 2 5 W

E

Vt V

¥

ILRt

V

A

545

546 \ A

FZNP 60 A

0.02

500 1800 60 2

8

E

FZNP 60 A

0.02

500 1800 60 8

8

Ia Pd

A¥ E Ia

¥

Ia Pd

A¥ E Ia

¥ ¥

1200 V

P

\ b

E

300 V

A \

¥

12 kW

A \ EX A MP L E W c

12 kW

16. 2 7 W

a

Solution a Ish

V Rsh

250 50

\

Ia E

b

IL + Ish ¥

V + IaRa

254 V

¥

I aRa

V ◊ Ish

¥

\

2050 W

c

P

VIL

¥

\

Pin 51.75 kW Pe

Pin Total watts generated in the armature Mechanical power supplied at the input

hm =

50800 51750

98.2 %

Total watts available to the load Total watts generated in the armature

he =

48750 50800

95.9 %

b

547 Total watts available to the load Mechanical power supplied 48750 = = 0.942 pu = 94.2 % 51750

hc =

EXA MP LE

1 6 .28

A 4-pole, dc shunt motor working on 250 V takes a current of 2 A when running light (i.e., at no load) at 1000 rpm. W and 250 W, respectively. (a) How much back emf is generated ? (b) What will be its back emf, speed and percentage speed drop if the motor takes 51 A at a certain load ?

Solution V 250 = =1A Rsh 250 Therefore, armature current, Ia1 = IL1 – Ish = 2 – 1 = 1 A Ish =

(a

\ Back emf, E1 = V – Ia1Ra = 250 – 1 ¥ 0.2 = 249.8 V (b) The armature current, Ia2 = IL2 – Ish = 51 – 1 = 50 A Back emf, E2 = V – Ia2Ra = 250 – 50 ¥ 0.2 = 240 V

\

Hence, the new speed is given as E2 240 N2 = N1 ¥ = 1000 ¥ = 961 rpm E1 249.8 N1 N2 1000 961 % speed drop = ¥ 100 = ¥ 100 = 3.9 % N1 1000

\ EXA MP LE

16.29

Calculate the value of the starting resistance for the following shunt motor : Output = 14 920 W; Armature resistance = 0.25 W

Supply = 240 V

The starting current is to be limited to 1.5 times full-load current. Ignore the current in shunt winding.

Solution \

The input power, Pin =

Po

=

14 920 = 17 349 W 0.86

Pin 17349 = = 72.3 A V 240 The permitted starting current, Ist = 1.5IL = 1.5 ¥ 72.3 = 108.45 A Full-load line current, IL =

Required total resistance in the armature circuit at the starting is given as V 240 = = 2.213 W Rt = Ist 108.45 Therefore, extra resistance required at starting is Rst = Rt – Ra = 2.213 – 0.25 = 1.963 W

548 EX A MP L E

16. 3 0

W

Solution

I

I

Ish E I E

\

I

Ish ¥

E μ N kN E2 480.2 ¥N ¥ E1 499.4

Pc

VI

¥

P

I Ra

¥

Pin

VI

¥

\

I

V I Ra

N

EX A MP L E

¥

V I Ra

h

Pin − ( Pv + Pc ) Pin

ª 962 rpm

50 000 − (2000 + 1960) 50 000

92 %

16. 3 1 W

W

Solution

Ish

IL E

V Rsh

250 50

50 kW 250 V V + I Ra

I

IL + Ish

¥ IL

EX A MP L E

I E

IL Ish V I Ra

N2 N1

E2 E1

¥

N

E2 E1

N1

246.1 ¥ 254.1

484 rpm

16. 3 2 W

W

549 Solution In a series motor, t μ I a2. But, here it is given that t μ N 2. Hence, we conclude that Ia μ N = kN Ia 2 N2 N \ = or Ia2 = 2 Ia1 N1 N1 The back emfs generated in the two cases are

Ia1 =

350 ¥ 20 = 28 A 250

E1 = V1 – Ia1(Ra + Rse) = 400 – 20 ¥ (0.6 + 0.4) = 380 V E2 = V2 – Ia2(Ra + Rse) = V2 – 28 ¥ (0.6 + 0.4) = (V2 – 28) V

and

F μ Ia , the back emf generated is given as E=

FZNP μ FN μ Ia N 60 A

Therefore, we must have E2 I N = a2 2 E1 I a1N1 V2 28 28 = 380 20

or

350 250

fi

V2 = 772.8 V

SUMMARY TE RM S

A N D

C ON CE PT S

There are two types of armature windings in a dc machine : ( ) Lap winding (A = P), and ( ) Wave winding (A = 2); A is the number of parallel paths. Types of dc machines : (1) Permanent magnet, (2) , and (3) : (a) Series, (b) Shunt, and (c) Compound : ( ) Short-shunt, and ( ) Long-shunt. Armature reaction It weakens and Different losses in dc machines : (1) Copper losses : ( ) Armature copper loss, ( ) Field copper loss, and ( ) Brush contact loss; (2) : ( ) Hysteresis loss, and ( ) Eddy-current loss; (3) :() Air friction (or windage) loss, and ( ) Bearing friction loss. constant losses ( ) must be equal to (copper losses). Important characteristics of a dc generator : (1) Open-circuit, Magnetic, or No-load Characteristic, (2) Load, or External Characteristic, (3) Internal Characteristic. which the voltage build-up is possible. Critical speed is possible. Due to armature voltage drop and demagnetisation effect of the armature reaction, the terminal voltage of a dc shunt generator slightly decreases as the load is increased. In a dc series generator, the terminal voltage increases proportionately to the load current. Ideally, we prefer a generator whose terminal voltage remains constant even on increasing the load. This can be achieved in a Depending upon whether the series ampere-turns are less than or more than the shunt ampere-turns, the generator may be or . this, we use a starter. A dc shunt motor is almost a constant speed motor.

550

IMP O R TA N T The emf

F O RM U LAE E

FZNP 60 A

number of parallel paths

A A

P

generator V Eg IaRa motor V E + IaRa Torque

FZ Ê P ˆ Á ˜ Ia , 2p Ë A ¯ td μ Ia td μ I a

td μ IaF

td

CHECK YOUR UNDERSTANDING two one

minus

12

Ia Ia

Your Score

DC Machines

551

A N SW E RS

1. 6.

2. 7.

3. 8.

4. 9.

5. 10.

REVIEW QUESTIONS 1. 2. 3.

13.

4.

14. 15.

5. 6.

16.

7.

17. 18.

8.

a b

19. 9.

20.

10. 21. a b

22.

c 23. 11. 24. 12. 25.

MULTIPLE CHOICE QUESTIONS c

2.

1. a b

a b

552

Basic Electrical Engineering

(c) to help in changing the direction of rotation of the armature ( ) None of the above

9.

3. The number of parallel paths in the armature winding of a four-pole wave-connected dc machine having 28 coil-sides is (a) 28 (c) 4

(b) 14 ( ) 2

4. Which one of the following equations does not apply to a shunt-wound dc generator ? V (a) Ish = (b) IL = Ia – Ish Rsh (c) E = V + IaRa ( ) V = E + IaRa 5. A dc series motor should always be started with load-connected, because (a) at no-load it will rotate at dangerously high speed (b) at no-load it will not develop high starting torque (c) it cannot start without load ( ) it draws a small amount of current at no load 6. The emf generated by a given dc generator depends upon (a (b) the speed only (c ( ) the terminal voltage 7. The speed of a dc motor is (a) directly proportional to both its back emf and

10. The emf induced in each conductor of the armature in a dc machine is (a) alternating in nature (b) direct in nature (c) pulsating in nature ( ) has a random waveshape 11. A dc machine having an armature-resistance of 0.1 W is connected to a 230-V dc supply. What should be the emf generated in the machine so that it may feed a current of 80 A to the supply ? (a) 8 V (b) 222 V (c) 230 V ( ) 238 V 12. The yoke of a dc machine

13.

14.

(b) inversely proportional to both its back emf (c proportional to its back emf ( ) directly proportional to its back emf but

15.

8. For reversing the direction of rotation of a dc motor, (a windings are required to be reversed (b) the connections of either the armature or the

16.

(c ( ) only the armature current need be reduced

current of a dc generator keeps on increasing, the emf generated (a (b) increases till the winding is burnt (c) increases till the magnetic saturation takes place ( then starts decreasing till it reduces to zero

(a) is always laminated (b) is never laminated (c) may or may not be laminated ( ) None of the above The back emf in a dc motor is given as (a) V + IaRa (b) V – IaRa (c) V ( ) IaRa Normally, a large number of commutator segments are used in a dc generator to (a) increase the magnitude of the output voltage (b) increase the output current (c) increase the kW power output ( ) make the dc output wave more smooth The residual magnetism is necessary for the operation of (a) a dc shunt generator (b) a dc shunt motor (c) a dc series motor ( ) a separately excited generator At the moment of starting a dc motor, its back emf is (a) zero (b) maximum (c) minimum ( ) optimum

553

DC Machines 19.

17. W

a c

b

a c

b

a c

b

a c

b

20.

18.

A N SW E RS

1. b 11.

2. a 12. c

3. 13. b

5. a 15. a

4. 14.

6. c 16. a

7. 17. c

8. b 18. a

9. c 19. c

10. a 20. b

PROBLEMS ( A )

S IM P L E

PRO BL EMS

1. Ans. a

b

5. [Ans. 2. W

[Ans.

[Ans.

3.

6. W

W Ans. 4. a b ( B)

TRI C KY

[Ans.

PR O BL EM S

7. [Ans.

554 W and

8. W

W

Ans. 14. Ans.

W

W

9. W

Ans. 15. W

W

W [Ans. W

10. W

Ra

Rsh

a

W

b c

W

[Ans. a

b

c

t

16. [Ans. W

11.

W

W

W

W

a b

[Ans. a

b

17. W

[Ans.

W

12.

Ans. 18. W

[Ans. 13.

( C )

a Ans. a C HA L L E NGI NG

W b

b W

PROBLEMS 20.

19.

I W

W [Ans.

[Ans.

555 21.

series with the armature, (b) a 30-W resistor were

of 250 W running at a speed of 1500 rpm on no load, it takes 5 A current from the mains. Calculate its speed when loaded so that it takes 50 A current from the mains. [Ans. 1364 rpm] 22. A 220-V, dc shunt motor takes a full-load current of 32 A while running at 850 rpm. It has an armature resistance of 0.5 W 110 W. Calculate the speed at which the machine runs, if (a) a 1.5-W resistor were introduced in

that the torque remains constant throughout and the [Ans. (a) 663 rpm; (b) 1061 rpm] 23. A 200-V, dc series motor runs at 1000 rpm and takes 20 A. Combined resistance of armature and series W. Calculate the resistance to be inserted in series to reduce the speed to 800 rpm, assuming that the torque varies as square of the speed. [Ans. 4.42 W]

EXPERIMENTAL EXERCISE 16.1 MA GN E TI S ATI O N CHARAC TERISTICS OF A D C S HU N T GEN ERATOR

Objectives 1. To plot the magnetisation characteristics of a dc shunt generator. 2.

Apparatus supply; Three-phase, 440-V ac supply; One 80-W, 5-A rheostat; Two dc voltmeters (0-300 V); One dc ammeter (0-2 A).

Circuit Diagram The circuit arrangement is shown in Fig. 16.20. Brief Theory

F (per pole) versus the Eg = FZNP 60A

Since Z, P and A are constant for a machine, if we keep the speed N constant, the emf Eg μ F. Further, if the number If. Hence, a plot between the induced emf E If represents the magnetisation characteristic. That is why this curve is often called (OCC) curve.

terminals are left open to measure the induced emf E by a high-resistance voltmeter V2. voltmeter V1.

1. Make the connections as shown in Fig. 16.20. The dc generator is mechanically coupled to the three-phase induction motor. Field winding terminals F-F1 are connected to 220-V dc supply through a rheostat. The voltmeter V2 is connected across the armature terminals A-A1. 2. By putting on the switch S2, connect the induction motor to three-phase ac supply. The induction motor drives the generator at a constant speed.

556

Basic Electrical Engineering

emf, Eg

3. Note the reading of voltmeter V2. This small voltage gives the measure of the residual magnetism. 4. Now, put on switch S1 by keeping the rheostat Rh at the minimum. If (as measured by ammeter A), by increasing the rheostat Rh, and note the corresponding values of induced emf E (as given by the voltmeter V2 winding (as given by the voltmeter V1). 6. Plot the magnetisation curve between V2 and If 1 and If. 7. Switch OFF Magnetisation curve

A

Field resistance line a O

Field current, If

DC Machines

Observations I

S. No.

V

V

4 5

Results

Precautions ON

Viva Voce 1. Ans. : 2. Ans. : E

3. Ans. : E

4. Ans. : E I

5. Ans. : 6. Ans. : 7. Ans. : 8. Ans. : 9. Ans. : 10. Ans. : 11. Ans. :

E EμN

557

558

Basic Electrical Engineering

12. Name the material of the brushes in dc machines. Why do you use this material? Ans. : It is carbon or graphite. We choose this material because it is soft, self-lubricating and highly conductive.

EXPERIMENTAL EXERCISE 16.2 S P E ED

C O N T ROL

O F

A

D C

SHUNT

MOTOR

Objectives 1. 2. To plot the speed versus armature voltage characteristic curve for a dc shunt motor. W, 2-A rheostat; One 25-W, 20-A rheostat; One dc voltmeter (0-300 V); One dc ammeter (0-2 A); One dc ammeter (0-20 A); One tachometer.

Apparatus

Circuit Diagram The circuit arrangement is shown in Fig. 16.22.

Brief Theory The equation governing the speed of a dc shunt motor is given as V - Ia Ra F where IaRa is the voltage drop across the armature, which is usually not more than 5 % of the terminal voltage V. Hence, we can say that V N∝ Nμ

Φ F

terminal voltage V. armature terminal voltage, and plot the speed control characteristics.

1. Make the connections as shown in Fig. 16.22.

If) and almost directly proportional to the

DC Machines Rac

559

R

ON

Rac R

I

R

I Rac

V

Speed, N

Speed, N

OFF

Rated value

Rated value 0 If (a) Speed versus field-current characteristic.

S. No.

I

0

V (b) Speed versus voltage characteristic.

N

V

N

Results

I

F V

Precautions ON

N

N

a b

560 Viva Voce 1. Ans. :

ON

2. N

Ans. :

ON

3. Ans. : Rac 4. Ans. :

5. Ans. : 6. Ans. :

t

kFIa

7. Ans. : 8. t

Ans. : E

kFIa

FRACTIONAL HORSE POWER MOTORS

17

OB JE CT IV ES

(i) (ii)

(iii)

( )

: (i) (PM)

( )

(VR)

(ii)

(iii)

In the previous chapters, we have considered the three-phase motors at some length. But it is not always convenient to have three-phase supplies available. The most common situation is in a house where almost universally we have single-phase supplies available.

Single-Phase Motors used in fractional horse-power ranges. In 1/8 hp to 1/4 hp range, these motors are widely used in fans, washing machines, refrigerators, blowers, centrifugal pumps, etc. Motors of very small sizes (1/300 hp to 1/20 hp) used in toys, hair dryers, vending machines, etc., are also single-phase motors. The ac series motors, also known as universal motors are widely used in portable tools, vacuum cleaners, and kitchen motors.

562

In this chapter, we shall also discuss the motors that are used in control systems. The control systems can be divided into two main categories : 1. Regulators. 2. Remote position control (RPC) systems.

Regulators Here, we seek to control the speed of the motor with high degree of accuracy. A dc shunt motor, described in Chapter 16, has good torque-speed characteristic for such applications. For better control, Fig. 17.1. DC supply

Control signal

Control amplifier

Fig. 17.1 The dc motor is relatively more expensive and less robust than the ac induction motor. Therefore, some engineers consider using a cage-rotor three-phase induction motor a better alternative for regulators. Its speed can be electronically controlled by just varying the frequency of the supply. There are other regulator motors such as the brushless dc motor, the switched reluctance motor, and the variable-frequency synchronous motor. But none of these proves to be better than the two considered above.

System An RPC motor moves its load to a position determined by the control system. Following two limitations are experienced by this system. 1. There is a limit to the mechanical ability of a motor to just move its load by the required angle. This limit occurs when the required angle of rotation becomes too small. can no longer drive the motor. These limitations have been overcome by following two arrangements, which we shall discuss in this chapter : 1. Geneva cam. 2. Stepper motor.

In physical appearance, a single-phase induction motor is similar to a three-phase squirrel cage induction motor. The stator has distributed single-phase winding. The rotor has a short-circuited squirrel cage winding. There is uniform air gap between the stator and the rotor.

Fractional Horse Power Motors

563

The problem with such a motor is that it has no starting torque. This can be seen from Fig. 17.2. For clarity

sinusoidal manner. Consider an instant when terminal B is positive with respect to terminal A and the current i action, emfs are induced in the rotor bars. Since the bars are shorted at the ends, according to Lenz’s law the half experience rightward forces. Moreover, the forces on the bars in the right half are leftward. As a result, the net torque developed is zero. in polarity and vary in strength sinusoidally. A

i

N AC supply

–

Bars Rotor

+ S

B

i

Fig. 17.2

three-phase induction motor does. B (along the vertical axis in Fig. 17.2) can be shown by a sinusoidal curve (Fig. 17.3a) and can be represented as B = –Bm sin wt where Bm

Double-Field Revolving Theory

B, sinusoidally pulsating along Br1 and Br2, each of magnitude Bm/2, but rotating in opposite directions. This is illustrated in Fig. 17.3. As shown in Fig. 17.3b, at wt Br1 and Br2 are shown by two equal and oppositely directed arrows along horizontal Br1 Br2 in clockwise direction, both at angular speed w. At later instants, when wt = 30°, 90° and 225°, the positions of Br1 and Br2 are shown in Figs. 17.3c, d and e, respectively. In each case, the horizontal components are equal and opposite and therefore cancel each other. The resultant is always along the vertical direction and has the same B.

564 B B3

B3 Br2 0

Br1

30° 90°

Br2

wt

225° 30°

225°

B1

w

30° Br2

Br1

w

Br1 225°

90° 90° Br1 Br2

B1 B2

B2 (a)

(b) wt = 0°

(c) wt = 30°

(d) wt = 90°

(e) wt = 225°

Fig. 17.3 Therefore, produce induction-motor action in the opposite direction. If the motor is moving, it must be moving in the direction) and rotating in the opposite direction to direction). In the forward direction, we have a motor action with a value of slip between 0 and 1. However, in the reverse direction, we have a plugging or braking action with a value of slip between 2 and 1. The total action is sum of these two actions. Fortunately, the motor action is stronger than the plugging action and therefore the device acts as a motor.

(Forward)

Torque

are similar to that of a three-phase induction motor, except that these are extended beyond the synchronous speed. The resultant of these two components is shown by the solid-line curve. It is seen to have zero value

(Forward)

(Backward)

–50 0 50 Percent synchronous speed

100

(Backward)

–100

Fig 17.4

(

)

(

)

Fractional Horse Power Motors

565

There remains only the problem of how to start a single-phase motor. Once the rotation commences, Fig. 17.4 indicates that a torque exists which speeds up the motor. TH E

RE M EDY

The remedy to the problem lies in somehow converting the single-phase motor into a two-phase motor, at least at the starting. But, before we go in for solving the problem, let us see whether a two-phase motor is capable of providing a starting torque.

Two-Phase Motor Consider a two-phase motor having two identical windings M-M1 and A-A1 placed at right angles on the stator (shown separately in Fig. 17.5a for clarity). These windings are excited by currents iM = IMp sin wt and iA = IAp sin (wt – p/2), respectively. That is, iA lags iM by 90° as shown in Figs. 17.5b and c. Further, let the two currents be of same magnitude. That is, IMp = IAp. The mmfs M and A produced by the peak values IMp and IAp will have same value Fp but displaced by 90°, as shown in Fig. 17.5a. Let us now see how the resultant of these two mmfs changes from instant to instant.

Fig. 17.5

566 At wt = 0, the currents iM = 0 and iA = –IAp. Therefore, M = 0 and A = Fp. Hence, the resultant mmf, = Fp in the vertically downward direction, as shown in Fig. 17.5d(i). At wt = p/4, the currents, iM = IMp/ 2 and iA = –IAp / 2 . Therefore, M = Fp/ 2 and A = –Fp / 2 , as shown in Fig. 17.5d(ii). The resultant mmf will be R

R

The resultant mmf

R

=

( Fp / 2 ) 2 + ( Fp / 2 ) 2 = Fp

will be directed at an angle p/4 from vertically downward direction.

At wt = p/3, the currents, iM = ( 3/2) IMp and iA = –IAp/2. Therefore, shown in Fig. 17.5d(iii). The resultant mmf will be R

=

M

= ( 3 / 2) Fp and

A

= –Fp/2, as

( 3/2) 2 Fp2 + ( Fp /2) 2Fp2 = Fp

The resultant mmf R will be directed at an angle p/3 from vertically downward direction. At wt = p/2, the currents, iM = IMp and iA = 0. Therefore, M = Fp and A = 0. Hence, the resultant mmf, = R Fp in the horizontal rightward direction, as shown in Fig. 17.5d(iv). Thus, we conclude that the resultant mmf in a two-phase motor of Fig. 17.5a rotates counterclockwise at speed w. Even if the two currents iM and iA are phase displaced by an angle less than 90°, there would be a . There are different ways of achieving the two-phase or nearly two-phase system from the single-phase. This gives rise to a variety of single-phase induction motors, described next.

modify the motor so that it actually becomes a two-phase motor at the time of starting. In most cases, the stator is provided with two windings, called main winding and auxiliary winding. The axes of these windings are displaced form each other by 90 electrical degrees. The currents to excite these two windings are derived currents is near about 90°, so that a starting torque may be developed. The direction of rotation of such motors can be reversed by interchanging the connections to either the main winding or the auxiliary winding.

(1) Split-Phase Motor Construction The stator has two windings—main and auxiliary (or starting) windings. The main winding occupies two-third of the stator slots and the auxiliary winding is placed in the remaining one third slots. Thus, a space displacement of 90° is obtained between the two windings. The main winding is made of thick wire and has low resistance and high reactance. But the auxiliary winding is made of thin wire and has high resistance and low reactance. Thus, the impedances of the two windings have different phase angles. Such a motor is schematically shown in Fig. 17.6a.

Starting Torque The main winding being mainly reactive, draws current IM that lags voltage V by an angle a little less than 90°. But the auxiliary winding being mainly resistive, draws current IA that lags the voltage by a small angle. Thus, as shown in Fig. 17.6b, the current IM lags the current IA by about 45°. This

Fractional Horse Power Motors

IA

I +

Auxiliary winding

IM

45°

Main winding

V

567

Centrifugal switch

V

IA

IM

I

– (a) Schematic diagram.

(b) Phasor diagram at the starting.

Fig. 17.6 way, the motor becomes equivalent to an unbalanced two-phase motor, and a starting torque is developed. Because of the high resistance of the auxiliary winding, this motor is also called resistance-start motor. When the motor picks up about 75 % of the synchronous speed, the auxiliary winding is disconnected by result in overheating which may even damage the motor.

Applications These motors have moderate starting torque with starting current 6 to 8 times the full-load current. Their ratings are from 1/20 hp to 1/2 hp. They are cheap and are used in washing machines, fans, blowers, centrifugal pumps, refrigerators, duplicating machines, grinders, etc.

This is an improved form of split-phase motor. Here, the time displacement between the currents of the main and the auxiliary windings is achieved by connecting a capacitor in series with the auxiliary winding (see Fig. 17.7a). The current IA in the auxiliary winding leads the applied voltage by some angle, and the current IM in the main winding lags the applied voltage by some angle, as shown in Fig. 17.7b. IA

I +

Auxiliary winding

IA

IM C V

Main winding

V

90°

I

Centrifugal switch

–

IM (a) Schematic diagram.

(b) Phasor diagram at the starting.

Fig. 17.7

Starting Torque By selecting the capacitor of suitable value, the current IA can be made to lead the current IM by about 90°. Thus, this motor develops a much larger starting torque than that developed by a resistance-start motor. The capacitor used is of electrolytic type, which is smaller and cheaper than a paper capacitor. However, an electrolytic capacitor and the auxiliary winding are designed for short time duty. Therefore, when the motor attains a speed of about 75 % of the rated speed, the auxiliary winding circuit is disconnected by a centrifugal switch.

568

Applications This motor gives an unusually high starting torque without drawing excessively high starting current. In addition, it is quite simple in construction. These motors are manufactured in various sizes from 1/8 hp to 5 hp to suit heavy-duty general purpose applications such as compressors, large fans, blowers, small portable hoists, jet pumps, etc. The size of the capacitor depends on the motor rating and varies from 50 mF to a few hundred mF.

Construction This motor uses the advantages of using a capacitor during its normal running as well as using a high-value capacitor at the starting. The capacitor required for optimum running performance is of low value, whereas we need a high-value capacitor for starting. Moreover, the capacitor for normal running should have continuous-duty rating, but that used for starting may have short-duty rating. As shown in Fig. 17.8, the capacitor C1 is a small-value (a few mF), continuous-duty, oil-impregnated paper capacitor. The other capacitor C2 is a much larger-value (50 mF to a few hundred mF), short-duty, electrolytic capacitor. I +

IA

Auxiliary winding

IM C2 Main winding

V

C1

Centrifugal switch

–

Fig. 17.8 Initially, when the motor is started, both the capacitors are in the circuit. After the motor has picked up enough speed, the centrifugal switch opens and disconnects the capacitor C2 from the circuit. The capacitor C1 remains in the circuit even during the normal running of the motor. Thus, this motor behaves more like a power factor.

Applications These motors are manufactured in a number of sizes from 1/8 hp to 5 hp. Though more costly, these are splendid machines where the load requirement is severe such as compressors, conveyer pumps, etc. Because of their quieter running, these motors are preferred in places like hospitals, studios, etc.

Construction In applications where the motor starts practically under zero loads, the cost can be reduced by eliminating one capacitor and the centrifugal switch. The permanent-capacitor motor uses only one oil-impregnated paper capacitor*, as shown in Fig. 17.9. Both the auxiliary winding and the capacitor remain in the circuit during starting as well as during normal running. Both the main winding and the auxiliary winding are made of equal rating. *

We cannot use electrolytic capacitor, as it is not suitable for continuous duty.

Fractional Horse Power Motors

IA

I +

569

Auxiliary winding

IM Main winding

V

C

–

Fig. 17.9

Applications Since this motor employs the same capacitor for starting as well as for running, it has neither optimum starting nor optimum running performance. Nevertheless, it has improved power factor and hp to 1 hp, and are used for table and ceiling fans, blowers, room coolers, portable drills and other loads requiring low starting torque.

Construction The shaded-pole motor is one of the most common motors to be found; the reason being its lowest cost of manufacture. The rotor is squirrel-cage type, as usual. But the stator has salient poles. There is only one winding, and that is a simple coil wound round the laminated salient poles. The coil does not have to be inserted into the slots in the stator, hence the low cost of manufacture. Low resistance copper band, called shaded rings, are placed to surround almost half of each pole, as shown in Fig. 17.10. Main winding + Single phase supply – Shaded ring

Fig. 17.10

Second Field *

*

The shaded ring acts like a short-circuited secondary winding of a transformer.

570

unshaded to the shaded portion of the pole. As the motor picks up speed, additional torque is created due to the single-phase induction motor action.

Performance The motor is However, since the shading rings are not taken out of the circuit after the machine has attained normal speed, an additional power loss occurs. It suffers from the drawbacks such as—(i) poor starting torque (about 50 % of the full-load torque), (ii) poor power-factor, (iii) very little overload capacity, (iv as 5 % in tiny motors and 35 % in larger motors), and (v) nonreversibility of the direction of rotation.

Applications These motors are manufactured in small sizes ranging from 5 W to 50 W. Some are available with built-in gear reducers to give any speed (even as low as 1 rpm). Because of their low starting as toys, small fans, electric clocks, hair dryers, slide projectors, desk fans, advertising displays, etc. EXA M P L E

17 . 1

a) its slip, and (b

Solution (a) The synchronous speed, Ns = \

Slip, s =

120 50 120 f = = 1500 rpm 4 P Ns N 1500 1410 = = 0.06 = 6 % Ns 1500

(b) The output power, Po The input power, Pin = VI cos f = 230 ¥ P 375 \ h = o = Pin 473.6 EXA M P L E

¥ 79.2 %

17 . 2

A 230-V, 50-Hz, split-phase induction motor has main-winding resistance of 5 W and inductive reactance of 12 W, and starting-winding resistance of 12 W and inductive reactance of 5 W a) the current in the main winding, (b) the current in the starting winding, (c) the line current, (d) the phase displacement between the two winding currents, and (e) the power factor.

Solution ZM = (5 + j12) W = 13–

W

and ZA = (12 + j5) W = 13–

(a) The current in the main winding, IM = \

230 ∠ 0° V = ZM 13 ∠ 67.38°

IM = 17.7 A

and

– fM

(b) The current in the starting winding, 230 ∠ 0° 13 ∠ 22.62° IA = 17.7 A and

IA = \

V ZA

=

– fA

W

Fractional Horse Power Motors

571

(c) The line current, IL = IM + IA = 17.7–– 67.38° + 17.7––22.62° = 32.73–– 45° A \ IL = 32.72 A (d) The phase displacement between the two winding currents, f = fA – fM = –22.62° – (– 67.38°) = 44.76° ª 45° (e) The power factor, = cos f = cos 45° = 0.707 (lagging) EX A MP L E

17. 3

A 230-V, 50-Hz, split-phase induction motor has the main-winding resistance of 2 W and inductive reactance of 20 W, and the auxiliary-winding resistance of 25 W and inductive reactance of 5 W. Determine the value of capacitance to be connected in series with the auxiliary winding to obtain maximum starting torque.

Solution For obtaining maximum torque, the phase angle between the main winding current IM and auxiliary winding current IA should be 90°. That is, fM – fA = 90°. Here, 20 X fM = tan–1 M = tan–1 = 84.3° RM 2 \ fA = fM – 90° = 84.3° – 90° = –5.7° 5 XC X XC Now, tan fA = A or tan (–5.7°) = or RA 25 Or –2.495 = 5 – XC fi XC = 7.495 W 1 1 \ C = = = 424.7 mF 2 f XC 2π × 50 × 7.495

– 0.0998 =

5

XC 25

A dc shunt or series motor can operate on both the dc and ac supplies. When ac is fed to a series motor, the direction* lag the armature current by such a large angle that a very low torque is developed. Hence, compared to a dc series motor, a dc shunt motor is not very suitable for ac operation. If an ordinary dc series motor is connected to an ac supply, it runs, but very unsatisfactorily because of the following reasons :

excessive heat. 3. Due to large voltages and currents induced in the short-circuited armature coils, vicious sparking occurs at the brushes during commutation period. armature circuits. **. *

**

To reduce the reactance of the armature

However, the torque developed is not of constant magnitude (as in a dc series motor) but pulsates between zero and maximum value each half-cycle. The dc series motor has only its rotor laminated.

572

Basic Electrical Engineering

winding, a compensating winding is added in the pole faces and connected in series with the armature winding, as shown in Fig. 17.11. The number of turns per pole in the compensating winding is so chosen that the mmf produced by it is equal and opposite to that of the armature.

Fig. 17.11 For making the power f

given core. So, we got to use more number of poles to produce required torque. The use of numerous poles lowering the frequency of the ac supply.

2

/3 The fractional kilowatt ac series motors have high speed (20 000 rpm) and large starting torque. Hence, etc.

A series motor specially designed to operate either on ac or dc at almost the same speed at the same output is called a universal motor

drive. These motors are widely used for sewing machines, table fans, vacuum cleaners, portable drills, dryers, blowers, electric shavers, projectors, etc. Their speed can be controlled by varying the voltage applied. It is to the armature as in a dc series motor. However, they are usually wound to run in only one direction.

Fractional Horse Power Motors

573

This is a simple ‘remote position control (RPC)’ mechanism used for moving the load to a position determined by the control system. It permits the motor to move without moving the load beyond the desired position. The basic Geneva cam is shown in Fig. 17.12. Here, it is assumed that the striker arm attached to the motor rotates clockwise. In order to see the action of the mechanism, we have marked one of the edges of the cam by a star. In Fig. 17.12a, the striker arm has just engaged in one of the slots on the cam. As the motor rotates clockwise, the cam rotates anticlockwise as long as the striker remains engaged in the slot. In Fig. 17.12b, the striker arm has rotated through 60° and the cam too has rotated through 60°. Further rotation of the striker brings the system to the position shown in Fig. 17.12c. Here, the striker has disengaged, leaving the cam rotated through exactly 90°. Cam

Striker leaving cam

Motor

* Identification mark

*

*

Striker entering cam (a)

(b)

Identification mark (c)

Fig. 17.12 the slots of the cam and rotates the load through exactly 90°. For the remaining part of the rotation of the motor, there is no engagement. So, here it does not matter if we cannot stop the motor in exactly the required position. Rather we can stop the motor anywhere when it is not engaged. Thus, the Geneva cam provides mechanical equivalent of a digital bit. The Geneva cams have now been largely superseded by the stepper motors except in systems that require large torques, and where the positioning of the load in steps of quite large angles is acceptable.

These motors, also called stepping motors or step motors angular step in response to each input current pulse from its controller. The industrial motors cannot be used for precise positioning of an object or precise control of speed without using closed-loop feedback system. The stepper motors are able to provide very precise position and/or speed control without needing any feedback loop. coil. The unique feature of a stepper motor is that its shaft rotates in a series of discrete steps in response to command pulses received from an electronic control circuit. By counting the pulses, we can determine the exact rotation achieved. This basic control is illustrated in Fig. 17.13. Note that the control has two elements— the number of pulses which determine the angle of rotation, and the direction data which determines the order in which the stator poles are excited.

574 Step pulses Electronic control circuit

Drive controller

Stepper motor

Direction

Fig. 17.13 Stepper motors are designed to develop torques ranging from 1 mNm (in tiny wrist watch motor of 3 mm diameter) up to 40 Nm in a motor of 15 cm diameter used for machine tool applications. Their output power ranges from about 1 W to about 2500 W. The only moving part in a stepper motor is its rotor which has no winding. Hence, it needs no commutator and brushes. This feature makes the stepper motor quite robust and reliable.

Based on their construction and working, the stepper motors can be divided into the following three basic categories : 1. Variable reluctance (VR) stepper motors. 2. Permanent magnet (PM) stepper motors. 3. Hybrid stepper motors.

It has wound stator poles but the rotor poles are made of a ferromagnetic material. It is called variable reluctance motor because the reluctance of the magnetic circuit formed by the stator and rotor poles varies with the angular position of the rotor. Figure 17.14 explains the working of a stepper motor with the stator having six poles and rotor having four poles. The rotor pole-width is same as the stator pole-width. The coils on the diametrically opposite statorpoles are connected in series to form one circuit or phase such that on energising it, one pole becomes N-pole and the other S-pole. Thus, there are three independent circuits or phases A, B and C. Each of these phases arrangement to supply current to the three phases in proper sequence is shown in Fig. 17.14e. When none of the coils is energised, the rotor is completely free to rotate. On energising one or more coils, the rotor assumes a position that forms a path of least reluctance with the magnetised stator poles. The motor has following four modes of operation.

Suppose that the switch S1 (in Fig. 17.14e) has been closed for energising the phase A. The nearest rotor poles are attracted by N and S poles of the stator so that the rotor takes the position shown in Fig. 17.14a. Let us mark by a star the rotor pole just under the N-pole of the stator. In this position, the rotor poles at right angles to the marked pair have no function whatsoever since they are effectively neutral. Let us refer to these poles as unmarked pair.

575

Fractional Horse Power Motors 30°

B

N

B

C¢

B

C¢

N

*

*

A

60°

S

N

*

C¢

A

S

A

B¢

C

S

B¢

C

B¢

A¢ (a)

C

A¢

A¢

(b)

(c)

A 90° C¢

N

B

A

A¢

B

B¢

C

*

Truth Table 1

S2

C¢

S3 B¢

S

A

B

C

q

A

+

0

0

0°

B

0

+

0

30°

C

0

0

+

60°

A

+

0

0

90°

S1

C V

A¢ (d)

(f ) One-phase oN mode.

(e)

Fig. 17.14

VR

Now, let us open the switch S1 and close the switch S2 to energise the next phase B. Note that the next phase is selected by going around the stator. The rotor poles of unmarked pair are now nearest to the new position of alignment. The rotor therefore rotates by 30° to align along the path of least reluctance (see Fig. 17.14b). Again, let us open the switch S2 and close switch S3 to energise the phase C. Once again, the nearest rotor poles to align are from the marked pair. These poles align themselves by rotating a further 30° (see Fig. 17.14c). If we again close switch S1 to energise phase A, the rotor further rotates by another 30° (see Fig. 17.14d closing the switches in sequence 1-2-3-1-2-…, that is, by energising the phases in sequence A-B-C-A-B-…, the rotor rotates anticlockwise in steps of 30°. Now, if the phases are energised in the reverse sequence, A-C-B-A-C-…, the rotor will rotate in steps of 30°. Since only one phase is energised at a time, this mode of operation is called one-phase ON operation. The stator phase switching truth table for this operation is given in Fig. 17.14 .

576 Note that the stator pole of either magnetic polarity attracts the rotor pole (by inducing opposite polarity).

because

b) The angle through which the shaft of the stepper motor rotates for each command pulse is called its step angle. Obviously, the smaller the step angle, the greater is the number of steps per revolution and the higher is the resolution or accuracy of positioning obtained. There are two equivalent ways of expressing the value of step angle, b. If Ns and Nr are the number of poles (also called teeth) on the stator and on the rotor, respectively, and m is the number of stator phases, then 1. b =

( Ns ~ N r ) ¥ 360° Ns ◊ N r

(17.1)

2. b =

360∞ mNr

(17.2)

For example, for the stepper motor shown in Fig. 17.14, we have Ns = 6, Nr = 4, and m = 3. Therefore, the value of step angle, using above two formulae, is (6 − 4 ) ( Ns ~ N r ) b= ¥ 360° = ¥ 360° = 30° N s ◊ Nr 6×4 360° 360∞ and b= = = 30° 3×4 mNr

Resolution The resolution of a stepper motor is given by the number of steps needed to complete one revolution of the rotor shaft. That is, Number of steps 360° Resolution = = (17.3) Number of revolutions β Speed A stepper motor has the extraordinary ability to operate at very high stepping rate (up to 20 000 steps per second in some motors), and yet to remain fully in synchronism with the command pulses. The shaft rotation appears continuous, if the pulse rate is high*. If is the stepping frequency (or the pulse rate) in pulses per second (pps) and b is the step angle, the shaft speed n is given by b Speed, n = rps (17.4) 360∞ EX A MP L E

17. 4

The step angle of a stepper motor is 2.5°. If the stepping frequency is 3600 pps, determine (a) the resolution, (b) the number of steps required for the shaft to make 25 revolutions, and (c) the shaft speed.

Solution 360° 360 = = 144 steps per revolution β 2.5 (b) The number of steps required for the shaft to make 25 revolutions, = 144 ¥ 25 = 3600 (a) The resolution =

*

The operation of stepper motor at high speed is called ‘slewing’. While slewing, the motor generally gives out an audible whining sound of fundamental frequency same as the stepping rate.

Fractional Horse Power Motors

(c) The shaft speed, n = EXA MP LE

577

2.5° × 3600 b rps = = 25 rps 360∞ 360°

1 7 . 5

A three-phase VR motor has a step angle of 15°. Find the number of its rotor and stator poles.

Solution Using Eq. 17.2, we have b=

360∞ mNr

Nr =

or

360° 360∞ = =8 mb 3 × 15°

For determining the number of stator poles, we use Eq. 17.1. Two possibilities, as given below, must be considered. ( N − 8) ( Ns - N r ) (i) When Ns > Nr : b = ¥ 360° or 15° = s ¥ 360° fi Ns = 12 N s ◊ Nr Ns × 8 (ii) When Ns < Nr : b =

( Nr - N s ) ¥ 360° N s ◊ Nr

or

15° =

(8 − Ns) ¥ 360° fi Ns × 8

Ns = 6

In this mode of operation, two stator phases are energised simultaneously. When phases A and B are energised together, the rotor comes to rest at a point midway between the two adjacent full-step positions (say, the marked pair will be at 15° anticlockwise from the reference given in Fig. 17.14). Now, if the stator phases are switched in the sequence AB, BC, CA, AB…, etc., the rotor will take full steps of 30° anticlockwise each time (as in one-phase on mode), as shown in the truth table No. 2 in Fig. 17.15a. Note that the equilibrium positions in this case are interleaved between those of the one-phase ON mode. Truth Table 3

Truth Table 2 A

B

C

A

B

C

q

0

0

0°

AB

+ +

+

0

15°

B

0

+

0

30°

BC

0

+

+

45°

C

0

0

+

60°

CA

+

0

+

75°

A

+

0

0

90°

q A

AB

+

+

0

15°

BC

0

+

+

45°

CA

+

0

+

75°

AB

+

+

0

105°

(a) Two-phase ON mode.

(b) Half-step operation.

Fig. 17.15

It is possible to have ‘half-stepping’ or ‘half-step operation’ by alternately having ‘one-phase ON’ and ‘two-phase ON’ modes. That is, by energising the three phases in the sequence A, AB, B, BC, C, CA, A…, etc. This operation is illustrated in Fig. 17.16, where the rotor moves by only 15° with each switching. Note that here the phase sequence is selected by going around the stator. The truth table for this operation is given in Fig. 17.15b.

578 15°

q = 0°

B

6

A

A

A

1

1

1

C

2

1 4

5

30°

C¢

B

6

C¢

B

6

2 3

B¢

C

5

2

1

C¢

4

4

2 Rotor 3

2

1

3

3

B¢

C

5

2

3

3

4

4

4

A¢

A¢

A¢

(a) Phase A excited.

(b) Phases A and B excited.

(c) Phase B excited.

B¢

Fig. 17.16 Note that in half-stepping operation, the step angle is halved (or, in other words, the resolution is doubled). This improvement has been done without increasing the number of poles, but by applying different scheme of switching.

It is also known as ministepping. This operation requires excitation of two phases simultaneously as in twophase ON operation, but with one difference. Here, the current in one phase is held constant at its maximum, while the current in the other phase is increased in very small steps till it reaches its maximum. After this, Using this technique, the resulting step becomes very small and hence rightly called a microstep. For instance, consider a VR stepper motor with a resolution of 200 steps/rev. This corresponds to a step angle, b = 360°/200 = 1.8°. With good electronic circuits available, it is possible to accurately increase or decrease the exciting current (say, 1 A) in 100 steps of 10 mA each. By employing the microstepping technique, the resolution of the VR motor can be increased to 20 000 steps/rev. Consequently, the step angle becomes as small as b = 0.018°. Thus, the microstepping provides smooth low speed operation with high resolution. possible disadvantage is the absence of detent torque needed to retain the rotor at the step position in case of power failure.

All the VR stepper motors discussed above are single-stack. Another technique for increasing the resolution is to have multistack.

Construction Figure 17.17a shows a three-stack VR motor. It has three magnetically isolated sections or stacks which can be excited by separate windings or phases. The three stators have common frame. The three rotors are assembled on the same shaft such that the rotors are aligned with the corresponding stators. Both the stators and the rotors have the same number of teeth.

Fractional Horse Power Motors

579

Stator A

iA

iB

B

C

iC Stator Teeth

Stack A Rotor

Stack B Rotor

Stator stack A

Stack C Rotor

Stator stack B

Rotor Teeth

Stator stack C

Stator (a) Construction.

(b) Alignment of rotor teeth with teeth of three stacks, when phase A is excited.

VR

Working b. It may be seen here that the teeth of rotor are

by another step. Three-stack motors are most common, although stepper motors with stacks up to seven are available. step angle, b EXA M P L E

¥ 17 . 6

Solution A four-stack stepper motor has four phases. That is, m b=

360∞ mNr

or

Nr =

360° 360∞ = = 50 mb 4 × 1.8°

Since in a multistack stepper motor, the number of stator teeth is the same as that of rotor teeth, Ns = Nr = 50.

580 the rotor is cylindrical and has radially magnetised permanent magnet. Unlike a direction of rotation depends on the polarity of stator winding current.

stepper motor, here the

energised by one winding only, the motor has only two windings or phases (marked as phase A and phase B). Hence, the step angle is 4−2 N - Nr b= s ¥ ¥ N s ◊ Nr 4×2 360∞ 360° or b= = 2×2 m ◊Nr A

A q = 0°

iA

S

90°

N N S

B

B¢ B

N

S

S B¢

iB

N

A¢

A¢ (a)

A

(b) A

q = 180°

q = 270°

N

S S N

B

S

N

B¢ B

iB

S iA A¢

A¢ (c)

Fig. 17.18

N

(d)

B¢

Fractional Horse Power Motors

581

Working There are two positions of the rotor in which this alignment can occur depending upon the polarity of the ¢ as A+ current iA is positive (that is, when it is entering terminal A and leaving terminal A¢ a), and A– iA is negative (that is, when it is entering terminal A¢ and leaving terminal A, as c). 1. 2. 3. 4.

One-phase ON mode, full-step operation. Two-phase ON mode, full-step operation. Alternate one-phase ON and two-phase ON mode, half-step operation. Microstepping.

mode Suppose that we start with A+

One-phase

a. The magnetic

rotor with q b, c and d sequence of A+, B+, A–, B–, A+…, the rotor rotates clockwise a. The rotation can be reversed just by changing the polarity anticlockwise operation, the required switching would be in the sequence A+, B–, A–, B+, A+…, etc. Truth Table 4

Truth Table 6

Truth Table 5

A

B

q

A

B

q

+

+

0

0°

A B

+ +

+

+

45°

+

0

+

90°

B A

+ –

–

+

135°

A

B

A

–

–

0

180°

– –

A B

–

–

225°

+

0

0°

+

+

45°

+

0 –

+

90°

+ –

+

135°

–

–

180°

– –

–

0 –

–

0

–

270°

– +

+

–

315°

+

+

0

0°

A B B

B A

A A B

B A

–

0

+

+

– 0

270°

– +

B A

+ +

0°

A B

(a) One-phase ON mode, full-step operation.

+ +

– +

315° 45°

q

+ +

A A

B

+

B B A

A

(b) Two-phase ON mode, full-step operation.

225°

(c) Alternate one-phase ON and two-phase ON mode, half-step operation.

Fig. 17.19

Two-phase

b. Note

it is possible to have full-step positions of the rotor midway between the two adjacent positions of one-phase ON

(i) For clockwise (ii) For anticlockwise

+

B+, B+ A–, A– B–, B– A+, A+ B+, … + – B , B– A–, A– B+, B+ A+, A+ B–, …

Alternate One-phase and Two-phase c represents half-stepping when one-phase ON and two-phase ON modes are used alternately. The sequences of switching +

(i) For clockwise

, A+ B+, B+, B+ A–, A–, A– B–, B–, B– A+, A+, … +

(ii) For anticlockwise

, A+ B–, B–, B– A–, A–, A– B+, B+, B+ A+, A+, …

Microstepping Here, both phases are energised simultaneously as illustrated in Truth Table No. 5 the current in the other phase is increased or decreased in very small steps. This technique greatly reduces the step size.

small permanent magnet rotors with large number of poles, it has relatively large step size and hence low resolution.

motors, this motor can also be a single-stack or multistack motor, and can operate in one-phase two-phase ON mode, or alternate one-phase ON and two-phase ON modes.

ON

mode,

Construction b. x

y

A

A

S N B

N

N N

N N A¢

N

S

N

S

B¢

S B

S

S

S S N A¢

Stator Outer x¢ casting

(a)

S

Stator winding Shaft

Air gap

(b)

y¢ (c)

B¢

Fractional Horse Power Motors

- ¢ and y-y¢

a and c, respectively. Note that both the

Thus, all the teeth of the end-cap on the left are N-poles, and all the teeth of the end-cap on the right are tooth-pitch with respect to the other. It means the tooth of one end-cap coincides with the slot of the other. The step angle of such a motor is (5 − 4) ( N r ~ Ns ) b= ¥ ¥ N s ◊ Nr 5×4

Working Here, we shall discuss one-phase

ON

mode only* +

), the top stator pole becomes S-pole and the bottom pole becomes N-pole. The rotor aligns so that one of the N-teeth of left-hand end-cap and one of ¢ de-energised and phase B is energised positively, the left-hand stator pole becomes S-pole and the right-hand pole becomes N-pole. The rotor has to rotate anticlockwise cap is aligned along the B-B¢ axis. Truth Table 7 A

B

q

A+

+

0

0°

B+

0

+

18°

A–

–

0

36°

B–

0

–

54°

A+

+

0

72°

17.20 Thus, we can rotate the rotor in full steps, either clockwise or anticlockwise by the phase sequences as (i) For anticlockwise (ii) For clockwise

+

, B+, A–, B–, A+, B+, … + , B–, A–, B+, A+, B–, A–, …

resolution. The resolution is further increased by providing more stacks.

Advantages The hybrid motor achieves much smaller step sizes easily with a simple magnetic structure.

*

ON

mode.

mode and the alternate one-phase ON and two-phase ON

SUMMARY TE R MS

A N D

CO NC E PTS -phase winding, and its rotor is squirrel cage.

magnitude but rotating in opposite directions with synchronous speed. This is known as motor is not self-starting.

split-phase motor, capacitor-start motor, capacitor-start capacitor-run motor, and shaded-pole motor. reactance of the armature winding. Furthermore, they are operated at low frequencies, such as 16 2/3 Hz and 25 Hz is an ac series motor designed to work on either ac or dc at very high speeds. is a digital electromagnetic device where each input pulse results in a discrete output (i.e., the ). (VR which minimises the magnetic circuit reluctance. The direction of rotation is independent of the polarity of excitation. (PM) stepper motor has a permanently magnetised cylindrical rotor. The direction of rotation depends on the polarity of the excitation. The direction of rotation depends on the polarity of the excitation. ON ON

positions are midway between the two consecutive positions of one-phase ON mode. can be achieved by alternate one-phase ON mode and two-phase ON mode. very small steps, using an electronic controller. IM P O RTA N T

F O RM U L AE

,b = where,

( Ns ~ N r ) ¥ N s ◊ Nr

b=

360∞ . mN r

Ns = the number of stator poles, Nr = the number of rotor poles. m = the number of phases in single-stack, or the number of stacks in multistack.

Fractional Horse Power Motors

585

CHECK YOUR UNDERSTANDING Before you proceed to the next chapter, take this test. Give yourself two marks for each correct answer and minus one for each wrong answer. If your score is 12 or more, go to the next chapter; otherwise study this chapter again. S. No.

Statement

1.

A single-phase induction motor can be thought of as a composite of two motors mechanically coupled. If one of these has a slip of s, the slip of other motor is 1 – s.

3. 4. 5. 6. 7. 8. 9. 10.

True

False

Marks

synchronous speeds. The cheapest motor in the fractional horse power motors is the shaded-pole motor. In a split-phase single-phase induction motor, the main winding and the auxiliary winding occupy equal number of stator slots. The auxiliary winding in a split-phase induction motor has higher X/R ratio as compared to the main winding. A split-phase motor is also known as a resistance-start motor. A capacitor-start capacitor-run motor has two capacitors, one is electrolytic capacitor, and the other is oil-impregnated paper capacitor. The starting torque of a shaded-pole motor is low. To provide very precise position control of an object, the stepper motors require a feedback loop. While switching the phases of a VR stepper motor, if the selection of phases is done clockwise around the stator, the rotor moves anticlockwise with each switching. Your Score A N SW E RS

1. False

2. True

3. True

4. False

5. False

6. False

7. True

8. True

9. False

10. True

REVIEW QUESTIONS 1. Explain why a single-phase motor does not develop a starting torque. 2. What are the main differences between a threephase and single-phase induction motors ? With the help of neat diagrams, explain the working principle of split-phase induction motor. 3. phase motors.

4. Prepare a table showing the horse power rating and applications of each type of single-phase induction motor. 5. Explain why the auxiliary winding in a capacitorstart motor should be disconnected after the motor has picked up speed. 6. Describe the construction and principle of working of a capacitor-start capacitor-run single-phase motor.

586 7. Explain in what ways does a capacitor-start motor differs from a capacitor-start capacitor-run motor. 8. Explain the working principle and application of single-phase shaded-pole motor. Is it possible to reverse the direction of rotation of such a motor ? If yes, how ? If not, why not ? 9. Explain why ac series motors are preferred for electric locomotives. 10. What is a universal motor ? Where is it used ? 11. Explain the working of a universal motor. What

motor so that it can operate as a universal motor ? 12. What is a stepper motor ? Why is it named so ? Does it run on ac or dc ? 13. State the different types of stepper motors and 14. Explain the working of a VR stepper motor in its half-step operation. Is there any other name given to this type of operation ? 15. What is a hybrid stepper motor ? In what way is it better than the VR stepper motor ?

MULTIPLE CHOICE QUESTIONS Here are some incomplete statements. Four alternatives completes the statement correctly. 1. In a resistance split-phase single-phase induction motor, a time-phase difference between the currents in the main and auxiliary windings is achieved by (a) placing the two windings at an angle of 90° electrical in the stator slots (b) applying two-phase supply across the two windings (c) having different ratio of resistance to inductive reactance for the two windings supplied from a single-phase supply (d) connecting the two windings in series across a single-phase supply 2. The direction of rotation of an ordinary shadedpole single-phase induction motor (a) can be reversed by interchanging the supply terminals connected to the stator winding (b) can be reversed by open-circuiting the shading rings (c) can be reversed by short-circuiting the shading rings (d) cannot be reversed 3. The direction of rotation of split-phase single-phase induction motor can be reversed by (a) interchanging the supply terminals connected to the stator winding (b) reversing the terminals of only the auxiliary winding across the supply (c) reversing the terminals of only the main winding across the supply

(d) reversing the terminals of either the main winding or the auxiliary winding across the supply 4. If the centrifugal switch of a resistance split-phase single-phase induction motor fails to close when de-energised, then (a) no starting torque will be developed when an attempt is made to restart (b the main winding when an attempt is made to restart (c enable the motor to run when an attempt is made to restart (d) the motor will develop extremely high torque when an attempt is made to restart 5. The motor used in the ceiling fans in homes is (a) the split-phase motor (b) capacitor-start motor (c) the shaded-pole motor (d) the ac series motor 6. In a capacitor-start capacitor-run motor, the two capacitors (a) have similar construction (b) are of different type (c) have equal capacitance (d) are disconnected when the motor attains its rated speed 7.

(a

Fractional Horse Power Motors (b)

3 / 2 times the amplitude of the pulsating

(c (d) one-fourth the amplitude of the pulsating 8. In a resistance split-phase single-phase motor, (a same X/R ratio (b X/R ratio than the main winding (c X/R ratio than the main winding (d) Any of the above 9. In a capacitor-start single-phase induction motor, the capacitor is connected (a) in series with the main winding (b (c (d) in parallel with the main winding 10. As compared to a resistance split-phase motor, the capacitor-start motor has (a) higher starting torque (b) lower starting torque (c) higher running torque (d) lower running torque 11. The main purpose of providing a compensating winding in an ac series motor is (a) to improve commutation (b) to reduce the reactance of the armature winding (c) to provide starting torque (d) All of the above 12. The electric motor used in toys which run on ac mains, is usually (a) capacitor-start motor (b) split-phase motor (c) capacitor-start capacitor-run motor (d) shaded-pole motor

587

13. If the centrifugal switch of the split-phase induction motor fails to open on running, (a) the motor will come to a stop (b) the main winding will get overheated (c (d) All of the above 14. The capacitor-start motor uses (a) a paper capacitor (b) an electrolytic capacitor (c) an air capacitor (d) Any of the above 15. Which of the following motors does not use a

16.

17.

18.

19.

(a) split-phase motor (b) capacitor-start capacitor-run motor (c) capacitor-start motor (d) shaded pole motor The step angle of a four-phase, 6-pole variable reactance stepper motor is b (a (c d The step angle of a permanent magnet stepper motor having 8 stator poles and 4 rotor poles is (a b (c d The operation of stepper motors at high speeds is referred to as (a) fast forward (b) slewing (c) jogging (d) inching Which one of the following phase-switching

(a b) (c d) 20. The rotor of a stepper motor has no (a) winding (b) commutator (c) brushes (d) All of the above A N SW E R S

1. c

2. d

3. d

4. a

5. b

6. b

7. a

8. c

9. b

10. a

11. b

12. d

13. c

14. b

15. d

16. a

17. c

18. c

19. d

20. d

588

PROBLEMS ( A )

S IM P L E

PRO BL EMS

1. Determine step angle of a single-stack, four-phase, 8/6-pole, VR stepper motor. What is its resolution ? [Ans. 15°, 24 steps/rev] 2. A stepper motor has a step angle of 1.8°. What number should be fed into the encoder of its driver ( B)

TRI C KY

P RO B LE MS

4. A 4-pole, 500-W, 230-V, 50-Hz capacitor-start induction motor takes a full-load line current of 5.6 A at a power factor of 0.62 (lagging) while running at 1440 rpm. Determine (a) its slip at fullload, (b c) its full-load torque. [Ans. (a) 0.04; (b) 62.6 %; (c) 3.316 Nm] 5. A stepper motor has a step angle of 1.8° and is driven at 4000 pulse per second. Determine (a) the ( C)

system if it is desired to turn the shaft by twelve complete revolutions ? [Ans. 2400] 3. Calculate the pulse rate required to obtain rotor speed of 2400 rpm for a stepper motor having a resolution of 200 steps/rev. [Ans. 8000 pps]

C H A LL EN GI NG

resolution, (b) the speed, (c) the number of pulses needed to rotate the shaft by 54°. [Ans. (a) 200 steps/rev.; (b) 1200 rpm; (c) 30] 6. A stepper motor is required to rotate the shaft by 180° in 0.025 s. If the pulse frequency available is 6000 pps, what resolution in steps/rev should the stepper motor have ? [Ans. 300 steps/rev]

PROBLEM S

7. A 230-V, 50-Hz, split-phase induction motor has the main winding resistance of 5 W and inductive reactance of 12 W, and the auxiliary winding resistance of 12 W and inductive reactance of 5 W. Determine the value of capacitance to be connected in series with the auxiliary winding to obtain maximum starting torque. [Ans. 318.3 mF]

8. A 230-V, 50-Hz, split-phase induction motor has the main-winding resistance of 1 W and inductive reactance of 10 W, and the auxiliary-winding resistance of 20 W and inductive reactance of 5 W. Determine the value of capacitance to be connected in series with the auxiliary winding to obtain maximum starting torque. [Ans. 455 mF]

ELECTRICAL MEASURING INSTRUMENTS

18

OB JE CT IV ES :

(i)

: (i)

(ii)

: (i)

(ii)

(ii)

(iii) (PMMC)

(i) Absolute instruments (ii

Absolute Instruments tangent galvanometer

Secondary Instruments

(iii) integrating

590

Basic Electrical Engineering

(i (ii (iii

Indicating Instruments

Recording Instruments

Integrating Instruments

No.

Effect

Instrument

Suitable for measurement of

Electrical Measuring Instruments

591

t I

restraining torque or restoring torque

i (ii tc

(i) Spring Control

tc a tc is

A

b

592

Basic Electrical Engineering

50

0

B

Graduated scale A

10

0

S

Pivot Spindle

Pointer

L

P

Control spring Balance weight

Pivot (a) Single spring.

I

(b) Double springs.

tc t = tc qμI disadvantages

q

t is

(ii) Gravity Control

a q

W sin q

b tc = (W sin q) ¥ L = WL sin q t tc = t

I or

WL sin q = kI fi

Ê WL ˆ ˜ sin q k ¯

I= Á Ë

or

I μ sin q

Electrical Measuring Instruments Volts Pointer deflected through angle, q

Pointer at zero position

q

L

0

q r q

Balance weight

q

W sin q

Pivoted spindle r

Control weight

(a) The arrangement.

W (b) The gravity force acting in deflected position.

advantages (i (ii (iii

annoying wait for some time

593

594

Basic Electrical Engineering

Fig. 18.3 RE M E DY The irritation and the time delay in taking the reading can be avoided if an extra torque is provided to kill (or ‘damp’) important to provide correct amount of damping torque time delay in taking the reading becomes unnecessarily long. This condition is described as overdamped. the system remains underdamped increasing the delay-time to take the reading. The instrument satisfying this condition is said to be critically damped or ‘dead beat’ (see Fig. 18.3b).

Methods of Obtaining Damping Torques The purpose of damping torque is to o the . Different methods of obtaining damping torque

(1) Air Friction Damping A ligh a the pressure falls and the motion is again opposed. Damping due to air friction can also be achieved by attaching to the moving system a vane b).

Electrical Measuring Instruments Piston Air chamber

Air chamber

Spindle Pointer

Vane Spindle

(a) Piston in an air chamber.

(b) Vane in an air chamber.

(2) Fluid Friction Damping

(3) a

N D

S M Pointer

Permanent magnet

Damping vane M D (a) Eddy currents in a metal disc.

b

(b) Eddy currents in a metal wave.

595

596

Basic Electrical Engineering

(1) Permanent Magnet Type : (2) Dynamometer Type :

a

as d’Arsonval movement

0

1

2

3

4

5

6

7

8

9

M

Scale

10

N Face plate

S P

P

X

X

Pointer C Permanent magnet

B A

Coil

C Fixed iron core

P

B

P

Pole pieces (a) The d¢Arsonval movement.

A (b) The front view and sectional plan along X-X.

(PMMC)

Electrical Measuring Instruments

b

F-F.

d

t = F ¥ (d/2) + F ¥ (d/2) = Fd F I acting on the coil is given as t = kI k tc = cq

c

or cq = kI fi

tc = t

q=

k I c

F

P N

S Q

F

q

q

597

598

Basic Electrical Engineering

The main advantages of the permanent magnet moving coil (PMMC) instrument are : (i (ii) Uniform scale. (iii (iv (v) Effective and reliable eddy-current damping. The main disadvantages are : (i) Cannot be used for ac measurement. (ii) More expensive compared to moving-iron type. (iii) Ageing of control springs and of the permanent magnets might cause errors.

Fig. 18.8 EXA M P L E

18 . 1 1

Solution

1

F1 = IBl ¥ n

¥

¥

¥

¥

–6

N

F = IBl ¥ n

¥

¥

¥

¥

–6

N

td = t1 + t = F1r + F r = (F1 + F )r

¥

–6

¥

63.75 ¥ 10 –6 Nm

Electrical Measuring Instruments

599

*

a

square-law the mean value of the square of the current

b

c

The main advantages (i (ii *

600

Basic Electrical Engineering

(iii (iv

transfer

The main disadvantages (i (ii

(iii (iv (v (vi

i) the attraction type

ii) the repulsion type

Pointer

Air damping chamber

Coil winding

Balanced weight

Moving iron Control weight

Electrical Measuring Instruments

instruments are unpolarised

10

15

0

5 20

Pointer Control spring Air damping device

Fixed coil

The main advantages (i

Moving iron

Fixed iron

601

602

Basic Electrical Engineering

(ii (iii (iv The main disadvantages (i (ii (iii (iv (v

on th

An ammeter series low resistance Ideally, an ammeter should have zero resistance A voltmeter parallel have very high resistance

very

Rm)

current sensitivity W Im voltage across its terminals is given as Vm = Im ¥ Rm

¥

W

Electrical Measuring Instruments

W

Im

W

Rsh

603

a Rsh I

Rm

Rm = 500 W

Im = 0.1 mA a

Im

b Im

Rsh

Ifsd

c (I – Im)

d 500 W

I

I

–

+ (a) Doubling the range.

Rsh

Ish Shunt, Rsh Potential terminals Current terminals

(b) Range extended many times.

b current terminals

potential terminals. I range multiplier*

I /Im = N resistance Rm

(a current much larger than the meter current Im Rsh I Im

Im Rm = (I

– Im)Rsh

\

Rsh =

I m Rm ( I fsd - I m )

or

Rsh =

Rm I m Rm Rm = = ( I fsd I m ) ( I fsd /I m - 1) ( N 1)

Multi-Range Ammeter Rsh

*

I

– Im = Ish

Ifsd

604

Basic Electrical Engineering 0.1 mA, 500 W

55.55 W

1 mA

I

I –

+ 10 mA 5.0505 W 0.5005 W

100 mA

EX A MP L E

18. 2 m

W

Solution Ish = I

Im Rm = Ish Rsh

m

– Im

or Rsh =

100 × 100 × 10 − 6 I m Rm = = 1.010101 W Ish 9.9 × 10 − 3

m

W

W

W

universal-shunt or ring-shunt

Rm universal-shunt R1 R2 R3 R4

R5

ring

Im Rm Rm R5

R1 R1 R2 R3

R4

Electrical Measuring Instruments

+

605

– Im, Rm

R1

B

R2

C

R3

R4

D

E

R5

A

10 mA 100 mA 500 mA 1 mA

1A

Range switch

–

+

EX A MP L E

18. 3 m W

Solution Im

m −6

Rsh =

100 × 100 × 100 I m Rm = ( I fsd - I m) (1 × 10 − 3 − 0.1 × 10 − 3)

W

resistor RT W

RT W

Rm

W

RT preset adjustment

W W

W

RT W RT Rm

(a) Range-switch at Rm1 = Rm W Ish1 Rsh1 = R1 + R2 + R3 + R4 + R5 = R

606

Basic Electrical Engineering

100 mA, 900 W + R1

RT

– R2

R3

R4

R5

10 mA 100 mA 500 mA 1 mA

1A Range switch

+

Rsh1 Rsh1 =

–

900 × 100 × 10 − 6 R m1I m = I sh1 0.9 × 10 − 3

W

(b) Range switch at R m2 = Rm + R1 R1) W; Ish2 Rsh2 = R2 + R3 + R4 + R5 = R – R1 Rsh2 = \ fi

Rm2 I m (900 + R1) × 100 × 10 = Ish2 9.9 × 10 − 3

900 + R1 99 9900 900 R1 = = 90 W 100 R1 =

R1

R1 −6

=

900 + R1 99 R1

(c) Range switch at Rm3 = Rm + R1 + R2 R2) W R2) W Ish2 Rsh3 = R3 + R4 + R5 = R – R1 – R2 R2 R2 −6 (990 + R2 ) × 100 × 10 990 + R2 R I Rsh3 = m3 m = = 999 I sh3 99.9 × 10 − 3 \ fi

990 + R2 999 9990 990 R2 = =9W 1000 R2 =

R2

R2

Electrical Measuring Instruments (d) Range switch at Rm4 = Rm + R1 + R2 + R3 R3) W R3) W Ish2 Rsh4 = R4 + R5 = R – R1 – R2 – R3 R3 = 1 – R3 (999 + R3) × 100 × 10 − 6 999 + R3 Rm4I m Rsh4 = = = 4999 I sh 4 499.9 × 10 − 3 999 + R3 1 – R3 = R3 R3 4999 4999 999 R3 = = 0.8 W 5000

\ fi (e) Range switch at

R4) W

Rm5 = Rm + R1 + R2 + R3 + R4 Ish2 Rsh5 = R5 = R – R1 – R2 – R3 – R4 Rsh5 =

(999.8 + R4 ) × 100 × 10 R m5 I m = I sh5 999.9 × 10 − 3

999.8 + R4 9999 1999.8 999.8 R4 = = 0.1 W 10000 R5 = R – R1 – R2 – R3 – R4 0.1 W

\

R4 =

fi \

R4) W

R4

−6

=

R4

999.8 + R4 9999

R4

R4

R4

Im

W Rs

0.1 mA, 500 W

Rs

RT

Im

–

+ +

Rs

–

V

RT

607

608

Basic Electrical Engineering

V = Im RT or \

V 10 V = W Im 0.1 mA W W = 99.5 kW

RT =

Rs = RT – Rm

V V

Im V The series resistor Rs

= Im(Rm + Rs)

or

Rs =

Vf sd Im

Rm

range multiplier

Multi-Range Voltmeter Switch position at

Net series resistance, Rs W W W W W W W W W

W multimeters

cales

Im = 0.1 mA 99.5 kW Multiplier

10 V

10

0

400 kW

50

50 V 500 kW 100 V

0.1 mA, 500 W

4000 kW 500 V

+

–

W W W W

Electrical Measuring Instruments

609

ideal voltmeter

resistance is very high Second

AC Voltage Measurement range switch DC meter Range switch AC voltage to be measured

EX A MP L E

R1

–

+ Bridge rectifier circuit

R2

R3

18. 4 m

W

Solution

Im

m

Rs = EX A MP L E

W

Rm

Rs

100 V Vfsd - Rm = 100 mA Im

18. 5

m

W b

(a

Solution

999.9 kW

Im

m

Rm

W

(a Rs =

Vf sd Im

Rm =

50 V 50 A

999 kW

610

Basic Electrical Engineering

(b n=

Vfsd 50 V = fsd = = 1000 Vm I m Rm 50 × 10 − 6 × 1 × 103

ohmmeter

(i) Shunt-Type Ohmmeter

R R R

R = R1

low-value X Im

R1 Rm

E1

Rx

S Y

(ii) Series-Type Ohmmeter

R

a

R medium-value

(iii)

R high-value W

a

W

E

RT Rm

preset resistor R

a zero adjust

Electrical Measuring Instruments

611

E

Rs limits

W series resistor Rs

m zero-adjust resistor R Resistance scale

100 mA, 500 W

100

RT

R0

50 20

25

15

40

10

5

60

80

0 kW

0

100 m A

Current scale

Rs

E

Pointer X +

Rx

– Y

(a) The basic construction.

(b) The current and resistance scales.

R

b resistance is inverted disadvantages (i (ii I = V/R.

Rs (iii R caution connected to an energised circuit.

The test leads of an ohmmeter are never

612

Basic Electrical Engineering

The sensitivity of a W/V rating can m

meter is measured in ohms-per-volt (W m

W

m

V W

1 Current sensitivity

W

W W does not depend on the range of the voltmeter.

W

W

ohms-per-volt rating

for a given range the internal resistance of the voltmeter

loading effect

EX A MP L E

18. 6 W W

W

a

b

100 kW 150 V

A

X 50 kW

V

Ri

Y B

Solution

true Vt

¥

50 k Ω 100 k Ω + 50 k Ω

W

Electrical Measuring Instruments

613

(a ¥

R

W

¥

W W

W 25 k Ω = 30 V 100 k Ω + 25 k Ω 30 V ¥

V

¥

R

W

¥

W

W ¥

V

W

47.6 k Ω = 48.36 V 100 k Ω + 47.6 k Ω

(b % error =

% error =

A multimeter current (in A

V

Vt

V1 Vt

Vt

V2 Vt

¥

¥

50

30 50

50

¥

48.36 ¥ 50

O

40 %

3.28 %

‘AVO meter’

p = vi P = VI

614

Basic Electrical Engineering

*

P = VI cos f f is the power factor wattmeter

a current coil

voltage coil or pressure coil R Wattmeter

Supply

*

Voltage coil

Current coil

R

Load

Electrical Measuring Instruments

615

(i) Voltage-Coil Inductance R) has I

b

V

f¢ f

a I

I

f¢ = f – b f¢

f I

b V

f f¢

Ip

f¢

f

V

b

Ip

I (a) Lagging power factor.

(b) Leading power factor.

b

f¢

angle f f¢ = f + b f R

(ii) R

(iii) Mutual Inductance

(iv)

616

Basic Electrical Engineering

(v) Stray Magnetic Fields

¥ Time motor meters ampere-hour meter integrating

watt-hour meter

*

meter constant.

Pressure coil

Aluminium disc

Shunt magnet

Current coil

Series magnet

I Line

*

Load

617

Electrical Measuring Instruments

(i) Driving System

series

shunt

voltage or pressure coil

current coil

power factor compensator or compensating loop

(ii) registering or counting system.

(iii)

(iv)

ADDITIONAL SOLVED EXAMPLES EXA M P L E

18 . 7

a b

Solution Given that t μ I = k I tc μ q = k q

(a t = tc or

k I=k q

fi qμI

or

q = kI

618

Basic Electrical Engineering q k=

60 20 A

60 ¥ 12 A = 36° 20 A tc μ sin q = k 2 sin q

q = kI = (b t = tc or

k1I = k2 sin q fi

sin q μ I

or sin q = kI

q k=

sin q = kI = EX A MP L E

sin 60 0.866 = 20 A 20 A

0.866 ¥ 20 A

fi q = sin –1

31.3°

18. 8 ¥ q

Solution

t t = Wl sin q q = sin–1

or EX A MP L E

tc −4

1.05 × 10 t = sin–1 = 61° Wl 0.005 × 0.024

18. 9

a (b

Solution (a

t μ I2 tc μ q t = tc

q μ I2

2 1

2 2 ⎛ 5⎞ ⎛I ⎞ q 2 = ⎜ 2 ⎟ ¥ q1 = ¥ ⎝ 10 ⎠ ⎝ I1 ⎠

(b

tc μ sin q t = tc

qμI

2

⎛I ⎞ = ⎜ 2⎟ ⎝ I1 ⎠

2

22.5° sin sin

2 1

⎛I ⎞ = ⎜ 2⎟ ⎝ I1 ⎠

2

2 ⎡ ⎤ ⎡⎛ I ⎞ 2 ⎤ −1 ⎛ 5 ⎞ × sin 90°⎥ q 2 = sin –1 ⎢⎜ 2 ⎟ × sin θ 1⎥ = sin ⎢ ⎣⎝ 10 ⎠ ⎦ ⎢⎣⎝ I1 ⎠ ⎥⎦

= sin–1

14.5°

–4

619

Electrical Measuring Instruments EXA M P L E

1 8.1 0 ¥ A =w¥l tc ¥ t = BINA

Solution

¥ ¥ ¥I¥

¥ ¥

¥ Nm ¥

m Nm

t = tc I =

3 × 10 − 3 0.1 × 80 × 2 × 10 −3

= 0.1875 A

SUMMARY TE R MS

A N D

C O NC EPT

Absolute Instruments Secondary Instruments

(i (ii (iii indicating instrument tc

tc μ sin q

i

air friction permanent magnet moving coil t μI

controlling damping. tc μ q ii

eddy currents. d’Arsonval movement) is

tc μ q

i ammeter voltmeter universal-shunt or ring-shunt

ii

620

Basic Electrical Engineering i

ohmmeters

ii iii

multimeter resistance (in O

A V ‘AVO meter’ wattmeter current coil pressure coil energy meter i) Driving System ii) Moving System iii) Braking System iv) Registering or Counting System IMP O RTA N T

F O R MU LAE Im Rm ( Ifsd Im ) Vfsd Rm Rs = Im Rsh =

1 Current sensitivity

CHECK YOUR UNDERSTANDING two one S. No.

minus

12 Statement

True

False

Marks

Electrical Measuring Instruments

621

Your Score A N SW E R S

1. True 6.

2. 7.

4. True 9. True

3. 8. True

5. True 10.

REVIEW QUESTIONS 1.

12.

2. 13. a

3. (b

c) an integrat 14.

4.

5.

15.

6. 16. 7. 8. 17. 9. a) the current

18. b 10. 11.

19.

m

622

Basic Electrical Engineering

20.

23.

21. 24. W in

22.

25. 26.

W

MULTIPLE CHOICE QUESTIONS Here are some incomplete statements. Four alternatives are provided below each. Tick the alternative that completes the statement correctly. not

1. instrument (a) Voltmeter (c

(a (b (c (d 6.

(b) Ammeter d) Tangent galvanometer

(a (b (c

2. (a (b) an integrating instrument (c (d) an absolute instrument

(d 7. W

in an instrument (a

(b) 5 mA (d

(a) 1 mA (c mA

3. 8.

m

W is

(b

(d

5.

W

(b (d

W W m

(b (d

W/V W/V

(b (d

W W

9. W

4.

(a (c (a (b (c (d

W

(a (c

(c

W/V W/V

10. W

(a (c

W W

623

Electrical Measuring Instruments

A N SW E RS

1. d

2. b

3. a

4. d

5. d

6. b

7. c

8. a

9. a

10. d

PROBLEMS ( A )

S I M P L E

PR O BL EMS 3.

1.

a) currents ¥

[Ans.

b [Ans. (a

Nm]

W; (b

W] W

4.

2.

a b [Ans.

( B)

T R IC KY

W]

[Ans. (a

W; (b

¥

–4

W]

PR OB LE MS

5. W W [Ans. 248 V]

( C )

C HA L L E NGI NG

PROBLEMS

6. [Ans.

ELECTRICAL INSTALLATION AND ILLUMINATION

19

OB JE CT IV E S :

(i)

(ii)

(iii)

( )

( ) (i) ( )

( )

MCBs (ii)

(iii)

(i) ) ( )

Following supply systems (i (ii (iii

( )

( )

(ii) ( )

( )

(iii) CFL (

625

systems of wiring

(i (ii (iii (iv (v (vi (vii

service connection

electronic energy meters

consumer’s terminal double-pole iron-clad

626

Basic Electrical Engineering To fan circuit To power circuit Lamp

L

Energy meter

Distribution fuse board

N Supply authority’s cut-out

Lighting circuit Consumer’s main switch

ing, the most

(i (ii (iii

(i) Vulcanised Indian Rubber (VIR)

(ii) Butyl Rubber Coatings

*

Incoming cable

copper or aluminium*

627

(iii) Ethylene-Propylene Rubber Coatings

(iv) Silicon Rubber Coatings (v) Polychloroprene Coatings

strands, twisted

switched in and switched off

switch to

OFF

628

Basic Electrical Engineering

single-pole single-throw ON

or

OFF

(i) Tumbler Switch

(ii) Flush Switch piano switches

(iii) Toggle Switch *

it ON and OFF ON

(iv) Push-to-On Switch

and OFF ON

OFF

(v) Bed Switch ON

or OFF

(vi) Call-Bell Switch

ON

singlepole double-throw

and S terminal A *

629 ON OFF

ON

the light at

OFF

B1

B2

S 1 C1

C2 S 2

A1

Live

A2 Lamp

Neutral

OFF ON

and S3 for a threeL

ON

L

OFF

ON OFF

3

3

(Two-way switch) S2

(One-way switch) S1

(Two-way switch) S3

Live

Neutral L1

L2

Room 1

Room 2

Room 3

(3 3 OFF

ON

L3

)

3 OFF

,

ON

S

OFF

630

Basic Electrical Engineering

a

A

Live A

C

C

Mains supply

Input

C

B

D

Output

D

B

Neutral

A

D B

Circuit 1

Circuit 2 (a)

(b)

(c)

b

b swapping over c

c) is found in verandah

S2

S1

S3

S4

Live Lamp Neutral

and S4

3 and OFF

ON

OFF

S3

631

ii

disadvantages (i iii

iv pandal

i

advantages ii

iii v

faults, (iv vi

vii

i of installation, (ii other material used, (v (viii (xi

ix

iii) nature of load, (iv vi vii x

632

Basic Electrical Engineering

a tinned brass clips b conduit Clip

Clip

Cable

Screw or nail

Plug

Holes for nailing in batten

Wall/ ceiling

(a)

Folding portion (female)

Folding portion (male) (b)

conduits. Nails Capping Wooden casing

Wall/ceiling Channels Wooden plug

Wires

633

conduits.

4

earthed

For concealed conduit wiring system

*, a

surf *

634

Basic Electrical Engineering Clip Cable Conduit Screw

Screw

Wall/Ceiling

aluminium

OFF

itself ON

blender, a

635

636

Basic Electrical Engineering

(i) (ii) (iii) (iv)

in live (red

OFF

ON

or OFF 3

4

Lamp

Fan regulator Ceiling can

0 5 4

Neutral link

Tube light

1 2 3

NL

S 1 S2 S3 S4 E

N

L

Control switch-board

637

e

earthing earthing

earthing

vide safety

earthing

through earth

lightning conductor

earth electrode. Galvanised iron

(1) Plate Earthing (i) Copper Plate (ii) GI Plate

a ¥ ¥

¥ ¥

copper

638

Basic Electrical Engineering

The plate is buried to a depth not less than 2 m and at least 0.6 m away from the foundation of any building. The layers of common salt and charcoal are 30-mm and 80-mm thick, respectively.

(2) Pipe Earthing As shown in Fig. 19.10b, a GI pipe with a few holes at its lower end is buried to a depth not less than 2 m and at least 0.6 m away from the foundation of any building. Normally, the size of pipe is either 2 m long and 38 mm diameter or 1.37 m and 51 mm diameter. However, for dry and rocky soil, we use longer pipes. Alternate layers of common salt and charcoal have thickness of 30 mm and 80 mm, respectively.

Fig. 19.10 In both the methods shown in Fig. 19.10, to maintain good conductivity of the soil, an arrangement is made for pouring water into the earth pit surrounding the earth electrode. This is especially needed during summer. As the pipe has much larger contact area with soil, it can handle larger leakage currents than the plate earthing of same electrode size. The earth wire (made of copper) is tightly fastened to the earth electrode by means of nut and bolt.

19.10

SAFETY PRECAUTIONS IN HANDLING ELECTRICAL APPLIANCES

The only safeguard against an injury or death while handling electrical appliances is to strictly follow the required precautions at every stage. If accidentally, the live wire comes in contact with the body of a person, system, the heart and the respiratory system, and as a result the muscular functions of the body are paralysed. This may cause severe burns, stoppage of breathing, and even death. Following safety precautions must be taken while working with electrical installations or while handling electrical appliances. 1. Make sure that all metallic parts of the electrical equipments are effectively earthed.

639 2. 3. 4. 5. 6. 7. Use rubber-sole dry-wooden 8. Use rubber gloves 9. insulated tools 10. two different terminals 11.

put off damaged or frayed

rating of fuse wire

12. 13.

OFF

leaky insulation

can give serious shock. 14. 15. Do not throw water on live conductors or throw sand

16. equipments. 17. 18.

W megger

640

Basic Electrical Engineering

(i (ii (iii (iv (v (vi

OFF OFF, ON, ON

W (i (ii (iii (iv (v

OFF OFF, ON, ON

earth continuity tester (i (ii (iii (iv (v

OFF OFF, ON, ON

W

641

OFF

simple test lamp (i (ii (iii (iv

ON, ON,

and

OFF

light bulbs) have been in

T T

range (l

i ii

iii iv v

vi tungsten

i) the support wires iv) the glass rod vi) the exhaust tube

stem tube vii) the fuse lead-in-wire

ii) the iii) the button (made v) the viii) the ix) the gas,

642

Basic Electrical Engineering

Pins Base

Exhaust tube Stem tube Glass rod

Fuse

Button Support wire

Lead-in-wire

Filament

Gas Bulb

x) the base xi) the bulb thin (diameter about one-hundredth of a millimetre

arcing

diffused light

At the rated voltage, the life WW ON

643

advantages i iii v) gives light with no glare or glitter, (vi vii viii

(ii

iv) gives light that does

disadvantages i

choke or starter ii

iii pf iv

(v

ON

and

OFF

stroboscopic effect

ON

vi vii viii

ix

phosphor substance

tungstate,

of absorbing ultraviolet radiation form of visible light wavelength (or

heating

) and a

644

Basic Electrical Engineering

Starter

L

S

C2 Choke

AC Supply

Tube

C1

N

Electrodes (Heating filaments)

neighbourhood

Note stabilising impedance ballast C C

A a

[

Electrical Installation and Illumination

645

Fig. 19.13 The purchase price of a CFL is higher (about Rs. 100) than that of an incandescent lamp (about Rs. 10) of the same luminous output. But this cost is recovered in energy savings and replacement costs over the lamp’s lifetime. Due to the potential to reduce electric consumption and hence pollution, various organisations have undertaken measures to encourage the adoption of CFLs. Efforts range from publicity to encourage awareness and make CFLs more widely available, to direct measures to provide CFLs to the public. Some state governments have subsidised CFLs or provided them free to customers as a means of reducing electric demand (and thereby delaying additional investments in generation).

It is an electric discharge lamp in which light is produced by electrical conduction through mercury vapour. It emits a very bright greenish-blue light, in which true colours of objects cannot be seen. It has an average life various sizes up to 2 kW. Compared to an incandescent lamp, the cost of a mercury lamp is quite high (about

As shown in Fig. 19.14, it consists of two tubes, one inside the other. The smaller inner tube is made of hard glass (or quartz) and is surrounded by the larger glass tube or bulb. The space between the two tubes is completely evacuated to prevent heat loss. The outer tube also absorbs harmful ultraviolet radiations. tube, it is also called arc tube. It houses two main electrodes A and B, and an auxiliary electrode C. The main electrodes are made of tungsten and are coated with barium oxide. The auxiliary electrode C is placed near the main electrode A, and it is connected to the other main electrode B through a resistance R. A choke is connected in series to limit the current. The capacitor C1 is connected across the main lines to improve the power factor.

646

Basic Electrical Engineering AC main supply C1

Capacitor for power-factor improvement Main electrodes

Choke Argon and mercury Fluorescent coating A

B

C

Inner hard glass tube R Resistance Auxiliary electrode

Outer glass tube (or bulb)

vertical position only

yellow light

neon gas

647 Insulation

AC main supply

Inner U-tube

C Outer double-walled envelope High leakage-reactance auto-transformer

Electrodes

sodium envelope

yellow light

C is used for only in horizontal position

helium gas iron electrodes W Series resistance

Helium gas

advertising and decoration

Iron filament

648

Basic Electrical Engineering

AC supply

C

metre C is used

SUMMARY

i

ii

CHECK YOUR UNDERSTANDING minus one

S. No.

12

Statement

True

ON

and OFF

False

Marks

649

A N SW E RS

2. False 7. False

1. 6.

4. False 9. False

3. 8. False

5. False 10.

REVIEW QUESTIONS 13.

1. installation ? 2.

14.

15. 3. 16. 4. 5. 17. 6. 18. installation to earth, what instrument is used and

7. 8.

19. 20.

9. 10.

21.

11. 22. 12.

23.

650

Basic Electrical Engineering

Supplementary Exercise—Part A : DC Circuits

S U P P L E M E N TA R Y E X E R C I S E S A.1 Solved Problems A.2 Practice Problems

PA R T

A : DC

651

A

CIRCUITS

Assemblage of

N OT E This set of exercises provides practice to those who wish to attain a higher standard of learning the basic principles of Electrical Engineering. The real key to success is practice.

652

Basic Electrical Engineering

A.1. SOLVED PROBLEMS P RO B L EM

A - 1

The single loop circuit in Fig. A-1 has a current source of I amperes connected to an independent voltage source of through it. Determine the power P1 absorbed by the independent voltage source and the power P2 absorbed by the dependent voltage source for (a) I = 4 A, (b) I = – 3 A, and (c) I = 5 mA. 8V

I

P1 P2

I

2I

Fig. A-1

Fig. A-2

Solution

I is directed away from the positive terminal of the 8-V source. Hence, the formula for power absorbed has a negative sign : P1 = – 8I and current references corresponds to the passive sign convention. Hence, P2 = 2I(I) = 2I 2. Therefore, (a) P1 = – 8I = – 8(4) = –32 W and P2 = 2I 2 = 2(4)2 = 32 W. The negative sign for P1 shows that independent source delivers power instead of absorbing it. Furthermore, the independent source of 8 V is delivering power of 32 W and the dependent source is absorbing power of 32 W. Hence, the current source is neither absorbing nor delivering any power. 2 2 (b) P1 = – 8I = – 8(– 3) = 24 W and P2 = 2I = 2(–3) = 18 W. The power absorbed by the dependent source is still

(c) P1 = – 8I = – 8(5 ¥ 10–3) = – 40 mW and P2 = 2I 2 = 2 (5 ¥ 10–3)2 = 50 mW . P RO B L EM

A - 2

Show that the total power absorbed by components is equal to the total power delivered in the circuit of Fig. A.2.

Solution Using passive sign convention, let us calculate the power absorbed by each component : P1 = – 16 (10) = –160 W P3 = 22 (6) = 132 W

and

P2 = – 6 (10) = – 60 W P4 = 22(0.4 ¥ 10) = 22 (4) = 88 W

The negative sign indicates that it is power delivered and not power absorbed. Thus, Power absorbed, Pa = 132 + 88 = 220 W Power delivered, Pd = 160 + 60 = 220 W The two are seen to be equal.

654

Basic Electrical Engineering

P RO B L EM

A -3 a, determine the voltage vx.

Fig. A-3 Solution

b). Applying KVL in

60 – v8 – v10 = 0 or 60 – (5 A ¥ 8 W) – v10 = 0 fi v10 = 20 V 20 V \ i10 = =2A 10 We note that vx appears across the 2-W resistor as well as across the current source ix. Now applying KVL in the central loop, we get v10 – v4 – vx = 0 Writing KCL for the node 1, we get v4 +2 fi 4 Substituting this value of v4 and knowing that v10 = 20 V, we get 5 = i4 + i10 =

20 –12 – vx = 0 P ROBLEM

fi

v4 = 12 V

vx = 8 V

A - 4

Determine the voltage, current, and power associated with each element in the circuit of Fig. A-4a.

Fig. A-4 v with an arbitrary polarity, as shown in Fig. A-4b. Also, we mark currents i1 and Solution i2 in the two resistors in conformance with the passive sign convention. Next, we apply KCL at the upper node to get v v + 30 + =0 fi v=2V – 120 + 1/ 30 1/15

Supplementary Exercise—Part A : DC Circuits v = 60 A 1/ 30 The absorbed power in each element is

\

i1 =

P

PROBLEM

and

i2 =

= vi = 2(–120) = – 240 W PR1 = vi1 = 2(60) = 120 W and

655

v = 30 A 1/15 P30A = vi = 2(30) = 60 W PR2 = vi2 = 2(30) = 60 W

A -5

Determine the value of v and the power delivered by the independent current source in the circuit of Fig. A-5.

Fig. A-5

Fig. A-6

Solution Applying KCL for the upper node, we get i6 – 2ix – 0. 024 – ix = 0 Applying Ohm’s law to each resistor, i6 = fi and

v 6000

and

ix =

v 2000

v ⎛ −v ⎞ ⎛ −v ⎞ −2 – 0.024 – = 0 fi v = 600 ¥ 0.024 = 14.4 V ⎝ 2000 ⎠ ⎝ 2000 ⎠ 6000 P24mA = iv = (24 mA)(14.4 V) = 345.6 mW

PR OB L EM

A- 6

Determine vx and i3 in the circuit of Fig. A-6.

Solution Since vx appears across the parallel combination of the 6-W and 3-W resistors, we can simplify the circuit by combining them without losing vx. 6×3 6 W || 3 W = =2W 6+3 Now, applying the principle of voltage division, we get 2 = 4 sin t V vx = (12 sin t) ¥ 4+2 Now, the current i3 can easily be determined as follows. 4 v i3 = x = sin t A 3 3W P ROBLE M

A- 7

a) the voltage of the dependent source, (b) the power supplied by the dependent source, and (c) the power dissipated by the 15-W resistor.

656

Basic Electrical Engineering

Fig. A-7

Solution (a) The voltage across the dependent source is same as that across the 3-W resistor, v. To determine this voltage, we leave the dependent source alone and simplify the remaining circuit. The two current sources can be combined into a single 2-A source. Two 6-W resistors are parallel and can be replaced by a single 3-W resistor (6 W || 6 W = 3 W). This 3-W is now combined with the series 15-W resistor to give a single 18-W resistor. The resulting circuit is given in Fig. A-8a.

Fig. A-8 W, 9-W and 18-W resistors. However, doing so loses i3 only 9-W and 18-W resistors, as shown in Fig. A-8b. Applying KCL to the top node of Fig. A-8b, we get v =0 – 0.9i3 – 2 + i3 + 6 Applying Ohm’s law to the 3-W resistor branch, we have v = 3i3. Putting this value of v in above equation, we get 10 3i – 0.9i3 – 2 + i3 + 3 = 0 fi i3 = A 6 3 Thus, the voltage across the dependent source (which is the same as the voltage across the 3-W resistor) is 10 v = 3i3 = 3 ¥ = 10 V 3 (b) The power supplied by the dependent source is 10 Pdep = v ¥ 0.9i3 = 10 ¥ 0.9 ¥ = 30 W 3 (c) To determine the power dissipated by the 15-W resistor, we must return to the original circuit. This resistor is in series with an equivalent 3-W resistor, and a voltage of 10 volts appears across this branch of total resistance 18 W.

Supplementary Exercise—Part A : DC Circuits

657

Therefore, the current through the 15-W resistor is 10 5 i15 = = A 18 9 \ P15W = (5/9)2 ¥ 15 = 4.63 W P ROBLEM

A- 8

What is the maximum voltage that can be applied across the series combination of a 150-W, 2-W resistor and a 100-W, 1-W resistor without exceeding the power rating of either resistor ?

Solution

From P = I 2R, the maximum safe current for the 150-W resistor is I1 =

P/ R =

2 /150 = 0.115 A

P/ R =

1/100 = 0.1 A

The maximum safe current for the 100-W resistor is I2 =

The maximum current cannot exceed the lesser of these two currents. Hence, the maximum voltage that can be applied is Vmax =I(R1 + R2) = 0.1 ¥ (150 + 100) = 25 V P ROBLE M

A -9

For the circuit of Fig. A-9, (a) calculate the current I and the power absorbed by the dependent source, and (b) determine the resistance ‘seen’ by the independent voltage source. 4W

2W

I

+ V1 – 4.5V1

24 V

Fig. A-9

Fig. A-10

Solution (a) Applying Ohm’s law to the 4-W resistor, we get V1 = 4 I. Now, writing KVL equation, we get 24 – 4I – 2I + 4.5V1 = 0

or 24 – 4I – 2I + 4.5(4I) = 0

fi

I = 24/(– 12) = – 2 A

direction for I. Since the current and voltage references for the dependent source are not the same as the passive sign convention, the formula for power absorbed has a negative sign : P = – 4.5V1(I) = – 4.5(4I)(I) = – 18I 2 = – 18(– 2)2 = – 72 W The negative sign here means that the dependent source is supplying power instead of absorbing it. (b positive terminal of the source. Thus, R=

24 24 = = –12 W -2 I

658

Basic Electrical Engineering

The negative sign here indicates that the remaining circuit supplies power to the independent source. Actually, it is the dependent source alone that supplies this power, as well as to the two resistors. P RO B L EM

A -1 0 a) if vx

Solution

Is; (b) if Is

vx; (c) calculate the ratio vx/Is.

Let us mark the currents and nodes in the circuit as shown in Fig. A-11a.

Fig. A-11 (a) Given : vx = 10 V. Therefore, i4 = (10 V)/(5 W) = 2 A Thus, using Ohm’s law, v 18 vb = i4 (4 + 5) = 2 (4 + 5) = 18 V fi i3 = b = =6A 3 3 Using KCL : i2 = i3 + i4 = 6 + 2 = 8 A Using Ohm’s law : va – vb = i2 ¥ 2 fi va = 18 + 8 ¥ 2 = 34 V fi i1 = 34/1 = 34 A Again, using KCL : Is = i1 + i2 = 34 + 8 = 42 A (b combination of 4 W and 5 W is in parallel with 3 W. Hence, the equivalent resistance from node b to ground is 9×3 Req = (4 + 5) || 3 = = 2.25 W 9+3 b. From this circuit, since Is = 50 A, it is obvious that 1 i2 = (50 A) = 9.52 A 1 + (2 + 2.25) This current again divides into i3 and i4 (Fig. A-11a). Thus, 3 i4 = (9.52 A) = 2.38 A 3 + ( 4 + 5) vx as vx = i4 (5 W) = (2.38 A)(5 W) = 11.90 V (c

Rt is Rt =

10 V vx = = 0.238 W 42 A Is

For the second case, the ratio Rt is Rt =

vx 11.9 V = = 0.238 W Is 50 A

vx/Is and it is called transfer resistance of the circuit.

Supplementary Exercise—Part A : DC Circuits P RO B L EM

659

A -1 1 V1.

Fig. A-12

Fig. A-13

Solution to the left-hand half of the circuit, we get 24 –16I1 – 4V1 = 0

(i)

Applying Ohm’s law to the right-hand-half of the circuit, we get V1 = – 0.5I1(4) = – 2I1

or

I1 = – 0.5V1

(ii)

Substituting this value of I1 in Eq. (i), gives 24 – 16(– 0.5V1) – 4V1 = 0 P ROBLE M

fi V1 = – 6 V

A -12

Determine voltage V1 and V2 in the circuit of Fig. A-13. Solution We observe that the dependent current source is 3 I I in terms of V1, using Ohm’s law : I =V1/5. Thus, the dependent current directed source is 3I = (V1/5) = 0.6V1, directed downward. Applying KCL at the top node, we get V1 V1 + + 0.6V1 = 9 fi V1 = 10 V 5 10 Finally, applying KVL around the outside loop gives V2 – 12(9) – V1 = 0

or

V2 = 12(9) + 10 = 118 V

Note that the 12-W resistor has no effect on V1, but it does have an effect on V2. P ROBLEM

A -13

Determine the power absorbed by the 5-W resistor in the circuit of Fig. A-14a.

Solution

redraw the given circuit in a simpler way as in Fig. A-14b. the upper node with respect to the lower node in terms of v1. The current through the 2-W resistor is v1/2. Therefore, the voltage across the series combination of 1-W and 2-W resistors is (v1/2)(1 + 2)= 1.5v1. Applying KCL at the upper node, we get –5 + fi

1.5v1 1.5v1 – 5v1 + =0 3 5

v1 = –

or

0.5v1 – 5v1 + 0.3v1 = 5

5 V = – 1.19 V 4.2

660

Basic Electrical Engineering

Fig. A-14 The current through the 5-W resistor is i=

1.5 × ( − 119 . ) 1.5v1 = 5 5

Hence, the power absorbed by the 5-W resistor is 2 P =i R

P ROBLEM

2

(5) = 0.637 W

A - 14 a) if R = 80 W

Req; (b) if Req = 80 W

R; (c) if R = Req 6W

X 10 W

40 W

R

100 W

30 W

Q

6W 6W

6W Req

R.

6W

20 W P

Fig. A-15

6W

Y

Fig. A-16

Solution (a

R = 80 W, the equivalent resistance is Req = 10 + [100 || {80 + (30 || (40 + 20))}] ⎡ ⎧ 30 × 60 ⎫⎤ = 10 + ⎢100 || ⎨80 + ⎬⎥ = 10 + [100 || {80 + 20}] = 10 + [100 || 100] 30 + 60 ⎭⎥⎦ ⎢⎣ ⎩ = 10 + [50] = 60 W

(b

Req = 80 W, we have 80 = 10 + 100 || (R + 20) or 80 = 10 +

100( R + 20) 100 + ( R + 20)

Dividing by 10, we get 10( R + 20) =8–1 100 + ( R + 20)

or

10R

R + 840

or 3R = 640

Supplementary Exercise—Part A : DC Circuits fi

R=

661

640 = 213.3 W 3

R = Req, we have

(c

R =10 + 100 || (R +120),

or

R = 10 +

100 R + 2000 R + 120

Multiplying throughout by (R + 120), we get 2 2 R + 120R = 10R + 1200 + 100R + 2000 or R + 10R – 3200 = 0

fi

W

R

W

Since a resistance cannot have negative value, R = 51.79 W P ROBLEM

A -15

Six resistors, each of 6 W, are connected as shown in Fig. A-16. Determine the equivalent resistance across any two diagonal points.

Solution We shall discuss below three methods of doing it. First Method We convert the delta-connection (represented by dotted lines) across points PYQ into its star-equivalent a). This introduces an additional node N as the star-point. Using a delta with 6 W as each side is converted into 2 W as each leg of the star. b between X and N, there are two parallel branches each having 6-W and 2-W resistors in series. Point N is connected to point Y through a 2-W resistor. One more 6-W resistor is connected between X and Y. Hence, the equivalent circuit is as c. Thus, RXY = 6 || {8 || 8) + 2} = 6 || {4 + 2} = 3 W 6W

X

Q

6W

X

6W

Q

2W

6W 2W

6W

Q

2W

N

6W

N

2W

2W

6W

2W

N 2W

Y

6W Y

(a)

2W

P

2W P

6W X

P

Y (b)

(c)

Fig. A-17 Second Method We convert the star-connection (represented by dotted lines) across points XYQ into its delta-equivalent (see Fig. A-18a). Using W as each leg is converted into 18 W as each side of the delta. The 18 W and 6 W in parallel gives 4.5 W. Therefore, the equivalent network between X and Y is as shown in Fig. A-18b. Thus, RXY = 4.5 || (4.5 + 4.5) = 3 W

662

Basic Electrical Engineering 6W

X

Q

18 W 6W

4.5 W Q 4.5 W 18 W

X

6W

Y

18 W 4.5 W

P

Y (a)

(b)

Fig. A-18 Third Method We break the two 6-W resistors in the centre of the network, each into a series combination of two 3-W resistors (Fig. A-19a). The network is now seen to be symmetrical about the dotted line along PQ. connected across XY, due to symmetry, all points along the dotted line (such as points Q, N and P), shall be at the same potential. Hence, all these points can be shorted without affecting the equivalent resistance of the network. The resulting b RXY = 2 ¥ {3 || (6 || 6) } = 2 ¥ {3 || 3} = 2 ¥ 1.5 = 3 W Note Obviously, the third method is simplest of all. 6W

X

Q 6W

3W 3W 6W

3W

X

6W

6W P Q

3W

N

3W

6W

6W (a)

P

3W

6W

Y (b)

Fig. A-19 P RO B L EM

A -1 6

Find voltage V1 in the circuit of Fig. A-20, using voltage division twice. 8W 16 W

54 W 36 A 20 W

80 V

+ V2

36 W

–

Fig. A-20

18 W

+ V1 –

I2

6W I1

Fig. A-21

5W

Y

Supplementary Exercise—Part A : DC Circuits

663

Solution

V1 can be found by voltage division if V2 V2 by voltage division of the source voltage of 80 V. For this, we need the equivalent resistance to the right of 16-W resistor. This resistance is (36)(54 + 18) Req = = 24 W 36 + 54 + 18 By applying the voltage division principle, we get

24 = 48 V 24 + 16 Note A common mistake committed by students is to ignore the loading effect of the resistors to the right of the V2 node. Again by voltage division, we get 18 V1 = 48 ¥ = 12 V 18 + 54 V2 = 80 ¥

P ROBLEM

A -1 7

Find current I1 in the circuit of Fig. A-21, using current division twice.

Solution

Obviously, the current I1 can be found by current division, if we know current I2. For this, we need the equivalent resistance of the bottom three branches : (20) (5) Req = 6 + = 10 W 20 + 5 By current division principle, we get 8 20 = 16 A and I1 = 16 ¥ = 12.8 A I2 = 36 ¥ 8 + 10 20 + 5 –110 V

R

–+

5A

40 V + –

G

6A

Fig. A-22 P RO B L EM

Fig. A-23

A -1 8

Find R and G in the circuit of Fig. A-22 if the 5-A source supplies 100 W and the 40-V source supplies 500 W.

Solution

Voltage across 5-A source, V5A =

P 100 = = 20 V (lower terminal ‘+’ marked) I 5

I40V =

P 500 = = 12.5 A (upwards) V 40

Current through 40-V source,

664

Basic Electrical Engineering

Current through 110-V source, I110V = 12.5 + 6 = 18.5 A (towards left) IG = I110V – 5 = 18.5 – 5 = 13.5 A (downwards) Voltage across G, VG = 40 – (– 110) = 150 V

\

G=

Thus,

13.5 A IG = = 0.09 S = 90 mS VG 150 V

Now, the voltage across R, VR = V5A + VG R=

Thus, P ROBLEM

VR 170 = = 34 W IR 5

A - 19

For the circuit shown in Fig. A-23, determine ix and compute the power absorbed by the 15-kW resistor.

Solution

Equivalent resistance as faced by the 5-V source, W

Req 5V Current supplied by the source, i = 20.78 k

= 240.6 mA

13.686 = 139 mA 10 + 13.686 2 Power, P15kW = (240.6 mA) (15 kW) = 868 mW Current, i x = (240.6 mA) ¥

P ROBLE M

A - 20

Find the power absorbed by each element of the circuit shown in Fig. A-24a.

6W

7A

12 W 4W

8A

6W

12 W

7A

(a)

4W

(b)

Fig. A-24 Solution

The given circuit can be redrawn in the simple way as in Fig. A-24b V is the voltage of the upper node with respect to the lower node, we can write the KCL equation as V 6

\

V V + +8 =0 4 12

P6 W =

or

V

1⎞ ⎛1 1 + + = –5 fi ⎝ 6 12 4 ⎠

( 30) V2 = 6 6

V = – 30 V

2

= 150 W; P

= VI

210 W

8A

665

Supplementary Exercise—Part A : DC Circuits ( 30) V2 = 12 12

P12W =

2

P4W =

= 75 W;

( 30) V2 = 4 4

2

= 225 W

P8A = VI = – 30 (8) = – 240 W

and

Comment : The sum of powers absorbed by all the elements of a circuit must add up to zero :

P RO B L EM

A -2 1

Nine resistors, R1 to R9, are connected in a network as shown in Fig. A-25a. Determine the current delivered by a 30-V source when connected across nodes A and B of the network.

R2 2W 2W

A

R3

4W

A

C

2W

R9 R1

30 V + –

2W2W

R8

R7

2W

2W 1W

1W

B

R6

R4

D

R12 R7

R1

30 V + –

C

2W

A

1.2 W R13

2W

30 V + –

R1

R14

2W

1W

B

R5

(a)

R6

(b)

1.23 W

1W

D

B

R6

(c)

Fig. A-25 Solution

R2 and R3 is in parallel with R9. Thus, as shown in Fig. A-25b, these three resistors can be replaced by a single resistor R12 across the nodes A and C : R12 = (2 + 2) || 4 = 2 W Similarly, the three resistors R4, R5 and R8 can be replaced by a single resistor R13 across the nodes A and C : R13 = (2 + 1) || 2 = 1.2 W

Now, we can replace the three resistors R12, R13 and R by a single resistor R14 across the nodes A and D, as shown in Fig. A-25c R14 = (2 + 1.2) || 2 = 1.23 W Further, the three resistors R14, R6 and R1 can be replaced by a single resistor across the nodes A and B to give total resistance : RAB = (1 + 1.23) || 2 = 1.054 W Finally, the current delivered by the 30-V source is given as I=

V 30 = = 28.46 A RAB 1.054

666

Basic Electrical Engineering

P RO B L EM

A - 22

The circuit of Fig. A-26 has two current sources and six resistors. Determine the output voltage Vo.

Fig. A-26 Solution tempted to go in for star-delta conversions. However, if you look at the circuit with proper perspective by redrawing it as a, it can easily be solved by series-parallel combinations of resistors. + R1

R4

10 A

Vo

1A R2

R3

+

R5 10 A

R11

10 W

1 A R12

5W

Vo

R6 – (a)

– (b)

Fig. A-27 R2 and R3 is in series with R1. Similarly, the parallel combination of R4 and R5 is in series with R6. Thus, 3×6 4×6 = 10 W and R12 = R6 + (R4 || R5) = 2.6 + =5W 3+6 4+6 b) simply has a single node-pair. Applying KCL, gives ⎛ 1 1⎞ V V + – 10 + o + 1 + o = 0 or Vo = 9 fi Vo = 30 V ⎝ 10 5 ⎠ 10 5

R11 = R1 + (R2 || R3) = 8 +

P R O B LE M

A- 2 3

For the circuit of Fig. A-28, determine the node-to-reference voltages.

Solution assign v1 to v4 to the remaining nodes. We mark supernode about each voltage to write KCL equations only at node 2 and at the supernode containing the dependent voltage source. No equation is needed for node 1, since it is obvious that v1 = –12 V. The KCL equations at node 2, v2 v1 v2 v3 + = 14 0.5 2

Supplementary Exercise—Part A : DC Circuits

667

v2

vx –

+ 0.5 W

–+

v1

14 A

2W

0.5vx

12 V Ref.

i2

1W

v3

2W

3W 2.5 W

– vy

7V

1W

+ –

i1 + –

0.2vy

+

6V

2W

i3

1W

v4

Fig. A-28

Fig. A-29

and at the 3-4 supernode, v3 v2 v v v + 4 1 + 4 = 0.5vx 2 2.5 1 Next, we relate the source voltages to the node voltages, v3 – v4 = 0.2vy

and

v4 – v1 = vy

Finally, we express the dependent current source in terms of the node voltages, 0.5vx = 0.5(v2 – v1) Eliminating vx and vy from the above equations, we get – 2v1 + 2.5v2 – 0.5v3 =14 0.1v1 – v2 + 0.5v3 + 1.4v4 = 0 v1 = – 12 0.2v1 + v3 – 1.2v4 = 0 Solving these equations, we get v1 = –12 V; P ROBLEM

v2 = – 4 V;

v3 = 0 V

and v4 = – 2 V

A -24

Figure A-29 shows a three-mesh circuit. Use mesh analysis to determine the three mesh currents.

Solution

The three mesh currents are assigned clockwise directions, as shown in Fig. A-29. We can either apply KVL around the three meshes, or to save time and avoid errors, we can directly write the mesh equations by inspection as follows : -1 -2 ˘ ⎡ i1 ⎤ ⎡7 − 6⎤ È1 + 2 È 3 -1 -2˘ È i1 ˘ ⎡1 ⎤ Í ˙⎢ ⎥ ⎢ ⎥ -3 ˙ ⎢i2 ⎥ = ⎢ 0 ⎥ or ÍÍ -1 6 -3˙˙ ÍÍi2 ˙˙ = ⎢⎢0 ⎥⎥ Í -1 1 + 2 + 3 ÍÎ -2 ÍÎ-2 -3 6 ˙˚ ÍÎi3 ˙˚ ⎣⎢6 ⎦⎥ 2 + 3 + 1˙˚ ⎣⎢i3 ⎦⎥ ⎣⎢ 6 ⎦⎥ -3

668

Basic Electrical Engineering

i1 = 3 A; P RO B L EM

i2 = 2 A and i3 = 3 A

A -2 5

For the circuit shown in Fig. A-30a, the current I. 5W

5W

5W

I

20 W

5W

+

+ 37.5 V

4W

I

V = 3I –

2

5W

7.5 A

(a)

20 W

I +

2

2

30 V

V = 3I –

(b)

V = 3I – (c)

Fig. A-30 Solution parallel with a 5-W resistor (see Fig. A-30b). The 5-W resistor and 20-W resistor can be combined into a (5 ¥ 20)/(5 + 20) = 4-W W resistor is transformed into a voltage source of ¥ 4 = 30 V in series with a 4-W resistor (Fig. A-30c). For this single mesh circuit, the KVL equation is 3I 2 + 9I – 30 = 0, which can be solved to get I = 2 A or – 5 A positive terminal. Hence, I = 2 A. P RO B L EM

A -2 6

Obtain the mesh currents and hence the voltage drop across 6-W resistor in the circuit of Fig. A- 31. 1 kW

8W

6W 0.5Vx

I1

4W I2

120 V

Fig. A-31 Solution

2 kW

2W + Vx – 60 V

4 kW 5 kW

26 V

I2 I1 13 kW

I3

500 W

Fig. A-32

Before we write KVL equations for the two meshes, we express the controlling quantity Vx in terms of the mesh current I2. Using Ohm’s law, Vx = 4 I2. The voltage of the dependent source is then 0.5 Vx = 0.5 (4 I2) = 2 I2. Note that because of the presence of the dependent source we cannot write the mesh equations by inspection. Thus, writing

Supplementary Exercise—Part A : DC Circuits

669

KVL equations for the two meshes, we have and

(8 + 6)I1 – 6I2 – 2I2 = – 120 (6 + 2 + 4)I2 = 120 – 60

⎡ 14 − 8⎤ ⎡ I1 ⎤ È-120˘ ⎢− 6 12 ⎥ ⎢ I ⎥ = Í 60˙ ˚ ⎣ ⎦ ⎣ 2⎦ Î Using a calculator, the above matrix equation can be solved to give I1 = – 8 A PROBLEM

and

I2 = 1 A

A -27 W resistor in the circuit of

Fig. A-32. W resistor so that only one current needs to be solved for. We can select the path of other two loop currents as shown. For convenience, let us take kW as the unit of resistance and mA as the unit of current. The KVL equations for the three loops are 13(I1 + I3 – I2) + 5I1 + 0.5(I1 + I3) = 0 or 18.5I1 – 13I2 + 13.5I3 = 0 or – 13I1 + 16I2 – 15I3 = 26 1(I2) + 2(I2 – I3) + 13(I2 – I3 – I1) = 26 13(I1 + I3 – I2) + 2(I3 – I2) + 4I3 + 0.5(I1 + I3) = 0 or 13.5I – 15I2 + 19.5I3 = 0 Using a calculator, we get

Solution

I1 = 2 mA P ROBL EM

A -28

Use the technique of mesh analysis to determine the three mesh-currents in the circuit of Fig. A-33a.

Fig. A-33 Solution the meshes 1 and 3. For mesh analysis, we use the concept of supermesh. The mesh currents i1, i2 and i3 are assigned as

670

Basic Electrical Engineering

shown in Fig. A-33b of meshes 1 and 3 as shown. Applying KVL to this loop, 1(i1 – i2) + 3(i3 – i2) + 1i3

i1 – 4i2 + 4i3

Applying KVL around mesh 2, 1(i2 – i1) + 2i2 + 3(i2 – i3) = 0

or – i1 + 6i2 – 3i3 = 0

The third equation is obtained by relating the source current to the assumed mesh currents, i1 – i3 Solving the three equations, we get i1 = 9 A; i2 = 2.5 A P RO B L EM

and i3 = 2 A

A -2 9

Use the technique of node analysis to determine the node voltage v1 in the circuit of Fig. A-34.

Solution independent voltage source between the nodes 2 and 3. For node analysis, we treat node 2, node 3 and the voltage source together as a supernode (indicated by the region enclosed by the broken line). Applying KCL to this supernode, v2 v1 v v v v + 3 1 + 3 + 2 = – (–3) – (–25) 3 4 5 1 or – 0.5833v1 + 1.3333v2 + 0.45v3 = 28

4W –3 A v2

v1

22 V

3W –8 A

1W

–+

5W –25 A

KCL equation at node 1 is Reference node

v1 v2 v v + 1 3 = –8 – 3 3 4 0.5833v1 – 0.3333v2 + 0.2500v3 = –11

or

v3

Fig. A-34

Since we have three unknowns, we need one more equation. This is obtained from the fact that there is a 22-V voltage source between node 2 and node 3 : v3 – v2 = 22 Solving these three equations, we get v1 = 1.071 V P RO B L EM

Solution

A -3 0

We note that the 15-A current source is located on the perimeter of the circuit, and hence clearly i1 = 15 A.

KVL : 1(i2 – i1) + 2i2 + 3(i2 – i3) = 0

or

6i2 – 3i3 = 15 fi

The controlling voltage vx is vx = 3(i3 – I2)

2i2 – i3 = 5

Supplementary Exercise—Part A : DC Circuits

i2

1W

671

2W 8A

3W 15 A

i1

i3

2W

ix

8W

+ v – 1v x 9 x

2W

1W

100 V + –

4W

Fig. A-35

10 W 3W

5W

Fig. A-36

The dependent current source being common between mesh 1 and 3, we have vx 3 (i3 i2 ) = i3 – i1 or = i3 – i1 fi i2 + 2i3 = 45 9 9 Solving above two equations, we get i2 = 11 A and i3 = 17 A. P ROBLEM

A- 3 1

A planar circuit with four meshes is given in Fig. A-36. Determine the current ix by (a) mesh analysis, and (b) nodal analysis.

Solution (a) By mesh analysis a therefore need to write three more equations. Writing KVL equations for meshes 1, 2 and 3 :

i4 = – 8 A; we

8A 8A i4

ix

8W

8W

2W 100 V + –

i1

4W

i2

ix v2

v1

10 W 3W

i3

2W + –

5W

100 V

(a)

3W

(b)

Fig. A-37 8i1 + 4(i1 – i2) = 100 4(i2 – i1) + 2i2 + 3(i2 – i3) = 0 3(i3 – i2) + 10(i3 + 8) + 5i3 = 0

4W

or or or

12i1 – 4i2 = 100 – 4i1 + 9i2 – 3i3 = 0 – 3i2 + 18i3 = – 80

v3 10 W 5W

672

Basic Electrical Engineering

Solving these equations for i2¸ we get ix = i2 = 2.79 A (b) By nodal analysis: Since the largest number of branches are incident on the bottom node, we choose this as the b. This leaves us with four nodes. However, since the voltage of the node between 100-V source and 8-W resistor is clearly known to be 100 V, we label the remaining nodes with voltages v1, v2, and v3. Applying KCL to these three nodes, we write v1 100 v v v + 1 + 1 2 4 2 8 v2 v2 v1 v2 v3 + + =8 3 2 10 v3 v2 v + 3 = –8 10 5 Solving these equations, we get

v1 – 0.5v2 = 12.5 or

– 0.5v1 + 0.9333v2 – 0.1v3 = 8

or

v1 = 25.89 V

–0.1v2 + 0.3v3 = – 8

and

v2 = 20.31 V

Thus, by Ohm’s law, we get v1 v2 = 2.79 A 2 Note that for this particular problem, the mesh analysis proves to be simpler. ix =

P ROBLEM

A -3 2

13 W

14 V I3 20Ix

15 W

20 V

18 W 7V

10 V 20 W

35 W

I1

Ix 16 V

I2

11 W

Fig. A-38 Solution is

The controlling quantity for the dependent source is Ix = I1 – I2. Hence the voltage of the dependent source 20Ix = 20(I1 – I2)

20I1 + 15(I1 – I3) + 20(I1 – I2) + 35(I1 – I2) = 10 + 16

or

90I1 – 55I2 – 15I3 = 26

Supplementary Exercise—Part A : DC Circuits 35(I2 – I2) + 18(I2 – I3) + 11I2 13I + 18(I3 – I2) + 15(I3 – I1) – 20(I1 – I2) = – 14 + 20

or

673

I1 + 64I2 – 18I3 = – 29 – 35I1 + 2I2 + 46I3 = 6

È 90 -55 -15˘ È I1 ˘ È 26˘ Í ˙Í ˙ Í ˙ 35 64 18 I = 2 Í ˙Í ˙ Í-29˙ ÍÎ-35 2 ÍÎ 6 ˙˚ 46 ˙˚ ÍÎ I 3 ˙˚ Note that due to the presence of the dependent voltage source, the resistance matrix has lost its symmetry. We could not The solutions of the above equations are i1 = 0.148 A, I2 = – 0.3 A, and I3 = 0.256 A. Finally, the power absorbed by the dependent voltage source is equal to the product of its voltage and the current P = [20(I1 – I2)](I1 – I3) = [20(0.148 + 0.3)](0.148 – 0.256) = – 0.968 W P RO B L EM

A -3 3

Consider the circuit of Fig. A-39a. Determine the maximum positive current to which the source Ix can be set before any resistor exceeds its power rating and overheats. i¢100 W

100 W 1 W 4 + –

6V

i¢¢100 W

100 W

100 W 64 W 1 W 4

Ix

+ –

64 W

6V

(a)

i¢¢64 W i¢64 W

(b)

64 W

Ix

(c)

Fig. A-39 Solution Both the resistors are rated to a maximum of

1 4

Ix and the

maximum current through each resistor. Since the power rating of each resistor is 0.25 W, the maximum current the 100-W resistor can tolerate is Pmax 0.25 = = 50 mA R 100 Similarly, the maximum current the 64-W resistor can tolerate is i100W(max) =

Pmax 0.25 = = 62.5 mA R 64 Next, to solve the problem we could apply mesh or nodal analysis. However, applying superposition principle gives i64W(max) =

6-V source to the currents in the two resistors by redrawing the circuit as in Fig. A-39b, 6 = 36.59 mA 100 + 64 Since this current is less than the maximum permissible current of either resistor, we are sure that the 6-V source acting alone will not pose any overheating problem. i¢100W = i¢64W =

674

Basic Electrical Engineering

Ix, we consider the circuit of Fig. A-39c. We note that i¢¢100 W is oposite in direction to i 100 ¢ W, but i≤64W adds to i ¢64W. Therefore, the currents that can safely be contributed by the source Ix to the two resistors, respectively, are i ≤100W(safe) = i100 W(max) – (– i¢100 W) = 50 – (– 36.59) = 86.59 mA i≤64W(safe) = i64W(max) – (i¢64W) = 62.5 – (36.59) = 25.91 mA

and

The current Ix must be limited to a value that does not produce currents greater than the above safe values for the two resistors. The limiting value of Ix for the 100-W resistor is given by i ≤100 W(safe) = I

W

Ix(lim)100 W = i ≤100W(safe)

or

⎛ 64 ⎞ ⎜ ⎟ ⎝ 100 + 64 ⎠ ⎛ 100 + 64 ⎞ ⎛ 100 + 64 ⎞ = 86.59 = 221.9 mA ⎝ 64 ⎠ ⎝ 64 ⎠

Similarly, the limiting value of Ix for the 64-W resistor is given by Ix(lim)64W = i≤64W(safe)

⎛ 100 + 64 ⎞ ⎛ 100 + 64 ⎞ = 25.91 = 42.49 mA ⎝ 100 ⎠ ⎝ 100 ⎠

Obviously, if none of the resistors is to be overheated, Ix must be less than 42.49 mA Ix is increased, the 64-W resistor will overheat long before the 100-W resistor does. Note that originally we found that the 100-W resistor has a smaller maximum current. So, it might appear reasonable to expect it to limit Ix. However, because Ix opposes the current sent by 6-V source through the 100-W resistor but adds to the current sent by 6-V source through the 64-W resistor, it turns out that it is 64-W resistor which sets the limit on Ix. PROBL EM

A-34

Use superposition principle to determine the current ix in the circuit of Fig. A-40a. 2W

10 V

+ –

1W

ix + v –

3A

(a)

2W

2ix

10 V

+ –

1W

2W

i¢x 2i¢x

(b)

1W

i¢¢x 3A

+ v¢¢ –

2i¢¢x

(c)

Fig. A-40 Solution that the dependent source has a nonzero value. Thus, we need to apply the superposition principle for the independent sources only. First, we consider the 10-V source acting alone (as in Fig. A-40b), and then the 3-A source acting alone (as in Fig. A-40c). Writing the single-mesh equation for the circuit of Fig. A-40b, 2i ¢x + 1ix + 2i ¢x = 10 fi Next, we write the single-node equation for the circuit of Fig. A-40c, v v¢¢ - 2i x¢¢ + =3 1 2

i¢x = 2 A

Supplementary Exercise—Part A : DC Circuits

675

and relate the dependent-source-controlling quantity to v ≤ : v≤ = 2(– i≤x) Solving above two equations, we get i x≤ ix = i¢x + i≤x = 2 + (– 0.6) = 1.4 A P ROBLEM

A -35

Find the Thevenin’s equivalent of the circuit of Fig. A-41. 30 V

10 W

8W a

+ 4W 100 V

5W

40 W

V 20 A –

b

Fig. A-41 Solution

Writing the single-node equation,

V - 100 V + = – 20 fi V = – 80 V 40 10 Since 8-W resistor has no current, Thevenin’s voltage is given as VTh = V + 30 = – 80 + 30 = – 50 V Note that the 5-W resistor (in parallel with the 100-V source) and the 4-W resistor (in series with the 20-A source) have no effect on VTh. For calculating RTh, we turn off all the sources in the given circuit to get the circuit of Fig. A-42a. Here, the voltage source is replaced by a short circuit and the current source by an open circuit. Thevenin’s resistance is given as RTh = Rab = 8 + 40 || 10 = 16 W Note again that the 5-W resistor (in parallel with the 100-V source) and the 4-W resistor (in series with the 20-A source) have no effect on RTh. Fig. A-42b shows the Thevenin’s equivalent circuit.

Fig. A-42 P RO B L EM

A -3 6

Compute the current through 2-W resistor in Fig. A-43a, by making use of source transformations to simplify the circuit to a single-loop circuit.

676

Basic Electrical Engineering

Fig. A-43 b). We can then combine resistances to Solution simplify the circuit. However, we must be careful to retain the 2-W resistor for two reasons. First, the dependent-source controlling voltage Vx W and 9-W resistors, we get a single 26-W resistor; and combining 3-W and 4-W resistors, we get a single W W resistor into its equivalent current source, we get the circuit of Fig. A-44a.

Fig. A-44

Supplementary Exercise—Part A : DC Circuits

677

W resistors (in parallel) into a single 3.5-W resistor. This resistor can now be used to b). We note that the controlling voltage Vx = (2)I = 2I. Applying KVL, we get I + 51Vx – 26I – 2I – 9 = 0 P RO B L EM

or – 1.5 – 31.5I + 51(2I) = 0 fi

I = 21.28 mA

A -3 7

Find the Thevenin’s equivalent of the circuit shown in Fig. A-45a. 2 kW

2 kW

3 kW

3 kW

+ vx 4000

4V + –

vx

+ 4V + –

vx 4000

–

vx

+ –

4V

–

(a)

(b)

Fig. A-45 Solution

VTh. We note that the voltage VTh is the same as the opencircuit voltage Voc = vx. With terminals open-circuited, the dependent source current must pass through the 2-kW resistor, W resistor. Applying KVL around the outer loop, we get vx + (3 ¥ 103)(0) – (2 ¥ 103) Hence,VTh = vx = 8 V

⎛ vx ⎞ –4=0 fi ⎝ 4000 ⎠

vx = 8 V

RTh. Here, because of the presence of the dependent source, we cannot RTh is needed : First Method ratio of this voltage to the current gives us RTh. Here, it is convenient to connect a 4-V source (Fig. A-45b). We then have vx = 4 V and the dependent current source becomes vx 4 = = 1 mA 4000 4000 The 4-V source in series with 2-kW resistor can be transformed into an equivalent 2-mA current source in parallel with 2-kW resistor (as shown in Fig. A-46a). Combining the two current sources and transforming it to an equivalent voltages source, we get the circuit of Fig. A-46b. The current I is given as I=

2 ( 6 - 4) V = mA = 0.4 mA (2 + 3) kW 5

Thus, RTh is given as 4V = 10 kW 0.4 mA Second Method Isc, so that RTh is given by the ratio of Voc to Isc. Upon short-circuiting the output terminals in Fig. A-45a, it is obvious that vx = 0 and consequently the dependent current source is dead. Hence, Voc VTh 8V 4V Isc = = 0.8 mA; RTh = = = = 10 kW Isc Isc 0.8 mA (2 + 3) kW The Thevenin’s equivalent is shown in Fig. A-46c. RTh =

678

Basic Electrical Engineering

Fig. A-46 P RO B L EM

A -3 8 a.

Fig. A-47 Solution

The given circuit has a dependent source but no independent source. Since the terminals are already opencircuited, i = 0. Consequently, the dependent source is dead and hence VTh = Voc = 0. RTh. Since the circuit has no independent source, both Voc and Isc are zero. Hence, we cannot determine RTh by taking the ratio of Voc and Isc. Let us, therefore, be a little tricky. We apply a 1-A source externally (as in b Vt across the terminals. The Thevenin’s resistance RTh is then given by the ratio Vt /1. Applying nodal analysis, we get V Vt - 1.5i + 2t = 1 3

\

P ROBLE M

RTh =

Vt 1

Vt - 1.5 (-1) V + 2t = 1 3

or =

0.6 = 0.6 W 1 c

fi Vt = 0.6 V

W.

A -39 RL will absorb maximum power and what is this power ? a 10I 8A

10 W

10 W

RL

I b

Fig. A-48

Supplementary Exercise—Part A : DC Circuits

679

Solution RL by an open circuit and then applying current division, we get 40 = 6.4 A 40 + 10

I=8¥

Consequently, the dependent voltage source becomes 10I = 64 V. Now, applying KVL, we get VTh = Voc = Vab = 10I + 64 = 10 ¥ 6.4 + 64 = 128 V Thevenin’s resistance RTh can be conveniently determined by the short-circuit current approach. When a short circuit is placed across terminals a and b, all the components of the circuit of Fig. A-48 become in parallel. Therefore, the voltage drop (from top to bottom) across the 10-W resistor must be equal to voltage across the dependent voltage source. That is, 10I = –10I This is possible only if I Isc = 8 A. Therefore, V 128 RTh = oc = = 16 W Isc 8 Thus, for maximum power absorption, RL = RTh = Pmax = PR OBLEM

. Finally, the maximum power absorbed is 2 VTh

4RTh

=

(128)2 = 256 W 4 16

A -4 0

Refer to the circuit shown in Fig. A-49. (a) Find current I by using a D-Y transformation. (b) What resistor R replacing the 20-W resistor causes the bridge to be balanced ? Also, what is the current I then ? I

8W 14 W

10 W 6W

196 V

1.6 W

20 W

Fig. A-49 Solution (a) We transform the top D to a Y. Figure A-50a shows the top D enclosing a Y as a memory aid for this transformation. All three Y formulae have the same denominator: D = 14 + 10 + 6 = 30 W D resistors : RA =

10 14 30

W; RB =

14 6 6 10 = 2.8 W; RC = =2W 30 30

680

Basic Electrical Engineering

Fig. A-50 b. Note that in this circuit all the resistors are in series-parallel. Thus, I=

196 = 12 A 8 + 4.67 + (2.8 + 1.6) || (2 + 20)

(b) We know that for the balance of the Wheatstone bridge, the product of opposite arms must be equal. That is 16 = 1.14 W 14 Under balance condition, the centre arm of the bridge carries no current. Hence, we can consider this as an open circuit, so that the bridge becomes a series-parallel arrangement. The current I is then R ¥ 14 = 1.6 ¥ 10

I= P ROBLEM

fi

R=

196 = 13.5 A 8 + (14 + 1.6) || (10 + 114 . )

A -41

Using a D

I1, I2, and I3 for the circuit shown in Fig. A-51. I1

4W

15 W 30 V I2

6W

20 W

20 W

15 W 20 W 40 V

15 W I3

8W

Fig. A-51

Supplementary Exercise—Part A : DC Circuits

681

The D of 15-W resistors transforms to a Y of 15/3 = 5-W resistors. These resistors are in parallel with the Y of 20-W resistors, as shown in Fig. A-52a.

Solution

4W

5W

4W

20 W

I1

30 V

4W 4W

20 W

5W

4W

I2

4W

6W I3

40 V 20 W

4W

5W (a)

8W (c)

(b)

Fig. A-52 Note parallel if all the resistances of each Y were not having the same value. However, when the resistances have the same a wire were connected between them. Comment What would you do to solve such a problem, if all the resistances of each Y were not having the same value ? D. Then the resistors of the two D D, which would again be converted to a Y. Let us now proceed with the solution. The two Y’s can be reduced to a single Y shown in Fig. A-52b, in which each Y resistance is 5 || 20 = 4 W. Replacing the D-Y combination in Fig. A-51 by this Y, we get a circuit shown in Fig. A-52c. Writing KVL equations for the two loops, we get 18I1 + 10I3 = 30

and

10I1 + 22I3 = 40

Solving these, we get I1 = 0.88 A and I3 = 1.42 A. Then, KCL applied to the right hand node gives I2 = –I1 – I3 = – 2.3 A.

A. 2. PRACTICE PROBLEMS ( A )

S I M P L E

P RO B LEMS

A-1. Find the energy stored in a 12-V car battery rated at 650 Ah. [Ans. 28.08 MJ] A-2. for 4 s causes it to give off 240 J of light and heat energy, what must be the voltage drop across the light bulb ? [Ans. 240 V] A-3. Find the power absorbed by each element in the circuit of Fig. A-53. [Ans. (from left to right) –56 W, 16 W, –60 W, 160 W, –60 W]

A-4. The circuit of Fig. A-54 shows a voltage source of V volts connected to a current source of I amperes. Find the power absorbed by the voltage source for (a) V = 2 V, I = 4 A (b) V = 3 V, I = –2 A (c) V = – 6 V, I = –8 A [Ans. (a) 8 W; (b) – 6 W; (c) 48 W] A-5. Determine the resistivity of platinum if a cube of 1 cm along each edge has a resistance of

682

Basic Electrical Engineering

+ –

I

Fig. A-53

Fig. A-54

W across opposite faces.

A-6.

V

[Ans. Wm] Vx = 1 V and VR = 9 V, calculate (a) the power absorbed by the element A, (b) the power supplied by each of the two sources. [Ans. (a) 45 W; (b) 5 W, 40 W]

A-9. Determine the current i and the power delivered by the 80-V source and by the 20-V source in the circuit of Fig. A-58. [Ans. 3 A, 240 W, –60W] i

10 W

7W

5W

30 V 80 V + –

+ –

20 V

8W

Fig. A-55

Fig. A-58

A-7. For the circuit shown in Fig. A-56, if v2 = 1000i2 and i2 = 5 mA, determine vs. [Ans. – 1 mV]

A-10. Determine the current i in the circuit of Fig. A-59. [Ans. – 0.333 A] 5V

vs + –

+ v1 –

i2 5v1

+ v2 –

15 W

25 W

i

–+

5V

+ –

5V

– +

5W

Fig. A-56 Fig. A-59

A-8. absorbed by each component. [Ans. P120V = –240 W, P30V = 60 W, P30 W = 120 W, P15W = 60 W]

Fig. A-57

A-11. Find the internal resistance of a 2-kW water heater that draws 8.33 A current. [Ans. 28.8 W] A-12. Find the voltage v in the circuit given in Fig. A-60. [Ans. A-13. Find the current and unknown voltages in the circuit shown in Fig. A-61. [Ans. 0.3 A, 3 V, 4.5 V, – 1.8 V, 2.4 V, – 3.3 V] A-14. Find the voltage Vab in the circuit shown in Fig. A-61. [Ans. A-15. A string of Deepawali lights consists of forty 3-W,

Supplementary Exercise—Part A : DC Circuits 10 W 10 W

v

+ V – 1

5V

+ V – 2

+

2W

+ 3A

15 W

a

4W

10 W

4A

–

Vab

12 V

683

– V3

I

2W +

V5 –

– V4 +

–

11 W

b

Fig. A-60

8W

8V

Fig. A-61

when the string is plugged into a 240-V socket, and what is the net hot resistance of each bulb. [Ans. 0.5 A, 12 W] A-16. A resistance R in series with an 8-W resistor absorbs 100 W when the two are connected across 60-V line. Determine the unknown resistance R. [Ans. 16 W or 4 W] A-17. terminal of a 180-V source through two resistors— one has 30 W resistance and the other of which has 45 V across it. Find the current and the unknown resistance. [Ans. 4.5 A, 10 W] A-18. Three parallel resistors have a total conductance of W and 5 kW, Ans. 1.25 kW] A-19. Refer to the circuit of Fig. A-62. (a) Determine the resistance R that results in the 12-V source delivering 3.6 mW to the circuit. (b) Determine the resistance R that results in the 25-kW resistor absorbing 1 mW. (c) Replace the resistor R with a voltage source such that no power is absorbed by either resistor.

A-20. Find the voltages across the various components in the circuit of Fig. A-63. [Ans. 5 V, 2 V, 3 V] 5W

2W

3.1 A

1W

–1.4 A

Fig. A-63 A-21. Determine the current in 4-W resistor in the circuit of Fig. A-64. [Ans. 0.83 A]

Fig. A-64

[Ans. (a) R = 0 W; (b) R = 20 kW; (c) 12 V]

6W

4W

i1

i2

15 kW 12 V

6W

+

+ –

R 25 kW

Fig. A-62

42 V + –

3W

Fig. A-65

– + 10 V

684

Basic Electrical Engineering

Fig. A-66 A-22. Use the technique of mesh analysis to determine i1 and i2 the power absorbed by the 3-W resistor. [Ans. 12 W] A-23. Determine the mesh currents in the circuit of Fig. A-66. [Ans. 5 A, – 8 A, 2 A]

A-30. ix in 15-W resistor. [Ans. 0.66 A]

7A

I3

8W

2W

6W

I1

8A

I2

40 V

Fig. A-67

Fig. A-68

A-24. Two 12-V batteries are being charged from a 16-V generator with an internal resistance of 2 W. The internal resistances for the two batteries are 0.8 W and 0.5 W positive terminals of the batteries. A-25. A-26. A-27. A-28. A-29.

[Ans. Find the node voltages in the circuit shown in Ans. V1 V2 Find the mesh currents in the circuit shown in Fig. A-68. [Ans. I1 = 3 A, I2 = – 8 A, I3 Use repeated source transformation to obtain current I in the circuit of Fig. A-69. [Ans. 2 A] Solve for the mesh currents in the circuit of Ans. I1 = 5 mA, I2 = – 2 mA] Determine the mesh currents in the circuit of source of – 1 V.

[Ans. I1

I2 = mA]

3W

4W

I +

18 V

6W

4A

V = 2I

2

–

Fig. A-69 7 kW

2 kW

20 V

4 kW

I1 18 V

Fig. A-70

I2

24 V

Supplementary Exercise—Part A : DC Circuits 5 kW

ix 15 W

5V + –

7W + –

5W

2A

685

Ix

1 mA

47 kW

3.5 V

3W

Fig. A-74 A-34. Find the voltage V across the 1-MW resistor in the

Fig. A-71 transformations.

A-31. What resistor draws a current of 5 A when connected across terminals a and b of the circuit of Ans. 6 W] 5W

40 mA

6 MW

6W

+ V – 1 MW

a

75 mA

4 MW

+ –

200 kW

3V

20 W

100 V

Fig. A-75

b

A-35. Using repeated source transformations, determine the Norton’s equivalent of network inside the

Fig. A-72 A-32. Compute current I transforming the 9-mA current source into an equivalent voltage source. [Ans. 3 kW

4.7 kW

RL if (a) RL = 15 W, (b) RL = 10 W, and (c) RL = 5 W. [Ans. IN = 1 A, RN = 5 W; (a) 0.25 A, (b) 0.333 A; (c) 0.5 A] 8W

I 9 mA

[Ans.

5 kW

+ 3V –

5A

2W

10 W

RL

Fig. A-73 A-33.

Ix W resistor after transforming the 5-V source into an equivalent current source. [Ans.

Fig. A-76 A-36. Use Thevenin’s theorem to determine the current I2W [Ans. 260.8 mA]

686

Basic Electrical Engineering 4W

5W

A-38. Find the Thevenin’s and Norton’s equivalents of I2 W

9V + –

4W

2W

6W

[Ans. W] A-39. Find the Thevenin’s equivalent of the circuit shown in Fig. A-80. [Ans. 12 V, 12 W] 8W

Fig. A-77

a

A-37. Find Thevenin’s and Norton’s equivalents for the network faced by 1-kW (i.e., treating 1-kW resistor as load). [Ans. 8 V, 1.6 mA, 5 kW] 2 kW

4V + –

6W

12 W

5A

48 V

3 kW

b

Fig. A-80 2 mA

1 kW

a + –

Fig. A-78 2 kW

15 V 8W

1 kW

4W + 10 V – b

3V + –

7 mA

5 kW

Fig. A-81 A-40. Convert the circuit shown in Fig. A-81 to a single voltage source in series with a single resistor. [Ans. 35/3 V, 8/3 W]

Fig. A-79

( B)

T R IC KY

PR OB LE MS

A-41. For the circuit shown in Fig. A-82, determine the power absorbed P1, P2 and P3 for (a) I = 2 A, and (b) I = –3 A. P1

I

8V P2

I P3 10 V

Fig. A-82

6I

[Ans. (a) P1 = 16 W, P2 = – 24 W, P3 = – 20 W; (b) P1 = – 24 W, P2 = – 54 W, P3 = 30 W] A-42. A dc circuit shown in Fig. A-83 has a voltage source V, a current source I an experiment, it was found that a particular resistor R dissipates 4 W when the voltage source V alone is active, and it dissipates 9 W when the current source I alone is active. Determine the power dissipated by R when both sources are active. [Ans. 25 W] A-43. vx and ix. [Ans. 12.8 V, 5.8 A]

Supplementary Exercise—Part A : DC Circuits

687

I

V

+ –

Resistive network

R

Fig. A-86 Fig. A-83

A-49. v1 and iy. (b

a ix v1 ix and iy. [Ans. (a) 25 V, 2.5 A; (b) 0.6 A, 0.3 A]

Fig. A-84 A-44. Determine voltage v in the circuit of Fig. A-85. [Ans. 50 V]

Fig. A-87 A-50. Find the total resistance RT of the resistor ladder network given in Fig. A-88. [Ans. 34 W] 16 W

RT

3W

8W

5W

24 W

4W

Fig. A-85 A-45. A short circuit across a current source draws 20 A. When the current source has an open circuit across it, the terminal voltage is 600 V. Find the internal resistance of the source. [Ans. 30 W] A-46. A short circuit across a current source draws 15 A. W resistor across the source draws 13 A, what is the internal resistance of the source ? [Ans. 65 W] A-47. Resistors R1, R2 and R3 are in series with a 100-V source. The total voltage drop across R1 and R2 is 50 V, and that across R2 and R3 is 80 V. Find the three resistances, if the total resistance is 50 W. [Ans. 10 W, 15 W and 25 W] A-48. i1, i2 and v3 using resistance combination method and current division. [Ans. 100 mA, 50 mA, 0.8 V]

14 W

9W

Fig. A-88 A-51. resistance RT with the terminals a and b (a) opencircuited, and (b) short-circuited. [Ans. (a) 45.5 W; (b) 33 W] 40 W a 90 W

RT

60 W b 10 W

Fig. A-89

688

Basic Electrical Engineering

A-52. A 12-V battery with a 0.3-W internal resistance is

A-57. Find R1 and R2 in the circuit of Fig. A-93. [Ans. 20 W and 5 W]

current should not exceed 2 A, what is the minimum resistance of a series resistor which will limit the current to this safe value ? [Ans. 1.2 W] A-53. Find the total resistance RT of the resistor ladder network shown in Fig. A-90. [Ans. 26.6 kW] 15 kW

6 kW

R2 2W 0.4 A

3 kW

R1

2A 12 V

RT

10 kW

8 kW

4 kW

Fig. A-93 6 kW

2 kW

5 kW

V in

A-58.

Fig. A-90

the circuit of Fig. A-94. [Ans. 36 V]

A-54. resistance RT with the terminals a and b (a) opencircuited, and (b) short-circuited. [Ans. (a) 18.2 W; (b) 18.1 W]

16 W

7W

RT

+ V –

45 W

8W a

10 W

8W

40 W

80 V 3W

15 W

b

Fig. A-94 5W

4W

Fig. A-91 A-55. The equivalent resistance of three parallel resistors is 10 W 40 W and 60 W, determine the resistance of the third resistor. [Ans. W] A-56. Calculate voltage V1 in the circuit of Fig. A-92. [Ans. 96 V]

A-59. current I in the load resistor RL for (a) RL = 0 W, (b) RL = 5 W, (c) RL = 25 W. [Ans. (a) 16 A; (b) 9.96 A; (c) 3.96 A] 8W 28 A 6W

20 W RL

I

Fig. A-95 A-60. Find i1 in the circuit of Fig. A-96. [Ans.

Fig. A-92

Supplementary Exercise—Part A : DC Circuits 10 W

90 V

A-64.

40 W

each current source.

+– + v2 –

i

+

20 V

[Ans.

R1 5W

1.8v3

v3

689

–

I + 100 V –

Fig. A-96

A

I1 I2

50 V B 30 V

R2

A-61. Find the power absorbed by each of the resistors in

Fig. A-99

[Ans. P2.5W = 250 W, P30 W P6 W P5W =180 W, P20 W = 45 W.] 2.5 W

1 W 3

6W

+–

100 V + –

30 W

5W

20 W

5V 4A

Fig. A-97

1 W 2

1 W 6

9A

Reference node

A-62. 10-W resistor.

[Ans.

Fig. A-100 A-65. Determine the node voltages V1 and V2 in the circuit of Fig. A-101. [Ans. 3.38 V, –5 V]

A

20 W

10 A

B

10 W

5W

15 A

Fig. A-98 A-63.

a R1 and R2 such that I1 = 1 A and I2 = 5 A, and (b R2 such that I1 = 0. [Ans. (a) R1 = 20 W, R2 = 8 W; (b) R2 = 2 W]

Fig. A-101 A-66. Determine the current Ix in the circuit of Fig. A-102. [Ans. – 4.86 mA]

690

Basic Electrical Engineering A-68. Find the Thevenin’s equivalent for the network of Fig. A-104. [Ans. –502.5 mV, –100.5 W]

Fig. A-102 A-67. Find the input resistance Rin in the circuit of Fig. A-103. [Ans. 33.3 W]

Fig. A-104 A-69. absorbed by the 12-W resistor in the circuit shown in Fig. A-105. [Ans. 685 W]

I

6W

Rin

25 W

1.5I

50 W 100 V

Fig. A-103

( C )

C HA L L E NGI NG

I

6A

12 W

Fig. A-105

PROBLEMS

A-70. A wire, 50 m in length and 2 mm2 in cross section, has a resistance of 0.56 W. A 100-m length of the same wire has a resistance of 2 W at the same temperature. Find the diameter of this wire. [Ans. 1.19 mm] A-71. Compute the power absorbed by each element in the circuit of Fig. A-106. [Ans. P120V = –960 W, P30 W =1920 W, Pdep = –1920 W, P15 W = 960 W]

Fig. A-106 A-72. Determine iA, iB, and iC [Ans. 3 A, –5.4 A, 6 A]

Fig. A-107 A-73. A wire-wound resistor is to be made from a 0.2-mm diameter constantan wire wound around a cylinder of 1-cm diameter. How many turns are needed for a resistance of 50 W. (Take resistivity of constantan –8 [Ans. 32 turns] as 49 ¥ 10 Wm) A-74. W resistor in the circuit of Fig. A-108. [Ans. A-75. A resistor in series with a 100-W resistor absorbs 80 W when the two are connected across a 240-V line. Find the unknown resistance. [Ans. 20 W or 500 W]

Supplementary Exercise—Part A : DC Circuits

691

Fig. A-108 A-76. The light bulb shown in the circuit of Fig. A-109. has a rating of 120 V, 60 W. What must be the supply voltage Vs for the light bulb to operate at rated values. [Ans. 285 V] 40 W

(b) P1 = –1.62 kW, P2 = 180 W, P3 = 360 W, P4 P5 = 405 W, S Pabs = 0] 100 kW

60 W

3 kW

25 kW 60 kW 10 kW

10 W

225 V

55 W

30 kW

Vs

I

Fig. A-111 Fig. A-109

i3 + v3 –

A-77. Find the current I in the circuit of Fig. A-110. [Ans. 4 A]

+ v2 –

i4

5i2

v1 4

i5 + v4 –

5W

+ v5 –

48 W 40 W

240 V

i2

+ v 60 V + – 1 20 W –

12 W 8W

i1

Fig. A-112

60 W

A-80

Fig. A-110 I in the circuit of Fig. A-111. [Ans. 4 mA] A-79. (a) Use Ohm’s law and Kirchhoff’s laws to evaluate all the currents and voltages in the circuit of Fig. A-112. (b) Calculate the power absorbed by each of the

A-78.

zero. [Ans. (a) v1 = v2 = 60 V, v3 = 15 V, v4 = v5 = 45 V, i1 i2 = 3 A, i3 = 24 A, i4 = 15 A, i5 = 9 A;

circuit. Without knowing the current-voltage relationship for the device, we can still solve this circuit since it obeys Ohm’s law and Kirchhoff’s laws. (a) Determine VDS, if ID = 1.5 mA. (b) Determine VGS, if ID = 1 mA and VG = 3 V. [Ans. (a) 1.5 V; (b) 1 V] A-81. Use both resistance and source combinations, as well as current division, in the circuit of Fig. A-114 W, 12-W and 15-W resistors. [Ans. P3W = 55.21 W, P12W = 0 W, P15W

692

Basic Electrical Engineering 300 W

vs

+ vp

50 kW

gmvp

+ vo

1 kW

–

–

Fig. A-116 A-84. Determine the current i1 [Ans. – 1.93 A]

Fig. A-113 10 W

5‡ 15 W 3W

2W

40 W 150 V + –

4‡ 10 V + –

60 W

i1

1W 6W

9‡

10 ‡ 1‡

3A

13 W

7‡ –+

2W 2W

30 V

12 W

Fig. A-114 A-82. Find the power absorbed by each circuit element of Fig. A-115, if the control for the dependent current source is (a) 0.8ix, and (b) 0.8iy. [Ans. (a) P5A = –1389 W, P10ms P40ms = 3086.9 W, Pdep = – 2469.5W; (b) P5A P10 ms = 240.8 W, P40ms = 963.5 W, Pdep = – 428.3 W]

Fig. A-117 A-85. Determine the mesh currents in the circuit of Fig. A-118. [Ans. I1 = 2 mA, I2 = – 3 mA, I3 = 4 mA] 3 kW

2 kW

I1

24 V

4 kW

I2

1

7 mA 2

I3

8V 3

Fig. A-118 10 mS

5A

iy 40 mS

A-86. For the circuit of Fig. A-119, (a) use nodal analysis to determine v1 and v2, and (b) compute the power absorbed by the 6-W resistor. [Ans. (a) 58.54 V, 64.4 V (b) 542.64 W] 10 A

ix 3W

Fig. A-115 vs = 10 cos 5t

12 W + v2 –

+ v1 –

gm = 25 ¥ 10–3 S and

A-83.

6W

240 V + –

30 W

vo (t). [Ans. vo(t) = – 248.5 cos 5t mV]

Fig. A-119

+ 60 V –

Supplementary Exercise—Part A : DC Circuits 6W

A-87. Use nodal analysis to obtain the value of vx in the circuit of Fig. A-120. [Ans. – 60.41 V]

90 V

693

a + Vx –

4W 0.125Vx

RL

8W

b

Fig. A-122 A-89.

Fig. A-120 A-88. Use superposition principle to obtain the voltage across each current source in the circuit of Fig. A-121. [Ans. v1 = 11.402 V; v2 = –0.5924 V]

Fig. A-121

RL will absorb maximum power and what is this power ? [Ans. 4.8 W

694

Basic Electrical Engineering

Supplementary Exercises—Part B : Electromagnetic Circuits

S U P P L E M E N TA R Y E X E R C I S E S B.1 Solved Problems B.2 Practice Problems

PA R T

B : ELECTROMAGNETIC

695

B

CIRCUITS

Assemblage of

N OT E This set of exercises provides practice to those who wish to attain a higher standard of learning the basic principles of Electrical Engineering. The real key to success is practice.

696

Basic Electrical Engineering

Supplementary Exercises—Part B : Electromagnetic Circuits

697

B.1. SOLVED PROBLEMS P RO B L EM

B - 1

Given the waveform of the current in a 3-H inductor as shown in Fig. B-1a, determine the inductor voltage and sketch it.

Fig. B-1 Solution Since the current is zero for t < –1 s, the voltage is zero in this interval. During the interval –1 s < t < 0, the current increases at a linear rate of 1 A/s, and hence the voltage produced is constant and is given by di L =3¥1=3V dt During the interval 0 < t < 2 s, the current is constant at 1 A. That is, di/dt = 0, and hence the voltage produced is zero. During the interval 2 s < t < 3 s, the current linearly decreases and results in di/dt = –1 A/s. Hence the voltage produced is constant and is given by di L = 3 ¥ (–1) = – 3 V dt Fort t > 3 s, the current is constant at zero, and hence the voltage v(t) is zero. The complete voltage waveform is shown in Fig. B-1b. P ROB L EM

B -2

A choke has an inductance of 3 H and a resistance (of the wire) of 0.1 W. As shown in Fig. B-2a, it is connected to an ac source of 12 sin (pt/6) A. (a) Find the maximum energy stored in the inductor. (b) Calculate how much energy is dissipated in its resistance in the time during which the energy is being stored in and then recovered from the inductor. i

0.1 W + vR –

12 sin

Choke L = 3 H, R = 0.1 W

pt A 6

(a)

12 sin

+

pt A 6

3H

–

(b)

Fig. B-2

vL

698

Basic Electrical Engineering

Solution (a) The energy stored in the inductor becomes maximum when the current through it becomes maximum. The maximum value of current is 12 A, and occurs at a time when sin (pt/6) = ±1. Therefore, the maximum energy stored is 1 1 w L = Li L2 = ¥ 3 ¥ (±12)2 = 216 J 2 2 (b) Figure B-2b shows the circuit representation of the choke. The voltage across the resistor is given by vR = Ri = 0.1 ¥ 12 sin (pt/6) = 1.2 sin (pt/6) V and the voltage across the inductance is given as d Ê di pt ˆ Ê pt ˆ Ê pt ˆ = 3 Á12 sin ˜ = 3 ¥ 12 ¥ (p/6) cos Á ˜ = 6pcos Á ˜ V ¯ Ë ¯ Ë dt dt 6 6 Ë 6¯ The current increases from zero to 12 A from t = 0 to t = 3 s; and again decreases from 12 A to zero in next 3 s. The vL = L

in the resistor is given by pR = Ri 2 = 0.1 ¥ (12sin(pt/6))2 = 14.4sin2 (pt/6) W Therefore, the energy converted into heat in the resistor within this 6 s is given as 6

wR =

Ú

6

0

P RO B L EM

Êp ˆ t dt = Ë 6 ˜¯

6

14.4 sin2 Á

pR dt = 0

p ˆ Ê 1ˆ Ê 14.4 Á ˜ Á1 - cos t ˜ dt = 43.2 J Ë 2¯ Ë 3 ¯

0

B - 3 2

. Assuming that there is a) the relative permeability of the metal, (b) the self-inductance of the coil, and (c) the average value in 1.5 ms.

Solution (a NI 3500 ¥ 0.6 = l 0.25 The relative permeability of the metal is given as B 0.45 mr = = = 42.63 m 0H 4 π × 10 −7 × 8400 H=

2

–2

2

–4

2

(b) Area of cross section of the metal ring, A = pr = p(2.5 ¥ 10 /2) = 4.911 ¥ 10 m \ F = BA = 0.45 ¥ 4.911 ¥ 10 – 4 m2 = 2.21 ¥ 10 – 4 Wb \

L =

Self-inductance,

NF 3500 ¥ 2.21 ¥ 10-4 = = 1.289 H 0.6 I

F1 = 2.21 ¥ 10 – 4 Wb F2 ¥ 2.21 ¥ 10 – 4

(c

\

e = –N

¥ 10 – 4 Wb

(0.1768 - 2.21) ¥ 10 - 4 dF = –3500 ¥ = 474.4 V 0.0015 dt

Supplementary Exercises—Part B : Electromagnetic Circuits P RO B L EM

699

B - 4 2

2

is produced

2

1.2 Wb/m . (a) Find the mean value of inductance between these limits of current. (b

Solution (a) The mean value of inductance, L =N

dB d (BA) dF =N = NA dI dI dI

= NA

B2 - B1 I 2 - I1

(b) The induced emf, e = –L P RO B L EM

–4

¥ (15 ¥ 10 ) ¥

1.2 - 0.8 = 0.096 H = 96 mH 10 - 5

di 5 - 10 = –0.096 ¥ = 12 V dt 0.04

B - 5

30 ms.

Solution We know that the inductance of a solenoid is given as N 2m A N 2m r m0 A = l l 2 (1200) ¥ 1 ¥ 4p ¥ 10 - 7 ¥ [p ¥ (2.5 ¥ 10 -2) 2 ] = = 4.44 ¥ 10 –3 H = 4.44 mH 0.8 The induced emf in the coil due to the change in current is given as [( -10) - 10] di e = –L = – 4.44 ¥ 10 –3 ¥ = 2.96 V 0.03 dt L =

P RO B L EM

B - 6

of 25 cm2 is placed centrally in the solenoid. Calculate (a) the mutual inductance between the two coils, and (b) the emf induced in the search coil when the current in the solenoid is changing uniformly at the rate of 200 A/s.

Solution I is given as NI –7 1000 I Bc = m0 = 4p ¥ 10 ¥ = 15.715 ¥ 10 – 4 I T 0.8 l

(a)

F2 = Bc ¥ Area = (15.715 ¥ 10–4I) (25 ¥ 10–4 Therefore, the mutual inductance between the coils is given as M=

¥ 10 I Wb

N 2F2 400 ¥ 392.875 ¥ 10 - 8I = = 1.57 ¥ 10 –3 H = 1.57 mH I1 I

700

Basic Electrical Engineering

(b) The emf induced in the search coil due to a change in current in the solenoid, e2 = –M P RO B L EM

di1 –3 = –1.57 ¥ 10 ¥ 200 = – 0.314 V di

B - 7

2 milliseconds, the voltage induced in a nearby coil is 60 V. Calculate the self-inductance of each coil and the mutual

Solution L1 = N1

F 0.3 ¥ 10 -3 = 250 ¥ = 37.5 mH I1 2 e2 = M

Hence, we have 60 = M ¥

(2 - 0)

di1 dt

= M ¥ 103 fi

M=

60

= 60 mH 2 ¥ 10 The self-inductance of the second coil can now be calculated from the relation, -3

2

M = k L1L2 P RO B L EM

103

2

Ê M ˆ 1 Ê 60 ˆ 1 fi L2 = Á ˜ = mH = 195.9 mH Ë k ¯ L1 ÁË 0.7 ˜¯ 37.5

B - 8

A solenoid of 500 turns is wound on a former of length 1 m and of diameter 3 cm. This is placed co-axially within and material is copper, determine the inductance and resistance of each solenoid. (r for copper = 0.5 mW m)

Solution Given: l1 = l2 = 1 m; N1 = N2 = 500;

A1 = p(1.5 ¥ 10 –2)2 = 7.07 ¥ 10 – 4 m2; A2 = p(3 ¥ 10 –2)2

¥ 10 – 4 m2

The self-inductances of the two solenoids are m A N 2 4 p ¥ 10 - 7(7.07 ¥ 10 - 4)(500) 2 L1 = 0 1 1 = = 0.222 mH l1 1 m A N 2 4 p ¥ 10 - 7 (28.28 ¥ 10 - 4)(500) 2 and L2 = 0 2 2 = = 0.888 mH l2 1 Now, the total lengths of the wire used in the two windings, lw1 = pd1N1 = p ¥ 3 ¥ 10 –2 ¥ 500 = 47.124 m and lw2 = pd2N2 = p ¥ 6 ¥ 10 –2 ¥ Area of cross section of the wire, Awl = Aw2 = pr 2 = p ¥ (0.5 ¥ 10 –3)2 ¥ 10–6 m2 Therefore, the resistances of the two solenoids are l 47.124 R1 = r w1 = 0.5 ¥ 10–6 ¥ = 30.0 W Aw1 0.785 ¥ 10 - 6 and

R2 = r

lw 2 94.248 = 0.5 ¥ 10 – 6 ¥ = 60.0 W Aw 2 0.785 ¥ 10 - 6

Supplementary Exercises—Part B : Electromagnetic Circuits P RO B L EM

701

B - 9

A long solenoid of cross-sectional area 2 cm2 has primary winding with 25 turns per cm. At the middle of the solenoid is wound a secondary winding of 100 turns. The primary winding is carrying a current of 2 A. The supply to the primary winding is suddenly disconnected so that its current falls to zero in 1 ms. Determine the average emf induced in the secondary winding.

Solution Assuming the length of the solenoid to be l metres, we have N1 = 2500l, N2 = 100; A = 2 cm2 = 2 ¥ 10 – 4 m2 Therefore, the mutual inductance is given as m0 AN1N 2 4p ¥ 10 -7 (2 ¥ 10 - 4) (2500 l ) (100) = = 62.83 mH l l The induced emf in the secondary is given as M=

e2 = M P RO B L EM

di1 dt

¥ 10 – 6 ¥

(2 - 0) 1 ¥ 10 -3

= 125.66 mV

B - 1 0

(a) the inductance of the coil, (b) the emf induced in the coil when a current of 5 A is switched off so that it uniformly reduces to zero in 1 ms, and (c) the mutual inductance between the coils if the second coil of 600 turns is uniformly

Solution (a

L1 = N1

dF (0.5 - 0) ¥ 10- 3 = 900 ¥ = 0.15 H di1 (3 - 0)

(b e1 = L1

di1 (5 - 0) = 0.15 ¥ = 750 V dt (1 - 0) ¥ 10 - 3

(c

0.5 5 3 There are following two ways of obtaining the induced emf in the second coil : F=

e2 = M

di1 dt

and

e2 = N2

dF dt

Hence, the mutual inductance between the two coils is given by dF dF dt dF di ◊ M 1 = N2 fi M = N2 = N2 di1 dt di1 dt dt \

M = N2

dF [(2.5 / 3) − 0] × 10 − 3 = 600 ¥ ª 0.1 H (5 − 0 ) di1

702

Basic Electrical Engineering

P RO B L EM

B - 1 1

The combined inductance of two coils connected in series is either 0.75 H or 0.25 H, depending on the relative (a) the mutual inductance, and (b

Solution (a) The equivalent inductance of the series aiding combination is given as Lsa = L1 + L2 + 2M 0.75 = 0.15 + L2 + 2M

or

(i)

And the equivalent inductance of the series opposing combination is given as Lso = L1 + L2 – 2M 0.25 = 0.15 + L2 – 2M

or Adding Eqs. (i) and (ii), we get

1.0 = 0.3 + 2L2

fi

(ii)

L2 = 0.35 H

Thus, from Eq. (i) we get 0.75 = 0.15 + 0.35 + 2M

fi M = 0.125 H

(b k=

P ROBLEM

M = L1L2

0.125 = 0.5455 0.15 0.35

B -12

Two coils A and B, having 100 and 150 turns respectively, are wound side-by-side on a closed iron circuit of cross section 125 cm2 and mean length 2 m. Determine (a) the self-inductance of each coil, (b) the mutual inductance between the coils, and (c) the emf induced in coil B, when the current in coil A is changed from zero to 5 A in 0.02 s. Take the relative permeability of iron as 2000.

Solution (a) The reluctance of the closed iron circuit is given as l 2 R= = = 63.662 kAt/Wb 7 m0 m r A 4p ¥ 10 ¥ 2000 ¥ 125 ¥ 10 - 4 The self-inductances of the two coils are LA =

2 NA (100)2 = = 0.157 H R 63.662 103

and

LB =

NB2 (150)2 = = 0.353 H 63.662 103 R

(b) The mutual inductance between the coils is M=

NANB 100 150 = = 0.2356 H R 63.662 103

(c) The emf induced in coil B is eB = M

diA 5 0 = 0.2356 ¥ = 58.9 V 0.02 dt

Supplementary Exercises—Part B : Electromagnetic Circuits P RO B L EM

703

B - 1 3

co-axially within another solenoid B of same length and same number of turns but of diameter 6 cm. Find (a) the mutual inductance, and (b

Solution Given : Na = Nb = N

la = lb = l = 0.9 m

(a) The mutual inductance of the arrangement is given by M=

N aF a Ib

(i) Ib is given by

B=

m0 N b I b lb

Fa = BAa =

(ii)

m0 N b I b Aa lb

(iii)

Substituting Eq. (iii) into Eq. (i), we get M=

m0 N a N b I b Aa m0 N a N b Aa = lb I b lb

(iv)

Thus, the mutual inductance of the arrangement is given as m0 N a N b Aa 4 π × 10 −7 × (800)2 × [( π / 4) (0.03)2 ] = = 0.632 ¥ 10 –3 = 0.632 mH lb 0.9 (b) The self-inductances of the two solenoids are given as M=

La =

m0 N a2Aa la

and

Lb =

m0 N b2 Ab lb

(v)

Using Eqs (iv) and (v m N N A M = 0 a b a ◊ lb L aLb

k =

= P RO B L EM

la lb m0 N a2 Aa m0 N b2 Ab

m0 N 2Aa l = ◊ l m0 N 2 Aa Ab

Aa = Aa Ab

=

Aa = Ab

m0 N 2 Aa ◊ l

l2 m02 N 4 Aa Ab

( / 4)(0.03)2 ( / 4)(0.06)2

=

0.03 = 0.5 0.06

B - 1 4 2

and relative permeability of 500, is 2 . (a) Determine

the exciting current, the inductance and the stored energy. (b above quantities.

Solution (a H=

B 1.2 = = 1910 At/m m0 m r 4p ¥ 10 - 7 ¥ 500

704

Basic Electrical Engineering

li = pD = p ¥ 0.1 = 0.3142 m Thus, total ampere turns required, At = Hli = 1910 ¥ 0.3142 = 600 At 600 Therefore, the exciting current, I = = =2A N 300 The inductance of the coil can be calculated as L=

NF NBA 300 × 1.2 × (8 × 10 − 4 ) = = = 0.144 H = 144 mH 2 I I

W=

1 2 1 2 LI = ¥ 0.144 ¥ (2) = 0.288 J 2 2

(b) After an air gap of 2 mm is cut in the ring, there are two kinds of magnetic paths in series. The iron path length, li = 0.3142 – 0.002 = 0.3122 m The air gap path length, lg = 0.002 m 2 , as calculated above, the ampere turns per metre of iron path length required is Hi = 1910 At/m. Hence, the total ampere turns required by the iron path is

Ati = Hli = 1910 ¥ 0.3122 = 596.3 And the ampere turns required by the air gap is Atg = Hglg =

B

lg =

0

1.2 4 π × 10 − 7

¥ 0.002 = 1910

Therefore, the total ampere turns required, At = Ati + Atg = 596.3 + 1910 = 2506.3 Hence, due to the presence of air gap, the exciting current is much higher and is given as At 2506.3 = = 8.35 A I= N 300 The inductance, due to the presence of air gap, is much reduced and is given as L=

NF NBA 300 × 1.2 × (8 × 10 − 4 ) = = = 0.03449 = 34.49 mH 8.35 I I

The stored energy, W=

1 2 1 LI = ¥ 0.03449 ¥ 2 2

2

= 1.202 J

Note magnetic path. P RO B L EM

B - 15

An iron ring of mean length 100 cm and circular cross-sectional area of 10 cm2 has an air gap of 1 mm and a winding of 100 turns. Assuming the relative permeability of iron as 500, determine the inductance of the coil.

Solution For a magnetic circuit, the mmf F magnetic path. That is, F = FR . Or, NI = FR

F produced and the reluctance R of the fi

F=

NI R

Supplementary Exercises—Part B : Electromagnetic Circuits

705

Therefore, the inductance of the winding is given by N NI N 2 NF L= = = ◊ I R I R

Rt = Ri + Rg = = \

L =

P RO B L EM

lg 1 li = (li + mrlg) + m0 m r A m0 A m0 m r A 1

4p ¥ 10

-7

¥ 500 ¥ 10 ¥ 10 - 4

((1 – 0.001) + 500 ¥

¥ 106 At/Wb

N2 (100)2 = = 4.2 mH R 2.38 10 6

B -1 6

A coil of 100 turns is wound on a toroidal core having a reluctance of 104 At/Wb. (a) When the coil current is 5 A and coil, assuming coil resistance to be zero. (b) How are your answers affected if the coil resistance is 2 W ?

Solution The inductance of the coil is given by L=

N2 (100)2 = =1H R 10 4

(a) The energy stored is W=

1 2 1 LI = ¥ 1 ¥ 52 = 12.5 J 2 2

The voltage applied across the coil is same as the emf induced in the coil, di v = e = L = 1 ¥ 200 = 200 V dt (b) The energy stored in the coil would remain the same. However, there would be an additional energy loss in the coil due to its resistance, WR = I 2R = 52 ¥ 2 = 50 W. Furthermore, the voltage across the coil would be increased by VR = IR = 5 ¥ 2 = 10 V. Therefore, the net voltage across the coil would be v = e + VR = 200 + 10 = 210 V P ROBLE M

B -1 7

A 0.5-m long single-layer solenoid has an effective diameter of 10 cm and is wound with 2500 turns. A small concentrated co-axial coil of diameter 12 cm is wound with 160 turns in the middle of the solenoid. Calculate the mutual inductance between the two coils.

Solution Let I1 B = m0H =

0 N1I1

l1

=

4p ¥ 10 - 7 ¥ 2500I1 0.5

The area of cross section of the solenoid, A1 =

4

¥ (0.1)2

¥ 10 –3 m2

¥ 10–3I1 Wb/m2

706

Basic Electrical Engineering

¥ 10 –3I1 ¥

F1 = BA1

¥ 10 –3 = 49.32 ¥ 10–6I1 Wb

F1, irrespective of its area of cross section. That is, –6

F2 = F1 = 49.32 ¥ 10 I1 Wb Hence, the mutual inductance between the two coils is M= P RO B L EM

N2 I1

2

=

160 ¥ 49.32 ¥ 10- 6I1 I1

¥ 10 –3 = 7.891 mH

B - 1 8 2

cross section. The relative

of 1200 turns on each limb, calculate the exciting current needed. Assume that the car makes close contact with the magnet.

Solution The force required to lift the car, Fr = 1190 ¥ The force that has to be exerted by one pole of the electromagnet on the car is Fr 11662 = 2 2 The force exerted by a pole of the electromagnet is given as B 2A F = 2m0 F=

\

F ¥ 2m0 = A

B =

H= The total ampere turns needed,

5831 ¥ 2 ¥ 4p ¥ 10 -7 80 ¥ 10 - 4

= 1.3535 Wb

B 1.3535 = = 1346.35 At/m m0 m r 4p ¥ 10 -7 ¥ 800 At = Hl = 1346.35 ¥ 1.2 = 1615.62

Since both the limbs have coils, the total number of turns, N = 2 ¥ 1200 = 2400 Finally, the exciting current required, At 1615.62 I= = = 0.6732 A 2400 N P RO B L EM

B - 1 9

The following particulars are taken from the magnetic circuit of a relay : Mean length of the iron circuit = 20 cm;

Length of air gap

= 2 mm;

Relative permeability of iron

Area of the core

= 0.5 cm2;

Solution The mmf, F = NI

= 500;

¥ 50 ¥ 10 –3 = 400 At

Supplementary Exercises—Part B : Electromagnetic Circuits

707

The total reluctance of magnetic path is given as lg 1 li R = Ri + Rg = = (li + mrlg) + m0 m r A m0 A m0 m r A =

1 4p ¥ 10

-7

¥ 500 ¥ 0.5 ¥ 10

-4

¥ (20 ¥ 10

–2

+ 500 ¥ 2 ¥ 10

–3

7

¥ 10 At/Wb

400 At F F = = = 1.047 ¥ 10 –5 Wb R 3.82 ¥ 107 At / Wb 1.047 × 10 −5

= 0.2094 Wb/m2 A 0.5 × 10 − 4 The pull on the armature is given as B 2A 0.2094 ¥ 0.5 ¥ 10 - 4 F= = = 4.166 N 2m0 2 ¥ 4p ¥ 10 - 7 B =

P RO B L EM

=

B - 2 0

The hysteresis loss of iron weighing 12 kg is equivalent to 300 J/m3/cycle. Find the loss of energy/hour at 50 Hz. The 3 . m 12 = ¥ 10 –3 m3 d 7800 Hysteresis loss in iron per cycle = (300 J/m3/cycle) ¥ ¥ 10 –3 m3) = 0.4614 J/cycle Hysteresis loss per hour = (0.4614 J/cycle) ¥ (50 cycles/s) (3600 s) 83.052 kJ

Solution The volume of 12 kg iron, V = \ \

P RO B L EM

B - 2 1

The hysteresis loop of an iron ring was found to have an area of 10 cm2 on a scale of 1 cm = 1000 At/m (x axis) and 1 cm = 0.2 Wb/m2 (y axis). The ring has a mean length of 100 cm and a cross-sectional area of 5 cm2. Compute the hysteresis loss in watts for a frequency of 50 Hz.

Solution Hysteresis loss per cycle, Whc = (10 cm2) ¥ (1000 At/m)/cm ¥ (0.2 Wb/m2)/cm = 2000 At-Wb/m3 = 2000 J/m3 The volume of the ring, V = (1 m) ¥ (5 ¥ 10 – 4 m2) = 5 ¥ 10 – 4 m3 Therefore, the hysteresis loss for the ring in watts, Wh = (2000 J/m3/cycle) ¥ (50 cycles/s) ¥ (5 ¥ 10 – 4 m3) = 50 J/s = 50 W P RO B L EM

B - 2 2

B-H of 0.3 Wb/m2 requires the following magnetising forces : Sheet steel, Hss = 200 At/m; Cast iron, Hci the air gap.

708

Basic Electrical Engineering

Solution Here, we have N = 100. For air gap, Fg = 0.3 ¥ 10 –3 Wb; Ag = (3 ¥ 2) cm2 = 6 ¥ 10 –4 m2; lg = 2 mm = 2 ¥ 10 –3 m For sheet steel, Fss = 0.3 ¥ 10 –3 Wb; Hss = 200 At/m; lss For cast iron,

Fci = 0.3 ¥ 10 –3 Wb; Hci g

lci = 10 cm = 0.1 m

0.3 × 10 −3

= 0.5 Wb/m2 Ag 6 × 10 − 4 Thus, the total ampere turns (or mmf) required, Bglg 0.5 ¥ 2 ¥ 10 - 3 F = + Hsslss + Hcilci = + 200 ¥ m0 4p ¥ 10 - 7 Bg =

=

¥ 0.1

Hence, the exciting current required, I=

2 mm air gap

Steel shell

A

F b

c

10 cm

B

l2

Cast steel

a

F 1040.4 = = 10.4 A N 100

l1

12

12

N= 3600

54 0.15

12 3 cm 2 cm e

1 10 cm

12 E

D

C 54

1 cm

Fig. B-3 P RO B L EM

30

d

Fig. B-4

B - 2 3

A magnetic circuit has two parallel branches, which are similar, as shown in Fig. B-4. All dimensions are given in cm. The relative permeability of steel at the operating point is 700. A coil of 3600 turns is wound on the central limb.

Solution Let B steel-limbs is also B. The area of cross section of the side limb, As = (12 ¥ 12) cm2 = 144 ¥ 10 –4 m2 –4

Fs = As B = 144 ¥ 10 B Wb Fc = 2Fs

¥ 10 –4B Wb

Supplementary Exercises—Part B : Electromagnetic Circuits The area of cross section of the central limb, Ac = (12 + 12) ¥ 12 cm2 = 288 ¥ 10 – 4 m2

Bc =

c

Ac

=

288 × 10 − 4 B 288 × 10 − 4

= B Wb/m2

Now, the mean length of one side steel-limb, ls = lABCD = (54 – 6 – 6) + (54 – 6 – 6 – 0.15) + (54 – 6 – 6) = 125.85 cm = 1.2585 m The mean length of the central limb, lc = 54 – 6 – 6 = 42 cm = 0.42 m F = Fc + Fs + Fg F = Hclc + Hsls + Hglg =

or

= =

B 4p ¥ 10 -7 ¥ 700 B 4p ¥ 10 -7 ¥ 700

B B B B (lc + ls + mr lg) lc + ls + lg = m0 m r m0 m r m0 0 r ¥ (0.42 + 1.2585 + 700 ¥ 0.15 ¥ 10 –2)

¥ (2.7285) At

F = NI = 3600 ¥ 1.6 = 5760 At ¥ (2.7285) = 5760 4p ¥ 10 -7 ¥ 700 5760 ¥ 4p ¥ 10 - 7 ¥ 700 B = = 1.857 Wb/m2 2.7285 B

\ or

P R O B LE M

B -2 4

The length of the iron path is 50 cm and the area of each pole shoe is 3 cm2 300. Find (a b

Solution (a) Let Bg or

F = Fi + Fg

or

2000 At = Hili + Hglg =

li +

Bg m0

lg =

Bg m0 m r

(li + mr lg)

Bg

¥ (0.5 + 300 ¥ 1.5 ¥ 10 –3) = 2520Bg At 4p ¥ 10 -7 ¥ 300 2000 2 Bg = = 0.794 Wb/m 2520 Bg2 A (0.794) 2 ¥ 3 ¥ 10 - 4 F = = = 75.2 N 2m0 2 ¥ 4p ¥ 10-7 =

\

Bg m0 m r

Ft = 2F = 2 ¥ 75.2 = 150.4 N

709

710

Basic Electrical Engineering

(b) With lg = 0.2 mm, we have 2000 At = =

Bg m0 m r

(li + mr lg) Bg

4p ¥ 10 - 7 ¥ 300

¥ (0.5 + 300 ¥ 0.2 ¥ 10

–3

Bg At

2000 2 = 1.3463 Wb/m 1845.44 The force needed at each pole to pull the armature away is given as

\

Bg =

F =

Bg2 A 2m0

=

(1.3463) 2 ¥ 3 ¥ 10 - 4

Due to the two poles, the total force needed Ft = 2F = 2 ¥ PROBLEM

2 ¥ 4p ¥ 10 - 7 432.76 N

B -25

and B, with relative permeabilities of 1000 and 1200 respectively, separated by two air gaps, each 2 mm wide. The part B is wound with a total of 1000 turns of wire on the two side limbs and carries a current of 1 A. Calculate (a) the reluctance of part A, (b) the reluctance of part B, (c) the reluctance of the air gaps, (d) the total reluctance of the complete magnetic circuit, (e) the mmf, (f g

Solution (a) The reluctance of part A, Ra =

la 17 ¥ 10 - 2 = = 0.1503 ¥ 106 At/Wb ma m ra A 4p ¥ 10 - 7 ¥ 1000 ¥ 32 ¥ 10 -4

(b) The reluctance of part B, Rb =

lb (17 + 8.5 + 8.5) ¥ 10 - 2 = = 0.2505 ¥ 106 At/Wb m0 m rb A 4p ¥ 10 - 7 ¥ 1200 ¥ 32 ¥ 10 - 4

(c) The reluctance of the air gaps, Rg =2¥

lg m0 A

=2¥

2 ¥ 10 - 3 4p ¥ 10 - 7 ¥ 32 ¥ 10 - 4

= 3.537 ¥ 106 At/Wb

17 cm Part A 2 mm F

10 cm Part B 20 cm

Fig. B-5

Fig. B-6

Supplementary Exercises—Part B : Electromagnetic Circuits

711

(d) The total reluctance of the complete magnetic circuit, 6

6

R t = R a + R b + R g = (0.1503 + 0.2505 + 3.537) ¥ 10 = 3.938 ¥ 10 At/Wb (e) The mmf, F = NI = 1000 ¥ 1 = 1000 At mmf F 1000 –3 (f F= = = = 0.2539 ¥ 10 = 0.2539 mWb reluctance R 3.938 10 6 B=

(g

P RO B L EM

A

=

0.2539 × 10 − 3 (3 × 3) × 10 − 4

2

= 0.2821 Wb/m

B - 2 6 2

2

sectional area of the central limb is 6 cm and of each outer limb is 4 cm wrought iron gives B (Wb/m2)

1.125

1.5

H (At/m)

500

2000

Solution For the central limb : B1 =

F1 0.9 × 10 − 3 = = 1.5 Wb/m2 6 × 10 − 4 A1 2

H = 2000 At/m.

Therefore, the ampere turns required for the central limb, At1 = H1l1 = 2000 ¥ 0.1 = 200 At For outer limb 1 1 F1 = ¥ 0.9 = 0.45 mWb 2 2 0.45 × 10 − 3 B2 = 2 = = 1.125 Wb/m2 A2 4 × 10 − 4

F2 =

2

H = 500 At/m.

Therefore, the ampere turns required for the outer limb, At2 = H2l2 = 500 ¥ 0.2 = 100 At 2

For the air gap : Bg = B2 = 1.125 Wb/m ; lg = 1 ¥ 10 –3 Hg =

Bg

=

1.125

4p ¥ 10 - 7 Therefore, the ampere turns required for the air gap,

¥ 105 At/m

0

Atg = Hglg

\

Exciting current, I =

¥ 105 ¥ 10 –3

At 1195.2 = = 2.39 A N 500

712

Basic Electrical Engineering

B. 2. PRACTICE PROBLEMS ( A )

S I M P L E

P RO B LEMS

B-1.

of motion is (a (b

2

Ans. (a) 28.8 V; (b) 20.4 V] Ans. 12 mV]

B-9.

B-2.

of the coil is 100 W emf when the plane of the coil is (a) at right angles b c) in the

Ans. 3 V, 30 mA] B-10.

Ans. (a) 0 V; (b) 45.4 V; (c) 52.4 V] 1 Wb/m2 when the plane of the coil is at (a) 30°, and (b) 45°

B-3. B at a point on the road if the line is 8 m above the road ? Ans. 2.5 mT] B-4.

Ans. (a) 90.67 V; (b) 74 V] B-11.

–1

of mean diameter 140 mm and the cross-sectional area of 750 mm2 Ans. 21.469 mH] B-12. A solenoid 80 cm in length and 8 cm in diameter Ans. 100 V]

(a

b

B-5. Ans. (a) 0.126 H; (b) 0.252 J] B-13. ¥ 10

W. When

–5

Ans. 9.4 mV] B-6.

Ans. 1.043 H, 11.04 J] B-14. An iron ring of mean diameter 30 cm having a ¥ 2

Ans. 50 At, 250 At/m] B-7.

mr = 800. Ans. 68.3 mH]

m B-15.

Ans. 4.5 V] B-8. A wire of length 80 cm moves at right angles to its 2

. Determine the

Ans. 0.15 H, 300 V]

Supplementary Exercises—Part B : Electromagnetic Circuits

mWb.

B-23. A solenoid of length 1 m and diameter 10 cm a) the approximate b

Ans. 0.09 H, 4.5 J, 180 V]

Ans. (a) 0.25 H; (b) 0.5 J]

B-16.

B-17.

713

B-24.

105 b) the voltage

(a

changed from 2 A to –2 A in 0.5 second, an emf of a b

c t = 1 s is 1 A. Neglect the resistance of the coil. Ans. (a) 0.4 H; (b) 80 V; (c) 0.2 J] B-18.

Ans. (a) 1 mH; (b) 20 mWb] B-25.

placed near each other. With coil B

m Ans. 28.8 V] B-19.

Ans. 1.25 mH] B-26. coil B

10 –5 Ans. 1.25 mH] mH

B-20. and 40 m

mWb in coil B.

is Ans. 1.44 V] B-27. 120 mH and 300 m

a

mWb-

coils, (b (c

a between the coils, (b (c

Ans. (a) 2.5 mH; (b) 0.125; (c) 50 mV] B-21. Two identical coils A and B, each having 1000

Ans. (a) 100 mH; (b) 0.527; (c) 2 mV] B-28.

2

mWb.

A changes form 6 A to – 6 A in 0.01 s, what will be ? Ans. 0.01 H, 0.005 H, 6.0 V] B-22.

1.0 Wb/m2 Ans. 2 B-29. An iron ring 15 cm in diameter and 10 cm in cross 2

Ans. 3.75 A] B-30.

a cross-sectional area of 3 cm2 mH. Find (a) the b Ans. (a) 60.32 mH; (b) 0.0083]

contact area

2

Ans. 3001 At]

714 ( B)

Basic Electrical Engineering

T R IC KY

PR OB LE MS

B-31. A conducting circular loop having a radius of

4 0.04 mWb in it, and 70 coil B. (a) Calculate the self-inductance of each coil, and the mutual inductance between them. (b A in 20 ms, what will be the emf induced in coil B ? [Ans. (a) 15 mH, 10.5 mH; (b) 4.2 V]

the average emf produced in the loop during this time. [Ans. B-32. A conducting square loop of side 5.0 cm is placed time, as B = B0 sin wt, where B0 = 0.40 T and –1 w = 300 s . The normal to the coil makes an a) the maximum emf induced in the coil, (b) the emf induced at t = (p/900) s, and (c) the emf induced at t = (p/600) s. [Ans. (a) 0.15 V; (b) 0.075 V; (c) zero] B-33. A 20-cm long conducting rod is set into pure translational motion with uniform velocity of 10 cm/s perpendicular to its length. A uniform direction perpendicular to the plane of motion. (a) Find the average magnetic force on the free electrons of the rod. (b inside the rod, the electric force on a free electron will balance the magnetic force ? (c) Find the motional emf induced in the rod across its ends. [Ans. (a) 1.6 ¥ 10 –21 N; (b) 0.01 V/m; (c) 2 mV] B-34. The current through a 0.2 H inductor is shown in Fig. B-7. Assuming the passive sign convention, vL at t equal to (a) 0, (b) 2 ms, and (c) 6 ms. [Ans. (a) 0.4 V, (b) 0.2 V, (c) –0.267 V]

B-36.

a at which the inductor is absorbing the maximum power, (b maximum power, and (c the inductor at t = 40 ms. [Ans. (a) 40 – ms; (b) 20+ ms and 40+ ms; (c) 2.5 J] iL

iL (A) 5

0.2 H + vL –

10

20 30

40

50 60 t (ms)

–5

Fig. B-8 is = 0.4t 2 A for t > 0 in the circuit Fig. B-9a, v in (t) for t > 0. (b vs = 40t V for t > 0 in the circuit of Fig. B-9b and iL iin (t) for t > 0. [Ans. (a) (4t 2 + 4t) V; (b) (4t 2 + 4t + 5) A]

B-37. (a

iL (mA)

iL

iin

10 W 4

is

2 –3

–2

–1

1

2

3

4

5

6

+ vin –

5H

vs + –

10 W

5H

7 t (ms)

(a)

–2

(b)

–4

Fig. B-9

–6

B-38. A nonmagnetic ring having a mean diameter of 30 cm and a cross-sectional area of 4 cm2 is uniformly wound with two coils A and B one over the other. Coil A has 90 turns and coil B has 240 turns. Calculate the mutual inductance between the two coils. Also, calculate the emfs induced in coil

Fig. B-7 B-35. Two identical coils A and B consisting of 1500 turns each lie in parallel planes. A current of

Supplementary Exercises—Part B : Electromagnetic Circuits is reversed in 0.02 s. Ans. 11.52 mH, 2.592 mV, 6.912 mV] B-39. An iron rod, 2 cm in diameter and 20 cm long, is

0.5 Wb/m2

715

B-44. 2

cross-sectional area 100 cm , and mean length of

Ans. 301 V] W

B-45. 2

(a) the B/H ratio for iron, and (b

30 cm and cross-sectional area 9 cm

a) the magnetising force, c) the mmf, and (d Ans. (a) 2000 At/m; (b) 1.7684 ¥ 105 At/Wb; (c) 600 At; (d) 3.393 mWb] B-46. An iron ring of 20 cm diameter having crosssectional area of 100 cm2 of wire. (a 2 . (b (b

Ans. (a) 6.67 ¥ 10 –5 H/m; (b) 0.94 H; 424 V] B-40.

of 20 cm2

2 2

Ans. 0.16 H, 16 V]

Ans. (a) 1.25 A; (b) 2500 J] k

B-47.

B-41. area of cross section of 100 cm2 and mean length

changes from 3 mA to 15 mA in 4 ms ? Given that one coil is twice as long as other and it has twice

Ans. 96 mV, 45.26 mH, 22.63 mH] B-48. Ans. 0.226 H, 22.6 V]

series is 0.6 H or 0.1 H depending on the relative

B-42. b)

0.2 H, determine (a

Ans. (a) 0.125 H; (b) 0.72] per ampere

B-49.

Ans. 7.5 mH, 20 m B-43.

a and (b

Ans. 32 mH, 48 mH] B-50. An iron ring of 20 cm mean diameter having a cross section of 100 cm2 2

Ans. (a) 1.15 H; (b) 0.586] Ans. 1.25 A, 2.5 J]

716

Basic Electrical Engineering 2

B-51. The pole face area of an electromagnet is 0.5 m /pole. pole faces are parallel to the surface of the ingot at a distance of 1 mm, determine the coil mmf required. [Ans. 250 At] B-52. Find the pull in kilograms exerted on the plunger

table, 2

B (Wb/m H (At/m) 900

1050

1200

1450

1650

Calculate the current required in the magnetising 1.25 mWb in the cast steel bar.

[Ans. 6.014 A] 2

6.25 cm

Ans. B-53. A cast-steel electromagnet has an air gap 3 mm wide and an iron path 40 cm long. Find the number of 0.7 Wb/m2 B-H curve for cast steel gives H = 660 At/m corresponding to B = 0.7 Wb/m2. Neglect leakage and fringing. [Ans. 1935 At] B-54. An iron ring consists of three parts, as shown in

Magnetising coil

100 turns

10 cm

2

21 cm

l1 = 10 cm, A1 = 5 cm2 l2

A2 = 3 cm2

l3 = 6 cm,

A3 = 2.5 cm2

Fig. B-11

2

in the ring. Given the relative permeabilities of the three parts as m1 = 2670, m2 = 1050 and m3 = 650. [Ans. 2

A1 = 5 cm l1 = 10 cm

B-56. An iron ring of mean length 100 cm with an air gap of 2 mm has a winding of 500 turns. The relative

the winding. Neglect fringing. [Ans. 0.5145 Wb/m2 ] B-57. An iron ring has 100-cm mean diameter, 10-cm2 cross section, and 2-mm wide air gap. The relative permeability of the material is 1500. There are 1000 turns of copper wire uniformly wound on it. 2

2

A3 = 2.5 cm l3 = 6 cm

3

A2 = 3 cm l2 = 8 cm

Fig. B-10 B-55. with a cast steel bar, as shown in Fig. B-11. The magnetising curve for cast steel is given by the

a) the reluctance of ring, (b) the c) the mmf required, (d) the current in the winding. Neglect leakage and fringing. [Ans. (a) 3.257 ¥ 106 At/Wb; (b) 1 mWb; (c) 3257 At; (d) 3.257 A] B-58. A circular iron ring having a cross-sectional area of 10 cm2 and a length of 4p cm in iron, has an air gap of 0.4p mm made by a saw cut. The relative permeability of iron is 1000. The ring is wound with a coil of 2000 turns and carries 2-mA current. fringing. [Ans. 3.636 mWb] B-59. An iron ring of mean circumference 1.0 m is uniformly wound with 400 turns of wire. When a

Supplementary Exercises—Part B : Electromagnetic Circuits

717

B-60. An iron ring of mean length 60 cm has an air gap

2

Ans. 1907]

2

Ans. 66.1 mWb/m ] ( C )

C HA L L E NGI NG

PROBLEMS

2

B-61.

in a steel ring of diameter 40 cm and cross-sectional area 15 cm2. The coil on a b

Ans. 0.3 H, 0.9 H. 0.4 H] B-66. connected in series is 0.5 H or 0.2 H, depending on

Ans. (a) 10 A, 59.4 mH; (b) 26.2 A, 22.67 mH]

a between the two coils, (b the coil B, (c two coils, and (d

B-62. 2

and mean

a between the coils, and (b

combination decreases at 1000 A/s. Ans. (a) 0.075 H; (b) 0.15 H; (c) 0.433; (d) 275 V, 125 V]

Ans. (a) 0.419 H; (b) 335.2 V]

B-67.

2

B-63.

in cross

Ans. 0.592 A] 10-mF capacitor were connected across the switch

B-68. air gap is 1.25 mm. The length of the iron path is 40 cm and the area of each pole shoe is 2 cm2. The a) the b) the force needed to

Ans. 0.5 H, 500 V, 224 V] B-64. a of the coil, (b

0.1 mm, the excitation remaining the same. Neglect ms, and (c

Ans. (a) 251.9 N; (b) 1940 N] B-69.

Ans. (a) 0.55 H; (b) 5500 V; (c) 0.4 H]

section 3 cm in diameter has an air gap 1.5 mm

B-65. are connected in series so as to magnetise (a) in the same direction, (b) in the opposite direction.

force, (b

a) the magnetomotive c

718

Basic Electrical Engineering

(d) the total reluctance, and (e) the relative permeability of iron. [Ans. (a) 1400 At; (b) 0.7621 Wb/m2; 6 (c d) 2.599 ¥ 10 At/Wb; (e) 972] B-70. A circular soft iron ring with a cross bar of silicon

B-72. A magnetic core made up of an alloy has the dimensions as shown in Fig. B-14. Calculate of 1 mWb in the air gap, neglecting leakage and fringing. The B-H curve is as follows :

relevant data mentioned. Estimate the required ampere turns to be applied to one-half of the ring 2 in the other half. [Ans. 2393 At] mr = 765

A 2

1

1.2

1.42

1.513

1.6

H (At/m)

200

400

1792

3300

6000

[Ans. 1.65 A]

50 cm

D

50 cm 600 turns

0.1 cm air gap

C

2 cm

B (T)

30 cm

2

10 cm

2

16 cm

28 cm

50 cm

2

12 cm mr = 442

10 cm

mr = 500

50 cm

Fig. B-14 B-73. An electromagnet of the form shown in Fig. B-15 is excited by two coils, each having 500 turns. When

Fig. B-12 B-71. the cast steel frame of square cross section, shown in Fig. B-13. All the dimensions are given in cm. Assuming the permeability of cast steel to be 1200, calculate the value of the required exciting current. [Ans. 1.697 A]

4

density gives the permeability of 1250. Calculate (a) the total reluctance, and (b 0.5-cm air gap. [Ans. (a ¥ 106 At/Wb; (b 0.5 cm air gap

500 turns

500 turns

0.1

4 Outer path = 60

48 cm 24

2

10 cm

Fig. B-15

Fig. B-13

Supplementary Exercise—Part C : AC Circuits

S U P P L E M E N TA R Y E X E R C I S E S C.1 Solved Problems C.2 Practice Problems

PA R T

C : AC

719

C

CIRCUITS

Assemblage of

N OT E This set of exercises provides practice to those who wish to attain a higher standard of learning the basic principles of Electrical Engineering. The real key to success is practice.

720

Basic Electrical Engineering

Supplementary Exercise—Part C : AC Circuits

721

C.1. SOLVED PROBLEMS P RO B L EM

C - 1

Find the angle by which the current i1 lags the voltage v1, if v1 = 80 cos (100 pt – 40°) V and i1 is equal to (a) i1 = 2.5 cos (100pt + 20°) A; (c) i1 = – 0.8 cos (100pt – 110°) A;

(b) i1 = 21.4 sin (100pt –70°) A (d) i1 = –5.5 sin (100pt + 50°) A

Solution (a) Since both v1 and i1 have positive amplitude, and are expressed as cosine function, the angle by which i1 lags the voltage v1 is given as f = (– 40°) – (20°) = – 60° (b) First, we express i1 as a cosine function, i1 = 21.4 sin (100pt – 70°) = 21.4 cos (100pt – 70° – 90°) = 21.4 cos (100pt – 160°) We can now say that i1 lags v1 by f = (– 40°) – (– 160°) = 120° (c) First, we express i1 with a positive amplitude, i1 = – 0.8 cos (100pt – 110°) = 0.8 cos (100pt – 110° + 180°) = 0.8 cos (100pt + 70°) Therefore, i1 lags v1 by f = (– 40°) – (70°) = – 110° (d) First, we express i1 as a cosine function with a positive amplitude, i1 = – 5.5 sin (100pt + 50°) = 5.5 cos (100pt + 50° + 90°) = 5.5 cos (100pt + 140°) Thus, i1 lags v1 by f = (– 40°) – (140°) = – 180°. That is, i1 and v1 are out of phase. P ROBLE M

C -2

A series combination of two impedances Z1 and Z2 is connected across an ac voltage source v, so that the voltages v1 and v2 across them, as shown in Fig. C-1a, are given as v1 = 40 cos (100pt – 40°) V The supply voltage v can be written as

and

v2 = 20 sin (100pt + 170°) V

v = (A cos 100pt + B sin 100pt) V = C cos (100pt + f) V Find A, B, C, and f.

Fig. C-1

722 Solution

Basic Electrical Engineering Applying KVL, we get v = v1 + (– v2) = {40 cos (100pt – 40°)} + {– 20 sin (100pt + 170°)} V

The voltage v can more conveniently be determined by using the phasor diagram drawn on a complex plane. First, all sinusoids should be expressed in sine-wave form. Hence, we express v1 as v1 = 40 cos (100pt – 40°) = 40 sin (100pt – 40° + 90°) = 40 sin (100pt + 50°) V As shown in Fig. C-1b, the phasor V1 has a magnitude of 40 V (peak or maximum value) and makes an angle of 50° with the reference axis. Similarly, phasor V2 has a magnitude of 20 V and makes an angle of 170° with the reference axis. Reversing its direction gives us phasor –V2 = 20 sin (100pt – 10°) V, which also has a magnitude of 20 V but makes an angle of – 10° with the reference axis. Thus, V = V1 + (– V2) = 40 – 50° + 20–– 10° = (45.41 + j 27.17) V Writing Eq. (i) as summation of two phasors, we get V = (45.41 – 0° + 27.17 – 90°) V which can be written as v = 27.17 sin (100pt + 90°) + 45.41 sin 100pt = (27.17 cos 100pt + 45.41 sin 100pt) V A = 27.17 V

Hence, we have

and B = 45.41 V

Now, we can re-write Eq. (i) in the polar form as V = (45.41 + j 27.17) = 52.92 – 30.9° V which can be written as v = 52.92 sin (100pt + 30.9°) = 52.92 cos (100pt + 30.9° – 90°) = 52.92 cos (100pt – 59.1°) V C = 52.92 V and f = –59.1°

Therefore, we have

P R O B L EM

C - 3

Find the average values of the periodic waves shown in Fig. C-2.

Fig. C-2

(i)

Supplementary Exercise—Part C : AC Circuits

723

Solution (a) It is a voltage sinusoid of peak value 5 – 3 = 2 V, riding over a constant voltage of 3 V. Since the average value of the sinusoid is zero, the average of the given waveform is 3 V. (b curve for one cycle is A = 8 ¥ (T/2) + 1 ¥ (T/2) = 4.5T Therefore, the average value of the wave is Area 4.5T V2(av) = = = 4.5 V T Period (c) It is a triangular wave of height 10 V. The area underneath the curve for one cycle is 1 A = ¥ 10 ¥ T = 5T 2 Therefore, the average value of the wave is Area 5T = =5V V3(av) = T Period P ROBLEM

C -4

Determine the period of the following waves : (a) 9 – 3.2 cos (400t + 30°),

(b) 5 sin2 4t

and

(c)10 sin 3t cos 3t

Solution (a) The expression is a sinusoid riding over a constant 9. Since, the time period is contributed by sinusoid only, we have 2p 2p = = 15.7 ¥ 10–3 s = 15.7 ms T= 400 w (b È 1 - cos (2 ¥ 4t ) ˘ 5 sin2 4t = 5 Í ˙ = 2.5(1 – cos 8t) 2 Î ˚ For the cosine-wave portion, the period is 2p 2p T= = = 0.785 s w 8 (c) Before the period can be determined, we must simplify the product of two sinusoids, by using the identity sin 2q = 2 sin q cos q. Thus, the given expression can be written as

È sin (2 ¥ 3t ) ˘ 10 sin 3t cos 3t = 10 Í ˙ = 5 sin 6t 2 Î ˚ From this, the time period is 2p 2p T= = = 1.05 s w 6

P ROBLE M

C - 5

An alternating current varying sinusoidally with a frequency of 50 Hz has an rms value of 10 A. Write the equation of a) 0.0025 s after passing through positive maximum value, and (b) 0.0075 s after passing through zero and increasing negatively.

724

Basic Electrical Engineering

Im = 2 ¥ 10 = 14.14 A; w = 2pf = 2p ¥ 50 = 100p rad/s Hence, the equation for instantaneous current is

Solution

i = 14.14 sin 100pt = 14.14 sin 314t A

(i)

(a) If we are to take t = 0 at the positive maximum, we should shift the wave represented by Eq. (i) to the left by p/2 to get i = 14.14 sin (314t + p/2) A = 14.14cos 314t A \

i (0.0025 s) = 14.14 cos (314 ¥ 0.0025) A = 10 A

(b) If we are to take t = 0 at the instant when the wave is passing through zero and increasing negatively, we should shift the wave represented by Eq. (i) to the left by p to get i = 14.14 sin (314t + p) = –14.14 sin 314t A \

i (0.0075 s) = –14.14 sin (314 ¥ 0.0075) = –10 A

P R O B L E M

C - 6

A current wave consists of two components : (i) a dc current of 10 A, and (ii) a sinusoidal ac current of peak value 5 A and frequency 50 Hz. (a) Write the expression of the current wave, taking t = 0 at an instant where the ac component is zero and increasing in positive direction. (b) Find the average value of the current wave. (c) Find the rms value of the current wave.

Solution (a) The resultant current, i = i1(dc) + i2(ac) = 10 + 5 sin 2p ¥ 50t = 10 + 5 sin 314t i = 10 + 5 sin q,

or

where q = 314t

(b) The dc component remains constant at 10 A, and hence its average value is 10 A. The average value of the ac component over full cycle is zero. Hence, the average of the given current wave, Iav = 10 + 0 = 10 A Alternatively, the average value can be determined using standard procedure, as follows. 1 Iav = 2p

2p

1

2p

1

Ú idq = 2p Ú (10 + 5 sin q) dq = 2p [10q – 5cos q] 0

2p

0

0

=

1 [10 ¥ 2p – 5 + 5] 2p

= 10 A

(c) Using the standard procedure, the rms value of the current wave is given as 2p

Irms =

1 i 2dq = 2p

Ú 0

=

=

1 2p

1 2p

2p

È

1 2p

2p

Ú

2p

(10 + 5 sin q ) 2 dq =

2p

Ê 1 - cos 2q ˆ ˘ ˜¯ ˙ dq 2 ˚

Ú [112.5 + 100 sin q - 12.5 cos 2q ] dq 0

Ú 0

0

Ú ÍÎ100 + 100 sin q + 25 ÁË 0

1 (100 + 2 ¥ 10 ¥ 5 sin q + 52 sin 2 q ) dq 2p

Supplementary Exercise—Part C : AC Circuits

725

2p

=

12.5 sin 2q ˘ 1 È 112.5q - 100 cos q Í ˙ 2 2p Î ˚0

=

1 [112.5 ¥ 2p - 100 + 100 + 0] = 2p

112.5 = 10.61 A

Alternatively, the rms value of the given current wave can easily or rms value of the given current wave is Irms R, the power delivered is I 2rms R. The same amount of power is delivered jointly by the two components of the current wave. That is, 2

2

2

2

fi

Irms R = I1(rms) R + I2(rms) R

2

2

I rms = I1(rms) + I 2(rms)

Hence, the rms value of the current wave is simply given as

PROBLEM

2 ⎛ 5 ⎞ 10 2 + ⎜ = ⎟ ⎝ 2⎠

I12(rms) + I 22(rms) =

Irms =

100 + 12.5 = 10.61 A

C -7

Find the rms value of an ac voltage wave of peak value Vm, which can be represented by alternate positive and negative semicircles of radius Vm.

Solution

Figure C-3 shows the given ac voltage wave, taking voltage v along y-axis and angle wt along x-axis. For Vm is same as the radius R of the circle. Any point (x, y) on the semicircle is then given by x 2 + y 2 = R2

or

2 v2 = V m – q2

Fig. C-3 While determining the rms value, we take the mean of squares of voltage. Since, the square of negative quantities is positive, we need to consider only half of the wave. Hence, we consider only the positive semicircle, for which q varies from – Vm to +Vm, or equivalently from –R to + R. Thus, the rms value of the voltage can now be determined as

Vrms =

1 2Vm

+Vm

Ú

-Vm

2

v ◊ dq =

1 2Vm

+V

+Vm

Ú

-Vm

(Vm2

2

- q ) ◊ dq =

m 1 È 2 q3 ˘ ÍVmq ˙ 2Vm ÍÎ 3 ˚˙ -Vm

726

Basic Electrical Engineering

P RO B L EM

ÈÊ 2 (V )3 ˆ Ê ( - Vm )3 ˆ ˘ Í ÁVmVm - m ˜ - Á - Vm2Vm ˜˙ = 3 ¯ Ë 3 ¯ ˙˚ ÍÎ Ë

=

1 2Vm

=

2 2 Vm = 0.818Vm 3

1 2Vm

Ê 3 2(Vm )3 ˆ ÁË 2Vm ˜ 3 ¯

C -8

Find iS for the circuit shown in Fig. C-4, if vS = 150 sin (2500t – 34°) V.

Solution

The current iS can be determined from iS = iR + iL + iC. From

iR =

vS 150 sin (2500t - 34∞) = = 15 sin (2500t – 34°) V 10 R

For the inductor branch, XL = wL = 2500 ¥ 6 ¥ 10 –3 = 15 W. Thus, the peak value of the current is given as V 150 Im = m = = 10 A XL 15 The current iL lags the applied voltage by 90°. Hence,

Fig. C-4

iL = 10 sin (2500t – 34° – 90°) = 10 sin (2500t –124°) A –6 For the capacitor branch, XC = 1/wC = 1/(2500 ¥ 20 ¥ 10 ) = 20 W. The peak value of the current is given as

Vm 150 = = 7.5 A XC 20 The current iC leads the applied voltage by 90°. Hence, Im =

iC = 7.5 sin (2500t – 34° + 90°) = 7.5 sin (2500t + 56°) A To add the sinusoids directly and then to simplify it by using trigonometrical identities is a very tedious job. Instead, phasors based on peak values can be used to add sinusoids. Thus, IS = IR + IL + IC = 15––34° + 10––124° + 7.5–56° = 15.2–– 43.46° A The above result can be expressed as iS = 15.2 sin (2500t – 43.46°) A P R O B L EM

C - 9

Transform the series ac circuit given in Fig. C-5a to its equivalent in phasor-domain, and then determine the current i.

Supplementary Exercise—Part C : AC Circuits

727

Solution

While drawing the circuit in phasor-domain, the current and voltages are replaced by corresponding phasors, the inductance is replaced by ZL = jwL = j4(2) = j8 W

and the capacitance is replaced by ZC = – j

1 j = = – j4 W C 4 (1/16)

The resistance, of course, is not changed. The given ac circuit is redrawn in phasor-domain in Fig. C-7b. Next, we apply KVL to get VS = V R + V L + V C 40–20° = 6I + j8I – j4I = (6 + j4)I

or from which

I=

40–20∞ 40– 20∞ = = 5.547––13.7° A 7.211–33.7∞ 6 + j4

i = 5.547 2 sin (4t – 13.7°) = 7. 84 sin (4t – 13.7°) A P ROBLE M

C -10

Determine the input impedance at w = 5 krad/s of the circuit shown in Fig. C-6a. 2 mF

100 W

v

–

100 W + –

+ 1–0° A Vin

3v 12 mH

V

Zin

+ 3V

j60 W

–

(a)

–j100 W

(b)

Fig. C-6 Solution by their corresponding impedances, ZC =

-j -j = = – j100 W, w C 5 ¥ 103 ¥ 2 ¥ 10 - 6

and

ZL = jwL = j5 ¥ 103 ¥ 12 ¥ 10–3 = j60 W

Zin. The best choice here is a current source of 1–0° A, because with it, Zin = Vin/1–0° = Vin. The resulting phasor-domain circuit is shown in Fig. C-6b. Note that the controlling voltage for the dependent source is the voltage drop across the combination of the resistor and the capacitor : V = –(1–0°) (100 – j100) = (–100 + j100) V Applying KVL, we get

Therefore,

Vin = (1–0°)(100) + (1–0°)(– j100) + 3(–100 + j100) + (1–0°)( j60) = 100 – j100 – 300 + j300 + j60 = –200 + j260 = 328–127.6° V V Zin = in = 328–127.6° W = (– 200 + j260) W 1 ∠ 0°

728

Basic Electrical Engineering

P RO B L EM

C -11

In the circuit of Fig. C-7a, w = 1200 rad/s, IC = 1.2–28° A, and IL = 3–53° A. Find (a) Is; (b) iR(t); (c) Vs.

Fig. C-7 Solution redraw the circuit in phasor-domain. We have ZL = jwL = j(1200)(10 ¥ 10–3) = j12 W

and

ZC =

-j -j = = – j 33.33 W w C (1200)(25 ¥ 10 - 6 )

The circuit is redrawn in phasor-domain in Fig. C-7b. (a) The voltage across 20-W resistance is V20W = VC = IC ZC = (1.2–28°)(33.33–– 90°) = 40––62° V \

I20W =

V20 20

=

40–- 62∞ = 2––62° A 20

Is = I20W + IC = 2–– 62° + 1.2–28° = 2.33––31.0° A (b) The current through the 10-W resistance is IR = Is + IL = 2.33––31.0° + 3–53° = 3.99–17.46° A Thus, the current iR in time-domain is iR (t) = 3.99 2 sin (1200t + 17.46°) A = 5.64 sin (1200t + 17.46°) A (c) Vs = IR (10) + IL ( j12) = (3.99–17.46°)(10) + (3–53°)(12–90°) = 34.9–74.52° V P R O B L E M

C - 12 I1 and I2 in the circuit shown in Fig. C-8a.

Solution The mesh currents as marked in Fig. C-8b have the same value as the branch currents marked in Fig. C-8a. Writing the mesh equations in matrix form by inspection, we get -3 ˘ È I1 ˘ È3 + j5 = Í -3 3 - j4 ˙˚ ÍÎ I 2 ˙˚ Î

È 10 - j15 ˘ Í-20 + j15˙ Î ˚

Using Casio fx-991ES, we get I1 = 4.87––164.6° A

and

I2 = 7.17––144.9° A

Supplementary Exercise—Part C : AC Circuits

729

Fig. C-8 PROBLEM

C -13

Determine the time-domain node voltages v1(t) and v2(t) in the circuit shown in Fig. C-9. –j5 W V1

V2 j10 W

1–0° A

5W

–j10 W

j5 W

10 W

0.5––90° A

Fig. C-9 Solution

V1 and V2 and then transform the results into time-domain.

Applying KCL at the left node, gives V1 V V - V2 V1 - V2 + 1 + 1 + = 1–0° 5 - j10 - j5 j10

or

Ê1 Ê 1 1 1 1 ˆ 1 ˆ + V1 Á - V2 Á + =1 Ë 5 j10 j 5 j10 ˜¯ Ë j 5 j10 ˜¯

or

(0.2 + j0.2)V1 – j0.1V2 = 1

(i)

Applying KCL at the right node, V2 - V1 V2 - V1 V2 V2 = – 0.5––90° + + + - j5 j10 j 5 10

or

Ê 1 Ê 1 1 ˆ 1 1 1ˆ - V1 Á + + V2 Á + + + = j0.5 j10 ˜¯ Ë - j5 Ë j 5 j10 j 5 10 ˜¯

or

– j 0.1V1 + (0.1 – j0.1)V2 = j0.5

Solving Eqs (i) and (ii) for V1 and V2 we get V1 = 1 – j2 = 2.24–– 63.4° V

and

V2 = –2 + j4 = 4.47–116.6° V

(ii)

730

Basic Electrical Engineering

Thus, the time-domain solutions are or

v1 (t) = 2.24 2 sin (w t – 63.4°) V

and

v2 (t) = 4.47 2 sin (wt + 116.6°) V

v1 (t) = 3.168 sin (wt – 63.4°) V

and

v2 (t) = 6.322 sin (wt + 116.6°) V

P ROBLE M

C -1 4 V1 for the circuit of Fig. C-9.

Solution

First we reduce each pair of parallel impedances by a single equivalent impedance. Thus, (5 W) || (– j10 W) =

1 = (4 – j2) W; 0.2 + j 0.1

( j5 W) || (10 W) =

1 = (2 + j4) W - j 0.2 + 0.1

V1

1–0° A

( j10 W) || (– j5 W) =

V2

–j10 W

(4 – j2) W

1 = – j10 W - j 0.1 + j 0.2

(2 + j4) W

0.5––90° A

Ref.

Fig. C-10 To use the superposition, we consider only one source at a time. First, activating only the left source and deactivating the right source (i.e., open-circuiting it), the equivalent admittance across the current source is given as Y1 =

1 1 = (0.2 + j0.1) + (0.05 + j0.15) = (0.25 + j0.25) S + 4 - j 2 (- j10) + (2 + j 4)

Hence, the partial response V1L due to only the left source is given as V1L = I/Y1 = (1–0°)/(0.25 + j0.25) = (2 – j2) V Now, we activate only the right source and deactivate the left source. Using current division concept, the current through (4 – j2) W impedance is given as I1 = (– 0.5–– 90°) ¥

( 2 + j 4) = (– 0.2 – j0.1) A (2 + j 4) + (- j10 + 4 - j 2)

V1R due to only the right source as V1R = I1Z = (– 0.2 – j0.1) (4 – j2) = –1.0 V Finally, adding the two partial responses, we get V1 = V1L + V1R = (2 – j2) – 1 = (1 – j2) V PR OBL EM

C -15

A non-inductive load takes 20 A at 200 volts. Calculate the inductance of a reactor to be connected in series in order that the same current be supplied from a 230-V, 50-Hz supply. Also, determine the power factor of this circuit.

Supplementary Exercise—Part C : AC Circuits

Solution

731

The resistance of the non-inductive load is

V 200 = = 10 W I 20 When the reactor is connected in series with the load, the total impedance should be R=

V1 230 = = 11.5 W I 20 Thus, if XL is the reactance of the reactor, we should have Z=

2

2

2

Z = R + X L fi XL = Z 2 Since, XL = wL, the inductance of the reactor is given as

11.52 10 2 = 5.679 W

5.679 = 0.018 H = 18 mH 2p ¥ 50 R = 0.8696 (lagging) pf = cos f = Z

L=

The power factor, PR OBL EM

R2 =

XL

=

C -16

A sinusoidal ac voltage having rms value of 200 V and frequency of 50 Hz is applied to an ac circuit. At t = 0, it has an instantaneous value of 200 2 V. The circuit draws a current of 5 A at 0.5 lagging power factor. Determine (a) the expression for instantaneous value of voltage and current, (b) the values of circuit parameters in ohms and impedance in complex number form, and (c) the values of instantaneous voltage and current at t = 0.0125 s.

Solution (a) If Vrms = 200 V, its peak value should be 200 2 V. Thus, in general, it can be expressed as v = 200 2 sin (2p ¥ 50t + q) V. At t = 0, it has a value of 200 2 V. Thus, 200 2 = 200 2 sin (0 + q) Hence, the expression for instantaneous voltage is

fi

q = 90°

v = 200 2 sin (100pt + 90°) V If rms current, Irms = 5 A, its peak value should be 5 2 A. Since, the power factor is 0.5 lagging, we should have f = cos–1 (0.5) = 60°, and the phase angle for current should be q i = 90° – 60° = 30° Hence, the expression for instantaneous current is i = 5 2 sin (100pt + 30°) A V 200 (b) The impedance, Z = = = 40 W I 5 \ R = Z cos f = 40 cos 60° = 20 W; XL = Z sin f = 40 sin 60° = 34.66 W or Z = (20 + j 34.66) W (c) At t = 0.0125 s, v = 200 2 sin {100p(0.0125) + 90°} = 200 2 sin (225° + 90°) = – 200 V and

i = 5 2 sin {100p(0.0125) + 30°} = 5 2 sin (225° + 30°) = – 6.83 A

732

Basic Electrical Engineering

P RO B L EM

C -17

An inductive coil, when connected across a 200-V, 50-Hz supply, draws a current of 6.25 A and a power of 1000 W. Another coil, when connected across the same supply, draws a current of 10.75 A and a power of 1155 W. Find the current drawn and the power consumed when the two coils are connected in series across the same supply.

Solution

For coil 1, we have

R1 =

P1 I12

=

1000 (6.25)

2

= 25.6 W; Z1 =

V 200 = = 32 W, XL1 = I1 6.25

Z12

R12 =

32 2

25.62 = 19.2 W

Similarly, for coil 2, we have R2 =

P2 I 22

=

1155 (10.75)

2

= 10 W; Z2 =

V 200 = = 18.6 W, XL2 = I2 10.75

Z22

R22 =

18.62 10 2 = 15.7 W

When the two coils are connected in series, the net impedance is given as Z = Z1 + Z2 = (25.6 + j19.2) + (10 + j15.7) = (35.6 + j34.9) W = 49.85–44.43° W Therefore, the current drawn, I =

P = I 2R = (4.01)2 ¥ 35.6 = 572.5 W

The power drawn, P RO BL EM

V 200 = = 4.01 A Z 49.85

C - 18

A 50-Hz sinusoidal voltage v = 141.4 sin wt V is applied to a series RL circuit consisting of R = 3 W and L = 0.01272 H. Compute (a) the effective value of the steady-state current and its phase angle, (b) the expression for the instantaneous current, (c) the effective value and phase angle of the voltage drops across each element, (d) the average power and the power factor of the circuit, and (e) the reactive power.

Solution

V=

Vm 141.4 = = 100 V; XL = wL = 2p ¥ 50 ¥ 0.01272 = 4 W 2 2

(a) The impedance of the circuit, Z = 3 + j4 = 5–53.13° W V 100– 0∞ = = 20–– 53.13° A 5 – 53.13∞ Z The effective (rms) current, I = 20 A The phase angle, f = 53.13° (lagging)

Hence, the current,

I =

(b) Peak value of current, Im = I 2 = 20 2 = 28.28 A The expression for the instantaneous current can be written as i(t) = 28.28 sin (wt – 53.13°) A (c) The voltage drop across the resistor, VR = IR = (20–– 53.13°) ¥ 3 = 60–– 53.13° V The voltage drop across the inductor, VL = I (jXL) = (20–– 53.13°) ¥ (4–90°) = 80–(90° – 53.13°) = 80–36.87° V (d) The average power, P = VI cos f = 100 ¥ 20 ¥ cos 53.13° = 1200 W The power factor, pf = cos 53.13° = 0.6 (lagging) (e) The reactive power, Q = VI sin f = 100 ¥ 20 ¥ sin 53.13° = 1600 VAR

Supplementary Exercise—Part C : AC Circuits P RO B L EM

733

C -19

A circuit consists of R = 35 W in series with an unknown impedance Zu. For a sinusoidal current of 2 A, the observed voltages are 200 V across R and Zu together, and 150 V across impedance Zu. Find (a) the unknown impedance Zu, and (b) the phase angle of the circuit. Also, draw the phasor diagram.

Fig. C-11. Solution

The given voltages are shown in the circuit given in Fig. C-11a. Let the unknown impedance be Zu = (Ru ± jXu) W, the + sign for inductive reactance, and – sign for capacitive reactance. Since the current through the circuit is known to be 2 A, the voltage drop across resistor R is VR = 2 ¥ 35 V. And, the magnitudes of the unknown impedance and the total circuit impedance are given as 150 200 Zu = = 75 W and Z= = 100 W 2 2 (a) Let us consider inductive reactance, and refer to the phasor diagram given in Fig. C-11b. The voltage phasor VR is in same phase as the current phasor I. However, as the impedance Zu is inductive, the voltage Vu leads the current I by an angle q V across the circuit is the sum of phasors Vu and VR. That is, V = Vu + VR. Using the law of parallelogram of addition of phasors (or vectors), we have 2 2 2 V = VR + Vu + 2VRVu cos q 2 2 2 (200) = (70) + (150) + 2(70) (150)cos q

or

(200)2 ( 70)2 (150)2 = 0.6 or q = 53.13° 2 ( 70) (150) We can now calculate the resistance and the reactance of the unknown impedance, fi

cos q =

Ru = Zu cos q = 75 ¥ 0.6 = 45 W

and

Xu = ± Z u2

Ru2 =

752

452 = 60 W

Thus, the unknown impedance is given as Zu = Ru ± jXu = (45 ± j60) W (b) Now, the total resistance, Rt = Ru + R = 45 + 35 = 80 W ⎛ Rt ⎞ ⎛ 80 ⎞ The phase angle of the circuit, f = cos–1 = cos–1 = 36.87° ⎝ Z⎠ ⎝ 100 ⎠ The phasor diagram for the capacitive unknown impedance is drawn in Fig. C-11c. P ROBLE M

C -20

An iron-cored coil takes 3 A at a pf of 0.7, when connected across a 110-V, 50 Hz supply. When the iron core is removed from the coil and the voltage is reduced to 30 V, the current taken by the coil becomes 4.5 A at a pf of 0.9. Calculate (a) the inductance in each case, and (b) the iron losses in the core.

734

Basic Electrical Engineering

Solution V1 110 = = 36.67 W. I1 3 f1 = cos –1 0.7 = 45.57° fi sin f1 = 0.7141 XL1 = Z1 sin f1 = 36.67 ¥ 0.7141 = 26.19 W

(a) Case 1 (Coil with iron core) : Impedance, Z1 = The phase angle, \

Hence, the inductance, L1 =

26.19 X L1 = = 0.08336 H = 83.36 mH 2p f 2p ¥ 50

V2 30 = = 6.667 W. 4.5 I2 –1 f2 = cos 0.9 = 25.84° fi sin f = 0.4359 XL2 = Z2 sin f2 = 6.667 ¥ 0.4359 = 2.906 W

Case 2 (Coil without iron core) : Impedance, Z2 = The phase angle, \

X L2 2.906 = = 0.00925 H = 9.25 mH 2 f 2 π × 50 (b) In Case 1 (Coil with iron core): the equivalent resistance responsible for losses, Hence, the inductance, L2 =

R1 = Z1 cos f1 = 36.67 ¥ 0.7 = 25.67 W In Case 2 (Coil without iron core): the equivalent resistance responsible for losses, R2 = Z2 cos f2 = 6.667 ¥ 0.9 = 6.0 W Hence, the equivalent resistance responsible for iron losses, Ri = R1 – R2 = 25.67 – 6.0 = 19.67 W 2 2 Thus, the iron loss at 3 A current, Pi = I Ri = 3 ¥ 19.67 = 177 W

P ROBLEM

C -21

A coil having a resistance of 15 W and an inductance of 0.2 H is connected in series with another coil having a resistance of 25 W and an inductance of 0.04 H to a 230-V, 50-Hz supply, as shown in Fig. C-12. Determine (a) the voltage across each coil, (b) the power dissipated in each coil, and (c) the power factor of the whole circuit.

I

+

15 W

25 W

0.2 H

V1

V2

– + +

230 V, 50 Hz

0.04 H

–

–

Fig. C-12 Solution R = R1 + R2 = 15 + 25 = 40 W; L = L1 + L2 = 0.2 + 0.04 = 0.24 H The impedances of each coil and of entire circuit are given as Z1 = R1 + jXL1 = 15 + j(2p ¥ 50 ¥ 0.2) = (15 + j62.83) = 64.60–76.57° W Z2 = R2 + jXL2 = 25 + j(2p ¥ 50 ¥ 0.04) = (25 + j12.57) = 27.98–26.69° W Z = R + jXL = 40 + j(2p ¥ 50 ¥ 0.24) = (40 + j75.4) = 85.35–62.05° W Therefore, the magnitude of the current through the circuit is given as, V 230 = = 2.695 A I= Z 85.35

Supplementary Exercise—Part C : AC Circuits

735

For convenience sake, let us take current I as the reference phasor. (a) Voltage across coil 1, V1 = IZ1 = (2.695–0°)(64.60–76.57°) = 174.1–76.57° V Voltage across coil 2, V2 = IZ2 = (2.695–0°)(27.98–26.69°) = 75.41–26.69° V (b) The power factor of coil 1, pf1 = cos f1 = cos 76.57° = 0.2323 \ Power dissipated, P1 = V1I cos f1 = 174.1 ¥ 2.695 ¥ 0.2323 = 109 W The power factor of coil 2, pf2 = cos f2 = cos 26.69° = 0.8934 \ Power dissipated, P2 = V2 Icos f2 = 75.41 ¥ 2.695 ¥ 0.8934 = 181.6 W (c) The power factor of the circuit, pf = cos f = cos 62.05° = 0.4686 P ROBLE M

C - 2 2

A 120-V, 60-W bulb (lamp) is to be operated on 220-V, 50-Hz supply mains. Calculate the value of (a) the noninductive resistance, (b) the pure inductance, and (c) the pure capacitance that would be required to be connected in series with the lamp, so that it works on the correct voltage. Which method is preferable and why ? Draw the relevant phasor diagrams for each case. P 60 = = 0.5 A V 120 (a) Let R be the required non-inductive resistance to be connected in series with the lamp, as shown in Fig. C-13a. Both the voltages VB and VR are in phase with the current I, as shown in phasor diagram of Fig. C-13b.

Solution The rated current of the lamp, I =

Fig. C-13 The voltage across R,

VR = V – VB = 220 – 120 = 100 V

VR 100 = = 200 W I 0.5 (b) Let L be the required inductance to be connected in series with the lamp, as shown in Fig. C-14a. The voltage VL leads the current I by 90°, as shown in phasor diagram of Fig. C-14b. \ The required resistance, R =

Fig. C-14

736

Basic Electrical Engineering

The voltage across the inductance,

\ The required inductive reactance, XL = Hence, the inductance,

V2

VL =

L =

VB2 =

220 2 120 2 = 184.4 V

VL 184.4 = = 368.8 W I 0.5 368.8 XL = = 1.174 H 2p f 2p ¥ 50

(c) Let C be the required capacitance to be connected in series with the lamp, as shown in Fig. C-15a. The voltage VC lags the current I by 90°, as shown in phasor diagram of Fig. C-15b.

Fig. C-15 The voltage across the capacitance,

VC =

V2

VB2 =

220 2 120 2 = 184.4 V

VC 184.4 = = 368.8 W I 0.5 1 1 = = 8.63 mF C = 2pf XC 2 π × 50 × 368.8

\ The required capacitive reactance, XC = Hence, the capacitance, COMMEN TS

Ploss = I 2R = (0.5)2 ¥ 200 = 50 W). Using L or C is a better alternative, as no power loss occurs. However, using a capacitor is preferred, as a capacitor costs much less than an inductor. PROBLEM

C -23

Find the simplest parallel circuit that has the same impedance at 400 Hz as the series combination of a 300-W resistor, a 0.25-H inductor and a 1-mF capacitor.

Solution Z = 300 + j2p (400)(0.25) – j1/[2p(400) (1 ¥ 10 –6)] = 300 + j230 = 378–37.5° W Thus, the admittance of the parallel circuit should be 1 1 = = 2.64 ¥ 10 –3––37.5° S = (2.096 ¥ 10 –3 – j1.61 ¥ 10 –3) S Y= Z 378∠ 37.5° The simplest parallel circuit which has this admittance is a resistor and an inductor. From the above expression for admittance, we have 1 1 R = = = 477 W G 2.096 × 10 −3

Supplementary Exercise—Part C : AC Circuits BL = –1.61 ¥ 10 –3 S

and

PROBLEM

fi

L=

737

−1 -1 = H = 247 mH ωBL 2p(400)( - 1.61 ¥ 10 - 3)

C -24

Find Iin and IL for the circuit shown in Fig. C-16.

Fig. C-16 Solution The three branches on the right are in parallel and have a total admittance of Y=

1 1 1 = 0.416––16° S + + 5 + j4 6 - j3 6 –30∞

Therefore, the total input impedance of the circuit is 1 1 Zin = 2 + =2+ = 4.36–8.72° W Y 0.416–-16∞ V 120 ∠ 30° = = 27.5–21.3° A Hence, the current, Iin = Zin 4.36 ∠8.72° The current IL can be found from the load voltage VL and its impedance ZL. The load voltage VL is equal to the current Iin divided by the total admittance of the three parallel branches, VL =

I in 27.5∠ 21.3° = = 66.2–37.3° V Y 0.416 ∠− 16°

PROBLEM

and

IL =

66.2 ∠ 37.3° VL = = 11–7.3° A ZL 6 ∠ 30°

C -25 j10 W impedance in the circuit shown in Fig. C-17, and hence

1

1–0° A

–j10 W

(4 – j2) W I1

2

(2 + j4) W 3

Fig. C-17

0.5––90° A

738

Basic Electrical Engineering

Solution

a, and is given as VTh = Voc = V12 = V13 + V32 = I1 (4 – j2) + I2 (2 + j4) = (1–0°)(4 – j2) + (0.5––90°)(2 + j4) = (6 – j3) V

Fig. C-18b. Thus, ZTh = (2 + j4) + (4 – j2) = (6 + j2) W

Fig. C-18 Thus, on connecting the load impedance – j10 W I12 =

P ROBLEM

c.

VTh 6 - j3 = = (0.6 + j0.3) A = 0.67–26.6° A ZTh + Z L 6 + j2 - j10

C - 26

Determine the average power dissipated by the 10-W resistor in the circuit of Fig. C-19. 10 W

5 sin 3t A

C1

0.2 F

C2

Fig. C-19

0.5 F

2 sin 5t A

Supplementary Exercise—Part C : AC Circuits

739

Solution across the 10-W resistor. However, we just cannot do this, since the two sources are operating at different frequencies. In such a situation, there is no way to compute the impedance of any capacitor or inductor in a circuit, as we are in a dilemma as to which frequency to use. The only way out of this dilemma is to use principle of superposition. We can consider one source at a time and I ¢due to the current source 2 sin 5t A acting alone. The other current source is deactivated (open-circuited), and all impedances calculated at w = 5 rad/s. The resulting subcircuit in phasor domain is shown in Fig. C-20a. - j1 -j1 I 2 -j -j Z C1 = = = – j W; Z C2 = = = – j0.4 W; I = m = = 1.414 A w C1 5 ¥ 0.2 w C2 5 ¥ 0.5 2 2 I¢

I¢¢

10 W

10 W

–j W

–j0.4 W

1.414–0° A

3.536–0° A

–j1.667 W

(a)

–j0.6667 W

(b)

Fig. C-20 I¢ from the circuit of Fig. C-20a, Ê ˆ - j0.4 I¢ = (1.414–0°) Á = 56––82.03° mA Ë 10 - j - j 0.4 ˜¯

Similarly, considering the current source 5 sin 3t convert the subcircuit in phasor domain at w = 3 rad/s, as shown in Fig. C-20b. ZC1 =

- j1 -j = = –j1.667 W; w C1 3 ¥ 0.2

ZC2 =

- j1 -j = = –j0.6667 W; w C2 3 ¥ 0.5

I≤through 10-W resistor. We

I=

Im 5 = = 3.536 A 2 2

I≤ from the circuit of Fig. C-20b, Ê ˆ - j1.667 I≤ = (3.536–0°) Á = 574––76.86° mA Ë 10 - j 0.6667 - j166.7 ˜¯

N O T E At this point, no matter how tempted we might be to add the two phasor currents I¢ and I≤, this would be incorrect, two currents to get the net heating power of the resistor. Hence the power dissipated by the 10-W resistor is given as P = (I¢)2R + (I≤)2R = [(0.056)2 + (0.574)2] ¥ 10 = 3.326 W P RO B L EM

C -27

Transform the practical ac voltage source given in Fig. C-21a into its equivalent current source.

740

Basic Electrical Engineering

Fig. C-21 Solution Total series impedance, Z = 3 + j4 + {6 || ( –j5)} = 3 + j4 +

6 ¥ (- j5) = 3 + j4 + (2.46 – j2.95) (6 - j5)

= (5.46 + j1.05) = 5.56–10.9° W The current source of the equivalent circuit is given as V 20 ∠ 30° = = 3.6–19.1° A 5.56 ∠10.9° Z Thus, the equivalent practical current source is as shown in Fig. C-21b. Is =

P ROBLEM

C -28

An inductive choke coil having a resistance of 5 W and a self-inductance of 0.06 H is connected to a 200-V, 50-Hz supply mains. Estimate the current taken and its power factor. Find also what value of capacitance must be arranged in series with it to give 2000 V across the capacitor with minimum applied potential difference. Estimate the necessary pd and also the pd across the inductive coil.

Solution Refer to Fig. C-22a. \

XL = j(2pf L) = j(100p ¥ 0.06) = j18.85 W ZCh = (5 + j18.85) = 19.5–75.14° W

Choke impedance,

Therefore, the current taken from the supply mains, I1 =

V1 200 ∠ 0° = = 10.26––75.14° A ZCh 19.5∠ 75.14°

Power factor, pf = cos 75.14° = 0.256 (lagging)

Fig. C-22

Supplementary Exercise—Part C : AC Circuits

741

Refer to Fig. C-22b. A capacitor of C farad has been connected in series with the choke coil. We wish to get a pd of 2000 V across this capacitor by applying minimum supply voltage V2. This would happen only when the series circuit resonates, for which we must have XC = XL

1/wC = 18.85 fi C =

or

The circuit current I2 is given as I2 =

1 = 168.9 mF 100p ¥ 18.85

VC 2000 = = 106.1 A XC 18.85

Under resonant condition, the impedance of the circuit, Z = R. Therefore, the supply voltage must be V2 = I2 Z2 = 106.1 ¥ 5 = 530.5 V The pd across the choke coil, VCh = I2 ZCh = 106.1 ¥ 19.5 = 2069 V PROB LE M

C -2 9

Find the voltages across the two impedances in the circuit shown in Fig. C-23a. Then transform the voltage source and the 10–30°-W in the two cases ? 10–30° W + V1 – 50–20° V

+ V2 –

8–20° W

5––10° A

(a)

10–30° W

+ V –

8–20° W

(b)

Fig. C-23 Solution Using voltage division concept, the voltage across 10–30°-W impedance is V1 = (50–20°) ¥

10 ∠ 30° 500 ∠ 50° = = 27.9–24.4° V 10 ∠ 30° + 8∠ 20° 17.9 ∠ 25.6°

By KVL, the voltage across 8–20°-W impedance is V2 = (50–20°) – (27.9–24.4°) = 22.3–14.4° V Now, the equivalent current source is given as I = (50–20°)/(10–30°) = 5––10° A so that the equivalent circuit becomes as shown in Fig. C-23b. As can be seen, the same voltage V exists across all the three parallel components. This voltage is found by multiplying current and the total impedance, V = (5––10°) ¥

(10–30∞) (8– 20∞) 400– 40∞ = = 22.3–14.4° V (10–30∞ + 8– 20∞) 17.9– 25.6∞

Note that voltage across the 8–20°-W impedance is the same as for the original circuit. However, the voltage across the 10–30°-W impedance is different in the two cases. This shows that the transformed source produces the same voltages and currents outside the source, but usually not inside it.

742

Basic Electrical Engineering

P RO B L EM

C -30

Determine the mesh currents in the circuit shown in Fig. C-24. 4W

j15 W

10–20° V

6W 15––30° V

I1

I2

3––13° A

–j7 W

Fig. C-24 Solution For writing the mesh equations, we can apply the self-impedance and mutual impedance approach. For mesh 1, the self-impedance, Z11 = 4 + j15 + 6 – j7 = (10 + j8) W and the mutual impedance, Z12 = (6 – j7) W The source voltage 15––30° V is aiding the current I1, but the source voltage 10–20° V is opposing it. Hence, the net source voltage for mesh 1 is 15––30° V – 10–20° V = 11.5––71.8° V Hence, the KVL equation for mesh 1 is (10 + j8)I1 – (6 – j7)I2 = 11.5––71.8° V No KVL equation is needed for mesh 2, since I2 is the only mesh current through the current source of 3––130° A. Since the direction of this source current is opposite to the mesh current I2, we have I2 = –3––13° A Substituting this value of I2 into the mesh 1 equation gives fi

(10 + j8)I1 – (6 – j7) (–3––13°) = 11.5––71.8° 11.5∠− 71.8° + (6 − j 7) ( − 3∠− 13° ) I1 = = 1.28–85.5° A (10 + j 8)

PROBLEM

C -31

In the circuit shown in Fig. C-25, the load ZL is a resistor with resistance R. Determine the value of R that will cause a current of 0.1 A through the load.

Solution voltage is simply the voltage drop across the j8-W impedance, j8 VTh = Voc = (1–30°) ¥ = 0.8–66.87° V 6 + j8 voltage source replaced by a short-circuit. Thus, ZTh = –j4 +

6 ( j 8) = (3.84 – j1.12) W = 4––16.26° W 6 + j8

Supplementary Exercise—Part C : AC Circuits Now, for a given load ZL, the load current is given as VTh IL = from which Z Th + Z L | ZTh + ZL | =

VT h 0.8 = =8W IL 0.1

fi

(3.84 + R)2 + (112 . )2 = 8

or

R2 + 7.86R – 48 = 0

or

or

ZTh + ZL =

743

VTh IL

|3.84 – j1.12 + R| = 8

2

2

(3.84 + R) + (1.12) = 64

fi R = 4.08 W or

–11.85 W

Since a resistance cannot have a negative value, R = 4.08 W j3 W

4W

6–30° V

Fig. C-25 P RO B L EM

+ V –

5––50° V

2W

Fig. C-26

C -32 V in the circuit of Fig. C-26.

Solution The voltage V can be considered to have two components, V1 and V2. The component V1 is due the source 6–30° V acting alone, and the component V2 is due the source 5––50° V acting alone. Thus, V1 = (6–30°) ¥

V2 = (5––50°) ¥

and Therefore, P R O B L E M

2 + j3 = 3.22–59.7° V 4 + 2 + j3 4 = 2.98––76.6° V 4 + 2 + j3

V = V1 + V2 = 3.22–59.7° + 2.98––76.6° = 2.32––2.96° V C -3 3

Find ZTh, VTh, and IN Fig. C-27.

Solution

ZTh, the current source is replaced by an open circuit and the voltage source is replaced by a short

circuit. Thus, ZTh = 4 + j3 = 5–36.87° W Isc = IN VTh. When we short circuit the terminals a and b, the voltage source of 40–60° V gets applied across the series combination of 4-W and j3-W impedances. Hence, the current (from left to the right) through the j3-W impedance is I1 =

40 ∠ 60° = 8–23.1° A 4 + j3

744

Basic Electrical Engineering 100-W impedance is the same as the current source, I2 = 6–20° A

Applying KCL at the terminal a, we get Isc = IN = 6–20° – 8–23.1° = 2.04––147.6° A = –2.04–32.4° A VTh = IN ZTh = (–2.04–32.4°)(5–36.87°) = –10.2–69.27° V

3W

j3 W

100 W

a

4W

6–20° A 40–60° V

b

Fig. C-27 P RO B L EM

120–30° V

a

–j4 W

Is

2W

I

j3 W

j5 W

b

4W

Fig. C-28

C -34 a) the current I if Is = 0 A, (b) the current I if Is = 10––50° A.

Solution (a) Since the current source produces 0 A, it is equivalent to an open circuit and can be removed form the circuit. I. Hence, we treat the impedance (2 + j3) W as ZTh, we short circuit the voltage source so that the 3-W and j5-W impedances, and as well as the – j4-W and 4-W impedances, come in parallel. These two parallel combinations are in series between a and b. Hence, 3 ( j 5) 4 (- j 4) ZTh = 3|| ( j5) + 4 || (–j4) = = 4.26–– 9.14° W + 3 + j5 4 - j 4 j5-W and 4-W impedances. These voltage drops can be found by applying voltage division. Thus, j5 4 VTh = (120–30°) ¥ – (120–30°) ¥ = 29.1–16° V 3 + j5 4 j4 I through the load, as VTh 29.1∠16° I= = = 4.39–– 4.5° A ZTh + Z L 4.26 ∠− 9.14° + (2 + j 3) (b) The current source does not affect ZTh, and its value remains the same as found above : ZTh = 4.26––9.14° W. the source-current multiplied with the impedance between terminals a and b (with the load replaced by an open circuit). This impedance is ZTh. Hence, the voltage contribution of the current source Is is Vs = Is ZTh = (10––50°)(4.26––9.14°) = 42.6––59.14° V

Supplementary Exercise—Part C : AC Circuits

745

Since, the direction of the source current I is into the terminal b, the above voltage drop is from terminal b to a. VTh = 29.1–16° – 42.6–– 59.14° = 45–82.14° V Finally, the current I through the load is given as I=

P ROBLEM

45∠82.14° VTh = = 6.79–61.6° A Z Th + Z L 4.26 ∠− 9.14° + 2 + j 3

C -35 i in the circuit shown in Fig. C-29. 2 mH

i

4W

4÷2 sin 1000t A

10÷2 cos (1000t – 25°) V

Fig. C-29 Solution We redraw the given circuit in phasor domain, for which XL = jwL = j1000 ¥ 2 ¥ 10

–3

= j2 W; I = 4–0° A; V = 10–(90 – 25)° = 10–65° V

The phasor-domain circuit is drawn in Fig. C-30. The current I can be considered to have a component I1 from the current source and another component I2 from the voltage source, so that I = I1 + I2 I1, we replace the voltage source by a short-circuit and then apply current division, 4 = 3.58––26.6° A I1 = (4–0°) ¥ 4 + j2 The component I2 10 ∠ 65° = –2.24–38.4° A 4 + j2 The negative sign appears because the current due to the voltage source has a direction opposite to that of the current I. Adding the two currents, we get I2 =

I = I1 + I2 = 3.58––26.6° – 2.24–38.4° = 3.32––64.2° A Finally, the current i in time-domain is i = 3.32 2 sin (1000t – 64.2°) A = 4.7 sin (1000t – 64.2°) A j2 W

4–0° A

4W

Fig. C-30

I

10–65° V

4W

30 mH

Fig. C-31

15 W

746

Basic Electrical Engineering

P RO B L EM

C -36

For the circuit shown in Fig. C-31, what is the power factor at 60 Hz ? What capacitor connected across the input terminals causes the power factor to be unity ? What additional capacitor causes the overall power factor to be 0.85 lagging ?

Solution XL = jwL = j(120p)(0.03) = j11.3 W. Therefore, the impedance of the circuit is Z=4+

15 ( j11.3) = 11.9–37.38° W 15 + j11.3

Hence, the power factor of the circuit is given as pf = cos 37.38° = 0.795 (lagging) Since the capacitor is to be connected in parallel, let us determine the admittance of the circuit. Before the capacitor is connected, the admittance is Y=

1 1 = = 0.0842––37.38° S = (66.9 – j51.1) mS Z 11.9 ∠ 37.38°

For unity power factor, the imaginary part of the total admittance should be zero. It means that the capacitor to be connected must have a susceptance of 51.1 mS. Consequently, its capacitance must be -3 = 51.1 ¥ 10 = 136 ¥ 10 –6 = 136 mF 120p We need a different capacitor to make the overall power factor to be 0.85 lagging. New power factor angle is given as q = cos–1 0.85 = 31.79°

C=

B

By adding a parallel capacitor to a circuit, the conductance G of the circuit does not change. That is, the new conductance, G = 66.9 mS. The new susceptance B can be determined from the admittance triangle as –3 –3 B = G tan (–q) = (66.9 ¥ 10 ) tan (–31.79°) = –41.5 ¥ 10 S

Because, it is the added capacitor that brings the change in susceptance of the circuit, its capacitance is given as C=

PROBLE M

DB 51.1 ¥ 10 - 3 - 41.5 ¥ 10 - 3 = = 25.6 ¥ 10 –6 F = 25.6 mF w 120p

C -37

maximum real power that the alternator can deliver ?

Solution The maximum volt-amperes or apparent power is the limitation on the capacity of an alternator. Since the real power is the product of the apparent power and the power factor, the maximum apparent power of the given alternator is P 35 Smax = = pf 0.7 At unity power-factor, all of the apparent power would be the real power. This means that the maximum real power that the alternator can deliver is Pmax = Smax ¥ 1 = 50 MW

Supplementary Exercise—Part C : AC Circuits P RO B L EM

747

C -38

A non-inductive resistor in series with an iron-cored coil and a capacitor is connected to an ac supply. The voltages and the current are as shown in the circuit of Fig. C-32. Determine the power loss in the coil and the voltage of the ac supply.

Fig. C-32 Solution Since the iron-cored choke coil has some losses, let its impedance be represented as Z2 = R2 + jXL2 = Z2 –q2 As shown in the phasor diagram in Fig. C-33, the voltage VR is in phase with current I, but the voltage V2 across the choke coil leads the current by angle q2. Since the phasor sum of the voltages VR and V2 has a magnitude of 50 V, we have 502 = 252 + 402 + 2(25)(40) cos q2 fi cos q2 = 0.1375 or q2 = 82.1° and sin q2 = 0.9905 V2 40 = = 116 W I 0.345 The resistance of the choke coil, R2 = Z2 cos q2 = 116 ¥ 0.1375 = 15.95 W

The impedance of the choke coil, Z2 =

Hence, the loss in the coil,

2 2 P1 = I R2 = (0.345) ¥ 15.95 = 1.9 W

Now, the supply voltage is the phasor sum of all the component voltages, V = VR + V2 + VC = 25 + (40 cos q2 + j40 sin q2) + (– j55) = 25 + (5.5 + j39.6) – j55 = 30.5 – j15.4 = 34.17––26.8° V IC

V = 440 V

25.8 42° 45. 573 °

C

B

I2 = 31.11 A

I = 40 A A 1

Fig. C-34

748

Basic Electrical Engineering

P R O B LE M

C - 39

A single-phase induction motor takes a current of 40 A at a power factor of 0.7 lagging from a 440-V, 50-Hz supply. What capacitor is to be connected across the input terminals to raise the power factor to 0.9 lagging ?

Solution The power factor angle of the motor is given as –1

q1 = cos 0.7 = 45.573° After connecting the capacitor in parallel, the new power-factor angle is given as –1 q2 = cos 0.9 = 25.842°

The parallel connected capacitor takes a current IC leading the applied voltage by 90°, and changes the input current I1 (= 40 A) to I2, as shown in the phasor diagram of Fig. C-34. Since the power drawn by the circuit remains the same, the active component of the current does not change by connecting a capacitor. That is, I1 cos q1 = I2 cos q2 I2 =

or

40

or

40 ¥ 0.7 = I2 ¥ 0.9

0.7 = 31.11 A 0.9

From the phasor diagram, we have IC = AB = AC – BC = I1 sin q1 – I2 sin q2 = 40 sin 45.573° – 31.11sin 25.842° = 15 A V 440 The reactance of the capacitor, XC = = = 29.33 W IC 15 1 1 = = 108.5 mF Thus, the capacitance, C = w X C 100p ¥ 29.33 PR OBLEM

C -4 0

A series RLC circuit has L = 50 mH, C = 2 mF, and R = 10 W. (a) Calculate the Q-factor of the circuit. (b) If the inductance is doubled, determine the new value of C required for resonance at the same frequency, and the new Qfactor.

Solution (a) The Q-factor, Q =

1 R

50 × 10 − 6

L 1 = C 10

(b) The resonant frequency, f0 =

2 × 10 − 6 1

=

2p LC

= 0.5

1 2p 50 ¥ 10 - 6 ¥ 2 ¥ 10 - 6

=

10 5 Hz 2

The changed inductance, L1 = 2L = 2 ¥ 50 ¥ 10 – 6 = 10 –4 H. If C1 is the required capacitance for keeping the resonant frequency same, we must have 105 1 = 2p 2p L1C1

The new Q-factor, Q1 =

1 R

or

L1 1 = C1 10

C1 =

1 L1 1010

100 × 10 − 6 1 × 10 − 6

= 1.0

=

1 10 − 4 × 1010

= 10

–6

F = 1 mF

Supplementary Exercise—Part C : AC Circuits P RO B L EM

749

C -41

A series RLC circuit with R = 25 W and L = 0.6 H results in a leading phase angle of 60° at a frequency of 40 Hz. At what frequency will the circuit resonate ?

Solution The reactance of the inductance at 40 Hz is XL = 2pfL = 2p ¥ 40 ¥ 0.6 = 150.8 W If XC is the reactance of the capacitor in the circuit, its total impedance is given as Z = R + j(XL – XC) = 25 + j(150.8 – XC) W Since, the phase angle is 60° leading, we should have 150.8 XC = – tan 60° fi XC = 193.3 W 25 Therefore, the value of the capacitance must be 1 1 C= = = 20.58 mF wX C 2p ¥ 40 ¥ 19.3 We can now calculate the frequency of resonance of the circuit as 1 1 f0 = = = 45.29 Hz 2p LC 2p 0.6 ¥ 20.58 ¥ 10 - 6 P ROBLEM

C -42

A coil of resistance 40 W and inductance 0.75 H forms a part of a series circuit for which the resonant frequency is a) the line current, (b) the power factor, (c) the power consumed, and (d) the voltage across the coil.

Solution (a) For a series resonant circuit, the resonant frequency is given as 1 1 1 f0 = fi C= = = 11.16 mF (2p ¥ 55) 2 ¥ 0.75 2p LC (2pf0)2L The reactances of the inductance and capacitance at 50 Hz are given as XL = 2pfL = 2p ¥ 50 ¥ 0.75 = 235.6 W 1 1 and XC = = = 285.2 W . × 10 − 6 2pf C 2 π × 50 × 1116 Therefore, the impedance of the circuit is Z = R + j(XL – XC) = 40 + j(235.6 – 285.2) = (40 – j49.6) W = 63.72––51.12° W We can now calculate the line current as I=

V 250 ∠ 0° = = 3.92–51.13° A 63.73∠− 5113 . ° Z

(b) The power factor, pf = cos 51.13° = 0.6276 (leading) (c) The power consumed, P = VI cos f = 250 ¥ 3.92 ¥ cos 51.13° = 615 W (d) The voltage across the coil, Vcoil = IZcoil = I(R + jXL) = (3.92–51.13°)(40 + j235.6) = 936.8–131. 5° V

750

Basic Electrical Engineering

P RO B L EM

C -43

An inductive circuit of resistance 2 W and inductance 0.01 H is connected to a 250-V, 50-Hz. What value of capacitance at resonance.

Solution The circuit diagram is shown in Fig. C-35a. The impedance of the inductive branch is given as Z1 = 2 + j(2p ¥ 50 ¥ 0.01) = (2 + j3.142) W = 3.725–57.52° W

Fig. C-35 At resonance, the net current I should be in phase with the applied voltage V, as shown in Fig. C-35b This can happen only if the reactive component of the inductive-circuit current I1 is equal and opposite in phase to the current I2 through the capacitor. Hence, I2 = I1 sin q1. V V XL = XC Z1 Z1

or

C =

Hence,

L Z12

The current at resonance, I = I1 cos q1 = P ROBLE M

=

1

2

fi

Z 1 = XL XC = (w 0 L)

0.01 (3.725)2

(

0C)

=

L C

= 720.7 mF

V R VR 250 2 = 2 = = 36.03 A Z1 Z1 Z1 (3.725)2

C -4 4

A coil having a resistance of 50 W and an inductance of 0.2 H is connected in series with a capacitor C1 across an ac voltage source of variable frequency, as shown in Fig. C-36a. The current through the circuit is found to be maximum when the frequency is 50 Hz. Calculate the capacitance C2 of the condenser to be connected in parallel (Fig. C-36b) with the above series combination so that the total current taken from the ac supply is a minimum when the frequency is 100 Hz. Taking the supply voltage as reference, draw the phasor diagram representing all the currents at 100 Hz.

Solution When the current is maximum at 50 Hz, we must have XL1 = XC1

fi

C1 =

1 2

(2pf ) L1

=

1 (2 π × 50)2 × 0.2

= 50.66 mF

At a frequency of 100 Hz, we have

and

XL1 ¢ = 2pf ¢L1 = 2p ¥ 100 ¥ 0.2 = 125.7 W 1 1 XC1 ¢ = = = 31.42 W 2p f ¢L1 2 π × 100 × 50.66 × 10 − 6

Supplementary Exercise—Part C : AC Circuits R1 = 50 W

L1 = 0.2 H

C1

L1

R1

751

50.66 mF

C2 f1 = 50 Hz

f2 = 100 Hz

(a)

(b)

Fig. C-36 Hence,

Z1 = 50 + j(125.7 – 31.42) = (50 + j94.28) W = 106.7–62° W

and

Z2 = 0 – jXC2 = XC2 ––90° W

The corresponding admittances of the parallel branches are 1 1 = = 9.372––62° mS and Y1 = Z1 106.7–62∞

Y2 =

1 1 = = wC2 –90° S Z2 X C 2–- 90∞

Therefore, the equivalent admittance of the parallel combination, Y = Y1 + Y2 = 9.372 ¥ 10–3––62° + wC2 –90° = 4.4 ¥ 10 –3 – j(8.275 ¥ 10 –3 – wC2) Since, I = VY, for minimum supply current (that is, for resonant condition), we must have minimum Y. This can happen only when the imaginary component of the Y is made zero, 8.275 ¥ 10 - 3 –3 = 13.17 mF 8.275 ¥ 10 – jwC2 = 0 fi C2 = 2p ¥ 100 At resonant condition, the overall admittance is Y = 4.4 ¥ 10 –3 S, the admittance of the second branch is –6 Y2 = wC2–90° = (2p ¥ 100 ¥ 13.17 ¥ 10 )–90° = 0.008275–90° S

Let the supply voltage be V–0° volts. Then, the supply current and the branch currents are given as –3 I = YV = (4.4 ¥ 10 )(V–0°) = 0.0044V A

I1 = Y1V = (9.372 ¥ 10–3–– 62°)(V–0°) = 0.009372V–– 62° A I2 = Y2V = (0.008275–90°)(V–0°) = 0.008275V–90° A Thus, the phasor diagram of the circuit of Fig. C-36b at resonance can be drawn as shown in Fig. C-37. and

I2 = 0.008275V A

I = 0.0044V A O

V 62°

I1 = 0.009372V A

Fig. C-37

752

Basic Electrical Engineering

P RO B L EM

C -45

Each phase of a star-connected load consists of a resistance of 100 W in parallel with a capacitance of 32 mF. Calculate the line current, the power absorbed, the total kVA, and the power factor, when connected to a 415-V, 50-Hz, threephase supply.

Solution For each phase : Z1 = 100 W; Z2 = – j/(2p ¥ 50 ¥ 32 ¥ 10 –6) = –j99.5 W Therefore, the load impedance per phase, Z Z 100 (- j 99.5) Zph = 1 2 = = 70.53–– 45.14° W 100 - j 99.5 Z1 + Z2 For star-connected load, Vph = VL/ 3 = 415/ 3 = 239.6 V. We can now calculate the phase current, Vph 239.6 Iph = = = 3.397 A Zph 70.53 \ Line current, IL = Iph = 3.397 A The power factor, pf = cos 45.14° = 0.7054 (leading) The power absorbed, P = 3VLIL cos f = 3 ¥ 415 ¥ 3.397 ¥ cos 45.14° = 1722 W The total kVA, kVA = 3VLIL = 3 ¥ 415 ¥ 3.397 = 2442 kVA P R O B L EM

C - 46

A 440-V, 50-Hz, three-phase supply drives a three-phase, delta-connected induction motor operating at rated load in parallel with another star-connected load having impedance of (20 + j15.65) W per phase, as shown in Fig. C-38. The motor draws a power of 75 kW at a power factor of 0.8 lagging and it can be considered as three impedances. Determine (a) the line and phase currents in the motor, (b) the line and phase currents in the impedance load, (c) total line current, and (d) the total power consumed.

Fig. C-38 Solution (a) For the delta-connected motor, Vph1 = VL = 440–0° V. The magnitude of the line current for the motor can be found from the power drawn by it, 75000 P IL1 = = = 123 A 3 440 0.8 3VL cos f

Supplementary Exercise—Part C : AC Circuits

753

The phase angle of the motor, f1 = cos –1 0.8 = 36.87°. As can be seen from the phasor diagram given in Fig. C-39a, the current Iph1 lags the voltage Vph1 by 36.87°, and the line current IL1 leads the phase current Iph1 by 30°. Hence, the phase angle of line current IL1, fL1 = 30° – 36.87° = – 6.87° Thus, the line current and the phase current for the motor are given as 123 and Iph1 = ––36.87° = 71––36.87° A IL1 = 123––6.87° A 3

Fig. C-39 (b) For the star-connected impedance load, as can be seen from the phasor diagram shown in Fig. C-39b, the phase voltage is given as 440 V Vph2 = L ––30° = ––30° = 254––30° V 3 3 The impedance of the load per phase, Zph2 = (20 + j15.65) W = 25.4–38.04° W Hence, the line current and phase current for the impedance load are given as IL2 = Iph2 =

Vph 2 Z ph 2

=

254 ∠− 30° = 10–– 68.04° A 25.4 ∠ 38.04°

(c) The total line current, IL = IL1 + IL2 = (123––6.87°) + (10––68.04°) = 128.1––10.79° A (d) The total power consumed, P = P1 + P2 = 3Vph1Iph1 cos f1 + 3Vph2Iph 2 cos f2 = 3 ¥ 440 ¥ 71 ¥ cos 36.87° + 3 ¥ 254 ¥ 10 ¥ cos 38.04° = 75000 + 6001 ª 81 kW P RO B LE M

C -4 7

A three-phase balanced supply system with phase voltage of 120 V and phase-sequence of abc is connected to an unbalanced delta-connected load, as shown in Fig. C-40. Find (a) IaA, (b) IbB, (c) IcC, and (d) the total complex power supplied by the source.

754

Basic Electrical Engineering b

B 10 W a

n

A

j5 W

120–0° V –j10 W c

C

Fig. C-40 Solution The phase voltages of the 3-phase supply are given as Van = 120–0° V; Vbn = 120––120° V;

and

Vcn = 120–120° V

Therefore, the line voltages are VAB = Van – Vbn = 120–0° – 120––120° = 207.8–30° V VBC = Vbn – Vcn = 120––120° – 120–120° = 207.8––90° V and

VCA = Vcn – Van = 120–120° – 120–0° = 207.8–150° V

and

IAB =

VAB 207.8∠30° = = 20.78–30° A ZAB 10

IBC =

VBC 207.8∠− 90° = = –41.4–0° A ZBC j5

ICA =

VCA 207.8–150∞ = = 20.78––120° A ZCA - j10

(a) IaA = IAB – ICA = 20.78–30° – 20.78––120° = 40.07–45° A (b) IbB = IBC – IAB = – 41.4–0° – 20.78–30° = 60.22––170.1° A (c) IcC = ICA – IBC = 20.78––120° – (– 41.4–0°) = 35.85––30° A (d) The total complex power is given as S = SAB + SBC + SCA = V I * + V I* + V I* AB AB

BC BC

CA CA

= (207.8–30°)(20.78––30°) + (207.8––90°)(– 41.4–0°) + (207.8–150°)(20.78–120°) = (4318 + j0) + (0 + j8603) + (0 – j4318) = (4318 + j4285) VA C O M ME N TS Note that, as it should be, the active power is contributed only by the impedance ZAB (= 10 W); both the impedances ZBC (= j5 W) and ZCA (= –j10 W) contribute to the reactive power.

755

Supplementary Exercise—Part C : AC Circuits

C. 2. PRACTICE PROBLEMS ( A )

S I M P L E

PR O BL EMS

C-1. Find the instantaneous value of i = 70 sin 400pt A at t = 3 ms. [Ans. –41.1 A] C-2. A current wave has a peak value of 58 mA and a radian frequency of 90 rad/s. Find the instantaneous current at t = 23 ms. [Ans. 50.9 mA] C-3. What is the shortest time required for a 2.1 krad/s peak value ? [Ans. 0.442 ms] C-4. Evaluate (a) v = 200 sin (339t + p/7) V (b) i = 67 cos (301t – 42°) mA at t = 11 ms. [Ans. (a) –172 V; (b) –56.63 mA] C-5. Find the angle by which i1 lags v1, if i1 = 4.5 sin (wt – 20°) A and v1 is equal to (a) v1 = 125 sin (wt + 60°) V (b) v1 = 125 sin (wt – 120°) V [Ans. (a) 80°; (b) –100°] C-6. A 30-W resistor has a voltage across it given as v = 170 sin (314t – 20°) V Find the average power dissipation of the resistor. [Ans. 482 W] C-7. Find the effective value of a periodic voltage that has a value of 20 V for one half-period and a value of –10 V for the other half-period. [Ans. 15.8 V] C-8. Find the frequencies at which a 250-mH inductor has reactances of 30 W and 50 kW. [Ans. 19.1 Hz, 31.8 kHz] C-9. Find the voltage across a 2-H inductor for the following currents, assuming passive-component references for voltage and current : (a) 10 A, (b) 10 sin (314.2t + 10°) A, and (c) 10 cos (10 4t – 20°) A [Ans. (a) 0 V; (b) 6.284 cos (314.2t + 10°) kV; (c) 0.2 cos (104t C-10. Find the capacitances of capacitors that have a reactance of 500 W at (a) 377 rad/s, (b) 10 kHz, and (c [Ans. (a) 5.31 mF; (b) 31.8 nF; (c) 14.1 pF]

C-11. What is the voltage across a 30-mH inductor that

C-12. C-13.

C-14.

C-15.

[Ans. 0.377 V] mF capacitor that has 200 V at 400 Hz across it ? [Ans. 50.3 mA] The voltage across and the current through a circuit-component are v = 20.3 sin (314t – 10°) V i = 15.6 sin (314t – 30°) mA Find the average power absorbed by the component. [Ans. 148.8 mW] mF capacitor for voltages of (a) v = 5 sin (314t – 30°) V and (b) v = 75 cos (10 4t + 60°) V? [Ans. (a) i = 3.14 sin (314t + 60°) mA; (b) i = –1.5 sin (104t + 60°) A] I1, I2, and I3. [Ans. 28.3–45° A, 20–90° A, 20–0° A] I1 –j5 W

100–0° V

I2

I3

5W

j5 W

Fig. C-41 C-16.

VR, VL, and VC in the circuit shown in Fig. C-42. [Ans. 100–30° V, 5000–120° V, 5000––60° V] 20 W

j1000 W

+ V – R

+ VL –

Vs = 100–30° V

+ –j1000 W –

Fig. C-42

VC

756

Basic Electrical Engineering

C-17. A coil is energised by a 120-V, 50-Hz source. It draws a current of 2 A that lags the voltage by 40°. What are the coil resistance and inductance ? [Ans. 46 W, 0.123 H] C-18. A load has a voltage of V = 120–30° V and a current of I = 30–50° A at a frequency of 400 Hz. Find the two-element series circuit that could represent the load. [Ans. R = 3.76 W, C = 290 mF] C-19. A capacitor is in series with a coil that has 1.5 H of inductance and 5 W of resistance. Find the capacitance that makes the combination purely resistive at 50 Hz. [Ans. 6.755 mF] C-20. Find the total impedance at 1 krad/s of a 1-H inductor and a 1-mF capacitor connected (a) in series, and (b) in parallel. [Ans. (a) 0 W; (b) W] C-21. Find the impedance across the terminals A and B for the circuit shown in Fig. C-43. The operating frequency is 50 Hz. [Ans. (10 – j30.87) W]

Fig. C-43 C-22.

IL for the circuit shown in Fig. C-44. [Ans. –2.22––36.3° A]

Fig. C-44 C-23. Three alternating currents : i1 = 150 sin (wt + p/4) A i2 = 40 sin (wt + p/2) A i3 = 80 sin (wt – p/6) A are fed simultaneously to a common conductor. Find the equation of the resultant current and its rms value.

[Ans. i = 204.9 sin (wt + 31.17°) A, 144.9 A] C-24. Three sinusoidal voltages acting in series are given by v1 = 10 sin 400t V, v2 = 10 2 sin (400t – 45°) V, and v3 = 20 cos 400t V Determine (a) an expression for the resultant voltage, and (b) the frequency and the rms value of the resultant voltage. [Ans. (a) v = 22.36 sin (400t + 26.57°) V; (b) 63.66 Hz, 15.81 V] C-25. A circuit consists of a resistance R and a capacitive reactance of 60 W connected in series. Determine the value of R for which the power factor of the circuit is 0.8. [Ans. 80 W] C-26. An iron choke takes 4 A when connected to a 20-V dc supply, and takes 5 A when connected to a 65-V, 50-Hz supply. Determine (a) the resistance and the inductance of the coil, (b) the power drawn by the coil, and (c) the power factor. [Ans. (a) 5 W, 38.2 mH; (b) 125 W; (c) 0.3846 (lagging)] C-27. circuit are v = 100 sin (wt + 30°) V i = 15 sin (wt + 60°) A Determine the impedance, the resistance, the reactance, the power factor and the power. [Ans. 6.67 W, 5.776 W, 3.335 W, 0.866 (leading), 649 W] C-28. The current in a circuit is given as (4.5 + j12) A, when the applied voltage is (100 + j150) V. Determine (a) the complex expression for the impedance, stating whether it is inductive or capacitive, (b) the power, and (c) the power factor. [Ans. (a) (13.7 – j3.196) W, Capacitive; (b) 2.251 kW; (c) 0.9728 (leading)] C-29. A series circuit consists of a non-inductive resistance of 5 W and an inductive reactance of 10 W. When connected to a single-phase ac supply, it draws a current i(t) = 27.89 sin (628.3t – 45°) A Find (a) the voltage applied to the series circuit in the form Vm sin (wt + q°), (b) the inductance, and (c) the power drawn by the circuit. [Ans. (a) v = 311.8 sin (628.3t + 18.43°) V; (b) 15.92 mH; (c) 1945 W]

Supplementary Exercise—Part C : AC Circuits C-30. A coil takes 2.5 A when connected across 200-V, 50-Hz mains. The power consumed by the coil is found to be 400 W. Find the inductance and the power factor of the coil. [Ans. 0.1528 H, 0.8 (lagging)] C-31. A pure inductive coil is connected in series with a 10-W resistor to a 50-Hz ac source. The voltages across the resistor and the inductor are found to be 30 V and 40 V respectively. Find the value of the inductive reactance and the supply voltage. [Ans. 13.33 W, 50 V] C-32. A series circuit with 10-W resistor and 20-mH inductor has a current i = 2 sin 500t total voltage across the circuit and the angle by which the current lags the voltage. [Ans. v = 28.28 sin (500t + 45°) V, 45°] C-33. A resistor and a capacitor are connected in series across a 150-V ac supply. When the frequency of the supply is 40 Hz, the current drawn is 5 A, and when the frequency is 50 Hz, the current is 6 A. Find the resistance and the capacitance of the circuit. [Ans. 11.7 W, 144 mF] C-34. A circuit consists of a resistance of 8 W and a series capacitive reactance of 6 W. A voltage v = 141 sin 314t V is applied to the circuit. Find (a) the complex impedance, (b) the rms and instantaneous values of the current, (c) the power delivered to the circuit, (d) the equation for the voltage appearing across the capacitor, and (e) the value of the capacitance. [Ans. (a) (8 – j6) = 10––36.87° W; (b) 10 A, 14.1 sin (314t + 36.87°) A; (c) 800 W; (d) 84.85 sin (314t – 53.13°) V; (e) 530.8 mF] C-35. A voltage 230–30° V is applied across two series components, one of which is a 20-W resistor and the other is a coil with an impedance of 40–20° W. Find the individual component voltage drops. [Ans. VR = 77.7–16.6° V; VZ = 155.2–36.6° V] C-36. A phasor-domain circuit has 200–15° V applied across three series-connected components having impedances of 20–15° W, 30––40° W, and 40–50° W. Use the concept of voltage division V across the component with impedance of 40–50° W. [Ans. V = 114–51.3° V] C-37. What is the total impedance of three parallel components that have impedances Z1 = 2.5–75° W,

757

Z2 = 4––50° W, and Z3 = 5–45° W ? [Ans. ZT = 1.9–39.7° W] C-38. A load has a voltage of V = 120–20° V and a current of I = 48–60° A, both at 2 kHz. Find the two-element parallel circuit which could represent the load. [Ans. R = 3.26 W, C = 20.5 mF] C-39. A resistor and a parallel 1-mF capacitor draw 0.48 A when a 120-V, 400-Hz source is applied. Find the admittance of the circuit in polar form. [Ans. 4–38.9° mS] C-40. Find the total impedance of two parallel components which have identical impedances of 100–60° W. [Ans. 50–60° W] C-41. What is the total impedance to two components which have impedances of 80––30° W and 60–40° W ? [Ans. 41.65–10.71° W] a) the C-42. open-circuit voltage Vab, (b) the downward current in a short circuit between a and b, and (c) Zab in parallel with the current source. [Ans. (a) 16.77––33.4° V; (b) 3–30° A; (c) (2.5 – j5) W] a 10 W –j5 W

3–30° A

j5 W b

Fig. C-45 C-43. Find Zab for the circuit shown in Fig. C-46 if w is equal to (a) 800 rad/s, and (b) 1600 rad/s. [Ans. (a) (477.9 + j175.6) W; (b) (587.6 + j119.8) W] 2 mF a 300 W

600 W

b

Fig. C-46

0.6 H

758

Basic Electrical Engineering

C-44. If w (a) Zin, and (b) Zin if a short circuit is connected between x and y. [Ans. (a) (22 – j6) W; (b) (9.6 + j2.8) W] x

20 W

10 W y

Zin

2 mF

C-50.

Zin at w = 4 rad/s if terminals a and b are (a) open-circuited, and (b) short-circuited. [Ans. (a) (10.56 – j1. 92) W; (b) (9.97 + j0.246) W] 8W

Zin

4H

C-49.

0.1 H

Fig. C-47 C-45.

(c) the phase angle, and (d) the voltages across each element. [Ans. (a) 1.918 A; (b) 0.3835 (leading); (c) 67.45°; (d) 38.36–67.45° V, 60.26–157.5° V, 152.6––22.55° V]

C-51.

a

C-52.

b

C-53.

1 F 8

2W

Fig. C-48 C-46. An ac circuit consists of a resistance of 5 W, an inductance of 0.1 H, and a capacitance of 100 mF, all in series. Determine for this circuit (a) the total reactance, (b) the impedance, (c) the admittance, (d) the susceptance, and (e) the conductance, at a frequency of 60 Hz. [Ans. (a) j11.17 W; (b) 12.24–65.9° W; (c) 0.0817––65.9° S; (d) j0.0746 S; (e) 0.0334 S] C-47. Find the active and reactive components of the current taken by a series circuit consisting of an inductance of 0.1 H and resistance of 8 W and a capacitor of 120 mF, when connected to a 240-V, 50-Hz supply. [Ans. 21.84 A, 13.35 A] C-48. A coil of resistance 20 W and inductance 100 mH is connected in series with a capacitance of 40 mF across a 100-V, 50-Hz supply. Calculate (a) the magnitude of the current, (b) the power factor,

C-54.

of R = 10 W, L = 0.1 H, and C = 100 mF. Find the impedance, the power factor and the power, if the mains frequency is 50 Hz. [Ans. 10.01––2.348° W, 0.9992 (leading), 1000 W] What power is consumed by a circuit that has an input admittance of (0.4 + j0.5) S and an input current of 30 A ? [Ans. 878 W)] through a load of 30–40° W. A current of 4 A Determine the resulting power components, i.e., the real, the reactive, and the apparent powers. [Ans. 368 W, 309 VAR, 480 VA] Find the power components of a load that draws 20––30° A when connected to a source of 240–20° V. [Ans. 3.09 kW, 3.68 kVAR, 4.8 kVA ] A coil of resistance 5 W and inductance 10 mH is connected in series with a condenser of 100 mF capacity across a single-phase, 230-V, 50-Hz supply. Find (a) the total power consumed, (b) the power factor, (c) the frequency at which resonance occurs, and (d) the current at resonance. [Ans. (a) 312 W; (b) 0.1719 (leading); (c) 159.2 Hz; (d) 46 A] A choke coil of resistance 5 W and an inductance 0.6 H, is in series with a capacitor of 10 mF. When 200 V ac is applied to it and the frequency is

the capacitor. [Ans. 9794 V] C-55. A circuit consisting of a coil of inductance 10 mH and resistance 300 W connected in parallel with a certain capacitor resonates at 48 kHz. Determine the capacitance of the capacitor and the dynamic impedance of the circuit. [Ans. 1.09 nF, 30.58 kW] C-56. If the bandwidth of a resonant circuit is 10 kHz and the lower half-power frequency is 110 kHz, what is its upper half-power frequency ? What is the value of its quality factor ? [Ans. 120 kHz, 11.5]

Supplementary Exercise—Part C : AC Circuits C-57. A series RLC circuit has the following parameters : R = 10 W, L = 0.01 H, and C = 100 mF. Compute (a) the resonant frequency, (b) the quality factor, and (c) the bandwidth. [Ans. (a) 159.2 Hz; (b) 1; (c) 159.2 Hz] C-58. An RLC series circuit consists of a resistance of 1000 W, an inductance of 100 mH, and a capacitance of 10 pF. Determine (a) the resonant frequency, (b) the quality factor of the circuit at resonance, and (c) the half-power frequencies. [Ans. (a) 15.92 kHz; (b) 10; (c) 15.12 kHz, 16.72 kHz] C-59. Three similar coils, each of resistance 20 W and inductance 0.5 H, are connected in (a) star, and then (b) delta, to a three-phase, 400-V, 50-Hz ( B)

TRI C KY

759

supply. Calculate the line current, and the total power absorbed, in both cases. [Ans. (a) 1.459 A, 127.7 W; (b) 4.3 77 A, 383.2 W] C-60. A 3-phase motor operating on a 400-V supply power factor of 0.82. Calculate (a) the line current, and (b) the phase current, if the windings are delta connected. [Ans. 37.73 A, 21.78 A] C-61. A 3-phase, three-wire, RYB system with effective voltage of 150 V has a balanced Y-connected load of 5––30° W and the power supplied to the load. [Ans. 17.32 A, 3897 W]

PR O BL EM S

C-62. Find the phase relationship for the following pairs of sinusoids : (a) v = 40 sin (377t + 50°) V and i = 4.7 sin (754t – 25°) A (b) v1 = 47 sin (7.1pt + 30°) V and v2 = 51 sin (7.1pt – 10°) V (c) v = 42.9 sin (400t + 60°) V and i = –3.7 sin (400t – 50°) A [Ans. (a) No phase relation; (b) v1 lead v2 by 40°; (c) v lead i by –70°] C-63. Find the angle by which v1 = 47 sin (628t – 20°) V [Ans. 40°] leads v2 = –69 cos (628t + 30°) V C-64. A 60-Hz sinusoidal current has an rms value of 10 2 A, and an instantaneous value of 7.07 A at t = 0. Assuming the current wave to enter positive half at t = 0, determine (a) the expression for the current, (b) the magnitude of the current at t = 0.0125 s, and t = 0.025 s after t = 0. [Ans. (a) i = 20 sin (120pt + 20.7°) A; (b) –18.71 A, –7.07 A] C-65. In an ac circuit supplied from 50-Hz mains, the potential difference has a maximum value of 500 V, and the current has a maximum value of 10 A. At t = 0, the instantaneous values of pd and current are 400 V and 4 A, respectively, both increasing positively. Assuming sinusoidal variations, obtain the expressions for pd and current. Determine

the instantaneous values of pd and current at t them. [Ans. (a) v = 500 sin (100 pt + 53.13°) V; i = 10 sin (100pt + 23.58) A; (b) –300 V, – 9.165 A, 29.55°] C-66. An alternating current varying sinusoidally with a frequency of 50 Hz has an rms value of 20 A. (a) Write down the equation for the instantaneous i) 0.0025 s, (ii) 0.0125 s, after it passes through a positive maximum value. (b) At what time, measured from positive maximum value, will the instantaneous current will be 14.14 A ? [Ans. (a) i = 28.28 sin (100pt) A, (i) 20 A, –3 (ii) –20 A; (b) 3.333 ¥ 10 s] C-67. Find the average values and the rms values of the waveforms shown in Fig. C-49. [Ans. I1(av) = 2.5 A, I2(av) = 0.4 A, I1(rms) = 6.042 A, I2 (rms) = 4.648 A] C-68. Find the average value of the periodic voltage waveforms shown in Fig. C-50. [Ans. (a) 3.5 V; (b) 4 V; (c) 15 V] C-69. Determine the equivalent impedance of the network shown in Fig. C-51, given an operating frequency of 5 rad/s. [Ans. 6.511–49.2° W]

760

Basic Electrical Engineering

Fig. C-49

Fig. C-50 200 mF 2 H

10 W

6W

500 mF

Fig. C-51 C-70. Find the current i(t) in the circuit shown in Fig. C-52, if vs (t) is 40 sin 3000t V. [Ans. 16 sin (3000t – 36.87°) mA]

16-mH inductor, what are the two other elements ? [Ans. R = 2.87, C = 238 mF] C-72. What two circuit-elements connected in series have the same total impedance at 4 krad/s as the parallel combination of a 50-mF capacitor and a 2-mH coil with a 10-W winding resistance. [Ans. R = 2.29, C = 44 mF] C-73. For the circuit shown in Fig. C-53, determine the indicated unknown phasors and the corresponding average power delivered by the source. [Ans. 6––53.1° A, 72––53.1° V, 96–36.9° V; 8.49 sin (314t – 53.1°) A, 102 sin (314t – 53.1°) V, 136 sin (314t + 36.9°) V, 432 W] I

12 W

+ VR – V = 120–0° V

j16 W + VL –

Fig. C-52 C-71. Three circuit-elements in series draw a current of 10 sin (400t + 70°) A in response to an applied voltage of 50 sin (400t + 15°) V. If one element is a

Fig. C-53 C-74. Find the current and unknown voltages in the

Supplementary Exercise—Part C : AC Circuits circuit of Fig. C-54. [Ans. 36.1sin (4000t + 11.9°) mA, 130 sin (4000t + 102°) V, 225sin (4000t – 78.1°) V] i

12 W + vR –

j16 W + vL –

v = 140 sin (4000t – 10°) V

+

vC

0.04 mF

–

Fig. C-54 C-75. A circuit consisting of a variable resistor in series with a capacitance of 80 mF is connected across a 120-V, 50-Hz supply, Calculate the value of the resistance so that the power absorbed is 100 W. [Ans. 132 W or 12 W] C-76. An adjustable resistor R in series with a capacitance of 25 mF draws a current of 0.8 A, when connected across a 50-Hz supply. Calculate (a) the value of resistor so that the voltage across the capacitor is half of the supply voltage, (b) the power, and (c) the power factor. [Ans. (a) 220.7 W; (b) 141.3 W; (c) 0.866 (leading)] C-77. A coil of 0.8 pf is connected in series with a 100-mF condenser. The frequency of supply is 50 Hz. The potential drop across the coil is found to be equal to the potential drop across the capacitor. Calculate the resistance and the inductance of the coil. [Ans. 25.46 W, 60.8 mH] C-78. The voltage and current for a series RLC circuit are v = 141.4 sin (314t + 45°) V i = 28.28 sin (314t – 15°) A Find (a) the rms values of voltage and current, (b) the power factor and the power consumption of the circuit, (c) the time period of the ac supply, and (d) the resistance of the circuit. [Ans. (a) 100 V, 20 A; (b) 0.5 (lagging), 1000 W; (c) 0.02 s; (d) 2.5 W] C-79. The voltage v = 141.42 sin (157.08t + p/12) V is applied to an ac circuit, and an ac ammeter, a wattmeter and a power-factor meter are connected to measure the respective quantities. The reading of the ammeter is 5 A and that of the power factor

761

meter is 0.5 lagging. Find (a) the expression for the instantaneous value of the current, (b) the wattmeter reading, (c) the impedance of the circuit in rectangular form. [Ans. (a) i = 7.07sin (157.08t – p/4) A; (b) 250 W; (c) (10 + j17.32) W] C-80. When an alternating voltage (80 + j60) V is applied to a circuit, the resulting current is (– 4 + j10) A. Find (a) the impedance of the circuit, (b) the phase angle, (c) the power consumed, and (d) the apparent power. [Ans. (a) (2.414 – j8.966) W; (b) 74.93°; (c) 280 W; (d) 1077 VA] C-81. When an iron-core coil is connected to a 12-V dc supply, it draws a current of 2.5 A, and when it is connected to a 230-V, 50-Hz supply, it draws 2 A current and consumes 50 W power. Determine for this value of current, (a) the power loss in the iron core, (b) the inductance of the coil, (c) the power factor, (d) the value of resistance which is equivalent to the effect of iron loss. [Ans. (a) 20 W; (b) 0.3657 H; (c) 0.1087 (lagging); (d) 7.7 W] C-82. (a) Find the equation for the instantaneous current, when a voltage represented by v = 141.4 sin 314t V is applied to a circuit consisting of R = 50 W and L = 0.2 H. (b) Calculate the value of the capacitance to be connected in series with this circuit to obtain minimum impedance of the circuit. (c) Find the power drawn by the circuit under this condition. [Ans. (a) i = 1.762 sin (314t – 51.47°) A; (b) 50.7 mF; (c) 200 W] C-83. An ac load draws 2 A at a power factor of 0.8 (lagging) from a 230-V, 50-Hz mains. If a 15-mF capacitor is connected across the load terminals, what will be the net current supplied by the mains and its power factor ? [Ans. 1.604 A, 0.9974 (lagging)] C-84. Two impedances, one inductive and the other capacitive, are connected in series. When a voltage of 120–30° V and frequency of 50 Hz is impressed across the combination, the current is 3––15°A. If one of the impedances is (10 + j48.3) W L and C in the impedances. [Ans. Z2 = (18.28 – j20.02) W; 0.1537 H, 159 mF]

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Basic Electrical Engineering

C-85. In the circuit shown in Fig. C-55, (a current I, and (b V1. [Ans. (a) 2.32––28.6°A; (b) 77.3–5° V]

I

–j20 W

j50 W

200–30° V

40 W

+ V –

20 W

Fig. C-58 C-91. Use current division twice to I for the circuit shown in Fig. C-59. [Ans. 1.41––19.5° A]

Fig. C-55

j3 W

C-86. A 0.5-W resistor is in parallel with a 10-mH inductor. At what radian frequency, do the circuit voltage and current have a phase angle difference of 40° ? [Ans. 59.6 rad/s] C-87. The current source i = 4 2 sin (400t – 10°) A in current iL in the inductor for the circuit. [Ans. 5.18 sin (400t – 57.1°) A]

8–40° A

10 W

20 mH

2W

3W

ZL = 4–30° W

Fig. C-59 C-92. Find the admittance Y of the circuit shown in Fig. C-60. [Ans. 2.29–– 42.2° S]

iL i

I

–j5 S

4S

j5 S

2S

80 mF Y

1S

–j1 S

–j2 S

Fig. C-56 C-88.

IR for the circuit shown in Fig. C-57. [Ans. 6.85––7° A] I

j3 W IR

20–45° A

2W

–j5 W

4W

Fig. C-57 C-89. A 2-mH coil with a 10-W winding resistance is in parallel with a 10-mF capacitor. Find its equivalent circuit at 8 krad/s, which has two circuit-elements in series. [Ans. R = 13.9 W, C = 7.2 mF] C-90. V for the circuit shown in Fig. C-58. [Ans. 81.22–6.03° V]

Fig. C-60 C-93.

I, VR and VC, and the corresponding sinusoidal quantities if delivered by the source. [Ans. I = 7.5–81.3° A; VR = 150–81.3° V; VC = 187–– 8.66° V; i = 10.6 sin (314t + 81.3°) A; vR = 212 sin (314t + 81.3°) V; vC = 265 sin (314t – 8.66°) V; 1125.4 W] I

20 W

–j25 W

+ VR –

+ VC –

240–30° V

Fig. C-61

Supplementary Exercise—Part C : AC Circuits C-94. Find the input admittance Yab of the network shown in Fig. C-62 and draw it as a parallel combination of a resistance R and an inductance L, giving values for R and L if w = 1 rad/s. [Ans. (0.5 – j0.5) S, (2 W) || (2 H)]

3W

763

j2 W a

–j4 W

3–60° A

4W

b

Fig. C-65 C-98. Find ZN, and IN circuit shown in Fig. C-66. [Ans. 7.92––16.48° W, 2.78–4.81° A]

Fig. C-62 C-95.

4W

equivalent across a and b for the circuit given in Fig. C-63. [Ans. 12.3––19.3° V, 3.07–15.7° W]

6––40° V a

4–30° A

5W

–j8 W

a b j3 W 4––35° A

Fig. C-66

6W

C-99. Determine the power factor of a fully loaded 5W b

Fig. C-63 C-96. Use superposition power absorbed by the 5-W resistor in the circuit shown in Fig. C-64. [Ans. 2.36 W]

Fig. C-64 C-97. Find ZTh, VTh, and IN ivalents of the circuit shown in Fig. C-65. [Ans. 1.35–10.9° W, 4.05–70.9° V, 3–60° A]

[Ans. 0.694 (lagging)] C-100. A resistor in parallel with a capacitor absorbs 20 W when the combination is connected to a 240-V, 50-Hz supply. If the power factor is 0.7 leading, what are the resistance and capacitance ? [Ans. 2.88 kW, 1.127 mF] C-101. A resistor in series with a capacitor absorbs 10 W when the combination is connected to a 120-V, 400-Hz supply. If the power factor is 0.6 leading, what are the resistance and capacitance ? [Ans. 518 W, 0.576 mF] C-102. An induction motor delivers 50 hp while operating factor is 0.6, what current does the motor draw ? If the power factor is 0.9, instead, what current does this motor draw ? [Ans. 176.5 A, 117.7 A] C-103. What resistor and capacitor in parallel present the same load to a 440-V, 50-Hz supply as a fully loaded 20-hp synchronous motor that operates at [Ans. 9.73 W, 245.2 mF]

764

Basic Electrical Engineering

C-104. A 20-mF capacitor and a parallel 200-W resistor draw 4 A at 50 Hz. Find the power components. [Ans. 1.99 kVA, 1.24 kW, –1.56 kVAR] C-105. A series RLC circuit is connected to 250-V, 50-Hz supply. It is found that the current in the circuit is unaltered when the capacitance is short-circuited. If R = 40 W and the current is 5 A, determine L and C. [Ans. 95.5 mH, 53 mF] C-106. A 230-V, 50-Hz voltage is applied across a coil of L = 0.5 H and R = 200 W in series with a capacitor. What value must the capacitor have in order that the total voltage across the coil shall be 250-V ? [Ans. 89.66 mF] C-107. The series combination of an inductive impedance Za consisting of R = 6 W and L = 25.5 mH and an unknown capacitive impedance Zb is connected across 240-V, 50-Hz supply, as shown in Fig. C-67. It is found that the voltage Va across Za is three time the voltage Vb across Zb, and are in quadrature. Determine the values of R and C. [Ans. 2.667 W, 1.592 mF]

Fig. C-67 C-108. An inductor coil, when connected to a 250-V, 50-Hz supply, absorbs 160 W taking a current of 2 A. (a) Determine the resistance, the inductance and the power factor of the coil. (b) If a 50-mF is connected in series with the coil, how much current will the circuit draw from the same supply ? [Ans. (a) 40 W, 0.377 H, 0.32 (lagging); (b) 3.687 A] C-109. Two circuits have the same numerical ohmic impedance, but the power factor of one circuit is 0.8 and the other 0.6, both lagging. Calculate the power factor of the circuit obtained by joining these two circuits in parallel. [Ans. 0.707 (lagging)] C-110. An ac circuit connected across 200-V, 50-Hz supply has two parallel branches A and B. The current in branch A is 4 A at 0.8 lagging power factor, and the total current is 5 A at unity power factor. Find

for the branch B, the power consumed and the impedance. [Ans. 360 W, (40 – j53.34) W] C-111. Two impedances Z1 = (10 + j15) W Z2 = (6 – j8) W are connected in parallel. The total current supplied is 15 A. What is the power taken by each branch ? [Ans. 738 W, 1438 W] C-112. When two impedances 20–– 45° W 30–30° W are connected in series across a certain ac supply, the resulting current is 10 A. If the supply voltage remains unaltered, calculate the supply current when the two impedances are connected in parallel. [Ans. 26.84–17.45° A] C-113. Z1 = (2 + j4) W and the second branch has Z2 = (6 + j0) W. To what value must the 6-W resistor be changed to result in the overall power factor of 0.9 lagging ? [Ans. 0.8 (lagging), 3.195 W] C-114. Three impedances Z1, Z2 and Z3 connected in parallel have an equivalent impedance of (6 + j8) W. A sinusoidal voltage Vm sin (wt ± q°) V is applied to it. The currents drawn by these impedances are i1 = 20 sin wt A, i2 = 40 sin (wt + 30°) A, and i3 = im sin (wt ± q°) A, respectively. The total current drawn is i = 25 sin (wt ± 30°) A. Determine (a) the current i3, (b) expression for the supply voltage, (c) the total power drawn from the supply, and (d) the impedance Z1. [Ans. (a) 33.83sin (wt – 167.2°) A; (b) 250sin (wt + 83.13°) V; (c) 1875 W; (d) (1.49 + j12.4) W] C-115. a) the current delivered by the source, and (b) the power factor. [Ans. (a) 6.935–56.31° A; (b) 0.5547 (leading)] 4W

100–0° V

8W

8W

Fig. C-68

8W

Supplementary Exercise—Part C : AC Circuits C-116. For the circuit shown in Fig. C-69 determine (a) the impedance, (b) the current I1, (c) the current I2, (d) the current I3, (e) the voltage VBC, (f) the power consumed by the whole circuit. [Ans. (a) 632.4–80.9° W; (b) 0.393––80.9° A; (c) 0.776–– 80.9° A; (d) 0.383–99.1° A; (e) 121.9–9.1° V; ( f ) 15.54 W] 100 W

A

1H

765

C-117. A balanced star-connected load of (8 + j6) W per phase is connected to a three-phase, 230-V supply. Find the line current, pf, active, reactive and total volt-amperes. [Ans. 13.28––36.87° A; 0.8 (lagging); 4232 W; 3174 VAR; 5290 VA]

B

I1 250–0° V 50 Hz

I2 I3

0.5 H

D

10 mF

C

Fig. C-69 ( C ) C-118.

C HA L L E NGI NG

PROBLEMS

vS across the current source, if current iS is given as iS = 0.234sin (3000t – 10°) A. [Ans. vS = 95.2 sin (3000t + 38.4°) V]

C-120. Use nodal analysis for the circuit of Fig. C-72 to V1 and V2. [Ans. 1.062–23.3° V, 1.593––50.0° V]

Fig. C-70 C-119. If the independent voltage source in the circuit of Fig. C-71 is v = 10 sin 1000t V, obtain the expressions for the time-domain currents i1 and i2. [Ans. i1 = 1.24 sin (1000t + 29.7°) A, i2 = 2.77sin (1000t + 56.3°) A]

Fig. C-72 C-121. A choke coil is connected to 240-V, variablefrequency ac supply, as shown in Fig. C-73. When the frequency of the supply is 50 Hz, an ammeter Choke coil

v

3W

500 mF

i1

i2 4 mH

R

L Ammeter

2i1

240 V

Fig. C-71

Fig. C-73

766

Basic Electrical Engineering

connected in series with the choke reads 60 A. ammeter reads 40 A. Calculate the resistance and the inductance of the choke. [Ans. 3.055 W, 8.218 mH] C-122. Two impedances ZA and ZB are connected in series across a 240-V, 50-Hz ac supply. The total current drawn is 3 A. The impedance ZA has 0.8 lagging power factor and the voltage across it is twice that across ZB and in quadrature to it. Analytically determine (a) the value of ZA in complex form and the power consumed in ZA, (b) the power consumed by ZB and its pf. [Ans. (a) (57.25 + j42.94) W, 515.3 W; (b) 193.1 W; 0.6 (leading)] C-123. An ohmic resistance is connected in series with an unknown choke coil across a 230-V, 50-Hz supply. The current drawn by the circuit is 1.5 A, and the voltages across the resistance and the choke coil are 90 V and 180 V, respectively. Calculate the resistance and the inductance of the choke coil, and the phase difference between the current and the supply voltage. [Ans. 45.93 W, 0.353 H, 46.31°] C-124. When a voltage of 100 V, 50 Hz is applied to a coil A, the current taken is 8 A and the power consumed is 120 W. An impedance ZB, whose power factor is 0.8 leading at 50 Hz, is connected in series with the coil A, and the combination is connected across a variable-frequency source of 200 V. The current taken from the supply is maximum, when the frequency is 100 Hz. Find the components of ZB and the value of maximum current taken by the circuit. [Ans. RB = 16.48 W, CB = 64.38 mF, Imax = 10.9 A] C-125. A 100-W resistance is connected in series with a choke coil. When a 440-V, 50-Hz, single-phase ac voltage is applied to this combination, the voltage across the resistance and the choke coil are 200 V and 300 V, respectively. Find the power consumed by the choke coil. Sketch a neat phasor diagram, indicating the current and all voltages. [Ans. 318 W] C-126. A sinusoidal ac voltage of 200 V is applied to a series circuit consisting of a resistor, a capacitor and an inductor choke. The voltages across the components are 170 V, 100 V and 150 V,

respectively, and the current drawn by the circuit is 4 A. Determine the power factor of the inductor choke and of the circuit. [Ans. 0.161 (lagging), 0.97 (lagging)] C-127. Find the input admittance at 50 krad/s of the circuit shown in Fig. C-74. [Ans. 2.83––135° S] i

2i

20 mH

0.5 W

20 mF

Fig. C-74 C-128. Find the input admittance at 1 krad/s of the circuit shown in Fig. C-75. [Ans. 4 S] I

0.25 W

3I

1H

1 mF

Fig. C-75 C-129. A certain industrial load has an impedance of 0.6–30° W at a frequency of 50 Hz. What capacitor connected in parallel with this load causes the angle of the total impedance to decrease to 15° ? Also, if the load voltage is 230 V, what is the decrease in the line current produced by adding the capacitor ? [Ans. 1.42 mF, 39.8 A] C-130. Determine the current i through the 4-W resistor in the circuit shown in Fig. C-76. [Ans. i = 0.1755 sin (2t – 20.56°) + 0.547 sin (5t – 43.15°) A] 3H 3 sin 2t V

1H i

4W

Fig. C-76

4 sin 5t V

Supplementary Exercise—Part C : AC Circuits 20–15° A

C-131. frequency at which (a) Rin =550 W, and (b) Xin = 50 W. [Ans. (a) 100 rad/s; (b) 100 rad/s] 500 W

0.25 W

V1

1H

1

V2 2

–j0.2 W

30–40° A

0.2 W

0.4 W

15–20° A

100 W

Zin

Fig. C-80

Fig. C-77

–j14 W

8W

j8 W

3W

C-132. Determine the mesh currents I1 and I2 in the circuit shown in Fig. C-78. [Ans. 0.974–41.5° A, – 0.63–– 48.2° A]

4W

6W

240–30° V

j2 W

4W 10––40° V

767

1

I1

I1

2

–j8 W

j10 W

12–10° V

R and L so that maximum power is absorbed by the

C-133. Find the node voltages in the circuit shown in Fig. C-79. [Ans. –3.59––5° V, –12––15° V]

20–10° A

0.25 W 2 j0.5 W

[Ans. R = 0 W, L = 3.9 mH, 207.5 W]

this power.

L

R

15––30° A

1

Fig. C-81 C-136.

Fig. C-78

V1

ZL

V2

6

45 cos (10 t + 30°) V

0.1 mF

8W

12––15° V

Fig. C-82 C-137. A fully loaded 10-hp induction motor operates from

Fig. C-79 C-134. Solve the node voltages in the circuit shown in Fig. C-80. [Ans. 5.13–47.3° V, 8.18–15.7° V] C-135. What load impedance ZL in the circuit shown in Fig. C-81 absorbs maximum average power and what is this power ? [Ans. 8.46–2.81° W, 1.67 kW]

a 0.8 lagging power factor. Find the overall power factor when a 33.3 mF capacitor is connected in parallel with the motor. [Ans. 0.8873] C-138. In the series circuit shown in Fig. C-83, the voltage and the current are as indicated. Find the values of R, r, L and the frequency of the applied voltage and its magnitude. [Ans. 7.143 W, 1.57 W, 7.28 mH, 247.5 Hz, 30.97 V]

768

Basic Electrical Engineering (b) the power factor of the circuit, and (c) the total power taken from the source. Draw the phasor diagram. [Ans. (a) 1.27–– 60.17° A, 1.44–90° A, 0.718–28.47° A; (b) 0.879; (c) 145.2 W] 90 W

Fig. C-83 C-139. A series resonant circuit has an impedance of 500 W at resonant frequency and its cutoff frequencies are 10 kHz and 100 kHz. Determine the resonant frequency, the quality factor at resonant frequency, and the values of R, L and C. [Ans. 31.62 kHz, 0.35, 500 W, 0.884 mH, 28.76 nF] C-140. Two circuits, with the impedances Z1 = (12 + j15) W and Z1 = (8 – j4) W, are connected in parallel. If the potential difference across the impedances is V = (230 + j0) V, calculate (a) the total current and the currents supplied to each branch, (b) the total power and power consumed by each branch, (c) the overall power factor and power factor of each branch. [Ans. (a) 30.56–4.04° A, 12––51.34° A, 25.7–26.56° A; (b) 7014 W, 1728 W; 5284 W; (c) 0.99 75 (leading), 0.6247 (lagging), 0.9021(leading)] C-141. Two circuits A and B are connected in parallel across 200-V, 50-Hz power mains, as shown in Fig. C-84. Calculate (a) the source current, (b) the current in each branch, and (c) the power factor. Draw the phasor diagram. [Ans. (a) 3.27––53.75° A; (b) 5.18––75° A, 2.44–75.9° A; (c) 0.59 (lagging)] I1

10 W

A 0.12 H

I2

20 W

40 mF

I

B +

200 V, 50 Hz

–

Fig. C-84 C-142. For the circuit shown in Fig. C-85, calculate (a) the current in each branch and the total rms current,

0.5 H 20 mF

+

230 V, 50 Hz

–

Fig. C-85 C-143. Two impedances Z1 and Z2 are connected in parallel across 200-V, 50-Hz, single-phase ac supply. The impedance Z1 carries a current of 2 A at 0.8 leading power factor, and Z2 consumes 668 W. If the total current is 5 A at 0.985 lagging power factor, determine (a) the values of Z1 and Y1, (b) the current through Z2 and its phase angle, and (c) the power consumed by Z1. [Ans. (a) 100–– 36.87° W, 0.01–36.87° S; (b) 3.92––31.88° A; (c) 320 W] C-144. (a) The load connected to a three-phase supply comprises three similar coils connected in star. The line currents are 25 A, and the kVA and kW inputs are 20 and 11, respectively. Find the line and the phase voltages, the resistance and the reactance of each coil. (b) If the coils are now connected in delta to the same three-phase supply, what would be line currents and the power taken ? [Ans. (a) 462 V, 267 V, 5.87 W, 8.92 W; (b) 75 A, 33 kW] C-145. (a) A balanced delta-connected load consumes 2 kW of power, when connected to 400-V, 50Hz, 3-phase supply. The same load takes 2 A at a lagging power factor when connected to a 230-V, 50-Hz, 3-phase supply. Determine the resistance and the inductance per phase and load power factor. (b) If the load is now connected in star to the same [Ans. (a) 166 W, 0.355 H; (b) 664.2 W]

Supplementary Exercise—Part D : Electrical Machines

S U P P L E M E N TA R Y E X E R C I S E S D.1 Solved Problems D.2 Practice Problems

PA R T

D : ELECTRICAL

769

D

MACHINES

Assemblage of

N OT E This set of exercises provides practice to those who wish to attain a higher standard of learning the basic principles of Electrical Engineering. The real key to success is practice.

770

Basic Electrical Engineering

Supplementary Exercise—Part D : Electrical Machines

771

D. 1. SOLVED PROBLEMS P RO B L EM

D -1

A single-phase transformer has 1000 turns on the primary and 200 turns on the secondary. The no-load current is 3 A at a pf of 0.2 lagging. Calculate the primary current and the power factor when the secondary current is 280 A at a pf of 0.8 lagging. –1

Solution The no-load phase angle, f0 = cos 0.2 = 78.46° (lagging). Therefore, taking the primary voltage as the reference, the no-load current is given as I0 = 3––78.46° A = (0.6 – j2.94) A When load is connected across the secondary, the secondary current is 280 A at a pf of 0.8 lagging. The phase angle of the secondary current, –1 f2 = cos 0.8 = 36.87° (lagging)

The corresponding primary balancing current is I¢1 = {I2 (N2/N1)}–f2 = {280(200/1000)}––36.87° = 56––36.87° A = (44.8 – j33.6) A Hence, the total primary current, I1 = I0 + I¢1 = (0.6 – j2.94) + (44.8 – j33.6) = (45.4 – j36.54) A = 58.28––38.83° A The power factor, pf = cos 38.83° = 0.779 (lagging) P R O B L EM

D - 2

A single-phase, 100-kVA, 2000-V/200-V, 50-Hz transformer has impedance drop of 10 %, and resistance drop of 5 %. Calculate (a) the regulation at full-load 0.8 pf lagging, and (b) the value of the pf at which the regulation is maximum.

Solution Given: \

cos f = 0.8, sin f = the % impedance drop = I2 Ze2 =

Again, given the % resistance drop = \

I2 Re2 =

1 − cos2 φ = 0.6.

I2 Ze2 ¥ 100 = 10 E2 10E2 10 200 = = 20 V 100 100 I2 Re 2 ¥ 100 = 5 E2 5E2 5 200 = = 10 V 100 100

Therefore, the reactance drop is given as I2 Xe2 =

( I2 Ze 2 )2 ( I2 Re 2 )2 =

20 2 10 2 = 17.32 V

772

Basic Electrical Engineering

(a) The percentage voltage regulation is given as (10 × 0.8 + 17.32 × 0.6) ( I 2Re 2 cos f + I 2 X e 2 sin f ) ¥ 100 = ¥ 100 = 9.2 % 200 E2 (b) For regulation to be zero, the power factor must be leading, and we must have VR =

\ Therefore, P R O B LE M

I R 10 ( I 2Re 2 cos f - I 2 X e 2 sin f ) ¥ 100 = 0 fi tan f = 2 e 2 = = 0.5774 I2 Xe 2 17.32 E2 –1 f = tan 0.5774 = 30° the power factor = cos f = cos 30° = 0.866 (leading) D -3

Solution Let Po

Pi be the iron loss and Pc be the full-load copper Po Po = Pin Po + Pi + Pc Po = (kVA)(pf ) = 50 000 ¥ h=

40 000 40 000 + Pi + Pc

⎧ 1 ⎫ − 1⎬ fi Pi + Pc = 40 000 ⎨ ⎩ 0.98 ⎭

In the second case, the output power, Po = (1/4)(kVA)(pf ) = 0.25 ¥ 50 000 ¥ The copper loss, P¢c = (1/4)2Pc Pc

12500 12500 + Pi + 0.0625Pc Solving the above two equations, we get

fi Pi

Pi = 372.1 W P R O B LE M

and

⎧ 1 ⎫ Pc = 12 500 ⎨ − 1⎬ ⎩ 0.969 ⎭ Pc = 444.2 W

D -4

a

b

c) the load in kVA at which the maximum

Solution (a

pf, Po = (kVA)( pf ) = 40 000 ¥ h=

Po 32 000 = = 96.1 % Po + Pi + Pc 32 000 + 450 + 850 x times the full load. Since, the copper loss must be equal to the iron loss at

(b Pc = Pi or

x2 ¥

fi

x=

450/850

Supplementary Exercise—Part D : Electrical Machines ¥ 40 000 ¥

Po hmax =

\

773

Po 23280 = = 94.71 % Po + Pi + Pc 23280 + 450 + 450

(c) The load in kVA = (kVA)(x) = 40 ¥ P RO B L EM

29.1 kVA

D -5

Solution 50 000 × 0.8 50 000 × 0.8 + Pi + Pc fi

Pi + Pc = 40 000

⎛ 1 ⎞ −1 ⎝ 0.98 ⎠

i) 2

.

50 000 × (1/ 2) × 0.8 50 000 × (1/ 2) × 0.8 + Pi + (1/ 2)2 Pc fi

Pi + (0.25)Pc = 20 000

⎛ 1 ⎞ −1 ⎝ 0.97 ⎠

ii)

Solving the above two equations, we get Pc = 263.69 W

and Pi = 552.64 W x times the full load. At this load, the copper loss becomes x2 times the

That is, x2

fi x=

(552.64)/(263.69) = 1.4477

= x(kVA) = 1.4477 ¥ 50 = 72.385 kVA

hmax = CO MM E N T S

72385 × 1 = 98.496 % 72385 × 1 + 552.64 + 552.64

774

Basic Electrical Engineering

P RO B L EM

D -6

The open-circuit and short-circuit tests on a 5-kVA, 200-V/400-V, 50-Hz, 1-phase transformer gives the following readings : OC Test : SC Test :

200 V 15 V

1A 10 A

100 W 85 W

(with secondary open-circuited) (with primary short-circuited)

(a) Draw the equivalent circuit referred to the primary. (b) Calculate the approximate regulation of the transformer at full load with (i) pf 0.8 lagging, and (ii) pf 0.8 leading.

Solution (a) In an open-circuit test, with secondary open-circuited and rated voltage applied to the primary, the power drawn meets the iron loss of the core. Phase angle of the no-load current of 1 A is given by cos f0 =

Poc 100 = = 0.5 fi sin f0 = Voc I0 200 1

1 (0.5)2 = 0.866

The resistance R0 representing the iron-loss and the magnetising reactance X0 are given by R0 =

Voc I0 cos

=

200 = 400 W; 1 0.5

X0 =

and

Voc I0 sin

=

200 = 231 W 1 0.866

In a short-circuit test, we are short-circuiting the primary and applying 15 V to the secondary to get a current of 10 A and power of 85 W. Thus, the equivalent resistance and leakage reactance of the two windings as referred to the secondary are given by P 85 V 15 Re2 = sc 2 = 2 = 0.85 W; Ze2 = sc = = 1.5 W; Isc 10 10 ( Isc) fi

Xe2 =

Ze22

Re22 =

1.52

0.852 = 1.236 W

We can now calculate the equivalent resistance and leakage reactance of the two windings as referred to the primary as follows. N E 400 =2 K = 2 = 2 = N1 E1 200 0.85 X 1.236 = 2 = 0.2125 W and Xe1 = e2 = 2 = 0.309 W K2 2 K2 2 Thus, the equivalent circuit of the transformer as referred to the primary is shown in Fig. D-1. \

Re1 =

Re2

Re1 = 0.2125 W Xe1 = 0.309 W

I1 + Iw 200 V

+

I0 R0 400 W

Im X0

L O A D

V¢2

231 W

–

–

Fig. D-1

Supplementary Exercise—Part D : Electrical Machines

775

(b) The full-load secondary current is given as I2(FL) =

kVA 5000 = = 12.5 A 400 E2

(i) The percentage regulation at full load with pf 0.8 lagging is given as 12.5 (0.85 × 0.8 + 1.236 × 0.6) = = 4.44 % E2 400 (ii) The percentage regulation at full load with pf 0.8 leading is given as % Regn =

I 2(FL) ( Re 2 cos f + X e 2 sin f )

% Regn = PROBLE M

I 2(FL) ( Re 2 cos f - X e 2 sin f ) E2

=

12.5 (0.85 × 0.8 + 1.236 × 0.6) = – 0.193 % 400

D -7

under : For 12 hours : For 6 hours : For 6 hours :

2 kW at a power factor of 0.5 lagging 12 kW at a power factor of 0.8 lagging 18 kW at a power factor of 0.9 lagging

Solution The input power, \

Po = (kVA)(pf ) = 15 ¥ 1 = 15 kW. Pin = Po + Pi + Pc = 15 + 2Pi (since, at hmax, we have Pc = Pi) 15 0.98 = fi Pi = Pc = 0.153 kW 15 + 2Pi

The total energy output for all-day, Wo = 2 ¥ 12 + 12 ¥ 6 + 18 ¥ 6 = 204 kW h For a transformer, the iron loss remains constant, but the copper loss is proportional to the square of the current. As the

kVA1 (for 2 kW load at 0.5 pf ) =

2 kW = 4 kVA 0.5

kVA2 (for 12 kW load at 0.8 pf ) =

12 kW = 15 kVA 0.8

kVA3 (for 18 kW load at 0.9 pf ) =

18 kW = 20 kVA 0.9

We can now calculate the copper losses in kW h for different durations : 2

⎛ 4 kVA ⎞ Wc1 = 0.153 ¥ ⎜ ⎟ ¥ 12 = 0.13056 kW h ⎝ 15 kVA ⎠ 2

⎛ 15 kVA ⎞ Wc2 = 0.153 ¥ ⎜ ⎟ ¥ 6 = 0.918 kW h ⎝ 15 kVA ⎠ 2

⎛ 20 kVA ⎞ Wc3 = 0.153 ¥ ⎜ ⎟ ¥ 6 = 1.632 kW h ⎝ 15 kVA ⎠

776

Basic Electrical Engineering

Thus, the total copper losses for all-day, Wc = (0.13056 + 0.918 + 1.632) kW h = 2.68 kW h the total iron loss for all-day, Wi = 0.153 ¥ 24 = 3.672 kW h Therefore, the total energy input for all-day, Win = Wo + Wi + Wc = 204 + 2.68 + 3.672 = 210.35 kW h hall-day = P ROBLEM

Wo 204 kWh = = 0.9698 = 96.98 % Win 210.35 kWh

D -8

A single-phase, 50-Hz, step-up transformer, having turns-ratio N2/N1 = 2, has no-load losses 38.4 W and no-load power factor 0.22 lag, when the OC test was carried out at LV side of the transformer. The transformer has 10 A as the secondary rated (full load) current. When the transformer is loaded to its full-load current, it is found that (i) the secondary terminal-voltage drops by 12.5 V from its no-load value at 0.8 pf (lag), and the voltage regulation (approx.) at this load is 2.5 %, and (ii) the secondary terminal-voltage drops by 6 V at unity power factor. (a) Determine the primary and the secondary no-load voltages and the rating of the transformer. (b) If the SC test is carried out at 50 % of its full-load current by short-circuiting LV side, what will be the voltmeter, ammeter and wattmeter readings ? (c pf load. (d) Find the no-load current, the magnetising component, and iron-loss component, if the OC test is carried out on LV side. (e) Draw the equivalent circuit of the transformer referred to the primary.

Solution (a) The percentage voltage regulation is given as % VR = \

E2 V2 ¥ 100 E2

fi

2.5 =

12.5 ¥ 100 E2

fi

E2 = 500 V

⎛N ⎞ ⎛ 1⎞ E1 = E2 ⎜ 1 ⎟ = 500 ¥ = 250 V ⎝ 2⎠ ⎝ N2 ⎠

The rating of the transformer = E2 I2(FL) = 500 ¥ 10 = 5000 VA = 5 kVA (b) On loading the transformer, the secondary voltage drop is given as \ or and fi

Voltage drop 12.5 12.5 6 Re2

= I2 (Re2 cos f + Xe2 sin f) = 10(Re2 ¥ 0.8 + Xe2 ¥ 0.6) = 8Re2 + 6Xe2 = 10(Re2 ¥ 1.0 + Xe2 ¥ 0) = 0.6 W

(For 0.8 lag pf ) (i) (For unity pf )

Substituting the value of Re2 into Eq. (i), we get Xe2 = 1.28 W When SC test on LV side is carried out at 50 % full-load, the ammeter reading = 50 % of 10 A = 5 A the voltmeter reading = Isc Ze2 = 5 ¥

0.62 + 1.282 = 7.06 V

2 2 the wattmeter reading = I sc Re2 = 5 ¥ 0.6 = 15 W

Supplementary Exercise—Part D : Electrical Machines

777

(c) The iron loss, Pi equals the iron loss. Hence, Pc = Pi = 38.4 W. 2 Since, Pc = I 2 Re2 Pc = Re2

I2 =

hmax =

38.4 =8A 0.6

Po 500 × 8 × 0.8 = = 97.66 % + + × Po Pi Pc 500 8 × 0.8 + 38.4 + 38.4

(d) When OC test is carried out on LV side, E1 = V1 = 250 V; Pi = 38.4 W; cos f 0 = 0.22 (lag); sin f0 =

and \

I0 =

The no-load current,

Poc cos V1

1 (0.22)2 = 0.9755. = 0

38.4 = 0.698 A 250 0.22

The iron-loss component, Iw = I0 cos f0 = 0.698 ¥ 0.22 = 0.1536 A The magnetising component, Im = I0 sin f0 = 0.698 ¥ 0.9755 = 0.68 A V 250 V 250 (e) R01 = 1 = = 1628 W; X01 = 1 = = 368 W 0.1536 0.68 Iw Im Re1 = Re2 (N1/N2)2 = 0.6 ¥ (1/2)2 = 0.15 W; Xel = Xe2 (N1/N2)2 = 1.28 ¥ (1/2)2 = 0.32 W Thus, the equivalent circuit is shown in Fig. D-2. Re1 = 0.15 W

I1 + Iw 250 V

+

I0 R01 1628 W

Xe1 = 0.32 W

Im X01

V¢2

368 W

–

L O A D

–

Fig. D-2 P RO B L EM

D -9

A 5-kVA, 200-V/400-V, 50-Hz, single-phase transformer has a leakage impedance of (0.12 + j0.32) W as referred to the low voltage side. Calculate the per unit values of resistance, reactance and impedance taking the rated quantities as the base values. Check that the per unit values are the same on the two sides.

Solution Base volt-ampere, VAbase = 5000 VA; Zlv = (0.12 + j0.32) W. For low voltage side : VA 5000 V 200 Vbase = 200 V; Ibase = = = 25 A; Zbase = = = 8 W; V 200 I 25 Therefore, 0.12 R Per unit resistance, Rpu = lv = = 0.015 pu Z base 8

778

Basic Electrical Engineering

Per unit reactance,

Xpu =

X lv 0.32 = Z base 8

= 0.04 pu

Zlv (0.12 + j 0.32) Ω Per unit impedance, Zpu = = = (0.015 + j0.04) pu Z base 8Ω For high voltage side : VA 5000 V 400 = = 12.5 A; Zbase = = = 32 W; Vbase = 400 V; Ibase = V 400 I 12.5 2 ⎛ 400 ⎞ = (0.48 + j1.28) W Zhv = (0.12 + j0.32) ¥ ⎝ 200 ⎠ Therefore, 0.48 R = 0.015 pu Per unit resistance, Rpu = lv = Z base 32

Per unit reactance,

Xpu =

Xhv 1.28 = Zbase 32

= 0.04 pu

Z hv (0.48 + j1.28) Ω = = (0.015 + j0.04) pu Zbase 32 Ω Note that the per unit resistance, reactance and impedance are the same on the two sides. Per unit impedance, Zpu =

P ROBLEM

D - 10

Determine the phase emf induced in a 4-pole, three-phase, 50-Hz star-connected alternator with 36 slots and

Solution Number of conductors per phase, Zph = (36 ¥ 30)/3 = 360 Number of turns per phase, T = Zph/2 = 360/2 = 180 Number of slots per pole = 36/4 = 9 Number of slots per pole per phase, q = 9/3 = 3 Slot angle, a = 180°/(36/4) = 20° The distribution factor for the winding is given as sin ( q / 2) sin (3 × 20° /2) kd = = = 0.9598 q sin ( / 2) 3 sin (20° /2) The emf induced per phase, Eph = 4.44kp kd fFT = 4.44 ¥ 0.95 ¥ 0.9598 ¥ 50 ¥ 0.05 ¥ 180 = 1821.8 V P R O B L EM

D - 11

Find the number of armature conductors in series per phase required for the armature of a three-phase, 50-Hz, 10-pole

Solution Number of slots per pole = 90/10 = 9 Number of slots per pole per phase, q = 9/3 = 3 Slot angle, a = 180°/9 = 20° \

kd =

sin ( q / 2) sin (3 × 20°/ 2) = = 0.9598 q sin ( / 2) 3 sin (20°/ 2)

Supplementary Exercise—Part D : Electrical Machines

779

Since nothing is mentioned about the pitch factor, we can take kp = 1. Now, the voltage per phase is given as E 11000 Eph = L = = 6351 V 3 3 However, the emf induced per phase is given as Eph = 4.44kp kd fFT \

T =

Eph 4.44kp kd f F

=

6351 4.44 1 0.9598

50

0.16

= 186.3

Thus, the number of conductors in series per phase, Zph = 2T = 2 ¥ 186.3 ª 373 P ROBLE M

D - 12

A three-phase, 20-pole, 50-Hz, salient-pole alternator with star-connected stator winding has 180 slots on the stator, Calculate (a) the speed, (b) the generated emf per phase, and (c) the line emf.

Solution (a) The speed, N =

120 50 120 f = = 300 rpm 20 P

(b) Number of slots per pole is 180/20 = 9. Number of slots per pole per phase, q = 9/3 = 3 Slot angle, a = 180°/9 = 20° \

kd = Total turns per phase,

T =

sin (qa /2) sin (3 ¥ 20∞/2) = = 0.9598 3 sin (20∞/2) q sin (a /2) Zph 2

=

(180

8)/ 3 2

= 240

The generated emf per phase, Eph = 4.44kp kd fFT = 4.44 ¥ 1 ¥ 0.9598 ¥ 50 ¥ 0.025 ¥ 240 = 1278.5 V (c) The line emf, EL = 3Eph = 3 ¥ 1278.5 = 2214.3 V P R O B L EM

D - 13

A 400-V, 10-kVA, 50-Hz, star-connected alternator has an effective armature resistance of 1.0 W. It generates an open a) the synchronous impedance, (b) the synchronous reactance. (c) If the alternator is supplying a load current of 15 A at 0.8 power factor lagging, to what value would the terminal voltage rise if the load is thrown off. (d) Calculate the percentage regulation at (i) 0.8 pf lagging, (ii) 0.8 pf leading, and (iii) unity pf.

Solution (a) Synchronous impedance, (b) Synchronous reactance, (c) Terminal voltage per phase,

Zs = Xs = V =

Voc I sc

= If same

Zs2

R2 =

90 V =6W 15 A 62 1.0 2 = 5.92 W

VL 400 = = 231 V 3 3

780

Basic Electrical Engineering –1 6 = tan R 1.0 Vz = IZs = 15 ¥ f = cos –1 = –1 Zs

q = tan

The power-factor angle, The angle between V and Vz, a = (q – f

a = 0.723

The open-circuit voltage or the terminal voltage when the load is thrown off,

V 2 + Vz2 + 2VVz cos a

E =

2312 + 90 2 + 2 × 231 × 90 × 0.723 = 302.5 V

=

Alternatively : The open-circuit voltage is given as (V cos f + IR ) 2 + (V sin f + IX s) 2

E = = (d) (i) At

(231 × 0.8 + 15 × 1)2 + (231 × 0.6 + 15 × 5.92)2 = 302.7 V ª 302.5 V

pf lagging : E V 302.5 231 ¥ 100 = ¥ 100 = 30.9 % V 231 pf leading : The angle between V and Vz, VR =

(ii) At

E

a = (q + f

a

The open-circuit voltage, E = =

V 2 + Vz2 + 2VVz cos a 2312 + 90 2 − 2 × 231 × 90 × 0.46 = 205.75 V

Therefore, percentage regulation is E V 205.75 231 ¥ 100 = ¥ 100 = –10.9 % V 231 –1 (iii) At unity pf : The power-factor angle, f = cos 1.0 = 0° The angle between V and Vz, a = (q + f a The open-circuit voltage, E = =

V 2 + Vz2 + 2VVz cos a 2312 + 90 2 + 2 × 231 × 90 × 0.164

Therefore, percentage regulation is VR = P RO B L EM

E

V V

¥ 100 =

261.3 231 ¥ 100 = 13.1 % 231

D -1 4 W per

801

Supplementary Exercise—Part D : Electrical Machines 0.75 W

W

D-19. A 4-pole, three-phase, star-connected alternator has pf

lagging. [Ans. D-14. A 40-kVA, 2000-V/250-V transformer has a W winding resistance of 0.0155 W. Calculate (a) the

when the alternator is driven at 1500 rpm. [Ans. D-20. alternator has 12 slots per pole and 4 conductors

(b) the total resistance drop on full load, and (c) the total power loss on full load. [Ans. (a) 0.0335 W; (b c D-15.

Ans. D-21. Calculate the speed and open-circuit line and phase

transformer has the following resistances and leakage reactances : R1 W; X1 = 3.2 W;

R2 W X2 = 0.03 W

Calculate the equivalent resistance and reactance

distributed.

[Ans.

D-22.

[Ans. 0.017 W W D-16. Open-circuit and short-circuit tests were conducted [Ans. D-23. A three-phase, star-connected alternator with 12 poles generates 1100 V on open circuit at a speed of

observations during these tests are : OC Test

Ans. SC Test

D-24. 5 ¥ 10 –2 pf

[Ans. [Ans.

D-25.

D-17. transformer on test gives the following results with force.

OC Test SC Test

[Ans.

D-26. of the generated emf. pf.

[Ans.

[Ans.

D-27.

D-18. winding transformer has a resistance of 0.01 W and W as referred to low-voltage side. Find (a) per unit resistance and reactance, (b

conductors are connected in series, calculate the generated line emf. [Ans. D-28.

[Ans. (a) 0.0125 pu, 0.045 pu; (b

782

Basic Electrical Engineering

Solution V =

(a) Applied voltage per phase,

VL 500 = = 288.68–0° V 3 3

Power output of the motor at full load, Po = 100 hp = 100 ¥ 746 W = 74 600 W = 74.6 kW P 74.6 Input power to the motor, Pin = o = = 83.82 kW 0.89 However, in terms of phase voltage and current, the input power is given as Pin = 3VI cos f. Hence, for the star-connected winding, the line current drawn from the supply, IL = I =

83.82 103 Pin = = 120.98 A 3 288.68 0.8 3V cos f

The phase angle, f = cos–1 0.8 = 36.87° (leading). Hence, the current drawn from the supply is given as 120.98–36.87° A. (b) The excitation voltage per phase, E = V – IZs = 288.68–0° – (120.98–36.87°)(0.06 + j0.6) = 332.34––10.83° V (c) The power angle, d R = 10.83° (d) The total mechanical power developed is equal to the power input less the copper losses. Thus, Pm = Pin – Pc = 83 820 – 3 ¥ (120.98)2 ¥ 0.06 = 81 185 W = 81.185 kW P ROBLEM

D - 1 7

A three-phase, 400-V, 10-kW, star-connected synchronous motor has a synchronous reactance of 4 W per phase and negligible resistance of the armature winding. The motor draws 8.5 kW at 0.5 pf lagging from the supply for a certain load. (a) Calculate (i) the armature current, and (ii) the power angle. (b) If the power developed by the motor remains the same and the excitation is increased by 50 %, calculate (i) the new power angle, (ii) the new armature current, and (iii) the new power factor.

Solution V =

(a) (i) Applied voltage per phase,

VL 400 = = 231–0° V 3 3

8500 = 2833.3 W 3 Pin 2833.3 = = 15.3 A I = 231 0.8 V cos

Input power to the motor per phase, Pin = \ Armature current per phase,

–1 Since, power factor is 0.8 lagging, f = cos 0.8 = 36.87°, and the armature current is

I = 15.3––36.87° A The excitation voltage per phase is given as E = V – IZs = 231–0° – (15.3––36.87°)(0 + j4) = 200.35––14.14° V (ii) The power angle, d R = 14.14°

Supplementary Exercise—Part D : Electrical Machines

783

(b) (i) Keeping the developed power the same, the excitation is increased by 50 %. Thus, the new excitation, E¢f = 1.5Ef = 1.5 ¥ 200.35 = 300.53 V. Since the power developed remains the same, we must have Ef sin d R = E¢f sin d¢R. Hence, E sin δ R 200.35 sin 14.14 sin d¢R = f = = 0.1628 Ef′ 300.53 –1

d¢R = sin 0.1628 = 9.37°

fi

(ii) The new armature current per phase is given as V − E′f 231∠ 0° − 300.35∠− 9.37° = = 20.4–53.19° A j4 Zs (iii) The new power factor, pf ¢ = cos f¢ = cos 53.19 = 0.599 (leading) I¢ =

P RO B L EM

D -1 8

A three-phase, 400-V, 50-Hz, 4-pole, 50-hp, star-connected induction motor is delivering rated output power while running at a slip of 4 per cent. The stator to rotor turns ratio is 1 : 0.6. Find (a) the synchronous speed, (b) the speed of c) the speed of the motor, (d) the slip in rpm, (e) the frequency of rotor induced emf, (f) the rotor induced emf at standstill, (g) the rotor induced emf while the motor is running on full load, and (h) the speed of i) the stator structure, (ii iii) the rotor structure.

Solution (a) The synchronous speed, Ns = (b (c) (d) (e) (f)

120 50 120 f = = 1500 rpm 4 P

1500 rpm The speed of the motor, N = (1 – s)Ns = (1 – 0.04) ¥ 1500 = 1440 rpm The slip in rpm, s = sNs = 0.04 ¥ 1500 = 60 rpm The frequency of rotor induced emf, fr = sf = 0.04 ¥ 50 = 2 Hz Assuming no voltage drop in the stator windings, the induced emf per phase in the stator is given as E1 =

VL 400 = = 231 V 3 3

T2 = 231 ¥ 0.6 = 138.6 V T1 (g) The rotor induced emf while the motor is running on full load, The rotor induced emf at standstill, E20 = E1 ¥

E2 = sE20 = 0.04 ¥ 138.6 = 5.54 V (h) (i structure. Hence, the speed = 1500 rpm. (ii zero. sNs rpm with respect to the rotor. Hence, the speed

(iii = sNs = 0.04 ¥ 1500 = 60 rpm P RO B L EM

D -1 9

A three-phase, 50-Hz, 500-V induction motor develops 20 hp at a slip of 5 %. The mechanical losses are 1 hp. Find

784

Basic Electrical Engineering

Solution The power output,

Po = 20 hp = 20 ¥ 746 W = 14 920 W The mechanical losses, Pm = 1 hp = 1 ¥ 746 W = 746 W Rotor gross output or power developed = Rotor shaft output + Mechanical losses Pd = Po + Pm = 14 920 + 746 = 15 666 W

Pd 15666 = = 16 491 W (1 s) 1 0.05 The input power to the stator, Pin = Rotor input power + Stator loss or Pin = 16 491 + 1000 = 17 491 W

Therefore, the rotor input power, Pg =

h= P ROBLEM

Po 14 920 = = 0.853 = 85.3 % Pin 17 491

D -20

A three-phase, 400-V, 50-Hz, 100-hp, star-connected induction motor has a star-connected slip-ring rotor with a transformation ratio T1/T2 of 2.5. The rotor has 0.02 W resistance and 0.6 mH inductance per phase. Assuming the stator losses to be negligible, calculate (a) At starting (s = 1) : (i) the rotor starting current per phase with slip-rings short-circuited when switched on normal voltage, (ii) the rotor power factor, and (b) At a slip of 3 % (s = 0.03): (i) the rotor current per phase, and (ii) the rotor power factor.

Solution As the stator winding is star-connected, the normal applied voltage per phase, V1 =

VL 400 = = 231 V 3 3

(a) At standstill (s = 1) : (i) Since the turns ratio T1/T2 = 2.5, the induced emf per phase in the rotor winding is given as T 1 E20 = V1 ¥ 2 = 231 ¥ = 92.4 V 2.5 T1 Rotor reactance per phase, X20 = 2psfL = 2p ¥ 1 ¥ 50 ¥ 0.6 ¥ 10 –3 = 0.1885 W Rotor impedance per phase,

Z20 =

Rotor current,

I20 =

(ii) The rotor power factor,

2 R22 + X20 =

(0.02)2 + (0.1885)2 = 0.1896 W

E20 92.4 = = 487 A 0.1896 Z20 R 0.02 = 0.105 pf = 2 = 0.1896 Z20

(b) At a slip of 3 % (s = 0.03) : (i) The voltage induced per phase in the rotor, Rotor reactance,

E2 = sE20 = 0.03 ¥ 92.4 = 2.77 V X2 = sX20 = 0.03 ¥ 0.1885 = 0.00566 W

Rotor impedance,

Z2 =

Rotor current,

I2 =

E2 2.77 = = 133 A 0.0208 Z2

(ii) Rotor power factor,

pf =

R2 0.02 = = 0.96 0.0208 Z2

R22 + X22 =

(0.02)2 + (0.00566)2 = 0.0208 W

Supplementary Exercise—Part D : Electrical Machines P RO B L EM

785

D -2 1

A three-phase, 4-pole, 400-V, 50-Hz, star-connected induction motor has the following values of various parameters with reference to its per-phase equivalent circuit as referred to the stator, shown in Fig. D-3. Stator impedance, Z1 = (0.15 + j0.4) W Rotor impedance referred to stator, Z2 = (0.16 + j0.4) W Resistance to account for core loss, Ri = 200 W Air-gap magnetising reactance, Xag = 20 W If the motor is running at a slip of 3 %, calculate (a) the stator current and power factor, (b) the rotor current, (c) the net mechanical output power if the friction and windage losses are 800 W, (d) the input power, and (e of the motor.

Fig. D-3 Solution (a) For s = 0.03, the effective scaled-up rotor resistance referred to the stator, which accounts for both the rotor copper loss and the mechanical power developed, is given as R 0.16 R2e = 2 = = 5.3 W 0.03 s Therefore, effective scaled-up rotor impedance referred to the stator is given as Z2e = R2e + jX2 = (5.3 + j0.4) W As can be seen from the equivalent circuit, Ri, jXag and Z2e are in parallel, and the equivalent impedance Zp of this parallel combination is given by 1 1 = Zp = −1 −1 -1 -1 -1 (200) + ( j 20) + (5.3 + j 0.4)−1 ( Ri) + ( jX ag ) + (Z 2e) = (4.67 + j1.56) W Hence, the total input impedance per phase for the stator is given as Zin = Z1 + Zp = (0.15 + j0.4) + (4.67 + j1.56) = (4.82 + j1.96) W The applied voltage per phase, V1 =

VL 400 = = 231 V 3 3

V1 231∠ 0° = = 44.4––22.13° A Zin ( 4.82 + j1.96) The power factor, pf = cos f1 = cos 22.13° = 0.926 (lagging)

Hence, the stator current, I1 =

786

Basic Electrical Engineering

(b) The induced emf per phase in the stator is given as E1 = V1 – I1 (R1 + jX1) = 231–0° – (44.4––22.13°)(0.15 + j0.4) = 218.6––3.66° V The core loss component of no-load current, E 218.6 ∠− 3.66° = 1.093––3.66° A Iw = 1 = Ri 200 The magnetising current,

Im =

E1 218.6∠− 3.66° = = 10.93–– 93.66° A Xag j 20

Thus, the rotor current per phase, as referred to the stator, I¢2 = I1 – (Iw + Im) = (44.4––22.13°) – {(1.093––3.66°) + (10.93––93.66°)} = 41.14––8.03° A (c) The mechanical power developed per phase, Pd = (I 2¢ )2

⎛1 − s⎞ ⎛ 1 − 0.03 ⎞ R¢ = (41.14)2 ¥ 0.16 = 8756 W ⎝ s ⎠ 2 ⎝ 0.03 ⎠

The total mechanical power developed, Pdt = 3Pd = 3 ¥ 8756 = 26 268 W = 26.268 kW Thus, the net output power, Po = Pdt – Mechanical losses = 26.268 – 0.8 = 25.468 kW (d) The input power drawn by the motor, Pin = 3V1 I1 cos f1 = 3 ¥ 231 ¥ 44.4 ¥ 0.926 = 28 492 W = 28.492 kW h=

(e

P ROBLEM

Po 25.468 = = 0.894 = 89.4 % Pin 28.492

D -2 2

A three-phase, 1-kV, 50-Hz, 6-pole, star-connected induction motor has wound rotor with slip rings. It has a resistance of 0.02 W and standstill reactance of 0.3 W per phase. Full-load torque is obtained at a speed of 970 rpm. Determine (a) the speed at which maximum torque occurs, (b) the ratio of maximum torque to full-load torque, and (c) the ratio of starting torque to full-load torque.

Solution 120 f 120 50 = = 1000 rpm P 6 N N 1000 970 = = 0.03 The slip, s = s Ns 1000 Torque exerted on the rotor is given as

(a) The synchronous speed, Ns =

t=

KV12sR2 R22

2 + s 2X 20

,

where K t=

K1sR2 2 R22 + s 2X 20

(i)

where K1 is another constant. The maximum torque occurs when R2 = sX20. Therefore, the slip at which maximum

Supplementary Exercise—Part D : Electrical Machines

787

torque occurs is given as

R2 0.02 = = 0.0667 X20 0.3 Hence, the speed at which the maximum torque occurs is given as sm =

N = (1 – sm) Ns = (1 – 0.0667) ¥ 1000 = 933.3 rpm (b) The maximum torque is obtained by putting the condition R2 = sX20 into Eq. (i), tm =

K1sR2 R22 + R22

=

0.0667 K1 = 1.6675K1 2 R2

(ii)

Given that full-load torque is obtained at a speed of 970 rpm or a slip of 0.03. Hence, from Eq. (i), we have tFL =

K1sR2 2 2 R2 + s 2X 20

=

K1 × 0.03 × 0.02 (0.02)2 + (0.03 × 0.3)2

= 1.2474K1

(iii)

Hence, using Eqs. (ii) and (iii), the ratio, m FL

=

1.6675 K1 = 1.337 1.2474 K1

(c) At starting, s = 1. Hence, from Eq. (i), the starting torque is given as 0.02 K1 KR tst = 2 1 2 2 = = 0.221K1 R2 + X 20 (0.02)2 + (0.3)2 Hence, using Eqs. (iii) and (iv), the ratio, 0.221 K1 st = = 0.177 1 .2474 K1 FL P ROBLEM

(iv)

D -2 3

A 3-phase, 220-V, 50-Hz, 4-pole, star-connected, squirrel-cage induction motor has rotor resistance of 0.1 W per phase and rotor standstill reactance of 0.8 W per phase. The ratio of rotor to stator turns is 0.65. (a) At a slip of 5 %, calculate (i) the speed, (ii) the rotor current, (iii) the rotor input power, (iv) the total mechanical power developed, and (v) the total torque exerted on the rotor. (b) Determine the slip at which maximum torque occurs and then at this slip, calculate (i) the speed, (ii) the rotor current, (iii) the rotor input power, (iv) the total maximum mechanical power developed, and (v) the total maximum torque exerted on the rotor. VL 220 = = 127 V 3 3 T The emf induced in rotor per phase at standstill, E20 = V1 ¥ 2 = 127 ¥ 0.65 = 82.55 V T1 120 50 120 f = = 1500 rpm The synchronous speed, Ns = 4 P (a) At 5 % slip :

Solution

Stator per phase voltage, V1 =

(i) The actual speed, (ii) Rotor impedance,

The rotor current,

N = (1 – s)Ns = (1 – 0.05) ¥ 1500 = 1425 rpm Z2 = (R2 + jsX20) = (0.1 + j0.05 ¥ 0.8) = (0.1 + j0.04) W = 0.1077–21.8° W E sE 0.05 82.55 = 38.32 A I2 = 2 = 20 = Z2 Z2 0.1077

788

Basic Electrical Engineering

(iii) The rotor copper loss per phase, PR = I 22 R2 = 38.322 ¥ 0.1 = 146.8 W PR 146.8 = = 2936 W s 0.05 Hence, the total rotor input, Pgt = 3Pg = 3 ¥ 2936 = 8808 W (iv) The total mechanical power, Pdt = Pgt – PRt = 8808 – 3 ¥ 146.8 = 8367.6 W (v) Since, the mechanical power developed for given torque is given as The rotor input per phase

Pg =

(i) The actual speed,

2pNt 60 8367.6 × 60 Pdt ¥ 60 tt = = = 56.07 Nm 2 π × 1425 2pN R 0.1 = 0.125 sm = 2 = 0.8 X20 Nm = (1 – sm)Ns = (1 – 0.125) ¥ 1500 = 1312.5 rpm

(ii) The rotor current,

I2m =

Pd =

\ (b) The slip for maximum torque,

sm E20 R22

+ ( sm X 20)

2

=

0.125 × 82.55 (0.1)2 + (0.125 × 0.8)2

= 72.96 A

2 2 (iii) The rotor copper loss per phase, PR = I 2m R2 = 72.96 ¥ 0.1 = 532.3 W

PR 532.3 = = 4258.4 W 0.125 sm Hence, the total rotor input, Pgt= 3Pg = 3 ¥ 4258.4 = 12 775 W (iv) The total mechanical power, Pdt = Pgt – PRt = 12 775 – 3 ¥ 532.3 = 11 178 W The rotor input per phase, Pg=

(v) The total maximum torque, P ROBLEM

tm =

11178 ¥ 60 Pdt ¥ 60 = = 81.3 Nm 2p ¥ 1312.5 2pN

D -24

A three-phase, 4-pole, 400-V, 50-Hz induction motor has standstill rotor impedance per phase of (0.03 + j0.15) W. Determine the speed at which maximum torque is developed. What resistance is to be added to the rotor circuit per phase so as to achieve 3/4th of maximum torque at the time of starting the motor ? 120 50 Solution The synchronous speed, Ns = 120 f = = 1500 rpm

4 P Maximum torque is obtained at speed when R2 = sX20. Hence, the slip at maximum torque, sm =

R2 0.03 = = 0.2 0.15 X20

Therefore, the speed at which maximum torque is developed is given as N = (1 – sm)Ns = (1 – 0.2) ¥ 1500 = 1200 rpm For a constant applied voltage, the torque developed is given as t=

K1sR2 R22

2 + s 2X 20

where K1 is a constant. Therefore, since R2 = sX20 the maximum torque is given as tm =

K1sm R2 R22 + R22

=

K 0.2 K K1sm = 1 = 1 2R2 2 0.03 0.3

Supplementary Exercise—Part D : Electrical Machines

789

At starting, s = 1, hence the starting torque is given as tst =

K1R2 2 R22 + X 20

Let the total resistance required in the rotor circuit per phase be x for making tst = (3/4)tm, then we must have K1x 3 ⎛ K1 ⎞ = or 1.2x = 3x2 + 0.0675 4 ⎝ 0.3 ⎠ x 2 + (0.15)2 x = 0.3323 or 0.0677

fi

Hence, the additional resistance required is r = x – R2 = (0.3323 – 0.03) or (0.0677 – 0.03) = 0.3023 W or 0.0377 W For starting an induction motor, an additional resistance is connected in the rotor circuit through the slip-rings so as to short-circuited. Hence, we select the lower value of r, that is, r = 0.0377 W. P RO B LE M

D - 25

A three-phase, 400-V, 50-Hz, 5-kW, delta-connected induction motor is started using a star-delta starter. The short0.84. Calculate (a) the starting current drawn by the motor, and (b) the ratio of the starting to the full-load current.

Solution (a) At the time of starting, the stator winding is connected in star. Hence, per phase voltage applied, V 400 V1 = L = = 231 V 3 3 For delta-connected motor, the short-circuit line current for an applied voltage of 100 V is given as 15 A. Hence, per phase short-circuit current at 100 V is 15 Isc1 = A 3 At the start, the stator winding is connected in star. Hence, per phase short-circuit current for an applied line voltage of 400 V is 15 231 Isc = = 20 A 3 100 Since the motor is connected in star, the line current at starting is also 20 A. (b) On full load, the input power is P 5000 Pin = o = = 6097 W 0.82 6097 Pin = = 10.47 A The line current, IFL = 3 400 0.84 3VL cos f 20 = 1.91 Hence, the ratio of starting current to full-load current = 10.47 P ROBLE M

D -26

A dc shunt generator supplies a load of 10 kW at 220 V through feeders of resistance 0.1 W and the armature resistance are 100 W and 0.05 W, respectively. Determine (a) the terminal voltage, (b current, and (c) the generated emf.

790

Basic Electrical Engineering

Solution (a) The voltage across the load terminals is 220 V and power supplied is 10 kW. Therefore, the load current, IL =

PL 10 103 = = 45.5 A VL 220 ¥ 0.1 = 4.55 V. Hence, the voltage

across the terminals of the generator, V = VL + 4.55 = 220 + 4.55 = 224.55 V V 224.55 = = 2.25 A Rsh 100 Ia = IL + Ish = 45.5 + 2.25 = 47.75 A

Ish =

(b (c) The armature current,

E = V + Ia Ra = 224.55 + 47.75 ¥ 0.05 = 226.94 V

The generated emf, P RO B L EM

D -2 7

A 4-pole, dc generator has 60 slots with 6 conductors per slot on its armature. (a generator gives an open-circuit voltage of 230 V, when the conductors are wave-connected and the generator is run at 750 rpm. (b produce an open-circuit voltage of 115 V, when it is run at the same speed.

Solution (a) For wave-connected armature, A = 2. Total number of conductors, Z = 60 ¥ 6 = 360. Since, Eg =

FZNP 60 A

fi F=

60Eg A NZP

=

60 230 2 = 0.0256 Wb 750 360 4

(b) To meet the requirement of producing 115 V at the same speed, the simplest way is to change the number of conductors per path (Z/A), if it is feasible. As P, N and F are to remain constant, from the expression for induced emf, we have E 115 (Z/A) μ Eg fi (Z/A)2 = (Z/A)1 ¥ 2 = (360/2) ¥ = 90 E1 230 It is possible to get 90 conductors per path on the armature if the 360 conductors are connected in lap winding (for which A = P P RO B L EM

D -2 8

W and 50 W, respectively. Determine the speed of the machine when supplying a load current of 450 A at a terminal voltage of 250 V.

Solution

W

is given as Ish =

V 250 = =5A Rsh 50

Therefore, the armature current, Ia = Ish + IL = 5 + 450 = 455 A \ The generated emf, E = V + Ia Ra = 250 + 455 ¥ 0.05 = 272.75 V

Supplementary Exercise—Part D : Electrical Machines

791

Due to the rotation of armature at N rpm, the generated emf is given as FZNP 60 A

N = 60EA F ZP Here, E = 272.75 V as calculated above, A = P = 4 as the armature is lap-wound, and the total number of conductors is E=

fi

Z = Number of slots ¥ Conductors per slot = 120 ¥ 4 = 480 Hence, the speed of rotation of the armature is given as 60 272.75 4 = 681.88 ª 682 rpm N = 60EA = 0.05 480 4 F ZP P ROBLEM

D -2 9

A long-shunt compound, dc generator delivers a load current of 50 A at 500 V. The resistances of the armature, seriesW, 0.03 W and 250 W, respectively. Calculate the generated emf and the armature current. Allow 1.0 V per brush for contact drop.

Solution Ish =

Vsh V 500 = L = =2A Rsh Rsh 250

A long-shunt compound dc generator. Therefore, the armature current, Ia = IL + Ish = 50 + 2 = 52 A \ The emf generated, Eg = VL + Ia (Rse + Ra) + 2 ¥ Vbrush = 500 + 52 ¥ (0.03 + 0.05) + 2 ¥ 1 = 506.16 V P RO B LE M

D - 30

In a long-shunt compound, dc generator, the terminal voltage is 230 V when it delivers 150 A of current. The armature, W, 0.015 W, 0.03 W and 92 W, respectively. Determine (a) the induced emf, and (b) the total power generated by the armature.

Solution (a Ish =

Vsh V 230 = L = = 2.5 A Rsh Rsh 92

792

Basic Electrical Engineering

Fig. D-5 The armature current, Ia = IL + Ish = 150 + 2.5 = 152.5 A 0.015 × 0.03 = 0.01 W 0.015 + 0.03 \ The emf induced, Eg = VL + Ia (Rp + Ra) = 230 + 152.5 ¥ (0.01 + 0.032) = 236.41 V (b) The total power generated by the armature, Pa = Eg Ia = 236.41 ¥ 152.5 = 36 052 W = 36.052 kW Rp = Rse || Rdi =

P RO B L EM

D -3 1

(a emf should not exceed 1000 V.

W b) the induced emf, and (c) the value of discharge resistance so that the induced

Solution (a) 600 × 50 × 10 − 3 NF = = 3.0 H 10 I Change in flux linkage 50 × 10 − 3 × 600 FN (b) The induced emf, E = = = = 1500 V t 0.02 Time L=

(c) To limit the induced emf to 1000 V, the total resistance needed, 1000 Rt = = 100 W 10 Therefore, the additional discharge resistor needed, Rd = Rt – Rsh = 100 – 50 = 50 W P RO B L EM

D -3 2

W, 0.3 W and 0.4 W, respectively. Calculate (a) the induced emf, and (b) the load resistance.

Supplementary Exercise—Part D : Electrical Machines

793

Solution The circuit diagram of the generator is shown in Fig. D-6. Rse = 0.3 W

Ish +

V¢

IL

Ise

Ia +

+

Rsh

Ra

100 W

0.4 W

VL 230 V

7.5 kW

– –

–

Fig. D-6 3

(a) The load current, IL =

P 7.5 10 = = 32.6 A V 230 Ise = 32.6 A. Therefore, the voltage across V¢ = VL + Ise Rse = 230 + 32.6 ¥ 0.3 = 239.78 V

Ish =

V 239.78 = ª 2.4 A 100 Rsh

Hence, the armature current, Ia = Ish + IL = 2.4 + 32.6 = 35 A The induced emf in the generator, E = V¢ + Ia Ra = 239.78 + 35 ¥ 0.4 = 253.78 V (b) The load resistance, RL = P ROB L EM

V2 230 2 = = 7.05 W P 7.5 103

D - 3 3

W and armature resistance of 0.03 W. Determine the speed of the machine running as a shunt motor and taking 40 kW input at 240 V. Assume the contact drop of 1 V per brush. P 40 103 = = 166.7 A V 240 V 240 = = =4A Rsh 60 = Ish + IL = 4 + 166.7 = 170.7 A = V + Ia Ra + Voltage drop at the brushes = 240 + 170.7 ¥ 0.03 + 2 ¥ 1 = 247.12 V

Solution As a generator : The load current, IL = Ish The armature current, Ia The generated emf, Eg

As a motor :

P 40 103 = = 166.7 A V 240 V 240 Ish = = =4A Rsh 60 The armature current, Ia = IL – Ish = 166.7 – 4 = 162.7 A The back emf, Eb = V – Ia Ra – Voltage drop at the brushes = 240 – 162.7 ¥ 0.03 – 2 ¥ 1 = 233.12 V

The line current drawn, IL =

794

Basic Electrical Engineering

The induced emf in the dc machine (a generator or a motor) is given E =

FZNP 60 A

Em N = m Eg Ng

\

P RO B L EM

or E μ N = kN (as other parameters are constant) fi

Nm = Ng

Em 23312 . = 450 ¥ = 424.5 rpm Eg 247.12

D -3 4

The open-circuit characteristic (OCC) of a separately excited dc generator driven at 750 rpm is given below. Field current, If (A) : Generated emf, Eg (V) :

0.2 40

0.4 66

0.6 86

0.8 101

1.0 112

1.2 121

1.4 128

1.6 133

a) the open-circuit voltage when W, (b voltage to 110 V, and (c

Solution (a) W. For this, we take a suitable value of Vsh = If Rsh = 1 ¥ 94 = 94 V. When the machine is used as a shunt generator, the voltage Vsh is the same as the open-circuit voltage Eg. We mark the on Eg-axis, we get point

Generated emf, Eg

C which gives the open-circuit voltage as 126 V.

160 Rt

G 140

Rsh = 94 W

C B

120

D E

100

A

80 70 60

F

40 20 0 0

0.2

0.4

0.6

0.8

1.0

1.2 1.4 1.6 Field current, If

The (b) Corresponding to Eg = 110 V (point D), a horizontal line is drawn which intersects OCC curve at point E. The line Rt needed to generate Eg of 110 V. The resistance Rt can be determined

Supplementary Exercise—Part D : Electrical Machines

795

by selecting a suitable point (say, point F) on the line OE and reading the corresponding values of voltage and current. Thus, V 70 Rt = = = 116.7 W I 0.6 \ The additional resistance needed, Radd = Rt – Rsh = 116.7 – 94 = 22.7 W (c point G, we get Rfc = P RO B L EM

V 150 = = 200 W I 0.75

D -3 5

The open-circuit characteristic (OCC) of a separately excited dc generator driven at 1000 rpm is given below. Field Current, If (A) : Generated emf, Eg (V) :

0 10

1 112

2 198

3 232

4 252

5 266

W. The machine is connected as a dc shunt generator and is driven at 1250 rpm. Find (a) its open-circuit voltage, (b voltage to 270 V, (c) the open-circuit voltage if an additional resistance of 130 W (d

Solution The OCC data given is at a speed of 1000 rpm, whereas the dc shunt generator is run at 1250 rpm. As the speed is increased, the generated emf also increases according to the following expression : Eg = Thus,

Eg 2 Eg1

=

FZNP 60 A

N2 N1

Field Current, If (A) : Generated emf, Eg (V) :

or

or

0 12.5

Eg μ N

Eg2 = Eg1 ¥

1 140

N2 1250 = Eg1 ¥ = Eg1 ¥ 1.25 N1 1000 2 247.5

3 290

4 315

5 332.5

Based on the above data, the OCC curve for the dc shunt generator running at 1250 rpm is drawn in Fig. D-8. (a) To draw the resistance line for Rsh = 70 W resistance line intersects the OCC curve at point B. The corresponding point on Eg-axis is C. Thus, The open-circuit voltage, Eg = 330 V (b) From point D(Eg = 270 V), we draw a horizontal line which intersects OCC curve at point E. Joining O to E gives Rt. The coordinates of the point E give the value of Rt as 270 V Rt = = 112.5 W 2.4 A

(c

Ra = Rt – Rsh = 112.5 – 70 = 42.5 W Rt1 = R a1 + Rsh = 130 + 70 = 200 W. To draw the line corresponding to this Eg as Eg = If Rt = 1 ¥ 200 = 200 V

796

Basic Electrical Engineering H

Generated emf, Eg

G

Rt

350 C

B

300

OCC

A

D 250 200

Rsh = 70 W

E F

150 100 50 O 1

2

3

4 5 Field current, If

The Next, we mark the point F Rt1 = 200 W that this line does not intersect the OCC curve anywhere. It means that in this case, the generator fails to build up voltage. So, Eg ª 0 V. (d resistance. Thus, Rfc = P ROBLEM

155 V = 155 W 1A

D -3 6

A 200-V dc shunt motor takes a total current of 100 A and runs at 750 rpm. The resistances of the armature winding W and 40 W, respectively. The friction and iron losses amount to 1500 W. Calculate (a) the torque developed by the armature, (b) the copper losses, (c) the shaft power, (d) shaft torque, and (e) the

Solution V 200 = =5A Rsh 40 The armature current, Ia = IL – Ish = 100 – 5 = 95 A The back emf, Eb = V – Ia Ra = 200 – 95 ¥ 0.1 = 190.5 V Now, we have Electrical input power = Mechanical power developed Ish =

(a

Eb Ia =

or \

Torque developed,

td =

2pNt d 60 60 × 190.5 × 95 60Eb I a = = 230.4 Nm 2 π × 750 2p N

Supplementary Exercise—Part D : Electrical Machines

797

(b) The back emf developed is given as Eb = V – Ia Ra. Multiplying this equation by Ia, and then rearranging the terms, we get 2 The armature copper loss = I a Ra = VIa – Eb Ia = 200 ¥ 95 – 190.5 ¥ 95 = 902.5 W 2

\

2

I sh Rsh = 5 ¥ 40 = 1000 W Total copper losses = 902.5 + 1000 = 1902.5 W

(c) Total losses = Total copper losses + (Friction losses + Iron losses) = 1902.5 + 1500 = 3402.5 W Input power, Pin = VI = 200 ¥ 100 = 20 000 W Output shaft power, Po = Pin – losses = 20 000 – 3402.5 = 16 598 W ª 16.6 kW (d) The shaft torque, tsh = (e

P R O B L EM

h=

60Po 60 × 16 598 = = 211.25 Nm 2pN 2 π × 750

Po 16 598 = = 0.8299 = 82.99 % Pin 20 000

D - 37

0.22 W and 0.13 W, respectively. Windage loss, friction loss and iron losses amount to 750 W. Find (a) the shaft power, (b) the total torque, and (c) the shaft torque.

Solution (a) The armature current, Ia = IL = 45 A. Therefore, the back emf is given as Eb = V – Ia (Ra + Rse) = 100 – 45 ¥ (0.22 + 0.13) = 84.25 V The mechanical power developed, Pd = Eb Ia = 84.25 ¥ 45 = 3791.3 W The output power or shaft power is given as Psh = Pd – Plosses = 3791.3 – 750 = 3041.3 W (b) If the total torque developed at the armature is td, then we can write 60Pd 60 ¥ 3791.3 2pNt d Pd = fi td = = = 48.27 Nm 2pN 2p ¥ 750 60 (c) In terms of the shaft torque (tsh), the output power or shaft power is given as Psh = PROBLEM

2pNt sh 60

fi

tsh =

60 Psh 60 ¥ 3041.3 = = 38.72 Nm 2pN 2p ¥ 750

D -38

W and 0.15 W, respectively. The iron and friction losses amount to 900 W. Find (a) the speed, (b) the total torque developed, (c) the shaft power, (d) the shaft torque, and (e

Solution (a) The armature current, Ia = IL = 46 A. Therefore, the back emf is given as Eb = V – Ia (Ra + Rse) = 220 – 46 ¥ (0.25 + 0.15) = 201.6 V

798

Basic Electrical Engineering Eb =

Since,

FZNP 60 A

60 × 2 × 201.6 N = 60AEb = = 252 rpm 20 × 10 − 3 × 1200 × 4 F ZP

fi

(b) The total torque developed is given as td =

20 ¥ 10 - 3 ¥ 1200 ¥ 4 ¥ 46 FZPI a = = 351.4 Nm 2p ¥ 2 2pA

(c) The mechanical power developed, Pd = Eb Ia = 201.6 ¥ 46 = 9273.6 W The output power or shaft power is given as Psh = Pd – Plosses = 9273.6 – 900 = 8373.6 W = 8.374 kW 60 ¥ 8373.6 60Psh = = 317.3 Nm 2p ¥ 252 2pN

(d) The shaft torque, tsh = h=

(e P RO B L EM

Po P 8373.6 = sh = = 0.8274 = 82.74 % Pin VIL 220 46

D -3 9

A 230-V, dc series motor has an armature-circuit resistance of 0.2 W W. At rated voltage, the motor draws a line current of 40 A and runs at a speed of 1000 rpm. Find the speed of the motor for a line current

Solution For line current of 40 A : The back emf induced in the motor is given as Eb1 = V – Ia1 (Ra + Rse) = 230 – 40 ¥ (0.2 + 0.1) = 218 V For line current of 20 A : The back emf induced, Eb2 = V – Ia2 (Ra + Rse) = 230 – 20 ¥ (0.2 + 0.1) = 224 V F2 = 0.6F1. Now, the back emf induced for a motor is given as Eb =

FZNP 60 A

fi

Eb μ FN = kFN

(as other parameters are constant)

Therefore, taking the ratio of emfs in the two cases, we get Eb 2 k = Eb1 k fi

N2 =

P RO B L EM

2 N2 1 N1

=

0.6F1N 2 0.6 N2 = F1N1 N1

N1 Eb 2 1000 = 0.6 Eb1 0.6

224 = 1712.5 rpm 218

D -4 0

A 200-V, dc shunt motor has an armature resistance of 0.4 W

Solution

Ish1 =

V 200 = =1A Rsh1 200

The back emf at 750 rpm, Eb1 = V – Ia1Ra = 200 – 25 ¥ 0.4 = 190 V

W. The motor runs at

Supplementary Exercise—Part D : Electrical Machines

799

Let R as Ish2 =

V 200 = A Rsh2 R

The torque developed by the motor, t μ FIa = kIsh Ia. Since the torque remains constant in the two cases, we must have kIsh1 Ia1 = kIsh2 Ia2 fi Ia2 =

Ish1Ia1 1 25 = = 0.125R A Ish2 200/ R

The back emf at 500 rpm, Eb2 = V – Ia2Ra = 200 – 0.125R ¥ 0.4 = (200 – 0.05R) V We know that the back emf, Eb μ FN = kIsh N. We can therefore write Eb 2 I N = sh2 2 Eb1 Ish1N1

or

(200

0.05 R) (200 / R) 500 200 = = 1 750 190 1.5R

2

fi

0.075R – 300R + 38 000 = 0

Solving this quadratic equation, we get R = 130.95 W

or

3869 W

R = 130.95 W. Hence, the reduction in the shuntThe value of 3869 W being greater than 200 W resistance, DRsh = 200 – 130.95 = 69.05 W P RO B L EM

D -4 1

A 100-kW, belt-driven, dc shunt generator running at 320 rpm on 220 V supply system, continues to run as a motor when the belt breaks, taking a power of 10 kW. Find the speed if the armature resistance is 0.03 W resistance is 55 W. Take the contact drop as 1 V per brush.

Solution As a generator : The line current, IL =

P 100 000 = = 454.5 A V 220

V 220 = =4A Rsh 55 The armature current, Ia = IL + Ish = 454.5 + 4 = 458.5 A The generated emf, Eg = V + Ia Ra + brush drop = 220 + 458.5 ¥ 0.03 + 2 ¥ 1 = 235.76 V Ish =

As a motor : The line current, IL = Ish =

P 10 000 = = 45.45 A V 220 V 220 = =4A Rsh 55

The armature current, Ia = IL – Ish = 45.45 – 4 = 41.45 A The generated emf, Eb = V – Ia Ra – brush drop = 220 – 41.45 ¥ 0.03 – 2 ¥ 1 = 216.76 V E μ N. Hence, Em N = m Eg Ns

fi

E 216.76 Nm = Ng m = 320 ¥ = 294.2 rpm 235.76 Eg

800

Basic Electrical Engineering

D. 2. PRACTICE PROBLEMS ( A )

S I M P L E

PR O BL EMS

D-1. A 50-kVA, single-phase transformer has 600 turns on primary and 40 turns on secondary. The primary winding is connected to 2.2-kV, 50-Hz supply. Determine (a) the secondary voltage at no load, (b) the primary and secondary currents at full load. [Ans. (a) 146.67 V; (b) 22.72 A, 340.9 A] D-2. The design requirement of a 11-kV/415-V, 50-Hz, single-phase, core-type transformer are approxiof 1.5 T. Find a suitable number of primary and secondary turns and the net cross-sectional area of the core. [Ans. 734, 28, 450 cm2] D-3. The required no-load voltage ratio in a single-phase, 50-Hz, core-type transformer is 6000-V/250-V. Find the turns per limb on the high and low voltage [Ans. 287, 12] D-4. A 80-kVA, 3200-V/400-V, 50-Hz transformer has 111 turns on its secondary. Calculate (a) the number of turns on primary winding, (b) the secondary current, and (c) the cross-sectional area of the core, [Ans. (a) 888; (b) 200 A; (c) 135 cm2] D-5. A 120-V/27.5-V, 400-Hz, step-down transformer is to be operated at 60 Hz. Find (a) the highest safe input voltage, and (b) the transformation ratio in both the frequency situations. [Ans. (a) 18 V; (b) 4.36] D-6. The following data refers to a single-phase transformer : Primary turns Secondary turns Output Secondary voltage

: 200, : 40 : 100 kVA, : 400 V

Neglecting the losses, calculate (a) the primary applied voltage, (b) the normal primary and secondary currents, (c) the secondary current when the load is 25 kW at 0.8 power factor. [Ans. (a) 2000 V; (b) 50 A, 250 A; (c) 78.125 A]

D-7. A 1000-V/200-V transformer takes 0.3 A at a pf of 0.2 lagging on open circuit. Find the magnetising and iron loss components of no-load primary current. [Ans. 0.06 A, 0.294 A] D-8. A 230-V/110-V, single-phase transformer takes an input of 350 VA at no load and rated voltage. The core loss is 110 W. Find (a) the no-load power factor, (b) the iron-loss component of the no-load current, and (c) the magnetising component of the no-load current. [Ans. (a) 0.314; (b) 0.477 A; (c) 1.44 A] D-9. Calculate the voltage regulation of a transformer in which the ohmic drop is 10 % of the output voltage, and the reactive drop is 5 % of the output voltage, when the power factor is (a) 0.8 lagging, and (b) 0.8 leading. [Ans. (a) 11 %; (b) 5 %] D-10. A 25-kVA, 1910-V/240-V, single-phase, 50-Hz transformer is employed to step-down the voltage, keeping the voltage at low-voltage side constant. Calculate (a) the turns ratio, (b) the full-load current on low-voltage side, (c) the load current referred to the high-voltage side, (d) the load impedance on low-voltage side for full load, and (e) the load impedance referred to high-voltage side. [Ans. (a) 7.95; (b) 104.16 A; (c) 13.1 A; (d) 2.3 W; (e) 145.36 W] D-11. The open-circuit test readings on a 400-V/200-V, single-phase transformer conducted on lowvoltage side are : voltage = 200 V, current = 0.7 A, power = 95 W. Calculate the no-load circuit parameters referred to the high-voltage side. [Ans. R01 = 1684 W, X01 = 1556 W] D-12. A 5-kVA, 200-V/100-V, 50-Hz, single-phase transformer has rated secondary voltage on full load. When the load is removed, the secondary voltage is found to be 110 V. Determine the percentage regulation. [Ans. 9.091 %] D-13. A 230-V/460-V, single-phase transformer has a primary resistance of 0.2 W and reactance of 0.5 W. The corresponding values for the secondary are

801

Supplementary Exercise—Part D : Electrical Machines 0.75 W

W

D-19. A 4-pole, three-phase, star-connected alternator has pf

lagging. [Ans. D-14. A 40-kVA, 2000-V/250-V transformer has a W winding resistance of 0.0155 W. Calculate (a) the

when the alternator is driven at 1500 rpm. [Ans. D-20. alternator has 12 slots per pole and 4 conductors

(b) the total resistance drop on full load, and (c) the total power loss on full load. [Ans. (a) 0.0335 W; (b c D-15.

Ans. D-21. Calculate the speed and open-circuit line and phase

transformer has the following resistances and leakage reactances : R1 W; X1 = 3.2 W;

R2 W X2 = 0.03 W

Calculate the equivalent resistance and reactance

distributed.

[Ans.

D-22.

[Ans. 0.017 W W D-16. Open-circuit and short-circuit tests were conducted [Ans. D-23. A three-phase, star-connected alternator with 12 poles generates 1100 V on open circuit at a speed of

observations during these tests are : OC Test

Ans. SC Test

D-24. 5 ¥ 10 –2 pf

[Ans. [Ans.

D-25.

D-17. transformer on test gives the following results with force.

OC Test SC Test

[Ans.

D-26. of the generated emf. pf.

[Ans.

[Ans.

D-27.

D-18. winding transformer has a resistance of 0.01 W and W as referred to low-voltage side. Find (a) per unit resistance and reactance, (b

conductors are connected in series, calculate the generated line emf. [Ans. D-28.

[Ans. (a) 0.0125 pu, 0.045 pu; (b

802

Basic Electrical Engineering

is employed. Flux per pole is 0.05 Wb. Calculate the line voltage on open circuit. [Ans. 928 V] D-29. A three-phase, star-connected, 2-pole alternator runs at 3600 rpm. If there are 500 conductors per phase in series on the armature winding and the magnitude and the frequency of the generated emf. [Ans. 11.53 kV, 60 Hz] D-30. current of 40 A produces a full-load current of 200 A on short circuit, and 1160 V on open circuit. Find the value of synchronous reactance if the armature resistance is 0.75 W. [Ans. 5.75 W] D-31. The effective armature resistance of a 2200-V, 50-Hz, 440-kVA, single-phase alternator is 0.5 W. full-scale current. The emf on open circuit for synchronous impedance and reactance. [Ans. 5.8 W, 5.78 W] D-32. A three-phase, 2300-V, star-connected synchronous motor has a resistance of 0.2 W per phase and synchronous reactance of 2.2 W per phase. The motor is operating at 0.5 power factor leading with a line current of 200 A. Determine the value of excitation voltage per phase and the power angle. [Ans. 1708.1 V, 8.57°] D-33. A 400-V, 50-Hz, three-phase, star-connected synchronous motor has per phase synchronous impedance of (0.5 + j4.0) W. It takes a current of 15 A at unity pf excitation voltage and the power angle. [Ans. 231.4 V, 15°] D-34. A 500-V, 50-Hz, single-phase synchronous motor takes 50 A at a pf of 0.8 lagging. It has a synchronous reactance of 2 W and negligible resistance. The armature has 120 full-pitch coils in series, with a distribution factor of 0.96. Assuming a sinusoidal pole. [Ans. 0.01748 Wb] D-35. An 8-pole alternator running at 750 rpm supplies power to a 6-pole, three-phase induction motor. The motor is on full load and running at a slip of 3 %. Find the full-load speed of the motor and the frequency of its rotor emf. [Ans. 970 rpm, 1.5 Hz]

D-36. A 6-pole, squirrel-cage induction motor is connected to a supply of 400 V, 50 Hz. It runs at a speed of 960 rpm. Calculate (a) the synchronous speed, and (b) the slip. [Ans. (a) 1000 rpm; (b) 4 %] D-37. If the electromotive force in the stator of an 8-pole induction motor has a frequency of 50 Hz, and that in the rotor 1.5 Hz, at what speed is the motor running and what is the slip ? [Ans. 727.5 rpm, 3 %] D-38. A three-phase induction motor is wound for 4 poles and is supplied from a 50-Hz voltage source. Calculate (a) the synchronous speed, and (b) the speed of the rotor when slip is 4 %. [Ans. (a) 1500 rpm; (b) 1440 rpm] D-39. A 6-pole, 50-Hz, squirrel-cage induction motor runs on full load at a shaft speed of 970 rpm. Calculate (a) the percentage slip, and (b) the rotor-voltage frequency. [Ans. (a) 3 %; (b) 1.5 Hz] D-40. A three-phase, 6-pole, 50-Hz induction motor runs at 4 % slip at a certain load. Determine (a) the synchronous speed, (b) the rotor speed, (c) the frequency of rotor current, (d) the speed of rotor’s

D-41.

D-42.

D-43.

D-44.

D-45.

[Ans. (a) 1000 rpm; (b) 960 rpm; (c) 2 Hz; (d) 40 rpm] A 2-pole, 50-Hz, three-phase induction motor is running on load at a slip of 4 %. Calculate (a) the synchronous speed of the machine, and (b) the actual running speed. [Ans. (a) 3000 rpm; (b) 2880 rpm] A 12-pole, 50-Hz induction motor runs at 475 rpm. Find (a) the slip, and (b) the frequency of the currents in the rotor winding. [Ans. (a) 5 %; (b) 2.5 Hz] At what speed will the shaft of a three-phase, 4-pole, 50-Hz induction motor run, if it is loaded to have a slip of 0.03 ? [Ans. 1455 rpm] A three-phase, 4-pole, 440-V, 50-Hz induction motor is running with a slip of 4 %. Find the rotor speed and frequency of induced currents. [Ans. 1440 rpm, 2 Hz] The rotor of a three-phase induction motor has 0.06 W resistance and 0.3 W standstill reactance per phase. Find the additional resistance required in the

803

Supplementary Exercise—Part D : Electrical Machines rotor circuit to make the starting torque equal to the maximum torque of the motor. [Ans. 0.24 W D-46.

load, the voltage is 120 V. Find the load current, if W and the armatureresistance is 0.02 W. Neglect armature reactance. [Ans. D-54. A 500-V, 10-pole dc shunt generator runs at

power developed and the rotor copper losses, if the If the armature current is 200 A and the armature resistance is 0.15 W [Ans.

[Ans. D-47. D-55. is the slip and at what speed is the motor running ? [Ans. D-48. A 4-pole, dc shunt generator with a lap-wound armature has an armature resistance of 0.1 W and W. The generator is

conductors, each of resistance 0.002 W, and is current is 50 A, calculate the terminal voltage. [Ans. D-56. A dc shunt generator gives full-load output of

the armature current, and the generated emf. The brush contact drop is 1 V per brush. [Ans.

W and 50 W, a) the emf generated, (b) the copper losses, and (c [Ans. (a) 207.7V; (b c

D-49. (a) If the armature is lap-wound and the generator is running at 400 rpm, calculate the induced emf. (b) If the armature were wave-wound, at what speed should it be driven so as to generate 400 V ? [Ans. (a b

D-57. at a terminal voltage of 200 V. The armature W and 40 W, a) the output of the engine driving the generator, and (b) the

D-50. 1000 rpm, when the armature is (a) lap-wound, and (b) wave-wound. [Ans. (a b D-51. A 4-pole, dc generator has 51 slots, each having 24 a) If the armature is wave-wound, at what speed must the armature rotate to give an induced emf of 220 V ? (b) If the armature is lap-wound, what voltage will be induced if the armature rotates at the same speed ? [Ans. (a b D-52. is required to give the open-circuit voltage of 220 V. The total number of armature-conductors is 2000 the speed at which the generator should be driven. [Ans. D-53. A dc shunt generator has an induced voltage of

[Ans. (a

b

D-58. 10 conductors per slot on its armature. It generates an emf of 400 V at no load, when running at 1000 rpm. Find (a b) the speed at which it should be run so as to generate a voltage of 220 V on open circuit. [Ans. (a b D-59. A 4-pole, dc shunt generator with lap-connected armature supplies 100 A to a load at 200 V. The are 0.1 W

W

a) the armature current, (b) the current per armature path, and (c) the emf generated. [Ans. (a) 102.5 A; (b c D-60. A short-shunt compound dc generator supplies

804

Basic Electrical Engineering

W, 0.03 W and 60 W, respectively. Determine the emf generated. [Ans. 114.07 V] D-61. A 4-pole, 500-V, dc shunt motor has 720 waveconnected conductors on its armature. The fullis 0.03 Wb. The armature resistance is 0.2 W and the contact drop is 1 V per brush. Calculate the fullload speed of the motor. [Ans. 675 rpm] D-62. Calculate the ohmic value of the starting resistance for the following dc shunt motor : Supply = 240 V Output = 14 920 W Armature resistance = 0.25 W

( B)

T R IC KY

The starting current is to be limited to 1.5 times the full-load current. Ignore the current in the shunt winding. [Ans. 1.9632 W] D-63. resistances of 0.06 W and 0.04 W, respectively, is connected across 220 V mains. The motor runs at 900 rpm and the armature takes 40 A current. Determine its speed when the armature takes 75 A, saturation. [Ans. 770 rpm] D-64. A 4-pole, 250-V, dc series motor has a waveconnected armature with 1254 conductors. The

each 0.2 W, calculate the speed.

[Ans. 250 rpm]

PR OB LE MS

D-65. A single-phase, 4-kVA transformer has 400 primary-turns and 1000 secondary-turns. The net cross-sectional area of the core is 60 cm2. When the primary winding is connected to 500-V, 50-Hz supply, calculate (a) the maximum value of b) the voltage induced in the secondary winding, and (c) the secondary fullload current. [Ans. (a) 0.938 T; (b) 1250 V; (c) 3.2 A] D-66. A 4600-V/230-V, 60-Hz step-down transformer has core dimensions of 76.2 mm by 111.8 mm. A 2 is to be used. Assuming 9 % loss of area due to stacking factor of laminations, calculate (a) the primary turns required, (b) the turns per volt, (c) the secondary turns required, and (d) the transmission ratio. [Ans. (a) 2422; (b) 0.5265; (c) 121; (d) 0.05] D-67. A 400-V/200-V, single-phase transformer is supplying a load of 25 A at a power factor of 0.866 lagging. On no load, the current and pf are 2.0 A and 0.208, respectively. Calculate the current taken from the supply and specify its phase. [Ans. 13.92 A, 36.13° (lagging)] D-68. An iron-cored transformer has 200 turns on the primary and 100 turns on the secondary. A supply of 400 V, 50 Hz is given to the primary and an impedance of (4 + j3) W is connected across the secondary. Assume ideal behaviour and calculate

(a) the voltage and current through the load, (b) the primary current, (c) the power taken from the supply, and (d) the input impedance of the transformer. [Ans. (a) 200 V, 40––36.87° A; (b) 20––36.87° A; (c) 6.4 kW; (d) (16 + j12) W] D-69. The iron loss of a 80-kVA, 1000-V/250-V, 1-phase, 50-Hz transformer is 800 W. When the primary winding carries a current of 50 A, the copper loss of the transformer is 400 W. Estimate (a) the area of is 1 tesla and there are 1000 turns on primary, (b) the current ratio (primary to secondary), (c) the pf, and (d for a load when the copper loss becomes equal to the iron loss and pf remains 0.8 lagging. [Ans. (a) 45 cm2; (b) 0.25; (c) 97.23 %; (d) 97.25 %] D-70. The SC test on a single-phase transformer with primary winding short-circuited and 30 V applied to the secondary gives a wattmeter reading of 60 W and secondary current of 10 A. If the normal primary voltage is 200 V, the transformer ratio is 1 : 2 and the full-load secondary current is 10 A, calculate the secondary potential difference at fullload current for (a) unity pf, (b) 0.8 pf lagging. [Ans. (a) 394 V; (b) 377.56 V]

Supplementary Exercise—Part D : Electrical Machines D-71. In a 25-kVA transformer, the iron loss and full-load copper loss are 350 W and 400 W, respectively. (a) half full-load at unity power factor, (b) threefourth full-load at 0.8 pf lagging. [Ans. (a) 96.525 %; (b) 96.308 %] D-72. The following test data is obtained on a 5-kVA, 220-V/400-V, single-phase transformer : OC Test : 220 V 2 A 100 W (on LV side) SC Test : 40 V 11.4 A 200 W (on HV side) at full load 0.9 power factor lag. [Ans. 93 %, 8.6 %] D-73. A 250-V/500-V, single-phase transformer gives the following results : SC Test : 20 V (with LV side shorted) OC Test : 250 V

12 A

100 W

1A

80 W

at 0.8 pf lagging. [Ans. 96.4 %] D-74. A 50-kVA, 1000-V/200-V, 1-phase, transformer takes 4 A at a pf of 0.2 lagging, when the secondary is open-circuited. Calculate the primary current and the power factor when a load taking 250 A at a lagging pf of 0.8 is connected across the secondary. Assume the voltage drops in the windings to be negligible. [Ans. 53.06––39.7° A, 0.7694 (lag)] D-75. A 550-V, 55-kVA, 50-Hz, single-phase alternator has an effective resistance of 0.2 W of 10 A produces an armature current of 200 A on short circuit and an emf of 4350 V on open circuit. Calculate (a) the synchronous reactance, and (b) the full-load regulation with power factor 0.8 lagging. [Ans. (a) 2.24 W; (b) 30.9 %] D-76. In a 60-kVA, 1-phase, 220-V alternator, the effective armature resistance and leakage reactance are 0.016 W and 0.07 W respectively. Calculate the emf induced in the armature when the alternator is delivering rated current at a pf of (a) unity, (b) 0.7 lagging, and (c) 0.7 leading. [Ans. (a) 225.2 V; (b) 234.5 V; (c) 208.1 V] D-77. A three-phase, 3300-V, 50-Hz, 500-kVA alternator has a star-connected stator. The effective resistance per phase of the stator winding is 0.4 W. Calculate

805

the regulation at full load (a) unity pf, and (b) 0.8 pf lagging. [Ans. (a) 3.32 %; (b) 12.67 %] D-78. A three-phase, 10-pole alternator has 2 slots per pole per phase on its stator with 10 conductors per equals to 0.05 Wb. The stator has a double layer winding with a coil span of 150 electrical degrees. If the alternator is running at 600 rpm, calculate the emf generated per phase at no load. [Ans. 1035.8 V] D-79. A three-phase, 8-pole, 50-Hz, star-connected alternator has 4 slots per pole per phase on its stator distributed sinusoidally and is 0.04 Wb. The stator has a double layer winding with a full pitch coil. Calculate (a) the pitch factor, (b) the distribution factor, (c) the emf generated per phase, and (d) the line voltage at no load. [Ans. (a) 1; (b) 0.957; (c) 1360 V; (d) 2354 V] D-80. A 4-pole, three-phase, star-connected alternator generates at no load a line voltage of 6.0 kV. The on its periphery 5 slots per pole per phase with 2 conductors per slot. The stator coils are short pitched by 3 slots, in order to reduce the harmonics in the generated emf wave. Determine the speed of the alternator. [Ans. 1500 rpm] D-81. A three-phase, 2500-kVA, 6600-V, star-connected alternator has an armature resistance of 0.073 W per phase and synchronous reactance of 10.7 W per phase. Calculate the voltage regulation at full load 0.8 pf lagging. [Ans. 45.63 %] D-82. A three-phase, 600-kVA, 11-kV, star-connected alternator has resistance of 1.5 W per phase and synchronous reactance of 25 W per phase. Find the percentage regulation for a load of 600 kW at 0.8 leading power factor. [Ans. –7.65 %] D-83. A three-phase, 400-V, 10-kW, star-connected synchronous motor has per phase synchronous impedance of (0.4 + j3.0) W. Determine the angle of retard and the voltage to which the motor must be excited to give a full-load output at 0.8 pf load to be 85 %. [Ans. 12°, 268.3 V] D-84. The input current to a three-phase, 11-kV, 50-Hz, star-connected synchronous motor is 60 A at a pf of 0.8 leading. It has an effective resistance of 1 W

806

D-85.

D-86.

D-87.

D-88.

Basic Electrical Engineering

and synchronous reactance of 30 W. Find the power supplied to the motor and the induced emf. [Ans. 914.5 kW, 13 041 V] A three-phase, 400-V, 50-Hz line supplies power to a three-phase, star-connected induction motor which runs at almost 1000 rpm on no load and at 960 rpm on full load. The motor has stator to rotor turns ratio of 1 : 0.4. Find (a) the number of poles, (b) the slip at full load, (c) the frequency of rotor emf, (d) the speed of rotor mmf with respect to the rotor, (e) the speed of rotor mmf with respect to the stator, (f) the speed of rotor mmf with respect to the stator mmf, (g) the rotor induced emf at standstill, and (h) the rotor induced emf while the motor is running on full load. [Ans. (a) 6; (b) 0.04; (c) 2 Hz; (d) 40 rpm; (e) 1000 rpm; (f) 0 rpm; (g) 92.4 V; (h) 3.69 V] A 6-pole, 410-V, 50-Hz, three-phase induction motor has a speed of 950 rpm on full load. Calculate the slip. How many complete alternations will the rotor-voltage make per minute ? [Ans. 5 %, 150 c/min] A three-phase, 12-pole generator driven at a speed of 500 rpm, supplies power to an 8-pole, threephase induction motor. If the slip of the motor is 3 % on full load, calculate the full-load speed of the motor. [Ans. 727.5 rpm] A three-phase, 400-V, 6-pole, 50-Hz induction motor develops 5 hp at 950 rpm. What is the stator

[Ans. 4226.3 W; 88.27 %] D-89. The power input to a three-phase induction motor is 60 kW. The stator losses total to 1 kW. Find the total mechanical power developed and the rotor copper loss per phase, if the motor is running at 3 % slip. [Ans. 57.231 kW; 590 W] D-90. A three-phase, 4-pole, 400-V, 50-Hz, 30-hp inducpower factor of 0.85 lagging. Calculate the current drawn by the induction motor from the mains. [Ans. 47.5 A] D-91. In a certain power station, the alternator is a 6-pole machine driven at 1000 rpm. A three-phase, 140-hp, 16-pole induction motor supplied from this alternator runs with a slip of 2.5 % . What is the actual speed of the motor ? If the motor delivers full

load at a speed of 360 rpm, what is the percentage slip of the motor ? [Ans. 365.6 rpm, 4 %] D-92. A three-phase, 6-pole, 400-V, 50-Hz induction motor develops 3.85 kW (including friction and windage losses) at a speed of 950 rpm. If the stator [Ans. 4.313 kW] D-93. A three-phase, 6-pole induction motor develops maximum torque when running at 1000 rpm and fed from a 60-Hz supply. The rotor resistance per phase is 1.1 W. Calculate the speed at which the motor will develop maximum torque when operated from a 50-Hz supply. [Ans. 800 rpm] D-94. An 8-pole, three-phase alternator is coupled to an engine running at 750 rpm. The alternator supplies power to an induction motor which has a full-load speed of 1425 rpm. Find the percentage slip and the number of poles on the motor. [Ans. 5 %, 4] D-95. A 110-V, 6-pole, lap-wound, dc shunt generator delivers a load current of 50 A. The armature resistance is 0.2 W is 55 W. The armature has 360 conductors and is rotating at 1800 rpm. Calculate (a) the no load voltage at the armature, and (b [Ans. (a) 120.4 V; (b) 11.148 mWb] D-96. The armature of a 4-pole, lap-wound dc shunt generator has 120 slots with 4 conductors per resistance is 0.05 W is 50 W. Find the speed of the machine when supplying 450 A at a terminal voltage of 250 V. [Ans. 681.88 rpm] D-97. A 4-pole, dc generator has a total of 500 conductors 0.01 Wb crossing the air gap with normal excitation. (a) What voltage will be generated at a speed of 900 rpm, if the armature is (i) wave-wound, (ii) lap-wound. (b) If the allowable current is 5 A per path, what will be the power in kW generated in each case ? [Ans. (a) (i) 150 V, (ii) 75 V; (b) (i) 1.5 kW, (ii) 1.5 kW] D-98. A 20-kW, 440-V, short-shunt compound, dc

W, 0.25 W and 240 W, respectively, calculate the combined bearing, windage and core

Supplementary Exercise—Part D : Electrical Machines loss of the machine. [Ans. D-99. A dc shunt generator load through a pair of feeders of total resistance of 0.05 W resistance are 0.1 W and 100 W the terminal voltage and the generated emf of the generator. [Ans. D-100. A dc shunt motor, having an armature resistance of 0.02 W W, draws a current of 200 A at 120 V from dc mains. Find the back emf induced. If the lap-wound armature has

motor runs.

when the input current is 100 A. [Ans. D-106. A 230-V, dc shunt motor runs at 1000 rpm on full load, drawing a current of 10 A. The armature resistance is 0.5 W is 230 W. Calculate the value of the resistance to be inserted in series with the armature so that the Ans. 1.253 W D-107. A series dc motor takes 20 A at 400 V and runs at W and 0.4 W voltage and the current taken to run the motor at 350 rpm, if the torque required varies as the square of the speed. [Ans.

[Ans.

D-101.

807

D-108. motor takes a current of 300 A at a speed of 400 rpm. The number of armature turns is 500 and

Ra = 0.7 W

windage, friction and iron losses can be assumed as 2.5 per cent. Calculate (a the armature, (b) the shaft torque, and (c) the shaft

A load torque is then applied to the motor shaft which causes Ia to rise to 40 A and the speed to fall

b (c D-102. A 120-V, dc shunt motor has an armature resistance of 0.2 W W. It runs at

[Ans. D-109. A 200-V, dc series motor runs at 1000 rpm and takes 20 A current. The combined resistance of the W. Find the resistance to be inserted in series so as to reduce the

[Ans. (a

Find the speed of the motor when it is operating at half the full load. [Ans. D-103. of 250 W and an armature resistance of 0.5 W running on no load, it takes 5 A from the lines and its speed is 1500 rpm. Calculate its speed on fullload when taking 50 A from the lines. [Ans. D-104. A 220-V, dc shunt motor runs at 1000 rpm when it A. The resistance of the armature-circuit is 0.3 W 220 W. Calculate the additional resistance required in the armature circuit to reduce the speed of the motor to 750 rpm, assuming that the motor then Ans. 2.215 W D-105. A 500-V, dc shunt motor takes 4 A on no load when running at 1000 rpm. The armature resistance (including that of the brushes) is 0.2 W current is 1 A. Estimate th

as the square of the speed. [Ans. 4.42 W D-110. A 4-pole, 240-V, lap-wound, dc series motor has W and 0.25 W

[Ans. D-111. A 240-V, dc series motor takes 40 A when giving its rated output at 1500 rpm. Its resistance is 0.3 W. Calculate the value of the resistance that must be added to obtain rated torque (a) at starting, and (b) at 1000 rpm. [Ans. (a) 5.7 W; (b W D-112. A dc shunt machine connected to 250 V mains has an armature resistance (including the brushes) of 0.12 W W. Find the ratio of the speed of the generator to the speed of [Ans.

808 ( C )

Basic Electrical Engineering

C HA L L EN G IN G

PROBLEMS

D-113. A single-phase transformer takes 10 A on no load at a power factor of 0.1 lagging. Its turns ratio is

current and power factor. Neglect the internal voltage drops in the transformer.

terminal voltage of full load, when the power factor is (a (b [Ans. (a (b D-120. Open-circuit and short-circuit tests on a 4-kVA, give the following results :

[Ans. 57.25– D-114.

OC Test SC Test

pf pf (lagging). [Ans.

pf lagging. [Ans. D-121. The test data on a single-phase, 250-V/500-V,

D-115. OC Test SC Test

pf (lagging). [Ans. D-116. In a 25-kVA, 2000-V/200-V transformer, the

[Ans.

(a b) Find the output power to obtain pf. (c answer in (b) above differ, if the pf is (i or (ii [Ans. (b c) (i) Same, (ii D-122. The following test data is obtained for a 5-kVA, 220-V/440-V, single-phase transformer : OC Test

D-117.

SC Test

a pf lag.

(b

Find iron loss and full-load copper loss of the

[Ans. (a [Ans.

b

D-123. a three-phase, star-connected alternator in which a

D-118. transformer has (i pf, and (ii pf a) the iron and copper losses at full load, (b) the load at which the c pf lagging. [Ans. (a b (c D-119. transformer gives the following readings on SC

of 250 A on short circuit and a generated emf of 1500 V on open circuit. The armature resistance is 2W pf lagging is switched off. [Ans. W W D-124. A three-phase, 2-pole, turbo-generator with starper pole per phase with 4 conductors per slot. The generator generates 3.3 kV between the lines when run at the rated speed. The stator winding is short-

809

Supplementary Exercise—Part D : Electrical Machines calculate (a) the distribution factor, (b) the pitch factor, and (c [Ans. (a b D-125. produces an armature current of 200 A on short circuit and 50 V on open circuit. The armature resistance is 0.1 W. To what voltage must the alternator be excited, if it delivers a load of 100 A pf lagging, with a terminal voltage of 200 V ? [Ans.

and (d emf per minute. [Ans. (a

(d D-131. The input power to the rotor of a three-phase, The rotor emf is found to be making 120 complete alternations per minute. Calculate (a) the slip, (b) the rotor speed, (c) the rotor copper loss per phase, (d) the total mechanical power developed, and (e) the rotor resistance per phase, if the rotor

D-126. has armature resistance of 1 W reactance of 20 W. Calculate the regulation for a pf of (a b and (c [Ans. (a b c D-127. chronous motor has an effective winding resistance of 0.5 W

b

[Ans. (a

b (d

e

W

D-132. squirrel-cage induction motor draws a starting a) The line on line. (b) The line and phase currents when started with an auto-transformer with a tapping of 55 percent. (c) The line and phase currents, when

power. If iron and mechanical losses amount to 500 a) the armature current, (b) the power supplied to the motor, and (c [Ans. (a) 14.5 A; (b

[Ans. (a) 43.3 A; (b (c) 14.4 D-133. (a having 4 poles and a lap-wound armature with

D-128. impedance of (0.5 + j W operating with a certain load and a certain value of

W and 300 W, b) Find the terminal voltage and the [Ans. (a

b

the new pf and mechanical torque developed. [Ans.

D-134.

motor runs at 1440 rpm with a load torque of

a) the total current, (b) the emf, and (c) the power generated in the armature; if the armature is (i) wave-wound, (ii) lap-wound. [Ans. (i) (a) 120 A; (b c (ii) (a) 240 A; (b) 150.4 V; (c D-135. A 4-pole, 250-V, long-shunt compound, dc generator

D-129.

the rotor copper loss.

[Ans.

D-130.

(b

a) the rotor copper losses, c) the line current,

are 0.1 W, 0.15 W and 250 W armature is lap-wound with 50 slots, each slot

810

Basic Electrical Engineering

calculate the speed of the generator. [Ans. D-136. A 440-V, compound generator has armature, W, 1 W and 200 W voltage while delivering 40 A to the external circuit for (a) the long-shunt, and (b) short-shunt connections. [Ans. (a) 503.3 V; (b D-137. A short-shunt compound, dc generator supplies a current of 100 A at a voltage of 220 V. The resistances 50 W, 0.025 W and 0.05 W brush drop is 2 V, and the iron and friction losses a) the generated emf, (b) the copper losses, (c) the output of the primemover driving the generator, and (d) the generator [Ans. (a b (c d D-138. A dc long-shunt compound generator has an armature resistance of 0.03 W resistance of 0.04 W 55 W. It supplies a load at 110 V through a pair of feeders of total resistance 0.04 W. The load consists a) the load current, (b) the terminal voltage, and (c) the emf generated. [Ans. (a b c D-139. The open-circuit characteristic (OCC) of a dc generator at 1000 rpm is given below: Field current, : 0.5 1.0 1.5 2.0 2.5 3.0 3.5 If (A) Generated emf, Eg The machine is connected as a dc shunt generator and is driven at 1000 rpm. The armature resistance W W, a) the open-circuit voltage, (b c) the terminal voltage when a load of 4.0 W is connected, and (d) the load current for which the terminal voltage is 100 V. [Ans. (a) 150 V; (b) 120 W; (c d D-140. motor has 450 conductors on its armature and

W and 200 W, a) the speed, (b) the armature torque, and (c) the useful torque. [Ans. (a b) 1055.5 Nm; (c D-141. A 220-V, dc shunt motor draws 4.5 A on no load and runs at 1000 rpm. The resistances of the armature W and 157 W Calculate the speed when loaded and drawing a current of 30 A. Assume that the armature reaction Ans. D-142. A dc shunt motor draws 230 A at 110 V when running at a speed of 450 rpm. It has an armature resistance of 0.03 W 45 W. (a) b at which it would have to run as a generator to give 140 A at 200 V. [Ans. (a b D-143. A 220-V, dc shunt motor has an armature resistance of 0.05 W W.

motor runs, if (a) a 1.5-W resistor is connected in series with the armature, and (b) a 30-W resistor is torque remains the same throughout. [Ans. (a b D-144. A 20-hp, 230-V, 1150-rpm, 4-pole, dc shunt motor

of 0.2 W current of 3.0 A. Calculate (a (b) the torque developed, (c) the rotational losses, and (d) the total losses expressed as a percentage of power. [Ans. (a b (c d D-145. A 250-V, dc shunt motor has an armature resistance of 0.5 W At full load, the motor runs at 400 rpm and takes an armature current pf 30 A. If a resistance of 1 W a) the

Supplementary Exercise—Part D : Electrical Machines speed at full-load torque, (b) the speed at double the full-load torque, (c) the stalling torque in terms of full-load torque. [Ans. (a) 348.9 rpm; (b) 272.3 rpm; (c) 5.56 times full-load torque] D-146. A 4-pole, 250-V, dc shunt motor has a lap-connected

811

0.12 W and 125 W, respectively. The rotational losses amount to 825 W. Calculate (a) the torque developed by the armature, (b) the useful torque when the current taken by the motor is 30 A. [Ans. (a) 85.6 Nm; (b) 75.34 Nm]

812

Basic Electrical Engineering

Supplementary Exercise—Part E : Miscellaneous

S U P P L E M E N TA R Y E X E R C I S E S E.1 Solved Problems E.2 Practice Problems

PA R T

813

E

E : MISCELLANEOUS Assemblage of

N OT E This set of exercises provides practice to those who wish to attain a higher standard of learning the basic principles of Electrical Engineering. The real key to success is practice.

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Basic Electrical Engineering

Supplementary Exercise—Part E : Miscellaneous

815

E. 1. SOLVED PROBLEMS P RO B L EM

E- 1

The capacitors in the circuit shown in Fig. E-1 are initially uncharged. Find the indicated voltages and currents at t = 0+ closes.

Fig. E-1 Solution Since the voltages across capacitors cannot change suddenly, we have + – v1 (0 ) = v1 (0 ) = 0 V

and

v4 (0+) = v4 (0–) = 0 V

+ + Further, with 0 V across them, the capacitors act like short circuits at t = 0 . The equivalent circuit at t = 0 is as shown in Fig. E-2a. It means that the 100-V of the source appears across both the 25-W and 50-W resistors. Therefore, + – v2 (0 ) = v3(0 ) = 100 V

The equivalent circuits. Obviously, the currents at t = 0+ are given as 0 100 100 = 0 A; i2(0+) = i3(0+) = = 4 A and i 4 (0+) = =2A 10 25 50 A long time after the switch closes, the capacitors are fully charged and their voltages are constant; they act like open circuits. The equivalent circuit at t = is as shown in Fig. E-2b. It means that i2( ) = i4 ( ) = 0 A. The 10-W and 25-W resistors are seen in series across 100-V source. Hence, the currents, i1 (0+) =

i1( ) = i3( ) =

100 = 2.86 A 10 + 25

816

Basic Electrical Engineering

Now, the voltages are given as v1( ) = i1( ) ¥ 10 = 2.86 ¥ 10 = 28.6 V; v3 ( ) = i4 ( ) ¥ 50 = 0 ¥ 50 = 0 V; PROBLE M

v2 ( ) = i3 ( ) ¥ 25 = 2.86 ¥ 25 = 71.5 V v4 ( ) = 100 – v3 ( ) = 100 – 0 = 100 V

E -2

In the circuit shown in Fig. E-3a, the capacitor is charged to 100 V. The switch closes at t = 0 s. Find vC and i for t > 0 s.

Fig. E-3 v and i are vC (0+), vC ( ), i(0+), i( ) and t = RThC. Because Solution the capacitor voltage cannot jump, we have vC (0+) = 100 V. A long time after the switch is put on, the capacitor gets fully charged and the current iC reduces to zero because the capacitor acts like an open circuit, as shown in Fig. E-3b. The voltage vC ( ) across the capacitor is the same as the voltage across the 60-W resistor. So, by voltage division, 60 vC ( ) = 300 ¥ = 180 V 60 + 40 300 =3A 40 + 60 v(0+), we write a node voltage equation for the middle top node, i( ) =

We can easily get i(0+) from v(0+

vC (∞) − 300 vC (∞) vC (∞) − 100 + + =0 40 60 16 Solving the above equation, we get v(0+) = 132 V. Therefore, i(0+) = 132/60 = 2.2 A. Now, the Thevenin resistance at the capacitor terminals is 16 + 60||40 = 40 W. Therefore, the time constant for charging of the capacitor is –3

t = RThC = 40 ¥ 2.5 ¥ 10

= 0.l s

We can now write the formulas for vC and i, +

and PROBLEM

–t/t

–10t

–10t

= 180 – 80 e vC (t) = vC ( ) + [vC (0 ) – vC ( )]e = 180 + (100 – 180) e i(t) = i( ) + [i(0+) – i( )] e–t/t = 3 + (2.2 – 3) e–10t = 3 – 0.8 e–10t A

V

E-3

In the circuit shown in Fig. E-4 the capacitor is initially uncharged and the switch is closed at t = 0 s. Find the current i for t > 0 s.

Supplementary Exercise—Part E : Miscellaneous

817

Fig. E-4 Solution For writing the expression of i for t > 0 s, we need i(0+), i( ) and t. At t = 0 s, the uncharged capacitor acts as a short circuit, and hence the equivalent circuit becomes as shown in Fig. E-5a. Obviously, the 20-mA current does not affect i(0+) Furthermore, the 60-kW and 6-kW resistors are placed in parallel. The current i1(0+) is given as i1(0+) =

100 100 V = = 2.2 mA Req ( 40 + 60 || 6) kΩ

By current division, we get 6 ⎞ 6 + + ⎛ = 0.2 mA i(0 ) = i1(0 ) ⎜ ⎟ = (2.2 mA) ¥ ⎝ 60 + 6 ⎠ 66

(a)

i

v2 20 mA

60 kW

40 kW

2.5 mA

20 mA

i

46 kW

v1

20 kW

40 kW v2

6 kW

20 kW

60 kW

40 kW v1

100 V

i1

(b)

Fig. E-5 After 5 time-constants, the capacitor no longer conducts current. It acts as an open circuit. The 6-kW and 40-kW resistors come in series. Transforming the 100-V source in series with 40-kW resistor into a 2.5-mA [= (100 V)/(40 kW)] source in parallel with 40-kW resistor, we get the circuit as shown in Fig. E-5b. Writing nodal analysis equations, by inspection, we get

ÈÊ 1 1 1ˆ ÍÁË 40 + 60 + 46 ˜¯ Í Í Ê 1ˆ Í- Á ˜ Î Ë 46 ¯

˘ ˙ Èv1 ( ) ˘ È2.5 ˘ ˙ Í ˙ ˙Í ˙=Í Í ˙ ˙ 1 ˆ Ív ( )˙ Í- 20˙ Ê 1 + ˙ ÁË 20 46 ˜¯ Î 2 ˚ ˚ Î ˚ Ê 1ˆ -Á ˜ Ë 46 ¯

Note that, for convenience, we have taken volt-milliampere-kilohm units. Using calculator, the above equations can be solved to give v1( ) = – 62.67 V. So, i( ) =

v1( ) = 60

62.67 = – 1.04 mA 60

818

Basic Electrical Engineering

Now, Thevenin resistance across the capacitor terminals is RTh = (6 + 40 || 60) || (40 + 20) = 20 kW t = RThC = (20 kW) (50 mF) = 1 s

so that

+ Having known i(0 ), i( ), and t, we can now write the expression for the current i as +

–t/t

i = i( ) + [i(0 ) – i( )] e PROBL EM

–t/1

= – 1.04 + [0.2 – (– 1.04)] e

–t

= – 1.04 + 1.24 e mA

E-4

Find the voltage across the inductor in the circuit of Fig. E-6a, after the switch is opened at t = 0 s.

Fig. E-6 Solution The current in an inductor cannot change abruptly. Hence, we must have iL(0+) = iL(0–). To determine iL(0–), we note that the switch had been closed for a time long enough to establish steady state. Hence, it acts as a short circuit. Furthermore, the closed switch short circuits the 50-W resistor, and the closed switch and the inductor short circuits the 150-W resistor. Thus, the current from the battery is (10 V)/(100 W) = 0.1 A, and this is the current iL(0–) through the closed switch and the inductor. Hence, iL(0+) = iL(0–) = 0.1 A. But, this does not give us the initial value of vL(t). For vL(0+), we resort to nodal analysis as follows. Immediately after the switch is opened, the inductor acts as current source of 0.1 A. Therefore, the equivalent circuit at t = 0+ is as shown in Fig. E-6b. Note that we have not shown the switch in this equivalent circuit as it is now open. Writing nodal equation for the node a, we get

v a − 10 v a + = – 0.1 fi va = 0 V 100 150 vL(0+) = – (50 W) (0.1 A) = – 5 V

Therefore,

RTh. As can be seen from the circuit of Fig. E-16, RTh = 50 + 150 || 100 = 110 W t = L/RTh = 0.1/110 = 909 ms

\

vL (0+) = vL( ) + [vL(0+) – vL ( )] e–t/t = 0 + [(– 5) – 0] e–t/909ms = – 5 e–t/909ms V PROBL EM

E-5

Find the time constant of the circuit shown in Fig. E-7, andtermine the energy stored in the inductor.

Supplementary Exercise—Part E : Miscellaneous

50 kW

819

14 kW

100 V

20 kW

30 kW

75 kW

150 kW

50 mH

Fig. E-7 Solution Thevenin resistance of the circuit across the inductor terminals, RTh = (50 + 30) || 20 + 14 + 75 || 150 = 80 kW Hence, the time constant of the circuit, –3

3

t = L/RTh = (50 ¥ 10 )/(80 ¥ 10 )s = 0.625 ms W resistor. Thevenin’s voltage is the voltage across the 20-kW resistor. By voltage division, we have 20 = 20 V 20 + 50 + 30 Since the inductor behaves as a short circuit, the inductor current is VTh 20 V = = 0.25 mA IL = RTh + 0 80 k VTh = (100 V) ¥

Therefore, the stored energy in the inductor, 1 2 1 W = LI L = ¥ 50 ¥ 10–3 ¥ (0.25 ¥ 10–3)2 J = 1.56 nJ 2 2 PROBLEM

E-6

In the circuit shown in Fig. E-8a, the switch has been in position 1 for a long time. (a) Find the indicated currents. (b) Find the indicated currents and voltage immediately after the switch has been thrown to position 2. 1 6.8 W 140 V

i2

18 W

18 W

v

2 12 W

6W

i3

50 V

+

20 W

2H i1

6W

+ 20 W

v –

(a)

i1

50 V

v

4A

–

(b)

Fig. E-8 Solution (a) The inductor acts as a short circuit, and shorts out the 20-W resistor. As a result, the current through 20-W resistor, i1 = 0 A. This short circuit also places the 18-W resistor in parallel with the 12-W resistor. Thus, the total resistance across the 140-V source is R = 6.8 + 12 || 18 = 14 W

820

Basic Electrical Engineering

Hence, the current drawn from the 140-V source, i =

140 = 10 A 14

By current division, we get 12 18 =4A and i3 = 10 ¥ =6A 12 + 18 12 + 18 (b) As soon as the switch leaves position 1, the left-hand side of the circuit is isolated. It becomes just a series circuit in which the current, 140 i3 = = 7.45 A 6.8 + 12 In the right-hand part of the circuit, the current in the inductor cannot suddenly change, and is 4 A, as it was found above. Hence, i2 = 4 A. Since this is a known current, it can be considered to be from a current source, as shown in Fig. E-8b. By nodal analysis, v v − 50 + + 4 = 0 fi v = – 20.9 V 20 6 + 18 i2 = 10 ¥

\

i1 =

P ROBLEM

v = 20

20.9 = – 1.05 A 20

E - 7

W. Calculate the resistance required (a) in parallel to enable the instrument to read up to 1 A, (b) in series to enable it to read up to 10 V.

Solution

Here, Im = 15 mA = 0.015 A, Rm = 5 W

Vm = Im Rm = (15 ¥ 10–3) ¥ 5 = 75 mV (a) The current through the shunt, Ish = Ifsd – Im = 1 – 0.015 = 0.985 A Vm 0.075 The resistance of the shunt required, Rsh = = = 0.07614 W Ish 0.985 (b) The required full-scale voltage, Vfsd = 10 V The voltage drop across the series resistor, Vs = Vfsd – Vm = 10 – 0.075 = 9.925 V Vs 9.925 \ The series resistance, Rs = = = 661.7 W Im 0.015 P ROBLEM

E-8

A voltage of 100 V is applied to a circuit comprising two 50-kW resistors in series. A voltmeter, with an fsd of 50 V W/V, is used to measure voltage across the 50-kW resistors. Calculate the percentage error in this measurement.

Solution

By potential division, the true value of the voltage across the 50-kW resistor (when the voltmeter is not

connected), Vt = 100 ¥

50 kΩ = 50 V (50 + 50) kΩ

The resistance of the voltmeter, Rm = (50 V) ¥ (1 kW/V) = 50 kW When the voltmeter is connected in the circuit, it shunts the 50-kW resistor. The equivalent resistance of this parallel combination, Re = (50 kW) || (50 kW) = 25 kW

Supplementary Exercise—Part E : Miscellaneous

821

The network effectively changes into an equivalent resistance of 25 kW in series with a 50-kW resistor. The voltmeter indicates the voltage that exists across the equivalent resistance of 25 kW, Vi = (100 V) ¥ \ The error,

Vi – Vt = 33.3 – 50 = – 16.7 V

and the percentage error,

PROBLEM

25 kΩ = 33.3 V (25 + 50) kΩ

Eroor ¥ 100 = Vt

16.7 ¥ 100 = – 33.4 % 50

E -9

A uniform potentiometer wire AB is 4 m long and has a resistance of 8 W. End A is connected to the negative terminal of a 2-V cell of negligible internal resistance, and end B is connected to the positive terminal. An ammeter of resistance 5 W has its negative terminal connected to A and its positive terminal to a point on the wire 3 m away from A. What current will the ammeter indicate ?

Solution

The arrangement described in the Problem is shown in Fig. E-9a, and its electrical-circuit equivalent is shown in Fig. E-9b.

Fig. E-9 The resistance RAC between end A and point C of the wire is given as RAC = RAB ¥ \

3 AC =8¥ =6W AB 4

RCB = RAB – RAC = 8 – 6 = 2 W

The net resistance connected across the 2-V cell, Rt = RCB + RAC || Ra = 2 + 6 || 5 = 4.73 W Therefore, the current I supplied by the cell, I=

V 2 = = 0.432 A Rt 4.73

By current division, the current through the ammeter, RAC 6 I1 = I ¥ = 0.423 ¥ = 0.231 A 6+5 RAC + Ra

822

Basic Electrical Engineering

P RO B L EM

E-1 0 W W having a negligible

reads correctly at 15 °C, determine the percentage error when the temperature is 25 °C.

Solution

The connections for a moving-coil instrument to work as an ammeter is shown in Fig. E-10. To extend its range, a shunt with resistance Rsh is put across the instrument. The shunt is made of a material such as manganin (alloy of copper, manganese and nickel) Also, a ‘swamping’ resistor, with resistance Rsw is put in series with the moving-coil instrument. It is also made of a

Rm Im

Ifsd

Rsw

Ish Shunt, Rsh

Ifsd

Moving-coil instrument as an ammeter.

resistance. The resistance Rsw is usually three times the meter resistance Rm, thereby reducing a possible error of, say, 4 % to about 1 %, due to variation in temperature.

At a temperature of 15 °C : Rm1 = 1.5 W; Rsw = 3.5 W Im = 15 mA; Ifsd = 20 A. Rms1 = Rm1 + Rsw = 1.5 + 3.5 = 5.0 W; Ish = Ifsd – Im = 20 – 0.015 = 19.985 A \

Rsh =

Im Rm1 0.015 5 = = 0.003753 W Ish 19.985

At a temperature of 25 °C : The resistance of the moving-coil increases and is calculated as follows : Rm1 = R0 (1 + at1) and Rm2 = R0(1 + at2) 1 + (1/ 234.5) 25 (1 + α t2 ) \ Rm2 = Rm1 = 1.5 ¥ = 1.56 W 1 + (1/ 234.5)15 (1 + α t1) Rms2 = Rm2 + Rsw = 1.56 + 3.5 = 5.06 W W to 5.06 W. The shunt resistance Rsh remaining the same, the current going to the meter branch reduces to Rsh 0.003753 Im2 = Ifsd = (20 A) ¥ = 0.014823 A Rsh + Rms2 0.003753 + 5.06 \

percentage error =

0.014823 0.015 ¥ 100 = – 1.18 % 0.015

E. 2. PRACTICE PROBLEMS ( A )

S I M P L E

PR O BL EMS

E-1. A 2-mF capacitor, initially charged to 300 V, is discharged through a 270-kW resistor. What is its voltage 0.25 s after the capacitor starts to discharge ? [Ans. 189 V]

E-2. A series combination of an uncharged 0.1-mF capacitor and a 2-MW resistor is connected across a 200-V source through a switch. Find the capacitor voltage and current at 0.1 s after the switch is closed. [Ans. 78.7 V, 60.7 mA]

Supplementary Exercise—Part E : Miscellaneous E-3. required for the capacitor voltage to reach 50 V. increase another 50 V, from 50 V to 100 V. [Ans. 57.5 ms, 81.1 ms] E-4. Find the time constant of the circuit shown in Fig. E-11. [Ans. 60 ms] 60 W

6V

capacitor 50 mJ ? (c) At what time is the power into the capacitor 1 W ? [Ans. (a) 0.4 t A for 0 < t < 0.1 s, 0 elsewhere; (b) 0.0841 s; (c) 0.063 s] iC(t) Vs + –

6W

40 W

C = 20 mF

Fig. E-14

2 mF

0.1 A

823

E-8. To move the bright spot of a CRT smoothly across plates must be increased linearly, as shown in Fig. E-15 If the capacitance of the plates is 1 pF,

Fig. E-11 E-5. The switch in the circuit shown in Fig. E-12 is closed at t a) the voltage vL across the inductor, and (b) the initial rate of increase of the current i(t). [Ans. (a) 10 e–t/0.4 V; (b) 5 A/s] 5W

t=0

[Ans. 4 mA when increasing, – 20 A when decreasing]

v (V)

Drops in 1 ps

20

+ vR – + 10 V

i(t)

vL 2 H

0 Rises in 5 ms

t

–

Fig. E-15 Fig. E-12 E-6. A 50-mH inductor is connected to a current source of is(t) = 0, for t < 0, and of is(t) = 150t3 A, for t > 0. Calculate the voltage across the inductor, vL(t), with the polarity shown in Fig. E-13. [Ans. 22.5t2 V, for t > 0]

+ is(t)

vL(t)

L = 50 mH

–

Fig. E-13 E-7. Figure E-14 shows a 20-mF capacitor that has a voltage vs(t) = 0 V, t < 0; vs(t) = 104t2 V, 0 < t < 0.1 s; vs(t) = 100 V, t > 0.1 s. (a) Find the current, iC (t). (b) At what time is the energy in the

E-9. A 0.1-mF capacitor, initially charged to 230 V, is discharged through a 3-MW resistor. (a) Find the capacitor voltage 0.2 s after the capacitor starts to discharge. (b) How long does it take the capacitor to discharge to 40 V ? [Ans. (a) 118 V; (b) 0.525 s] E-10. A current i = 0.32t inductor. Find the energy stored at t = 4 s. [Ans. 0.123 J] E-11. Closing a switch connects in series a 20-V source, a 2-W resistor and a 3.6-H inductor. How long does it take the current to get to its maximum value, and what is this value ? [Ans. 9 s, 10 A] E-12. A short is placed across a coil carrying a current of 0.5 A at that instant. If the coil has an inductance of 0.5 H and a resistance of 2 W, what is the coil current 0.1 s after the short is applied ? [Ans. 0.335 A] E-13. with 15 mA and has a resistance of 5 W. Calculate

824

Basic Electrical Engineering

the resistance of the necessary components in order that the instrument may be used as (a) a 2-A ammeter, and (b) a 100-V voltmeter. [Ans. (a) Rsh = 0.0378 W; (b) Rs = 6662 W] E-14. A moving-coil milliammeter has a coil of resistance 15 W of 5 mA. This instrument is to be adapted to operate (a 100 V, and (b) as an ammeter with a full-scale nents to be used to meet the above requirements. [Ans. (a) Rs = 19.985 kW; (b) Rsh = 0.0376 W] E-15. A moving-coil galvanometer of resistance 5 W, gives full-scale reading when a current of 15 mA

( B)

T R IC KY

passes through it. Explain how its range could be altered so as to read up to (a) 5 A, and (b) 150 V. [Ans. (a) Rsh = 0.015045 W; (b) Rs = 9.995 kW] E-16. read the rms value of a sinusoidal voltage, by what factor must the scale readings be multiplied when it is used to measure the rms value of (a) a squarewave voltage, and (b) a voltage having a form factor of 1.15 ? [Ans. (a) 0.9; (b) 1.036] E-17. A dc voltmeter has a resistance of 28 600 W. When connected in series with an external resistor across a 480 V dc supply, the instrument reads 220 V. What is the value of the external resistor ? [Ans. 33.8 kW]

PRO BLE MS

E-18. How long does it take a 10-mF capacitor charged to 200 V to discharge through a 160-kW resistor, and what is the total energy dissipated in the resistor ? [Ans. 8 s, 0.2 J] E-19. t = 0 s connects in series a 150-V source, a 1.6-kW resistor, and the parallel combination of a 1-kW resistor and an uncharged 0.2-mF capacitor. Find (a) the initial capacitor current, (b W resistor currents, (c (d) the time required for the capacitor to reach its [Ans. (a) 93.8 mA; (b) 0 A, 57.7 mA; (c) 57.7 V; (d) 0.615 ms] E-20. Repeat Prob. E-19 for a 200-V source and an initial capacitor voltage of 50 V opposed in polarity to that of the source. [Ans. (a) 43.8 mA; (b) 50 mA, 76.9 mA; (c) 76.9 V; (d) 0.615 ms] E-21. The switch in the circuit shown in Fig. E-16 is closed at t = 0 s. Determine the current i(t) through the capacitor. [Ans. 0.183e–t/1.1 ms]

Fig. E-16

E-22. Assuming that the switch in the circuit shown in Fig. E-17 has been in position a for a long time before it is switched to position b at t vL(t), the voltage across the inductor with the polarity shown. [Ans. –10 e–t/1.5 ms V]

Fig. E-17 E-23. A current source is placed in series with a resistor and an inductor, as shown in Fig. E-18. During this period, the switch is open. The switch is closed at t = 0 s, so that the circuit is separated into two independent loops that share a common short circuit but do not interact. Find the current in the right-hand loop after the switch closes. [Ans. 10 e–t/0.02 ms mA]

Fig. E-18

Supplementary Exercise—Part E : Miscellaneous

825

E-24. For the circuit shown in Fig. E-19, assume that the switch has been in position a for along time and then is changed to position b at t = 0. Find the current i(t). [Ans. 1e–t/0.1 ms A]

Fig. E-22

Fig. E-19 E-25. The circuit shown in Fig. E-20 has attained a steady state with the switch closed. The switch is now opened at t 5-kW resistor after the switch is opened. [Ans. 3.35 + 6.65e–t/0.067 ms V]

E-28. After being closed for a long time, the switch in the circuit of Fig. E-23 is opened at t = 0. (a) Find the voltage across the switch as a function of time for t > 0. (b) In what period of time is half the initial stored energy in the capacitor lost to the resistor ? [Ans. (a) vsw (t) = 10(1– e–t/6 ms)V; (b) 2.08 ms]

Fig. E-23

Fig. E-20 E-26. The switch in the circuit of Fig. E-21 has been in position a for a long time. It is moved to b at t = 0, and then to c at t vC across the capacitor for 0 < t < 30 ms. [Ans. 103 t V, for 0 < t < 10 ms; 10 e–(t–10)/5 ms V, for 10 ms < t < 30 ms]

10 mA

a

c b

500 W

+

vC

10 mF

–

Fig. E-21 E-27. The switch in the circuit of Fig. E-22 is closed at t = 0. Find the time at which the power into the inductor is maximum. [Ans. 0.347 s]

E-29. A capacitor is charged to 100 V and then disconnected. Leakage reduces its voltage to 35 V in 45 min. Estimate the additional time required for the voltage to drop to 8 V. [Ans. 63 min 16 s] E-30. Closing a switch connects in series a 21-V source, a 3-W resistor and a 2.4-H inductor. Find (a) the initial b and (c) the initial rate of current increase. [Ans. (a) 0 A, 7 A; (b) 21 V, 0 V; (c) 8.75 A/s] E-31. A coil for a relay has an inductance of 2 H and a resistance of 30 W. If the relay requires a current of 250 mA to operate, how soon will it operate after 12 V is applied to the coil ? [Ans. 65.4 ms] E-32. with 15 mA and has a resistance of 5 W. (a) Calculate the resistance of the necessary components in order that the instrument may be used as a 15-A ammeter. (b to be made of manganin strip having a resistivity of 0.5 mW m, as thickness of 0.6 mm and a length of 50 mm, calculate the width of the strip. [Ans. (a) 5005 mW; (b) 8.325 mm]

826

Basic Electrical Engineering

E-33. The coil of a moving-coil instrument is wound with 40½ turns. The mean width of the coil is 4 cm and

torque in Nm when the coil is carrying a current of 10 mA. [Ans. 81 mNm] E-34. A rectangular moving coil of a milliammeter is wound with 30½ turns. The effective length of the magnetic controlling torque of the hairspring is 0.5 ¥ 10–6 Nm/ Ans. 51.2 mA] E-35. The resistance of a coil is measured by the ammetervoltmeter method. With the voltmeter connected ( C )

C HA L L E NGI NG

across the coil, the readings on the ammeter and voltmeter are 0.4 A and 3.2 V, respectively. The resistance of the voltmeter is 500 W. Calculate (a) the true value of the resistance, (b) the percentage error had the resistance of the milliammeter been neglected. [Ans. (a) 8.13 W ; (b) – 1.6 %] E-36. A voltmeter is connected across a circuit consisting of a milliammeter in series with an unknown resistor R. If the readings on the instruments are 0.8 V and 12 mA, respectively, and if the resistance of the milliammeter is 6 W, calculate (a) the true resistance of R, and (b) the percentage error had the resistance of the milliammeter been neglected. [Ans. (a) 60.67 W ; (b) 9.9 %]

PROBLEMS

E-37. After a long time in position 1, the switch in the circuit shown in Fig. E-24 is thrown in position 2 at t = 0 s for duration of 30 s and then returned to position 1. (a) Find the equation for v for t 0 s. (b) Find v at t = 5 s and t = 40 s. (c) Make a sketch for v for 0 s t 80 s. [Ans. (a) v = 70 – 50 e–0.025t for 0 £ t £ 30 s, v = 20 + 26.4e–0.1(t–30) for t ≥ 30 s; (b) 25.9 V, 29.7 V] 5 MW 1 20 V + v –

2

20 MW

2 mF

70 V

Fig. E-24 E-38. In the circuit shown in Fig. E-25, the current source of 1 A is active. Find the indicated voltages and currents at t = 0+ s, immediately after the switch is closed. [Ans. v1(0+) = v2(0+) =20 V, i1(0+) = 1 A, i2(0+) = 0.106 A, i3 (0+) = – 0.106 A, i4(0+) = 0.17 A, i5(0+) = 0.064 A] E-39. voltages and currents a long time after the switch is closed. [Ans. v1( ) = 22.2 V, v2( ) = 25.6 V, i1( ) = 1.11 A, i2( )= 0 A, i3 ( ) = – 0.111 A, i4( ) = 0.111 A, i5( ) = 0 A]

Fig. E-25 E-40. In the circuit shown in Fig. E-26 the switch has been open for a long time. It is closed for 1 second and then opened again for a long time. (a the current in the inductor. (b) How much energy comes out of the battery in this process ? [Ans. (a) iL (t) = 10 t A, for 0 < t < l s; iL(t) = 10 e–(t–1)/0.ls for t > l s; (b) 60 J]

Fig. E-26 E-41. The switch in the circuit shown in Fig. E-27 is closed at t = 0. Find di/dt at the instant of time when the stored energy in the inductor is 0.5 mJ. [Ans. 3.568 mA/s]

Supplementary Exercise—Part E : Miscellaneous

827

+

E-45. Repeat Prob. E-44 for v(0 ) = 20 V, and for the 60-kW resistor replaced by a 70-kW resistor. [Ans. 63 – 43 e–1.96t V, 0.9 – 0.253 e–1.96t mA] E-46. After a long time in position 1, the switch in the circuit shown in Fig. E-30 is thrown at t = 0 s to position 2 for 2 s, after which it is returned to position 1. Find v for t > 0 s.

Fig. E-27

[Ans. v1 (t) = – 200 + 300 e–0.1t V, for 0 £ t £ 2 s; v2(t) = 100 – 81.13 e–0.2t V, for t ≥ 2 s]

E-42. For the circuit in Fig. E-28, the original energy stored in the capacitor is 500 J and the switch is open. The switched is closed for 1 second and then capacitor.

0.5 MW 1

2

1 MW

[Ans. 409.3 J] 100 V

+

v –

10 mF

200 V

Fig. E-30 Fig. E-28 E-43. Closing a switch connects in series a 300-V source, a 2.7-MW resistor, and a 2-mF capacitor charged to 50 V with its positive plate towards the positive terminal of the source. (a) Find the capacitor current 3 s after the switch closes. (b) Find the time required for the capacitor voltage to increase to 250 V. (Ans. (a) 53.1 mA; (b) 8.69 s] E-44. The switch in the circuit shown in Fig. E-29 is closed at t = 0 s. Find v and i for t > 0 s. The capacitor is initially uncharged. [Ans. 60( 1 – e –2t) V, (1 – 0.4 e–2t) mA]

Fig. E-29

E-47. After a long time in position 2, the switch in the circuit shown in Fig. E-30 is thrown at t = 0 s to position 1 for 4 s, after which it is returned to position 2. Find v for t > 0 s. [Ans. v1(t) = 100 – 300 e–0.2t V, for 0 £ t £ 4 s; v2(t) = – 200 + 246 e–0.1t V, for t ≥ 4 s] E-48. A moving-coil instrument, used as a voltmeter, has a coil of 150 turns with a width of 3 cm and 0.15 T. If the full-scale reading is 150 V and the total resistance of the instrument is 100 000 W the torque exerted by the control spring at the full scale. [Ans. 30.4 mNm]

828

Basic Electrical Engineering

GLOSSARY

A : An abbreviation for ampere. Absolute instrument : It measures a quantity directly in terms of absolute parameters, without needing prior calibration. Absolute permeability : m = B/H henry/meter. AC or ac : 1. Abbreviation for alternating current. Electric current which alternates at regular intervals. Its direction is reversed 50 times per second in India. 2. Pertaining to, or utilising ac. AC generator : Also called alternator. It produces alternating emf that delivers power to circuits. Accuracy of measurement : The ratio of the error in the indicated value to the true value. Active component : Also called active element. An independent source which can deliver or absorb energy continuously. Active current : The component of the current phasor along the voltage phasor. Active power : Also called actual power, real power, or true power. In an ac circuit, the product of the voltage and the current that is in phase with the voltage. It is measured in watts. Adapter : A device used to connect different types of electrical terminals. Admittance : Y = G + jB conductance, and its imaginary susceptance. It is the reciprocal of impedance Air capacitor : A capacitor in which the dielectric is nearly all air. Used for tuning electrical circuits with minimum energy loss. The ratio of energy output in 24 hours to the energy output plus energy losses in 24 hours. Alternating current : ac. Alternator : A device which produces ac. Also called synchronous generator. Ammeter : Abbreviation of ampere meter. An instrument used for measuring electric current. Ampere : Ampere-hour meter : Ammeter shunt : A resistor connected in parallel to an ammeter, in order to increase the current range that can be measured. Ampere-turn :

830

Basic Electrical Engineering

Amplitude : alternating current. Amplitude factor : Also called peak factor quantity. Angular velocity : Also called angular speed or angular frequency, represented by the symbol w Apparent power : Armature : 1. The rotating, or moving part, in an electric generator or motor. 2. in an electromagnetic device, the moving Armature coil : A coil of insulated copper wire wrapped around the armature core. It is a part of the armature winding. Armature core : The assembly of laminations forming the magnetic circuit of the armature of an electric machine. Armature reactance : A reactance associated with the armature winding of a machine. It is caused by the armature Armature reaction : Asynchronous machine : An ac machine which does not run at a speed which is synchronised to the frequency of the power supply, as opposed to a synchronous machine which does. Autotransformer : the other part as the secondary winding. Auxiliary winding : A special winding on a machine, in addition to the main winding. Average value : The adding together of many instantaneous values of an amplitude taken at equal time intervals divided by the number of measurements taken. For a sine wave, the average value is 0.637 times its peak amplitude value. Ayrton shunt : A high-resistance shunt utilised to reduce the sensitivity of a measuring instrument, such as a galvanometer. This increases its range. It is also called universal shunt. B : Symbol for susceptance and . Back emf : The emf which arises in an inductance (because of the rate of change in current) or in an electric motor Balanced current : A term used in connection with a polyphase circuit to denote the current which is equal to the sum of currents in all phases. Balanced load : A load connected to a polyphase system such that each phase current has the same magnitude and all phase currents add up to zero. Balanced voltage : A term used in polyphase circuits to denote voltages which are equal in all phases. Battery : A source of dc power that incorporates two or more cells. B-H curve : Bimetallic strip : switch. Blowing current : The current (ac or dc) which will cause a fuse link to melt. Branch : Part of a circuit that lies between two nodes. Breadth factor : A factor used in the calculation of the emf generated in the winding of an ac machine to allow for the fact that the emfs in each of the individual coils are not in phase with one another. Also called or Brush : A conductor which slides to maintain contact between stationary and moving parts of an electrical device, such as a motor. It is usually made of graphite, or a metal.

Glossary

831

Bus bar : A length of a constant-voltage conductor in a power circuit. Also called power line. C : Symbol for capacitance and . Cable : An insulated electrical conductor, often in strands. Or a combination of electrical conductors insulated from one another. Cage rotor : A form of rotor which is used in induction motors. Calibration : Candela : The fundamental SI unit of luminous intensity. If, in a given direction, a source emits monochromatic radiations of frequency 540 ¥ 1012 Hz, and the radiant intensity in that direction is 1/683 watt per steradian, the luminous intensity of the source is 1 candela (Cd). Capacitance : The ability of a to store charge. It is equal to the quantity of charge which is required to be given to the capacitor so as to increase the potential difference across its terminals by one volt. Its symbol is C, and it is measured in farads (F). Capacitive reactance : The impedance associated with a capacitance. Its symbol is XC, and is measured in ohms (W). Its value is frequency dependent and is given as XC = 1/wC = 1/2p . Capacitor : A component having capacitance. Also called a . Carbon brush : A small block of carbon used in electrical equipment to make contact with a moving surface. CFL (Compact Fluorescent Lamp) : It holders). Charge : Its symbol is Q negative charge, positive charge. Choke : dissipation. Circuit : An arrangement of passive and active components joined by conductors so as to provide path or paths for electric current. Circulating current : Also called mesh current circuit. produced by the primary. Coil : A length of an insulated conductor wound around a core of iron, ferrite, or air. Coil span : The distance between one side of a coil and the other, measured around the periphery of the armature. Commutation : 1. The transfer of current from one path of a circuit to another path within the same circuit. 2. The reversals of current through the windings of an armature, to provide dc at the brushes. Commutator : 1. A device, such as a switch, used to reverse current. 2. In a dc motor or generator, the section that causes the direction of the electric current to be reversed in the armature windings. 3. In a dc motor or generator, the section which maintains electrical continuity between the rotor and the stator. Compound generator : A dc generator having both series and shunt windings. Compound motor : A dc motor having both series and shunt windings. Compound winding : A winding which has both series and shunt windings. Conductivity : s It is the reciprocal of resistivity. Conductor Conduit : Conduit box :

832

Basic Electrical Engineering

Constant current source : Constant voltage source : Continuity : A continuous electrical connection between two points. Controlling torque : Also called restoring torque torque in an instrument. Copper loss : Also called I 2R them, in transformers and electrical machines. Core losses : The losses occurring in the iron core of the transformers and electrical machines due to hysteresis and eddy Core-type transformer : A transformer in which most of the core is enclosed by the windings. Crest factor : peak factor or amplitude factor cannot build up voltage. Critical speed of a shunt generator : The lowest speed below which the generator fails to build up voltage. Cumulatively compound machine : Current : i or I Current coil : A term used with wattmeter, energy meter, etc. to denote the coil connected in series and hence carrying main current. Cycle : One complete oscillation. Damped oscillation : diminishing until the amplitude becomes zero. Also called damped vibration. Damping torque : This torque acts on the moving system of an instrument, and it comes into action only when the system is moving so as to oppose the motion. D’Arsonval movement : DC or dc : Abbreviation of direct current. DC generator : A machine which converts mechanical energy into dc electrical energy. DC machine : A machine which converts mechanical energy into dc electrical energy and vice versa. DC meter : An instrument which responds only to dc component of a signal, e.g., moving coil instrument. DC motor : A machine which converts dc electrical energy into mechanical energy. Delta connection : A three terminal having a combination of three circuit elements, such as resistors, connected in series, D. Demagnetisation : Reducing the residual magnetism of a previously magnetised material by applying a magnetising force in the opposite direction. Dielectric : Insulating material which separates the two plates of a capacitor. Dielectric Constant : It denotes the ratio of the insulating quality of a material to that of air. Differentially wound motor : oppose each other. Direct Current : Distribution factor : A factor used in ac machines which accounts for the fact that the emfs produced in each of the individual coils are not in phase. Also called breadth factor, or

Glossary

833

Double throw switch : A switch which enables a connection to be made with either of the two contacts. Also called twoDynamo : A large capacity generator. Dynamometer-type instrument : E or e : voltage or Earth plate : A plate buried in the earth for the purpose of providing an electrical connection between an electric system and the earth. Earth terminal : Earthing : Eddy currents : rise to power loss in electrical machines. Effective value : Also called A value obtained by squaring multiple instantaneous values, averaging these

The ratio of the useful output power of a device or system, to its total input power. Electric Discharge Lamp : A lamp in which the light is obtained from the discharge between the electrodes in an evacuated glass tube. Electric potential : Electric switch : A switch which serves for opening, closing, or changing connections in electric circuits. Electrodynamometer : Electrolyte : The chemical ingredient between the plates of a voltage cell. Electromagnet : in the coil. Electromagnetism : 1. The science that studies electromagnetic phenomena. 2. Electro-motive force : Its abbreviation is emf Equivalent circuit : An arrangement of simple circuit elements having the same characteristics as the more complicated Excitation loss : I 2R External circuit : The circuit to which current is supplied from a generator or a battery or other source of electrical energy. F: farad, the unit of measuring capacitance. f: frequency Field coil : Field control : Filament : Fluorescent bulb : ultraviolet radiations from the discharge are converted to light of acceptable color. Fluorescent tube : inside the tube.

834

Basic Electrical Engineering

Form factor : It is the ratio of the effe Frequency : hertz Friction and windage loss : The losses in an electric machine due to friction of sliding parts and also due to air resistance. Fringing : a magnetic circuit. Full load : Fuse : circuit. G: conductance, which is the reciprocal of resistance. Galvanometer : Generator : A machine or device which converts mechanical energy into electrical energy. Governor : Gravity-controlled instrument : An electrical measuring instrument in which the controlling torque is provided solely by action of gravity. Ground : A point of reference and/or a common return path for current. H: henry, the unit of inductance. H: . Henry : Hertz : Horse power : Hysteresis : 1. A phenomenon on which an induced or observed effect remains after the inducing cause is removed. In this way, two physical quantities are related in a manner that depends on whether one is increasing or decreasing with respect to the other. 2. A phenomenon in which changes in the magnetisation induced in a ferromagnetic material lag behind the changes in the magnetising force. 3. A phenomenon in which there is a lag between the cause and the introduced or observed effect. Hysteresis loop : Hysteresis loss : The energy dissipated, usually in the form of heat, in each complete magnetising cycle of the material. It is equal to the area enclosed by the hysteresis loop. Hz : I electric current. Impedance : Its symbol is Z R + jX Impedance triangle : of an ac circuit, and whose hypotenuse is proportional to the impedance of the circuit. Incandescent lamp : light bulb. Inductance : 1 induced in the same circuit, it is called self-inductance is called mutual inductance M

L

2. The

Glossary

835

Induction : 1. of nearby entities. 2. The generation of an electromotive force in a circuit or conductor caused by a change in the Induction motor : in its stator. Inductive reactance : symbol is XL, and is measured in ohms. XL = wL = 2p f L. Inductor : In-phase : A state in which two or more periodic quantities having the same frequency and wave shape pass through periodic quantities do not pass through corresponding values at the same instant at all times. Inertia : The tendency of a body to preserve its state of rest or of uniform motion. Instantaneous value : The value of varying quantity, such as voltage or current, at a particular instant within a cycle. Insulated neutral : connected to earth directly or through low impedance. Insulated wire : A solid conductor insulated throughout its length. Insulator : A material having high electrical resistivity. Integrating instrument : An instrument whose indications are the total over time of a measured quantity, such as power. Iron core : A core, such as that of a transformer or armature, which is made of solid or laminated iron. Iron loss : The energy dissipated by an iron core, due to eddy currents and hysteresis loss, as found in a transformer or inductor. j: 1 , used by electrical engineers in place of mathematician’s i rotation. J: Junction : A point in a circuit where two or more conductors are connected. 3 k: 3 kilowatt-hour : kV : Abbreviation for kilovolt. kVA : Abbreviation for kVAR : Abbreviation for kilovolt-ampere-reactive. kW : Abbreviation for kilowatt. kW h : Abbreviation for kilowatt-hour. L: inductance. Lag : The amount, measured as a time interval or the angle in electric degrees, by which one periodically varying wave is delayed in phase with respect to another wave of same period. Lagging current : produces it. Lagging load : lagging current with respect to the voltage across its terminals. Lamination : A sheet steel stamping so shaped that a number of them can be put together to form the magnetic circuit of electric machines, transformers, etc.

836

Basic Electrical Engineering

Lap winding : A adjacent to it. Lead : 1. The amount, measured as a time interval or the angle in electric degree, by which one periodically varying wave is advanced in phase with respect to another wave of same period. 2. A term used to denote an electric wire or cable connected to a terminal. Leading current : produces it. Leading load : leading current with respect to the voltage across its terminals. Leakage : leakage factor. Leakage inductance : . Leakage reactance : coil but not the other. Lenz’s law : It states that the induced emf in a circuit opposes the very cause due to which it has been induced. Lifting magnet : A large electromagnet on a crane or hoist, used to lift iron objects. Lightning conductor : earthed. Linear resistor : A resistor which obeys ohm’s law. Also called ohmic resistor. Linear element : Line current : Line voltage : The voltage between the lines of an electric power supply. Live : Live wire : The wire which is at a higher potential than the neutral wire of the supply. Load : The rate at which energy is fed into a process or removed from it. Load characteristic : Load factor : Loss : Loss angle : The difference between 90° and the angle of lead of current over the voltage in a capacitor. Lumen : is at a unit distance from the source of one candela. 2 Lux : The unit of illumination M: mega –3 m: milli M: mutual inductance. Magnetic circuit : H) : Also called magnetic intensity. F

Glossary B) : A quantitative It is the number of lines of force p

B = F/A

837

tesla

2

Magnetic leakage : affecting nearby apparatus. Magnetic susceptibility : The amount by which the relative permeability of a medium differs from unity. It equals to the Magnetising coil : generator or motor. Magnetising current : The current in a coil for magnetisation of ferromagnetic material in its core. Magnetomotive force, mmf electromotive force, emf. Its unit is ampere-turns Mains : The term refers to the source of electric power, usually the electric supply system. Male and female : Maximum power transfer theorem : A theorem stating that the greatest possible amount of power is transferred from a source to the load when they have conjugate impedances. Maximum rating : or a device can operate safely. times. Megger : An instrument used for measuring insulation resistance. Mercury vapour lamp : A lamp producing light with a high ultraviolet content by means of an electric arc in mercury vapour. Mesh : A loop which does not contain any other loops inside it. Mesh current : Meter : A general term for any measuring instrument. times. –3

times. Motor : An electrical machine which converts electrical energy to mechanical energy. Moving-coil meter : Moving-iron meter : A meter in which the moving element is a small bar of soft iron which is suspended or pivoted in Multiplier : A resistor in series with the coil of a voltmeter to increase the range. Mutual inductance : is M. N: newton. N: number of turns. –9 n: nano times. –9 times. Natural frequency : driving force. Neon lamp : A gas discharge tube operated in the glow discharge region. It contains neon gas and emits a red glow.

838

Basic Electrical Engineering

Network : An interconnection of electrical components. If it contains only passive components, such as resistances, inductances, and capacitances, it is called passive network. components, it is called Network analysis : Node : Nodal analysis : Nonlinear circuit : A circuit whose output is not linearly proportional to its input. Nonplanar circuit : A circuit which cannot be drawn on a plane without crossovers. Ohm : W Ohmic contact : Ohmic heating : material or conductor. Ohmic loss : Ohmic resistance : The resistance to dc, offered by a device, circuit, or material. Ohm metre : Ohms per volt : A measure of the sensitivity of an instrument, such as a voltmeter. It refers to the resistance, in ohm, Open circuit : Open-circuit characteristic : The term used for the curve obtained by plotting the emf generated by an electric generator Output : The power, voltage, or current delivered by any circuit or machine. Output impedance : Output regulation : Of the power supply, the variation of voltage with load current. Overload : P: P: pico times. Pd : Abbreviation for potential difference. Panel : A sheet of metal, plastic or other material upon which instruments, switches, etc. are mounted. Paper capacitor : A capacitor in which thin paper acts as dielectric separating aluminium foil electrodes, rolled together. Parallel circuit : A circuit whose components are connected such that each has the same terminal voltage. Passive component : A component or device, such as a resistor or a transformer, that cannot operate on an applied electrical signal, as in amplifying, rectifying, or switching, and is not a source of energy. Peak factor : crest factor or Peak-to-peak value : Peak value : Periodic : Occurring, appearing, or characterised by regular and repetitive intervals or cycles.

Glossary

839

Permeability : Its symbol is m B

H m = B/H. Permeability of free space m0 = 4p ¥ –7 Permeance : The reciprocal of reluctance of a magnetic circuit. Permittivity : symbol is e e 0 = 8.85 ¥ Phase : reference point. Phase difference : Phasor : It is the representation of alternating current or voltage by a rotating arrow. This representation allows Phasor diagram : A graphical representation of phasors of sinusoidal quantities at the same frequency with proper phase relationship. Phase sequence : times. Pitch factor of an alternator : Planar circuit : nonplanar circuit, dimensional. Pole pitch : The distance between the central line of two adjacent poles in an electric machine. Polyphase : Potentiometer : adjusted potential provided by a current from a steady source. Power : Power factor : power factor is equal to the cosine of the phase angle between them. Primary voltage : The voltage which is applied to the input side of the transformer. Q: charge, and also for quality factor Quadrature : Refers to the phasors having a phase difference of 90°. Quality factor : Its symbol is Q of a circuit. R: Radian frequency : w = 2p f frequency or Ramp voltage : Rating : such equipments. Reactance : The imaginary number component of impedance Z reactance XL capacitive reactance XC Reactive current : does not add to the real power.

angular

inductive

840

Basic Electrical Engineering

Reactive load : A load in which the current lags or leads the voltage applied to its terminals. Reactive power : The power in an ac circuit which cannot perform work. It is given by VI sin f. Real power : Same as active power, or true power. Relative permeability : m r = m/m0. Relay : An electromagnetic switch. Reciprocity theorem : A theorem stating that if a voltage source at one point produces a given current at another point Reluctance : permeance. Residual magnetism : The magnetism which remains in a material after the magnetising force is removed. Resistance : R W). Resistance box : switches or keys. Resistive load : emf. Resistivity : The resistance between the opposite faces of a cube of the material having side of 1 metre. It is symbolised by r ohm metres W m). It is the reciprocal of conductivity. Resistor : Electric component designed to introduce known resistance into a circuit. Resonance : The condition in an ac circuit when XL = XC . Resonant frequency : etc. Retentivity : The ability of a material to hold magnetism. Rheostat : A variable resistor designed for the control of current in a circuit. Root mean square value : Also called effective value. Rotor : or armature of a motor. A stationary part in such a device is called stator.

rotate.

Its magnitude remains constant but its direction keeps on rotating at synchronous speed. S: Saturation : density. Sawtooth : Self-excited : Self-inductance : The property of a coil that induces an emf in itself when the current through it changes. Series circuit : A circuit whose components carry the same current. Series motor : armature. Shell-type transformer : A transformer in which most of the windings are enclosed by the core. Short circuit : with higher resistance.

Glossary

841

Short circuit test : A low voltage test carried out on an electrical machine with its output terminals short circuited and Shunt : galvanometer. Shunt motor : Sinusoidal wave : A periodic wave whose amplitude follows the values of a sine curve. Slip : Slip ring motor : Slip rings : the current into or away from the rotor winding. Slot : carrying conductors forming the winding are embedded. Sodium vapour lamp : An electrical lamp in which sodium vapour is activated by current passing between the two electrodes producing a yellow glareless light. Solenoid : Speed control : The method by which the speed of an electric motor may be varied. Speed-torque characteristic : The curve showing the relationship between the speed of a motor and the torque developed. Spring control : Controlling the movement of an indicating instrument with a spring. Square wave : A wave which switches abruptly back and forth between two stable levels of voltage. Squirrel-cage rotor : A winding consisting of a number of copper bars distributed in slots around the periphery of the Stabilising choke : A choke coil inserted in series with an electric discharge lamp to compensate for its negative resistance characteristic. Stalling torque : Standstill : A term pertaining to the electrical behaviour of a machine which is at rest. Standstill torque : The load torque which brings an electric motor to a standstill. Star connection : neutral point Starter : A device used for starting an electric motor and accelerating it to its normal speed. Starting current : The current drawn by a motor when starting up. Starting resistance : limits the current to a safe value. Starting torque : The torque developed by a motor at starting. Stator : Stator winding : The part of the winding of a machine accommodated in the stator. Steady state : Step-down transformer : One in which energy is transferred from a high voltage to a low voltage. It has N1 > N2. Step-up transformer : One in which energy is transferred from a low voltage to a high voltage. It has N1 < N2. and which may interfere with the operation of the device.

842

Basic Electrical Engineering

Superposition theorem : It states that in a linear circuit containing more than one source, the resultant current in any Surge : A sudden change in current or voltage. B) : Y Y = G + jB. Switch : The device for opening and closing an electrical circuit. Switch board : An assembly of switch panel. Switch box : Switch panel : An insulating panel on which a switch is mounted. Symmetrical winding : Synchronous motor : An ac motor which runs at a speed which is synchronous to the frequency of the power supply. Synchronous speed : 2 T: tera times. Also, symbol for tesla . An incremental change in the resistance of a component, device, or material as a function of temperature. Tera : times. Terminal : A point in an electrical circuit at which an electrical element may be connected. Terminal voltage : The voltage at the terminals of a piece of electrical equipment or machine. Test board : A switch board carrying instruments and sometimes an electric lamp, used for testing the equipments and appliances. Thermistor : Thevenin’s theorem : be represented by a single voltage source and a series impedance. Three-phase four-wire system : lines neutral Three-phase induction motor : ability to start without requiring any additional device. Three-way switch : Time constant : The time required for an electrical quantity, such as a current or a voltage, to rise from zero or another e Toroid : Torque : Total loss : The power loss in an electrical machine, equal to the difference between the input and the output powers. Transformer : A device in which ac power is transferred from one voltage level to another, without any change in its frequency. Transformer core : The structure, usually of laminated iron or ferrite, forming the magnetic circuit. Transformer oil : A mineral oil of high dielectric strength, forming the cooling and insulating medium of electric power transformers. Transient response : The speed with which a component, circuit, device, piece of equipment, or system changes its output in response to a sudden and intense change in its input. Transmission line : Transmission voltage : The nominal voltage at which electric power is transmitted from one place to another. Tree : True power : It is the average power consumed by a circuit.

Glossary

843

Tungsten lamp : Turns ratio : N2 N Two-wire system : Unbalanced system : Unidirectional current : The current whose amplitude may vary but its direction never changes. Universal motor : V: volt V: VA : volt-ampere apparent power. VAR : volt-ampere-reactive reactive power. Voltage : The difference in potential of two points, measured in volts. Voltage divider : loads. Voltage drop : Voltage gradient : The difference in potential per unit length of a conductor. Voltage regulation : The percentage variation in the output voltage of a power supply from no load to full load. Voltmeter : An instrument for measuring potential difference. W : Symbol for watt, the SI unit of . Watt : The SI unit of power. It is equal to the power rate of one joule of work per second. Watt-hour : One watt used continuously for one hour. Watt-hour meter : An integrating meter for the measurement of total electrical energy consumed in a circuit. Wattless voltage : The voltage component which is in quadrature with the current of an ac circuit. Such component does not add to power. Wattmeter : An instrument for measuring electrical power. Wave : A periodic variation in voltage or current. Waveform : A graph of a wave. Wavelength : The length of one period of a wave. Wave winding : A type of armature winding in which there are only two parallel circuits, irrespective of the number of poles in a dc machine. Winding : A complete group of interconnected conductors in an electrical machine or a transformer, that is designed Wound rotor : Same as slip-ring rotor. X : Symbol for Y : Symbol for . Y-connection : Same as star-connection. Yoke : It forms a part of the magnetic circuit of a machine, and provides mechanical support to the entire assembly. Z : Symbol for

844

Index

Index

845

INDEX A

Batten Holder 636

Absolute Instruments 589 Absolute Potential 4 AC Current 3 Acceptor Circuit 312 Actual Power 254 Additions of Phasors 249 Advantages of Three-Phase System 336 Air Gaps in Magnetic Circuits 172

Blocked Rotor 483 Blocked Rotor Reactance 487 Bracket Holder 636 Braking System 617 Branch Currents 73 Brushes 516 Building Up of Voltage 531 C

Alternating Current 3 Ammeters 12, 602 Amplitude Factor 242 Analogy between Electric and Magnetic Circuits 167 Angle of Lag 236 Angle of Lead 236 Angular Frequency 235 Apparent Power 276 Approximate Equivalent Circuit 389 Armature 515 Armature Reaction 526 Armature Winding 430, 516 Autotransformers 397 Auxiliary Winding 566, 567–569 Available Power 114 Average Power 240 Average Value 239 B Bandwidth 318 Base Units 7

Capacitance 51 Capacitances Connected in Parallel 54 Capacitances Connected in Series 54 Capacitive Reactance 260 Carbon-moulded Resistors 29 Cartesian Form 247 Charging of a Capacitor 213 Chokes 186, 395 Cleat Wiring 631

Coil Span Factor 438 Combination of Sources 62 Comments on Resonance 320 Commutator 515 Comparison between Series and Parallel Resonance 324 Complex Admittance 278 Complex Conjugate 140, 248, 276

846

Index

Complex Impedance 274, 275, 280 Complex Numbers 246 Complex Plane 243, 246 Complex Power 276, 454 Composite Magnetic Circuit 168 Compound Motor 539 Concept of Three-Phase Voltages 337 Condition for Maximum Regulation 393 Condition for Zero Regulation 393 Conditions for Ideal Transformer 375 Conductance 14, 28, 278 Conductance Matrix 84 Constant Speed Operation 453 Constrained Node 82, 83 Construction of Transformer 383 Controlled Sources 66 Copper Loss 386, 429 Core Losses 380, 395 Core of Transformer 383 Core-type Transformer 383 Core-loss Current 380 Coulomb 1, 4 Counting Independent Nodes 82 Counting Independent Loops 74 Coupled Coils in Parallel 193 Coupled Coils in Series 191 Crest Factor 242 Critical Field Resistance 531 Current at Resonance 315 Current Controlled Current Source 66 Current Controlled Voltage Source 66 Current Density 4 Current Divider 19 Current Transformer 401 Cutoff Frequencies 318 Cycle 234 Cylindrical or Non-salient Type 434 D D.H. Bernard Tellegen 117 Damper Winding 453, 458

Datum Node 80 DC Current 3, 58 Decaying Current 210, 223 Delta (D)-Connected Three-Phase System 342 Delta-Connected System 345 Delta-to-Star Transformation 22 Dependent Sources 66 Diamagnetic Materials 165 Direct Current 3 Discharging of a Capacitor 213 Distribution Factor 439 Dot Convention 190 Double-Field Revolving Theory 563 Double-Layer Armature Winding 516 Double-Subscript Notation for Currents 337 Double-Subscript Notation for Voltages 335 Drift Current 2 Drift Velocity 2 Duality 28 Dynamically Induced EMF 150 Dynamo 152 E Eddy Currents 172, 382, 395 Eddy-Current Loss 382 Effect of Air Gap 172 Effect of Change in Excitation 457 Effect of Change in Mechanical Load 457 Effect of Core Losses 380 Effect of Flux Leakage 386 Effect of Magnetisation 379 Effect of Winding Resistance 386 Effective Value 240

Elastomer Insulated Cables 626 Electric Current 2, 5 Electric Field 5 Electrical Degrees 432 Electromagnetic Induction 148 Electromechanical Energy Conversion 426 Electromotive Force 7, 58

Index Electrons 1 EMF Equation of Transformer 374 Energy 5 Energy Meter 616 Energy Sources 57 Energy Stored in an Inductor 53 Energy Stored in Magnetic Field 195 Equivalent Sources 60, 62 Exciting Circuit 380 Exciting Current 516 F Ferromagnetic Materials 165, 167, 382 Field Cols 514 Field Current 516

Floating Neutral 335 Flush Switch 628 Force on Current-Carrying Conductor 144 Form Factor 242 Four-Wire Three-Phase Voltage System 340, 342 Free Electrons 1–3, 30, 153 Frequency 234 Frequency of Resonance 311 Fringing 169 Fringing Flux 169

G Generation of Three-Phase Voltages 338 Generator Action 152, 427, 432 Geneva Cam 562, 573 Gravity Control 592 Ground Node 80 Growth of Current in Series RL Circuit 209

Horse Power 6 Hunting 458 Hybrid Stepper Motors 574, 582 Hysteresis Loss 381 I Ideal Current Source 58 Ideal Parallel Resonant Circuit 322 Ideal Sources 58 Ideal Transformer 374 Ideal Voltage Source 57 Imaginary 246 Imaginary Numbers 246 Impedance Match 114, 377 Impedance Transformation 376 Impedance Triangle 272, 274, 276 Importance of Maximum Power Transfer 114 Independent Loops 72, 74, 85 Independent Node 81 Independent Source 66 Inductance 50 Inductance from Geometrical Viewpoint 184 Inductances Connected in Parallel 54 Inductances Connected in Series 54 Induction Motor 477 Inductors 183 Inductive Reactance 258 Initial and Final Values 217 Instantaneous Value 211 Instrument Transformers 401 Internal Resistance 7 Iron losses 381, 395 J Joule 4, 8 K Kilocalories 6

H Half-Power Frequencies 310, 318, 320, 323 Henry 52, 141 High-Q Coils 315 Holes 2

847

848

Index

L Laminations 172 Lap Winding 517 Large System 448 Leakage Flux 169, 386 Leakage Reactance 386, 445 Leakage Transformer 402 Lifting Power of a Magnet 196 Load Component of Primary Current 384 Load Test on a Single-Phase Transformer 418 Loading Devices 13 Locked Rotor 483 Loop 72 Loop Current 72 Loop Moved in a Magnetic Field 153 Loop-Current Analysis 72 Lorentz Force 153 Low-Q Coils 315 M Magnetic Circuit 164-167 Magnetic Coupling 187 Magnetic Field due to a Circular Loop 143 Magnetic Field due to a Coil 142 Magnetic Field due to a Long Straight Wire 143 Magnetic Field due to a Solenoid 143 Magnetic Field due to a Toroid 144 Magnetic Field due to Electric Current 142 Magnetic Field Intensity 165 Magnetic Field Strength 164 Magnetic Permeability 164, 165 Magnetising Current 397 Magnetisation Characteristics 165, 173 Magnetising Force 165 Magnetomotive Force 164 Magnetostriction Losses 395 Maximum Power Transfer Theorem 113 Meaning of Time Constant 207 Measurement of M 192 Measurement of Power 352, 354, 613, 615 Mechanical Degrees 432

Mechanical Losses 395, 485 Mesh 21, 75 Mesh Analysis 75 Mesh Current Matrix 77 Michael Faraday 148 Microstepping 578 Ministepping 578 Mutual Inductance 186 Mutual Resistances 77 Mutual Conductance 84 N Network Components 50 Neutral Shift 342 Neutrons 1 Nodal Analysis 84 Node 67 Node-Current Source Matrix 84 Node-Voltage Analysis 80 No-load Current 379 Nonplanar 75, 85 Nonsinusoidal Waveforms 251 Normal Excitation 458 Normalised slip speed 478

O Ohm 14 Ohm Metre 16 One-Wattmeter Method 352 Open-Circuit 15 Open-Circuit and Short-Circuit Tests on a Transformer 423 Open-Circuit Test 372, 402, 403, 407, 421, 424, 451, 512, 774, 800 Optimisation 114 Over Excitation 458 P Parallel Combination of Resistances 17 Parallel RC Circuit 278 Parallel RLC Circuit 282

Index Paramagnetic Materials 165 Passband 318 Passive Sign Convention 6 Peak Factor 242 Peak Value 234 Pendant Holder 636 Periodic Time 234 Permanent Magnet Generators 522, 523 Permanent-Capacitor Motor 561 Permeability 53, 143 Permeance 166, 167 Phase 235 Phase Angle 235 Phase Difference 236, 243 Phase Shift 236 Phasor Diagram 243, 279 Phasors 242 Pitch Factor 438 Planar 75 Polar Form 247 Poles 514 Potential Difference 4 Power 6 Power Absorbed 6 Power Factor 255 Power in Three-Phase System 350 Power Supply Systems 11, 459 Practical Current Source 59 Practical Parallel Resonance Circuit 321 Practical Current Source 59 Practical Transformer at No Load 379 Practical Transformer on Load 386 Practical Voltage Source 58 Primary Winding 377 Primary Balancing Current 384 Primary Leakage Flux 386 Principle of Duality 28 Protons 1 Q Q Factor for Series Resonant Circuit 315 Q Gain 315

Q-Factor 325, 748 Quality Factor 314 R Random 1, 2 Range-Multiplier 603, 608 Rate of Growth of Current 210 RC Timers 219 Reactance 258–260 Real Power 255 Reciprocity Theorem 116 Reference Node 80 Regulation Down 391 Regulation Up 391 Reluctance 166 Resistance Colour Code 29 Resistance Matrix 76 Resistance Measurement 610 Resistivity 16 Resonance 281, 310 Resonance Curve 310, 318 Resonance Frequency 317, 318 Resonant Transformer 402 Rheostat 11, 13 Right-hand Thumb Rule 141 RMS Value 240 Root Mean Square 241 Rotary Machines 430 Rotation of Phasor by 90° 248 Rotor Equivalent Circuit 487 S Salient or Projected Pole 435 Saving in Copper 400 Secondary Winding 374 Secondary Leakage Flux 386 Selectivity 312, 319 Selectivity of Tuned Circuit 318 Self-inductance 183 Self-conductance 84 Self-excited Generators 522, 523 Self-resistances 77 Separately Excited Generators 522, 523 Series RC Circuit 228, 275

849

850

Index

Series RL Circuit 209, 273 Series RLC Circuit 280 Shell-type Transformer 385 Short Circuit 15 Short-Circuit Test 403 SI Units 7 Siemens 14 Single-Phase Motors 561 Single-Phase Supply 11, 336 Skewed Rotor 477 Slip of Induction Motor 478 Socket Outlets 634 Source Matrix 77 Source Resistance 58 Source Transformation 60 Split-Phase Motor 566 Squirrel Cage 477, 480, 562 Stacking Factor 173 Staircase Wiring 628 Stand-alone System 448 Star (Y)-Connected Three-phase System 340 Star and Delta Connections 21 Star-Connected System 340, 343 Star-to-Delta Transformation 23 Stator Magnetic Structure 514 Step Down 374 Step-down Transformer 376 Stepper Motors 573 Step Up 373 Step-up Transformer 376 Stray Losses 395, 527 Supermesh Method 79 Supernode 82 Superposition Theorem 104 Susceptance 278 Switches 13, 15, 627 Synchronous Impedance 445 Synchronous Machine 431 Synchronous Reactance 444 Synchronous Speed 432 T Tachometer 12, 511, 558

Terminal Voltage 7

Three-Ammeter Method 305 Three-Phase Loads 339 Three-Phase Supply 11, 336, 340 Three-Phase System 336 Three-Phase Transformers 400 Three-Voltmeter Method 307 Three-Wattmeter Method 352 Time Constant 206, 210 Time Period 234 Tinned Brass 632 Toggle Switch 628 Torque Angle 457 Torque experienced by a Coil 146 Torque-Slip Characteristic Curve 492, 493, 504 Transfer Resistance 116 Transformation Ratio 375 Transformer on Load 384-386 Transformer Testing 402 Transient Response 204, 219 Transient Time 204 Tumbler Switch 628 Turning off the Sources 105, 113 Turns-ratio 375 Two-Phase System 335, 566 Two-Wattmeter Method 353 U Unbalanced Three-Phase System 341 Under Excitation 458 Universal Motor 561, 572, 584, 586 Useful Flux 169 V Variacs 399 Variation of Resistance With Temperature 30 Vector Diagram 246 Voltage Controlled Current Source 67 Voltage Controlled Voltage Source 66 Voltage Divider 18

Index

Voltage Regulation 390 Voltage Regulation of a Transformer 390 Voltage Resonant Circuit 315 Voltage Transformer 402 Voltage-Coil Inductance 615 Volt-Amperes 276, 376 Voltmeters 12, 602

Wattmeter 12, 614 Wattmeter Errors 615 Wave Winding 517 Wire Wound Resistor 29 Y Yoke 514

W

Z

Watt 6, 8

Zig-zag 1

851