Network Analysis And Synthesis [1 ed.] 9781259062957

Table of contents :
Title
Contents
Chapter 01
Chapter 02
Chapter 03
Chapter 04
Chapter 05
Chapter 06
Chapter 07
Chapter 08
Chapter 09
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16
Appendix
Index

Citation preview

Network Analysis and Synthesis

About the Author Ravish R. Singh is presently Vice-Principal at Thakur College of Engineering and Technology, Mumbai. He obtained a BE degree from University of Mumbai in 1991 and an MTech degree from IIT Bombay in 2001. He is currently involved in doctoral research at the Faculty of Technology, University of Mumbai. He has published three books, namely Electrical Networks, Engineering Mathematics and Basic Electrical and Electronics Engineering with McGraw Hill Education (India). He is a member of IEEE, ISTE, IETE, CSI and has published several acclaimed research papers in national and international journals. His fields of interest include Circuits, Signals and Systems and Engineering Mathematics.

Network Analysis and Synthesis

Ravish R Singh Thakur College of Engineering and Technology Mumbai

McGraw Hill Education (India) Private Limited NEW DELHI

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

McGraw Hill Education (India) Private Limited Published by the McGraw Hill Education (India) Private Limited P-24, Green Park Extension, New Delhi 110 016. Network Analysis and Synthesis Copyright© 2013, by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. ISBN (13): 978-1-25-906295-7 ISBN (10): 1-25-906295-3 Vice President and Managing Director: Ajay Shukla Head—Higher Education Publishing and Marketing: Vibha Mahajan Publishing Manager (SEM & Tech. Ed.): Shalini Jha Editorial Executive: Koyel Ghosh Manager —Production Systems: Satinder S Baveja Assistant Manager—Editorial Services: Sohini Mukherjee Production Executive: Anuj K Shriwastava Assistant General Manager (Marketing)—Higher Education: Vijay Sarathi Senior Product Specialist: Tina Jajoriya Senior Graphic Designer—Cover: Meenu Raghav General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) 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 McGraw Hill Education (India) 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 BeSpoke Integrated Solutions, Puducherry, India 605 008 and printed at Cover Printer:

Dedicated to My Father Late Shri Ramsagar Singh and My Mother Late Shrimati Premsheela Singh

Contents Preface

xiii

1. BASIC NETWORK CONCEPTS 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12

1.1

Introduction 1.1 Resistance 1.1 Inductance 1.2 Capacitance 1.7 Sources 1.14 Some Definitions 1.15 Series and Parallel Combination of Resistors 1.16 Series and Parallel Combination of Inductors 1.26 Series and Parallel Combination of Capacitors 1.28 Star-Delta Transformation 1.32 Source Transformation 1.51 Source Shifting 1.58 Exercises 1.60 Objective-Type Questions 1.64 Answers to Objective-Type Questions 1.65

2. ELEMENTARY NETWORK THEOREMS 2.1 2.2 2.3 2.4 2.5 2.6

Introduction 2.1 Kirchhoff’s Laws 2.1 Mesh Analysis 2.20 Supermesh Analysis 2.35 Node Analysis 2.43 Supernode Analysis 2.63 Exercises 2.69 Objective-Type Questions 2.73 Answers to Objective-Type Questions

2.75

3. NETWORK THEOREMS (APPLICATION TO dc NETWORKS) 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

2.1

Introduction 3.1 Superposition Theorem 3.1 Thevenin’s Theorem 3.30 Norton’s Theorem 3.64 Maximum Power Transfer Theorem 3.91 Reciprocity Theorem 3.112 Millman’s Theorem 3.116 Tellegen’s Theorem 3.121 Substitution Theorem 3.124

3.1

viii Contents 3.10 Compensation Theorem 3.126 Exercises 3.128 Objective-Type Questions 3.133 Answers to Objective-Type Questions 3.134

4. SINGLE-PHASE ac CIRCUITS 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15

Introduction 4.1 Generation of Alternating Voltages 4.1 Terms Related to Alternating Quantities 4.2 Root Mean Square (rms) or Effective Value 4.3 Average Value 4.4 Phasor Representations of Alternating Quantities 4.23 Mathematical Representations of Phasors 4.27 Behaviour of a Pure Resistor in an ac Circuit 4.34 Behaviour of a Pure Inductor in an ac Circuit 4.36 Behaviour of a Pure Capacitor in an ac Circuit 4.38 Series RL Circuit 4.41 Series RC Circuit 4.61 Series RLC Circuit 4.68 Parallel ac Circuits 4.81 Locus Diagram 4.100 Exercises 4.103 Objective-Type Questions 4.111 Answers to Objective-Type Questions 4.114

5. RESONANCE 5.1 5.2 5.3 5.4

5.1

Introduction 5.1 Series Resonance 5.1 Parallel Resonance 5.18 Comparison of Series and Parallel Resonant Circuits 5.21 Exercises 5.38 Objective-Type Questions 5.39 Answers to Objective-Type Questions 5.40

6. NETWORK THEOREMS (APPLICATION TO ac CIRCUITS) 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9

4.1

Introduction 6.1 Mesh analysis 6.1 Node Analysis 6.9 Superposition Theorem 6.14 Thevenin’s Theorem 6.27 Norton’s Theorem 6.41 Maximum Power Transfer Theorem 6.51 Reciprocity Theorem 6.64 Millman’s Theorem 6.68

6.1

Contents ix

6.10 Tellegen’s Theorem 6.72 6.11 Substitution Theorem 6.73 6.12 Compensation Theorem 6.76 Exercises 6.78 Objective-Type Questions 6.80 Answers to Objective-Type Questions

6.81

7. COUPLED CIRCUITS 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10

Introduction 7.1 Self-Inductance 7.1 Mutual Inductance 7.2 Coefficient of Coupling (k) 7.2 Inductances in Series 7.3 Inductances in Parallel 7.4 Dot Convention 7.9 Coupled Circuits 7.15 Conductively Coupled Equivalent Circuits 7.41 Tuned Circuits 7.44 Exercises 7.51 Objective-Type Questions 7.54 Answers to Objective-Type Questions 7.55

8. THREE-PHASE CIRCUITS 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13

7.1

8.1

Introduction 8.1 Advantages of a Three-Phase System 8.2 Some Definitions 8.2 Interconnection of Three Phases 8.2 Star, or WYE, Connection 8.3 Delta, or Mesh, Connection 8.4 Voltage, Current and Power Relations in a Balanced Star-Connected Load 8.4 Voltage, Current and Power Relations in a Balanced Delta-Connected Load 8.6 Balanced Y/Δ And Δ/Y Conversions 8.8 Relation between Power in Delta and Star Systems 8.8 Comparison between Star and Delta Connections 8.9 Three-Phase Unbalanced Circuits 8.28 Measurement of Three-Phase Power 8.38 Exercises 8.50 Objective-Type Questions 8.53 Answers to Objective-Type Questions 8.55

9. NETWORK TOPOLOGY 9.1 Introduction 9.1 9.2 Graph of a Network 9.1 9.3 Definitions Associated with a Graph 9.2

9.1

x Contents 9.4 9.5 9.6 9.7 9.8 9.9 9.10

Incidence Matrix 9.6 Loop Matrix or Circuit Matrix 9.8 Cutset Matrix 9.10 Relationship Among Submatrices of A, B and Q 9.12 Kirchhoff’s Voltage Law 9.24 Kirchhoff’s Current Law 9.24 Relation Between Branch Voltage Matrix Vb, Twig Voltage Matrix Vt and Node Voltage Matrix Vn 9.25 9.11 Relation Between Branch Current Matrix Ib and Loop Current Matrix Il 9.26 9.12 Network Equilibrium Equation 9.26 9.13 Duality 9.52 Exercises 9.58 Objective-Type Questions 9.60 Answers to Objective-Type Questions 9.61

10. TRANSIENT ANALYSIS 10.1 10.2 10.3 10.4 10.5

Introduction 10.1 Initial Conditions 10.1 Resistor–Inductor Circuit 10.27 Resistor–Capacitor Circuit 10.49 Resistor–Inductor–Capacitor Circuit 10.66 Exercises 10.79 Objective-Type Questions 10.82 Answers to Objective-Type Questions 10.85

11. LAPLACE TRANSFORM AND ITS APPLICATION 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10 11.11 11.12 11.13 11.14 11.15

10.1

Introduction 11.1 Laplace Transformation 11.1 Laplace Transforms of Some Important Functions 11.2 Properties of Laplace Transform 11.4 Laplace Transform of Periodic Functions 11.15 Waveform Synthesis 11.21 Inverse Laplace Transform 11.30 Solution of Differential Equations with Constant Coefficients 11.36 Solution of a System of Simultaneous Differential Equations 11.39 The Transformed Circuit 11.42 Resistor–Inductor Circuit 11.43 Resistor–Capacitor Circuit 11.49 Resistor–Inductor–Capacitor Circuit 11.54 Response of RL Circuit to Various Functions 11.60 Response of RC Circuit to Various Functions 11.68 Exercises 11.79 Objective-Type Questions 11.81 Answers to Objective-Type Questions 11.83

11.1

Contents xi

12. NETWORK FUNCTIONS

12.1

12.1 12.2 12.3 12.4 12.5 12.6 12.7

Introduction 12.1 Driving-Point Functions 12.1 Transfer Functions 12.2 Analysis of Ladder Networks 12.5 Analysis of Non-Ladder Networks 12.15 Poles and Zeros of Network Functions 12.20 Restrictions on Pole and Zero Locations for Driving-Point Functions [Common Factors in N(s) and D(s) Cancelled] 12.21 12.8 Restrictions on Pole and Zero Locations for Transfer Functions [Common Factors in N(s) and D(s) Cancelled] 12.21 12.9 Time-Domain Behaviour from the Pole-Zero Plot 12.39 12.10 Graphical Method for Determination of Residue 12.42 Exercises 12.50 Objective-Type Questions 12.53 Answers to Objective-Type Questions 12.55

13. TWO-PORT NETWORKS 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 13.10 13.11 13.12 13.13 13.14

Introduction 13.1 Open-Circuit Impedance Parameters (Z Parameters) 13.2 Short-Circuit Admittance Parameters (Y Parameters) 13.8 Transmission Parameters (ABCD Parameters) 13.18 Inverse Transmission Parameters (A′B′C′D′ Parameters) 13.24 Hybrid Parameters (h Parameters) 13.28 Inverse Hybrid Parameters (g Parameters) 13.33 Inter-relationships Between the Parameters 13.37 Interconnection of Two-Port Networks 13.63 T-Network 13.79 Pi (p )-Network 13.79 Lattice Networks 13.84 Terminated Two-Port Networks 13.87 Image Parameters 13.97 Exercises 13.101 Objective-Type Questions 13.104 Answers to Objective-Type Questions 13.107

14. FOURIER ANALYSIS 14.1 14.2 14.3 14.4 14.5 14.6 14.7

13.1

Introduction 14.1 Trigonometric Fourier Series 14.1 Waveform Symmetry 14.9 Exponential Fourier Series 14.25 Average Value of a Periodic Complex Wave 14.33 RMS Value of Periodic Complex Wave 14.33 Power Supplied by Complex Wave 14.34

14.1

14.8 14.9 14.10 14.11 14.12

Fourier Transform 14.37 Fourier Transforms of Some Useful Functions 14.38 Fourier Transform of Periodic Function 14.47 Properties of Fourier Transform 14.50 Energy Density Spectrum 14.58 Exercises 14.60 Objective-Type Questions 14.62 Answers to Objective-Type Questions 14.64

15. FILTERS AND ATTENUATORS 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9 15.10 15.11 15.12 15.13 15.14 15.15 15.16 15.17 15.18 15.19

Introduction 15.1 Classification of Filters 15.1 T-Network 15.1 p Network 15.4 Characteristic of Filters 15.6 Constant-k Low Pass Filter 15.7 Constant-k High-pass Filter 15.14 Band-pass Filter 15.18 Band-stop Filter 15.22 m-Derived Filters 15.25 m-Derived Low-Pass Filter 15.28 m-Derived High-Pass Filter 15.31 Terminating Half Sections 15.34 Composite Filter 15.37 Attenuator 15.40 Lattice Attenuator 15.41 T-Type Attenuator 15.42 p -Type Attenuator 15.45 Ladder-Type Attenuator 15.47 Exercises 15.48 Objective-Type Questions 15.49 Answers to Objective-Type Questions

15.1

15.50

16. NETWORK SYNTHESIS 16.1 16.2 16.3 16.4 16.5 16.6 16.7

Index I.1

Introduction 16.1 Hurwitz Polynomials 16.1 Positive Real Functions 16.16 Elementary Synthesis Concepts 16.24 Realisation of LC Functions 16.30 Realisation of RC Functions 16.47 Realisation of RL Functions 16.63 Exercises 16.72 Objective-Type Questions 16.74 Answers to Objective-Type Questions 16.76

16.1

Preface Overview Network Analysis and Synthesis is one of the core subjects for students of Electrical Engineering, Electronics Engineering, Electronics and Telecommunication Engineering, Instrumentation Engineering, Biomedical Engineering and related branches of engineering in the third/fourth semester course of almost all universities in India. This subject is also one of the important topics of competitive examinations such as IAS, IES, etc. and examinations conducted by various public-sector undertakings in this field.

About the Book Network Analysis and Synthesis is a complete solution for the undergraduate student who wishes to prepare for university and competitive examinations. This highly illustrative text makes the subject matter interesting and easy for students. As per the course requirement, five major topics that are mandatory to be covered by any book on the subject are dc and ac circuits, transient analysis, network function, two-port networks and network synthesis. Generally, numerical problems are expected in university examinations in this subject. The weightage given to problems is more than 70-80% in examinations. This book attempts to cover almost all the topics and solved problems on these important topics. The simplistic presentation style sequentially takes the readers from basic to advance concepts and the crisp theory helps in ready understanding. Numerous solved examples and graded problems have been provided in this book, especially in important chapters from the examination viewpoint. Objective-type questions from various papers of competitive examinations are included in each chapter. This will help students in sharpening their knowledge about core concepts. Covering both analysis and synthesis of networks, the text provides a simple explanation about the concepts of electrical networks with brief theory and a large number of problems. Numerous examples and exercise problems have been included to help the reader develop an intuitive grasp of the contents.

Salient Features ● ● ● ● ●

Span of coverage meets curriculum requirements Simplistic presentation and logical sequencing of topics Graded problem-solving approach Numerous solved examples from frequently asked examination questions Rich pedagogy * * *

Illustrations: 1320 Solved Examples: 1065 Practice Problems: 340

Chapter Organisation The entire text is organised into 16 chapters, primarily based on the electrical and electronics engineering syllabi in BE/B.Tech undergraduate courses of universities around the country. The summary of the book can be organised in the following major parts: ● ● ●

Circuit Elements and Network Analysis Network Theorems for dc and ac Circuits Resonance and Coupled Circuits

xiv Preface ● ● ● ●

Network Topology/Graph Theory Transient Analysis and Laplace Transforms Two-Port Networks Network Functions

Within this structure, a more in-depth and detailed breakdown can be obtained from the table of contents. The fluid flow of text dealing with different topics as well as illustrations, examples, questions and numerical problems are extremely relevant and appropriate for students as well as instructors. The main strength of the book thus is its strong focus towards students’ readability and understanding with the scope for independent study and problem-solving skill development. I am confident that the depth of this book has been judiciously developed so that students not only treat this as a textbook, but, in addition, can also gather enough practical and theoretical knowledge to appear in national-level competitive examinations and interviews.

Web Supplements The text is supplemented with an exhaustive Online Learning Center, which can be accessed at https://www.mhhe.com/singh/nas1

Acknowledgements I would like to thank all the staff at McGraw Hill Education (India), especially Shalini Jha, Koyel Ghosh, Sagar Divekar, Satinder Singh Baveja, Sohini Mukherjee and Anuj Kr. Shrivastava for coordinating with me during the editorial, copyediting and production stages of this book. My acknowledgements would be incomplete without a mention of the contribution of my family members. I feel indebted to my father and mother for their lifelong inspiration. I also send a heartfelt thanks to my wife, Nitu; son, Aman; and daughter, Aditri, for always motivating and supporting me during the preparation of the project. Finally, I would like to appreciate the role of all those reviewers who, time and again have provided valuable inputs on the subject of Network Analysis and Synthesis which have significantly enriched this text and provide their constructive comments and inputs. Their names are given below: Gagan Deep Yadav

Yamuna Institute of Engineering and Technology, Yamuna Nagar, Haryana

Mukesh Kumawat

Star Academy Technology and Management, Indore, Madhya Pradesh

Kshipra Kshingwekar

Jai Narain College of Technology (JNCT), Bhopal, Madhya Pradesh

Santosh Sharma

Rajasthan Technical University, Kota, Rajasthan

Nikhil Gupta

Malaviya National Institute of Technology, Jaipur, Rajasthan

Prithviraj Purkait

Haldia Institute of Technology, Haldia, West Bengal

K V V S R Chowdary

Kalinga Institute of Industrial Technology (KIIT) University, Bhubaneswar, Odisha

Dhananjay Rao

Kalinga Institute of Industrial Technology (KIIT) University, Bhubaneswar, Odisha

Pranita Joshi

Atharva College of Engineering, Mumbai, Maharashtra

Shamala Rajaram Mahadik Dr. J J Magdum College of Engineering, Kolhapur, Maharshtra

Preface xv

Ch. Venkata Ramesh

NITTE Meenakshi Institute of Technology, Bangalore, Karnataka

Rupesh K C

Siddaganga Institute of Technology, Tumkur, Karnataka

G V Chalapathi Rao

Maturi Venkata Subba Rao (MVSR) Engineering College, Hyderabad, Andhra Pradesh

P Sravani

Mathrusri Engineering College, Hyderabad, Andhra Pradesh

R Dhanasekaran

Syed Ammal Engineering College, Ramnad, Tamil Nadu

A Pon Raj

Easwari Engineering College, Chennai, Tamil Nadu

R Arun Chithra

KLN Engineering College, Madurai, Tamil Nadu

R Manimekalai

Selvam College of Technology, Namakkal, Tamil Nadu

S Ramesh

KSR College of Engineering, Nammakal, Tamil Nadu

Suggestions for improvements will always be welcome.

Ravish R. Singh Publisher’s Note Do you have any further request or a suggestion? We are always open to new ideas (the best ones come from you!). You may send your comments to [email protected] Piracy-related issues may also be reported!

1 1.1

Basic Network Concepts INTRODUCTION

We know that like charges repel each other whereas unlike charges attract each other. To overcome this force of attraction, a certain amount of work or energy is required. When the charges are separated, it is said that a potential difference exists and the work or energy per unit charge utilised in this process is known as voltage or potential difference. The phenomenon of transfer of charge from one point to another is termed current. Current (I) is defined as the rate of flow of electrons in a conductor. It is measured by the number of electrons that flow in unit time. Energy is the total work done in the electric circuit. The rate at which the work is done in an electric circuit is called electric power. Energy is measured in joules (J) and power in watts (W).

1.2

RESISTANCE

Resistance is the property of a material due to which it opposes the flow of electric current through it. Certain materials offer very little opposition to the flow of electric current and are called conductors, e.g., metals, acids and salt solutions. Certain materials offer very high resistance to the flow of electric current and are called insulators, e.g., mica, glass, rubber, Bakelite, etc. The practical unit of resistance is ohm and is represented by the symbol Ω. A conductor is said to have resistance of one ohm if a potential difference of one volt across its terminals causes a current of one ampere to flow through it. The resistance of a conductor depends on the following factors. (i) (ii) (iii) (iv)

It is directly proportional to its length. It is inversely proportional to the area of cross section of the conductor. It depends on the nature of the material. It also depends on the temperature of the conductor.

Hence, R∝

l A

R=ρ

l A

where l is length of the conductor, A is the cross-sectional area and r is a constant known as specific resistance or resistivity of the material.

1.2 Network Analysis and Synthesis 1. Power Dissipated in a Resistor We know that v = R i When current flows through any resistor, power is absorbed by the resistor which is given by p=vi The power dissipated in the resistor is converted to heat which is given by t

t

∫ v i dt

∫ R i i dt = i

0

Example 1.1

2

Rt

0

A 25 W resistor has a voltage of 150 sin 377 t. Find the corresponding current i and

power p. R = 25 Ω v 150 si 377 t v 150 sin 377 t i= = = 6 sin 377 t R 25

Solution

p

Example 1.2

vi

sin 2 337 t si

(150 si 377 t ) (6 sin i 377 t))

A current waveform shown in Fig. 1.1 is applied to a 2 W resistor. Draw the voltage

waveform. i (A )

5

0

1

2

3

4

t(s )

5

Fig. 1.1 v ( V)

Solution For the resistor, (a) For 0 (b) For 1 (c) For 3

1, 3, 4,

v

Ri

v v v

R i = 2 × 5 = 10 V Ri = 2 × 0 = 0 R i = 2 × 5 = 10 V

The voltage waveform is shown in Fig. 1.2.

1.3

10

0

1

2

3

4

t( s)

Fig. 1.2

INDUCTANCE

Inductance is the property of a coil that opposes any change in the amount of current flowing through it. If the current in the coil is increasing, the self-induced emf is set up in such a direction so as to oppose the rise of current. Similarly, if the current in the coil is decreasing, the self-induced emf will be in the same direction as the applied voltage. Inductance is defined as the ratio of flux linkage to the current flowing through the coil. The practical unit of inductance is henry and is represented by the symbol H. A coil is said to have an inductance of one henry if a current of one ampere when flowing through it produces flux linkages of one weber-turn in it.

1.3 Inductance 1.3

The inductance of an inductor depends on the following factors. (i) (ii) (iii) (iv)

It is directly proportional to the square of the number of turns. It is directly proportional to the area of cross section. It is inversely proportional to the length. It depends on the absolute permeability of the magnetic material.

Hence, L∝

N 2A l

L=μ

N 2A l

where l is the mean length, A is the cross-sectional area and m is the absolute permeability of the magnetic material. 1. Current–Voltage Relationships in an Inductor We know that v

L

di dt

Expressing inductor current as a function of voltage, di =

1 v dt L

di =

l v dt L ∫0

Integrating both the sides, i(t )

∫

t

i(0)

t

i( t ) =

1 v dt i(0) L ∫0

The quantity i(0) denotes the initial current through the inductor. When there is no initial current through the inductor, t

i( t ) =

1 v dt L ∫0

2. Energy Stored in an Inductor Consider a coil of inductance L carrying a changing current I. When the current is changed from zero to a maximum value I, every change is opposed by the self-induced emf produced. To overcome this opposition, some energy is needed and this energy is stored in the magnetic field. The voltage v is given by v

L

di dt

Energy supplied to the inductor during interval dt is given by dE = v i dt = L

di i dt dt

L i dt

1.4 Network Analysis and Synthesis Hence, total energy supplied to the inductor when current is increased from 0 to I amperes is I

I

∫ ddEE

E

∫ L i di

0

0

5(1 − e

Example 1.3

An inductance of 3 mH has a current i age and the maximum stored energy. Solution

1 2 LI 2 5000t

). Find the corresponding volt-

L = 3 mH i v I max

5(1 e −5000 t ) di d 5000 t ⎤ 5000 5000t = 3 × 10 −3 ⎡⎣5(1 − e −5000 ) ⎦ = 3 × 10 −3 (5 × 5000 0 e −5000t ) dt dt i(∞) = 5(( − e −∞∞ ) = 5 A L

75 e

5000 t

1 1 2 L I max = × 3 × 10 −3 × ( ) 2 = 37.5 mJ 2 2

Emax

Example 1.4

A current waveform is applied to a 2 H inductor. Draw the voltage waveform for

Fig. 1.3. i (A)

6

0

2

4

6

8

t( s)

Fig. 1.3 Solution (a) For 0

v

L

di dt

2, v = 2×

(6 − 0 ) =6V ( 2 − 0)

6

(b) For 2 < t < 6, v = 2× (c) For 6

v( V)

(6 − 6 ) =0 ( 6 − 2) 0

8, v = 2×

( 0 − 6) = −6 V (8 − 6)

The voltage waveform is shown in Fig 1.4.

2

4

6

8

−6

Fig. 1.4

t ( s)

1.3 Inductance 1.5

Example 1.5

A voltage waveform shown in Fig. 1.5 is applied across a 1 H inductor. Draw the

current waveform. v ( V)

1

0

1

2

t ( s)

3

Fig. 1.5 t

i=

Solution (a) For 0

1 v dt i(0) L ∫0

1, v =1 t

i

1 1dt d i(0) = [t ]t0 dt 1 ∫0

0

t

i(1) = 1 A (b) For 1 t

2, v=0 t

i

1 0 ddt i(1) = 0 + 1 = 1 A 1 ∫1

i ( A)

i( 2) = 1 A (c) For 2

3, v =1

2 t

i

1 dt i( 2) = [t ]t2 1 t − 2 + 1 = t − 1 1 ∫2

1 t ( s)

i(3) = 3 − 1 = 2 A

0

1

The current waveform is shown in Fig. 1.6.

Example 1.6

2

3

4

Fig. 1.6

A voltage waveform shown in Fig. 1.7 is applied across a 2 H inductor. Draw the

current waveform. v ( V) 3 2

0 −1

2

4

Fig. 1.7

6

8

t ( s)

1.6 Network Analysis and Synthesis t

1 v ( t ) + i( 0) L ∫0

i=

Solution (a) For 0

2, v=2

i ( A) t

i

1 2 ddt i(0) = 0.5[2tt ]t0 2 ∫0

0

t 3.5

i( 2) = 2 A (b) For 2

8,

1.5

3 t + 3 = − 0.5t + 3 6

v

0

2

4

6

t ( s)

8

t

i

1 ( 0.5t + 3) dt dt + i( 2) 2 ∫2

Fig. 1.8

t

⎡ ⎤ t2 = 0.5 ⎢ − 0.5 + 3 ⎥ + 2 = 0.5 5[[ −0 0.25 5 2 + 3t + 0.5 × 2 − 3 × 2] + 2 = 0.5[ −0.25 2 ⎣ ⎦2 i(8) 0.5 5[[ 0.25(64) 3(8) 5] 2 = 3 5 A

2

3

5] 2

The current waveform is shown in Fig. 1.8.

Example 1.7 R

2 Ω and L

The following current waveform i(t) is passed through a series RL circuit with 2 mH . Find the voltage across each element and sketch the same. i ( A) 5

0

1

2

3

4

5

6

7

8

t (ms)

−5

Fig. 1.9 Solution (a) Voltage across the resistor of 2 Ω vR Ri Voltage across the resistor is a trapezoidal waveform with a peak value of 10 V. (b) Voltage across the inductor of 2 mH vL

L

di dt

1.4 Capacitance 1.7

For 0

i ( A)

1 ms,

⎛ 5−0 ⎞ vL = 2 × 10 −3 ⎜ = 10 V ⎝ 1 × 10 −3 − 0 ⎟⎠ For 1 s

5

3 ms,

0

−5 − 5 ⎛ ⎞ vL = 2 × 10 −3 ⎜ =0 − 3 − 3 ⎝ 3 × 10 − 1 × 10 ⎟⎠ For 3

s

2

3

1

2

2

4

5

6

7

8

3

5

6

7

8

3

5

6

t (ms)

−5 vR ( V) 10

5 ms,

5−5 ⎛ ⎞ vL = 2 × 10 −3 ⎜ = −10 V ⎝ 5 × 10 −3 − 3 × 10 −3 ⎟⎠ For 5 ms < t < 7 ms, −5 + 5 ⎛ ⎞ vL = 2 × 10 −3 ⎜ =0 ⎝ 7 × 10 −3 − 5 × 10 −3 ⎟⎠ For 7 ms < t < 8 ms,

0

The voltage waveforms are shown in Fig 1.10.

t (ms)

−10 vL ( V) 10

0

0+5 ⎛ ⎞ vL = 2 × 10 −3 ⎜ = 10 V ⎝ 8 × 10 −3 − 7 × 10 −3 ⎟⎠

1.4

1

1

7

8

t (ms)

−10

Fig. 1.10

CAPACITANCE

Capacitance is the property of a capacitor to store an electric charge when its plates are at different potentials. If Q coulombs of charge is given to one of the plates of a capacitor and if a potential difference of V volts is applied between the two plates then its capacitance is given by C=

Q V

The practical unit of capacitance is farad and is represented by the symbol F. A capacitor is said to have capacitance of one farad if a charge of one coulomb is required to establish a potential difference of one volt between its plates. The capacitance of a capacitor depends on the following factors. (i) It is directly proportional to the area of the plates. (ii) It is inversely proportional to the distance between two plates. (iii) It depends on the absolute permittivity of the medium between the plates. Hence, A d A C=ε d where d is the distance between two plates, A is the cross-sectional area of the plates and e is absolute permittivity of the medium between the plates. C∝

1.8 Network Analysis and Synthesis 1. Current–Voltage Relationships in a Capacitor The charge on a capacitor is given by q = Cv where q denotes the charge and v is the potential difference across the plates at any instant. We know that dq d dv i= = Cv = C dt dt dt Expressing capacitor voltage as a function of current, dv =

1 i dt C

dv

1 i dt C ∫0

Integrating both the sides, v(t )

∫

t

v( )

t

v(t ) =

1 i dt v( ) C ∫0

The quantity v (0) denotes the initial voltage across the capacitor. When there is no initial voltage on the capacitor, t

v(t ) =

1 i dt C ∫0

2. Energy Stored in a Capacitor Let a capacitor of capacitance C farads be charged from a source of V volts. Then current i is given by i

C

dv dt

Energy supplied to the capacitor during interval dt is given by dE = v i dt = v C

dv dt dt

Hence, total energy supplied to the capacitor when potential difference is increased from 0 to V volts is E

Example 1.8

V

V

0

0

∫ ddEE ∫ C v dv

A voltage is defined by ⎧0 ⎪ v(t ) = ⎨2t ⎪4 e −(t( t ⎩

and is applied to the 10 μ F capacitor. Find i(t). Solution

1 CV 2 2

i

C

dv dt

)

t 2s

1.4 Capacitance 1.9

For t < 0, i=0 For 0

2 s, v

2t

i = 10 × 10 −6 For t

d ( 2tt ) = 20 μ A dt

2 s, v

4 e − ( t − 2)

i = 10 × 10 −6

Example 1.9

d [4ee − ( t dt

2)

]

0

0 6[

e

( t − 2)

] = −40 e − ( t −2) μ A

A voltage waveform shown in Fig. 1.11 is applied to the capacitor. Draw the current

waveform. v (V ) C=2F

4 2 0

1

2

3

4

t (s)

−2

Fig. 1.11 Solution (a) 0 At t At t

i

dv C dt

i

C

dv ⎛ 2 − 2⎞ = 2⎜ =0 ⎝ 1 − 0 ⎟⎠ dt

i

C

dv ⎛ 4 − 2⎞ = 2⎜ =4A ⎝ 2 − 1 ⎟⎠ dt

1 0, v = 2 V 1, v = 2 V

(b) 1 t 2 At t 1, v = 2 V At t 2, v = 4 V i (A )

4

(c) 2 t 4 At t 2, v = 4 V At t 4, v = −2 V

0

i

C

dv ⎛ −2 − 4 ⎞ ⎛ −6 ⎞ = 2⎜ ⎟⎠ = 2 ⎜⎝ ⎟⎠ = −6 A ⎝ dt 4−2 2

The current waveform is shown in Fig. 1.12.

1

2

3

−6

Fig. 1.12

4

t (s)

1.10 Network Analysis and Synthesis

Example 1.10

A voltage waveform shown in Fig. 1.13 is applied to the capacitor. Draw the current

waveform. v (V)

C = 1000 μF

50

≈ 2

0

≈

4 5

t (s)

6

−50

Fig. 1.13 Solution (a) 0 At t At t i (b) 2 At t At t

2 0, v = 0 V 2, v = 50 V

i (A) 25 × 10−3

dv ⎛ 50 − 0 ⎞ ⎛ 50 ⎞ C = 10 −3 ⎜ = 10 −3 ⎜ ⎟ = 25 × 10 −3 A ⎝ 2 − 0 ⎟⎠ ⎝ 2⎠ dt 0 5 2, v = 50 V −33.33 × 10−3 5, v = −50 V

2

dv ⎛ −50 − 50 ⎞ ⎛ −100 ⎞ = 10 −3 ⎜ = 10 −3 ⎜ = −33.33 × 10 −3 A ⎝ 5 − 2 ⎟⎠ ⎝ 3 ⎟⎠ dt (c) 5 6 At t 5, v = −50 V At t 6, v = −50 V i

C

i

C

4

5

6

t (s)

Fig. 1.14

dv ⎛ −50 + 50 ⎞ = 10 −3 ⎜ =0 ⎝ 6 − 5 ⎟⎠ dt

The current waveform is shown in Fig. 1.14.

Example 1.11

A current waveform shown in Fig. 1.15 is applied to a 0.1 F capacitor. Draw the

voltage waveform. i (A) 2 1 0

1

2

3

−2

Fig. 1.15

4

t (s)

1.4 Capacitance 1.11 t

v=

Solution (a) For 0 i=2

1 i dt v(0) C ∫0

1, t

v

1 2 dt d v(0) = 10[2tt ]t0 0 1 ∫0

0

20 t

v(1) = 20 V (b) For 1 t i =1

2, vC (V)

t

v

1 dt v(1) = 10[t ]1t 0 1 ∫1

20 10(t − 1) + 20 30

v( 2) = 10( 2 − 1) + 20 = 30 3 V (c) For 2 t 4, i = −2

20

t

v

1 0 1 ∫2

dt v( 2) = 10[ −2tt ]t2

30 10( 2t + 4) + 30

1

0

2

3

4

t (s)

−10

v( 4) = 10( −2 × 4 + 4) + 30 = −10 V Fig. 1.16

The voltage waveform is shown in Fig. 1.16.

Example 1.12 circuit with R = 1 W, L

A current with periodic waveform shown in Fig. 1.17 is applied to a series RLC 1 mH and C μ F. F Sketch the voltage across each element. i (A) 2

0

1

2

3

4

t (ms )

Fig. 1.17 Solution (a) Voltage across the resistor of 1 Ω vR R i Voltage across the resistor is a triangular waveform with peak value of 2 A and slope 2 103. (b) Voltage across inductor of 1 mH di vL L dt For 0 1 ms, ⎛ 2−0 ⎞ vL = 1 × 10 −3 ⎜ =2V ⎝ 1 × 10 −3 − 0 ⎟⎠ For 1 s 2 ms, 0−2 ⎛ ⎞ vL = 1 × 10 −3 ⎜ = −2 V ⎝ 2 × 10 −3 − 1 × 10 −3 ⎟⎠

1.12 Network Analysis and Synthesis Voltage across the inductor is a square waveform. (c) Voltage across capacitor of 100 μF t

vC = For 0

1 i dt v(0) C ∫0

1 ms, 2

i

100 × 10 −6 7

vC (

2⎞ ⎛ 4 2000t 7 2 2000 t dt + 0 = 1 0 ⎜ 2 ⎟ = 10 t ∫ ⎝ ⎠ 0 t

1

vC

For 1 s

t = 2000t

10 −3

((1 10 3 ) 2

10 V

2 ms, 2

i

10 −3

t + 4 = −2000t + 4

1

vC

t

∫

100 × 10 −6 1 ms = 10 4 ( −1000 1000

2

( 2000 000t + 4)dt + v(1 ms) = 10 4 ( −1000

2

+ 4 − 3 × 10 −3 ) + 10

+ 4 ) − 20

⎡ −1000( 2 × 10 −3 ) 2 + 4( 2 × 10 1 −3 ) ⎤⎦ − 20 = 20 V ⎣ Voltage across the capacitor keeps rising continuously. The voltage waveforms are shown in Fig. 1.18. vC ( 2

=

4

i (A) 2

0 v R (V) 2

1

2

3

4

0 v i (V) 2

1

2

3

4

0

1

2

3

4

t(ms)

t(ms)

t(ms)

−2 v c (V) 20 10

0

1

2

3

Fig. 1.18

4

t(ms )

1.4 Capacitance 1.13

Example 1.13 Draw the waveform for iR iL , iC for the network shown in Fig. 1.19 (a) when it is excited by a voltage source having a waveform shown in Fig. 1.19 (b). v (V)

iR v (t)

iC

iL

10 Ω

2F

5H

10

0

2

4

t (s)

6

−10 (a)

Fig. 1.19 Solution (a) Current through the resistor of 10 Ω v R Current through the resistor is a rectangular waveform with a peak value of 1 A. (b) Current through the inductor of 5 H iR =

t

iL = For 0

1 v dt i(0) L ∫0

2, v = 10

v (V)

t

iL

1 0 dt i(0) = 0.2[10tt ]t0 5 ∫0

0

2t

0

iL ( 2) = 2( 2) = 4 A For 2 t 4, v = −10 iL

0 dt i( 2) =

6

2

4

6

2

4

6

t (s)

1

0.2[ −10t ]t2

+4

⎛ 10 − 10 ⎞ iC = 2 ⎜ =0 ⎝ 2 − 0 ⎟⎠ t

4

i R (A)

= 0.2( −10t 20) 4 2t + 8 iL ( 4) 2( 4) 8 = 0 (c) Current through the capacitor of 2 F dv iC C dt For 0 2,

For 2

2

−10

t

1 5 ∫2

10

4,

⎛ −10 + 10 ⎞ iC = 2 ⎜ =0 ⎝ 4 − 2 ⎟⎠ The current waveforms are shown in Fig. 1.20.

0

t (s)

−1 i L (A) 4

0

t (s)

i C (A) t (s)

0

Fig. 1.20

1.14 Network Analysis and Synthesis

1.5

SOURCES

Source is a basic network element which supplies energy to the networks. There are two classes of sources, namely, 1. Independent sources 2. Dependent sources

1.5.1 Independent Sources Output characteristics of an independent source are not dependent on any network variable such as a current or voltage. Its characteristics, however, may be time-varying. There are two types of independent sources: 1. Independent voltage source 2. Independent current source 1. Independent Voltage Source An independent voltage source is a two-terminal network element that establishes a specified voltage across its terminals. The value of this voltage at any instant is independent of the value or direction of the v (t) V current that flows through it. The symbols for such voltage sources are shown in Fig. 1.21. The terminal voltage may be a constant, or it may be some (a) (b) specified function of time. 2. Independent Current Source An independent current Fig. 1.21 Symbols for independent source is a two-terminal network element which produces a voltage source specified current. The value and direction of this current at any instant is independent of the value or direction of the voltage that appears across the terminals of the source. The symbols for such current sources are shown in Fig. 1.22. i (t) I The output current may be a constant or it may be a function of time.

1.5.2 Dependent Sources

(a)

(b)

If the voltage or current of a source depends in turn upon some other voltage or current, it is called as dependent or controlled source. The Fig. 1.22 Symbols for independent dependent sources are of four kinds, depending on whether the current source control variable is voltage or current and the controlled source is a voltage source or current source. 1. Voltage-Controlled Voltage Source (VCVS) A voltage-controlled voltage source is a four-terminal network component that establishes a voltage vcd between c two points c and d in the circuit that is proportional to a a + + + voltage vab between two points a and b. mvab vcd vab − The symbol for such a source is shown in − − d b Fig. 1.23. The (+) and (−) sign inside the diamond of the component symbol identifies the component as a Fig. 1.23 Symbol for VCVS voltage source. vcd = m vab The voltage vcd depends upon the control voltage vab and the constant m, a dimensionless constant called voltage gain.

1.6 Some Definitions 1.15

2. Voltage-Controlled Current Source (VCCS) A voltage-controlled current source is a four-terminal network component that establishes a current icd in a branch of the circuit that is proportional to the voltage vab between two points a and b. The symbol for such a source is shown in Fig. 1.24. The arrow inside the diamond of the component symbol identifies the component as a current source. icd = gm vab

icd

a

gmvab

vab b

c

+

+

vcd −

−

d

Fig. 1.24 Symbol for VCCS

The current icd depends only on the control voltage vab and the constant gm, called the transconductance or mutual conductance. The constant gm has dimension of ampere per volt or siemens (S). 3. Current-Controlled Voltage Source (CCVS) A current-controlled voltage source is a four-terminal network component that establishes a voltage vcd between two points c and d in the circuit that is proportional to the current iab in some branch of the circuit. The symbol for such a source is shown in Fig. 1.25. vcd = r iab

a

b

iab +

+

+ r iab −

−

c

vcd −

d

Fig. 1.25 Symbol for CCVS

The voltage vcd depends only on the control current iab and the constant r called the transresistance or mutual resistance. The constant r has dimension of volt per ampere or ohm (Ω). a

4. Current-Controlled Current Source (CCCS) A current-controlled current source is a four-terminal network component that establishes a current icd in one branch of a circuit that is proportional to the current iab in some branch of the network. The symbol for such a source is shown in Fig. 1.26. icd = b iab

icd

iab

+

+

c

b iab b

−

−

Fig. 1.26

d

Symbol for CCCS

The current icd depends only on the control current iab and the dimensionless constant b, called the current gain.

1.6

SOME DEFINITIONS

1. Network and Circuit The interconnection of two or more circuit elements (viz., voltage sources, resistors, inductors and capacitors) is called an electric network. If the network contains at least one closed path, it is called an electric circuit. Every circuit is a network, but all networks are not circuits. Figure 1.27(a) shows a network which is not a circuit and Fig. 1.27(b) shows a network which is a circuit.

L

R

C

(a)

Fig. 1.27

L

R

V

C

(b)

(a) Network which is not a circuit (b) Network which is a circuit

1.16 Network Analysis and Synthesis 2. Linear and Non-linear Elements If the resistance, inductance or capacitance offered by an element does not change linearly with the change in applied voltage or circuit current, the element is termed as linear element. Such an element shows a linear relation between I voltage and current as shown in Fig. 1.28. Ordinary resistors, capacitors and inductors are examples of linear elements. A non-linear circuit element is one in which the current does not change linearly with the change in t applied voltage. A semiconductor diode operating in the en t m curved region of characteristics as shown in Fig. 1.28 is en le m E e l common example of non-linear element. ar rE ne ea Li Other examples of non-linear elements are voltagen Li ndependent resistor (VDR), voltage-dependent capacitor No (varactor), temperature-dependent resistor (thermistor), lightV 0 dependent resistor (LDR), etc. Linear elements obey Ohm’s law whereas non-linear elements do not obey Ohm’s law. 3. Active and Passive Elements An element Fig. 1.28 V-I characteristics of linear and non-linear elements which is a source of electrical signal or which is capable of increasing the level of signal energy is termed as active element. Batteries, BJTs, FETs or OP-AMPs are treated as active elements because these can be used for the amplification or generation of signals. All other circuit elements, such as resistors, capacitors, inductors, VDR, LDR, thermistors, etc., are termed passive elements. The behaviour of active elements cannot be described by Ohm’s law. 4. Unilateral and Bilateral Elements If the magnitude of current flowing through a circuit element is affected when the polarity of the applied voltage is changed, the element is termed unilateral element. Consider the example of a semiconductor diode. Current flows through the diode only in one direction. Hence, it is called an unilateral element. Next, consider the example of a resistor. When the voltage is applied, current starts to flow. If we change the polarity of the applied voltage, the direction of the current is changed but its magnitude is not affected. Such an element is called a bilateral element. 5. Lumped and Distributed Elements A lumped element is the element which is separated physically, like resistors, inductors and capacitors. Distributed elements are those which are not separable for analysis purposes. Examples of distributed elements are transmission lines in which the resistance, inductance and capacitance are distributed along its length. 6. Active and Passive Networks A network which contains at least one active element such as an independent voltage or current source is an active network. A network which does not contain any active element is a passive network. 7. Time-invariant and Time-variant Networks A network is said to be time-invariant or fixed if its input–output relationship does not change with time. In other words, a network is said to time-invariant, if for any time shift in input, an identical time-shift occurs for output. In time-variant networks, the input–output relationship changes with time.

1.7

SERIES AND PARALLEL COMBINATIONS OF RESISTORS

Let R1 R2 and R3 be the resistances of three resistors connected in series across a dc voltage source V as shown in Fig. 1.29. Let V1 V2 and V3 be the voltages across resistances R1 R2 and R3 respectively. In series combination, the same current flows through each resistor but voltage across each resistor is different.

I

R1

R2

R3

V1

V2

V3

V

Fig. 1.29 Series combination of resistors

1.7 Series and Parallel Combinations of Resistors 1.17

V V1 + V2 V3 RT I = R1 I + R2 I + R3 I R1 + R2

RT

R3

Hence, when a number of resistors are connected in series, the equivalent resistance is the sum of all the individual resistance. 1. Voltage Division and Power in a Series Circuit I= V1 V2 V3 Total power

PT

V R1 + R2 + R3 R1 R1 I = V R1 + R2 + R3 R2 R2 I = V R1 + R2 + R3 R3 R3 I = V R1 R2 + R3 P1 + P2 P3

= I 2 R1 + I 2 R2 + I 2 R3 =

I

V12 V22 V32 + + R1 R2 R3

Figure 1.30 shows three resistors connected in parallel across a dc voltage source V. Let I1 I 2 and I 3 be the current flowing through resistors R1 R2 and R3 respectively. In parallel combination, the voltage across each resistor is same but current through each resistor is different.

I1

R1

I2

R2

I3

R3

V

Fig. 1.30

Parallel combination of resistors

I = I1 + I 2 I 3 V V V V = + + RT R1 R2 R3 1 1 1 1 = + + RT R1 R2 R3 R1 R2 R3 RT = R2 R3 + R3 R1 + R1 R2 Hence, when a number of resistors are connected in parallel, the reciprocal of the equivalent resistance is equal to the sum of reciprocals of individual resistances. 2. Current Division and Power in a Parallel Circuit V

RT I

R1 I1

R2 I 2

R3 I 3

V R I R2 R3 = T = I R1 R1 R1 R2 + R2 R3 + R3 R1 V R I R1 R3 I2 = = T = I R2 R2 R1 R2 + R2 R3 + R3 R1 V R I R1 R2 I3 = = T = I R3 R3 R1 R2 + R2 R3 + R3 R1 I1 =

1.18 Network Analysis and Synthesis Total power

P1 + P2

PT = =

+

P3 I 22 R2 + I 32 R3 2 3

V V V + + R1 R2 R3

R1 R2 R1 R2 V RT I R1 I1 R2 I 2 V R I R2 I1 = = T = I R1 R1 R1 R2 V R I R1 I2 = = T = I R2 R2 R1 R2

RT =

Note: For two branch circuits,

Example 1.14

I12 R1 2

Find an equivalent resistance between A and B in the network of Fig. 1.31. B

A 10 Ω

10 Ω

10 Ω

Fig. 1.31 Solution Marking all the junctions and redrawing the network (Fig. 1.32), A

C

10 Ω

D

10 Ω

B 10 Ω

(a) D

A 10 Ω

10 Ω

10 Ω

B (b)

C

Fig. 1.32 RAB = 10 � 10 � 10 = 3.33 Ω

Example 1.15

Find an equivalent resistance between A and B in the network of Fig. 1.33. 6Ω A 8Ω

10 Ω

B 12 Ω

Fig. 1.33

1.7 Series and Parallel Combinations of Resistors 1.19

Solution Marking all the junctions and redrawing the network (Fig. 1.34), 6Ω A 8Ω

D

A

C

8Ω

10 Ω

B

D

6Ω

10 Ω

12 Ω

B

12 Ω

C (b)

(a)

Fig. 1.34 RAB = 8 � 6 � 10 � 12 = 2.11 Ω

Example 1.16

Find the equivalent resistance between A and B in the network of Fig. 1.35. 2Ω

15 Ω

B A 10 Ω

8Ω

10 Ω

20 Ω 30 Ω

40 Ω

Fig. 1.35 Solution Marking all the junctions and redrawing the network (Fig. 1.36), 2Ω B

C

15 Ω 2Ω

A

10 Ω B

8Ω

25 Ω

D

F

10 Ω

20 Ω

30 Ω

E

40 Ω

A

8Ω

2Ω

C, E

D, F (b)

B 50 Ω

12.5 Ω

50 Ω A

8Ω (c)

(a)

Fig. 1.36 RAB = 22.5 Ω

Example 1.17

What is the resistance between the terminals A and B in network of Fig. 1.37 when the potential difference between C and D is zero?

1.20 Network Analysis and Synthesis C

R R

R A R

B

D R

Fig. 1.37 Solution When the potential difference between C and D is zero, the resistor connected between C and D is shorted. Hence, the points C and D are at same potential. Redrawing the network (Fig. 1.38), Simplifying the network (Fig. 1.39), R C, D

A

R 2 A

R R

R

R 2

B

B

Fig. 1.38

Fig. 1.39 RAB =

Example 1.18

R R + =RΩ 2 2

Determine the current delivered by the source in the network of Fig. 1.40.

2Ω

2Ω

4Ω 2Ω 30 V

2Ω

1Ω

Fig. 1.40

3Ω

1Ω

2Ω

1.7 Series and Parallel Combinations of Resistors 1.21

Solution The network can be simplified by series–parallel reduction technique (Fig. 1.41). 4Ω 2Ω

30 V

4Ω 2Ω

2Ω

3Ω

2Ω

30 V

2Ω

1.5 Ω

3Ω 1Ω

1Ω

(b)

(a) 3.5 Ω 30 V

2Ω

2Ω

2Ω

30 V

1.27 Ω

1Ω

1Ω (d)

(c)

2.27 Ω

2Ω

30 V

1.06 Ω

30 V

I (f)

(e)

Fig. 1.41 I=

Example 1.19

30 = 28.3 A 1 06

Find the current delivered by the source in the network of Fig. 1.42. 2Ω

2Ω 2Ω

4Ω

2Ω

4Ω

4Ω

50 V

5Ω

2Ω

2Ω

1Ω

Fig. 1.42

3Ω

1Ω

2Ω

1.22 Network Analysis and Synthesis Solution The network can be simplified by series–parallel reduction technique (Fig. 1.43). 4Ω 4Ω

5Ω

50 V

4Ω 2Ω

2Ω

4Ω

4Ω

2Ω 2Ω

5Ω

50 V

3Ω

2Ω

3Ω

4Ω 2Ω

1.5 Ω

1Ω

1Ω (a)

(b)

5Ω

50 V

2Ω

2Ω 2Ω

1.5 Ω

1Ω (c) 3.5 Ω 50 V

2Ω

5Ω

1.27 Ω

5Ω

50 V

2Ω

2Ω

1Ω

1Ω

(d)

50 V

5Ω

(e)

0.88 Ω

2.27 Ω 50 V

2Ω

I (f)

(g)

Fig. 1.43 I=

50 = 56.82 82 A 0 88

Example 1.20 Three equal resistors of 30 W each are connected in parallel across a 120 V dc supply as shown in Fig. 1.44. What is the current through each of them (a) if one of the resistors burns out, or (b) if one of the resistors gets shorted? + 120 V

I1

I2 30 Ω

−

Fig. 1.44

I3 30 Ω

30 Ω

1.7 Series and Parallel Combinations of Resistors 1.23

Solution (a) If one of the resistors burns out, it will act as an open circuit (Fig. 1.45). +

I3 = 0 I1

I2 =

120 =4A 30

I1

I2 30 Ω

120 V

I3 30 Ω

−

Fig. 1.45 (b) If one of the resistors gets shorted, the effective resistance becomes zero (Fig. 1.46). +

I1

I2 = 0

I1

I2 30 Ω

120 V

I3 30 Ω

−

Fig. 1.46

Example 1.21

A lamp rated at 100 V, 75 W is to be connected across a 230 V supply. Find the value of resistance to be connected in series with the lamp. Also find the power loss occuring in the resistor. Solution The network is shown in Fig. 1.47. V1

P1 = 75 W, V = 230 V

100 I

R

L

100 V 230 V

Fig. 1.47 (a) Value of resistance Rated current of the lamp I=

P1 75 = = 0.75 75 A V1 100

Lamp will operate normally on 230 V supply if the current flowing through the lamp remains the rated current, i.e., 0.75 A. Voltage across the resistor R V2 = 230 − 100 = 130 V R=

130 = 173.33 Ω 0 75

P2 =

V22 (130) 2 = = 97.55 W R 173.33

(b) Power loss occurring in the resistor

1.24 Network Analysis and Synthesis

Example 1.22 A 100 V, 60 W lamp is connected in series with a 100 V, 100 W lamp and the combination is connected across 200 V mains. Find the value of the resistance that should be connected in parallel with the first lamp so that each lamp may get the rated current at rated voltage. Solution The network is shown in Fig. 1.48. V1 = 100 V

R

P1 = 60 W L1

V2 = 100 V P2 = 100 W

L2

100 V

100 V

V = 200 V Rated current of the lamp L1

200 V

P 60 I1 = 1 = = 0.6 6A V1 100

Fig. 1.48

Rated current of the lamp L2 I2 =

P2 100 = =1A V2 100

Let R be the value of resistance that should be connected in parallel with the lamp L1 so that rated current flows through the lamp L1. Current through resistor R I = 1 − 0.6 = 0.4 A R=

V1 100 = = 250 Ω I 04

Example 1.23 A 100 V, 60 W lamp is connected in series with a 100 V, 100 W lamp across 200 V supply. What will be the current drawn by the lamps? What will be the power consumed by each lamp and will such a combination work? Solution The network is shown in Fig. 1.49. V1 V2

L1

P1 = 60 W P2 = 100 W

100 100

V = 200 V

L2

100 V

100 V 200 V

(a) Current drawn by the lamps Resistance of the lamp L1

Fig. 1.49 R1 =

V12 (100) 2 = = 166.67 67 Ω P1 60

R2 =

V22 (100) 2 = = 100 Ω P2 100

Resistance of the lamp L2

Current drawn by the lamps I=

V R1

R2

=

200 = 0.75 A 166.67 + 100

1.7 Series and Parallel Combinations of Resistors 1.25

(b) Power consumed by the lamps Power consumed by the lamp

L1

Power consumed by the lamp L2

I 2 R1 = (0.75) 2 ×166 166.67 = 93.75 W I 2 R2 = (0.75) 2 × 100 = 56.25 W

If a 100 V, 60 W lamp draws a power of 93.75 W, its filament will be overheated and will burn out. Hence, such a combination will not work.

Example 1.24

A resistor R is connected in series with two lamps of 12 V, 9 W across a 300 V supply. Find the value of R so that both the lamps operate at rated conditions. If one of the lamps is short circuited, find the current through the circuit and the power dissipited in each lamp. Solution The network is shown in Fig. 1.50.

L1

V1

V2 = 12 V

P1

P2 = 9 W

L2

12 V

12 V

V = 300 V

R

v3

300 V

(a) Value of R Rated current of lamps L1 and L2

Fig. 1.50 I=

P1 P2 9 = = = 0.75 A V1 V2 12

Resistance of the lamps L1 and L2 R1

R2 =

V12 V22 (12) 2 = = = 16 Ω P1 P2 9

Voltage across the resistor R V3 = 300 − 12 − 12 = 276 V R=

V3 276 = = 368 Ω I 0 75

(b) Current through the circuit if the lamp L2 is short circuited If the lamp L2 is short circuited, resistance across terminals of L2 is zero. If the new current is I′ then = I ′(16 + 368) I ′ = 0.78 A (c) Power dissipated in each lamp if the lamp L2 is short circuited ( I ′ ) 2 R1 = (0.78) 2 16

Power dissipated in the lamp

L1

Power dissipated in the lamp

L2 = 0

9.73 W

30 . It is connected in I series with a resistor across a 100 V supply. If the voltage across the arc and the resistor are equal, find the ohmic value of the resistor.

Example 1.25

A dc arc has voltage/current relation expressed by V

44 +

1.26 Network Analysis and Synthesis Solution The network is shown in Fig 1.51.

R

I

V = 100 V V1 V2 V V1 + V2 = 100 V1 V2 = 50

v1

v2 100 V

For a dc arc,

Fig. 1.51 30 V1 = 44 + I 30 50 = 44 + I I =5A

Ohmic value of the resistor R=

Example 1.26

V2 50 = = 10 Ω I 5

Find the voltage VAB in the network as shown in Fig. 1.52. 10 Ω

10 V

30 Ω

A

B

20 Ω

40 Ω

Fig. 1.52 Solution The resistors of 10 Ω and 20 Ω are in series. Similarly, the resistors of 30 Ω and 40 Ω are in series. By voltage-division rule, 20 VA = 10 × = 6 67 V 10 + 20 40 VB = 10 × = 5 71 V 30 + 40 VAB VA VB = 6.67 − 5.71 0.96 V

1.8

SERIES AND PARALLEL COMBINATION OF INDUCTORS

Let L1 L2 and L3 be the inductances of three inductors connected in series across an ac voltage source v as shown in Fig. 1.53. Let v1 v2 and v3 be the voltages across inductances L1 L2 and L3 respectively. In series combination, the same current flows through each inductor but the voltage across each inductor is different. v = v1 + v1 + v3 di di di di LT = L1 + L2 + L3 dt dt d dt dt LT L1 + L2 L3

L1

i v1

L2 v2

L3 v3

v

Fig. 1.53

Series connection of inductors

1.8 Series and Parallel Combination of Inductors 1.27

Hence, when a number of inductors are connected in series, the equivalent inductance is the sum of all the individual inductances. L1 i1 Figure 1.54 shows three inductors connected in parallel across an ac voltage source v. Let i1 i2 and i3 be the current L2 i2 i through each inductance L1 L2 and L3 respectively. In parallel combination, the voltage across each inductor L3 i3 is same but the current through each inductor is different. i 1 LT

i1 + i2 1

i3 1

1

∫ v ddt = L1 ∫ v dt + L2 ∫ v ddt + L3 ∫ v dt

v

LT =

Parallel connection of inductors

Fig. 1.54

1 1 1 1 = + + LT L1 L2 L3 L1 L2 L3 L1 L2 + L2 L3 + L3 L1

Hence, when a number of inductors are connected in parallel, the reciprocal of the equivalent inductance is equal to the sum of reciprocals of individual inductances.

Example 1.27

Find an equivalent inductance between terminals A and B in the network of Fig. 1.55. 8H 2H

1H

A

B 8H

Fig. 1.55 Solution The network can be simplified by series–parallel reduction technique (Fig. 1.56). 2H

4H

1H B

A (a) 7H

B

A (b)

Fig. 1.56 LAB = 7 H

Example 1.28

Find an equivalent inductance of the network shown in Fig 1.57. 4H 5H

4H

2H

3H

3H B

A 5H

4H

Fig. 1.57

1.28 Network Analysis and Synthesis Solution The network can be simplified by series–parallel reduction technique (Fig. 1.58). 5H

1.33 H

2H

3H

3H

A

B 5H (a) 5H

5H

1.33 H

3H

A

B 5H

(b) 5H

2.5 H

1.33 H

3H B

A (c) 11.83 H

B

A (d)

Fig. 1.58 LAB = 11.83 H

1.9

SERIES AND PARALLEL COMBINATION OF CAPACITORS

Let C1 C2 and C3 be the capacitances of three capacitors connected in series across an ac voltage source v as shown in Fig 1.59. Let v1 v2 and v3 be the voltages across capacitances C1 C2 C3 C1 C2 and C3 respectively. In series combination, the charge on each capacitor is same but v1 v2 v3 voltage across each capacitor is different. v = v1 + v2 + v3 1 1 1 1 v id dt = i dt + i ddt + i dt ∫ ∫ ∫ CT C1 C2 C3 ∫ Fig. 1.59 Series combination of 1 1 1 1 capacitors = + + CT C1 C2 C3 C1C2C3 CT = C1C2 + C2C3 + C3C1 Hence, when a number of capacitors are connected in series, the reciprocal of the equivalent capacitance is equal to the sum of reciprocals of individual capacitances. 1. Voltage Division in a Series Circuit Q

CT V C1V1 C2V2 C3V3 Q CT V C2 C3 V1 = = = V C1 C1 C1C2 + C2C3 + C3C1 Q CT V C1 C3 V2 = = = V C2 C2 C1C2 + C2C3 + C3C1 Q CT V C1 C2 V3 = = = V C3 C3 C1C2 + C2C3 + C3C1

1.9 Series and Parallel Combination of Capacitors 1.29

Figure 1.60 shows three capacitors connected in parallel across an ac voltage source v. Let i1 i2 and i3 be the current through each capacitance C1 C2 and C3 respectively. In parallel combination, the voltage across each capacitor is same but current through each capacitor is different. i i1 + i2 i3 dv dv dv dv CT = C1 + C2 + C3 dt dt d dt dt CT C1 + C2 C3 Hence, when a number of capacitors are connected in parallel, the equivalent capacitance is the sum of all the individual capacitance.

Example 1.29

i

i1

C1

i2

C2

i3

C3

v

Fig. 1.60

Parallel combination of capacitors

What is the equivalent capacitance between terminals A and B in the network of

Fig. 1.61. 2 μF 1 μF 2 μF 2 μF

A

B

Fig. 1.61 Solution The network can be simplified by series–parallel reduction technique (Fig. 1.62). 2 μF 1 μF

3 μF

1 μF

A

1 μF

A

B

B (b)

(a) 0.75 μF B

A (c)

Fig. 1.62 C AB = 0.75 μF

Example 1.30

Find the equivalent capacitance for the network of Fig. 1.63. 4 μF 2 μF

3 μF

A

3 μF

5 μF B

2 μF

Fig. 1.63

1.30 Network Analysis and Synthesis Solution The network can be simplified by series–parallel reduction technique (Fig. 1.64). 4 μF 2 μF

1.5 μF

5 μF

2 μF

A

B

5.5 μF

5 μF

A

B

2 μF

2 μF (a)

(b)

1.47 μF

5 μF

3.47 μF

A

5 μF

A

B

B

2 μF (c)

(d) 2.05 μF A

B (e)

Fig. 1.64 C AB = 2.05 05 μF

Example 1.31

What is equivalent capacitance between terminals A and B in the network of

Fig. 1.65?

6 μF

4 μF

A 4 μF 2 μF

B 6 μF

Fig. 1.65 Solution The network can be simplified by series–parallel reduction technique (Fig. 1.66). 6 μF

8 μF

A

6 μF

6 μF

A

2 μF

A

2 μF

3.429 μF

5.429 μF

6 μF B

B (a)

B (b)

Fig. 1.66

(c)

1.9 Series and Parallel Combination of Inductors 1.31

CAB = 2.85 μF

Example 1.32

What is equivalent capacitance between terminals A and B in the network of Fig. 1.67? 3 μF A 6 μF 3 μF

3 μF

6 μF

B

Fig. 1.67 Solution The network can be simplified by series–parallel reduction technique (Fig. 1.67). 3 μF A

A

9 μF

3 μF

2.25 μF

3 μF

6 μF

6 μF

B

B (a)

(b) A

8.25 μF

3 μF B (c)

Fig. 1.68 CAB = 11.25 μF

Example 1.33 A combination of four capacitors is shown in Fig. 1.69. Find the value of C to obtain an equivalent capacitance of 0.5 mF. C

0.2 μF

0.6 μF

0.8 μF

Fig. 1.69

1.32 Network Analysis and Synthesis Solution (C + .6)(0.2 + 0.8) = 0.5 C + 0.6 + 0.2 + 0.8 C C + 0.8 0.5 C = 0.22 Ceq =

C = 0.4 μF F

1.10

STAR-DELTA TRANSFORMATION

When a circuit cannot be simplified by normal series–parallel reduction technique, the star-delta transformation can be used. Figure 1.70 (a) shows three resistors RA, RB and RC connected in delta. Figure 1.70 (b) shows three resistors R1, R2 and R3 connected in star. 1

1

R1 RC

RB R3

R2 2

2

3

3

RA (a)

Fig. 1.70

(b)

(a) Delta networks

(b) Star Network

These two networks will be electrically equivalent if the resistance as measured between any pair of terminals is the same in both the arrangements.

1.10.1

Delta to Star Transformation

Referring to delta network shown in Fig. 1.70 (a), the resistance between terminals 1 and 2 RC ( RA RB ) = RC � ( RA RB ) = RA RB + RC Referring to the star network shown in Fig. 1.70 (b), the resistance between terminals 1 and 2 = R1 Since the two networks are electrically equivalent, R1 Similarly,

R2

and

R3

RC ( RA + RB ) RA RB + RC R ( R + RC ) R3 = A B RA RB + RC R ( R + RC ) R1 = B A RA RB + RC

R2 =

R2 .

...(1.1) ...(1.2) ...(1.3)

Subtracting Eq. (1.2) from Eq. (1.1), R1

R3 =

RB RC − RA RB RA RB + RC

...(1.4)

1.10 Star-delta Transformation 1.33

Adding Eq. (1.4) and Eq. (1.3), R1 = R2 =

Similarly,

R3 =

RA

RB RC RB + RC

RA

RA RC RB + RC

RA RB RA RB + RC

Thus, star resistor connected to a terminal is equal to the product of the two delta resistors connected to the same terminal divided by the sum of the delta resistors.

1.10.2

Star to Delta Transformation

Multiplying the above equations, R1 R2 = R2 R3 = R3 R1 =

RA RB RC2

...(1.5)

RB + RC ) 2

( RA

RA2 RB RC ( RA

...(1.6)

RB + RC ) 2 RA RB2 RC

( RA

...(1.7)

RB + RC ) 2

Adding Eqs (1.5), (1.6) and (1.7), R1 R2 + R2 R3 + R3 R1 =

RA RB RC ( RA + RB ( RA

RB + RC )

RC ) 2

=

RA RB RC RA RB + RC

= RA R1 = RB R2 = RC R3 Hence, RA =

R1 R2 + R2 R3 + R3 R1 = R2 R1

R3 +

R2 R3 R1

RB =

R1 R2 + R2 R3 + R3 R1 = R1 R2

R3 +

R3 R1 R2

RC =

R1 R2 + R2 R3 + R3 R1 = R1 R3

R2 +

R1 R2 R3

Thus, delta resistor connected between the two terminals is the sum of two star resistors connected to the same terminals plus the product of the two resistors divided by the remaining third star resistor. Note: (1) When three equal resistors are connected in delta (Fig. 1.71), the equivalent star resistance is given by RY =

RΔ RΔ R = Δ RΔ RΔ RΔ 3

RΔ

3RY

1.34 Network Analysis and Synthesis A

A

RY RΔ

RΔ RY

RY B

C

B

C

RΔ

Fig. 1.71 Equivalent star resistance for three equal delta resistors (2) Star-delta transformation can also be applied to network containing inductors and capacitors.

Example 1.34

Find an equivalent resistance between A and B in the network of Fig. 1.72. 4Ω 4.5 Ω

4.5 Ω 3Ω

3Ω

A 7.5 Ω

B 7.5 Ω

3Ω

Fig. 1.72 Solution Converting the two delta networks formed by resistors of 4.5 Ω, 3 Ω and 7.5 Ω into equivalent star networks (Fig. 1.73 and Fig. 1.74), 4Ω

R4

R3

R6

R1

B

A R5

R2 3Ω

Fig. 1.73 4Ω

R1 R2 R3

4.5 × 7.5 = 2.25 5Ω 4.5 + 7.5 + 3 7 5×3 R5 = =1 5Ω 4.5 + 7.5 + 3 4 5×3 R4 = =09Ω 4.5 + 7.5 + 3

0.9 Ω

0.9 Ω

R6 =

2.25 Ω

2.25 Ω

B

A 1.5 Ω

1.5 Ω 3Ω

Fig. 1.74

1.10 Star-delta Transformation 1.35

Simplifying the network (Fig. 1.75), 5.8 Ω

A

B 2.25 Ω

2.25 Ω 6Ω (a) 2.95 Ω

2.25 Ω

2.25 Ω

A

B (b) 7.45 Ω B

A (c)

Fig. 1.75 RAB = 7 45 Ω

Example 1.35

Find an equivalent resistance between A and B in the network of Fig. 1.76. 10 Ω A 10 Ω

C

10 Ω

D

B 10 Ω

10 Ω

Fig. 1.76 Solution Redrawing the network (Fig. 1.77), C 10 Ω

10 Ω

10 Ω

A

10 Ω

B

10 Ω D

Fig. 1.77 Converting the delta network formed by three resistors of 10 Ω into an equivalent star network (Fig. 1.78 and Fig. 1.79),

1.36 Network Analysis and Synthesis C

C

R2

10 Ω 3

10 Ω

10 Ω

10 Ω 3

R1 A

A

B

B 10 Ω 3

10 Ω

R3

10 Ω D

D

Fig. 1.79

Fig. 1.78

R2 = R3 =

R1

10 × 10 10 = Ω 10 + 10 + 10 3

Simplifying the network (Fig. 1.80), 40 Ω 3

10 Ω 3

10 Ω 3

A

B

20 Ω 3

A

B (b)

40 Ω 3 (a) 10 Ω A

B (c)

Fig. 1.80 RAB = 0 Ω

Example 1.36

Find an equivalent resistance between A and B in the network of Fig. 1.81. A

6Ω

9Ω

1.5 Ω

3Ω

4Ω

C

B 1Ω

Fig. 1.81

1.10 Star-delta Transformation 1.37

Solution Converting the star network formed by resistors of 3 Ω, 4 Ω and 6 Ω into an equivalent delta network (Fig. 1.82 and Fig. 1.83), A

A

1.5 Ω

9Ω

1.5 Ω

9Ω

13

R2

18 Ω

.5 9Ω

R3 B

C

Ω

R1

B

C

1Ω

1Ω

Fig. 1.82

Fig. 1.83 6 4 = 188 Ω 3 6 3 R2 = 6 3 + = 133.5 Ω 4 4 3 R3 = 4 3 + =9Ω 6 R1 = 6 4 +

Simplifying the network (Fig. 1.84), A

1.35 Ω

6Ω

RAB = 6 � (1.35 + 0.9) = 6 � 2.25 B

= 6 Ω

C 0.9 Ω

Fig. 1.84

Example 1.37

Find an equivalent resistance between A and N by solving outer delta ABC in Fig. 1.85. A

2Ω

12 Ω

12 Ω

N 2Ω

2Ω

C

B 12 Ω

Fig. 1.85

1.38 Network Analysis and Synthesis Solution Converting outer delta ABC into a star network (Fig. 1.86), A

A

4Ω

RY =

12 × 12 = 12 + 12 + 12

4Ω

Ω

2Ω 4Ω

N

2Ω

B

C

2Ω

M

B

C

Fig. 1.86 Simplifying the network (Fig. 1.87), 2Ω

M

2Ω

A

M

A 2Ω

2Ω 4Ω

4Ω 4Ω

2Ω

6Ω

6Ω

N

N (a)

(b)

2Ω

M

A

A A 4Ω

3Ω

N

4Ω

5Ω

2.22 Ω N

N (c)

(d)

(e)

Fig. 1.87 RAN = .

Example 1.38

Ω

Find an equivalent resistance between A and B in the network of Fig. 1.88. 4Ω 2Ω

41 Ω 6Ω

6Ω

17 Ω

A

15 Ω 15 Ω

4Ω

11 Ω B

Fig. 1.88

1.10 Star-delta Transformation 1.39

Solution The resistors of 2 Ω and 4 Ω and the resistors of 4 Ω and 11 Ω are connected in series (Fig. 1.89). 15 Ω

41 Ω 6Ω 15 Ω

6Ω

6Ω

15 Ω

17 Ω

A

B

Fig. 1.89 Converting the two outer delta networks into equivalent star networks (Fig. 1.90), 41 Ω 5Ω

2Ω

6 6 =2Ω 6 6 6 15 × 15 RY2 = =5Ω 15 + 15 + 15 RY1 =

2Ω

5Ω

A

B 17 Ω

2Ω

5Ω

Fig. 1.90 Simplifying the network (Fig. 1.91),

48 Ω

A

B 2Ω

5Ω 24 Ω (a)

2Ω

5Ω

16 Ω

B

A (b) 23 Ω A

B (c)

Fig. 1.91 RAB = 3 Ω

Example 1.39

Find an equivalent resistance between A and B in the network of Fig. 1.92. 15 Ω

20 Ω

A 25 Ω 30 Ω 45 Ω

35 Ω

B 40 Ω

Fig. 1.92

1.40 Network Analysis and Synthesis Solution Drawing the resistor of 30 Ω from outside (Fig. 1.93), 15 Ω

20 Ω

A 25 Ω

30 Ω

35 Ω

45 Ω B 40 Ω

Fig. 1.93 Converting the delta network formed by resistors of 20 Ω, 25 Ω and 35 Ω into an equivalent star network (Fig. 1.94), 15 Ω

A

R1 =

20 × 35 = 8 75 Ω 20 + 35 + 25

R2 =

20 × 25 = 6 25 5Ω 20 + 35 + 25

R3 =

35 × 25 = 100.94 9 Ω 20 + 35 + 25 2

R2

R1

45 Ω

30 Ω R3 B 40 Ω

Fig. 1.94

Redrawing the network (Fig. 1.95), 15 Ω

6.25 Ω

8.75 Ω

A 10.94 Ω 45 Ω

30 Ω 40 Ω

B

Fig. 1.95 Simplifying the network (Fig. 1.96), 15 Ω

6.25 Ω

A 50.94 Ω

45 Ω

38.75 Ω

B (a) 15 Ω

6.25 Ω

15 Ω

A

A 45 Ω

22.01 Ω

45 Ω B

B

(c)

(b)

Fig. 1.96 Continued

28.26 Ω

1.10 Star-delta Transformation 1.41 15 Ω

A

A

32.36 Ω

17.36 Ω B

B

(e)

(d)

Fig. 1.96 RAB = 32.36 Ω

Example 1.40

Find an equivalent resistance between A and B in the network of Fig. 1.97. 10 Ω

A

20 Ω

5Ω

5Ω

15 Ω 25 Ω

10 Ω 2Ω

B 5Ω

30 Ω

Fig. 1.97 Solution The resistors of 5 Ω and 25 Ω and the resistors of 10 Ω and 5 Ω are connected in series (Fig. 1.98). 10 Ω

20 Ω

A 5Ω

15 Ω

30 Ω

15 Ω 2Ω B 30 Ω

Fig. 1.98 Converting the delta network formed by the resistors of 20 Ω, 5 Ω and 15 Ω and 15 Ω into an equivalent star network (Fig. 1.99), 10 Ω A

20 × 5 = 25Ω 20 + 5 15 20 × 15 R2 = =75Ω 20 + 5 15 5 15 R3 = = 1.8875 5Ω 20 + 5 15 R1 =

R1

R2 30 Ω

15 Ω R3 2Ω

B 30 Ω

Fig. 1.99

1.42 Network Analysis and Synthesis Redrawing the network (Fig. 1.100), 12.5 Ω

2.5 Ω

10 Ω A

A 7.5 Ω 1.875 Ω 30 Ω

15 Ω

15 Ω

37.5 Ω

3.875 Ω

2Ω B

B

30 Ω

30 Ω

(a)

(b)

Fig. 1.100 Converting the delta network formed by the resistors of 3.875 Ω, 37.5 Ω and 30 Ω into an equivalent star network (Fig. 1.101), 12.5 Ω A

3.875 × 37.5 R4 = = 2 004 Ω 3.875 + 37.5 30 R5 = R6 =

R4

3.875 × 30 = 1.63 Ω 3.875 + 37.5 30

B

15 Ω

R6

37.5 30 = 15.76 Ω 3.875 + 37.5 30

R5

Fig. 1.101 Simplifying the network (Fig. 1.102), 12.5 Ω

2.04 Ω

A

14.54 Ω A

15.76 Ω

15.76 Ω B

B 16.63 Ω

1.63 Ω

15 Ω (a)

(b)

7.76 Ω

23.52 Ω

15.76 Ω B

A

A

(c)

B (d)

Fig. 1.102 RAB = 23.52 Ω

1.10 Star-delta Transformation 1.43

Example 1.41

Find an equivalent resistance between A and B in the network of Fig. 1.103. A

6Ω

4Ω

3Ω

5Ω 8Ω

5Ω

4Ω

B

Fig. 1.103 Solution Converting the star network formed by the resistors of 3 Ω, 5 Ω and 8 Ω into an equivalent delta network (Fig. 1.104 and Fig. 1.105), A

6Ω

4Ω

A

6Ω

9.875 Ω

R1

R2

4Ω

15.8 Ω

R3

26.33 Ω

5Ω

4Ω

5Ω

4Ω

B

B

Fig. 1.104

Fig. 1.105

3 5 = 9.875 Ω 8 3 8 R2 = 3 8 + = 15.8 Ω 5 5 8 R3 = 5 8 + = 26.33 Ω 3 The resistors of 15.8 Ω and 5 Ω and the resistors of 26.33 Ω and 4 Ω are connected in parallel (Fig. 1.106), R1 = 3 5 +

6Ω

A

4Ω

9.875 Ω

3.47 Ω

3.8 Ω

B

Fig. 1.106

1.44 Network Analysis and Synthesis Converting the delta network into a star network (Fig. 1.107), A 6Ω

A

4Ω

6Ω

4Ω

2.19 Ω R4

2Ω

R6

0.77 Ω R5 B B

Fig. 1.107 R4 =

3.8 × 9.875 = 2.199 Ω 3.8 + 9.875 + 3 47

R5 =

3.8 × 3.47 = 0 77 Ω 3.8 + 9.875 + 3 47

R6 =

3.47 9.875 =2Ω 3.8 + 9.875 + 3 47

Simplifying the network (Fig. 1.108), 4Ω

2Ω 0.77 Ω

A

B 6Ω

2.19 Ω (a) 6Ω 0.77 Ω

A

B 8.19 Ω (b) 3.46 Ω

0.77 Ω

A

B (c) 4.23 Ω A

B (d)

Fig. 1.108 RAB = . 3 Ω

1.10 Star-delta Transformation 1.45

Example 1.42

Find an equivalent resistance between A and B in the network of Fig. 1.109. 6Ω

4Ω

A 3Ω

5Ω 4Ω

8Ω

B

Fig. 1.109 Solution Converting the star network formed by the resistors of 3 Ω, 4 Ω and 5 Ω into an equivalent delta network (Fig. 1.110), 5Ω

3Ω R3

5 4 R1 = 5 4 + = 15 5.67 Ω 3 3 4 R2 = 3 4 + =94Ω 5 5 3 R3 = 5 3 + = 11.75 Ω 4

4Ω

⇒

R2

R1

(a)

(b)

Fig. 1.110

Similarly, converting the star network formed by the resistors of 4 Ω, 6 Ω and 8 Ω into an equivalent delta network (Fig. 1.111), 6Ω

4Ω

R6

6 8 = 26 6Ω 4 4 8 R5 = 4 8 + = 17 7.33 Ω 6 6 4 R6 = 6 4 + = 13 Ω 8 R4 = 6 8 +

8Ω

⇒

R5

R4

(a)

(b)

Fig. 1.111

These two delta networks are connected in parallel between points A and B (Fig. 1.112). 13 Ω A 11.75 Ω 26 Ω

15.67 Ω

9.4 Ω

17.33 Ω

B

A 6.17 Ω

Fig. 1.112 The resistors of 9.4 Ω and 17.33 Ω are in parallel with a short. Hence, the equivalent resistance of this combination becomes zero.

9.78 . Ω B

Fig. 1.113

1.46 Network Analysis and Synthesis Simplifying the parallel networks (Fig. 1.113), RAB = 6.17 17 � 9.78 = 3 78 Ω

Example 1.43

Determine the current supplied by the battery in the network of Fig. 1.114.

4Ω

6Ω 6Ω

4Ω

6Ω

50 V

2Ω

Fig. 1.114 Solution Converting the delta network formed by resistors of 6 Ω, 6 Ω and 6 Ω into an equivalent star network (Fig. 1.115), 4Ω

2Ω 2Ω

4Ω

50 V

2Ω

2Ω

Fig. 1.115 Simplifying the network (Fig. 1.116), 6Ω 2Ω

3Ω

2Ω

5Ω

I

6Ω 2Ω

50 V (a)

2Ω

50 V (b)

Fig. 1.116

50 V

2Ω (c)

1.10 Star-delta Transformation 1.47

50 = 7.14 A 5+ 2

I=

Example 1.44

Calculate the current flowing the 10 W resistor in Fig. 1.117. 180 V F 8Ω

30 Ω

4Ω

17 Ω

34 Ω A D

B 12 Ω

30 Ω

12 Ω C

13 Ω E

10 Ω

Fig. 1.117 Solution Between terminals A and B resistors of 8 Ω and 4 Ω are connected in series. Similarly, between terminals F and E, resistors of 17 Ω and 13 Ω are connected in series (Fig. 1.118), 180 V

F 30 Ω

12 Ω A

34 Ω B 12 Ω

12 Ω

30 Ω

D 30 Ω

C

E

10 Ω

Fig. 1.118 Converting delta ABC and DEF into an equivalent star network (Fig. 1.119), 180 V 34 Ω 10 Ω

4Ω 4Ω

10 Ω

10 Ω

4Ω 10 Ω

Fig. 1.119

1.48 Network Analysis and Synthesis Simplifying the network (Fig. 1.120), 180 V

180 V

48 Ω

I

4Ω

I′

I

10 Ω

4Ω

16 Ω

10 Ω

24 Ω (a)

(b)

Fig. 1.120 I=

180 =6A 4 + 16 + 10

By current division rule, I′

Example 1.45

I 24 Ω = I

0Ω

48 =4A 24 + 48

= 6×

Find the current supplied by the battery in the network of Fig. 1.121. 40 Ω

20 Ω

50 Ω

10 Ω

30 Ω

5Ω

15 V

Fig. 1.121 Solution Converting the star network formed by resistors of 40 Ω, 20 Ω and 50 Ω into an equivalent delta network (Fig. 1.122 and Fig. 1.123), R1 = 40 + 20 +

40 × 20 = 76 Ω 50

20 × 50 = 95 9 Ω 40

190 Ω

R2

40 × 50 R2 = 40 + 50 + = 190 90 Ω 20 R3 = 20 + 50 +

76 Ω

R1

95 Ω

R3 10 Ω

15 V

30 Ω

5Ω

Fig. 1.122

10 Ω

15 V

30 Ω

5Ω

Fig. 1.123

1.10 Star-delta Transformation 1.49

The resistors of 190 Ω and 10 Ω and the resistors of 95 Ω and 30 Ω are connected in parallel (Fig. 1.124). 76 Ω

9.5 Ω

22.8 Ω

5Ω

15 V

Fig. 1.124 Simplifying the network (Fig. 1.125),

I=

76 Ω

15 = 0.542 A 22.67 + 5

22.67 Ω

I

32.3 Ω

5Ω

15 V

5Ω

15 V

(a)

(b)

Fig. 1.125

Example 1.46

Find an equivalent inductance between terminals A and B in the network of Fig. 1.126. 6H A 2H

2H 6H

6H 2H B

Fig. 1.126 Solution Converting the star network formed by inductors of 2 H, 2 H and 2 H into an equivalent delta network (Fig. 1.127), 6H A 6H

6H 6H

6H

6H

B

Fig. 1.127 Simplifying the network (Fig. 1.128), 3H A

A 3H

A

3H

B

3H B

(a)

6H

2H B

(b)

Fig. 1.128

(c)

1.50 Network Analysis and Synthesis LAB = 2 H

Example 1.47

Determine the capacitance between terminals A and B in the network of Fig. 1.129. 2 μF

4 μF

3 μF

A

B

5 μF

1 μF

Fig. 1.129 Solution Converting delta network formed by capacitors of 2 μF, 1 μF and 3 μF into an equivalent star network [Fig. 1.130 (a)], C1 = 1 + 2 +

2 μF

1 2 = 3.667 μ F 3

C1 C3

1 3 C2 = 1 3 + = 5.5 μ F 2

A

C2

3 μF

B

5 μF

1 μF

2 3 = 11 μ F 1

C3 = 2 3 +

4 μF

(a)

Simplifying the network (Fig. 1.130), 11 μF

4 μF

3.67 μF

2.93 μF B

A

3.67 μF B

A 5.5 μF

5 μF

2.62 μF (c)

(b) 3.67 μF

2.21 μF

5.55 μF A

B

A (d)

B (e)

Fig. 1.130 C AB = 2.21 μF

Example 1.48 network of Fig. 1.31.

If the combined capacitance of network shown is 5 μF, find capacitance C in the

1.11 Source Transformation 1.51

1 μF

2 μF

3 μF

3 μF

2 μF

C

4 μF 1 μF

1 μF

Fig. 1.131 Solution The capacitors of 1 μF and 1 μF and the capacitors of 1 μF and 3 μF are in parallel (Fig. 1.132), 4 μF

2 μF 4 μF

3 μF 2 μF C

2 μF

Fig. 1.132 Replacing star network formed by capacitors of 4 μF, 4 μF and 2 μF into an equivalent delta network (Fig. 1.133), 4 4 C1 = = 1 6 μF 4 4 2 4 2 C2 C3 = = 0 8 μF 4 4 2

1.6 μF

2 μF

C1 2 μF

3 μF C3 0.8 μF

C

0.8 μF

C2

Fig. 1.133

Simplifying the network (Fig. 1.134), 3 μF

3.6 μF 0.6 μF

2.8 μF

1.575 μF

C

(a)

0.8 μF

3C 3+C

(b)

Fig. 1.134 1.575 + 0.8 +

1.11

3C =5 3+C 3C = 2.625 3+C C = 21 μF

SOURCE TRANSFORMATION

A voltage source with a series resistor can be converted into a equivalent current source with a parallel resistor. Conversely, a current source with a parallel resistor can be converted into a voltage source with a series resistor as shown in Fig. 1.135.

1.52 Network Analysis and Synthesis R ⇔I=

V

V R

R

(a)

(b)

Fig. 1.135 Source transformation Source transformation can be applied to dependent sources as well. The controlling variable, however must not be tampered with any way since the operation of the controlled sources depends on it.

Example 1.49

Replace the given network of Fig. 1.136 with a single current source and a resistor. A 10 A

6Ω

20 V

5Ω B

Fig. 1.136 Solution Since the resistor of 5 Ω is connected in parallel with the voltage source of 20 V it becomes redundant. Converting parallel combination of current source and resistor into equivalent voltage source and resistor (Fig. 1.137), By source transformation (Fig. 1.138), A

A 60 V

A 80 V

6Ω

6Ω

20 V B

6Ω

13.33 A

B

Fig. 1.137 B

Fig. 1.138

Example 1.50

Reduce the network shown in Fig. 1.139 into a single source and a single resistor between terminals A and B. A 2Ω

1A

2Ω 4V

3Ω

1Ω 3V

6V B

Fig. 1.139

1.11 Source Transformation 1.53

Solution Converting all voltage sources into equivalent current sources (Fig. 1.140), A 1A

2Ω

2Ω

2A

2A

3Ω

1Ω

3A

B

Fig. 1.140 Adding the current sources and simplifying the network (Fig. 1.141), A 1Ω

3A

0.75 . Ω

1A

B

Fig. 1.141 Converting the current sources into equivalent voltage sources (Fig. 1.142), A

A

3V 3.75 V

1Ω

1.75 Ω

0.75 Ω 0.75 V

B

B

Fig. 1.142

Example 1.51

Replace the circuit between A and B in Fig. 1.143 with a voltage source in series

with a single resistor. A

3A

5Ω 30 Ω

50 Ω

6Ω

20 V B

Fig. 1.143

1.54 Network Analysis and Synthesis Solution Converting the series combination of voltage source of 20 V and a resistor of 5 Ω into equivalent parallel combination of current source and resistor (Fig. 1.144), A

50 Ω

30 Ω

3A

4A

5Ω

6Ω

B

Fig. 1.144 Adding the two current sources and simplifying the circuit (Fig. 1.145), A

30 || 50 || 5 || 6 = 2.38 Ω

7A

B

Fig. 1.145 By source transformation (Fig. 1.146), 2.38 Ω A

16.67 V B

Fig. 1.146

Example 1.52

Find the power delivered by the 50 V source in the network of Fig. 1.147. 3Ω

5Ω 50 V

2Ω

10 A

10 V

Fig. 1.147 Solution Converting the series combination of voltage source of 10 V and resistor of 3 Ω into equivalent current source and resistor (Fig. 1.148),

1.11 Source Transformation 1.55

5Ω 2Ω

10 A

3Ω

3.33 A

50 V

Fig. 1.148 Adding the two current sources and simplifying the network (Fig. 1.149),

5Ω

1.2 Ω

13.33 A 50 V

Fig. 1.149 By source transformation (Fig. 1.150),

1.2 Ω

5Ω I 50 V

16 V

Fig. 1.150 I=

50 − 16 = 5.48 A 5 + 1.2

Power delivered by the 50 V source = 50 × 5.48 = 274 W

Example 1.53

Find the current in the 4 W resistor shown in network of Fig. 1.151. 6V

5A

2Ω

2A

4Ω

Fig. 1.151 Solution Converting the parallel combination of the current source of 5 A and the resistor of 2 Ω into an equivalent series combination of voltage source and resistor (Fig. 1.152),

1.56 Network Analysis and Synthesis 2Ω

6V

10 V 2A

4Ω

2A

4Ω

2A

4Ω

Fig. 1.152 Adding two voltage sources (Fig. 1.153), 2Ω

4V

Fig. 1.153 Again by source transformation (Fig. 1.154),

2Ω

2A

Fig. 1.154 Adding two current sources (Fig. 1.155),

2Ω

4A

4Ω

Fig. 1.155 By current-division rule, I4 Ω = 4 ×

Example 1.54

2 = 1.33 A 2 4

Find the voltage across the 4 W resistor shown in network of Fig. 1.156. 3Ω

6V

2Ω

6Ω

Fig. 1.156

1Ω

3A

4Ω

1.11 Source Transformation 1.57

Solution Converting the series combination of the voltage source of 6 V and the resistor of 3 Ω into equivalent current source and resistor (Fig. 1.157), 2Ω

3Ω

2A

1Ω

6Ω

4Ω

3A

Fig. 1.157 By series–parallel reduction technique (Fig. 1.158), 2Ω

1Ω

2Ω

2A

3A

4Ω

Fig. 1.158 By source transformation (Fig. 1.159), 2Ω

1Ω

1Ω

2Ω

4Ω

4Ω

3A

4Ω

3A

4V

4V (a)

(b)

Fig. 1.159 (continued) 1Ω

1Ω

4Ω

1Ω I

1A

4Ω

3A

4Ω

4Ω 4A

(c)

(d)

Fig. 1.159 I=

16 = 1.78 A 4 +1+ 4

Voltage across the 4 Ω resistor = 4 = 4 × 1.78 = 7.12 V

4Ω

4 Ω 16 V

(e)

1.58 Network Analysis and Synthesis

Example 1.55

Find the voltage at Node 2 of the network shown in Fig. 1.160. 50 Ω

1

2 I

100 Ω

15 V

100 Ω

+ 10 I −

Fig. 1.160 Solution We cannot change the network between nodes 1 and 2 since the controlling current I, for the controlled source, is in the resistor between these nodes. Applying source transformation to series combination of controlled source and the 100 Ω resistor (Fig. 1.161), 1

I

50 Ω

1 I

2

100 Ω

0.1 I

15 V

1

I

100 Ω

50 Ω

50 Ω

2

0.1 I

15 V

2

50 Ω

50 Ω

+ −

15 V

5I

Fig. 1.161 Applying KVL to the mesh, 15 − 50 I − 50 I − 5

0 15 I= = 0.143 A 105

Voltage at Node 2 = 15 − 50 I = 15 − 50 × 0.143 = 7.86 V

1.12

SOURCE SHIFTING

Source shifting is the simplification technique used when there is no resistor in series with a voltage source or a resistor in parallel with a current source.

Example 1.56 Calculate the voltage across the 6 Ω resistor in the network of Fig. 1.162 using source-shifting technique.

1.12 Source Shifting 1.59 3Ω 2

4Ω

1Ω

3

4

1

2Ω

18 V

+ Va

6Ω

−

Fig. 1.162 Solution Adding a voltage source of 18 V to the network and connecting to Node 2 (Fig. 1.163), we have 2

3Ω

4Ω

1Ω

3

1

4

18 V 2Ω

6Ω

18 V

+ Va −

Fig. 1.163 Since nodes 1 and 2 are maintained at the same voltage by the sources, the connection between nodes 1 and 2 is removed. Now the two voltage sources have resistors in series and source transformation can be applied (Fig. 1.164). 18 V

18 V

3Ω

4Ω

1Ω + 2Ω

6Ω

Va −

Fig. 1.164 Simplifying the network (Fig. 1.165),

1.60 Network Analysis and Synthesis 18 V 18 V

3Ω

3Ω 4.5 A 4.5 A

1Ω

1.33 Ω

+

4Ω 2Ω

6Ω

1Ω

Va −

6Ω −

(a) 18 V

1.33 Ω

+

Va

3Ω

5.985 V

Va

(b)

3Ω

1Ω

2.33 Ω

18 V

6Ω

5.985 V

6Ω −

+

Va (c)

(d)

Fig. 1.165 Applying KCL at the node, Va 18 Va − 5.985 Va + + =0 3 2 33 6 Va = 9.23 V

Exercises 1.1

Find the resistance between terminals A and B in the network of Fig. 1.166.

1.2

Find the resistance between terminals A and B in the network of Fig. 1.167.

24 Ω

4Ω 4Ω

4Ω 8Ω

A B 8Ω

A

Fig. 1.166 [4 Ω]

4Ω

8Ω

4Ω 8Ω

8Ω

4Ω

B

Fig. 1.167 [2.67 Ω]

Exercises 1.61

Find the equivalent resistance between terminals A and B in the network of Fig. 1.168.

1.3

1Ω

A

2Ω

2Ω

3Ω 12 Ω

40 Ω

A

60 Ω

B

2Ω

3Ω

B 2Ω

2Ω

Fig. 1.172

Fig. 1.168

[3 Ω]

[8 Ω] What is the equivalent resistance between terminals A and B of the networks shown in Fig. 1.169?

1.4

1.8

6Ω

A

1Ω

A

A

4Ω

B

6Ω

6Ω

4Ω B (a)

(b)

[(a) 0 (b) 0] Find the equivalent resistance between terminals A and B in the network of Fig. 1.170. 4Ω

6Ω

5Ω B

3Ω

Fig. 1.173 1.9

[5 Ω] Find the equivalent resistance between A and B in the network of Fig. 1.174. 40 Ω

4Ω A

3Ω

3Ω

Fig. 1.169 1.5

Find the equivalent resistance between A and B in the network of Fig. 1.173.

12 Ω

60 Ω

B

4Ω

4Ω

20 Ω

30 Ω

A

B 100 Ω

Fig. 1.174

Fig. 1.170 1.6

[4 Ω] Find the equivalent resistance between terminals A and B in the network of Fig. 1.171. A

[25 Ω] 1.10 Find the equivalent resistance between A and B in the network of Fig. 1.175. A

4Ω 4Ω

2R

R

R

4Ω

2R

2R

B 4Ω

Fig. 1.171 1.7

[4 Ω] Find the equivalent resistance between terminals A and B in the network of Fig. 1.172.

B

R

Fig. 1.175

⎡4 ⎤ ⎢ 7 R⎥ ⎣ ⎦

1.62 Network Analysis and Synthesis 1.11 Find the equivalent resistance between A and B in the network of Fig. 1.176.

A 3Ω

6Ω

20 Ω

3Ω

10 Ω

45 Ω

20 Ω

A

B 3.25 Ω

20 Ω

2.5 Ω

B

Fig. 1.176

Fig. 1.179

[17 Ω] 1.12 Find RAB by solving the outer delta (X-B-Y) only in the network of Fig. 1.177. Y

5Ω

1Ω

[3.5 Ω] 1.15 Find the equivalent resistance between terminals A and B in the network of Fig. 1.180.

1Ω

7Ω

A

5Ω

5Ω 3Ω

2Ω

3Ω X

3Ω

B 9Ω 3Ω

Fig. 1.177

1Ω

1Ω

[1.41 Ω]

5Ω

1.13 Find the equivalent resistance between A and B in the network of Fig. 1.178.

Fig. 1.180 [1.82 Ω]

3Ω 2Ω

2Ω

2Ω

8Ω

A

1.16 Find the equivalent resistance between the terminals A and B in the network of Fig. 1.181. 3Ω

6Ω 3Ω

B

8Ω

6Ω

9Ω 9Ω

B 5Ω

A

Fig. 1.178

6Ω

[2.625 Ω] 1.14 Find the equivalent resistance between the terminals A and B in the network of Fig. 1.179.

4Ω

4Ω

Fig. 1.181 [10.32 Ω] 1.17 Find the equivalent resistance between A and B in the network of Fig. 1.182.

Exercises 1.63 6Ω

3Ω

A

B 6Ω

1.21 Determine the power supplied to the network in the network of Fig. 1.186.

6Ω

6Ω 6Ω

9Ω

6Ω

3Ω

6Ω

100 V

Fig. 1.182 [4.59 Ω] 1.18 Find the equivalent resistance between A and B in the network of Fig. 1.183. 5Ω

3Ω

6Ω 9Ω

2Ω

A

Fig. 1.186

4Ω 4Ω

2Ω

4Ω

[4705.88 Ω] 1.22 Replace the given network with a single voltage source and a resistor.

2Ω

4Ω

2Ω

B 2Ω

5A

2Ω

Fig. 1.183 [6.24 Ω] 1.19 Determine the current I in the network of Fig. 1.184. 4Ω

I

6Ω

50 V

3Ω

10 A

2Ω

2Ω

10 V

3Ω 5Ω

4Ω

2Ω

Fig. 1.187 [8.6 V, 0.43 Ω] 1.23 Use source transformation to simplify the network until two elements remain to the left of terminals A and B. 3.5 kΩ

6 kΩ

8Ω

A

2Ω 2 kΩ

3 kΩ

20 mA

12 kΩ

Fig. 1.184 [8.59 A] 1.20 Find the voltage between terminals A and B in the network of Fig. 1.185.

B

Fig. 1.188 [88.42 V, 7.92 k Ω] 1.24 Determine the voltage Vx in the network of Fig. 1.189 by source-shifting technique.

A

3Ω

1Ω 1Ω

2Ω

1A

2Ω

1Ω Vx

1Ω

2Ω

2V B

2Ω

5Ω

1Ω

Fig. 1.185 [0.56 V]

Fig. 1.189 [1.129 V]

1.64 Network Analysis and Synthesis

Objective-Type Questions 1.1

1.2

A network contains linear resistors and ideal voltage sources. If values of all the resistors are doubled then the voltage across each resistor is (a) halved (b) doubled (c) increased by four times (d) not changed Four resistances 80 Ω, 50 Ω, 25 Ω, and R are connected in parallel. Current through 25 Ω resistor is 4 A. Total current of the supply is 10 A. The value of R will be (a) 66.66 Ω (b) 40.25 Ω (c)

1.3

36.36 Ω

A

A

(a) 5 V source in series with a 10 Ω resistor (b) 1 V source in series with a 2.4 Ω resistor (c) 15 V source in series with a 2.4 Ω resistor (d) 1 V source in series with a 10 Ω resistor Consider the star network shown in Fig. 192. The resistance between terminals A and B with C open is 6 Ω, between terminals B and C with A open is 11 Ω and between terminals C and A with B open is 9 Ω, then resistacne RA, RB and RC will be RA

R2 B

C

B

R3 C

RB

15 Ω

1.4

(b) (d)

3, 9, 1.5 3, 1.5, 9

C

Fig. 1.192

If each branch of a delta network has resistance 3 R, then each branch of the equivalent wye network has resistance (a)

R 3

3 3R

(a) RΑ = 4 Ω, (b) RΑ = 2 Ω, (c) RΑ = 3 Ω, (d) RΑ = 5 Ω,

(b) 3 R

R 3 Viewed from the terminal AB, the network of Fig.1.191 can be reduced to an equivalent network of a single voltage source in series with a single resistor with the following parameters (c)

1.5

1.5, 3, 9 9, 3, 1.5

RC

B

Fig. 1.190 (a) (c)

4Ω

A

R1

30 Ω

5Ω

10 Ω

Fig. 1.191

A delta connected network with its wyeequivalent is shown in Fig.190. The resistances R1, R2 and R3 (in ohms) are respectively. A

5V

B

1.6

(d) 76.56 Ω

10 V

(d)

1.7

RΒ = 2 Ω, RC = 5 Ω RΒ = 4 Ω, RC = 7 Ω RΒ = 3 Ω, RC = 4 Ω RΒ = 1 Ω, RC = 10 Ω

A 10 V battery with an internal resistance of 1 Ω is connected across a nonlinear load whose V-I characteristic is given by 7 V 2 2V . The current delivered by the battery is (a) 0 (b) 10 A (c) 5 A (d) 8 A

Answers to Objective-Type Questions 1.65

1.8

1.9

If the length of a wire of resistance R is uniformly stretched to n times its original value, its new resistance is (a) nR

(b)

(c) n2R

(d)

R n R

4 8 A (d) A 15 15 1.10 The current waveform in a pure resistor at 10 Ω is shown in Fig. 1.194. Power dissipated in the resistor is (c)

i

n2

All the resistances in Fig. 1.193 are 1 Ω each. The value of I will be

9

0

I

3

t

6

Fig. 1.194 (a) (c)

1V

Fig. 1.193 (a)

1 A 15

(b)

2 A 15

7.29 W 135 W

(b) (d)

52.4 W 270 W

1.11 Two wires A and B of the same material and length L and 2L have radius r and 2r respectively. The ratio of their specific resistance will be (a) 1 : 1 (b) 1 : 2 (c)

1:4

(d)

1:8

Answers to Objective-Type Questions 1.1. (d)

1.2. (c)

1.3. (d)

1.4. (a)

1.8. (c)

1.9. (d)

1.10. (d)

1.11. (b)

1.5. (b)

1.6. (b)

1.7. (c)

2 2.1

Elementary Network Theorems INTRODUCTION

In Chapter 1, we have studied basic network concepts. In network analysis, we have to find currents and voltages in various parts of networks. In this chapter, we will study elementary network theorems like Kirchhoff’s laws, mesh analysis and node analysis. These methods are applicable to all types of networks. The first step in analyzing networks is to apply Ohm’s law and Kirchhoff’s laws. The second step is the solving of these equations by mathematical tools.

2.2

KIRCHHOFF’S LAWS

The entire study of electric network analysis is based mainly on Kirchhoff’s laws. But before discussing this, it is essential to familiarise ourselves with the following terms: Node A node is a junction where two or more network elements are connected together. Branch An element or number of elements connected between two nodes constitute a branch. Loop A loop is any closed part of the circuit. Mesh A mesh is the most elementary form of a loop and cannot be further divided into other loops. All meshes are loops but all loops are not meshes. 1. Kirchhoff’s Current Law (KCL) The algebraic sum of currents meeting at a junction or node in an electric circuit is zero. Consider five conductors, carrying currents I1, I2, I3, I4 and I5 meeting at a point O as shown in Fig. 2.1. Assuming the incoming currents to be positive and outgoing currents negative, we have I1 ( I 2 ) + I 3 ( I 4 ) + I 5 = 0 I1 − I 2 + I 3 I 4 + I 5 = 0 I1 I 3 + I 5 I 2 + I 4

I1 I2 O I3 I5

Fig. 2.1

I4

Kirchhoff’s current law

Thus, the above law can also be stated as the sum of currents flowing towards any junction in an electric circuit is equal to the sum of the currents flowing away from that junction. 2. Kirchhoff’s Voltage Law (KVL) The algebraic sum of all the voltages in any closed circuit or mesh or loop is zero. If we start from any point in a closed circuit and go back to that point, after going round the circuit, there is no increase or decrease in potential at that point. This means that the sum of emfs and the sum of voltage drops or rises meeting on the way is zero.

2.2 Network Analysis and Synthesis 3. Determination of Sign A rise in potential can be assumed to be positive while a fall in potential can be considered negative. The reverse is also possible and both conventions will give the same result. (i) If we go from the positive terminal of the battery or source to the negative terminal, there is a fall in potential and so the emf should be assigned a negative sign (Fig. 2.2a). If we go from the negative terminal of the battery or source to the positive terminal, there is a rise in potential and so the emf should be given a positive sign (Fig. 2.2b).

(a) Fall in potential

(b) Rise in potential

Fig. 2.2 Sign convention (ii) When current flows through a resistor, there is a voltage drop across it. If we go through the resistor in the same direction as the current, there is a fall in the potential and so the sign of this voltage drop is negative (Fig. 2.3a). If we go opposite to the direction of the current flow, there is a rise in potential and hence, this voltage drop should be given a positive sign (Fig. 2.3b). I

+

+

I

(b) Rise in potential

(a) Fall in potential

Fig. 2.3 Sign convention

Example 2.1 In Fig. 2.4, the voltage drop across the 15 W resistor is 30 V, having the polarity indicated. Find the value of R. 2A 5Ω

5A +

− 15 Ω

−

30 V I

+

Fig. 2.4 Solution Current through the 15 Ω resistor I= Current through the 5 Ω resistor resisto i

30 =2A 15 5 2 7A

Applying KVL to the closed path, −5 (7) − R ( 2) + 100 − 30 = 0 −35 − 2 R + 100 − 30 = 0 R = 17.5 Ω

R

−

100 V

3A

+

2.2 Kirchhoff’s Laws 2.3

Example 2.2

Determine the currents I1, I2 and I3 in Fig. 2.5. I2

I1

12 Ω

I3

9A

8Ω

4A

16 Ω

Fig. 2.5 Solution Assigning currents to all the branches (Fig. 2.6), I2

I1

12 Ω

I3

9A

8Ω

4A

16 Ω

(I1 − I2)

(I1 − I2 + 9 + I3 + 4)

(I3 + 4)

Fig. 2.6 From Fig. 2.6,

Also,

I1 I1 − I 2 I 2 I 3 = 13 −12 I1 − 8 ( 1 − 2 ) = 0 −20 I1 + 8 2 = 0

9 I3

4 …(i) …(ii)

−12 I1 −16 16 I 3 = 0

and Solving Eqs (i), (ii) and (iii),

…(iii)

I1 = 4 A I 2 = 10 A I 3 = −3 A

Example 2.3

Find currents in all the branches of the network shown in Fig. 2.7. 80 A 0.02 Ω

0.02 Ω 60 A

30 A 0.01 Ω 0.01 Ω 60 A 70 A

0.01 Ω

0.03 Ω 120 A

Fig. 2.7

2.4 Network Analysis and Synthesis Solution Let Then

I AF

x

I FE

x − 30

I ED

x + 40

I DC

x − 80

I CB

x − 20

I BA

x − 80

80 A (x − 80) 0.02 Ω

0.02 Ω 30 A

A

60 A B (x − 20)

x F

0.01 Ω

0.01 Ω C 60 A (x − 80)

(x − 30) E 70 A

0.01 Ω

Applying KVL to the closed path AFEDCBA (Fig. 2.8),

D (x + 40)

0.03 Ω 120 A

Fig. 2.8 −0.02 − 0.01( − 30) − 0.01( x + 40) − 0.03 ( − 80) − 0.01( x − 2200) − 0.02 ( x − 80 8 )=0 x = 41 A I AF = 41 A I FE = 41 − 30 = 11 A I ED = 41 + 40 = 81 A I DC = 41 − 80 = −39 A I CD = 39 A I CB = 41 − 20 = 21 A I BA = 41 − 80 = −39 A I AB = 39 A

Example 2.4

Find currents in all the branches of the network shown in Fig. 2.9. B

4Ω

C

3Ω

5Ω

2Ω

1A O

1A

A

1Ω

Fig. 2.9 Solution Assigning currents to all the branches (Fig. 2.10), Applying KVL to the closed path OBAO, −2(1 − x ) − 3 y + 1( ) = 0 3x − 3 y = 2 y ) 5( x + y ) = 0 9 x 12 1 y 4

4Ω

C

1A (x + y)

y

3Ω

…(i)

Applying KVL to the closed path ABCA, 3 y 4(1 x

(1 −x − y) B

5Ω

2Ω (1 −x )

1A O

…(ii)

1Ω

x

Fig. 2.10

A

2.2 Kirchhoff’s Laws 2.5

Solving Eqs (i) and (ii), = 0 57 A y = −0.095 A I OA = 0 57 A I OB = 1 − 0.57 = 0.43 A I AB = 0.095 A I AC = 0.57 − 0.095 = 0.475 A I BC = 1 − 0.57 + 0.095 = 0.525 A

Example 2.5

What is the potential difference between points x and y in the network shown in Fig. 2.11? 2Ω −

4V

x

+ +

4V

−

I1

−

3Ω −

3Ω

2V +

5Ω I2

y

Fig. 2.11 2 =04A 2 3 4 I2 = =05A 3 5 Potential difference between points x and y = Vxy V x − Vy I1 =

Solution

Writing KVL equation for the path x to y, Vx Vx

3I + 4 − 3I 3I

Vy = 0

3 (0 4) 4 3 (0 5) Vy = 0 Vx V y = −3 7 Vxy = −3.7 V

Example 2.6

Find the voltage between points A and B in Fig. 2.12. B

10 Ω A

I1

5Ω

20 V 5 V 12 Ω

Fig. 2.12

6Ω

4Ω

I2

15 V

+

2.6 Network Analysis and Synthesis 20 = 1 33 A 10 + 5 15 I2 = =1 5A 4 6 I1 =

Solution

From Fig. 2.13, Voltage between points A and B = VAB

VA VB B

Writing KVL equation for the path A to B, VA

VA I (1.33)

10 Ω

−

I1 −

+

VA VB = 17.65 VAB = 17.65 V

Example 2.7

−

+

5 15 + 6 I VB = 0 1 6 (1. ) VB = 0

A

5Ω

20 V

+

+

5 V 12 Ω

− 15 V

Fig. 2.13

Determine the potential difference VAB for the given network in Fig. 2.14. 2A

3Ω

A

10 Ω 2Ω

4Ω

5Ω

8V

5V

B

Fig. 2.14 Solution The resistor of 3 Ω is connected across a short circuit. Hence, it gets shorted (Fig. 2.15). 2A A

10 Ω

+ 2Ω

5V I1

+ 5Ω −

8V

−

I2

B

Fig. 2.15 5 = 2.5 A 2 I2 = 2 A VAB VA VB I1 =

Potential difference

4Ω

Writing KVL equation for the path A to B, VA 2 I + 8 5 I VB = 0 VA 2 ( 2.5) 8 5 ( 2) VB = 0

6Ω

I2

4Ω

2.2 Kirchhoff’s Laws 2.7

VA VB = 7 VAB = 7 V

Example 2.8

Find the voltage of the point A w.r.t. B in Fig. 2.16. 5A

5Ω

A

+

−

8V

3Ω

3Ω

10 V

+

−

I1

4Ω I2

B

Fig. 2.16 10 = 1.25 A 5 3 I2 = 5 A Applying KVL to the path from A to B, VA 3 I 8 + 3 I VB = 0 VA 3 (1.25) 8 3 (5) VB = 0 I1 =

Solution

VA VB = −3 25 VAB = −3.225 V

Example 2.9

In Fig. 2.17, what values must R1 and R2 have

(a) when I1 = 4 A and I2 = 6 A both charging? (b) when I1 = 2 A discharging and I2 = 20 A charging? (c) when I1 = 0? b (I1 + I2)

2Ω

c

d I1

I2

R1

R2

110 V 80 V a

f

50 V e

Fig. 2.17 Solution Applying KVL to the closed path abcfa, 110 − 2 ( 1 2 ) 1 1 80 0 110 − 2

1

2 I 2 − R1 I1 − 80 = 0 ( 2 1 ) 1 2 I 2 = 30

Applying KVL to the closed path fcdef, 80 + R1 I1 − R2 I 2 − 50 = 0 R1 I1 − R2 I 2 = −30

…(i)

…(ii)

2.8 Network Analysis and Synthesis Case (a) I1 = 4 A and I2 = 6 A both charging i.e. I1 = 4 A and I2 = 6 A Substituting I1 and I2 in Eq. (i), (2 + R1) 4 + 2 (6) = 30 R1 = 2.5 Ω Substituting R1, I1 and I2 in Eq. (ii), 2.5 (4)−R2 (6) = −30 R2 = 6.67 Ω Case (b) I1 = 2 A discharging and I2 = 20 A charging i.e. I1 = −2 A and I2 = 20 A Substituting I1 and I2 in Eq. (i), (2 + R1) (−2) + 2 (20) = 30 R1 = 3 Ω Substituting R1, I1 and I2 in Eq. (ii), 3 (−2) − R2 (20) = −30 R2 = 1.2 Ω

Case (c) I1 = 0

Substituting in Eq. (i) (2 + R1) (0) + 2 I2 = 30 I2 = 15 A Substituting I1 and I2 in Eq. (ii), 0 − 15 R2 = −30 R2 = 2 Ω

Example 2.10

In Fig. 2.18, find I1 and I2 when (a) R = 2.3 W, (b) R = 0.5 W, and (c) for what values

of R is I1 = 0? b (I1 + I2)

0.2 Ω

c

e I1

I2

0.2 Ω 130 V

R

110 V a

d

f

Fig. 2.18 Solution Applying KVL to closed path abcda, 130 − 0.2 ( I1 + I 2 ) − 0.2 I1 − 110 = 0 0.44 I1 + 0 2 I 2 = 20

…(i)

Applying KVL to the closed path dcefd, 110 + 0 2 I1 − R I 2 = 0 0 2 I1 R I 2 = −110

…(ii)

2.2 Kirchhoff’s Laws 2.9

Case (a) R = 2.3 Ω Substituting R in Eq. (ii), 0.22 I1 2 3 I 2 = −110

…(iii)

Solving Eqs (i) and (iii), I1 = 25 A I 2 = 50 A Case (b) R = 0.5 Ω Substituting R in Eq. (i), 0.22 I1 0 5 I 2 = −110

…(iv)

Solving Eqs (i) and (iv), I1 = −50 A I 2 = 200 A Case (c)

I1 = 0 Substituting I1 in Eq. (i),

0 2 I 2 = 20 I 2 = 100 A

Substituting I1 and I2 in Eq. (ii), 0.2 (0)

(100) = −110 R = 1.1 Ω

Example 2.11

In Fig. 2.19, find the value of R. 10 Ω 3A 14 Ω

80 V

R

Fig. 2.19 Solution Assigning currents to all the branches (Fig. 2.20), Applying KVL to the closed path abcda, b

I

10 Ω

c

80 −10 10 I − 14 ( I − 3) 0 I = 5.08 A Applying KVL to the closed path dcefd, 14 ( 3) 3R = 0 14 (5.08 3) 3) − 3 0 R = 9 71 7 Ω

3A

e

(I − 3) 14 Ω

80 V

a

d

Fig. 2.20

R

f

2.10 Network Analysis and Synthesis

Example 2.12

Determine current drawn by the ammeter shown in Fig. 2.21. 10 Ω

5Ω

9V 30 Ω

A

5Ω

Fig. 2.21 Solution Assigning currents to all the branches (Fig. 2.22), Applying KVL to the closed path abcda, (I1 + I2) b 10 Ω −5 ( 1 + 2 ) + 9 − 10 ( 1 + 2 ) − 30 I1 = 0 45 I1 + +15 155 I 2 = 9 …(i)

I2

30 Ω

0 …(ii)

2

e

I1

9V

Applying KVL to the closed path dcefd, 30 I1 5

5Ω

c

A

5Ω

Solving Eqs (i) and (ii), I 2 = 0.4 A Current drawn by ammeter = 0.4 A

a

d

f

Fig. 2.22

Example 2.13

Find branch currents in the various branches of Fig. 2.23. 10 Ω

20 Ω

0.1 Ω

0.2 Ω 5Ω

2V

4V

Fig. 2.23 Solution Assigning currents to various branches (Fig. 2.24), Applying KVL to the closed path abcda, 2 0.1

1

10 0

5 ( I1 I 2 ) = 0 15.1 1 5 I 2 = 2

b

1

1

2)

20

02 2 4=0 5 1 25 25 2 I 2 = 4

10 Ω

c

…(i)

Applying KVL to the closed path dcefd, 5(

I1

I2

20 Ω

e

(I1 − I2)

0.1 Ω

0.2 Ω 5Ω

2

Solving Eqs (i) and (ii), I1 = 0.086 A I 2 = −0.142 A

…(ii)

2V

4V

a

d

Fig. 2.24

f

2.2 Kirchhoff’s Laws 2.11

Example 2.14

In Fig. 2.25, find the value of R and current flowing through it when the current is

zero in the branch OA. A

480 Ω 1.5 Ω

1Ω O

4Ω

R

C

B

2Ω

10 V

Fig. 2.25 Solution Assigning currents to all branches (Fig. 2.26), Applying KVL to the closed path OACO, 480 I 3 1 5 ( I1 − I 3 ) R ( I 2

I3 ) = 0

But current in the branch OA is zero, i.e. I3 = 0 −1 5 I1

A I3

R I2 = 0

(I1 − I3)

480 Ω

…(i)

1.5 Ω

1Ω

Applying KVL to the closed path BOCB, −4

2

− (

2

+

3 ) − 2 ( I1

+ I 2 ) + 10 = 0

B

−4 4 I2 Substituting I1 in Eq. (i) and (ii),

10 V

2Ω

(I1 + I2)

C

Fig. 2.26

I1 I1

0 4 I2

−6 2 + 2 = 0 −14 I 2 − R I 2 = −10

From Eq. (iv) and (v), I 2 = 0.5 A

Current in branch OC = I 2

(I2 + I3)

I2

I3 = 0

But

Substituting I2 in Eq. (iv),

R

I1

I3 = 0 −2 2 I1 (6 R) I 2 = −10 ...(ii) Applying KVL to the closed path BOAB, −4 I 2 + 480 I 3 + I1 = 0

But

and

O

4Ω

−6 (0.5) + (0.5) = 0 R=6Ω I 3 = 0.5 A

…(iii) …(iv) …(v)

2.12 Network Analysis and Synthesis

Example 2.15

In Fig. 2.27, find the current supplied by the battery. 2Ω 6Ω

2Ω

50 V

5Ω 4Ω

3Ω

Fig. 2.27 Solution Assigning currents to all the branches (Fig. 2.28), Applying KVL to the closed path OABEDO, (I1 + I2) 50 − 2 ( 1 2 ) 6 I1 4 ( 1 3 ) 0 A 12 I1 + 2 2 4 I 3 = 50 …(i) Applying KVL to the closed path BCEB, −2 2 + 5 I 3 + 6 1 = 0 6 1 − 2 I2 + 5 3 = 0 …(ii)

2Ω I1

3

− 3( I 2 + I3 ) + 4 ( 1 − 3 ) = 0 4 1 − 3 I 2 − 12 I 3 = 0

50 V

E (I1 − I3)

I3

2Ω

3Ω

…(iii) O

D

Fig. 2.28

I1 = 2.817 A I 2 = 6.647 A I 3 = −0.723 A Current supplied by the battery = I1 + I 2 = 2.8 817 7 + 6.647 = 9.464 A

In Fig.2.29, find the current flowing through the 2 W resistor. 16 Ω

16 Ω

32 Ω

20 V

20 V

2Ω

Fig. 2.29

C (I2 + I3)

5Ω

4Ω

Solving Eqs (i), (ii) and (iii),

Example 2.16

I2

6Ω

Applying KVL to the closed path ECDE, −5

B

2.2 Kirchhoff’s Laws 2.13

Solution Assigning currents to all the branches (Fig. 2.30), Applying KVL to the closed path OABFGO, −

I 16 I

A

(I1 + I )

B I

…(i)

0

32 Ω

16 Ω

Applying KVL to the closed path BCFB, 20 V 0 1− 1 48 − + 32 I = 0 …(ii)

(I1 − I )

= 20

20 V

I3

F

Applying KVL to the closed path GFCDEG, −

(I1 + I − I ) D

C

2Ω O

…(iii)

G

E

Fig. 2.30

Solving Eqs (i), (ii) and (iii), I1 = 1 05 A I 2 = 6 32 A I 3 = 1 58 A Current through th

resistor

Example 2.17

A

In Fig. 2.31, find the current through the 4 W resistor. 2Ω

2Ω 12 Ω

12 V

10 V

3Ω

1Ω

4Ω

24 V

Fig. 2.31 Solution Assigning currents to all the branches (Fig. 2.32), Applying KVL to the closed path ABEDA, 1

+ 12 = 0 + 12 I 3 = −12 …(i)

Applying KVL to the closed path BCFEB, −

−I

+

− I3 −

0 = 0 … (ii)

)−

3( −

+ 4 0 + = −24 …(iii)

C

(I2 I3) 12 Ω

12 V

(I1 − I ) D

Applying KVL to the closed path DEFHGD, 1

2Ω

B I

A

10 V

I

(I1 I3) 3Ω

1Ω

F

I G

24 V

4Ω

Fig. 2.32

H

2.14 Network Analysis and Synthesis Solving Eqs (i), (ii) and (iii), I1 = 4.11 A I1 = 2 72 A I 3 = 2 06 A Current through the 4 Ω resisto resistor i

Example 2.18

.

1

A

In Fig. 2.33, find the current through the 10 W resistor. 5Ω

10 Ω

12 Ω

15 Ω

4V

8Ω

6V

Fig. 2.33 Solution Assigning currents to all the branches (Fig. 2.34), Applying KVL to the closed path ABGHA, −5 ( 1 +

2 ) − 15 1

(I1 + I2) 5 Ω B A

+4=0

−20 I1 − 5

2

= −4

…(i)

I2 I1

I3

15 Ω

8Ω

4V

Applying KVL to the closed path BCFGB, −10 I 2 − 8

3

+ 15

15 I1 −10 10 I 2 − 8

1

=0

3

=0

H

10 Ω C 12 Ω (I2 − I3)

G

F

Fig. 2.34

…(ii)

Applying KVL to the closed path CDEFC, −12 (

2

−

3 ) − 6 + 8 I3

=0

−12 I 2 + 20 0 I3 = 6

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 0 19 A I 2 = 0.032 A I 3 = 0 32 A Current through the 10 Ω resistor resisto =

Example 2.19

2

0. 03 A

In Fig. 2.35, determine the current supplied by each battery. 20 Ω

1Ω

2Ω

40 Ω 10 Ω

8V

12V

Fig. 2.35

D 6V

E

2.2 Kirchhoff’s Laws 2.15

Solution Assigning currents to all the branches (Fig. 2.36), Applying KVL to the closed path ADEA, 40 I 2 + 8 − 1I1 = 0 I1 + 40 I 2 = 8 Applying KVL to the closed path ABDA, −20 (

1

−

2 ) − 10 ( 1

−

2

+

0 I2 3 ) + 40

…(i)

B (I1 − I2 + I3)

I2

I1

10 Ω

−30 I1 + 70 I 2 − 10 I 3 = 0 Applying KVL to the closed path BCDB, 2 3 12 10 0( 1 2 3 ) 0 10 I1 100 I 2 + 12 I 3 = 12

…(ii)

2Ω

40 Ω

1Ω

=0

20 Ω

(I1 − I2)

A

I3

8V

12 V C D

E

Fig. 2.36

…(iii)

Solving Eqs (i), (ii), and (iii), I1 = 0.1005 A I 2 = 0.197 A I 3 = 1.081 A

Current supplied by the 8 V battery 1 0.1005 A Current supplied by th t e 12 V battery = I 3 = 1.081 A

Example 2.20

In Fig. 2.37, find the value of the unknown resistance R such that 2 A current flows

through it.

R

2A 2Ω

10 V

4Ω

3Ω

5Ω

Fig. 2.37 Solution Assigning currents to all the branches (Fig. 2.38), Applying KVL to the closed path ABCDEA, −2 R + 4 ( I1 − I 2 − 2) + 2 ( I1 − 2) = 0 6 1 − 4 I 2 − 2 = 12 Applying KVL to the closed path HEDGH, 10 − 2 ( 1 2) 3 2 0 2 1 3 I 2 = 14

A

…(i)

E

2) 5 ( I1 − I 2 ) = 0 9 1 12 2 8 Solving Eqs (i), (ii) and (iii), 2

2Ω

…(ii)

H

4 Ω (I1 − I2 − 2)

D I2

3Ω

10 V G

Fig. 2.38

4 ( I1 − I 2

…(iii) I1 = 3 76 A I 2 = 2.16 A R = 0 98 Ω

Unknown resistance R = 0 98 Ω

(I1 − 2)

B

I1

Applying KVL to the closed path GDCFG, 3

R

2A

C (I1 − I2) 5Ω F

2.16 Network Analysis and Synthesis

Example 2.21

In Fig. 2.39, find the current delivered by the 12 V battery. 3Ω 5Ω

2Ω

4Ω

4Ω

12 V

4Ω

Fig. 2.39 Solution Assigning currents to all the branches (Fig. 2.40), Applying KVL to the closed path ABCDEA, 3(

1

2)

2 I2 5( 2 3 ) 0 3 1 10 0 2 5 I3 = 0

…(i) E

Applying KVL to the closed path HEDGH, 4(

1

…(ii)

3

2 I 2 − 12 + 4 4 1 2 I2 + 4

1 3

0 12

5Ω

G

…(iii)

I1

Fig. 2.40

Current delivered by the 12 V battery = I1 = 1 66 A

EXAMPLES WITH DEPENDENT SOURCES In the network of Fig. 2.41, find I1, I2 and V. V I2

I1

Fig. 2.41

I2 C I1

12 V

H

3Ω

B

4Ω

I1 = 1 66 A I 2 = 0 93 A I 3 = 0 87 A

2A

2Ω

I3

Solving Eqs (i), (ii), and (iii),

Example 2.22

(I1 − I2)

D

4Ω

Applying KVL to the closed path GDCFG, 4

(I2 − I3)

(I1 − I3)

5( 2 3 ) − 4 3 0 4 1 5 I 2 − 13 I 3 = 0

3)

3Ω

A

4V

6Ω

F 4Ω

2.2 Kirchhoff’s Laws 2.17

Solution Applying KCL at the node, 2+ 4V =

V V + 3 6

V V + − 4V 2 3 6 7 − V =2 2 4 V 7 4 =− A 21 2 =− A 21

V =− V 3 V I2 = 6 I1 =

Example 2.23

Find voltages V1 and V2 in the network of Fig. 2.42. 2Ω

I

4I − +

+ V1 −

6V

+ 4 Ω V2 −

Fig. 2.42 Solution Applying KVL to the loop, 6 2I + 4 I

4I = 0 I =3A

From Fig. 2.42, V1 = 2 I = 2 (3) = 6 V V2 = 4 I = 4 (3) = 12 V

Example 2.24

Find the power delivered by the dependent source in the network of Fig. 2.43. I

1Ω

3 Vx − + + Vx −

30 V

3Ω

Fig. 2.43 From Fig. 2.43, Vx

0 5I

Applying KVL to the loop, 30 −1 1 I 3 Vx − 0 5 I − 3 0 30 − I + 3 (0 (0.5 ) 0.5 3 I = 0

0.5 Ω

2.18 Network Analysis and Synthesis 30 − 3

0

I = 10 A Vx = 0.5 (10) = 5 V Power supplied by the dependent source = (3Vx) (I) = 3 × 5 × 10 = 150 W

Example 2.25

Find the current I2 in the network of Fig. 2.44. 8V

+

4V

V1

10 V1

I2 16 V

−

Fig. 2.44 Solution Applying KVL to the left loop, −4 + 8 − V1 = 0 V1 = 4 V Applying KCL to the right part, 10 V1 I 2 = 0 10 ( 4) 2 0 I 2 = −40 A

Example 2.26

Find the voltage Vx in the network of Fig. 2.45. + 2A

−

6Ω Vx

+ Vy −

3Ω

1V 6 y

Fig. 2.45 Solution Applying KCL at the node, 1 V 2 + Vy = x 6 9

…(i)

⎛V ⎞ V Vy = 3 ⎜ x ⎟ = x ⎝ 9⎠ 3

…(ii)

From Fig. 2.45,

Substituting Eq. (ii) in Eq. (i), 1 ⎛V ⎞ V 2+ ⎜ x ⎟ = x 6⎝ 3 ⎠ 9 V V 2+ x − x = 0 18 9 Vx = 36 V

2.2 Kirchhoff’s Laws 2.19

Example 2.27

In the network of Fig. 2.46, find the current I1 and power dissipated in the 500 W

resistor.

I1 300 Ω 50 V

0.4 I1

500 Ω

Fig. 2.46 Solution Assigning currents to all the branches as shown in Fig. 2.47. b

I1 300 Ω

e

c 0.6 I1

50 V

0.4 I1

500 Ω

a

d

f

Fig. 2.47 Applying KVL to the closed loop abcda, 50 − 300 I1 500 (0 0 6 I1 ) = 0 I1 = 0.083 A Power dissipated in the 500 Ω resistor = 500 (0.6 I1)2 = 500 (0.6 × 0.083)2 = 1.24 W

Example 2.28

Find the current I in the network shown in Fig. 2.48. 2A

3Ω

5Ω 4V

+ Vx −

2Ω

I + −

3 Vx

Fig. 2.48 Solution Assigning currents in all the branches as shown in Fig. 2.49. 2A a

5Ω

c

(I − 2) 4V

3Ω

+ 2Ω

Vx − b

d

e

I

+ −

3 Vx

f

Fig. 2.49 From Fig. 2.49, Vx Applying KVL to the closed loop fecdf, 3Vx 5 I − 4 2 ( I − 2) 3 [2 ( I 2)] 5 I − 4 2 ( I − 2)

2( I − 2) 0 0

…(i)

2.20 Network Analysis and Synthesis 6

12 5 I − 4 2 I + 4

0

− I = 12 I = −12 A

2.3

MESH ANALYSIS

A mesh is defined as a loop which does not contain any other loops within it. Mesh analysis is applicable only for planar networks. A network is said to be planar if it can be drawn on a plane surface without crossovers. In this method, the currents in different meshes are assigned continuous paths so that they do not split at a junction into branch currents. If a network has a large number of voltage sources, it is useful to use mesh analysis. Basically, this analysis consists of writing mesh equations by Kirchhoff’s voltage law in terms of unknown mesh currents.

Steps to be Followed in Mesh Analysis 1. 2. 3.

Identify the mesh, assign a direction to it and assign an unknown current in each mesh. Assign the polarities for voltage across the branches. V2 Apply KVL around the mesh and use Ohm’s law to express the branch voltages in terms of unknown mesh currents and the resistance. R1 V 4. Solve the simultaneous equations for unknown mesh 1 I1 I2 currents. Consider the network shown in Fig. 2.50 which has three meshes. R2 R4 I3 Let the mesh currents for the three meshes be I1, I2, and I3 and all the three mesh currents may be assumed to flow in the clockwise R5 V3 direction. The choice of direction for any mesh current is arbitrary. Fig. 2.50 Applying KVL to Mesh 1, V1

R3

I 2 ) − R2 ( I1 − I 3 ) = 0

R1 ( I1

V1

…(i)

R3 I 2 R4 ( I 2 I 3 ) − R1 ( I 2 − I1 ) = 0 − R1 I1 + ( R1 R3 + R4 ) I 2 − R4 I 3 = V2

…(ii)

( R1

R2 ) I1

R1 I 2

R2 I 3

Applying KVL to Mesh 2, V2 Applying KVL to Mesh 3, − R2 ( I 3 − I1 ) − R4 ( I 3 − I 2 ) − R5 I 3 + V3 = 0 − R2 I1 − R4 I 2 + ( R2 + R4 + R5 ) I 3 = V3 Writing Eqs (i), (ii), and (iii) in matrix form, ⎡ R1 R2 ⎢ − R1 ⎢ R2 ⎣ −R

− R1 R1 + R3 R4 R4

R2

R2 ⎤ ⎡ I1 ⎤ ⎡V1 ⎤ ⎥ ⎢ I 2 ⎥ = ⎢V2 ⎥ − R4 ⎥⎢ ⎥ ⎢ ⎥ R4 + R5 ⎦ ⎣ I 3 ⎦ ⎣V3 ⎦

In general, ⎡ R11 ⎢ R21 ⎢ ⎣ R31

R12 R22 R32

R13 ⎤ ⎡ I1 ⎤ ⎡V1 ⎤ R23 ⎥ ⎢ I 2 ⎥ = ⎢V2 ⎥ ⎥⎢ ⎥ ⎢ ⎥ R33 ⎦ ⎣ I 3 ⎦ ⎣V3 ⎦

…(iii)

2.3 Mesh Analysis 2.21

where, R11 = Self-resistance or sum of all the resistance of mesh 1 R12 = R21 = Mutual resistance or sum of all the resistances common to meshes 1 and 2 R13 = R31 = Mutual resistance or sum of all the resistances common to meshes 1 and 3 R22 = Self-resistance or sum of all the resistance of mesh 2 R23 = R32 = Mutual resistance or sum of all the resistances common to meshes 2 and 3 R33 = Self-resistance or sum of all the resistance of mesh 3 If the directions of the currents passing through the common resistance are the same, the mutual resistance will have a positive sign, and if the direction of the currents passing through common resistance are opposite then the mutual resistance will have a negative sign. If each mesh current is assumed to flow in the clockwise direction then all self-resistances will always be positive and all mutual resistances will always be negative. The voltages V1, V2 and V3 represent the algebraic sum of all the voltages in meshes 1, 2 and 3 respectively. While going along the current, if we go from negative terminal of the battery to the positive terminal then its emf is taken as positive. Otherwise, it is taken as negative.

Example 2.29

Find the current through the 5 W resistor is shown in Fig. 2.51. 1Ω

2Ω 3Ω

5Ω

10 V 5V

6Ω

4Ω

20 V

Fig. 2.51 Solution Assigning clockwise currents in three meshes as shown in Fig. 2.52. Applying KVL to Mesh 1, 10 −1 1 1 3 ( I1

I2 ) − 6 ( 1 3 ) 0 10 I1 3 2 6 I 3 = 10

1Ω

2Ω

…(i)

−3 (

2

− 1) − 2 I2 − 5 2 − 5 = 0 −3 1 + 1 10 0 2 = −5

I2

…(ii)

3

− 1 ) + 5 − 4 3 + 20 = 0 −6 1 + 110 0 3 = 25

Writing Eqs (i), (ii) and (iii) in matrix form, 3 6 ⎤ ⎡ I1 ⎤ ⎡10 ⎤ ⎡10 −3 ⎢ −3 10 0 ⎥ ⎢ I 2 ⎥ = ⎢ −5⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ −6 0 10 ⎦ ⎣ I 3 ⎦ ⎣ 25 ⎦ We can write matrix equation directly from Fig. 2.52, ⎡ R11 ⎢ R21 ⎢ ⎣ R31

10 V I1

Applying KVL to Mesh 3, −6 (

R12 R22 R32

R13 ⎤ ⎡ I1 ⎤ ⎡V1 ⎤ R23 ⎥ ⎢ I 2 ⎥ = ⎢V2 ⎥ ⎥⎢ ⎥ ⎢ ⎥ R33 ⎦ ⎣ I 3 ⎦ ⎣V3 ⎦

5Ω

3Ω

Applying KVL to Mesh 2,

5V 6Ω

4Ω I3

…(iii)

20 V

Fig. 2.52

2.22 Network Analysis and Synthesis R11 = Self-resistance of Mesh 1 = 1 + 3 + 6 = 10 Ω R12 = Mutual resistance common to meshes 1 and 2 = −3 Ω Here, negative sign indicates that the current through common resistance are in opposite direction. R13 = Mutual resistance common to meshes 1 and 3 = −6 Ω Similarly, R21 = −3 Ω R22 = 3 2 5 = 10 Ω R23 = 0 where

R31 = −6 Ω R32 = 0 R33 = 6 4 = 100 Ω For voltage matrix, V1 = 10 V V2 = −5 V V3 = algebraic sum of all the voltages in mesh 3 = 5 + 20 = 25 V Solving Eqs (i), (ii) and (iii), I1 = 4.27 A I 2 = 0 78 A I 3 = 5 06 A I5 I 2 = 0 78 A

Example 2.30

Find the current through the 2 W resistor of the network shown in Fig. 2.53. 6Ω

2Ω

1Ω

10 V

3Ω

10 Ω

20 V

Fig. 2.53 Solution Assigning clockwise currents in three meshes as shown in Fig. 2.54, Applying KVL to Mesh 1, 6Ω 10 − 6 1 1( I1 − I 2 ) = 0 …(i) 7 1 2 10 Applying KVL to Mesh 2, −1( 2 − 1 ) − 2 I 2 − 3 ( 2 − 3 ) = 0 − I1 + 6 2 − 3 I 3 = 0

2Ω

1Ω

10 V

I1

3Ω I2 20 V

…(ii)

Applying KVL to Mesh 3, −3 ( 3 − 2 ) − 10 3 − 20 = 0 −3 2 + 113 3 3 = −20 Solving Eqs (i), (ii) and (iii), I1 = 1 34 A I 2 = −0 62 A

10 Ω I3

Fig. 2.54

…(iii)

2.3 Mesh Analysis 2.23

I 3 = −1 68 A I 2 = −0 62 A

I2

Example 2.31

Determine the current through the 5 W resistor of the network shown in Fig. 2.55. 10 V 1Ω

8V 2Ω

4Ω

3Ω 5Ω

12 V

Fig. 2.55 Solution Assigning clockwise currents in three meshes as shown in Fig. 2.56. Applying KVL to Mesh 1, 8 1(

2 ) − 2( 1

1

3

1

2

3)

0

2 I3 = 8

…(i)

2

2Ω

3

− 1) − 3(

3Ω I3

…(ii) 12 V

Applying KVL to Mesh 3, −2 (

I2

I1

3 ( I 2 − I 3 ) 1( 2 1 ) 0 − I1 + 8 2 3 I 3 = 10

4Ω

1Ω

8V

Applying KVL to Mesh 2, 10 − 4

10 V

5Ω

Fig. 2.56

− 2 ) − 5 3 + 12 = 0 −2 1 − 3 I 2 + 10 I 3 = 12 3

…(iii)

Solving Eqs (i), (ii), and (iii), I1 = 6 01 A I 2 = 3 27 A I 3 = 3 38 A I5 I 3 = 3 38 A

Example 2.32

Find the current supplied by the battery of the network shown in Fig. 2.57. 3Ω 1Ω I2

2Ω

4V I1

4Ω

Fig. 2.57

5Ω I3

6Ω

2.24 Network Analysis and Synthesis Solution Applying KVL to Mesh 1, 4 3 I1 1( 1

2 ) 4 ( I1 − I 3 ) = 0 8 1 2 4 I3 = 4

…(i)

Applying KVL to Mesh 2, −2 2 − 5 ( I 2 − I 3 ) −1 1( 2 − 1 ) = 0 − I1 + 8 2 − 5 I 3 = 0

…(ii)

Applying KVL to Mesh 3, −6 3 − 4 ( I 3 − I1 ) − 5 ( 3 − 2 ) = 0 −4 1 − 5 I 2 + 15 I 3 = 0

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 0 66 A I 2 = 0 24 A I 3 = 0 26 A Current supplied by the battery = I1 = 0.66 A.

Example 2.33

Find the current through the 4 W resistor in the network of Fig. 2.58. 2Ω

2Ω 2Ω 6V

8V

I2 4Ω

I1 2Ω

2Ω I3

Fig. 2.58 Solution Applying KVL to Mesh 1, 8 2 I1 2 (

2)

1

6

1

2 ( I1 − I 3 ) = 0 2 I2 − 2 3 8

…(i)

Applying KVL to Mesh 2, −2 (

2

− 1) − 2 I2 − 4 ( 2 − 3 ) − 6 = 0 −2 1 + 8 I 2 − 4 3 = −6

…(ii)

Applying KVL to Mesh 3, −2 (

3

− 1 ) + 6 − 4 ( 3 − 2 ) − 2 I3 = 0 −2 1 − 4 I 2 + 8 3 = 6

Solving Eqs (i), (ii) and (iii), I1 = 2 A I2 = 0 5 A I3 = 1 5 A I4 I 3 − I 2 = 1.5 − 0.5 = 1 A

…(iii)

2.3 Mesh Analysis 2.25

Example 2.34

Determine the voltage V which causes the current I1 to be zero in the network

of Fig. 2.59. 6Ω

2Ω

20 V

3Ω

I2 1Ω

5Ω

I1

V

4Ω I3

Fig. 2.59 Solution Applying KVL to Mesh 1, 20 − 6

1

2 ( I1 − I 2 ) 5 (

3)

1

V +13 13 I1 2

2

V

0

5 I 3 = 20

…(i)

Applying KVL to Mesh 2, −2 (

2

− 1 ) − 3 I 2 −1 1(

2

−

3)

=0

− 6 I 2 + I3 = 0

…(ii)

+ V − 5( 3 − 1) = 0 V + 5 I1 + 2 − 10 3 = 0

…(iii)

2

1

Applying KVL to Mesh 3, −1(

3

−

2 ) − 4 I3

Putting I1 = 0 in Eqs (i), (ii) and (iii), V

2 I 2 − 5 I 3 = 20 −6 I 2 I 3 = 0

V

I 2 −10 − 10 I 3 = 0

…(iv)

Solving Eqs (i), (ii) and (iii), V = 43.7 V

Example 2.35

Find the current through the 2 W resistor in the network of Fig. 2.60. 3Ω

12 Ω

6A I1

36 V

2Ω

6Ω I2

I3

9V

Fig. 2.60 Solution Mesh 1 contains a current source of 6 A. Hence, we can write current equation for mesh 1. Since direction of current source and mesh current I1 are same, I1 = 6 …(i)

2.26 Network Analysis and Synthesis Applying KVL to Mesh 2, 36 −12 12 ( 2 36 −12 12 (

2

1) − 6 ( 2

3)

0

3

0

6) 6 I 2 + 6

6 I 3 = 108

18 I 2 Applying KVL to Mesh 3, −6 ( 3 − 2 ) − 3 I 3 − 2 6

2

3

…(ii)

−9 = 0

− 11 11

3

=9

…(iii)

Solving Eqs (ii) and (iii), I3 = 3 A I2 I3 = 3 A

Example 2.36

Determine the mesh currents I1, I2 and I3 in the network of Fig. 2.61. 30 V I1 I2

I3

1A

6Ω

15 Ω

50 V

10 Ω 5 Ω

Fig. 2.61 Solution Applying KVL to Mesh 1, −30 − 6 1 − 15 15 ( 1 − 2 ) = 0 21 I1 − 15 I 2 = −30

…(i)

Applying KVL to Mesh 2, −10 ( 2 − 3 ) − 15 ( 2 − 1 ) + 50 − 5 2 = 0 −15 I1 + 30 I 2 − 10 I 3 = 50

…(ii)

For Mesh 3, I3 = 1

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 0 I2 = 2 A I3 = 1 A

Example 2.37

Find the current through the 5 W resistor in the network of Fig. 2.62. 3A 2Ω I 2 4A I1

5Ω I3

Fig. 2.62

2Ω 2V

I4

3A

2.3 Mesh Analysis 2.27

Solution Writing current equations for Meshes 1, 2 and 4, I1 = 4

(i)

I2 = 3

… (ii)

I 4 = −3

…(iii)

Applying KVL to Mesh 3, −5 (

3

− 1) − 2 (

3

−

2 ) − 2( 3

−

4) − 2

=0

…(iv)

Substituting Eqs (i), (ii) and (iii) in Eq. (iv), −5 (

3

− 4) − 2 (

3

− 3) − 2 (

3

+ 3) − 2 = 0 I3 = 2 A I 5 Ω = I1 − I 3 = 4 − 2 = 2 A

EXAMPLES WITH DEPENDENT SOURCES Example 2.38

Obtain the branch currents in the network shown in Fig. 2.63. IA

5Ω

10IB + −

5Ω

IB

10 Ω 5V

10 V

+ −

5IA

Fig. 2.63 Solution Assigning clockwise currents in two meshes as shown in Fig. 2.64, From Fig. 2.64, 10IB 5Ω IA …(i) I A I1 + − I B I 2 …(ii)

5Ω

Applying KVL to Mesh 1,

IB

10 Ω

5 5 I1 10 I 10 ( I1 − I 2 ) − 5 5 5 I1 10 0 I 2 −10 10 I1 + 10 0 I2 − 5

A 1

0 0

−20 I1 = −5 1 I1 = = 0.25 A 4 Applying KVL to Mesh 2, 5 A 10 ( 2 1 ) 5 I 2 − 10 = 0 5 1 10 0 2 10 1 5 I 2 = 10 15 I1 15 1 I 2 = 10

5V

10 V I1

+ −

5IA

I2

Fig. 2.64 …(iii)

…(iv)

2.28 Network Analysis and Synthesis Putting I1 = 0.25 A in Eq. (iv), 15 (0.25) 15

Example 2.39

10 I 2 = −0.416 A 2

Find the mesh currents in the network shown in Fig. 2.65. 2V2

4Ω

+ − 5Ω

2Ω

+ V2 −

+ V1 −

1Ω 10 V

− 2V1 +

5V

Fig. 2.65 Solution Assigning clockwise currents in the two meshes as shown in Fig. 2.66, From Fig. 2.66, …(i) V1 5 I1 V2 2 I 2 …(ii) Applying KVL to Mesh 1,

2V2 + − 5Ω

4Ω

2Ω

+ V2 −

+ V1 −

1Ω 10 V

5V

− 2V1 +

I1

−5 − 5 I1 − 2 2 − 4 I1 −1 1( 1 − 2 ) + 2V1 = 0 −5 − 5 I1 − 2 ( 2 I 2 ) − 4 1 − 1 + 2 + 2 ( −5 I1 ) = 0 20 I1 + 3 2 5

I2

Fig. 2.66 …(iii)

Applying KVL to Mesh 2, −2V1 − 1( I 2 − I1 ) − 2

2

− 10 = 0

−2 ( −5 I1 ) − I 2 + I1 − 2 11 I1 − 3

2 2

= 10 = 10

…(iv)

Solving Eqs (iii) and (iv), I1 = 0.161 A I 2 = −2.742 A

Example 2.40

Find currents Ix and Iy of the network shown in the Fig. 2.67. 2Ix + − 5Ω

4 Ω Iy

2Ω

1Ω 10 V

5V

− 2Iy + Ix

Fig. 2.67

2.3 Mesh Analysis 2.29

Solution Assigning clockwise currents in the two Meshes as shown in Fig. 2.68. From Fig. 2.68, 2Ix 4 Ω Iy I y I1 + − I1 − I 2

Ix

5Ω

1Ω

Applying KVL to Mesh 1, −5 − 5 I1 − 2 −5 − 5 I1 − 2 (

1

x

− 4 I1 −1 1(

−

2 ) − 4 I1

−5 − 5 I1 − 2 I1 + 2 I 2 − 4

1

1

−

2) + 2Iy

=0

− 2Iy + Ix

I1

5V

− I1 + I 2 + 2 I1 = 0

−

2Ω

10 V I2

Fig. 2.68

+ 2 + 2 I1 = 0 −10 I1 + 3 2 = 5 1

…(iii)

Applying KVL to Mesh 2, −2

y

− 1( I 2 − I1 ) − 2 I 2 − 10 = 0 −2

1

−

2

+ 1 − 2 I 2 = 10 − I1 − 3 2 = 10

...(iv)

Solving Eqs (iii) and (iv), 15 = −1.364 A 11 I 2 = −2.878 A I1 = −

I y = −1.364 A Ix

Example 2.41

I1 − I 2 = −1.364 + 2.878 = 1.514 A

Find the currents in the three meshes of the network shown in Fig. 2.69. Ix

1Ω

1Ω

+ −

1Ω

Ix

+ I − y 1Ω

5V

1A

1Ω

Iy

Fig. 2.69 Solution Assigning clockwise currents in the three meshes is shown in Fig. 2.70. Ix

1Ω

1Ω

+ −

1Ω

Ix

+ I − y 1Ω

5V I1

1Ω

I2

Iy

Fig. 2.70

1A I3

2.30 Network Analysis and Synthesis From Fig. 2.70, Ix = I1 I y I 2 − I3

...(i) …(ii)

Applying KVL to Mesh 1, 5 1 I1 − I y − 1( 5

1

(

3)

2

( −22

1

2)

0

1

2)

0

1

5

3

…(iii)

Applying KVL to Mesh 2, −1( −(

2

2

− 1 ) + I y − 1 I 2 − I x −11(

− 1) + (

2

−

3 ) − I2

2

−

3)

=0

− I1 − ( I 2 − I 3 ) = 0 −22

2

…(iv)

0

For Mesh 3, I 3 = −1

…(v)

Solving Eqs (iii), (iv) and (v), I1 = 2 A I2 = 0 I 3 = −1 A

Example 2.42

For the network shown in Fig. 2.71, find the power supplied by the dependent voltage

source. 50 Ω 20 Ω

5A

+ V1 −

30 Ω

+ −

0.4 V1

0.01 V1

Fig. 2.71 Solution Assigning clockwise currents in three Meshes as shown in Fig. 2.72. 50 Ω

20 Ω

5A

+ V1 −

I1

30 Ω

I3

+ −

0.4 V1

Fig. 2.72

I2

0.01 V1

2.3 Mesh Analysis 2.31

From Fig. 2.72, V1 20 ( I1 − I 3 ) 0 4 V1 = 0 20 I1 − 20 I 3 33 33 I1 − 33.33 I 3

0 6 V1 V1

…(i)

For Mesh 1, I1 = 5

…(ii)

For Mesh 2, 0.33 I1

I2

I2 0 01 V1 0.33 I 3 = 0

0.01(33.33 I1 33.33 I 3 ) …(iii)

Applying KVL to Mesh 3, −50 I 3 − 30 ( I 3 − I 2 ) − 20 ( I 3 − I1 ) = 0 −20 I1 − 30 I 2 + 100 I 3 = 0

…(iv)

Solving Eqs (ii), (iii) and (iv), I1 = 5 A I 2 = −1.47 A I 3 = 0 56 A V1

33 33 I1 − 33.33 I 3 = 33.33( 33 (5) 33.33 (0.56) = 148 V

Power supplied by the dependent voltage source = 0.4 V1 (I1 − I2) = 0.4 (148) (5 + 1.47) = 383.02 W

Example 2.43

Find the voltage Vx in the network shown in Fig. 2.73. 0.45 A

16.67 Ω

33.33 Ω

+ Vx − 25 Ω

30 V

− +

2 Vx

10 V

`

Fig. 2.73 Solution Assigning clockwise currents in the three meshes as shown in Fig. 2.74. 0.45 A

16.67 Ω

I3 33.33 Ω

+ Vx −

25 Ω

− +

30 V I1

Fig. 2.74

10 V

I2

2 Vx

2.32 Network Analysis and Synthesis From Fig. 2.74, 16 67 I1

Vx Applying KVL to Mesh 1, −30 − 16.67 67 I1 − 33 33 (

1

−

25 ( 1 3 ) − 25

−

2 ) − 10

…(i)

=0

−30 − 16.67 67 I1 − 33 33 I1 + 33.33 I 3 − 25 5 I1 + 25 I 2 − 10 = 0 −75 I1 + 25 I 2 + 33.33 I 3 = 40

…(ii)

Applying KVL to Mesh 2, 10 − 25 ( 2 1 ) + 2 Vx 10 − 25 ( 2 1 ) + 2 (16.67 1 )

0 0

10 − 25 I 2 + 25 25 I1 + 33.34 I1 = 0 58.34 I1 255 I 2 = −10

…(iii)

I 3 = 0 45

…(iv)

For Mesh 3, Solving Eqs (ii), (iii) and (iv), I1 = −0 9 A I 2 = −1 7 A I 3 = 0 45 A Vx

Example 2.44

16 67 I1 = 16.67 ( −0.9) = −15 V

…(v)

For the network shown in Fig. 2.75, find the mesh currents I1, I2 and I3. 15 A

2Ω

I3 0.111 Vx

1Ω

+ Vx −

I1

3Ω

1Ω

I2 2Ω

Fig. 2.75 Solution From Fig. 2.75, Vx

3( I1 − I 2 )

…(i)

I 3 = 15

…(ii)

Writing current equation for the two current sources,

and

0.111 Vx I1 0.111 [3 ( 1 2 ) = I1 0.333 333 I1 0 333 I 2 − I1 I 3 = 0 −0 0.667 667 I1 0 333 I 2 + I 3 = 0

I3 I3 …(iii)

2.3 Mesh Analysis 2.33

Applying KVL to Mesh 2, −3 ( 2 − 1 ) − 1( 2 − 3 ) − 2 2 = 0 −3 1 + 6 I 2 − I 3 = 0

…(iv)

Solving Eqs (ii), (iii) and (iv), I1 = 17 A I 2 = 11 A I 3 = 15 A

Example 2.45

For the network shown in Fig. 2.76, find the magnitude of V0 and the current supplied by it, given that power loss in RL = 2 W resistor is 18 W. 10 Ω

5Ω

5Ω

+ Vx − V0

2Ω

4Ω

RL = 2 Ω

2Ω

2Vx

Fig. 2.76 Solution Assigning clockwise currents in meshes is shown in Fig. 2.77. 10 Ω

5Ω

5Ω

+ Vx − V0

2Ω I1

4Ω

2Ω

2Vx

I2

I3

RL = 2 Ω I4

Fig. 2.77 From Fig. 2.77, Vx

…(i)

I1

Also, I 4 2 RL = 18 I 4 2 ( 2) 18 I4 = 3 A

…(ii)

Applying KVL to Mesh 1, V0

5 I1 2 ( I1 I 2 ) = 0 7 I1 2 I 2 = V0

…(iii)

Applying KVL to Mesh 2, −2 (

2

− 1) − 4 I2 = 0 −2 1 + 6 I 2 = 0

…(iv)

For Mesh 3, 10 I1

I 3 2Vx I3 = 0

2 (5 I1 ) 10 I1 …(v)

2.34 Network Analysis and Synthesis Applying KVL to Mesh 4, −2 (

4

−

3 ) − 5 I4

−2

4

=0

−2 3 + 9 I 4 = 0 −2 3 + 9 (3) = 0 I 3 = 13.5 A From Eq. (v), I1 =

I 3 13.5 = = 1.35 A 10 10

From Eq. (iv), −2 (1.35) + 6

=0 I 2 = 0 45 A 2

From Eq. (iii), 7 (1.35) 2 (0.45 45)) = V0 V0 = 8 55 V Current supplied by voltage source V0 = I1 = 1.35 A

Example 2.46

In the network shown in Fig. 2.78, find voltage V2 such that Vx = 0. + − 0 1 Vx

2A 10 Ω

24 V

3A

5Ω + Vx −

20 Ω

V2

Fig. 2.78 Solution Assigning clockwise currents in four meshes as shown in Fig. 2.79. From Fig. 2.79, …(i) Vx 20 ( I 3 − I 4 ) 2A

Writing current equations for Meshes 1 and 2, I1 = 2 I2 = 3 Applying KVL to Mesh 3, 24 −10 10 ( 3 1 ) − 20 ( 3 4 ) = 0 24 −10 10 ( 3 2) 20 0( 3 4) 0 −30 30 I 3 + 200 I 4 = −44 Applying KVL to Mesh 4, −20 ( 4 − 3 ) − 5 ( 4 − 2 ) + V2 = 0 −20 ( 4 − 3 ) − 5 ( 4 − 3) + V2 = 0 20 I 3 − 25 I 4 = −V2 − 15 But Vx = 0 20 ( I 3 I 4 ) = 0 I3 I 4

I2 + − 0.1 Vx

I1 10 Ω

…(ii) …(iii) 24 V

I3

3A

5Ω + Vx −

V2

20 Ω I4

Fig. 2.79 …(iv)

…(v)

2.4 Supermesh Analysis 2.35

From Eq. (iv), 30 I 3 + 20 I 3 = −44 I 3 = 4.4 A I 4 = 4.4 A From Eq. (v), 20 ( 4.4) 25 ( 4.4 4)) = − V2 − 15 V2 = 7 V

2.4

SUPERMESH ANALYSIS

Meshes that share a current source with other meshes, none of which contains a current source in the outer loop, form a supermesh. A path around a supermesh doesn’t pass through a current source. A path around each mesh contained within a supermesh passes through a current source. The total number of equations required for a supermesh is equal to the number of meshes contained in the supermesh. A supermesh requires one mesh current equation, that is, a KVL equation. The remaining mesh current equations are KCL equations.

Example 2.47

Find the current through the 10 W resistor of the network shown in Fig. 2.80. 5Ω

1Ω 2V

10 Ω I1

15 Ω

4A I2

I3

Fig. 2.80 Solution Applying KVL to Mesh 1, 2 1 I1 10 ( I1 − I 2 ) = 0 11 I1 10 I 2 = 2

…(i)

Since meshes 2 and 3 contain a current source of 4 A, these two meshes will form a supermesh. A supermesh is formed by two adjacent meshes that have a common current source. The direction of the current source of 4 A and current (I3 − I2) are same, i.e., in the upward direction. Writing current equation to the supermesh,1 I2 = 4

…(ii)

Applying KVL to the outer path of the supermesh, −10 ( 2 − 1 ) − 5 2 − 15 15 3 = 0 10 I1 −15 15 I 2 − 15 I 3 = 0 15

…(iii)

I3

Solving Eqs (i), (ii) and (iii), I1 = −2 35 A I 2 = −2 78 A I 3 = 1.22 A Current through the 10 Ω resistor = I1 − I2 = −(2.35) − (−2.78) = 0.43 A

…(iv)

2.36 Network Analysis and Synthesis

Example 2.48

Find the current in the 3 W resistor of the network shown in Fig. 2.81.

2Ω

1Ω

I2

7V I1

7A

3Ω 1Ω

2Ω

I3

Fig. 2.81 Solution Meshes 1 and 3 will form a supermesh. Writing current equation for the supermesh, I3 = 7

I1

…(i)

Applying KVL to the outer path of the supermesh, 7 1(

2 ) − 3( 3

1

− I1 + 4

1I 3 = 0

2) 2

− 4 I 3 = −7

…(ii)

Applying KVL to Mesh 2, −1(

2

− 1) − 2 I2 − 3( I1 − 6

2

−

3)

=0

2 + 3 I3 = 0

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 9 A I2 = 2 5 A I3 = 2 A Current through the 3 Ω resistor = I2− I3 = 2.5 − 2 = 0.5 A

Example 2.49

Find the current in the 5 W resistor of the network shown in Fig. 2.82.

10 Ω

50 V

I1

2Ω

I2

3Ω

2A

1Ω 5Ω

Fig. 2.82

I3

2.4 Supermesh Analysis 2.37

Solution Applying KVL to Mesh 1, 50 −10 10 ( 1 2 ) − 5 ( 1 3 ) 0 15 I1 10 I 2 − 5 3 50 Meshes 2 and 3 will form a supermesh as these two meshes share a common current source of 2 A. Writing current equation for the supermesh, I 2 I3 = 2

…(i)

…(ii)

Applying KVL to the outer path of the supermesh, −10 (

2

− 1) − 2

2

− 1I 3 − 5 (

3

− 1) = 0

−15 I1 +12 12 I 2 + 6

3

=0

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 20 A I 2 = 17.33 A I 3 = 15.33 A Current through the 5 Ω resistor = I1 − I3 = 20 − 15.33 = 4.67 A

Example 2.50

Determine the power delivered by the voltage source and the current in the 10 W resistor of the network shown in Fig. 2.83. 5Ω 3A

5Ω

I2 10 Ω

50 V

I1 1Ω

10 A I3

Fig. 2.83 Solution Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I1 I 2 = 3

…(i)

Applying KVL to the outer path of the supermesh, 50 − 5 1 5 I 2 −10 10 ( I 2 I 3 ) −1 1( 1 3 ) 0 −6 6 1 15 5 2 11 3 50

…(ii)

For Mesh 3, I 3 = 10 Solving Eqs (i), (ii) and (iii), I1 = 9 76 A I 2 = 6 76 A I 3 = 10 A Power delivered by the voltage source = 50 I1 = 50 × 9.76 = 488 W I10 Ω I 3 I 2 = 10 − 6.76 3.24 A

…(iii)

2.38 Network Analysis and Synthesis

Example 2.51

For the network shown in Fig. 2.84, find current through the 8 W resistor. 6Ω I2

2Ω 3A

4Ω 5V

I1

11 Ω

8Ω I4

7A

12 A

I3

10 V

Fig. 2.84 Writing current equations for Meshes 1 and 4, I1 = −3

...(i)

I 4 = −12

...(ii)

Meshes 2 and 3 will form a supermesh. Writing current equation for the supermesh, I2 = 7

I3

…(iii)

Applying KVL to the outer path of the supermesh, 5 4 ( 2 1 ) − 6 2 8 ( 3 − 4 ) + 10 = 0 5 4 ( 2 + 3) 6 I 2 − 8 ( 3 12) 10 0 −10 I 2 8 I 3 93

…(iv)

Solving Eqs (iii) and (iv), I 2 = −8 28 A I 3 = −1.28 A I 3 − I 4 = −1.28 12 1 10.72 A

I8

EXAMPLES WITH DEPENDENT SOURCES Example 2.52

In the network of Fig. 2.85, find currents I1 and I2. 8Ω

3V0

V0

+ − 2Ω

−10 V

10 Ω

−3 A I 2

I1

Fig. 2.85 Solution From Fig. 2.85, −10 − 8

1

−

0 =0 V0 = −10 − 8 I1

…(i)

Meshes 1 and 2 will form a supermesh. Writing current equations for the supermesh, I2

I1 = −3

…(ii)

2.4 Supermesh Analysis 2.39

Applying KVL to the outer path of the supermesh, 10 8 I1 − 3V0 − 10 I 2 = 0 − 10 − 8 1 ) − 10 2 = 0 1 3 ( −10

−10 − 8

16 I1 −10 10 I 2 = −20

…(iii)

Solving Eqs (ii) and (iii), I1 = −8 33 A I 2 = −11.33 A

Example 2.53

In the network of Fig. 2.86, find the current through the 3 W resistor. −4 V

3Ω

+ Vx −

2Ω

1Ω

+ −

2A

5 Vx

Fig. 2.86 Solution Assigning clockwise currents in two meshes as shown in Fig. 2.87. From Fig. 2.87, −4 V Vx 2 I1 …(i) Meshes 1 and 2 will form a supermesh. Writing current equations for the supermesh,

2Ω

I1 = 2

I2

…(ii)

3Ω

+ Vx −

1Ω 2A

I1

Applying KVL to the outer path of the supermesh,

1

− 3 I2 = 4

…(iii)

Solving Eqs (ii) and (iii), I1 = 2 A I2 = 4 A I3

Example 2.54

I2 = 4 A

Find the currents I1 and I2 at the network shown in Fig. 2.88. 10 Ω 14 Ω

4Ω I3

110 V

I2

Fig. 2.87

−2 1 − 4 − 3 2 − 5Vx = 0 −2 1 − 4 − 3 2 − 5 ( −2 1 ) = 0 8

+ −

2Ω

+ V1 −

I1

6Ω

0.5 V1 I2

Fig. 2.88

5 Vx

2.40 Network Analysis and Synthesis Solution From Fig. 2.88, V1 = 2 (I1 − I2) Meshes 2 and 3 will form a supermesh. Writing current equation for the supermesh, I3

I 2 = 0.5 V1 = 0 5 × 2 ( I1 I3

I 2 ) = I1

I2

I1

Applying KVL to outer path of the supermesh, −2 ( 2 − 1 ) − 10 3 − 6 I 2 = 0 −2 2 + 2 I1 − 10 I1 − 6 2 = 0 I1 = − I 2 Applying KVL to Mesh 1, 110 −14 14

1

4 1 2 ( I1 I 2 ) = 0 110 − 20 I1 + 2 2 0 110 + 20 I1 + 2

0

2

I 2 = −5 A I1 = I 2 = 5 A

Example 2.55

For the network of Fig. 2.89, find current through the 8 W resistor. 5 Iy

+ −

10 Ω

6Ω

Iy 8Ω

50 V Ix

52 V

0.5 Ix

Fig. 2.89 Solution Assigning clockwise currents to the three meshes as shown in Fig. 2.90. From Fig. 2.90, Ix Iy

I1 I 2 − I3

…(i) 10 Ω

…(ii)

I3

5 Iy 6Ω

Iy 8Ω

Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I 2 I1 = 0.5 I x −0 0 5 I1 I 2 = 0

+ −

50 V Ix

0.5 ( I1 ) …(iii)

I1

0.5 Ix

52 V I2

Fig. 2.90

Applying KVL to the outer path of the supermesh, 50 −10 10 (

1

3) − 6( 2

−10 10 I1

8 I 2 − 52 = 0 14 I 2 +16 16 I 3 = 2

3)

…(iv)

2.4 Supermesh Analysis 2.41

Applying KVL to Mesh 3, 5 y − 6 ( I 3 − I 2 ) −10 10 ( I 3 − I1 ) = 0 −5 (

2

−

3) − 6

( I 3 − I 2 ) − 10 ( I 3 − I1 ) = 0 10 I1 + I 2 − 11 I 3 = 0

…(v)

Solving Eqs (iii), (iv) and (v), I1 = −1 56 A I 2 = −0 58 A I 3 = −1.11 A I8 I 2 = −0.58 A

Example 2.56

For the network shown in Fig. 2.91, find the current through the 10 W resistor. 10 Ω

20 V 5 Ω

15 V

4Ω

2A I1

2 I1 I2

40 V I3

Fig. 2.91 Solution Meshes 1, 2 and 3 will form a supermesh. Writing current equations for the supermesh, I1 I 2 = 2

…(i)

I 2 = 2 I1 I3 0

I3 I2

…(ii)

2 I1 and Applying KVL to the outer path of the supermesh, 15 −10 10 I1 20 0 5 2 4 I 3 + 40 = 0 10 I1 + 5 2 4 I 3 = 35

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 1 96 A I 2 = −0 04 A I 3 = 3 89 A I10 Ω

I1 = 1.96 A

Example 2.57

In the network shown in Fig. 2.92, find the power delivered by the 4 V source and voltage across the 2 W resistor. 2Ω

+ V2 − 5A 5Ω 4V

6Ω 4Ω

I1 1Ω

I2

I3

Fig. 2.92

V2 2

2.42 Network Analysis and Synthesis Solution From Fig. 2.92, V2

…(i)

2 I1

Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I2

I1 = 5

…(ii)

Applying KVL to the outer path of the supermesh, 4 5 I 2 2 1 6 I1 4 ( 1 3 ) 1( I 2 − I 3 ) = 0 −12 12 I1 6 2 5 I 3 = −4

…(iii)

For Mesh 3, V2 2 I1 = = I1 2 2 I3 = 0 I3 =

I1

…(iv)

Solving Eqs (ii), (iii) and (iv), I1 = −2 A I2 = 3 A I 3 = −2 A Power delivered by the 4 V source = 4 I2 = 4 (3) = 12 W V2 2 I1 2 ( −2 2) 4V

Example 2.58

Find currents I1, I2, I3, I4 of the network shown in Fig. 2.93. 1 Ω 10 1 Ω 5 6V

1 Ω 20 − Vx +

1 Ω 15 5 Vx

I2

I3

1 Ω 2 40 A

1 Ω 6

I1

I4

Fig. 2.93 From Fig. 2.93, Vx

1 ( I 2 − I1 ) 5

…(i)

For Mesh 4, I 4 = 40 Applying KVL to Mesh 1, 1 1 1 −6 − I1 − ( I1 − I 2 ) − ( I1 − I 4 ) = 0 10 5 6 1 1 1 1 1 −6 − I1 − I1 + I 2 − I1 + (40) 0 10 5 5 6 6 7 1 2 − I1 + I 2 = − 15 5 3

…(ii)

…(iii)

2.5 Node Analysis 2.43

Mesh 2 and 3 will form a supermesh. Writing current equation for the supermesh, I3 I1 2 I 2

I 2 = 5 Vx

⎡1 ⎤ ( I 2 − I1 ) 5 ⎣ ⎦

I 2 − I1 …(iv)

0

I3

Applying KVL to the outer path of the supermesh, 1 1 1 1 − ( 2 − 1) − I 2 − I3 − ( 3 − 4 ) = 0 5 20 15 2 1 1 1 1 1 1 − I 2 + I1 − I 2 − I 3 − I 3 + ( 40) = 0 5 5 20 15 2 2 1 1 17 I1 I 2 − I 3 = −20 5 4 30

…(v)

Solving Eqs (iii), (iv) and (v), I1 = 10 A I 2 = 20 A I 3 = 30 A I 4 = 40 A

2.5

NODE ANALYSIS

Node analysis is based on Kirchhoff’s current law which states that the algebraic sum of currents meeting at a point is zero. Every junction where two or more branches meet is regarded as a node. One of the nodes in the network is taken as reference node or datum node. If there are n nodes in any network, the number of simultaneous equations to be solved will be (n − 1).

Steps to be followed in Node Analysis 1. 2. 3. 4.

Assuming that a network has n nodes, assign a reference node and the reference directions, and assign a current and a voltage name for each branch and node respectively. Apply KCL at each node except for the reference node and apply Ohm’s law to the branch currents. Solve the simultaneous equations for the unknown node voltages. Using these voltages, find any branch currents required.

Example 2.59

Calculate the current through 2 W resistor for the network shown in Fig. 2.94. VA 1Ω

0.5 Ω 1Ω

20 V

VB

2Ω

1Ω 20 V

Fig. 2.94 Solution Assume that the currents are moving away from the nodes.

2.44 Network Analysis and Synthesis Applying KCL at Node A, 20 VA VA VB + + =0 1 1 05 1 20 ⎛1 1 1 ⎞ VB = ⎜⎝ + + ⎟⎠ VA 1 1 0.5 0.5 1 4 VA 2 VB = 20 2

…(i)

VB VA VB VB − 20 + + =0 05 2 1 1 20 ⎛ 1 1 1⎞ − VA VB = 0.5 ⎝ 0.5 2 1⎠ 1 −2 VA + 3.5 VB = 20

…(ii)

VA

Applying KCL at Node B,

Solving Eqs (i) and (ii), VA = 11 V VB = 12 V Current through the 2 Ω resisto resistor =

Example 2.60

VB 2

12 2

6A

Find the voltage at nodes 1 and 2 for the network shown in Fig. 2.95. 1Ω

V1

2Ω

1 1A

V2 2

2Ω

1Ω

2A

Fig. 2.95 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V V −V 1= 1 + 1 2 2 2 1 ⎛ 1 1⎞ V2 = 1 ⎜⎝ + ⎟⎠ V1 2 2 2 V1 0 5 V2 = 1

…(i)

Applying KCL at Node 2, 2=

V2 V2 − V1 + 1 2

1 ⎛ 1⎞ − V1 + 1 + V2 = 2 ⎝ 2⎠ 2 −0 0.5 5V1 + 1 5 V2 = 4

…(ii)

2.5 Node Analysis 2.45

Solving Eqs (i) and (ii), V1 = 2 V V2 = 2 V

Example 2.61

Find the current in the 100 W resistor for the network shown in Fig. 2.96. 20 Ω

30 Ω

V1

50 Ω

1A

60 V

V2

100 Ω

40 V

Fig. 2.96 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V1 60 V1 − V2 + =1 20 30 1⎞ 1 60 ⎛ 1 V2 = +1 ⎜⎝ + ⎟⎠ V1 20 30 30 20 …(i)

0.083 V1 − 0.033 V2 = 4 Applying KCL at Node 2, V2 V1 V 2 − 40 V2 + + =0 30 50 100 1 1 1 ⎞ 40 ⎛ 1 − V1 V2 = ⎝ ⎠ 30 30 50 100 50 −0.0 033 V1 0 063 V2 = 0.8

…(ii)

Solving Eqs (i) and (ii), V1 = 67.25 V V2 = 48 V Current through the 100 Ω resistor =

Example 2.62

V2 48 = = 0 48 A 100 100

Find VA and VB for the network shown in Fig. 2.97. 1Ω

VA

2A

2Ω

2Ω 1V

Fig. 2.97

VB

2Ω

1A

2V

2.46 Network Analysis and Synthesis Solution Assume that the currents are moving away from the nodes. Applying KCL at Node A, 2= ⎛1 1 ⎞ ⎜⎝ + + 1⎠ VA VB 2 2 2VA VB

VA VA − 1 VA − VB + + 2 2 1 1 2+ 2 …(i)

2.5

Applying KCL at Node B, VB VA VB − 2 + =1 1 2 2 ⎛ 1⎞ −V VA 1 VB = 1 + ⎝ 2⎠ 2 −V VA 1 VB = 2

…(ii)

Solving Eqs (i) and (ii), VA = 2.875 V V B = 3 25 V

Example 2.63

Find currents I1, I2 and I3 for the network shown in Fig. 2.98. 10 Ω

V1

2Ω

V2

5Ω

2Ω

50 V

4Ω I1

25 V

I2

I3

Fig. 2.98 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V1 V1 − 25 V1 V2 + + =0 2 5 10 1 25 ⎛1 1 1 ⎞ V2 = ⎜⎝ + + ⎟⎠ V1 2 5 10 10 5 0.8 V1 0 1 V2 = 5

…(i)

V2 V1 V2 V2 − ( −50) + + =0 10 4 2 1 50 ⎛ 1 1 1⎞ − V1 V2 = − ⎝ 10 4 2 ⎠ 10 2 −0 1 V1 + 0 85 V2 = −25

…(ii)

Applying KCL at Node 2,

2.5 Node Analysis 2.47

Solving Eqs (i) and (ii), V1 = 2 61 V V2 = −29.1 V V 2 61 I1 = − 1 = − = −1 31 A 2 2 V V2 2.61 ( 2.91) I2 = 1 = = 3 17 A 10 10 V + 50 5 −29 29.1 50 I3 = 2 = = 10.45 A 2 2

Example 2.64

Find currents I1, I2 and I3 and voltages Va and Vb for the network shown in Fig. 2.99. I1

0.2 Ω

I2 0.3 Ω + a Va

120 V

−

I3 0.1 Ω + b Vb −

30 A

110 V 20 A

Fig. 2.99 Solution Applying KCL at Node a, I1 30 I 2 120 − Va V − Vb = 30 + a 0.2 0.3 36 − 0.3 3 Va = 1.8 + 0.22Va − 0.2 Vb 0.55 Va − 0.2 Vb = 34.2

…(i)

Applying KCL at Node b,

0.1 1 Va

I 2 I 3 = 20 Va Vb 110 − Vb + = 20 0.3 0.1 0.1 Vb + 33 − 0 3 Vb = 20 0 03 0.11 Va 0.4 Vb = −32.4

Solving Eqs (i) and (ii), Va = 112 V Vb = 109 V 120 − Va 120 − 112 = = 40 A 0.2 0.2 V Vb 112 − 109 I2 = a = = 10 A 0.3 0.3 110 − Vb 110 − 109 I3 = = = 10 A 0.1 0.1 I1 =

…(ii)

2.48 Network Analysis and Synthesis

Example 2.65

Calculate the current through the 5 W resistor for the network shown in Fig. 2.100. 3Ω

V1

4A

5Ω

V2

V3

2Ω

4Ω

2Ω

8A

20 V

Fig. 2.100 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V1 V1 − V2 + =0 2 3 1 ⎛ 1 1⎞ V2 = 4 ⎜⎝ + ⎟⎠ V1 2 3 3 4+

0.83 83 V1 0 33 V2 = −4

…(i)

Applying KCL at Node 2, V1 V2 − ( −20) V2 V3 + + =0 3 2 5 1 1 20 ⎛ 1 1 1⎞ − V1 V − V =− ⎝ 3 2 5⎠ 2 5 3 3 2 −0 333 3 V1 1 03 V2 − 0 2 V3 = −10

V2

…(ii)

Applying KCL at Node 3, V3 V2 V3 + =8 5 4 1 ⎛ 1 1⎞ − V2 V3 = 8 ⎝ 5 4⎠ 5 −0.2 V2 0 45 V3 8

…(iii)

Solving Eqs (i), (ii) and (iii), V1 = −8 76 V V2 = −9 92 V V3 = 13.37 V Current through the 5 Ω resistor resisto =

V3 V2 5

13.37 − ( −9.92) 5

4 66 A

2.5 Node Analysis 2.49

Example 2.66

Find the voltage across the 5 W resistor for the network shown in Fig. 2.101. 9A 4Ω 2Ω

V1

5Ω

V2

4Ω

V3

20 Ω

100 Ω

12 V

Fig. 2.101 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V1 12 V1 − V2 V1 − V3 + + +9 4 2 4 ⎛ 1 1 1⎞ ⎜⎝ + + ⎟⎠ V1 4 2 4

0

1 1 12 V2 − V3 = −9 + 2 4 4

V1 0 5 V2 − 0.25 V3

6

…(i)

Applying KCL at Node 2, V2 V1 V2 V2 − V3 + + =0 2 100 5 1 1 ⎛1 − V1 ⎝ 2 100 2

1⎞ 1 V2 − V3 = 0 5⎠ 5

−0 5 V1 + 0.7 71 V2 − 0 2

3

=0

…(ii)

Applying KCL at Node 3, V3 V2 V3 V3 − V1 + + =9 5 20 4 1 − V1 4

1 ⎛ 1 1 1⎞ V2 + + + V3 = 9 ⎝ 5 20 4 ⎟⎠ 5 −0. 0 25 2 V1 − 0.22 V2

0 5 V3 = 9

Solving Eqs (i), (ii) and (iii), V1 = 6 35 V V2 = 11.76 V V3 = 25.88 V Voltages across the 5 Ω resistor esisto V3

2

25.88 11.76 14.12 V

…(iii)

2.50 Network Analysis and Synthesis

Example 2.67

Determine the current through the 5 W resistor for the network shown in Fig. 2.102. 4Ω

36 V 2Ω

V1

4Ω

3A

5Ω

V2

100 Ω

V3

20 Ω

Fig. 2.102 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V1 V1 − V2 V1 − 36 − V3 + + =3 4 2 4 1 1 36 ⎛ 1 1 1⎞ V2 − V3 = 3 + ⎜⎝ + + ⎟⎠ V1 4 2 4 2 4 4 V1 − 0.5 V2 0.25 2 V3 = 12

…(i)

Applying KCL at Node 2, V2 V1 V2 V2 − V3 + + =0 2 100 5 1 1 1⎞ 1 ⎛1 − V1 V2 − V3 = 0 ⎝ ⎠ 2 2 100 5 5 −0 5 V1 + 0.771 V2 − 0 2 3 = 0

…(ii)

Applying KCL at Node 3, V3 V2 V3 V3 − ( −36) − V1 + + =0 5 20 4 1 1 ⎛ 1 1 1⎞ − V1 V2 + + + V3 = −9 ⎝ 5 20 4 ⎟⎠ 4 5 −0.25 25 V1 0 2 2 0.5 V3 9

…(iii)

Solving Eqs (i), (ii) and (iii), V1 = 13.41 V V2 = 7 06 V V3 = −8 47 V Current through the 5 Ω resistor resisto =

V2 V3 5

7.06 ( 8.47) 5

3

A

2.5 Node Analysis 2.51

Example 2.68

Find the voltage drop across the 5 W resistor in the network shown in Fig. 2.103. 4Ω

5Ω

2A

V2

V1

1Ω

1A

V3

2Ω

Fig 2.103 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, 1+

V1 V2 V1 V3 + =0 5 4

⎛ 1 1⎞ ⎜⎝ + ⎟⎠ V1 5 4 0.45 45 V1 0 2

1 1 V2 − V3 = −1 5 4 2

0.2 25 V3 = −1

…(i)

Applying KCL at Node 2, V2 V1 V2 + =2 5 1 1 ⎛1 ⎞ − V1 1 V2 ⎝5 ⎠ 5

2

−0.2 V1 1 2 V2 = 2

…(ii)

Applying KCL at Node 3, V3 V3 − V1 + +2 2 4 1 ⎛1 − V1 ⎝2 4 −0 0.25 5 V1 0

0

1⎞ V3 = −2 4⎠ V3 = −2 .

Solving Eqs (i), (ii) and (iii), V1 = −4 V V2 = 1 V V3 = −4 V V5

V2 − V1 = 1 ( 4) = 5 V

…(iii)

2.52 Network Analysis and Synthesis

Example 2.69

Find the power dissipated in the 6 W resistor for the network shown in Fig. 2.104. V1

3Ω 20 V

1Ω

6Ω

V2

2Ω V3

5A

5Ω

Fig. 2.104 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V1 20 V1 − V2 V1 − V3 + + =0 3 1 2 1⎞ 1 20 ⎛1 ⎜⎝ + 1 + ⎟⎠ V1 V2 − V3 = 3 2 2 3 1 83 V1 V2 − 0.5 5 V3 = 6.67

…(i)

Applying KCL at Node 2, V2 V1 V2 V3 + =5 1 6 1 ⎛ 1⎞ 1 V2 − V3 = 5 ⎝ 6⎠ 6 1 17 V2 0.17 V3 = 5

…(ii)

V3 V1 V3 V3 − V2 + + =0 2 5 6 1 1 ⎛ 1 1 1⎞ − V1 V2 + + + V3 = 0 ⎝ 2 5 6 ⎟⎠ 2 6 −0 0.5 V1 − 0.17 V2 0 87 8 V3 = 0

…(iii)

− V1 −V V1 Applying KCL at Node 3,

Solving Eqs (i), (ii) and (iii), V1 = 23.82 V V2 = 27.4 V V3 = 19.04 V I6 Ω =

V2 V3 27.4 19.04 = = 1.39 A 6 6

Power dissipated in the 6 Ω resistor = (1.39)2 × 6 = 11.59 W

Example 2.70 the 10 W resistor zero.

Find the voltage V in the network shown in Fig. 2.105 which makes the current in

2.5 Node Analysis 2.53 3Ω

10 Ω

V1

2Ω

V

7Ω

V2

5Ω

50 V

Fig. 2.105 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V1 V V1 V1 − V2 + + =0 3 2 10 1 1 ⎛1 1 1 ⎞ V2 − V = 0 ⎜⎝ + + ⎟⎠ V1 3 2 10 10 3 0.93 V1 0.1 V2 0 33 V 0

…(i)

Applying KCL at Node 2, V2 V1 V2 V2 − 50 + + =0 10 5 7 1 50 ⎛ 1 1 1⎞ − V1 V2 = ⎝ 10 5 7 ⎠ 10 7 −0.1 V1 + 0.44 V2 7.14

…(ii)

V1 V2 =0 10 V2 = 0

I10 Ω = V1 Solving Eqs (i), (ii) and (iii),

Example 2.71

V = 52.82 V

Find V1 and V2 for the network shown in Fig. 2.106. 50 Ω

Va

2A 10 Ω

80 V

20 Ω

Vb

+ V − 1

+

V2 50 Ω

−

Fig. 2.106 Solution

…(iii)

Assume that the currents are moving away from the nodes.

Vc

20 V

2.54 Network Analysis and Synthesis Applying KCL at Node a, 80 Va − Vb + +2=0 50 10 1⎞ 1 80 ⎛ 1 Vb = −2 ⎜⎝ + ⎟⎠ Va 50 10 10 50 0.12 12 Va 0 1 Vb = −0 4 Va

…(i)

Applying KCL at Node b,

−

1 Va 10

Vb Va Vb Vb − Vc + + =0 10 50 20 1 1⎞ 1 ⎛ 1 Vb − Vc = 0 ⎝ 10 50 20 ⎠ 20 −0..1 Va 0..17 17 Vb − 0.05 Vc 0

…(ii)

Node c is directly connected to a voltage source of 20 V. Hence, we can write voltage equation at Node c. Vc = 20

…(iii)

Solving Eqs (i), (ii), and (iii), Va = 3 08 V Vb = 7 69 V V1 Va Vb = 3.08 − 7.69 = −4 61 V V2 Vb Vc = 7.69 20 12.31 V

Example 2.72

Find the voltage across the 100 W resistor for the network shown in Fig. 2.107. 50 Ω

VA

20 Ω

VB

20 Ω 12 V

60 V

0.6 A 50 Ω

20 Ω 50 Ω

VC

100 Ω

Fig. 2.107 Solution Node A is directly connected to a voltage source of 20 V. Hence, we can write voltage equation at Node A. …(i) VA = 60 Assume that the currents are moving away from the nodes. Applying KCL at Node B,

−

1 VA 20

VB VA VB VC VB + + =06 20 20 20 1 1⎞ 1 ⎛ 1 VB − VC = 0..6 ⎝ 20 20 20 ⎠ 20 −0 0.0 05 5 VA 0.15 1 VB − 0.05 05 VC 0.6

…(ii)

2.5 Node Analysis 2.55

Applying KCL at Node C,

−

1 VA 50

VC VA VC VB VC 12 VC + + + =0 50 20 50 100 1 1 1 1 ⎞ 12 ⎛ 1 VB + + + + ⎟ VC = ⎝ 50 20 50 100 20 0 ⎠ 50 −0 0.02 02 VA 0 05 0 VB + 0.1V 1 VC = 0.24

…(iii)

Solving Eqs (i), (ii), and (iii), VC = 31.68 V Voltages across the 100 Ω resistor = 31.68 V

EXAMPLES WITH DEPENDENT SOURCES Example 2.73

Find the voltage across the 5 W resistor in the network shown in Fig. 2.108. I1

V1

20 Ω

10 Ω

10 Ω

5Ω

2A

− +

30I1

50 V

Fig. 2.108 Solution From Fig. 2.108, I1 =

V1 50 V1 − 50 = 20 + 10 30

…(i)

Assume that the currents are moving away from the node. Applying KCL at Node 1, 2=

V1 V1 + 30 I1 V1 − 50 + + 5 10 30

V1 5 V 2= 1 5 2=

⎛ V − 50 ⎞ V1 + 30 ⎜ 1 ⎝ 30 ⎟⎠ V1 − 50 + + 10 30 2V1 − 50 V1 − 50 + + 10 30

Solving Eq. (ii), V1 = 20 V Voltage across the 5 Ω resistor = 20 V

…(ii)

2.56 Network Analysis and Synthesis

Example 2.74

For the network shown in Fig. 2.109, find the voltage Vx. 40 Ω

Vx Iy 100 Ω

0.6 A

+ −

50 Ω 25 Iy

0.2 Vx

Fig. 2.109 Solution From Fig. 2.109, Iy =

Vx 100

…(i)

Assume that the currents are moving away from the node. Applying KCL at Node x, Vx Vx Vx 0 2 Vx + + 100 50 40 Vx Vx 0 8 Vx ⎛ Vx ⎞ 25 ⎜ +0 6 = + + ⎝ 100 ⎟⎠ 100 50 40 25 I y + 0 6 =

1 1 0 8⎞ ⎛1 − − ⎜⎝ − ⎟ Vx = −0 6 4 100 50 40 ⎠ 0.2 2 Vx = −0.6 Vx = −3 V

Example 2.75

For the network shown in Fig. 2.110, find voltages V1 and V2. V1

5Ω

0.5 V1 +−

V2 10 Ω

20 Ω 2A

4A

2Ω 20 V

40 V

Fig. 2.110 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V 20 V1 − 0 5 V1 V2 2= 1 + 20 5 1 ⎛ 1 1 0 5⎞ V2 = 3 ⎜⎝ + − ⎟ V1 20 5 5 ⎠ 5 0.15 15 V1 0.2 V2 = 3

…(i)

2.5 Node Analysis 2.57

Applying KCL at Node 2, 0 V1 − V1 V2 V2 − 40 + + =4 5 2 10 40 ⎛ 0 5 1⎞ ⎛1 1 1 ⎞ − ⎟ V1 V2 = 4 + ⎜⎝ ⎠ ⎝ ⎠ 5 5 5 2 10 10 −0 0.1V 1 V1 + 0 8 V2 = 8 V2

…(ii) Solving Eqs (i) and (ii), V1 = 40 V V2 = 15 V

Example 2.76

Determine the voltages V1 and V2 in the network of Fig. 2.111. 0.5 Ω

2V

V1

1Ω

0.25 Ω

V2

V1

1Ω

Fig. 2.111 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V1 − 2 V1 V1 V2 + + =0 0.5 0.25 1 1 2 ⎛ 1 ⎞ + + 1⎟ V1 V2 = ⎜⎝ ⎠ 0.5 0.25 05 7 V7 V2 = 4

…(i)

Applying KCL at Node 2, V2 V1 V 2 + + V1 = 0 1 1 2 V2 0 V2 = 0 From Eq. (i), V1 =

4 V 7

…(ii)

2.58 Network Analysis and Synthesis

Example 2.77

In the network of Fig. 2.112, find the node voltages V1, V2 and V3. 1Ω

V1

1Ω

V2

V3

2V

− Vx + 2A

2Ω

2Ω

2Ω

+ −

4 Vx

Fig. 2.112 Solution From Fig. 2.112, V2 − V1

Vx

…(i)

Assume that the currents are moving away from the nodes. Applying KCL at Node 1,

⎛1 ⎞ ⎜⎝ + 1⎠ V1 2

2=

V1 V1 − V2 + 2 1

2

2

1 5 V1 V2 = 2

…(ii)

Applying KCL at Node 2, V2 V1 V2 V2 − V3 + + =0 1 2 1 −V V1

⎛ 1 ⎞ 1 1 V2 V3 ⎝ 2 ⎠

0

−V V1 2 5 V2 − V3 = 0

…(iii)

At Node 3, 4 Vx

2

4(V2 V1 )

2

V3 V3

4 V1 4 V2 + V3 = 2 Solving Eqs (ii), (iii) and (iv), V1 = −1 33 V V2 = −4 V V3 = −8 67 V

…(iv)

2.5 Node Analysis 2.59

Example 2.78

For the network shown in Fig. 2.113, find the node voltages V1 and V2. 3A

0.5 Ω

V1

3V2

1A

Ix

1Ω

V2

0.25 Ω

2A

2 Ix

Fig. 2.113 Solution From Fig. 2.113, Ix =

V1 V2 05

…(i)

Assume that the currents are moving away from the nodes. Applying KCL at Node 1, 3 V2 1 =

V1 V1 − V2 + +3 1 0.5

1 ⎞ 1 ⎞ ⎛ ⎛ V2 = −2 ⎜⎝1 + ⎟⎠ V1 ⎝ 3 0.5 5 0.5 ⎠ 3 V1 5 V2 = −2

…(ii)

Applying KCL at Node 2, 3 2= 5=

V2 V1 V2 + + 2 Ix 0.5 0.25 V2 V1 V2 ⎛ V −V ⎞ + + 2⎜ 1 2 ⎟ ⎝ 05 ⎠ 0.5 0.25

2 ⎞ 1 2 ⎞ ⎛ 1 ⎛ 1 + + − ⎜⎝ − ⎟⎠ V1 + ⎝ ⎟ V2 = 5 0.5 0.5 0.5 0.25 0 5 ⎠ 2 V1 + 2 V2 = 5 Solving Eqs (ii) and (iii), V1 = 1 31 V V2 = 1.19 V

…(iii)

2.60 Network Analysis and Synthesis

Example 2.79

Find voltages V1 and V2 in the network shown in Fig. 2.114. V1 V1

I1

2Ω

V2

1Ω

5A

1Ω

2I1

Fig. 2.114 Solution From Fig. 2.114, V1 V2 2 Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V V −V 5 = 1 + 1 2 + V1 1 2 1 ⎛ 1 ⎞ V2 = 5 ⎜⎝1 + + 1⎟⎠ V1 2 2 2 5 V1 0 5 V2 = 5 Applying KCL at Node 2, V2 V1 V2 + = 2 I1 V1 1 1 ⎛V V ⎞ V 2 V1 + V2 = 2 ⎜ 1 2 ⎟ + V1 ⎝ 2 ⎠ 3 V1 3 V2 V1 V2 I1 =

…(i)

…(ii)

…(iii)

Solving Eqs (ii) and (iii), V1 = 2 5 V V2 = 2 5 V

Example 2.80

Find the power supplied by the 10 V source in the network shown in Fig. 2.115. 2Ω

V1

10 V

4Ω

V2

+ V3 − 10 A

2Ω

Fig. 2.115

4V3

1Ω

2.5 Node Analysis 2.61

Solution Assume that the currents are moving away from the nodes. From Fig. 2.115, V3 V1 + 10 − V2

…(i)

Applying KCL at Node 1, 10 +

V1 V1 + 10 − V2 V1 − V2 + + =0 2 4 2

10 ⎛ 1 1 1⎞ ⎛ 1 1⎞ ⎜⎝ + + ⎟⎠ V1 ⎝ + ⎠ V2 = −10 − 2 4 2 4 2 4 1 25

…(ii)

0 75 V2 = −12 5

Applying KCL at Node 2, V2 10 V1 V2 V1 V2 + + = 4 V3 4 2 1 V2 10 V1 V2 V1 V2 + + = 4 (V1 10 V2 ) 4 2 1 10 ⎛ 1 1 ⎞ ⎛1 1 ⎞ + 40 ⎜⎝ − − − 4⎟⎠ V1 + ⎝ + + 1 + 4⎟⎠ V2 = 4 2 4 2 4 −4.75 75 V1 + 5 75 V2 = 42.5

…(iii)

Solving Eqs (ii) and (iii), V1 = −11.03 V V2 = −1 72 V V 10 V2 −11 11.03 + 10 − ( −1.72) I10 V = 1 = = 0.173 A 4 4 Power supplied by the 10 V source = 10 × 0.173 = 1.73 W

Example 2.81

For the network shown in Fig. 2.116, find voltages V1 and V2. 0.03 Vx

V1

20 Ω + Vx −

0.4 A

40 Ω

V2

100 Ω

V3

Iy 40 Ω

+ −

80 Iy

Fig. 2.116 Solution From Fig. 2.116, Vx

V1 − V 2

V Iy = 2 40

…(i) …(ii)

2.62 Network Analysis and Synthesis Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V V −V 0.4 = 1 + 1 2 + 0.03 Vx 100 20 V1 V1 − V2 0.4 = + + 0.03 ( 1 100 20 1 ⎛ 1 ⎞ ⎛ 1 ⎞ + + 0.03 03⎟ V1 + 0 03 2 0.4 ⎜⎝ ⎠ ⎝ ⎠ 100 20 20 0.09 09 V1 0 08 V2 = 0.4

2)

…(iii)

Applying KCL at Node 2, V2 V1 V2 V2 − V3 + + =0 20 40 40 1 1 1⎞ 1 ⎛ 1 − V1 V2 − V3 = 0 ⎝ 20 40 40 ⎠ 20 40 − 0.05 V1 0 0..1 V2 − 0.025 V3 = 0

…(iv)

For Node 3, V3 2 V2 V3

⎛V ⎞ 80 I y = 80 ⎜ 2 ⎟ = 2 V2 ⎝ 40 ⎠ …(v)

0

Solving Eqs (iii), (iv) and (v), V1 = 40 V V2 = 40 V V3 = 80 V

Example 2.82

Find voltages Va , Vb and Vc in the network shown in Fig. 2.117.

Va

4A

2V

2Ω

2Ω

I1

Vb

3Ω

1Ω

Vc

2I1

Fig. 2.117 Solution From Fig. 2.117, Va Vc 2 Assume that the currents are moving away from the nodes. I1 =

5Ω

2.6 Supernode Analysis 2.63

Applying KCL at Node a, 4=

Va Va − Vc Va − 2 − Vb + + 1 2 2

1 1 ⎛ 1 1⎞ ⎜⎝1 + + ⎟⎠ Va − Vb − Vc = 5 2 2 2 2 2 Va

0.5 Vb − 0 5 Vc = 5

…(i)

Applying KCL at Node b, Vb Vb

2 Va Vb Vc + = 2 I1 2 3 2 Va Vb Vc ⎛ V Vc ⎞ + = 2⎜ a ⎝ 2 ⎟⎠ 2 3

2 Va Vb Vc + = Va Vc 2 3 ⎛ 1 ⎞ ⎛ 1 1⎞ ⎛ 1⎞ Vb + 1 − ⎟ Vc = −1 ⎜⎝ − − 1⎟⎠ Va ⎝ ⎠ ⎝ 3⎠ 2 2 3 −1.5 Va + 0 83 Vb + 0.67 Vc = −1 Vb

…(ii)

Applying KCL at Node c, Vc Vb Vc + = I1 3 5 Vc Vb Vc Va − Vc + = 3 5 2 1 1 ⎛ 1 1 1⎞ − Va − Vb + + + ⎟ Vc = 0 ⎝ 3 5 2⎠ 2 3 −0.55 Va − 0 33 Vb + 1.033 Vc = 0

…(iii)

Solving Eqs (i), (ii), and (iii), Va = 4.303 V V b = 3 88 V V c = 3 33 V

2.6

SUPERNODE ANALYSIS

Nodes that are connected to each other by voltage sources, but not to the reference node by a path of voltage sources, form a supernode. A supernode requires one node voltage equation, that is, a KCL equation. The remaining node voltage equations are KVL equations.

2.64 Network Analysis and Synthesis

Example 2.83

Determine the current in the 5 W resistor for the network shown in Fig. 2.118. 2Ω

V1

V3

5Ω

3Ω

10 A

20 V

V2

1Ω

2Ω 10 V

Fig. 2.118 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, 10 =

V1 V1 − V2 + 3 2

1 ⎛ 1 1⎞ V2 = 10 ⎜⎝ + ⎟⎠ V1 3 2 2 0.83 83 V1 0 5 2 10 …(i) Nodes 2 and 3 will form a supernode. Writing voltage equation for the supernode, V2

V3 = 20

…(ii)

Applying KCL at the supernode, V2 V1 V2 V3 − 10 V3 + + + =0 2 1 5 2 1 ⎛1 ⎞ ⎛ 1 1⎞ − V1 1 V2 + + ⎟ V3 = 2 ⎝2 ⎠ ⎝ 5 2⎠ 2 −0 0.5 5 V1 1 5 V2 + 0 7 V3 = 2 …(iii) Solving Eqs (i), (ii) and (iii), V1 = 19.04 V V2 = 11.6 V V3 = −8 4 V

I5 Ω =

V3 − 10 −8 4 − 10 = = −3.68 A 5 5

2.6 Supernode Analysis 2.65

Example 2.84

Find the power delivered by the 5 A current source in the network shown in Fig. 2.119. 10 V

V1

V2

3Ω 1Ω 2A

5A

V3

5Ω

2Ω

Fig. 2.119 Solution Assume that the currents are moving away from the nodes. Nodes 1 and 2 will form a supernode. Writing voltage equation for the supernode, V1 V2 = 10

…(i)

Applying KCL at the supernode, V1 V3 V2 V2 − V3 + + =5 3 5 1 1 ⎛1 ⎞ ⎛1 ⎞ V1 + + 1 V2 − + 1 V3 3 ⎝5 ⎠ ⎝3 ⎠ 3 2+

0 33 V1 +1 1 2 V2 − 1.33 V3

3

…(ii)

V3 V1 V3 V2 V3 + + =0 3 1 2 1 1⎞ ⎛1 − V1 V2 + + 1 + ⎟ V3 = 0 ⎝3 3 2⎠ −0 0.33V 33 V1 V2 1.83 8 V3 = 0

…(iii)

Applying KCL at Node 3,

Solving Eqs (i), (ii) and (iii), V1 = 13.72 V V2 = 3 72 V V3 = 4 51 V Power delivered by the 5 A source = 5 V2 = 5 × 3.72 = 18.6 W

2.66 Network Analysis and Synthesis

Example 2.85

In the network of Fig. 2.120, find the node voltages V1, V2 and V3. 0.5 Ω 0.33 Ω

V1

V2

5V

V3

0.2 Ω

4A

1Ω

Fig. 2.120 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V V V V 4= 1 2 + 1 3 0.33 0. 5 1 ⎞ 1 1 ⎛ 1 + V2 − V3 = 4 ⎜⎝ ⎟ V1 0.33 0.5 ⎠ 0.33 0.5 5.003 3 V1 3 03 V2 − 2 V3 4

…(i)

Nodes 2 and 3 will form a supernode. Writing voltage equation for the supernode, V3 V2 = 5

…(ii)

V2 V1 V2 V3 V3 V1 + + + =0 0.33 0.2 1 05 1 1 ⎞ 1 ⎞ 1 ⎞ ⎛ ⎛ 1 ⎛ − + V3 = 0 ⎜⎝ − ⎟ V1 + ⎝ ⎟ V2 + ⎝1 + 0.33 0.5 ⎠ 0 33 0.2 ⎠ 0.5 ⎠ −5 5.03 V1 8.03 V2 + 3 V3 0

…(iii)

Applying KCL at the supernode,

Solving Eqs (i), (ii) and (iii), V1 = 2 62 V V2 = −0 17 V V3 = 4 83 V

EXAMPLES WITH DEPENDENT SOURCES Example 2.86

For the network shown in Fig. 2.121, determine the voltage Vx. 7Ω 6V

V1

8V

11 Ω

V2 +

3Ω

4Ω

10 Ω

2A

5V

+ 2 Vx −

Vx −

V3

12 A

Fig. 2.121

2.6 Supernode Analysis 2.67

Solution From Fig. 2.121, V2 − V3

Vx

Assume that the currents are moving away from the nodes. Node 1 and 2 will form a supernode. Writing voltage equations for the supernode, V1 V2 = 6

…(i)

Applying KCL at the supernode, V1 + 5 V2 − 2Vx V2 8 V3 V2 V3 + + + 4 10 7 11 V1 + 5 V2 − 2(V2 V3 ) V2 − 8 − V3 V2 − V3 2= + + + 4 10 7 11 2=

1 ⎛ 1 V1 ⎝ 10 4

1 1⎞ 5 8 ⎛1 1 1 ⎞ V2 + − − ⎟ V3 = 2 − + ⎝ 5 7 11⎠ 7 11⎠ 4 7 0 25 V1 + 0.133V 133 2 − 0 033 V3 = 1 89

1 5

…(ii)

Applying KCL at Node 3, V3 V2 V3 8 V2 + + 12 = 0 11 7 8 ⎛ 1 1⎞ ⎛ 1 1⎞ V3 = −12 − ⎜⎝ − − ⎟⎠ V2 ⎝ ⎠ 11 7 11 7 7 −0.233 33 V2 + 0 233 V3 = −13.14

…(iii)

Solving Eqs (i), (ii) and (iii), V1 = 1 8 V V2 = −4.2 V V3 = −60.6 V

Example 2.87

Find the node voltages in the network shown in Fig. 2.122. 6Ω

V1

6A

5Ω

ix

V2 5ix

10 Ω

−+

20 Ω

V3

15 Ω

2Ω

V4

40 V

Fig. 2.122 Solution From Fig. 2.122, Ix =

V2 V1 5

…(i)

For Node 4, V4 = 40

…(ii)

2.68 Network Analysis and Synthesis Applying KCL at Node 1, V1 V1 − V2 V1 − V4 + + =0 10 5 6 V V − V V − 40 6+ 1 + 1 2 + 1 =0 10 5 6 1 40 ⎛ 1 1 1⎞ V2 = −6 ⎜⎝ + + ⎟⎠ V1 10 5 6 5 6 7 1 2 V1 V2 = 15 5 3 Nodes 2 and 3 will form a supernode, Writing voltage equation for the supernode, 6+

V3 V2 = 5 I x V1 2V2 V3

…(iii)

⎛V V ⎞ 5 ⎜ 2 1 ⎟ = V2 V1 ⎝ 5 ⎠ …(iv)

0

Applying KCL to the supernode, V2 V1 V2 V3 V3 V4 + + + 5 20 15 2 V2 V1 V2 V3 V3 − 40 + + + 5 20 15 2 1 1 1 1 1 ⎛ ⎞ ⎛ ⎞ − V1 + ⎜ + ⎟ V2 + + V3 ⎝ 5 20 ⎠ ⎝ 15 2 ⎟⎠ 5 1 1 17 − V1 + V2 + V3 5 4 30 Solving Eqs (iii), (iv) and (v), V1 V2

=0 =0 = 20 = 20

…(v)

= 10 V = 20 V

V3 = 30 V V4 = 40 V

Example 2.88

Find the node voltages in the network shown in Fig. 2.123. V2 Vx −

V1

+ 14 A

0.5 Ω

2Ω

0.5 Vx

12 V

1Ω

V4

Fig. 2.123

+

− 2.5 Ω Vy +

− 0.5 Vy

V3

Exercises 2.69

Solution Selecting the central node as reference node, V1 = −12 V

…(i)

Applying KCL at Node 2,

−

V2 V1 V2 V3 + = 14 05 2 1 ⎛ 1 1⎞ V2 − V3 = 14 ⎝ 0.5 2 ⎠ 2

1 V1 0.55 −2 2 V1 2 V2

0 5 V3 = 14

…(ii)

Nodes 3 and 4 will form a supernode, Writing voltage equation for the supernode, V3 V4 = 0.2Vy

0 2 (V4 V1 )

0.2V1 V3 − 1.2V4 = 0

…(iii)

Applying KCL to the supernode, V3 V2 V V −V − 0 5Vx + 4 + 4 1 = 0 2 1 2.5 V3 V2 V4 V1 − 0.5( 5 (V2 V1 ) + V4 + =0 2 25 1 ⎞ 1 1 ⎞ ⎛ ⎛1 ⎞ ⎛ 1 V4 = 0 ⎜⎝ 0 5 − ⎟ V1 ⎝ + 0 5⎠ V2 + V3 + ⎝1+ 2 5⎠ 2 2 2 5⎠ 0 1 V1 V2 + 0 5 V3 +1 1 4 V3 = 0

…(iv)

Solving Eqs (i), (ii), (iii) and (iv), V1 = −12 V V2 = −4 V V3 = 0 V4 = −2 V

Exercises KIRCHHOFF’S LAWS 2.1

Find Ix and Vx in the network shown in Fig. 2.124. 4A 10 V

a

2.2

Find V1 and V2 in the network shown in Fig. 2.125.

2A

b

9V

+ 1A −

5Ω

Vx

− V2 +

12 V

c

d

3A

18 V

V1

10 V

3V

Fig. 2.125

Ix

[

Fig. 2.124 [

, −15 V ]

,,5 V ]

2.70 Network Analysis and Synthesis 2.3

Find the values of unknown currents in the network shown in Fig. 2.126.

2.7

In the network shown in Fig. 2.130, find the voltage between points A and B. 5Ω

5A

7A

I1

10 Ω

1 A I2

A

15 Ω

5V 4Ω

2.4

2 A, I 3

4 A, I 4

[

40 A

Find the current through the 10 Ω resistor in the network shown in Fig. 2.131. 6Ω

0.1 Ω

2 V 10 Ω

Y

0.1 Ω

4Ω

70 A

20 V

[ .6 . 8 A]

[ ] for the network as shown in

Find I and VAB Fig. 2.128.

15 Ω

Fig. 2.131

Fig. 2.127 2.5

5Ω

0.05 Ω

5V

50 A

]

MESH ANALYSIS 2.8

0.05 Ω

0.05 Ω

20 V

10 A ]

Find the current in the branch XY as shown in Fig. 2.127. 160 A X

10 Ω

Fig. 2.130

Fig. 2.126 , I2

6Ω

30 V

I4

6A 8A

[

B

12 A

I3 10 A

4Ω

2.9

Find the current through the 20 Ω resistor in the network shown in Fig. 2.132. 15 Ω

60 V

A 6A

1A

2Ω

5Ω

I

3Ω

12 V

B

+ − 4Ω

20 Ω

20 V

15 Ω

10 Ω

5Ω

40 V

5Ω

1A

Fig. 2.128 2.6

[ , 19 V ] In the network shown in Fig. 2.129, find the voltage between points A and B. 5Ω

3Ω

A

Fig. 2.132 [ .4 . 6 A] 2.10 Find the current through the 10 Ω resistor in the network shown in Fig. 2.133. 10 Ω

2Ω

2A

5Ω

6Ω

30 V

6V 3Ω

B

30 Ω

2Ω

10 A

B

5Ω

20 Ω

100 V

Fig. 2.129 [

]

Fig. 2.133 [ .3 . 7 A]

Exercises 2.71

2.11 Find the current through the 1 Ω resistor in the network shown in Fig. 2.134.

10 Ω

i1

20 Ω

1Ω

30 Ω A

2Ω

3Ω

2Ω

1A

80 V

10 V

+

40 Ω 30 Ω

Fig. 2.134 [ .9 . 5 A] [ .4 V,

V,

V,

V, 9977.3399 V]

2.15 Find currents I1, I2, and I3 in the network shown in Fig. 2.138.

6V

2Ω

1Ω

4Ω

2A

3Ω

I1

Fig. 2.135

2Ω

Fig. 2.138

2.13 Find currents Ix and Iy in the network shown in Fig. 2.136.

[

I3

2Ω

Ix

+−

20 Ω

4Ix

I1

+ − 10IY

2A

I2

25 Ω

Iy

2Ω

5V

, 11 A, 17 A]

2.16 Find currents Ix in the network shown in Fig. 2.139.

2IY 2Ix

1Ω

I3

[ .3 . 3 A]

+−

2Ω

I2

+ Vx −

1 V 9 x

15 A

5Ω

V3

Fig. 2.137

2.12 Find the current through the 4 Ω resistor in the network shown in Fig. 2.135.

5A

−

2Ω

10 Ω

1.5 Ix

5Ω

5A

Ix

Fig. 2.139

Fig. 2.136

[ .3 . 3 A]

[ .5 A, 0.1 A] 2.14 In the network shown in Fig. 2.137, find V3 if element A is a (i) short circuit (ii) 5 Ω resistor (iii) 20 V independent voltage source, positive reference on the right (iv) dependent voltage source of 1.5 i1, with positive reference on the right (v) dependent current source 5 i1, arrow directed to the right

2.17 Find currents I1 in the network shown in Fig. 2.140. 5 Ω I1

30 V

19 V

2Ω

4A

4Ω

1.5I1

25 V

Fig. 2.140 [−12 A]

2.72 Network Analysis and Synthesis

NODE ANALYSIS

[ .62 A ]

2.18 Find the current Ix in the network shown in Fig. 2.141. Ix

5Ω

2.22 Find the current through the branch ab in the network shown in Fig. 2.145.

20 Ω

10 Ω

1Ω a 2Ω

24 V

20 Ω

2A

36 V

10 Ω

10 V 2Ω b

Fig. 2.141

1Ω

[ .93 A ] 2.19 Find VA and VB in the network shown in Fig. 2.142. 1Ω

VA

[ .038 A ] 2.23 Find the current through the 4 Ω resistor in the network shown in Fig. 2.146.

2Ω

2Ω

2A

2Ω

VB

Fig. 2.145

4Ω 1A

10 V

2V

1V

2Ω

Fig. 2.142 [ .88 V, 3.25 V ] 2.20 Find the current through the 6 Ω resistor in the network shown in Fig. 2.143.

2Ω

2Ω 2Ω

2Ω

5A

6V

10 V 2 Ω

Fig. 2.146 10 Ω

4A

5Ω

3A

[ .34 A ]

6Ω

2.24 Find the current through the 4 Ω resistor in the network shown in Fig. 2.147. Fig. 2.143

2A

[ . 04 A ] 2.21 Calculate the current through the 10 Ω resistor in the network shown in Fig. 2.144.

2Ω

2Ω 2V

6V

10 Ω 2Ω

4Ω 25 V

+ −

2Ω

4Ω

3A

− +

7Ω

3Ω

12 V

Fig. 2.147 [1 A]

2Ω

3Ω

Fig. 2.144

2.25 Find the voltage Vx in the network shown in Fig. 2.148.

Objective–Type Questions 2.73

1Ω 1Ω

4I1

+ Vx −

2Ω

2.28 Determine V1 in the network shown in Fig. 2.151. 50 Ω

I1

20 Ω

8V

4V

Fig. 2.148

5A

[ ..31 V] 2.26 Find the currents Vx in the network shown in Fig. 2.149.

20 Ω

+ V1 −

+ 0.4V1 −

0.01V1

Fig. 2.151

3V1

[140 V] 2.29 Find the voltage Vy in the network shown in Fig. 2.152.

5Ω + Vx −

0.5Vy

4Vx 7A

+−

6Ω

4Ω

15 Ω

+

3Ω

4A

20 Ω Vy

6A

Fig. 2.149

−

[ .0 . 9 V] 2.27 Find the voltage Vx in the network shown in Fig. 2.150. 1 Ω 3 1Ω

Fig. 2.152 [−10 V] 2.30 Find the voltage V2 in the network shown in Fig. 2.153. 0.8V2 +−

0.5 Ω

Vx

2Ω

1V

3I

0.5 Ω 5A

5Ω

+ V2 −

8A 2.5 Ω

Fig. 2.150 Fig. 2.153

[ .2 . V]

[

Objective–Type Questions 2.1

Two electrical sub-networks N1 and N2 are connected through three resistors as shown in Fig. 2.154. The voltages across the 5 Ω resistor and 1 Ω resistor are given to be 10 V and 5 V respectively. Then the voltage across the 15 Ω resistor is (a) −105 V (b) 105 V (c) −15 V (d) 15 V

+

5Ω − 10 V

N1

N2

15 Ω 1Ω +

5V

−

Fig. 2.154

.9 V]

2.74 Network Analysis and Synthesis 2.2

2.3

1Ω

The nodal method of circuit analysis is based on (a) KVL and Ohm’s law (b) KCL and Ohm’s law (c) KCL and KVL (d) KCL, KVL and Ohm’s law The voltage across terminals a and b in Fig. 2.155 is 2Ω

a

4Ω

(a) 6 A (c) 12 A

1Ω

2Ω

+ Vx −

b

+ 8V −

(a) 0.5 V (b) 3 V (c) 3.5 V (d) 4 V The voltage V0 in Fig. 2.156 is

+ Vy −

+ 2V −

+ Vz −

+ 1V −

(a) −6, 3, −3 (c) 6, 3, 3

16 V 12 Ω

10 Ω

2.5 A none of these

+ 2V −

Fig. 2.159

2Ω

8A

(b) (d)

The value of Vx,Vy and Vz in Fig. 2.159 shown are

3A

Fig. 2.155

2.4

2Ω V

Fig. 2.158

2.7 + 1V −

2Ω

IS

6Ω

+ V0 −

2.8

(b) (d)

−6, −3, 1 6, 1, 3

The circuit shown in Fig. 2.160 is equivalent to a load of 2Ω

I

Fig. 2.156

2.5

5Ω

V1 5

2.9 Fig. 2.157 (a) (b) (c) (d) 2.6

delivers 80 W absorbs 80 W delivers 40 W absorbs 40 W

If V = 4 in Fig. 2.158, the value of Is is given by

2I

Fig. 2.160

5Ω

V1 = 20 V

+ −

4Ω

(a) 48 V (b) 24 V (c) 36 V (d) 28 V The dependent current source shown in Fig. 2.157. (a)

4 Ω 3

(b)

8 Ω 3

(b)

4Ω

(d)

2Ω

In the network shown in Fig. 2.161, the effective resistance faced by the voltage source is i 4 i

4Ω

V

Fig. 2.161

Answers to Objective-Type Questions 2.75

2.10

(a)

4Ω

(b)

3Ω

(c)

2Ω

(d)

1Ω

A network contains only an independent current source and resistors. If the values of

all resistors are doubled, the value of the node voltages will (a) become half (b) remain unchanged (c) become double (d) none of these

Answers to Objective-Type Questions 2.1. (a)

2.2. (b)

2.3. (c)

2.4. (d)

2.5. (a)

2.6. (d)

2.7. (a)

2.8. (a)

2.9. (d)

2.10. (b)

2.11. (b)

2.12. (b)

2.13. (c)

2.14. (b)

3 3.1

Network Theorems (Application to dc Networks) INTRODUCTION

In Chapter 2, we have studied elementary network theorems like Kirchhoff’s laws, mesh analysis and node analysis. There are some other methods also to analyse circuits. In this chapter, we will study superposition theorem, Thevenin’s theorem, Norton’s theorem, maximum power transfer theorem, reciprocity theorem, Millman’s theorem, Tellegen’s theorem, substitution theorem and compensation theorem. We can find currents and voltages in various parts of the circuits with these methods.

3.2

SUPERPOSITION THEOREM

It states that ‘in a linear network containing more than one independent source and dependent source, the resultant current in any element is the algebraic sum of the currents that would be produced by each independent source acting alone, all the other independent sources being represented meanwhile by their respective internal resistances.’ The independent voltage sources are represented by their internal resistances if given or simply with zero resistances, i.e., short circuits if internal resistances are not mentioned. The independent current sources are represented by infinite resistances, i.e., open circuits. The dependent sources are not sources but dissipative components—hence they are active at all times. A dependent source has zero value only when its control voltage or current is zero. A linear network is one whose parameters are constant, i.e., they do not change with voltage and current. Explanation resistor R4.

Consider the network shown in Fig. 3.1. Suppose we have to find current I4 through R1

V

R3

R2

R4

I

Fig. 3.1 Network to illustrate superposition theorem

3.2 Network Analysis and Synthesis R1 R3 The current flowing through resistor R4 due to constant voltage source V is found to be say IÄ4 (with I4′ proper direction), representing constant current source R4 R2 V with infinite resistance, i.e., open circuit. The current flowing through resistor R4 due to constant current source I is found to be say I4″ (with proper direction), representing the constant voltage Fig. 3.2 When voltage source V is acting alone source with zero resistance or short circuit. R1 R3 The resultant current I4 through resistor R4 is found by superposition theorem. I4′′ I 4 I 4′ + I 4″ R2

I

R4

Steps to be followed in Superposition Theorem 1. Find the current through the resistance when only one independent source is acting, replacing all Fig. 3.3 When current source I is acting alone other independent sources by respective internal resistances. 2. Find the current through the resistance for each of the independent sources. 3. Find the resultant current through the resistance by the superposition theorem considering magnitude and direction of each current.

Example 3.1

Find the current through the 2 W resistor in Fig. 3.4. 20 V

5Ω

2Ω

10 Ω

40 V

10 V

Fig. 3.4 Solution Step I

When the 40 V source is acting alone (Fig. 3.5) I

5Ω

I′

2Ω

10 Ω

40 V

Fig. 3.5 By series parallel reduction technique (Fig. 3.6), 40 I= =6A 5 + 1.67 From Fig. 3.5, by current-division rule, 10 I′ = 6× =5A(→) 10 + 2

I

5Ω

1.67 Ω

40 V

Fig. 3.6

3.2 Superposition Theorem 3.3

Step II

When the 20 V source is acting alone (Fig. 3.7) I

20 V

5Ω

2Ω

I ′′

10 Ω

Fig. 3.7 By series–parallel reduction technique (Fig. 3.8), I=

5Ω

I

20 V

20 =3A 5 + 1.67

1.67 Ω

From Fig. 3.7, by current-division rule, I ′′ = 3 × Step III

10 = 2 5 A( ← ) = − 2 5 A ( → ) 10 + 2

Fig. 3.8

When the 10 V source is acting alone (Fig. 3.9) 5Ω

2Ω

10 Ω

10 V

Fig. 3.9 I ′′′

By series–parallel reduction technique (Fig. 3.10), I ′′′ = Step IV

10 = 1 88 A( → ) 3 33 + 2

2Ω 10 V

3.33 Ω

I ′′′

By superposition theorem, I

Example 3.2

I ′ + I ′′ I ′′′ = 5 − 2.5 + 1.88 = 4.38 A ( → )

Fig. 3.10

Find the current through the 1 W resistor in Fig. 3.11. 4Ω

100 V

50 V

40 V

1Ω

2Ω

Fig. 3.11

3.4 Network Analysis and Synthesis Solution Step I When the 100 V source is acting alone (Fig. 3.12) I

4Ω I′

1Ω

100 V

2Ω

Fig. 3.12 I

By series–parallel reduction technique (Fig. 3.13), I=

100 = 21.41 A 4 + 0.67

4Ω

0.67 Ω

100 V

From Fig. 3.12, by current-division rule, I ′ = 21.41 × Step II

2 = 2 +1

Fig. 3.13

(↓)

When the 50 V source is acting alone (Fig. 3.14) 4Ω 50 V

2Ω

1Ω

Fig. 3.14 By series–parallel reduction technique (Fig. 3.15), I ′′ = Step III

50 = 1 + 1.33

I ′′

.46 A( ↑ )= − 21.46 A( ↓ )

50 V

1.33 Ω

1Ω

When the 40 V source is acting alone (Fig. 3.16)

Fig. 3.15 4Ω

I I ′′′

1Ω

Fig. 3.16

40 V

2Ω

3.2 Superposition Theorem 3.5

By series–parallel reduction technique (Fig. 3.17), 40 I= = 14.29 A 0 8+ 2 From Fig. 3.16, by current-division rule, 4 I ′′′ = 14.29 × = 4 +1 Step IV By superposition theorem, I

I 40 V

0.8 Ω

2Ω

(↓ ) Fig. 3.17

I ′ + I ′′ I ′′′ = 14.27 − 21.46 + 11.43 = 4.24 A(↓)

Example 3.3

Find the current through the 8 W resistor in Fig. 3.18. 5Ω

10 Ω

15 Ω

4V

12 Ω

8Ω

6V

Fig. 3.18 Solution Step I When the 4 V source is acting alone (Fig. 3.19) I

5Ω

I1

10 Ω

12 Ω I′

15 Ω

4V

8Ω

Fig. 3.19 By series–parallel reduction technique (Fig. 3.20), I

5Ω

I1

10 Ω

15 Ω

4V

I

4.8 Ω

5Ω

I1

15 Ω

4V

14.8 Ω

(b)

(a)

I=

I

4 = 0.32 A 5 + 7.45

From Fig. 3.20(b), by current-division rule, 15 I1 = 0 32 × = 0.16 A 15 + 14.8 From Fig. 3.19, by current-division rule, 12 I ′ = 0.16 × = (↓) 12 + 8

5Ω

7.45 Ω

4V

(c)

Fig. 3.20

3.6 Network Analysis and Synthesis When the 6 V source is acting alone (Fig. 3.21)

Step II

5Ω

10 Ω

12 Ω

15 Ω

8Ω

6V

Fig. 3.21 By series–parallel reduction technique (Fig. 3.22), 10 Ω

12 Ω

12 Ω

I

I ′′

3.75 Ω

8Ω

13.75 Ω

6V

8Ω

(a)

(b)

I=

12 Ω

6 = 0.35 A 12 + 5 06

From Fig. 3.22(b), by current division rule,

5.06 Ω

13.75 I ′′ = 0 35 × = 13.75 + 8 By superposition theorem,

Step III

I

Example 3.4

I ′ + I ′′ = 0.096 + 0.22 =

(c)

(↓)

Fig. 3.22

4Ω

5Ω

40 V

8A

3Ω

Fig. 3.23 Solution When the 40 V source is acting alone (Fig. 3.24) I

40 V

12 Ω

I′

5Ω

Fig. 3.24

4Ω

3Ω

I

6V

(↓)

Find the current through the 4 W resistor in Fig. 3.23. 12 Ω

Step I

6V

3.2 Superposition Theorem 3.7

By series–parallel reduction technique (Fig. 3.25), I

12 Ω

I′

5Ω

40 V

12 Ω

I

7Ω

2.92 Ω

40 V

(b)

(a)

Fig. 3.25 I=

40 = 2.68 A 12 + 2 92

From Fig. 3.25(a), by current-division rule, I ′ = 2.68 ×

5 = 1 12 A( → ) 5+ 7

− 1 12 A ( ← ) −1

Step II When the 8 A source is acting alone (Fig. 3.26) 12 Ω

4Ω

5Ω

8A

3Ω

Fig. 3.26 By series–parallel reduction technique (Fig. 3.27), 4Ω

I ′′

I ′′

3.53 Ω

3Ω

8A

7.53 Ω

(a)

(b)

Fig. 3.27 I″ = 8× Step III

3 = 2 28 A( ← ) 7 53 + 3

By superposition theorem, I

3Ω

I ′ + I ′′ = −1.12 + 2.28 = 1.16 A ( ← )

8A

3.8 Network Analysis and Synthesis

Example 3.5

Find the current in the 10 W resistor in Fig. 3.28. 2Ω

1Ω 10 Ω

5Ω

4A

10 V

Fig. 3.28 Solution Step I

When the 10 V source is acting alone (Fig. 3.29) 2Ω

1Ω

5Ω

10 Ω 10 V

Fig. 3.29 By series–parallel reduction technique (Fig. 3.30), I

I I ′′

1Ω

1Ω

7Ω

10 Ω 10 V

4.12 Ω

10 V

(a)

(b)

Fig. 3.30

I=

10 = 1.95 A 1 + 4.12

From Fig. 3.30 (a), by current-division rule, I ′ = 1.95 ×

7 = 7 + 10

(↓ )

3.2 Superposition Theorem 3.9

Step II

When the 4 A source is acting alone (Fig. 3.31) 2Ω

I

I ′′

1Ω

10 Ω

5Ω

4A

Fig. 3.31 By series–parallel reduction technique (Fig. 3.32), 2Ω

I

I

0.91 Ω

4A

5Ω

2.91 Ω

4A

(a)

(b)

Fig. 3.32

I = 4×

5 = 2.53 A 2 91 + 5

From Fig. 3.31, by current-division rule, I ′′ = 2.53 × Step III

1 − 1 + 10

(↓)

By superposition theorem, I

Example 3.6

I ′ + I ′′ = 0.8 + 0.23 =

(↓)

Find the current through the 8 W resistor in Fig. 3.33. 8Ω

5A

12 Ω

Fig. 3.33

30 Ω

25 A

5Ω

3.10 Network Analysis and Synthesis Solution Step I

8Ω

I′

When the 5 A source is acting alone (Fig. 3.34) I′ = 5×

12 = 1.2 12 + 8 + 30

(→)

30 Ω

12 Ω

5A

When the 25 A source is acting alone (Fig. 3.35)

Step II

I ′′ = 25 ×

Fig. 3.34

30 = 15 A (→) 30 + 12 + 8

I ′′

8Ω

By superposition theorem,

Step III

I

12 Ω

I ′ + I ′′ = 1.2 +15 + 15 = 16 2 A(→)

30 Ω

Fig. 3.35

Example 3.7

Find the current through the 4 W resistor in Fig. 3.36. 2Ω

5A

6Ω

5Ω

4Ω 6Ω

20 V

Fig. 3.36 Solution Step I

When the 5 A source is acting alone (Fig. 3.37) 2Ω

5A

6Ω

5Ω

4Ω 6Ω

Fig. 3.37 By series–parallel reduction technique (Fig. 3.38),

25 A

3.2 Superposition Theorem 3.11 2Ω 2Ω

I′

6Ω 4Ω

5A

8.73 Ω

5A

2.73 Ω

(a)

4Ω

(b)

Fig. 3.38 I′ = 5× Step II

8 73 = 8 73 + 4

(↓ )

When the 20 V source is acting alone (Fig. 3.39) 2Ω

6Ω 5Ω

I

I ′′

4Ω

6Ω

20 V

Fig. 3.39 By series–parallel reduction technique (Fig. 3.40), I

5Ω

I ′′

I

6Ω

20 V

10 Ω

3.75 Ω

20 V

(b)

(a)

Fig. 3.40 I=

20 = 2.29 A 5 + 3.75

From Fig. 3.40 (a), by current-division rule, I ′′ = 2.29 × Step III

6 = 6 + 10

(↓ )

By superposition theorem I

5Ω

I ′ + I ′′ = 3.43 3 + 0.86 =

(↓ )

3.12 Network Analysis and Synthesis

Example 3.8

Find the current through the 3 W resistor in Fig. 3.41. 2Ω

5Ω

3Ω

10 Ω

4Ω

5A 20 V

Fig. 3.41 Solution Step I

When the 5 A source is acting alone (Fig. 3.42) 2Ω

5Ω

3Ω

10 Ω

4Ω

5A

Fig. 3.42

I′

By series–parallel reduction technique (Fig. 3.43), I′ = 5× Step II

15 = 15 + 2 + 3

(↓ )

2Ω

3Ω

15 Ω

5A

When the 20 V source is acting alone (Fig. 3.44) 2Ω

Fig. 3.43 3Ω

5Ω

I

I ′′

10 Ω

4Ω

20 V

Fig. 3.44 By series–parallel reduction technique (Fig. 3.45), I ′′

I

20 Ω

4Ω

I

20 V

(a)

3.33 Ω

20 V

(b)

Fig. 3.45

3.2 Superposition Theorem 3.13

I=

20 =6A 3 33

From Fig. 3.45(a), by current-division rule, I ′′ = 6 ×

4 = 20 + 4

(↑) = −1 A(↓)

By superposition theorem,

Step III

I

Example 3.9

I ′ + I ′′ = 3.75 75 − 1 =

(↓)

Find the current in the 1 W resistors in Fig. 3.46. 1A

2Ω

3Ω

1Ω

4V

3A

Fig. 3.46 2Ω

Solution Step I

3Ω I′

When the 4 V source is acting alone (Fig. 3.47)

1Ω

4V

4 I′ = = 2 +1

(↓) Fig. 3.47

Step II When the 3 A source is acting alone (Fig. 3.48) By current-division rule, 2 I ′′ = 3 × = 1+ 2 Step III

2Ω

(↓ )

3Ω I ′′

1Ω

When the 1 A source is acting alone (Fig. 3.49) 1A

2Ω

Fig. 3.48 3Ω

1Ω

Fig. 3.49

3A

3.14 Network Analysis and Synthesis Redrawing the network (Fig. 3.50), By current-division rule,

I ′′′

3Ω

2 I ′′′ = 1 × = 2 +1

2Ω

(↓)

1Ω

1A

Step IV By superposition theorem, I

I ′ + I ′′ I ′′′ = 1.33 + 2 + 0.66 = 4 A (↓)

Example 3.10

Fig. 3.50

Find the voltage VAB in Fig. 3.51. +

A

5Ω

6V

VAB 10 V

5A

−

B

Fig. 3.51 Solution Step I When the 6 V source is acting alone (Fig. 3.52) VABB′ = 6 V

+

A

5Ω

6V

VAB′

−

B

+

Fig. 3.52

A

5Ω VAB′′

Step II When the 10 V source is acting alone (Fig. 3.53) Since the resistor of 5 Ω is shorted, the voltage across it is zero.

10 V

VAB ″ = 10 V

−

B

Fig. 3.53 Step III When the 5 A source is acting alone (Fig. 3.54) Due to short circuit in both the parts,

+

VAB ″′ = 0

5Ω VAB′′′

Step IV By superposition theorem, 5A

VAB

VAB′ + VAB ″

A

−

VAB ″′ = 6 + 10 + 0 = 16 V Fig. 3.54

B

3.2 Superposition Theorem 3.15

Example 3.11

Find the current through the 5 W resistor in Fig. 3.55. 5Ω

10 Ω

20 Ω

2A

24 V

36 V

Fig. 3.55 Solution Step I

When the 24 V source is acting alone (Fig. 3.56) 5Ω

10 Ω

20 Ω

24 V

Fig. 3.56

I′

5Ω

By series–parallel reduction technique (Fig. 3.57), I′ =

24 = 2.06 5 + 6.67

(→) = −2.06 A(←)

6.67 Ω

24 V

Step II When the 2 A source is acting alone By series–parallel reduction technique (Fig. 3.58), 5Ω

Fig. 3.57 5Ω

10 Ω

I ′′

20 Ω

2A

6.67 Ω

2A

(a)

(b)

Fig. 3.58 By current-division rule, I″ = 2×

6 67 = 1.14 (←) 5 + 6.67

Step III When the 36 V source is acting alone (Fig. 3.59) By series–parallel reduction technique, 5Ω

10 Ω

I ′′′

20 Ω

10 Ω

I

36 V

(a)

4Ω

36 V

(b)

Fig. 3.59

I

3.16 Network Analysis and Synthesis I=

36 = 2.57 A 10 + 4

By current-division rule, I ′′′ = 2.57 ×

20 = 2 06 A(←) 20 + 5

Step IV By superposition theorem, I ′ + I ′′ I ′′′ = −2.06 + 1.14 + 2.06 06 = 1 14 A (←)

I

Example 13.12

Find the current through the 4 W resistor in Fig. 3.60. 6V

2Ω

5A

4Ω

2A

Fig. 3.60 I′

Solution

I′ = 5×

2 = 2+4

2Ω

5A

Step I When the 5 A source is acting alone (Fig. 3.61) By current-division rule, (↓ )

Fig. 3.61 I ′′

Step II When the 2 A source is acting alone (Fig. 3.62) By current-division rule, I″ = 2×

2 = 2+4

4Ω

2Ω

2A

4Ω

(↓ )

Step III When the 6 V source is acting alone (Fig. 3.63) Applying KVL to the mesh,

Fig. 3.62

−2l ′′′ − 6 − 4l ′′′ = 0

6V

I ′′′ = −1 A (↓) 2Ω

Step IV By superposition theorem,

4Ω I ′′′

I

I ′ + I ′′ I ′′′ = 1.67 + 0.67 − 1 = 1.34

(↓ ) Fig. 3.63

3.2 Superposition Theorem 3.17

EXAMPLES WITH DEPENDENT SOURCES Example 3.13

Find the current through the 6 W resistor in Fig. 3.64. 6Ω

I

8Ω

3I

15 V

10 V

Fig. 3.64 Solution Step I When the 15 V source is acting alone (Fig 3.65) From Fig. 3.65, I ′ = I1

I′

…(i)

4

1

I1 = 3I ′ 2

8Ω

3I ′

15 V I1

Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I2

6Ω

I2

Fig. 3.65

3I1

0

…(ii)

Applying KVL to the outer path of the supermesh, 15 − 6

1

8I 2 = 0

6

1

88II 2 = 15

…(iii)

Solving Eqs (ii) and (iii), I1 = 0.39A I 2 = 1.59A I ′ = I1 = 0.39

(→)

Step II When the 10 V source is acting alone (Fig 3.66) From Fig. 3.66, I ′′ = I1 Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I 2 I1 = 3I ′′ 3I1 41 2 0 Applying KVL to the outer path of the supermesh, −6 1 − 88I 2 + 10 = 0 6 1 88II 2 = 10

I ′′ 6 Ω

8Ω

...(i) 3I ′′ I1

10 V I2

…(ii) Fig. 3.66

…..(iii)

3.18 Network Analysis and Synthesis Solving Eqs (ii) and (iii), I1 = 0 26 A

Step III

I 2 = 1 05 A I ′′ I1 = 0.26 By superposition theorem,

(→)

I = I ′ + I ″ = 0.39 + 0.26 = 0.65

Example 3.14

(→)

Find the current Ix in Fig. 3.67. Ix 5 Ω

1Ω

+ −

30 A

20 V

4Ix

Fig. 3.67 Solution Step I When the 30 A source is acting alone (Fig. 3.68) From Fig. 3.68, I ′x

I1

Ix′ 5 Ω

…(i) 30 A

Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I1

1Ω

I1

I 2 = 30

I2

Fig. 3.68

−5 1 − 1I 2 − 4 ′x = 0 −5 1 − 2 − 4 I1 = 0 91 2 0 Solving Eqs (ii) and (iii),

…(iii)

I1 = 3 A I 2 = −27 A I ′x I1 = 3 A (→)

Step II When the 20 V source is acting alone (Fig. 3.69) Applying KVL to the mesh,

Step III

4Ix′

…(ii)

Applying KVL to the outer path of the supermesh,

20 − 5 x″

+ −

1I x″ − 4 ″x 0 1I I x″ = 2 A( → )

Ix′′ 5 Ω

1Ω

+ −

20 V

By superposition theorem, Ix

I x′ + I ″x

3+ 2

5 Α(→) Fig. 3.69

4Ix′′

3.2 Superposition Theorem 3.19

Example 3.15

Find the current I1. in Fig. 3.70. Vx

+ −

4Vx

I1

10 Ω

2A

2Ω

5V

Fig. 3.70 Solution Step I When the 5 V source is acting alone (Fig 3.71) From the figure, Vx = 5 – 10I1′ Applying KVL to the mesh,

Vx

2Ω

5V

5 – 10I1′ – 4 (5 – 10I1′) – 2I1′ = 0

Fig. 3.71

5 – 10I1′ – 20 + 40I1′ – 2I1′ = 0 15 = 0. 0 54 28

(↑ ) Vx

Step II When the 2 A source is acting alone (Fig 3.72) From Fig . 3.72, 10 I1′

Vx

I1′ = 2

+−

4Vx

I1′

…(i)

10 Ω

Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I2

4Vx

10 Ω

5 – 10I1′ – 4Vx – 2I1′ = 0

I1′ =

+ −

I1′

2A I1′

…(ii)

2Ω

I2

Fig. 3.72

Applying KVL to the outer path of the supermesh, −10 I1′ − 4 x − 2 I 2 = 0 −10 I1′ − 4( −10 10 1′ ) − 2 I 2 = 0 30 I1′

2

2

0

…(iii)

Solving Eqs (ii) and (iii), I1 = 0.14 (↑) I 2 = 2.14 A Step III

By superposition theorem, I1

I1′ + I1″ = 0.54 0.14 = 0.68

(↑ )

3.20 Network Analysis and Synthesis

Example 3.16

Determine the current through the 10 W resistor in Fig. 3.73. 10 Ω

− +

10Vx

+

5Ω

100 V

Vx

10 A

−

Fig. 3.73 Solution Step I When the 100 V source is acting alone (Fig. 3.74) From the figure, Vx = 5I′ Applying KVL to the mesh,

10 Ω

− +

I′

+ 5Ω

100 V

100 – 10I′ + 10Vx – 5I′ = 0 100 – 10I′ + 10(5I′) – 5I′ = 0

10Vx

Vx

−

I ′ = −2.86

(→)

Fig. 3.74

Step II When the 10 A source is acting alone (Fig. 3.75) From Fig. 3.75, Vx ( I1 − I 2 ) …(i) Applying KVL to Mesh 1, −10 I1 +10 10Vx − 5( 1 − 2 ) = 0 −10 I1 +10 10 0{5( 1 − 2 )} − 5( I1 − I 2 ) = 0 35 I1 45 I 2 = 0

10 Ω

I ′′

− +

10Vx

+

5Ω

Vx

I1

10 A I2

−

…(ii)

Fig. 3.75

For Mesh 2, I 2 = −10

…(iii)

Solving Eqs (ii) and (iii), I1 = −12.86 A I 2 = −10 A I ′′ = I1 = −12.86

(→) Step III By superposition theorem, I I ′ + I ′′ = −2.86 − 12 12.86 = −15.72 (→)

Example 3.17

Find the current I in the network of Fig. 3.76. 3Ω

17 V

I

2Ω

+ Vx −

4Ω

1A

Fig. 3.76

+ −

5Vx

3.2 Superposition Theorem 3.21

Solution Step I When the 17 V source is acting alone (Fig. 3.77) From the figure, Vx = –2I′ Applying KVL to the mesh, –2I′ – 17 – 3I′ – 5Vx = 0 –2I′ – 17 – 3I′ – 5(–2I′) = 0 I′ =3.4 A (→) Step II When the 1 A source is acting alone (Fig. 3.78) From Fig. 3.78, Vx 2 I1 …(i) Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I 2 I1 = 1 …(ii) Applying KVL to the outer path of the supermesh,

I′

2Ω

+ Vx −

2Ω

4Ω I1

1A

I2

Fig. 3.78

Find the voltage V1 in Fig. 3.79.

4I

I

4Ω

+

+ −

V1

5A

20 V

−

Fig. 3.79 Solution Step I When the 5 A source is acting alone (Fig. 3.80) From Fig. 3.80, I=

V1′ 4

Applying KCL at Node 1, V1′ − 4 I V1′ + =5 1 4

+ −

5Vx

I ′′

+ Vx −

…(iii)

1Ω

5Vx

3Ω

I ′′ I 2 = 1.6 (→) By superposition theorem, I = I′ + I′′ = 3.4 + 1.6 = 5 A (→)

Example 3.18

+ −

Fig. 3.77

−2 1 − 3I 2 − 5Vx = 0 −2 1 − 3I 2 − 5( −2 I1 ) = 0 81 3 3II 2 = 0 Solving Eqs (ii) and (iii), I1 = 0 6 A I2 = 1 6 A Step III

3Ω

17 V

1Ω

4I

V1′

+ −

I

5A

Fig. 3.80

4Ω

3.22 Network Analysis and Synthesis ⎛ V ′⎞ V ′ V1′− 4 ⎜ 1 ⎟ + 1 = 5 ⎝ 4⎠ 4 V1′ = 20 V 1Ω

Step II When the 20 V source is acting alone (Fig. 3.81) Applying KVL to the mesh, 4I – I – 4I – 20 = 0 I = –20 A V1′′ = 4I – 1(I) = 3I = 3 (–20) = –60 V

4I

4Ω

I

+ −

20 V

Step III By superposition theorem, V1 = V1′ + V1′′ = 20 – 60 = –40 V

Example 3.19

V1′′

Fig. 3.81

Find the current in the 6 W resistor in Fig. 3.82. 1Ω

− +

2Vx I

− Vx +

6Ω

3A

18 V

Fig. 3.82 1Ω

Solution Step I When the 18 V source is acting alone (Fig. 3.83) From Fig. 3.83, Vx = –I′ Applying KVL to the mesh, 18 – I′ + 2Vx – 6I′ = 0 18 – I′ – 2I′ – 6I′ = 0 I′ = 2 A (↓) Step II When the 3 A source is acting alone (Fig. 3.84) From Fig. 3.84, Vx

1 I1 = − I1

…(i)

I1 = 3

I′

6Ω

18 V

Fig. 3.83 1Ω

− +

− Vx +

I1

…(ii)

2Vx

− Vx +

2Vx I ′′

6Ω

3A

Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I2

− +

I2

Fig. 3.84

Applying KVL to the outerpath of the supermesh, −1I1 + 2Vx − 6 2 = 0 − I1 + 2( − 1 ) − 6 I 2 = 0 3 1 66II 2 = 0

…(iii)

3.2 Superposition Theorem 3.23

Solving Eqs (ii) and (iii), I1 = −2 A I2 = 1 A I ′′ = I 2 = 1 A (↓) Step III

By superposition theorem, I6 Ω = I′ + I′′ = 2 + 1 = 3 A (↓)

Example 3.20

Find the current Iy in Fig. 3.85. Iy

4Ω

+ −

120 V

10Iy 8 Ω

12 A

40 V

Fig. 3.85 I y′

Solution Step I When the 120 V source is acting alone (Fig. 3.86) Applying KVL to the mesh,

4Ω

+ −

10I y′

8Ω

120 V

120 – 4Iy′ – 10Iy′ – 8Iy′ = 0 Iy′ = 5.45 A (→)

Fig. 3.86

Step II When the 12 A source is acting alone (Fig. 3.87) From Fig. 3.87, I y ″ I1 …(i) Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I 2 I1 = 12 Applying KVL to the outer path of the supermesh, −4 1 − 1 10 0 y ″ − 8I 2 = 0

′′

10Iy′′ + −

Iy

4Ω

8Ω

12 A I1

I2

...(ii) Fig. 3.87

−4 1 − 10 10 1 − 8 I 2 = 0 14 I1 8 2 0

…(iii)

Solving Eqs (ii) and (iii), I1 = −4 36 A I 2 = 7 64 A Iy ″ = I1 = −4.36 A ( → ) Step III When the 40 V source is acting alone (Fig. 3.88) Applying KVL to the mesh, –4 Iy′′′ – 10Iy′′′ – 8Iy′′′ – 40 = 0 I y ″′ = −

40 = –1.82 A (→) 22

Iy′′′

10Iy′′′ + −

4Ω

8Ω

40 V

Fig. 3.88

3.24 Network Analysis and Synthesis Step IV By superposition theorem, Iy = Iy′ + Iy′′ + Iy′′′ = 5.45 – 4.36 – 1.82 = –0.73 A (→)

Example 3.21

Find the voltage Vx in Fig. 3.89. 3Ω + Vx

6Ω −

+ −

3Vx

24 Ω

18 V

36 V 5A

Fig. 3.89 Solution Step I When the 18 V source is acting alone (Fig. 3.90) From Fig. 3.90, Vx′ = 3I Applying KVL to the mesh, 18 – 3I – 6I – 3Vx′ = 0 18 – 3I – 6I – 3 (3I) = 0 I=1A Vx′ = 3 V

3Ω + Vx

6Ω

+ −

3Vx′

−

18 V

Step II When the 5 A source is acting alone (Fig. 3.91) From Fig. 3.91, Vx′′ = –3I1 Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I 2 I1 = 5 Applying KVL to the outer path of the supermesh,

I

Fig. 3.90 3Ω + Vx

…(i)

6Ω − 24 Ω

5A

I1

+ − I2

...(ii) Fig. 3.91

−3 1 − 6 I 2 − 3V″x = 0 −3 1 − 6 I 2 − 3(3I1 ) = 0 12 I1 6 2 0 Solving Eqs (ii) and (iii),

3Vx ′

…(iii)

I1 = −1 67A I 2 = 3 33 A Vx″

3I1 = 3( −1.67) = −5 V

Step III When the 36 V source is acting alone (Fig. 3.92) From the figure, Vx′′′ = –3I Applying KVL to the mesh,

3Ω

6Ω

+ −

3Vx′′′

+ Vx′′′ − 36 V I

36 + 3Vx′′′ – 6I – 3I = 0 Fig. 3.92

3.2 Superposition Theorem 3.25

⎛ −V ′′′⎞ ⎛ −V ′′′⎞ 36 + 3Vx′′′ 6 ⎜ x ⎟ − 3 ⎜ x ⎟ = 0 ⎝ 3 ⎠ ⎝ 3 ⎠ 36 + 3Vx′′′ 2Vx′′′ + Vx″′ = 0 Vx′′′ = −6 V Step IV By superposition theorem, Vx = Vx′ + Vx′′ + Vx′′′ = 3 – 5 – 6 = –8 V

Example 3.22

Find the voltage V in the network of Fig. 3.93. 8Ω − V

15 Ω

5Ω

+ + −

12 Ω

−5 A

10 V

8V

Fig. 3.93 Solution Step I When the 10 V source is acting alone (Fig. 3.94) From Fig. 3.94, V′ 8 I1 ...(i) Applying KVL to Mesh 1, −10 − 8 1 − 15 15

1

− 12 (

1

−

2)

8Ω

15 Ω

5Ω

− V′ + + −

12 Ω I2

10 V I1

8V′

=0

35 I1 12 I 2 = −10

...(ii)

Fig. 3.94

Applying KVL to Mesh 2, −12( 2 − 1 ) − 5 I 2 − 8V ′ = 0 −12 I 2 +12 12 I1 − 5 2 − 8( −8 8 1) = 0 76 I1 17 I 2 = 0

...(iii)

Solving Eqs (ii) and (iii), I1 = 0 54 A I 2 = 2.4 A V′ 8 I1 8(0.54) Step II When the –5 A source is acting alone (Fig. 3.95) From Fig. 3.95, V″ 8 I1 ...(i) Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I1

I 2 = −5

8Ω

15 Ω

5Ω

− V ′′ +

I1

...(ii)

4.32 V

−5 A I2

Fig. 3.95

12 Ω I3

+ −

8 V ′′

3.26 Network Analysis and Synthesis Applying KVL to the outer path of the supermesh, −8 1 − 15 15 2 − 12( 2 − 3 ) = 0 −8 1 − 27 27 2 + 12 3 = 0 Applying KVL to Mesh 3,

...(iii)

−12( 3 − 2 ) − 5 I 3 − 8V ″ = 0 −12 I 3 +12 12 I 2 − 5 3 − 8( −8 8 1) = 0 64 I1 12 I 2 − 17 I 3 = 0

...(iv)

Solving Eqs (ii), (iii) and (iv), I1 = 4 97 A I 2 = 9 97 A I 3 = 25.74 A V″ 8 I1 8( −4 4.97) Step III

39.76 V

By superposition theorem, V V ′ + V ″ = −4.32 − 39 39.76 = −44.08 V

Example 3.23

For the network shown in Fig. 3.96, find the voltage V0. 50 Ω

200 Ω + 40 Ω

+ 1A

V0 −

V1

25 V

+ −

0.5V1

−

Fig. 3.96 Solution Step I When the 1 A source is acting alone (Fig. 3.97) From Fig. 3.97, V1 200 I 2 ...(i) For Mesh 1, I1 = 1 ...(ii) Applying KVL to Mesh 2, 0.5 1 40 ( I 2 − I1 ) 200 00 I 2 = 0 0.5 ( 200 I 2 ) 40 I 2 + 40 I1 200 I 2 = 0 40 I1 140 I 2 = 0 Solving Eqs (ii) and (iii), I1 = 1 A I 2 = 0 29 A V0 ′ 50 I1 − 200 I 2 = 0 V0 ′ − 50(1) − 200(0.29) 0 V0 ′ = 108 V

50 Ω

200 Ω +

1A

+ V0′ −

40 Ω V1 I1

+ −

0.5V1

I2

−

Fig. 3.97 ...(iii)

3.2 Superposition Theorem 3.27 50 Ω

Step II When the 25 V source is acting alone (Fig. 3.98) From Fig. 3.98, V1 200 I − 25 = 0 200 I + 25

V1

...(i)

200 Ω + 40 Ω

V0′′

25 V

V1

+ −

Applying KVL to Mesh 1,

0.5V1

I

−

0 5V1 40 I − 200 I − 25 = 0 0.5( 200 I 25) 25) 40 I − 200 I − 25 = 0

Fig. 3.98

I = −0 09 A V0″ V1 = 200 I + 25 = 200( 200 −0.09) 25

7V

By superposition theorem,

Step III

V0 = V0′ + V0″ = 108 + 7 = 115 V

Example 3.24

For the network shown in Fig. 3.99, find the voltage Vx. 10 A

2Ω

4Ω + Vx

20 V

Vx 2

6Ω

−

Fig. 3.99 2Ω

Solution Step I When the 20 V source is acting alone (Fig. 3.100) From Fig. 3.100, Vx′

6( I1 − I 2 )

...(i)

Applying KVL to Mesh 1, 20 − 2

6( I1 − I 2 ) = 0 81 6 6II 2 = 20

1

4Ω +

20 V

Vx′

6Ω

Vx′ I1

2 I2

−

Fig. 3.100 ...(ii)

For Mesh 2, Vx′ 6( I1 I 2 ) = = 3I 3I1 3I 2 2 2 4II 2 = 0 4 I2 =

3

1

Solving Eqs (ii) and (iii), I1 = 5 71 A I 2 = 4.29 A Vx′

6( I1

I2 )

6(5 (5.71 4.29) = 8 52 V

...(iii)

3.28 Network Analysis and Synthesis Step II When the 10 A source is acting alone (Fig. 3.101) From Fig. 3.101, Vx″ = 6(II1 I 2 ) Applying KVL to Mesh 1, −2( 1 − 3 ) − 6( 1 − 2 ) = 0 8 1 6I2 − 2 3 0 For Mesh 2, V ″ 6( I1 I 2 ) I2 = x = = 3I 3I1 3I 2 2 2 31 4 4II 2 = 0 For Mesh 3, I 3 = −10 Solving Eqs (ii), (iii) and (iv), I1 = −5 71 A I 2 = −4.29 A I 3 = −10 A Vx ″ Step III

10 A

...(i)

2Ω

I3

4Ω +

...(ii)

Vx′′ 2

6Ω

Vx′′ I1

I2

−

Fig. 3.101 ...(iii) ...(iv)

6( I1 − I 2 ) = 6( 5.71 + 4.29)

8.52 V

By superposition theorem, Vx

Example 3.25

Vx ′ + Vx ″ = 8.52 − 8.52 = 0

Calculate the current I in the network shown in Fig. 3.102. 4Ω 2I1

20 Ω

70 V

− + 2Ω

50 V I 10 Ω

Fig. 3.102 Solution Step I When the 70 V source is acting alone (Fig. 3.103) From Fig. 3.103, I ′ I3 ...(i) Applying KVL to Mesh 1, −4

1

− 2 I1 − 20 ( I1 − I 2 ) = 0 26 I1 20 I 2 = 0

Applying KVL to Mesh 2, 70 − 20( 2 1 ) − 2( 2 3 ) 0 −20 I1 + 22 I 2 − 2 3 = 70

...(ii)

4Ω

I1 20 Ω

2I1

I1

− + 2Ω

70 V I2

I3

10 Ω

...(iii)

Fig. 3.103

I′

3.2 Superposition Theorem 3.29

Applying KVL to Mesh 3, −2(

3

− 2

2 ) + 2 I1

−10 10 I 3 = 0

2 I 2 − 12 I 3 = 0

1

...(iv)

Solving Eqs (ii), (iii) and (iv), I1 = 8 94 A I 2 = 11.62 A I 3 = 3 43 A I′

I 3 = 3.43 (←)

Step II When the 50 V source is acting alone (Fig. 3.104) From Fig. 3.104,

4Ω

I1

I ′′ = I 3 …(i)

20 Ω

2I1

I1

− +

Applying KVL to Mesh 1, 2Ω

−4 1 − 2 I1 − 20( I1 − I 2 ) = 0 26 I1 20 0 I 2 = 0 …(ii)

I2

50 V I3

10 Ω

Applying KVL to Mesh 2, −20(

2

− 1 ) − 2(

2

−

−20 I1 + 22 I 2 − 2

3)

=0

3

=0

I′′

Fig. 3.104 …(iii)

Applying KVL to Mesh 3, −2(

3

−

2 ) + 2 I1

2

1

+ 50 − 10 0I3 = 0 2 I 2 − 12 I 3 = −50

…(iv)

Solving Eqs (ii), (iii), and (iv), I1 = 1 06 A I 2 = 1 38 A I 3 = 4 57 A I ′′ Step III

I 3 = 4.57

(←)

By superposition theorem, I

Example 3.26

I ′ + I ′′ = 3.43 + 4.57 = 8 A (←)

Find the voltage V0 in the network of Fig. 3.105. 4V

5Ω + 10 V

1A

V0 −

Fig. 3.105

2Ω

1Ω

+ −

V0 2

3.30 Network Analysis and Synthesis Solution 5Ω

Step I When the 10 V source is acting alone (Fig. 3.106) Applying KCL at the node,

1Ω + V0′

10 V

V′ V0′ − 0 V0′ − 10 V0′ 2 =0 + + 5 2 1 ⎛ 1 1 1⎞ ⎜⎝ + + ⎟⎠ V0′ = 2 5 2 2 V0′ = 1.667 V Step II When the 1A current source is acting alone (Fig. 3.107) Applying KCL at the node,

−

Fig. 3.106

5Ω

1Ω + 1A

V ′′ V ′′− 0 V0′′ V0′′ 0 2 + + + 5 2 1

4V

5Ω + V0″′

2

1Ω

+ V0″′ 2 −

−

Fig. 3.108

By superposition theorem, V0 = V0′ V0′′ V0′′′ = 1.67 0.83 + 3.33

3.3

+ V0″ − 2

2Ω

Fig. 3.107

V ′′′ V ′′′− 4 − 0 V0′′′ V0′′′ 0 2 =0 + + 5 2 1 ⎛ 1 1 1⎞ ⎜⎝ + + ⎟⎠ V0′′′ = 4 5 2 2 V0′′′ = 3.33 33 V Step IV

V0″ −

⎛ 1 1 1⎞ ⎜⎝ + + ⎟⎠ V0′′= −1 5 2 2 V0′′= −0.83 V Step III When the 4 V source is acting alone (Fig. 3.108) Applying KCL at the node,

+ V0′ − 2

2Ω

4.17 V

THEVENIN’S THEOREM

It states that ‘any two terminals of a network can be replaced by an equivalent voltage source and an equivalent series resistance. The voltage source is the voltage across the two terminals with load, if any, removed. The series resistance is the resistance of the network measured between two terminals with load removed and constant voltage source being replaced by its internal resistance (or if it is not given with zero resistance, i.e., short circuit) and constant current source replaced by infinite resistance, i.e., open circuit.’

3.3 Thevenin’s Theorem 3.31 RTh IL Network

RL

IL VTh

RL

(a)

(b)

Fig. 3.109 Network illustrating Thevenin’s theorem Explanation

Consider a simple network as shown in Fig. 3.110. R1

R3 A

R2

V

RL B

Fig. 3.110

Network R1

For finding load current through RL, first remove the load resistor RL from the network and calculate open circuit voltage VTh across points A and B as shown in Fig. 3.111. VTh =

R2 R1

R2

R3 +

+

−

Fig. 3.111 Calculation of VTh R1

R3 A

R1 R2 R1 R2

R2

R Th

Thevenin’s equivalent network is shown in Fig. 3.113. IL =

VTh RTh

RL

B

Fig. 3.112 Calculation of RTh RTh A

If the network contains both independent and dependent sources, Thevenin’s resistance RTh is calculated as, RTh

V = Th IN

where I N is the short-circuit current which would flow in a short circuit placed across the terminals A and B. Dependent sources are active at all times. They have zero values only when the control voltage or current is zero. RTh may be negative in

VTh

A

VTh

R2

V

V

For finding series resistance RTh , replace the voltage source by a short circuit and calculate resistance between points A and B as shown in Fig. 3.112. RTh

R3

RL IL B

Fig 3.113 Thevenin’s equivalent network

B

3.32 Network Analysis and Synthesis some cases which indicates negative resistance region of the device, i.e., as voltage increases, current decreases in the region and vice-versa. If the network contains only dependent sources then VTh = 0 IN = 0 For finding RTh in such a network, a known voltage V is applied across the terminals A and B and current is calculated through the path AB. V RTh = I RTh or a known current source I is connected across the A terminals A and B and voltage is calculated across the terminals A and B. V RTh = I B Thevenin’s equivalent network for such a network is shown in Fig. 3.114. Fig. 3.114 Thevenin’s equivalent network Steps to be Followed in Thevenin’s Theorem 1. 2. 3. 4. 5.

Remove the load resistance RL. Find the open circuit voltage VTh across points A and B. Find the resistance RTh as seen from points A and B. Replace the network by a voltage source VTh in series with resistance RTh. Find the current through RL using Ohm’s law. IL =

Example 3.27

VTh RTh

RL

Find the current through the 2 W resistor in Fig. 3.115. 5Ω

40 V

20 V

2Ω

10 Ω

10 V

Fig. 3.115 Solution

20 V

5Ω

Step I Calculation of VTh (Fig. 3.116) Applying KVL to the mesh, 40 − 5 20 10 0 15 I = 20 I = 1.33 A

+

−

A +

+

10 Ω

40 V

I

−

Fig. 3.116

B VTh

− 10 V

3.3 Thevenin’s Theorem 3.33

Writing the VTh equation,

5Ω

A

B RTh

10 I VTh + 10 = 0 VTh Step II

10 I + 10 = 10(1 (1.33) 10

10 Ω

23.33 V

Calculation of RTh (Fig. 3.117) Fig. 3.117 RTh = 5 10 = 3.33 Ω

3.33 Ω A

Step III

Calculation of IL (Fig. 3.118)

2Ω

23.33 V

IL =

IL

23.33 = 4.38 A 3 33 + 2

B

Fig. 3.118

Example 3.28

Find the current through the 8 W resistor in Fig. 3.119. 5Ω

10 Ω

5Ω

250 V

8Ω

75 V

Fig. 3.119 5Ω

I

+

Solution

−

+ 5Ω

250 V

VTh

−

250 = 25 A 5+5

− 75 V

Writing the VTh equation, 250 − 5

VTh

75

Fig. 3.120 5Ω

0

VTh = 175 1 − 5 I = 175 − 5( 25) Step II

Calculation of RTh (Fig. 3.121) RTh = (

A

+

Step I Calculation of VTh (Fig. 3.120) I=

10 Ω

10 Ω A

50 V 5Ω

) + 10 = 12.5 Ω

RTh

B

Fig. 3.121

B

3.34 Network Analysis and Synthesis 12.5 Ω

Calculation of IL (Fig. 3.122)

Step III

IL =

A 8Ω

50 V

50 = 2.44 A 12.5 + 8

IL B

Fig. 3.122

Example 3.29

Find the current through the 2 W resistor connected between terminals A and B in

the Fig. 3.123. 2Ω

1Ω

3Ω A

12 Ω

2V

2Ω

4V

2Ω

1Ω

B

Fig. 3.123 Solution Step I Calculation of VTh (Fig. 3.124) Applying KVL to Mesh 1,

+

2 2 I1 12( I1 − I 2 ) = 0 14 I1 12 I 2 = 2 ...(i)

−

+ −

+

−

12 Ω

2V

− +

I1

3Ω A

+

+

−

V I2

B

4V

−

Applying KVL to Mesh 2, −12(

2

− 1 ) − 1 2 − 3I 2 − 4 = 0 −12 I1 +16 16 I 2 = −4

Fig. 3.124 ...(ii)

Solving Eqs (i) and (ii), I 2 = −0.4 A

2Ω

1Ω A

Writing the VTh equation, VTh

Step II

B

Fig. 3.125

Calculation of RTh (Fig. 3.125)

Calculation of IL (Fig. 3.126) IL =

28 = 0.82 A 1.43 + 2

RTh

12 Ω

3I 2 − 4 = 0 VTh 4 3I 2 = 4 + 3( −0.4) = 2.8 V

1.43 Ω

RTh = [( 2 12)) + ] 3 = 1. 3 Ω Step III

3Ω

A 2Ω

2.8 V IL

B

Fig. 3.126

3.3 Thevenin’s Theorem 3.35

Example 3.30

Find the current through the 10 W resistor in Fig. 3.127. 6Ω

2Ω

1Ω

10 V

3Ω

10 Ω

20 V

Fig. 3.127 6Ω

Solution Step I Calculation of VTh (Fig. 3.128) Applying KVL to Mesh 1, 10 − 6

1

1( I1 − I 2 ) = 0 7

1

2

10

+

2Ω −

+

−

+ − I1

3Ω I2

− +

2

VTh

−

−

− 1 ) − 2I 2 − 3 − I1 + 6

2 2

=0 =0

Fig. 3.128

…(ii) 6Ω

2Ω A

Solving Eqs (i) and (ii), I 2 = 0.24 A

…(ii) 1Ω

Writing the VTh equation, 3

2

VTh

20 VTh

Step II Step III

3Ω

R Th

0 B

3I 2 20 3(0 (0.24) 20 19.28 V =19.28 V (terminal B is positive w.r.t A)

Fig. 3.129 1.47 Ω

Calculation of RTh (Fig. 3.129) RTh = [(6 1) + ] 3 = . 7 Ω

A IL

Calculation of IL (Fig. 3.130) IL =

10 Ω

19.28 V

19.28 = .68 ( ↑ ) 1.47 + 10

B

Fig. 3.130

Example 3.31

B

20 V

Applying KVL to Mesh 2, −1(

A

+

1Ω

10 V

…..(i)

+

Find the current through the 10 W resistor in Fig. 3.131. 10 Ω

10 A

5Ω

Fig. 3.131

30 Ω

20 Ω

100 V

3.36 Network Analysis and Synthesis Solution

A

Step I Calculation of VTh (Fig. 3.132) For Mesh 1, 1

+

+

5Ω

10 A

= 10

Th

I1

30 Ω

B −

+

20 Ω

100 V

−

I2

−

Applying KVL to Mesh 2, 100 − 30 I 2

Fig. 3.132

20 I 2 = 0

R Th

I2 = 2 A

A

30 Ω

B

Writing the VTh equation, 5

1

20

VTh Step II

5Ω

0

2

5 I1 20

2

20 Ω

5(10) − 20( 2) 10 V Fig. 3.133

Calculation of RTh (Fig. 3.133)

17 Ω

RTh = 5 + ( 20 30) = 17 Ω Step III

A

Calculation of IL (Fig. 3.134)

10 Ω

10 V IL

10 = 0.37 A 17 + 10

IL =

B

Fig. 3.134

Example 3.32

Find the current through the 40 W resistor in Fig. 3.135. 50 Ω

25 V

20 Ω

10 Ω

40 Ω

30 Ω

10 V

Fig. 3.135 Solution Step I Calculation of VTh (Fig. 3.136) Since the 20 Ω resistor is connected across the 25 V source, the resistor becomes redundant.

50 Ω +

25 V

20 Ω

10 Ω −

+ + A VTh I − B

V20 Ω = 25 V Fig. 3.136

−

10 V

30 Ω

3.3 Thevenin’s Theorem 3.37

Applying KVL to the mesh, 25 − 50 I − 10 I + 10 = 0 I = 0.58 A Writing the VTh equation, VTh − 10I + 10 = 0 VTh

Step II

10( I ) − 10 = 10( .58) .5 ) .2 V = 4.2 V(terminal B is positive w . r .t )

Calculation of RTh (Fig. 3.137) 50 Ω

10 Ω

A RTh B

20 Ω

30 Ω

A RTh B

50 Ω

(a)

10 Ω

(b)

Fig. 3.137

8.33 Ω A

RTh = 50 10 = 8 33 Ω

IL 40 Ω

4.2 V

Step III

Calculation of IL (Fig. 3.138) IL =

Example 3.33

B

4.2 = 0.09 (↑) 8 33 + 40

Fig. 3.138

Find the current through the 10 W resistor in Fig. 3.139. 2V

6Ω

10 Ω

5Ω

15 Ω

4Ω

50 V

20 V

Fig. 3.139 Solution Step I Calculation of VTh (Fig. 3.140) 50 =5A 6 4 20 I2 = =1A 5 15 I1 =

6Ω +

2V

−

A +

+

V Th

5Ω

B −

−

4Ω

50 V I1

+

+ 15 Ω

−

−

Fig. 3.140

20 V I2

3.38 Network Analysis and Synthesis Writing the VTh equation, 4 I1 2 V

h

− 15I 15 I 2 = 0 VTh 4 I1 2 − 15I 15 I 2 = 4(5) + 2 − 15(1) = 7 V

Calculation of RTh (Fig. 3.141)

Step II

6Ω

RTh = (

)+(

A

) = 6.15 Ω

R Th

5Ω

B

4Ω

15 Ω

Fig. 3.141 Step III

Calculation of IL (Fig. 3.142) 6.15 Ω A

IL =

7 = 0.43 A 6 15 + 10

10 Ω

7V IL

B

Fig. 3.142

Example 3.34

Determine the current through the 24 W resistor in Fig. 3.143. 30 Ω

20 Ω

220 V 24 Ω 50 Ω

5Ω

Fig. 3.143 Solution Step I Calculation of VTh (Fig. 3.144)

220 I2 = =88A 20 + 5 Writing the VTh equation, VTh

30 I1 − 20 I 2 = 0 VTh 20 I 2 − 30 I1 = 20(8.8) − 30( 2.75) = 93 5 V

+

30 Ω +

220 I1 = = 2 75 A 30 + 50

I1 − +

220 V

A 50 Ω

Fig. 3.144

V Th

−

20 Ω − I2

B 5Ω

3.3 Thevenin’s Theorem 3.39

Step II

Calculation of RTh (Fig. 3.145) 30 Ω

20 Ω

A

R Th

B

50 Ω

5Ω

Fig. 3.145 Redrawing the circuit (Fig. 3.146),

RTh = (

)+(

30 Ω

) = 2 .7 5 Ω

20 Ω

A

B 50 Ω

5Ω

Fig. 3.146 Step III

Calculation of IL (Fig. 3.147) 22.75 Ω A

IL =

24 Ω

93.5 V

93.5 =2A 22.75 + 24

IL B

Fig. 3.147

Example 3.35

Find the current through the 3 W resistor in Fig. 3.148. 4Ω

5Ω 3Ω

8Ω

1Ω 2Ω

50 V

Fig. 3.148

3.40 Network Analysis and Synthesis Solution Step I Calculation of VTh (Fig. 3.149) Applying KVL to Mesh 1, 50 − 2 1 1( I1 − I 2 ) 8( 1 2 ) 11I1 9 2

4Ω

+

0 50

…(i) +

Applying KVL to Mesh 2, −4

2

− 5 I 2 − 8(

2

−

− 1 ) − 1( I 2 − I1 ) = 0 −9 1 + 1 18 8 2 =0

2Ω

...(ii)

−

I2

Step II

+ −

+ I1 50 V

Writing the VTh equation, 5 I 2 − 8(I (I

−

B −

− 1 Ω −+

I1 = 7 69 A I 2 = 3 85 A I )=0 VTh 5 I 2 + 8(I ( I I ) = 5(( .85) ((33.85 − 7.69) = −11.47 4 V = 11.47 V(the terminal B is positive w . r .t . )

Calculation of RTh (Fig. 3.150) 4Ω

5Ω

A R Th B 8Ω

1Ω 2Ω

Fig. 3.150 Redrawing the network (Fig. 3.151), A

4Ω

5Ω

2Ω 1Ω

8Ω

B

Fig. 3.151

5Ω

A + V Th

Solving Eqs (i) and (ii),

VTh

+

Fig. 3.149

+

−

8Ω

3.3 Thevenin’s Theorem 3.41

Converting the upper delta into equivalent star network (Fig. 3.152), R1 =

4 2 = 0 73 Ω 4+2+5

R2 =

4 5 = 1 82 Ω 4+2+5

R3 =

5 2 = 0 91 Ω 4+2+5

A

A

1.82 Ω

R2

R1

1Ω

R3

0.73 Ω

8Ω

1Ω

0.91 Ω

8Ω

B

B

Fig. 3.152

Fig. 3.153

Simplifying the network (Fig. 3.153), RTh = 1.82 + (1.73 73 8.91) = 3.27 Ω

A

1.82 Ω

1.73 Ω

8.91 Ω

B

Fig. 3.154 Step III

Calculation of IL (Fig. 3.155) IL =

3.27 Ω

11.47 = .83 ( ↑ ) 3 27 + 3

A 3Ω

11.47 V IL

B

Fig. 3.155

3.42 Network Analysis and Synthesis

Example 3.36

Find the current through the 20 W resistor in Fig. 3.156. 120 V

45 V

20 Ω

10 Ω

15 Ω

5Ω

5Ω

20 V

Fig. 3.156 120 V

Solution Step I Calculation of VTh (Fig. 3.157) Applying KVL to Mesh 1, 45 − 120 − 15 I1 5(

2)

1

45 V

10 0( 1 2 ) 0 30 I1 15 I 2 = −75

...(i)

Applying KVL to Mesh 2, 20 − 5

2

10(

2

1)

5( I 2 − I1 ) = 0

I1 − +

10 Ω

+

− B

−

+ −

−

Solving Eqs (i) and (ii),

− +

5Ω

Writing the VTh equation, 45 − VTh − 10( I1 − I 2 ) = 0

Step II

I 2 ) = 45 −10 10[ −33.2 − ( −11.4)] = 63 V

Calculation of RTh (Fig. 3.158) A R Th

15 Ω

B 10 Ω

5Ω

5Ω

Fig. 3.158

5Ω

+

Fig. 3.157

I1 = −3 2 A I 2 = −1.4 A

VTh = 45 − 10( I1

15 Ω

I2

...(ii)

−15 15 I1 + 20 I 2 = 20

+ A V Th

20 V

+ −

3.3 Thevenin’s Theorem 3.43

Converting the delta formed by resistors of 10 Ω, 5 Ω and 5 Ω into an equivalent star network (Fig. 3.159), R1 =

10 × 5 = 2 5Ω 20

R2 =

10 × 5 = 2 5Ω 20

R3 =

5 5 = 1.25 Ω 20

Simplifying the network (Fig. 3.160),

A RTh B

A B R1

R2

2.5 Ω

15 Ω

R3

15 Ω

1.25 Ω

2.5 Ω

Fig. 3.159

Fig. 3.160

RTh = ( .25 | | 2.. ) + 2.5 = 4.67 Ω

A RTh B

2.5 Ω

16.25 Ω

2.5 Ω

Fig. 3.161 Step III

Calculation of IL IL =

4.67 Ω

63 = 2.55 A 4 67 + 20

A 20 Ω

63 V IL

B

Fig. 3.162

Example 3.37

Find the current through the 3 W resistor in Fig. 3.163.

12 Ω 3Ω

6Ω

6A 42 V

Fig. 3.163 Solution Step I Calculation of VTh (Fig. 3.164) For Mesh 1, I1 = 6 Applying KVL to Mesh 2, 42 −12 12(

2

1) − 6 2

+

+ −

+

12 Ω

…(i)

− +

6A I1

6Ω 42 V

I2

V Th

− −

0

−12 12 I1 + 18 I 2 = 42

A

...(ii) Fig. 3.164

B

3.44 Network Analysis and Synthesis Solving Eqs (i) and (ii),

A

I 2 = 6.33 A

6Ω

12 Ω

R Th

Writing the VTh equation, VTh Step II

6 I 2 = 38 V

B

Calculation of RTh (Fig. 3.165)

Fig. 3.165 4Ω

RTh = 6 12 = 4 Ω Step III

A

Calculation of IL (Fig. 3.166) IL =

3Ω

38 V IL

38 = 5.43 A 4+3

B

Fig. 3.166

Example 3.38

Find the current through the 30 W resistor in Fig. 3.167. 15 Ω

60 Ω

30 Ω

40 Ω

13 A

150 V

50 V

Fig. 3.167 Solution Step I Calculation of VTh (Fig. 3.168) Meshes 1 and 2 form a supermesh. Writing current equation for supermesh, I 2 I1 = 13 ...(i) Writing voltage equation for supermesh,

15 Ω +

60 Ω −

+

A −

+

I1

V Th

B −

40 Ω

13 A

150 V

+

I2

50 V

−

150 −15 15 I1 60 I 2 − 40 I 2 = 0 Fig. 3.168

15 I1 +100 100 I 2 = 150 ...(ii) Solving Eqs (i) and (ii), I1 = −10 A I2 = 3 A

15 Ω

60 Ω

Writing the VTh equation, 40 I 2 VTh − 50 = 0 VTh 40 I 2 − 50 = 40(3) 50 Step II

40 Ω

70 V

Calculation of RTh (Fig. 3.169) RTh = 75 40 = 26.09 Ω

A

Fig. 3.169

R Th

B

3.3 Thevenin’s Theorem 3.45

Step III

26.09 Ω

Calculation of IL (Fig. 3.170)

A

70 = 1.25 A 26.09 + 30

IL =

30 Ω

70 V IL

B

Fig. 3.170

Example 3.39

Find the current through the 20 W resistor in Fig. 3.171. 10 Ω

5Ω

5A

20 Ω

100 V

Fig. 3.171 Solution Step I Calculation of VTh (Fig. 3.172) 10 Ω

VTh = 100 V

+ A V Th − B

5Ω

5A

100 V

Fig. 3.172 Step II

Calculation of RTh (Fig.3.173) 10 Ω A R Th B

5Ω

A R Th B

5Ω

(a)

(b)

Fig. 3.173 RTh = 0 Step III

Calculation of I L (Fig. 3.174) IL =

100 = 5A 20

A 20 Ω

100 V IL

B

Fig. 3.174

3.46 Network Analysis and Synthesis

Example 3.40

Find the current through the 20 W resistor in Fig. 3.175. 10 Ω

20 Ω

5Ω

4Ω

10 V

8Ω

2A

Fig. 3.175 Solution Step I

10 Ω

Calculation of VTh (Fig. 3.176) +

10 = 0.71 A 10 + 4 I2 = 2 A

A −

I1 =

+

+

V Th

5Ω

B +

−

4Ω

10 V I1

−

+ 8Ω

−

−

2A I2

Writing the VTh equation, 4

1

Step II

8I 2 = 0 VTh = 4(0 (0.71) + 8( 2) = 18.84 V

Fig. 3.176

Calculation of RTh (Fig. 3.177). 10 Ω

R Th A

4Ω

R Th

5Ω

B

A

8Ω

B

2.86 Ω

(a)

8Ω

(b)

Fig. 3.177 RTh = 10.86 Ω Step III

Calculation of I L (Fig. 3.178) 10.86 Ω A

IL =

18.84 = 0.61 A 10.86 + 20

20 Ω

18.84 V IL

B

Fig. 3.178

3.3 Thevenin’s Theorem 3.47

Example 3.41

Find the current through the 5 W resistor in Fig. 3.179. 10 Ω

2Ω

50 V

50 V

5Ω

100 V

2Ω

3Ω

Fig. 3.179 Solution Step I Calculation of VTh (Fig. 3.180) Applying KVL to Mesh 1, 100 −10 10 I1 + 500 2 1 2( I1 I 2 ) = 0 14 I 2 2 I 2 = 150 Applying KVL to Mesh 2, −2( 2 − 1 ) + 50 − 3I 2 = 0 …(ii) −2 1 + 5 I 2 = 50 Solving Eqs (i) and (ii), I1 = 12.88 A I 2 = 15.15 A

...(i)

10 Ω +

2Ω

50 V

−

+

50 V

−

A + 100 V

I1

2Ω

V Th B −

Writing the VTh equation,

+

+ − − +

3Ω I2

Fig. 3.180

100 −10 10 I1 − VTh = 0

VTh = 100 − 10(12.88) = −28.8 V = 28.8 V( terminal B is positiv s e w.r.t. A) Step II

Calculation of RTh (Fig. 3.181) 10 Ω

10 Ω

2Ω

A

A 2Ω

R Th

3Ω

R Th B

B

(b)

(a) 10 Ω

RTh = 10 3.2 = 2.42 Ω

2Ω

3.2 Ω A R Th B (c)

Fig. 3.181

1.2 Ω

−

3.48 Network Analysis and Synthesis Step III

2.42 Ω

Calculation of I L (Fig. 3.182) IL =

28.8 = 3.88 2.42 + 5

A

(↑ )

IL

28.8 V

5Ω

B

Fig. 3.182

Example 3.42

Find the current through the 10 W resister in Fig. 3.183. 10 Ω 2Ω

2Ω

1Ω

15 V

1Ω

10 V

1Ω

Fig. 3.183 A

Solution Step I Calculation of VTh (Fig. 3.183) Applying KVL to Mesh 1,

+

2

= −25

+

...(i)

2

2Ω −

2

0

− I1 + 4

2

10

...(ii)

+

−

1Ω − +

I1 1 ) 2I 2 1

+ −

15 V

Applying KVL to Mesh 2, 10 −1 1(

−

2Ω

−15 − 2 1 − 1( I1 − I 2 ) − 100 − 1 1 = 0 4 1−

B VTh

−

10 V

+ 1Ω

Fig. 3.184

Solving Eqs (i) and (ii), I1 = −6 A I 2 = 1A Writing the VTh equation, −VTh + 2 I 2 + 2 I1 = 0 VTh = 2 I1 + 2 I 2 = 2( −6) + 2(1) = −10 V

= 10 V( the terminal B is positive w.r.t. A)

+ 1Ω

I2

−

3.3 Thevenin’s Theorem 3.49

Step II

Calculation of RTh (Fig. 3.185)

2Ω

A

RTh

B

2Ω

1Ω

1Ω

1Ω

Fig. 3.185 Converting the star network formed by resistors of 2 Ω, 2 Ω and 1 Ω into an equivalent delta network (Fig. 3.186), A

2Ω

2 2 R1 = 2 + 2 + =8Ω 1 2 1 R2 = 2 + 1 + =4Ω 2 2 1 R3 = 2 + 1 + =4Ω 2

RTh

B

2Ω

1Ω

1Ω

Fig. 3.186 Simplifying the network (Fig. 3.187), RTh A

RTh

A

B

B 8Ω

8Ω

4Ω

4Ω

0.8 Ω

0.8 Ω (b)

RTh A 1Ω

(a)

1Ω

B 1.33 Ω (c)

Fig. 3.187 RTh = 1 33 Ω

1Ω

3.50 Network Analysis and Synthesis Step III

Calculation of I L (Fig. 3.188) 1.33 Ω A

IL =

10 = 0.88 1 33 + 10

(↑ )

IL

10 V

10 Ω

B

Fig. 3.188

Example 3.43

Find the current through the 1 W resistor in Fig. 3.189. 1A

2Ω

3Ω

4V

1Ω

3A

Fig. 3.189 Solution Step I Calculation of VTh (Fig. 3.190) 1A

− +

2Ω

+ I2 − A +

4V

− +

3Ω

+ −

V Th

I1

3A

B − 2Ω

Fig. 3.190

A

Writing the current equations for Meshes 1 and 2,

Writing the VTh equation, 4 2(

1

VTh Step II Step III

R Th

I1 = −3 I2 = 1 2 ) − VTh

=0 = 4 2(

3Ω

B

Fig. 3.191 1

−

Calculation of RTh (Fig. 3.191) RTh = 2 Ω Calculation of I L (Fig. 3.192) 12 IL = = 4A 2 +1

2)

2Ω

= 4 2( −4) 12 V

A 1Ω

12 V IL

B

Fig. 3.192

3.3 Thevenin’s Theorem 3.51

Example 3.44

Find the current through the 3 W resistor in Fig. 3.193. 2Ω

1Ω

2Ω

10 V

3Ω

10 A

Fig. 3.193 Solution Step I Calculation of VTh (Fig. 3.194) 2Ω

1Ω + 2Ω

10 V

10 A

A

V Th −

B

Fig. 3.194 By source transformation (Fig. 3.195), 2Ω +

1Ω −

2Ω −

10 V I

A

+

+

20 V

Fig. 3.195

V Th −

B 2Ω

1Ω A

Applying KVL to the mesh, 10 − 2

2

20 4

0

2Ω

10

R Th

I = −2.5 A B

Writing the VTh equation, 10 − 2 Step II

VTh 0 VTh = 10 − 2 I

2Ω

10 2( −2 2.5) = 15 V

A

Calculation of RTh (Fig. 3.196) RTh = (

Step III

Fig. 3.196

) +1 = 1+1 = 2 Ω

Calculation of IL (Fig. 3.197) IL =

15 = 3A 2+3

3Ω

15 V IL

B

Fig. 3.197

3.52 Network Analysis and Synthesis

EXAMPLES WITH DEPENDENT SOURCES Example 3.45

Obtain the Thevenin equivalent network for the given network of Fig. 3.198 at

terminals A and B. A

I1

4Ω

2I1

8V B

Fig. 3.198 Solution Step I Calculation of VTh (Fig. 3.199) From Fig. 3.199, I1

+

I1

4Ω

2 I1

3I1 0 I1 = 0

2I1

V Th

8V

−

Writing the VTh equation,

A

B

Fig. 3.199 8 0 − VTh = 0 VTh = 8 V

Step II Calculation of IN (Fig. 3.200), Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I2 3

1

4Ω

I1 = 2 I1 2

0

A

I1

2I1

I1

...(i)

IN

I2

8V B

Applying KVL to the outer path of the supermesh, Fig. 3.200

8 4I 4 I1 = 0 I1 = 2

...(ii)

Solving Eqs (i) and (ii), I2 = 6 A IN I2 = 6 A Step III

Calculation of RTh

1.33 Ω A

8V

RTh =

VTh 8 = = 1 33 Ω IN 6

Step IV Thevenin’s Equivalent Network (Fig. 3.201)

B

Fig. 3.201

3.3 Thevenin’s Theorem 3.53

Example 3.46

Find Thevenin’s equivalent network of Fig. 3.202. 2Ω

3Ω +

4V

0.1 Vx

A

Vx −

B

Fig. 3.202 2Ω

Solution Step I Calculation of VTh (Fig. 3.203) Vx

+

VTh

I1

3Ω −

4V

+ 0.1 Vx

A

Vx = VTh

I1

0 1 Vx

−

Writing the VTh equation,

B

Fig. 3.203 4 2 I1 − Vx = 0

2Ω

3Ω

4 2( −0.1Vx ) − Vx = 0 4 0.8Vx

+

0

4V

0.1 Vx

A

Vx IN

Vx = 5 V −

Vx = VTh = 5 V

B

Fig. 3.204

Step II Calculation of IN (Fig. 3.204) From Fig. 3.204,

2Ω

3Ω A

Vx = 0 The dependent source 0.1 Vx depends on the controlling variable Vx. When Vx = 0, the dependent source vanishes, i.e., 0.1Vx = 0 as shown in Fig. 3.205. IN = Step III

4V

IN

B

4 = 0.8 A 2+3

Fig. 3.205 6.25 Ω A

Calculation of RTh RTh =

VTh 5 = = 6 25 Ω IN 08

5V

B

Step IV Thevenin’s Equivalent Network (Fig. 3.206) Fig. 3.206

3.54 Network Analysis and Synthesis

Example 3.47

Obtain the Thevenin equivalent network of Fig. 3.207 for the terminals A and B. 1Ω

Vx

2Ω

4Vx − +

A

2A

2V B

Fig. 3.207 Solution Step I Calculation of VTh (Fig. 3.208) From Fig. 3.208, 2 2 I1 − Vx = 0 Vx = 2 2 I1 For Mesh 1, I1 = −2 A Vx = 2 2( −2 2) Writing the VTh equation, 2 2 I1 − 0 4Vx − VTh = 0 2 2( −2) 0 + 4(6) − VTh = 0 VTh = 30 V

1Ω

Vx

4Vx − +

−

2Ω 2V

−

6V

Step II Calculation of IN (Fig. 3.209) From Fig. 3.209, Vx 2 2 I1 Meshes 1 and 2 will form a supermesh, Writing current equation for the supermesh I 2 I1 = 2 Applying KVL to the outer path of the supermesh,

Vx

…(i)

1Ω

4Vx − +

IN

2A 2V

I1

I2

…(ii)

B

Fig. 3.209 ...(iii)

I1 = 0 73 A I 2 = 2 73 A I N I 2 = 2 73 A

10.98 Ω A

30 V

Calculation of RTh VTh 30 = = 10.98 Ω IN 2 73

Step IV Thevenin’s Equivalent Network (Fig. 3.210)

A

2Ω

Solving Eqs (ii) and (iii),

RTh =

B

Fig. 3.208

2 2 I1 1 2 + 4Vx = 0 2 2 I1 I 2 + 4( 2 − 2 1 ) 0 10 I1 + I 2 = 10

Step III

A

VTh

2A

+ I 1

+

B

Fig. 3.210

3.3 Thevenin’s Theorem 3.55

Example 3.48

Find the Thevenin equivalent network of Fig. 3.211 for the terminals A and B. 8I1 − +

1Ω A

I1

10 Ω

10 Ω

5V B

Fig. 3.211 Solution

8I1 − +

Step I Calculation of VTh (Fig. 3.212) Applying KVL to the mesh, 5 10

1

10

I1 =

10 Ω

5 = 0.25 A 20

10 Ω

8

1

1

1

2

0 − VTh = 0 VTh = 5 2 I1 = 5 2(0.25)

4.5 V 8I1 − +

10 0( 1 2 ) 0 20 I1 10 I 2 = 5

...(i)

− 1 ) + 8 1 −1 − 1I 2 = 0

1Ω A

IN

10 Ω 5V

18 I1 −11 11I 2 = 0

10 Ω I1

I2 B

...(ii)

Fig. 3.213

Solving Eqs (i) and (ii), I1 = 1.375 A I 2 = 2.25 A I N I 2 = 2.25 A Step III

B

Fig. 3.212

Applying KVL to Mesh 2, −10(

VTh −

Step II Calculation of IN (Fig. 3.213) Applying KVL to Mesh 1, 5 10

A

5V

Writing the VTh equation, 5 10

+

I1

0

1

1Ω

2Ω A

Calculation of RTh

4.5 V

RTh

V 45 = Th = = IN 2.25

Ω

Step IV Thevenin’s Equivalent Network (Fig. 3.214)

B

Fig. 3.214

3.56 Network Analysis and Synthesis

Example 3.49

Find VTh and RTh between terminals A and B of the network shown in Fig. 3.215 . 1Ω

2Ω

12 V

Ix

A

1Ω

2Ix

B

Fig. 3.215 1Ω

Solution Step I

2Ω +

A

Calculation of VTh (Fig. 3.216) 12 V

Ix = 0

VTh

1Ω

The dependent source 2Ix depends on the controlling variable Ix. When Ix = 0, the dependent source vanishes, i.e., 2 x 0 as shown in Fig. 3.216.

−

B

Fig. 3.216

Writing the VTh equation, VTh = 12 ×

1 =6V 1+1 1Ω

Step II Calculation of IN (Fig. 3.217) From Fig. 3.217, Ix =

V1 2

12 V

2Ω

V1

2Ix

1Ω

Ix

A

IN

Applying KCL at Node 1, B

V1 12 V1 V1 + + = 2I x 1 1 2 V1 ⎛V ⎞ V1 V1 + − 12 = 2 ⎜ 1 ⎟ ⎝ 2⎠ 2

Fig. 3.217

V1 = 8 V IN = Step III

V1 8 = =4A 2 2

Calculation of RTh RTh =

Example 3.50 terminals a and b .

VTh 6 = = 5Ω IN 4

Obtain the Thevenin equivalent network of Fig. 3.218 for the given network at

3.3 Thevenin’s Theorem 3.57 3Ω

4Ω

Vx

5Vx + −

a

2Ω

2A

b

Fig. 3.218 3Ω

Solution Step I Calculation of VTh (Fig. 3.219) Applying KCL at Node x,

Vx

+

Vx − 5 Vx

a

VTh −

Writing the VTh equation, = −16 V

5Vx + −

2Ω

2A

V 2= x 2 Vx = 4 V VTh

4Ω

b

Fig. 3.219 4 Vx

( the terminal a s

3Ω

ega ve w. . . b )

Step II Calculation of IN (Fig. 3.220) Applying KCL at Node x,

4Ω

Vx

5Vx + −

a IN

2Ω

2A

Vx Vx − 5 Vx + 2 4 V V 2 = x − Vx = − x 2 2 2=

b

Fig. 3.220

Vx = −4 V V Vx IN = x = −Vx = 4 A 4 Step III

−4 Ω a

−16 V

Calculation of RTh RTh =

VTh −16 = = −4 Ω IN 4

b

Step IV Thevenin’s Equivalent Network (Fig. 3.221)

Example 3.51

Fig. 3.221

Obtain the Thevenin equivalent network of Fig. 3.222 for the given network. 150 V 10 Ω

30 Ω

5A

+ Vx

15 Ω

−

Fig. 3.222

+ −

1 V 3 x

3.58 Network Analysis and Synthesis Solution Step I Calculation of VTh (Fig. 3.223) From Fig. 3.223 Vx VTh Applying KCL at the node, 1 Vx 1 0 Vx 3 + Vx + 5 = 0 10 15

150 V 10 Ω

30 Ω A

+

VTh

+ Vx

5A

− B

30 Ω

Step IV

IN

1 V 3 x

15 Ω

5A

B

Fig. 3.224 37.5 Ω A

V 60 = x = =2A 30 30

75 V

Calculation of RTh VTh 75 = = 37.5 Ω IN 2

B

Thevenin’s Equivalent Network (Fig. 3.225)

Example 3.52

+ −

150 V 10 Ω

Vx

A

Step II Calculation of IN (Fig. 3.224) Applying KCL at Node x, 1 Vx 1 0 Vx Vx Vx 3 +5+ + =0 30 15 10 Vx Vx Vx Vx + + − = 15 − 5 30 15 10 30 Vx = 60 V

RTh =

1 V 3 x

Fig. 3.223

VTh = 75 V

Step III

+ −

−

Vx = 75 V

IN

15 Ω

Fig. 3.225

Find the Thevenin’s equivalent network of the network to the left of A-B in the

Fig. 3.226. 10 Ix

10 V

+ −

A

Ix 1A

5Ω

10 Ω

B

Fig. 3.226 10 Ix

Solution Step I Calculation of VTh (Fig. 3.227) From Fig. 3.227, I x I1 − I 2 For Mesh 1, I1 = 1

10 V

+ −

+

Ix

…(i)

+

5Ω

1A I1

I2

10 Ω

VTh

−

−

…(ii) Fig. 3.227

A

B

3.3 Thevenin’s Theorem 3.59

Applying KVL to Mesh 2, −5( −5(

2

2

− 1 ) − 10

− 1 ) − 10(

1

−

2

=0

2 ) − 10 2

=0

x

5

−1 10 0

5II 2 = 0 5

1

…(iii)

Solving Eqs (ii) and (iii), I1 = 1 A I 2 = −1 A I x I1 I 2 = 1 ( 1) = 2 A Writing the VTh equation, 10 I 2 10 VTh = 0 10( −11) 10 − VTh = 0 VTh = −20 V Step II Calculation of IN (Fig. 3.228) From Fig. 3.228, I x I1 − I 2 For Mesh 1, I1 = 1 Applying KVL to Mesh 2, −5( −5(

2

2

− 1 ) − 10

− 1 ) − 10(

1

−

2

−

3)

=0

2 ) − 10( 2

−

3)

=0

x

−1 10 0(

10 Ix

10 V

+ −

…(i) …(ii)

A

Ix

5Ω

1A I1

10 Ω I2

IN I3 B

Fig. 3.228

−5 1 − 5 I 2 + 10 I 3 = 0

…(iii)

Applying KVL to Mesh 3, −10(

3

−

2 ) − 10

=0

10 I 2 10 0 I 3 = 10

…(iv)

Solving Eqs (ii), (iii) and (iv), I1 = 1 A I2 = 3 A I3 = 2 A I N I3 = 2 A Step III

−20 Ω A

−20 V

Calculation of RTh

VTh −20 = = −10 Ω IN 2 Step IV Thevenin’s Equivalent Network (Fig. 3.229) RTh =

Example 3.53 Fig. 3.230.

B

Fig. 3.229

Find Thevenin’s equivalent network at terminals A and B in the network of

3.60 Network Analysis and Synthesis 2Ω

4Ω A

4Vx

+ Vx −

− +

5Ω

B

Fig. 3.230 Solution Since the network does not contain any independent source, VTh = 0 IN = 0 But the RTh can be calculated by applying a known voltage source of 1 V at the terminals A and B as shown in Fig. 3.231. V 1 RTh = = I I From Fig. 3.231, Vx ( I1 − I 2 ) Applying KVL to Mesh 1, −4Vx − 2 I1 − 5( 1 − 2 ) = 0 −4 [

−

] − 2 1 − 5I1 + 5

2

2Ω

4Ω A

4Vx

− +

I1

+ Vx −

5Ω

1V I2 B

Fig. 3.231 ...(i)

=0

−27 I1 + 25 I 2 = 0

…(ii)

Applying KVL to Mesh 2, −5(

2

− 1) − 4I2 − 1 = 0 5

1

9II 2 = 1 9

…(iii)

Solving Eqs (ii) and (iii), A

I1 = −0 21 A I 2 = −0 23 A

4.35 Ω

Hence, current supplied by voltage source of 1 V is 0.23 A. 1 RTh = = 4 35 Ω 0 23 Hence, Thevenin’s equivalent network is shown in Fig. 3.232.

Example 3.54

B

Fig. 3.232

Find the current in the 9 W resistor in Fig. 3.233. − +

6Ix Ix

4Ω 6Ω 20 V

Fig. 3.233

9Ω

3.3 Thevenin’s Theorem 3.61

Solution Step I Calculation of VTh (Fig. 3.234) Applying KVL to the mesh, −4 x 6

− +

Ix

4Ω

6I x = 0

x

Ix = 5 A x

VTh

+ 6Ω −

20 V

Writing the VTh equation, 6

6Ix

−

− +

6Ix Ix

Ix = 0

A

4Ω 6Ω

The dependent source 6Ix depends on the controlling variable Ix. When I x = 0, the dependent source vanishes, i.e., 6 x 0 as shown in Fig. 3.236. 20 IN = =5A 4

IN

20 V B

Fig. 3.235

6Ix A

4Ω IN 20 V

A

⇒

4Ω IN 20 V

B

B

(a)

(b)

Fig. 3.236

6Ω A

Calculation of RTh RTh =

VTh 30 = =6Ω IN 5

9Ω

30 V IL

Step IV Calculation of IL (Fig. 3.237)

B

30 IL = =2A 6+9

Example 3.55

B

Fig. 3.234

Step II Calculation of IN (Fig. 3.235). From Fig. 3.235,

− +

A

VTh

0

6(5) − VTh = 0 VTh = 30 V

Step III

+

Fig. 3.237

Determine the current in the 16 W resistor in Fig. 3.238. 10 Ω

6Ω Ix

40 V

0.8Ix

Fig 3.238

16 Ω

3.62 Network Analysis and Synthesis 10 Ω

Solution Step I Calculation of VTh (Fig. 3.239) From Fig. 3.239, Ix = 0 The dependent source 0.8Ix depends on the controlling variable Ix. When Ix = 0, the dependent source vanishes, as shown in Fig. 3.240. i.e., 0 8I x = 0 VTh = 40 V

I1

+ 40 V

−

2 2

−

…(ii) 10 Ω

6Ω

…(iii)

40 V

0.8Ix

I1 = 3 A 5 I2 = A 3

IN I2 B

Fig. 3.241 5 A 3

24 Ω

Calculation of RTh

A

RTh =

VTh 40 = = 5 IN 3

Ω

16 Ω

40 V IL

Step IV Calculation of IL (Fig. 3.242)

B

40 IL = =1A 24 + 16

Example 3.56

Ix A

0 40

I2 =

Fig. 3.242

Find the current in the 6 W resistor in Fig. 3.243. 1Ω −

18 V

B

Fig. 3.240

I1

IN

A

VTh

Solving Eqs (ii) and (iii),

Step III

6Ω

40 V

Applying KVL to the outer path of the supermesh, 40 −10 10 I1 6 10 I1 + 6

B

Fig. 3.239 10 Ω

…(i)

0.8 I 2

A

VTh

0.8Ix

+

Step II Calculation of IN (Fig. 3.241) From Fig. 3.241, I x I2 Meshes 1 and 2 will form a supermesh, Writing current equation for the supermesh, I1 I 2 = 0.8 I x 1 8 I2 = 0

6Ω

Vx

− +

+

3A

Fig. 3.243

2Vx

6Ω

3.3 Thevenin’s Theorem 3.63

Solution Step I Calculation of VTh (Fig. 3.244) From Fig. 3.244, Vx 1 1 = − I1 For Mesh 1, I1 = −3 A Vx = 3 V

1Ω −

…(i)

Vx

− +

+

18 V

2Vx +

3A

…(ii)

A

VTh

I1 −

Writing the VTh equation,

B

Fig. 3.244

18 −1 1 I1 2 Vx − VTh = 0 18 + 3 2(3) VTh 0 VTh = 27 V Step II Calculation of IN (Fig. 3.245) From Fig. 3.245, Vx I1 Meshes 1 and 2 will form a supermesh, Writing current equation for supermesh, I 2 I1 = 3 Applying KVL to the outer path of the supermesh, 18 −1 1I1 2 Vx = 0 18 − I1 + 2( 1 ) 0

1Ω

…(i)

−

Vx

− +

+

2Vx

IN

18 V

3A

…(ii)

I1

I2 B

Fig. 3.245

…(iii)

I1 = 6 A Solving Eqs (ii) and (iii), I2 = 9 A Step III

IN

I2 = 9 A

RTh =

VTh 27 = = 3Ω IN 9

3Ω A

Calculation of RTh

6Ω

27 V IL

Step IV Calculation of IL (Fig. 3.246)

B

27 IL = =3A 3+ 6

Example 3.57

Fig. 3.246

Find the current in the 10 W resistor. 10 Ω

100 V

A

− +

10Vx

+ Vx −

Fig. 3.247

5Ω

10 A

3.64 Network Analysis and Synthesis Solution Step I Calculation of VTh (Fig. 3.248) From the figure, Vx = 10 × 5 = 50 V

+ VTh −

A

− +

10Vx

B + 5Ω −

100 V

Writing the VTh equation,

Vx

10 A

5Ω

10 A

100 − VTh + 10Vx − Vx = 0 100 − VTh + 9Vx 100 − VTh + 9(50)

0 0

Fig. 3.248

VTh = 550 V Step II Calculation of IN (Fig. 3.249) From Fig. 3.249, Vx 5( I N + 10) Applying KVL to Mesh 1,

A

B

− +

10Vx

IN + Vx −

100 V

100 +10 10Vx − Vx = 0 Vx = − 100 − =5 9 IN = − Step III

100 9 N

Fig. 3.249

+ 50

550 A 45 −45 Ω

Calculation of RTh

A

550 RTh = = −45 Ω 550 − 45 Step IV Calculation of IL (Fig. 3.250) IL =

3.4

10 Ω

550 V IL

B

550 110 =− A −45 + 10 7

Fig. 3.250

NORTON’S THEOREM

It states that ‘any two terminals of a network can be replaced by an equivalent current source and an equivalent parallel resistance.’ The constant current is equal to the current which would flow in a short circuit placed across the terminals. The parallel resistance is the resistance of the network when viewed from these open-circuited terminals after all voltage and current sources have been removed and replaced by internal resistances.

IL Network

(a)

RL

IL

IN or IN

RN

(b)

Fig. 3.251 Network illustrating Norton’s theorem

RL

3.4 Norton’s Theorem 3.65

Explanation

Consider a simple network as shown in Fig.3.252 R1

R3 A

V

R2

RL

B

Fig. 3.252

Network

For finding load current through RL , first remove the load resistor RL from the network and calculate short circuit current ISC or I N which would flow in a short circuit placed across terminals A and B as shown in Fig. 3.253. For finding parallel resistance RN , replace the voltage source by a short circuit and calculate resistance between points A and B as shown in Fig. 3.254. RN

R3 +

R1

R3 A

V

R2

IN

B

R1 R2 R1 R2

Fig. 3.253

Calculation of IN

R1

R3 A

Norton’s equivalent network is shown in Fig.3.255. IL

IN

RN RN RL

R2

RN

If the network contains both independent and dependent sources, Norton’s resistances RN is calculated as

B

Fig. 3.254

V RN = Th IN

Calculation of RN IL

where VTh is the open-circuit voltage across terminals A and B. If the network contains only dependent sources, then

IN

RN

RL

VTh = 0 IN = 0

Fig. 3.255

To find RTh in such network, a known voltage V or current I is applied across the terminals A and B, and the current I or the voltage V is calculated respectively. RN =

V I

Norton’s equivalent network for such a network is shown in Fig. 3.256.

Norton’s equivalent network A

RN

B

Fig. 3.256

Norton’s equivalent network

3.66 Network Analysis and Synthesis Steps to be followed in Norton’s Theorem 1. 2. 3. 4. 5.

Remove the load resistance RL and put a short circuit across the terminals. Find the short-circuit current ISC or IN. Find the resistance RN as seen from points A and B. Replace the network by a current source IN in parallel with resistance RN. Find current through RL by current–division rule. IL =

Example 3.58

I N RN RN RL

Find the current through the 10 W resistor in Fig. 3.257. 5Ω 1Ω

10 Ω

15 Ω

4A

2V

Fig. 3.257

2 1I1 = 0 I1 = 2

1Ω IN

...(i)

2V

Meshes 2 and 3 will form a supermesh. Writing the current equation for the supermesh, I2 = 4

I3

I1

2

− 15 15

3

B

Fig. 3.258

...(ii)

=0

5Ω

...(iii)

Solving Eqs (i), (ii) and (iii),

A

1Ω

I1 = 2 A

Fig. 3.259 A

Calculation of RN (Fig. 3.259) RN = 1 (5 + 15) = 0 95 Ω

Step III

15 Ω

RN B

I 2 = −3 A I3 = 1A I N I1 I 2 = 2 ( 3) = 5 A Step II

15 Ω

4A I3

I2

Applying KVL to the supermesh, −5

5Ω

A

Solution Step I Calculation of IN (Fig. 3.258) Applying KVL to Mesh 1,

IL 5A

0.95 Ω

10 Ω

Calculation of IL (Fig.3.260) IL = 5 ×

0 95 = 0.43 A 0 95 + 10

B

Fig. 3.260

3.4 Norton’s Theorem 3.67

Example 3.59

Find the current through the 10 W resistor in Fig. 3.258. 8Ω

12 V

V

2Ω

5Ω

10 Ω

Fig. 3.261 12 V

8Ω

A

Solution Step I Calculation of IN (Fig. 3.262) Applying KVL to Mesh 1,

20 V

5Ω

−5 1 + 2 20 0 − 2( I1 − I 2 ) = 0 7 1 − 2 I 2 = 20

...(i)

2Ω I1

B

Applying KVL to Mesh 2, −2(

2

IN I2

Fig. 3.262

− 1 ) − 8 I 2 − 12 = 0 − 2 1 +1 10 0 2 = −12

8Ω

Solving Eqs (i) and (ii),

5Ω

2Ω

...(ii) RN

I 2 = −0 67 A IN Step II

Fig. 3.263

Calculation of RN (Fig. 3.263) RN = (

Step III

I 2 = −0 67 A A

) + 8 = 9.43 Ω

I L = 0 67 ×

9.43 Ω

0.67 A

Calculation of IL (Fig. 3.264) 9 43 = 0.33 9 43 + 10

IL

(↑ )

B

Fig. 3.264

Example 3.60

Find the current in the 10 W resistor in Fig. 3.265. 50 Ω

50 V

40 Ω

20 Ω 10 V

Fig. 3.265

10 Ω

10 Ω

3.68 Network Analysis and Synthesis 50 Ω

Solution Step I Calculation of IN (Fig. 3.266) The resistance of Ω becomes redundant as it is connected across the 50 V source (Fig. 3.267). Applying KVL to Mesh 1, 50 − 50 I1 20( I1 − I 2 ) − 10 = 0 70 I1 20 I 2 = 40

40 Ω

50 V

20 Ω

IN

10 V

...(i) Fig. 3.266

Applying KVL to Mesh 2, 10 − 20(

2

1)

=0

−20 20 I1 + 20 I 2 = 0 Solving Eqs (i) and (ii),

Step II

50 Ω

...(ii) 40 Ω

20 Ω

50 V I1

I1 = 1 A I2 = 1 5 A IN I2 = 1 5 A

10 V

IN I2

Fig. 3.267

Calculation of RN (Fig. 3.268) 50 Ω A RN

40 Ω

RN = 50 20 = 14.28 Ω

20 Ω

B

Fig. 3.268 Step III

Calculation of IL (Fig. 3.269) A IL

IL = 1 5 ×

14.28 Ω

1.5 A

14.28 = 0 88 A 14.28 + 10

10 Ω

B

Fig. 3.269

Example 3.61

Find the current through the 10 W resistor in Fig. 3.270. 6Ω

10 V

2Ω

1Ω

3Ω

20 V

Fig. 3.270

10 Ω

3.4 Norton’s Theorem 3.69 6Ω

Solution Step I Calculation of IN (Fig. 3.271) Applying KVL to Mesh 1, −6

7

1

2

10

A

1Ω

10 V

1( I1 − I 2 ) = 0

1

2Ω

I1

...(i)

3Ω I2

B

Applying KVL to Mesh 2, −1(

2

20 V

− 1 ) − 2 I 2 − 3(

2

−

− I1 + 6

2

− 3I 3 = 0

3)

=0

Fig. 3.271 ...(ii) 6Ω

Applying KVL to Mesh 3, −3(

3

IN I3

2Ω A

− 2 ) − 20 = 0 3 2 − 3I 3 = 20

1Ω

...(iii)

3Ω

RN

Solving Eqs (i), (ii) and (iii),

B

I 3 = −13.17 A IN Step II

Fig. 3.272

I 3 = −13.17 A

2Ω

Calculation of RN (Fig. 3.272)

A IL

RN = [(6 1) + ] 3 = 1.46 Ω Step III

1.46 Ω

13.17 A

10 Ω

Calculation of IL (Fig. 3.273) I L = 13.17 ×

1.46 = .68 (↑) 1.46 + 10

Example 3.62

B

Fig. 3.273

Find the current through the 10 W resistor in Fig. 3.274. 10 Ω

20 Ω

30 Ω

20 Ω

20 Ω

50 V

100 V

40 V

Fig. 3.274 Solution Step I Calculation of IN (Fig. 3.274) Applying KVL to Mesh 1, 50 − 20( 1 2 ) − 40 = 0 20 I1 20 I 2 = 10

A

...(i)

Applying KVL to Mesh 2, 40 − 20(

2

1 ) − 20 I 2

20( 2 − 3 ) = 0 −20 20 I1 + 60 I 2 − 20 I 3 = 40 ...(ii)

IN

B

20 Ω

20 Ω

20 Ω

50 V I1

30 Ω

I2 40 V

Fig. 3.275

100 V I3

3.70 Network Analysis and Synthesis Applying KVL to Mesh 3, −20( 3 −

2 ) − 30 I 3

− 100 = 0

...(iii)

−20 I 2 + 50 I 3 = −100 Solving Eqs (i), (ii) and (iii), I1 = 0 81 A IN Step II

I1 = 0 81 A

Calculation of RN (Fig. 3.276) 20 Ω

RN A

RN = [( 20 30) +

30 Ω

B

20 Ω

] 20 = 12.3 Ω

20 Ω

Fig. 3.276 Step III

Calculation of IL (Fig. 3.277) A IL

I L = 0 81 ×

12.3 Ω

0.81 A

12.3 = 0 45 A 12.3 + 10

10 Ω

B

Fig. 3.277

Example 3.63

Obtain Norton’s equivalent network as seen by RL in Fig. 3.278. 30 Ω

40 V

10 Ω

30 Ω

60 Ω

120 V

RL

10 V

Fig. 3.278 30 Ω

Solution Step I Calculation of IN (Fig. 3.279) Applying KVL to Mesh 1,

90 I1 60 I 2 = 120

10 Ω

I1

A

IN

30 Ω

60 Ω

120 V

120 − 30 I1 60( I1 − I 2 ) = 0

40 V

I2

...(i) Fig. 3.279

B

10 V I3

3.4 Norton’s Theorem 3.71

Applying KVL to Mesh 2, −60(

2

10 Ω

30 Ω

− 1 ) + 40 − 10 I 2 − 30( 30(

2

−

3)

A

RN

B

=0

−60 I1 +100 100 I 2 − 30 I 3 = 40

...(ii) 30 Ω

60 Ω

Applying KVL to Mesh 3, −30(

3

−

2 ) + 10

=0

30 I 2 − 30 I 3 = −10

...(iii)

Fig. 3.280

Solving Eqs (i), (ii) and (iii), A

I 3 = 4 67 A IN Step II

I 3 = 4 67 A

RN = [(30 60) + Step III

15 Ω

4.67 A

Calculation of RN (Fig. 3.280) ] 30 = 15 Ω

RL

B

Norton’s equivalent network (Fig. 3.281) Fig. 3.281

Example 3.64

Find the current through the 8 W resistor in Fig. 3.282. 5V

5A

12 Ω

2A

4Ω

8Ω

Fig. 3.282 Solution Step I Calculation of IN (Fig. 3.283) 5V A

4Ω

12 Ω

5A

2A

IN B

Fig. 3.283 The resistor of the 4 Ω gets shorted as it is in parallel with the short circuit. Simplifying the network by source transformation (Fig. 3.284), 12 Ω

5V A

60 V

2A I1

I2

IN B

Fig. 3.284

3.72 Network Analysis and Synthesis Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh, I 2 I1 = 2 Applying KVL to the supermesh,

A

...(i)

60 −12 12 I1 − 5 0 12 I1 = 55

12 Ω

4Ω

B

...(ii)

Fig. 3.285

Solving Eqs (i) and (ii), I1 = 4 58 A I 2 = 6 58 A I N I 2 = 6 58 A Step II

A IL

3Ω

6.58 A

8Ω

Calculation of RN (Fig. 3.285) RN = 12 4 = 3 Ω

Step III

RN

B

Calculation of IL (Fig. 3.286) I L = 6.58 ×

Example 3.65

Fig. 3.286 3 = 1.79 A 3+8

Find the current through the 1 W resistor in Fig. 3.287.

2Ω

1Ω 1A

3Ω

1V

2Ω

2Ω

Fig. 3.287 Solution Step I Calculation of IN (Fig. 3.288) A

2Ω

IN 1A

3Ω

1V

2Ω

B

2Ω

Fig. 3.288

3.4 Norton’s Theorem 3.73

By source transformation (Fig. 3.289), A

2Ω I3

3Ω 3V

IN I1

1V

2Ω

I2

B

2Ω

Fig. 3.289 A

Applying KVL to Mesh 1, −3 − 3I1 − 2(

− 3) +1 = 0 5 1 − 2 I 3 = −2

1

2Ω

...(i)

RN 3Ω

2Ω

Applying KVL to Mesh 2, −1 − 2(

2

−

3) − 2 2

4

2

=0 − 2 I 3 = −1

B 2Ω

...(ii)

(a) 2Ω

Applying KVL to Mesh 3, −2(

3

− 1 ) − 2( 3 − 2 ) = 0 −2 1 − 2 I 2 + 4 3 = 0

...(iii)

2Ω

A

B

Solving Eqs (i), (ii) and (iii), 3Ω

I1 = −0 64 A I 2 = −0 55 A I 3 = −0 59 A I N I 3 = −0 59 A Step II

2Ω (b)

1.2 Ω

1Ω

A

B (c)

Calculation of RN (Fig. 3.290)

Fig. 3.290

RN = 2.2 Ω Step III

Calculation of IL (Fig. 3.291) A

I L = 0 59 ×

2.2 = 0 41 A 2.2 + 1

0.59 A

2.2 Ω

1Ω IL B

Fig. 3.291

3.74 Network Analysis and Synthesis

EXAMPLES WITH DEPENDENT SOURCES Example 3.66

Find Norton’s equivalent network across terminals A and B of Fig. 3.292. 3I1 I2

I1

10 Ω

5Ω

5V

+ 10I2 −

10 Ω

− +

A

B

Fig. 3.292 Solution Step I Calculation of VTh (Fig. 3.293) From Fig. 3.293, I2 I x I1

3I1 I2

I1

10 Ω

Ix

5 10

Ix

5V

5 I x − 10 I 2 = 0

x

A

VTh

+ 10I2 −

−

5 I x − 10 I x = 0 Ix = 0 2 A

x

+

5Ω

Applying KVL to the mesh, 5 10

10 Ω

− +

B

Fig. 3.293

I1 = −0 2 A Writing the VTh equation, 5 10

3 I1 − VTh = 0

x

5 10 (0.2) 3 ( −0.2) − VTh = 0 VTh = 2.4 V Step II Calculation of IN (Fig. 3.294) From Fig. 3.294, I2 I x I1 I y − I x Applying KVL to Mesh 1, 5 10 x 5( I x − I y ) − 10 I 2 = 0 5 10

5 Ix + 5

x

y

25 I x

10

x

0

5

y

5

3I1

…(i) …(ii)

10

x

5

y

5( 5I x

x)

y

3(

y

10

y

0

12 I x 12 I y = 0

A

5Ω

Ix

+ 10I2 −

IN Iy B

…(iii)

3I1 − 10 I y = 0 x)

I1

10 Ω

5V

Applying KVL to Mesh 2, 10 I 2

I2

10 Ω

− +

…(iv)

Fig. 3.294

3.4 Norton’s Theorem 3.75

Solving Eqs (iii) and (iv),

I x = 0 25 A

A

I y = 0 25 A Step III

IN

I y = 0.25 A

RN =

VTh 2.4 = =96Ω IN 0 25

9.6 Ω

0.25 A

Calculation of RN B

Step IV Norton’s Equivalent Network (Fig. 3.295)

Example 3.67

Fig. 3.295

For the network shown in Fig. 3.296, find Norton’s equivalent network. 20 Ω A

5Ω

3Vx 15 Ω + Vx

2A

2Ω

− B

Fig. 3.296 20 Ω

Solution Step I Calculation of VTh (Fig. 3.297) From Fig. 3.297, Vx

+

−

5Ω

2I 2

+

…(i)

−

I1 15 Ω

For Mesh 1, I1

3Vx

3( 2II 2 )

6I2

…(ii)

For Mesh 2,

I1

6I2

6( 2)

+

Vx I2

I2 = 2

3Vx

VTh

+ 2A

A

2Ω

−

−

…(iii)

B

Fig. 3.297

12 A

20 Ω

Writing the VTh equation,

A

VTh 0 I1 +15 + 15( I1 I 2 ) − 2I 2I 2 = 0 VTh + 5( −12) + 15( −12 − 2) − 2( 2) = 0 VTh = 274 V

5Ω

IN +

Step II Calculation of IN (Fig. 3.298) From Fig. 3.298, Vx

3Vx

I1 15 Ω

Vx

2A

2( I 2 − I 3 )

…(i)

For Mesh 2, I2 = 2

…(ii)

I2

I3

2Ω

− B

Fig. 3.298

3.76 Network Analysis and Synthesis Meshes 1 and 3 will form a supermesh. Writing the current equation for the supermesh, I 3 I1 = 3Vx 3 [ 2 I 2

]

I3

6I 2 − 6I3

I1 6 I 2 7 I 3 = 0 Applying KVL to the outer path of the supermesh, −5 1 − 20 20 3 − 2( I 3 − I 2 ) − 15( I1 − I 2 ) = 0

…(iii)

−20 I1 +17 17 I 2 − 22 I 3 = 0

…(iv)

Solving Eqs (ii), (iii) and (iv), I1 = −0 16 A I2 = 2 A I 3 = 1 69 A I N I 3 = 1 69 A Step III

A

Calculation of RN

162.13 Ω

1.69 A

V 274 RN = Th = = 162.13 Ω I N 1 69

B

Step IV Norton’s Equivalent Network (Fig. 3.299)

Example 3.68

Fig. 3.299

Obtain Norton’s equivalent network across A-B in the network of Fig. 3.300. 2Ω

5Ω

A I2

+ + 10I2 −

8Ω

V1

15 V

−

0.6V1

15 Ω

B

Fig. 3.300

Step I Calculation of VTh (Fig. 3.301) From Fig. 3.301, V1

2Ω

5Ω

Solution

8( I x − I y )

+ V1

15 V

+

−

Ix

+ 10I2 −

8Ω Iy

I2

15 Ω

0.6V1 I2

+ VTh

− −

…(i)

Applying KVL to Mesh 1,

A

B

Fig. 3.301 15 − 5

x

8( I x − I y ) = 0 13I x

8

y

15

…(ii)

Applying KVL to Mesh 2, −8(

y

− 8

) − 2 I y −10 10 I 2 = 0 x

10

y

10

2

0

…(iii)

3.4 Norton’s Theorem 3.77

For Mesh 3, I2

0 6V1 = 0.6 ⎡⎣8( I x

I y ) ⎤⎦

4.8 I y − I 2 = 0

4.8I 8I x

…(iv)

Solving Eqs (ii), (iii) and (iv), I x = 3 28 A 2Ω

5Ω

I y = 3 45 A

A

I 2 = −0 83 A

+ V1

15 V

Writing the VTh equation,

+ 10I2 −

8Ω

−

15 I 2 VTh = 0 15( −0 0.83) VTh 0 VTh = −12.45 V

0.6V1

15 Ω

IN

B

Fig. 3.302

Step II Calculation of IN (Fig. 3.302) From Fig. 3.302,

The dependent source of 10 I2 depends on the controlling variable I2. When I 2 = 0, the dependent source vanishes, i.e. 10 I 2 = 0 as shown in Fig. 3.303. From Fig. 3.303,

2Ω

5Ω

A

I2 = 0

V1

I2

+ V1

15 V Ix

−

8Ω

0.6V1

IN

Iy B

Fig. 3.303

8( I x − I y )

…(i)

Applying KVL to Mesh 1, 15 − 5

8( I x − I y ) = 0

x

13I x

8

y

…(ii)

15

Applying KVL to Mesh 2, −8(

y

−

−8

) − 2I y = 0 x

+1 10 0

y

=0

…(iii)

Solving Eqs (ii) and (iii), I x = 2.27 A I y = 1 82 A Iy)

8( 2.27 1.82) = 3.66 V

V1

8( I x

IN

0 6 V1 = 0.6(3.6) = 2.16 A

A

For Mesh 3, Step III

Calculation of RN V −12.45 RN = Th = = −5 76 Ω IN 2.16

Step IV Norton’s Equivalent Network (Fig. 3.304)

2.16 A

−5.76 Ω

B

Fig. 3.304

3.78 Network Analysis and Synthesis

Example 3.69

Find Norton’s equivalent network of Fig. 3.305. A I1

0.5I1 + −

1Ω 2Ω 2V B

Fig. 3.305 0.5I1 + −

Solution Step I Calculation of VTh (Fig. 3.306) Applying KVL to the mesh, 2 − 2 I1 + 0 5 I1 −1 11 2 2.5 1

I1

0 0

VTh

−

2V

I1 = 0.8 8A

A

+

1Ω 2Ω

+

−

B

Fig. 3.306

Writing the VTh equation,

A

1 1 VTh 1(0.8) VTh

0 0

0.5I1 + −

I1

1Ω

VTh = 0 8 V Step II Calculation of IN (Fig. 3.307) When a short circuit is placed across the 1 Ω resistor, it gets shorted. I1 = 0

2V B

The dependent source of 0 5 I1 depends on the controlling variable I1. When I1 = 0, the dependent source vanishes, i.e. 0.5 I1 = 0 as shown in Fig. 3.308. 2 IN = = 1 A 2 Step III Calculation of RN

Fig. 3.307 A

2Ω

IN

2V

V 08 RN = Th = =08Ω IN 1

B

Fig. 3.308

Step IV Norton’s Equivalent Network (Fig. 3.309) A

1A

0.8 Ω

B

Fig. 3.309

IN

2Ω

3.4 Norton’s Theorem 3.79

Example 3.70

Find Norton’s equivalent network at the terminals A and B of Fig. 3.310. 6Ix 3Ω

2Ω A

Ix 9V

6Ω B

Fig. 3.310 6Ix

Solution Step I Calculation of VTh (Fig. 3.311) From Fig. 3.311, I x I1 Applying KVL to Mesh 1, 9 3(

3Ω +

…(i)

9

−

+

6Ω

+

A

VTh

I1

0 33II 2 = 9

1

Ix

9V

2) −6 1

1

2Ω

I2 −

−

…(ii)

B

Fig. 3.311

For Mesh 2, I2 6

1

2

6I x 0

6 I1 …(iii)

Solving Eqs (ii) and (iii), I1 = −1 A I 2 = −6 A Writing the VTh equation, 9 3( 1 2 ) + 2 2 VTh 0 9 3( −1 6) + 2( 6) − VTh = 0 VTh = −18 V Step II Calculation of IN (Fig. 3.312) From Fig. 3.312,

6Ix

Ix

I1 − I 3

…(i)

3Ω

Applying KVL to Mesh 1, 9 3(

1

2 ) − 6( 1

9

3)

3I 2 − 6

1

3

2Ω

I2

A

Ix

0 9

…(ii)

9V

6Ω I1

IN I3

For Mesh 2,

B

6

1

2

I2 6I x 6I3 = 0

Applying KVL to Mesh 3, −6( 3 − 1 ) − 2( 3 − 2 ) = 0 −6 1 − 2 I 2 + 8 3 = 0

6( I1

I3 ) …(iii)

Fig. 3.312

…(iv)

3.80 Network Analysis and Synthesis Solving Eqs (ii), (iii) and (iv),

A

I1 = 5 A I2 = 3 A I3 = 4 5 A I N I3 = 4 5 A Step III

−4 Ω

4.5 A

B

Calculation of RN

Fig. 3.313

V −18 RN = Th = = −4 Ω IN 45 Step IV Norton’s Equivalent Network (Fig. 3.313)

Example 3.71

Find Norton’s equivalent network to the left of terminal A-B in Fig. 3.314. A

6Ω

0.5I

4Ω

I B

Fig. 3.314 V

Solution Since the network does not contain any independent source, VTh = 0 IN = 0 But RN can be calculated by applying a known current source of 1 A at the terminals A and B as shown in Fig. 3.315. From Fig. 3.315, I=

A

6Ω

0.5I

4Ω

1A

I B

Fig. 3.315

V 6

Applying KCL at the node, V V + 0 5I + = 1 6 4 V ⎛V ⎞ V + 0 5⎜ ⎟ + = 1 ⎝ 6⎠ 4 6 ⎛ 1 0 5 1⎞ + ⎟V =1 ⎜⎝ + 6 6 4⎠ V =2 V 2 RN = = = 2 Ω 1 1 Hence, Norton’s equivalent network is shown in Fig. 3.316.

A

2Ω

B

Fig. 3.316

3.4 Norton’s Theorem 3.81

Example 3.72

Find the current through the 2 W resistor in the network shown in Fig. 3.317. 2Ω

−10 V

−2Ix

Ix

+ −

4Ω

2A

10 Ω

Fig. 3.317 Solution Step I Calculation of VTh (Fig. 3.318) From Fig. 3.318, Ix = 0

−2Ix

A B Ix V + Th − + −

−10 V

The dependent source of −2 Ix depends on the controlling variable Ix. When I x = 0, the dependent source vanishes, i.e. −2 x = 0 as shown in Fig. 3.319. I1 = 2

2A

10 Ω

Fig. 3.318 A B Ix VTh + −

Writing the VTh equation, −10 − VTh − 4 I1 = 0 −10 V

−10 − VTh − 4( 2) = 0

+ −

10 Ω

I1

Step II Calculation of IN (Fig. 3.320) From Fig. 3.320, Ix

4Ω

2A

VTh = −18 V

Fig. 3.319

I1 = 2

−2Ix

A IN B Ix

…(i)

I1

Mesh 1 and 2 will form a supermesh. Writing the current equation for supermesh, I2

4Ω

the

−10 V

+ −

…(ii)

4Ω

2A I1

I2

Applying KVL to the outer path of the supermesh,

10 Ω I3

Fig. 3.320

−10 − 4(

2

−

−4

2

+ 4 I 3 = 10

3)

=0 …(iii)

For Mesh 3, I3 2

1

3

( 2II x ) 0

2I x

2 I1 …(iv)

3.82 Network Analysis and Synthesis Solving Eqs (ii), (iii) and (iv), I1 = 4 5 A I2 = 6 5 A I3 = 9 A I N I1 = 4 5 A Step III

A IL

Calculation of RN RN =

VTh −18 = = −4 Ω IN 45

−4 Ω

4.5 A

2Ω

Step IV Calculation of IL (Fig. 3.321)

B

−4 I L = 4.5 × =9A −4 + 2

Example 3.73

Fig. 3.321

Find the current through the 2W resistor in the network of Fig. 3.322. −

Vi

+

1Ω 2Ω 5V

− +

4Vi

Fig. 3.322 −

Solution Step I Calculation of VTh (Fig. 3.322) From Fig. 3.323, 5 Vi 4Vi = 0 Vi = −1 V

Vi

+ +

1Ω 5V

− +

Writing the VTh equation,

VTh

4Vi −

−4Vi − VTh = 0 VTh = −4 4Vi = −4( −1) = 4 V

Applying KVL to the mesh, −4Vi − 1I N = 0 I N = −4 4Vi = −4( −5) = 20 A

B

Fig. 3.323 −

Step II Calculation of IN (Fig. 3.324) From Fig. 3.324, 5 Vi 0 Vi = −5 V

A

Vi

+

A

1Ω 5V

IN − +

4Vi B

Fig. 3.324

3.4 Norton’s Theorem 3.83

Step III

Calculation of RN

A

RN =

VTh 4 = =02Ω IN 20

2Ω

0.2 Ω

20 A

Step IV Calculation of IL (Fig. 3.325)

B

02 I L = 20 × = 1.82 82 A 0 2+2

Example 3.74

Fig. 3.325

Find the current in the 2 W resistor in the network of Fig. 3.326. 1Ω

Ix

10 V

2Ix

3Ω

1A

2Ω

Fig. 3.326 Solution Step I Calculation of VTh (Fig. 3.327) Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I 2 I1 = 1 …(i) Applying KVL to the outer path of the supermesh, 10 −1 1 1 3I 2 = 0 I1 3I 2 10 …(ii) Solving Eqs (i) and (ii),

1Ω

Ix

2Ix +

+ 10 V

3Ω

1A I1

VTh

−

I2

A

−

B

Fig. 3.327

I1 = 1 75 A I 2 = 2 75 A Writing the VTh equation, 3 2 VTh 0 3( 2.75) − VTh = 0 VTh = 8 25 V Step II Calculation of IN (Fig. 3.328) From Fig. 3.328, I x I1 …(i) Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh, I 2 I1 = 1 …(ii) Applying KVL to the outer path of the supermesh, 10 −1 1 1 3( I 2 − I 3 ) = 0

1Ω

Ix

2Ix

1A

10 V I1

I2

Fig. 3.328

3Ω

IN I3

3.84 Network Analysis and Synthesis 3I 3 = 10

I1 3I 2

…(iii)

For Mesh 3, I3 2

1

3

2I x 0

2 I1 …(iv)

Solving Eqs (ii), (iii) and (iv), I1 = −3 5 A I 2 = −2 5 A I 3 = −7 A IN

I 3 = −7 A

Calculation of RN

Step III

RN =

A IL

VTh 8 25 = = −1.18 Ω IN −7

−1.18 Ω

−7 A

Step IV Calculation of IL (Fig. 3.329)

B

−1.18 I L = −7 × = 10.07 07 A −1.18 + 2

Example 3.75

2Ω

Fig. 3.329

Find the current through the 10 W resistor for the network of Fig. 3.330. 3Ix − +

2Ω

Ix

5Ω

10 Ω

10 V

Fig. 3.330 3Ix

Solution

− +

Step I Calculation of VTh (Fig. 3.331) Applying KVL to the mesh, 10 − 2

x

3I x − 5 3I

x

0

+

2Ω

+

5Ω

VTh

−

10 V

−

I x = 2.55 A Writing the VTh equation, 5

0

3Ix

5( 2.5) VTh

0

− +

VTh = 12.5 V Step II Calculation of IN (Fig. 3.332) From Fig. 3.332, Ix = 0

B

Fig. 3.331

VTh

x

A

Ix

2Ω

Ix

5Ω

10 V

Fig. 3.332

IN

3.4 Norton’s Theorem 3.85

The dependent source of 3 Ix depends on the controlling variable Ix. When I x = 0, the dependent source 3 Ix vanishes, i.e. 3 Ix = 0 as shown in Fig. 3.333. N

Step III

A

2Ω

10 = =5A 2

IN

10 V B

Calculation of RN

Fig. 3.333 V 12.5 RN = Th = =25Ω IN 5

A IL

Step IV Calculation of IL (Fig. 3.334) IL = 5 ×

10 Ω

2.5 Ω

5A

25 =1A 2 5 + 10

B

Fig. 3.334

Example 3.76

Find the current through the 5 W resistor in the network of Fig. 3.335. 2Ω

12 V

4Ω

Ix

+ −

5Ω

4Ix

Fig. 3.335 Solution Step I Calculation of VTh (Fig. 3.336) Applying KVL to the mesh, 12 − 2 I x 4 I x − 4 x 0 12 V 12 − 10 I x = 0 I x = 1.2 2A Writing the VTh equation, 12 − 2 x VTh 0 12 − 2(1.2) − VTh = 0 VTh = 9 6 V Step II Calculation of IN (Fig. 3.337) From Fig. 3.337, I x I1 …(i) Applying KVL to Mesh 1, 12 − 2 1 0 12 V I1 = 6 A …(ii) Applying KVL to Mesh 2, −4 2 − 4 I x = 0 −4 2 − 4 I1 = 0 …(iii) Solving Eqs (ii) and (iii), I 2 = −6 A I N I1 − I 2 = 6 − ( −6) = 12 A

2Ω +

4Ω

Ix −

+ A V Th − B

+ −

4Ix

+ −

4Ix

Fig. 3.336

2Ω

4Ω

Ix

IN I1

I2

Fig. 3.337

3.86 Network Analysis and Synthesis Step III

Calculation of RN

IL

VTh 9 6 = =08Ω IN 12

RN =

12 A

5Ω

0.8 Ω

Step IV Calculation of IL (Fig. 3.338) I L = 12 ×

Example 3.77

08 = 1.66 66 A 0 8+5

Fig. 3.338

Find the current through the 10 W resistor for the network of Fig. 3.339. 5Ω +

4Ω

0.5Vx

Vx

10 Ω −

5V

Fig. 3.339 Solution Step I Calculation of VTh (Fig. 3.340) For the mesh, 0 5Vx = −0.5VTh

I

5Ω +

4Ω 5V

0.5Vx

VTh = Vx

I −

Writing the VTh equation, 5 4 I − 0 VTh

5Ω

5 4( −0.5VTh ) − VTh = 0

Step II Calculation of IN (Fig. 3.341) From Fig. 3.341, Vx = 0

+

4Ω

IN Step III

0.5Vx

5V

Vx −

A

IN

B

Fig. 3.341

The dependent source of 0 5V Vx depends on the controlling variable Vx. When Vx = 0, the dependent source vanishes, i.e. 0.5 Vx = 0 as shown in Fig. 3.342. 5 5 = = A 4+5 9

B

Fig. 3.340

0

VTh = −5 V

A

5Ω A

4Ω

IN

5V B

Calculation of RN RN =

VTh −5 = = −9 Ω 5 IN 9

Step IV Calculation of IL (Fig. 3.343) IL =

5 −9 × = −5 A 9 −9 + 10

Fig. 3.342 A IL 5A 9

−9 Ω

10 Ω

B

Fig. 3.343

3.4 Norton’s Theorem 3.87

Example 3.78

Find the current through the 10 W resistor in the network shown in Fig. 3.344. 1000 Ω

I1 + + 2Vx −

12 V

25 Ω

Vx

5I1

10 Ω

−

Fig. 3.344 Solution Step I Calculation of VTh (Fig. 3.345) From Fig. 3.345, Vx

2 ( I1 ) = −125 125 I1 …(i)

1000 Ω

I1

+ −

12 V

+

2Vx

5I1

Vx

A

+

25 Ω

VTh

−

Applying KVL to Mesh 1,

−

12 −1000I 1000 1 − 2Vx 0 12 −1000 1000 I1 2( 125 I1 ) = 0

B

Fig. 3.345

…(ii)

I1 = 0.016 A Vx 12 I1 = −125(0.016) = −2 V Writing the VTh equation, VTh

1000 Ω

I1

A

Vx = −2 V

Step II Calculation of IN (Fig. 3.346) From Fig. 3.346,

+ −

12 V

+

2Vx

5I1

Vx

25 Ω

IN

−

Vx = 0

B

The dependent source of 2Vx depends on the controlling variable Vx. When Vx = 0, the dependent source vanishes, i.e. 2 Vx = 0 as shown in Fig. 3.347. I1 = IN Step III

12 = 0.012 A 1000 5 I1 5(0.012) = −0.06 A

Fig. 3.346 1000 Ω

A

12 V

5I1

IN

B

Fig. 3.347

Calculation of RN RN =

I1

VTh −2 = = 33.33 Ω IN −0 06

A IL

−0.06 A

33.33 Ω

10 Ω

Step IV Calculation of IL (Fig. 3.348) I L = −0 06 ×

33.33 = −0.046 046 A 33.33 + 10

B

Fig. 3.348

3.88 Network Analysis and Synthesis

Example 3.79

Find the current through the 5 W resistor for the network of Fig. 3.349. 4V

1Ω

2Ω +

1Ω

Vx

− + −

5Ω 4Vx

2A

Fig. 3.349 Solution Step I Calculation of VTh (Fig. 3.350) From Fig. 3.350, Vx 2 I For the mesh, I =2 Vx = 2( 2) = 4 V Writing the VTh equation, 4Vx

2 I + 1I

I 2 )] − 2

2

2 I1 − I 2

I1 + 4

Step III

…(i)

2Ω +

Vx

1Ω

− + −

…(ii) I

2A

4Vx

+ A VTh − B

Fig. 3.350

4V

…(i) …(ii)

1Ω

I2

2Ω +

1Ω

Vx

−

A + −

I1

2A

0

11I1 11I 2 = −4 Solving Eqs (ii) and (iii),

1Ω

4 VTh = 0

4( 4) + 2( 2) + 2 + 4 − VTh = 0 VTh = 26 V Step II Calculation of IN (Fig. 3.351) From Fig. 3.351, Vx 2( I1 − I 2 ) For Mesh 1, I1 = 2 Applying KVL to Mesh 2, 4Vx 2( I 2 I1 ) 1( 2 1 ) + 4 = 0 4[2( I1

4V

IN

4Vx B

…(iii) Fig. 3.351

I1 = 2 A I 2 = 2 36 A I N I 2 = 2 36 A

Calculation of RN RN =

VTh 26 = = 11.02 Ω IN 2 36

A IL

2.36 A

11.02 Ω

Step IV Calculation of IL (Fig. 3.352) 11.02 I L = 2 36 × = 1.62 A 11.02 + 5

5Ω

B

Fig. 3.352

3.4 Norton’s Theorem 3.89

Example 3.80

Find the current through the 1 W resistor in the network of Fig. 3.353. 6Ω

Ix

3Ω

3Ix

12 V

1Ω

Fig. 3.353 Solution Step I Calculation of VTh (Fig. 3.354) From Fig. 3.354,

6Ω

Ix

I1

1

2

0

+

+

…(i)

3Ω

3Ix

12 V

Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh, I 2 I1 = 3I x 3I1 4

Ix

I1

I2

−

VTh −

Fig. 3.354

…(ii)

Applying KVL to the outer path of the supermesh, 12 − 6 1 3I 2 = 0 6

33II 2 = 12

1

…(iii)

Solving Eqs (ii) and (iii), I1 = 0 67 A I 2 = 2 67 A

6Ω

Ix A

Writing the VTh equation, 3

VTh

2

0

I = 3I x

3I1

4

0

1

2

IN B

Step II Calculation of IN (Fig. 3.355) When a short circuit is placed across a 3 Ω resistor, it gets shorted as shown in Fig. 3.356. From Fig. 3.356, I x I1 …(i) Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh, I2

3Ω

3Ix

12 V

3( 2.67) − VTh = 0 VTh = 8 V

Fig. 3.355 6Ω

Ix A 3Ix

12 V I1

IN I2 B

…(ii)

Fig. 3.356

Applying KVL to the outer path of the supermesh, 12 − 6

1

0

I1 = 2

…(iii)

3.90 Network Analysis and Synthesis Solving Eqs (ii) and (iii), I1 = 2 A I2 = 8 A IN Step III

A IL

I2 = 8 A

Calculation of RN

1Ω

1Ω

8A

V 8 RN = Th = = 1 Ω IN 8

B

Step IV Calculation of IL (Fig. 3.357)

Fig. 3.357 1 IL = 8 × =4A 1+1

Example 3.81

Find the current through the 1.6 W resistor in the network of Fig. 3.358. 3Ix − + Ix

1Ω

10 A

1.6 Ω

6Ω

Fig. 3.358 Solution Step I Calculation of VTh (Fig. 3.359) From Fig. 3.359,

3Ix − + Ix

Ix

I1 − I 2 …(i)

I1

For Mesh 1, I1 = 10

2

6Ω I2

VTh

− −

…(ii)

A

B

Fig. 3.359

Applying KVL to Mesh 2, −1(

+

1Ω

10 A

+

− 1 ) + 3I x − 6

− I 2 + I1 + 3(

2

=0

1 − 2 ) − 6I2 = 0 4 1 10 0 2 0

Solving Eqs (ii) and (iii), I1 = 10 A I2 = 4 A Writing the VTh equation, 6 2 VTh 0 6( 4) − VTh = 0 VTh = 24 V

…(iii)

3.5 Maximum Power Transfer Theorem 3.91

Step II Calculation of IN (Fig. 3.360) When a short circuit is placed across the 3 Ω resistor, it gets shorted as shown in Fig. 3.361. From Fig. 3.361, I1 − I 2

Ix

…(i)

3Ix − +

A

Ix

1Ω

10 A

6Ω

IN

For Mesh 1, I1 = 10

B

…(ii) Fig. 3.360

Applying KVL to Mesh 2,

3Ix

−1( 2 − 1 ) + 3I x = 0 − I 2 + I1 + 3( 1 − 2 ) = 0 4 1 4I2 = 0 Solving Eqs (ii) and (iii), I1 = 10 A

− +

…(iii)

1Ω

10 A

IN

I1

I2 B

I 2 = 10 A I N I 2 = 10 A Step III

A

Ix

Fig. 3.361

Calculation of RN

A IL

V 24 RN = Th = = 2.4 Ω IN 10

2.4 Ω

10 A

1.6 Ω

Step IV Calculation of IL (Fig. 3.362) I L = 10 ×

3.5

2.4 =6A 2.4 + 1.6

B

Fig. 3.362

MAXIMUM POWER TRANSFER THEOREM

It states that ‘the maximum power is delivered from a source to a load when the load resistance is equal to the source resistance.’ Proof From Fig. 3.363,

RS

I=

Power delivered to the load RL

V Rs

RL 2

P = I RL =

V

V 2 RL

( Rs + RL ) 2 To determine the value of RL for maximum power to be transferred Fig. 3.363 to the load, dP =0 dR RL dP d V2 = RL dR RL dRL ( Rs RL ) 2 =

V 2 [( Rs + RL ) 2

( 2 RL )( Rs

( Rs + RL )

4

RL I

RL )]

Network illustrating maximum power transfer theorem

3.92 Network Analysis and Synthesis ) 2 − 2 RL (

( Rs 2

)=0

RL 2 + 2 Rs RL − 2 RL Rs − 2 RL2 = 0 Rs

RL

Hence, the maximum power will be transferred to the load when load resistance is equal to the source resistance. Steps to be followed in Maximum Power Transfer Theorem 1. 2. 3. 4.

RTh A

Remove the variable load resistor RL. Find the open circuit voltage VTh across points A and B. Find the resistance RTh as seen from points A and B. Find the resistance RL for maximum power transfer. RL

RL = RTh

VTh IL

B

RTh

Thevenin’s equivalent network

Fig. 3.364

5. Find the maximum power (Fig. 3.364). IL =

VTh V = Th RTh RL 2 RTh I L2 RL =

Pmax

2 VTh 2 4 RTh

× RTh =

2 VTh 4 RTh

Example 3.82 Find the value of resistance RL in Fig. 3.365 for maximum power transfer and calculate maximum power. RL

2Ω

2Ω

3V

10 V

6V 2Ω

Fig. 3.365 Solution Step I Calculation of VTh (Fig. 3.366) Applying KVL to the mesh, 3 2I − 2

2Ω +

6=0 I = −0.75 A

+

+

VTh

2Ω

3V I

Writing the VTh equation,

2Ω

5.5 V

= 5.5 V( terminal B is positive w.r.t A)

B − 10 V

− 6V

6 2 I − VTh − 10 = 0 VTh = 6 2 I − 10 = 6 2( −0.75) 10

A −

Fig. 3.366

3.5 Maximum Power Transfer Theorem 3.93

Step II

2Ω

Calculation of RTh (Fig. 3.367) RTh = (

A

RTh

B

) + 2 = 3Ω 2Ω

Step III Calculation of RL For maximum power transfer, RL

RTh = 3 Ω 2Ω

Step IV Calculation of Pmax (Fig. 3.368)

Fig. 3.367 3Ω A

Pmax =

2 VTh

4 RTh

2

=

( .5) = 2.52 W 4 3

3Ω

5.5 V

B

Fig. 3.368

Example 3.83 Find the value of resistance RL in Fig. 3.369 for maximum power transfer and calculate maximum power. 5Ω

1Ω RL

5Ω

4A

10 V

8V

Fig. 3.369 Solution Step I Calculation of VTh (Fig. 3.370) I 2 I1 = 4 Applying KVL to the outer path, 8 1I1 5 1 5 I 2 − 10 = 0 −6 6 1 5I 2 = 2

5Ω −

...(i)

1Ω + 8V

+

+ A V Th − B I1

4A I2

5Ω − 10 V

Fig. 3.370

I1 = −2 A I2 = 2 A

5Ω

...(ii)

Writing the VTh equation,

Step II

−

...(ii)

Solving Eqs (i) and (ii),

8 1I1 − VTh = 0 VTh = 8 I1

+

1Ω

8 − ( −2 2) 10 V

A R Th B

Calculation of RTh (Fig. 3.371) RTh = 10 1 = 0.91 Ω

Fig. 3.371

5Ω

3.94 Network Analysis and Synthesis 0.91 Ω

Step III Calculation of RL For maximum power transfer,

A

RTh = 0 91 Ω

RL

0.91 Ω

10 V

Step IV Calculation of Pmax Pmax =

2 VTh ( )2 = = 27.47 W 4 RTh 4 0.91

B

Fig. 3.372

Example 3.84 Find the value of the resistance RL in Fig. 3.373 for maximum power transfer and calculate the maximum power. 10 Ω

50 A

2Ω

5Ω

RL

3Ω

Fig. 3.373 10 Ω

Solution

+

Step I Calculation of VTh (Fig. 3.374) For Mesh 1, I1 = 50

2Ω −

+

−

+ −

5Ω

50 A I1

+

+

3Ω I2

− +

VTh

−

−

Applying KVL to Mesh 2, −5(

2

− 1 ) − 2I 2 − 3 5 1 − 10 10

2

Fig. 3.374

=0

2 =0 I1 = 2 I 2 I 2 = 25 A

10 Ω

Step II

2Ω A

5Ω

VTh = 3I 2 = 3( 25) = 75 V

3Ω

RTh = ( + ) 3 = 2.1 Ω

Fig. 3.375 2.1 Ω

Step III Calculation of RL For maximum power transfer, RL

RTh = 2.1 Ω

A

2.1 Ω

75 V

Step IV Calculation of Pmax (Fig. 3.376) Pmax =

2 VTh ( )2 = = 669.64 W 4 RTh 4 2.1

RTh B

Calculation of RTh (Fig. 3.375)

B

Fig. 3.376

A

B

3.5 Maximum Power Transfer Theorem 3.95

Example 3.85 Find the value of resistance RL in Fig. 3.377 for maximum power transfer and calculate maximum power. 3Ω

5Ω

RL

2Ω

6A

10 V

4Ω

Fig. 3.377 3Ω

Solution

−

Step I Calculation of VTh (Fig. 3.378) Writing the current equation for the supermesh, I2

I1 = 6

5Ω +

...(i)

10 V

+

+

I1

6A I2

2Ω

1

2I 2 = 0

5

1

2 I 2 = 10

VTh

− −

Applying KVL to the supermesh, 10 − 5

A

B

4Ω

...(ii)

Fig. 3.378 3Ω

Solving Eqs (i) and (ii), I1 = −0 29 A I 2 = 5 71 A

A

5Ω

2Ω

RTh

Writing the VTh equation, 2 I 2 = 11.42 V

VTh Step II

B

4Ω

Calculation of RTh (Fig. 3.379) RTh = (

Fig. 3.379

) + 3 + 4 = 8.43 Ω

Step III Calculation of RL For maximum power transfer, RL

8.43 Ω A

RTh = 8 43 Ω 8.43 Ω

11.42 V

Step IV

Calculation of Pmax (Fig. 3.380) Pmax =

2 VTh ( .42) 2 = = 3.87 W 4 RTh 4 8.43

B

Fig. 3.380

Example 3.86 Find the value of resistance RL in Fig. 3.381 for maximum power transfer and calculate the maximum power.

3.96 Network Analysis and Synthesis 10 Ω

RL

120 V

6A

5Ω

Fig. 3.381 10 Ω

Solution Step I Calculation of VTh (Fig. 3.382) Applying KVL to Mesh 1, 120 −10 10 I1 5(

1

2)

+

−

120 V

0

I1

15 I1 5 I 2 = 120

...(i)

Writing current equation for Mesh 2, I 2 = −6 Solving Eqs (i) and (ii),

5Ω − + I2

B −

6A

10 Ω

Writing the VTh equation,

A R Th B

120 −10 10 I1 − VTh = 0 Step II

+ −

V Th

Fig. 3.382

...(ii)

I1 = 6 A

VTh = 120 − 10(6)

A +

5Ω

60 V

Calculation of RTh (Fig. 3.383)

Fig. 3.383

RTh = 10 5 = 3.33 Ω 3.33 Ω

Step III Calculation of RL For maximum power transfer, RL

A

RTh = 3 33 Ω

3.33 Ω

60 V

Step IV Calculation of Pmax (Fig. 3.384) Pmax =

2 VTh ( )2 = = 270.27 W 4 RTh 4 3.33

B

Fig. 3.384

Example 3.87 Find the value of resistance RL in Fig. 3.385 for maximum power transfer and calculate the maximum power. 10 Ω

RL

3A

20 V

Fig. 3.385

25 Ω

6Ω

3.5 Maximum Power Transfer Theorem 3.97 10 Ω

Solution Step I

+

Calculation of VTh (Fig. 3.386) I1 = 3

A + V Th B −

…(i)

Applying KVL to Mesh 2, −25( 2 − 1 ) − 10 I 2 − 6 2 = 0 −25 I1 + 41I 2 = 0

−

+ − 25 Ω

3A I1

20 V

…(ii)

+

− +

6Ω −

I2

Fig. 3.386

Solving Eqs (i) and (ii),

10 Ω

I 2 = 1.83 A Writing the VTh equation, 20 + VTh − 10 I 2

6 2 0 VTh = −20 +10 10 (1.83) 6 (1.83)

R Th

A

25 Ω

9.28 V

Calculation of RTh (Fig. 3.387)

Step II

6Ω

B

Fig. 3.387

RTh = 25 || (10 + 6) = 9.76 Ω

9.76 Ω

Step III Calculation of RL For maximum power transfer, RL RTh = 9 76 Ω

A

9.76 Ω

9.28 V

Step IV Calculation of Pmax (Fig. 3.388) Pmax =

2 VTh ( .28) 2 = = 2.21 W 4 RTh 4 × 9.76

B

Fig. 3.388

Example 3.88 Find the value of resistance RL in Fig. 3.389 for maximum power transfer and calculate maximum power. 1Ω

2Ω

1A

5V

5Ω

RL

3Ω

10 Ω

Fig. 3.389 Solution Step I

Calculation of VTh (Fig. 3.390) 1Ω +

2Ω −

5V I1

+ − 1A I2

+

5Ω −

3Ω

10 Ω − +

Fig. 3.390

+

+

I3

−

A

VTh −

B

3.98 Network Analysis and Synthesis Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh, I2

I1 = 1

…(i)

Writing the voltage equation for the supermesh, 5 1I1 10 0( I 2 − I 3 ) = 0

…(ii)

I1 +10 10 I 2 − 10 I 3 = 5 Applying KVL to Mesh 3, −10(

3

−

2) − 2 3

− 3I 3 = 0 −10 I 2 +15 15 I 3 = 0

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 0 38 A I 2 = 1 38 A I 3 = 0 92 A Writing the VTh equation, VTh Step II

3I 3 = 2.76 V

Calculation of RTh (Fig. 3.391) 1Ω

2Ω

5Ω A

3Ω

10 Ω

RTh B

(a) 2Ω

5Ω

5Ω A

3Ω

0.91 Ω

RTh

A

1.48 Ω

RTh

B

B

(b)

(c)

Fig. 3.391 RTh = 6 8 Ω

6.48 Ω A

Step III Calculation of RL For maximum power transfer, RL

RTh = 6 48 Ω

6.48 Ω

2.76 V

Step IV Calculation of Pmax (Fig. 3.392) Pmax =

2 VTh ( .76) 2 = = 0.29 W 4 RTh 4 × 6.48

B

Fig. 3.392

3.5 Maximum Power Transfer Theorem 3.99

Example 3.89 For the network shown in Fig. 3.393, find the value of the resistance RL for maximum power transfer and calculate the maximum power. 2Ω

RL

1Ω

2Ω 2A

8V

3A 6V

Fig. 3.393 Solution Step I Calculation of VTh (Fig. 3.394) 2Ω +

1Ω −

+

A −

+

VTh

B −

2Ω 8V I1

2A I2

3A 6V

Fig. 3.394 I1 = 2 I 2 = −3 A

I2

Solving Eqs (i), and (ii), I1 = −5 A

…(i) …(ii) 2Ω

1Ω

RTh A

Writing the VTh equation, 8 2 I1 1

Step II

2

VTh

B

2Ω

6=0 VTh = 8 2( −5 5) ( 3) − 6 15 V Fig. 3.395

Calculation of RTh (Fig. 3.395) RTh = 5 Ω

Step III Calculation of RL For maximum power transfer, RL

5Ω A

RTh = 5 Ω 5Ω

15 V

Step IV Calculation of Pmax (Fig. 3.396) Pmax =

2 VTh ( )2 = = 11.25 W 4 RTh 4 × 5

B

Fig. 3.396

3.100 Network Analysis and Synthesis

Example 3.90 For the value of resistance RL in Fig. 3.397 for maximum power transfer and calculate the maximum power. RL

15 Ω

5Ω

18 Ω

15 Ω

10 Ω

27 Ω

20 Ω

9Ω

27 Ω

100 V

Fig. 3.397 Solution Step I Calculation of VTh (Fig. 3.398) 15 Ω

A +

5Ω

18 Ω

B

VTh

−

15 Ω

10 Ω

27 Ω

20 Ω

9Ω

27 Ω

100 V

Fig. 3.398 By star-delta transformation (Fig. 3.399), I=

A

100 = 2.08 A 5 + 5 + 20 + 9 + 9

VTh

= 100 − 14( 2.08) = 70.88 V

B

−

9Ω

+

− 5Ω

9I = 0 VTh = 100 − 14 I

VTh

5Ω

+

Writing the VTh equation, 100 − 5

5Ω

+

9Ω 20 Ω

I 100 V

Fig. 3.399

9Ω

−

3.5 Maximum Power Transfer Theorem 3.101

Step II

Calculation of RTh (Fig. 3.400) A

RTh

B 9Ω

5Ω

9Ω

5Ω

5Ω

9Ω 20 Ω (a)

5Ω

A

RTh

B

5Ω 5Ω

20 Ω

5Ω

9Ω

A

RTh

9Ω

B

9Ω

14 Ω

9Ω

34 Ω

5Ω

A

RTh

B

9Ω

9.92 Ω (c)

(b)

(d)

Fig. 3.400 RTh = 23.92 Ω 23.92 Ω

Step III Calculation of RL For maximum power transfer, RL

A

RTh = 23.92 Ω

23.92 Ω

70.88 V

Step IV Calculation of Pmax (Fig. 3.401) Pmax =

2 VTh ( .88) 2 = = 52.51 W 4 RTh 4 × 23.92

B

Fig. 3.401

Example 3.91 For the value of resistance RL in Fig. 3.402 for maximum power transfer and calculate the maximum power. 2A

5Ω 10 Ω

20 Ω

80 V

20 V RL

Fig. 3.402

3.102 Network Analysis and Synthesis Solution Step I Calculation of VTh (Fig. 3.403) Applying KVL to Mesh 1, 80 − 5

1

10 0(

1

2)

20 0(

1

2)

2A

5Ω +

−

20 0

− +

35 I1 30 I 2 = 60 …(i)

10 Ω

+ −

80 V I1

Writing the current equation for Mesh 2, I2 = 2

I2

A +

− +

20 Ω

+ − 20 V

V Th B −

…(ii)

Solving Eqs (i) and (ii),

Fig. 3.403

I1 = 3.43 A

5Ω

Writing the VTh equation, VTh

10 Ω

20 ( I1 − I 2 ) − 20 = 0 VTh = 20(3.43 − 2) + 20 = 48.6 V

Step II

20 Ω A R Th

Calculation of RTh (Fig. 3.404)

B

RTh = 15 || 20 = 8 57 Ω Fig. 3.404 Step III Calculation of RL For maximum power transfer, RL

8.57 Ω A

RTh = 8 57 Ω

Pmax =

8.57 Ω

48.6 V

Step IV Calculation of Pmax (Fig. 3.405) 2 VTh ( .6) 2 = = 68.9 W 4 RTh 4 × 8.57

B

Fig. 3.405

Example 3.92 For the value of resistance RL in Fig. 3.406 for maximum power transfer and calculate the maximum power. 10 Ω

20 Ω

100 V RL 30 Ω

Fig. 3.406

40 Ω

3.5 Maximum Power Transfer Theorem 3.103 I2

I1 10 Ω + −

Solution Step I Calculation of VTh (Fig. 3.407) 100 =25A 10 + 30 100 I2 = = 1 66 A 20 + 40 I1 =

+

100 V

A

+ 30 Ω

Writing the VTh equation,

V Th

+

− B

+ −

−

20 Ω −

40 Ω

Fig. 3.407

VTh 10 I1 − 20 I 2 = 0 VTh 20 I 2 −10 − 10 I1 = 20(1.66) − 10( 2.5) = 8.2 V Step II

Calculation of RTh (Fig. 3.408) 10 Ω

20 Ω

A

R Th

30 Ω

B 40 Ω

Fig. 3.408 Redrawing the network (Fig. 3.409), 10 Ω

20 Ω

A

RTh = (

||

)+(

||

B

) = 0.83 Ω 30 Ω

40 Ω

Fig. 3.409 Step III Value of RL For maximum power transfer, RL

20.83 Ω A

RTh = 20.83 Ω

Step IV Calculation of Pmax (Fig. 3.410) Pmax =

20.83 Ω

8.2 V

2 VTh ( .2) 2 = = 0.81 W 4 RTh 4 × 20.83

B

Fig. 3.410

3.104 Network Analysis and Synthesis

Example 3.93 For the value of resistance RL in Fig. 3.411 for maximum power transfer and calculate the maximum power. 6Ω

RL

72 V 2Ω 3Ω

4Ω

Fig. 3.411 Solution Step I Calculation of VTh (Fig. 3.412) Applying KVL to Mesh 1, 72 − 6

1

3( I1 − I 2 ) = 0 9

1

72 V

3I 2 = 72 …(i)

+ I1

+

Applying KVL to Mesh 2, −3(

2

A

6Ω + −

3Ω

− 1 ) − 2I 2 − 4

2

=0

−3 1 + 9 I 2 = 0 …(ii)

2Ω

− I2

−

Fig. 3.412

Solving Eqs (i) and (ii), I1 = 9 A I2 = 3 A Writing the VTh equation, VTh

Step II

6 I1 − 2 I 2 = 0 VTh 6 I1 + 2 I 2 = 6(9) + 2(3) = 60 V

Calculation of RTh (Fig. 3.413) A

6Ω

2Ω

R Th B

6Ω

A

3Ω 4Ω

2Ω 3Ω

R Th

4Ω

B

Fig. 3.413 RTh = [(6 || 3) + ] || 4 = 2 Ω

−

+ V Th

−

B

+ 4Ω

3.5 Maximum Power Transfer Theorem 3.105 2Ω

Step III Calculation of RL For maximum power transfer, RL

A

RTh = 2 Ω

2Ω

60 V

Step IV Calculation of Pmax (Fig. 3.414) Pmax =

2 VTh ( )2 = = 450 W 4 RTh 4 × 2

B

Fig. 3.414

Example 3.94 For the network shown in Fig. 3.415 find the value of the resistance RL for maximum power transfer and calculate maximum power. 10 Ω

5Ω

25 A

2Ω

10 A

RL

10 Ω

30 V

10 Ω

30 V

Fig. 3.415 Solution Step I Calculation of VTh (Fig. 3.416) 10 Ω

5Ω

25 A

2Ω + A V Th

10 A

− B

Fig. 3.416 By source transformation, the current source of 25 A and the 5 Ω resistor is converted into an equivalent voltage source of 125 V and a series resistor of 5 Ω. Also the voltage source of 30 V is connected across the 10 Ω resistor. Hence, the 10 Ω resistor becomes redundant (Fig. 3.417). 2Ω

10 Ω

+

5Ω

10 A

A V Th −

B

125 V

Fig. 3.417

30 V

10 Ω

3.106 Network Analysis and Synthesis 10 Ω

Applying KCL at the node, VTh 12 V − 30 − 10 + Th =0 15 2

2Ω

A 5Ω

B

VTh = 58.81 V Step II

10 Ω

R Th

Calculation of RTh (Fig. 3.418)

Fig. 3.418

RTh = 15 || 2 = 1.76 Ω Step III Value of RL For maximum power transfer RL

A R Th

15 Ω

2Ω

B

RTh = 1 76 Ω

Step IV Calculation of Pmax (Fig. 3.420)

Fig. 3.419 1.76 Ω

Pmax

A 1.76 Ω

58.81 V

V2 ( .81) 2 = Th = = 491.28 W 4 RTh 4 × 1.76

B

Fig. 3.420

Example 3.95 For the network shown in Fig. 3.421, find the value of the resistance RL for maximum power transfer and calculate maximum power. 12 V

2Ω 4Ω 10 V

2Ω

6V

5Ω

10 Ω

RL 4A

8V

Fig. 3.421 Solution Step I Calculation of VTh (Fig. 3.422) 12 V

2Ω +

2Ω

−

4Ω

6V

+

A

+ 5Ω

10 Ω

4 A 10 V I

V Th

− 8V

−

Fig. 3.422

B

3.5 Maximum Power Transfer Theorem 3.107

Applying KVL to the outer path, 10 − 2 I 12 5 I − 8

0

I=−

10 = −1.43 A 7

Writing the VTh equation, 8 5 I + 6 VTh 0 VTh = 8 6 + 5 I Step II

8 + 6 5( −1 1.43)

6.85 V

Calculation of RTh (Fig. 3.423) 2Ω

2Ω A

4Ω 10 Ω

RTh = ( || ) + 2 = 3.43 Ω

5Ω

R Th

B

Fig. 3.423 Step III Value of RL For maximum power transfer,

3.43 Ω

RL

RTh = 3 43 Ω

Step IV Calculation of Pmax (Fig. 3.424) Pmax =

A 3.43 Ω

6.85 V

2 VTh ( .85) 2 = = 3.42 W 4 RTh 4 × 3.43

B

Fig. 3.424

EXAMPLES WITH DEPENDENT SOURCES Example 3.96 For the network shown in Fig. 3.425, find the value of RL for maximum power transfer. Also, calculate maximum power. I

20 Ω

10 I

40 Ω + −

RL 50 V

Fig. 3.425

3.108 Network Analysis and Synthesis Solution Step I Calculation of VTh (Fig. 3.426) Applying KVL to the mesh, 10 I

20 I

I

20 Ω

40 I − 50 = 0 I = −1 A

VTh

+ −

10 I

Writing the VTh equation,

40 Ω

A + V Th B

−

50 V

40 I − 50 = 0

VTh − 40( −1) − 50 = 0

Fig. 3.426

VTh = 10 V I

Step II Calculation of IN (Fig. 3.427) From Fig. 3.427, I Applying KVL to Mesh 1, 10 I 10 I 2

I2

20 I1 = 0 20 I1 = 0

20 Ω

…(i)

+ −

10I

IN B

I1

−40 I 2 − 50 = 0 I 2 = −1.25 25 A Solving Eqs (i), (ii) and (iii), I1 = −0.625 A I N I1 − I 2 = −0.625 + 1.25 = 0.625 A Step III Calculation of RN V 10 RTh = Th = = 16 Ω IN 0.625 Step IV Calculation of RL For maximum power transfer, RL

50 V

Fig. 3.427 …(iii)

16 Ω

A

RTh = 16 Ω

16 Ω

10 V

Calculation of Pmax (Fig. 3.428) Pmax =

I2

…(ii)

Applying KVL to Mesh 2,

Step V

40 Ω

A

VTh ( )2 = = 1.56 W 4 RTh 4 × 16

B

Fig. 3.428

Example 3.97 For the network shown in Fig. 3.429, calculate the maximum power that may be dissipated in the load resistor RL. − +

3Ω

2Ix Ix

10 A

4Ω

Fig. 3.429

6Ω

RL

3.5 Maximum Power Transfer Theorem 3.109 2Ix

Solution Step I Calculation of VTh (Fig. 3.430) From Fig. 3.430, Ix

3Ω

− +

+

…(i)

I2

6Ω

4Ω

10 A I1

+

Ix

I2

For Mesh 1,

V Th

− −

I1 = 10

…(ii)

A

B

Fig. 3.430

Applying KVL to Mesh 2, −4( 2 − 1 ) + 2 I x − 6 2 = 0 −4 2 + 4 I1 + 2 2 − 6 I 2 = 0 4

8II 2 = 0 8

1

…(iii)

Solving Eqs (ii) and (iii), I1 = 10 A I2 = 5 A Writing the VTh equation, 0 VTh = 0

6I2

VTh

6I2

6(5)

30 V

Step II Calculation of IN (Fig. 3.431) From Fig. 3.431, I x I 2 − I3 For Mesh 1, I1 = 10

2Ix

6Ω

4Ω

10 A

…(ii)

−4

2

2

I1

I2

2

−

−

3)

=0

+6

3

=0

8I 2 + 4

3

0

2 ) − 3I 3

=0

2

3 ) − 6I2

4

IN

I3

B

− ) + 2 I x − 6(

+ 4 I + 2(

A Ix

…(i)

Applying KVL to Mesh 2, −4(

3Ω

− +

1

Fig. 3.431 …(iii)

Applying KVL to Mesh 3, −6(

3

− 6

2

9II 3 = 0 9

Solving Eqs (ii), (iii) and (iv), I1 = 10 A I2 = 7 5 A I3 = 5 A IN

I3 = 5 A

…(iv)

3.110 Network Analysis and Synthesis Step III

6Ω

Calculation of RTh

A

RTh

V 30 = Th = =6Ω IN 5

6Ω

30 V

Step IV Calculation of RL For maximum power transfer,

B

RTh = 6 Ω

RL Step V

Fig. 3.432

Calculation of Pmax Pmax =

2 VTh ( )2 = = 37.5 W 4 RTh 4 × 16

Example 3.98

For the network shown in Fig. 3.433, find the value of RL for maximum power transfer. Also, find maximum power. 1Ω −

2Vx

Vx

2V +

+ −

1Ω RL 1A

Fig. 3.433 Solution Step I Calculation of VTh (Fig. 3.434) From Fig. 3.434, Vx 1I = − I For Mesh 1, I = −1 Vx = 1 V Writing the VTh equation, 2Vx 1I + 2 VTh

1Ω −

…(i)

2Vx

Vx

2V +

+

A

1Ω

+ −

V Th 1A

I

…(ii)

−

B

Fig. 3.434

0

2(1) − ( −1) + 2 − VTh = 0 VTh = 5 V Step II Calculation of IN (Fig. 3.435) From Fig. 3.435, Vx 1I1 = − I1 Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh, I 2 I1 = 1 Applying KVL to the outer path of the supermesh, 2Vx 1I1 + 2 0 2( 1 ) 1 + 2 = 0 31 0

1Ω −

…(i) 2Vx

…(ii)

+ −

Vx

2V

A

+

1Ω IN

I1

1A

I2

B

Fig. 3.435 …(iii)

3.5 Maximum Power Transfer Theorem 3.111

Solving Eqs (ii) and (iii), I1 = 0 67 A I 2 = 1 67 A I N I 2 = 1 67 A Step III

Calculation of RTh RTh =

VTh 5 = =3Ω I N 1 67

Step IV Calculation of RL For maximum power transfer,

3Ω

A

RL Step V

RTh = 3 Ω

Pmax =

Example 3.99

3Ω

5V

Calculation of Pmax ( Fig. 3.436) 2 VTh ( )2 = = 2.08 W 4 RTh 4 × 3

B

Fig. 3.436

What will be the value of RL in Fig. 3.437 to get maximum power delivered to it?

What is the value of this power? 0.5 V − +

4Ω

3A

+

4Ω

RL

V −

Fig. 3.437 Solution Step I Calculation of VTh (Fig. 3.438) By source transformation, From Fig. 3.438, VTh 4 I

0.5 VTh − +

4Ω

3A

+ 4Ω

VTh

−

Applying KVL to the mesh, 12 − 4 I

0 5 VTh

B

Fig. 3.438

4I = 0

0.5 VTh

12 − VTh + 0 5 VTh − VTh = 0

− +

VTh = 8 V Step II Calculation of IN (Fig. 3.439) If two terminals A and B are shorted, the 4 Ω resistor gets shorted. =0

A

4Ω 12 V

+

4Ω

A

VTh

I −

Fig. 3.439

B

3.112 Network Analysis and Synthesis Dependent source 0.5 V depends on the controlling variable V . When V = 0, the dependent source vanishes, ie. 0.5 V = 0 as shown in Fig. 3.441. 12 IN = =3A 4 4Ω

0.5 V − +

4Ω

4Ω

A

+

IN

V

B

−

A

12 V IN

12 V

B

Fig. 3.440 Step III

Fig. 3.441 2.67 Ω

Calculation of RTh RTh

V 8 = Th = = 2 67 Ω IN 3

A 2.67 Ω

8V IL

Step IV Calculation of RL For maximum power transfer, RL RTh = 2 67 Ω

B

Fig. 3.442

Step V Calculation of Pmax (Fig. 3.442) Pmax =

3.6

2 VTh ( )2 = =6W 4 RTh 4 × 2.67

RECIPROCITY THEOREM

It states that ‘in a linear, bilateral, active, single source network, the ratio of excitation to response remains same when the positions of excitation and response are interchanged.’ In other words, it may be stated as ‘if a single voltage source Va in the branch ‘a’ produces a current Ib in the branch + N t k I ‘b’ then if the voltage source Va is removed and inserted in V − the branch ‘b’, it will produce a current Ib in branch ‘a’’. Explanation Consider a network shown in Fig. 3.443. Fig. 3.443 Network When the voltage source V is applied at the port 1, it produces a current I at the port 2. If the positions of the excitation (source) and response are interchanged, i.e., if + I Network − V the voltage source is applied at the port 2 then it produces a current I at the port 1. The limitation of this theorem is that it is applicable only Fig. 3.444 Network when excitation to a single-source network. This theorem is not applicable in and response are the network which has a dependent source. This is applicable interchanged only in linear and bilateral networks. In the reciprocity theorem, position of any passive element (R, L, C) do not change. Only the excitation and response are interchanged.

3.6 Reciprocity Theorem 3.113

Steps to be followed in Reciprocity Theorem 1. Identify the branches between which reciprocity is to be established. 2. Find the current in the branch when excitation and response are not interchanged. 3. Find the current in the branch when excitation and response are interchanged.

Example 3.100

Calculate current I and verify the reciprocity theorem for the network shown in

Fig. 3.445. 5Ω

4Ω I

10 Ω

20 V

6Ω

Fig. 3.445 Solution Case I Calculation of current I when excitation and response are not interchanged (Fig. 3.446) Applying KVL to Mesh 1, 20 − 5

10 1 2 ) 0 10( 15 I1 10 I 2 = 20 Applying KVL to Mesh 2, −10(

5Ω

4Ω I

10 Ω

20 V I1

6Ω I2

1

…(i)

Fig. 3.446

− 1) − 4 2 − 6I2 = 0 −10 I1 + 20 I 2 = 0 Solving. Eqs (i) and (ii), I1 = 2 A I2 = 1 A I I2 = 1 A 2

…(ii)

Case II Calculation of current I when excitation and response are interchanged (Fig. 3.447). Applying KVL to Mesh 1, −5 1 − 10 10( 0 1 − 2) = 0 15 I1 10 I 2 = 0

5Ω

4Ω 20 V

I

10 Ω

…(i)

I1

I2

6Ω

Applying KVL to Mesh 2, −10(

2

− 1) − 4 2 − 2 20 0 − 6I2 = 0 −10 I1 + 200 I 2 = −20

Fig. 3.447

Solving Eqs (i) and (ii), I1 = −1 A I 2 = −1 5 A I I1 = 1 A Since the current I remains the same in both the cases, reciprocity theorem is verified.

…(ii)

3.114 Network Analysis and Synthesis

Example 3.101

Find the current I and verify reciprocity theorem for the network shown in Fig. 3.446. 2Ω

2Ω

3Ω

5V

3Ω

4Ω

I 4Ω

Fig. 3.448 Solution Case I Calculation of the current I when excitation and response are not interchanged (Fig. 3.449) Applying KVL to Mesh 1, 5V

4( I1 − I 3 ) = 0 9 1 3I 2 − 4 3 5 Applying KVL to Mesh 2, 5 2 I1 3(

−3(

2)

1

2

− 1 ) − 2I 2 − 3

2

2Ω

…(i)

2Ω

3Ω

I1 4Ω

=0

I3

−3 1 + 8 8I 2 = 0 Applying KVL to Mesh 3, −4( 3 − 1 ) − 4 I 3 = 0

…(ii)

3Ω

I2

I

4Ω

Fig. 3.449

−4 1 + 8 8I 3 = 0

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 0 85 A I 2 = 0 32 A I 3 = 0 43 A I

I 3 = 0 43 A

Case II Calculation of current I when excitation and response are interchanged (Fig. 3.450). Applying KVL to Mesh 1, −2 1 − 3( I1 − I 2 ) − 4( 9

1

−

3I 2 − 4

1

3)

=0

3

0

2Ω I

2

− 1 ) − 2I 2 − 3

3Ω

I1 4Ω

I2

…(i)

Applying KVL to Mesh 2, −3(

2Ω

I3 2

=0

−3 1 + 8 8I 2 = 0

…(ii)

4Ω

5V

Fig. 3.450

3Ω

3.6 Reciprocity Theorem 3.115

Applying KVL to Mesh 3, −4(

3

− 1) + 5 − 4

3

=0

−4 1 + 88I 3 = 5

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 0 43 A I 2 = 0 16 A I 3 = 0 84 A I

I1 = 0 43 A

Since the current I remains the same in both the cases, reciprocity theorem is verified.

Example 3.102

Find the voltage V and verify reciprocity theorem for the network shown in

Fig. 3.451. 4Ω

2Ω 5Ω −

10 A

V

+

6Ω

8Ω

Fig. 3.451 Solution Case I Calculation of the voltage V when excitation and response are not interchanged (Fig. 3.451) For Mesh 1, I1 = 10 …(i) Applying KVL to Mesh 2, −4(

2

− 1 ) − 2 I 2 − 5(

2

−

−4 1 + 1 11 1

2

− 5I 3 = 0

3)

4Ω 5Ω −

10 A I1

6Ω

=0 …(ii)

2Ω

I2

V I3

+ 8Ω

Fig. 3.452

Applying KVL to Mesh 3, −6(

3

− 1 ) − 5(

3

−

2) −8 3

=0

−6 1 − 5 I 2 + 19 I 3 = 0

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 10 A I 2 = 5 76 A I 3 = 4 67 A V ( I 2 I3 )

5(5.76 4.67) = 5 45 V

3.116 Network Analysis and Synthesis Case II Calculation of voltage V when excitation and response are interchanged (Fig. 3.453). +

4Ω

2Ω 5Ω

V 6Ω

10 A

8Ω

−

Fig. 3.453 By source transformation (Fig. 3.454), Applying KVL to Mesh 1,

+

−4 1 − 2 I1 − 50 − 5( 1 − 2 ) = 0 11I1 5 2 50 …(i) Applying KVL to Mesh 2, −6

2

− 5( I 2 − I1 ) + 50 − 8 −5 1 + 1 19 9

2 2

4Ω

6Ω

…(ii)

50 V

V

=0 = 50

2Ω I1 5Ω

I2

8Ω

−

Solving Eqs (i) and (ii),

Fig. 3.454 I1 = −3 8 A I 2 = 1 63 A

From Fig. 3.454, V

4 I1 + 6 I 2 = 0

V + 4( −3.8) + 6(1.63) = 0 V = 5.42 42 V Since the voltage V is same in both the cases, the reciprocity theorem is verified.

3.7

MILLMAN’S THEOREM

It states that ‘if there are n voltage sources V1 V2 ,… ,Vn with internal resistances R1 R2 ,… , Rn respectively connected in parallel then these voltage sources can be replaced by a single voltage source Vm and a single series resistance Rm, ’(Fig. 3.455). A R1 V1

R2

Rn

V2

Vn

A

⇒

Rm Vm

B

Fig. 3.455 Millman’s network where

Vm =

V1G1 + V2G2 + … + VnGn G1 G2 + … + Gn

B

3.7 Millman’s Theorem 3.117

Rm =

and

1 1 = Gm G1 + G

Gn

Explanation By source transformation, each voltage source in series with a resistance can be converted to a current source in parallel with a resistance as shown in Fig. 3.456. A I1

R1

I2

R2

In

Rn

B

Fig. 3.456 Equivalent network Let Im be the resultant current of the parallel current sources and Rm be the equivalent resistance as shown in Fig. 3.457. Im

Im

Rm

V V V I n = 1 + 2 + … + n = V1 G1 + V2 G2 + …Vn Gn R1 R2 Rn

I +I

1 1 1 1 = + +…+ Rm R1 R2 Rn Gm G1 + G2 Gn

Fig. 3.457 Equivalent network A Rm

By source transformation, the parallel circuit can be converted into a series circuit as shown in Fig. 3.458. I V G + V G + … + Vn Gn Vm I m Rm = m = 1 1 2 2 Gm G1 G2 + … + Gn

Vm B

Fig. 3.458 Millman’s equivalent network

Dual of Millman’s Theorem

It states that ‘if there are n current sources I 1 I 2 ,… , I n with internal resistances R1 R2 ,… , Rn respectively, connected in series then these current sources can be replaced by a single current source Im and a single parallel resistance Rm ’ (Fig. 3.459). I1

I2

In

Im

⇒ R1

R2

Rn

A

Rm B

Fig. 3.459 Millman’s network where

Im = Rm

I1 R1 + I 2 R2 + … + I n Rn R1 + R2 + … + Rn R +R Rn

A

B

3.118 Network Analysis and Synthesis Steps to be followed in Millman’s Theorem 1. Remove the load resistance RL. 2. Find Millman’s voltage across points A and B. Vm =

V1 G1 + V2 G2 + … + Vn Gn G1 G2 + … + Gn

3. Find the resistance Rm between points A and B. Rm =

G1

1 G2 + … + Gn

4. Replace the network by a voltage source Vm in series with the resistance Rm. 5. Find the current through RL using ohm’s law. IL =

Example 3.103

Vm Rm

RL

Find Millman’s equivalent for the left of the terminals A-B in Fig. 3.460. A 4Ω

2Ω

1A

6V B

Fig. 3.460 A

Solution By source transformation, the network is redrawn as shown in Fig. 3.461. Step I Calculation of Vm ⎛ 1⎞ ⎛ 1⎞ 6⎜ ⎟ + 2⎜ ⎟ ⎝ 4⎠ ⎝ 2⎠ V G +V G Vm = 1 1 2 2 = = 3.33 33 V 1 1 G1 G2 + 4 2 Step II

4Ω

2Ω

6V

2V B

Fig. 3.461 A

Calculation of Rm Rm =

1 1 1 = = = 1.33 Ω 1 1 Gm G1 + G2 + 4 2

Step III Millman’s Equivalent Network (Fig. 3.462)

1.33 Ω 3.33 V B

Fig. 3.462

3.7 Millman’s Theorem 3.119

Example 3.104

Find the current through the 10 W resistor in the network of Fig. 3.463.

2Ω

4Ω

6Ω

5V

10 V

15 V

10 Ω

Fig. 3.463 Solution Step I

Calculation of Vm ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ 5 ⎜ ⎟ − 10 ⎜ ⎟ + 15 ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ 6⎠ V G +V G +V G 2 4 Vm = 1 1 2 2 3 3 = = 2.73 73 V 1 1 1 G1 G2 + G3 + + 2 4 6

Step II

Calculation of Rm 1 1 = = 1.09 Ω Gm 1 1 1 + + 2 4 6 Calculation of IL (Fig. 3.464) Rm =

Step III

IL =

Example 3.105

1.09 Ω 2.73 V

2 73 = 0.25 25 A 1 09 + 10

10 Ω IL

Fig. 3.464

Find the current through the 10 W resistor in the network of Fig. 3.465. 50 Ω

2Ω 50 V

40 Ω

10 Ω 100 V

Fig. 3.465 Solution Since the 40 Ω resistor is connected in parallel with the 50 V source, it becomes redundant. The network can be redrawn as shown in Fig. 3.466. Step I Calculation of Vm V G +V G Vm = 1 1 2 2 = G1 G2

⎛ 1⎞ ⎛ 1⎞ 50 ⎜ ⎟ − 100 ⎜ ⎟ ⎝ 50 ⎠ ⎝ 20 ⎠ = −57.15 15 V 1 1 + 50 20

50 Ω

20 Ω 10 Ω

50 V

40 Ω

100 V

Fig. 3.466

3.120 Network Analysis and Synthesis Step II

Calculation of Rm 1 1 1 = = = 14.29 Ω 1 1 Gm G1 + G2 + 50 20 Calculation of IL (Fig. 3.467) Rm =

Step III

IL =

Example 3.106

IL

14.29 Ω 57.15 V

57.15 = 2.35 35 A 14.29 + 10

Fig. 3.467

Draw Millman’s equivalent network across terminals AB in the network of

Fig. 3.468.

C

2Ω

4A

8V

6V

2Ω

1Ω

6V

2Ω

3V

1Ω A

B D

5V

5Ω

Fig. 3.468 Step I

By source transformation, the network is redrawn as shown in Fig. 3.469.

C 8V

8V

6V

2Ω

2Ω

1Ω

6V

2Ω

3V

1Ω A

B D

5V

5Ω

Fig. 3.469 Step II

10 Ω

Applying Millman’s theorem at terminals CD,

Vm1

V G +V G +V G = 1 1 2 2 3 3 = G1 G2 + G3

⎛ 1⎞ ⎛ 1⎞ −8 ⎜ ⎟ + 8 ⎜ ⎟ + 6(1) ⎝ 2⎠ ⎝ 2⎠ =3V 1 1 + +1 2 2

3.8 Tellegen’s Theorem 3.121

1 1 1 = = = 0.5 Ω Gm1 G1 + G2 G3 1 1 + +1 2 2 Applying Millman’s theorem at terminals CA,

4V

Rm1 =

Step III

Vm2

Rm2

A 3V

0.5 Ω

⎛ 1⎞ 6 ⎜ ⎟ + 3(1) ⎝ 2⎠ V4 G4 + V5 G5 = = =4V 1 G4 G5 +1 2 1 1 1 = = = = 0.67 Ω Gm2 G4 + G4 1 +1 2

B 5V

5Ω

Fig. 3.470 A 6.17 Ω

Step IV Millman’s Equivalent Network (Fig. 3.470) Simplifying Fig. 3.470 further, the Millman’s equivalent network is shown in fig. 3.471.

3.8

0.67 Ω

2V B

Fig. 3.471

TELLEGEN’S THEOREM

It states that ‘the algebraic sum of the powers in all branches of the network at any instant is zero’. n

∑ Vk

Ik = 0

k =1

This condition is valid for the network which obeys kirchhoff’s voltage and current laws. Explanation

Consider a network shown in Fig. 3.472. I3 +

V3

−

I2 I1 V1

I5 +

+ −

V2

−

I4 + V − 5 + V4 −

I6 + V6 −

Fig. 3.472 Network illustrating Tellegen’s theorem According to Tellegen’s theorem, 6

∑ Vk

I k = V1 I1 + V2 I 2 + V3 I 3 + V4 I 4 + V5 I 5 + V6 I 6

k =1

= V1 I1 + (V1 V4 ) I 2 (V1 − V6 ) I 3 + V4 I 4 + (V4 − V6 ) I 5 + V6 I 6 = V1 ( I1 + I 2 + I 3 ) + V4 ( − I 2 + I 4 + I 5 ) + V6 (− − I3 − I5 + I6 ) Applying KCL at Node 1, I1 I 2 + I 3 = 0 Applying KCL at Node 4, I 2 I 4 + I5

3.122 Network Analysis and Synthesis Applying KCL at Node 6, I3 = I6

I5 6

∑ Vk

I k = V1 (0) V4 (0) +V V6 (0) = 0

k =1

Hence, Tellegen’s theorem is verified.

Example 3.107

Determine the power supplied by the source in the network of Fig. 3.473. I5 +

V5

I1

− I3

+

V1

−

V

I2 + V − 3 +

I4 +

V2

V4 −

−

V1 V2

5 10

I1 2 A I2 = 0 8 A

V3 V4

3 7

I3 I4

6A 3A

5

4A

V5 = 8 V,

Fig. 3.473 Solution By Tellegen’s theorem, Power supplied by the source = V1 I1 + V2 I 2 + V3 I 3 + V4 I 4 + V5 I 5 = 5 × 2 + 10 × 0.8 + 3 × 6 + 7 × 3 + 8 × 4 = 89 W

Example 3.108

Prove the Tellegen’s theorem for the network shown in Fig. 3.474. I6 +

+ I1

−

III

2 I 4

I2

1

V6

V2

−

+ I3 +

V1

V4

−

V3 −

I

II

3 I5 + V5 −

Fig. 3.474 V1 I1

8 4A

V2 I2

4 V, 2 A,

Solution Applying KVL to Mesh 1, V1 V2 − V3 = 0 V3 V1 − V2 = 8 4 = 4 V Applying KVL to Mesh 2, V3 V4 − V5 = 0 V5 V3 − V4 = 4 2 = 2 V

V4 = 2 V I3 = 1 A

3.8 Tellegen’s Theorem 3.123

Applying KVL to Mesh 3, −V6 + V4 + V2 = 0 V6 = V2 + V4 = 4 + 2 = 6 V Applying KCL at Node 1, I1

I 2 + I6 = 0 I6

I1 − I 2 = −44 2 = −6 A

Applying KCL at Node 2, I2 I4

I3 + I 4 I 2 − I3 = 2 1 = 1 A

Applying KCL at Node 3, I5

I 4 + I 6 = 1 ( 6) = −5 A

6

∑ Vb I b = V1 I1 + V2 I 2 + V3 I 3 + V4 I 4 + V5 I 5 + V6 I 6 = 8 × 4 + 4 × 2 + 4 × 1 + 2 × 1 + 2 × (

)+6×(

)=0

b=1

Hence, Tellegen’s theorem is verified.

Example 3.109

Find the value of source V2 in Fig. 3.475, using Tellegen’ theorem if the power

absorbed by V2 is 20 W. 10 Ω

5Ω

10 Ω

100 V

5Ω V2

Fig. 3.475 Solution Finding Thevenin’s equivalent network across voltage source V2 (Fig. 3.476). 10 Ω

10 Ω

5Ω

5Ω

5Ω 10 Ω

100 V

10 Ω

+ A VTh − B

I1

Fig. 3.477

Fig. 3.476 I1 = VTh

100 =5A 10 + 10 10 I1 = 10(5)

5Ω

RTh = ( 50 V

) + 5 + 5 = 15 Ω

A RTh B

3.124 Network Analysis and Synthesis Thevenin’s equivalent network is shown in Fig. 3.478.

I

15 Ω A

V2 I = 20 Applying Tellegen’s theorem,

50 V

V2

2

−50 I +15 15 I + V2 I = 0

B

15 I 2 − 50 I + 20 = 0 I = 2 87 or 0.46 A 20 20 V2 = = = 6 97 V I 2 87 20 V2 = = 43.48 V 0 46

or

3.9

Fig. 3.478

SUBSTITUTION THEOREM

It states that ‘any branch in a network can be substituted by a different branch without disturbing the voltages and currents in the entire network provided the new branch has the same set of terminal voltage and current as the original network.’ Explanation Any branch can be replaced by any combination of elements that will have the same voltage across it and same current through it as the original branch (Fig. 3.479). A

A

A

+

VAB

⇒

VAB

⇒

IAB

−

IAB

B

B

(a) Branch of a (b) Branch replaced by a voltage network whose VAB and IAB are source VAB known

B (c) Branch replaced by a current source IAB

Fig. 3.479 Substitution theorem Steps to be followed in Substitution Theorem 1. Find the branch voltage and branch current in the network. 2. Replace the branch by an independent voltage source and find the voltage and current in the network. 3. Replace the branch by an independent current source and find the voltage and current in the network.

Example 3.110 In the network of Fig. 3.480 verify substitution theorem by replacing the 6 W resistor by (a) a voltage source, and (b) a current source.

3.9 Substitution Theorem 3.125 2Ω

4Ω

6Ω

8Ω

100 V

Fig. 3.480 Solution Step I Calculation of current through 6 Ω resistor (Fig. 3.481) Applying KVL to Mesh 1, 100 − 2 1 8( I1 − I 2 ) = 0 10 I1 8 I 2 = 100 …(i) Applying KVL to Mesh 2, −8(

2

− 1) − 4I2 − 6

2

=0

−8 1 + 1 18 8

2

=0

2Ω

4Ω

6Ω

8Ω

100 V I1

I2

Fig. 3.481 …(ii)

Solving Eqs (i) and (ii), I1 = 15.52 A I2 = 6 9 A I6 V6

I2 = 6 9 A 6 I 2 6(6.9) = 41.4 V

Step II When the 6 Ω resistor is replaced by the voltage source of 41.4 V (Fig. 3.482) Applying KVL to Mesh 1, 100 − 2 1 8( I1 − I 2 ) = 0 10 I1 8 I 2 = 100 …(i) Applying KVL to Mesh 2, −8(

2

2Ω

8Ω

100 V I1

− 1 ) − 4 I 2 − 41.4 = 0 −8 1 + 1 12 2

2

4Ω

41.4 V I2

Fig. 3.482

= −41.4 …(ii)

Solving Eqs (i) and (ii),

I1 = 51.51 A I2 = 6 9 A By replacing the 6 Ω resistor by a voltage source has not affected the currents I1 and I2. Hence, the substitution theorem is verified. 2Ω 4Ω Step III When the 6 Ω resistor is replaced by current source of 6.9 A (Fig. 3.483) Applying KVL to Mesh 1, 100 − 2 1 8( I1 − I 2 ) = 0 10 I1 8 I 2 = 100

8Ω

100 V I1

6.9 A I2

…(i) Fig. 3.483

3.126 Network Analysis and Synthesis For Mesh 2, I2 = 6 9

…(ii)

Solving Eqs (i) and (ii), I1 = 15.52 A I2 = 6 9 A By replacing the 6 Ω resistor by a current source has not affected the currents I1 and I2. Hence, the substitution theorem is verified.

3.10

COMPENSATION THEOREM

It states that ‘in any linear bilateral active network, if any branch carrying a current I has its resistance R changed by an amount d R, the resulting changes that occur in the other branches are the same as those which would have been produced by an opposing voltage source of value Vc (Id R) introduced into the modified branch.’ Explanation

Consider a network shown in Fig. 3.484 (a), having load resistance R. RTh

RTh R

VTh

R

VTh

I

(a)

I′

dR

(b)

Fig. 3.484 Network illustrating compensation theorem I=

VTh RTh

R

If the load resistance R be changed to R δ R as shown in Fig. 3.484(b) then the current flowing in the circuit is I′ =

RTh

VTh + R+δR

The change in the current is

δI

VTh VTh − RTh + R + δ R RTh R VTh δ R =− ( RTh + R + δ R) ( RTh + R)

I′ − I =

VTh δR RTh + R RTh + R + δ R I δR =− RTh + R + δ R Vc =− RTh + R + δ R =−

where Vc

I δ R and is called the compensation voltage.

3.10 Compensation Theorem 3.127

d I has the same direction as I. This shows that the change in current d I due to a change in any branch in a linear network can be calculated by determining the current in that branch in a network obtained from the original network by removing all the independent sources and placing a voltage source called compensation source in series with the branch whose value is Vc I δ R, whose I is the current through the branch before its resistance is changed and d R is the change in resistance. (Fig. 3.485).

Example 3.111

RTh R +dR δI

Vc = Id R

Fig. 3.485 Compensation network

In the network of Fig. 3.486, the resistance of 4 W is changed to 2 W . Verify the

compensation theorem. 1Ω

4Ω

1V

8Ω

Fig. 3.486 Solution Step I Calculation of change in current Finding Thevenin’s equivalent network across the 4 Ω resistor, 8 VTh = 1 × = 0 89 V 8 +1 1× 8 RTh = = 0 89 Ω 1+ 8 Thevenin’s equivalent network is shown in Fig. 3.487. 0 89 IL = = 0 18 A 0 89 + 4 When resistance of 4 Ω is changed to 2 Ω, 0 89 I L′ = = 0 31 A 0 89 + 2 δ I L I L′ − I L = 0.31 − 0.18 = 0.13 A Step II Calculation of change of current by compensation theorem δ R = 2 − 4 = −2 Ω R δR = 4 − 2 = 2 Ω Vc I L δ R = 0.18 × ( −2) = −0.36 V The compensating network is shown in Fig. 3.489. 0 36 δ IL = = 0.12 12 A 0 89 + 2 Since change in current is same in both the steps, the compensation theorem is verified.

0.89 Ω

4Ω

0.89 V IL

Fig. 3.487 0.89 Ω

0.89 V IL′

2Ω

Fig. 3.488 0.89 Ω 2Ω δIL

Fig. 3.489

0.36 V

3.128 Network Analysis and Synthesis

Exercises 3.4

Superposition Theorem 3.1

Find the current through the 10 Ω resistor in Fig. 3.490. 10 Ω

10 Ω

30 Ω

5Ω

10 A

Calculate the current through the 10 Ω resistor in Fig. 3.493.

4Ω

20 Ω

100 V

25 V

2Ω

[0.37 A ] Find the current through the 8 Ω resistor in Fig. 3.491.

12 Ω

30 Ω

3Ω

[1.62 A ] 3.5

Find the current through the 1Ω resistor in Fig. 3.494.

2Ω

25 A

1Ω

3Ω

1A

2Ω 1V

Fig. 3.491

[16.2 A ] 3.3

2Ω

Find the potential across the 3 Ω resistor in Fig. 3.492.

Fig. 3.494

[0.41 A ] 3.6

9Ω

Find the current through the 4 Ω resistor in Fig. 3.495.

15 A

7Ω 2Ω

5Ω 5A

12 V

Fig. 3.493

8Ω

5A

− +

7Ω

Fig. 3.490

3.2

2Ω

+ −

3Ω

2Ω

4Ω

10 V

2Ω

2Ω 2Ω

2Ω

5A

6V 4V

Fig. 3.495

Fig. 3.492

[3.3 V ]

[1.33 A ]

Exercises 3.129

3.7

Find the current Ix in Fig. 3.496. Ix

5Ω

50 V

10 Ω

20 Ω 5Ω

100 V

20 Ω

2A

24 V

50 V

2Ω

3Ω

2Ω

36 V

Fig. 3.500 Fig. 3.496

[− 3.8

]

[3.87 A ] 3.12 Find the current through the 6 Ω resistor in Fig. 3.501.

Find the voltage Vx in Fig. 3.497. 2Ω

4V

+ 50 V

Vx

− 3A

3Ω

]

Determine the voltages V1 and V2 in Fig. 3.498.

10 A

V1

4Ω

+ −

0.5V2

V2

6Ω

Fig. 3.501

[− 4Ω

5Ω

2A

100 V

0.1Vx

Fig. 3.497 3.9

4Ω

4Ω

[1.26 A ] 3.13 Find the current through the 6 Ω resistor in Fig. 3.502. 10 V 2 Ω

I1 4 Ω

0.25I1

20 V

4A

10 Ω

5Ω

6Ω

3A

Fig. 3.498

[6 V,12 V ] 3.10

Find the voltage Vx in Fig. 3.499. 10 Ω

4A

20 Ω

30 Ω

+ Vx −

[ 2.04 A ] 3.14 Find the current through the 2 Ω resistor connected between terminals A and B in Fig. 3.503.

60 V

I1

Fig. 3.502

2 Ω 12 V

0.4I1

10 V

Fig. 3.499

10 Ω

4Ω

5Ω

4A

8V

[7.5 V ]

Thevenin’s Theorem 3.11 Find the current through the 5 Ω resistor in Fig. 3.500.

2Ω 6V A

2Ω

B

Fig. 3.503

[1.26 A ]

3.130 Network Analysis and Synthesis 3.15 Find the current through the 5 Ω resistor in Fig. 3.504.

(i) 4Vx

10 V

5Ω

+ −

4Ω 40 V

100 V

20 Ω

20 V

6Ω

30 V 50 V

7Ω

6Ω

+ Vx −

4A

4Ω

Fig. 3.507 20 V

10 Ω

[ 4.67 A ] 3.16 Find the current through the 20 Ω resistor in Fig. 3.505.

8Ω

40 Ω

10 V

36 Ω

9I1

Vx

21 Ω

15 Ω

Ω]

I1 100 Ω

Fig. 3.504

15 Ω

[−

(ii)

15 Ω

100 Ω

Fig. 3.508

[9.09 7Ω

20 Ω

36 Ω

, 9.09 Ω ]

(iii) 0.5Vx − +

100 V

+

Fig. 3.505

[1.54 A ]

3A

4Ω

4Ω

3.17 Calculate the current through the 10 Ω resistor in Fig. 3.506.

Vx −

Fig. 3.509

10 Ω

[8 V, 2.66 Ω] (iv)

4Ω 25 V

+ −

3Ω

2Ω − +

7Ω

12 V 2Ω

2Ω

3Ω

+ Vx −

5Ω

10 A

Vx 4

Fig. 3.506

[1.62 A ] 3.18 Determine Thevenin’s equivalent network for figures 3.507 to 3.510 shown below.

Fig. 3.510

[150 V, 20 Ω]

Exercises 3.131

3.23 Find the current through the 2 Ω resistor in Fig. 3.515.

3.19 Find the current Ix in Fig. 3.511. Ix 5 Ω

2Ω

10 V

10 Ω 10Ix

+ −

5Ω

4A

1A

10 V

5Ω

5A

3Ω

2Ω

Fig. 3.511

[4 A ]

20 V

3.20 Find the current in the 24 Ω resistor in Fig. 3.512.

Fig. 3.515

[5 A ]

Ix 1000 Ω

48 V

3Vx

+ −

10Ix

+ Vx −

13 Ω

24 Ω

3.24 Find the current through the 5 Ω resistor in Fig. 3.516. 3Ω

6Ω

5Ω

6A

Fig. 3.512

10 V

Norton’s Theorem

4Ω

5V

5Ω

15 Ω

20 V

4Ω

2Ω

Fig. 3.516

3.21 Find the current through the 10 Ω resistor in Fig. 3.513. 2 V 10 Ω

3Ω

10 Ω 2 A

[0.225 A ]

6Ω

6Ω

[ 4.13 A ] 3.25 Find Norton’s equivalent circuit for the portion of network shown in Fig. 3.517 to the left of ab. Hence obtain the current in the 10 Ω resistor. a

Fig. 3.513

[0.68 A ] 3.22 Find the current through the 20 Ω resistor in Fig. 3.514. 10 Ω

10 V

20 Ω

4Ω

4Ω 4Ω 7V

2Ω

10 Ω

12 V

8A

5Ω

8Ω

9Ω

6Ω

b

2A

Fig. 3.514

[0.61 A ]

Fig. 3.517

[0.053 A ] 3.26 Find Norton’s equivalent network and hence find the current in the 10 Ω resistor in Fig. 3.518.

3.132 Network Analysis and Synthesis 2V

3.30 Find the value of the resistance RL in Fig. 3.522 for maximum power transfer and calculate the maximum power.

3Ω

2Ω I1

3Ω

10 Ω

2I1

8A 2A

Fig. 3.518

[0.25 A ]

4Ω

3.27 Find Norton’s equivalent network In Fig. 3.519. 2Ω

12 V

4 V/2 Ω

2Ω

8Ω

2 V/1 Ω

I2

+ V1 −

8I2

+ −

0.1 V1

2Ω

2Ω RL

20 Ω

Fig. 3.522

[ 2.18 Ω, 29 35 ]

Fig. 3.519

[0.533 A, 31 Ω] Maximum Power Transfer Theorem

3.31 Find the value of the resistance RL in Fig. 3.523 for maximum power transfer and calculate the maximum power.

3.28 Find the value of the resistance RL in Fig. 3.520 for maximum power transfer and calculate the maximum power. 3Ω

3Ω 2Ω

1Ω

6A

2Ω

3V 2Ω

6V

1Ω

2A

RL

RL

2Ω

Fig. 3.523

[3 Ω 2 52 ]

Fig. 3.520

[1.75 Ω,1.29 ] 3.29 Find the value of the resistance RL in Fig. 3.521 for maximum power transfer and calculate the maximum power.

25 A

9Ω

RL

10 Ω

2Ω

6Ω

3Ω 110 V

3.32 Find the value of the resistance RL in Fig. 3.524 for maximum power transfer and calculate the maximum power.

3Ω

5Ω

10 A

RL

10 Ω

30 A

Fig. 3.524

[1.76 Ω, 490.187 ]

Fig. 3.521

[ 2.36 Ω, 940 ]

Objective-Type Questions 3.133

Objective-Type Questions 3.1

The value of the resistance R connected across the terminals A and B in Fig. 3.525, which will absorb the maximum power is

3.5

The maximum power that can be transferred to the load RL from the voltage source in Fig. 3.528 is 100 Ω

3 kΩ

4 kΩ R

V

A

B

6 kΩ

RL

10 V

4 kΩ

Fig. 3.528

Fig. 3.525 (a) (c) 3.2

4 kΩ 8 kΩ

(b) (d)

(a) (c)

4.11 kΩ 9kΩ

Superposition theorem is not applicable to networks containing

3.6

(a) nonlinear elements (b) dependent voltage source (c) dependent current source (d) transformers 3.3

(b) (d)

10 W 0.5 W

For the circuit shown in Fig. 3.529, Thevenin’s voltage and Thevenin’s equivalent resistance at terminals a-b is 1A

0.5I1

The value of R required for maximum power transfer in the network shown in Fig. 3.526 is 5Ω

1W 0.25 W

+ −

a I1

5Ω

5Ω

4Ω

10 V

b

Fig. 3.529 20 Ω

25 V

3A

R

(a) 5 V and 2 Ω (c) 4 V and 2 Ω 3.7

Fig. 3.526 3.4

The value of RL in Fig. 3.530 for maximum power transfer is

(a) 2 Ω (b) 4 Ω (c) 8 Ω (d) 16 Ω In the network of Fig. 3.527, the maximum power is delivered to RL if its value is

9Ω

6Ω

I1

20 Ω

0.5I1

40 Ω RL

(b) 7.5 V and 2.5 Ω (d) 3 V and 2.5 Ω

2V

6Ω

+ −

6Ω

1A

9Ω

RL

V

Fig. 3.530

Fig. 3.527 (a)

16 Ω

(c)

60 Ω

40 Ω 3 (d) 20 Ω (b)

(a) (c)

3Ω 4.1785 Ω

(b) (d)

1.125 Ω none of these

3.134 Network Analysis and Synthesis

Answers to Objective-Type Questions 3.1 (a)

3.2 (a)

3.3 (c)

3.4 (a)

3.5 (c)

3.6 (b)

3.7 (a)

4 4.1

Single-Phase ac Circuits INTRODUCTION

An alternating waveform changes its magnitude and direction periodically. Many times alternating voltages and currents are represented by a sinusoidal waveform. A sinusoidal waveform is more useful than other waveforms, like square, triangular, sawtooth etc. Any waveform can be written in terms of sinusoidal function with the help of Fourier series. The derivative and integration of sinusoidal waveform result in sinusoidal waveform only. The sinusoidal waveform can be easily analyzed and generated.

4.2

GENERATION OF ALTERNATING VOLTAGES

An alternating voltage can be generated either by rotating a coil in a stationary magnetic field or by a rotating a magnetic field within a stationary coil. In both the cases, the magnetic field is cut by the conductors or coils and an emf is induced in the coil according to Faraday’s laws of electromagnetic induction. The magnitude of the induced emf depends upon the number of turns of the coil, the strength of the magnetic field and the speed at which the coil or magnetic field rotates. Consider a rectangular coil of N turns of area A m2 and rotating in anti-clockwise direction with angular velocity of w radians per second in a uniform magnetic field as shown in Fig. 4.1. Let fm be the maximum flux cutting the coil when its axis coincides with XX ′ axis (reference position of the coil). Thus when the coil is along XX ′, the flux linking with it is maximum, i.e., fm. When the coil is along YY ′, i.e., parallel to the lines of flux, the flux linking with it is zero. The coil rotates through an angle θ ω t at any instant t. At this instant, the flux linking with the coil is

φ

φm cos ω t w rad/s

N

S

(a)

4.2 Network Analysis and Synthesis N Y

X′

w rad/s Coil

q = wt

0

X

fm cosw wt Y′ S (b)

Fig. 4.1 Generation of alternating voltage According to Faraday’s laws of electromagnetic induction, dφ d e N = − N (φm c ω t ) N φm ω sin ω t Em i ω t dt dt e E m = Nφ m ω where = maximum value of induced emf Em When ω t 0, i ω t = 0, e = 0 π π q = wt When ωt , si 1 e = Em p p 3p 0 2p 2 2 2 2 If the induced emf is plotted against time, a sinusoidal −E m waveform is obtained as shown in Fig. 4.2. Fig. 4.2 Sinusoidal wave form

4.3

TERMS RELATED TO ALTERNATING QUANTITIES

1. Waveform A waveform is a graph in which the instantaneous value of any quantity is plotted against time. Figure 4.3 shows a few waveforms. 2. Cycle One complete set of positive and negative values of an alternating quantity is termed a cycle. 3. Frequency The number of cycles per second of an alternating quantity is known as its frequency. It is denoted by f and is measured in hertz (Hz) or cycles per second (c/s). 4. Time Period The time taken by an alternating quantity to complete one cycle is called its time period. It is denoted by T and is measured in seconds. 1 T= f

v

v

t

0

t

0

v

0

Fig. 4.3

t

Alternating waveforms

5. Amplitude The maximum positive or negative value of an alternating quantity is called the amplitude. 6. Phase The phase of an alternating quantity is the time that has elapsed since the quantity has last passed through zero point of reference. 7. Phase Difference This term is used to compare the phases of two alternating quantities. Two alternating quantities are said to be in phase when they reach their maximum and zero values at the same time. Their maximum value may be different in magnitude.

4.4 Root Mean Square (rms) or Effective Value 4.3

A leading alternating quantity is one which reaches its maximum or zero value earlier compared to the other quantity. A lagging alternating quantity is one which attains its maximum or zero value later than the other quantity. A plus (+) sign, when used in connection with the phase VB VA difference, denotes ‘lead’ whereas a minus (−) sign denotes ‘lag’. Vm sin ω t A vB

Vm sin (ω t + φ )

Here, the quantity B leads A by a phase angle f as shown in Fig. 4.4.

4.4

t

f

Fig. 4.4 Phase difference

ROOT MEAN SQUARE (rms) OR EFFECTIVE VALUE

i Normally, the current is measured by the amount of work it will do or the amount of heat it will produce. Hence, rms or effective value i2 of alternating current is defined as that value of steady current(direct current) which will do the same amount of work in the same time or i1 would produce the same heating effect as when the alternating current is applied for the same time. Figure 4.5 shows the positive half cycle of a non-sinusoidal 0 t1 t2 alternating current waveform. The waveform is divided in m Fig. 4.5 equal intervals of time each of duration t . Let the average m values of instantaneous currents during these intervals be respectively i1, i2,

i3 im t3

t m

t

Mid-ordinate method …, im. This waveform

is applied to a circuit consisting of a resistance of R ohms. The work done in different intervals will be t⎞ ⎛ 2 t⎞ t⎞ ⎛ 2 ⎛ 2 ⎜⎝ i1 R × ⎟⎠ , ⎜⎝ i2 R × ⎟⎠ , …, ⎝ im R × ⎟⎠ joules. m m m Thus, the total work done in t seconds on applying an alternating current waveform to a resistance i 2 i 2 + …+ + im2 R= 1 2 × Rt joules m Let I be the value of the direct current that while flowing through the same resistance does the same amount of work in the same time t. i 2 i 2 + …im2 I 2 Rt = 1 2 × Rt m i 2 i 2 + …+ im2 I2 = 1 2 m Hence, rms value of the alternating current is given by I rms =

i12

i22 + …+ + im2 = Mean value of (i ) 2 m

The rms value of any current i(t) over the specified interval t1 to t2 is expressed mathematically as I rms =

1 t2

t2

t1 ∫t 1

i 2 (t )dt

4.4 Network Analysis and Synthesis The rms value of an alternating current is of considerable importance in practice because the ammeters and voltmeters record the rms value of alternating current and voltage respectively.

The rms Value of a Sinusoidal Waveform

v

Figure 4.6 shows a sinusoidal waveform. Vm si θ

v

1 2π

Vrms

=

=

=

1 2π Vm2 2π

Vm

< θ < 2π

2π

∫v

2

( ) dθ

p

0

2p

q

0

2π

∫ Vm

2

2

Fig. 4.6 Sinusoidal waveform

θ dθ

0

2π

∫ sin

2

θ dθ =

0

Vm2 2π

2π

2π

Vm2 ⎡ θ i θ⎤ ⎛ 1 − cos2θ ⎞ − ⎟⎠ dθ = 2 2π ⎢⎣ 2 4 ⎥⎦ 0

∫ ⎜⎝ 0

Vm2 ⎡ 2 ⎤ V − 0 − 0 + 0 ⎥ = m = 0.707 Vm ⎢ 2π ⎣ 2 2 ⎦

Crest or Peak or Amplitude Factor It is defined as the ratio of maximum value to rms value of the given quantity. Maximum u value Peak factor ( p ) = rms value

4.5

AVERAGE VALUE

The average value of an alternating quantity is defined as the arithmetic mean of all the values over one complete cycle. In case of a symmetrical alternating waveform (whether sinusoidal or non-sinusoidal), the average value over a complete cycle is zero. Hence, in such a case, the average value is obtained over half the cycle only. Referring to Fig. 4.5, the average value of the current is given by I avg =

i1 i2 + …+ im m

The average value of any current i(t) over the specified interval t1 to t2 is expressed mathematically as I avg =

t2

1 t2

t1

∫ i(t ) dt

v Vm

t1

p

0

2p

Average Value of a Sinusoidal Waveform Figure 4.7 shows a sinusoidal waveform. v

Vm si θ

Fig. 4.7

Sinusoidal waveform

< θ < 2π

Since this is a symmetrical waveform, the average value is calculated over half the cycle.

q

4.5 Average Value 4.5 π

Vavg

=

π

1 v π ∫0

d

Vm [+ π

]=

π

1 V V Vm sin θ dθ = m ∫ sin θ dθ = m [ − ∫ π0 π 0 π

]π0

2Vm = 0.637 Vm π

Form Factor It is defined as the ratio of rms value to the average value of the given quantity. Form factor (

f

)=

rms value Average value

Example 4.1

An alternating current takes 3.375 ms to reach 15 A for the first time after becoming instantaneously zero. The frequency of the current is 40 Hz. Find the maximum value of the alternating current. i = 15 A, t = 3.375 ms, f = 40 Hz i = Im sin 2p ft 15 = Im sin (2 × 180 × 40 × 3.375 × 10−3) Im = 20 A

Solution

(angle in degrees)

Example 4.2

An alternating current of 50 c/s frequency has a maximum value of 100 A. (a) Calculate 1 its value second after the instant the current is zero. (b) In how many seconds after the zero value will 600 the current attain the value of 86.6 A? f = 50 c/s, Im = 100 A 1 (a) Value of current second after the instant the current is zero 600 i = Im sin 2p ft Solution

1 ⎞ ⎛ = 100 sin 2 × 180 × 50 × ⎟ ⎝ 600 ⎠ = 50 A (b) Time at which current will attain the value of 86.6 A after the zero value i = Im sin 2pft 86.6 = 100 sin (2 × 180 × 50 × t) 1 t= second 300

Example 4.3

(angle in degrees)

(angle in degrees)

An alternating current varying sinusoidally with a frequency of 50 c/s has an rms value of 20 A. Write down the equation for the instantaneous value and find this value at (a) 0.0025 s, and (b) 0.0125 s after passing through zero and increasing positively. (c) At what time, measured from zero, will the value of the instantaneous current be 14.14 A? Solution

f = 50 c/s, Irms = 20 A I m I rms × 2 = 20 2 = 28.28 A

4.6 Network Analysis and Synthesis Equation of current, i = Im sin 2p ft = 28.28 sin (100p × t) = 28.28 sin (100 × 180 × t) (a) Value of current at t = 0.0025 second i = 28.28 sin (100 × 180 × 0.0025) = 20 A (b) Value of current at t = 0.0125 second

(c)

(angle in degrees) (angle in degrees)

i = 28.28 sin (100 × 180 × 0.0125) = − 20 A Time at which value of instantaneous current will be 14.14 A i = 28.28 sin (100 × 180 × t) 14.14 = 28.28 sin 18000 t t = 1.66 ms

(angle in degrees)

(angle in degrees)

Example 4.4

An alternating current of 60 Hz frequency has a maximum value of 110 A. Calculate (a) time required to reach 90 A after the instant current is zero and increasing positively, and (b) its value 1 second after the instant current is zero and its value decreasing thereafter. 600 i Solution f = 60 Hz, Im = 110 A (a) Time required to reach 90 A after the instant current is zero 110 A and increasing positively i = Im sin 2p ft 90 = 110 sin (2 × 180 × 60 × t) (angle in degrees) 0 t = 2.54 ms 1 (b) Value of current second after the instant current is 600 zero and decreasing thereafter From Fig. 4.8, T 1 1 1 t= + = + 2 600 2 f 600

=

T

T 2

t

Fig. 4.8

1 1 + = 0.01 s 2 × 60 600

i = Im sin 2p ft = 110 sin (2 × 180 × 60 × 0.01) = − 64.66 A

(angle in degrees)

Example 4.5 A sinusoidal wave of 50 Hz frequency has its maximum value of 9.2 A. What will be its value at (a) 0.002 s after the wave passes through zero in the positive direction, and (b) 0.0045 s after the wave passes through the positive maximum. Solution f = 50 Hz, Im = 9.2 A (a) Value of current at 0.002 s after the wave passes through zero in the positive direction

4.5 Average Value 4.7

i = Im sin 2p ft = 9.2 sin (2 × 180 × 50 × 0.002) = 5.41 A (b)

(angle in degrees)

i Value of current at 0.0045 s after the wave passes through the positive maximum From Fig. 4.9, 9.2 A

t= =

T 1 + 0.0045 = + 0.0045 4 4f

0

T 4

1 + 0.0045 = 9.5 ms 4 × 50

t

T

T 2

Fig. 4.9

i = Im sin 2p ft = 9.2 sin (2 × 180 × 50 × 9.5 × 10−3) = 1.44 A

(angle in degrees)

Example 4.6 An alternating current varying sinusoidally with a frequency of 50 Hz has an rms value of current of 20 A. At what time measured from negative maximum value will the instantaneous current be 10 2 A? f = 50 Hz, Irms = 20 A

Solution

Time at which instantaneous current will be 10 2 A i = 10 2 A = 14.14 A Im = Irms × 2 = 20 2 = 28.28 A i = Im sin 2π ft 14.14 = 28.28 sin (2 × 180 × 50 × t) t = 1.67 ms i (b) Time, measured from negative maximum value, at which instantaneous current will be 10 2 A 28.28 A From Fig. 4.10, 14.14 A T 1 t = + 1.67 × 10 −3 = + 1.67 × 10 −3 0 4 4f (a)

=

1 + 1.67 × 10 −3 = 6.67 ms 4 × 50

(angle in degrees)

T 4

T 2

3T 4

T

t

Fig. 4.10

Example 4.7 An alternating current is given by i = 14.14 sin 377 t. Find (a) rms value of the current, (b) frequency, (c) instantaneous value of the current when t = 3 ms, and (d) time taken by the current to reach 10 A for first time after passing through zero. i = 14.14 sin 377 t

Solution (a)

The rms value of the current I rms =

Im 2

=

14.14 2

= 10 A

4.8 Network Analysis and Synthesis (b)

Frequency 2π f = 377 f =

(c)

377 = 60 Hz 2π

Instantaneous value of the current when t = 3 ms i = 14.14 sin (377 × 3 × 10−3) = 12.79 A

(angle in radians)

(d) Time taken by the current to reach 10 A for the first time after passing through zero i = 14.14 sin 377 t (angle in radians) 10 = 14.14 sin 377 t t = 2.084 ms

Example 4.8 An alternating current varying sinusoidally at 50 Hz has its rms value of 10 A. Write down an equation for the instantaneous value of the current. Find the value of the current at (a) 0.0025 second after passing through the positive maximum value, and (b) 0.0075 second after passing through zero value and increasing negatively. f = 50 Hz,

Solution (a)

Irms = 10 A

Equation for instantaneous value of the current Im

I rms × 2 = 10 2 = 14.14 A

i = Im sin 2p ft = 14.14 sin (2 × 180 × 50 × t) = 14.14 sin (18000 t) (b)

(angle in degrees) i

Value of the current at 0.0025 s after passing through 14.14 A the positive maximum value From Fig. 4.11, t=

T 1 + 0.0025 = + 0.0025 4 4f

0

T 4

1 + 0.0025 = 7.5 ms 4 × 50 i = 14.14 sin(18000 × 7.5 × 10−3) = 10 A =

T 1 + 0.0075 = + 0.0075 2 2f

1 + 0.0075 = 17.5 ms 2 × 50 i = 14.14 sin (18000 × 17.5 × 10−3) = −10 A =

T

t

Fig. 4.11 (angle in degrees)

(c) Value of the current 0.0075 s after passing through zero value and increasing negatively. 14.14 A From Fig. 4.12, t=

T 2

0

i

T 2

T

t

Fig. 4.12 (angle in degrees)

4.5 Average Value 4.9

Example 4.9 Draw a neat sketch in each case of the waveform and write expressions of instantaneous value for the following: π (a)) Sinusoidal current of amplitude 10 A, 50 Hz passing through its zero value at ω t = and increasing 3 positively π (b) Sinusoidal current of amplitude 8 A, 50 Hz passing through its zero value at ω t = − and increasing 6 positively. Solution i

(a) The current waveform is lagging in nature. From Fig. 4.13, I m sin( 2π ft f − φ) =

i

π⎞ ⎛ si sin 2π × 50 × t − ⎟ ⎝ 3⎠

10 A

π⎞ ⎛ = 10 sin 100π t − ⎟ ⎝ 3⎠ (b)

3

The current waveform is leading in nature. From Fig. 4.14, i

i

π⎞ ⎛ I m sin ( 2π ft ft + φ ) = sin 2π × 50 × t + ⎟ ⎝ 6⎠ = 8 si

q = wt

p

0

Fig. 4.13

8A

π⎞ ⎛ 100π t + ⎟ ⎝ 6⎠ −

q = wt

p 0 6

Fig. 4.14

Example 4.10

π⎞ ⎛ The instantaneous current is given by l = 7.071 sin ⎜ 157.08t − ⎟ . Find its effective ⎝ 4⎠

value, periodic time and the instant at which it reaches its positive maximum value. Sketch the waveform from t = 0 over one complete cycle. Solution (a)

i

Effective value I eff

(b)

π⎞ ⎛ 157.08t − ⎟ ⎝ 4⎠

7.707 7071si

I rms =

Im 2

=

7.071 2

= 5A

Periodic time 2p f = 157.08 f = 25 Hz T=

(c)

1 1 = = 0.04 s f 25

Instant at which the current reaches its positive maximum value i.e., i = 7.071 A

4.10 Network Analysis and Synthesis From Fig. 4.14,

π⎞ ⎛ 7.071 = 7.071sin 071sin 157.08t − ⎟ ⎝ 4⎠ 1 = sin (157.08t − 0.785)

i 7.071 A

t = 0.015 s (d)

0

The waveform is shown in Fig. 4.15.

p 4

5p 4

9p 4

wt

Fig. 4.15

Example 4.11

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) expression for instantaneous current, (b) magnitude of the current at t = 0.0125 second, and (c) magnitude of the current at t = 0.025 second after t = 0. Solution (a)

i(0) = 7.07 A,

f

I rms = 10 2 A

Expression for instantaneous current Im

I rms × 2 = 10 2 × 2 = 20 A

Since i = 7.07 A at t = 0, the current is leading in nature. At t = 0 I m sin ( 2π ft f + φ) 7.07 = 20 sin i ( × × + φ)

φ = 20.7° i 20 sin (120 ( π t + 20.7 )

i 20 A

Magnitude of current at t = 0.0125 second i = 20 sin (120 × 180 × 0.0125 + 20.7°) 0 (angle in degrees) = −18.7 A (c) Magnitude of current at t = 0.025 second after t = 0 Fig. 4.16 From Fig. 4.16, 20.7° 20.7° 1 20 7° 1 Time corresponding to a phase shift of 20.7° = ×T = × = × = 0.958 ms 360° 360° f 360° 60 Time 0.025 second after t = 0 (b)

t = 0.958 × 10−3 + 0.025 = 25.958 ms i = 20 sin (120 × 180 × 25.958 × 10–3 + 20.7°) = −13.22 A

Example 4.12

(

t

(angle in degrees)

A 50 Hz sinusoidal voltage applied to a single-phase circuit has an rms value of 200 V.

)

Its value at t = 0 is 2 × 200 V positive. The current drawn by the circuit is 5 A (rms) and lags behind the voltage by one-sixth of a cycle. Write the expressions for the instantaneous values of voltage and current. Sketch their waveforms and find their values at t = 0.0125 second.

4.5 Average Value 4.11

Solution

Vrms = 200 V,

f

I rms = 5 A

v (0) = 2 × 200 = 282.84 V (a)

Instantaneous value of voltage Vm v

Vrms × 2 = 200 2 = 282.84 V Vm sin ( 2π ft f + φ)

At t = 0, 282.84 = 282 8 .84 sin (0 + φ ) sin φ = 1

φ = 90° v = 282.8 84 4 si ( 2π × 50 × + 90°) (b)

Instantaneous value of current The current lags behind the voltage by one-sixth of a cycle.

φ=

1 × 360° = 60° 6

I m = I rms × 2 = 5 2 = 7.07 A i = I m sin ( 2π ft + 90° − 60°) = 7.07 sin s (2π 50

30 )

(c)

Voltage and current waveforms are shown in Fig. 4.17.

(d)

Value of voltage at t = 0.0125 s v = 282.84 sin (100 × 180 × 0.0125 + 90°) (angle in degrees) = 200 V

(e)

282 84 i (100π t + 90°)

7 07 sin 7.07 si (100 ( 00π t + 30 )

v i p − 2

Value of current at t = 0.0125 s i = 7.07 sin (100 × 180 × 0.0125 + 30°) (angle in degrees) = −6.83 A

0

Fig. 4.17

At the instant t = 0, the instantaneous value of a 50 Hz sinusoidal current is 5 A and increases in magnitude further. Its rms value is 10 A. (a) Write the expression for its instantaneous value. (b) Find the current at t = 0.01 s and t = 0.015 s. (c) Sketch the waveform indicating these values.

Example 4.13

Solution (a)

i(0) = 5 A,

f

I rms = 10 A

Expression for instantaneous value of current Im Since

i

I rms × 2 = 10 2 = 14.14 A

5 A a t = 0, the current is leading in nature.

4.12 Network Analysis and Synthesis i

I m sin ( 2π ft f + φ)

At t = 0, 5 14.14 si ( 2 × 180 × 50 × 0 + φ ) 5 14.14 sin φ

φ = 20.7° i = 14.14 ssii (100 (100 (b)

(angle in degrees)

+ 20.7°)

Current at t = 0 01 s i = 14.14 sin (100 × 180 × 0.01 + 20.7°) = −5 A

(angle in degrees)

(c) Current at t = 0.015 s i = 14.14 sin (100 × 180 × 0.015 + 20.7°) = −13.23 A i

(angle in degrees)

(d) Waveform

0.015

0 −5 A

0.01

14.14 A

The waveform is shown in Fig. 4.18.

t(s)

−13.23 A

Fig. 4.18

Example 4.14

In a certain 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 the instant t = 0, the instantaneous values of potential difference and current are 400 V and 4 A respectively, both increasing in the positive direction. State expressions for instantaneous values of potential difference and current at time t. Calculate the instantaneous values at time t = 0.015 second. Find the phase angle between potential difference and current. Solution (a)

f

Vm = 500 V

Im

10 A v(0) = 400 V,

i( 0) = 4 A

Expression for instantaneous value of potential difference and current Since v = 400 V and i = 4 A at t = 0, the voltage and current waveforms are leading in nature. ) At t = 0

v

Vm sin ( 2π ft f + φ1 )

400 = 500 sin ( 2π 50 0 + φ1 ) φ1 = 53.13° v = 500 sin(100π t + 53.13°) (ii) At t = 0

i = Im sin (2p ft + f2) 4 10 in ( 2π × 500 × 0 + φ2 )

φ2 = 23.58° i = 10 sin (100π t + 23.58°)

4.5 Average Value 4.13

(b)

Instantaneous values of potential difference and current at t = 0.015 second (i)

v = 500 sin (100 × 180 × 0.015 + 53.13°) = −300 V

(angle in degrees)

i = 10 sin (100 × 180 × 0.00155 + 23.58°)

(ii)

= −9.17 A

(angle in degrees)

(c) Phase angle between potential difference and current φ φ1 − φ2 = 53.13° − 23.58° = 29.55°

Example 4.15 Find the following parameters of a voltage v = 200 sin 314 t: (a) frequency (b) form factor, and (c) crest factor. Solution (a)

v

200 si 314 t

v

Vm sin 2π ft

Frequency 2π f = 314 314 f = = 50 Hz 2π

For a sinusoidal waveform, Vavg = Vrms = (b)

2Vm π Vm 2

Form factor Vm Vrms kf = = 2 = 1.11 Vavg 2Vm π

(c)

Crest factor kp =

Vm V = m = 1.414 Vrms Vm 2

Example 4.16 A non-sinusoidal voltage has a form factor of 1.2 and peak factor of 1.5. If the average value of the voltage is 10 V, calculate (a) rms value, and (b) maximum value. Solution

kf

k p = 1.5, Vavg = 10 V

4.14 Network Analysis and Synthesis (a)

rms value kf =

Vrms Vavg

Vrms 10 = 12 V

1.2 = Vrms (b)

Maximum value Vm Vrms V 15= m 12 Vm = 18 V kp =

Example 4.17 The waveform of a voltage has a form factor of 1.15 and a peak factor of 1.5. If the maximum value of the voltage is 4500 V, calculate the average value and rms value of the voltage. Solution (a)

k p = 1.5, Vm = 4500 V

kf

rms value of the voltage Vm Vrms 4500 15= Vrms Vrms = 3000 V kp =

(b)

Average value of the voltage kf = 1.15 =

Vrms Vavg 3000 Vavg

Vavg = 2608.7 V

Example 4.18

A 50 Hz sinusoidal current has a peak factor of 1.4 and a form factor of 1.1. Its average value is 20 A. The instantaneous value of the current is 15 A at t = 0. Write the equation of the current and draw its waveform. Solution (a)

kp = 1 4

f

Equation of current kf =

I rms I avg

1.1 =

I rms 20

kf

. , I avg = 20 A, i(0) = 15 15 A

4.5 Average Value 4.15

I rms = 22 A Im I rms I 1.4 = m 22 I m = 30.8 A kp =

Since i = 15 Α at t = 0, the current is leading in nature as shown in Fig. 4.19.

π fftt + φ )

i At t = 0 15 = 30.8

( 2π × 50 0 × 0 + φ)

φ = 29.14° i = 30.8 8 sin (100π (b)

29.14°)

The waveform is shown in Fig. 4.19. i 30.8 A wt −29.14°

Fig. 4.19

Example 4.19

Find the average value and rms value of the waveform shown in Fig. 4.20. v Vm

q

p

0

3p

2p

Fig. 4.20 Solution v = Vm sin q (a) Average value of the waveform

0 k3

w The variation of output voltage V0 with angular 0 w0 frequency w for different values of coefficient of coupling k is shown in Fig. 7.91. Fig. 7.91 Variation of V0 with w for different values of k

7.10.2 Double-Tuned Circuit Consider a double-tuned circuit with tuning capacitors placed both in primary and secondary circuits as shown in Fig. 7.92. − −j

1 wC1

R2

M

R1

+

jwL1

+

jwL2

− −j

V1 −

1 wC2

V2 −

Fig. 7.92

Double-tuned circuit

The conductively coupled equivalent circuit is shown in Fig. 7.93. − −j

1 wC1

jw (L1 − M )

jw (L2 − M )

R2 +

R1 − −j

jw M

+

I1

V1 −

1 wC2

V2 −

Fig. 7.93 Conductively coupled equivalent circuit Applying KVL to Mesh 1, V1

R1I1 + j

1 I1 ω C2

j (

1

M )I1

jω M ( I1 I 2 ) = 0

1 ⎞ ⎛ ⎜⎝ R1 + j L1 − j ωC ⎟⎠ I1 1

j

I 2 = V1

...(7.19)

7.48 Network Analysis and Synthesis Applying KVL to Mesh 2, −

−

L −

−

1 I2 C2

0

1 ⎤ I2 C2

0

+j

+

−

...(7.20)

From Eq. (7.20), 1 C

−

2

I j M

I2

Substituting Eq. (7.21) in Eq. (7.19), ⎡ ⎢

1 C2

−

1

I2

1 C2

−

1 C2

−

2

j M

I2

j M

⎤

1 C

+ V1

j M j MV1

I2 = −

1

1 C

1 + C2

−

2

2

2

1. Output Voltage 1

1 I2 C2 MV1

=

⎡

C

+

−

⎤⎡

1 C

R2 +

2

1 ⎤ + C2

2

2

2. Voltage Amplification A=

V = V1

M R1 +

−

1 C2

2

−

If the primary and secondary circuits resonate at the same frequency w0, 1 0

1

1 C2

2

2

...(7.21)

7.10 Tuned Circuits 7.49

Then, at the resonant frequency w 0,

ω 0 MV V1 R1 R2 + ω 02 M 2 MV V1 V2 = 2 2 C2 ( R1 R2 0M ) M A= 2 2 C2 ( R1 R2 0M ) I2 =

The maximum amplification or maximum output voltage can be obtained by, dA =0 dM 2 2 0 M ) − 2C2

C2 ( R1 R2

(R R

⎡C2 ( R1 R2 ⎣

1 2

2 2 0M

+

)

2 2 0M

2 2 2 0 M )⎤ ⎦

2ω 02 M 2

ω 02 M 2

=0

0 R1 R2

M=

R1 R2 ω0 V0

where M = inductance.

R1 R2 ω0

is the critical value of mutual

k = k1

Hence, maximum value of I2, V2 and A is I 2max = V2max = Am max =

k1 = kc k2 > k1 k3 < k1

k = k2

V1 2 R1 R2 V1

k = k3 w

w0

0

2ω 0C2 R1 R2 1

Fig. 7.94

2ω 0C2 R1 R2

Variation of V0 with w for different values of k

The variation of the output voltage V0 with the angular frequency w for different values of coefficient of coupling k is shown in Fig. 7.94.

Example 7.39 For the single-tuned circuit shown in Fig. 7.95, determine (a) the resonant frequency, (b) Output voltage at resonance, and (c) maximum output voltage. Assume R1>>w0L1 and k = 0.8. 10 Ω

5Ω

M

+

+ 60 μH

1 μF

50∠0°V

0.1 μF

V2

− −

Fig. 7.95

7.50 Network Analysis and Synthesis Solution k L1 L2 = 0.8 (1 10 −66 )(660 0

M

0 6 ) = 6. 2 μ H

(a) Resonant frequency 1

ω0 =

f0 =

L2C2

1

=

−6

−6

(60 10 )(0 (0.1 × 10 )

= 0 41 106 rad/s

ω 0 0 41 106 = = 0. 0 0653 MHz = 65.3 kHz 2π 2π

(b) Output voltage at resonance V0 =

(

MV V1

)

2 2 0M

C2 R1 R2

=

6 2 × 10 −6 × 50

(0 42

⎡ 0.1 1 × 10 −6 (10 × 5) ⎣

106

) (6 2 ×10 ) ⎤⎥⎦ 2

−6 2

= 54.6 V

(c) Maximum output voltage V2max =

V1 2ω 0C2 R1 R2

=

50 6

2(0.42 × 10 )(0.1 10 6 ) 10 × 5

= 84.18 V

Example 7.40

The resonant frequency of the tuned circuit shown in Fig. 7.96 is 1000 rad/s. Calculate the self-inductances of two coils and critical value of mutual inductance. 1 μF

5Ω

3Ω

M

+

L1

Vs

L2

−

Fig. 7.96 Solution

ω0 = L1 = L2 =

Mc =

1 L1C1 1

ω 02C1 1

ω 02C2

= = =

1 L2C2 1

(1000)

2

(1 1

(1000)

2

(

10

6

)

=1H

)

=05H

R1 R2 5 3 = = 3 87 mH ω0 1000

2 μF

Exercises 7.51

Example 7.41

The tuned frequency of a double-tuned circuit shown in Fig. 7.97 is 104 rad/s. If the source voltage is 2 V and has a resistance of 0.1 W, calculate the maximum output voltage. C1

0.01 Ω

0.1 Ω

M

0.1 Ω 2 μH

25 μH

C2

2V

Fig. 7.97 Solution MC =

ω0 = 10 4 =

RS ) R2

( R1

ω0

=

(0.01 + 0.1) (0.1) 10 4

= 10.49 μH

1 L2C2 1

(25

)C

2

C2 = 0.4 mF V0 max =

V1 2ω 0C2

( RS + R1 ) R2

=

2 2

0

4

0.4 10

3

(0.1

0.01)( 0.1)

= 2 38 V

Exercises 7.1

Two coupled coils have inductances of 0.8 H and 0.2 H. The coefficient of coupling is 0.90. Find the mutual inductance and the turns ratio N1 . N2 [0.36 H, 2]

7.2

Two coils with coefficient of coupling 0.5 are connected in such a way that they magnetise (i) in the same direction, and (ii) in opposite directions. The corresponding equivalent inductances are 1.9 H and 0.7 H. Find selfinductances of the two coils and the mutual inductance between them. [0.4 H, 0.9 H, 0.3 H] Two coils having 3000 and 2000 turns are wound on a magnetic ring. 60% of the flux

7.3

produced in the first coil links with the second coil. A current of 3 A produce a flux of 0.5 mwb in the first coil and 0.3 mwb in the second coil. Determine the mutual inductance and coefficient of coupling. [0.2 H, 0.63] 7.4

Find the equivalent inductance of the network shown in Fig. 7.98. 7H 2H

3H

5H 4H

6H

Fig. 7.98 [10 H]

7.52 Network Analysis and Synthesis Find the effective inductance of the network shown in Fig. 7.99.

7.5

(ii) A

2H 3H

2Ω

2Ω

4H

5H

j5 Ω

2H B

Fig. 7.99

Fig. 7.102

[4.8 H] (iii)

Write mesh equations of the network shown in Fig. 7.100.

7.6

R1

L2

A

R3

2Ω

j4 Ω

L1 −

i1

R2 i2

−3Ω −j

j5 Ω

+

v(t)

j5 Ω j2 Ω

L3 j8 Ω

i3 B

Fig. 7.100

4Ω

Fig. 7.103 d dii dii ⎤ ⎡ (i1 i2 ) + M12 2 + M13 3 ⎥ ⎢ v i1 R1 L1 d dt dt dt ⎢ ⎥ di i d ⎢ R (i − i ) R i L 3 + M (i1 i2 ) ⎥ 33 3 13 ⎢ 2 3 2 ⎥ dt dt ⎢ ⎥ dii ⎢ −M ⎥ M 23 2 = 0 ⎢⎣ ⎥⎦ dt 7.7

⎡(a) ( ⎢ 2 ⎢⎣(c) (6.22 7.8

Find the input impedance at terminals AB of the coupled circuits shown in Fig. 7.101 to 7.103. (i)

j 4.65) Ω

(b) (1

5Ω

j2 Ω

2Ω

+

+ j8 Ω

j2 Ω

−

j4 Ω

j1.5) Ω ⎤ ⎥ ⎥⎦

In the coupled circuit shown in Fig. 7.104, find V2 for which I1 = 0. What voltage appears at the 8 Ω inductive reactance under this condition?

100∠0°V

3Ω

)

V2 −

A

Fig. 7.104 j3 Ω

j5 Ω

−8Ω −j

[ 7.9

B

Fig. 7.101

.5

45°V, 100∠0°V]

For the coupled circuit shown in Fig. 7.105, find the components of the current I2 resulting from each source V1 and V2.

Exercises 7.53 2Ω

V1 = 10∠0°V

+

j2 Ω

j4 Ω

+

j3 Ω

−

−

I2

V2 = 10∠0°V

7.13 For the circuit shown in Fig. 7.109, what load connected to terminals A and B absorbs maximum power and what is this power? 30 Ω

j 25 Ω A

Fig. 7.105

+

[ .77∠112.6°Α,1.72 6 ∠86.05°Α]

k = 0.5

j40 Ω

100∠15° V −

7.10 Find the voltage across the 5 Ω resistor in the network shown in Fig. 7.106. j2 Ω

j10 Ω

B

Fig. 7.109

j1 Ω

56.3° Ω, 83.3 W]

[ .1

7.14 Find the current I in the circuit of Fig. 7.110.

+ −3Ω −j

10∠0° V

5Ω

− 15 Ω 14 Ω −j

− + −

[ .2

33..

j2 Ω

j3 Ω

j4 Ω

5Ω

Fig. 7.110 [ .3 . 8∠39 39..4° A] 7.15 Obtain a conductively coupled circuit for the circuit shown in Fig. 7.111. 2Ω

3Ω −

j5 Ω

j3 Ω

j4 Ω

+ −

Fig. 7.107 [

.16 W] Fig. 7.111

7.12 Obtain Thevenin’s and Norton’s equivalent network at terminals AB of the coupled circuit shown in Fig. 7.108. 4Ω

j8 Ω

j5 Ω

j8 Ω

−2Ω −j

3Ω

100∠0°V

2Ω

4Ω

+

A

+

−

B

−

10∠0°V

70∠−30° V −

° ]

+ 100∠0°V

j 12 Ω

I

7.11 Find the power dissipated in the 5 Ω resistor in the network of Fig. 7.107. 2Ω

j3 Ω

10 Ω

+

j 10 Ω

100∠20° V

Fig. 7.106

j4 Ω

j2 Ω

−1Ω −j

+

3Ω

−

j3 Ω

−2Ω −j

100∠0°V

10∠90°V

Fig. 7.108 [ .0 . 7 45° V, 1.04∠ − 27.9°A, 6.8∠72.9° Ω]

Fig. 7.112

7.54 Network Analysis and Synthesis

Objective-Type Questions 7.1

7.2

Two coils are wound on a common magnetic core. The sign of mutual inductance M for finding out effective inductance of each coil is positive if the (a) two coils are wound in the same sense. (b) fluxes produced by the two coils are equal (c) fluxes produced by the coils act in the same direction (d) fluxes produced by the two coils act in opposition When two coils having self-inductances of L1 and L2 are coupled through a mutual inductance M, the coefficient of coupling k is given by k= (a) k= (c)

7.3

k= (b)

2M L1 L2

L1 + L2

M

L1 L2 M

(b) L1 − L2

1 1 ( 1 ( 1 2) 2) (d) 4 2 Consider the following statements: The coefficient of coupling between two oils depends upon 1. Orientation of the coils 2. Core material 3. Number of turns on the two coils 4. Self-inductance of the two coils of these statements, (a) 1, 2 and 3 are correct (b) 1 and 2 are correct (c) 3 and 4 are correct (d) 1, 2 and 4 are correct (c)

7.7

7.8 k=

(d)

7.6

L1 L2

The overall inductance of two coils connected in series, with mutual inductance aiding selfinductance is L1; with mutual inductance opposing self-inductance, the overall inductance is L2. The mutual inductance M is given by (a)

7.4

M 2 L1 L2

7.5

Two coupled coils connected in series have an equivalent inductance of 16 mH or 8 mH depending on the inter connection. Then the mutual inductance M between the coils is (a) 12 mH (b) 8 2 mH (c) 4 mH (d) 2 mH Two coupled coils with L1 = L2= 0.6 H have a coupling coefficient of k = 0.8. The turns N1 ratio is N2 (a) 4 (b) 2 (c) 1 (d) 0.5 The coupling between two magnetically coupled coils is said to be ideal if the coefficient of coupling is (a) zero (b) 0.5 (c) 1 (d) 2 The mutual inductance between two coupled coils is 10 mH . If the turns in one coil are doubled and that in the other are halved then the mutual inductance will be (a) 5 mH (b) 10 mH (c) 14 mH (d) 20 mH

Two perfectly coupled coils each of 1 H selfinductance are connected in parallel so as to aid each other. The overall inductance in henrys is (a) 2 (b) 1 1 (c) (d) Zero 2 7.10 The impedance Z as shown in Fig.7.113 is

7.9

j5 Ω

j2 Ω

j 10 Ω

j10 Ω

j2 Ω

Fig. 7.113 (a) (c)

j 29 Ω j19 Ω

(b) (d)

j9 Ω j39 Ω

Answers to Objective-Type Questions 7.55

Answers to Objective-Type Questions 1. (c) 7. (c)

2. 8.

(b) (b)

3. 9.

(c) (b)

4. (d) 10. (b)

5.

(d)

6.

(c)

8 8.1

Three-Phase Circuits INTRODUCTION

A system which generates a single alternating voltage and current is termed a single-phase system. It utilises only one winding. A polyphase system utilises more than one winding. It will produce as many induced voltages as the number of windings. A three-phase system consists of three separate but identical windings displaced by 120 electrical degrees from each other as shown in Fig. 8.1. When these three windings are rotated in an anticlockwise direction with constant angular velocity in a uniform magnetic field, the emfs induced in each winding have the same magnitude and frequency but are displaced 120° from one another. w rad/s R Y1

B1 S

N Y

B R1

Fig. 8.1 Three-phase system The instantaneous values of generated voltage in windings RR1, YY1 and BB1 are given by eR eY eB

Em sin θ Em sin (θ − 120°) Em sin (θ − 240°)

where Em is the maximum value of the induced voltage in each winding. The waveforms of these three voltages are shown in Fig. 8.2. e eR

eY

eB q

0

Fig. 8.2 Voltage waveforms

8.2 Network Analysis and Synthesis Figure 8.3 shows the phasor diagram of these three induced voltages. ER

120°

120° 120°

EB

EY

Fig. 8.3 Phasor diagram

8.2

ADVANTAGES OF A THREE-PHASE SYSTEM

1. In a single-phase system, the instantaneous power is fluctuating in nature. However, in a three-phase system, it is constant at all times. 2. The output of a three-phase system is greater than that of a single-phase system. 3. Transmission and distribution of a three-phase system is cheaper than that of a single-phase system. 4. Three-phase motors are more efficient and have higher power factors than single-phase motors of the same frequency. 5. Three-phase motors are self-starting whereas single-phase motors are not self-starting.

8.3

SOME DEFINITIONS

1. Phase Sequence The sequence in which the voltages in the three phases reach the maximum positive value is called the phase sequence or phase order. From the phasor diagram of a three-phase system, it is clear that the voltage in the coil R attains maximum positive value first, next in the coil Y and then in the coil B. Hence, the phase sequence is R-Y-B. 2. Phase Voltage The voltage induced in each winding is called the phase voltage. 3. Phase Current The current flowing through each winding is called the phase current. 4. Line Voltage The voltage available between any pair of terminals or lines is called the line voltage. 5. Line Current The current flowing through each line is called the line current. 6. Symmetrical or Balanced System A three-phase system is said to be balanced if the (a) voltages in the three phases are equal in magnitude and differ in phase from one another by 120°, and (b) currents in the three phases are equal in magnitude and differ in phase from one another by 120°. 7. Balanced Load The load is said to be balanced if loads connected across the three phases are identical, i.e., all the loads have the same magnitude and power factor.

8.4

INTERCONNECTION OF THREE PHASES

In a three-phase system, there are three windings. Each winding has two terminals, viz., ‘start’ and ‘finish’. If a separate load is connected across each winding as shown in Fig. 8.4, six conductors are required to transmit and distribute power. This will make the system complicated and expensive.

8.5 Star, or WYE, Connection 8.3 F Load S F Load S F Load S

Fig. 8.4 Non-interlinked three-phase system In order to reduce the number of conductors, the three windings are connected in the following two ways: 1. Star, or Wye, connection 2. Delta, or Mesh, connection

8.5

STAR, OR WYE, CONNECTION

In this method, similar terminals (start or finish) of the three windings are joined together as shown in Fig. 8.5. The common point is called star or neutral point. Figure 8.6 shows a three-phase system in star connection. R

IR

R

S Zph F F S

IR + IY + IB

O

N F Y

S

B

Fig. 8.5

N Zph

Three-phase star connection

Fig. 8.6

IY

Y

IB

B

Zph

Three-phase, four-wire system

This system is called a three-phase, four-wire system. If three identical loads are connected to each phase, the current flowing through the neutral wire is the sum of the three currents IR, IY and IB. Since the impedances are identical, the three currents are equal in magnitude but differ in phase from one another by 120°. iR iY iR

iB iY + iB

I m sin θ I m sin (θ − 120°) I m sin(θ − 240°) I m sin θ + I m i (

°) I m sin (θ − 240°) = 0

8.4 Network Analysis and Synthesis Hence, the neutral wire can be removed without any way affecting the voltages or currents in the circuit as shown in Fig. 8.7. This constitutes a three-phase, three-wire system. If the load is not balanced, the neutral wire carries some current. IR

R

Zph N

O

Zph

IY

Y

IB

B

Zph

Fig. 8.7 Three-phase, three-wire system

8.6

DELTA, OR MESH, CONNECTION

In this method, dissimilar terminals of the three windings are joined together, i.e., the ‘finish’ terminal of one winding is connected to the ‘start’ terminal of the other winding, and so on, as shown in Fig. 8.8. This system is also called a three-phase, three-wire system. For a balanced system, the sum of the three-phase voltages round the closed mesh is zero. The three emfs are equal in magnitude but differ in phase from one another by 120°. R

eY eB eR + eY + eB

8.7 8.7.1

R F

S

F S

Y

F

B

Three-phase delta connection

Fig. 8.8

Em sin θ Em sin (θ − 120°) Em sin (θ − 240°) Em si θ + Em i (

°) Em sin (θ − 240°) = 0

VOLTAGE, CURRENT AND POWER RELATIONS IN A BALANCED STAR-CONNECTED LOAD Relation between Line Voltage and Phase Voltage

Figure 8.9 shows a balanced star-connected load. Since the system is balanced, the three-phase voltages VRN, VYN, and VBN are equal in magnitude and differ in phase from one another by 120°. VRN VYN = VBN V ph where V ph indicates the rms value of phase voltage. Let

IR R

IY

VYB = VBR

IR IB

Y VBR

Zph

N IY

VYB IB

VYN = V pph ∠− 120° VBN = V pph ∠− 240° VRY

Zph

VRY

VRN = V pph ∠0°

Let

S

B

VL

Fig. 8.9 Star connection

Zph

8.7 Voltage, Current and Power Relations in a Balanced Star-Connected Load 8.5

where VL indicates the rms value of line voltage. Applying Kirchhoff’s voltage law, VRN + VNY

VRY

= VRN

VYN

= V ph ∠0° V pph ∠− 120° =(

p

+

) − (− ( .5 V p − j 0.

p

)

= 1.5 V p + j 0.866 V ph = 3 V ph ∠30° Similarly,

VYB

VYN + VNB = 3 V pph ∠30°

VBR

VBN + VNR = 3 V pph ∠30°

Thus, in a star-connected, three-phase system, VL voltages by 30°.

8.7.2

3 V ph and line voltages lead respective phase

Relation between Line Current and Phase Current

From Fig. 8.9, it is clear that line current is equal to the phase current. IL = Iph

8.7.3

Phasor Diagram (Lagging Power Factor)

Figure 8.10 shows the phasor diagram of a balanced star-connected inductive load. VRY VRN

VNY

f

IB f VBN

VNB

IR

30°

f

30°

VYB

30°

IY

VYN

VNR VBR

Fig. 8.10 Phasor diagram

8.7.4

Power

The total power in a three-phase system is the sum of powers in the three phases. For a balanced load, the power consumed in each load phase is the same.

8.6 Network Analysis and Synthesis Total active power P = 3 × power in each phase = 3 Vph Iph cos f In a star-connected, three-phase system, V ph =

VL

I ph

IL

3

P = 3×

VL 3

φ

× IL

3 VL I L cos φ

where f is the phase difference between phase voltage and corresponding phase current. Similarly, total reactive power Q = 3Vph Iph sin f = 3 VL I L sin φ Total apparent power S

V phh I ph

3 VL I L

8.8

VOLTAGE, CURRENT AND POWER RELATIONS IN A BALANCED DELTA-CONNECTED LOAD

8.8.1

Relation between Line Voltage and Phase Voltage

Figure 8.11 shows a balanced delta-connected load. IR

R

R IBR

VRY

Zph

Zph

Y VYB B

IRY

VBR IB

Zph

B IYB

Y

IY

Fig. 8.11 Delta connection From Fig. 8.11, it is clear that line voltage is equal to phase voltage. VL

8.8.2

V ph

Relation between Line Current and Phase Current

Since the system is balanced, the three-phase currents IRY, IYB, and IBR are equal in magnitude but differ in phase from one another by 120°. Let I RY IYB = I BR I ph where Iph indicates rms value of the phase current. I RY = I ph ∠0° IYB = I ph ∠− 120° I BR = I ph ∠− 240° Let

IR

IY = I B

IL

8.8 Voltage, Current and Power Relations in a Balanced Delta-Connected Load 8.7

where IL indicates rms value of the line current. Applying Kirchhoff’s current law, IR

I BBRR = I RY I RY − I BR

IR

= I ph ∠0° I ph ∠− 240° =(

p

+

) − (− ( .5 I p + j 0.

p

)

= 1.5 I p − j 0.866 I ph = 3 Similarly,

ph

∠− − 30°

IY

IYB

I RY = 3 I ph ∠− 30°

IB

I BR

IYB = 3 I ph ph ∠− 30°

Thus, in a delta-connected, three-phase system, I L phase currents by 30°.

8.8.3

3 I ph and line currents lag behind the respective

Phasor Diagram (Lagging Power Factor)

Figure 8.12 shows the phasor diagram of a balanced delta connected inductive load. VRY IR

IRY −IYB

f

30°

IB −IBR

30°

IBR f −IRY

VBR

f 30°

IYB

IY

Fig. 8.12 Phasor diagram

8.8.4

Power

P 3V ph I ph cos φ In a delta-connected, three-phase system, V ph I ph = P

VL IL 3 3 VL ×

IL 3

× cos φ = 3 VL I L cos φ

VYB

8.8 Network Analysis and Synthesis Total reactive power Q

V phh I ph i φ

Total apparent a power S

V phh I ph

8.9

3 VL I L sin i φ

3 VL I L

BALANCED Y/D AND D/Y CONVERSIONS

Any balanced star-connected system can be converted into the equivalent delta-connected system and vice versa. For a balanced star-connected load, Line voltage = VL Line current = IL Impedance/phase = ZY V ph = I ph ZY =

VL 3 IL V ph I ph

=

VL 3 IL

For an equivalent delta-connected system, the line voltages and currents must have the same values as in the star-connected system, i.e., Line voltage = VL Line current = IL Impedance/phase = ZΔ V ph I ph = ZΔ =

ZY

VL IL 3 V ph I ph

=

VL V = 3 L = 3ZY IL IL 3

1 ZΔ 3

Thus, when three equal phase impedances are connected in delta, the equivalent star impedance is one-third of the delta impedance.

8.10

RELATION BETWEEN POWER IN DELTA AND STAR SYSTEMS

Let a balanced load be connected in star having impedance per phase as Zph. For a star-connected load, V ph = I ph = IL

VL 3 V ph Z ph

=

I ph =

VL 3 Z ph VL 3 Z ph

8.11 Comparison between Star and Delta Connections 8.9

Now

PY

φ

3 VL I L

3 VL ×

VL 3Z ph

× cos φ =

VL2 cos φ Z ph

For a delta-connected load,

Now

V ph

VL

I ph =

V ph V = L Z ph Z ph

IL

3 I pphh = 3

VL Z ph

PΔ

3 VL I L

φ

PY

3 VL × 3

VL ×c Z ph

φ=3

VL2 cos φ = 3 PY Z ph

1 PΔ 3

Thus, power consumed by a balanced star-connected load is one-third of that in the case of a deltaconnected load.

8.11

COMPARISON BETWEEN STAR AND DELTA CONNECTIONS Star Connection

1. VL 2. I L

Delta Connection

1. VL

3 V ph I ph

2. I L

V ph 3 I ph

3. Line voltage leads the respective phase 3. Line current lags behind the respective voltage by 30°. phase current by 30° 4. Power in star connection is one-third of 4. Power in delta connection is 3 times of power in delta connection. the power in star connection. 5. Three-phase, three-wire and three- 5. Only three-phase, three-wire system is phase, four-wire systems are possible. possible. 6. The phasor sum of all the phase currents 6. The phasor sum of all the phase voltages is zero. is zero.

Example 8.1 Three equal impedances, each of (8 + j10) ohms, are connected in star. This is further connected to a 440 V, 50 Hz, three-phase supply. Calculate (a) phase voltage, (b) phase angle, (c) phase current, (d) line current, (e) active power, and (f ) reactive power. Solution For a star-connected load, (a) Phase voltage

Z ph = ( + j ) Ω

V ph =

VL 3

=

440 3

VL = 440 V,

= 254.03 V

f = 50 Hz

8.10 Network Analysis and Synthesis (b) Phase angle Z ph = 8 + j10 = 12.81 ∠51.34° Ω Z ph =

.8 Ω

φ = 51.34° (c) Phase current I ph =

V ph

IL

I ph = 19.83 A

Z ph

=

254.03 = 19.83 A 12.81

(d) Line current (e) Active power P (f)

3 VL I L

φ = 3 × 440 × 19.83 × cos (51.34°) = 9 44 kW

Reactive power Q

VL I L i φ = 3 × 440 × 19.83 × sin (51.34°) = 11.81 kVAR

Example 8.2 A balanced delta-connected load of impedance (8 − j6) ohms per phase is connected to a three-phase, 230 V, 50 Hz supply. Calculate (a) power factor, (b) line current, and (c) reactive power. Solution For a delta-connected load, (a) Power factor

Z ph = ( − j ) Ω

VL = 230 V,

f = 50 Hz

Z ph = 8 − j 6 = 10 ∠ − 36.87° Ω Z ph = 10 Ω

φ = 36.87° pf = cos φ = cos (36.87°) = 0.8 (

)

(b) Line current V ph

VL = 230 V

I ph =

V ph

IL

Z ph

=

230 = 23 A 10

3 I pphh = 3 × 23 = 39.84 A

(c) Reactive power Q

VL I L i φ = 3 × 230 × 39.84 × sin (36.87°) = 9 52 kVAR

Example 8.3 Three coils, each having a resistance and an inductance of 8 W and 0.02 H respectively, are connected in star across a three-phase, 230 V, 50 Hz supply. Find the (a) power factor, (b) line current, (c) power, (d) reactive volt-amperes, and (e) total volt-amperes. Solution

R

8 Ω, L = 0.02 H, V

f = 50 Hz

8.11 Comparison between Star and Delta Connections 8.11

For a star-connected load, (a) Power factor

=

×

×

Z ph =

2= 38.13° Ω

28

Z ph . = cos (38.13° = 0 786 (

pf

)

(b) Line current V ph = I ph

V

=

230

132 79 V 3 V ph 132 79 = = 13.05 A Z ph 10 17 = 13 A 3

(c) Power ×

×

.05 ×

×

×

05 ×

=

088 kW

(d) Reactive volt-amperes =

° = 3 21 kVAR

(e) Total volt-amperes = 3×

× 13

198 kVA

Example 8.4 Three coils, each having a resistance of 8 W and an inductance of 0.02 H, are connected in delta to a three-phase, 400 V, 50 Hz supply. Calculate the (a) line current, and (b) power absorbed. Solution For a delta-connected load, (a) Line current

R

50 Hz

= 400 V = π× R

Z ph

×

2= 38 13° Ω

28

L

Z ph = I ph = I

° V ph Z ph

=

400 10.17 =

39.33 A ×

=

12 A

×

×

(b) Power absorbed =

.12 ×

°

37 12 kW

8.12 Network Analysis and Synthesis

Example 8.5 The three equal impedances of each of 10 Æ 60°W, are connected in star across a three-phase, 400 V, 50 Hz supply. Calculate the (a) line voltage and phase voltage, (b) power factor and active power consumed. (c) If the same three impedances are connected in delta to the same source of supply, what is the active power consumed? Solution

Z ph = 10 ∠60° Ω, VL

f = 50 Hz

For a star-connected load, (a) Line voltage and phase voltage VL = 400 V V 400 V ph = L = = 230.94 V 3 3 (b) Power factor and active power consumed

φ = 60° pf = cos φ = cos (60°) = 0.5 (lagging) I ph = IL P

V ph Z ph

230.94 = 23.094 A 10

=

I ph = 23.094 A

φ = 3 × 400 × 23.094 × 0 5 = 8 kW

3 VL I L

(c) Active power consumed for delta-connected load VL = 400 V Z ph = 10 Ω V ph

VL = 400 V

I ph =

V ph Z ph

400 = 40 A 10

=

IL

3 I pphh = 3 × 40 = 69.28 A

P

3 VL I L

φ = 3 × 400 × 69.28 28 c (60 (60 )

Example 8.6

24 kW

Three similar coils A,B, and C are available. Each coil has a 9 W resistance and a 12 W reactance. They are connected in delta to a three-phase, 440 V, 50 Hz supply. Calculate for this load, the (a) phase current, (b) line current, (c) power factor, (d) total kVA, (e) active power, and (f) reactive power. If these coils are connected in star across the same supply, calculate all the above quantities. Solution

R

9 Ω,

XL =

Ω

V

For a delta-connected load, (a) Phase current VL

V ph = 440 V

Z ph = R

jjX

Z ph = 15 Ω

9

j12 = 15 ∠53 13° Ω

f = 50 Hz

8.11 Comparison between Star and Delta Connections 8.13

φ = 53.13° V ph 440 I ph = = = 29.33 A Z ph 15 (b) Line current IL (c) Power factor

3 I pphh = 3 × 29.33 = 50.8 A

pf = cos φ = cos (

.13°) = .6 (lagging)

(d) Total kVA VL I L = 3 × 440 × 50.8 = 38 38.71 kVA

S (e) Active power P (f)

φ = 3 × 440 × 50.8 × 0.6 = 23 23.23 kW

3 VL I L

Reactive power VL I L i φ = 3 × 440 × 50.8 × in (53.13°°) = 30 97 kVAR

Q

If these coils are connected in star across the same supply, (a) Phase current VL = 440 V Z ph = 5 Ω V ph =

VL

I ph =

V ph

=

3 Z ph

440

= 254.03 V 3 254.03 = = 16.94 A 15

(b) Line current IL

I ph = 16.94 A

(c) Power factor pf = 0.6 (lagging) (d) Total kVA S

VL I L = 3 × 440 × 16.94 = 12.91 kVA

(e) Active power P (f)

3 VL I L

φ = 3 × 440 × 16.94 × 0.6 = 7.74 kW

Reactive power Q

VL I L i φ = 3 × 440 × 16.94 × sin (53.13°) = 12.33 kVAR

Example 8.7

A 415 V, 50 Hz, three-phase voltage is applied to three star-connected identical impedances. Each impedance consists of a resistance of 15 W , a capacitance of 177 μF and an inductance of 0.1 henry in series. Find the (a) power factor, (b) phase current, (c) line current, (d) active power, (e) reactive power, and (f) total VA. Draw a neat phasor diagram. If the same impedances are connected in delta, find the (a) line current, and (b) power consumed. Solution

V

f = 50 Hz, R

Ω

C = 177 1 μF, L = 0.1 H

8.14 Network Analysis and Synthesis For a star-connected load, (a) Power factor ffL = 2π × 50 × 0.1 = 3 .

X XC =

Ω

1 1 = = 17.98 Ω 2π fC 2π × 50 × 177 × 10 −6

Z ph = R

jjX L

Z ph = 0.

jjX XC

j 31.42 − j17 98 15

15

j13.44 = 20.14 ∠41.86° Ω

Ω

φ = 41.86° pf = cos φ = cos ( 41.86°) = 0.744 (lagging) (b) Phase current V ph =

VL

I ph =

V ph

=

3 Z ph

415

= 239.6 V 3 239.6 = = 11.9 A 20.14

(c) Line current I ph = 11.9 A

IL (d) Active power P

φ = 3 × 415 × 11.9 × 0.744 = 6.36 kW

3 VL I L

(e) Reactive power (f)

Q

VL I L i φ = 3 × 415 × 11.9 × in ( 41.86°°) = 5.71 kVAR

S

VL I L = 3 × 415 × 11.9 = 8.55 kVA

Total VA

The phasor diagram is shown in Fig. 8.13. VRY VRN

VNY

IR

30°

VNB

f

VYB IB

30°

f f 30°

IY

VBN

VYN

VNR VBR

Fig. 8.13

f = 41.86°

8.11 Comparison between Star and Delta Connections 8.15

If the same impedances are connected in delta, (a) Line current VL V ph = 415 V Z ph = 0. Ω I ph = IL

V ph Z ph

=

415 = 20.61 A 20.14

3 I ph ph = 3 × 20.61 = 35.69 A

(b) Power consumed P

3 VL I L

φ = 3 × 415 × 35.69 × 0.774 = 19.09 kW

Example 8.8 Each phase of a delta-connected load consists of a 50 mH inductor in series with a parallel combination of a 50 W resistor and a 50 μF capacitor. The load is connected to a three-phase, 550 V, 800 rad/s ac supply. Find the (a) power factor, (b) phase current, (c) the line current, (d) power drawn, (e) reactive power, and (f) kVA rating of the load. Solution

L

50 0

, R = 500 Ω

C

μF, VL = 5500 V, ω = 800 rad/s

For a delta-connected load, (a) Power factor

ω L = 800 × 50 × 10 −3 = 0 Ω 1 1 XC = = = 25 Ω ωC 800 × 50 × 10 −6 R( jX j C) ( − j 25) Z ph = jX L + = j 0+ = 10 + j 20 = 22.36 ∠63.43° Ω R jX j C 50 − j 25 Z ph = .36 Ω φ = 63.43° pf = cos φ = cos (63.43°) = 0.447 ( ) XL

(b) Phase current VL I ph

V ph = 550 V V ph 550 = = = 24.6 A Z ph 22.36

(c) Line current IL

3 I pphh = 3 × 24.6 = 42 42.61 A

P

3 VL I L

(d) Power drawn

φ = 3 × 550 × 42.61 × 0.447 = 18.14 kW

(e) Reactive power

(f)

Q

VL I L i φ = 3 × 550 × 42.61 × sin (63.43°) = 36.3 kVAR

S

VL I L = 3 × 550 × 42.61 = 40.59 kVA

kVA rating of the load

8.16 Network Analysis and Synthesis

Example 8.9 A balanced star-connected load is supplied from a symmetrical three-phase 400 volts, 50 Hz system. The current in each phase is 30 A and lags 30° behind the phase voltage. Find the (a) phase voltage, (b) resistance and reactance per phase, (c) load inductance per phase, and (d) total power consumed. Solution

f = 50 Hz, I ph = 30 A, φ = 30°

VL

For a star-connected load, (a) Phase voltage V ph =

VL 3

=

400 3

= 230.94 V

(b) Resistance and reactance per phase

Z ph

V ph

230.94 = =77Ω I ph 30 = Z ph ∠φ = 7.7∠30° = (6.67 + j 3.85) Ω

Z ph =

R ph = 6 67 Ω X ph = 3 85 Ω (c) Load inductance per phase Xp ffL L ph 3 85 = 2π × 50 × L ph L ph = 0.01225 H (d) Total power consumed P

φ = 3 × 230.94 × 30 × cos (30°) = 18 kW

3V phh I phh

Example 8.10

A symmetrical three-phase 400 V system supplies a basic load of 0.8 lagging power factor and is connected in star. If the line current is 34.64 A, find the (a) impedance, (b) resistance and reactance per phase, (c) total power, and (d) total reactive voltamperes. Solution

VL

400

V ph =

VL

pf

0.8 (lagging), I L = 34.64 A

For a star-connected load, (a) Impedance

I ph Z ph

=

400

= 230.94 V 3 3 I L = 34.64 A V ph 230.94 = = = 6.67 6 Ω I ph 34.64

(b) Resistance and reactance per phase pf = cos φ = 0.8 (lagging)

φ = cos −1 (0.. ) = 36.87° Z ph = Z ph ∠φ = 6.67∠36.87° = (5.33

j 4) Ω

8.11 Comparison between Star and Delta Connections 8.17

R ph = 5 33 Ω X ph = Ω (c) Total power P

φ = 3 × 400 × 34.64 × 0.8 = 19.19 kW

3 VL I L

(d) Total reactive volt-amperes VL I L i φ = 3 × 400 × 34.64 × sin (36.87 ) = 14.4 kVAR

Q

Example 8.11 A balanced star-connected load is supplied by a 415 V, 50 Hz three-phase system. Current in each phase is 20 A and lags 30° behind its phase voltage. Find the (a) phase voltage, (b) power, and (c) circuit parameters. Also, find power consumed when the same load is connected in delta across the same supply. Solution

f = 50 Hz, I ph = 20 A, φ = 30°

VL

For a star-connected load, (a) Phase voltage V ph =

VL

=

3

415 3

= 239.6 V

(b) Power IL

I ph = 20 A

P

3 VL I L

φ = 3 × 415 × 20 × cos (30 ) = 12.45 kW

(c) Circuit parameters Z ph =

V ph I ph

=

239.6 = 20

Z ph = Z p ∠φ

.98 Ω

11.98∠30

j 6) Ω

R ph = 0.37 Ω X ph = 6 Ω Since

Xp 6

fL ph fL 2π × 50 × L ph

L ph = 19.1 mH (d) Power consumed by same delta load across the same supply PΔ

3PY = 3 × 12.45 × 103 = 37.35 kW

Example 8.12

Three identical coils connected in delta to a 440 V, three-phase supply take a total power of 50 kW and a line current of 90 A. Find the (a) phase current, (b) power factor, and (c) apparent power taken by the coils. Solution

VL

440

P = 50 kW, I L = 90 A

8.18 Network Analysis and Synthesis For a delta-connected load, (a) Phase current I ph =

IL

=

3

90

= 51.96 A

3

(b) Power factor 3 VL I L cos φ

P

50 × 103 = 3 × 440 × 90 × cos φ pf = cos φ = 0.73 ( ) (c) Apparent power VL I L = 3 × 440 × 90 = 68.59 kVA

S

Example 8.13 Three similar choke coils are connected in star to a three-phase supply. If the line current is 15 A, the total power consumed is 11 kW and the volt-ampere input is 15 kVA, find the line and phase voltages, the VAR input and the reactance and resistance of each coil. If these coils are now connected in delta to the same supply, calculate phase and line currents, active and reactive power. Solution

15 A, P = 11 kW, S = 15 kVA

IL

For a star-connected load, (a) Line voltage S

VL I L

3

15 × 10 = 3 × VL × 15 VL = 577.35 V (b) Phase voltage V ph =

VL

=

3

577.35 3

= 333.33 V

(c) VAR input P 11 × 103 S 15 × 103 φ = 42.86°

cos φ

Q=

VL I L sin i φ

0.733

3 × 577.35 × 15 × si s n ( 42.86°) = 10.2 kVAR

(d) Reactance and resistance of coil I ph

I L = 15 A

Z ph =

V ph I ph

=

333.33 = 15

.

Ω

R = Z phh cos φ = 22.22 × 0.733 = 16.29 Ω XL If these coils are now connected in delta,

Z ph sin φ = 22.22 × sin ( 42.

) = 155 11 Ω

8.11 Comparison between Star and Delta Connections 8.19

(a) Phase current V ph VL = 577.35 V Z ph = . Ω V ph 577.35 I ph = = = 25.98 A Z ph 22.22 (b) Line current IL

3 I pphh = 3 × 25.98 = 45 A

(c) Active power P

3 VL I L

φ = 3 × 577.35 × 45 × 0.733 = 32.98 kW

(d) Reactive power VL I L i φ = 3 × 577.35 × 45 × sin ( 42.86 ) = 30.61 kVAR

Q

Example 8.14

A three-phase, star-connected source feeds 1500 kW at 0.85 power factor lag to a balanced mesh-connected load. Calculate the current, its active and reactive components in each phase of the source and the load. The line voltage is 2.2 kV. Solution

P

1 00 k , pf

P

3 VL I L cos φ

0 8 (lagging), VL = 2.2 kV

For a mesh or delta-connected load, (a) Line current 3

1500 × 10 = 3 × 2.2 × 103 × I L × 0.85 I L = 463.12 A (b) Active component of current in each phase of the load I ph =

IL

=

463.12

= 267.38 A 3 3 I ph cos φ = 267.38 × 0.85 = 227.27 A (c) Reactive component of current in each phase of the load I ph sin φ = 267.38 × sin (cos −1 0.85)

.38 × 0.526 = 140.85 A

For a star-connected source, the phase current in the source will be the same as the line current drawn by the load. (d) Active component of this current in each phase of the source I L cos φ = 463.12 × 0.85 = 393.65 A (e) Reactive component of this current in each phase of the source I L sin φ = 463.12 × 0.526 = 243.6 A

Example 8.15

A three-phase, 208-volt generator supplies a total of 1800 W at a line current of 10 A when three identical impedances are arranged in a wye connection across the line terminals of the generator. Compute the resistive and reactive components of each phase impedance. Solution

VL

208

P = 1800 W

I L = 10 A

8.20 Network Analysis and Synthesis For a wye-connected load, VL

V ph =

=

208

3 3 I L = 10 A

I ph

= 120.09 V

Z ph =

V ph 120.09 = = I ph 10

P

3 VL I L cos φ

Ω

1800 = 3 × 208 × 10 × cos φ cos φ = 0.5 φ = 60° R ph = Z ph cos φ = 12 × 0.5 = 6 Ω X ph = Z ph sin φ = 12 × sin (60°) = 10.39 Ω

Example 8.16

A balanced, three-phase, star-connected load of 100 kW takes a leading current of 80 A, when connected across a three-phase, 1100 V, 50 Hz supply. Find the circuit constants of the load per phase. Solution

P

100 00 kW, I L = 80 A

f = 500 Hz

V

For a star-connected load, V ph = I ph Z ph = P

VL

=

1100

=

635.08 =79 Ω 80

3 3 I L = 80 A V ph I ph

= 635.08 V

3 VL I L cos φ

100 × 10 = 3 × 1100 × 80 × cos φ cos φ = 0.656 (leading) φ = 49° 3

R ph = Z ph cos φ = 7.94 × 0.656 = 5.21 Ω X ph = Z ph sin φ = 7.94 × sin ( 49°) = 6 Ω This reactance will be capacitive in nature as the current is leading. 1 2π fC 1 6= 2π × 50 × C C = 530.52 μF

XC =

8.11 Comparison between Star and Delta Connections 8.21

Example 8.17 Three identical impedances are connected in delta to a three-phase supply of 400 V. The line current is 34.65 A, and the total power taken from the supply is 14.4 kW. Calculate the resistance and reactance values of each impedance. Solution

I L = 34.65 65 A, P = 14.4 kW

VL

400

VL

V ph = 400 V

For a delta-connected load, I ph =

IL

Z ph =

V ph I ph

P

3

=

34.65

= 20 A 3 400 = = 20 Ω 20

3 VL I L cos φ

14.4 × 10 = 3 × 400 × 34.65 × cos φ cos φ = 0.6 φ = 53.13° R ph = Z ph cos φ = 20 × 0.6 = 12 Ω 3

X ph = Z ph sin φ = 20 × sin (53.13°) = 16 Ω

Example 8.18 A balanced, three-phase load connected in delta draws a power of 10.44 kW at 200 V at a power factor of 0.5 lead. Find the values of the circuit elements and the reactive volt-amperes drawn. Solution

P

10 44 k , VL = 200 V, pf = 0.5 (

)

For a delta-connected load, (a) Values of circuit elements VL P

V ph = 200 V 3 VL I L cos φ

3

10.44 × 10 = 3 × 200 × I L × 0.5 I L = 60.28 A I 60.28 I ph = L = = 34.8 A 3 3 V ph 200 Z ph = = = 5 75 Ω I ph 34.8 R ph

Z phh cos φ = 5.75 × 0.5 = 2.875 Ω

X ph

Z phh sin φ = 5.75 × sin in (cos ( −1 0.5) = 5.747 × 0.866 = 4 98 Ω

(b) Reactive volt-amperes drawn Q

VL I L i φ = 3 × 200 × 60.28 × 0.866 = 18.08 kVAR

8.22 Network Analysis and Synthesis

Example 8.19 Each leg of a balanced, delta-connected load consists of a 7 W resistance in series with a 4 W inductive reactance. The line-to-line voltages are 2360 0 ° V , Ebc = 2360 ∠ 120° V , Eca = 2360 ∠120° V

Eab Determine

(a) phase current Iab, Ibc, and Ica both magnitude and phase) (b) each line current and its associated phase angle (c) the load power factor

Solution

7 Ω,

R

XL =

Ω

VL = 2360 V

For a delta-connected load, (a) Phase current V ph VL = 2360 V Z ph = 7 + j 4 = 8.06 ∠29.74° Ω Eab 2360 ∠0° = = 292.8∠ ∠− 29.74° A Z ph 8.06 ∠29.74° E 2360 ∠− 120° = bc = = 292.8∠− 149.71° A Z ph 8.06 29.74° E 2360 ∠120° = ca = = 292.8 90.26° A Z ph 8 06 29 74°

Iab = Ibc Ica

(b) Line current In a delta-connected, three-phase system, line currents lag behind respective phase currents by 30°. IL 3 I pphh = 3 × 292.8 = 507.14 A I La = 507.14 ∠ − 59.71°A I Lb = 507.14 ∠− ∠ 179.71°A I Lc = 507.14 ∠ ∠60 60.26°A (c) Load power factor pf = cos ( 29.

) = 0.

(lagging)

Example 8.20

A three-phase, 200 kW, 50 Hz, delta-connected induction motor is supplied from a three-phase, 440 V, 50 Hz supply system. The efficiency and power factor of the three-phase induction motor are 91% and 0.86 respectively. Calculate (a) line currents, (b) currents in each phase of the motor, (c) active, and (d) reactive components of phase current. Solution

Po

200 kW

VL = 440 V,

For a delta-connected load (induction motor), (a) Line current

η=

Output power Po = Input power Pi

200 × 103 Pi Pi = 219.78 kW

0 91 =

f = 50 Hz, η = 91%, pf = 0 86

8.11 Comparison between Star and Delta Connections 8.23

3 VL I L cos φ

Pi 3

219.78 × 10 = 3 × 440 × I L × 0.86 I L = 335.3 A (b) Currents in each phase of motor I ph =

IL 3

=

335.3 3

= 193.6 A

(c) Active component of phase current I ph cos φ = 193.6 × 0.86 = 166.5 A (d) Reactive component of phase current I ph sin φ = 193.6 × sin in (cos ( −1 0.86)

.6 × 0.51 = 98.7 A

Example 8.21 A three-phase, 400 V, star-connected alternator supplies a three-phase, 112 kW, mesh-connected induction motor of efficiency and power factor 0.88 and 0.86 respectively. Find the (a) current in each motor phase, (b) current in each alternator phase, and (c) active and reactive components of current in each motor phase, (d) active and reactive components of current in each alternator phase. Solution

VL

Po = 112 kW, η = 0 88

400

pf = 0.86

For a mesh-connected load (induction motor), (a) Current in each motor phase V ph

η=

VL = 400 V Output power Po = Input power Pi

112 × 103 Pi Pi = 127.27 kW

0 88 =

Pi

3 VL I L cos φ

3

127.27 × 10 = 3 × 400 × I L × 0.86 I L = 213.6 A I 213.6 I ph = L = = 123.32 A 3 3 Current in a star-connected alternator phase will be same as the line current drawn by the motor. (b) Current in each alternator phase I L = 213.6 A (c) Active component of current in each phase of a motor I ph cos φ = 123.32 × 0.86 = 105.06 A Reactive component of current in each phase of the motor I ph sin φ = 123.32 × sin (cos −1 0.86)

.32 × 0.51 = 62.89 A

8.24 Network Analysis and Synthesis (d) Active component of current in each alternator phase I L cos φ = 213.6 × 0.86 = 183.7 A Reactive component of current in each alternator phase I L sin φ = 213.6 × sin in (cos ( −1 0.86)

.6 × 0.51 = 108.94 A

Example 8.22

Three similar resistors are connected in star across 400 V, three-phase lines. The line current is 5 A. Calculate the value of each resistor. To what value should the line voltage be changed to obtain the same line current with the resistors connected in delta? Solution

VL

IL = 5 A

400

For a star-connected load, V ph = I ph Z ph

VL

=

400

= 230.94 V 3 3 IL = 5 A V ph 230.94 = R phh = = = 46.19 Ω I ph 5

For a delta-connected load, IL = 5 A R ph = 6. 9 Ω I 5 I ph = L = A 3 3 5 V ph I phh R phh = × 46.19 = 133.33 V 3 VL = 133.33 V Voltage needed is one-third of the star value.

Example 8.23

Three 100 W, non-inductive resistors are connected in (a) star, and (b) delta across a 400 V, 50 Hz, three-phase supply. Calculate the power taken from the supply in each case. If one of the resistors is open-circuited, what would be the value of total power taken from the mains in each of the two cases? Solution VL For a star-connected load, V ph =

Z ph = 100 Ω

400 VL

400

= 230.94 V 3 V ph 230.94 I ph = = = 2 31 A Z ph 100 I L I ph = 2 31 A 3

=

cos φ = 1 P = 3 VL I L cos φ = 3 × 400 × 2.31 × 1 = 1600.41 W

( For pure resistor, pf = 1)

8.11 Comparison between Star and Delta Connections 8.25

For a delta-connected load, V ph

VL = 400 V

I ph =

V ph 400 = =4A Z ph 100

IL

3 I pphh = 3 × 4 = 6.93 A

P

3 VL I L

φ = 3 × 400 × 6.93 × 1 = 4801.24 W A

When One of the Resistors is Open-circuited

100 Ω 400 V

1. Star Connection The circuit consists of two 100 Ω resistors in series across a 400 V supply as shown in Fig. 8.14.

100 Ω

C

400 Currents in lines A and dC A 200 Power taken from the mains 400 0 × 2=800 W

B

Fig. 8.14 Delta connection

Hence, when one of the resistors is open circuited, the power consumption is reduced by half. 2. Delta Connection In this case, currents in A and C remains as usual 120° out of phase with each other as shown in Fig. 8.15.

A 100 Ω

400 V

C 100 Ω

400 Currents in each phase = =4A 100 Power taken from the mains 2 × 4 × 400 = 3200 W

B

Fig. 8.15 Delta connection

Hence, when one of the resistors is open-circuited, the power consumption is reduced by one-third. Three identical impedances of 10 ∠30° Ω each are connected in star and another set of three identical impedances of 18 ∠60° Ω are connected in delta. If both the sets of impedances are connected across a balanced, three-phase 400 V, supply, find the line current, total volt-amperes, active power and reactive power.

Example 8.24

Solution

ZY

10 ∠30° Ω

Z = 18∠

Ω

VL = 400 V

Three identical delta impedances can be converted into equivalent star impedances. Z′Y =

Z Δ 18 ∠60° = = 6 ∠60° Ω 3 3

Now two star-connected impedances of 10 ∠30 Ω and 6 ∠60 Ω are connected in parallel across a threephase supply. Zeq =

( ∠ )(6 )(6 ∠60°) = 3.87∠48.83° Ω 10 ∠30° + 6 ∠60°

8.26 Network Analysis and Synthesis For a star-connected load, (a) Line current V

=

I ph = IL =

V 3 V ph Z ph ph

400

= =

3 V ph

= 230 94 V 230 94 3 87

Z

59 67 A

59.67 A

(b) Total volt-amperes 3×

× 59.

41.34 kVA

(c) Active power =

×

×

.67 ×

= 27 21 kW

=

×

×

.67 ×

= 31 12 kVAR

(d) Reactive power

Example 8.25

Three star-connected impedances ZY = + j37.7 per phase are connected in parallel with three delta-connected impedances Z Δ − per phase. The line voltage is 398 V. Find the line current, pf, active and reactive power taken by the combination. Solution

V

Z

39

Three identical delta-connected impedances can be converted into equivalent star impedances. Z′Y =

162 ∠− 9 3 3

−

Now two star-connected impedances of 42 across the three-phase supply. ( .68 Zeq = 42 68

Ω and 54.03∠− 79.3 Ω are connected in parallel

62 05

=

3°

9 88

For a star-connected load, (a) Line current V ph I ph IL (b) Power factor

pf =

V

398 3

V ph Z ph ph

3 V ph Zeq 3 36

= 229.79 V 229 79 68.33

= os .88 )

3 36 A

99 (lagging)

(c) Active power ×

×

×

×

×

6×

=

9 kW

(d) Reactive power =

= 397.43 VAR

8.11 Comparison between Star and Delta Connections 8.27

Example 8.26 Three coils, each having a resistance of 20 W and a reactance of 15 W, are connected in star to a 400 V, three-phase, 50 Hz supply. Calculate(a) line current, (b) power supplied, and (c) power factor. If three capacitors, each of same capacitance, are connected in delta to the same supply so as to form a parallel circuit with the above coils, calculate the capacitance of each capacitor to obtain a resultant power factor of 0.95 lagging. Solution

R ph

20 Ω,

X ph = 15 Ω, VL = 400 V

For a star-connected load, (a) Line current Z ph = R ph p V pph =

VL

I ph =

V ph

IL

jX ph p =

3 Z ph

20

j15 = 25 ∠36.87° Ω

400

= 230.94 V 3 230.94 = = 9 24 A 25

I ph = 9 24 A

(b) Power supplied P1

3 VL I L

φ1

3 × 400 × 9.24

s (36.87 )

5.12 kW

(c) Power factor pf = cos φ1 = cos (

.87°) = .8 (lagging)

(d) Value of capacitance of each capacitor Q

VL I L i φ1

3 × 400 × 9.24

i (36.87 )

3.84 kVAR

When capacitors are connected in delta to the same supply, pf = 0 95

φ2 = cos −11 (0.95) = 18 19° tan φ2 = tan (18.19°) = 0.33 Since capacitors do not absorb any power, power remains the same even when capacitors are connected. But reactive power changes. P2 = 5.12 kW Q P2 tan φ2 = 5.12 0.33 = 1.69 kVAR Difference in reactive power is supplied by three capacitors. From Fig. 8.16, Q Q 2.15 103

Q1 − Q2 = 3.84 − 1.69 = 2.15 kVAR VL I L sin φ

P1 = P2 = P f2 f1

Q2 S2 S1

3 × 400 × I L × sin (90 )

IL = 3 1 A

Fig. 8.16

Q1

8.28 Network Analysis and Synthesis I ph =

IL

I ph =

V ph

C=

8.12

3

= 1 79 A = Vp

XC

fC

I ph Vp

f

=

1 79 = 14 1 .24 μF 400 × 2π × 50

THREE-PHASE UNBALANCED CIRCUITS

If the loads connected across the three phases are not identical to each other, i.e., the loads have different magnitude and power factors, the loads are said to be unbalanced. The phase currents in delta and phase or line current in star connection differ in unbalanced loading giving rise to flow of current, in a neutral wire. I R + IY

IN

IB

There may be three cases of unbalanced loads: (i) Unbalanced delta-connected load (ii) Unbalanced three-wire star-connected load (iii) Unbalanced four-wire star-connected load

8.12.1 Unbalanced Delta-connected Load Figure 8.17 shows an unbalanced delta-connected load connected to a balanced three-phase supply. For a delta-connected load, Let

VL

V ph

VRRY = V ph ∠0° VYB = V ph ∠− 120°

IR R

VBR = V ph ∠− 240°

IBR

Phase currents are given by V = RRY Z RY

Y

IYB =

VYYB ZYB

B

I BR =

VBBR Z BR

I RY

IY

ZRY

ZBR ZYB

IB IYB

Fig. 8.17 Unbalanced delta-connected load

Line currents are given by IR

I RY − I BR

IY

IYB

I RY

IB

I BR

IYB

IRY

8.12 Three-Phase Unbalanced Circuits 8.29

8.12.2

Unbalanced Four-wire Star-connected Load

Figure 8.18 shows an unbalanced star-connected load connected to a balanced 3-phase, 4-wire supply. The neutral point N of the load is connected to the neutral point O of the supply. Hence, voltage across the three load impedances is equal to the phase voltage of the supply.

IR R ZR IN

N

For a star-connected load,

N

IY

VL

VRRN = V ph ∠0°

Let

ZB

Y

3 V ph

ZY

IB

VYN = V pph ∠− 120°

B

VBN = V pph ∠− 240° Fig. 8.18 Phase currents are given by

Unbalanced four-wire star-connected load

VRN ZR V IY = YN ZY VBN IB = ZB IR =

For a star-connected load, phase currents are equal to the line currents. The current in a neutral wire is given by IN

I R + IY

IB

8.12.3 Unbalanced Three-wire Star-connected Load Figure 8.19 shows an unbalanced three-wire star-connected load. R

ZR N

O ZB

ZY

Y B

Fig. 8.19

Unbalanced three-wire star-connected load

If the neutral point N of the load is not connected to the neutral point O of the supply, there exists a voltage between the supply neutral point and the load neutral point. The load phase voltage is not the same as the

8.30 Network Analysis and Synthesis supply phase voltage. There are many methods to solve such unbalanced star-connected loads. Two most commonly used methods are (i) Star-delta transformation (ii) Millman’s theorem, In star-delta transformation, the star-connected loads, are replaced by an equivalent delta-connected load. Z R ZY ZB ZY Z B = ZY + Z B + ZR Z Z = ZB + ZR + B R ZY

Z RY = Z R + ZY + ZYB Z BR

The problem is then solved as an unbalanced delta-connected load. The line currents so calculated are equal to the line currents of the original star-connected load. An unbalanced three-wire star-connected load can also be solved by Millman’s theorem. Let VRO , VYO and VBO be the phase voltages of the supply which are equal in magnitude but differ in phase from one another by 120°. Let VRN , VYN and VBN be the load phase voltages which are unequal in magnitude as well as differ in phase by unequal angles. The voltage between the neutral points, N and O, i.e, VNO , is calculated using Millman’s theorem. VNO =

VRO YR + VYO YY + VBO YB YR YY + YB

where YR YY and YB are the admittances of the branches of the unbalanced star-connected load. The voltage across phase impedance, i.e., load phase voltage can be calculated as, VRN VYN

VRO − VNO VYO − VNO

VBN

VBO − VNO

The phase currents in the load are given by VRN ZR V IY = YN ZY V I B = BR ZB IR =

The line currents are equal to the phase currents for a star-connected load.

Example 8.27

A three-phase supply with a line voltage of 250 V has an unbalanced delta-connected load as shown in Fig. 8.20.

8.12 Three-Phase Unbalanced Circuits 8.31 IA

A

A V

CA

25 V

ZCA 25 ∠90° Ω

IB

B

AB = 20

∠0° Ω

IAB IC

C

BC

B

C ZBC 15 ∠20° Ω

Fig. 8.20 Determine (a) phase currents, (b) line currents, (c) power, if phase sequence is ABC.

=

Solution Let

Z

=1

Z

∠−

total active power and (d)

°

total reactive

° V, 5 90° Ω

20

(a) Phase currents VAB 250 ∠0° = ∠ ° Z AB 20 ∠0° VBC 250 ∠− 120 .67∠− 140 A Z BC 15 ∠20 V 250 ∠− 240° = = = 10 ∠30°A ZCA 25 ∠90°

I AB = I BC I (b) Line currents

A

∠ °− ∠− 14

− .48°A 27.45∠− 157 02°

B

=

C

= 10 ∠30° −16.67∠− 140° = 2

0

∠36 25

(c) Total active power =

×

=2 = × +

=

× co

13 kW

× os ( 20 ) × cos ( ) 0 +

3 92 kW

.05 kW

(d) Total reactive power

+

Example 8.28

= × × sin °) 0 = 250 × 16 67 × 43 kVAR = × × sin ( ) 6.25 kVAR = 0 + 1 43 + .68 kVAR

In the circuit of Fig. 8.21, a 400 V, 50 Hz, 3-phase supply of phase sequence ABC is supplied to a delta-connected load consisting of a 100 W resistor between lines A and B, a 378 mH inductor between lines B and C, and a 37.8 μF capacitor between lines C and A. Determine phase and line currents.

8.32 Network Analysis and Synthesis IA

A

ICA

VAB B VCA B

ZCA

IB

ZAB IAB

VBC

ZBC

IC

C

IBC

Fig. 8.21 VL = 400 V,

Solution

f = 50 Hz, R 100 Ω, L = 318 mH, C = 31.8 μF

VAB = 400 ∠0° V VBC = 400 ∠ − 120° V VCA = 400 ∠ − 240° V

Let

ffL = 2π × 50 × 318 × 10 −3 = 99.9 Ω 1 1 XC = = = 103. 35 Ω 2π fC 2π × 50 × 31 8 × 10 −6 X

Z AB = R = 100 ∠0 Ω Z BC = jX L = j 99.9 = 99 99.9 ∠90° Ω ZCA = − jX C = − j103.35 = 103.35 3 ∠− 90° Ω (a) Phase currents VAB 400 ∠0° = = 4 ∠0°A Z AB 100 ∠0° V 400 ∠− 120° = BC = = 4 ∠150°A Z BC 99.9∠90° V 400 ∠− 240° = CA = = 3 87∠− 150°A ZCA 103.35∠− 90°

I AB = I BC ICA (b) Line currents

I A I AB − ICA = 4 ∠0° − 3.87∠− 150° = 7.6 ∠14.75°°A I B I BC − I AB = 4 ∠150° − 4 ∠0° = 7.73∠165°A IC = ICA − I BC = 3.87 87 8 ∠−150° 1500° − 4 ∠150 150 50° = 3.94 ∠ ∠− 88.36°A

Example 8.29 Z A (10 conductor.

A three-phase, 400 V, 4-wire system of Fig. 8.22 has a star-connected load with j0 ) Ω , Z B = (15 + j10 j10 ) , ZC (0 + j5 ) Ω . Find the line currents and current through neutral

8.12 Three-Phase Unbalanced Circuits 8.33 IA A VL = 400 0V

ZA IB

IN

B

N

ZC

ZB

IC C

Fig. 8.22 Solution

VL = 400 V Z A = (10 + j10) Ω = 10 ∠0 Ω Z B = ( + j ) Ω = 18.03∠ .69 Ω Z C = ( + j ) Ω = 5∠ 9 0 Ω

For a star-connected load, V ph =

VL 3

=

400 3

= 230.94 V

The phase sequence is assumed as A-B-C. VAN = 230.94 ∠0° V VBN = 230.94 ∠− 120° V VCN = 230.94 ∠− 240° V Phase currents VAN 230.94 ∠0° = = 23.09∠0°A ZA 10 ∠0° V 230.94 ∠− 120° I B = BN = = 12.81 8 ∠− 153.69°A ZB 18.03 33 69° V 230.94 ∠− 240° IC = CN = = 46.19∠30 3 °A ZC 5 90° IA =

Line currents are equal to the phase currents in a star-connected load. I L1

I A = 23.09∠0°A

I L2

I B = 12.81∠ − 153.69°A

I L3

IC = 46.19∠30°A

Current through neutral conductor IN

IA + IB

IC = 23.09∠0° + 12.81∠ ∠− 153.69° + 46.19∠30° = 544.447∠18.65°A

8.34 Network Analysis and Synthesis

Example 8.30

A 3-phase, 4 wire, 208 V, CBA system as shown in Fig. 8.23 has a star-connected

load with Z A 5 0° Ω Z B = 5 30° Ω , ZC current through neutral wire.

10 10

60° Ω . Obtain the phase currents, line currents and

IC C VL = 208 8V

ZC IB

B

N ZA

ZB

IA A

Fig. 8.23 Solution

VL = 208 V

Z A = 5∠0 Ω, Z B = 5∠30° Ω

C

= 10 ∠− 60° Ω

For a star-connected load, VL

V ph = VCCN VBN VCN

=

208

= 120.09 V 3 3 = 120.09∠0°V = 120.09∠− 120° V = 120.09∠− 240° V

(a) Phase currents VCN 120.09∠0° = = 12∠60° A ZC 10 ∠− 60° V 120.09∠− 120° I B = BN = = 24.02∠− 150° A ZB 5∠30° V 120.09∠− 240° I A = AN = = 24.02∠120° A ZA 5∠0°

IC =

(b) Line currents are equal to the phase currents in a star-connected load. I L1

IC = 12∠60°A

I L2

I B = 24.02∠− 150°A

I L3

I A = 24.02∠120°A

(c) Current through neutral wire IN

IC + I B

I A = 12∠60° + 24.02∠ ∠− 150° + 24.02∠120°

32.97∠144.42°° A

8.12 Three-Phase Unbalanced Circuits 8.35

Example 8.31 A symmetrical 440 V, 3-phase system supplied a star-connected load with the branch j Ω , Z B = − j5 Ω as shown in Fig. 8.24. Calculate line currents and voltage impedances Z R 10 Ω ZY = j5 across each phase impedance by Millman’s theorem. The phase sequence is RYB. IR

R

ZR

VRY O Y

ZB

VBR IY

N ZY

VYB IB

B

Fig. 8.24 Solution

V

10 Ω

440

j5 Ω

B

= − j5 Ω

For a star-connected load, V ph =

VL 3

=

440 3

= 254.03 V

Let O be the neutral point of the supply system. Let

VRO = 254.03∠0° V VYO = 254.03∠ − 120° V VBO = 254.03∠ − 240° V YR =

1 1 1 = = = 0 1∠0° � Z R 10 10 ∠0°

YY =

1 1 1 = = = 0 2∠− 90° � ZY j 5 5∠ 9 0°

YB =

1 1 1 = = = 0 2∠90° � Z B − j 5 5∠− 90°

By Millman’s theorem, VRO YR + VYO YY + YBO YB YR YY + YB ( .03∠0°)(0 )(0. ∠ ) + ( .03∠− 120°)(0.2∠ ∠− 90 90°) + ( 254.03∠− 240°)(0.2∠90 90°) = 0.1∠0° + 0.2∠ ∠− 90° + 0.2 90° = 625.96 ∠180° V

VNO =

8.36 Network Analysis and Synthesis Voltage across phase impedance − VNO − VNO 254 0 ∠− 120 − = 254.03∠− 24 °

0 V = 29∠− 23 79 V ° 545 29 23 79

Phase currents/ line currents IR IY IB

880 ∠0 = ∠0 A 10 ∠0° 545.29∠− .79 5∠90° 545. ∠23.79° 109 5 90°

VRN ZR VYN ZY VBN ZB

−

79 A

113 79 A

Example 8.32

In the circuit of Fig. 8.25, a symmetrical 3-phase 100 V, three-wire supply feeds an unbalanced star-connected load, with impedances of the load as Z = =4 . Find the line currents and voltage across the impedance using star-delta transformation method. IR

R

5Ω N

−j4 Ω

IY

Y

j2 Ω

IB

C

Fig. 8.25 Solution

= ∠

V

∠− 90 Ω

The unbalanced star-connected load can be converted into equivalent delta-connected load by star-delta transformation technique. + + +

ZB ZR ZY

= ∠

+

= ∠ =4

− 9

The equivalent delta-connected load is shown in Fig. 8.26. For a delta-connected load, Let

VRY

= 100 V 0 V

∠ ° 4 ∠− 90 ∠ + 5∠0° 4 90 + 2 90°

38 66° °

=

∠−

34° Ω

6.4 ∠− 141.34° Ω

8.12 Three-Phase Unbalanced Circuits 8.37

VYB = VBR =

∠ 120° V ∠ 240° V

IR

R

IBR

Phase currents I RY IYB I BR

100 ∠0° ∠38 66 100 ∠− 6 5 4 100 ∠− 240 6 4 ∠− 1 34°

VRY Z RY VYB ZYB VBR Z BR

ZBR

IY

Y

38 66 A

ZRY IRY

39 06 − 68 66° A

ZYB

IB

B

IYB

98 66° A

15

Fig. 8.26

Line currents °

R

∠− −

39. −

−

°

.66

19 7 −

.

−

.

5 14° Α 36° Α

These line currents are equal to the line currents of the original star-connected load. For a star-connected load, I I RN

IL 5 A

IYN I BN

=

−121 14° A 128 36° A = ∠ ( 2∠9 = 4 90

− . − 26 69∠

= 135.3∠− 8.65 V 39 4 1 14° V 36 . V

Example 8.33

A three-phase three-wire unbalanced load is star-connected as shown in Fig. 8.27. V = V Calculate voltage The phase voltage of two of the arms are between star point of the load and the supply neutral. R

O

N

Y

B

Solution

V

Let

VRO VYO

= V 0 V 120°

Fig. 8.27 °V

8.38 Network Analysis and Synthesis From Fig. 8.27, VRO VNO

VRN + VNO VRO − VRN = V∠0° − 100°∠− 10°

…(i)

VNO

VYO − VYN = V∠ − 120° − 11500 ∠100°

…(ii)

Similarly, From Eqs. (i) and (ii), V∠0° 100 ∠ 10° = V∠−120 V∠ 120° −150 ∠100° V∠0° V∠ V∠− 120° = 100 ∠− 10° −150 ∠100° 3 V 30° 206.79 52 97° V = 119.39∠− 82.97 VNO = 119.39∠ ∠− 82.97 − 100 ∠− 10 = 131.88∠ ∠− 129.67° V

8.13

MEASUREMENT OF THREE-PHASE POWER

In a three-phase system, total power is the sum of powers in three phases. The power is measured by wattmeter. It consists of two coils: (i) Current coil, and (ii) Voltage coil. Current coil is connected in series with the load and it senses current. Voltage coil is connected across supply terminals and it senses voltages. There are three methods to measure three-phase power: 1. Three-wattmeter method 2. Two-wattmeter method 3. One-wattmeter method

8.13.1 Three-Wattmeter Method This method is used for balanced as well as unbalanced loads. Three wattmeters are inserted in each of the three phases of the load whether star connected or delta connected as shown in Fig. 8.28. Each wattmeter will measure the power consumed in each phase. Forr balanced load , W1 W2 = W3 Forr unbalanced load , W1 W2 ≠ W3 Total t power P = W1 W2 + W3 W1 R Zph R

N

N Zph

Zph

Zph Y

W2 Y

W1

W3 Zph

Zph

B W2

B

W3

Fig. 8.28

Three-wattmeter method

8.13 Measurement of Three-Phase Power 8.39

8.13.2 Two-Wattmeter Method This method is used for balanced as well as unbalanced loads. The current coils of the two wattmeters are inserted in any two lines and the voltage coil of each wattmeter is joined to a third line. The load may be star or delta connected as shown in Fig. 8.29. The sum of the two wattmeter readings gives three-phase power. R

R

W1 Zph

Y

W2

Y

N

Zph

W1

Zph

W2

Zph Zph

Zph

B

B (b)

(a)

Two-wattmeter method

Fig. 8.29 Total power P

W1 + W2

8.13.3 One-Wattmeter Method This method is used for balanced loads only. When the load is balanced, total power is given by 3 V ph I ph cos φ

P

Hence, one wattmeter is used to measure power in one phase. The wattmeter reading is then multiplied by three to obtain three-phase power. The load may be star or delta connected as shown in Fig. 8.30. R

W1

R

Y Zph

N

N Zph

Zph

Zph

Zph

Zph

B W

B

Fig. 8.30

One-wattmeter method

8.13.4 Measurement of Power by Two-wattmeter Method Figure 8.31 shows a balanced star-connected load and this load may be assumed to be inductive. Let VRN, VYN and VBN be the three phase voltages and IR, IY and IB be the phase currents. The phase currents will lag behind their respective phase voltages by angle f. Current through current coil of W1 I R Voltage across voltage coil of W1 VRB = VRN

VNB = VRN

VBN

8.40 Network Analysis and Synthesis Figure 8.31 shows the phasor diagram of a balanced star-connected inductive load. VRB VRN

IR

VNB

° 30

f

VYB 30° f

VBN

VYN

IY

Fig. 8.31 Phasor diagram

φ ). )

From the phasor diagram, it is clear that the phase angle between VRB and IR is ( W1 Current through current coil of W2 Voltage across voltage coil of W2

VRB I R cos (30° − φ ) IY VYB = VYN

VNB = VYN

VBN

From the phasor diagram, it is clear that phase angle between VYB and IY is ( But

W2 IR VRB W1 W2

φ )).

VYB IY cos (30° + φ ) IY = I L VYB = VL VL I L cos (30° − φ ) VL I L cos (30° + φ )

W1 W2 = VL I L [cos (30

φ ) + cos (30° φ ))] = VL I L ( 2 c 30°

Thus, the sum of two wattmeter readings gives three-phase power.

8.13.5 Measurement of Power Factor by Two-wattmeter Method 1. Lagging Power Factor W1 W2

VL I L cos (30° − φ ) VL I L cos (30° + φ ) ∴W W1 W2

W1 W2 = 3 VL I L cos φ W1 W2 = VL I L [cos (30 φ ) − cos (30° φ ) = VL I L sin φ W1 − W2 V I sin φ = L L W1 + W2 3VL I L cos φ W − W2 tan φ = 3 1 W1 + W2

s φ)

3 VL I L cos o φ

8.13 Measurement of Three-Phase Power 8.41

⎛ W − W2 ⎞ φ = tan −1 ⎜ 3 1 ⎝ W1 + W2 ⎟⎠ ⎧ ⎛ W − W2 ⎞ ⎫ pf = cos φ = cos ⎨tan −1 ⎜ 3 1 ⎬ ⎝ W1 + W2 ⎟⎠ ⎭ ⎩ 2. Leading Power Factor Figure 8.32 shows the phasor diagram of a balanced star-connected capacitive load.

VRB VRN

W1 VL I L cos (30° + φ ) W2 VL I L cos (30° − φ ) ∴W W1 < W2

f

VBN

− +

)⎞ ) ⎟⎠

IB

30°

f

f

W1 W2 = 3 VL I L cos φ W1 W2 = −VL I L sin φ ( ) tan φ = − 3 ( ) ⎛ ( φ = tan −1 − 3 ⎝ (

VNB

IR

3 ° 30

VYB IY VYN

Fig. 8.32 Phasor diagram

⎧ W − W2 ⎞ ⎫ ⎛ pf = cos φ = cos ⎨tan −1 ⎜ − 3 1 ⎬ ⎝ W1 + W2 ⎟⎠ ⎭ ⎩

Example 8.34

Two wattmeters are used to measure power in a three-phase balanced load. Find the power factor if (a) two readings are equal and positive, (b) two readings are equal and opposite, and (c) one wattmeter reads zero. Solution (a)

Power factor if two readings are equal and positive W1

W2

tan φ = 3

W1 W2 W1 W2

3 ( 0)

0

φ = 0° Power factor = cos φ = cos (0°) = 1 (b) Power factor if two readings are equal and opposite W1

W2 W W2 tan φ = 3 1 =∞ W1 W2

φ = 90° Power factor = cos φ = cos (90°) = 0

8.42 Network Analysis and Synthesis (c) Power factor if one wattmeter reads zero W2 = 0 W1 W2 W1 W2 φ = 60°

tan φ = 3

⎛W ⎞ 3⎜ 1⎟ = 3 ⎝ W1 ⎠

Power factor = cos φ = cos (60°) = 0 5

Example 8.35 What will be the relation between readings on the wattmeter connected to measure power in a three-phase balanced circuit with (a) unity power factor, (b) zero power factor, and (c) power factor = 0.5. Solution (a) Relation between wattmeter readings with power factor = 1 cos φ = 1 φ = 0° tan φ tan ( ) = 0 W1 − W2 W1 + W2 W − W2 0= 3 1 W1 + W2 W1 = W2

tan φ = 3

(b) Relation between wattmeter readings with power factor = 0 cos φ = 0 φ = 90° tan φ tan ( tan φ = 3

)=∞

W1 − W2 W1 + W2

W1 + W2 = 0 W1 = −W2 (c) Relation between wattmeter readings with power factor = 0.5 cos φ = 60° tan φ tan (60°) = 1.732 W − W2 tan φ = 3 1 W1 + W2 W1 W2 W1 + W2 W2 = W1 + W2 W2 = 0

1.732 = 3 W1

8.13 Measurement of Three-Phase Power 8.43

Example 8.36

In a balanced three-phase system, the power is measured by two-wattmeter method and the ratio of two-wattmeter readings is 4:1. The load is inductive. Determine the load power factor. Solution

W1 4 = W2 1

Load power factor W1

4 W2

tan φ = 3

W1 W2 W1 W2

3

(3W2 ) ⎛ 3⎞ = 3 ⎜ ⎟ = 1.039 ⎝ 5⎠ (5W2 )

φ = 46.1° pf = cos φ = cos ( 46.1°) = 0.0693 (

)

Example 8.37

Find the power and power factor of the balanced circuit in which the wattmeter readings are 5 kW and 0.5 kW, the latter being obtained after the reversal of the current coil terminals of the wattmeter. Solution W1 5 k , W2 0 5 kW (a) Power When the latter reading is obtained after the reversal of the current coil terminals of the wattmeter, W1 = 5 kW W2 = −0 5 kW Power = W1 W2 = 5 ( 0.5) 4.5 kW (b) Power factor W1 − W2 ( = 3 W1 + W2 ( φ = 64.72°

.5) = 2.12 .5)

tan φ = 3

Power factor = cos o φ = cos (64.

) = 0.43

Example 8.38

Two wattmeters connected to measure the input to a balanced, three-phase circuit indicate 2000 W and 500 W respectively. Find the power factor of the circuit (a) when both readings are positive and (b) when the latter is obtained after reversing the connection to the current coil of one instrument. Solution

W1

2000

W2 = 500 W

(a) Power factor of the circuit when both readings are positive W1 = 2000 W W2 = 500 W W − W2 ( tan φ = 3 1 = 3 W1 + W2 ( φ = 46.102° Power factor = cos f = cos (46.102°) = 0.693

) = 1.039 )

8.44 Network Analysis and Synthesis (b) Power factor of the circuit when the latter reading is obtained after reversing the connection to the current coil of one instrument W1 = 2000 W W2 = −500 W W W2 tan φ = 3 1 W1 W2 φ = 70.89°

3

( 2000 + 500) = 2.887 ( 2000 − 500)

Power factor = cos f = cos (70.89°) = 0.33

Example 8.39 A three-phase, 10 kVA load has a power factor of 0.342. The power is measured by the two-wattmeter method. Find the reading of each wattmeter when the (a) power factor is leading, and the (b) power factor is lagging. Solution

S = 10 kVA, pf = 0.342 S

VL I L

3

10 × 10 = 3 VL I L VL I L = 5.77 kVA cos φ = 0.342 φ = 72° (a) Reading of each wattmeter when the power factor is leading W1

VL I L cos (30° + φ ) = 5.77 × 103

W2

VL I L cos (30° − φ )

s (30° 70°)

3

..77 7 × 100 coss (30° 70°)

1 kW .42 kW

(b) Reading of each wattmeter when the power factor is lagging W1 W2

VL I L cos (30° φ ) = 4.42 kW VL I L cos (30° + φ ) = −1 kW

Example 8.40

A three-phase, star-connected load draws a line current of 20 A. The load kVA and kW are 20 and 11 respectively. Find the readings on each of the two wattmeters used to measure the three-phase power. Solution

IL S

, S = 20 kVA, P = 11 kW VL I L

3

20 × 10 = 3 VL I L VL I L = 11.55 kVA P

3 VL I L cos φ

11 × 10 = 20 × 103 × cos φ cos φ = 0.55 φ = 56.63° 3

W1 = VL I L cos (30° φ ) = 11.55 × 103 × cos (30° 56.63 ) = 10.32 kW W W2

VL I L cos (30° + φ ) = 11.55 × 103 × cos (30° + 56 56.63 .63 ) = 0 68 kW

8.13 Measurement of Three-Phase Power 8.45

Example 8.41 Calculate the total power and readings of the two wattmeters connected to measure power in three-phase balanced load, if the reactive power is 15 kVAR and load pf is 0.8 lagging. Solution

Q = 15 kVAR, pf = 0.8 (lagging)

(a) Reading of the two wattmeters cos φ 0.8 φ = 36.87° Q = 3 VL I L sin φ 15 × 103 = 3 VL L in (36.87 ) VL I L = 14.43 kVA W1 W2

VL I L cos (30° φ ) = 14.43 × 103 × cos (30° − 36.87 . ) 14 03 kW VL I L co c s (30° + φ ) .43 × 103 × cos (30 .43 (30° + 36. 36.. ) 5.67 kW

(b) Total power P

W1 + W2 = 14.03 + 5.67 = 19.7 kW

Example 8.42 Two wattmeters are connected to measure power in a three-phase circuit. The reading of one of the wattmeters is 5 kW when the load power factor is unity. If the power factor of the load is changed to 0.707 lagging without changing the total input power, calculate the readings of the two wattmeters. Solution

W1 = 5 kW

(a) When the power factor is unity, W1 W2 = 5 kW Power = W1 W2 = 5 + 5 = 10 kW

…(i)

(b) When the power factor is changed to 0.707 lagging, pf = cos φ = 0.707 (lagging) φ = 45° tan φ = tan ( 45°) = 1 W1 − W2 W1 + W2 W − W2 1= 3 1 10 10 W1 − W2 = = 5.77 kW 3 tan φ = 3

…(ii)

Solving Eq. (i) and (ii), W1 = 7 89 kW W2 = 2.11 kW

Example 8.43

A three-phase RYB system has effective line voltage of 173.2 V. Wattmeters in line R and Y read 301 W and 1327 W respectively. Find the impedance of the balanced star-connected load. Solution

VL

173 2 , W1 = 301 W, W2 = 1327 W

8.46 Network Analysis and Synthesis If the load is capacitive and pf is leading then W1 W2 . W − W2 ( ) tan φ = − 3 1 =− 3 = −1.09 W1 + W2 ( ) φ = 47.47° P = W1 + W2 = 301 + 1327 = 1628 W 3 VL I L cos φ

P

1628 = 3 × 173.2 × I L × cos ( 47.47°) I L = 8 03 A For a star-connected load, V ph = I ph Z ph

VL

=

173.2

= 100 V 3 3 I L = 8 03 A V ph 100 = = = .45 Ω I ph 8 03

Example 8.44

Three coils each with a resistance of 10 W and reactance of 10 W are connected in star across a three-phase, 50 Hz, 400 V supply. Calculate (a) line current, and (b) readings on the two wattmeters connected to measure the power. Solution For a star-connected load, (a) Line current

R

V ph =

0 Ω,

VL 3

=

XL =

400 3

Ω

VL = 400 V

= 230.94 V

Z ph = R jjX L 10 j10 = 14.14 ∠45° Ω Z ph = . Ω φ = 45° Power factor = cos φ = cos ( 45°) = 0.707 (lagging) I ph = IL

V ph

230.94 = = 16.33 A Z ph 14.14 I ph = 16.33 A

(b) Readings on the two wattmeters

Also,

P 3 VL I L φ = 3 × 400 × 16.33 × 0.707 = 7998.83 W W1 W2 = 7998.83 W W − W2 tan φ = 3 1 W1 + W2 W − W2 tan 45° = 3 1 7998.83 W1 − W2 = 4618.13 W

Solving Eqs. (i) and (ii), W1 = 6308.48 W W2 = 1690.35 W

...(i)

...(ii)

8.13 Measurement of Three-Phase Power 8.47

Example 8.45 Three coils each having a resistance of 20 W and reactance of 15 W are connected in (a) star, and (b) delta, across a three-phase, 400 V, 50 Hz supply. Calculate in each case, the readings on two wattmeters connected to measure the power input. Solution

0 Ω,

R

XL =

Ω

VL = 400 V

(a) Readings on two wattmeters for a star-connected load V ph = Z ph

VL

=

400

= 230.94 V 3 3 = 20 + j 15 = 25∠36.87° Ω

Z ph = 25 Ω

φ = 36.87° I ph =

V ph Z ph

=

230.94 = 9 24 A 25

IL

I ph = 9 24 A

W1

VL I L cos (30° φ ) = 400 × 9.24

W2

VL I L co c s (30° + φ )

s (30° 36.87 . ) = 3669.46 W

.24 coss (30° 36 36.

) = 1451.86 W

(b) Readings on two wattmeters for a delta-connected load V ph

VL = 400 V

Z ph = 25 Ω

φ = 36.87° I ph = IL

V ph Z ph

=

400 = 16 A 25

3 I pphh = 3 × 16 = 27.72 A

W1

VL I L cos (30° φ ) = 400 × 27.72 × cos (30° − 36.87 . ) = 11008.39 W

W2

VL I L cos (30° + φ )

.72 × cos (30° (30° + 36. 36.. 36

) = 4355.57 W

Example 8.46 Two wattmeters connected to measure three-phase power for star-connected load reads 3 kW and 1 kW. The line current is 10 A. Calculate (a) line and phase voltage (b) resistance and reactance per phase. Solution

W1

3 k , W2

1 kW, I L = 10 A

(a) Line and phase voltage W1 − W2 ( = 3 W1 + W2 ( φ = 40.89°

tan φ = 3

) = 0.866 )

Power factor = cos φ = cos ( 40.89°) = 0.756 P

W1 + W2 = 3 + 1 = 4 kW

8.48 Network Analysis and Synthesis 3 VL I L cos φ

P 3

4 × 10 = 3 × VL × 10 × 0.756 VL = 305.48 V For a star-connected load, V ph =

VL 3

=

305.48 3

= 176.37 V

(b) Resistance and reactance per phase I ph Z ph

I L = 10 A V ph 176.37 = = = 7.637 Ω I ph 10

R XL

Z phh cos φ = 17.637 × 0.756 = 13.33 Ω Z ph sin φ = 17.637 × sin ( 40. ) = .55 Ω

Example 8.47 The power input to a 2000 V, 50 Hz, three-phase motor running on full load at an efficiency of 90% is measured by two wattmeters which indicate 300 kW and 100 kW respectively. Calculate the (a) input power, (b) power factor, and (c) line current. Solution

VL

2000 V, η

90%, W1 = 300 kW

W2 = 100 kW

(a) Input power P

W1 + W2 = 300 + 100 = 400 kW

(b) Power factor W1 − W2 ( = 3 W1 + W2 ( φ = 40.89° pf = cos φ = co c s ( 40. ) = 0..

tan φ = 3

) = 0.866 ) (lagging)

(c) Line current P

3 VL I L cos φ

3

400 × 10 = 3 × 2000 × I L × 0 76 I L = 151.93 A

Example 8.48 A three-phase, 400 V, 50 Hz induction motor has a full load output of 14.9 kW at which the efficiency and power factor are 0.88 and 0.8 respectively. What is the line current? Find the readings on the two wattmeters connected to measure the power input to the motor. Solution

V

(a) Line current Po Pi 14.9 × 103 0 88 = Pi Pi = 16.93 kW

η=

f = 50 Hz, Po = 14 9 kW, pf = 0.8, η = 0.88

8.13 Measurement of Three-Phase Power 8.49

pf = cos φ = 0 8 φ = 36.87° Pi = 3 VL I L cos φ 16.93 × 103 3 × 400 × I L × 0.8 I L = 30.55 A (b) Readings on the two wattmeters W1 W2

VL I L cos (30° φ ) = 400 × 30.55 × cos (30° − 36.87 . ) 12.13 kW VL I L co c s (30° + φ ) .55 × cos (30° (30° + 36. 36.. ) 4.8 kW 36

Example 8.49

A three-phase, 220 V, 50 Hz, 11.2 kW induction motor has a full load efficiency of 88 per cent and draws a line current of 38 A under full load, when connected to a three-phase, 220 V supply. Determine power factor at which the motor is operating. Find the reading on two wattmeters connected in the circuit to measure the input to the motor. Solution

VL

220

Po = 11.2 kW, k , η = 88

I L = 38 A

(a) Power factor at which the motor is operating Po Pi 11.2 × 103 0 88 = Pi Pi = 12.73 kW

η=

But

Pi

3 VL I L cos φ

12.73 × 10 = 3 × 220 × 38 × cos φ pf = cos φ = 0.88 (lagging) 3

(b) Reading on two wattmeters

φ = 28.36° W1 = VL I L cos (30° φ ) = 220 × 38 × cos (30 W2 = VL I L cos (30° + φ ) = 220 × 38 × cos (30

. 36° ) = 8356.58 W . 36° ) = 4385.49 W

Example 8.50 Find the reading of a wattmeter when the network shown in Fig. 8.33 is connected to a symmetrical 440 V, 3-phase supply. The phase sequence is RYB. W R

IR −j 40 Ω

IRY −j 53 Ω

50 Ω IBR B

Y

Fig. 8.33

8.50 Network Analysis and Synthesis Solution

VL = 440 V VRRY = 440 ∠0°V

Let

IRY

VYB = 440 ∠− 120°V

VBR

VBR = 440 ∠− 240°V

IBR

Z RY = − j 53 = 53∠− 90 Ω Z BR = 50 + j 40 = 64.03∠38.66° Ω V 440 ∠0° I RY = RRY = = 8 3∠90° A Z RY 53∠− 90° V 440 ∠− 240° I BR = BBR = = 6.87∠81.34° A Z BR 64.03∠38.66° I R = I RY − I BR = 8.3∠90° − 6.87∠81.34° = 1.83∠124 2 .44° A

IR 115.56°

VYB

The phasor diagram of the system is shown in Fig. 8.34. From the phasor diagram, it is clear that angle between VYB and IR is 115.56° W

VRY

Fig. 8.34

VYB I R cos (115.56°) = 440 × 1.83 × cos (115.56°) = 347.41 W

Exercises 8.1

Three coils, each of 5 Ω resistance, and 6 Ω inductive reactance are connected in closed delta and supplied from a 440 V, three-phase system. Calculate the line and phase currents, the power factor of the system and the intake in watts. [ .58 A, 56.33 A, 0.64 ( ), 47.61 kW]

8.2

Three coils, each having a resistance of 10 Ω and inductance of 0.02 H are connected (i) in star, (ii) in delta to a three-phase, 50 Hz supply, the line voltage being 500 volts. Calculate for each case the line current and the total power taken from the supply. [(a)) : . A, 17.94 kW, ( ) delta : 73.39 A, .83 kW]

8.3

A balanced delta-connected load of (8 + j6) ohms per phase is connected to a three-phase, 230 V supply with phase sequence R-Y-B. Find the line current, power factor, power, reactive volt-amperes and the total voltamperes. Draw the phasor diagram. [ .85 A,, 0.8 ( ), 12.74 kW, 9.52 k A , 15..

k

]

8.4

A balanced star-connected load with (6 + j8) Ω per phase is connected to a three-phase, 440 V supply. Find the line current, power factor, power, reactive volt-amperes and voltamperes total. [ .404 A, . (lagging), 11.616 kW,15.488 kVAR , 19.

8.5

8.6

8.7

k

]

Calculate the active and reactive components of the current in each phase of a star-connected 5000 V, three-phase, alternator supplying 3000 kW at a power factor of 0.8. [ , 150 50 A ] Three similar coils, connected in star, take a total power of 1.5 kW at a power factor of 0.2 lagging from a three-phase, 440 V, 50 Hz supply. Calculate the resistance and inductance of each coil. [ .16 Ω, 0.08 H] A balanced three-phase load connected in delta draws a power of 10 kW at 440 V at a power factor of 0.6 lead. Find the values of the circuit elements and the reactive volt-amperes drawn. [

.91 Ω, 27.88 Ω, 13.33 kVAR ]

Exercises 8.51

8.8

A balanced star-connected load, connected to a 400 V, 50 Hz, three-phase ac supply draws a phase current of 50 A at 0.6 power factor lagging. Calculate (a) phase voltage, (b) total power, and (c) parameters in the star-connected load. .94 V, 20.84 kW,( 2.77 + j 3.7)

[ 8.9

]

Three equal star-connected inductors, consume 8 kW power at 0.8 power factor when connected to 415 V, three-phase, three-wire, 50 Hz supply. Estimate the load parameters per phase and determine the line currents. [ .2 Ω, 0.0386 H,13.83 A ]

8.10 A balanced three-phase, star-connected load of 150 kW takes a leading current of 100 A with a line voltage of 1100 V, 50 Hz. Find the circuit constant of the load per phase. [ Ω, 813 μF] 8.11 Three pure elements connected in star draw x kVAR. What will be the value of elements that will draw the same kVAR, when connected in delta across the same supply? [

]

8.12 A balanced wye-connected load with (10 + j20) ohms per phase is connected to a three-phase, 400 V supply. Determine the voltage across, current through and power dissipated in each resistor. Also, determine the total power. [

.2 V,10 10.3 32 2 V, 1067 W,3201 , 3201 W ]

8.13

A delta-connected three-phase load is supplied from a 3- phase, 400 volts balanced supply system. The line current is 20 A and power taken by the load is 10 kW. Find (a) impedance in each branch, (b) line current, power factor and power consumed if the same load is connected in star. [( 24.95 + j 24.05) , 6.66 A, 0. (lagging), 3323. ]

8.14

A balanced star-connected load is supplied from a symmetrical three-phase, 400 V system. The current in each phase is 30 A and lags 30° behind the phase voltage. Find (a) phase voltage, (b) the circuit elements, and (c) draw the vector diagram showing the currents and the voltages. [

.94 V, 6.67 Ω, 3.849 8 9 Ω]

8.15 A three-phase, delta-connected load having a (3 + j4) ohms per phase is connected across a 230 V, three-phase source. Calculate the magnitude of the line current. [ .21 A ] 8.16 A 220 V, three-phase voltage is applied to a balanced delta-connected load. The rms value of the phase current is 20 ∠ −30°A. Determine (i) magnitude and phase of the line current (ii) total power received by the three-phase load (iii) value of the resistive portion of the phase impedance Also, draw the phasor diagram showing clearly the line voltages, phase current and line currents. [ .65 ∠− 60°A, 11.43 kW, 9.53 Ω] 8.17 A three-phase, 37.3 kW, 440 V, 50 Hz induction motor operates on full load with an efficiency of 89% and at a power factor of 0.85 lagging. Calculate total kVA rating of capacitance required to raise power factor at 0.95 lagging. What will be the value of capacitance/phase if capacitors are (a) delta connected (b) star connected? [ .19 kVA, 66.8 μF F, 200.4 200 4 μF] 8.18 Three coils each having impedance (4 + j3) ohms are connected in star to a 440 V, three-phase, 50 Hz supply. Calculate the line current and active power. Now if three pure capacitors, each of C farads, connected in delta, are connected across the same supply, it is found that the total power factor of the circuit becomes 0.96 lag. Find the value of C. Also, find the total line current. [ .8 A, 30.976 kW, 77.75 μF, 42.34 A] 8.19 Three identical coils each having an resistance of 8 Ω and inductance of 0.02 H are connected in (i) star, and (ii) delta across a 3f, 400 V, 50 Hz supply. Draw a neat phasor diagram and calculate the reading of two wattmeters connected to measure power. Also, calculate pf of the circuit. .99 kW, 3.3 kW, 0.7866 ( ), 26.98 kW,10.14 kW, 0. (lagging g)]

8.52 Network Analysis and Synthesis 8.20 Three identical coils each having a reactance of 20 Ω and resistance of 10 Ω are connected in (i) star, (ii) delta across a 440 V, three-phase line. Calculate for each method of connection the line current and readings of each of the two wattmeters connected. 36 A, 4.17 kW, 299.07 W, 34.08 A, 12.52 kW,, 897 89 .

]

8.21 A 3-phase motor load has a pf of 0.397 lagging. Two wattmeters connected to measure power show the input as 30 kW. Find the reading on each wattmeter. [ k , − 5 kW ] 8.22 Each of the wattmeters connected to measure the input to a 3-phase induction motor reads 10 kW. If the power factor of the motor be changed to 0.866 lagging, determine the readings of the two wattmeters, the total input power remaining unchanged. [ .67 kW, 13.33 kW ] 8.23 A 3f, star-connected load draws a line current of 25 A. The load kVA and kW are 20 and 16 respectively. Find the readings on each of the two wattmeters used to measure the 3f power. [ .46 kW, 4.54 kW ] 8.24 Three similar coils are star-connected to a 3f, 50 Hz supply. The line current taken is 25 A and the two wattmeters connected to measure the input indicate 5.185 kW and 10.37 kW respectively. Calculate (a) the line and phase voltages, and (b) the resistance and reactance of each coil. [ , 240 V, 5.31 Ω, 4.8 Ω] 8.25 A three-phase, 500 V motor load has a power factor of 0.4. Two wattmeters connected to measure power show the input to be 30 kW. Find the reading on each instrument. [ k , − 5 kW ] 8.26 The power in a three-phase circuit is measured by two wattmeters. If the total power is 100

kW and power factor is 0.66 leading, what will be the reading of each wattmeter? [ .26 kW, 82.74 kW ] 8.27 Two wattmeters are connected to measure the input to a 400 V, three-phase connected motor outputting 24.4 kW at a power factor of 0.4 lag and 80% efficiency. Calculate (i) resistance and reactance of motor per phase, (ii) reading of each wattmeter. [ .55 Ω, 5.58 Ω, 34915 W, − 4850 48 0 W ] 8.28 In a balanced three-phase, 400 V circuit, the line current is 115.5 A. When power is measured by the two wattmeter method, one meter reads 40 kW and the other, zero. What is the power factor of the load? If the power factor were unity and the line current the same, what would be the reading of each wattmeter? [ .5, 40 kW, 40 kW ] 8.29 A 440 V, three-phase, delta-connected induction motor has an output of 14.92 kW at pf of 0.82 and efficiency of 85%. Calculate the readings on each of the two wattmeters connected to measure the input. If another starconnected load of 10 kW at 0.85 pf lagging is added in parallel to the motor, what will be the current drawn from the line and the power taken from the line? [ .35 kW, 5.26 kW, 43.56 A, 27.6 kW ] 8.30 Balanced delta-connected impedances, each of 10 ∠30 Ω are connected across threephase 400 V mains. Determine the twowattmeter readings if the current coils of the two wattmeters are connected in lines R and Y and the pressure coils are connected between R and B and Y and B lines respectively. [ .7 kW, 13.86 kW ] 8.31 A balanced star-connected load, each phase having a resistance of 10 Ω and the inductive reactance of 30 Ω is connected to 400 V, 50 Hz supply. The phase rotation is R, Y and B. Wattmeters connected to read total power have their currents coils in the red and blue

Objective-Type Questions 8.53

voltage, current in each phase, load parameter, power in each phase, total power, readings of the wattmeters connected in the load circuit to measure the total power. Draw a neat circuit diagram and vector/phase diagram.

lines respectively. Calculate the reading on each wattmeter. [ , 583 W ] 8.32 A balanced star-connected load is supplied from a symmetrical three-phase 400 V, 50 Hz supply system. The current in each phase is 20 A and lags behind its phase voltage by an angle of 2p/9 radians. Calculate line voltage, phase

, 254.034 V, 20 A,, 9.73 Ω, 0.026 H,

[

3.892 kW, 11.676 kW, 8.666 kW, 3.009 kW ]

Objective-Type Questions Choose the correct alternative in the following questions: 8.1 In a three-phase system, voltages differ in phase by (a) 30° (b) 60° (c) 90° (d) 120° 8.2 The rated voltage of a three-phase power system is given as (a) rms phase voltage (b) peak phase voltage (c) rms line to line voltage (d) peak line to line voltage 8.3

8.4

Total instantaneous power supplied by a threephase ac supply to a balanced R-C load is (a) zero (b) constant (c) pulsating with zero average (d) pulsating with non-zero average Which of the following equations is valid for a three-phase, balanced star-connected system? (a) (b) (c) (d)

IR IR

R

Y B

Fig. 8.35 (a) (c) 8.7

8.8

IY + I B = 0 I B = IY

A three-phase load is balanced if all the three phases have the same (a) impedance (b) power factor (c) impedance and power factor (d) none of the above

8.6

The phase sequence of a three-phase system shown in Fig. 8.35 is

(b) RBY (d) YBR

In a 3-phase system, VYN = 100∠−120° V and VBN = 100∠120° V. Then VYB will be ∠90° V

(a) (b)

173∠ − 90° V

(c) (d)

200 ∠60° V none of the above.

If a balanced delta load has an impedance of (6 + j9) ohms per phase then the impedance of each phase in the equivalent star load is (a) (6 + j9) ohms (b) (2 + j3) ohms (c) (2 + j8) ohms (d) (3 + j4.5) ohms

I R IY − I B = 0 VR VB + VY = IN Z

8.5

RYB BRY

8.9

Three equal impedances are first connected in delta across a three-phase balanced supply. If the same impedances are connected in star across the same supply (a) phase current will be one-third (b) line current will be one-third (c) power consumed will be one-third (d) none of the above.

8.54 Network Analysis and Synthesis 8.10

8.11

Three identical resistors connected in star carry a line current of 12 A. If the same resistors are connected in delta across the same supply, the line current will be (a) 12 A (b) 4 A (c) 8 A (d) 36 A

If one of the resistors in Fig. 8.38 is opencircuited, power consumed in the circuit is 10 Ω 400 V

60 kW 40 kW

Fig. 8.38

The power consumed in the star-connected load shown in Fig. 8.36 is 690 W. The line current is

(a) (c) 8.15

R R

400 V

Y 400 V

(b) (d)

4000 W none of the above

IR

R R

R1 IB

R

B R1

Fig. 8.36 (b) 1 A

(a)

2.5 A

(c)

1.725 A

(d)

none of the above

ZL

Fig. 8.39

ZL

90 ∠32.44°

(c)

80 ∠− 32.44° (d)

(b)

80 ∠32.44° 90 ∠− 32.44°

1:1: 3

(b)

1:1: 2

(c)

1:1: 0

(d)

1:1:

3 2

The phase sequence RYB denotes that (a) emf of phase Y lags behind that of phase R by 120° (b) emf of phase Y leads that of phase R by 120° (c) emf of phase Y and phase R are in phase (d) none of the above

8.17

In the two wattmeter method of measurement, if one of the wattmeters reads zero, then power factor will be (a) zero (b) unity (c) 0.5 (d) 0.866

ZL

(a)

(a)

8.16 400 V

Fig. 8.37

IY

Y

If the three-phase balanced source in Fig. 8.37 delivers 1500 W at a leading power factor of 0.844 then the value of ZL (in ohms) is approximately

3f balanced source

8000 W 16000 W

For the three-phase circuit shown in Fig. 8.39, the ratio of the currents I R IY : I B is given by

B

8.13

10 Ω

400 V

(b) 20 kW (d) 180 kW

400 V

10 Ω

400 V

Three delta-connected resistors absorb 60 kW when connected to a 3-phase line. If the resistors are connected in star, the power absorbed is (a) (c)

8.12

8.14

Answers to Objective-Type Questions 8.55

8.18 Two wattmeters, which are connected to measure the total power on a three-phase system, supplying a balanced load, read 10.5 kW and −2.5 kW, respectively. The total power and the power factor, respectively are (a) 13 kW, 0.334 (b) 13 kW, 0.684 (c) 8 kW, 0.52 (d) 8 kW, 0.334

8.19

The minimum number of wattmeter(s) required to measure three-phase, three-wire balanced or unbalanced power is (a) 1 (b) 2 (c) 3 (d) 4

8.20

One of the two wattmeters has read zero in the two-wattmeter method of power measurement. This indicated that the load phase angle is (a) 0° (b) 30° (c) 60° (d) 90°

Answers to Objective-Type Questions 8.1 (d) 8.5 (c) 8.9 (c)

8.2 (c) 8.6 (b) 8.10 (d)

8.3 (b) 8.7 (b) 8.11 (b)

8.4 (a) 8.8 (b) 8.12 (b)

8.13 (d) 8.17 (c)

8.14 (a) 8.18 (d)

8.15 (a) 8.19 (b)

8.16 (a) 8.20 (c).

9 Network Topology

9.1

INTRODUCTION

The purpose of network analysis is to find voltage across and current through all the elements. When the network is complicated and has a large number of nodes and closed paths, network analysis can be done conveniently by using ‘Network Topology’. This theory does not make any distinction between different types of physical elements of the network but makes the study based on a geometric pattern of the network. The basic elements of this theory are nodes, branches, loops and meshes. Node Branch Loop Mesh

9.2

It is defined as a point at which two or more elements have a common connection. It is a line connecting a pair of nodes, the line representing a single element or series connected elements. Whenever there is more than one path between two nodes, there is a circuit or loop. It is a loop which does not contain any other loops within it.

GRAPH OF A NETWORK

A linear graph is a collection of nodes and branches. The nodes are joined together by branches. The graph of a network is drawn by first marking the nodes and then joining these nodes by lines which correspond to the network elements of each branch. All the voltage and current sources are replaced by their internal impedances. The voltage sources are replaced by short circuits as their internal impedances are zero whereas current sources are replaced by open circuits as their internal impedances are infinite. Nodes and branches are numbered. Figure 9.1 shows a network and its associated graphs. Each branch of a graph may be given an orientation or a direction with the help of an arrow head which represents the assigned reference direction for current. Such a graph is then referred to as a directed or oriented graph. Branches whose ends fall on a node are said to be incident at that node. Degree of a node is defined as the number of branches incident to it. Branches 2, 3 and 4 are incident at Node 2 in Fig. 9.1(c). Hence, the degree of Node 2 is 3.

9.2 Network Analysis and Synthesis (6)

(6)

1

(2)

3

3

(3)

2

1

(2)

3

(3)

(3)

(2) (1) V

2

1

2

(6)

(5)

(4)

(1)

(4)

(5)

(1)

(4)

(5)

+ − 4

4

4

(a) Network

(b) Undirected graph

(c) Directed or oriented graph

Fig. 9.1 Network and its graphs

9.3

DEFINITIONS ASSOCIATED WITH A GRAPH

1. Planar Graph A graph drawn on a two-dimensional plane is said to be planar if two branches do not intersect or cross at a point which is other than a node. Figure 9.2 shows such graphs. (2) 2

1

(3)

(1) (5)

(8)

3

3

(4)

(1)

2

(2)

1

(6)

(3)

(6)

(7)

4

(9)

6 (4)

4

5

(5)

Fig. 9.2 Planar graphs 2. Non-planar Graph A graph drawn on a two-dimensional plane is said to be non-planar if there is intersection of two or more branches at another point which is not a node. Figure 9.3 shows non-planar graphs. 1

(2)

2

(2)

2

1 (5)

(3)

3

(6)

(1)

(4) 3

4 (3)

(1)

(7)

(8)

4 (5)

5

(6)

(4) 6

Fig. 9.3 Non-planar graphs 3. Sub-graph It is a subset of branches and nodes of a graph. It is a proper sub-graph if it contains branches and nodes less than those on a graph. A sub-graph can be just a node or only one branch of the graph. Figure 9.4 shows a graph and its proper sub-graph.

9.3 Definitions Associated with a Graph 9.3 (2)

(2)

2

1

(3) (1)

(4)

(5)

2

3

(5)

3 3

1 (6)

(4)

4

4 (a)

(b)

Fig. 9.4 (a) Graph (b) Proper sub-graph 4. Path It is an improper sub-graph having the following properties: 1. At two of its nodes called terminal nodes, there is incident only one branch of sub-graph. 2. At all remaining nodes called internal nodes, there are incident two branches of a graph. In Fig. 9.5, branches 2, 5 and 6 together with all the four nodes, constitute a path. (3) 2

1

(2)

3

(4)

(1)

(5)

(6)

4

Fig. 9.5

Path

5. Connected Graph A graph is said to be connected if there exists a path between any pair of nodes. Otherwise, the graph is disconnected. 6. Rank of a Graph

If there are n nodes in a graph, the rank of the graph is (n − 1).

7. Loop or Circuit A loop is a connected sub-graph of a connected graph at each node of which are incident exactly two branches. If two terminals of a path are made to coincide, it will result in a loop or circuit. Figure 9.6 shows two loops. (2) 1

2

(3) 3

(1)

(4)

(2) 1

4 Loops: {1, 2, 3, 4}

Fig. 9.6

(1)

2

Loops: {1, 2}

Loops

Loops of a graph have the following properties: 1. 2. 3.

There are at least two branches in a loop. There are exactly two paths between any pair of nodes in a circuit. The maximum number of possible branches is equal to the number of nodes.

9.4 Network Analysis and Synthesis 8. Tree A tree is a set of branches with every node connected to every other node in such a way that removal of any branch destroys this property. Alternately, a tree is defined as a connected sub-graph of a connected graph containing all the nodes of the graph but not containing any loops. Branches of a tree are called twigs. A tree contains (n − 1) twigs where n is the number of nodes in the graph. Figure 9.7 shows a graph and its trees. (6) 2

1

(1)

3

(2)

(3)

(4)

1

2

(4)

(5)

2 (2)

1

3

(1)

(3)

(5)

(5)

4

4

4 (a) Graph

3

(b)

(c)

Twigs: {1, 4, 5}

Twigs: {2, 3, 5}

Fig. 9.7 Graph and its trees Trees have the following properties: 1. 2. 3. 4. 5. 6.

There exists only one path between any pair of nodes in a tree. A tree contains all nodes of the graph. If n is the number of nodes of the graph, there are (n − 1) branches in the tree. Trees do not contain any loops. Every connected graph has at least one tree. The minimum terminal nodes in a tree are two.

9. Co-tree Branches which are not on a tree are called links or chords. All links of a tree together constitute the compliment of the corresponding tree and is called the co-tree. A co-tree contains b − (n − 1) links where b is the number of branches of the graph. In Fig. 9.7 (b) and (c) the links are {2, 3, 6} and {1, 4, 6} respectively.

Example 9.1

Draw directed graph of the networks shown in Fig. 9.8. R2

R1

C1

L

L1 L2

v C1

R1 R2 V

C2

+ −

C3 C2

(a)

R3 (b)

Fig. 9.8

9.3 Definitions Associated with a Graph 9.5

Solution For drawing the directed graph, 1. replace all resistors, inductors and capacitors by line segments, 2. replace the voltage source by a short-circuit, 3. assume directions of branch currents, and 4. number all the nodes and branches. The directed graph for the two networks are shown in Fig. 9.9. 1 (1)

1

(2)

4

(2)

(1)

2

5

(5)

(3)

(3)

(4)

(7) (8)

(6)

2

3

(a)

(b)

Fig. 9.9

Example 9.2

Figure 9.10 shows a graph of the network. Show all the trees of this graph. (1) 1

(2)

2

3 (4) (5)

(3) 4

Fig. 9.10 Solution A graph has many trees. A tree is a connected sub-graph of a connected graph containing all the nodes of the graph but not containing any loops. Figure 9.11 shows various trees of the given graph. (2)

2

3

1

(1)

2

3

1 (1) 2

3

1 (4)

(3)

(4)

4 1

2

4

(3)

4 3

(4) (5)

(3)

(5)

1

2

(3)

(5)

1 (1) 2 (2) 3

(3)

4

4

Fig. 9.11

3

1 (1) 2 (2) 3

(4)

(5)

(5)

4 (2) 3

1 (1) 2 (2)

4

4

9.6 Network Analysis and Synthesis

9.4

INCIDENCE MATRIX

A linear graph is made up of nodes and branches. When a graph is given, it is possible to tell which branches are incident at which nodes and what are its orientations relative to the nodes.

9.4.1 Complete Incidence Matrix (Aa) For a graph with n nodes and b branches, the complete incidence matrix is a rectangular matrix of order n × b. Elements of this matrix have the following values: aij = 1, if branch j is incident at node i and is oriented away from node i. = −1, if branch j is incident at node i and is oriented towards node i. = 0, if branch j is not incident at node i. For the graph shown in Fig. 9.12, branch 1 is incident at nodes 1 and 4. It is oriented away from Node 1 and oriented towards Node 4. Hence, a11 = 1 and a41 = −1. Since branch 1 is not incident at nodes 2 and 3, a21 = 0 and a31 = 0. Similarly, other elements of the complete incidence matrix are written. (6) 2

1

(2)

3

(4)

(1)

(3)

(5)

Nodes ↓

Branches → 1

2

3

4

5

6

1

1

1

0

0

0

1

2

0 −1

1 −1

0

0

3

0

0

1 −1

4

−1

0

0 −1

1

0 −1

0

4

Fig. 9.12 Graph The complete incidence matrix is 1⎤ ⎡ 1 1 0 0 0 ⎢ 0 −1 1 −1 0 0 ⎥ Aa = ⎢ ⎥ 1 1 −1⎥ ⎢ 0 0 0 ⎢⎣ −1 0 −1 0 −1 0 ⎥⎦ It is seen from the matrix Aa that the sum of the elements in any column is zero. Hence, any one row of the complete incidence matrix can be obtained by the algebraic manipulation of other rows.

9.4.2 Reduced Incidence Matrix (A) The reduced incidence matrix A is obtained from the complete incidence matrix Aa by eliminating one of the rows. It is also called incidence matrix. It is of order (n − 1) × b. Eliminating the third row of matrix Aa, ⎡ 1 1 0 0 0 1⎤ A = ⎢ 0 −1 1 −1 0 0 ⎥ ⎢ ⎥ ⎣ −1 0 −1 0 −1 0 ⎦ When a tree is selected for the graph as shown in Fig. 9.13, the incidence matrix is obtained by arranging a column such that the first (n − 1) column corresponds to twigs of the tree and the last b − (n − 1) branches corresponds to the links of the selected tree.

9.4 Incidence Matrix 9.7 2

1

Twigs

3

(2)

(4)

2

Twigs: {2, 3, 4} Links: {1, 5, 6}

(3)

3

⎡ 1 0 A = ⎢ −1 1 ⎢ ⎣ 0 −11

4

Links 4

1

0 1 1 0 0 −11

5 6 0 1⎤ 0 0⎥ ⎥ 1 0⎦

Fig. 9.13 Tree The matrix A can be subdivided into submatrices At and Al. A [ At : Al ] Where At the is twig matrix and Al is the link matrix.

9.4.3 Number of Possible Trees of a Graph Let the transpose of the reduced incidence matrix A be AT. It can be shown that the number of possible trees of a graph will be given by Number of possible trees = |AAT| For the graphs shown in Fig. 9.12, the reduced incidence matrix is given by ⎡ 1 1 0 0 0 1⎤ A = ⎢ 0 −1 1 −1 0 0 ⎥ ⎢ ⎥ ⎣ −1 0 −1 0 −1 0 ⎦ Then transpose of this matrix will be ⎡ 1 0 −1⎤ ⎢ 1 −1 0 ⎥ ⎢ ⎥ 0 1 −1⎥ AT = ⎢ ⎢0 −1 0 ⎥ ⎢0 0 −1⎥ ⎢ ⎥ ⎣ 1 0 0⎦ Hence, number of all possible trees of the graph ⎡ 1 0 −1⎤ ⎢ 1 −1 0 ⎥ ⎥ 3 −1 −1⎤ ⎡ 1 1 0 0 0 1⎤ ⎢0 1 −1⎥ ⎡ AAT = ⎢ 0 −1 1 −1 0 0 ⎥ ⎢ = ⎢ −1 3 −1⎥ ⎢ ⎥ ⎢0 −1 0 ⎥ ⎢ ⎥ ⎣ −1 0 −1 0 −1 0 ⎦ ⎢ ⎣ −1 −1 3⎦ ⎥ 0 0 −1 ⎢ ⎥ ⎣ 1 0 0⎦ |

T

3 −1 −1 | = −1 3 −1 = 3(( −1 −1 3

Thus, 16 different trees can be drawn.

) + ( )(

) − 1((

) = 16

9.8 Network Analysis and Synthesis

9.5

LOOP MATRIX OR CIRCUIT MATRIX

When a graph is given, it is possible to tell which branches constitute which loop or circuit. Alternately, if a loop matrix or circuit matrix is given, we can reconstruct the graph. For a graph having n nodes and b branches, the loop matrix Ba is a rectangular matrix of order b columns and as many rows as there are loops. Its elements have the following values: bij = 1, if branch j is in loop i and their orientations coincide. = − 1, if branch j is in loop i and their orientations do not coincide. = 0, if branch j is not in loop i. A graph and its loops are shown in Fig. 9.14. (6) 2

1

3

(2)

(4) (3)

(1)

(5)

Loop 1: {1, 2, 3} Loop 2: {3, 4, 5} Loop 3: {2, 4, 6} Loop 4: {1, 2, 4, 5} Loop 5: {1, 5, 6} Loop 6: {2, 3, 5, 6} Loop 7: {1, 3, 4, 6}

4

Fig. 9.14

Graph

All the loop currents are assumed to be flowing in a clockwise direction. Loops ↓ 1 2 3 4 5 6 7

Branches → 1 2 3 4 −1 1 1 0 0 0 −1 −1 0 −1 0 1 −1 1 0 −1 −1 0 0 0 0 −1 −1 0 −1 0 1 1

⎡ −1 1 1 0 ⎢ 0 0 −1 −1 ⎢ 1 ⎢ 0 −1 0 Ba = ⎢ −1 1 0 −1 ⎢ −1 0 0 0 ⎢ 0 −1 −1 0 ⎢ 1 1 ⎣ −1 0

5 0 1 0 1 1 1 0 0 1 0 1 1 1 0

6 0 0 1 0 1 1 1

0⎤ 0⎥ ⎥ 1⎥ 0⎥ 1⎥ 1⎥⎥ 1⎦

9.5.1 Fundamental Circuit (Tieset) and Fundamental Circuit Matrix When a graph is given, first select a tree and remove all the links. When a link is replaced, a closed loop or circuit is formed. Circuits formed in this way are called fundamental circuits or f-circuits or tiesets. Orientation of an f-circuit is given by the orientation of the connecting link. The number of f-circuits is same as the number of links for a graph. In a graph having b branches and n nodes, the number of f-circuits or tiesets will be (b − n + 1). Figure 9.15 shows a tree and f-circuits (tiesets) for the graph shown in Fig. 9.14.

9.5 Loop Matrix or Circuit Matrix 9.9 (6) 1

2 (2)

3 (4) (3)

4 (a) Tree

(2) (1)

(4)

(2)

(4)

(3)

(3)

tieset 1

(2)

(4) (3)

(5)

tieset 5

tieset 1: {1, 2, 3} tieset 5: {5, 3, 4} tieset 6: {6, 2, 4}

tieset 6

(b) f-circuits (tiesets)

Fig. 9.15

Tree and f-circuits

Here, b = 6 and n = 4. Number of tiesets = b − n + 1 = 6 − 4 + 1 = 3 f-circuits are shown in Fig. 9.15. The orientation of each f-circuit is given by the orientation of the corresponding connecting link. The branches 1, 2 and 3 are in the tieset 1. Orientation of tieset 1 is given by orientation of branch 1. Since the orientation of branch 1 coincides with orientation of tieset 1, b11 = 1. The orientations of branches 2 and 3 do not coincide with the orientation of tieset 1, Hence, b12= − 1 and b13 = − 1. The branches 4, 5 and 6 are not in tieset 1. Hence, b14 = 0, b15 = 0 and b16 = 0. Similarly, other elements of the tieset matrix are written. Then, the tieset schedule will be written as Branches →

Tiesets ↓

1

3

4 5 6

1

1 −1 −1

0 0 0

5

0

6

0 −1

2

0 −1 −1 1 0 0

1 0 1

Hence, an f-circuit matrix or tieset matrix will be given as ⎡ 1 −1 −1 0 0 0 ⎤ B = ⎢0 0 −1 −1 1 0 ⎥ ⎢ ⎥ 1 0 1⎦ ⎣0 −1 0 Usually, the f-circuit matrix B is rearranged so that the first (n − 1) columns correspond to the twigs and b − (n − 1) columns to the links of the selected tree. Twigs gs 2 3 4 1 ⎡−1 B=⎢ 0 ⎢ ⎣−1

1 0 1 −1 0 1

Links 1 5 6 1 0 0

0 1 0

0⎤ 0⎥ ⎥ 1⎦

The matrix B can be partitioned into two matrices Bt and Bl. B [ Bt : Bl ] [ Bt : U ] where Bt is the twig matrix, Bl is the link matrix and U is the unit matrix.

9.10 Network Analysis and Synthesis

9.5.2

Orthogonal Relationship between Matrix A and Matrix B

For a linear graph, if the columns of the two matrices Aa and Ba are arranged in the same order, it can be shown that Aa BaT = 0 Ba AaT = 0 The above equations describe the orthogonal relationship between the matrices Aa and Ba. If the reduced incidence matrix A and the f-circuit matrix B are written for the same tree, it can be shown that A BT = 0 or

B AT = 0

or

These two equations show the orthogonal relationship between matrices A and B.

9.6

CUTSET MATRIX

1

(2)

2

Consider a linear graph. By removing a set of branches without affecting the nodes, (5) (3) two connected sub-graphs are obtained and the original graph becomes unconnected. (1) The removal of this set of branches which results in cutting the graph into two parts (4) 4 3 are known as a cutset. The cutset separates the nodes of the graph into two groups, each being in one of the two groups. Fig. 9.16 Graph Figure 9.16 shows a graph. Branches 1, 3 and 4 will form a cutset. This set of branches separates the graph into two parts. One having an isolated node 4 and other part having branches 2 and 5 and nodes 1, 2 and 3. Similarly, branches 1 and 2 will form a cutset. Each branch of the cutset has one of its terminals incident at a node in one part and its other end incident at other nodes in the other parts. The orientation of a cutset is made to coincide with orientation of defining branch. For a graph having n nodes and b branches, the cutset matrix Qa is a rectangular matrix of order b columns and as many rows as there are cutsets. Its elements have the following values: qij = 1, if the branch j is in the cutset i and the orientation coincide. = −1, if the branch j is in the cutset i and the orientations do not coincide. = 0, if the branch j is not in the cutset i. Figure 9.17 shows a directed graph and its cutsets. (6) 1

3

2 (2)

(4) (3)

(1)

(5)

Cutset 1: {1, 2, 6} Cutset 2: {2, 3, 4} Cutset 3: {3, 1, 5} Cutset 4: {4, 5, 6} Cutset 5: {5, 2, 3, 6} Cutset 6: {6, 1, 3, 4}

Cutsets ↓ 1 2 3 4 5 6

Branches → 1 2 3 4 5 6 1 1 0 0 0 1 0 1 −1 1 0 0 1 0 1 0 1 0 0 0 0 1 1 −1 0 −1 1 0 1 −1 1 0 1 −1 0 1

4

Fig. 9.17 Directed graph For the cutset 2, which cuts the branches 2, 3 and 4 and is shown by a dotted circle, the entry in the cutset schedule for the branch 2 is 1, since the orientation of this cutset is given by the orientation of the branch 2 and hence it coincides. The entry for branch 3 is −1 as orientation of branch 3 is opposite to that of cutset 2,

9.6 Cutset Matrix 9.11

i.e., branch 2 goes into cutset while branch 3 goes out of cutset. The entry for branch 4 is 1 as the branch 2 and the branch 4 go into the cutset. Thus their orientations coincide. Hence, the cutset matrix Qa is given as ⎡1 1 0 0 ⎢0 1 −1 1 ⎢ 1 0 1 0 Qa = ⎢ 0 0 0 1 ⎢ ⎢0 −1 1 0 ⎢ 1 0 1 −1 ⎣

0 1⎤ 0 0⎥ ⎥ 1 0⎥ 1 −1⎥ 1 −1⎥ 0 1⎥⎦

9.6.1 Fundamental Cutset and Fundamental Cutset Matrix When a graph is given, first select a tree and note down its twigs. When a twig is removed from the tree, it separates a tree into two parts (one of the separated part may be an isolated node). Now, all the branches connecting one part of the disconnected tree to the other along with the twig removed, constitutes a cutset. This set of branches is called a fundamental cutset or f-cutset. A matrix formed by these f-cutsets is called an f-cutset matrix. The orientation of the f-cutset is made to coincide with the orientation of the defining branch, i.e., twig. The number of f-cutsets is the same as the number of twigs for a graph. Figure 9.18 shows a graph, selected tree and f-cutsets corresponding to the selected tree. (6) 1

(6) 3

2 (2)

(2)

(4)

(1)

(3)

(5)

(2)

(4) (3)

(1)

(4) (3)

(5)

f-cutset 2: {2, 1, 6} f-cutset 3: {3, 1, 5} f-cutset 4: {4, 5, 6}

4 (a) Graph

(b) Tree

(c) f-cutsets

Fig. 9.18 Graph, selected tree and f-cutsets The branches 2, 1, and 6 are in the f-cutset 2. Orientation of f-cutset 2 is given by orientation of the branch 2 which is moving away from the f-cutset 2. Since the orientations of branches 2, 1 and 6 coincide with the orientation of the f-cutset 2, q11 = 1, q12 = 1 and q16 = 0. The branches 3, 4 and 5 are not in the f-cutset 2. Hence, q13 = 0, q14 = 0 and q15 = 0. Similarly, other elements of the f-cutset matrix are written. The cutset schedule is f-cutsets ↓ 2 3 4 Hence, the f-cutset matrix Q is given by ⎡ 1 Q=⎢ 1 ⎢ ⎣ 0

1 1 1 0

Branches → 2 3 4 5 6 1 0 0 0 1 0 1 0 1 0 0 0 1 1 −1

0 1⎤ 1 0⎥ ⎥ 1 −1⎦ The f-cutset matrix Q is rearranged so that the first (n − 1) columns correspond to twigs and b − (n − 1) columns to links of the selected tree. 1 0 0

0 1 0

0 0 1

9.12 Network Analysis and Synthesis Twigs

Links

2

3

4

1

5

6

⎡ 1 Q=⎢ 0 ⎢ ⎣ 0

0 1 0

0 0 1

1 1 0

0 1⎤ 1 0⎥ ⎥ 1 −11⎦

The matrix Q can be subdivided into matrices Qt and Ql. Q

Qt : Q

U : Ql ]

where Qt is the twig matrix, Ql is the link matrix and U is the unit matrix.

9.6.2

Orthogonal Relationship between Matrix B and Matrix Q

For a linear graph, if the columns of two matrices Ba and Qa are arranged in the same order, it can be shown that Q BaT = 0 B QaT = 0

or

If the f-circuit matrix B and the f-cutset matrix Q are written for the same selected tree, it can be shown that B QT = 0 or Q BT = 0 These two equations show the orthogonal relationship between matrices A and B.

9.7

RELATIONSHIP AMONG SUBMATRICES OF A, B AND Q

Arranging the columns of matrices A, B and Q with twigs for a given tree first and then the links, we get the partitioned forms as A [ At : Al ] B [ Bt : Bl ] [ Bt : U ] Q

Qt : Q

U : Ql ]

From the orthogonal relation, ABT = 0, ⎡ Bt T ⎤ AB = [ At : Al ] ⎢ … ⎥ ⎢ ⎥ ⎢⎣ Bl T ⎥⎦ T

At Bt T + Al Bl T = 0 At Bt T = − Al Bl T Since At is non-singular, i.e., |A| ≠ 0, At−1 exists. Premultiplying with At−1, Bt T At−1 Al Bl T Since Bl is a unit matrix

Bt

Bl ( At 1 Al )T

Bt

( At−1. Al )T

9.7 Relationship Among Submatrices of A, B and Q 9.13

Hence, matrix B is written as B [ ( At−1 Al )T : U ]

…(9.1)

ABT = 0

We know that

Al Bl T = − At Bt T Postmultiplying with (

T −1

) , At Bt T ( Bl T ) −1 = − At Bt T (B ( Bl 1 )T

Al

At ( Bl 1 Bt )T

Hence matrix A can be written as A [ Al : At ( Bl−1 Bt )T ]

…(9.2)

= At [U : − ( Bl 1 Bt )T ] Similarly we can prove that Q

U : − ( Bl 1 Bt )T ]

A

At Q

Q

At−1 A

…(9.3)

From Eqs (9.2) and (9.3), we can write

We have shown that

At−1[ At : Al ] = [U : At−11 Al ]

Bt

( At−1. Al )T

Bt T

( At−1 Al )

Q

U : − Bt T ]

Hence, Q can be written as Bt T

Q

Example 9.3 For the circuit shown in Fig. 9.19, draw the oriented graph and write the (a) incidence matrix, (b) tieset matrix, and (c) f-cutset matrix. R5 R2 R1 V

L1

R4 R6

+ −

C

R3

I

(4)

Fig. 9.19 Solution For drawing the oriented graph, 1. replace all resistors, inductors and capacitors by line segments, 2. replace the voltage source by short circuit and the current source by an open circuit, 3. assume the directions of branch currents arbitrarily, and 4. number all the nodes and branches. The oriented graph is shown in Fig. 9.20.

1

2 (3) (1)

(2)

3

Fig. 9.20

9.14 Network Analysis and Synthesis (a)

Incidence Matrix (A) Nodes Branches → ⎡ −1 0 −1 1⎤ ↓ 1 2 3 4 Aa = ⎢ 0 1 1 −1⎥ ⎢ ⎥ 1 −1 0 −1 1 ⎣ 1 −1 0 0 ⎦ 2 0 1 1 −1 3 1 −1 0 0 Eliminating the third row from the matrix Aa, we get the incidence matrix A. ⎡ −1 0 −1 1⎤ A= ⎢ ⎣ 0 1 1 −1⎥⎦

(b) Tieset Matrix (B) The oriented graph, selected tree and tiesets are shown in Fig. 9.21. (4) 1

2

Links: {3, 4} Tieset 3: {3, 1, 2} Tieset 4: {4, 1, 2}

(3) (1)

(2)

1

2

3 ⎡ −11 B= ⎢ 4⎣ 1

3 4

1 1 0⎤ 1 0 1⎥⎦

3

(c)

Fig. 9.21 f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 9.22. (4) 1

2 (3)

(1)

Twigs: {1, 2} f-cutset 1: {1, 3, 4} f-cutset 2: {2, 3, 4}

1 2 3 1⎡1 0 1 Q= ⎢ 2 ⎣0 1 1

(2)

4 1⎤ 1⎥⎦

3

Fig. 9.22

Example 9.4 For the network shown in Fig. 9.23, draw the oriented graph and write the (a) incidence matrix, (b) tieset matrix, and (c) f-cutset matrix. 2Ω 2Ω 4Ω 1F

2Ω V1

10 Ω

+ −

I 1H

2Ω + − V2

Fig. 9.23

4F

9.7 Relationship Among Submatrices of A, B and Q 9.15

Solution For drawing the oriented graph, 1. replace all resistors, inductors and capacitors by line segments, 2. replace all voltage sources by short circuits and current source by an open circuit, 3. assume directions of branch currents arbitrarily, and 4. number all the nodes and branches. The oriented graph is shown in Fig. 9.24.

(6)

(7)

5 2

1

3

(1)

(2) (3)

(4)

(5)

4

(a) Incidence Matrix (A)

Fig. 9.24 Nodes ↓ 1 1 1 2 −1 3 0 4 0 5 0

2 0 −1 1 0 0

Branches → 3 4 5 6 7 0 1 0 1 0 1 0 0 0 0 0 0 1 0 1 −1 −1 −1 0 0 0 0 0 −1 −1

1 0 1 0⎤ ⎡ 1 0 0 ⎢ −1 −1 1 0 0 0 0 ⎥ ⎢ ⎥ Aa = ⎢ 0 1 0 0 1 0 1⎥ ⎢ 0 0 −1 −1 −1 0 0 ⎥ ⎢⎣ 0 0 0 0 0 −1 −1⎥⎦

Eliminating the last row from the matrix Aa, we get the incidence matrix A. 1 0 ⎡ 1 0 0 ⎢ −1 −1 1 0 0 A= ⎢ 1 0 0 1 ⎢ 0 ⎢⎣ 0 0 −1 −1 −1 (b)

1 0 0 0

0⎤ 0⎥ ⎥ 1⎥ 0 ⎥⎦

Tieset Matrix (B) The oriented graph, selected tree and tiesets are shown in Fig. 9.25. (6)

(7)

5 2

1 (1)

3 Links: {2, 4, 7} Tieset 2: {2, 3, 5} Tieset 4: {4, 1, 3} Tieset 7: {7, 6, 1, 3, 5}

(2) (3)

(4)

(5)

1 2

3

4

5

1 0 −1 1 1 0 1 0 −11

2⎡ 0 1 B = 4 ⎢ −1 0 ⎢ 7⎣ 1 0

6 7 0 0⎤ 0 0⎥ ⎥ 1 1⎦

4

(c)

Fig. 9.25 f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 9.26. 5

(6)

(7) 2

1

3

(2)

(1) (3) (4)

(5)

4

Fig. 9.26

Twigs: {1, 3, 5, 6} f-cutset 1: {1, 4, 7} f-cutset 3: {3, 4, 7, 2} f-cutset 5: {5, 2, 7} f-cutset 6: {6, 7}

1

2

3

4 5

6

1⎡1 0 3 ⎢0 −1 Q= ⎢ 5 ⎢0 1 6 ⎢⎣0 0

0 1 0 0

1 1 0 0

0 −1⎤ 0 −1⎥ ⎥ 0 1⎥ 1 1⎥⎦

0 0 1 0

7

9.16 Network Analysis and Synthesis

Example 9.5 For the circuit shown in Fig. 9.27, draw the oriented graph and write (a) incidence matrix, (b) tieset matrix, and (c) cutset matrix. i L1

r2

r1

r5

r4

r6

r3

+ V

C1

−

L2

Fig. 9.27

1. 2. 3. 4.

replace all resistors, inductors and capacitors by line segments, replace voltage source by short circuit and current source by an open circuit, assume directions of branch currents arbitrarily, and number the nodes and branches.

(1)

3

⎡ −1 1 0 −1⎤ Aa = ⎢ 0 0 1 1⎥ ⎢ ⎥ ⎣ 1 −1 −1 0 ⎦

Eliminating the third row from the matrix Aa, we get the incidence matrix A. ⎡ −1 1 0 −1⎤ A= ⎢ ⎣ 0 0 1 1⎥⎦ Tieset Matrix (B) The oriented graph, selected tree and tiesets are shown in Fig. 9.29.

(1)

(4) 2

(2)

(3)

Links: {1, 3} Tieset 1: {1, 2} Tieset 3: {3, 2, 4}

1 1⎡1 B= ⎢ 3 ⎣0

3

Fig. 9.29 (c)

(3)

Fig. 9.28

Nodes Branches → ↓ 1 2 3 4 1 −1 1 0 −1 2 0 0 1 1 3 1 −1 −1 0

1

2

(2)

The oriented graph is shown in Fig. 9.28. (a) Incidence Matrix (A)

(b)

(4)

1

Solution For drawing the oriented graph,

f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 9.30.

2

3

1 0 1 1

4 0⎤ 1⎥⎦

9.7 Relationship Among Submatrices of A, B and Q 9.17 (4)

1

(1)

2 Twigs: {2, 4} f-cutset 2: {2, 1, 3} f-cutset 4: {4, 3} (2)

1 2 3 4

(3)

2 ⎡ −1 1 1 0 ⎤ Q= ⎢ 4 ⎣ 0 0 1 1⎥⎦

3

Fig. 9.30

Example 9.6 For the circuit shown in Fig. 9.31, (a) draw its graph, (b) draw its tree, and (c) write the fundamental cutset matrix. 2Ω 1F

1F 1Ω

1Ω

I

1Ω 1H (3)

Fig. 9.31 Solution (a) For drawing the oriented graph, 1. replace all resistors, inductors and capacitors by line segments, 2. replace the current source by an open circuit, 3. assume directions of branch currents, and 4. number all the nodes and branches. The oriented graph is shown in Fig. 9.32. (b) Tree The oriented graph and its selected tree are shown in Fig. 9.33. (3) 2

1

3

(2)

(4) (5)

(1)

(6)

Twigs: {2, 4, 5} f-cutset 2: {2, 1, 3} f-cutset 4: {4, 3, 6} f-cutset 5: {5, 1, 6}

4

Fig. 9.33 (c) Fundamental Cutset Matrix (Q) 1

2

3 4 5 6

2 ⎡ 1 1 1 0 0 0⎤ Q = 4 ⎢0 0 −1 1 0 1⎥ ⎢ ⎥ 5 ⎣ 1 0 0 0 1 1⎦

2

1

3

(2)

(4) (5)

(1)

4

Fig. 9.32

(6)

9.18 Network Analysis and Synthesis

Example 9.7 The graph of a network is shown in Fig. 9.34. Write the (a) incidence matrix, (b) tieset matrix, and (c) f-cutset matrix. (2) 1

3 (4)

(5) 2

(1)

(3)

(6)

4

Fig. 9.34 Solution (a) Incidence Matrix (A) 1 1 ⎡ −1 2⎢ 0 Aa = ⎢ 3⎢ 0 4 ⎢⎣ 1

2

3

4

5

6

1 0 1 0 0 0 −1 1 1 1 0 −1 0 −11 0 0

0⎤ 1⎥ ⎥ 0⎥ 1⎥⎦

The incidence matrix A is obtained by eliminating any row from the matrix Aa. 1 0 0⎤ ⎡ −1 1 0 A = ⎢ 0 0 0 −1 1 1⎥ ⎢ ⎥ ⎣ 0 −1 1 0 −1 0 ⎦ (b)

Tieset, Matrix (B) The oriented graph, selected tree and tiesets are shown in Fig. 9.35. (2) 1

3

(5)

(4) 2 (1)

(3)

(6)

Links: {1, 2, 3} Tieset 1: {1, 4, 6} Tieset 2: {2, 4, 5} Tieset 3: {3, 5, 6}

1 2 3

4

1⎡1 0 0 B = 2 ⎢0 1 0 ⎢ 3 ⎣0 0 1

5

1 0 1 −1 0 1

6 1⎤ 0⎥ ⎥ 1⎦

4

(c)

Fig. 9.35 f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 9.36. (2) 1

3

(5)

(4) 2 (1)

(6)

(3)

4

Fig. 9.36

Twigs: {4, 5, 6} f-cutset 4: {4, 1, 2} f-cutset 5: {5, 2, 3} f-cutset 6: {6, 1, 3}

1

2

3

4

5

6

4 ⎡ −1 1 0 1 0 0 ⎤ Q = 5 ⎢ 0 1 −1 0 1 0 ⎥ ⎢ ⎥ 6 ⎣ −1 0 1 0 0 1⎦

9.7 Relationship Among Submatrices of A, B and Q 9.19

Example 9.8

For the graph shown in Fig. 9.37, write the incidence matrix, tieset matrix and f-cutset

matrix. (5) 2

1

4

3

(2)

(3)

(4) (6)

(1) (7) 5

Fig. 9.37 Solution (a) Incidence Matrix (A) 1

2

1⎡ 1 2⎢ 0 ⎢ Aa = 3 ⎢ 0 4⎢ 0 5 ⎢⎣ −1

3

4

5

1 0 0 1 1 0 0 −1 1 0 0 −11 0 0 0

6

7 0⎤ 0⎥ ⎥ 0⎥ 1⎥ 1⎥⎦

0 0 1 0 0 0 1 1 0 −1

The incidence matrix A is obtained by eliminating any row from the matrix Aa. ⎡1 1 0 0 0 ⎢0 −1 1 0 1 A= ⎢ ⎢0 0 −1 1 0 ⎢⎣0 0 0 −1 −1 (b)

(c)

Tieset Matrix (B) Links: {5, 6, 7} Tieset 5: {5, 3, 4} Tieset 6: {6, 1, 2, 3, 4} Tieset 7: {7, 1, 2, 3, 4}

1 5⎡ 0 B = 6 ⎢ −1 ⎢ 7⎣ 1

2

3

4

0 0⎤ 0 0⎥ ⎥ 0 0⎥ 1 −1⎥⎦ 5

6 7

0 −1 −1 1 0 0 ⎤ 1 1 1 0 1 0⎥ ⎥ 1 1 1 0 0 1⎦

f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 9.38. (5) 2

1 (2)

3 (4)

(3) (6)

(1)

(7) 5

Fig. 9.38

4

1 Twigs: {1, 2, 3, 4} f-cutset 1: {1, 6, 7} f-cutset 2: {2, 6, 7} f-cutset 3: {3, 5, 6, 7} f-cutset 4: {4, 5, 6, 7}

1 ⎡1 2 ⎢0 Q= ⎢ 3 ⎢0 4 ⎢⎣0

2 3 4 5 0 1 0 0

0 0 1 0

0 0 0 1

6

0 1 0 −1 1 1 1 1

7 1⎤ 1⎥ ⎥ 1⎥ 1⎥⎦

9.20 Network Analysis and Synthesis

Example 9.9

For the graph shown in Fig. 9.39, write the incidence matrix, tieset matrix and

f-cutset matrix. (1) 1

2 (5)

(6) (2)

(4) 5 (8)

(7) 3

4 (3)

Fig. 9.39 Solution (a) Incidence Matrix (A) 1

2

1⎡ 1 2 ⎢ −1 ⎢ Aa = 3 ⎢ 0 4⎢ 0 5 ⎢⎣ 0

3

4

5

6

7

8

1 1 0 0 0⎤ 0 0 1 0 0⎥ ⎥ 0 0 0 1 0⎥ 1 0 0 0 1⎥ 0 −1 −1 −1 −1⎥⎦

0 0 1 0 1 1 0 −1 0 0

The incidence matrix is obtained by eliminating any one row. ⎡ 1 0 0 −1 ⎢ −1 1 0 0 A= ⎢ ⎢ 0 −1 1 0 ⎢⎣ 0 0 −1 1 (b) Tieset Matrix (B) Links: {2, 4, 6, 8} Tieset 2: {2, 7, 5, 1} Tieset 4: {4, 5, 7, 3} Tieset 6: {6, 5, 1} Tieset 8: {8, 7, 3} (c)

1 2 3 4 2⎡1 4 ⎢0 B= ⎢ 6 ⎢1 8 ⎢⎣0

1 0 0 0

0 1 0 1

1 0 0 0

0 1 0 0

5 6

0 −1 1 1 0 −1 0 0

0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦

0 0 1 0 7

0 1 0 −1 1 0 0 −1

8 0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦

f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 9.40. (1) 1

2 (5)

Twigs: {1, 3, 5, 7} f-cutset 1: {1, 6, 2} f-cutset 3: {3, 4, 8} f-cutset 5: {5, 4, 6, 2} f-cutset 7: {7, 2, 8, 4}

(6) (2)

(4) 5 (8)

(7)) 3

4 (3)

Fig. 9.40

1 1⎡1 3 ⎢0 Q= ⎢ 5 ⎢0 7 ⎢⎣0

2 3

4

5

6

7

8

1 0 1 1

0 1 1 1

0 0 1 0

1 0 1 0

0 0 0 1

0⎤ 1⎥ ⎥ 0⎥ 1⎥⎦

0 1 0 0

9.7 Relationship Among Submatrices of A, B and Q 9.21

Example 9.10

How many trees are possible for the graph of the network of Fig. 9.41. 2

3

+ −

1

Fig. 9.41 (2)

Solution To draw the graph, 1. replace all resistors, inductors and capacitors by line segments, 2. replace voltage source by short circuit and current source by an open circuit, 3. assume directions of branch currents arbitrarily, and 4. number all the nodes and branches. The oriented graph is shown in Fig. 9.42. The complete Incidence Matrix (Aa) is written as 3

2

3 (1)

(3) (4) 1

Fig. 9.42

4

1 ⎡ 1 0 −1 Aa = 2 ⎢ −1 1 0 ⎢ 3 ⎣ 0 −1 1 1

1⎤ 0⎥ ⎥ 1⎦

The reduced incidence matrix A is obtained by eliminating the last row from matrix Aa. ⎡ 1 0 −1 1⎤ A= ⎢ ⎣ −1 1 0 0 ⎥⎦ ⎡ 1 −1⎤ 1 0 − 1 1 1⎥ ⎡ 3 −1⎤ ⎡ ⎤⎢ 0 AAT = ⎢ ⎢ ⎥= ⎥ ⎣ −1 1 0 0 ⎦ ⎢ −1 0 ⎥ ⎢⎣ −1 2⎥⎦ ⎢⎣ 1 0 ⎥⎦ 3 −1 The number of possible trees = AAT = = 6 − 1 = 5. −1 2

Example 9.11

Draw the oriented graph from the complete incidence matrix given below;

↓ 1 2 3 4 5

(8)

Branches →

Nodes 1

2

3

4

5

6

7

8

1 0 0 0 1 0 0 1 0 1 0 0 −1 1 0 0 0 0 1 0 0 −1 1 −1 0 0 0 1 0 0 −1 0 −1 −1 −1 −1 0 0 0 0

1

(5)

(1)

2

(6)

(2)

(3)

3

(7)

4

(4) 5

Fig. 9.43

Solution First, note down the nodes 1, 2, 3, 4, 5 as shown in Fig. 9.43. From the complete incidence matrix, it is clear that the branch number 1 is between nodes 1 and 5 and it is going away from node 1 and towards node 5 as the entry against node 1 is 1 and that against 5 is −1. Hence, connect the nodes 1 and 5 by a line, point the arrow towards 5 and call it branch 1 as shown in Fig. 9.43. Similarly, draw the other oriented branches.

9.22 Network Analysis and Synthesis

Example 9.12

The reduced incidence matrix of an oriented graph is given below. Draw the

graph. ⎡ 0 −1 1 1 0 ⎤ A = ⎢ 0 0 −1 −1 −1⎥ ⎢ ⎥ 1⎦ ⎣ −1 0 0 0 Solution First, writing the complete incidence matrix from the matrix A such that the sum of all entries in each column of Aa will be zero, we have (4)

1

2

3

4

5

1 ⎡ 0 −1 1 2 ⎢ 0 0 −1 1 Aa = ⎢ 3 ⎢ −1 0 0 4 ⎢⎣ 1 1 0

1 1 0 0

0⎤ 1⎥ ⎥ 1⎥ 0 ⎥⎦

2

1 (3)

3 (5) (1)

(2)

4

Fig. 9.44 Now, the oriented graph can be drawn with matrix Aa as shown in Fig. 9.44.

Example 9.13

The reduced incidence matrix of an oriented graph is ⎡ 0 −1 1 0 0 ⎤ A = ⎢ 0 0 −1 −1 −1⎥ ⎢ ⎥ 1⎦ ⎣ −1 0 0 0

(a) Draw the graph. (b) How many trees are possible for this graph? (c) Write the tieset and cutset matrices. Solution (a) First, writing the complete incidence matrix Aa such that the sum of all the entries in each column of Aa is zero, we have 2

1

1 1⎡ 0 2⎢ 0 Aa = ⎢ 3 ⎢ −1 4 ⎢⎣ 1

2

3

4

5

1 1 0 −1 1 0 0 1 0

0 1 0 1

0⎤ 1⎥ ⎥ 1⎥ 0 ⎥⎦

(3)

3 (5)

(4)

(2)

(1)

4

Fig. 9.45 Now, the oriented graph can be drawn with the matrix Aa, as shown in Fig. 9.45. (b) The number of possible trees = |AAT| ⎡ 0 0 −1⎤ ⎡ 0 −1 1 0 0 ⎤ ⎢ −1 0 0 ⎥ ⎡ 2 −1 0 ⎤ ⎢ ⎥ AAT = ⎢ 0 0 −1 −1 −1⎥ ⎢ 1 −1 0 ⎥ = ⎢ −1 3 −1⎥ ⎢ ⎥ ⎢ ⎥ 1⎦ ⎢ 0 −1 0 ⎥ ⎣ 0 −1 2⎦ ⎣ −1 0 0 0 ⎢⎣ 0 −1 1⎥⎦

9.7 Relationship Among Submatrices of A, B and Q 9.23

2 −1 0 AAT = −1 3 −1 = 2(6 − 1) + 1( −2) = 8 0 −1 2 The number of possible trees = 8. (c) Tieset Matrix (B) The oriented graph, selected tree and tiesets are shown in Fig. 9.46. 2

1

3 (3)

(5) (4)

(2)

1

(1)

2 3

4 5

1 ⎡1 0 0 −1 1 ⎤ B= ⎢ 2 ⎣0 1 1 −1 0 ⎥⎦

Links: {1, 2} Tieset 1: {1, 4, 5} Tieset 2: {2, 3, 4}

Fig. 9.46 f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 9.47 2

1

3 (3)

(5) (4)

(2)

(1)

Twigs: {3, 4, 5} f-cutset 3: {3, 2} f-cutset 4: {4, 2, 1} f-cutset 5: {5, 1}

1 3⎡ 0 Q = 4⎢ 1 ⎢ 5 ⎣ −1

2 3 4 5 1 1 0 0⎤ 1 0 1 0⎥ ⎥ 0 0 0 1⎦

4

Fig. 9.47

Example 9.14

The fundamental cutset matrix of a network is given as follows; Twigs g a c e

Links b d f

1 0 0 0 1 0 0 0 1

1 0 1 0 1 1 1 1 1

Draw the oriented graph. Solution No. of links l = b − n + 1 No. of nodes n = b − l + 1 = 6 − 3 + 1 = 4 f-cutsets are written as, f-cutsets a: {a, b, f } f-cutsets c: {c, d, f } f-cutsets e: {e, b, d, f } The oriented graph is drawn as shown in Fig. 9.48.

(f)

(a) (b)

(d) (e)

Fig. 9.48

Twigs: {a, c, e} Links: {b, d, f} (c)

9.24 Network Analysis and Synthesis

Example 9.15

Draw the oriented graph of a network with the f-cutset matrix as shown below: Twigs g 1 1 0 0 0

2 0 1 0 0

3 0 0 1 0

Links k 4 5 0 −1 0 1 0 0 1 0

Solution No. of links l = b − n + 1 No. of nodes n = b − l + 1 = 7 − 3 + 1 = 5 f-cutsets are written as f-cutset 1: {1, 5} f-cutset 2: {2, 5, 7} f-cutset 3: {3, 6, 7} f-cutset 4: {4, 6} Then oriented graph can be drawn as shown in Fig. 9.49.

6 0 0 1 1

7 0 1 1 0

(5)

(2)

(7)

(3)

(1) (4)

Twigs: {1, 2, 3, 4} Links: {5, 6, 7}

(6)

Fig. 9.49

9.8

KIRCHHOFF’S VOLTAGE LAW

KVL states that if vk is the voltage drop across the kth branch, then

∑ vk = 0 the sum being taken over all the branches in a given loop. If l is the number of loops or f-circuits, then there will be l number of KVL equations, one for each loop. The KVL equation for the f-circuit or loop ‘l’ can be written as b

∑ bik

k

=0

(k = 1, 2, …, l)

k =1

where bik is the elements of the tieset matrix B, b being the number of branches. The set of l KVL equations can be written in matrix form. BV Vb = 0 ⎡ v1 ⎤ ⎢v ⎥ Vb = ⎢ 2 ⎥ is a column vector of branch voltages. ⎢ ⎥ ⎢⎣ vb ⎥⎦

where and B is the fundamental circuit matrix.

9.9

KIRCHHOFF’S CURRENT LAW

KCL states that if ik is the current in the kth branch then at a given node

∑ ik = 0 the sum being taken over all the branches incident at a given node. If there are ‘n’ nodes, there will ‘n’ such equations, one for each node

9.10 Relation Between Branch Voltage Matrix Vb, Twig Voltage Matrix Vt And Node Voltage Matrix Vn

9.25

b

∑ aik ik = 0

(k = 1, 2,… , n)

k =1

so that set of n equations can be written in matrix form. Aa I b = 0

…(9.4)

⎡ i1 ⎤ ⎢i ⎥ where I b = ⎢ 2 ⎥ is a column vector of branch currents. ⎢ ⎥ ⎢⎣ib ⎥⎦ and Aa is the complete incidence matrix. If one node is taken as reference node or datum node, we can write the Eq. (9.4) as, AI b = 0 where A is the incidence matrix of order (n − 1) × b. We know that

A

…(9.5)

At Q

Equation (9.5) can be written as AQ QII b = 0 Premultiplying with At−1, At 1 At Q I b = At−1. 0 I Q Ib = 0 Q Ib = 0 where Q is the f-cutset matrix.

9.10

RELATION BETWEEN BRANCH VOLTAGE MATRIX Vb, TWIG VOLTAGE MATRIX Vt AND NODE VOLTAGE MATRIX Vn BV Vb = 0

We know that

⎡Vt ⎤ : Bl ] ⎢ ⎥ = 0 ⎢ ⎥ ⎣Vl ⎦

[

Bt Vt + Bl Vl = 0 Bl Vl = − Bt Vt Premultiplying with Bl−1 . Vl Now

Bl−11 Bt Vt = − ( Bl 1 Bt )Vt

⎡Vt ⎤ Vb = ⎢ ⎥ ⎢ ⎥ ⎣Vl ⎦ ⎡ ⎢ =⎢ ⎢− ⎢⎣

(

Vt …

⎤ ⎡ ⎥ ⎢ U ⎥=⎢ … Vt ⎥⎥ ⎢⎢ − ⎦ ⎣

)

(

)

⎤ ⎥ ⎥ .Vt ⎥ ⎥⎦

…(9.6)

9.26 Network Analysis and Synthesis Also,

V

QT Vt

Q

At−1 A AT ( AtT ) −1

QT

AT ( At 1 )T

Vb

AT ( AtT ) 1Vt = AT

Vn

( AtT ) −1Vt is node voltage matrix.

Hence Eq. (9.6) can be written as

where

9.11

{A

T t

Vt

}

AT Vn

RELATION BETWEEN BRANCH CURRENT MATRIX Ib AND LOOP CURRENT MATRIX Il

We know that, A I b = 0 ⎡ It ⎤ : At ] ⎢…⎥ = 0 ⎢ ⎥ ⎣ Il ⎦

[

At I t + Al I l = 0 At I t = − Al I l Premultiplying with At−1 , At−11 Al I l = − ( At 1 Al ) I l

It

1 1 ⎡ I t ⎤ ⎡ − ( At Al ) I l ⎤ ⎡ − ( At Al ) ⎤ ⎢ ⎥ ⎢ I b = ⎢…⎥ = … = … ⎥ .I l ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ Il ⎥⎦ ⎢⎣ U ⎥⎦ ⎣ I l ⎦ ⎢⎣

Now

I b = BT I l

9.12 9.12.1 1.

NETWORK EQUILIBRIUM EQUATION KVL Equation

If there is a voltage source vsk in the branch k having impedance zk and carrying current ik, as shown in Fig. 9.50, vk

zk ik

vsk

Vb

Zb I b Vs

(k = 1, 2,…, b)

In matrix form,

vsk

Zk

+

−+

ik

where Zb is the branch impedance matrix, Ib is the column vector of branch currents and Vs is the column vector of source voltages. Hence, KVL equation can be written as BV Vb = 0 B ( Zb I b Vs ) = 0 B Zb I b BV Vs

vk

Fig. 9.50 Circuit diagram

−

9.12 Network Equilibrium Equation 9.27

Ib

Also,

BT I l

B Zb BT I l = BV Vs Z Il = E where E BV Vs and Z B Z b BT The matrix Z is called loop impedance matrix. 2.

If there is a voltage source in series with an impedance and a current source in parallel with the combination as shown in Fig. 9.51, (v + vsk ) ik = k − isk zk vk = zk ik zk isk vsk In matrix form, Vb Zb I b Zb I s Vs

vk ik +

Zk

−+

−

vsk isk

Fig. 9.51 Circuit diagram

KVL equation is BV Vb = 0. BV Vb = B ( Zb I b Zb I s Vs ) = 0 B Zb I b BV Vs B Zb I s Now

Ib

BT I l

B Zb BT I l = BV Vs − B Zb I s Z I l = BV Vs − B Zb I s where Z

9.12.2 1.

B Zb BT is the loop impedance matrix. This is the generalised KVL equation.

KCL Equation

If the branch k contains an input current source isk and an admittance yk as shown in Fig. 9.52, i

yk vk

isk

Ib

Yb Vb

Is

ik +

vk yk

−

(k = 1, 2,…, b)

In the matrix form, where Yb is the branch admittance matrix. Hence KCL equation is given by, A Ib = 0 A (Yb Vb I s ) = 0 AY Yb Vb A I s Also

Vb

AT Vn

AY Yb AT Vn = A I s Y Vn = I where and

Y I

AY Yb AT A Is

isk

Fig. 9.52 Circuit diagram

9.28 Network Analysis and Synthesis The matrix Y is called admittance matrix. This is the KCL equation in matrix form. In terms of f-cutset matrix, the KCL equation can be written as Q Ib = 0 Q (Yb Vb Vs ) = 0 QY Yb V Q I s Also

V

QT Vt

QY Yb QT Vt = Q I s Y Vt = I

2.

where Y QY Y QT and I Q Is This is the KCL equation in matrix form. If there is a voltage source in series with an impedance and a current source in parallel with the combination as shown in Fig. 9.53, 1 yk = zk i yk v + yk vsk isk In matrix form, I b Yb Vb Yb Vs I s

ik +

zk

vk

−+ vsk isk

Fig. 9.53 Circuit diagram

KCL equation will be given by, A Ib = 0 A (Yb Vb Yb Vs I s ) = 0 AY Yb Vb A I s Also

Vb

AY Yb Vs

AT Vn

AY Yb AT Vn = A I s − AY Yb Vs Y Vn = A I s − AV Vb Vs where Y AY Yb AT is the node admittance matrix. This is a generalised KCL equation. In terms of f-cutset matrix, the KCL equation can be written as Q Ib = 0 Q (Yb Vb Yb Vs I s ) = 0 QY Yb V Q I Also

V

QY Yb Vs

QT Vt

QY Yb QT Vt = Q I s − QY Yb Vs Y Vt = Q I s − QY Yb Vs This is a generalised KCL equation.

−

9.12 Network Equilibrium Equation 9.29

Note (i) For a graph having b branches, the branch impedance matrix Zb is a square matrix of order b, having branch impedances as diagonal elements and the mutual impedances between the branches as nondiagonal elements. For a network having no mutual impedances, only diagonal elements will be present in the branch impedance matrix. (ii) For a graph having b branches, the branch admittance matrix Yb is a square matrix of order b, having branch admittances as diagonal elements and the mutual admittances between the branches as nondiagonal elements. For a network having no mutual admittances, only diagonal elements will be present in the branch admittance matrix. (iii) For a graph having b branches, the voltage source matrix or vector Vs is a rectangular matrix of order b × 1, having the value of the voltage source in the particular branch. The value will be positive if there is a voltage rise in the direction of current and will be negative if there is a voltage fall in the direction of current. (iv) For a graph having b branches, the current source matrix or vector Is is a rectangular matrix of order b × 1, having the value of the current source in the particular parallel branch. The value will be positive if the direction of the current source and the corresponding parallel branch current are not same. The value will be negative if the directions of the current source and corresponding parallel branch current are same.

Example 9.16 Write the incidence matrix of the graph of Fig. 9.54 and express branch voltages in terms of node voltages. Write the tieset matrix and express branch currents in terms of loop currents. (8) (5)

(6)

2

1

3

(2)

(1)

5

(7)

(3)

(4)

Fig. 9.54 Solution (a) Incidence Matrix 1

2

3

4

5

6

7

8

1⎡ 1 0 0 0 1 0 2⎢ 0 1 0 0 −1 1 ⎢ Aa = 3 ⎢ 0 0 1 0 0 −11 4⎢ 0 0 0 1 0 0 5 ⎢⎣ −1 −1 −1 −1 0 0

0 0 1 1 0

1⎤ 0⎥ ⎥ 1⎥ 0⎥ 0 ⎥⎦

The incidence matrix is obtained by eliminating any one row. ⎡1 ⎢0 A= ⎢ ⎢0 ⎢⎣0

0 1 0 0

0 0 1 0

0 1 0 0 1⎤ 0 −1 1 0 0 ⎥ ⎥ 0 0 −1 1 −1⎥ 1 0 0 −1 0 ⎥⎦

9.30 Network Analysis and Synthesis (b)

Branch voltages in terms of node voltages AT Vn

Vb

⎡V1 ⎤ ⎡ 1 ⎢V2 ⎥ ⎢0 ⎢ ⎥ ⎢ ⎢V3 ⎥ ⎢0 ⎢V4 ⎥ = ⎢0 ⎢V5 ⎥ ⎢ 1 ⎢V ⎥ ⎢0 ⎢ 6⎥ ⎢ ⎢V7 ⎥ ⎢0 ⎢⎣V8 ⎥⎦ ⎢⎣ 1 (c)

0 0 1 0 0 1 0 0 1 0 1 −1 0 1 0 −1

0⎤ 0⎥ ⎥ 0 ⎥ ⎡Vn1 ⎤ 1⎥ ⎢Vn2 ⎥ ⎢ ⎥ 0 ⎥ ⎢Vn3 ⎥ 0 ⎥⎥ ⎢⎣Vn4 ⎥⎦ 1⎥ 0 ⎥⎦

Tieset Matrix Selected tree and tiesets are shown in Fig. 9.55. (8) (5) 1

(6)

2

(2)

(1)

1

3

(7)

(3)

Links: {1, 4, 6, 8} Tieset 1: {1, 2, 5} Tieset 4: {4, 3, 7} Tieset 6: {6, 2, 3} Tieset 8: {8, 2, 3, 5}

1 ⎡1 4 ⎢0 B= ⎢ 6 ⎢0 8 ⎢⎣0

2

3 4

1 0 0 −1 1 1 1 1

5 6 7 8

0 −1 1 0 0 0 0 −1

0 0 1 0

0 1 0 0

0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦

4

5

(4)

Fig. 9.55 (d) Branch currents in terms of loop currents Ib

BT I l

⎡ I1 ⎤ ⎡ 1 0 0 0 ⎤ ⎢ I 2 ⎥ ⎢ −1 0 −1 −1⎥ ⎢ ⎥ ⎢ ⎥ ⎡I ⎤ ⎢ I 3 ⎥ ⎢ 0 −1 1 1⎥ ⎢ l1 ⎥ 1 0 0 ⎥ I l4 ⎢I4 ⎥ = ⎢ 0 ⎢ I 5 ⎥ ⎢ −1 0 0 −1⎥ ⎢⎢ I l6 ⎥⎥ ⎢I ⎥ ⎢ 0 0 1 0 ⎥⎥ ⎢⎣ I l8 ⎥⎦ ⎢ 6⎥ ⎢ 1 0 0⎥ ⎢ I7 ⎥ ⎢ 0 ⎢⎣ I 8 ⎥⎦ ⎢⎣ 0 0 0 1⎥⎦

Example 9.17

Branch current and loop-current relationships are expressed in matrix form as

Draw the oriented graph.

⎡ I1 ⎤ ⎡ 1 0 0 ⎢I2 ⎥ ⎢ 0 1 0 ⎢ ⎥ ⎢ I 0 1 1 3 ⎢ ⎥ ⎢ 1 1 ⎢I4 ⎥ = ⎢ 0 ⎢ I 5 ⎥ ⎢ 1 −1 0 ⎢ I ⎥ ⎢ 0 0 −1 ⎢ 6⎥ ⎢ ⎢ I 7 ⎥ ⎢ −1 0 0 ⎢⎣ I 8 ⎥⎦ ⎢⎣ 0 0 0

1⎤ 1⎥ ⎥ 0 ⎥ ⎡ I l1 ⎤ 0 ⎥ ⎢ I l2 ⎥ ⎢ ⎥ 0 ⎥ ⎢ I l3 ⎥ 0 ⎥⎥ ⎢⎣ I l4 ⎥⎦ 0⎥ 1⎥⎦

9.12 Network Equilibrium Equation 9.31

Solution Writing the equation in matrix form, Ib

BT I l

⎡ 1 0 0 −1⎤ ⎢ 0 1 0 −1⎥ ⎢ ⎥ 0 1 1 0⎥ ⎢ 0 1 1 0⎥ BT = ⎢⎢ 1 −1 0 0 ⎥ ⎢ 0 0 −1 0 ⎥ ⎢ ⎥ ⎢ −1 0 0 0 ⎥ ⎢⎣ 0 0 0 1⎥⎦

1 ⎡ 1 ⎢ 0 ∴B = ⎢ ⎢ 0 1 ⎢⎣ −1

2 3 4 0 1 0 1

0 1 1 0

5

0 1 1 −1 1 0 0 0

6

7

0 −1 0 0 1 0 0 0

8 0⎤ 0⎥ ⎥ 0⎥ 1⎥⎦

(8) (1)

1

2

(2)

3

(5)

(7)

From tieset matrix, No. of links l=4 No of branches b = 8 No of nodes n= b − l + 1 = 8 − 4 + 1 = 1 The oriented graph is shown in Fig. 9.56.

(3)

4

5

(4)

(6)

Fig. 9.56

Example 9.18 For the given graph shown in Fig. 9.57, write down the basic tieset matrix and taking a tree of branches 2, 4, 5, write down KVL equations from the matrix. (2) 1 (5) (1)

2

(3)

3

4 (4)

(6)

Fig. 9.57 Solution Selecting branches 2, 4, and 5 as the tree as shown in Fig. 9.58, (2) 1

2

1 2 (5) (1)

(3)

Links: {1, 3, 6} Tieset 1: {1, 5, 4} Tieset 3: {3, 2, 5} Tieset 6: {6, 2, 5, 4}

3

4 (4)

(6)

Fig. 9.58

3

1⎡1 0 0 B = 3 ⎢0 1 1 ⎢ 6 ⎣0 1 0

4 5 6 1 1 0⎤ 0 1 0⎥ ⎥ 1 1 1⎦

9.32 Network Analysis and Synthesis The KVL equation in matrix form is given by BV Vb = 0 ⎡V1 ⎤ ⎢V ⎥ ⎡ 1 0 0 −1 1 0 ⎤ ⎢ 2 ⎥ ⎢0 1 1 0 1 0 ⎥ ⎢V3 ⎥ = 0 ⎢ ⎥ ⎢V4 ⎥ ⎣0 1 0 −1 1 1⎦ ⎢V ⎥ 5 ⎢V ⎥ ⎣ 6⎦ V1 V4 + V5 = 0 V2 V3 + V5 = 0 V4 + V5 V6 = 0

V2

Example 9.19 Obtain the f-cutset matrix for the graph shown in Fig. 9.59 taking 1, 2, 3, 4 as tree branches. Write down the network equations from the f-cutset matrix. 2

(5)

(6)

(3)

3 1 (2)

(7) (8)

(1)

(4)

4

Fig. 9.59 1 2 3

Solution Twigs: {1, 2, 3, 4} f-cutset 1 : {1, 6, 7, 8} f-cutset 2 : {2, 5, 6, 7, 8} f-cutset 3 : {3, 5, 6} f-cutset 4 : {4, 6, 7}

1 ⎡1 2 ⎢0 Q= ⎢ 3 ⎢0 4 ⎢⎣0

The KCL equation in matrix form is given by Q Ib = 0

⎡1 ⎢0 ⎢ ⎢0 ⎢⎣0

0 1 0 0

0 0 1 0

0 0 0 1

0 1 −1 1 −1 1 1 −1 0 0 1 −1

⎡ I1 ⎤ ⎢I2 ⎥ ⎢ ⎥ 1⎤ ⎢ I 3 ⎥ −1⎥ ⎢ I 4 ⎥ =0 ⎥ 0⎥ ⎢ I 5 ⎥ 0 ⎥⎦ ⎢⎢ I 6 ⎥⎥ ⎢ I7 ⎥ ⎢⎣ I 8 ⎥⎦

0 1 0 0

0 0 1 0

4

5

6

7

8

0 0 0 1

0 1 1 0

1 1 1 1

1 1 0 1

1⎤ 1⎥ ⎥ 0⎥ 0 ⎥⎦

9.12 Network Equilibrium Equation 9.33

I1 I2

Example 9.20

I6 − I7

I5 − I6 I3 I4

I8 = 0

I 7 − I8 = 0 I5 − I6 = 0 I6 − I7 = 0

The reduced incidence matrix of a graph is given as

⎡ 1 0 0 0 −1⎤ A = ⎢ −1 −1 −1 0 0 ⎥ ⎢ ⎥ 1 −1 0 ⎦ ⎣ 0 0 Express branch voltages in terms of node voltages. Solution For the given graph, No of branches b = 5 No of nodes n = 3 Branch voltages can be expressed in terms of node voltages by Vb

T

Vn

⎡V1 ⎤ ⎡ 1 −1 0 ⎤ ⎢V2 ⎥ ⎢ 0 −1 0 ⎥ ⎡Vn1 ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢V3 ⎥ = ⎢ 0 −1 1⎥ ⎢Vn2 ⎥ ⎢V4 ⎥ ⎢ 0 0 −1⎥ ⎣Vn3 ⎦ ⎢⎣V5 ⎥⎦ ⎢⎣ −1 0 0 ⎥⎦ V1 Vn1 − Vn2

Example 9.21

V2

Vn2

V3

Vn2 + Vn3

V4

Vn3

V5

Vn1

The fundamental cutset matrix of a graph is given as ⎡ −1 1 0 0 −1⎤ Q = ⎢ 0 0 1 0 −1⎥ ⎢ ⎥ ⎣1 0 0 1 0⎦

Express branch voltages in terms of twig voltages. Solution For the given graph, No. of branches b = 5 No. of twigs = 3 Branch voltages are expressed in terms of twig voltages by V

QT Vt

⎡V1 ⎤ ⎡ −1 0 ⎢V2 ⎥ ⎢ 1 0 ⎢ ⎥ ⎢ 1 ⎢V3 ⎥ = ⎢ 0 ⎢V4 ⎥ ⎢ 0 0 ⎢⎣V5 ⎥⎦ ⎢⎣ −1 −1

1⎤ 0 ⎥ ⎡Vt1 ⎤ ⎥ 0 ⎥ ⎢Vt2 ⎥ ⎢ ⎥ 1⎥ ⎣Vt3 ⎦ 0 ⎥⎦

9.34 Network Analysis and Synthesis Vt1 + Vt3

V1 V2

Vt1

V3

Vt2

V4

Vt3 Vt1 − Vt2

V5

Example 9.22 For this network shown in Fig. 9.60, write down the tieset matrix and obtain the network equilibrium equation in matrix form using KVL. Calculate the loop currents and branch currents.

2Ω

1Ω

1Ω

2V

+ −

2Ω

2Ω 1Ω

Fig. 9.60 Solution The oriented graph and one of its trees are shown in Fig. 9.61. 1

(1) (5)

3

(4) 2 (3)

1

(1)

(2)

(5)

(6)

4

3

(2)

(4)

Links: {1, 2, 3} Tieset 1: {1, 4, 5} Tieset 2: {2, 4, 6} Tieset 3: {3, 5, 6}

(6)

2

1 2 3 1⎡1 0 0 B = 2 ⎢0 1 0 ⎢ 3 ⎣0 0 1

4

(3)

Fig. 9.61 The KVL equation in matrix form is given by

Here,

B Zb BT I l = BV Vs − B Zb I s I s = 0, B Zb BT I l = BV Vs

Zb

⎡1 ⎢0 ⎢ ⎢0 ⎢0 ⎢0 ⎢0 ⎣

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 2 0 0

0 0 0 0 2 0

0⎤ ⎡ 1 0 0⎤ ⎡ 2⎤ ⎢0 ⎢0 ⎥ 0⎥ 1 0⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 0⎥ T ⎢0 0 1⎥ 0 ;B = ; Vs = ⎢ ⎥ 0⎥ 1 − 1 0 ⎢ ⎥ ⎢0 ⎥ ⎢ 1 0 −1⎥ ⎢0 ⎥ 0⎥ ⎥ ⎢ ⎥ ⎢0 ⎥ 2⎦ ⎣0 −1 1⎦ ⎣ ⎦

4

5

1 1 1 0 0 −1

6 0⎤ 1⎥ ⎥ 1⎦

9.12 Network Equilibrium Equation 9.35

1 1 ⎡1 0 0 B Zb = ⎢0 1 0 −1 0 ⎢ ⎣0 0 1 0 −1

2 2 ⎡1 0 0 B Zb BT = ⎢0 1 0 −2 0 ⎢ ⎣0 0 1 0 −2

1 1 ⎡1 0 0 BV Vs = ⎢0 1 0 −1 0 ⎢ ⎣0 0 1 0 −1

⎡1 0 0 0 0 0⎤ ⎢0 1 0 0 0 0⎥ 0⎤ ⎢ ⎥ ⎡1 0 0 1 0 0 0⎥ ⎢ −1⎥ ⎢ = 0 ⎥ ⎢0 0 0 2 0 0 ⎥ ⎢ 1⎦ ⎢ 0 0 0 0 0 2 0⎥ ⎣ ⎢ 0 0 0 0 0 2⎥ ⎣ ⎦ 1 0 0 ⎡ ⎤ ⎢0 1 0⎥ 0⎤ ⎢ ⎥ ⎡ 5 −2 0 0 1⎥ ⎢ −2⎥ ⎢ = −2 5 ⎥ 1 −1 0 ⎥ ⎢ 2⎦ ⎢⎢ − 2 − 2 1 0 −1⎥ ⎣ ⎢0 −1 1⎥ ⎣ ⎦ ⎡ 2⎤ ⎢0 ⎥ 0 ⎤ ⎢ ⎥ ⎡ 2⎤ 0 −1⎥ ⎢ ⎥ = ⎢0 ⎥ ⎥ 0 ⎢ ⎥ 1⎦ ⎢⎢ ⎥⎥ ⎣0 ⎦ 0 ⎢0 ⎥ ⎣ ⎦

The KVL equation in matrix form is given by ⎡ 5 ⎢ −2 ⎢ 2 ⎣ −2

2 −2⎤ ⎡ I l1 ⎤ ⎡ 2⎤ 5 −2⎥ ⎢ I l2 ⎥ = ⎢0 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 2 5⎦ ⎣ I l3 ⎦ ⎣0 ⎦

Solving this matrix equation, 6 A 7 4 I l2 = A 7 4 I l3 = A 7 I l1 =

The branch currents are given by Ib

BT I l

⎡ I1 ⎤ ⎡ 1 0 ⎢ I 2 ⎥ ⎢0 1 ⎢ ⎥ ⎢ I 0 0 ⎢ 3⎥ = ⎢ ⎢ I 4 ⎥ ⎢ 1 −1 ⎢ I5 ⎥ ⎢ 1 0 ⎢ I ⎥ ⎢0 −1 ⎣ 6⎦ ⎣

⎡6⎤ ⎢7⎥ ⎢4⎥ 0⎤ ⎡ 6 ⎤ ⎢ ⎥ 0⎥ ⎢ 7 ⎥ ⎢ 7 ⎥ ⎥ ⎢ ⎥ ⎢4⎥ 1⎥ ⎢ 4 ⎥ ⎢ ⎥ = 0⎥ ⎢ 7 ⎥ ⎢ 7 ⎥ 2 −1⎥ ⎢ 4 ⎥ ⎢ ⎥ ⎢7⎥ ⎢ ⎥ ⎥ 1⎦ ⎣ 7 ⎦ ⎢ ⎥ 2 ⎢ ⎥ ⎢7⎥ ⎢⎣ 0 ⎥⎦

0 0 1 0 0 1

−2⎤ −2⎥ ⎥ 5⎦

2 2 2 0 0 −2

0⎤ 2⎥ ⎥ 2⎦

9.36 Network Analysis and Synthesis

Example 9.23 For the network shown in Fig. 9.62, write down the tieset matrix and obtain the network equilibrium equation in matrix form using KVL. Calculate loop currents. 2Ω

+−

4Ω

6Ω

Il 3

6Ω

8V

2Ω

12 V + −

Il1

4Ω + − 6V

Il2

Fig. 9.62 Solution The oriented graph and its selected tree are shown in Fig. 9.63. (3)

(3)

2

1

(4)

3

(5) (6)

(1)

2

1

(2)

(4)

3

(5) (6)

(1)

4

Links: {1, 2, 3} Tieset 1: {1, 4, 6} Tieset 2: {2, 5, 6} Tieset 3: {3, 5, 4}

(2)

1 2

3

4

5

6

1⎡1 0 0 1 0 1⎤ B = 2 ⎢0 1 0 0 1 −1⎥ ⎢ ⎥ 3 ⎣0 0 1 −1 −1 0 ⎦

4

Fig. 9.63 The KVL equation in matrix form is given by B Zb BT I l = BV Vs − B Zb I s I s = 0,

Here, T

B Zb B I l = BV Vs ⎡6 ⎢0 ⎢ ⎢0 ⎢0 ⎢0 ⎢0 ⎣

0⎤ ⎡1 0 ⎥ ⎢0 0 1 ⎥ ⎢ 0⎥ T ⎢0 0 Zb ;B = 0⎥ ⎢1 0 ⎢0 0⎥ 1 ⎢ 1 −1 2⎥⎦ ⎣ ⎡6 0 0 ⎢0 4 0 1 0 1⎤ ⎢ ⎡1 0 0 0 0 2 B Zb = ⎢0 1 0 0 1 −1⎥ ⎢ ⎢ ⎥ ⎢0 0 0 ⎣0 0 1 −1 −1 0 ⎦ ⎢0 0 0 ⎢0 0 0 ⎣ 0 4 0 0 0 0

0 0 2 0 0 0

0 0 0 4 0 0

0 0 0 0 6 0

⎡1 0 ⎢0 1 4 0 2⎤ ⎢ ⎡6 0 0 0 0 BZb BT = ⎢0 4 0 0 6 −2⎥ ⎢ ⎢ ⎥ ⎢1 0 ⎣0 0 2 −4 −6 0 ⎦ ⎢0 1 ⎢ 1 −1 ⎣

0⎤ ⎡ 12⎤ ⎥ ⎢ −6 ⎥ 0 ⎥ ⎢ ⎥ 1⎥ −8 ; Vs = ⎢ ⎥ −1⎥ ⎢ 0⎥ ⎢ 0⎥ −1⎥ ⎢ 0⎥ 0 ⎥⎦ ⎣ ⎦ 0 0 0⎤ 0 0 0⎥ 4 0 ⎥ ⎡6 0 0 0 0 0⎥ ⎢ = 0 4 0 0 6 4 0 0⎥ ⎢ 0 0 2 −4 −6 0 6 0⎥ ⎣ 0 0 2⎥⎦ 0⎤ 0⎥ ⎥ ⎡ 12 −2 −4 ⎤ 1⎥ ⎢ = −1 12 −6 ⎥ −1⎥ ⎢ ⎥ −4 −6 12⎦ −1⎥ ⎣ 0 ⎥⎦

2⎤ 2⎥ ⎥ 0⎦

9.12 Network Equilibrium Equation 9.37

⎡1 0 0 BV Vs = ⎢0 1 0 ⎢ ⎣0 0 1

⎡ 12⎤ ⎢ −6 ⎥ 0 1⎤ ⎢ ⎥ ⎡ 12⎤ −8 1 −1⎥ ⎢ ⎥ = ⎢ −6 ⎥ ⎥ 0 ⎢ ⎥ 1 0 ⎦ ⎢⎢ ⎥⎥ ⎣ −8⎦ 0 ⎢ 0⎥ ⎣ ⎦

1 0 1

Hence, the KVL equation in matrix form is given by 4 ⎤ ⎡ I l1 ⎤ ⎡12 ⎤ 6 ⎥ ⎢ I l2 ⎥ = ⎢ −6 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 12⎦ ⎣ I l3 ⎦ ⎣ −8⎦

2 ⎡ 12 −2 ⎢ −2 2 12 ⎢ 4 6 ⎣ −4 Solving this matrix equation,

I l1 = 0 55A I l2 = −0.866 A I l3 = −0.916 A

Example 9.24 For the network shown in Fig. 9.64, draw the oriented graph. Write the tieset schedule and hence obtain the equilibrium equation on loop basis. Calculate the values of branch currents and branch voltages. + 1V −

1A 1Ω 1Ω

1Ω

1Ω

1Ω

Fig. 9.64 Solution The oriented graph and one of its trees are shown in Fig. 9.65. 1

1 (4)

(1) (2)

2

(6)

3

(4) (1)

(3)

(5)

(2)

4

2

(6)

3

Links: {1, 2, 3} Tieset 1: {1, 4, 5, 6} Tieset 2: {2, 4, 5} Tieset 3: {3, 4}

(3)

(5)

4

Fig. 9.65 1 2 3 1⎡1 0 0 B = 2 ⎢0 1 0 ⎢ 3 ⎣0 0 1

4

5

1 1 1 1 1 0

6 1⎤ 0⎥ ⎥ 0⎦

9.38 Network Analysis and Synthesis The KVL equation in matrix form is given by B Zb BT I l = BV Vs − B Zb I s ⎡1 ⎢0 ⎢ 0 Zb = ⎢ ⎢0 ⎢0 ⎢0 ⎣

0⎤ ⎡ 1 ⎢ 0 0⎥ ⎥ ⎢ 0⎥ T ⎢ 0 ;B = 0⎥ ⎢ −1 ⎢ 1 0⎥ ⎢ −1 1⎥⎦ ⎣ ⎡1 0 0 ⎢0 1 0 ⎡ 1 0 0 −1 1 −1⎤ ⎢ 0 0 1 B Zb = ⎢0 1 0 −1 1 0 ⎥ ⎢ ⎢ ⎥ ⎢0 0 0 ⎣0 0 1 1 0 0 ⎦ ⎢0 0 0 ⎢0 0 0 ⎣ 0 1 0 0 0 0

0 0 1 0 0 0

⎡1 0 0 BZb BT = ⎢0 1 0 ⎢ ⎣0 0 1

⎡1 0 0 BV Vs = ⎢0 1 0 ⎢ ⎣0 0 1

0 0 0 0 0 0

0 0 0 0 1 0

0 1 0 −1 1 0

⎡ 1 0 0⎤ ⎢ 0 1 0⎥ 0 1 −1⎤ ⎢ ⎥ ⎡3 1 0⎤ 0 0 1⎥ ⎢ 0 1 0⎥ ⎢ = 1 2 0⎥ ⎥ ⎢ −1 −1 1⎥ ⎢ ⎥ 0 0 0⎦ ⎢ 0 0 1⎦ 1 1 0⎥ ⎣ ⎢ −1 0 0 ⎥ ⎣ ⎦ ⎡0 ⎤ ⎢0 ⎥ −1 1 −1⎤ ⎢ ⎥ ⎡ −1⎤ 0 −1 1 0 ⎥ ⎢ ⎥ = ⎢ −1⎥ ⎥ ⎢1 ⎥ ⎢ ⎥ 1 0 0 ⎦ ⎢ ⎥ ⎣ 1⎦ 0 ⎢0 ⎥ ⎣ ⎦

⎡ −1⎤ ⎢ 0⎥ ⎡ 1 0 0 0 1 −1⎤ ⎢ ⎥ ⎡ −1⎤ 0 B Zb I s = ⎢0 1 0 0 1 0 ⎥ ⎢ ⎥ = ⎢ 0 ⎥ ⎢ ⎥ ⎢ 0⎥ ⎢ ⎥ ⎣0 0 1 0 0 0 ⎦ ⎢ 0 ⎥ ⎣ 0 ⎦ ⎢ 0⎥ ⎣ ⎦ Hence, the KVL equation in matrix form is given by ⎡ 3 1 0 ⎤ ⎡ I l1 ⎤ ⎡ −1⎤ ⎡ −1⎤ ⎡ 0 ⎤ ⎢1 2 0 ⎥ ⎢ I l ⎥ = ⎢ −1⎥ − ⎢ 0 ⎥ = ⎢ −1⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣0 0 1 ⎦ ⎣ I l3 ⎦ ⎣ 1⎦ ⎣ 0 ⎦ ⎣ 1⎦ Solving this matrix equation, 1 A 5 3 I l2 = − A 5 I l3 = 1 A I l1 =

0 0 0 0 0 0

0⎤ ⎡0 ⎤ ⎢0 ⎥ 0⎥ ⎥ ⎢ ⎥ 1⎥ 0 ; Vs = ⎢ ⎥ ; I s 1⎥ ⎢1 ⎥ ⎢0 ⎥ 0⎥ ⎢0 ⎥ 0 ⎥⎦ ⎣ ⎦ 0 0⎤ 0 0⎥ ⎥ ⎡1 0 0 0⎥ ⎢ = 0 1 0 0⎥ ⎢ 0 0 1 0⎥ ⎣ 0 1 ⎥⎦

⎡ −1⎤ ⎢ 0⎥ ⎢ ⎥ 0 =⎢ ⎥ ⎢ 0⎥ ⎢ 0⎥ ⎢ 0⎥ ⎣ ⎦ 0 0 1 0 0 1 1 0 0

1⎤ 0⎥ ⎥ 0⎦

9.12 Network Equilibrium Equation 9.39

The branch currents are given by I b = BT I l

Is

⎡ I1 ⎤ ⎡ 1 0 ⎢I2 ⎥ ⎢ 0 1 ⎢ ⎥ ⎢ ⎢ I3 ⎥ = ⎢ 0 0 ⎢ I 4 ⎥ ⎢ −1 −1 ⎢ I5 ⎥ ⎢ 1 1 ⎢ ⎥ ⎢ ⎣ I 6 ⎦ ⎣ −1 0

⎡ 4⎤ ⎢− 5 ⎥ ⎢ 3⎥ ⎡ −1⎤ ⎢ − ⎥ 0⎤ ⎢ 5⎥ ⎡ 1⎤ 0⎥ ⎢ ⎥ ⎢ 0⎥ ⎢ 1 ⎥ ⎥ ⎢ ⎥ 1⎥ ⎢ 5 ⎥ + ⎢ 0 ⎥ = ⎢ ⎥ 3 ⎢ 7⎥ 1⎥ ⎢ − ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢ 5⎥ 0⎥ ⎣ 1 ⎦ ⎢ 0⎥ ⎢ 5 ⎥ ⎥ ⎢ ⎥ 2 0⎦ ⎣ 0⎦ ⎢ − ⎥ ⎢ 5⎥ ⎢ 1⎥ ⎢− ⎥ ⎣ 5⎦

The branch voltages are given by Vb

Zb I b Vs

⎡V1 ⎤ ⎡1 ⎢V2 ⎥ ⎢0 ⎢ ⎥ ⎢ ⎢V3 ⎥ = ⎢0 ⎢V4 ⎥ ⎢0 ⎢V5 ⎥ ⎢0 ⎢ ⎥ ⎢ ⎣V6 ⎦ ⎣0

Example 9.25

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 0 0 0

0 0 0 0 1 0

⎡ 4⎤ ⎡ 4⎤ ⎢− 5 ⎥ ⎢− 5 ⎥ ⎢ 3⎥ ⎢ ⎥ 0⎤ ⎢ − ⎥ ⎡0 ⎤ ⎢ − 3 ⎥ ⎢ 5⎥ ⎢ ⎥ ⎥ ⎢ 5⎥ 0 ⎢ ⎥ 0 ⎥ 1 ⎢ ⎥ ⎢ ⎥ 0⎥ ⎢ 7 ⎥ − ⎢0 ⎥ = 1 ⎢ ⎥ 0 ⎥ ⎢ ⎥ ⎢1 ⎥ ⎢ −1⎥ ⎢ 5⎥ 0 ⎥ ⎢ 2 ⎥ ⎢0 ⎥ ⎢ − 2 ⎥ ⎥ − ⎢ ⎥ 1 ⎦ ⎢ ⎥ ⎣0 ⎦ ⎢ 5 ⎥ 5 ⎢ 1⎥ ⎢ ⎥ ⎢− ⎥ ⎢ − 1⎥ ⎣ 5⎦ ⎢⎣ 5 ⎥⎦

For the network shown in Fig. 9.66, obtain the loop equation in matrix form. R1 R2

I

V

R4 R3

Fig. 9.66 Solution The oriented graph and one of its trees are shown in Fig. 9.67. 1

1 (2)

(1)

2

(2) (4)

(1)

2

(3) 3

(3) 3

Fig. 9.67

Links: {1, 4} (4) Tieset 1: {1, 2, 3} Tieset 4: {4, 2, 3}

1 2 3 4 1 ⎡1 1 1 0 ⎤ B= ⎢ 4 ⎣0 1 1 1 ⎥⎦

9.40 Network Analysis and Synthesis The KVL equation in matrix form is given by B Zb BT I l = BV Vs − B Zb I s 0 0 0⎤ ⎡ R1 ⎢ 0 R2 0 0⎥ T Zb = ⎢ ⎥; B 0 0 R 0⎥ 3 ⎢ 0 0 R4 ⎦⎥ ⎢⎣ 0 0 ⎡ R1 ⎡1 1 1 0 ⎤ ⎢ 0 R2 B Zb = ⎢ ⎢ 0 ⎣0 1 1 1 ⎥⎦ ⎢ 0 0 ⎢⎣ 0

⎡1 0⎤ ⎡V ⎤ ⎡0 ⎤ ⎢1 1⎥ ⎢0⎥ ⎢I ⎥ =⎢ ⎥ ; Vs = ⎢ ⎥ ; I s = ⎢ ⎥ 1 1 0 ⎢ ⎥ ⎢ ⎥ ⎢0 ⎥ 0 1 0 ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ ⎢⎣0 ⎥⎦ 0 0⎤ 0 0 ⎥ ⎡ R1 R2 R3 0⎤ ⎥= R3 0 ⎥ ⎢⎣ 0 R2 R3 R4 ⎥⎦ 0 R4 ⎥⎦

⎡1 0⎤ 0 ⎤ ⎢1 1 ⎥ ⎡ R1 + R2 + R3 ⎢ ⎥= R4 ⎥⎦ ⎢1 1 ⎥ ⎢⎣ R2 R3 0 1 ⎢⎣ ⎥⎦ ⎡V ⎤ ⎡1 1 1 0 ⎤ ⎢ 0 ⎥ ⎡V ⎤ BV Vs = ⎢ ⎢ ⎥= ⎣0 1 1 1 ⎥⎦ ⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎡0 ⎤ 0 ⎤ ⎢ I ⎥ ⎡ R2 I ⎤ ⎡ R1 R2 R3 B Zb I s = ⎢ ⎢ ⎥= ⎣ 0 R2 R3 R4 ⎥⎦ ⎢0 ⎥ ⎢⎣ R2 I ⎥⎦ ⎢⎣0 ⎥⎦ Hence, the KVL equation in matrix form is given by ⎡ R R2 B Z b BT = ⎢ 1 ⎣ 0 R2

⎡ R1 ⎢⎣

R2 + R3 R2 R3

R3 R3

R2 + R3 ⎤ ⎡ I l1 ⎤ ⎡V ⎤ ⎡ R2 = − R2 R3 + R4 ⎥⎦ ⎢⎣ I l4 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ R2

R2 + R3 ⎤ R2 R3 + R4 ⎥⎦

I ⎤ ⎡V R2 I ⎤ = I ⎥⎦ ⎢⎣ − R2 I ⎥⎦

Example 9.26 For the network shown in Fig. 9.68, write down the tieset matrix and obtain the network equilibrium equation in matrix form using KVL. 2A

5Ω j5 Ω

5Ω

2Ω +

10 V

−4Ω −j

−

Fig. 9.68

2Ω

9.12 Network Equilibrium Equation 9.41

Solution The oriented graph and its selected tree are shown in Fig. 9.69. (3)

(3)

2

1

(4)

3

(5)

(1)

(6)

2

1

(4)

(2)

(1)

4

3

(5) (6)

(2)

Links: {1, 2, 3} Tieset 1: {1, 4, 6} Tieset 2: {2, 5, 6} Tieset 3: {3, 5, 4}

1 2

3

4

5

1⎡1 0 0 1 0 B = 2 ⎢0 1 0 0 1 ⎢ 3 ⎣0 0 1 −1 −1

4

Fig. 9.69 The KVL equation in matrix form is given by B Zb BT I l = BV Vs − B Zb I s ⎡2 ⎢0 ⎢ 0 Zb = ⎢ ⎢0 ⎢0 ⎢0 ⎣

0 0⎤ ⎡1 ⎢0 0 0⎥ ⎥ ⎢ 0 0 ⎥ T ⎢0 ;B = 0 0⎥ ⎢1 ⎢0 j5 0⎥ ⎢1 0 − j 4 ⎥⎦ ⎣ ⎡2 0 0 ⎢0 2 0 1 0 1⎤ ⎢ ⎡1 0 0 0 0 5 B Zb = ⎢0 1 0 0 1 −1⎥ ⎢ ⎢ ⎥ ⎢0 0 0 ⎣0 0 1 −1 −1 0 ⎦ ⎢0 0 0 ⎢0 0 0 ⎣ 0 2 0 0 0 0

0 0 5 0 0 0

0 0 0 5 0 0

⎡1 0 ⎢0 1 5 0 − j 4⎤ ⎢ ⎡2 0 0 0 0 B Z b BT = ⎢ 0 2 0 0 j5 j 4⎥ ⎢ ⎢ ⎥ 1 0 0 ⎦ ⎢⎢ ⎣0 0 5 −5 − j 5 0 1 ⎢ 1 −1 ⎣

0 0⎤ ⎡10 ⎤ ⎢ 0⎥ 1 0⎥ ⎥ ⎢ ⎥ 0 1⎥ 0 ; Vs = ⎢ ⎥ ; I s 0 −1⎥ ⎢ 0⎥ ⎢ 0⎥ 1 −1⎥ ⎢ 0⎥ 1 0 ⎥⎦ ⎣ ⎦ 0 0 0⎤ 0 0 0⎥ ⎥ ⎡2 0 0 0 0⎥ ⎢ = 0 2 5 0 0⎥ ⎢ 0 0 0 j5 0⎥ ⎣ 0 0 j 4 ⎥⎦ 0⎤ 0⎥ j4 ⎥ ⎡7 − j 4 1⎥ ⎢ = j 4 2 + j1 −1⎥ ⎢ −5 − j5 −1⎥ ⎣ ⎥ 0⎦

⎡10 ⎤ ⎢0⎥ 1 0 1⎤ ⎢ ⎥ ⎡100 ⎤ ⎡1 0 0 0 BV Vs = ⎢0 1 0 0 1 −1⎥ ⎢ ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢0 ⎥ ⎢ ⎥ ⎣0 0 1 −1 −1 0 ⎦ ⎢ 0 ⎥ ⎣ 0 ⎦ ⎢0⎥ ⎣ ⎦ ⎡2 0 0 B Zb I s = ⎢0 2 0 ⎢ ⎣0 0 5

5 0 0 j5 5 − j5

⎡ 0⎤ ⎢ 0⎥ j 4⎤ ⎢ ⎥ ⎡ 0⎤ −2 j 4⎥ ⎢ ⎥ = ⎢ 0⎥ ⎥ ⎢ 0⎥ ⎢ ⎥ 0 ⎦ ⎢ ⎥ ⎣ −10 ⎦ 0 ⎢ 0⎥ ⎣ ⎦

⎡ 0⎤ ⎢ 0⎥ ⎢ ⎥ −2 =⎢ ⎥ ⎢ 0⎥ ⎢ 0⎥ ⎢ 0⎥ ⎣ ⎦ 0 0 5

5 0 0 j5 5 − j5

−5⎤ − j 5⎥ ⎥ 10 + j 5⎦

j 4⎤ j 4⎥ ⎥ 0⎦

6 1⎤ 1⎥ ⎥ 0⎦

9.42 Network Analysis and Synthesis Hence, the KCL equation in matrix form is given by j4 5⎤ ⎡ I l1 ⎤ ⎡10 ⎤ ⎡ 0 ⎤ ⎡10 ⎤ ⎡7 j 4 ⎢ j 4 2 + j1 j 5⎥ ⎢ I l2 ⎥ = ⎢ 0 ⎥ − ⎢ 0 ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 5 j 5 10 + j 5⎦ ⎣ I l3 ⎦ ⎣ 0 ⎦ ⎣ −10 ⎦ ⎣10 ⎦ ⎣ −5

Example 9.27 For the network shown in Fig. 9.70, write down the tieset matrix and obtain the network equilibrium equation in matrix form using KVL. j5.66 Ω

j5 Ω

50 ∠0° V

j10 Ω

3Ω

+ −

−4Ω −j

I1

5Ω I2

Fig. 9.70 Solution The branch currents are so chosen that they assume the direction out of the dotted terminals. Because of this choice of current direction, the mutual inductance is positive. The oriented graph and its selected tree are shown in Fig. 9.71. 1

1

(1)

(2)

(3)

2

(1)

(2)

(3)

Links: {1, 3} Tieset 1: {1, 2} Tieset 3: {3, 2}

1 ⎡1 B=⎢ ⎣0

2

3

1 0⎤ 1 1⎥⎦

2

Fig. 9.71 The KVL equation in matrix form is given by

Here,

B Zb BT I l = BV Vs − B Zb I s I s = 0, B Zb BT I l = BV Vs 0 j 5 66 ⎤ ⎡ j5 ⎡ 1 0⎤ ⎡50 ∠0°⎤ Zb = ⎢ 0 3 − j4 0 ⎥ ; BT ⎢ 1 1⎥ ; Vs = ⎢ 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 5 j10 ⎦ 1⎦ ⎣ j 5 66 ⎣0 ⎣ 0 ⎦ 0 j 5 66 ⎤ ⎡ j5 3 − j4 j 5.66 ⎤ ⎡ 1 1 0⎤ ⎢ ⎡ j5 B Zb = ⎢ 0 3 − j 4 0 ⎥=⎢ ⎥ 0 − 1 1 j 5 66 3 + j 4 5 + j10 ⎥⎦ ⎢ ⎥ ⎣ ⎦ j 5 666 0 5 + j10 ⎦ ⎣ ⎣ ⎡ 1 0⎤ 3 j4 j 5.66 ⎤ ⎢ −3 + j 9.66 ⎤ ⎡ 3 + j1 1 −1⎥ = ⎢ ⎥ 3 j 4 5 + j10 ⎦ ⎢ −3 + j 9 66 8 + j 6 ⎥⎦ ⎥ 1⎦ ⎣ ⎣0 ⎡50 ∠0°⎤ ⎡ 1 1 0⎤ ⎢ ⎡50 ∠0°⎤ BV Vs = ⎢ 0 ⎥=⎢ ⎥ 0 − 1 1 ⎢ ⎥ ⎣ ⎦ ⎣ 0 ⎥⎦ ⎣ 0 ⎦

⎡ j5 B Z b BT = ⎢ ⎣ j 5 66

9.12 Network Equilibrium Equation 9.43

Hence, the KVL equation in matrix form is given by −3 j 9.66 ⎤ ⎡ I l1 ⎤ ⎡50 ∠0°⎤ ⎡ 3 j1 = ⎢⎣ −3 3 j 9 66 8 j 6 ⎥⎦ ⎢⎣ I l2 ⎥⎦ ⎢⎣ 0 ⎥⎦

Example 9.28 For the network shown in Fig. 9.72, write down the tieset matrix and obtain the network equilibrium equation in matrix form using KVL. j4 Ω

3Ω

j3 Ω

+ 50 ∠45° V

j5 Ω

−

−j8 Ω

I2

I1

Fig. 9.72 Solution The branch currents are so chosen that they assume the direction out of the dotted terminals. Because of this choice of current direction, the mutual inductance is positive. The oriented graph and its selected tree are shown in Fig. 9.73. 1

1

(1)

(2)

(3)

(1)

(2)

(3)

Links: {1, 3} Tieset 1: {1, 2} Tieset 3: {3, 2}

1 ⎡1 B=⎢ ⎣0

2 3 1 0⎤ 1 1⎥⎦

2

2

Fig. 9.73 The KVL equation in matrix form is given by B Zb BT I l = BV Vs − B Zb I s Here,

I s = 0, B Zb BT I l = BV Vs ⎡3 + j 4 Zb = ⎢ j 3 ⎢ ⎣ 0

j3 0 ⎤ ⎡ 1 0⎤ ⎡50 ∠45°⎤ j 5 0 ⎥ ; BT = ⎢ 1 −1⎥ ; Vs = ⎢ 0 ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 0 − j 8⎦ 1⎦ ⎣0 ⎣ 0 ⎦ ⎡3 + j 4 j 3 0 ⎤ 0 ⎤ ⎡ 1 1 0⎤ ⎢ ⎡3 + j 7 j 8 B Zb = ⎢ j3 j5 0 ⎥ = ⎢ ⎥ 0 − 1 1 − j 3 − j 5 − j8⎥⎦ ⎢ ⎥ ⎣ ⎦ 0 − j 8⎦ ⎣ ⎣ 0 ⎡ 1 0⎤ j8 0⎤ ⎢ ⎡3 + j15 1 −1⎥ = ⎢ j 5 − j8⎥⎦ ⎢ ⎥ ⎣ − j8 1⎦ ⎣0 ⎡50 ∠45°⎤ ⎡ 1 1 0⎤ ⎢ ⎡50 ∠45°⎤ BV Vs = ⎢ 0 ⎥=⎢ ⎥ 0 − 1 1 ⎢ ⎥ ⎣ ⎦ ⎣ 0 ⎥⎦ ⎣ 0 ⎦

⎡3 + j 7 B Z b BT = ⎢ ⎣ − j3

j 8⎤ j 3⎥⎦

9.44 Network Analysis and Synthesis Hence, the KVL equation in matrix form is given by, ⎡3 j15 ⎢⎣ − j8

Example 9.29

j8⎤ ⎡ I l1 ⎤ ⎡50 ∠45°⎤ = j 3⎥⎦ ⎢⎣ I l3 ⎥⎦ ⎢⎣ 0 ⎥⎦

For the network shown in Fig. 9.74, obtain branch voltages using KCL equation on

node basis. 8Ω

3Ω

4Ω

4Ω

5Ω 12 V

24 V

Fig. 9.74 Solution The oriented graph is shown in Fig. 9.75. (3)

1

2

(2)

(4)

(1)

(5) 3

Fig. 9.75 The complete incidence matrix for the graph is 1

2

3

4 5

1 ⎡ 1 −1 1 0 Aa = 2 ⎢ 0 0 −1 1 ⎢ 3 ⎣ −1 1 0 −1

0⎤ 1⎥ ⎥ 1⎦

Eliminating the last row from the matrix Aa, we get the incidence matrix A. 1

2

3

4 5

1 ⎡ 1 −1 1 0 0 ⎤ A= ⎢ 2 ⎣0 0 −1 1 1⎥⎦ The KCL equation in matrix form is given by AY Yb AT Vn = AI s − AY Yb Vs Is = 0 ,

9.12 Network Equilibrium Equation 9.45

AY Yb AT Vn = AY Yb Vs

Here, ⎡1 ⎢4 0 0 ⎢ ⎢0 1 0 ⎢ 3 ⎢ 1 Yb = ⎢ 0 0 8 ⎢ ⎢0 0 0 ⎢ ⎢ ⎢0 0 0 ⎣

⎤ 0⎥ ⎥ ⎡ 1 0⎤ ⎡ 0⎤ 0 0⎥ ⎥ ⎢ −1 0 ⎥ ⎢ 12⎥ ⎥ T ⎢ ⎥ ⎢ ⎥ − ⎥ ; Vs = ⎢ 0 ⎥ 0 0⎥ ; A = ⎢ ⎥ 1⎥ ⎢ 0 ⎢ 24 ⎥ 1 ⎢⎣ 0 ⎢⎣ 0 ⎥⎦ 1⎥⎦ 0⎥ ⎥ 4 1⎥ 0 ⎥ 5⎦ 0

⎡1 ⎢4 ⎢ ⎢0 ⎢ ⎡ 1 −1 1 0 0 ⎤ ⎢ AY Yb = ⎢ ⎢0 ⎣0 0 −1 1 1⎥⎦ ⎢ ⎢0 ⎢ ⎢ ⎢0 ⎣ 1 1 ⎡1 − ⎢ 3 8 AY Yb AT = ⎢ 4 1 ⎢0 0 − 8 ⎣ 1 1 ⎡1 ⎢4 − 3 8 AY Yb Vs = ⎢ 1 ⎢0 0 − 8 ⎣ Hence, the KCL equation in matrix form is given by ⎡ 17 ⎢ 24 ⎢ 1 ⎢− ⎣ 8

1⎤ − ⎥ ⎡V ⎤ n1 ⎡ −4 ⎤ ⎡ 4 ⎤ 8 = −⎢ ⎥ = ⎢ ⎥ 23 ⎥⎥ ⎢⎣Vn2 ⎥⎦ ⎣ 6 ⎦ ⎣ −6 ⎦ 40 ⎦

Solving this matrix equation, Vn1 = 3 96 V Vn2 = −9 57 V

0

0

0

1 3

0

0

0

1 8

0

0

0

1 4

0

0

0

⎤ 0⎥ ⎥ 0⎥ ⎥ ⎡1 ⎥ ⎢ 0⎥ = ⎢ 4 ⎥ ⎢0 0⎥ ⎣ ⎥ 1⎥ ⎥ 5⎦

⎡ 1 0⎤ ⎤ ⎡ 17 0 ⎥ ⎢ −1 0 ⎥ ⎢ ⎢ ⎥ 1 −1⎥ = ⎢ 24 1 1 ⎥⎥ ⎢ 1 1⎥ ⎢ − ⎢ 0 4 5⎦ ⎢ 0 8 ⎣ 1⎥⎦ ⎣ ⎡0⎤ ⎤⎢ ⎥ 0 0 ⎥ 12 ⎢ ⎥ ⎡ −4 ⎤ 0 = 1 1 ⎥⎥ ⎢ ⎥ ⎢⎣ 6 ⎥⎦ ⎢ 24 ⎥ 4 5⎦ ⎢ 0 ⎥ ⎣ ⎦ 0

1⎤ − ⎥ 8 23 ⎥⎥ 40 ⎦

−

1 3

0

1 8 1 − 8

0 1 4

⎤ 0⎥ 1 ⎥⎥ 5⎦

9.46 Network Analysis and Synthesis Branch voltages are given by, Vb

AT Vn

⎡V1 ⎤ ⎡ 1 0 ⎤ ⎡ 3 96 ⎤ ⎢V2 ⎥ ⎢ −1 0 ⎥ ⎢ −3 96 ⎥ ⎢ ⎥ ⎢ ⎥ ⎡ 3 96 ⎤ ⎢ ⎥ ⎢V3 ⎥ = ⎢ 1 −1⎥ ⎢⎣ −9 57⎥⎦ = ⎢ 13.53⎥ 1⎥ ⎢V4 ⎥ ⎢ 0 ⎢ −9 57⎥ ⎢⎣V5 ⎥⎦ ⎢⎣ 0 ⎢⎣ −9 57⎥⎦ 1⎥⎦

Example 9.30 For the network shown in Fig. 9.76, write down the f-cutset matrix and obtain the network equilibrium equation in matrix form using KCL. 1Ω

1Ω

1Ω 1Ω

10 V

2A

Fig. 9.76 Solution The oriented graph and its selected tree are shown in Fig. 9.77. (3)

1

(1)

(2)

1

2

(3) 2

(1)

(4)

(2)

3

Twigs: {2, 4} f-cutset 2: {2, 1, 3} f-cutset 4: {4, 3}

(4)

1 2

2 ⎡ −1 1 1 0 ⎤ Q= ⎢ 4 ⎣ 0 0 −1 1⎥⎦

3

Fig. 9.77 The KCL equation in matrix form is given by QY Yb QT Vt = Q I s − QY Yb Vs ⎡1 ⎢0 ⎢ ⎢0 ⎢⎣0

0⎤ ⎡0 ⎤ ⎢0 ⎥ 0⎥ Yb ⎥ I s = ⎢ ⎥ ; Vs 0⎥ ⎢0 ⎥ 1 ⎥⎦ ⎢⎣ 2⎥⎦ ⎡1 0 0 − 1 1 1 0 ⎡ ⎤ ⎢0 1 0 QY Yb = ⎢ ⎢ ⎣ 0 0 −1 1⎥⎦ ⎢0 0 1 ⎢⎣0 0 0 0 1 0 0

0 0 1 0

⎡ −1 ⎡ −1 1 1 0 ⎤ ⎢ 1 QY Yb Q = ⎢ ⎢ ⎣ 0 0 −1 1⎥⎦ ⎢ 1 ⎢⎣ 0 T

⎡10 ⎤ ⎡ −1 0 ⎤ ⎢ 0⎥ T ⎢ 1 0⎥ = ⎢ ⎥;Q = ⎢ ⎥ ⎢ 0⎥ ⎢ 1 1⎥ 1⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 0⎤ 0 ⎥ ⎡ −1 1 1 0 ⎤ ⎥= 0 ⎥ ⎢⎣ 0 0 −1 1⎥⎦ 1 ⎥⎦

0⎤ 0⎥ ⎡ 3 ⎥= 1⎥ ⎢⎣ −1 1⎥⎦

1⎤ 2⎥⎦

3 4

9.12 Network Equilibrium Equation 9.47

⎡0 ⎤ ⎡ −1 1 1 0 ⎤ ⎢0 ⎥ ⎡0 ⎤ Q Is = ⎢ ⎢ ⎥= ⎣ 0 0 −1 1⎥⎦ ⎢0 ⎥ ⎢⎣ 2⎥⎦ ⎢⎣ 2⎥⎦ ⎡10 ⎤ − 1 1 1 0 ⎡ ⎤ ⎢ 0 ⎥ ⎡ −10 ⎤ QY Yb Vs = ⎢ ⎢ ⎥= ⎣ 0 0 −1 1⎥⎦ ⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ Hence, the KCL equation is given by 1⎤ ⎡Vt2 ⎤ ⎡0 ⎤ ⎡ −10 ⎤ ⎡10 ⎤ = − = 2⎥⎦ ⎢⎣Vt4 ⎥⎦ ⎢⎣ 2⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 2 ⎥⎦

⎡ 3 ⎢⎣ −1 Solving this matrix equation,

Vt2 = 4.4 V Vt4 = 3 2 V

Example 9.31

Calculate the twig voltages using KCL equation for the network shown in Fig. 9.78. 2Ω

5Ω 5Ω

10 Ω

10 Ω

5Ω

+ 910 V −

Fig. 9.78 Solution The oriented graph and one of the trees are shown in Fig. 9.79. (5)

(3) (1)

(5)

(4) (6)

(3) (2)

(1)

Twigs: {1, 2, 3} f-cutset 1: {1, 4, 5, 6} f-cutset 2: {2, 4, 5} f-cutset 3: {3, 4, 6}

(4) (6))

(2)

Fig. 9.79 The network equilibrium equation on node basis can be written as

Here,

QY Yb QT Vt = QI s − QY Yb Vs I s = 0, QY Yb QT Vt = QY Yb Vs

1 2

3

4

5

6

1 ⎡ 1 0 0 −1 −1 1⎤ Q = 2 ⎢0 1 0 −1 −1 0 ⎥ ⎢ ⎥ 3 ⎣0 0 1 1 0 1⎦

9.48 Network Analysis and Synthesis 0 0 0 0 0⎤ ⎡0 2 ⎡ 1 ⎢ 0 0.2 ⎢ 0 0 0 0 0⎥ ⎢ ⎥ ⎢ 0 0 02 0 0 0⎥ T ⎢ 0 Yb = ⎢ ;Q = 0 0 01 0 0⎥ ⎢ 0 ⎢ −1 ⎢ 0 ⎢ −1 0 0 0 0.5 0⎥ ⎢ 0 ⎥ ⎢ 1 0 0 0 0 0 1 ⎣ ⎦ ⎣ 0 0 0 ⎡0 2 ⎢ 0 0.2 0 0 1 −1 1⎤ ⎢ ⎡1 0 0 0 0 0 2 0 QY Yb = ⎢0 1 0 1 −1 0 ⎥ ⎢ 0 0 01 ⎢ ⎥⎢ 0 ⎣0 0 1 1 0 1⎦ ⎢ 0 0 0 0 ⎢ 0 0 0 0 ⎣

0 0⎤ ⎡9110 ⎤ ⎢ 0⎥ 1 0⎥ ⎥ ⎢ ⎥ 0 1⎥ 0⎥ ; Vs = ⎢ −1 −1⎥ ⎢ 0⎥ ⎢ 0⎥ −1 0 ⎥ ⎢ 0⎥ 0 1⎥⎦ ⎣ ⎦ 0 0⎤ 0 0⎥ 0 0 −0.1 −0.5 0.1⎤ ⎥ ⎡0.2 0 0⎥ ⎢ = 0 0.2 0 −0.1 −0 5 0⎥ 0 0⎥ ⎢ ⎥ 0 0 0.2 −0.1 0 0.1⎦ 0.5 0⎥ ⎣ 0 0 1⎥⎦

0 0 −0.1 −0.5 ⎡0.2 QY Yb QT = ⎢ 0 0.2 0 −0.1 −0 5 ⎢ 0 0.2 −0.1 0 ⎣ 0

0 0 −0.1 −0.5 ⎡0.2 QY Yb Vs = ⎢ 0 0.2 0 −0.1 −0 5 ⎢ 0 0.2 −0.1 0 ⎣ 0

⎡ 1 0 0⎤ ⎢ 0 1 0⎥ 0.1⎤ ⎢ ⎥ ⎡ 0.9 0.6 0 2⎤ 0 0 1⎥ ⎢ 0⎥ ⎢ = 0.6 0.8 0 1⎥ ⎥ ⎢ −11 1 1⎥ ⎢ ⎥ 0.1⎦ ⎢ 0.2 0.1 0 3⎦ −11 1 0 ⎥ ⎣ ⎢ 1 0 1⎥⎦ ⎣ 1 ⎤ ⎡910 ⎢ 0⎥ 0.1⎤ ⎢ ⎥ ⎡182⎤ 0⎥ ⎢ ⎥ 0⎥ ⎢ = 0 0⎥ ⎢ ⎥ ⎥ 0.1⎦ ⎢⎢ 0 0⎥ ⎣ ⎦ ⎢ 0⎥ ⎣ ⎦

Hence, KCL equation can be written as, ⎡ 0.9 0.6 0 2⎤ ⎡ vt1 ⎤ ⎡ −182⎤ ⎢0.6 0.8 0 1⎥ ⎢ vt ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎣ 0.2 0.1 0 3⎦ ⎣ vt3 ⎦ ⎣ 0 ⎦ Solving this matrix equation, vt1 = −460 V vt2 = 320 V vt3 = 200 V For the network shown in Fig. 9.80, obtain equilibrium equation on node basis.

Ω

5

Fig. 9.80

5

Ω

10 A

10

Ω

5

Ω

Example 9.32

9.12 Network Equilibrium Equation 9.49

Solution The oriented graph and its selected tree are shown in Fig. 9.81. (2)

(2)

2

1

1

(3)

(1)

(3)

(1)

(4) 3

2 (4)

Twigs: {1, 3} f-cutset 1: {1, 2} f-cutset 3: {3, 2, 4}

1

2

3 4

1 ⎡1 −1 0 0 ⎤ Q= ⎢ 3 ⎣0 −1 1 1 ⎥⎦

3

Fig. 9.81 The KCL equation in matrix form is given by Here,

QY Yb QT Vt = Q I s − QY Yb Vs Vs = 0, QY Yb QT Vt = Q I s Y

⎡5 ⎢0 ⎢ ⎢0 ⎢⎣0

0 5 0 0

⎡1 −1 QY Yb = ⎢ ⎣0 −1

⎡ 5 −5 QY Yb QT = ⎢ ⎣0 −5

⎡1 −1 Q Is = ⎢ ⎣0 −1

0 0⎤ ⎡ 1 0⎤ ⎡ −10 ⎤ ⎥ ⎢ ⎥ ⎢ 0⎥ 0 0 −1 −1 T ⎥ Q =⎢ ⎥ ; Is = ⎢ ⎥ 5 0⎥ 1⎥ ⎢ 0 ⎢ 0⎥ 0 10 ⎥⎦ 1⎥⎦ ⎢⎣ 0 ⎢⎣ 0 ⎥⎦ ⎡ 5 0 0 0⎤ 0 0⎤ ⎢0 5 0 0 ⎥ ⎡ 5 5 0 0 ⎤ ⎢ ⎥= 1 1 ⎥⎦ ⎢0 0 5 0 ⎥ ⎢⎣0 5 5 10 ⎥⎦ 1 ⎥⎦ ⎢⎣0 0 0 10 ⎡ 1 0⎤ 0 0 ⎤ ⎢ −1 −1⎥ ⎡10 5⎤ ⎢ ⎥= 5 10 ⎥⎦ ⎢ 0 1⎥ ⎢⎣ 5 20 ⎥⎦ 1⎥⎦ ⎢⎣ 0 ⎡ −10 ⎤ 0 0 ⎤ ⎢ 0 ⎥ ⎡ −10 ⎤ ⎢ ⎥= 1 1 ⎥⎦ ⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦

Hence, KCL equation will be written as ⎡10 5⎤ ⎡ vt1 ⎤ ⎡ −10 ⎤ ⎢⎣ 5 20 ⎥⎦ ⎢ vt ⎥ = ⎢⎣ 0 ⎥⎦ ⎣ 3⎦ Solving this matrix equation, 8 vt1 = − V 7 2 vt3 = V 7

Example 9.33 For the network shown in Fig. 9.82, write down the f-cutset matrix and obtain the network equilibrium equation in matrix form using KCL and calculate v.

9.50 Network Analysis and Synthesis 2v 2Ω 2Ω 2V

+ v −

+ −

2Ω

2Ω

Fig. 9.82 Solution The oriented graph and its selected tree are shown in Fig. 9.83 Since voltage v is to be determined, Branch 2 is chosen as twig, 1

(4)

(2)

2

1

(2)

(3)

(1)

(4)

(3)

(1) 3

2 Twigs: {2, 4} f-cutset 2: {2, 1, 3} f-cutset 3: {4, 3}

1

2

3 4

⎡ −1 1 1 0 ⎤ Q=⎢ ⎣ 0 0 −1 1⎥⎦

3

Fig. 9.83 The KCL equation in matrix form is given by QY Yb QT Vt = Q I s − QY Yb Vs 0 0 0⎤ ⎡0 5 ⎡ −1 0 ⎤ ⎡ 0⎤ ⎡ 2⎤ ⎢ 0 0.5 ⎢ 0⎥ ⎢0 ⎥ 0 0⎥ T ⎢ 1 0⎥ Y ⎢ ⎥ Q =⎢ ⎥ ; Is = ⎢ ⎥ ; Vs = ⎢ ⎥ 0 05 0⎥ ⎢ 0 ⎢ 1 −1⎥ ⎢ 0⎥ ⎢0 ⎥ 0 0 0.5⎥⎦ 1⎥⎦ ⎢⎣ 0 ⎢⎣ 0 ⎢⎣ −2v ⎥⎦ ⎢⎣0 ⎥⎦ 0 0 0⎤ ⎡0 5 ⎢ 0 0 ⎥ ⎡ −0.5 0.5 0.5 0⎤ ⎡ −1 1 1 0 ⎤ 0 0.5 QY Yb = ⎢ ⎢ ⎥= ⎥ 0 05 0 ⎥ ⎢⎣ 0 0 −0.5 0.5⎥⎦ ⎣ 0 0 −1 1⎦ ⎢ 0 0 0 0 5⎥⎦ ⎢⎣ 0 ⎡ −1 0 ⎤ 0 ⎤ ⎢ 1 0 ⎥ ⎡ 1 5 −0 5⎤ ⎡ −0.5 0.5 0 5 T QY Yb Q = ⎢ ⎢ ⎥= 0 −0.5 0.5⎥⎦ ⎢ 1 −1⎥ ⎢⎣ −0 5 1⎥⎦ ⎣ 0 1⎥⎦ ⎢⎣ 0 ⎡ 0⎤ ⎡ −1 1 1 0 ⎤ ⎢ 0 ⎥ ⎡ 0 ⎤ Q Is = ⎢ ⎢ ⎥= ⎣ 0 0 −1 1⎥⎦ ⎢ 0 ⎥ ⎢⎣ −2v ⎥⎦ ⎢⎣ −2v ⎥⎦ ⎡ 2⎤ 0 ⎤ ⎢0 ⎥ ⎡ −1⎤ ⎡ −0.5 0.5 0 5 QY Yb Vs = ⎢ ⎢ ⎥= 0 −0.5 0.5⎥⎦ ⎢0 ⎥ ⎢⎣ 0 ⎥⎦ ⎣ 0 ⎢⎣0 ⎥⎦ ⎡ 1⎤ Q I s − QY Yb Vs = ⎢ ⎣ −2v ⎥⎦

9.12 Network Equilibrium Equation 9.51

Hence, the KCL equation can be written as ⎡ 1.5 −0.5⎤ ⎡ vt2 ⎤ ⎡ 1⎤ = ⎢⎣ −0 5 1⎥⎦ ⎢⎣ vt4 ⎥⎦ ⎢⎣ −2v ⎥⎦ From Fig. 9.82, vt2 v Solving this matrix equation, vt2 = 0 44 V vt4 = 0 66 V vt2 = 0 44 V

v

Example 9.34 For the network shown in Fig. 9.84, write down the f-cutset matrix and obtain the network equilibrium equation in matrix form using KCL and calculate v. 4Ω 0.5 A

2Ω −

+ v

1V + −

4Ω

2Ω

Fig. 9.84 Solution The voltage and current sources are converted into accompanied sources by source–shifting method as shown in Fig. 9.85. 4Ω 2Ω 1V

+ −

+ v − + 1V −

0.5 A 2Ω

4Ω

0.5 A

Fig. 9.85 The oriented graph and its selected tree are shown in Fig. 9.86. 2

(1) (2)

2 Twigs: {1, 2} f-cutset 1: {1, 4} f-cutset 2: {2, 3}

1

1 (4)

(1)

(3)

(2)

3

(4) (3)

3

Fig. 9.86

1

2

3

4

⎡ 1 0 0 −1⎤ Q=⎢ ⎣0 1 −1 0 ⎥⎦

9.52 Network Analysis and Synthesis The KCL equation in the matrix form is given by QY Yb QT Vt = Q I s − QY Yb Vs 0 0⎤ ⎡0 25 0 ⎡ 1 ⎢ 0 0.5 0 0⎥ T ⎢ 0 Yb = ⎢ ⎥;Q ⎢ 0 0 25 0 ⎥ ⎢ 0 ⎢ 0 0 0 0.5⎥⎦ ⎢⎣ 0 ⎢⎣ −1 0 ⎡0 25 0 0.5 0 ⎡ 1 0 0 −1⎤ ⎢ 0 QY Yb = ⎢ ⎢ 0 0 25 ⎣0 1 −1 0 ⎥⎦ ⎢ 0 0 0 ⎢⎣ 0

0⎤ ⎡ 0 ⎤ ⎡1 ⎤ ⎢ 0 ⎥ ⎢1 ⎥ 1⎥ ⎥; I = ⎢ ⎥; V = ⎢ ⎥ −1⎥ s ⎢ 0.5 ⎥ s ⎢0 ⎥ 0 ⎥⎦ ⎢⎣ −0.5⎥⎦ ⎢⎣0 ⎥⎦ 0⎤ 0 ⎥ ⎡0.25 0 0 −0.5⎤ ⎥= 0 ⎥ ⎢⎣ 0 0.5 −0.25 0 ⎥⎦ 0.5⎥⎦

⎡ 1 0⎤ ⎢ 0 0 . 25 0 0 − 0 . 5 1⎥ ⎡0 75 0 ⎤ ⎡ ⎤ QY Yb QT = ⎢ ⎢ ⎥=⎢ ⎥ 0 0 . 5 − 0 . 25 0 0 − 1 0 0 . 75⎥⎦ ⎣ ⎦⎢ ⎥ ⎣ − 1 0 ⎢⎣ ⎥⎦ ⎡ 0 ⎤ ⎡1 0 0 −1⎤ ⎢ 0 ⎥ ⎡ 0 5⎤ Q Is = ⎢ ⎢ ⎥= ⎣0 1 −1 0 ⎥⎦ ⎢ 0 5 ⎥ ⎢⎣ −0 5⎥⎦ ⎢⎣ −0 5⎥⎦ ⎡1 ⎤ 0 −0.5⎤ ⎢1 ⎥ ⎡0 25⎤ ⎡0.25 0 Q Yb Vs = ⎢ ⎢ ⎥= 0.5 −0.25 0 ⎥⎦ ⎢0 ⎥ ⎢⎣ 0 5 ⎥⎦ ⎣ 0 ⎢⎣0 ⎥⎦ ⎡0 25⎤ Q I s − Q Yb Vs = ⎢ ⎣ −1 ⎥⎦ Hence, the KCL equation can be written as 0 ⎤ ⎡ vt1 ⎤ ⎡0 25⎤ ⎡0 75 = ⎢⎣ 0 0.75⎥⎦ ⎢⎣ vt2 ⎥⎦ ⎢⎣ −1 ⎥⎦ Solving this matrix equation, vt1 = 0 33 V vt2 = −1 33 V From Fig. 9.85, v 1 + vt2 = −0.33 V

9.13

DUALITY

Two networks are said to be the dual of each other when the mesh equations of one network are the same as the node equations of the other. Kirchhoff ’s voltage law and current law are same, word for word, with voltage substituted for current, independent loop for independent node pair, etc. Similarly, two graphs are said to be dual of each other if the incidence matrix of any one of them is equal to the circuit matrix of the other. Only planar networks have duals.

9.13 Duality 9.53

Table 9.1 Conversion for dual electrical circuits Loop basis

Node basis

Current

Voltage

Resistance

Conductance

Inductance

Capacitance

Branch current

Branch voltage

Mesh

Node

Short circuit

Open circuit

Parallel path

Series path

The following steps are involved in constructing the dual of a network: 1. Place a node inside each mesh of the given network. These internal nodes correspond to the independent nodes in the dual network. 2. Place a node outside the given network. The external node corresponds to the datum node in the dual network. 3. Connect all internal nodes in the adjacent mesh by dashed lines crossing the common branches. Elements which are the duals of the common branches will form the branches connecting the corresponding independent node in the dual network. 4. Connect all internal nodes to the external node by dashed lines corresponding to all external branches. Duals of these external branches will form the branches connecting independent nodes and the datum node. 5. A clockwise current in a mesh corresponds to a positive polarity (with respect to the datum node) at the dual independent node. 6. A voltage rise in the direction of a clockwise mesh current corresponds to a current flowing towards the dual independent node.

Example 9.35

Draw the dual of the network shown in Fig. 9.87. C

R1

V

+ −

L

R2

Fig. 9.87 Solution The following steps are involved in constructing the dual of the network as shown in Fig. 9.88. 1. Place a node inside each mesh. 2. Place a node outside the mesh which will correspond to the datum node. 3. Connect two internal nodes through a dashed line. The element which is dual of the common branch (here capacitance) will form the branch connecting the corresponding independent node in the dual network. 4. Connect all internal nodes to the external node by dashed lines crossing all the branches. The dual of these branches will form the branches connecting the independent node and datum.

9.54 Network Analysis and Synthesis R1

V

+ −

C

L

R2

1

2

0

Fig. 9.88 The dual network is shown in Fig. 9.89. C

1

2

G1

I

G2

L

O

Fig. 9.89

Example 9.36

Draw the dual of the network of Fig. 9.90. C2

C4

R1

I

L3

R5

Fig. 9.90 Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in Fig. 9.91. C2

R1

I 1

C4

L3 2

3

0

Fig. 9.91

R5

9.13 Duality 9.55

The dual network is shown in Fig. 9.92. G1

1

V

C3

2

+ −

3

L2

L4

G5

O

Fig. 9.92

Example 9.37

Draw the dual of the network shown in Fig. 9.93. V +−

R1

R2

R3

R4

C

R5

L

Fig. 9.93 Solution For drawing the dual network, proceed in the same way as in Example 9.35, as shown in Fig. 9.94. V +−

R1

R2

1

R3

R4 3

2

C L

R5

0

Fig. 9.94 The dual network is shown in Fig. 9.95. G4

G3

1

G1

G2

L

2

3

C

I O

Fig. 9.95

G5

9.56 Network Analysis and Synthesis

Example 9.38

Draw the dual of the network shown in Fig. 9.96.

L2

V

C

+ −

R1

L1

R3

R2

I

Fig. 9.96 Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in Fig. 9.97.

C

1

V

L2

2

+ −

R1

L1

R3

R2 3 I

O

Fig. 9.97 The dual network is shown in Fig. 9.98. C1 L 1

I

G1

G2

2

C2

O

Fig. 9.98

G3

3

− +

V

9.13 Duality 9.57

Example 9.39

Draw the dual of the network shown in Fig. 9.99. 8F

3Ω

4H

10 sin w t

5Ω

Fig. 9.99 Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in Fig. 9.100. 8F

3Ω

1

2

4H

10 sin w t

5Ω

0

Fig. 9.100 The dual network is shown in Fig. 9.101. 4F 2

Ω

3

10 sin w t

8H

O

Fig. 9.101

Example 9.40

Draw the dual of the network shown in Fig. 9.102. 4F

2H

5V

+ −

1H

1A

2Ω

1F

Fig. 9.102

5

Ω

1

9.58 Network Analysis and Synthesis Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in Fig. 9.103. 4F

2H

5V

1

+ −

2

1H

1A

2Ω

1F 0

Fig. 9.103 The dual network is shown in Fig. 9.104. 1V

5A

2F

2

−+

1H

1F

4H

2

Ω

1

O

Fig. 9.104

Exercises 9.1

For the networks shown in Fig. 9.105–9.108, write the incidence matrix, tieset matrix and f-cutset matrix. (i)

C

R2

R4

(iii) 1Ω 2Ω

L2 1Ω

R1

2Ω

+ 10 V −

R3

L1 V + −

1Ω

Fig. 9.107 Fig. 9.105

(ii)

(iv)

I R1

L C

R1 V

I1

2Ω

20 A

R2

+ V 2 −

+ −

C1 L1 L2 R2

Fig. 9.108 Fig. 9.106

Exercises 9.59

For the graph shown in Fig. 9.109, write the incidence matrix, tieset matrix and f-cutset matrix.

9.2

Draw the dual networks for the circuits shown in Fig. 9.111–9.115. (i)

9.6

(4) 1

3

2 (2)

(1)

(6)

(5)

4 4A

(3)

4Ω

2F

3H

(7)

Fig. 9.111 (ii) 5

R1

R3

Fig. 9.109 +

The incidence matrix is given as follows:

9.3

1 2 −1 −1 0 1 0 0 1 0

Branches → 3 4 5 6 0 0 0 0 1 0 1 0 −1 −1 0 1 0 1 0 0

7 1 0 0 0

v (t )

8 0 0 0 1

Fig. 9.112 (iii) R

Draw oriented graph and write tieset matrix. 9.4 The incidence matrix is given below: 1

2

3

0 0 1 0 −1 −1 −1 1 0 1 0 0 9.5

Branches → 4 5 6 7 1 1 0 0 0 0 0 −1 1

8

9

V

1Ω

L1

L2

R1

R2

R3

C

Fig. 9.114 (v)

1V

2V 3Ω

C

+ −

V

2Ω 2Ω

+ −

(iv)

1 0 1 0 0 0 −1 0 0 −1 0 0 −1 −1 1 1 1 0 0 0

1Ω

L

Fig. 9.113

10

Draw the oriented graph. For the network shown in Fig. 9.110, draw the oriented graph and obtain the tieset matrix. Use this matrix to calculate the current i. i

C

R2

−

1Ω

2H V

3F

+ −

3Ω 3Ω

3Ω

4F

Fig. 9.110 1A

[0.91 A]

Fig. 9.115

9.60 Network Analysis and Synthesis Using the principles of network topology, write the loop/node equation in matrix form for the network shown in Fig. 9.116.

9.7

R1 R2 Vg

ig

+ −

R4 R3

Fig. 9.116

Objective-Type Questions The number of independent loops for a network with n nodes and b branches is (a) n − 1 (b) b − n (c) b − n + 1 (d) independent of the number of nodes

9.1

A network has 7 nodes and 5 independent loops. The number of branched in the network is (a) 13 (b) 12 (c) 11 (d) 10

9.2

Identify which of the following is NOT a tree of the graph shown in Fig. 9.117.

9.3

9.5

(a) 3 (b) 4 (c) 6 (d) 7 Consider the network graph shown in Fig. 9.119.

Fig. 9.119 Which one of the following is NOT a tree of this graph? (a)

(b)

a 2

1

d

f

e 4

3

c

b

h

(d)

(c)

g 5

Fig. 9.117 (a) begh (c) adfg

9.6

(b) defg (d) aegh

Figure 9.120 below shows a network and its graph is drawn aside. A proper tree chosen for analyzing the network will contain the edges.

The minimum number of equations required to analyze the circuit shown in Fig. 9.118 is

9.4

C

C

a

b

c

+ − R

R d

V

R

C

Fig. 9.118

R

Fig. 9.120 (a) (c)

ab, bc, ad ab, bd, ca?

(b) (d)

ab, bc, ca ac, bd, ad

Answers to Objective-Type Questions 9.61

9.7

9.8

The graph of an electrical network has n nodes and b branches. The number of links with respect to the choice of a tree is given by (a) b − n + 1 (b) b + n (c) n − b + 1 (d) n − 2b − 1.

(2)

2

1

(a) (c) 9.10

5

(5)

Fig. 9.122

8 7

(3)

(1)

In the graph shown in Fig. 9.121, one possible tree is formed by the branches 4, 5, 6, 7. Then one possible fundamental cutset is 6

(4)

4 5

(b) (d)

C 8

Which one of the following is a cutset of the graph shown in Fig. 9.123? (3)

4 (2)

3

(4)

Fig. 9.121 (a) (c) 9.9

1, 2, 3, 8 1, 5, 6, 8

(b) (d)

1, 2, 5, 6 1, 2, 3, 7, 8

(5)

(1)

Which one of the following represents the total number of trees in the graph given in Fig. 9.122?

Fig. 9.123 (a) (c)

1, 2, 3 and 4 1, 4, 5 and 6

Answers to Objective-Type Questions 9.1. (c) 9.6. (d)

9.2. (c) 9.7. (a)

9.3. (c) 9.8. (d)

(6)

9.4. (b) 9.9. (d)

9.5. (b) 9.10. (d)

(b) (d)

2, 3, 4 and 6 1, 2, 4 and 5

10 Transient Analysis

10.1

INTRODUCTION

Whenever a network containing energy storage elements such as inductor or capacitor is switched from one condition to another, either by change in applied source or change in network elements, the response current and voltage change from one state to the other state. The time taken to change from an initial steady state to the final steady state is known as the transient period. This response is known as transient response or transients. The response of the network after it attains a final steady value is independent of time and is called the steady-state response. The complete response of the network is determined with the help of a differential equation.

10.2

INITIAL CONDITIONS

In solving the differential equations in the network, we get some arbitary constant. Initial conditions are used to determine these arbitrary constants. It helps us to know the behaviour of elements at the instant of switching. To differentiate between the time immediately before and immediately after the switching, the signs ‘–’ and ‘+’ are used. The conditions existing just before switching are denoted as i (0–), v (0–), etc. Conditions just after switching are denoted as i (0+), v (0+). Sometimes conditions at t = ∞ are used in the evaluation of arbitrary constants. These are known as final conditions. In solving the problems for initial conditions in the network, we divide the time period in the following ways: 1. Just before switching (from t = –∞ to t = 0–) 2. Just after switching (at t = 0+) 3. After switching (for t > 0) If the network remains in one condition for a long time without any switching action, it is said to be under steady-state condition. 1. Initial Conditions for the Resistor For a resistor, current and voltage are related by v(t) = Ri(t). The current through a resistor will change instantaneously if the voltage changes instantaneously. Similarly, the voltage will change instantaneously if the current changes instantaneously.

10.2 Network Analysis and Synthesis 2. Initial Conditions for the Inductor

For an inductor, current and voltage are related by, v(t ) = L

di dt

Voltage across the inductor is proportional to the rate of change of current. It is impossible to change the current through an inductor by a finite amount in zero time. This requires an infinite voltage across the inductor. An inductor does not allow an abrupt change in the current through it. The current through the inductor is given by, t

i( t ) =

1 v(t )dt + i( ) L ∫0

where i(0) is the initial current through the inductor. If there is no current flowing through the inductor at t = 0–, the inductor will act as an open circuit at t = 0+. If a current of value I0 flows through the inductor at t = 0–, the inductor can be regarded as a current source of I0 ampere at t = 0+. 3. Initial Conditions for the Capacitor

For the capacitor, current and voltage are related by,

i( t ) = C

dv(t ) dt

Current through a capacitor is proportional to the rate of change of voltage. It is impossible to change the voltage across a capacitor by a finite amount in zero time. This requires an infinite current through the capacitor. A capacitor does not allow an abrupt change in voltage across it. The voltage across the capacitor is given by, t

v(t ) =

1 i(t )dt + v( ) C ∫0

where v(0) is the initial voltage across the capacitor. If there is no voltage across the capacitor at t = 0–, the capacitor will act as a short circuit at t = 0+. If the capacitor is charged to a voltage V0 at t = 0–, it can be regarded as a voltage source of V0 volt at t = 0+. These conditions are summarized in Fig. 10.1. Element with initial conditions

Equivalent circuit at t = 0+

R

R

L

OC

C

SC I0

I0

+

V0

−

+−

Fig. 10.1

Initial conditions

V0

10.2 Initial Conditions 10.3

Similarly, we can draw the chart for final conditions as shown in Fig. 10.2 Equivalent circuit at t = ∞

Element with initial conditions R

R

L

SC

C OC SC I0

+

V0

V0

−

+−

I0 OC

Fig. 10.2 Final conditions 4. Procedure for Evaluating Initial Conditions (a) Draw the equivalent network at t = 0–. Before switching action takes place, i.e., for t = –∞ to t = 0–, the network is under steady-state conditions. Hence, find the current flowing through the inductors iL (0–) and voltage across the capacitor vC(0–). (b) Draw the equivalent network at t = 0+, i.e., immediately after switching. Replace all the inductors with open circuits or with current sources iL(0+) and replace all capacitors by short circuits or voltage sources vC (0+). Resistors are kept as it is in the network. (c) Initial voltages or currents in the network are determined from the equivalent network at t = 0+. di + dv + d 2 i + d 2 v + ( )), (0 ), 2 ( )), 2 (0 ) are determined by writing integrodt dt dt ddt differential equations for the network for t > 0, i.e., after the switching action by making use of initial condition.

(d) Initial conditions, i.e.,

Example 10.1

In the given network of Fig. 10.3, the switch is closed at t = 0. With zero current in di d 2i the inductor, find i, and 2 at t = 0+. dt dt 10 Ω 1H

100 V i(t)

Fig. 10.3 Solution At t = 0–, no current flows through the inductor. i(

)

0

10.4 Network Analysis and Synthesis 10 Ω

At t = 0 + , the network is shown in Fig. 10.4. At t = 0 + , the inductor acts as an open circuit. 100 V +

i(

)

i( 0+)

0

For t > 0, the network is shown in Fig. 10.5. Writing the KVL equation for t > 0, 100 − 10i 10 − 1

di dt

At t = 0 + ,

+

10 Ω

0

…(i)

) = 100 − 10i(

1H

100 V i(t)

di = 100 − 10i dt di ( dt

Fig. 10.4

…(ii) Fig. 10.5 +

) = 100 − 10( ) = 100 A / s

Differentiating Eq. (ii), d 2i dt 2

d i

At t = 0 + ,

dt

2

2

= −10

(0 + )

10

di dt di + (0 ) = −10(100) = −1000 A / s 2 (0 dt

Example 10.2

In the network of Fig. 10.6, the switch is closed at t = 0. With the capacitor di d 2i uncharged, find value for i, and 2 at t = 0+. dt dt 1000 Ω 1 μF

100 V i(t)

Fig. 10.6 Solution At t = 0−, the capacitor is uncharged. vC (

)

0

i(

)

0

1000 Ω

At t = 0 + , the network is shown in Fig. 10.7. At t = 0+, the capacitor acts as a short circuit. vC (

+

)

0 100 i( + ) = = 0.1 A 1000

v C ( 0+)

100 V i( 0+)

Fig. 10.7

10.2 Initial Conditions 10.5

For t > 0, the network is shown in Fig. 10.8. Writing the KVL equation for t > 0, 100 −1000 1000i −

t

1 1 10

1000 Ω

−6

100 V

∫ i dt = 0

1 μF

…(i) i(t)

0

Differentiating Eq. (i),

Fig. 10.8

di 0 − 1000 − 106 i = 0 dt di 106 =− i dt 1000 + At t = 0 ,

di ( dt

+

)=−

…(ii)

106 i( 1000

+

)=−

106 ( .1) 1000

/s

Differentiating Eq. (ii), d 2i dt 2 At t = 0 + ,

=−

d 2i dt

(0 + ) = − 2

106 di 1000 dt 106 di + 106 (0 ) = − (0 ( 100) = 105 A / s 2 1000 dt 1000

Example 10.3

In the network shown in Fig. 10.9, the switch is closed. Assuming all initial condidi d 2i and 2 at t = 0+. tions as zero, find i, dt dt 10 Ω

1H

10 μF

10 V i(t)

Fig. 10.9 Solution At t = 0–, i(

)

0

vC (

)

0

At t = 0 + , the network is shown in Fig. 10.10. At t = 0+, the inductor acts as an open circuit and the capacitor acts as a short circuit. i( vC (

+ +

)

0

)

0

10 Ω

10 V

i( 0+)

Fig. 10.10

v C ( 0+)

10.6 Network Analysis and Synthesis 10 Ω

For t > 0, the network is shown in Fig. 10.11. Writing the KVL equation for t > 0, t

10 − 10i − 1 At t = 0 + ,

di 1 − i dt = 0 dt 10 × 10 −6 ∫0

10 −10 10i(0 + )

…(i)

1H

10 μF

10 V

i(t)

Fig. 10.11

di + (0 ) − 0 0 dt di + (0 ) 10 A / s dt

Differentiating Eq. (i), 0 10 At t = 0+,

0 10

di d 2 i 1 − − dt dt 2 10 × 10

dt

and

d 2v dt 2

i=0

di + d 2i 1 (0 ) − 2 (0 + ) − i( 0 + ) dt dt 10 5 d 2i

Example 10.4

6

2

(0 + )

0 10 10 = −100 A / s 2

In the network shown in Fig. 10.12, at t = 0, the switch is opened. Calculate v,

dv dt

at t = 0+. v(t)

1H

1A

100 Ω

Fig. 10.12 Solution

At t = 0 , the switch is closed. Hence, no current flows through the inductor. –

iL (

)

0 v( 0+)

At t = 0 + , the network is shown in Fig. 10.13. At t = 0+, the inductor acts as an open circuit. iL ( v(

+ +

)

0

i L ( 0+) 100 Ω

1A

) = 100 × 1 = 100 V Fig. 10.13

10.2 Initial Conditions 10.7

For t > 0, the network is shown in Fig. 10.14. Writing the KCL equation for t > 0,

v(t)

t

v 1 + v dt = 1 100 1 ∫0

1H

1A

…(i)

100 Ω

Differentiating Eq. (i), 1 dv +v =0 100 dt dv + ( ) = −100 v ( dt

At t = 0+,

Fig. 10.14

…(ii) +

) = −100 × 100 = −10000 V / s

Differentiating Eq. (ii), 1 d 2 v dv + =0 100 dt 2 dt d 2v

At t = 0+,

dt 2

Example 10.5 and

d 2v dt 2

(0 + ) = −100

dv + (0 ) = −100 × ( 10 4 ) = 106 V / s 2 (0 dt

In the given network of Fig. 10.15, the switch is opened at t = 0. Solve for v,

dv dt

at t = 0+. v(t)

10 A

1 kΩ

1 μF

Fig. 10.15 Solution At t = 0–, switch is closed. Hence, the voltage across the capacitor is zero. v(

)

)=0

vC (

v( 0+)

10 A

1 kΩ

V C ( 0+)

+

At t = 0 , the network is shown in Fig. 10.16. At t = 0+, the capacitor acts as a short circuit. v(

+

)

vC (

+

Fig. 10.16

)=0

For t > 0, the network is shown in Fig. 10.17. Writing the KCL equation for t > 0, v dv +10 10 −6 = 10 1000 dt At t = 0+,

v( + ) dv + 10 −66 ( 1000 dt

+

) 10

v(t)

…(i)

10 A

1 kΩ

Fig. 10.17

1 μF

10.8 Network Analysis and Synthesis dv ( dt

+

)=

10 10 −6

= 10 × 106 V / s

Differentiating Eq. (i), 1 dv d 2v + 10 −6 2 = 0 1000 dt dt 1 dv + d 2v (0 ) 10 −6 2 (0 + ) = 0 1000 dt dt

At t = 0+,

d 2v dt 2

Example 10.6 and

d 2v dt 2

(0 + ) = −

1 1000 × 10

6

× 10 × 106 = −10 × 109 V / s 2

For the network shown in Fig. 10.18, the switch is closed at t = 0, determine v,

dv dt

at t = 0+.

2Ω

10 A

1H

0.5 μF v (t)

Fig. 10.18 Solution At t = 0–, no current flows through the inductor and there is no voltage across the capacitor. iL (

)

0

v(

)

0

At t = 0 + , the network is shown in Fig. 10.19. At t = 0+, the inductor acts as an open circuit and the capacitor acts as a shot circuit. iL ( v(

+ +

)

0

)

0

v( 0+)

2Ω

10 A

i L ( 0+)

Fig. 10.19

For t > 0, the network is shown in Fig. 10.20. Writing the KCL equation for t > 0,

10 A

2Ω

1H

t

v 1 dv + ∫ vdt + 0 5 × 10 −6 = 10 2 11 dt At t = 0+,

v(

+

2

)

+ 0 0.

…(i) Fig. 10.20 dv + (0 ) = 10 dt dv ( dt

+

)

20 106 V / s 20

0.5 μF v (t)

10.2 Initial Conditions 10.9

Differentiating Eq. (i), 1 dv d 2v + v + 0 5 × 10 −6 2 = 0 2 dt dt 1 dv + (0 ) 2 dt

At t = 0+,

((0 0 + ) + 0.5 10 −6

d 2v ddt

2

d 2v dt 2

(0 + ) (0 + )

0 20 1012 V /s / s2

Example 10.7

In the network shown in Fig. 10.21, the switch is changed from the position 1 to the di d 2i position 2 at t = 0, steady condition having reached before switching. Find the values of i, and 2 at dt dt t = 0+. 10 Ω

1 2

20 V

1H

20 Ω i(t)

Fig. 10.21

10 Ω

20 V

Solution At t = 0–, the network attains steady-state condition. Hence, the inductor acts as a short circuit. i(

)=

i( 0−)

20 =2A 10

Fig. 10.22 10 Ω

At t = 0 + , the network is shown in Fig. 10.23. At t = 0+, the inductor acts as a current source of 2 A. i(

+

)

20 Ω

2A

For t > 0, the network is shown in Fig. 10.24. Writing the KVL equation for t > 0,

At t = 0 , +

di −20i −10 10i − 1 = 0 dt di −30i(0 + ) − (0 + ) = 0 dt di + (0 ) = −30 × 2 = −60 A / s dt

Differentiating Eq. (i), −30

2A i( 0+)

di d 2 i − =0 dt dt 2

Fig. 10.23 10 Ω

…(i) 20 Ω

1H i(t )

Fig. 10.24

2A

10.10 Network Analysis and Synthesis −30

At t = 0+,

di + d 2i (0 ) − 2 (0 + ) = 0 dt dt d 2i dt

2

(0 + ) = 1800 A / s 2

Example 10.8

In the network shown in Fig.10.25, the switch is changed from the position 1 to the di d 2i and 2 at position 2 at t = 0, steady condition having reached before switching. Find the values of i, dt dt t = 0+. 20 Ω

1 2

30 V

1 μF

10 Ω i(t)

Fig. 10.25 20 Ω

Solution At t = 0–, the network attains steady-state condition. Hence, the capacitor acts as an open circuit.

vC( 0−)

30 V

vC (

)

30 V

i(

)

0

i ( 0−)

Fig. 10.26

At t = 0 + , the network is shown in Fig. 10.27. At t = 0+, the capacitor acts as a voltage source of 30 V. vC ( i(

+ +

)

20 Ω

30 V

)=−

10 Ω i( 0+)

30 = −1 A 30

Fig. 10.27

For t > 0, the network is shown in Fig. 10.28. Writing the KVL equation for t > 0, −10i − 20i −

20 Ω

t

1 1 × 10 −6

∫ i dt − 30 = 0

…(i)

1 μF

10 Ω

0

Differentiating Eq. (i),

i(t)

Fig. 10.28 di −30 − 106 i = 0 dt

At t = 0+,

vC( 0+) 30 V

−30

…(ii)

di + ( 0 ) − 10 6 ( 0 + ) = 0 dt di + 106 ( −1) (0 ) = = 0.33 × 105 A / s dt 30

30 V

10.2 Initial Conditions 10.11

Differentiating Eq. (ii), −30 2

−30

At t = 0+,

d i dt

d 2i dt

2

− 106

di + (0 ) = 0 dt

( 0 + ) − 10 6

2

di =0 dt

d 2i dt

(0 + ) = − 3

106 × 0 33 × 105 = −1.1 × 109 A s 2 30

Example 10.9

In the network shown in Fig. 10.29, the switch is changed from the position 1 to the d 2i di position 2 at t = 0, steady condition having reached before switching. Find the values of i, and 2 at dt dt t = 0 +. 20 Ω

1 2

40 V

1 μF

1H i(t)

Fig. 10.29 20 Ω

Solution At t = 0–, the network attains steady state. Hence, the capacitor acts as an open circuit. vC (

)

40 V

i(

)

0

40 V i( 0−)

Fig. 10.30

At t = 0 + , the network is shown in Fig. 10.31. At t = 0+, the capacitor acts as a voltage source of 40 V and the inductor acts as an open circuit. vC ( i(

+ +

)

40 V

)

0

20 Ω v C ( 0+) 40 V i( 0+)

Fig. 10.31

For t > 0, the network is shown in Fig. 10.32. Writing the KVL equation fo t > 0,

20 Ω 1 μF

t

di 1 −1 − 20i − ∫ i dt − 40 = 0 dt 1 × 10 −6 0 At t = 0+,

−

di ( dt

+

) − 20i((

+

) − 0 − 40 = 0 di ( dt

v C ( 0−)

+

) = − 40 A / s

…(i)

1H i(t)

Fig. 10.32

40 V

10.12 Network Analysis and Synthesis Differentiating Eq. (i), − 2

−

At t = 0+,

d i dt

2

d 2i dt

2

(0 + ) − 20

− 20

di − 106 i − 0 = 0 dt

di + (0 ) − 106 i(0 + ) = 0 dt d 2i dt

2

(0 + ) = 800 A / s 2

Example 10.10

In the network of Fig. 10.33, the switch is changed from the position ‘a’ to ‘b’ at di d 2i and 2 at t = 0+. t = 0. Solve for i, dt dt 1 kΩ

a b

100 V

0.1 μF

1H i(t)

Fig. 10.33 Solution At t = 0–, the network attains steady condition. Hence, the inductor acts as a short circuit. i( vC (

)=

100 = 0.1 A 1000

)

0

100 V i ( 0−)

Fig. 10.34

At t = 0 + , the network is shown in Fig. 10.35. At t = 0+, the inductor acts as a current source of 0.1 A and the capacitor acts as a short circuit. i( vC (

+ +

)

0.1 A

)

0

1 kΩ

1 kΩ v C ( 0+)

Fig. 10.35

For t > 0, the network is shown in Fig. 10.36. Writing the KVL equation for t > 0, −

t

1 0 1 × 10

−6

di

1000i − 1 = 0 ∫ i dt −1000 dt

0.1 A i( 0+)

1 kΩ

…(i)

0.1 μF

1H i(t )

0

Fig. 10.36

0.1 A

10.2 Initial Conditions 10.13

−0 − 1000i(0 + ) −

At t = 0+,

di + (0 ) = 0 dt di + (0 ) = −1000 i (0 + ) = −1000 × 0.1 = −100 A /s / dt

Differentiating Eq. (i), − At t = 0+,

1 10

i − 1000 −7

−107 i(0 + ) − 1000

di d 2 i − =0 dt dt 2

di + d 2i (0 ) − 2 (0 + ) = 0 dt dt d 2i dt 2

(0 + ) = −107 (0.1) − 1000 0 ( 100) = −99 105 A / s 2

Example 10.11

The network of Fig. 10.37 attains steady-state with the switch closed. At t dv switch is opened. Find the voltage across the switch vK and K at t 0 + . dt K

0, the

1H

vK 1Ω

2V 0.5 F

i(t)

Fig. 10.37 Solution: At t = 0 − , the network is shown in Fig. 10.38. At t = 0 − , the network attains steady-state condition. The capacitor acts as an open circuit and the inductor acts as a short circuit. i( vC (

)=

2 =2A 1

)

0

At t = 0 + , the network is shown in Fig. 10.39. At t = 0 + , the capacitor acts as a short circuit and the inductor acts as a current source of 2 A. 2V i( vC ( vK (

+ + +

)

2A

)

0

)

0

vK( 0−) 2V

1Ω

vC( 0−)

i( 0−)

Fig. 10.38

2A

vC( 0+)

1Ω i( 0+)

Fig. 10.39

10.14 Network Analysis and Synthesis 1 i dt C∫ dvK i = dt C vK =

Also,

dvK ( dt

At t = 0 + ,

+

)=

+

i(

C

)

=

2 = 4 A /s 0.5

Example 10.12

In the network shown in Fig. 10.40, assuming all initial conditions as zero, find dii di d 2i d 2 i2 + i1 (0 + ) i (0 + ), 1 (0 + ), 2 (0 + ), 21 (0 + ) and ( ). dt dt d dt dt 2 R2

R1

V

L

C i1(t)

i2(t)

Fig. 10.40 Solution At t = 0–, all initial conditions are zero. vC (

)

0

i1 (

)

0

i2 (

)

0

At t = 0 + , the network is shown in Fig. 10.41. At t = 0+, the inductor acts as an open circuit and the capacitor acts as a short circuit. i1 (0 + ) =

V R1

i2 (0 + )

0

+

vC (0 )

i1( 0+)

Fig. 10.41

t

and

−

i2( 0+)

0

1 (i1 i2 )dt = 0 C ∫0

…(i)

1 di (i2 − i1 )dt − R2 i2 − L 2 = 0 C∫ dt

…(ii)

R1i1 −

vC( 0+)

V

For t > 0, the network is shown in Fig. 10.42. Writing the KVL equations for two meshes for t > 0, V

R2

R1

R1

R2

L

C

V i1(t)

i2(t)

Fig. 10.42

10.2 Initial Conditions 10.15

From Eq. (ii), at t = 0 + , 1 − C

0+

∫ (i2 − i1 )dt − R2i2 (0

+

di2 + (0 ) = 0 dt

)−

0

di2 + (0 ) = 0 dt Differentiating Eq. (i), di1 1 − (i1 i2 ) = 0 dt C di1 + 1 1 (0 ) − i1 (0 + ) + i2 (0 + ) 0 1 dt C C di 1 V R1 1 (0 + ) + =0 dt C R1 di1 + V (0 0 )=− 2 dt R1 C 0 − R1

At t = 0 + ,

0

…(iii)

Differentiating Eq. (iii), − R1 2

At t = 0+,

− R1

d i1 dt

2

(0 + ) −

d 2 i1 dt

2

−

1 dii1 1 di2 + =0 C dt C dt

1 dii1 + 1 di2 + (0 ) + (0 ) = 0 C dt C dt d 2 i1 dt

2

(0 + ) =

V R13C 2

Differentiating Eq. (ii), −

1 di d 2i (i2 − i1 ) − R2 2 − L 22 = 0 C dt dt d 2 i2

At t = 0+,

dt

2

R2 di2 + 1 V ((00 ) − [i2 (0 + ) i1 (0 + )] = [i L dt LC R1 LC

(0 + )

Example 10.13

In the network shown in Fig. 10.43, assuming all initial conditions as zero, find dii di d i d 2 i2 i1 , i2 1 , 2 , 21 and at t 0 + . dt dt dt dt 2 2

R2

C

V

L

R1 i2(t)

i1(t)

Fig. 10.43

10.16 Network Analysis and Synthesis Solution At t = 0 − , all initial conditions are zero. vC (

)

0

i1 (

)

0

i2 (

)

0

At t = 0 + , the network is shown in Fig. 10.44. At t = 0 + , the capacitor acts as a short circuit and the inductor acts as an open circuit. i1 (0 + ) =

V R1

i2 (0 + )

0

+

vC (0 )

vC( 0+)

R2

R1

V i1( 0+)

i2( 0+)

Fig. 10.44

0

For t > 0, the network is shown in Fig. 10.45.

C

R2

Writing the KVL equation for t > 0, V

t

1 V − ∫ i1 dt C0 and

R1 (i1 i2 ) = 0

…(i)

dii2 =0 dt

…(ii)

− R1 (i2 − i1 ) − R2 i2 − L

L

R1 i1(t)

i2(t)

Fig. 10.45

From Eq. (ii),

At t = 0 + ,

dii2 1 = [ R1i1 − ( R1 R2 )i2 ] dt L 1⎡ V R2 )i2 (0 + )] = ⎢ R1 − (R ( R1 L ⎣ R1

dii2 + 1 (0 ) = [ R1i1 (0 + ) ( R1 dt L

…(iii) ⎤ V R2 )0 ⎥ = ⎦ L

Differentiating Eq. (i), 0

At t = 0 + ,

i1 C

1

di1 dt

1

di2 0 dt di1 di2 i = − 1 dt dt R1C

...(iv)

di2 + i1 (0 + ) V V (0 ) − (0 = − 2 dt R1C L R1 C

dii1 (0 ) dt

Differentiating Eq. (iii), d 2 i2 dt At t = 0 + ,

2

d i2 dt 2

2

=

1 ⎡ dii1 R1 − ( R1 L ⎢⎣ dt

R ⎞ ⎛ 1 (0 + ) = −V ⎜ + 2 ⎝ R1 LC L2 ⎟⎠

R2 )

dii2 ⎤ dt ⎥⎦

10.2 Initial Conditions 10.17

Differentiating Eq. (iv), d 2 i1 dt 2

d i1

At t = 0 + ,

dt

2

(0 + )

2

d i2 dt

2

((0 0+ ) −

2

=

d 2 i2 dt

−

2

1 dii1 R1C dt

1 dii1 + V VR R 1 ⎛V V ⎞ V 2V VR R (0 ) = − − 22 − − 2 ⎟= 3 2− − 2 ⎜ R1C dt R1 LC L R1C ⎝ L R1 C ⎠ R1 C R1 LC L2

Example 10.14 In the network shown in Fig. 10.46, a steady state is reached with the switch open. At t = 0, the switch is closed. For the element values given, determine the value of va (0–), vb (0–), va (0+) and vb (0+). 10 Ω 10 Ω

va(t)

20 Ω

vb(t)

5V

2H 10 Ω

Fig. 10.46 Solution At t = 0 − , the network is shown in Fig. 10.47.

10 Ω 10 Ω

At t = 0–, the network attains steady-state condition. Hence, the inductor acts as a short circuit. 5

)=

vb (

)

0

)

20 5× = 3.33 V 30

va (

(

)

=

5 2 = A 7.5 3

iL (

+

) 5 va ( + ) va ( + + 10 10

20 Ω

vb(0−)

5V iL(0−)

Fig. 10.47

At t = 0 + , the network is shown in Fig. 10.48. At t = 0+, the inductor acts as a current source 2 of A. 3 2 iL ( + ) = A 3 Writing the KCL equations at t = 0+, va (

va(0−)

+

) vb ( 20

+

)

=0

10 Ω 10 Ω

5V

va(0+)

10 Ω

Fig. 10.48

20 Ω

vb(0+) 2 A 3

10.18 Network Analysis and Synthesis vb (

and

+

) va ( 20

+

)

+

vb (

+

) 5 2 + =0 10 3

Solving these two equations, va ( vb (

+ +

) 1.9 V )

0.477 V

Example 10.15 In the accompanying Fig. 10.49 is shown a network in which a steady state is reached with switch open. At t = 0, switch is closed. Determine va (0–), va (0+), vb(0–) and vb (0+). 10 Ω 10 Ω

va(t)

20 Ω vb(t)

5V

2F 10 Ω

Fig. 10.49 10 Ω

Solution At t = 0 − , the network is shown in Fig. 10.50. At t = 0–, the network attains steady-state condition. Hence, the capacitor acts as an open circuit.

10 Ω

va(0−)

20 Ω

vb(0−)

5V

va (

)

5V

vb (

)

5V Fig. 10.50

At t = 0 + , the network is shown in Fig. 10.51. At t = 0+, the capacitor acts as a voltage source of 5 V. vb (

+

)

10 Ω 10 Ω

5V

Writing the KCL equation at t = 0+, va (

5V

+

) 5 va ( + ) va ( + ) 5 + + =0 10 10 20 0..

va ( 0 + ) va (

Example 10.16

+

)

.75

va(0+)

10 Ω

20 Ω

vb(0+)

5V

Fig. 10.51

3V

The network shown in Fig. 10.52 has two independent node pairs. If the switch is dv dv opened at t = 0. Find v1 , v2 , 1 and 2 at t = 0+. dt dt

10.2 Initial Conditions 10.19 L

v1(t)

v2(t)

R2

R1

i(t)

C

Fig. 10.52 Solution At t = 0–, no current flows through the inductor and there is no voltage across the capacitor. iL (

)

0

vC (

)

v2 (

)=0

At t = 0 + , the network is shown in Fig. 10.53. At t = 0+, the inductor acts as an open circuit and the capacitor acts as a short circuit. iL ( v1 ( v2 (

+ + +

iL(0+) i( 0)

)

0

)

R1 ii((

)

0

+

v2(0+)

v1(0+)

R2

R1

) Fig. 10.53

For t > 0, the network is shown in Fig. 10.54. Writing the KCL equation at Node 1 for t > 0,

L

v1(t)

v2(t )

t

v1 1 + ( v1 − v2 )dt = i(t ) R1 L ∫0

…(i)

i(t)

R1

R2

C

Differentiating Eq. (i), Fig. 10.54

1 dv1 1 di + ( v1 − v2 ) = R1 dt L dt dv1 + (0 ) dt

At t = 0 + ,

R1

⎡ di + (0 ) (0 dt ⎣d

1 ⎤ R1i(0 + ) ⎥ L ⎦

Writing the KCL equation at Node 2 for t > 0, t

1 ( v2 L ∫0 At t = 0+,

0+

v1 )dt +

v2 dv +C 2 = 0 R2 dt

v2 ( 0 + ) dv + C 2 (0 + ) R2 dt dv2 + (0 ) dt

…(ii)

0 0

In the network shown in Fig. 10.55, the switch is closed at t = 0, with zero capacidv dv d 2 v2 tor voltage and zero inductor current. Solve for v1 , v2 , 1 , 2 and at t = 0+. dt dt dt 2

Example 10.17

10.20 Network Analysis and Synthesis R1 iC (t )

+ v1(t )

iL (t ) L

V

− + v2(t )

C R2

−

Fig. 10.55 Solution At t = 0–, no current flows through the inductor and there is no voltage across the capacitor. vC (

)

0

v1 (

)

0

v2 (

)

0

iL (

)

0

iC (

)

0

At t = 0 + , the network is shown in Fig. 10.56. At t = 0+, the inductor acts as an open circuit and the capacitor acts as a short circuit. vC ( v1 ( v2 ( iL ( iC (

+ + + + +

)

0

)

0

)

0

)

0

R1 iC (0+)

iL (0+)

v1 (0+)

V R2

v2 (0+)

Fig. 10.56

V )= R1 R1

For t > 0, the network is shown in Fig. 10.57.

iC (t )

Writing the KVL equation for t > 0, vC (t ) = v1 (t ) + v2 (t )

…(i)

V

C

L

v1(t )

R

v2(t )

vC (t )

Differentiating Eq. (i), dvC dv1 dv2 = + dt dt dt

iL (t )

…(ii) Fig. 10.57

10.2 Initial Conditions 10.21 t

vC =

Now,

1 iC dt C ∫0

…(iii)

dvC iC = dt C At t = 0+,

dvC ( dt

)=

Also

v1

iC ( + ) V = V /s C R1C L

diiL dt

…(iv)

diiL v1 = dt L At t = 0+,

diiL ( dt

Also,

+

)=

v2

…(v)

v1 ( + ) =0 L …(vi)

R2 iL

dv2 dii = R2 L dt dt At t = 0+,

…(vii)

dv2 + (0 ) dt

R2

diiL + ((00 ) = 0 dt

dvC + (0 ) dt

dv1 dv (0 ) + 2 (0 + ) dt dt

dv1 + V (0 ) = dt R1C

/s

Differentiating Eq. (vii), d 2 v2 dt 2 At t = 0+,

d 2 v2 dt 2

= R2

(0 + )

R2

d 2 iL dt 2 d 2 iL + ((00 ) dt

Differentiating Eq. (v), d 2 iL dt At t = 0+,

d 2 iL dt

2

d 2 v2 dt

2

2

=

1 dv1 L dt

(0 + ) =

1 dv1 + 1 V (0 ) = (0 L dt L R1C

(0 + ) =

R2V V / s2 R1 LC

10.22 Network Analysis and Synthesis

Example 10.18 In the network shown in Fig. 10.58, a steady state is reached with switch open. At t = 0, switch is closed. Find the three loop currents at t = 0+. 2Ω

4Ω

i2(t )

0.5 F

6V 1H

i1(t )

1F

i3(t )

Fig. 10.58 Solution At t = 0 − , the network is shown in Fig. 10.59. At t = 0–, the network attains steady-state condition. Hence, the inductor act as a short circuit and the capacitors act as open circuits. i4 (0 )

i1 (0 − ) =

i2 (0 )

0

i3 (0 )

0

v1 (0 ) v2 (0 ) = 6 ×

2Ω

4Ω

v1(0−)

6V

6 =1A 6

v2(0−)

i1(0−)

(0−)

i3

Fig. 10.59

4 =4V 6

Since the charges on capacitors are equal when connected in series, Q

Q2

C1v1 = C2 v2 v1 (0 − ) −

v2 ( 0 )

and

i2(0−)

=

C2 1 = =2 C1 0 5

v1 (0 − ) =

8 V 3

v2 ( 0 ) =

4 V 3

10.2 Initial Conditions 10.23

At t = 0 + , the network is shown in Fig. 10.60. At t = 0+, the inductor is replaced by a current source of 1 A and the capacitors are replaced by a 8 4 voltage source of V and V respectively. 3 3

2Ω

4Ω 6V

8 V 3 4 v2 ( 0 + ) = V 3

i1(0+)

v1 (0 + ) =

At t = 0

+,

1A

i2(0+)

8 V 3

i3(0+)

4 V 3

Fig. 10.60

8 4 6 2i1 (0 ) − − = 0 3 3 +

i1 (0 + ) 1 A Now,

i1 (0 + ) i3 (0 (0 + ) = 1 i3 (0 + )

0

Writing the KVL equation for Mesh 2, 8 =0 3 8 −4 2 (0 + ) + 4 − = 0 3

−4 [ 2 (0 + ) − 1 (0 + )] −

i2 (0 + ) =

1 A 3

Example 10.19

In the network shown in Fig. 10.61, the switch K is closed at t = 0 connecting a dii di voltage V0 sin ω t to the parallel RL-RC circuit. Find (a) i ((00 + ) andd i2 (0 + ) (b) 1 (0 + ) andd 2 (0 + ). dt dt K i1 (t )

i2 (t )

R

R

C

L

V0 sinwt

Fig. 10.61 Solution At t = 0 − , no current flows in the inductor and there is no voltage across the capacitor. vC (

)

0

i1 (

)

0

i2 (

)

0

10.24 Network Analysis and Synthesis At t = 0 + , the network is shown in Fig. 10.62. At t = 0 + , the inductor acts as an open circuit and the capacitor acts as a short circuit. The voltage source V0 i ω t acts as a short circuit. i1 (0 + )

0

i2 (0 + )

0

vC (0 + )

0

R

Fig. 10.62

and

V0 i

1 i1 dt = 0 C∫ dii R i2 − L 2 = 0 dt

R i1 −

t t

i1 (t )

…(i) …(ii)

ωt − R

i2 (t )

R

R

C

L

V0 sinwt

Differentiating Eq. (i), V0 ω

R

vC (0+ )

For t > 0, the network is shown in Fig. 10.63. Writing the KVL equation for t > 0, V0 i

i2 (0+ )

i1 (0+ )

dii1 i1 − =0 dt C

Fig. 10.63

dii1 V0ω i = cos ω t − 1 …(iii) dt R RC dii1 Vω i (0 + ) V0ω (0 ) = 0 cos ω t − 1 = dt R RC R t = 0+

At t = 0 + , From Eq. (ii),

dii2 V0 R = sin ω t − i2 dt L L dii2 (0 ) dt

At t = 0 + ,

V0 i L

t t = 0+

R i2 (0 + ) = 0 L

Example 10.20 In the network of Fig. 10.64, the switch K is changed from ‘a’ to ‘b’ at t (a steady state having been established at the position a). Find i1 , i2 and i3 at t = 0 + . a

C3

R2

L2

K V

R3

b

i3 C2

L1 R1

i1

Fig. 10.64

C1 i2

0

10.2 Initial Conditions 10.25

Solution At t = 0 − , the network is shown in Fig. 10.65. At t = 0 − , the network attains steadystate condition. Hence, the capacitors act as open circuits and inductors act as short circuits. V i1 (0 )

0

i2 (0 )

0

i3 (0 )

0

vC3(0−)

R3

i1(0−)

i2(0−)

0

vC1 (0 )

0 V

At t = 0 , the network is shown in Fig. 10.66. At t = 0 + , the capacitor C3 acts as a voltage source of V volts and capacitors C1 and C2 act as short circuts. The inductors act as open circuits.

vC3(0+)

R3

R1

i1 (0 )

i2 (0 (0 ) = −

i3 (0 + )

0

0. At t

vC2(0+)

i2(0+)

V R2 + R3

Fig. 10.66

Example 10.21 to t

R1

vC2(0−)

i3(0+) vC1(0+)

(0+)

i1 +

vC1(0−)

R2

+

+

i3(0−)

Fig. 10.65

vC3 (0 ) V vC2 (0 )

R2

In the network of Fig. 10. 67, the switch K1 has been closed for a long time prior

0, the switch K2 is closed. Find v (0 + ) andd iC (0 + ). K1

K2

i1 10 Ω iC C

10 V

i2 10 Ω

vC

Fig. 10.67 Solution At t = 0 − , the network is shown in Fig. 10.68. At t = 0 − , the network attains steady-state condition. Hence, the capacitor acts as an open circuit. i1 (0 )

0

i2 (0 )

0

iC (0 )

0

vC (0 ) 10 V

i1(0−)

10 Ω i2(0−)

iC(0−) 10 V

vC(0−)

Fig. 10.68

10 Ω

10.26 Network Analysis and Synthesis At t = 0 + , the network is shown in Fig. 10.69. At t = 0 + , the capacitor acts as a voltage source of voltage V. vC (

+

i1(0+) 10 Ω iC(0+) 10 V

vC(0+)

) 10 V

i2(0+) 10 Ω

10 V

Writing the KVL equation at t = 0 + , 10 −10 10 i1 (0 + ) 10

0

10 −10 10 i2 (0 + )

0

i1 (0 + )

0

and

+

i2 (0 )

Fig. 10.69

1A

+

iC (0 (0 + ) + i2 (0 + )

i1 (0 )

iC (0 + ) 1 A

Example 10.22

In the network shown in fig. 10.70, a steady state is reached with the switch open. dii di At t = 0, the switch is closed. Determine vC (0 ), i1 (0 + )), i2 (0 (0 + ), 1 (0 + ) andd 2 (0 + ). dt dt

10 Ω i2(t)

i1(t) 20 Ω

20 Ω

1H

1 μF

100 V

Fig. 10.70 Solution At t = 0 − , the network is shown in Fig. 10.71. At t = 0 − , the network is in steady-state. Hence, the inductor acts as a short circuit and the capacitor acts as an open circuit. vC (

) = 100 ×

10 Ω

i1 ( i2 (

)

20 Ω

Fig. 10.71

i1(0+)

0

At t = 0 + , the network is shown in Fig. 10.72. At t = 0 + , the inductor acts as a current source of 3.33 A and the capacitor acts as a voltage source of 66.67 V.

20 Ω

100 V

20 = 66.67 V 20 + 10

66.67 )= = 3.33 A 20

i2 (0−)

i1 (0−)

100 V

i2(0+) 20 Ω 3.33 A

Fig. 10.72

20 Ω 66.67 A

10.3 Resistor-Inductor Circuit 10.27

vC ( i1 (

+ +

)

66.67 V

3.33 A 100 − 66.67 i2 ( + ) = = 1.67 A 20 For t > 0 − , the network is shown in Fig. 10.73. Writing the KVL equations for t > 0, 100 − 20i1 − 1 and

di1 dt

0

)

20 Ω

20 Ω 100 V

1 μF 1H

3.33 A 66.67 V

…(i) Fig. 10.73

100 − 20 i2 −

1

∫ i2 dt − 66.67 = 0

10 −6

…(ii)

dii1 + (0 ) = 100 − 20i1 (0 (0 + ) = 100 − 20 (3.33) dt

At t = 0 + ,

i2(t)

i1(t)

33.3 A/s

Differentiating Eq. (ii), 0 20 At t = 0 + ,

di2 dt

106

20

di2 + (0 ) dt

2

0 106

(0 2 (0

+

)

di2 + 106 (0 ) = − × 1 67 = −83500 A/s 2 dt 20

10.3

RESISTOR-INDUCTOR CIRCUIT

Consider a series RL circuit as shown in Fig. 10.74. The switch is closed at time t = 0. The inductor in the circuit is initially un-energised. Applying KVL to the circuit for t > 0,

V

L i(t)

di Ri L = 0 dt

V

R

Fig. 10.74 Series RL circuit

This is a linear differential equation of first order. It can be solved if the variables can be separated. (

) dt L di L di = dt V Ri

Integrating both the sides, −

L ln (V − Ri ) = t + k R

10.28 Network Analysis and Synthesis where ln denotes that the logarithm is of base e and k is an arbitrary constant. k can be evaluated from the initial condition. In the circuit, the switch is closed at t = 0, i.e., just before closing the switch, the current in the inductor is zero. Since the inductor does not allow sudden change in current, at t = 0+, just after the switch is closed, the current remains zero. Setting i = 0 at t = 0, − − −

L [ R

L ln V = k R

L L ln (V − Ri ) = t − ln V R R −

]= t

−

R

− t V − Ri =e L V

V − Ri = Ve

R − t L

Ri = V − Ve

R − t t R

i=

V V − Lt − e R R

The complete response is composed of two parts, the steady-state V response or forced response or zero state response and transient R

for t > 0 i(t) V R

R

V − Lt e . t R O The natural response is a characteristic of the circuit. Its form may be found by considering the source-free circuit. The forced response Fig. 10.75 Current response of series RL circuit has the characteristics of forcing function, i.e., applied voltage. Thus, when the switch is closed, response reaches the steady-state value after some time interval as shown in Fig. 10.75. Here, the transient period is defined as the time taken for the current to reach its final or steady state value from its initial value. L The term is called time constant and is denoted by T. R L T= R V At one time constant, the current reaches 63.2 per cent of its final value . R response or natural response or zero input response

1

i(T ) =

V V − T t V V −1 V V V − e = − e = − 0.368 = 0.632 R R R R R R R

10.3 Resistor-Inductor Circuit 10.29

Similarly, i( T )

V V e R R

V V V 0.135 = 0.865 R R R

i( T )

V V e R R

i(5T ) =

V V −5 V V V − e = − 0.0067 = 0.993 R R R R R

2

V V V 0.0498 = 0.950 R R R

3

After 5 time constants, the current reaches 99.33 per cent of its final value. The voltage across resistor is vR

R×

Ri ⎛ = V ⎜1 ⎝

R − t⎞ V⎛ ⎜1 − e L ⎟ R⎝ ⎠

R − t⎞ e L

⎟ ⎠

f

v(t)

t>0

V

VL

Similarly, voltage across inductor is vL =

O

R − t⎞ di V d⎛ =L ⎜1 − e L ⎟ dt R dt ⎝ ⎠

= Ve

R − t L

VR

Fig. 10.76

t

Voltage response of series RL circuit

for t > 0

Note: 1. Consider a homogeneous equation, di + Pi = 0 dt

where P is a constant.

The solution of this equation is given by, i (t) = k e−Pt The value of k is obtained by putting t = 0 in the equation for i (t). 2. Consider a non-homogeneous equation, di + Pi = Q dt where P is a constant and Q may be a function of the independent variable t or a constant. The solution of this equation is given by, i(t) = e−Pt ∫Q ePt dt + k e−Pt The value of k is obtained by putting t = 0 in the equation of i(t ).

Example 10.23 In the network of Fig. 10.77, the switch is initially at the position 1. On the steady state having reached, the switch is changed to the position 2. Find current i(t).

10.30 Network Analysis and Synthesis R1

1 2

V

L

R2 i (t)

Fig. 10.77 Solution At t = 0 − , the network is shown in Fig. 10.78. At t = 0−, the network has attained steady-state condition. Hence, the inductor acts as a short circuit.

R1

V

V )= R1

i(

i ( 0−)

Fig. 10.78

Since the inductor does not allow sudden change in current, +

i(

)=

V R1 R1

For t > 0, the network is shown in Fig. 10.79. Writing the KVL equation for t > 0, − R2 i − R1i − L

i(t )

di ( R1 + R2 ) + i=0 dt L Comparing with the differential equation

Fig. 10.79

di + Pi = 0, dt

P=

R1

R2 L

The solution of this differential equation is given by, i(t ) = k e − Pt i( t ) = k e At t = 0, i( ) =

L

R2

di =0 dt

⎛R R ⎞ −⎜ 1 2 ⎟ t ⎝ L ⎠

V R1 V = k e0 = k R1 ⎛ R1 R2 ⎞ ⎟t L ⎠

V −⎜ i( t ) = e ⎝ R1

for t > 0

V R1

10.3 Resistor-Inductor Circuit 10.31

Example 10.24 In the network shown in Fig. 10.80, the switch is closed at t = 0, a steady state having previously been attained. Find the current i (t). R2 R1 V i(t)

L

Fig. 10.80 Solution At t = 0 − , the network is shown in Fig. 10.81. At t = 0−, the network has attained steady-state condition. Hence, the inductor acts as a short circuit. )=

i(

R2 R1 V

V R1

i( 0−)

R2

Since the current through the inductor cannot change instantaneously, i(

+

)=

Fig. 10.81

V R1

R2

For t > 0, the network is shown in Fig. 10.82. Writing the KVL equation for t > 0, V

R1 i L

R1 V

di =0 dt

L

i (t)

di R1 V + i= dt L L

Fig. 10.82

Comparing with the differential equation P=

di + Pi = Q, dt R1 V , Q= L L

The solution of this differential equation is given by, Pt

i( t ) = e =e

−

∫ QQe

R1 t L

Pt

R

R

− 1t V − L1 t L e dt k e ∫L R

=

dt k e − Pt

− 1t V +ke L R1

V R1 + R2

10.32 Network Analysis and Synthesis At t = 0, i( ) =

V R1

R2 V R1

=

R2

V +k R1

k=−

VR R2 R1 ( R1 + R2 ) R

i( t ) =

− 1t V VR R2 − e L R1 R1 ( R1 + R2 )

=

R − 1t ⎞ V ⎛ R2 e L ⎟ ⎜1 − R1 ⎝ R1 R2 ⎠

for t > 0

Example 10.25 t

In the network of Fig. 10.83, a steady state is reached with the switch K open. At 0, the switch K is closed. Find the current i(t ) for t 0. 30 Ω

20 Ω

20 V

K

10 V

1 H 2 i(t)

Fig. 10.83 Solution At t = 0 − , the network is shown in Fig. 10.84. At t = 0 − , the network has attained steady-state condition. Hence, the inductor acts as a short circuit i(

)=

20 + 10 = 0.6 A 30 + 20

30 Ω

20 V i( 0−)

10 V

Since the current through the inductor cannot change instantaneously, i(

+

)

1 di =0 2 dt

di + 40i = 20 dt

Fig. 10.84

0.6 A 20 Ω

For t > 0, the network is shown in Fig. 10.85. Writing the KVL equation for t > 0. 10 − 20i −

20 Ω

1 H 2

10 V i(t)

Fig. 10.85

0.6 A

10.3 Resistor-Inductor Circuit 10.33

Comparing with the differential equation

di + Pi = Q, dt P Q = 20

The solution of this differential equation is given by, − Pt

∫ Q e ddt k e = e −40 t ∫ 0 e 40 t ddt + k e −40 t Pt

i( t ) = e

= At t = 0, i( )

Pt

20 + k e −40 t 40

= 0..5 + k e −40 t

0.6 A

0.6 = 0.5 + k k=01 40 i(( ) = 0.5 + 0.1 e −40t

fo t > 0

Example 10.26 The network of Fig. 10.86 is under steady state with switch at the position 1. At t = 0, switch is moved to position 2. Find i (t). 40 Ω

1 2 50 V

20 mH

10 V i(t )

Fig. 10.86 40 Ω

Solution At t = 0 − , the network is shown in Fig. 10.87. At t = 0−, the network has attained steady-state condition. Hence, the inductor acts as a short circuit.

50 V i( 0−)

50 )= = 1.25 A 40

i(

Fig. 10.87 Since current instantaneously,

through i(

+

the

inductor

cannot

change

) 1.25 A 40 Ω

For t > 0, the network is shown in Fig. 10.88. Writing the KVL equation for t > 0, 10 − 40i − 20 × 10 −3

di =0 dt

di + 2000i = 500 dt

10 V

20 mH i(t)

Fig. 10.88

1.25 A

10.34 Network Analysis and Synthesis Comparing with the differential equation

di + Pi = Q, dt P

Q = 500

The solution of this differential equation is given by, − Pt

∫ Q e ddt k e = e −2000 t ∫ 500 e 2000 t + k e −2000 t

i( t ) = e

=

Pt

Pt

500 + k e −2000 t 2000

= 0 25 + k e −2000 t At t = 0, i( ) 1.25 A 1.25 0.25 + k k =1 2000 t

i(( ) = 0 25 Example 10.27

f

0

In the network of Fig. 10.89, the switch is moved from 1 to 2 at t = 0. Determine i(t). 1 5Ω

2 2Ω

20 V

0.5 H i(t)

40 V

Fig. 10.89 Solution At t = 0 − , the network is shown in Fig. 10.90. At t = 0−, the network has attained steady-state condition. Hence, the inductor acts as a short circuit. i(

5Ω 20 V

20 )= =4A 5

i( 0−)

Fig. 10.90

Since the current through the inductor cannot change instantaneously, i(

+

)

4A

For t > 0, the network is shown in Fig. 10.91. 2Ω

Writing the KVL equation for t > 0, di 40 − 2 0.5 dt di +4 dt

0.5 H

0 80

40 V

i(t)

Fig. 10.91

4A

10.3 Resistor-Inductor Circuit 10.35

Comparing with the differential equation

di + Pi = Q, dt P Q = 80

The solution of this differential equation is given by, − Pt

∫ Q e ddt k e = e −4tt ∫ 80 e 4 t ddt + k e −4 t

i( t ) = e

Pt

Pt

80 + k e −4 t 4 = 20 + k e −4 t =

At t

0, i(0) = 4 A 4 = 20 + k k = −16 4 i(t ) = 20 − 16 e −4t

fo t > 0

Example 10.28 For the network shown in Fig. 10.92, steady state is reached with the switch closed. The switch is opened at t = 0. Obtain expressions for iL (t) and vL (t). 100 Ω

iL(t)

15 V

3000 Ω

90 mH

vL(t)

Fig. 10.92 Solution At t = 0 − , the network is shown in Fig. 10.93. At t = 0−, the network has attained steady-state condition. Hence, the inductor acts as a short circuit. iL (

100 Ω

15 V

15 )= = 0.15 A 100

i L ( 0−)

Since current through the inductor cannot change instantaneously,

Fig. 10.93

iL (0+) = 0.15 A For t > 0, the network is shown in Fig. 10.94. Writing the KVL equation for t > 0, −3000iL − 90 × 10 −3

diiL =0 dt

diiL + 33.33 × 103 iL = 0 dt

3000 Ω

90 mH iL(t)

Fig. 10.94

0.15 A

10.36 Network Analysis and Synthesis di + Pi = 0, dt

Comparing with the differential equation

P = 33.33 × 103 The solution of this differential equation is given by, iL (t ) = k e − Pt iL (t ) = k e −33.33×10 At t = 0, iL ( )

3

t

0.15 A 0 15 = k iL ( ) = 0 15 vL (t ) = L

Also,

33 33×103 t

f

0

diiL dt

= 90 × 10 −3

3 d (0.15 e −33 33×10 t ) dt

= −90 × 10 −3 × 0.15 × 33.33 × 103 × e −33.33×10 = −450 e −33.33×10

Example 10.29

3

t

3

t

for t > 0

In the network of Fig. 10.95, the switch is open for a long time and it closes at

t = 0. Find i (t). 10 Ω

10 Ω

50 V

0.1 H 10 Ω

i(t)

Fig. 10.95

10 Ω

10 Ω

−

Solution At t = 0 , the network is shown in Fig. 10.96. At t = 0−, the network has attained steady-state condition. Hence, the inductor acts as a short circuit. i(

)=

50 V

i( 0−)

50 = 2.5 A 10 + 10

Fig. 10.96

Since current through the inductor cannot change instantaneously, i (0+) = 2.5 A

10 Ω

For t > 0, the network is shown in Fig. 10.97. For t > 0, representing the network to the left of the inductor by Thevenin’s equivalent network, 50 V

10 Veq = 50 × = 25 V 10 + 10 Req = (10 10) + 10 = 15 Ω

10 Ω

10 Ω i(t)

Fig. 10.97

0.1 H

2.5 A

10.3 Resistor-Inductor Circuit 10.37

For t > 0, Thevenin’s equivalent network is shown in Fig. 10.98. Writing the KVL equation for t > 0, 25 − 15i − 0 1

15 Ω

di =0 dt 25 V

di + 150i = 250 dt Comparing with the differential equation

0.1 H i(t)

di + Pi = Q, dt

Fig. 10.98 Q = 250

P The solution of this differential equation is given by, i( t ) = e

Pt

∫Q e

Pt

ddt k e − Pt

= e −150 t ∫ 50 e150 t ddt + k e − Pt =

250 + k e −150 t 150

= 1.667 + k e −150 t At t = 0, i( )

2 .5 A 2.5 = 1.667 + k k = 0.833 150 i(( ) = 1.667 .667 + 0.833 e −150t

Example 10.30

for t > 0

In Fig. 10.99, the switch is closed at t = 0. Find i (t) for t > 0. 2Ω

1Ω

10 A

2Ω i (t)

Fig. 10.99

Solution At t = 0–,

i(

)

0

Since current through inductor cannot change instantaneously, i(

+

)

0

1H

2.5 A

10.38 Network Analysis and Synthesis For t > 0, simplifying the network by source-transformation technique as shown in Fig. 10.100. 2Ω

1Ω

10 A

2Ω

2Ω

1H

0.67 Ω

10 A

i (t)

1H i (t)

(a)

(b)

Writing the KVL equation for t > 0, 6.67 2.67i − 1

2.67 Ω

di dt

0 6.67 V

di + 2.67 dt

1H

6.67

Comparing with the differential equation

i (t) (c)

di + Pi = Q, dt

Fig. 10.100

P = 2.67, Q = 6.67 The solution of this differential equation is given by, − Pt

∫ Q e ddt k e = e −2.67t ∫ 6.67 e 2.67t ddt + k e −2.67t

i( t ) = e

=

Pt

Pt

6 67 + k e −2 67t 2 67

= 2 5 + k e −2 67t At t = 0, i (0) = 0 0

2 .5 + k

k = −2 5 i(( ) = 2.5 − 2.5 e −2 67t 2 67 t

= 2.5(1

Example 10.31

)

f

0

Find the current i (t) for t > 0. 60 Ω

25 A

140 Ω

20 Ω

0.3 H i (t)

Fig. 10.101

10.3 Resistor-Inductor Circuit 10.39

Solution At t = 0–, the inductor acts as a short circuit. Simplifying the network as shown in Fig. 10.102. 60 Ω

20 Ω

140 Ω

25 A

25 A

140 Ω

i (0−)

60 Ω

i (0−)

(a)

(b)

Fig. 10.102 i(

)

25 × 25

140 = 17.5 A 140 + 60

Since current through the inductor cannot change instantaneously, i(

+

) 17.5 A

For t > 0, the network is shown in Fig. 10.103. 60 Ω

20 Ω

140 Ω

25 A

0.3 H

17.5 A

i (t)

Fig. 10.103 Simplifying the network by source transformation as shown in Fig. 10.104,

60 Ω

20 Ω

17.5 A

0.3 H

15 Ω

i (t)

0.3 H i (t)

(a)

(b)

Fig. 10.104 Writing the KVL equation for t > 0, −15i − 0 3

di =0 dt

di + 50i = 0 dt Comparing with the differential equation

di + Pi = 0, dt P = 50

17.5 A

10.40 Network Analysis and Synthesis The solution of this differential equation is given by, i(t ) = k e − Pt = k e −50 t At t = 0, i (0) = 17.5 A k = 17.5 50 t i(t ) = 17.5 e −50

f

t>0

Example 10.32 In the network of Fig. 10.105, the switch is in position ‘a’ for a long time. At t = 0, the switch is moved from a to b. Find v2 (t). Assume that the initial current in the 2 H inductor is zero. 1Ω

a

b +

1V

1 Ω 2

1H

v2(t)

2H −

Fig. 10.105 Solution At t = 0 , the switch is in the position a. The network has attained steady-state condition. Hence, the inductor acts as a short circuit. Current through the 1 H inductor is given by –

1 ) = =1A 1 ) 0

( v2 (

Since current through the inductor cannot change instantaneously, +

i(

+

v2 (

) 1A 1×

)

1 = −0.5 V 2

For t > 0, the network is shown in Fig. 10.106. Writing the KCL equation for t > 0, t

+ 1H

t

1 v 1 v2 dt + 1 + 2 + ∫ v2 d dtt = 0 ∫ 1 10 20 2 Differentiating Eq. (i), v2 + 2

…(i)

1 Ω 2

dv + Pv = 0, dt P=

3 4

v2(t)

2H −

Fig. 10.106

dv2 1 + v2 = 0 dt 2 dv2 3 + v2 = 0 dt 4

Comparing with the differential equation

1A

10.3 Resistor-Inductor Circuit 10.41

The solution of this differential equation is given by, v2 (t ) = K e − Pt = k e

3 − t 4

At t = 0, v2 (0) = –0.5 V −0 5 = k e° k = −0 5 v2 ( ) = −0 5 e

3 − t 4

fo t > 0

Example 10.33 In the network shown in Fig. 10.107, a steady-state condition is achieved with switch open. At t = 0 switch is closed. Find va (t). 10 Ω + 5Ω

3V

va(t) 0.5 H −

Fig. 10.107 Solution At t = 0–, the network has attained steady-state condition. Hence, the inductor acts as a short circuit. iL (

)

0

va (

)

3×

5 =1V 10 + 5

Since current through inductor cannot change instantaneously, iL ( va (

+ +

)

0

) 1V

For t > 0, the network is shown in Fig. 10.108. Writing the KCL equation for t > 0,

10 Ω +

t

1 v v −3 va dt + a + a =0 0 5 ∫0 5 10

3V

5Ω

−

Differentiating Eq. (i), 2

a

0.5 H va(t)

0.2

dva dv 0.1 a 0 dt dt dva 20 + va = 0 dt 3

Fig. 10.108

10.42 Network Analysis and Synthesis Comparing with the differential equation

dv + Pv = 0, dt P=

20 3

The solution of this differential equation is given by, va (t ) = k e − Pt = k e

−

20 t 3

At t = 0, va (0) = 1 V 1= k va ( t ) = e

Example 10.34 is closed at t = 0.

−

20 t 3

fo t > 0

In the network of Fig. 10.109, determine currents i1 (t) and i2 (t) when the switch

10 Ω 5Ω 100 V

i2(t)

5Ω

0.01 H

i1(t)

Fig. 10.109

Solution At t = 0–, At t = 0+,

i1 (0 )

i2 (0 ) = 0

i1 (0 + )

0

i2 (0 + ) =

100 = 6 67 A 15

For t > 0, the network is shown in Fig. 10.110. Writing the KVL equations for t > 0, di1 =0 dt 100 −10 10( 1 + 2 ) − 5 2 0

100 −10 10( 1 + 2 ) − 5 1 0.01 and

10 Ω

…(i) …(ii)

5Ω 100 V

i2(t) i1(t)

0.01 H

From Eq. (ii), i2 =

100 − 10i1 15

Substituting in Eq. (i), dii1 + 833i1 = 3333 dt

Fig. 10.110

5Ω

10.3 Resistor-Inductor Circuit 10.43

Comparing with the differential equation

di + Pi = Q, dt P Q = 3333

The solution of this differential equation is given by, − Pt

d ke ∫ Q e dt = e −833t ∫ 3333 e833t ddt + k e −833t Pt

i1 (t ) = e

=

Pt

3333 + k e −833t 833

= 4 + k e −833t At t = 0, i1 (0) = 0 4+k

0

k = −4 i1 (t ) = 4 4 e −833t 833t = 4(1 − e −833t ) 100 − 10i1 i2 (t ) = 15

=

100 − 10( 4 − 4 15

fo t > 0

−833t

= 4 + 2.67 e −833t

) fo t > 0

Example 10.35 The switch in the network shown in Fig. 10.111 is closed at t = 0. Find v (t) for all t > 0. Assume zero initial current in the inductor. 30 Ω

10 V

+ −

10 Ω i1(t) i2(t)

Fig. 10.111 Solution At t = 0 − ,

i1 (0 )

0

i2 (0 )

0

Since current through the inductor cannot change instantaneously, i2 (0 + )

0

i1 (0 + ) =

10 = 0 25 A 30 + 10

0.2 H

10.44 Network Analysis and Synthesis For t > 0, the network is shown in Fig. 10.112. Writing the KVL equations for t > 0,

and

30 Ω

10 − 30( 1 + 2 ) −10 10 i1 = 0

…(i)

di2 =0 dt

…(ii)

10 − 30( 1 + 2 ) − 0 2

10 V

+ −

10 Ω i1(t)

i2(t)

0.2 H

From Eq. (i), 10 − 30 i2 = 0.25 0.75 i2 …(iii) 40 Substituting Eq. (iii) into Eq. (ii), dii2 + 37.5 i2 2.5 dt di Comparing with the differential equation + Pi = Q, dt P

Fig. 10.112

i1 =

Q = 2.5

The solution of this differential equation is given by, − Pt

d ke ∫ Q e dt 37.5t 5t 37.5t −37.5t = e −37 ∫ 5 e ddt + k e

i2 (t ) = e

= At t

Pt

Pt

25 + k e −37.5t 37.5

= 0.067 + k e −37.5t

0, i2 (0) = 0

0 0.067 + k k = −0.067 i2 ( ) = 0.067 − 0.067 e −37.5t di v2 ( ) = 0 2 2 dt d = 0.2 (0.067 − 0.067 e dt = 0 5e 5 e −37 5t

37 5t

)

fo t > 0

Example 10.36 For the network shown in Fig. 10.113, find the current i(t) when the switch is changed from the position 1 to 2 at t = 0. 40 Ω

60 Ω

1 2

500 V

+ − 10i i(t)

Fig. 10.113

0.4 H

10.3 Resistor-Inductor Circuit 10.45

Solution At t = 0 − , the network is shown in Fig. 10.114. At t = 0 − , the network attains steady-state condition. Hence, the inductor acts as a short circuit.

Since current instantaneously,

i(

)

60 Ω

500 V

500 i( ) = =5A 40 + 60 through the inductor cannot +

40 Ω

i( 0−)

change

Fig. 10.114

5A

60 Ω

For t > 0, the network is shown in Fig. 10.115. Writing the KVL equation for t > 0, 10i 60i − 0 4

di =0 dt

+ −

10i

0.4 H

5A

i (t)

di + 125i 125i = 0 dt di Comparing with the differential equation + Pi = 0, dt P = 125 The solution of this differential equation is given by,

Fig. 10.115

i(t ) = k e − Pt = k e −125t At t

0, i(0) = 5 A

5=k i( t ) = 5

Example 10.37 the switch is opened at t

125t

f

0

For the network shown in Fig. 10.116, find the current in the 20 W resistor when 0. i 30 Ω 50 V

20 Ω

+ − 10i i2(t)

i1(t)

2H

Fig. 10.116 Solution At t = 0 − , the network is shown in Fig. 10.117. At t = 0 − , the network attains steady-state condition. Hence, the inductor acts as a short circuit.

i (0 − ) 30 Ω 50 V

i1(0 − )

+ − − − 10i (0 ) i2(0 )

i(0−) = i2(0−) Fig. 10.117

20 Ω

10.46 Network Analysis and Synthesis Writing the KVL equations at t = 0 − , 50 − 30( 1 10i2

2 ) − 10i2

=0

30(( 2 − 1 ) 200i2 = 0 30

Solving these equations, i1 (0 )

3.33 A

i2 (0 )

2.5 A

Since the current through the inductor cannot change instantaneously, i2 (0 + )

2.5 A

For t > 0, the network is shown in Fig. 10.118.

i(t) = i2(t) 30 Ω

Writing the KVL equation for t > 0, 10i2

30i2 − 20i2 − 2

di2 dt

20 Ω

+ − 10i2 i2(t)

0

di2 + 20i 20i2 = 0 dt

2H

2.5 A

Fig. 10.118

di Comparing with the differential equation + Pi = 0, dt P = 20 The solution of this differential equation is given by, i2 (t ) = k e − Pt = k e −20 t At t

0, i2 (0) = 2.5 A 25=k 20t i2 ( ) = 2 5 e −20

fo t > 0

In the network of Fig. 10.119, an exponential voltage v(t) = 4 e −3t is applied at 0. Find the expression for current i(t). Assume zero current through inductor at t 0.

Example 10.38 t

0.5 Ω

+ 4e − 3t −

0.25 H i(t)

Fig. 10.119 Solution At t = 0 − ,

i(

)

0

Since current through the inductor cannot change instantaneously, i(

+

)

0

10.3 Resistor-Inductor Circuit 10.47

Writing the KVL equation for t > 0, 4

3t

0.5

0.25

di =0 dt

di + 2i 16 e −3t dt Comparing with the differential equation

di + Pi = Q, dt Q = 16 e −3t

P

The solution of this differential equation is given by, − Pt

∫ Q e ddt k e = e −2tt ∫ 6 e 3t e 2t dt k e 2t = 16 e −2t ∫ e t dt k e −2t

i( t ) = e

Pt

= −16 e At t

Pt

3t

ke

2t

0, i(0) = 0 0 16 + k k = 16 i(t ) = −16 e −33 +16 166 e

2t

fo t > 0

Example 10.39 For the network shown in Fig. 10.120, a sinusoidal voltage source v = 150 sin(500t + θ ) volts is applied at a time when θ = 0. Find the expression for the current i(t). 50 Ω

150 sin(500t + q )

0.2 H i(t)

Fig. 10.120 Solution Writing the KVL equation for t > 0, 150 sin(500t θ ) 50

0.2

di =0 dt

di + 250i 250i dt Comparing with the differential equation P

750 sin(500t + θ )

di + Pi = Q, dt Q = 750 sin(500t + )

10.48 Network Analysis and Synthesis The solution of this differential equation is given by, i t =e

e dt

e

Pt

=e

e

=

⎡ e 250 t e −250 t ⎢ 2 ⎣ ( 250) +

=

+

2

Aco

.

and

A si

. =

tan −1

=1 8

12 = 06

.

it = i At t

= .

+ k e −250 t

t+

+ k e −250

.

.6 A 1 342

and

{250 sin(500t +

Let

sin

250 t

ke

°

°) sin

t

t

° +ke

sn

)+ke

250 t

250 t

i 0 =0 −63 43° + k k =1 2 i

=

+

° +1 2

>0

Example 10.40 For the network shown in Fig. 10.121, find the transient current when the switch is moved from the position 1 to 2 at t 0 The network is in steady state with the switch in the position 1. The voltage applied to the network is v = + 30 )V 1 2

200 Ω

150 cos(200t + 30 ) 0.5 H

Fig. 10.121 Solution At t = 0 − , the network is shown in Fig 10.122. At t = 0 − the network attains steady-state condition. V 150 ∠ ° I= = = Z 200 + × 200 × 0 5

200 Ω 150 cos(200t + 30 ) i t)

∠ 43 A Fig. 10.122

0.5

10.4 Resistor–Capacitor Circuit 10.49

The steady-state current passing through the network when the switch is in the position 1 is i

0.67 cos( 200t + 3.43°)

…(i)

For t > 0, the network is shown in Fig. 10.123. Writing the KVL equation for t > 0, −200 i − 0 5

200 Ω

di =0 dt

0.5 H

i(t)

di + 400 i = 0 dt

Fig. 10.123

Comparing with the differential equation

di + Pi = 0, dt P = 400

The solution of this differential equation is given by, p i(t ) = k e − pt = k e −400 t

...(ii)

From Eqs (i) and (ii), 0.67 cos( 200 3.43 3 ) k e −400 t 0.67 c (3.43°) = k k = 6 67

At t = 0,

i(( ) = 0 67

10.4

400 t

f

0

RESISTOR–CAPACITOR CIRCUIT

Consider a series RC circuit as shown in Fig. 10.124. The switch is closed at time t = 0. The capacitor is initially uncharged. Applying KVL to the circuit for t > 0, 1

V

Ri −

R

V

C i(t)

1 i dt = 0 C ∫0

Fig. 10.124

Series RC circuit

Differentiating the above equation, 0

di dt

i C

0

di 1 + i=0 dt RC This is a linear differential equation of first order. The variables may be separated to solve the equation. di dt =− i RC Integrating both the sides, ln i = −

1 t k RC

10.50 Network Analysis and Synthesis The constant k can be evaluated from initial condition. In the circuit shown, the switch is closed at t = 0. Since the capacitor never allows sudden change in voltage, it will act as short circuit at t = 0+. Hence, current in the V circuit at t = 0+ is . R V Setting i = at t = 0, R ln

V =k R

1 V t ln RC R V 1 ln i − l n = − t R RC ln i = −

ln

i 1 =− t RC ⎛V ⎞ ⎜⎝ ⎟⎠ R 1

− t i = e RC V R 1

V − t i = e RC fo t > 0 R When the switch is closed, the response decays with time as shown in Fig. 10.125(a). The term RC is called time constant and is denoted by T. T = RC After 5 time constants, the current drops to 99 per cent of its initial value. The voltage across the resistor is 1

vR

V − t R e RC R

Ri = Ve

−

1 t RC

for t > 0

i(t) V R

t

O

Fig. 10.125(a) Current response of series RC circuit

v(t)

Similarly, the voltage across the capacitor is V t

1 vC = ∫ i dt C0 t

=

1 V C ∫0 R

= −Ve

−

VR

1 − t e RC

1 t RC

VC

+k

O

t

Fig. 10.125(b) Voltage response of series RC circuit

10.4 Resistor–Capacitor Circuit 10.51

At t = 0, vC (0) = 0

k=V

Example 10.41

1 ⎞ ⎛ − t V ⎜1 − e RC ⎟ ⎝ ⎠

vC

Hence,

f

t>0

The switch in the circuit of Fig. 10.126 is moved from the position 1 to 2 at t = 0. Find vC (t). 5 kΩ

1 2 100 V

50 V

+ vC(t)

1 μF

−

Fig. 10.126 Solution At t = 0−, the network is shown in Fig. 10.127. At t = 0−, the network has attained steady-state condition. Hence, the capacitor acts as an open circuit.

5 kΩ

v C ( 0−)

100 V

vC (0 ) = 100 V −

Since the voltage across the capacitor cannot change instantaneously,

Fig. 10.127

vC (0 ) = 100 V +

5 kΩ

For t > 0, the network is shown in Fig. 10.128. Writing the KCL equation for t > 0,

dvC vC + 50 1 × 10 + =0 dt 5000 dvC + 200 vC = 10 4 dt dv Comparing with the differential equation + Pv = Q, dt −6

50 V

−

Fig. 10.128

Q = 10 4

P Solution of this differential equation is given by,

− Pt

∫ Q e ddt k e = e −200 t ∫ 0 4 e 200 t ddt + k e −200 t

vC (t ) = e

=

Pt

+ vC(t)

Pt

10 4 + k e −200 t 200

= −50 + k e −200 t

1 μF

10.52 Network Analysis and Synthesis At t = 0, vC (0) = 100 V

50 50 + k

1

k = 150 vC ( ) = −50 + 150 e −200 t

for t > 0

Example 10.42 In the network shown in Fig. 10.129, the switch closes at t = 0. The capacitor is initially uncharged. Find vC (t) and iC (t). 9 kΩ

10 V

4 kΩ iC(t)

1 kΩ

+ vC(t) −

3 μF

Fig. 10.129 Solution At t = 0−, the capacitor is uncharged. Hence, it acts as a short circuit. vC (

) 0

iC (

) 0

At t = 0+, the network is shown in Fig. 10.130.

iT (0 + ) 9 kΩ

Since voltage across the capacitor cannot change instantaneously, vC ( At t = 0+,

iT ( iC (

+

+

+

10 V

) 0

⎡ 10 )=⎢ ⎣9 k ( ) 1.02 m ×

⎤ 10 = = 1.02 mA ⎥ ) ⎦ 9.8 k

4 kΩ iC (0 + )

v C ( 0+)

1 kΩ

Fig. 10.130

1k = 0.204 mA 1k+4 k

For t > 0, the network is shown in Fig. 10.131.

9 kΩ

4 kΩ

For t > 0, representing the network to the left of the capacitor by Thevenin’s equivalent network, 10 V

1k =1V 9 k +1 k Req = (9 k 1 k) + 4 k = 4.9 kΩ

+ vC(t) −

1 kΩ

3 μF

Veq = 10 ×

Fig. 10.131

For t > 0, Thevenin’s equivalent network is shown in Fig. 10.132. Writing the KCL equation for t > 0, 3 10 −6

dvC v −1 + C =0 dt 4 9 × 103 dvC + 68.02 vC = 68.02 dt

4.9 kΩ

1V

3 μF

Fig. 10.132

vC(t)

10.4 Resistor–Capacitor Circuit 10.53

Comparing with the differential equation

dv + Pv = Q, dt Q = 68.02

P

The solution of this differential equation is given by, vC (t ) = e

Pt

∫Q e

Pt

ddt k e − Pt

= e −68.02t ∫ 68.0 e68.02t ddt + k e −68.02t = 1 + k e −68.02 t At t

0, vC (0) = 0 0 1+ k k = −1 vC (t ) = 1 iC (t ) = C

68 02 t

f

dvC dt

= 3 10 −6 = 3 10

d ( dt

6

e

68.02 02e

= 204.06 × 10 −66 e

Example 10.43

0

68 02 t

)

68.02 t

68.02 t

foor t > 0

For the network shown in Fig. 10.133, the switch is open for a long time and closes

at t = 0. Determine vC (t). 100 Ω

1200 V 300 Ω

+ vC(t) −

50 μF

Fig. 10.133

Solution At t = 0−, the network is shown in Fig. 10.134.

100 Ω

At t = 0–, the network has attained steady-state condition. Hence, the capacitor acts as an open circuit. vC (0−) = 1200 V

v C ( 0−)

1200 V

Since the voltage across the capacitor cannot change instantaneously, vC (0+) = 1200 V

Fig. 10.134

10.54 Network Analysis and Synthesis

For t > 0, the network is shown in Fig. 10.135.

100 Ω

Writing the KCL equation for t > 0, 50 × 10 −6

+

dvC vC vC − 1200 + + =0 dt 300 100 dvC + 266.67 67 vC = 0.24 106 dt

Comparing with the differential equation

300 Ω

1200 V

vC (t)

−

50 μF

Fig. 10.135

dv + Pv = Q, dt Q = 0 24 × 106

P

The solution of this differential equation is given by, d ∫ Q e dt = e −266.67t ∫ 0.

vC (t ) = e

=

Pt

Pt

k e − Pt 06 e 266.67t ddt + k e −266 67t

0 24 × 106 + k e −266.67 t 266.67

= 900 + k e −266.67 t At t

0, vC (0) = 1200 V 1200 = 900 + k k = 300 vC (t ) = 900 + 300 e −266 67 t

Example 10.44

for t > 0

In Fig. 10.136, the switch is closed at t = 0 Find vC (t) for t > 0. 100 Ω

5V

2Ω

1F

+ −

vC(t)

Fig. 10.136

Solution At t = 0−,

vC (0−) = 0 Since the voltage across the capacitor cannot change instantaneously, vC (0+) = 0 Since the resistor of 2 Ω is connected in parallel with the voltage source of 5 V, it becomes redundant. For t > 0, the network is as shown in Fig. 10.137. Writing KCL equation for t > 0, vC dv +1 C = 0 100 dt dvC 100 + vC = 5 dt

100 Ω 1F

5V

Fig. 10.137

vC (t)

10.4 Resistor–Capacitor Circuit 10.55

dvC + 0.01vC = 0.05 dt dv + Pv = Q, dt P Q = 0.05 The solution of this differential equation is given by, Comparing with the differential equation

− Pt

d ke ∫ Q e dt = e −0.01 ∫ 0.05 e 0.01t ddt + k e 0.01t Pt

vC (t ) = e

=

Pt

0 05 + k e −0 01t 0 01

= 5 + k e −0 01t At t

0, vC (0) = 0 0 5+ k k = −5 vC (t ) = 5 5 e −0 01t = 5(1 − e −0 01t )

Example 10.45

for t > 0

In the network shown, the switch is shifted to position b at t = 0. Find v (t) for t > 0. 1 F 4

a b

5V

2Ω

+ v (t) −

2Ω

Fig. 10.138 Solution At t = 0 , the network is shown in Fig. 10.139. At t = 0−, the network has attained steady-state condition. Hence, the capacitor acts as an open circuit.

vC( 0−)

−

5V

2Ω

vC (0−) = 5 V v (0−) = 0

Fig. 10.139

At t = 0+, the network is shown in Fig. 10.140.

5V

At t = 0+, the capacitor acts as a voltage source of 5 V. i( v(

+ +

)=− )

+ v( 0−) −

5 = −1.25 A 4

1.25 × 2

2.5 V

2Ω

2Ω i( 0+)

Fig. 10.140

+ v( 0+) −

10.56 Network Analysis and Synthesis

For t > 0, the network is shown in Fig. 10.141.

1F 4

5V

Writing the KVL equation for t > 0,

t

−2 − 5 −

1 i dt − 2 = 0 1∫ 0 4

…(i) 2Ω

2Ω

i(t )

Differentiating Eq. (i), −4

di − 4i = 0 dt di +i = 0 dt

Fig. 10.141

di + Pi = 0, dt P =1 The solution of this differential equation is given by,

Comparing with the differential equation

i(t ) = k e − Pt = k e − t At t

0, i(0) = −1.25 A k = −1.25 i(t ) = −1.25 e − t f v(t ) = 2i(t ) = −2 5e − t

Example 10.46

t>0

fo t > 0

In the network of Fig. 10.142, the switch is open for a long time and at t = 0, it is

closed. Determine v2 (t). 0.25 Ω + 6V

1 Ω 2

0.3 F

v2(t ) −

Fig. 10.142 Solution At t = 0−, the switch is open. v2 (0−) = 0 Since voltage across capacitor cannot change instantaneously,

0.25 Ω + 6V

0.3 F

v2 (0+) = 0

v2 dv v −6 +0 3 2 + 2 =0 1 dt 0 25 2

v2(t ) −

For t > 0, the network is shown in Fig. 10.143. Writing KCL equation for t > 0,

1 Ω 2

Fig 10.143

10.4 Resistor–Capacitor Circuit 10.57

dv2 + 20 v2 = 80 dt Comparing with the differential equation

dv + Pv = Q, dt Q = 80

P The solution of this differential equation is given by,

− Pt

∫ Q e ddt k e = e −20 t ∫ 80 e 20 t ddt + k e −20 t

v(t ) = e

=

Pt

Pt

80 + k e −20 t 20

v2 t = 4 + k e −20 t At t

0, v2 (0) = 0 0 4+k k = −4 v2 (t ) = 4 4 e −20 t 20 t = 4(1 − e −20t )

for t > 0.

Example 10.47 The switch is moved from the position a to b at t = 0, having been in the position a for a long time before t = 0. The capacitor C2 is uncharged at t = 0. Find i (t) and v2 (t) for t > 0. R

a

R1

b

+ V0

C2

C1 i(t)

v2 (t) −

Fig. 10.144 Solution At t = 0–, the network has attained steady-state condition. Hence, the capacitor C1 acts as an open circuit and it will charge to V0 volt. vC (0−) = V0 1

vC (0−) = 0 2

Since the voltage across the capacitor cannot change instantaneously, vC1 (0 + ) V0 vC2 (0 + ) i( 0 + ) =

0 V0 R1

10.58 Network Analysis and Synthesis For t > 0, the network is shown in Fig. 10.145. Writing the KVL equation for t > 0, t

1 V0 − ∫ i dt d C1 0

R1

+

t

1 R1i − i dt = 0 C2 ∫0

C1

...(i)

C2

V0

i(t )

−

Differentiating Eq. (i), −

i di i − R1 − =0 C1 dt C2

Fig. 10.145

di 1 ⎛ C1 + C2 ⎞ + i=0 dt R1 ⎜⎝ C1 C2 ⎟⎠ Comparing with the differential equation

di + Pi = 0, dt P=

1 ⎛ C1 C2 ⎞ R1 ⎜⎝ C1 C2 ⎟⎠

The solution of this differential equation is given by, i( t ) = k e At t

0, i(0) =

v2(t)

− Pt

= ke

−

1 ⎛ C1 C2 ⎞ t R1 ⎜⎝ C1 C2 ⎟⎠

V0 R1 k=

V0 R1 1 ⎛ C1 C2 ⎞ t C1 C2 ⎠⎟

V − ⎜ i(t ) = 0 e R1 ⎝ R1 1

V − t = 0 e RC R1

where, C =

C1C2 C1 + C2

t

1 v2 ( t ) = i dt C2 ∫0 t

=

t

1 V0 − R1C e dt C2 ∫0 R1

1 ⎞ ⎛ − t V0 R1C = R1C ⎜1 − e ⎟ ⎜⎝ ⎟⎠ R1C2

=

V0C1 C1 + C2

t ⎞ ⎛ − t ⎜1 − e R1C ⎟ ⎜⎝ ⎟⎠

for t > 0

10.4 Resistor–Capacitor Circuit 10.59

Example 10.48 t > 0.

For the network shown in Fig.10.146, the switch is opened at t = 0. Find v (t) for + K

1 F 2

1Ω

10 A 1 Ω 2

v (t) −

Fig. 10.146 Solution At t = 0 − , the network is shown in Fig. 10.147. At t = 0 − , the network attains steady-state condition. Hence, the capacitor acts as an open circuit. vC(0−) = 0

+ 1 Ω 2

10 A

v C (0 − ) v(0 − )

1Ω

−

−

Writing the KCL equation at t = 0 , v(

) 1

Fig. 10.147

) = 10 1 2 3v(( ) = 10

+

v(

v(

)

3.33 V

Since the voltage across the capacitor cannot change instantaneously, vC (

+

)

v( v(

+

) = 3.33 V

For t > 0, the network is shown in Fig. 10.148. Writing the KCL equation for t > 0,

+

1 dv v + = 10 2 dt 1

1Ω

10 A

+ −

1 F 2

v (t) −

dv + 2v dt

20 Fig. 10.148

dv Comparing with the differential equation + Pv = Q, dt P Q = 20 The solution of this differential equation is given by, − Pt

d ke ∫ Q e dt = e −2tt ∫ 0 e 2t dt + k e −2t

v(t ) = e

Pt

Pt

20 + k e −2t 2 = 10 + k e −2t =

10.60 Network Analysis and Synthesis At t

0, v(0) = 3.33 V 3 33 10 + k k = 6 67 v(( ) = 10 + 6 67 e −2t

Example 10.49 opened at t

For the network shown in Fig. 10.149, find the current i(t) when the switch is

0. 10 Ω i(t) + −

100 V

5i 10 Ω 4 μF

Fig. 10.149 10 Ω

Solution At t = 0 − , the network is shown in Fig. 10.150. At t = 0 − , the network attains steady-state condition. Hence, the capacitor acts as open circuit. i( vC (

)=

+ −

100 V

100 =5A 10 + 10

) 100 00 10 0ii((

i( 0−)

) 5i( i(

Fig. 10.150

At t = 0 + , the network is shown in Fig. 10.151. +

25 + 5 (0 (0 ) −10 10 0i ( 0 ) +

i( 0 )

v c ( 0+) = 2 5 V

0

i( 0+) + −

5A

For t > 0, the network is shown in Fig. 10.152. 25 −

1 4 10 −6

10 Ω v C ( 0−)

) = 100 − 10( ) 5( ) = 25 V

−

5i( 0−)

5i( 0+) 10 Ω 25 V

t

∫ i dt + 5

10

0

…(i)

0

Fig. 10.151

Differentiating Eq. (i), 0 0.25 × 106 i + 5

di di 10 dt dt

0

di + 50000 i = 0 dt

+ −

i(t ) 5i 4 μF 25 V

di Comparing with the differential equation + Pi = 0, dt P = 50000

Fig. 10.152

10 Ω

10.4 Resistor–Capacitor Circuit 10.61

The solution of this differential equation is given by, i(t ) = k e − Pt = k e −50000 t At t

0, i(0) = 5 A 5=k i( t ) = 5

Example 10.50 opened at t

50000 t

f

0

For the network shown in Fig. 10.153, find the current i(t) when the switch is

0. 10 Ω i(t) 20 Ω

10 Ω

20i

2 μF

100 V + −

Fig. 10.153 10 Ω

Solution At t = 0 − , the network is shown in Fig. 10.154. At t = 0 − , the network attains steady-state condition. Hence, the capacitor acts as an open circuit. Writing the KVL equation at t = 0 − , 100 −10 10 i(0 ) 20 (0 − ) − 20 i(0 ) 20 i(0 ) 20 (0 ) − 0 vC (

)

C (0

+ −

)=0

+

)

i2 (

+

10 Ω

20 Ω

+

20 i(0 (0 ) 20 0 (0 (0 ) + 10 i(0 ) 80 i( vC (

+ +

vc (0 −)

i(0 +)

)

+

+ −

80 V

20 i(0 + ) 20 0 2 ((0 0 + ) 10 i2 (0 + ) 80 +

20i (0−)

Fig. 10.154

At t = 0 + , the network is shown in Fig. 10.155. From Fig. 10.155, i(

10 Ω

2A

) = 40( )

40 i((

20 Ω 100 V

0

i( 0 ) Also,

i (0 −)

+ 20i(0 +) i 2 (0 +) −

0 0

) 1.6 A )

Fig. 10.155

80 V i(t)

For t > 0, the network is shown in Fig. 10.156.

10 Ω

20 Ω

From Fig. 10.156, i(t ) = −i2 (t ) Writing the KVL equation for t > 0, 20i 20i2 10i2

80 V

1 2 10 −6

t

∫ i2 dt − 80 = 0 0

+ 20i −

2 μF i 2 (t)

Fig. 10.156

80 V

10.62 Network Analysis and Synthesis 20i 20i2 + 10i2 50i

t

1 2 10 −6 1 2 10 −6

∫ i dt − 80 = 0 0

∫ i dt − 80 = 0

…(i)

Differentiating Eq. (i), 50

Comparing with the differential equation

di + 5 105 dt di + 1 × 10 4 dt

0 0

di + Pi = 0, dt P = 1 × 10 4

The solution of this differential equation is given by, i(t ) = k e − Pt = k e −1×10 At t

4

t

0, i(0) = 1.6 A 16=k i(( ) = 1 6 e −11

10 4 t

fo t > 0

In the network of Fig. 10.157, an exponential voltage 4 e −5t is applied at time 0. Find the expression for current i(t). Assume zero voltage across the capacitor at t 0.

Example 10.51 t

0.2 Ω

+ −

4e −5t

1F i(t)

Fig. 10.157 Solution At t = 0 − ,

vC (

)

0

i(

)

0

+

0.2 Ω

At t = 0 , the network is shown in Fig. 10.158. Since voltage across the capacitor cannot change instantaneously, vC (

+

)

0 4 i( + ) = = 20 A 0.2

Writing the KVL equation for t > 0,

4

+ −

i(0 + )

v c (0 + )

Fig. 10.158

t

4

5t

0.2i −

1 i dt = 0 1 ∫0

…(i)

10.4 Resistor–Capacitor Circuit 10.63

Differentiating Eq. (i), di −i = 0 dt di + 5i = −100 100 e −5t dt

−20 e −5t − 0 2

Comparing with the differential equation

di + Pi = Q, dt Q = −100 e −5t

P The solution of this differential equation is given by,

− Pt

∫ Q e ddt k e = e −5tt ∫ 00 e 5t e5t dt

i( t ) = e

Pt

Pt

ke

5t

= −100t e −5t + k e −5t At t

0, i(0) = 20 A 20 = k i(t ) = −100 t e −55 + 200 e

55t

fo t > 0

Example 10.52 source e

−t

In the network shown in Fig. 10.159, the switch is closed at t to the network. At t 0, 0 v (0) = 0.5 V. Determine v(t). 1Ω + e −t

1 Ω 0.5 V+ 2 −

+ −

1F

v (t ) −

Fig. 10.159 Solution At t = 0 − ,

v( )

) = 0.5 V

vC (

Since voltage across the capacitor cannot change instantaneously, v(

+

)

vC (

+

) = 0.5 V

Writing the KCL equation for t > 0, v e−t v dv + +1 = 0 1 1 dt 2 dv + 3v = e − t dt Comparing with the differential equation

dv + Pv = Q, dt P

Q = e−t

0 connecting a

10.64 Network Analysis and Synthesis The solution of this differential equation is given by, − Pt

∫ Q e ddt k e = e −3tt ∫ e t e3t dt k e 3t = e −3t ∫ e 2t dt + k e −3t Pt

v(t ) = e

1 −t e + k e −3t 2

= At t

0,

Pt

v(0) = 0.5 V 1 +k 2 k=0

05=

v(( ) = 0 5 e − t

Example 10.53 In the network shown in Fig. 10.160, a sinusoidal voltage v = 100 sin(500t + θ ) volts is applied to the circuit at a time corresponding to θ = 45°. Obtain the expression for the current i(t). 15 Ω

100 sin (500t + 45°)

100 μF i(t)

Fig. 10.160 Solution Writing the KVL equation for t > 0, 100 sin(500t 45 5°) 15i −

t

1 100 × 10 −6

∫ i dt = 0

…(i)

0

Differentiating Eq. (i), (

)(

) cos(500t + 45°) − 15

di − 10 4 i = 0 dt

di + 666.67i 67i = 3333.33 cos o( dt Comparing with the differential equation

)

di + Pi = Q, dt Q = 3333.33 cos(500t + 5°)

P The solution of this differential equation is given by, − Pt

∫ Q e ddt k e = e −666 67t ∫ 3333.33 cos(

i( t ) = e

Pt

Pt

t

) e666.67t + k e

666.67 t

10.4 Resistor–Capacitor Circuit 10.65

⎡ ⎤ e666.67t = 3333.33 e −666.67t ⎢ {666.67 cos(500t + 45°) + 500 sin(500 + 455°)}⎥ + k e −666.67t 2 2 ⎣ (666.67) + (500) ⎦ = 3.2 cos(500 + 45°) + 2.4 sin(500t 45 5 ) k e −666.67t Let and 2

A si

2

φ

Asin φ

3.2

A cos φ

2.4

A cos cos φ = ( .2) 2 2

2

( . ) 2 = 16 (2

A=4 ⎛ 3.2 ⎞ φ = tan −1 ⎜ = 53.13° ⎝ 2.4 ⎟⎠

and

i(t ) = 4 sin( i (53..

) cos(500t

5°)

c ( cos(

. 3°) si (

) + k e −666.67t

t

= 4 sin(500t 98. 3 ) k e −666.67t Putting t = 0, in Eq. (i), 100 sin( 45°)) 15 (0) 0 = 0 i(0) 4.71 4.71 = 4 sin(98 in((98.13°) + k k = 0 75 i(t (t )

si (

t

.13°))

.75 e −666 67t

f

t>0

Example 10.54 In the network of Fig. 10.161, the switch is moved from the position 1 to 2 at t 0. The switch is in the position 1 for a long time. Initial charge on the capacitor is 7 10 −4 coulombs. Determine the current expression i(t), when ω = 1000 rad / s. 1 50 Ω

2 100 sin (wt + 30°)

50 Ω i(t)

20 μF

Fig. 10.161 Solution At t = 0 − , the network is shown in Fig. 10.162. At t = 0 − , the network attains steady-state condition. I=

V = Z

100 ∠30° = 1.41 ∠75° A 1 50 − j 1000 × 20 × 10 −6

The steady-state current passing through the network when the switch is in the position 1 is i 1.41si (1000t + 75°)

50 Ω 100 sin (wt + 30°) i(∞)

20 μF

Fig. 10.162 …(i)

10.66 Network Analysis and Synthesis For t > 0, the network is as shown in Fig. 10.163. Writing the KVL equation for t > 0, −50i − 50i −

50 Ω 50 Ω

t

1

∫ i dt − vC (0) = 0 20 × 10 −6

…(ii)

i(t)

0

Differentiating Eq. (ii),

+ vC (0 −) −

20 μF

Fig. 10.163

di di 1 −50 − 50 − i=0 dt dt 20 × 10 −6 di + 500i = 0 dt di + Pi = 0, dt

Comparing with differential equation

P = 500 The solution of this differential equation is given by, i(t ) = k e − Pt = k e −500 t

…(iii)

From Eqs (i) and (ii), 1.41 sin(1000 At t = 0,

75 )

k e −500 t

1.41si (75 ) = k k = 1 36 i(( ) = 1 36

10.5

500 t

f

0

RESISTOR–INDUCTOR–CAPACITOR CIRCUIT

Consider a series RLC circuit as shown in Fig. 10.164. The switch is closed at time t = 0. The capacitor and inductor are initially uncharged. Applying KVL to the circuit for t > 0, Ri L

L V i(t)

t

V

R

di 1 − i dt = 0 dt C ∫0

C

Fig. 10.164 Series RLC circuit

Differentiating the above equation, 0−R

di d 2i 1 −L 2 − i=0 dt C dt

d 2i dt 2

+

R di 1 + i=0 L dt LC

This is a second-order differential equation. The auxiliary equation or characteristic equation will be given by, ⎛ R⎞ ⎛ 1 ⎞ s2 + ⎜ ⎟ s + ⎜ =0 ⎝ L⎠ ⎝ LC ⎟⎠

10.5 Resistor–Inductor–Capacitor Circuit 10.67

Let s1 and s2 be the roots of the equation. 2

s1 = −

R 1 ⎛ R⎞ + ⎜ ⎟ − = −α + α 2 − ω 02 = −α + β ⎝ 2L ⎠ 2L LC

s2 = −

R 1 ⎛ R⎞ − ⎜ ⎟ − = −α − α 2 − ω 02 = −α − β ⎝ ⎠ 2L 2L LC

2

α=

where

ω0 = β

and

R 2L 1 LC

α 2 − ω 02

The solution of the above second-order differential equation will be given by, i(t ) = k1e s t + k2 e s2t where k1 and k2 are constants to be determined and s1 and s2 are the roots of the equation. Now depending upon the values of a and w0, we have 3 cases of the response. Case I When a > w0 i.e.,

R > 2L

1

LC The roots are real and unequal and it gives an overdamped response. In this case, the solution is given by,

i(t)

t

i = e–a t (k1 eb t + k2 e–b t) k1 e s t

i

or

k2 e s2t

for t > 0

The current curve for an overdamped case is shown in Fig. 10.165.

i(t )

Case II When a = w0 i.e.,

Fig. 10.165 Overdamped response

R = 2L

1 LC

The roots are real and equal and it gives a critically damped response. In this case the solution is given by, e −α t ( k

i

t

Fig. 10.166

k t ) for t > 0

The current curve for critically damped case is shown in Fig. 10.166. Case III When a < w0 i.e.,

R < 2L

1 LC

The roots are complex conjugate and it gives an underdamped response.

Critically damped response

10.68 Network Analysis and Synthesis In this case, the solution is given by, it where Let

2 0

s1 2 2

2

where

= −

j

= j 1 2

and

2

−

i t =e

Hence,

d

d

−

⎥+

2 −

−

⎤

)

2

−

for t > 0

The current curve for an underdamped case is shown in Fig. 10.167. i t)

t

Fig. 10.167

Example 10.55

Underdamped response

In the network of Fig. 10.168, the switch is closed at t

0 Obtain the expression

for current i t for t > 9Ω

1H

0.05 F

20 V t)

Fig. 10.168 Solution At t = 0 − , the switch is open. i

)

0

vC

)

0

Since current through the inductor and voltage across the capacitor cannot change instantaneously, i

)

0

vC

)

0

10.5 Resistor–Inductor–Capacitor Circuit 10.69 9Ω

For t > 0, the network is shown in Fig. 10.169.

1H

Writing the KVL equation for t > 0, t

20 − 9

di 1 1 − i dt = 0 dt 0 05 ∫0

…(i)

0.05 F

20 V

i(t )

Fig. 10.169

Differentiating Eq. (i), 0 9

di d 2 i − − 20i = 0 dt dt 2

d 2i dt (

2

+9

2

di dt

20 )i )

0 0

D1

4, D2

5

The solution of this differential equation is given by, i(t ) = k1 e −4 t + k2 e

5t

…(ii)

Differentiating Eq. (ii), di = −4 k1 e −4 t − 5k2 e dt At t

5t

…(iii)

0, i(0) = 0 0 = k1 + k2 di ( ) dt

Putting t = 0 in Eq. (i), 20 − 9 (0 + ) −

…(iv) …(v)

4 k1 5k2

di + (0 ) 0 = 0 dt di + (0 ) dt

20 9i(0 + )

20 A / s

From Eq. (v), 20 = −4 k1 5k2

…(vi)

Solving Eqs (iv) and (vi), k1 = 20 k2 = 20 4t i(t ) = 20 e −4t

Example 10.56 at t

0e

5t

for t > 0

In the network shown in Fig. 10.170, the switch is moved from the position 1 to 2 0. The switch is in the position 1 for a long time. Determine the expression for the current i(t ).

10.70 Network Analysis and Synthesis 10 Ω

2H

1 2 20 V

3F 50 V

i(t )

Fig. 10.170 Solution At t = 0 − , the network is shown in Fig. 10.171. At t = 0 − , the network attains steady-state condition. Hence, the inductor acts as a short circuit and the capacitor acts as an open circuit. vC (

)

20 V

i(

)

0

10 Ω

20 V

vC (0−)

i(0 −)

Fig. 10.171

Since the current through the inductor and the voltage across the capacitor cannot change instaneously, vC ( i(

+ +

)

20 V

)

0

For t > 0, the network is shown in Fig. 10.172. Writing the KVL equation for t > 0,

10 Ω

2H

t

20 − 10i − 2

di 1 − i dt − 20 = 0 dt 3 ∫0

…(i)

3F 20 V

20 V

i(t)

Differentiating Eq. (i), 0 10

di dt

2

d 2i dt 2

d 2i

1 − i−0 3

Fig. 10.172

0

di 1 + i=0 dt 6 dt 1⎞ ⎛ 2 ⎜⎝ D + 5 D + ⎟⎠ i = 0 6 2

+5

D1

0 03

2

4.97

The solution of this differential equation is given by, 03t i(t ) = k1 e −0.03 + k2 e

4.97 t

…(ii)

Differentiating Eq. (ii),

At t

0, i(0) = 0

di = −0.03k1 e −0 03t − 4.97k2 e dt 0 = k1 + k2 di ( ) dt

0.03 k1

4.97 t

…(iii) …(iv)

.97 k2

…(v)

10.5 Resistor–Inductor–Capacitor Circuit 10.71

Putting t = 0 in Eq. (i), 20 −10 10 i(0 + ) 2

di + (0 ) 0 = 0 dt di + 20 − 10 i(0 + ) (0 ) = = 10 A / s dt 2

From Eq. (v), 10 = −0.03 03 k1 4 97 k2

…(vi)

Solving Eqs (iv) and (vi), k1 = 2 02 k2 = 2 02 i(t ) = 2.02 e

0 03t 03t

.0 e

4.97 t

for t > 0

Example 10.57 the network. At t

In the network of Fig. 10.173, the switch is closed and a steady state is reached in 0, the switch is opened. Find the expression for the current i (t) in the inductor. 10 Ω i2(t) 100 V

10 μF

1H

Fig. 10.173 Solution At t = 0 − , the network is shown in Fig. 10.174. At t = 0 − , the network attains steady-state condition. Hence, the inductor acts as a short circuit and the capacitor acts as an open circuit. i2 (0 ) = vC (0 )

10 Ω i2 (0 −) vC (0 −)

100 V

100 = 10 A 10

Fig. 10.174

0

Since current through the inductor and voltage across capacitor cannot change instantaneously, i2 (0 + ) 10 A vC (0 + )

0

For t > 0, the network is shown in Fig. 10.175. Writing the KVL equation for t > 0, 10 A

t

di 1 −1 2 − i dt = 0 dt 10 × 10 −6 ∫0

i2(t) 1H

…(i) Fig. 10.175

10 μF

10.72 Network Analysis and Synthesis Differentiating Eq. (i), d 2 i2

−

dt 2 d 2 i2 dt 2

− 105 i = 0 + 105 i = 0

( D 2 + 05 )i = 0 D = j 316 D = − j 316 The solution of this differential equation is given by, i2 (t ) = k1 cos 3 6t + k2 s 3 6t Differentiating Eq. (ii), dii2 = −316 k1 t + 316 k2 cos 3 6t dt At t 0, i2 (0) = 10 A 10 = k1 dii2 (0) = 316 k2 dt Putting t = 0 in Eq. (i), −

di ( dt

+

…(ii) …(iii) …(iv) …(v)

)−0 = 0

di ( dt

+

)=0

From Eq. (v), 0 = 316 k2 k2 = 0 i2 (t ) = 10 cos 316 316t

fo t > 0

Example 10.58 In the network of Fig. 10.176, capacitor C has an initial voltage vc (0 ) of 10 V and at the same instant, current in the inductor L is zero. The switch is closed at time t 0. Obtain the expression for the voltage v(t ) across the inductor L. v (t)

+ 10 V

−

1Ω 4

1F

Fig. 10.176 Solution At t = 0 − ,

iL (

)

0

v(

)

vC (

) = 10 V

1 H 2

10.5 Resistor–Inductor–Capacitor Circuit 10.73

Since current through the inductor and voltage across capacitor cannot change instantaneously, +

iL (

+

v(

)

0

)

vC (

+

) = 10 V

For t > 0, the network is shown in Fig. 10.177. Writing the KCL equation for t > 0,

v (t)

+

t

1

dv v 1 + + v dt = 0 dt 1 1 ∫0 4 2

…(i)

10 V

−

Differentiating Eq. (i),

1 H 2

Fig. 10.177

d 2v dt

1Ω 4

1F

2

(D2

+4

dv + 2v = 0 dt

4 D 2)v = 0 4D D1 1 D2

3

The solution of this differential equation is given by, v(t ) = k1 e −tt + k2 e

3t

…(ii)

Differentiating Eq. (iii), dv = − k1 e −tt − 3 k2 e dt At t

3t

…(iii)

0, v(0) = 10 V 10 = k1 + k2 dv ( ) dt

…(iv) …(v)

k1 3 k2

Putting t = 0 in Eq. (i), dv ( dt

+

) 4 v( v( ) 0 = 0 dv ( dt

+

)

40 V / s

From Eq. (v), −40 = − k1 − 3 k2

…(vi)

Solving Eqs (iv) and (vi), k1 = −5 k2 = 15 v(t ) = −5 e

t

15 e

3t

f

t

0

Example 10.59 In the network of Fig. 10.178, the switch is opened at t for v(t ). Assume zero initial conditions.

0 obtain the expression

10.74 Network Analysis and Synthesis v (t )

0.5 Ω

2A

0.5 H

1F

Fig. 10.178 Solution At t = 0 − ,

iL (

)

0

v(

)

vC (

)=0

Since current through the inductor and voltage across the capacitor can not change instantaneously, iL ( v(

+ +

)

0

)

vC (

+

)=0

For t > 0, the network is shown in Fig. 10.179. Writing the KCL equation for t > 0,

v (t)

t

v 1 dv + vdt + 1 ∫ 0.5 0.5 0 dt

2

…(i)

0.5 Ω

2A

0.5 H

1F

Differentiating Eq. (i), 2

dv d 2v + 2v 2v + 2 = 0 dt dt

d 2v dt

2

(D2

+2

Fig. 10.179

dv + 2v = 0 dt

2 D 2) v = 0 D

1

j1

D

1

j1

The solution of this differential equation is given by, v(t ) = e t ( k cos cos t k2 sin t )

… (ii)

Differentiating Eq. (ii), dv = −e t ( k cos cos t k2 sin t ) e t ( −kk s t + k2 cos t ) dt = e − t [ − k (cos (cos t + si t ) + k (cos t − sin t )]

…(iii)

At t = 0, v(0) = 0 0 = k1

…(iv)

dv ( ) = − k1 k2 dt

…(v)

Putting t = 0 in Eq. (i), 2 ( 0) + 0

dv ( 0) = 2 dt

10.5 Resistor–Inductor–Capacitor Circuit 10.75

dv ( ) dt

2 V /s

From Eq. (v), 2 = −k1 + k2

…(vi)

Solving Eq. (iv) and (vi), k1 = 0 k2 = 2 v(t ) = 2 e

t

i t

f

t

0

Example 10.60 The network shown in Fig. 10.180, a sinusoidal voltage v = 150 sin(200t + φ ) is applied at φ = 30°. Determine the current i(t). 10 Ω

0.5 H

200 μF

150 sin (200t + f) i(t )

Fig. 10.180 Solution Writing the KVL equation for t > 0, t

150 sin( 200t 30°) 10

di 1 − i dt = 0 dt 200 × 10 −6 ∫0

0.5

…(i)

Differentiating Eq. (i), di dt

30000 cos( 200t + 30°) 10

d 2i dt (

2

0.5 + 20

d 2i dt 2

− 5000i 5000i = 0

di + 10000 0 i dt

60000 cos( 200t + 30°)

)i

60000 cos( 200 00t + 30°)

2

…(ii)

The roots of the characteristic equation are D

10

j 99 5

D

10

j 99 5

The complimentary function is iC

e −10 t ( K1 c

iP

A cos( 200t 30°) B sin(

99 5t + K 2 i 99.5t )

Let the particular function be t+

°)

iP′

00 A sin( 200t + 30°) + 200 B cos( 200t + 30°)

iP″

0000 A cos( 200t 30°)

B sin(200 i ( 200t + 30°)

10.76 Network Analysis and Synthesis Substituting these values in Eq. (ii), 30° − 40000 B + (

200t +

+ 30 + B sin 200t + 30

° +

60000 cos( 200t + −

B 4000 A

=

−200 A

200t + 0° + 200

cos 200 + 30°

°)

+

+

200t +

)

°)

+ 30°)

Equating the coefficients, B 4000 A + + 4000 +

=0 = 60000

Solving these equations, A 1 97 B = 0 26 t+

i Let

Asin

and

A cos

° +

+ 30°

. 7 . 6 =

cos

= 3.95

A = 1.987 −1 97 = −82.48 0 26

= tan −1

and

t + 30° + 1.

i

. °) sin( 200 +

cos(

°)

t + 30° − 82 48 ) − 52 48°) The solution of the differential equation is given by, it

e

k

.

sin 9. t

t

…(iii)

48 )

Differentiating Eq. (iii), di dt

k 10

At t

99 5t

k1 os 99

n

. t

1

(

)

t

.

)

0 0

1

k1

in(

. °) …(iv)

1 58

di dt

−52. =

)

1 5 ) 242.03 226.23

…(v)

10.5 Resistor–Inductor–Capacitor Circuit 10.77

Putting t = 0 in Eq. (i), 150 sin(30°)) 10(0) 0.5

di ( 0) − 0 0 dt di (0) = 150 A / s dt

From Eq. (v), 150 = 99.5 k2 + 226.23 k2 = −0 77 i(( ) = e −10 t (1 (1.58

Example 10.61 (a) L =

s 99.5t − 0.77 si 99.5 )

.987 sin( 200t − 52. .987

) fo t > 0

The switch in the network of Fig. 10.181 is opened at t = 0. Find i (t) for t > 0 if,

1 H and C = 1 F 2

(b) L = 1 H and C = 1 F 2Ω

L

4V

(c) L = 5 H and C = 1 F

i(t)

+ C vC (t)

2Ω

−

Fig. 10.181 Solution At t = 0–, the network has attained steady-state condition. Hence, the inductor acts as a short circuit and the capacitor acts as an pen circuit. vC (

)

4×

i(

)

0

2 =2V 2 2

Since current through the inductor and voltage across the capacitor cannot change instantaneously, vC ( i( Case I

When R = 2 Ω, L =

+ +

)

2V

)

0

1 H, C = 1 F 2

α=

ω0 =

R = 2L 1 LC

α > ω0

2 1 2× 2 =

=2 1

1 ×1 2

=

1 05

= 1.414

10.78 Network Analysis and Synthesis This indicates an overdamped case. i(t ) = A1 e s1t − A2 e s2t where,

s1 = −α − α 2 − ω 02 = −2

4−2

and

s2 = −α + α 2 − ω 02 = −2

2 = −0.586

i(t ) = k1 e

−3 414 t

2− 2

3.414

0.586 t

+ k2 e

At t = 0, i (0) = 0 k1 k2 = 0

…(i)

Also vL(0+) + vC(0+) + vR(0+) = 0 vL ( vL ( di ( dt

+ +

+

)

+

vR ( di ( dt

) − vC (

+

+

)

2i(

)

L

)=

vL ( + ) 2 =− = −4 A / s L 0.5

+

) vC (

+

) = −2 V

…(ii)

)

Differentiating the equation of i (t) and putting the condition at t = 0, − 3.414 k1 − 0.586 k2 = −4

…(iii)

Solving Eqs (i) and (iii), we get k1 = 1.414

(e −3.414 t

(t ) = 1. Case II

k2 = − 1.414

and

e

0.586 t

)

When R = 2 Ω, L = 1 H, C =1 F

α=

R 2 2 = = =1 2L 2 × 1 2 1

ω0 = α

LC

=

1 1

=1

ω0

This indicates a critically damped case. i(t) = e–a t (k1 + k2 t) = e–t (k1 + k2 t)

At t = 0, i(0) = 0

k1 = 0 Also,

di + (0 ) dt

vL (0 + )

L

di ( dt

vL ( + ) 2 = − = −2 A / s L 1

+

)=

for t > 0

Exercises 10.79

Differentiating the equation of i(t) and putting the condition at t = 0, di dt

= − k1 k2 = −2 t =0

k2 = −2 i(t ) = −2t e − t Case III

f

t>0

When R = 2 Ω, L = 5 H, C = 1 F R 2 = =02 2 L 10 1 1 ω0 = = = 0.447 LC 5 α < ω0

α=

This indicates an underdamped case. i (t) = e–a t (B1 cos wd t + B2 sin wd t)

ωd

where,

ω o2 − α 2 = (0.447) 2 − (0.2) 2 = 0 4 ± jω d = −0.2 ± j 0.4

s, i( t ) = e

0 2t

( B1 cos 0 4t (B

B2 sin 0.4t )

Applying the initial condition, i(0+) = 0 di ( dt

and

+

vL ( + ) 2 =− L 5 i( 0) 0

)=−

B1

B2 = −1 i(t ) = −e −0 2t si 0.4t

for t > 0

Exercises 10.1 The switch in Fig. 10.182 is moved from the position a to b at t = 0, the network having been in steady state in the position a. Determine dii dii3 + i1 (0 + ), i2 (0 + ), i3 (0 + ), 2 (0 + ) (0 ). dt dt

1Ω

1 Ω i1(t)

a

i2(t) 10 V

10.2 The switch K is closed at t = 0 in the network shown in Fig. 10.183. Determine di d 2i + i( + )), (0 + ) (0 ). dt ddt 2

b

1Ω

1H

i3(t )

1Ω

2Ω

2H

1F

V0

1Ω i1(t)

1F i(t)

Fig. 10.183 Fig. 10.182 [1.66 A, 5 A, −3.33 A, −3.33 A/s, 2.22 A/s]

⎡ 1 ⎢0, 2 V0 ⎣

⎤ V0 ⎥ ⎦

10.80 Network Analysis and Synthesis 10.3 In the network of Fig. 10.184, the switch K is closed at t = 0. At t = 0−, all capacitor voltages and inductor currents are zero. Find dv dv dv3 v1 , 1 , v2 , 2 , v3 and at t = 0 + . dt d dt d dt

10.6 The network shown in Fig. 10.187 is under steady-state when the switch is closed. At t = 0, it is opened. Obtain an expression for i (t). 4 kΩ i(t)

K

R1

L1

V1

V3

10 kΩ

4A

4 mH

8 kΩ

R2 v (t)

+ −

C3

C1 C2

Fig. 10.187

V2 R

.857 e −2 ×10 t ] 6

[ (t )

Fig. 10.184 ⎡ 1 (0 + ) ⎤ , 0, 0, 0, 0 ⎥ ⎢0, ⎣ C R1 ⎦

10.7 The switch in Fig. 10.188 is open for a long time and closes at t = 0. Determine i (t) for t > 0. 6Ω

3Ω i(t)

10.4 In the network at Fig. 10.185, the capacitor C1 is charged to voltage 1000 V and the switch K d 2 i2 is closed at t = 0. Find at t = 0 + . dt 2

2H

Fig. 10.188

K

10 μF

30 Ω

240 V

2 MΩ

+ 1000 V −

10 μF

1 MΩ i1

[i(t) = 25(1 − e−4t)]

i2

10.8 In the network shown in Fig. 10.189, the steady state is reached with the switch open. At t = 0, the switch is closed Find vC (t) for t > 0. 3 kΩ

Fig. 10.185 ⎡ 17 2⎤ ⎢ 400000 A / s ⎥ ⎣ ⎦

5 kΩ

2 kΩ

vC (t) 1 kΩ

10.5 In the network shown in Fig. 10.186, switch is closed at t = 0. Obtain the current i2 (t). 10 Ω

+ −

5 μF

6V

Fig. 10.189 [vC(t) = 5e−20t] 10 Ω

50 V i1(t )

2 μF

i2(t)

Fig. 10.186 [i2(t) = 5 e–100000t]

10.9 The circuit shown in Fig. 10.190 has acquired steady state before switching at t = 0. (i) Obtain vC (0+), vC (0−), i (0+) and i (0−). (ii) Obtain time constant for t > 0. (iii) Find current i (t) for t > 0.

Exercises 10.81

+

vC (t)

10.13 In Fig. 10.194, the switch is open until time t = 100 seconds and is closed for all times thereafter. Find v (t) for all times greater than 100 if v (100) = −3 V.

10 kΩ i(t )

2 μF

5V

5 kΩ

12 Ω

8Ω

+

−

Fig. 10.190

2

20 V 20 Ω

1H i (t)

v (t ) −

Fig. 10.194 (t ⎡ − ⎢ v(t ) = 5 − 8e 160 ⎢⎣

10 Ω

1

2F

6F

5V

[(i) 5 V, 5 V, 1 mA, 0, (ii) 0.01 s, (iii) e−100 t mA] 10.10 In the network shown in Fig. 10.191, the switch is initially at the position 1 for a long time. At t = 0, the switch is changed to the position 2. Find current i (t) for t > 0.

)⎤

⎥ ⎥⎦

10.14 A series RL circuit shown in Fig. 10.195 has a constant voltage V applied at t = 0. At what time does vR = vL.

Fig. 10.191 [i (t) = 2 e−30t] 10.11 In the network shown in Fig. 10.192, the switch is closed at t = 0. Find v (t) for t > 0.

vR

10 Ω

vL

1H

10 V

+

Fig. 10.195 1 Ω 2

3A

1 H 3

1Ω

[0.0693 s]

v (t ) −

Fig. 10.192 [v(t) = e−t] 10.12 In the network shown in Fig. 10.193, the switch is in the position 1 for a long time and at t = 0, the switch is moved to the position . Find v (t) for t > 0. 1Ω

1V

1

2

1H

+ 0.5 Ω

10.15 In the circuit shown in Fig. 10.196, at time t = 0, the voltage across the capacitor is zero and the switch is moved to the position y. The switch is kept at position y for 20 seconds and then moved to position z and kept in that position thereafter. Find the voltage across the capacitor at t = 30 seconds. 10 kΩ

10 V

y

+ vC (t) −

z

0.1 μF

5 kΩ

2 H v(t ) −

Fig. 10.193 [v (t) = −0.5

−3 t e 4]

Fig. 10.196 [0] 10.16 Determine whether RLC series circuit shown in Fig 10.197 is underdamped, overdamped or

10.82 Network Analysis and Synthesis critically damped. Also, find vL ( and i (∞). 200 Ω

+

)),

di + (0 ) dt

or

critically damped. Also find 2 di d v vL ( + )), (0 + ), 2 ( + ) if v(t ) = u(t ). dt dt

0.1 H

2Ω

+ v (t) − L 200 u(t )

+ −

1H + v (t) − L

10 μF u(t)

i(t)

+ −

+

v(t) −

i(t)

Fig. 10.197

1 F 2

Fig. 10.198

[critically damped, 200 V, 2000 A/s, 0] 10.17 Determine whether RLC circuit of Fig 10.198 is underdamped, overdamped

[underdamped 1 V, 1 A/s, 2 V/s2]

Objective-Type Questions 10.1 The voltages vC1 vC2 and vC3 across the capacitors in the circuit in Fig. 10.199 under steady state are respectively 1H

10 kΩ

+ 100 V −

2F 2H + − vC 2

+

vC1 −

1F

2 kΩ

(a)

eat − ebt

(b)

eat + ebt

(c)

aedt − bebt

(d)

aeat + bebt

10.3 The differential equation for the current i(t) in the circuit of Fig. 10.201 is

40 kΩ

vC 3

−

2Ω

i(t)

+

2H

3F

1F

sin t

Fig. 10.199 Fig. 10.201

(a) 80 V, 32 V, 48 V (b) 80 V, 48 V, 32 V

(a)

(c) 20 V, 8 V, 12 V (d) 20 V, 12 V, 8 V 10.2 In the circuit of Fig. 10.200, the voltage v(t) is 1Ω

1Ω

eat

(b)

+ v ( t) −

(c) 1H

ebt

(d) Fig. 10.200

2

d 2i dt

d 2i dt 2 2

2

+2

d 2i dt 2

d 2i dt 2

+2

di + 2i ( t ) = cos t dt

+2

+2

di + i ( t ) = sin t dt

di + i ( t ) = cos t dt

di + 2i ( t ) = sin t dt

Objective-Type Questions 10.83

10.4 At t = 0+, the current i1 in Fig. 10.202 is 1

i(t )

(b)

C

1 0.63

2

V

R L

R i1(t)

i2(t )

(c)

Fig. 10.202 (a) (c)

V 2R V − 4R −

(b)

t

0.5

C i(t) 0.5 0.31

−

V R

t

0.5

(d) zero

(d)

10.5 For the circuit shown in Fig. 10.203, the time constant RC = 1 ms. The input voltage is vi(t) = 2 sin 103 t. The output voltage v0(t) is equal to

i(t ) 1 0.63

t

2

R

Fig. 10.205 vi (t )

C

vo (t)

10.7 The condition on R, L and C such that the step response v(t) in Fig. 10.206 has no oscillations is

Fig. 10.203 (a)

sin (103 − 45°)

(c)

sin (103 t − 53)

L

(b) sin (103 t + 45°) (d) sin (103 t + 53°)

10.6 For the RL circuit shown in Fig. 10.204, the input voltage vi(t) = u(t). The current i(t) is

+ v(t) −

C

2Ω

(a)

1 L 2 C

(b)

R≥

L C

L C

(d)

R=

1

(a)

R≥

(c)

R≥2

i(t)

Fig. 10.204

v(t)

Fig. 10.206

1H

vi (t )

R

LC

10.8 The switch S in Fig. 10.207 closed at t = 0. If v2 (0) = 10 V and vg(0) = 0 respectively, the voltages across capacitors in steady state will be

i(t ) 0.5 0.31

2

t

10.84 Network Analysis and Synthesis the terminal pair, the time constant of the system will be

v1 (t ) 8 μF v2 (t )

1 MΩ

2 μF

i(t )

Network of linear resistors and independent sources

Fig. 10.207 (a) v2 (∞) = v1 (∞) = 0 (b) v2(∞) = 2 V, v1 (∞) = 8 V (c) v2 (∞) = v1 (∞) = 8 V (d) v2(∞) = 8 V, v1 (∞) = 2 V

+

A

−

B

(a) i(t ) 4 mA

10.9 The time constant of the network shown in Fig. 10.208 is

8V

(0, 0)

v( t)

(b)

R

Fig. 10.210 2R

10 V

C

Fig. 10.208 (a)

2 RC

(b)

3 RC

1 2 RC (d) RC 2 3 10.10 In the series RC circuit shown in Fig. 10.209, the voltage across C starts increasing when the dc source is switched on. The rate of increase of voltage across C at the instant just after the switch is closed i.e., at t = 0+ will be (c)

C

(a) 3 μs (b) 12 s (c) 32 s (d)

unknown, unless actual network is specified 10.12 In the network shown in Fig. 10.211, the circuit was initially in the steady-state condition with the switch K closed. At the instant when the switch is opened, the rate of decay of current through inductance will be K

R

2Ω

2Ω

2V

2H 1V

Fig. 10.211 Fig. 10.209 (a)

zero

(b)

(c)

RC

(d)

infinity 1 RC

10.11 The v – i characteristic as seen from the terminal pair (A – B) of the network of Fig. 10.210(a) is shown in Fig. 10.210(b). If an inductance of value 6 mH is connected across

(a) (c)

zero 1 A/s

(b) (d)

0.5 A/s 2 A/s

10.13 A step function voltage is applied to an RLC series circuit having R = 2 Ω, L = 1 H and C = 1 F. The transient current response of the circuit would be (a) over damped (b) critically damped (c) under damped (d) none of these

Answers to Objective-Type Questions 10.85

Answers to Objective-Type Questions 10.1 (b)

10.2 (d)

10.3 (c)

10.4 (d)

10.5 (a)

10.6 (b)

10.8 (d)

10.9 (d)

10.10 (d)

10.11 (a)

10.12 (d)

10.13 (b)

10.7 (c)

11 11.1

Laplace Transform and Its Application INTRODUCTION

Time-domain analysis is the conventional method of analysing a network. For a simple network with firstorder differential equation of network variable, this method is very useful. But as the order of network variable equation increases, this method of analysis becomes very tedious. For such applications, frequency domain analysis using Laplace transform is very convenient. Time-domain analysis, also known as classical method, is difficult to apply to a differential equation with excitation functions which contain derivatives. Laplace transform methods prove to be superior. The Laplace transform method has the following advantages: (1) (2) (3)

Solution of differential equations is a systematic procedure. Initial conditions are automatically incorporated. It gives the complete solution, i.e., both complementary and particular solution in one step.

Laplace transform is the most widely used integral transform. It is a powerful mathematical technique which enables us to solve linear differential equations by using algebraic methods. It can also be used to solve systems of simultaneous differential equations, partial differential equations and integral equations. It is applicable to continuous functions, piecewise continuous functions, periodic functions, step functions and impulse functions. It has many important applications in mathematics, physics, optics, electrical engineering, control engineering, signal processing and probability theory.

11.2

LAPLACE TRANSFORMATION

The Laplace transform of a function f (t) is defined as F ( s) = L { f (t )}

∞

∫ f (t ) e

− st

dt

0

where s is the complex frequency variable. jω

s

The function f (t) must satisfy the following condition to possess a Laplace transform, ∞

∫ | f (t ) | e 0

−σ t

dt < ∞

11.2 Network Analysis and Synthesis where s is real and positive. The inverse Laplace transform L–1 {F ( s )} is σ + j∞

f (t ) =

11.3

1 F ( s) e st dds 2π j σ −∫j∞

LAPLACE TRANSFORMS OF SOME IMPORTANT FUNCTIONS

1. Constant Function k The Laplace transform of a constant function is ∞

∞

L{k} = ∫ ke

− st

0

⎡ e − st ⎤ k dt = k ⎢ ⎥ = s ⎣ −s ⎦ 0

2. Function tn The Laplace transform of f(t) is ∞

L{t n } = ∫ t n e − st dt 0

dx Putting st = x, dt = s ∞

1 ∞

n

dx ⎛ x⎞ L{t n } = ∫ ⎜ ⎟ e − x = s n+ ⎝ s⎠ s 0 n +1 n!

If n is a positive integer, L{t n } =

∫x

n

e x ddx =

0

n +1 s n +1

, s > 0, n + 1 > 0

n!

s n +1

3. Unit-Step Function The unit-step function (Fig 11.1) is defined by the equation, u(t) = 1 =0

t>0 ta t0 =0 ta =0 t a.

L{ f (t )}

s

F F(( s) =

2

s +1

By time-shifting theorem, L{cos(t a)} e − as

Example 11.12 Solution

s 2

s +1

Find the Laplace transform of e t

a

t > a.

Let f (t ) = e t L{ f (t )}

F( F ( s) =

1 s −1

By time-shifting theorem, L{e t a } = e − as

Example 11.13 Solution

By time-shifting theorem,

1 s −1

⎛ π⎞ Find the Laplace transform of sin t − ⎟ ⎝ 4⎠

t>

Let f (t ) = sin t L{ f (t )}

F( F ( s) =

1 s2 + 1

πs

− ⎧ ⎛ π⎞⎫ 1 L ⎨si t − ⎟ ⎬ = e 4 2 ⎝ ⎠ 4 s +1 ⎩ ⎭

Example 11.14

Find the Laplace transform of (

as

∫ f (t ) e 0

Let f(t) = cos t

Solution

e

)3

t

1.

π . 4

− st

as dt = e −as F ( s)

11.8 Network Analysis and Synthesis Let f(t) = t3

Solution

3!

f

s4

By time-shifting theorem, 3

e−s

L t

s4

11.4.5 Multiplication by t (Frequency-Differentiation Theorem) If

F ( s) then L t f )} = −

f t

d F s ds ∞

Proof

t e − st dt

f 0

Differentiating both the sides w.r.t s using DUIS, ∞

d F s ds

∞

d ds 0

=

t

t e − st dt

∂ ∂s 0

Example 11.15

at}

L

s

L t os t

t} = L

d L ds

L

ds

t}

t 4

3 d 1 4 ds s 2 −

s

1 d 1 s + 2 ds s s

si t

t =− 3s s 2

2

}=

1 1 s + 2 2 s s +4

=−

1

−

1 s

+

⎤

− ⋅

s s

=

Find the Laplace transform of t sin3 t . L

L t in t

2as

d a = 2 ds s a 2

at

1 co 2t 1 = 2 2 t =−

Example 11.17 Solution

a2

Find the Laplace transform of t cos 2 t .

Example 11.16 L

at.

a 2

L t in at

Solution

d F ds

Find the Laplace transform of t s

Solution

( )}

0

−

( )}

dt t e − st dt = − L

−

0

L

t e

−

1 3 4 s2 1 2

s +9 =

3s ⋅ 2 (

1 2

s +9 =−

3 1 4 s2 + 1

3 −2 s 4 ( s2 + 1

+

+9

2s 2

s + 9) ⎦

s +5 s

1 2

=

3s 2

24 s s + 5 s

s

1 s

+

s2 2

4

11.4 Properties of Laplace Transform 11.9

Example 11.18 Solution

Find the Laplace transform of t sin 2t cosh t .

L{sin 2t cosh t}

⎧⎪ ⎛ e t e t ⎞ ⎫⎪ 1 t L ⎨sin i 2t ⎜ i 2t e −tt sin t} ⎟ ⎬ = 2 L{e sin 2 ⎝ ⎠ ⎪⎭ ⎩⎪

⎤ 1⎡ 2 2 1 1 + + 2 ⎢ ⎥= 2 2 2 2 ⎣ ( − 1) + 4 ( + 1) + 4 ⎦ s − 2 + 5 s + 2 + 5

=

d d ⎛ L{sin 2t cosh t} = − ⎜ 2 ds d ds ⎝ s

L{{ sin 2t cosh t}

1 2s + 5

+

2s − 2 2s 2 ⎞ + 2 ⎟⎠ = 2 2 s + 2s + 5 ( s − 2 5) ( s + 2 5) 2 1

2

11.4.6 Division by t (Frequency-Integration Theorem) ∞

If L{ f (t )} Proof

f (t ) ⎫ F ( s), then L ⎧⎨ ⎬ = ∫ F ( s) ds ⎩ t ⎭ s ∞

L { f (t )}

∫ f (t ) e

F (s F(s ( s)

− st

dt

0

Integrating both the sides w.r.t s from s to ∞, ∞

∞

∫ F ( s) ds

∫ f (t )e

st

dt dds

s 0

s

Since s and t are independent variables, interchanging the order of integration, ∞

∞

∞ ⎡∞

∫ F ( s)ds = ⎢⎢ ∫ f (t )e 0

s

⎣s

st

∞ ∞ ⎤ f (t ) − st ⎡1 ⎤ ds d ⎥ dt = ∫ ⎢ f (t )e − st ⎥ dt = ∫ e dt − t t ⎦s ⎥⎦ 0⎣ 0 ∞

⎧ f (t ) ⎫ L⎨ ⎬ = ∫ F ( s)ds ⎩ t ⎭ s

Example 11.19 Solution

L{

Find the Laplace transform of

e t} =

1 e−t . t

1 1 − s s +1 ∞

∞

⎧1 − e − t ⎫ 1 ⎞ ⎛1 ∞ −t L⎨ g s + 1) s ⎬ = ∫ L 1 e ds = ∫ − ⎟⎠ ds = log s − log( ⎝ t s s + 1 ⎩ ⎭ s s ∞

∞

s ⎤ ⎡ = ⎢log s + 1⎦ s ⎣

⎡ ⎤ ⎢ 1 ⎥ log ⎢ ⎥ = log ⎢1 + 1 ⎥ ⎣ s ⎦s

⎛ ⎞ s s +1 ⎜ 1 ⎟ log ⎜ = − log = log ⎟ 1 s +1 s ⎜1+ ⎟ ⎝ s⎠

11.10 Network Analysis and Synthesis

Example 11.20 Solution L

e

e

∞

− bt

−e

t

⎩

Example 11.21

s

s a log s+b

1

= s

s

1 = log 2

Example 11.22 L

s b

1 ds = s b

et

t

=

1

e

L

2

1 s +1 −

+

=

1⎡

log

s 1 s +1

∞ s

cos h2t sin 2t . t

e 2t

⎞

t

=∫

sin 2t ⎭

1 e2 L 2 t

+L

e

t

t t

+4

n 2t = L t s

L

s a s+b = log s+b s a

2

t} =

L

t

− log

1 s 1 1 s +1 − log = log 2 s +1 s 1

Find the Laplace transform of cosh t t

1 1 2 s 1 log( s − 2

1⎞ s − og 1 1+ s

1 s⎥ = 1 1 2 1+ s ⎦s 1

1+

∞ s

+

sinh t . t

1 ds s +1

−

s

e 2

−

a s − og b 1+ s

1+

t =L

L

Solution

s a

Find the Laplace transform of

Solution sinh t = t

1

a⎤ s = log = b 1+ s ⎦s ⎣

s

.

1

∞

⎡

bt

e t

1 s a

∫

d

s

⎡

L

=

Le −

e

Find the Laplace transform of

s

2 s +

ds = tan −1

s⎤ = − tan 2 s 2

2

cot

By first shifting theorem, L

cosh t t

t 2

L

e t

+L

e t

1 cot 2

1

2

cot

11.4.7 Time-Differentiation Theorem: Laplace Transform of Derivatives If

f t

F ( s) then − f( )

f f

2

− sf

−

+2 2

2

11.4 Properties of Laplace Transform 11.11

In general, f

−

s ′′F s

t

Proof f ′ t e − st dt

L f′ t 0

Integrating by parts, f t e − st

f t

L fn t

Example 11.23

n

t e

(0) + sL {

d

t

}

0

′ + sL{ f t}

f t

s∫

0

Similarly,

In general,

t e − st dt = − f

s

0

s −

+

f′

f t

f )}

−

−

Find L f (t ) and L f ′(t ) of

sin t . t

t

Solution L f t L

L =

′

sin t = t

s

s 1

−

Example 11.24

∞

− lim t →0

f

e =

Example 11.25 Solution

−

∞ s

ta

2

−

−

s

s

s

− st

0

t t

5 5.

∞

5

t dt =

t=

d

st 5

e

s

0

3 0 = ⋅ (1 − e − s

+ = 0

−3 s

−

−

− s

− =

Find L f (t ) and L f ′(t ) of f(t) = e −5t sin t . L f t L f′ t

Example 11.26

s co

3 0

s

f t

s +1

an −1 s

ds

Find L f (t ) and L f (t ) of the following function: t

Solution

si t t

1

t}

Le −

⋅

1 +1

s 1

s +

s

0

+ 26

2

Find L f (t ) and L f ′(t ) of the following function: t

t 6

0

t t

3 3

+

26

=

−5

s

11.12 Network Analysis and Synthesis 3

e − st ⎤ ⋅ t⎥ ⎣ −s ⎦0 1

3

Solution

L f t

e 0

=−

3 s

t dt

e

t dt

e

dt

3

−

−

+

1

s

3

e

s2 ⎦0

+6

e

st

s

∞

3

3 − s s2

+ = 2 s s s 1 1 3s f ( 0) = + e 3 s s

=

L

st

11.4.8 Time-Integration Theorem: Laplace Transform of Integral t

If L f t

F s

L

∫

F s s

t dt ⎭

0

t

Proof

L

t

⎛ e − st

t

f t t

f t t

s

∞

− st

f t

⎩0

Integrating by parts, L

t

f t t

⎡ e − st ∫⎢

t

d dt ∫0

0

t dt t

Example 11.27

dt

⎭

0

Find the Laplace transform of

∫e

−

⎞

dt =

1 s 0

=

1 L f s

}=

F s

t 3 dt .

0

Solution t

L

Le

t

e−

t

3! s

6

=

4

4

s

1 6 L e −2 t 3 = s s s+2

0

4

t

Example 11.28

Find the Laplace transform of

∫ t cosh t dt . 0

Solution L t cosh t

et

L t t

t cosh t dt

L 0

e t⎞ 1 = 2 ⎭ 2 1 s

=

te s

t cosh t

1 1 2 ( s 1)

⎤

1 (

)

1

ss t

Example 11.29

Find the Laplace transform of the

∫ te

4t

sin 3t dt .

0

Solution L t in 3t L te

in t}

d L ds 6 [( s +

t =− 4

d 3 ds s 6 4 s

6s s

2

1 2 1 s2 + 1 = 2 −

11.4 Properties of Laplace Transform 11.13 t

L

t e−

1 s

t dt

0

−

=

6 s+4 s s2 + s +

)2

t

Example 11.30

t sin 3t dt .

Find the Laplace transform of e 0

Solution

d L ds

L t in 3t t

1 s

t sin 3t

L 0

2

s

s + 6

t

6s

6

t

L e −4 t ∫ t

d 3 ds s +

t

6

[s

0

11.4.9 Initial Value Theorem If

f t

F ( s) then lim ( = lim

( s)

s→

Proof We know that, − f( )

f

∞

L f t )} +

sF s

t e − dt +

0 ∞

lim

( s)

s

s →∞

e − st dt +

f′

lim

0

∞

lim[

0

s →∞

dt +

t

t →0

11.4.10 Final Value Theorem If

f t

F ( s) then lim (

lim

( s)

s →∞

t

Proof We know that − f( )

f

∞

sF s

L f t )} +

t e − dt +

0

lim

s

( s)

f′

lim

s→ 0

t

e

st

dt +

0

lim f ′

e

st

d

0 0

+

t →∞

=

t dt + 0

+ f (0

im (

t →∞

(t )

11.14 Network Analysis and Synthesis

Example 11.31

Verify the initial and final value theorems for f (t ) = e −tt (t ) 2 = e t (t (t 2 2 2 F ( s) = + + 3 (s ) ( s )2 2s 2s sF ( s) = + + 3 ( s 1) ( s 1) 2 lim f (t ) = 1

Solution

t

t + 1)2 .

t ) 1 s +1 s s +1

t →0

⎡ ⎤ 2 2 ⎢ ⎥ 2 1 ⎥ s lim sF ( s) = lim ⎢ s 3 + + =1 2 1⎥ s →∞ s →∞ ⎢ 1⎞ 1⎞ ⎛ ⎛ 1 + ⎢ ⎜1 + ⎟ ⎥ ⎜⎝1 + ⎟⎠ s⎥ s ⎢⎣ ⎝ s ⎠ ⎦ Hence, the initial value theorem is verified. lim f (t ) = 0 t →∞

lim sF ( ) = 0 s→ 0

Hence, the final value theorem is verified. Verify the initial and final value theorems for e t (t 2 + cos3t).

Example 11.32

f (t ) = e − t (t 2

Solution

F ( s) = sF ( s) =

2 (s

)

3

2s

(s lim f (t ) = 1

)3

cos 3t ) + +

s +1 ( s )2 + 9 s( s ) (s

)2 + 9

t →0

⎡ ⎛ 1⎞ ⎤ 2 ⎢ ⎜⎝1 + ⎟⎠ ⎥ 2 s ⎥ =1 lim sF ( s) = lim ⎢ s 3 + 2 ⎢ ⎥ s →∞ s →∞ 1 1 9 ⎛ ⎞ ⎢ ⎛⎜1 + ⎞⎟ ⎥ 1 + + ⎜⎝ ⎟ s⎠ ⎢⎣ ⎝ s ⎠ s 2 ⎥⎦ Hence, the initial value theorem is verified. lim f (t ) = lim( li ( 2 cos 3t )e − t = 0 t→

t →∞

⎡ 2s s( s + ) ⎤ ⎥=0 lim sF ( ) = lim ⎢ + s→ 0 s→ 0 ⎢ ( )3 ( s + 1)2 + 9 ⎥⎦ ⎣ Hence, the final value theorem is verified.

Example 11.33 F(s) =

2s+ 1 s3 + 6s 2 + 11s+ 6

Find the initial and final values of the function whose Laplace transform is .

11.5 Laplace Transform of Periodic Functions 11.15

F ( s) =

Solution

2 1 + s s2 sF ( s) = lim 3 = lim =0 1 6 s →∞ s →∞ s 6 s 2 11s 6 s→∞ 1 + 6 + 11 6s + s s 2 s3 2s2 + s sF ( s) = lim 3 =0 s →0 s → 0 s + 6 s 2 + 11s + 6

f

Find the final value of the function whose Laplace transform is I(s) s = I ( s) =

Solution

s+6 s( s ) s→0

s+ 6 . s(s+ ( 3)

s+6 =2 s+3

li s ( s) = lim

I( )

11.5

s + 6s 6 s 2 + 11s + 6 2s2 + s

f

Example 11.34

2s + 1 3

s→0

LAPLACE TRANSFORM OF PERIODIC FUNCTIONS

A function f(t) is said to be periodic if there exists a constant T(T > 0) such that values of t. f (t T ) f (t T T ) f (t (t T ) = f (t )

(t T )

f (t ), for all

In general, f (t nT ) = f (t ) for all t, where n is an integer (positive or negative) and T is the period of the function. If f (t) is a piecewise continuous periodic function with period T then L{ f (t )} =

1 1 − e −Ts ∞

Proof

L{ f (t

∫ f (t(t ) e

T

∫ f ( t )e

dt

0

− st

0

In the second integral, putting t When

st

∞

T

dt = ∫ f (t )e

st

∫ f ( t )e

ddt

0

T

x + T , dt dx d t T, x = 0 t→ , x→∞ T

L{ f (t

∫ 0

∞

f (t (t )e − st d dt + ∫ f ( x T )e − s( x 0

∞

T

= ∫ f (tt))e

st

0

0

T dt e −Ts d ∫ f ( x )e

dx

∞

st

dt e −Ts ∫ f (t )e 0

T

= ∫ f (t )e − st dt + e 0

sx

0

T

= ∫ f ( t )e

T)

Ts

L{ f (t )}

st

dt

dx

− st

dt

11.16 Network Analysis and Synthesis T

(

Ts

) L{ f (t )} = ∫ f (t )e − st dt 0

L{ f (t )} =

Example 11.35

T

1 1 − e −Ts

∫ f ( t )e

− st

dt

0

Find the Laplace transform of the waveform shown in Fig. 11.7. f (t )

A

T

0

2T

t

3T

Fig. 11.7 Solution The function f (t) is a periodic function with period T. At f (t ) = 0 0. 2Ω

a

b + 1Ω

2V

1H

2H

v (t) −

Fig. 11.45

2Ω

At t = 0−, the network is shown in Fig 11.46. At t = 0−, the network has attained steady-state condition. Hence, the 2V inductor of 2H acts as a short circuit. 2 i( ) = = 1 A 2 Since current through the inductor cannot change instantaneously, i (0+) = 1 A For t > 0, the transformed network is shown in Fig. 11.47. 2s Applying KCL at the node for t > 0,

Solution

Fig. 11.46 + s

1

V ( s) + 2 V ( s) V ( s) + + =0 2s 1 s 3⎞ 1 ⎛ V ( s) ⎜ 1 + ⎟ = − ⎝ 2s ⎠ s

2

V (s) −

Fig. 11.47

1 − s =− 2 =− 1 V ( s) = 2s + 3 2 3 s +1 5 2s Taking the inverse Laplace transform, v (t) = − e−1.5 t

Example 11.82

i (0−)

for t > 0

In the network of Fig. 11.48, the switch is opened at t = 0. Find i(t). 10 Ω 3Ω 36 V

6Ω 0.1 H

i (t)

Fig. 11.48

iT (0−) 10 Ω

Solution At t = 0 , the network is shown in Fig. 11.49. At t = 0 , the switch is closed and steady-state condition is reached. Hence, the inductor acts as a short circuit. 36 36 iT ( ) = = =3A 10 + ( ) 10 + 2 6 iL ( ) 3 × =2A 6 3 −

–

3Ω 36 V −

iL (0 )

Fig. 11.49

6Ω

11.46 Network Analysis and Synthesis Since current through the inductor cannot change instantaneously, iL(0+) = 2 A For t > 0, the transformed network is shown in Fig. 11.50. Applying KVL to the mesh for t > 0, −0. − 0.1s

3

I( ) =

6

0.1s

( ) − 3 ( ) − 6I ( s) = 0 0. ( ) + ( ) = 0.2

I(s) 0.2

02 2 = 0.1 9 s + 90

Taking inverse Laplace transform, i(t) = 2e−90 t

Fig. 11.50

for t > 0

Example 11.83 The network shown in Fig. 11.51 has acquired steady-state with the switch closed for t < 0. At t = 0, the switch is opened. Obtain i (t) for t > 0. 10 Ω

4Ω

4Ω

36 V

2H i (t)

Fig. 11.51

Solution At t = 0−, the network is shown in Fig 11.52. At t = 0−, the switch is closed and the network has acquired steady-state. Hence, the inductor acts as a short circuit. 36 36 iT ( ) = = =3A 10 + ( ) 10 + 2 36 V 4 i( ) 3 × = 1.5 A 4 4 Since current through the inductor cannot change instantaneously, i(0+) = 1.5 A For t > 0, the transformed network is shown in Fig. 11.53. Applying KVL to the mesh for t > 0,

4Ω

10 Ω iT (0− ) 4Ω

i (0− )

Fig. 11.52 4

−4I ( s ) − 4I (s ) − 2sI ( s ) + 3 = 0 8

2s

( ) + 2sI ( s) = 3 I ( s) =

Taking the inverse Laplace transform, i(t) = 1. 5e−4 t

4

3 15 = 2s + 8 s + 4

I (s)

3

Fig. 11.53 for t > 0

Example 11.84 In the network shown in Fig. 11.54, the switch is closed at t = 0, the steady-state being reached before t = 0. Determine current through inductor of 3 H.

11.11 Resistor–Inductor Circuit 11.47 2Ω

2H

2Ω

1V i1 (t)

3H i2 (t)

Fig. 11.54

Solution At t = 0−, the network is shown in Fig. 11.55. At t = 0−, steady-state condition is reached. Hence, the inductor of 2 H acts as a short circuit. 1 i1 (0 ) = A 2 i2 (0 ) 0 Since current through the inductor cannot change instantaneously, 1 i1 (0 + ) = A 2 i2 (0 + ) 0 For t > 0, the transformed network is shown in Fig. 11.56. Applying KVL to Mesh 1, 1 1 s − 2 s 1 ( ) 1 2[ 1 ( ) 2 ( )] 0 s 1 ( 2 + 2s) I1 ( s) 2 2 ( s) = 1 + s Applying KVL to Mesh 2, −2 [I2(s) − I1(s)] − 2I2(s) − 3s I2 (s) = 0 −2I1(s) + (4 + 3s) I2(s) = 0 By Cramer’s rule, 2 + 2s 1 +

2Ω

1V i1 (0−)

Fig. 11.55

1

2s

2

2 I1 (s)

3s I2 (s)

Fig. 11.56

1 s

2 1 ( s 1) ( s + 1) −2 0 s +1 s +1 s 3 I 2 ( s) = = = = = 2 + 2s −2 ( 2 2 s)(4 ( 4 3s) − 4 s(3s 2 + 7s 7 s 2 ) 3s ⎛ s + 1 ⎞ ( s + 2 ) s ( s + 2 ) ⎛ s + 1 ⎞ ⎟ ⎟ −2 4 + 3s ⎝ ⎝ 3⎠ 3⎠ By partial-fraction expansion, I 2 ( s) =

A

B

A B C + + 1 s s+2 s+ 3 1 ( s +1) 3 s I 2 ( s) s = 0 = 1⎞ ⎛ ( s + 2) s + ⎟ ⎝ 3⎠ ( s + ) I 2 ( s) S = −2

1 (s + ) = 3 1⎞ ⎛ s s+ ⎟ ⎝ 3⎠

=

1 2

s=0

=− s = −2

1 10

11.48 Network Analysis and Synthesis

C

1 ( s +1) 1⎞ ⎛ s + ⎟ I 2 ( s) s = − 1 = 3 ⎝ 3⎠ s ( s + 2) 3

=− s= −

I 2 ( s) =

2 5

1 3

11 1 1 2 1 − − 1 2 s 10 s + 2 5 s+ 3

Taking inverse Laplace transform 1

i2 (t ) =

1 1 −2t 2 − 3 t − e − e for t > 0 2 10 5

Example 11.85 In the network of Fig. 11.57, the switch is closed at t = 0 with the network previously unenergised. Determine currents i1(t). 10 Ω

1H

1H 10 Ω

100 V i1 (t)

10 Ω

i2 (t)

Fig. 11.57 Solution For t > 0, the transformed network is shown in Fig. 11.58. 10

s

s 100 s

10 I1 (s)

I2 (s)

10

Fig. 11.58 Applying KVL to Mesh 1, 100 −10I 10 I1 ( s) sI1 ( s) 10 [ I1 s s

I2 s ] = 0

( s 20) I1 ( s) 100 I 2 ( s) = Applying KVL to Mesh 2, −10 [ −

100 0 s

10 0 I 2 ( s) = 0 ] − s I 2 ( s) −10 −10 I1 ( ) + ( s + 20 0) I 2 ( s) = 0

By Cramer’s rule, 100 −10 100 s ( s 20) 0 s 20 100( I1 ( s) = = s = s + 20 −10 ( s 20) 2 − 100 s(( 2 −10 s + 20

) )

=

100( s((

) )(

)

11.12 Resistor–Capacitor Circuit 11.49

By partial-fraction expansion, I1 ( s) = A

A B C + + s s + 10 s + 30 100( s 20) 2 20 s I1 ( s) |s 0 = = ( s 10)(s ( s 30) s= 0 3

B

( s + 10) I1 ( s) |s

10 =

C

( s + 30) I1 ( ) |s=−30 =

100( s + 20) = −5 s( s + 30) s=−10 100( s((

) )

=− s =−30

5 3

20 1 5 5 1 I1 ( ) = − − 3 s s + 10 3 s + 30 Taking inverse Laplace transform, 20 5 30 t i1 (t ) = − 5e −10 t e 3 3 Similarly, 100 s + 20 1000 s −10 0 1000 1000 s I 2 ( s) = = = = s + 20 −10 ( s 20) 2 100 s( s 2 + 40 s + 300) s( s +10)( s + 30) −10 s + 20 By partial-fraction expansion, A B C I 2 ( s) = + + s s + 10 s + 30 10 1000 A sI 2 ( s) |s 0 = = ( s 10)(s ( s 30) s = 0 3 B=(

) I 2 ( ) |s= s −10 =

1000 = −5 s(( ) s = −10

(

) I 2 ( ) |s= −30 =

1 1000 5 = s( s 10) s = −30 3

C I 2 ( s) =

10 1 5 5 1 − + 3 s s + 10 3 s + 30

Taking inverse Laplace transform, i2 (t ) =

11.12

10 − 5e −10 t 3

5 e 3

30 t

fo t > 0 R

RESISTOR–CAPACITOR CIRCUIT

Consider a series RC circuit as shown in Fig. 11.59. The switch is closed at time t = 0. For t > 0, the transformed network is shown in Fig. 11.60. Applying KVL to the mesh, V 1 − RI ( s) − I ( s) = 0 s Cs

V

C i (t)

Fig. 11.59

RC circuit

11.50 Network Analysis and Synthesis 1⎞ V ⎛ ⎜⎝ R + ⎟⎠ I ( s) = Cs s

R

V V s R I ( s) = = = 1 RCs C +1 1 R+ s+ Cs Cs RC Taking the inverse Laplace transform, V s

V s

Fig. 11.60

1

i( t ) =

Example 11.86

I (s)

V − RC t e R

1 Cs

Transformed network

fo t > 0

In the network of Fig. 11.61, the switch is moved from a to b at t = 0. Determine

i (t) and vc (t).

1Ω

10 V

a

6F

1Ω

b

+

3F

vc (t)

i (t) −

Fig. 11.61 −

At t = 0 , the network is shown in Fig. 11.62. At t = 0–, the network has attained steady-state condition. Hence, the capacitor of 6 F acts as an open circuit. 1Ω v6 F (0−) = 10 V i (0−) = 0 v3 F (0−) = 0 v6F (0 − ) Since voltage across the capacitor cannot change 10 V instantaneously, i (0 − ) v6 F (0+) = 10 V v3 F (0+) = 0 Fig. 11.62 For t > 0, the transformed network is shown in 11.63. Applying KVL to the mesh for t > 0, 1 10 1 1 − I ( s) − I ( s) − I ( s) = 0 s 6s 3s 1 1 1 10 I ( s) + I ( s) + I ( s) = 6s 1 V (s) 6s 3s s c 3s 10 60 10 I (s) 10 I ( s) = = = 1 1 ⎞ 6 s + 3 s + 0.5 s ⎛ s ⎜1 + + ⎟ ⎝ 6 s 3s ⎠

Solution

Taking the inverse Laplace transform, i(t) = 10e for t > 0 Voltage across the 3 F capacitor is given by 1 10 Vc ( s) = I ( s) = 3s 3s(s ( .5) −0.5t

Fig. 11.63

11.12 Resistor–Capacitor Circuit 11.51

By partial-fraction expansion, Vc ( s) =

A B + s s + 0.5 10

A

sV Vc ( s) s = 0 =

B

( s + .5)Vc ( s) s = −0.5 =

3(s (

.5) s = 0 10 3s

=

20 3 =−

s = −0.5

20 3

20 1 20 1 Vc ( s) = − 3 s 3 s + 0.5 Taking the inverse Laplace transform, 20 20 −0.5t − e 3 3 20 = ( e 0 5t ) 3

vc (t ) =

f

t>0

Example 11.87 The switch in the network shown in Fig. 11.64 is closed at t = 0. Determine the voltage cross the capacitor. 10 Ω

10 Ω 2 F

10 V

vc (t)

Fig. 11.64

Solution At t = 0 , the capacitor is uncharged. −

vc(0−) = 0 Since the voltage across the capacitor cannot change instantaneously, vc(0+) = 0 For t > 0, the transformed network is shown in Fig. 11.65. Applying KCL at the node for t > 0, 10 10 s Vc ( s) − V ( s ) V ( s ) s + c + c =0 1 10 10 2s 1 2 Vc ( ) 0.2 2V Vc ( s) = s 1 05 Vc ( s) = = s( 2 0.2) s( s 0.1) By partial-fraction expansion, Vc ( s) = A

A B + s s + 0.1 sV Vc ( s) s = 0 =

0.5 0.5 = =5 s + 0.1 s = 0 0.1

10

10

Fig. 11.65

1 2s

Vc (s)

11.52 Network Analysis and Synthesis B Vc ( s) =

( s + .1)Vc ( s) s = −0.1 =

0.5 0.5 =− = −5 s s = −0.1 0.1

5 5 − s s + 0.1

Taking inverse Laplace transform, vc ( t ) = 5 − 5e −0 1t

for t > 0

Example 11.88 In the network of Fig. 11.66, the switch is closed for a long time and at t = 0, the switch is opened. Determine the current through the capacitor. v (t) i1 (t)

i2 (t) 0.5 F

2A 1Ω

1Ω

Fig. 11.66 At t = 0−, the network is shown in Fig. 11.67. At t = 0−, the switch is closed and steady-state condition is reached. Hence, the capacitor acts as an open circuit. vc (0−) = 0

Solution

v (0 − ) vc (0 − ) 1Ω

2A 1Ω

Fig. 11.67 Since voltage across the capacitor cannot change instantaneously, vc (0+) = 0 For t > 0, the transformed network is shown in Fig. 11.68. Applying KVL to two parallel branches, 2 I1 ( s) + I1 ( s) = I 2 ( s) s Applying KCL at the node for t > 0, 2 = I1 ( s) + I 2 ( s) s 2 2 I1 ( s) + I1 ( ss)) = − I1 ( s) s s 2 2 I1 ( s) + 2 I1 ( s) = s s 2 1 I1 ( s) = s = 2 s +1 +2 s

V (s) I1(s)

2 s

2 s

I2(s) 1

1

Fig. 11.68

11.12 Resistor–Capacitor Circuit 11.53

Taking the inverse Laplace transform, i1 ( t ) = e − t

for t > 0

In the network of Fig. 11.69, the switch is moved from a to b, at t = 0. Find v(t).

Example 11.89

4Ω

a b

+

6V

v (t) −

1F

2Ω

2Ω

Fig. 11.69

Solution At t = 0−, the network is shown in Fig 11.70. At t = 0−, steady-state condition is reached. Hence, the capacitor acts as an open circuit. v(

)

Since voltage instantaneously,

6×

2 4 2

across

4Ω

=2V the

capacitor

cannot

change

v (0+) = 2 V For t > 0, the transformed network is shown in Fig. 11.71. Applying KCL at the node for t > 0, 2 V ( s) − V ( s) s + V ( s) = 0 + 1 6 2 s ⎛2 ⎞ V ( s ) ⎜ + s⎟ = 2 ⎝3 ⎠ V ( s) =

+ v (0−) −

6V

2Ω

Fig. 11.70 4

V (s) 1 s

2

2 s

2

Fig. 11.71

2 s+

2 3

Taking the inverse Laplace transform, v ( t ) = 2e

2 − t 3

fo t > 0

Example 11.90 The network shown in Fig. 11.72 has acquired steady-state at t < 0 with the switch open. The switch is closed at t = 0. Determine v (t). 2Ω +

4V

2Ω

1F

1F

v (t)

−

Fig. 11.72

11.54 Network Analysis and Synthesis

Solution At t = 0−, the network is shown in Fig 11.73. At

2Ω

t = 0–, steady-state condition is reached. Hence, the capacitor of 1 F acts as an open circuit. v(

)

2

4×

2Ω

4V

=2V

2 2 Since voltage across the capacitor cannot change instantaneously, v(0+) = 2 V For t > 0, the transformed network is shown in Fig. 11.74.

Fig. 11.73

2 1 s

4 s

2

1 v (s) s

2 s

Fig. 11.74 Applying KCL at the node for t > 0, 4 2 V ( s) − V ( s ) s+ s + V ( s) = 0 + 1 1 2 2 s s 2 2 sV ( s) V ( s) = + 2 s 2 +2 2s + 2 2 2 2 1 V ( s) = s = = − = − 2 s + 1 s( s ) s 2 s + 1 s s + 0.5

V ( s) −

Taking the inverse Laplace transform, v ( t ) = 2 − e −0 5t

11.13

fo t > 0

RESISTOR–INDUCTOR–CAPACITOR CIRCUIT

Consider a series RLC circuit shown in Fig. 11.75. The switch is closed at time t = 0. R

L V i (t)

Fig. 11.75 RLC circuit For t > 0, the transformed network is shown in Fig. 11.76.

C

v (0−)

11.13 Resistor–Inductor–Capacitor Circuit 11.55

Applying KVL to the mesh,

R

V 1 − RI ( s) Ls I ( s) − I ( s) = 0 s Cs 1⎞ V ⎛ ⎜⎝ R + Ls + ⎟⎠ I ( s) = Cs s

Ls

V s I (s)

⎛ LCs C 2 RCs RCs + 1⎞ V I ( s) = ⎜ ⎟ Cs s ⎝ ⎠

Fig. 11.76 Transformed network

V s I ( s) = = LCs C 2 + RCs + 1 s 2 + Cs

V V L L = R 1 ( s − s )( s − s2 ) s+ L LC

⎛ R⎞ ⎛ 1 ⎞ where s1 and s2 are the roots of the equation s 2 + ⎜ ⎟ s + ⎜ = 0. ⎝ L⎠ ⎝ LC ⎟⎠ 2

s1 = −

R 1 ⎛ R⎞ + ⎜ ⎟ − = −α + α 2 − ω 02 = −α + β ⎝ 2L ⎠ 2L LC

s2 = −

R 1 ⎛ R⎞ − ⎜ ⎟ − = −α − α 2 − ω 02 = −α − β ⎝ 2L ⎠ 2L LC

2

α=

where

ω0 = and β By partial-fraction expansion, of I(s), I ( s) =

A

B I ( s) =

R 2L 1 LC

α 2 − ω 02 A B + s s1 s s2 ( s s ) I ( s) s

s1

( s − s ) I ( s) s

s2

=

=

V L s1

s2 V L

s2

s1

=−

V

1 ⎤ ⎡ 1 − ⎢ s ) ⎣ s − s1 s − s2 ⎥⎦

V

⎡e s t s )⎣

L( s

V L s1

s2

Taking the inverse Laplace transform, i( t ) =

1 Cs

L( s

e s2t ⎤⎦ = k1 e s t + k2 e s2t

where k1 and k2 are constants to be determined and s1 and s2 are the roots of the equation.

11.56 Network Analysis and Synthesis Now depending upon the values of s1 and s2, we have 3 cases of the response. Case I

When the roots are real and unequal, it gives an overdamped response. R 1 > 2L LC > 0 In this case, the solution is given by i t) = e or

e

t

for t > 0

it

When the roots are real and equal, it gives a critically damped response. R 1 2L LC = 0 In this case, the solution is given by i(t) = e−at (k1 + k2 t) for t > 0 Case III When the roots are complex conjugate, it gives an underdamped response. R 1 < 2L LC < 0 In this case, the solution is given by it Case II

2 0

s1 2 = −

where 2

Let

2

where

2

=

= j

d

1

j

2

and Hence it

2

−

2

dt

e e

⎡ 2

−

Example 11.91

+

−

−

2j

−

for t > 0

The switch in Fig. 11.77 is opened at time t = 0. Determine the voltage v(t) for t > 0. +

2A

0.5 Ω

Fig. 11.77

0.5 H

0.5 F

v()

11.13 Resistor–Inductor–Capacitor Circuit 11.57

At t = 0−, the network is shown in Fig. 11.78. At t = 0−, the network has attained steady-state condition. Hence, the inductor acts as a short circuit and the capacitor acts as an open circuit.

Solution

+

iL (0 − ) 2A

v (0 − )

0.5 Ω

−

Fig. 11.78 iL(0−) = 0 v(0−) = 0 Since current through the inductor and voltage across the capacitor cannot change instantaneously, iL(0+) = 0 v(0+) = 0 For t > 0, the transformed network is shown in Fig. 11.79. Applying KCL at the node for t > 0, V ( s) V ( s) V ( s) + + = 1 0.5 0.5s 0.5s 2 2V ( s) + V ( s) + 0.. sV ( s) = s 2 s V ( s) = = 2 + 0.5s + 2 s

+

2 s

2 s

0.5

0.5s

2 s

1 0.5s

V(s)

−

4 2

s + 4s + 4

=

Fig. 11.79

4 (s

)

2

Taking inverse Laplace transform, v ( t ) = 4tt e −2t

for t > 0

Example 11.92 In the network of Fig. 11.80, the switch is closed and steady-state is attained. At t = 0, switch is opened. Determine the current through the inductor. 2.5 . Ω

5V

200 μF

0.5 H

Fig. 11.80

Solution At t = 0−, the network is shown in Fig. 11.81. At t = 0–, the switch is closed and steady-state condition is attained. Hence, the inductor acts as a short circuit and the capacitor acts as an open circuit. Current through inductor is same as the current through the resistor.

11.58 Network Analysis and Synthesis iL (

)=

2.5 Ω

5 =2A 2.5

Voltage across the capacitor is zero as it is connected in parallel with a short. vc (0−) = 0

vc (0−)

5V

iL (0−)

Fig. 11.81

Since voltage across the capacitor and current through the inductor cannot change instantaneously, iL (0+) = 2 A vc (0+) = 0

−

1 200 × 10 −6 s

0.5s

1 200 × 10−6s

For t > 0, the transformed network is shown in Fig. 11.82. Applying KVL to the mesh for t > 0,

I (s)

1

I ( s) − 0.5s I ( s) + 1 = 0 Fig. 11.82

I ( s) 0.5s I ( s) − 1 + 5000 =0 s I ( s) =

1 5000 0.5 5s + s

=

2s 2

s + 10000

Taking inverse Laplace transform, i ( t ) = 2 cos 100t

fo t > 0

Example 11.93 In the network shown in Fig. 11.83, the switch is opened at t = 0. Steady-state condition is achieved before t = 0. Find i(t). 0.5 H

1V

1Ω

1F i (t)

Fig. 11.83

Solution At t = 0−, the network is shown in Fig 11.84. At

t = 0−, the switch is closed and steady-state condition is achieved. Hence, the capacitor acts as an open circuit and the inductor acts as a short circuit. vc (0−) = 1 V i (0−) = 1 A Since current through the inductor and voltage across the capacitor cannot change instantaneously,

1Ω

1V i (0−)

Fig. 11.84

11.13 Resistor–Inductor–Capacitor Circuit 11.59

vc(0+) = 1 V i(0+) = 1 A For t > 0, the transformed network is shown in Fig. 11.85. Applying KVL to the mesh for t > 0,

0.5s

0.5

1 s 1 1

I(s)

s 1 1 − I ( s) − 0.5s I ( s) 0.5 I ( s) = 0 s s Fig. 11.85 1 1 0.5 + = I ( s) + 0.5 s I ( s) I ( s) s s 1 ⎡ 1 ⎤ I ( s) ⎢1 + + 0.5s ⎥ = 0.5 + s ⎣ s ⎦ s+2 (s ) 1 s +1 1 I ( s) = 2 = = + 2 2 s + 2s 2 s 2 ( s ) 1 ( s +1) + 1 ( s + 1) 2 + 1

Taking the inverse Laplace transform, i ( t ) = e −tt cos t + e t si t

for t > 0

Example 11.94 In the network shown in Fig. 11.86, the switch is closed at t = 0. Find the currents i1(t) and i2(t) when initial current through the inductor is zero and initial voltage on the capacitor is 4 V. 1Ω 1Ω

10 V i1 (t)

1Ω + 4V −

i2 (t)

1H

1F

Fig. 11.86 Solution For t > 0, the transformed network is shown in Fig. 11.87. 1 10 s

1

I1 (s)

s

1 1 s 4 s

I2 (s)

Fig. 11.87 Applying KVL to Mesh 1, 10 − I1 ( s) (1 s

)[

1

( s 2) I1 ( ) (

2

]

1) I 2 ( s) =

0 10 s

11.60 Network Analysis and Synthesis Applying KVL to Mesh 2, −( + ) [

1 4 ) − I2( ) − = 0 s s 1 4 ⎛ ⎞ −( + ) I1 ( ) + s + 2 + ⎟ I 2 ( s) = − ⎝ s⎠ s −

] − I2(

By Cramer’s rule, 10 ( s 1) −(s s 2 ⎛ 10 ⎞ ⎛ s + 2 s 1⎞ ⎛ 4⎞ 10 4 4 1 ⎜⎝ ⎟⎠ ⎜ (ss + 1) 2 ( s + 1) − s+ 2+ ⎟ − ( s + 1) ⎜⎝ s ⎟⎠ 2 s s ⎝ ⎠ s s s s I1 ( s) = = = 2 s+2 ( s 1) ⎛ s 2 + 2s ⎞ ( s +1) 2s 1 2 ( s + 2) − ( s + 1) 2 ( s + 2) ⎜ ⎟ − ( s + 1) 1 s s ⎝ ⎠ −(s ( s 1) s + 2 + s 10 4 ( s +1) − 2 3s + 5 s s = = ( s +1) s(( ) ( s + 2) −( ) s By partial-fraction expansion, A B I1 ( s) = + s s +1 3s 5 A sI1 ( s) |s 0 = =5 s + 1 s=0 B I1 ( s) = Taking inverse Laplace transform, Similarly,

( s 1) I1 ( s) |s

1=

3s + 5 = −2 s s = −1

5 2 − s s +1

i1 ( t ) = 5 2e − t

fo t > 0

10 s 4 −(s ( s 1) − 3s 1 3s 3 − 2 3( + 1) − 2 3 2 s I 2 ( s) = = = = = − 2 2 2 s+2 ( s 1) s + 1 ( +1) 2 ( s 1) ( +1) ( +1) 1 −(s ( s 1) s + 2 + s Taking inverse Laplace transform, i2 (t ) = 3e 3e t 2te t fo t > 0 s+2

11.14

RESPONSE OF RL CIRCUIT TO VARIOUS FUNCTIONS

Consider a series RL circuit shown in Fig. 11.88. When the switch is closed at t = 0, i(

)

i( i(

+

) = 0.

11.14 Response of RL Circuit to Various Functions 11.61 R

v (t)

+ −

L i(t)

Fig. 11.88 RL circuit R

For t > 0, the transformed network is shown in Fig. 11.89. Applying KVL to the mesh, V ( s) − R I (s ( s) Ls I ( s) = 0 V ( s) 1 V ( s) I ( s) = = R R Ls L s+ L (a) When the unit step signal is applied, v ( t ) = u( t ) Taking Laplace transform, 1 V ( s) = s 1 1 s I ( s) = R L s+ L 1 1 = R⎞ L ⎛ s s+ ⎟ ⎝ L⎠ By partial-fraction expansion, ⎛ ⎞ 1⎜A B ⎟ I ( s) = ⎜ + R⎟ L s s+ ⎟ ⎜ ⎝ L⎠ =

1

V (s) + −

Fig. 11.89 Transformed network

=

L R

A

s I ( s)

B

R⎞ 1 L ⎛ s + ⎟ I ( s) s = − R = =− ⎝ s s= − R L⎠ R L

s=0

R s+ L

s=0

L

⎛ ⎞ 1⎜L1 L 1 ⎟ I ( s) = ⎜ − R⎟ L Rs R s+ ⎟ ⎜ ⎝ L⎠ ⎛ ⎞ 1 ⎜1 1 ⎟ = ⎜ − R⎟ R s s+ ⎟ ⎜ ⎝ L⎠

Ls I (s)

11.62 Network Analysis and Synthesis Taking inverse Laplace transform, i( t ) = (b)

1 [ R

e

⎛ R⎞ −⎜ ⎟ t ⎝ L⎠

When unit ramp signal is applied, v(t ) = r (t ) = t Taking Laplace transform, V ( s) =

]

f

t>0

for t > 0 1

s2 1 I ( s) = L

1 R⎞ ⎛ s2 s + ⎟ ⎝ L⎠

By partial-faction expansion, 1 L

1 R⎞ ⎛ s2 s + ⎟ ⎝ L⎠

=

A B C + 2+ R s s s+ L

1 R⎞ R⎞ ⎛ ⎛ = As s + ⎟ + B s + ⎟ + Cs 2 ⎝ ⎝ L L⎠ L⎠ Putting s = 0,

Putting s = −

R , L

Comparing coefficients of s2,

B=

1 R

C=

L R2

A C=0 A

C=−

L

R2 L 1 1 1 L 1 I ( s) = − 2 + + 2 2 R s Rs R s+ R L Taking inverse Laplace transform, i( t ) = −

L R2

⎛ R⎞

+

1 L −⎜ ⎟ t t + 2 e ⎝ L⎠ R R

1 L = t− 2[ R R (c) When unit impulse signal is applied, v(t ) = δ (t ) Taking Laplace transform, V ( s) = 1 1 1 I ( s) = R L s+ L

e

⎛ R⎞ −⎜ ⎟ t ⎝ L⎠

]

f

t>0

11.14 Response of RL Circuit to Various Functions 11.63

Taking inverse Laplace transform, ⎛ R⎞

1 −⎜ ⎟ t i( t ) = e ⎝ L ⎠ L

fo t > 0

Example 11.95 At t = 0, unit pulse voltage of unit width is applied to a series RL circuit as shown in Fig. 11.90. Obtain an expression for i(t). 1Ω

v (t )

+ −

v (t)

1

1H i (t)

0

t

1

Fig. 11.90 Solution

v(t ) = u ( t ) − u ( t − 1)

1 e −ss 1 − e s − = s s s For t > 0, the transformed network is shown in Fig. 11.91. Applying KVL to the mesh, V ( s) − I ( s) − sI ( s) = 0 V ( s) I ( s) = s +1 1 − e−s = s( s ) V ( s) =

1

V (s)

+ −

s I (s)

Fig. 11.91

−s

=

1 e − s( s ) s( s + 1)

=

1 1 e−s e−s − − + s s +1 s s +1

Taking inverse Laplace transform, i(t ) = u(t ) − e −tt u(t ) u(t u( t =(

t

e )u(t (t ) [

) e

e

( t −1)

(t

)

u(t (t

)

] (t − 1)

f

>0

Example 11.96 For the network shown in Fig. 11.92, determine the current i(t) when the switch is closed at t = 0. Assume that initial current in the inductor is zero. 5Ω

r (t − 3) + −

2H i(t)

11.64 Network Analysis and Synthesis Fig. 11.92

5

Solution For t > 0, the transformed network is shown in Fig. 11.93. Applying KVL to the mesh for t > 0, e

−3 s

s2

− 5 I ( s) − 2 s I ( s) = 0 5 I ( s) + 2 s I ( s) = I ( s) =

e−3s s2

+ −

e −3s

2s I (s)

Fig. 11.93

s2 e −3s

=

s 2 ( 2 s + 5)

0 5 e −3s s2 (

2.5)

By partial-fraction expansion, 05 2

s (

2.5)

=

A B C + 2+ s s s+2 5

0.5

((ss 2.5 5)) B( = As 2 + 2.5 A =(

2

)s2

2.5) Cs 2 2.5 B C Cs 2

( .5 .5 A + B))

.5 B

0

Comparing coefficients of s , s and s , A C=0 2 5A + B = 0 2.5 B = 0.5 Solving these equations,

A = −0 08 B=02 C = 0 08 ⎛ 0.08 0.2 0.08 ⎞ I ( s) = e −3s − + 2 + ⎟ ⎝ s s + 2 5⎠ s = −0.08

e −33 e 3s e −3s + 0.2 2 + 0.08 s s + 2.5 s

Taking inverse Laplace transform, i(t ) = −0.. u(t − 3)

.2r (t

) + 0. e

2 ( t 3)

u(t − 3)

Example 11.97 Determine the expression for vL (t) in the network shown in Fig. 11.94. Find vL(t) when (i) vs(t) = d (t), and (ii) vs(t) = e−t u(t). 5Ω + Vs (t )

+ −

1 v (t ) 2H L −

Fig. 11.94

11.14 Response of RL Circuit to Various Functions 11.65

Solution For t > 0, the transformed network is shown in Fig. 11.95. By voltage-division rule, VL ( s) = Vs ( s) × (a)

5

s 2 s +5 2

=

s Vs ( s) s + 10

+ Vs (s )

For impulse input, Vs ( s) = 1

+ −

s V (s ) 2 L −

s s + 10 − 10 10 = = 1− s + 10 s + 10 s + 10 Taking inverse Laplace transform, VL ( s) =

VL (t ) = δ (t (t ) 10e −10 t u(t ) (b)

Fig. 11.95

for t > 0

For vs (t ) = e − t u(t ), Vs ( s) = VL ( s) =

1 s +1 (s

s ))(s (s

)

By partial-fraction expansion, VL ( s) =

A B + s + 10 s + 1

A

(s

B

(s

)VL ( s) s = −10 = )VL ( s) s = −1 =

s 10 = s + 1 s = −10 9

s s + 10

=− s = −1

1 9

10 1 1 1 VL ( s) = − 9 s + 10 9 s + 1 Taking inverse Laplace transform, 10 −10 t 1 t e u(t u( t ) e u( t ) 9 9 ⎛ 10 −10 t 1 t ⎞ = e e ⎟ u( t ) ⎝ 9 9 ⎠

vL (t ) =

for t > 0

Example 11.98 For the network shown in Fig. 11.96, determine the current i (t) when the switch is closed at t = 0. Assume that initial current in the inductor is zero. 2Ω

2d (t − 3)

+ −

1H i (t)

Fig. 11.96

11.66 Network Analysis and Synthesis

Solution For t > 0, the transformed network is shown in Fig. 11.97.

2

Applying KVL to the mesh for t > 0, 2e

3s

2 I ( s) sI ( s) = 0 2 ( s) s ( s) I ( s) =

2e

2e−3s

−3 s

Fig. 11.97

Taking inverse Laplace transform,

Example 11.99

s I (s)

2e −3s s+2

i(t ) = 2ee −2(t(t(

+ −

)

u(t (

fo t > 0

)

Determine the current i(t) in the network shown in Fig. 11.98, when the switch is

closed at t = 0.

10 Ω

50 sin 25 t

5H i (t)

Fig. 11.98 10

Solution For t > 0, the transformed network is shown in Fig. 11.99. Applying KVL to the mesh for t > 0, 1250 − 10I 10 I ( s) − 5 ( ) 0 2 s + 625 I ( s) =

1250 s 2 + 625

5s I (s)

250 Fig. 11.99

2

( s + 625)( s + 2)

By partial-fraction expansion, I ( s) =

As + B 2

s + 625

+

250 = ( As B)( s = ( A C )s Comparing coefficients,

2

C s+2 ) + C ( s2 ( A + B) s ((2 B

) C)

A C=0 2A + B = 0 2B C = 250

Solving the equations, A = −0.397 B = 0.795 C = 0.397 −0.397 s + 0.795 0.397 0.397 s 0.795 0.397 I ( s) = + =− 2 + + s+2 s 2 + 625 s + 625 s 2 + 625 s + 2

11.14 Response of RL Circuit to Various Functions 11.67

Taking the inverse Laplace transform, i(t ) = −0.397 cos 25t 0.032 si 25t + 0.397e −2t

Example 11.100

fo t > 0

Find impulse response of the current i(t) in the network shown in Fig. 11.100. 1Ω

i1 (t)

i (t) d (t)

+ −

1Ω 2H

Fig. 11.100

Solution The transformed network is shown in Fig. 11.101. 1(( s ) 2 s + 1 = 2s + 1 + 1 2s + 2 V ( s) 1 2s 2 I1 ( s) = = = Z ( s) 2 s 1 2 s 1 2s 2 By current-division rule, 1 1 2s + 2 1 1 1 I ( s) = I1 ( s) × = × = = 2 s + 2 2 s + 2 2 s + 1 2 s + 1 2 s + 0.5 Taking inverse Laplace transform, 1 i(t ) = e −0.5t u(t (t ) fo t > 0 2

1

I1 (s)

I (s)

Z ( s) =

1

+ −

1 2s

Fig. 11.101

Example 11.101

The network shown in Fig. 11.102 is at rest for t < 0. If the voltage v(t ) = u(t ) cos t Aδ (t ) is applied to the network, determine the value of A so that there is no transient term in the current response i(t). 1Ω

v (t)

2H i (t)

Fig. 11.102 v(t ) = u(t ) cos t Aδ (t ) s V ( s) = 2 +A s +1

11.68 Network Analysis and Synthesis

Solution For t > 0, the transformed network is shown in Fig. 11.103.

1

Applying KVL to the mesh for t > 0, V ( s) = 2 sI (s ( s) I ( s) = I ( s) =

s s2 + 1

+A

V (s)

2s I (s)

s A A(s ( s2 ) K K s + K3 = 1 + 22 1 1⎞ ⎛ s +1 2 ⎜ s + ( s2 ) s + ⎝ 2 2⎠

Fig. 11.103

The transient part of the response is given by the first term. Hence, for the transient term to vanish, K1 = 0.

K1

When

11.15

−1 ⎛ 5⎞ + A⎜ ⎟ ⎝ 4⎠ 1 2 ⎛ ⎞ s + ⎟ I ( s) s = − 1 = ⎝ 2⎠ ⎛ 5⎞ 2 2⎜ ⎟ ⎝ 4⎠

K1 = 0 5 1 A= 4 2 2 A= =04 5

RESPONSE OF RC CIRCUIT TO VARIOUS FUNCTIONS

Consider a series RC circuit as shown in Fig. 11.104. R

v (t)

+ −

C i (t)

Fig. 11.104 RC circuit For t > 0, the transformed network is shown in Fig. 11.105. Applying KVL to the mesh, V ( s) − RI ( s) −

(a)

1 I ( s) = 0 Cs V ( s) sV ( s) I ( s) = = 1 1 ⎞ + R R⎛s + ⎟ ⎝ Cs RC ⎠

When unit step signal is applied, v ( t ) = u( t )

R

1 Cs

V (s) + − I (s)

Fig. 11.105 Transformed network

11.15 Response of RC Circuit to Various Functions 11.69

Taking Laplace transform, 1 s

V ( s) =

1 1 s I ( s) = = 1 ⎞ 1 ⎞ ⎛ ⎛ R s+ ⎟ R⎝s + ⎟ ⎝ RC ⎠ RC ⎠ s×

Taking inverse Laplace transform, 1

1 − RC t e R

i( t ) = (b)

fo t > 0

When unit ramp signal is applied, v(t ) = r (t ) = t Taking Laplace transform, V ( s) =

1 s2

1 1 2 s R I ( s) = = 1 ⎞ 1 ⎞ ⎛ ⎛ R s+ ⎟ s⎝s + ⎟ ⎝ RC ⎠ RC ⎠ s×

By partial-fraction expansion, I ( s) =

A

B

A + s

B s+

s I ( s)

1 RC

s=0

1 R

= s+

1 RC

=C s=0

1 1 ⎞ ⎛ R s+ ⎟ I ( s) s = − 1 = ⎝ RC ⎠ s R RC

= −C s= −

I ( s) =

C − s

1 RC

C s+

1 RC

Taking inverse Laplace transform, i(t ) = C Ce (c)

When unit inpulse signal is applied, v(t) = d (t)

−

1 t RC

fo t > 0

11.70 Network Analysis and Synthesis Taking Laplace transform, V ( s) = 1 1 1 1 ⎞ ⎛ − 1 ⎜ ⎟ RC RC = I ( s) = = 1 − RC ⎟ ⎜ 1 1 ⎞ 1 ⎞ R ⎛ ⎛ ⎜ s+ ⎟ R s+ R s+ ⎟ ⎟ ⎝ ⎝ ⎝ RC ⎠ RC ⎠ RC ⎠ s+

s

Taking inverse Laplace transform, 1⎡ 1 − RC t ⎤ ⎢δ (t ) − ⎥ e R⎢ RC ⎥⎦ ⎣ 1

i( t ) =

fo t > 0

Example 11.102 A rectangular voltage pulse of unit height and T-seconds duration is applied to a series RC network at t = 0. Obtain the expression for the current i(t). Assume the capacitor to be initially uncharged. v (t )

R

1

0

v (t) T

+ −

C i (t)

t

(a)

(b)

Fig. 11.106 v(t) = u(t) − u(t − T) sT 1 e −sT 1 − e − sT V ( s) = − = s s s

Solution

R

V (s) + −

I (s) For t > 0, the transformed network is shown in Fig. 11.107. Applying KVL to the mesh for t > 0, 1 Fig. 11.107 V ( s) − RI ( s) − I ( s) = 0 Cs 1 ⎡ ⎤ s V ( s) 1 − e − sT 1⎢ 1 e − sT ⎥ R I ( s) = = V ( s) = = ⎢ − ⎥ 1 1 1 1 ⎥ 1 ⎞ R⎢ ⎛ R+ s+ s + s + R s+ ⎟ ⎝ Cs RC RC RC ⎦ ⎣ R ⎠ RC

1 Cs

Taking inverse Laplace transform, ⎛ 1 ⎞ ⎡ ⎛ 1⎞ −⎜ ⎟ (t 1 ⎢ − ⎝⎜ RC ⎠⎟ t i( t ) = e u(t ) e ⎝ RC ⎠ u(t R⎢ ⎣

T)

⎤ u( t T ) ⎥ ⎥ ⎦

fo t > 0

Example 11.103 For the network shown in Fig. 11.108, determine the current i(t) when the switch is closed at t = 0 with zero initial conditions.

11.15 Response of RC Circuit to Various Functions 11.71 3Ω

+ −

2r (t − 2)

1F i (t)

Fig. 11.108 Solution For t > 0, the transformed network is shown in Fig. 11.109. Applying KVL to the mesh for t > 0, 2e −2 s s2

3

2e−2s + − s2

1 − 3 I ( s) − I ( s) = 0 s 1⎞ 2e −2 s ⎛ 3 + I ( s ) = ⎜⎝ ⎟ s⎠ s2 I ( s) =

I(t)

1 s

Fig. 11.109

2e −2 s 0 67e −2 s = 1 ⎞ s( s + 0.33) ⎛ s2 ⎜ 3 + ⎟ ⎝ s⎠

By partial-fraction expansion, 0 67 A = + s(( 0.33)

B 0.33

A=

0 67 =2 s + 0 33 s = 0

B=

0 67 = −2 s s = −0 33

I ( s) = e −2 s

2 ⎞ e −2 s e 2s ⎛2 − = 2 − 2 ⎝ s s + 0.33 ⎟⎠ s s + 0.33

Taking inverse Laplace transform, ) − 2ee −0.

i(t ) = 2u(t u( t

((tt − 2 )

u( t u(t

fo t > 0

)

Example 11.104 For the network shown in Fig. 11.110, determine the current i(t) when the switch is closed at t = 0 with zero initial conditions. 5Ω

d (t)

+ −

2F i (t)

Fig. 11.110 Solution For t > 0, the transformed network is shown in Fig. 11.111.

11.72 Network Analysis and Synthesis Applying KVL to the mesh for t > 0, 1 I ( s) = 0 2s 1⎞ ⎛ ⎜⎝ 5 + ⎟⎠ I ( s) = 1 2s

1 5 I ( s) −

5

1

I ( s) =

1 2s 2s = 10 s + 1 0 2s = s+0 1 0.2( s + 0.1 − 0.1) = s+0 1 01 ⎞ ⎛ = 0 2 1− ⎝ s + 0 1⎟⎠ 0 02 = 0 2− s + 0.1 5+

1

+ −

I(s)

1 2s

Fig. 11.111

Taking inverse Laplace transform, .02 e −0.1t u(t )

i(t ) = 0. 0. δ (t )

Example 11.105

For the network shown in Fig. 11.112, find the response v0 (t). 2Ω +

vs (t) =

1 cost u(t ) 2

+ −

1 F 4

vo (t) −

Fig. 11.112

Solution For t > 0, the transformed network is shown in Fig. 11.113. 1 s 2 s2 + 1 By voltage-division rule, 4 2V ( s) Vo ( s) = Vs ( s) × s = s = 2 4 s+2 (s 2+ s By partial-fraction expansion,

2

Vs ( s) =

Vo ( s) = s s

+ Vs (s)

+ −

4 s Vo (s)

s ))( )(s (s As + B 2

−

)

+

C s+2

s +1 ( As B)( s 2

Fig. 11.113

) + c(s ( s2

)

( A C ) s + ( 2 A B) s + (

)

11.15 Response of RC Circuit to Various Functions 11.73

Comparing coefficient of s2, s and s0, A C=0 2A + B = 1 2B C = 0 Solving the equations, A=04 B=02 C = −0 4 0.4 s + 0.2 0.4 0.4 s 02 04 Vo ( s) = − = 2 + 2 − 2 s + 2 s +2 s +1 s +1 s +1 Taking the inverse Laplace transform, i(t ) = 0.4 cos t 0.2 si t − 0.4e

2t

fo t > 0

Example 11.106 Find the impulse response of the voltage across the capacitor in the network shown in Fig. 11.114. Also determine response vc (t) for step input. 2Ω

v (t)

1H

+

+ −

−

1 F vc (t)

Fig. 11.114

Solution For t > 0, the transformed network is shown in Fig. 11.115. 2

By voltage-division rule, Vc ( s) = V ( s) ×

1 s

1 2+ s+ s V ( s) V ( s) = 2 = s + 2s + 1 ( s )2

(a)

For impulse input,

V (s)

1 )2

(s

Taking inverse Laplace transform, vc (t ) = te − t u(t ) (b)

for t > 0

For step input, V ( s) = Vc ( s) =

+ 1 Vc (s ) s −

+ −

Fig. 11.115

V ( s) = 1 Vc ( s) =

s

1 s 1 s( s

)2

11.74 Network Analysis and Synthesis By partial-fraction expansion, Vc ( s) =

A B C + + s s + 1 ( s )2 ) 2 + Bs(s Bs( s

1 = A( s

) Cs C

s

) B ( s 2 + s ) Cs

= ( A B) s2

( A B C )s A

= A( s

2

Comparing coefficient of s2, s1 and s0, A=1 A+ B = 0 B A = −1 2A + B + C = 0 A − B = −2 + 1 = −1 1 1 1 Vc ( s) = − − s s + 1 ( s )2 C

Taking inverse Laplace transform, vc (t ) = u(t ) − e −tt u(t ) te t u(t ) =(

e

t

t t )utt te

f

t>0

Example 11.107 For the network shown in Fig. 11.116, determine the current i(t) when the switch is closed at t = 0 with zero initial conditions. 5Ω

5r (t − 1)

1H

+ −

1 F 6

i (t)

Fig. 11.116

Solution For t > 0, the transformed network is shown in Fig. 11.117. Applying KVL to the mesh for t > 0, 5e − s s2

5

6 − 5I 5 I ( s) − sI ( s) − I ( s) = 0 s 5 I ( s) + sI ( s) +

6 5e − s I ( s) = 2 s s I ( s) =

5e−s s2

5e 2

−ss

s( s + s + )

s

+ − I (s)

s

=

5e s( s + )(s )( s + )

Fig. 11.117

6 s

11.15 Response of RC Circuit to Various Functions 11.75

By partial-fraction expansion, 1 s( s 3)(

A B C + + 2) s s + 3 2 1 1 A= = ( s 3)( 2) s = 0 6 =

B=

1 1 = s( s 2) s = −3 3

C=

1 1 =− s( s + 3) s = −2 2

⎡1 1 1 ⎤ 5 e−s 5 e−s 5 e−s I ( s) = 5e 5e − s ⎢ + − + − ⎥= 3 s+3 2 s+2 ⎣ 6 s 3( 3) 2( s 2) ⎦ 6 s Taking inverse Laplace transform, i( t ) =

5 u( t 6

5 ) + e −3(t(t( 3

)

u( t u(t

)

5 e 2

2(t (t

)

u( t

)

f

t>0

Example 11.108 For the network shown in Fig. 11.118, the switch is closed at t = 0. Determine the current i(t) assuming zero initial conditions. 2Ω

1H

sin t

0.5 F i(t)

Fig. 11.118 Solution For t > 0, the transformed network is shown in Fig. 11.119. Applying KVL to the mesh for t > 0, 1 2

s +1

− 2I 2 I ( s) − s I ( s) −

2 I ( s) = 0 s

1 s2 + 1

2⎞ 1 ⎛ ⎜⎝ 2 + s + ⎟⎠ I ( s) = 2 s s +1 I ( s) =

s 2

( s + 1)( s

s

2

I (s)

Fig. 11.119

2

2 s + 2) 2s

By partial-fraction expansion, I ( s) = s

As + B 2

s +1

+

Cs + D 2

s + 2s + 2

( As B)( s 2 = As3 + 2 As 2

s

) + (C ((Cs Cs D )( s 2 C

)

2 As + Bs 2 + 2 Bs + 2 B + Cs3 + Cs + Ds 2 + D

= ( A + C ) s3 + ( 2 A

)s2

(2 A + 2 B + C )s + (2 B + D)

2 s

11.76 Network Analysis and Synthesis Comparing coefficients of s3, s2, s1 and s0, A C=0 2A + B + D = 0 2A 2B + C = 1 2B + D = 0 Solving these equations, A = 0.2, B = 0.4, C = −0.2, D = −0.8 I ( s) =

0.2 s + 0.4

−

0.2 s + 0.8

s +1 +1 s2 + 2s + 2 0.2 s 0.4 0.2 s + 0.2 0.6 = 2 + − s + 1 s 2 + 1 ( +11) 2 (1) 2

=

2

0.2 s 2

s +1 +1

+

0.4 2

0.2( s + 1)

−

2

−

06

s +1 + 1 ( + 1) + 1 ( + 1) 2 + 1

Taking inverse Laplace transform, i(t ) = 0. 0.2 cos t 0.4 si t − 0.2 e −tt cos t − 0.6 e

t

si t

−t

= 0.2 cos t 0.4 si t − e ( .2 cos t + 0.6 sin t )

fo t > 0

Example 11.109 For the network shown in Fig. 11.120, the switch is closed at t = 0. Determine the current i(t) assuming zero initial conditions in the network elements. 5Ω

6e−2t

1H

+ −

0.25 F i(t)

Fig. 11.120 Solution For t > 0, the transformed network is shown in Fig. 11.121. Applying KVL to the mesh for t > 0, 6 4 − 5I 5 I ( s) − s I ( s) − I ( s) = 0 s+2 s 4⎞ 6 ⎛ ⎜⎝ 5 + s + ⎟⎠ I ( s) = s s+2 I ( s) = =

s

5

6 s+2

+ −

I (s)

6s ( s + 2)(

2

)

6s (

)( )(

)(

)

Fig. 11.121

4 s

11.15 Response of RC Circuit to Various Functions 11.77

By partial-fraction expansion, I ( s) =

A C B + + s + 2 s +1 s + 4

6s =6 ( s )( )( s ) s = −2 )(s 6s B ( s + 1) ( s) |s = −1 = = −2 ( + 2)( + 4) s = −1 6s C = ( + 4) I ( ) |s = −4 = = −4 ( + 2)( s + 1) s = −4 6 2 4 I ( s) = − − s + 2 s +1 s + 4 Taking inverse Laplace transform, A

(s

) I ( s) |s = −2 =

i(t ) = 6e −2t u(t uu((t (t ) 2e t u(t ) − 4 e −4 t u(t (t )

f

t>0

Example 11.110 The network shown has zero initial conditions. A voltage vi(t) = d (t) applied to two terminal network produces voltage vo(t) = [e−2 t + e−3 t] u(t). What should be vi(t) to give vo(t) = t e−2 t u(t)? + vi (t ) −

Network

+ vo (t ) −

Fig. 11.122

Solution For vi(t) = d (t),

Vi ( s) = 1 vo (t ) = [e [e

2t

3t

]u(t ) 1 1 Vo ( s) = + s+2 s+3 System function H ( s) =

e

Vo ( s) Vi ( s) =

1 1 2 +5 + = s + 2 s + 3 ( s + 2)( + 3)

For vo (t ) = te −2t u(t ), Vo ( s) =

1 (s

)2

From Eq. (i), Vi ( s) = By partial-fraction expansion,

Vo ( s) 1 ( s )(s )( ) (s ) = × = 2 H ( s) ( s ) 2s + 5 2(s ( s .5)( s + 2)

A B + s + 2 s + 2.5 A=1 B = −0.5 1 0.5 Vi ( s) = − s + 2 s + 2.5 Vi ( s) =

…(i)

11.78 Network Analysis and Synthesis Taking inverse Laplace transform, vi (t ) = e −22tt

0.5e

2.5t

for t > 0

Example 11.111 A unit impulse applied to two terminal black box produces a voltage vo(t) = 2e−t −e−3t. Determine the terminal voltage when a current pulse of 1 A height and a duration of 2 seconds is applied at the terminal. + is (t)

vo (t )

Black box

−

Fig. 11.123 Solution

vo (t ) = 2e −tt

e

3t

is (t )

2 1 Vo ( s) = − s +1 s + 3 When is (t ) = δ (t ),

1

I s ( s) = 1

0

Vo ( s) = Z ( s) I s ( s) Z ( s) =

…(i)

t

2

Fig. 11.124

Vo ( s) 2 1 = − I s ( s) s + 1 s + 3

When is (t) is a pulse of 1 A height and a duration of 2 seconds then, is (t ) = u(t ) − u(t I s ( s) =

)

−2 s

1 e − s s

From Eq. (i), 1 ⎤ ⎡ 1 e −2 s ⎤ ⎡ 2 Vo ( s) = ⎢ − ⎥⎢ − s ⎥ ⎣ s + 1 s + 3⎦ ⎣ s ⎦ =

2 1 2e −2 s e −2 s − − + s( s ) s( s ) s( ) s(( )

1 ⎤ 1 ⎡1 1 ⎤ 1 ⎤ e −2 s ⎡ 1 1 ⎤ ⎡1 −2 s ⎡ 1 = 2⎢ − − − − 2 e − + − ⎥ ⎢ ⎥ ⎢ ⎥ 3 ⎣ s s + 3 ⎥⎦ ⎣ s s + 1⎦ 3 ⎣ s s + 3 ⎦ ⎣ s s + 1⎦ Taking the inverse Laplace transform, v(t ) = 2[u [u(t ) e t u(t )]

1 [u u((t ) e 3

3t

u(t )] − 2[[u u ( t − 2) e

(t (t

)

1 u(t − 2)] + [ (t − 2) − 3

−33( − 2 )

(t − 2)] for t > 0

Exercises 11.79

Exercises ⎛ 1 − cos 2t ⎞ 11.1 Find L{ f ′(t )} of f (t ) = ⎜⎝ ⎟⎠ t ⎡ ⎛ s2 + 4 ⎞ ⎤ ⎢ s log ⎜ ⎟⎥ ⎜⎝ s ⎟⎠ ⎥ ⎢ ⎣ ⎦

`

11.2 Find Laplace transform of the follwoing function: 2 f (t ) = t + 1 0 t>2 =3 ⎡1 ⎢ s (1 e ⎣

1 Ω 8

v (t )

2s

⎤ )⎥ ⎦

11.3 For the network shown in Fig 11.125, the switch is closed at t = 0. Find the current i1(t) for t > 0.

1 Ω 2

10 A

1F

Fig. 11.127 [v(t) = 1 + 4 e−10t] 11.6 The circuit of Fig. 11.128, has been in the condition shown for a long time. At t = 0, switch is closed. Find v(t) for t > 0. 5Ω

v (t )

20 V

3Ω

2F

100 Ω

50 Ω

100 V

Fig. 11.128 [v(t) = 7.5 + 12.5 e−(4/15)t]

4H

i1 (t )

Fig. 11.125 [i1(t) = 3 − e−25 t]

11.7 Figure 11.129 shows a circuit which is in the steady-state with the switch open. At t = 0, the switch is closed. Determine the current i (t). Find its value at t = 0.114 μ seconds. 800 Ω

11.4 Determine the current i(t) in the network of Fig. 11.126, when the switch is closed at t = 0. The inductor is initially unenergized.

i (t )

12 V

2Ω

400 Ω

0.001 μF

200 Ω

i (t )

2Ω

24 V

2Ω

0.5 H

Fig. 11.129 6

[i(t) = 0.00857 + 0.01143 e−8.75 × 10 t, 0.013 A] 11.8 Find i(t) for the network shown in Fig. 11.130. i (t ) 10 Ω

Fig. 11.126 [i(t) = 4(1 − e−6t)] 11.5 In the network of Fig. 11.127, after the switch has been in the open position for a long time, it is closed at t = 0. Find the voltage across the capacitor.

1F 50 V

5Ω

Fig. 11.130

0.5 F 5Ω

11.80 Network Analysis and Synthesis [i(t) = 0.125 e−0.308t + 3.875 e−0.052t] 11.9 Determine v(t) in the network of Fig. 11.131 where iL(0−) = 15 A and vc(0−) = 5 V.

10 Ω

2H

1H

20 Ω

10 V

0.5 H

30 Ω

+ 0.33 Ω

10 V

1 F v (t )

Fig. 11.134 −

Fig. 11.131 [v(t) = 10 − 10e−t + 5e−2t] 11.10 The network shown in Fig. 11.132 has acquired steady state with the switch at position 1 for t < 0. At t = 0, the switch is thrown to the position 2. Find v(t) for t > 0. 2Ω

1

[i(t) = 0.1818 − 0.265 e−13.14t + 0.083 e−41.86t ] 11.13 The network shown in Fig. 11.135 is in steady state with s1 closed and s2 open. At t = 0, s1 is opened and s2 is closed. Find the current through the capacitor. 2Ω

2H

s1

3H

10 V

2

s2

1 μF

+ 3Ω 2V

v (t )

0.5 F

1H −

Fig. 11.135 [i(t) = 5 cos (0.577 × 103 t)] 11.14 In the network shown in Fig. 11.136, find currents i1(t) and i2(t) for t > 0. 10 Ω

Fig. 11.132 [v(t) = 4e−t − 2 e−2t] 11.11 In the network shown in Fig. 11.133, the switch is closed at t = 0. Find current i1(t) for t > 0. 3Ω

20 V

i1 (t )

1H

0.2 F

50 V i1 (t )

40 Ω

i2 (t )

1Ω

1 Ω i (t ) 2

1 F 3

Fig. 11.136 [i1(t) = 5 e−0.625t, i2(t) = 1 − e−0.625t ] 11.15 For the network shown in Fig. 11.137, find currents i1(t) and i2(t) for t > 0. 5Ω

Fig. 11.133 [i1(t) = 5 + 5e−2t − 10e−3t] 11.12 In the network shown in Fig. 11.134, the switch is closed at t = 0. Find the current through the 30 Ω resistor.

20 μF

50 V i1 (t )

Fig. 11.137

i2 (t )

5Ω 0.1 H

Objective-Type Questions 11.81 7Ω

99499.5t ⎤ ⎡ i1 (t ) = 0.101 .101e 101e −100.5t 0.05e 99 ⎢ ⎥ ⎢⎣i2 (t ) = 5 5.05e −100.5t + 0.05e −9949.5t ⎥⎦

11.16 In the network shown in Fig. 11.138, the switch is opened at t = 0, the steady state having been established previously. Find i(t) for t > 0. 3Ω i (t )

5Ω 10 V 0.1 F

1.8 H 3.5 Ω

5 cos 2t

1H

+ −

1 F v (t ) 2

Fig. 11.140 6 −tt ⎡ ⎢ v(t ) = − 5 e ⎣

9 e 10

6t

+

3 21 ⎤ cos 2t s i 2t ⎥ 10 10 ⎦

11.19 For the network shown in Fig. 11.141, determine v(t) when the input is (i) an impulse function [e−t u(t)] −t (ii) i(t) = 4e u(t) [4t e−t u(t)]

Fig. 11.138 [i(t) = 1.5124e−2.22t + 3.049e−2.5t] 11.17 Find the current i(t) in the network of Fig. 11.139, if the switch is closed at t = 0. Assume initial conditions to be zero. 20 Ω i (t )

10 Ω

15 Ω

+ 1Ω

i (t )

v (t )

1F

−

Fig. 11.141 11.20 For a unit-ramp input shown in Fig. 11.142, find the response vc(t) for t > 0.

5A

10 Ω 1H

2.5 H

10 F vc (t )

r (t )

Fig. 11.139 [i(t) = 3 + 0.57e−7.14 t] 11.18 In the network shown in Fig. 11.140, find the voltage v(t) for t > 0.

Fig. 11.142 [vc(t) = −100 u(t) + 100e−0.01t u(t) + tu(t)]

Objective-Type Questions 11.1 If the Laplace transform of the voltage across 1 a capacitor of value F is 2 1 Vc ( s) = 2 s +1 the value of the current through the capacitor at t = 0+ is (a) 0 (b) 2 A

(c)

1 A 2

(d)

1A

11.2 The response of an initially relaxed linear constant parameter network to a unit impulse applied at t = 0 is 4 e−2t u(t). The response of this network to a unit-step function will be

11.82 Network Analysis and Synthesis 2[1 − e −2t] u(t) 4[e−t − e–2t] u(t) sin 2t (1 − 4 e−4t) u(t) 11.3 The Laplace transform of a unit-ramp function starting at t = a is (a) (b) (c) (d)

(a)

(c)

1

(b)

)2

( e − as

3V

vi

(b)

( s + a) 2

11.7 A 2 mH inductor with some initial current can be represented as shown in Fig. 11.145. The value of the initial current is

a s2

I (S)

−+ 0.002s

1mV

Fig. 11.145 (a) 0.5 A (b) 2 A (c) 1 A (d) 0 11.8 A current impulse 5 d (t) is forced through a capacitor C. The voltage vc(t) across the capacitor is given by

( s − α )2 + α 2 s+α ( s − α )2 + α 2 1

( )2 (d) none of the above 11.5 The circuit shown in Fig. 11.143 has initial current i(0−) = 1 A through the inductor and an initial voltage vc(0−) = −1 V across the capacitor. For input v(t) = u(t), the Laplace transform of the current i(t) for t ≥ 0 is 1Ω

vo

Fig. 11.144

s −α

(c)

1 kΩ

t

2s

11.4 The Laplace transform of eat cos a t is equal to (a)

−2Ω −j

e − as

(d)

s2

0.1 μF

(a)

5t

(b)

5 u(t) − C

(c)

5 t C

(d)

5u(t ) C

11.9 In the circuit shown in Fig. 11.146, it is desired to have a constant direct current i(t) through the ideal inductor L. The nature of the voltage source v(t) must be

1H

i (t )

+ v (t )

1F

v (t )

L

i (t )

−

Fig. 11.143 (a) (c)

s 2

s + s +1 s−2 s2 + s + 1

(b) (d)

Fig. 11.146 s+2 2

s + s +1 s−2 s2 + s + 1

11.6 A square pulse of 3 volts amplitude is applied to an RC circuit shown in Fig. 11.144. The capacitor is initially uncharged. The output voltage v0 at time t = 2 seconds is (a) 3 V (b) −3 V (c) 4 V (d) −4 V

(a) a constant voltage (b) a linearly increasing voltage (c) an ideal impulse (d) as exponential increasing voltage 11.10 When a unit-impulse voltage is applied to an inductor of 1 H, the energy supplied by the source is (a) ∞ (b) 1 J (c)

1 J 2

(d)

0

Answers to Objective-Type Questions 11.83

Answers to Objective-Type Questions 11.1 (c) 11.7 (a)

11.2 (a) 11.8 (d)

11.3 (c) 11.9 (c)

11.4 (a) 11.10 (c)

11.5 (b)

11.6 (b)

12 Network Functions

12.1

INTRODUCTION

A network function gives the relation between currents or voltages at different parts of the network. It is broadly classified as driving point and transfer function. It is associated with terminals and ports. Any network may be represented schematically by a rectangular box. Terminals are needed to connect any network to any other network or for taking some measurements. Two such associated terminals are called terminal pair or port. If there is only one pair of terminals in the network, it is called a one-port network. If there are two pairs of terminals, it is called a two-port network. The port to which energy source is connected is called the input port. The port to which load is connected is known as the output port. One such network having only one pair of terminals (1 – 1′ ) is shown in Fig. 12.1 (a) and is called one-port network. Figure 12.1 (b) shows a two-port network with two pairs of terminals. The terminals 1 – 1′ together constitute a port. Similarly, the terminals 2 – 2′ constitute another port. 1 1′

I + V −

Oneport network

1 1′

(a)

I2

I1 + V1 −

Twoport network

2 + V2 − 2′

(b)

Fig. 12.1 (a) One-port network (b) Two-port network A voltage and current are assigned to each of the two ports. V1 and I1 are assigned to the input port, whereas V2 and I2 are assigned to the output port. It is also assumed that currents I1 and I2 are entering into the network at the upper terminals 1 and 2 respectively.

12.2

DRIVING-POINT FUNCTIONS

If excitation and response are measured at the same ports, the network function is known as the driving-point function. For a one-port network, only one voltage and current are specified and hence only one network function (and its reciprocal) can be defined.

12.2 Network Analysis and Synthesis 1. Driving-point Impedance Function It is defined as the ratio of the voltage transform at one port to the current transform at the same port. It is denoted by Z (s). Z ( s) =

V ( s) I ( s)

2. Driving-point Admittance Function It is defined as the ratio of the current transform at one port to the voltage transform at the same port. It is denoted by Y (s). Y ( s) =

I ( s) V ( s)

For a two-port network, the driving-point impedance function and driving-point admittance function at port 1 are Z11 ( s) =

V1 ( s) I1 ( s)

Y11 ( s) =

I1 ( s) V1 ( s)

Z 22 ( s) =

V2 ( s) I 2 ( s)

Y22 ( s) =

I 2 ( s) V2 ( s)

Similarly, at port 2,

12.3

TRANSFER FUNCTIONS

The transfer function is used to describe networks which have at least two ports. It relates a voltage or current at one port to the voltage or current at another port. These functions are also defined as the ratio of a response transform to an excitation transform. Thus, there are four possible forms of transfer functions. 1. Voltage Transfer Function It is defined as the ratio of the voltage transform at one port to the voltage transform at another port. It is denoted by G (s). V2 ( s) V1 ( s) V ( s) G21 ( s) = 1 V2 ( s) G12 ( s) =

2. Current Transfer Function It is defined as the ratio of the current transform at one port to the current transform at another port. It is denoted by a (s).

α12 ( ) =

I2 ( ) I1 ( )

α 21 ( ) =

I1 ( ) I2 ( )

12.3 Transfer Functions 12.3

3. Transfer Impedance Function It is defined as the ratio of the voltage transform at one port to the current transform at another port. It is denoted by Z (s). Z12 ( s) =

V2 ( s) I1 ( s)

Z 21 ( s) =

V1 ( s) I 2 ( s)

4. Transfer Admittance Function It is defined as the ratio of the current transform at one port to the voltage transform at another port. It denoted by Y (s).

Example 12. 1

Y12 ( s) =

I 2 ( s) V1 ( s)

Y21 ( s) =

I1 ( s) V2 ( s)

Determine the driving-point impedance function of a one-port network shown in Fig. 12.2. R

L

C

R

Fig. 12.2 Solution The transformed network is shown in Fig. 12.3. 1 R ( R Ls) s+ R Ls 1 Cs L Z ( s) = = = 2 1 1 C 2 R LCs C R RCs Cs + 1 + (R ( R Ls) s + s+ Cs L LC

Example 12.2

1 Cs

Z(s)

Ls

Fig. 12.3

Determine the driving-point impedance of the network shown in Fig. 12.4. 2s

2s

1 2s

Z (s)

1 2s

Fig. 12.4 Solution 1 ⎛ 1⎞ 1 ⎛ 1⎞ 1 1 2s + 4 s + 8s3 + 2 s + ⎜⎝ 2 s + ⎟⎠ ⎜⎝ 2 s + ⎟⎠ 4 + s2 + 1 2s 2s 2s 2s 2 s 2 s = 16 s +12 Z ( s) = 2 s + = 2s + = 2 s + = 1 1 2 + 4s2 2 + 4s2 2 + 4s2 8s3 + 4 s + 2s + 2s 2s 2s

12.4 Network Analysis and Synthesis

Example 12.3

Determine the driving-point impedance of the network shown in Fig. 12.5. 1 s

1 s

s

Z (s)

s

Fig. 12.5 Solution ⎛1 ⎞ s ⎜ + s⎟ ⎝ s ⎠ 1 ( s 2 ) s 1 s + s3 1 s4 + 3 2 1 Z ( s) = + = + = + = 1 s s 2s2 + 1 s 2s2 + 1 2 s3 + s s+ +s s

Example 12.4

Find the driving-point admittance function of the network shown in Fig. 12.6.

3s

s

1 5s

1 2s

Y (s)

Fig. 12.6 Solution ⎛ 1⎞ s⎜ ⎟ ⎝ 2s ⎠ 1 1 s 30 s 4 +15 + 5s 2 + 2 s 2 + 1 + 5s 2 30 s 4 + 22s 22 s 2 + 1 Z ( s ) = 3s + + = 3s + + 2 = = 1 5s 5s 2 s + 1 5 ( 2 s 2 1) 5 ( 2 s 2 1) s+ 2s 1 5s ( s 2 ) Y ( s) = = Z ( s) 30 s 4 + 22 s 2 + 1

Example 12.5

Find the transfer impedance function Z12(s) for the network shown in Fig. 12.7. I1 (s) +

V1 (s)

+ R

1 Cs

−

V2 (s) −

Fig. 12.7

12.4 Analysis of Ladder Networks 12.5

⎛ 1⎞ R⎜ ⎟ ⎝ Cs ⎠ V2 ( s) = I1 ( s) 1 R+ Cs V2 ( s) R = I1 ( s) RCs C +1 V2 ( s) 1 Z12 ( s) = = 1 ⎞ I1 ( s) ⎛ C s+ ⎟ ⎝ RC ⎠

Solution

Example 12.6

Find voltage transfer function of the two-port network shown in Fig. 12.8. I1(s)

R

I2(s)

+

V1(s)

+ 1 Cs

−

V2(s) −

Fig. 12.8 Solution By voltage division rule, 1 Cs

Voltage transfer function

12.4

1 1 V2 ( s) = V1 ( s) = V1 ( s) = V1 ( s) RC 1 1 RCs RC C +1 R+ s+ Cs RC 1 V2 ( s) = RC 1 V1 ( s) s+ RC

ANALYSIS OF LADDER NETWORKS

Z1 Z3 I1 I2 = 0 Va Ib Vb The network functions of a ladder network can be obtained by a simple method. This method depends + upon the relationships that exist between the branch Y4 Y2 currents and node voltages of the ladder network. V1 Consider the network shown in Fig. 12.9 where all the − impedances are connected in series branches and all the admittances are connected in parallel branches. Fig. 12.9 Ladder network Analysis is done by writing the set of equations. In writing these equations, we begin at the port 2 of the ladder and work towards the port 1. Vb = V2 Ib = Y4 V2 Va = Z3 Ib + V2 = (Z3Y4 + 1) V2 I1 = Y2 Va + Ib = [Y2 (Z3 Y4 + 1) + Y4] V2 V1 = Z1 I1 + Va = [Z1 {Y2 (Z3 Y4 + 1) + Y4} + (Z3 Y4 + 1)] V2

+ V2 −

12.6 Network Analysis and Synthesis Thus, each succeeding equation takes into account one new impedance or admittance. Except the first two equations, each subsequent equation is obtained by multiplying the equation just preceding it by imittance (either impedance or admittance) that is next down the line and then adding to this product the equation twice preceding it. After writing these equations, we can obtain any network function.

Example 12.7

For the network shown in Fig. 12.10, determine transfer function 1Ω

I1

1Ω

I2 = 0

+

V V

+

V

1F

V

1F

Fig. 12.10 Solution The transformed network is shown in Fig. 12.11. V Ib

1

I1

V V 1 s

1

Ib

V

I

0

+

+

sV +V

1 s

1 s

V −

V −

s +1 V I1

Hence,

Fig. 12.11

V 1 s

V +V V 1 = 2 V s s

Example 12.8

s

b

s

s

sV

s

sV s + )V

For the network shown in Fig. 12.12, determine the voltage transfer function I1

s

s

Va

I

V

I

+

+

V1

1

V

1

Fig. 12.12 Solution

Vb = V2 Ib

V 1

I1

V 1

0

V V = s+2 V

I s

V

)V

V . V

12.4 Analysis of Ladder Networks 12.7

V2 1 = V1 s 2 + 3s 3s 1

Hence,

Example 12.9

Find the network functions 1H

I1

V1 V2 V , and 2 for the network shown in Fig. 12.13. I1 V1 I1 1H

I2 = 0 +

+ V1

1F

V2

1F

−

−

Fig. 12.13 Solution The transformed network is shown in Fig. 12.14. I1

Vb

V2 V I b = 2 = sV V2 1 s Va s I b V2 = s ( sV V2 ) V2 = (s ( s 2 1)V2 I1 = V1 Hence,

s

s

Va

Ib

I2 = 0

Vb

+

+ 1 s

1 s

V1

−

−

Fig. 12.14

Va + I b sV Va I b = s ( s 2 + 1)V2 sV V2 ( s3 2 s)V2 1 s s I1 Va = s ( s3 + 2 s) ( s 2 + 1) ( s 4 2 s 2 + s 2 + 1)

((ss 4

3s 2 1)V2

V1 s 4 + 3s 3s 2 1 = I1 s3 + 2 s V2 1 = 4 V1 s + 3s 3s 2 1 V2 1 = 3 V1 s + 2 s

Example 12.10

Find the network functions

I1

1 F 4

V1 V2 V , , and 2 for the network in Fig. 12.15. I 1 V1 I1

3H

I2 = 0

+ V1

+ 1 H 2

−

2F

V2 −

Fig. 12.15

V2

12.8 Network Analysis and Synthesis Solution The transformed network is shown in Fig. 12.16. Vb

V2 V I b = 2 = 2 sV V2 1 2s

Va

4 s

I1

Ib

I2 = 0

Vb

+

3s I b V2 = 3s ( 2 sV V2 ) V2 = (6 s

2

1 2s

s 2

V1 −

1)V2

⎛ 14 s 2 + 2 ⎞ Va 2 + I b = (6 s 2 1)V2 + 2 sV V2 = ⎜ ⎟ V2 s s s ⎝ ⎠ 2 ⎛ 6s4 4 4 ⎛ 14 s 2 + 2 ⎞ 2 V1 = I1 Va = ⎜ V ( 6 s 1 ) V = 2 ⎟ 2 ⎜ s s⎝ s ⎠ ⎝

V2 −

I1 =

Hence,

3s

Va

+

Fig. 12.16 557s 7s2 8 ⎞ ⎟ V2 s2 ⎠

V1 6 s 4 557s 7s2 8 = I1 14 s3 + 2 s V2 s2 = 4 V1 6 s 57s 57 s 2 8 V2 s = I1 14 s 2 + 2

Example 12.11 For the ladder network of Fig. 12.17, find the driving point-impedance at the 1 – 1Ä terminal with 2 – 2Ä open. 1

I1

1Ω

1Ω

1H

1H

I2 = 0

+

+

V1 1′

1Ω

1H

1F

1F

1F

−

2

V2 −

2′

Fig. 12.17 Solution The transformed network is shown in Fig. 12.18. 1

1′

I1

1

s

1

a

s

1

Vb

+

Ia

Ib

V1

1 s

1 s

−

V2 V I b = 2 = s V2 1 s

Vc

I2 = 0 + 1 s

2

V2 −

Fig. 12.18 Vc

s

2′

12.4 Analysis of Ladder Networks 12.9

V Ia =

V + 1 s

V I1

V 1 s s +2

s

s

a

=s s s2

s

V

= s

s(

b

s

+ 2s V

sV +

s

+

s + s

s

+ +

+1 V

s + 2s V

s) V s

s

+

s

V

V Hence,

Z11

V I1

Example 12.12

s

s

s

s

s s

Determine the voltage transfer function 1Ω

I1

V for the network shown in Fig. 12.19. V

2

I

0

+

+

V

1F

V

1Ω

1F

−

Fig. 12.19 Solution The transformed network is shown in Fig. 12.20. I1

=V

1

Ia

V

2s

V

+

+I

V 1 s

s

+1 V

s +1 V

1

Ib

V

1 s

s

V = I1 =

V + 1 s

Fig. 12.20

a

s

V Hence,

V = V 2s

s +

1V

+1

s 1

s + 4s 2

+

s

Vc

+1 V

s

4s

V

I

0 +

I 1

V

12.10 Network Analysis and Synthesis

Example 12.13

For the network shown in Fig. 12.21, determine the transfer function 3Ω 2

I1

1Ω

I2

2F 3

+

I2 . V1

I=0 +

V1

1Ω 6

2F

V2

−

−

Fig. 12.21 Solution The transformed network is shown in Fig. 12.22. 3 2 1

I1

3 2s

+

I2

Va

Ia

I=0

Vb

1 6

1 2s

V1

+

Ib

V2

−

−

Fig. 12.22 Vb

Va = V2

V2 V2 + = 2 sV V2 + 6 V2 ( 2 s + 6)V2 1 1 2s 6 ⎛3 3 ⎞ × ⎡ ( 2 s 5)( s + 3) ⎤ (6 15) ⎜ 2 2s ⎟ ⎛ 9 ⎞ V1 = ⎜ + 1⎟ I1 V2 = ⎜ + 1⎟ I1 V2 = ( 2 6)V2 V2 = ⎢ + 1⎥ V2 3 3 ⎝6 6 ⎠ 6 6 ( s + 1) ⎣ ⎦ ⎜ + ⎟ ⎝ 2 2s ⎠ ⎛ 2 s 2 + 6 s + 5s + 15 + s + 1⎞ ⎛ 2 s 2 + 12 s + 16 ⎞ ⎡ ( 2 s 5)( s + 3) ( s 1) ⎤ =⎢ V2 = ⎜ V2 = ⎜ ⎟ ⎟ V2 ⎥ s +1 s +1 s +1 ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ 2(ss 2 + 6s 6 s 8) 2 ( s 4)( s + 2) = V2 = V2 s +1 s +1 I2 = −Ib = −6 V2 I1 = I a + I b =

Also,

I2 3 ( s 1) =− V1 ( s + 4)( s 2)

Example 12.14

For the network shown in Fig. 12.23, compute α 12 (s) s = 1Ω

I2

I=0

+ I1

V1

+ 1F

1F

V2 −

−

Fig. 12.23

I2 V and Z12 ((s) s) = 2 . I1 I1

12.4 Analysis of Ladder Networks 12.11

Solution The transformed network is shown in Fig. 12.24.

1

Va

Hence, and

V2 + V2 I1 I2 = = sV V2 V1 1 s − V1 1 I 2 + Va sV V2 V2 = (s 1)V2 V I1 = 1 + I 2 sV V1 I 2 = s (s + 1)V V + sV = ( s 2 + 2s)V2 1 s I2 1 α12 ( ) = = I1 s + 2 Z12 ( s) =

I2

Va

I=0 +

1 s

1 s

V2 −

Fig. 12.24

V2 1 = I1 s 2 + 2 s

Example 12.15

Determine the voltage ratio

V2 V2 I and , current ratio 2 , transfer impedance I1 V1 I1

V1 for the network shown in Fig. 12.25. I1

driving- point impedance

3Ω

3H

I1 +

I=0

I2

1F 3

+

2Ω

V1

1Ω

1F

1F 2

−

V2 −

Fig. 12.25 Solution The transformed network is shown in Fig. 12.26. 3

I1

Va

3 s

+

3s

Ia 1 s

V1 −

I2

Vb

I=0

Vc

+

Ib 2 2 s

1

V2 −

Fig. 12.26 Vb = V2 V I 2 = 2 = V2 1

Vc

V s ⎛ 3s + 2 ⎞ + 2 = V V2 = ⎜ V2 2 1 2s + 2 2 ⎝ 2 s + 2 ⎟⎠ 2+ s ⎛ 9 s 2 + 8s + 2 ⎞ 3s (3s + 2) Va = 3s I a + V2 = V2 + V2 = ⎜ ⎟ V2 2s + 2 ⎝ 2s + 2 ⎠ Ia

Ib + I2 =

V2

12.12 Network Analysis and Synthesis I1 =

V

a

s 3⎞ ⎛ 3× s

V

3+ ⎡ 27 ⎣ ⎛ 36

3 I + s +1

=

V = V 36 I2 = I1 s V I1 s

=

⎛ 9s

3s 2 s

3 ⎛ s +1

2

+5 2

⎞

8 2 s 2s 2

2

V +

⎛ 2

+

+ 15s +

s + +2

s + 25s + 8 ⎞

+

s+

s+2

(36

V

+

V

s + 25s + 8 2s 2 s + 5s 2 2s 2 + 5s 2

Example 12.16

s +2

36

s + 25s + 8

s+2

s

s+2

For the resistive two-port network of Fig. 12.27, find 2Ω

I1

2Ω

2Ω

+

V V I2 I , , and 2 V I1 V I1 I

+ 1Ω

V

1Ω

1Ω

1Ω

V

Fig. 12.27 Solution The network is redrawn as shown in Fig. 12.28. I1

+2 V 2

+ 25 + 8) V +2

s+2

s

V 36 = I1 s

V

s

+2 Hence,

s 8s + 2 V 2s 2

s

+

2

V

2

V

+

Ia

I

V

1

1

Fig. 12.28

V

I +

1

V

1

12.4 Analysis of Ladder Networks 12.13

V2 = −V2 1 3 I 2 3V2

I2 = − Vb

Vb Vb 4 + = Vb 4 V2 1 3 3 2 I b Vb 8 V2 + 3 V2 = 11 V2 V = a + I b = 11 V2 + 4 2 = 15 V2 1 = 2 a + a = 30 2 + 11 V2 = 4411 V2 V = 1 + I a = 41 V2 +15 155 V2 = 566 V2 1 1 = 41 1 = Ω 56 1 =− 41 1 =− 56

Ib = Va Ia V1 I1 V2 V1 V2 I1 I2 V1 I2 I1

Hence,

Example 12.17

Find the network function Ia

1Ω

1Ω

+

V2 for the network shown in Fig. 12.29. V1 2V1

2Ia

V1

I2 = 0

+ −

+ 1Ω

−

V2 −

Fig. 12.29 Solution The network is redrawn as shown in Fig. 12.30. Ia

1

3Ia

1

+

V1

2V1

I2 = 0

+ −

3Ia 2Ia

−

1

+

V2 −

Fig. 12.30

12.14 Network Analysis and Synthesis From Fig. 12.30, V2 = 1 (3 Ia) = 3 Ia Applying KVL to the outermost loop, V1 − 1 (Ia) − 1 (3 Ia) − 2 V1 − 1 (3 Ia) = 0 V1 = −7 Ia V2 3 =− V1 7

Hence,

Example 12.18

I2 for the network shown in Fig. 12.31. I1

Find the network function 2Ia

2Ω

I2

− +

+

Ia I1

1Ω

I1 2

1Ω

1Ω

V2 −

Fig. 12.31 Solution The network is redrawn as shown in Fig. 12.32. 2 Ia

a

Ia + I2 +

− + I

I1

I1 2 d

2

I2 +

I1 2

+

f

Ia I1 2

1

1

I2

1

V2 −

b

c

e

Fig. 12.32 From Fig. 12.32, I1 + I a + I 2 +

I =

3 I1 2

Ia + I2

I1 2 ...(i)

Applying KVL to the loop abcda, −1 I − 1 I a − 2 −I − 3 I +3

a a a

=0 =0 =0

...(ii)

12.5 Analysis of Non-Ladder Networks

12.15

Substituting Eq. (i) in Eq. (ii), 3 I1 2

Ia + I2 3 I1 2

3

a

0

I2 + 4

a

0

...(iii)

Applying KVL to the loop dcefd, 1

a

1I2

I ⎞ ⎛ 2 I2 + 1 ⎟ = 0 ⎝ 2⎠ Ia

3

2

1

Ia

0 3 I 2 + I1

...(iv)

Substituting Eq. (iv) in Eq. (iii), 3 I1 2

I 2 + 4 (3 I 2

3 I1 2

I 2 + 12 I 2

I1 ) = 0 4

1

0

11 I1 13 I 2 = 0 2 13 I 2 = −

I2 11 =− I1 26

Hence,

12.5

11 I1 2

ANALYSIS OF NON-LADDER NETWORKS

The above method is applicable for ladder networks. There are other network configurations to which the technique described is not applicable. Figure 12.33 shows one such network. Z4 I1

Z2

Z1

+ V1

I2 +

Z3

−

V2 −

Fig. 12.33 Non-ladder network For such a type of network, it is necessary to express the network functions as a quotient of determinants, formulated on KVL and KCL basis.

12.16 Network Analysis and Synthesis

Example 12.19

For the resistive bridged T network shown in Fig. 12.34, find

V2 V2 I 2 I , , and 2 . V1 I 1 V1 I1

2Ω

1Ω

I1

1Ω

I3

+

I2 +

0.5 Ω

V1 I1

−

V2 I2

1Ω

−

Fig. 12.34 Solution Applying KVL to Mesh 1, V1 = 1.5 I1 + 0.5 I2 − I3 Applying KVL to Mesh 2, 0 = 0.5 I1 + 2.5 I2 + I3 Applying KVL to Mesh 3, 0 = −I1 + I2 + 4 I3 Writing these equations in matrix form, 0 ⎡V1 ⎤ ⎡1 ⎢ 0 ⎥ = ⎢0.5 2.5 ⎢ ⎥ ⎢ 1 ⎣ 0 ⎦ ⎣ −1 V1 0 0 Δ1 I1 = = 1.5 Δ 0.5 −1

1⎤ ⎡ I1 ⎤ 1⎥ ⎢ I 2 ⎥ ⎥⎢ ⎥ 4 ⎦ ⎣ I3 ⎦ 0 5 −1 2 .5 1 1 4 V (10 − 1) = 1 = V1 0.5 −1 9 2.5 1 1 4

1 5 V1 −1 05 0 1 −1 0 4 Δ2 −V ( 2 + 1) 1 I2 = = = 1 = − V1 1.5 0.5 −1 Δ 9 3 0.5 2.5 1 −1 1 4 From Fig. 12.34, From Eq. (v), From Eqs. (iv) and (v),

Hence,

V2 = −1 (I2) = −I2 V1 = −3 I2 1 1 V1 = − I1 3 3 I1 3I 2 I2 1 =− � V1 3 I2

...(i) ...(ii) ...(iii)

...(iv)

...(iv)

...(v)

12.5 Analysis of Non-Ladder Networks

12.17

1 − V1 I2 1 = 3 =− I1 V1 3 V2 −I2 1 = = V1 −3 I 2 3 V2 −I2 1 = = Ω I1 −3 I 2 3

Example 12.20

For the network of Fig. 12.35, find Z11, Z12 and G12. Za

a

I1

b

+

+

V1

Zb

V2

Zb −

− d

c

Za

Fig. 12.35 Solution The network can be redrawn as shown in Fig. 12.36. Since the network consists of two identical impedances connected in parallel, the current in I1 divides equally in each branch. V1

( Z a + Zb )

I1

I1 2

Z11 =

V1 Z a Zb = I1 2

V2

Zb

Z12 =

V2 Zb − Z a = I1 2

+

⎛I ⎞ Z a ⎜ 1 ⎟ = ( Zb ⎝ 2⎠

I1 2

Za )

I1 2

V1

I1 2 Za

+

I1 2 Zb V2

−

Zb −

Fig. 12.36

By voltage-division rule, V2 = G12 =

Example 12.21

Zb Za

Zb

V2 Zb = V1 Z a

Za

V1

Za

Zb

V1 =

Zb Za

Za V1 Zb

Za Zb

For the network shown in Fig. 12.37, determine Z11 (s), G12 (s) and Z12 (s).

Za

12.18 Network Analysis and Synthesis 1F 1Ω

I1

1Ω

I2 = 0

+

+

V1

V2

1F

−

−

Fig. 12.37 Solution The transformed network is shown in Fig. 12.38. 1 s

1

1

I3

+

+ 1 s

V1 −

I2 = 0

I1

V2 −

Fig. 12.38 Applying KVL to Mesh 1, ⎛ 1⎞ V1 = 1 + I1 ⎝ s⎠

I3

...(i)

Applying KVL to Mesh 2, V2 =

1 I1 s

I3

...(ii)

Applying KVL to Mesh 3, 1⎞ ⎛ − I1 + ⎜ 2 + ⎟ I 3 = 0 ⎝ s⎠ ⎛ s ⎞ I3 = ⎜ I1 ⎝ 2 s + 1⎟⎠

...(iii)

Substituting Eq. (iii) in Eqs. (i) and (ii), ⎡ s 2 + 3s s ⎞ 3s 1 ⎤ ⎛ 1⎞ ⎛ s ⎞ ⎛ s +1 V1 = 1 + I1 − ⎜ I = I = ⎢ ⎥ I1 1 1 ⎟ ⎝ s⎠ ⎝ 2 s 1⎠ ⎝ s 2 s 1⎠ ⎣ s (22 s + 1) ⎦ ⎡ s 2 + 2 s + 1⎤ 1 s V2 = I1 + I1 = ⎢ ⎥ I1 s 2s + 1 ⎣ s ( 2 s + 1) ⎦ Hence,

Z11 ( s) =

V1 s 2 + 3s 3s 1 = I1 s ( 2 s 1)

Z12 ( s) =

V2 s 2 + 2s 2s 1 = I1 s ( 2 s 1)

G12 ( s) =

V2 s 2 + 2s 2s 1 = 2 V1 s + 3s 3s 1

12.5 Analysis of Non-Ladder Networks

Example 12.22

12.19

For the network shown in Fig.12.39, find the driving-point admittance Y11 and

transfer admittance Y12. 1Ω 1F

I1

V1

1F

+ −

I2

1Ω

1Ω

Fig. 12.39 Solution The transformed network is shown in Fig. 12.40. 1 1 s

I1 + −

V1

I3

1

I1

1 s

I2

I2

1

Fig. 12.40 Applying KVL to Mesh 1, V1 =

1 ⎛1 ⎞ + 1 I1 + I 2 − I 3 ⎝s ⎠ s

...(i)

1⎞ 1 ⎛ 2+ I 2 + I3 ⎝ s⎠ s

...(ii)

Applying KVL to Mesh 2, 0

1

Applying KVL to Mesh 3, 1 1 ⎛2 ⎞ 0 = − I1 + I 2 + ⎜ + 1⎟ I 3 ⎝s ⎠ s s Writing these equations in matrix from, ⎡1 1 ⎢ s +1 ⎡V1 ⎤ ⎢ 1 ⎢0⎥ = ⎢ 1 2+ ⎢ ⎥ ⎢ s ⎣0⎦ ⎢ 1 1 ⎢− s ⎣ s Δ I1 = 1 Δ

⎤ ⎥ ⎥ ⎡ I1 ⎤ ⎥ ⎢I2 ⎥ ⎥ ⎢I ⎥ 2 ⎥ ⎣ 3⎦ + 1⎥ s ⎦ 1 s 1 s

−

...(iii)

12.20 Network Analysis and Synthesis 1 +1 s Δ=

− =

2+

1 1 s

1 s

s2 + 5 s

1 s 1 1⎞ ⎛ 2 ⎞ 1 ⎤ ⎡ ⎛ 2 ⎞ 1 ⎤ 1 ⎡ ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ ⎤ ⎛ 1 ⎞ ⎡⎛ = + 1⎟ ⎢⎜ 2 + ⎟ ⎜ + 1⎟ − 2 ⎥ −11 (1) + 1⎟ + 2 ⎥ − ⎢(1) ⎜ ⎟ + ⎜ ⎟ ⎜ 2 + ⎟ ⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ s s s s s s s s⎠ ⎦ s ⎦ ⎣ s ⎦ s⎣ ⎣ 2 +1 s −

1 1 s

2

2

V1

1

Δ1 = 0

2+

0

1 s

I1 Hence,

1 s

2 +1 s

5s + 1⎞ 5s ⎟ s ⎠ 2

I1 2 s 2 5s + 1 = V1 s 2 + 5s 5s 2 Δ I2 = 2 Δ

Y11 =

Δ2 =

1 −

12.6

2

⎛ 2 s 2 5s + 1⎞ V1 ⎜ 2 ⎟ 5s 2 ⎠ ⎝ s + 5s

1 + 1 V1 s

Hence,

1 s ⎛2 ⎡⎛ 1 1⎞ ⎛ 2 ⎞ 1 ⎤ = V1 ⎢ 2 + ⎟ ⎜ + 1⎟ − 2 ⎥ = V1 ⎜ ⎝ ⎠ ⎝ ⎠ s s s s ⎦ ⎝ ⎣

−

1 s

0 0

1 s ⎛ s 2 2 s + 1⎞ 1 1⎞ ⎛2 = −V1 ⎜ + 1 + 2 ⎟ = −V1 ⎜ ⎟ ⎝s s s ⎠ s2 ⎝ ⎠

−

2 +1 s

I2

⎛ s 2 + 2s 2s 1 ⎞ V1 ⎜ 2 ⎟ ⎝ s + 5 2⎠

Y12 =

I2 s 2 + 2s 2s 1 =− 2 V1 s + 5s 5s 2

POLES AND ZEROS OF NETWORK FUNCTIONS

The network function F(s) can be written as ratio of two polynomials. F ( s) =

N ( s) an s n + an 1 s n −1 + … + a1 s + a0 = D( s) bm s m + bm 1 s m −1 + … + b1 s + b0

where a0, a1, …, an and b0, b1, …, bm are the coefficients of the polynomials N(s) and D(s). These are real and positive for networks of passive elements. Let N(s) = 0 have n roots as z1, z2, ……, zn and D(s) = 0 have m roots as p1, p2, ……, pm. Then F(s) can be written as

12.8 Restrictions on Pole and Zero Locations for Transfer Functions 12.21

F ( s) = H

( s z )(s )( s z ) …( s zn ) ( s p )(s )( s p ) …( s pm )

an is a constant called scale factor and z1, z2, …, zn, p1, p2, …, pm are complex frequencies. When bm the variable s has the values z1, z2, …, zn, the network function becomes zero; such complex frequencies are known as the zeros of the network function. When the variable s has values p1, p2, …, pm, the network function becomes infinite; such complex frequencies are known as the poles of the network function. A network function is completely specified by its poles, zeros and the scale factor. If the poles or zeros are not repeated, then the function is said to be having simple poles or simple zeros. If the poles or zeros are repeated, then the function is said to be having multiple poles or multiple zeros. When n > m, then (n – m) zeros are at s = ∞, and for m > n, (m − n) poles are at s = ∞. The total number of zeros is equal to the total number of poles. For any jw network function, poles and zeros at zero and infinity are taken into account in addition to finite poles and zeros. Poles and zeros are critical frequencies. The network function becomes infinite at poles, while the network function becomes zero at zeros. The s 0 network function has a finite, non-zero value at other frequencies. Poles and zeros provide a representation of a network function as shown in Fig. 12.41. The zeros are shown by circles and the poles by crosses. This diagram is referred to as pole-zero plot. Fig. 12.41 Pole-zero plot where

12.7 (1) (2) (3) (4) (5) (6) (7)

12.8 (1) (2) (3) (4)

RESTRICTIONS ON POLE AND ZERO LOCATIONS FOR DRIVING-POINT FUNCTIONS [COMMON FACTORS IN N(S) AND D(S) CANCELLED] The coefficients in the polynomials N(s) and D(s) must be real and positive. The poles and zeros, if complex or imaginary, must occur in conjugate pairs. The real part of all poles and zeros, must be negative or zero, i.e., the poles and zeros must lie in left half of s plane. If the real part of pole or zero is zero, then that pole or zero must be simple. The polynomials N(s) and D(s) may not have missing terms between those of highest and lowest degree, unless all even or all odd terms are missing. The degree of N(s) and D(s) may differ by either zero or one only. This condition prevents multiple poles and zeros at s = ∞. The terms of lowest degree in N(s) and D(s) may differ in degree by one at most. This condition prevents multiple poles and zeros at s = 0.

RESTRICTIONS ON POLE AND ZERO LOCATIONS FOR TRANSFER FUNCTIONS [COMMON FACTORS IN N(S) AND D(S) CANCELLED] The coefficients in the polynomials N(s) and D(s) must be real, and those for D(s) must be positive. The poles and zeros, if complex or imaginary, must occur in conjugate pairs. The real part of poles must be negative or zero. If the real part is zero, then that pole must be simple. The polynomial D(s) may not have any missing terms between that of highest and lowest degree, unless all even or all odd terms are missing.

12.22 Network Analysis and Synthesis (5) (6) (7) (8)

The polynomial N(s) may have terms missing between the terms of lowest and highest degree, and some of the coefficients may be negative. The degree of N(s) may be as small as zero, independent of the degree of D(s). For voltage and current transfer functions, the maximum degree of N(s) is the degree of D(s). For transfer impedance and admittance functions, the maximum degree of N(s) is the degree of D(s) plus one.

Example 12.23 (a)

Test whether the following represent driving-point immittance functions.

5s4 + 3s 3 2 − 2s + 1 s3 6s 6 20

(b)

s3 + s 2 + 5s+ 2 s4 + 6s3 + 9s 2

(c)

s 2 + 3s+ 2 s 2 + 6s+ 2

Solution (a) The numerator and denominator polynomials have a missing term between those of highest and lowest degree and one of the coefficient is negative in numerator polynomial. Hence, the function does not represent a driving-point immittance function. (b) The term of lowest degree in numerator and denominator polynomials differ in degree by two. Hence, the function does not represent a driving-point immittance function. (c) The function satisfies all the necessary conditions. Hence, it represents a driving-point immittance function.

Example 12.24 (a) G21 =

Test whether the following represent transfer functions.

3s+ 2 3

2

5s + 4s + 1

(b) a 12 =

2s 2 + 5s+ 1 s+7

(c) Z 21 =

1 3

s + 2s

Solution (a) The polynomial D(s) has a missing term between that of highest and lowest degree. Hence, the function does not represent a transfer function. (b) The degree of N(s) is greater than D(s). Hence the function does not represent a transfer function. (c) The function satisfies all the necessary conditions. Hence, it represents a transfer function.

Example 12.25

Obtain the pole-zero plot of the following functions.

(a) F(s) =

s(s+ ( 2) (s+ 1)(s 1 + 3) 3

(b) F(s) =

(c) F(s) =

s(s+ ( 2) (s+ 1+ j1)(s+ 1 - j1)

(d) F(s) =

(e) F(s) =

s2 + 4 (s+ 22)(s 2 + 9) 9

s(s+ ( 1) (s+ 22)2 (s+ 3) (s+ 1) 1 2 (s+ 5) (s+ 2)(s 2 + 3 + j2)(s+ 3 - j2)

12.8 Restrictions on Pole and Zero Locations for Transfer Functions

12.23

Solution (a) The function F(s) has zeros at s = 0 and s = − 2 and poles at s = − 1 and s = − 3. The pole-zero plot is shown in Fig. 12.42. jw

s

0

−3 −2 −1

Fig. 12.42 (b) The function F(s) has zeros at s = 0 and s = −1 and poles at s = −2, −2 and s = −3. The pole-zero plot is shown in Fig. 12.43. jw

s

0

−3 −2 −1

Fig. 12.43 (c) The function F(s) has zeros at s = 0 and s = −2 and poles at s = −1 −j1 and s = −1 + j1. The pole-zero plot is shown in Fig. 12.44. jw j1

−2

s

0

−1

−j 1

Fig. 12.44 (d) The function F (s) has zeros at s = −1, −1 and s = −5 and poles at s = −2, s = −3 + j2 and s = −3 − j2. The pole-zero plot is shown in Fig. 12.45. jw j2 j1 −5 −4 −3 −2

−1

0 −j 1

−j 2

Fig. 12.45

s

12.24 Network Analysis and Synthesis (e) The function F (s) has zeros at s = j2 and s = −j2 and poles at s = −2, s = j3 and s = −j3. The pole-zero plot is shown in Fig. 12.46. jw j3 j2 j1 −2

−1

s

0 −j 1

−j 2 −j 3

Fig. 12.46

Example 12.26

Find poles and zeros of the impedance of the network shown in Fig. 12.47 and plot

them on the s-plane. 1F

1 H 2

2Ω

Z (s)

Fig. 12.47 1 s

Solution The transformed network is shown in Fig. 12.48. s ×2 1 2 1 2s 2 s 2 + s + 4 2(s 2(( 2 .5s + 2) Z ( s) = + = + = = s s s s+4 s (s ) s (s ) +2 2 2 ( s + 0.25 + j1.4)( s + 0.25 − j1.4) = s ( + 4)

Z (s)

2

s 2

Fig. 12.48

The function Z (s) has zeros at s = −0.25 + j1.4 and s = −0.25 − j1.4 and poles at s = 0 and s = −4 as shown in Fig. 12.49. jw

j 1.4

−4

−3

−2

− 1 − 0.25

0

− j 1 .4

Fig. 12.49

s

12.8 Restrictions on Pole and Zero Locations for Transfer Functions

Example 12.27

12.25

Determine the poles and zeros of the impedance function Z (s) in the network

shown in Fig. 12.50. 1 Ω 2

Z (s)

4F

1 Ω 6

Fig. 12.50 Solution The transformed network is shown in Fig. 12.51. 1 2

1 4s

Z (s)

1 6

Fig. 12.51 1 1 × 1 4s 6 1 1 4s + 8 s + 2 0.. ( s + 2) Z ( s) = + = + = = = 2 1 1 2 4 s + 6 2(( s ) 2 s + 3 s +1 5 + 4s 6 The function Z(s) has zero at s = −2 and pole at s = −1.5.

Example 12.28

Determine Z(s) in the network shown in Fig. 12.52. Find poles and zeros of Z(s)

and plot them on s-plane. 1H

1 F 20

Z (s)

4Ω

Fig. 12.52 Solution The transformed network is shown in Fig. 12.53. s

Z (s)

20 s

Fig. 12.53

4

12.26 Network Analysis and Synthesis 20 ×4 80 20 s ( s ) + 20 s 2 + 5s + 20 Z ( s) = s + s = s+ = s+ = = 20 4 s + 20 s+5 s+5 s+5 +4 s ( s + 2.5 + j 3.71)( s + 2.5 − j 3.71) = s+5 The function Z(s) has zeros at s = –2.5 + j3.71 and s = –2.5 –j 3.71 and pole at s = –5. The pole-zero diagram is shown in Fig. 12.54. jw j 3.71

−5

−4

−3

−2

s

0

−1

−j 3.71

Fig. 12.54

Example 12.29

For the network shown in Fig. 12.55, plot poles and zeros of function

I0 . Ii

I0 4Ω Ii

0.5 F 2H

Fig. 12.55 Solution The transformed network is shown in Fig. 12.56. By current-division rule,

I0

⎛ ⎞ ⎜ 4 2s ⎟ Ii ⎜ 2⎟ ⎜ 4 + 2s + ⎟ ⎝ s⎠

I0 4 2 s

Ii

I0 s ( 4 2 s) s ( s + 2) s ( s + 2) = = = I i 4 s 2s2 + 2 s2 + 2s + 1 ( s + +1)( s + 1) The function has zeros at s = 0 and s = –2 and double poles at s = –1. The pole-zero diagram is shown in Fig. 12.57.

2s

Fig. 12.56

12.8 Restrictions on Pole and Zero Locations for Transfer Functions jw

−4 −3

−2

s

0

−1

Fig. 12.57

Example 12.30

Draw the pole-zero diagram of I 2 for the network shown in Fig. 12.58. I1 I2

10 H 250 μF

I1

200 Ω

Fig. 12.58 Solution The transformed network is shown in Fig. 12.59. I2

10s 1

I1

250 × 10−6s 200

Fig. 12.59 By current-division rule, 1 250 × 10 −6 s I 2 I1 1 + 10 s + 200 250 × 10 −6 s I2 400 400 = 2 = I1 s + 20 s + 400 ( s + 10 1 − j17 17 32)( s + 10 + j17.32) The function has no zero and poles at s = –10 + j17.32 and s = −10 −j17.32. The pole-zero diagram is shown in Fig. 12.60. jw j17.32

−10

0

− 17.32 −j

Fig. 12.60

s

12.27

12.28 Network Analysis and Synthesis

Example 12.31 I1

For the network shown in Fig. 12.61, draw pole-zero plot of

Vc . V1

1Ω

+

V1

− +

+ −

2Vc

1 F 2

1H

5I1

Vc

−

Fig. 12.61 Solution The transformed network is shown in Fig. 12.62. I1

1

c

+

V1

− +

+ −

2Vc

2 s

s

5I1

Vc

−

Fig. 12.62 Applying KVL to the left loop, V1 − 1I1 + 2Vc = 0 I1 = V1 + 2Vc Applying KCL at Node C, Vc Vc + 2 s s Vc s 5 (V1 2Vc ) + + Vc s 2 Vc s 5V1 10 Vc + + Vc s 2 ⎛ 20 s + 2 + s 2 ⎞ Vc ⎜ ⎟ 2s ⎝ ⎠ Vc V1 5 I1 +

=0

=0 =0 = −5V1 =−

10 s 2

s + 20 s + 2

=−

10 s ( s + 0.1)( s + 19.9)

jw

The function has zero at s = 0 and poles at s = − 0.1 and s = −19.9. The pole-zero diagram is shown in Fig. 12.63. −20

−2

− 1 − 0.1

Fig. 12.63

0

s

12.8 Restrictions on Pole and Zero Locations for Transfer Functions

Example 12.32

12.29

Find the driving point admittance function and draw pole-zero plot for the network

shown in Fig. 12.64. I1

2

2 +

V1

− +

+ −

0.1V2

10I1

0.5s

1

V2

−

Fig. 12.64 Solution Applying KVL to the left loop, V1 2 I1 0.1 V2

0 V 0 1 V2 I1 = 1 2

...(i)

Applying KCL at Node 2, 10 I1 +

V2 V2 + =0 0 5s 1

2 10 I1 + V2 V2 = 0 s 10 I1 +

⎛2 ⎞ +1 ⎝s ⎠

2

0

⎛ 2 + s⎞ 10 I1 + ⎜ V2 = 0 ⎝ s ⎟⎠ ⎛ s + 2⎞ ⎜⎝ ⎟ V2 s ⎠

10 I1

V2 = −

10 s I1 s+2

Substituting Eq. (ii) in Eq. (i), V1 + 0 1 I1 =

⎛ 10 s ⎞ I ⎝ s + 2⎠ 1 ⎛ 10 s ⎞ = 0.55V V1 + 0 05 I1 ⎝ s + 2⎠ 2

0 5s ⎞ ⎛ I1 ⎜1 + = 0 5V1 ⎝ s + 2 ⎟⎠ Hence,

Y11 =

I1 = V1

05 0.5 ( s + 2) 0 5s + 1 = = 0 5s s + 0 5s + 2 1 5s + 2 1+ s+2

...(ii)

12.30 Network Analysis and Synthesis The function has zero at s = −2 and pole at s = −1.33. The pole-zero diagram is shown in Fig 12.65. jw

− 3 − 2 −1.33 − 1

s

0

Fig. 12.65

Example 12.33

For the network shown in Fig. 12.66, determine

V2 V . Plot the pole-zero diagram of 2 . Ig Ig I2 = 0

1H

+ 1Ω

Ig

1F

1F

V2 −

Fig. 12.66 Solution The transformed network is shown in Fig. 12.67. s

Va

Ib Vb

I2 = 0

Vc

+ Ig

1

1 s

1 s

1

V2

−

Fig. 12.67

Hence,

Vc

Vb = V2

Ib =

Vb Vc + = sV V2 + V2 1 1 s

Va

s I b Vb = s ( s + 1)V2 V2 = ( s 2 + s + 1)V2

Ig =

Va Va + + I b = ( s 2 + s + 1)V2 1 1 s 1

V2 = I g s3

2s2

( s + 1)V2

s ( s2

s +11)V2 + ( s +11)V2 = ( s3 + 2

3s + 2

The function has no zeros. It has poles at s = −1, s = −0.5 + j1.32 and s = −0.5 –j1.32, The pole-zero diagram is shown in Fig. 12.68.

2

+ 3s + 2)V2

12.8 Restrictions on Pole and Zero Locations for Transfer Functions

12.31

jw j1.32

−2

−1

s

0

− 0.5

−j1.32

Fig. 12.68

Example 12.34

For the transfer function H(s) =

network shown in Fig. 12.69. Find L and C when R = 5 W.

V0 10 = 2 , realise the function using the Vi s + 3s+ 10

L + Vi

+ −

C

R

V0

−

Fig. 12.69 Solution The transformed network is shown in Fig. 12.70. Ls + Vi

+ −

1 Cs

R

V0

−

Fig. 12.70 Simplifying the network as shown in Fig. 12.71, Ls + Vi

+ −

Z (s)

V0

−

Fig. 12.71 1 R Cs Z ( s) = = 1 RCs C +1 R+ Cs R×

12.32 Network Analysis and Synthesis R C +1 Vi RCs R Ls + RCs C +1

V0

1 V0 R LC = = 1 1 Vi RLC s 2 + Ls + R s2 + s+ RC LC V0 10 = Vi s 2 + 3s 3s 10

But

...(i)

...(ii)

R=5Ω

and Comparing Eq. (ii) with Eq. (i),

1 =3 RC 1 = 10 LC Solving the above equations, L = 1.5 H 1 C= F 15

Example 12.35

Obtain the impedance function Z(s) for which pole-zero diagram is shown in Fig. 12.72. jw Z (∞ ) = 1

−3

−2

−1

0

s

Fig. 12.72 Solution The function Z(s) has poles at s = –1 and s = –3 and zeros at s = 0 and s = –2. s (s ) Z ( s) = H =H ( s )(s )( ) For s = ∞, Z( ) When

H

1 =H ( )( )

Z(∞) = 1, H=1 Z ( s) =

s (s ) ( s ))(s ( (s )

⎛ 2⎞ s 2 ⎜1 + ⎟ ⎝ s⎠ ⎛ 1⎞ ⎛ 3⎞ s 2 ⎜1 + ⎟ ⎜1 + ⎟ ⎝ s⎠ ⎝ s⎠

12.8 Restrictions on Pole and Zero Locations for Transfer Functions

Example 12.36

12.33

Obtain the admittance function Y(s) for which the pole-zero diagram is shown in

Fig. 12.73. jw

Y( ∞ ) = 1

j1

−2

s

0

−1

−j 1

Fig. 12.73 Solution The function Y(s) has poles at s = –1 + j1 and s = −1 −j1 and zeros at s = 0 and s = −2.

Y ( s) = H

s (s ) j )( s

(s

j)

=H

s (s (s

)2

⎛ 2⎞ s 2 ⎜1 + ⎟ ⎝ s⎠ =H 2 =H 2 ⎛ 2 2⎞ (j ) s + 2s + 2 s 2 ⎜1 + + 2 ⎟ ⎝ s s ⎠ )

s (s

)

For s = ∞, Y( ) When

H

() =H ()

Y (∞) = 1, H=1 s (s ) Y ( s) = 2 s + 2s + 2

Example 12.37 A network and its pole-zero configuration are shown in Fig. 12.74. Determine the values of R, L and C if Z (j0) = 1. jw j

√111

R 1 Cs

Z (s) Ls

−3

− 1.5 5

2

s

0

−j

√111 2

Fig. 12.74 1⎛ R⎞ 1 ⎜⎝ s + ⎟⎠ Ls + R C L Cs = Z ( s) = = 2 1 LCs C RCs RCs + 1 s 2 + R s + 1 ( Ls R) + Cs L LC ( Ls R)

Solution

...(i)

12.34 Network Analysis and Synthesis From the pole-zero diagram, zero is at s = –3 and poles are at s

15

j

111 2

ds

15

j

111 2

s+3 111 ⎞ ⎛ 111 ⎞ j s 15 j ⎟ ⎜ 2 ⎠⎝ 2 ⎟⎠ s+3 s+3 Z ( s) = H =H 2 2 s + 3 + 30 ⎛ 111 ⎞ ( s + 1.5)2 − ⎜ j 2 ⎟ ⎝ ⎠ Z ( s) = H

When

⎛ ⎜s 1 5 ⎝

Z (j0) = 1, ⎛ 3⎞ 1= H ⎜ ⎟ ⎝ 30 ⎠ H = 10 10 ( s + 3) Z ( s) = 2 s + 3 30

...(ii)

Comparing Eq. (ii) with Eq. (i), R =3 L 1 = 10 C 1 = 30 LC Solving the above equations, 1 F 10 1 L= H 3 R =1Ω

C=

Example 12.38

A network is shown in Fig. 12.75. The poles and zeros of the driving-point function Z(s) of this network are at the following places: 1 3 Poles at − ± j 2 2 Zero at –1 If Z (j0) = 1, determine the values of R, L and C. R Z (s)

1 Cs Ls

Fig. 12.75

12.35

12.8 Restrictions on Pole and Zero Locations for Transfer Functions

1⎛ R⎞ 1 ⎜⎝ s + ⎟⎠ Ls + R C L Cs = Z ( s) = = 2 1 R LCs C RCs RCs + 1 s 2 + s + 1 Ls + R + Cs L L LC ( Ls R)

Solution

...(i)

1 3 The poles are at − ± j and zero is at –1. 2 2 Z ( s) = H

⎛ 1 ⎜s 2 ⎝

s +1 3⎞ ⎛ 1 j s ⎟ ⎜ 2 ⎠⎝ 2

j

3⎞ 2 ⎟⎠

s +1

=H 1⎞ ⎛ ⎜⎝ s 2⎠

2

⎛ ⎝

j

3⎞ 2 ⎟⎠

2

=H

s +1 2

s + s +1

Z (j0) = 1,

When

1= H

(1) (1)

H =1 Z ( s) =

s +1

...(ii)

2

s + s +1

Comparing Eq. (ii) with Eq. (i), C =1 R =1 L 1 =1 LC Solving the above equations, C=1F L=1H R=1Ω

Example 12.39 The pole-zero diagram of the driving-point impedance function of the network of Fig. 12.76 is shown below. At dc, the input impedance is resistive and equal to 2 W. Determine the values of R, L and C. jw j4 Z (s)

1 Cs

R Ls

−2

−1

0

s

−j 4

Fig. 12.76 1⎛ R⎞ 1 ⎜⎝ s + ⎟⎠ Ls + R C L Cs = Z ( s) = = 2 1 R LCs C RCs + 1 s 2 + s + 1 RCs Ls + R + Cs L L LC ( Ls R)

Solution

From the pole-zero diagram, zero is at s = –2 and poles are at s = –1+ j4 and s = −1 − j4.

...(i)

12.36 Network Analysis and Synthesis Z ( s) = H

(s

s+2 j )( s

j )

=H

s+2 (s

)2

( j )2

=H

s+2 s 2 + 2 s + 17

w = 0, Z ( j0) = 2

At dc, i.e.,

2= H H = 17 Z ( s) = 17

2 17 s+2

...(ii)

s 2 + 2 s 17

Comparing Eq. (ii) with Eq. (i), 1 = 17 C R =2 L 1 = 17 LC Solving the above equations, 1 F 17 L =1H R=2Ω

C=

Example 12.40

The network shown in Fig. 12.77 has the driving-point admittance Y (s) of the form Y(s) ( =H

(s − s1 )(s s ) (s − s )

(a) Express s1, s2, s3 in terms of R, L and C. (b) When s1 = –10 + j104, s2 = –10 – j104 and Y (j0) = 10–1 mho, find the values of R, L and C and determine the value of s3.

Y (s)

Ls

1 Cs

R

Fig. 12.77 Solution

R 1 ⎞ ⎛ C s2 + s + ⎟ ⎝ 1 ( Ls R) Cs (Ls C + 1 LCs C RCs + 1 RCs L LC ⎠ Y ( s) = Cs + = = = R Ls + R Ls + R Ls + R s+ L H ( s s ))(s (s s ) Y ( s) = (s s ) 2

(a)

But

...(i)

12.8 Restrictions on Pole and Zero Locations for Transfer Functions

12.37

2

− s1 s2 =

where

R 4 ⎛ R⎞ ± ⎜ ⎟ − 2 ⎝ L⎠ L LC R 1 ⎛ R⎞ =− ± ⎜ ⎟ − ⎝ 2L ⎠ 2 2L LC

R L s1 = −10 + j104 s2 = − 10 − j104

s3 = − (b) When

Y ( s) = H

(s

j

4

))(( s s s3

j

4

)

s 2 + 200 s + 108 s s3

=H

...(ii)

Comparing Eq. (ii) with Eq. (i), R = 20 L s3 = −20 Y ( s) = H

( s 2 + 200 s + 108 ) ( s + 20)

At s = j0, (

8

) = 10 −1 20 H = 0.02 10 −6

Y( j ) = H

Y ( s) = 0.02 10 −6

( s2

s ( s + 20)

8

)

...(iii)

Comparing Eq. (iii) with Eq. (i), C = 0.02 0 × 10 −6 = 0.02 μF 1 = 108 LC 1 L= H 2 R = 20 L R = 10 Ω

Example 12.41

A network and pole-zero diagram for driving-point impedance Z(s) are shown in Fig. 12.78. Calculate the values of the parameters R, L, G and C if Z(j0) = 1. jw j3 j2 j1

R Z (s)

G

1 Cs

Ls

Fig. 12.78

−3

−2

−1

0

s

− j1 − j2 − j3

12.38 Network Analysis and Synthesis Solution It is easier to calculate Y(s) and then invert it to obtain Z(s). Y ( s) = G Cs C +

Z( ) =

1 (G Cs C )( Lss R) + 1 LCs C 2 = = Ls + R Ls + R

1 = Y ( ) LCs C 2

(

(GL G RC ) s + 1 + GR Ls + R 1⎛ R⎞ ⎜ s + ⎟⎠ Ls + R C⎝ L = G R ) s + 1 + GR s 2 + ⎛ + ⎞ s + ⎛ 1 + GR ⎞ ⎜⎝ ⎟ ⎝ C L ⎟⎠ LC ⎠

...(i)

From the pole-zero diagram, zero is at s = –2 and poles are at s = –3 ± j3. Z ( s) = H When

(s

(s ) j )( s

j )

=H

(s (s

)

2

) (j )

2

=H

s+2 2

s + 6 s + 18

Z (j0) = 1, 2 18

1= H H =9 Z( s) =

9( 2

(s + 6

2)

...(ii)

18)

Comparing Eq. (ii) with Eq. (i), 1 C R L G R + C L 1 + GR LC

=9 =2 =6 = 18

Solving the above equation, 1 F 9 9 L= H 10 4 G= � 9 9 R= Ω 5 C=

Example 12.42

A series R-L-C circuit has its driving-point admittance and pole-zero diagram is shown in Fig. 12.79. Find the values of R, L and C.

12.9 Time-Domain Behaviour from the Pole-Zero Plot 12.39 jw

j 25 Scale factor = 1 s

0

−1

−j 25

Fig. 12.79 Solution The function Y (s) has poles at s = –1 + j25 and s = –1 – j25 and zero at s = 0. Y ( s) = H

(s

j

s ))(s (s

j

)

=H

s (s

)

2

(j

)

2

=H

s 2

s + 2 s + 626

H=1

Scale factor

Y ( s) =

s

...(i)

s 2 + 2 s + 626

For a series RLC circuit, R 1 ⎞ ⎛ L s2 + s + ⎟ ⎝ 1 LCs C R RCs Cs + 1 C L LC ⎠ Z ( s) = R + Ls + = = Cs Cs s 1 s Y ( s) = = R 1 ⎞ Z ( s) ⎛ L ⎜ s2 + s + ⎟ ⎝ L LC ⎠ 2

...(ii)

Comparing Eq. (i) with Eq. (ii), L =1H 1 = 626 LC 1 C= F 626 R =2 L R=2Ω

12.9

TIME-DOMAIN BEHAVIOUR FROM THE POLE-ZERO PLOT

The time-domain behaviour of a system can be determined from the pole-zero plot. Consider a network function F ( s) = H

( s z )(s )( s z )… ( s zn ) ( s p )(s )( s p )… ( s pm )

The poles of this function determine the time-domain behaviour of f(t).The function f(t) can be determined from the knowledge of the poles, the zeros and the scale factor H. Figure 12.80 shows some pole locations and the corresponding time-domain response.

12.40 Network Analysis and Synthesis (i) When pole is at origin, i.e., at s = 0, the function f(t) represents steady-state response of the circuit i.e., dc value. (Fig. 12.80) jw

f (t)

0

s

t

0

Fig. 12.80 Pole at origin (ii)

When pole lies in the left half of the s-plane, the response decreases exponentially. (Fig. 12.81) jw

0

f (t)

s

0

t

Fig. 12.81 Pole in left half of the s-plane (iii)

When pole lies in the right half of the s-plane, the response increases exponentially. A pole in the right-half plane gives rise to unbounded response and unstable system. (Fig. 12.82) jw

0

f (t)

s

0

t

Fig. 12.82 Pole in right half of the s-plane (iv)

For s = 0 +jwn, the response becomes f (t) = Ae± jwnt = A(cos wnt ± j sin wnt).The exponential response e ± jwnt may be interpreted as a rotating phasor of unit length. A positive sign of exponential e jwnt indicates counterclockwise rotation, while a negative sign of exponential e−jwnt indicates clockwise rotation. The variation of exponential function e jwnt with time is thus sinusoidal and hence constitutes the case of sinusoidal steady state. (Fig. 12.83) f (t) jw

0

s

0

Fig. 12.83 Poles on jw -axis

t

12.9 Time-Domain Behaviour from the Pole-Zero Plot 12.41

(v)

For s = sn + jwn, the response becomes f (t) = Aest = Ae(sn+jwnt) = Aesnt ejwnt. The response esnt is an exponentially increasing or decreasing function. The response ejwnt is a sinusoidal function. Hence, the response of the product of these responses will be over damped sinusoids or under damped sinusoids (Fig. 12.84). f (t)

jw

s

0

t

0

(a)

f (t)

jw

s 0

t

0

(b)

Fig. 12.84 (a) Complex conjugate poles in left half of the S-plane (b) Complex conjugate poles in right half of the S-plane (vi)

The real part s of the pole is the displacement of the pole from the imaginary axis. Since s is also the damping factor, a greater value of s (i.e., a greater displacement of the pole from the imaginary axis) means that response decays more rapidly with time. The poles with greater displacement from the real axis correspond to higher frequency of oscillation (Fig. 12.85). f (t) jw 0 0

t

s

f (t)

0

t

(a)

Fig. 12.85 Nature of response with different positions of poles

12.42 Network Analysis and Synthesis f (t)

jw t

0 s

0

f (t)

t

0

(b)

Fig. 12.85 (Continued)

12.9.1 Stability of the Network Stability of the network is directly related to the location of poles in the s-plane. (i) (ii) (iii) (iv) (v)

When all the poles lie in the left half of the s-plane, the network is said to be stable. When the poles lie in the right half of the s-plane, the network is said to be unstable. When the poles lie on the jw axis, the network is said to be marginally stable. When there are multiple poles on the jw axis, the network is said to be unstable. When the poles move away from jw axis towards the left half of the s-plane, the relative stability of the network improves.

12.10

GRAPHICAL METHOD FOR DETERMINATION OF RESIDUE

Consider a network function, F ( s) = H

( s z )(s )( s z ) ( s p )(s )( s p )

( s zn ) ( s pm )

By partial fraction expansion, F ( s) =

K1 K2 + + (s p ) (s p )

+

Km ( s pm )

( s − p ) F ( s ) |s → p

(p (p

The residue Ki is given by Ki

i

H

z )( )( p p )( )( p

z ) p )

(p (p

zn ) pm )

Each term (pi – zi) represents a phasor drawn from zero zi to pole pi. Each term (pi – pk), i ≠ k, represents a phasor drawn from other poles to the pole pi. Ki

H

Product of phasors (polar form) from each zero to pi Product of phasor a s (polar form) from other poles to pi

12.10 Graphical Method for Determination of Residue 12.43

The residues can be obtained by graphical method in the following way: (1) (2) (3) (4) (5) (6)

Draw the pole-zero diagram for the given network function. Measure the distance from each of the other poles to a given pole. Measure the distance from each of the other zeros to a given pole. Measure the angle from each of the other poles to a given pole. Measure the angle from each of the other zeros to a given pole. Substitute these values in the required residue equation.

The graphical method can be used if poles are simple and complex. But it cannot be used when there are multiple poles. 2s The current I(s) in a network is given by I(s) s = . Plot the pole-zero (s + 1)(s + 2) pattern in the s-plane and hence obtain i(t).

Example 12.43

Solution Poles are at s = –1 and s = –2 and zero is at s = 0. The pole-zero plot is shown in Fig. 12.86. By partial-fraction expansion, I ( s) =

K1 K + 2 s +1 s + 2 jw

s 0

−2 −1

Fig. 12.86 The coefficients K1 and K2, often referred as residues, can be evaluated from he pole-zero diagram. From Fig. 12.87, Phasor from zero at origin to pole at A ⎛ 1∠180° ⎞ K1 H = 2⎜ = 2 ∠180° = −2 ⎝ 1∠0° ⎟⎠ Phasor from pole at B to pole at a A jw

B

A s

−2 −1

0

Fig. 12.87 From Fig. 12.88, K2

H

⎛ 2 ∠180 Phasor from zero at origin to pole at B 80° ⎞ = 2⎜ =4 Phasor from pole at A to pole at a B ⎝ 1 80° ⎟⎠

12.44 Network Analysis and Synthesis jw

B

A s 0

−2 −1

Fig. 12.88 2 4 + s +1 s + 2 Taking inverse Laplace transform, i (t) = –2e−t + 4e−2t I ( s) = −

Example 12.44

The voltage V(s) of a network is given by

V ( s) =

3s (s

)(s )( )( s 2

s

)

Plot its pole-zero diagram and hence obtain v (t). Solution

V ( s) =

3s (s

)( s )(s )(

2

s

)

=

(s

)(s )( )( s

3s j ))(s (s

j)

Poles are at s = –2 and s = –1 ± j1 and zero is at s = 0 as shown in Fig. 12.89. jw B

j1

A

−2

0

−1 C

s

−1 −j

Fig. 12.89 By partial-fraction expansion, V ( s) =

K1 K2 K 2* + + s + 2 s 1 j1 s 1 j1

The coefficients K1, K2 and K2* can be evaluated from the pole-zero diagram. From Fig. 12.90, K1 =

⎡ ⎤ 2∠180 80° = 3⎢ ⎥ = 3 180° = −3 ( BA) (CA) 35 ) ( 2 135 35 ) ⎦ ⎣ ( 2 − 135 3 (OA)

12.10 Graphical Method for Determination of Residue 12.45 jw B

j1 √2

135°

A

180°

−2

s

0

−1

135°

√2

−j 1 C

Fig. 12.90 From Fig. 12.91, ⎡ ( 2 135 35 ) = 3⎢ ( AB ) (CB) 5 ) ( 2 90 ⎣ ( 2 45 3 K 2* = 2 K2 =

3 (OB)

⎤ 3 ⎥= )⎦ 2

jw B

j1

√2

√2

A

45°

−2

−1

135° 0

s

90°

−1 −j C

Fig. 12.91

V ( s) = −

3 (s

)

+

3 2 (s

j)

+

3 2 (s

j)

Taking inverse Laplace transform, v(t ) = −3e −2tt

Example 12.45

3 ( ⎡e 2⎣

j )t )t

+ e(

j )t ⎤

⎦

3e

2t

⎛ e j1 + e − j1 ⎞ 3 + 2 × e−t ⎜ ⎟ = −3e 2 2 ⎝ ⎠

2t

3e t cos t

Find the function v(t) using the pole-zero plot of following function: V(s) =

(s + 2)(s + 6) (s + 1)(s + 5)

Solution If the degree of the numerator is greater or equal to the degree of the denominator, we have to divide the numerator by the denominator such that the remainder can be expanded into partial fractions.

12.46 Network Analysis and Synthesis V ( s) =

s 2 + 8 s + 12 2

s + 6s + 5

= 1+

2s + 7 2

s + 6s + 5

= 1+

2(s ( .5) ( s )(s )( )

By partial fraction expansion, V ( s) = 1 +

K1 K + 2 s +1 s + 5

K1 and K2 can be evaluated from the pole-zero diagram shown in Fig. 12.92 and Fig. 12.93. jw

jw

s

s

0

−5

− 3.5

0

−1

−5

−3.5

Fig. 12.92

−1

Fig. 12.93

From Fig. 12.92 ⎛ 2 5∠0° ⎞ 5 K1 = 2 ⎜ = ⎝ 4 ∠0° ⎟⎠ 4 From Fig. 12.93 80° ⎞ 3 ⎛ 1 5∠180 K2 = 2 ⎜ = ⎝ 4 ∠180° ⎟⎠ 4 5 3 4 4 V ( s) = 1 + + s +1 s + 5 Taking inverse Laplace transform, 5 −tt 3 e + e 4 4

v (t ) = δ (t (t )

Example 12.46

5t

The pole-zero plot of the driving-point impedance of a network is shown in

Fig. 12.94. Find the time-domain response. jw

j1

−1

Scale factor = 5

0

s

−1 −j

Fig. 12.94 Solution The function Z(s) has poles at s = −1 + j1 and s = −1 −j1 and zero at s = 0.

12.10 Graphical Method for Determination of Residue 12.47

Z ( s) = H

s j )( s

(s

j)

H=5

Scale factor

Z ( s) =

jw

5s j )( s

(s

j)

j1

A

By partial fraction expansion, Z ( s) =

K1 K1* + s 1 j1 s 1 j1

s

−1 −j

B

d K1* can be evaluated from the pole-zero

The coefficients K1

0

−1

Fig. 12.95

diagram. From Fig. 12.95, K1 =

5 (OA) ( BA)

=

5 ( 2 135 35 ) = 3 54 ∠455° 2∠90°

K1*

= 3 54 ∠ 45 5° 3.54 ∠45 5° 3.54 ∠ 455° Z ( s) = + s 1 j1 s 1 j1 Taking inverse Laplace transform, z(t) = 3.54 ∠ 45° e(−1−j1)t + 3.54 ∠−45°e(−1+j1)t Evaluate amplitude and phase of the network function F(s) =

Example 12.47 pole-zero plot at s = j2.

F ( s) =

Solution

4s s2 + 2s + 2

=

(s

4s j )( s

4s s

2

2s 2

from the

j)

The pole-zero plot is shown in Fig. 12.96. At s = j2, jw j2 j1 −1

s 0

−j 1

Fig. 12.96 |

|=

Product of phasor magnitudes from all zero to j 2 2 = = 0.447 Product of phaso h r magnitudes from all poles to j 2 ( 2 ) ( 10 )

12.48 Network Analysis and Synthesis

φ ( ) = tan −1

Example 12.48

⎛ 2⎞ ⎝ 0⎠

⎛ 3⎞ ⎛ 1⎞ tan −1 ⎜ ⎟ − tan −1 ⎝ 1⎠ ⎝ 1⎠

90° 71.56° − 45 4 ° = −26.56°

Using the pole-zero plot, find magnitude and phase of the function

Solution

F ( s) =

(s

)( )(s ) at s s (s )

F ( s) =

(s

)( )(s ) s (s )

j 4.

The pole-zero plot is shown in Fig. 12.97. At s = j4, jw j4

s

−3

−2

−1

0

Fig. 12.97 ⎪ ( j )⎪=

φ (ω ) = tan −1

⎛ 4⎞ ⎝ 1⎠

Example 12.49

Product of phasor magnitudes from all zeros to j 4 ( ) ( ) = = 1.15 product of phaso p r magnitudes from all poles to j 4 ( )( )

⎛ 4⎞ ⎛ 4⎞ tan −1 ⎜ ⎟ − tan1 ⎝ 3⎠ ⎝ 0⎠

⎛ 4⎞ tan −1 ⎜ ⎟ = 75.96° + 53.13° − 90° − 63.43° = −24.34° ⎝ 2⎠

Plot amplitude and phase response for F(s) =

s s 10

Solution F( j ) = ⎪F(( j )⎪=

jω jω + 10 ω

ω 2 + 100

12.10 Graphical Method for Determination of Residue 12.49

v

ÃF(jv)Ã

0

0

10

0.707

100

0.995

1000

1

1

The amplitude response is shown in Fig. 12.98.

φ (ω ) = tan −1

⎛ω⎞ ⎝ 0⎠

F(jw)

0.7

w 0

10

100

1000

Fig. 12.98

⎛ω⎞ ⎛ω⎞ tan −1 ⎜ ⎟ = 90° − tan −1 ⎜ ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ f (w)

v

e (v)

0

90°

10

45°

100

5.7°

1000

0°

90°

45° w 10

The phase response is shown in Fig. 12.99

100

1000

Fig. 12.99

Example 12.50

Sketch amplitude and phase response for F(s) =

s 10 s 10

Solution F(jw)

jω + 10 F( j ) = jω − 10 ⎪F ( fω )⎪=

1

ω 2 + 100 ω 2 + 100

w 0

For all w, magnitude is unity. The amplitude response is shown in Fig. 12.100.

φ (ω ) = tan −1

⎛ω⎞ ⎝ 10 ⎠

⎛ ω⎞ ⎛ω⎞ tan −1 ⎜ − ⎟ = 2 tan −1 ⎜ ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠

The phase response is shown in Fig. 12.101. v

e(v)

0

0°

10

90°

100

168.6°

1000

178.9°

Fig. 12.100

f (w) 180° 90°

w 0

10

100

1000

Fig. 12.101

12.50 Network Analysis and Synthesis

Exercises 12.1 Determine the driving-point impedance transfer impedance ratio I1

V1 , I1

V2 and voltage transfer I1

V2 for the network shown in Fig. 12.102. V1 5Ω

1

I1

2H

V1 1′

1 F 2

2H

V1

2F

2F

V2 −

−

2′

Fig. 12.104 1 1 ⎡ ⎤ , G21 = 4 ⎢ Z 21 = 3 ⎥ 2 2 s 3 s 4 s 7 s + 1 ⎣ ⎦

V2

12.4 For the two-port network shown in Fig. 12.105, determine Z11, Z21 and voltage transfer ratio G21(s).

−

−

2

+

+

+

I2 = 0

+

I2 = 0

2Ω

1H

Fig. 12.102

I1

⎡V1 7 s 7 s + 5 V2 2s ; = 2 ; ⎢ = s 2 s 1 I1 s + s + 1 ⎣ I1 V2 2s ⎤ = 2 V1 7 s 7 s + 5 ⎥⎦ 2

2Ω

2H

+

+ 1F

V1

1H

−

V2 −

Fig. 12.105 12.2 For the network shown in Fig. 12.103, V V determine 2 and 2 . V1 I1 2H I1 + V1

I2 = 0

1 F 2 1 F 2

+ 1 F 2

12.5 Draw the pole-zero diagram of the following network functions:

V2 −

−

⎡ 2ss3 s 2 + 3s 2 s , Z 21 = 2 , ⎢ Z11 = 2 s + 2s 2s 1 s + 2s 2s 1 ⎣ s ⎤ G21 = 3 ⎥ 2 2ss s + 3s 2 ⎦

(i) F ( s) =

Fig. 12.103

(ii) F ( s) =

⎡V2 s 2 + 1 V2 2s2 2 ⎤ ; = ⎢ = 2 ⎥ 2 ⎣ V1 2 s 1 I1 s (3s 2) ⎦

(iii) F ( s) =

12.3 Find the open-circuit transfer impedance Z21 and open-circuit voltage ratio G21 for the ladder network shown in Fig. 12.104.

(iv) F ( s) =

s2 + 4 s2 + 6s + 4 5 s − 12 2

s + 4 s + 13 s +1 (s

2

s ( s2 4

)2

s ) 2

s + s +1

Exercises 12.51 2

(v) F s

s2

(vi) F s =

Z (s

s

1

s

s+ 2

Fig. 12.108

s s

⎡ ⎢Z s ⎣

s

s

s

s

12.6 For the network shown in Fig. 12.106, draw the pole-zero plot of the impedance function Z(s).

)

V s V s

1Ω

I1

4Ω

s + 0.

s

V V and for the V I1 network shown in Fig. 12.109 and plot poles and zeros of

1 F 20

1. s s 2 + 0

12.9 Find network functions

2

Z (s)

H

s −s 2 s2

(viii) F s =

2F

+ s2

s

(vii) F s

1F

s 2

2

+

+

V

V

1

−

1Ω

Fig. 12.106 Z s

⎡ s

j1.

⎤

s s+5

⎡V

12.7 For the network shown in Fig. 12.107, draw the pole-zero plot of driving-point impedance function Z(s). 5Ω

Z (s)

Fig. 12.109

10 Ω

1

⎣V

2s

2s

s + 2s 1 ⎤

2s

s + 2s 1 ⎦

12.10 For the network shown in Fig. 12.110, V V determine and Plot the poles and I1 I1 zeros of

10

I1

V . I1 2H

1H

+

Fig. 12.107 Z s =

V s + 2 s 1 I1

5(s s s

s + 0. .03)

12.8 Find the driving-point impedance of the network shown in Fig. 12.108. Also, find poles and zeros.

V

+ 1

1 2

V

Fig. 12.110 V s s +2 V = , I1 I1 s s

2 s

⎤ ⎥ s⎦

12.52 Network Analysis and Synthesis 12.11 For the network shown in Fig. 12.111, V V determine 1 and 2 . Plot the pole and zeros I1 V1 V2 for . V1 I1

12.14 Obtain the impedance function for which the pole-zero diagram is shown in Fig. 12.114. jw j1

2F

1F

Z ( j 0) = 1 +

+ 1H

V1

1H

−2

0

−1

−j1

V2

−

−

Fig. 12.114

Fig. 12.111 ⎡V1 s + 3s 3s 1⎤ ⎢ = ⎥ 3 2s + s ⎦ ⎣ I1 4

1H

1H

+

+ 1Ω

1Ω

V1

2 (s (s ) ⎤ ⎡ ⎢ Z ( s) = 2 ⎥ s + 2s + 2 ⎦ ⎣

2

12.12 For the network shown in Fig. 12.112, plot the poles and zeros of transfer impedance function. I1

s

12.15 For the network shown in Fig. 12.115, poles and zeros of driving point function Z(s) are, Poles: (–1 ± j4); zero: –2 If Z (j0) = 1, find the values of R, L and C.

V2

−

−

R

1 Cs

Z (s)

Ls

Fig. 12.112 1 ⎤ ⎡V2 ⎢ I = s + 2⎥ ⎣ 1 ⎦ 12.13 For the network shown in Fig. 12.113, V V determine 1 and 2 . Plot the poles and I1 V1 zeros of transfer impedance function. I1

2H

4H

2 ⎡ ⎢1 Ω, 0.5 Η, 17 ⎣

R1

1H 2F

−

1 F V2

+

+ −

Fig. 12.113 ⎡V1 16 s 4 + 100 s 2 + 1 V2 1 , = 3 , ⎢ = 3 I I 8s 3s 8s 3s 1 ⎣ 1 V2 1 ⎤ = ⎥ 4 2 V1 16 s +10 0 s + 1⎦

⎤ F⎥ ⎦

12.16 For the two-port network shown in Fig. V 02 12.116, find R1, R2 and C. 2 = 2 V1 s + 3s 3s 2

+

+ V1

Fig. 12.115

V1

C

−

R2

V2 −

Fig. 12.116 1 ⎡3 ⎤ ⎢ 5 Ω, 15 Ω, 0.5 F⎥ ⎣ ⎦

Objective-Type Questions 12.53

12.17 For the given network function, draw the pole-zero diagram and hence obtain the time domain voltage. V ( s) =

5 (s ( ) ( s )(s )(s ( ) [v(t) = 3e−2t + 2e−7t]

12.18 A

transfer

function is given by 10 s Y ( s) = . Find time(s j )(s )( s j ) domain response using graphical method.

⎡5.26 18.4° e ⎣

(5 (5

15) 15 5)

+ 5.266

188.4° e

( 5 j 15) t ⎤

⎦

Objective-Type Questions 12.1 Of the four networks N1, N2, N3 and N4 of Fig. 12.117, the networks having identical drivingpoint functions are (a) N1 and N2 (b) N2 and N4 (c) N1 and N3 (d) N1 and N4

12.2 The driving-point impedance Z(s) of a network has the pole-zero locations as shown in Fig. 12.118. If Z(0) = 3, then Z(s) is jw j1

1Ω

2H

1F

2Ω

−3

1F

s

0

−1

−j 1 N1

Fig. 12.118 1Ω

2H

(a) 2Ω

3(

3)

2

s +2

(b)

3

2(

3)

2

s + 2s 2s 2

1F N2 1Ω

1Ω

1H 1F

(c)

3( s

2

3) 2s 2

1Ω

2( s

2

3

+ 100 μF

vi (t) −

v0 −

1F

Fig. 12.119

N4

Fig. 12.117

2

10 mH

+ 2H

3)

12.3 For the circuit shown in Fig. 12.119, the initial conditions are zero. Its transfer function V ( s) H ( s) = 0 is V1 ( s) 10 kΩ

N3

(d)

(a)

1 s

2

6

100 s + 10

6

(b)

106 s

2

1003 s + 106

12.54 Network Analysis and Synthesis 103

(c)

s 2 10 03 s + 106

106

(d)

s 2 10 06 s + 106

12.4 In Fig. 12.120, assume that all the capacitors are initially uncharged. If vi(t) = 10 u(t), then v0(t) is given by

12.7 A network has response with time as shown in Fig. 12.122. Which one of the following diagrams represents the location of the poles of this network? x (t)

1 kΩ + 4 μF

vi (t)

1 μF

4 kΩ

t

0

+ v0 (t)

Fig. 12.122

−

−

jw

Fig. 12.120 (a) 8 e–0.004 t (c) 8 u(t)

(b) 8 (1–e–0.004 t) (d) 8

12.5 A system is represented by the transfer 10 function . The dc gain of this ( 1 1) )( )( 2) system is

(a)

jw s

0

(b)

0 jw

jw (c)

s

s

0

(d)

0

s

(a) 1

(b) 2

Fig. 12.123

(c) 5

(d) 10

12.8 The transfer function of a low-pass RC network is

12.6 Which one of the following is the ratio

V24 V13

(a) (RCs) (1 + RCs)

(b)

1 1+ RCs C

(d)

s 1+ RCs C

of the network shown in Fig. 12.121. (c)

1Ω 1

2 1Ω

1Ω

3

RCs C 1+ RCs C

12.9 The driving-point admittance function of the network shown in Fig. 12.124 has a

4

R

1Ω

L

Fig. 12.121 (a)

1 3

(b)

2 3

(c)

3 4

(d)

4 3

Fig. 12.124 (a) (b) (c) (d)

pole at s = 0 and zero at s = ∞ pole at s = 0 and pole at s = ∞ pole at s = ∞ and zero at s = 0 pole at s = ∞ and zero at s = ∞

C

Answers to Objective-Type Questions 12.55

I 2 ( s) for the V1 ( s) network shown in Fig. 12.125 is

12.10 The transfer function Y12 ( s) =

(a)

(c) 1Ω

I2(s)

V1(s)

1H

1F

Fig. 12.125

s2

(b)

2

s + s +1 1 s +1

(d)

s s +1 s +1 s2 + 1

12.11 As the poles of a network shift away from the x axis, the response (a) remains constant (b) becomes less oscillating (c) becomes more oscillating (d) none of these

Answers to Objective-Type Questions 1. (c)

2. (b)

3. (d)

4. (c)

5. (c)

7. (d)

8. (b)

9. (a)

10. (a)

11. (c)

6. (a)

13 Two-Port Networks

13.1

INTRODUCTION

A two-port network has two pairs of terminals, one pair at the input known as input port and one pair at the output I1 I2 known as output port as shown in Fig. 13.1. There are + Two-port network four variables V1, V2, I1 and I2 associated with a two-port V1 − network. Two of these variables can be expressed in terms of the other two variables. Thus, there will be two dependent Fig. 13.1 Two-port network variables and two independent variables. The number of possible combinations generated by four variables taken two at a time is 4C2, i.e., six. There are six possible sets of equations describing a two-port network. Table 13.1 Two–port parameters Parameter

Variables

Equation

Express

In terms of

Open-Circuit Impedance

V1, V2

I1, I2

V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2

Short-Circuit Admittance

I1, I2

V1, V2

I1 = Y11 V1 + Y12 V2 I2 = Y21 V1 + Y22 V2

Transmission

V1, I1

V2, I2

V1 = AV2 − BI2 I1 = CV2 − DI2

Inverse Transmission

V2, I2

V1, I1

V2 I2

Hybrid

V1, I2

I1, V2

V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2

Inverse Hybrid

I1, V2

V1, I2

I1 = g11 V1 + g12 I2 V2 = g21 V1 + g22 I2

A′ V1 C ′ V1

B′ I1 D ′ I1

+ V2 −

13.2 Network Analysis and Synthesis

13.2

OPEN-CIRCUIT IMPEDANCE PARAMETERS (Z PARAMETERS)

The Z parameters of a two-port network may be defined by expressing two-port voltages V1 and V2 in terms of two-port currents I1 and I2. ( , V2 ) f ( I1 , ) V1 Z11 I1 Z12 I 2 V2

Z 21 I1 Z 22 I 2

In matrix form, we can write ⎡V1 ⎤ ⎡ Z11 ⎢⎣V2 ⎥⎦ = ⎢⎣ Z 21

[V ] [ Z ] [ I ]

Z12 ⎤ ⎡ I1 ⎤ Z 22 ⎥⎦ ⎢⎣ I 2 ⎥⎦

The individual Z parameters for a given network can be defined by setting each of the port currents equal to zero. Case 1 When the output port is open-circuited, i.e., I2 = 0 V Z11 = 1 I1 I 2 = 0 where Z11 is the driving-point impedance with the output port open-circuited. It is also called open-circuit input impedance. Similarly, V Z 21 = 2 I1 I 2 = 0 where Z21 is the transfer impedance with the output port open-circuited. It is also called open-circuit forward transfer impedance. Case 2 When input port is open-circuited, i.e., I1 = 0 Z12 =

V1 I2

I1 = 0

where Z12 is the transfer impedance with the input port open-circuited. It is also called open-circuit reverse transfer impedance. Similarly, V Z 22 = 2 I 2 I1 = 0 where Z22 is the open-circuit driving-point impedance with the input port open-circuited. It is also called open circuit output impedance. As these impedance parameters are measured with either the input or output port open-circuited, these are called open-circuit impedance parameters. The equivalent circuit of the two-port network in terms of Z parameters is shown in Fig. 13.2.

I1

+

+ Z11

V1 Z12I2 −

Fig. 13.2

13.2.1 Condition for Reciprocity A network is said to be reciprocal if the ratio of excitation at one port to response at the other port is same if excitation and response are interchanged. (a) As shown in Fig. 13.3, voltage Vs is applied at the input port with the output port short-circuited.

I2

Z22 + −

Fig. 13.3

−

Equivalent circuit of the two-port network in terms of Z parameter

I1 + Vs −

V2

+ Z I − 21 1

I2 Network

I2′

Network for deriving condition for reciprocity

13.2 Open-Circuit Impedance Parameters (Z Parameters) 13.3

i.e.,

V1 Vs V2 = 0 I2 I2 ′ From the Z-parameter equations, Vs Z11 I1 Z12 I 2 ′ 0 = Z 21 I1 − Z 22 I 2 ′ Z I1 = 22 I 2 ′ Z 21 Z 22 Vs Z11 I 2 ′ Z12 I 2 ′ Z 21 Vs Z Z − Z12 Z 21 = 11 22 I2 ′ Z 21 (b) As shown in Fig. 13.4, voltage Vs is applied at the output port with input port short-circuited. I1′ V2 Vs i.e., V1 = 0

i.e.,

I1

I1 I1 ′ Fig. 13.4 From the Z-parameter equations, 0 = − Z11 I1 ′ + Z12 I 2 Vs = − Z 221 I1 ′ + Z 222 I 2 Z I 2 = 111 I1 ′ Z112 Z Vs Z 21 I1 ′ Z 22 11 I1 ′ Z12 Vs Z11Z 22 − Z12 Z 21 = I1 ′ Z12 Hence, for the network to be reciprocal, Vs Vs = I1 ′ I 2 ′ Z12 Z 21

I2 Network

+ −

Vs

Network for deriving condition for reciprocity

13.2.2 Condition for Symmetry For a network to be symmetrical, the voltage-to-current ratio at one port should be the same as the voltageto-current ratio at the other port with one of the ports open-circuited. (a) When the output port is open-circuited, i.e., I2 = 0 From the Z-parameter equation, Vs Z11 I1 Vs = Z11 I1 (b) When the input port is open-circuited, i.e., I1 = 0 From the Z-parameter equation, Vs Z 22 I 2 Vs = Z 22 I2 Hence, for the network to be symmetrical,

13.4 Network Analysis and Synthesis Vs Vs = I1 I 2 Z11 Z 22

i.e.,

Example 13.1 Test results for a two-port network are (a) I1 = 0.1 Æ 0é A, V1 = 5.2 Æ 50é V, V2= 4.1 Æ-25é V with Port 2 open-circuited (b) I = 0.1 Æ 0é A, V = 3.1 Æ−80é V, V =4.2 Æ 60é V, with Port 1 2 1 2 open-circuited. Find Z parameters. Solution V 5 2∠50° Z11 = 1 = = 52 ∠50 ∠50° Ω, I1 I 2 = 0 0 1∠0° Z 221 =

V2 I1

= I2 =0

Z112 =

4.1∠ − 25° = 41 ∠ − 25° Ω, 0 1∠0°

Hence, the Z-parameters are ⎡ Z11 ⎢⎣ Z 21

Example 13.2

Z222 =

V2 I2

=

3 1∠ − 80° = 31 ∠ 80° Ω 0.1∠0°

=

4.2∠60° = 42 ∠60° Ω 0 1∠0°

I1 = 0

I1 = 0

Z12 ⎤ ⎡52∠50° 31∠ − 80°⎤ = Z 22 ⎥⎦ ⎢⎣ 41∠ − 25° 42∠60° ⎥⎦

Find the Z parameters for the network shown in Fig. 13.5. I1

Z1

Z3

I2 +

+

V2

Z2

V1

−

−

Fig. 13.5 Solution First Method Case 1 When the output port is open-circuited, i.e., I2 = 0. Applying KVL to Mesh 1, V1 ( Z1 + Z 2 ) I1 Z11 = Also

V1 I2

V2 Z 21 =

V1 I1

= Z1 Z 2 I2 = 0

Z 2 I1 V2 I1

= Z2 I2 = 0

Case 2 When the input port is open-circuited, i.e., I1 = 0. Applying KVL to Mesh 2, V2 ( Z 2 + Z3 ) I 2 Z 22 =

V2 I2

= Z2 I1 = 0

Z3

13.2 Open-Circuit Impedance Parameters (Z Parameters) 13.5

Also

V1

Z2 I 2

Z12 =

V1 I2

Hence, the Z-parameters are ⎡ Z11 ⎢⎣ Z 21

= Z2 I1 = 0

Z12 ⎤ ⎡ Z1 Z 2 = Z 22 ⎥⎦ ⎢⎣ Z 2

Z2 ⎤ Z 2 + Z3 ⎥⎦

Second Method The network is redrawn as shown in Fig. 13.6. Applying KVL to Mesh 1, V1

Z1 I1 = ( Z1

Z 2 ( I1

I2 )

Z 2 ) I1 Z 2 I 2

Z1

…(i)

Z3 +

+

Applying KVL to Mesh 2, V2

Z3 I 2

Z 2 ( I1

= Z 2 I1 + ( Z 2

Z3 ) I 2

I1

…(ii)

Z12 ⎤ ⎡ Z1 Z 2 = Z 22 ⎥⎦ ⎢⎣ Z 2

Example 13.3

I2

−

−

Comparing Eqs (i) and (ii) with Z-parameter equations, ⎡ Z11 ⎢⎣ Z 21

V2

Z2

V1

I2 )

Fig. 13.6

Z2 ⎤ Z 2 + Z3 ⎥⎦

Find Z-parameter for the network shown in Fig. 13.7. I1

2Ω

1Ω

1H

I2

+

+

V1

1Ω

2F

V2

−

−

Fig. 13.7 Solution The transformed network is shown in Fig. 13.8. Z1 = 1 ⎛ 1⎞ ⎜⎝ ⎟⎠ (1) 1 2s Z2 = = 1 2s 1 +1 2s Z3 s + 2

1 +

+ Z1

V1

Z 21 2 =

V1 I1 V2 I1

= Z1

Z2 = 1 +

I2 0

= Z2 = I2 = 0

1 2s 1

1 2s 2 = , 2s 1 2s 1

1 2s

1 Z 2

−

Z3

V2 −

Fig. 13.8

From definition of Z-parameters, Z11 =

s

2

Z12 = Z 222 =

V1 I2 V2 I2

= Z2 = I1 = 0

1 2s 1

= Z 2 + Z3 = I1 = 0

1 2 2 5s + 3 +s+2 = 2s 1 2s 1

13.6 Network Analysis and Synthesis

Example 13.4

Find Z-parameters for the network shown in Fig. 13.9. I1

1Ω

1Ω

I2

+

+

2Ω

V1

V2

2Ω

−

−

Fig. 13.9 Solution The network is redrawn as shown in Fig. 13.10. Applying KVL to Mesh 1, …(i)

V1 3I1 2 I 3 Applying KVL to Mesh 2, V2 2 I 2 2 I 3 Applying KVL to Mesh 3, −2 1 + 2 I 2 + 5

3

…(ii)

1Ω

1Ω

+

+

2Ω

V1

=0

I1

2 2 I1 − I 2 5 5 Substituting Eq. (iii) in Eq. (i),

…(iii)

I3 =

2Ω I3

V2 I2

−

−

Fig. 13.10

4 4 I1 I2 5 5 11 4 = I1 I2 5 5

V1 = 3I1

…(iv)

Substituting Eq. (iii) in Eq. (ii), 2I 2 +

V2 =

4 I1 5

4 4 I1 I2 5 5 6 I2 5

…(v)

Comparing Eqs (iv) and (v) with Z-parameter equations, ⎡ Z11 ⎢⎣ Z 21

⎡11 4 ⎤ Z12 ⎤ ⎢ 5 5 ⎥ = Z 22 ⎥⎦ ⎢⎢ 4 6 ⎥⎥ ⎣ 5 5⎦

Example 13.5

Find the Z-parameters for the network shown in Fig. 13.11. I1

1Ω

1Ω

I2

+ V1

+

1H

−

1H

V2 −

Fig. 13.11

13.2 Open-Circuit Impedance Parameters (Z Parameters) 13.7

Solution The transformed network is shown in Fig. 13.12. Applying KVL to Mesh 1, 1 V1 ( s + 1) I1 sI 3 …(i) + Applying KVL to Mesh 2, V2 sI sI 2 sI 3 …(ii) V1 I1 Applying KVL to Mesh 3,

+ s

s I3

−

− sI1 + sI 2 + ( 2 s + 1) I 3 = 0 I3 =

1

s s I1 − I2 … 2s 1 2s 1 (iii)

V2 I2

−

Fig. 13.12

Substituting Eq. (iii) in Eq. (i), V1

( s + 1) I1

⎛ s s⎜ I1 ⎝ 2s 1

s ⎞ I2 ⎟ 2s 1 ⎠

⎛ s 2 + 3s ⎛ s2 ⎞ 3s 1⎞ =⎜ I + 1 ⎟ ⎜ 2 s + 1⎟ I 2 ⎝ 2s 1 ⎠ ⎝ ⎠

…(iv)

Substituting Eq. (iii) in Eq. (ii), V2

s 2 sI

⎛ s s⎜ I1 ⎝ 2s 1

s ⎞ I2 ⎟ 2s 1 ⎠

⎛ s2 ⎞ ⎛ s2 + s ⎞ =⎜ I1 + ⎜ ⎟ ⎟ I2 ⎝ 2 s 1⎠ ⎝ 2s 1⎠

…(v)

Comparing Eqs (iv) and (v) with Z-parameter equations, ⎡ Z11 ⎢⎣ Z 21

⎡ s 2 + 3s 3s 1 Z12 ⎤ ⎢ 2 s +1 1 =⎢ Z 22 ⎥⎦ ⎢ s 2 ⎢⎣ 2 s 1

s2 ⎤ ⎥ 2s 1 ⎥ 2s s2 + s ⎥ 2 s 1 ⎥⎦

Example 13.6 Find the open-circuit impedance parameters for the network shown in Fig. 13.13. Determine whether the network is symmetrical and reciprocal. 4Ω

I1

1Ω

3Ω

+ V1

I2 +

2Ω

−

V2 −

Fig. 13.13

13.8 Network Analysis and Synthesis Solution The network is redrawn as shown in Fig. 13.14. Applying KVL to Mesh 1, V I + I2 0 V …(i) Applying KVL to Mesh 2, V I + I2 0 V Applying KVL to Mesh 3, − − − = 2+ I − + I 0 1 8 Substituting Eq. (iii) in Eq. (i),

1Ω

I3

3Ω

…(ii) +

+

I1

3 − I2 8

…(iii)

1 8

V

2Ω

V

V 23 8

4Ω

I

Fig. 13.14

3 − I2 8

19 8

…(iv)

Substituting Eq. (iii) in Eq. (ii), 1

V

3 − I2 8

19 31 8 8 Comparing Eqs (iv) and (v) with Z-parameter equations, 23 19 ⎤ Z Z12 ⎤ 8 8 Z Z 22 19 31 8 8⎦ Since Z11 ≠ Z22, the network is not symmetrical. Since Z12 = Z21, the network is reciprocal.

13.3

…(v)

SHORT-CIRCUIT ADMITTANCE PARAMETERS (Y PARAMETERS)

The Y parameters of a two-port network may be defined by expressing the two-port currents I1 and I2 in terms of the two-port voltages V1 and V2. 2

( , V V

In matrix form, we can write I1 I2 I

⎡Y Y Y V

Y ⎤ ⎡V ⎤ Y V

13.3 Short-Circuit Admittance Parameters (Y Parameters) 13.9

The individual Y parameters for a given network can be defined by setting each of the port voltages equal to zero. Case 1 When the output port is short-circuited, i.e., V2 = 0 I Y11 = 1 V1 V2 = 0 where Y11 is the driving-point admittance with the output port short-circuited. It is also called short-circuit input admittance. Similarly, I Y21 = 2 V1 V2 = 0 where Y21 is the transfer admittance with the output port short-circuited. It is also called short-circuit forward transfer admittance. Case 2 When the input port is short-circuited, i.e., V1 = 0 I Y12 = 1 V2 V1 = 0 where Y12 is the transfer admittance with the input port short-circuited. It is also called short-circuit reverse transfer admittance. Similarly, I Y22 = 2 V2 V1 = 0 where Y22 is the short-circuit driving-point admittance with the input port short-circuited. It is also called the short circuit output admittance. As these admittance parameters are measured with either input or output port short-circuited, these are called short-circuit admittance parameters. The equivalent circuit of the two-port network in terms of Y parameters is shown in Fig. 13.15. I2

I1 + V1

+ Y11

Y12V2

Y22

V2 −

−

Fig. 13.15

Y21V1

2

Equivalent circuit of the two-port network in terms of Y-parameters

Condition for Reciprocity (a) As shown in Fig. 13.16, voltage Vs is applied at input port with the output port short-circuited. i.e, V1 Vs I1 I2 V2 = 0 I2 I 2′ + Network I2′ From the Y-parameter equation, − Vs − I 2′ Y21 Vs I2 = −Y21 Vs

Fig. 13.16 Network for deriving condition for reciprocity

13.10 Network Analysis and Synthesis (b) As shown in Fig. 13.17, voltage Vs is applied at output port with the input port short-circuited. i.e, V2 Vs V1 = 0 I1 I2 I1 I1′ From the Y-parameter equation, + I ′ Network

1

− I1′

−

Y12 Vs

I1′ = −Y12 Vs Hence, for the network to be reciprocal,

Fig. 13.17

Vs

Network for deriving condition for reciprocity

I 2′ I1′ = Vs Vs Y12 Y21

i.e,

13.3.2 Condition for Symmetry (a) When the output port is short-circuited, i.e., V2 = 0. From the Y-parameter equation, I1 Y11 Vs Vs 1 = I1 Y11 (b) When the input port is short-circuited, i.e., V1 = 0. From the Y-parameter equation, I 2 Y22 Vs Vs 1 = I 2 Y22 Hence, for the network to be symmetrical, Vs Vs = I1 I 2 Y11 Y22 i.e.,

Example 13.7

Test results for a two-port network are

(a) Port 2 short-circuited: V1 (b) Port 1 short-circuited: V2 Find Y-parameters.

50 0 ∠0° V I 1 50 0 ∠0° V I 2

2.1 3

30° A, I 2 15° A, I 1

1. .

0° A 20° A.

Solution Y11 =

I1 2.1∠ − 30° = = 0.042 0 ∠ − 30° , V1 V2 = 0 50 ∠0°

I Y221 = 2 V1

V2 = 0

−1.1∠ − 20° = = −0.022∠ − 20 2 ° , 50 ∠0°

Y12 =

I1 V2

2 22

I2 V2

V1 = 0

−1.1∠ − 20 2 ° = −0.022∠ − 20° 50 ∠0°

V1 = 0

3 15° 50 ∠0°

=

Hence, the Y-parameters are 0 ∠ − 20°⎤ ⎡ Y11 Y12 ⎤ ⎡ 0.042∠ − 30° −0.022 ⎢Y 21 Y22 ⎥ = ⎢⎣ −0.022 0 ∠ − 20 0 ° 0 . 06 15° ⎥⎦ ⎣ ⎦

0.06

155°

13.3 Short-Circuit Admittance Parameters (Y Parameters) 13.11

Example 13.8

Find Y-parameters for the network shown in Fig. 13.18. I1

3

1Ω

3Ω

I2

+

+ V2

2Ω

V1 −

−

Fig. 13.18 Solution First Method Case 1 When the output port is short-circuited, i.e., V2 = 0 as shown in Fig. 13.19, Req = 1 + Now,

Also,

2×3 6 11 = 1+ = Ω 2+3 5 5

2 2 5 ( I1 ) = − × V1 5 5 11 I2 2 Y21 = =− V1 V2 = 0 11 I2

1Ω

I1

11 I1 5 I 5 Y11 = 1 = V1 V2 = 0 11 V1

3Ω

I2

+

2Ω

V1

2 V1 11

−

Fig. 13.19

Case 2 When the input port is short-circuited, i.e., V1 = 0 as shown in Fig. 13.20, Req = 3 + Now

V2 Y22

Also

1× 2 2 11 = 3+ = Ω 1+ 2 3 3

I1

2 2 3 ( I 2 ) = − × V2 3 3 11 I1 2 Y12 = =− V2 V1 = 0 11

3Ω

I2 +

11 I2 3 I 3 = 2 = V2 V1 = 0 11

I1

1Ω

2Ω

V2 −

2 V2 11

Fig. 13.20

Hence, the Y-parameters are 2⎤ ⎡ 5 − ⎥ ⎢ Y Y ⎡ 11 12 ⎤ 11 11 ⎥ ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 2 3 ⎢− ⎥ ⎣ 11 11 ⎦ Second Method (Refer Fig. 13.18) V V I1 = 1 3 1 = V1 V3

…(i)

13.12 Network Analysis and Synthesis V2 V3 3 V2 V3 = − 3 3

I2 =

…(ii)

Applying KCL at Node 3, I1

I2 =

V3 2

…(iii)

Substituting Eqs (i) and (ii) in Eq. (iii), V V V V1 V3 + 2 − 3 = 3 3 3 2 V2 11 V1 + = V3 3 6 6 2 V3 V1 + V 2 11 11 Substituting Eq. (iv) in Eq. (i), 6 2 I1 = V1 V1 V2 11 11 5 2 = V1 V2 11 11 Substituting Eq. (iv) in Eq. (ii), V 1⎛ 6 2 ⎞ I 2 = 2 − ⎜ V1 V2 ⎟ ⎝ 3 3 11 11 ⎠ 2 3 = − V1 V2 11 11 Comparing Eqs (v) and (vi) with Y-parameter equations, 2⎤ ⎡ 5 − ⎥ ⎡Y11 Y12 ⎤ ⎢ 11 11 ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 2 3 ⎥⎥ ⎢− ⎣ 11 11 ⎦

Example 13.9

…(iv)

…(v)

…(vi)

Find Y-parameters of the network shown in Fig. 13.21. I1

2Ω

2Ω

V3

+

I2 +

1Ω V1

V2

3V2 −

−

Fig. 13.21 Solution From Fig. 13.21, V1 V3 2 1 1 = V1 V3 2 2

I1 =

…(i)

13.3 Short-Circuit Admittance Parameters (Y Parameters) 13.13

V2 V3 2 1 1 = V2 V3 2 2

I2 =

…(ii)

Applying KCL at Node 3, I1 I 2 + 3 V2 Substituting Eqs (i) and (ii) in Eq. (iii), V1 V3 V2 V3 + + 3V 3 V2 2 2 2 V3

…(iii)

0 0 V1 7 V2 1 7 V1 + V2 2 2

V3

…(iv)

Substituting Eq. (iv) in Eq. (i), 1 1 ⎛1 V1 − ⎜ V1 2 2 ⎝2 1 7 = V1 V2 4 4

I1

7 ⎞ V2 ⎟ 2 ⎠ …(v)

Substituting Eq. (iv) in Eq. (ii), 1 1 ⎛1 7 ⎞ V2 − ⎜ V1 V2 ⎟ 2 2 ⎝2 2 ⎠ 1 5 = − V1 V2 4 4 Comparing Eqs (v) and (vi) with Y-parameter equations, 7⎤ ⎡ 1 − ⎥ ⎡Y11 Y12 ⎤ ⎢ 4 4 ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1 5⎥ ⎢− − ⎥ ⎣ 4 4⎦ I2

…(vi)

Example 13.10 Determine Y-parameters for the network shown in Fig. 13.22. Determine whether the network is symmetrical and reciprocal. I1

3

1Ω

2

2Ω

+ V1

I2 +

2Ω

4Ω

−

V2 −

Fig. 13.22 Solution From Fig. 13.31, V1 V3 1 = V1 V3

I1 =

…(i)

Applying KCL at Node 3, V3 V3 − V2 + 2 2 V2 = V3 − 2

I1 =

…(ii)

13.14 Network Analysis and Synthesis Applying KCL at Node 2, V2 V2 − V3 + 4 2 3 V = V2 − 3 4 2

I2 =

…(iii)

Substituting Eq. (i) in Eq. (ii), V2 2 V V V3 = 1 + 2 2 4

V1 V3 = V3 −

…(iv)

Substituting Eq. (iv) in Eq. (ii), V1 V2 V2 + − 2 4 2 V1 V2 = − 2 4

I1 =

Substituting Eq. (iv) in Eq. (iii),

…(v)

3 1 ⎛V V ⎞ V2 − ⎜ 1 + 2 ⎟ 4 2⎝ 2 4 ⎠

I2

V1 V + 2 4 8 Comparing Eqs (v) and (vi) with Y-parameter equations, 1⎤ ⎡ 1 − ⎥ ⎡Y11 Y12 ⎤ ⎢ 2 4 ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1 5 ⎥ ⎢− ⎥ ⎣ 4 8 ⎦ Since Y11 ≠ Y22, the network is not symmetrical. Since Y12 = Y21, the network is reciprocal.

…(vi)

=−

Example 13.11

Determine the short-circuit admittance parameters for the network shown in Fig. 13.23. I1

1Ω

1Ω

I2 +

+ V1

V2

1F

1F

−

−

Fig. 13.23 Solution The transformed network is shown in Fig. 13.24. From Fig. 13.24, V1 V3 1 = V1 V3

I1

…(i)

3

1

2

I2

+ V1

I1 =

1

+ 1 s

1 s

V2 −

−

Fig. 13.24

13.3 Short-Circuit Admittance Parameters (Y Parameters) 13.15

Applying KCL at Node 3, V3 (V3 V2 ) + 1 1 s = ( s 1) V3 V2

I1 =

…(ii)

Applying KCL at Node 2, I2 =

V2 (V2 V3 ) + 1 1 s …(iii)

= ( s 1) V2 V3 Substituting Eq. (i) in Eq. (ii), V1 V3 = ( s 1) V3 V2 ( s 2) V3 = V1 V2 V3 =

1 1 V1 + V2 s+2 s+2

I1

( s + 1)

…(iv)

Substituting Eq. (iv) in Eq. (ii),

=

1 ⎛ 1 ⎞ V1 + V2 V2 ⎝s+2 s+2 ⎠

s +1 V1 s+2

1 V2 s+2

…(v)

Substituting Eq. (iv) in Eq. (iii), I2

1 ⎛ 1 ⎞ ( s + 1) V2 − ⎜ V1 + V2 ⎟ ⎝s+2 s+2 ⎠

1 s 2 + 3s 3s 1 V1 + V2 s+2 s+2 Comparing Eqs (v) and (vi) with Y-parameter equations, 1 ⎤ ⎡ s +1 − ⎡Y11 Y12 ⎤ ⎢ s + 2 s+2 ⎥ ⎥ ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1 s 2 + 3s 3s 1 ⎥ ⎢− ⎢⎣ s + 2 s + 2 ⎥⎦

…(vi)

=−

Example 13.12

Determine Y-parameters for the network shown in Fig. 13.25. I1

1 F 2

1 F 2

I2

+ V1

+

1H

−

1H

V2 −

Fig. 13.25

13.16 Network Analysis and Synthesis Solution The transformed network as shown in Fig. 13.26. From Fig. 13.26, V V I1 = 1 3 + 2 s V1 s s …(i) − = V1 − V3 2 2 Applying KCL at Node 3,

I1

2 s

2 s

3

2

I2 +

s

s

V2 −

Fig. 13.26

s V s (V1 V3 ) = 3 + (V (V3 V2 ) 2 s 2 s 1 s s s V3 + V3 + V3 = V1 + V2 2 s 2 2 2 V3 =

s2 2( s 2 1)

V1 +

s2 ((ss 2 1)

V2

…(ii)

Substituting Eq. (ii) in Eq. (i), ⎤ s s ⎡ s2 s2 I1 = V1 − ⎢ 2 V1 + V2 ⎥ 2 2 2 ⎣ 2( s 1) 2( s 1) ⎦ ⎡s s3 ⎤ s3 =⎢ − V − V2 ⎥ 1 2 4( s 2 1) ⎣ 2 4( s 1) ⎦ =

s3 + 2 s 4( s 2 1)

V1 −

s3 4( s 2 1)

V2

…(iii)

Applying KCL at Node 2, I2 = =

V2 s + (V2 V3 ) s 2 s2 + 2 V2 2s

s V3 2

…(iv)

Substituting Eq. (ii) in Eq. (iv), I2 =

⎤ s2 + 2 s ⎡ s2 s2 V2 − ⎢ 2 V1 + V2 ⎥ 2 2s 2 ⎣ 2( s 1) 2( s 1) ⎦

=−

⎡ s2 + 2 s3 s3 ⎤ V + − ⎢ ⎥ V2 1 4( s 2 1) 4( s 2 1) ⎦ ⎣ 2s

s3 s4 + 6 2 4 V + V2 1 4( s 2 1) 4 s( s 2 1) Comparing Eqs (iii) and (v) with Y-parameter equation, =−

⎡Y11 ⎢⎣Y21

⎡ s3 + 2 s ⎢ Y12 ⎤ ⎢ 4( s 2 1) = Y22 ⎥⎦ ⎢ s3 ⎢− 2 ⎢⎣ 4( s 1)

⎤ s3 ⎥ 2 4( s 1) ⎥ s4 + 6s2 + 4 ⎥ ⎥ 4 s( s 2 1) ⎥⎦ −

…(v)

13.3 Short-Circuit Admittance Parameters (Y Parameters) 13.17

Example 13.13

Obtain Y-parameters of the network shown in Fig. 13.27. 1 s

I1

1

3

1

2

1

+

I2 +

1 s

V1

V2

−

−

Fig. 13.27 Solution Applying KCL at Node 1, V1 V3 V1 V2 + 1 1 s = ( s 1) V1 s V2 − V3

…(i)

V2 V3 V2 V1 + 1 1 s = ( s 1) V2 s V1 − V3

…(ii)

I1 =

Applying KCL at Node 2, I2 =

Applying KCL at Node 3, V3 V3 V1 V3 V2 + + =0 1 1 1 s ( s 2) V3 V1 V2 0 V3 =

1 1 V1 + V2 s+2 s+2

I1

( s + 1) V1

…(iii)

Substituting Eq. (iii) in Eq. (i), sV V2

1 ⎛ 1 ⎞ V1 + V2 ⎟ ⎝s+2 s+2 ⎠

⎡ ( s +11)( s 2) − 1⎤ ⎡ s ( s + 2) 1 ⎤ =⎢ V1 ⎢ ⎥ ⎥ V2 ( s + 2) ⎣ ⎦ ⎣ ( s + 2) ⎦ ⎛ s 2 + 3s ⎛ s 2 + 2s 3s 1⎞ 2 s 1⎞ =⎜ V1 − ⎜ ⎟ ⎟ V2 ⎝ s+2 ⎠ ⎝ s+2 ⎠

…(iv)

13.18 Network Analysis and Synthesis Substituting Eq. (iii) in Eq. (ii), I2

1 ⎛ 1 ⎞ V1 + V2 ⎟ ⎝s+2 s+2 ⎠ ⎡ s ( s + 2) 1 ⎤ ⎡( )( )( ) − 1⎤ = −⎢ V1 + ⎢ ⎥ ⎥ V2 ( ) ⎣ ( s + 2) ⎦ ⎣ ⎦ ( s + 1) V2

s V1

⎛ s 2 + 2 s + 1⎞ ⎛ s 2 + 3s + 1⎞ = −⎜ V1 + ⎜ ⎟ ⎟ V2 ⎝ s+2 ⎠ ⎝ s+2 ⎠ Comparing Eqs (iv) and (v) with Y-parameter equations, ⎡Y11 ⎢⎣Y21

13.4

⎡ s 2 + 3s 3s 1 Y12 ⎤ ⎢ s+2 =⎢ Y22 ⎥⎦ ⎢ ( s 2 + 2ss 1) ⎢⎣ − s+2

−

…(v)

( s 2 + 2s 2 s 1) ⎤ ⎥ s+2 ⎥ s2 + 3 1 ⎥ ⎥⎦ s+2

ΤRANSMISSION PARAMETERS (ABCD PARAMETERS)

The transmission parameters or chain parameters or ABCD parameters serve to relate the voltage and current at the input port to voltage and current at the output port. In equation form, ( , I1 ) f (V2 , ) V1

AV AV2

B 2 BI

I1 CV V2 DI D 2 Here, the negative sign is used with I2 and not for parameters B and D. The reason the current I2 carries a negative sign is that in transmission field, the output current is assumed to be coming out of the output port instead of going into the port. In matrix form, we can write ⎡V1 ⎤ ⎡ A B ⎤ ⎡ V2 ⎤ ⎢⎣ I1 ⎥⎦ = ⎢⎣C D ⎥⎦ ⎢⎣ − I 2 ⎥⎦ ⎡A B⎤ is called transmission matrix. where matrix ⎢ ⎣C D ⎥⎦ For a given network, these parameters are determined as follows: Case 1 When the output port is open-circuited, i.e., I2 = 0 V A= 1 V2 I 2 = 0 where A is the reverse voltage gain with the output port open-circuited. I Similarly, C= 1 V2 I 2 = 0 where C is the transfer admittance with the output port open-circuited. Case 2 When output port is short-circuited, i.e., V2 = 0 V B=− 1 I 2 V2 = 0 where B is the transfer impedance with the output port short-circuited. I Similarly, D=− 1 I 2 V2 = 0 where D is the reverse current gain with the output port short-circuited.

13.4 Τransmission Parameters (ABCD Parameters) 13.19

13.4.1 Condition for Reciprocity (a) As shown in Fig. 13.28, voltage Vs is applied at the input port with the output port short-circuited. i.e., V V 1

s

I1

V2 = 0 I 2′ From the transmission parameter equations,

+ − Vs

I2

I2

I2′

Network

Vs B I 2′ Fig. 13.28 Network for deriving condition for reciprocity Vs =B ′ I2 (b) As shown in Fig. 13.29, voltage Vs is applied at the output port with the input port short-circuited. i.e.,

V2 Vs V1 = 0 I1′

I1 I1′

I1

0 = AV Vs − BI 2 − I1 ′ = CVs D DI 2 A I 2 = Vs B AD −II1 ′ CV Vs − Vs B Vs B = I1 ′ AD − BC Hence, for the network to be reciprocal, Vs Vs = I 2 ′ I1 ′

i.e.,

Network

+ −

Vs

Fig. 13.29 Network for deriving condition for reciprocity

From the transmission parameter equations,

i.e.,

I2

B AD − BC AD − BC = 1 B=

13.4.2 Condition for Symmetry (a) When the output port is open-circuited, i.e., I2 = 0. From the transmission-parameter equations, Vs AV V2 I1 CV V2 Vs A = I1 C (b) When the input port is open-circuited, i.e., I1 = 0. From the transmission parameter equation, CV Vs = DI 2 Vs D = I2 C

13.20 Network Analysis and Synthesis Hence, for network to be symmetrical, Vs Vs = I1 I 2 A D

i.e.,

Example 13.14

Find the transmission parameters for the network shown in Fig. 13.30. I1

1Ω

2Ω

+

I2 +

5Ω

V1 −

V2 −

Fig. 13.30 Solution First Method Case 1 When the output port is open-circuited, i.e., I2 = 0. V1 6 I1 V2 I1 and A=

V1 V2

I C= 1 V2 Case 2

Now and

=

6 I1 6 = 5 I1 5

=

1 5

I2 = 0

I2 = 0

When the output port is short-circuited, i.e., V2 = 0, as shown in Fig. 13.31, 1Ω 2Ω I1 5× 2 10 17 Req = 1 + = 1+ = Ω + 5+ 2 7 7 17 V1 I1 V1 5Ω 7 5 5 − I2 ( I1 ) = − I1 7 7 Fig. 13.31 17 I1 V1 17 B=− =− 7 = Ω 5 I 2 V2 = 0 5 − I1 7 I1 7 D=− = I 2 V2 = 0 5

Hence, the transmission parameters are ⎡ 6 17 ⎤ ⎡A B⎤ ⎢5 5 ⎥ ⎢⎣C D ⎥⎦ = ⎢ 1 7 ⎥ ⎢ ⎥ ⎣5 5 ⎦

I2

13.4 Τransmission Parameters (ABCD Parameters) 13.21

Second Method (Refer Fig. 13.37) Applying KVL to Mesh 1, V1 6 I1 5 I 2 Applying KVL to Mesh 2, V2 I1 7 I 2 Hence, 51 2 7I 2 1 7 I1 V2 − I 2 5 5 Substituting Eq. (iii) in Eq. (i), 7 ⎞ ⎛1 V1 6 V2 − I 2 I2 ⎝5 5 ⎠ 6 17 = V2 I2 5 5 Comparing Eqs (iii) and (iv) with transmission parameter equations,

…(i) …(ii)

…(iii)

…(iv)

⎡ 6 17 ⎤ ⎡A B⎤ ⎢5 5 ⎥ ⎢⎣C D ⎥⎦ = ⎢ 1 7 ⎥ ⎢ ⎥ ⎣5 5 ⎦

Example 13.15

Obtain ABCD parameters for the network shown in Fig 13.32. I1

1Ω

1Ω

I2

+ V1

+

2Ω

V2

2Ω

−

−

Fig. 13.32 Solution The network is redrawn as shown in Fig 13.33. Applying KVL to Mesh 1, V1 3I1 2 I 3 …(i) + Applying KVL to Mesh 2, V2 2 I 2 2 I 3 …(ii) V 1 Applying KVL to Mesh 3, −2 ( 3 − 1 ) − 3 − 2 ( 3 + 2 ) = 0 − 5 3 = 2 I1 − 2 I 2 2 2 I 3 = I1 − I 2 …(iii) 5 5 Substituting Eq. (iii) in Eq. (i), 2 ⎞ ⎛2 V1 3I1 2 ⎜ I1 I2 ⎟ ⎝5 5 ⎠ 11 4 = I1 I2 5 5

1Ω

1Ω +

2Ω I1

2Ω I3

V2 I2

−

Fig. 13.33

…(iv)

13.22 Network Analysis and Synthesis Substituting Eq. (iii) in Eq. (ii), ⎛2 2 ⎜ I1 ⎝5 4 6 = I1 I2 5 5 4 6 I1 V2 − I 2 5 5 5 3 I1 V2 − I 2 4 2 V2

2I 2

2 ⎞ I2 ⎟ 5 ⎠

…(v)

Substituting Eq. (v) in Eq. (iv), 11 ⎛ 5 3 ⎞ V2 − I 2 5 ⎝4 2 ⎠ 11 5 = V2 I2 4 2 Comparing Eqs (v) and (vi) with ABCD parameter equations, V1

⎡11 ⎡A B⎤ ⎢ 4 ⎢⎣C D ⎥⎦ = ⎢ 5 ⎢ ⎣4

Example 13.16

4 I2 5 …(vi)

5⎤ 2⎥ 3 ⎥⎥ 2⎦

Determine the transmission parameters for the network shown in Fig. 13.34. 1

I1

1

2

I2

+

+ s

V1

1 s

V2 −

−

Fig. 13.34 Solution Applying KCL at Node 1, V1 + (V1 V2 ) s s +1 = V1 V2 s

I1 =

…(i)

Applying KCL at Node 2, V2 + (V2 V1 ) 1 s = ( s 1) V2 V1 V1 ( s + 1) V2 I 2 I2 =

…(ii)

13.4 Τransmission Parameters (ABCD Parameters) 13.23

Substituting Eq. (ii) in Eq. (i), s +1 s ⎡ ( s + 1) 2 ⎢ ⎣ s

I1

2

=

−I ⎤ 1 V ⎦

s +1 s

s 1 s +1 V − s s

…(iii)

2

Comparing Eqs (ii) and (iii) with ABCD parameter equations, s+ − ⎡A B⎤ = 2 s 1 s +1 C D s s

Example 13.17

Find transmission parameters for the two-port network shown in Fig. 13.35. 1.5 V

10 Ω

I1

I

+

+

25 Ω

V I1

20 Ω

V

I

I

−

Fig. 13.35 Solution Applying KVL to Mesh 1, + 25 (

)

=

…(i)

Applying KVL to Mesh 2, + I3 I

V Applying KVL to Mesh 3, − (

− )+

+

− +

−

−

2

+

…(ii)

=0

) − 20 I 2 −

0 =0

20

− 20 I 2

82 I3

…(iii)

0 94

Substituting Eq. (iii) in Eq. (i), − 25 0 9 I

I2 …(iv)

Substituting Eq. (iii) in Eq. (ii), 18

I + 20 0 9 I I

I2 …(v)

From Eq. (v), V − 0 81I 2

…(vi)

13.24 Network Analysis and Synthesis Substituting Eq. (vi) in Eq. (iv), V1

11. (0 0 3 V2 − 0.81I 2 ) 6 I 2

= 0.61V 61 V2 3 32 I 2 Comparing Eqs (vi) and (vii) with ABCD parameter equations, ⎡ A B ⎤ ⎡ 0.61 −3.32⎤ ⎢⎣C D ⎥⎦ = ⎢⎣0.053 −0.81⎥⎦

13.5

…(vii)

INVERSE TRANSMISSION PARAMETERS (A�B�C�D� PARAMETERS)

The inverse transmission parameters serve to relate the voltage and current at the outport port to the voltage and current at the input port. In equation form, ( , I 2 ) f (V1 , ) V2 A′ V1 B′ I1 I2

C ′ V1

D ′ I1

In matrix form, we can write ⎡V2 ⎤ ⎡ A′ ⎢⎣ I 2 ⎥⎦ = ⎢⎣C ′

B′ ⎤ ⎡ V1 ⎤ D ′ ⎥⎦ ⎢⎣ − I1 ⎥⎦

⎡ A′ B′ ⎤ where matrix ⎢ is called the inverse transmission matrix. ⎣C ′ D ′ ⎥⎦ For a given network, these parameters are determined as follows: Case 1 When the input port is open-circuited, i.e., I1 = 0 V A′ = 2 V1 I1 = 0 where A′ is the forward voltage gain with the input port open-circuited. I Similarly, C′ = 2 V1 I1 = 0 where C ′ is the transfer admittance with the input port open-circuited. Case 2 When the input port is short-circuited, i.e., V1 = 0 V B′ = − 2 I1 V1 = 0 where B′ is the transfer impedance with the input port short-circuited. Similarly, I D′ = − 2 I1 V1 = 0 where D′ is the forward current gain with the input port short-circuited.

13.5.1 Condition for Reciprocity (a) As shown in Fig 13.36, voltage Vs is applied at input port with the output port short-circuited. i.e.,

V1 Vs V2 = 0 I2 I2 ′

I1 + Vs −

I2 Network

I2′

Fig. 13.36 Network for deriving condition for reciprocity

13.5 Inverse Transmission Parameters (A′B′C′D′ Parameters) 13.25

From the inverse transmission parameter equations, 0 = A′ Vs − B′ I1 − I 2 ′ = C ′ Vs D ′ I1 I 2 ′ A′ D ′ − B ′C ′ = Vs B′ (b) As shown in Fig 13.37, voltage Vs is applied at the output port with the input port short-circuited. V2 Vs i.e., I1 I2 V1 = 0 + I1′ I1 I1 ′ Network − Vs From the inverse transmission parameter equations, Vs B′ I1 ′ Fig. 13.37 Network for deriving condition for A′ I2 = Vs reciprocity B′ I1 ′ 1 = Vs B′ Hence, for the network to be reciprocal, I 2 ′ I1 ′ = Vs Vs i.e., A′ D ′ − B′C ′ = 1

13.5.2 Condition for Symmetry The condition for symmetry is obtained from the Z-parameters. V D′ Z11 = 1 =0= I1 I 2 = 0 C′ Similarly, V A′ Z 22 = 2 =0= I 2 I1 = 0 C′ For symmetrical network Z11 Z 22 A′ D ′

Example 13.18

Find the inverse transmission parameters for the network shown in Fig. 13.38. I1

1Ω

2Ω

+ V1

+

3Ω

−

V2 −

Fig. 13.38 Solution First method Case 1 When the input port is open-circuited, i.e., I1 = 0 V1 3I 2 V2 I2

13.26 Network Analysis and Synthesis A′ =

V2 V1

I C′ = 2 V1

=

5 3

=

1 3

I1 = 0

I1 = 0

Case 2 When the input port is short-circuited, i.e., V1 = 0 as shown in Fig. 13.39. By current division rule, 1Ω 2Ω I1 3 I1 I2 4 I 4 3Ω D′ = − 2 = I1 V1 = 0 3 3 1 3 11 Req = 2 + = 2+ = Ω 3 1 4 4 Fig. 13.39 11 11 ⎛ 4 ⎞ 11 V2 I 2 = ⎜ − ⎟ I1 I1 4 4 ⎝ 3⎠ 3 V 11 B′ = − 2 = Ω I1 V1 = 0 3 Second Method Refer Fig. 13.38. Applying KVL to Mesh 1, V1 = 4I1 + 3I2 Applying KVL to Mesh 2, V2 = 3I1 + 5I2 Hence, 3I2 = V1 − 4I1 1 4 I2 V1 − I1 3 3 Substituting Eq. (iii) in Eq. (ii), 4 ⎞ ⎛1 V2 3I1 5 ⎜ V1 I1 ⎟ ⎝3 3 ⎠ 5 11 = V1 I1 3 3 Comparing Eqs (iii) and (iv) with inverse transmission parameter equations, ⎡ 5 11⎤ ⎡ A′ B′ ⎤ ⎢ 3 3 ⎥ ⎢⎣C ′ D ′ ⎥⎦ = ⎢ 1 4 ⎥ ⎢ ⎥ ⎣3 3 ⎦

Example 13.19

I2 + V2 −

...(i) ...(ii) …(iii)

…(iv)

Find the inverse transmission parameters for the network shown in Fig. 13.40 2Ω

I1

I2

+ V1

+

2Ω

2Ω

−

V2 −

Fig. 13.40

13.5 Inverse Transmission Parameters (A′B′C′D′ Parameters) 13.27

Solution The network is redrawn as shown in Fig 13.41. Applying KVL to Mesh 1, I1 …(i) + V1 2 I1 2 I 3 Applying KVL to Mesh 2, V2 Applying KVL to Mesh 3,

…(ii)

V1

=0 1 1 I 3 = I1 − I 2 …(iii) 3 3 Substituting Eq. (iii) in Eq. (i),

−

−2 1 + 2 I 2 + 6

2I 2

2I3

2Ω

I2 +

2Ω I1

2Ω

V2 I2

I3

−

3

Fig. 13.41

⎛1 2 I1 2 ⎜ I1 ⎝3 4 2 = I1 I2 3 3 Substituting Eq. (iii) in Eq. (ii),

1 ⎞ I2 ⎟ 3 ⎠

⎛1 2 ⎜ I1 ⎝3 4 I2 3

1 ⎞ I2 ⎟ 3 ⎠

V1

V2

2I 2 =

2 I1 3

...(iv)

...(v)

Rewriting Eq. (iv), 2 I2 3

V1 −

4 I1 3

3 V1 − 2 I1 2 Substituting Eq. (vi) in Eq. (v), I2

V2

...(vi)

2 4⎛3 ⎞ I1 + ⎜ V1 2 I1 ⎟ ⎝ ⎠ 3 3 2 = 2V 2V1 2 I1

...(vii)

Comparing Eq. (vi) and (vii) with inverse transmission parameter equations, ⎡ A′ ⎢⎣C ′

Example 13.20

⎡2 B′ ⎤ ⎢ = 3 D ′ ⎥⎦ ⎢ ⎣2

2⎤ ⎥ 2⎥ ⎦

Find the inverse transmission parameters for the network shown in Fig. 13.42. I1

1Ω

1Ω

I2

+ V1

+

2Ω

−

1Ω

V2 −

Fig. 13.42

13.28 Network Analysis and Synthesis Solution The network is redrawn as shown in Fig. 13.43. Applying KVL to Mesh 1, I1 1Ω …(i) V1 3I1 2 I 3 + Applying KVL to Mesh 2, V2 I 2 + I 3 …(ii) V1 Applying KVL to Mesh 3, I1 −2 1 + 2 + 4 I 3 = 0 − 1 1 I 3 = I1 − I 2 …(iii) 2 4 Substituting Eq. (iii) in Eq. (i), 1 ⎞ ⎛1 V1 3I1 2 ⎜ I1 I2 ⎟ ⎝2 4 ⎠ = 2 I1

1Ω

I2

+

2Ω

1Ω I3

V2 I2

−

Fig. 13.43

1 I2 2

…(iv)

Substituting Eq. (iii) in Eq. (ii), V2

Rewriting Eq. (iv), 1 I2 2 I2

1 1 I1 I2 2 4 1 3 = I1 I2 2 4 I2 +

…(v)

V1 − 2 I1 2

1

4 I1

...(vi)

Substituting the Eq. (vi) in Eq. (v), 1 3 I1 + ( 2V1 4 1 ) 2 4 3 5 = V1 I1 2 2 Comparing Eqs (vi) and (vii) with inverse transmission parameter equations, V2

⎡ A′ ⎢⎣C ′

13.6

⎡3 B′ ⎤ ⎢ = D ′ ⎥⎦ ⎢ 2 ⎣2

...(vii)

5⎤ 2⎥ 4 ⎥⎦

HYBRID PARAMETERS (h PARAMETERS)

The hybrid parameters of a two-port network may be defined by expressing the voltage of input port V1 and current of output port I2 in terms of current of input port I1 and voltage of output port V2. ( , I 2 ) f ( I1 , ) V1 h11 I1 h12 V2 I 2 h21 I1 h22 V2 In matrix form, we can write ⎡V1 ⎤ ⎡ h11 h12 ⎤ ⎡ I1 ⎤ ⎢⎣ I 2 ⎥⎦ = ⎢⎣ h21 h22 ⎥⎦ ⎢⎣V2 ⎥⎦ The individual h parameters can be defined by setting I1 = 0 and V2 = 0.

13.6 Hybrid Parameters (h Parameters) 13.29

Case 1 When the output port is short-circuited i.e., V2 = 0 V h11 = 1 I1 V2 = 0 where h11 is the short-circuit input impedance. h21 =

I2 I1 V2 = 0

where h21 is the short-circuit forward current gain. Case 2 When the input port is open-circuited, i.e., I1 = 0 V h12 = 1 V2 I1 = 0 where h12 is the open-circuit reverse voltage gain. I h22 = 2 V2

I1 = 0

where h22 is the open-circuit output admittance. Since h parameters represent dimensionally an impedance, an admittance, a voltage gain and a current gain, these are called hybrid parameters. The equivalent circuit of a two-port network in terms of hybrid parameters is shown in Fig 13.44. I1

h11

I2

+ V1

+ h12V2

+ −

h21I1

h22

V2 −

−

Fig. 13.44 Equivalent circuit of the two-port network in terms of h-parameters

13.6.1 Condition for Reciprocity (a) As shown in Fig. 13.45, voltage Vs is applied at the input port and the output port is short-circuited. i.e., V1 Vs V2 = 0 I2 ′ I2 From the h-parameter equations,

I1 + Vs −

Vs h11 I1 − I 2 ′ h21 I1 Fig. 13.45 Vs h = − 11 I2 ′ h21 I1 (b) As shown in Fig. 13.46, voltage Vs is applied at the output port with the input port shortI1′ circuited. i.e., V1 = 0 V2 Vs Fig. 13.46 I1 I1 ′

I2 I2′

Network

Network for deriving condition for reciprocity I2 Network

+ −

Vs

Network for deriving condition for reciprocity

13.30 Network Analysis and Synthesis From the h-parameter equations, 0 = h11 1 I1 + h1122 Vs h12 Vs = − h11 1 I1 = h11 11 I1 ′ Vs h111 = I1 ′ h112 Hence, for the network to be reciprocal, Vs Vs = I 2 ′ I1 ′ i.e., h21 h12

13.6.2 Condition for Symmetry The condition for symmetry is obtained from the Z-parameters. V h I +h V V Z11 = 1 = 11 1 12 2 = h11 h12 2 I1 I 2 0 I1 I1 I2 = 0 But with I2 = 0, 0 = h21 2 I1 + h2222 V2 V2 h = − 221 I1 h22 h h h h −h h Δh Z11 h11 − 12 21 = 11 22 12 21 = h22 h22 h22 where Δh = h11h22 − h12 h21 Similarly, V Z 22 = 2 I 2 I1 = 0 With I1 = 0, I 2 h22 V2

For a symmetrical network,

V2 I2

Z11

Z 22

= I1 = 0

1 h22

Δh 1 = h22 h22 Δh = 1

i.e., i.e., i.e.,

Z 22 =

h11 h22 − h12 h21 = 1

Example 13.21 In the two-port network shown in Fig. 13.47, compute h-parameters from the following data: (a) With the output port short-circuited:V V1 2 V I 1 1 A, I 2 A (b) With the input port open-circuited:V V1 10 V V2 50 V , I 2 A I1 + V1 −

I2 Two-port network

Fig. 13.47

+ V2 −

13.6 Hybrid Parameters (h Parameters) 13.31

Solution V1 I1 V

h11 =

=

2=

I2 I1

h21 =

0

= V2 = 0

25 = 25 5 Ω, 1

h12 =

2 = 2, 1

h222 =

V1 V2 I2 V2

=

10 = 0.2 50

=

2 = 0.04 50

I1 = 0

I1 = 0

Hence, the h-parameters are ⎡ h11 h12 ⎤ ⎡ 25 0 2 ⎤ ⎢⎣ h 21 h22 ⎥⎦ = ⎢⎣ 2 0.04 ⎥⎦

Example 13.22 Determine hybrid parameters for the network of Fig. 13.48. Determine whether the network is reciprocal. 1Ω

I1

2Ω

I2

+

+

2Ω

V1 I1

4Ω

V2

I3

I2

−

−

Fig. 13.48 Solution First Method Case 1 When Port 2 is short-circuited, i.e., V2 = 0 as shown in Fig. 13.49, 2×2 = 2+2

Req = 1 + Now,

Also,

I1 ×

2 2 2

V2 = 0

I2

2Ω

V1

=−

=−

2Ω

+

2I1 V h11 = 1 =2Ω I1 V2 = 0 I2

1Ω

I1

V1

I h21 = 2 I1 Case 2

Ω

I1 2

4Ω

−

Fig. 13.49

1 2

When Port 1 is open-circuited, i.e., I1 = 0 as shown in Fig. 13.50, Req = V1

( + )×4 = 2+2+4 2I y

I Iy = 2 2 V2 4 I x I Ix = 2 2

Ω

I1 = 0

+ V1

1Ω

1Ω ly

2Ω

−

I2

lx

4Ω

+ V2 −

Fig. 13.50

13.32 Network Analysis and Synthesis

I1 = 0

I2 V2

I1 = 0

h22 =

I2 2 =1 = = I 4I x 2 4× 2 2 2I 1 = x = 4I x 2 2I y

V h12 = 1 V2

2×

Hence, the h-parameters are ⎡ 2 h12 ⎤ ⎢ = ⎢ h22 ⎥⎦ ⎢ 1 − ⎣ 2

⎡ h11 ⎢⎣ h21

1⎤ 2⎥ 1 ⎥⎥ 2⎦

Second Method (Refer Fig. 13.48) Applying KVL to Mesh 1, V1

3I1 2 I 3

…(i)

V2

4I2

…(ii)

Applying KVL to Mesh 2, Applying KVL to Mesh 3, −2 (

3

− 1 ) − 2I3 − 4 (

3

+

4I3

=0 = 2 I1 − 4 I 2 I I I3 = 1 − 2 4 2 2)

8

3

…(iii)

Substituting Eq. (iii) in Eq. (i), V1 =

3I1

I ⎞ ⎛I 2⎜ 1 − 2⎟ ⎝4 2⎠

5 I1 2

I2

...(iv)

Substituting Eq. (iii) in Eq. (ii), I2⎞ ⎛I 4⎜ 1 − ⎟ ⎝4 2⎠ = 4I 4 I 2 I1 2 I 2 = I1 + 2 I 2 1 1 I 2 = − I1 + V2 2 2

V2

4I2

...(v)

Substituting Eq. (v) in Eq. (iv), 5 1 I1 − I1 2 2 1 = 2 I1 V2 2 Comparing Eqs (v) and (vi) with h-parameter equations, V1

⎡ h11 ⎢ h21 ⎣ Since h12 = − h21, the network is reciprocal.

⎡ 2 h12 ⎤ ⎢ =⎢ ⎥ h22 ⎦ ⎢ 1 − ⎣ 2

1⎤ 2⎥ 1 ⎥⎥ 2⎦

1 V2 2 ...(vi)

13.7 Inverse Hybrid Parameters (g Parameters) 13.33

Example 13.23

Find h-parameters for the network shown in Fig. 13.51. 1 F 2

I1

1 F 2

I2

+

+

1H

V1

1H

V2

−

−

Fig. 13.51 Solution As solved in Example 13.12, derive the equations for I1 and I2 in terms of V1 and V2. I1 =

s3 + 2 s s3 V − V2 1 4( s 2 1) 4( s 2 1)

I2 = −

s3 4( s

2

1)

s 4 + 6s 6s2

V1 +

4ss( s

2

…(i) 4

1)

V2

…(ii)

From Eq. (i), V1 = Substituting Eq. (iii) in Eq (ii),

4( s 2 1) s2 I + V2 1 s ( s 2 + 2) s2 2

I2 = − =−

s3 4( s

2

s2 2

⎡ 4( s 2 1) ⎤ s 4 + 6s s2 6s2 4 I1 + 2 V2 ⎥ + V2 ⎢ 2 1) ⎣ s( s + 2) s 2 ⎦ 4 s( s 2 + 1) I1 +

2

2(

2

+ 1)

s +2 s(( + 2) Comparing Eqs (iii) and (iv) with h-parameter equations, ⎡ h11 ⎢ h21 ⎣

13.7

…(iii)

⎡ 4( s 2 1) ⎢ h12 ⎤ ⎢ s( s 2 + 2) = h22 ⎥⎦ ⎢ s2 ⎢− 2 ⎢⎣ s + 2

V2

…(iv)

⎤ ⎥ s 2 ⎥ 2( s 2 1) ⎥ ⎥ s( s 2 + 2) 2 ⎥⎦ s2

2

INVERSE HYBRID PARAMETERS (g PARAMETERS)

The inverse hybrid parameters of a two-port network may be defined by expressing the current of the input port I1 and voltage of the output port V2 in terms of the voltage of the input port V1 and the current of the output port I2. ( , V2 ) f (V1 , ) I g11 V g12 I 2 V g21 V g22 I 2 In matrix form, we can write g12 ⎤ ⎡V1 ⎤ ⎡ I1 ⎤ ⎡ g ⎢⎣V2 ⎥⎦ = ⎢⎣ g g22 ⎥⎦ ⎢⎣ I 2 ⎥⎦ The individual g parameters can be defined by setting V1 = 0 and I2 = 0.

13.34 Network Analysis and Synthesis Case 1

When the output port is open-circuited, i.e., I2 = 0 g11 =

I1 V1

I2 = 0

where g11 is the open-circuit input admittance. V g21 = 2 V1 I 2 = 0 where g21 is the open-circuit forward voltage gain. Case 2 When the input port is short-circuited, i.e., V1 = 0 g12 =

I1 I 2 V1 = 0

where g12 is the short-circuit reverse current gain. V g22 = 2 I 2 V1 = 0 where g22 is the short-circuit output impedance. The equivalent circuit of a two-port network in terms of inverse hybrid parameters is shown in Fig. 13.52. g22

I2 + V1

I2 +

g11

g12I2

−

+ g V − 21 1

V2 −

Fig. 13.52 Equivalent circuit of two-port network in terms of g-parameters

13.7.1 Condition for Reciprocity (a) As shown in Fig. 13.53, voltage Vs is applied at the input port and the output port is short-circuited, i.e., V1 Vs I1 I2 V2 = 0 + I2 ′ I2 Network Vs I2′ − From the g-parameter equation, g Vs = g I 2 ′ Fig. 13.53 Network for deriving condition for Vs g = 22 reciprocity I 2 ′ g21 (b) As shown in Fig. 13.54, voltage Vs is applied at the output port with the input port short-circuited, i.e., V1 = 0 I1 V2 Vs I2 I1 I1 ′ + I1′ From the g-parameter equations, Network − Vs − I ′ g12 I 2 V g22 I 2 Fig. 13.54 Network for deriving condition for Vs g reciprocity = − 22 I1 ′ g12

13.7 Inverse Hybrid Parameters (g Parameters) 13.35

Hence, for the network to be reciprocal, Vs Vs = I 2 ′ I1 ′ g g12

i.e.,

13.7.2 Condition for Symmetry The condition for symmetry is obtained from the Z-parameters. V 1 Z11 = 1 = I I 2 = 0 g11 Similarly, Z 22 =

V2 I2

I1 = 0

0 = g V1 + g I 2 g V1 = − 12 I 2 g11 ⎛ g ⎞ V g21 − 12 ⎟ I ⎝ g11 ⎠ Z 22 = For a symmetrical network, Z11

V2 I2

= I1 = 0

g22 I 2

g g22 − g g21 g11

Z 22 .

1 g g − g12 g21 = 11 22 g11 g11 g g22 − g g21 = 1

i.e., i.e.,

Table 13.2 Conditions for reciprocity and symmetry Parameter

Condition for Reciprocity

Condition for Symmetry

Z

Z12

Z 21

Z11

Z 22

Y

Y12

Y21

Y11

Y22

A

D

T

AD − BC = 1

T′ h

A′ D ′ − B′C ′ = 1 h12 h21

g

Example 13.24

g

A′ D ′ h11 h22 − h12 h21 = 1 g g22 − g

g21

Find g-parameters for the network shown in Fig. 13.55. I1

1Ω

2Ω

+ V1

I2 +

3Ω

−

V2 −

Fig. 13.55

g21 = 1

13.36 Network Analysis and Synthesis Solution First Method Case 1 When the output port is open-circuited, i.e., I2 = 0 Applying KVL to Mesh 1,

Also

V1

4 I1

g11 =

I1 V1

V2

3 I1

g21 =

V2 V1

=

1 4

=

3 I1 3 = 4 I1 4

I2 = 0

I2 = 0

Case 2 When the input port is short-circuited, i.e., V1 = 0 as shown in Fig. 13.56. By current division rule, I1 g12 =

3 I2 4

I1

I1 3 =− I 2 V1 = 0 4

1Ω

2Ω

I2 +

3Ω

3 ×1 3 11 Req = 2 + = 2+ = Ω 3 +1 4 4 11 V2 I2 4 V 11 g22 = 2 = Ω I 2 V1 = 0 4

V2 −

Fig. 13.56

Second Method (Refer Fig. 13.55). Applying KVL to Mesh 1, V1

4 I1 3 I 2

…(i)

V2 4 1

3 I1 5 I 2 1 3 I2

…(ii)

1 3 V1 − I 2 4 4

…(iii)

Applying KVL to Mesh 2, Hence,

I1 Substituting Eq. (iii) in Eq. (ii),

3 ⎞ ⎛1 V1 − I 2 ⎝4 4 ⎠ 3 11 = V1 I2 4 4 Comparing Eqs (iii) and (iv) with g-parameter equations, V2

⎡g ⎢⎣ g

3

⎡1 g12 ⎤ ⎢ 4 = g22 ⎥⎦ ⎢⎢ 3 ⎣4

3⎤ − ⎥ 4 11 ⎥⎥ 4 ⎦

I2 …(iv)

13.8 Inter-relationships between the Parameters 13.37

Example 13.25

Find g-parameters for the network shown in Fig. 13.57. 5I1

I1

1Ω

I2

+

1Ω

V2

1Ω

V1

+

−

−

Fig. 13.57 Solution The network is redrawn using source transformation as shown in Fig. 13.58. Applying KVL to Mesh 1, 1Ω 1Ω I1 V1 1 I1 1 ( I1 I 2 ) = 0 V1 2 I1 + I 2 …(i) + Applying KVL to Mesh 2, V1 1Ω V2 5 I1 1 I 2 1 ( I 2 I1 ) 0 −

V2 6 I1 2 I 2 …(ii) 2 I1 V1 − I 2

Hence,

I1

1 1 V1 − I 2 2 2

V2

6

Substituting Eq (iii) in Eq. (ii),

1 ⎞ ⎛1 V1 − I 2 ⎝2 2 ⎠

− +

I2 + V2 −

Fig. 13.58 …(iii) 2 I2

= 3 V1 − I 2 Comparing Eqs (iii) and (iv) the g- parameter equations, 1⎤ ⎡1 g12 ⎤ ⎢ − ⎥ ⎡g = 2 2 ⎢⎣ g g22 ⎥⎦ ⎢ ⎥ 3 1 ⎣ ⎦

13.8

5I1

…(iv)

INTER-RELATIONSHIPS BETWEEN THE PARAMETERS

When it is required to find out two or more parameters of a particular network then finding each parameter will be tedious. But if we find a particular parameter then the other parameters can be found if the interrelationship between them is known.

13.8.1 Z-parameters in Terms of Other Parameters 1. Z-parameters in Terms of Y-parameters We know that 1

By Cramer’s rule,

I2

Y11 11 V1 Y12 V2 Y2211 V1 Y22 V2

I1 I2 V1 = Y11 Y21

Y12 Y22 Y I Y I Y Y = 22 1 12 2 = 22 I1 − 12 I 2 Y12 Y11Y22 − Y12Y21 ΔY ΔY Y22

13.38 Network Analysis and Synthesis where Comparing with

Also, Comparing with

ΔY = Y11Y22 − Y12Y21 V1 Z11I1 Z12 , I 2 , Y Z11 = 22 ΔY Y Z12 = − 12 ΔY Y11 I1 Y21 I 2 Y Y V2 = = 11 I 2 − 21 I1 ΔY ΔY ΔY V2 Z 21 I1 Z 22 I 2 , Y Z 22 = 11 ΔY Y Z 21 = − 21 ΔY

2. Z-parameter in Terms of ABCD Parameters We know that V1 AV AV2 BI B 2 I1 CV V2 D DI 2 Rewriting the second equation, V2 = Comparing with

V2

1 D I1 + I 2 C C Z 21 I1 Z 22 I 2 ,

1 C D Z 22 = C D ⎤ ⎡1 V1 A I1 + I 2 C ⎦ ⎣C V1 Z11 I1 Z12 I 2 , Z 21 =

Also, Comparing with

B 2= BI

A ⎡ AD I1 + C ⎣ C

A ⎤ ⎡ AD − BC ⎤ B I 2 = I1 + ⎢ ⎥ I2 C C ⎦ ⎣ ⎦

A C AD − BC Z12 = C Z11 =

3. Z-parameters in Terms of A�B�C�D� Parameters We know that V2 A′ V1 B′ I1 I 2 C ′ V1 D ′ I1 Rewriting the second equation, V1 = Comparing with

V1 Z11 =

D′ 1 I1 + I 2 C′ C′ Z11 I1 Z12 I 2 , D′ C′

13.8 Inter-relationships between the Parameters 13.39

Z12 = Also,

V2

Comparing with

V2

1 C′ 1 ⎤ ⎡ D′ I1 + I2 C ′ C ′ ⎦ ⎣ Z 21 I1 Z 22 I 2 , A′

A′ ⎡ A′ D ′ − B′C ′ ⎤ B′ I1 = ⎢ ⎥ I1 + C ′ I 2 C ′ ⎣ ⎦

A′ D ′ − B′C ′ ΔT ′ = C′ C′ A′ Z 22 = C′ 4. Z-parameters in Terms of Hybrid Parameters We know that Z 21 =

V1

h11 I1 h12 V2

I 2 h21 I1 h22 V2 Rewriting the second equation, h 1 V2 = − 21 I1 + I2 h22 h22 Comparing with V2 Z 21 I1 Z 22 I 2 , h21 h22 1 = h22

Z 21 = − Z 22

1 h h h h112 ⎡ h ⎤ ⎡ h1 h222 − h12 1 h221 ⎤ h11 I1 h12 ⎢ − 21 I1 + I2 h11 I1 + 12 I 2 − 12 21 I1 = ⎢ 11 ⎥ I1 + h I 2 h h h h h ⎣ 22 22 ⎦ 22 22 ⎣ 222 ⎦ 222 Comparing with V1 Z11 I1 Z12 I 2 , Also,

V1

h11h22 − h12 h21 Δh = h22 h22 h12 Z12 = h22 Z11 =

5. Z-parameters in Terms of Inverse Hybrid Parameters We know that I V Rewriting the first equation,

g11 V g 21 V

V1 = Comparing with

V1

g12 I 2 g22 I 2

1 g I1 − 12 I 2 g11 g11 Z11 I1 Z12 I 2 ,

1 g11 g = − 12 g11

Z11 = Z12 Also

V

⎡ 1 ⎤ g g21 ⎢ I1 − 12 I g11 ⎦ ⎣ g11

g 22 I 2 =

⎡ g g22 − g g21 ⎤ g21 I1 + ⎢ ⎥ I2 g11 g11 ⎣ ⎦

13.40 Network Analysis and Synthesis Comparing with

V2

Z 21 I1 Z 22 I 2 , g21 g11 g g22 − g = g11

Z 21 = Z 22

g21

=

Δg g11

13.8.2 Y-parameters in Terms of Other Parameters 1. Y-parameters in terms of Z-parameters We know that

By Cramer’s rule,

V1 V2

Z11 I1 Z12 I 2 Z 21 I1 Z 22 I 2

V1 Z12 V2 Z 22 Z V Z12 V2 Z Z I1 = = 22 1 = 22 V1 − 12 V2 Z11 Z12 Z11 Z 22 − Z12 Z 21 ΔZ ΔZ Z 21 Z 22 where Comparing with

ΔZ = Z11 Z 22 − Z12 Z 21 I1 Y1111 V1 Y12 V2 , Z 22 ΔZ Z Y12 = − 12 ΔZ Z11 V1 Z 21 V2 Z V Z12V1 Z Z I2 = = 11 2 = − 21 V1 + 11 V2 ΔZ ΔZ ΔZ ΔZ I 2 Y21 21 V1 Y22 V2 , Y11 =

Also, Comparing with

Y21 = − Y22 =

Z 21 ΔZ

Z11 ΔZ

2. Y-parameters in Terms of ABCD Parameters We know that V1 A AV V2 B BI 2 I1

CV V2

D DI 2

I2 =

1 A V1 + V2 B B Y21 21 V1 Y22 V2 ,

Rewriting the first equation,

Comparing with

I2

Y21 = − Y22 = Also,

I1

1 B

A B CV V2

A ⎤ D ⎡ 1 ⎡ BC − AD ⎤ D ⎢ − V1 + V2 ⎥ = V1 + ⎢ ⎥ V2 B B B B ⎣ ⎦ ⎣ ⎦

13.8 Inter-relationships between the Parameters 13.41

Comparing with

I1

Y1111 V1 Y12 V2 ,

D B BC − AD AD − BC ΔT Y12 = =− =− B B B Y11 =

3. Y-parameters in Terms of A�B�C�D� Parameters We know that V2 A′V1 B′ I1 C ′V1

I2

D ′ I1

Rewriting the first equation, I1 = Comparing with

I1

1 A′ V1 − V2 B′ B′ Y1111 V1 Y12 V2 ,

A′ B′ 1 Y12 = − B′ Y11 =

Also,

I2

C ′V1

Comparing with

I2

Y2211 V1

and

⎡ A′ D ′ ⎢ V1 ⎣ B′ Y22 V2 ,

1 ⎤ D′ ⎡ A′D ′ − B′C ′ ⎤ V2 = − ⎢ ⎥ V1 + B′ V2 B′ ⎥⎦ B ′ ⎣ ⎦

ΔT ′ ⎡ A′D ′ − B′C ′ ⎤ Y21 = − ⎢ =− ⎥ B′ B′ ⎣ ⎦ D′ Y22 = B′

4. Y-parameters in Terms of Hybrid Parameters We know that V1

h11 I1 h12 V2

I2

h21 I1 h22 V2

I1 =

1 h V1 − 12 V2 h11 h11 Y1111 V1 Y12 V2 ,

Rewriting the first equation,

Comparing with

I1

1 h11 h Y12 = − 12 h11 Y11 =

Also

I2

Comparing with

I2

h h ⎡ 1 ⎤ ⎡h h − h h ⎤ h21 ⎢ V1 − 12 V2 h22V2 = 21 V1 + ⎢ 11 22 12 21 ⎥ V2 h h h h11 ⎣ 11 11 ⎦ 11 ⎣ ⎦ Y2211 V1 Y22 V2 , h21 h11 h h −h h Δh = 11 22 12 21 = h11 h11

Y21 = Y22

13.42 Network Analysis and Synthesis 5. Y-parameters in Terms of Inverse Hybrid Parameters We know that I

g11 V

g12 I 2

V g21 V g22 I 2 Rewriting the second equation, g 1 I 2 = − 21 V1 + V2 g22 g22 Comparing with I 2 Y2211 V1 Y22 V2 , g21 g22 1 = g22

Y21 = − Y22 Also,

I

g11 V

Comparing with

I1

Y1111 V1

⎡ g ⎤ ⎡ g g22 − g g21 ⎤ 1 g g12 ⎢ − 21 V1 + V2 ⎥ = ⎢ V1 + 12 V2 ⎥ g22 ⎦ ⎣ g22 g22 ⎣ g22 ⎦ Y12 V2 ,

g g22 − g g21 Δg = g22 g22 g Y12 = 12 g22 Y11 =

13.8.3 ABCD Parameters in Terms of Other Parameters 1. ABCD Parameters in Terms of Z-parameters We know that V1 Z11 I1 Z12 I 2 V2 Z 21 I1 Z 22 I 2 Rewriting the second equation, 1 Z I1 = V2 − 22 I 2 Z 21 Z 21 Comparing with I1 CV V2 D DI 2 , 1 C= Z 21 Z 22 D= Z 21 Also,

V1

Comparing with

V1

Z Z11 Z Z ⎡ 1 ⎤ Z11 ⎢ V2 − 22 I 2 Z12 V2 − 22 11 I 2 12 I 2 = Z Z Z Z 21 ⎣ 21 21 ⎦ 21 Z ⎡ Z Z − Z12 Z 21 ⎤ = 11 V2 − ⎢ 11 22 ⎥ I2 Z 21 Z 21 ⎣ ⎦

AV AV2 B BI 2 , Z11 A= Z 21 Z Z − Z12 Z 21 ΔZ B = 11 22 = Z 21 Z 21

Z12 I 2

13.8 Inter-relationships between the Parameters 13.43

2. ABCD Parameters in terms of Y-parameters We know that I1

Y11 11 V1 Y12 V2

I2

Y2211 V1 Y22 V2

Rewriting the second equation, Y22 1 V2 + I2 Y21 Y21 V1 A AV V2 B BI 2 , Y A = − 22 Y21 1 B=− Y21

V1 = − Comparing with

Also, Comparing with

1 Y ⎡ Y ⎤ ⎡Y Y − Y Y ⎤ Y11 ⎢ − 22 V2 + I 2 Y1122 V2 = ⎢ 12 21 11 22 ⎥ V2 + 111 I 2 Y21 ⎦ Y21 Y221 ⎣ Y21 ⎣ ⎦ I1 CV V2 D DI 2 , Y12 Y21 − Y11 Y22 ΔY C= =− Y21 Y21 Y D = − 11 Y21 I1

3. ABCD Parameters in terms of A�B�C�D� Parameters We know that V2 A′V1 B′ I1 I2

C ′V1

D ′ I1

By Cramer’s rule, V2 B′ I 2 D′ D′ V1 = = V2 A′ B′ ΔT ′ C ′ D′ where Comparing with

Also,

Comparing with

B′ I2 ΔT ′

ΔT ′ = A′ D ′ − B′C ′ V1 AV AV2 BI B 2, D′ A= ΔT ′ B′ B= ΔT ′ A′ V2 C′ I2 C′ A′ − I1 = =− V2 + I2 ΔT ′ ΔT ′ ΔT ′ C′ A′ I1 = V2 − I2 ΔT ′ ΔT ΔT ′ I1 CV V2 D DI 2 , C′ C= ΔT ′ A′ D= ΔT ′

13.44 Network Analysis and Synthesis 4. ABCD Parameters in Terms of Hybrid Parameters We know that V1 h11 I1 h12 V2 I 2 h21 I1 h22 V2 Rewriting the second equation, h 1 I1 = − 22 V2 + I2 h21 h21 I1 CV V2 D DI 2 , Comparing with h C = − 22 h21 1 D=− h21 Also, Comparing with

h h ⎡ 1 ⎤ ⎡h h − h h ⎤ h11 ⎢ I 2 − 22 V2 h12 V2 = ⎢ 12 21 11 22 ⎥ V2 + 11 I 2 h21 ⎦ h21 h21 ⎣ h21 ⎣ ⎦ V1 A AV V2 B BI 2 , h12 h21 − h11 h22 Δh A= =− h21 h21 h B = − 11 h21

V1

5. ABCD Parameters in Terms of Inverse Hybrid Parameters We know that I g11 V g12 I 2 V g21 V g22 I 2 Rewriting the second equation, 1 g V1 = V2 − 22 I 2 g21 g21 Comparing with V1 A AV V2 B BI 2 , 1 A= g21 g B = 22 g21 Also, Comparing with

⎡ 1 g g11 ⎢ V2 − 22 I g g21 ⎣ 21 I1 CV V2 D DI 2 , g C = 11 g21 g g22 − g g21 D= = g21 I

⎤ ⎦

g12 I 2 =

⎡ g g22 − g g21 ⎤ g11 V2 − ⎢ ⎥ I2 g21 g21 ⎣ ⎦

Δg g21

13.8.4 A�B�C�D� Parameters in Terms of Other Parameters 1. A�B�C �D� Parameters in Terms of Z-parameters We know that V1 V2

Z11 I1 Z12 I 2 Z 21 I1 Z 22 I 2

13.8 Inter-relationships between the Parameters 13.45

Rewriting the first equation,

Comparing with

I2 =

1 Z V1 − 11 I1 Z12 Z12

I2

C ′V1

D ′ I1 ,

1 Z12 Z D ′ = 11 Z12 C′ =

Also,

V2

Comparing with

V2

Z Z Z Z ⎡ 1 ⎤ Z 22 ⎢ V1 − 11 I1 Z1122 I1 + 22 V1 − 22 11 I1 Z12 ⎦ Z12 Z12 ⎣ Z12 Z ⎡ Z Z − Z12 Z 21 ⎤ = 22 V1 − ⎢ 11 22 ⎥ I1 Z12 Z12 ⎣ ⎦ Z 21 I1

A′ V1

B′ I1 ,

Z 22 Z12 Z Z − Z12 Z 21 ΔZ B′ = 11 22 = Z12 Z12 A′ =

2. A�B�C �D� Parameters in Terms of Y-parameters We know that I1

Y1111 V1 Y12 V2

I2

Y21 21 V1 Y22 V2

Rewriting the first equation, Y11 1 V1 + I1 Y12 Y12 V2 A′V1 B′ I1 , Y A′ = − 11 Y12 1 B′ = − Y12

V2 = − Comparing with

Also, Comparing with

1 ⎤ ⎡ Y12 Y21 − Y11 Y22 ⎤ Y ⎡ Y Y2211 V1 Y22 ⎢ − 11 V1 + I1 ⎥ = ⎢ V1 + 222 I1 ⎥ Y12 ⎦ ⎣ Y12 Y112 ⎣ Y12 ⎦ I 2 C ′V1 D ′ I1 , Y Y −Y Y ΔY C ′ = 12 21 11 22 = − Y12 Y12 Y D ′ = − 22 Y12 I2

3. A� B� C � D� Parameters in Terms of ABCD Parameters We know that V1 I1

AV AV2 CV V2

B BI 2 DI D 2

13.46 Network Analysis and Synthesis By Cramer’s rule,

where Comparing with

Also,

Comparing with

V1 B I1 D D B V2 = = V2 I2 A B ΔT ΔT C D ΔT = AD − BC V2 A′V1 B′ I1 , D A′ = ΔT B B′ = ΔT A V1 C I1 C A −I2 = =− V1 + I1 ΔT ΔT ΔT C A I2 = V1 − I1 ΔT ΔT I 2 C ′ V1 D ′ I1 , C ΔT A D′ = ΔT C′ =

4. A� B� C � D� Parameters in Terms of Hybrid Parameters We know that V1

h11 I1 h12 V2

I2

h21 I1 h22 V2

V2 =

1 h V1 − 11 I1 h12 h12 A′ V1 B′ I1 ,

Rewriting the first equation,

Comparing with

V2

1 h12 h B′ = 11 h12 A′ =

Also,

I2

Comparing with

I2 C′ =

h ⎡ 1 ⎤ h ⎡h h − h h ⎤ h21 I1 h22 ⎢ V1 − 11 I1 ⎥ = 22 V1 + ⎢ 11 22 12 21 ⎥ I1 h12 ⎦ h12 h12 ⎣ h12 ⎣ ⎦ C ′ V1 D ′ I1 , h22 h12

⎡ h h − h h ⎤ Δh D ′ = ⎢ 11 22 12 21 ⎥ = h12 ⎣ ⎦ h12 5. A� B� C� D� parameters in Terms of Inverse Hybrid Parameters We know that I V

g11 V g 21 V

g12 I 2 g22 I 2

13.8 Inter-relationships between the Parameters 13.47

Rewriting the first equation, g11 1 V1 + I1 g12 g12 C ′ V1 D ′ I1 ,

I2 = − Comparing with

I2

g11 g12 1 D′ = − g12 C=−

Also,

V

g21 V

Comparing with

⎡ g ⎡ g g22 − g g21 ⎤ 1 ⎤ g22 g22 ⎢ − 11 V1 + I1 ⎥ = − ⎢ ⎥ V2 + g I1 g g g 12 12 12 ⎣ 12 ⎦ ⎣ ⎦ V2 A′ V1 B′ I1 , ⎡ g g22 − g g21 ⎤ Δg A′ = − ⎢ =− ⎥ g12 g12 ⎣ ⎦ g22 B′ = g12

13.8.5 Hybrid Parameters in Terms of Other Parameters 1. Hybrid Parameters in terms of Z-parameters We know that V1 V2 Rewriting the second equation,

Z11 I1 Z12 I 2 Z 21 I1 Z 22 I 2 Z 21 1 I1 + V2 Z 22 Z 22 h21 I1 h22 V2 ,

I2 = − Comparing with

I2

Z 21 Z 22 1 h22 = Z 22 h21 = −

Also,

V1

Z11 I1

Comparing with

V1

h11 I1

1 Z112 ⎡ Z ⎤ ⎡ Z Z − Z12 Z 21 ⎤ Z12 ⎢ − 21 I1 + V2 ⎥ = ⎢ 11 22 ⎥ I1 + Z V2 Z Z Z ⎣ 22 22 ⎦ ⎣ 22 ⎦ 222 h12 V2 ,

Z11 Z 22 − Z12 Z 21 ΔZ = Z 22 Z 22 Z12 h12 = Z 22 h11 =

2. Hybrid Parameters in terms of Y-parameters We know that I1 Y11 11 V1 Y12 V2 I 2 Y2211 V1 Y22 V2 Rewriting the first equation, 1 Y V1 = I1 − 12 V2 Y11 Y11

13.48 Network Analysis and Synthesis Comparing with

V1

h11 I1 h12 V2 ,

1 Y11 Y h12 = − 12 Y11 h11 =

Also,

I2

Comparing with

I2

Y Y ⎡ 1 ⎤ ⎡Y Y − Y Y ⎤ Y21 ⎢ I1 − 12 V2 Y22 V2 = ⎢ 11 22 12 21 ⎥ V2 + 21 I1 Y11 ⎦ Y11 Y11 ⎣ Y11 ⎣ ⎦ h21 I1 h22 V2 ,

Y22 Y11 Y Y −Y Y ΔY h22 = 11 22 12 21 = Y11 Y11 h21 =

3. Hybrid Parameters in Terms of ABCD Parameters We know that V1 AV AV2 BI B 2 I1 CV V2 DI D 2 Rewriting the second equation, 1 C I2 = I1 + V2 D D Comparing with I 2 h21 I1 h22 V2 , h21 = −

1 D

h22 =

C D

Also,

V1

AV2 AV

Comparing with

V1

h11 I1

C ⎤ B ⎡ 1 ⎡ AD − BC ⎤ B ⎢ − I1 + V2 ⎥ = I1 + ⎢ ⎥ V2 D D D D ⎣ ⎦ ⎣ ⎦ h12 V2 ,

B D AD − BC ΔT h12 = = D D h11 =

4. Hybrid Parameters in Terms of A�B�C�D� Parameters We know that V2 A′ V1 B′ I1 I 2 C ′ V1 D ′ I1 Rewriting the first equation, B′ 1 V1 = I1 + V2 A′ A′ Comparing with V1 h11 I1 h12 V2 , B′ A′ 1 h12 = A′ h11 =

13.8 Inter-relationships between the Parameters 13.49

Also,

I2

Comparing with

I2

1 ⎤ C′ ⎡ B′ ⎡ A′ D ′ − B′ C ′ ⎤ I1 + V2 D ′ I1 = − ⎢ ⎥ I1 + A′ V2 A ′ A ′ A ′ ⎣ ⎦ ⎣ ⎦ h21 I1 h22 V2 ,

C′

ΔT ′ ⎡ A′ D ′ − B′C ′ ⎤ h21 = − ⎢ =− ⎥ A′ A′ ⎣ ⎦ C′ h22 = A′ 5. Hybrid Parameters in Terms of Inverse Hybrid Parameters We know that I

g11 V

g12 I 2

V

g 21 V

g22 I 2

By Cramer’s rule, I V V1 = g g where Comparing with

g12 g 22 g g = 22 I1 − 12 V2 g12 Δg Δg g22

Δg = g11 g − g12 g21 V1 h11 I1 h12 V2 , g22 Δg g h12 = − 12 Δg h11 =

Also,

I2 =

Comparing with

I2

g g

g g = − 21 I1 + 11 V2 Δg Δg Δg h21 I1 h22 V2 ,

h21 = − h22 =

I1 V2

g21 Δg

g11 Δg

13.8.6 Inverse Hybrid Parameters in Terms of Other Parameters 1. Inverse Hybrid Parameters in Terms of Z-parameters We know that V1 V2

Z11 I1 Z12 I 2 Z 21 I1 Z 22 I 2

I1 =

1 Z V1 − 12 I 2 Z11 Z11 g11 V g12 I 2 ,

Rewriting the first equation,

Comparing with

I g11 =

1 Z11

13.50 Network Analysis and Synthesis g12 = − Also,

V2

Comparing with

V

Z12 Z11

Z Z 21 ⎡ 1 ⎤ ⎡ Z Z − Z12 Z 21 ⎤ Z 21 ⎢ V1 − 12 I 2 Z 22 V1 + ⎢ 11 22 22 I 2 = ⎥ I2 Z11 ⎦ Z11 Z11 ⎣ Z11 ⎣ ⎦ g21 V g22 I 2 , Z 21 Z11 Z Z − Z12 Z 21 ΔZ = 11 22 = Z11 Z11

g21 = g22

2. Inverse Hybrid Parameters in Terms of Y-parameters We know that I1 Y1111 V1 Y12 V2 I2 Rewriting the second equation,

Y21 21 V1 Y22 V2 Y21 1 V1 + I2 Y22 Y22 g21 V g22 I 2 ,

V2 = − Comparing with

V

Y21 Y22 1 = Y22

g21 = − g22 Also,

I1

Comparing with

I

1 Y ⎡ Y ⎤ ⎡Y Y − Y Y ⎤ Y1111 V1 Y12 ⎢ − 21 V1 + I 2 ⎥ = ⎢ 11 22 12 21 ⎥ V1 + 112 I 2 Y Y Y Y ⎣ 22 22 ⎦ ⎣ 22 ⎦ 222 g11 V g12 I 2 ,

Y11 Y22 − Y12 Y21 ΔY = Y22 Y22 Y g12 = 12 Y22 g11 =

3. Inverse Hybrid Parameters in Terms of ABCD Parameters We know that V1 A AV V2 B BI 2 I1 CV V2 DI D 2 Rewriting the first equation, 1 B V2 = V1 + I 2 A A Comparing with V g21 V g22 I 2 , 1 A B = A

g21 = g22 Also,

I1

C

B ⎤ C ⎡1 V1 + I 2 D 2 = V1 DI A A A ⎣ ⎦

⎡ AD − BC ⎤ ⎢ ⎥ I2 A ⎣ ⎦

13.8 Inter-relationships between the Parameters 13.51

Comparing with

I

g11 V

g12 I 2 ,

C g11 = A ΔT ⎡ AD − BC ⎤ g12 = − ⎢ =− ⎥ A A ⎣ ⎦ 4. Inverse Hybrid Parameters in Terms of A�B�C�D� Parameters We know that V2

A′ V1

B′ I1

I2

C ′ V1

D ′ I1

Rewriting the second equation, I1 = Comparing with

I

C′ 1 V1 − I2 D′ D′ g11 V g12 I 2 ,

C′ D′ 1 g12 = − D′ g11 =

Also,

V2

Comparing with

V

⎡ C′ B′ ⎢ V1 ⎣ D′ g21 V g22 I 2 , A′ V1

1 ⎤ ⎡ A′ D ′ − B′C ′ ⎤ B′ I2 = ⎥ V1 + D ′ I 2 D ′ ⎥⎦ ⎢⎣ D′ ⎦

⎡ A′ D ′ − B′C ′ ⎤ ΔT ′ g21 = ⎢ ⎥ = D′ D′ ⎣ ⎦ B′ g22 = D′ 5. Inverse Hybrid Parameters in Terms of Hybrid Parameters We know that

By Cramer’s rule,

V1

h11 I1 h12 V2

I2

h21 I1 h22 V2

V1 h12 I 2 h22 h h I1 = = 22 V1 − 12 I 2 h11 h12 Δh Δh h21 h22 where Comparing with

Δh = h11 h22 − h12 h21 I g11 V g12 I 2 , h22 Δh h g12 = − 12 Δh h11 V1 h21 I 2 h h V2 = = − 21 V1 + 11 I 2 Δh Δh Δh V g21 V g22 I 2 , g11 =

Also, Comparing with

13.52 Network Analysis and Synthesis g21 = − g22 =

h21 Δh

h11 Δh

Table 13.3 Inter-relationship between parameters ΔX = X11 X 22 − X12 X 21 In terms of [Z]

[Z]

[Y] −

[T]

[T ′]

[h] −

[g]

[Y]

ΔT C

D′ C′

Y11 ΔY

1 C

D C

ΔT ′ C′

A′ C′

Y11 Y12

D B

−

A′ B′

−

1 B

A B

Z12

Z 21

Z 22

Z 22 ΔZ

−

Z 21 Δ

Z11 ΔZ

Z11 Z 21

ΔZ Z 21

−

Y22 Y21

−

1 Y21

A

1 Z 221

Z 222 Z 221

−

ΔY Y21

−

Y11 Y21

Z 22 Z12

ΔZ Z12

−

Y11 Y12

−

1 Z112

Z111 Z112

−

ΔY Y12

ΔZ Z 22

Z12 Z 22

Z 21 Z 22

1 Z 22

1 Z111

−

Z 21 Z11

ΔZ Z11

Z12 ΔZ

Y22 ΔY

−

Y21 Δ

Y12 ΔY

−

ΔT B

.

1 C′ −

1 B′

[g]

Δh h22

h12 h22

1 g11

−

h21 h22

1 h22

g21 g11

Δg g11

1 h111

−

Δg Δ g22

g12 g22

h21 h11

Δh h11

g21 g

1 g22

h112 h111

g12 g11

ΔT ′ B′

D′ B′

B

D′ ΔT ′

B′ ΔT ′

−

Δh h21

−

h11 h21

1 g21

g22 g21

C

D

C′ ΔT ′

A′ ΔT ′

−

h22 h21

−

1 h21

g11 g21

Δg g21

1 Y12

D ΔT

B ΔT

A′

B′

1 h112

h111 h112

−

Δg g12

−

g22 g12

−

Y22 Y12

C ΔT

A ΔT

C′

D′

h22 h12

Δh h12

−

g11 g

−

1 g12

1 Y111

−

Y112 Y111

B D

ΔT D

B′ A′

1 A′

h11 h12

g22 Δg

−

g12 Δg

Y21 Y11

ΔY Y11

1 D

C D

ΔT ′ A′

C′ A′

h21 h22

g21 Δg

g11 Δg

ΔY Y22

Y12 Y22

C A

−

C′ D′

−

g

g12

Y21 Y22

1 Y22

1 A

B A

g

g22

Y21 Y22

Z112 Z111

[h]

A C

Z11

−

[T �]

[T]

−

−

ΔT A

−

ΔT ′ D′

1 D′

B′ D′

−

h22 Δh

−

h12 Δh

h21 Δh

h11 Δh

−

−

Example 13.26 The Z parameters of a two-port network are Z11 = 20 W, Z22 = 30 W, Z12 = Z21 = 10 W. Find Y and ABCD parameters.

13.8 Inter-relationships between the Parameters 13.53

Solution ΔZ = Z11 Z 22 − Z12 Z 21 = ( 20)(30) − (10)(10) = 500 Y-parameters Z 22 30 3 = = , ΔZ 500 50 Z 10 1 Y21 = − 21 = − =− , ΔZ 5 500 50 Hence, the Y-parameters are 1⎤ ⎡ 3 − ⎥ ⎡Y11 Y12 ⎤ ⎢ 50 50 ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1 2 ⎥⎥ ⎢− ⎣ 50 50 ⎦ ABCD parameters Y11 =

Z12 10 1 =− =− ΔZ 500 50 Z 20 2 = 111 = = ΔZ 500 50

Y12 = − 222

ΔZ 500 = = 50 Z 21 10 Z 30 D = 22 = =3 Z 21 10 1

Z11 20 = = 2, Z 21 10 1 1 C= = = 0.1, Z 21 10 A=

B=

Hence, the ABCD parameters are ⎡ A B ⎤ ⎡ 2 50 ⎤ ⎢⎣C D ⎥⎦ = ⎢⎣0 1 3 ⎥⎦

Example 13.27 Currents I1 and I2 entering at Port 1 and Port 2 respectively of a two-port network are given by the following equations: I 1 0 V1 − 0.2V2 I2 0 2V1 + V2 Find Y, Z and ABCD parameters for the network. Solution Y11 =

I1 V1 V2

= 0.5 ,

Y12 =

0

I2 = −0.2 , V1 V2 = 0 Hence, the Y-parameters are ⎡Y11 Y12 ⎤ ⎡ 0.5 −0.2⎤ ⎢⎣Y21 Y22 ⎥⎦ = ⎢⎣ −0 2 1 ⎥⎦ Y21 =

Y22 =

I1 V2

V1 = 0

I2 V2

V1 = 0

= −0 2 =1

Z-parameters ΔY = Y11 Y22 − Y12 Y21 = (0.5)(1) − ( −0.2)( −0.2) = 0.46 Y22 1 = = 2.174 Ω, ΔY 0 46 Y2 ( −0.2) Z 21 = − 21 =− = 0.434 Ω, ΔY 0 46 Z12 ⎤ ⎡ 2.174 0.434 ⎤ = Z 22 ⎥⎦ ⎢⎣ 0.434 1.087 ⎥⎦ Z11 =

⎡ Z11 ⎢⎣ Z 21

Y12 ( −0.2) =− = 0.434 3 Ω ΔY 0 46 Y 05 = 111 = = 1.087 Ω ΔY 0 46

Z12 = − Z 222

13.54 Network Analysis and Synthesis ABCD parameters Y22 1 =− = 5, Y21 −0 2 ΔY 0 46 C=− =− = 2.3, Y21 −0 2 Hence, the ABCD parameters are 5⎤ ⎡A B⎤ ⎡ 5 ⎢⎣C D ⎥⎦ = ⎢⎣ 2.3 2.5⎥⎦

1 1 =− =5 Y21 −0 2 Y 05 D = − 11 = − =25 Y21 −0 2

A= −

Example 13.28

B=−

Using the relation Y = Z−1, show that | |=

1 ⎛ Z 22 Z ⎞ 2 + 111 ⎟ . 2 ⎝ Y111 Y222 ⎠

Solution We know that Z −1 Z ⎤ ⎡ Z 22 − 12 ⎥ Y12 ⎤ ⎢ ΔZ ΔZ = Y22 ⎥⎦ ⎢⎢ Z 21 Z11 ⎥⎥ − ⎣ ΔZ ΔZ ⎦ | Z | Z11 Z 22 Z12 Z 21 Y

⎡Y11 ⎢⎣Y21

i.e.,

⎛ ⎞ 1 ⎛ Z 222 Z111 ⎞ 1 ⎜ Z 222 Z111 ⎟ 1 + = + = ( ΔZ + ΔZ Z) 2 ⎜⎝ Y111 Y222 ⎟⎠ 2 ⎜ Z 222 Z111 ⎟ 2 ⎜⎝ ⎟ ΔZ ΔZ ⎠ |

Example 13.29

|=

1 ( ΔZ ) = ΔZ 2

Z11 Z12

Z12 Z 21

1 ⎛ Z 22 Z11 ⎞ + 2 ⎝ Y11 Y22 ⎟⎠

For the network shown in Fig 13.59, find Z and Y-parameters. 2Ω

I1

I2

+ V1

+

1Ω

2Ω

−

3I2

V2 −

Fig. 13.59 Solution The network is redrawn by source transformation technique as shown in Fig 13.60. Applying KVL to Mesh 1, 2Ω V1 I1 − I 3 …(i) Applying KVL to Mesh 2, + V2 2( I 2 I 3 ) 6 I 2 2Ω = −4 I 2 2 I 3 …(ii) V 1Ω 1 Applying KVL to Mesh 3, I1 I3 − 6I2 −( − ) − 2 I 3 − 2( + ) + 6 I 2 = 0 + − 5 I 3 = I1 + 4 I 2 1 4 …(iii) Fig. 13.60 I 3 = I1 + I 2 5 5

+ V2 I2

−

13.8 Inter-relationships between the Parameters 13.55

Substituting Eq. (iii) in Eq. (i), V1

1 4 I1 − I1 I2 5 5 4 4 = I1 I2 5 5

…(iv)

Substituting Eq. (iii) in Eq. (ii), 4I2 + 2

V2

⎛1 I1 ⎝5

4 ⎞ I2 ⎟ 5 ⎠

2 12 I1 I2 5 5 Comparing Eqs (iv) and (v) with Z-parameter equations, 4⎤ ⎡4 − ⎥ ⎡ Z11 Z12 ⎤ ⎢ 5 5 ⎢⎣ Z 21 Z 22 ⎥⎦ = ⎢ 2 12 ⎥⎥ ⎢ − 5⎦ ⎣5 Y-parameters 40 8 ⎛ 4 ⎞ ⎛ 12 ⎞ ⎛ 4 ⎞ ⎛ 2 ⎞ ΔZ = Z11 Z 22 − Z12 Z 21 = ⎜ ⎟ ⎜ − ⎟ − ⎜ − ⎟ ⎜ ⎟ = − =− ⎝ 5⎠ ⎝ 5 ⎠ ⎝ 5⎠ ⎝ 5⎠ 255 5 12 4 − − Z 22 3 Z 1 Y11 = = 5 = , Y12 = − 12 = 5 = − 8 8 ΔZ 2 ΔZ 2 − − 5 5 2 4 − Z 21 1 Z111 1 5 Y21 = − = = , = 5 =− 222 = 8 4 8 ΔZ ΔZ 2 − − 5 5 Hence, the Y-parameters are 1⎤ ⎡3 − ⎥ ⎡Y11 Y12 ⎤ ⎢ 2 2 ⎥ ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1 1 ⎢ − ⎥ ⎣4 2⎦ =

Example 13.30

…(v)

Find Z and h-parameters for the network shown in Fig 13.61. 4I1

2Ω

I1

+

2Ω

V1

2Ω

+−

I1

I2

+

2Ω I3

−

V2 I2

−

Fig. 13.61 Solution Applying KVL to Mesh 1, V1 2 I1 2( I1 I 3 ) = 4I 4 I1 2 I 3 Applying KVL to Mesh 2, V2 2 I 2 2( I 2 I 3 ) = 4I 4 I 2 2I3

…(i) …(ii)

13.56 Network Analysis and Synthesis Applying KVL to Mesh 3, −2( 3 − 1 ) − 4 I1 − 2(

+ 2) = 0 I1 + I 2 = −2 I 3

3

…(iii)

Substituting Eq. (iii) in Eq. (i), V1 = 4 I1

I1

= 5 I1

I2

Substituting Eq. (iii) in Eq. (ii), V2

I2

4 I 2 − I1

…(iv)

I2

…(v)

= − I1 3I 2 Comparing Eqs (iv) and (v) with Z-parameter equations, ⎡ Z11 ⎢⎣ Z 21

Z12 ⎤ ⎡ 5 1⎤ = Z 22 ⎥⎦ ⎢⎣ −1 3⎥⎦

h-parameters ΔZ = Z11 Z 22 − Z12 Z 21 = (5) (3) − (1) ( −1) = 15 + 1 = 16 ΔZ 16 = Ω, Z 22 3 Z 1 h21 = − 21 = , Z 22 3 Hence, the h-parameters are

Z12 1 = Z 22 3 1 1 h22 = = Z 22 3

h11 =

⎡16 h12 ⎤ ⎢ 3 = h22 ⎥⎦ ⎢⎢ 1 ⎣3

⎡ h11 ⎢⎣ h22

Example 13.31

h12 =

1⎤ 3⎥ 1 ⎥⎥ 3⎦

Find Y and Z-parameters for the network shown in Fig 13.62. 1Ω 2

I1

3

1Ω

2

I2

+

+ 1Ω 2

2V1

V1 −

V2 −

Fig. 13.62 Solution Applying KCL at Node 3, 2(

Now,

1

3)

2V1 ( V2 V3 = 3 I1

3

2)

2V1 + (V3 V2 ) V = 2V1 + 2 − V2 3 2 = 2V1 V2 3

…(i)

…(ii)

13.8 Inter-relationships between the Parameters 13.57

I2

2V2 + (V2 V3 ) V = 3V2 − 2 3 8 = V2 3

…(iii)

Comparing Eqs (ii) and (iii) with Y-parameter equations, 2⎤ ⎡ 2 − ⎥ ⎡Y11 Y12 ⎤ ⎢ 3 ⎥ ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 8 ⎢0 ⎥ 3 ⎦ ⎣ Z-parameters 16 ⎛ 8⎞ ΔY = Y11 Y22 − Y12 Y21 = ( 2) ⎜ ⎟ − 0 = ⎝ 3⎠ 3 ⎛ 2⎞ ⎜⎝ − ⎟⎠ 1 Y 3 Z12 = − 12 = − = Ω 16 ΔY 8 3 Y 2 3 Z 222 = 111 = = Ω ΔY 16 8 3

8 Y22 1 Z11 = = 3 = Ω, ΔY 16 2 3 Y21 0 Z 21 = − =− = 0, 16 ΔY 3 Hence, the Z-parameters are ⎡ Z11 ⎢⎣ Z 21

Example 13.32

⎡1 Z12 ⎤ ⎢ 2 = Z 22 ⎥⎦ ⎢⎢ 0 ⎣

1⎤ 8⎥ 3 ⎥⎥ 8⎦

For the network shown in Fig 13.63, find Y and Z-parameters. I1

1

1Ω

+ V1

3 V1

2

I2

+−

2Ω

+

1Ω

−

V2 −

Fig. 13.63 Solution Applying KCL at Node 1, V1 V1 − 3V1 V2 + 2 1 3 = − V1 V2 2

I1 =

…(i)

Applying KCL at Node 2, V2 V2 + 3V1 V1 + 1 1 = 2V 2V1 2V2

I2 =

…(ii)

13.58 Network Analysis and Synthesis Comparing Eqs (i) and (ii) with Y-parameter equations, ⎡ 3 ⎡Y11 Y12 ⎤ ⎢ − = ⎢⎣Y21 Y22 ⎥⎦ ⎢ 2 ⎣ 2

⎤ −1⎥ 2 ⎥⎦

Z-parameters ⎛ 3⎞ ΔY = Y11 Y22 − Y12 Y21 = ⎜ − ⎟ ( 2) − ( −1) ( 2) = −3 + 2 = −1 ⎝ 2⎠ Z11 =

Y22 2 = = −2 Ω, ΔY −1

Z 21 = −

Y21 2 =− = ΔY ( −1)

Y12 ( −1) =− = −1 Ω ΔY ( −1) 3 − Y111 3 = = 2= Ω ΔY −1 2

Z12 = −

Ω,

Z 222

Hence, the Z-parameters are ⎡ Z11 ⎢⎣ Z 21

Example 13.33

2 ⎡ −2 Z12 ⎤ ⎢ = Z 22 ⎥⎦ ⎢ 2 ⎣

1⎤ 3⎥ ⎥ 2⎦

Find Z-parameters for the network shown in Fig 13.64. Hence, find Y and h-parameters. 0.9 I1 I1

1Ω

I2

+

+

10 Ω

V2

1Ω

V1 −

−

Fig. 13.64 Solution The network is redrawn by source transformation technique as shown in Fig 13.65. Applying KVL to Mesh 1, 9I1 1Ω 10 Ω I1 V1 2 I1 + I 2 …(i) +

Applying KVL to Mesh 2, 9 I1 10 I 2 1( I1 + I 2 )

V2

−+

= 10 I1 11I 2

…(ii)

Comparing Eqs (i) and (ii) with Z-parameter equations,

1Ω

V1 −

Z 22 11 = , ΔZ 12 Z 10 5 Y21 = − 21 = − = − ΔZ 12 6

Y22

Fig. 13.65

Z12 1 =− ΔZ 12 Z 2 1 = 11 = = ΔZ 12 6

Y12 = − ,

+ V2 −

⎡ Z11 Z12 ⎤ ⎡ 2 1 ⎤ ⎢⎣ Z 21 Z 22 ⎥⎦ = ⎢⎣10 11⎥⎦ Y-parameters ΔZ = Z11 Z 22 − Z12 Z 21 = ( 2) (11) − (1) (10) = 22 − 10 = 12 Y11 =

I2

13.8 Inter-relationships between the Parameters 13.59

Hence, the Y-parameters are ⎡ 11 ⎡Y11 Y12 ⎤ ⎢ 12 ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 5 ⎢− ⎣ 6

−

1⎤ 12 ⎥ 1 ⎥⎥ 6 ⎦

h-parameters ΔZ 12 = Ω, Z 22 11 Z 10 h21 = − 21 = − , Z 22 11

Z12 1 = Z 22 11 1 1 h22 = = Z 22 11 1

h11 =

Hence, h-parameters are

h12 =

1⎤ ⎡ 12 ⎡ h11 h12 ⎤ ⎢ 11 11⎥ ⎢⎣ h21 h22 ⎥⎦ = ⎢ 10 1 ⎥ ⎢− ⎥ ⎣ 11 11⎦

Example 13.34

Find Y and Z-parameters of the network shown in Fig 13.66. 1

I1

1Ω

+

1Ω

V1

2V2

2V1

2

I2

−+

+

2Ω

V2

−

−

Fig. 13.66 Solution Applying KCL at Node 1, I1 2V2 = I1

V1 3V1 − V2 + 1 1 4V1 3V2

…(i)

Applying KCL at Node 2, V2 V2 − 2V1 V1 + 2 1 = −3V1 1 V2 Comparing Eqs (i) and (ii) with Y-parameter equations, 3⎤ ⎡Y11 Y12 ⎤ ⎡ 4 ⎢⎣Y21 Y22 ⎥⎦ = ⎢⎣ −3 1.5⎥⎦ Z-parameters ΔY = Y11Y22 − Y12Y21 = ( 4) (1.5) − ( −3)( −3) = −3 I2 =

Y22 15 =− = −0.5 Ω, ΔY 3 Y ( −3) = − 21 = − = −1 Ω, ΔY −3

Z11 = Z 21 Hence, the Z-parameters are ⎡ Z11 ⎢⎣ Z 21

⎡ −0 5 −1 ⎤ Z12 ⎤ ⎢ = 4⎥ Z 22 ⎥⎦ ⎢ −1 − ⎥ 3⎦ ⎣

…(ii)

Y12 ( −3) =− = −1 Ω ΔY −3 Y 4 4 = 111 = − = − Ω ΔY 3 3

Z12 = − 222

13.60 Network Analysis and Synthesis

Example 13.35

Determine Y and Z-parameters for the network shown in Fig 13.67. 1

I1

2

2Ω

I2

+

+

1Ω

V1

2Ω

V2

3I1

−

−

Fig. 13.67 Solution Applying KCL at Node 1, V1 V1 − V2 + 1 2 = 1.5V1 0 5V2

I1 =

…(i)

Applying KCL at Node 2, V2 V V + 3I1 + 2 1 2 2 V2 V V = + 3 (1.5V1 0. V2 ) + 2 1 2 2 = 0.5 5V2 + 4 5V1 − 1.55V2 + 0 5V2 − 0 5V1 = 4 V1 − 0 5V2 Comparing Eqs (i) and (ii) with the Y-parameter equation, ⎡Y11 Y12 ⎤ ⎡1.5 −0.5⎤ ⎢⎣Y21 Y22 ⎥⎦ = ⎢⎣ 4 0.5⎥⎦ Z-parameters ΔY = Y11Y22 − Y12Y21 = (1.5)( −0.5) − ( −0.5)( 4) = 1.25 I2 =

Y22 05 =− = −0.4 Ω, ΔY 1.25 Y 4 = − 21 = − = −3.2 Ω, ΔY 1.25

Z11 = Z 21 Hence, the Z-parameters are ⎡ Z11 ⎢⎣ Z 21

Example 13.36

Y12 05 = =04Ω ΔY 1.25 Y 15 = 111 = = 1.2 Ω ΔY 1.25

Z12 = − 22

Z12 ⎤ ⎡ −0.4 0.4 ⎤ = Z 22 ⎥⎦ ⎢⎣ −3.2 1.2 ⎥⎦

Determine the Y and Z-parameters for the network shown in Fig 13.68. I1

0.5 Ω

1

3

2

1Ω

+ V1

…(ii)

I2 +

1Ω

2V1

−

0.5 Ω

V2

−

Fig. 13.68

13.8 Inter-relationships between the Parameters 13.61

Solution Applying KCL at Node 1, V1 V1 − V3 + 1 0.5 = 3V 3V1 2V3

…(i)

V2 V2 − V3 + 05 1 = 3V2 V3

…(ii)

I1 = Applying KCL at Node 2,

I2 =

Applying KCL at Node 3, V3 V1 V V + 2V1 + 3 2 = 0 05 1 1 V3 V2 3 Substituting Eq. (iii) in Eqs (i) and (ii),

…(iii)

2 3V1 − V2 3 8 I 2 0V1 + V2 3 Comparing Eqs (iv) and (v) with Y-parameter equations, 2⎤ ⎡ 3 − ⎥ ⎡Y11 Y12 ⎤ ⎢ 3 ⎥ ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 8 ⎢0 ⎥ 3 ⎦ ⎣ Z-parameters I1

…(iv) …(v)

⎛ 8⎞ Y11Y22 Y12Y21 = (3) ⎜ ⎟ − 0 = 8 ⎝ 3⎠ 8 Y 1 Z11 = 22 = 3 = Ω, ΔY 8 3 2 Y12 1 Z 21 = − = 3= Ω, ΔY 8 12 ΔY

Z12

2 Y12 3 1 =− = = Ω ΔY 8 12

Z 22 =

Y11 3 = Ω YΔ Δ 8

Hence, the Z-parameters are ⎡ Z11 ⎢⎣ Z 21

Example 13.37

⎡1 1 ⎤ Z12 ⎤ ⎢ 3 12 ⎥ = Z 22 ⎥⎦ ⎢⎢ 3 ⎥⎥ 0 8⎦ ⎣

Determine Z and Y-parameters of the network shown in Fig. 13.69. I1

2Ω

4Ω

+ V1

I2 +

0.05I2

+ −

− 10V1 +

V2 −

−

Fig. 13.69

13.62 Network Analysis and Synthesis Solution Applying KVL to Mesh 1, V1 − 4I1 − 0.05 I2 = 0 V1 = 4I1 + 0.05I2 Applying KVL to Mesh 2, V2 − 2I2 + 10V1 = 0 V2 = 2I2 − 10V1 Substituting Eq. (i) in Eq. (ii), V2 = 2I2 − 40I1 − 0.5I2 = −40I1 + 1.5I2 Comparing Eqs (i) and (iii) with Z-parameter equations, ⎡ Z11 Z12 ⎤ ⎡ 4 0.05⎤ ⎢⎣ Z 21 Z 22 ⎥⎦ = ⎢⎣ −40 1 5 ⎥⎦ Y-parameters ΔZ Z11Z 22 Z12 Z 21 = ( 4)(1.5) − (0.05)( −40) = 8 Z 22 1 5 = , ΔZ 8 Z ( −40) 40 Y221 = − 221 = − = ΔZ 8 8 Y11 =

Hence, the Y-parameters are

⎡1.5 ⎡Y11 Y12 ⎤ ⎢ 8 ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 40 ⎢ ⎣ 8

Example 13.38

−

…(i)

...(ii)

...(iii)

Z112 0 05 =− ΔZ 8 Z111 4 = = ΔZ 8

Y12 = − ,

Y222

0.05 ⎤ 8 ⎥ 4 ⎥⎥ 8 ⎦

Determine Z and Y-parameters of the network shown in Fig. 13.70. 2V3

1Ω

I1

+

3I2

I3

+−

I2

+

2Ω

V1

2Ω

I1

−

+ V3 −

V2 I2

−

Fig. 13.70 Solution Applying KVL to Mesh 1, V1 − 1I1 − 3I2 − 2(I1 + I2) = 0 V1 = 3I1 + 5I2 Applying KVL to Mesh 2, V2 − 2(I2 − I3) − 2(I1 + I2) = 0 V2 − 2I2 + 2I3 − 2I1 − 2I2 = 0 V2 = 2I1 + 4I2 − 2I3

...(i)

…(ii)

13.9 Interconnection of Two-Port Networks 13.63

Writing equation for Mesh 3, I3 = 2V3

...(iii)

From Fig. 13.70, V3 = 2(I1 + I2) I3 = 2V3 = 4I1 + 4I2

...(iv)

Substituting Eq. (iv) in Eq. (ii), V2 = −6I1 − 4I2 Comparing Eqs (i) and (v) with Z-parameter equations, 5⎤ ⎡ Z11 Z12 ⎤ ⎡ 3 ⎢⎣ Z 21 Z 22 ⎥⎦ = ⎢⎣ −6 6 4 ⎥⎦ Y-parameters ΔZ = Z111 Z22 − Z112 Z221 = ( )(− )( ) − ( )(− )( ) = 18 Z 22 4 2 Z 5 Y11 = =− =− , Y112 = − 12 = − ΔZ ΔZ 18 9 18 Z ( −66) 1 Z11 3 Y21 = − 21 = − = , = 222 = ΔZ 18 3 ΔZ 18 Hence, Y-parameters are 5⎤ ⎡ 2 − − ⎥ ⎡Y11 Y12 ⎤ ⎢ 9 18 ⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1 3 ⎥⎥ ⎢ 18 ⎦ ⎣ 3

13.9

...(v)

INTERCONNECTION OF TWO-PORT NETWORKS

We shall now discuss the various types of interconnections of two-port networks, namely, cascade, parallel, series, series-parallel and parallel-series. We shall derive the relation between the input and output quantities of the combined two-port networks.

13.9.1 Cascade Connection 1. Transmission Parameter Representation Figure 13.71 shows two-port networks connected in cascade. In the cascade connection, the output port of the first network becomes the input port of the second network. Since it is assumed that input and output currents are positive when they enter the network, we have I1 ′ I1

−

N1

I2′

I1′

I2

+ V1

I2 +

+

V2

V1′

−

−

+ N2

V2′ −

Fig. 13.71 Cascade Connection Let A1,B1,C1,D1 be the transmission parameters of the network N1 and A2,B2,C2,D2 be the transmission parameters of the network N2. For the network N1, ⎡V1 ⎤ ⎡ A1 B1 ⎤ ⎡ V2 ⎤ ...(13.1) ⎢⎣ I1 ⎥⎦ = ⎢⎣C1 D1 ⎥⎦ ⎢⎣ − I 2 ⎥⎦

13.64 Network Analysis and Synthesis For the network N2,

Since V1′

V2 and I 2′

⎡V1 ′ ⎤ ⎡ A2 ⎢⎣ I1 ′ ⎥⎦ = ⎢⎣C2

B2 ⎤ ⎡ V2 ′ ⎤ D2 ⎥⎦ ⎢⎣ − I 2 ′ ⎥⎦

...(13.2)

I 2 , we can write

⎡ V2 ⎤ ⎡ A2 B2 ⎤ ⎡ V2 ′ ⎤ ⎢⎣ − I 2 ⎥⎦ = ⎢⎣C2 D2 ⎥⎦ ⎢⎣ − I 2 ′ ⎥⎦ Combining Eqs (13.1) and (13.3), ⎡V1 ⎤ ⎡ A1 B1 ⎤ ⎡ A2 B2 ⎤ ⎡ V2 ′ ⎤ ⎡ A B ⎤ ⎡ V2 ′ ⎤ ⎢⎣ I1 ⎥⎦ = ⎢⎣C1 D1 ⎥⎦ ⎢⎣C2 D2 ⎥⎦ ⎢⎣ − I 2 ′ ⎥⎦ = ⎢⎣C D ⎥⎦ ⎢⎣ − I 2 ′ ⎥⎦

...(13.3)

⎡ A B ⎤ ⎡ A1 B1 ⎤ ⎡ A2 B2 ⎤ ...(13.4) ⎢⎣C D ⎥⎦ = ⎢⎣C1 D1 ⎥⎦ ⎢⎣C2 D2 ⎥⎦ Equation 13.4 shows that the resultant ABCD matrix of the cascade connection is the product of the individual ABCD matrices. 2. Inverse Transmission Parameter Representation Figure 13.72 shows two-port networks connected in cascade. Since it is assumed that input and output currents are positive when they enter the network, we have Hence,

−I1′ = I2 I1 N1

V1 −

I2′

I1′

I2

+

+

+

V2

V1′

−

−

+ N2

V2′ −

Fig. 13.72 Cascade connection A1′

B1′ , C1′

D1′

Let be the transmission parameters of the network N1 and A2′ B2′ , C2′ D2′ be the transmission parameters of the network N2. For the network N1, ⎡V2 ⎤ ⎡ A1 ′ B1 ′ ⎤ ⎡ V1 ⎤ ...(13.5) ⎢⎣ I 2 ⎥⎦ = ⎢⎣C1 ′ D1 ′ ⎥⎦ ⎢⎣ − I1 ⎥⎦ For the network N2, ⎡V2 ′ ⎤ ⎡ A2 ′ B2 ′ ⎤ ⎡ V2 ′ ⎤ ...(13.6) ⎢⎣ I 2 ′ ⎥⎦ = ⎢⎣C2 ′ D2 ′ ⎥⎦ ⎢⎣ − I1 ′ ⎥⎦ Since V1′ V2 and − I1′ I 2 , we can write ⎡V2 ′ ⎤ ⎡ A2 ′ B2 ′ ⎤ ⎡V2 ⎤ ⎢⎣ I 2 ′ ⎥⎦ = ⎢⎣C2 ′ D2 ′ ⎥⎦ ⎢⎣ I 2 ⎥⎦ Combining equations (13.5) and (13.7), ⎡V2 ′ ⎤ ⎡ A2 ′ B2 ′ ⎤ ⎡ A1 ′ B1 ′ ⎤ ⎡ V1 ⎤ ⎡ A′ ⎢⎣ I 2 ′ ⎥⎦ = ⎢⎣C2 ′ D2 ′ ⎥⎦ ⎢⎣C1 ′ D1 ′ ⎥⎦ ⎢⎣ − I1 ⎥⎦ = ⎢⎣C ′ ⎡ A′ B′ ⎤ ⎡ A2 ′ B2 ′ ⎤ ⎡ A1 ′ B1 ′ ⎤ Hence, ⎢⎣C ′ D ′ ⎥⎦ = ⎢⎣C2 ′ D2 ′ ⎥⎦ ⎢⎣C1 ′ D1 ′ ⎥⎦

...(13.7) B′ ⎤ ⎡ V1 ⎤ D ′ ⎥⎦ ⎢⎣ − I1 ⎥⎦ ...(13.8)

Equation 13.8 shows that the resultant A′B′C′D′ matrix of the cascade connection is the product of the individual A′B′C′D′ matrices.

13.9 Interconnection of Two-Port Networks 13.65

Example 13.39 Two identical sections of the network shown in Fig. 13.73 are connected in cascade. Obtain the transmission parameters of the overall connection. 2Ω

I1

2Ω

I2

+

+

1Ω

V1

2Ω

V2

−

−

Fig. 13.73 Solution The network is redrawn as shown in Fig. 13.74. Applying KVL to Mesh 1, V1 = 3I1 − I3 …(i) Applying KVL to Mesh 2, + V2 = 2I2 + 2I3 …(ii) Applying KVL to Mesh 3, V1 −I1 + 2I2 + 5I3 = 0 1 2 I3 I1 − I 2 …(iii) − 5 5 Substituting Eq. (iii) in Eq. (i), 2 ⎞ ⎛1 V1 3I1 − ⎜ I1 I2 ⎟ ⎝5 5 ⎠ =

14 I1 5

2 I2 5

2I 2

⎛1 2 ⎜ I1 ⎝5

2Ω

2Ω +

1Ω I1

2Ω I3

V2 I2

−

Fig. 13.74

…(iv)

Substituting Eq. (iii) in Eq. (ii), V2 = I1

2 ⎞ I2 ⎟ 5 ⎠

2 6 I1 I2 5 5 5 V2 − 3I 2 2

…(v)

Substituting Eq. (v) in Eq. (iv), V1

14 ⎛ 5 ⎞ V2 − 3I 2 ⎝ ⎠ 5 2 = 7V 7V2 8 I 2

2 I2 5

Comparing the Eqs (vi) and (v) with ABCD parameter equations, ⎡ A1 B1 ⎤ ⎡ 7 8⎤ ⎢⎣C1 D1 ⎥⎦ = ⎢⎣ 2 5 3⎥⎦ Hence, transmission parameters of the overall cascaded network are ⎡ A B ⎤ ⎡ 7 8⎤ ⎡ 7 8⎤ ⎡69 80 ⎤ ⎢⎣C D ⎥⎦ = ⎢⎣ 2.5 3⎥⎦ ⎢⎣ 2.5 3⎥⎦ = ⎢⎣ 25 29⎥⎦

…(vi)

13.66 Network Analysis and Synthesis

Example 13.40

Determine ABCD parameters for the ladder network shown in Fig. 13.75. 2Ω

I1

2H

I2 +

+ V1

1F

2F

1Ω

−

V2 −

Fig. 13.75 Solution The above network can be considered as a cascade connection of two networks N1 and N2. The network N1 is shown in Fig. 13.76. Applying KVL to Mesh 1, 2s 2 I2 I1 1⎞ 1 ⎛ V1 = 2 + I1 + I 2 …(i) + + ⎝ 2s ⎠ 2s Applying KVL to Mesh 2, 1 V2 V1 2s 1 1⎞ ⎛ V2 = I1 2 s + ⎟ I 2 …(ii) ⎝ 2s 2s ⎠ − − From Eq. (ii), Fig. 13.76 I1 = 2s V2 − (4s2 + 1) I2 ...(iii) Substituting Eq. (iii) in Eq. (i), 1⎞ 1 ⎛ V1 = 2 + [2ss V2 ( 4 s 2 1) I 2 ] + I 2 ⎝ 2s ⎠ 2s = ( 4s 4 s 1)V2 − (8s 8s 2 2 s + 2) I 2 Comparing Eqs (iv) and (iii) with ABCD parameter equations, 8s 2 2 s + 2 ⎤ ⎡ A1 B1 ⎤ ⎡ 4 s 1 8s = ⎢ ⎥ ⎢⎣C1 D1 ⎥⎦ 4s2 + 1 ⎦ ⎣ 2s The network N2 is shown in Fig. 13.77. I2′

I1′ +

1 s

1

V2′

−

−

I2′

I1′ +

V1′

…(iv)

+

V1′

+ Z(s) =

1 s+1

−

−

(a)

V2′

(b)

Fig. 13.77 Applying KVL to Mesh 1, V1 ′ =

1 1 I1 ′ + I2 ′ s +1 s +1

…(i)

V2 ′ =

1 1 I1 ′ + I2 ′ s +1 s +1

…(ii)

Applying KVL to Mesh 2,

13.9 Interconnection of Two-Port Networks 13.67

From Eq. (ii), 2′

I

…(iii) …(iv)

Also, Comparing Eqs (iv) and (iii) with ABCD parameter equations, ⎤

0⎤ ⎡ 1 s +1 1

Hence, overall ABCD parameters are A B⎤ ⎡ s = C D s

s + s + 2⎤ ⎡ 1 0⎤ s = 2 s + 1 1 +1 4

s + 2⎤

s + s +

13.9.2 Parallel Connection Figure13.78 shows two-port networks connected in parallel. In the parallel connection, the two networks have the same input voltages and the same output voltages. I

I1′

I + V −

I + V −

N

I

I1 N1

Fig. 13.78 Parallel connection Let of the network N2. For the network N1,

be the Y-parameters of the network N1 and ⎡ I1 ′ ⎤ I2 ′

⎡

2

be the Y-parameters

⎤ ⎡V ⎤ V

For the network N2, I1 ″ ⎤ = I2 ″

⎤ V⎤ V

For the combined network, Hence,

⎡ I1 ⎤ ⎡ = I2

I

=

⎡Y

Y ⎤ V⎤ = ″ V Y

Y ⎤ V⎤ Y V

Thus, the resultant Y-parameter matrix for parallel connected networks is the sum of Y matrices of each individual two-port networks.

Example 13.41

Determine Y-parameters for the network shown in Fig. 13.79.

13.68 Network Analysis and Synthesis 3Ω

I1

2Ω

2Ω

+ V1

I2 +

2Ω

−

V2 −

Fig. 13.79 Solution The above network can be considered as a parallel connection of two networks, N1 and N2. The network N1 is shown in Fig. 13.80. Applying KCL at Node 3, 3 2Ω 2Ω I2′ I1′ V3 I1 ′ I 2 ′ = …(i) + + 2 From Fig. 13.80, V2 V1 2Ω V1 V3 I1 ′ = …(ii) − − 2 V2 V3 Fig. 13.80 I2 ′ = …(iii) 2 Substituting Eqs (ii) and (iii) in Eq (i), V1 V3 V2 V3 V3 + = 2 2 2 3V3 V1 + V2 V V V3 = 1 + 2 ...(iv) 3 3 Substituting Eq. (iv) in Eq. (ii), V 1 ⎛ V1 V2 ⎞ I1 ′ = 1 − + ⎟ 2 2⎝ 3 3 ⎠ 1 1 …(v) = V1 V2 3 6 Substituting Eq. (iv) in Eq. (iii), V 1 ⎛ V1 V2 ⎞ I2 ′ = 2 − + ⎟ 2 2⎝ 3 3 ⎠ 1 1 …(vi) = − V1 V2 6 3 Comparing Eqs (v) and (vi) with Y-parameter equations, 1⎤ ⎡ 1 − ⎥ 3Ω I2′′ I1′′ ⎡Y11 ′ Y12 ′ ⎤ ⎢ 3 6 = ⎢ ⎥ ⎢⎣Y21 ′ Y22 ′ ⎥⎦ + + ⎢− 1 1 ⎥ ⎣ 6 3 ⎦ V1 V2 The network N2 is shown in Fig. 13.81. − − V V 1 1 I1 ″ I 2 ″ = 1 2 = V1 V2 3 3 3 Fig. 13.81

13.9 Interconnection of Two-Port Networks 13.69

Hence, the Y-parameters are 1⎤ ⎡ 1 − ⎥ ⎢ Y ″ Y ″ ⎡ 11 12 ⎤ 3 3 ⎢⎣Y21 ″ Y22 ″ ⎥⎦ = ⎢ 1 1 ⎥ ⎢− ⎥ ⎣ 3 3 ⎦ The overall Y-parameters of the network are ⎡ 1 1 + ⎡Y11 Y12 ⎤ ⎡ Y11 ′ Y11 ″ Y12 ′ Y12 ″ ⎤ ⎢ 3 3 = = ⎢ ⎢⎣Y21 Y22 ⎥⎦ ⎢⎣Y21 ′ Y21 ″ Y22 ′ Y22 ″ ⎥⎦ ⎢− 1 − 1 ⎣ 6 3

Example 13.42

1 1⎤ ⎡ 2 − − ⎥ ⎢ 6 3 = 3 1 1 ⎥⎥ ⎢⎢ 1 + − 3 3 ⎦ ⎣ 2

1⎤ − ⎥ 2 2 ⎥⎥ 3 ⎦

Find Y-parameters for the network shown in Fig. 13.82 2Ω

I1

I2

+

+

1Ω

V1

0.5 Ω

0.5 Ω

V2

0.5 Ω 2Ω

−

−

Fig. 13.82 Solution The above network can be considered as a parallel combination of two networks N1 and N2. The network N1 is shown in Fig. 13.83. Applying KCL at Node 1, I1 ′ =

V1 V1 − V2 + 1 2

3 = V1 2

1 V2 2

…(i)

V2 V2 − V1 + 05 2 1 5 …(ii) = − V1 V2 2 2 Comparing Eqs (i) and (ii) with Y-parameter equation, I2 ′ =

⎡ 3 ⎡Y11 ′ Y12 ′ ⎤ ⎢ 2 ⎢⎣Y21 ′ Y22 ′ ⎥⎦ = ⎢ 1 ⎢− ⎣ 2

1⎤ − ⎥ 2 5 ⎥⎥ 2 ⎦

The network N2 is shown in Fig. 13.84.

2

I2′

+

+

1Ω

V1

Applying KCL at Node 2,

2Ω

1

I1′

0.5 Ω

V2

−

−

Fig. 13.83 I1′′

0.5 Ω

0.5 Ω

3

+ V1

I2′′ +

2Ω

−

V2 −

Fig. 13.84

13.70 Network Analysis and Synthesis Applying KCL at Node 3, V3 2 V1 V3 I1 ″ = = 2V 2V1 2V3 where 05 V V I 2 ″ = 2 3 = 2V 2V2 2V3 05 Substituting I1″ and I2″ in Eq (i), I1 ″ I 2 ″ =

2V1 2V3 + 2

…(i)

2V3 = 0 5V3

2

4 5V3

and

2

1

2V2

V3

4 4 V1 + V2 9 9

I1 ″

2V1 2V3

2V1 − 2

I2 ″

2V2

2V2 − 2

2V3

…(ii) ⎛4 V1 ⎝9

4 ⎞ 10 V2 ⎟ = V1 9 ⎠ 9

⎛4 V1 ⎝9

4 ⎞ V2 ⎟ 9 ⎠

8 V2 9

…(iii)

8 10 = − V1 + V2 9 9 Comparing Eqs (iii) and (iv) with Y-parameter equations, 8⎤ ⎡ 10 − ⎥ ⎡Y11 ″ Y12 ″ ⎤ ⎢ 9 9 ⎢⎣Y21 ″ Y22 ″ ⎥⎦ = ⎢ 8 10 ⎥ ⎢− ⎥ ⎣ 9 9 ⎦ Hence, overall Y-parameters of the network are

…(iv)

⎡ 3 10 + ⎡Y11 Y12 ⎤ ⎡ Y11 ′ Y11 ″ Y12 ′ Y12 ″ ⎤ ⎢ 2 9 = = ⎢ ⎢⎣Y21 Y22 ⎥⎦ ⎢⎣Y21 ′ Y21 ″ Y22 ′ Y22 ″ ⎥⎦ ⎢− 1 − 8 ⎣ 2 9

Example 13.43

1 8 ⎤ ⎡ 47 − − ⎥ ⎢ 2 9 = 18 5 10 ⎥⎥ ⎢⎢ 25 + − 2 9 ⎦ ⎣ 18

Find Y-parameters for the network shown in Fig. 13.85. 1F

I1

1F

2Ω

2Ω

+ V1

I2 +

1Ω

−

2F

V2 −

Fig. 13.85

25 ⎤ 18 ⎥ 65 ⎥⎥ 1 ⎦ 18

−

13.9 Interconnection of Two-Port Networks 13.71

Solution The above network can be considered as a parallel connection of two networks, N1 and N2. The network N1 is shown in Fig. 13.86. I1 ′

2Ω

2Ω

I2′

+ V1

2F

−

I1′ +

+

V2

V1

−

−

2

(a)

3

2

I2′ +

1 2s

V2 −

(b)

Fig. 13.86 Applying KCL at Node 3, I1 ′ I 2 ′ = 2 s V3 From Fig. 13.86,

I1 ′ =

…(i)

V1 V3 2

1 1 = V1 − V3 2 2 V2 − V3 I2 ′ = 2 1 1 = V2 − V3 2 2 Substituting Eq. (ii) and Eq. (iii) in Eq. (i), V1 V3 V2 V3 − + − = ( 2 s) V3 2 2 2 2 V V ( 2 s 1)V3 = 1 + 2 2 2 1 1 V3 = V1 + V2 2( 2 s + 1) 2( 2 s + 1) Substituting Eq. (iv) in Eq. (ii), ⎤ V 1⎡ 1 1 I1 ′ = 1 − ⎢ V1 + V2 2 2 ⎣ 2( 2 s + 1) 2( 2 s + 1) ⎥⎦ ⎛ 4s 1⎞ ⎛ 1 ⎞ =⎜ V1 − ⎜ V2 ⎝ 8s 4 ⎟⎠ ⎝ 8s + 4 ⎟⎠ Substituting Eq. (iv) in Eq. (iii), ⎤ V 1⎡ 1 1 I2 ′ = 2 − ⎢ V1 V2 2 2 ⎣ 2( 2 s + 1) 2( 2 s + 1) ⎥⎦ ⎛ 1 ⎞ ⎛ 4s 1⎞ = −⎜ V1 + ⎜ V2 ⎝ 8s 4 ⎟⎠ ⎝ 8s + 4 ⎟⎠ Comparing Eqs (v) and (vi) with Y-parameter equations, 1 ⎤ ⎡ 4s 1 − ⎡Y11 ′ Y12 ′ ⎤ ⎢ 8s 4 8 s 4⎥ ⎢⎣Y21 ′ Y22 ′ ⎥⎦ = ⎢ 4 s 1 ⎥⎥ ⎢− 1 ⎣ 8s 4 8s 4 ⎦

…(ii)

…(iii)

…(iv)

…(v)

…(vi)

13.72 Network Analysis and Synthesis The network N2 is shown in Fig. 13.87. I1′′

1F

1F

I2′′

+

1Ω

V1 −

1 s

I1′′ +

+

V2

V1

−

−

1 s

3

I2′′ +

1

V2 −

(a)

(b)

Fig. 13.87 Applying KCL at Node 3, I1 ″ I 2 ″ = V3

…(i)

From Fig. 13.87, V1 V3 1 s = sV V1 − sV V3 V V I2 ′ = 2 3 1 s = sV V2 − sV V3 Substituting Eqs (ii) and (iii) in Eq. (i), sV V1 − sV V3 + sV V2 − sV V3 = V3 ( 2 s 1)V3 = sV V1 + sV V2 ⎛ s ⎞ ⎛ s ⎞ V3 = V1 + ⎜ V2 ⎝ 2 s 1⎠ ⎝ 2 s 1⎟⎠ Substituting Eq. (iv) in Eq. (ii), ⎡⎛ s ⎞ ⎤ s I1 ″ sV V1 s ⎢⎜ V2 ⎥ ⎟⎠ V1 + ⎝ ( 2 s 1) ⎦ ⎣ 2s 1 I1 ″ =

⎛ s2 ⎡ s( s + 1) ⎤ =⎢ V1 − ⎜ ⎥ ⎝ 2s ⎣ 2s 1 ⎦

⎞ V2 1⎟⎠

…(ii)

…(iii)

…(iv)

…(v)

Substituting Eq. (iv) in Eq. (iii), ⎡⎛ s ⎞ ⎛ s ⎞ ⎤ s ⎢⎜ V2 ⎟⎠ V1 + ⎝ ⎝ 2 s 1 2 s 1⎠ ⎥⎦ ⎣ ⎛ s2 ⎞ ⎡ s( s + ) ⎤ = −⎜ V1 + ⎢ ⎟ ⎥ V2 ⎝ 2 s 1⎠ ⎣ 2s + 1 ⎦ Comparing Eqs (v) and (vi) with Y-parameter equations, ⎡ s( s + 1) ⎛ s2 ⎞ ⎤ −⎜ ⎢ ⎟⎥ ⎝ 2 s 1⎠ ⎥ ⎡Y11 ″ Y12 ″ ⎤ ⎢ 2 s 1 ⎥ ⎢⎣Y21 ″ Y22 ″ ⎥⎦ = ⎢ ⎛ 2 ⎞ s( s + ) ⎥ ⎢− s ⎢⎣ ⎜⎝ 2 s 1⎟⎠ 2 s + 1 ⎥⎦ I2 ″

sV V2

…(vi)

13.9 Interconnection of Two-Port Networks 13.73

Hence, the overall Y-parameters of the network are ⎡Y Y

Y ⎤ Y

+ 4 2s + 1

⎢

⎤

Y Y

⎢−

+ 4 2s + 1 + 42 1

−

4 2s + 1

13.9.3 Series Connection Figure 13.88 shows two-port networks connected in series. In a series connection, both the networks carry the same input current. Their output currents are also equal. I1

+

+

I2

+

V

+

V ′

N

V

V +

+ N

V ′′

Fig. 13.88

V ′′

Series connection

Let be the Z-parameters of the network N1 and 22 Z-parameters of the network N2. For the network N1, V′ = V ′

22

be the

I1 I2

12 22

For the network N2, V″ V ″

12 22

+

For the combined network Hence,

⎡V ⎤ V

I1 I2 =

⎡

⎤

⎤ ⎡ I1 I2 22 ″

Z

12 21

⎡Z Z

Z12 ⎤ ⎡ I1 ⎤ Z 22 I 2

Thus, the resultant Z-parameter matrix for the series-connected networks is the sum of Z matrices of each individual two-port network.

Example 13.44 Two identical sections of the network shown in Fig. 13.89 are connected in series. Obtain Z-parameters of the overall connection. I1

2Ω

2Ω

+ V

I +

1Ω

Fig. 13.89

V

13.74 Network Analysis and Synthesis Solution Applying KVL to Mesh 1, V1 = 3I1 + I2

…(i)

Applying KVL to Mesh 2, V2 = I1 + 3I2 Comparing Eqs (i) and (ii) with Z-parameter equations,

…(ii)

⎡ Z11 ″ Z12 ″ ⎤ ⎡3 1⎤ ⎢⎣ Z 21 ″ Z 22 ″ ⎥⎦ = ⎢⎣1 3⎥⎦ Hence, Z-parameters of the overall connection are ⎡ Z11 ⎢⎣ Z 21

Example 13.45

Z12 ⎤ ⎡3 1⎤ ⎡3 1⎤ ⎡6 2⎤ = + = Z 22 ⎥⎦ ⎢⎣1 3⎥⎦ ⎢⎣1 3⎥⎦ ⎢⎣ 2 6 ⎥⎦

Determine Z-parameters for the network shown in Fig. 13.90. L

L

+

+

C

V2

V1 L −

C

C

−

Fig. 13.90 Solution The above network can be considered as a series connection of two networks, N1 and N2. The network N1 is shown in Fig. 13.91. Ls Ls I1 I2 Applying KVL to Mesh 1, V1 ′

1⎞ ⎛ ⎛ 1⎞ Lss + L I1 + ⎜ ⎟ I 2 ⎝ ⎝ Cs ⎠ Cs ⎠ C

+

+

…(i)

1 Cs

V1′

Applying KVL to Mesh 2, V2 ′ =

1⎞ ⎛ 1⎞ ⎛ I1 Ls + ⎟ I 2 L ⎝C ⎠ ⎝ Cs Cs ⎠

−

1⎞ ⎛ Lss + L I1 ( Ls) I 2 ⎝ Cs ⎠ C

Fig. 13.91

…(i)

Applying KVL to Mesh 2, V2 ″

1⎞ ⎛ ( Ls) I1 + Ls + ⎟ I 2 ⎝ Cs ⎠

−

…(ii)

Comparing Eqs (i) and (ii) with Z-parameter equations, 1 1 ⎤ ⎡ Ls + ⎥ ⎡ Z11 ′ Z12 ′ ⎤ ⎢ Cs Cs ⎢⎣ Z 21 ′ Z 22 ′ ⎥⎦ = ⎢ 1 1⎥ ⎢ Ls + ⎥ L Cs ⎦ ⎣ Cs The network N2 is shown in Fig. 13.92. Applying KVL to Mesh 1, V1 ″

V2′

+

+ Ls

V1′′ −

…(ii)

I2

I1

1 Cs

V2′′

1 Cs

Fig. 13.92

−

13.9 Interconnection of Two-Port Networks 13.75

Comparing Eqs (i) and (ii) with Z-parameter equations, 1 ⎡ Ls + ⎡ Z11 ″ Z12 ″ ⎤ ⎢ Cs ⎢⎣ Z 21 ″ Z 22 ″ ⎥⎦ = ⎢ ⎢ Ls ⎣

⎤ ⎥ 1⎥ Ls + ⎥ Cs ⎦ Ls

Hence, the overall Z-parameters of the network are, ⎡ Z11 ⎢⎣ Z 21

2 ⎡ 2 Ls + Z12 ⎤ ⎡ Z11 ′ Z11 ″ Z12 ′ Z12 ″ ⎤ ⎢ Cs = = Z 22 ⎥⎦ ⎢⎣ Z 21 ′ Z 21 ″ Z 22 ′ Z 22 ″ ⎥⎦ ⎢⎢ 1 Ls + Cs ⎣

1 ⎤ Cs ⎥ = ⎛ Ls + 1 ⎞ ⎡ 2 1 ⎤ ⎟ 2⎥ ⎝ Cs ⎠ ⎢⎣1 2⎥⎦ 2 Ls + ⎥ Cs ⎦ Ls +

13.9.4 Series-Parallel Connection Figure 13.93 shows two networks connected in series-parallel. Here, the input ports of two networks are connected in series and the output ports are connected in parallel. I1 +

I2′

+ V1′ −

I2′′

+ V1′′

+

N1

V1

−

I2

N2

V2

−

−

Fig. 13.93 Series-parallel connection Let h11 ′, h12 ′, h21 ′, h22 ′ be the h-parameters of the network N1 and h11 ″, h12 ″, h21 ″, h22 ″ be the h-parameters of the network N2. For the network N1, ⎡V1 ′ ⎤ ⎡ h11 ′ h12 ′ ⎤ ⎡ I1 ⎤ ⎢⎣V2 ′ ⎥⎦ = ⎢⎣ h21 ′ h22 ′ ⎥⎦ ⎢⎣V2 ⎥⎦ For the network N2, ⎡V1 ″ ⎤ ⎡ h11 ″ h12 ″ ⎤ ⎡ I1 ⎤ ⎢⎣ I 2 ″ ⎥⎦ = ⎢⎣ h21 ″ h22 ″ ⎥⎦ ⎢⎣V2 ⎥⎦ For the combined network, V1 Hence,

V1 ′ + V1 ″

d I2 = I2 ′ I2 ″

⎡V1 ⎤ ⎡ V1 ′ V1 ″ ⎤ ⎡ h11 ′ h11 ″ h12 ′ h12 ″ ⎤ ⎡ I1 ⎤ ⎡ h11 ⎢⎣ I 2 ⎥⎦ = ⎢⎣ I 2 ′ I 2 ″ ⎥⎦ = ⎢⎣ h21 ′ h21 ″ h22 ′ h22 ″ ⎥⎦ ⎢⎣V2 ⎥⎦ = ⎢⎣ h21

h12 ⎤ ⎡ I1 ⎤ h22 ⎥⎦ ⎢⎣V2 ⎥⎦

Thus, the resultant h-parameter matrix is the sum of h-parameter matrices of each individual two-port networks.

Example 13.46

Determine h-parameters for the network shown in Fig. 13.94.

13.76 Network Analysis and Synthesis I1

1Ω

1Ω

+

I2 +

2Ω

V1

1Ω

1Ω

1Ω

1Ω

V2

2Ω −

−

1Ω

1Ω

Fig. 13.94 Solution The above network can be considered as a series-parallel connection of two networks N1 and N2. The network N1 is shown in Fig. 13.95. Applying KVL to Mesh 1, 1Ω 1Ω I2 I1 V1 = 4I1 + 2I2 …(i) + + Applying KVL to Mesh 2, V2 = 2I1 + 4I2 …(ii) Rewriting Eq. (ii) V2 V1 2Ω 4 2 2 I1 + V2 1 1 …(iii) I2 I1 + V2 2 4 − − Substituting Eq. (iii) in Eq. (i), 1Ω 1Ω 1 ⎞ ⎛ 1 V1 4 I1 2 ⎜ − I1 V2 ⎟ Fig. 13.95 ⎝ 2 4 ⎠ 1 V2 …(iv) 2 Comparing Eqs (iii) and (iv) with h-parameters equations, 1⎤ ⎡ 3 ⎡ h11 ′ h12 ′ ⎤ ⎢ 2⎥ ⎢⎣ h21 ′ h22 ′ ⎥⎦ = ⎢ 1 1 ⎥ ⎢− ⎥ ⎣ 2 4⎦ For network N2, h-parameters will be same as the two networks are identical. 1⎤ ⎡ 3 ⎢ h ″ h ″ ⎡ 11 12 ⎤ 2⎥ ⎢⎣ h21 ″ h22 ″ ⎥⎦ = ⎢ 1 1 ⎥ ⎢− ⎥ ⎣ 2 4⎦ Hence, the overall h-parameters of the network are 1⎤ ⎡ 1⎤ ⎡ 3 ⎥ ⎢ 3 2⎥ ⎡ 6 1⎤ ⎡ h11 h12 ⎤ ⎡ h11 ′ h12 ′ ⎤ ⎡ h11 ″ h12 ″ ⎤ ⎢ 2 ⎢ ⎥ ⎢⎣ h21 h22 ⎥⎦ = ⎢⎣ h21 ′ h22 ′ ⎥⎦ + ⎢⎣ h21 ″ h22 ″ ⎥⎦ = ⎢ 1 1 ⎥ + ⎢ 1 1 ⎥ = ⎢ −1 1 ⎥ ⎢− ⎥ ⎢− ⎥ ⎣ 2⎦ ⎣ 2 4⎦ ⎣ 2 4⎦ = 3I1

13.9 Interconnection of Two-Port Networks 13.77

13.9.5 Parallel–Series Connection Figure 13.96 shows two networks connected in parallel–series. Here the input ports of two networks are connected in parallel and the output ports are connected in series. I1′

I1

+

+

+

V2′

N1

V1

I2

− V2

I1′′

+ N2

−

V2′′

−

−

Fig. 13.96 Parallel–series connection Let g ′ g12 ′, g ′ g22 ′ be the g-parameters of the network N1 and g ″ g12 ″, g ″ g22 ″ be the g-parameters of the network N2, For the network N1, ⎡ I1 ′ ⎤ ⎡ g ′ ⎢⎣V2 ′ ⎥⎦ = ⎢⎣ g ′

g12 ′ ⎤ ⎡V1 ⎤ g22 ′ ⎥⎦ ⎢⎣ I 2 ⎥⎦

⎡ I1 ″ ⎤ ⎡ g ″ ⎢⎣V2 ″ ⎥⎦ = ⎢⎣ g ″

g12 ″ ⎤ ⎡V1 ⎤ g22 ″ ⎥⎦ ⎢⎣ I 2 ⎥⎦

For the network N2,

For the combined network,

I1

I1 ′ + I1 ″

d V2 = V2 ′ V2 ″

⎡ I1 ⎤ ⎡ I1 ′ I1 ″ ⎤ ⎡ g ′ + g11 ″ ⎢⎣V2 ⎥⎦ = ⎢⎣V2 ′ V2 ″ ⎥⎦ = ⎢⎣ g ′ + g21 ″ g12 ⎤ ⎡V1 ⎤ ⎡g =⎢ ⎥⎢ ⎥ g g 22 ⎦ ⎣ I 2 ⎦ ⎣

Hence,

g ′ + g12 ″ ⎤ ⎡V1 ⎤ g ′ + g22 ″ ⎥⎦ ⎢⎣ I 2 ⎥⎦

Thus, the resultant g-parameter matrix is the sum of the g-parameter matrices of each individual two-port network.

Example 13.47

Determine the g-parameters for the network shown in Fig. 13.97. I1

3Ω

3Ω

I2

+

+

1Ω

V1

3Ω

1Ω

V2

3Ω

1Ω −

1Ω −

Fig. 13.97

13.78 Network Analysis and Synthesis Solution The above network can be considered as a parallel series connection of two networks N1 and N2. The network N1 is shown in Fig. 13.98. Applying KVL to Mesh 1, V1 = 4I1 − I3 …(i) I1 I2 3Ω 3Ω Applying KVL to Mesh 2, V2 = I2 + I3 …(ii) + + Applying KVL to Mesh 3, − I1 + I 2 + 5 I 3 = 0 I3 =

1Ω

V1

1 1 I1 − I 2 5 5

I1

…(iii)

1Ω I3

V2 I2

−

−

Fig. 13.98

Substituting Eq (iii) in Eq (i), V1

1 ⎞ ⎛1 4 I1 − ⎜ I1 I2 ⎟ ⎝5 5 ⎠ 19 1 = I1 I2 5 5

…(iv)

Substituting Eq (iii) in Eq (ii), V2

1 ⎞ ⎛1 I 2 + ⎜ I1 I2 ⎟ ⎝5 5 ⎠ 1 4 = I1 I2 5 5

…(v)

Rewriting Eq (iv), 19 I1 5 I1

1 V1 − I 2 5 5 1 V1 − I 2 19 19

…(vi)

Substituting Eq (vi) in Eq (v), 1⎛ 5 1 ⎞ 4 V1 − I 2 I2 5 ⎝ 19 19 ⎠ 5 1 15 = V1 I2 19 19 Comparing Eqs (vi) and (vii) with g-parameters equation, 1⎤ ⎡5 − ⎥ ⎡ g ′ g12 ′ ⎤ ⎢19 19 ⎢⎣ g ′ g22 ′ ⎥⎦ = ⎢ 1 15 ⎥⎥ ⎢ ⎣19 19 ⎦ For the network N2, g-parameters will be same as the two networks are identical. 1⎤ ⎡5 − ⎥ ⎡ g ″ g12 ″ ⎤ ⎢19 19 ⎢⎣ g ″ g22 ″ ⎥⎦ = ⎢ 1 15 ⎥⎥ ⎢ ⎣19 19 ⎦ Hence, the overall g-parameters of the network are 1⎤ ⎡5 ⎡5 − ⎥ ⎢ g12 ⎤ ⎡ g ′ g12 ′ ⎤ ⎡ g ″ g12 ″ ⎤ ⎢19 ⎡g 19 = + = + 19 ⎢⎣ g g22 ⎥⎦ ⎢⎣ g ′ g22 ′ ⎥⎦ ⎢⎣ g ″ g22 ″ ⎥⎦ ⎢⎢ 1 15 ⎥⎥ ⎢⎢ 1 ⎣19 19 ⎦ ⎣19 V2

…(vii)

1 ⎤ ⎡10 19 ⎥ = ⎢19 15 ⎥⎥ ⎢⎢ 2 19 ⎦ ⎣19

−

2⎤ 19 ⎥ 30 ⎥⎥ 19 ⎦

−

13.11 PI (p )-Network 13.79

13.10

T-NETWORK

Any two-port network can be represented by an equivalent T network as shown in Fig 13.99. The elements of the equivalent T network may be expressed in terms of Z-parameters. Applying KVL to Mesh 1, V1

ZAI = (Z A

ZC ( I1 ZC ) I1

ZC I 2

...(13.9)

Applying KVL to Mesh 2, V2

ZA

I1

I2 )

ZB

+

+

...(13.10)

V2

ZC

V1

Z B I ZC ( I 2 I1 ) = ZC I + ( Z B ZC ) I 2

I2

−

−

Comparing Eqs (13.9) and (13.10) with Z-parameter equations, Z11 Z A + ZC Z12 ZC

Fig. 13.99 T-Network

Z 21 Z 22

ZC Z B + ZC

ZA ZB ZC

Z11 Z12 = Z11 Z 21 Z 22 − ZZ 21 = Z 22 Z12 Z12 = Z 21

Solving the above equations,

13.11

PI (o )-NETWORK

Any two-port network can be represented by an equivalent pi (p) network as shown in Fig. 13.100. Applying KCL at Node 1, I1

YAV1 YB (V1 V2 ) = (YA YB )V1 YBV2 Applying KCL at Node 2, I2

I1

...(13.11)

YB

1

2

I2 +

+

YCV YB (V2 V1 ) = −YBV1 + (YB YC )V2

...(13.12) Comparing Eqs (13.11) and (13.12) with Y-parameter equations, Y11 YA + YB Y12 YB Y21 YB Y22 YB + YC

YA

V1

YB YC

Fig. 13.100 p-network

Y11 Y12 = Y11 Y21 Y12 = −Y21 Y22 + Y12

V2 −

−

Solving the above equations, YA

YC

Y22 + Y21

13.80 Network Analysis and Synthesis

Example 13.48 The Z-parameters of a two-port network are: Z11 = 10 W, Z12 = Z21 = 5 W, Z22 = 20 W. Find the equivalent T-network. I1

Solution The T-network is shown in Fig 13.101 Applying KVL to Mesh 1, V1 = (Z1 + Z2)I1 + Z2 I2 …(i) Applying KVL to Mesh 2, V2 = Z2I1 + (Z2 + Z3)I2 …(ii) Comparing Eqs (i) and (ii) with Z parameter equations, Z11 Z1 + Z 2 = 10 Z12

Z1

Z3

I2 +

+ Z2

V1

V2 −

−

Z2 = 5

Fig. 13.101

Z 21 Z 2 = 5 Z 22 Z 2 + Z3 = 20 Solving the above equations, Z1 = 5 Ω Z2 = 5 Ω Z3 = 15 Ω

Example 13.49 Y22 = 0.07

Admittance parameters of a pi network are Y11 = 0.09 , Y12 = Y21 = −0.05

and

. Find the values of Ra, Rb and Rc.

Solution The pi network is shown in Fig 13.102. Applying KCL at Node, 1, I1 = =

V1 V1 V2 + Ra Rb 1⎞ ⎛ 1 + V1 ⎝ Ra Rb ⎟⎠

1 V2 Rb

Applying KCL at Node 2,

+ Ra

−

1 1⎞ ⎛ 1 V1 + + V2 ⎝ RB Rc ⎟⎠ Rb

…(ii)

Comparing Eqs (i) and (ii) with Y-parameter equations, 1 1 + = 0 09 Ra Rb 1 Y12 = − = −0 05 Rb 1 Y21 = − = −0 05 Rb 1 1 Y22 = + = 0.007 Rb Rc Y11 =

I2

+ V1

V V V I2 = 2 + 2 1 Rc Rb =−

Rb

I1

…(i)

Rc

V2 −

Fig. 13.102

13.11 PI (p )-Network 13.81

Solving the above equations, Ra = 25 Ω Rb = 20 Ω Rc = 50 Ω

Example 13.50 Find the parameters YA, YB and YC of the equivalent p network as shown in Fig. 13.103 to represent a two-terminal pair network for which the following measurements were taken: (a) With terminal 2 short-circuited, a voltage of 10 ∠ 0° V applied at terminal pair I resulted in I1 = 2.5 ∠ 0° A and I2 = −0.5 ∠ 0° A. (b) With terminal 1 short-circuited, the same voltage at terminal pair 2 resulted in I2 = 1.5 ∠ 0° A and I1 = −1.1 ∠−20° A. 1

I1

YB

2

I2

+ V1

1′

1

+ YA

YC

−

2

V2 −

2′

Fig. 13.103 Solution Since measurements were taken with either of the terminal pairs short-circuited, we have to calculate Y-parameters first. I 2 5∠0° Y11 = 1 = = 0 25 V1 V2 = 0 10 ∠0° Y21 =

I2 −0 5∠0° = = −0.05 V1 V2 = 0 10 ∠0°

Y222 =

I2 1.5∠0° = = 0.15 V2 V1 = 0 10 ∠0°

Applying KCL at Node 1, I1

YAV1 YB (V1 V2 ) = (YA YB )V1 YBV2

…(i)

Applying KCL at Node 2, I2

YCV YB (V2 V1 ) = −YBV1 + (YB YC )V2 Comparing Eqs (i) and (ii) with the Y-parameter equation, Y11 YA + YB = 0 25 Y12 Y21 = −YB = −0 05 Y22 YB + YC = 0 15 Solving the above equation, YA = 0 20 YB = 0 05 YC = 0 10

…(ii)

13.82 Network Analysis and Synthesis

Example 13.51 A network has two input terminals (a, b) and two output terminals (c, d) as shown in Fig. 13.104. The input impedance with c and d open-circuited is (250 + j100) ohms and with c and d short-circuited is (400 + j 300) ohms. The impedance across c and d with a and b open-circuited is 200 ohms. Determine the equivalent T-network parameters. ZA

ZB

a

c

ZC

b

d

Fig. 13.104 Solution The input impedance with c and d open-circuited is ZA + ZC = 250 + j100 The input impedance with c and d short-circuited is, Z Z Z A + B C = 400 + j 300 Z B ZC

…(i)

…(ii)

The impedance across c and d with a and b open-circuited is ZB + ZC = 200 Subtracting Eq. (i) from (ii), Z B ZC −Z Z B ZC

1 0

…(iii)

j 200

…(iv)

From Eq. (iii), ZB = 200 − ZC

…(v)

Subtracting the value of ZB in the equation (iv) and simplifying, ZC = (100 − j200) Ω

…(vi)

From Eqs (i) and (vi), ZA = (150 + j300) Ω From Eqs (iii) and (vi), ZB = (100 + j200) Ω

Example 13.52

Find the equivalent p-network for the T-network shown in Fig. 13.105. I1

2Ω

2.5 Ω

+ V1

I2 +

5Ω

−

V2 −

Fig. 13.105

13.11 PI (p )-Network 13.83

Solution Figure 13.106 shows T-network and p-network. ZA

ZB

Z3

ZC

Z1

Z2

p-network (Delta network)

T-network (Star network)

Fig. 13.106 For converting a T-network (star network) into an equivalent p-network (delta network), we can use stardelta transformation technique. 5.5 Ω

Z A ZC 2 5 = 2 5+ = 11 Ω ZB 25 Z Z 2 2.5 Z3 Z A + Z B + A B = 2 + 2.5 + =5 5Ω ZC 5 Z Z 2 5×5 Z 2 = Z B + ZC + B C = 2 5 + 5 + = 13.75 Ω ZA 2 Z1

Z A + ZC +

11 Ω

The equivalent p-network is shown in Fig. 13.107.

Example 13.53

13.75 Ω

Fig. 13.107

For the network shown in Fig. 13.108. Find the equivalent T-network. 8Ω

1Ω

I1

I3

4Ω

I2

+

+

2Ω

V1 −

V2

I1

I2

−

Fig. 13.108 Solution Applying KVL to Mesh 1, V1

3I1 2 I 2

I3

…(i)

V2

2 I1 6 I 2

4I3

…(ii)

Applying KVL to Mesh 2, Applying KVL to Mesh 3, 13I 3

I1 + 4

2

I3

0 1 4 I1 − I 2 13 13

…(iii)

13.84 Network Analysis and Synthesis Substituting the Eq. (iii) in Eq. (i), V1

3I1 2 I 2 =

38 I1 13

1 4 I1 + I 2 13 13

30 I2 13

…(iv)

Substituting the Eq. (iii) in Eq. (ii), V2

2 I1 6 I 2 =

30 I1 13

The T-network is shown in Fig. 13.109. Applying KVL to Mesh 1, V1 ( Z A + ZC ) I1 + ZC I 2 Applying KVL to Mesh 2, V2 ZC I ( Z B + ZC ) I 2 Comparing Eqs (iv) and (v) with Eqs (vi) and (vii), ZA

ZB

4

4 ⎞ ⎛1 I1 − I 2 ⎟ ⎝ 13 13 ⎠

62 I2 13

…(v)

I1

…(vi)

ZA

ZB

+

+

…(vii) V 1

38 13 30 ZC = 13 62 ZC = 13

ZC

−

ZC =

I2

V2 −

Fig. 13.109

Solving the above equations, 8 Ω 13 32 ZB = Ω 13 30 ZC = Ω 13 ZA =

13.12

LATTICE NETWORKS

A lattice network is one of the common two-port networks, shown in Fig. 13.110. It is used in filter sections and is also used as attenuator. This network can be represented in terms of z-parameters. ZA

I1

I2

+

V1

+ ZB

ZB

−

V2

− ZA

Fig. 13.110 Lattice network

13.12 Lattice Networks 13.85

The lattice network can be redrawn as a bridge network as shown in Fig. 13.111. This lattice network is symmetric and reciprocal. The current I1 divides equally between the two arms of the bridge. When the output port is open-circuited, i.e., I2 = 0 I1 I1 I1 + 2 V1 = ( Z A Z B ) I1 ZA 2 ZB 2 V1 ZA ZB Z11 = = I1 I 2 = 0 2 Also

I I I V2 = 1 Z B − 1 Z A = 1 ( Z B 2 2 2 V Z ZA Z 21 = 2 = B I1 I 2 = 0 2

Z12

Z 22 =

ZA

Z 21 =

ZB

V2

−

I2

ZA )

ZB

ZA

−

Since the network is symmetric, Z11

+

V1

Fig. 13.111 Bridge network

ZB 2 ZA 2

Solving the above equations, ZA

Z11 Z12

Z B Z11 + Z12 The lattice network can be represented in terms of other two-port network parameters, with the help of inter-relationship formulae of various parameters.

Example 13.54

Find the lattice equivalent of a symmetrical T network shown in Fig. 13.112. I1

1Ω

1Ω

+

I2 +

2Ω

V1 −

V2 −

Fig. 13.112 Solution Applying KVL to Mesh 1, V1

3I1 2 I 2

…(i)

V2 2 I1 3I 2 Comparing Eqs (i) and (ii) with Z-parameter equations, Z11 = 3 Ω Z12 = 2 Ω Z 21 = 2 Ω Z 22 = 3 Ω

…(ii)

Applying KVL to Mesh 2,

13.86 Network Analysis and Synthesis Since Z11 = Z22 and Z12 = Z21, the network is symmetric and reciprocal. The parameters of lattice network are Z A Z11 Z12 = 3 − 2 = 1 Ω Z11 + Z12 = 3 + 2 = 5 Ω

ZB The lattice network is shown in Fig. 13.113.

1Ω

I1

I2

+

+

5Ω

V1

5Ω

V2

−

−

1Ω

Fig. 13.113

Example 13.55

Find the lattice equivalent of a symmetric p-network shown in Fig. 13.114. 5Ω

I1

I2

+

+

10 Ω

V1

10 Ω

V2

−

−

Fig. 13.114 Solution The network is redrawn as shown in Fig. 13.115. Applying KVL to Mesh 1, I1 V1 10 I1 − 10 I 3 …(i) + Applying KVL to Mesh 2, V2 10 I 2 + 10 I 3 …(ii) V1 I1 Applying KVL to Mesh 3, −10 I1 +10 10 I 2 + 25 I 3 = 0 2 2 I − I2 5 5 Substituting Eq (iii) in Eq (i), I3 =

5Ω

+

10 Ω

10 Ω I3

−

…(iii)

⎛2 V1 10 I1 − 10 ⎜ I1 ⎝5

I2

V2 I2

−

Fig. 13.115

2 ⎞ I2 ⎟ 5 ⎠

= 6 I1 4 I 2

…(iv)

Substituting Eq (iii) in Eq (ii), V2

⎛2 10 I 2 + 10 ⎜ I1 ⎝5 = 4 I1 6 I 2

2 ⎞ I2 ⎟ 5 ⎠ …(v)

13.13 Terminated Two-Port Networks 13.87

Comparing the Eqs (iv) and (v) with Z-parameter equations, Z11 = 6 Ω Z12 = 4 Ω

+

10 Ω

V1

Z11 Z12 = 6 − 4 = 2 Ω Z11 + Z12 = 6 + 4 = 10 Ω

10 Ω

−

V2

−

2Ω

Fig. 13.116

The lattice network is shown in Fig. 13.116.

13.13

I2

+

Since Z11 = Z22 and Z12 = Z21, the network is symmetric and reciprocal. The parameters of lattice network are ZA ZB

2Ω

I1

Z 21 = 4 Ω Z 22 = 6 Ω

TERMINATED TWO-PORT NETWORKS

13.13.1 Driving-Point Impedance at Input Port A two-port network is shown in Fig. 13.117. The output port of the network is terminated in load impedance ZL. The input impedance of this network can be expressed in terms of parameters of two-port network parameters. I1

I2

+

+

Twoport network

V1

V2

−

ZL

−

Fig. 13.117 Terminated two-port network 1. Input Impedance in Terms of Z-parameters We know that

From Fig. 13.117,

V1

Z11 I1 Z12 I 2

V2

Z 21 I1 Z 22 I 2

V2 ZL I2 − I 2 Z L = Z 21 I1 + Z 22 I 2 Z 21 I2 = − I1 Z 22 Z L Zin =

V1 = Z11 I1

Z 21 ⎛ ⎞ Z11Z 222 + Z11Z L − Z12 Z 221 2 Z12 − = ⎝ Z 22 Z L ⎟⎠ Z 22 Z L

If the output port is open-circuited, i.e., ZL = ∞, Z11Z 22 − Z12 Z 21 + Z11 ZL Zin = lim = Z11 Z 22 Z L→∞ +1 ZL

13.88 Network Analysis and Synthesis If the output port is short-circuited, i.e., ZL = 0, Zin =

Z11Z 22 − Z12 Z 21 Z 22

2. Input Impedance in Terms of Y-parameters We know that I1

Y1111V1 Y12V2

I2

Y21 21V1 Y22V2

From Fig. 13.117, V2

ZL I 2

I2 = −

V2 = −YLV2 ZL

where YL =

1 ZL

−YLV2 = Y21V1 + Y22V2 Y21 V2 = − V1 Y22 + YL Y21 ⎞ Y Y Y Y −Y Y +Y Y ⎛ I1 = Y11V1 + Y12 ⎜ − V1 = Y11V1 − 21 12 V1 = 11 22 12 21 11 L V1 ⎟ ⎝ Y22 + YL ⎠ Y22 + YL Y22 + YL Zin =

V1 Y22 YL = I1 Y11Y22 − Y12Y21 + Y11YL

When output port is open-circuited, i.e., YL = 0 Zin =

Y22 Y11Y22 − Y12Y21

When output port is short-circuited, i.e., YL = ∞, Y22 +1 1 YL Zin = lim = YL →∞ Y11Y22 − Y12Y21 Y11 + Y11 YL 3. Input Impedance in Terms of Transmission Parameters We know that V1 A AV V2 BV V2 I1 CV V2 DI D 2 From Fig. 13.117, V2 I1

ZL I 2 CZ Z L I 2 DI 2 = (CZ L + D ) I 2 I1 I2 = − CZ L + D I1 ⎞ ⎛ AZ L + B ⎞ ⎛ V1 AZ AZ L I 2 − B = − = I1 ⎝ CZ L D ⎟⎠ ⎜⎝ CZ L + D ⎟⎠ V AZ L + B Zin = 1 = I1 CZ L + D

13.13 Terminated Two-Port Networks 13.89

If the output port is open-circuited, i.e., ZL = ∞, A Zin = C If the output port is short-circuited, i.e., ZL = 0, B Zin = D 4. Input Impedance in Terms of Hybrid Parameters We know that V1 h11 I1 h12V2 I 2 h21 I1 h22V2 V2 ZL I2 I 2 h21 I1 h22 Z L I 2 h21 I2 = IL 1 + h22 2 ZL h Z V2 = − 21 L I1 1 + h22 Z L Substituting the value of V2 in V1, ⎡ − h21Z L ⎤ ⎡ (h h − h h )Z h11 I1 h12 ⎢ I L ⎥ = ⎢ 11 22 12 21 L 1 h Z 1 + h22 ⎣ 22 L ⎦ ⎣ 2 ZL V ( h1 h222 − h12 h111 1 h221 ) Z L Zin = 1 = 11 I1 1 + h22 2 ZL V1

h11 ⎤ ⎥ I1 ⎦

If the output port is open-circuited, i.e., ZL = ∞, Zin =

h11h22 − h12 h21 h22

If the output port is short-circuited, i.e., ZL = 0, Zin = h11

13.13.2 Driving-Point Impedance at Output Port A two-port network is shown in Fig. 13.118. The input port is terminated in load impedance ZL. The output impedance of this network can be expressed in terms of two port network parameters. I1 + ZL

V1

I2 Twoport network

−

+ V2 −

Fig. 13.118 Terminated two-port network 1. Output Impedance in terms of Z-parameters We know that V1 V2

Z11I1 Z12 I 2 Z 21I1 Z 22 I 2

13.90 Network Analysis and Synthesis From Fig. 13.118, V1 Z L I1 − I1Z1 = Z11 I1 + Z12 I 2 Z12 ⎞ ⎛ I1 = − I2 ⎝ Z L Z11 ⎟⎠ V2

Z12 ⎞ Z Z ⎞ ⎛ Z Z − Z12 Z 21 + Z 22 Z L ⎞ ⎛ ⎛ Z 21 − I 2 + Z 22 I 2 = I 2 ⎜ Z 22 − 21 12 ⎟ = ⎜ 11 22 ⎟ ⎟⎠ I 2 ⎝ Z L + Z11 ⎠ ⎝ Z L + Z11 ⎠ ⎝ Z11 + Z L

Z0 =

V2 Z11Z 22 − Z12 Z 21 + Z 22 Z L = I2 Z11 + Z L

If the input port is open-circuited, i.e., ZL = ∞, Z0 = Z22 If the input port is short-circuited, i.e., ZL= 0, Z0 =

Z11Z 22 − Z12 Z 21 Z11

2. Output Impedance in Terms of Y-parameters We know that I1

Y1111V1 Y12V2

I2

Y2211V1 Y22V2

From Fig. 13.118, V1

Z L I1 V I1 = − 1 = −YLV1 ZL −YLV1 = Y11V1 + Y12V2 Y12 ⎞ ⎛ V1 = − V2 ⎝ YL Y11 ⎟⎠ I2

Y12 ⎞ ⎛ Y21 − V2 Y22V2 ⎝ YL Y11 ⎟⎠

Z0 =

V2 YL + Y11 = I 2 Y11Y22 − Y12Y21 21 Y Y222

Y Y ⎤ ⎡ ⎡Y Y − Y Y + Y Y ⎤ V2 ⎢Y22 − 21 12 ⎥ = V2 ⎢ 11 22 12 21 L 22 ⎥ YL Y11 ⎦ YL + Y11 ⎣ ⎣ ⎦

If input port is open-circuited, i.e., YL = 0, Z0 =

Y11 Y11Y22 − Y12Y21

If input port is short-circuited, i.e., YL = ∞, Z0 =

1 Y22

3. Output Impedance in Terms of ABCD Parameters We know that V1 AV AV2 BI B 2 I1 CV V2 D DI 2

13.13 Terminated Two-Port Networks 13.91

From Fig. 13.118,

V2 (CZ L

V1 Z L I1 V1 AV V2 BI 2 = −ZL = I1 CV V2 DI 2 A) = I ( DZ L B) V D L+B DZ Z0 = 2 = I 2 CZ L + A

If input port is open-circuited, i.e., ZL = ∞, Z0 = If input port is short-circuited, i.e., ZL = 0, Z0 =

D C B A

4. Output Impedance in Terms of h-parameters We know that V1

h11I1 h12V12

I2

h21I1 h22V2

From Fig.13.118, V1 Z L I1 − I1Z L = h11 I1 + h22V2 h12 ⎞ ⎛ I1 = − V2 ⎝ h11 Z L ⎟⎠ h12 ⎞ ⎛ ⎡ h1 h222 − h12 1 h221 + h22 2 ZL ⎤ I 2 h21 − V2 + h222V2 = V2 ⎢ 11 ⎥ ⎝ h11 + Z L ⎟⎠ h + Z ⎣ 1 11 L ⎦ V2 h11 Z L Z0 = = I 2 h11h22 − h12 h21 + h22 Z L If input port is open-circuited, i.e., ZL = ∞, Z0 =

1 h22

If input port is short-circuited i.e., ZL = 0, Z0 =

Example 13.56

h11 h11h22 − h12 h21

Measurements were made on a two-terminal network shown in Fig. 13.119. 1

I1

I2

+ V1

2 +

Network

V2

−

−

1′

2′

Fig. 13.119

RL

13.92 Network Analysis and Synthesis (a) With terminal pair 2 open, a voltage of 100 ∠0° V applied to terminal pair 1 resulted in °A

I1

2 ∠0° V

V2

(b) With terminal pair 1 open, the same voltage applied to terminal pair 2 resulted in I2 ° A V1 50 ∠0° V Write mesh equations for this network. What will be the voltage across a 10-W resistor connected across Terminal pair 2 if a 100 ∠0° V is connected across terminal pair 1? Solution Since measurements were done with either of the terminal pairs open-circuited, we have to calculate Z-parameters first. Z11 = Z 21 = Z 22 = Z112 =

V1 I1 V2 I1 V2 I2 V1 I2

=

100 ∠0° = 10 Ω 10 ∠0°

=

25∠0° = 2 5Ω 10 ∠0°

=

100 ∠0° =5Ω 20 ∠0°

=

50 ∠0° = 2 5Ω 20 ∠0°

I 2 =0

I 2 =0

I1 = 0

I1 = 0

Putting these values in Z-parameter equations, V1 10 I1 + 2 5 I 2

…(i)

V2 2 I1 + 5 I 2 When a 10-Ω resistor is connected across terminal pair 1, V1 100 ∠0° V V2 RL I 2 10 I 2

…(ii)

Substituting values of V1 and V2 in Eqs (i) and (ii), 100 = 10I1 + 2.5I2 and −10I2 = 2.5I1 + 5I2 2.5I1 = −15I2 I1 = −6I2 100 = −60I2 + 2.5I2 I2 = −

100 = −1.74 A 57.5

Voltage across the resistor = −I2RL = −10(−1.74) = 17.4 V

Example 13.57 The Z-parameters of a two-port network shown in Fig. 13.120 are Z11 = Z22 = 10 W, Z21 = Z22 = 4 W. If the source voltage is 20 V, determine I1,I2,V2 and input impedance. I1

I2

+ Vs

+ −

V1

+ Network

−

V2 −

Fig. 13.120

5Ω

13.13 Terminated Two-Port Networks 13.93

Solution V1 Vs = 20 V V2 20 I 2 The two-port network can be represented in terms of Z-parameters. V1 10 I1 + 4 I 2

…(i)

V2 4 I1 10 I 2 −20 I 2 = 4 1 + 10 I 2 4

1

…(ii)

= −30 30 I 2

I1 = −7 7 5I 2 Substituting the value of I1 in Eq. (i), V1 10(

. I 2 ) + 4I 4I2

71 I 2

20 = −71 I 2 I 2 = −0.28 A I1 = −7.5( −00.28) V2 = −20( −00.28) Input impedance Zi =

2 .1 A 56 V

V1 20 = = 9 52 Ω I1 2.1

Example 13.58

The Z-parameters of a two-port network shown in Fig. 13.121 are, Z11 = 2 W, V I V Z12 = 1 W, Z21 = 2 W, Z22 = 5 W. Calculate the voltage ratio 2 , current ratio − 2 and input impedance 1 . Vs I1 I1 I1

I2

+ Vs

+ −

+ Network

V1 −

V2

5Ω

−

Fig. 13.121 Solution The two-port network can be represented in terms of Z-parameters. V1 V2

2 I1 + I 2 2 I1 5 I 2

…(i) …(ii)

When the 5 Ω resistor is connected across port-2, V2

I2

…(iii)

Applying KVL to the input port, Vs 1I1 − V1 = 0 V1

Vs + I1

…(iv)

Substituting values of V1 and V2 in Eqs (i) and (ii), Vs

I1 = 2 I1 Vs

I2

3I1 + I 2

…(v)

13.94 Network Analysis and Synthesis −5

and

2

= 2 I1 + 5 I 2

0 = 2 I1 +10 10 0I2

…(vi)

Solving Eqs (v) and (vi), we get Vs 0 I1 = 3 2 3 0 I2 = 3 2 I2 1 − = I1 5 V2

1 10 5 = Vs 1 14 10 Vs 0 1 = − Vs 1 14 10

2 I1 5 I 2

2

⎛ 5 ⎞ ⎛ 1 ⎞ Vs + 5 − Vs ⎝ 14 ⎟⎠ ⎝ 14 ⎠

5 Vs 14

V2 5 = Vs 14 V1

2 I1 + I 2

2

⎛ 5 ⎞ 1 Vs − Vs ⎝ 14 ⎟⎠ 14

9 Vs 14

V1 9 = Ω I1 5

Example 13.59 The following equations give the voltages V1 and V2 at the two ports of a two-port network shown in Fig. 13.122. V1 5 I 1 2 I 2 V2 2 I 1 + I 2 A load resistor of 3 W is connected across port 2. Calculate the input impedance. I1 + V1

I2 Twoport network

−

+ V2

3Ω

−

Fig. 13.122 Solution From Fig. 13.122, V2 = −3I2 Substituting Eq. (i) in the given equation, −3 2 = 2I1 + I 2 I I2 = − 1 2 Substituting the Eq. (ii) in the given equation. V1 5 I1 I1 4 I1 V Input impedance Zi = 1 = 4 Ω I1

...(i)

(ii)

13.13 Terminated Two-Port Networks 13.95

Example 13.60 The y-parameters for a two-port network shown in Fig. 13.123 are given as, Y11 = 4 , Y22 = 5 , Y12 = Y21 = 4 . If a resistor of 1 W is connected across port-1 of the network then find the output impedance. I2

I1 +

+

1Ω

V1

Network

V2

−

−

Fig. 13.123 Solution The two-port network can be represented in terms of Y-parameters. I1 4V1 4V2 I 2 4V1 5V2 When the 1-Ω resistor is connected across port-1 of the network, V1 1I1 = − I1 I1 V1 Substituting value of I1 in Eq (i), −V1 = 4V1 + 4V2 −5V1 = 4V2 4 V1 = − V2 5 Substituting value of V1 in Eq (ii), ⎛ 4 ⎞ V2 + 5V2 ⎝ 5 ⎟⎠ V 5 Output impedance Z0 = 2 = Ω I2 9 I2

4

…(i) …(ii)

9 V2 5

Example 13.61 The following equation gives the voltage and current at the input port of a twoport network shown in Fig. 13.124. V1 5V2 3I 2 I1 6V2 2 I 2 A load resistance of 5 W is connected across the output port. Calculate the input impedance. I1 + V1

I2 Twoport network

−

+ V2 −

Fig. 13.124 Solution From Fig. 13.124, V2 = −5I2 Substituting the value of V2 in the given equations, V1 ( I 2 ) − 3I 3I 2 28 I 2 I1 6( I 2 ) − 2I 2I 2 32 I 2 V −28 I 2 7 Input impedance Zi = 1 = = Ω I1 −32 I 2 8

5Ω

13.96 Network Analysis and Synthesis

Example 13.62 The ABCD parameters of a two-port network shown in Fig. 13.125 are A = 2.5, B = 4 W, C = 1 , D = 2. What must be the input voltage V1 applied for the output voltage V2 to be 10 V across the load of 10 W connected at Port 2? I1

I2

+

+

V1

V2

Network

−

10 Ω

−

Fig. 13.125 Solution The two-port network can be represented in terms of ABCD parameters. V1 = 2.5 V2 − 4I2 I1 = V2 − 2I2 When the 10 Ω resistor is connected across Port 2, V2 10 I 2 = 10

…(i) …(ii) ...(iii)

I 2 = −1A V1 = 2.5( 5(10) − 4( 1) = 29 V

Example 13.63

The h-parameters of a two-port network shown in Fig. 13.126 are h11 = 4 W, h12 = 1, h21 = 1, h22 = 0.5 . Calculate the output voltage V2 when the output port is terminated in a 3 W resistance and a 1 V is applied at the input port. I1

I2

+

1V

+ −

+ Network

V1 −

V2

3Ω

−

Fig. 13.126 Solution V1 = 1 V V2

3 I2

The two-port network can be represented in terms of h-parameters. V1 4 I1 + V2 I 2 I1 + 0 5 V2 I 2 I1 + 0.5( −3I 2 ) 2 5 I 2 I1 Substituting the value of V1 and I1 in Eq (i), 1 4( 2.5 I 2 ) 3 I 2 1 7I 2 I2 =

1 A 7

3 ⎛ 1⎞ V2 = −3 ⎜ ⎟ = − V ⎝ 7⎠ 7

...(i) …(ii)

13.14 Image Parameters 13.97

Example 13.64 The h-parameters of a two-port network shown in Fig. 13.127 are h11 = 1 W, h12 = −h21 = 2, h22 = 1 . The power absorbed by a load resistance of 1 W connected across port-2 is 100W. The network is excited by a voltage source of generated voltage Vs and internal resistance 2 W. Calculate the value of Vs. 2Ω

I1

I2

+ Vs

+ −

+ Network

V1

V2

−

2Ω

−

Fig. 13.127 Solution The two-port network can be represented in terms of h-parameters. V1 I1 + 2 V2 I2 2 I1 + V2 When the 1 Ω resister is connected across port-2,

…(i) …(ii)

V22 = 100 1 V2 = 10 V V2

1I2

10

I 2 = −10 A Substituting values of I2 and V2 in Eq (ii), −10 = −2 1 + 10 I1 = 10 A Applying KVL to the input port, Vs Vs

2 I1 − V1 = 0

2 I1 − ( I1 2V2 ) = 0 Vs

3 I1 − 2V2 = 0 Vs

13.14

3 I1 + 2V2 = 3(10) + 2( ) = 50 V

IMAGE PARAMETERS

If the driving-point impedance at port 1, with impedance Zi2 connected across port 2, is Zi1 and driving-point impedance at port 2, with impedance Zi1 connected across the port 1, is Zi2 then Zi1 and Zi2 are known as image impedances of the network. These are also known as image parameters. The image parameters can be expressed in terms of ABCD parameters. Figure 13.128 shows a two-port network terminated in Zi2 at Port 2. I1

V1

AV A V2

B I2

+

I1

CV V2

D I2

V1

V2

Zi 2 I 2

V AV V2 Zi1 = 1 = I1 CV V2

−

B I2 − A Zi 2 I 2 = D I 2 −C C Zi2 i2 I2

B I2 − A Zi2 i2 − B = D I 2 −C Zi2 i2 − D

I2

+ Network

V2

Zi 2

−

Fig. 13.128 Terminated two-port network

13.98 Network Analysis and Synthesis =

A Zi 2 + B C Zi 2 + D

Similarly, if the two-port network is terminated in Zi1 at port 1 as shown in Fig. 13.129 then V1

I1

...(13.13)

Zi1 I1

D B ⎛ ⎞ ⎛ ⎞ V1 I ⎝ AD − BC ⎠ 1 ⎜⎝ AD − BC ⎟⎠ V2 Zi 2 = = C A I2 ⎛ ⎞ ⎛ ⎞ V1 ⎜ −I ⎝ AD − BC ⎟⎠ 1 ⎜⎝ AD − BC ⎟⎠

I2

+ Zi1

+ Network

V1 −

V2 −

Fig. 13.129 Terminated two-port network

…(from Cramer’s rule)

D B ⎛ ⎞ ⎛ ⎞ − Zi1 I1 ⎜ −I ⎝ AD − BC ⎟⎠ 1 ⎜⎝ AD − BC ⎟⎠ − DZi1 − B = = C A ⎛ ⎞ ⎛ ⎞ −CZi1 − A − Zi1 I1 ⎜ −I ⎝ AD − BC ⎟⎠ 1 ⎜⎝ AD − BC ⎟⎠ =

DZi1 + B CZi1 + A

…(13.14)

Solving Eqs (13.13) and (13.14), Zi1 =

AB CD

Zi 2 =

BD AC

The image impedances Zi1 and Zi2 do not define a network completely. A third parameter image transfer constant f is also used to define the network. f can also be expressed in terms of ABCD parameters. Let the network be terminated in impedance Zi2 at Port 2. V2

Zi 2 I 2

V1

AV2 AV

B I2

I1

CV V2

D I2

AV V2

B ⎞ ⎛ V ⎞ ⎛ B⎜ 2 ⎟ = A+ V2 ⎝ − Zi 2 ⎠ ⎝ Zi 2 ⎟⎠

C Zi I 2 − D I 2 = −(C Zi 2 + D ) I 2

V1 B AC = A+ = A+ B = A+ V2 Zi 2 BD − Hence,

I1 BD = C Zi 2 + D = C +D= I2 AC

ABCD C AD + ABCD C = D D ABCD C +D= A

−

V1 I1 ( AD + ABCD C )2 = = ( AD + BC ) 2 V2 I 2 AD

−

V1 I1 = V2 I 2 =

ABCD C + AD AD + ABCD C = A A

AD + BC AD + AD − 1

(∵ AD − BC = 1)

13.14 Image Parameters 13.99

cosh h φ , then h

Let tanh hf =

AD − 1 AD

f = tanh h −1 Also

−

=

A − 1 = sinh AD i hφ BC AD

BC AD

V1 I1 = cosh f + si h f = ef V2 I 2 f = log

−

V1 I1 1 ⎛ V I ⎞ = log loge − 1 1 ⎟ ⎝ V2 I 2 ⎠ V2 I 2 2

Relationship between Image parameters and the Open-circuit and Short-Circuit Impedances Open and short-circuit impedance parameters can be expressed in terms of ABCD parameters. V1 I1

AV AV2 CV V2

B I2 D I2

When the output port is open-circuited, i.e., I2 = 0, V1

AV A V2

I1

CV V2

Hence, the impedance measured at input port with output port open-circuited is Zoc1 =

V1 A = I1 C

When the output port is short-circuited, i.e., V2 = 0, V1 = −BI2 I1 = −DI2 Hence, the impedance measured at input port with output port short-circuited is Z sc 1 =

V1 B = I1 D

Similarly the impedance measured at output port with input port open-circuited and short-circuited are D C B = A

Zoc2 = Z sc2

The ratio of short-circuit to open-circuit impedance at the two ports is Z sc1 Z sc2 BC = = Zoc1 Zoc2 AD

13.100 Network Analysis and Synthesis The image parameters can also be expressed in terms of open circuit and short-circuit impedances. Zi1 =

AB = Zoc1 Z sc1 CD

Zi 2 =

BD = Zoc2 Z sc2 AC

f = tan nh −11

Example 13.65

BC C = tannh AD

1

Z sc1 Z ssc2 = tanh −1 Zoc Zooc2 o1

The ABCD parameters of a two-port network are given as, A

7 D = . Find the image parameters. 5

6 17 1 ,B= ,C = , 5 5 5

Solution

AB Zi1 = = CD

⎛ 6 ⎞ ⎛ 17 ⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ 5 5 = 3 82 Ω ⎛ 1⎞ ⎛ 7⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ 5 5

BD = AC

⎛ 17 ⎞ ⎛ 7 ⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ 5 5 = 4.45 Ω ⎛ 6 ⎞ ⎛ 1⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ 5 5

Zi 2 =

f = tan nh −11

Example 13.66

BC = tannh AD

1

1 ⎞ ⎛ 1⎞ ⎛ 17 ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ 17 5 5 = tannh −1 = 0.75° 42 ⎛ 6⎞ ⎛ 7⎞ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ 5 5

Find the image parameters of the network shown in Fig. 13.130. I1

30 Ω

20 Ω

+ V1

+

10 Ω

−

V2 −

Fig. 13.130 Solution

I2

Zoc1 = 30 + 10 = 40 Ω 10 × 20 Z sc1 = 30 + = 36.67 Ω 10 + 20 Zi Zoc1 Z sc1 = 40 × 36.67 6 = 38.3 Ω

Exercises 13.101

Zoc2 = 20 + 10 = 30 Ω 10 × 30 Z sc2 = 20 + = 27.5 Ω 10 + 30 Zi Zoc2 Z sc2 = 30 × 27.5 = 28.72 Ω f = tan nh −11

Example 13.67

Z 1 36.67 = tan anh −1 = 1 91° Zoc1 40

Find image parameters for the lattice two-port network shown in Fig. 13.131. 2Ω

4Ω

4Ω

2Ω

Fig. 13.131 Solution The network is redrawn as shown in Fig. 13.132. Zoc1 Z sc1

Zoc2 = ( 2 + 4) || ( 2 + 4) = 3 Ω Z sc2 = ( 2 || 4) + ( 2 || 4) = 2 67 Ω

Zi1 = Zoc1 Z sc1 = (3)(2 ( 2.67) = 2.83 Ω Zoc2 Z sc2 = (3)( 2.67) = 2.83 Ω

Zi

f = tan nh

1

Z Zoc1

ta h −1

2

4

4

2

2 67 = 1 76° 3

Fig. 13.132

Exercises 13.1 Determine Z-parameters for the network shown in Fig. 13.133. I1

1Ω

1Ω

I2

+ V1

2Ω

13.2 Find Z-parameters for the network shown in Fig. 13.134.

0.5 Ω

−

I1 +

+

V2

V1

1F

2F

I2 +

2H

2H

−

−

V2 −

Fig. 13.134

Fig. 13.133 ⎡13 ⎢ Z=⎢7 ⎢2 ⎣7

2⎤ 7⎥ 3 ⎥⎥ 7⎦

⎡ 4s4 + 6s2 + 1 ⎢ 3 Z = ⎢ 4 s +3 s ⎢ 4s ⎢ ⎣ 4s2 + 1

4 s3 ⎤ ⎥ 4s2 + 1 ⎥ 4 s3 + 2 s ⎥ ⎥ 4s2 + 1 ⎦

13.102 Network Analysis and Synthesis 13.3 Find Y-parameters of the network shown in Fig 13.135. I1

2Ω

2Ω

2Ω

2i

I2

+ 1Ω

V1

1Ω

+

V2

−

8Ω

i +

V2

2Ω

V1

−

−

Fig. 13.135 ⎡ 0.36 −0.033⎤ Y =⎢ ⎣ −0.033 −0.36 ⎥⎦ 13.4 Find Y-parameters for the network shown in Fig. 13.136. 3Ω

I1

1H

I2

+

+

V1

2Ω

2F

−

Fig. 13.138 1⎤ ⎡ 3 ⎢ 20 − 20 ⎥ Y =⎢ 1 ⎥⎥ ⎢− 1 ⎣ 4 4 ⎦ 13.7 Show the ABCD parameters of the network shown in Fig. 13.139. I1

V2

1F

−

1F

1F

−

+ 1H

V1

⎡10 s + 13s + 2 ⎢ 5s + 6 Y =⎢ 2 ⎢ ⎢⎣ − 5s + 6

−

⎤ 2 ⎥ 5s + 6 ⎥ 5s 2 + 6 s + 5 ⎥ 5s + 6 ⎥⎦ −

Fig. 13.139 ⎡1 + s 2 ⎢ ⎡ A B ⎤ ⎢ s2 = ⎢⎣C D ⎥⎦ ⎢ 1 ⎢ ⎣ s

13.5 Find Y-parameters for the network shown in Fig. 13.137. 1F 4

I2

+

+ 2Ω

1F 3

V2

−

2

I1

I2

+

Fig. 13.136

V1

+

4Ω

2H

−

1Ω

⎡ 7s + 6 ⎢ 12 Y =⎢ ⎢ −s ⎢⎣ 4

13.8 Find ABCD parameters for the network shown in Fig. 13.140. I1

V2

+

−

V1

Fig. 13.137 s ⎤ − ⎥ 4 ⎥ 2 s + 4s + 2 ⎥ ⎥⎦ 4s

13.6 Find Y-parameters for the network shown in Fig. 13.138.

1 + 2s2 ⎤ ⎥ s3 ⎥ 1 + s2 ⎥ ⎥ s2 ⎦

4Ω

8Ω

6Ω

I2 +

1Ω

2Ω

−

V2 −

Fig. 13.140 ⎡ 27 206 ⎤ ⎡A B⎤ ⎢ ⎥ = ⎢⎣C D ⎥⎦ ⎢11 42 ⎥ ⎣2 ⎦ 13.9 For the network shown in Fig. 13.141, determine parameter h21.

Exercises 13.103 2I1 1Ω

I1

I2

+

+

1F

V2

1Ω

V1 −

−

Fig. 13.141

2Ω

I1

I2

+

+

2Ω

V1

2Ω

2Ω

2Ω

I2

+

+

1Ω

V1

V2

1Ω

−

−

Fig. 13.144 −( 2 + s ) ⎤ ⎡ ⎢ h21 = 1 + s ⎥ ⎣ ⎦

Determine Y and Z-parameters for the network shown in Fig. 13.142.

13.10

I1

2Ω

1Ω

1⎤ ⎡ 3 ⎢ 4 − 4⎥ Y =⎢ ⎥ ⎢− 1 3 ⎥ ⎣ 4 4 ⎦ 13.13 Two identical sections of the network shown in Fig. 13.145 are connected in parallel. Obtain Y-parameters of the connection.

V2

2I1

I1

1Ω

2Ω

I2

+ −

+

−

[Y11

1 , Y12

Z11

0.5 , Y21

2Ω

V1

Fig. 13.142 1.5 , Y22

2 2 6 Ω, Z12 = Ω, Z 21 = − Ω, 5 5 5

22

−

0.5

−

4 ⎤ Ω 5 ⎥⎦

Fig. 13.145 ⎡ ⎢ 1 Y =⎢ ⎢− 1 ⎣ 2

13.11 For the bridged T, R-C network shown in Fig. 13.143 determine Y-parameters using interconnections of two-port networks. 1 2F

I1

1Ω 2

1Ω

+ V1 −

1⎤ − ⎥ 2 5 ⎥⎥ 4 ⎦

13.14 Determine Y-parameters using interconnection of two-port networks for the network shown in Fig. 13.146.

I2

1Ω

+

1 F 2

V2

4Ω

V2 −

Fig. 13.143 ⎡ s 2 + 8s + 8 ( s 2 + 6 s + 8) ⎤ − ⎢ ⎥ 2( s + 6) 2( s + 6) ⎥ Y =⎢ ⎢ ( s 2 + 6 s + 8) s 2 + 10 0s + 8 ⎥ ⎢− ⎥ 2( s + 6) 2( s + 6) ⎦ ⎣ 13.12 For the network of Fig. 13.144, find Y-parameters using interconnection of two-port networks.

1Ω 2Ω

I1 + V1 −

I2 + V2 −

2Ω 1Ω 2

1Ω 2

Fig. 13.146 ⎡ 3.1 −0.9⎤ Y =⎢ ⎣ −0.9 3.1 ⎥⎦ 13.15 Determine the transmission parameters of the network shown in Fig. 13.147 using the concept of interconnection of two two-port networks.

13.104 Network Analysis and Synthesis 1H

1H

Determine the h-parameters of the overall network.

1H

3Ω 1F

3Ω

1F

1Ω

1Ω

Fig. 13.147 ⎡1 3s 2 + s 4 ⎢ 3 ⎣ 2s s

3 4 s3 + s5 ⎤ ⎥ 1 + 3s 2 + s 4 ⎦

Fig. 13.149 ⎡ 15 ⎢ 2 ⎢ 1 ⎢− ⎣ 2

13.16 Two networks shown in Fig. 13.148 are connected in series. Obtain the Z-parameters of the resulting network. 1Ω

1Ω

13.18 The h-parameters of a two-port network shown in Fig. 13.150 are h11 = 2 Ω, h12 = 4, h21 = −5, h22 = 2 . Determine the supply voltage Vs if the power dissipated in the load resistor of 4 Ω is 25 W and Rs = 2 Ω.

2Ω

Rs = 2 Ω

(a) 10 Ω

1⎤ − ⎥ 2 5 ⎥⎥ 2 ⎦

20 Ω

Vs

+ −

I1 + V1 −

I2 + V2 −

Network

4Ω

Fig. 13.150

5Ω

(b)

Fig. 13.148 ⎡18 7 ⎤ ⎢⎣ 7 28⎥⎦ 13.17 Two identical sections of the network shown in Fig. 13.149 are connected in series-parallel.

[58 V] 13.19 The Z-parameters of a two-port network are Z11 = 2.1 Ω, Z12 = Z21 = 0.6 Ω, Z22 = 1.6 Ω. A resistor of 2 Ω is connected across port 2. What voltage must be applied at port 1 to produce a current of 0.5 A in the 2 Ω resistor. [6 V] 13.20 If a two-port network has Z11 = 25 Ω, Z12 = Z21 = 20 Ω, Z22 = 50 Ω, find the equivalent T-network. [10 Ω, 30 Ω, 20 Ω]

Objective-Type Questions 13.1 The open-circuit impedance matrix of the two-port network shown in Fig. 13.151 is 2Ω

1Ω

Fig. 13.151

3I1

⎡ −2 1⎤ (a) ⎢ ⎣ −8 3⎥⎦

⎡ −2 −8⎤ (b) ⎢ 3 ⎥⎦ ⎣1

⎡0 1⎤ (c) ⎢ ⎣1 0 ⎥⎦

⎡2 (d) ⎢ ⎣ −1

1⎤ 3 ⎥⎦

13.2 Two two-port networks are connected in cascade. The combination is to be represented as a single two-port network. The parameters are obtained by multiplying the individual

Objective-Type Questions 13.105

(a) (b) (c) (d)

(a) −0.2 mho (c) −0.05 mho

z-parameter matrix h-parameter matrix y-parameter matrix ABCD parameter matrix

13.7 The Z-parameters Z11 and Z21 for the twoport network in Fig. 13.154 are,

13.3 For a two-port network to be reciprocal

I1

(a) z11 = z22

(b) y21 = y12

(c) h21 = −h12

(d) AD − BC = 0

non-reciprocal and passive non-reciprocal and active reciprocal and passive reciprocal and active

+

V1

R

−

−

Fig. 13.154 6 16 Ω, Ω 11 11

(b)

6 4 Ω, Ω 11 11

6 16 Ω, Ω 11 11

(d)

4 4 Ω, Ω 11 11

13.8 The impedance parameters Z11 and Z12 of a two-port network in Fig. 13.155. 2Ω

2Ω

I2

+

3Ω

1Ω

+

V1

V2

− 10V1 +

(c)

13.5 A two-port network is shown in Fig. 13.152. The parameter h21 for this network can be given by R

I2

4Ω

(a) −

I1

2Ω

+

13.4 The short-circuit admittance matrix of a two1⎤ ⎡ ⎢0 − 2⎥ port network is ⎢ ⎥ . The two-port ⎢1 0 ⎥ ⎣2 ⎦ network is (a) (b) (c) (d)

(b) 0.1 mho (d) 0.05 mho

1Ω

V2

R

Fig. 13.155 −

−

Fig. 13.152 1 (a) − 2 (c) −

(d)

(b) 3 Ω, 0.5 Ω (d) 2.25 Ω, 0.5 Ω

13.9 The h parameters of the circuit shown in Fig. 13.156.

1 (b) 2

3 2

(a) 2.75 Ω, 0.25 Ω (c) 3 Ω, 0.25 Ω

10 Ω

3 2 20 Ω

13.6 The admittance parameter Y12 in the two-port network in Fig. 13.153. 20 Ω

Fig. 13.156 5Ω

Fig. 13.153

10 Ω

⎡ 0.1 0.1⎤ (a) ⎢ ⎣ −0.1 0.3⎥⎦

⎡10 −1 ⎤ (b) ⎢ ⎣ 1 0.05⎥⎦

⎡30 20 ⎤ (c) ⎢ ⎣ 20 20 ⎥⎦

1 ⎤ ⎡10 (d) ⎢ ⎣ −1 0.05⎥⎦

13.106 Network Analysis and Synthesis 13.10 A two-port network is represented by ABCD ⎡V ⎤ ⎡ A B ⎤ ⎡ V2 ⎤ . parameters given by ⎢ 1 ⎥ = ⎢ ⎣ I1 ⎦ ⎣C D ⎥⎦ ⎢⎣ − I 2 ⎥⎦ If port 2 is terminated by RL, then the input impedance seen at port 1 is given by (a)

A BR BRL C DR DRL

(b)

AR RL + C BR RL + D

(c)

DR RL + A BR RL + C

(d)

B AR RL D C CR RL

13.11 In the two-port network shown in Fig. 13.157, Z12 and Z21 are respectively

re

bI 1

ro

(a)

A C

(b)

AD BC

AB D (d) DC C 13.15 A two-port network is symmetrical if (a) Z11 Z22 − Z12 Z21 = 1 (b) AD − BC = 1 (c) h11 h22 − h12 h21 = 1 (d) Y11 Y22 − Y12 Y21 = 1 (c)

13.16 For the network shown in Fig. 13.158 admittance parameters are Y11 = 8 mho, Y12 = Y21 = −6 mho and Y22 = 6 mho. The value of YA, YB and YC (in mho) will be respectively (a) 2, 6, −6 (c) 2,0,6

(b) 2,6,0 (d) 2,6,8 YC

Fig. 13.157 (a) re and b r0

(b) 0 and −b r0

(c) 0 and b r0

(d) re and −b r0

13.12 If a two-port network is passive, then we have, with the usual notation, the following relationship for symmetrical network (a) (b) (c) (d)

h12 = h21 h12 = −h21 h11 = h22 h11h22 − h12h21 = 1

13.13 A two-port network is defined by the following pair of equations I1 = 2V1 +V2 and I2 = V1 + V2. Its impedance parameters (Z11, Z12, Z21, Z22) are given by (a) 2, 1, 1, 1 (b) 1, −1, −1, 2 (c) 1, 1, 1, 2 (d) 2, −1, −1, 1 13.14 A two-port network has transmission ⎡A B⎤ . The input impedance parameters ⎢ ⎣C D ⎥⎦ of the network at port 1 will be

YA

YB

Fig. 13.158 13.17 The impedance matrices of two two-port ⎡ 3 2⎤ ⎡15 5 ⎤ and ⎢ . networks are given by ⎢ ⎥ 2 3 ⎣ ⎦ ⎣ 3 25⎥⎦ If these two networks are connected in series, the impedance matrix of the resulting twoport network will be ⎡3 5 ⎤ (a) ⎢ ⎣ 2 25⎥⎦

⎡18 7 ⎤ (b) ⎢ ⎣ 7 28⎥⎦

⎡15 2⎤ (c) ⎢ (d) inderminate ⎣ 5 3⎥⎦ 13.18 If the p network and T network are equivalent, then the values of R1, R2 and R3 (in ohms) will be respectively (a) 6, 6, 6 (c) 9, 6, 9

(b) 6, 6, 9 (d) 6, 9, 6

Answers to Objective-Type Questions 13.107 16 Ω

1 3F

1Ω

24 Ω

2H

24 Ω

Fig. 13.160 R1

2s ⎤ ⎡ 2s 1 (a) ⎢ 3⎥ 2s + ⎥ ⎢ 2s s⎦ ⎣

R2

R3

Fig. 13.159 13.19 For a two-port symmetrical bilateral network, if A = 3 and B = 1, the value of the parameter C will be (a) 4 (b) 6 (c) 8 (d) 16 13.20 The impedance matrix for the network shown in Fig. 13.160 is

2s ⎤ ⎡ 2s 1 (b) ⎢ 3⎥ ⎢ −2 s 2 s + ⎥ s⎦ ⎣ 3 ⎡ ⎤ 2s ⎤ 2s ⎥ ⎡ 2s 1 ⎢ 2s + 2 (c) ⎢ 3 ⎥ (d) ⎢ 3⎥ ⎢ −2 s 2 s + ⎥ ⎢ 2s 2s + ⎥ s⎦ ⎣ s⎦ ⎣ 13.21 With the usual notations, a two-port resistive network satisfies the conditions 3 4 A D= B C . The Z11 of the network is 2 3 5 4 (a) (b) 3 3 2 1 (c) (d) 3 3

Answers to Objective-Type Questions 1. (a)

2. (d)

8. (a)

9. (d)

15. (c)

16. (c)

3. (b) , (c)

4. (b)

5. (a)

6. (c)

7. (c)

10. (d)

11. (b)

12. (d)

13. (b)

14. (a)

17. (b)

18. (a)

19. (c)

20. (a)

21. (b)

14 Fourier Analysis

14.1

INTRODUCTION

Any arbitrary periodic function can be represented by an infinite series of sinusoids of harmonically related frequencies. This representation is known as Fourier-series representation. The Fourier representation of periodic functions is extended to nonperiodic functions by letting the fundamental period T tend to infinity, and this Fourier method of representing nonperiodic functions as a function of frequency is called Fourier transform. The Fourier representation of functions is also known as frequency-domain representation. The Fourier-series representation can be obtained only for periodic functions, but the Fourier-transform technique can be applied to both periodic and nonperiodic functions to obtain the frequency-domain representation of functions. The information about magnitude and phase of various frequency components can be obtained by Fourier analysis. The plot of magnitude versus frequency is called magnitude spectrum, and the plot of phase versus frequency is called phase spectrum.

14.2

TRIGONOMETRIC FOURIER SERIES

Any periodic function f (t) can be expressed into an infinite trigonometric series if it satisfies the following conditions: 1. It is well defined and single valued, except possibly at a finite number of points. 2. It has a finite number of discontinuities in the period T. 3. It has a finite number of positive and negative maxima in the period T. These conditions are known as Dirichlet’s conditions. Any function f (t) which is periodic and satisfies the Dirichlet’s condition, can be represented in terms of sine and cosine functions. f (t ) = a0 + a1 cos ω t a2 cos cos 2 t … an cos n t + b ∞

∞

n =1

n =1

t+b

t+

+ bn sin i n ωt + … …(14.1)

= a0 + ∑ an cos ω t + ∑ bn sin si n ω t where the coefficients a0, an and bn are given by, T

a0 =

1 f (t ) dt T ∫0

14.2 Network Analysis and Synthesis T

an =

2 f (t ) cos n ω t dt T ∫0

bn =

2 f (t ) sin n ω t dt T ∫0

T

14.2.1

Evaluation of Fourier Coefficients

1. Evaluation of a0

Integrating Eq. (14.1) w.r.t. t,

T

∫

T

f (t ) dt = a

0

∫ dt 0

T ∞ ⎞ ⎛ a cos n ω t dt + ⎟ ⎜ ∑ bn s ∫⎝∑ n ∫ ⎠ 0 n =1 0 ⎝ n =1

T

⎛

∞

⎞ n ωt ⎟ dt ⎠

= a0 T + 0 + 0 T ⎡ ⎤ ⎢∵ ∫ sin n ω t dt = 0 for all n ⎥ ⎢ ⎥ 0 ⎢ ⎥ T ⎢ ⎥ and cos n ω t dt = 0 f ≠ 0 ∫ ⎢ ⎥ 0 ⎣ ⎦ T

1 f (t ) dt T ∫0

a0 =

2. Evaluation of an Multiplying Eq. (14.1) by cos n wt and integrating w.r.t t between the limits 0 and T, T

∫

T

T

∞

T

∞

0

0

n =1

0

n =1

0

= 0+ 2 T

an = 3. Evaluation of bn

0

n ω t dt

T an + 0 2

T

∫ f (t ) cos n ω t dt 0

Multiplying Eq. (14.1) by sin n wt and integrating w.r.t t between the limits 0 and T,

T

∫

∫ cos n ω t ∑ an cos n ω t dt + ∫ cos n ω t ∑ bn

∫ a0 cos n ω t dt

f (t ) cos n ω t dt

f (t ) sin n ω t dt

T

T

∞

T

∞

0

0

n =1

0

n =1

i n ω t dt ∫ a0 sin

= 0+0+ bn =

2 T

∫ sin n ω t ∑ an cos n ω t dt + ∫ sin n ω t ∑ bn

n ω t dt

T bn 2

T

∫ f (t ) sin n ω t dt 0

14.2.2 Complex Fourier Spectrum A Fourier series expansion of a periodic function is equivalent to resolving the function in terms of its components of various frequencies. A periodic function with period T has frequency components of angular

14.2 Trigonometric Fourier Series 14.3

2π . Thus, the periodic function f (t) possesses its spectrum of frequencies w, 2w, 3w,… nw where ω = T frequencies, known as line spectrum. The spectrum exists only at w, 2w, 3w,… etc. Thus, the spectrum is not a continuous curve but exists only at some discrete values of w. The line spectrum is the plot which shows the variation of magnitude and phase of the function w.r.t n. (i) Magnitude Spectrum It is a plot of magnitude cn versus nw where cn = an2 + bn2 . (ii) Phase Spectrum It is a plot of phase angle fn versus nw where φn = tan −1

bn . an

Example 14.1 Find the trigonometric Fourier series of the waveform shown in Fig. 14.1. Draw the magnitude and phase spectrum. f(t)

A

t p

0

2p

3p

Fig. 14.1 Solution

T = 2π 2π 2π ω= = =1 T 2π f (t ) = A 0 < t < π = 0 π < < 2π

The trigonometric Fourier series of f (t) is ∞

f (t ) = a0 + ∑ an cos n n =1

∞

t + ∑ bn s

n ωt

n 1 n=

T

a0 =

1 f (t ) dt T ∫0

π 2π ⎤ 1 ⎡ ⎢ ∫ A dt + ∫ 0 dt ⎥ 2π ⎢⎣ 0 ⎥⎦ π A π = [t ]0 2π A = (π − 0) 2π A = 2

=

T

an = =

2 f (t ) cos n ω t dt T ∫0 π 2 ⎡ ⎢∫ A 2π ⎢⎣ 0

n t dt d

2π

⎤

π

⎦

d⎥ ∫ 0 dt ⎥

[ ω = 1]

14.4 Network Analysis and Synthesis π

=

A ⎡ si n t ⎤ π ⎢⎣ n ⎥⎦ 0

A ⎡1 ⎤ (sin nπ − sin ) ⎥ ⎢ π ⎣n ⎦ =0 [ sin si nπ = 0] =

T

bn = =

2 f (t ) sin n ω t dt T ∫0 π 2 ⎡ ⎢∫ A 2π ⎢⎣ 0

n t dt d

2π

⎤

π

⎦

d⎥ ∫ 0 dt ⎥

[ ω = 1]

π

=

A ⎡ cos n t ⎤ − π ⎢⎣ n ⎥⎦ 0

A + [− nπ A = [ −( −1)) n + ] nπ ⎧ 0 if n is even ⎪ bn = ⎨ 2 A ⎪⎩ nπ if n is odd =

f (t ) = =

A + 2

∞

]

A

∑ nπ sin n ω t

n=1 n= odd

A 2A 2A + sin i ωt + ssin i 3 ω t + …… 2 π 3π

The magnitude and phase spectrum are shown in Fig. 14.2. cn

fn 2A p

A 2

2A 3p

p 2

nw 0

w

2w

3w

4w

nw 0

w

2w

3w

4w

Fig. 14.2

Example 14.2 Find the trigonometric Fourier series of the waveform shown in Fig. 14.3. Draw the magnitude and phase spectrum.

14.2 Trigonometric Fourier Series 14.5 f(t)

A

t

0

2p

4p

Fig. 14.3 Solution

T = 2π 2π 2π = =1 T 2π A f (t ) = t 0 2π

ω=

t < 2π

The trigonometric Fourier series of f (t) is ∞

f (t ) = a0 + ∑ an cos n n =1

∞

t + ∑ bn s

nω t

n 1 n=

T

a0 = =

1 f (t ) dt T ∫0 1 2π

2π

A

∫ 2π t dt 0

2π

A ⎡t2 ⎤ = ⎢ ⎥ 4π 2 ⎣ 2 ⎦ 0

⎤ A ⎡ 4π 2 − 0 ⎢ ⎥ 4π 2 ⎣ 2 ⎦ A = 2 =

T

an = = =

2 f (t ) cos n ω t dt T ∫0 2 2π

2π

A

A

∫ 2π t cos n t ddt

2π 2

[∵ ω = 1]

0 2π

∫ t cos n t dt 0

2π

A ⎡ ⎛ sin n t ⎞ ⎛ cos n t ⎞ ⎤ = t ⎟⎠ − (1) ⎜⎝ − ⎟⎥ 2 ⎢ ⎜ ⎝ n 2π ⎣ n2 ⎠ ⎦ 0

14.6 Network Analysis and Synthesis =

A ⎡ ⎛ sin nπ ⎞ ⎟+ ⎢ 2π ⎜ n ⎠ 2π 2 ⎣ ⎝

=

A ⎡ 1 1⎤ 0+ 2 − 2⎥ 2π 2 ⎣ n n ⎦

2 nπ n

2

−

cos 0 ⎤ ⎥ n2 ⎦ ⎡∵sin 2 π = 0 ⎤ ⎢ ⎥ ⎣cos 2 nπ = 1 ⎦

=0 T

bn = = =

2 f (t ) sin n ω t dt T ∫0 2 2π

2π

A

d ∫ 2π t sin n dt

[∵ ω = 1]

0

A 2π 2

2π

∫ t si

n t dt

0

2π

=

A ⎡ ⎛ cos n t ⎞ ⎛ sin n t ⎞ ⎤ t − ⎟⎠ − (1) ⎜⎝ − 2 ⎟⎠ ⎥ 2 ⎢ ⎝ n 2π ⎣ n ⎦0

=

A ⎡ ⎛ cos nπ ⎞ i 2 nπ sin 0 ⎤ −2π ⎜ − 2 ⎥ ⎟+ 2 ⎢ ⎝ n ⎠ 2π ⎣ n2 n ⎦

=

A ⎡ 2π ⎤ ⎢− ⎥ 2π 2 ⎣ n ⎦

=− f (t ) = =

⎡∵ cos 2 nπ 1⎤ ⎢⎣ sin 2 π 0 ⎥⎦

A nπ

A ∞ A +∑− sin n ω t 2 n =1 n π A A − sin ω t 2 π

A sin 2 ω t − …… 2π

The magnitude and phase spectrum are shown in Fig. 14.4. fn

cn A 2

0

A p

w

A 2p

2w

p 2

nw

Fig. 14.4

0

w

2w

nw

14.2 Trigonometric Fourier Series 14.7

Example 14.3

Find the trigonometric Fourier series of the waveform shown in Fig. 14.5. f (t )

A

p

0

t

3p

2p

Fig. 14.5 Solution

T = 2π 2π 2π ω= = =1 T 2π f (t ) = A si ω t f (t ) = A si t 0