Circuit Theory and Transmission Lines [2 ed.] 9789352602704, 9352602706

Table of contents :
Title
Contents
1 Basic Network Concepts
2 Elementary Netowrk Theorems
3 Network Theorems (Application to dc Networks)
4 Cpupled Circuits
5 Resonance
6 Transient Analysis
7 Laplace Transform and its Applicaiton
8 Netowrk Functions
9 Network Synthesis
10 Two-port Newtworks
11 Filters and Attenuators
12 Transmission Lines
Question Papers
Index

Citation preview

Circuit Theory and Transmission Lines Second Edition

About the Author Ravish R Singh is presently Academic Advisor at Thakur Educational Trust, Mumbai. He obtained a BE degree from University of Mumbai in 1991, an MTech degree from IIT Bombay in 2001, and a PhD degree from Faculty of Technology, University of Mumbai, in 2013. He has published several books with McGraw Hill Education (India) on varied subjects like Engineering Mathematics (I and II), Applied Mathematics, Electrical Engineering, Electrical and Electronics Engineering, etc., for all-India curricula as well as regional curricula of some universities like Gujarat Technological University, Mumbai University, Pune University, Jawaharlal Nehru Technological University, Anna University, Uttarakhand Technical University, and Dr A P J Abdul Kalam Technical University. Dr Singh is a member of IEEE, ISTE, and IETE, and has published research papers in national and international journals. His fields of interest include Circuits, Signals and Systems, and Engineering Mathematics.

Circuit Theory and Transmission Lines Second Edition

Ravish R Singh Academic Advisor Thakur Educational Trust Mumbai, Maharashtra

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. Circuit Theory and Transmission Lines, 2e Copyright© 2016, 2014, 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): 978-93-5260-270-4 ISBN (10): 93-5260-270-6 Managing Director: Kaushik Bellani Director—Products (Higher Education and Professional): Vibha Mahajan Manager—Product Development: Koyel Ghosh Specialist—Product Development: Piyali Chatterjee Head––Production (Higher Education and Professional): Satinder S Baveja Senior Production Executive: Jagriti Kundu AGM—Product Management (Higher Education and Professional): Shalini Jha Manager—Product Management: Ritwick Dutta General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar

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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 1.3 Sources 1.4 Some Definitions 1.6 Series and Parallel Combination of Resistors 1.7 Series and Parallel Combination of Inductors 1.9 Series and Parallel Combination of Capacitors 1.10 Star-Delta Transformation 1.10 Source Transformation 1.13 Source Shifting 1.19 Exercises 1.21 Objective-Type Questions 1.22 Answers to Objective-Type Questions 1.23

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 2.2 Supermesh Analysis 2.17 Node Analysis 2.26 Supernode Analysis 2.46 Exercises 2.52 Objective-Type Questions 2.56 Answers to Objective-Type Questions

2.57

3. NETWORK THEOREMS (APPLICATION TO DC NETWORKS) 3.1 3.2 3.3 3.4 3.5 3.6

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 Millman’s Theorem 3.112 Exercises 3.117 Objective-Type Questions 3.122 Answers to Objective-Type Questions 3.123

3.1

viii�Contents

4. COUPLED CIRCUITS 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10

5. RESONANCE 5.1 5.2 5.3 5.4

6.1

Introduction 6.1 Initial Conditions 6.1 Resistor–Inductor Circuit 6.27 Resistor–Capacitor Circuit 6.49 Resistor–Inductor–Capacitor Circuit 6.66 Exercises 6.79 Objective-Type Questions 6.82 Answers to Objective-Type Questions 6.85

7. LAPLACE TRANSFORM AND ITS APPLICATION 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

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. TRANSIENT ANALYSIS 6.1 6.2 6.3 6.4 6.5

4.1

Introduction 4.1 Self-Inductance 4.1 Mutual Inductance 4.2 Coefficient of Coupling (k) 4.2 Inductances in Series 4.3 Inductances in Parallel 4.4 Dot Convention 4.9 Coupled Circuits 4.15 Conductively Coupled Equivalent Circuits 4.37 Tuned Circuits 4.40 Exercises 4.47 Objective-Type Questions 4.49 Answers to Objective-Type Questions 4.50

Introduction 7.1 Laplace Transformation 7.1 Laplace Transforms of Some Important Functions 7.2 Properties of Laplace Transform 7.4 Inverse Laplace Transform 7.7 The Transformed Circuit 7.12 Resistor–Inductor Circuit 7.13 Resistor–Capacitor Circuit 7.19

7.1

���������ix

7.9 Resistor–Inductor–Capacitor Circuit 7.25 7.10 Response of RL Circuit to Various Functions 7.31 7.11 Response of RC Circuit to Various Functions 7.39 Exercises 7.49 Objective-Type Questions 7.52 Answers to Objective-Type Questions 7.53

8. NETWORK FUNCTIONS

8.1

8.1 8.2 8.3 8.4 8.5 8.6 8.7

Introduction 8.1 Driving-Point Functions 8.1 Transfer Functions 8.2 Analysis of Ladder Networks 8.5 Analysis of Non-Ladder Networks 8.15 Poles and Zeros of Network Functions 8.20 Restrictions on Pole and Zero Locations for Driving-Point Functions [Common Factors in N(s) and D(s) Cancelled] 8.21 8.8 Restrictions on Pole and Zero Locations for Transfer Functions [Common Factors in N(s) and D(s) Cancelled] 8.21 8.9 Time-Domain Behaviour from the Pole-Zero Plot 8.39 8.10 Graphical Method for Determination of Residue 8.42 Exercises 8.50 Objective-Type Questions 8.53 Answers to Objective-Type Questions 8.55

9. NETWORK SYNTHESIS 9.1 9.2 9.3 9.4 9.5 9.6 9.7

Introduction 9.1 Hurwitz Polynomials 9.1 Positive Real Functions 9.16 Elementary Synthesis Concepts 9.24 Realisation of LC Functions 9.30 Realisation of RC Functions 9.47 Realisation of RL Functions 9.63 Exercises 9.72 Objective-Type Questions 9.74 Answers to Objective-Type Questions 9.76

10. TWO-PORT NETWORKS 10.1 10.2 10.3 10.4 10.5 10.6 10.7

9.1

Introduction 10.1 Open-Circuit Impedance Parameters (Z Parameters) 10.2 Short-Circuit Admittance Parameters (Y Parameters) 10.8 Transmission Parameters (ABCD Parameters) 10.18 Inverse Transmission Parameters (A�B�C�D� Parameters) 10.24 Hybrid Parameters (h Parameters) 10.28 Inverse Hybrid Parameters (g Parameters) 10.33

10.1

x�Contents 10.8 10.9 10.10 10.11 10.12 10.13

Inter-relationships between the Parameters 10.37 Interconnection of Two-Port Networks 10.63 T-Network 10.79 Pi (� )-Network 10.79 Lattice Networks 10.84 Terminated Two-Port Networks 10.87 Exercises 10.97 Objective-Type Questions 10.100 Answers to Objective-Type Questions 10.103

11. FILTERS AND ATTENUATORS 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 11.16 11.17 11.18 11.19

Introduction 11.1 Classification of Filters 11.1 T-Network 11.1 � Network 11.4 Characteristic of Filters 11.6 Constant-k Low Pass Filter 11.7 Constant-k High-pass Filter 11.14 Band-pass Filter 11.18 Band-stop Filter 11.22 m-Derived Filters 11.25 m-Derived Low-Pass Filter 11.28 m-Derived High-Pass Filter 11.31 Terminating Half Sections 11.34 Composite Filter 11.37 Attenuator 11.40 Lattice Attenuator 11.41 T-Type Attenuator 11.42 � -Type Attenuator 11.45 Ladder-Type Attenuator 11.47 Exercises 11.48 Objective-Type Questions 11.49 Answers to Objective-Type Questions

11.1

11.50

12. TRANSMISSION LINES 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9

Introduction 12.1 Types of Transmission Lines 12.1 Primary Constants of a Transmission Line 12.2 Equivalent Circuit Representation of Transmission Line 12.2 Secondary Constants of a Transmission Line 12.3 Infinite Line 12.10 Terminated Transmission Lines 12.11 General Solution of a Transmission Line 12.11 Reflection 12.15

12.1

���������xi

12.10 12.11 12.12 12.13 12.14 12.15

Reflection Coefficient 12.15 Standing Waves 12.16 Standing Wave Ratio 12.16 Scattering parameters or S-parameters 12.18 Impedance Smith Chart 12.19 Impedance Matching Using Smith Chart 12.36 Exercises 12.42 Objective-Type Questions 12.42 Answers to Objective-Type Questions 12.43

Solved Question Paper December 2015 (EXC 304) Solved Question Paper December 2015 (ETC 304) Solved Question Paper May 2016 (EXC 304) Solved Question Paper May 2016 (ETC 304) Index

SQP.1 SQP.1 SQP.1 SQP.1 I.1

Preface Overview Circuit Theory and Transmission Lines is an important subject for third-semester students of Electronics Engineering and Electronics & Telecommunication Engineering. With lucid and brief theory, this textbook provides thorough understanding of the topics of this subject. Following a problem-solving approach and discussing both analysis and synthesis of networks, it offers good coverage of dc circuits, network theorems, two-port networks, network synthesis, and transmission lines. Generally, numerical problems are expected in university examinations in this subject. The weightage given to problems in examinations is more than 70–80%. Questions from important topics of this subject are part of competitive examinations such as IAS, IES, etc. Hence, numerous solved examples and exercise problems are included in each chapter of this book to help students develop and master problem-solving skills required to ace any examination with confidence. Objective-type questions from various competitive examinations are also included at the end of each chapter for easy revision of core concepts.

Salient Features � � � � � � � �

Up-to-date and full coverage of the latest revised syllabus of University of Mumbai Covers both analysis and synthesis of networks Uses problem-solving approach to explain topics Lucid coverage of network theorems, transient analysis, two-port networks, network synthesis Separate chapter on transmission lines Extensively supported by illustrations Solution of 2015 and 2016 University of Mumbai question papers is provided at the end of the book. Examination-oriented excellent pedagogy: � Illustrations: 1500+ � Solved Examples within chapters 570 � Unsolved Problems: 185 � Objective Type Questions: 135

Chapter Organisation This text is organised into 12 chapters. Chapter 1 covers basic circuit elements and basic laws comprising of networks. Further, dc network theorems are elucidated in Chapters 2 and 3. Chapters 4 and 5 discuss coupled circuits and resonance, respectively. Chapters 6 and 7 discuss transient analysis in time domain and frequency domain, respectively. Chapters 8 and 9 cover network functions and network synthesis. Chapter 10 elucidates two-port networks. Chapter 11 describes filters and attenuators. Lastly, transmission lines and radio frequency is covered in Chapter 12.

Acknowledgements 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

xiv�Preface project. I appreciate the support extended by the team at McGraw Hill Education (India), especially Koyel Ghosh, Piyali Chatterjee, Satinder Singh Baveja, Anuj Shrivastava and Jagriti Kundu during the editorial, copyediting and production stages of this book. I am grateful to the reviewers mentioned below for taking out time to review certain chapters of the book and sharing their valuable suggestions: Amit Bagade Gajraj Singh Reena Sonkusare

Ramrao Adik Institute of Technology, Nerul, Navi Mumbai, Maharashtra Ramrao Adik Institute of Technology, Nerul, Navi Mumbai, Maharashtra Sardar Patel Institute of Technology, Mumbai, Maharashtra

Suggestions for improvements will always be welcome.

Ravish R Singh Publisher’s Note Remember to write to us. We look forward to receiving your feedback, comments, and ideas to enhance the quality of this book. You can reach us at [email protected]. Please mention the title and authors’ name as the subject. In case you spot piracy of this book, please do let us know.

Roadmap to the Syllabus (As per latest revised syllabus of University of Mumbai) This text is useful for Electronics Engineering Circuit Theory—EXC304 Module 1: Analysis of Electrical Circuits 1.1 Analysis of DC Circuits: Analysis of circuits with and without controlled sources using generalized loop, node matrix, superposition, Thevenin, Norton, Millman theorems 1.2 Analysis of Coupled Circuits: Self and mutual inductances, coefficient of coupling, dot convention, equivalent circuit, solution using loop analysis 1.3 Series and Parallel Resonance Circuits: Selectivity, bandwidth, quality factor

GO TO: CHAPTER 1. BASIC NETWORK CONCEPTS CHAPTER 2. ELEMENTARY NETWORK THEOREMS CHAPTER 3. NETWORK THEOREMS (APPLICATION TO dc NETWORKS) CHAPTER 4. COUPLED CIRCUITS CHAPTER 5. RESONANCE

Module 2: Time and Frequency Domain Analysis 2.1 Time Domain Analysis of R-L and R-C Circuits: Forced and natural response, time constant, initial and final values Solution Using First Order Equation for Standard Input Signals: Transient and steady state time response, solution using universal formula 2.2 Time Domain Analysis of R-L-C Circuits: Forced and natural response, effect of damping Solution Using Second Order Equation for Standard Input Signals: Transient and steady state time response 2.3 Frequency Domain Analysis of RLC Circuits: S-domain representation, applications of Laplace transform in solving electrical networks, driving point and transfer function, poles and zeros, calculation of residues by analytical and graphical method, frequency response

GO TO: CHAPTER 6. TRANSIENT ANALYSIS CHAPTER 7. LAPLACE TRANSFORM AND ITS APPLICATION CHAPTER 8. NETWORK FUNCTIONS

xvi�Roadmap to the Syllabus Module 3: Synthesis of RLC Circuits 3.1 Positive Real Functions: Concept of positive real function, testing for Hurwitz polynomials, testing for necessary and sufficient conditions for positive real functions 3.2 Synthesis of RC, RL, LC Circuits: Concepts of synthesis of RC, RL, LC driving point functions

GO TO: CHAPTER 9. NETWORK SYNTHESIS

Module 4: Two-Port Networks 4.1 Parameters: Open circuit, short circuit, transmission and hybrid parameters, relationships among parameters, reciprocity and symmetry conditions 4.2 Series/Parallel Connection: T and Pi representations, interconnection of two-port networks

GO TO: CHAPTER 10. TWO-PORT NETWORKS Module 5: Filters and Attenuators 5.1 Basic Filter Circuits: Low pass, high pass, band pass and band stop filters, transfer function, frequency response, cutoff frequency, bandwidth, quality factor, attenuation constant, phase shift, characteristic impedance 5.2 Concept of Design and Analysis of Filters: Constant K, M derived and composite filters 5.3 Attenuators: Basic concepts, classification, attenuation in dB, K factor (impedance factor) and design concepts

GO TO: CHAPTER 11. FILTERS AND ATTENUATORS

Module 6: Transmission Lines 6.1 Power Frequency Lines: Representation, losses and efficiency in power lines, effect of length, calculation of inductance and capacitance 6.2 Radio Frequency Lines: Representation, propagation constant, attenuation constant, phase constant, group velocity, input impedance, characteristic impedance, reflection coefficient, standing wave ratio, VSWR, ISWR, S-parameters 6.3 Smith Chart: Impedance locus diagram, impedance matching

GO TO: CHAPTER 12. TRANSMISSION LINES

������������������������xvii

This text is useful for Electronics and Telecommunication Engineering Circuits and Transmission Lines—ETC304

Module 1: Electrical Circuit Analysis 1.1 Analysis of DC Circuits: Analysis of circuits with and without controlled sources using generalized loop and node matrix methods and source transformation, superposition, Thevenin, Norton, Millman theorems 1.2 Magnetic Circuits: Self and mutual inductances, coefficient of coupling, dot convention, equivalent circuit, solution using loop analysis 1.3 Tuned Coupled Circuits: Analysis of tuned coupled circuits

GO TO: CHAPTER 1. BASIC NETWORK CONCEPTS CHAPTER 2. ELEMENTARY NETWORK THEOREMS CHAPTER 3. NETWORK THEOREMS (APPLICATION TO dc NETWORKS) CHAPTER 4. COUPLED CIRCUITS

Module 2: Time and Frequency Domain Analysis 2.1 Time Domain Analysis of R-L and R-C Circuits: Forced and natural response, time constant, initial and final values Solution Using First Order Equation for Standard Input Signals: Transient and steady state time response, solution using universal formula 2.2 Time Domain Analysis of R-L-C Circuits: Forced and natural response, effect of damping Solution Using Second Order Equation for Standard Input Signals: Transient and steady state time response 2.3 Frequency Domain Analysis of RLC Circuits: S-domain representation, applications of Laplace transform in solving electrical networks, driving point and transfer function, poles and zeros, calculation of residues by analytical and graphical method, analysis of ladder and lattice network Response to Standard Signals: Transient and steady state time response of R-L-C circuits

GO TO: CHAPTER 6. TRANSIENT ANALYSIS CHAPTER 7. LAPLACE TRANSFORM AND ITS APPLICATION CHAPTER 8. NETWORK FUNCTIONS

xviii�Roadmap to the Syllabus Module 3: Synthesis of RLC circuits 3.1 Positive Real Functions: Concept of positive real function, testing for Hurwitz polynomials, testing for necessary and sufficient conditions for positive real functions 3.2 Synthesis of RC, RL, LC and RLC Circuits: Properties and synthesis of RC, RL, LC driving point functions

GO TO: CHAPTER 9. NETWORK SYNTHESIS

Module 4: Two-Port Circuits 4.1 Parameters: Open circuits, short circuit, transmission and hybrid parameters, relationship among parameters, reciprocity and symmetry conditions 4.2 Interconnections of two-port circuits, T & � representation 4.3 Terminated two-port circuits

GO TO: CHAPTER 10. TWO-PORT NETWORKS

Module 5: Radio Frequency Transmission Lines 5.1 Transmission Line Representation: T and � representations, terminated transmission line, infinite line 5.2 Parameters of Radio Frequency Lines: Propagation constant, attenuation constant, phase constant, group velocity, input impedance, characteristic impedance, reflection coefficient, standing wave ratio, VSWR, ISWR, S-parameters 5.3 Smith Chart: Impedance locus diagram, impedance matching

GO TO: CHAPTER 12. TRANSMISSION LINES

1

Basic Network Concepts

  1.1     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 W. 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 Circuit Theory and Transmission Lines 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

0

0

E = ∫ v i dt = ∫ R i i dt = i 2 Rt

  1.3     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. 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

1.4 Capacitance 1.3

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.  E   nergy 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 = L i dt dt

Hence, total energy supplied to the inductor when current is increased from 0 to I amperes is I

I

0

0

E = ∫ dE = ∫ L i di =

1 2 LI 2

  1.4     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 C∝

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.

1.4 Circuit Theory and Transmission Lines 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 i=

dq d dv = 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(0)

t

v(t ) =

1 i dt + v(0) 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 V

V

0

0

E = ∫ dE = ∫ C v dv =

1 CV 2 2

  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 Sources 1.5

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 v (t) V at any instant is independent of the value or direction of the current that flows through it. The symbols for such voltage sources are shown in Fig. 1.1. The terminal voltage may be a constant, or it may be some (a) (b) specified function of time. Fig. 1.1 Symbols for independent 2.  Independent  Current  Source  An independent current voltage source source is a two-terminal network element which produces a 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 i (t) current sources are shown in Fig. 1.2. 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 Fig. 1.2 Symbols for independent voltage or current, it is called as dependent or controlled source. The current source dependent sources are of four kinds, depending on whether the 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 two points c and d in the circuit that is proportional to a voltage vab between two points a and b. The symbol for such a source is shown in Fig. 1.3. The (+) and (−) sign inside the diamond of the component symbol identifies the component as a voltage source. vcd = m vab The voltage vcd depends upon the control voltage vab and the constant m, a dimensionless constant called voltage gain. 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. 1.4.

a

c +

+ + mvab −

vab b

−

Fig. 1.3

a

+

+ gmvab

Symbol for VCCS

c

vcd −

−

Fig. 1.4

d

Symbol for VCVS

icd

vab b

vcd

−

d

1.6 Circuit Theory and Transmission Lines The arrow inside the diamond of the component symbol identifies the component as a current source. icd = gm vab 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. 1.5. vcd = r iab

iab a

c +

+ + r iab − b

−

Fig. 1.5

icd

iab

a

d

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 (W). 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. 1.6. icd = b iab

vcd

−

c +

+ b iab b

−

−

Fig. 1.6

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 interconnecL L R R tion of two or more circuit elements (viz., voltage sources, resistors, inductors and capacitors) C V is called an electric network. If the network C contains at least one closed path, it is called an electric circuit. Every circuit is a network, (b) (a) but all networks are not circuits. Figure 1.7(a) shows a network which is not a circuit and Fig. Fig. 1.7 (a) Network which is not a circuit 1.7(b) shows a network which is a circuit. (b) Network which is a circuit 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 voltage and current as shown in Fig. 1.8. Ordinary resistors, capacitors and inductors are examples of linear elements.

1.7 Series and Parallel Combinations of Resistors 1.7

A non-linear circuit element is one in which the current does not change linearly with the change in applied voltage. A semiconductor diode operating in I the curved region of characteristics as shown in Fig. 1.8 is common example of non-linear element. Other examples of non-linear elements are voltagedependent resistor (VDR), voltage-dependent capacitor t (varactor), temperature-dependent resistor (thermistor), lighten t m en dependent resistor (LDR), etc. Linear elements obey Ohm’s le m E le ar law whereas non-linear elements do not obey Ohm’s law. rE ne a

Li

ne

i 3.  Active  and  Passive  Elements  An element -L on N which is a source of electrical signal or which is capable of increasing the level of signal energy is termed as V 0 active element. Batteries, BJTs, FETs or OP-AMPs are treated as active elements because these can be used Fig. 1.8 V-I characteristics of linear and for the amplification or generation of signals. All other non-linear elements 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. 1.9. 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. 1.9

Series combination of resistors

1.8 Circuit Theory and Transmission Lines V = V1 + V2 + V3 RT I = R1 I + R2 I + R3 I RT = R1 + R2 + 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 V R1 + R2 + R3 R1 V1 = R1 I = V R1 + R2 + R3 R2 V2 = R2 I = V R1 + R2 + R3 R3 V3 = R3 I = V R1 + R2 + R3 PT = P1 + P2 + P3 I=

Total power

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

I

V12 V22 V32 + + R1 R2 R3

Figure 1.10 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. 1.10 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 I1 = = 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 I I3 = = T = R3 R3 R1 R2 + R2 R3 + R3 R1

1.8 Series and Parallel Combination of Inductors 1.9

PT = P1 + P2 + P3

Total power

= I12 R1 + I 22 R2 + I 32 R3 = Note: For two branch circuits,

V2 V2 V3 + + 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 =

  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. 1.11. 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 dt dt LT = L1 + L2 + L3

L2

L1

i

v2

v1

L3 v3

v

Fig. 1.11 Series connection of inductors

Hence, when a number of inductors are connected in series, the equivalent inductance is the sum of all the individual inductances. L1 i1 Figure 1.12 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 = i1 + i2 + i3 1 LT

1

1

1

∫ v dt = L1 ∫ v dt + L2 ∫ v dt + L3 ∫ v dt 1 1 1 1 = + + LT L1 L2 L3 LT =

v

Fig. 1.12

Parallel connection of inductors

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.

1.10 Circuit Theory and Transmission Lines

  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 1.13. 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 i dt = i dt + i dt + i dt CT ∫ C1 ∫ C2 ∫ C3 ∫

v

Fig. 1.13

1 1 1 1 = + + CT C1 C2 C3 CT =

Series combination of capacitors

C1C2C3 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 = = V C2 C2 C1C2 + C2C3 + C3C1 Q CT V C1 C2 V V3 = = = C3 C3 C1C2 + C2C3 + C3C1

V2 =

i

i1

C1

i2

C2

i3

C3

v

Figure 1.14 shows three capacitors connected in parallel across Fig. 1.14 Parallel combination an ac voltage source v. Let i1 , i2 and i3 be the current through each of capacitors 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 CT

dv dv dv dv = C1 + C2 + C3 dt dt 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.

  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 1.15 (a) shows three resistors RA, RB and RC connected in delta. Figure 1.15 (b) shows three resistors R1, R2 and R3 connected in star.

1.10 Star-delta Transformation 1.11 1

1

R1 RC

RB R3

R2 2

2

3

3

RA (a)

Fig. 1.15 (a)

(b)

Delta network

(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. 1.15 (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. 1.15 (b), the resistance between terminals 1 and 2 = R1 + R2 . Since the two networks are electrically equivalent, R1 + R2 =

RC ( RA + RB ) RA + RB + RC

...(1.1)

Similarly,

R2 + R3 =

RA ( RB + RC ) RA + RB + RC

...(1.2)

and

R3 + R1 =

RB ( RA + RC ) RA + RB + RC

...(1.3)

R1 − R3 =

RB RC − RA RB RA + RB + RC

...(1.4)

R1 =

RB RC RA + RB + RC

R2 =

RA RC RA + RB + RC

R3 =

RA RB RA + RB + RC

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

Adding Eq. (1.4) and Eq. (1.3),

Similarly,

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.

1.12 Circuit Theory and Transmission Lines

1.10.2 Star to delta Transformation Multiplying the above equations, R1 R2 = R2 R3 = R3 R1 =

RA RB RC2

...(1.5)

( RA + RB + RC ) 2 RA2 RB RC

...(1.6)

( RA + RB + RC ) 2 RA RB2 RC

...(1.7)

( RA + RB + RC ) 2

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

RA RB RC ( RA + RB + RC ) ( RA + RB + RC )

2

=

RA RB RC RA + RB + RC

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

R1 R2 + R2 R3 + R3 R1 RR = R2 + R3 + 2 3 R1 R1

RB =

R1 R2 + R2 R3 + R3 R1 RR = R1 + R3 + 3 1 R2 R2

RC =

R1 R2 + R2 R3 + R3 R1 RR = R1 + R2 + 1 2 R3 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. 1.16), the equivalent star resistance is given by RY =

R∆ R∆ R = ∆ 3 R∆ + R∆ + R∆

R∆ = 3RY A

A

RY R∆

R∆ RY

B

C

B

RY C

R∆

Fig. 1.16

Equivalent star resistance for three equal delta resistors

(2) Star-delta transformation can also be applied to network containing inductors and capacitors.

1.11 Source Transformation 1.13

  1.11     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. 1.17. R ⇔I=

V

V R

R

(a)

(b)

Fig. 1.17 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 1.1 

Replace the given network of Fig. 1.18 with a single current source and a resistor. A 10 A

6Ω

20 V

5Ω B

Fig. 1.18

A

A

60 V

Solution Since the resistor of 5 W 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. 1.19), By source transformation (Fig. 1.20), A

13.33 A

80 V 6Ω 6Ω 20 V B

B

Fig. 1.19

6Ω

B

Fig. 1.20

  example 1.2  Reduce the network shown in Fig. 1.21 into a single source and a single resistor between terminals A and B.

1.14 Circuit Theory and Transmission Lines A 2Ω

1A

2Ω 4V

3Ω

1Ω 3V

6V B

Fig. 1.21 Solution Converting all voltage sources into equivalent current sources (Fig. 1.22), A 1A

2Ω

2Ω

2A

2A

3Ω

1Ω

3A

B

Fig. 1.22 Adding the current sources and simplifying the network (Fig. 1.23), A 1Ω

3A

0.75 Ω

1A

B

Fig. 1.23 Converting the current sources into equivalent voltage sources (Fig. 1.24), A

A

3V 3.75 V

1Ω

1.75 Ω

0.75 Ω 0.75 V

B

B

Fig. 1.24

  example 1.3  a single resistor.

Replace the circuit between A and B in Fig. 1.25 with a voltage source in series with

1.11 Source Transformation 1.15 A 5Ω

3A 30 Ω

50 Ω

6Ω

20 V B

Fig. 1.25 Solution Converting the series combination of voltage source of 20 V and a resistor of 5 W into equivalent parallel combination of current source and resistor (Fig. 1.26), A

50 Ω

30 Ω

3A

4A

5Ω

6Ω

B

Fig. 1.26 Adding the two current sources and simplifying the circuit (Fig. 1.27), A

30 || 50 || 5 || 6 = 2.38 Ω

7A

B

Fig. 1.27 By source transformation (Fig. 1.28), 2.38 Ω A

16.67 V

B

Fig. 1.28

  example 1.4 

Find the power delivered by the 50 V source in the network of Fig. 1.29. 3Ω

5Ω 50 V

2Ω

10 A

Fig. 1.29

10 V

1.16 Circuit Theory and Transmission Lines Solution Converting the series combination of voltage source of 10 V and resistor of 3 W into equivalent current source and resistor (Fig. 1.30),

5Ω 2Ω

10 A

3Ω

3.33 A

50 V

Fig. 1.30 Adding the two current sources and simplifying the network (Fig. 1.31),

5Ω

1.2 Ω

13.33 A 50 V

Fig. 1.31 By source transformation (Fig. 1.32),

1.2 Ω

5Ω

I

50 V

16 V

Fig. 1.32 I=

50 − 16 = 5.48 A 5 + 1.2

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

  example 1.5 

Find the current in the 4 W resistor shown in network of Fig. 1.33. 6V

5A

2Ω

2A

4Ω

Fig. 1.33 Solution Converting the parallel combination of the current source of 5 A and the resistor of 2 W into an equivalent series combination of voltage source and resistor (Fig. 1.34),

1.11 Source Transformation 1.17 2Ω

6V

10 V 2A

4Ω

2A

4Ω

2A

4Ω

Fig. 1.34 Adding two voltage sources (Fig. 1.35), 2Ω

4V

Fig. 1.35 Again by source transformation (Fig. 1.36),

2A

2Ω

Fig. 1.36 Adding two current sources (Fig. 1.37),

4A

2Ω

4Ω

Fig. 1.37 By current-division rule, I4 W = 4 ×

  example 1.6 

2 = 1.33 A 2+4

Find the voltage across the 4 W resistor shown in network of Fig. 1.38. 3Ω

6V

2Ω

6Ω

Fig. 1.38

1Ω

3A

4Ω

1.18 Circuit Theory and Transmission Lines Solution Converting the series combination of the voltage source of 6 V and the resistor of 3 W into equivalent current source and resistor (Fig. 1.39), 2Ω

6Ω

3Ω

2A

1Ω

4Ω

3A

Fig. 1.39 By series–parallel reduction technique (Fig. 1.40), 2Ω

1Ω

2Ω

2A

3A

4Ω

Fig. 1.40 By source transformation (Fig. 1.41), 2Ω

1Ω

1Ω

2Ω

4Ω

4Ω

3A

4Ω

3A

4V

4V (a)

(b)

1Ω

1Ω

4Ω

1Ω I

1A

4Ω

3A

4Ω

4Ω 4A

(c)

(d)

Fig. 1.41 I=

16 = 1.78 A 4 +1+ 4

Voltage across the 4 W resistor = 4 I = 4 × 1.78 = 7.12 V

4Ω

4 Ω 16 V

(e)

1.12 Source Shifting 1.19

  example 1.7 

Find the voltage at Node 2 of the network shown in Fig. 1.42. 50 Ω

1

2 I

100 Ω 100 Ω

15 V + 10 I −

Fig. 1.42 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 W resistor (Fig. 1.43), 1

I

50 Ω

1 I

2

15 V

0.1 I

100 Ω

1

I

100 Ω

50 Ω

50 Ω

2

0.1 I

15 V

2

50 Ω

50 Ω

+ −

15 V

5I

Fig. 1.43 Applying KVL to the mesh, 15 − 50 I − 50 I − 5 I = 0 I=

15 = 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.

1.20 Circuit Theory and Transmission Lines

  example 1.8 

Calculate the voltage across the 6 W resistor in the network of Fig. 1.44 using source-

shifting technique. 3Ω 2

4Ω

1Ω

3

1

4

2Ω

18 V

+ Va

6Ω

−

Fig. 1.44 Solution Adding a voltage source of 18 V to the network and connecting to Node 2 (Fig. 1.45), we have 2

3Ω

3

4Ω

1Ω

1

4

18 V 2Ω

6Ω

18 V

+ Va −

Fig. 1.45 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. 1.46). 18 V

18 V

3Ω

4Ω

1Ω + 2Ω

6Ω

Va −

Fig. 1.46

Exercises 1.21

Simplifying the network (Fig. 1.47), 18 V

18 V

3Ω

3Ω

4.5 A 1Ω

4.5 A 1Ω

1.33 Ω +

4Ω 2Ω

6Ω

Va −

6Ω −

18 V

+

Va

3Ω

5.985 V

Va

(b)

(a)

`

1.33 Ω

3Ω

1Ω

2.33 Ω

18 V

6Ω

5.985 V

6Ω −

+

Va (c)

(d)

Fig. 1.47 Applying KCL at the node, Va − 18 Va − 5.985 Va + + =0 3 2.33 6 Va = 9.23 V

Exercises 1.1

Use source transformation to simplify the network until two elements remain to the left of terminals A and B.

Determine the voltage Vx in the network of Fig. 1.49 by source-shifting technique.

1.2

3Ω

3.5 kΩ

6 kΩ

A 2Ω

1Ω Vx

2 kΩ

3 kΩ

20 mA

12 kΩ 2V

2Ω

5Ω

B

Fig. 1.48 [88.42 V, 7.92 k W]

Fig. 1.49 [1.129 V]

1.22 Circuit Theory and Transmission Lines

Objective-Type Questions 1.1

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) (b) (c) (d)

1.2

1.3

halved doubled increased by four times not changed

1.6

Four resistances 80 W, 50 W, 25 W, and R are connected in parallel. Current through 25 W resistor is 4 A. Total current of the supply is 10 A. The value of R will be (a) 66.66 W (b) 40.25 W (c) 36.36 W (d) 76.56 W Viewed from the terminal AB, the network of Fig. 1.50 can be reduced to an equivalent network of a single voltage source in series with a single resistor with the following parameters 5V

10 Ω

4Ω

nR

(b)

(c)

n2R

(d)

All the resistances in Fig. 1.51 are 1 W each. The value of I will be

Fig. 1.51 1 2 A A (b) 15 15 4 8 A A (c) (d) 15 15 The current waveform in a pure resistor at 10 W is shown in Fig. 1.52. Power dissipated in the resistor is (a)

1.7

Fig. 1.50

i

(a) 5 V source in series with a 10 W resistor (b) 1 V source in series with a 2.4 W resistor (c) 15 V source in series with a 2.4 W resistor 1.4

1.5

9

1 V source in series with a 10 W resistor

A 10 V battery with an internal resistance of 1 W is connected across a nonlinear load whose V-I characteristic is given by 7 I = V 2 + 2V . The current delivered by the battery is (a) 0 (b) 10 A (c) 5 A (d) 8 A If the length of a wire of resistance R is uniformly stretched to n times its original value, its new resistance is

n2

1V

B

(d)

R n R

I

A 10 V

(a)

0

3

t 6

Fig. 1.52 (a) (c) 1.8

7.29 W 135 W

(b) (d)

52.4 W 270 W

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) (c)

1:1 1:4

(b) (d)

1:2 1:8

Answers to Objective-Type Questions 1.23

Answers to Objective-Type Questions 1.1. (d) 1.8. (b)

1.2. (c)

1.3. (b)

1.4. (c)

1.5. (c)

1.6. (d)

1.7. (d)

2

Elementary Network Theorems

  2.1     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

I4

Fig. 2.1  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 Circuit Theory and Transmission Lines 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

+

−

(a) Fall in potential

−

+

I

(b) Rise in potential

Fig. 2.3  Sign convention

  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 4. Solve the simultaneous equations for unknown mesh V1 I1 I2 currents. Consider the network shown in Fig. 2.4 which has three meshes. R2 R4 Let the mesh currents for the three meshes be I1, I2, and I3 and all I3 the three mesh currents may be assumed to flow in the clockwise R5 direction. The choice of direction for any mesh current is arbitrary. V3

R3

Fig. 2.4  Circuit for mesh analysis

Applying KVL to Mesh 1, V1 − R1 ( I1 − I 2 ) − R2 ( I1 − I 3 ) = 0 ( R1 + R2 ) I1 − R1 I 2 − R2 I 3 = V1

…(2.1)

2.3  Mesh Analysis  2.3

Applying KVL to Mesh 2, V2 − R3 I 2 − R4 ( I 2 − I 3 ) − R1 ( I 2 − I1 ) = 0 − R1 I1 + ( R1 + R3 + R4 ) I 2 − R4 I 3 = V2

…(2.2)

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

…(2.3)

Writing Eqs (2.1), (2.2), and (2.3) in matrix form,  R1 + R2  − R1   − R2

− R1 R1 + R3 + R4 − R4

− R2   I1  V1    I 2  = V2  − R4     R2 + R4 + R5   I 3  V3 

In general,  R11  R21   R31

R12 R22 R32

R13   I1  V1  R23   I 2  = V2      R33   I 3  V3 

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 2.1 

Find the current through the 5 W resistor is shown in Fig. 2.5. 1Ω

2Ω 3Ω

5Ω

10 V 6Ω

5V

4Ω

20 V

Fig. 2.5 Solution Assigning clockwise currents in three meshes as shown in Fig. 2.6. Applying KVL to Mesh 1, 10 − 1I1 − 3 ( I1 − I 2 ) − 6 ( I1 − I 3 ) = 0 10 I1 − 3 I 2 − 6 I 3 = 10

…(i)

2.4 Circuit Theory and Transmission Lines Applying KVL to Mesh 2, 1Ω

−3 ( I 2 − I1 ) − 2 I 2 − 5 I 2 − 5 = 0

2Ω

…(ii)

−3 I1 + 10 I 2 = −5

5Ω

3Ω

Applying KVL to Mesh 3,

I2

−6 ( I 3 − I1 ) + 5 − 4 I 3 + 20 = 0 −6 I1 + 10 I 3 = 25

…(iii)

10 V 5V

I1

Writing Eqs (i), (ii) and (iii) in matrix form,

6Ω

4Ω I3

10 −3 −6   I1  10   −3 10 0   I 2  =  −5       −6 0 10   I 3   25 

20 V

Fig. 2.6

We can write matrix equation directly from Fig. 2.6,  R11  R21   R31

R12 R22 R32

R13   I1  V1  R23   I 2  = V2      R33   I 3  V3 

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 where

R22 R23 R31 R32 R33

= 3 + 2 + 5 = 10 Ω =0 = −6 Ω =0 = 6 + 4 = 10 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 I 5 Ω = I 2 = 0.78 A

  Example 2.2 

Find the current through the 2 W resistor of the network shown in Fig. 2.7. 6Ω

10 V

2Ω

1Ω

3Ω

20 V

Fig. 2.7

10 Ω

2.3  Mesh Analysis  2.5

Solution Assigning clockwise currents in three meshes as shown in Fig. 2.8, Applying KVL to Mesh 1, 6Ω 10 − 6 I1 − 1( I1 − I 2 ) = 0 …(i) 7 I1 − I 2 = 10 10 V

2Ω

1Ω

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

3Ω I2

I1

− I1 + 6 I 2 − 3 I 3 = 0

10 Ω I3

20 V

…(ii) Fig. 2.8

Applying KVL to Mesh 3, −3 ( I 3 − I 2 ) − 10 I 3 − 20 = 0 −3 I 2 + 13 I 3 = −20

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 1.34 A I 2 = −0.62 A I 3 = −1.68 A I 2 Ω = I 2 = −0.62 A

  Example 2.3 

Determine the current through the 5 W resistor of the network shown in Fig. 2.9. 10 V 1Ω

8V

2Ω

4Ω

3Ω 5Ω

12 V

Fig. 2.9 Solution Assigning clockwise currents in three meshes as shown in Fig. 2.10. Applying KVL to Mesh 1, 8 − 1( I1 − I 2 ) − 2 ( I1 − I 3 ) = 0 3 I1 − I 2 − 2 I 3 = 8

…(i)

10 V

8V

I2

I1

Applying KVL to Mesh 2, 10 − 4 I 2 − 3 ( I 2 − I 3 ) − 1( I 2 − I1 ) = 0 − I1 + 8 I 2 − 3 I 3 = 10

2Ω

12 V

−2 ( I 3 − I1 ) − 3 ( I 3 − I 2 ) − 5 I 3 + 12 = 0 −2 I1 − 3 I 2 + 10 I 3 = 12 Solving Eqs (i), (ii), and (iii), I1 = 6.01 A I 2 = 3.27 A

3Ω I3

…(ii)

Applying KVL to Mesh 3,

5Ω

Fig. 2.10 …(iii)

4Ω

1Ω

2.6 Circuit Theory and Transmission Lines I 3 = 3.38 A I 5 Ω = I 3 = 3.38 A

  Example 2.4 

Find the current supplied by the battery of the network shown in Fig. 2.11. 3Ω 1Ω I2

2Ω

4V I1

5Ω I3

4Ω

6Ω

Fig. 2.11 Solution Applying KVL to Mesh 1, 4 − 3 I1 − 1( I1 − I 2 ) − 4 ( I1 − I 3 ) = 0 8 I1 − I 2 − 4 I 3 = 4

…(i)

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

…(ii)

Applying KVL to Mesh 3, −6 I 3 − 4 ( I 3 − I1 ) − 5 ( I 3 − I 2 ) = 0 −4 I1 − 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 2.5 

Find the current through the 4 W resistor in the network of Fig. 2.12. 2Ω

2Ω 2Ω 6V

I2 4Ω

8V I1 2Ω

2Ω I3

Fig. 2.12 Solution Applying KVL to Mesh 1, 8 − 2 I1 − 2 ( I1 − I 2 ) − 2 ( I1 − I 3 ) = 0 6 I1 − 2 I 2 − 2 I 3 = 8

…(i)

2.3  Mesh Analysis  2.7

Applying KVL to Mesh 2, −2 ( I 2 − I1 ) − 2 I 2 − 4 ( I 2 − I 3 ) − 6 = 0 −2 I1 + 8 I 2 − 4 I 3 = −6

…(ii)

Applying KVL to Mesh 3, −2 ( I 3 − I1 ) + 6 − 4 ( I 3 − I 2 ) − 2 I 3 = 0 −2 I1 − 4 I 2 + 8 I 3 = 6

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = 2 A I 2 = 0.5 A I 3 = 1.5 A I 4 Ω = I 3 − I 2 = 1.5 − 0.5 = 1 A

  Example  2.6 

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

of Fig. 2.13. 6Ω 2Ω I 2

20 V

3Ω

1Ω I1

5Ω V

4Ω I3

Fig. 2.13 Solution Applying KVL to Mesh 1, 20 − 6 I1 − 2 ( I1 − I 2 ) − 5 ( I1 − I 3 ) − V = 0 V + 13 I1 − 2 I 2 − 5 I 3 = 20

…(i)

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

…(ii)

Applying KVL to Mesh 3, −1( I 3 − I 2 ) − 4 I 3 + V − 5 ( I 3 − I1 ) = 0 V + 5 I1 + I 2 − 10 I 3 = 0

…(iii)

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 I 3 = 0 Solving Eqs (i), (ii) and (iii), V = 43.7 V

…(iv)

2.8 Circuit Theory and Transmission Lines

  Example 2.7 

Find the current through the 2 W resistor in the network of Fig. 2.14. 3Ω

12 Ω

6A I1

36 V

2Ω

6Ω I2

I3

9V

Fig. 2.14 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) Applying KVL to Mesh 2, 36 − 12 ( I 2 − I1 ) − 6 ( I 2 − I 3 ) = 0 36 − 12 ( I 2 − 6) − 6 I 2 + 6 I 3 = 0 18 I 2 − 6 I 3 = 108

…(ii)

Applying KVL to Mesh 3, −6 ( I 3 − I 2 ) − 3 I 3 − 2 I 3 − 9 = 0 6 I 2 − 11 I 3 = 9

…(iii)

Solving Eqs (ii) and (iii), I3 = 3 A I 2 Ω = I3 = 3 A

 Example 2.8 

Determine the mesh currents I1, I2 and I3 in the network of Fig. 2.15. 30 V I1

15 Ω

I2 1A

I3

6Ω 50 V

10 Ω 5 Ω

Fig. 2.15 Solution Applying KVL to Mesh 1, −30 − 6 I1 − 15 ( I1 − I 2 ) = 0 21 I1 − 15 I 2 = −30

…(i)

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

…(ii)

For Mesh 3, I3 = 1 Solving Eqs (i), (ii) and (iii),

…(iii)

2.3  Mesh Analysis  2.9

I1 = 0 I2 = 2 A I3 = 1 A

 Example 2.9 

Find the current through the 5 W resistor in the network of Fig. 2.16. 3A

2Ω

4A I1

I2

5Ω I3

2Ω 2V

3A I4

Fig. 2.16 Solution Writing current equations for Meshes 1, 2 and 4, I1 = 4 I2 = 3

… (i) … (ii)

I 4 = −3

…(iii)

−5 ( I 3 − I1 ) − 2 ( I 3 − I 2 ) − 2 ( I 3 − I 4 ) − 2 = 0

…(iv)

Applying KVL to Mesh 3,

Substituting Eqs (i), (ii) and (iii) in Eq. (iv), −5 ( I 3 − 4) − 2 ( I 3 − 3) − 2 ( I 3 + 3) − 2 = 0 I3 = 2 A I 5 Ω = I1 − I 3 = 4 − 2 = 2 A

EXAMPLES WItH dEPEndEnt SourcES    Example 2.10 

Obtain the branch currents in the network shown in Fig. 2.17. IA

5Ω

10IB

5Ω

IB

+ − 10 Ω

5V

10 V

+ −

5IA

Fig. 2.17 Solution We assign clockwise currents in two meshes as shown in Fig. 2.18.

2.10 Circuit Theory and Transmission Lines From Fig. 2.18, 5Ω

IA

10IB

5Ω

+ −

IB

10 Ω 5V

10 V

+ −

I1

I2

5IA

Fig. 2.18 I A = I1 I B = I2

…(i) …(ii)

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

…(iii)

Applying KVL to Mesh 2, 5 I A − 10 ( I 2 − I1 ) − 5 I 2 − 10 = 0 5 I1 − 10 I 2 + 10 I1 − 5 I 2 = 10 15 I1 − 15 I 2 = 10

…(iv)

Putting I1 = 0.25 A in Eq. (iv), 15 (0.25) − 15 I 2 = 10 I 2 = −0.416 A

 Example 2.11 

Find the mesh currents in the network shown in Fig. 2.19. 2V2

4Ω

2Ω

+ − 5Ω

+ V2 −

+ V1 −

1Ω 10 V

− 2V1 +

5V

Fig. 2.19 Solution Assigning clockwise currents in the two meshes as shown in Fig. 2.20, From Fig. 2.20, …(i) V1 = −5 I1 …(ii) V2 = 2 I 2 Applying KVL to Mesh 1,

5Ω 5V

2V2 + −

4Ω

2Ω

+ V2 −

+ V1 −

1Ω 10 V I1

− 2V1 +

Fig. 2.20

I2

2.3  Mesh Analysis  2.11

−5 − 5 I1 − 2V2 − 4 I1 − 1( I1 − I 2 ) + 2V1 = 0 −5 − 5 I1 − 2 ( 2 I 2 ) − 4 I1 − I1 + I 2 + 2 ( −5 I1 ) = 0 20 I1 + 3 I 2 = −5

…(iii)

Applying KVL to Mesh 2, −2V1 − 1( I 2 − I1 ) − 2 I 2 − 10 = 0 −2 ( −5 I1 ) − I 2 + I1 − 2 I 2 = 10 11 I1 − 3 I 2 = 10

…(iv)

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

 Example 2.12 

Find currents Ix and Iy of the network shown in Fig. 2.21. 2Ix

4 Ω Iy

+ − 5Ω

2Ω

1Ω 10 V

− 2Iy + Ix

5V

Fig. 2.21 Solution Assigning clockwise currents in the two Meshes as shown in Fig. 2.22. From Fig. 2.22, 2Ix 4Ω I y

+ −

I y = I1

...(i)

I x = I1 − I 2

...(ii)

Applying KVL to Mesh 1,

5Ω

2Ω

1Ω 10 V

5V

− 2Iy + Ix

I1

−5 − 5 I1 − 2 I x − 4 I1 − 1( I1 − I 2 ) + 2 I y = 0 −5 − 5 I1 − 2 ( I1 − I 2 ) − 4 I1 − I1 + I 2 + 2 I1 = 0

I2

Fig. 2.22

−5 − 5 I1 − 2 I1 + 2 I 2 − 4 I1 − I1 + I 2 + 2 I1 = 0 −10 I1 + 3 I 2 = 5 Applying KVL to Mesh 2,

…(iii)

−2 I y − 1( I 2 − I1 ) − 2 I 2 − 10 = 0 −2 I1 − I 2 + I1 − 2 I 2 = 10 − I1 − 3 I 2 = 10 Solving Eqs (iii) and (iv), 15 = −1.364 A 11 I 2 = −2.878 A I1 = −

I y = −1.364 A I x = I1 − I 2 = −1.364 + 2.878 = 1.514 A

...(iv)

2.12 Circuit Theory and Transmission Lines

 Example 2.13 

Find the currents in the three meshes of the network shown in Fig. 2.23. Ix

1Ω

1Ω

+ −

1Ω

Ix

+ I − y 5V

1A

1Ω 1Ω

Iy

Fig. 2.23 Solution Assigning clockwise currents in the three meshes is shown in Fig. 2.24. Ix

1Ω

1Ω

+ −

1Ω

Ix

+ I − y 5V

1A

1Ω I1

1Ω

I2

I3

Iy

Fig. 2.24 From Fig. 2.24, Ix = I1 I x = I1 I y = I 2 − I3

...(i) …(ii)

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

…(iii)

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

…(iv)

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

…(v)

2.3  Mesh Analysis  2.13

 Example 2.14 

For the network shown in Fig. 2.25, find the power supplied by the dependent voltage

source. 50 Ω 20 Ω

5A

+ V1 −

30 Ω

+ −

0.4 V1

0.01 V1

Fig. 2.25 Solution Assigning clockwise currents in three meshes as shown in Fig. 2.26. 50 Ω

5A

+ V1 −

20 Ω

I3

I1

+ −

30 Ω

0.4 V1

I2

0.01 V1

Fig. 2.26 From Fig. 2.26, V1 − 20 ( I1 − I 3 ) − 0.4 V1 = 0 0.6 V1 = 20 I1 − 20 I 3 V1 = 33.33 I1 − 33.33 I 3

…(i)

I1 = 5

…(ii)

For Mesh 1, For Mesh 2, I 2 = −0.01 V1 = − 0.01(33.33 I1 − 33.33 I 3 ) 0.33 I1 + I 2 − 0.33 I 3 = 0

…(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 (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

2.14 Circuit Theory and Transmission Lines

 Example 2.15 

Find the voltage Vx in the network shown in Fig. 2.27. 0.45 A

16.67 Ω

33.33 Ω

+ Vx − 25 Ω

30 V

− +

2 Vx

10 V

`

Fig. 2.27 Solution Assigning clockwise currents in the three meshes as shown in Fig. 2.28. 0.45 A

16.67 Ω

I3 33.33 Ω

+ Vx −

25 Ω

− +

30 V I1

10 V

2 Vx

I2

Fig. 2.28 From Fig. 2.28, Vx = 16.67 I1 Applying KVL to Mesh 1, −30 − 16.67 I1 − 33.33 ( I1 − I 3 ) − 25 ( I1 − I 2 ) − 10 = 0 −30 − 16.67 I1 − 33.33 I1 + 33.33 I 3 − 25 I1 + 25 I 2 − 10 = 0 −75 I1 + 25 I 2 + 33.33 I 3 = 40

…(i)

…(ii)

Applying KVL to Mesh 2, 10 − 25 ( I 2 − I1 ) + 2 Vx = 0 10 − 25 ( I 2 − I1 ) + 2 (16.67 I1 ) = 0 10 − 25 I 2 + 25 I1 + 33.34 I1 = 0 58.34 I1 − 25 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 = 16.67 I1 = 16.67 ( −0.9) = −15 V

…(v)

2.3  Mesh Analysis  2.15

 Example 2.16 

For the network shown in Fig. 2.29, find the mesh currents I1, I2 and I3. 15 A I3 2Ω

0.111 Vx

1Ω

+ Vx −

I1

3Ω

1Ω

I2 2Ω

Fig. 2.29 Solution From Fig. 2.29, Vx = 3 ( I1 − I 2 )

…(i)

Writing current equation for the two current sources, I 3 = 15

…(ii)

0.111 Vx = I1 − I 3 0.111 [3 ( I1 − I 2 ) = I1 − I 3

and

0.333 I1 − 0.333 I 2 − I1 + I 3 = 0 −0.667 I1 − 0.333 I 2 + I 3 = 0

…(iii)

Applying KVL to Mesh 2, −3 ( I 2 − I1 ) − 1( I 2 − I 3 ) − 2 I 2 = 0 −3 I1 + 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 2.17  For the network shown in Fig. 2.30, 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

Fig. 2.30

2Vx

2Ω

RL = 2 Ω

2.16 Circuit Theory and Transmission Lines Solution Assigning clockwise currents in meshes is shown in Fig. 2.31. 10 Ω

5Ω

5Ω

+ Vx − V0

2Ω I1

2Ω

2Vx

4Ω I2

I3

RL = 2 Ω I4

Fig. 2.31 From Fig. 2.31, V x = 5 I1

…(i)

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)

−2 ( I 2 − I1 ) − 4 I 2 = 0 −2 I1 + 6 I 2 = 0

…(iv)

Applying KVL to Mesh 2,

For Mesh 3, I 3 = 2Vx = 2 (5 I1 ) = 10 I1 10 I1 − I 3 = 0 Applying KVL to Mesh 4, −2 ( I 4 − I 3 ) − 5 I 4 − 2 I 4 = 0 −2 I 3 + 9 I 4 = 0 −2 I 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 I 2 = 0 I 2 = 0.45 A From Eq. (iii), 7 (1.35) − 2 (0.45) = V0 V0 = 8.55 V Current supplied by voltage source V0 = I1 = 1.35 A

…(v)

2.4  Supermesh Analysis  2.17

 Example 2.18 

In the network shown in Fig. 2.32, find voltage V2 such that Vx = 0. + − 0.1 Vx

2A 10 Ω

24 V

3A

5Ω + Vx −

20 Ω

V2

Fig. 2.32 Solution Assigning clockwise currents in four meshes as shown in Fig. 2.33. From Fig. 2.33, …(i) Vx = 20 ( I 3 − I 4 ) Writing current equations for Meshes 1 and 2, I1 = 2 I2 = 3 Applying KVL to Mesh 3, 24 − 10 ( I 3 − I1 ) − 20 ( I 3 − I 4 ) = 0 24 − 10 ( I 3 − 2) − 20 ( I 3 − I 4 ) = 0 −30 I 3 + 20 I 4 = −44 Applying KVL to Mesh 4, −20 ( I 4 − I 3 ) − 5 ( I 4 − I 2 ) + V2 = 0 −20 ( I 4 − I 3 ) − 5 ( I 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 V x −

2A I1 10 Ω

…(ii) …(iii) 24 V

I3

3A

5Ω + Vx −

V2

20 Ω I4

Fig. 2.33 …(iv)

…(v)

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) = − 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.

2.18 Circuit Theory and Transmission Lines

 Example 2.19 

Find the current through the 10 W resistor of the network shown in Fig. 2.34. 5Ω

1Ω 2V

10 Ω I1

15 Ω

4A I2

I3

Fig. 2.34 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 I3 − I 2 = 4

…(ii)

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

…(iii)

Solving Eqs (i), (ii) and (iii), I1 = −2.35 A I 2 = −2.78 A I 3 = 1.22 A

…(iv)

Current through the 10 Ω resistor = I1 − I2 = −(2.35) − (−2.78) = 0.43 A

 Example 2.20 

Find the current in the 3 W resistor of the network shown in Fig. 2.35.

2Ω

1Ω

I2

7V

I1

7A

3Ω 1Ω

2Ω

Fig. 2.35 Solution Meshes 1 and 3 will form a supermesh. Writing current equation for the supermesh,

I3

2.4  Supermesh Analysis  2.19

I1 − I 3 = 7

…(i)

Applying KVL to the outer path of the supermesh, 7 − 1( I1 − I 2 ) − 3 ( I 3 − I 2 ) − 1I 3 = 0 − I1 + 4 I 2 − 4 I 3 = −7

…(ii)

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

…(iii)

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

 Example 2.21 

Find the current in the 5 W resistor of the network shown in Fig. 2.36.

10 Ω

50 V

I1

2Ω

I2

3Ω

2A

1Ω 5Ω

I3

Fig. 2.36 Solution Applying KVL to Mesh 1, 50 − 10 ( I1 − I 2 ) − 5 ( I1 − I 3 ) = 0 15 I1 − 10 I 2 − 5 I 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 ( I 2 − I1 ) − 2 I 2 − 1I 3 − 5 ( I 3 − I1 ) = 0 −15 I1 + 12 I 2 + 6 I 3 = 0 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

…(iii)

2.20 Circuit Theory and Transmission Lines

 Example 2.22 

Determine the power delivered by the voltage source and the current in the 10 W resistor of the network shown in Fig. 2.37. 5Ω 3A

5Ω

I2 10 Ω

50 V

I1 1Ω

10 A I3

Fig. 2.37 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 I1 − 5 I 2 − 10 ( I 2 − I 3 ) − 1( I1 − I 3 ) = 0 −6 I1 − 15 I 2 + 11 I 3 = −50

…(ii)

For Mesh 3, I 3 = 10

…(iii)

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

 Example 2.23 

For the network shown in Fig. 2.38, find current through the 8 W resistor. 6Ω I2

2Ω

4Ω

11 Ω

8Ω I4

3A

5V I1

7A I3

12 A 10 V

Fig. 2.38 Writing current equations for meshes 1 and 4, I1 = −3 I 4 = −12

...(i) ...(ii)

Meshes 2 and 3 will form a supermesh. Writing current equation for the supermesh, I3 − I 2 = 7

…(iii)

2.4  Supermesh Analysis  2.21

Applying KVL to the outer path of the supermesh, 5 − 4 ( I 2 − I1 ) − 6 I 2 − 8 ( I 3 − I 4 ) + 10 = 0 5 − 4 ( I 2 + 3) − 6 I 2 − 8 ( I 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 8 Ω = I 3 − I 4 = −1.28 + 12 = 10.72 A

EXAMPLES WItH dEPEndEnt SourcES  Example 2.24 

In the network of Fig. 2.39, find currents I1 and I2. 8Ω

3V0

V0

+ − 2Ω

−10 V

10 Ω

−3 A I 2

I1

Fig. 2.39 Solution From Fig. 2.39, −10 − 8 I1 − V0 = 0 V0 = −10 − 8 I1

…(i)

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

…(ii)

Applying KVL to the outer path of the supermesh, −10 − 8 I1 − 3V0 − 10 I 2 = 0 −10 − 8 I1 − 3 ( −10 − 8 I1 ) − 10 I 2 = 0 16 I1 − 10 I 2 = −20

…(iii)

Solving Eqs (ii) and (iii), I1 = −8.33 A I 2 = −11.33 A

 Example 2.25 

In the network of Fig. 2.40, find the current through the 3 W resistor. −4 V

2Ω

+ Vx −

3Ω 1Ω 2A

Fig. 2.40

+ −

5 Vx

2.22 Circuit Theory and Transmission Lines Solution Assigning clockwise currents in two meshes as shown in Fig. 2.41. From Fig. 2.41, −4 V Vx = −2 I1 …(i) Meshes 1 and 2 will form a supermesh. Writing current equations for the supermesh,

2Ω

I 2 − I1 = 2

…(ii)

3Ω 1Ω

+ Vx −

2A

I1

Applying KVL to the outer path of the supermesh, −2 I1 − 4 − 3 I 2 − 5 ( −2 I1 ) = 0 8 I1 − 3 I 2 = 4

…(iii)

Solving Eqs (ii) and (iii), I1 = 2 A I2 = 4 A I3 Ω = I 2 = 4 A Find the currents I1 and I2 at the network shown in Fig. 2.42. 10 Ω 14 Ω

4Ω I3

110 V

I2

Fig. 2.41

−2 I1 − 4 − 3 I 2 − 5Vx = 0

 Example 2.26 

+ −

2Ω

+ V1 −

I1

6Ω

0.5 V1 I2

Fig. 2.42 Solution From Fig. 2.42, V1 = 2 (I1 − I2) Meshes 2 and 3 will form a supermesh. Writing current equation for the supermesh, I 3 − I 2 = 0.5 V1 = 0.5 × 2 ( I1 − I 2 ) = I1 − I 2 I 3 = I1 Applying KVL to outer path of the supermesh, −2 ( I 2 − I1 ) − 10 I 3 − 6 I 2 = 0 −2 I 2 + 2 I1 − 10 I1 − 6 I 2 = 0 I1 = − I 2 Applying KVL to Mesh 1, 110 − 14 I1 − 4 I1 − 2 ( I1 − I 2 ) = 0 110 − 20 I1 + 2 I 2 = 0 110 + 20 I1 + 2 I 2 = 0 I 2 = −5 A I1 = − I 2 = 5 A

5 Vx

2.4  Supermesh Analysis  2.23

 Example 2.27 

For the network of Fig. 2.43, find current through the 8 W resistor. 5 Iy

+ −

10 Ω

6Ω

Iy 8Ω

50 V Ix

52 V

0.5 Ix

Fig. 2.43 Solution Assigning clockwise currents to the three meshes as shown in Fig. 2.44. From Fig. 2.44, I x = − I1

…(i)

I y = I 2 − I3

…(ii)

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

5 Iy + −

10 Ω

I3

Iy

6Ω

8Ω

50 V Ix

I 2 − I1 = 0.5 I x = 0.5 ( − I1 ) −0.5 I1 + I 2 = 0

0.5 Ix

I1

52 V I2

Fig. 2.44

…(iii)

Applying KVL to the outer path of the supermesh, 50 − 10 ( I1 − I 3 ) − 6 ( I 2 − I 3 ) − 8 I 2 − 52 = 0 −10 I1 − 14 I 2 + 16 I 3 = 2

…(iv)

Applying KVL to Mesh 3, −5 I y − 6 ( I 3 − I 2 ) − 10 ( I 3 − I1 ) = 0 −5 ( I 2 − I 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 I 8 Ω = I 2 = −0.58 A

 Example 2.28 

For the network shown in Fig. 2.45, find the current through the 10 W resistor. 10 Ω 15 V

20 V 5 Ω

2A I1

4Ω

2 I1 I2

Fig. 2.45

40 V I3

2.24 Circuit Theory and Transmission Lines Solution Meshes 1, 2 and 3 will form a supermesh. Writing current equations for the supermesh, I1 − I 2 = 2

...(i)

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

...(ii)

…(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 2.29  In the network shown in Fig. 2.46, 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

V2 2

Fig. 2.46 Solution From Fig. 2.46, V2 = 2 I1

…(i)

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, 4 − 5 I 2 − 2 I1 − 6 I1 − 4 ( I1 − I 3 ) − 1( I 2 − I 3 ) = 0 −12 I1 − 6 I 2 + 5 I 3 = −4

…(ii)

…(iii)

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

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

…(iv)

2.4  Supermesh Analysis  2.25

Power delivered by the 4 V source = 4 I2 = 4 (3) = 12 W V2 Ω = 2 I1 = 2 ( −2) = −4 V

 Example 2.30 

Find currents I1, I2, I3, I4 of the network shown in Fig. 2.47. 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. 2.47 From Fig. 2.47, 1 Vx = ( I 2 − I1 ) 5

…(i)

I 4 = 40

…(ii)

For Mesh 4, 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

…(iii)

Mesh 2 and 3 will form a supermesh. Writing current equation for the supermesh, 1  I 3 − I 2 = 5 Vx = 5  ( I 2 − I1 )  = I 2 − I1 5  I1 − 2 I 2 + I 3 = 0

…(iv)

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

…(v)

2.26 Circuit Theory and Transmission Lines 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 2.31 

Calculate the current through 2 W resistor for the network shown in Fig. 2.48. VA 1Ω

0.5 Ω 1Ω

VB

2Ω

20 V

1Ω 20 V

Fig. 2.48 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node A, VA − 20 VA VA − VB + + =0 1 1 0.5 1 20 1 1 1  VB =  + +  VA − 1 1 0.5  0.5 1 4 VA − 2 VB = 20

…(i)

Applying KCL at Node B, VB − VA VB VB − 20 + + =0 0.5 2 1 1 20  1 1 1 − VA  + +  VB = 0.5  0.5 2 1 1 −2 VA + 3.5 VB = 20 Solving Eqs (i) and (ii),

…(ii)

2.5  Node Analysis  2.27

VA = 11 V VB = 12 V Current through the 2 Ω resistor =

 Example 2.32 

VB 12 = =6A 2 2

Find the voltage at nodes 1 and 2 for the network shown in Fig. 2.49. 1Ω

2Ω

V1

V2 2

1 2Ω

1A

1Ω

2A

Fig. 2.49 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  +  V1 − V2 = 1 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 −0.5V1 + 1.5 V2 = 4

…(ii)

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

 Example 2.33 

Find the current in the 100 W resistor for the network shown in Fig. 2.50. 20 Ω

60 V

V1

30 Ω

1A

V2

50 Ω 40 V

Fig. 2.50

100 Ω

2.28 Circuit Theory and Transmission Lines 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 +1  +  V1 − V2 = 20 30 30 20 0.083 V1 − 0.033 V2 = 4

…(i)

Applying KCL at Node 2, V2 − V1 V 2 − 40 V2 + + =0 30 50 100 1 1 1  40  1 − V1 +  + + V2 =  30 50 100  30 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 2.34 

48 V2 = = 0.48 A 100 100

Find VA and VB for the network shown in Fig. 2.51. 1Ω

VA

2A

2Ω

2Ω

VB

2Ω

1A

2V

1V

Fig. 2.51 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node A, VA VA − 1 VA − VB + + 2 2 1 1 1 1   + + 1 VA − VB = 2 + 2 2 2 2VA − VB = 2.5 2=

Applying KCL at Node B,

…(i)

2.5  Node Analysis  2.29

VB − VA VB − 2 + =1 1 2 2  1 −VA + 1 +  VB = 1 +  2 2 −VA + 1.5 VB = 2

…(ii)

Solving Eqs (i) and (ii), VA = 2.875 V V B = 3.25 V

 Example 2.35 

Find currents I1, I2 and I3 for the network shown in Fig. 2.52. 10 Ω

V1

2Ω

V2

5Ω 2Ω

50 V

4Ω I1

25 V

I2

I3

Fig. 2.52 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   + +  V1 − V2 = 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,

Solving Eqs (i) and (ii), V1 = 2.61 V V2 = −29.1 V V 2.61 = −1.31 A I1 = − 1 = − 2 2 V − V2 2.61 − ( −2.91) = = 3.17 A I2 = 1 10 10 V + 50 −29.1 + 50 I3 = 2 = = 10.45 A 2 2

2.30 Circuit Theory and Transmission Lines

 Example 2.36 

Find currents I1, I2 and I3 and voltages Va and Vb for the network shown in Fig. 2.53. I1

0.2 Ω

I2 0.3 Ω

I3 0.1 Ω + b

+ a Va

120 V

Vb −

−

110 V 20 A

30 A

Fig. 2.53 Solution Applying KCL at Node a, I1 120 − Va 0.2 36 − 0.3 Va 0.5 Va − 0.2 Vb

= 30 + I 2 V − Vb = 30 + a 0.3 = 1.8 + 0.2Va − 0.2 Vb = 34.2

…(i)

Applying KCL at Node b, I 2 + I 3 = 20 Va − Vb 110 − Vb + = 20 0.3 0.1 0.1 Va − 0.1 Vb + 33 − 0.3 Vb = 20 0.03 0.1 Va − 0.4 Vb = −32.4

…(ii)

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 = = 10 A I2 = a 0.3 0.3 110 − Vb 110 − 109 = = 10 A I3 = 0.1 0.1 I1 =

 Example 2.37 

Calculate the current through the 5 W resistor for the network shown in Fig. 2.54. 3Ω

V1

4A

5Ω

V2

V3

2Ω 4Ω

2Ω 20 V

Fig. 2.54

8A

2.5  Node Analysis  2.31

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  +  V1 − V2 = 4 2 3 3 4+

0.83 V1 − 0.33 V2 = −4

…(i)

Applying KCL at Node 2, V2 − V1 V2 − ( −20) V2 − V3 + + =0 3 2 5 1 1 20  1 1 1 − V1 +  + +  V2 − V3 = −  3 2 5 3 5 2 −0.33 V1 + 1.03 V2 − 0.2 V3 = −10

…(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 =

 Example 2.38 

V3 − V2 13.37 − ( −9.92) = = 4.66 A 5 5

Find the voltage across the 5 W resistor for the network shown in Fig. 2.55. 9A

4Ω V1

4Ω

2Ω

V2

5Ω

100 Ω

12 V

Fig. 2.55

V3

20 Ω

2.32 Circuit Theory and Transmission Lines Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V1 − 12 V1 − V2 V1 − V3 + + +9= 0 4 2 4 1 1 12  1 1 1  + +  V1 − V2 − V3 = −9 + 4 2 4 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 1 1 − V1 +  + + V2 − V3 = 0  2 100 5  2 5 −0.5 V1 + 0.71 V2 − 0.2 V3 = 0

…(ii)

Applying KCL at Node 3, V3 − V2 V3 V3 − V1 + + =9 5 20 4 1 1  1 1 1 − V1 − V2 +  + + V3 = 9  5 20 4  4 5 −0.25 V1 − 0.2 V2 + 0.5 V3 = 9

…(iii)

Solving Eqs (i), (ii) and (iii), V1 = 6.35 V V2 = 11.76 V V3 = 25.88 V Voltages across the 5 Ω resistor = V3 − V2 = 25.88 − 11.76 = 14.12 V

 Example 2.39 

Determine the current through the 5 W resistor for the network shown in Fig. 2.56. 36 V

2Ω

V1

3A

4Ω

4Ω 5Ω

V2

100 Ω

Fig. 2.56

V3

20 Ω

2.5  Node Analysis  2.33

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  + +  V1 − V2 − V3 = 3 + 4 2 4 2 4 4 V1 − 0.5 V2 − 0.25 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 + 0.71 V2 − 0.2 V3 = 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 V1 − 0.2 V2 + 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 =

 Example 2.40 

V2 − V3 7.06 − ( −8.47) = = 3.11 A 5 5

Find the voltage drop across the 5 W resistor in the network shown in Fig. 2.57. 4Ω

5Ω V1

1A

2A

V2

1Ω

Fig 2.57

V3

2Ω

2.34 Circuit Theory and Transmission Lines 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  1 1  +  V1 − V2 − V3 = −1 5 4 5 4 0.45 V1 − 0.2 V2 − 0.2 25 V3 = −1

…(i)

Applying KCL at Node 2, V2 − V1 V2 + =2 5 1 1 1  − V1 +  + 1 V2 = 2 5  5 −0.2 V1 + 1.2 V2 = 2

…(ii)

Applying KCL at Node 3, V3 V3 − V1 + +2=0 2 4 1  1 1 − V1 +  +  V3 = −2  2 4 4 −0.25 V1 + 0.75 V3 = −2 …(iii)

. Solving Eqs (i), (ii) and (iii), V1 = −4 V V2 = 1 V V3 = −4 V V5 Ω = V2 − V1 = 1 − ( −4) = 5 V

 Example 2.41 

Find the power dissipated in the 6 W resistor for the network shown in Fig. 2.58. V1

3Ω 20 V

1Ω

2Ω 6Ω

V2

V3

5A

Fig. 2.58 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1,

5Ω

2.5  Node Analysis  2.35

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 V3 = 6.67

…(i)

Applying KCL at Node 2, V2 − V1 V2 − V3 + =5 1 6 1  1 − V1 + 1 +  V2 − V3 = 5  6 6 −V1 + 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 −0.5 V1 − 0.17 V2 + 0.87 V3 = 0

…(iii)

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 2.42 

Find the voltage V in the network shown in Fig. 2.59 which makes the current in the

10 W resistor zero. 3Ω

V

10 Ω

V1

2Ω

V2

7Ω

5Ω

Fig. 2.59 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1,

50 V

2.36 Circuit Theory and Transmission Lines V1 − V V1 V1 − V2 + + =0 3 2 10 1 1 1 1 1   + +  V1 − V2 − V = 0 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 10 5 7 7 −0.1 V1 + 0.44 V2 = 7.14

…(ii)

V1 − V2 =0 10 V1 − V2 = 0 I10 Ω =

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

 Example 2.43 

…(iii)

V = 52.82 V

Find V1 and V2 for the network shown in Fig. 2.60. 50 Ω

Va

2A 10 Ω

80 V

+ V − 1

20 Ω

Vb

+

V2 50 Ω

−

Vc

20 V

Fig. 2.60 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node a, Va − 80 Va − Vb + +2=0 50 10 1 1 80  1 −2  +  Va − Vb = 50 10 10 50 0.12 Va − 0.1 Vb = −0.4

…(i)

Applying KCL at Node b, Vb − Va Vb Vb − Vc + + =0 10 50 20 1 1 1 1  1 − Va +  + + Vb − Vc = 0  10 50 20  10 20 −0..1 Va + 0.17 Vb − 0.05 Vc = 0

…(ii)

2.5  Node Analysis  2.37

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 2.44 

Find the voltage across the 100 W resistor for the network shown in Fig. 2.61. 50 Ω

VA

20 Ω

VB

20 Ω

VC

12 V 60 V

0.6 A

20 Ω

50 Ω

50 Ω

100 Ω

Fig. 2.61 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, VB − VA VB − VC VB + + = 0.6 20 20 20 1 1 1 1  1 − VA +  + + VB − VC = 0.6  20 20 20  20 20 −0.05 VA + 0.15 VB − 0.05 VC = 0.6

…(ii)

Applying KCL at Node C, VC − VA VC − VB VC − 12 VC + + + =0 50 20 50 100 1 1 1 1 1  12  1 − VA − VB +  + + +  VC =  50 20 50 20 50 10 00 50 −0.02 VA − 0.05 VB + 0.1 VC = 0.24 Solving Eqs (i), (ii), and (iii), VC = 31.68 V Voltages across the 100 Ω resistor = 31.68 V

…(iii)

2.38 Circuit Theory and Transmission Lines

EXAMPLES WItH dEPEndEnt SourcES  Example 2.45 

Find the voltage across the 5 W resistor in the network shown in Fig. 2.62. 20 Ω

I1

V1

10 Ω

10 Ω

5Ω

2A

− +

50 V

30I1

Fig. 2.62 Solution From Fig. 2.62, I1 =

V1 − 50 V1 − 50 = 20 + 10 30

…(i)

Assume that the currents are moving away from the node. Applying KCL at Node 1, V1 V1 + 30 I1 V1 − 50 + + 5 10 30  V1 − 50  V1 + 30   30  V1 − 50 V + 2= 1 + 5 10 30 − 2 50 V V V − 50 1 + 1 2= 1 + 5 10 30 2=

…(ii)

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

 Example 2.46 

For the network shown in Fig. 2.63, find the voltage Vx. 40 Ω

Vx Iy 0.6 A

100 Ω

50 Ω 25 Iy

Fig. 2.63

+ −

0.2 Vx

2.5  Node Analysis  2.39

Solution From Fig. 2.63, 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 Vx = −0.6 Vx = −3 V

 Example 2.47 

For the network shown in Fig. 2.64, find voltages V1 and V2. V1

5Ω

0.5 V1 +−

V2

10 Ω

20 Ω 2A

4A

2Ω 20 V

40 V

Fig. 2.64 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, 2=

V1 − 20 V1 − 0.5 V1 − V2 + 20 5

1  1 1 0.5   + −  V1 − V2 = 3 20 5 5 5 0.15 V1 − 0.2 V2 = 3

…(i)

Applying KCL at Node 2, V2 + 0.5 V1 − V1 V2 V2 − 40 + + =4 5 2 10 40  0.5 1  1 1 1  −  V1 +  + +  V2 = 4 +   5 2 10  10 5 5 −0.1V1 + 0.8 V2 = 8 Solving Eqs (i) and (ii), V1 = 40 V V2 = 15 V

…(ii)

2.40 Circuit Theory and Transmission Lines

 Example 2.48 

Determine the voltages V1 and V2 in the network of Fig. 2.65. 0.5 Ω

1Ω

V1

V2

V1

0.25 Ω

2V

1Ω

Fig. 2.65 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  0.5 7 V7 − V2 = 4

…(i)

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

…(ii)

From Eq. (i), V1 =

 Example 2.49 

4 V 7

In the network of Fig. 2.66, find the node voltages V1, V2 and V3. 1Ω

V1

1Ω

V2

V3

2V

− Vx + 2A

2Ω

2Ω

2Ω

+ −

4 Vx

Fig. 2.66 Solution From Fig. 2.66, Vx = V2 − V1

…(i)

2.5  Node Analysis  2.41

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

V1 V1 − V2 + 2 1

1   + 1 V1 − V2 = 2 2 1.5 V1 − V2 = 2

…(ii)

Applying KCL at Node 2, V2 − V1 V2 V2 − V3 + + =0 1 2 1  1  −V1 + 1 + + 1 V2 − V3 = 0  2  −V1 + 2.5 V2 − V3 = 0

…(iii)

At Node 3, V3 − 4 Vx = 2 V3 − 4(V2 − V1 ) = 2 4 V1 − 4 V2 + V3 = 2

…(iv)

Solving Eqs (ii), (iii) and (iv), V1 = −1.33 V V2 = −4 V V3 = −8.67 V

 Example 2.50 

For the network shown in Fig. 2.67, find the node voltages V1 and V2. 3A

0.5 Ω

V1

3 V2

1A

Ix

1Ω

Fig. 2.67

V2

0.25 Ω

2A

2 Ix

2.42 Circuit Theory and Transmission Lines Solution From Fig. 2.67, Ix =

V1 − V2 0.5

…(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    1 +  V1 −  3 +  V2 = −2 0.5  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   0.5  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

…(iii)

Solving Eqs (ii) and (iii), V1 = 1.31 V V2 = 1.19 V

  Example 2.51 

Find voltages V1 and V2 in the network shown in Fig. 2.68. V1 V1

I1

2Ω

1Ω

5A

V2

1Ω

2I1

Fig. 2.68 Solution From Fig. 2.68, V1 − V2 2 Assume that the currents are moving away from the nodes. Applying KCL at Node 1, I1 =

…(i)

2.5  Node Analysis  2.43

5=

V1 V1 − V2 + + V1 1 2

1  1  1 + + 1 V1 − V2 = 5 2 2 2.5 V1 − 0.5 V2 = 5

…(ii)

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

…(iii)

Solving Eqs (ii) and (iii), V1 = 2.5 V V 2 = 2.5 V

  Example 2.52 

Find the power supplied by the 10 V source in the network shown in Fig. 2.69. 2Ω

V1

10 V

4Ω

V2

+ V3 − 10 A

2Ω

4V3

1Ω

Fig. 2.69 Solution Assume that the currents are moving away from the nodes. From Fig. 2.69, V3 = V1 + 10 − V2

…(i)

Applying KCL at Node 1, 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 V1 − 0.75 V2 = −12.5

10 +

Applying KCL at Node 2,

…(ii)

2.44 Circuit Theory and Transmission Lines 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 V1 + 5.75 V2 = 42.5

…(iii)

Solving Eqs (ii) and (iii), V1 = −11.03 V V2 = −1.72 V V + 10 − V2 −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 2.53 

For the network shown in Fig. 2.70, find voltages V1 and V2. 0.03 Vx

20 Ω

V1

+ Vx − 0.4 A

40 Ω

V2

100 Ω

V3

Iy 40 Ω

+ −

80 Iy

Fig. 2.70 Solution From Fig. 2.70, Vx = V1 − V 2

…(i)

V2 40

…(ii)

Iy =

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 (V1 − V2 ) 100 20 1  1   1  + + 0.03 V1 −  + 0.03 V2 = 0.4    20  100 20 0.09 V1 − 0.08 V2 = 0.4

…(iii)

2.5  Node Analysis  2.45

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 + 0.1 V2 − 0.025 V3 = 0

…(iv)

For Node 3, V  V3 = 80 I y = 80  2  = 2 V2  40  …(v)

2 V2 − V3 = 0 Solving Eqs (iii), (iv) and (v), V1 = 40 V V2 = 40 V V3 = 80 V

  Example 2.54 

Find voltages Va , Vb and Vc in the network shown in Fig. 2.71.

Va

4A

2V

2Ω

2Ω

I1

Vb

3Ω

Vc

2I1

1Ω

5Ω

Fig. 2.71 Solution From Fig. 2.71, Va − Vc 2 Assume that the currents are moving away from the nodes. Applying KCL at Node a, I1 =

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 Applying KCL at Node b,

…(i)

2.46 Circuit Theory and Transmission Lines Vb + 2 − Va Vb − Vc + = 2 I1 2 3 Vb + 2 − Va Vb − Vc  V − Vc  + = 2 a  2  2 3 Vb + 2 − Va Vb − Vc + = Va − Vc 2 3  1   1 1  1  − − 1 Va +  +  Vb + 1 −  Vc = −1 2 2 3 3 −1.5 Va + 0.83 Vb + 0.67 Vc = −1

…(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 −0.5 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.

Example 2.55 

Determine the current in the 5 W resistor for the network shown in Fig. 2.72. 2Ω

V1

10 A

20 V

V2

V3

5Ω 3Ω

1Ω

2Ω 10 V

Fig. 2.72

2.6  Supernode Analysis  2.47

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  +  V1 − V2 = 10 3 2 2 0.83 V1 − 0.5 V2 = 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 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 =

  Example 2.56 

V3 − 10 −8.4 − 10 = = −3.68 A 5 5

Find the power delivered by the 5 A current source in the network shown in Fig. 2.73. 10 V

V1

V2

3Ω 1Ω 2A

5A

V3 2Ω

Fig. 2.73

5Ω

2.48 Circuit Theory and Transmission Lines 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 + 1.2 V2 − 1.33 V3 = 3

…(ii)

Applying KCL at Node 3, V3 − V1 V3 − V2 V3 + + =0 3 1 2 1 1 1 − V1 − V2 +  + 1 +  V3 = 0 3 3 2 −0.33 V1 − V2 + 1.8 83 V3 = 0

…(iii)

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

  Example 2.57 

In the network of Fig. 2.74, find the node voltages V1, V2 and V3. 0.5 Ω 0.33 Ω

V1 4A

V2

5V 0.2 Ω

V3 1Ω

Fig. 2.74 Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, 4=

V1 − V2 V1 − V3 + 0.33 0.5

1  1 1  1 + V2 − V3 = 4   V1 − 0.33 0.5  0.33 0.5 5.03 V1 − 3.03 V2 − 2 V3 = 4

…(i)

2.6  Supernode Analysis  2.49

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 0.5 1  1  1   1  1  − +  −  V1 +   V2 + 1 +  V3 = 0 0.33 0.5  0.33 0.2  0.5  −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 2.58 

For the network shown in Fig. 2.75, determine the voltage Vx. 7Ω 6V

V1

8V

11 Ω

V2 +

3Ω

4Ω

10 Ω

2A

5V

+ 2 Vx −

Vx −

V3

12 A

Fig. 2.75 Solution From Fig. 2.75, Vx = V2 − V3 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 V + 5 V2 − 2(V2 − V3 ) V2 − 8 − V3 V2 − V3 + 2= 1 + + 7 11 4 10 2=

1 5 8  1 1 1 1 1 1 1  V1 +  − + +  V2 +  − −  V3 = 2 − +  10 5 7 11  5 7 11 4 4 7 0.25 V1 + 0.133 V2 − 0.033 V3 = 1.89

…(ii)

2.50 Circuit Theory and Transmission Lines Applying KCL at Node 3, V3 − V2 V3 + 8 − V2 + + 12 = 0 11 7 8  1 1  1 1  − −  V2 +  +  V3 = −12 − 11 7 11 7 7 −0.233 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 2.59 

Find the node voltages in the network shown in Fig. 2.76. 6Ω

V1

6A

5Ω

ix

V2 5ix

10 Ω

−+

20 Ω

V3

15 Ω

2Ω

V4

40 V

Fig. 2.76 Solution From Fig. 2.76, Ix = For Node 4,

V2 − V1 5

V4 = 40

…(i) …(ii)

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 6+

1 40  1 1 1 −6  + +  V1 − V2 = 10 5 6 5 6 7 1 2 V1 − V2 = 15 5 3

…(iii)

2.6�Supernode Analysis�2.51

Nodes 2 and 3 will form a supernode, Writing voltage equation for the supernode, �V �V � V3 � V2 � 5 I x � 5 � 2 1 � � V2 � V1 � 5 � V1 � 2V2 � V3 � 0

…(iv)

Applying KCL to the supernode, V2 � V1 V2 V3 V3 � V4 � � � �0 5 20 15 2 V2 � V1 V2 V3 V3 � 40 � � � �0 5 20 15 2 1 �1 1 � � 1 1� � V1 � � � � V2 � � � � V3 � 20 � 5 20 � � 15 2 � 5 1 1 17 � V1 � V2 � V3 � 20 5 4 30

…(v)

Solving Eqs (iii), (iv) and (v), V1 � 10 V V2 � 20 V V3 � 30 V V4 � 40 V

���������������

Find the node voltages in the network shown in Fig. 2.77. V2 Vx �

V1

� 14 A

0.5 �

2�

0.5 Vx

12 V

1�

��

� 2.5 � Vy �

V3

� 0.2 Vy

V4

��������� Solution Selecting the central node as reference node, V1 � �12 V

…(i)

Applying KCL at Node 2,

Chapter 02.indd 51

5/30/2016 3:58:09 PM

2.52 Circuit Theory and Transmission Lines V2 − V1 V2 − V3 + = 14 0.5 2 1 1  1 1 − V1 +  +  V2 − V3 = 14  0.5 2  0.5 2 −2 V1 + 2.5 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 (V2 − V1 ) + V4 + =0 2 2.5 1  1 1   1    0.5 −  V1 −  + 0.5 V2 + V3 + 1 +  V4 = 0 2.5  2 2 2.5  0.1 V1 − V2 + 0.5 V3 + 1.4 V3 = 0

…(iv)

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

Exercises MESH ANALYSIS 2.1

Find the current through the 10 Ω resistor in the network shown in Fig. 2.78.

2.2

Find the current through the 20 Ω resistor in the network shown in Fig. 2.79. 15 Ω

6Ω

2 V 10 Ω

20 Ω

20 V 5V

4Ω

60 V

5Ω

15 Ω

20 V

Fig. 2.78

10 Ω

5Ω

5Ω

40 V

[0.68 A]

15 Ω

Fig. 2.80 [1.46 A]

Exercises  2.53

2.3

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

2IY 5Ω

10 Ω

+−

30 Ω

2Ix

I3

20 Ω

4Ix Iy

2Ω 2Ω

I2

+ − 10IY

I1 5Ω

+−

Ix

5V

10 A

2Ω

100 V

Fig. 2.83 [0.5 A, 0.1 A] 2.7

Fig. 2.80 [0.37 A] 2.4

Find the current through the 1 Ω resistor in the network shown in Fig. 2.81. 1Ω

2Ω 3Ω

1A

In the network shown in Fig. 2.84, 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

2Ω

i1

10 Ω

10 V 20 Ω

30 Ω A

Fig. 2.81 80 V

[0.95 A]

+

40 Ω

2.5

30 Ω

Find the current through the 4 Ω resistor in the network shown in Fig. 2.82.

−

V3

Fig. 2.84 6V

[69.4 V, 72.38 V, 73.68 V, 70.71 V, 97.39 V] 2.8

5A

2Ω

2A

Find currents I1, I2, and I3 in the network shown in Fig. 2.85.

4Ω

1Ω

Fig. 2.82

15 A I1

[1.33 A] 2.6

Find currents Ix and Iy in the network shown in Fig. 2.83.

1 V 9 x

3Ω

I2

+ Vx −

2Ω

2Ω

1Ω

I3

Fig. 2.85 [15 A, 11 A, 17 A]

2.54 Circuit Theory and Transmission Lines Find currents Ix in the network shown in Fig. 2.86.

2.9

20 Ω

25 Ω

10 Ω

2A

2.10 Find currents I1 in the network shown in Fig. 2.87. 5 Ω I1

5Ω

1.5 Ix

5A

30 V

19 V

2Ω

4Ω

Fig. 2.86

25 V

1.5I1

4A

Ix

Fig. 2.87 [8.33 A]

[−12 A]

nodE AnALYSIS 2.11 Find the current Ix in the network shown in Fig. 2.88. Ix

24 V

5Ω

2.14 Calculate the current through the 10 Ω resistor in the network shown in Fig. 2.91.

20 Ω

10 Ω

20 Ω

2A

36 V

4Ω 25 V

2Ω

+ −

− +

7Ω

12 V

Fig. 2.88 [ −0.93 A ] 2.12 Find VA and VB in the network shown in Fig. 2.89. 1Ω

VA

VB

3Ω

Fig. 2.91 [1.62 A ]

2Ω

2.15 Find the current through the branch ab in the network shown in Fig. 2.92.

2Ω 2A

2Ω

2Ω

1A

2V

10 Ω

1V

1Ω a 2Ω

Fig. 2.89 [2.88 V, 3.25 V ] 2.13 Find the current through the 6 Ω resistor in the network shown in Fig. 2.90.

10 Ω

10 V 2Ω b

10 V 2 Ω 1Ω

4A

10 Ω

5Ω

3A

6Ω

Fig. 2.92 [0.038 A ]

Fig. 2.90 [2.04 A ]

2.16 Find the current through the 4 Ω resistor in the network shown in Fig. 2.93.

Exercises  2.55

2.20 Find the voltage Vx in the network shown in Fig. 2.97.

10 V

4Ω 2Ω

2Ω

1 Ω 3

2Ω 2Ω

2Ω

1Ω

5A

0.5 Ω

Vx

6V

0.5 Ω

3I

1V

Fig. 2.93 [1.34 A ] 2.17 Find the current through the 4 Ω resistor in the network shown in Fig. 2.94. 2A

2Ω

Fig. 2.97 [6.2 V] 2.21 Determine V1 in the network shown in Fig. 2.98.

2Ω

50 Ω

2V

6V

20 Ω 4Ω

3Ω

2Ω

5A

20 Ω

+ V1 −

+ 0.4V1 −

0.01V1

3A

Fig. 2.98 Fig. 2.94 [1 A] 2.18 Find the voltage Vx in the network shown in Fig. 2.95. 1Ω 1Ω

4I1

+ Vx −

[140 V] 2.22 Find the voltage Vy in the network shown in Fig. 2.99. 0.5Vy +−

2Ω 4A

15 Ω

I1

6A

3Ω

20 Ω Vy

8V

4V

−

Fig. 2.99

Fig. 2.95 [ −4.31 V] 2.19 Find the currents Vx in the network shown in Fig. 2.96.

[−10 V] 2.23 Find the voltage V2 in the network shown in Fig. 2.100.

3V1

0.8V2 +− 2Ω

5Ω

+ Vx − 7A

+

4Ω

4Vx 6Ω

5A

5Ω

+ V2 −

8A 2.5 Ω

Fig. 2.100

Fig. 2.96 [2.09 V]

[25.9 V]

2.56 Circuit Theory and Transmission Lines

Objective–Type Questions 2.1

Two electrical sub-networks N1 and N2 are connected through three resistors as shown in Fig. 2.101. 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

2.5

(b) (d)

24 V 28 V

The dependent current source shown in Fig. 2.104. 5Ω

V1 = 20 V

5Ω −

+

(a) 48 V (c) 36 V

V1 5

5Ω

10 V N1

N2

15 Ω 1Ω +

5V

Fig. 2.104 (a) delivers 80 W (b) absorbs 80 W (c) delivers 40 W (d) absorbs 40 W

−

Fig. 2.101 2.2

2.3

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. 2.102 is 2Ω

a

2.6

If V = 4 in Fig. 2.105, the value of Is is given by 1Ω

4Ω

2Ω

2Ω V

IS

1Ω

Fig. 2.105 1V

+ −

2Ω

(a) 6 A (c) 12 A

3A

2.7

b

Fig. 2.102

2.4

+ Vx −

+ 8V −

16 V 10 Ω

Fig. 2.103

12 Ω 6Ω

2.5 A none of these

The value of Vx,Vy and Vz in Fig. 2.106 shown are

(a) 0.5 V (b) 3 V (c) 3.5 V (d) 4 V The voltage V0 in Fig. 2.103 is 2Ω

8A

(b) (d)

+ 2V −

+ V0 −

+ Vy −

+ Vz −

+ 1V −

+ 2V −

Fig. 2.106 (a) −6, 3, −3  (c) 6, 3, 3 

(b) (d)

−6, −3, 1 6, 1, 3

Answers to Objective-Type Questions  2.57

2.8

The circuit shown in Fig. 2.107 is equivalent to a load of

i 4 i

2Ω

I

4Ω + −

4Ω

2I

V

Fig. 2.108 Fig. 2.107

2.9

(a)

4 Ω  3

(b)

8 Ω 3

(b)

4Ω 

(d)

2Ω

2.10

In the network shown in Fig. 2.108, the effective resistance faced by the voltage source is

(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

Network Theorems (Application to dc Networks)

  3.1     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 and Millman’s 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 Circuit Theory and Transmission Lines 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 = I ′ + I ′′ + I ′′′ = 5 − 2.5 + 1.88 = 4.38 A ( → )

  example 3.2 

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 Circuit Theory and Transmission Lines 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 = 14.27 A( ↓ ) 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

I ′′

50 = 21.46 A( ↑ )= − 21.46 A( ↓ ) 1 + 1.33

When the 40 V source is acting alone (Fig. 3.16)

50 V

1.33 Ω 1Ω

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 × = 11.43 A(↓) 4 +1 Step IV By superposition theorem,

I 40 V

0.8 Ω

2Ω

Fig. 3.17

I = 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

8Ω

15 Ω

4V

12 Ω

6V

Fig. 3.18 Solution Step I When the 4 V source is acting alone (Fig. 3.19) I

I1

5Ω

10 Ω

12 Ω I′

8Ω

15 Ω

4V

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, I1 = 0.32 ×

15 = 0.16 A 15 + 14.8

5Ω

7.45 Ω

4V

From Fig. 3.19, by current-division rule, I ′ = 0.16 ×

12 = 0.096 A( ↓ ) 12 + 8

(c)

Fig. 3.20

3.6 Circuit Theory and Transmission Lines Step II

When the 6 V source is acting alone (Fig. 3.21) 5Ω

10 Ω

12 Ω

8Ω

15 Ω

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,

Step III

5.06 Ω

(c)

I = I ′ + I ′′ = 0.096 + 0.22 = 0.316 A( ↓ )

Fig. 3.22

Find the current through the 4 W resistor in Fig. 3.23. 12 Ω

4Ω

8A

3Ω

5Ω

40 V

Fig. 3.23 Solution Step I 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

13.75 I ′′ = 0.35 × = 0.22 A( ↓ ) 13.75 + 8 By superposition theorem,

  example 3.4 

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

2.92 Ω

40 V

7Ω

(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( → ) = − 1.12 A ( ← ) 5+ 7

Step II When the 8 A source is acting alone (Fig. 3.26) 12 Ω

4Ω

8A

3Ω

5Ω

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

3Ω

3 = 2.28 A( ← ) 7.53 + 3

By superposition theorem, I = I ′ + I ′′ = −1.12 + 2.28 = 1.16 A ( ← )

8A

3.8 Circuit Theory and Transmission Lines

  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 = 0.8 A(↓) 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

0.91 Ω

I

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 − 0.23 A( ↓ ) 1 + 10

By superposition theorem, I = I ′ + I ′′ = 0.8 + 0.23 = 1.03 A( ↓ )

  example 3.6 

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 Circuit Theory and Transmission Lines Solution

8Ω

I′

Step I When the 5 A source is acting alone (Fig. 3.34) I′ = 5× Step II

12 = 1.2 A (→) 12 + 8 + 30

30 Ω

12 Ω

When the 25 A source is acting alone (Fig. 3.35) I ′′ = 25 ×

Step III

5A

Fig. 3.34

30 = 15 A (→) 30 + 12 + 8

I ′′

8Ω

By superposition theorem, 12 Ω

I = I ′ + I ′′ = 1.2 + 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 5A

4Ω

5A

8.73 Ω

4Ω

2.73 Ω

(a)

(b)

Fig. 3.38 I′ = 5× Step II

8.73 = 3.43 A(↓) 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

5Ω

6 = 0.86 A(↓) 6 + 10

By superposition theorem I = I ′ + I ′′ = 3.43 + 0.86 = 4.29 A (↓)

3.12 Circuit Theory and Transmission Lines

  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 = 3.75 A(↓) 15 + 2 + 3

5A

2Ω

3Ω

15 Ω

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 × Step III

4 = 1 A(↑) = −1 A(↓) 20 + 4

By superposition theorem, I = I ′ + I ′′ = 3.75 − 1 = 2.75 A ( ↓ )

  example 3.9 

Find the current in the 1 W resistors in Fig. 3.46. 1A

2Ω

3Ω

1Ω

4V

3A

Fig. 3.46 2Ω

Solution

3Ω I′

Step I When the 4 V source is acting alone (Fig. 3.47)

1Ω

4V

4 I′ = = 1.33 A( ↓ ) 2 +1 Fig. 3.47

Step II When the 3 A source is acting alone (Fig. 3.48) By current-division rule,

2Ω

2 I ′′ = 3 × = 2 A (↓ ) 1+ 2 Step III

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 Circuit Theory and Transmission Lines Redrawing the network (Fig. 3.50), By current-division rule,

I ′′′

3Ω 2Ω

2 I ′′′ = 1 × = 0.66 A( ↓ ) 2 +1 Step IV

1Ω

1A

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. + 6V

A

5Ω VAB 10 V

5A

−

B

Fig. 3.51 Solution Step I When the 6 V source is acting alone (Fig. 3.52) VAB′ = 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 Step IV

A

5Ω VAB′′′

By superposition theorem, 5A

−

VAB = VAB′ + VAB ″ + 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 A(→) = −2.06 A(←) 5 + 6.67

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 A(←) 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)

36 V

4Ω

(b)

Fig. 3.59

I

3.16 Circuit Theory and Transmission Lines I=

36 = 2.57 A 10 + 4

By current-division rule, I ′′′ = 2.57 × Step IV

20 = 2.06 A(←) 20 + 5

By superposition theorem, I = I ′ + I ′′ + I ′′′ = −2.06 + 1.14 + 2.06 = 1.14 A (←)

  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 Step I When the 5 A source is acting alone (Fig. 3.61) By current-division rule, I′ = 5×

2Ω

5A

2 = 1.67 A(↓) 2+4

Fig. 3.61

Step II When the 2 A source is acting alone (Fig. 3.62) By current-division rule, 2 I″ = 2× = 0.67 A(↓) 2+4

4Ω

I ′′

2Ω

Step III When the 6 V source is acting alone (Fig. 3.63) Applying KVL to the mesh,

2A

4Ω

Fig. 3.62

−2l ′′′ − 6 − 4l ′′′ = 0

6V

I ′′′ = −1 A (↓) Step IV

By superposition theorem, I = I ′ + I ′′ + I ′′′ = 1.67 + 0.67 − 1 = 1.34 A (↓)

2Ω

4Ω I ′′′

Fig. 3.63

3.2�Superposition Theorem�3.17

EXAMPLES WITH DEPENDENT SOURCES ��������������

Find the current through the 6 � resistor in Fig. 3.64. I

6�

8�

3I

15 V

10 V

��������� Solution Step I When the 15 V source is acting alone (Fig 3.65) From Fig. 3.65, I � � I1

I�

…(i)

8�

3I �

15 V I1

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

I2

���������

I 2 � I1 � 3I � � 3I1 4 I1 � I 2 � 0

6�

…(ii)

Applying KVL to the outer path of the supermesh, 15 � 6 I1 � 8 I 2 � 0 6 I1 � 8 I 2 � 15

…(iii)

Solving Eqs (ii) and (iii), I1 � 0.39A I 2 � 1.58A I � � I1 � 0.39 A (�) 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 4 I1 � I 2 � 0 Applying KVL to the outer path of the supermesh, �6 I1 � 8 I 2 � 10 � 0 6 I1 � 8 I 2 � 10

Chapter 03.indd 17

I �� 6 �

8�

...(i) 3I �� I1

10 V I2

…(ii) ���������

…..(iii)

5/30/2016 5:47:18 PM

3.18 Circuit Theory and Transmission Lines Solving Eqs (ii) and (iii), I1 = 0.26 A

Step III

I 2 = 1.05 A I ′′ = I1 = 0.26 A (→) By superposition theorem, I = I ′ + I ″ = 0.39 + 0.26 = 0.65 A (→)

  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

1Ω

…(i) 30 A

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

I1

I1 − I 2 = 30

I2

4Ix′

…(ii) Fig. 3.68

Applying KVL to the outer path of the supermesh, −5 I1 − 1I 2 − 4 I′x = 0 −5 I1 − I 2 − 4 I1 = 0 9 I1 + I 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, 20 − 5 I x″ − 1I x″ − 4 I ″x = 0 I x″ = 2 A( → ) Step III

+ −

Ix′′ 5 Ω

1Ω

+ −

20 V

By superposition theorem, I x = 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′

Vx

+ −

I1′

4Vx

10 Ω

2Ω

Applying KVL to the mesh, 5V

5 – 10I1′ – 4Vx – 2I1′ = 0 5 – 10I1′ – 4 (5 – 10I1′) – 2I1′ = 0

Fig. 3.71

5 – 10I1′ – 20 + 40I1′ – 2I1′ = 0 I1′ = Step II

15 = 0.54 A(↑) 28 Vx

When the 2 A source is acting alone (Fig 3.72)

From Fig . 3.72,

I1

Vx = −10 I1′

…(i)

Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I 2 − I1′ = 2

+−

4Vx

′

10 Ω

2A I1′

…(ii)

2Ω

I2

Fig. 3.72

Applying KVL to the outer path of the supermesh, −10 I1′ − 4Vx − 2 I 2 = 0 −10 I1′ − 4( −10 I1′ ) − 2 I 2 = 0 30 I1′ − 2 I 2 = 0 Solving Eqs (ii) and (iii), I1 = 0.14 A (↑) I 2 = 2.14 A Step III

By superposition theorem, I1 = I1′ + I1″ = 0.54 + 0.14 = 0.68 A (↑)

…(iii)

3.20 Circuit Theory and Transmission Lines

  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 A (→)

Fig. 3.74

Step II When the 10 A source is acting alone (Fig. 3.75) From Fig. 3.75, Vx = 5( I1 − I 2 ) …(i) Applying KVL to Mesh 1, −10 I1 + 10Vx − 5( I1 − I 2 ) = 0 −10 I1 + 10{5( I1 − I 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 A (→) Step III By superposition theorem, I = I ′ + I ′′ = −2.86 − 12.86 = −15.72 A (→)

  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

I1

1A

I2

Fig. 3.78

4I

4Ω

+

+ −

5A

V1

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

4Ω

Find the voltage V1 in Fig. 3.79. I

+ −

I ′′

+ Vx −

…(iii)

1Ω

5Vx

3Ω

I ′′ = I 2 = 1.6 A (→) By superposition theorem, I = I′ + I′′ = 3.4 + 1.6 = 5 A (→)

  example 3.18 

+ −

Fig. 3.77

−2 I1 − 3I 2 − 5Vx = 0 −2 I1 − 3I 2 − 5( −2 I1 ) = 0 8 I1 − 3I 2 = 0 Solving Eqs (ii) and (iii), I1 = 0.6 A I 2 = 1.6 A Step III

3Ω

17 V

1Ω

4I

V1′

+ −

I

5A

Fig. 3.80

4Ω

3.22 Circuit Theory and Transmission Lines  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 4I

I = –20 A

V1′′

4Ω

I

+ −

20 V

V1′′ = 4I – 1(I) = 3I = 3 (–20) = –60 V Step III By superposition theorem,

Fig. 3.81

V1 = V1′ + V1′′ = 20 – 60 = –40 V

  example 3.19 

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)

− +

I 2 − I1 = 3

6Ω

18 V

Fig. 3.83 1Ω

− +

− Vx +

I1

…(ii)

I′

− Vx +

2Vx I ′′

6Ω

3A

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

2Vx

I2

Fig. 3.84

Applying KVL to the outerpath of the supermesh, −1I1 + 2Vx − 6 I 2 = 0 − I1 + 2( − I1 ) − 6 I 2 = 0 3I1 + 6 I 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

4Ω + −

Step I When the 120 V source is acting alone (Fig. 3.86) Applying KVL to the mesh,

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 I1 − 10 I y ″ − 8 I 2 = 0

Iy′′

4Ω

10Iy′′ + −

8Ω

12 A I1

I2

...(ii) Fig. 3.87

−4 I1 − 10 I1 − 8 I 2 = 0 14 I1 + 8 I 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,

10Iy′′′ + −

4Ω

–4 Iy′′′ – 10Iy′′′ – 8Iy′′′ – 40 = 0 I y ″′ = −

Iy′′′

40 = –1.82 A (→) 22

8Ω

40 V

Fig. 3.88

3.24 Circuit Theory and Transmission Lines 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

6Ω + −

+ Vx

−

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

I

Fig. 3.90 3Ω + Vx

…(i)

− 24 Ω

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,

6Ω

I1

5A

+ −

I2

...(ii) Fig. 3.91

−3I1 − 6 I 2 − 3V″x = 0 −3I1 − 6 I 2 − 3(3I1 ) = 0 12 I1 + 6 I 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,

8Ω

15 Ω

5Ω

− V′ + + −

12 Ω I2

10 V I1

8V′

−10 − 8 I1 − 15 I1 − 12 ( I1 − I 2 ) = 0 35 I1 − 12 I 2 = −10

...(ii)

Fig. 3.94

Applying KVL to Mesh 2, −12( I 2 − I1 ) − 5 I 2 − 8V ′ = 0 −12 I 2 + 12 I1 − 5 I 2 − 8( −8 I1 ) = 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) = −4.32 V 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

5Ω

− V ′′ +

I1

...(ii)

15 Ω

−5 A I2

Fig. 3.95

12 Ω I3

+ −

8 V ′′

3.26 Circuit Theory and Transmission Lines Applying KVL to the outer path of the supermesh, −8 I1 − 15 I 2 − 12( I 2 − I 3 ) = 0 −8 I1 − 27 I 2 + 12 I 3 = 0 Applying KVL to Mesh 3,

...(iii)

−12( I 3 − I 2 ) − 5 I 3 − 8V ″ = 0 −12 I 3 + 12 I 2 − 5 I 3 − 8( −8 I1 ) = 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( −4.97) = −39.76 V Step III

By superposition theorem, V = V ′ + V ″ = −4.32 − 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, 0.5V1 − 40 ( I 2 − I1 ) − 200 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

Step II When the 25 V source is acting alone (Fig. 3.98) From Fig. 3.98,

50 Ω

200 Ω +

V1 − 200 I − 25 = 0

40 Ω

V1 = 200 I + 25

...(i)

25 V

V1

V0′′

+ −

Applying KVL to Mesh 1, 0.5V1 − 40 I − 200 I − 25 = 0 0.5( 200 I + 25) − 40 I − 200 I − 25 = 0

0.5V1

I

−

Fig. 3.98 I = −0.09 A V0″ = V1 = 200 I + 25 = 200( −0.09) + 25 = 7 V

Step III

By superposition theorem, 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 +

20 V

Vx

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 I1 − 6( I1 − I 2 ) = 0 8 I1 − 6 I 2 = 20

4Ω +

20 V

Vx′

6Ω

Vx′ I1

2 I2

−

Fig. 3.100 ...(ii)

For Mesh 2, 6( I1 − I 2 ) Vx′ = = 3I1 − 3I 2 2 2 3I1 − 4 I 2 = 0 I2 =

Solving Eqs (ii) and (iii), I1 = 5.71 A I 2 = 4.29 A Vx′ = 6( I1 − I 2 ) = 6(5.71 − 4.29) = 8.52 V

...(iii)

3.28 Circuit Theory and Transmission Lines Step II When the 10 A source is acting alone (Fig. 3.101) From Fig. 3.101, Vx″ = 6(I1 − I 2 ) Applying KVL to Mesh 1, −2( I1 − I 3 ) − 6( I1 − I 2 ) = 0 8 I1 − 6 I 2 − 2 I 3 = 0 For Mesh 2, V ″ 6( I1 − I 2 ) I2 = x = = 3I1 − 3I 2 2 2 3I1 − 4 I 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

10 A

...(i)

2Ω

4Ω

I3

+

...(ii)

I1

Vx′′ 2

6Ω

Vx′′

I2

−

Fig. 3.101 ...(iii) ...(iv)

Vx ″ = 6( I1 − I 2 ) = 6( −5.71 + 4.29) = −8.52 V Step III

By superposition theorem, Vx = Vx ′ + Vx ″ = 8.52 − 8.52 = 0

  example 3.25 

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 I1 − 2 I1 − 20 ( I1 − I 2 ) = 0 26 I1 − 20 I 2 = 0

...(ii)

4Ω

I1 20 Ω

2I1

I1

− + 2Ω

70 V I2

I3

Applying KVL to Mesh 2, 70 − 20( I 2 − I1 ) − 2( I 2 − I 3 ) = 0 −20 I1 + 22 I 2 − 2 I 3 = 70

10 Ω

...(iii)

Fig. 3.103

I′

3.2  Superposition Theorem  3.29

Applying KVL to Mesh 3, −2( I 3 − I 2 ) + 2 I1 − 10 I 3 = 0 2 I1 + 2 I 2 − 12 I 3 = 0

...(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 A (←) 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 I1 − 2 I1 − 20( I1 − I 2 ) = 0 26 I1 − 20 I 2 = 0 …(ii)

I2

50 V I3

10 Ω

Applying KVL to Mesh 2, −20( I 2 − I1 ) − 2( I 2 − I 3 ) = 0

I′′

Fig. 3.104

−20 I1 + 22 I 2 − 2 I 3 = 0

…(iii)

Applying KVL to Mesh 3, −2( I 3 − I 2 ) + 2 I1 + 50 − 10 I 3 = 0 2 I1 + 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 ′′ = I 3 = 4.57 A (←) Step III

By superposition theorem, I = I ′ + I ′′ = 3.43 + 4.57 = 8 A (←)

  example 3.26 

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 Circuit Theory and Transmission Lines Solution 5Ω

Step I When the 10 V source is acting alone (Fig. 3.106) Applying KCL at the node,

1Ω +

10 V

V0′

V′ V0′ − 0 V0′ − 10 V0′ 2 =0 + + 5 2 1  1 1 1  + +  V0′ = 2 5 2 2 V0′ = 1.67 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 +1+ + 5 2 1

+ V0″ − 2

2Ω

Fig. 3.107

4V

5Ω

V ′′′ V ′′′− 4 − 0 V0′′′ V0′′′ 0 2 =0 + + 1 5 2  1 1 1  + +  V0′′′ = 4 5 2 2 V0′′′ = 3.33 V Step IV

V 0″ −

 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

+ 2

V0″′

1Ω

+ V0″′ 2 −

−

Fig. 3.108

By superposition theorem, V0 = V0′ + V0′′+ V0′′′ = 1.67 − 0.83 + 3.33 = 4.17 V

  3.3     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 =

R3 +

VTh

R2

V

R2 V R1 + R2

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 +

−

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 =

VTh 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

RL IL B

Fig 3.113  Thevenin’s equivalent  network

B

3.32 Circuit Theory and Transmission Lines 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 I − 20 − 10 I = 0 15 I = 20 I = 1.33 A

+

−

A +

+

10 Ω

40 V

I

−

Fig. 3.116

VTh

B − 10 V

3.3  Thevenin’s Theorem  3.33

Writing the VTh equation,

5Ω

A

RTh

B

10 I − VTh + 10 = 0 10 Ω

VTh = 10 I + 10 = 10(1.33) + 10 = 23.33 V Step II

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 Step I Calculation of VTh (Fig. 3.120) I=

10 Ω −

+ + 5Ω

250 V

VTh

−

250 = 25 A 5+5

− 75 V

Writing the VTh equation, 250 − 5 I − VTh − 75 = 0

Fig. 3.120 5Ω

10 Ω A

VTh = 175 − 5 I = 175 − 5( 25) = 50 V Step II

Calculation of RTh (Fig. 3.121)

A

5Ω

RTh = (5  5) + 10 = 12.5 Ω

RTh

B

Fig. 3.121

B

3.34 Circuit Theory and Transmission Lines Calculation of IL (Fig. 3.122)

Step III

IL =

12.5 Ω 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 2Ω

12 Ω

2V

4V

B

Fig. 3.123 Solution 2Ω

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)

1Ω −

+

+ −

+ 12 Ω

2V

− +

I1

3Ω −

A

+

+

−

VTh I2

B

4V

−

Applying KVL to Mesh 2, −12( I 2 − I1 ) − 1I 2 − 3I 2 − 4 = 0 −12 I1 + 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,

B

Fig. 3.125

Calculation of RTh (Fig. 3.125)

1.43 Ω

RTh = [( 2  12) + 1]  3 = 1.43 Ω Step III

Calculation of IL (Fig. 3.126) IL =

2.8 = 0.82 A 1.43 + 2

RTh

12 Ω

VTh − 3I 2 − 4 = 0 VTh = 4 + 3I 2 = 4 + 3( −0.4) = 2.8 V Step II

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. 2Ω

6Ω

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 I1 − 1( I1 − I 2 ) = 0 7 I1 − I 2 = 10

+

2Ω −

+

−

+ − I1

3Ω I2

− +

VTh

− −

−1( I 2 − I1 ) − 2 I 2 − 3I 2 = 0 − I1 + 6 I 2 = 0

Fig. 3.128

…(ii) 6Ω

2Ω A

Solving Eqs (i) and (ii), I 2 = 0.24 A

…(ii) 1Ω

Writing the VTh equation,

3Ω

R Th

3I 2 − VTh − 20 = 0 B

VTh = 3I 2 − 20 = 3(0.24) − 20 = −19.28 V =19.28 V (terminal B is positive w.r.t A)

Step III

Fig. 3.129 1.47 Ω

Calculation of RTh (Fig. 3.129) RTh = [(6  1) + 2]  3 = 1.47 Ω

A IL

Calculation of IL (Fig. 3.130) IL =

10 Ω

19.28 V

19.28 = 1.68 A( ↑ ) 1.47 + 10

B

Fig. 3.130

    example 3.31   

B

20 V

Applying KVL to Mesh 2,

Step II

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 Circuit Theory and Transmission Lines Solution

30 Ω

A V Th B +

Step I Calculation of VTh (Fig. 3.132) For Mesh 1,

−

+

+ 5Ω

10 A

I1 = 10

I1

20 Ω

100 V

−

I2

−

Applying KVL to Mesh 2, Fig. 3.132

100 − 30 I 2 − 20 I 2 = 0

R Th

I2 = 2 A

A

30 Ω

B

Writing the VTh equation, 5Ω

5 I1 − VTh − 20 I 2 = 0

20 Ω

VTh = 5 I1 − 20 I 2 = 5(10) − 20( 2) = 10 V Step II

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 = 10( I ) − 10 = 10(0.58) − 10 = −4.2 V = 4.2 V(terminal B is positive w . r .t A) Step II

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 A(↑) 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 =1A I2 = 5 + 15 I1 =

2V

6Ω +

−

A +

V Th

5Ω

B −

−

4Ω

50 V I1

+

+

+

15 Ω

−

−

Fig. 3.140

20 V I2

3.38 Circuit Theory and Transmission Lines Writing the VTh equation, 4 I1 + 2 − VTh − 15 I 2 = 0 VTh = 4 I1 + 2 − 15I 2 = 4(5) + 2 − 15(1) = 7 V Calculation of RTh (Fig. 3.141)

Step II

6Ω

A

RTh = (6  4) + (5  15) = 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 = 8.8 A I2 = 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 = 2.75 A I1 = 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

R Th A

B

50 Ω

5Ω

Fig. 3.145 Redrawing the circuit (Fig. 3.146), 30 Ω

RTh = (30  50) + ( 20  5) = 22.75 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 Circuit Theory and Transmission Lines Solution Step I Calculation of VTh (Fig. 3.149) Applying KVL to Mesh 1, 50 − 2 I1 − 1( I1 − I 2 ) − 8( I1 − I 2 ) = 0 11I1 − 9 I 2 = 50

4Ω

−

+

…(i) +

Applying KVL to Mesh 2,

−

−1Ω

−4 I 2 − 5 I 2 − 8( I 2 − I1 ) − 1( I 2 − I1 ) = 0 −9 I1 + 18 I 2 = 0

I2

A + V Th B −

+ −

+ −

+

Solving Eqs (i) and (ii),

I1

I1 = 7.69 A I 2 = 3.85 A

50 V

Writing the VTh equation, VTh − 5 I 2 − 8( I 2 − I1 ) = 0 VTh = 5 I 2 + 8( I 2 − I1 ) = 5(3.85) + 8(3.85 − 7.69) = −11.47 V = 11.47 V(the terminal B is positive w . r .t . A) Step II

−

2Ω

...(ii)

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

+

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

R2

1.82 Ω

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  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 = 1.83 A( ↑ ) 3.27 + 3

A 3Ω

11.47 V IL

B

Fig. 3.155

3.42 Circuit Theory and Transmission Lines

  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 V

I1

45 − 120 − 15 I1 − 5( I1 − I 2 ) − 10( I1 − I 2 ) = 0 30 I1 − 15 I 2 = −75

...(i) − +

Applying KVL to Mesh 2, 20 − 5 I 2 − 10( I 2 − I1 ) − 5( I 2 − I1 ) = 0

10 Ω

+

− B

−

+ −

−

− +

Fig. 3.157

I1 = −3.2 A I 2 = −1.4 A Writing the VTh equation, 45 − VTh − 10( I1 − I 2 ) = 0

VTh = 45 − 10( I1 − I 2 ) = 45 − 10[ −3.2 − ( −1.4)] = 63 V Calculation of RTh (Fig. 3.158) A R Th

15 Ω

B 10 Ω

5Ω

5Ω

Fig. 3.158

5Ω

+ 5Ω

Solving Eqs (i) and (ii),

Step II

15 Ω

I2

...(ii)

−15I1 + 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 = 2.5 Ω 20

R2 =

10 × 5 = 2.5 Ω 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 = (16.25 | | 2.5) + 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 6A

3Ω

6Ω 42 V

Fig. 3.163 Solution Step I Calculation of VTh (Fig. 3.164) For Mesh 1, I1 = 6 Applying KVL to Mesh 2,

+

+ − +

12 Ω

…(i)

6A

6Ω

− + I1

42 V

I2

42 − 12( I 2 − I1 ) − 6 I 2 = 0 −12 I1 + 18 I 2 = 42

A

V Th

− −

...(ii) Fig. 3.164

B

3.44 Circuit Theory and Transmission Lines Solving Eqs (i) and (ii),

A

I 2 = 6.33 A 6Ω

12 Ω

Writing the VTh equation,

R Th

VTh = 6 I 2 = 38 V Step II

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)

15 Ω +

60 Ω −

+

I1

+

+

V Th

B −

40 Ω

13 A

150 V

Writing voltage equation for supermesh,

A −

I2

50 V

−

150 − 15 I1 − 60 I 2 − 40 I 2 = 0 Fig. 3.168

15 I1 + 100 I 2 = 150 ...(ii) Solving Eqs (i) and (ii), I1 = −10 A I2 = 3 A Writing the VTh equation,

15 Ω

60 Ω

40 I 2 − VTh − 50 = 0 VTh = 40 I 2 − 50 = 40(3) − 50 = 70 V Step II

40 Ω

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

Calculation of IL (Fig. 3.170)

IL =

26.09 Ω A

70 = 1.25 A 26.09 + 30

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

5A

5Ω

20 Ω

100 V

Fig. 3.171 Solution Step I Calculation of VTh (Fig. 3.172) VTh = 100 V

10 Ω

5A

+ A V Th − B

5Ω

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 =

A

100 = 5A 20

20 Ω

100 V IL

B

Fig. 3.174

3.46 Circuit Theory and Transmission Lines

  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 10 Ω

Step I 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 I1 − VTh + 8 I 2 = 0 VTh = 4(0.71) + 8( 2) = 18.84 V Step II

Fig. 3.176

Calculation of RTh (Fig. 3.177).

10 Ω

R Th A

B

4Ω

R Th

5Ω

A

B

2.86 Ω

8Ω

(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. 50 V

10 Ω

50 V

2Ω

5Ω

100 V

2Ω

3Ω

Fig. 3.179 Solution Step I Calculation of VTh (Fig. 3.180) Applying KVL to Mesh 1, 100 − 10 I1 + 50 − 2 I1 − 2( I1 − I 2 ) = 0 14 I 2 − 2 I 2 = 150 Applying KVL to Mesh 2, −2( I 2 − I1 ) + 50 − 3I 2 = 0 …(ii) −2 I1 + 5 I 2 = 50

...(i)

50 V

10 Ω +

50 V

2Ω

−

+

−

A +

Solving Eqs (i) and (ii), I1 = 12.88 A I 2 = 15.15 A

100 V

I1

V Th

2Ω

B −

Writing the VTh equation,

+

+ − − +

3Ω I2

Fig. 3.180

100 − 10 I1 − VTh = 0

VTh = 100 − 10(12.88) = −28.8 V = 28.8 V( terminal B is positive w.r.t. A) Step II

Calculation of RTh (Fig. 3.181)

A

A R Th

2Ω

R Th

3Ω

B

B

(b)

(a) 3.2 Ω

10 Ω

RTh = 10  3.2 = 2.42 Ω

2Ω

10 Ω

2Ω

10 Ω

A R Th B (c)

Fig. 3.181

1.2 Ω

−

3.48 Circuit Theory and Transmission Lines Step III

2.42 Ω

Calculation of I L (Fig. 3.182) IL =

A

28.8 = 3.88 A (↑) 2.42 + 5

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,

+

VTh

B −

2Ω

−15 − 2 I1 − 1( I1 − I 2 ) − 10 − 1I1 = 0 4 I1 − I 2 = −25

+

...(i)

2Ω −

Applying KVL to Mesh 2, − I1 + 4 I 2 = 10

...(ii)

− +

1Ω

15 V

− +

I1

10 − 1( I 2 − I1 ) − 2 I 2 − 1I 2 = 0

+ −

+

−

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), 2×2 =8Ω R1 = 2 + 2 + 1 2 ×1 =4Ω R2 = 2 + 1 + 2 2 ×1 =4Ω R3 = 2 + 1 + 2

A

2Ω

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 Circuit Theory and Transmission Lines Step III

Calculation of I L (Fig. 3.188) 1.33 Ω A

IL =

10 = 0.88 A (↑) 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

− + 4V

2Ω

+ I2 − A + I1

− +

3Ω

+ −

V Th

3A

B − 2Ω

3Ω

Fig. 3.190 A

Writing the current equations for Meshes 1 and 2,

Writing the VTh equation,

R Th

I1 = −3 I2 = 1

B

Fig. 3.191

4 − 2( I1 − I 2 ) − VTh = 0 VTh = 4 − 2( I1 − I 2 ) = 4 − 2( −4) = 12 V Step II Step III

Calculation of RTh (Fig. 3.191) RTh = 2 Ω Calculation of I L (Fig. 3.192) 12 IL = = 4A 2 +1

2Ω 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

3Ω

10 A

2Ω

10 V

Fig. 3.193 Solution Step I Calculation of VTh (Fig. 3.194) 2Ω

1Ω + 10 A

2Ω

10 V

A

V Th −

B

Fig. 3.194 By source transformation (Fig. 3.195), 2Ω +

1Ω −

2Ω 10 V

− I

A

+

+

V Th

20 V

−

Fig. 3.195

B 2Ω

1Ω A

Applying KVL to the mesh, 10 − 2 I − 2 I − 20 = 0 4 I = −10

2Ω

R Th

I = −2.5 A B

Writing the VTh equation, Fig. 3.196

10 − 2 I − VTh = 0 VTh = 10 − 2 I = 10 − 2( −2.5) = 15 V Step II

A

Calculation of RTh (Fig. 3.196) RTh = ( 2  2) + 1 = 1 + 1 = 2 Ω

Step III

2Ω

Calculation of IL (Fig. 3.197) IL =

15 = 3A 2+3

3Ω

15 V IL

B

Fig. 3.197

3.52 Circuit Theory and Transmission Lines

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 = −2 I1

4Ω

3I1 = 0 I1 = 0

2I1

A

V Th

8V

−

Writing the VTh equation,

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,

4Ω

I 2 − I1 = 2 I1 3I1 − I 2 = 0

A

I1

2I1 I1

...(i)

IN

I2

8V B

Applying KVL to the outer path of the supermesh, Fig. 3.200

8 − 4 I1 = 0 I1 = 2

...(ii)

Solving Eqs (i) and (ii), I2 = 6 A IN = I2 = 6 A Step III

Calculation of RTh

A

8V

RTh = Step IV

1.33 Ω

VTh 8 = = 1.33 Ω IN 6

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

3Ω −

4V

+ 0.1 Vx

I1 = −0.1 Vx

A

Vx = VTh

I1 −

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., 0.1 Vx = 0 as shown in Fig. 3.205. IN

4V

IN

B

4 = = 0.8 A 2+3

Fig. 3.205 6.25 Ω

Step III

A

Calculation of RTh RTh =

Step IV

VTh 5 = = 6.25 Ω IN 0.8

5V

B

Thevenin’s Equivalent Network (Fig. 3.206) Fig. 3.206

3.54 Circuit Theory and Transmission Lines

  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) = 6 V Writing the VTh equation, 2 − 2 I1 − 0 + 4Vx − VTh = 0 2 − 2( −2) − 0 + 4(6) − VTh = 0 VTh = 30 V 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,

1Ω

Vx

4Vx − +

−

2Ω 2V

−

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

Step IV

10.98 Ω A

30 V

Calculation of RTh VTh 30 = = 10.98 Ω IN 2.73

Thevenin’s Equivalent Network (Fig. 3.210)

A

2Ω

Solving Eqs (ii) and (iii),

RTh =

B

Fig. 3.208

2 − 2 I1 − 1I 2 + 4Vx = 0 2 − 2 I1 − I 2 + 4( 2 − 2 I1 ) = 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,

+

I1

10 Ω

5 − 10 I1 − 10 I1 = 0 I1 =

1Ω

5 = 0.25 A 20

10 Ω

A

VTh

5V −

Writing the VTh equation,

B

Fig. 3.212

5 − 10 I1 + 8 I1 − 0 − VTh = 0 VTh = 5 − 2 I1 = 5 − 2(0.25) = 4.5 V 8I1 − +

Step II Calculation of IN (Fig. 3.213) Applying KVL to Mesh 1,

A

IN

5 − 10 I1 − 10( I1 − I 2 ) = 0

...(i)

20 I1 − 10 I 2 = 5

10 Ω 5V

Applying KVL to Mesh 2, −10( I 2 − I1 ) + 8 I1 − 1I 2 = 0 18 I1 − 11I 2 = 0

1Ω

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

Calculation of RTh

A

4.5 V

RTh Step IV

2Ω

V 4.5 = Th = =2Ω IN 2.25

Thevenin’s Equivalent Network (Fig. 3.214)

B

Fig. 3.214

3.56 Circuit Theory and Transmission Lines

  example 3.49 

Find VTh and RTh between terminals A and B of the network shown in Fig. 3.215 . 1Ω

2Ω

12 V

2Ix

Ix

A

1Ω

B

Fig. 3.215 1Ω

Solution

2Ω +

Step I 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 I x = 0 as shown in Fig. 3.216.

A

−

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 V V  V1 + V1 + 1 − 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 = = 1.5 Ω IN 4

Obtain the Thevenin equivalent network of Fig. 3.218 for the given network at

3.3  Thevenin’s Theorem  3.57 3Ω

2A

4Ω

Vx

5Vx + −

a

2Ω

b

Fig. 3.218 3Ω

Solution Step I Calculation of VTh (Fig. 3.219) Applying KCL at Node x,

Vx

2A

4Ω

5Vx + −

VTh

2Ω

V 2= x 2 Vx = 4 V

a +

−

Writing the VTh equation,

b

Fig. 3.219

VTh = Vx − 5 Vx = − 4 Vx = −16 V

( the terminal a is negative w.r.t. b)

Step II Calculation of IN (Fig. 3.220) Applying KCL at Node x,

3Ω

4Ω

Vx

5Vx + −

a IN

2A

2Ω

Vx Vx − 5 Vx + 2 4 Vx V 2= − Vx = − x 2 2 2=

b

Fig. 3.220

Vx = −4 V V − 5 Vx IN = x = −Vx = 4 A 4 Step III

a

−16 V

Calculation of RTh RTh =

Step IV

−4 Ω

VTh −16 = = −4 Ω IN 4

b

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 Circuit Theory and Transmission Lines Solution Step I Calculation of VTh (Fig. 3.223) From Fig. 3.223 Vx = VTh Applying KCL at the node, 1 Vx − 150 − Vx 3 + Vx + 5 = 0 10 15

150 V 10 Ω

30 Ω A

+

VTh

+ Vx

5A

− B

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

30 Ω A

Step II Calculation of IN (Fig. 3.224) Applying KCL at Node x, 1 Vx − 150 − 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

A

+

5Ω

1A

…(i)

I1

10 Ω I2

VTh

−

−

…(ii) Fig. 3.227

B

3.3  Thevenin’s Theorem  3.59

Applying KVL to Mesh 2, −5( I 2 − I1 ) − 10 I x − 10 I 2 = 0 −5( I 2 − I1 ) − 10( I1 − I 2 ) − 10 I 2 = 0 5 I1 + 5 I 2 = 0

…(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( −1) − 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,

10 Ix

10 V

+ −

…(ii)

A

Ix

…(i)

5Ω

1A I1

IN

10 Ω I2

I3 B

−5( I 2 − I1 ) − 10 I x − 10( I 2 − I 3 ) = 0 −5( I 2 − I1 ) − 10( I1 − I 2 ) − 10( I 2 − I 3 ) = 0

Fig. 3.228

−5 I1 − 5 I 2 + 10 I 3 = 0

…(iii)

Applying KVL to Mesh 3, −10( I 3 − I 2 ) − 10 = 0 10 I 2 − 10 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

Calculation of RTh

Step IV

VTh −20 = = −10 Ω IN 2 Thevenin’s Equivalent Network (Fig. 3.229) RTh =

  example  3.53  Fig. 3.230.

−20 Ω A

−20 V

B

Fig. 3.229

Find Thevenin’s equivalent network at terminals A and B in the network of

3.60 Circuit Theory and Transmission Lines 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 = 5( I1 − I 2 ) Applying KVL to Mesh 1, −4Vx − 2 I1 − 5( I1 − I 2 ) = 0

2Ω

4Ω A

4Vx

− +

I1

+ Vx −

1V

5Ω I2

B

Fig. 3.231 ...(i)

−4 [5( I1 − I 2 ] − 2 I1 − 5 I1 + 5 I 2 = 0 −27 I1 + 25 I 2 = 0

…(ii)

Applying KVL to Mesh 2, −5( I 2 − I1 ) − 4 I 2 − 1 = 0 5 I1 − 9 I 2 = 1

…(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, 20 − 4 I x + 6 I x − 6 I x = 0

6Ix − +

Ix

4Ω

Ix = 5 A

+ 6Ω −

20 V

Writing the VTh equation,

+

VTh −

6 I x − VTh = 0 6(5) − VTh = 0 VTh = 30 V

A

B

Fig. 3.234 6Ix

Step II Calculation of IN (Fig. 3.235). From Fig. 3.235,

− +

Ix = 0 The dependent source 6Ix depends on the controlling variable Ix. When I x = 0, the dependent source vanishes, i.e., 6 I x = 0 as shown in Fig. 3.236. 20 IN = =5A 4

Ix

A

4Ω 6Ω

IN

20 V B

Fig. 3.235

6Ix A

− +

A

4Ω

4Ω IN

20 V

⇒

IN 20 V

B

B

(a)

(b)

Fig. 3.236

6Ω A

Step III Calculation of RTh RTh = Step IV

VTh 30 = =6Ω IN 5

9Ω

30 V IL

Calculation of IL (Fig. 3.237)

B

30 IL = =2A 6+9

  example 3.55   

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 Circuit Theory and Transmission Lines Solution

10 Ω

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., 0.8 I x = 0 VTh = 40 V

+ 40 V

−

10 Ω

40 V

…(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

VTh 40 = = 24 Ω 5 IN 3

16 Ω

40 V IL

Calculation of IL (Fig. 3.242)

B

40 IL = =1A 24 + 16

  example 3.56 

Ix A

Solving Eqs (ii) and (iii),

RTh =

Fig. 3.242

Find the current in the 6 W resistor in Fig. 3.243. 1Ω −

18 V

B

Fig. 3.240

I1

IN = I2 =

A

VTh

Applying KVL to the outer path of the supermesh,

Step IV

6Ω

−

40 − 10 I1 − 6 I 2 = 0 10 I1 + 6 I 2 = 40

B

Fig. 3.239

…(i)

I1 − I 2 = 0.8 I x = 0.8 I 2 I1 − 1.8 I 2 = 0

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,

Step III

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 = −1I1 = − I1

1Ω −

…(i)

For Mesh 1,

Vx

− +

+

18 V

A +

3A

…(ii)

I1 = −3 A

2Vx

VTh

I1 −

Vx = 3 V Writing the VTh equation,

B

Fig. 3.244

18 − 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,

1Ω

Vx = − I1 Meshes 1 and 2 will form a supermesh, Writing current equation for supermesh, I 2 − I1 = 3

…(i)

−

Vx

− +

+

2Vx

IN

18 V

3A

…(ii)

I1

I2

Applying KVL to the outer path of the supermesh, 18 − 1I1 + 2 Vx = 0 18 − I1 + 2( − I1 ) = 0

B

Fig. 3.245

…(iii)

I1 = 6 A Solving Eqs (ii) and (iii), I2 = 9 A IN = I2 = 9 A Step III

Step IV

3Ω A

Calculation of RTh RTh =

VTh 27 = = 3Ω IN 9

6Ω

27 V IL

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 Circuit Theory and Transmission Lines 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 = 0 100 − VTh + 9(50) = 0

Fig. 3.248

VTh = 550 V Step II

Calculation of IN (Fig. 3.249)

A

From Fig. 3.249,

B IN

Vx = 5( I N + 10)

− +

10Vx

+ Vx −

100 V

Applying KVL to Mesh 1, 100 + 10Vx − Vx = 0 Vx = −

100 9

Fig. 3.249

100 − = 5 I N + 50 9 550 A IN = − 45 −45 Ω

Step III

Calculation of RTh

Step IV

550 RTh = = −45 Ω 550 − 45 Calculation of IL (Fig. 3.250) IL =

A

10 Ω

550 V IL

B

550 110 =− A −45 + 10 7

Fig. 3.250

  3.4     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

IN

R2

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 RN =

B

Fig. 3.254  Calculation of RN

VTh IN

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  Norton’s equivalent network

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.

A

RN

B

Fig. 3.256  Norton’s equivalent network

3.66 Circuit Theory and Transmission Lines 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, I3 − I 2 = 4

I1

B

Fig. 3.258 5Ω

Applying KVL to the supermesh, ...(iii)

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

A

15 Ω

B

I 2 = −3 A I3 = 1A I N = I1 − I 2 = 2 − ( −3) = 5 A

Fig. 3.259 A

Calculation of RN (Fig. 3.259) RN = 1  (5 + 15) = 0.95 Ω

Step III

RN

1Ω

I1 = 2 A

Step II

15 Ω

4A I3

I2

...(ii)

−5 I 2 − 15 I 3 = 0

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

20 V

5Ω

2Ω

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 I1 + 20 − 2( I1 − I 2 ) = 0 7 I1 − 2 I 2 = 20

...(i)

2Ω I1

IN I2 B

Applying KVL to Mesh 2,

Fig. 3.262

−2( I 2 − I1 ) − 8 I 2 − 12 = 0 − 2 I1 + 10 I 2 = −12

8Ω

Solving Eqs (i) and (ii),

5Ω

...(ii) RN

2Ω

I 2 = −0.67 A I N = I 2 = −0.67 A Step II

Fig. 3.263

Calculation of RN (Fig. 3.263)

A

RN = (5  2) + 8 = 9.43 Ω Step III

0.67 A

Calculation of IL (Fig. 3.264) I L = 0.67 ×

9.43 Ω

IL

9.43 = 0.33 A (↑) 9.43 + 10

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 Circuit Theory and Transmission Lines Solution Step I Calculation of IN (Fig. 3.266) The resistance of 40 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

50 Ω

50 V

40 Ω

20 Ω

IN

10 V

...(i) Fig. 3.266

Applying KVL to Mesh 2, 10 − 20( I 2 − I1 ) = 0 −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 I 2 = 1.5 A I N = I 2 = 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

I L = 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

Solution Step I Calculation of IN (Fig. 3.271) Applying KVL to Mesh 1,

6Ω

2Ω

1Ω

10 V

10 − 6 I1 − 1( I1 − I 2 ) = 0 7 I1 − I 2 = 10

A

I1

...(i)

3Ω I2

IN I3 B

Applying KVL to Mesh 2,

20 V

−1( I 2 − I1 ) − 2 I 2 − 3( I 2 − I 3 ) = 0

Fig. 3.271

− I1 + 6 I 2 − 3I 3 = 0

...(ii) 6Ω

Applying KVL to Mesh 3,

2Ω A

−3( I 3 − I 2 ) − 20 = 0 3I 2 − 3I 3 = 20

1Ω

...(iii)

3Ω

RN

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

B

I 3 = −13.17 A

Fig. 3.272

I N = I 3 = −13.17 A Step II

2Ω

Calculation of RN (Fig. 3.272)

A IL

RN = [(6  1) + 2]  3 = 1.46 Ω Step III

1.46 Ω

13.17 A

10 Ω

Calculation of IL (Fig. 3.273) I L = 13.17 ×

  example 3.62 

1.46 = 1.68 A (↑) 1.46 + 10

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( I1 − I 2 ) − 40 = 0 20 I1 − 20 I 2 = 10

A

...(i)

Applying KVL to Mesh 2, 40 − 20( I 2 − I1 ) − 20 I 2 − 20( I 2 − I 3 ) = 0 −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 Circuit Theory and Transmission Lines Applying KVL to Mesh 3, −20( I 3 − I 2 ) − 30 I 3 − 100 = 0

...(iii)

−20 I 2 + 50 I 3 = −100 Solving Eqs (i), (ii) and (iii), I1 = 0.81 A I N = I1 = 0.81 A Step II

Calculation of RN (Fig. 3.276) RN A

20 Ω

30 Ω

B

20 Ω

RN = [( 20  30) + 20]  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

120 V

40 V

10 Ω

RL

30 Ω

60 Ω

10 V

Fig. 3.278 30 Ω

Solution Step I Calculation of IN (Fig. 3.279) Applying KVL to Mesh 1, 120 V

120 − 30 I1 − 60( I1 − I 2 ) = 0 90 I1 − 60 I 2 = 120

40 V

10 Ω

IN

30 Ω

60 Ω I1

A

I2

...(i) Fig. 3.279

B

10 V I3

3.4  Norton’s Theorem  3.71

Applying KVL to Mesh 2,

A

10 Ω

30 Ω

RN

B

−60( I 2 − I1 ) + 40 − 10 I 2 − 30( I 2 − I 3 ) = 0 −60 I1 + 100 I 2 − 30 I 3 = 40

...(ii) 30 Ω

60 Ω

Applying KVL to Mesh 3, −30( I 3 − I 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 I N = I 3 = 4.67 A Step II

15 Ω

4.67 A

Calculation of RN (Fig. 3.280) RN = [(30  60) + 10]  30 = 15 Ω

Step III

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

12 Ω

5A

4Ω

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 Circuit Theory and Transmission Lines Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh,

A

I 2 − I1 = 2

...(i)

12 Ω

4Ω

RN

Applying KVL to the supermesh, 60 − 12 I1 − 5 = 0 12 I1 = 55

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

Calculation of RN (Fig. 3.285) RN = 12  4 = 3 Ω

Step III

8Ω

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

IN I1

1V

2Ω

3V I2

B

2Ω

Fig. 3.289 A

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

2Ω

...(i)

RN 3Ω

2Ω

Applying KVL to Mesh 2, −1 − 2( I 2 − I 3 ) − 2 I 2 = 0 4 I 2 − 2 I 3 = −1

B 2Ω

...(ii)

(a) 2Ω

Applying KVL to Mesh 3, −2( I 3 − I1 ) − 2( I 3 − I 2 ) = 0 −2 I1 − 2 I 2 + 4 I 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 Circuit Theory and Transmission Lines

eXAmpLeS WIth dependent SourceS   example 3.66 

Find Norton’s equivalent network across terminals A and B of Fig. 3.292. 3I1 I1

I2

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

3I1 I2

I1

10 Ω

I1 = − I x

+

A

5Ω

Applying KVL to the mesh, Ix

5V

5 − 10 I x − 5 I x − 10 I 2 = 0

10 Ω

− +

VTh

+ 10I2 −

−

5 − 10 I x − 5 I x − 10 I x = 0 I x = 0.2 A

B

Fig. 3.293

I1 = −0.2 A Writing the VTh equation, 5 − 10 I x + 3 I1 − VTh = 0 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 I x − 5( I x − I y ) − 10 I 2 = 0

3I1

…(i) …(ii)

I2

I1

10 Ω

IN 5V

Ix

+ 10I2 −

Iy B

…(iii)

Applying KVL to Mesh 2, 10 I 2 − 5( I y − I x ) + 3I1 − 10 I y = 0 10 I x − 5 I y + 5 I x + 3( I y − I x ) − 10 I y = 0 12 I x − 12 I y = 0

A

5Ω

5 − 10 I x − 5 I x + 5 I y − 10 I x = 0 25 I x − 5 I y = 5

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 I N = I y = 0.25 A Step III

RN = Step IV

9.6 Ω

0.25 A

Calculation of RN VTh 2.4 = = 9.6 Ω IN 0.25

B

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 = 2 I 2

+

−

5Ω

…(i)

+ −

I1 15 Ω

3Vx +

VTh

For Mesh 1, I1 = −3Vx = −3( 2 I 2 ) = −6 I 2

…(ii)

+ Vx

2A I2

For Mesh 2, I2 = 2

A

2Ω

−

−

…(iii)

B

Fig. 3.297

I1 = −6 I 2 = −6( 2) = −12 A

20 Ω

Writing the VTh equation,

A

VTh − 0 + 5 I1 + 15( I1 − I 2 ) − 2 I 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 = 2( I 2 − I 3 )

Vx

2A

…(i)

For Mesh 2, I2 = 2

3Vx

I1 15 Ω

…(ii)

I2

I3

2Ω

− B

Fig. 3.298

3.76 Circuit Theory and Transmission Lines Meshes 1 and 3 will form a supermesh. Writing the current equation for the supermesh, I 3 − I1 = 3Vx = 3 [ 2( I 2 − I 3 ) ] = 6 I 2 − 6 I 3 I1 + 6 I 2 − 7 I 3 = 0 Applying KVL to the outer path of the supermesh, −5 I1 − 20 I 3 − 2( I 3 − I 2 ) − 15( I1 − I 2 ) = 0

…(iii)

−20 I1 + 17 I 2 − 22 I 3 = 0

…(iv)

Solving Eqs (ii), (iii) and (iv), I1 I2 I3 IN Step III

= −0.16 A =2A = 1.69 A = I 3 = 1.69 A

A

Calculation of RN

162.13 Ω

1.69 A

V 274 RN = Th = = 162.13 Ω I N 1.69 Step IV

B

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 + V1

15 V

A

I2 + 10I2 −

8Ω

−

0.6V1

15 Ω

B

Fig. 3.300

Step I Calculation of VTh (Fig. 3.301) From Fig. 3.301, V1 = 8( I x − I y )

2Ω

5Ω

Solution

+

+ V1

15 V Ix

−

8Ω Iy

+ 10I2 −

…(i)

I2

15 Ω

0.6V1 I2

+ VTh

− −

Applying KVL to Mesh 1,

A

B

Fig. 3.301 15 − 5 I x − 8( I x − I y ) = 0 13I x − 8 I y = 15

…(ii)

Applying KVL to Mesh 2, −8( I y − I x ) − 2 I y − 10 I 2 = 0 8 I x − 10 I y − 10 I 2 = 0

…(iii)

3.4  Norton’s Theorem  3.77

For Mesh 3, I 2 = 0.6V1 = 0.6 8( I x − I y )  4.8 I x − 4.8 I y − I 2 = 0

…(iv)

Solving Eqs (ii), (iii) and (iv), I x = 3.28 A 2Ω

5Ω

I y = 3.45 A I 2 = −0.83 A

I2

+ V1

15 V

Writing the VTh equation,

+ 10I2 −

8Ω

−

15 I 2 − VTh = 0 15( −0.83) − VTh = 0 VTh = −12.45 V

0.6V1

15 Ω

A IN

B

Fig. 3.302

Step II Calculation of IN (Fig. 3.302) From Fig. 3.302,

A

I2 = 0 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 + V1

15 V Ix

−

8Ω

0.6V1

IN

Iy B

Fig. 3.303

V1 = 8( I x − I y )

…(i)

Applying KVL to Mesh 1, 15 − 5 I x − 8( I x − I y ) = 0 13I x − 8 I y = 15

…(ii)

Applying KVL to Mesh 2, −8( I y − I x ) − 2 I y = 0 −8 I x + 10 I y = 0

…(iii)

Solving Eqs (ii) and (iii), I x = 2.27 A I y = 1.82 A V1 = 8( I x − I y ) = 8( 2.27 − 1.82) = 3.6 V

A

For Mesh 3, I N = 0.6 V1 = 0.6(3.6) = 2.16 A Step III

Step IV

Calculation of RN V −12.45 RN = Th = = −5.76 Ω IN 2.16 Norton’s Equivalent Network (Fig. 3.304)

2.16 A

−5.76 Ω

B

Fig. 3.304

3.78 Circuit Theory and Transmission Lines

  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,

I1

2 − 2 I1 + 0.5 I1 − 1I1 = 0 2 − 2.5 I1 = 0

VTh

−

2V −

I1 = 0.8 A

B

Fig. 3.306

Writing the VTh equation,

A

1I1 − VTh = 0 1(0.8) − VTh = 0

0.5I1 + −

VTh = 0.8 V

I1

2Ω 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 0.8 RN = Th = = 0.8 Ω IN 1

B

Fig. 3.308

Norton’s Equivalent Network (Fig. 3.309) A

1A

IN

1Ω

Step II Calculation of IN (Fig. 3.307) When a short circuit is placed across the 1 Ω resistor, it gets shorted. I1 = 0

Step IV

A

+

1Ω 2Ω

+

0.8 Ω

B

Fig. 3.309

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

3Ω

…(i)

Applying KVL to Mesh 1,

+

Ix −

9V

9 − 3( I1 − I 2 ) − 6 I1 = 0 9 I1 − 3I 2 = 9

2Ω

I2 −

+

6Ω

+

A

VTh

I1

−

…(ii)

B

Fig. 3.311

For Mesh 2, I 2 = 6 I x = 6 I1 6 I1 − I 2 = 0

…(iii)

Solving Eqs (ii) and (iii), I1 = −1 A I 2 = −6 A Writing the VTh equation, 9 − 3( I1 − I 2 ) + 2 I 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, I x = I1 − I 3

6Ix

…(i)

3Ω

Applying KVL to Mesh 1, 9 − 3( I1 − I 2 ) − 6( I1 − I 3 ) = 0 9 I1 − 3I 2 − 6 I 3 = 9

2Ω

I2

A

Ix

…(ii)

9V

IN

6Ω I1

I3

For Mesh 2,

B

I 2 = 6 I x = 6( I1 − I 3 ) 6 I1 − I 2 − 6 I 3 = 0 …(iii)

Fig. 3.312

Applying KVL to Mesh 3, −6( I 3 − I1 ) − 2( I 3 − I 2 ) = 0 −6 I1 − 2 I 2 + 8 I 3 = 0

…(iv)

3.80 Circuit Theory and Transmission Lines Solving Eqs (ii), (iii) and (iv),

A

I1 I2 I3 IN

=5A =3A = 4.5 A = I 3 = 4.5 A

B

Step III

Calculation of RN

Step IV

V −18 RN = Th = = −4 Ω IN 4.5 Norton’s Equivalent Network (Fig. 3.313)

  example 3.71   

−4 Ω

4.5 A

Fig. 3.313

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 + 0.5 I + = 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 VTh + −

The dependent source of −2 Ix depends on the controlling variable Ix. When I x = 0, the dependent source vanishes, i.e. −2 I x = 0 as shown in Fig. 3.319. I1 = 2

+ −

−10 V

2A

10 Ω

Fig. 3.318 A B Ix VTh + −

Writing the VTh equation, −10 − VTh − 4 I1 = 0 −10 − VTh − 4( 2) = 0

−10 V

+ −

4Ω

2A

10 Ω

I1

VTh = −18 V Step II Calculation of IN (Fig. 3.320) From Fig. 3.320,

Fig. 3.319

I x = I1

−2Ix

A IN B Ix

…(i)

Mesh 1 and 2 will form a supermesh. Writing the current equation for supermesh, I 2 − I1 = 2

4Ω

the −10 V

…(ii)

+ −

4Ω

2A I1

Applying KVL to the outer path of the supermesh,

I2

10 Ω I3

Fig. 3.320

−10 − 4( I 2 − I 3 ) = 0 −4 I 2 + 4 I 3 = 10

…(iii)

For Mesh 3, I 3 = −( −2 I x ) = 2 I x = 2 I1 2 I1 − I 3 = 0

…(iv)

3.82 Circuit Theory and Transmission Lines Solving Eqs (ii), (iii) and (iv), I1 I2 I3 IN Step III

A IL

Calculation of RN RN =

Step IV

= 4.5 A = 6.5 A =9A = I1 = 4.5 A

VTh −18 = = −4 Ω IN 4.5

4.5 A

2Ω

−4 Ω

Calculation of IL (Fig. 3.321)

B

I L = 4.5 ×

  example 3.73 

−4 =9A −4 + 2

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 VTh

5V − +

Writing the VTh equation,

4Vi −

−4Vi − VTh = 0 VTh = −4Vi = −4( −1) = 4 V −

Applying KVL to the mesh,

Vi

+

A

1Ω IN

5V − +

4Vi B

−4Vi − 1I N = 0 I N = −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

Fig. 3.324

3.4  Norton’s Theorem  3.83

Step III

Calculation of RN

A

RN = Step IV

VTh 4 = = 0.2 Ω IN 20

2Ω

0.2 Ω

20 A

Calculation of IL (Fig. 3.325)

B

0.2 I L = 20 × = 1.82 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

2Ix

10 V

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 − 1I1 − 3I 2 = 0 I1 + 3I 2 = 10 …(ii) Solving Eqs (i) and (ii),

1Ω

Ix

2Ix +

+ 10 V

3Ω

1A I1

I2

A

VTh

−

−

B

Fig. 3.327

I1 = 1.75 A I 2 = 2.75 A Writing the VTh equation, 3I 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 − 1I1 − 3( I 2 − I 3 ) = 0

1Ω

Ix

2Ix

1A

10 V I1

I2

Fig. 3.328

3Ω

IN I3

3.84 Circuit Theory and Transmission Lines I1 + 3I 2 − 3I 3 = 10

…(iii)

For Mesh 3, I 3 = 2 I x = 2 I1 2 I1 − I 3 = 0

…(iv)

Solving Eqs (ii), (iii) and (iv), I1 = −3.5 A I 2 = −2.5 A I 3 = −7 A I N = 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

Calculation of IL (Fig. 3.329)

Step IV

I L = −7 ×

B

−1.18 = 10.07 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 I x + 3I x − 5 I x = 0

+

5Ω

VTh

+

2Ω

−

10 V

I x = 2.5 A

−

Writing the VTh equation,

3Ix

5( 2.5) − VTh = 0

− +

Step II Calculation of IN (Fig. 3.332) From Fig. 3.332, Ix = 0

B

Fig. 3.331

5 I x − VTh = 0 VTh = 12.5 V

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. IN Step III

A

2Ω

10 = =5A 2

IN

10 V B

Calculation of RN

Fig. 3.333 V 12.5 RN = Th = = 2.5 Ω IN 5

Step IV

A IL

Calculation of IL (Fig. 3.334) IL = 5 ×

10 Ω

2.5 Ω

5A

2.5 =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

4Ω

Ix

+ −

5Ω

12 V

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 I x 12 − 10 I x Ix Writing the VTh equation, 12 − 2 I x − VTh 12 − 2(1.2) − VTh VTh

=0 =0 = 1.2 A

2Ω +

−

+ A V Th − B

12 V

+ −

4Ix

+ −

4Ix

Fig. 3.336

=0 =0 = 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 I1 = 0 12 V I1 = 6 A …(ii) Applying KVL to Mesh 2, −4 I 2 − 4 I x = 0 −4 I 2 − 4 I1 = 0 …(iii) Solving Eqs (ii) and (iii), I 2 = −6 A I N = I1 − I 2 = 6 − ( −6) = 12 A

4Ω

Ix

2Ω

4Ω

Ix

IN I1

I2

Fig. 3.337

3.86 Circuit Theory and Transmission Lines Step III

Calculation of RN

Step IV

IL

VTh 9.6 = = 0.8 Ω IN 12

RN =

12 A

5Ω

0.8 Ω

Calculation of IL (Fig. 3.338) I L = 12 ×

  example 3.77 

0.8 = 1.66 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 5Ω

Solution Step I Calculation of VTh (Fig. 3.340) For the mesh, I = −0.5Vx = −0.5VTh

+

4Ω 5V

0.5Vx

VTh = Vx

I −

Writing the VTh equation, 5Ω

5 − 4( −0.5VTh ) − VTh = 0

Step II Calculation of IN (Fig. 3.341) From Fig. 3.341, Vx = 0

+

4Ω

Step III

Vx −

IN

B

Fig. 3.341 5Ω A

4Ω

IN

5V B

Calculation of RN RN =

Step IV

5 5 = A 4+5 9

0.5Vx

A

5V

The dependent source of 0.5Vx 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. IN =

B

Fig. 3.340

5 − 4 I − 0 − VTh = 0 VTh = −5 V

A

VTh −5 = = −9 Ω 5 IN 9

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

Vx

5I1

25 Ω

10 Ω

−

Fig. 3.344 Solution Step I Calculation of VTh (Fig. 3.345) From Fig. 3.345, Vx = −25(5 I1 ) = −125 I1 …(i)

1000 Ω

I1 A

+ + + −

12 V

2Vx

5I1

Vx

VTh

25 Ω

−

Applying KVL to Mesh 1,

−

12 − 1000 I1 − 2Vx = 0 12 − 1000 I1 − 2( −125 I1 ) = 0

B

Fig. 3.345

…(ii)

I1 = 0.016 A Vx = −125 I1 = −125(0.016) = −2 V Writing the VTh equation,

1000 Ω

I1 A

VTh = 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. 12 = 0.012 A 1000 = −5 I1 = −5(0.012) = −0.06 A

I1 = IN Step III

Fig. 3.346 1000 Ω

A 12 V

5I1

B

VTh −2 = = 33.33 Ω IN −0.06

A IL

−0.06 A

Step IV

IN

Fig. 3.347

Calculation of RN RN =

I1

33.33 Ω

10 Ω

Calculation of IL (Fig. 3.348) I L = −0.06 ×

33.33 = −0.046 A 33.33 + 10

B

Fig. 3.348

3.88 Circuit Theory and Transmission Lines

  example 3.79   

Find the current through the 5 W resistor for the network of Fig. 3.349. 4V

1Ω

2Ω +

1Ω

Vx

− + −

2A

5Ω 4Vx

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,

4V

1Ω

…(i)

Solving Eqs (ii) and (iii),

−

…(ii)

+ −

I

2A

4Vx

4V

…(i)

1Ω

I2

2Ω +

…(ii) 1Ω

Vx

− + −

I1

2A

A IN

4Vx B

…(iii) Fig. 3.351

I1 = 2 A I 2 = 2.36 A I N = I 2 = 2.36 A

Step III

Calculation of RN RN =

Step IV

+ A VTh − B

Fig. 3.350

4[2( I1 − I 2 )] − 2 I 2 + 2 I1 − I 2 + I1 + 4 = 0 11I1 − 11I 2 = −4

Vx

1Ω

4Vx + 2 I + 1I + 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( I 2 − I1 ) + 4 = 0

2Ω +

VTh 26 = = 11.02 Ω IN 2.36

A IL

2.36 A

11.02 Ω

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

3Ix

12 V

3Ω

1Ω

Fig. 3.353 Solution Step I Calculation of VTh (Fig. 3.354) From Fig. 3.354,

6Ω

I x = I1

+

+

…(i)

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 I1 − I 2 = 0

Ix

I1

I2

3Ω −

VTh −

Fig. 3.354

…(ii)

Applying KVL to the outer path of the supermesh, 12 − 6 I1 − 3I 2 = 0 6 I1 + 3I 2 = 12

…(iii)

Solving Eqs (ii) and (iii), I1 = 0.67 A I 2 = 2.67 A

6Ω

Ix

A

Writing the VTh equation, 3I 2 − VTh = 0

3Ix

12 V

3( 2.67) − VTh = 0 VTh = 8 V

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,

Fig. 3.355 6Ω

Ix

A 3Ix

12 V I1

IN I2 B

I 2 − I1 = 3I x = 3I1 4 I1 − I 2 = 0

3Ω

…(ii)

Fig. 3.356

Applying KVL to the outer path of the supermesh, 12 − 6 I1 = 0 I1 = 2

…(iii)

3.90 Circuit Theory and Transmission Lines Solving Eqs (ii) and (iii), I1 = 2 A I2 = 8 A

A IL

IN = I2 = 8 A Step III

Calculation of RN RN =

Step IV

1Ω

1Ω

8A

VTh 8 = =1Ω IN 8

B

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

I x = I1 − I 2 …(i) I1 = 10

+

A

+

1Ω

10 A I1

For Mesh 1,

− +

6Ω I2

VTh

− −

…(ii)

B

Fig. 3.359

Applying KVL to Mesh 2, −1( I 2 − I1 ) + 3I x − 6 I 2 = 0 − I 2 + I1 + 3( I1 − I 2 ) − 6 I 2 = 0 4 I1 − 10 I 2 = 0 Solving Eqs (ii) and (iii), I1 = 10 A I2 = 4 A Writing the VTh equation,

6 I 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, I x = I1 − I 2

…(i)

I1 = 10

…(ii)

3Ix A

− +

Ix

1Ω

10 A

IN

6Ω

For Mesh 1,

B

Fig. 3.360

Applying KVL to Mesh 2,

3Ix

−1( I 2 − I1 ) + 3I x = 0 − I 2 + I1 + 3( I1 − I 2 ) = 0 4 I1 − 4 I 2 = 0 Solving Eqs (ii) and (iii), I1 = 10 A

…(iii) I1

I2 B

Fig. 3.361

Calculation of RN V 24 RN = Th = = 2.4 Ω IN 10

Step IV

IN

1Ω

10 A

I 2 = 10 A I N = I 2 = 10 A Step III

A

− +

Ix

A IL

2.4 Ω

10 A

1.6 Ω

Calculation of IL (Fig. 3.362) I L = 10 ×

2.4 =6A 2.4 + 1.6

B

Fig. 3.362

  3.5     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

V I= Rs + RL V 2

Power delivered to the load RL = P = I RL =

V 2 RL

RL I

( Rs + RL ) 2 To determine the value of RL for maximum power to be transferred Fig. 3.363  N   etwork illustrating  to the load, maximum power transfer  dP theorem =0 dRL dP d V2 RL = dRL dRL ( Rs + RL ) 2 =

V 2 [( Rs + RL ) 2 − ( 2 RL )( Rs + RL )] ( Rs + RL ) 4

3.92 Circuit Theory and Transmission Lines ( Rs + RL ) 2 − 2 RL ( Rs + RL ) = 0 Rs 2 + 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 = RTh

VTh IL

B

RL = RTh

Fig. 3.364  Thevenin’s equivalent network

5. Find the maximum power (Fig. 3.364). IL =

VTh V = Th RTh + RL 2 RTh

Pmax = I L2 RL =

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 − 2I − 6 = 0 I = −0.75 A Writing the VTh equation, 6 + 2 I − VTh − 10 = 0 VTh = 6 + 2 I − 10 = 6 + 2( −0.75) − 10 = −5.5 V = 5.5 V( terminal B is positive w.r.t A)

2Ω +

A −

+

+

VTh

2Ω

3V I

B − 10 V

− 6V 2Ω

Fig. 3.366

3.5  Maximum Power Transfer Theorem  3.93

Step II

2Ω

Calculation of RTh (Fig. 3.367)

A

RTh = ( 2  2) + 2 = 3 Ω

RTh

B

2Ω

Step III Calculation of RL For maximum power transfer, RL = RTh = 3 Ω Step IV

2Ω

Calculation of Pmax (Fig. 3.368)

Fig. 3.367 3Ω A

Pmax =

2 VTh

4 RTh

2

=

(5.5) = 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 5Ω

Solution Step I Calculation of VTh (Fig. 3.370) I 2 − I1 = 4 Applying KVL to the outer path, 8 − 1I1 − 5 I1 − 5 I 2 − 10 = 0

...(i)

−6 I1 − 5 I 2 = 2

...(ii)

−

1Ω + 8V

Step II

+

+ A V Th − B I1

...(ii)

4A I2

1Ω

A R Th B

Calculation of RTh (Fig. 3.371) RTh = 10  1 = 0.91 Ω

5Ω − 10 V

5Ω

Writing the VTh equation, 8 − 1I1 − VTh = 0 VTh = 8 − I1 = 8 − ( −2) = 10 V

−

Fig. 3.370

Solving Eqs (i) and (ii), I1 = −2 A I2 = 2 A

+

Fig. 3.371

5Ω

3.94 Circuit Theory and Transmission Lines Step III Calculation of RL For maximum power transfer,

0.91 Ω A

RL = RTh = 0.91 Ω 0.91 Ω

10 V

Step IV

Calculation of Pmax Pmax =

2 (10) 2 VTh = = 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

RL

3Ω

5Ω

Fig. 3.373 10 Ω

Solution

+

Step I Calculation of VTh (Fig. 3.374) For Mesh 1, I1 = 50

2Ω −

+

−

+ −

+

3Ω

5Ω

50 A I1

+

I2

− +

VTh

− −

Applying KVL to Mesh 2, Fig. 3.374

−5( I 2 − I1 ) − 2 I 2 − 3I 2 = 0 5 I1 − 10 I 2 = 0 I1 = 2 I 2 I 2 = 25 A

10 Ω

2Ω A

5Ω

VTh = 3I 2 = 3( 25) = 75 V

3Ω

RTh = (5 + 2)  3 = 2.1 Ω

Fig. 3.375 2.1 Ω

Step III Calculation of RL For maximum power transfer, RL = RTh = 2.1 Ω Step IV

A

2.1 Ω

75 V

Calculation of Pmax (Fig. 3.376)

Pmax =

2 (75) 2 VTh = = 669.64 W 4 RTh 4 × 2.1

RTh B

Calculation of RTh (Fig. 3.375)

Step II

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, I 2 − I1 = 6

+

5Ω +

...(i)

10 V

+

I1

6A I2

2Ω

5 I1 + 2 I 2 = 10

VTh

− −

Applying KVL to the supermesh, 10 − 5 I1 − 2 I 2 = 0

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, VTh = 2 I 2 = 11.42 V Step II

B

4Ω

Calculation of RTh (Fig. 3.379) Fig. 3.379

RTh = (5  2) + 3 + 4 = 8.43 Ω Step III Calculation of RL For maximum power transfer,

8.43 Ω A

RL = RTh = 8.43 Ω 8.43 Ω

11.42 V

Step IV

Calculation of Pmax (Fig. 3.380) Pmax =

2 (11.42) 2 VTh = = 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 Circuit Theory and Transmission Lines 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 V

120 − 10 I1 − 5( I1 − I 2 ) = 0

I1

15 I1 − 5 I 2 = 120 Writing current equation for Mesh 2, I 2 = −6 Solving Eqs (i) and (ii),

−

...(i)

+ −

V Th

5Ω − + I2

B −

6A

Fig. 3.382

...(ii) 10 Ω

I1 = 6 A Writing the VTh equation,

A R Th B

120 − 10 I1 − VTh = 0 VTh = 120 − 10(6) = 60 V Step II

A +

Calculation of RTh (Fig. 3.383)

5Ω

Fig. 3.383

RTh = 10  5 = 3.33 Ω 3.33 Ω

Step III Calculation of RL For maximum power transfer, RL = RTh = 3.33 Ω Step IV

A

3.33 Ω

60 V

Calculation of Pmax (Fig. 3.384) Pmax =

2 (60) 2 VTh = = 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( I 2 − I1 ) − 10 I 2 − 6 I 2 = 0 −25 I1 + 41I 2 = 0

−

+ − 3A I1

20 V

…(ii)

+

25 Ω

6Ω

I2

− +

−

Fig. 3.386

Solving Eqs (i) and (ii),

10 Ω

I 2 = 1.83 A Writing the VTh equation,

R Th

20 + VTh − 10 I 2 − 6 I 2 = 0 VTh = −20 + 10 (1.83) + 6 (1.83) = 9.28 V Step II

A 25 Ω

Calculation of RTh (Fig. 3.387)

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 Step IV

6Ω

B

A

9.76 Ω

9.28 V

Calculation of Pmax (Fig. 3.388) Pmax =

2 (9.28) 2 VTh = = 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 Circuit Theory and Transmission Lines Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh, I 2 − I1 = 1

…(i)

Writing the voltage equation for the supermesh, 5 − 1I1 − 10( I 2 − I 3 ) = 0

…(ii)

I1 + 10 I 2 − 10 I 3 = 5 Applying KVL to Mesh 3, −10( I 3 − I 2 ) − 2 I 3 − 3I 3 = 0 −10 I 2 + 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 = 3I 3 = 2.76 V Step II

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 = 6.48 Ω

6.48 Ω A

Step III Calculation of RL For maximum power transfer, RL = RTh = 6.48 Ω Step IV

6.48 Ω

2.76 V

Calculation of Pmax (Fig. 3.392) Pmax =

2 ( 2.76) 2 VTh = = 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 3A

2A

8V

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 I 2 − I1 = 2 I 2 = −3 A

…(i) …(ii)

Solving Eqs (i), and (ii), I1 = −5 A

2Ω

RTh

1Ω A

Writing the VTh equation,

B

2Ω

8 − 2 I1 − 1I 2 − VTh − 6 = 0 VTh = 8 − 2( −5) − ( −3) − 6 = 15 V Step II

Fig. 3.395

Calculation of RTh (Fig. 3.395) RTh = 5 Ω

Step III Calculation of RL For maximum power transfer,

5Ω A

RL = RTh = 5 Ω 5Ω

15 V

Step IV

Calculation of Pmax (Fig. 3.396) Pmax =

2 (15) 2 VTh = = 11.25 W 4 RTh 4 × 5

B

Fig. 3.396

3.100 Circuit Theory and Transmission Lines

  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

27 Ω

15 Ω

10 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

+

−

27 Ω

15 Ω

10 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

5Ω

= 100 − 14( 2.08) = 70.88 V

B

−

9Ω

+

− 5Ω

100 − 5 I − VTh − 9 I = 0 VTh = 100 − 14 I

VTh

5Ω

+

Writing the VTh equation,

+

9Ω 20 Ω

I 100 V

Fig. 3.399

9Ω

−

3.5  Maximum Power Transfer Theorem  3.101

Calculation of RTh (Fig. 3.400)

Step II

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

(b)

(d)

(c)

Fig. 3.400 RTh = 23.92 Ω 23.92 Ω

Step III Calculation of RL For maximum power transfer,

A

RL = RTh = 23.92 Ω

23.92 Ω

70.88 V

Step IV

Calculation of Pmax (Fig. 3.401) Pmax =

2 (70.88) 2 VTh = = 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 Circuit Theory and Transmission Lines Solution Step I Calculation of VTh (Fig. 3.403) Applying KVL to Mesh 1,

2A

5Ω +

−

80 − 5 I1 − 10( I1 − I 2 ) − 20( I1 − I 2 ) − 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, 10 Ω

VTh − 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,

8.57 Ω A

RL = RTh = 8.57 Ω Step IV

Pmax =

8.57 Ω

48.6 V

Calculation of Pmax (Fig. 3.405) 2 ( 48.6) 2 VTh = = 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 = 2.5 A 10 + 30 100 = 1.66 A I2 = 20 + 40 I1 =

+

100 V

A

+ 30 Ω

Writing the VTh equation,

V Th

+

20 Ω −

− B

+ −

−

40 Ω

Fig. 3.407

VTh + 10 I1 − 20 I 2 = 0 VTh = 20 I 2 − 10 I1 = 20(1.66) − 10( 2.5) = 8.2 V Step II

Calculation of RTh (Fig. 3.408) 10 Ω

20 Ω

R Th A

B

30 Ω

40 Ω

Fig. 3.408 Redrawing the network (Fig. 3.409), 10 Ω

RTh = (10 || 30) + ( 20 || 40) = 20.83 Ω

20 Ω

A

B

30 Ω

40 Ω

Fig. 3.409 Step III Value of RL For maximum power transfer,

20.83 Ω A

RL = RTh = 20.83 Ω Step IV

Calculation of Pmax (Fig. 3.410) Pmax =

20.83 Ω

8.2 V

2 (8.2) 2 VTh = = 0.81 W 4 RTh 4 × 20.83

B

Fig. 3.410

3.104 Circuit Theory and Transmission Lines

  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,

A

6Ω + −

72 − 6 I1 − 3( I1 − I 2 ) = 0

72 V

9 I1 − 3I 2 = 72 …(i)

+ I1

+

Applying KVL to Mesh 2,

3Ω

−3( I 2 − I1 ) − 2 I 2 − 4 I 2 = 0 −3I1 + 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 − 6 I1 − 2 I 2 = 0 VTh = 6 I1 + 2 I 2 = 6(9) + 2(3) = 60 V Step II

Calculation of RTh (Fig. 3.413) A

6Ω

2Ω

R Th B

6Ω

A

3Ω 4Ω

2Ω

R Th 3Ω

4Ω

B

Fig. 3.413 RTh = [(6 || 3) + 2] || 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,

A

RL = RTh = 2 Ω Step IV

2Ω

60 V

Calculation of Pmax (Fig. 3.414) Pmax =

2 (60) 2 VTh = = 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

25 A

5Ω

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

25 A

5Ω

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 Circuit Theory and Transmission Lines Applying KCL at the node,

10 Ω

VTh − 125 V − 30 − 10 + Th =0 15 2

2Ω

A R Th

5Ω

VTh = 58.81 V Step II

10 Ω

B

Calculation of RTh (Fig. 3.418)

Fig. 3.418

RTh = 15 || 2 = 1.76 Ω Step III Value of RL For maximum power transfer

A R Th

15 Ω

RL = RTh = 1.76 Ω Step IV

2Ω

B

Calculation of Pmax (Fig. 3.420)

Fig. 3.419 1.76 Ω

Pmax =

A 1.76 Ω

58.81 V

2 (58.81) 2 VTh = = 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 + 5I + 6 − VTh = 0 VTh = 8 + 6 + 5 I = 8 + 6 + 5( −1.43) = 6.85 V Step II

Calculation of RTh (Fig. 3.423) 2Ω

2Ω A

4Ω

RTh = ( 2 || 5) + 2 = 3.43 Ω

10 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 (6.85) 2 VTh = = 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 Circuit Theory and Transmission Lines Solution Step I Calculation of VTh (Fig. 3.426) Applying KVL to the mesh,

I

20 Ω

10 I − 20 I − 40 I − 50 = 0 I = −1 A Writing the VTh equation,

40 Ω

A + V Th B

+ −

10 I

−

50 V

VTh − 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 = I2

20 Ω

…(i)

Applying KVL to Mesh 1, 10 I − 20 I1 = 0 10 I 2 − 20 I1 = 0

40 Ω

A + −

10I

B

I1

IN I2

50 V

…(ii)

Applying KVL to Mesh 2, −40 I 2 − 50 = 0 I 2 = −1.25 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,

Fig. 3.427 …(iii)

16 Ω

A

RL = RTh = 16 Ω Step V

Pmax =

16 Ω

10 V

Calculation of Pmax (Fig. 3.428) (10) 2 VTh = = 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,

3Ω

− +

+

I x = I2

…(i)

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( I 2 − I1 ) + 2 I x − 6 I 2 = 0 −4 I 2 + 4 I1 + 2 I 2 − 6 I 2 = 0 4 I1 − 8 I 2 = 0

…(iii)

Solving Eqs (ii) and (iii), I1 = 10 A I2 = 5 A Writing the VTh equation, 6 I 2 − 0 − VTh = 0 VTh = 6 I 2 = 6(5) = 30 V Step II Calculation of IN (Fig. 3.431) From Fig. 3.431,

2Ix

I x = I 2 − I3

6Ω

4Ω

10 A

…(ii)

A Ix

…(i)

For Mesh 1, I1 = 10

3Ω

− +

I1

I2

IN

I3

B

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

Fig. 3.431

−4 I 2 + 4 I1 + 2( I 2 − I 3 ) − 6 I 2 + 6 I 3 = 0 4 I1 − 8 I 2 + 4 I 3 = 0

…(iii)

Applying KVL to Mesh 3, −6( I 3 − I 2 ) − 3I 3 = 0 6 I 2 − 9I3 = 0 Solving Eqs (ii), (iii) and (iv), I1 = 10 A I 2 = 7.5 A I3 = 5 A I N = I3 = 5 A

…(iv)

3.110 Circuit Theory and Transmission Lines Step III

Calculation of RTh

6Ω

RTh

A

V 30 = Th = =6Ω IN 5

6Ω

30 V

Step IV Calculation of RL For maximum power transfer,

B

RL = RTh = 6 Ω Step V

Fig. 3.432

Calculation of Pmax Pmax =

2 (30) 2 VTh = = 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. 2V

1Ω −

2Vx

Vx

+

+ −

1Ω RL 1A

Fig. 3.433 Solution Step I Calculation of VTh (Fig. 3.434) From Fig. 3.434, Vx = −1I = − I

−

…(i)

2Vx

For Mesh 1, I = −1 Vx = 1 V

2V

1Ω Vx

+

+

A

1Ω

+ −

V Th 1A

I

…(ii)

−

B

Fig. 3.434

Writing the VTh equation, 2Vx − 1I + 2 − VTh = 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( − I1 ) − I1 + 2 = 0 3I1 = 0

2V

1Ω −

…(i) 2Vx

…(ii)

+ −

Vx

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 = RTh = 3 Ω Step V

Pmax =

  example 3.99 

3Ω

5V

Calculation of Pmax ( Fig. 3.436) 2 (5) 2 VTh = = 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 − +

3A

4Ω

+

4Ω

V

RL −

Fig. 3.437 0.5 VTh

Solution Step I Calculation of VTh (Fig. 3.438) By source transformation, From Fig. 3.438, VTh = 4 I

− +

4Ω

3A

+ 4Ω

VTh

−

Applying KVL to the mesh,

B

Fig. 3.438

12 − 4 I + 0.5 VTh − 4 I = 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. V =0

A

4Ω 12 V

+

4Ω

A

VTh

I −

Fig. 3.439

B

3.112 Circuit Theory and Transmission Lines 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 0.5 V

4Ω

− +

4Ω

4Ω

A

+

IN

V

B

−

A 12 V IN

12 V

B

Fig. 3.440

Fig. 3.441

Step III Calculation of RTh RTh

2.67 Ω

A

V 8 = Th = = 2.67 Ω IN 3

2.67 Ω

8V IL

Step IV Calculation of RL For maximum power transfer, RL = RTh = 2.67 Ω Step V

B

Fig. 3.442

Calculation of Pmax (Fig. 3.442) Pmax =

2 (8) 2 VTh = =6W 4 RTh 4 × 2.67

  3.6     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. 3.443). A R1 V1

R2

Rn

V2

Vn

A Rm

⇒ Vm B

Fig. 3.443  Millman’s network where

Vm =

V1G1 + V2G2 + … + VnGn G1 + G2 + … + Gn

and

Rm =

1 1 = Gm G1 + G2 + … + Gn

B

3.6  Millman’s Theorem  3.113

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. 3.444. A I1

R1

I2

R2

In

Rn B

Fig. 3.444  Equivalent network Let Im be the resultant current of the parallel current sources and Rm be the equivalent resistance as shown in Fig. 3.445. I m = I1 + I 2 + … + I n =

Im

Rm

V1 V2 V + + … + n = V1 G1 + V2 G2 + …Vn Gn R1 R2 Rn

1 1 1 1 = + +…+ Rm R1 R2 Rn Gm = G1 + G2 + … + Gn

Fig. 3.445  Equivalent network A Rm

By source transformation, the parallel circuit can be converted into a series circuit as shown in Fig. 3.446.

Vm

I V G + V G + … + Vn Gn Vm = I m Rm = m = 1 1 2 2 Gm G1 + G2 + … + Gn

B

Fig. 3.446  M   illman’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. 3.447). I1

I2

In

Im

⇒ R1

R2

Rn

A

Rm B

Fig. 3.447  Millman’s network

where

I1 R1 + I 2 R2 + … + I n Rn R1 + R2 + … + Rn Rm = R1 + R2 + … + Rn Im =

A

B

3.114 Circuit Theory and Transmission Lines 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 =

1 G1 + 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 3.100 

Vm Rm + RL

Find Millman’s equivalent for the left of the terminals A-B in Fig. 3.448. A 4Ω 2Ω

1A

6V B

Fig. 3.448 A

Solution By source transformation, the network is redrawn as shown in Fig. 3.449. Step I Calculation of Vm  1  1 6  + 2   2  4 V G +V G = 3.33 V Vm = 1 1 2 2 = 1 1 G1 + G2 + 4 2 Step II

2Ω

6V

2V B

Fig. 3.449 A

Calculation of Rm Rm =

Step III

4Ω

1 1 1 = = = 1.33 Ω 1 1 Gm G1 + G2 + 4 2

Millman’s Equivalent Network (Fig. 3.450)

1.33 Ω 3.33 V B

Fig. 3.450

3.6  Millman’s Theorem  3.115

  example 3.101 

Find the current through the 10 W resistor in the network of Fig. 3.451.

2Ω

4Ω

6Ω

5V

10 V

15 V

10 Ω

Fig. 3.451 Solution Step I Calculation of Vm  1  1  1 5   − 10   + 15    6  4  2 V1 G1 + V2 G2 + V3 G3 Vm = = = 2.73 V 1 1 1 G1 + G2 + G3 + + 2 4 6 Step II

Calculation of Rm 1 1 = = 1.09 Ω 1 1 1 Gm + + 2 4 6 Calculation of IL (Fig. 3.452) Rm =

Step III

IL =

  example 3.102 

1.09 Ω 2.73 V

2.73 = 0.25 A 1.09 + 10

10 Ω IL

Fig. 3.452

Find the current through the 10 W resistor in the network of Fig. 3.453. 50 Ω

2Ω 50 V

40 Ω

10 Ω 100 V

Fig. 3.453 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. 3.454. Step I Calculation of Vm  1  1 50   − 100    20   50  V1 G1 + V2 G2 = −57.15 V Vm = = 1 1 G1 + G2 + 50 20

50 Ω

20 Ω 10 Ω

50 V

40 Ω

100 V

Fig. 3.454

3.116 Circuit Theory and Transmission Lines Step II

Calculation of Rm 1 1 1 = = = 14.29 Ω 1 1 Gm G1 + G2 + 50 20 Calculation of IL (Fig. 3.455) Rm =

Step III

10 Ω

57.15 V

57.15 IL = = 2.35 A 14.29 + 10

  example  3.103 

IL

14.29 Ω

Fig. 3.455

Draw Millman’s equivalent network across terminals AB in the network of

Fig. 3.456.

C

6V

2Ω

3V

1Ω A

2Ω

4A

8V

6V

2Ω

1Ω B D

5V

5Ω

Fig. 3.456 Step I By source transformation, the network is redrawn as shown in Fig. 3.457.

C

6V

2Ω

3V

1Ω A

8V

8V

6V

2Ω

2Ω

1Ω B D

5V

5Ω

Fig. 3.457 Step II Applying Millman’s theorem at terminals CD,

Vm1

 1  1 −8   + 8   + 6(1)  2  2 V G +V G +V G = 1 1 2 2 3 3 = =3V 1 1 G1 + G2 + G3 + +1 2 2

   Exercises  3.117

1 1 1 = = = 0.5 Ω Gm1 G1 + G2 + G3 1 1 + +1 2 2 Applying Millman’s theorem at terminals CA,

0.67 Ω

4V

Rm1 =

Step III

Vm2

Rm2

A 3V

0.5 Ω

 1 6   + 3(1)  2 V G + V5 G5 =4V = 4 4 = 1 G4 + G5 +1 2 1 1 1 = = = = 0.67 Ω Gm2 G4 + G4 1 +1 2

B

5Ω

5V

Fig. 3.458 A 6.17 Ω

Step IV Millman’s Equivalent Network (Fig. 3.458) Simplifying Fig. 3.458 further, the Millman’s equivalent network is shown in Fig. 3.459.

2V B

Fig. 3.459

Exercises 9Ω

Superposition theorem 3.1

Find the current through the 10 Ω resistor in Fig. 3.460. 10 Ω

15 A

30 Ω

5Ω 5A

7Ω 2Ω 10 A

20 Ω

5Ω

3Ω

100 V 4V

Fig. 3.462

Fig. 3.460

[3.3 V ]

[0.37 A ] 3.2

Find the current through the 8 Ω resistor in Fig. 3.461.

3.4

Calculate the current through the 10 Ω resistor in Fig. 3.463. 10 Ω

8Ω

5A

12 Ω

30 Ω

4Ω

25 A 25 V

+ −

2Ω − +

7Ω

12 V

Fig. 3.461

[16.2 A ] 3.3

Find the potential across the 3 Ω resistor in Fig. 3.462.

2Ω

3Ω

Fig. 3.463

[1.62 A ]

3.118 Circuit Theory and Transmission Lines Find the current through the 1Ω resistor in Fig. 3.464.

3.5

3.9

Determine the voltages V1 and V2 in Fig. 3.468. 4Ω

V1

4Ω

+ −

0.5V2

V2

I1 4 Ω

2Ω 1Ω 3Ω

1A

10 A

0.25I1

20 V

2Ω 1V

Fig. 3.468

2Ω

[6 V,12 V ]

Fig. 3.464

[0.41 A ]

3.10

Find the voltage Vx in Fig. 3.469.

4Ω

10 V

2Ω

2Ω

2Ω

I1 4A

20 Ω

30 Ω

+ Vx −

0.4I1

Fig. 3.469

5A

2Ω

2Ω

60 V

10 Ω

Find the current through the 4 Ω resistor in Fig. 3.465.

3.6

[7.5 V ]

6V

thevenin’s theorem

Fig. 3.465

[1.33 A ] 3.7

3.11 Find the current through the 5 Ω resistor in Fig. 3.470.

Find the current Ix in Fig. 3.466. Ix

5Ω

50 V

10 Ω

20 Ω

5Ω

100 V

24 V

2A

20 Ω

50 V

2Ω

3Ω

2Ω

36 V

Fig. 3.470 Fig. 3.466

[ −1.143 A ] 3.8

Find the voltage Vx in Fig. 3.467. 2Ω

4V

4Ω +

50 V

[3.87 A ] 3.12 Find the current through the 6 Ω resistor in Fig. 3.471.

Vx

0.1Vx

4Ω

− 100 V

Fig. 3.467

3A

2A

3Ω

5Ω

6Ω

Fig. 3.471

[ −38.5 V ]

[1.26 A ]

   Exercises  3.119

3.13 Find the current through the 6 Ω resistor in Fig. 3.472.

3A

5Ω

10 Ω

40 Ω

8Ω

10 V 2 Ω

4A

15 Ω

36 Ω

15 Ω

7Ω

6Ω

21 Ω

20 Ω

15 Ω

36 Ω

100 V

Fig. 3.475

Fig. 3.472

[1.54 A ]

[ 2.04 A ] 3.14 Find the current through the 2 Ω resistor connected between terminals A and B in Fig. 3.473. 2 Ω 12 V

10 V

10 Ω

3.17 Calculate the current through the 10 Ω resistor in Fig. 3.476. 10 Ω

2Ω 6V A

4Ω

5Ω

4A

8V

2Ω

4Ω 25 V

+ −

2Ω − +

7Ω

12 V

B

Fig. 3.473

2Ω

[1.26 A ] 3.15 Find the current through the 5 Ω resistor in Fig. 3.474.

3Ω

Fig. 3.476

[1.62 A ] 3.18 Determine Thevenin’s equivalent network for figures 3.477 to 3.480 shown below.

5Ω

(i) 40 V

100 V 20 Ω

6Ω

20 V 7Ω

50 V

+ −

4Ω

4Ω 20 V

4Vx

10 V

30 V

10 Ω

+ Vx −

6Ω

4A

Fig. 3.474

[ 4.67 A ]

3.16 Find the current through the 20 Ω resistor in Fig. 3.475.

Fig. 3.477

[ −58 V,12 Ω]

3.120 Circuit Theory and Transmission Lines (ii)

Ix 1000 Ω

I1 100 Ω 48 V

3Vx

10 V 9I1

Vx

+ −

+ Vx −

13 Ω

10Ix

24 Ω

100 Ω

Fig. 3.482

[0.225 A ]

Fig. 3.478

[9.09 V, 9.09 Ω] norton’s theorem 3.21 Find the current through the 10 Ω resistor in Fig. 3.483.

(iii) 0.5Vx − +

3A

6Ω

+

4Ω

4Ω

Vx

2 V 10 Ω

4Ω

5V

5Ω

20 V

15 Ω

−

Fig. 3.483

[0.68 A ]

Fig. 3.479

[8 V, 2.66 Ω] (iv) 3Ω

2Ω

3.22 Find the current through the 20 Ω resistor in Fig. 3.484.

5Ω

+ Vx −

10 Ω

Vx 4

10 A

20 Ω

5Ω

4Ω

10 V

8Ω

2A

Fig. 3.484

[0.61 A ]

Fig. 3.480

[150 V, 20 Ω] 3.19 Find the current Ix in Fig. 3.481. Ix 5 Ω

10Ix

+ −

4A

3.23 Find the current through the 2 Ω resistor in Fig. 3.485. 10 Ω

2Ω

5Ω

1A

10 V

5A

10 V

5Ω

3Ω

Fig. 3.481

2Ω

20 V

[4 A ] 3.20 Find the current in the 24 Ω resistor in Fig. 3.482.

Fig. 3.485

[5 A ]

   Exercises  3.121

3.24 Find the current through the 5 Ω resistor in Fig. 3.486.

2Ω

2Ω I2

3Ω

6Ω

6A

6Ω

5Ω

12 V

8Ω

+ V1 −

8I2

+ −

0.1 V1

20 Ω

3Ω

10 Ω 2 A

Fig. 3.489 10 V

[0.533 A, 31 Ω]

4Ω

2Ω

Fig. 3.486

[ 4.13 A ] maximum power transfer theorem 3.25 Find Norton’s equivalent circuit for the portion of network shown in Fig. 3.487 to the left of ab. Hence obtain the current in the 10 Ω resistor.

3.28 Find the value of the resistance RL in Fig. 3.490 for maximum power transfer and calculate the maximum power. 3Ω

1Ω

a 6V

4Ω 4Ω

9Ω

6Ω

7V

2Ω

2Ω

1Ω

2A

RL

10 Ω

12 V

8A

Fig. 3.490

b

[1.75 Ω,1.29 W ] Fig. 3.487

[0.053 A ] 3.26 Find Norton’s equivalent network and hence find the current in the 10 Ω resistor in Fig. 3.488.

3.29 Find the value of the resistance RL in Fig. 3.491 for maximum power transfer and calculate the maximum power.

6Ω

3Ω

2V

3Ω

2Ω

9Ω 110 V

I1

3Ω

2I1

10 Ω

RL

3Ω

Fig. 3.491 Fig. 3.488

[0.25 A ] 3.27 Find Norton’s equivalent network In Fig. 3.489.

[ 2.36 Ω, 940 W ] 3.30 Find the value of the resistance RL in Fig. 3.492 for maximum power transfer and calculate the maximum power.

3.122 Circuit Theory and Transmission Lines 8A

3Ω 2Ω

2A

6A

2Ω

RL

3V 4Ω

4 V/2 Ω

2Ω 2Ω

Fig. 3.493

[3 Ω, 2.52 W ] 2Ω

2 V/1 Ω

3.32 Find the value of the resistance RL in Fig. 3.494 for maximum power transfer and calculate the maximum power.

RL

10 Ω

2Ω

Fig. 3.492

[ 2.18 Ω, 29.35 W ]

25 A

5Ω

3.31 Find the value of the resistance RL in Fig. 3.493 for maximum power transfer and calculate the maximum power.

10 A

RL

10 Ω

30 A

Fig. 3.494

[1.76 Ω, 490.187 W ]

Objective-Type Questions 3.1

The value of the resistance R connected across the terminals A and B in Fig. 3.495, which will absorb the maximum power is

The value of R required for maximum power transfer in the network shown in Fig. 3.496 is 5Ω

4Ω

4 kΩ

3 kΩ

25 V

R V

3.3

A

20 Ω

3A

R

B 6 kΩ

4 kΩ

Fig. 3.496 Fig. 3.495 (a) 4 kΩ (c) 8 kΩ 3.2

(b) (d)

4.11 kΩ 9kΩ

Superposition theorem is not applicable to networks containing (a) nonlinear elements (b) dependent voltage source (c) dependent current source (d) transformers

3.4

(a) 2 Ω (b) 4 Ω (c) 8 Ω (d) 16 Ω In the network of Fig. 3.497, the maximum power is delivered to RL if its value is I1

40 Ω 0.5I1

20 Ω

RL 50 V

Fig. 3.497

   Answers to Objective-Type Questions  3.123

3.5

(a) 16 Ω

(b)

40 Ω 3

(c) 60 Ω

(d)

20 Ω

1A

a I1

0.5I1 + −

5Ω

10 V

5Ω

The maximum power that can be transferred to the load RL from the voltage source in Fig. 3.498 is

b

Fig. 3.499 (a) 5 V and 2 Ω (c) 4 V and 2 Ω

100 Ω

3.7 RL

10 V

(b) 7.5 V and 2.5 Ω (d) 3 V and 2.5 Ω

The value of RL in Fig. 3.500 for maximum power transfer is 9Ω 6Ω

6Ω

Fig. 3.498

3.6

(a) 1 W

(b)

10 W

(c) 0.25 W

(d)

0.5 W

2V

For the circuit shown in Fig. 3.499, Thevenin’s voltage and Thevenin’s equivalent resistance at terminals a-b is

+ −

6Ω

1A

9Ω

Fig. 3.500 (a) (c)

3Ω 4.1785 Ω

(b) (d)

1.125 Ω none of these

Answers to Objective-Type Questions 3.1 (a)

3.2 (a)

3.3 (c)

3.4 (a)

RL

3.5 (c)

3.6 (b)

3.7 (a)

4 Coupled Circuits

  4.1     IntroductIon Two circuits are said to be coupled circuits when energy transfer takes place from one circuit to the other without having any electrical connection between them. Such coupled circuits are frequently used in network analysis and synthesis. Common examples of coupled circuits are transformer, gyrator, etc. In this chapter, we will discuss self and mutual inductance, magnetically coupled circuits, dot conventions and tuned circuits.

  4.2     Self-Inductance Consider a coil of N turns carrying a current i as shown in Fig. 4.1. When current flows through the coil, a flux f is produced in the coil. The flux produced by the coil links with the coil itself. If the current flowing through the coil changes, the flux linking the coil also changes. Hence, an emf is induced in the coil. This is known as self-induced emf. The direction of this emf is given by Lenz’s law. We know that φ ∝i φ = k , a constant i φ=ki

i v

Fig. 4.1  Coil carrying current

Hence, rate of change of flux = k × rate of change of current dφ di =k dt dt According to Faraday’s laws of electromagnetic induction, a self-induced emf can be expressed as v = −N

φ di dφ di di = − Nk = − N = −L dt dt i dt dt

Nφ and is called coefficient of self-inductance. i The property of a coil that opposes any change in the current flowing through it is called self-inductance or inductance of the coil. If the current in the coil is increasing, the self-induced emf is set up in such a direction so where L =

4.2 Circuit Theory and Transmission Lines as to oppose the rise in current, i.e., the direction of self-induced emf is opposite to that of the applied voltage. Similarly, if the current in the coil is decreasing, the self-induced emf will be in the same direction as the applied voltage. Self-inductance does not prevent the current from changing, it serves only to delay the change.

  4.3     Mutual Inductance If the flux produced by one coil links with the other coil, placed closed to the first coil, an emf is induced in the second coil due to change in the flux produced by the first coil. This is known as mutually induced emf. Consider two coils 1 and 2 placed adjacent to each other as shown in Fig. 4.2. Let Coil 1 has N1 turns while Coil 2 has N2 turns. Mutual flux

i1 +

+ L1

v1 −

L2

Coil 1

v2 Coil 2 −

Fig. 4.2  Two adjacent coils If a current i1 flows in Coil 1, flux is produced and a part of this flux links Coil 2. The emf induced in Coil 2 is called mutually induced emf. We know that

φ2 ∝ i1 φ2 = k , a constant i1 φ2 = k i1 Hence, rate of change of flux = k × rate of change of current i1 dφ2 di =k 1 dt dt According to Faraday’s law of electromagnetic induction, the induced emf is expressed as v2 = − N 2 where M =

φ di dφ 2 di di = −N2 k 1 = −N2 2 1 = −M 1 dt dt i1 dt dt

N 2 φ2 and is called coefficient of mutual inductance. i1

  4.4     coeffIcIent of couPlInG (k) The coefficient of coupling (k) between coils is defined as fraction of magnetic flux produced by the current in one coil that links the other. Consider two coils having number of turns N1 and N2 respectively. When a current i1 is flowing in Coil 1 and is changing, an emf is induced in Coil 2. N φ M= 2 2 i1

4.5  Inductances in Series  4.3

φ2 φ1 φ2 = k1 φ1 N k φ M= 2 1 1 i1 k1 =

Let

…(4.1)

If the current i2 is flowing in Coil 2 and is changing, an emf is induced in Coil 1, M=

N1 φ1 i2

φ1 φ2 φ1 = k2 φ2 N k φ M= 1 2 2 i2

k2 =

Let

…(4.2)

Multiplying Eqs (4.1) and (4.2), M 2 = k1k2 ×

N1φ1 N 2φ2 × = k 2 L1 L2 i1 i2

M = k L1 L2 k = k1k2

where

  4.5     InductanceS In SerIeS 1. Cumulative Coupling Figure 4.3 shows two coils 1 and 2 connected in series, so that currents through the two coils are in the same direction in order to produce flux in the same direction. Such a connection of two coils is known as cumulative coupling. Let L1 = coefficient of self-inductance of Coil 1 L2 = coefficient of self-inductance of Coil 2 M = coefficient of mutual inductance If the current in the coil increases by di amperes in dt seconds then

Coil 1

Coil 2

i

Fig. 4.3  C   umulative  coupling 

di dt di Self-induced emf in Coil 2 = −L2 dt Self-induced emf in Coil 1 = −L1

di dt di Mutually induced emf in Coil 2 due to change of current in Coil 1 = −M dt

Mutually induced emf in Coil 1 due to change of current in Coil 2 = −M

Total induced emf

v = −( L1 + L2 + 2 M )

di dt

…(4.3)

4.4 Circuit Theory and Transmission Lines If L is the equivalent inductance then total induced emf in that single coil would have been v = −L

di dt

…(4.4)

Equating Eqs (4.3) and (4.4), L = L1 + L2 + 2 M 2. Differential Coupling Figure 4.4 shows the coils connected in series but the direction of current in Coil 2 is now opposite to that in 1. Such a connection of two coils is known as differential coupling. Hence, total induced emf in coils 1 and 2. v = − L1

di di di di − L2 + 2 M = −( L1 + L2 − 2 M ) dt dt dt dt

Coil 1

Coil 2

i

Fig. 4.4  D   ifferential  coupling

Coils 1 and 2 connected in series can be considered as a single coil with equivalent inductance L. The induced emf in the equivalent single coil with same rate of change of current is given by, v = −L −L

di dt

di di = −( L1 + L2 − 2 M ) dt dt L = L1 + L2 − 2 M

  4.6     InductanceS In Parallel 1. Cumulative Coupling Figure 4.5 shows two coils 1 and 2 connected in parallel such that fluxes produced by the coils act in the same direction. Such a connection of two coils is known as cumulative coupling. i1 Coil 1 Let L1 = coefficient of self-inductance of Coil 1 L2 = coefficient of self-inductance of Coil 2 M = coefficient of mutual inductance

l

If the current in the coils changes by di amperes in dt seconds then di1 dt di Self-induced emf in Coil 2 = −L2 2 dt Self-induced emf in Coil 1 = −L1

i2

Fig. 4.5  Cumulative coupling

di2 dt di Mutually induced emf in Coil 2 due to change of current in Coil 1 = −M 1 dt Mutually induced emf in Coil 1 due to change of current in Coil 2 = −M

Total induced emf in Coil 1 = − L1

di1 di −M 2 dt dt

Coil 2

4.6  Inductances in Parallel  4.5

Total induced emf in Coil 2 = − L2

di2 di −M 1 dt dt

As both the coils are connected in parallel, the emf induced in both the coils must be equal. di1 di di di − M 2 = − L2 2 − M 1 dt dt dt dt di1 di1 di2 di2 L1 −M = L2 −M dt dt dt dt di di ( L1 − M ) 1 = ( L2 − M ) 2 dt dt

− L1

di1  L2 − M  di2 = dt  L1 − M  dt

...(4.5)

i = i1 + i2

Now,

di di1 di2 = + dt dt dt  L − M  di2 di2 = 2 +  L1 − M  dt dt  L − M  di2 = 2 +1  L1 − M  dt  L + L2 − 2 M  di2 = 1  L1 − M  dt

...(4.6)

If L is the equivalent inductance of the parallel combination then the induced emf is given by di dt Since induced emf in parallel combination is same as induced emf in any one coil, di di di L = L1 1 + M 2 dt dt dt di 1  di1 di  =  L1 +M 2 dt L  dt dt  v = −L

di  1   L2 − M  di2 +M 2  L1   dt  L L1 − M  dt  di 1  L −M =  L1  2 + M 2  L   L1 − M   dt =

Substituting Eq. (4.6) in Eq. (4.7),  di2  L1 + L2 − 2 M  di2 1   L2 − M   L − M  dt = L  L1  L − M  + M  dt 1 1   L −M +M L1  2  L1 − M  L= L1 + L2 − 2 M L1 − M

...(4.7)

4.6 Circuit Theory and Transmission Lines =

L1 L2 − L1 M + L1 M − M 2 L1 + L2 − 2 M

=

L1 L2 − M 2 L1 + L2 − 2 M

2. Differential Coupling Figure 4.6 shows two coils 1 and 2 connected in parallel such that fluxes produced by the coils act in the opposite direction. Such a connection of two coils is known as differential coupling. di1 dt di Self-induced emf in Coil 2 = −L2 2 dt

i1

Coil 1 Coil 2

i i2

Self-induced emf in Coil 1 = −L1

Fig. 4.6  Differential coupling

di2 dt di Mutually induced emf in Coil 2 due to change of current in Coil 1 = M 1 dt Mutually induced emf in Coil 1 due to change of current in Coil 2 = M

di1 di +M 2 dt dt di di Total induced emf in Coil 2 = − L2 2 + M 1 dt dt Total induced emf in Coil 1 = − L1

As both the coils are connected in parallel, the emf induced in the coils must be equal. di1 di di di + M 2 = − L2 2 + M 1 dt dt dt dt di1 di1 di2 di2 L1 +M = L2 +M dt dt dt dt di1 di2 ( L1 + M ) = ( L2 + M ) dt dt

− L1

di1  L2 + M  di2 = dt  L1 + M  dt Now,

...(4.8)

i = i1 + i2 di di1 di2 = + dt dt dt  L + M  di2 di2 = 2 +  L1 + M  dt dt  L + M  di2 = 2 +1  L1 + M  dt  L + L2 + 2 M  di2 = 1  L1 + M  dt

...(4.9)

4.6  Inductances in Parallel  4.7

If L is the equivalent inductance of the parallel combination then the induced emf is given by v = −L

di dt

Since induced emf in parallel combination is same as induced emf in any one coil, L

di di di = L1 1 − M 2 dt dt dt di 1  di1 di  =  L1 −M 2 dt L  dt dt  =

di  1   L2 + M  di2 −M 2  L1   dt  L L1 + M  dt

=

 di 1   L2 + M  − M 2  L1   L L1 + M   dt

...(4.10)

Substituting Eq. (4.9) in Eq. (4.10),  di2  L1 + L2 + 2 M  di2 1   L2 + M   L + M  dt = L  L1  L + M  − M  dt 1 1   L +M −M L1  2  L1 + M  L= L1 + L2 + 2 M L1 + M =

L1 L2 + L1 M − L1 M − M 2 L1 + L2 + 2 M

=

L1 L2 − M 2 L1 + L2 + 2 M

  example 4.1  The combined inductance of two coils connected in series is 0.6 H or 0.1 H depending on relative directions of currents in the two coils. If one of the coils has a self-inductance of 0.2 H, find (a) mutual inductance, and (b) coefficient of coupling. Solution

L1 = 0.2 H, Ldiff = 0.1 H, Lcum = 0.6 H

(a) Mutual inductance Lcum = L1 + L2 + 2 M = 0.6 Ldiff = L1 + L2 − 2 M = 0.1 Adding Eqs (i) and (ii), 2( L1 + L2 ) = 0.7 L1 + L2 = 0.35 L2 = 0.35 − 0.2 = 0.15 H

...(i) ...(ii)

4.8 Circuit Theory and Transmission Lines Subtracting Eqs (ii) from Eqs (i), 4 M = 0.5 M = 0.125 H (b) Coefficient of coupling k=

M L1 L2

=

0.125 0.2 × 0.15

= 0.72

  example 4.2  Two coils with a coefficient of coupling of 0.6 between them are connected in series so as to magnetise in (a) same direction, and (b) opposite direction. The total inductance in the same direction is 1.5 H and in the opposite direction is 0.5 H. Find the self-inductance of the coils. k = 0.6, Ldiff = 0.5 H, Lcum = 1.5 H

Solution

Ldiff = L1 + L2 − 2 M = 0.5 Lcum = L1 + L2 + 2 M = 1.5

...(i) ...(ii)

Subtracting Eq. (i) from Eq. (ii), 4M = 1 M = 0.25 H Adding Eq. (i) and (ii), 2( L1 + L2 ) = 2 L1 + L2 = 1

...(iii) M

k= 0.6 =

L1 L2 0.25 L1 L2

L1 L2 = 0.1736

...(iv)

Solving Eqs (iii) and (iv), L1 = 0.22 H L2 = 0.78 H

  example 4.3  Two coils having self-inductances of 4 mH and 7 mH respectively are connected in parallel. If the mutual inductance between them is 5 mH, find the equivalent inductance. L1 = 4 mH, L2 = 7 mH,

Solution For cumulative coupling,

L=

M = 5 mH

4 × 7 − (5) 2 L1 L2 − M 2 = = 3 mH L1 + L2 − 2 M 4 + 7 − 2(5)

For differential coupling, L=

4 × 7 − (5) 2 L1 L2 − M 2 = = 0.143 mH L1 + L2 + 2 M 4 + 7 + 2(5)

4.7  Dot Convention  4.9

  example  4.4  Two inductors are connected in parallel. Their equivalent inductance when the mutual inductance aids the self-inductance is 6 mH and it is 2 mH when the mutual inductance opposes the self-inductance. If the ratio of the self- inductances is 1:3 and the mutual inductance between the coils is 4 mH, find the self-inductances. Solution Lcum = 6 mH, Ldiff = 2 mH,

L1 = 1.3, M = 4 mH L2

For cumulative coupling, L=

L1 L2 − M 2 L1 + L2 − 2 M

6=

L1 L2 − ( 4) 2 L1 + L2 − 2( 4)

6=

L1 L2 − 16 L1 + L2 − 8

L=

L1 L2 − M 2 L1 + L2 + 2 M

2=

L1 L2 − ( 4) 2 L1 + L2 + 8

2=

L1 L2 − 16 L1 + L2 + 8

...(i)

For differential coupling,

...(ii)

From Eqs (i) and (ii), 2( L1 + L2 + 8) = 6( L1 + L2 − 8) L1 + L2 + 8 = 3L1 + 3L2 − 24 L1 + L2 = 16 But

L1 = 1.3 L2 1.3 L2 + L2 = 16 2.3 L2 = 16 L2 = 6.95 mH L1 = 1.3 L2 = 9.035 mH

  4.7     dot conVentIon Consider two coils of inductances L1 and L2 respectively connected in series as shown in Fig. 4.7. Each coil will contribute the same mutual flux (since it is in a series connection, the same current flows through L1 and L2) and hence, same mutual inductance (M). If the mutual fluxes of the two coils aid each other as

4.10 Circuit Theory and Transmission Lines shown in Fig 4.7 (a), the inductances of each coil will be increased by M, i.e., the inductance of coils will become (L1 + M) and (L2 + M). If the mutual fluxes oppose each other as shown in Fig. 4.7 (b), inductance of the coils will become (L1 − M) and (L2 − M). Whether the two mutual fluxes aid to each other or oppose will depend upon the manner in which coils are wound. The method described above is very inconvenient because we have to include the pictures of the coils in the circuit. There is another simple method of defining the directions of currents in the coils. This is known as dot convention. L2

L1

L1

L2

I

I (a) I

L1

(b) L2

I

L1

(c)

L2

(d)

Fig. 4.7  Dot convention Figure 4.7 shows the schematic connection of the two coils. It is not possible to state from Fig. 4.7(a) and Fig. 4.7(b) whether the mutual fluxes are additive or in opposition. However dot convention removes this confusion. If the current enters from both the dotted ends of Coil 1 and Coil 2, the mutual fluxes of the two coils aid each other as shown in Fig. 4.7(c). If the current enters from the dotted end of Coil 1 and leaves from the dotted end of Coil 2, the mutual fluxes of the two coils oppose each other as shown in Fig. 4.7(d). When two mutual fluxes aid each other, the mutual inductance is positive and polarity of the mutually induced emf is same as that of the self-induced emf. When two mutual fluxes oppose each other, the mutual inductance is negative and polarity of the mutually induced emf is opposite to that of the selfinduced emf.

  example 4.5 

Obtain the dotted equivalent circuit for Fig. 4.8 shown below. R L1 +

C

v (t) −

i(t) L2

Fig. 4.8 Solution The current in the two coils is shown in Fig. 4.9. The corresponding flux due to current in each coil is also drawn with the help of right-hand thumb rule.

4.7  Dot Convention  4.11 f1

R

L1 + C

v (t) − L2 f2

Fig. 4.9 From Fig. 4.9, it is seen that, the flux f1 is in upward direction in Coil 1, and flux f2 is in downward direction in Coil 2. Hence, fluxes are opposing each other. The mutual inductances are negative and mutually induced emfs have opposite polarities as that of self-induced emf. The dots are placed in two coils to illustrate these conditions. Hence, current i(t) enters from the dotted end in Coil 1 and leaves from the dotted end in Coil 2. The dotted equivalent circuit is shown in Fig. 4.10.

R

L1 + C

v (t) −

i(t)

L2

Fig. 4.10

  example 4.6 

Obtain the dotted equivalent circuit for the circuit of Fig. 4.11. j4 Ω j2 Ω

j3 Ω

j3 Ω

j5 Ω

j6 Ω

Fig. 4.11 Solution The current in the three coils is shown in Fig. 4.12. The corresponding flux due to current in each coil is also drawn with the help of right-hand thumb rule. From Fig. 4.12, it is seen that the flux is towards the left in Coil 1, towards the right in Coil 2 and towards the left in Coil 3. Hence, fluxes f1 and f2 oppose each other in coils 1 and 2, fluxes f2 and f3 oppose each other in coils 2 and 3, and fluxes f1 and f3 aid each other in coils 1 and 3. The dots are placed in three coils to illustrate these conditions. Hence, current enters from the dotted end in Coil 1, leaves from the dotted end in Coil 2 and enters from the dotted end in Coil 3. The dotted equivalent circuit is shown in Fig. 4.13.

j4 Ω j2 Ω

j3 Ω

f1

f 2f 3

i j3 Ω

j5 Ω

j6 Ω

Fig. 4.12 j4Ω

j3 Ω

j2 Ω

j5 Ω

j3 Ω

Fig. 4.13

j6 Ω

4.12 Circuit Theory and Transmission Lines

  example 4.7 

Obtain the dotted equivalent circuit for the circuit shown in Fig. 4.14.

j6Ω

A j5Ω

j4Ω

B j3Ω

j1Ω

j2Ω

Fig. 4.14 Solution The current in the three coils is shown in Fig. 4.15. The corresponding flux due to current in each coil i A is also drawn with the help of right-hand thumb rule. From Fig. 4.15, it is seen that all the three fluxes f1, f2, f3 j5 Ω aid each other. Hence, all the mutual reactances are positive and mutually induced emfs have same polarities as that of self-induced emfs. The dots are placed in three coils to illustrate these conditions. Hence, currents enter from the dotted end in each of the three coils. The dotted equivalent circuit is shown in Fig. 4.16.

f1 j6 Ω j4 Ω

B j3 Ω

j1 Ω f3

f2 j2 Ω

Fig. 4.15

j6 Ω j5 Ω

j4 Ω

j2 Ω

j1 Ω

j3 Ω

A

B

Fig. 4.16

  example 4.8 

Obtain the dotted equivalent circuit for the coupled circuit of Fig. 4.17.

j3 Ω

j5 Ω

−j8 Ω

10 Ω + − 50∠ 0°V

Fig. 4.17 Solution The current in the two coils is shown in Fig. 4.18 . The corresponding flux due to current in each coil is also drawn with the help of right-hand thumb rule.

4.7  Dot Convention  4.13 f2

f1

j3 Ω

j5 Ω

− j8 Ω

10 Ω + − 50∠ 0°V

Fig. 4.18 j5 Ω

From Fig. 4.18, it is seen that the flux f1 is in clockwise direction in Coil 1 and in anti-clockwise direction in Coil 2. Hence, fluxes are opposing each other. The dots are placed in two coils to illustrate these conditions. Hence, current enters from the dotted end in Coil 1 and leaves from the dotted end in Coil 2. The dotted equivalent circuit is shown in Fig. 4.19.

  example 4.9 

10 Ω

j3 Ω

M12

+ − 50∠ 0°V

Fig. 4.19

Find the equivalent inductance of the network shown in Fig. 4.20. 1H 1H

0.5 H i

1H

5H

2H

Fig. 4.20 Solution L = ( L1 + M12 + M13 ) + ( L2 + M 23 + M 21 ) + ( L3 + M 31 + M 32 ) = (1 + 0.5 + 1) + ( 2 + 1 + 0.5) + (5 + 1 + 1) = 13 H

  example 4.10 

Find the equivalent inductance of the network shown in Fig. 4.21. 1H 2H i

10 H

1H 5H

6H

Fig. 4.21 Solution L = ( L1 + M12 − M13 ) + ( L2 − M 23 + M 21 ) + ( L3 − M 31 − M 23 ) = (10 + 2 − 1) + (5 − 1 + 2) + (6 − 1 − 1) = 21 H

−j8 Ω

4.14 Circuit Theory and Transmission Lines

  example 4.11 

Find the equivalent inductance of the network shown in Fig. 4.22. k3 = 0.65

i

12 H

k2 = 0.37 k1 = 0.33 14 H 14 H

Fig. 4.22 Solution M12 = M 21 = k1 L1 L2 = 0.33 (12)(14) = 4.28 H M 23 = M 32 = k2 L2 L3 = 0.37 (14)(14) = 5.18 H M 31 = M13 = k3 L3 L1 = 0.65 (12)(14) = 8.42 H L = ( L1 − M12 + M13 ) + ( L2 − M 23 − M 21 ) + ( L3 + M 31 − M 32 ) = (12 − 4.28 + 8.42) + (144 − 5.18 − 4.28) + (14 + 8.42 − 5.18) = 37.92 H

  example 4.12 

Find the equivalent inductance of the network shown in Fig. 4.23. 8H 16 H

15 H

A

B

Fig. 4.23 Solution

For Coil A, LA = L1 − M12 = 15 − 8 = 7 H

For Coil B, LB = L2 − M12 = 16 − 8 = 8 H 1 1 1 1 1 15 = + = + = L LA LB 7 8 56 56 L= = 3.73 H 15

  example 4.13 

Find the equivalent inductance of the network shown in Fig. 4.24. 10 H

10 H

15 H 30 H

25 H

A

B

Fig. 4.24

35 H

C

4.8  Coupled Circuits  4.15

Solution

For Coil A, LA = L1 + M12 − M13 = 25 + 10 − 10 = 25 H

For Coil B, LB = L2 − M 23 + M 21 = 35 − 15 + 10 = 25 H For Coil C, LC = L3 − M 32 − M 31 = 35 − 15 − 10 = 10 H 1 1 1 1 1 1 1 9 = + + = + + = L LA LB LC 25 25 10 50 50 = 5.55 H L= 9

  example 4.14 

Find the equivalent impedance across the terminals A and B in Fig. 4.25. 5Ω A 2Ω

3Ω j2 Ω

j4 Ω

j3 Ω

B

Fig. 4.25 Solution

Z1 = 5 Ω, Z 2 = ( 2 + j 4) Ω, Z3 = (3 + j 3) Ω, Z M = j 2 Ω Z = Z1 +

( 2 + j 4) (3 + j 3) − ( j 2) 2 Z 2 Z3 − Z 2M = 6.9 ∠ 24.16° Ω = 5+ 2 + j 4 + 3 + j 3 − 2( j 2) Z 2 + Z 3 − 2Z M

  4.8     couPled cIrcuItS Consider two coils located physically close to one another as shown in Fig. 4.26. When current i1 flows in the first coil and i2 = 0 in the second coil, flux f1 is produced in the coil. A fraction of this flux also links the second coil and induces a voltage in this coil. The voltage v1 induced in the first coil is di v1 = L1 1 dt i2 = 0 The voltage v2 induced in the second coil is di v2 = M 1 dt

i2 = 0

i1

i2 M

+ v1

L1

+ L2

v2 −

−

Fig. 4.26  Coupled circuit

4.16 Circuit Theory and Transmission Lines The polarity of the voltage induced in the second coil depends on the way the coils are wound and it is usually indicated by dots. The dots signify that the induced voltages in the two coils (due to single current) have the same polarities at the dotted ends of the coils. Thus, due to i1, the induced voltage v1 must be positive at the dotted end of Coil 1. The voltage v2 is also positive at the dotted end in Coil 2. The same reasoning applies if a current i2 flows in Coil 2 and i1 = 0 in Coil 1. The induced voltages v2 and v1 are di v2 = L2 2 dt i1 = 0 v1 = M

and

di2 dt

i1 = 0

The polarities of v1 and v2 follow the dot convention. The voltage polarity is positive at the doted end of inductor L2 when the current direction for i2 is as shown in Fig. 4.26. Therefore, the voltage induced in Coil 1 must be positive at the dotted end also. Now if both currents i1 and i2 are present, by using superposition principle, we can write di1 di +M 2 dt dt di1 di2 v2 = M + L2 dt dt v1 = L1

This can be represented in terms of dependent sources, as shown in Fig. 4.27. i1

i2 +

+ L1

L2 v2

v1 + −

Mdi2 dt

+ −

Mdi1 dt −

−

Fig. 4.27  Equivalent circuit Now consider the case when the dots are placed at the opposite ends in the two coils, as shown in Fig. 4.28. i1

i2 M

+ v1

L1

+ L2

v2 −

−

Fig. 4.28  Coupled circuit Due to i1, with i2 = 0, the dotted end in Coil 1 is positive, so the induced voltage in Coil 2 is positive at the dot, which is the reverse of the designated polarity for v2. Similarly, due to i2, with i1 = 0, the dotted ends have negative polarities for the induced voltages. The mutually induced voltages in both cases have polarities that are the reverse of terminal voltages and the equations are

4.8  Coupled Circuits  4.17

di1 di −M 2 dt dt di1 di v2 = − M + L2 2 dt dt v1 = L1

This can be repressed in terms of dependent sources as shown in Fig. 4.29. i1

i2 +

+ L1

L2 v2

v1 − +

Mdi2 dr

− +

Mdi1 dt −

−

Fig. ☻4.29  Equivalent circuit

The various cases are summarised in the table shown in Fig. 4.30. Coupled circuit

Time-domain equivalent circuit

i1

i2 M

+ L1

v1

L2

i1

+

v2

v1

L1

v1

L2

i1

+

+

v2

v1 −

i2 M

+ L1

L2

i1

+

+

v2

v1 −

i2 M

+ L1

L2

L1 Mdi2 dt

L2 + −

+ −

Mdi1 dt

L1 Mdi2 dt

L2 + −

+ −

Mdi1 dt

+

v2

v1

−

−

+

+

v2

v1

−

−

L1 Mdi2 dt

L2 − +

− +

Mdi1 dt

+

+

v2

v1

−

−

+

+

v2

v1

−

−

Mdi2 dt

L2 − +

− +

Mdi1 dt

jwL2

j wM i2 + −

+ j wM i 1 −

v2

+

v2

v1

−

−

+

jwL1

jwL2

j wM i2 + −

+ j wM i 1 −

v2 −

i2 +

jwL1

jwL2

j wM i2 − +

− j wM i 1 +

v2 −

i2 jwL1

jwL2

j wM i2 − +

− j wM i 1 +

Fig. 4.30  Coupled circuits for various cases

−

i2

i1 +

+

jwL1

i1

i2 L1

i2

i1

i2

i1 +

i1

i2

i1

−

−

i2

i1

−

−

v1

−

i2 M

+

−

+

−

−

v1

i1

Frequency-domain equivalent circuit

+ v2 −

4.18 Circuit Theory and Transmission Lines

  example 4.15 

Write mesh equations for the network shown in Fig. 4.31. R1

L1

L3 R2

+

M12

v1(t ) −

M23

L2

i1

R3

i2

Fig. 4.31 Solution Coil 1 is magnetically coupled to Coil 2. Similarly, Coil 2 is magnetically coupled with Coil 1 and Coil 3. By applying dot convention, the equivalent circuit is drawn with the dependent sources. The equivalent circuit in terms of dependent sources is shown in Fig. 4.32. L1

R1

M12 d (i1 − i2) dt + −

L3

M23

d (i − i ) dt 1 2

− +

R2

L2 + v1(t )

R3 −

− di M23 2 i + dt 2

i1

+ di M12 1 − dt

Fig. 4.32 (a) In Coil 1, there is a mutually induced emf due to current (i1 − i2) in Coil 2. The polarity of the mutually induced emf is same as that of self-induced emf because currents i1 and (i1 − i2) enter in respective coils from the dotted ends. (b) In Coil 2, there are two mutually induced emfs, one due to current i1 in Coil 1 and the other due to current i2 in Coil 3. The polarity of the mutually induced emf in Coil 2 due to the current i1 is same as that of the self-induced emf because currents i1 and (i1 − i2) enter in respective coils from dotted ends. The polarity of the mutually induced emf in Coil 2 due to the current i2 is opposite to that of the self-induced emf because current (i1 − i2) leaves from the dotted end in Coil 2 and the current i2 enters from the dotted end in Coil 3. (c) In Coil 3, there is a mutually induced emf due to the current (i1 − i2) in Coil 2. The polarity of the mutually induced emf is opposite to that of self-induced emf because the current (i1 − i2) leaves from the dotted end in Coil 2 and the current i2 enters from the dotted end in Coil 3. Applying KVL to Mesh 1, v1 (t ) − R1 i1 − L1

di1 d d di di − M12 (i1 − i2 ) − R2 (i1 − i2 ) − L2 (i1 − i2 ) + M 23 2 − M12 1 = 0 dt dt dt dt dt

4.8  Coupled Circuits  4.19

( R1 + R2 ) i1 + ( L1 + L2 + 2 M12 )

di1 di − R2 i2 − ( L2 + M12 + M 23 ) 2 = v1 (t ) dt dt

…(i)

Applying KVL to Mesh 2, M12

di1 di d di d − M 23 2 − L2 (i2 − i1 ) − R2 (i2 − i1 ) − L3 2 + M 23 (i1 − i2 ) − R3i2 = 0 dt dt dt dt dt di1 di − R2 i1 − ( L2 + M12 + M 23 ) + ( R2 + R3 ) i2 + ( L2 + L3 + 2 M 23 ) 2 = 0 dt dt

  example 4.16 

…(ii)

Write KVL equations for the circuit shown in Fig. 4.33. R2

R1

R3

i3

C

M + v1(t ) −

i1

L1

+ v2(t ) −

L2

i2

Fig. 4.33 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.34. R2

R3

R1 L1 + v1(t) −

C

i3

L2

i1 Mdi2 dt

+ −

+ −

Mdi1 dt

i2

+ v2 −

Fig. 4.34 Applying KVL to Mesh 1, di1 di −M 2 =0 dt dt di1 di2 R1 i1 + L1 +M = v1 (t ) dt dt

v1 (t ) − R1 i1 − L1

Applying KVL to Mesh 2, t

M

1 di1 di − L2 2 − R3 (i2 − i3 ) − ∫ (i2 − i3 ) dt − v2 (t ) = 0 dt dt C0

…(i)

4.20 Circuit Theory and Transmission Lines t

M

1 di1 di − L2 2 − R3 (i2 − i3 ) − ∫ (i2 − i3 ) dt = v2 (t ) dt dt C0

…(ii)

Applying KVL to Mesh 3, t

− R2 i3 −

  example 4.17 

1 (i3 − i2 )dt − R3 (i3 − i2 ) = 0 C ∫0

…(iii)

Write down the mesh equations for the network shown in Fig. 4.35. Z1

M L2

L1

+

ZL

V1 −

I1

Z2 I2

Fig. 4.35 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.36. Z1 jwL1

jwL2 ZL

+

j wM I2

V1

+ −

+ −

j wM I1

−

I1

Z2

I2

Fig. 4.36 Applying KVL to Mesh 1, V1 − Z1I1 − jω L1I1 − jω MI 2 − Z 2 ( I1 − I 2 ) = 0 ( Z1 + jω L1 + Z 2 ) I1 − ( Z 2 − jω M ) I 2 = V1

…(i)

Applying KVL to Mesh 2, −Z 2 ( I 2 − I1 ) + jω MI1 − jω L2 I 2 − Z L I 2 = 0 −( Z 2 − jω M ) I1 + ( Z 2 + jω L2 + Z L ) I 2 = 0

…(ii)

4.8  Coupled Circuits  4.21

  example 4.18 

Write mesh equations for the network shown in Fig. 4.37. R

L3

+

L1

v(t)

L2

i1

−

i2

i3

C

Fig. 4.37 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.38. R

L2

L1

+ v1 −

i1

+ −

M12 d (i2 − i3) dt

− +

M13 di3 dt

i2

L3

+ −

M21 d (i1 − i2) dt

+ −

M23 di3 dt

i3

− +

M31 d (i1 − i2) dt

+ −

M32 d (i2 − i3) dt C

Fig. 4.38 Applying KVL to Mesh 1, d (i1 − i2 ) − M12 dt d Ri1 + L1 (i1 − i2 ) + M12 dt

v(t ) − Ri1 − L1

d (i2 − i3 ) + M13 dt d (i2 − i3 ) − M13 dt

di3 =0 dt di3 = v(t ) dt

…(i)

Applying KVL to Mesh 2, − M13 M13

di3 + M12 dt di3 − M12 dt

d d (i2 − i3 ) − L1 (i2 − i1 ) − L2 dt dt d d (i2 − i3 ) + L1 (i2 − i1 ) + L2 dt dt

d d di (i2 − i3 ) − M 21 (i1 − i2 ) − M 23 3 = 0 dt dt dt d d di3 (i2 − i3 ) + M 21 (i1 − i2 ) + M 23 =0 dt dt dt

…(ii)

Applying KVL to Mesh 3, di3 d + M 21 (i1 − i2 ) − L2 dt dt di d − M 23 3 − M 21 (i1 − i2 ) + L2 dt dt M 23

1 d di d d (i3 − i2 ) − L3 3 + M 31 (i1 − i2 ) − M 32 (i2 − i3 ) − ∫ i3 dt = 0 dt dt dt dt C 1 d di d d (i3 − i2 ) + L3 3 − M 31 (i1 − i2 ) + M 32 (i2 − i3 ) + ∫ i3 dt = 0 …(iii) dt dt dt dt C

4.22 Circuit Theory and Transmission Lines

  example 4.19 

Write KVL equations for the network shown in Fig. 4.39. R

L3

+

L1

v(t) i1

−

L2 i2

i3

C

Fig. 4.39 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.40. R

L2

L1

+ v1 −

i1

+ −

M12 di2 dt

+ −

L3

M21 di1 dt

i2 − +

M13 di3 dt

− +

M31 di1 dt

+ −

M32 di2 dt

i3 + −

M23 di3 dt

C

Fig. 4.40 Applying KVL to Loop 1, di1 − M12 dt di R(i1 + i2 + i3 ) + L1 1 + M12 dt

di2 + M13 dt di2 − M13 dt

di3 =0 dt di3 = v(t ) dt

di2 di − M 21 1 − M 23 dt dt di2 di1 + M 21 + M 23 R(i1 + i2 + i3 ) + L2 dt dt

di3 =0 dt di3 = v(t ) dt

v(t ) − R(i1 + i2 + i3 ) − L1

…(i)

Applying KVL to Loop 2, v(t ) − R(i1 + i2 + i3 ) − L2

…(ii)

Applying KVL to Loop 3, di3 di + M 31 1 − M 32 dt dt di di R(i1 + i2 + i3 ) + L3 3 − M 31 1 + M 32 dt dt

v(t ) − R(i1 + i2 + i3 ) − L3

di2 1 − i3 dt = 0 dt C ∫ di2 1 + i3 dt = v(t ) dt C ∫

…(iii)

4.8  Coupled Circuits  4.23

  example 4.20 

In the network shown in Fig. 4.41, find the voltages V1 and V2. M=2H

3H +

v1

5H

i3 −

+

v2

−

i2 = 10e−t A

i1 = 5e−t A

Fig. 4.41 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.42. 2 3H +

di3 dt

2 5H

i3

+ −

−

v1

i1 = 5e−t A

+

di1 dt + −

v2

−

i2 = 10e−t A

Fig. 4.42 From Fig. 4.42, i3 = i1 + i2 = 5 e − t + 10 e − t = 15 e − t A di1 di d d + 2 3 = 3 (5 e − t ) + 2 (15 e − t ) = −15 e − t − 30 e − t = −45 e − t V dt dt dt dt di3 di1 d d v2 = 5 +2 = 5 (15 e − t ) + 2 (5 e − t ) = −75 e − t − 10 e − t = −85 e − t V dt dt dt dt v1 = 3

  example 4.21 

In the network shown in Fig. 4.43, find the voltages V1 and V2. M=2H 2H +

v1

4H

i3 −

+

v2

i2 = 10e−t A

i1 = 10e−t A

Fig. 4.43

−

4.24 Circuit Theory and Transmission Lines Solution

The equivalent circuit in terms of dependent sources is shown in Fig. 4.44. di3 dt

2 2H +

2 4H

i3

− +

v1

−

i1 = 10e−t A

di1 dt − +

+

−

i2 = 10e−t A

Fig. 4.44 From Fig. 4.44, i3 = i1 + i2 = 10 e − t + 10 e − t = 20 e − t A v1 = 2 v2 = 4

di1 di d d − 2 3 = 2 (10 e − t ) − 2 ( 20 e − t ) = −20 e − t + 40 e − t = 20 e − t A dt dt dt dt

di3 di d d − 2 1 = 4 ( 20 e − t ) − 2 (10 e − t ) = −80 e − t + 20 e − t = −60 e − t A dt dt dt dt

  example 4.22 

Calculate the current i2(t) in the coupled circuit of Fig. 4.45. i1(t)

i2(t)

0.1 H

+

0.2 H

30 sint

0.2 H

−

Fig. 4.45 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.46. i1(t)

+

i2(t)

0.2 H

0.2 H

30 sint −

0.1

di2 dt

− +

− +

0.1

di1 dt

Fig. 4.46 Applying KVL to Mesh 1, 30 sin t − 0.2

di1 di + 0.1 2 = 0 dt dt

…(i)

4.8  Coupled Circuits  4.25

Applying KVL to Mesh 2, − 0.2

di2 di + 0.1 1 = 0 dt dt di1 di =2 2 dt dt

…(ii)

Substituting Eq. (ii) in Eq. (i), di  di  30 sin t − 0.2  2 2  + 0.1 2 = 0  dt  dt 0.3

di2 = 30 sin t dt di2 = 100 sin t dt di2 = 100 sin t dt

Integrating both the sides, t

i2 (t ) = 100∫ sin t dt 0

= 100 [ − cos t ]0 t

= 100 (1 − cos t )

  example 4.23 

Find the voltage V2 in the circuit shown in Fig. 4.47 such that the current in the left-hand loop (Loop 1) is zero. 2Ω

1Ω

j2 Ω

+

+ j4 Ω

5∠0° V

V2

j3 Ω

I1

−

I2

Loop 1

−

Loop 2

Fig. 4.47 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.48. 2Ω

1Ω j4 Ω

+

j3 Ω

+

V2

5∠0° V −

I1

j 2 I2

− +

− +

Fig. 4.48

j 2 I1

I2

−

4.26 Circuit Theory and Transmission Lines Applying KVL to Loop 1, 5 ∠ 0° − 2 I1 − j 4I1 + j 2I 2 = 0 ( 2 + j 4) I1 − j 2I 2 = 5 ∠ 0°

…(i)

Applying KVL to Loop 2, − j 2I1 − j 3I 2 − 1I 2 − V2 = 0 …(ii)

− j 2I1 − (1 + j 3) I 2 = V2 Writing Eqs (i) and (ii) in matrix form, − j 2   I1  5 ∠ 0° 2 + j 4  − j 2 −(1 + j 3)   I 2  =  V2  By Cramer’s rule, 5 ∠ 0° − j2 −(1 + j 3) V2 I1 = 2 + j4 − j2 − j 2 −(1 + j 3) But I1 = 0. ∴ −(5 ∠ 0°)(1 + j 3) + j 2 V2 = 0 V2 =

  example 4.24 

Determine the ratio

(5 ∠ 0°)(1 + j 3) = 7.91 ∠ − 18.43° V j2

V2 in the circuit of Fig. 4.49, if I1 = 0. V1

8Ω

2Ω

j2 Ω

+

+ j8 Ω

V1

j2 Ω

V2

I1

−

I2

−

Fig. 4.49 Solution The equivalent circuit in terms of dependent sources is as shown in Fig. 4.50. 8Ω

2Ω j8 Ω

+

j2 Ω

+

V1

V2 −

I1

j 2 I2

+ −

+ −

j 2 I1

I2

−

Fig. 4.50 Applying KVL to Mesh 1, V1 − 8I1 − j8I1 − j 2I 2 = 0 (8 + j8) I1 + j 2I 2 = V1

…(i)

4.8  Coupled Circuits  4.27

Putting I1 = 0 in Eq (i),

j2 I 2 = V1 Applying KVL to Mesh 2, V2 − 2I 2 − j 2I 2 − j 2I1 = 0 j 2I1 + ( 2 + j 2) I 2 = V2 Putting I1 = 0 in Eq (iii),

…(ii)

…(iii)

( 2 + j 2) I 2 = V2

…(iv)

From Eqs (ii) and (iv), V2 ( 2 + j 2)I 2 2 + j 2 = = = 1.41 ∠ − 45° V V1 j 2I 2 j2

  example 4.25 

For the coupled circuit shown in Fig. 4.51, find input impedance at terminals A and B. A

j4 Ω

3Ω

j3 Ω

+

j5 Ω

V1

−j8 Ω

−

B

Fig. 4.51 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.52. A

j4 Ω

3Ω

j 3 (I1 − I2) + − j5 Ω

+ −j8 Ω

V1 I1

−

+ j3 I I 1 2 −

B

Fig. 4.52 Applying KVL to Mesh 1, V1 − 3I1 − j 4 I1 − j 3 ( I1 − I 2 ) − j 5 ( I1 − I 2 ) − j 3I1 = 0 (3 + j15) I1 − j8 I 2 = V1

…(i)

Applying KVL to Mesh 2, j 3 I1 − j 5 ( I 2 − I1 ) + j8 I 2 = 0 j8 I1 + j 3 I 2 = 0 I2 = −

j8 I1 = −2.67 I1 j3

…(ii)

4.28 Circuit Theory and Transmission Lines Substituting Eq (ii) in Eq (i), (3 + j15)I1 − j8 ( −2.67 I1 ) = V1 (3 + j 36.36) I1 = V1 Zi =

  example 4.26 

V1 = (3 + j 36.36) Ω = 36.48 ∠ 85.28° Ω I1

Find equivalent impedance of the network shown in Fig. 4.53. j2 Ω

2Ω

j6 Ω j4 Ω

j4 Ω

j3 Ω

Zeq −j5 Ω

5Ω

Fig. 4.53 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.54. j2 Ω

2Ω

j 4 (I1 − I2) − + j3 Ω

+

− +

V1 −

I1

j 4 I1

5Ω − +

j4 Ω

j 6 I2

I2

− j 6 (I1 − I2) +

−j 5 Ω

Fig. 4.54 Applying KVL to Mesh 1, V1 − 2I1 − j 2I1 + j 4 ( I1 − I 2 ) − j 3 ( I1 − I 2 ) + j 4 I1 − 5( I1 − I 2 ) + j 6 I 2 = 0 (7 − j 3) I1 − (5 + j 5) I 2 = V1

…(i)

4.8  Coupled Circuits  4.29

Applying KVL to Mesh 2, − j 6 I 2 − 5( I 2 − I1 ) − j 4 I1 − j 3( I 2 − I1 ) − j 4 I 2 + j 6( I1 − I 2 ) + j 5I 2 = 0 (5 + j 5)I1 = (5 + j 4) I 2  5 + j5  I2 =  I1  5 + j 4 

…(ii)

Substituting Eq. (ii) in Eq. (i),  5 + j5  (7 − j 3)I1 − (5 + j 5)  I1 = V1  5 + j 4  Zi =

  example 4.27 

V1 (5 + j 5) (5 + j 5) = 7 − j3 − = 5.63 ∠ − 47.15° Ω I1 5 + j4

Find the voltage across the 5 W resistor in Fig. 4.55 using mesh analysis. j5.66 Ω j5 Ω

+

j10 Ω

3Ω

50∠0° V

I1

−

−j4 Ω

5Ω

I2

Fig. 4.55 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.56. j5 Ω

j5.66 I2

j10 Ω

j5.66 I1

+ − +

3Ω

50∠0° V −

+ −

I1

−j 4 Ω

5Ω

I2

Fig. 4.56 Applying KVL to Mesh 1, 50 ∠ 0° − j 5 I1 − j 5.66 I 2 − (3 − j 4) ( I1 − I 2 ) = 0 (3 + j 1) I1 − (3 − j 9.66) I 2 = 50 ∠ 0°

…(i)

Applying KVL to Mesh 2, −(3 − j 4) ( I 2 − I1 ) − j 10 I 2 − j 5.66 I1 − 5I 2 = 0 −(3 − j 9.66) I1 + (8 + j 6) I 2 = 0

…(ii)

4.30 Circuit Theory and Transmission Lines Writing Eqs (i) and (ii) in matrix form, −(3 − j 9.66)   I1  50 ∠ 0°   3+ j 1 =  −(3 − j 9.66) 8 + j 6   I 2   0  By Cramer’s rule, 3+ j 1 50 ∠ 0° −(3 − j 9.66) 0 = 3.82 ∠ − 112.14° Α I2 = 3+ j 1 −(3 − j 9.66) −(3 − j 9.66) 8 + j6 V5 Ω = 5 I 2 = 5 (3.82 ∠ − 112.14°) = 19.1 ∠ − 112.14° V

  example 4.28 

Find the voltage across the 5 W resistor in Fig. 4.57 using mesh analysis. k = 0.8 j5 Ω

+

j10 Ω

3Ω

50∠0° V

I1

−

−j4 Ω

5Ω

I2

Fig. 4.57 Solution For a magnetically coupled circuit, X M = k X L1 X L2 = 0.8 (5) (10) = 5.66 Ω The equivalent circuit in terms of dependent sources is shown in Fig. 4.58. j5 Ω

j5.66 I2

j10 Ω

j5.66 I1

− + +

3Ω

50∠0° V −

− +

I1

−j 4 Ω

5Ω

I2

Fig. 4.58 Applying KVL to Mesh 1, 50 ∠ 0° − j 5 I1 + j 5.66 I 2 − (3 − j 4) ( I1 − I 2 ) = 0 (3 + j 1) I1 − (3 + j 1.66) I 2 = 50 ∠ 0°

…(i)

4.8  Coupled Circuits  4.31

Applying KVL to Mesh 2, −(3 − j 4) ( I 2 − I1 ) − j 10 I 2 + j 5.66 I1 − 5I 2 = 0 −(3 + j 1.66) I1 + (8 + j 6) I 2 = 0

…(ii)

Writing Eqs (i) and (ii) in matrix form, −(3 + j 1.66)   I1  50 ∠ 0°  3+ j 1 =  −(3 + j 1.66) 8 + j 6   I 2   0  By Cramer’s rule, 3+ j 1 50 ∠ 0° −(3 + j 1.66) 0 = 8.62 ∠ − 24.79° Α I2 = 3+ j 1 −(3 + j 1.66) −(3 + j 1.66) 8 + j6 V5 Ω = 5 I 2 = 5 (8.62 ∠ − 24.79°) = 43.1 ∠ − 24.79° Α

  example 4.29 

Find the current through the capacitor in Fig. 4.59 using mesh analysis. j4 Ω

3Ω

j3 Ω

+

j5 Ω

50∠45° V −

I1

−j8 Ω

I2

Fig. 4.59 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.60. 3Ω

j4 Ω

j 3 (I1 − I2) + − j5 Ω

+ 50∠45° V −

I1

+ −

−j8 Ω j 3 I1 I2

Fig. 4.60 Applying KVL to Mesh 1, 50 ∠ 45° − (3 + j 4) I1 − j 3( I1 − I 2 ) − j 5 ( I1 − I 2 ) − j 3 I1 = 0 (3 + j 15) I1 − j 8 I 2 = 50 ∠ 45°

…(i)

4.32 Circuit Theory and Transmission Lines Applying KVL to Mesh 2, j 3 I1 − j 5 ( I 2 − I1 ) + j8 I 2 = 0 …(ii)

− j8 I1 − j 3 I 2 = 0 Writing Eqs (i) and (ii) in matrix form, 3 + j 15 − j8  I1  50 ∠ 45° =   − j8 − j 3  I 2   0 By Cramer’s rule, 3 + j 15 50 ∠ 45° − j8 0 = 3.66 ∠ − 310.33° Α I2 = 3 + j 15 − j8 − j8 − j3 IC = I 2 = 3.66 ∠ − 310.33°Α

  example 4.30 

Find the voltage across the 15 Ω resistor in Fig. 4.61 using mesh analysis. j10 Ω

20 Ω

j15 Ω j20 Ω

+ 120∠0° V −

15 Ω

I2

I1

Fig. 4.61 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.62. j10 Ω

20 Ω

j 5 (I1 − I2) + −

j20 Ω

+

15 Ω

120∠0° V −

I1

+ −

j 5 I2

I2

Fig. 4.62 Applying KVL to Mesh 1, 120 ∠ 0° − 20 I1 − j 20( I1 − I 2 ) − j 5 I 2 = 0 ( 20 + j 20) I1 − j 15 I 2 = 120 ∠ 0°

…(i)

4.8  Coupled Circuits  4.33

Applying KVL to Mesh 2, j 5 I 2 − j 20 ( I 2 − I1 ) − j10 I 2 − j 5 ( I1 − I 2 ) − 15 I 2 = 0 − j15 I1 + (15 + j 20) I 2 = 0

…(ii)

Writing Eqs (i) and (ii) in matrix form, − j15   I1  120 ∠ 0°  20 + j 20 =   − j15 15 + j 20   I 2   0 By Cramer’s rule, 20 + j 20 120 ∠ 0° − j15 0 = 2.53 ∠ 10.12°Α I2 = 20 + j 20 − j15 − j15 15 + j 20 V15 Ω = 15 I 2 = 15 ( 2.53 ∠ 10.12°) = 37.95 ∠ 10.12°V

  example 4.31 

Find the current through the 6 W resistor in Fig. 4.63 using mesh analysis. 4Ω j2 Ω + j3 Ω

120∠0° V −

j8 Ω

I1

I2

6Ω

Fig. 4.63 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.64. 4Ω

j3 Ω

+

j8 Ω

200∠30° V −

I1

− +

j 2 I2

I2

− + j 2 (I1 − I2)

6Ω

Fig. 4.64 Applying KVL to Mesh 1, 120 ∠ 0° − 4 I1 − j 3( I1 − I 2 ) + j 2 I 2 = 0 ( 4 + j 3) I1 − j 5 I 2 = 120 ∠ 0°

…(i)

4.34 Circuit Theory and Transmission Lines Applying KVL to Mesh 2, − j 2 I 2 − j 3 ( I 2 − I1 ) − j8 I 2 + j 2 ( I1 − I 2 ) − 6 I 2 = 0 − j 5 I1 + (6 + j15) I 2 = 0

…(ii)

Writing Eqs (i) and (ii) in matrix form, − j 5   I1  120 ∠ 0° 4 + j 3   − j 5 6 + j 15  I 2  =  0 By Cramer’s rule, 4 + j 3 120 ∠ 0° − j5 0 = 7.68 ∠ 2.94°Α I2 = 4+ j 3 − j5 − j 5 6 + j 15

  example 4.32 

Determine the mesh current I3 in the network of Fig. 4.65. 12 Ω

j 16 Ω

j5 Ω 4Ω

7Ω

I3

−j8 Ω

+

6Ω

j4 Ω

200∠30° V −

I1

I2 −j4 Ω

Fig. 4.65 Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.66. j 16 Ω

j 5 (I1 − I2)

12 Ω

− +

4Ω

7Ω

I3 j4 Ω

+

−j8 Ω

6Ω

200∠30° V −

I1

− +

j 5 I3

Fig. 4.66

I2 −j4 Ω

4.8  Coupled Circuits  4.35

Applying KVL to Mesh 1, 200 ∠ 30° − 4 ( I1 − I3 ) − j 4( I1 − I 2 ) + j 5 I3 = 0 ( 4 + j 4) I1 − j 4 I 2 − ( 4 + j 5) I3 = 200 ∠ 30°

…(i)

Applying KVL to Mesh 2, − j 5 I3 − j 4 ( I 2 − I1 ) − (7 − j8) ( I 2 − I3 ) − (6 − j 4) I 2 = 0 − j 4 I1 + (13 − j8) I 2 − (7 − j13) I3 = 0

…(ii)

− j 16 I3 + j 5 ( I1 − I 2 ) − 12 I3 − (7 − j 8) ( I3 − I 2 ) − 4( I3 − I1 ) = 0 −( 4 + j 5) I1 − (7 − j13) I 2 + ( 23 + j8) I3 = 0

…(iii)

Applying KVL to Mesh 3,

Writing Eqs. (i), (ii) and (iii) in matrix form, − j4 −( 4 + j 5)   I1   200 ∠ 30°  4 + j4  − j4  13 − j8 −(7 − j13)   I 2  =  0      23 + j8   I3   0  −( 4 + j 5) −(7 − j13)  By Cramer’s rule, − j4 4 + j4 200 ∠ 30° − j4 13 − j8 0 −( 4 + j 5) −(7 − j13) 0 = 16.28 ∠ 16.87°Α I3 = − j4 −( 4 + j 5) 4 + j4 − j4 13 − j8 −(7 − j13) −( 4 + j 5) −(7 − j13) 23 + j8

  example 4.33 

Obtain the dotted equivalent circuit for the coupled circuit shown in Fig. 4.67 and find mesh currents. Also find the voltage across the capacitor.

5Ω j5 Ω

5Ω j2 Ω

j5 Ω

+

+

−

−

10∠ 0° V

10∠ 90° V

−j10 Ω

Fig. 4.67 Solution The currents in the coils are as shown in Fig. 4.68. The corresponding flux due to current in each coil is also drawn with the help of right-hand thumb rule.

4.36 Circuit Theory and Transmission Lines 5 Ω I1

f1

I2 5 Ω j2 Ω

j5 Ω

j5 Ω

f2

+ 10∠ 0°V

I1

−

+ 10∠ 90°V

I2

−

−j10 Ω

Fig. 4.68 From Fig. 4.68, it is seen that two fluxes f1 and f2 aid each other. Hence, dots are placed at the two coils as shown in Fig. 4.69. 5Ω

j2 Ω

j5 Ω

j5 Ω

5Ω

+

+ −j10 Ω

10∠0° V

10∠90° V

I1

−

−

I2

Fig. 4.69 The equivalent circuit in terms of dependent sources is shown in Fig. 4.70. 5Ω

j5 Ω

j 2 I2

+

− +

10∠0° V −

I1

j 2 I1

j5 Ω

+ −

−

+

− +

5Ω −

+ +

−j 10 Ω

10∠90° V

I2

−

Fig. 4.70 Applying KVL to Mesh 1, 10 ∠ 0° − (5 + j 5) I1 − j 2 I 2 + j10 ( I1 + I 2 ) = 0 (5 − j 5) I1 − j 8 I 2 = 10 ∠ 0°

…(i)

Applying KVL to Mesh 2, − j10 ( I 2 + I1 ) + j 5 I 2 − j 2 I1 + 5 I 2 − 10 ∠ 90° = 0 − j 8 I1 + (5 − j 5) I 2 = 10 ∠ 90° Writing Eqs. (i) and (ii) in matrix form, 5 − j 5 − j8   I1   10 ∠ 0°   − j8 5 − j 5  I 2  = 10 ∠ 90°

…(ii)

4.9  Conductively Coupled Equivalent Circuits  4.37

By Cramer’s rule, 10 ∠ 0° − j8 10 ∠ 90° 5 − j 5 = 0.72 ∠ − 82.97°A I1 = 5 − j 5 − j8 − j8 5 − j 5 5 − j 5 10 ∠ 0° − j8 10 ∠ 90° = 1.71 ∠ 106.96°A I2 = 5 − j 5 − j8 − j8 5 − j 5 VC = − j 10 ( I1 + I 2 ) = ( − j 10) (0.72 ∠ − 82.97° + 1.71 ∠ 106.96° Α) = 10.08 ∠ 24.03°V

  4.9     conductIVely couPled equIValent cIrcuItS For simplifying circuit analysis, it is desirable to replace a magnetically coupled circuit with an equivalent circuit called conductively coupled circuit. In this circuit, no magnetic coupling is involved. The dot convention is also not needed in the conductively coupled circuit. Consider a coupled circuit as shown in Fig. 4.71. I1

I2

jw M

+

+

jwL2

jwL1

V1

V2

−

−

Fig. 4.71  Coupled circuit The equivalent circuit in terms of dependent sources is shown in Fig. 4.72. I1

+

I2 jwL1

jwL2

+

V1

V2 −

j wM I2

+ −

+ jwM I1 −

−

Fig. 4.72  Equivalent circuit Applying KVL to Mesh 1, V1 − jω L1 I1 − jω M I 2 = jω L1 I1 + jω M I 2 =

1

…(4.11)

Applying KVL to Mesh 2, V2 − jω L2 I 2 − jω M I1 = 0 jω M I1 + jω L2 I 2 = V2

…(4.12)

4.38 Circuit Theory and Transmission Lines Writing Eqs (4.11) and (4.12) in matrix form,  jω L1 jω M   I1   V1   jω M jω L2   I 2  =  V2 

…(4.13)

Consider a T-network as shown in Fig. 4.73. Z1

I1

Z2

I2

+

+

Z3

V1

V2 −

−

Fig. 4.73  T-network Applying KVL to Mesh 1, V1 − Z1 I1 − Z3 ( I1 + I 2 ) = 0 ( Z1 + Z3 ) I1 + Z3 I 2 = V1

… (4.14)

RTh = [( 2  12) + 1]  3 = 1.43 Ω

… (4.15)

Applying KVL to Mesh 2,

Writing Eqs (4.14) and (4.15) in matrix form,  Z1 + Z3  Z3

Z3   I1   V1  = Z 2 + Z3   I 2   V2 

Comparing matrix equations, Z1 + Z3 = j ω L1 Z3 = j ω M Z 2 + Z3 = j ω L2 Solving these equations, Z1 = j ω L1 − j ω M = j ω ( L1 − M ) Z 2 = j ω L2 − j ω M = j ω ( L2 − M ) Z3 = j ω M Hence, the conductively coupled circuit of a magnetically coupled circuit is shown in Fig. 4.74. I1

jw (L1 − M )

jw (L2 − M ) I2

+

+ jw M

V1 −

V2 −

Fig. 4.74  Conductively coupled equivalent circuit

4.9  Conductively Coupled Equivalent Circuits  4.39

  example 4.34 

Find the conductively coupled equivalent circuit for the network shown in Fig. 4.75. −j 4 Ω

j6 Ω

j2 Ω +

V1

j3 Ω −

j5 Ω

10 Ω

I1

I2

2Ω

Fig. 4.75 Solution The current I1 leaves from the dotted end and I2 enters from the dotted end. Hence, mutual inductance M is negative. In the conductively coupled equivalent circuit, Z1 = jω (L1 − M ) = jω L1 − jω M = j 3 − j 2 = j1 Ω Z2 = jω ( L2 − M ) = jω L2 − jω M = j 5 − j 2 = j 3 Ω Z 3 = jω M = j 2 Ω The conductively coupled equivalent circuit is shown in Fig. 4.76. −j4 Ω

j1 Ω

j3 Ω

j6 Ω

+ j2 Ω

V1

10 Ω

I2

I1

−

2Ω

Fig. 4.76

  example 4.35 

Draw the conductively coupled equivalent circuit of Fig. 4.77. j6 Ω j5 Ω

+

3Ω

V1 −

j10 Ω

I1

−j 4 Ω

5Ω

I2

Fig. 4.77 Solution The current I1 enters from the dotted end and I2 leaves from the dotted end. Hence, the mutual inductance M is negative.

4.40 Circuit Theory and Transmission Lines In the conductively coupled equivalent circuit,

−j 1 Ω

j4 Ω

Z1 = jω ( L1 − M ) = jω L1 − jω M = j 5 − j 6 = − j1 Ω Z2 = jω ( L2 − M ) = jω L2 − jω M = j10 − j 6 = j 4 Ω Z 3 = jω M = j 6 Ω

j6 Ω

+

V1

5Ω

−

(3 − j 4) Ω I2

I1

The conductively coupled equivalent circuit is shown in Fig. 4.78. Fig. 4.78

  example 4.36 

Find the conductively coupled equivalent circuit of the network in Fig. 4.79. j 1Ω

4Ω

2Ω

+

V1

j4 Ω

j2 Ω −

I1

Fig. 4.79 Solution The currents I1 and I2 leave from the dotted terminals. Hence, mutual inductance is positive. In the conductively coupled equivalent circuit, Z1 = jω ( L1 + M ) = jω L1 + jω M = j 4 + j 2 = j 6 Ω Z2 = jω (L2 + M ) = jω L2 + jω M = j 2 + j 2 = j 4 Ω Z 3 = − jω M = − j 2 Ω The conductively coupled equivalent circuit is shown in Fig. 4.80. (4 − j2) Ω

j4 Ω

j6 Ω

+

V1

−j 2 Ω −

I1

(2 + j 4) Ω

I2

Fig. 4.80

  4.10     tuned cIrcuItS At radio frequencies, the amplification of signals is done by tuned circuits. There are two types of tuned circuits—single-tuned circuit and double-tuned circuit. In the single-tuned circuit, only the secondary is tuned. In the double-tuned circuits, both primary and secondary of the coupled circuits are tuned. The tuned circuits are used in TV receivers to couple the antenna to the RF stage. In RF circuit design, the tuned circuits are generally employed for obtaining maximum power transfer to the load connected to secondary or for obtaining maximum possible value of the secondary voltage.

4.10  Tuned Circuits  4.41

4.10.1 Single-Tuned Circuit Consider a single-tuned circuit as shown in Fig. 4.81. I1

I2

M

R2 +

R1 jwL1

+

jwL2

−j

V1 −

1 wC2

V2 −

Fig. 4.81  Single-tuned circuit The conductively coupled equivalent circuit is shown in Fig. 4.82. jw (L1 − M )

jw (L2 − M )

R2 +

R1 j wM

+

I1

V1 −

−j

I2

1 wC2

V2 −

Fig. 4.82  Conductively coupled equivalent circuit Applying KVL to Mesh 1, V1 − R1I1−jw(L1−M)I1− jwM(I1−I2) = 0 (R1 + jw L1)I1 − jw M I2 = V1 Applying KVL to Mesh 2, 1 − jω M ( I 2 − I1 ) − jω ( L2 − M )I 2 − R2 I 2 + j I2 = 0 ωC2  − jω M I1 +  R2 + 

1   j  ω L2 −  I2 = 0  ωC2  

…(4.16)

…(4.17)

From Eq. (4.17), 1   R2 + j  ω L2 −  ωC2  I1 = I2 jω M Substituting Eq. (4.18) in Eq. (4.16),   R2 + ( R1 + jω L1 )   

1   j  ω L2 −  ωC2   I 2 − jω MI 2 = V1  jω M  

   ( R1 + jω L1 )  R2 +   I2   

 1   j  ω L2 − + ω2M 2     ω C2     = V1 jω M  

…(4.18)

4.42 Circuit Theory and Transmission Lines jω M V1

I2 =



1  2 2 +ω M 2

( R1 + jω L1 )  R2 + j  ω L2 − ωC 

1. Output Voltage 1  1  V2 = − j  I2 = I2  ωC2  jω C 2 =

MV1   C2 ( R1 + jω L1 )  R2 +  

 1   + ω2M 2 j  ω L2 −    ω C2   

2. Voltage Amplification A=

V2 = V1

M   C2 ( R1 + jω L1 )  R2 +  

 1   j  ω L2 − + ω2M 2    ω C2   

If R1 >> jwL1 and secondary circuit resonates at

ω0 =

1 L2C2

then, at the resonant frequency w0,

ω 0 MV1 R1 R2 + ω 02 M 2 MV1 V2 = C2 ( R1 R2 + ω 02 M 2 ) M A= C2 ( R1 R2 + ω 02 M 2 ) I2 =

Output current, output voltage and voltage amplification depends on the mutual inductance M at resonant frequency. The maximum amplification or maximum output voltage can be obtained by dA =0 dM C2 ( R1 R2 + ω 02 M 2 ) − 2C2ω 02 M 2 C2 ( R1 R2 + ω 02 M 2 )   

2

=0

R1 R2 + ω 02 M 2 − 2ω 02 M 2 = 0

ω 02 M 2 = R1 R2 M= where M =

R1 R2 ω0

R1 R2 is the critical value of mutual inductance. ω0

4.10  Tuned Circuits  4.43

Hence, maximum value of I2, V2 and A is I 2max = V2max = Amax =

V0

V1

k = k1

2 R1 R2 V1 2ω 0C2 R1 R2

k = k2

1

k = k3

k1 > k2 > k3

2ω 0C2 R1 R2

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. 4.83. Fig. 4.83  V   ariation of V0 with w  for  different values of k

4.10.2

Double-Tuned Circuit

Consider a double-tuned circuit with tuning capacitors placed both in primary and secondary circuits as shown in Fig. 4.84. −j

1 wC1

R2

M

+

R1

jwL1

jwL2

+

−j

V1 −

1 wC2

V2 −

Fig. 4.84  Double-tuned circuit The conductively coupled equivalent circuit is shown in Fig. 4.85. −j

1 wC1

jw (L1 − M )

jw (L2 − M )

R2 +

R1 j wM

+

V1 −

I1

−j

1 wC2

V2 −

Fig. 4.85  Conductively coupled equivalent circuit Applying KVL to Mesh 1, V1 − R1I1 + j

1 I1 − jω ( L1 − M )I1 − jω M ( I1 − I 2 ) = 0 ω C2 1    R1 + jω L1 − j ωC  I1 − jω MI 2 = V1 1

...(4.19)

4.44 Circuit Theory and Transmission Lines Applying KVL to Mesh 2, − jω M ( I 2 − I1 ) − jω ( L2 − M )I 2 − R2 I 2 + j  − jω MI1 +  R2 + 

1 I2 = 0 ω C2

1   j  ω L2 −  I2 = 0  ωC2  

...(4.20)

From Eq. (4.20), 1   R2 + j  ω L2 −  ωC2  I1 = I2 jω M Substituting Eq. (4.21) in Eq. (4.19),  1   R2 + j  ω L2 −   ωC2   1   + − R j ω L j I 2 − jω MI 2 = V1 1  1  jω M ωC2          R1 +  I2     I2 =

 1   j  ω L2 − + ω2M 2     ω C2     = V1 jω M  

1   j  ω L1 − R2 +  ωC2   

jω MV1  1  1    2 2  R1 + j  ω L1 − ωC    R2 + j  ω L2 − ωC  + ω M  2  2  

1. Output Voltage 1  1  V2 = − j  I2 = I2  ωC2  jω C 2 =

MV1  1  1      C2   R1 + j  ω L1 − R2 + j  ω L2 − + ω2M 2          ω C2   ω C2    

2. Voltage Amplification A=

V2 = V1

M   C2   R1 +  

1   j  ω L1 − R2 +  ωC2   

 1   j  ω L2 − + ω2M 2    ω C2   

If the primary and secondary circuits resonate at the same frequency w0,

ω0 =

1 L1C1

=

1 L2C2

...(4.21)

4.10  Tuned Circuits  4.45

Then, at the resonant frequency w 0,

ω 0 MV1 R1 R2 + ω 02 M 2 MV1 V2 = C2 ( R1 R2 + ω 02 M 2 ) M A= C2 ( R1 R2 + ω 02 M 2 ) I2 =

The maximum amplification or maximum output voltage can be obtained by, dA =0 dM C2 ( R1 R2 + ω 02 M 2 ) − 2C2ω 02 M 2

(R R

C2 ( R1 R2 + ω 02 M 2 )   

1 2

2

=0

)

+ ω 02 M 2 − 2ω 02 M 2 = 0

ω 02 M 2 = R1 R2 M= where M = inductance.

R1 R2 ω0

R1 R2 ω0

V0

is the critical value of mutual

k = k1

Hence, maximum value of I2, V2 and A is I 2max = V2max =

k1 = kc k2 > k1 k3 < k1

k = k2

V1 2 R1 R2

k = k3

V1

w 0

2ω 0C2 R1 R2

Amax =

1

w0

Fig. 4.86  V   ariation of V0 with w  for  different values of k

2ω 0C2 R1 R2

The variation of the output voltage V0 with the angular frequency w for different values of coefficient of coupling k is shown in Fig. 4.86.

  example 4.37  For the single-tuned circuit shown in Fig. 4.87, 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. 4.87

4.46 Circuit Theory and Transmission Lines Solution M = k L1 L2 = 0.8 (1 × 10 −6 )(60 ×10 −6 ) = 6.2 µH (a) Resonant frequency 1

ω0 =

f0 =

L2C2

1

=

−6

−6

(60 × 10 )(0.1 ×10 )

= 0.41 × 106 rad/s

ω 0 0.41 × 106 = = 0.0653 MHz = 65.3 kHz 2π 2π

(b) Output voltage at resonance V0 =

MV1

(

)

C2 R1 R2 + ω 02 M 2

=

6.2 × 10 −6 × 50

(

0.1 × 10 −6 (10 × 5) + 0.42 × 106 

)( 2

)

2 6.2 ×10 −6  

= 54.6 V

(c) Maximum output voltage V2max =

V1 2ω 0C2 R1 R2

=

50 2(0.42 × 10 )(0.1 × 10 −6 ) 10 × 5 6

= 84.18 V

  example 4.38 

The resonant frequency of the tuned circuit shown in Fig. 4.88 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. 4.88 Solution

ω0 = L1 = L2 =

Mc =

1 L1C1 1

ω 02C1 1

ω 02C2

= = =

1 L2C2 1

(1000)

2

(1 × 10 ) −6

1

(1000)

2

=1H

(2 × 10 ) −6

= 0.5 H

R1 R2 5×3 = = 3.87 mH ω0 1000

2 µF

Exercises  4.47

  example 4.39 

The tuned frequency of a double-tuned circuit shown in Fig. 4.89 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. 4.89 Solution MC =

ω0 = 10 4 =

( R1 + RS ) R2 ω0

=

(0.01 + 0.1) (0.1) 10 4

= 10.49 µH

1 L2C2 1

(25 ×10 ) C −6

2

C2 = 0.4 mF V0 max =

V1 2ω 0C2

( RS + R1 ) R2

=

2 4

2 ×10 × 0.4 × 10

−3

×

(0.1 + 0.01)(0.1)

= 2.38 V

Exercises 4.1

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]

4.2

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

4.3

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] 4.4

Find the equivalent inductance of the network shown in Fig. 4.90. 7H 3H 2H

5H 4H

6H

Fig. 4.90 [10 H]

4.48 Circuit Theory and Transmission Lines Find the effective inductance of the network shown in Fig. 4.91.

4.5

(ii) A

2H 2Ω

3H

j5 Ω

4H

5H

2Ω

j5 Ω j2 Ω

2H B

Fig. 4.94

Fig. 4.91

(iii)

[4.8 H]

A

Write mesh equations of the network shown in Fig. 4.92.

4.6

L2

R1

2Ω

j4 Ω +

L1

v(t) i1

−

R2 i2

−j 3 Ω

j5 Ω

R3

j8 Ω

L3 i3 B

4Ω

Fig. 4.92

Fig. 4.95

d di2 di3    v = i1 R1 + L1 dt (i1 − i2 ) + M12 dt + M13 dt     R (i − i ) + R i + L di3 + M d (i − i )  33 3 13 1 2   2 3 2 dt dt     − M 23 di2 = 0   dt 4.7

(a) (3 + j 36.3) Ω (b) (1 + j1.5) Ω    (c) (6.22 + j 4.65) Ω  4.8

In the coupled circuit shown in Fig. 4.96, find V2 for which I1 = 0. What voltage appears at the 8 Ω inductive reactance under this condition?

Find the input impedance at terminals AB of the coupled circuits shown in Fig. 4.93 to 4.95.

5Ω

+ j8 Ω

j2 Ω

−

3Ω

2Ω

+ 100∠0°V

(i)

j2 Ω

V2 −

j4 Ω

A

Fig. 4.96 j3 Ω

j5 Ω

[141.5∠ − 45°V, 100∠0°V]

−j 8 Ω

4.9

B

Fig. 4.93

For the coupled circuit shown in Fig. 4.97, find the components of the current I2 resulting from each source V1 and V2.

Objective-Type Questions  4.49 2Ω

V1 = 10∠0°V

4.12 Find the current I in the circuit of Fig. 4.100.

j2 Ω

+

+ j4 Ω

j3 Ω

I2

−

−

14 Ω −j15 Ω

V2 = 10∠0°V

j4 Ω

j3 Ω

10 Ω

+

+ j 10 Ω

100∠20° V

Fig. 4.97

−

j 12 Ω

70∠−30° V −

I

[0.77∠112.6°Α, 1.72 ∠86.05°Α] 4.10 Find the voltage across the 5 Ω resistor in the network shown in Fig. 4.98. j2 Ω

k = 0.5

Fig. 4.100

[7.07∠45° V, 1.04∠ − 4.13 Obtain a conductively coupled circuit for the circuit shown in Fig. 4.101.

j1 Ω

+ −j 3 Ω

10∠0° V

5Ω

2Ω

j5 Ω

j3 Ω

j4 Ω

− + 100∠0°V

Fig. 4.98 [19.2∠ − 33.02°V ] 4.11 Find the power dissipated in the 5 Ω resistor in the network of Fig. 4.99. 2Ω

j2 Ω

j3 Ω

j4 Ω

−j 2 Ω

3Ω −

Fig. 4.101

5Ω

2Ω

j2 Ω

−j 1 Ω

+ 3Ω

100∠0°V −

+

3Ω

−

j3 Ω

100∠0°V

−j2 Ω

Fig. 4.99 [668.16 W]

Fig. 4.102

Objective-Type Questions 4.1

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

4.2

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 (a)

(c)

k= k=

M 2 L1 L2 2M L1 L2

(b)

(d)

k= k=

M L1 L2 L1 L2 M

4.50 Circuit Theory and Transmission Lines 4.3

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) L1 + L2

4.4

4.5

4.6

(b) L1 − L2

4.7

4.8

1 1 ( L1 − L2 ) ( L1 + L2 ) (c) (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 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

4.9

(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 4.10 The impedance Z as shown in Fig.4.103 is j5 Ω

j2 Ω

j 10 Ω

j10 Ω

Fig. 4.103 (a) (c)

j 29 Ω j19 Ω

(b) (d)

Answers to Objective-Type Questions 4.1 4.7

(c) (c)

4.2 (b) 4.8 (b)

4.3 4.9

(c) (b)

4.4 4.10

(d) (b)

j2 Ω

4.5

(d)

4.6

(c)

j9 Ω j39 Ω

5 Resonance

  5.1     IntroductIon Resonance is a very important phenomenon in many electrical applications. The study of resonance is very useful in the telecommunication field. A circuit containing reactance is said to be in resonance if the voltage across the circuit is in phase with the current through it. At resonance, the circuit thus behaves as a pure resistor and the net reactance is zero.

  5.2     SerIeS reSonance

R

Consider the series RLC circuit as shown in Fig. 5.1. The impedance of the circuit is

L

i

Z = R + jX L − jX C = R + j ( X L − X C ) At resonance, the circuit is purely resistive. v = Vm sin wt

X L − XC = 0

Fig. 5.1  Series circuit

X L = XC

ω0 L =

1 ω 0C

ω 02 =

1 LC

ω0 =

1

f0 =

LC 1 2π LC

where f 0 is called the resonant frequency of the circuit.

C

5.2 Circuit Theory and Transmission Lines 1.  Power Factor Power factor = cos φ = At resonance Z = R

R Z

R =1 R

Power factor =

2.   Current Since impedance is minimum, the current is maximum at resonance. Thus, the circuit accepts more current and as such, an RLC circuit under resonance is called an accepator circuit. V V = Z R

I0 = 3.  Voltage At resonance,

1 ω 0C

ω0 L = ω 0 L I0 =

1 I0 ω 0C

VL0 = VC0 Thus, potential difference across inductor equal to potential difference across capacitor being equal and opposite cancel each other. Also, since I 0 is maximum, VL 0 and VC 0 will also be maximum. Thus, voltage magnification takes place during resonance. Hence, it is also referred to as voltage magnification circuit. 4.  Phasor Diagram Figure 5.2 shows the phasor diagram of a series resonant circuit. VL

I VR VC I VR

Fig. 5.2  Phasor diagram 5.   Behaviour  of  R,  L  and  C  with  Change  in  Frequency Resistance remains constant with the change in frequencies. Inductive reactance X L is directly proportional to frequency f. It can be drawn as a straight line passing through the origin. Capacitive reactance X C is inversely proportional to the frequency f. It can be drawn as a rectangular hyperbola in the fourth quadrant. Figure 5.3 shows the behaviour of R, L and C with change in frequency. Total impedance Z = R + j ( X L − X C )

XL

Z

R

0 XC

Fig. 5.3  Behaviour of R, L and C with       change in frequency

5.2  Series Resonance  5.3

(a) When f < f 0 , impedance is capacitive and decreases up to f0. The power factor is leading in nature. (b) At f =  f0, impedance is resistive. The power factor is unity. (c) When f > f0, impedance is inductive and goes on increasing beyond f0. The power factor is lagging in nature. Figure 5.4 shows the impedance curve of series resonant circuit. Z Capacitive

0

Inductive

f

f0

Fig. 5.4  Impedance curve 6.   Bandwidth For the series RLC circuit, bandwidth is defined as the range of frequencies for which the P power delivered to R is greater than or equal to 0 2 where P0 is the power delivered to R at resonance.

I

V R 0.707 I0 I0 =

From the shape of the resonance curve shown in Fig. 5.5, it is clear that there are two frequencies for which the power delivered to R is half the power at resonance. For this reason, these frequencies are referred as those corresponding to the half-power points. The magnitude of the current at each halfpower point is the same.

Bandwidth = w 2 − w 1

w

0

w1 w0 w2

Fig. 5.5  Resonance curve

1 2 I 0 R = I 22 R 2 where the subscript 1 denotes the lower half point and the subscript 2, the higher half point. It follows then that I12 R =

Hence,

I1 = I 2 =

I0

= 0.707 I 0

2

Accordingly, the bandwidth may be identified on the resonance curve as the range of frequencies over which the magnitude of the current is equal to or greater than 0.707 of the current at resonance. In Fig. 5.5, the bandwidth is w 2 −w1. Expression for Bandwidth Generally, at any frequency w, I=

V = Z

V 2

R + ( X L − XC )

2

=

V 1   R2 +  ω L −   ωC 

…(i) 2

5.4 Circuit Theory and Transmission Lines At half-power points, I=

I0

2 V I0 = R V I= 2R

But

…(ii)

From Eqs (i) and (ii), V 1   R2 +  ω L −   ωC 

2

=

V 2R

2

1   R2 +  ω L −  = 2R  ωC  Squaring both the sides, 2

1   2 R2 +  ω L −  = 2R  ωC  2

1   2  ω L −  =R ωC  1 ±R=0 ωC 1 R =0 ω2 ± ω − L LC

ωL −

ω=±

R R2 1 ± + 2 2L LC 4L

 R2  1 For low values of R, the term   can be neglected in comparison with the term LC .  4 L2  Then w is given by,

ω=±

R 1 R 1 ± =± ± 2L LC 2L LC

The resonant frequency for this circuit is given by

ω0 =

1

LC R + ω0 ω=± 2L R ω1 = ω 0 − 2L

(considering only posititve sign of w0)

5.2  Series Resonance  5.5

R 2L R f1 = f 0 − 4π L R f 2 = f0 + 4π L

ω2 = ω0 +

and

and

R L R Bandwidth = f 2 − f1 = 2π L Bandwidth = ω 2 − ω1 =

or

7.   Quality Factor It is measure of voltage magnification in the series resonant circuit. It is also a measure of selectivity or sharpness of the series resonant circuit. Q0 = Substituting values of VL0 and V, Q0 =

Voltage across inductor or capacitor VL0 VC0 = = Voltage at resonance V V I 0 X L0 I0 R

=

X L0 R

=

ω0 L 1 = R ω 0CR

Substituting values of w0,  1   L 1 L LC  = Q0 = R R C

 example  5.1  A series RLC circuit has the following parameter values: R = 10 W  , L = 0.01 H, C = 100 mF. Compute the resonant frequency, bandwidth, and lower and upper frequencies of the bandwidth. Solution

R = 10 Ω,  L = 0.01 H, C = 100 µF

(a) Resonant frequency f0 =

1 2π LC

=

1 2π 0.01 × 100 ×10 −6

= 159.15 Hz

(b) Bandwidth R 10 = = 159.15 Hz 2π L 2π × 0.01 (c) Lower frequency of bandwidth BW =

f1 = f0 −

159.15 BW = 159.15 − = 79.58 Hz 2 2

(d) Upper frequency of bandwidth f2 = f0 +

159.15 BW = 159.15 + = 238.73 Hz 2 2

5.6 Circuit Theory and Transmission Lines

 example  5.2 

An RLC series circuit with a resistance of 10 W, inductance of 0.2 H and a capacitance of 40 µF is supplied with a 100 V supply at variable frequency. Find the following w.r.t. the series resonant circuit: (a) (b) (c) (d) (e) (f) (g) (h)

frequency of which resonance takes place current power power factor voltage across RLC at that time quality factor half-power points resonance and phasor diagrams R = 10  Ω,  L = 0.2 H, C = 40 µF,

Solution

V = 100 V

(a) Resonant frequency f0 =

1 2π LC

=

1 2π 0.2 × 40 × 10 −6

= 56.3 Hz

(b) Current I0 =

V 100 = = 10 A R 10

(c) Power P0 = I 02 R = (10)2 × 10 = 1000 W (d) Power factor pf = 1 (e) Voltage across R, L, C VR0 = R I0 = 10 × 10 = 100 V VL0 = X L0 I0 = 2π f0 LI 0 = 2π × 56.3 × 0.2 × 10 = 707.5 V VC0 = XC0 I 0 = (f)

1 1 × 10 = 707.5 V I0 = 2π f0C 2π × 56.3 × 40 × 10 −6

Quality factor Q=

1 L 1 0.2 = = 7.07 R C 10 40 × 10 −6

(g) Half-power points 10 R = 56.3 − = 52.32 Hz 4π L 4π × 0.2 10 R = 56.3 + = 60.3 Hz f2 = f0 + 4π L 4π × 0.2 f1 = f0 −

5.2  Series Resonance  5.7

(h) Resonance and phasor diagram are shown in Fig. 5.6. I

I0 I0 √2

0

f1

f0

f

f2

V

I

(b)

(a)

Fig. 5.6

 example 5.3 

A series RLC circuit is connected to a 200 V ac supply. The current drawn by the circuit at resonance is 20 A. The voltage drop across the capacitor is 5000 V at series resonance. Calculate resistance and inductance if capacitance is 4 µF. Also, calculate the resonant frequency. Solution

V = 200 V, I 0 = 20 A, VC0 = 5000 V, C = 4 µF

(a) Resistance R=

V 200 = = 10 Ω 20 I0

(b) Resonant frequency X C0 = X C0 = 250 =

VC0 I0

=

5000 = 250 Ω 20

1 2π f0C 1 2π × f0 × 4 × 10 −6

f0 = 159.15 Hz (c) Inductance At resonance XC0 = X L0 = 250 Ω X L0 = 2π f0 L 250 = 2π × 159.15 × L L = 0.25 H

5.8 Circuit Theory and Transmission Lines

 example 5.4 

A resistor and a capacitor are connected in series with a variable inductor. When the circuit is connected to a 230 V, 50 Hz supply, the maximum current obtained by varying the inductance is 2 A. The voltage across the capacitor is 500 V. Calculate the resistance, inductance and capacitance of the circuit. Solution

V = 230 V,

f0 = 50 Hz, I 0 = 2 A, VC0 = 500 V I0 = 2 A VC0 = 500 V

(a) Resistance R=

V 230 = = 115 Ω 2 I0

(b) Capacitance X C0 = X C0 =

VC0 I0

=

500 = 250 Ω 2

1 2π f0C

1 2π × 50 × C C = 12.73 µF

250 =

(c) Inductance At resonance XC0 = X L0 = 250 Ω X L0 = 2π f0 L 250 = 2π × 50 × L L = 0.795 H

 example 5.5  A coil of 2 W resistance and 0.01 H inductance is connected in series with a capacitor across 200 V mains. What must be the capacitance in order that maximum current occurs at a frequency of 50 Hz? Find also the current and the voltage across the capacitor. Solution

R = 2 Ω, L = 0.01 H, V = 200 V,

(a) Capacitance f0 = 50 =

1 2π LC 1 2π 0.01 × C

C = 1013.2 µF (b) Current I0 =

V 200 = = 100 A 2 R

f0 = 50 Hz

5.2  Series Resonance  5.9

(c) Voltage across capacitor VC0 = I 0 X C0 = I 0

1 1 = 100 × = 314.16 V 2π f 0C 2π × 50 × 1013.2 × 10 −6

 example 5.6  A series RLC circuit has a quality factor of 5 at 50 rad/s. The current flowing through the circuit at resonance is 10 A and the supply voltage is 100 V. Find the circuit constants. Solution

Q0 = 5, ω 0 = 50 rad / s, I 0 = 10 A, V = 100 V V R 100 10 = R R = 10 Ω I0 =

ω0 L R 50 × L 5= 10 L =1H

Q0 =

Q0 =

1 L R C

1 1 10 C C = 400 µF 5=

 example 5.7  A series RLC circuit has R = 2.5 W, C = 100 µF and a variable inductance. The applied voltage is 50 V at 800 rad/s. The inductance is varied till the voltage across the resistance is maximum. Find (a) value of inductance under this condition, (b) Q, (c) current, and (d) voltage across resistance, capacitance and inductance. Solution

R = 2.5 Ω, C = 100 µF, V = 50 V, ω = 800 rad/s

(a) Value of inductance

ω0 = 800 =

1 LC 1

L × 100 × 10 −6 L = 0.0156 H

(b) Quality factor Q0 =

ω 0 L 800 × 0.0156 = =5 2.5 R

5.10 Circuit Theory and Transmission Lines (c) Current I0 =

V 50 = = 20 A R 2.5

(d) Voltage across resistance, capacitance and inductance VR0 = I 0 R = 20 × 2.5 = 50 V VL0 = Q0 V = 5 × 50 = 250 V VC0 = Q0 V = 5 × 50 = 250 V

 example 5.8  A series RLC circuit which resonates at 500 kHz has R = 25 W, L = 100 µH and C = 1000 pF. Determine the quality factor, new value of C required to resonate at 500 kHz when the value of L is doubled, and the new quality factor. f 0 = 500 kHz, R = 25 Ω, L = 100 µH, C = 1000 pF

Solution (a) Quality factor

Q0 =

1 L 1 100 × 10 −6 = = 12.65 R C 25 1000 × 10 −12

(b) New value of C when L = 200 µH f0 =

1 2π LC 1

3

500 × 10 =

2π 200 × 10 −6 × C C = 506.61 pF

(c) New Q factor Q0 =

1 L 1 200 × 10 −6 = = 25.13 R C 25 506.61 × 10 −12

 example 5.9  Find the values of R,L and C in a series RLC circuit that resonates at 1.5 kHz and consumes 50 W from a 50 V ac source operating at the resonant frequency. The bandwidth is 0.75 kHz. Solution

f 0 = 1.5 kHz, P = 50 W, V = 50 V, B W = 0.75 kHz P=

V2 R

(50) 2 R R = 50 Ω

50 =

5.2  Series Resonance  5.11

R 2π L 50 0.75 × 103 = 2π × L L = 0.011 H BW =

f0 = 1.5 × 103 =

1 2π LC 1

2π 0.011 × C C = 1.023 µF

 example 5.10  A 50 W resistor is connected in series with an inductor having internal resistance, a capacitor and 100 V variable frequency supply as shown in Fig. 5.7. At a frequency of 200 Hz, a maximum current of 0.7 A flows through the circuit and voltage across the capacitor is 200 V. Determine the circuit constants. 50 Ω

r

L

C

100 V

Fig. 5.7

Solution

V = 100 V, R = 50 Ω, VC0 = 200 V VC0 = I 0 X C 0 200 = 0.7 X C 0 X C 0 = 285.71 Ω X C0 =

1 2π f 0C

1 2π × 200 × C C = 2.79 µF

285.71 =

f0 = 200 =

1 2π LC 1

2π L × 2.79 × 10 −6 L = 0.23 H

f 0 = 200 Hz, I 0 = 0.7 A

5.12 Circuit Theory and Transmission Lines At resonance, total resistance is V 100 = = 142.86 Ω I 0.7 RT = R + r 142.86 = 50 + r RT =

r = 92.86 Ω

 example 5.11  In the circuit shown in Fig. 5.8, a maximum current of 0.1 A flows through the circuit when the capacitor is at 5 µF with a fixed frequency and a voltage of 5 V. Determine the frequency at which the circuit resonates, the bandwidth, quality factor Q and the value of resistance at resonant frequency. R

0.1 H

5 µF

0.1 A

5V

Fig. 5.8 I 0 = 0.1 A, C = 5 µF, V = 5 V, L = 0.1 H

Solution

R= f0 =

5 V = = 50 Ω I 0 0.1 1 2π LC

=

1 2π 0.1 × 5 × 10 −6

= 225.08 Hz

1 L 1 0.1 = = 2.83 R C 50 5 × 10 −6 R 50 BW = = = 79.58 Hz 2π L 2π × 0.1 Q0 =

 example 5.12  A series RLC circuit has R = 10 W and L = 60 mH. At a frequency of 25 Hz, the pf of the circuit is 45° lead. At what frequency will the circuit be resonant? Solution

R = 10 Ω,  L = 60 mH XL = 2πf L = 2π × 25 × 60 × 10−3 = 9.42 Ω

For a series RLC circuit, XC − X L R X − 9.42 tan 45° = C 10 tan φ =

5.2  Series Resonance  5.13

X C − 9.42 10 X C = 19.42 Ω 1=

XC =

1 2π fC

1 2π × 25 × C C = 327.82 µF

19.42 =

f0 =

1

=

2π LC

1 2π 60 × 10

−3

× 327.82 × 10 −6

= 35.89 Hz

A voltage v(t ) = 10 sin ω t is applied to a series RLC circuit. At the resonant frequency of the circuit, the voltage across the capacitor is found to be 500 V. The bandwidth of the circuit is known to be 400 rad/s and the impedance of the circuit at resonance is 100 W  . Determine inductance and capacitance, resonant frequency, upper and lower cut-off frequencies.

 example 5.13 

Solution

v(t ) = 10 sin ω t , VC0 = 500 V, BW = 400 rad / s, R = 100 Ω

(a) Inductance and capacitance V=

Vm

10

= 7.07 V 2 V 7.07 I0 = = = 0.0707 A R 100 R BW = L 100 400 = L 2

=

L = 0.25 H VC 500 = 70.72 Q0 = 0 = 7.07 V Q0 =

1 L R C

1 0.25 100 C C = 4.99 nF

70.72 =

(b) Resonant frequency f0 =

1 2π LC

=

1 2π × 0.25 × 4.99 × 10 −9

= 4506.09 Hz

5.14 Circuit Theory and Transmission Lines (c) Lower cut-off frequency f1 = f 0 −

100 R = 4506.09 − = 4474.26 Hz 4π L 4π × 0.25

f 2 = f0 +

100 R = 4506.09 + = 4537.92 Hz 4π L 4π × 0.25

(d) Upper cut-off frequency

 example 5.14  A series resonant circuit has an impedance of 500 W at resonant frequency. Cut-off frequencies are 10 kHz and 100 Hz. Determine (a) resonant frequency, (b) value of L, C, and (c) quality factor at resonant frequency. R = 500 Ω,

Solution

f1 = 100 Hz,

f 2 = 10 kHz

(a) Resonant frequency BW = f 2 − f1 = 10000 − 100 = 9900 Hz R 4π L R f 2 = f0 + 4π L f1 = f 0 −

Adding Eqs (i) and (ii), f1 + f 2 = 2 f 0 f + f 2 100 + 10000 f0 = 1 = = 5050 Hz 2 2 (b) Values of L and C R 2π L 500 9900 = 2π L L = 8.038 mH BW =

X L0 = 2π f 0 L = 2π × 5050 × 8.038 × 10 −3 = 255.05 Ω At resonance

X L0 = X C0 = 255.05 Ω X C0 =

1 2π f 0C

1 2π × 5050 × C C = 0.12 µF

255.05 =

(c) Quality factor Q0 =

1 L 1 8.038 × 10 −3 = = 0.5176 R C 500 0.12 × 10 −6

…(i) …(ii)

5.2  Series Resonance  5.15

 example 5.15 

Impedance of a circuit is observed to be capacitive and decreasing from 1 Hz to 100 Hz. Beyond 100 Hz, the impedance starts increasing. Find the values of circuit elements if the power drawn by this circuit is 100 W at 100 Hz, when the current is 1 A. The power factor of the circuit at 70 Hz is 0.707. f 0 = 100 Hz, P0 = 100 W, I 0 = 1 A, ( pf )70 Hz = 0.707

Solution

The impedance of the circuit is capacitive and decreasing from 1 Hz to 100 Hz. Beyond 100 Hz, the impedance starts increasing. f 0 = 100 Hz P0 = I 02 R 100 = (1) 2 × R R = 100 Ω 1 f0 = 2π LC 100 =

1 2π LC

LC = 2.53 × 10 −6

…(i)

Power factor at 70 Hz is 0.707. R = 0.707 Z 100 = 0.707 Z Z = 141.44 Ω Impedance at 70 Hz = Z70 = R 2 + ( X C − X L ) 2 1   141.44 = (100) 2 +  − 2π × 70 × L  2π × 70 × C  2.27 × 10 −3 − 439.82 L = 100.02 C

2

…(ii)

Solving Eqs (i) and (ii), L = 0.2187 H C = 11.58 µF

 example 5.16  A constant voltage at a frequency of 1 MHz is applied to an inductor in series with a variable capacitor. When the capacitor is set to 500 pF, the current has its maximum value while it is reduced to one-half when the capacitance is 600 pF. Find resistance, inductance and Q-factor of inductor.

5.16 Circuit Theory and Transmission Lines Solution

f 0 = 1 MHz, C1 = 500 pF, C2 = 600 pF

(a) Resistance and inductance of inductor C = 500 pF

At resonance

f0 = 106 =

1 2π LC 1 2π L × 500 × 10 −12

L = 0.05 mH X L = 2π f 0 L = 2π × 106 × 0.05 × 10 −3 = 314.16 Ω When capacitance is 600 pF, the current reduces to one-half of the current at resonance. 1 1 = = 265.26 Ω 2π fC 2π × 106 × 600 × 10 −12 1 I = I0 2 V 1V = Z 2R Z = 2R

XC =

R2 + ( X L − X C )2 = 2 R R 2 + (314.16 − 265.26) 2 = 4 R 2 3R 2 = 2391.21 R = 28.23 Ω (b) Quality factor Q0 =

1 L 1 0.05 × 10 −3 = = 11.2 R C 28.23 500 × 10 −12

  example  5.17 

In the circuit shown in Fig. 5.9, determine the circuit constants when the circuit draws a maximum current at 10 mF with a 10 V, 100 Hz supply. When the capacitance is changed to 12 mF, the current that flows through the circuit becomes 0.707 times its maximum value. Determine Q of the coil at 900 rad/s. Also find the maximum current that flows through the circuit. R

L

10 V, 100 Hz

Fig. 5.9

C

5.2  Series Resonance  5.17

C = 10 µF, V = 10 V,

Solution

f0 =

f 0 = 100 Hz

1 2π LC 1

100 =

2π L × 10 × 10 −6 L = 0.25 H

When C = 12 µF, w  = 900 rad/s I = 0.707 I 0 V V = 0.707 Z R R Z= 0.707 R R2 + ( X L − X C )2 = 0.707 2

1 R   R +  900 × 0.25 − = −6    0 . 707 900 × 12 × 10 2

R = 132.37 Ω

ω L 900 × 0.25 = = 1.7 R 132.37 V 10 I0 = = = 0.076 A R 132.37

Q0 =

 example 5.18 

A series RLC circuit is excited from a constant voltage variable frequency source. 600 Hz and falls to half the maximum value The current in the circuit becomes maximum at a frequency of 2π 400 Hz. If the resistance in the circuit is 3W, find L and C. at 2π Solution

f0 = f0 =

600 Hz, R = 3 Ω 2π 1

2π LC 600 1 = 2π 2π LC LC = 2.78 × 10 −6 At f =

400 Hz, 2π

π I0 I= 2

…(i)

5.18 Circuit Theory and Transmission Lines V V = Z 2R Z = 2R R2 + ( X L − X C )2 = 2 R R2 + ( X L − X C )2 = 4 R2 ( X L − X C ) 2 = 3R 2 = 3(3) 2 = 27 2

  400 1    2π × 2π × L −  = 27 400  2π × ×C   2π 2

1    400 L −  = 27 400C 

…(ii)

2

 2.78 × 10 −6 1  − = 27  400 × C C  400  C = 267.12 µF L=

2.78 × 10 −6 2.78 × 10 −6 = 10.41 mH = C 267.12 × 10 −6

  5.3     ParaLLeL reSonance Consider a parallel circuit consisting of a coil and a capacitor as shown in Fig. 5.10. The impedances of two branches are Z1 = R + jX L Z 2 = − jX C

XC

i

R − jX L 1 1 = = 2 Z1 R + jX L R + X L2 j 1 1 = = Y2 = XC Z 2 − jX C Y1 =

Y = Y1 + Y2 =

Admittance of the circuit

R − jX L 2

R +

X L2

+

v = Vm sin wt

Fig. 5.10  Parallel circuit

 X j R 1  = 2 − j 2 L 2 − 2 XC R + X L  R + X L X C 

At resonance, the circuit is purely resistive. Hence, the condition for resonance is XL

−

R 2 + X L2

1 =0 XC

XL 2

R +

X L2

=

1 XC

X L X C = R 2 + X L2

ω0 L

1 = R 2 + ω 02 L2 ω0c

XL

R

5.3    Parallel Resonance  5.19

L = R 2 + ω 02 L2 C

ω 02 L2 = ω02 = ω0 = f0 =

L − R2 C 1 R2 − 2 LC L 1 R2 − 2 LC L 1 2π

1 R2 − 2 LC L

where f0 is called the resonant frequency of the circuit. If R is very small as compared to L then 1

ω0 = f0 =

LC 1 2π LC

1.   Dynamic Impedance of a Parallel Circuit At resonance, the circuit is purely resistive. The real R part of admittance is 2 . Hence, the dynamic impedance at resonance is given by R + X L2 R 2 + X L2 ZD = R At resonance, R 2 + X L2 = X L X C = ZD =

L C

L CR

2.  Current Since impedance is maximum at resonance, the current is minimum at resonance. I0 =

V V VCR = = L ZD L CR

3.   Phasor Diagram At resonance, power factor of the circuit is unity and the total current drawn by the circuit is in phase with the voltage. This will happen only when the current I C is equal to the reactive component of the current in the inductive branch, i.e., I C = I L sin φ

IC

IL cos f f

V

I

IL sin f IL

Fig. 5.11  Phasor diagram

V

5.20 Circuit Theory and Transmission Lines Hence, at resonance I C = I L sin φ I = I L cos φ

and

The phasor diagram of parallel resonant circuit is shown in Fig. 5.11. 4.   Behaviour of Conductance G, Inductive Susceptance  BL  and Capacitive Susceptance  with Change in Frequency Conductance remains constant with the change in frequencies. Inductive susceptance BL is BL =

1 1 1 =−j =−j jX L XL 2π fL

It is inversely proportional to the frequency. Thus, it decreases with the increase in the frequency. Hence, it can be drawn as a rectangular hyperbola in the fourth quadrant. Capacitive susceptance BC is BC

Y

1 1 BC = =−j = j 2π fC − jX C XC It is directly proportional to the frequency. It can be drawn as a straight line passing through the origin. Figure 5.12 shows the behaviour of G, BL and BC with change in frequency.

G

0 BL

Fig. 5.12  Behaviour of G, BL and Bc with        change in frequency

(a) When f < f 0 , inductive susceptance predominates. Hence, the current lags behind the voltage and the power factor is lagging in nature. (b) When f = f 0 , net susceptance is zero. Hence, the admittance is minimum and impedance is maximum. At f 0 , the current is in phase with the voltage and the power factor is unity.

Z

ZD =

L CR

0 f0 (c) When f > f 0 , capacitive susceptance predominates. Hence, the current leads the voltage and power factor Fig. 5.13  Impedance curve is leading in nature. Figure 5.13 shows the impedance curve of a parallel resonant circuit.

f

I

5.   Bandwidth The bandwidth of a parallel resonant circuit is defined in the same way as that for a series resonant circuit. Figure 5.14 shows the resonance curve of a parallel resonant circuit. 6.   Quality Factor It is a measure of current magnification in a parallel resonant circuit. Current through inductor or capacitor I C0 Q0 = = I0 Current at resonance

I0 =

V ZD 0

f0

Fig. 5.14  Resonance curve

f

5.4  Comparison of Series and Parallel Resonant Circuits  5.21

Substituting values of I C0 and I 0 , V 1 X C0 ω 0 C ω 0 L X C0 Q0 = = = = CR CR VCR R L L L Neglecting the resistance R, the resonant frequency ω 0 is given by 1

ω0 =

LC

 1   L 1 L LC  Q0 = = R R C

  5.4     comParISon of SerIeS and ParaLLeL reSonant cIrcuItS Parameter

Series Circuit

Current at resonance

I=

Impedance at resonance Power factor at resonance

Parallel Circuit

V and is maximum R

I=

VCR and is minimum L

Z = R and is minimum

Z=

L and is maximum CR

Unity

Unity 1

Resonant frequency

f0 =

Q-factor

Q=

It magnifies

Voltage across L and C

2π LC 2π f 0 L R

1 R2 − 2 LC L

f0 =

1 2π

Q=

2π f 0 L R

Current through L and C

 example 5.19 

A coil having an inductance of L henries and a resistance of 12 W is connected in parallel with a variable capacitor. At ω = 2.3 × 10 6 rad /s, resonance is achieved and at this instant, capacitance C = 0.021 µ F . Find the inductance of the coil. Solution

R = 12 Ω, ω 0 = 2.3 × 106 rad/s, C = 0.021 µF

ω0 =

1 R2 − 2 LC L

5.22 Circuit Theory and Transmission Lines 1

2.3 × 106 =

L × 0.021 × 10 −6 L = 89.7 µΗ

−

(12) 2 L2

 example 5.20 

A coil of 20 W resistance has an inductance of 0.2 H and is connected in parallel with a condenser of 100 µF capacitance. Calculate the frequency at which this circuit will be have as a noninductive resistance. Find also the value of dynamic resistance. Solution

R = 20 Ω, L = 0.2 H, C =100 µF

(a) Resonant frequency 1 2π

f0 =

1 1 R2 − 2 = 2π LC L

1 0.2 × 100 × 10 −6

2

 20  − = 31.83 Hz  0.2 

(b) Dynamic resistance ZD =

0.2 L = = 100 Ω CR 100 × 10 −6 × 20

 example 5.21  In the circuit shown in Fig. 5.15, an inductance at 0.1 H having a Q0 of 5 is in parallel with a capacitor. Determine the value of capacitance and coil resistance at a resonant frequency of 500 rad/s. R

L C

Fig. 5.15 Solution

L = 0.1 H, Q0 = 5, ω 0 = 500 rad/s ω L Q0 = 0 R 500 × 0.1 5= R R = 10 Ω

ω0 =

R2 1 − 2 LC L

500 =

(10) 2 1 − 0.1 × C (0.1) 2

C = 38.46 µF

5.4  Comparison of Series and Parallel Resonant Circuits  5.23

  example  5.22 

A coil of 10 W resistance and 0.2 H inductance is connected in parallel with a variable capacitor across a 220 V, 50 Hz supply. Calculate (a) the capacitance of the capacitor for resonance, (b) the dynamic impedance of the circuit, and (c) supply current. R = 10 Ω, L = 0.2 H, V = 220 V,

Solution

f 0 = 50 Hz

(a) Capacitance of the capacitor f0 =

1 2π

1 R2 − 2 LC L

50 =

1 2π

1 (10) 2 − 0.2 × C (0.2) 2

C = 49.41 µF (b) Dynamic impedance of the circuit ZD =

0.2 L = = 404.78 Ω CR 49.41 × 10 −6 × 10

(c) Supply current I=

V 220 = = 0.543 A Z D 404.78

  example  5.23  Find the value of the dynamic impedance of the circuit shown in Fig. 5.16 at a frequency of 500 kHz and for bandwidth of operation equal to 20 kHz. The resistance of the coil is 5 W.. L

5Ω C

Fig. 5.16 Solution

f 0 = 500 kHz, BW = 20 kHz, R = 5 Ω R 2π L 5 20 × 103 = 2π L L = 39.79 µΗ BW =

f0 =

1 2π

1 R2 − 2 LC L

500 × 103 =

1 2π

1 39.79 × 10 −6 × C

−

(5) 2 (39.79 × 10 −6 ) 2

5.24 Circuit Theory and Transmission Lines C = 2.54 nF ZD =

L 39.79 × 10 −6 = = 3.13 k Ω CR 2.54 × 10 −9 × 5

 example 5.24  A coil having a resistance of 20 W and an inductance of 200 µH is connected in parallel with a variable capacitor. This parallel combination is connected in series with a resistance of 8000 W. A voltage of 230 V at a frequency of 106 Hz is applied across the circuit as shown in Fig. 5.17. Calculate (a) the value of capacitance at resonance, (b) Q factor of the circuit, (c) dynamic impedance of the circuit, and (d) total circuit current. 200 µH

20 Ω 8000 Ω

C

230 V, 106 Hz

Fig. 5.17 R = 20 Ω, L = 200 µH,

Solution

f 0 = 106 Hz, V = 230 V, RS = 8000 Ω

(a) Value of capacitance f0 =

1 2π

106 =

1 2π

1 R2 − 2 LC L 1 200 × 10 −6 × C

−

( 20) 2 ( 200 × 10 −6 ) 2

C = 126.65 pF (b) Quality factor Q0 =

2π f 0 L 2π × 106 × 200 × 10 −6 = = 62.83 20 R

ZD =

200 × 10 −6 L = = 78958 Ω CR 126.65 × 10 −12 × 20

(c) Dynamic impedance

(d) Current Total impedance of the circuit at resonance = Z D + RS = 78958 + 8000 = 869558 Ω I=

230 = 2.65 mA 86958

5.4  Comparison of Series and Parallel Resonant Circuits  5.25

  example  5.25 

Derive the expression for resonant frequency for the parallel circuit shown in Fig. 5.18. Also calculate the impedance and current at resonance. R L C

v = Vm sin w t

Fig. 5.18 Z1 = R, Z 2 = jX L , Z3 = − jX C

Solution

(a) Resonant frequency of the circuit For a parallel circuit, 1 1 1 1 = + + Z Z1 Z 2 Z3 Y=

1 1 1 1 1 1 1 + + = −j +j = − R jX L − jX C R XL XC R

1   1 j −  X L X C 

At resonance, the circuit is purely resistive. Hence, the condition for resonance is 1 1 − =0 X L XC X L = XC 1 ω 0C 1 ω 02 = LC 1 ω0 = LC

ω0 L =

f0 =

1 2π LC

where f 0 is called the resonant frequency of the circuit. (b) Impedance at resonance. At resonance, the circuit is purely resistive. Hence, the imaginary part of Y is zero. 1 R ZD = R YD =

5.26 Circuit Theory and Transmission Lines (c) Current at resonance V V = ZD R

I0 =

 example 5.26 

Derive the expression for the resonant frequency of the parallel circuit as shown in

Fig. 5.19. R1

L

R2

C

v = Vm sin w t

Fig. 5.19 Solution Z1 = R1 + jX L Z 2 = R2 − jX C For a parallel circuit, 1 1 1 = + Z Z1 Z 2 Y=

 X  1 1 R − jX L R2 + jX C R R X + = 12 + 2 = 2 1 2 + 2 2 2 − j 2 L 2 − 2 C 2  2 2 R1 + jX L R2 − jX C  R1 + X L R2 + X C  R1 + X L R2 + X C R1 + X L R2 + X C

At resonance, the circuit is purely resistive. Hence, the condition for resonance is XL R12 + X L2

−

XC R22 + X C2 XL R12

+

X L2

=0 =

XC R22

+ X C2

1 ω0 L ω 0C = R12 + ω 02 L2 R 2 + 1 2 ω 02C 2

ω 02 LC ω 2C 2 = 2 20 2 2 2 + ω0 L R2 ω 0 C + 1

R12

LC ( R22ω 02C 2 + 1) = C 2 ( R12 + ω 02 L2 )

ω 02 R22 LC 3 + LC = C 2 R12 + ω 02 L2C 2 ω 02 R22 LC 3 − ω 02 L2C 2 = C 2 R12 − LC

5.4  Comparison of Series and Parallel Resonant Circuits  5.27

ω 02 LC 2 (CR22 − L) = C (CR12 − L) ω 02 LC (CR22 − L) = CR12 − L ω 02 = ω0 = f0 =

CR12 − L LC (CR22 − L) CR12 − L LC (CR22 − L)

=

1

CR12 − L

2π LC

CR22 − L

1

CR12 − L

LC

CR22 − L

where f 0 is called the resonant frequency of the circuit.

  example  5.27 

Derive the expression for resonant frequency for the parallel circuit shown in

Fig. 5.20. r

L R C

v = Vm sin w t

Fig. 5.20 Z1 = r + jX L , Z 2 = R, Z3 = − jX C

Solution For a parallel circuit,

1 1 1 1 = + + Z Z1 Z 2 Z3 Y=

 1 1 1 1 r − jX L 1 1 r + + + + = 2 + +j = 2 2 2 R r + jX L R − jX C r + X L R XC  r + X L

 1  X j − 2 L 2  XC r + X L 

At resonance, the circuit is purely resistive, i.e., the imaginary part of Y is equal to zero. 1 X − 2 L 2 =0 XC r + X L 1 XL = X C r 2 + X L2 X L X C = r 2 + X L2

5.28 Circuit Theory and Transmission Lines 1 = r 2 + ω 0 2 L2 ω 0C L = r 2 + ω 0 2 L2 C L ω 0 2 L2 = − r 2 C

ω0 L

ω02 = ω0 = f0 =

1 r2 − 2 LC L 1 r2 − 2 LC L 1 r2 − 2 LC L

1 2π

where f0 is called the resonant frequency of the circuit.

 example 5.28 

Find the value of R in the circuit shown in Fig. 5.21 to achieve resonance.

R

−j 2 Ω

10 Ω

j10 Ω

Fig. 5.21 Solution

Z1 = 10 + j10 Ω, Z 2 = R − j 2 Ω

For a parallel circuit, 1 1 1 = + Z Z1 Z 2 Y=

1 1 10 − j10 R + j 2  10 R  2   10 + = + = + −  − j   10 + j10 R − j 2 100 + 100 R 2 + 4  200 R 2 + 4  200 R 2 + 4 

At resonance, the circuit is purely resistive, i.e., the imaginary part of Y is equal to zero. 10 2 − =0 200 R 2 + 4 2 1 = R 2 + 4 20 R 2 + 4 = 40 R 2 = 36 R=6Ω

5.4  Comparison of Series and Parallel Resonant Circuits  5.29

 example 5.29 

Find the value of R1 such that the circuit given in Fig. 5.22 is in resonance. R1

j6 Ω

10 Ω

−j 4 Ω

Fig. 5.22 Z1 = R1 + j 6 Ω, Z 2 = 10 − j 4 Ω

Solution For a parallel circuit, 1 1 1 = + Z Z1 Z 2 Y=

 6 1 1 R − j 6 10 + j 4  R1 10  4  + = 12 + = 2 + − j 2 −  R1 + j 6 10 − j 4 R1 + 36 100 + 16  R1 + 36 116   R1 + 36 116 

At resonance, the circuit is purely resistive, i.e., the imaginary part of Y is equal to zero. 6 R12

4 =0 + 36 116 6 4 = 2 116 R1 + 36 −

116 × 6 = 174 4 R12 = 138

R12 + 36 =

R1 = 11.75 Ω

 example 5.30 

Find the value of C in the circuit shown in Fig. 5.23 to give resonance. 20 Ω 10 Ω

j 37.7 Ω C

230 V, 50 Hz

Fig. 5.23

Solution

Z1 = 20 + j 37.7 Ω, Z 2 = 10 − jX C

5.30 Circuit Theory and Transmission Lines For a parallel circuit, 1 1 1 = + Z Z1 Z 2 1 1 20 − j 37.7 10 + jX C Y= + = + 20 + j 37.7 10 − jX C 400 + 1421.29 100 + X C2  20   37.7 10 XC  = + − j −  2  1821.29 100 + X C   1821.29 100 + X C2  At resonance, the circuit is purely resistive, i.e., the imaginary part of Y is equal to zero. 37.7 XC − =0 1821.29 100 + X C2 XC 100 + X C2

= 0.021

0.021 X C2 − X C + 2.1 = 0 Solving the quadratic equation, X C = 45.41 Ω

X C = 2.2 Ω 1 = 2.2 2π × 50 × C C = 1.44 mF

or

1 = 45.41 2π × 50 × C C = 70.1 µF

or or

 example 5.31 

Find the value of L at which the circuit resonates at a frequency of 1000 rad/s in the circuit shown in Fig. 5.24. 5Ω

10 Ω

L

−j 12 Ω

Fig. 5.24 Z1 = 5 + jX L Ω, Z 2 = 10 − j12 Ω

Solution For a parallel circuit, 1 1 1 = + Z Z1 Z 2 Y=

 XL 1 1 5 − jX L 10 + j12  5 10  12  + = + = + − j −  2 2 2 5 + jX L 10 − j12 25 + X L 100 + 144  25 + X L 244   25 + X L 244 

At resonance, the circuit is purely resistive, i.e., the imaginary part of Y is equal to zero.

5.4  Comparison of Series and Parallel Resonant Circuits  5.31

XL 25 +

X L2

−

12 =0 244

XL X L2

25 + 12 X L2

=

12 244

− 244 X L + 300 = 0

Solving the quadratic equation, X L = 19.01 2π × 50 × L = 19.01 L = 19.01 mH

or or

X L = 1.31 Ω 2π × 50 × L = 1.31

or

L = 1.31 mH

 example 5.32  Two impedances Z1 = (20 + j10) Ω and Z2 = (10 − j30) Ω are connected in parallel and this combination is connected in series with Z3 = (30 + jX) Ω . Find the value of X which will produce resonance. Z1 = 20 + j10 Ω , Z 2 = 10 − j 30 Ω, Z3 = 30 + jX Ω

Solution Z=

( 20 + j10) (10 − j 30) Z1Z 2 + Z3 = + (30 + jX ) = 19.23 − j 3.85 + 30 + jX = 49.23 + j ( X − 3.85) 20 + j10 + 10 − j 30 Z1 + Z 2

At resonance, the circuit is purely resistive, i.e., imaginary part of Z is equal to zero. X − 3.85 = 0 X = 3.85 Ω

 example 5.33 

Impedances Z 2 and Z3 in parallel are in series with an impedance Z1 across a

100 V, 50 Hz ac supply. Z1 = (6.25+ j1.25) Ω, Z 2 = 5 Ω , Z3 = (5 − jX C ) Ω . Determine the value of capacitance of XC such that the total current of the circuit will be in phase with the total voltage. Z1 = 6.25 + j1.25 Ω, Z 2 = 5 Ω, Z3 = 5 − jX C Ω

Solution Z= = =

5(5 − jX C ) Z 2 Z3 + Z1 = + (6.25 + j1.25) 5 + 5 − jX C Z 2 + Z3 25 − j 5 X C 25 − j 5 X C 10 + jX C + 6.25 + j1.25 = × + 6.25 + j1.25 10 − jX C 10 − jX C 10 + jX C 250 + 5 X C 2 100 + X C 2

−j

25 X C 100 + X C 2

  25 X C  250 + 5 X C2  + 6.25 + j1.25 =  6.25 + − j − 1.25 2  2  100 + X C  100 + X C  

At resonance, the circuit is purely resistive, i.e., the imaginary part of Z is equal to zero. 25 X C

− 1.25 = 0 100 + X C 2 25 X C = 1.25 100 + X C 2 1.25 X C 2 − 25 X C + 125 = 0

5.32 Circuit Theory and Transmission Lines X C 2 − 20 X C + 100 = 0 X C = 10 1 = 10 2π × 50 × C C = 318.31 µF

 example 5.34 

Find the frequency at which the circuit shown in Fig. 5.25 will be at resonance. If the capacitor and inductors are interchanged, what would be the value of resonant frequency? −

j Ω 2w

jw Ω

1Ω

Fig. 5.25

Z1 = jω Ω, Z 2 = −

Solution

j Ω, Z3 = 1 Ω 2ω

 j   −  (1) j j ( 2ω + j ) Z Z 1 2ω  Z = Z1 + 2 3 = jω + = jω − = jω − = + 2 2 j Z 2 + Z3 2ω − j 4 ω + 1 4 ω +1 − +1 2ω At resonance, the circuit is purely resistive, i.e., imaginary part of Z is equal to zero.

ω−

2ω 4ω 2 + 1

2ω   j ω −   4ω 2 + 1

=0

ω=

2ω 4ω 2 + 1 1

1 = 2 4ω 2 + 1 4ω 2 + 1 = 2 4ω 2 − 1 = 0 ω = ± 0.5 rad/s Considering positive values of w,

ω = 0.5 rad/s

If the capacitor and inductor are interchanged, Z=−  

ω2 j (1)( jω ) j jω (1 − jω ) j (ω 2 + jω ) ω   1 − j − + =− + =− + =  2 2 2  ω 2 2ω 1 + jω 2ω 2 ω 1+ ω2  1+ ω 1+ ω 1+ ω                                  

5.4  Comparison of Series and Parallel Resonant Circuits  5.33

At resonance, the circuit is purely resistive, i.e., the imaginary part of Z is equal to zero. 1 ω − =0 2ω 1 + ω 2 1 ω = 2ω 1 + ω 2 2ω 2 − ω 2 − 1 = 0

ω2 −1 = 0 ω = ± 1 rad / s Considering positive values of w, 

ω = 1 rad / s

 example 5.35  A 400 V, 50 Hz supply feeds energy to a parallel circuit consisting of 10 ∠30° Ω branch and 10 ∠− 60° Ω branch. Determine the impedance of a circuit element such that if connected in series with the source, the system comes at resonance. Zp =

Solution

(10∠30°) (10∠− 60°) = 7.07 ∠− 15° = 6.83 − j1.83 Ω 10 ∠30° + 10∠− 60°

Zeq = Z + Z p = Z + 6.83 − j 1.83 At resonance, the circuit is purely resistive, i.e., the imaginary part of Z is equal to zero. Hence Z must be equal to j1.83 Ω. Z = j1.83 Ω = jX L X L = 1.83 2π × 50 × L = 1.83 L = 5.82 mH

 example 5.36 

A coil of 400 W resistance and 318 µH inductance is connected in parallel with a capacitor and the circuit resonates at 1 MHz. If a second capacitor of 23.5 pF capacitance is connected in parallel with the first capacitor, find the frequency at which the circuit resonates. Solution

R = 400 Ω, L = 318 µH, f0 =

1 2π

106 =

1 2π

f 0 = 1 MHz, C2 = 23.5 pF

1 R2 − 2 LC1 L 1 318 × 10 −6 × C1

−

( 400) 2 (318 × 10 −6 ) 2

59 pF C1 = 76.5 When capacitor of 23.5 pF is connected in parallel with C1 = 76.59 pF , CT = C1 + C2 = 76.59 + 23.5 = 100.09 pF

5.34 Circuit Theory and Transmission Lines The new resonant frequency f 0′ =

1 2π

1 1 R2 − 2 = 2π LCT L

1 318 × 10 −6 × 100.09 × 10 −12

−

( 400) 2 (318 × 10 −6 ) 2

= 869.34 kHz

 example 5.37  A series circuit consisting of a 12 W resistor, 0.3 H inductor and a variable capacitor is connected across a 100 V, 50 Hz ac supply as shown in Fig. 5.26. The capacitance value is adjusted to obtain maximum current. Find the capacitance value. Now, the supply frequency is raised to 60 Hz, the voltage remaining same at 100 V. Find the value of the capacitance C2 to be connected across the above series circuit so that current drawn from the supply is minimum. R

C1

L

100 V, 50 Hz

Fig. 5.26 R = 12 Ω, L = 0.3 H, V = 100 V,

Solution (a) Value of capacitance C1 The resonance occurs at

f = 50 Hz

f = 50 Hz f0 = 50 =

1 2π LC1 1 2π 0.3 × C1

C1 = 33.77 µF (b) Value of capacitance C2 The resonance occurs at 60 Hz for the circuit shown in Fig. 5.27. R

L

C1

C2

100 V, 60 Hz

Fig. 5.27 X L = 2π f 0 L = 2π × 60 × 0.3 = 113.1 Ω 1 1 X C1 = = = 78.555 Ω 2π f 0C1 2π × 60 × 33.77 × 10 −6 Z1 = 12 + j113.1 − j 78.55 = 36.57 ∠70.85° Ω Y1 =

1 1 = = 0.027∠70.85° = 8.86 × 10 −3 − j 0.0255  Z1 36.57 ∠70.85°

Yreq = Y1 + Y2 = 8.86 × 10 −3 − j 0.0255 + Y2

5.4  Comparison of Series and Parallel Resonant Circuits  5.35

At resonance, the imaginary part of Yreq becomes zero. Y2 = j 0.0255  Y2 =

1 = 0.0255  X C2

X C2 = 39.22 Ω X C2 =

1 2π f 0C2

1 2π × 60 × C2 C2 = 67.63 µF

39.22 =

 example 5.38 

Find active and reactive components of the current taken by a series circuit consisting of a coil of inductance 0.1 H and a resistance of 8 W and a capacitor of 120 µF connected to a 240 V, 50 Hz supply. Find the value of the capacitor that has to be connected in parallel with the above series circuit so that the pf of the entire circuit is unity. L = 0.1 H, R = 8 Ω, C = 120 µF, V = 240 V,

Solution

f = 50 Hz

X L = 2π fL = 2π × 50 × 0.1 = 31.42 Ω 1 1 XC = = = 26.53 Ω 2π fC 2π × 50 × 120 × 10 −6 Z1 = R + jX L − jX C = 8 + j 31.42 − j 26.53 = 9.38 ∠31.44° Ω I=

V 240 = = 25.4 45 A Z1 9.43

Active component of current = I cos φ = 25.45 cos(31.44°) = 21.71 A Reactive component of current = I sin φ = 25.45 sin(31.44°) = 13.27 A When a capacitor is connected in parallel with the above series circuit, to make pf of the entire circuit unity, Z1 = R + jX L − jX C = 8 + j 31.42 − j 26.53 = 8 + j 4.89 Ω Z 2 = − jX C For a parallel circuit, 1 1 1 = + Z Z1 Z 2 Y=

1 1 8 − j 4.89 1 8 + = +j = + 8 + j 4.89 − jX C 64 + 23.91 X C 87.91

At resonance, pf is unity and the imaginary part of Y is equal to zero. 1 4.89 − =0 X C 87.91 1 = 0.056 XC

4.89   1 j −  X C 87.91

5.36 Circuit Theory and Transmission Lines X C = 17.86 1 = 17.86 2π × 50 × C C = 178.23 µF

 example 5.39  A coil of 10 W  resistance and 0.5 H inductance is connected in series with a capacitor. On applying a sinusoidal voltage the current is maximum when the frequency is 50 Hz. A second capacitor is connected in parallel with the circuit. What capacitance should it have so that the combination acts as a non-inductive resistor at 100 Hz? R = 10 Ω, L = 0.5 H,

Solution

f0 = 50 =

f 0 = 50 Hz

1 2π LC 1

2π 0.5 × C C = 20.26 µF

When a second capacitor is connected in parallel with the circuit so that the combination acts as a non-inductive resistor at 100 Hz, X L = 2π × 100 × 0.5 = 314.16 Ω 1 XC = = 78.56 Ω 2π × 100 × 20.26 × 10 −6 Z1 = 10 + j 314.16 − j 78.56 = 10 + j 235.6 Ω Z 2 = − jX C For a parallel circuit, 1 1 1 = + Z Z1 Z 2 Y=

1 1 10 − j 235.6 1 10 + = +j = + 10 + j 235.6 − jX C 100 + 55507.36 X C 55607.36

235.6   1 j −  X C 55607.36 

At resonance, the circuit acts as a non-inductive resistor, i.e., the imaginary part of Y is equal to zero. 1 235.6 − =0 X C 55607.36 X C = 236.02 1 = 236.02 2π × 100 × C C = 6.7 µF

 example 5.40  A coil having a resistance and inductance of 15 W and 8 mH respectively is connected in parallel with another coil having a resistance and inductance of 4 W and 18 mH as shown in Fig. 5.28. If this parallel combination is to be replaced by a single coil, calculate the value of resistance and inductance

5.4  Comparison of Series and Parallel Resonant Circuits  5.37

of that coil. What value of capacitance should be connected in parallel with this coil in order to get unity power factor? Assume operating frequency to be 50 Hz. R1

L1

R2

L2

Fig. 5.28 R1 = 15 Ω, L1 = 8 mH, R2 = 4 Ω, L2 = 18 mH,

Solution

f = 50 Hz

(a) Value of resistance and inductance of the coil X L1 = 2π fL1 = 2π × 50 × 8 × 10 −3 = 2.51 Ω X L2 = 2π fL2 = 2π × 50 × 18 × 10 −3 = 5.65 Ω Z1 = R1 + jX L1 = 15 + j 2.51 = 15.21 ∠9.5° Ω Z 2 = R2 + jX L2 = 4 + j 5.65 = 6.92 ∠54.77° Ω Z=

Z 1Z 2

=

Z1 + Z 2 R = 3.84 Ω X L = 3.34 Ω X L = 2π fL 3.34 = 2π × 50 × L L = 10.63 mH

(15.21 ∠9.5°)(6.92 ∠54.7° ) = 5.09∠ 40.96° Ω = 3.84 + j 3.34 Ω 15.21 ∠9.5° + 6.92 ∠54.7°

(b) Value of capacitance When a capacitance C is connected in parallel with this coil as shown in Fig. 5.29, power factor becomes unity. Hence, the circuit behaves like a parallel resonant circuit at 50 Hz. R

L

C

Fig. 5.29 1 R2 − 2 LC L

f0 =

1 2π

50 =

1 1 (3.84) 2 − 2π 10.63 × 10 −3 × C (10.63 × 10 −3 ) 2

C = 410.46 µF

5.38 Circuit Theory and Transmission Lines

Exercises 5.1

A series RLC circuit has the following parameter values: R = 10 Ω, L =  0.014 H, C = 100 µF. Compute the following. (i) resonant frequency in rad/s (ii) quality factor of the circuit (iii) bandwidth (iv) lower and upper frequency points of the bandwidth [845.2 rad/s, 1.185, 714.3 rad/s

5.4

5.5

487.9 rad/s, 1202.1 rad/s] 5.2

5.3

A series RLC circuit as R = 500 Ω,  L = 60 mH and C = 0.053 µF. Estimate (i) circuit impedance at 50 Hz (ii) resonance frequency in rad/s (iii) quality factor of the circuit (iv) bandwidth (v) current at resonance frequency with an applied voltage of 30 V [0.0041 Ω, 17733 rad/s 2.128, 8333.33 rad/s 60 mA ] A choking coil of 10 Ω  resistance and 0.1 H inductance is connected in series with a capacitor of 200 µF capacitance across a supply voltage of 230 V, 50 Hz ac. Find the following w.r.t. the resonant circuit. (i) Neat circuit diagram indicating all the relevant parameters. (ii) At what frequency will the circuit resonate. (iii) Value of reactances at resonance condition (iv) Impedance of the coil at resonance condition (v) Voltage across the coil and voltage across the capacitance (vi) Half-power frequencies f1 and f2 (vii) Q factor of the circuit (vii) Power and power factor [35.59 Hz, 22.36 Ω,10 Ω, 744.28 V, 514.28 V, 27.63 Hz, 43.55 Hz, 2.24, 5.29 kW, 1]

5.6

5.7

5.8

A coil of 40 Ω resistance and 0.75 H inductance forms part of a series circuit for which the resonant frequency is 55 Hz. If the supply is 250 V, 50 Hz, find (i) the line current (ii) the power factor (iii) the voltage across the coil [3.85 A, 0.616 (leading), 920.07 V ] An RLC series circuit of 10 Ω  resistance should be designed to have a bandwidth of 100 Hz. Determine the values of L and C so that the circuit resonates at 250 Hz. [0.0159 H, 25 µF] A resistor and capacitor are connected in series with a variable inductor. When the circuit is connected to a 220 V, 50 Hz supply, the maximum current obtained by varying the inductance is 0.314 A. The voltage across the capacitor is 800 V. Calculate the resistance, inductance and capacitance of the circuit. [700.63 Ω, 1.25 µF, 8.10 H] A coil of 2 Ω resistance and 0.01 H inductance is connected in series with a capacitor across 230 V mains. What must be the capacitance, in order that maximum current occurs at a frequency of 50 Hz? Find also the current and the voltage across the capacitor. [1 mF, 115 A, 361.28 V ] A choking coil of 10-ohm resistance and 0.1 H inductance is connected in series with a capacitor of 200 µF capacitance. Calculate the current, the coil voltage and the capacitor voltage. The supply voltage is 230 V at 50 Hz. At what frequency will the circuit resonate? Calculate the voltages at resonant frequency across the coil and capacitor. For this, assume that supply voltage is 230 V of variable frequency. [12.47 A, 411.17, 198.47 V, 35.6 Hz, 563.3 V, 514.2 V ]

5.9

An inductive coil having an inductance of 0.04 H and a resistance of 25 Ω has been connected in series with another inductive coil of 0.2 H

Objective-Type Questions  5.39

inductance and 15 Ω  resistance. The whole circuit has been energized from 230 V, 50 Hz mains. Calculate power dissipation in each coil and power factor of the whole circuit. Draw the phasor diagram. Suggest a suitable capacitor for the above circuit to resonate at 50 Hz. [181.57 W,108.95 W, 0.468 (lagging), 42.22 µF] 5.10 A voltage v (t ) = 10 2 sin ω t is applied to a series RLC circuit. At resonant frequency, voltage across the capacitor is found to be 500 V. The bandwidth of the circuit is known to

be 400 rad/s and impedance at resonance, is 10 Ω.  Determine resonant frequency, upper and lower cut-off frequencies, L and C. [3.183 kHz, 3.151 kHz, 3.214 kHz, 0.025 H, 0.1 µF] 5.11 An inductive coil having a resistance of 20 Ω and an inductance of 0.2 H is connected in parallel with a 20 µF capacitor with variable frequency, 230 V supply. Find the resonant frequency at which the total current taken from the supply is in phase with the supply voltage. Also, find the value of this current. [19.49 Hz, 4.6 A ]

Objective-Type Questions 5.1

5.2

For a series RLC resonant circuit, the total reactance at the lower half power frequency is (a) 2 R∠45° (b) 2 R∠− 45° (c) R (d) −R In a series RLC high Q circuit, the current peaks at a frequency (a) equal to the resonant frequency (b) greater than the resonant frequency (c) less than the resonant frequency (d) none of the above.

In a series RLC circuit, R = 2 k Ω, L = 1 H, 1 L =1H C = µF. The resonant frequency is 400 1 (a) 2 × 10 4 Hz (b) × 10 4 Hz π (c) 10 4 Hz (d) 2π × 10 4 Hz 5.3

5.4

5.5

For a series resonant circuit at low frequency, circuit impedance is ______ and at high frequency circuit impedance is ______. (a) capacitive, inductive (b) inductive, capacitive (c) resistive, inductive (d) capacitive, resistive A circuit with a resistor, inductor and capacitor in series in resonant of f 0 Hz. If all the component values are now doubled, the

new resonant frequency is (a)

2 f0

(b)

f0

f0 (d) f 0 4 2 In a series RLC circuit at resonance, the magnitude of the voltage developed across the capacitor (a) is always zero (b) can never be greater than the input voltage (c) can be greater than the input voltage, F however it is 90° out of phase with the input voltage (d) can be greater than the input voltage and is in phase with the input voltage The power in a series RLC circuit will be half of that at resonance when the magnitude of the current is equal to V V (a) (b) 2R 3R (c)

5.6

=

5.7

2V R 2R A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q of 100. If each of R, L and C is doubled from its original value, the new Q of the circuit is (c)

5.8

V

(d)

5.40 Circuit Theory and Transmission Lines (a) 25 (b) 50 (c) 100 (d) 200 5.9 A series RLC circuit has a Q of 100 and an impedance of (100 + j 0) Ω at its resonant angular frequency of 107 rad/s. The values of R and L are (a) 100 Ω, 103 H (b) 100 Ω, 10 −3 H (c) 10 Ω, 10 H (d) 10 Ω, 0.1 H 5.10 The following circuit in Fig. 5.30 resonates at 4H

(a) all frequencies (b) 0.5 rad/s (c) 5 rad/s (d) 1 rad/s 5.11 At resonance, the parallel circuit shown in Fig. 5.31 behaves like

R C L

1F

10 Ω

Fig. 5.31 (a) an open circuit (b) a short circuit (c) a pure resistor of value R (d) a pure resistor of value much higher than R

1F

Fig. 5.30

Answers to Objective-Type Questions 5.1 (d) 5.7 (c)

5.2 (a) 5.8 (b)

5.3 (b) 5.9 (b)

5.4 (a) 5.10 (b)

5.5 (d) 5.11 (d)

5.6 (c)

6 Transient Analysis

6.1

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.

6.2

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.

6.2 Circuit Theory and Transmission Lines 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(0) 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(0) 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. 6.1. Element with initial conditions

Equivalent circuit at t = 0+

R

R

L

OC

C SC I0

I0

V0 +

V0

−

+−

Fig. 6.1 Initial conditions

6.2 Initial Conditions 6.3

Similarly, we can draw the chart for final conditions as shown in Fig. 6.2 Equivalent circuit at t = ∞

Element with initial conditions R

R

L

SC

C OC SC I0

V0

V0 +

−

+−

I0 OC

Fig. 6.2 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 ), (0 ), 2 (0 ), 2 (0 ) are determined by writing integrodt dt dt dt 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 6.1

In the given network of Fig. 6.3, the switch is closed at t = 0. With zero current in the di d 2i inductor, find i, and 2 at t = 0+. dt dt 10 Ω 1H

100 V i(t)

Fig. 6.3 Solution At t = 0–, no current flows through the inductor. i(0 - ) = 0

6.4 Circuit Theory and Transmission Lines At t = 0 + , the network is shown in Fig. 6.4.

10 Ω

At t = 0 + , the inductor acts as an open circuit. 100 V +

i(0 ) = 0

i( 0+)

For t > 0, the network is shown in Fig. 6.5. Writing the KVL equation for t > 0, 100 - 10i - 1

Fig. 6.4 10 Ω

di =0 dt

…(i)

1H

100 V i(t)

di = 100 - 10i dt At t = 0 + ,

…(ii) Fig. 6.5

di + (0 ) = 100 - 10i(0 + ) = 100 - 10(0) = 100 A / s dt

Differentiating Eq. (ii), d 2i dt 2 d 2i

At t = 0 + ,

dt

2

= -10

(0 + ) = -10

di dt di + (0 ) = -10(100) = -1000 A / s 2 dt

Example 6.2

In the network of Fig. 6.6, the switch is closed at t = 0. With the capacitor und 2i di charged, find value for i, and 2 at t = 0+. dt dt 1000 Ω 1 µF

100 V i(t)

Fig. 6.6 Solution At t = 0-, the capacitor is uncharged. vC (0 - ) = 0 i( 0 - ) = 0 1000 Ω

At t = 0 + , the network is shown in Fig. 6.7. At t = 0+, the capacitor acts as a short circuit. +

vC (0 ) = 0 100 i( 0 + ) = = 0.1 A 1000

v C ( 0+)

100 V i( 0+)

Fig. 6.7

6.2 Initial Conditions 6.5

For t > 0, the network is shown in Fig. 6.8. Writing the KVL equation for t > 0, 100 - 1000i -

t

1 1 × 10

1000 Ω

-6

∫ i dt = 0

100 V

1 µF

…(i) i(t)

0

Differentiating Eq. (i),

Fig. 6.8

di 0 - 1000 - 106 i = 0 dt di 106 =i dt 1000 + At t = 0 ,

…(ii)

di + 106 106 (0 ) = i( 0 + ) = (0.1) = -100 A / s dt 1000 1000

Differentiating Eq. (ii), d 2i dt 2 At t = 0 + ,

=-

d 2i dt

(0 + ) = 2

106 di 1000 dt 106 di + 106 (0 ) = ( -100) = 105 A / s 2 1000 dt 1000

Example 6.3

In the network shown in Fig. 6.9, the switch is closed. Assuming all initial conditions d 2i di as zero, find i, and 2 at t = 0+. dt dt 10 Ω

1H

10 µF

10 V i(t)

Fig. 6.9 Solution At t = 0–, i( 0 - ) = 0 vC (0 - ) = 0 At t = 0 + , the network is shown in Fig. 6.10. At t = 0+, the inductor acts as an open circuit and the capacitor acts as a short circuit. i( 0 + ) = 0 vC (0 + ) = 0

10 Ω

10 V

i ( 0+)

Fig. 6.10

v C ( 0+)

6.6 Circuit Theory and Transmission Lines For t > 0, the network is shown in Fig. 6.11. Writing the KVL equation for t > 0,

10 Ω

t

di 1 10 - 10i - 1 i dt = 0 dt 10 × 10 -6 ∫0 At t = 0 + ,

10 - 10i(0 + ) -

…(i)

1H

10 µF

10 V

i(t)

Fig. 6.11

di + (0 ) - 0 = 0 dt di + (0 ) = 10 A / s dt

Differentiating Eq. (i), 0 - 10 0 - 10

At t = 0+,

di d 2 i 1 i=0 dt dt 2 10 × 10 -6

di + d 2i 1 (0 ) - 2 (0 + ) - -5 i(0 + ) = 0 dt dt 10 d 2i dt 2

Example 6.4 d 2v dt 2

(0 + ) = -10 × 10 = -100 A / s 2

In the network shown in Fig. 6.12, at t = 0, the switch is opened. Calculate v,

dv and dt

at t = 0+. v(t)

1H

1A

100 Ω

Fig. 6.12 Solution

At t = 0–, the switch is closed. Hence, no current flows through the inductor. iL (0 - ) = 0 v( 0+)

At t = 0 + , the network is shown in Fig. 6.13. At t = 0+, the inductor acts as an open circuit. iL (0 + ) = 0

i L ( 0+) 1A

100 Ω

v(0 + ) = 100 × 1 = 100 V Fig. 6.13

6.2 Initial Conditions 6.7

For t > 0, the network is shown in Fig. 6.14. 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), Fig. 6.14 1 dv +v =0 …(ii) 100 dt dv + (0 ) = -100 v (0 + ) = -100 × 100 = -10000 V / s dt

At t = 0+, Differentiating Eq. (ii),

1 d 2 v dv + =0 100 dt 2 dt d 2v

At t = 0+,

dt

Example 6.5 d 2v dt 2

2

(0 + ) = -100

dv + (0 ) = -100 × ( -10 4 ) = 106 V / s 2 dt

In the given network of Fig. 6.15, the switch is opened at t = 0. Solve for v,

dv and dt

at t = 0+. v(t)

10 A

1 kΩ

1 µF

Fig. 6.15 Solution At t = 0–, switch is closed. Hence, the voltage across the capacitor is zero. v(0 - ) = vC (0 - ) = 0

v( 0+)

10 A

1 kΩ

V C ( 0+)

+

At t = 0 , the network is shown in Fig. 6.16. At t = 0+, the capacitor acts as a short circuit. v(0 + ) = vC (0 + ) = 0

Fig. 6.16

For t > 0, the network is shown in Fig. 6.17. Writing the KCL equation for t > 0, v dv + 10 -6 = 10 1000 dt At t = 0+,

v(0 + ) dv + 10 -6 (0 + ) = 10 1000 dt

v(t)

…(i)

10 A

1 kΩ

Fig. 6.17

1 µF

6.8 Circuit Theory and Transmission Lines dv + 10 (0 ) = -6 = 10 × 106 V / s dt 10 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

Example 6.6 and

d 2v dt 2

2

(0 + ) = -

1 1000 × 10 -6

× 10 × 106 = -10 × 109 V / s 2

For the network shown in Fig. 6.18, the switch is closed at t = 0, determine v,

dv dt

at t = 0+.

2Ω

10 A

1H

0.5 µF v (t)

Fig. 6.18 Solution

At t = 0–, no current flows through the inductor and there is no voltage across the capacitor. iL (0 - ) = 0 v(0 - ) = 0

At t = 0 + , the network is shown in Fig. 6.19. At t = 0+, the inductor acts as an open circuit and the capacitor acts as a short circuit.

10 A

v( 0+)

2Ω i L ( 0+)

iL (0 + ) = 0 v(0 + ) = 0

Fig. 6.19

For t > 0, the network is shown in Fig. 6.20. Writing the KCL equation for t > 0,

10 A

2Ω

t

v 1 dv + ∫ vdt + 0.5 × 10 -6 = 10 2 11 dt At t = 0+,

…(i) Fig. 6.20

+

v(0 ) dv + 0 + 0.5 × 10 -6 (0 + ) = 10 2 dt dv + (0 ) = 20 × 106 V / s dt

1H

0.5 µF v (t )

6.2 Initial Conditions 6.9

Differentiating Eq. (i), 1 dv d 2v + v + 0.5 × 10 -6 2 = 0 2 dt dt 1 dv + d 2v (0 ) + v(0 + ) + 0.5 × 10 -6 2 (0 + ) = 0 2 dt dt

At t = 0+,

d 2v dt 2

(0 + ) = -20 × 1012 V / s 2

Example 6.7

In the network shown in Fig. 6.21, 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+. 10 Ω

1 2

20 V

1H

20 Ω i(t)

Fig. 6.21

10 Ω

20 V

Solution At t = 0–, the network attains steady-state condition. Hence, the inductor acts as a short circuit. i(0 - ) =

i( 0−)

20 =2A 10

Fig. 6.22 10 Ω

At t = 0 + , the network is shown in Fig. 6.23. At t = 0+, the inductor acts as a current source of 2 A.

20 Ω

2A i( 0+)

+

i(0 ) = 2 A For t > 0, the network is shown in Fig. 6.24. Writing the KVL equation for t > 0,

At t = 0+,

di -20i - 10i - 1 = 0 dt di -30i(0 + ) - (0 + ) = 0 dt di + (0 ) = -30 × 2 = -60 A / s dt

Differentiating Eq. (i), -30

di d 2 i =0 dt dt 2

Fig. 6.23 10 Ω

…(i) 20 Ω

1H i(t)

Fig. 6.24

2A

6.10 Circuit Theory and Transmission Lines -30

At t = 0+,

di + d 2i (0 ) - 2 (0 + ) = 0 dt dt d 2i dt 2

(0 + ) = 1800 A / s 2

Example 6.8

In the network shown in Fig.6.25, 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

30 V

1 µF

10 Ω i(t)

Fig. 6.25 20 Ω

Solution At t = 0–, the network attains steady-state condition. Hence, the capacitor acts as an open circuit. i( 0−)

i( 0 - ) = 0

Fig. 6.26

At t = 0 + , the network is shown in Fig. 6.27. At t = 0+, the capacitor acts as a voltage source of 30 V.

20 Ω

vC (0 + ) = 30 V

i( 0+)

30 i( 0 ) = = -1 A 30

Fig. 6.27

For t > 0, the network is shown in Fig. 6.28. Writing the KVL equation for t > 0,

20 Ω 1 µF

t

1 1 × 10 -6

∫ i dt - 30 = 0

…(i)

10 Ω i(t)

0

Differentiating Eq. (i),

Fig. 6.28 di -30 - 106 i = 0 dt

At t = 0+,

vC( 0+) 30 V

10 Ω

+

-10i - 20i -

vC( 0−)

30 V

vC (0 - ) = 30 V

-30

…(ii)

di + (0 ) - 106 i(0 + ) = 0 dt di + 106 ( -1) (0 ) = = 0.33 × 105 A / s dt 30

30 V

6.2 Initial Conditions 6.11

Differentiating Eq. (ii), -30 -30

At t = 0+,

d 2i dt

2

d 2i dt 2

- 106

di =0 dt

di + (0 ) = 0 dt

(0 + ) - 106

d 2i dt

3

(0 + ) = -

106 × 0.33 × 105 = -1.1 × 109 A / s 2 30

Example 6.9

In the network shown in Fig. 6.29, 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. 6.29 20 Ω

Solution At t = 0–, the network attains steady state. Hence, the capacitor acts as an open circuit.

v C ( 0−)

40 V

-

vC (0 ) = 40 V

i(

i( 0 - ) = 0

0−)

Fig. 6.30

At t = 0 + , the network is shown in Fig. 6.31. At t = 0+, the capacitor acts as a voltage source of 40 V and the inductor acts as an open circuit.

20 Ω v C ( 0+) 40 V

vC (0 + ) = 40 V

i( 0+)

i( 0 + ) = 0

Fig. 6.31

For t > 0, the network is shown in Fig. 6.32. 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 + (0 ) - 20i(0 + ) - 0 - 40 = 0 dt di + (0 ) = - 40 A / s dt

…(i)

1H i(t)

Fig. 6.32

40 V

6.12 Circuit Theory and Transmission Lines 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 6.10

In the network of Fig. 6.33, the switch is changed from the position ‘a’ to ‘b’ at d 2i di t = 0. Solve for i, and 2 at t = 0+. dt dt 1 kΩ

a b 100 V

1H

0.1 µF i(t)

Fig. 6.33 Solution At t = 0–, the network attains steady condition. Hence, the inductor acts as a short circuit. i( 0 - ) =

1 kΩ

100 V

100 = 0.1 A 1000

i ( 0−)

vC (0 - ) = 0

Fig. 6.34

At t = 0 + , the network is shown in Fig. 6.35. At t = 0+, the inductor acts as a current source of 0.1 A and the capacitor acts as a short circuit.

1 kΩ

v C ( 0+)

0.1 A

+

i(0 ) = 0.1 A

i( 0+)

vC (0 + ) = 0

Fig. 6.35

For t > 0, the network is shown in Fig. 6.36. Writing the KVL equation for t > 0, -

t

1 0.1 × 10

-6

di

∫ i dt - 1000i - 1 dt = 0

1 kΩ

…(i)

0.1 µF

1H i(t)

0

Fig. 6.36

0.1 A

6.2 Initial Conditions 6.13

-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( -100) = -9 × 105 A / s 2

The network of Fig. 6.37 attains steady-state with the switch closed. At t = 0 , the dv switch is opened. Find the voltage across the switch vK and K at t = 0 + . dt

Example 6.11

K

1H

vK 2V

1Ω 0.5 F

i(t)

Fig. 6.37 Solution: At t = 0 - , the network is shown in Fig. 6.38. 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( 0 - ) =

vK( 0−) 2V

1Ω

vC( 0−)

2 =2A 1

i( 0−)

Fig. 6.38

vC (0 - ) = 0 At t = 0 + , the network is shown in Fig. 6.39. At t = 0 + , the capacitor acts as a short circuit and the inductor acts as a current source of 2 A. 2V +

i( 0 ) = 2 A vC (0 + ) = 0 +

vK (0 ) = 0

2A

vC( 0+)

1Ω i( 0+)

Fig. 6.39

6.14 Circuit Theory and Transmission Lines 1 i dt C∫ dvK i = dt C vK =

Also,

dvK + i( 0 + ) 2 (0 ) = = = 4 A /s dt C 0.5

At t = 0 + ,

Example 6.12

In the network shown in Fig. 6.40, assuming all initial conditions as zero, find d 2 i2 + di di d 2i i1 (0 + ), i2 (0 + ), 1 (0 + ), 2 (0 + ), 21 (0 + ) and (0 ). dt dt dt 2 dt R2

R1

V

L

C i1(t)

i2(t)

Fig. 6.40 Solution At t = 0–, all initial conditions are zero. vC (0 - ) = 0 i1 (0 - ) = 0 i2 (0 - ) = 0 At t = 0 + , the network is shown in Fig. 6.41. At t = 0+, the inductor acts as an open circuit and the capacitor acts as a short circuit.

R2

R1 vC( 0+) V

i1 (0 + ) =

V R1

i1( 0+)

i2 (0 + ) = 0

i2( 0+)

Fig. 6.41

vC (0 + ) = 0 For t > 0, the network is shown in Fig. 6.42. Writing the KVL equations for two meshes for t > 0, t

1 (i1 - i2 )dt = 0 C ∫0

…(i)

1 di (i2 - i1 )dt - R2 i2 - L 2 = 0 ∫ C dt

…(ii)

V - R1i1 -

and

-

R1

R2

L

C

V i1(t)

i2(t)

Fig. 6.42

6.2 Initial Conditions 6.15

From Eq. (ii), at t = 0 + , 1 C

0+

∫ (i2 - i1 )dt - R2i2 (0

+

)-L

0

di2 + (0 ) = 0 dt di2 + (0 ) = 0 dt

Differentiating Eq. (i), di1 1 - (i1 - i2 ) = 0 dt C di 1 1 0 - R1 1 (0 + ) - i1 (0 + ) + i2 (0 + ) = 0 dt C C di1 + 1 V R1 (0 ) + =0 dt C R1 di1 + V (0 0 )=- 2 dt R1 C 0 - R1

At t = 0 + ,

…(iii)

Differentiating Eq. (iii), - R1 At t = 0+,

- R1

d 2 i1 dt

2

(0 + ) -

d 2 i1 dt 2

-

1 di1 1 di2 + =0 C dt C dt

1 di1 + 1 di2 + (0 ) + (0 ) = 0 C dt C dt d 2 i1 dt 2

(0 + ) =

V R13C 2

Differentiating Eq. (ii), -

di d 2i 1 (i2 - i1 ) - R2 2 - L 22 = 0 C dt dt d 2 i2

At t = 0+,

dt

2

(0 + ) = -

R2 di2 + V 1 (0 ) [i2 (0 + ) - i1 (0 + )] = L dt LC R1 LC

Example 6.13

In the network shown in Fig. 6.43, assuming all initial conditions as zero, find d 2 i2 di1 di2 d 2 i1 i1 , i2 , , 2 and at t = 0 + . dt dt dt dt 2 R2

C

V

L

R1 i2(t)

i1( t)

Fig. 6.43

6.16 Circuit Theory and Transmission Lines Solution At t = 0 - , all initial conditions are zero. vC (0 - ) = 0 i1 (0 - ) = 0 i2 (0 - ) = 0 At t = 0 + , the network is shown in Fig. 6.44. At t = 0 + , the capacitor acts as a short circuit and the inductor acts as an open circuit.

R1

V

V R1

i1 (0 + ) =

R2

vC( 0+)

i1( 0+)

i2 (0 + ) = 0

i2( 0+)

Fig. 6.44

vC (0 + ) = 0 For t > 0, the network is shown in Fig. 6.45.

C

R2

Writing the KVL equation for t > 0, V

t

and

1 V - ∫ i1 dt - R1 (i1 - i2 ) = 0 C0

…(i)

di2 =0 dt

…(ii)

- R1 (i2 - i1 ) - R2 i2 - L

L

R1 i1(t)

i2(t)

Fig. 6.45

From Eq. (ii),

At t = 0 + ,

di2 1 = [ R1i1 - ( R1 + R2 )i2 ] dt L di2 + 1 1 V  V (0 ) = [ R1i1 (0 + ) - ( R1 + R2 )i2 (0 + )] =  R1 - ( R1 + R2 )0  = dt L L  R1  L

…(iii)

Differentiating Eq. (i), 0-

At t = 0 + ,

i1 di di - R1 1 + R1 2 = 0 C dt dt di1 di2 i = - 1 dt dt R1C di1 + di i (0 + ) V V (0 ) = 2 (0 + ) - 1 = - 2 dt dt R1C L R1 C

Differentiating Eq. (iii), d 2 i2 dt At t = 0 + ,

2

2

=

1  di1 di  R1 - ( R1 + R2 ) 2   L  dt dt 

R   1 (0 + ) = -V  + 22   R1 LC L  dt

d i2 2

...(iv)

6.2 Initial Conditions 6.17

Differentiating Eq. (iv), d 2 i1 dt At t = 0 + ,

2

d i1 dt

2

(0 + ) =

2

d i2 dt

2

(0 + ) -

2

=

d 2 i2 dt

2

-

1 di1 R1C dt

V VR V  V 2V VR 1 di1 + 1 V (0 ) = - 22 - 2 = 3 2- 22  R1C dt R1 LC L R1C  L R1 C  R1 C R1 LC L

Example 6.14 In the network shown in Fig. 6.46, 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. 6.46 Solution Fig. 6.47.

At t = 0 - , the network is shown in

10 Ω

At t = 0–, the network attains steady-state condition. Hence, the inductor acts as a short circuit. iL (0 - ) =

5 5 2 = = A (10  30) 7.5 3

10 Ω

va(0−)

20 Ω

vb(0−)

5V iL(0−)

vb (0 - ) = 0

Fig. 6.47

20 va ( 0 ) = 5 × = 3.33 V 30 -

At t = 0 + , the network is shown in Fig. 6.48. At t = 0+, the inductor acts as a current source 2 of A. 3 2 iL (0 + ) = A 3 5V Writing the KCL equations at t = 0+, va (0 + ) - 5 va (0 + ) va (0 + ) - vb (0 + ) + + =0 10 10 20

10 Ω 10 Ω

va(0+)

10 Ω

Fig. 6.48

20 Ω

vb(0+) 2 A 3

6.18 Circuit Theory and Transmission Lines vb (0 + ) - va (0 + ) vb (0 + ) - 5 2 + + =0 20 10 3

and Solving these two equations,

va (0 + ) = 1.9 V v b (0 + ) = -0.477 V

Example 6.15 In the accompanying Fig. 6.49 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. 6.49 10 Ω

Solution At t = 0 - , the network is shown in Fig. 6.50. At t = 0–, the network attains steady-state condition. Hence, the capacitor acts as an open circuit. va ( 0 - ) = 5 V

10 Ω

va(0−)

20 Ω

vb(0−)

5V

vb (0 - ) = 5 V Fig. 6.50

At t = 0 + , the network is shown in Fig. 6.51. At t = 0+, the capacitor acts as a voltage source of 5 V.

10 Ω 10 Ω

vb (0 + ) = 5 V

va(0+)

20 Ω

vb(0+)

Writing the KCL equation at t = 0+, va ( 0 + ) - 5 va ( 0 + ) va ( 0 + ) - 5 + + =0 10 10 20 0.25va (0 + ) = 0.75

5V

10 Ω

5V

Fig. 6.51

va ( 0 + ) = 3 V

Example 6.16

The network shown in Fig. 6.52 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

6.2 Initial Conditions 6.19 L

v1(t)

v2(t)

R2

R1

i(t)

C

Fig. 6.52 Solution At t = 0–, no current flows through the inductor and there is no voltage across the capacitor. iL (0 - ) = 0 vC (0 - ) = v2 (0 - ) = 0 At t = 0 + , the network is shown in Fig. 6.53. At t = 0+, the inductor acts as an open circuit and the capacitor acts as a short circuit.

v2(0+)

v1(0+) iL(0+) i( 0)

iL (0 + ) = 0

R2

R1

v1 (0 + ) = R1 i(0 + ) v2 ( 0 + ) = 0

Fig. 6.53

For t > 0, the network is shown in Fig. 6.54. 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), 1 dv1 1 di + ( v1 - v2 ) = R1 dt L dt

Fig. 6.54

dv1 + 1  di  (0 ) = R1  (0 + ) - R1i(0 + )  dt L  dt 

At t = 0 + ,

Writing the KCL equation at Node 2 for t > 0, t

1 v dv ( v2 - v1 )dt + 2 + C 2 = 0 L ∫0 R2 dt At t = 0+,

0+

…(ii)

v2 ( 0 + ) dv + C 2 (0 + ) = 0 R2 dt dv2 + (0 ) = 0 dt

In the network shown in Fig. 6.55, the switch is closed at t = 0, with zero capacitor d 2 v2 dv dv voltage and zero inductor current. Solve for v1 , v2 , 1 , 2 and at t = 0+. dt dt dt 2

Example 6.17

6.20 Circuit Theory and Transmission Lines R1 iC (t ) V

L

+ v1(t )

R2

− + v2(t )

iL (t )

C

−

Fig. 6.55 Solution At t = 0–, no current flows through the inductor and there is no voltage across the capacitor. vC (0 - ) = 0 v1 (0 - ) = 0 v2 ( 0 - ) = 0 iL (0 - ) = 0 iC (0 - ) = 0 At t = 0 + , the network is shown in Fig. 6.56. At t = 0+, the inductor acts as an open circuit and the capacitor acts as a short circuit.

R1 iC (0+)

vC (0 + ) = 0

iL (0+)

v1 (0+)

V

+

v1 (0 ) = 0

R2

+

v2 (0+)

v2 ( 0 ) = 0 iL (0 + ) = 0 iC (0 + ) =

Fig. 6.56

V R1 R1

For t > 0, the network is shown in Fig. 6.57.

iC (t )

Writing the KVL equation for t > 0, vC (t ) = v1 (t ) + v2 (t )

…(i)

V

C

L

v1(t )

R

v2(t )

v C (t )

Differentiating Eq. (i), dvC dv1 dv2 = + dt dt dt

iL (t )

…(ii) Fig. 6.57

6.2 Initial Conditions 6.21 t

vC =

Now,

1 iC dt C ∫0

…(iii)

dvC iC = dt C At t = 0+,

dvC + i (0 + ) V (0 ) = C = V /s dt C R1C v1 = L

Also

diL dt

diL v1 = dt L At t = 0+,

…(v)

diL + v (0 + ) (0 ) = 1 =0 dt L

Also,

At t = 0+,

v2 = R2 iL

…(vi)

dv2 di = R2 L dt dt

…(vii)

dv2 + di (0 ) = R2 L (0 + ) = 0 dt dt dvC + dv dv (0 ) = 1 (0 + ) + 2 (0 + ) dt dt dt V dv1 + V /s (0 ) = R1C dt

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 + (0 ) dt

Differentiating Eq. (v), d 2 iL dt 2 At t = 0+,

…(iv)

d 2 iL dt 2 d 2 v2 dt

2

=

1 dv1 L dt

(0 + ) =

1 dv1 + 1 V (0 ) = L dt L R1C

(0 + ) =

R2V V / s2 R1 LC

6.22 Circuit Theory and Transmission Lines

Example 6.18 In the network shown in Fig. 6.58, 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. 6.58 Solution At t = 0 - , the network is shown in Fig. 6.59. 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 W (0 - ) = i1 (0 - ) =

2Ω

4Ω

v2(0−)

i1(0−) (0−)

i3

i3 (0 - ) = 0 Fig. 6.59

4 =4V 6

Since the charges on capacitors are equal when connected in series, Q1 = Q2 C1v1 = C2 v2 v1 (0 - ) -

v2 ( 0 )

and

v1(0−)

6V

6 =1A 6

i2 (0 - ) = 0

v1 (0 - ) + v2 (0 - ) = 6 ×

i2(0−)

=

1 C2 = =2 C1 0.5

v1 (0 - ) =

8 V 3

v2 ( 0 - ) =

4 V 3

6.2 Initial Conditions 6.23

At t = 0 + , the network is shown in Fig. 6.60. 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 v1 (0 + ) =

At t = 0

+,

i1(0+)

i2(0+)

8 V 3

i3(0+)

4 V 3

1A

Fig. 6.60

8 4 6 - 2i1 (0 ) - - = 0 3 3 +

i1 (0 + ) = 1 A Now,

i1 (0 + ) - i3 (0 + ) = 1 i3 (0 + ) = 0

Writing the KVL equation for Mesh 2, 8 =0 3 8 -4i2 (0 + ) + 4 - = 0 3

-4 [i2 (0 + ) - i1 (0 + )] -

i2 (0 + ) =

1 A 3

Example 6.19

In the network shown in Fig. 6.61, the switch K is closed at t = 0 connecting a voltdi di age V0 sin ω t to the parallel RL-RC circuit. Find (a) i1 (0 + ) and i2 (0 + ) (b) 1 (0 + ) and 2 (0 + ). dt dt K i1 (t )

i2 (t )

R

R

C

L

V0 sinwt

Fig. 6.61 Solution At t = 0 - , no current flows in the inductor and there is no voltage across the capacitor. vC (0 - ) = 0 i1 (0 - ) = 0 i2 (0 - ) = 0

6.24 Circuit Theory and Transmission Lines At t = 0 + , the network is shown in Fig. 6.62. At t = 0 + , the inductor acts as an open circuit and the capacitor acts as a short circuit. The voltage source V0 sin ω t acts as a short circuit.

i2 (0+ )

i1 (0+ ) R

i1 (0 + ) = 0

R

vC (0+ )

i2 (0 + ) = 0 vC (0 + ) = 0 Fig. 6.62

For t > 0, the network is shown in Fig. 6.63. Writing the KVL equation for t > 0, 1 i1 dt = 0 C∫ di V0 sin ω t - R i2 - L 2 = 0 dt

V0 sin ω t - R i1 and

i1 (t )

…(i) …(ii)

R

R

V0 sinwt C

Differentiating Eq. (i), V0 ω cos ω t - R

i2 (t )

di1 i1 - =0 dt C

L

Fig. 6.63

di1 V0ω i = cos ω t - 1 …(iii) dt R RC di1 + Vω i (0 + ) V0ω (0 ) = 0 cos ω t - 1 = dt R RC R t = 0+

At t = 0 + , From Eq. (ii),

di2 V0 R = sin ω t - i2 dt L L di2 + V R (0 ) = 0 sin ω t - i2 (0 + ) = 0 + dt L L t =0

At t = 0 + ,

Example 6.20 In the network of Fig. 6.64, the switch K is changed from ‘a’ to ‘b’ at t = 0 (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

C1 i1

Fig. 6.64

i2

6.2 Initial Conditions 6.25

Solution At t = 0 - , the network is shown in Fig. 6.65. At t = 0 - , the network attains steadystate condition. Hence, the capacitors act as open circuits and inductors act as short circuits. V

vC3(0−)

R2 R3

i1 (0 - ) = 0 i2 (0 - ) = 0

i3(0−)

vC1(0−)

i1(0−)

vC2(0−)

i2(0−)

-

i3 (0 ) = 0 vC3 (0 - ) = V

Fig. 6.65

vC2 (0 - ) = 0 vC1 (0 - ) = 0 V

At t = 0 + , the network is shown in Fig. 6.66. 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.

R2

vC3(0+)

R3

R1

vC1(0+)

i1(0+)

vC2(0+)

i2(0+)

V i1 (0 ) = i2 (0 ) = R1 + R2 + R3 +

i3(0+)

+

Fig. 6.66

i3 (0 + ) = 0

Example 6.21

In the network of Fig. 6. 67, the switch K1 has been closed for a long time prior to

t = 0. At t = 0 , the switch K2 is closed. Find vC (0 + ) and iC (0 + ). K1

K2

i1 10 Ω iC

10 V

C

i2 vC

10 Ω

Fig. 6.67 Solution At t = 0 - , the network is shown in Fig. 6.68. At t = 0 - , the network attains steadystate condition. Hence, the capacitor acts as an open circuit.

i1(0−)

10 Ω i2(0−)

iC(0−) 10 V

vC(0−)

i1 (0 - ) = 0 i2 (0 - ) = 0 iC (0 - ) = 0 vC (0 - ) = 10 V

Fig. 6.68

10 Ω

6.26 Circuit Theory and Transmission Lines At t = 0 + , the network is shown in Fig. 6.69. At t = 0 + , the capacitor acts as a voltage source of voltage V.

i1(0+) 10 Ω iC(0+) 10 V

vC (0 + ) = 10 V

i2(0+) 10 V

10 Ω

vC(0+)

Writing the KVL equation at t = 0 + , Fig. 6.69

10 - 10 i1 (0 + ) - 10 = 0 10 - 10 i2 (0 + ) = 0

and

i1 (0 + ) = 0 i2 (0 + ) = -1 A i1 (0 + ) = iC (0 + ) + i2 (0 + ) iC (0 + ) = 1 A

Example 6.22

In the network shown in Fig. 6.70, a steady state is reached with the switch open. At di di t = 0, the switch is closed. Determine vC (0 - ), i1 (0 + ), i2 (0 + ), 1 (0 + ) and 2 (0 + ). dt dt

10 Ω i2(t)

i1(t) 20 Ω

20 Ω

1H

1 µF

100 V

Fig. 6.70 Solution At t = 0 - , the network is shown in Fig. 6.71. 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 (0 - ) = 100 ×

10 Ω i2 (0−)

i1 (0−)

20 Ω

100 V 20 Ω

20 = 66.67 V 20 + 10

Fig. 6.71

66.67 i1 (0 - ) = = 3.33 A 20 i2 (0 - ) = 0 At t = 0 + , the network is shown in Fig. 6.72. 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.

i1(0+) 100 V

i2(0+) 20 Ω 3.33 A

Fig. 6.72

20 Ω 66.67 A

6.3 Resistor-Inductor Circuit 6.27

vC (0 + ) = 66.67 V i1 (0 + ) = 3.33 A 100 - 66.67 i2 (0 + ) = = 1.67 A 20 For t > 0 - , the network is shown in Fig. 6.73. Writing the KVL equations for t > 0, di 100 - 20i1 - 1 1 = 0 dt and

100 - 20 i2 -

i2(t)

i1(t)

20 Ω

20 Ω 100 V

1 µF 1H

3.33 A 66.67 V

…(i) Fig. 6.73 1 10 -6

∫ i2 dt - 66.67 = 0

…(ii)

di1 + (0 ) = 100 - 20i1 (0 + ) = 100 - 20 (3.33) = 33.3 A/s dt

At t = 0 + , Differentiating Eq. (ii),

0 - 20 At t = 0 + ,

di2 - 106 i2 = 0 dt di 20 2 (0 + ) = -106 i2 (0 + ) dt di2 + 106 (0 ) = × 1.67 = -83500 A/s 2 dt 20

6.3

rEsIstor-Inductor cIrcuIt

Consider a series RL circuit as shown in Fig. 6.74. The switch is closed at time t = 0. The inductor in the circuit is initially un-energised.

R

V

L

Applying KVL to the circuit for t > 0, di V - Ri - L = 0 dt

i(t)

Fig. 6.74

Series RL circuit

This is a linear differential equation of first order. It can be solved if the variables can be separated. (V - Ri ) dt = L di L di = dt V - Ri

Integrating both the sides, -

L ln (V - Ri ) = t + k R

6.28 Circuit Theory and Transmission Lines 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 ln V = k R

L L ln (V - Ri ) = t - ln V R R

L [ln (V - Ri) - ln V ] = t R 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 response or forced response or zero state response

V 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. 6.75 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. 6.75. 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

6.3 Resistor-Inductor Circuit 6.29

Similarly, i( 2T ) =

V V -2 V V V - e = - 0.135 = 0.865 R R R R R

i(3T ) =

V V -3 V V V - e = - 0.0498 = 0.950 R R R R R

i(5T ) =

V V -5 V V V - e = - 0.0067 = 0.993 R R R R R

After 5 time constants, the current reaches 99.33 per cent of its final value. The voltage across the resistor is R - t V vR = Ri = R × 1 - e L  R  R  - t = V 1 - e L   

v(t)

for t > 0

V

VR VL

Similarly, the voltage across the inductor is R - t di V d vL = =L 1 - e L  dt R dt  

= Ve

R - t L

t

O

Fig. 6.76

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 6.23 In the network of Fig. 6.77, 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).

6.30 Circuit Theory and Transmission Lines R1

1 2

V

L

R2 i (t)

Fig. 6.77 Solution At t = 0 - , the network is shown in Fig. 6.78. At t = 0-, the network has attained steady-state condition. Hence, the inductor acts as a short circuit.

R1

V

V i( 0 ) = R1 -

i( 0−)

Fig. 6.78

Since the inductor does not allow sudden change in current, i( 0 + ) =

V R1 R1

For t > 0, the network is shown in Fig. 6.79. 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. 6.79

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(0) =

 R +R  - 1 2  t  L 

V R1 V = k e0 = k R1 i( t ) =

L

R2

di =0 dt

 R1 + R2  t L 

V -  e R1

for t > 0

V R1

6.3 Resistor-Inductor Circuit 6.31

Example 6.24 In the network shown in Fig. 6.80, 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. 6.80 Solution At t = 0 - , the network is shown in Fig. 6.81. At t = 0-, the network has attained steady-state condition. Hence, the inductor acts as a short circuit. i( 0 - ) =

R2 R1 V

V R1 + R2

i( 0−)

Since the current through the inductor cannot change instantaneously, i( 0 + ) =

Fig. 6.81

V R1 + R2

For t > 0, the network is shown in Fig. 6.82. 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. 6.82

Comparing with the differential equation P=

di + Pi = Q, dt R1 V , Q= L L

The solution of this differential equation is given by, i(t ) = e - Pt ∫ Qe Pt dt + k e - Pt =e

-

R1 t L

R

R

=

R

- 1t V - L1 t L e dt + k e ∫L

- 1t V +ke L R1

V R1 + R2

6.32 Circuit Theory and Transmission Lines At t = 0, i(0) =

V R1 + R2 V V = +k R1 + R2 R1 k=-

VR2 R1 ( R1 + R2 ) R

i(t ) =

- 1t V VR2 e L R1 R1 ( R1 + R2 )

=

R - 1t  V  R2 e L  1 R1  R1 + R2 

for t > 0

Example 6.25 In the network of Fig. 6.83, a steady state is reached with the switch K open. At t = 0 , the switch K is closed. Find the current i(t ) for t > 0. 30 Ω

20 V

20 Ω

K 1 H 2

10 V

i(t)

Fig. 6.83 Solution At t = 0 - , the network is shown in Fig. 6.84. At t = 0 - , the network has attained steady-state condition. Hence, the inductor acts as a short circuit. i(0 - ) =

20 + 10 = 0.6 A 30 + 20

30 Ω

20 Ω

20 V i( 0−)

10 V

Since the current through the inductor cannot change instantaneously,

Fig. 6.84

i(0 + ) = 0.6 A For t > 0, the network is shown in Fig. 6.85. Writing the KVL equation for t > 0. 10 - 20i -

1 di =0 2 dt

di + 40i = 20 dt

20 Ω

1 H 2

10 V i(t)

Fig. 6.85

0.6 A

6.3 Resistor-Inductor Circuit 6.33

Comparing with the differential equation

di + Pi = Q, dt P = 40, Q = 20

The solution of this differential equation is given by, i(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -40 t ∫ 20 e 40 t dt + k e -40 t =

20 + k e -40 t 40

= 0..5 + k e -40 t

At t = 0, i(0) = 0.6 A

0.6 = 0.5 + k k = 0.1 i(t ) = 0.5 + 0.1 e -40 t

for t > 0

Example 6.26 The network of Fig. 6.86 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. 6.86 40 Ω -

Solution At t = 0 , the network is shown in Fig. 6.87. At t = 0-, the network has attained steady-state condition. Hence, the inductor acts as a short circuit.

50 V i( 0−)

50 i(0 ) = = 1.25 A 40 -

Fig. 6.87 Since current instantaneously,

through

the

inductor

cannot

change

i(0 + ) = 1.25 A 40 Ω

For t > 0, the network is shown in Fig. 6.88. 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. 6.88

1.25 A

6.34 Circuit Theory and Transmission Lines Comparing with the differential equation

di + Pi = Q, dt P = 2000, Q = 500

The solution of this differential equation is given by, i(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -2000 t ∫ 500 e 2000 t + k e -2000 t =

500 + k e -2000 t 2000

= 0.25 + k e -2000 t At t = 0, i(0) = 1.25 A 1.25 = 0.25 + k k =1 i(t ) = 0.25 + e -2000 t

Example 6.27

for t > 0

In the network of Fig. 6.89, the switch is moved from 1 to 2 at t = 0. Determine i(t). 1 5Ω

2 0.5 H

2Ω

20 V

i(t) 40 V

Fig. 6.89 Solution At t = 0 - , the network is shown in Fig. 6.90. At t = 0-, the network has attained steady-state condition. Hence, the inductor acts as a short circuit. i(0 - ) =

5Ω 20 V

20 =4A 5

i( 0−)

Fig. 6.90

Since the current through the inductor cannot change instantaneously, i(0 + ) = 4 A For t > 0, the network is shown in Fig. 6.91. 2Ω

Writing the KVL equation for t > 0, 40 - 2i - 0.5

0.5 H

di =0 dt

di + 4i = 80 dt

40 V

i(t)

Fig. 6.91

4A

6.3 Resistor-Inductor Circuit 6.35

Comparing with the differential equation

di + Pi = Q, dt P = 4, Q = 80

The solution of this differential equation is given by, i(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -4 t ∫ 80 e 4 t dt + k e -4 t 80 + k e -4 t 4 = 20 + k e -4 t =

At t = 0, i(0) = 4 A 4 = 20 + k k = -16 i(t ) = 20 - 16 e -4 t

for t > 0

Example 6.28 For the network shown in Fig. 6.92, steady state is reached with the switch closed. The switch is opened at t = 0. Obtain expressions for iL (t) and vL (t). iL(t)

100 Ω

15 V

3000 Ω

90 mH

vL(t)

Fig. 6.92 Solution At t = 0 - , the network is shown in Fig. 6.93. At t = 0-, the network has attained steady-state condition. Hence, the inductor acts as a short circuit.

100 Ω

15 V

15 iL (0 ) = = 0.15 A 100

i L ( 0−)

-

Since current through the inductor cannot change instantaneously,

Fig. 6.93

iL (0 ) = 0.15 A +

For t > 0, the network is shown in Fig. 6.94. Writing the KVL equation for t > 0, -3000iL - 90 × 10

-3

diL =0 dt

diL + 33.33 × 103 iL = 0 dt

3000 Ω

90 mH iL(t)

Fig. 6.94

0.15 A

6.36 Circuit Theory and Transmission Lines 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

3

t

At t = 0, iL (0) = 0.15 A 0.15 = k iL (t ) = 0.15 e -33.33×10 vL (t ) = L

Also,

3

t

for t > 0

diL 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 6.29

3

t

3

t

for t > 0

In the network of Fig. 6.95, 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. 6.95

10 Ω

10 Ω

-

Solution At t = 0 , the network is shown in Fig. 6.96. At t = 0-, the network has attained steady-state condition. Hence, the inductor acts as a short circuit. i(0 - ) =

50 V

i( 0−)

50 = 2.5 A 10 + 10

Fig. 6.96

Since current through the inductor cannot change instantaneously, i (0+) = 2.5 A 10 Ω

For t > 0, the network is shown in Fig. 6.97. 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 W

10 Ω

10 Ω i(t)

Fig. 6.97

0.1 H

2.5 A

6.3 Resistor-Inductor Circuit 6.37

For t > 0, Thevenin’s equivalent network is shown in Fig. 6.98. 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 P = 150,

Fig. 6.98 Q = 250

The solution of this differential equation is given by, i(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -150 t ∫ 250 e150 t dt + k e - Pt =

250 + k e -150 t 150

= 1.667 + k e -150 t At t = 0, i(0) = 2.5 A 2.5 = 1.667 + k k = 0.833 i(t ) = 1.667 + 0.833 e -150 t

Example 6.30

for t > 0

In Fig. 6.99, the switch is closed at t = 0. Find i (t) for t > 0. 2Ω

10 A

1Ω

2Ω i (t)

Fig. 6.99

Solution At t = 0–,

i(0 - ) = 0

Since current through inductor cannot change instantaneously, i(0 + ) = 0

1H

2.5 A

6.38 Circuit Theory and Transmission Lines For t > 0, simplifying the network by source-transformation technique as shown in Fig. 6.100. 2Ω

1Ω

10 A

2Ω

2Ω

1H

10 A

1H

0.67 Ω

i (t)

i (t)

(a)

(b)

Writing the KVL equation for t > 0, 6.67 - 2.67i - 1

2.67 Ω

di =0 dt

6.67 V

di + 2.67i = 6.67 dt Comparing with the differential equation

1H i (t) (c)

di + Pi = Q, dt

Fig. 6.100

P = 2.67, Q = 6.67 The solution of this differential equation is given by, i(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -2.67t ∫ 6.67 e 2.67t dt + k e -2.67t =

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(t ) = 2.5 - 2.5 e -2.67t = 2.5(1 - e -2.67t )

Example 6.31

for t > 0

Find the current i (t) for t > 0. 60 Ω

25 A

20 Ω

140 Ω

0.3 H i (t)

Fig. 6.101

6.3 Resistor-Inductor Circuit 6.39

Solution At t = 0–, the inductor acts as a short circuit. Simplifying the network as shown in Fig. 6.102. 60 Ω

25 A

140 Ω

25 A

20 Ω

140 Ω

i (0−)

60 Ω

i (0−)

(a)

(b)

Fig. 6.102 i(0 - ) = 25 ×

140 = 17.5 A 140 + 60

Since current through the inductor cannot change instantaneously, i(0 + ) = 17.5 A For t > 0, the network is shown in Fig. 6.103. 60 Ω

25 A

20 Ω

140 Ω

0.3 H

17.5 A

i (t)

Fig. 6.103 Simplifying the network by source transformation as shown in Fig. 6.104,

60 Ω

20 Ω

17.5 A

0.3 H

15 Ω

i (t)

0.3 H i (t)

(a)

(b)

Fig. 6.104 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

6.40 Circuit Theory and Transmission Lines 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 i(t ) = 17.5 e -50 t

for t > 0

Example 6.32 In the network of Fig. 6.105, 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. 6.105 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 i( 0 - ) = = 1 A 1 v2 ( 0 ) = 0 Since current through the inductor cannot change instantaneously, i( 0 + ) = 1 A v2 (0 + ) = -1 ×

1 = -0.5 V 2

For t > 0, the network is shown in Fig. 6.106. Writing the KCL equation for t > 0, t

+ 1H

t

1 1 v v2 dt + 1 + 2 + ∫ v2 dt = 0 1 2 1 ∫0 0 2 Differentiating Eq. (i), v2 + 2

…(i)

1 Ω 2

v2(t)

2H −

Fig. 6.106

dv2 1 + v2 = 0 dt 2 dv2 3 + v2 = 0 dt 4

Comparing with the differential equation

1A

dv + Pv = 0, dt P=

3 4

6.3 Resistor-Inductor Circuit 6.41

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 (t ) = -0.5 e

3 - t 4

for t > 0

Example 6.33 In the network shown in Fig. 6.107, a steady-state condition is achieved with switch open. At t = 0 switch is closed. Find va (t). 10 Ω + 3V

va(t)

5Ω 0.5 H

−

Fig. 6.107 Solution circuit.

At t = 0–, the network has attained steady-state condition. Hence, the inductor acts as a short iL (0 - ) = 0 va ( 0 - ) = 3 ×

5 =1V 10 + 5

Since current through inductor cannot change instantaneously, iL (0 + ) = 0 va ( 0 + ) = 1 V For t > 0, the network is shown in Fig. 6.108. 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), 2va + 0.2

0.5 H va(t)

dva dv + 0.1 a = 0 dt dt dva 20 + va = 0 dt 3

Fig. 6.108

6.42 Circuit Theory and Transmission Lines 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

= ke

-

20 t 3

At t = 0, va (0) = 1 V 1= k va ( t ) = e

Example 6.34

-

20 t 3

for t > 0

In the network of Fig. 6.109, determine currents i1 (t) and i2 (t) when the switch is

closed at t = 0.

10 Ω 5Ω 100 V

i2(t)

5Ω

0.01 H

i1(t)

Fig. 6.109 i1 (0 - ) = i2 (0 - ) = 0

Solution At t = 0–, At t = 0+,

i1 (0 + ) = 0 i2 (0 + ) =

100 = 6.67 A 15

For t > 0, the network is shown in Fig. 6.110. Writing the KVL equations for t > 0, di1 =0 dt 100 - 10(i1 + i2 ) - 5i2 = 0

100 - 10(i1 + i2 ) - 5i1 - 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), di1 + 833i1 = 3333 dt

Fig. 6.110

5Ω

6.3 Resistor-Inductor Circuit 6.43

Comparing with the differential equation

di + Pi = Q, dt P = 833, Q = 3333

The solution of this differential equation is given by, i1 (t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -833t ∫ 3333 e833t dt + k e -833t =

3333 + k e -833t 833

= 4 + k e -833t At t = 0, i1 (0) = 0 0 = 4+k k = -4 i1 (t ) = 4 - 4 e -833t = 4(1 - e -833t ) 100 - 10i1 i2 (t ) = 15 =

for t > 0

100 - 10( 4 - 4 e -833t ) 15

= 4 + 2.67 e -833t

for t > 0

Example 6.35 The switch in the network shown in Fig. 6.111 is closed at t = 0. Find v2 (t) for all t > 0. Assume zero initial current in the inductor. 30 Ω

10 V

+ −

10 Ω i1(t) i2(t)

Fig. 6.111 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

6.44 Circuit Theory and Transmission Lines For t > 0, the network is shown in Fig. 6.112. Writing the KVL equations for t > 0,

30 Ω

10 - 30(i1 + i2 ) - 10 i1 = 0

…(i)

di and 10 - 30(i1 + i2 ) - 0.2 2 = 0 dt From Eq. (i),

…(ii)

10 V

+ −

10 Ω i1(t)

i2(t)

0.2 H

Fig. 6.112

10 - 30 i2 = 0.25 - 0.75 i2 …(iii) 40 Substituting Eq. (iii) into Eq. (ii), di2 + 37.5 i2 = 2.5 dt di + Pi = Q, Comparing with the differential equation dt P = 37.5, Q = 2.5 i1 =

The solution of this differential equation is given by, i2 (t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -37.5t ∫ 2.5 e37.5t dt + k e -37.5t =

2.5 + k e -37.5t 37.5

= 0.067 + k e -37.5t

At t = 0, i2 (0) = 0

0 = 0.067 + k k = -0.067 i2 (t ) = 0.067 - 0.067 e -37.5t di2 dt d = 0.2 (0.067 - 0.067 e -37.5t ) dt

v2 (t ) = 0.2

= 0.5 e -37.5t

for t > 0

Example 6.36 For the network shown in Fig. 6.113, 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. 6.113

0.4 H

6.3 Resistor-Inductor Circuit 6.45

Solution At t = 0 - , the network is shown in Fig. 6.114. At t = 0 - , the network attains steady-state condition. Hence, the inductor acts as a short circuit.

40 Ω

500 V

500 =5A 40 + 60 the inductor cannot

i(0 - ) = Since current through instantaneously,

60 Ω

i( 0−)

change

Fig. 6.114

i(0 + ) = 5 A 60 Ω

For t > 0, the network is shown in Fig. 6.115. Writing the KVL equation for t > 0, 10i - 60i - 0.4

di =0 dt

10i

+ −

0.4 H

5A

i (t)

di + 125i = 0 dt di + Pi = 0, Comparing with the differential equation dt P = 125 The solution of this differential equation is given by,

Fig. 6.115

i(t ) = k e - Pt = k e -125t At t = 0, i(0) = 5 A

5=k i(t ) = 5 e -125t

for t > 0

Example 6.37 For the network shown in Fig. 6.116, find the current in the 20 W resistor when the switch is opened at t = 0. i 30 Ω 50 V

20 Ω

+ − 10i i2(t)

i1(t)

2H

Fig. 6.116 i (0 − )

-

Solution At t = 0 , the network is shown in Fig. 6.117. At t = 0 - , the network attains steady-state condition. Hence, the inductor acts as a short circuit.

30 Ω 50 V i1(0 − )

+ − − − 10i (0 ) i2(0 )

i(0-) = i2(0-) Fig. 6.117

20 Ω

6.46 Circuit Theory and Transmission Lines Writing the KVL equations at t = 0 - , 50 - 30(i1 - i2 ) - 10i2 = 0 10i2 - 30(i2 - i1 ) - 20i2 = 0 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. 6.118.

i(t ) = i2(t) 30 Ω

Writing the KVL equation for t > 0, 10i2 - 30i2 - 20i2 - 2

di2 =0 dt

20 Ω

+ − 10i2 i2(t)

di2 + 20i2 = 0 dt

2H

2.5 A

Fig. 6.118

di + Pi = 0, Comparing with the differential equation 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 2.5 = k i2 (t ) = 2.5 e -20 t

for t > 0

In the network of Fig. 6.119, an exponential voltage v(t) = 4 e -3t is applied at t = 0. Find the expression for current i(t). Assume zero current through inductor at t = 0.

Example 6.38

0.5 Ω

+ 4e − 3t −

0.25 H i(t)

Fig. 6.119 Solution

At t = 0 - ,

i(0 - ) = 0

Since current through the inductor cannot change instantaneously, i(0 + ) = 0

6.3 Resistor-Inductor Circuit 6.47

Writing the KVL equation for t > 0, 4 e -3t - 0.5i - 0.25

di =0 dt

di + 2i = 16 e -3t dt Comparing with the differential equation

di + Pi = Q, dt P = 2, Q = 16 e -3t

The solution of this differential equation is given by, i(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -2t ∫ 16 e -3t e 2t dt + + k e -2t = 16 e -2t ∫ e - t dt + k e -2t = -16 e -3t + k e -2t At t = 0, i(0) = 0 0 = -16 + k k = 16 i(t ) = -16 e -3t + 16 e -2t

for t > 0

Example 6.39 For the network shown in Fig. 6.120, 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. 6.120 Solution Writing the KVL equation for t > 0, 150 sin(500t + θ ) - 50i - 0.2

di =0 dt

di + 250i = 750 sin(500t + θ ) dt Comparing with the differential equation

di + Pi = Q, dt

P = 250, Q = 750 sin(500t + θ )

6.48 Circuit Theory and Transmission Lines The solution of this differential equation is given by, i(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -250 t ∫ 750 sin(500t + θ ) e 250 t + k e -250 t   e 250 t = 750 e -250 t  {250 sin(500t + θ ) - 500 cos(500t + θ )} + k e -250t 2 2  ( 250) + (500)  = 0.6 sin(500t + θ ) - 1.2 cos(500t + θ ) + k e -250 t Let

Acos φ = 0.6

and

A sin φ = 1.2 A2 cos 2 φ + A2 sin 2 φ = (0.6) 2 + (1.2) 2 = 1.8 A = 1.342  1.2  φ = tan -1   = 63.43°  0.6 

and

i(t ) = 1.342 cos(63.43°) sin(500t + θ ) - 1.342 sin(63.43°) cos(500t + θ ) + k e -250 t i(t ) = 1.342 sin(500t + θ - 63.43°) + k e -250 t At t = 0, θ = 0, i(0) = 0 0 = 1.342 sin( -63.43°) + k k = 1.2 i(t ) = 1.342 sin(500t + θ - 63.43°) + 1.2 e -250 t

for t > 0

Example 6.40 For the network shown in Fig. 6.121, 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 = 150 cos(200t + 30° ) V . 1 2

200 Ω

150 cos(200t + 30°) 0.5 H

Fig. 6.121 Solution At t = 0 - , the network is shown in Fig 6.122. At t = 0 - the network attains steady-state condition.

200 Ω 150 cos(200t + 30°) i(t)

V 150 ∠30° I= = = 0.67∠3.43° A Z 200 + j × 200 × 0.5 Fig. 6.122

0.5 H

6.4 Resistor–Capacitor Circuit 6.49

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. 6.123. 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. 6.123

Comparing with the differential equation

di + Pi = 0, dt P = 400

The solution of this differential equation is given by, i(t ) = k e - pt = k e -400 t

...(ii)

From Eqs (i) and (ii), 0.67 cos( 200t + 3.43°) = k e -400 t 0.67 cos(3.43°) = k k = 6.67

At t = 0,

i(t ) = 0.67 e -400 t

6.4

for t > 0

rEsIstor–capacItor cIrcuIt

Consider a series RC circuit as shown in Fig. 6.124. 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. 6.124

Series RC circuit

Differentiating the above equation, di i - =0 dt C di 1 + i=0 dt RC

0-R

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

6.50 Circuit Theory and Transmission Lines 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 - ln = t R RC ln i = -

ln

1 i =t RC V    R 1

t i = e RC V R 1

i=

V - RC t for t > 0 e R

When the switch is closed, the response decays with time as shown in Fig. 6.125(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

V - t vR = Ri = R e RC R = Ve

-

1 t RC

for t > 0

i(t) V R

t O

Fig. 6.125(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

t

O

Fig. 6.125(b) Voltage response of series RC circuit

6.4 Resistor–Capacitor Circuit 6.51

At t = 0, vC (0) = 0

k=V 1   t vC = V 1 - e RC   

Hence,

Example 6.41

for t > 0

The switch in the circuit of Fig. 6.126 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. 6.126 Solution At t = 0-, the network is shown in Fig. 6.127. 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 instantaneously,

across

the

capacitor cannot

change Fig. 6.127

vC (0+) = 100 V 5 kΩ

For t > 0, the network is shown in Fig. 6.128.

Writing the KCL equation for t > 0,

dvC vC + 50 1 × 10 + =0 dt 5000 dvC + 200vC = 10 4 dt dv + Pv = Q, Comparing with the differential equation dt -6

50 V

P = 200, Q = 10 4 Solution of this differential equation is given by, vC (t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -200 t ∫ 10 4 e 200 t dt + k e -200 t =

10 4 + k e -200 t 200

= -50 + k e -200 t

+ vC(t) −

Fig. 6.128

1 µF

6.52 Circuit Theory and Transmission Lines At t = 0, vC (0) = 100 V

100 = –50 + k k = 150 vC (t ) = -50 + 150 e -200 t

for t > 0

Example 6.42 In the network shown in Fig. 6.129, 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. 6.129 Solution

At t = 0-, the capacitor is uncharged. Hence, it acts as a short circuit. vC (0 ) = 0 iC (0 ) = 0

At t = 0+, the network is shown in Fig. 6.130.

iT (0 + ) 9 kΩ

Since voltage across the capacitor cannot change instantaneously, vC (0 + ) = 0 At t = 0+,

10 V

  10 10 iT (0 + ) =   = 9.8 k = 1.02 mA 9 + ( 4 1 ) k k  k   1 k iC (0 + ) = 1.02 m × = 0.204 mA 1k+4 k

4 kΩ iC (0 + )

v C ( 0+)

1 kΩ

Fig. 6.130

For t > 0, the network is shown in Fig. 6.131.

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 kW

+ vC(t) −

1 kΩ

3 µF

Veq = 10 ×

Fig. 6.131

For t > 0, Thevenin’s equivalent network is shown in Fig. 6.132. 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. 6.132

vC(t)

6.4 Resistor–Capacitor Circuit 6.53

Comparing with the differential equation

dv + Pv = Q, dt

P = 68.02, Q = 68.02 The solution of this differential equation is given by, vC (t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -68.02t ∫ 68.02 e68.02t dt + k e -68.02t = 1 + k e -68.02 t At t = 0, vC (0) = 0 0 = 1+ k k = -1 vC (t ) = 1 - e -68.02 t iC (t ) = C

for t > 0

dvC dt

= 3 × 10 -6

d (1 - e -68.02 t ) dt

= 3 × 10 -6 × 68.02e -68.02 t = 204.06 × 10 -6 e -68.02 t

foor t > 0

Example 6.43 For the network shown in Fig. 6.133, 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. 6.133

Solution At t = 0-, the network is shown in Fig. 6.134.

At t = 0–, the network has attained steady-state condition. Hence, the capacitor acts as an open circuit. vC (0-) = 1200 V

100 Ω

v C ( 0−)

1200 V

Since the voltage across the capacitor cannot change instantaneously, vC (0+) = 1200 V

Fig. 6.134

6.54 Circuit Theory and Transmission Lines

For t > 0, the network is shown in Fig. 6.135. Writing the KCL equation for t > 0, 50 × 10

-6

100 Ω +

dvC vC vC - 1200 + + =0 dt 300 100 dvC + 266.67 vC = 0.24 × 106 dt

vC (t)

300 Ω

1200 V

50 µF −

Fig. 6.135

dv + Pv = Q, Comparing with the differential equation dt P = 266.67, Q = 0.24 × 106 The solution of this differential equation is given by, vC (t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -266.67t ∫ 0.24 × 106 e 266.67t dt + 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

for t > 0

In Fig. 6.136, the switch is closed at t = 0 Find vC (t) for t > 0.

Example 6.44

100 Ω

5V

1F

2Ω

+ −

vC(t)

Fig. 6.136

Solution At t = 0-,

vC (0-) = 0 Since the voltage across the capacitor cannot change instantaneously, vC (0+) = 0

Since the resistor of 2 W is connected in parallel with the voltage source of 5 V, it becomes redundant. For t > 0, the network is as shown in Fig. 6.137. Writing KCL equation for t > 0, vC - 5 dvC +1 =0 100 dt dv 100 C + vC = 5 dt

100 Ω 1F

5V

Fig. 6.137

vC (t)

6.4 Resistor–Capacitor Circuit 6.55

dvC + 0.01vC = 0.05 dt dv + Pv = Q, dt P = 0.01, Q = 0.05 The solution of this differential equation is given by, Comparing with the differential equation

vC (t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -0.01 ∫ 0.05 e 0.01t dt + k e -0.01t =

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 )

for t > 0

In the network shown, the switch is shifted to position b at t = 0. Find v (t) for t > 0.

Example 6.45

1 F 4

a b

+ v (t) −

5V 2Ω

2Ω

Fig. 6.138 vC( 0−)

Solution At t = 0 , the network is shown in Fig. 6.139. At t = 0-, the network has attained steady-state condition. Hence, the capacitor acts as an open circuit. -

2Ω

+ v( 0−) −

2Ω

+ v( 0+) −

5V

vC (0-) = 5 V v (0-) = 0

Fig. 6.139

At t = 0+, the network is shown in Fig. 6.140.

5V

At t = 0+, the capacitor acts as a voltage source of 5 V. i( 0 + ) = -

5 = -1.25 A 4

v(0 + ) = -1.25 × 2 = -2.5 V

2Ω i( 0+)

Fig. 6.140

6.56 Circuit Theory and Transmission Lines

For t > 0, the network is shown in Fig. 6.141.

1F 4

5V

Writing the KVL equation for t > 0,

t

-2i - 5 -

1 i dt - 2i = 0 1∫ 0 4

…(i) 2Ω

2Ω

i(t)

Differentiating Eq. (i), -4

di - 4i = 0 dt di +i = 0 dt

Fig. 6.141

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 for t > 0 v(t ) = 2i(t ) = -2.5e - t

Example 6.46

for t > 0

In the network of Fig. 6.142, 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. 6.142 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 0.25 dt 2

v2(t ) −

For t > 0, the network is shown in Fig. 6.143. Writing KCL equation for t > 0,

1 Ω 2

Fig 6.143

6.4 Resistor–Capacitor Circuit 6.57

dv2 + 20 v2 = 80 dt Comparing with the differential equation

dv + Pv = Q, dt P = 20, Q = 80

The solution of this differential equation is given by, v(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -20 t ∫ 80 e 20 t dt + k e -20 t =

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 = 4(1 - e -20 t )

for t > 0.

Example 6.47 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

v2 (t)

i(t) −

Fig. 6.144 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 + ) = 0 i( 0 + ) =

V0 R1

6.58 Circuit Theory and Transmission Lines For t > 0, the network is shown in Fig. 6.145. Writing the KVL equation for t > 0, t

V0 -

R1

+

t

1 1 i dt - R1i i dt = 0 ∫ C1 0 C2 ∫0

C1

...(i)

C2

V0

i(t )

Differentiating Eq. (i),

−

-

i di i - R1 =0 C1 dt C2

Fig. 6.145

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=

i( t ) =

V0 R1 1  C1 + C2  t C1 C2 

V0 - R1  e R1 1

V - t = 0 e R1C R1

where, C =

C1C2 C1 + C2

t

v2 ( t ) =

1 i dt C2 ∫0 t

t

=

1 V0 - R1C e dt C2 ∫0 R1

=

1   t V0 R1C 1 - e R1C    R1C2

=

V0 C1 C1 + C2

t   t 1 - e R1C   

for t > 0

6.4 Resistor–Capacitor Circuit 6.59

Example 6.48 t > 0.

For the network shown in Fig.6.146, the switch is opened at t = 0. Find v (t) for + K 10 A

1Ω 1 Ω 2

1 F 2

v (t)

−

Fig. 6.146 Solution At t = 0 - , the network is shown in Fig. 6.147. At t = 0 - , the network attains steady-state condition. Hence, the capacitor acts as an open circuit. vC(0-) = 0

+ 10 A

1 Ω 2

v C (0 − ) v(0 − )

1Ω

−

-

Writing the KCL equation at t = 0 , -

Fig. 6.147 -

v(0 ) v(0 ) + = 10 1 1 2 3v(0 - ) = 10 v(0 - ) = 3.33 V Since the voltage across the capacitor cannot change instantaneously, vC (0 + ) = v(0 + ) = 3.33 V For t > 0, the network is shown in Fig. 6.148. Writing the KCL equation for t > 0, 1 dv v + = 10 2 dt 1

+ 10 A

dv + 2v = 20 dt dv + Pv = Q, Comparing with the differential equation dt P = 2, Q = 20 The solution of this differential equation is given by, v(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -2t ∫ 20 e -2t dt + k e -2t 20 + k e -2t 2 = 10 + k e -2t =

1Ω

+ −

1 F 2

v (t)

−

Fig. 6.148

6.60 Circuit Theory and Transmission Lines At t = 0, v(0) = 3.33 V 3.33 = 10 + k k = 6.67 v(t ) = 10 + 6.67 e -2t

Example 6.49

For the network shown in Fig. 6.149, find the current i(t) when the switch is opened

at t = 0. 10 Ω i(t) + −

100 V

5i 10 Ω 4 µF

Fig. 6.149 10 Ω

Solution At t = 0 - , the network is shown in Fig. 6.150. At t = 0 - , the network attains steady-state condition. Hence, the capacitor acts as open circuit. i( 0 - ) =

i( 0−) + −

100 V

100 =5A 10 + 10

Fig. 6.150

At t = 0 + , the network is shown in Fig. 6.151.

v c ( 0+) = 2 5 V

25 + 5i(0 - ) - 10i(0 + ) = 0

i( 0+)

+

i( 0 ) = 5 A

+ −

For t > 0, the network is shown in Fig. 6.152. 1 4 × 10 -6

10 Ω v C ( 0−)

vC (0 - ) = 100 - 10i(0 - ) - 5i(0 - ) = 100 - 10(5) - 5(5) = 25 V

25 -

5i( 0−)

5i( 0+) 10 Ω 25 V

t

∫ i dt + 5i - 10i = 0

…(i)

0

Fig. 6.151

Differentiating Eq. (i), 0 - 0.25 × 106 i + 5

di di - 10 = 0 dt dt

di + 50000 i = 0 dt

i(t) + −

5i

4 µF 25 V

di + Pi = 0, Comparing with the differential equation dt P = 50000

Fig. 6.152

10 Ω

6.4 Resistor–Capacitor Circuit 6.61

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 e -50000 t

Example 6.50

for t > 0

For the network shown in Fig. 6.153, find the current i(t) when the switch is opened

at t = 0. 10 Ω i(t ) 20 Ω

10 Ω

20i

2 µF

100 V + −

Fig. 6.153 10 Ω

Solution At t = 0 - , the network is shown in Fig. 6.154. At t = 0 - , the network attains steady-state condition. Hence, the capacitor acts as an open circuit. Writing the KVL equation at t = 0 - ,

i( 0 −) 20 Ω + −

100 - 10 i(0 - ) - 20 i(0 - ) - 20 i(0 - ) = 0

20i (0−)

i( 0 - ) = 2 A Also,

20 i(0 - ) + 20 i(0 - ) - 0 - vC (0 - ) = 0 -

10 Ω

100 V + −

vc (0 −)

Fig. 6.154

-

vC (0 ) = 40 i(0 ) = 40( 2) = 80 V At t = 0 + , the network is shown in Fig. 6.155.

i(0 +)

From Fig. 6.155, i(0 + ) = -i2 (0 + )

10 Ω

20 Ω

20 i(0 + ) - 20 i2 (0 + ) - 10 i2 (0 + ) - 80 = 0

+ 20i(0 +) i 2 (0 +) −

20 i(0 + ) + 20 i(0 + ) + 10 i(0 + ) - 80 = 0 i(0 + ) = 1.6 A

80 V

Fig. 6.155

+

vC (0 ) = 80 V i(t)

For t > 0, the network is shown in Fig. 6.156.

10 Ω 20 Ω

From Fig. 6.156, i(t ) = -i2 (t ) Writing the KVL equation for t > 0, 20i - 20i2 - 10i2 -

2 µF

1 2 × 10 -6

t

∫ i2 dt - 80 = 0 0

+ 20i −

i 2 (t)

Fig. 6.156

80 V

6.62 Circuit Theory and Transmission Lines 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 i = 0 dt di + 1 × 10 4 i = 0 dt

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

4

t

At t = 0, i(0) = 1.6 A 1.6 = k i(t ) = 1.6 e -1×10

4

t

for t > 0

In the network of Fig. 6.157, an exponential voltage 4 e -5t is applied at time t = 0. Find the expression for current i(t). Assume zero voltage across the capacitor at t = 0.

Example 6.51

0.2 Ω

4e −5t

+ −

1F i(t)

Fig. 6.157 vC (0 - ) = 0

Solution At t = 0 - ,

i( 0 - ) = 0 At t = 0 + , the network is shown in Fig. 6.158. Since voltage across the capacitor cannot change instantaneously, vC (0 + ) = 0 4 i( 0 + ) = = 20 A 0.2 Writing the KVL equation for t > 0, 4 e -5t - 0.2i -

0.2 Ω

4

+ −

i(0 + )

v c (0 + )

Fig. 6.158

t

1 i dt = 0 1 ∫0

…(i)

6.4 Resistor–Capacitor Circuit 6.63

Differentiating Eq. (i), di -i = 0 dt di + 5i = -100 e -5t dt

-20 e -5t - 0.2

Comparing with the differential equation

di + Pi = Q, dt P = 5, Q = -100 e -5t

The solution of this differential equation is given by, i(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -5t ∫ -100 e -5t e5t dt + k e -5t = -100t e -5t + k e -5t At t = 0, i(0) = 20 A 20 = k i(t ) = -100 t e -5t + 20 e -5t

for t > 0

In the network shown in Fig. 6.159, the switch is closed at t = 0 connecting a to the network. At t = 0, vC (0) = 0.5 V. Determine v(t).

Example 6.52 source e

-t

1Ω + e −t

1 Ω 0.5 V+ 2 −

+ −

1F

v (t ) −

Fig. 6.159 v(0) = vC (0 - ) = 0.5 V

Solution At t = 0 - ,

Since voltage across the capacitor cannot change instantaneously, v(0 + ) = vC (0 + ) = 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 = 3,

Q = e-t

6.64 Circuit Theory and Transmission Lines The solution of this differential equation is given by, v(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -3t ∫ e - t e3t dt + k e -3t = e -3t ∫ e 2t dt + k e -3t 1 -t e + k e -3t 2

= At t = 0,

v(0) = 0.5 V 1 +k 2 k=0

0.5 =

v(t ) = 0.5 e - t

Example 6.53 In the network shown in Fig. 6.160, 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. 6.160 Solution Writing the KVL equation for t > 0, 100 sin(500t + 45°) - 15i -

t

1 100 × 10 -6

∫ i dt = 0 0

Differentiating Eq. (i), (100)(500) cos(500t + 45°) - 15

di - 10 4 i = 0 dt

di + 666.67i = 3333.33 cos(500t + 45°) dt Comparing with the differential equation

di + Pi = Q, dt P = 666.67, Q = 3333.33 cos(500t + 45°)

The solution of this differential equation is given by, i(t ) = e - Pt ∫ Q e Pt dt + k e - Pt = e -666.67t ∫ 3333.33 cos(500t + 45°) e666.67t + k e -666.67t

…(i)

6.4 Resistor–Capacitor Circuit 6.65

  e666.67t = 3333.33 e -666.67t  {666.67 cos(500t + 45°) + 500 sin(500t + 45°)} + k e -666.67t 2 2  (666.67) + (500)  = 3.2 cos(500t + 45°) + 2.4 sin(500t + 45°) + k e -666.67t Asin φ = 3.2

Let

A cos φ = 2.4

and

A sin φ + A cos 2 φ = (3.2) 2 + ( 2.4) 2 = 16 2

2

2

A=4  3.2  φ = tan -1  = 53.13°  2.4 

and

i(t ) = 4 sin(53.13°) cos(500t + 45°) + 4 cos(53.13°) sin(500t + 45°) + k e -666.67t = 4 sin(500t + 98.13°) + k e -666.67t Putting t = 0, in Eq. (i), 100 sin( 45°) - 15 i(0) - 0 = 0 i(0) = 4.71 4.71 = 4 sin(98.13°) + k k = 0.75 i(t ) = 4 sin(500t + 98.13°) + 0.75 e -666.67t

for t > 0

In the network of Fig. 6.161, 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.

Example 6.54

1 2 100 sin (wt + 30° )

50 Ω 50 Ω i(t)

20 µF

Fig. 6.161 Solution At t = 0 - , the network is shown in Fig. 6.162. 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 = 1.41sin(1000t + 75°)

50 Ω 100 sin (wt + 30° ) i(∞)

20 µF

Fig. 6.162 …(i)

6.66 Circuit Theory and Transmission Lines For t > 0, the network is as shown in Fig. 6.163. Writing the KVL equation for t > 0, -50i - 50i -

50 Ω 50 Ω

t

1 20 × 10

-6

∫ i dt - vC (0) = 0

…(ii)

20 µF

vC (0 −) −

0

Differentiating Eq. (ii), -50

+

i(t)

Fig. 6.163 di di 1 i=0 - 50 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(1000t + 75°) = k e -500 t At t = 0,

1.41sin(75°) = k k = 1.36 i(t ) = 1.36 e -500 t

6.5

for t > 0

rEsIstor–Inductor–capacItor cIrcuIt

Consider a series RLC circuit as shown in Fig. 6.164. The switch is closed at time t = 0. The capacitor and inductor are initially uncharged. Applying KVL to the circuit for t > 0,

L V i(t)

t

V - Ri - L

R

di 1 i dt = 0 dt C ∫0

Fig. 6.164

C

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  =0 s2 +   s +   L  LC 

6.5 Resistor–Inductor–Capacitor Circuit 6.67

Let s1 and s2 be the roots of the equation. 2

s1 = -

1 R  R +   = -α + α 2 - ω 02 = -α + β   2L 2L LC

s2 = -

1 R  R -   = -α - α 2 - ω 02 = -α - β  2L  2L LC

2

α=

where

ω0 =

R 2L 1 LC

β = α 2 - ω 02

and

The solution of the above second-order differential equation will be given by, i(t ) = k1e s1t + 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) i = k1 e s1t + k2 e s2t

or

for t > 0

Fig. 6.165

The current curve for an overdamped case is shown in Fig. 6.165. Case II

i(t )

When a = w0 i.e.,

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,

t

Fig. 6.166

i = e -α t ( k1 + k2 t ) for t > 0 The current curve for critically damped case is shown in Fig. 6.166. Case III When a < w0 i.e.,

Overdamped response

R < 2L

1 LC

The roots are complex conjugate and it gives an underdamped response.

Critically damped response

6.68 Circuit Theory and Transmission Lines In this case, the solution is given by, i(t ) = k1e s1t + k2 e s2t s1, 2 = -α ± α 2 + ω 02

where Let

α 2 - ω 02 = -1 ω 02 - α 2 = jω d j = -1

where and

ω d = ω 02 - α 2

Hence,

i(t ) = e -α t ( k1e jω d t + k2 e - jω d t )   e jω d t + e - j ω d t   e jω d t - e - jω d t   = e -α t ( k1 + k2 )    + j ( k1 - k2 )  2 j2       = e -α t [( k1 + k2 ) cos ω d t + j ( k1 - k2 ) sin ω d t ] for t > 0

The current curve for an underdamped case is shown in Fig. 6.167. i(t )

t

Fig. 6.167 Underdamped response

Example 6.55 current i(t ) for t > 0.

In the network of Fig. 6.168, the switch is closed at t = 0. Obtain the expression for 9Ω

1H

0.05 F

20 V i(t)

Fig. 6.168 Solution

At t = 0 - , the switch is open. i( 0 - ) = 0 vC (0 - ) = 0

Since current through the inductor and voltage across the capacitor cannot change instantaneously, i( 0 + ) = 0 vC (0 + ) = 0

6.5 Resistor–Inductor–Capacitor Circuit 6.69

For t > 0, the network is shown in Fig. 6.169.

9Ω

1H

Writing the KVL equation for t > 0, t

di 1 20 - 9i - 1 i dt = 0 dt 0.05 ∫0

…(i)

20 V

0.05 F i(t )

Fig. 6.169

Differentiating Eq. (i), 0-9

di d 2 i - 20i = 0 dt dt 2

d 2i dt

2

+9

di + 20i = 0 dt

( D 2 + 9 D + 20)i = 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 -5t dt

…(iii)

At t = 0, i(0) = 0

Putting t = 0 in Eq. (i), 20 - 9i(0 + ) -

0 = k1 + k2

…(iv)

di (0) = -4 k1 - 5k2 dt

…(v)

di + (0 ) - 0 = 0 dt di + (0 ) = 20 - 9i(0 + ) = 20 A / s dt

From Eq. (v), 20 = -4 k1 - 5k2

…(vi)

Solving Eqs (iv) and (vi), k1 = 20 k2 = 20 i(t ) = 20 e -4 t - 20 e -5t

for t > 0

Example 6.56 In the network shown in Fig. 6.170, the switch is moved from the position 1 to 2 at t = 0. The switch is in the position 1 for a long time. Determine the expression for the current i(t ).

6.70�Circuit Theory and Transmission Lines 10 �

2H

1 2 20 V

3F 50 V

i(t)

���������� Solution At t � 0 � , the network is shown in Fig. 6.171. 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.

10 �

20 V

vC (0�)

i(0 �)

vC (0 � ) � 20 V i( 0 � ) � 0

����������

Since the current through the inductor and the voltage across the capacitor cannot change instaneously, vC (0 � ) � 20 V i( 0 � ) � 0 For t � 0, the network is shown in Fig. 6.172. Writing the KVL equation for t � 0,

10 �

2H

t

50 � 10i � 2

di 1 � i dt � 20 � 0 dt 3 �0

…(i)

3F 50 V

20 V

i(t)

Differentiating Eq. (i), 0 � 10

di d 2i 1 �2 2 � i�0 � 0 dt 3 dt d 2i

����������

di 1 � i�0 dt 6 dt 1� � 2 �� D � 5 D � �� i � 0 6 2

�5

D1 � �0.03, D2 � �4.97 The solution of this differential equation is given by, i(t ) � k1 e �0.03t � k2 e �4.97t

…(ii)

Differentiating Eq. (ii), di � �0.03k1 e �0.03t � 4.97k2 e �4.97t dt At t � 0, i(0) � 0

0 � k1 � k2 di (0) � �0.03 k1 � 4.97 k2 dt

Chapter 06.indd 70

…(iii) …(iv) …(v)

5/30/2016 5:52:35 PM

6.5 Resistor–Inductor–Capacitor Circuit 6.71

Putting t = 0 in Eq. (i), 20 - 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 k1 - 4.97 k2

…(vi)

Solving Eqs (iv) and (vi), k1 = 2.02 k2 = 2.02 i(t ) = 2.02 e -0.03t - 2.02 e -4.97t

for t > 0

Example 6.57 In the network of Fig. 6.173, the switch is closed and a steady state is reached in the network. At t = 0 , the switch is opened. Find the expression for the current i2 (t) in the inductor. 10 Ω i2(t) 100 V

1H

10 µF

Fig. 6.173 Solution At t = 0 - , the network is shown in Fig. 6.174. 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 - ) =

10 Ω i2 (0 −) vC (0 −)

100 V

100 = 10 A 10

Fig. 6.174

vC (0 - ) = 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. 6.175. Writing the KVL equation for t > 0,

i2(t) 1H 10 A

t

-1

di2 1 i dt = 0 dt 10 × 10 -6 ∫0

…(i) Fig. 6.175

10 µF

6.72 Circuit Theory and Transmission Lines Differentiating Eq. (i), d 2i2

-

dt 2 d 2i2 dt

2

- 105 i = 0 + 105 i = 0

( D 2 + 105 )i = 0 D1 = j 316, D2 = - j 316 The solution of this differential equation is given by, i2 (t ) = k1 cos 316t + k2 sin 316t Differentiating Eq. (ii), di2 = -316 k1 sin 316t + 316 k2 cos 316t dt At t = 0, i2 (0) = 10 A 10 = k1 di2 (0) = 316 k2 dt Putting t = 0 in Eq. (i), -

…(ii) …(iii) …(iv) …(v)

di + (0 ) - 0 = 0 dt di + (0 ) = 0 dt

From Eq. (v), 0 = 316 k2 k2 = 0 i2 (t ) = 10 cos 316t

for t > 0

In the network of Fig. 6.176, 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.

Example 6.58

v (t)

+ 10 V

−

1Ω 4

1F

Fig. 6.176 Solution

At t = 0 - ,

iL (0 - ) = 0 v(0 - ) = vC (0 - ) = 10 V

1 H 2

6.5 Resistor–Inductor–Capacitor Circuit 6.73

Since current through the inductor and voltage across capacitor cannot change instantaneously, iL (0 + ) = 0 v(0 + ) = vC (0 + ) = 10 V For t > 0, the network is shown in Fig. 6.177. 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

−

1Ω 4

1F

Differentiating Eq. (i),

1 H 2

Fig. 6.177 2

d v dt 2

+4

dv + 2v = 0 dt

( D 2 + 4 D + 2)v = 0 D1 = -1, D2 = -3 The solution of this differential equation is given by, v(t ) = k1 e - t + k2 e -3t

…(ii)

Differentiating Eq. (iii), dv = - k1 e - t - 3 k2 e -3t dt

…(iii)

10 = k1 + k2

…(iv)

dv (0) = - k1 - 3 k2 dt

…(v)

At t = 0, v(0) = 10 V

Putting t = 0 in Eq. (i), dv + ( 0 ) + 4 v ( 0) + 0 = 0 dt dv + (0 ) = -40 V / s dt 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

for t > 0

Example 6.59 In the network of Fig. 6.178, the switch is opened at t = 0 obtain the expression for v(t ). Assume zero initial conditions.

6.74 Circuit Theory and Transmission Lines v (t )

0.5 H

0.5 Ω

2A

1F

Fig. 6.178 Solution

At t = 0 - ,

iL (0 - ) = 0 v(0 - ) = vC (0 - ) = 0

Since current through the inductor and voltage across the capacitor can not change instantaneously, iL (0 + ) = 0 v(0 + ) = vC (0 + ) = 0 For t > 0, the network is shown in Fig. 6.179. Writing the KCL equation for t > 0,

v (t)

t

v 1 dv + vdt + 1 = 2 0.5 0.5 ∫0 dt

…(i)

2A

0.5 Ω

0.5 H

1F

Differentiating Eq. (i), 2

dv d 2v + 2v + 2 = 0 dt dt

d 2v dt

2

+2

Fig. 6.179

dv + 2v = 0 dt

( D 2 + 2 D + 2) v = 0 D1 = -1 + j1, D2 = -1 - j1 The solution of this differential equation is given by, v(t ) = e - t ( k1 cos t + k2 sin t )

… (ii)

Differentiating Eq. (ii), dv = -e - t ( k1 cos t + k2 sin t ) + e - t ( - k1 sin t + k2 cos t ) dt = e - t [ - k1 (cos t + sin t ) + k2 (cos t - sin t )]

…(iii)

At t = 0, v(0) = 0 0 = k1

…(iv)

dv (0) = - k1 + k2 dt

…(v)

Putting t = 0 in Eq. (i), 2v ( 0) + 0 +

dv ( 0) = 2 dt

6.5 Resistor–Inductor–Capacitor Circuit 6.75

dv ( 0 ) = 2 V /s dt From Eq. (v), 2 = - k1 + k2

…(vi)

Solving Eq. (iv) and (vi), k1 = 0 k2 = 2 v(t ) = 2 e - t sin t

for t > 0

Example 6.60 The network shown in Fig. 6.180, 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. 6.180 Solution Writing the KVL equation for t > 0, t

di 1 i dt = 0 dt 200 × 10 -6 ∫0

150 sin( 200t + 30°) - 10i - 0.5

…(i)

Differentiating Eq. (i), 30000 cos( 200t + 30°) - 10

di d 2i - 0.5 2 - 5000i = 0 dt dt d 2i

dt 2

+ 20

di + 10000i = 60000 cos( 200t + 30°) dt

( D 2 + 20 D + 10000)i = 60000 cos( 200t + 30°)

…(ii)

The roots of the characteristic equation are D1 = -10 + j 99.5, D2 = -10 - j 99.5 The complimentary function is iC = e -10 t ( K1 cos 99.5t + K 2 sin 99.5t ) Let the particular function be iP = A cos( 200t + 30°) + B sin( 200t + 30°) iP′ = -200 A sin( 200t + 30°) + 200 B cos( 200t + 30°) iP″ = -40000 A cos( 200t + 30°) - 40000 B sin( 200t + 30°)

6.76 Circuit Theory and Transmission Lines Substituting these values in Eq. (ii), -40000 A cos( 200t + 30°) - 40000 B sin( 200t + 30°) + 20[ -200 A sin( 200t + 30°) + 200 B cos( 200t + 30°)] +10000[ A cos( 200t + 30°) + B sin( 200t + 30°)] = 60000 cos( 200t + 30°) ( -40000 B - 4000 A + 10000 B) sin(200t + 30°) + ( -40000 A + 4000 B + 10000 A) cos( 200t + 30°) = 60000 cos( 200t + 30°) Equating the coefficients, -40000 B - 4000 A + 10000 B = 0 -40000 A + 4000 B + 10000 A = 60000 Solving these equations, A = -1.97 B = 0.26 iP = -1.97 cos( 200t + 30°) + 0.26 sin( 200t + 30°) Let

Asin φ = -1.97

and

A cos φ = 0.26 A2 sin 2 φ + A2 cos 2 φ = ( -1.97) 2 + (0.26) 2 = 3.95 A = 1.987

and

 -1.97  φ = tan -1  = -82.48°  0.26  iP = 1.987 sin( -82.48°) cos( 200t + 30°) + 1.987 cos( -82.48°) sin( 200t + 30°) = 1.987 sin( 200t + 30° - 82.48°) = 1.987 sin( 200t - 52.48°)

The solution of the differential equation is given by, i(t ) = e -10 t ( k1 cos 99.5t + k2 sin 99.5t ) + 1.987 sin( 200t - 52.48°)

…(iii)

Differentiating Eq. (iii), di = e -10 t ( -99.5 k1 sin 99.5t + 99.5 k2 cos 99.5t ) dt - 10 e -10 t ( k1 cos 99.5t + k2 sin 99.5t ) + (1.987)( 200) cos( 200t - 52.48°) At t = 0, i(0) = 0 0 = k1 + 1.987 sin( -52.48°) k1 = 1.58

…(iv)

di (0) = 99.5 k2 - 10 k1 + (1.987)( 200) cos( -52.48°) dt = 99.5 k2 - 10(1.58) + 242.03 = 99.5 k2 + 226.23

…(v)

6.5 Resistor–Inductor–Capacitor Circuit 6.77

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(t ) = e -10 t (1.58 cos 99.5t - 0.77 sin 99.5t ) + 1.987 sin( 200t - 52.48°) for t > 0

Example 6.61 (a) L =

The switch in the network of Fig. 6.181 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 L

2Ω

(c) L = 5 H and C = 1 F

i(t) +

4V

C vC (t)

2Ω

−

Fig. 6.181 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 (0 - ) = 4 ×

2 =2V 2+2

i( 0 - ) = 0 Since current through the inductor and voltage across the capacitor cannot change instantaneously, vC (0 + ) = 2 V i( 0 + ) = 0 Case I  When R = 2 W, L =

1 H, C = 1 F 2

α=

ω0 =

R = 2L 1 LC

α > ω0

2 2× =

1 2

=2 1

1 ×1 2

=

1 0.5

= 1.414

6.78 Circuit Theory and Transmission Lines This indicates an overdamped case. i(t ) = A1 e s1t - A2 e s2t where,

s1 = -α - α 2 - ω 02 = -2 - 4 - 2 = -2 - 2 = -3.414

and

s2 = -α + α 2 - ω 02 = -2 + 2 = -0.586 i(t ) = k1 e -3.414 t + k2 e -0.586 t

At t = 0, i (0) = 0 k1 + k2 = 0

…(i)

Also vL(0+) + vC(0+) + vR(0+) = 0 vL (0 + ) = - vR (0 + ) - vC (0 + ) = -2i(0 + ) - vC (0 + ) = -2 V vL (0 + ) = L

…(ii)

di + (0 ) dt

di + v (0 + ) 2 (0 ) = L == -4 A / s dt L 0.5 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 and

k2 = - 1.414

i(t ) = 1.414(e -3.414 t - e -0.586 t ) Case II  When R = 2 W, L = 1 H, C =1 F

α=

R 2 2 = = =1 2L 2 × 1 2

ω0 =

1 LC

=

1 1

=1

α = ω0 This indicates a critically damped case. At t = 0, i(0) = 0

i(t) = e–a t (k1 + k2 t) = e–t (k1 + k2 t) k1 = 0

Also,

vL (0 + ) = L

di + (0 ) dt

di + v (0 + ) 2 (0 ) = L = - = -2 A / s dt L 1

for t > 0

Exercises 6.79

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

for t > 0

Case III  When R = 2 W, L = 5 H, C = 1 F R 2 = = 0.2 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)

ω d = ω o2 - α 2 = (0.447) 2 - (0.2) 2 = 0.4

where,

s1, 2 = -α ± jω d = -0.2 ± j 0.4 i(t ) = e -0.2t ( B1 cos 0.4t + B2 sin 0.4t ) Applying the initial condition, i(0+) = 0 di + v (0 + ) 2 (0 ) = - L =dt L 5 B1 = i(0) = 0

and

B2 = -1 i(t ) = -e -0.2t sin 0.4t

for t > 0

Exercises 6.1

The switch in Fig. 6.182 is moved from the position a to b at t = 0, the network having been in steady state in the position a. Determine di di i1 (0 + ), i2 (0 + ), i3 (0 + ), 2 (0 + ) and 3 (0 + ). dt dt

The switch K is closed at t = 0 in the network shown in Fig. 6.183. Determine di d 2i + i(0 + ), (0 + ) and (0 ). dt dt 2 1Ω

1H

1Ω

1 Ω i1(t )

a

i2(t) 10 V

6.2

b

1Ω

i3(t) 2Ω

V0

2H

1F

1Ω i1(t)

i(t)

1F

Fig. 6.183 Fig. 6.182 [1.66 A, 5 A, -3.33 A, -3.33 A/s, 2.22 A/s]

 1  0, 2 V0 , - V0   

6.80 Circuit Theory and Transmission Lines 6.3

In the network of Fig. 6.184, 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 dt dt

6.6

The network shown in Fig. 6.187 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. 6.187

V2

[i(t ) = 2.857 e -2 ×10 t ] 6

R

6.7

Fig. 6.184  1 v(0 + )  , 0, 0, 0, 0   0,  C R1 

The switch in Fig. 6.188 is open for a long time and closes at t = 0. Determine i (t) for t > 0. 6Ω

3Ω i(t)

6.4

In the network at Fig. 6.185, 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

240 V

30 Ω

Fig. 6.188

K 2 MΩ

10 µF

[i(t) = 25(1 - e-4t)] 6.8

+ 1000 V −

10 µF

1 MΩ i1

i2

In the network shown in Fig. 6.189, 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. 6.185  17 2  400000 A / s   

5 kΩ

2 kΩ

vC (t) 1 kΩ

6.5

2H

In the network shown in Fig. 6.186, switch is closed at t = 0. Obtain the current i2 (t).

+ −

5 µF

6V

Fig. 6.189

10 Ω

[vC(t) = 5e-20t] 50 V

10 Ω i1(t )

2 µF

i2(t )

Fig. 6.186 [i2(t) = 5 e–100000t]

6.9

The circuit shown in Fig. 6.190 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 6.81

+

6.13 In Fig. 6.194, 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)

vC (t )

2 µF

5V

5 kΩ

12 Ω

8Ω

+

−

Fig. 6.190

2F

6F

5V

v ( t)

-100 t

[(i) 5 V, 5 V, 1 mA, 0, (ii) 0.01 s, (iii) e mA] 6.10 In the network shown in Fig. 6.191, 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.

Fig. 6.194 ( t -100 )    v(t ) = 5 - 8e 160   

10 Ω

1 2

20 V

−

1H

20 Ω i (t)

6.14 A series RL circuit shown in Fig. 6.195 has a constant voltage V applied at t = 0. At what time does vR = vL.

Fig. 6.191 [i (t) = 2 e-30t]

vR

10 Ω

vL

1H

10 V

6.11 In the network shown in Fig. 6.192, the switch is closed at t = 0. Find v (t) for t > 0. +

Fig. 6.195 1 Ω 2

3A

1 H 3

1Ω

[0.0693 s]

v (t )

−

Fig. 6.192 [v(t) = e-t] 6.12 In the network shown in Fig. 6.193, 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

6.15 In the circuit shown in Fig. 6.196, 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. 6.196 −

Fig. 6.193 [v (t) = -0.5

-3 t e 4]

[0] 6.16 Determine whether RLC series circuit shown in Fig 6.197 is underdamped, overdamped or

6.82 Circuit Theory and Transmission Lines or

critically damped. Also find 2 di d v vL (0 + ), (0 + ), 2 (0 + ) if v(t ) = u(t ). dt dt

di critically damped. Also, find vL (0 + ), (0 + ) dt and i (∞). 200 Ω

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. 6.197

1 F 2

Fig. 6.198

[critically damped, 200 V, 2000 A/s, 0] 6.17 Determine whether RLC circuit of Fig 6.198 is underdamped, overdamped

[underdamped 1 V, 1 A/s, 2 V/s2]

Objective-Type Questions The voltages vC1 , vC2 and vC3 across the capacitors in the circuit in Fig. 6.199 under steady state are respectively

6.1

1H

10 kΩ

+ 100 V −

2F 2H + −

vC1 −

1F

6.3

40 kΩ

+ vC 3 −

(b)

eat + ebt

(d)

aeat + bebt

The differential equation for the current i(t) in the circuit of Fig. 6.201 is 2Ω

i(t)

vC 2

+

2 kΩ

(a) eat - ebt (c) aedt - bebt

2H

3F

1F

sin t

Fig. 6.199 Fig. 6.201

(a) 80 V, 32 V, 48 V (b) 80 V, 48 V, 32 V (c) 20 V, 8 V, 12 V (d)

(a)

20 V, 12 V, 8 V

In the circuit of Fig. 6.200, the voltage v(t) is

6.2

+ v ( t) −

(c) 1H

Fig. 6.200

ebt

(d)

d 2i dt 2

d 2i dt

1Ω

1Ω

eat

(b)

2

2

2

+2

d 2i dt

d 2i dt

2

+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 6.83

6.4

At t = 0+, the current i1 in Fig. 6.202 is 1

(b)

C

1 0.63

2

V

R L

R i1(t)

t 0.5

C

i2(t)

(c)

Fig. 6.202 (a) (c) 6.5

i(t )

V 2R V 4R -

(b)

i(t) 0.5 0.31

-

V R

t 0.5

(d) zero

(d)

For the circuit shown in Fig. 6.203, 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. 6.205 vi (t )

C

vo (t)

6.7

The condition on R, L and C such that the step response v(t) in Fig. 6.206 has no oscillations is

Fig. 6.203 (a) sin (103 - 45°) (c) sin (103 t - 53) 6.6

L

(b) sin (103 t + 45°) (d) sin (103 t + 53°)

+ v(t) −

For the RL circuit shown in Fig. 6.204, the input voltage vi(t) = u(t). The current i(t) is

C

2Ω

Fig. 6.204 6.8 i(t ) 0.5 0.31

t 2

1 L 2 C

(b)

R≥

L C

(d)

R=

(a)

R≥

(c)

R≥2

i(t)

(a)

v(t )

Fig. 6.206

1H

vi (t )

R

L C 1 LC

The switch S in Fig. 6.207 closed at t = 0. If v2 (0) = 10 V and vg(0) = 0 respectively, the voltages across capacitors in steady state will be

6.84 Circuit Theory and Transmission Lines 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

B −

Fig. 6.207 (a) v2 (∞) = v1 (∞) = 0

(a)

(b) v2(∞) = 2 V, v1 (∞) = 8 V

i(t )

(c) v2 (∞) = v1 (∞) = 8 V (d) v2(∞) = 8 V, v1 (∞) = 2 V 6.9

A +

4 mA

The time constant of the network shown in Fig. 6.208 is

8V

(0, 0)

R

v(t)

(b)

Fig. 6.210 2R

10 V

C

Fig. 6.208 (a)

2 RC

(b)

3 RC

1 2 RC RC (d) 2 3 6.10 In the series RC circuit shown in Fig. 6.209, 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 ms (b) 12 s (c) 32 s (d)

unknown, unless actual network is specified 6.12 In the network shown in Fig. 6.211, 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. 6.211 Fig. 6.209 (a) zero

(b)

(c) RC

(d)

infinity 1 RC

6.11 The v – i characteristic as seen from the terminal pair (A – B) of the network of Fig. 6.210(a) is shown in Fig. 6.210(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

6.13 A step function voltage is applied to an RLC series circuit having R = 2 W, 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 6.85

Answers to Objective-Type Questions 6.1 (b)

6.2 (d)

6.3 (c)

6.4 (d)

6.5 (a)

6.6 (b)

6.8 (d)

6.9 (d)

6.10 (d)

6.11 (a)

6.12 (d)

6.13 (b)

6.7 (c)

7 7.1

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.

7.2

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. s = σ + jω The function f (t) must satisfy the following condition to possess a Laplace transform, ∞

∫ | f (t ) | e 0

−σ t

dt < ∞

7.2 Circuit Theory and Transmission where s is real and positive. The inverse Laplace transform L–1 {F ( s )} is σ + j∞

f (t ) =

7.3

1 F ( s) e st ds 2π j σ −∫j∞

laplace transforms of some Important functIons

  1.   Constant Function k The Laplace transform of a constant function is ∞

∞  e − st  k L{k} = ∫ ke − st dt = k   = − s s  0 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 n +1  x L{t n } = ∫   e − x = s n +1 ∫ x n e − x dx = n +1 , s > 0, n + 1 > 0  s s s 0 0 n + 1 = n! n!

If n is a positive integer, n

L{t } =

s n +1

  3.   Unit-Step Function The unit-step function (Fig 7.1) is defined by the equation, u(t) = 1 t>0 =0 ta t0 =0 ta =0 t 0, the transformed network is shown in Fig. 7.11. Applying KVL to the mesh, V − RI ( s) − Ls I ( s) = 0 s V L I ( s) = R  ss +   L By partial-fraction expansion, A B I ( s) = + R s s+ L V L A = sI ( s) |s = 0 = s × R  ss +   L

R

V s

Ls I (s)

Fig. 7.11 Transformed network

= s=0

V R

7.14 Circuit Theory and Transmission V R R   L B =  s +  I ( s) = s+  ×    R R L L  s= − ss +  L  L

=− s= −

V R

R L

V −V    R I ( s) = R + R s s+ L Taking the inverse Laplace transform, R

V V − Lt − e R R R − t V = 1 − e L  R  

i( t ) =

for t > 0

example 7.11 In the network of Fig. 7.12, the switch is moved from the position 1 to 2 at t = 0, steady-state condition having been established in the position 1. Determine i (t) for t > 0. 1Ω

1 2

10 V

1Ω

1H i (t)

Fig. 7.12 At t = 0−, the network is shown in Fig 7.13. At t = 0−, the network has attained steady-state condition. Hence, the inductor acts as a short circuit. 1Ω

Solution 

10 = 10 A 1 Since the current through the inductor cannot change instantaneously, i(0 − ) =

10 V i (0−)

i (0 ) = 10 A +

Fig. 7.13

For t > 0, the transformed network is shown in Fig. 7.14. Applying KVL to the mesh for t > 0, − I ( s) − I ( s) − sI ( s) + 10 = 0 I ( s) ( s + 2) = 10 10 I ( s) = s+2 Taking inverse Laplace transform, i (t) = 10e−2t for t > 0

1

s

1 I (s)

10

Fig. 7.14

7.7 Resistor–Inductor Circuit 7.15

example 7.12 The network of Fig. 7.15 was initially in the steady state with the switch in the position a. At t = 0, the switch goes from a to b. Find an expression for voltage v (t) for t > 0. 2Ω

a

b + 1Ω

2V

1H

2H

v (t)

−

Fig. 7.15

2Ω

At t = 0 , the network is shown in Fig 7.16. At t = 0−, the network has attained steady-state condition. Hence, the 2V inductor of 2H acts as a short circuit. 2 i(0 − ) = = 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. 7.17. 2s Applying KCL at the node for t > 0,

Solution 

−

i (0−)

Fig. 7.16 + s

1

V ( s) + 2 V ( s) V ( s) + + =0 2s 1 s 3 1  V ( s)  1 +  = −  2s  s

V (s)

2 −

Fig. 7.17

1 s =− 2 =− 1 V ( s) = 2s + 3 2s + 3 s + 1.5 2s Taking the inverse Laplace transform, v (t) = − e−1.5 t −

example 7.13

for t > 0

In the network of Fig. 7.18, the switch is opened at t = 0. Find i(t). 10 Ω 3Ω 36 V

6Ω 0.1 H

i (t)

Fig. 7.18

iT (0−) 10 Ω

Solution  At t = 0−, the network is shown in Fig. 7.19. At t = 0–, the switch is closed and steady-state condition is reached. Hence, the inductor acts as a short circuit. 36 36 = =3A 10 + (3  6) 10 + 2 6 iL (0 − ) = 3 × =2A 6+3

iT (0 − ) =

3Ω 36 V iL

(0−)

Fig. 7.19

6Ω

7.16�Circuit Theory and Transmission Since current through the inductor cannot change instantaneously, iL(0+) = 2 A For t > 0, the transformed network is shown in Fig. 7.20 Applying KVL to the mesh for t > 0,

3

I(s)

0.1sI � s � � 9I ( s) � � 0.2 I ( s) �

6

0.1s

�0.2 � 0.1s I � s � � 3I � s � � 6I � s � � 0

0.2

�2 � 0.2 � 0.1s � 9 s � 90

Taking inverse Laplace transform, i(t) = –2e−90 t

���������

for t > 0

Example 7.14 The network shown in Fig. 7.21 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 �

4�

36 V

4�

2H i (t)

���������

Solution At t = 0−, the network is shown in Fig 7.22. At t = 0−, the switch is closed and the network has acquired steady-state. Hence, the inductor acts as a 4� 10 � short circuit. � iT (0 ) 36 36 iT (0 � ) � � �3A 10 � ( 4 � 4) 10 � 2 4� 36 V 4 i( 0 � ) � 3 � � 1.5 A i (0� ) 4�4 Since current through the inductor cannot change ��������� instantaneously, i(0+) = 1.5 A 4 For t > 0, the transformed network is shown in Fig. 7.23. Applying KVL to the mesh for t > 0, �4I � s � � 4I �s � � 2sI � s � � 3 � 0

2s

8I � s � � 2sI � s � � 3 I ( s) � Taking the inverse Laplace transform, i(t) = 1. 5e−4 t

4

3 1.5 � 2s � 8 s � 4

I (s)

3

��������� for t > 0

�Example 7.15� In the network shown in Fig. 7.24, the switch is closed at t = 0, the steady-state being reached before t = 0. Determine current through inductor of 3 H.

Chapter 07.indd 16

5/30/2016 5:57:55 PM

7.7 Resistor–Inductor Circuit 7.17 2H

2Ω

2Ω

1V i1 (t)

3H i2 (t)

Fig. 7.24

Solution At t = 0 , the network is shown in Fig. 7.25. 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. 17.26. Applying KVL to Mesh 1, 1 1 s − 2 s I1 ( s) + 1 − 2[ I1 ( s) − I 2 ( s)] = 0 s 1 ( 2 + 2s) I1 ( s) − 2 I 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. 7.25

1

2s

2

2 I1 (s)

3s I2 (s)

Fig. 7.26

1 s

2 1 ( s + 1) ( s + 1) −2 0 + 1 s s + 1 s 3 = = = I 2 ( s) = = −2 2 + 2s 1 1 ( 2 + 2 s)( 4 + 3s) − 4 s(3s 2 + 7 s + 2)   3s  s +  ( s + 2) s( s + 2)  s +  −2 4 + 3s   3 3 By partial-fraction expansion, A B C + + 1 s s+2 s+ 3 1 ( s + 1) 3 A = s I 2 ( s) s = 0 = 1  ( s + 2)  s +   3

I 2 ( s) =

B = ( s + 2) I 2 ( s) S = −2

1 ( s + 1) = 3 1  ss +   3

=

1 2

s=0

=− s = −2

1 10

7.18 Circuit Theory and Transmission 1 ( s + 1) 1   C =  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 7.16 In the network of Fig. 7.27, the switch is closed at t = 0 with the network previously unenergised. Determine currents i1(t). 10 Ω

1H

1H 100 V

10 Ω i1 (t)

i2 (t)

10 Ω

Fig. 7.27 Solution For t > 0, the transformed network is shown in Fig. 7.28. 10

s

s 100 s

10 I1 (s)

I2 (s)

10

Fig. 7.28 Applying KVL to Mesh 1, 100 − 10 I1 ( s) − sI1 ( s) − 10 [ I1 ( s) − I 2 ( s) ] = 0 s 100 ( s + 20) I1 ( s) − 10 I 2 ( s) = s Applying KVL to Mesh 2, −10 [ I 2 ( s) − I1 ( s) ] − s I 2 ( s) − 10 I 2 ( s) = 0 −10 I1 ( s) + ( s + 20) I 2 ( s) = 0 By Cramer’s rule,

100 −10 100 s ( s + 20) 0 s + 20 100( s + 20) 100( s + 20) = s = I1 ( s) = = s + 20 −10 ( s + 20) 2 − 100 s( s 2 + 40 s + 300) s( s + 10)( s + 30) −10 s + 20

7.8 Resistor–Capacitor Circuit 7.19

By partial-fraction expansion, A B C + + s s + 10 s + 30 100( s + 20) 20 = A = s I1 ( s) |s= 0 = ( s + 10)( s + 30) s= 0 3

I1 ( s) =

B = ( s + 10) I1 ( s) |s=−10 =

100( s + 20) = −5 s( s + 30) s=−10

C = ( s + 30) I1 ( s) |s=−30 =

100( s + 20) 5 =− s( s + 10) s=−30 3

I1 ( s) =

20 1 5 5 1 − − 3 s s + 10 3 s + 30

Taking inverse Laplace transform, i1 (t ) = Similarly,

20 5 − 5e −10 t − e −30 t 3 3

100 1000 s 0 −10 1000 1000 s I 2 ( s) = = = = 2 2 s + 20 −10 ( s + 20) − 100 s( s + 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 1000 10 A = sI 2 ( s) |s = 0 = = ( s + 10)( s + 30) s = 0 3 s + 20

B = ( s + 10) I 2 ( s) |s = −10 =

1000 = −5 s( s + 30) s = −10

C = ( s + 30) I 2 ( s) |s = −30 =

1000 5 = s( s + 10) s = −30 3

I 2 ( s) =

10 1 5 5 1 − + 3 s s + 10 3 s + 30

Taking inverse Laplace transform, i2 (t ) =

7.8

10 5 − 5e −10 t + e −30 t 3 3

for t > 0 R

resIstor–capacItor cIrcuIt

Consider a series RC circuit as shown in Fig. 7.29. The switch is closed at time t = 0. For t > 0, the transformed network is shown in Fig. 7.30. Applying KVL to the mesh,

V

C i (t)

Fig. 7.29

RC circuit

7.20 Circuit Theory and Transmission V 1 − RI ( s) − I ( s) = 0 s Cs 1 V   R +  I ( s) = Cs s

R

V V s R = = I ( s) = 1 RCs + 1 1 s+ R+ Cs RC Cs Taking the inverse Laplace transform, V s

V s

example 7.17

Transformed network

Fig. 7.30

1

i( t ) =

I (s)

V − RC t e R

1 Cs

for t > 0

In the network of Fig. 7.31, the switch is moved from a to b at t = 0. Determine

i (t) and vc (t).

1Ω

a

1Ω

b

+

10 V

6F

3F

vc (t)

i (t) −

Fig. 7.31 −

Solution  At t = 0 , the network is shown in Fig. 7.32. At t = 0–, the network has attained steady-state condition. Hence, the capacitor of 6 F acts as an open circuit. v6 F (0−) = 10 V i (0−) = 0 v3 F (0−) = 0 Since voltage across the capacitor cannot change instantaneously, v6 F (0+) = 10 V v3 F (0+) = 0 For t > 0, the transformed network is shown in 7.33. Applying KVL to the mesh for t > 0, 10 1 1 − I ( s) − I ( s) − I ( s) = 0 s 6s 3s 1 1 10 I ( s) + I ( s) + I ( s) = 6s 3s s 10 60 10 I ( s) = = = 1 1  6 s + 3 s + 0.5  s 1 + +   6 s 3s  Taking the inverse Laplace transform, i(t) = 10e

−0.5t

1Ω

v6F (0 − )

10 V i (0 − )

Fig. 7.32 1 1 6s 10 s

1 V (s) c 3s

I (s)

Fig. 7.33 for t > 0

7.8 Resistor–Capacitor Circuit 7.21

Voltage across the 3 F capacitor is given by Vc ( s) =

1 10 I ( s) = 3s 3s( s + 0.5)

Vc ( s) =

A B + s s + 0.5

By partial-fraction expansion,

A = sVc ( s) s = 0 =

10 20 = 3( s + 0.5) s = 0 3

B = ( s + 0.5)Vc ( s) s = −0.5 = Vc ( s) =

10 20 =− 3s s = −0.5 3

20 1 20 1 − 3 s 3 s + 0.5

Taking the inverse Laplace transform, 20 20 −0.5t − e 3 3 20 = (1 − e −0.5t ) 3

vc (t ) =

example 7.18

for t > 0

The switch in the network shown in Fig. 7.34 is closed at t = 0. Determine the voltage

cross the capacitor. 10 Ω

10 Ω 2 F

10 V

vc (t)

Fig. 7.34

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. 7.35. Applying KCL at the node for t > 0, 10 Vc ( s) − s + Vc ( s) + Vc ( s) = 0 1 10 10 2s 2 sVc ( s) + 0.2Vc ( s) = Vc ( s) =

1 s 1 0.5 = s( 2 s + 0.2) s( s + 0.1)

10 s

10

10

Fig. 7.35

1 2s

Vc (s)

7.22 Circuit Theory and Transmission By partial-fraction expansion, Vc ( s) =

A B + s s + 0.1

A = sVc ( s) s = 0 =

0.5 0.5 = =5 s + 0.1 s = 0 0.1

B = ( s + 0.1)Vc ( s) s = −0.1 = Vc ( s) =

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 7.19 In the network of Fig. 7.36, 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. 7.36 At t = 0−, the network is shown in Fig. 7.37. 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. 7.37 Since voltage across the capacitor cannot change instantaneously, vc (0+) = 0 For t > 0, the transformed network is shown in Fig. 7.38. 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

V (s) I1(s)

2 s

2 s

I2(s) 1

1

Fig. 7.38

7.8 Resistor–Capacitor Circuit 7.23

2 2 I1 ( s) + I1 ( s) = − I1 ( s) s s 2 2 I1 ( s) + 2 I1 ( s) = s s 2 1 I1 ( s) = s = 2 s +1 +2 s Taking the inverse Laplace transform, i1 ( t ) = e − t

example 7.20

for t > 0

In the network of Fig. 7.39, the switch is moved from a to b, at t = 0. Find v(t). 4Ω

a b

+

6V

v (t) −

1F

2Ω

2Ω

Fig. 7.39

Solution At t = 0−, the network is shown in Fig 7.40. At t = 0−, steady-state condition is reached. Hence, the capacitor acts as an open circuit.

4Ω

2 v(0 ) = 6 × =2V 4+2 −

Since voltage instantaneously,

across

the

capacitor

cannot

change

+ v (0−) −

6V

v (0+) = 2 V For t > 0, the transformed network is shown in Fig. 7.41. Applying KCL at the node for t > 0, V ( s) + 6

Fig. 7.40 4

V (s) 1 s

2 s + V ( s) = 0 1 2 s 2  V ( s )  + s = 2 3 

V ( s) −

V ( s) =

2

2 s

Fig. 7.41 2

s+

2 3

Taking the inverse Laplace transform, v ( t ) = 2e

2 − t 3

2Ω

for t > 0

2

7.24 Circuit Theory and Transmission

example 7.21 The network shown in Fig. 7.42 has acquired steady-state at t < 0 with the switch open. The switch is closed at t = 0. Determine v (t). 2Ω +

2Ω

4V

1F

1F

v (t)

−

Fig. 7.42

Solution  At t = 0 , the network is shown in Fig 7.43. At t = 0–, steady-state condition is reached. Hence, the capacitor of 1 F acts as an open circuit. −

v(0 − ) = 4 × Since voltage across instantaneously,

the

2Ω

2 =2V 2+2

capacitor

2Ω

4V

cannot

change Fig. 7.43

v(0+) = 2 V For t > 0, the transformed network is shown in Fig. 7.44. 2

4 s

1 s 2

1 v (s) s

2 s

Fig. 7.44 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

V ( s) −

2 +2 2s + 2 2 2 2 1 V ( s) = s = = − = − 2 s + 1 s( 2 s + 1) s 2 s + 1 s s + 0.5 Taking the inverse Laplace transform, v ( t ) = 2 − e −0.5t

for t > 0

v (0−)

7.9 Resistor–Inductor–Capacitor Circuit 7.25

7.9

resIstor–Inductor–capacItor cIrcuIt

Consider a series RLC circuit shown in Fig. 7.45. The switch is closed at time t = 0. R

L V i (t)

C

Fig. 7.45 RLC circuit For t > 0, the transformed network is shown in Fig. 7.46. 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 2 + RCs + 1 V   I ( s) = s Cs  

Fig. 7.46

V s I ( s) = = LCs 2 + RCs + 1 s 2 + Cs

Transformed network

V V L L = R 1 ( s − s1 )( s − s2 ) s+ L LC

 R  1  = 0. where s1 and s2 are the roots of the equation s 2 +   s +   L  LC  2

s1 = −

1 R  R +   − = −α + α 2 − ω 02 = −α + β   2L 2L LC

s2 = −

1 R  R −   − = −α − α 2 − ω 02 = −α − β  2L  2L LC

2

where

α=

R 2L

ω0 =

1 LC

and β = α 2 − ω 02 By partial-fraction expansion, of I(s),

1 Cs

7.26 Circuit Theory and Transmission I ( s) =

A B + s − s1 s − s2

A = ( s − s1 ) I ( s) s = s

V = L s1 − s2

B = ( s − s2 ) I ( s) s = s

V V L = =− L s1 − s2 s2 − s1

1

2

I ( s) =

1  V  1 −  L( s1 − s2 )  s − s1 s − s2 

Taking the inverse Laplace transform, i( t ) =

V e s1t − e s2t  = k1 e s1t + k2 e s2t  L( s1 − s2 ) 

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 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 α > ω0 In this case, the solution is given by

(

i ( t ) = e −α t k1e βt + k2e − βt or

)

i(t ) = k1 e s1t + k2 e s2t

for t > 0

When the roots are real and equal, it gives a critically damped response. R 1 = 2L LC α = ω0 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 α < ω0 In this case, the solution is given by i(t ) = k1 e s1t + k2 e s2t Case II

where Let where and

s1,2 = −α ± α 2 − ω 02

α 2 − ω 02 = −1 ω 02 − α 2 = jω d j = −1

ω d = ω 02 − α 2

7.9 Resistor–Inductor–Capacitor Circuit 7.27

Hence

(

i(t ) = e −α t k1 e jω d t + k2 e − jω d t

)

  e jω d t + e − jω d t   e jω d t − e − jω d t   = e −α t ( k1 + k2 )   + j ( k1 − k2 )   2 2j       = e −α t [( k1 + k2 ) cos ω d t + j ( k1 − k2 )sin ω d t ]

example 7.22

for t > 0

The switch in Fig. 7.47 is opened at time t = 0. Determine the voltage v(t) for t > 0. +

2A

0.5 Ω

0.5 H

0.5 F

v (t)

−

Fig. 7.47 At t = 0−, the network is shown in Fig. 7.48. 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. 7.48 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. 7.49. 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) + 0.5sV ( s) = s 2 s V ( s) = = 2 + 0.5s + 2 s

+

2 s

2 s

0.5

0.5s

2 s

V(s)

−

4 2

1 0.5s

s + 4s + 4

=

Fig. 7.49

4 ( s + 2)

2

7.28 Circuit Theory and Transmission Taking inverse Laplace transform, v ( t ) = 4t e −2t

for t > 0

example 7.23 In the network of Fig. 7.50, 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

0.5 H

200 µF

Fig. 7.50 At t = 0−, the network is shown in Fig. 7.51. 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.

Solution 

iL (0 − ) =

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

5V

vc (0−) iL (0−)

Fig. 7.51

Since voltage across the capacitor and current through the inductor cannot change instantaneously, iL (0+) = 2 A vc (0+) = 0

0.5s

For t > 0, the transformed network is shown in Fig. 7.52. Applying KVL to the mesh for t > 0, −

1 200 × 10 −6 s

1 200 × 10−6s I (s)

I ( s) − 0.5s I ( s) + 1 = 0 Fig. 7.52

I ( s) 0.5s I ( s) − 1 + 5000 =0 s I ( s) =

1 5000 0.5 5s + s

=

2s s 2 + 10000

Taking inverse Laplace transform, i ( t ) = 2 cos 100t

for t > 0

1

7.9 �Resistor–Inductor–Capacitor Circuit�7.29

Example 7.24 In the network shown in Fig. 7.53, the switch is opened at t � 0. Steady-state condition is achieved before t = 0. Find i(t). 0.5 H

1V

1�

1F i (t)

���������

Solution At t = 0−, the network is shown in Fig 7.54. 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. 1V vc (0−) = 1 V − i (0 ) = 1 A Since current through the inductor and voltage across the capacitor cannot change instantaneously, vc(0+) = 1 V i(0+) = 1 A For t > 0, the transformed network is shown in Fig. 7.55. Applying KVL to the mesh for t > 0, 1 1 � I ( s) � 0.5s I ( s) � 0.5 � I ( s) � 0 s s 1 1 0.5 � � I ( s) � 0.5 s I ( s) � I ( s) s s

1� i (0�)

��������� 0.5s

0.5

1 s 1 1 s

I(s)

��������� 1 � 1 � I ( s) �1 � � 0.5s � � 0.5 � s � s � s �1 1 s�2 ( s � 1) � 1 � I ( s) � 2 � � 2 2 s � 2 s � 2 ( s � 1) � 1 ( s � 1) � 1 ( s � 1) 2 � 1 Taking the inverse Laplace transform, i � t � � e � t cos t � e � t sin t

for t � 0

�������������� In the network shown in Fig. 7.56, 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. 1� 1�

10 V i1 (t)

1H

1� i2 (t)

� 4V �

1F

���������

Chapter 07.indd 29

6/1/2016 3:49:43 PM

7.30 Circuit Theory and Transmission Solution For t > 0, the transformed network is shown in Fig. 7.57. 1 10 s

1

I1 (s)

s

1 1 s 4 s

I2 (s)

Fig. 7.57 Applying KVL to Mesh 1, 10 − I1 ( s) − (1 + s) [ I1 ( s) − I 2 ( s) ] = 0 s 10 ( s + 2) I1 ( s) − ( s + 1) I 2 ( s) = s Applying KVL to Mesh 2, 1 4 −( s + 1) [ I 2 ( s) − I1 ( s) ] − I 2 ( s) − I 2 ( s) − = 0 s s 1 4  −( s + 1) I1 ( s) +  s + 2 +  I 2 ( s) = −  s s By Cramer’s rule, 10 −( s + 1) s 2  10   s + 2 s + 1  4 10 4 4 1    (ss + 1) 2 − ( s + 1) −  − ( s + 1)  s  s+ 2+ 2 s s   s s s s = = I1 ( s) = 2 2 −( s + 1) s+2  s + 2 s + 1 ( s + 1) 2 ( s + 2) − ( s + 1) 2 ( s + 2)   − ( s + 1) 1 s s   −( s + 1) s + 2 + s 10 4 ( s + 1) − 2 3s + 5 s s = = ( s + 1) s( s + 1) ( s + 2) − ( s + 1) 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 = ( s + 1) I1 ( s) |s = −1 = I1 ( s) = Taking inverse Laplace transform,

3s + 5 = −2 s s = −1

5 2 − s s +1

i1 ( t ) = 5 − 2e − t

for t > 0

7.10 Response of RL Circuit to Various Functions 7.31

Similarly, 10 s 4 −( s + 1) − 3 2 3s + 1 3s + 3 − 2 3( s + 1) − 2 s = = − I 2 ( s) = = = 2 2 2 s+2 −( s + 1) + 1 s ( s + 1) ( s + 1) ( s + 1) 2 ( s + 1) 1 −( s + 1) s + 2 + s Taking inverse Laplace transform, i2 (t ) = 3e − t − 2te − t for t > 0 s+2

7.10

response of rl cIrcuIt to varIous functIons

Consider a series RL circuit shown in Fig. 7.58. When the switch is closed at t = 0, i(0 − ) = i(0 + ) = 0. R

v (t)

L

+ −

i(t)

Fig. 7.58

RL circuit R

For t > 0, the transformed network is shown in Fig. 7.59. Applying KVL to the mesh,

(a)

V ( s) − R I ( s) − Ls I ( s) = 0 V ( s) 1 V ( s) I ( s) = = R R + Ls L s+ L When the unit step signal is applied, v ( t ) = u( t ) Taking Laplace transform, V ( s) =

1 s

1 I ( s) = L =

By partial-fraction expansion,

1 s s+

R L 1

1 R L  ss +   L

V (s) + −

Ls I (s)

Fig. 7.59 Transformed network

7.32 Circuit Theory and Transmission   1A B  I ( s) =  + R L s s+    L A = s I ( s)

s=0

=

1

=

R s+ L

L R

s=0

1 L R  =− B =  s +  I ( s) s = − R = R  s R L L s= − L

  1L1 L 1  I ( s) =  − R L Rs R s+    L   1  1 1 =  − R R s s+    L Taking inverse Laplace transform,  R

i( t ) = (b)

−  t 1 [1 − e  L  ] R

When unit ramp signal is applied, v(t ) = r (t ) = t Taking Laplace transform, V ( s) =

for 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

7.10 Response of RL Circuit to Various Functions 7.33

A+C = 0 A = −C = −

Taking inverse Laplace transform,

L

R2 L 1 1 1 L 1 I ( s) = − 2 + + 2 2 s R R s R s+ R L  R

1 L −  t i( t ) = − 2 + t + 2 e  L  R R R L

 R

−  t 1 L t − 2 [1 − e  L  ] R R

= (c)

for t > 0

When unit impulse signal is applied, v(t ) = δ (t ) Taking Laplace transform,

V ( s) = 1 I ( s) =

1 L

1 s+

R L

Taking inverse Laplace transform,  R

1 −  t i( t ) = e  L  L

for t > 0

example 7.26 At t = 0, unit pulse voltage of unit width is applied to a series RL circuit as shown in Fig. 7.60. Obtain an expression for i(t). v (t)

1Ω

v (t)

1

+ −

1H i (t)

0

t

1

Fig. 7.60 Solution

1

v(t ) = u ( t ) − u ( t − 1)

1 e−s 1 − e−s − = s s s For t > 0, the transformed network is shown in Fig. 7.61. Applying KVL to the mesh, V ( s) =

V (s)

+ −

s I (s)

Fig. 7.61

7.34 Circuit Theory and Transmission V ( s) − I ( s) − sI ( s) = 0 V ( s) I ( s) = s +1 1 − e−s = s( s + 1) =

1 e−s − s( s + 1) s( s + 1))

=

1 1 e−s e−s − − + s s +1 s s +1

Taking inverse Laplace transform, i(t ) = u(t ) − e − t u(t ) − u(t − 1) + e − ( t −1) u(t − 1) = (1 − e − t )u(t ) − [1 − e − ( t −1) ]u(t − 1)

for t > 0

example 7.27 For the network shown in Fig. 7.62, 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)

Fig. 7.62 Solution For t > 0, the transformed network is shown in Fig. 7.63. Applying KVL to the mesh for t > 0, e −3s s2

− 5 I ( s) − 2 s I ( s) = 0 5 I ( s) + 2 s I ( s) = I ( s) =

e

5

e−3s s2

−3 s

0.5 s ( s + 2.5)

=

2s I (s)

s2 e −3s 2

s ( 2 s + 5)

=

0.5 e −3s 2

s ( s + 2.5)

By partial-fraction expansion, 2

+ −

A B C + 2+ s s s + 2.5

0.5 = As( s + 2.5) + B( s + 2.5) + Cs 2 = As 2 + 2.5 As + Bs + 2.5 B + Cs 2 = ( A + C ) s 2 + ( 2.5 A + B ) s + 2.5 B Comparing coefficients of s2, s and s0, A+C = 0 2.5 A + B = 0 2.5 B = 0.5

Fig. 7.63

7.10 Response of RL Circuit to Various Functions 7.35

Solving these equations,

A = −0.08 B = 0.2 C = 0.08  0.08 0.2 0.08  I ( s) = e −3s  − + 2 +   s s + 2.5  s = −0.08

e −3s e −3s e −3s + 0.2 2 + 0.08 s s + 2.5 s

Taking inverse Laplace transform, i(t ) = −0.08u(t − 3) + 0.2r (t − 3) + 0.08e −2.5( t − 3) u(t − 3)

example 7.28 Determine the expression for vL (t) in the network shown in Fig. 7.64. Find vL(t) when (i) vs(t) = d (t), and (ii) vs(t) = e−t u(t). 5Ω + 1 v (t ) 2H L

+ −

Vs (t )

−

Fig. 7.64

Solution For t > 0, the transformed network is shown in Fig. 7.65. By voltage-division rule, VL ( s) = Vs ( s) ×

(a)

s 2 s +5 2

5

=

s Vs ( s) s + 10

+ Vs (s)

s V (s ) 2 L

+ −

For impulse input,

−

Vs ( s) = 1 VL ( s) =

s s + 10 − 10 10 = = 1− s + 10 s + 10 s + 10

Taking inverse Laplace transform, VL (t ) = δ (t ) − 10e −10 t u(t ) (b)

for t > 0

For vs (t ) = e − t u(t ), Vs ( s) =

1 s +1

VL ( s) =

s ( s + 10)( s + 1)

Fig. 7.65

7.36 Circuit Theory and Transmission By partial-fraction expansion, VL ( s) =

A B + s + 10 s + 1 10 s = s + 1 s = −10 9 1 s = =− 9 s + 10 s = −1

A = ( s + 10)VL ( s) s = −10 = B = ( s + 1)VL ( s) s = −1 VL ( s) =

10 1 1 1 − 9 s + 10 9 s + 1

Taking inverse Laplace transform, 10 −10 t 1 e u( t ) − e − t u( t ) 9 9  10 −10 t 1 − t  = e − e  u( t )  9 9 

vL (t ) =

for t > 0

example 7.29 For the network shown in Fig. 7.66, 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. 7.66

Solution  For t > 0, the transformed network is shown in Fig. 7.67. Applying KVL to the mesh for t > 0, 2e −3s − 2 I ( s) − sI ( s) = 0 2 I ( s) + sI ( s) = 2e −3s 2e −3s I ( s) = s+2 Taking inverse Laplace transform, i ( t ) = 2e

example 7.30

−2( t − 3)

2

2e−3s

+ −

s I (s)

Fig. 7.67 u(t − 3)

for t > 0

Determine the current i(t) in the network shown in Fig. 7.68, when the switch is

closed at t = 0. 10 Ω

50 sin 25 t

5H i (t)

Fig. 7.68

7.10 Solution of Differential Equations with Constant Coefficients 11.37

Solution  For t > 0, the transformed network is shown in Fig. 7.69. Applying KVL to the mesh for t > 0, 1250 − 10 I ( s) − 5sI ( s) = 0 s 2 + 625 1250 250 s 2 + 625 I ( s) = 2 ( s + 625)( s + 2)

10

5s I (s)

By partial-fraction expansion, I ( s) =

As + B 2

s + 625

+

C s+2

Fig. 7.69

250 = ( As + B )( s + 2) + C ( s 2 + 625) = ( A + C ) s 2 + ( 2 A + B) s + ( 2 B + 625C ) Comparing coefficients,

A+C = 0 2A + B = 0 2 B + 625 C = 250

Solving the equations,

A = −0.397 B = 0.795 C = 0.397 I ( s) =

−0.397 s + 0.795 2

s + 625

+

0.397 0.397 s 0.795 0.397 =− 2 + 2 + s+2 s + 625 s + 625 s + 2

Taking the inverse Laplace transform, i(t ) = −0.397 cos 25t + 0.032 sin 25t + 0.397e −2t

example 7.31

for t > 0

Find impulse response of the current i(t) in the network shown in Fig. 7.70. 1Ω

i1 (t)

i (t) d (t)

+ −

1Ω 2H

Fig. 7.70 1

I1 (s)

Solution The transformed network is shown in Fig. 7.71. 1( 2 s + 1) 2 s + 1 Z ( s) = = 2s + 1 + 1 2s + 2 1 2s + 2 V ( s) = = I1 ( s) = Z ( s) 2 s + 1 2 s + 1 2s + 2

I (s) 1

+ −

1 2s

Fig. 7.71

7.38 Circuit Theory and Transmission By current-division rule, 1 1 2s + 2 1 1 1 = × = = 2 s + 2 2 s + 2 2 s + 1 2 s + 1 2 s + 0.5 Taking inverse Laplace transform, I ( s) = I1 ( s) ×

i( t ) =

1 −0.5t e u( t ) 2

for t > 0

example 7.32 The network shown in Fig. 7.72 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. 7.72 v(t ) = u(t ) cos t + Aδ (t ) s V ( s) = 2 +A s +1

Solution For t > 0, the transformed network is shown in Fig. 7.73.

1

Applying KVL to the mesh for t > 0, V ( s) = 2 sI ( s) + I ( s) = I ( s) =

s 2

s +1

+A

K s + K3 s + A( s 2 + 1) K = 1 + 22 1 1 2  s +1 2  s +  ( s + 1) s +   2 2

V (s)

2s I (s)

Fig. 7.73

The transient part of the response is given by the first term. Hence, for the transient term to vanish, K1 = 0. −1  5 + A   4 1 2  K1 =  s +  I ( s) s = − 1 =   2  5 2 2   4 When

K1 = 0 5 1 A= 4 2 2 A = = 0.4 5

7.11 Response of RC Circuit to Various Functions 7.39

7.11

response of rc cIrcuIt to varIous functIons

Consider a series RC circuit as shown in Fig. 7.74. R

+ −

v (t)

C i (t)

Fig. 7.74 RC circuit For t > 0, the transformed network is shown in Fig. 7.75. 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 

R

V (s) + −

Taking Laplace transform, 1 s

1 1 s I ( s) = = 1  1    Rs +   R  s +  RC  RC  s×

Taking inverse Laplace transform, 1

i( t ) = (b)

1 − RC t e R

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×

I (s)

Fig. 7.75 Transformed network

When unit step signal is applied, v ( t ) = u( t )

V ( s) =

1 Cs

for t > 0

7.40 Circuit Theory and Transmission By partial-fraction expansion, A B I ( s) = + 1 s s+ RC A = s I ( s)

s=0

1 R

= s+

1 RC

=C s=0

1 1   R B = s+  I ( s) s = − 1 =  s RC  RC

= −C s= −

I ( s) =

C − s

1 RC

C

1 RC Taking inverse Laplace transform, s+

i(t ) = C − Ce (c)

−

1 t RC

for t > 0

When unit inpulse signal is applied, v(t) = d (t) Taking Laplace transform, V ( s) = 1 1 1 1   − 1   RC RC = = 1 − RC  I ( s) =  1 1 1 R      s+  Rs + Rs +      RC  RC  RC  Taking inverse Laplace transform, s+

s

i( t ) =

1 1 1 − RC t  δ (t ) −  e R RC  

for t > 0

example 7.33 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. 7.76

7.11�Response of RC Circuit to Various Functions�7.41

Solution

v(t) = u(t) − u(t − T) V ( s) �

1 e � sT 1 � e � sT � � s s s

For t � 0, the transformed network is shown in Fig. 7.77. Applying KVL to the mesh for t > 0,

R

1 Cs

V (s ) + −

I (s)

��������� V ( s) � RI ( s) �

1 I ( s) � 0 Cs 1 � � s � sT � sT � � 1 1 1 e V ( s) � e � � � I ( s) � � R V ( s) � � 1 1 1 1 � 1 � R� � s� s� R� s� R�s � � � RC RC � Cs RC � RC �

Taking inverse Laplace transform, i( t ) �

� 1 � � � 1� � �� � ( t �T ) 1 � � �� RC �� t e u(t ) � e � RC � u( t � T ) � � R� � �

for t � 0

�������������� For the network shown in Fig. 7.78, determine the current i(t) when the switch is closed at t = 0 with zero initial conditions. 3�

2r (t � 2)

� �

1F i (t)

��������� 3

Solution For t > 0, the transformed network is shown in Fig. 7.79. Applying KVL to the mesh for t > 0, 2e �2 s s2

1 � 3 I ( s) � I ( s) � 0 s 1� 2e �2 s � �� 3 � �� I ( s) � 2 s s I ( s) �

2e−2s + − s2

1 s

I(s)

���������

0.67e �2 s 2e �2 s � 1 � s( s � 0.33) � s2 � 3 � � � s�

By partial-fraction expansion,

Chapter 07.indd 41

6/13/2016 3:21:57 PM

7.42 Circuit Theory and Transmission 0.67 A B = + s( s + 0.33) s s + 0.33 A=

0.67 =2 s + 0.33 s = 0

B=

0.67 = −2 s s = −0.33

2  e −2 s e −2 s 2 I ( s) = e −2 s  − =2 −2   s s + 0.33  s s + 0.33 Taking inverse Laplace transform, i(t ) = 2u(t − 2) − 2e −0.33( t − 2) u(t − 2)

for t > 0

example 7.35 For the network shown in Fig. 7.80, determine the current i(t) when the switch is closed at t = 0 with zero initial conditions. 5Ω

d (t)

+ −

2F i (t)

Fig. 7.80 Solution For t > 0, the transformed network is shown in Fig. 7.81. Applying KVL to the mesh for t > 0, 1 I ( s) = 0 2s 1   5 +  I ( s) = 1 2s

1 − 5 I ( s) −

I ( s) =

5

1

1 2s 2s = 10 s + 1 0.2 s = s + 0.1 0.2( s + 0.1 − 0.1) = s + 0.1 0.1   = 0.2 1 −  s + 0.1 0.02 = 0.2 − s + 0 .1 5+

Taking inverse Laplace transform, i(t ) = 0.2δ (t ) − 0.02 e −0.1t u(t )

1

+ −

I(s)

Fig. 7.81

1 2s

7.11 Response of RC Circuit to Various Functions 7.43

example 7.36

For the network shown in Fig. 7.82, find the response v0 (t). 2Ω + vs (t) =

1 cost u(t ) 2

1 F 4

+ −

vo (t)

−

Fig. 7.82

Solution For t > 0, the transformed network is shown in Fig. 7.83. 2 1 s 2 2 s +1 By voltage-division rule, + Vs (s) − 4 2V ( s) s Vo ( s) = Vs ( s) × s = s = 2 4 s+2 ( s + 1)( s + 2) 2+ Fig. 7.83 s By partial-fraction expansion, As + B C Vo ( s) = 2 + s +1 s + 2 s = ( As + B)( s + 2) + c( s 2 + 1) s = ( A + C ) s 2 + ( 2 A + B) s + ( 2B + C )

Vs ( s) =

+ 4 s Vo (s)

−

Comparing coefficient of s2, s and s0, A+C = 0 2A + B = 1 2B + C = 0 Solving the equations, A = 0.4, B = 0.2, C = −0.4 0.4 s + 0.2 0.4 0.4 s 0.2 0.4 Vo ( s) = − = 2 + 2 − 2 s + 2 s +1 s +1 s + 2 s +1 Taking the inverse Laplace transform, i(t ) = 0.4 cos t + 0.2 sin t − 0.4e −2t

for t > 0

example 7.37 Find the impulse response of the voltage across the capacitor in the network shown in Fig. 7.84. Also determine response vc (t) for step input. 2Ω

v (t)

1H

+

+ −

−

Fig. 7.84

1 F vc (t)

7.44 Circuit Theory and Transmission

Solution For t > 0, the transformed network is shown in Fig. 7.85. By voltage-division rule,

Vc ( s) = V ( s) ×

2

1 s

1 2+ s+ s V ( s) V ( s) = 2 = s + 2 s + 1 ( s + 1) 2

(a)

V (s)

+ 1 Vc (s) s −

+ −

Fig. 7.85

For impulse input, V ( s) = 1 Vc ( s) =

1 ( s + 1) 2

Taking inverse Laplace transform, vc (t ) = te − t u(t ) (b)

s

for t > 0

For step input, V ( s) = Vc ( s) =

1 s 1 s( s + 1) 2

By partial-fraction expansion, Vc ( s) =

A B C + + s s + 1 ( s + 1) 2

1 = A( s + 1) 2 + Bs( s + 1) + Cs = A( s 2 + 2 s + 1) + B( s 2 + s) + Cs = ( A + B) s2 + ( 2 A + B + C ) s + A Comparing coefficient of s2, s1 and s0, A=1 A+ B = 0 B = − A = −1 2A + B + C = 0 C = −2 A − B = −2 + 1 = −1 Vc ( s) =

1 1 1 − − s s + 1 ( s + 1) 2

Taking inverse Laplace transform, vc (t ) = u(t ) − e − t u(t ) − te − t u(t ) = (1 − e − t − te − t )ut

for t > 0

7.11 Response of RC Circuit to Various Functions 7.45

example 7.38 For the network shown in Fig. 7.86, 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. 7.86

Solution For t > 0, the transformed network is shown in Fig. 7.87. Applying KVL to the mesh for t > 0, 5e − s s

2

5

6 − 5 I ( s) − sI ( s) − I ( s) = 0 s 5 I ( s) + sI ( s) +

5e−s s

6 5e − s I ( s) = 2 s s I ( s) =

5e − s 2

s( s + 5 s + 6 )

=

2

5e − s s( s + 3)( s + 2)

s + − I (s)

6 s

Fig. 7.87

By partial-fraction expansion, 1 A B C = + + s( s + 3)( s + 2) s s + 3 s + 2 A=

1 1 = ( s + 3)( s + 2) s = 0 6

B=

1 1 = s( s + 2) s = −3 3

C=

1 1 =− 2 s( s + 3) s = −2

1 1 1  5 e−s 5 e−s 5 e−s + − I ( s) = 5e − s  + − = 3 s+3 2 s+2  6 s 3( s + 3) 2( s + 2)  6 s Taking inverse Laplace transform, i( t ) =

5 5 5 u(t − 1) + e −3( t −1) u(t − 1) − e −2( t −1) u(t − 1) 6 3 2

for t > 0

example 7.39 For the network shown in Fig. 7.88, the switch is closed at t = 0. Determine the current i(t) assuming zero initial conditions.

7.46 Circuit Theory and Transmission 1H

2Ω

sin t

0.5 F i(t)

Fig. 7.88 Solution For t > 0, the transformed network is shown in Fig. 7.89. Applying KVL to the mesh for t > 0, 1 s +1 2

− 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) =

2

s

2 s I (s)

Fig. 7.89

s ( s + 1)( s + 2 s + 2) 2

2

By partial-fraction expansion, I ( s) =

As + B 2

Cs + D

+

2

s + 1 s + 2s + 2 s = ( As + B)( s 2 + 2 s + 2) + (Cs + D )( s 2 + 1) = As3 + 2 As 2 + 2 As + Bs 2 + 2 Bs + 2 B + Cs3 + Cs + Ds 2 + D = ( A + C ) s3 + ( 2 A + B + D ) s 2 + ( 2 A + 2 B + C ) s + ( 2 B + D )

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 s 2 + 2s + 2 0.2s 0.4 0.2s + 0.2 + 0.6 = 2 + 2 − s + 1 s + 1 ( s + 1) 2 + (1) 2

=

2

0.2 s 2

s +1

+

0.4 2

−

0.2( s + 1) 2

−

0.6

s + 1 ( s + 1) + 1 ( s + 1) 2 + 1

Taking inverse Laplace transform, i(t ) = 0.2 cos t + 0.4 sin t − 0.2 e − t cos t − 0.6 e − t sin t = 0.2 cos t + 0.4 sin t − e − t (0.2 cos t + 0.6 sin t )

for t > 0

7.11 Response of RC Circuit to Various Functions 7.47

example 7.40 For the network shown in Fig. 7.90, 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. 7.90 Solution For t > 0, the transformed network is shown in Fig. 7.91. Applying KVL to the mesh for t > 0, 6 4 − 5 I ( s) − s I ( s) − I ( s) = 0 s+2 s 4 6   5 + s +  I ( s) = s s+2

5

6 s+2

+ −

s

4 s I (s)

6s

I ( s) =

Fig. 7.91

( s + 2)(( s 2 + 5s + 4) 6s = ( s + 2)( s + 1)( s + 4)

By partial-fraction expansion, I ( s) =

A B C + + s + 2 s +1 s + 4

6s =6 ( s + 1)( s + 4) s = −2 6s = −2 B = ( s + 1) I ( s) |s = −1 = ( s + 2)( s + 4) s = −1 6s = −4 C = ( s + 4) I ( s) |s = −4 = ( s + 2)( s + 1) s = −4 6 2 4 I ( s) = − − s + 2 s +1 s + 4 Taking inverse Laplace transform, i(t ) = 6e −2t u(t ) − 2e − t u(t ) − 4 e −4 t u(t ) A = ( s + 2) I ( s) |s = −2 =

for t > 0

example 7.41 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

Fig. 7.92

+ vo (t ) −

7.48 Circuit Theory and Transmission

Solution For vi(t) = d (t), Vi ( s) = 1 vo (t ) = [e −2t + e −3t ]u(t ) 1 1 Vo ( s) = + s+2 s+3 System function H ( s) =

Vo ( s) Vi ( s) =

1 1 2s + 5 + = s + 2 s + 3 ( s + 2)( s + 3)

…(i)

For vo (t ) = te −2t u(t ), Vo ( s) =

1 ( s + 2) 2

From Eq. (i), Vi ( s) =

Vo ( s) 1 ( s + 2)( s + 3) ( s + 3) = × = 2 H ( s) ( s + 2) 2s + 5 2( s + 2.5)( s + 2)

By partial-fraction expansion, A B + s + 2 s + 2.5 A=1 B = −0.5 1 0.5 Vi ( s) = − s + 2 s + 2.5

Vi ( s) =

Taking inverse Laplace transform, vi (t ) = e −2t − 0.5e −2.5t

for t > 0

example 7.42

A unit impulse applied to two terminal black box produces a voltage vo(t) = 2e −e . Determine the terminal voltage when a current pulse of 1 A height and a duration of 2 seconds is applied at the terminal. −t

−3t

+ is (t )

Black box

vo (t )

−

Fig. 7.93 is (t )

Solution

−t

−3t

vo (t ) = 2e − e 2 1 Vo ( s) = − s +1 s + 3

When is (t ) = δ (t ),

1 0

I s ( s) = 1

t 2

Fig. 7.94

Exercises 7.49

Vo ( s) = Z ( s) I s ( s) Z ( s) =

…(i)

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 − 2) I s ( s) =

1 e −2 s − s s

From Eq. (i), 1   1 e −2 s   2 Vo ( s) =  −  − s   s + 1 s + 3  s  =

e −2 s 2 1 2e −2 s + − − s( s + 1) s( s + 3) s( s + 1) s( s + 3)

1  1  1 1 1  1  e −2 s  1 1 −2 s  1 = 2 − 2 e + − − − − −       3  s s + 3   s s + 1  s s + 1 3  s s + 3  Taking the inverse Laplace transform, 1 1 v(t ) = 2[u(t ) − e − t u(t )] − [u(t ) − e −3t u(t )] − 2[u(t − 2) − e − ( t − 2) u(t − 2)] + [u(t − 2) − e −3( t − 2) u(t − 2)] 3 3 for t > 0

Exercises 7.1

For the network shown in Fig 7.95, the switch is closed at t = 0. Find the current i1(t) for t > 0.

2Ω i (t )

100 Ω 2Ω 50 Ω

100 V

4H 24 V

i1 (t )

0.5 H

Fig. 7.96

Fig. 7.95 [i1(t) = 3 − e−25 t] 7.2

2Ω

Determine the current i(t) in the network of Fig. 7.96, when the switch is closed at t = 0. The inductor is initially unenergized.

[i(t) = 4(1 − e−6t)] 7.3

In the network of Fig. 7.97, 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.

7.50 Circuit Theory and Transmission + 0.33 Ω

10 V

1 Ω 2

10 A

0.5 H

1 Ω 8

v (t )

1 F v (t)

1F

−

Fig. 7.97 [v(t) = 1 + 4 e−10t]

7.8

The circuit of Fig. 7.98, has been in the condition shown for a long time. At t = 0, switch is closed. Find v(t) for t > 0.

7.4

5Ω

Fig. 7.101 [v(t) = 10 − 10e−t + 5e−2t] The network shown in Fig. 7.102 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

2 +

v (t)

20 V

3Ω

3Ω

2F

2V

v (t )

0.5 F

7.5

Fig. 7.98 [v(t) = 7.5 + 12.5 e−(4/15)t] Figure 7.99 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 m seconds.

1H −

Fig. 7.102 7.9

800 Ω

[v(t) = 4e−t − 2 e−2t] In the network shown in Fig. 7.103, the switch is closed at t = 0. Find current i1(t) for t > 0. 1H

3Ω

1Ω

i (t )

12 V

7.6

400 Ω

0.001 µF

200 Ω

1 Ω i (t ) 2

i1 (t)

1 F 3

Fig. 7.99 6 [i(t) = 0.00857 + 0.01143 e−8.75 × 10 t, 0.013 A]

Fig. 7.103 [i1(t) = 5 + 5e−2t − 10e−3t]

Find i(t) for the network shown in Fig. 7.100.

7.10 In the network shown in Fig. 7.104, the switch is closed at t = 0. Find the current through the 30 W resistor.

i (t) 10 Ω

1F 50 V

20 V

5Ω

10 Ω

0.5 F

1H

2H

5Ω 20 Ω

10 V

30 Ω

Fig. 7.100 [i(t) = 0.125 e−0.308t + 3.875 e−0.052t] 7.7

Determine v(t) in the network of Fig. 7.101 where iL(0−) = 15 A and vc(0−) = 5 V.

Fig. 7.104 [i(t) = 0.1818 − 0.265 e−13.14t + 0.083 e−41.86t ]

Exercises 7.51

7.11 The network shown in Fig. 7.105 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.

having been established previously. Find i(t) for t > 0. 3Ω i (t )

2Ω

2H

s1

s2

5Ω

1.8 H

0.1 F

3.5 Ω

10 V

3H

10 V

1 µF

Fig. 7.108 [i(t) = 1.5124e−2.22t + 3.049e−2.5t]

Fig. 7.105 [i(t) = 5 cos (0.577 × 103 t)] 7.12 In the network shown in Fig. 7.106, find currents i1(t) and i2(t) for t > 0.

7.15 Find the current i(t) in the network of Fig. 7.109, if the switch is closed at t = 0. Assume initial conditions to be zero. 20 Ω

10 Ω

i (t )

10 Ω

0.2 F

50 V i1 (t)

40 Ω

5A

i2 (t )

1H

[i1(t) = 5 e−0.625t, i2(t) = 1 − e−0.625t ] 7.13 For the network shown in Fig. 7.107, find currents i1(t) and i2(t) for t > 0.

[i(t) = 3 + 0.57e−7.14 t] 7.16 In the network shown in Fig. 7.110, find the voltage v(t) for t > 0. 7Ω

5Ω

20 µF i1 (t)

2.5 H

Fig. 7.109

Fig. 7.106

50 V

15 Ω

i2 (t )

5Ω

5 cos 2t

1H

+ −

1 F v (t ) 2

0.1 H

Fig. 7.110 Fig. 7.107  i1 (t ) = 0.101e −100.5t + 10.05e −9949.5t    i2 (t ) = 5 − 5.05e −100.5t + 0.05e −9949.5t  7.14 In the network shown in Fig. 7.108, the switch is opened at t = 0, the steady state

6 − t 9 −6 t 3 21    v(t ) = − 5 e + 10 e + 10 cos 2t + 10 sin 2t    7.17 For the network shown in Fig. 7.111, determine v(t) when the input is (i) an impulse function [e−t u(t)] (ii) i(t) = 4e−t u(t) [4t e−t u(t)]

7.52 Circuit Theory and Transmission 10 Ω

+ 1Ω

i (t)

1F

v (t)

10 F vc (t )

r (t )

−

Fig. 7.111

Fig. 7.112

7.18 For a unit-ramp input shown in Fig. 7.112, find the response vc(t) for t > 0.

[vc(t) = −100 u(t) + 100e−0.01t u(t) + tu(t)]

Objective-Type Questions 7.1

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)

7.2

7.3

(d)

(c) (d) 7.5

1A

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 (a) 2[1 − e −2t] u(t) (b) 4[e−t − e–2t] u(t) (c) sin 2t (d) (1 − 4 e−4t) u(t) The Laplace transform of a unit-ramp function starting at t = a is (a)

(c) 7.4

1 A 2

(b)

1 ( s + a) 2

(b)

(d)

s2

The Laplace transform of eat cos a t is equal to (a)

s −α ( s − α )2 + α 2

The circuit shown in Fig. 7.113 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 1H

1F

v (t ) i (t)

−

Fig. 7.113 (a) (c)

( s + a) 2 s2

( s − α )2 none of the above

+

e − as

a

1

1Ω

7.6

e − as

s+α ( s − α )2 + α 2

s 2

s + s +1 s−2 2

s + s +1

(b) (d)

s+2 2

s + s +1 s−2 2

s + s +1

A square pulse of 3 volts amplitude is applied to an RC circuit shown in Fig. 7.114. The capacitor is initially uncharged. The output voltage v0 at time t = 2 seconds is (a) 3 V (c) 4 V

(b) −3 V (d) −4 V

Answers to Objective-Type Questions 7.53 0.1 µF

3V

7.9

−j2 Ω

vi

vo

1 kΩ

In the circuit shown in Fig. 7.116, 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

t

2s

i (t)

Fig. 7.114 7.7

A 2 mH inductor with some initial current can be represented as shown in Fig. 7.115. The value of the initial current is

v (t)

L

I (S)

−+ 0.002s

Fig. 7.116

1mV

(a) a constant voltage (b) a linearly increasing voltage (c) an ideal impulse (d) as exponential increasing voltage

Fig. 7.115

7.8

(a) 0.5 A (b) 2 A (c) 1 A (d) 0 A current impulse 5 d (t) is forced through a capacitor C. The voltage vc(t) across the capacitor is given by (a)

5t

(b)

5 u(t) − C

(c)

5 t C

(d)

5u(t ) C

7.10

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)

Answers to Objective-Type Questions 7.1 (c) 7.7 (a)

7.2 (a) 7.8 (d)

7.3 (c) 7.9 (c)

7.4 (a) 7.10 (c)

7.5 (b)

7.6 (b)

0

8 Network Functions

8.1

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. 8.1 (a) and is called one-port network. Figure 8.1 (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. 8.1 (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.

8.2

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.

8.2 Circuit Theory and Transmission Lines 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,

8.3

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 ( s) =

I 2 ( s) I1 ( s)

α 21 ( s) =

I1 ( s) I 2 ( s)

8.3 Transfer Functions 8.3

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 8. 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. 8.2. R

L

C

R

Fig. 8.2 Solution The transformed network is shown in Fig. 8.3. 1 R (R + L Lss) s+ R + Ls L 1 L Z ( s) = Cs = = 2 1 1 C 2 R LC LCs s + RCs R Cs + 1 + (R + L Lss) s + s+ Cs L LC

Example 8.2

1 Cs

Z((s)

Ls

Fig. 8.3

Determine the driving-point impedance of the network shown in Fig. 8.4. 2s

2s

1 2s

Z (s)

1 2s

Fig. 8.4 Solution 1  1 1  1 1 1 2s + 4 s + 8s3 + 2 s +  2 s +   2 s +  4 2 2s  2s 2s  2s 2s = 2 s = 16 s + 12 s + 1 Z ( s) = 2 s + = 2s + = 2 s + 1 1 2 + 4s2 2 + 4s2 2 + 4s2 8s3 + 4 s + 2s + 2s 2s 2s

8.4 Circuit Theory and Transmission Lines

Example 8.3

Determine the driving-point impedance of the network shown in Fig. 8.5. 1 s

1 s

s

Z (s)

s

Fig. 8.5 Solution 1  s  + s  s  1 (1 + s 2 ) s 1 s + s3 1 s 4 + 3s 2 + 1 Z ( s) = + = + = + = 1 s s 2s2 + 1 s 2s2 + 1 2 s3 + s s+ +s s

Example 8.4

Find the driving-point admittance function of the network shown in Fig. 8.6.

3s

s

1 5s

1 2s

Y (s)

Fig. 8.6 Solution  1 s   2s  1 1 s 30 s 4 + 15s 2 + 2 s 2 + 1 + 5s 2 30 s 4 + 22 s 2 + 1 Z ( s ) = 3s + + = 3s + + 2 = = 2 1 5s 5s 2 s + 1 5 s ( 2 s + 1 ) 5s ( 2 s 2 + 1) s+ 2s 1 5s ( 2 s 2 + 1) Y ( s) = = Z ( s) 30 s 4 + 22 s 2 + 1

Example 8.5

Find the transfer impedance function Z12(s) for the network shown in Fig. 8.7. I1 (s) +

V1 (s)

+ R

−

1 Cs

V2 (s) −

Fig. 8.7

8.4 Analysis of Ladder Networks 8.5

 1 R   Cs  V2 ( s) = I1 ( s) 1 R+ Cs V2 ( s) R = I1 ( s) RCs RC + 1 V ( s) 1 Z12 ( s) = 2 = 1  I1 ( s)  Cs+   RC 

Solution

Example 8.6

Find voltage transfer function of the two-port network shown in Fig. 8.8. I1(s)

R

I2(s)

+

V1(s)

+ 1 Cs

−

V2(s) −

Fig. 8.8 Solution By voltage division rule, 1 Cs

Voltage transfer function

8.4

1 1 RC V2 ( s) = V1 ( s) = V1 ( s) = V1 ( s) 1 1 RCss + 1 RC 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. 8.9 where all the − impedances are connected in series branches and all the admittances are connected in parallel branches. Fig. 8.9 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 −

8.6 Circuit Theory and Transmission Lines 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 8.7

For the network shown in Fig. 8.10, determine transfer function 1Ω

I1

1Ω

I2 = 0

+

V2 . V1

+

V1

1F

V2

1F

−

−

Fig. 8.10 Solution The transformed network is shown in Fig. 8.11. 1

I

Hence,

Example 8.8

I1

V1

s

s

Va

Ib

Vb

I2 = 0 +

1

1

−

V2 −

Fig. 8.12 Vb = V2 V2 = V2 1 Va = s I b + V2 = sV2 + V2 = ( s + 1)V2 V I1 = a + I b = ( s + 1)V2 + V2 = ( s + 2)V2 1 V1 = s I1 + Va = s ( s + 2))V V2 + ( s + 1))V V2 = ( s 2 + 3s + 1)V2 Ib =

Ib

Vb

I2 = 0 + 1 s

For the network shown in Fig. 8.12, determine the voltage transfer function +

Solution

1

V

a 1 Vb = V2 + V I b = 2 = sV2 1 1 V1 s s − Va = 1 I b + V2 = sV2 + V2 = ( s + 1)V2 Fig. 8.11 V I1 = a + I b = sV sVa + I b = s( s + 1)V2 + ssV V2 = ( s 2 + 2 s)V2 1 s V1 = 1 I1 + Va = ( s 2 + 2 s)V2 + ( s + 1)V2 = ( s 2 + 3s + 1)V V2 V2 1 = V1 s 2 + 3s + 1

V2 −

V2 . V1

8.4 �Analysis of Ladder Networks�8.7

V 1 � 2 V s s

Hence,

�������������

Find the network functions

V V V , and for the network shown in Fig. 8.13. I1 V I1

1H

I1

1H

I

0 +

+ V1

1F

V

1F

−

Fig. 8.13 Solution The transformed network is shown in Fig. 8.14. V V Ib � � sV 1 s

V

I =0

s

s

V −

−

V

Fig. 8.14

� s s �1 � s s � 2s

s

Ib

+

V

V � I1 � 1 s V I1 V V V I1

s

+

V � s

Hence,

s

I1

V

s

s �

2s V

s

s

s

s

V

s s3

s 1

s

s 1

s3

��������������

s

Find the network functions

I1

1 4

V V I1 V

and

V for the network in Fig. 8.15. I1

3H

I =0

+ V1

+ 1 2

2

V

Fig. 8.15

Chapter 08.indd 7

5/30/2016 6:01:40 PM

8.8 Circuit Theory and Transmission Lines Solution The transformed network is shown in Fig. 8.16. Vb = V2

4 s

I1

V I b = 2 = 2 sV2 1 2s

3s

Va

Ib

Vb

I2 = 0

+

+

V1

Va = 3s I b + V2 = 3s ( 2 sV sV2 ) + V2 = (6 s 2 +1 + 1)V2

1 2s

s 2

−

V2 −

 14 s + 2  Va 2 + I b = (6 s 2 + 1)V2 + 2 sV2 =   V2 s s s   2  6 s 4 + 57 s 2 + 8  4 4  14 s 2 + 2  2 V1 = I1 + Va =  V + ( 6 s + 1 ) V = 2 2    V2 s s s s2    2

I1 =

Hence,

Fig. 8.16

V1 6 s 4 + 57 s 2 + 8 = I1 14 s3 + 2 s V2 s2 = 4 V1 6 s + 57 s 2 + 8 V2 s = 2 I1 14 s + 2

Example 8.11

For the ladder network of Fig. 8.17, 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. 8.17 Solution The transformed network is shown in Fig. 8.18. I1 1

1′

1

s

1

Va

s

1

Vb

+

Ia

Ib

V1

1 s

1 s

−

Vc

I2 = 0 + 1 s

2

V2 −

Fig. 8.18 Vc = V2 V I b = 2 = s V2 1 s

s

2′

8.4 Analysis of Ladder Networks 8.9

Ia =

V + 1 s

b

= s(

+2

V I1

1

V 1 s

2

s

s2

2

s) V V V

Hence,

Z11

V I1

Example 8.12

s

Determine the voltage transfer function 2

1Ω

I1

V for the network shown in Fig. 8.19. V I =0

+

+

V

1

F

V

1Ω

Fig. 8.19 Solution The transformed network is shown in Fig. 8.20. I1

V Ic

V 1 s

1V

1

V 1 s

V

2s

V

a

=s

V = V

1 s

s

1

s 1 4s 2

V Ib

1 s

Fig. 8.20

1 Hence,

Ia

Va

−

+1 V I1

1

+

V

I =0 Ic 1

+ V

8.10 Circuit Theory and Transmission Lines

Example 8.13

For the network shown in Fig. 8.21, determine the transfer function

I2 . V1

3Ω 2

I1 +

I2

1Ω

2F 3

I=0 +

V1

1Ω 6

2F

V2

−

−

Fig. 8.21 Solution The transformed network is shown in Fig. 8.22. 3 2 1

I1 +

I2

Va

3 2s

Ia

Vb

I=0

1 6

1 2s

V1

+

Ib

V2

−

−

Fig. 8.22 Vb = Va = V2 V2 V2 + = 2 sV2 + 6 V2 = ( 2 s + 6)V2 1 1 2s 6 3 3  ×  ( 2 s + 5)( s + 3)  (6 s + 1155)  2 2s   9  V1 =  + 1 I1 + V2 =  + 1 I1 + V2 = ( 2 s + 6)V2 + V2 =  + 1 V2 3 3  6s + 6  6s + 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)  = V = V = 2 2     V2  s +1 s +1 s +1       2 2(ss + 6 s + 8) 2 ( s + 4)( 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 8.14

For the network shown in Fig. 8.23, compute α 12 (s) (s = 1Ω

I2

I=0

+ I1

V1

+ 1F

1F

V2 −

−

Fig. 8.23

I2 V and Z12 ((s) s) = 2 . I1 I1

8.4 Analysis of Ladder Networks 8.11

Solution The transformed network is shown in Fig. 8.24.

Hence, and

1

Va = V2 + V2 I1 I2 = = sV2 V1 1 s − V1 = 1 I 2 + Va = sV sV2 + V2 = (s + 1)V2 V I1 = 1 + I 2 = sV sV1 + I 2 = s (s + 1)V2 + sV2 = ( s 2 + 2s)V2 1 s I2 1 α12 ( s) = = I1 s + 2 Z12 ( s) =

I2

Va

I=0 +

1 s

1 s

V2 −

Fig. 8.24

V2 1 = I1 s 2 + 2 s

Example 8.15

Determine the voltage ratio

V2 V2 I and , current ratio 2 , transfer impedance I1 V1 I1

V1 for the network shown in Fig. 8.25. I1

driving- point impedance

3Ω

3H

I1

I2

1F 3

+

I=0 +

2Ω

V1

1Ω

1F

1F 2

−

V2 −

Fig. 8.25 Solution The transformed network is shown in Fig. 8.26. 3 Va

I1 +

3 s

3s

Ia 1 s

V1 −

I2

Vb

Vc

I=0

Ib 2 2 s

+ 1

V2 −

Fig. 8.26 Vc = Vb = V2 V I 2 = 2 = V2 1 Ia = Ib + I2 =

V2

+

V2 s  3s + 2  = V2 + V2 =  V2  2 s + 2  1 2s + 2

2 s  9 s 2 + 8s + 2  3s ( 3s + 2 ) Va = 3s I a + V2 = V2 + V2 =   V2 2s + 2  2s + 2  2+

8.12 Circuit Theory and Transmission Lines I1 =

V + 1 s 3×

V 3+ = = Hence,

Example 8.16

3 s

s 2 V 2s 2

=

3 I + s +1

=

3s 2 2s 2

3 s +1

2

8 2 2s 2

9s

=

2

5 2

V +

V

2 2

2

15s

+ 2

s 2

 25s + 8 

36

2



25 + 8) V 2

(36

V

V

2 25s + 8 2s 2 5s 2 2s 2 5s 2

36

2

25s + 8

36 2

2 V V I I , , and 2 . V I1 V I1

For the resistive two-port network of Fig. 8.27, find 2Ω

I1

2Ω

2Ω

+

I +

V

1Ω

1Ω

V

1Ω

−

1Ω

−

Fig. 8.27 Solution The network is redrawn as shown in Fig. 8.28. I1

V

s

 27

V = V 36 I2 = I1 V I1 V = I1

a

2

V

+

Ia

V

1

2

V I

V

2

I +

V

Fig. 8.28

1

8.4 Analysis of Ladder Networks 8.13

V2 = −V2 1 Vb = −3 I 2 = 3V2 I2 = −

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 V2 = 15 V2 1 = 2 Ia + +V Va = 30 V2 + 11 V2 = 4411 V2 V = 1 + I a = 41 V2 + 15 V2 = 56 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 8.17

Find the network function Ia

1Ω

1Ω

V2 for the network shown in Fig. 8.29. V1 2V1

I2 = 0

+ −

+ 2Ia

V1

+

1Ω

−

V2 −

Fig. 8.29 Solution The network is redrawn as shown in Fig. 8.30. Ia

1

3Ia

1

2V1

I2 = 0

+ − +

V1

2Ia

−

3Ia

+

1

V2 −

Fig. 8.30

8.14 Circuit Theory and Transmission Lines From Fig. 8.30, 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 8.18

I2 for the network shown in Fig. 8.31. I1

Find the network function 2Ia

2Ω

I2

− +

+

Ia I1

1Ω

I1 2

1Ω

1Ω

V2 −

Fig. 8.31 Solution The network is redrawn as shown in Fig. 8.32. 2 Ia

a

Ia + I2 +

− + I I1

I1 2 d

2

I2 +

I1 2

I2 f

+

Ia I1 2

1

1

1

V2 −

b

c

e

Fig. 8.32 From Fig. 8.32, I = I1 + I a + I 2 + =

3 I1 + I a + I 2 2

I1 2 ...(i)

Applying KVL to the loop abcda, −1 I − 1 I a − 2 I a = 0 −I − 3 Ia = 0 I + 3 Ia = 0

...(ii)

8.5 Analysis of Non-Ladder Networks 8.15

Substituting Eq. (i) in Eq. (ii), 3 I1 + I a + I 2 + 3 I a = 0 2 3 I1 + I 2 + 4 I a = 0 2

...(iii)

Applying KVL to the loop dcefd, I   1Ia −1I2 − 2  I2 + 1  = 0  2 I a − 3 I 2 − I1 = 0 I a = 3 I 2 + I1

...(iv)

Substituting Eq. (iv) in Eq. (iii), 3 I1 + I 2 + 4 (3 I 2 + I1 ) = 0 2 3 I1 + I 2 + 12 I 2 + 4 I1 = 0 2 11 I1 + 13 I 2 = 0 2 13 I 2 = −

I2 11 =− I1 26

Hence,

8.5

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 8.33 shows one such network. Z4 I1

Z2

Z1

+ V1

I2 +

Z3

−

V2 −

Fig. 8.33 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.

8.16 Circuit Theory and Transmission Lines

Example 8.19

For the resistive bridged T network shown in Fig. 8.34, find

V2 V2 I 2 I , , and 2 . V1 I 1 V1 I1

2Ω

1Ω

I1

1Ω

I3

+

+

V1 −

I2

V2

0.5 Ω

I1

I2

1Ω

−

Fig. 8.34 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, V1  1.5 0.5 −1  I1   0  = 0.5 2.5 1  I 2       1 4   I3   0   −1 V1 0.5 −1 0 2 .5 1 0 1 4 ∆1 V (10 − 1) I1 = = = 1 = V1 1.5 0.5 −1 ∆ 9 0.5 2.5 1 −1 1 4 1.5 V1 −1 0.5 0 1 −1 0 4 ∆2 −V ( 2 + 1) 1 I2 = = = 1 = − V1 1.5 0.5 −1 ∆ 9 3 0.5 2.5 1 −1 1 4 From Fig. 8.34, From Eq. (v), From Eqs. (iv) and (v),

Hence,

V2 = −1 (I2) = −I2 V1 = −3 I2 1 1 I 2 = − V1 = − I1 3 3 I1 = −3I 2 I2 1 =−  V1 3

...(i) ...(ii) ...(iii)

...(iv)

...(iv)

...(v)

8.5 Analysis of Non-Ladder Networks 8.17

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 8.20

For the network of Fig. 8.35, find Z11, Z12 and G12. a

I1

Za

b

+

+

V1

Zb

V2

Zb −

− d

c Za

Fig. 8.35 Solution The network can be redrawn as shown in Fig. 8.36. 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 2

I1 +

V1 Z a + Zb = I1 2

Z11 =

I I I  V2 = Zb 1 − Z a  1  = ( Zb − Z a ) 1   2 2 2 V2 Zb − Z a = I1 2

Z12 =

V1

I1 2 Za

+

I1 2 Zb V2

−

Zb −

Fig. 8.36

By voltage-division rule, V2 = G12 =

Example 8.21

Zb Za Z − Za V1 − V1 = b V1 Z a + Zb Z a + Zb Z a + Zb V2 Zb − Z a = V1 Z a + Zb For the network shown in Fig. 8.37, determine Z11 (s), G12 (s) and Z12 (s).

Za

8.18 Circuit Theory and Transmission Lines 1F

I1

1Ω

1Ω

I2 = 0

+

+

V1

V2

1F

−

−

Fig. 8.37 Solution The transformed network is shown in Fig. 8.38. 1 s

I3

1

1

+

I2 = 0 +

1 s

V1

V2

I1 −

−

Fig. 8.38 Applying KVL to Mesh 1,  1 V1 = 1 +  I1 − I 3  s

...(i)

1 I1 + I 3 s

...(ii)

Applying KVL to Mesh 2, V2 = 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 + 1  s   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 + 1 = I1 s ( 2 s + 1)

Z12 ( s) =

V2 s 2 + 2 s + 1 = I1 s ( 2 s + 1)

G12 ( s) =

V2 s 2 + 2 s + 1 = V1 s 2 + 3s + 1

8.5 Analysis of Non-Ladder Networks 8.19

Example 8.22

For the network shown in Fig.8.39, find the driving-point admittance Y11 and trans-

fer admittance Y12. 1Ω 1F

I1

V1

1F

+ −

I2

1Ω

1Ω

Fig. 8.39 Solution The transformed network is shown in Fig. 8.40. 1 1 s

I1

V1

+ −

I3

1

1 s

I2

I2

1

I1

Fig. 8.40 Applying KVL to Mesh 1, 1 1  V1 =  + 1 I1 + I 2 − I 3 s  s

...(i)

Applying KVL to Mesh 2, 1 1  0 = I1 +  2 +  I 2 + I 3   s s

...(ii)

1 1 2  0 = − I1 + I 2 +  + 1 I 3 s  s s

...(iii)

Applying KVL to Mesh 3,

Writing these equations in matrix from, 1 1  s +1 V1   1 0 =  1 2+    s 0  1 1 − s  s ∆ I1 = 1 ∆

    I1   I2   I  2   3 + 1 s  1 s 1 s

−

8.20 Circuit Theory and Transmission Lines 1 +1 s ∆=

− =

2+

1 1 s

1 s

1 s 1 1  2  1    2  1  1   1  1  1   1   =  + 1  2 +   + 1 − 2  − 1 (1)  + 1 + 2  − (1)   +    2 +    s   s s  s  s    s  s  s   s  s  s  2 +1 s −

1 1 s

s 2 + 5s + 2 s2 V1

1

∆1 = 0

2+

0

1 s

1 s  2 s 2 + 5s + 1  1 1  2  1  = V1  2 +   + 1 − 2  = V1   s s  s  s  s2   

− 1 s

2 +1 s

 2 s 2 + 5s + 1 I1 = V1  2   s + 5s + 2  Hence,

I1 2 s 2 + 5s + 1 = V1 s 2 + 5s + 2 ∆ I2 = 2 ∆

Y11 =

1 + 1 V1 s ∆2 =

1 −

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

 s2 + 2s + 1  I 2 = −V1  2   s + 5s + 2  Hence,

8.6

Y12 =

I2 s2 + 2s + 1 =− 2 V1 s + 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

8.8 Restrictions on Pole and Zero Locations for Transfer Functions 8.21

F ( s) = H

( s − z1 )( )( s − z2 ) …( s − zn ) ( s − p1 )( )( s − p2 ) …( 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. 8.41. The zeros are shown by circles and the poles by crosses. This diagram is referred to as pole-zero plot. Fig. 8.41 Pole-zero plot where

RESTRICTIONS ON POLE AND ZERO LOCATIONS FOR DRIVING-POINT FUNCTIONS [COMMON FACTORS IN N(S) AND D(S) CANCELLED]

8.7 (1) (2) (3) (4) (5) (6) (7)

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]

8.8 (1) (2) (3) (4)

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.

8.22 Circuit Theory and Transmission Lines (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 8.23 (a)

Test whether the following represent driving-point immittance functions.

5s4 + 3s 2 − 2s + 1 3

s +6 6ss + 20

(b)

s3 + s 2 + 5s+ 2 4

3

s + 6s + 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 8.24 (a) G21 =

Test whether the following represent transfer functions. 3s+ 2

3

2

5s + 4s 4s + 1

(b) a 12 =

2s 2 + 5s+ 1 s+7

(c) Z 21 =

1 s3 + 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 8.25

Obtain the pole-zero plot of the following functions.

(a) F(s) =

s( 2) s(s+ (s+ +1 1)(s + 3 3)

(b) F(s) =

(c) F(s) =

s( 2) s(s+ (s+ 1+ j1)(s+ 1 - j1)

(d) F(s) =

(e) F(s) =

s2 + 4 (s+ +2 2)(s 2 + 9 9)

s( 1) s(s+ (s+ + 22))2 ((s+ s+ 3) (s+ + 11))2 ((s+ s+ 5) (s+ + 22)(s + 3 + j2)(s+ 3 - j2)

8.8 Restrictions on Pole and Zero Locations for Transfer Functions

8.23

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. 8.42. jw

s 0

−3 −2 −1

Fig. 8.42 (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. 8.43. jw

s 0

−3 −2 −1

Fig. 8.43 (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. 8.44. jw j1

s

−2

0

−1

−j 1

Fig. 8.44 (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. 8.45. jw j2 j1 −5 −4 −3 −2

−1

0 −j 1

−j 2

Fig. 8.45

s

8.24 Circuit Theory and Transmission Lines (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. 8.46. jw j3 j2 j1 s

−2

−1

0 −j 1

−j 2 −j 3

Fig. 8.46

Example 8.26

Find poles and zeros of the impedance of the network shown in Fig. 8.47 and plot

them on the s-plane. 1F

Z (s)

1 H 2

2Ω

Fig. 8.47 1 s

Solution The transformed network is shown in Fig. 8.48. s ×2 1 2 1 2s 2 s 2 + s + 4 2 ( s 2 + 0.5s + 2) Z ( s) = + = + = = s s s s+4 s ( s + 4) s ( s + 4) +2 2 2 ( s + 0.25 + j1.4)( s + 0.2 25 5 − j1.4) = s ( s + 4)

Z (s)

2

s 2

Fig. 8.48

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. 8.49. jw

j 1 .4 0

−4

−3

−2

− 1 − 0.25 − j 1 .4

Fig. 8.49

s

8.8 Restrictions on Pole and Zero Locations for Transfer Functions

Example 8.27

8.25

Determine the poles and zeros of the impedance function Z (s) in the network shown

in Fig. 8.50. 1 Ω 2

Z (s)

4F

1 Ω 6

Fig. 8.50 Solution The transformed network is shown in Fig. 8.51. 1 2

1 4s

Z (s)

1 6

Fig. 8.51 1 1 × 1 4s 6 1 1 4s + 8 s + 2 0.5 ( s + 2) Z ( s) = + = + = = = 2 1 1 2 4 s + 6 2 ( 4 s + 6) 2 s + 3 s +11.5 + 4s 6 The function Z(s) has zero at s = −2 and pole at s = −1.5.

Example 8.28

Determine Z(s) in the network shown in Fig. 8.52. Find poles and zeros of Z(s) and

plot them on s-plane. 1H

1 F 20

Z (s)

4Ω

Fig. 8.52 Solution The transformed network is shown in Fig. 8.53. s

Z (s)

20 s

Fig. 8.53

4

8.26 Circuit Theory and Transmission Lines 20 ×4 80 20 s ( s + 5) + 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 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. 8.54. jw j 3.71

s

−5

−4

−3

−2

0

−1

−j 3.71

Fig. 8.54

Example 8.29

For the network shown in Fig. 8.55, plot poles and zeros of function

I0 . Ii

I0 4Ω Ii

0.5 F 2H

Fig. 8.55 Solution The transformed network is shown in Fig. 8.56. By current-division rule,    4 + 2s  I0 = Ii  2  4 + 2s +   s

I0 4 2 s

Ii

I0 s ( 4 + 2 s) s ( s + 2) s ( s + 2) = = 2 = 2 I i 4 s + 2 s + 2 s + 2 s + 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. 8.57.

2s

Fig. 8.56

8.8 Restrictions on Pole and Zero Locations for Transfer Functions jw

s

−4 −3

−2

0

−1

Fig. 8.57

Example 8.30

Draw the pole-zero diagram of I 2 for the network shown in Fig. 8.58. I1 I2

10 H I1

250 µF 200 Ω

Fig. 8.58 Solution The transformed network is shown in Fig. 8.59. I2

10s 1

I1

250 × 10−6s 200

Fig. 8.59 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 + 1 10 0 − 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. 8.60. jw j17.32

−10

0

− 17.32 −j

Fig. 8.60

s

8.27

8.28 Circuit Theory and Transmission Lines

Example 8.31

For the network shown in Fig. 8.61, draw pole-zero plot of I1

Vc . V1

1Ω

+

V1

− +

+ −

2Vc

1 F 2

1H

5I1

Vc

−

Fig. 8.61 Solution The transformed network is shown in Fig. 8.62. I1

1

c

+

V1

− +

+ −

2Vc

2 s

s

5I1

Vc

−

Fig. 8.62 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 10 Vc + + Vc s 2  20 s + 2 + s 2  Vc   2s   5 I1 +

=0

=0 =0 = −5V1

Vc 10 s 10 s =− 2 =− V1 ( s + 0.1)( s + 19.9) s + 20 s + 2

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. 8.63. s

−20

−2

− 1 − 0 .1

Fig. 8.63

0

8.8 Restrictions on Pole and Zero Locations for Transfer Functions

Example 8.32

8.29

Find the driving point admittance function and draw pole-zero plot for the network

shown in Fig. 8.64. I1

2

2 +

V1

− +

+ −

0.1V2

10I1

0.5s

1

V2

−

Fig. 8.64 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 2  10 I1 +  + 1 V2 = 0 s   2 + s 10 I1 +  V2 = 0  s   s + 2   V2 = −10 I1 s  V2 = −

10 s I1 s+2

Substituting Eq. (ii) in Eq. (i),  10 s  V1 + 0.1  − I  s + 2  1  10 s  I1 = = 0.5V1 + 0.05  − I1  s + 2  2 0.5s 5s   I1 1 + = 0.5V1  s + 2  Hence,

Y11 =

I1 = V1

0.5 0.5 ( s + 2) 0.5s + 1 = = 0.5s s + 0.5s + 2 1.5s + 2 1+ s+2

...(ii)

8.30 Circuit Theory and Transmission Lines The function has zero at s = −2 and pole at s = −1.33. The pole-zero diagram is shown in Fig 8.65. jw

s

− 3 − 2 −1.33 − 1

0

Fig. 8.65

Example 8.33

For the network shown in Fig. 8.66, determine

V2 V . Plot the pole-zero diagram of 2 . Ig Ig I2 = 0

1H

+

Ig

1F

1Ω

1F

V2 −

Fig. 8.66 Solution The transformed network is shown in Fig. 8.67. s

Va

Ib Vb

I2 = 0

Vc

+ Ig

1

1 s

1 s

1

V2 −

Fig. 8.67 Vc = Vb = V2 Ib =

Vb Vc + = sV2 + V2 = ( s + 1)V2 1 1 s

Va = s I b + Vb = s ( s + 1)V2 + V2 = ( s 2 + s + 1)V2 Ig =

Hence,

Va Va + + I b = ( s 2 + s + 1)V2 + s ( s 2 + s + 1)V2 + ( s + 1)V2 = ( s3 + 2 s 2 + 3s + 2)V2 1 1 s 1

V2 = I g s3 + 2 s 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. 8.68.

8.8 Restrictions on Pole and Zero Locations for Transfer Functions

8.31

jw j1.32

−2

−1

s

0

− 0.5

−j1.32

Fig. 8.68

Example 8.34

For the transfer function H(s) =

network shown in Fig. 8.69. 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. 8.69 Solution The transformed network is shown in Fig. 8.70. Ls + Vi

1 Cs

+ −

R

V0 −

Fig. 8.70 Simplifying the network as shown in Fig. 8.71, Ls + Vi

+ −

Z (s)

V0 −

Fig. 8.71 1 R Cs Z ( s) = = 1 RCss + 1 R+ Cs R×

8.32 Circuit Theory and Transmission Lines R RC + 1 V0 = Vi RCs R Ls + RC + 1 RCs 1 V0 R LC = = 1 1 Vi RLC s 2 + Ls + R s2 + s+ RC LC V0 10 = Vi s 2 + 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 8.35

Obtain the impedance function Z(s) for which pole-zero diagram is shown in Fig. 8.72. jw Z (∞ ) = 1 s

−3

−2

−1

0

Fig. 8.72 Solution The function Z(s) has poles at s = –1 and s = –3 and zeros at s = 0 and s = –2. s ( s + 2) Z ( s) = H =H ( s + 1)( s + 3) For s = ∞, Z (∞) = H When

1 =H (1)( )(1)

Z(∞) = 1, H=1 Z ( s) =

s ( s + 2) ( s + 1)( s + 3)

 2 s 2 1 +   s  1  3 s 2 1 +  1 +   s  s

8.8 Restrictions on Pole and Zero Locations for Transfer Functions

Example 8.36

8.33

Obtain the admittance function Y(s) for which the pole-zero diagram is shown in

Fig. 8.73. jw

Y( ∞ ) = 1

j1

s

−2

0

−1

−j 1

Fig. 8.73 Solution The function Y(s) has poles at s = –1 + j1 and s = −1 −j1 and zeros at s = 0 and s = −2.  2 s 2 1 +   s ( s + 2) s ( s + 2) s ( s + 2) s Y ( s) = H =H =H 2 =H 2 2 ( s + 1 + j1)( s + 1 − j1)  2 2 ( s + 1) − ( j1) s + 2s + 2 s 2 1 + + 2   s s  For s = ∞, Y (∞) = H When

(1) =H (1)

Y (∞) = 1, H=1 s ( s + 2) Y ( s) = 2 s + 2s + 2

Example 8.37

A network and its pole-zero configuration are shown in Fig. 8.74. Determine the values of R, L and C if Z (j0) = 1. jw j

√111

R 1 Cs

Z (s) Ls

−3

− 1.5

2 s

0

−j

√111 2

Fig. 8.74 1 R 1  s +  Ls + R C L Cs = Z ( s) = = 2 1 R LCs + R LCs RCs Cs + 1 s 2 + s + 1 ( Ls + R) + Cs L LC ( Ls + R)

Solution

...(i)

8.34 Circuit Theory and Transmission Lines From the pole-zero diagram, zero is at s = –3 and poles are at s = −1.5 + j

111 111 and s = − 1.5 − j 2 2

s+3  111   111   s + 1.5 + j 2   s + 1.5 − j 2     s+3 s+3 Z ( s) = H =H 2 2 s + 3s + 30  111  ( s + 1.5)2 −  j 2    Z ( s) = H

When

Z (j0) = 1,  3 1= H    30  H = 10 10 ( s + 3) Z ( s) = 2 s + 3s + 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 8.38 A network is shown in Fig. 8.75. 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. 8.75

8.8 Restrictions on Pole and Zero Locations for Transfer Functions

1 R 1 s+   Ls + R C L Cs = Z ( s) = = 2 1 R LCs + R LCs RCs Cs + 1 s 2 + s + 1 Ls + R + Cs L LC L C ( Ls + R)

Solution

8.35

...(i)

1 3 The poles are at − ± j and zero is at –1. 2 2 Z ( s) = H

s +1 s +1 s +1 =H =H 2 2 2  s + s +1 1 3  1 3  1 3  s+  − j s+ 2 + j 2  s+ 2 − j 2        2  2 

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 8.39

The pole-zero diagram of the driving-point impedance function of the network of Fig. 8.76 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. 8.76 1 R 1 s+   Ls + R C L Cs = Z ( s) = = 2 1 R LCs + R LCs RCs Cs + 1 s 2 + s + 1 Ls + R + Cs L LC L C ( 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)

8.36 Circuit Theory and Transmission Lines s+2 s+2 s+2 =H =H 2 2 2 ( s + 1 + j 4)( s + 1 − j 4) ( s + 1) − ( j 4) s + 2 s + 17 w = 0, Z ( j0) = 2

Z ( s) = H At dc, i.e.,

2= H H = 17 Z ( s) = 17

2 17 s+2

...(ii)

2

s + 2 s + 117

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 8.40

The network shown in Fig. 8.77 has the driving-point admittance Y (s) of the form Y( = H Y(s)

(s − s1 )(s − s2 ) (s − s3 )

(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. Ls Y (s)

1 Cs R

Fig. 8.77 Solution 1 ( Ls + R) C Css + 1 LCs LCs 2 + R RCs Cs + 1 Y ( s) = Cs + = = = Ls + R Ls + R Ls + R

(a)

But

Y ( s) =

H ( s − s1 )( s − s2 ) ( s − s3 )

R 1   C  s2 + s +   L LC  R s+ L

...(i)

8.8 Restrictions on Pole and Zero Locations for Transfer Functions

8.37

2

− s1 , s2 =

where

R 4  R ±   − 2  L L LC L R 1  R =− ±   −  2L  2 2L LC L

R L s1 = −10 + j104 s2 = − 10 − j104

s3 = − (b) When

Y ( s) = H

( s + 10 − j10 4 )( )( s + 10 + j10 4 ) s 2 + 20 s + 108 =H s − s3 s − s3

...(ii)

Comparing Eq. (ii) with Eq. (i), R = 20 L s3 = −20 Y ( s) = H

( s 2 + 20 s + 108 ) ( s + 20)

At s = j0, (108 ) = 10 −1 20 H = 0.02 × 1 10 −6

Y ( j 0) = H

Y ( s) = 0.02 × 1 10 −6

( s 2 + 20 s + 108 ) ( s + 20)

...(iii)

Comparing Eq. (iii) with Eq. (i), C = 0.02 02 × 10 −6 F = 0.02 µF 1 = 108 LC 1 L= H 2 R = 20 L R = 10 Ω

Example 8.41 A network and pole-zero diagram for driving-point impedance Z(s) are shown in Fig. 8.78. 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. 8.78

−3

−2

−1

0

s

− j1 − j2 − j3

8.38 Circuit Theory and Transmission Lines Solution

It is easier to calculate Y(s) and then invert it to obtain Z(s). 1 (G + C Css)( Ls Ls + R) + 1 LC LCss 2 + (G GL L + RC ) s + 1 + GR = = Ls + R Ls + R Ls + R 1 R s+   1 Ls + R C L Z ( s) = = = 2 G R Y ( s) LC    1 + GR  LCss + (GL + RC ) s + 1 + GR s 2 +  +  s +   C L LC 

Y ( s) = G + C Cs +

...(i)

From the pole-zero diagram, zero is at s = –2 and poles are at s = –3 ± j3. ( s + 2) ( s + 2) s+2 =H =H 2 2 2 ( s + 3 − j 3)( s + 3 + j 3) ( s + 3) − ( j 3) s + 6 s + 18

Z ( s) = H When

Z (j0) = 1, 2 18

1= H H =9 Z ( s) =

9 ( s + 2) 2

(s + 6s + 1 18 8)

...(ii)

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 8.42

A series R-L-C circuit has its driving-point admittance and pole-zero diagram is shown in Fig. 8.79. Find the values of R, L and C.

8.9 Time-Domain Behaviour from the Pole-Zero Plot 8.39 jw

j 25 Scale factor = 1 s

−1

0

−j 25

Fig. 8.79 Solution The function Y (s) has poles at s = –1 + j25 and s = –1 – j25 and zero at s = 0. Y ( s) = H

s s s =H =H 2 ( s + 1 + j 25)( )( s + 1 − j 25) ( s + 1) 2 − ( j 25) 2 s + 2 s + 626

H=1

Scale factor

Y ( s) =

s

...(i)

2

s + 2 s + 626

For a series RLC circuit, 1 LCss 2 + R LC RCs Cs + 1 Z ( s) = R + L Ls + = = Cs Cs 1 s Y ( s) = = 1  Z ( s)  2 R Ls + s +   L LC 

R 1   L  s2 + s +   L LC  s ...(ii)

Comparing Eq. (i) with Eq. (ii), L =1H 1 = 626 LC 1 C= F 626 R =2 L R=2Ω

8.9

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 − z1 )( )( s − z2 )… ( s − zn ) ( s − p1 )( )( s − p2 )… ( 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 8.80 shows some pole locations and the corresponding time-domain response.

8.40 Circuit Theory and Transmission Lines (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. 8.80) jw

f (t)

s

0 t

0

Fig. 8.80 Pole at origin (ii)

When pole lies in the left half of the s-plane, the response decreases exponentially. (Fig. 8.81) jw

0

f (t)

s

0

t

Fig. 8.81 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. 8.82) jw

0

f (t)

s

0

t

Fig. 8.82 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. 8.83) f (t) jw

0

s

0

Fig. 8.83 Poles on jw -axis

t

8.9 Time-Domain Behaviour from the Pole-Zero Plot 8.41

(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. 8.84). f (t)

jw

s

0

t

0

(a)

f (t)

jw

s 0

t

0

(b)

Fig. 8.84 (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. 8.85). f (t) jw 0

t

s 0

f (t)

0

(a)

Fig. 8.85

Nature of response with different positions of poles

t

8.42 Circuit Theory and Transmission Lines f (t)

jw t

0 s

0

f (t)

0

t

(b)

Fig.8.85

(Continued)

8.9.1 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.

8.10

GRAPHICAL METHOD FOR DETERMINATION OF RESIDUE

Consider a network function, F ( s) = H

( s − z1 )( )( s − z2 ) ( s − zn ) ( s − p1 )( )( s − p2 ) ( s − pm )

By partial fraction expansion, F ( s) =

K1 K2 Km + ++ ( s − p1 ) ( s − p2 ) ( s − pm )

The residue Ki is given by Ki = ( s − pi ) F ( s) |s→ p = H i

( pi − z1 )( pi − z2 ) ( pi − zn ) ( pi − p1 )( pi − p2 ) ( pi − 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 asors (polar form) from other poles to pi

8.10 Graphical Method for Determination of Residue 8.43

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) I(s = . Plot the pole-zero pat(s + 1)(s + 2) tern in the s-plane and hence obtain i(t).

Example 8.43

Solution Poles are at s = –1 and s = –2 and zero is at s = 0. The pole-zero plot is shown in Fig. 8.86. By partial-fraction expansion, I ( s) =

K1 K + 2 s +1 s + 2 jw

s 0

−2 −1

Fig. 8.86 The coefficients K1 and K2, often referred as residues, can be evaluated from he pole-zero diagram. From Fig. 8.87, Phasor from zero at origin to pole at A  1∠180°  K1 = H = 2 = 2 ∠180° = −2  1∠0°  Phasor from pole att B to pole aatt A jw

B

A s

−2 −1

0

Fig. 8.87 From Fig. 8.88, K2 = H

 2 ∠180°  Phasor from zero at origin to pole at B = 2 =4 Phasor from pole att A to pole aatt B  1 ∠180° 

8.44 Circuit Theory and Transmission Lines jw

B

A s 0

−2 −1

Fig. 8.88 2 4 + s +1 s + 2 Taking inverse Laplace transform, i (t) = –2e−t + 4e−2t I ( s) = −

Example 8.44

The voltage V(s) of a network is given by V ( s) =

3s ( s + 2))(( s 2 + 2 s + 2)

Plot its pole-zero diagram and hence obtain v (t). Solution

V ( s) =

3s ( s + 2))(( s 2 + 2 s + 2)

=

3s ( s + 2))(( s + 1 + j1)( s + 1 − j1)

Poles are at s = –2 and s = –1 ± j1 and zero is at s = 0 as shown in Fig. 8.89. jw B

j1

A s

−2

0

−1

−1 −j C

Fig. 8.89 By partial-fraction expansion, V ( s) =

K1 K2 K 2* + + s + 2 s + 1 − jj1 1 s + 1 + j1 j

The coefficients K1, K2 and K2* can be evaluated from the pole-zero diagram. From Fig. 8.90, K1 =

  2∠180° = 3  = 3 180° = −3 ( BA) (CA)  ( 2∠ − 135°) ( 2∠135°)  3 (OA)

8.10 Graphical Method for Determination of Residue 8.45 jw B

j1 √2

135°

A

180° s

−2

0

−1 135° √2

−j 1 C

Fig. 8.90 From Fig. 8.91,   3 ( 2∠135°) = 3 = ( AB ) (CB) 45°) ( 2∠90°)  2  ( 2∠45 3 K 2* = 2 K2 =

3 (OB)

jw B

j1

√2

√2

A

45°

−2

−1

135° s 0 90°

−1 −j C

Fig. 8.91 3 3 3 2 2 V ( s) = − + + ( s + 2) ( s + 1 − j1) ( s + 1 + j1) Taking inverse Laplace transform, v(t ) = −3e −2t +

Example 8.45

3 ( −1+ j1 3 − t  e j1 + e − j1  j1))tt ( −1− j1)t )t  −2 t e + e = − 3 e + 2 × e  = −3e −2t + 3e − t cos t    2 2 2  

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.

8.46 Circuit Theory and Transmission Lines V ( s) =

s 2 + 8 s + 12 2

s + 6s + 5

= 1+

2s + 7 2

s + 6s + 5

= 1+

2( s + 3.5 . ) ( s + 1)( s + 5)

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. 8.92 and Fig. 8.93. jw

jw

s

s

0

−5

− 3.5

0

−1

−5

−3.5

Fig. 8.92

−1

Fig. 8.93

From Fig. 8.92  2.5∠0°  5 K1 = 2  =  4 ∠0°  4 From Fig. 8.93  1.5∠180°  3 K2 = 2  =  4 ∠180°  4 5 3 V ( s) = 1 + 4 + 4 s +1 s + 5 Taking inverse Laplace transform, 5 3 v (t ) = δ (t ) + e − t + e −5t 4 4

Example 8.46

The pole-zero plot of the driving-point impedance of a network is shown in

Fig. 8.94. Find the time-domain response. jw

j1

Scale factor = 5

s

−1

0

−1 −j

Fig. 8.94 Solution The function Z(s) has poles at s = −1 + j1 and s = −1 −j1 and zero at s = 0.

8.10 Graphical Method for Determination of Residue 8.47

Z ( s) = H

s ( s + 1 + j1)( s + 1 − j1)

H=5

Scale factor

Z ( s) =

jw

5s ( s + 1 + j1)( s + 1 − j1)

j1

A

By partial fraction expansion, Z ( s) =

K1 K1* + s + 1 + j1 s + 1 − j1

s 0

−1

−j1

B

The coefficients K1 and K1* can be evaluated from the pole-zero

Fig. 8.95

diagram. From Fig. 8.95, K1 =

5 (OA) ( BA)

=

5 ( 2∠135°) = 3.5544 ∠45 45 ° 2∠90 90°

K1* = 3.54 ∠ − 45° 3.54 ∠45° 3.54 ∠ − 45° 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 8.47 pole-zero plot at s = j2. Solution

F ( s) =

4s 2

s + 2s + 2

=

4s 2

s + 2s 2s + 2

from the

4s ( s + 1 + j1)( s + 1 − j1)

The pole-zero plot is shown in Fig. 8.96. At s = j2, jw j2 j1 s

−1

0

−j 1

Fig. 8.96 | F ( j 2) | =

Product of phasor magnitudes from all zero too j 2 2 = = 0.447 Product of phaso hasor magnitudes from all poles too j 2 ( 2 ) ( 10 10 )

8.48 Circuit Theory and Transmission Lines  2  3  1 φ (ω ) = tann −1   − ttan an −1   − tann −1   = 90° − 71.56 .56° − 4455° = −26.56°  0  1  1

Example 8.48

Solution

Using the pole-zero plot, find magnitude and phase of the function F ( s) =

( s + 1)( s + 3) at s = j 4. s ( s + 2)

F ( s) =

( s + 1)( s + 3) s ( s + 2)

The pole-zero plot is shown in Fig. 8.97. At s = j4, jw j4

s 0

−3

−2

−1

Fig. 8.97 F ( j 4)=

Product of phasor magnitudes from all zeros to j 4 (5) ( 17 ) = = 1.15 product of p phaso hasor magnitudes from all poles to j 4 ( 20 ) ( 4)

 4  4  4  4 φ (ω ) = tann −1   + ttan an −1   − tan n1   − ttan an −1   = 75.96° + 53.13° − 90° − 63.43° = −24.34°  1  3  0  2

Example 8.49

Plot amplitude and phase response for F(s) =

s s+1 10

Solution F ( jω ) = F ( jω )=

jω jω + 10 ω

ω 2 + 100

8.10 Graphical Method for Determination of Residue 8.49

v

ÃF(jv)Ã

0

0

10

0.707

100

0.995

1000

1

F(jw))

1

The amplitude response is shown in Fig. 8.98.

0.7

w 0

10

100

1000

Fig. 8.98

ω ω ω φ (ω ) = tann −1   − ttan an −1   = 90° − tan −1    0  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. 8.99

100

1000

Fig. 8.99

Example 8.50

Sketch amplitude and phase response for F(s) =

Solution

s + 110 s − 110 F(jw))

jω + 10 F ( jω ) = 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. 8.100. ω  ω ω φ (ω ) = tann −1   − ttan an −1  −  = 2 tan −1    10   10   10  The phase response is shown in Fig. 8.101. v

e(v)

0

0°

10

90°

100

168.6°

1000

178.9°

Fig. 8.100

f (w) 180° 90°

w 0

10

100

1000

Fig. 8.101

8.50 Circuit Theory and Transmission Lines

Exercises Determine the driving-point impedance

8.1

transfer impedance ratio I1

V1 , I1

5Ω

2H

V1 1′

2F

1 F 2

2H

V1

2

V2 −

2′

Fig. 8.104 1 1   , G21 = 4  Z 21 = 3  2 s + 3s 4 s + 7 s 2 + 1 

V2

For the two-port network shown in Fig. 8.105, determine Z11, Z21 and voltage transfer ratio G21(s).

8.4

−

−

2F

−

+

+

I2 = 0 +

I2 = 0

2Ω

1H

+

V2 and voltage transfer I1

V2 for the network shown in Fig. 8.102. V1

I1

1

Fig. 8.102 I1

V1 7 s 2 + 7 s + 5 V2 2s ; = 2 ;  = I s 2 + s + 1 I s + s +1 1  1 V2 2s  = 2 V1 7 s + 7 s + 5 

2Ω

2H

+

+ 1F

V1 −

1H

V2 −

Fig. 8.105 8.2

For the network shown in Fig. 8.103, V V determine 2 and d 2. V1 I1

 2 s 3 + 4 s 2 + 3s + 2 s , Z 21 = 2 ,  Z11 = 2 s + 2s + 1 s + 2s + 1  s  G21 = 3  2 2 s + 4 s + 3s + 2 

2H I1 + V1

I2 = 0

1 F 2 1 F 2

+ 1 F 2

8.5

Draw the pole-zero diagram of the following network functions:

V2

(i) F ( s) = −

−

8.3

Fig. 8.103

(ii) F ( s) =

V2 s 2 + 1 V2 2s2 + 2  ; =  = 2  2  V1 2 s + 1 I1 s (3s + 2) 

(iii) F ( s) =

Find the open-circuit transfer impedance Z21 and open-circuit voltage ratio G21 for the ladder network shown in Fig. 8.104.

(iv) F ( s) =

s2 + 4 s2 + 6s + 4 5 s − 12 2

s + 4 s + 13 s +1 2

( s + 2 s + 2) 2 s ( s 2 + 5) s4 + 2s2 + 1

Exercises 8.51

(v) F ( s) = (vi) F ( s) = (vii) F ( s) =

1F

s2 + s + 2 s 4 + 5s3 + 6 s 2 Z (s)

s2 − s 3

1H

8.6

s + 2s − s − 2 s 2 + 3s + 2

Fig. 8.108

s 2 + 3s

  1.5s ( s 2 + 0.33)  Z ( s) = 2  2 ( s + 1.707 .707)( s + 0..293)  

( s + 4))(( s + 1) 2

( s + 1)( )( s 2 + 2 s + 5)

For the network shown in Fig. 8.106, draw the pole-zero plot of the impedance function Z(s).

V2 V and 1 for the V1 I1 network shown in Fig. 8.109 and plot poles

Find network functions

8.9

2H

and zeros of 1 F 20

Z (s)

I1

4Ω

Fig. 8.106

2H

+

For the network shown in Fig. 8.107, draw the pole-zero plot of driving-point impedance function Z(s).

Z (s)

10 Ω

10 F

1F

+ V2 −

1Ω

Fig. 8.109 V2 1 V1 2( s3 + 2 s 2 + 2 s + 1)  = ,  =  3 2 2 s3 + 2 s 2 + 2 s + 1   V1 2( s + 2 s + 2 s + 1) I1 8.10 For the network shown in Fig. 8.110, V V determine 1 and 2 . Plot the poles and I1 I1 zeros of

5F

I1

Fig. 8.107  5 ( s + 0.01)( s + 0.04)   Z ( s) =  s ( s + 0.03)   8.8

1F

−

1.94)( )( s + 2.5 + j1.94)   ( s + 2.5 − jj1 Z ( s) =   s+5  

5Ω

V2 ( s) . V1 ( s)

1Ω

V1

8.7

1H

2

2

(viii) F ( s) =

2F

Find the driving-point impedance of the network shown in Fig. 8.108. Also, find poles and zeros.

V2 . I1 2H

1H

+ V1

+ 1F

−

1 V F 2 2 −

Fig. 8.110 V1 2 s 4 + 5s 2 + 2 V2 2  , = 3  =  3 I1 2 s + 3s  2 s + 3s  I1

8.52 Circuit Theory and Transmission Lines 8.11 For the network shown in Fig. 8.111, determine V1 V V and 2 . Plot the pole and zeros for 2 . I1 V1 V1 I1

8.14 Obtain the impedance function for which the pole-zero diagram is shown in Fig. 8.114. jw

2F

1F

j1 +

+

Z ( j 0) = 1 s

1H

V1

1H

V2

−

−2

0

−1

−j1

−

Fig. 8.111

Fig. 8.114 V1 s 4 + 3s 2 + 1  =  2 s3 + s   I1

8.12 For the network shown in Fig. 8.112, plot the poles and zeros of transfer impedance function. I1

1H

1H

+

+ V1

1Ω

1Ω

V2

−

2 ( s + 1)    Z ( s) = 2  s + 2s + 2   8.15 For the network shown in Fig. 8.115, 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.

−

Fig. 8.112

Ls

1  V2  I = s + 2  1 

Fig. 8.115

8.13 For the network shown in Fig. 8.113, determine V1 V and d 2 . Plot the poles and zeros of transfer I1 V1 impedance function. I1

2H

4H +

+ V1

2  1 Ω, 0.5 Η, 17 

 F 

8.16 For the two-port network shown in Fig. 8.116, V 0.2 find R1, R2 and C. 2 = 2 V1 s + 3s + 2

1 F V2

2F

R1

1H −

−

Fig. 8.113 V1 16 s + 10 s + 1 V2 1 , = 3 ,  = 3 I1 8s + 3s 8s + 3s  I1 V2 1  = 4 2 V1 16 s + 10 s + 1 4

R

1 Cs

Z (s)

+

+ V1

C

R2

V2

2

−

−

Fig. 8.116 1 3  Ω, 0.5 F  5 Ω, 15 Ω,  

Objective-Type Questions 8.53

8.17 For the given network function, draw the pole-zero diagram and hence obtain the time domain voltage. V ( s) =

5 ( s + 5) ( s + 2)( s + 7) [v(t) = 3e−2t + 2e−7t]

8.18 A

transfer

function is given by 10 s Y ( s) = . Find time( s + 5 + j15)( )( s + 5 − j15) domain response using graphical method.

15) t 5.26 18.4° e − ( 5+ j 15 + 5.26 26 ∠ − 1188.4° e − ( 5 − j 15)t   26 ∠18

Objective-Type Questions 8.1

Of the four networks N1, N2, N3 and N4 of Fig. 8.117, the networks having identical drivingpoint functions are (a) N1 and N2 (b) N2 and N4 (c) N1 and N3 (d) N1 and N4 2H

8.2

The driving-point impedance Z(s) of a network has the pole-zero locations as shown in Fig. 8.118. If Z(0) = 3, then Z(s) is jw j1

1Ω 1F

2Ω

s

−3

1F

0

−1

−j 1 N1 2H

Fig. 8.118 1Ω

(a)

3 ( s + 3)

(b)

2

s + 2s + 3

2 ( s + 3) 2

s + 2s + 2

1F

2Ω

(c)

N2 1Ω 1Ω

1H 1F

8.3

3 ( s − 3) s − 2s − 2

1Ω

2 ( s − 3) 2

s − 2s − 3

For the circuit shown in Fig. 8.119, the initial conditions are zero. Its transfer function V ( s) H ( s) = 0 is V1 ( s) 10 kΩ

N3

(d)

2

10 mH

+ 2H

+ 100 µF

vi (t) −

−

1F

Fig. 8.119

N4

Fig. 8.117

v0

(a)

1 s 2 + 106 s + 106

(b)

106 s 2 + 103 s + 106

8.54 Circuit Theory and Transmission Lines 103

(c)

s 2 + 103 s + 106

8.7

106

(d)

s 2 + 106 s + 106

In Fig. 8.120, assume that all the capacitors are initially uncharged. If vi(t) = 10 u(t), then v0(t) is given by

8.4

A network has response with time as shown in Fig. 8.122. Which one of the following diagrams represents the location of the poles of this network? x (t)

1 kΩ

+ 4 µF

vi (t)

4 kΩ

1 µF

v0 (t)

Fig. 8.122

−

−

jw

Fig. 8.120 (a) 8 e–0.004 t

(b) 8 (1–e–0.004 t)

(c) 8 u(t)

(d) 8

A system is represented by the transfer 10 function . The dc gain of this ( s + 1 ))(( s + 2) system is

8.5

(a)

jw

s

0

(b)

0 jw

jw (c)

(b) 2

(c) 5

(d) 10

s

0

(d)

0

8.8

s

V24 V13

The transfer function of a low-pass RC network is (a) (RCs) (1 + RCs)

(b)

1 1+ RCss

(d)

s 1+ RCss

of the network shown in Fig. 8.121. (c) 1Ω 1

2 1Ω

s

Fig. 8.123

(a) 1

Which one of the following is the ratio

8.6

t

0

+

1Ω

3

8.9

RCss 1+ RCss

The driving-point admittance function of the network shown in Fig. 8.124 has a

4

R

L

1Ω

Fig. 8.121 (a)

1 3

(b)

2 3

(c)

3 4

(d)

4 3

Fig. 8.124 (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 8.55

I 2 ( s) for the V1 ( s) network shown in Fig. 8.125 is

8.10 The transfer function Y12 ( s) =

(a)

(c) 1Ω

I2(s )

s2

(b)

2

s + s +1 1 s +1

(d)

s s +1 s +1 s2 + 1

8.11 As the poles of a network shift away from the x axis, the response V1(s)

1H

(a) (b) (c) (d)

1F

Fig. 8.125

remains constant becomes less oscillating becomes more oscillating none of these

Answers to Objective-Type Questions 8.1. (c)

8.2. (b)

8.3. (d)

8.4. (c)

8.8. (b)

8.9. (a)

8.10. (a)

8.11. (c)

8.5. (c)

8.6. (a)

8.7. (d)

9 Network Synthesis

9.1

INTRODUCTION

In the study of electrical networks, broadly there are two topics: ‘Network Analysis’ and ‘Network Synthesis’. Any network consists of excitation, response and network function. In network analysis, network and excitation are given, whereas the response has to be determined. In network synthesis, excitation and response are given, and the network has to be determined. Thus, in network synthesis we are concerned with the realisation of a network for a given excitation-response characteristic. Also, there is one major difference between analysis and synthesis. In analysis, there is a unique solution to the problem. But in synthesis, the solution is not unique and many networks can be realised. The first step in synthesis procedure is to determine whether the network function can be realised as a physical passive network. There are two main considerations; causality and stability. By causality we mean that a voltage cannot appear at any port before a current is applied or vice-versa. In other words, the response of the network must be zero for t < 0. For the network to be stable, the network function cannot have poles in the right half of the s-plane. Similarly, a network function cannot have multiple poles on the jw axis.

9.2

HURwITz POlyNOmIals

A polynomial P(s) is said to be Hurwitz if the following conditions are satisfied: (a) P(s) is real when s is real. (b) The roots of P(s) have real parts which are zero or negative.

Properties of Hurwitz Polynomials 1. All the coefficients in the polynomial P ( s) = an s n + an −1s n −1 + … + a1s + a0 are positive. A polynomial may not have any missing terms between the highest and the lowest order unless all even or all odd terms are missing. 2. The roots of odd and even parts of the polynomial P(s) lie on the jw -axis only. 3. If the polynomial P(s) is either even or odd, the roots of polynomial P(s) lie on the jw -axis only. 4. All the quotients are positive in the continued fraction expansion of the ratio of odd to even parts or even to odd parts of the polynomial P(s).

9.2 Circuit Theory and Transmission Lines 5. If the polynomial P(s) is expressed as W (s) P1(s), then P(s) is Hurwitz if W (s) and P1(s) are Hurwitz. 6. If the ratio of the polynomial P(s) and its derivative P′(s) gives a continued fraction expansion with all positive coefficients then the polynomial P(s) is Hurwitz. This property helps in checking a polynomial for Hurwitz if the polynomial is an even or odd function because in such a case, it is not possible to obtain the continued fraction expansion.

Example 9.1

State for each case, whether the polynomial is Hurwitz or not. Give reasons in each case.

(a) s + 4s3 + 3s + 2 4

(b) s6 + 5s5 + 4s4 − 3s3 + 2 s 2 + s + 3 Solution (a) In the given polynomial, the term s2 is missing and it is neither an even nor an odd polynomial. Hence, it is not Hurwitz. (b) Polynomial s6 + 5s5 + 4 s 4 − 3s3 + 2 s 2 + s + 3 is not Hurwitz as it has a term (−3s3) which has a negative coefficient.

Example 9.2

Test whether the polynomial P ( s) = s4 + s3 + 5 s 2 + 3 s + 4 is Hurwitz.

Solution Even part of P ( s) = m( s) = s 4 + 5s 2 + 4 Odd part of P ( s) = n( s) = s3 + 3s m( s) n( s) By continued fraction expansion, Q ( s) =

s 3 + 3s ) s 4 + 5 s 2 + 4 ( s s 4 + 3s 2 1  2 s 2 + 4 s3 + 3s  s  2 s3 + 2 s s) 2 s 2 + 4 ( 2 s 2s2  1 4 s  s  4 s 0 Since all the quotient terms are positive, P(s) is Hurwitz.

Example 9.3

Test whether the polynomial P ( s) = s3 + 4 s 2 + 5 s + 2 is Hurwitz.

Solution Even part of P ( s) = m( s) = 4 s 2 + 2 Odd part of P ( s) n( s) = s3 + 5s

9.2 Hurwitz Polynomials 9.3

The continued fraction expansion can be obtained by dividing n(s) by m(s) as n(s) is of higher order than m(s). Q ( s) =

n( s) m( s)

1  4 s 2 + 2 s3 + 5s  s  4 s3 +

2 s 4 9  2 8 s 4 s + 2  s  9 2 4s2  9 9 2 s  s  2 4 9 s 2 0

Since all the quotient terms are positive, P(s) is Hurwitz.

Example 9.4

Test whether the polynomial P ( s) = s4 + s3 + 3 s 2 + 2 s + 12 is Hurwitz.

Solution Even part of P ( s) = m( s) = s 4 + 3s 2 + 12 Odd part of P ( s) = n( s) = s3 + 2 s Q ( s) =

m( s) n( s)

By continued fraction expansion,   s3 + 2 s s 4 + 3s 2 + 12  s   s4 + 2s2   s 2 + 12 s3 + 2 s  s   s3 + 12 s  1  −10 s s 2 + 12  − s   10 s2  10  12 − 10 s  − s   12 − 10 s 0 Since two quotient terms are negative, P(s) is not Hurwitz.

9.4 Circuit Theory and Transmission Lines

Example 9.5 Solution

Prove that polynomial P ( s) = s4 + s3 + 2 s 2 + 3 s + 2 is not Hurwitz.

Even part of P ( s) = m( s) = s 4 + 2 s 2 + 2 Odd part of P ( s) = n( s) = s3 + 3s Q ( s) =

m( s) n( s)

By continued fraction expansion, s 3 + 3s ) s 4 + 2 s 2 + 2 ( s s 4 + 3s 2 − s 2 + 2) s3 + 3s( − s s3 − 2 s  1  5 s − s 2 + 2  − s   5 − s2  5 2 5s  s  2 5s 0 Since two quotient terms are negative, P(s) is not Hurwitz.

Example 9.6 Solution

Prove that polynomial P ( s) = 2 s4 + 5 s3 + 6 s 2 + 3 s + 1 is Hurwitz.

Even part of P ( s) = m( s) = 2s 4 + 6 s 2 + 1 Odd part of P ( s) = n( s) = 5s3 + 3s Q ( s) =

m( s) n( s)

By continued fraction expansion, 2  5s3 + 3s 2 s 4 + 6 s 2 + 1  s  5 6 2s4 + s2 5 24 2  3  25 s + 1 5s + 3s  s   24 5 25 s 5s3 + 24

9.2 Hurwitz Polynomials 9.5

47  24 2  576 s s + 1 s  235 24  5 24 2 s 5  47  24 1 s s  24  47 47 s 24 0 Since all the quotient terms are positive, the polynomial P(s) is Hurwitz.

Example 9.7

Test whether the polynomial P ( s) = s4 + 7 s3 + 6 s 2 + 21s + 8 is Hurwitz.

Solution Even part of P ( s) = m( s) = s 4 + 6 s 2 + 8 Odd part of P ( s) = n( s) = 7 s3 + 21s m( s) n( s) By continued fraction expansion, Q ( s) =

1  7 s3 + 21s s 4 + 6 s 2 + 8  s  7 s 4 + 3s 2 7  3s 2 + 8 7 s3 + 21s  s  3 56 7 s3 + s 3 7  2 9 s  3s + 8  s 7 3  3s 2 7  7 8 s  s  3  24 7 s 3 0 Since all the quotient terms are positive, the polynomial P(s) is Hurwitz.

Example 9.8

Check whether P ( s) = s4 + 5 s3 + 5 s 2 + 4 s + 10 is Hurwitz.

Solution Even part of P ( s) = m( s) = s 4 + 5s 2 + 10 Odd part of P ( s) = n( s) = 5s3 + 4 s

9.6 Circuit Theory and Transmission Lines Q ( s) =

m( s) n( s)

By continued fraction expansion, 1  5s3 + 4 s s 4 + 5s 2 + 10  s  5 s4 +

4 2 s 5 21 2   25 s + 10 5s3 + 4 s  s   21 5 250 s 5s3 + 21 −

166  21 2  441 s s + 10  − s  830 21  5 21 2 s 5  166  166 10 − s − s  21  210 166 − s 21 0

Since the last two quotient terms are negative, the polynomial P(s) is not Hurwitz.

Example 9.9

Test whether the polynomial s5 + 3s3 + 2s is Hurwitz.

Solution Since the given polynomial contains odd functions only, it is not possible to perform a continued fraction expansion. d P ( s) = 5s 4 + 9 s 2 + 2 ds P ( s) Q ( s) = P ′( s)

P ′( s) =

By continued fraction expansion, 1  5s 4 + 9 s 2 + 2 s5 + 3s3 + 2 s  s  5 9 2 s5 + s3 + s 5 5 6 3 8  4  25 s + s 5 s + 9 s 2 + 2  s  6 5 5 

9.2 Hurwitz Polynomials 9.7

5s 4 +

20 2 s 3 7 2 8  18 6 s + 2 s3 + s  s 5 3 5  35 6 3 36 s + s 5 35 20  7 2  49 s s + 2  s  12 35  3 7 2 s 3  20  10 2 s s  35  35 20 s 35 0

Since all the quotient terms are positive, the polynomial P(s) is Hurwitz.

Example 9.10

Test whether the polynomial P(s) is Hurwitz. P ( s) = s5 + s 3 + s

Solution Since the given polynomial contains odd functions only, it is not possible to perform continued fraction expansion. d P ′( s) = P ( s ) = 5 s 4 + 3s 2 + 1 ds P ( s) Q ( s) = P ′( s) By continued fraction expansion, 1  5s 4 + 3s 2 + 1 s5 + s3 + s  s  5 3 1 s5 + s3 + s 5 5 2 3 4  4  25 s + s 5s + 3s 2 + 1  s   2 5 5 5s 4 + 10 s 2 4  2  2 − 7 s 2 + 1 s3 + s  − s  5 5  35 2 3 2 s − s 5 35

9.8 Circuit Theory and Transmission Lines 26   245 s − 7 s 2 + 1  − s  26 35  − 7s 2  26  26 1 s s  35  35 26 s 35 0 Since the third and fourth quotient terms are negative, P(s) is not Hurwitz.

Example 9.11

Test the polynomial P(s) of Hurwitz property. P(s) = s6 + 3s5 + 8s4 + 15s3 + 17s 2 + 12s + 4

Solution Even part of P ( s) = m( s) = s6 + 8s 4 + 17 s 2 + 4 Odd part of P ( s) = n( s) = 3s5 + 15s3 + 12 s m( s) n( s) By continued fraction expansion, Q ( s) =

1  3s5 + 15s3 + 12 s s6 + 8s 4 + 17 s 2 + 4  s  3 s6 + 5s 4 + 4 s 2   3s 4 + 13s 2 + 4 3s5 + 15s3 + 12 s  s   3s5 + 13s3 + 4 s 3  2 s3 + 8s 3s 4 + 13s 2 + 4  s  2 3s 4 + 12 s 2   s 2 + 4 2 s 3 + 8 s  2 s   2 s3 + 8s 0 The division has terminated abruptly (i.e., the number of partial quotients (that is four) is not equal to the order of polynomial (that is six) with common factor (s2 + 4). P ( s) = s6 + 3s5 + 8s 4 + 15s3 + 17 s 2 + 12 s + 4 = ( s 2 + 4)( s 4 + 3s3 + 4 s 2 + 3s + 1)

9.2 Hurwitz Polynomials 9.9

If both the factors are Hurwitz, P(s) will be Hurwitz. P1 ( s) = s 2 + 4

Let

Since it contains only even functions, we have to find the continued fraction expansion of

P1 ( s) . P1′( s)

P1′( s) = 2 s P1 ( s) s 2 + 4 s 2 4 s 1 = = + = + 2s 2s 2s 2 s P1′( s) 2 Since all the quotient terms are positive, P1(s) is Hurwitz. P2 ( s) = s 4 + 3s3 + 4 s 2 + 3s + 1

Now, let

m2 ( s) = s 4 + 4 s 2 + 1 n2 ( s) = 3s3 + 3s By continued fraction expansion, 1  3s3 + 3s s 4 + 4 s 2 + 1  s  3 s4 + s2 3s 2 + 1) 3s3 + 3s ( s 3s3 + s 3  2 s 3s 2 + 1  s  2 3s 2 1) 2 s ( 2 s 2s 0 Since all the quotient terms are positive, P2(s) is Hurwitz. Hence, P ( s) = ( s 2 + 4)( s 4 + 3s3 + 4 s 2 + 3s + 1) is Hurwitz.

Example 9.12

Test whether the polynomial P ( s) = s7 + 2 s6 + 2 s5 + s4 + 4 s3 + 8 s 2 + 8 s + 4 is Hur-

witz. Solution Even part of P ( s) = m( s) = 2s6 + s 4 + 8s 2 + 4 Odd part of P ( s) = n( s) = s7 + 2 s5 + 4 s3 + 8s Q ( s) =

n( s) m( s)

9.10 Circuit Theory and Transmission Lines By continued fraction expansion, 1  2 s 6 + s 4 + 8 s 2 + 4 s 7 + 2 s 5 + 4 s 3 + 8 s  s  2 s7 +

1 5 s + 4 s3 + 2 s 2 3 5 s 2

 4 + 6 s 2 s 6 + s 4 + 8 s 2 + 4  s  3 + 8s 2

2 s6 s4

3 3 + 4 s 5 + 6 s  s 2 2 3 5 s + 6s 2 0

Since the division has terminated abruptly it indicates a common factor s4 + 4. The polynomial can be written as P ( s) = ( s 4 + 4)( s3 + 2 s 2 + 2 s + 1) If both the factor are Hurwitz, P(s) will be Hurwitz. In the polynomial (s4 + 4), the terms s3, s2 and s are missing. Hence, it is not Hurwitz. Therefore, P(s) is not Hurwitz.

Example 9.13

Test whether the polynomial 2 s6 + s5 + 13 s4 + 6 s3 + 56 s 2 + 25 s + 25 is Hurwitz.

Solution Even part of P ( s) = m( s) = 2 s6 + 13s 4 + 56 s 2 + 25 Odd part of P ( s) = n( s) = s5 + 6 s3 + 25s Q ( s) =

m( s) n( s)

By continued fraction expansion, s5 + 6 s3 + 25s)2 s6 + 13s 4 + 56 s 2 + 25( 2 s 2 s6 + 12 s 4 + 50 s 2 s 4 + 6 s 2 + 25) s5 + 6 s3 + 25s( s s5 + 6 s3 + 25s 0 The division has terminated abruptly. P ( s) = 2 s6 + s5 + 13s 4 + 6 s3 + 56 s 2 + 25s + 25 = ( s 4 + 6 s 2 + 25) ( 2 s 2 + s + 1) Let

P1 ( s) = s 4 + 6 s 2 + 25

9.2 Hurwitz Polynomials 9.11

Since P1(s) contains only even functions, we have to find the continued fraction expansion of

P1 ( s) . P1′( s)

P1′( s) = 4 s3 + 12 s By continued fraction expansion, 1  4 s3 + 12 s s 4 + 6 s 2 + 25  s  4 s 4 + 3s 2 4  3s 2 + 25 4 s3 + 12 s  s  3 100 4 s3 + s 3 −

64  2  9 s 3s + 25  − s  64 3  3s 2  64  64 25 − s  − s  3  75 64 − s 3 0

Since two of the quotient terms are negative, P1(s) is not Hurwitz. We need not test the other factor (2s2 + s + 1) for being Hurwitz. Hence, P(s) is not Hurwitz. There is another method to test a Hurwitz polynomial. In this method, we construct the Routh–Hurwitz array for the required polynomial. Let

P ( s) = an s n + an −1s n −1 + an − 2 s n − 2 + … + a1s + a0

The Routh–Hurwitz array is given by, sn s n −1 sn− 2 sn−3 . . . . s1 s0

an an − 2 an −1 an − 3 bn bn −1 cn cn −1 . . . . . .

an − 4 … an − 5 … bn − 2 … …

9.12 Circuit Theory and Transmission Lines The coefficients of sn and sn − 1 rows are directly written from the given equation. where

bn =

an −1an − 2 − an an − 3 an −1

bn −1 =

an −1an − 4 − an an − 5 an −1

bn − 2 =

an −1an − 6 − an an − 7 an −1

cn =

bn an − 3 − an −1bn −1 bn

cn −1 =

bn an − 5 − an −1bn − 2 bn

Hence, for polynomial P(s) to be Hurwitz, there should not be any sign change in the first column of the array.

Example 9.14 Solution

Test whether P ( s) = s4 + 7 s3 + 6 s 2 + 21s + 8 is Hurwitz.

The Routh array is given by, s4 s3 s2 s1 s0

1 6 8 7 21 3 8 7 3 0 8

Since all the elements in the first column are positive, the polynomial P(s) is Hurwitz.

Example 9.15

Determine whether P ( s) = s4 + s3 + 2 s 2 + 3 s + 2 is Hurwitz.

Solution The Routh array is given by, s4 s3 s2 s1 s0

1 1 −1 5 2

2 3 2 0

2

Since there is a sign change in the first column of the array, the polynomial P(s) is not Hurwitz.

Example 9.16

Test whether P ( s) = s5 + 2 s4 + 4 s3 + 6 s 2 + 2 s + 5 is Hurwitz.

9.2 Hurwitz Polynomials 9.13

Solution The Routh array is given by, s5 s4 s3 s2 s1 s0

1 2 1 7 −1.21 5

4 2 6 5 −0.5 5

Since there is a sign change in the first column of the array, the polynomial is not Hurwitz.

Example 9.17

Test whether the polynomial P(s) = s5 + s3 + s is Hurwitz.

Solution The given polynomial contains odd functions only. P′(s) = 5s4 + 3s2 + 1

The Routh array is given by, s5 s4 s3 s2 s1 s0

1 5 0.4 −7 0.86 1

1 1 3 1 0.8 1

Since there is a sign change in the first column of the array, the polynomial is not Hurwitz.

Example 9.18

Test whether the polynomial P(s) = s8 + 5s6 + 2s4 + 3s2 + 1 is Hurwitz.

Solution The given polynomial contains even functions only.

P ′(s) = 8s7 + 30s5 + 8s3 + 6s

The Routh array is given by, s8 s7 s6 s5 s4 s3 s2 s1 s0

1 8 1.25 23.6 1.33 −46.6 1.75 8.49 1

5 2 30 8 1 2.25 −6.4 −0.4 2.27 1 0 −18.14 1

3 1 6 0 1 0

Since there is a sign change in the first column of the array, the polynomial is not Hurwitz.

Example 9.19

Test whether P(s) = s5 + 12s4 + 45s3 + 60s2 + 44s + 48 is Hurwitz.

9.14 Circuit Theory and Transmission Lines

Solution The Routh array is given by, s5 s4 s3 s2 s1 s0

1 12 40 48 0

45 44 60 48 40 48 0

Notes: When all the elements in any one row is zero, the following steps are followed: (i) Write an auxiliary equation with the help of the coefficients of the row just above the row of zeros. (ii) Differentiate the auxiliary equation and replace its coefficient in the row of zeros. (iii) Proceed for the Routh test. Auxiliary equation, A( s) = 48s 2 + 48 A′( s) = 96s s5 s4 s3 s2 s1 s0

1 12 40 48 96 48

45 44 60 48 40 48 0

Since there is no sign change in the first column of the array, the polynomial P(s) is Hurwitz.

Example 9.20

Check whether P(s) = 2s6 + s5 + 13s4 + 6s3 + 56s2 + 25s + 25 is Hurwitz.

Solution The Routh array is given by, s6 s5 s4 s3 s2 s1 s0

2 1 1 0

13 6 6 0

56 25 25 25 0

A(s) = s4 + 6s2 + 25 A′(s) = 4s3 + 12s

9.2 Hurwitz Polynomials 9.15

Now, the Routh array will be given by, s6 s5 s4 s3 s2 s1 s0

2 1 1 4 3 −21.3 25

13 56 25 6 25 6 25 12 25

Since there is a sign change in the first column of the array, the polynomial P(s) is not Hurwitz.

Example 9.21

Determine the range of values of ‘a’ so that P(s) = s4 + s3 + as2 + 2s + 3 is Hurwitz.

Solution The Routh array is given by, s4 s3 s2

1 1 a−2

s1

2a − 7 a−2

a 3 2 3

s0 3 For the polynomial to be Hurwitz, all the terms in the first column of the array should be positive, i.e., a−2>0 a>2 2a − 7 >0 and a−2 7 a> 2 7 Hence, P(s) will be Hurwitz when a > . 2

Example 9.22

Determine the range of values of k so that the polynomial P(s) = s3 + 3s2 + 2s + k is

Hurwitz.

Solution The Routh array is given by, s3 s2 s1 s0

1 3 6−k 3 k

2 k 0

9.16 Circuit Theory and Transmission Lines For the polynomial to be Hurwitz, all the terms in the first column of the array should be positive, 6−k >0 3 6−k >0

i.e.,

i.e., k < 6 and k > 0 Hence, P(s) will be Hurwitz for 0 < k < 6.

9.3

POsITIvE REal FUNCTIONs

A function F(s) is positive real if the following conditions are satisfied: (a) F(s) is real for real s. (b) The real part of F(s) is greater than or equal to zero when the real part of s is greater than or equal to zero, i.e., Re F(s) ≥ 0 for Re(s) ≥ 0

9.3.1

Properties of Positive Real Functions

1 is also positive real. F ( s) The sum of two positive real functions is positive real. The poles and zeros of a positive real function cannot have positive real parts, i.e., they cannot be in the right half of the s plane. Only simple poles with real positive residues can exist on the jw-axis. The poles and zeros of a positive real function are real or occur in conjugate pairs. The highest powers of the numerator and denominator polynomials may differ at most by unity. This condition prevents the possibility of multiple poles and zeros at s = ∞. The lowest powers of the denominator and numerator polynomials may differ by at most unity. Hence, a positive real function has neither multiple poles nor zeros at the origin.

1. If F(s) is positive real then 2. 3. 4. 5. 6. 7.

9.3.2

Necessary and Sufficient Conditions for Positive Real Functions

The necessary and sufficient conditions for a function with real coefficients F(s) to be positive real are the following: 1. F(s) must have no poles and zeros in the right half of the s-plane. 2. The poles of F(s) on the jw-axis must be simple and the residues evaluated at these poles must be real and positive. 3. Re F (jw) ≥ 0 for all w. Testing of the Above Conditions Condition (1) requires that we test the numerator and denominator of F(s) for roots in the right half of the s-plane, i.e., we must determine whether the numerator and denominator of F(s) are Hurwitz. This is done through a continued fraction expansion of the odd to even or even to odd parts of the numerator and denominator. Condition (2) is tested by making a partial-fraction expansion of F(s) and checking whether the residues of the poles on the jw-axis are positive and real. Thus, if F(s) has a pair of poles at s = ± jw0, a partial-fraction expansion gives terms of the form

9.3 Positive Real Functions 9.17

K1 K1* + s − jω 0 s + jω 0 Since residues of complex conjugate poles are themselves conjugate, K1 = K1* and should be positive and real. Condition (3) requires that Re F(jw) must be positive and real for all w. Now, to compute Re F(jw) from F(s), the numerator and denominator polynomials are separated into even and odd parts. F ( s) =

m1 ( s) + n1 ( s) m1 + n1 = m2 ( s) + n2 ( s) m2 + n2

Multiplying N(s) and D(s) by m2 − n2, F ( s) =

m1 + n1 m2 − n2 m1m2 − n1n2 m2 n1 − m1n2 = + m2 + n2 m2 − n2 m22 − n22 m22 − n22

But the product of two even functions or odd functions is itself an even function, while the product of an even and odd function is odd. Ev F ( s) = Od F ( s) =

m1m2 − n1n2 m22 − n22 m2 n1 − m1n2 m22 − n22

Now, substituting s = jw in the even polynomial gives the real part of F(s) and substituting s = jw into the odd polynomial gives imaginary part of F(s). Ev F ( s) s = jω = Re F ( jω ) Od F ( s) s = jω = j Im F ( jω ) We have to test Re F(jw) ≥ 0 for all w. The denominator of Re F(jw) is always a positive quantity because m22 − n22

s = jω

≥0

Hence, the condition that Ev F(jw) should be positive requires m1m2 − n1n2

s = jω

= A(ω 2 )

should be positive and real for all w ≥ 0.

Example 9.23

Test whether F ( s) =

s+3 is a positive real function. s +1

jw

Solution (a) F ( s) =

N ( s) s + 3 = D ( s) s + 1

The function F(s) has pole at s = −1 and zero at s = −3 as shown in Fig. 9.1. Thus, pole and zero are in the left half of the s-plane. (b) There is no pole on the jw axis. Hence, the residue test is not carried out.

s −3 −2 −1 0

Fig. 9.1

9.18 Circuit Theory and Transmission Lines (c) Even part of N ( s ) = m1 = 3 Odd part of N ( s ) = n1 = s Even part of D ( s ) = m2 = 1 Odd part of D ( s ) = n2 = s A(ω 2 ) = m1m2 − n1n2 |s = jω = (3)(1) − ( s)( s) |s = jω = 3 − s 2 |s = jω = 3 + ω 2 A(w2) is positive for all w ≥ 0. Since all the three conditions are satisfied, the function is positive real.

Example 9.24

Test whether F ( s) =

s2 + 6 s + 5 is positive real function. s 2 + 9 s + 14

Solution (a) F ( s) =

jw

N ( s) s 2 + 6 s + 5 ( s + 5)( s + 1) = 2 = D( s) s + 9 s + 14 ( s + 7)( s + 2)

The function F (s) has poles at s = −7 and s = −2 and zeros at s = −5 an s = −1 as shown in Fig. 9.2. Thus, all the poles and zeros are in the left half of the s plane. (b) Since there is no pole on the jw axis, the residue test is not carried out.

s 0 −7 −6 −5 −4 −3 −2

−1

Fig. 9.2

(c) Even part of N ( s) = m1 = s 2 + 5 Odd part of N ( s) = n1 = 6s Even part of D( s) = m2 = s 2 + 14 Odd part of D( s) = n2 = 9s A(ω 2 ) = m1m2 − n1n2 |s = jω = (s 2 + 5) (s 2 + 14) − (6s)(9s) |s = jω = s 4 − 35s 2 + 70 |s = jω = ω 4 + 35ω 2 + 70 A(w2) is positive for all w ≥ 0. Since all the three conditions are satisfied, the function is positive real.

Example 9.25

Test whether F ( s) =

s( s + 3)( s + 5 ) is positive real function. ( s + 1)( s + 4 )

Solution (a) F ( s) =

N ( s) s( s + 3)( s + 5) s3 + 8s 2 + 15s = = 2 D( s) ( s + 1)( s + 4) s + 5s + 4

9.3 Positive Real Functions 9.19

The function F(s) has poles at s = −1 and s = −4 and zeros at s = 0, s = −3 and s = −5 as shown in Fig. 9.3. Thus, all the poles and zeros are in the left half of the s plane. (b) There is no pole on the jw axis, hence the residue test is not carried out.

jw

s 0 −5

−4

−3

−2

−1

(c) Even part of N ( s) = m1 = 8s 2 Odd part of N ( s) = n1 = s3 + 15s

Fig. 9.3

Even part of D( s) = m2 = s 2 + 4 Odd part of D( s) = n2 = 5s A(ω 2 ) = m1m2 − n1n2 |s = jω = (8s 2 )( s 2 + 4) − ( s3 + 15s)(5s) |s = jω = 3s 4 − 43s 2 |s = jω = 3ω 4 + 43ω 2 A(w2) is positive for all w ≥ 0. Since all the three conditions are satisfied, the function is positive real.

Example 9.26

Test whether F ( s) =

s2 + 1 is positive real function. s3 + 4 s

Solution (a) F ( s) =

N ( s) s2 + 1 ( s + j1)( s − j1) = 3 = D( s) s + 4 s s( s + j 2)( s − j 2)

jw j2

The function F(s) has poles at s = 0, s = −j2 and s = j2 and zeros at s = −j1 and s = j1 as shown in Fig. 9.4. Thus, all the poles and zeros are on the jw axis. (b) The poles on the jw axis are simple. Hence, residue test is carried out. F ( s) =

s2 + 1 s3 + 4 s

=

s( s 2 + 4 )

The constants K1, K2 and K2* are called residues. s2 + 1 s +4 2

= s=0

K 2 = ( s + j 2) F ( s ) | s = − j 2 = K 2* = K 2 =

3 8

Thus, residues are real and positive.

−j1

Fig. 9.4

K1 K2 K 2* + + s s + j2 s − j2

K1 = s F ( s) |s = 0 =

0

−j2

s2 + 1

By partial-fraction expansion, F ( s) =

j1

1 4

s2 + 1 s ( s − j 2)

= s= − j 2

3 −4 + 1 = ( − j 2)( − j 2 − j 2) 8

s

9.20�Circuit Theory and Transmission Lines (c) Even part of N ( s) � m1 � s 2 � 1 Odd part of N ( s) � n1 � 0 Even part of D( s) � m2 � 0 Odd part of D( s) � n2 � s3 � 4s

� �

A � 2 � m1m2 � n1n2 |s � j� � ( s 2 � 1)(0) � (0)(s3 � 4s) |s � j� � 0 A(�2) is zero for all � � 0. Since all the three conditions are satisfied, the function is positive real.

�Example 9.27�

Test whether F ( s) �

2 s3 � 2 s 2 � 3s � 2 is positive real function. s2 � 1

Solution (a) F ( s) �

N ( s) 2 s3 � 2 s 2 � 3s � 2 2s3 � 2s 2 � 3s � 2 � � D ( s) ( s � j1)( s � j1) s2 � 1

Since numerator polynomial cannot be easily factorized, we will prove whether N(s) is Hurwitz. Even part of N ( s) � m( s) � 2s 2 � 2 Odd part of N ( s) � n( s) � 2s3 � 3s By continued fraction expansion, � � 2 s 2 � 2� 2 s3 � 3s � s � � 2 s3 � 2 s � � s� 2 s 2 � 2 � 2 s � � 2s2 �1 � 2 � s � s � �2 s 0 Since all the quotient terms are positive, N(s) is Hurwitz. This indicates that zeros are in the left half of the s plane. The function F(s) has poles at s = −j1 and s = j1. Thus, all the poles and zeros are in the left half of the s plane. (b) The poles on the j� axis are simple. Hence, residue test is carried out. F ( s) �

2 s3 � 2 s 2 � 3s � 2 s2 � 1

As the degree of the numerator is greater than that of the denominator, division is first carried out before partial-fraction expansion.

Chapter 09.indd 20

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9.3 �Positive Real Functions�9.21

� � s 2 � 1� 2 s3 � 2 s 2 � 3s � 2 � 2 s � 2 � � 2 s3

� 2s 2s2 � s � 2 2s2

�2 s

F ( s) � 2 s � 2 �

s s2 � 1

By partial-fraction expansion, F ( s) � 2 s � 2 �

K1 K* � 1 s � j1 s � j1

K1 � ( s � j1) F ( s) |s � � j1 � K1* � K1 �

� j1 1 � � j1 � j1 2

1 2

Thus, residues are real and positive. (c) Even part of N ( s) � m1 � 2s 2 � 2 Odd part of N ( s) � n1 � 2s3 � 3s Even part of D( s) � m2 � s 2 � 1 Odd part of D( s) � n2 � 0 A(� 2 ) � m1m2 � n1n2 |s � j� � ( 2s 2 � 2)( s 2 � 1) � ( 2s3 � 3s)(0) |s � j� � 2s 4 � 4s 2 � 2 |s � j� � 2(� 4 � 2� 2 � 1) � 2(� 2 � 1) 2 A(�2) � 0 for all � � 0. Since all the three conditions are satisfied, the function is positive real.

��������������

Test whether F ( s) �

s3 � 6 s 2 � 7 s � 3 is positive real function. s2 � 2s � 1

Solution (a) F ( s) �

N ( s) s3 � 6 s 2 � 7 s � 3 s3 � 6 s 2 � 7 s � 3 � � D ( s) ( s � 1)( s � 1) s2 � 2s � 1

Since a numerator polynomial cannot be easily factorized, we will test whether N(s) is Hurwitz. Even part of N � s � � m � s � � 6s 2 � 3 Odd part of N � s � � n � s � � s3 � 7s

Chapter 09.indd 21

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9.22 Circuit Theory and Transmission Lines By continued fraction expansion,  1 6 s 2 + 3 s3 + 7 s  s 6  s3 + 0.5s   6.5s 6 s 2 + 3  0.92 s   6s2   3 6.5s  2.17 s   6.5s 0 Since all the quotient terms are positive, N(s) is Hurwitz. This indicates that the zeros are in the left half of the s plane. The function F(s) has a double pole at s = −1. Thus, all the poles and zeros are in the left half of the s plane. (b) There is no pole on the jw axis. Hence, the residue test is not carried out. (c) Even part of N ( s) = m1 = 6s 2 + 3 Odd part of N ( s) = n1 = s3 + 7s Even part of D( s) = m2 = s 2 + 1 Odd part of D( s) = n2 = 2s A(ω 2 ) = m1m2 − n1n2 |s = jω = (6s 2 + 3)(s 2 + 1) − ( s3 + 7s)( 2s) |s = jω = 4s 4 − 5s 2 + 3 |s = jω = 4ω 4 + 5ω 2 + 3 A(w2) is positive for all w ≥ 0. Since all the three conditions are satisfied, the function is positive real. Test whether F ( s) =

Example 9.29

s2 + s + 6 is a positive real function. s2 + s + 1

Solution

(a) F ( s) =

 1 23   1 23  s+ 2 + j 2  s+ 2 − j 2    

N ( s) s + s + 6 = = D ( s) s 2 + s + 1  1 3  1 3 s+ 2 + j 2  s+ 2 − j 2     2

1 23 1 3 The function F(s) has zeros at s = − ± j and poles at s = − ± j . 2 2 2 2 (b) There is no pole on the jw axis. Hence, the residue test is not carried out. (c) Even part of N ( s) = m1 = s 2 + 6 Odd part of N ( s) = n1 = s Even part of D( s) = m2 = s 2 + 1 Odd part of D( s) = n2 = s

9.3 Positive Real Functions 9.23

A(ω 2 ) = m1m2 − n1n2 |s = jω = ( s 2 + 6)(s 2 + 1) − ( s)(s) |s = jω = s 4 + 6s 2 + 6 |s = jω = ω 4 − 6ω 2 + 6 For w = 2, A(w2) = 16 − 24 + 6 = −2 This condition is not satisfied. Hence, the function F(s) is not positve real.

Example 9.30

Test whether F ( s) =

s2 + 4 is positive real function. s3 + 3s 2 + 3s + 1

Solution (a) F ( s) =

N ( s) s2 + 4 ( s + j 2)( s − j 2) = 3 = D( s) s + 3s 2 + 3s + 1 ( s + 1)3

The function F(s) has two zeros at s = ± j2 and three poles at s = −1. Thus, all the poles and zeros are in the left half of the s plane. (b) There is no pole on the jw axis. Hence, the residue test is not carried out. (c) Even part of N ( s) = m1 = s 2 + 4 Odd part of N ( s) = n1 = 0 Even part of D( s) = m2 = 3s 2 + 1 Odd part of D( s) = n2 = s3 + 3s A(ω 2 ) = m1m2 − n1n2 |s = jω = ( s 2 + 4)(3s 2 + 1) − (0)( s3 + 3s) |s = jω = 3s 4 + 13s 2 + 4 |s = jω = 3ω 4 − 13ω 2 + 4 For w = 1, A(w)2 = 3 − 13 + 4 = −6 This condition is not satisfied. Hence, the function F(s) is not positive real.

Example 9.31

Test whether F ( s) =

s3 + 5 s is positive real function. s4 + 2 s 2 + 1

Solution (a) F ( s) =

N ( s) s3 + 5s s( s 2 + 5) s( s + j 5 )( s − j 5 ) = 4 = = D( s) s + 2 s 2 + 1 ( s 2 + 1) 2 ( s ± j1)( s ± j1)

The function F(s) has zeros at s = 0, s = ± j 5 and two poles at s = j1 and two poles at s = −j1. Thus, poles on the jw axis are not simple. Hence, the function F (s) not positive real.

Example 9.32

Test whether F ( s) =

s4 + 3s3 + s 2 + s + 2 is positive real function. s3 + s 2 + s + 1

Solution F ( s) =

N ( s) s 4 + 3s3 + s 2 + s + 2 = D ( s) s3 + s 2 + s + 1

Here, it is easier to prove that N(s) and D(s) are Hurwitz.

9.24 Circuit Theory and Transmission Lines By Routh array, s4 s3

1 1 2 3 1 2 s2 3 2 s1 −8 s0 2 Since there is a sign change in the first column of the array, N(s) is not Hurwitz. Thus, all the zeros are not in the left half of the s plane. The remaining two tests need not be carried out. Hence, the function F(s) is not positive real.

9.4

ElEmENTaRy syNTHEsIs CONCEPTs

We know that impedances and admittances of passive networks are positive real functions. Hence, addition of impedances of the two passive networks gives a function which is also a positive real function. Thus, Z(s) = Z1(s) + Z2(s) is a positive real function, if Z1(s) and Z2(s) are positive real functions. Similarly, Y(s) = Y1(s) + Y2(s) is a positive real function, if Y1(s) and Y2(s) are positive real functions. There is a special terminology for synthesis procedure. We have, Z1(s) Z ( s) = Z1 ( s) + Z 2 ( s) Z 2 ( s) = Z ( s) − Z1 ( s)

Z(s)

Here, Z1(s) is said to have been removed from Z(s) in forming the new function Z2(s) as shown in Fig. 9.5. If the removed network is associated with Y(s) the pole or zero of the original network impedance then that pole or zero is also said to have been Fig. 9.5 removed. There are four important removal operations.

9.4.1

Z2(s)

Z(s)

Y(s)

Y1(s)

Y2(s)

Network interpretation of the removal of impedance and admittance

Removal of a Pole at Infinity

Consider an impedance function Z(s) having a pole at infinity which means that the numerator polynomial is one degree greater than the degree of the denominator polynomial. Z ( s) =

where Let

an +1s n +1 + an s n + … + a1s + a0

bn s + bn −1s a H = n +1 bn Z1 ( s) = Hs

n −1

+ … + b1s + b0

cn s n + cn −1s n −1 + … + c1s + c0

= Hs +

cn s n + cn −1s n −1 + … + c1s + c0 bn s n + bn −1s n −1 + … + b1s + b0

= Z ( s) − Hs bn s n + bn −1s n −1 + … + b1s + b0 Z1(s) = Hs represents impedance of an inductor L of value H. Hence, the removal of a pole at infinity Y(s) C Z2(s) Z2(s) corresponds to the removal of an inductor from the Z(s) network of Fig. 9.6(a). If the given function is an admittance function (a) (b) Y(s), then Y1(s) = Hs represents the admittance of a capacitor YC(s) = Cs. The network for Y1(s) is a Fig. 9.6 Network interpretation of the removal of a pole at infinity capacitor of value C = H as shown in Fig. 9.6(b). and

Z 2 ( s) =

n

9.4 Elementary Synthesis Concepts 9.25

9.4.2

Removal of a Pole at Origin

If Z(s) has a pole at the origin then it may be written as Z ( s) = K0 =

where Z1 ( s) =

a0 + a1s + … + an −1s n −1 + an s n b1s + b2 s 2 + … + bm s m

=

K 0 d1 + d2 s + … + dn s n −1 + = Z1 ( s) + Z 2 ( s) s b1 + b2 s + … + bm s m −1

a0 b1

1 K0 represents the impedance of a capacitor of value . K0 s

K If the given function is an admittance function Y(s) then removal of Y1 ( s) = 0 corresponds to an inductor s 1 of value . K0 Z(s)

C

Thus, removal of a pole from the impedance function Z(s) at the origin corresponds to the removal of a capacitor, and from admittance function Y(s) corresponds to removal of an Fig. 9.7 inductor as shown in Fig. 9.7.

9.4.3

Z2(s) (a)

Y(s)

L

Z2(s) (b)

Network interpretation of the removal of a pole at origin

Removal of Conjugate Imaginary Poles

If Z(s) contains poles on the imaginary axis, i.e., at s = ± jw1 then Z(s) will have factors (s + jw1) (s − jw1) = s2 + w12 in the denominator polynomial Z ( s) =

p( s) ( s + ω12 ) q1 ( s) 2

By partial-fraction expansion, Z ( s) =

K1 K1* + + Z 2 ( s) s + jω1 s − jω1

For a positive real function, jw axis poles must themselves be conjugate and must have equal, positive and real residues. K1 = K1* 2K s Z ( s) = 2 1 2 + Z 2 ( s) s + ω1 2K s 1 1 = Z1 ( s) = 2 1 2 = 2 Ya + Yb s + ω1 s ω + 1 2 K1 2 K1s

Hence, Thus,

where Ya = and Yb =

s 1 is the admittance of a capacitor of value C = 2 K1 2 K1

ω12 2K is the admittance of an inductor of value L = 21 2 K1s ω1

9.26 Circuit Theory and Transmission Lines If the given function is an admittance function Y(s) then 2K s 1 = Y1 ( s) = 2 1 2 = + Z Zb s + ω1 a where Z a =

1 s ω2 + 1 2 K1 2 K1s

ω2 s 1 is the impedance of an inductor of value L = and Zb = 1 is the impedance of 2 K1 2 K1 2 K1s 2 K1

a capacitor of value C =

. L ω12 Thus, removal of conjugate imaginary poles from impedance function Z(s) corresponds to the Z(s) C Z2(s) Y(s) Z2(s) L C removal of the parallel combination of L − C and from admittance function Y(s) corresponds to removal of series combination of L − C as shown Fig. 9.8 Network interpretation of the removal of in Fig. 9.8. conjugate imaginary poles

9.4.4

Removal of a Constant

If a real number R1 is subtracted from Z (s) such that Z 2 ( s) = Z ( s) − R1 Z ( s) = R1 + Z 2 ( s) then R1 represents a resistor. If the given function is an admittance function Y(s), then removal of Y1(s) = R1 represents a conductance of value R1. Thus, removal of a constant from impedance function Z(s) corresponds to the removal of a resistance, and from admittance function Y(s) corresponds to removal of a conductance. Synthesize the impedance function Z ( s) =

Example 9.33 Solution By long division of Z(s),

s3 + 4 s s2 + 2

.

  s 2 + 2 s3 + 4 s  s   s3 + 2 s 2s Z ( s) = s +

2s

= Z1 ( s) + Z 2 ( s) s +2 Z1(s) = s represents impedance of an inductor of value 1 H. Y2 ( s) =

2

1 1 s2 + 2 s2 2 1 = = + = s + = Y3 ( s) + Y4 ( s) 2s 2s 2s 2 Z 2 ( s) s

1 1 s represents the admittance of a capacitor of value F. 2 2 1 Y4 ( s) = represents the admittance of an inductor of value 1 H. s The impedances are connected in the series branches whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.9. Y3 ( s) =

1H

Z(s)

1 F 2

Fig. 9.9

1H

9.4 Elementary Synthesis Concepts 9.27

Example 9.34

Realise the network having impedance function Z ( s) =

s 2 + 2 s + 10 s( s + 5 )

Solution By long division of Z(s),  2 s 2 + 5s s 2 + 2 s + 10  s  2 s + 10 s2 Z ( s) = Z1 ( s) =

2 2 s2 s + 2 = + = Z1 ( s) + Z 2 ( s) s s + 5s s s + 5

2 1 represents the impedance of capacitor of value F. 2 s Y2 ( s) =

1 5 s+5 = = 1 + = Y3 ( s) + Y4 ( s) Z 2 ( s) s s

1 F 2

Y3(s) = 1 represents the admittance of a resistor of value 1 Ω. 5 1 represents the admittance of an inductor of value H. 5 s The impedances are connected in the series branches whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.10. Y4 ( s) =

Example 9.35

Z(s)

Realise the network having impedance function Z ( s) =

Solution By long division of Z(s),   2 s 3 + 2 s 6 s 3 + 5 s 2 + 6 s + 4  3   + 6s

6 s3 5s 2 Z ( s) = 3 +

5s + 4 2

2 s3 + 2 s

+4

= Z1 (ss) + Z 2 ( s)

Z1(s) = 3 represents the impedance of a resistor of value 3 Ω. Y2 ( s) =

1 2 s3 + 2 s = 2 Z 2 ( s) 5s + 4

1 H 5

1Ω

Fig. 9.10 6 s3 + 5 s 2 + 6 s + 4 2 s3 + 2 s

.

9.28�Circuit Theory and Transmission Lines By long division of Y2(s), � �2 5 s 2 � 4� 2 s 3 � 2 s � s �5 � 8 2 s3 � s 5 2 s 5 2 s 2 5 Y2 ( s) � s � 2 � Y3 ( s) � Y4 (ss) 5 5s � 4 Y3 ( s) �

2 2 s represents the admittance of a capacitor of value F. 5 5 Z 4 ( s) �

1 5s 2 � 4 25s 2 � 20 25 10 � � � � Z5 ( s) � Z6 ( s) s� 2 2s 2 Y4 ( s) s s 5

25 25 H. s represents the impedance of an inductor of value 2 2 10 1 F. Z6 ( s) � represents the impedance of a capacitor of value 10 s The impedances are connected in the series branches, whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.11. Z5 ( s) �

��������������

3�

Z(s)

25 H 2

1 F 10

2 F 5

���������

Realise the network having impedance function Z ( s) �

s4 � 10 s 2 � 7 s3 � 2 s

Solution By long division of Z(s), � � s3 � 2 s� s 4 � 10 s 2 � 7 � s � � s4 � 2s2 8s 2 � 7 Z ( s) � s �

8s 2 � 7 s3 � 2 s

� Z1 ( s) � Z 2 ( s)

Z1(s) = s represents the impedance of an inductor of value 1 H. Y2 ( s) �

Chapter 09.indd 28

1 s3 � 2 s � 2 Z 2 ( s) 8s � 7

5/30/2016 6:14:17 PM

9.4 Elementary Synthesis Concepts 9.29

By long division of Y2(s),  1 8s 2 + 7 s3 + 2 s  s 8  7 s3 + s 8 9 s 8 9 s 1 8 Y2 ( s) = s + 2 = Y3 ( s) + Y4 ( s) 8 8s + 7 Y3 ( s) =

1 1 s represents the admittance of a capacitor of value F. 8 8 Z 4 ( s) =

1 8s 2 + 7 64 56 = = = Z5 ( s) + Z6 ( s) s+ 9 9 9s Y4 ( s) s 8

64 64 H. s represents the impedance of an inductor of value 9 9 56 9 F. Z6 ( s) = represents the impedance of a capacitor of value 9s 56 The impedances are connected in the series branches, whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.12. Z5 ( s) =

Example 9.37

  s + 1 4 s 2 + 6 s  4 s   4s2 + 4s 2s Y ( s) = 4 s +

2s = Y1 ( s) + Y2 ( s) s +1

Y1(s) = 4s represents the admittance of a capacitor of value 4 F. Z 2 ( s) =

1 s +1 1 1 = = + = Z3 ( s) + Z 4 ( s) 2s 2 2s Y2 ( s)

1 F 8

Z(s)

Realise the network having admittance function Y ( s) =

Solution By long division of Y(s),

64 H 9

1H

Fig. 9.12 4 s2 + 6 s . s+1

9 F 56

9.30 Circuit Theory and Transmission Lines 1 1 Ω. represents the impedance of a resistor of value 2 2 1 Z 4 ( s) = represents the impedance of a capacitor of value 2 F. 2s The impedances are connected in the series branches, whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.13.

1 Ω 2

Z3 ( s) =

Example 9.38

Realise the admittance function Y ( s) =

Z(s)

2F

4F

Fig. 9.13

3 + 5s . 4 + 2s

Solution By long division,  3 4 + 2 s 3 + 5 s  4  3+

3 s 2 7 s 2

7 s 3 Y ( s) = + 2 = Y1 ( s) + Y2 ( s) 4 4 + 2s Y1 ( s) =

3 4 represents the admittance of a resistor of value Ω. 4 3

1 4 + 2s 8 + 4 s 8 4 = = = + = Z3 ( s) + Z 4 ( s) 7 7s 7s 7 Y2 ( s) s 2 8 7 Z3 ( s) = represents the impedance of a capacitor of value F. 7s 8 4 4 Z 4 ( s) = represents the impedance of a resistor of value Ω. 7 7 The impedances are connected in the series branches, whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.14. Z 2 ( s) =

9.5

7 F 8

4 Ω 7

4 Ω 3

Fig. 9.14

REalIsaTION OF LC FUNCTIONs

LC driving point immitance functions have the following properties. 1. 2. 3. 4. 5.

It is the ratio of odd to even or even to odd polynomials. The poles and zeros are simple and lie on the jw-axis. The poles and zeros interlace on the jw-axis. There must be either a zero or a pole at the origin and infinity. The difference between any two successive powers of numerator and denominator polynomials is at most two. There cannot be any missing terms.

9.5 Realisation of LC Functions 9.31

6. The highest powers of numerator and denominator polynomials must differ by unity; the lowest powers also differ by unity. There are a number of methods of realising an LC function. But we will study only four basic forms—Foster I, Foster II, Cauer I and Cauer II forms. The Foster forms are obtained by partial-fraction expansion of F(s), and the Cauer forms are obtained by continued fraction expansion of F(s).

9.5.1

Foster Realisation

Consider a general LC function F(s) given by F ( s) =

H ( s 2 + ω12 )( s 2 + ω 32 )... s( s 2 + ω 22 )( s 2 + ω 42 )...

where 0 ≤ ω12 < ω 22 < ω 32 … and H is positive. By partial-fraction expansion of F(s), F ( s) =

K0 K2 K2 + + + ... + K ∞ s s s + jω 2 s − jω 2

Combining terms with conjugate poles, F ( s) =

K0 2K 2 s + ... + K ∞ s s s + ω 22

where K0, Ki and K∞ are the residues of F(s) at the origin, at jwi and at infinity respectively. These residues are given by K 0 = sF ( s)s = 0 Ki

(s =

K∞ = Foster I Form

2

)

+ ω i2 F ( s) 2s

s 2 = − ω i2

F ( s) s s→∞

If F(s) represents an impedance function, it gives a series connection of impedances. 2K s K F ( s ) = Z ( s ) = 0 + 2 2 2 … + K ∞ s = Z1 ( s) + Z 2 ( s) + … + Z n ( s) s s + ω2

K0 1 represents the impedance of a capacitor of farad. s K0 The last term K∞ s represents the impedance of an inductor of K∞ henry. 2K s The remaining terms, i.e., 2 i 2 represent the impedance of a parallel combination of capacitor Ci and s + ωi inductor Li. For parallel combination of Li and Ci, The first term

 1  C  s 2K s i Z ( s) = = = 2 i 2 1 1 s + ωi Ci s + s2 + Li s Li Ci 1

9.32 Circuit Theory and Transmission Lines Ci =

1 2K and Li = 2i 2 Ki ωi

Table 9.1 Realisation of Foster-I form of LC network Impedance function

Element

K0 1 = s C0 s

C0 =

1 K0

Li =

2 Ki

Ci =

1 2 Ki

 1  C  s 2 Ki s i = 2 2 1 2 s + ωi s + Li Ci

K∞ s = Ls

ω i2

L∞ = K ∞

The network corresponding to Foster I form is shown in Fig. 9.15. C0

Z(s)

Fig. 9.15

L1

Li

C1

Ci

L∞

Foster-I form of LC network

If Z(s) has no pole at the origin then capacitor C0 is not present in the network. Similarly, if there is no pole at ∞, inductor L∞ is not present in the network. Foster II Form admittances.

If F(s) represents an admittance function, it gives the parallel combination of F ( s) = Y ( s) =

The first term

2K s K0 + 2 2 2 + …+ K ∞ s = Y1 ( s ) + Y2 ( s ) + …Yn ( s ) s s + ω2

K0 1 represents the admittance of an inductor of henry. s K0

The last term K∞s represents the admittance of a capacitor of K∞ farad. The remaining terms, i.e., a capacitor Ci.

2 Ki s s + ω i2 2

represent the admittance of a series combination of an inductor Li and

9.5 Realisation of LC Functions 9.33

For series combination of Li and Ci,  1  L  s 2K s i Y ( s) = = = 2 i 2 1 1 s + ωi Li s + s2 + Ci s Li Ci 1

Li =

1 2 Ki

and Ci =

2 Ki

ω i2

Table 9.2 Realisation of Foster-II form of LC network Admittance function

Element

K0 1 = s L0 s

 1  L  s 2 Ki s i = 2 2 1 s + ωi s2 + Li Ci

L0 =

1 K0

Li =

1 2 Ki 2 Ki

Ci =

K∞ s = Cs

ω i2

C∞ = K ∞

The network corresponding to the Foster II form is shown L0 L1 Li Y(s) C∞ in Fig. 9.16. If Y (s) has no pole at the origin then inductor L0 is not C1 Ci present. Similarly, if there is no pole at infinity, capacitor C∞ Fig. 9.16 Foster-II form of LC network is not present.

9.5.2 Cauer Realisation or Ladder Realisation Cauer I Form Since the numerator and denominator polynomials of an LC function always differ in degrees by unity, there is always a zero or a pole at s = ∞. The Cauer I Form is obtained by successive removal of a pole or a zero at infinity from the function. Consider an impedance function Z(s) having a pole at infinity. By removing the pole at infinity, we get Z 2 ( s ) = Z ( s ) − L1s Now, Z2(s) has a zero at s = ∞. If we invert Z2(s), Y2(s) will have a pole at s = ∞. By removing this pole, Y3 ( s) = Y2 ( s) − C2 s Now Y3(s) has a zero at s = ∞, which we can invert and remove. This process continues until the remainder is zero. Each time we remove a pole, we remove an inductor or a capacitor depending on whether the function is an impedance or an admittance. The impedance Z(s) can be written as a continued fraction expansion.

9.34 Circuit Theory and Transmission Lines Z ( s ) = L1s +

1 C2 s +

1 L3 s +

1 C4 s + ...

Thus, the final structure is a ladder network whose series arms are inductors and shunt arms are capacitors. The Cauer I network is shown in Fig. 9.17. L1

Z1

L3

Z(s)

C2

C4

Z3 Y2

Z(s)

(a)

Y4

(b)

Fig. 9.17 Cauer I form of LC network If the impedance function has zero at infinity, i.e., if degree of numerator is less than that of its denominator by unity, the function is first inverted and continued fraction expansion proceeds as usual. In this case, the first element is a capacitor as shown in Fig. 9.18. L2

Z(s)

Z2

L4

C1

C3

C5

(a)

Fig. 9.18

Z(s)

Z4

Y1

Y3

Y5

(b)

Cauer-I form of LC network

Cauer II Form Since the lowest degrees of numerator and denominator polynomials of LC function must differ by unity, there is always a zero or a pole at s = 0. The Cauer II form is obtained by successive removal of a pole or a zero at s = 0 from the function. In this method, continued fraction expansion of Z(s) is carried out in terms of poles at the origin by removal of the pole at the origin, inverting the resultant function to create a pole at the origin which is removed and this process is continued until the remainder is zero. To do this, we arrange both numerator and denominator polynomials in ascending order and divide the lowest power of the denominator into the lowest power of the numerator. Then we invert the remainder and divide again. The impedance Z (s) can be written as a continued fraction expansion. Z ( s) =

1 1 + 1 1 C1s + 1 1 L2 s + 1 C3 s + ... L4 s

C3

C1

Z(s)

Thus, the final structure is a ladder network whose first element is a series capacitor and second element is a shunt inductor as Fig. 9.19 shown in Fig. 9.19.

L2

L4

Cauer II form of LC network

9.5 Realisation of LC Functions 9.35

If the impedance function has a zero at the origin then the first element is a shunt inductor and the second element is a series capacitor as shown in Fig. 9.20. Z(s) Thus, the LC function F(s) can be realised in four different forms. All these forms have the same number of elements and the number is equal to the number of poles and zeros of F(s) includng any at infinity. Fig. 9.20

Example 9.39

L1

jw

5 s( s 2 + 4 ) ( s + 1)( s + 3) 2

2

(b) Z ( s) =

2( s 2 + 1)( s 2 + 9 )

j3

s( s 2 + 2 )

j2 j√3 j1

5 s( s 2 + 4 ) ( s 2 + 1)( s 2 + 3)

−3 −2 −1

s

−j √3 −j2 −j3

2( s + 1)( s + 9) 2

0 −j1

The function Z(s) has poles at s = ± j1 and s = ± j 3 and zeros at s = 0 and s = ± j 2 as shown in Fig. 9.21. Since the poles and zeros do not interlace on the jw axis, the function Z(s) is not an LC impedance function. (b) Z ( s) =

L5

State whether the following functions are driving point immittance of LC

Solution (a) Z ( s) =

L3

Cauer–II form of LC network

networks are not: (a) Z ( s) =

C4

C2

2

Fig. 9.21

s ( s 2 + 2)

The function Z(s) has poles at s = 0 and s = ± j 2 and zeros at s = ± j1 and s = ± j 3 as shown in Fig. 9.22. The poles and zeros are simple and lie on the jw-axis. The poles and zeros interlace on the jw axis. Hence, the function Z(s) is an LC impedance function. jw j3 j2 j√2 j1 s −3 −2 −1

0 −j1 −j√2 −j2 −j3

Fig. 9.22

9.36 Circuit Theory and Transmission Lines

Example 9.40

Realise the Foster and Cauer forms of the following impedance function Z ( s) =

4( s 2 + 1)( s 2 + 9 ) s( s 2 + 4 )

Solution The function Z(s) has poles at s = 0 and s = ± j2 and zeros

jw

at s = ± j1 and s = ± j3 as shown in Fig. 9.23. From the pole-zero diagram, it is clear that poles and zeros are simple and lie on the jw axis. Poles and zeros are interlaced. Hence, the given function is an LC function.

j3

Foster I Form

0

j2 j1 s

The Foster I form is obtained by partial-fraction expansion of the impedance function Z(s). But degree of numerator is greater than degree of denominator. Hence, division is first carried out. Z ( s) =

4( s 2 + 1)( s 2 + 9) s( s + 4 ) 2

4 s 4 + 40 s 2 + 36

=

4

−j2 −j3

s + 4s 3

s + 4 s) 4 s + 40 s + 36 ( 4 s 3

−j1

2

Fig. 9.23

4 s 4 + 16 s 2 24 s 2 + 36 24 s 2 + 36 24 s 2 + 36 Z ( s) = 4 s + 3 = 4s + s + 4s s( s 2 + 4 ) By partial-fraction expansion, K0 K1 K1* K 2K s + + = 4s + 0 + 2 1 s s + j2 s − j2 s s +4 4(1)(9) =9 K 0 = sZ ( s) s = 0 = 4

Z ( s) = 4 s + where

K1 =

( s 2 + 4) Z ( s) 2s

Z ( s) = 4 s +

= s 2 = −4

4( −4 + 1)( −4 + 9) 15 = 2( −4) 2

9 15s + s s2 + 4

The first term represents the impedance of an inductor of 4 H. The second term represents the impedance 1 of a capacitor of F. The third term represents the impedance of a parallel LC network. 9 For a parallel LC network,  1   s C Z LC ( s) = 1 2 s + LC By direct comparison, C=

1 F 15

9.5 Realisation of LC Functions 9.37

L=

15 H 4

1 F 9

4H

The network is shown in Fig. 9.24.

Foster II Form The Foster II form is obtained by partial-

15 H 4

fraction expansion of the admittance function Y(s). Y ( s) =

1 F 15

s( s 2 + 4 ) 4( s 2 + 1)( s 2 + 9) Fig. 9.24

By partial-fraction expansion, Y ( s) =

where

2K s 2K s K1 K* K2 K 2* + 1 + + = 2 1 + 2 2 s + j1 s − j1 s + j 3 s − j 3 s + 1 s + 9

K1 =

( s 2 + 1) Y ( s) 2s

K2 =

( s + 9) Y ( s) 2s

Y ( s) =

s2 + 1

+

( −1 + 4) 3 = 8( −1 + 9) 64

=

( −9 + 4) 5 = 8( −9 + 1) 64

s 2 = −1

2

 3   s 32

=

s 2 = −9

 5   s 32 s2 + 9

These two terms represent admittance of a series LC network. For a series LC network,  1   s L YLC ( s) = 1 s2 + LC By direct comparison, 32 H 3 32 H L2 = 5 L1 =

3 F 32 5 F C2 = 288

C1 =

The network is shown in Fig. 9.25.

3 F 32

5 F 288

32 H 3

32 H 5

Fig. 9.25

Cauer I Form The Cauer I form is obtained from continued fraction expansion about the pole at infinity. Z ( s) =

4 s 4 + 40 s 2 + 36 s3 + 4 s

Since the degree of the numerator is greater than the degree of the denominator by one, it indicates the presence of a pole at infinity.

9.38 Circuit Theory and Transmission Lines By continued fraction expansion,   s3 + 4 s 4 s 4 + 40 s 2 + 36  4 s ← Z   4 s 4 + 16 s 2  1  24 s 2 + 36 s3 + 4 s  s ← Y   24 s3 +

3 s 2 5   48 s 24 s 2 + 36  s ← Z  5 2  24 s 2

 5  5 36 s  s ← Y  2  72 5 s 2 0 The impedances are connected in the series branches whereas the admittances are connected in the parallel branches in a Cauer or ladder realisation. The network is shown in Fig. 9.26.

48 H 5

4H

1 F 24

5 F 72

Fig. 9.26

Cauer II Form The Cauer II form is obtained from partial-fraction expansion about pole at origin. Z ( s) =

4( s 2 + 1)( s 2 + 9) s( s 2 + 4 )

=

4 s 4 + 40 s 2 + 36 s3 + 4 s

The function Z(s) has a pole at origin. Arranging the numerator and denominator polynomials in ascending order of s, 36 + 40 s 2 + 4 s 4 Z ( s) = 4 s + s3 By continued fraction expansion, 9  4 s + s3  36 + 40 s 2 + 4 s 4  ← Ζ  s 36 + 9 s 2  4  ←Y 31s 2 + 4 s 4  4 s + s3    31s 4s +

16 3 s 31 15 3   961 s  31s 2 + 4 s 4  ←Z  15s 31  31s 2

9.5 �Realisation of LC Functions�9.39

� 15 3 � 15 �Y 4s4 � s � 31 �� 124 s

1 F 9

15 3 s 31 0

31 H 4

The impedances are connected in the series branches whereas the admittances are connected in the parallel branches in a Cauer or ladder realisation. The network is shown in Fig. 9.27.

���������������

15 F 961

124 H 15

���������

Realise Foster forms of the LC impedance function Z ( s) �

( s 2 � 1)( s 2 � 3) s( s 2 � 2 )

Solution Foster I Form The Foster I form is obtained by continued-fraction expansion of the impedance function Z(s). Since the degree of the numerator is greater than the degree of the denominator, division is first carried out. Z ( s) �

( s 2 � 1)( s 2 � 3) 2

s( s � 2)

s4 � 4s2 � 3

�

s3 � 2 s

� � s 3 � 2 s� s 4 � 4 s 2 � 3 � s � � s 4 � 2s 2 2s2 � 3 Z ( s) � s �

2s 2 � 3 3

s � 2s

� s�

2s2 � 3 s(ss 2 � 2)

By partial-fraction expansion, Z ( s) � s � where

2K s K0 K1 K1* K � � � s� 0 � 2 1 s s � j2 s � j2 s s �2

(1)(3) 3 � 2 2 2 ( s � 2) ( �2 � 1)( �2 � 3) 1 � K1 � Z ( s) � �2) 4 2s 2(� 2

K 0 � sZ ( s)|s � 0 �

s � �2

� 3� � 1� �� �� �� �� s 2 2 � 2 Z ( s) � s � s s �2 The first term represents the impedance of an inductor of 1 H. The second term represents the impedance 2 of a capacitor of F. The third term represents the impedance of a parallel LC network. 3

Chapter 09.indd 39

6/13/2016 3:18:36 PM

9.40 Circuit Theory and Transmission Lines For a parallel LC network,  1   s C Z LC ( s) = 1 s2 + LC

By direct comparison,

1H

2 F 3

C = 2F

1 H 4

2F

1 L= H 4 Fig. 9.28

The network is shown in Fig. 9.28.

Foster II Form The Foster II form is obtained by partial-fraction expansion of the admittance function Y(s). Y ( s) =

s ( s 2 + 2) ( s 2 + 1)( s 2 + 3)

By partial-fraction expansion, Y ( s) = where

2K s 2K s K1 K* K2 K 2* + 1 + + = 2 1 + 2 2 s + j1 s − j1 s + j 3 s − j 3 s + 1 s + 3

K1 =

( s 2 + 1) Y ( s) 2s

K2 =

( s + 3) Y ( s) 2s

Y ( s) =

( −1 + 2) 1 = 2( −1 + 3) 4

=

1 −3 + 2 = 2( −3 + 1) 4

s 2 = −1

2

 1   s 2

=

s 2 = −3

 1   s 2

+ s2 + 1 s2 + 3 These two terms represent admittance of a series LC network. For a series LC network,  1   s L YLC ( s) = 1 s2 + LC By direct comparison, 1 F 2 1 L2 = 2H, C2 = F 6 L1 = 2H, C1 =

1 F 6

2H

2H

Fig. 9.29

The network is shown in Fig. 9.29.

Example 9.42

1 F 2

Realise Foster forms of the following LC impedance function: Z ( s) =

( s 2 + 1)( s 2 + 3) s( s 2 + 2)( s 2 + 4 )

Solution Foster I Form The Foster I form is obtained by partial-fraction expansion of the impedance function Z (s).

9.5 Realisation of LC Functions 9.41

By partial-fraction expansion, Z ( s) =

2K s 2K s K0 K1 K1* K2 K 2* K + + + + = 0+ 2 1 + 2 2 2 2 s s + j s − j s s +2 s +4 s+ j 2 s− j 2

K 0 = sZ ( s) s = 0 =

where

(1)(3) 3 = ( 2)( 4) 8

K1 =

( s 2 + 2) Z ( s) 2s

K2 =

( s + 4) Z ( s) 2s

=

( −2 + 1)( −2 + 3) 1 = 2( −2)( −2 + 4) 8

=

( −4 + 1)( −4 + 3) 3 = 2( −4)( −4 + 2) 16

s 2 = −2

2

3 Z ( s) = 8 + s

 1   s 4 s2 + 2

+

s 2 = −4

 3   s 8 s2 + 4

8 The first term represents the impedance of a capacitor of F. The other two terms represent the impedance 3 of a parallel LC network. For a parallel LC network,  1   s C Z LC ( s) = 1 s2 + 1H 3 H LC 8F 8 32 By direct comparison, 3 1 C1 = 4 F, L1 = H 8 4F 8F 3 8 3 C2 = F, L2 = H 3 32 Fig. 9.30 The network is shown in Fig. 9.30. Foster II Form The Foster II form is obtained by partial-fraction expansion of the admittance function Y(s). Y ( s) =

s( s 2 + 2)( s 2 + 4) ( s 2 + 1)( s 2 + 3)

=

s5 + 6 s3 + 8s s4 + 4s2 + 3

Since the degree of the numerator is greater than the degree of the denominator, division is first carried out. s 4 + 4 s 2 + 3) s5 + 6 s3 + 8s( s s5 + 4 s3 + 3s 2 s3 + 5s Y ( s) = s +

2 s3 + 5s s4 + 4s2 + 3

= s+

2 s3 + 5s ( s + 1)( s 2 + 3) 2

By partial-fraction expansion, Y ( s) = s +

2K s 2K s K1 K* K2 K 2* + 1 + + = s+ 2 1 + 2 2 s + j1 s − j1 s + j 3 s − j 3 s +1 s + 3

9.42 Circuit Theory and Transmission Lines where

K1 =

( s 2 + 1) Y ( s) 2s

K2 =

( s + 3) Y ( s) 2s

=

( −1 + 2)( −1 + 4) 3 = 2( −1 + 3) 4

=

( −3 + 2)( −3 + 4) 1 = 2( −3 + 1) 4

s 2 = −1

2

Y ( s) = s +

 3   s 2

s 2 = −3

 1   s 2

+ s2 + 1 s2 + 3 The first term represents the admittance of capacitor of 1 F. The other two terms represent admittance of a series LC network. For a series LC network,  1   s L YLC ( s) = 1 s2 + LC By direct comparison, 2H 2H 2 3 3 L1 = H, C1 = F 1F 3 2 3F 1F 1 2 6 L2 = 2 H, C2 = F 6 Fig. 9.31 The network is shown in Fig. 9.31.

Example 9.43

Realise Cauer forms of the following LC impedance function: Z ( s) =

10 s 4 + 12 s 2 + 1 2 s3 + 2 s

Solution Cauer I Form The Cauer I form is obtained from continued fraction expansion about the pole at infinity. Z ( s) =

10 s 4 + 12 s 2 + 1

2 s3 + 2 s Since the degree of the numerator is greater than the degree of the denominator by one, it indicates the presence of a pole at infinity. By continued fraction expansion, 2 s3 + 2 s) 10 s 4 + 12 s 2 + 1 (5s ← Z 10 s 4 + 10 s 2

)

2 s 2 + 1 2 s3 + 2 s ( s ← Y 2 s3 + s s ) 2 s 2 + 1( 2 s ← Z 2s2 1) s ( s ← Y s 0

9.5 Realisation of LC Functions 9.43

The impedances are connected in the series branches whereas the admittances are connected in the parallel branches in a Cauer or ladder realisation. The network is shown in Fig. 9.32. 5H

2H

1F

1F

Fig. 9.32 Cauer II Form The Cauer II form is obtained from continued fraction expansion about the pole at the origin. 10 s 4 + 12 s 2 + 1 Z ( s) = 2 s3 + 2 s The function Z(s) has a pole at the origin. Arranging the numerator and denominator polynomials in ascending order of s, Z ( s) =

1 + 12 s 2 + 10 s 4

2 s + 2 s3 By continued fraction expansion of Z(s),  1  2 s + 2 s3  1 + 12 s 2 + 10 s 4  ← Ζ   2s 1+

s2  2  ←Y 11s 2 + 10 s 4  2 s + 2 s3    11s 2s +

20 3 s 11 2 3  121 s  11s 2 + 10 s 4  ←Z  2s 11 

11s 2  2  2 10 s 4  s3  ←Y  11  110 s 2 3 s 11 0 The impedances are connected in the series branches whereas the admittances are connected in the parallel branches in Cauer or ladder realisation. The network is shown in Fig. 9.33.

2F

2 F 121

11 H 2

Fig. 9.33

110 H 2

9.44 Circuit Theory and Transmission Lines

Example 9.44

Realise the following network function in Cauer I form: 6 s4 + 42 s 2 + 48

Z ( s) =

s5 + 18 s3 + 48 s Solution The Cauer I form is obtained by continued fraction expansion of Z(s) about the pole at infinity. In the above function, the degree of the numerator is less than the degree of the denominator which indicates the presence of a zero at infinity. The admittance function Y(s) has a pole at infinity. Hence, the continued fraction expansion of Y(s) is carried out. Y ( s) = By continued fraction expansion

s5 + 18s3 + 48s 6 s 4 + 42 s 2 + 48

1  6 s 4 + 42 s 2 + s 2  s5 + 18s3 + 48s  s ← Y  6 s 5 + 7 s 3 + 8s 6  11s3 + 40 s 6 s 4 + 42 s 2 + 48  s ← Z   11 6s4 +

240 2 s 11 222 2   121 s + 30 11s3 + 40 s  s ←Y   222 11 5808 11s3 + s 222 3072  222 2  49284 s s + 48  s←Z  33792 222  11 222 2 s 11

 3072  128 s s ←Y 48  222  444 3072 s 222 0 The impedances are connected in the series branches whereas the admittances are connected in the parallel branches in a Cauer or ladder realisation. The network is shown in Fig. 9.34. 6 H 11

1F 6

49284 H 33792

121 F 222

Fig. 9.34

128 F 444

9.5 Realisation of LC Functions 9.45

Example 9.45

Realise Cauer II form of the function: Z LC ( s) =

s( s4 + 3 s 2 + 1) 3s4 + 4 s 2 + 1

Solution The Cauer II form is obtained by continued fraction expansion about the pole at the origin. The given function has a zero at the origin. The admittance function Y(s) has a pole at origin. Hence, the continued fraction expansion of Y(s) is carried out. Arranging the polynomials in ascending order of s, YLC ( s) =

3s 4 + 4 s 2 + 1

s5 + 3s3 + s By continued fraction expansion of Y(s), we have

=

1 + 4 s 2 + 3s 4 s + 3s 3 + s 5

1  s + 3s3 + s5  1 + 4 s 2 + 3s 4  ← Y  s 1 + 3s 2 + s 4 1  s 2 + 2 s 4  s + 3s 3 + s 5  ← Z  s s + 2 s3 1  s3 + s5  s 2 + 2 s 4  ← Y  s s2 + s4 1  s 4  s3 + s5  ← Z  s s3  1 s5  s 4  ← Y  s

1F

1F

4

s 0

1H

The impedances are connected in the series branches whereas the admittances are connected in the parallel branches in a Cauer or ladder realisation. The network is shown in Fig. 9.35.

Example 9.46

1H

1H

Fig. 9.35

Obtain the Cauer I form of realisation for the function Z LC ( s) =

s5 + 7 s3 + 10 s s4 + 5 s 2 + 4

Solution The Cauer I form is obtained by continued fraction expansion of ZLC (s) about pole at infinity.

9.46 Circuit Theory and Transmission Lines   s 4 + 5s 2 + 4 s5 + 7 s3 + 10 s  s ← Z   s5 + 5s3 + 4 s 1  2 s 3 + 6 s s 4 + 5 s 2 + 4  s ← Y  2 s 4 + 3s 2   2 s 2 + 4 2 s 3 + 6 s  s ← Z   2 s3 + 4 s   2 s 2 s 2 + 4  s ← Y   2s 2  1 4 2 s  s ← Z  2 2s 0 The impedances are connected in the series branches, whereas the admittances are connected in the parallel branches in a Cauer or ladder realisation. The network is shown in Fig. 9.36. 1H

1H 2

1H

1F 2

1F

Fig. 9.36

Example 9.47

Synthesize the following LC impedance function in Cauer II form: Z ( s) =

s3 + 2 s s4 + 4 s 2 + 3

Solution The Cauer II form is obtained by continued fraction expansion about the pole at the origin. Z ( s) =

s ( s 2 + 2) ( s 2 + 3)( s 2 + 1)

But Z(s) has a zero at the origin. Hence, the continued fraction expansion of Y (s) is carried out. Arranging the polynomials in ascending order of s, Y ( s) =

3 + 4s2 + s4 2 s + s3

9.6 Realisation of RC Functions 9.47

By continued fraction expansion of Y (s),  3  2 s + s3  3 + 4 s 2 + s 4  ← Y   2s 3+

3 2 s 2 5 2 4 4 s + s  2 s + s3  ← Z   5s 2 4 2 s + s3 5 s3  5 2 4  25 ←Y s +s   2s 5  2 5 2 s 2 1  1 s 4  s3  ← Z  5  5s 1 3 s 5 0

The impedances are connected in the series branches whereas the admittances are connected in the parallel branches in a Cauer or ladder realisation. The network is shown in Fig. 9.37.

9.6

5F 4

2H 3

2 H 25

5F

Fig. 9.37

REalIsaTION OF RC FUNCTIONs

RC driving point immittance functions have following properties: 1. The poles and zeros are simple and are located on the negative real axis of the s plane. 2. The poles and zeros are interlaced. 3. The lowest critical frequency nearest to the origin is a pole. 4. The highest critical frequency farthest to the origin is a zero. 5. Residues evaluated at the poles of ZRC (s) are real and positive. d Z RC is negative. 6. The slope dσ 7. Z RC (∞) < Z RC ( 0 ) . RC functions can also be realised in four different ways. The impedance function of RC networks is given by, Z ( s) =

H ( s + σ1 )( s + σ 3 )... s( s + σ 2 )...

9.6.1 Foster Realisation Foster I Form

The Foster I form is obtained by partial-fraction expansion of Z(s). Z ( s) =

K0 K1 K2 + + + ... + K ∞ s s + σ1 s + σ 2

9.48 Circuit Theory and Transmission Lines where K0, K1, K2, … K∞ are residues of Z(s).

K o = sZ ( s) s = 0

K i = ( s + σ i ) Z ( s) s = −σ i Z ( s) K∞ = s s→∞ The first term

K0 1 represents the impedance of a capacitor of farads. s Ko

The last term K∞ represents the impedance of a resistor of K∞ ohms. Ki The remaining terms, i.e., represent the impedance of the parallel combination of resistor Ri and s + σi capacitor Ci. For parallel combinaton of Ri and Ci,  1  Ri   Ci s  Ki Z ( s) = = 1 s + σi Ri + Ci s Ri =

Ki 1 and Ci = σi Ki

Table 9.3 Realisation of Foster-I form of RC network Impedance function

K0 1 = s C0 s

Ki = s + σi

 1  ( Ri )   Ci s  1 Ri + Ci s

K ∞ = R∞

Element

1 K0 Ki Ri = σi

C0 =

Ci =

R∞ = K∞

The network corresponding to the Foster-I form is shown in Fig. 9.38. Foster II Form

The Foster-II form is obtained by partial fraction

1 Ki

C0

Z(s)

R1

Ri

C1

Ci

R∞

1 expansion of Y(s). Since Y ( s) = has negative residue at its Z ( s) Y ( s) Fig. 9.38 Foster-I form of RC network . pole, Foster II form is obtained by expanding s Y ( s) K o n Ki = +∑ + K∞ s s i =1 ( s + σ i )

9.6 Realisation of RC Functions 9.49

Multiplying this equation by s, n

Ki s + K∞ s i =1 ( s + σ i )

Y ( s) = K o + ∑

The first term Ko represents the conductance of a resistor of

1 ohms. Ko

The last term K∞ s represents the admittance of a capacitor of K∞ farads. Ki s The remaining terms, i.e., represent the admittance of series combination of resistor Ri and capacitor s + σi Ci with Ri =

1 K ohms and Ci = i farads. σi Ki Table 9.4 Realisation of Foster II form of RC network Admittance function

K0 =

Element

1 R0

Ro =

1 Ko

1 Ki K Ci = i σi

 1  R  s Ki i = 1 s + ωi s+ Ri Ci

Ri =

C∞ = K ∞

K∞ s = C∞ s The network corresponding to the Foster II form is shown in Fig. 9.39. Z(s)

R0

R1

Ri

C1

C∞

Ci

Fig. 9.39 Foster II form of RC network

9.6.2 Cauer Realisation Cauer I Form The Cauer I form is obtained by removal of the pole from the impedance function Z(s) at s = ∞. This is the same as a continued fraction expansion of the impedance function about infinity. The impedance Z(s) can be written as a continued fraction expansion. Z ( s) = R1 +

The network is shown in Fig. 9.40.

1 C2 s +

1 R3 +

1 C4 s + ...

9.50 Circuit Theory and Transmission Lines In the network shown in Fig. 9.40, if Z(s) has a zero at s = ∞, the first element is the capacitor C1. If Z(s) is a constant at s = ∞, the first element is R1. If Z(s) has a pole at s = 0, the last element is Cn. If Z(s) is a constant at s = 0, the last element is Rn.

R1

R3

C2

Z(s)

Rn − 1 C4

Cn

Cauer II Form The Cauer II form is obtained by removal of the pole from the impedance function at the origin. This is Fig. 9.40 Cauer-I form of RC network the same as a continued fraction expansion of an impedance function about the origin. If the given impedance function has a pole at the origin, it is removed as a capacitor C1. The reciprocal of the remainder function has a minimum value at s = 0 which is removed as a constant of resistor R2. If the original impedance has no pole at the origin, then the first capacitor is absent and the process is repeated with the removal of the constant corresponding to the resistor R2. The impedance Z(s) can be written as a continued fraction expansion. 1 1 C1 C3 Cn − 1 + 1 1 C1s + 1 1 R2 + Z(s) R2 R4 Rn C3 s 1 + ... R4 The network is shown in Fig. 9.41. In the network shown in Fig. 9.41, if Z (s) has a pole at s = 0, the Fig. 9.41 Caver-II form of RC network first element is C1. If Z (s) is a constant at s = 0, the first element is R2. If Z (s) has a zero at s = ∞, the last element is Cn. If Z (s) is constant at s = ∞, the last element is Rn. Z ( s) =

Example 9.48 (a) Z ( s) =

Determine whether following functions are RC impedance function or not.

3( s + 2)( s + 4 ) s( s + 3)

(b)

2( s + 1)( s + 3) ( s + 2)( s + 6 )

Solution (a) Z ( s) =

3( s + 2)( s + 4) s( s + 3)

The function Z(s) has poles at s = 0 and s = −3 and zeros at s = −2 and s = −4 as shown in Fig. 9.42. The poles and zeros are simple and located on the negative real axis of the s plane. The poles and zeros are interlaced. The lowest critical frequency nearest to the origin is a pole. Hence, the function Z(s) is an RC impedance function. (b) Z ( s) =

jw

s −5

−4

−3

−2

−1

Fig. 9.42

2( s + 1)( s + 3) ( s + 2)( s + 6)

The function Z(s) has poles at s = −2 and s = −6 and zeros at s = −1 and s = −3 as shown in Fig. 9.43. The poles and zeros are simple and located on the negative real axis of the s-plane. The poles and zeros are interlaced. But the lowest critical frequency nearest to the origin is not a pole, but zero. Hence, the function Z(s) is not an RC impedance function.

0

jw

s −6 −5

−4 −3

−2

−1

Fig. 9.43

0

9.6 Realisation of RC Functions 9.51

Example 9.49

Realise the Foster and Cauer forms of the impedance function Z ( s) =

( s + 1)( s + 3) s( s + 2 )

Solution The function Z(s) has poles at s = 0 and s = –2 and zeros at s = –1 and s = –3 as shown in Fig. 9.44. From the pole-zero diagram, it is clear that poles and zeros are simple and lie on the negative real axis. The poles and zeros are interlaced and the lowest critical frequency nearest to the origin is a pole. Hence, the function Z(s) is an RC function. Foster I Form The Foster I form is obtained by partial fraction expansion of impedance function Z(s). Since the degree of the numerator is greater than the degree of the denominator, division is first carried out. s2 + 4s + 3 Z ( s) = 2 s + 2s   s 2 + 2 s s 2 + 4 s + 3  1  

jw

s −3

−2

−1

0

Fig. 9.44

s2 + 2s 2s + 3 Z ( s) = 1 +

2s + 3 s + 2s 2

= 1+

2s + 3 s( s + 2)

By partial-fraction expansion, K1 K 2 + s s+2 (1)(3) 3 = where K1 = sZ ( s) s = 0 = 2 2 ( −2 + 1)( −2 + 3) 1 K 2 = ( s + 2) Z ( s) s = −2 = = −2 2 3 1 Z ( s) = 1 + 2 + 2 s s+2 The first term represents the impedance of a resistor of 1 Ω. The second term represents the impedance of 2 a capacitor of F. The third term represents the impedance of parallel RC circuit for which 3 1 2F 3 1Ω Ci Z RC ( s) = 1 s+ Ri Ci 1 Ω By direct comparison, 2F 4 1 R= Ω 4 C=2F The network is shown in Fig. 9.45. Fig. 9.45 Z ( s) = 1 +

9.52 Circuit Theory and Transmission Lines Foster II Form The Foster II form is obtained by the partial-fraction expansion of admittance function Y ( s) . s 1 s( s + 2) Y ( s) = = Z ( s) ( s + 1)( s + 3) Y ( s) s+2 = s ( s + 1)( s + 3) By partial-fraction expansion, Y ( s) K K = 1 + 2 s s +1 s + 3 Y ( s) ( −1 + 2) 1 where K1 = ( s + 1) = = s s = −1 ( −1 + 3) 2 Y ( s) ( −3 + 2) 1 = K 2 = ( s + 3) = −3 + 1) 2 s s = −3 (− 1 1 Y ( s) = 2 + 2 s s +1 s + 3 1 1 s s 2 Y ( s) = + 2 s +1 s + 3 These two terms represent the admittance of a series RC circuit. For a series RC circuit.  1  R  s i YRC ( s) = 1 s+ Ri Ci By direct comparison, 1 F 2 1 R2 = 2 Ω, C2 = F 6 R1 = 2 Ω,

C1 =

The network is shown in Fig. 9.46.

2Ω

2Ω

1F 2

1F 6

Fig. 9.46

Cauer I Form The Cauer I form is obtained by continued fraction expansion about the pole at infinity. Z ( s) =

s2 + 4s + 3 s2 + 2s

By continued fraction expansion,   s 2 + 2 s s 2 + 4 s + 3  1 ← Z   s 2 + 2s 1  2 s + 3 s 2 + 2 s  s ← Y  2

9.6 Realisation of RC Functions 9.53

s2 +

3 s 2 1   s 2 s + 3  4 ← Z  2  2s

 1 1 3 s  s ← Y  2 6 1 s 2 0 The impedances are connected in the series branches whereas admittances are connected in the parallel branches. The network is shown in Fig. 9.47.

1Ω

4Ω 1F 2

1F 6

Fig. 9.47

Cauer II Form The Cauer II form is obtained from continued fraction expansion about the pole at the origin. Arranging the numerator and denominator polynomials of Z(s) in ascending order of s, Z ( s) =

3 + 4s + s2 2s + s2

By continued fraction expansion,  3  2s + s2  3 + 4 s + s2  ← Z   2s 3+

3 s 2 5  4 s + s2  2s + s 2  ← Y  5 2 4 2s + s 2 5 1 2 5  25 ←Z s  s + s2    2s 2 5 5 s 2  1 1 s2  s2  ← Y  5 5 1 2 s 5 0

2F 3

2 25 F

5 Ω 4

The impedances are connected in the series branches whereas admittances are connected in the parallel branches. The network is shown in Fig. 9.48.

Fig. 9.48

5Ω

9.54 Circuit Theory and Transmission Lines

Example 9.50

Determine the Foster form of realisation of the RC impedance function. Z ( s) =

( s + 1)( s + 3) s( s + 2)( s + 4 )

Solution Foster I Form The Foster I form is obtained by the partial-fraction expansion of the impedance function Z(s). By partial-fraction expansion, Z ( s) = where

K0 K K + 1 + 2 s s+2 s+4

K 0 = sZ ( s) s = 0 =

(1)(3) 3 = ( 2)( 4) 8

K1 = ( s + 2) Z ( s) s = −2 =

( −2 + 1)( −2 + 3) ( −1)(1) 1 = = ( −2)( −2 + 4) ( −2)( 2) 4

K 2 = ( s + 4) Z ( s) s = −4 =

( −4 + 1)( −4 + 3) ( −3)( −1) 3 = = ( −4)( −4 + 2) ( −4)( −2) 8

3 3 1 Z ( s) = 8 + 4 + 8 s s+2 s+4 8 The first term represents the impedance of a capacitor of F. The remaining terms represent the impedance 3 of a parallel RC circuit for which 1 Ci Z RC ( s) = 1 s+ Ri Ci 8F 3

By direct comparison, 1 Ω, C1 = 4 F 8 3 8 Ω, C2 = F R2 = 32 3 R1 =

1Ω 8

4F

3 Ω 32

8F 3

Fig. 9.49

The network is shown in Fig. 9.49.

Foster II Form The Foster II form is obtained by partial-fraction expansion of admittance function Y ( s) =

s( s + 2)( s + 4) ( s + 1)( s + 3)

Y ( s) ( s + 2)( s + 4) s 2 + 6 s + 8 = = s ( s + 1)( s + 3) s 2 + 4 s + 3

Y ( s) . s

9.6 Realisation of RC Functions 9.55

Since the degree of the numerator is equal to the degree of the denominator, division is carried out first.

)

s 2 + 4 s + 3 s 2 + 6 s + 8 (1 s2 + 4s + 3 2s + 5 Y ( s) 2s + 5 2s + 5 = 1+ 2 = 1+ s ( s + 1)( s + 3) s + 4s + 3 By partial-fraction expansion, Y ( s) K K = 1+ 1 + 2 s s +1 s + 3 Y ( s) ( −1 + 2)( −1 + 4) (1)(3) 3 K1 = ( s + 1) = = = where s s = −1 ( −1 + 3) 2 2 K 2 = ( s + 3)

( −3 + 2)( −3 + 4) ( −1)(1) 1 Y ( s) = = = s s= −3 ( −3 + 1) ( −2) 2

1 3 Y ( s) 2 = 1+ + 2 s s +1 s + 3 1 3 s s Y ( s) = s + 2 + 2 s +1 s + 3 The first term represents the admittance of a capacitor of 1 F. The other two terms represent the admittance of a series RC circuit. For a series RC circuit,  1  R  s i YRC ( s) = 1 s+ Ri Ci By direct comparison, 2 3 Ω, C1 = F 3 2 1 R2 = 2 Ω, C2 = F 6 R1 =

1F

2Ω

3F 2

1F 6

Fig. 9.50

The network is shown in Fig. 9.50.

Example 9.51

2Ω 3

Realise Foster forms of the following RC impedance function Z ( s) =

2( s + 2)( s + 4 ) ( s + 1)( s + 3)

Solution Foster I Form The Foster I form is obtained by the partial-fraction expansion of the impedance function Z(s). Since the degree of the numerator is equal to the degree of the denominator, division is carried out first.

9.56 Circuit Theory and Transmission Lines Z ( s) =

2 s 2 + 12 s + 16 s2 + 4s + 3

  s 2 + 4 s + 3 2 s 2 + 12 s + 16  2   2 s 2 + 8s + 6 4 s + 10 Z ( s) = 2 +

4 s + 10 s2 + 4s + 3

= 2+

4 s + 10 ( s + 1)( s + 3)

By partial-fraction expansion, Z ( s) = 2 +

K1 K + 2 s +1 s + 3

2( −1 + 2)( −1 + 4) =3 ( −1 + 3) 2( −3 + 2)( −3 + 4) =1 = ( −3 + 1)

K1 = ( s + 1) Z ( s) s = −1 =

where

K 2 = ( s + 3) Z ( s) s = −3

3 1 + s +1 s + 3 The first term represents the impedance of a resistor of 2 Ω. The remaining terms represent the impedance of a parallel RC circuit for which 1 Ci Z RC ( s) = 1Ω 1 3 3Ω s+ 2Ω Ri Ci By direct comparison, 1 1F 1F R1 = 3 Ω, C1 = F 3 3 1 R2 = Ω, C2 = 1 F 3 Fig. 9.51 The network is shown in Fig. 9.51. Y ( s) . Foster II Form The Foster II form is obtained by partial-fraction expansion of admittance function s ( s + 1)( s + 3) Y ( s) = 2( s + 2)( s + 4) Y ( s) ( s + 1)( s + 3) = s 2 s( s + 2)( s + 4) Z ( s) = 2 +

By partial-fraction expansion,

where

Y ( s) K 0 K K = + 1 + 2 s s s+2 s+4 (1)(3) 3 Y ( s) K0 = s = = s s = 0 ( 2)( 2)( 4) 16 K1 = ( s + 2)

( −1)(1) 1 ( −2 + 1)( −2 + 3) Y ( s) = = = 2( −2)( −2 + 4) 2( −2)( 2) 8 s s = −2

9.6 Realisation of RC Functions 9.57

K 2 = ( s + 4)

3 Y ( s) ( −4 + 1)( −4 + 3) ( −3)( −1) = = = s s = −4 2( −4)( −4 + 2) 2( −4)( −2) 16

3 3 1 Y ( s) 16 16 8 = + + s s s+2 s+4 3 1 s s 3 Y ( s) = + 8 + 16 16 s + 2 s + 4 16 Ω. The other two terms represent the admittance The first term represents the admittance of a resistor of 3 of a series RC circuit. For a series RC circuit.  1  R  s i YRC ( s) = 1 s+ Ri Ci By direct comparison, 1 F 16 16 3 R2 = Ω, C2 = F 3 64 The network is shown in Fig. 9.52. R1 = 8 Ω,

Example 9.52

C1 =

16 Ω 3

8Ω 1 F 16

16 Ω 3

3 F 64

Fig. 9.52

Obtain the Cauer forms of the RC impedance function Z ( s) =

( s + 2)( s + 6 ) 2( s + 1)( s + 3)

Solution Cauer I Form The Cauer I form is obtained by continued fraction expansion about the pole at infinity. Z ( s) =

( s + 2)( s + 6) s 2 + 8s + 12 = 2( s + 1)( s + 3) 2 s 2 + 8s + 6

By continued fraction expansion, 1  2 s 2 + 8s + 6 s 2 + 8s + 12  ← Z  2 s2 + 4s + 3 1  4 s + 9 2 s 2 + 8s + 6  s ← Y  2 9 2s 2 + s 2 7  8 s + 6 4 s + 9  ← Z  7 2 48 4s + 7

9.58 Circuit Theory and Transmission Lines 15  7  49  s + 6  s ← Y 72 30 7 s 2  15  5 6  ← Z  7  14 15 7 0

1 Ω 2

The impedances are connected in the series branches whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.53.

8 Ω 7

1 F 2

5 Ω 14

49 F 30

Fig. 9.53

Cauer II Form The Cauer II form is obtained by continued fraction expansion about the pole at the origin. Arranging the polynomials in ascending order of s, Z ( s) =

12 + 8s + s 2 6 + 8s + 2 s 2

By continued fraction expansion,   6 + 8s + 2 s 2  12 + 8s + s 2  2   12 + 16 s + 4 s 2 − 8 s − 3s 2 Since negative term results, continued fraction expansion of Y(s) is carried out. Y ( s) =

6 + 8s + 2 s 2 12 + 8s + s 2

By continued fraction expansion, 1  12 + 8s + s 2  6 + 8s + 2 s 2  ← Y  2 1 6 + 4s + s2 2 3  3 4 s + s 2  12 + 8s + s 2  ← Z s 2  9 12 + s 2 7 3  8 s + s2  4s + + s2  ← Y  7 2 2 8 2 4s + s 7 5 2 7  49 s  s + s2  ←Z   5s 14 2

9.6 Realisation of RC Functions 9.59

7 s 2  5  5 s2  s2  ← Y  14  14 5 s 14 0 The impedances are connected in the series branches, whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.54.

Example 9.53

1 F 3

5 F 49

14 Ω 5

7 Ω 8

2Ω

Fig. 9.54

Realise the RC impedance in Cauer I and Foster I forms s+4 Z ( s) = ( s + 2)( s + 6 )

Solution Cauer I Form The Cauer I form is obtained by continued fraction expansion of Z(s) about the pole at infinity. In the above function, the degree of the numerator is less than the degree of the denominator which indicates presence of a zero at infinity. Hence, the admittance function Y(s) has a pole at infinity. Y ( s) =

s 2 + 8s + 12 s+4

By continued fraction expansion,   s + 4 s 2 + 8s + 12  s ← Y   s2 + 4s 1  4 s + 12 s + 4  ← Z  4 s+3   1 4 s + 12  4 s ← Y   4s

1

Ω

4  1 12 1  ← Z   12 1 1F 0 The impedances are connected in series branches, whereas the admittances are connected in parallel branches. The network is shown Fig. 9.55 in Fig. 9.55. Foster I Form The Foster I form is obtained by partial fraction expansion of Z(s). s+4 Z ( s) = ( s + 2)( s + 6) By partial-fraction expansion, K K Z ( s) = 1 + 2 s+2 s+6

1 Ω 12

4F

9.60 Circuit Theory and Transmission Lines ( −2 + 4) 1 = ( −2 + 6) 2 ( −6 + 4) 1 = = ( −6 + 2) 2

K1 = ( s + 2) Z ( s) s = −2 =

where

K 2 = ( s + 6) Z ( s) s = −6

1 1 Z ( s) = 2 + 2 s+2 s+6 These two terms represent the impedance of a parallel RC circuit for which Z RC ( s) =

1 Ci s+

1 Ri Ci

1Ω 4

1 Ω 12

By direct comparison, 1 Ω, C1 = 2 F 4 1 Ω, C2 = 2 F R2 = 12 R1 =

2F

2F

Fig. 9.56

The network is shown in Fig. 9.56.

The RC driving-point impedance function is given as Z ( s) = H Realise the impedance function in the ladder form, given Z( −2) = 1.

Example 9.54

Solution Putting s = − 2, Z ( −2) = H

( s + 1)( s + 4) s( s + 3)

( −2 + 1)( −2 + 4) =H ( −2)( −2 + 3)

H =1 ( s + 1)( s + 4) s( s + 3) Cauer I Form The Cauer I form is obtained by continued fraction expansion of Z(s) about the pole at infinity. By continued fraction expansion of Z(s),   s 2 + 3s  s 2 + 5 s + 4  1 ← Z   2 s + 3s 1  2 s + 4  s 2 + 3s  s ← Y  2 2 s + 2s   s 2 s + 4  2 ← Z   2s  1 4 s  s ← Y  4 s 0 Z ( s) =

9.6 Realisation of RC Functions 9.61

The impedances are connected in the series branches whereas admittances are connected in the parallel branches. The network is shown in Fig. 9.57.

1Ω

1 F 2

Cauer II Form The Cauer II form is obtained by continued fraction expansion about the pole at the origin. Arranging the polynomials in ascending order of s, Z ( s) =

1 F 4

Fig. 9.57

4 + 5s + s 2 3s + s

2Ω

2

By continued fraction expansion, 4  3s + s 2  4 + 5s + s 2  ← Z   3s 4+

4 s 3 11  9 s + s 2  3s + s 2  ← Y   11 3 9 3s + s 2 11 2 2  11  121 s  s + s2  ←Z   6s 11 3 11 s 3  2 2 s2  s2  ← Y  11  11

3 F 4

2 2 s 11 0

6 F 121

11 Ω 9

The impedances are connected in the series branches whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.58.

Example 9.55

11 Ω 2

Fig. 9.58

An impedance function has the pole-zero diagram as shown in Fig. 9.59. Find the 3 impedance function such that Z(−4) = and realise in Cauer I and Foster II forms. 4 jw

s −3

−2

−1

0

Fig. 9.59

9.62 Circuit Theory and Transmission Lines Solution The function Z(s) has poles at s = 0 and s = −2 and zeros at s = −1 and s = −3. Z ( s) = H

( s + 1)( s + 3) s( s + 2)

Putting s = −4, Z ( −4) = H

( −4 + 1)( −4 + 3) ( −3)( −1) 3 =H = H ( −4)( −4 + 2) ( −4)( −2) 8

3 3 = H 4 8 H =2 Z ( s) =

2( s + 1)( s + 3) 2 s 2 + 8s + 6 = s( s + 2) s2 + 2s

Cauer I Form The Cauer I form is obtained by continued fraction expansion of Z(s) about the pole at infinity. By continued fraction expansion of Z(s),   s 2 + 2 s 2 s 2 + 8 s + 6  2 ← Z   2s2 + 4 s 1  4 s + 6 s 2 + 2 s  s ← Y  4 s2 +

3 s 2 1   s 4 s + 6  8 ← Z   2 4s 1 1 6 s  s ← Y  2  12 1 s 2 0

The impedances are connected in the series branches whereas admittances are connected in the parallel branches. The network is shown in Fig. 9.60. Foster II Form The Foster II form is obtained by partial fraction Y ( s) . expansion of s

2Ω

8Ω 1 F 4

Fig. 9.60

1 F 12

9.7 Realisation of RL Functions 9.63

Y ( s) s+2 = s 2( s + 1)( s + 3) By partial fraction expansion, Y ( s) K K = 1 + 2 s s +1 s + 3 K1 = ( s + 1)

where

Y ( s) ( −1 + 2) 1 = = s s = −1 2( −1 + 3) 4

K 2 = ( s + 3)

−1 Y ( s) ( −3 + 2) 1 = = = s s = −3 2( −3 + 1) 2( −2) 4

1 1 Y ( s) 4 = + 4 s s +1 s + 3 1 1 s s Y ( s) = 4 + 4 s +1 s + 3 Two terms represent the admittance of a series RC circuit. For a series RC circuit,  1  R  s i YRC ( s) = 1 s+ Ri Ci 4Ω

By direct comparison, R1 = 4 Ω, C1 =

1 F 4

R2 = 4 Ω, C2 =

1 F 12

1 F 4

4Ω 1 F 12

Fig. 9.61

The network is shown in Fig. 9.61.

9.7

REalIsaTION OF RL FUNCTIONs

RL driving point immittance functions have following properties: 1. 2. 3. 4. 5.

The poles and zeros are simple and are located on the negative real axis of the s plane. The poles and zeros are interlaced. The lowest critical frequency is a zero which may be at s = 0. The highest critical frequency is a pole which may be at infinity. Z ( s) Residues evaluated at the poles of ZRL(s) are real and negative while that of RL are real and s positive. d Z RL is positive. 6. The slope dσ 7. Z RL ( 0 ) < Z RL (∞) .

9.64 Circuit Theory and Transmission Lines The admittance of an inductor is similar to the impedance of a capacitor. Hence, properties of an RL admittance are identical to those of an RC impedance and vice-versa, i.e., Z RC ( s ) = YRL ( s )

Z RL ( s ) = YRC ( s )

An RL admittance can be considered as the dual of an RC impedance and vice-versa.

Example 9.56

Indicate which of the following functions are either RL, RC or LC impedance functions.

(a) Z ( s) =

4( s + 1)( s + 3) s( s + 2 )

(b) Z ( s) =

s( s + 4 )( s + 8 ) ( s + 1)( s + 6 )

(c) Z ( s) =

( s + 1)( s + 4 ) s( s + 2 )

(d) Z ( s) =

2( s + 1)( s + 3) ( s + 2)( s + 16 )

4( s + 1)( s + 3) s( s + 2) This is an RC impedance function since (i) poles and zeros are on the negative real axis, (ii) they are interlaced, and (iii) critical frequency nearest to the origin is a pole. Solution (a) Z ( s) =

(b) Z ( s) =

s( s + 4)( s + 8) ( s + 1)( s + 6)

This is an RL impedance function as (i) poles and zeros are on the negative real axis, (ii) they are interlaced, and (iii) critical frequency nearest to the origin is a zero. (c) Z ( s) =

( s + 1)( s + 4) s ( s + 2)

This is an RC impedance function since (i) poles and zeros are on the negative real axis, (ii) they are interlaced, and (iii) critical frequency nearest to the origin is a pole. (d) Z ( s) =

2( s + 1)( s + 3) ( s + 2)( s + 6)

This is an RL impedance function as (i) poles and zeros are on the negative real axis, (ii) they are interlaced, and (iii) critical frequency nearest to the origin is a zero.

Example 9.57

Realise following RL impedance function in Foster-I and Foster-II form. Z ( s) =

2( s + 1)( s + 3) ( s + 2)( s + 6 )

Solution Foster I Form The Foster I form is obtained by partial-fraction expansion of the impedance function Z(s). By partial-fraction expansion, Z ( s) =

K1 K + 2 s+2 s+6

9.7 Realisation of RL Functions 9.65

where

K1 = ( s + 2) Z ( s) s = −2 =

2( −2 + 1)( −2 + 3) 1 =− ( −2 + 6) 2

K 2 = ( s + 6) Z ( s) s = −6 =

2( −6 + 1)( −6 + 3) 15 =− ( −6 + 2) 2

Since residues of Z (s) are negative, partial fraction expansion of

Z ( s) is carried out. s

Z ( s) 2( s + 1)( s + 3) = s s( s + 2)( s + 6) By partial fraction expansion, Z ( s) K 0 K K = + 1 + 2 s s s+2 s+6 2(1)(3) 1 Z ( s) = = s s = 0 ( 2)(6) 2 2( −2 + 1)( −2 + 3) Z ( s) = K1 = ( s + 2) = ( 2)( −2 + 6) s s = −2 Z ( s) 2( −6 + 1)( −6 + 3) K 2 = ( s + 6) = = s s = −6 ( −6)( −6 + 2) K0 = s

where

1 4 5 4

5 1 1 Z ( s) 2 4 = + + 4 s s s+2 s+6 5 1 s s 1 4 Z ( s) = + + 4 2 s+2 s+6 1 The first term represents the impedance of the resistor of Ω . The other two terms represent the impedance 2 of the parallel RL circuit for which Z RL ( s) =

Ri s R s+ i Li

By direct comparison, 1 1 R1 = Ω, L1 = H 4 8 5 5 H R2 = Ω, L2 = 4 24

1 Ω 2

1Ω 4

5 Ω 4

1 H 8

5 H 24

Fig. 9.62

The network is shown in Fig. 9.62. Foster II Form The Foster II form is obtained by partial fraction expansion of Y (s). Since the degree of the numerator is equal to the degree of the denominator, division is first carried out. Y ( s) =

( s + 2)( s + 6) s 2 + 8s + 12 = 2( s + 1)( s + 3) 2 s 2 + 8s + 6

9.66 Circuit Theory and Transmission Lines 1  2 s 2 + 8s + 6 s 2 + 8s + 12   2 s2 + 4s + 3 4s + 9 Y ( s) =

1 4s + 9 1 4s + 9 + 2 = + 2 2 s + 8s + 6 2 2( s + 1)( s + 3)

By partial-fraction expansion, 4s + 9 K K = 0 + 1 2( s + 1)( s + 3) s + 1 s + 3 ( −4 + 9) 5 K 0 = ( s + 1)Y1 ( s) s = −1 = = where 2( −1 + 3) 4 ( −12 + 9) 3 = K1 = ( s + 3)Y1 ( s) s = −3 = 2( −3 + 1) 4 3 5 1 4 4 Y ( s) = + + 2 s +1 s + 3 The first term represents the admittance of a resistor of 2 Ω. The other two terms represent the admittance of a series RL circuit. For a series RL circuit, 1 Li YRL ( s) = R s+ i Li 4Ω By direct comparison, 4Ω 5 4 4 2Ω R1 = Ω, L1 = H 4 5 5 4 H H 5 3 4 R2 = 4 Ω, L2 = H 3 Fig. 9.63 The network is shown in Fig. 9.63. Y1 ( s) =

Example 9.58

Find the Foster forms of the RL impedance function: Z ( s) =

( s + 1)( s + 4 ) ( s + 5 )( s + 3)

Solution Z ( s) . Foster I Form The Foster I form is obtained by partial-fraction expansion of impedance function s Z ( s) ( s + 1)( s + 4) = s s( s + 5)( s + 3) By partial-faction expansion,

where

Z ( s) K 0 K K = + 1 + 2 s s s+3 s+5 (1)( 4) 4 Z ( s) = = K0 = s s s = 0 (5)(3) 15

9.7 Realisation of RL Functions 9.67

Z ( s) = s s = −3 Z ( s) K 2 = ( s + 5) = s s = −5 2 4 1 Z ( s) 15 5 3 = + + s s s+3 s+5 2 1 s s 4 5 3 Z ( s) = + + 15 s + 3 s + 5 K1 = ( s + 3)

( −3 + 1)( −3 + 4) ( −2)(1) 1 = = ( −3)( −3 + 5) ( −3)( 2) 3 ( −5 + 1)( −5 + 4) ( −4)( −1) 2 = = ( −5)( −5 + 3) ( −5)( −2) 5

The first term represents the impedance of the resistor of impedance of a parallel RL circuit for which Ri s Z RL ( s) = R s+ i Li By direct comparison, 1 1 R1 = Ω, L1 = H 3 9 2 2 H R2 = Ω, L2 = 5 25 The network is shown in Fig. 9.64.

4 Ω . The other two terms represent the 15

4 Ω 15

1Ω 3

2 Ω 5

1 H 9

2 H 25

Fig. 9.64

Foster II Form The Foster II form is obtained by partial fraction expansion of Y (s). Since the degree of the numerator is equal to the degree of the denominator, division is first carried out. ( s + 5)( s + 3) s 2 + 8s + 15 = ( s + 1)( s + 4) s 2 + 5s + 4 s 2 + 5s + 4) s 2 + 8s + 15(1

Y ( s) =

s 2 + 5s + 4 3s + 11 3s + 11 Y ( s) = 1 + ( s + 1)( s + 4) By partial-fraction expansion, K K Y1 ( s) = 0 + 1 s +1 s + 4 where

( −3 + 11) 8 = ( −1 + 4) 3 ( −12 + 11) 1 = = ( −4 + 1) 3

K 0 = ( s + 1)Y1 ( s) s = −1 = K1 = ( s + 4)Y1 ( s) s = −4 8 1 Y1 ( s) = 3 + 3 s +1 s + 4 1 8 Y ( s) = 1 + 3 + 3 s +1 s + 4

9.68 Circuit Theory and Transmission Lines The first term represents the admittance of a resistor of 1 Ω. The other two terms represent the admittance of a series RL circuit. For a series RL circuit, 1 Li YRL ( s) = R s+ i Li 3Ω By direct comparison, 12 Ω 8

1Ω

3 3 R1 = Ω, L1 = H 8 8 R2 = 12 Ω, L2 = 3 H

3 H 8

Fig. 9.65

The network is shown in Fig. 9.65.

Example 9.59

3H

Find the Cauer forms of the RL impedance function: Z ( s) =

2( s + 1)( s + 3) ( s + 2)( s + 6 )

Solution Cauer I Form The Cauer I form is obtained by a continued fraction expansion of Z(s) about the pole at infinity. Z ( s) =

2( s + 1)( s + 3) 2 s 2 + 8s + 6 = ( s + 2)( s + 6) s 2 + 8s + 12

By continued fraction expansion, s 2 + 8s + 12)2 s 2 + 8s + 6( 2 ← Z 2 s 2 + 16 s + 24 − 8s − 18 Since a negative term results, continued fraction expansion of Y (s) is carried out. Y ( s) =

s 2 + 8s + 12 2 s 2 + 8s + 6

By continued fraction expansion, 1  2 s 2 + 8s + 6 s 2 + 8s + 12  ← Y  2 s2 + 4s + 3 1  4 s + 9 2 s 2 + 8s + 6  s ← Z  2 2s2 +

9 s 2 7  8 s + 6 4 s + 9  ← Y  7 2 48 4s + 7

9.7 Realisation of RL Functions 9.69

15  7  49  s + 6  s ← Z 72 30 7 s 2  15  15 6  ←Y  7  42

1 H 2

15 7 0

7Ω 8

2Ω

The impedances are connected in the series branches whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.66.

49 H 30

42 Ω 15

Fig. 9.66

Cauer II Form The Cauer II form is obtained from a continued fraction expansion about the pole at the origin. Arranging the numerator and denominator polynomials of Z(s) in ascending order of s, Z ( s) =

6 + 8s + 2 s 2 12 + 8s + s 2

By continued fraction expansion, 1  12 + 8s + s 2  6 + 8s + 2 s 2  ← Z  2 6 + 4s + 4s +

1 2 s 2 3 2 3 s  12 + 8s + s 2  ← Y s 2  12 +

9 s 2 7 3 8  s + s2  4s + s2  ← Z  2 2 7 8 4s + s2 7 5 2 7  98 s  s + s2  ←Y  10 s 14  2 7 s 2  5  5 s2  s2  ← Z  14  14 5 s 14 0

9.70 Circuit Theory and Transmission Lines The impedances are connected in the series branches whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.67. 1Ω 2

5 Ω 14

8Ω 7

1 H 3

10 H 98

Fig. 9.67

Example 9.60

Obtain the Foster I and Cauer I forms of the RL impedance function. Z ( s) =

s( s + 4)( s + 8) ( s + 1)( s + 6 )

Solution Foster I Form The Foster I form is obtained by partial fraction expansion of

Z ( s) . s

Z ( s) ( s + 4)( s + 8) = s ( s + 1)( s + 6) Since the degree of the numerator is equal to the degree of the denominator, division is first carried out. s 2 + 7 s + 6) s 2 + 12 s + 32(1 s2 + 7s + 6 5s + 26 Z ( s) 5s + 26 5s + 26 = 1+ 2 = 1+ ( s + 1)( s + 6) s s + 7s + 6 By partial-fraction expansion, Z ( s) K K = 1+ 0 + 1 s s +1 s + 6 where

K0 =

5s + 26 −5 + 26 21 = = −1 + 6 5 s + 6 s = −1

K1 =

−30 + 26 4 5s + 26 = = −6 + 1 5 s + 1 s = −6

24 21 Z ( s) = 1+ 5 + 5 s s +1 s + 6 21 4 s s 5 + 5 Z ( s) = s + s +1 s + 6

9.7 Realisation of RL Functions 9.71

The first term represents the impedance of the inductor of 1 H. The other two terms represent the impedance of a parallel RL circuit for which 21 Ω 4 Ω 5 5 Ri s Z RL ( s) = 1H R s+ i Li 21 4 H H By direct comparison, 5 30 21 21 Ω, L1 = H R1 = 5 5 4 4 Fig. 9.68 H R2 = Ω, L2 = 5 30 The network is shown in Fig. 9.68. Cauer I Form The Cauer I form is obtained by continued fraction expansion of Z(s) about the pole at infinity. Z ( s) =

s3 + 12 s 2 + 32 s s2 + 7s + 6

By continued fraction expansion,   s 2 + 7 s + 6 s3 + 12 s 2 + 32 s  s ← Z   s3 + 7 s 2 + 6 s 1  5s 2 + 26 s s 2 + 7 s + 6  ← Y  5 s2 +

26 s 5 9   25 s + 6 5s 2 + 26 s  s ← Z   9 5 50 5s 2 + s 3 28  9  27 ←Y s s + 6   1400 3 5 9 s 5  28  28 6 s s←Z  3  18 28 s 3 0

The impedances are connected in the series branches, whereas the admittances are connected in the parallel branches. The network is shown in Fig. 9.69.

9.72 Circuit Theory and Transmission Lines 25 H 9

1H

28 H 18

140 Ω 27

5Ω

Fig. 9.69

Exercises Test the following polynomials for Hurwitz property: (i) s3 + s2 + 2s + 2 (ii) s4 + s2 + s + 1 (iii) s3 + 4s2 + 5s + 2 (iv) s4 + 7s3 + 6s2 + 21s + 8 (v) s4 + s3 + s + 1 (vi) s7 + 3s6 + 8s5 + 15s4 + 17s3 + 12s2 + 4s (vii) s7 + 2s6 + 2s5 + s4 + 4s3 + 8s2 + 8s + 4 (viii) s7 + 3s5 + 2s3 + s (ix) s5 + 2s3 + s (x) s3 + 2s2 + 4s + 2 (xi) s4 + s3 + 4s2 + 2s + 3 (xii) s5 + 8s4 + 24s3 + 28s2 + 23s + 6 (xiii) s7 − 2s6 + 2s5 + 9s2 + 8s + 4 (xiv) s7 + 3s5 + 2s3 + 3 (xv) s5 + s3 + s (xvi) s6 + 7s4 + 5s3 + s2 + s (xvii) s4 + s3 + 2s2 + 3s + 2 9.2 Determine whether the following functions are positive real: 9.1

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

s3 + 5s s4 + 2s2 + 1 s( s + 3)( s + 5) ( s + 1)( s + 4) 2s2 + 2s + 1 s + 2s2 + s + 2 3

s 4 + 3s3 + s 2 + s + 2 s3 + s 2 + s + 1 2 s 3 + 2 s 2 + 3s + 2 s2 + 1

(vi) (vii) (viii) (ix) (x) (xi) (xii)

s2 + 2s + 1 s2 + 4s + 4 s3 + 2 s 2 + 3s + 1 s3 + 2 s 2 + s + 2 s3 + 2 s 2 + s + 1 s2 + s + 1 s+4 s2 + 2s + 1 s2 + 4s + 3 s2 + 6s + 8 s2 + 1 s3 + 4 s s 4 + 2 s3 + 3s 2 + 1

s 4 + s3 + 3s 2 + 2 s + 1 s2 + 2s + 4 (xiii) ( s + 1)( s + 3) 2s + 4 (xiv) s+5 s2 + 2s (xv) s2 + 1 s2 + 4 (xvi) 3 s + 3s 2 + 3s + 1 9.3 Determine whether the following functions are LC, RC or RL function: 2( s + 1)( s + 3) (i) F ( s) = ( s + 2)( s + 6) (ii)

Z ( s) =

3( s + 2)( s + 4) s( s + 3)

Exercises 9.73

(iii) Z ( s) =

( s + 1)( s + 8) ( s + 2)( s + 4)

(iv) Z ( s) = (v) Z ( s) =

Ks( s + 4) 2

9.5

( s + 1)( s + 3) 2

2

2( s 2 + 1)( s 2 + 9) s ( s 2 + 2)

4( s + 1)( s + 3) (vi) Z ( s) = s( s + 2) Z ( s) =

(viii)

F ( s) =

( s + 1)( s + 2) s( s + 3)

(ix) Z ( s) =

( s + 1)( s + 3) ( s + 2)( s + 4)

(x) Z ( s) =

( s + 2)( s + 4) ( s + 1)

(xi) Y ( s) = (xii) Y ( s) = (xiii) (xiv) 9.4

( s 2 + 1)( s 2 + 3)

(vii)

Z ( s) = Z ( s) =

s( s + 2) 2

(ii) Z ( s) = (iii)

(ii)

Z ( s) =

(iii)

Z ( s) =

s2 + 2s + 2

s2 + s + 1 Realise the following functions in Foster II form: 3( s + 2)( s + 4) (i) Z ( s) = s( s + 3)

(iv) Y ( s) =

2( s 2 + 1)( s 2 + 9) s( s 2 + 4 ) 4( s + 1)( s + 3) ( s + 2)( s + 6) 4( s 2 + 4)( s 2 + 25) s( s 2 + 16)

( s + 2)( s + 5) s( s + 4)( s + 6) Realise the following functions in Cauer I form: Z ( s) =

4( s + 3) ( s + 1)( s + 5)

(i)

F ( s) =

2( s + 1)( s + 3) ( s + 2)( s + 6)

(ii)

Z ( s) =

(iii)

Z ( s) =

(iv)

Z ( s) =

(v)

F ( s) =

( s + 1)( s + 3) s( s + 2)

(vi)

Z ( s) =

s+4 ( s + 2)( s + 6)

(vii)

Z ( s) =

6( s + 2)( s + 4) s( s + 3)

(viii)

Z ( s) =

(ix)

Z ( s) =

(x)

Z ( s) =

9.6

s( s 2 + 4)( s 2 + 16) ( s 2 + 9)( s 2 + 25) ( s 2 + 1)( s 2 + 8) s( s 2 + 4 )

3( s + 2)( s + 4) s( s + 3) 2( s + 1)( s + 9) 2

2

s( s 2 + 4 )

4( s + 1)( s + 3) F ( s) = ( s + 2)( s + 6)

s+4 (iv) Z ( s) = ( s + 2)( s + 6) ( s + 1)( s + 4) (v) Z ( s) = s ( s + 2) (vi) Y ( s) =

Z ( s) =

(v)

Realise the following functions in Foster I form: (i) Z ( s) =

(vii)

( s + 2)( s + 5) s( s + 4)( s + 6)

2( s + 1)( s + 3) ( s + 2)( s + 6) ( s 2 + 1)( s 2 + 3) s ( s 2 + 2) ( s + 1)( s + 3) ( s + 2)( s + 4) 2( s 2 + 1)( s 2 + 9) s( s 2 + 4 )

s3 + 2 s s4 + 4s2 + 3 s( s 2 + 2)( s 2 + 5) ( s 2 + 1)( s 2 + 3) s2 + 2s + 2 s2 + s + 1

9.74 Circuit Theory and Transmission Lines 9.7

Realise the following function in Cauer II form: (i)

F ( s) =

(ii) Z ( s) = (iii) Z ( s) = (iv) Z ( s) =

(v)

F ( s) =

(vi) Z ( s) = (vii) Z ( s) = (viii) Z ( s) = 9.8

jw s

( s 2 + 1)( s 2 + 3)

−5 −4 −3 −2 −1

0

s( s 2 + 2) ( s + 1)( s + 3) ( s + 2)( s + 4)

Fig. 9.70 9.9

2( s 2 + 1)( s 2 + 9) s( s 2 + 4 )

An impedance function has the pole-zero diagram as shown in Fig. 9.71. Find the 8 impedance function such that Z( −4) = and 3 realise in Cauer forms.

( s + 1)( s + 3) s( s + 2)

jw s

s3 + 12 s 2 + 32 s

−3

−2

−1

0

s2 + 7s + 6 2( s + 1)( s + 3) ( s + 2)( s + 6)

Fig. 9.71 9.10 For the realisation of a given function F(s).

s( s 2 + 2)( s 2 + 5)

F ( s) =

( s 2 + 1)( s 2 + 3)

n K0 sK + ∑ 2 i 2 + sK ∞ s i =1 ( s + ω i )

where K0, Ki (i = 1, 2, 3, … n) and K∞ are constants. (i) Mention the type of function (RC, RL or LC) (ii) Given that K0 = 6, K1 = 8, w1 = 4, K2 = 10, w2 = 8, K∞ = 5, find the component values of realised network for F(s) = Z(s) and F(s) = Y(s). Draw neat diagrams.

s + 2s + 2 2

s2 + s + 1

An impedance function has the pole-zero diagram as shown in Fig. 9.70 below. If Z (−2) = 3, synthesise the impedance function in Foster and Cauer forms.

Objective-Type Questions 9.1

9.2

The necessary and sufficient condition for a rational function F(s) to be the driving-point impedance of an RC network is that all poles and zeros should be (a) simple and lie on the negative real axis in the s-plane (b) complex and lie in the left half of s-plane (c) complex and lie in the right-half of s-plane (d) simple and lie on the positive real axis of the s-plane The number of roots of s3 + 5s2 + 7s + 3 = 0 in the left half of s-plane is

(a) zero (c) two 9.3

(b) one (d) three

The first and the last critical frequencies of a driving-point impedance function of a passive network having two kinds of elements, are a pole and a zero respectively. The above property will be satisfied by (a) RL network only (b) RC network only (c) LC network only (d) RC as well as RL network

Objective-Type Questions 9.75

9.4

The pole-zero pattern of a particular network is shown in Fig. 9.72. It is that of an

(a) (b) (c) (d)

jw j2

9.9

j1 s

−j2

9.6

9.7

(b) RC network (d) none of these

The first critical frequency nearest to the origin of the complex frequency plane for an RL driving-point impedance function will be (a) a zero in the left-half plane (b) a zero in the right-half plane (c) a pole in the left-half plane (d) a pole in the right-half plane Consider the following polynomials: P1 = s8 + 2s6 + 4s4 P2 = s6 − 3s2 + 2s2 + 1 P3 = s4 + 3s3 + 3s2 + 2s + 1 P4 = s7 + 2s6 + 2s4 + 4s3 + 8s2 + 8s + 4 which one of these polymials is not Hurwitz? (a) P1 (c) P3

(b) P2 (d) P4

For very high frequencies, the driving-point 4( s + 1)( s + 3) admittance function, Y ( s) = s ( s + 2)( s + 4) behaves as 3 (a) a resistance of Ω 2 (b) a capacitance of 4 F 1 (c) an inductance of H 4 (d) an inductance of 4 H

9.8

(b) s = −1, s = −3 (d) s = −3, s = −4

9.10 Consider the following from the point of view of possible realisation as driving-point impedances using passive elements:

Fig. 9.72

9.5

An RC driving-point impedance function has zeros at s = −2 and s = −5. The admissible poles for the functions would be (a) s = 0, s = −6 (c) s = 0, s = −1

−j1

(a) LC network (c) RL network

a resistance of 0.75 Ω at low frequencies a resistance of 1 Ω at high frequencies both (a) and (b) above none of the above

s+3 The driving-point impedance Z ( s) = s +4 behaves as

1.

3.

1 s( s + 5) s2 + 3

2.

4.

s+3 s ( s + 5) 2

s+5 s( s + 5)

s ( s + 5) Of these, the realisable are (a) 1, 2 and 4 (b) 1, 2 and 3 (c) 3 and 4 (d) none of these 9.11 The poles and zeros of a driving-point function of a network are simple and interlace on the negative real axis with a pole closest to the origin. It can be realised (a) by an LC network (b) as an RC driving point impedance (c) as an RC driving point admittance (d) only by an RLC network 9.12 If F1(s) and F2(s) are two positive real functions then the function which is always positive real, is F ( s) (a) F1(s) F2(s) (b) 1 F2 ( s) (c)

2

2

F1 ( s) F2 ( s) F1 ( s) + F2 ( s)

(d) F1(s) − F2(s)

9.13 The circuit shown in Fig. 9.73 is

Fig. 9.73 (a) Cauer I form (b) Foster I form (c) Cauer II form (d) Foster II form

9.76 Circuit Theory and Transmission Lines 9.14 For an RC driving-point impedance function, the poles and zeros

(c) should alternate on the imaginary axis (d) can lie anywhere on the left half-plane

(a) should alternate on the real axis (b) should alternate only on the negative real axis

Answers to Objective-Type Questions 9.1 (a)

9.2 (a)

9.3 (b)

9.4 (a)

9.5 (a)

9.6 (b)

9.7 (c)

9.8 (c)

9.9 (b)

9.10 (a)

9.11 (b)

9.12 (c)

9.13 (b)

9.14 (b)

10 Two-Port Networks

  10.1     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. 10.1. 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. 10.1  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 10.1  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

V 1, I 1

V2, I2

V1 = AV2 − BI2 I1 = CV2 − DI2

Inverse Transmission

V 2, I 2

V1, I1

V2 = A′ V1 − B ′ I1 I 2 = C ′ V1 − D ′ I1

Hybrid

V 1, I 2

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

+ V2 −

10.2 Network Analysis and Synthesis

  10.2     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. (V1 , V2 ) = f ( I1 , I 2 ) V1 = Z11 I1 + Z12 I 2 V2 = Z 21 I1 + Z 22 I 2 In matrix form, we can write V1   Z11 Z12   I1  V2  =  Z 21 Z 22   I 2  [V ] = [ Z ][ I ] 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. 10.2.

I1

+ Z11

V1 Z12I2 −

Z22 + −

V2

+ Z I − 21 1

−

Fig. 10.2  E   quivalent circuit of the two-port  network in terms of Z parameter

10.2.1 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. 10.3, voltage Vs is applied at the input port with the output port short-circuited.

I2

+

I1 + Vs −

I2 Network

I2′

Fig. 10.3  N   etwork for deriving condition  for reciprocity

10.2  Open-Circuit Impedance Parameters (Z Parameters)  10.3

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 I 2 ′ − Z12 I 2 ′ Vs = Z11 Z 21 Vs Z Z − Z12 Z 21 = 11 22 I2 ′ Z 21 (b) As shown in Fig. 10.4, voltage Vs is applied at the output port with input port short-circuited. I1′ V2 = Vs i.e., V1 = 0

i.e.,

I1

I2 Network

+ −

Vs

I1 = − I1 ′ Fig. 10.4   Network for deriving condition for  From the Z-parameter equations, reciprocity 0 = − Z11 I1 ′ + Z12 I 2 Vs = − Z 21 I1 ′ + Z 22 I 2 Z I 2 = 11 I1 ′ Z12 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

10.2.2 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,

10.4 Network Analysis and Synthesis Vs Vs = I1 I 2 Z11 = Z 22

i.e.,

  example  10.1  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 5.2∠50° V = = 52 ∠50° Ω, Z11 = 1 0.1∠0° I1 I 2 = 0 Z 21

V = 2 I1

I2 =0

Z12 =

4.1∠ − 25° = = 41 ∠ − 25° Ω, 0.1∠0°

Hence, the Z-parameters are  Z11  Z 21

  example 10.2 

Z22

V = 2 I2

=

3.1∠ − 80° = 31 ∠ − 80° Ω 0.1∠0°

=

4.2∠60° = 42 ∠60° Ω 0.1∠0°

I1 = 0

I1 = 0

31∠ − 80° Z12  52∠50° = Z 22   41∠ − 25° 42∠60° 

Find the Z parameters for the network shown in Fig. 10.5. I1

Z1

Z3

I2 +

+ Z2

V1

V2

−

−

Fig. 10.5 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

V1 I1

= Z1 + Z 2 I2 = 0

V2 = Z 2 I1 Z 21 =

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

= Z 2 + Z3 I1 = 0

10.2  Open-Circuit Impedance Parameters (Z Parameters)  10.5

V1 = Z 2 I 2

Also

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. 10.6. Applying KVL to Mesh 1, V1 = Z1 I1 + Z 2 ( I1 + I 2 ) = ( Z1 + Z 2 ) I1 + Z 2 I 2

Z1

Z3

…(i)

+

+

Applying KVL to Mesh 2, V2 = Z3 I 2 + Z 2 ( I1 + I 2 ) = Z 2 I1 + ( Z 2 + Z3 ) I 2

I1

…(ii)

Z12   Z1 + Z 2 = Z 22   Z 2

  example 10.3 

I2

−

−

Comparing Eqs (i) and (ii) with Z-parameter equations,  Z11  Z 21

V2

Z2

V1

Fig. 10.6

Z2  Z 2 + Z3 

Find Z-parameter for the network shown in Fig. 10.7. I1

2Ω

1Ω

1H

I2

+

+

V1

2F

V2

1Ω

−

−

Fig. 10.7 Solution The transformed network is shown in Fig. 10.8. Z1 = 1

1

 1   (1) 1 2s = Z2 = 1 2s + 1 +1 2s Z3 = s + 2

+

+ Z1

V1

Z 21 =

V1 I1 V2 I1

= Z1 + Z 2 = 1 + I2= 0

= Z2 = I2 = 0

1 , 2s + 1

1 2s + 2 = , 2s + 1 2s + 1

1 2s

1 Z 2

−

Z3

V2 −

Fig. 10.8

From definition of Z-parameters, Z11 =

s

2

Z12 = Z 22 =

V1 I2 V2 I2

= Z2 = I1 = 0

1 2s + 1

= Z 2 + Z3 = I1 = 0

1 2 s 2 + 5s + 3 +s+2 = 2s + 1 2s + 1

10.6 Network Analysis and Synthesis

  example 10.4 

Find Z-parameters for the network shown in Fig. 10.9. I1

1Ω

1Ω

I2

+

+

V1

2Ω

V2

2Ω

−

−

Fig. 10.9 Solution The network is redrawn as shown in Fig. 10.10. Applying KVL to Mesh 1, V1 = 3I1 − 2 I 3 …(i) 1Ω Applying KVL to Mesh 2, V2 = 2 I 2 + 2 I 3 Applying KVL to Mesh 3,

…(ii)

1Ω

+

+

2Ω

V1

−2 I1 + 2 I 2 + 5 I 3 = 0

I1

2 2 I1 − I 2 5 5 Substituting Eq. (iii) in Eq. (i),

…(iii)

I3 =

V2

2Ω I3

I2

−

−

Fig. 10.10

4 4 I1 + I 2 5 5 11 4 = I1 + I 2 5 5

V1 = 3I1 −

…(iv)

Substituting Eq. (iii) in Eq. (ii), 4 4 I1 − I 2 5 5 4 6 = I1 + I 2 5 5

V2 = 2 I 2 +

…(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 10.5 

Find the Z-parameters for the network shown in Fig. 10.11. I1

1Ω

1Ω

I2

+ V1

+

1H

−

1H

V2 −

Fig. 10.11

10.2  Open-Circuit Impedance Parameters (Z Parameters)  10.7

Solution The transformed network is shown in Fig. 10.12. Applying KVL to Mesh 1, V1 = ( s + 1) I1 − sI 3

…(i)

Applying KVL to Mesh 2, V2 = sI 2 + sI 3

…(ii)

1

+

+ s

V1 I1

Applying KVL to Mesh 3,

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. 10.12

Substituting Eq. (iii) in Eq. (i), s  s  V1 = ( s + 1) I1 − s  I1 − I2   2s + 1 2s + 1   s2   s 2 + 3s + 1 I1 +  =  I2   2 s + 1  2s + 1 

…(iv)

Substituting Eq. (iii) in Eq. (ii), s  s  V2 = sI 2 + s  I1 − I2   2s + 1 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 + 1 Z12   2 s + 1 = Z 22   s 2  2 s + 1

s2   2s + 1  s2 + s  2 s + 1 

  example 10.6  Find the open-circuit impedance parameters for the network shown in Fig. 10.13. Determine whether the network is symmetrical and reciprocal. 4Ω

I1

1Ω

3Ω

+ V1

I2 +

2Ω

−

V2 −

Fig. 10.13

10.8 Network Analysis and Synthesis Solution The network is redrawn as shown in Fig. 10.14. Applying KVL to Mesh 1, V1 − 1( I1 − I 3 ) − 2 ( I1 + I 2 ) = 0 V1 = 3I1 + 2 I 2 − I 3 …(i) Applying KVL to Mesh 2, V2 − 3 ( I 2 + I 3 ) − 2 ( I1 + I 2 ) = 0 V2 = 2 I1 + 5 I 2 + 3I 3 Applying KVL to Mesh 3, −4 I 3 − 3 ( I 2 + I 3 ) − 1( I 3 − I1 ) = 0 I1 − 3I 2 + 8 I 3 = 0 1 3 I 3 = I1 − I 2 8 8 Substituting Eq. (iii) in Eq. (i),

4Ω

1Ω

I3

3Ω

…(ii) +

+

V1 I1

I2

−

…(iii)

V2

2Ω

−

Fig. 10.14

3  1 V1 = 3I1 + 2 I 2 −  I1 − I 2  8 8  =

23 19 I1 + I 2 8 8

…(iv)

Substituting Eq. (iii) in Eq. (ii), 3  1 V2 = 2 I1 + 5 I 2 + 3  I1 − I 2  8 8  19 31 I1 + I 2 8 8 Comparing Eqs (iv) and (v) with Z-parameter equations,  23 19   Z11 Z12   8 8  Z 21 Z 22  =  19 31   8 8 Since Z11 ≠ Z22, the network is not symmetrical. Since Z12 = Z21, the network is reciprocal. =

…(v)

  10.3     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. ( I1 , I 2 ) = f (V1 ,V2 ) I1 = Y11 V1 + Y12 V2 I 2 = Y21 V1 + Y22 V2 In matrix form, we can write  I1  Y11 Y12  V1   I 2  = Y21 Y 22  V2  [ I ] = [Y ] [V ]

10.3  Short-Circuit Admittance Parameters (Y Parameters)  10.9

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. 10.15. I2

I1 + V1

+ Y11

Y12V2

Y21V1

Y22

2

V2 −

−

Fig. 10.15  Equivalent circuit of the two-port network in terms of Y-parameters

10.3.1 Condition for Reciprocity (a) As shown in Fig. 10.16, voltage Vs is applied at input port with the output port short-circuited. i.e, V1 = Vs I1 I2 V2 = 0 I 2 = − I 2′ + Network I2′ From the Y-parameter equation, − Vs − I 2′ = Y21 Vs I2 = −Y21 Vs

Fig. 10.16  N   etwork for deriving condition  for reciprocity

10.10 Network Analysis and Synthesis (b) As shown in Fig. 10.17, voltage Vs is applied at output port with the input port short-circuited. V2 = Vs V1 = 0 I1 = − I1′

i.e,

From the Y-parameter equation,

I1

I2

I1′

Network

− I1′ = Y12 Vs I1′ = −Y12 Vs Hence, for the network to be reciprocal,

+ −

Vs

Fig. 10.17  N   etwork for deriving condition for  reciprocity

I 2′ I1′ = Vs Vs Y12 = Y21

i.e,

10.3.2 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 10.7 

Test results for a two-port network are

(a) Port 2 short-circuited: V1 = 50 ∠0° V , I 1 = 2.1 ∠ − 30° A, I 2 = −1.1 ∠ − 20° A (b) Port 1 short-circuited: V2 = 50 ∠0° V , I 2 = 3 ∠ − 15° A, I 1 = −1.1 ∠ − 20° A. Find Y-parameters. Solution Y11 =

2.1∠ − 30° I1 = = 0.042∠ − 30° , 50 ∠0° V1 V2 = 0

I Y21 = 2 V1

V2 = 0

−1.1∠ − 20° = = −0.022∠ − 20° , 50 ∠0°

Y12 = Y22

I1 V2

I = 2 V2

=

−1.1∠ − 20° = −0.022∠ − 20°  50 ∠0°

=

3∠ − 15° = 0.06 ∠ − 15°  50 ∠0°

V1 = 0

V1 = 0

Hence, the Y-parameters are  Y11 Y12   0.042∠ − 30° −0.022∠ − 20° Y 21 Y22  =  −0.022∠ − 20° 0.06 ∠ − 15°   

10.3  Short-Circuit Admittance Parameters (Y Parameters)  10.11

  example 10.8 

Find Y-parameters for the network shown in Fig. 10.18. I1

1Ω

3

3Ω

I2

+

+

V1

V2

2Ω

−

−

Fig. 10.18 Solution First Method Case 1 When the output port is short-circuited, i.e., V2 = 0 as shown in Fig. 10.19, Req = 1 + Now,

Also,

Case 2

11 I1 5 I 5 Y11 = 1 =  V1 V2 = 0 11 V1 =

2 2 5 2 I 2 = ( − I1 ) = − × V1 = − V1 5 5 11 11 I 2 Y21 = 2 =−  V1 V2 = 0 11

3Ω

I2

+ V1

2Ω

−

Fig. 10.19

1× 2 2 11 = 3+ = Ω 1+ 2 3 3

11 I2 3 3 I = 2 =  V2 V1 = 0 11

I1

1Ω

3Ω

2 2 3 2 ( − I 2 ) = − × V2 = − V2 3 3 11 11 2 I Y12 = 1 =−  11 V2 V1 = 0 I1 =

I2 +

V2 = Y22

Also

1Ω

I1

When the input port is short-circuited, i.e., V1 = 0 as shown in Fig. 10.20, Req = 3 +

Now

2×3 6 11 = 1+ = Ω 2+3 5 5

2Ω

V2 −

Fig. 10.20

Hence, the Y-parameters are 2  5 −  Y11 Y12   11 11 Y21 Y22  =  2 3  −  11 11  Second Method (Refer Fig. 10.18) V −V I1 = 1 3 1 = V1 − V3

…(i)

10.12 Network Analysis and Synthesis V2 − V3 3 V2 V3 = − 3 3

I2 =

…(ii)

Applying KCL at Node 3, I1 + I 2 =

V3 2

…(iii)

Substituting Eqs (i) and (ii) in Eq. (iii), V V V V1 − V3 + 2 − 3 = 3 3 3 2 V2 11 = V3 V1 + 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), 2  V 1 6 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 10.9 

…(iv)

…(v)

…(vi)

Find Y-parameters of the network shown in Fig. 10.21. I1

V3

2Ω

2Ω

+

I2 +

1Ω V1

V2

3V2 −

−

Fig. 10.21 Solution From Fig. 10.21, V1 − V3 2 1 1 = V1 − V3 2 2

I1 =

…(i)

10.3  Short-Circuit Admittance Parameters (Y Parameters)  10.13

V2 − V3 2 1 1 = V2 − V3 2 2

I2 =

…(ii)

Applying KCL at Node 3, I1 + I 2 + 3 V2 = 0 Substituting Eqs (i) and (ii) in Eq. (iii), V1 − V3 V2 − V3 + + 3 V2 = 0 2 2 2 V3 = V1 + 7 V2 1 7 V3 = V1 + V2 2 2 Substituting Eq. (iv) in Eq. (i), 1 1 1 7  I1 = V1 −  V1 + V2   2 2 2 2  1 7 = V1 − V2 4 4 Substituting Eq. (iv) in Eq. (ii), 1 1 1 7  I 2 = 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

…(iii)

…(iv)

…(v)

…(vi)

  example 10.10  Determine Y-parameters for the network shown in Fig. 10.22. Determine whether the network is symmetrical and reciprocal. I1

3

1Ω

2

2Ω

+ V1

I2 +

2Ω

4Ω

−

V2 −

Fig. 10.22 Solution From Fig. 10.31, V1 − V3 1 = V1 − V3

I1 =

…(i)

Applying KCL at Node 3, V3 V3 − V2 + 2 2 V2 = V3 − 2

I1 =

…(ii)

10.14 Network Analysis and Synthesis Applying KCL at Node 2, V2 V2 − V3 + 4 2 3 V3 = V2 − 4 2

I2 =

…(iii)

Substituting Eq. (i) in Eq. (ii), V2 2 V1 V2 V3 = + 2 4

V1 − V3 = V3 −

…(iv)

Substituting Eq. (iv) in Eq. (ii), V1 V2 V2 + − 2 4 2 V V = 1− 2 2 4

I1 =

Substituting Eq. (iv) in Eq. (iii),

…(v)

3 1 V V  I 2 = V2 −  1 + 2  4 2 2 4  V1 5V2 + 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 10.11 

Determine the short-circuit admittance parameters for the network shown in Fig. 10.23. I1

1Ω

1Ω

I2 +

+ V1

V2

1F

1F

−

−

Fig. 10.23 Solution The transformed network is shown in Fig. 10.24. From Fig. 10.24, V1 − V3 1 = V1 − V3

I1

…(i)

3

1

2

I2

+ V1

I1 =

1

+ 1 s

1 s

V2 −

−

Fig. 10.24

10.3  Short-Circuit Admittance Parameters (Y Parameters)  10.15

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

…(iv)

Substituting Eq. (iv) in Eq. (ii), 1  1  I1 = ( s + 1)  V1 + V2  − V2 s+2 s+2  =

s +1 1 V1 − V2 s+2 s+2

…(v)

Substituting Eq. (iv) in Eq. (iii), 1  1  I 2 = ( s + 1) V2 −  V1 + V2   s+2 s+2  1 s 2 + 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 + 1 −  s + 2 s + 2 

…(vi)

=−

  example 10.12 

Determine Y-parameters for the network shown in Fig. 10.25. I1

1 F 2

1 F 2

I2

+ V1

+

1H

−

1H

V2 −

Fig. 10.25

10.16�Network Analysis and Synthesis Solution The transformed network is shown in Fig. 10.26. From Fig. 10.26, 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 −

����������

s V s (V1 � V3 ) � 3 � (V3 � V2 ) 2 s 2 s 1 s s s V3 � V3 � V3 � V1 � V2 2 s 2 2 2 V3 �

s2 2

2( s � 1)

V1 �

s2 2( s 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 s V2 � V3 2s 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 � 6s2 � 4 V � V2 1 4( s 2 � 1) 4 s( s 2 � 1) Comparing Eqs (iii) and (v) with Y-parameter equation, ��

�Y11 ��Y21

Chapter 10.indd 16

� s3 � 2 s � Y12 � � 4( s 2 � 1) � Y22 �� � s3 �� 2 � 4( s � 1)

…(v)

� s3 � 2 4( s � 1) � s4 � 6s2 � 4 � � 4 s( s 2 � 1) � �

6/13/2016 3:25:08 PM

10.3  Short-Circuit Admittance Parameters (Y Parameters)  10.17

  example 10.13 

Obtain Y-parameters of the network shown in Fig. 10.27. 1 s

I1

1

3

1

1

+

2

I2 +

1 s

V1 −

V2 −

Fig. 10.27 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

…(iii)

Substituting Eq. (iii) in Eq. (i), 1  1  I1 = ( s + 1) V1 − sV2 −  V1 + V2  s+2 s+2   s( s + 2) + 1  ( s + 1)( s + 2) − 1 V1 −  =  V2  ( s + 2)  ( s + 2)     s 2 + 3s + 1  s 2 + 2 s + 1 = V − 1   s + 2  V2  s+2   

…(iv)

10.18 Network Analysis and Synthesis Substituting Eq. (iii) in Eq. (ii), 1  1  I 2 = ( s + 1) V2 − s V1 −  V1 + V2   s+2 s+2   ( s + 1)( s + 2) − 1  s( s + 2) + 1  V1 +  = −  V2  ( s + 2)    ( s + 2)   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

 s 2 + 3s + 1 Y12   s+2 = Y22   ( s 2 + 2 s + 1)  − s+2

−

…(v)

( s 2 + 2 s + 1)   s+2  s 2 + 3s + 1   s+2

  10.4     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, (V1 , I1 ) = f (V2 , − I 2 ) V1 = AV2 − BI 2 I1 = CV2 − DI 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 where matrix  is called transmission 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.

10.4  Τransmission Parameters (ABCD Parameters)  10.19

10.4.1 Condition for Reciprocity (a) As shown in Fig. 10.28, voltage Vs is applied at the input port with the output port short-circuited. i.e., V =V 1

s

I1

V2 = 0

+ − Vs

I 2′ = − I 2 From the transmission parameter equations,

I2

I2′

Network

Vs = B I 2′ Fig. 10.28   Network for deriving condition for  reciprocity Vs =B ′ I2 (b) As shown in Fig. 10.29, voltage Vs is applied at the output port with the input port short-circuited. V2 = Vs V1 = 0

i.e.,

I1′ = − I1

I1 I1′

0 = AVs − BI 2 − I1 ′ = CVs − DI 2 A I 2 = Vs B AD − I1 ′ = CVs − Vs B Vs B = I1 ′ AD − BC Hence, for the network to be reciprocal, Vs Vs = I 2 ′ I1 ′

i.e.,

Network

+ −

Vs

Fig. 10.29  N   etwork for deriving condition for  reciprocity

From the transmission parameter equations,

i.e.,

I2

B AD − BC AD − BC = 1 B=

10.4.2 Condition for Symmetry (a) When the output port is open-circuited, i.e., I2 = 0. From the transmission-parameter equations, Vs = AV2 I1 = CV2 Vs A = I1 C (b) When the input port is open-circuited, i.e., I1 = 0. From the transmission parameter equation, CVs = DI 2 Vs D = I2 C

10.20 Network Analysis and Synthesis Hence, for network to be symmetrical, Vs Vs = I1 I 2 A= D

i.e.,

  example 10.14 

Find the transmission parameters for the network shown in Fig. 10.30. I1

1Ω

2Ω

+

I2 +

V1

5Ω

−

V2 −

Fig. 10.30 Solution First Method Case 1 When the output port is open-circuited, i.e., I2 = 0. V1 = 6 I1 V2 = 5 I1 and A= C= Case 2

Now and

V1 V2 I1 V2

=

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. 10.31, 1Ω 2Ω I1 5× 2 10 17 Req = 1 + = 1+ = Ω + 5+ 2 7 7 17 V1 = I1 V1 5Ω 7 5 5 − I 2 = ( − I1 ) = − I1 7 7 Fig. 10.31 17 I1 17 V1 B=− =− 7 = Ω 5 5 I 2 V2 = 0 − I1 7 7 I1 D=− = I 2 V2 = 0 5

Hence, the transmission parameters are  6 17  A B 5 5  C D  =  1 7    5 5 

I2

10.4  Τransmission Parameters (ABCD Parameters)  10.21

Second Method (Refer Fig. 10.37) Applying KVL to Mesh 1, V1 = 6 I1 + 5 I 2 Applying KVL to Mesh 2, V2 = 5 I1 + 7 I 2 Hence, 5 I1 = V2 − 7 I 2 1 7 I1 = V2 − I 2 5 5 Substituting Eq. (iii) in Eq. (i), 7  1 V1 = 6  V2 − I 2  + 5 I 2 5 5  6 17 I2 = V2 − 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 10.15 

Obtain ABCD parameters for the network shown in Fig.10.32. I1

1Ω

1Ω

I2

+ V1

+

2Ω

V2

2Ω

−

−

Fig. 10.32 Solution The network is redrawn as shown in Fig.10.33. 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 ( I 3 − I1 ) − I 3 − 2 ( I 3 + I 2 ) = 0 − 5 I 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 − I 2  5 5  11 4 = I1 + I 2 5 5

1Ω

1Ω +

2Ω I1

V2

2Ω I3

I2

−

Fig. 10.33

…(iv)

10.22 Network Analysis and Synthesis Substituting Eq. (iii) in Eq. (ii), 2  2 V 2 = 2 I 2 + 2  I1 − I 2  5 5  4 6 = I1 + I2 5 5 4 6 I1 = V2 − I 2 5 5 5 3 I1 = V2 − I 2 4 2

…(v)

Substituting Eq. (v) in Eq. (iv), 11  5 3  4  V2 − I 2  + I 2 5 4 2 5 11 5 = V2 − I 2 4 2 Comparing Eqs (v) and (vi) with ABCD parameter equations, V1 =

11 A B  4 C D  =  5  4

  example 10.16 

…(vi)

5 2 3  2

Determine the transmission parameters for the network shown in Fig. 10.34. 1

I1

1

2

I2

+

+ s

V1

1 s

V2 −

−

Fig. 10.34 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)

10.4  Τransmission Parameters (ABCD Parameters)  10.23

Substituting Eq. (ii) in Eq. (i), s +1 [( s + 1) V2 − I 2 ] − V2 s  ( s + 1) 2  s +1 I2 = − 1 V2 − s s  

I1 =

 s 2 + s + 1  s + 1 V2 −  I2 =   s  s   Comparing Eqs (ii) and (iii) with ABCD parameter equations, −1   s +1 A B  2 = + + + 1 1 s s s C D    s s  

  example 10.17 

…(iii)

Find transmission parameters for the two-port network shown in Fig. 10.35. I1

10 Ω

1.5 V1

I2

−+

+

25 Ω

V1 I1

+ V2

20 Ω

I3

−

I2

−

Fig. 10.35 Solution Applying KVL to Mesh 1, V1 = 10 I1 + 25 ( I1 − I 3 ) = 35 I1 − 25I 3

…(i)

V2 = 20( I 2 + I 3 ) = 20 I 2 + 20 I 3

…(ii)

Applying KVL to Mesh 2,

Applying KVL to Mesh 3, −25 ( I 3 − I1 ) + 1.5 V1 − 20 ( I 2 + I 3 ) = 0 −25 I 3 + 25 I1 + 1.5 (35 I1 − 25 I 3 ) − 20 I 2 − 20 I 3 = 0 −25 I 3 + 25 I1 + 52.5 I1 − 37.5 I 3 − 20 I 2 − 20 I 3 = 0 82.5 I 3 = 77.5 I1 − 20 I 2 I 3 = 0.94 I1 − 0.24 I 2

…(iii)

V1 = 35 I1 − 25 (0.94 I1 − 0.24 I 2 ) = 11.5 I1 + 6 I 2

…(iv)

V2 = 20 I 2 + 20 (0.94 I1 − 0.24 I 2 ) = 18.8 I1 + 15.2 I 2

…(v)

I1 = 0.053 V2 − 0.81I 2

…(vi)

Substituting Eq. (iii) in Eq. (i),

Substituting Eq. (iii) in Eq. (ii),

From Eq. (v),

10.24 Network Analysis and Synthesis Substituting Eq. (vi) in Eq. (iv), V1 = 11.5 (0.053 V2 − 0.81I 2 ) + 6 I 2 = 0.61 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

…(vii)

  10.5     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, (V2 , I 2 ) = f (V1 , − I1 ) V2 = A′ V1 − B ′ I1 I 2 = 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.

10.5.1 Condition for Reciprocity (a) As shown in Fig. 10.36, 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. 10.36  N   etwork for deriving condition  for reciprocity

10.5  Inverse Transmission Parameters (A′B′C′D′ Parameters)  10.25

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. 10.37, voltage Vs is applied at the output port with the input port short-circuited. i.e., V2 = Vs I1 I2 V1 = 0 + I1′ I1 = − I1 ′ Network − Vs From the inverse transmission parameter equations, Vs = B′ I1 ′ Fig. 10.37   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

10.5.2 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 10.18 

Find the inverse transmission parameters for the network shown in Fig. 10.38. I1

1Ω

2Ω

+ V1

+

3Ω

−

V2 −

Fig. 10.38 Solution First method Case 1 When the input port is open-circuited, i.e., I1 = 0 V1 = 3I 2 V2 = 5 I 2

10.26 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. 10.39. By current division rule, 1Ω 2Ω I1 3 I1 = − I 2 4 4 I 3Ω D′ = − 2 = I1 V1 = 0 3 3 ×1 3 11 Req = 2 + = 2+ = Ω 3 +1 4 4 Fig. 10.39 11 11  4  11 V2 = I 2 =  −  I1 = − I1 4 4  3 3 11 V2 B′ = − = Ω I1 V1 = 0 3 Second Method Refer Fig. 10.38. Applying KVL to Mesh 1, V1 = 4I1 + 3I2 Applying KVL to Mesh 2, V2 = 3I1 + 5I2 Hence, 3I2 = V1 − 4I1 1 4 I 2 = 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 10.19 

I2 + V2 −

...(i) ...(ii) …(iii)

…(iv)

Find the inverse transmission parameters for the network shown in Fig. 10.40 2Ω

I1

I2

+ V1

+

2Ω

2Ω

−

V2 −

Fig. 10.40

10.5  Inverse Transmission Parameters (A′B′C′D′ Parameters)  10.27

Solution  The network is redrawn as shown in Fig. 10.41. Applying KVL to Mesh 1, I1 V1 = 2 I1 − 2 I 3 Applying KVL to Mesh 2,

…(i)

+

V2 = 2 I 2 + 2 I 3

…(ii)

V1

−2 I1 + 2 I 2 + 6 I 3 = 0 1 1 I 3 = I1 − I 2 …(iii) 3 3 Substituting Eq. (iii) in Eq. (i),

−

2Ω

+

2Ω I1

Applying KVL to Mesh 3,

I2

V2

2Ω I2

I3

−

Fig. 10.41

1  1 V1 = 2 I1 − 2  I1 − I 2  3 3  4 2 = I1 + I 2 3 3 Substituting Eq. (iii) in Eq. (ii),

...(iv)

1  1 V2 = 2 I 2 + 2  I1 − I 2  3 3  2 4 = I1 + I 2 3 3

...(v)

Rewriting Eq. (iv), 2 4 I 2 = V1 − I1 3 3 3 I 2 = V1 − 2 I1 2 Substituting Eq. (vi) in Eq. (v),

...(vi)

2 43  I1 +  V1 − 2 I1   3 32 = 2V1 − 2 I1

V2 =

...(vii)

Comparing Eq. (vi) and (vii) with inverse transmission parameter equations,  A′ C ′

  example 10.20 

2 B′   = 3 D ′   2

2  2 

Find the inverse transmission parameters for the network shown in Fig. 10.42. I1

1Ω

1Ω

I2

+ V1

+

2Ω

−

1Ω

V2 −

Fig. 10.42

10.28 Network Analysis and Synthesis Solution The network is redrawn as shown in Fig. 10.43. 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 I1 + I 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 − I 2  2 4  = 2 I1 +

I2

1Ω

+

2Ω

V2

1Ω I3

I2

−

Fig. 10.43

1 I2 2

…(iv)

Substituting Eq. (iii) in Eq. (ii), 1 1 I1 − I 2 2 4 1 3 = I1 + I 2 2 4

V2 = I 2 +

Rewriting Eq. (iv),

1 I 2 = V1 − 2 I1 2 I 2 = 2V1 − 4 I1

…(v)

...(vi)

Substituting the Eq. (vi) in Eq. (v), 1 3 I1 + ( 2V1 − 4 I1 ) 2 4 3 5 = V1 − I1 2 2 Comparing Eqs (vi) and (vii) with inverse transmission parameter equations, V2 =

 A′ C ′

3 B′   = D ′   2 2

...(vii)

5 2 4 

  10.6     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. (V1 , I 2 ) = f ( I1 , V2 ) 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.

10.6  Hybrid Parameters (h Parameters)  10.29

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. 10.44. I1

h11

I2

+ V1

+ h12V2

+ −

h21I1

h22

V2 −

−

Fig. 10.44  Equivalent circuit of the two-port network in terms of h-parameters

10.6.1 Condition for Reciprocity (a) As shown in Fig. 10.45, 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 −

I2 Network

I2′

Vs = h11 I1 − I 2 ′ = h21 I1 Fig. 10.45   Network for deriving condition  for reciprocity Vs h = − 11 I2 ′ h21 I1 I2 (b) As shown in Fig. 10.46, voltage Vs is applied at the output port with the input port short+ I1′ Network circuited. − Vs i.e., V1 = 0 V2 = Vs Fig. 10.46   Network for deriving condition for  reciprocity I1 = − I1 ′

10.30 Network Analysis and Synthesis From the h-parameter equations, 0 = h11 I1 + h12 Vs h12 Vs = − h11 I1 = h11 I1 ′ Vs h11 = I1 ′ h12 Hence, for the network to be reciprocal, Vs Vs = I 2 ′ I1 ′ i.e., h21 = − h12

10.6.2 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 I1 + h22 V2 V2 h = − 21 I1 h22 h h h h −h h ∆h Z11 = h11 − 12 21 = 11 22 12 21 = h22 h22 h22 where ∆h = h11h22 − h12 h21 Similarly, V Z 22 = 2 I 2 I1 = 0 With I1 = 0, I 2 = h22 V2 Z 22 =

= I1 = 0

1 h22

Z11 = Z 22

For a symmetrical network,

∆h 1 = h22 h22 ∆h = 1

i.e., i.e., i.e.,

V2 I2

h11 h22 − h12 h21 = 1

  example  10.21  In the two-port network shown in Fig. 10.47, compute h-parameters from the following data: (a) With the output port short-circuited:V1 = 25 V , I 1 = 1 A, I 2 = 2 A (b) With the input port open-circuited:V1 = 10 V , V2 = 50 V , I 2 = 2 A I1 + V1 −

I2 Two-port network

Fig. 10.47

+ V2 −

10.6  Hybrid Parameters (h Parameters)  10.31

Solution V1 I1 V

h11 =

=

2=

I2 I1

h21 =

0

= V2 = 0

25 = 25 Ω, 1

h12 =

2 = 2, 1

h22 =

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 10.22 

Determine hybrid parameters for the network of Fig. 10.48. Determine whether the network is reciprocal. I1

1Ω

I2

2Ω

+

+

2Ω

V1 I1

V2

4Ω I3

I2

−

−

Fig. 10.48 Solution First Method Case 1 When Port 2 is short-circuited, i.e., V2 = 0 as shown in Fig. 10.49, Req = 1 + Now,

Also,

Case 2

2×2 =2Ω 2+2

V1 = 2I1 V h11 = 1 =2Ω I1 V2 = 0 2 I =− 1 2+2 2 1 I2 h21 = =− 2 I1 V2 = 0 I 2 = − I1 ×

I1

1Ω

I2

2Ω

+

2Ω

V1

4Ω

−

Fig. 10.49

When Port 1 is open-circuited, i.e., I1 = 0 as shown in Fig. 10.50, ( 2 + 2) × 4 =2Ω 2+2+4 V1 = 2 I y

Req =

I Iy = 2 2 V2 = 4 I x I Ix = 2 2

I1 = 0

+ V1

1Ω

I2

1Ω ly

2Ω

−

lx

4Ω

+ V2 −

Fig. 10.50

10.32 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  h11  h21

 2 h12   =  h22   1 −  2

1 2 1  2

Second Method (Refer Fig. 10.48) Applying KVL to Mesh 1, V1 = 3I1 − 2 I 3

…(i)

Applying KVL to Mesh 2, V2 = 4 I 2 + 4 I 3 Applying KVL to Mesh 3, −2 ( I 3 − I1 ) − 2 I 3 − 4 ( I 3 + I 2 ) = 0 8 I 3 = 2 I1 − 4 I 2 I I I3 = 1 − 2 4 2 Substituting Eq. (iii) in Eq. (i), I  I V1 = 3I1 − 2  1 − 2  4 2 5 = I1 + I 2 2 Substituting Eq. (iii) in Eq. (ii), I2 I V2 = 4 I 2 + 4  1 −  4 2 = 4 I 2 + I1 − 2 I 2 = I1 + 2 I 2 1 1 I 2 = − I1 + V2 2 2

…(ii)

…(iii)

...(iv)

...(v)

Substituting Eq. (v) in Eq. (iv), 5 1 1 I1 − I1 + V2 2 2 2 1 = 2 I1 + V2 2 Comparing Eqs (v) and (vi) with h-parameter equations, V1 =

 h11 h  21 Since h12 = − h21, the network is reciprocal.

 2 h12   =  h22   1 −  2

1 2 1  2

...(vi)

10.7  Inverse Hybrid Parameters (g Parameters)  10.33

  example 10.23 

Find h-parameters for the network shown in Fig. 10.51. 1 F 2

I1

1 F 2

I2

+

+

1H

V1

1H

V2

−

−

Fig. 10.51 Solution As solved in Example 10.12, derive the equations for I1 and I2 in terms of V1 and V2. I1 =

s3 + 2 s s3 V1 − V2 2 4( s + 1) 4( s 2 + 1)

I2 = −

s3 2

4( s + 1)

V1 +

s4 + 6s2 + 4 4 s( s 2 + 1)

…(i) V2

…(ii)

From Eq. (i), V1 = Substituting Eq. (iii) in Eq (ii),

s2 4( s 2 + 1) I1 + 2 V2 2 s( s + 2) s +2

I2 = − =−

 4( s 2 + 1)  s4 + 6s2 + 4 s2 V2 I V +  + 1 2 s 2 + 2  4 s( s 2 + 1) 4( s 2 + 1)  s( s 2 + 2) s3

s2 2

I1 +

2( s 2 + 1)

s +2 s ( s 2 + 2) Comparing Eqs (iii) and (iv) with h-parameter equations,  h11  h21 

…(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)  s2

2

  10.7     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. ( I1 , V2 ) = f (V1 , I 2 ) I1 = g11 V1 + g12 I 2 V2 = g21 V1 + g22 I 2 In matrix form, we can write  I1   g11 g12  V1  V2  =  g21 g22   I 2  The individual g parameters can be defined by setting V1 = 0 and I2 = 0.

10.34 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. 10.52. g22

I2 + V1

I2 +

g11

g12I2

−

+ g V − 21 1

V2 −

Fig. 10.52  Equivalent circuit of two-port network in terms of g-parameters

10.7.1 Condition for Reciprocity (a) As shown in Fig. 10.53, 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, g21 Vs = g22 I 2 ′ Fig. 10.53   Network for deriving condition for  Vs g = 22 reciprocity I 2 ′ g21 (b) As shown in Fig. 10.54, 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 − I1 ′ = g12 I 2 Vs = g22 I 2 Fig. 10.54   Network for deriving condition for  Vs g reciprocity = − 22 I1 ′ g12

10.7  Inverse Hybrid Parameters (g Parameters)  10.35

Hence, for the network to be reciprocal, Vs Vs = I 2 ′ I1 ′ g21 = − g12

i.e.,

10.7.2 Condition for Symmetry The condition for symmetry is obtained from the Z-parameters. V 1 Z11 = 1 = I1 I 2 = 0 g11 Similarly, Z 22 =

V2 I2

I1 = 0

0 = g11 V1 + g12 I 2 g V1 = − 12 I 2 g11  g  V2 = g21  − 12  I 2 + g22 I 2  g11  Z 22 =

V2 I2

= I1 = 0

g11 g22 − g12 g 21 g11

For a symmetrical network, Z11 = Z 22 . 1 g g − g12 g21 = 11 22 g11 g11 g11 g22 − g12 g21 = 1

i.e., i.e.,

Table 10.2  Conditions for reciprocity and symmetry Parameter

Condition for Reciprocity

Condition for Symmetry

Z

Z12 = Z 21

Z11 = Z 22

Y

Y12 = Y21

Y11 = Y22

T

AD − BC = 1

A= D

T′ h

A′ D ′ − B′C ′ = 1 h12 = − h21

A′ = D ′ h11 h22 − h12 h21 = 1

g

g12 = − g21

g11 g22 − g12 g21 = 1

  example 10.24 

Find g-parameters for the network shown in Fig. 10.55. I1

1Ω

2Ω

+ V1

I2 +

3Ω

−

V2 −

Fig. 10.55

10.36 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, V1 = 4 I1 g11 =

I1 V1

=

1  4

=

3 I1 3 = 4 I1 4

I2 = 0

V2 = 3 I1

Also

g21 =

V2 V1

I2 = 0

Case 2 When the input port is short-circuited, i.e., V1 = 0 as shown in Fig. 10.56. By current division rule, I1 = −

3 I2 4

I1

I 3 g12 = 1 =− 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. 10.56

Second Method (Refer Fig. 10.55). Applying KVL to Mesh 1, V1 = 4 I1 + 3 I 2

…(i)

V2 = 3 I1 + 5 I 2 4 I1 = V1 − 3 I 2

…(ii)

Applying KVL to Mesh 2, Hence,

I1 = Substituting Eq. (iii) in Eq. (ii),

1 3 V1 − I 2 4 4

3  1 V2 = 3  V1 − I 2  + 5 I 2 4 4  3 11 = V1 + I 2 4 4 Comparing Eqs (iii) and (iv) with g-parameter equations,  g11  g21

1 g12   4 = g22   3 4

3 −  4 11  4 

…(iii)

…(iv)

10.8  Inter-relationships between the Parameters  10.37

  example 10.25 

Find g-parameters for the network shown in Fig. 10.57. 5I1

I1

1Ω

I2

+

1Ω

V1

1Ω

−

+ V2 −

Fig. 10.57 Solution The network is redrawn using source transformation as shown in Fig. 10.58. 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) Hence, 2 I1 = V1 − I 2 Fig. 10.58 I1 = Substituting Eq (iii) in Eq. (ii),

1 1 V1 − I 2 2 2

5I1 − +

I2 + V2 −

…(iii)

1  1 V2 = 6  V1 − I 2  + 2 I 2 2 2  = 3 V1 − I 2 Comparing Eqs (iii) and (iv) the g- parameter equations, 1 1 −   g11 g12   = 2 2  g21 g22     3 −1 

…(iv)

  10.8     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.

10.8.1 Z-parameters in Terms of Other Parameters 1. Z-parameters in Terms of Y-parameters We know that

By Cramer’s rule,

I1 = Y11 V1 + Y12 V2 I 2 = Y21 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 ∆Y ∆Y Y22

10.38 Network Analysis and Synthesis where Comparing with

Also, Comparing with

∆Y = Y11Y22 − Y12Y21 V1 = Z11I1 + Z12 , I 2 , Y Z11 = 22 ∆Y Y Z12 = − 12 ∆Y Y11 I1 Y21 I 2 Y Y V2 = = 11 I 2 − 21 I1 ∆Y ∆Y ∆Y V2 = Z 21 I1 + Z 22 I 2 , Y Z 22 = 11 ∆Y Y Z 21 = − 21 ∆Y

2. Z-parameter in Terms of ABCD Parameters We know that V1 = AV2 − BI 2 I1 = CV2 − DI 2 Rewriting the second equation, D 1 I1 + I 2 C C V2 = Z 21 I1 + Z 22 I 2 , V2 = Comparing with

1 C D Z 22 = C D  A A 1  AD   AD − BC  − B  I 2 = I1 +  V1 = A  I1 + I 2  − BI 2 = I1 +   I2 C C C C C C       V1 = Z11 I1 + Z12 I 2 , Z 21 =

Also, Comparing with

A C AD − BC Z12 = C Z11 =

3. Z-parameters in Terms of ABCD Parameters V2 = A′ V1 − B ′ I1 I 2 = C ′ V1 − D ′ I1 Rewriting the second equation, D′ 1 I1 + I 2 C′ C′ V1 = Z11 I1 + Z12 I 2 , V1 = Comparing with

Z11 =

D′ C′

We know that

10.8  Inter-relationships between the Parameters  10.39

Z12 = Also, Comparing with

1 C′

1  A′  D′  A′ D ′ − B′C ′  V2 = A′  I1 + I 2  − B′ I1 =  I1 + I2  C′  C′ C′  C′   V2 = Z 21 I1 + Z 22 I 2 ,

A′ D ′ − B′C ′ ∆T ′ = 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, 1 h 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 h  h  h h − h h  V1 = h11 I1 + h12  − 21 I1 + I 2  = h11 I1 + 12 I 2 − 12 21 I1 =  11 22 12 21  I1 + 12 I 2 h h h h h h 22 22  22  22 22 22   Comparing with V1 = Z11 I1 + Z12 I 2 , Also,

h11h22 − h12 h21 ∆h = h22 h22 h Z12 = 12 h22 Z11 =

5. Z-parameters in Terms of Inverse Hybrid Parameters We know that I1 = g11 V1 + g12 I 2 V2 = g21 V1 + g22 I 2 Rewriting the first equation, g 1 V1 = I1 − 12 I 2 g11 g11 Comparing with V1 = Z11 I1 + Z12 I 2 , 1 g11 g = − 12 g11

Z11 = Z12 Also

 1   g g − g12 g 21  g g V2 = g21  I1 − 12 I 2  + g22 I 2 = 21 I1 +  11 22  I2 g11  g11 g11  g11  

10.40 Network Analysis and Synthesis Comparing with

V2 = Z 21 I1 + Z 22 I 2 , g21 g11 g g − g12 g21 ∆g = 11 22 = g11 g11

Z 21 = Z 22

10.8.2 Y-parameters in Terms of Other Parameters 1. Y-parameters in terms of Z-parameters We know that

By Cramer’s rule,

V1 = Z11 I1 + Z12 I 2 V2 = 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 ∆Z ∆Z Z 21 Z 22

where Comparing with

∆Z = Z11 Z 22 − Z12 Z 21 I1 = Y11 V1 + Y12 V2 , Z 22 ∆Z Z Y12 = − 12 ∆Z Z11 V1 Z 21 V2 Z V − Z12V1 Z Z I2 = = 11 2 = − 21 V1 + 11 V2 ∆Z ∆Z ∆Z ∆Z I 2 = Y21 V1 + Y22 V2 , Y11 =

Also, Comparing with

Y21 = − Y22 =

Z 21 ∆Z

Z11 ∆Z

2. Y-parameters in Terms of ABCD Parameters We know that V1 = AV2 − BI 2 I1 = CV2 − DI 2 Rewriting the first equation,

Comparing with

A 1 I 2 = − V1 + V2 B B I 2 = Y21 V1 + Y22 V2 , Y21 = − Y22 =

Also,

1 B

A B

A  D  1  BC − AD  I1 = CV2 − D  − V1 + V2  = V1 +   V2 B  B B  B  

10.8  Inter-relationships between the Parameters  10.41

Comparing with

I1 = Y11 V1 + Y12 V2 , D B BC − AD AD − BC ∆T Y12 = =− =− B B B Y11 =

3. Y-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, 1 A′ V1 − V2 B′ B′ I1 = Y11 V1 + Y12 V2 , I1 = Comparing with

A′ B′ 1 Y12 = − B′ Y11 =

Also, Comparing with

and

1  D′  A′  A′D ′ − B′C ′  I 2 = C ′V1 − D ′  V1 − V2  = −   V1 + B′ V2 ′ ′ B B ′ B     I 2 = Y21 V1 + Y22 V2 , ∆T ′  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 I 2 = h21 I1 + h22 V2 Rewriting the first equation, h 1 V1 − 12 V2 h11 h11 I1 = Y11 V1 + Y12 V2 , I1 = Comparing with

1 h11 h Y12 = − 12 h11 Y11 =

Also Comparing with

h h h h − h h   1  I 2 = h21  V1 − 12 V2  + h22V2 = 21 V1 +  11 22 12 21  V2 h11  h11 h11    h11 I 2 = Y21 V1 + Y22 V2 , h21 h11 h h −h h ∆h = 11 22 12 21 = h11 h11

Y21 = Y22

10.42 Network Analysis and Synthesis 5. Y-parameters in Terms of Inverse Hybrid Parameters We know that I1 = g11 V1 + g12 I 2 V2 = g21 V1 + g22 I 2 Rewriting the second equation, g 1 I 2 = − 21 V1 + V2 g22 g22 Comparing with I 2 = Y21 V1 + Y22 V2 , g21 g22 1 Y22 = g22 Y21 = −

Also, Comparing with

 g   g g − g12 g21  1 g I1 = g11 V1 + g12  − 21 V1 + V2  =  11 22 V1 + 12 V2  g22   g22 g22  g22  I1 = Y11 V1 + Y12 V2 , g11 g22 − g12 g21 ∆g = g22 g22 g Y12 = 12 g22 Y11 =

10.8.3 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, Z 1 I1 = V2 − 22 I 2 Z 21 Z 21 Comparing with I1 = CV2 − DI 2 , 1 C= Z 21 Z D = 22 Z 21 Also,

Comparing with

Z Z Z Z  1  V1 = Z11  V2 − 22 I 2  + Z12 I 2 = 11 V2 − 22 11 I 2 + Z12 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   V1 = AV2 − BI 2 , Z A = 11 Z 21 Z Z − Z12 Z 21 ∆Z B = 11 22 = Z 21 Z 21

10.8  Inter-relationships between the Parameters  10.43

2. ABCD Parameters in terms of Y-parameters We know that I1 = Y11 V1 + Y12 V2 I 2 = Y21 V1 + Y22 V2 Rewriting the second equation, 1 Y22 V2 + I2 Y21 Y21 V1 = AV2 − BI 2 , Y A = − 22 Y21 1 B=− Y21

V1 = − Comparing with

Also, Comparing with

1 Y  Y  Y Y − Y Y  I1 = Y11  − 22 V2 + I 2  + Y12 V2 =  12 21 11 22  V2 + 11 I 2 Y21  Y21 Y21  Y21   I1 = CV2 − DI 2 , Y Y −Y Y ∆Y C = 12 21 11 22 = − Y21 Y21 Y11 D=− Y21

3. ABCD Parameters in terms of ABCD Parameters V2 = A′V1 − B ′ I1

We know that

I 2 = C ′V1 − D ′ I1 By Cramer’s rule, V2 B′ I 2 D′ D′ B′ V1 = = V2 − I2 A′ B′ ∆T ′ ∆T ′ C ′ D′ where Comparing with

Also,

Comparing with

∆T ′ = A′ D ′ − B′C ′ V1 = AV2 − BI 2 , D′ A= ∆T ′ B′ B= ∆T ′ A′ V2 C′ I2 C′ A′ − I1 = =− V2 + I2 ∆T ′ ∆T ′ ∆T ′ C′ A′ I1 = V2 − I2 ∆T ′ ∆T ′ I1 = CV2 − DI 2 , C′ C= ∆T ′ A′ D= ∆T ′

10.44 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 Comparing with I1 = CV2 − DI 2 , h C = − 22 h21 1 D=− h21 Also, Comparing with

h h  1  h h − h h  V1 = h11  I 2 − 22 V2  + h12 V2 =  12 21 11 22  V2 + 11 I 2 h21  h21 h21  h21   V1 = AV2 − BI 2 , h h −h h ∆h A = 12 21 11 22 = − h21 h21 h11 B=− h21

5. ABCD Parameters in Terms of Inverse Hybrid Parameters We know that I1 = g11 V1 + g12 I 2 V2 = g21 V1 + g22 I 2 Rewriting the second equation, g 1 V1 = V2 − 22 I 2 g21 g21 Comparing with V1 = AV2 − BI 2 , 1 A= g21 g B = 22 g21 Also, Comparing with

 1   g g − g12 g21  g g I1 = g11  V2 − 22 I 2  + g12 I 2 = 11 V2 −  11 22  I2 g21  g21 g21  g21   I1 = CV2 − DI 2 , g C = 11 g21 g g − g12 g21 ∆g D = 11 22 = g21 g21

10.8.4 ABCD Parameters in Terms of Other Parameters 1. ABC D Parameters in Terms of Z-parameters V1 = Z11 I1 + Z12 I 2 V2 = Z 21 I1 + Z 22 I 2

We know that

10.8  Inter-relationships between the Parameters  10.45

Rewriting the first equation, I2 = Comparing with

Z 1 V1 − 11 I1 Z12 Z12

I 2 = C ′V1 − D ′ I1 , 1 Z12 Z D ′ = 11 Z12 C′ =

Also,

Comparing with

Z Z Z Z  1  V2 = Z 21 I1 + Z 22  V1 − 11 I1  = Z12 I1 + 22 V1 − 22 11 I1 Z Z Z Z12  12  12 12 Z  Z Z − Z12 Z 21  = 22 V1 −  11 22  I1 Z12 Z12   V2 = A′ V1 − B ′ I1 , Z 22 Z12 Z11 Z 22 − Z12 Z 21 ∆Z B′ = = Z12 Z12 A′ =

2. ABC D Parameters in Terms of Y-parameters We know that I1 = Y11 V1 + Y12 V2 I 2 = Y21 V1 + Y22 V2 Rewriting the first equation, 1 Y11 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 I 2 = Y21 V1 + Y22  − 11 V1 + I1  =  V1 + 22 I1  Y12   Y12 Y12  Y12  I 2 = C ′V1 − D ′ I1 , ∆Y Y Y −Y Y C ′ = 12 21 11 22 = − Y12 Y12 Y22 D′ = − Y12

3. A B C  D Parameters in Terms of ABCD Parameters We know that V1 = AV2 − BI 2 I1 = CV2 − DI 2

10.46 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 ∆T ∆T C D ∆T = AD − BC V2 = A′V1 − B ′ I1 , D A′ = ∆T B B′ = ∆T A V1 C I1 C A −I2 = =− V1 + I1 ∆T ∆T ∆T C A I2 = V1 − I1 ∆T ∆T I 2 = C ′ V1 − D ′ I1 , C ∆T A D′ = ∆T C′ =

4. A B C  D Parameters in Terms of Hybrid Parameters We know that V1 = h11 I1 + h12 V2 I 2 = h21 I1 + h22 V2 Rewriting the first equation, h 1 V1 − 11 I1 h12 h12 V2 = A′ V1 − B ′ I1 , V2 = Comparing with

1 h12 h11 B′ = h12 A′ =

Also, Comparing with

h h h − h h   1  h I 2 = h21 I1 + h22  V1 − 11 I1  = 22 V1 +  11 22 12 21  I1 h h h h12    12  12 12 I 2 = C ′ V1 − D ′ I1 , C′ =

h22 h12

 h h − h h  ∆h D ′ =  11 22 12 21  = h12  h12  5. A B C D parameters in Terms of Inverse Hybrid Parameters We know that I1 = g11 V1 + g12 I 2 V2 = g21 V1 + g22 I 2

10.8  Inter-relationships between the Parameters  10.47

Rewriting the first equation, g11 1 V1 + I1 g12 g12 I 2 = C ′ V1 − D ′ I1 , I2 = − Comparing with

g11 g12 1 D′ = − g12 C=−

 g  g g − g12 g21  1  g22 V2 = g21 V1 + g22  − 11 V1 + I1  = −  11 22  V2 + g I1 g g g 12 12 12  12    Comparing with V2 = A′ V1 − B ′ I1 , Also,

 g g − g12 g21  ∆g A′ = −  11 22 =−g g 12 12   g22 B′ = g12

10.8.5 Hybrid Parameters in Terms of Other Parameters 1. Hybrid 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 I 2 = − 21 I1 + V2 Z 22 Z 22 Comparing with I 2 = h21 I1 + h22 V2 , Z 21 Z 22 1 h22 = Z 22 h21 = −

Also, Comparing with

1 Z12  Z   Z Z − Z12 Z 21  V1 = Z11 I1 + Z12  − 21 I1 + V2  =  11 22  I1 + Z V2 Z Z Z  22   22 22 22  V1 = h11 I1 + h12 V2 , Z11 Z 22 − Z12 Z 21 ∆Z = Z 22 Z 22 Z h12 = 12 Z 22 h11 =

2. Hybrid Parameters in terms of Y-parameters We know that I1 = Y11 V1 + Y12 V2 Rewriting the first equation,

I 2 = Y21 V1 + Y22 V2 V1 =

Y 1 I1 − 12 V2 Y11 Y11

10.48 Network Analysis and Synthesis Comparing with

V1 = h11 I1 + h12 V2 , 1 Y11 Y h12 = − 12 Y11 h11 =

Also, Comparing with

Y Y  1  Y Y − Y Y  I 2 = Y21  I1 − 12 V2  + Y22 V2 =  11 22 12 21  V2 + 21 I1 Y Y Y Y11  11  11 11   I 2 = h21 I1 + h22 V2 , Y22 Y11 Y11 Y22 − Y12 Y21 ∆Y h22 = = Y11 Y11 h21 =

3. Hybrid Parameters in Terms of ABCD Parameters We know that V1 = AV2 − BI 2 I1 = CV2 − DI 2 Rewriting the second equation, 1 C I 2 = − I1 + V2 D D Comparing with I 2 = h21 I1 + h22 V2 , h21 = − h22 = Also, Comparing with

1 D

C D

C  B  1  AD − BC  V1 = AV2 − B  − I1 + V2  = I1 +   V2 D D D D     V1 = h11 I1 + h12 V2 , B D AD − BC ∆T h12 = = D D h11 =

4. Hybrid Parameters in Terms of ABCD Parameters 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 =

We know that

10.8  Inter-relationships between the Parameters  10.49

Also, Comparing with

1  C′  B′  A′ D ′ − B ′ C ′  I 2 = C ′  I1 + V2  − D ′ I1 = −   I1 + A′ V2 ′ ′ ′ A A A     I 2 = h21 I1 + h22 V2 , ∆T ′  A′ D ′ − B ′C ′  h21 = −   = − A′ A ′   C′ h22 = A′

5. Hybrid Parameters in Terms of Inverse Hybrid Parameters We know that I1 = g11 V1 + g12 I 2 V2 = g21 V1 + g22 I 2 By Cramer’s rule, I1 g12 V2 g22 g g V1 = = 22 I1 − 12 V2 g11 g12 ∆g ∆g g21 g22 where Comparing with

∆g = g11 g22 − g12 g21 V1 = h11 I1 + h12 V2 , g22 ∆g g h12 = − 12 ∆g h11 =

Also, Comparing with

g11 I1 g21 V2 g g I2 = = − 21 I1 + 11 V2 ∆g ∆g ∆g I 2 = h21 I1 + h22 V2 , h21 = − h22 =

g21 ∆g

g11 ∆g

10.8.6 Inverse Hybrid Parameters in Terms of Other Parameters 1. Inverse Hybrid 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 first equation, Z 1 V1 − 12 I 2 Z11 Z11 I1 = g11 V1 + g12 I 2 , I1 = Comparing with

g11 =

1 Z11

10.50 Network Analysis and Synthesis g12 = − Also, Comparing with

Z12 Z11

Z Z  Z Z − Z12 Z 21   1  V2 = Z 21  V1 − 12 I 2  + Z 22 I 2 = 21 V1 +  11 22  I2 Z Z Z Z11    11  11 11 V2 = g21 V1 + g22 I 2 , Z 21 Z11 Z11 Z 22 − Z12 Z 21 ∆Z = = Z11 Z11

g21 = g22

2. Inverse Hybrid Parameters in Terms of Y-parameters We know that I1 = Y11 V1 + Y12 V2 I 2 = Y21 V1 + Y22 V2 Rewriting the second equation, 1 Y V2 = − 21 V1 + I2 Y22 Y22 Comparing with V2 = g21 V1 + g22 I 2 , Y21 Y22 1 = Y22

g21 = − g22 Also, Comparing with

1 Y  Y  Y Y − Y Y  I1 = Y11 V1 + Y12  − 21 V1 + I 2  =  11 22 12 21  V1 + 12 I 2 Y22   Y22 Y22  Y22  I1 = g11 V1 + g12 I 2 , Y11 Y22 − Y12 Y21 ∆Y = Y22 Y22 Y g12 = 12 Y22 g11 =

3. Inverse Hybrid Parameters in Terms of ABCD Parameters We know that V1 = AV2 − BI 2 I1 = CV2 − DI 2 Rewriting the first equation, B 1 V2 = V1 + I 2 A A Comparing with V2 = g21 V1 + g22 I 2 , 1 A B = A

g21 = g22 Also,

B  C 1  AD − BC  I1 = C  V1 + I 2  − DI 2 = V1 −   I2 A A A A    

10.8  Inter-relationships between the Parameters  10.51

I1 = g11 V1 + g12 I 2 ,

Comparing with

C A ∆T  AD − BC  g12 = −  =−  A A   g11 =

4. Inverse 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 second equation, C′ 1 V1 − I2 D′ D′ I1 = g11 V1 + g12 I 2 , I1 = Comparing with

C′ D′ 1 g12 = − D′ g11 =

Also, Comparing with

1   A′ D ′ − B′C ′  B′  C′ V2 = A′ V1 − B′  V1 − I2  =   V1 + D ′ I 2 ′ ′ ′ D D D     V2 = g21 V1 + g22 I 2 ,

 A′ D ′ − B′C ′  ∆T ′ g21 =   = D′ D′   B′ g22 = D′ 5. Inverse Hybrid Parameters in Terms of Hybrid Parameters We know that V1 = h11 I1 + h12 V2 By Cramer’s rule,

I 2 = h21 I1 + h22 V2 V1 h12 I 2 h22 h h I1 = = 22 V1 − 12 I 2 h11 h12 ∆h ∆h h21 h22

where Comparing with

∆h = h11 h22 − h12 h21 I1 = g11 V1 + g12 I 2 , h22 ∆h h g12 = − 12 ∆h h11 V1 h21 I 2 h h V2 = = − 21 V1 + 11 I 2 ∆h ∆h ∆h V2 = g21 V1 + g22 I 2 , g11 =

Also, Comparing with

10.52 Network Analysis and Synthesis g21 = − g22 =

h21 ∆h

h11 ∆h

Table 10.3  Inter-relationship between parameters ∆X = X 11 X 22 − X12 X 21 In terms of [Z]

[Z]

[Y] −

[T]

[T ′]

[h] −

[g]

[Y]

∆T C

D′ C′

Y11 ∆Y

1 C

D C

∆T ′ C′

A′ C′

Y11 Y12

D B

−

A′ B′

−

1 B

A B

Z12

Z 21

Z 22

Z 22 ∆Z

−

Z 21 ∆Z

Z11 ∆Z

Z11 Z 21

∆Z Z 21

−

Y22 Y21

−

1 Y21

A

1 Z 21

Z 22 Z 21

−

∆Y Y21

−

Y11 Y21

Z 22 Z12

∆Z Z12

−

Y11 Y12

−

1 Z12

Z11 Z12

−

∆Y Y12

∆Z Z 22

Z12 Z 22

Z 21 Z 22

1 Z 22

1 Z11

−

Z 21 Z11

∆Z Z11

Z12 ∆Z

Y22 ∆Y

−

Y21 ∆Y

Y12 ∆Y

−

∆T B

.

1 C′ −

1 B′

[g]

∆h h22

h12 h22

1 g11

−

h21 h22

1 h22

g21 g11

∆g g11

1 h11

−

∆g g22

g12 g22

h21 h11

∆h h11

g21 g22

1 g22

h12 h11

g12 g11

∆T ′ B′

D′ B′

B

D′ ∆T ′

B′ ∆T ′

−

∆h h21

−

h11 h21

1 g21

g22 g21

C

D

C′ ∆T ′

A′ ∆T ′

−

h22 h21

−

1 h21

g11 g21

∆g g21

1 Y12

D ∆T

B ∆T

A′

B′

1 h12

h11 h12

−

∆g g12

−

g22 g12

−

Y22 Y12

C ∆T

A ∆T

C′

D′

h22 h12

∆h h12

−

g11 g12

−

1 g12

1 Y11

−

Y12 Y11

B D

∆T D

B′ A′

1 A′

h11 h12

g22 ∆g

−

g12 ∆g

Y21 Y11

∆Y Y11

1 D

C D

∆T ′ A′

C′ A′

h21 h22

g21 ∆g

g11 ∆g

∆Y Y22

Y12 Y22

C A

−

C′ D′

−

g11

g12

Y21 Y22

1 Y22

1 A

B A

g21

g22

Y21 Y22

Z12 Z11

[h]

A C

Z11

−

[T ]

[T]

−

−

∆T A

−

∆T ′ D′

1 D′

B′ D′

−

h22 ∆h

−

h12 ∆h

h21 ∆h

h11 ∆h

−

−

  example 10.26    The Z parameters of a two-port network are Z11 = 20 W, Z22 = 30 W, Z12 = Z21 = 10 W. Find Y and ABCD parameters.

10.8  Inter-relationships between the Parameters  10.53

Solution ∆Z = Z11 Z 22 − Z12 Z 21 = ( 20)(30) − (10)(10) = 500 Y-parameters 30 3 Z 22 = = , ∆Z 500 50 1 10 Z =− , Y21 = − 21 = − 500 50 ∆Z Hence, the Y-parameters are 1  3 −  Y11 Y12   50 50 Y21 Y22  =  1 2  −  50 50  ABCD parameters Y11 =

10 1 Z12 =− =−  ∆Z 500 50 20 2 Z = 11 = =  ∆Z 500 50

Y12 = − Y22

∆Z 500 = = 50 Z 21 10 30 Z D = 22 = =3 Z 21 10

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 10.27  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 = 0.5V1 − 0.2V2 I 2 = −0.2V1 + V2 Find Y, Z and ABCD parameters for the network. Solution Y11 =

I1 = 0.5 , V1 V2 = 0

Y12 =

I2 = −0.2 , V1 V2 = 0 Hence, the Y-parameters are Y11 Y12   0.5 −0.2 Y21 Y22  =  −0.2 1 

Y22 =

Y21 =

I1 V2

V1 = 0

I2 V2

V1 = 0

= −0.2  = 1

Z-parameters ∆Y = Y11 Y22 − Y12 Y21 = (0.5)(1) − ( −0.2)( −0.2) = 0.46 1 Y22 = = 2.174 Ω, ∆Y 0.46 ( −0.2) Y = 0.434 Ω, Z 21 = − 21 = − ∆Y 0.46 Z12   2.174 0.434  = Z 22   0.434 1.087  Z11 =

 Z11  Z 21

( −0.2) Y12 =− = 0.434 Ω ∆Y 0.46 Y 0.5 = 11 = = 1.087 Ω ∆Y 0.46

Z12 = − Z 22

10.54 Network Analysis and Synthesis ABCD parameters 1 Y22 =− = 5, Y21 −0.2 0.46 ∆Y 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 0.5 Y D = − 11 = − = 2.5 Y21 −0.2

A= −

  example 10.28   

B=−

Using the relation Y = Z−1, show that |Z |=

1  Z 22 Z11  + . 2  Y11 Y22 

Solution We know that

i.e.,

Y11 Y21

Y = Z −1 Z   Z 22 − 12  Y12   ∆Z ∆Z = Y22   Z 21 Z11  −  ∆Z ∆Z  | Z | = Z11 Z 22 − Z12 Z 21

  1 1  Z 22 Z11  1  Z 22 Z11  1 =  + + Z ) = ( 2 ∆Z ) = ∆Z = Z11 Z12 − Z12 Z 21 = ( ∆Z + ∆Z    2 2  Y11 Y22  2 Z 22 Z11 2   ∆Z ∆Z  | Z |=

  example 10.29   

1  Z 22 Z11  + 2  Y11 Y22 

For the network shown in Fig. 10.59, find Z and Y-parameters. 2Ω

I1

I2

+ V1

+

1Ω

2Ω

−

3I2

V2

−

Fig. 10.59 Solution The network is redrawn by source transformation technique as shown in Fig. 10.60. 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 −( I 3 − I1 ) − 2 I 3 − 2( I 2 + I 3 ) + 6 I 2 = 0 + − 5 I 3 = I1 + 4 I 2 1 4 …(iii) Fig. 10.60 I 3 = I1 + I 2 5 5

+ V2 I2

−

10.8  Inter-relationships between the Parameters  10.55

Substituting Eq. (iii) in Eq. (i), 1 4 V1 = I1 − I1 − I 2 5 5 4 4 = I1 − I 2 5 5 Substituting Eq. (iii) in Eq. (ii), 4  1 V2 = −4 I 2 + 2  I1 + I 2  5 5  2 12 = I1 − I 2 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  =− ∆Z = Z11 Z 22 − Z12 Z 21 =    −  −  −    = −  5  5   5  5 255 5 12 5 = 3 , 8 2 − 5 2 − 1 Z 21 = 5 = , Y21 = − 8 4 ∆Z − 5 Hence, the Y-parameters are 1 3 −  Y11 Y12   2 2  Y21 Y22  =  1 1  −  4 2 Z Y11 = 22 = ∆Z

  example 10.30   

−

…(v)

4 5 = −1  8 2 − 5 4 1 Z = 11 = 5 = −  8 2 ∆Z − 5

Z Y12 = − 12 = ∆Z

Y22

…(iv)

−

Find Z and h-parameters for the network shown in Fig. 10.61. I1

4I1

2Ω

I2

2Ω

+−

+

2Ω

V1 I1

+ V2

2Ω I3

−

I2

−

Fig. 10.61 Solution Applying KVL to Mesh 1, V1 = 2 I1 + 2( I1 − I 3 ) = 4 I1 − 2 I 3 Applying KVL to Mesh 2, V2 = 2 I 2 + 2( I 2 + I 3 ) = 4 I 2 + 2I3

…(i) …(ii)

10.56 Network Analysis and Synthesis Applying KVL to Mesh 3, −2( I 3 − I1 ) − 4 I1 − 2( I 3 + I 2 ) = 0 I1 + I 2 = −2 I 3 Substituting Eq. (iii) in Eq. (i), V1 = 4 I1 + I1 + I 2

…(iii) …(iv)

= 5 I1 + I 2

Substituting Eq. (iii) in Eq. (ii),

V2 = 4 I 2 − I1 − I 2

…(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 ∆Z = Z11 Z 22 − Z12 Z 21 = (5)(3) − (1)( −1) = 15 + 1 = 16 ∆Z 16 = Ω, 3 Z 22 1 Z h21 = − 21 = , Z 22 3 Hence, the h-parameters are

Z12 1 = Z 22 3 1 1 =  h22 = Z 22 3

h11 =

 h11  h22

  example 10.31 

h12 =

16 h12   3 = h22   1 3

1 3 1  3

Find Y and Z-parameters for the network shown in Fig. 10.62. I1

1Ω 2

3

1Ω

2

I2

+

+

V1

2V1

−

1Ω 2

V2 −

Fig. 10.62 Solution Applying KCL at Node 3,

Now,

2(V1 − V3 ) = 2V1 + (V3 − V2 ) V V3 = 2 3 I1 = 2V1 + (V3 − V2 ) V = 2V1 + 2 − V2 3 2 = 2V1 − V2 3

…(i)

…(ii)

10.8  Inter-relationships between the Parameters  10.57

I 2 = 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 ∆Y = Y11 Y22 − Y12 Y21 = ( 2)   − 0 =  3 3  2  −  1 Y12 3 = Ω =− Z12 = − 16 ∆Y 8 3 Y 2 3 Z 22 = 11 = = Ω ∆Y 16 8 3

8 1 Y22 = 3 = Ω, Z11 = ∆Y 16 2 3 Y21 0 Z 21 = − =− = 0, 16 ∆Y 3 Hence, the Z-parameters are  Z11  Z 21

  example 10.32 

1 Z12   2 = Z 22   0 

1 8 3  8

For the network shown in Fig. 10.63, find Y and Z-parameters. I1

1

1Ω

2

I2

+−

+ V1

3 V1

2Ω

+

1Ω

−

V2 −

Fig. 10.63 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 = 2V1 + 2V2

I2 =

…(ii)

10.58 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 ∆Y = Y11 Y22 − Y12 Y21 =  −  ( 2) − ( −1)( 2) = −3 + 2 = −1  2 Z11 =

2 Y22 = = −2 Ω, ∆Y −1

Z 21 = −

( −1) Y12 =− = −1 Ω ∆Y ( −1) 3 − 3 Y11 = = 2= Ω ∆Y −1 2

Z12 = −

2 Y21 =− = 2 Ω, ∆Y ( −1)

Z 22

Hence, the Z-parameters are  Z11  Z 21

  example 10.33 

 −2 −1 Z12   = 3 Z 22   2   2

Find Z-parameters for the network shown in Fig. 10.64. Hence, find Y and h-parameters. 0.9 I1 I1

1Ω

I2

+ V1

+

10 Ω

V2

1Ω

−

−

Fig. 10.64 Solution The network is redrawn by source transformation technique as shown in Fig. 10.65. Applying KVL to Mesh 1, 9I1 1Ω 10 Ω I1 V1 = 2 I1 + I 2 …(i) +

Applying KVL to Mesh 2, V2 = 9 I1 + 10 I 2 + 1( I1 + I 2 ) = 10 I1 + 11I 2

−+

…(ii)

Comparing Eqs (i) and (ii) with Z-parameter equations,

V1

1Ω

−

Z 22 11 = , ∆Z 12 10 5 Z Y21 = − 21 = − = − , ∆Z 12 6

+ V2 −

 Z11 Z12   2 1   Z 21 Z 22  = 10 11 Y-parameters ∆Z = Z11 Z 22 − Z12 Z 21 = ( 2)(11) − (1)(10) = 22 − 10 = 12 Y11 =

I2

1 Z12 =−  ∆Z 12 Z11 2 1 = = =  ∆Z 12 6

Y12 = − Y22

Fig. 10.65

10.8  Inter-relationships between the Parameters  10.59

Hence, the Y-parameters are  11 Y11 Y12   12 Y21 Y22  =  5 −  6

−

1 12  1  6 

h-parameters ∆Z 12 = Ω, Z 22 11 10 Z h21 = − 21 = − , 11 Z 22

1 Z12 = Z 22 11 1 1 =  h22 = Z 22 11

h11 =

Hence, h-parameters are

h12 =

1  12  h h  11 12  11 11  h21 h22  =  10 1   −  11 11

  example 10.34 

Find Y and Z-parameters of the network shown in Fig. 10.66. 1

I1

1Ω

2V1

2

I2

−+

+

1Ω

V1

2V2

+ V2

2Ω

−

−

Fig. 10.66 Solution Applying KCL at Node 1, V1 3V1 − V2 + 1 1 I1 = 4V1 − 3V2

I1 + 2V2 =

…(i)

Applying KCL at Node 2, V2 V2 − 2V1 − V1 + 2 1 = −3V1 + 1.5V2 Comparing Eqs (i) and (ii) with Y-parameter equations, Y11 Y12   4 −3  Y21 Y22  =  −3 1.5 Z-parameters ∆Y = Y11Y22 − Y12Y21 = ( 4)(1.5) − ( −3)( −3) = −3 I2 =

1.5 Y22 =− = −0.5 Ω, ∆Y 3 ( −3) Y = −1 Ω, = − 21 = − −3 ∆Y

…(ii)

( −3) Y12 =− = −1 Ω ∆Y −3 Y 4 4 = 11 = − = − Ω ∆Y 3 3

Z11 =

Z12 = −

Z 21

Z 22

Hence, the Z-parameters are  Z11  Z 21

 −0.5 −1  Z12   = 4 Z 22   −1 −  3 

10.60 Network Analysis and Synthesis

  example 10.35 

Determine Y and Z-parameters for the network shown in Fig. 10.67. 1

I1

2

2Ω

I2

+

+

V1

1Ω

V2

3I1

2Ω

−

−

Fig. 10.67 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 − 0.5V2 ) + 2 1 2 2 = 0.5V2 + 4.5V1 − 1.5V2 + 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 ∆Y = Y11Y22 − Y12Y21 = (1.5)( −0.5) − ( −0.5)( 4) = 1.25 I2 =

Z12 = −

Z 21

Z 22

 Z11  Z 21

Z12   −0.4 0.4  = Z 22   −3.2 1.2 

Determine the Y and Z-parameters for the network shown in Fig. 10.68. I1

1

0.5 Ω

3

2

1Ω

+ V1

0.5 Y12 = = 0.4 Ω ∆Y 1.25 Y 1.5 = 11 = = 1.2 Ω ∆Y 1.25

Z11 =

Hence, the Z-parameters are

  example 10.36 

0.5 Y22 =− = −0.4 Ω, ∆Y 1.25 4 Y = − 21 = − = −3.2 Ω, ∆Y 1.25

…(ii)

I2 +

1Ω

2V1

−

0.5 Ω

V2

−

Fig. 10.68

10.8  Inter-relationships between the Parameters  10.61

Solution Applying KCL at Node 1, V1 V1 − V3 + 1 0.5 = 3V1 − 2V3

…(i)

V2 V2 − V3 + 0.5 1 = 3V2 − V3

…(ii)

I1 = Applying KCL at Node 2,

I2 =

Applying KCL at Node 3, V3 − V1 V −V + 2V1 + 3 2 = 0 0.5 1 1 V3 = V2 3 Substituting Eq. (iii) in Eqs (i) and (ii),

…(iii)

2 I1 = 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

…(iv) …(v)

 8 ∆Y = Y11Y22 − Y12Y21 = (3)   − 0 = 8  3 8 Y 1 Z11 = 22 = 3 = Ω, ∆Y 8 3 2 Y12 1 Z 21 = − Ω, = 3= ∆Y 8 12

Z12

2 Y12 3 1 =− = = Ω ∆Y 8 12

Z 22 =

Y11 3 = Ω Y∆ 8

Hence, the Z-parameters are  Z11  Z 21

  example 10.37 

1 1  Z12   3 12  = 3  Z 22   0 8 

Determine Z and Y-parameters of the network shown in Fig. 10.69. I1

2Ω

4Ω

+ V1

I2 +

0.05I2

+ −

− 10V1 +

V2 −

−

Fig. 10.69

10.62 Network Analysis and Synthesis Solution Applying KVL to Mesh 1, V1 − 4I1 − 0.05 I2 = 0 V1 = 4I1 + 0.05I2 …(i) Applying KVL to Mesh 2, V2 − 2I2 + 10V1 = 0 V2 = 2I2 − 10V1 ...(ii) Substituting Eq. (i) in Eq. (ii), V2 = 2I2 − 40I1 − 0.5I2 = −40I1 + 1.5I2 ...(iii) Comparing Eqs (i) and (iii) with Z-parameter equations,  Z11 Z12   4 0.05  Z 21 Z 22  =  −40 1.5  Y-parameters ∆Z = Z11Z 22 − Z12 Z 21 = ( 4)(1.5) − (0.05)( −40) = 8 Z 1.5 Z 0.05 Y11 = 22 = , Y12 = − 12 = −  ∆Z 8 ∆Z 8 Z ( −40) 40 Z 4 Y21 = − 21 = − = , Y22 = 11 =  ∆Z 8 8 ∆Z 8 Hence, the Y-parameters are 0.05  1.5 −  Y Y  11 12  8 8  Y21 Y22  =  40 4   8   8

  example 10.38 

Determine Z and Y-parameters of the network shown in Fig. 10.70. 2V3

I1

1Ω

3I2

I3

I2

2Ω

+−

+

+

2Ω

V1 I1

−

+ V3 −

V2 I2

−

Fig. 10.70 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)

10.9  Interconnection of Two-Port Networks  10.63

Writing equation for Mesh 3, I3 = 2V3

...(iii)

From Fig. 10.70, 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 −4  Y-parameters ∆Z = Z11 Z22 − Z12 Z21 = (3)( −4) − (5)( −6) = 18 5 Z 4 2 Z Y11 = 22 = − = − , Y12 = − 12 = −  ∆Z 18 9 ∆Z 18 ( −6) 1 3 Z Z Y21 = − 21 = − Y22 = 11 =  = , ∆Z 18 3 ∆Z 18 Hence, Y-parameters are 5  2 − −  Y11 Y12   9 18 Y21 Y22  =  1 3   18   3

...(v)

  10.9     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.

10.9.1 Cascade Connection 1. Transmission Parameter Representation Figure 10.71 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 ′ = − I 2 I1 + V1 −

N1

I2′

I1′

I2 +

+

V2

V1′

−

−

+ N2

V2′ −

Fig. 10.71  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  ...(10.1)  I1  = C1 D1   − I 2 

10.64 Network Analysis and Synthesis For the network N2,

V1 ′   A2  I1 ′  = C2

B2   V2 ′  D2   − I 2 ′ 

...(10.2)

Since V1′ = V2 and I 2′ = − I 2 , we can write  V2   A2 B2   V2 ′   − I 2  = C2 D2   − I 2 ′  Combining Eqs (10.1) and (10.3), V1   A1 B1   A2 B2   V2 ′   A B   V2 ′   I1  = C1 D1  C2 D2   − I 2 ′  = C D   − I 2 ′   A B   A1 C D  = C1

Hence,

B1   A2 D1  C2

...(10.3)

B2  D2 

...(10.4)

Equation 10.4 shows that the resultant ABCD matrix of the cascade connection is the product of the individual ABCD matrices. 2. Inverse Transmission Parameter Representation Figure 10.72 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 −I1′ = I2 I1 + V1 −

N1

I2′

I1′

I2 +

+

V2

V1′

−

−

+ N2

V2′ −

Fig. 10.72  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, V2   A1 ′ B1 ′   V1  ...(10.5)  I 2  = C1 ′ D1 ′   − I1  For the network N2, V2 ′   A2 ′ B2 ′   V2 ′  ...(10.6)  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 (10.5) and (10.7), 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 ′ 

...(10.7) B′   V1  D ′   − I1  ...(10.8)

Equation 10.8 shows that the resultant A′B′C′D′ matrix of the cascade connection is the product of the individual A′B′C′D′ matrices.

10.9  Interconnection of Two-Port Networks  10.65

  example 10.39  Two identical sections of the network shown in Fig. 10.73 are connected in cascade. Obtain the transmission parameters of the overall connection. 2Ω

I1

2Ω

I2

+

+

V1

1Ω

2Ω

V2

−

−

Fig. 10.73 Solution The network is redrawn as shown in Fig. 10.74. 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 I 3 = I1 − I 2 …(iii) − 5 5 Substituting Eq. (iii) in Eq. (i), 2  1 V1 = 3I1 −  I1 − I 2  5 5  14 2 = I1 + I 2 5 5

2Ω

2Ω +

1Ω I1

V2

2Ω I3

I2

−

Fig. 10.74

…(iv)

Substituting Eq. (iii) in Eq. (ii), 2  1 V2 = 2 I 2 + 2  I1 − I 2  5 5  2 6 I1 + I 2 5 5 5 I1 = V2 − 3I 2 2 Substituting Eq. (v) in Eq. (iv), =

14  5  2  V2 − 3I 2  + I 2 5 2 5 = 7V2 − 8 I 2

…(v)

V1 =

Comparing the Eqs (vi) and (v) with ABCD parameter equations, B1   7 8 = D1   2.5 3 Hence, transmission parameters of the overall cascaded network are  A1 C1

 A B   7 8  7 8 69 80  C D  =  2.5 3  2.5 3 =  25 29

…(vi)

10.66 Network Analysis and Synthesis

  example 10.40 

Determine ABCD parameters for the ladder network shown in Fig. 10.75. 2H

2Ω

I1

I2 +

+ V1

1F

2F

1Ω

−

V2 −

Fig. 10.75 Solution The above network can be considered as a cascade connection of two networks N1 and N2. The network N1 is shown in Fig. 10.76. 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. 10.76 I1 = 2s V2 − (4s2 + 1) I2 ...(iii) Substituting Eq. (iii) in Eq. (i), 1 1  V1 =  2 +  [2 s V2 − ( 4 s 2 + 1) I 2 ] + I 2   2s 2s = ( 4 s + 1)V2 − (8s 2 + 2 s + 2) I 2 Comparing Eqs (iv) and (iii) with ABCD parameter equations, 2  A1 B1   4 s + 1 8s + 2 s + 2 =   C1 D1  4s2 + 1   2s The network N2 is shown in Fig. 10.77. I2′

I1′ +

1 s

1

V2′

−

−

I2′

I1′ +

V1′

…(iv)

+

V1′

+ Z(s) =

1 s+1

−

−

(a)

V2′

(b)

Fig. 10.77 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,

10.9  Interconnection of Two-Port Networks  10.67

From Eq. (ii), I1 ′ = ( s + 1)V2 ′ − I 2 ′ V1 ′ = V2 ′

Also,

…(iii) …(iv)

Comparing Eqs (iv) and (iii) with ABCD parameter equations,  A2 C2

B2   1 0 = D2   s + 1 1 

Hence, overall ABCD parameters are 2 0  8s3 + 10 s 2 + 8s + 3 8s 2 + 2 s + 2  A B   4 s + 1 8s + 2 s + 2   1   s + 1 1 =  3  2 C D  =  2s 4s + 1     4 s + 4 s 2 + 3s + 1 4s2 + 1  

10.9.2 Parallel Connection Figure 10.78 shows two-port networks connected in parallel. In the parallel connection, the two networks have the same input voltages and the same output voltages. I1

I2′

I1′

+ V1 −

I2 + V2 −

N1

I2′′

I1′′ N1

Fig. 10.78  Parallel connection Let Y11 ′, Y12 ′, Y21 ′, Y22 ′ be the Y-parameters of the network N1 and Y1 ″, Y12 ″, Y21 ″, Y22 ″ be the Y-parameters of the network N2. For the network N1,  I1 ′  Y11 ′ Y12 ′  V1   I 2 ′  = Y21 ′ Y22 ′  V2  For the network N2,  I1 ″  Y11 ″ Y12 ″  V1   I 2 ″  = Y21 ″ Y22 ″  V2  For the combined network, Hence,

I1 = I1 ′ + I1 ″ and I 2 = I 2 ′ + I 2 ″.  I1   I1 ′ + I1 ″   Y11 ′ + Y11 ″ Y12 ′ + Y12 ″  V1  Y11 Y12  V1   I 2  =  I 2 ′ + I 2 ″  = Y21 ′ + Y21 ″ Y22 ′ + Y22 ″  V2  = Y21 Y22  V2 

Thus, the resultant Y-parameter matrix for parallel connected networks is the sum of Y matrices of each individual two-port networks.

  example 10.41 

Determine Y-parameters for the network shown in Fig. 10.79.

10.68 Network Analysis and Synthesis 3Ω

I1

2Ω

2Ω

+ V1

I2 +

2Ω

−

V2 −

Fig. 10.79 Solution The above network can be considered as a parallel connection of two networks, N1 and N2. The network N1 is shown in Fig. 10.80. Applying KCL at Node 3, 3 2Ω 2Ω I2′ I1′ V3 I1 ′ + I 2 ′ = …(i) + + 2 From Fig. 10.80, V2 V1 2Ω V1 − V3 I1 ′ = …(ii) − − 2 V −V Fig. 10.80 I2 ′ = 2 3 …(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 V V  I1 ′ = 1 −  1 + 2  2 2 3 3  1 1 …(v) = V1 − V2 3 6 Substituting Eq. (iv) in Eq. (iii), 1 V V  V I2 ′ = 2 −  1 + 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. 10.81. − − 1 1 V −V I1 ″ = − I 2 ″ = 1 2 = V1 − V2 3 3 3 Fig. 10.81

10.9  Interconnection of Two-Port Networks  10.69

Hence, the Y-parameters are 1  1 −  Y11 ″ Y12 ″   3 3 Y21 ″ Y22 ″  =  1 1  −   3 3  The overall Y-parameters of the network are 1 1  2  1 1 + − −   Y11 Y12   Y11 ′ + Y11 ″ Y12 ′ + Y12 ″   3 3 6 3 3 Y21 Y22  = Y21 ′ + Y21 ″ Y22 ′ + Y22 ″  =  1 1 1 1  =  1 − − +  −  6 3 3 3   2

  example 10.42 

1 −  2 2  3 

Find Y-parameters for the network shown in Fig. 10.82 2Ω

I1

I2

+

+

1Ω

V1

0.5 Ω

0.5 Ω

V2

0.5 Ω 2Ω

−

−

Fig. 10.82 Solution The above network can be considered as a parallel combination of two networks N1 and N2. The network N1 is shown in Fig. 10.83. Applying KCL at Node 1, I1 ′ =

V1 V1 − V2 + 1 2

3 1 = V1 − V2 2 2

…(i)

V2 V2 − V1 + 0.5 2 1 5 …(ii) = − V1 + V2 2 2 Comparing Eqs (i) and (ii) with Y-parameter equation,  3 Y11 ′ Y12 ′   2 Y21 ′ Y22 ′  =  1 −  2

1 −  2 5  2 

The network N2 is shown in Fig. 10.84.

2

2Ω

I2′

+

+

1Ω

V1

Applying KCL at Node 2, I2 ′ =

1

I1′

0.5 Ω

V2

−

−

Fig. 10.83 I1′′

0.5 Ω

3

0.5 Ω

+ V1

I2′′ +

2Ω

−

V2 −

Fig. 10.84

10.70 Network Analysis and Synthesis Applying KCL at Node 3, V3 2 V1 − V3 = 2V1 − 2V3 where I1 ″ = 0.5 V −V I 2 ″ = 2 3 = 2V2 − 2V3 0.5 Substituting I1″ and I2″ in Eq (i), I1 ″ + I 2 ″ =

…(i)

2V1 − 2V3 + 2V2 − 2V3 = 0.5V3 4.5V3 = 2V1 + 2V2 4 4 V3 = V1 + V2 9 9

…(ii)

4  10 8 4 I1 ″ = 2V1 − 2V3 = 2V1 − 2  V1 + V2  = V1 − V2 9 9  9 9

…(iii)

4  4 I 2 ″ = 2V2 − 2V3 = 2V2 − 2  V1 + V2  9 9 

and

8 10 = − V1 + V2 9 9 Comparing Eqs (iii) and (iv) with Y-parameter equations, 8  10 −   Y Y ″ ″  11 12  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 10.43 

1 8   47 − −   2 9 = 18 5 10   25 + − 2 9   18

Find Y-parameters for the network shown in Fig. 10.85. 1F

I1

1F

2Ω

2Ω

+ V1

I2 +

1Ω

−

2F

V2 −

Fig. 10.85

25  18  65  18 

−

10.9  Interconnection of Two-Port Networks  10.71

Solution The above network can be considered as a parallel connection of two networks, N1 and N2. The network N1 is shown in Fig. 10.86. I1 ′

2Ω

2Ω

I2′

+ V1

2F

−

I1′ +

+

V2

V1

−

−

2

3

2

I2′ +

1 2s

V2 −

(b)

(a)

Fig. 10.86 Applying KCL at Node 3, I1 ′ + I 2 ′ = 2 s V3 From Fig. 10.86,

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),  1 1 V 1 I1 ′ = 1 −  V1 + V2 2 2  2( 2 s + 1) 2( 2 s + 1)   1   4s + 1 V1 −  V2 =  8s + 4   8s + 4 

…(ii)

…(iii)

…(iv)

…(v)

Substituting Eq. (iv) in Eq. (iii),  1 1 V2 1  −  V1 − V2  2 2  2( 2 s + 1) 2( 2 s + 1)   4s + 1  1  V1 +  V2 = −  8s + 4   8s + 4  Comparing Eqs (v) and (vi) with Y-parameter equations, 1   4s + 1 −  Y Y ′ ′  11 12  8s + 4 8s + 4  Y21 ′ Y22 ′  =  4 s + 1  − 1  8s + 4 8s + 4  I2 ′ =

…(vi)

10.72 Network Analysis and Synthesis The network N2 is shown in Fig. 10.87. I1′′

1F

1F

I2′′

+ V1

1Ω

−

I1′′ +

+

V2

V1

−

−

1 s

(a)

1 s

3

I2′′ +

1

V2 −

(b)

Fig. 10.87 Applying KCL at Node 3, I1 ″ + I 2 ″ = V3

…(i)

From Fig. 10.87, V1 − V3 1 s = sV1 − sV3 V −V I2 ′ = 2 3 1 s = sV2 − sV3 Substituting Eqs (ii) and (iii) in Eq. (i), sV1 − sV3 + sV2 − sV3 = V3 ( 2 s + 1)V3 = sV1 + sV2  s   s  V3 =  V1 +  V2   2 s + 1  2 s + 1 Substituting Eq. (iv) in Eq. (ii),  s   s I1 ″ = sV1 − s  V2   V1 +  2 + 1 2 1 s s ( + )   I1 ″ =

 s2   s( s + 1)  V = −  1  2 s + 1 V2    2s + 1 

…(ii)

…(iii)

…(iv)

…(v)

Substituting Eq. (iv) in Eq. (iii),  s   s   I 2 ″ = sV2 − s  V1 +  V2   2 s + 1   2 s + 1  s2   s( s + 1)  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 + 1)  − s    2 s + 1 2 s + 1 

…(vi)

10.9  Interconnection of Two-Port Networks  10.73

Hence, the overall Y-parameters of the network are  4 s 2 + 8s + 1   Y11 Y12   Y11 ′ + Y11 ″ Y12 ′ + Y12 ″   4( 2 s + 1) = = Y21 ′ Y22  Y21 ′ + Y21 ″ Y22 ′ + Y22 ″   ( 4 s 2 + 1) −  4( 2 s + 1)

( 4 s 2 + 1)   4( 2 s + 1)  4 s 2 + 8s + 1   4( 2 s + 1)  −

10.9.3 Series Connection Figure 10.88 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

+

+

V1′ −

I2

+

+

V2′

N1

− V2

V1 + V1′′

−

+ N2

V2′′

−

−

−

Fig. 10.88  Series connection Let Z11 ′, Z12 ′, Z 21 ′, Z 22 ′ be the Z-parameters of the network N1 and Z11 ″, Z12 ″, Z 21 ″, Z 22 ″ be the Z-parameters of the network N2. For the network N1, V1 ′   Z11 ′ Z12 ″   I1  V2 ′  =  Z 21 ′ Z 22 ″   I 2  For the network N2,

For the combined network

V1 ″   Z11 ′ Z12 ″   I1  V2 ″  =  Z 21 ′ Z 22 ″   I 2  V1 = V1 ′ + V1 ″ and V2 = V2 ′ + V2 ″.

V1   V1 ′ + V1 ″   Z11 ′ + Z11 ″ Z12 ′ + Z12 ″   I1   Z11 Z12   I1  V2  = V2 ′ + V2 ″  =  Z 21 ′ + Z 21 ″ Z 22 ′ + Z 22 ″   I 2  =  Z 21 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.

Hence,

  example 10.44  Two identical sections of the network shown in Fig. 10.89 are connected in series. Obtain Z-parameters of the overall connection. I1

2Ω

2Ω

+ V1

I2 +

1Ω

−

V2 −

Fig. 10.89

10.74 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 10.45 

Z12  3 1 3 1 6 2 = + = Z 22  1 3 1 3  2 6 

Determine Z-parameters for the network shown in Fig. 10.90. L

L

+

+

C

V2

V1 L −

C

C

−

Fig. 10.90 Solution The above network can be considered as a series connection of two networks, N1 and N2. The network N1 is shown in Fig. 10.91. Ls Ls I1 I2 Applying KVL to Mesh 1, 1   1 V1 ′ =  Ls +  I1 +   I 2   Cs  Cs 

+

+

…(i)

1 Cs

V1′

V2′

Applying KVL to Mesh 2, 1  1  V2 ′ =   I1 +  Ls +  I 2  Cs   Cs 

−

Fig. 10.91

Comparing Eqs (i) and (ii) with Z-parameter equations, 1 1   Ls +   Z11 ′ Z12 ′  Cs Cs   Z 21 ′ Z 22 ′  =  1 1  Ls +   Cs Cs  The network N2 is shown in Fig. 10.92. Applying KVL to Mesh 1, 1  V1 ″ =  Ls +  I1 + ( Ls) I 2  Cs 

…(i)

Applying KVL to Mesh 2, 1  V2 ″ = ( Ls) I1 +  Ls +  I 2  Cs 

−

…(ii)

+

+ Ls

V1′′ −

…(ii)

I2

I1

1 Cs

V2′′

1 Cs

Fig. 10.92

−

10.9  Interconnection of Two-Port Networks  10.75

Comparing Eqs (i) and (ii) with Z-parameter equations, 1  Ls +  Z Z ″ ″  11 12  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 +

10.9.4 Series-Parallel Connection Figure 10.93 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 +

N1

V2

V1 I2′′

+ −

V1′′

N2

−

−

Fig. 10.93  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 = V1 ′ + V1 ″ and I 2 = I 2 ′ + I 2 ″ Hence,

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 10.46 

Determine h-parameters for the network shown in Fig. 10.94.

10.76 Network Analysis and Synthesis I1

1Ω

1Ω

+

I2 +

2Ω

V1

1Ω

1Ω

1Ω

1Ω

V2

2Ω −

−

1Ω

1Ω

Fig. 10.94 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. 10.95. 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 I 2 = −2 I1 + V2 1 1 …(iii) I 2 = − I1 + V2 2 4 − − Substituting Eq. (iii) in Eq. (i), 1Ω 1Ω 1   1 V1 = 4 I1 + 2  − I1 + V2  Fig. 10.95  2 4  1 = 3I1 + V2 …(iv) 2 Comparing Eqs (iii) and (iv) with h-parameters equations, 1  3  ′ ′ h h  11 12  2  h21 ′ h22 ′  =  1 1  −   2 4 For network N2, h-parameters will be same as the two networks are identical. 1  3  h11 ″ h12 ″   2  h21 ″ h22 ″  =  1 1  −   2 4 Hence, the overall h-parameters of the network are 1  1  3 3  6 1    ′ ′ ″ ″ h h h h h h  11 12   11 12   11 12  2 + 2 =  + = = 1      h21 h22   h21 ′ h22 ′   h21 ″ h22 ″   − 1 1   − 1 1   −1 2   2 4  2 4

10.9  Interconnection of Two-Port Networks  10.77

10.9.5 Parallel–Series Connection Figure 10.96 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

+

+

I2 +

V2′

N1N1

− V2

V1 I1′′

+ N2N2

−

V2′′

−

−

Fig. 10.96  Parallel–series connection Let g11 ′, g12 ′, g21 ′, g22 ′ be the g-parameters of the network N1 and g11 ″, g12 ″, g21 ″, g22 ″ be the g-parameters of the network N2, For the network N1,  I1 ′   g11 ′ V2 ′  =  g21 ′

g12 ′  V1  g22 ′   I 2 

For the network N2,

For the combined network,

 I1 ″   g11 ″ g12 ″  V1  V2 ″  =  g21 ″ g22 ″   I   2 I1 = I1 ′ + I1 ″ and V2 = V2 ′ + V2 ″  I1   I1 ′ + I1 ″   g11 ′ + g11 ″ V2  = V2 ′ + V2 ″  =  g21 ′ + g21 ″

Hence,

g =  11  g21

g12 ′ + g12 ″  V1  g22 ′ + g22 ″   I 2 

g12  V1  g22   I 2 

Thus, the resultant g-parameter matrix is the sum of the g-parameter matrices of each individual two-port network.

  example 10.47 

Determine the g-parameters for the network shown in Fig. 10.97. I1

3Ω

3Ω

I2

+

+

1Ω

V1

3Ω

1Ω

V2

3Ω

1Ω −

1Ω −

Fig. 10.97

10.78 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. 10.98. 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 =

1 1 I1 − I 2 5 5

1Ω

V1 I1

…(iii)

V2

1Ω I3

I2

−

−

Fig. 10.98

Substituting Eq (iii) in Eq (i), 1  1 V1 = 4 I1 −  I1 − I 2  5 5  19 1 = I1 + I 2 5 5

…(iv)

Substituting Eq (iii) in Eq (ii), 1  1 V2 = I 2 +  I1 − I 2  5 5  1 4 = I1 + I 2 5 5

…(v)

Rewriting Eq (iv), 19 1 I1 = V1 − I 2 5 5 5 1 I1 = V1 − I 2 19 19 Substituting Eq (vi) in Eq (v), 1 5 1  4 V2 =  V1 − I 2  + I 2  5 19 19  5 1 15 = V1 + I 2 19 19 Comparing Eqs (vi) and (vii) with g-parameters equation, 1 5 −   g g ′ ′  11 12  19 19   g21 ′ g22 ′  =  1 15   19 19  For the network N2, g-parameters will be same as the two networks are identical. 1 5 −   g11 ″ g12 ″  19 19  g21 ″ g22 ″  =  1 15   19 19  Hence, the overall g-parameters of the network are 1 5 5 −    g11 g12   g11 ′ g12 ′   g11 ″ g12 ″  19 19 19 + 1  g21 g22  =  g21 ′ g22 ′  +  g21 ″ g22 ″  =  1 15    19 19  19

…(vi)

…(vii)

1  10 19  = 19 15   2 19  19

−

2 19  30  19 

−

10.11  PI (p )-Network  10.79

  10.10     T-NeTwORk Any two-port network can be represented by an equivalent T network as shown in Fig. 10.99. The elements of the equivalent T network may be expressed in terms of Z-parameters. Applying KVL to Mesh 1, ZA

I1

V1 = Z A I1 + ZC ( I1 + I 2 ) = ( Z A + ZC ) I1 + ZC I 2

...(10.9)

Applying KVL to Mesh 2,

ZB

+

+

...(10.10)

V2

ZC

V1

V2 = Z B I 2 + ZC ( I 2 + I1 ) = ZC I1 + ( Z B + ZC ) I 2

I2

−

−

Comparing Eqs (10.9) and (10.10) with Z-parameter equations, Z11 = Z A + ZC Z12 = ZC

Fig. 10.99  T-Network

Z 21 = ZC Z 22 = Z B + ZC Solving the above equations, Z A = Z11 − Z12 = Z11 − Z 21 Z B = Z 22 − ZZ 21 = Z 22 − Z12 ZC = Z12 = Z 21

  10.11     PI (o )-NeTwORk Any two-port network can be represented by an equivalent pi (p) network as shown in Fig. 10.100. Applying KCL at Node 1, I1 = YAV1 + YB (V1 − V2 ) = (YA + YB )V1 − YBV2 Applying KCL at Node 2, I 2 = YCV2 + YB (V2 − V1 ) = −YBV1 + (YB + YC )V2

I1

...(10.11)

YB

1

2

I2 +

+

...(10.12) Comparing Eqs (10.11) and (10.12) with Y-parameter equations, Y11 = YA + YB Y12 = −YB Y21 = −YB Y22 = YB + YC

V1

YA

YC

−

−

Solving the above equations, YA = Y11 + Y12 = Y11 + Y21 YB = −Y12 = −Y21 YC = Y22 + Y12 = Y22 + Y21

V2

Fig. 10.100  p-network

10.80 Network Analysis and Synthesis

  example 10.48  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. 10.101 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

Z1

Z3

I2 +

+ Z2

V1

V2

−

−

Z12 = Z 2 = 5

Fig. 10.101

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 10.49 

Admittance parameters of a pi network are Y11 = 0.09 , Y12 = Y21 = −0.05  and

Y22 = 0.07 . Find the values of Ra, Rb and Rc. Solution The pi network is shown in Fig. 10.102. Applying KCL at Node, 1, I1 =

V1 V1 − V2 + Ra Rb

1 1  1 V1 − V2 = +  Ra Rb  Rb Applying KCL at Node 2, I2 =

Rb

I1

…(i)

+ V1

V2 V2 − V1 + Rc Rb

+ 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 = −0.05 Y12 = − Rb 1 = −0.05 Y21 = − Rb 1 1 + = 0.007 Y22 = Rb Rc Y11 =

I2

Rc

V2

−

Fig. 10.102

10.11  PI (p )-Network  10.81

Solving the above equations, Ra = 25 Ω Rb = 20 Ω Rc = 50 Ω

  example  10.50  Find the parameters YA, YB and YC of the equivalent p network as shown in Fig. 10.103 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. 10.103 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°

Y22 =

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, I 2 = YCV2 + 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)

10.82 Network Analysis and Synthesis

  example 10.51  A network has two input terminals (a, b) and two output terminals (c, d) as shown in Fig. 10.104. The input impedance with c and d open-circuited is (250 + j100) ohms and with c and d shortcircuited 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. ZB

ZA a

c

ZC

b

d

Fig. 10.104 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),

…(iii)

Z B ZC − ZC = 150 + j 200 Z B + ZC

…(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 10.52 

Find the equivalent p-network for the T-network shown in Fig. 10.105. I1

2Ω

2.5 Ω

+ V1

I2 +

5Ω

−

V2

−

Fig. 10.105

10.11  PI (p )-Network  10.83

Solution Figure 10.106 shows T-network and p-network. ZA

ZB

Z3

ZC

Z1

Z2

p-network (Delta network)

T-network (Star network)

Fig. 10.106 For converting a T-network (star network) into an equivalent p-network (delta network), we can use stardelta transformation technique. Z A ZC 2×5 = 2+5+ = 11 Ω ZB 2.5 Z Z 2 × 2.5 Z3 = Z A + Z B + A B = 2 + 2.5 + = 5.5 Ω ZC 5 Z Z 2.5 × 5 Z 2 = Z B + ZC + B C = 2.5 + 5 + = 13.75 Ω ZA 2

5.5 Ω

Z1 = Z A + ZC +

11 Ω

The equivalent p-network is shown in Fig. 10.107.

  example 10.53 

13.75 Ω

Fig. 10.107

For the network shown in Fig. 10.108. Find the equivalent T-network. 8Ω

I1

1Ω

I3

I2

4Ω

+

+ V2

2Ω

V1 I1

I2

−

−

Fig. 10.108 Solution Applying KVL to Mesh 1, V1 = 3I1 + 2 I 2 − I 3

…(i)

V2 = 2 I1 + 6 I 2 + 4 I 3

…(ii)

Applying KVL to Mesh 2, Applying KVL to Mesh 3, 13I 3 − I1 + 4 I 2 = 0 1 4 I 3 = I1 − I 2 13 13

…(iii)

10.84 Network Analysis and Synthesis Substituting the Eq. (iii) in Eq. (i), V1 = 3I1 + 2 I 2 − =

1 4 I1 + I 2 13 13

38 30 I1 + I2 13 13

…(iv)

Substituting the Eq. (iii) in Eq. (ii), 4  1 V2 = 2 I1 + 6 I 2 + 4  I1 − I 2   13 13  30 62 = I1 + I 2 13 13 The T-network is shown in Fig. 10.109. Applying KVL to Mesh 1, V1 = ( Z A + ZC ) I1 + ZC I 2 Applying KVL to Mesh 2, V2 = ZC I1 + ( Z B + ZC ) I 2 Comparing Eqs (iv) and (v) with Eqs (vi) and (vii),

…(vi)

I1

…(v)

ZA

ZB

+

+

…(vii) V 1

38 13 30 ZC = 13 62 Z B + ZC = 13

ZC

−

Z A + ZC =

I2

V2

−

Fig. 10.109

Solving the above equations, 8 Ω 13 32 ZB = Ω 13 30 ZC = Ω 13 ZA =

  10.12     laTTICe NeTwORks A lattice network is one of the common two-port networks, shown in Fig. 10.110. 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. 10.110  Lattice network

10.12  Lattice Networks  10.85

The lattice network can be redrawn as a bridge network as shown in Fig. 10.111. 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 = = 2 I1 I 2 = 0 Also

I I I V2 = 1 Z B − 1 Z A = 1 ( Z B − Z A ) 2 2 2 V2 ZB − ZA Z 21 = = I1 I 2 = 0 2

+

V1

I2 ZB

V2

−

ZA

−

Since the network is symmetric,

Fig. 10.111  Bridge network

ZA + ZB 2 ZB − ZA Z12 = Z 21 = 2 Solving the above equations, Z11 = Z 22 =

Z A = 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 10.54 

Find the lattice equivalent of a symmetrical T network shown in Fig. 10.112. I1

1Ω

1Ω

+

I2 +

V1

2Ω

−

V2

−

Fig. 10.112 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

…(ii)

Applying KVL to Mesh 2,

Z 21 = 2 Ω Z 22 = 3 Ω

10.86 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 Z B = Z11 + Z12 = 3 + 2 = 5 Ω The lattice network is shown in Fig. 10.113. 1Ω

I1

I2

+

+

5Ω

V1

5Ω

V2

−

−

1Ω

Fig. 10.113

  example 10.55   

Find the lattice equivalent of a symmetric p-network shown in Fig. 10.114. 5Ω

I1

I2

+

+

V1

10 Ω

V2

10 Ω

−

−

Fig. 10.114 Solution The network is redrawn as shown in Fig. 10.115. 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

+

10 Ω I1

−10 I1 + 10 I 2 + 25 I 3 = 0 2 2 I − I2 5 5 Substituting Eq (iii) in Eq (i),

I2

+

Applying KVL to Mesh 3,

I3 =

5Ω

−

…(iii)

V2

10 Ω I3

I2

−

Fig. 10.115

2  2 V1 = 10 I1 − 10  I1 − I 2  5 5  = 6 I1 + 4 I 2

…(iv)

Substituting Eq (iii) in Eq (ii), 2  2 V2 = 10 I 2 + 10  I1 − I 2  5 5  = 4 I1 − 6 I 2

…(v)

10.13  Terminated Two-Port Networks  10.87

Comparing the Eqs (iv) and (v) with Z-parameter equations, Z11 = 6 Ω Z12 = 4 Ω

2Ω

I1

Z 21 = 4 Ω Z 22 = 6 Ω

I2

+

Since Z11 = Z22 and Z12 = Z21, the network is symmetric and reciprocal. The parameters of lattice network are Z A = Z11 − Z12 = 6 − 4 = 2 Ω Z B = Z11 + Z12 = 6 + 4 = 10 Ω

+

10 Ω

V1

10 Ω

−

V2

−

2Ω

Fig. 10.116

The lattice network is shown in Fig. 10.116.

  10.13     TeRmINaTeD TwO-pORT NeTwORks 10.13.1 Driving-Point Impedance at Input Port A two-port network is shown in Fig. 10.117. 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. 10.117  Terminated two-port network 1. Input Impedance in Terms of Z-parameters We know that V1 = Z11 I1 + Z12 I 2 From Fig. 10.117,

V2 = Z 21 I1 + Z 22 I 2 V2 = − Z L I 2 − I 2 Z L = Z 21 I1 + Z 22 I 2 I2 = − Zin =

Z 21 I1 Z 22 + Z L

V1 Z 21  Z11Z 22 + Z11Z L − Z12 Z 21  = = Z11 + Z12  −  Z 22 + Z L  Z 22 + Z L I1

If the output port is open-circuited, i.e., ZL = ∞, Z11Z 22 − Z12 Z 21 + Z11 ZL = Z11 Zin = lim Z 22 Z L→∞ +1 ZL

10.88 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 = Y11V1 + Y12V2 I 2 = Y21V1 + Y22V2 From Fig. 10.117, V2 = − Z L 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 = AV2 − BV2 I1 = CV2 − DI 2 From Fig. 10.117, V2 = − Z L I 2 I1 = −CZ L I 2 − DI 2 = −(CZ L + D ) I 2 I1 I2 = − CZ L + D I1   AZ L + B   = I1 V1 = AZ L I 2 − B =  −  CZ L + D   CZ L + D  V AZ L + B Zin = 1 = I1 CZ L + D

10.13  Terminated Two-Port Networks  10.89

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 = − Z L I 2 I 2 = h21 I1 − h22 Z L I 2 h21 I2 = IL 1 + h22 Z L h Z V2 = − 21 L I1 1 + h22 Z L Substituting the value of V2 in V1,  − h21Z L   (h h − h h )Z + h  V1 = h11 I1 + h12  I L  =  11 22 12 21 L 11  I1 1 + h22 Z L  1 + h22 Z L   V (h h − h h )Z + h Zin = 1 = 11 22 12 21 L 11 I1 1 + h22 Z L 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

10.13.2 Driving-Point Impedance at Output Port A two-port network is shown in Fig. 10.118. 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. 10.118  Terminated two-port network 1. Output Impedance in terms of Z-parameters We know that V1 = Z11I1 + Z12 I 2 V2 = Z 21I1 + Z 22 I 2

10.90 Network Analysis and Synthesis From Fig. 10.118, V1 = − Z L I1 − I1Z1 = Z11 I1 + Z12 I 2 Z12   I1 =  − I2  Z L + Z11  Z Z   Z Z − Z12 Z 21 + Z 22 Z L  Z12    I 2 + Z 22 I 2 = I 2  Z 22 − 21 12  =  11 22 V2 = Z 21  −   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 = Y11V1 + Y12V2 I 2 = Y21V1 + Y22V2 From Fig. 10.118, V1 = − Z L I1 V I1 = − 1 = −YLV1 ZL −YLV1 = Y11V1 + Y12V2 Y12   V1 =  − V2  YL + Y11  Y12  Y Y    Y Y − Y Y + Y Y  I 2 = Y21  − V2 + Y22V2 = V2 Y22 − 21 12  = V2  11 22 12 21 L 22    YL + Y11  YL + Y11  YL + Y11    Z0 =

V2 YL + Y11 = I 2 Y11Y22 − Y122Y21 + YLY22

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 = AV2 − BI 2 I1 = CV2 − DI 2

10.13  Terminated Two-Port Networks  10.91

From Fig. 10.118, V1 = − Z L I1 V1 AV2 − BI 2 = −ZL = I1 CV2 − DI 2 V2 (CZ L + A) = I 2 ( DZ L + B) V DZ L + B 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 I 2 = h21I1 + h22V2 From Fig. 10.118, V1 = − Z L I1 − I1Z L = h11 I1 + h22V2 h12   I1 =  − V2  h11 + Z L  h12   h h − h h + h Z  I 2 = h21  − V2 + h22V2 = V2  11 22 12 21 22 L   h11 + Z L  h11 + Z L   h11 + Z L V2 = 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 10.56   

h11 h11h22 − h12 h21

Measurements were made on a two-terminal network shown in Fig. 10.119. 1

I1

I2

+ V1

2 +

Network

V2

−

−

1′

2′

Fig. 10.119

RL

10.92 Network Analysis and Synthesis (a) With terminal pair 2 open, a voltage of 100 ∠0° V applied to terminal pair 1 resulted in I 1 = 10 ∠0° A, V2 = 25 ∠0° V (b) With terminal pair 1 open, the same voltage applied to terminal pair 2 resulted in I 2 = 20 ∠0° 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 = Z12 =

V1 I1 V2 I1 V2 I2 V1 I2

=

100 ∠0° = 10 Ω 10 ∠0°

=

25∠0° = 2.5 Ω 10 ∠0°

=

100 ∠0° =5Ω 20 ∠0°

=

50 ∠0° = 2.5 Ω 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 = 2.5 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 10.57  The Z-parameters of a two-port network shown in Fig. 10.120 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. 10.120

5Ω

10.13  Terminated Two-Port Networks  10.93

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 I1 + 10 I 2

…(ii)

4 I1 = −30 I 2 I1 = −7.5 I 2 Substituting the value of I1 in Eq. (i), V1 = 10( −7.5 I 2 ) + 4 I 2 = −71 I 2 20 = −71 I 2 I 2 = −0.28 A I1 = −7.5( −0.28) = 2.1 A V2 = −20( −0.28) = 56 V Input impedance Zi =

V1 20 = = 9.52 Ω I1 2.1

  example 10.58 

The Z-parameters of a two-port network shown in Fig. 10.121 are, Z11 = 2 W, I V V Z12 = 1 W, Z21 = 2 W, Z22 = 5 W. Calculate the voltage ratio 2 , current ratio − 2 and input impedance 1 . I1 Vs I1 I1

I2

+ Vs

+ −

+ Network

V1 −

V2

5Ω

−

Fig. 10.121 Solution The two-port network can be represented in terms of Z-parameters. V1 = 2 I1 + I 2 V2 = 2 I1 + 5 I 2

…(i) …(ii)

When the 5 Ω resistor is connected across port-2, V2 = −5 I 2

…(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 + I 2 Vs = 3I1 + I 2

…(v)

10.94 Network Analysis and Synthesis −5 I 2 = 2 I1 + 5 I 2

and

0 = 2 I1 + 10 I 2

…(vi)

Solving Eqs (v) and (vi), we get Vs 0 I1 = 3 2 3 0 I2 = 3 2 I2 1 − = I1 5

1 10 5 = Vs 1 14 10 Vs 0 1 = − Vs 1 14 10

 5   1  5 V2 = 2 I1 + 5 I 2 = 2  Vs  + 5  − Vs  = Vs  14   14  14 V2 5 = Vs 14 9  5  1 V1 = 2 I1 + I 2 = 2  Vs  − Vs = Vs  14  14 14 V1 9 = Ω I1 5

  example 10.59  The following equations give the voltages V1 and V2 at the two ports of a two-port network shown in Fig. 10.122. 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. 10.122 Solution From Fig. 10.122, V2 = −3I2 Substituting Eq. (i) in the given equation, −3I 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)

10.13  Terminated Two-Port Networks  10.95

  example 10.60  The y-parameters for a two-port network shown in Fig. 10.123 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 +

+

V1

1Ω

Network

V2

−

−

Fig. 10.123 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),

…(i) …(ii)

9  4  I 2 = 4  − V2  + 5V2 = V2  5  5 V2 5 Output impedance Z0 = = Ω I2 9

  example 10.61  The following equation gives the voltage and current at the input port of a twoport network shown in Fig. 10.124. 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. 10.124 Solution From Fig. 10.124, V2 = −5I2 Substituting the value of V2 in the given equations, V1 = 5( −5 I 2 ) − 3I 2 = −28 I 2 I1 = 6( −5 I 2 ) − 2 I 2 = −32 I 2 V −28 I 2 7 Input impedance Zi = 1 = = Ω I1 −32 I 2 8

5Ω

10.96 Network Analysis and Synthesis

  example 10.62  The ABCD parameters of a two-port network shown in Fig. 10.125 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

Network

−

V2

10 Ω

−

Fig. 10.125 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(10) − 4( −1) = 29 V

  example 10.63  The h-parameters of a two-port network shown in Fig. 10.126 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. 10.126 Solution V1 = 1 V V2 = −3 I 2 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)

Exercises  10.97

  example  10.64  The h-parameters of a two-port network shown in Fig. 10.127 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. 10.127 Solution The two-port network can be represented in terms of h-parameters. V1 = I1 + 2 V2

…(i)

I 2 = −2 I1 + V2 When the 1 Ω resister is connected across port-2, V22 1 V2 V2 I2

…(ii)

= 100 = 10 V = −1 I 2 = 10 = −10 A

Substituting values of I2 and V2 in Eq (ii), −10 = −2 I1 + 10 I1 = 10 A Applying KVL to the input port, Vs − 2 I1 − V1 = 0 Vs − 2 I1 − ( I1 + 2V2 ) = 0 Vs − 3 I1 − 2V2 = 0 Vs = 3 I1 + 2V2 = 3(10) + 2(10) = 50 V

Exercises 10.1 Determine Z-parameters for the network shown in Fig. 10.128. I1

1Ω

I2

1Ω

+ V1

2Ω

10.2 Find Z-parameters for the network shown in Fig. 10.129.

0.5 Ω

I1 +

+

V2

V1

1F

2F

I2 +

2H

2H

− −

V2 −

−

Fig. 10.129

Fig. 10.128 13  Z=7 2 7

2 7 3  7

 4s4 + 6s2 + 1  3 Z =  4 s +3 s  4s  2 4 s +1 

4 s3   4s2 + 1  4 s3 + 2 s   4s2 + 1 

10.98 Network Analysis and Synthesis 10.3 Find Y-parameters of the network shown in Fig. 10.130. I1

2Ω

2Ω

1Ω

1Ω

−

8Ω

i

I2

2Ω

+ V1

2i

+

+

V2

V1

−

−

−

Fig. 10.133

 0.36 −0.033 Y =  −0.033 −0.36  10.4 Find Y-parameters for the network shown in Fig. 10.131. 3Ω

1H

I2

+

1  3  20 − 20  Y = 1  − 1  4 4  10.7 Show the ABCD parameters of the network shown in Fig. 10.134.

+

V1

2Ω

2F

1F

−

I1

V2

+

−

V1

10 s 2 + 13s + 2  5s + 6 Y = 2   − 5s + 6

1F 4

I2

1H

V2 −

Fig. 10.134 1 + s 2   A B   s2 = C D   1   s

10.8 Find ABCD parameters for the network shown in Fig. 10.135. I1

+

4Ω

8Ω

6Ω

+

1F 3

1 + 2s2   s3  1 + s2   s2 

I2

+ 2Ω

1F

+

 2 −  5s + 6  2 5s + 6 s + 5  5s + 6 

10.5 Find Y-parameters for the network shown in Fig. 10.132.

V1

1F

−

Fig. 10.131

I1

V2

2Ω

Fig. 10.130

I1

+

4Ω

2H

1Ω

+

V2 V1

−

I2

1Ω

2Ω

V2

− −

Fig. 10.132  7s + 6  12 Y =  −s  4

s   4  2 s + 4s + 2   4s −

10.6 Find Y-parameters for the network shown in Fig. 10.133.

−

Fig. 10.135  27 206  A B   = C D  11 42  2  10.9 For the network shown in Fig. 10.136, determine parameter h21.

Exercises  10.99 2I1 1Ω

I1

I2

+

+

1F

V1

V2

1Ω

−

−

Fig. 10.136

2Ω

I1

I2

+

+

V1

2Ω

1Ω

2Ω

2Ω

−

+

V1

1Ω

−

1  3  4 − 4 Y =  − 1 3   4 4  10.13 Two identical sections of the network shown in Fig. 10.140 are connected in parallel. Obtain Y-parameters of the connection. I1

1Ω

I2

2Ω

+ V1

[Y11 = 1 , Y12 = −0.5 , Y21 = 1.5 , Y22 = 0.5 

−

+

2Ω

Fig. 10.140   1 Y = − 1  2

10.11 For the bridged T, R-C network shown in Fig. 10.138 determine Y-parameters using interconnections of two-port networks. 1 2F 1Ω 2

+ V1 −

I2

1Ω

V2

I1

−

+ V1 −

1Ω 2Ω

−

( s 2 + 6 s + 8)   2( s + 6)  s 2 + 10 s + 8   2( s + 6) 

10.12 For the network of Fig. 10.139, find Y-parameters using interconnection of two-port networks.

I2 + V2 −

2Ω 1Ω 2

Fig. 10.138  s 2 + 8s + 8  2( s + 6) Y =  ( s 2 + 6 s + 8) − 2( s + 6) 

1 −  2 5  4 

10.14 Determine Y-parameters using interconnection of two-port networks for the network shown in Fig. 10.141. +

1 F 2

V2

4Ω

−

2 2 6 4  Ω, Z12 = Ω, Z 21 = − Ω, Z 22 = Ω  5 5 5 5 

1Ω

V2

1Ω

−

Fig. 10.137

I1

I2

+

V2

2I1

−

Z11 =

2Ω

Fig. 10.139 −( 2 + s)    h21 = 1 + s   

Determine Y and Z-parameters for the network shown in Fig. 10.137.

10.10

I1

2Ω

1Ω 2

Fig. 10.141  3.1 −0.9 Y =  −0.9 3.1  10.15 Determine the transmission parameters of the network shown in Fig. 10.142 using the concept of interconnection of two two-port networks.

10.100 Network Analysis and Synthesis 1H

1H

Determine the h-parameters of the overall network.

1H

3Ω 1F

3Ω

1F

1Ω

1Ω

Fig. 10.142 1 + 3s 2 + s 4  3  2s + s

3s + 4 s 3 + s 5   1 + 3s 2 + s 4 

Fig. 10.144  15  2  1 −  2

10.16 Two networks shown in Fig. 10.143 are connected in series. Obtain the Z-parameters of the resulting network. 1Ω

1Ω

10.18 The h-parameters of a two-port network shown in Fig. 10.145 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. 10.145

5Ω

(b)

Fig. 10.143 18 7   7 28 10.17 Two identical sections of the network shown in Fig. 10.144 are connected in series-parallel.

[58 V] 10.19 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] 10.20 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 10.1 The open-circuit impedance matrix of the two-port network shown in Fig. 10.146 is 2Ω

1Ω

Fig. 10.146

3I1

 −2 1 (a)   −8 3

 −2 −8 (b)  3  1

0 1  (c)  1 0 

 2 −1 (d)   −1 3 

10.2 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  10.101

(a) (b) (c) (d)

(a) −0.2 mho

z-parameter matrix h-parameter matrix y-parameter matrix ABCD parameter matrix

10.3 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

4Ω V2

V1

− 10V1 + −

−

Fig. 10.149

(c)

10.5 A two-port network is shown in Fig. 10.147. The parameter h21 for this network can be given by R

R

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

10.8 The impedance parameters Z11 and Z12 of a two-port network in Fig. 10.150. 2Ω

2Ω

I2

+

I2 +

(a) −

I1

2Ω

+

10.4 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

(c) −0.05 mho (d) 0.05 mho 10.7 The Z-parameters Z11 and Z21 for the twoport network in Fig. 10.149 are,

3Ω

1Ω

1Ω

+

V1

V2

R

−

−

Fig. 10.147 (a) −

1 2

(b)

1 2

(c) −

3 2

(d)

3 2

Fig. 10.150 (a) 2.75 Ω, 0.25 Ω (c) 3 Ω, 0.25 Ω

(b) 3 Ω, 0.5 Ω (d) 2.25 Ω, 0.5 Ω

10.9 The h parameters of the circuit shown in Fig. 10.151. 10 Ω

20 Ω

10.6 The admittance parameter Y12 in the two-port network in Fig. 10.148. 20 Ω

Fig. 10.151 5Ω

Fig. 10.148

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

10.102 Network Analysis and Synthesis 10.10 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 + BRL C + DRL

(b)

ARL + C BRL + D

(c)

DRL + A BRL + C

(d)

B + ARL D + CRL

10.11 In the two-port network shown in Fig. 10.152, Z12 and Z21 are respectively

re

bI 1

(c)

AB DC

(a) (b) (c) (d)

Z11 Z22 − Z12 Z21 = 1 AD − BC = 1 h11 h22 − h12 h21 = 1 Y11 Y22 − Y12 Y21 = 1

10.16 For the network shown in Fig. 10.153 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

(b) 2,6,0

(c) 2,0,6

(d) 2,6,8

ro

(a) re and b r0

(b) 0 and −b r0

(c) 0 and b r0

(d) re and −b r0

10.12 If a two-port network is passive, then we have, with the usual notation, the following relationship for symmetrical network (a) h12 = h21 (b) h12 = −h21 (c) h11 = h22 (d) h11h22 − h12h21 = 1 10.13 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 10.14 A two-port network has transmission A B . The input impedance parameters  C D  of the network at port 1 will be AD (b) BC

D C

10.15 A two-port network is symmetrical if

YC

Fig. 10.152

A (a) C

(d)

YA

YB

Fig. 10.153 10.17 The impedance matrices of two two-port 15 5   3 2 and  . networks are given by   3 25  2 3 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)   5 3

(d) inderminate

10.18 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

(b) 6, 6, 9

(c) 9, 6, 9

(d) 6, 9, 6

Answers to Objective-Type Questions  10.103 16 Ω

1 3F

1Ω

24 Ω

2H

24 Ω

Fig. 10.155 R1

2s   2s + 1 (a)  3 2s +   2s s 

R2

R3

Fig. 10.154 10.19 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 10.20 The impedance matrix for the network shown in Fig. 10.155 is

 2 s + 1 −2 s  (b)  3  −2 s 2 s +  s  3   2s  −2 s   2s + 1  2s + 2 (c)  3  (d)  3  −2 s 2 s +   2s 2s +  s  s  10.21 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 10.1. (a)

10.2. (d)

10.8. (a)

10.9. (d)

10.15. (c)

10.16. (c)

10.3. (b) , (c)

10.4. (b)

10.5. (a)

10.6. (c)

10.7. (c)

10.10. (d)

10.11. (b)

10.12. (d)

10.13. (b)

10.14. (a)

10.17. (b)

10.18. (a)

10.19. (c)

10.20. (a)

10.21. (b)

11 Filters and Attenuators

  11.1     IntroductIon Filters are frequency-selective networks that attenuate signals at some frequency and allow others to pass with or without attenuation. A filter is constructed from purely reactive elements. Ideally, filters should produce no attenuation in the desired band, called pass band and should provide attenuation at all other frequencies called attenuation band or stop band. The frequency which separates the pass band and the stop band is called cut-off frequency. Filter networks are widely used in communication systems to separate various channels in carrier-frequency telephone circuits.

  11.2     classIfIcatIon of fIlters On the basis of frequency characteristics, filters are classified into four categories: (i) Low-pass filter (ii) High-pass filter (iii) Band-pass filter (iv) Band-stop filter A low-pass filter allows all frequencies up to a certain cut-off frequency to pass through it and attenuates all the other frequencies above the cut-off frequency. A high-pass filter attenuates all the frequency below the cut-off frequency and allows all other frequencies above the cut-off frequency to pass through it. Z1 Z1 A band-pass filter allows a limited band of frequencies to 2 2 pass through it and attenuates all other frequencies below or 1 2 above the frequency band. A band-stop filter attenuates a limited band of frequencies but allows all other frequencies to pass through it. Z2 Z0

  11.3     t-netWorK Figure 11.1 shows a T network.

1′

2′

Fig. 11.1   T-network

11.2 Circuit Theory and Transmission Lines

11.3.1 Characteristic Impedance For a T network, the value of input impedance when it is terminated by characteristic impedance Z0, is given by Z  Z 2  1 + Z0    Z 2 Zin = 1 + Z1 2 + Z 2 + Z0 2 Zin = Z0

But

Z  2Z 2  1 + Z 0   2  Z1 Z0 = + 2 Z1 + 2Z 2 + 2Z0 Z Z Z + 2Z 2 Z 0 = 1+ 1 2 2 Z1 + 2Z 2 + 2Z0 =

Z12 + 2Z1Z 2 + 2Z1Z 0 + 2Z1Z 2 + 4 Z 2 Z 0 2(Z1 + 2Z 2 + 2Z 0 )

2Z1Z0 + 4 Z 2 Z0 + 4Z0 2 = Z12 + 2Z1Z 2 + 2Z1Z 0 + 2Z1Z 2 + 4 Z 2 Z 0 4Z 02 = Z12 + 4Z1Z 2 Z02 = Z0 =

Z12 + Z1Z2 4 Z12 + Z1Z2 4

Hence, characteristic impedance for symmetrical T section is given by, Z 0T =

Z12 + Z1Z 2 4

Characteristic impedance can also be expressed in terms of open-circuit impedance Zoc and short-circuit impedance Zsc. Open-circuit impedance Zoc =

Short-circuit impedance Z sc

Z + 2Z2 Z1 + Z2 = 1 2 2

Z1 Z2 Z Z1Z 2 Z 2 + 4 Z1Z 2 Z1 = + 2 = 1+ = 1 2 Z1 2 Z1 + 2Z 2 2Z1 + 4 Z 2 + Z2 2

2 2  Z + 2Z 2   Z1 + 4 Z1Z 2  Z1 Zoc Z sc =  1 = + Z1Z 2 = Z02T       2Z1 + 4 Z 2  2 4

Z0T = Zoc Z sc

11.3.2

Propagation Constant

The propagation constant γ of the network in Fig. 11.2 is given by,

11.3  T-network  11.3

γ = loge

I1 I2

Z1

Z1

2

2

1

Applying KVL to Mesh1,

2

+ Z2

Vs

Z Vs − 1 I1 − Z 2 ( I1 − I 2 ) = 0 2 Z  Vs =  1 + Z 2  I1 − Z 2 I 2  2 

−

I1

Z0 I2

1′

2′

Fig. 11.2  T-network Applying KVL to Mesh 2, Z1 I 2 − Z0 I 2 = 0 2 Z  − Z 2 I1 +  1 + Z 2 + Z0  I 2 = 0  2 

− Z 2 ( I 2 − I1 ) −

Z  Z 2 I1 =  1 + Z 2 + Z 0  I 2  2  Z1 + Z 2 + Z0 I1 = 2 = eγ I2 Z2 Z1 + Z 2 + Z 0 = Z 2 eγ 2 Z0 = Z 2 (eγ − 1) −

Z1 2

The characteristic impedance of the T-network is given by, Z0 =

Z12 + Z1Z 2 4

Z12 Z + Z1Z 2 = Z 2 (eγ − 1) − 1 4 2 Z12 Z2 + Z1Z 2 = Z 22 (eγ − 1) 2 + 1 − Z1Z 2 (eγ − 1) 4 4 Z22 (eγ − 1) 2 − Z1Z2 (1 + eγ − 1) = 0 Z22 (eγ − 1) 2 − Z1Z2 eγ = 0 Z2 (eγ − 1) 2 − Z1eγ = 0 Z (eγ − 1) 2 = 1 eγ Z2 e 2γ − 2eγ + 1 = e − γ ( e 2γ

Z1

Z 2 e −γ Z − 2eγ + 1) = 1 Z2

11.4 Circuit Theory and Transmission Lines eγ + e − γ − 2 = Dividing both the sides by 2,

Z1 Z2

eγ + e − γ Z = 1+ 1 2 2Z 2 Z cosh γ = 1 + 1 2Z 2 2

Z  Z Z2  sinh γ = cosh 2 γ − 1 = 1 + 1  − 1 = 1 + 1 + 12 − 1 =  2Z 2  Z 2 4Z 2

Z1  Z1  + Z 2  2Z 2 

2

1 Z2 Z Z1Z 2 + 1 = 0T Z2 4 Z2 sinh γ Z 0T Z 0T tan nh γ = = = Z1 cosh γ Z   Z 2 1 + 1  Z 2 + 2  2Z 2  =

But

Z0T = Zoc Z sc

and

Zoc = tanh γ =

Also,

sinh

Z1 + Z2 2 Z sc Zoc

γ 1 1 Z  1 + 1 − 1 = = (cosh γ − 1) =  2 2 2  2Z 2 

Z1 4Z 2

  11.4     p netWorK Figure 11.3 shows a p-network. Z1 1

2

2 Z2

2 Z2

1′

Z0

2′

Fig. 11.3  p-network

11.4.1 Characteristic Impedance For a p network, the value of input impedance when it is terminated by impedance Z0, is given by, 2Z 2 Z0   2Z 2  Z1 +  2 Z 2 + Z 0  Zin = 2Z 2 Z0 Z1 + + 2Z 2 2Z 2 + Z0

11.4  p Network  11.5

Zin = Z0

But

2Z 2 Z0   2Z 2  Z1 +  2 Z 2 + Z 0  Z0 = 2Z 2 Z0 + 2Z 2 Z1 + 2Z 2 + Z0 Z 0 Z1 +

2 Z 2 Z 02 2 Z 2 ( 2Z1Z 2 + Z0 Z1 + 2Z0 Z 2 ) + 2Z 0 Z 2 = 2Z 2 + Z0 2Z 2 + Z 0

2Z0 Z1Z 2 + Z1Z 02 + 2Z 0 Z 22 + 4 Z 2 Z 02 + 2Z 2 Z 02 = 4 Z1Z 22 + 2Z 0 Z1Z 2 + 4 Z 0 Z 22 Z1Z02 + 4 Z 2 Z02 = 4 Z1Z 22 Z 02 ( Z1 + 4 Z 2 ) = 4 Z1Z 22 Z02 = Z0 =

4 Z1Z 22 Z1 + 4 Z 2 4 Z1Z 22 Z1 + 4 Z 2

Dividing both the sides by 4Z2, Z0 =

Z1Z 2 Z 1+ 1 4Z 2

Hence, characteristic impedance of a symmetrical p network is given by Z0π =

Z1Z2 = Z 1+ 1 4Z2

Z1Z2 Z1Z2 +

Z12 4

Z12 + Z1Z2 4 ZZ = 1 2 Z 0T

Z 0T =

But

Z0π

Characteristic impedance can also be expressed in terms of open-circuit impedance Zoc and short-circuit impedance Zsc. 2Z 2 ( Z1 + 2Z 2 ) 2Z 2 ( Z1 + 2Z 2 ) = 2Z 2 + Z1 + 2Z 2 Z1 + 4 Z 2 2Z1Z 2 = 2Z 2 + Z1

Open-circuit impedance Zoc = Short-circuit impedance Z sc

Zoc Z sc =

4Z1Z 22 ZZ = 1 2 Z Z1 + 4 Z 2 1+ 1 4Z 2

Z 0π = Z oc Z sc

11.6 Circuit Theory and Transmission Lines

11.4.2 Propagation Constant The propagation constant of a symmetrical p network is same as that of a symmetrical T network.

  11.5     characterIstIc of fIlters A filter transmits or passes desired range of frequencies without loss and attenuates all undesired frequencies. The propagation constant γ = α + jβ where a is attenuation constant and b is the phase constant. We know that sinh

γ = 2

Z1 4Z2

α β α β  α + jβ  sinh  = sinh cos + j cosh sin =  2  2 2 2 2

Z1 4Z2

Depending upon the type of Z1 and Z2, there are two cases: Case (i) If Z1 and Z2 are same type of reactances then cosh

Z1 γ and sinh are real. 4Z2 2

α β sin = 0 2 2 β α   if β = 0 or nπ where n = 0,1, 2, … ∵ cosh cannot be zero  sin = 0 2 2   β cos = 1 2 α Z1 sinh = 2 4Z 2 α = 2 sinh −1

Z1 4Z 2

Z Z Case (ii) If Z1 and Z2 are opposite types of reactances then 1 is negative, i.e. 1 < 0 and 4Z2 4Z2 is purely imaginary.

Z1 4Z2

α β cos = 0 2 2 α β Z1 j cosh sin = j 2 2 4Z 2 sinh

cosh

α β sin = 2 2

Z1 4Z 2

The two conditions of operation of the filter, viz. the pass band and stop band are mathematically obtained from these equations. Both the above equations must be satisfied simultaneously by a and b. Two conditions may arise

11.6  Constant-k Low Pass Filter  11.7

(a)

α = 0 i.e a = 0 when b ≠ 0 2 Z β Since 1 is negative and sin is real, 4Z2 2 sinh

sin

β = 2

Z1 4Z 2

as cosh

α =1 2

This signifies the region of zero attenuation or pass band which is limited by the upper limit of sine terms. β sin = 1 2 Z −1 < 1 < 0 4Z2 The phase angle in the pass band is given by,

β = sin −1 2

Z1 4Z2

β = 2 sin −1 (b)

Z1 4Z2

β =0 2 β sin = ±1 2 Z1 α Since is negative and cosh is real, 4Z2 2 α Z1 cosh = 2 Z2 cos

This signifies a stop band since a ≠ 0.

α = cosh −1 2 α = 2cosh −1 and

Z1 4Z2 Z1 4 Z2

Z1 < −1 4Z 2

  11.6     constant-k loW Pass fIlter A T or p network is said to be of the constant k type if Z1 and Z2 are opposite types of reactances satisfying the relation Z1Z2 = k 2

11.8 Circuit Theory and Transmission Lines where k is a constant, independent of frequency. k is often referred to as design impedance or nominal impedance of the constant-k filter. The constant-k, T or p-type filter is also known as the prototype filter because other complex networks can be derived from it. Figure 11.4 shows constant-k, T and p-section filters. Z1

Z1

2

2

Z1

2Z2

Z2

(a) T-section

2Z2

(b) p-section

Fig. 11.4  Constant–k Filter In constant-k low pass filter, Z1 = jω L 1 1 Z2 = − j = ω c jω c 1. Nominal Impedance  1  L k = Z1Z 2 = ( jω L)  =  jω c  C 2. Cut-off Frequency The cut-off frequencies are obtained when (i)

When

(ii)

When

Z1 Z = 0 and 1 = −1 4Z 2 4Z 2

Z1 =0 4Z 2 Z1 = 0 jω L = 0 ω=0 f =0 Z1 = −1 4Z2 Z1 = −4 Z 2 4j jω L = ωC 2 ω LC = 4 4 ω2 = LC

ω = ωc = f = fc =

2 LC 1

π LC

11.6  Constant-k Low Pass Filter  11.9

Hence, the pass band starts at f = 0 and continues up to the cut-off frequency fc. All the frequencies above fc are in the attenuation or stop band. 3. Attenuation Constant In pass band, a = 0 In stop band,

α = 2 cosh −1

Z1 ω 2 LC ω2 = 2 cosh −1 = 2 cosh −1 4Z 2 4 ωc2 a

 f  ω = 2 cosh −1   = 2 cosh −1    ωc   fc  The attenuation constant a is zero throughout the pass band but increases gradually from the cut-off frequency. The variation of a is plotted in Fig. 11.5.

4. Phase Constant In pass band,

Fig. 11.5  Variation of a with frequency

β = 2 sin

−1

b

 f  Z1 = 2 sin −1   4Z 2  fc 

p

In stop band, β = π . The phase constant b is zero at zero frequency and increases gradually through the pass band, reaches p at cutoff frequency fc and remains at p for all frequency beyond fc. The variation of b is plotted in Fig. 11.6.

Z12 Z   + Z1Z 2 = Z1Z 2 1 + 1   4 4Z 2 

Z0

2

 f  L  ω 2 LC  ω = 1− = k 1−   = k 1−       ωc 4  C  fc  Z 0π

ZZ = 1 2 = Z 0T

f

fc

0

Fig. 11.6  Variation of b with frequency

5. Characteristic Impedance Z 0T =

f

fc

0

Z0p

2

k

k  f  1−    fc 

Z0T

2

0

fc

f

The plot of characteristic impedance versus frequency is Fig. 11.7    Variation of characteristic  impedance with frequency shown in Fig. 11.7.

11.10 Circuit Theory and Transmission Lines The characteristic impedance Z0T is real when f < fc . At f = fc, Z0T = 0. For f > fc, Z0T is imaginary in the stop band, rising to infinite reactance at infinite frequency. Z0p is real when f < fc. At f = fc, Z0p is infinite and for f > fc, Z0p is imaginary. 6. Design of Filter L C 1

k= fc =

π LC

Solving these two equations, L=

k π fC

C=

1 π fC k

The constant-k, T and p section filters are shown in Fig. 11.8. L 2

L 2

L

C 2

C

(a) T-section

C 2

(b) p-section

Fig. 11.8  Constant-k low-pass filter

  example 11.1 

Find the nominal impedance, cut-off frequency and pass band for the network shown

in Fig. 11.9. 25 mH

25 mH

0.2 µF

Fig. 11.9 Solution The network is a low-pass filter. L = 25 mH, C = 0.2 µF 2 L = 50 mH

11.6  Constant-k Low Pass Filter  11.11

(a)

Nominal Impedance

(b)

0.2 × 10 −6

= 500 Ω

Cut-off frequency fc =

(c)

50 × 10 −3

L = C

k=

1

=

π LC

1

π 50 × 10

−3

× 0.2 × 10 −6

= 3.18 kHz

Pass band The pass band is from zero to 3.18 kHz.

  example 11.2  A low-pass filter is composed of a symmetrical p section. Each series branch is a 0.02 H inductor and shunt branch is a 2 µF capacitor. Find (a) cut-off frequency, (b) nominal impedance, (c) characteristic impedance at 200 Hz and 2000 Hz, (d) attenuation at 200 Hz and 2000 Hz, and (e) phase shift constant at 200 Hz and 2000 Hz.

(a)

C = 4 µF

Cut-off frequency

fc = (b)

1

=

π LC

1

= 1125.4 Hz

π 0.02 × 4 × 10 −6

Nominal impedance k=

(c)

C = 2 µF 2

L = 0.02 H,

Solution

0.02

L = C

4 × 10 −6

= 70.71 Ω

Characteristic impedance at 200 Hz Z0 =

k  f  1−    fc 

2

=

70.71  200  1−   1125.4 

2

= 71.85 Ω

Characteristic impedance at 2000 Hz 70.71 k Z0 = = j 48.13 Ω = 2 2  f   2000  1−  1−    1125.4   fc  (d)

Attenuation at 200 Hz and 2000 Hz The 200 Hz frequency lies in the pass band. α=0 At 2000 Hz,  f   2000  α = 2 cosh −1   = 2 cosh −1  = 2.35  1125.4   fc 

(e)

Phase shift constant at 200 Hz and 2000 Hz At 200 Hz,  f   200  β = 2 sin −1   = 2 sin −1  = 20.47°  1125.4   fc 

11.12 Circuit Theory and Transmission Lines 2000 Hz frequency lies in the attenuation band. β = 180°

  example  11.3  A p section filter network consists of a series arm inductor of 20 mH and two shunt-arm capacitors of 0.16 µF each.Calculate the cut-off frequency, and attenuation and phase shift at 15 kHz. What is the value of the nominal impedance in the pass band? Solution

L = 20 mH,

C = 0.16 µF 2

C = 0.32 µF (a)

Cut-off frequency fc =

(b)

1

=

π LC

1

π 20 × 10

−3

× 0.32 × 10 −6

= 3.98 kHz

Attenuation at 15 kHz  15 × 103   f  α = 2 cosh −1   = 2 cosh −1   =4  fc   3.98 × 103 

(c)

Phase shift at 15 kHz 15 kHz frequency lies in the attenuation band.

β = 180° (d)

Nominal impedance k=

20 × 10 −3

L = C

0.32 × 10 −6

= 250 Ω

  example 11.4  Each of the two series elements of a T-section low-pass filter consists of an inductor of 60 mH having negligible resistance and a shunt element having a capacitance of 0.2 µF. Calculate (a) the cut-off frequency, (b) nominal impedance, and (c) characteristic impedance at frequencies of 1 kHz and 5 kHz. Solution (a)

L = 60 mH, C = 0.2 µF

Cut-off frequency fc =

(b)

1

π LC

1

π 60 × 10

−3

× 0.2 × 10 −6

= 2.91 kHz

Nominal impedance k=

(c)

=

L = C

60 × 10 −3 0.2 × 10 −6

= 547.72 Ω

Characteristic impedance at 1 kHz 2

2

 1 × 103   f  Z0 = k 1 −   = 547.72 1 −   = 514.36 Ω  fc   2.91 × 103 

11.6  Constant-k Low Pass Filter  11.13

Characteristic impedance at 5 kHz 2

2

 5 × 103   f  Z0 = k 1 −   = 547.72 1 −   = j 765.29 Ω  fc   2.91 × 103 

  example  11.5  Design a constant-k low-pass T and p section filters having cut-off frequency of 4 kHz and nominal characteristic impedance of 500 W. f c = 4 kHz, k = 500 Ω

Solution

500 k = = 39.79 mH π f c π × 4 × 103 1 1 C= = = 0.16 µF π f c k π × 4 × 103 × 500 L=

L , i.e. 19.9 mH in each series branch and a capacitor of 0.16 µF 2 in the shunt branch as shown in Fig. 11.10 (a). The p-section consists of an inductor of 39.79 mH in the series C branch and a capacitor of , i.e. 0.08 µF in each shunt branch as shown in Fig. 11.10 (b). 2 The T section consists of an inductor of

19.9 mH

19.9 mH

39.79 mH

0.16 µF

0.08 µF

(a)

0.08 µF

(b)

Fig. 11.10

  example 11.6  Design a constant-k, T section low-pass filter is having 2.5 kHz cut-off frequency and nominal impedance of 700 W. Also find the frequency at which this filter produces an attenuation of 19.1 dB. Find its characteristic impedances and phase constant at pass band and stop or attenuation band. Solution 

f c = 2.5 kHz, k = 700 Ω 700 k = = 89.13 mH π f C π × 2.5 × 103 1 1 C= = = 0.18 µF π f C k π × 2.5 × 103 × 700 L=

L The T-section filter consists of an inductor of , i.e. 44.57 mH in each series arm and a capacitor of 0.18 2 µF in the shunt arm as shown in Fig. 11.11.

11.14 Circuit Theory and Transmission Lines

α = 19.1 dB = 2.197 nepers

44.57 mH

44.57 mH

 f  α = 2 cosh −1    fC  −1 

0.18 µF

 2.197 = 2 cosh   2.5 × 103  f = 4.17 kHz b is imaginary in pass band. In attenuation band,

f

Fig. 11.11

β = 180°

  11.7     constant-k hIgh-Pass fIlter A constant-k high-pass filter is obtained by changing the positions of series and shunt arm of the constant-k low-pass filter, Figure 11.12 shows a constant-k, T and p section, high-pass filter. 2C

C

2C

L

2L

(a) T-section

(b) p-section

Fig. 11.12  Constant-k high-pass filter In a constant-k high-pass filter 1 1 = ω C jω C Z 2 = jω L Z1 = − j

1. Nominal Impedance  1  L k = Z1Z 2 =  ( jω L ) =   jω C  C 2. Cut-off Frequency The cut-off frequencies are obtained when (i)

When

Z1 =0 4Z2 Z1 = 0 1 =0 jω C ω=∞ f =∞

(ii) When

Z1 = −1 4Z2

Z1 Z = 0 and 1 = −1 4Z2 4Z2

2L

11.7  Constant-k High-pass Filter  11.15

Z1 = −4 Z 2 1 = 4 jω L ωc 1 ω 2 LC = 4 1 ω2 = 4 LC

−j

ω = ωc = f = fc =

1 a

2 LC 1

4π LC Hence, the filter passes all the frequencies beyond fc. The pass band starts at f = fc and continues up to infinite frequency. All the frequencies below the cut-off frequency lie in the attenuation or stop band. 0

3. Attenuation Constant In pass band, α = 0  f  Z1 = 2 cosh −1  c  4Z 2  f 

In stop band, α = 2 cosh −1

fc

Fig. 11.13  Variation of a with frequency b

The attenuation constant a decreases gradually to zero at the cut-off frequency and remains at zero through the pass band. The variation of a is plotted in Fig. 11.13.

p

4. Phase Constant

fc

0

In pass band, β = 2 sin −1

 f  Z1 = 2 sin −1  c  4Z 2  f 

In stop band, β = π . The phase constant b remains at constant value p in the stop band and decreases to −p in at fc and reaches zero value gradually as f increases in the pass band. The variation of b is plotted in Fig. 11.14.

f

f

−p

Fig. 11.14  Variation of b  with frequency Z0

5. Characteristic Impedance Z0p

Z 0T = = Z 0π =

Z12

Z   + Z1Z 2 = Z1Z 2 1 + 1   4Z 2  4

 fc  1  L 1 −  = k 1 −   2  f  C 4ω LC Z1Z 2 = Z 0T

k

2

Z0T 0

fc

k  f  1−  c   f 

2

Fig. 11.15  V   ariation of characteristic  impedance with frequency

The variation of characteristic impedance is plotted in Fig. 11.15.

f

11.16 Circuit Theory and Transmission Lines 6. Design of Filter k= fc =

L C 1 4π LC

Solving these two equations, k 4π f c 1 C= 4π f c k L=

  example 11.7 

Find the characteristic impedance, cut-off frequency and pass band for the network

shown in Fig. 11.16. 0.4 µF

0.4 µF

50 mH

Fig. 11.16 Solution The network is a high-pass filter. 2C = 0.4 µF, L = 50 mH C = 0.2 µF (a)

Characteristic impedance k=

(b)

50 × 10 −3 0.2 × 10 −6

= 500 Ω

Cut-off frequency fc =

(c)

L = C 1 4π LC

=

1 4π 50 × 10

−3

× 0.2 × 10 −6

= 795.77 Hz

Pass band The pass band is from 795.77 Hz to infinite frequency.

  example 11.8  A high-pass filter section is constructed from two capacitors of 1 µF each and a 15 mH inductance. Find (a) cut-off frequency, (b) infinite frequency characteristic impedance, (c) characteristic impedance at 200 Hz and 2000 Hz, (d) attenuation at 200 Hz and 2000 Hz, and (e) phase constant at 200 Hz and 2000 Hz. Solution

L = 15 mH, 2C = 1 µF C = 0.5 µF

11.7  Constant-k High-pass Filter  11.17

(a)

Cut-off frequency fc =

(b)

=

1 4π 15 × 10

−3

= 918.88 Hz

× 0.5 × 10 −6

Infinite-frequency characteristic impedance At f = ∞,

(c)

1 4π LC

Z0 = k =

1.5 × 10 −3 L = = 173.21 Ω C 0.5 × 10 −6

Characteristic impedance at 200 Hz 2

2

2

2

 f   918.88  Z0T = k 1 −  c  = 173.21 1 −  = j 776.72 Ω  2000   f  Characteristic impedance at 2000 Hz  f   918.88  Z0T = k 1 −  c  = 173.21 1 −  = 153.85 Ω  2000   f  (d)

Attenuation at 200 Hz  f   918.88  α = 2 cosh −1  c  = 2 cosh −1  = 4.41  200   f  The frequency of 2000 Hz lies in the pass band. α=0

(e)

Phase constant The frequency 200 Hz lies in the attenuation band. β = π radians At 2000 Hz,  f   918.88  β = 2 sin −1  C  = 2 sin −1  = 54.7°  2000   f 

  example 11.9  Design a constant-k high-pass T and p sections filters having a cut-off frequency of 2000 Hz and infinite frequency characteristic impedance of 300 W. Solution

f c = 2000 Hz, k = 300 Ω L=

k 300 = = 11.94 mH 4π f c 4π × 2000

C=

1 1 = = 0.13 µF 4π f c k 4π × 2000 × 300

The T-section filter consists of a capacitor of 2C, i.e 0.26 µF in each series arm and an inductor of 11.94 mH in shunt arm as shown in Fig. 11.17 (a). The p-section filter consists of a capacitor of 0.13 µF in the series arm and an inductor of 2L, i.e. 23.88 mH in each shunt arm as shown in Fig. 11.17 (b).

11.18 Circuit Theory and Transmission Lines 0.26 µF

0.26 µF

0.13 µF

11.94 mH

23.88 mH

(a)

23.88 mH

(b)

Fig. 11.17

  example 11.10  Design a T-section constant-k high-pass filter having a cut-off frequency of 10 kHz and a design impedance of 600 W. Find its characteristic impedance and phase constant at 25 kHz. Solution

f c = 10 kHz, k = 600 Ω k 600 = = 4.77 mH 4π f c 4π × 10 × 103 1 1 C= = = 0.013 µF 4π f c k 4π × 10 × 103 × 600 L=

0.026 µF

0.026 µF

4.77 mH

The T-section filter consists of a capacitor of 2C, i.e. 0.026 µF in each series arm and an inductor of 4.77 mH in shunt arm as shown in Fig. 11.18. (a)

Fig. 11.18

Characteristic impedance at 25 kHz 2

2

 10 × 103   f  Z0 = k 1 −  c  = 600 1 −   = 549.91 Ω  f   25 × 103  (b)

Phase constant at 25 kHz  10 × 103   f  β = 2 sin −1  C  = 2 sin −1   = 47.16°  f   25 × 103  The frequency of 25 kHz lies in the attenuation band. In the attenuation band, β = 180° = π radians

  11.8     Band-Pass fIlter A band-pass filter attenuates all the frequencies below a lower cut-off frequency and above an upper cut-off frequency. It passes a band of frequencies without attenuation. A band-pass filter is obtained by using a low pass filter followed by a high-pass filter. Figure 11.19 shows a band-pass filter. The series arm is a series resonant circuit comprising L1 and C1 while its shunt arm is formed by a parallel resonant circuit L2 and C2. The resonant frequency of series arm and shunt arm are made equal.

11.8  Band-pass Filter  11.19 L1

L1 2C1

2

2C1

L2

L1

2

2L2

C2

(a) T-section

C2 2

C1

2L2

C1

(b) p-section

Fig. 11.19  Band-pass filter For series arm, L1 1 = 2 2ω 0C1 1 ω 02 = L1C1 1 = ω 0 L2 ω 0C 2 1 ω 02 = L2C2 L1C1 = L2C2

ω0

For shunt arm,

For series arm, Z1 = jω L1 −

 ω 2 L1C1 − 1 j = j ω c1 ω c1  

For shunt arm, 1 jω L2 jω c2 Z2 = = 1 1 − ω 2 L2C2 jω L2 + jω c2 jω L2

 ω 2 L1C1 − 1  jω L2  L2  ω 2 L1C1 − 1  L2 L1 Z1Z 2 = j  = − = = = k2      2 2 ω c1   1 − ω L2C2  C1  1 − ω L2C2  C1 C2  where k is constant. For constant k filter, at cut-off frequency, Z1 = −4 Z2 Z12 = −4 Z1Z2 = −4 k 2 Z1 = ± j 2k i.e. the value of Z1 at lower cut-off frequency is equal to the negative value of Z1 at the upper cut-off frequency.  1  1 + jω1 L1 = −  + jω 2 L1  jω1C1  jω 2C1  1 − ω12 L1C1 =

ω1 2 ω 2 L1C1 − 1 ω2

(

)

11.20 Circuit Theory and Transmission Lines

ω 02 =

But 1−

1 L1C1

ω12 ω1  ω 22  = − 1  ω 02 ω 2  ω 02 

(ω 02 − ω12 )ω 2 = ω1 (ω 22 − ω 02 )

ω 02ω 2 − ω12ω 2 = ω1ω 22 − ω1ω 02 ω 02 (ω1 + ω 2 ) = ω1ω 2 (ω 2 + ω1 ) ω 02 = ω1ω 2 ω 0 = ω1ω 2 f0 =

f1 f 2

Thus, resonant frequency is the geometric mean of the cutoff frequencies. The variation of attenuation, phase constant and characteristic impedance with frequency are shown in Fig. 11.20. a

p

0

f1

f0

f2

f

Z0p

Z0

b

k

0

f1

f0

f2

f

0

f1

f0

Z0T f2

f

−p (a)

(b)

(c)

Fig. 11.20  V   ariation of (a) attenuation, (b) phase constant, and (c) characteristic impedance with  frequency for constant–k band-pass filter Design of Filter If the filter is terminated in a load resistance R = k then at lower cut-off frequency, Z1 = −2 jk 1 + jω1 L1 = −2 jk jω1C1 1 − ω1 L1 = 2k ω1C1 1 − ω12 L1C1 = 2kω1C1 1−

ω12 = 2kω1C1 ω 02 2

 f  1 −  1  = 4π kf1C1  f0 

1   2 ∵ ω 0 = L C  1 1

11.8  Band-pass Filter  11.21

1−

f12 = 4π kf1C1 f1 f 2

(∵ f

0

=

f1 f 2

)

f 2 − f1 = 4π kf1 f 2C1 f −f C1 = 2 1 4π kf1 f 2 1 4π kf1 f 2 4π kf1 f 2 k = L1 = 2 = 2 = 2 2 ω 0 C1 ω 0 ( f 2 − f1 ) 4π f 0 ( f 2 − f1 ) π ( f 2 − f1 ) For shunt arm, L2 L1 = = k2 C1 C2 ( f − f )k L2 = C1k 2 = 2 1 4π f1 f 2 L 1 C2 = 12 = π ( f k 2 − f1 ) k

Z1Z 2 =

  example 11.11  In a constant-k band-pass filter, the ratio of capacitances in the shunt and series arms is 100:1. The resonant frequency of both arms is 1000 Hz. Find the bandwidth, Solution

C2 100 = = 100, f 0 = 1000 Hz C1 1 1 C2 = π k ( f 2 − f1 ) f −f C1 = 2 1 4π kf1 f 2 1 4π kf1 f 2 C2 = C1 π k ( f 2 − f1 ) f 2 − f1 4 f1 f 2 100 = ( f 2 − f1 ) 2 100 =

4 f 02 ( f 2 − f1 ) 2

2 × 1000 = = 200 100 100 BW = f 2 − f1 = 200 Hz

f 2 − f1 =

2 f0

  example 11.12  Design a band pass constant-k filter with cut-off frequency of 4 kHz and 10 kHz and nominal characteristic impedance of 500 W . Solution

f1 = 4 kHz, f 2 = 10 kHz, k = 500 Ω 500 k = = 26.53 mH L1 = π ( f 2 − f1 ) π (10 × 103 − 4 × 103 ) C1 =

10 × 103 − 4 × 103 f 2 − f1 = = 0.024 µF 4π kf1 f 2 4π × 500 × 4 × 103 × 10 × 103

11.22 Circuit Theory and Transmission Lines ( f 2 − f1 )k (10 × 103 − 4 × 103 ) 500 = = 5.97 mH 4π f1 f 2 4π × 4 × 103 × 10 × 103 1 1 = = 0.11 µF C2 = 3 π ( f 2 − f1 )k π × (10 × 10 − 4 × 103 ) × 500 L2 =

L1 , i.e. 13.27 mH and a capacitor of 2 2 C1, i.e. 0.048 µF in each series arm and a parallel combination of an inductor of 5.97 mH and a capacitor

The T-section filter consists of series combination of an inductor of of 0.11 µF in shunt arm as shown in Fig. 11.21. 13.27 mH

0.048 µF

13.27 mH 0.048 µF

5.97 mH

0.11 µF

Fig. 11.21 The p-section filter consists of a series combination of an inductor 26.53 mH and a capacitor of 0.024 µF in C the series arm and a parallel combination of an inductor of 2L2, i.e. 11.94 mH and a capacitor of 2 , i.e. 2 0.055 µF in each shunt arm as shown in Fig. 11.22. 26.53 mH

11.94 mH

0.055 µF

0.024 µF

11.94 mH

0.055 µF

Fig. 11.22

  11.9     Band-stoP fIlter A band-stop filter attenuates a specified band of frequencies and allows all frequencies below and above this band. A band-stop filter is realised by connecting a low-pass filter in parallel with a high-pass filter. Figure 11.23 shows a band-stop filter. As in the band-pass filter, the series and shunt arms are chosen to resonate at same frequency w0. For series arm, ω 0 L1 1 = 2 2ω 0C1 1 ω 02 = L1C1

11.9  Band-stop Filter  11.23 L1

L1

2

2

2C1

L1

2C1

C1

2L2

L2 C2

2L2

C2

C2

2

2 (b) p-section

(a) T-section

Fig. 11.23  Band-stop filter For shunt arm, 1 ω 0 C2 1 ω 02 = L2C2 L1C1 = L2C2 L L Z1Z2 = 1 = 2 = k 2 C2 C1

ω 0 L2 =

Similarly, and

f0 =

At cut-off frequencies,

Z1 = −4 Z2

f1 f 2

Z1Z2 = −4 Z22 = k 2 k Z2 = ± j 2 The variation of attenuation, phase shift and characteristic impedance with frequency are shown in Fig. 11.24. b

a

Z0

Z0p

p

0

0

f1

f0 (a)

f2

f

f1

f0

f2

f

k

Z0T 0

−p (b)

f1 f0 f2

f

(c)

Fig 11.24  V   ariation of (a) attenuation, (b) phase constant, and (c) characteristic impedance with  frequency for constant-k band-stop filter. Design of Filter If the load is terminated in load resistance R = k then at lower cut-off frequency,

11.24 Circuit Theory and Transmission Lines k  1  − ω1 L2  = j Z2 = j   ω1C2  2 1 k − ω1 L2 = ω1C2 2 1 − ω12 L2C2 = ω1C2 1−

k 2

ω12 k = ω1C2 ω 02 2

1   2 ∵ ω 0 = L C  2 2

2

 f  1 −  1  = kπ f1C2  f0  C2 = L2 =

2 1   f1   1  1 1  1  f 2 − f1  f −f 1 −    = − = = 2 1    kπ f1   f 0   kπ  f1 f 2  kπ  f1 f 2  π k f1 f 2  

1

ω 02C2

=

π kf1 f 2 2 ω 0 ( f 2 − f1 )

=

k π kf1 f 2 = 2 4π f 0 ( f 2 − f1 ) 4π ( f 2 − f1 ) 2

L1 L2 = C2 C1 k ( f 2 − f1 ) L1 = k 2C2 = π f1 f 2 1 L C1 = 22 = 4 π ( k f 2 − f1 ) k

k2 =

  example 11.13  Design a constant-k band-stop filter having cut-off frequencies at 2000 Hz and 5000 Hz and characteristic resistance of 600 W. Solution

f1 = 2000 Hz,

f 2 = 5000 Hz, k = 600 Ω

k ( f 2 − f1 ) 600(5000 − 2000) = = 57.3 mH π f1 f 2 π × 2000 × 5000) 1 1 = = 0.044 µF C1 = 4π k ( f 2 − f1 ) 4π × 600 × (5000 − 2000) L1 =

600 k = = 15.92 mH 4π ( f 2 − f1 ) 4π × (5000 − 2000) 5000 − 2000 f −f = 0.16 µF C2 = 2 1 = π kf1 f 2 π × 600 × 2000 × 5000 L2 =

L1 , i.e. 28.65 mH and a capacitor of 2 C1, i.e. 2 0.088 µF in each series arm and a series combination of an inductor of 11.92 mH and a capacitor of 0.16 µF in shunt arm as shown in Fig. 11.25.

The T-section filter consists of a parallel combination of

11.10  m-derived Filters  11.25 28.65 mH

28.65 mH

57.3 mH

0.088 µF

0.088 µF

0.044 µF 31.84 mH

31.84 mH

0.08 µF

0.08 µF

15.92 mH

0.16 µH

Fig. 11.26

Fig. 11.25

The p-section filter consists of a parallel combination of an inductor of 57.3 mH and a capacitor of 0.044 µF C in the series arm and a series combination of an inductor of 2L2, i.e. 31.84 mH and a capacitor of 2 , i.e. 2 0.08 µF in each shunt arm as shown in Fig. 11.26.

  11.10     m-derIved fIlters There are two disadvantages of constant-k filters. (i) (ii)

The attenuation is not sharp in the stop band. The characteristic impedance varies widely in the pass band.

If the constant-k filter is regarded as prototype, it is possible to design a filter to have rapid attenuation in the stop band and the same characteristic impedance as the prototype throughout the pass band and stop band. Such a filter is called an m- derived filter. Constant-k and m-derived T sections are shown in Fig. 11.27. The constant-k T-network is identical to that of the m-derived T-network except that the series arm is multiplier by m. Z1

Z1

2

2

m

Z1

m

2

Z2

Z2

Fig. 11.27  Constant-k and m-derived T sections For the constant-k T section, ZOT =

Z12 + Z1 Z 2 4

ZOT =

m2 Z12 + m Z1 Z′2 4

For the m-derived T section,

Z1 2

11.26 Circuit Theory and Transmission Lines Equating the characteristic impedance of the networks, Z12 + Z1 Z 2 = 4

m2 Z12 + mZ1 Z ′2 4

Z12 m2 Z12 + Z1 Z 2 = + mZ1 Z ′2 4 4 Z2 mZ1 Z′2 = 1 (1 − m2 ) + Z1 Z 2 4  1 − m2  Z Z′2 =  Z1 + 2  m  4m 

m

Z1

m

2

Z1 2

1 - m2 4m

Z1

Hence, Z′2 must be an impedance composed of two series Z2  1 − m2  Z2 1 − m2 m impedances   Z1 and m . The term 4 m should be 4 m   positive to realise the impedance Z′2 physically, i.e. 0 < m < 1. Thus, an m-derived filter can be obtained from constant-k filter by Fig. 11.28  m-derived T-section filter modifying its series and shunt impedances as shown in Fig. 11.28. Similarly, the constant-k p-section filter can be modified to m-derived p section filter. A constant-k and m-derived p sections are shown in Fig. 11.29. Z1

2Z2

Z1

2 Z2 m

2Z2

(a)

2 Z2 m

(b)

Fig. 11.29  (a) Constant-k section (b) m-derived p section For the constant-k p section, ZOπ =

Z1 Z 2 Z 1+ 1 4Z 2

For the m-derived p section, Z2 m ZOπ = Z′ 1+ 1 Z 4 2 m Equating the characteristic impedance of the networks, Z1′

Z Z1′ 2 Z1 Z 2 m = ′ Z1 Z 1+ 1+ 1 4Z 2 Z 4 2 m

11.10  m-derived Filters  11.27

Z Z1′ 2 Z1 Z 2 m = ′ Z1 Z 1+ 1+ 1 4Z 2 Z 4 2 m   Z′  Z  Z   Z1 Z 2 1 + 1  = Z1′ 2 1 + 1  Z2  4Z 2  m  4    m Z1 Z 2 +

mZ1 Z1′ Z1′ Z 2 Z1Z1′ + = m 4m 4

4 Z1 Z 2 + m Z1 Z1′ =

4 Z1′ Z 2 + Z1 Z1′ m

 Z 4Z 2  Z1′  1 + − mZ1  = 4 Z1 Z 2 m  m Z1 Z 2 Z1′ = Z1 Z 2 mZ1 + − 4m 4m 4 Z1 Z 2

=

4 m2 (1 − m2 )

Z1 Z 2 = Z1 Z2 2 4 m2 (1 − m ) + Z2 mZ1 + 4m m m(1 − m)

4m Z 2 2 1 − m = 4m mZ1 + Z2 1 − m2 mZ1

 4m  Hence, Z1′ must be an impedance composed of two parallel impedances mZ1 and  Z 2 . Thus, an  1 − m2  m-derived filter can be obtained from a constant-k filter by modifying its series and shunt impedances as shown in Fig. 11.30. mZ1

2 Z2 m

4m 1 - m2

Z2

Fig. 11.30  m-derived p-section filter

2 Z2 m

11.28 Circuit Theory and Transmission Lines

  11.11     m-derIved loW-Pass fIlter Figure 11.31 shows an m-derived low-pass T and p filter. At a particular frequency, the shunt arm of the T section or series arm of the p section will be in resonance giving a short circuit to the input for the T section and open circuit for the p section causing infinite attenuation. mZ1

mZ1

2

2

1 - m2 4m

mZ1

Z1

4m 2Z2 m

Z2

1 - m2

Z2

2Z2 m

m

(b) p-section

(a) T-section

Fig. 11.31  m-derived low-pass filter Let

Z1 = jω L Z2 =

1 −j = jω C ω C

At resonant frequency,  1 − m2  1  4m  ω 0 L = m ω C   0

ω02 = ω0 = f0 =

4 (1 − m2 ) LC 2 LC (1 − m2 ) 1

π LC (1 − m2 )

= f∞

For low pass filter, fc = f∞ =

1

π LC fc

mL 2

1 − m2

 f  m = 1−  c   f∞ 

mL 2 1 - m2

2

For sharp cut-off, f∞ should be near to fC . For smaller value of m, f∞ comes close to fc. The m-derived T-section low-pass filter is shown in Fig. 11.32.

L

4m mC

Fig. 11.32  m   -derived T-section   low-pass filter

11.11  m-derived Low-pass Filter  11.29

Similarly, for m-derived p-section, the inductance and capacitance in the series arm constitute a resonant circuit. At resonant frequency, 1 m ω0 L =  1 − m2   4m  ω 0C  

ω 02 = ω0 = f0 =

4 LC (1 − m2 ) 4 LC (1 − m2 ) 1

π LC (1 − m2 )

= f∞

For low-pass filter, fc = f∞ =

1

π LC fc 1 − m2

For both m-derived low-pass networks, f ∞ > f c for positive values of m (0 < m < 1). The m-derived p section low-pass filter is shown in Fig. 11.33. mL

mC 2

1 - m2

mC 2

C

4m

Fig. 11.33  m-derived p-section low-pass filter a

The variation of attenuation with frequency is shown in Fig. 11.34. It is seen that the attenuation rises abruptly becomes infinite at f∞ and then decreases for the m-derived filter. Thus, an m-derived filter gives a sharp cut-off. The decrease in attenuation for frequencies higher than f∞ can be avoided by connecting an m-derived filter and a constant-k filter in cascade. The variation of b with frequency is shown in Fig. 11.35.

0

fc

f∞

f

Fig. 11.34  Variation of a with frequency 

11.30 Circuit Theory and Transmission Lines Z0 Z0p b

k p

Z0T 0 fÉ

fc

0

f

Fig. 11.35  Variation of b with frequency        

fc

fÉ

f

Fig. 11.36  V   ariation in characteristic  impedance with frequency.

The variation of characteristic impedance with frequency is shown in Fig. 11.36. The characteristic impedance of an m-derived filter also varies with frequency. The impedance matching can be achieved by using half-section terminations.

  example 11.14  Design an m-derived T-section low-pass filter having cut-off frequency of 800 Hz, design impedance of 500 W and frequency of attenuation of 1000 Hz. Solution k = 500 Ω, f c = 800 Hz, For a constant-k T-section low-pass filter 500 k L= = = 198.94 mH π f c π × 800 1 1 C= = = 0.8 µF π f c k π × 800 × 500

f ∞ = 1000 Hz

For an m-derived T-section low-pass filter 2

2

 f   800  m = 1−  c  = 1−  = 0.6  1000   f∞  mL 0.6 × 198.94 × 10 −3 = = 59.68 mH 2 2 mC = 0.6 × 0.8 × 10 2

−6

59.68 mH

59.68 mH

53.05 mH

= 0.48 µF

1− m 1 − (0.6)  −3  4 m  L =  4(0.6)  × 198.94 × 10 = 53.05 mH     The m-derived T-section low-pass filter is shown in Fig. 11.37. 2

0.48 µF

Fig. 11.37

  example 11.15  Design an m-derived p-section low-pass filter having cut-off frequency of 1500 Hz, design impedance of 500 Ω and infinite attenuation frequency of 2000 Hz. Solution

fC = 1500 Hz, k = 500 Ω,

f ∞ = 2000 Hz

For constant-k p-section low-pass filter, 500 k L= = = 106.1 mH π f c π × 1500

11.12  m-derived High-pass Filter  11.31

C=

1 1 = = 0.424 µF π f c k π × 1500 × 500

For m-derived p-section low-pass filter, 2

2

 f   1500  m = 1−  c  = 1−  = 0.66  2000   f∞ 

70.03 mH

mC 0.66 × 0.424 × 10 −6 = = 0.14 µF 2 2

0.09 µF

mL = 0.66 × 106.1 × 10 −3 = 70.03 mH 0.14 µF

0.14 µF

 1 − m2  1 − (0.66) 2 −6  4 m  C = 4 × 0.66 × 0.424 × 10 = 0.09 µF  

Fig. 11.38

The m-derived low-pass filter is shown in Fig. 11.38.

  11.12     m-derIved hIgh-Pass fIlter Figure 11.39 shows m-derived high-pass T and p filter. mZ 1

mZ 1

2

2

mZ 1

1 − m2 Z 1 4m

2Z 2 m

Z2 m

4m Z 2 1 − m2

(b) p-section

(a) T-section

Fig. 11.39  m-derived high-pass filter Let

Z1 =

1 jω c

Z 2 = jω L At resonant frequency,

ω0

L = m

ω 02 =

ω0 = f0 =

1  4m  ω0  C  1 − m2  1 1 − m2 = L  4m  4 LC  C m  1 − m2  1 − m2 2 LC 1 − m2 4π LC

= f∞

2Z 2 m

11.32 Circuit Theory and Transmission Lines For a high-pass filter, fc =

2C m

2C m

1 4π LC

f∞ = fc 1 − m

4m C 1−m 2

2

 f  m = 1−  ∞   fc 

L m

2

The m-derived T-section high-pass filter is shown in Fig. 11.40. Fig. 11.40   m-derived  T-section  highpass filter Similarly, for an m-derived p section, the resonant circuit is formed by the series arm inductance and capacitance. At resonant frequency, 1  4m  ω0 L =  2 ω  0 1− m C m

ω 02 = ω0 = f0 =

1 − m2 4 LC 1 − m2 2 LC 1 − m2 4π LC

C m

= f∞

For a high-pass filter, fc =

1

4m L 1−m 2

2L m

4π LC

2L m

f ∞ = f c 1 − m2  f  m = 1−  ∞   fc 

2

Fig. 11.41  m   -derived p-section high-pass  filter

The m-derived p section high-pass filter is shown in Fig. 11.41. The variation of attenuation phase constant and characteristic impedance with frequency are shown in Fig. 11.42 for an m-derived high-pass filter. The attenuation rises abruptly, becomes infinite at f∞ and then decreases, thus giving a sharp cut-off. Z0

b

a

Z0p

k 0

0

f∞ (a)

f

f∞

fc

f Z0T

−p 0

fc (b)

fc

f

(c)

Fig. 11.42  V   ariation of (a) attenuation, (b) phase constant, (c) characteristic impedance of m-derived  high-pass filter.

11.12  m-derived High-pass Filter  11.33

The use of m-derived and constant-k high pass filters in series gives a sharp cut-off and high attenuation in the entire stop band. The half-section termination provides impedance matching.

  example 11.16  Design an m-derived T-section high-pass filter with a cutoff frequency of 2 kHz, design impedance of 700 W and m = 0.6. Solution

fC = 2 kHz, k = 700 Ω, m = 0.6

For constant-k T-section high-pass filter, k 700 = = 27.85 mH 4π f c 4π × 2 × 103 1 1 C= = = 0.057 µF 4π f c k 4π × 2000 × 700 L=

For an m-derived T-section high-pass filter, L 27.85 × 10 −3 = = 46.42 mH m 0.6

0.19 µF

2C 2 × 0.057 × 10 −6 = = 0.19 µF m 0.6 4 × 0.6  4m  C= × 0.057 × 10 −6 = 0.21 µF  2 2  1− m 1 − (0.6) The m-derived high-pass filter is shown in Fig. 11.43.

0.19 µF

0.21 µF 46.42 mH

Fig. 11.43

  example 11.17  Design the p-section of an m-derived high-pass filter having a design impedance of 600 W, cut-off frequency of 4 kHz and an infinite attenuation at 3.6 kHz. k = 600 Ω,

Solution

f c = 4 kHz,

f ∞ = 3.6 kHz

For a constant-k p-section high-pass filter, k 600 = = 11.94 mH 4π f c 4π × 4 × 103 1 1 C= = = 0.033 µF F 4π f c k 4π × 4 × 103 × 600 L=

For an m-derived p-section high-pass filter, 2

2

 3.6 × 103   f  m = 1−  ∞  = 1−   = 0.436  fc   4 × 103  2 L 2 × 11.94 × 10 −3 = 54.77 mH = 0.436 m C 0.033 × 10 −6 = = 0.076 µF 0.436 m

11.34 Circuit Theory and Transmission Lines 0.076 µF

4 × 0.436  4m  × 11.94 × 10 −3 = 25.71 mH  L= 1 − m2  1 − (0.436) 2 The m-derived p-section high-pass filter is shown in Fig. 11.44.

25.71 mH 54.77 mH

54.77 mH

  11.13     termInatIng half sectIons A filter is composed of a number of sections. As the characteristic Fig. 11.44 impedance of an equivalent T or p section does not match with each other, a half section is used for impedance matching between T and p sections. The input impedance of the half section is same as the characteristic impedance of a T section, while the output impedance of the half section is same as the characteristic impedance of the p section. Hence, matching can be obtained if a half section is connected in between a T and p section.

11.13.1

Constant-k Half Sections

Figure 11.45(a) shows a constant-k T-section filter. If this section is bisected longitudinally, a half section is obtained as shown in Fig. 11.45 (b). Z1

Z1

Z1

Z1

2

2

2

2

2Z 2

Z2

(a)

2Z 2

(b)

Fig. 11.45  (a) Constant-k T-section (b) constant-k half T-section Similarly, a constant-k p section can be bisected into two half sections as shown in Fig. 11.46. Z1

2Z 2

2Z 2

(a)

Z1

Z1

2

2

2Z 2

2Z 2

(b)

Fig. 11.46  (a) Constant-k p-section (b) constant-k half p-section Figure 11.47 shows a constant-k half section. The image impedance of the section as seen from terminals 1-2 and terminals 3-4 can be found from the open circuit and short-circuit impedances.

11.13  Terminating Half Sections  11.35

Z12 = Z oc Z sc =

Z34 = Zoc Z sc

Z   ( 2Z 2 )  2Z 2 1   2 = Z 2Z 2 + 1 2

Z1

4Z1 Z 22 = Z 0π Z1 + 4Z 2

Z Z  =  1 + 2Z 2   1   2  2 

2 1

3 2Z 2 4

2

Z   = Z1 Z 2 1 + 1  = Z 0T  4Z 2 

Fig. 11.47  Constant-k half section

Thus, the half section has the impedance characteristics of a p section between terminals 1-2 and that of a T section between terminals 3-4. Hence, this half section can be used to match a p section to a T section. It can also be used to match a filter section to a terminating impedance which differs from the characteristic impedance of a p section.

11.13.2

m-derived Half Sections

Figure 11.48 shows and m-derived half section. The image impedance of the section as seen from terminals 1-2 and terminals 3-4 can be found from the open circuit and short-circuit impedances.

mZ 1 2 1

=

 1 − m2 2Z 2   1 − m2 2Z 2   Z1  + Z Z1 + 1 m   2m   m   2m m   2   1 − m2 2Z 2 mZ1 + Z1 + 2m 2 m 2

=

 1 − m2 2Z 2   mZ1   2m Z1 + m   2    1 − m2 2Z 2 mZ1 + Z1 + 22mm 22 mm 2

=

2  1 − m2 2Z 2   mZ1  + Z    2m 1 m   2  

 1 − m2 2Z 2 mZ1   mZ1   2m Z1 + m + 2   2    2

=

=

  1 − m2  2  Z1 + Z1 Z 2     4   2 Z1 m2 Z12 m2 Z12 − + Z1 Z 2 + 4 4 4 Z1 Z 2 +

(1 − m2 ) 2 Z1 4

Z1 Z 2 +

3 1−m 2 2Z 1 4m

Z12 = Zoc Z sc

Z12 4

2

2Z 2 m

4

Fig. 11.48  m-derived half section

11.36 Circuit Theory and Transmission Lines =

Z1  2 1 + 4Z (1 − m)  2   Z1   2 1 + 4Z (1 − m)  2  

Z1 Z 2 Z 0T

= Z 0π

Z34 = Zoc Z sc  mZ  1 − m2  2Z 2   mZ1  =  1 + Z1 +    m   2   2m   2 =

m2 Z12 Z12 m2 Z12 + − + Z1 Z 2 4 4 4

Z   = Z1 Z 2 1 + 1   4Z 2  = Z 0T Hence, the terminals 3-4 of the m-derived half section can be used to match the impedance of a constant-k T-section filter.

  example 11.18  Find the values of shunt and series element of each half section of a constant-k T-section high-pass filter. The termination is 600 Ω and the filter is cut off below 20000 Hz. Solution

k = 600 Ω,

f c = 20000 Hz

k 600 = = 2.39 mH 4π f c 4π × 20000 1 1 C= = = 0.0066 µF F 4π f c k 4π × 20000 × 600 The half section consists of a capacitor of 2C, i.e. 0.0133 µF in the series branch and an inductor of 2L, i.e. 4.78 mH in the shunt branch as shown in Fig. 11.49. L=

0.0133 µF

4.78 mH

Fig. 11.49

  example 11.19  A prototype high-pass filter with L = 4.77 mH and C = 0.0133 µ F has a cut-off frequency of 10 kHz and design impedance of 600 Ω. Design a suitable terminating half section with m = 0.6 such as to operate with same cut-off frequency. Solution

L = 4.77 mH, C = 0.0133 µF,

f c = 10 kHz, k = 600 Ω

For m-derived T-section high-pass filter L 4.77 × 10 −3 = = 7.95 mH m 0.6 2C 2 × 0.0133 × 10 −6 = = 0.044 µF m 0.6 4 × 0.6  4m  C= × 0.0133 × 10 −6 = 0.05 µF  2  1− m 1 − (0.6) 2

0.044 µF

15.9 mH 0.025 µF

Fig. 11.50

11.14  Composite Filter  11.37

The m-derived T-half section is shown in Fig. 11.50. For m-derived p-section high-pass filter, 8.95 mH

2 L 2 × 4.77 × 10 −3 = = 15.9 mH m 0.6 C 0.0133 × 10 −6 = = 0.022 µF m 0.6 4 × 0.6  4m  L= × 4.77 × 10 −3 = 17.89 mH  2 2  1− m 1 − (0.6)

0.044 µF

15.9 mH

Fig. 11.51

The m-derived p-half section is shown in Fig. 11.51.

  11.14      comPosIte fIlter A constant-k filter does not give a sharp cut-off. An m-derived filter section has a sharp cut off, but its attenuation decreases for frequencies beyond f∞. Moreover, two terminating half sections, one at each end, are needed for impedance matching. A filter composed of all the above sections is known as a composite filter. Figure 11.52 shows a composite low-pass filter. L 2

mL 2

L 2

1−m 2 L 2m

mL 2

C

mC 2 Terminating half section

Constant-k T-section

mL 2

mL 2

1−m 2 L 4m

1−m 2 L 2m

mC

mC 2

m-derived T-section

Terminating half section

Fig. 11.52  Composite low-pass filter A composite filter consists of the following components: (i) One or more constant-k sections to produce cut-off frequency, i.e. transition from pass band to the stop band at a specified frequency fc. (ii) One or more m-derived sections to give infinite attenuation at a frequency f∞ near to fc. (iii) Two terminating m-derived half sections with m = 0.6 to provide matching with the source and the load. Figure 11.53 shows the attenuation versus frequency characteristics of a constant-k low-pass filter, m-derived low-pass filter and a composite filter. a

Composite filter Constant-k filter m-derived filter

f 0

fc

Fig. 11.53  Attenuation characteristics

11.38 Circuit Theory and Transmission Lines For a composite filter, the attenuation rises very rapidly with frequency in the range fc to f∞ and falls only marginally with frequency.

  example 11.20  Design a composite low-pass filter to have a cut-off frequency of 1000 Hz and a characteristic impedance of 600 W. Use one constant-k T section, one m-derived T section and two terminating half sections with m = 0.6. The frequency of infinite attenuation is 1050 Hz. Solution

f c = 1000 Hz, k = 600 Ω,

f ∞ = 1050 Hz

For constant-k T section, 600 k = = 0.19 H π f c π × 1000 1 1 C= = = 0.53 µF π f c k π × 1000 × 600 L=

L Hence, the constant-k T section filter consists of an inductor of 1 , i.e. 0.095 H in series arm and a 2 capacitor C1, i.e. 0.53 µF in the shunt arm. For the m-derived full T section, 2

2

 f   1000  m = 1−  c  = 1−  = 0.305  1050   f∞  mL 0.305 × 0.19 = = 0.029 H 2 2 mC = 0.305 × 0.53 × 10 −6 = 0.162 µF  1 − m2  1 − (0.305) 2 L = × 0.19 = 0.14 H  4m  4 × 0.305   For each terminating half section with m = 0.6, mL 0.6 × 0.19 = = 0.057 H 2 2 mC 0.6 × 0.53 × 10 −6 = = 0.159 µF 2 2  1 − m2  1 − (0.6) 2  2m  L = 2 × 0.6 × 0.19 = 0.101 H   The composite low-pass filter is shown in Fig. 11.54.

11.14  Composite Filter  11.39 0.057 H

0.095 H 0.095 H

0.029 H 0.029 H

0.53 µF

0.101 H

0.14 H

0.101 H 0.159 µF

0.162 µF

0.159 µF Terminating half section

0.057 H

Constant-k T-section

m-derived T-section

Terminating half section

Fig. 11.54

  example 11.21  Design a composite high-pass filter having a characteristic impedance of 600 W and a cut-off frequency of 1000 Hz. Use one a constant-k T section, one m-derived T section with m = 0.2 and two m-derived terminating half sections with m = 0.6. Solution

k = 600 Ω,

f c = 1000 Hz

For a constant-k T section, L=

k 600 = = 0.0477 H 4π f c 4π × 1000

C=

1 1 = = 0.133 µF 4π f c k 4π × 1000 × 600

Hence, the constant-k T section filter consists of an inductor of 0.0477 H in the shunt branch and a capacitor of 2C, i.e. 0.266 µF in each series branch. For an m-derived T section with m = 0.2, 2C 2 × 0.133 × 10 −6 = = 1.33 µF m 0.2 L 0.0477 = = 0.239 H m 0.2 4 × 0.2  4m  C= × 0.133 × 10 −6 = 0.11 µF  2  1− m 1 − (0.2) 2 For each terminating half section with m = 0.6, 2C 2 × 0.133 × 10 −6 = = 0.44 µF m 0.6 2 L 2 × 0.0477 = = 0.159 H m 0.6 2 × 0.6  2m  × 0.133 × 10 −6 = 0.25 µF  C = 1 − m2  1 − (0.6) 2 The composite high-pass filter is shown in Fig. 11.55.

11.40 Circuit Theory and Transmission Lines 0.44 µF

0.266 µF 0.266 µF

1.33 µF 1.33 µF

0.0477 H

0.159 H

0.239 H 0.11 µF

0.25 µF Termenating half section

Constant-k T-section

m-derived T-section

0.44 µF

0.159 H 0.25 µF Termenating half section

Fig. 11.55

  11.15     attenuator An attenuator is a two-port resistive network. It is used to reduce the signal level when used between a generator and load. Attenuators may be symmetrical or asymmetrical. They may provide fixed or variable attenuation. A fixed attenuator is also called a pad. The attenuation is expressed in decibels (dB). Attenuation in dB = 10 log10

P1 P2

where P1 is the input power and P2 is the output power. For a properly matched network, P1 = I12 R0 =

V12 R0

P2 = I 22 R0 =

V22 R0

where R0 is the characteristic resistance of the network. Hence,

attenuation in dB = 20 log10

V1 I = 20 log10 1 V2 I2

Attenuation is also expressed in nepers as the natural logarithm of the voltage or current ratio. Attenuation in nepers = ln

1 P V1 I = ln 1 = ln 1 V2 I 2 2 P2

V1 I1 P = = N then 1 = N 2 V2 I 2 P2

If

Attenuation in dB = 20 log10 N  dB  N = antilog    20  where N is the attenuation in nepers.

11.16  Lattice Attenuator  11.41

Relation between Decibel and Neper V Attenuation in nepers = ln 1 v2 Attenuation in dB = 20 log10

V1 V2

Changing the base of the algorithm, V1 20 V V2 = ln 1 = 8.686 (attenuation in neper) attenuation in dB = 20 ln 10 2.303 V2 ln

Attenuation in nepers =

1 (attenuation in dB ) = 0.115 (attenuation in dB) 8.686

  11.16     lattIce attenuator A lattice is a general type of symmetrical and balanced network. Any symmetrical balanced or unbalanced network can be transformed into an equivalent lattice network. Figure 11.56 shows a lattice network terminated through a characteristic resistance R0. The elements of a lattice attenuator can be specified in terms of characteristic impedance and propagation constant. We know that

RA

RB

RB

Z0 = Zoc Z sc Zoc Z sc

R0

RA

R + RB = A 2 2 RA RB = RA + RB

Fig. 11.56  Lattice attenuator

 R + RB   2 RA RB  = RA RB Z0 = R0 = Zoc Z sc =  A   2   RA + RB  Redrawing the network as shown in Fig. 11.57, Applying KVL to the loop, V1 = R0 I1 = RA ( I1 − I ) + R0 I 2 + RA ( I + I 2 ) R0 I1 = RA ( I1 + I 2 ) + R0 I 2 ( R0 − RA ) I1 = ( RA + R0 ) I 2 R 1+ A I1 RA + R0 R0 = = RA I 2 R0 − RA 1− R0 RA RA 1+ R0 I R0 N = eα = 1 = = R I2 1 − A 1 − RA R0 R0 1+

I1

I1 − I

I

+ RB

RA

V1 R0

I + I2

R0 RA

I2

I1 − I − I2

RB

− I1

Fig. 11.57  Modified network

11.42 Circuit Theory and Transmission Lines  RA  1+ R  0  α = log   RA  1− R   0  We can express RA and RB in terms of attenuation. R  R  N 1 − A  = 1 + A  R0  R0  N − 1 RA = R0   N + 1  N + 1 Similarly, RB = R0   N − 1 These equations can be used to find the equivalent lattice network for T and p networks using the bisection theorem.

Bisection Theorem A network is said to have been bisected when the open-circuited and short-circuited input impedances of the two bisected networks are equivalent. Also, the square root of the product of these impedances is the characteristic impedance of the original whole network. The bisection theorem states that ‘any symmetrical balanced or unbalanced network can be transformed into an equivalent lattice network. The series arm of the lattice is equal to the short-circuited impedance of the bisected network and the diagonal arm is equal to the open-circuited impedance of the bisected network’.

  example  11.22 

Design a lattice attenuator if the characteristic resistance is 200 W and the

attenuation is 20 dB. Solution

N ( dB) = 20 dB, R0 = 200 Ω N ( dB) = 20 log10 N 163.6 Ω

20 = 20 log10 N N = 10  N − 1  10 − 1 RA = R0  = 200  = 163.6 Ω  N + 1  10 + 1  10 + 1  N + 1 RB = R0  = 200  = 244.5 Ω  10 − 1  N − 1 The lattice attenuator is shown in Fig. 11.58.

244.5 Ω

244.5 Ω

200 Ω

163.6 Ω

Fig. 11.58

  11.17     T-tyPe attenuator A T-type attenuator is a symmetrical but unbalanced network and is shown in Fig. 11.59 (a). Its bisected network is shown in Fig. 11.59 (b).

11.17  T-Type Attenuator  11.43 R1 2

R1 2

R1 2

R1 2

2R2

R2

2R2

(b)

(a)

Fig. 11.59  (a) T-type attenuator (b) Bisection of T-type attenuator From the bisection theorem, the series arm of the lattice network is equal to the short-circuited input resistance of the bisected network and the diagonal arm is equal to the open-circuited input resistance of the bisected network. R RA = 1 2 R1 RB = + 2 R2 2 From lattice network,  N − 1 RA = R0   N + 1  N + 1 and RB = R0   N − 1 Solving these equations,  N − 1 R1 = 2 R0   N + 1  N + 1 N − 1 2 R2 = R0  −  N − 1 N + 1 R  4 N  2N R R2 = 0  2  = 2 0 2  N − 1 N − 1 1 If the ratio of R1 and R2 of a symmetrical T network is , find the ratio of the input 4 current to the output current. Also, calculate the attenuation in dB.

  example 11.23 

Solution

R1 1 = R2 4

For a T-type attenuator,

Dividing Eq. (ii) by Eq. (i),

 N − 1 R1 = 2 R0   N + 1

(i)

 2N  R2 = R0  2   N − 1

(ii)

2N N 2 −1 = 1  N − 1 4 2  N + 1

11.44 Circuit Theory and Transmission Lines 2N = N 2 −1 2N = ( N + 1)( N − 1)

1  N − 1   2  N + 1 1  N − 1   2  N + 1

4 N = ( N − 1) 2 N 2 − 6N + 1 = 0 N = 5.83 or N = 0.172 N in dB = 20 log10 5.83 = 15.31 dB or N in dB = 20 log10 0.172 = −15.29 dB Since attenuation cannot be negative, N = 5.83 nepers = 15.31 dB I1 = N = 5.83 I2

  example 11.24  An attenuator is composed of symmetrical T-section having a series arm of 175 W and a shunt arm of 350 W. Find the characteristic impedance and attenuation in dB. Solution

R1 = 175 Ω, R1 = 350 Ω, R2 = 350 Ω 2  N − 1 R1 = 2 R0  = 350  N + 1

…(i)

 2N  R2 = R0  2  = 350  N − 1

…(ii)

Dividing Eq. (ii) by Eq. (i), 2N N 2 −1 = 1  N − 1 2  N + 1  N − 1 = 2  N + 1 N −1 N N −1 = ( N + 1)( N − 1) N + 1 2N 2

N = ( N − 1) 2 N 2 − 3N + 1 = 0 N = 2.618 or N = 0.3819 N in dB = 20 log10 2.618 = 8.36 dB or N in dB = 20 log10 0.3819 = −8.36 dB Since attenuation cannot be negative, N = 2.618 nepers = 8.36 dB

11.18  o-Type Attenuator  11.45

Putting N = 2.618 in Eq. (i),  2.618 − 1 350 = 2 R0   2.618 + 1 R0 = 391.32 Ω

  example  11.25 

Design a T-type attenuator to give an attenuation of 60 dB and characteristic

resistance of 500 W . Solution

N ( dB ) =  60 dB, R0 = 500 Ω N ( dB) = 20 log10 N 60 = 20 log10 N N = 1000  N − 1  1000 − 1 R1 = 2 R0  = 2 × 500  = 998 Ω  N + 1  1000 + 1

499 Ω

499 Ω

1Ω

500 Ω

 2 × 1000   2N  R2 = R0  2  = 500   =1Ω 2  N − 1  (1000) − 1 Fig. 11.60

The T-type attenuator is shown in Fig. 11.60.

  11.18     o-tyPe attenuator A symmetrical p-type attenuator is shown in Fig. 11.61 (a). Its bisected network is shown in Fig. 11.61 (b). R1 2

R1

2R2

2R2

(a)

R1 2

2R2

2R2

(b)

Fig. 11.61  (a) p-attenuator (b) Bisection of p-type attenuator From the bisection theorem, the series arm of the lattice network is equal to the short-circuited input resistance of the bisected network and the diagonal arm is equal to the open-circuit input resistance of the bisected network. R1 R2 RA = R1 + 2 R2 2 RB = 2 R2 From lattice network,  N − 1 RA = R0   N + 1  N + 1 RB = R0   N − 1

11.46 Circuit Theory and Transmission Lines Solving these equations, R1 =

 N 2 − 1 2 RA RB = R0   RB − RA  2N 

R2 =

RB R0  N + 1 =   2 2  N − 1

  example 11.26  An attenuator is composed of a symmetrical p-section having series arm of 275 W and each shunt arm of 450 W. Find the characteristic impedance and attenuation in dB. R1 = 275 Ω, 2 R2 = 450 Ω, R2 = 225 Ω

Solution

 N 2 − 1 R1 = R0   = 275  2N  R2 =

R0  N + 1   = 225 2  N − 1

…(i) …(ii)

Dividing Eq. (ii) by Eq. (i), N 2 −1 275 2N = 1  N + 1 225   2  N − 1 ( N + 1)( N − 1)  N + 1 = 0.611   N − 1 2N ( N − 1) 2 = 2 N (0.611) N 2 − 3.222 N + 1 = 0 N = 2.872 or N = 0.348 N in dB = 20 log10 2.872 = 9.16 dB N in dB = 20 log10 0.348 = −9.16 dB

or Since attenuation cannot be negative,

N = 2.872 neper = 9.16 dB Putting N = 2.872 in Eq. (i),  ( 2.872) 2 − 1 275 = R0    2( 2.872)  R0 = 217.92 Ω

  example  11.27 

Design a p-type attenuator to give attenuation of 20 dB and characteristic

resistance of 500 W . N ( dB) = 20 dB, R0 = 500 Ω

Solution N ( dB) = 20 log10 N 20 = 20 log10 N N = 10

11.19  Ladder-Type Attenuator  11.47 2475 Ω

 N 2 − 1  (10) 2 − 1 500 R1 = R0  =   = 2475 Ω   2N   2 × 10  R R2 = 0 2

611 Ω

 N + 1 500  10 + 1  =   = 305.5 Ω N − 1 2  10 − 1

The p-type attenuator is shown in Fig. 11.62.

611 Ω

500 Ω

Fig. 11.62

  11.19     ladder-tyPe attenuator When a number of symmetrical T-or is p-type attenuators are connected in cascade to provide the attenuation in steps, the resultant network is termed ladder-type attenuator. Figure 11.63 shows a ladder-type attenuator with three identical and symmetrical p-sections connected in cascade. R1

R0

R1

2R2

2R2

R1

2R2

2R2

2R2

2R2

R0

Fig. 11.63  Ladder-type attenuator The output terminals are terminated with resistance R0 which is the characteristic resistance of the network. The source end also has resistance R0 so that it is a matched network. The net attenuation is the sum of the attenuation of individual sections between the input and output terminals.

  example 11.28 

Design a ladder-type attenuator for a load resistance of 500 Ω and an attenuation

of 3 dB per step. Solution

N ( dB) = 3 dB, R0 = 500 Ω N ( dB) = 20 log10 N 3 = 20 log10 N N = 1.413  N 2 − 1  (1.413) 2 − 1 500 R1 = R0  =   = 176.32 Ω   2N   2 × 1.413  R2 = 176.32 Ω

426.93 Ω

R0 2

 N + 1 500  1.413 + 1  =   = 1460.65 Ω N − 1 2  1.413 − 1 176.32 Ω

1460.65 Ω

176.32 Ω

1460.65 Ω

Fig. 11.64

2921.31 Ω

500 Ω

11.48 Circuit Theory and Transmission Lines

Exercises 11.1 A T-section low-pass filter has a series inductance of 80 mH and a shunt capacitance of 0.022 µF. Find the cut-off frequency and nominal impedance. Also design the equivalent p-section.  f c = 7.587 kHz, k = 1.907 kΩ,L = 80 mH,   C  = 0.011 µF 2   11.2 Calculate the frequency of a constant-k, T-section low-pass filter having a cut-off frequency of 1000 Hz at which it has an attenuation of 10 dB. [ f = 1170 Hz] 11.3 Design a low-pass constant-k (a) T-section, and (b) p-section filter with f c = 6 kHz and RO = 500 Ω. Calculate a and b for the filters for f = 10 kHz. Also determine the frequency at which the attenuation is 10 dB.  α = 2.2 neper, β = 1.46 radians,   β = π at 10 kHzT → L = 13.25 mH, 2    C = 0.106 µFπ → L = 26.5 mH,   C   = 0.053 µF  2   f = 7 kHz   11.4 Design a p-section constant-k high-pass filter having a cut-off frequency f c = 8 kHz and nominal characteristic impedance k = 600 Ω. Find (a) its characteristic impedance at a frequency of 12 kHz, (b) phase constant at frequency of 12 kHz, (c) attenuation at frequency of 800 Hz. L = 5.971 mH, C = 0.016 µF,   Z = 805 Ω, β = 83.6°, α = 5 nepers   Oπ 11.5 Design a band-pass with f1 = 2 kHz and k = 500 Ω.

constant-k filter f 2 = 3 kHz and

 L1 = 159 mH, C1 = 0.026 µF,   L = 20.8 mH, C = 0.637 µF 2   2

11.6 Design a band-stop constant-k filter with cut-off frequencies of 8 kHz and 12 kHz and nominal characteristic impedance of 500 Ω.  L1 = 6.6 mH, C1 = 0.0398 µF,   L = 9.9 mH, C = 0.0265 µF 2   2 11.7 Design an m-derived T-section low-pass filter with cut-off frequency f c = 1 kHz, design impedance of 600 Ω and frequency of infinite attenuation of 1050 Hz.

[ L = 191.08 mH,

C = 0.53 µF, m = 0.3]

11.8 Design an m-derived high-pass T-section filter having a design impedance of 600 Ω, cut-off frequency of 5 kHz and m = 0.35. Also determine the frequency of infinite attenuation. L  2C   m = 0.148 µF, m = 27.28 mH,     4 m C = 0.0414 µF, f = 4.684 kHz  ∞  1 − m2 11.9 In a constant-k band-pass filter, the ratio of capacitance in the shunt and series arms is 50:1 and the resonant frequency of both arms is 1000 Hz. Find the bandwidth of the filter. [282.84 Hz ] 11.10 Design a symmetrical lattice attenuator to have characteristic impedance of 800 Ω and attenuation of 20 dB. [654.55 Ω, 977.78 Ω] 11.11 Design a T-type attenuator to have a characteristic resistance of 200 Ω and attenuation of 20 dB. [163.6 Ω, 40.4 Ω] 11.12 Design a p-type attenuator to give 20 dB attenuation and characteristic impedance of 100 Ω. [495 Ω, 122.22 Ω]

Objective–Type Questions  11.49

Objective–Type Questions 11.1 A 500 Ω is resistance attenuator is to have an attenuation of 20 db. The resistances of the series and shunt arms of the lattice network, respectively are (a) 400 Ω, 500 Ω approxi (b) 410 Ω, 510 Ω approx (c) 410 Ω, 610 Ω approx (d) none of the above 11.2 2. For the design of low-pass prototype filter Fig. 11.65 of with load resistance RL = 1 Ω and angular frequency w = 1 rad/s the values of L and C would be L 2

L 2

R1

R0

R1

Fig. 11.66 (a)

Fig. 11.65 1 H and I F 1 H and 2 F 2 H and I F 2 H and 2 F

11.3 The passband of a typical filter network with Z1 and Z2 as the series and shunt-arm impedances is characerised by Z1 Z1 (a) −1 < 4 Z < 0 (b) −1 < 4 Z < 1 2 2 0
10 7 5

1 1

1.1

∞ 30 0

0.1

0.01

0 0

1.1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 1

0.99

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

CENTER 1

1.1

1.3 0.95

0.4

0.6 1.4

W A V EL E N

0.

30

0.2

40

0.49

3

31

o)

IV IT

CE

7

0.3

0.

C PA CA

R ,O

AN PT CE US ES

0.1

60

) / Yo (+ j B

19

06

0 13

4

35

0.

5 65

43

0.

6

0.3

2.

0

0.1

70

1.8

7 .0

1.0

0.8

55

0.6 60

0 12

40 1.4

2

0.35

80

0.7

0

0.4

0.9

110

1

0.4

0.15

0.36

90

50

9

.08

0.39 100

0.4

0.0

0.37

0.38

0.1

ORIGIN

Fig. 12.15 Complete Smith chart Generally, the circumference of the Smith chart is marked with the length normalised to the wavelength. Arrows indicating ‘towards generator’ and ‘towards load’ are also indicated. A clockwise rotation on the Smith chart indicates movement ‘towards generator from the load’and an anticlockwise rotation on the Smith

12.14 Impedance Smith Chart 12.25

chart indicates a movement ‘towards load from the generator’. One complete rotation around the Smith chart λ λ corresponds to a phase change of 2p and represents a distance on the line. The distance on the line 2 2 corresponds to a movement of 360° on the chart. Hence, the distance λ corresponds to a 720° movement on the chart. λ = 720°

12.14.4 Constant VSWR Circles Consider Eq. (12.17). 2

r  1  2  u −  +v = 1+ r  (1 + r ) 2 If v = 0, the centres of all these circles lie on the horizontal real axis of the Smith chart. 2

1 r    u −  = 1+ r  (1 + r ) 2 u−

1 r =± 1+ r 1+ r u=

1 r r ±1 ± = 1+ r 1+ r 1+ r

u=

r +1 r −1 or u = 1+ r 1+ r

i.e.,

u = 1 or u =

For

u=

r −1 1+ r

r −1 , 1+ r

By componendo-dividendo, 1 + u (1 + r ) + ( r − 1) 2r = = =r 1 − u (1 + r ) − ( r − 1) 2 But

u + jv = Γ u= Γ r=

1+ Γ 1− Γ

(∵ v = 0) =S

Thus, constant VSWR circles can be drawn with radius equal to the distance between centre of the Smith chart and point r = S. The constant VSWR circles have the following characteristics: 1. All the circles have the same centre, i.e. the origin of the complex Γ plane.

12.26 Circuit Theory and Transmission Lines 2. The origin in the complex Γ plane represents Γ = 0 or S = 1. As we move radially outwards, the Γ and S increases. For the outermost unity circle, Γ = 1 and S = ∞. 3. The origin corresponds to the condition Γ = 0 and S = 1, i.e. no reflection on the transmission line. For

+0

.5

+2.0

+1.0

the outermost circle, i.e. Γ = 1, the entire power is reflected on the transmission line.

2.0 VSWR circle

.0

2 −0.

50

5

2

0

∞

−1.0

−2

−0 .

.0

5

−5 .0

−

1.0

0.5

0

0.2

+0.2

+

+5

5.0 VSWR circle

Fig. 12.16 Smith chart with VSWR circles added

12.14.5 Applications of Smith Chart The Smith chart is a very useful graphical tool for solving transmission-line problems. The Smith chart finds application in finding the following quantities on a transmission line. 1. 2. 3. 4.

Normalised impedance VSWR Reflection coefficient Input impedance of the line

1. Plotting Normalised Impedance Any complex normalised impedance can be represented by a point on the Smith chart which is intersection R X of constant resistance circle, i.e. r = and constant reactance circle, i.e. x = . Z0 Z0

12.14 Impedance Smith Chart 12.27

2. Measurement of VSWR VSWR can be measured by drawing a constant VSWR circle with centre of the chart (1, 0) and radius equal to distance between the centre and point indicating the normalised impedance. The point of intersection of VSWR circle with the real axis at the right side of the centre gives VSWR for given line. 3. Measurement of Reflection Coefficient The angle of reflection coefficient Γ can be measured by extending a line from the centre of the chart to the outer circumference of the chart through the point which indicates the normalised impedance. The point at which the extended line cuts the outer circumference gives the angle of reflection coefficient Γ. The magnitude of reflection coefficient Γ can be obtained from Γ-scale provided at the bottom of the chart. An arc is cut by intersecting the Γ-scale with radius equal to the distance between the centre of the chart and a point indicating normalised impedance. The distance from centre of the Γ-scale to the point of intersection on the horizontal Γ-scale gives the magnitude of reflection coefficient. 4. Measurement of Input Impedance of the Line To find the input impedance of the line, first locate the normalised impedance and extend the line up to the outer circumference of the chart. To find the input impedance, move along the outer circumference in a clockwise direction toward the generator with a distance equal to the length of the transmission line. Mark the point on the outer circumference and draw a line from the centre of the chart to the generator point. This line cuts the circle drawn corresponding to the VSWR of the line. This point gives the normalised input impedance and the actual input impedance can be obtained by multiplying this normalised impedance by Z0.

  example 12.13  A lossless 75 W transmission line is terminated by an impedance of 150 + j150 W. Using the Smith chart, find (a) VSWR, and (b) reflection coefficient. Solution Z0 = 75 Ω, Z L = 150 + j150 Ω 1. The normalised load impedance is given by, ZL =

150 + j150 = 2 + j2 75

Mark the normalised load impedance point A at the intersection of the r = 2 circle and the x = 2 circle. 2. Draw the VSWR circle with centre O and radius equal to distance OA. The VSWR circle cuts the real axis at r = 4.2 on the right side. VSWR = 4.2 3. The magnitude of reflection coefficient is obtained from the linear scale shown at the bottom of the chart. Mark a point from centre at a distance equal to the distance OA. This gives the value of Γ as 0.62. To find the angle of reflection coefficient, draw a line OA and extend it up to A′ on the outer circumference of the chart. This line intersects the scale for the angle of reflection coefficient at 30°. Γ = 0.62 ∠ 30°

12.28 Circuit Theory and Transmission Lines 0.13

0.12

0.14

0.11

0.

06 0.

44 70

0. 0

45

1.2

1.0

1.6

0

X/

PO NE N

75

T

(+ j

0.4

4

NC EC OM

0

6

0.0

15

0.3

0.4

Z

5

14 5

0.0

4

0.

0.8

0 1.

IVE

85

5.0 10

0.25 0.26 0.24 0.27 0.23 0.25 0.24 0.26 0.23 O 0.27 REFL ECTI ON C EFFI CI EN T I N D EG REES L E OF ANG ISSI ON CO EFFI CI EN T IN TRA N SM DEGR L E OF EES

0.2

U CT

A

8

0.

A NG

IN D

20

160

8 0.2

0.6

90

15 4.0

0

1.

2 0.2

0.4 7

RE AC TA

80

9

0.48

A′ f = 30°

1

0.2

0.2

OR —>

20 0

3.

0.6

30

WA R DG EN E RA T

25 0.4

0.2

170

10 0.1

0.49

0.4

GTH S TO

18

32

0.3

20

0.2 D OW A R -170

50

20

10

5.0

4.0

3.0

2.0

1.8

1.6

1.4

1.0

1.2

0.9

0.8

0.7

0.6

0.5

0.4

20

0.4

0 1.

PTA CE US ES

IV CT DU IN R

0 2.

1.8

0

9

-110

0.0

1

0.13

0.12

0.36

0.4

0.1

0.11 -100

-90

0.4

0.39

0.37

0.38

1

0.9

0.8 0.9

15 2

0.7

0.6 0.8

2.5

6

7

10

3

4

0.5

0.4

0.7

3

5

8

0.3 0.6

2 6 8

0.2 0.5

1.8

9

5 10

0.1 0.4

0.3

1.6

1.4

4

3

12

14

0.05 0.2

1.2 1.1 1 2

1 20

0.01

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

TOW A RD L OA D —> 10 7 5

0.1

0 0

1.1

0.1

0 1

0.99

0.9

CENTER 1

1.1

Γ = 0.62 0.0

15

1 1 30 ∞ 0

4

1.2

1.3 1.4

0.4

0.6

1.2

1.3 0.95

1.4

10 7 5

1 1

1.1

� 30 0

0.1

0 0

1.1

0.1

0 1

0.99

0.9

CENTER 1

1.1

0.2 1.2

1.3 0.95

4 1.3 1.4

0.4

0.6 1.4

10 7 5

1 1

1.1

∞ 30 0

0.1

0.01

0 0

1.1

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 1

0.99

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

CENTER 1

1.1

0.95

0.4

0.6 1.4

0.8

1.5

0.9

1.6 1

1.8 1.5

2

3 2

0.8

0.7

4

3

1.6 1.7 1.8 1.9 2

5

4

0.6

0.5

3 0.4

20 ∞

10

5

2.5

10 ∞15

6 4

0.3

0.2

5

∞10 0.1

0

SM

0.9

1.3

10 7 5

1 1

1.1

∞ 30 0

0.1

0 0

1.1

0.1

0 1

0.99

0.9

CENTER 1

1.1

0.2 1.2

1.3 0.95

4 1.3 1.4

0.4

0.6 1.4

10 7 5

1 1

1.1

∞ 30 0

0.1

0.01

0 0

1.1

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 1

0.99

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

CENTER 1

1.1

0.95

0.4

0.6 1.4

0.8

1.5

0.9

1.6 1

1.8 1.5

2

3 2

0.8

0.7

4

3

1.6 1.7 1.8 1.9 2

5

4

0.6

0.5

10

5

2.5

3 0.4

6

0.3

20 ∞ 10 ∞15

4 0.2

5

∞10 0.1

0

SM

0.9

1.3

0

Check whether the polynomial is Hurwitz or not by continued fraction method. F(s) = s4 + s3 + 4s2 + 2s + 3 Solution: F(s) = s4 + s3 + 4s2 + 2s + 3 Even part of F(s) = m(s) = s4 + 4s2 + 3 Odd part of

F(s) = n(s) = s3 + 2s m( s ) Q(s) = n(s)

(5)

����������������������������������������������SQP.3

By continued fraction expansion, s3 + 2s s4 + 4s2 + 3 s s4 + 2s2 2s3 + 3 s3 + 2s 1 s 3s 2 s3 + 2 s 2s2 + 3 4s 2 2s2 3 s 1s 2 6 s 2 0 Since all the quotient terms are positive, F(s) is Hurwitz. (d)

Find out

V2 for the following network shown in Fig. 3. V1

(5)

Fig. 3 Solution: Refer Example 8.7 on page 8.6. 2.

(a)

Find the voltage across 5 � resistor in the network shown below, if k = 0.8 is coefficient of coupling. (8)

Fig. 4 Solution: For a magnetically coupled circuit, XM � k X L1 X L2 � 0.8 2(1) � 1.13 �

SQP.4�Solved Question Paper December 2015 (EXC 304) The equivalent circuit in terms of dependent sources is shown in Fig. 5.

Fig. 5 Applying KVL to Mesh 1, 10 �0� � j2 I1 � j1.13I 2 � 3 j ( I1 � I 2 ) � 0 jI1 � j1.87 I 2 � �10 �0� � jI1 � j1.87 I 2 � 10 �0�

(i)

Applying KVL to Mesh 2, 3 j ( I 2 � I1 ) � jI 2 � j1.13I1 � 5I 2 � 0 �1.87 I1 � I 2 ( j 2 � 5) � 0 1.87 I1 � (5 � j 2)I 2 � 0

(ii)

Writing Eqs (i) and (ii) in matrix form, � � j 1.87 � � I1 � �10 �0�� �1.87 5 � j 2 � � I � � � 0 � � � � 2� � � By Cramer’s rule, �j 1.87 I2 � �j 1.87

10 �0� 0 � �2.14 �13.17 A 1.87 5 � j2

V5 � = 5I2 � 5( �2.14 �13.17�) � �10.7 �13.17� V (b)

In the circuit shown in Fig. 6, find out the expression for voltage v(t) across capacitor for t > 0. At t = 0, the switch is closed. (8)

Fig. 6 –

Solution: At t = 0 , the network is shown in Fig. 7. At t = 0, the network has attained steady state condition. Hence, the capacitor acts as open circuit.

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Fig. 7 vc(0–) = 20 V Since the voltage across capacitor cannot change instantaneously, vc(0+) = 20 V For t > 0, the network is shown in Fig. 8.

Fig. 8 Writing the KCL equation for t > 0, vc � 20 2 dvc vc � � �0 5 dt 3 dvc 4 vc � �2 dt 15 dv � Pv � Q, dt 4 P � ,Q � 2 15 The solution of this differential equation is given by Q vc (t ) � � ke � Pt P Comparing the equation with

4

� t 2 � 15 � ke 15 4 vc (t ) � 7.5 � ke �0.26 t

�

At t = 0,

vc(0) = 20 V 20 = 7.5 + k k = 12.5 vc (t ) � 7.5 � 12.5 e �0.26 t

for t > 0

SQP.6�Solved Question Paper December 2015 (EXC 304) (c) Define ABCD parameters for the two port network. Hence obtain condition for symmetry. Solution: Refer Section 10.4 on page 10.18. 3.

(a)

(4)

di d 2i and 2 at t � 0 � in the circuit shown in Fig. 9. Switch is changed from position dt dt 1 to 2 at t = 0. (6) Find i,

Fig. 9 Solution: Refer Example 6.8 on page 6.10. (b)

Compare and obtain Foster I and Foster II of the following RC impedance function. 2(s � 2)(s � 4) Z ( s) � (s � 1)(s � 3)

(8)

Solution: Refer Example 9.51 on page 9.55. (c)

(6)

Obtain the Cauer from I of LC network. (s2 � 4)(s2 � 16) Z ( s) � s(s2 � 9)

Solution: Z (s) �

s 4 � 20 s2 � 64

s3 � 9s The Cauer I form is obtained from continued fraction expansion about the pole at infinity and since the degree of numerator is greater than denominator by one, it indicates presence of poles at infinity. By continued fraction expansion, s3 + 9s s4 + 20s2 + 64 s � Z s4 + 9s2 11s2 + 64 s3 + 9s

1 s �Y 11

64 s 11 2 35 11s + 64 121 s�Z s 11s2 35 11 35 s �Y 64 35s 704 11

s3 +

35s 11 0

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The network is shown in Fig. 10.

Fig. 10 4.

Derive the characteristic equation of a transmission line, also obtain ����� of the transmission line. (8) Solution: Refer Section 12.5 on page 12.3. (a)

(b)

Derive the relation for nominal impedance and cut off frequency for a constant k low pass filter. (4) Solution: Refer Section 11.6 on page 11.7. (c)

A network and its pole zero diagram are shown in Fig. 11. Determine the values of R, L, C if Z(0) = 1. (8)

Fig. 11 Solution: Refer Example 8.37 on page 8.33. 5.

(a)

Check whether the following functions are PRF or not. s(s � 3)(s � 5) (i) F (s) � (s � 1)(s � 4)

(8)

Solution: Refer Example 9.25 on page 9.18. (ii)

F ( s) �

s3 � 6 s 2 � 7s � 3 s2 � 2s � 1

Solution: Refer Example 9.28 on page 9.21. (b)

Find the current Ix using superposition theorem.

(6)

SQP.8�Solved Question Paper December 2015 (EXC 304)

Fig. 12 Solution: Refer Example 3.14 on page 3.18. (c)

The current I(s) is given by I (s) �

2s . Plot the pole zero pattern in s plane and hence, (s � 1)(s � 2)

obtain i(t) by finding out residues by graphical method. Solution: Refer Example 8.43 on page 8.43. 6.

(6)

The characteristic impedance of a high frequency line is 100 �. If it is terminated by a load impedance of 100 + j100 �. Using Smith chart find out (i) VSWR, (ii) Reflection coefficient, (iii) impedance at �/10th of wavelength away from load, (iv) VSWR minimum and VSWR maximum away from the load. (8) Solution: Z0 = 100 �, ZL = 100 + j 100 � 1. The normalised load impedance is given by 100 � j100 ZL � � 1 � j1 100 Mark the normalised load impedance point A at the intersection of the r = 1 circle and the x = 1 circle. 2. Draw the VSWR with centre O and radius equal to distance OA. VSWR = 2.52 3. Draw a line OA and extend it upto the point A� on the outer circumference of chart. This line intersects the scale for the angle of reflection coefficient at 63°. This is the angle of reflection coefficient. The magnitude of reflection coefficient is obtained from the linear scale shown at the bottom of the chart. Mark C point from centre at a distance equal to the distance OA. This gives the value of |�| as 0.445. � = 0.445�63°

(a)

4.

1 � away from the load, move in a clockwise direction 10 � i.e. 0.1� on outer circumference and locate it as point from the point A� through a distance of 10 B� at 0.162� + 0.1� = 0.262�. Draw the line OB� which intersects the constant VSWR circle at the point B. The point B is at the intersection of the r = 2.6 circle and the x = 0.47 circle. Hence, normalised input impedance at the point B is To find input impedance of the line

Zs � 2.6 � j 0.47 The actual impedance is Zs � Z0 Zs � 100(2.6 � j 0.47) � 260 � j 47

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The first voltage maximum occurs to the right of the centre on the real axis at s = 2.6 and the first voltage minimum occurs to the left of the centre on the real axis at s = 0.38. To find the location of voltage maximum from the load, move in clockwise direction from the point A� to the point P (voltage maximum point). Total distance is distance AQ + distance QP = 0.162� + 0.25� = 0.412 �. Hence, the first voltage maximum is located at 0.412� from the load. To find the location for voltage minimum for the load, move in a clockwise direction from the point A� to the point Q (voltage minimum point). Total distance is distance AQ = 0.162�. Hence, the first voltage minimum is located at 0.162� from the load. The complete Smith chart is shown in Fig. 13.

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0, where r(t) is a ramp signal (Fig. 12).

Fig. 12 Solution: For t > 0, the transformed network is shown in Fig. 13.

Fig. 13 Applying KVL to the Mesh for t > 0, 2e �2 s s

2

� 5I (s) � sI (s) � 5I (s) � sI (s) �

6 I ( s) � 0 s 6 2e �2 s I ( s) � 2 s s I ( s) �

2e �2 s s(s2 � 5s � 6)

�

2e �2 s s(s � 3)(s � 2)

By partial-fraction expansion, 1 A B C � � � s(s � 3)(s � 2) s s � 3 s � 2 A� B� C�

1 (s � 3)(s � 2) 1 s(s � 2) 1 s(s � 3)

� s�0

� s ��3

1 3

�� s ��2

1 6

1 2

�1 1 1 � I (s) � 2e �2 s � � � 6 3( 3) 2( s s s � � 2) �� � �

1 e �2 s 2 e �2 s e �2 s � � � 3 s 3 s�3 s�2

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Taking inverse Laplace transform, 1 2 i(t ) � u(t � 2) � e �3(t � 2)u(t � 2) � e �2(t � 2)u(t � 2) for t � 0 3 3 (b)

Derive an expression for characteristic equation of a transmission line. Also obtain ‘�’, ‘�’ (10) and ‘�’ of the line. Solution: Refer Section 12.5 on page 12.3. 6.

(a)

Find Thevenin’s equivalent of the network shown in Fig. 14.

(8)

Fig. 14 Solution: Refer Example 3.52 on page 3.58. (b) Explain various types of filters. Solution: Refer Section 11.2 on page 11.1.

(4)

(c)

(8)

Obtain Z11(s), Z12(s), G12(s) for the ladder network shown in Fig. 15.

Fig. 15 Solution: The transformed network is shown in Fig. 16.

Fig. 16

SQP.14�Solved Question Paper May 2016 (EXC 304) Vb � V2 Ib �

V2 s � V 5 5 2 s

� 3s 2 � �s � Va � 3sI b � Vb � 3s � V2 � � V2 � � � 1� V2 �5 � � 5 � Va Va � � Ib 1 2 2 s s �s � � 3s 2 � � � � 2� � � 1� V2 � V2 �2 �� 5 5 �

I1 �

� � s � 4 � � 3s 2 � 5 � s � � �� �� � � � � V2 � �� 2 � 5 � 5 �� s� �1 � � (3s3 � 12 s2 � 5s � 20) � � V2 5� � 10 1 � (3s3 � 12 s2 � 7s � 20) V2 10 V1 � 2sI1 � Va �1 � � 3s2 � � 2s � (3s3 � 12s2 � 7s � 20) V2 � � � � 1� V2 � � 10 � � 5 � Hence,

1 (3s 4 � 12s3 � 10 s2 � 20 s � 5) V2 5

V1 2(3s 4 � 12 s3 � 10 s2 � 20 s � 5) � I1 3s3 � 12s2 � 7s � 20 V 10 Z12 (s) � 2 � 3 I1 3s � 12 s2 � 7s � 20 Z11 (s) �

G12 (s) �

V2 5 � 4 3 V1 3s � 12s � 10s2 � 20s � 5

Index

A Active and Passive Elements 1.7 Active and Passive Networks 1.7 Analysis of Ladder Networks 8.5 Analysis of Non-Ladder Networks 8.15 Applications of Smith Chart 12.26 Attenuation Band 11.1 Attenuator 11.40

B Band-Pass Filter 11.1, 11.18 Band-Stop Filter 11.1, 11.22 Bandwidth 5.3 Branch 2.1

C Capacitance 1.3 Cascade Connection 10.63 Cauer I Form 9.33 Cauer II Form 9.34 Cauer Realisation or Ladder Realisation 9.33 Characteristic Impedance 11.2, 12.3 Characteristic of Filters 11.6 Coefficient of Coupling (k) 4.2 Composite Filter 11.37

Condition for Reciprocity 10.2 Condition for Symmetry 10.3 Conductively Coupled Equivalent Circuits 4.37 Constant Reactance Circles 12.22 Constant Resistance Circles 12.21 Constant VSWR Circles 12.25 Constant-k Half Sections 11.34 Constant-k High-Pass Filter 11.14 Constant-k Low Pass Filter 11.7 Cosine Function 7.3 Coupled Circuits 4.15 Critical Value of Mutual Inductance 4.42 Critically Damped Response 6.67 Cumulative Coupling 4.3 Current 1.1 Current Transfer Function 8.2 Current-Controlled Current Source (CCCS) 1.6 Current-Controlled Voltage Source (CCVS) 1.6 Cut-off Frequency 11.1

Dependent Sources 1.5 Differential Coupling 4.4 Dot Convention 4.9 Double-Tuned Circuit 4.43 Driving Point 8.1 Driving-Point Admittance Function 8.2 Driving-Point Functions 8.1 Driving-Point Impedance at Input Port 10.87 Driving-Point Impedance at Output Port 10.89 Driving-Point Impedance Function 8.2 Dynamic Impedance of a Parallel Circuit 5.19

D

F

Delayed or Shifted Unit-Step Function 7.2 Delayed Unit-Ramp Function 7.3

Farad 1.3 Final Value Theorem 7.7 Forced Response 6.28 Foster I Form 9.31

E Elementary Synthesis Concepts 9.24 Energy 1.1 Equivalent Circuit Representation of Transmission Line 12.2 Exponential Function (eat) 7.3 Exponentially Damped Function 7.4

I.2 Index Foster II Form 9.32 Foster Realisation 9.31

G General Solution of a Transmission Line Graphical Method for Determination of Residue 8.42

12.11

H Henry 1.2 High-Pass Filter 11.1 Hurwitz Polynomials 9.1 Hybrid Parameters (h Parameters) 10.28 Hyperbolic Cosine Function 7.4 Hyperbolic Sine Function 7.4

I Impedance Matching Using Smith Chart 12.36 Impedance Smith Chart 12.19 Independent Current Source 1.5 Independent Sources 1.5 Independent Voltage Source 1.5 Inductance 1.2 Inductances in Parallel 4.4 Inductances in Series 4.3 Infinite Line 12.10 Initial Conditions 6.1 Initial Conditions for the Capacitor 6.2 Initial Conditions for the Inductor 6.2 Initial Conditions for the Resistor 6.1 Initial Value Theorem 7.7 Input Port 8.1 Interconnection of Two-Port Networks 10.63 Inter-Relationships Between the Parameters 10.37 Inverse Hybrid Parameters (g Parameters) 10.33 Inverse Laplace Transform 7.7

Inverse Transmission Parameters (A′ B′ C′ D′ Parameters) 10.24

Norton’s Theorem 3.64

K Kirchhoff’s Current Law (KCL) 2.1 Kirchhoff’s Laws 2.1 Kirchhoff’s Voltage Law (KVL) 2.1

ohm 1.1 One-Port Network 8.1 Open-Circuit Impedance Parameters (Z Parameters) 10.2 Output Port 8.1 Overdamped Response 6.67

L

P

Ladder-Type Attenuator 11.47 Laplace Transformation 7.1 Laplace Transforms of Some Important Functions 7.2 Lattice Attenuator 11.41 Lattice Networks 10.84 Linear and Non-linear Elements 1.6 Loop 2.1 Low-Pass Filter 11.1 Lumped and Distributed Elements 1.7

p Network 11.4 p-Type Attenuator 11.45 Parallel Connection 10.67 Parallel Resonance 5.18 Parallel–Series Connection 10.77 Partial Fraction Expansion 7.8 Pass Band 11.1 Phasor Diagram 5.2 PI (p )-Network 10.79 Poles and Zeros of Network Functions 8.20 Positive Real Functions 9.16 Potential Difference 1.1 Power Factor 5.2 Port 8.1 Primary Constants of a Transmission Line 12.2 Propagation Constant 11.2, 12.4 Properties of Hurwitz Polynomials 9.1 Properties of Laplace Transform 7.4 Properties of Positive Real Functions 9.16

M Maximum Power Transfer Theorem 3.91 m-Derived Filters 11.25 m-Derived Half Sections 11.35 m-Derived High-Pass Filter 11.31 m-Derived Low-Pass Filter 11.28 Mesh 2.1 Mesh Analysis 2.2 Millman’s Theorem 3.112 Mutual Inductance 4.2 Mutually Induced emf 4.2

N Natural Response 6.28 Nepers 11.40 Network and Circuit 1.6 Node 2.1 Node Analysis 2.26

O

Q Quality Factor

5.5

R Realisation of LC Functions 9.30

Index  I.3

Realisation of RC Functions 9.47 Realisation of RL Functions 9.63 Reflection 12.15 Reflection Coefficient 12.15 Relation between Decibel and Neper 11.41 Resistance 1.1 Resistor–Capacitor Circuit 6.49, 7.19 Resistor-Inductor Circuit 6.27, 7.13 Resistor–Inductor–Capacitor Circuit 6.66, 7.25 Response 6.28 Response of RC Circuit to Various Functions 7.39 Response of RL Circuit to Various Functions 7.31

Series-Parallel Connection 10.75 Short-Circuit Admittance Parameters (Y Parameters) 10.8 Sine Function 7.3 Single-Tuned Circuit 4.41 Source Shifting 1.19 Source Transformation 1.13 Sources 1.4 Stability of the Network 8.42 Standing Wave Ratio 12.16 Standing Waves 12.16 Star-Delta Transformation 1.10 Steady-State Response 6.1, 6.28 Stop Band 11.1 Supermesh Analysis 2.17 Supernode Analysis 2.46 Superposition Theorem 3.1

S

T

Scattering Parameters or S-Parameters 12.18 Secondary Constants of a Transmission Line 12.3 Self-Induced emf 4.1 Self-Inductance 4.1 Series and Parallel Combination of Capacitors 1.10 Series and Parallel Combination of Inductors 1.9 Series and Parallel Combinations of Resistors 1.7 Series Connection 10.73 Series Resonance 5.1

Terminal Pair 8.1 Terminated Transmission Lines 12.11 Terminated Two-Port Networks 10.87 Terminating Half Sections 11.34 The Smith Chart 12.23 The Transformed Circuit 7.12 Thevenin’s Theorem 3.30 Time-Domain Behaviour from the Pole-Zero Plot 8.39 Time-Invariant and Time-Variant Networks 1.7 T-Network 10.79, 11.1

Transfer Function 8.1, 8.2 Transfer Admittance Function 8.3 Transfer Impedance Function 8.3 Transient Period 6.1 Transient Response 6.1, 6.28 Transients 6.1 Transmission Lines 12.1 Transmission Parameters (ABCD Parameters) 10.18 T-type Attenuator 11.42 Tuned circuits 4.40 Types of Transmission Lines 12.1

U Underdamped Response 6.67 Unilateral and Bilateral Elements 1.7 Unit-Impulse Function 7.3 Unit-Ramp Function 7.3 Unit-Step Function 7.2

V Voltage 1.1 Voltage Amplification 4.42 Voltage Transfer Function 8.2 Voltage-Controlled Current Source (VCCS) 1.5 Voltage-Controlled Voltage Source (VCVS) 1.5

Z Zero Input Response 6.28 Zero State Response 6.28