Circuit Theory and Networks Analysis and Synthesis [2 ed.]
 9789353161736, 9353161738

Table of contents :
Title
Contents
1 Basic Circuit Concepts
2 Analysis of DC Circuits
3 Analysis of AC Circuits
4 Magnetic Circuits
5 Graph Theory
6 Time Domain Analysis of RLC Circuits
7 Frequency Domain Analysis of RLC Circuits
8 Network Functions
9 Two-Port Networks
10 Synthesis of RLC Circuits
11 Filters
APPENDIX
Index

Citation preview

Circuit Theory and Networks Analysis and Synthesis ��������������

About the Author Ravish R Singh is presently Director at Thakur Ramnarayan College of Arts and Commerce, 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, Applied Mathematics, Electrical Networks, Network Analysis and Synthesis, Electrical Engineering, Basic 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 Networks Analysis and Synthesis ��������������

Ravish R Singh Director Thakur Ramnarayan College of Arts and Commerce Mumbai, Maharashtra

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

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai 600 116 Circuit Theory and Networks—Analysis and Synthesis, 2e (MU 2018) Copyright © 2019 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. 1 2 3 4 5 6 7 8 9

D103074

22 21 20 19 18

Printed and bound in India. ISBN (13): 978-93-5316-173-6 ISBN (10): 93-5316-173-8 Director—Science & Engineering Portfolio: Vibha Mahajan Senior Portfolio Manager—Science & Engineering: Hemant K Jha Portfolio Manager—Science & Engineering: Navneet Kumar Production Head: Satinder S Baveja Copy Editor: Taranpreet Kaur Assistant Manager—Production: Anuj K Shriwastava General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought.

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

Contents Preface Roadmap to the Syllabus

���� ����������������������� 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12

���

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. ANALYSIS OF DC CIRCUITS 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12

xiii xv

2.1

Introduction 2.1 Kirchhoff’s Laws 2.1 Mesh Analysis 2.2 Supermesh Analysis 2.15 Node Analysis 2.23 Supernode Analysis 2.36 Superposition Theorem 2.42 Thevenin’s Theorem 2.62 Norton’s Theorem 2.82 Maximum Power Transfer Theorem 2.106 Reciprocity Theorem 2.118 Millman’s Theorem 2.122 Exercises 2.127 Objective-Type Questions 2.130 Answers to Objective-Type Questions 2.132

3. ANALYSIS OF AC CIRCUITS 3.1 Introduction 3.1 3.2 Mesh analysis 3.1

3.1

viii�Contents 3.3 3.4 3.5 3.6 3.7 3.8 3.9

Node Analysis 3.9 Superposition Theorem 3.14 Thevenin’s Theorem 3.27 Norton’s Theorem 3.41 Maximum Power Transfer Theorem 3.51 Reciprocity Theorem 3.64 Millman’s Theorem 3.68 Exercises 3.72 Objective-Type Questions 3.74 Answers to Objective-Type Questions 3.75

���� ������������������ 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9

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 Exercises 4.41 Objective-Type Questions 4.43 Answers to Objective-Type Questions 4.44

���� ������������� 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10

���

Introduction 5.1 Graph of a Network 5.1 Graph Terminologies 5.2 Incidence Matrix 5.6 Loop Matrix or Circuit Matrix 5.8 Cutset Matrix 5.10 Relationship Among Submatrices of A, B and Q 5.12 Kirchhoff’s Voltage Law 5.24 Kirchhoff’s Current Law 5.24 Relation Between Branch Voltage Matrix Vb, Twig Voltage Matrix Vt and Node Voltage Matrix Vn 5.25 5.11 Relation Between Branch Current Matrix Ib and Loop Current Matrix Il 5.26 5.12 Network Equilibrium Equation 5.26 Exercises 5.53 Objective-Type Questions 5.54 Answers to Objective-Type Questions 5.55

���

���������ix

���� ������������������������������������� 6.1 6.2 6.3 6.4 6.5

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

���� ������������������������������������������ 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11

�.1

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 Frequency Domain Representaion of RLC Circuits 7.12 Resistor–Inductor Circuit 7.13 Resistor–Capacitor Circuit 7.19 Resistor–Inductor–Capacitor Circuit 7.25 Response of RL Circuit to Various Functions 7.31 Response of RC Circuit to Various Functions 7.39 Exercises 7.49 Objective-Type Questions 7.52 Answers to Objective-Type Questions 7.53

���� ������������������ 8.1 8.2 8.3 8.4 8.5 8.6 8.7

�.1

�.1

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

x�Contents

���� ������������������ 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11

Introduction 9.1 Open-Circuit Impedance Parameters (Z Parameters) 9.2 Short-Circuit Admittance Parameters (Y Parameters) 9.8 Transmission Parameters (ABCD Parameters) 9.18 Hybrid Parameters (h Parameters) 9.24 Inter-relationships between the Parameters 9.29 Interconnection of Two-Port Networks 9.47 T-Network 9.61 Pi (� )-Network 9.61 Lattice Networks 9.66 Terminated Two-Port Networks 9.69 Exercises 9.79 Objective-Type Questions 9.82 Answers to Objective-Type Questions 9.85

������������������������������� 10.1 10.2 10.3 10.4 10.5 10.6 10.7

���1

Introduction 10.1 Hurwitz Polynomials 10.1 Positive Real Functions 10.16 Elementary Synthesis Concepts 10.24 Realisation of LC Functions 10.30 Realisation of RC Functions 10.47 Realisation of RL Functions 10.63 Exercises 10.72 Objective-Type Questions 10.74 Answers to Objective-Type Questions 10.76

������������� 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10

���

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 Terminating Half Sections 11.25 Exercises 11.27 Objective-Type Questions 11.27 Answers to Objective-Type Questions 11.28

��.1

���������xi

����������������������������������������������������������� ���������

Index

Circuit Theory and Networks (May 2018) Circuit Theory and Networks (December 2017)

A.1–A.14 A.15–A.27

Electrical Network Analysis and Synthesis (May 2018) Electrical Network Analysis and Synthesis (December 2017)

A.28–A.35 A.36–A.51 I.1

������� Overview Circuit Theory and Networks (Analysis and Synthesis) is an important subject for third-semester students of of Electronics & Telecommunication Engineering and Electronics 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, and network synthesis. 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 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 Extensively supported by illustrations Examination-oriented excellent pedagogy: � Illustrations: 1500+ � Solved Examples within chapters 539 � Unsolved Problems: 195 � Objective Type Questions: 130

Chapter Organisation This text is organised into 11 chapters. Chapter 1 covers basic circuit elements and laws comprising networks. Chapter 2 elucidates DC network theorems while AC network theorems are covered in Chapter 3. Chapter 4 discusses about magnetic circuits. Further, Chapter 5 discusses the concepts of graph theory. Chapters 6 and 7 elaborate upon transient analysis in time domain and frequency domain, respectively. Chapters 8 and 9 cover network functions and two-port networks. Chapter 10 deals with network synthesis. Lastly, Chapter 11 describes filters.

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 project. I appreciate the support extended by the team at McGraw Hill Education (India), especially Hemant K Jha,

xiv�Preface Navneet Kumar, Satinder Singh Baveja, Anuj Shrivastava and Jagriti Kundu during the editorial, copyediting and production stages of this book. Suggestions for improvements will always be welcome.

Ravish R Singh ���������������� 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 Circuit Theory and Networks—ECC304 Module 1: Electrical Circuit Analysis 1.1 Analysis of DC and AC Circuits: Analysis of circuits with and without controlled sources using generalized loop and node matrix methods, circuit theorems: Superposition, Thevenin’s, Norton’s, Maximum Power Transfer and Reciprocity theorems 1.2 Magnetic Circuits: Concept of self and mutual inductances, coefficient of coupling, dot convention, equivalent circuit, coupled circuit-solution using mesh analysis

GO TO: CHAPTER 1. BASIC CIRCUIT CONCEPTS CHAPTER 2. ANALYSIS OF DC CIRCUITS CHAPTER 3. ANALYSIS OF AC CIRCUITS CHAPTER 4. MAGNETIC CIRCUITS

Module 2: Graph Theory 2.1 Objectives of graph theory, linear oriented graphs, graph terminologies, matrix representation of a graph: incidence matrix, circuit matrix, cut-set matrix, reduced incident matrix, tieset matrix, f-cutset matrix. 2.2 Relationship between sub matrices A, B and Q. 2.3 KVL and KCL using matrix

GO TO: CHAPTER 5. GRAPH THEORY

Module 3: Time and Frequency Domain Analysis 3.1 Time Domain Analysis of R-L and R-C Circuits: Forced and natural response, initial and final values solution using first order differential equation for impulse, step, ramp, exponential and sinusoidal signals 3.2 Time Domain Analysis of R-L-C Circuits: Forced and natural response, effect of damping factor, solution using second order equation for step, ramp, exponential and sinusoidal signals.

xvi�Roadmap to the Syllabus 3.3 Frequency Domain Analysis: Frequency-domain representation of R, L, C, initial value theorem and final value theorem, applications of Laplace Transform in analyzing electrical circuits

GO TO: CHAPTER 6. TIME DOMAIN ANALYSIS OF RLC CIRCUITS CHAPTER 7. FREQUENCY DOMAIN ANALYSIS OF RLC CIRCUITS

Module 4: Network Functions 4.1 Network functions for the one port and two port networks, driving point and transfer functions, poles and zeros of network functions, necessary condition for driving point functions, necessary condition for transfer functions, calculation of residues by analytical and graphical methods, time domain behavior as related to the pole-zero plot, stability and causality, testing for Hurwitz polynomial 4.2 Analysis of ladder and symmetrical lattice network

GO TO: CHAPTER 8. NETWORK FUNCTIONS Module 5: Two Port Networks 5.1 Parameters: Open circuits, short circuit, transmission and hybrid parameters, relationship among parameters, conditions for reciprocity and symmetry 5.2 Interconnections of two port network T and � representation 5.3 Terminated two port network

GO TO: CHAPTER 9. TWO-PORT NETWORKS Module 6: Synthesis of RLC circuits 6.1 Positive Real Functions: Concept of positive real function, testing for necessary and sufficient conditions for positive real functions 6.2 Synthesis of LC, RC and RL Circuits: Properties of LC, RC and RL driving point functions, LC, RC and RL network synthesis in Cauer-I and Cauer-II , Foster-I and Foster-II forms

GO TO: CHAPTER 10. SYNTHESIS OF RLC CIRCUITS

������������������������xvii

This text is useful for Electrical Network Analysis and Synthesis—ELX304 Module 1: Analysis of DC Circuits 1.1 DC Circuit Analysis: Analysis of DC circuits with dependent sources using generalized loop, node matrix analysis. 1.2 Application of Network Theorems to DC Circuits: Superposition, Thevenin, Norton, Maximum Power Transfer and Millman theorems.

GO TO: CHAPTER 1. BASIC CIRCUIT CONCEPTS CHAPTER 2. ANALYSIS OF DC CIRCUITS Module 2: Analysis of AC Circuits 2.1 Analysis of Steady State AC circuits: Analysis of AC circuits with independent sources using generalized loop, node matrix analysis. 2.2 Application of Network Theorems to AC Circuits: Superposition, Thevenin, Norton, Maximum Power Transfer and Millman theorems. 2.3 Analysis of Coupled Circuits: Self and mutual inductances, coefficient of coupling, dot convention, equivalent circuit, solution using loop analysis.

GO TO: CHAPTER 3. ANALYSIS OF AC CIRCUITS CHAPTER 4. MAGNETIC CIRCUITS Module 3: Time and Frequency Domain Analysis of Electrical Networks 3.1 Time Domain Analysis of R-L and R-C Circuits: Forced and natural responses, time constant, initial and final values. 3.2 Solution using First Order Equation for Standard Input Signals: Transient and steady state time response, solution using universal formula. 3.3 Frequency Domain Analysis of RLC Circuits: S-domain representation, concept of complex frequency, applications of Laplace Transform in solving electrical networks, driving point and transfer function, poles and zeros, calculation of residues by analytical and graphical method. GO TO: CHAPTER 6. TIME DOMAIN ANALYSIS OF RLC CIRCUITS CHAPTER 7. FREQUENCY DOMAIN ANALYSIS OF RLC CIRCUITS CHAPTER 8. NETWORK FUNCTIONS

xviii�Roadmap to the Syllabus 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 9. TWO-PORT NETWORKS

Module 5: Synthesis of RLC Circuits 5.1 Positive Real Functions: Concept of positive real function, testing for Hurwitz polynomials, testing for necessary and sufficient conditions for positive real functions. 5.2 Synthesis of RC, RL, LC Circuits: Concepts of synthesis of RC, RL, LC driving point functions.

GO TO: CHAPTER 10. SYNTHESIS OF RLC CIRCUITS

Module 6: Filters 6.1 Basic Filter Circuits: Low pass, high pass, band pass and band stop filters, transfer function, frequency response, cut-off frequency, bandwidth, quality factor, attenuation constant, phase shift, characteristic impedance. 6.2 Design and Analysis of Filters: Constant K filters

GO TO: CHAPTER 11. FILTERS

1

Basic Circuit Concepts

�1.1���������������� 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 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 �. 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��

l A

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

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

t

0

0

E � � v i dt � � R i i dt � i 2 Rt

�1.3������������� 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��

N 2A l

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

di dt

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

di i dt � 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 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�� d C�

where d is the distance between two plates, A is the cross-sectional area of the plates and � is absolute permittivity of the medium between the plates.

1.4�Circuit Theory and Networks—Analysis and Synthesis 1. Current–Voltage Relationships in a Capacitor The charge on a capacitor is given by q � Cv where q denotes the charge and v is the potential difference across the plates at any instant. We know that 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���������� 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. ���������������������������������� 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 ������������������ ���� ������������ 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 c voltage-controlled voltage source is a four-terminal a + + network component that establishes a voltage vcd between + mvab vcd v − two points c and d in the circuit that is proportional to a ab − − voltage vab between two points a and b. d 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 � � vab The voltage vcd depends upon the control voltage vab and the constant �, 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

icd

vab b

+

+ gmvab

vcd −



������������������������

c

d

1.6�Circuit Theory and Networks—Analysis and Synthesis 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

a

b

iab +

c

vcd −



d

������������������������

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 (�). 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 � � iab

+

+ r iab −

a

icd

iab

+

+

c

b iab b





d

������������������������

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

�1.6�������������������� 1. Network and Circuit The interconnection L L R R 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. ��������������������������������������������� 1.7(b) shows a 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��������������������������������������������������

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 e dependent resistor (LDR), etc. Linear elements obey Ohm’s El em l r a law whereas non-linear elements do not obey Ohm’s law. rE ne Li

a

ne

i 3. Active and Passive Elements An element which is a -L on N source of electrical signal or which is capable of increasing the level of signal energy is termed as active element. V 0 Batteries, BJTs, FETs or OP-AMPs are treated as active elements because these can be used for the amplification �������������������������������������������� or generation of signals. All other circuit elements, such as ������������������� resistors, capacitors, inductors, VDR, LDR, thermistors, etc., are termed passive elements. The behaviour of active elements cannot be described by Ohm’s law.

4. Unilateral and Bilateral Elements If the magnitude of current flowing through a circuit element is affected when the polarity of the applied voltage is changed, the element is termed unilateral element. Consider the example of a semiconductor diode. Current flows through the diode only in one direction. Hence, it is called an unilateral element. Next, consider the example of a resistor. When the voltage is applied, current starts to flow. If we change the polarity of the applied voltage, the direction of the current is changed but its magnitude is not affected. Such an element is called a bilateral element. 5. Lumped and Distributed Elements A lumped element is the element which is separated physically, like resistors, inductors and capacitors. Distributed elements are those which are not separable for analysis purposes. Examples of distributed elements are transmission lines in which the resistance, inductance and capacitance are distributed along its length. 6. Active and Passive Networks A network which contains at least one active element such as an independent voltage or current source is an active network. A network which does not contain any active element is a passive network. 7. Time-invariant and Time-variant Networks A network is said to be time-invariant or fixed if its input–output relationship does not change with time. In other words, a network is said to time-invariant, if for any time shift in input, an identical time-shift occurs for output. In time-variant networks, the input–output relationship changes with time.

�1.7������������������������������������������������� 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. V � V1 � V2 � V3 RT I � R1 I � R2 I � R3 I RT � R1 � R2 � R3

I

R1

R2

R3

V1

V2

V3

V

����������������������������������������

����Circuit Theory and Networks—Analysis and Synthesis 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 I�

Total power

PT � P1 � P2 � P3 � 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

�������������������������������� ������������

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 � T � I3 � R3 R3 R1 R2 � R2 R3 � R3 R1

1.8�������������������������������������������������

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����� ������������������������������������������� 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

�������������������������������� 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

������������������������������������������

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.

�����Circuit Theory and Networks—Analysis and Synthesis

�1.9������������������������������������������������� 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 � 1 1 1 1 � � � CT C1 C2 C3 CT �

v

������������������ ������������ ��� 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 �������������������������������� an ac voltage source v. Let i1 , i2 and i3 be the current through each ������������� capacitance C1 , C2 and C3 respectively. In parallel combination, the voltage across each capacitor is same but current through each capacitor is different. i � i1 � i2 � i3 dv dv dv dv CT � C1 � C2 � C3 dt dt 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���������������������������� 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���������������������������1.11 1

1

R1 RC

RB R3

R2 2

3

2

3

RA (a)

(b)

��������������������������������������������� 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), 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, the resistance between terminals 1 and 2 � RC � ( RA � RB ) �

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 Networks—Analysis and Synthesis

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 RΔ

RΔ RY

B

C

B

RY C



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

1.11�����������������������1.13

�1.11������������������������ 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)

������������������������������� 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.

�������������

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

6�

20 V

5� B

���������

A 60 V

Solution Since the resistor of 5 � is connected in parallel with the voltage source of 20 V it becomes redundant. Converting parallel combination 6 � of current source and resistor into equivalent voltage source and resistor (Fig. 1.19), 20 V By source transformation (Fig. 1.20), A

13.33 A

A 80 V 6�

B

B

���������

6�

B

���������

������������� 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 Networks—Analysis and Synthesis A 2�

1A

2� 4V

3�

1� 3V

6V B

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

2�

2�

2A

2A

3�

1�

3A

B

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

3A

0.75 �

1A

B

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

A

3V 3.75 V

1�

1.75 �

0.75 � 0.75 V

B

B

���������

������������� a single resistor.

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

1.11�����������������������1.15 A 5�

3A 30 �

50 �

6�

20 V B

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

50 �

30 �

3A

4A

5�

6�

B

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

30 || 50 || 5 || 6 � 2.38 �

7A

B

��������� By source transformation (Fig. 1.28), 2.38 � A

16.67 V B

���������

�������������

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

5� 50 V

2�

10 A

���������

10 V

1.16�Circuit Theory and Networks—Analysis and Synthesis Solution Converting the series combination of voltage source of 10 V and resistor of 3 � into equivalent current source and resistor (Fig. 1.30),

5� 2�

10 A

3�

3.33 A

50 V

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

5�

1.2 �

13.33 A 50 V

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

5�

1.2 �

I

50 V

16 V

��������� I�

50 � 16 � 5.48 A 5 � 1.2

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

�������������

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

5A

2�

2A

4�

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

1.11��������������������������� 2�

6V

10 V 2A

4�

2A

4�

2A

4�

��������� Adding two voltage sources (Fig. 1.35), 2�

4V

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

2A

2�

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

4A

2�

4�

��������� By current-division rule, I4 � � 4 �

�������������

2 � 1.33 A 2�4

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

6V

2�

6�

���������

1�

3A

4�

�����Circuit Theory and Networks—Analysis and Synthesis Solution Converting the series combination of the voltage source of 6 V and the resistor of 3 � into equivalent current source and resistor (Fig. 1.39), 2�

6�

3�

2A

1�

4�

3A

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

1�

2�

2A

3A

4�

��������� By source transformation (Fig. 1.41), 2�

1�

1�

2�

4�

4�

3A

4�

3A

4V

4V (a)

(b)

1�

1�

4�

1� I

1A

4�

3A

4�

4� 4A

(c)

(d)

��������� I�

16 � 1.78 A 4 �1� 4

Voltage across the 4 � resistor � 4 I � 4 � 1.78 � 7.12 V

4�

4 � 16 V

(e)

1.12���������������������

�������������

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

1

2 I

100 � 100 �

15 V

� 10 I �

��������� 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 � resistor (Fig. 1.43), 1

I

50 �

1 I

2

15 V

0.1 I

100 �

1

I

100 �

50 �

50 �

2

0.1 I

15 V

2

50 �

50 �

� �

15 V

5I

��������� 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 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.

�����Circuit Theory and Networks—Analysis and Synthesis

�������������

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

shifting technique. 3� 2

4�

1�

3

1

4

2�

18 V

� Va

6�



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

3�

3

4�

1�

1

4

18 V 2�

6�

18 V

� Va �

��������� 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

3�

4�

1� � 2�

6�

Va �

���������

����������1.21

Simplifying the network (Fig. 1.47), 18 V

18 V

3�

3�

4.5 A 1�

4.5 A 1�

1.33 � �

4� 2�

6�

Va �

6� �

18 V



Va

3�

5.985 V

Va

(b)

(a)

`

1.33 �

3�

1�

2.33 �

18 V

6�

5.985 V

6� �



Va (c)

(d)

��������� 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

3�

3.5 k�

6 k�

A 2�

1� Vx

2 k�

3 k�

20 mA

12 k� 2V

2�

5�

B

��������� [88.42 V, 7.92 k �]

��������� [1.129 V]

1.22�Circuit Theory and Networks—Analysis and Synthesis

Objective-Type Questions 1.1

1.2

1.3

A network contains linear resistors and ideal voltage sources. If values of all the resistors are doubled then the voltage across each resistor is (a) halved (b) doubled (c) increased by four times (d) not changed Four resistances 80 �, 50 �, 25 �, and R are connected in parallel. Current through 25 � resistor is 4 A. Total current of the supply is 10 A. The value of R will be (a) 66.66 � (b) 40.25 � (c) 36.36 � (d) 76.56 � 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

(a)

(c) n2R 1.6

5V

10 �

4�

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

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

(a)

1.7

���������

i

(a) 5 V source in series with a 10 � resistor (b) 1 V source in series with a 2.4 � resistor (c) 15 V source in series with a 2.4 � resistor (d) 1 V source in series with a 10 � resistor A 10 V battery with an internal resistance of 1 � 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) 1.5

5A

(d)

n2

1V

B

1.4

(d)

R n R

I

A 10 V

(b)

nR

8A

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

9

0

3

t

6

��������� (a) (c) 1.8

7.29 W 135 W

(b) 52.4 W (d) 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) 1 : 1 (b) 1 : 2 (c)

1:4

(d)

1:8

������������������������������������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

Analysis of DC Circuits

������������������� In Chapter 1, we have studied basic circuit 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. There are some other methods also to analyse circuits. We will also study superposition theorem, Thevenin’s theorem, Norton’s theorem, maximum power transfer theorem, Reciprocity theorem and Millman’s theorem. We can find currents and voltages in various parts of the circuits with these methods.

����������������������� 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 I2

O

I3 I5

I4

I1 � I 3 � I 5 � I 2 � I 4 ���������������������� 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. 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

������������������������ (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

������������������������

�������������������� 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.

4.

Identify the mesh, assign a direction to it and assign an unknown current in each mesh. Assign the polarities for voltage across the branches. V2 Apply KVL around the mesh and use Ohm’s law to express the branch voltages in terms of unknown mesh currents and the resistance. R1 V Solve the simultaneous equations for unknown mesh 1 I1 I2 currents.

Consider the network shown in Fig. 2.4 which has three meshes. Let the mesh currents for the three meshes be I1, I2, and I3 and all the three mesh currents may be assumed to flow in the clockwise direction. The choice of direction for any mesh current is arbitrary.

R2

V3

I3

R3

R4

R5

����������������������������������

2.3������������������

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)

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)

� 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)

Applying KVL to Mesh 2,

Applying KVL to Mesh 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.

��������������

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

2� 3�

5�

10 V 6�

5V

20 V

��������

4�

������������������������������������������������������ Solution Assign 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)

Applying KVL to Mesh 2, �3 ( I 2 � I1 ) � 2 I 2 � 5 I 2 � 5 � 0 �3 I1 � 10 I 2 � �5

1�

2�

…(ii)

5�

3�

Applying KVL to Mesh 3, �6 ( I 3 � I1 ) � 5 � 4 I 3 � 20 � 0 �6 I1 � 10 I 3 � 25

I2

…(iii)

Writing Eqs (i), (ii) and (iii) in matrix form, �10 �3 �6 � � I1 � �10 � � �3 10 0 � � I 2 � � � �5� � �� � � � � �6 0 10 � � I 3 � � 25 �

10 V I1

5V 6�

4� I3 20 V

��������

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 � R12 � Mutual resistance common to meshes 1 and 2 � �3 � Here, negative sign indicates that the current through common resistance are in opposite direction. R13 � Mutual resistance common to meshes 1 and 3 � �6 � Similarly, R21 � �3 � where

R22 R23 R31 R32 R33

� 3 � 2 � 5 � 10 � �0 � �6 � �0 � 6 � 4 � 10 �

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 � � I 2 � 0.78 A

2.3������������������

��������������

Determine the current through the 5 � resistor of the network shown in Fig. 2.7. 10 V 1�

8V 2�

4�

3� 5�

12 V

�������� Solution Assign clockwise currents in three meshes as shown in Fig. 2.8. 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

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

4�

1�

8V

2�

3� I3

…(ii) 12 V

Applying KVL to Mesh 3,

5�

��������

�2 ( I 3 � I1 ) � 3 ( I 3 � I 2 ) � 5 I 3 � 12 � 0 �2 I1 � 3 I 2 � 10 I 3 � 12

…(iii)

Solving Eqs (i), (ii), and (iii), I1 � 6.01 A I 2 � 3.27 A I 3 � 3.38 A I 5 � � I 3 � 3.38 A

��������������

Find the current through the 2 � resistor in the network of Fig. 2.9. 3�

12 �

6A I1

36 V

2�

6� I2

I3

9V

�������� 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 � � I3 � 3 A

�������������

Find the current through the 5 � resistor in the network of Fig. 2.10. 3A 2�

4A I1

I2

5� I3

2� 2V

3A I4

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

...(i)

I2 � 3

...(ii)

I 4 � �3

...(iii)

Applying KVL to Mesh 3, �5 ( I 3 � I1 ) � 2 ( I 3 � I 2 ) � 2 ( I 3 � I 4 ) � 2 � 0 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 � � I1 � I 3 � 4 � 2 � 2 A

…(iv)

2.3������������������

�������������������������������� ��������������

Obtain the branch currents in the network shown in Fig. 2.11. IA

5�

10IB

5�

����

IB

10 � 5V

10 V

�� �

5IA

��������� Solution Assign clockwise currents in two meshes as shown in Fig. 2.12. From Fig. 2.12, IA

5�

10IB

5�

����

IB

10 � 5V

10 V I1

�� �

5IA

I2

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

…(iv)

������������������������������������������������������

�������������

Find the mesh currents in the network shown in Fig. 2.13. 2V2 ���� 5�

4�

2�

��V2��

� V1 ��

1� 10 V

�� 2V1 �

5V

��������� Solution Assign clockwise currents in the two meshes as shown in Fig. 2.14. From Fig. 2.14, …(i) V1 � �5 I1 …(ii) V2 � 2 I 2 Applying KVL to Mesh 1,

2V2 ���� 5�

4�

2�

��V2��

� V1 ��

1� 10 V

5V

�� 2V1 �

I1

�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

I2

���������

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

�������������

Find currents Ix and Iy of the network shown in Fig. 2.15. 2Ix ���� 5�

4 � Iy

2�

1� 10 V

5V

�� 2Iy � Ix

���������

2.3������������������

Solution Assign clockwise currents in the two meshes as shown in Fig. 2.16. From Fig. 2.16, 2Ix 4� I I y � I1

...(i)

I x � I1 � I 2

...(ii)

Applying KVL to Mesh 1,

2�

y

����

5�

1�

�� 2Iy � Ix

I1

5V

�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

10 V I2

���������

�5 � 5 I1 � 2 I1 � 2 I 2 � 4 I1 � I1 � I 2 � 2 I1 � 0 �10 I1 � 3 I 2 � 5

…(iii)

Applying KVL to Mesh 2, �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

...(iv)

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

�������������

Find the currents in the three meshes of the network shown in Fig. 2.17. Ix

1�

1�

����

1�

Ix

�� I � y 1�

5V

1A

1�

Iy

��������� Solution Assign clockwise currents in the three meshes as shown in Fig. 2.18. Ix

1�

1�

����

1�

Ix

�� I � y 1�

5V I1

1�

I2

Iy

���������

1A I3

������������������������������������������������������� From Fig. 2.18, 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

…(v)

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

�������������

For the network shown in Fig. 2.19, find the power supplied by the dependent voltage

source. 50 � 20 �

5A

� V1 ��

30 �

�� 0.4 V1 �

0.01 V1

��������� Solution Assign clockwise currents in three meshes as shown in Fig. 2.20. 50 �

20 �

5A

� V1 ��

I1

I3

30 �

�� 0.4 V1 � I2

���������

0.01 V1

2.3�������������������

From Fig. 2.20, 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)

For Mesh 1, I1 � 5

…(ii)

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 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

��������������

Find the voltage Vx in the network shown in Fig. 2.21. 0.45 A

16.67 �

33.33 �

��Vx �� 25 �

30 V

�� 2 Vx �

10 V

`

��������� Solution Assign clockwise currents in the three meshes as shown in Fig. 2.22. 0.45 A

16.67 �

I3 33.33 �

��Vx ��

25 �

�� 2 Vx �

30 V I1

���������

…(iii)

10 V

I2

������������������������������������������������������� From Fig. 2.22, 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)

For the network shown in Fig. 2.23, find the mesh currents I1, I2 and I3. 15 A I3 2�

0.111 Vx

1�

� Vx �

I1

3�

1�

I2 2�

��������� Solution From Fig. 2.23, Vx � 3 ( I1 � I 2 )

…(i)

Writing current equation for the two current sources, I 3 � 15 and

0.111 Vx � I1 � I 3 0.111 [3 ( I1 � I 2 )] � I1 � I 3

…(ii)

2.3�������������������

0.333 I1 � 0.333 I 2 � I1 � I 3 � 0 �0.667 I1 � 0.333 I 2 � I 3 � 0

…(iii)

�3 ( I 2 � I1 ) � 1( I 2 � I 3 ) � 2 I 2 � 0 �3 I1 � 6 I 2 � I 3 � 0

…(iv)

Applying KVL to Mesh 2,

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

��������������

For the network shown in Fig. 2.24, find the magnitude of V0 and the current supplied by it, given that power loss in RL = 2 � resistor is 18 W. 10 �

5�

5�

��Vx �� V0

2�

4�

RL ��2 �

2�

2Vx

��������� Solution Assign clockwise currents in meshes as shown in Fig. 2.25. 10 �

5�

5�

��Vx �� V0

2� I1

4� I2

2�

2Vx I3

RL ��2 � I4

��������� From Fig. 2.25, 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)

������������������������������������������������������� Applying KVL to Mesh 2, �2 ( I 2 � I1 ) � 4 I 2 � 0 �2 I1 � 6 I 2 � 0

…(iv)

For Mesh 3, I 3 � 2Vx � 2 (5 I1 ) � 10 I1 10 I1 � I 3 � 0

…(v)

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

��������������

In the network shown in Fig. 2.26, find voltage V2 such that Vx = 0. � � 0.1 Vx

2A 10 �

24 V

3A

5� � Vx �

20 �

V2

��������� Solution Assign clockwise currents in four meshes as shown in Fig. 2.27. From Fig. 2.27, Vx � 20 ( I 3 � I 4 )

2A

…(i)

I1 10 �

Writing current equations for Meshes 1 and 2, 24 V

I1 � 2 I2 � 3

…(ii) …(iii)

I2 � 0.1 V x �

I3

3A

5� � Vx �

V2

20 �

���������

I4

2.4������������������������

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

…(iv)

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

…(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

������������������������� 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.

��������������

Find the current in the 3 � resistor of the network shown in Fig. 2.28.

2�

1�

I2

7V I1

7A

3� 1�

2�

I3

��������� Solution Meshes 1 and 3 will form a supermesh. Writing current equation for the supermesh, 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 � resistor � I2� I3 ��2.5 � 2 � 0.5 A

��������������

Find the current in the 5 � resistor of the network shown in Fig. 2.29.

10 �

50 V

I1

2�

I2

3�

2A

1� 5�

I3

��������� 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

…(iii)

2.4������������������������

Current through the 5 � resistor � I1 � I3 ��20 � 15.33 � 4.67 A

��������������

Determine the power delivered by the voltage source and the current in the 10 � resistor of the network shown in Fig. 2.30. 5� 3A

5�

I2 10 �

50 V

I1 1�

10 A I3

��������� 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 � � I 3 � I 2 � 10 � 6.76 � 3.24 A

��������������

For the network shown in Fig. 2.31, find current through the 8 � resistor. 6� I2

2� 3A

4� I1

11 �

8�

5V

I4

7A I3

���������

12 A 10 V

������������������������������������������������������� Writing current equations for meshes 1 and 4, I1 � �3

...(i)

I 4 � �12

...(ii)

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

…(iii)

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 I8 � � I 3 � I 4 � �1.28 � 12 � 10.72 A

������������������������������� ��������������

In the network of Fig. 2.32, find currents I1 and I2. 8�

3V0

V0

� � 2�

�10 V

10 �

�3 A I 2

I1

��������� Solution From Fig. 2.32, �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 Solving Eqs (ii) and (iii), I1 � �8.33 A I 2 � �11.33 A

…(iii)

2.4������������������������

��������������

In the network of Fig. 2.33, find the current through the 3 � resistor. �4 V

2�

3� 1�

� Vx �

� �

2A

5 Vx

��������� Solution Assign clockwise currents in two meshes as shown in Fig. 2.34. From Fig. 2.34, �4 V Vx � �2 I1 …(i) Meshes 1 and 2 will form a supermesh. Writing current equations for the supermesh,

2�

…(ii)

I 2 � I1 � 2

3� 1�

� Vx �

2A

I1

Applying KVL to the outer path of the supermesh,

…(iii)

Solving Eqs (ii) and (iii), I1 � 2 A I2 � 4 A I3 � � I 2 � 4 A Find the currents I1 and I2 at the network shown in Fig. 2.35. 10 � 14 �

4� I3

110 V

I2

���������

�2 I1 � 4 � 3 I 2 � 5Vx � 0 �2 I1 � 4 � 3 I 2 � 5 ( �2 I1 ) � 0 8 I1 � 3 I 2 � 4

��������������

� �

2�

� V1 �

I1

6�

0.5 V1 I2

��������� Solution From Fig. 2.35, 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

5 Vx

������������������������������������������������������� 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

��������������

For the network of Fig. 2.36, find current through the 8 � resistor. 5 Iy

� �

10 �

6�

Iy 8�

50 V Ix

52 V

0.5 Ix

��������� Solution Assign clockwise currents to the three meshes as shown in Fig. 2.37. From Fig. 2.37, I x � � I1 I y � I 2 � I3 Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I 2 � I1 � 0.5 I x � 0.5 ( � I1 ) �0.5 I1 � I 2 � 0

5 Iy � �

…(i) …(ii)

10 �

I3

Iy

6�

8�

50 V Ix

…(iii)

I1

0.5 Ix

52 V I2

���������

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)

2.4������������������������

Solving Eqs (iii), (iv) and (v), I1 � �1.56 A I 2 � �0.58 A I 3 � �1.11 A I 8 � � I 2 � �0.58 A

��������������

For the network shown in Fig. 2.38, find the current through the 10 � resistor. 10 �

20 V 5 �

15 V

4�

2A

2 I1 I2

I1

40 V I3

��������� 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

...(ii)

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

…(iii)

Solving Eqs (i), (ii) and (iii), I1 � 1.96 A I 2 � �0.04 A I 3 � 3.89 A I10 � � I1 � 1.96 A

�������������� In the network shown in Fig. 2.39, find the power delivered by the 4 V source and voltage across the 2 � resistor. 2�

� V2 � 5A 5� 4V

6�

4�

I1 1�

I2

I3

V2 2

��������� Solution From Fig. 2.39, V2 � 2 I1

…(i)

������������������������������������������������������� Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I 2 � I1 � 5

…(ii)

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

…(iii)

For Mesh 3, V2 2 I1 � � I1 2 2 I1 � I 3 � 0 I3 �

…(iv)

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

��������������

Find currents I1, I2, I3, I4 of the network shown in Fig. 2.40. 1 � 10 1 � 5

6V I1

1 � 20 � Vx �

I2

1 � 15 5 Vx

I3

1 � 2 40 A

1 � 6

I4

��������� From Fig. 2.40, 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 Mesh 2 and 3 will form a supermesh.

…(iii)

2.5�������������������

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

…(v)

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

�������������������� 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.

��������������

Determine the current through the 5 � resistor for the network shown in Fig. 2.41. 4�

36 V 2�

V1

3A

4�

5�

V2

100 �

���������

V3

20 �

������������������������������������������������������� 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 � 2 100 5 �� 2 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� V3 � �9 � V1 � V2 � � � � � 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 � resistor �

��������������

V2 � V3 7.06 � ( �8.47) � � 3.11 A 5 5

Find the power dissipated in the 6 � resistor for the network shown in Fig. 2.42. V1

3� 20 V

1� V2

6�

2� V3

5A

���������

5�

2.5�������������������

Solution Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V1 � 20 V1 � V2 V1 � V3 � � �0 3 1 2 1� 1 20 �1 �� 3 � 1 � 2 �� V1 � V2 � 2 V3 � 3 1.83 V1 � V2 � 0.5 V3 � 6.67

…(i)

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 2,

Applying KCL at Node 3,

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

V2 � V3 27.4 � 19.04 � � 1.39 A 6 6

Power dissipated in the 6 � resistor � (1.39)2 ��6 ��11.59 W

��������������

Find V1 and V2 for the network shown in Fig. 2.43. 50 �

Va

2A 10 �

80 V

� V � 1

���������

20 �

Vb



V2 50 �



Vc

20 V

������������������������������������������������������� 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 10 50 20 20 �0..1 Va � 0.17 Vb � 0.05 Vc � 0

…(ii)

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

…(iii)

Solving Eqs (i), (ii), and (iii), Va � 3.08 V Vb � 7.69 V V1 � Va � Vb � 3.08 � 7.69 � �4.61 V V2 � Vb � Vc � 7.69 � 20 � �12.31 V

��������������

Find the voltage across the 100 � resistor for the network shown in Fig. 2.44. 50 �

VA

20 �

VB

20 � 12 V

60 V

0.6 A 50 �

20 � 50 �

VC

100 �

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

2.5�������������������

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 1000 � 50 20 50 �0.02 VA � 0.05 VB � 0.1 VC � 0.24

…(iii)

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

������������������������������� ��������������

Find the voltage across the 5 � resistor in the network shown in Fig. 2.45. I1

V1

10 �

20 �

10 �

5�

2A

� �

30I1

50 V

��������� Solution From Fig. 2.45, I1 �

V1 � 50 V1 � 50 � 20 � 10 30

Assume that the currents are moving away from the node.

…(i)

������������������������������������������������������� 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 V1 2V1 � 50 V1 � 50 � 2� � 5 10 30 2�

…(ii)

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

��������������

For the network shown in Fig. 2.46, find the voltage Vx. 40 �

Vx Iy 0.6 A

100 �

50 � 25 Iy

� �

0.2 Vx

��������� Solution From Fig. 2.46, Iy �

Vx 100

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

…(i)

2.5�������������������

��������������

For the network shown in Fig. 2.47, find voltages V1 and V2. V1

0.5 V1

5�

���

V2 10 �

20 � 2A

4A

2� 20 V

40 V

��������� 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� �0.1 V1 � 0.8 V2 � 8

…(ii)

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

��������������

Determine the voltages V1 and V2 in the network of Fig. 2.48. 0.5 �

2V

V1

1�

0.25 �

���������

V2

V1

1�

������������������������������������������������������� 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 �

��������������

4 V 7

In the network of Fig. 2.49, find the node voltages V1, V2 and V3. 1�

V1

1�

V2

V3

2V

��Vx � 2A

2�

2�

2�

� �

4 Vx

��������� Solution From Fig. 2.49, Vx � V2 � V1

…(i)

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)

2.5�������������������

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

��������������

For the network shown in Fig. 2.50, find the node voltages V1 and V2. 3A

0.5 �

V1

3V2

1A

Ix

1�

V2

0.25 �

2A

2 Ix

��������� Solution From Fig. 2.50, Ix �

V1 � V2 0.5

Assume that the currents are moving away from the nodes.

…(i)

������������������������������������������������������� Applying KCL at Node 1, 3 V2 � 1 �

V1 V1 � V2 � �3 1 0.5

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

��������������

Find voltages V1 and V2 in the network shown in Fig. 2.51. V1 V1

I1

2�

1�

5A

V2

1�

2I1

��������� Solution From Fig. 2.51, V1 � V2 2 Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V V �V 5 � 1 � 1 2 � V1 1 2 1 � 1 � ��1 � � 1�� V1 � V2 � 5 2 2 2.5 V1 � 0.5 V2 � 5 I1 �

…(i)

…(ii)

2.5�Node Analysis�2.33

Applying KCL at Node 2, V2 � V1 V2 � � 2 I1 � V1 2 1 V 2 � V1 � V �V � � V2 � 2 � 1 2 � � V1 � 2 � 2 5 V1 � 5 V2 V1 � V2

…(iii)

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

���������������

Find the power supplied by the 10 V source in the network shown in Fig. 2.52. 2�

V1

4�

10 V

V2

� V3 � 10 A

2�

4V3

1�

��������� Solution From Fig. 2.52, V3 � V1 � 10 � V2

…(i)

Assume that the currents are moving away from the nodes. 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, V2 � 10 � V1 V2 � V1 V2 � � � 4 V3 4 2 1 V2 � 10 � V1 V2 � V1 V2 � � � 4 (V1 � 10 � V2 ) 4 2 1

…(ii)

������������������������������������������������������� 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

���������������

For the network shown in Fig. 2.53, find voltages V1 and V2. 0.03 Vx

20 �

V1

� Vx � 0.4 A

40 �

V2

100 �

V3

Iy 40 �

� �

80 Iy

��������� Solution From Fig. 2.53, 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�������������������

Applying KCL at Node 2, V2 � V1 V2 V2 � V3 � � �0 20 40 40 1 1 1� 1 � 1 V2 � V3 � 0 � V1 � � � � � 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

���������������

Find voltages Va , Vb and Vc in the network shown in Fig. 2.54.

Va

4A

2V

2�

2�

I1

Vb

3�

1�

Vc

2I1

5�

��������� Solution From Fig. 2.54, 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

…(i)

������������������������������������������������������� Applying KCL at Node b, 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

������������������������� 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.

��������������

Determine the current in the 5 � resistor for the network shown in Fig. 2.55. 2�

V1

10 A

3�

20 V

V2

V3

5� 1�

2� 10 V

���������

2.6������������������������

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� �� 3 � 2 �� V1 � 2 V2 � 10 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 �00.5 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 � �

���������������

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. 2.56. 10 V

V1

V2

3� 1� 2A

5A

V3 2�

���������

5�

������������������������������������������������������� 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 � � �3 � 3 5 0.33 V1 � 1.2 V2 � 1.33 V3 � 3

…(ii)

V3 � V1 V3 � V2 V3 � � �0 3 1 2 1 1� �1 � V1 � V2 � � � 1 � � V3 � 0 �3 3 2� �0.33 V1 � V2 � 1.883 V3 � 0

…(iii)

2�

Applying KCL at Node 3,

Solving Eqs (i), (ii) and (iii), V1 � 13.72 V V2 � 3.72 V V3 � 4.51 V Power delivered by the 5 A source � 5 V2 ��5 ��3.72 � 18.6 W

���������������

In the network of Fig. 2.57, find the node voltages V1, V2 and V3. 0.5 � 0.33 �

V1 4A

V2

5V 0.2 �

V3 1�

��������� 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 �� 0.33 � 0.5 �� V1 � 0.33 V2 � 0.5 V3 � 4 5.03 V1 � 3.03 V2 � 2 V3 � 4

…(i)

2.6������������������������

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 � �� � 0.33 � 0.5 �� V1 � �� 0.33 � 0.2 �� V2 � ��1 � 0.5 �� V3 � 0 �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

������������������������������� ���������������

For the network shown in Fig. 2.58, determine the voltage Vx. 7� 6V

V1

8V

11 �

V2 �

3�

4�

10 �

2A

5V

�� 2 Vx �

Vx �

V3

12 A

��������� Solution From Fig. 2.58, 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 Applying KCL at the supernode, V1 � 5 V2 � 2Vx V2 � 8 � V3 V2 � V3 � � � 4 10 7 11 V1 � 5 V2 � 2(V2 � V3 ) V2 � 8 � V3 V2 � V3 � 2� � � 7 11 4 10 2�

…(i)

������������������������������������������������������� 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)

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

���������������

Find the node voltages in the network shown in Fig. 2.59. 6�

V1

6A

5�

ix

V2 5ix

10 �

���

20 �

V3

15 �

2�

V4

40 V

��������� Solution From Fig. 2.59, 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������������������������

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. 2.60. 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, 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 V2 V3 V4

� �12 V � �4 V �0 � �2 V

���������������������������� 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.

2.7���������������������������

Explanation resistor R4.

Consider the network shown in Fig. 2.61. Suppose we have to find current I4 through R1

R3

R2

V

I

R4

����������������������������������������������������� The current flowing through resistor R4 due to constant voltage source V is found to be say I�4 (with proper direction), representing constant current source 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 source with zero resistance or short circuit. R1

R3 I4� R4

R2

V

����������������������������������������������� R1

R3 I4�� R2

R4

I

����������������������������������������������� The resultant current I4 through resistor R4 is found by superposition theorem. I 4 � I 4� � I 4� Steps to be followed in Superposition Theorem 1. 2. 3.

Find the current through the resistance when only one independent source is acting, replacing all other independent sources by respective internal resistances. Find the current through the resistance for each of the independent sources. Find the resultant current through the resistance by the superposition theorem considering magnitude and direction of each current.

�������������������������������������������������������

��������������

Find the current through the 4 � resistor in Fig. 2.64. 2�

5A

6�

5�

4� 6�

20 V

��������� Solution Step I When the 5 A source is acting alone (Fig. 2.65) 2�

5A

6�

5�

4� 6�

��������� By series-parallel reduction technique (Fig. 2.66), 2� 2�

I�

6� 5A

4�

5A

8.73 �

2.73 �

(a)

(b)

��������� I� � 5�

8.73 � 3.43 A(�) 8.73 � 4

4�

2.7���������������������������

Step II

When the 20 V source is acting alone (Fig. 2.67) 2�

6� 5�

I

I ��

4�

6�

20 V

��������� By series-parallel reduction technique (Fig. 2.68), I

5�

I ��

I

6�

20 V

10 �

5�

3.75 �

20 V

(b)

(a)

��������� I�

20 � 2.29 A 5 � 3.75

From Fig. 2.68(a), by current-division rule, I �� � 2.29 �

6 � 0.86 A(�) 6 � 10

Step III By superposition theorem, I � I � � I �� � 3.43 � 0.86 � 4.29 A (�)

��������������

Find the current through the 3 � resistor in Fig. 2.69. 2�

5�

3�

10 �

4�

5A

���������

20 V

������������������������������������������������������� Solution Step I

When the 5 A source is acting alone (Fig. 2.70) 2�

5�

3�

10 �

4�

5A

��������� I�

By series-parallel reduction technique (Fig. 2.71), I� � 5� Step II

15 � 3.75 A(�) 15 � 2 � 3

5A

2�

3�

15 �

When the 20 V source is acting alone (Fig. 2.72) 2�

��������� 3�

5�

I

I ��

10 �

4�

20 V

��������� By series-parallel reduction technique (Fig. 2.73), I ��

I

20 �

I

4�

20 V

3.33 �

(a)

20 V

(b)

��������� I�

20 �6A 3.33

From Fig. 2.73(a), by current-division rule, I �� � 6 �

4 � 1 A(�) � �1 A(�) 20 � 4

Step III By superposition theorem, I � I � � I �� � 3.75 � 1 � 2.75 A ( � )

2.7���������������������������

��������������

Find the current in the 1 � resistors in Fig. 2.74. 1A

2�

3�

1�

4V

3A

��������� 2�

Solution

3� I�

Step I When the 4 V source is acting alone (Fig. 2.75)

1�

4V

4 I� � � 1.33 A( � ) 2 �1 Step II When the 3 A source is acting alone (Fig. 2.76) By current-division rule,

��������� 2�

2 I �� � 3 � � 2 A (� ) 1� 2

3� I ��

1�

3A

Step III When the 1 A source is acting alone (Fig. 2.77) 1A

2�

��������� 3�

1�

��������� Redrawing the network (Fig. 2.78), By current-division rule, I ��� � 1 � Step IV

2 � 0.66 A( � ) 2 �1

I ���

3� 2� 1A

By superposition theorem, I � I � � I �� � I ��� � 1.33 � 2 � 0.66 � 4 A (�)

���������

1�

�������������������������������������������������������

��������������

Find the voltage VAB in Fig. 2.79. A

� 5�

6V

VAB 10 V

5A



B

��������� Solution Step I When the 6 V source is acting alone (Fig. 2.80) VAB� � 6 V



A

5�

6V

VAB�



B

���������



A

5�

Step II When the 10 V source is acting alone (Fig. 2.81) Since the resistor of 5 � is shorted, the voltage across it is zero.

VAB�� 10 V

VAB � � 10 V

B

���������

Step III When the 5 A source is acting alone (Fig. 2.82) Due to short circuit in both the parts,



VAB���

By superposition theorem,

5A �

VAB � VAB� � VAB � � VAB �� � 6 � 10 � 0 � 16 V

���������

������������������������������� ��������������

Find the current through the 6 � resistor in Fig. 2.83. I

15 V

A

5�

VAB �� � 0 Step IV



6�

8�

3I

���������

10 V

B

2.7���������������������������

Solution Step I When the 15 V source is acting alone (Fig. 2.84) From Fig. 2.84,

I�

I � � I1

…(i)

6�

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

…(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 � 1.57A I � � I1 � 0.39 A (�) Step II When the 10 V source is acting alone (Fig. 2.85) From Fig. 2.85,

I �� 6 �

...(i) 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 …(ii) Applying KVL to the outer path of the supermesh, �6 I1 � 8 I 2 � 10 � 0 6 I1 � 8 I 2 � 10 Solving Eqs (ii) and (iii), I1 � 0.26 A I 2 � 1.05 A I �� � I1 � 0.26 A (�) Step III By superposition theorem,

3I �� I1

…..(iii)

Find the current Ix in Fig. 2.86. Ix 5 �

20 V

1�

30 A

���������

10 V I2

���������

I � I � � I � � 0.39 � 0.26 � 0.65 A (�)

��������������

8�

� �

4Ix

������������������������������������������������������� Solution Step I When the 30 A source is acting alone (Fig. 2.87) From Fig. 2.87,

Ix� 5 �

I �x � I1

1�

…(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) ���������

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. 2.88) Applying KVL to the mesh,

Ix�� 5 �

20 � 5 I x� � 1I x� � 4 I �x � 0 I x� � 2 A( � )

1�

� �

20 V

4Ix��

Step III By superposition theorem, I x � I x� � I �x � � � ��� � ����

��������������

���������

Find the current I1. in Fig. 2.89. Vx

����

4Vx

I1

10 �

2A

2�

5V

��������� Vx

Solution Step I When the 5 V source is acting alone (Fig. 2.90) From Fig. 2.90, Vx = 5 – 10I1� Applying KVL to the mesh, 5 – 10I1� – 4Vx – 2I1� = 0

I1�

10 �

����

4V x

2�

5V

���������

2.7���������������������������

5 – 10I1� – 4 (5 – 10I1�) – 2I1� = 0 5 – 10I1� – 20 + 40I1� – 2I1� = 0 I1� �

15 � 0.54 A(�) 28 Vx

Step II When the 2 A source is acting alone (Fig. 2.91) From Fig. 2.91, Vx � �10 I1�

���

…(i)

10 �

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

2�

2A I1�

I 2 � I1� � 2

4 Vx

I1�

…(ii)

I2

���������

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

…(iii)

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 (�)

��������������

Determine the current through the 10 � resistor in Fig. 2.92. 10 � ����

100 V

10Vx

� 5�

Vx

10 A



��������� Solution Step I When the 100 V source is acting alone (Fig. 2.93) From Fig. 2.93, Vx = 5I� Applying KVL to the mesh,

10 � I�

����

10Vx

� 5�

100 V

Vx



100 – 10I����10Vx – 5I� = 0 100 – 10I� + 10(5I�) – 5I� = 0 ���������

������������������������������������������������������� I � � �2.86 A (�) Step II When the 10 A source is acting alone (Fig. 2.94) From Fig. 2.94, 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 �

I ��

����

10Vx

� 5�

Vx

I1

10 A I2



…(ii)

���������

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 (�)

��������������

Find the current I in the network of Fig. 2.95. 3�

17 V

I

2�

� Vx �

4� � �

5Vx

1A

���������

I�

2�

I 2 � I1 � 1

…(i)

…(ii)

� Vx �

� �

5Vx

� �

5 Vx

��������� 3�

Step II When the 1 A source is acting alone (Fig. 2.97) From Fig. 2.97, Vx � �2 I1 Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh,

3�

17 V

Solution Step I When the 17 V source is acting alone (Fig. 2.96) From Fig. 2.96, Vx = –2I� Applying KVL to the mesh, –2I� – 17 – 3I� – 5Vx = 0 –2I� – 17 – 3I� – 5(–2I�) = 0 I� =3.4 A (�)

I ��

2�

� Vx �

4� I1

1A

���������

I2

2.7���������������������������

Applying KVL to the outer path of the supermesh, �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 I �� � I 2 � 1.6 A (�) Step III By superposition theorem, I = I� + I�� = 3.4 + 1.6 = 5 A (�)

��������������

…(iii)

Find the voltage V1 in Fig. 2.98. 1�

4I

I

4�



� �

5A

V1

20 V



��������� Solution Step I When the 5 A source is acting alone (Fig. 2.99) From Fig. 2.99, I�

V1�

1�

V1� 4

4I

Applying KCL at Node 1,

I

� �

4�

5A

V1�� 4 I V1� � �5 1 4

���������

� V �� V � V1�� 4 � 1 � � 1 � 5 � 4� 4 V1� = 20 V 1�

Step II When the 20 V source is acting alone (Fig. 2.100) Applying KVL to the mesh, 4I – I – 4I – 20 = 0 I = –20 A

4I

V1��

I

4�

� �

20 V

V1�� = 4I – 1(I) = 3I = 3 (–20) = –60 V Step III

By superposition theorem, V1 = V1� + V1���= 20 – 60 = –40 V

����������

�������������������������������������������������������

��������������

Find the current in the 6 � resistor in Fig. 2.101. 1�

����

2Vx I

� Vx �

6�

3A

18 V

���������� 1�

Solution Step I When the 18 V source is acting alone (Fig. 2.102) From Fig. 2.102, Vx = –I� Applying KVL to the mesh, 18 – I� + 2Vx – 6I� = 0 18 – I� – 2I� – 6I� = 0 I� = 2 A (�)

6�

���������� 1�

����

� Vx �

I1

…(ii)

I ��

I2

����������

…(iii)

Solving Eqs (ii) and (iii), I1 � �2 A I2 � 1 A I �� � I 2 � 1 A (�) Step III By superposition theorem, I6 � = I� + I�� = 2 + 1 = 3 A (�) Find the current Iy in Fig. 2.104. Iy

120 V

2Vx

6�

3A

�1I1 � 2Vx � 6 I 2 � 0 � I1 � 2( � I1 ) � 6 I 2 � 0 3I1 � 6 I 2 � 0

��������������

I�

18 V

…(i)

Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I 2 � I1 � 3 Applying KVL to the outerpath of the supermesh,

2Vx

� Vx �

Step II When the 3 A source is acting alone (Fig. 2.103) From Fig. 2.103, Vx � �1 I1 � � I1

����

4�

����

10Iy 8 �

12 A

����������

40 V

2.7���������������������������

Solution

I y�

Step I When the 120 V source is acting alone (Fig. 2.105) Applying KVL to the mesh,

4�

����

10I y�

8�

120 V

120 – 4Iy� – 10Iy� – 8Iy� = 0 Iy� = 5.45 A (�)

����������

Step II When the 12 A source is acting alone (Fig. 2.106) From Fig. 2.106, I y � � I1 …(i)

Iy��

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

10Iy�� ����

4�

12 A I1

I 2 � I1 � 12 Applying KVL to the outer path of the supermesh,

8�

I2

...(ii) ����������

�4 I1 � 10 I y � � 8 I 2 � 0 �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. 2.107) Applying KVL to the mesh,

Iy���

10Iy��� ����

8�

–4 Iy��� – 10Iy��� – 8Iy��� – 40 = 0 4�

40 = –1.82 A (�) 22 By superposition theorem, I y �� � �

Step IV

40 V

����������

Iy = Iy� + Iy�� + Iy��� = 5.45 – 4.36 – 1.82 = –0.73 A (�)

��������������

Find the voltage Vx in Fig. 2.108. 3� � Vx

6� �

����

3Vx

24 �

18 V

36 V 5A

����������

������������������������������������������������������� Solution Step I When the 18 V source is acting alone (Fig. 2.109) From Fig. 2.109, Vx� = 3I Applying KVL to the mesh, 18 – 3I – 6I – 3Vx� = 0 18 – 3I – 6I – 3 (3I) = 0 I=1A Vx � = 3 V

3� � Vx

����

3Vx�



18 V I

���������� 3�

Step II When the 5 A source is acting alone (Fig. 2.110) From Fig. 2.110, …(i) Vx�� = –3I1 Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I 2 � I1 � 5 Applying KVL to the outer path of the supermesh,

6�

� Vx

6� � 24 �

I1

5A

� �

I2

...(ii) ����������

�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. 2.111) From Fig. 2.111, Vx��� = –3I

3�

6�

����

� Vx��� � 36 V

Applying KVL to the mesh, 36 + 3Vx��� – 6I – 3I = 0 � �V ���� � �V ���� 36 � 3Vx���� 6 � x � � 3 � x � � 0 � 3 � � 3 � 36 � 3Vx��� � 2Vx��� � Vx�� � 0 Vx��� � �6 V Step IV

3Vx���

I

����������

By superposition theorem, Vx = Vx� + Vx�� + Vx��� = 3 – 5 – 6 = –8 V

2.7���������������������������

��������������

Find the voltage V in the network of Fig. 2.112. 8� � V

15 �

5�

� �5 A

10 V

� �

12 �

8V

���������� Solution Step I When the 10 V source is acting alone (Fig. 2.113) From Fig. 2.113, V � � �8 I1 ...(i) 10 V Applying KVL to Mesh 1, �10 � 8 I1 � 15 I1 � 12 ( I1 � I 2 ) � 0 35 I1 � 12 I 2 � �10

8�

15 �

5�

� V� � 12 � I2

I1

...(ii)

� �

8V�

����������

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. 2.114) 8�

15 �

5�

� V �� �

I1

�5 A I2

12 � I3

� �

8 V ��

���������� From Fig. 2.114, V � � �8 I1

...(i)

Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I1 � I 2 � �5

...(ii)

������������������������������������������������������� 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

...(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)

Applying KVL to Mesh 3,

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

��������������

For the network shown in Fig. 2.115, find the voltage V0. 50 �

200 � � 40 �

� 1A

V0 �

V1

25 V

� �

0.5V1



���������� Solution Step I When the 1 A source is acting alone (Fig. 2.116) From Fig. 2.116, 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

50 �

200 � �

1A

� V0� �

40 � V1 I1

� �

0.5V1



...(iii) ����������

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 � � ��� V

I2

2.7���������������������������

Step II When the 25 V source is acting alone (Fig. 2.117) From Fig. 2.117,

50 �

200 � �

V1 � 200 I � 25 � 0

40 �

V1 � 200 I � 25

...(i)

V1

V0��

Applying KVL to Mesh 1,

25 V

� �

I

0.5V1



0.5V1 � 40 I � 200 I � 25 � 0 0.5( 200 I � 25) � 40 I � 200 I � 25 � 0 ���������� I � �0.09 A V0� � V1 � 200 I � 25 � 200( �0.09) � 25 � 7 V Step III By superposition theorem, V0 � V0� � V0� � ��� � � � ��� V

��������������

For the network shown in Fig. 2.118, find the voltage Vx. 10 A

2�

4� �

20 V

Vx

Vx 2

6�



���������� 2�

Solution Step I When the 20 V source is acting alone (Fig. 2.119) From Fig. 2.119, 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

4� �

20 V

Vx�

6�

Vx� I1

2 I2



���������� ...(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)

������������������������������������������������������� Step II When the 10 A source is acting alone (Fig. 2.120) From Fig. 2.120,

10 A

Vx� � ��I1 � I 2 ) ...(i) 2 � I3 4� � Applying KVL to Mesh 1, �2( I1 � I 3 ) � 6( I1 � I 2 ) � 0 6� Vx�� ...(ii) 8 I1 � 6 I 2 � 2 I 3 � 0 I1 I2 � 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 Vx � � �� I1 � I 2 ) � 6( �5.71 � 4.29) � �8.52 V

Vx�� 2

...(iii) ...(iv)

Step III By superposition theorem, Vx � Vx � � Vx � � ���� � ���� � 0

��������������

Calculate the current I in the network shown in Fig. 2.121. 4� 2I1

20 �

����� 70 V

2�

50 V I 10 �

���������� Solution Step I When the 70 V source is acting alone (Fig. 2.122) From Fig. 2.122, I � � I3 ...(i) Applying KVL to Mesh 1, �4 I1 � 2 I1 � 20 ( I1 � I 2 ) � 0 26 I1 � 20 I 2 � 0

...(ii)

4�

I1 20 �

2I1

I1

���� 2�

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 �

...(iii)

����������

I�

2.7���������������������������

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. 2.123) From Fig. 2.123,

4�

I1

I �� � I 3 …(i)

20 �

2I1

I1

�����

Applying KVL to Mesh 1, 2�

�4 I1 � 2 I1 � 20( I1 � I 2 ) � 0 26 I1 � 20 I 2 � 0 …(ii)

I2

50 V I3

10 �

Applying KVL to Mesh 2, �20( I 2 � I1 ) � 2( I 2 � I 3 ) � 0 �20 I1 � 22 I 2 � 2 I 3 � 0

I��

���������� …(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 (�)

��������������

Find the voltage V0 in the network of Fig. 2.124. 4V

5� � 10 V

1A

V0 �

����������

2�

1�

� �

V0 2

������������������������������������������������������� Solution 5�

Step I When the 10 V source is acting alone (Fig. 2.125) Applying KCL at the node,

1� �

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. 2.126) Applying KCL at the node,



����������

5�

1� � 1A

V �� V ��� 0 V0�� V0�� 0 2 �1� � 5 2 1

V 0�

� V0� � 2

2�



����������

� 1 1 1� �� � � �� V0��� �1 5 2 2 V0��� �0.83 V Step III When the 4 V source is acting alone (Fig. 2.127) Applying KCL at the node,

� V0� � 2

2�

4V

5�

V ��� V ���� 4 � 0 V0��� V0��� 0 2 �0 � � 1 5 2 � 1 1 1� �� � � �� V0��� � 4 5 2 2

� 2

V0��

1�

� V0�� 2 �



����������

V0��� � 3.33 V Step IV

By superposition theorem, V0 � V0� � V0��� V0��� � 1.67 � 0.83 � 3.33 � 4.17 V

������������������������� 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.’

2.8������������������������ RTh IL Network

RL

IL VTh

RL

(a)

(b)

�������������������������������������������������� Explanation

Consider a simple network as shown in Fig. 2.129. R1

R3 A

R2

V

RL B

������������������ 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. 2.130. 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. 2.131. RTh � R3 �



R1

R3 A

R1 R2 R1 � R2

VTh RTh � RL

If the network contains both independent and dependent sources, Thevenin’s resistance RTh is calculated as, RTh �

VTh IN

where IN 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 some

B

�����������������������������

R2

R Th

Thevenin’s equivalent network is shown in Fig. 2.132. IL �

A

B

����������������������������� RTh A VTh

RL IL B

��������������������������������� �������

������������������������������������������������������� 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. 2.133. ���������������������������������������� 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 �

��������������

VTh RTh � RL

Determine the current through the 24 � resistor in Fig. 2.134. 30 �

20 �

220 V 24 � 50 �

5�

���������� Solution Step I Calculation of VTh (Fig. 2.135) 220 � 2.75 A I1 � 30 � 50 I2 �

220 � 8.8 A 20 � 5



30 � �

I1 � �

220 V

A 50 �

����������

V Th



20 � � I2

B 5�

2.8������������������������

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 Step II

Calculation of RTh (Fig. 2.136) 30 �

20 �

A

R Th

B

50 �

5�

���������� Redrawing the circuit (Fig. 2.137),

30 �

20 �

A

RTh � (30 � 50) � ( 20 � 5) � 22.75 �

B

50 �

5�

���������� Step III

Calculation of IL (Fig. 2.138) 22.75 � A

IL �

93.5 �2A 22.75 � 24

24 �

93.5 V IL

B

����������

��������������

Find the current through the 20 � resistor in Fig. 2.139. 120 V

45 V

20 �

10 �

15 �

5�

5�

����������

20 V

������������������������������������������������������� Solution Step I Calculation of VTh (Fig. 2.140) Applying KVL to Mesh 1,

120 V

45 V

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

I1

...(i) � �

Applying KVL to Mesh 2, 20 � 5 I 2 � 10( I 2 � I1 ) � 5( I 2 � I1 ) � 0 �15I1 � 20 I 2 � 20

10 �

� A V Th



� B



15 �

� �

...(ii)

� �

5�

� �

I2



Solving Eqs (i) and (ii), I1 � �3.2 A I 2 � �1.4 A

5�

� 20 V

����������

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 Step II

Calculation of RTh (Fig. 2.141) A R Th

15 �

B 10 �

5�

5�

���������� Converting the delta formed by resistors of 10 �, 5 � and 5 � into an equivalent star network (Fig. 2.142), R1 �

10 � 5 � 2.5 � 20

R3 �

R1

5�5 � 1.25 � 20

Simplifying the network (Fig. 2.143 and Fig. 2.144),

A RTh B

A B

10 � 5 R2 � � 2.5 � 20

R2

2.5 �

15 �

R3

����������

2.5 �

1.25 �

����������

15 �

2.8������������������������

RTh � (16.25 | | 2.5) � 2.5 � 4.67 �

A RTh B

2.5 �

16.25 �

2.5 �

���������� Step III

Calculation of IL (Fig. 2.145) IL �

4.67 �

63 � 2.55 A 4.67 � 20

A 20 �

63 V IL

B

����������

��������������

Find the current through the 10 � resister in Fig. 2.146. 10 � 2�

2�

1�

15 V

1�

10 V

1�

���������� Solution Step I Calculation of VTh (Fig. 2.147) Applying KVL to Mesh 1, �15 � 2 I1 � 1( I1 � I 2 ) � 10 � 1I1 � 0 4 I1 � I 2 � �25

A � �

2� �

� �

Solving Eqs (i) and (ii),



� �

I1

...(ii) �

10 V

� 1�

I1 � �6 A I 2 � 1A

� �

1�

15 V

10 � 1( I 2 � I1 ) � 2 I 2 � 1I 2 � 0 � I1 � 4 I 2 � 10



2�

...(i)

Applying KVL to Mesh 2,

B VTh

����������

1� I2



������������������������������������������������������� 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) Step II

Calculation of RTh (Fig. 2.148) 2�

A

RTh

B

2�

1�

1�

1�

���������� Converting the star network formed by resistors of 2 �� 2 � and 1 � into an equivalent delta network (Fig. 2.149), A

2�

2�2 �8� 1 2 �1 �4� R2 � 2 � 1 � 2 2 �1 �4� R3 � 2 � 1 � 2 R1 � 2 � 2 �

RTh

B

2�

1�

1�

���������� Simplifying the network (Fig. 2.150), RTh A

RTh

A

B

B 8�

8� 4�

4�

0.8 �

0.8 �

(b)

RTh A 1�

(a)

1�

B 1.33 � (c)

����������

1�

2.8������������������������

RTh � 1.33 � Step III Calculation of IL (Fig. 2.151) 1.33 � A

IL �

10 � 0.88 A (�) 1.33 � 10

IL

10 V

10 �

B

����������

��������������

Find the current through the 1 � resistor in Fig. 2.152. 1A

2�

3�

4V

1�

3A

���������� Solution Step I Calculation of VTh (Fig. 2.153) 1A

� �

2�

� I2 � A �

4V I1

� �

V Th

3�

� � 3A

B � 2�

3�

���������� A

Writing the current equations for Meshes 1 and 2, I1 � �3 I2 � 1 Writing the VTh equation,

R Th B

����������

4 � 2( I1 � I 2 ) � VTh � 0 VTh � 4 � 2( I1 � I 2 ) � 4 � 2( �4) � 12 V Step II

Calculation of RTh (Fig. 2.154) RTh � 2 �

Step III Calculation of IL (Fig. 2.155) IL �

12 � 4A 2 �1

2� A 1�

12 V IL

B

����������

�������������������������������������������������������

������������������������������� ��������� �����

Obtain the Thevenin equivalent network for the given network of Fig. 2.156 at

terminals A and B. A

I1

4�

2I1

8V B

���������� Solution Step I Calculation of VTh (Fig. 2.157) From Fig. 2.157,



I1

4�

I1 � �2 I1 3I1 � 0 I1 � 0

A

V Th

2I1

8V



Writing the VTh equation,

B

���������� 8 � 0 � VTh � 0 VTh � 8 V

Step II Calculation of IN (Fig. 2.158), Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh,

4�

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, ����������

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 �

VTh 8 � � 1.33 � IN 6

Thevenin’s Equivalent Network (Fig. 2.159)

B

����������

2.8������������������������

��������������

Find Thevenin’s equivalent network of Fig. 2.160. 2�

3� �

4V

0.1 Vx

A

Vx �

B

���������� 2�

Solution Step I Calculation of VTh (Fig. 2.161)



Vx � VTh

3� �

4V

� 0.1 Vx

A

Vx � VTh

I1

I1 � �0.1 Vx



Writing the VTh equation,

B

���������� 4 � 2 I1 � Vx � 0

2�

3�

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

����������

Step II Calculation of IN (Fig. 2.162) From Fig. 2.162,

2�

3� 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. 2.163. IN �

4V

IN

B

4 � 0.8 A 2�3

���������� 6.25 �

Step III Calculation of RTh

A

RTh �

Step IV

VTh 5 � � 6.25 � IN 0.8

5V

Thevenin’s Equivalent Network (Fig. 2.164)

B

����������

�������������������������������������������������������

��������������

Obtain the Thevenin equivalent network of Fig. 2.165 for the terminals A and B. 1�

Vx

2�

4Vx ����

A

2A

2V B

���������� Solution Step I Calculation of VTh (Fig. 2.166) From Fig. 2.166, 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. 2.167) From Fig. 2.167, 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,

1�

Vx

4Vx � �



2� 2V



Vx

…(i)

1�

4Vx � �

IN

2A 2V

I1

I2

…(ii)

B

���������� ...(iii)

10.98 � A

30 V

Step III Calculation of RTh

A

2�

I1 � 0.73 A I 2 � 2.73 A I N � I 2 � 2.73 A VTh 30 � � 10.98 � IN 2.73

Thevenin’s Equivalent Network (Fig. 2.168)

B

����������

Solving Eqs (ii) and (iii),

Step IV

A

VTh

2A � I 1

2 � 2 I1 � 1I 2 � 4Vx � 0 2 � 2 I1 � I 2 � 4( 2 � 2 I1 ) � 0 10 I1 � I 2 � 10

RTh �



B

����������

2.8������������������������

�������������

Find the Thevenin equivalent network of Fig. 2.169 for the terminals A and B. 8I1 � �

1� A

I1

10 �

10 �

5V B

���������� Solution

8I1 � �

Step I Calculation of VTh (Fig. 2.170) Applying KVL to the mesh,

1� �

I1

10 �

5 � 10 I1 � 10 I1 � 0 5 I1 � � 0.25 A 20

10 �

A

VTh

5V �

Writing the VTh equation,

B

����������

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. 2.171) Applying KVL to Mesh 1,

A

IN

5 � 10 I1 � 10( I1 � I 2 ) � 0 20 I1 � 10 I 2 � 5

...(i)

10 � 5V

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

1�

10 � I1

I2 B

...(ii)

����������

Solving Eqs (i) and (ii), I1 � 1.375 A I 2 � 2.25 A

2�

I N � I 2 � 2.25 A Step III Calculation of RTh

4.5 V

RTh Step IV

A

V 4.5 � Th � �2� IN 2.25

Thevenin’s Equivalent Network (Fig. 2.172)

B

����������

�������������������������������������������������������

��������������

Find VTh and RTh between terminals A and B of the network shown in Fig. 2.173. 1�

2�

12 V

Ix

A

1�

2Ix

B

���������� 1�

Solution Step I

2� �

Calculation of VTh (Fig. 2.174) 12 V

Ix � 0

VTh

1�

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. 2.174.

A



B

����������

Writing the VTh equation, VTh � 12 �

1 �6V 1�1 1�

Step II Calculation of IN (Fig. 2.175) From Fig. 2.175, Ix �

V1 2

12 V

2�

V1

2Ix

1�

Ix

A

IN

Applying KCL at Node 1, B

V1 � 12 V1 V1 � � � 2I x 1 1 2 V1 �V � V1 � V1 � � 12 � 2 � 1 � � 2� 2

����������

V1 � 8 V IN �

V1 8 � �4A 2 2

Step III Calculation of RTh RTh �

��������� ����� terminals a and b .

VTh 6 � � 1.5 � IN 4

Obtain the Thevenin equivalent network of Fig. 2.176 for the given network at

2.8������������������������ 3�

2A

4�

Vx

5Vx � �

a

2�

b

���������� 3�

Solution Step I Calculation of VTh (Fig. 2.177) Applying KCL at Node x,

Vx

2A

4�

5Vx � �

VTh

2�

V 2� x 2 Vx � 4 V

a �



Writing the VTh equation,

b

����������

VTh � Vx � 5 Vx = � 4 Vx � �16 V

3�

� the terminal a is negative w.r.t. b�

Step II Calculation of IN (Fig. 2.178) Applying KCL at Node x,

4�

Vx

5Vx � �

a IN

2A

2�

Vx Vx � 5 Vx � 2 4 Vx V 2� � Vx � � x 2 2 2�

b

����������

Vx � �4 V V � 5 Vx IN � x � �Vx � 4 A 4

�4 � a

�16 V

Step III Calculation of RTh RTh � Step IV

VTh �16 � � �4 � IN 4

b

Thevenin’s Equivalent Network (Fig. 2.179)

��������������

����������

Obtain the Thevenin equivalent network of Fig. 2.180 for the given network. 150 V 10 �

30 �

5A

� Vx

15 �



����������

� �

1 V 3 x

������������������������������������������������������� Solution Step I Calculation of VTh (Fig. 2.181) From Fig. 2.181, Vx � VTh Applying KCL at the node, 1 Vx � 150 � Vx 3 � Vx � 5 � 0 10 15

150 V 10 �

30 � A � VTh

� Vx

5A

� B

IN

1 V 3 x

15 �

5A

B

���������� 37.5 � A

V 60 � x � �2A 30 30

75 V

Step III Calculation of RTh VTh 75 � � 37.5 � IN 2

B

Thevenin’s Equivalent Network (Fig. 2.183)

��������� �����

� �

150 V 10 �

Vx

30 � A

Step II Calculation of IN (Fig. 2.182) 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

Step IV

1 V 3 x

����������

VTh � 75 V

RTh �

� �



Vx � 75 V

IN

15 �

����������

Find the Thevenin’s equivalent network of the network to the left of A-B in the

Fig. 2.184. 10 V

10 Ix ����

A

Ix 1A

5�

10 �

B

����������

10 Ix

10 V

����

Solution Step I Calculation of VTh (Fig. 2.185) From Fig. 2.185, I x � I1 � I 2 For Mesh 1, I1 � 1



Ix

A



5�

1A I1

10 � I2

…(i) …(ii)

VTh





����������

B

2.8������������������������

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. 2.186) From Fig. 2.186, I x � I1 � I 2 For Mesh 1, I1 � 1 Applying KVL to Mesh 2,

10 Ix

10 V

����

…(ii)

A

Ix

…(i)

5�

1A I1

IN

10 � 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

����������

�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 I2 I3 IN

�1A �3A �2A � I3 � 2 A

�10 � A

�20 V

Step III Calculation of RTh VTh �20 � � �10 � IN 2 Thevenin’s Equivalent Network (Fig. 2.187) RTh �

Step IV

��������� ����� Fig. 2.188.

B

����������

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

������������������������������������������������������� 2�

4� A

4Vx

� �

� Vx �

5�

B

���������� 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 4Vx in Fig. 2.189. V 1 RTh � � I I From Fig. 2.189, Vx � 5( I1 � I 2 ) Applying KVL to Mesh 1, �4Vx � 2 I1 � 5( I1 � I 2 ) � 0

2�

4� A

� �

I1

� Vx �

1V

5� I2

B

���������� ...(i)

�4 �5( I1 � I 2 � � 2 I1 � 5 I1 � 5 I 2 � 0 �27 I1 � 25 I 2 � 0

…(ii)

�5( I 2 � I1 ) � 4 I 2 � 1 � 0 5 I1 � 9 I 2 � 1

…(iii)

Applying KVL to Mesh 2,

Solving Eqs (ii) and (iii), A

I1 � �0.21 A I 2 � �0.23 A

4.35 �

Hence, current supplied by voltage source of 1 V is 0.23 A. 1 RTh � � 4.35 � 0.23 Hence, Thevenin’s equivalent network is shown in Fig. 2.190.

��������������

B

����������

Find the current in the 9 � resistor in Fig. 2.191. 6Ix ����

Ix

4� 6� 20 V

����������

9�

2.8������������������������

Solution Step I Calculation of VTh (Fig. 2.192) Applying KVL to the mesh, 20 � 4 I x � 6 I x � 6 I x � 0 Ix � 5 A Writing the VTh equation,

6Ix ����

Ix

4�

� 6� �

20 V



VTh �

6 I x � VTh � 0 6(5) � VTh � 0 VTh � 30 V

A

B

���������� 6Ix

Step II Calculation of IN (Fig. 2.193). From Fig. 2.193,

����

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. 2.194. 20 IN � �5A 4

Ix

A

4� 6�

IN

20 V B

����������

6Ix A

����

A

4�

4� IN

20 V



IN 20 V

B

B

(a)

(b)

���������� Step III

A

Calculation of RTh RTh �

Step IV

6�

VTh 30 � �6� IN 5

9�

30 V IL

Calculation of IL (Fig. 2.195) IL �

B

30 �2A 6�9

��������������

����������

Determine the current in the 16 � resistor in Fig. 2.196. 10 �

6� Ix

40 V

0.8Ix

����������

16 �

������������������������������������������������������� Solution

10 �



Step I Calculation of VTh (Fig. 2.197) From Fig. 2.197, 40 V Ix � 0 The dependent source 0.8Ix depends on the controlling variable Ix. When Ix = 0, the dependent source vanishes, as shown in Fig. 2.198. 0.8 I x � 0 i.e., VTh � 40 V



6�

10 �

6�

…(iii)

40 V

0.8Ix I1

I1 � 3 A 5 I2 � A 3

IN I2 B

���������� 5 A 3

24 �

Step III Calculation of RTh

A

RTh

V 40 � Th � � 24 � 5 IN 3

16 �

40 V IL

Calculation of IL (Fig. 2.200)

B

40 IL � �1A 24 � 16

��������������

Ix A

Solving Eqs (ii) and (iii),

IN � I2 �

����������

Find the current in the 6 � resistor in Fig. 2.201. 1� ����

����������� Vx

18 V

B

����������

…(ii)

Applying KVL to the outer path of the supermesh, 40 � 10 I1 � 6 I 2 � 0 10 I1 � 6 I 2 � 40

A

VTh �

I1 � I 2 � 0.8 I x � 0.8 I 2 I1 � 1.8 I 2 � 0

B

���������� 10 �

40 V

…(i)

A

VTh

0.8Ix



Step II Calculation of IN (Fig. 2.199) From Fig. 2.199, I x � I2 Meshes 1 and 2 will form a supermesh, Writing current equation for the supermesh,

Step IV

6�

3A

����������

2Vx

6�

2.8������������������������

Solution Step I Calculation of VTh (Fig. 2.202) From Fig. 2.202, Vx � �1I1 � � I1 For Mesh 1, I1 � �3 A Vx � 3 V

1� ����

����������� Vx

…(i)

2Vx

18 V

3A

…(ii)

A �

VTh

I1 �

Writing the VTh equation,

B

����������

18 � 1 I1 � 2 Vx � VTh � 0 18 � 3 � 2(3) � VTh � 0 VTh � 27 V Step II Calculation of IN (Fig. 2.203) From Fig. 2.203, Vx � � I1 Meshes 1 and 2 will form a supermesh, Writing current equation for supermesh, I 2 � I1 � 3 Applying KVL to the outer path of the supermesh, 18 � 1I1 � 2 Vx � 0 18 � I1 � 2( � I1 ) � 0 I1 � 6 A

1�

…(i)

����

����������� Vx

2Vx

IN

18 V

3A I1

…(ii)

I2 B

���������� …(iii)

Solving Eqs (ii) and (iii), I2 � 9 A IN � I2 � 9 A Step III

Step IV

3� A

Calculation of RTh RTh �

VTh 27 � � 3� IN 9

6�

27 V IL

Calculation of IL (Fig. 2.204)

B

27 IL � �3A 3� 6

��������������

����������

Find the current in the 10 � resistor. 10 � ����

100 V

A

10Vx

� Vx �

����������

5�

10 A

������������������������������������������������������� Solution Step I Calculation of VTh (Fig. 2.206) From Fig. 2.206, Vx � 10 � 5 � 50 V

VTh ����������� A

����

10Vx

B � 5� �

100 V

Writing the VTh equation, 100 � VTh � 10Vx � Vx � 0 100 � VTh � 9Vx � 0 100 � VTh � 9(50) � 0 VTh � 550 V

Vx

10 A

5�

10 A

����������

Step II Calculation of IN (Fig. 2.207) From Fig. 2.207, Vx � 5( I N � 10) Applying KVL to Mesh 1,

A

B IN

����

10Vx

� Vx �

100 V

100 � 10Vx � Vx � 0 Vx � �

100 9

����������

100 � � 5 I N � 50 9 550 A IN � � 45 �45 �

Step III Calculation of RTh

Step IV

A

550 RTh � � �45 � 550 � 45 Calculation of IL (Fig. 2.208) IL �

10 �

550 V IL

B

550 110 �� A �45 � 10 7

����������

����������������������� 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)

������������������������������������������������

RL

2.9����������������������

Explanation

Consider a simple network as shown in Fig. 2.210. R1

R3 A

V

R2

RL

B

������������������ 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. 2.211. 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. 2.212. RN � R3 �

R1

R3 A

V

IN

R2

B

R1 R2 R1 � R2

���������������������������� R1

R3 A

Norton’s equivalent network is shown in Fig. 2.213. 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

����������������������������

VTh IN

where VTh is the open-circuit voltage across terminals A and B. If the network contains only dependent sources, then

IL IN

RN

RL

VTh � 0 IN � 0

��������������������������������������

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. 2.214.

A

RN

B

��������������������������������������

������������������������������������������������������� 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 �

��������������

I N RN RN � RL

Find the current through the 10 � resistor in Fig. 2.215. 5� 1�

10 �

15 �

4A

2V

����������

2 � 1I1 � 0 I1 � 2

1� IN 2V

I1

15 �

4A I3

I2

...(i) B

Meshes 2 and 3 will form a supermesh. Writing the current equation for the supermesh, I3 � I 2 � 4

5�

A

Solution Step I Calculation of IN (Fig. 2.216) Applying KVL to Mesh 1,

���������� ...(ii)

5�

Applying KVL to the supermesh, �5 I 2 � 15 I 3 � 0 Solving Eqs (i), (ii) and (iii), I1 I2 I3 IN Step II

A

...(iii)

15 �

B

� 2A � �3 A � 1A � I1 � I 2 � 2 � ( �3) � 5 A

Calculation of RN (Fig. 2.217)

RN

1�

���������� A IL 5A

0.95 �

10 �

RN � 1 � (5 � 15) � 0.95 � Step III Calculation of IL (Fig. 2.218) 0.95 IL � 5 � � 0.43 A 0.95 � 10

B

����������

2.9����������������������

��������������

Find the current through the 10 � resistor in Fig. 2.219. 10 �

20 �

30 �

20 �

20 �

50 V

100 V

40 V

���������� Solution Step I Calculation of IN (Fig. 2.220) Applying KVL to Mesh 1, 50 � 20( I1 � I 2 ) � 40 � 0 20 I1 � 20 I 2 � 10

A

...(i)

IN

B

20 �

30 �

20 �

20 �

50 V I1

I2

100 V I3

40 V

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)

����������

Applying KVL to Mesh 3, �20( I 3 � I 2 ) � 30 I 3 � 100 � 0 �20 I 2 � 50 I 3 � �100

...(iii)

Solving Eqs (i), (ii) and (iii), I1 � 0.81 A I N � I1 � 0.81 A Step II

Calculation of RN (Fig. 2.221) RN A

20 �

30 �

B

20 �

RN � [( 20 � 30) � 20] � 20 � 12.3 �

20 �

���������� Step III

Calculation of IL (Fig. 2.222) A IL

I L � 0.81 �

12.3 � 0.45 A 12.3 � 10

0.81 A

12.3 �

10 �

B

����������

�������������������������������������������������������

��������������

Find the current through the 8 � resistor in Fig. 2.223. 5V

5A

12 �

2A

4�

8�

���������� Solution Step I Calculation of IN (Fig. 2.224) 5V A

12 �

5A

4�

2A

IN B

���������� The resistor of the 4 � gets shorted as it is in parallel with the short circuit. Simplifying the network by source transformation (Fig. 2.225), 12 �

5V A

60 V

2A I1

I2

IN B

���������� Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh, I 2 � I1 � 2 Applying KVL to the supermesh,

A

...(i) 12 �

60 � 12 I1 � 5 � 0 12 I1 � 55

4�

...(ii) B

Solving Eqs (i) and (ii), ����������

I1 � 4.58 A I 2 � 6.58 A I N � I 2 � 6.58 A Step II

RN

Calculation of RN (Fig. 2.226)

A IL 6.58 A

3�

8�

RN � 12 � 4 � 3 � B

Step III Calculation of IL (Fig. 2.227) I L � 6.58 �

3 � 1.79 A 3�8

����������

2.9����������������������

��������������

Find the current through the 1 � resistor in Fig. 2.228. 2�

1� 1A

3�

2�

1V

2�

���������� Solution Step I Calculation of IN (Fig. 2.229) A

2�

IN 1A

3�

2�

1V

B

2�

���������� By source transformation (Fig. 2.230), A

2� I3

3� 3V

IN I1

1V

2�

I2

B

2�

���������� Applying KVL to Mesh 1, �3 � 3I1 � 2( I1 � I 3 ) � 1 � 0 5I1 � 2 I 3 � �2

...(i)

������������������������������������������������������� Applying KVL to Mesh 2,

A 2�

�1 � 2( I 2 � I 3 ) � 2 I 2 � 0 4 I 2 � 2 I 3 � �1

...(ii)

RN 3�

Applying KVL to Mesh 3,

B

�2( I 3 � I1 ) � 2( I 3 � I 2 ) � 0 �2 I1 � 2 I 2 � 4 I 3 � 0

2�

...(iii)

(a) 2�

Solving Eqs (i), (ii) and (iii), I1 I2 I3 IN Step II

2�

� �0.64 A � �0.55 A � �0.59 A � I 3 � �0.59 A

2�

A

B

3�

2� (b)

Calculation of RN (Fig. 2.231)

1.2 �

1�

A

RN � 2.2 �

B (c)

Step III Calculation of IL (Fig. 2.232)

���������� A

I L � 0.59 �

2.2 �

0.59 A

2.2 � 0.41 A 2.2 � 1

IL B

����������

������������������������������� ��������������

Find Norton’s equivalent network across terminals A and B of Fig. 2.233. 3I1 I2

1�

I1

10 �

5�

5V

� 10I2 �

����

10 � A

B

����������

2.9����������������������

Solution Step I Calculation of VTh (Fig. 2.234) From Fig. 2.234, I2 � I x I1 � � I x Applying KVL to the mesh, 5 � 10 I x � 5 I x � 10 I 2 5 � 10 I x � 5 I x � 10 I x Ix I1

3I1 I1

I2

10 �



A

5�

Ix

5V

�0 �0 � 0.2 A � �0.2 A

10 �

����

VTh

� 10I2 �



B

����������

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. 2.235) From Fig. 2.235, 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)

I1

I2

10 �

10 � A

����

5� IN

5V

Ix

� 10I2 �

Iy

5 � 10 I x � 5 I x � 5 I y � 10 I x � 0 25 I x � 5 I y � 5

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

…(iv)

Solving Eqs (iii) and (iv), I x � 0.25 A I y � 0.25 A I N � I y � 0.25 A Step III Calculation of RN RN � Step IV

A

0.25 A

9.6 �

2.4 VTh � � 9.6 � 0.25 IN

Norton’s Equivalent Network (Fig. 2.236)

B

����������

�������������������������������������������������������

��������������

For the network shown in Fig. 2.237, find Norton’s equivalent network. 20 � A

3V x

5� 15 � � Vx

2A

2�

� B

���������� Solution Step I Calculation of VTh (Fig. 2.238) From Fig. 2.238, Vx � 2 I 2

20 � �



5�

…(i)

� �

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

I1 15 �



VTh Vx

2A I2

I2 � 2

3Vx



…(ii)

For Mesh 2,

A

2�





…(iii)

B

����������

I1 � �6 I 2 � �6( 2) � �12 A

20 �

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

5�

IN �

Step II Calculation of IN (Fig. 2.239) From Fig. 2.239, Vx � 2( I 2 � I 3 )

3Vx

I1 15 �

Vx

2A I2

I3

2�



…(i)

For Mesh 2,

B

����������

I2 � 2 Meshes 1 and 3 will form a supermesh. Writing the current equation for the supermesh,

…(ii)

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 �20 I1 � 17 I 2 � 22 I 3 � 0

…(iii)

…(iv)

2.9����������������������

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

� �0.16 A �2A � 1.69 A � I 3 � 1.69 A

A

162.13 �

1.69 A

Step III Calculation of RN RN �

VTh 274 � � 162.13 � I N 1.69

B

����������

Norton’s Equivalent Network (Fig. 2.240)

Step IV

��������������

Obtain Norton’s equivalent network across A-B in the network of Fig. 2.241. 2�

5�

A I2

� V1

15 V

8�



� 10I2 �

0.6V1

15 �

B

���������� Solution Step I

Calculation of VTh (Fig. 2.242) 2�

5�



� V1

15 V Ix



8� Iy

� 10I2 �

I2

15 �

0.6V1 I2



A

VTh

� �

B

���������� From Fig. 2.242, 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 � 10 I 2 � 0 8 I x � 10 I y � 10 I 2 � 0

…(iii)

������������������������������������������������������� 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 I y � 3.45 A

2�

5�

A

I 2 � �0.83 A

� V1

15 V

Writing the VTh equation,

I2 � 10I2 �

8�



15 I 2 � VTh � 0 15( �0.83) � VTh � 0 VTh � �12.45 V

0.6V1

15 �

IN

B

����������

Step II Calculation of IN (Fig. 2.243) From Fig. 2.243,

2�

5�

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. 2.244. From Fig. 2.244,

V1

15 V Ix



8�

0.6V1

IN

Iy B

����������

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 Calculation of RN V �12.45 RN � Th � � �5.76 � 2.16 IN Step IV

Norton’s Equivalent Network (Fig. 2.245)

2.16 A

�5.76 �

B

����������

2.9����������������������

��������������

Find Norton’s equivalent network of Fig. 2.246. A I1

0.5I1 � �

1� 2� 2V B

���������� 0.5I1 � �

Solution Step I Calculation of VTh (Fig. 2.247) Applying KVL to the mesh,

I1

2 � 2 I1 � 0.5 I1 � 1I1 � 0 2 � 2.5 I1 � 0 I1 � 0.8 A

VTh



2V



B

����������

Writing the VTh equation,

A

1I1 � VTh � 0 1(0.8) � VTh � 0 VTh � 0.8 V

0.5I1

� �

I1

2V B

The dependent source of 0.5I1 depends on the controlling variable I1. When I1 � 0, the dependent source vanishes, i.e. 0.5 I1 � 0 as shown in Fig. 2.249. 2 IN � � 1 A 2 Step III Calculation of RN

���������� A

2�

IN

2V

V 0.8 RN � Th � � 0.8 � IN 1

B

����������

Norton’s Equivalent Network (Fig. 2.250) A

1A

IN

1� 2�

Step II Calculation of IN (Fig. 2.248) When a short circuit is placed across the 1 � resistor, it gets shorted. I1 � 0

Step IV

A



1� 2�



0.8 �

B

����������

�������������������������������������������������������

��������������

Find Norton’s equivalent network at the terminals A and B of Fig. 2.251. 6Ix 3�

2� A

Ix 9V

6� B

���������� Solution Step I Calculation of VTh (Fig. 2.252) From Fig. 2.252, I x � I1 Applying KVL to Mesh 1, 9 � 3( I1 � I 2 ) � 6 I1 � 0 9 I1 � 3I 2 � 9

6Ix

3� �

…(i)

2�

I2 �

Ix �

9V





A

VTh

6� I1



…(ii)

B

����������

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. 2.253) From Fig. 2.253, I x � I1 � I 3

6Ix

…(i) 3�

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

2�

I2 Ix

9V

…(ii)

IN

6� I1

A

I3

For Mesh 2,

B

I 2 � 6 I x � 6( I1 � I 3 ) 6 I1 � I 2 � 6 I 3 � 0 …(iii) 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)

2.9����������������������

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

A

I1 I2 I3 IN

�5A �3A � 4.5 A � I 3 � 4.5 A

�4 �

4.5 A

B

Step III Calculation of RN

����������

VTh �18 � � �4 � 4.5 IN Norton’s Equivalent Network (Fig. 2.254) RN �

Step IV

��������������

Find Norton’s equivalent network to the left of terminal A-B in Fig. 2.255. A

6�

0.5I

4�

I B

���������� 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. 2.256. From Fig. 2.256, I�

A

6�

0.5I

4�

1A

I B

����������

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 � 1 1 Hence, Norton’s equivalent network is shown in Fig. 2.257.

A

2�

B

����������

�������������������������������������������������������

��������������

Find the current through the 2 � resistor in the network shown in Fig. 2.258. 2�

�10 V

�2Ix

Ix

� �

4�

2A

10 �

���������� Solution Step I Calculation of VTh (Fig. 2.259) From Fig. 2.259, Ix � 0

�2Ix

A B Ix V � Th �

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. 2.260.

� �

�10 V

2A

4�

10 �

����������

I1 � 2 A B Ix V � Th �

Writing the VTh equation, �10 � VTh � 4 I1 � 0 �10 V

�10 � VTh � 4( 2) � 0

� �

4�

2A

10 �

I1

VTh � �18 V Step II Calculation of IN (Fig. 2.261) From Fig. 2.261,

����������

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

�2Ix

A IN B Ix

…(i)

I x � I1

the �10 V

� �

4�

2A I1

I2

10 � I3

…(ii)

Applying KVL to the outer path of the supermesh,

����������

�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)

2.9����������������������

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

� 4.5 A � 6.5 A �9A � I1 � 4.5 A

A IL

Step III Calculation of RN RN � Step IV

VTh �18 � � �4 � 4.5 IN

4.5 A

2�

�4 �

Calculation of IL (Fig. 2.262)

B

I L � 4.5 �

��������������

�4 �9A �4 � 2

����������

Find the current through the 2� resistor in the network of Fig. 2.263. �

Vi



1� 2� 5V

� �

4Vi

���������� Solution Step I Calculation of VTh (Fig. 2.264) From Fig. 2.264, 5 � Vi � 4Vi � 0 Vi � �1 V



Vi

� �

1� VTh

5V

� �

Writing the VTh equation,

4Vi �

�4Vi � VTh � 0 VTh � �4Vi � �4( �1) � 4 V

Applying KVL to the mesh, �4Vi � 1I N � 0 I N � �4Vi � �4( �5) � 20 A

B

���������� �

Step II Calculation of IN (Fig. 2.265) From Fig. 2.265, 5 � Vi � 0 Vi � �5 V

A

Vi



A

1� 5V

IN � �

4Vi B

����������

������������������������������������������������������� Step III Calculation of RN

A

RN � Step IV

4 VTh � � 0.2 � 20 IN

2�

0.2 �

20 A

Calculation of IL (Fig. 2.266)

B

0.2 I L � 20 � � 1.82 A 0.2 � 2

��������������

����������

Find the current in the 2 � resistor in the network of Fig. 2.267. 1�

Ix

10 V

2Ix

3�

1A

2�

���������� Solution Step I Calculation of VTh (Fig. 2.268) 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),

Ix

1�

2Ix �

� 10 V

3�

1A I1

I2

A

VTh





B

����������

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. 2.269) From Fig. 2.269, 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

1�

Ix

2Ix

1A

10 V I1

I2

����������

3�

IN I3

2.9����������������������

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 Step III Calculation of RN RN �

A IL

VTh 8.25 � � �1.18 � IN �7

�1.18 �

�7 A

Calculation of IL (Fig. 2.270)

Step IV

B

�1.18 I L � �7 � � 10.07 A �1.18 � 2

��������������

2�

����������

Find the current through the 10 � resistor for the network of Fig. 2.271. 3Ix ����

2�

Ix

5�

10 �

10 V

���������� 3Ix

Solution

����

Step I Calculation of VTh (Fig. 2.272) Applying KVL to the mesh, 10 � 2 I x � 3I x � 5 I x � 0



5�

VTh



2�



10 V



I x � 2.5 A Writing the VTh equation,

3Ix ����

5( 2.5) � VTh � 0

Step II Calculation of IN (Fig. 2.273) From Fig. 2.273, Ix � 0

B

����������

5 I x � VTh � 0

VTh � 12.5 V

A

Ix

2�

Ix

5�

10 V

����������

IN

�������������������������������������������������������� The dependent source of 3 Ix depends on the controlling variable Ix. When Ix � 0, the dependent source 3 Ix vanishes, i.e. 3 Ix � 0 as shown in Fig. 2.274. IN

A

2�

10 � �5A 2

IN

10 V B

Step III Calculation of RN

���������� 12.5 V RN � Th � � 2.5 � 5 IN

Step IV

A IL

Calculation of IL (Fig. 2.275) IL � 5 �

10 �

2.5 �

5A

2.5 �1A 2.5 � 10

B

����������

��������������

Find the current through the 5 � resistor in the network of Fig. 2.276. 2�

12 V

Ix

4� � �

5�

4Ix

���������� Solution Step I Calculation of VTh (Fig. 2.277) Applying KVL to the mesh, 12 � 2 I x � 4 I x � 4 I x � 0 12 V 12 � 10 I x � 0 I x � 1.2 A Writing the VTh equation, 12 � 2 I x � VTh � 0 12 � 2(1.2) � VTh � 0 VTh � 9.6 V Step II Calculation of IN (Fig. 2.278) From Fig. 2.278, 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

Ix

2� �



4� � A V Th � B

� �

4Ix

� �

4Ix

����������

2�

Ix

4�

IN I1

I2

����������

2.9�����������������������

Step III Calculation of RN

Step IV

IL

VTh 9.6 � � 0.8 � 12 IN

RN �

12 A

5�

0.8 �

Calculation of IL (Fig. 2.279) I L � 12 �

��������������

0.8 � 1.66 A 0.8 � 5

����������

Find the current through the 10 � resistor for the network of Fig. 2.280. 5� �

4�

0.5Vx

Vx

10 � �

5V

���������� 5�

Solution Step I Calculation of VTh (Fig. 2.281) For the mesh, I � �0.5Vx � �0.5VTh



4� 5V

0.5Vx

VTh ��Vx

I �

Writing the VTh equation, 5�

5 � 4( �0.5VTh ) � VTh � 0

Step II Calculation of IN (Fig. 2.282) From Fig. 2.282, Vx � 0



4�

5 5 � A 4�5 9

0.5Vx

5V

Vx �

IN

B

5� A

4�

IN

5V B

Step III Calculation of RN RN �

Step IV

A

����������

The dependent source of 0.5 Vx� depends on the controlling variable Vx. When Vx � 0, the dependent source vanishes, i.e. 0.5 Vx � 0 as shown in Fig. 2.283. IN �

B

����������

5 � 4 I � 0 � VTh � 0

VTh � �5 V

A

VTh �5 � � �9 � 5 IN 9

Calculation of IL (Fig. 2.284) IL �

5 �9 � � �5 A 9 �9 � 10

���������� A IL 5A 9

�9 �

10 �

B

����������

��������������������������������������������������������

��������������

Find the current through the 10 � resistor in the network shown in Fig. 2.285. 1000 �

I1 � � 2Vx �

12 V

Vx

5I1

25 �

10 �



���������� Solution Step I Calculation of VTh (Fig. 2.286) From Fig. 2.286, Vx � �25(5 I1 ) � �125 I1 …(i)

1000 �

I1 �

A

� � �

12 V

2Vx

5I1

Vx

VTh

25 �



Applying KVL to Mesh 1,



12 � 1000 I1 � 2Vx � 0 12 � 1000 I1 � 2( �125 I1 ) � 0

B

����������

…(ii)

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

1000 �

I1

A

VTh � Vx � �2 V Step II Calculation of IN (Fig. 2.287) From Fig. 2.287,

� � �

12 V

2Vx

5I1

Vx

25 �

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. 2.288. 12 � 0.012 A 1000 � �5 I1 � �5(0.012) � �0.06 A

I1 � IN

���������� 1000 �

5I1

IN

B

���������� A IL

VTh �2 � � 33.33 � IN �0.06 �0.06 A

Step IV

A

12 V

Step III Calculation of RN RN �

I1

33.33 �

10 �

Calculation of IL (Fig. 2.289) I L � �0.06 �

33.33 � �0.046 A 33.33 � 10

B

����������

2.9�����������������������

��������������

Find the current through the 5 � resistor for the network of Fig. 2.290. 4V

1�

2� �

1�

Vx

� � �

2A

5�

4Vx

���������� Solution Step I Calculation of VTh (Fig. 2.291) From Fig. 2.291, Vx � 2 I For the mesh, I �2 Vx � 2( 2) � 4 V Writing the VTh equation, 4Vx � 2 I � 1I � 4 � VTh � 0 4( 4) � 2( 2) � 2 � 4 � VTh � 0 VTh � 26 V Step II Calculation of IN (Fig. 2.292) From Fig. 2.292, Vx � 2( I1 � I 2 ) For Mesh 1, I1 � 2 Applying KVL to Mesh 2,

4V

1�

…(i)

Vx

1�

� � �

…(ii) I

2A

4Vx

4V

…(i)

1�

I2

2� �

…(ii) 1�

Vx



…(iii)

A � �

I1

2A

IN

4Vx B

����������

Solving Eqs (ii) and (iii), I1 � 2 A I 2 � 2.36 A

A IL

I N � I 2 � 2.36 A Step III Calculation of RN

Step IV

V 26 RN � Th � � 11.02 � IN 2.36

2.36 A

11.02 �

11.02 � 1.62 A 11.02 � 5

5�

B

Calculation of IL (Fig. 2.293) I L � 2.36 �

� A VTh � B

����������

4Vx � 2( I 2 � I1 ) � 1( I 2 � I1 ) � 4 � 0 4[2( I1 � I 2 )] � 2 I 2 � 2 I1 � I 2 � I1 � 4 � 0 11I1 � 11I 2 � �4

2� �

����������

��������������������������������������������������������

��������������

Find the current through the 1 � resistor in the network of Fig. 2.294. 6�

Ix

3Ix

12 V

3�

1�

���������� Solution Step I Calculation of VTh (Fig. 2.295) From Fig. 2.295,

Ix

6�

I x � I1

4 I1 � I 2 � 0

3Ix

12 V

Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh, I 2 � I1 � 3I x � 3I1





…(i) I1

I2

3� �

VTh �

���������� …(ii)

Applying KVL to the outer path of the supermesh, 12 � 6 I1 � 3I 2 � 0 …(iii)

6 I1 � 3I 2 � 12 Solving Eqs (ii) and (iii), I1 � 0.67 A I 2 � 2.67 A

Ix

6�

A

Writing the VTh equation, 3I 2 � VTh � 0 3( 2.67) � VTh � 0 VTh � 8 V Step II Calculation of IN (Fig. 2.296) When a short circuit is placed across a 3 � resistor, it gets shorted as shown in Fig. 2.297. From Fig. 2.297, I x � I1 …(i) Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh,

3Ix

12 V

IN B

���������� 6�

Ix A 3Ix

12 V I1

IN I2 B

I 2 � I1 � 3I x � 3I1 4 I1 � I 2 � 0

3�

…(ii)

����������

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

…(iii)

2.9�����������������������

Solving Eqs (ii) and (iii), I1 � 2 A I2 � 8 A IN � I2 � 8 A

A IL

Step III Calculation of RN RN � Step IV

1�

1�

8A

VTh 8 � �1� 8 IN

B

Calculation of IL (Fig. 2.298)

���������� 1 IL � 8 � �4A 1�1

��������������

Find the current through the 1.6 � resistor in the network of Fig. 2.299. 3Ix Ix

����

1�

10 A

1.6 �

6�

���������� Solution Step I Calculation of VTh (Fig. 2.300) From Fig. 2.300,

3Ix Ix

I x � I1 � I 2 …(i)



6� I2

VTh

� �

I1 � 10

A



1�

10 A I1

For Mesh 1,

����

…(ii)

B

����������

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)

�������������������������������������������������������� Step II Calculation of IN (Fig. 2.301) When a short circuit is placed across the 3 � resistor, it gets shorted as shown in Fig. 2.302. From Fig. 2.302, I x � I1 � I 2

…(i)

I1 � 10

…(ii)

3Ix A

����

Ix

1�

10 A

IN

6�

For Mesh 1,

B

����������

Applying KVL to Mesh 2, �1( I 2 � I1 ) � 3I x � 0 � I 2 � I1 � 3( I1 � I 2 ) � 0 4 I1 � 4 I 2 � 0 …(iii) Solving Eqs (ii) and (iii), I1 � 10 A I 2 � 10 A I N � I 2 � 10 A Step III Calculation of RN 24 V RN � Th � � 2.4 � 10 IN Step IV

3Ix A

����

Ix

IN

1�

10 A I1

I2 B

���������� A IL

2.4 �

10 A

1.6 �

Calculation of IL (Fig. 2.303) I L � 10 �

2.4 �6A 2.4 � 1.6

B

����������

�������������������������������������� 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. 2.304,

RS

I�

V Rs � RL 2

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

V 2 RL

V

RL I

( Rs � RL ) 2 To determine the value of RL for maximum power to be transferred ��������������������������������� to the load, ����������������������� dP ������� �0 dRL dP d V2 RL � dRL dRL ( Rs � RL ) 2 �

V 2 [( Rs � RL ) 2 � ( 2 RL )( Rs � RL )] ( Rs � RL ) 4

2.10�������������������������������������

( 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 5.

Find the maximum power (Fig. 2.305). IL �

VTh V � Th RTh � RL 2 RTh

Pmax � I L2 RL �

2 VTh 2 4 RTh

� RTh �

2 VTh 4 RTh

��������� ������ For the value of resistance RL in Fig. 2.306 for maximum power transfer and calculate the maximum power. RL

15 �

5�

15 �

10 �

20 �

100 V

����������

18 �

27 �

27 �

9�

�������������������������������������������������������� Solution Step I Calculation of VTh (Fig. 2.307) 15 �

A �

5�

18 �

B

VTh



27 �

15 �

10 �

20 �

9�

27 �

100 V

���������� A

By star-delta transformation (Fig. 2.308), I�

100 � 2.08 A 5 � 5 � 20 � 9 � 9



5�

VTh

B



9�

5�



Writing the VTh equation,



� 5�

100 � 5 I � VTh � 9 I � 0 VTh � 100 � 14 I

9�

9� 20 �

I

� 100 � 14( 2.08) 100 V

� 70.88 V

���������� Step II

Calculation of RTh (Fig. 2.309) A

RTh

B 9�

5�

9�

5�

5�

9� 20 � (a)

5�

A

RTh

B

5� 5�

20 � (b)

9�

5�

A

RTh

B

9�

14 �

9�

34 �

9� 5�

A

RTh

B

9.92 � (c)

����������

(d)

9�



2.10�������������������������������������

RTh � 23.92 � Step III Calculation of RL For maximum power transfer,

23.92 �

A

RL � RTh � 23.92 � Step IV

Pmax �

23.92 �

70.88 V

Calculation of Pmax (Fig. 2.310) 2 (70.88) 2 VTh � � 52.51 W 4 RTh 4 � 23.92

B

����������

��������� ������ For the value of resistance RL in Fig. 2.311 for maximum power transfer and calculate the maximum power. 2A

5�

10 �

20 �

80 V

20 V RL

���������� Solution Step I Calculation of VTh (Fig. 2.312) 2A

5� �

� � �

10 �

80 V I1

� �

I2

A � V Th

� �

20 �

� � 20 V

B �

���������� Applying KVL to Mesh 1, 80 � 5 I1 � 10( I1 � I 2 ) � 20( I1 � I 2 ) � 20 � 0 35 I1 � 30 I 2 � 60

…(i)

Writing the current equation for Mesh 2, I2 � 2

…(ii)

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

5�

I1 � 3.43 A 10 �

Writing the VTh equation,

20 � A R Th

VTh � 20 ( I1 � I 2 ) � 20 � 0 VTh � 20(3.43 � 2) � 20 � 48.6 V Step II

B

Calculation of RTh (Fig. 2.313)

����������

RTh � 15 || 20 � 8.57 � Step III Calculation of RL For maximum power transfer,

8.57 �

A

RL � RTh � 8.57 � Step IV

Pmax �

8.57 �

48.6 V

Calculation of Pmax (Fig. 2.314) 2 ( 48.6) 2 VTh � � 68.9 W 4 RTh 4 � 8.57

B

����������

��������� ������ For the value of resistance RL in Fig. 2.315 for maximum power transfer and calculate the maximum power. 10 �

20 �

100 V RL 30 �

40 �

���������� Solution Step I Calculation of VTh (Fig. 2.316) 100 � 2.5 A 10 � 30 100 � 1.66 A I2 � 20 � 40 I1 �

Writing the VTh equation, VTh � 10 I1 � 20 I 2 � 0 VTh � 20 I 2 � 10 I1 � 20(1.66) � 10( 2.5) � 8.2 V

I2

I1 10 � � � �

100 V

A

V Th



20 � �

� B





30 �

40 �



����������



2.10�������������������������������������

Step II

Calculation of RTh (Fig. 2.317) 10 �

20 �

A

R Th

B

30 �

40 �

���������� Redrawing the network (Fig. 2.318), 10 �

20 �

A

B

RTh � (10 || 30) � ( 20 || 40) � 20.83 � 30 �

40 �

���������� 20.83 �

Step III Value of RL For maximum power transfer, RL � RTh � 20.83 � Step IV

20.83 �

8.2 V

Calculation of Pmax (Fig. 2.319) Pmax �

A

2 (8.2) 2 VTh � � 0.81 W 4 RTh 4 � 20.83

B

����������

��������� ������ For the value of resistance RL in Fig. 2.320 for maximum power transfer and calculate the maximum power. 6�

RL

72 V 2� 3�

����������

4�

�������������������������������������������������������� Solution Step I

Calculation of VTh (Fig. 2.321)

A

6� � �

Applying KVL to Mesh 1, 72 � 6 I1 � 3( I1 � I 2 ) � 0 9 I1 � 3I 2 � 72 …(i)

72 V

� I1

Applying KVL to Mesh 2,

� 3�

�3( I 2 � I1 ) � 2 I 2 � 4 I 2 � 0 �3I1 � 9 I 2 � 0 …(ii)

2� I2







� V Th



B

� 4�

����������

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. 2.322) A 6�

2�

R Th B

6�

A

3� 4�

2�

R Th 3�

4�

B

���������� RTh � [(6 || 3) � 2] || 4 � 2 � 2�

Step III Calculation of RL For maximum power transfer, RL � RTh � 2 � Step IV

A

2�

60 V

Calculation of Pmax (Fig. 2.323) Pmax �

2 (60) 2 VTh � � 450 W 4 RTh 4 � 2

B

����������

������������������������������� ���������������

For the network shown in Fig. 2.324, find the value of RL for maximum power transfer. Also, calculate maximum power.

2.10������������������������������������� I

20 �

40 � RL

10 I

� �

50 V

���������� I

Solution Step I Calculation of VTh (Fig. 2.325) Applying KVL to the mesh,

20 �

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

10 I

B

� �

VTh � 40 I � 50 � 0 VTh � 40( �1) � 50 � 0 VTh � 10 V Step II Calculation of IN (Fig. 2.326) From Fig. 2.326, I � I2 Applying KVL to Mesh 1,



50 V

����������

I

…(i)

20 �

40 �

A IN

10 I � 20 I1 � 0 10 I 2 � 20 I1 � 0

40 �

A � V Th

10I

� �

B

I1

I2

…(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 � IN 0.625 Step IV Calculation of RL For maximum power transfer, RL � RTh � 16 �

…(iii)

16 �

A

16 �

10 V

Step V

50 V

Calculation of Pmax (Fig. 2.327) Pmax �

(10) 2 VTh � � 1.56 W 4 RTh 4 � 16

B

����������

��������������������������������������������������������

��������������� For the network shown in Fig. 2.328, calculate the maximum power that may be dissipated in the load resistor RL. ����

3�

2Ix Ix

10 A

6�

4�

RL

���������� 2Ix

Solution Step I Calculation of VTh (Fig. 2.329) From Fig. 2.329,

3�

����



…(i)

I x � I2

6�

4�

10 A I1



Ix

I2

V Th

� �

For Mesh 1, I1 � 10

…(ii)

A

B

����������

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. 2.330) From Fig. 2.330, I x � I 2 � I3 For Mesh 1, I1 � 10

2Ix

6�

4�

10 A

…(ii)

4 I1 � 8 I 2 � 4 I 3 � 0

A Ix

…(i) I1

I2

IN

I3

B

Applying KVL to Mesh 2, �4( I 2 � I1 ) � 2 I x � 6( I 2 � I 3 ) � 0 �4 I 2 � 4 I1 � 2( I 2 � I 3 ) � 6 I 2 � 6 I 3 � 0

3�

����

���������� …(iii)

2.10�������������������������������������

Applying KVL to Mesh 3, �6( I 3 � I 2 ) � 3I 3 � 0 6 I 2 � 9I3 � 0

…(iv)

Solving Eqs (ii), (iii) and (iv), I1 � 10 A I 2 � 7.5 A I3 � 5 A I N � I3 � 5 A Step III Calculation of RTh

6�

A

RTh �

VTh 30 � �6� IN 5

6�

30 V

Step IV Calculation of RL For maximum power transfer,

B

RL � RTh � 6 � Step V

����������

Calculation of Pmax (Fig. 2.331) Pmax �

2 (30) 2 VTh � � 37.5 W 4 RTh 4 � 16

���������������

For the network shown in Fig. 2.332, find the value of RL for maximum power transfer. Also, find maximum power. 2V

1� �

2Vx

Vx

� �



1� RL 1A

���������� Solution Step I Calculation of VTh (Fig. 2.333)



From Fig. 2.333, Vx � �1I � � I

…(i)

2V x

Vx

� �

Vx � 1 V





A

1� V Th

I

For Mesh 1, I � �1

2V

1�

1A



…(ii) ����������

B

�������������������������������������������������������� Writing the VTh equation, 2Vx � 1I � 2 � VTh � 0 2(1) � ( �1) � 2 � VTh � 0 VTh � 5 V Step II Calculation of IN (Fig. 2.334) From Fig. 2.334, 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,

2V

1� �

…(i) 2Vx

� �

Vx

A



1� IN

I1

…(ii)

1A

I2

B

����������

2Vx � 1I1 � 2 � 0 2( � I1 ) � I1 � 2 � 0 3I1 � 0

…(iii)

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 � �3� I N 1.67

Step IV Calculation of RL For maximum power transfer,

3�

A

RL � RTh � 3 � Step V

Pmax �

3�

5V

Calculation of Pmax (Fig. 2.335) 2 (5) 2 VTh � � 2.08 W 4 RTh 4 � 3

B

����������

��������������� What will be the value of RL in Fig. 2.336 to get maximum power delivered to it? What is the value of this power? 0.5 V ���� 3A

4�



4�

V

RL �

����������

2.10�������������������������������������

Solution Step I Calculation of VTh (Fig. 2.337) By source transformation, From Fig. 2.337, VTh � 4 I

0.5 VTh ����

4�

3A

� 4�

VTh



Applying KVL to the mesh, 12 � 4 I � 0.5 VTh � 4 I � 0

A

B

����������

12 � VTh � 0.5 VTh � VTh � 0 VTh � 8 V

0.5 VTh

Step II Calculation of IN (Fig. 2.338) If two terminals A and B are shorted, the 4 � resistor gets shorted. V �0 Dependent source 0.5 V depends on the controlling variable V . When V = 0, the dependent source vanishes, i.e. 0.5 V = 0 as shown in Fig. 2.339 and Fig. 2.340. IN �

����



4�

4�

12 V



����������

0.5 V

4� A



IN

V

B



A

12 V

4�

4�

IN

12 V

B

����������

����������

Step III Calculation of RTh RTh

V 8 � Th � � 2.67 � IN 3

2.67 �

A

2.67 �

8V IL

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

Calculation of Pmax (Fig. 2.341) Pmax �

2 (8) 2 VTh � �6W 4 RTh 4 � 2.67

VTh

I

12 �3A 4

����

A

B

����������

B

��������������������������������������������������������

��������������������������� It states that ‘in a linear, bilateral, active, single source network, the ratio of excitation to response remains same when the positions of excitation and response are interchanged.’ In other words, it may be stated as ‘if a single voltage source Va in the branch ‘a’ produces a current Ib in the branch + N t k I ‘b’ then if the voltage source Va is removed and inserted in V − the branch ‘b’, it will produce a current Ib in branch ‘a’’. Explanation Consider a network shown in Fig. 2.342. ������������������ When the voltage source V is applied at the port 1, it produces a current I at the port 2. If the positions of the + excitation (source) and response are interchanged, i.e., if I Network − V the voltage source is applied at the port 2 then it produces a current I at the port 1 (Fig. 2.343). The limitation of this theorem is that it is applicable only ������������������������������������ to a single-source network. This theorem is not applicable in ����������������� the network which has a dependent source. This is applicable ������������� only in linear and bilateral networks. In the reciprocity theorem, position of any passive element (R, L, C) do not change. Only the excitation and response are interchanged. Steps to be followed in Reciprocity Theorem 1. 2. 3.

Identify the branches between which reciprocity is to be established. Find the current in the branch when excitation and response are not interchanged. Find the current in the branch when excitation and response are interchanged.

���������������

Calculate current I and verify the reciprocity theorem for the network shown in

Fig. 2.344. 5�

4� I

20 V

10 �

6�

���������� Solution Case I Calculation of current I when excitation and response are not interchanged (Fig. 2.345) Applying KVL to Mesh 1, 20 � 5 1 10 10( 1 2 ) 0 15 I1 10 I 2 � 20 Applying KVL to Mesh 2, �10( 2 � 1 ) � 4 2 � 6 I 2 � 0 �10 I1 � 20 I 2 � 0

…(i)

5�

4� I

10 �

20 V I1

6� I2

����������

…(ii)

2.10��������������������������

Solving. Eqs (i) and (ii), I1 � 2 A I2 � 1 A I2 � 1 A

I

Case II Calculation of current I when excitation and response are interchanged (Fig. 2.346). Applying KVL to Mesh 1,

5�

4� 20 V

I

10 �

�5 1 � 10 10( 0 1 � 2) � 0 15 I1 100 I 2 � 0

I1

…(i)

6�

I2

Applying KVL to Mesh 2, �10(

2

����������

� 1 ) � 4 2 � 20 20 � 6 I 2 � 0 �10 I1 � 20 I 2 � �20

…(ii)

Solving Eqs (i) and (ii), I1 � �1 A I 2 � �1 5 A I I1 � 1 A Since the current I remains the same in both the cases, reciprocity theorem is verified.

��������� ������

Find the current I and verify reciprocity theorem for the network shown in

Fig. 2.347. 2�

2�

3�

5V

3�

4�

I 4�

���������� Solution Case I Calculation of the current I when excitation and response are not interchanged (Fig. 2.348) Applying KVL to Mesh 1, 5 2 I1 3( 1 2 ) 4( I1 � I 3 ) � 0 9

1

3I 2 � 4

3

5

2

� 1 ) � 2I 2 � 3

5V

2�

3�

I1 4�

I3 2

�0

�3 1 � 88I 2 � 0

3�

I2

…(i)

Applying KVL to Mesh 2, �3(

2�

4�

…(ii)

����������

I

�������������������������������������������������������� Applying KVL to Mesh 3, �4(

� 1 ) � 4I3 � 0 �4 1 � 88I 3 � 0 Solving Eqs (i), (ii) and (iii), 3

…(iii)

I1 � 0 85 A I 2 � 0 32 A I 3 � 0 43 A I I 3 � 0 43 A Case II Calculation of current I when excitation and response are interchanged (Fig. 2.349). Applying KVL to Mesh 1, �2 1 � 3( I1 � I 2 ) � 4( 9

1



3I 2 � 4

1

3)

�0

3

0

2�

…(i) I

Applying KVL to Mesh 2, �3(

2

� 1 ) � 2I 2 � 3

2

�0

4�

� 1) � 5 � 4

3

I2

3�

I3

…(ii)

Applying KVL to Mesh 3, 3

3�

I1 4�

�3 1 � 88I 2 � 0

�4(

2�

�0

5V

����������

�4 1 � 88I 3 � 5

…(iii)

Solving Eqs (i), (ii) and (iii), I1 � 0 43 A I 2 � 0 16 A I 3 � 0 84 A I

I1 � 0 43 A

Since the current I remains the same in both the cases, reciprocity theorem is verified.

��������� ������

Find the voltage V and verify reciprocity theorem for the network shown in

Fig. 2.350. 4�

2� 5�

10 A

� 6�

����������

V

� 8�

2.10��������������������������

Solution Case I Calculation of the voltage V when excitation and response are not interchanged (Fig. 2.351) For Mesh 1, I1 � 10 …(i) Applying KVL to Mesh 2, �4(

2

� 1 ) � 2 I 2 � 5( �4 1 � 11 11

3)

4� 5�

10 A





2

� 5I 3 � 0

3

� 1 ) � 5(

3



I3

6�

8�

…(ii)

Applying KVL to Mesh 3, �6(



V

I1

�0

2

2�

I2

����������

2) �8 3

�0

�6 1 � 5 I 2 � 19 I 3 � 0

…(iii)

Solving Eqs (i), (ii) and (iii), I1 � 10 A I 2 � 5 76 A I 3 � 4 67 A V ( I 2 I3 )

5(5.76 4.67) � 5 45 V

Case II Calculation of voltage V when excitation and response are interchanged (Fig. 2.352). �

4�

2� 5�

V

10 A

6�

8�



���������� By source transformation (Fig. 2.353), Applying KVL to Mesh 1, �4 1 � 2 I1 � 50 � 5(

1



2)



4�

11I1 5 I 2

50 …(i)

2� I1 5�

�0

50 V

V

Applying KVL to Mesh 2, �6

2

� 5( I 2 � I1 ) � 50 0 �8 �5 1 � 1199

2

�0

2

� 50

6�

…(ii)

Solving Eqs (i) and (ii),

I2



���������� I1 � �3 8 A I 2 � 1 63 A

8�

�������������������������������������������������������� From Fig. 2.353, 4 I1 � 6 I 2 � 0

V

V � 4( �3.8) � 6(1.63) � 0 V � 5 42 V Since the voltage V is same in both the cases, the reciprocity theorem is verified.

������������������������� 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. 2.354). A R1 V1

R2

A Rm

Rn

V2

� Vm

Vn B

B

���������������������������� where

Vm �

V1G1 � V2G2 � � � VnGn G1 � G2 � � � Gn

and

Rm �

1 1 � Gm G1 � G2 � � � Gn

Explanation By source transformation, each voltage source in series with a resistance can be converted to a current source in parallel with a resistance as shown in Fig. 2.355. A I1

R1

I2

R2

In

Rn

B

����������������������������� Let Im be the resultant current of the parallel current sources and Rm be the equivalent resistance as shown in Fig. 2.356.

Im

Rm

V V V I m � I1 � I 2 � � � I n � 1 � 2 � � � n � V1 G1 � V2 G2 � �Vn Gn R1 R2 Rn ��������������������� �������

����������������������������� A

1 1 1 1 � � ��� Rm R1 R2 Rn Gm � G1 � G2 � � � Gn

Rm Vm

By source transformation, the parallel circuit can be converted into a series circuit as shown in Fig. 2.357. Vm � I m Rm �

B

I m V1 G1 � V2 G2 � � � Vn Gn � Gm G1 � G2 � � � Gn

���������������������������������� �������

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. 2.358). I1

I2

In

Im

� R1

R2

Rn

A

Rm B

A

���������������������������� where

I1 R1 � I 2 R2 � � � I n Rn R1 � R2 � � � Rn Rm � R1 � R2 � � � Rn Im �

Steps to be followed in Millman’s Theorem 1. 2.

Remove the load resistance RL. Find Millman’s voltage across points A and B. Vm �

3.

Find the resistance Rm between points A and B. Rm �

4. 5.

V1 G1 � V2 G2 � � � Vn Gn G1 � G2 � � � Gn

1 G1 � G2 � � � Gn

Replace the network by a voltage source Vm in series with the resistance Rm. Find the current through RL using ohm’s law. IL �

Vm Rm � RL

B

��������������������������������������������������������

���������������

Find Millman’s equivalent for the left of the terminals A-B in Fig. 2.359. A 4�

2�

1A

6V B

���������� A

Solution By source transformation, the network is redrawn as shown in Fig. 2.360. Step I Calculation of Vm � 1� � 1� 6� � � 2� � � 2� � 4� V1 G1 � V2 G2 � 3.33 V Vm � � 1 1 G1 � G2 � 4 2

4�

2�

6V

2V B

���������� A

Calculation of Rm

Step II

Rm �

1.33 �

1 1 1 � � � 1.33 � Gm G1 � G2 1 1 � 4 2

3.33 V B

Step III Millman’s Equivalent Network (Fig. 2.361)

���������������

����������

Find the current through the 10 � resistor in the network of Fig. 2.362.

2�

4�

6�

5V

10 V

15 V

10 �

���������� Solution Step I

Calculation of Vm � 1� � 1� � 1� 5 � � � 10 � � � 15 � � � 6� � � � � V G �V G �V G 2 4 Vm � 1 1 2 2 3 3 � � 2.73 V 1 1 1 G1 � G2 � G3 � � 2 4 6

�����������������������������

Step II

Calculation of Rm 1.09 �

1 1 � � 1.09 � Gm 1 1 1 � � 2 4 6 Step III Calculation of IL (Fig. 2.363) Rm �

IL �

���������������

2.73 V

10 � IL

����������

2.73 � 0.25 A 1.09 � 10

Find the current through the 10 � resistor in the network of Fig. 2.364. 50 �

2� 40 �

50 V

10 � 100 V

���������� Solution Since the 40 � resistor is connected in parallel with the 50 V source, it becomes redundant. The network can be redrawn as shown in Fig. 2.365. Step I

Calculation of Vm

50 �

10 � 50 V

� 1� � 1� 50 � � � 100 � � � 20 � � 50 � V1 G1 � V2 G2 � �57.15 V Vm � � 1 1 G1 � G2 � 50 20 Step II

20 �

40 �

100 V

����������

Calculation of Rm Rm �

1 1 1 � � � 14.29 � 1 1 Gm G1 � G2 � 50 20

Step III Calculation of IL (Fig. 2.366) IL �

��������� ������ Fig. 2.367.

57.15 � 2.35 A 14.29 � 10

14.29 �

IL

10 �

57.15 V

����������

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

��������������������������������������������������������

C

6V

2�

3V

1� A

2�

4A

8V

6V

2�

1� B D

5V

5�

���������� Step I

By source transformation, the network is redrawn as shown in Fig. 2.368.

C

6V

2�

3V

1� A

8V

8V

6V

2�

2�

1� B D

5V

5�

���������� Step II

Applying Millman’s theorem at terminals CD,

Vm1

� 1� � 1� �8 � � � 8 � � � 6(1) � 2� � � V G �V G �V G 2 � 1 1 2 2 3 3 � �3V 1 1 G1 � G2 � G3 � �1 2 2

1 1 1 � � � 0.5 � Gm1 G1 � G2 � G3 1 1 � �1 2 2 Step III Applying Millman’s theorem at terminals CA, Rm1 �

Vm2

Rm2

� 1� 6 � � � 3(1) � 2� V4 G4 � V5 G5 �4V � � 1 G4 � G5 �1 2 1 1 1 � � � � 0.67 � Gm2 G4 � G4 1 �1 2

�����������������

Step IV Millman’s Equivalent Network (Fig. 2.369) Simplifying Fig. 2.369 further, the Millman’s equivalent network is shown in Fig. 2.370. 4V

0.67 � A A

3V 6.17 �

0.5 �

2V

B 5V



5�

B

�����������

����������

Exercises ������������� 2.1

i1

20 �

Find currents Ix and Iy in the network shown in Fig. 2.371.

5�

���

2Ix

I3

30 � A

2IY

80 V

2�

Ix

���

4Ix

I1

� � 10IY

I2



40 �

Iy

30 �



2�

5V

2�

V3

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

���������� [0.5 A, 0.1 A] 2.2

10 �

In the network shown in Fig. 2.372, find V3 if element A is a (i) short circuit (ii) 5 � resistor (iii) 20 V independent voltage source, positive reference on the right (iv) dependent voltage source of 1.5 i1, with positive reference on the right (v) dependent current source 5 i1, arrow directed to the right

2.3

Find currents I1, I2, and I3 in the network shown in Fig. 2.373. 1�

15 A I1

1 V 9 x 2�

3�

I2

� Vx �

2�

1�

I3

���������� [15 A, 11 A, 17 A]

�������������������������������������������������������� Find currents Ix in the network shown in Fig. 2.374.

2.4

20 �

Find the voltage Vx in the network shown in Fig. 2.378. 1 � 3

25 �

10 �

2A

2.8

1.5 Ix

1�

5�

0.5 �

Vx

5A

Ix

0.5 �

3I

1V

���������� [8.33 A] 2.5

Find currents I1 in the network shown in Fig. 2.375. 5 � I1

19 V

2�

4�

���������� [6.2 V] 2.9

Determine V1 in the network shown in Fig. 2.379. 50 �

30 V

20 �

25 V

1.5I1

4A

5A

����������

20 �

� V1 �

� 0.4V1 �

0.01V1

[�12 A] ���������� [140 V]

������������� 2.6

Find the voltage Vx in the network shown in Fig. 2.376. 1� 1�

4I1

� Vx �

2.10 Find the voltage Vy in the network shown in Fig. 2.380. 0.5Vy ���

2�

I1

4A

15 �

� 6A

3�

20 � Vy

8V

4V



���������� 2.7

[ �4.31 V] Find the currents Vx in the network shown in Fig. 2.377.

���������� [�10 V] 2.11 Find the voltage V2 in the network shown in Fig. 2.381.

3V1

0.8V2 ���

5�

2�

� Vx � 7A

4�

4Vx 6�

5A

5�

� V2 �

8A 2.5 �

����������

���������� [2.09 V]

[25.9 V]

�����������������

(ii)

�������������������� 2.12 Find the voltage Vx in Fig. 2.382. 2�

I1 100 �

4� �

50 V

Vx



10 V 9I1

Vx

100 V

0.1Vx

����������

����������

� �38.5 V �

�9.09 V, 9.09 ��

2.13 Determine the voltages V1 and V2 in Fig. 2.383. 4�

4�

V1

V2

100 �

(iii)

I1 4 �

0.5Vx ����

� �

10 A

0.25I1

0.5V2



20 V

3A

4�

4�

����������



�6 V,12 V � ����������

2.14 Find the voltage Vx in Fig. 2.384. 10 �

Vx

�8 V, 2.66 ��

60 V

(iv)

I1

20 �

4A

30 �

� Vx �

3�

5�

0.4I1

2�

����������

� Vx �

Vx 4

10 A

�7.5 V � ������������������

����������

2.15 Determine Thevenin’s equivalent network for figures 2.385 to 2.388 shown below. (i) 4Vx

10 V

�150 V, 20 �� 2.16 Find the current Ix in Fig. 2.389.

����

4�

� Vx �

Ix 5 �

2�

6� 10Ix

� �

4A

5�

1A

10 V

4A

���������� ����������

� �58 V,12 ��

�4 A �

�������������������������������������������������������� 2.17 Find the current in the 24 � resistor in Fig. 3.390.

2V

3�

2� I1

Ix 1000 �

2I1

3� 48 V

3Vx

� �

13 �

10Ix

� Vx �

10 �

24 �

����������

�0.25 A �

���������� [0.225 A]

2.19 Find Norton’s equivalent network in Fig. 2.392. 2�

���������������� 2.18 Find Norton’s equivalent network and hence find the current in the 10 � resistor in Fig. 2.391.

2� I2

8�

12 V

� V1 �

8I2

� �

0.1 V1

20 �

����������

�0.533 A, 31 ��

Objective-Type Questions 2.1

Two electrical sub-networks N1 and N2 are connected through three resistors as shown in Fig. 2.393. The voltages across the 5 � resistor and 1 � resistor are given to be 10 V and 5 V respectively. Then the voltage across the 15 � resistor is (a) �105 V (b) 105 V (c) �15 V (d) 15 V �

2.3

The voltage across terminals a and b in Fig. 2.394 is 2�

1V

a

� �

1�

2�

3A

b

5� �

����������

10 V N1

N2 15 � 1� �

5V

2.4 �

(a) (a) 0.5 V (b) 3 V (c) 3.5 V (d) 4 V The voltage V0 in Fig. 2.395 is 2�

���������� 2.2

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

16 V 8A

12 �

10 �

6�

���������� (a) (c)

48 V 36 V

(b) 24 V (d) 28 V

� V0 �

��������������������������������

2.5

The dependent current source shown in Fig. 2.396. 5�

V1 ��20 V

2.9

V1 5

5�

(a)

4 � 3

(b)

8 � 3

(c)

4�

(d)

2�

In the network shown in Fig. 2.400, the effective resistance faced by the voltage source is i

����������

4

(a) (c) 2.6

delivers 80 W delivers 40 W

(b) (d)

absorbs 80 W absorbs 40 W

i

If V � 4 in Fig. 2.397, the value of Is is given by

4�

1�

V

���������� 4�

2�

2� V

IS

����������

(a)

4�

(b)

3�

(c)

2�

(d)

1�

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 2.11 The value of the resistance R connected across the terminals A and B in Fig. 2.401, which will absorb the maximum power is

2.10 (a) (c) 2.7

6A 12 A

(b) (d)

2.5 A none of these

The value of Vx,Vy and Vz in Fig. 2.398 shown are + Vx −

+ Vy −

+ 2V −

+ 8V −

+ Vz −

+ 1V −

+ 2V −

4 k�

3 k�

���������� (a) �6, 3, �3 (c) 6, 3, 3

(b) (d)

R

�6, �3, 1 6, 1, 3

V

A

B 6 k�

2.8

4 k�

The circuit shown in Fig. 2.399 is equivalent to a load of ����������

2�

I

(a) (c) 4�

����������

� �

2I

4 k� 8 k�

(b) (d)

4.11 k� 9k�

2.12 Superposition theorem is not applicable to networks containing (a) nonlinear elements (b) dependent voltage source

�������������������������������������������������������� (c) dependent current source (d) transformers

100 �

2.13 The value of R required for maximum power transfer in the network shown in Fig. 2.402 is 5�

4�

���������� (a) (c)

25 V

20 �

3A

R

1W 0.25 W

2� 8�

(b) (d)

1A

(b) 4 � (d) 16 �

0.5I1 � �

5�

10 V

b

����������

I1

20 �

a I1

5�

2.14 In the network of Fig. 2.403, the maximum power is delivered to RL if its value is

0.5I1

10 W 0.5 W

2.16 For the circuit shown in Fig. 2.405, Thevenin’s voltage and Thevenin’s equivalent resistance at terminals a-b is

���������� (a) (c)

RL

10 V

(a) 5 V and 2 � (b) 7.5 V and 2.5 � (c) 4 V and 2 � (d) 3 V and 2.5 � 2.17 The value of RL in Fig. 2.406 for maximum power transfer is

40 �

RL

50 V

9�

���������� 6�

(a)

16 �

(b)

40 � 3

(c)

60 �

(d)

20 �

2V

2.15 The maximum power that can be transferred to the load RL from the voltage source in Fig. 2.404 is

6�

� �

6�

1A

9�

RL

���������� (a) (c)

3� 4.1785 �

(b) (d)

1.125 � 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. (a)

2.12. (a)

2.13. (c)

2.14. (a)

2.15. (c)

2.16. (b)

2.17. (a)

3

Analysis of AC Circuits

�3.1��������������� We have discussed the network theorems with reference to resistive load and dc sources. Now, all the theorems will be discussed when a network consists of ac sources, resistors, inductors and capacitors. All the theorems are also valid for ac sources.

�3.2���������������� Mesh analysis is useful if a network has a large number of voltage sources. In this method, currents are assigned in each mesh. We can write mesh equations by Kirchhoff’s voltage law in terms of unknown mesh currents,

�������������

Find mesh currents I1 and I2 in the network of Fig. 3.1. 3�

j4 �

� j10 �

100�45� V �

I1

� 10 � �j

I2

�������� Solution Applying KVL to Mesh 1, 100�45° − (3 � j4)I1 − j10(I1 − I2) � 0 (3 � j14)I1 − j10I2 � 100 �45° Applying KVL to Mesh 2, −j10 (I2 − I1) � j10 (I2) � 0 j10I1 � 0 I1 � 0 Substituting I1 in Eq. (i), � j10 I 2 � 100 � 5� I2 �

100 �45� � 10 �135� A � j10

…(i)

…(ii)

3.2�Circuit Theory and Networks—Analysis and Synthesis

�������������

Find mesh current I1, I2 and I3 in the network of Fig. 3.2. 5�

�2� �j

j5 �

� 3�

10�30� V �

I1

I2

5�

2�

I3

�2� �j

Fig. 3.2 Solution Applying KVL to Mesh 1, 10 �30° − (5 − j2) I1 − 3 (I1 − I2) � 0 (8 − j2) I1 − 3I2 � 10�30° Applying KVL to Mesh 2, −3 (I2 − I1) − j5I2 − 5 (I2 − I3) � 0 −3I1 � (8 � j5) I2 − 5I3 � 0 Applying KVL to Mesh 3, –5 (I3 – I2) – (2 – j2) I3 � 0 –5I2 + (7 – j2) I3 � 0 Writing Eqs (i), (ii) and (iii) in matrix form, �3 0 � � I1 � �10�30�� �8 j 2 � �3 8 � j5 5 � �I2 � � � 0 � � �� � � � 5 7 j 2� � I3 � � 0 � � 0 By Cramer’s rule, �3 10 �30� 0 0 8 � j5 5 0 5 7 j2 � 1.43 43 38 7� A I1 � �3 8 j2 0 �3 8 � j5 5 0 5 7 j2 j 2 10 �30� 0 �3 �5 0 0 0 7 j2 � 0.693� � 2.2� A I2 � �3 8 j2 0 �3 8 � j5 5 0 5 7 j2 8

j2 �3 10 �30� �3 8 � j5 0 0 5 0 � 0.476 �13.8� A I3 � �3 8 j2 0 �3 8 � j5 5 0 5 7 j2 8

… (i)

… (ii)

… (iii)

3.2�Mesh Analysis����

�������������In the network of Fig. 3.3, find the value of V2 so that the current through (2 + j3) ohm impedance is zero. 5�

j3 �

2�

4� �

� j5 �

30�0��V �

6� I2

I1

I3



V2

��������

��������Applying KVL to Mesh 1, 30�0° − 5I1 − j5 (I1 − I2) � 0 (5 � j5) I1 − j5I2 � 30 �0°

…(i)

Applying KVL to Mesh 2, −j5 (I2 − I1) − (2 � j3) I2 − 6 (I2 − I3) � 0 −j5I1 � (8 � j8) I2 − 6I3 � 0

…(ii)

Applying KVL to Mesh 3, −6(I3 − I2) − 4I3 − V2 � 0 −6I2 � 10I3 � −V2 Writing Eqs (i), (ii) and (iii) in matrix form, 0 � � I1 � �30�0�� �5 j 5 � j 5 � � j 5 8 � j8 6 � � I 2 � � � 0 � � �� � � � 6 10 � � I3 � � V2 � � 0 By Cramer’s rule, 5 � j 5 30 �0� 0 � j5 �6 0 0 10 2 �0 I2 � 5 j5 � j5 0 � j 5 8 � j8 6 0 6 10 (

)(

) (

)(

) V2 �

0 j1500 � 35.36� 45� V 30 � j 30

�������������Find the value of the current I3 in the network shown in Fig. 3.4. 10�30� V

�4� �j

4�





� 20�0� V

j10 � �

I1

20 � I2

10 �

4� I3 20 �

��������

�4� �j

…(iii)

����Circuit Theory and Networks—Analysis and Synthesis

��������Applying KVL to Mesh 1, 20�0° − (4 − j4) I1 − j10 (I1 − I2) − 10 (I1 − I3) � 0 (14 � j6) I1 − j10I2 − 10I3 � 20 �0° Applying KVL to Mesh 2, −j10 (I2 − I1) − 10�30° − 20I2 − (4 − j4) (I2 − I3) � 0 −j10I1 � (24 � j6) I2 − (4 − j4) I3 � −10�30° Applying KVL to Mesh 3, −10(I3 − I1) − (4 − j4) (I3 − I2) − 20I3 � 0 −10I1 − (4 − j4) I2 � (34 − j4) I3 � 0 Writing Eqs (i), (ii) and (iii) in matrix form, j10 �14 � j 6 � � j10 24 � j 6 � �( 4 j4 j 4) � �10

… (i)

… (ii)

… (iii)

10 � � I1 � � 20 �0� � ( 4 � j 4) � �I 2 � � � �10 �30�� �� � � � 34 � j 4 � � I 3 � � 0 �

By Cramer’s rule, j10 14 � j 6 20 �0� � j10 24 � j 6 10 30� �10 �( 4 j 4) 0 � 0 44 � 14� A I3 � 14 � j 6 10 10 � j10 24 j 6 � ( 4 j 4 ) �10 �( 4 j4 j 4) 34 � j 4

�������������Find the voltage VAB in the network of Fig. 3.5. 1�

� 50 � �j

100 � A B

I2 4�

100 �

j 200 �

96 � I1 � � 10�0� V

��������

��������Applying KVL to Mesh 1, − 96 I1 − (100 � 4 � j200) (I1 − I2) � 10 �0° � 0 (200 � j200) I1 − (104 � j200) I2 � 10 �0°

…(i)

Applying KVL to Mesh 2, − (1 − j50 − 100) I2 − (100 � 4 � j200) (I2 − I1) � 0 − (104 � j200) I1 � (205 � j150) I2 � 0 Writing Eqs (i) and (ii) in matrix form, �(104 � j 200) � � I1 � �10�0�� � 200 � j 200 � �� �(104 � j200 j 200) 205 � j150 �� ��I 2 �� �� 0 ��

…(ii)

3.2�Mesh Analysis����

By Cramer’s rule, 10 �0� � 200 0 205 � j150 200 � �(104 � 200 � � 200 � j150

I1

0 051

200 � 200 10 �0 � � j 200 0 200 � � 104 � j 200 � � j 200 � j150 � 2

I2

A

045 26 34� A

0 � � � 0.045�

26 34� � ( 4 � j 200) ( 0.051 V

������������

10



For the network shown in Fig. 3.6, find the voltage across the capacitor.

1�



j2 �



3�

I2 j2 �

1�

1



j3 �

5�0� V

3

j1 �



�������� Solution Applying KVL to Mesh 1, 0 � 2)

0 5� 0

…(i)

Applying KVL to Mesh 2, 2





� 2I 3

Applying KVL to Mesh 3, � ) � � j2 �

� �

)



0 �0 �0

Writing Eqs (i), (ii) and (iii) in matrix form, �2 2 �

5 3

2 2

…(ii)

� j 3 � � I1 � 5 � � I2 � 0 � 2 � j2 I 0

…(iii)

����Circuit Theory and Networks—Analysis and Synthesis By Cramer’s rule, 5

2

j2 2� 2 � j3

0 0 �2

3

I2 �

� �1

5

2

5

2 � j2

( 2 �

I3

�2 V

�2 5�0� 5 j2 0 2 0 � 0 91 �2 � j3 5� 2 2



������������

130 51� A

A

2 0 91

.

�130 51�

3 03

.

V

Find the voltage across the 2 � resistor in the network of Fig. 3.7. j �

3�

2�

� 8�45 �V

2�30� A I1

j �



I2

�������� Solution For Mesh 1, I1 � 2

0

…(i)

Applying KVL to Mesh 2, �

)(

)� j

� �45� � 0 � 45

…(ii)

Substituting I1 in Eq. (i), 8�45� I2 � V

2 2 2



1 0

9

3 19� � 65�

65 A 78

84 37 V

3.2�Mesh Analysis����

������������

Find the current through 3 � resistor in the network of Fig. 3.8. 3�

5�

j2 � �

1�0� A I3

1� I2

I1

10�0� V

j1 �



�������� Solution Applying KVL to Mesh 1, 10 �0�

j 2I1 3 (

1

1(I1 I 2 ) � 0 ) I1 � I 2 � 10 �0�

…(i)

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

I 2 � 1�0�

…(ii)

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

� 1 ) � 5I 3 � j1 3 � 0 I1 I 2 � (5 1) I 3 � 0

2

…(iii)

Writing Eqs (i), (ii) and (iii) in matrix form, 0 � 4 j 2 �1 � � I1 � �10 �0�� � 0 � �I 2 � � � 1�0� � 1 1 � �� � � � 1 1 � 5 1 ( j ) � � � I3 � � 0 � By Cramer’s rule, 10 �0� 1 0 1�0� �1 1 0 1 �(5 1) � 2.11 I1 � j 4 2 �1 0 0 1 1 1 1 �(5 1) I 3 � � I1 � 2.11

������������

28.01� A

28.01� A

Find the currents I1 and I2 in the network of Fig. 3.9. 2Vx

6�

���



Vx

9�0� V �

I1

� �

3�

�3 � �j

��������

I2

����Circuit Theory and Networks—Analysis and Synthesis Solution From Fig. 3.9, Vx � � j 3( 1 �

2)

…(i)

Applying KVL to Mesh 1, 9�0� � 6

1

j 3 ( I1 � I 2 ) � 0 ) I1 � j 3

(

2

9 � 0�

…(ii)

Applying KVL to Mesh 2, j3 ( j3

2

1)

2Vx � 3

j 3I1 � 2 [ jj33 ( I1 I 2 )] � 3 j3 j9 (3 j9 j 99))

2 2 2

0 0 0

…(iii)

Writing Eqs (ii) and (iii) in matrix form, j 3 � � I1 � �9�0�� �6 j 3 � �� j 9 3 � j 9�� �� I 2 �� �� 0 �� By Cramer’s rule, j3 9�0� 0 3 � j9 � 1 3�2.49� A I1 � j3 6 j3 j9 3 � j9 j 3 9 0� j9 0 I2 � � 1.24 4 � 15 95� A j3 6 j3 j9 3 � j9 6

�������������

Find the voltage across the 4 � resistor in the network of Fig. 3.10. 6�30��V � �

j2 �

Ix

�1� �j

2�

� �

I1

2Ix

4�

I2

��������� Solution From Fig. 3.10, Ix

I1

…(i)

Applying KVL to Mesh 1, �2 1 � 6 �30� � j1( 1 � 2 ) � 2I x � 0 �2 1 � 6 �30� � j1 1 � j1I 2 � 2 1 � 0 ( ) I1 � j1 2 6 �30�

…(ii)

3.3�Node Analysis����

Applying KVL to Mesh 2, �0

2 �4

0 �0

)

…(iii)

Writing Eqs (ii) and (iii) in matrix form, 1 1



� � I1 � I2

1 4

30 0

By Cramer’s rule, 2

I2 �

1 6 0 1 0 1 1 1 �4

4

.

(0 74 � � 2.

V

2

A

�� . 1 V

3.3���������������� Node analysis uses Kirchhoff’s current law for finding currents and voltages in a network. For ac networks, Kirchhoff’s current law states that the phasor sum of currents meeting at a point is equal to zero.

��������������In the network shown in Fig. 3.11, determine Va and Vb. j6 �

3�

Va

Vb

j5 �

� j4 �

j6 �

10�0��V

j4 �

���������

��������Applying KCL at Node a, � 1 j

1 j6

3

1 3

1

�0 10 � � j6

. V � 1 67

90

…(i)

Applying KCL at Node b, V

� �

1 3

V 1 3

� 1 j

.



1

1 V j1 b

�0 0 �0

…(ii)

�����Circuit Theory and Networks—Analysis and Synthesis Adding Eqs (i) and (ii), � j1.25 25Vb � 1.67�� 90� Vb �

1 67�� 90� � 1 34 �0� V � j1.25

Substituting Vb in Eq. (i), 0.33(1 (1.34 0�)) 1.67 90� 1 73�75 5 17� Va � � 5 24 4 0 33

0 33Va

75.17 1 �V

��������������For the network shown in Fig. 3.12, find the voltages V1 and V2. 5� � 50�0� V

4�

V1

V2

2� �

j2 �

�2� �j



50�90� V �

��������� Applying KCL at Node 1, 0�0� V1 V1 V2 � � �0 5 j2 4

V1

� 1 1 1� �� 5 � j 2 � 4 �� V1

1 V2 � 10 �0� 4

(0.45 � j 0.5)V1 0 25

2

10 �0�

…(i)

Applying KCL at Node 2, V2

V1 4



V2 V2 � 50 �90� � �0 j2 2

�1 1 � V1 4 �4

1 � j2

�0.25V1 ( .7755

1� V2 � 25�90� 2� j 0. )

2

25 90� 25

Writing Eqs (i) and (ii) in matrix form, 0.25 � � V1 � � 10�0� � �0 45 j 0.5 � �� �0 255 0.75 � j 0 5�� �� V2 �� �� 25�90��� By Cramer’s rule, 10 �0� 0.25 j 25 0 75 j 0.5 � 24.7�7 . 5� V V1 � 0 45 j 0.5 0.25 �0 25 0.75 � j 0 5 0 45 j 0.5 10 0� �00 25 25 90� � 34 34 34 �52 5 82 8 �V V2 � 0 45 j 0.5 0.25 �0 25 0.75 � j 0 5

…(ii)

3.3�Node Analysis�����

�������������� Find the voltage VAB in the network of Fig. 3.13. j5 � 2�

I � 10�0� A

2

A

j 10 � I

1

B

3�

j4 �

���������

��������Applying KCL at Node 1, 10 �0� � �1 1 � �� 2 � 3 j 4 �� V1

V1 V2 V1 � 2 3 � j4

1 V2 � 10 �0� 2

j 0.16)V1 0 V2 � 10 �0�

(0.62

…(i)

Applying KCL at Node 2, V2

V1 2

�1 1 � V1 2 �2



1 j5

V2 V2 � �0 j 5 j10 1 � V2 � 0 j10 �

�00.5V1 � (0 5 � j 0.3)V2 � 0 Writing Eqs (i) and (ii) in matrix form, �0 5 � � V1 � �10�0�� �0 62 j 0.16 � �� �0 5 0 5 � j 0 3�� �� V2 �� �� 0 �� By Cramer’s rule, 10 �0� 0.5 0 0.5 j 0.3 � 21.8 8 56 42� V V1 � 0 62 j 0.16 �0 5 �0 5 0 5 � j0 3 0 62 j 0.16 10�0� �0 5 0 � 18.7 87.42 4 �V V2 � 0 62 j 0.16 �0 5 �0 5 0 5 � j0 3 VA � V2 � 18.77�877 42� V 21.88� �56 4 � V1 ( j4) � ( j4) � 17 45 5�93 32� V VB � 3 � j4 3 � j4 VAB VA VB � ( .7�87.. �) � ( .45�93.. �) � 2.23�34.1� V

…(ii)

�����Circuit Theory and Networks—Analysis and Synthesis

�������������

Find the node voltages V1 and V2 in the network of Fig. 3.14. �1 � �j

V1

2�

V2

j2 �

2V2

2�30� A

��������� Solution Applying KCL at Node 1, V1 V1 � V2 � � 2V2 2 j1 �1 � 1 � V1 2 �� 2 � � j1� 1�� � ( .5 � j1) j1) 1)

1� V2 � 0 j1�

( 2 � j1) 1)

2

0

…(i)

Applying KCL at Node 2, V2 V1 V2 � � 2�30� � j1 j2 � 1 1 V1 j1 � � j1 � j1 Writing Eqs (i) and (ii) in matrix form, �0.5 1 �� � j1

1� V2 � 2 30� j2�

� j05

2

� 2�30�

…(ii)

( 2 � j1) � � V1 � � 0 � � j 0.5 �� �� V2 �� �� 2�30���

By Cramer’s rule, ( 2 � 1) j 0.5 � 2.46 6 �130.62� V ( 2 � j1) j 0.5

0 2�30� V1 � 0.5 � 1 � j1

0.5 j1 0 � j1 2�30� V2 � � 1.2 23 3�167.49� V 0.. ((2 2 � j1) � j1 j 0.5

�������������

In the network of Fig. 3.15, find the voltage V2 which results in zero current through

4 � resistor. 5�

V1

4�

V3

2�



� j2 �

50�0� V �

�2� �j

V2 �

���������

3.3�Node Analysis�����

Solution Applying KCL at Node 1, �

5



2

1 1 1 � � 5 j2 4 .45

0

4 1

0 )

10 �0�

.

10 0

…(i)

Applying KCL at Node 3, �

4 1 4

�V �0 2

2

1 4

1 j

1 2

� 0 5V � 0 5V

…(ii)

Writing Eqs (i) and (ii) in matrix form, 5

� � V1 V

0 25 j

.

�10 �0�

By Cramer’s rule, 10 0 V V1 � 04 5 � . .

V � I4 � 10 0 7

0. 5

1

�������������

0.125 5� � �



4

0 55

�V 01 5 5 V �

5

.25 05 10 .7 0. 0 125 V � 0 25 5 1 5 � j0 5

5 10 5V 05 � 5 0.25 0 55 � j0 5

0 1

25 5

�0

25 15 95 � 0 25V 0 ) 5

26.26

2� V

Find the voltage across the capacitor in the network of Fig. 3.16. V1

2�60� A

12�30� V V2 �

j1 �

���������

2�

j2 �

�����Circuit Theory and Networks—Analysis and Synthesis Solution Nodes 1 and 2 will form a supernode. Writing the voltage equation for the supernode, V1 V2 � 12�30�

…(i)

Applying KCL to the supernode,

(

)

V1 V2 V2 � � � 2 60� j1 2 j2 � ( .5 � j 0. ) 2 � 2�60�

…(ii)

Writing Eqs (i) and (ii) in matrix form, 1 � � V1 � �12�30�� � 1 �� � j1 0 5 j 0.5�� �� V2 �� � �� 2�60� �� By Cramer’s rule, 1 12�30� � j1 2�60� � 18.55�157.42� V V2 � 1 1 � j1 0 5 j 0.5 Vc

57.42� V V2 � 18.55�157

��������������������������� The superposition theorem can be used to analyse an ac network containing more than one source. The superposition theorem states that in a network containing more than one voltage source or current source, the total current or voltage in any branch of the network is the phasor sum of currents or voltages produced in that branch by each source acting separately. As each source is considered, all of the other sources are replaced by their internal impedances. This theorem is valid only for linear systems.

��������������Find the current through the 3 � j4 ohm impedance. 5�

j5 �



3�



j4 �



50�90��V

50�0��V �

��������� Solution

j5 �

5�

Step I When the 50 �90° V source is acting alone (Fig. 3.18) (3 � j 4)( jj5) � 6 35� 23 2� � 3 � j9 50 �90� IT � � 7 87�66 8� A 6 35 35� �23 2�

ZT � 5 �

I� �

3�



j4 �

50�90��V

���������

3.4�Superposition Theorem�����

By current division rule, � j5 � I � � ( .87�66.. �) � � 4.15�85.3� (�) � 3 � j 9 �� j5 �

5�

When the 50�0° V source is acting alone (Fig. 3.19)

Step II

I�

5(3 � j 4) ZT � j 5 � � 6 74 �68 2� � 8 � j4 50 �0� IT � � 7 42� �� � 68 2� A 6 74 74� �68 2�

3�



j4 �



50�0� V

���������

By current division rule, � 5 � I� � ( .42�� 68.. �) � � 4.15�� �� 94.77� A (�) � 4.15� .3� A (�) � 8 � j 4 ��

Step III By superposition theorem, I � I� � I�� � 4.15 �85.3° � 4.15 �85.3° � 8.31 �85.3°A (�)

��������������Determine the voltage across the (2 + j5) ohm impedance for the network shown in Fig. 3.20. j4 �

�3� �j



2�



j5 �

50�0� V

20�30� A

��������� Solution j4 �

Step I When the 50�0° V source is acting alone (Fig. 3.21) I�

50 �0� � 5 422� � 77 47� A 2 � j 4 � j5

Voltage cross (2 � j5) � impedance

2�

� 50�0� V �

I�(

� j4 � � �) � � 8.68�42.53� A � 2 � j 9 ��

Voltage across (2 � j5) � impedance V�� � (2 � j5) (8.68 �42.53°) � 46.69 �110.72° V

j5 �

I

V� � (2 � j5) (5.42 �− 77.47°) � 29.16 �− 9.28° V Step II When the 20�30° A source is acting alone (Fig. 3.22) By current division rule,

�3� �j

��������� �3� �j

j4 � I

2� 20�30��A j5 �

���������

�����Circuit Theory and Networks—Analysis and Synthesis Step III By superposition theorem, V � V� � V�� � 29.16 �−9.28° � 46.69 �110.72° � 40.85 �72.53° V

�������������� Determine the voltage VAB

for the network shown in Fig. 3.23.

j5 � �2� �j

A

4�0� A

� 50�0� V

5�



B

��������� Solution Step I When the 50�0° V source is acting alone (Fig. 3.24) j5 � A

�2� �j

� 5�

50�0��V �

B

��������� � VAB � 50 �0� V

Step II

When the 4�0° A source is acting alone (Fig. 3.25) j5 � �2� �j

A

4�0� A

5� B

��������� � VAB �0

Step III By superposition theorem, VAB

� � VAB � VAB � 50 �0� � 0 � 50 �0� V

3.4�Superposition Theorem�����

��������������Find the current I in the network shown in Fig. 3.26. �5� �j

j3 �

I 4�

2�





13 �25� V

20 ��30� V

3 �50� A �



��������� Solution Step I

When the 13�25° V source is acting alone (Fig. 3.27) j3 �

4�

�5� �j

2�

� 13 �25��V �

I�

��������� I� �

Step II

13� 5� � 2.057 0 �43.43� 6 � j2

(� )

When the 20�−30° V source is acting alone (Fig. 3.28) 4�

j3 �

�5� �j

2� � 20 ��30� V �

I�

��������� I �� �

20 � � 30� V � 3 16� � 11 57�� A(�) � 3.16 16 �168.43� A( � ) 6 � j2

Step III When the 3�50° A source is acting alone (Fig. 3.29) 4�

I��� j3 �

�5� �j

3 �50� A

���������

2�

�����Circuit Theory and Networks—Analysis and Synthesis By current division rule, 2 � j5 � 2 56 �0.23� 6 � j2 By superposition theorem, I ��� � 3�50� �

Step IV

(�) � 2 56 � � 179.77� A( � )

I � I� � I�� � I��� � 2.057 �43.13° � 3.16 �168.43° � 2.56 �−179.77° A � 4.62 �153.99° A (�)

�������������

Find the current through the j3 � reactance in the network of Fig. 3.30. �5 � �j



� j5 �

�2 � �j

5�30� V �

10�60� V �

j3 �

��������� Solution Step I When the 5�30° V source is acting alone (Fig. 3.31) �5 � �j

� j5 �

�2 � �j

5�30� V �

j3 �

��������� When a short circuit is placed across j15 � reactance, it gets shorted as shown in Fig. 3.32. I�

�5 � �j

� �2 � �j

5�30� V �

j3 �

��������� I� �

5�30� � 2.5�120� ( � ) � j5 � j3

3.4�Superposition Theorem�����

Step II When the 10�60° V source is acting alone (Fig. 3.33) �5 � �j

� j5 �

�2 � �j

10�60� V �

j3 �

��������� When a short circuit is placed across the −j2 � reactance, it gets shorted as shown in Fig. 3.34 �5 � �j

I�� � j5 �

10�60��V �

j3 �

��������� I� �

10 �60� � 5�150� A ( � ) � 5� � 30� A (�) � j5 � j3

Step III By superposition theorem, I I � � I � � 2 5�120� � 5� �� 30� � 3.1�� 6.21�

�������������

(�)

Find the current I0 in the network of Fig. 3.35. 2�0��A �2� �j

6�

I0 8�

� j4 �

10�30��A �

��������� �2� �j

Solution Step I When the 10�30° V source is acting alone (Fig. 3.36) j 4(8 j 2) � 8 664 �24.12� � ZT � 6 � j4 8 j2 10 �30� � 1.16 �5 88 8 �A IT � 8 64� � . �

6�

I 0� 8�

IT �

j4 �

10�30� V �

���������

�����Circuit Theory and Networks—Analysis and Synthesis By current division rule, I �0 � 1.166 5.88� � Step II

j4 � 0 56 81 84� j2 � j4

8

(�)

When the 2�0° A source is acting alone (Fig. 3.37) 2�0� A 6�

�2� �j I0��

j4 �

8�

��������� The network can be redrawn as shown in Fig. 3.38. �2� �j

�2� �j

I0�� 8�

I0�� j4 �

6�

2�0��A

8�

2�0��A (1.85 � j2.77) �

(a)

(b)

��������� By current division rule, I 0� � 2 0� �

1 85 j 2.77 � 0 67 6 �51 83� 1 85 j 2.77 � 8 j 2

(�)

Step III By superposition theorem, I0

�������������

I �0 � I 0� � 0 56 81 84� 0.67�51 �51.83� � 1.19 9 65 46�

(�)

Find the current through the j5 � branch for the network shown in Fig. 3.39.

�4� �j

3�

j5 � �



� 15 �90� V

10 �0��V �



���������

20 �0� V �

3.4�Superposition Theorem�����

Solution Step I

When the 10�0° V source is acting alone (Fig. 3.40) I�

j5 �

�4� �j

3� �

10 �0��V �

��������� 3( � j4 j ) � 4 04 �61 66� � 3 � j4 10 �0� I� � � 2.48 8� � 61.66 6 �A ( � ) 4 04 04� �61 66�

ZT � j 5 �

I��

Step II

When the 15�90° V source is acting alone (Fig. 3.41) ( j5 j )( j4 j ) � 20.22� �� 81.47� � ZT � 3 � j5 j 4 15�90� IT � � 0.74 �171 7 .47 7� A 20.22� � 81.47�

IT

j5 � �

15 �90� V �

By current division rule, I�

3�

���������

� j4 0 7 �171.47 � � j 4 � j5

2 96 9 � 8.53 A ( � ) � 2 96�171.47�

(� )

Step III When the 20 �0° V source is acting along (Fig. 3.42) IT

I���

j5 �

�4� �j

3� �

20 �0� V �

��������� 3( j5 j ) � 3 447�� � 50 51� � 3 � j5 20 �0� IT � � 5 76 �50.51 0. � A 3 47� �� 50 51�

ZT � � j 4 �

By current division rule, I�� � 5 76 �50 51� �

3 � 2 96 � �� � 8.53� A ( � ) � 2 96�171.47� 3 � j5

(� )

�4� �j

�����Circuit Theory and Networks—Analysis and Synthesis Step IV

By superposition theorem, I

I � � I � I �� � 2.48 48� �� 61 66� � 2.96 96 � 7 . 7� � 2 96 �171.47� � 4.86 8 �� 164 6 .41� A

�������������

Find the voltage drop across the capacitor for the network shown in Fig. 3.43. 20 �45� V � 2�

j5 �





2�



10 �0� V �2� �j

4�

��������� Solution

I�

Step I When the 10�0° V source is acting alone (Fig. 3.44) 2� ( 2 � j 5)( 2 j 2) ZT � 4 � 2 � j5 2 j 2 j5 � � 7� � 5.91� � 10 �0� � 1.433�5 91� A IT � 7� � 5.91�

2�

� �

10 �0� V

4�

���������

By current division rule, � � 2 � j5 I� � ( .43�5. �) � � 1.54�37.24� A ( � ) � 2 � j 5 � 2 � j 2 �� Step II

When the 20�45° V source is acting alone (Fig. 3.45) 20 �45� V �

2�

I��



2� 4�

�2� �j

j5 �

��������� ZT � ( � j ) � I �� �

4(( � j ) � 4.48� � 8.84� � 4 � 2 � j5

20 � 5� � 4.46 46 �533 84� A ( � ) 4.48� � 8.84�

4.46 .46 �53.84� A ( � )

�2� �j

3.4�Superposition Theorem�����

Step III By superposition theorem, I Vc

�������������

I � � I � � 1 54 �37 24 � 4.46 6 �53.84� � 3 01� � 117.78� A (

2) I � ( � j 2) (3.01� � 117.78�))

6.02�152 5 .22� V

Find the node voltage V2 in the network of Fig. 3.46. 5 �30� V

V1

V2

5�

10 �0� A

2�

j10 �

5 �0� V

��������� Solution Step I When the 10 �0� A source is acting alone (Fig. 3.47) 5 �30� V

V1�

5�

10 �0��A

V2�

2�

j10 �

��������� Applying KCL at Node 1, V1� V1� � V2� � � 10 �0� 5 5�30� 1 � � �1 �� � � V1 5 5�30� � ( .37

j 0. )

� 1

1 V2� � 10 �0� 5 30�

( .17 1

j0 0.. )V2� � 10 �0�

…(i)

Applying KCL at Node 2, V2� V1� V2� V2� � � �0 5�30� 2 j10 �

1 V1� 5�30�

� 1 � 5�30�

�( .17 � j 0. )

� 1

1 2

1 � � V2 � 0 j10 �

� ( .67 6 � j 0. 0. )V2� � 0

…(ii)

�����Circuit Theory and Networks—Analysis and Synthesis Writing Eqs (i) and (ii) in matrix form, � 0 37 j 0.1 �� �(0 (0.17 j 0.1) (0

(0.17 � j 0.1) � � V1� � �10 �0�� � 0.67 67 � j 0 2 �� �� V2� �� �� 0 ��

By Cramer’s rule, 0.37 j 0.1 10 0� 0.17 j 0.1) 0 � ( (0 � 8 57� �� � 3.36� V V2� � 0 37 j 0.1 ((00.17 � j 0.1) 0 67 � j 0.2 �(0.17 � j 0 1) Step II

When the 5�0� A source is acting alone (Fig. 3.48) 5 �30� V

V2��

2�

5�

j10 �

5 �0��A

��������� V2� V �� V � � 2 � 2 � 5 0� 5�30� 5 2 j10 (0.61

11.93�)V2�

5�0�

V2�

� 8.2� 2 111.93� V Step III By superposition theorem, V2 V2� � V2� � 8 57 � 3.36� � 8 2�11.93� � 16.62�4 62� 4.12� V

�������������

Find current through inductor in the network of Fig. 3.49. 8�135� V � j2 �

2�0��A

� �1 � �j

2�

2�90� A

8�135��V

��������� Solution

j2 �

Step I When the 8�135� V source is acting alone (Fig. 3.50) Applying KVL to the mesh, 8�135 35� � ( � 1) � j 2I � � 0 8�135� I� � � 8 45� j1

(�)

8� � 135� A (�)



� I�

�1 � �j

2�

���������

3.4�Superposition Theorem�����

Step II

When the 2�0� A source is acting alone (Fig. 3.51) �1 � �j

j2 �

2�

2�0� A

�1 � �j

��������� The network can be redrawn as shown in Fig. 3.52. By current division rule,

j2 � I��

� � j1 � � � j1� I �� � 2�0� � � 2 � 0� � � 2�180� A(�) � � � j1� 1 � j2� � j1 ��

2�

2�0� A

Step III When the 2�90� A source is acting alone (Fig. 3.53) ��������� �1 � �j

j2 �

2�

2�90� A

���������

j2 �

The network can be redrawn as shown in Fig. 3.54. By current division rule,

I���

�1 � �j

� � j1 � I ��� � 2�90� � � 2� � 90� A (�) � 2�90� A (�) � � j1� 1 � j 2 ��

2�

2�90� A

Step III By superposition theorem, I

I � � I �� I ��� � 8 �� 135� � 2�180 80� � 2�90� � 8.49 49� �� 154.47�A

�������������

���������

Determine the source voltage VS so that the current through 2 � resistor is zero in

the network of Fig. 3.55. 3� � Vs

2�

j3 �

4�

�3� �j



� 20�90��V �

���������

�����Circuit Theory and Networks—Analysis and Synthesis Solution Step I

When the voltage source Vs is acting alone (Fig. 3.56) 3�

2�

4�

j3 �

Vs �

I1�

�3� �j I2�

I3�

��������� Appling KVL to Mesh 1, 3I1� � j3 j 3(I1� I �2 ) � 0

Vs

)I1� � j 3I �2 � Vs

( Appling KVL to Mesh 2, � j 3( � � � ) � 2I � � jj33( �

� 2





� 3)

…(i)

�0

� 3

� j 3 � 2I � j 3 � 0 Appling KVL to Mesh 3, � j3( 3� � �2 ) � 4I 3� � 0

…(ii)

j 3 � ( 4 j3 j 33)) 3� 0 Writing Eqs (i), (ii) and (iii) in matrix form,

…(iii)

� �� 0 � � I1 � � Vs � �3 j 3 � j 3 � � j3 2 j 3 � � I 2� � � � 0 � � � � � 0 3 4 j j 3� � I � � � 0 � � �� 3 �� By Cramer’s rule, 3 j 3 Vs 0 j3 � j3 0 0 0 4 j3 (9 � I 2� � 3 j3 � j3 0 j3 � j3 2 j3 4 j3 0 Step II

j12) Vs �

When the 20 �90° V source is acting alone (Fig. 3.57) 3�

2�



j3 � I1��

4�

�3� �j I2��

20�90��V I3��



��������� Applying KVL to Mesh 1, �3 (

� 1

� j 3 ( I1� � I 2� ) � 0 ) I1� � j 3

� 2

0

…(i)

3.5�Thevenin’s Theorem�����

Applying KVL to Mesh 2, �

j3 (





) � 2I � � jj33 ( � j3



� 2



� 3)

�0

� 3

�0



� 2I � j 3

…(ii)

Applying KVL to Mesh 3, j3 (

� 3

� 2)

j3

4 I3� � 20 �90� � 0 �

(4

� 3

j 3) j3

20 �90�

…(iii)

Writing Eqs (i), (ii) and (iii) in matrix form, � 0 � � I1 � � 0 �3 j 3 � j 3 � � � � � j3 � 2 j 3 � � I 2� � � � 0 � � � � � j 3 4 j 3� � I3 � � �20 �90�� � 0 � �

By Cramer’s rule, 3 j3 0 0 j3 � j3 0 0 20 90� 4 � j 3 �180 � j180 � I �2 � 3 j3 � j3 0 � j3 � j3 2 j3 4 j3 0 Step III By superposition theorem, I2 (9

I�2 � I �2 �

(9

112 12) 2)V � ( �180 � 180) �0 �

12)V � ( �180 � 180) � 0 12) ( )V � 180 � j180 9 � � 8.13 13� V Vs � 16.97

������������������������ Thevenin’s theorem gives us a method for simplifying a network. In Thevenin’s theorem, any linear network can be replaced by a voltage source VTh in series with an impedance ZTh.

��������������Obtain Thevenin’s equivalent network for the terminals A and B in Fig. 3.58. 3�

�4� �j

j5

�j �4� A



4�



j6 �

50�0� V

B

���������

�����Circuit Theory and Networks—Analysis and Synthesis Solution Step I

Calculation of VTh (Fig. 3.59) �4� �j

3�

j5

�j �4� �

A

4�



VTh

50�0� V �

j6 �

I



B

��������� Applying KVL to the mesh, 50 �0° − (3 − j4) I − (4 � j6) I � 0 I�

50 �0� � 6 87� � 15 95� A (3 � 4) � (4 ( 4 � 6)

VTh � (4 � j6) I � (4 � j6) (6.87 �−15.95°) � 49.5 �40.35° V Step II

Calculation of ZTh (Fig. 3.60) ZTh � (

� 3�

)�

( � )( � ) � 4.83� � 1.13� � ( � )�( � )

�4� �j

j5

�j �4� A

4� ZTh j6 � B

��������� Step III

Thevenin’s Equivalent Network (Fig. 3.61) 4.83 ��1.13� � A � 49.5 �40.35� V

� B

���������

��������������Find Thevenin’s equivalent network for Fig. 3.62. 5�

j5 �

�2� �j

A � 3�

10 �30� V

5�

� B

���������

3.5�Thevenin’s Theorem�����

Solution Step I

Calculation of VTh (Fig. 3.63) j5 �

�2� �j

5�



A

� 3�

10 �30��V �

I1

5� I2

VTh �

B

��������� Applying KVL to Mesh 1, 10 �30° − (5 − j2) I1 − 3(I 1 − I2) � 0 (8 − j2) I1 − 3I2 � 10 �30° Applying KVL to Mesh 2, −3 (I2 − I1) − j5 I2 − 5 I2 � 0 −3I1 � (8 � j5) I2 � 0 Writing Eqs (i) and (ii) in matrix form;

…(i)

…(ii)

j2 �3 � � I1 � �10�30�� � � �� �3 8 � j 5�� ��I 2 �� �� 0 �� By Cramer’s rule, j 2 10 �30� �3 0 � 0.433 33�9 7� A I2 � 8 j2 �3 �3 8 � j5 8

VTh Step II

5I 2

5(0.433�9.7�))

Calculation of ZTh (Fig. 3.64)

�2� �j

� � ( � )3 � � ZTh � � � � � j 5� � 5 � � 5 j 2 3 � �� � � [ .94 � j 0. � j ] � 5 � ( .94 j 4.735) � 5 (1.94 j 4.735)5 � � 3 04 33 4� � 6 94 j 4.735

5�

3�

5�

ZTh B

���������

3.04 �33.4� � A � � B

���������

j5 � A

Step III Thevenin’s equivalent Network (Fig. 3.65)

2.16 �9.7��V

.16 �9.7� V

�����Circuit Theory and Networks—Analysis and Synthesis

��������������Obtain Thevenin’s equivalent network for Fig. 3.66. 4�





2�

10 �0� V A



5 �90� V �

j6 �

�4� �j B

��������� Solution Step I Calculation of VTh 4�

j6 �





2�

10 �0� V �



A

5 �90� V � �

I



VTh �4� �j



B

��������� Applying KVL to the mesh, ( ) 5�90� � 0 I�

5�90� � 1 77�45� A 2 � j2

VTh � (−j4) I � 5 �90° − 10 � 0° � (4 �−90°) (1.77 �45°) � 5 �90° − 10 �0° � 18 �146.31° V Step II

Calculation of ZTh (Fig. 3.67) 4� A 2� �4� �j

j6 �

ZTh

B

��������� ZTh � 4 �

( 2 � 6)(� ( 4) � 11.3 3� �� � 44 93� � 2 � j2

Step III Thevenin’s Equivalent Network 11.3 ��44.93� � A � 18 �146.31��V

� B

���������

3.5�Thevenin’s Theorem�����

�������������

Obtain Thevenin’s equivalent network for Fig. 3.70.

j15 �

10 �0� A

2� A �5� �j

3�

B

���������

I

Solution j15 �

10 �0� A

Step I Calculation of VTh (Fig. 3.71) By current division rule,

2�



I�

VTh

�5� �j

3�

( � �)( ) � 13.42�26.57� A 5 � j 5 � j15



VTh � (−j5) I � (5 �−90°) (13.42 �26.57°) � 67.08 �−63.43° V

���������

Step II Calculation of ZTh (Fig. 3.72) ZTh �

j 15 �

( � )( )( � ) � 7.07� � 81.86� � � j 5 � 5 � j15

2� A

Step III Thevenin’s Equivalent Network

3�

7.07 ��81.86� �

�5� �j

A

67.08 ��63.43� V

���������

� B

���������

��������������Obtain Thevenin’s equivalent network for Fig. 3.74. 50 �

21 � � A

20 �0� V �

12 � j 24 �

���������

ZTh B



B 30 � j60 �

A

B

�����Circuit Theory and Networks—Analysis and Synthesis Solution Step I

Calculation of VTh (Fig. 3.75) I2

I1

50 �

21 � �

A

20 �0� V �

VTh �



B

12 �

30 �

j 24 �

j60 �

��������� 20�0� � 0 49 � 36 02� A 21 � 12 � j 24 20 �0� � 0 2� � 36.86� A I2 � 80 � j 60 I1 �

VTh � (12 � j24) I1 − (30 � j60) I 2 � (26.83 �63.43°) (0.49 �−36.02°) − (67.08 �63.43°) (0.2 �−36.86°) � 0.33 �171.12° V Step II

Calculation of ZTh (Fig. 3.76) 21

50 B

A 12

j 24

30

j60

���������

ZTh �

21(12 � 24) 50(30 � 60) � � 47.4 �2 �26 8� � 33 � j 24 80 � j 60

Step III Thevenin’s Equivalent Network 47.4 �26.8��� A � 0.33 �171.12� V

� B

���������

3.5�Thevenin’s Theorem�����

�������������

Find Thevenin’s equivalent network across terminals A and B for Fig. 3.78. A 1�

5�

2 �45� A



j2 �

10 �90� V



B

��������� Solution Step I Calculation of VTh (Fig. 3.79) �

1�

5�

2 �45� A

VTh



j2 �

10 �90� V �



��������� Applying KCL at the node, VTh V � 10�90� � Th � 2 45� 1 j2 5 � 1 1� �� 1 j 2 � 5 �� VTh � 2 45� 2�90� ( .57

45�)) VTh

Step II

.7�67.5� � 6.49�112.5� V

Calculation of ZTh (Fig. 3.80) A 1� 5�

ZTh

j2 � B

��������� ZTh �

A

5(1 � 2) � 1 77�45� � 5 � 1� j2

B

�����Circuit Theory and Networks—Analysis and Synthesis Step III Thevenin’s Equivalent Network (Fig. 3.81) 1.77�45� � A � 6.49 �112.5� V

� B

���������

�������������

Find the current through the (

) � impedance in the network of Fig. 3.82.

5� �

3�

2�

20 �0��V �

j2 �

5�

20 �0� A �2� �j

��������� Solution Step I Calculation of VTh (Fig. 3.83) 5�

V1

3� �

� VTh B �

2�

A

20 �0��V �

20 �0� A �2� �j I2

��������� Applying KCL at the node, V1 20 �0� V1 � � 20 �0� 5 2 � j2 �1 1 � �� 5 � 2 j 2 �� V1 � 20 �0� 4 �0� 0 51�299 05� V1 24 0� V1 � 47.06 .06 �� 29.05� V VTh � V1 � 47.06�� 29.05� V

3.5�Thevenin’s Theorem�����

Step II

Calculation of ZTh (Fig. 3.84) 5� 3� ZTh

2�

A B

�2� �j

��������� ZTh � 3 �

5 ( 2 � 2) � 4 79� �� � 11 35� � 5 � 2 � j2

Step III Calculation of IL (Fig. 3.85) 4.79��11.35� � A

5�

� 47.06 ��29.05� V

IL



j2 � B

��������� IL �

�������������

47.06 � �� 29.05� � 4 73� �� � 39 96� A 4 79� �� 11 35� � 5 � j 2

Find the current through the 5 � resistor in the network of Fig. 3.86. j5 �

6 �0� A

5�

4�

�j �2�

4 �0� A

��������� Solution Step I Calculation of VTh (Fig. 3.87) j5 �

V1

� 6 �0��A

A Vth B

4�

���������

V2

� j2 �

4 �0� A

�����Circuit Theory and Networks—Analysis and Synthesis Applying KCL at Node 1, V1 V1 � V2 � � 6 0� � 0 4 j5 �1 1 � V1 �� 4 � j 5� 5 �� ( .25

j 0. )

j 0.2

1

1 V2 � �6 6 0� j5 6 �0�

2

…(i)

Applying KCL at Node 2, V2

V1 j5

� 1� � 1 V1 �� � j5� 5 �� � j5



V2 � 4 0� j2

1� V2 � 4 0� j2�

j 0 2 V j0 j 0 3 V2 � 4 �0� Writing Eqs (i) and (ii) in matrix form, �0 25 j 0.2 �� j0 2

…(ii)

j 0.2� � V1 � � �6 �0�� � j 0 3�� �� V2 �� �� 4 �0� ��

By Cramer’s rule, �66 0� j 0 2 4 � 0� j 0 3 � 20.8�� 126.87� V V1 � 0 25 j 0 2 j 0.2 j0 2 j0 3 �� 126.87� V VThh � V1 � 20.88� Step II

Calculation of ZTh (Fig. 3.88) j5 �

A ZTh

4�

�2� �j

B

��������� ZTh �

4( � 2 � 5) � 2.4 �53 �53.13� � 4 � j 2 � j 5)

Step III Calculation of IL (Fig. 3.89)

2.4 �53.13��� A

IL �

20.8� �� 126.87� � 3 1� �� 143.47� � 2.4 �53. 3� � 5

� 20.8 ��126.87� V

5� �

IL B

���������

3.5�Thevenin’s Theorem�����

�������������

In the network of Fig. 3.90, find the current through the 10 � resistor. 5 �30��V �



2�



10 �0� V

1�

10 �

� �2� �j

5�

��������� Solution 5 �30��V

Step I Calculation of VTh (Fig. 3.91) Applying KVL to the mesh, j2

1I � 10 �0� 5I 5I � 0 ( j2 6)I � 10 �0� I � 1 58� � 161.57� A Writing VTh equation,



2�







10 �0� V

1�

VTh

� I

�2� �j

5� �

5I 10 0� 5�30� 0 VTh � 0 5(1.58� � 161.57�) � 10 �0� 5�30� � VTh � 0 VTh � 5 32 3 � � 110.06� V Step II

A

B

���������

Calculation of ZTh (Fig. 3.92)

2� A

ZTh � 2 �

5(1 � 2) � 3 488� � 21 04� � 5 � 1 � j2

1� 5�

ZTh

�2� �j

Step III Calculation of IL (Fig. 3.93)

B

3.48 ��21.04� � A � 5.32 ��110.06� V

10 � �

IL B

��������� IL �

5 32� �� 110.06� � 0 4� �� 104.67� A 3 48� �� 21 04� � 10

���������

�����Circuit Theory and Networks—Analysis and Synthesis

�������������

) � impedance in the network of Fig. 3.94.

Find the current through ( 2�

j5 �

3�

4�

� 100 �0� V

�5� �j

� 50 �90� V





j6 �

��������� Solution Step I Calculation of VTh (Fig. 3.95) 2�

j5 �

� 100 �0� V

� I

3�

�5� �j

� A VTh � B

� 50 �90� V �

��������� Applying KVL to the mesh, 100 �0� 2I � j 5

3I � j 5I � 50 90� 0 I � 22.36 �� 26.57� A

Writing VTh equation, VTh

3I � j 5I � 50 �90� � 0

VTh � (3 � j 5)( 22.36 �� 26.57�) � 50 �90� � 0 VTh � 80 661 1 Step II

82 88� V

Calculation of ZTh (Fig. 3.96) 2�

j5 �

3�

�5� �j

A ZTh

B

��������� ZTh �

( � )( � ) � 6.28�9.16� � 2 � j5 � 3 � j5

3.5�Thevenin’s Theorem�����

Step III Calculation of IL (Fig. 3.97) 6.28�9.16� � A

4�

� 80.61 ��82.88��V

IL



j6 � B

��������� IL �

�������������

80.61� � 82.88� � 6 52 5 � � 117.34� A 6 28 28� �9.16� � 4 � j 6

Obtain Thevenin’s equivalent network across terminals A and B in Fig. 3.98. I

4�

j2 �

A

� j1 �

� 10 �0� V

� �



2I B

��������� Solution

I

Step I Calculation of VTh (Fig. 3.99) Applying KVL to the mesh, 10 �0� 4I � j1

2I � 0 I � 1 64 9.46� A

4�

� 10 �0��V

� �



Applying KVL to Mesh 2, 2I j1(I � I ) j 2I 2 � 0 2I1

j1I � j1I1

j2 I 2 � 0

A

VTh

2I



B

���������

10 �0� 4I � 0 VTh 0 10 �0� 4(1 (1.64 9.46�) � VTh � 0 VTh � 3 69 � 17� V Calculation of IN (Fig. 3.100) From Fig. 3.100, I I1 Applying KVL to Mesh 1, � 10 �0� 4I1 � j1( 1 2 ) 2I � 0 100 �0�V � 10 �0� 4I1 � j1 I1 j1I 1I 2 2 1 0 ( ) I1 � j1 2 10 0� …(i)



� j1 �

Writing VTh equation,

Step II

j2 �

I

4�

j2 �

A

� j1 �

I1

� �

2I

IN I2

B

����������

�����Circuit Theory and Networks—Analysis and Synthesis ( ) 1 j1 2 0 )I Writing Eqs (i) and (ii) in matrix form, �6 j1 j1 � � I1 � �10 �0�� �� 2 j1 � j1�� ��I 2 �� � �� 0 �� By Cramer’s rule, 6 j1 10�0� 2 j1 0 � 2 71 7 �� 102.53� A I2 � 6 j1 j1 2 j1 � j1

…(ii)

I 2 � 2 71�� 102 1 .53� A

IN Step III Calculation of ZTh ZTh �

3 69� � 17� VTh � � 1 36 �85 53� � 2 71� � 102.53� IN

Thevenin’s Equivalent Network (Fig. 3.101)

Step IV

1.36 �85.53� � A � 3.69 ��17� V � B

����������

�������������

Find Thevenin’s equivalent network across terminals A and B for Fig. 3.102. 2�

j4 � �



A

5 �0��V �

0.2 Vx

Vx

1� �

B

���������� Solution Step I

2�

Calculation of VTh (Fig. 3.103)



� �

… (i)

1�

0.2 Vx

Vx � VTh

I

Writing VTh equation, � I � � � � 0 � Vx � 0

A

5 �0��V

From Fig. 3.103, I � �0 2 Vx

j4 �



����������

B

3.6�Norton’s Theorem�����

0 2Vx

5 0� Vx � 0 Vx � 6 25 0� V VTh Vx � 6 25�0� V

Step II

Calculation of IN (Fig. 3.104) 2�

j4 � �



A

5 �0� V �

Vx

0.2 Vx

IN

1� �

B

���������� From Fig. 3.104, Vx � 0

2�

A

The dependent source depends on the controlling variable Vx. When Vx � 0, the dependent source vanishes, i.e. 0 2 Vx � 0 as shown in Fig. 3.105. IN �

j4 �

� 5 �0� V �

IN

1�

5�0� � 1�� � 53 13� A 1� 2 � j4

B

����������

Step III Calculation of ZTh ZTh � Step IV

6 25�0� VTh � � 6 225�53 13� � 1�� 53 13� IN

Thevenin’s Equivalent Network (Fig. 3.106) 6.25 �53.13� � A � 6.25 �0��V � B

����������

���������������������� Norton’s theorem states that any linear network can be replaced by a current source IN parallel with an impedance ZN where IN is the current flowing through the short-circuited path placed across the terminals.

�����Circuit Theory and Networks—Analysis and Synthesis

��������������Obtain Norton’s equivalent network between terminals A and B as shown in Fig. 3.107. 3�

j4 � A



4�

25 �0� V �

�j5 � B

���������� Solution Step I

Calculation of IN (Fig. 3.108) When a short circuit is placed across (4 − j4) � impedance, it gets shorted as shown in Fig. 3.109. 3�

3�

j4 � A



4�

25 �0��V

j4 �



A



IN

�j5 �

25 �0��V

B IN

����������

B

���� IN Step II

3�

����������

j4 � A

25�0� � � 5�� 53 13� A 3 � j4

4�

Calculation of ZN (Fig. 3.110) ZN �

( � j )( 3 � j4 4

j ) � 4.53�9.92� � j5

B

����������

Step III Norton’s Equivalent Network A

5 ��53.13� A

4.53 �9.92� � B

����������

�������������

ZN

�j5 �

Obtain Norton’s equivalent network at the terminals A and B in Fig. 3.112. 5� A 1�

4�

j2 �

j4 �

10 �30� A

B

����������

3.6�Norton’s Theorem�����

Solution Step I

Calculation of IN (Fig. 3.113) 5�

1�

4�

j2 �

j4 �

A

10 �30��A

IN

B

���������� By series-parallel reduction technique (Fig. 3.114) 5�

10 �30� A

A

IN

1.62 �58.24� �

B

���������� �58 24� � 62� � 1 62 I N � �10 �30�� � � � 2 69�75� A � 1 62 �58 24� � 5 � 62� Step II

Calculation of ZN (Fig. 3.115) 5� A 1�

4�

j2 �

j4 �

ZN

B

���������� ZN � 5 �

(1 � j 2) ( 4 1 � j2 4

j 4) � 6 01 0 �13 24� � j4

�����Circuit Theory and Networks—Analysis and Synthesis Step III Norton’s Equivalent Network (Fig. 3.116) A

2.69 �75� A

6.01 �13.24� �

B

����������

�������������

Find Norton’s equivalent network across terminals A and B in Fig. 3.117. A 10 �

j4 � 4 �45� A

� 3�

25 �90� V � B

���������� Solution Step I

Calculation of IN (Fig. 3.118) A

10 �

j4 �

IN

4 �45� A

� 3�

25 �90� V � B

���������� When a short circuit is placed across the (

) � impedance, it gets shorted as shown in Fig. 3.119. A 10 �

4 �45��A

IN

� 25 �90��V � B

����������

3.6�Norton’s Theorem�����

By source transformation, the network is redrawn as shown in Fig. 3.120. A

A

4 �45� A

2.5 �90��A

10 �

IN

4 �45� A

2.5 �90� A

B

B (a)

(b)

���������� I N � 4 �45� � 2.5 5�90� Step II

�����62 04� A

Calculation of ZN (Fig. 3.121) A j4 � ZN

10 � 3�

B

����������� ZN �

10 (3 � j 4) � 3 68�36 03� � 10 � 3 � j 4

Step III Norton’s Equivalent Network (Fig. 3.122) A

6.03 �62.04� A

3.68 �36.03���

B

����������

�������������

Obtain the Norton’s equivalent network for Fig. 3.123.

10 �0� A

IN

5�

j3 � j5 � A

j2 �

�5� �j B

����������

�����Circuit Theory and Networks—Analysis and Synthesis Solution Step I

Calculation of IN (Fig. 3.124) 5�

10 �0��A

j3 � j5 � A �5� �j

j2 �

IN B

���������� By source transformation, the network can be redrawn as shown in Fig. 3.125. Writing KVL equations in matrix form, �5 �� j 5



j 5� � I1 � �50�0�� � 0 �� ��I 2 �� �� 0 ��

50 �0� V

A I1

5 50 �0� j5 0 � 10 �� 90� A I2 � 5 j5 j5 0

Step II

j5 �

5�

By Cramer’s rule,

IN

j3 �



IN

�5� �j

j2 �

I2 B

I 2 � 10 �� 90� A

����������

Calculation of ZN (Fig. 3.126) 5�

j3 � j5 �

j2 �

ZN

�5� �j

���������� Z N � j5 �

(5 � j 5) ( jj5) �5� 5 � j5 j5

Step III Norton’s Equivalent Network (Fig. 3.127) A

10 ��90��A

5� B

����������

3.6�Norton’s Theorem�����

�������������

Obtain the Norton’s equivalent network for Fig. 3.128. 5�

10 � 10 �45��V



�2� �j



A 5�

B

10 �

j2 �

���������� Solution Step I

� 15 � j 2 � �10 � j 2 � � �5



15 � j 2 10 45� 5 �10 � j 2 0 0 �5 0 15 � j 2 � 1 41.28� A I2 � �5 15 � j 2 10 j 2 �10 � j 2 15 j 2 0 �5 0 15 � j 2

Step II

IN A 5 � I3

I1

B 10 � j2 �

By Cramer’s rule,

IN

� 2 � I2 �j

� 10 �45��V

10 j 2 �5 � � I1 � �10 � 5�� 15 j 2 0 � �I 2 � � � 0 � �� � � � 0 15 � j 2� � I 3 � � 0 �

15 � j 2 �10 � j 2 �5 I3 � 15 � j 2 �10 � j 2 �5

5�

10 �

Calculation of IN (Fig. 3.129) Writing KVL equations in matrix form,

����������

10 j 2 10� 45� 15 j 2 0 0 0 � 0 49 37 41� A �5 10 j 2 15 � j 2 0 0 15 � j 2

I 3 � I 2 � 0 49�37 41 � 1�41.28 28� � 0 51�� 135� A

Calculation of ZN (Fig. 3.130) 5�

10 � �2� �j ZN A 5�

�j2 �

10 �

5�

B

A

B 10 �

5�

10 �

j2 � (a)

(b)

����������

j2 �

�����Circuit Theory and Networks—Analysis and Synthesis ZN �

5(10 � j 2) 5(10 � j 2) � � 6 72 � 5 � 10 � j 2 5 � 10 � j 2

Step III Norton’s Equivalent Network (Fig. 3.131) A 0.51 ��135� A

6.72 � B

����������

�������������

Find the current through the 8 � resistor in the Network of Fig. 3.132. 5�

10 �

8�



20 �0� V

5 �0� A �

j4 �

���������� Solution Step I

Calculation of IN (Fig. 3.133) 5� A

10 �



IN

20 �0��V

5 �0� A



j4 �

B

���������� When a short circuit is placed across the (

) � impedance, it gets shorted as shown in Fig. 3.134.

5� A �

IN

20 �0��V �

B

����������

5 �0� A

3.6�Norton’s Theorem�����

By source transformation, the network is redrawn as shown in Fig. 3.135. A IN

5�

4 �0� A

5 �0��A

B

���������� I N � 4 �0� � 5�0� � 9�0� A Step II

Calculation of ZN (Fig. 3.136) 5� A 10 � ZN j4 �

B

���������� ZN �

5(10 � j 4) � 3 47�6.87� � 5 � 10 � j 4

A IL 9 �0� A

Step III Calculation of IL (Fig. 3.137)

3.47 �6.87 �

8� B

9 � 0� IL � � 0 79� �� � 2.08� A 3 47 47� �6.87� � 8

�������������

����������

Obtain Norton’s equivalent network across the terminals A and B in Fig. 3.138. 5I

I

100 �

− 5� −j

A



10 �0� V

j 10 � �

B

����������

�����Circuit Theory and Networks—Analysis and Synthesis Solution Step I

Calculation of VTh (Fig. 3.139) 5I

100 �

I



−5� −j �







A



10 �0� V

j10��

VTh





B

���������� I�

10 �0� � 0 1�� � 5 71� A 100 � j10

Writing VTh equation, 10 �00� �� � j )(5 ) VTh 0 10 �0� � 100 (0.1� �� 5.71� 71 ) ( j 5)(5) (0.1 5.71 ) � VTh � 0 VTh � 3 5�85.1� V Step II

Calculation of IN (Fig. 3.140) 5I

100��

I

−5� −j �

A





10 �0��V

j10 �

IN



B

���������� By source transformation, the network is redrawn as shown in Fig. 3.141. − 25 I −j

− 5�� −j

100��

I

� �

A



j10 �

10 �0� V �

I1

IN I2 B

���������� From Fig. 3.141, I � I1

…(i)

�����Maximum Power Transfer Theorem�����

Applying KVL to Mesh 1, 10 �0� � 100 I1

j100 ( I1 � I 2 ) � 0

(100 � j10 j10) I1 � j10 I 2 � 10 �0�

…(ii)

Applying KVL to Mesh 2, � j10 ( � ) � j 5 2 � j 25 � 0 � j10 I � j100 I1 � j 5 � j 225 5 1�0 j 35I � jj5 5I 2 � 0 …(iii)

Writing Eqs (ii) and (iii) in matrix form, �100 � j10 � j10 � � I1 � �10 �0�� � �� j 35 � j 5 �� �� I 2 �� �� 0 �� By Cramer’s rule, 100 � j10 10�0� j 35 0 � 0 6 �30.96� A I2 � 100 � j10 � j10 j 35 � j5 I 2 � 0 6 �30.96� �

IN Step III Calculation of ZN ZN � Step IV

3 5�85.1� VTh � � 5 83�54.14� � 0 6 �30.96� IN

Norton’s Equivalent Network (Fig. 3.142) A

0.6 �30.96� A

5.83�54.14� �

B

����������

Zs

������������������������������������� This theorem is used to determine the value of load impedance for which the source will transfer maximum power. Consider a simple network as shown in Fig. 3.143. There are three possible cases for load impedance ZL.

+ Vs

ZL −

����������

�����Circuit Theory and Networks—Analysis and Synthesis Case (i)

When the load impedance is variable resistance (Fig. 3.144) IL �

V

V



ZS = RS + j X S s

L

V

IL �

+ Vs

s

L=

The power delivered to the load is 2

Vs



P

RL

L

L

�������������������������������� s

For power to be maximum, dP dRL Vs

2

0

{( s

2

R

0 2 L



R

�0 � X2

R

Hence, load resistance RL should be equal to the magnitude of the source impedance for maximum power transfer. Case (ii) When the load impedance is a complex impedance with variable resistance and variable reactance (Fig. 3.145) IL �

Vs

S=

RS + j X S

+ Vs

L=

RL + j X L



IL �

Vs �

����������������������������������

The power delivered to the load is PL

L

VS

R

2 L

For maximum value of PL, denominator of the equation should be small, ie. P �

VS (

2 L

)2

.

�����Maximum Power Transfer Theorem�����

Differentiating the above equation w.r.t. RL and equating to zero, dP

2

� Vs



�2 R (



L

s

) �2

R

2

L

� RL � � �

0 �

( �2

2

0 0

Hence, load resistance RL should be equal to source resistance RL and load reactance XL should be equal to negative value of source reactance for maximum power transfer. �



s

s

i.e. load impedance should be a complex conjugate of the source impedance. Case (iii) When the load impedance is a complex impedance with variable resistance and fixed reactance (Fig. 3.146) S = RS + jX S IL

Vs + Vs

Vs

IL

���������������������������������

The power delivered to the load is Vs



P

ZL = RL + jX L

2 L

For maximum power, dP

0

L

Vs

�2 R

2

s

� RL �



{(

�0

�2 R



0 �0

� 2

R

0 )2

�(

R

� �

L



XL





L

�����Circuit Theory and Networks—Analysis and Synthesis Hence, load resistance RL should be equal to the magnitude of the impedance Z s � jX L for maximum power transfer.

s

� jX L , i.e.

������������� For maximum power transfer, find the value of ZL in the network of Fig. 3.147 if (i) ZL is an impedance, and (ii) ZL is pure resistance. 6�

�j 8 �

� Vs

ZL



���������� Zs � ( � j ) �

Solution (i)

If ZL is an impedance

(ii)

For maximum power transfer, Z L

Z*s � ( � j ) �

If ZL is a resistance For maximum power transfer, Z L

Z S � 6 � j8 � 10 �

�������������

For the maximum power transfer, find the value of ZL in the network of Fig. 3.148

for the following cases: (i) ZL is variable resistance, (ii) ZL is complex impedance, with variable resistance and variable reactance, and (iii) ZL is complex impedance with variable resistance and fixed reactance of j5 �. A 3�

2� 10 A

5 �0� V

j5 �

B

���������� Solution Thevenin’s impedance can be calculated by replacing voltage source by a short circuit and current source by an open circuit. A 2� 3�

ZTh

j5 � B

����������

�����Maximum Power Transfer Theorem�����

ZTh �

3 ( 2 � 5) � ( 2.1 � j 0.9) � 3 � 2 � j5

For maximum power transfer, value of ZL will be, (i)

ZL is variable resistance ZL

(ii)

ZTh � 2.1 � j 0 9 � 2.28 �

ZL is complex impedance with variable resistance and variable reactance ZL

(iii)

Z*Th � ( .1 � j 0.. ) �

ZL is complex impedance with variable resistance and fixed reactance of j5 � ZL

ZTh � j 5

21

j 0.9 � j 5 � 6 26 �

��������������Find the impedance ZL so that maximum power can be transferred to it in the network of Fig. 3.150. Find maximum power. 3�

3�

� j3 �

5 �0� V

ZL

� j3 �



���������� Solution Step I

Calculation of VTh (Fig. 3.151) IT 3 �

3� �

I

� j3 �

5 �0��V

� j3 �

A

VTh

� �

���������� ZT � 3 � IT �

j 3(3 j 3) �67 71�26 57� � 3 � j3 j3

5�0� � 0 75�� 26.57� A 6 71� � 6 57�

By current division rule, j3 � 0 75�63 43� A 3 � j3 � j3 75�633 43�) 2.24 �� 26.57� V � ( � 3)(0 75

I � 0 75� �� � 26 57� � VTTh

�����Circuit Theory and Networks—Analysis and Synthesis Calculation of ZTh (Fig. 3.152) ZTh � [(3 || j3) � 3] || (−j3) � 3 �−53.12° � � (1.8 − j2.4) �

Step II

3�

3�

A

� j3 �

j3 �

B

Step III Calculation of ZL For maximum power transfer, the load impedance should be a complex conjugate of the source impedance.

����������

ZL � (1.8 � j2.4) � Calculation of Pmax (Fig. 3.153)

Step IV

(1.8 � j 2.4) �

A � (1.8 � j 2.4) �

2.24 ��26.57� V �

B

���������� Pmax �

|

Th T

|2

4 RL



| .24 |2 � 0.7 W 4 � 1.8

��������������Find the value of ZL for maximum power transfer in the network shown and find maximum power. j10 �

5� 100 �0� V � 7�

ZL

� � 20 � �j

���������� Solution Step I

I1

Calculation of VTh (Fig. 3.155) I1 �

100 �0� � 8 94 � 63 43� A 5 j10

I2 �

100 �0� � 4 72 70 7� A 7 j 20

I2

5�

7�

� 100 �0� V

A



� VTh � B

j10 �

����������

� 20 � �j

�����Maximum Power Transfer Theorem�����

VTh Step II

VA � VB � ( .94� � 63.. �) ( j ) � (

2�70 �) ( � j

) � 71.76�97..3 3� V

Calculation of ZTh (Fig. 3.156) 5�

7�

A

B j10 �

� 20 � �j

����������

ZTh �

Step III

5( 10) 7( � 20) 50 �90� 140 �� 90� � ( .23 � j 0.. ) � � � � 5 � j10 7 � j 20 11.18�63. 3� 21.19�� 70.7�

For maximum power transfer, the load impedance should be complex conjugate of the source impedance. ZL � (10.23 � j0.18) �

Step IV

Calculation of Pmax (Fig. 3.157) (10.23 � j0 .18)�� A � (10.23 � j 0 .18)��

71.76 �97.3� V �

B

����������

Pmax �

|

Th T

4 RL

|2



| .76 |2 � 125.84 W 4 � 10.23

��������������Find the value of load impedance ZL so that maximum power can be transferred to it in the network of Fig. 3.158. Find maximum power. 3� 2� �

50 �45� V



ZL j10 �

����������

�����Circuit Theory and Networks—Analysis and Synthesis Solution Step I

Calculation of VTh (Fig. 3.159) 3� 2� �

50 �45� V

� A VTh T � B



j10 �

I

���������� I� VTh Step II

50 � 5� � 4.477� �� � 18 43� A 3 � 2 � j10 1 .43 ) � 45.6 6 �60 0 26� V ( 2 j10) I � ( 2 � j10) ( 4.47�� 18

Calculation of ZTh (Fig. 3.160) 3� 2� A B

ZTh T j10 �

���������� ZTh �

3( 2 � 10) � ( 2.64 � j 0.72) � 3 � 2 � j10

Step III Calculation of ZL For maximum power transfer, the load impedance should be complex conjugate of the source impedance. ZL � (2.64 − j0.72) � Step IV

Calculation of Pmax (Fig. 3.161) (2.64 ��j 0.72) � A � (2.64 � j 0.72)��

45.6 �60.26� V � B

���������� Pmax �

|

Th T

4 RL

|2



| .6 |2 � 196.91 W 4 � 2.64

�����Maximum Power Transfer Theorem�����

��������������Determine the load ZL required to be connected in the network of Fig. 3.162 for maximum power transfer. Determine the maximum power drawn. j1 �

2�

4 �0� A

4�

ZL

���������� Solution Step I

Calculation of VTh (Fig. 3.163) j1 � I1



I2 2�

4 �0� A

4�

A

VTh �

B

���������� I 2 � 4 0� � VTh

4I2

2 6

j1

� 1.3 315�� 9 46� A 315

4 (1.315�� 9.46�)

5.26 �� 9 46� V j1 �

Step II

A

Calculation of ZTh (Fig. 3.164) ZTh

2�

4( 2 � 1) � � 1.477�17 1� � (1.41 � j 0.43) � 4 � 2 � j1

ZL � (1.41 − j0.43) � Calculation of Pmax (Fig. 3.165) (1.41 ��j 0.43) � A � (1.41 � j 0.43) �

5.26 ��9.46��V �

B

���������� Pmax �

|

Th T

4 RL

|2



ZTh B

Step III Calculation of ZL For maximum power transfer, the load impedance should be the complex conjugate of the source impedance.

Step IV

4�

| .26 |2 � 4.91 W 4 � 1.41

����������

�����Circuit Theory and Networks—Analysis and Synthesis

�������� ����� In the network shown in Fig. 3.166, find the value of ZL for which the power transferred will be maximum. Also find maximum power. 5 �60� �

10 ��30� �



10 �0� V



ZL



5 �90� V �

���������� Solution Step I

Calculation of VTh (Fig. 3.167) 5 �60� � �

10 ��30� � �

A � VTh B �

10 �0� V �

5 �90� V �

I

���������� Applying KVL to the mesh, 10 �00� �� 60�) � � 30�) � 90� � � ������� 26.57� � (11.188�� 3 43�) 0 I � 1� �� 23 14� A Writing VTh equation, 10�00� �� 60�� VTh 0 10 �00� �� 60�) (1� � 23.14�) � VTh � 0 VTh � 6 71� �26.56� V Step II

Calculation of ZTh (Fig. 3.168) 5 �60� �

10 ��30� �

A ZTh B

���������� ZTh �

(5�60�������30�� � 4.47�33.43� � � ( .73 � j 2.46) � 5�60� � ��30�

�����Maximum Power Transfer Theorem�����

Step III Calculation of ZL For maximum power transfer, the load impedance should be the complex conjugate of the source impedance. Z*Th � ( .73 � j 2.

ZL

)�

Calculation of Pmax (Fig. 3.169)

Step IV

(3.73 � j2.46) �

A



(3.73 � j2.46) �

6.71 ��26.56� V �

B

����������

Pmax �

VTh 4 RL

2



( .71) 2 � 3.02 W 4 � ����

������������� In the network shown in Fig. 3.170, find the value of ZL so that power transfer from the source is maximum. Also find maximum power. j9 �

j9 � �

j9 �

10 �0� V � 8�

ZL

���������� Solution Step I

Calculation of VTh (Fig. 3.171) Applying Star-delta transformation (Fig. 3.172) j9 �

j9 �

Z1

( 9)( 9) � j3 � Z 2 � Z3 � j9 j9 j9

VTh � Voltage drop across (8 + j3)� impedence �( �

10 �0� ) � 8.54 �� � 16.31� V 8 � j3 � j3

j9 � 10 �0� V A

� VTh � B

����������

8�

�����Circuit Theory and Networks—Analysis and Synthesis

Z1 Z2

Z3

10 �0� V

A

� VTh � B

8�

���������� Step II

Calculation of ZTh (Fig. 3.173) j3 �

j3 � j3 �

ZTh 8�

���������� ZTh � j 3 �

j 3(8 � j 3) � 5 51�82.49� � � (0.72 � j 5.46) � j 3 � 8 � j 3)

Step III Calculation of ZL For maximum power transfer, the load impedance should be the complex conjugate of the source impedance. Z L Z�Th � ( .72 � j 5. ) � Step IV

Calculation of Pmax (Fig. 3.174) (0.72 � j5.46) �

A



(0.72 � j5.46) �

8.54 ��16.31� V �

B

���������� Pmax �

VTh 4 RL

2



( .54) 2 � 25.32 W 4 � 0.72

�����Maximum Power Transfer Theorem�����

������������� For the network shown in Fig. 3.175, find the value of ZL that will transfer maximum power from the source. Also find maximum power. 4�

� Vx �

j 10 � ZL



� �

100 �0� V

5Vx



���������� �

Solution Step I

4�

Calculation of VTh (Fig. 3.176) From Fig. 3.176, Vx 4 I Applying KVL to the mesh, 100 �00� 4I � j 10 I � 5Vx 0 100�00� ( 4 � j10) I � 5 ( 4I 4 )�0 100 �0� I� � 3 885 5 24 � j10

Vx �

� 100 �0� V

j 10 �



A

VTh B

� I



� �

5Vx

����������

22 62� �

Writing VTh equation, 100 � 0� 4I � VTh � 0 100 �00� 4(3.85 22.62� 62 ) VTh 0 VTh � 86 �3 95� V Step II

Calculation of IN (Fig. 3.177) From Fig. 3.177, Vx

4�

4 I1

Applying KVL to Mesh 1, 100 �00� �

� Vx �

IN

� 100 �0� V

B

I1

I2



0 1 I1 � 25 A

Applying KVL to Mesh 2, � j10 I 2 � 5Vx � 0 � j10 I 2 � 5( 4 I1 ) � 0 � j10 I 2 � 5(100) � 0 I 2 � 50 �90�A I N � I1 � I 2 � 25 � 50�90�

����������

������ 63.43��

Step III Calculation of ZTh ZTh �

j 10 �

A

VTh 86 �3 95� � � 1 54 �67 38� � � (0.59 � j . IN 55.99� �� 63 43�

)�

� �

5Vx

�����Circuit Theory and Networks—Analysis and Synthesis Calculation of ZL

Step IV

Z�Th � ( .59 � j1.

For maximum power transfer, Z L Step V

)�

Calculation of Pmax (Fig. 3.178) (0.59 � j1.42) �

A � 86 �3.95� V

(0.59 � j1.42) �



B

���������� Pmax �

VTh

2



4 RL

( )2 � 3133.9 W 4 � 0.59

�������������������������� The Reciprocity theorem states that ‘In a linear, bilateral, active, single-source network, the ratio of excitation to response remains same when the positions of excitation and response are interchanged.’

�������������

Find the current through the 6 � resistor and verify the reciprocity theorem. �1� �j

1�

I � 2�

j1 �

5 �0� V �

���������� Solution Case I

Calculation of current I when excitation and response are not interchanged (Fig. 3.180) �1� �j

1�

I �

j1 �

5 �0� V �

2� I2

I1

���������� Applying KVL to Mesh 1, 5�0� �11 1 j1( I1 � I 2 ) � 0 (1 jj1 1) I1 � j1 2 5�0�

…(i)

�����Reciprocity Theorem�����

Applying KVL to Mesh 2, � j1(

� ) � j1I 2 � 2 2 � 0 � j1 1 � 2I 2 � 0

…(ii)

Writing Eqs (i) and (ii) in matrix form, �1 j1 � j1� � I1 � �5 0�� � �� � j1 2 �� ��I 2 �� �� 0 �� By Cramer’s rule, 1 j1 5 0� � j1 0 � 1 39 56 31� � I2 � 1 j1 � j1 � j1 2 I 2 � 1 39 56 31� �

I

Case II Calculation of current I when excitation and response are interchanged (Fig. 3.181) �1� �j

1�

I

2� j1 �



I2

I1

5 �0� V �

���������� Applying KVL to Mesh 1, ��I1 j1( I1 I 2 ) (1 jj1 1) I1 j1I 2

0 0

…(i)

Applying KVL to Mesh 2, � j1(

� ) � j1I 2 � 2 2 � 5�0� � 0 � j1 1 � 2I 2 � �5�0�

Writing Eqs (i) and (ii) in matrix form, �1 j1 � j1� � I1 � � 0 � � �� � j1 2 �� ��I 2 �� �� �5�0��� By Cramer’s rule, 0 j1 �55 0� 2 � 1 38�� 123.69� � I1 � 1 j1 � j1 � j1 2 I

I1 � 1 39 56 31� �

Since the current I is same in both the cases, the reciprocity theorem is verified.

…(ii)

�����Circuit Theory and Networks—Analysis and Synthesis

�������������

In the network of Fig. 3.182, find the voltage Vx and verify the reciprocity theorem.

10 �

j5 �

j5 �

�2� �j

20 �90� A



Vx �

���������� Solution Case I

Calculation of voltage Vx when excitation and response are interchanged. (Fig. 3.183) Ix 10 �

j5 �

j5 �

�2� �j

20 �90� A



Vx �

���������� By current division rule, Ix � ( Vx



)

( �j ) �1 17.46 �77.91 91� � � j ) (j � j )

(

( j )I x � ( �

) ( .46 �77.91��

����� �� � 12.09� V

Case II Calculation of voltage Vx when excitation and response are interchanged (Fig. 3.184) �

Ix 10 �

j5 �

Vx j5 �

�2� �j

20 �90� A



���������� (� j ) � 3.12�� � 38.66�� (� j ) ( j �j ) j ) Ix ( j ) ( .12�� 38.66�� � 34.88�� 12.09�V

I x � ( 20�90�� Vx

(

Since the voltage Vx is same in both the cases, the reciprocity theorem is verified.

�����Reciprocity Theorem�����

�������������

Find I and verify the reciprocity theorem for the network shown in Fig. 3.185. j2 �

4�



1�

2� I

� 3�

4�

10 �45 V

2�

���������� Solution Case I Calculation of I when excitation and response are not interchanged (Fig. 3.186) j2 �

4�

3�

1�

2�

� 3�

4�

10 �45 V

2

1

j � 3

���������� Applying KVL to Mesh 1, 10 �45 � 10

5

…(i)

Applying KVL to Mesh 2, �



� 1� 2 �



0 � 3I 3 � 0





3

…(ii)

Applying KVL to Mesh 3, � �

�0

Writing Eqs (i), (ii) and (iii) in matrix form, �4 � 1



0 � I1 10� 45 � I2 � 0 I3 0

By Cramer’s rule, 4 I3

0 4

0 704

�4 10 45 j1 0 3 0 � 0 704 � .72� � �4 0 1 72��

…(iii)

�����Circuit Theory and Networks—Analysis and Synthesis Case II Calculation of I when excitation and response are interchanged (Fig. 3.187) 2�

j4 �

3�

1�

2� 2� 3�

4� 1

� 10 �45 V

I3

2



���������� Applying KVL to Mesh 1, � �

I � �

� �

…(i)

Applying KVL to Mesh 2, 1

2

3

� 3I 3 Applying KVL to Mesh 3, � �







� 5



0 0

0 � 10 �45

…(ii)

…(iii)

Writing Eqs (i), (ii) and (iii) in matrix form, �4 � 1

� � �

0 � I1 0 � I2 � 0 I3 10 �45

By Cramer’s rule,

I1 �

0 0 10 � 45 �



4 5� � j3 �4 � 1

0 2 � j5 0

0 704 30.72��

� 0 704 �30 72��

Since the current I is same in both the cases, the reciprocity theorem, is verified.

����������������������� Millman’s theorem states that ‘If there are n voltage sources V1, V2. … Vn with internal impedances Z1, Z2, … Zn respectively connected in parallel then these voltage sources can be replaced by a single voltage source Vm and a single series impedance Zm. �

Vm Zm

1

��� ��� Y 1

3.9�Millman’s Theorem�����

�������������

Find the current through the 40 � resistor for the network shown in Fig. 3.188. 10 �

30 � � 20 � �j

j 20 � �







10 �0� V

40 �

20 �0� V

���������� Solution Step I

Calculation of Vm � � � � 1 1 (10 �0 ) � 0 �0 ) � � ( 20 � � � 30 � j 20 � � 10 j 20 � V Y � V2 Y2 � Vm � 1 1 � 18.2� �� 15 95�V 1 1 Y1 Y2 � 30 � j 20 10 � j 20

Step II

Calculation of Zm Zm �

1 1 � � Ym Y1 � Y2

1 � 20.15 1 � 29.74� � 1 1 � 30 � j 20 10 j 20

Step III Calculation of IL (Fig. 3.189) 20.15 �29.74� � 40 �

� 18.2 ��15.95� V

IL



���������� 18.22� �� 15 95� IL � � 0 31 3 � �� 25 81� � 20.15�29.74� � ��

�������������

Find the current I in the network shown in Fig. 3.190. � 20 � �j

j20 �

I

10 �

30 � � 10 �0� V

40 �

� �



����������

20 �0� V

�����Circuit Theory and Networks—Analysis and Synthesis Solution The network is redrawn as shown in Fig. 3.191. I

� 20�� �j

j20�� 40��

30��

10��





10 �0� V

20 �0� V

0V �



���������� Step I

Calculation of Vm � � 1 � 1� 0 � � � ( 20 �0 ) � � � � 10 � j 20 �� 40 V Y � V2 Y2 Vm � 1 1 � 14.86 �� 21.8� V � 1 1 Y1 � Y2 � 40 10 � j 20 2

Step II

Calculation of Zm Zm �

1 1 1 � � � 16.6 61�41.63� � 1 1 Ym Y1 � Y2 � 40 10 � j 20

Step III Calculation of I (Fig. 3.192) I � 20�� �j 16.61 �41.63� �

30 �

10 �0� V









14.86 ��21.8� V

���������� 10 � 14.86 �� 21.8� I= � 0 15 15�136.47� � 30 � j 20 � 16.61�41.63�

�������������

Apply the dual Millman theorem and find the power loss in the (2 + j2) � impedance. 10 A

20 A

4�

5�

2�

j2 �

����������

3.9�Millman’s Theorem�����

Solution Step I

Calculation of Im Im �

I1Z1 � I 2 Z 2 (10) ( 4) � ( 20) (5) � � 15.56 � 4�5 Z1 Z 2

Calculation of Zm

Step II

Zm

Z1 � Z 2 � 4 � 5 � 9 �

Step III Calculation of P (Fig. 3.194) IL 2�

15.56 A

9� j2 �

���������� By current division rule, I L � 15.56 �

9 � 12.53� � 10.3� � 9 � 2 � j2

I 2L RL � (12.53) 2 � 2 � 314 W

P

������������� In the network shown in Fig. 3.195, what load ZL will receive the maximum power. Also find maximum power. 5�

j5 �

3�

� 50 �0� V

�4� �j



ZL





25 �90� V

���������� Solution Step I

Calculation of Vm V Y � V2 Y2 � Vm � 1 1 Y1 Y2

Step II

� 1 � � 1 � (50 �0 ) � � ( 25�90 ) � � 5 � j 5 �� � 3 j 4 �� � 9 81� �� 78 69�V 1 1 � 5 � j5 3 � j 4

Calculation of Zm Zm �

1 1 = � Ym Y1 Y2

1 � 4 39� 39� � 15 26 � � ( 4.23 � j1.15) � 1 1 � 5 � j5 3 j 4

�����Circuit Theory and Networks—Analysis and Synthesis Step III Calculation of ZL For maximum power transfer Step IV

Z*Th � ( .23 � j1. ) �

ZL

Calculation of Pmax (Fig. 3.196) A (4.23 � j1.15) � (4.23 � j1.15) �

� 9.81 ��78.69� V



B

���������� Pmax �

Vm

2



4 RL

( .81) 2 � 5.69 W 4 � ����

Exercises ������������� 3.1

I1 j2 �

Find the current through the 3 � j4 � impedance in Fig. 3.197.

�4� �j

I3

� 50 �45� V

5�

� 3�

20 �

10 �0� V �

j4 �

���������� [11.6 �113.2° A]

j2.5 �

3.4 ����������

In the network of Fig. 3.200, find V2 which results in zero current through the 4 � resistor.

[0] 3.2

5�

In the network of Fig. 3.198, find V0. 2�

2�

j2 �

j5 �



10 �

5�

4�

2�



2� �

� j2 �

50 �0� V �

� �j 2 �

V2 �

� j2 �

10 �0� V

j2 �

j2 � V0



���������� �

���������� [1.56 �128.7° V] 3.3

Find the current I3 in the network of Fig. 3.199.

[26.3��113.2° V]

������������� 3.5

For the network shown in Fig. 3.201, find the voltage VAB.

Exercises����� 5�

3.9

A



5�



j20 �

In the network of Fig. 3.205, find the current through capacitance.

20 �

100 �45� V

j1 � 5�



B 20 �0� V

���������� Find the voltages at nodes 1 and 2 in the network of Fig. 3.202. 10 �

1

���������� [4.86 �80.8° A]

2

2�

������������������

3�



�j10 �

j5 �

50 �0� V

10 �0� V



[75.4��55.2° V] 3.6

�j5 �







j4 �

3.10 Obtain Thevenin’s equivalent network for the network shown in Fig. 3.206. j20 �

���������� [15.95 �49.94° V, 12.9 �55.5° V] 3.7

In the network of Fig. 3.203, find the current in the 10 �30° V source. 5�

�2� �j

� 50 �0� V

50 �

40 � A



B 1000 �

1000 �

� 400 � �j

j5 �

���������� 2�

� 10 �30� V

3�

5�



�2� �j

[0.192 �−43.4° V, 88.7 �11.55° �] 3.11 Obtain Thevenin’s equivalent network for Fig. 3.207. 10 �

����������

j16 �

j5 �

[1.44 �38.8° A]

10 �

� 10 �0� V

��������������������� 3.8

For the network shown in Fig. 3.204, find the current in the 10 � resistor. 10 �



� j4 �



10 �90� V

����������

5� �5� �j

3�

[11.17 �−63.4° V, 10.6 �45° �]

� 100 �0� V



� j4 �

50 �30� V �

���������������� 3.12 Find Norton’s equivalent network for Fig. 3.208.

���������� [73.4 �−21.84° A]

�����Circuit Theory and Networks—Analysis and Synthesis

���������������������� �������

50 �0�V �



j5 �

j5 �

5�

j10 � A

3.14 Determine the maximum power delivered to the load in the network shown in Fig. 3.210.

B

5�

10 �

j15 �

5 � �j6 �

� � 50 �90�V

3�

����������

j4 �

[2.77 �–33.7° A, 2.5 � j12.5 �] 3.13 Find the current through the (3 � j4) � impedance in the network of Fig. 3.209. j5 �

5�







���������� [1032.35 W] 3.15 For the network shown in Fig. 3.211, find the value of ZL that will receive the maximum power. Determine also this power.

3� �

50 �0�V

50 �90�V j4 �

ZL

�j10 �

50 �0� A

j5 �

2� ZL

� 50 �0�V �

���������� [8.3 �85.2° A]

4�

�j3 �

���������� [3.82 − j1.03 �, 54.5 W]

Objective-Type Questions 3.1

In Fig. 3.212, the equivalent impedance seen across terminals a, b, is 2�

3.2

The Thevenin equivalent voltage VTh appearing between the terminals A and B of the network shown in Fig. 3.213 is given by

4�

3�

j3 �



Zeq �j4 � 2�

100 �0� V

16 � 3 �8 � (C) � +j12� � �3 �

�j6

j4

VTh �

���������� (a)

j2

4�

A

B

����������

(b)

8 � 3

(d)

none of the above

(a) 3.3

j16(3 − j4)

(b)

j16(3 � j4)

(c) 16(3 � j4) (d) 16 (3 − j4) A source of angular frequency of 1 rad/s has a source impedance consisting of a 1 � resistance

����������������������������������������

3.4

in series with a 1 H inductance. The load that will obtain the maximum power transfer is (a) 1 � resistance (b) 1 � resistance in parallel with 1 H inductance (c) 1 � resistance in series with 1 F capacitance (d) 1 � resistance in parallel with 1 F capacitance For the network shown in Fig. 3.214, the instantaneous current I1is

(a) (b) (c) (d) 3.6

28 A 4A 20 A cannot be determined

In the network of Fig. 3.216, the magnitudes of VL and VC are twice that of VR. The inductance of the coil is VR 5�

L

� j2 �

j2 � I1

VC VL

���������� 10 �60��A

3�

5 �0��A

(a) (c) 3.7

����������

3.5

C

5 �0�

(a)

10 3 �90� A 2

(b)

(c)

5 �60° A

(d) 5�−60° A

2.14 mH 31.8 mH

(b) (d)

5.3 H 1.32 H

Phase angle of the current I with respect to the voltage V1 in the circuit shown in Fig. 3.217.

10 3 �� 90� A 2

I V1 � 100 (1 � j ) 10 � V2 � 100 (1 � j )

In the network shown in Fig. 3.215, the current supplied by the sinusoidal current source I is

���������� (a) (c)

I 12 A

16 A

0° −45°

����������

Answers to Objective-Type Questions 3.1 (b)

3.2 (d)

3.3 (c)

3.4 (a)

3.5 (c)

3.6 (c)

3.7 (d)

(b) (d)

45° −90°

j10 �

4 Magnetic Circuits

�4.1��������������� 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������������������ Consider a coil of N turns carrying a current i as shown in Fig. 4.1. When current flows through the coil, a flux � 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 � �i � � k , a constant i ��ki

i v

������������������������������

Hence, rate of change of flux � k �� rate of change of current d� di �k dt dt According to Faraday’s laws of electromagnetic induction, a self-induced emf can be expressed as v � �N

� di d� di di � � Nk � � N � �L dt dt i dt dt

N� 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 �

������������������������������������������������������ 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�������������������� 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 −

��������������������������� 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

�2 � i1 �2 � k , a constant i1 �2 � k i1 Hence, rate of change of flux � k � rate of change of current i1 d� 2 di �k 1 dt dt According to Faraday’s law of electromagnetic induction, the induced emf is expressed as v2 � � N 2 where M �

� di d� 2 di di � �N2 k 1 � �N2 2 1 � �M 1 dt dt i1 dt dt

N 2 �2 and is called coefficient of mutual inductance. i1

�4.4����������������������������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 � M� 2 2 i1

4.5��������������������������

�2 �1 �2 � k1 �1 N k � 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 �1 i2

�1 �2 �1 � k2 �2 N k � M� 1 2 2 i2

k2 �

Let

…(4.2)

Multiplying Eqs (4.1) and (4.2), M 2 � k1k2 �

N1�1 N 2�2 � � k 2 L1 L2 i1 i2

M � k L1 L2 k � k1k2

where

�4.5������������������������ 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

��������������������� ���������

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)

������������������������������������������������������ 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

����������� ������������ ��������

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�������������������������� 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

����������������������������

di2 dt di1 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 in Coil 1 � � L1

di1 di �M 2 dt dt

Coil 2

4.6����������������������������

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 di1 di2 � ( L2 � M ) ( L1 � M ) 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)

������������������������������������������������������ �

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

������������������������������

di2 dt di1 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

di1 di �M 2 dt dt di2 di Total induced emf in Coil 2 � � L2 �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����������������������������

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

������������� 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)

������������������������������������������������������ 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

������������� 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

�������������

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, M � 5 mH

Solution For cumulative coupling,

L�

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�������������������

��������� ���� 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 L1 � 1.3 L2

But

1.3 L2 � L2 � 16 2.3 L2 � 16 L2 � 6.95 mH L1 � 1.3 L2 � 9.035 mH

�4.7����������������� 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

������������������������������������������������������� 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

L2

L1

(c)

(d)

����������������������� 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.

�������������

Obtain the dotted equivalent circuit for Fig. 4.8 shown below. R L1 +

C

v (t) −

i(t ) L2

�������� 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�������������������� f1

R

L1 + C

v (t ) − L2 f2

�������� From Fig. 4.9, it is seen that, the flux �1 is in upward direction in Coil 1, and flux �2 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

���������

�������������

Obtain the dotted equivalent circuit for the circuit of Fig. 4.11. j4 � j2 �

j3 �

j3 �

j5 �

j6 �

��������� 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 �1 and �2 oppose each other in coils 1 and 2, fluxes �2 and �3 oppose each other in coils 2 and 3, and fluxes �1 and �3 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 � j2 �

j3 �

f1

f 2f 3

i j3 �

j5 �

j6 �

��������� j4 �

j3 �

j2 �

j5 �

j3 �

���������

j6 �

�������������������������������������������������������

�������������

Obtain the dotted equivalent circuit for the circuit shown in Fig. 4.14.

j6 �

A j 5�

j 4�

B j3�

j1�

j 2�

��������� 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 �1, �2, �3 j5 � 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 � j4 �

B j3 �

j1 � f3

f2 j2 �

���������

j6 � j5 �

j4 �

j2 �

j1 �

j3 �

A

B

���������

�������������

Obtain the dotted equivalent circuit for the coupled circuit of Fig. 4.17.

j3 �

j5 �

� j8 �

10 � � � 50��0�V

��������� 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�������������������� f2

f1

j3 �

j5 �

� j8 �

10 � � � 50��0�V

��������� j5 �

From Fig. 4.18, it is seen that the flux �1 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.

�������������

10 �

j3 �

M12 � � 50��0�V

���������

Find the equivalent inductance of the network shown in Fig. 4.20. 1H 1H

0.5 H i

1H

5H

2H

��������� 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

��������������

Find the equivalent inductance of the network shown in Fig. 4.21. 1H 2H i

10 H

1H 5H

6H

��������� 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

�j 8 �

�������������������������������������������������������

��������������

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

��������� 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

��������������

Find the equivalent inductance of the network shown in Fig. 4.23. 8H 16 H

15 H

A

B

��������� 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

��������������

Find the equivalent inductance of the network shown in Fig. 4.24. 10 H

10 H

15 H 30 H

25 H

A

B

���������

35 H

C

4.8����������������������

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

��������������

Find the equivalent impedance across the terminals A and B in Fig. 4.25. 5� A 2�

3� j2 �

j4 �

j3 �

B

��������� Solution

Z1 � 5 �, Z 2 � ( 2 � j 4) �, Z3 � (3 � j 3) �, Z M � j 2 � Z � Z1 �

( 2 � j 4) (3 � j 3) � ( j 2) 2 Z 2 Z3 � Z 2M � 6.9 � 24.16� � � 5� 2 � j 4 � 3 � j 3 � 2( j 2) Z 2 � Z 3 � 2Z M

����������������������� 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 �1 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 −



�������������������������

������������������������������������������������������� 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 −



���������������������������� 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 −



������������������������� 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����������������������

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 −



����������������������������

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 −

+ L1

L2

Mdi2 dt

L2 + −

+ −

Mdi1 dt

+

+

v2

v1





i1 +

+

v2

v1





i2 L1

Mdi2 dt

L2 + −

+ −

Mdi1 dt

L1 Mdi2 dt

L2 − +

− +

Mdi1 dt

+

+

v2

v1





+

+

v2

v1





Mdi2 dt

L2 − +

− +

Mdi1 dt

jwL2

j wM i2 + −

+ jwM i 1 −

v2

+

v2

v1







i2 +

jwL1

jwL2

j wM i2 + −

+ jwM i 1 −

v2 −

i2 +

jwL1

jwL2

j wM i2 − +

− jwM i 1 +

v2

i1 +

+

jwL1

i1

i2 L1

i2

i1

i2

i1

i2 M

L1

i1





i2

i1





v1



i2 M

+



+





v1

i1 +

Frequency-domain equivalent circuit



i2 jwL1

jwL2

j wM i2 − +

− j wM i 1 +

+ v2

��������������������������������������������



�������������������������������������������������������

��������������

Write mesh equations for the network shown in Fig. 4.31. R1

L1

L3 R2

+

M12

v1(t ) −

M23

L2

i1

R3

i2

��������� 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

��������� (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����������������������

( 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

��������������

�(ii)

Write KVL equations for the circuit shown in Fig. 4.33. R2

R1 + v1(t ) −

R3

i3

C

M

i1

L1

+ v2(t) −

L2

i2

��������� 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 −

��������� Applying KVL to Mesh 1, di1 di �M 2 �0 dt dt di di R1 i1 � L1 1 � M 2 � 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)

������������������������������������������������������� 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 �

��������������

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

��������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.36. Z1 jwL1

jwL2 ZL

+

jwM I2

V1

+ −

+ −

j wM I1



I1

Z2

I2

��������� Applying KVL to Mesh 1, V1 � Z1I1 � j� L1I1 � j� MI 2 � Z 2 ( I1 � I 2 ) � 0 ( Z1 � j� L1 � Z 2 ) I1 � ( Z 2 � j� M ) I 2 � V1

�(i)

�Z 2 ( I 2 � I1 ) � j� MI1 � j� L2 I 2 � Z L I 2 � 0 �( Z 2 � j� M ) I1 � ( Z 2 � j� L2 � Z L ) I 2 � 0

�(ii)

Applying KVL to Mesh 2,

4.8����������������������

��������������

Write mesh equations for the network shown in Fig. 4.37. R

L3

+

L1

v(t )

L2

i1



i2

i3

C

��������� 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

i2 M13

di3 dt

L3

+ −

M21 d (i1 − i2) dt

+ −

M23 di3 dt

i3

− +

M31 d (i1 − i2) dt

+ −

M32 d (i2 − i3) dt C

��������� 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

�������������������������������������������������������

��������������

Write KVL equations for the network shown in Fig. 4.39. R

L3

+

L1

v(t ) i1



L2 i2

i3

C

��������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.40. R

L2

L1

+ v1 −

i1

+ −

M12 di2 dt

− +

di3 dt

L3

+ −

M21 di1 dt

+ −

di3 dt

i2 M13

i3 M23

− +

M31 di1 dt

+ −

M32 di2 dt C

��������� 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 di3 di1 � M 31 � M 32 R(i1 � i2 � i3 ) � L3 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����������������������

��������������

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

��������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.42. 2

3H +

di3 dt

i3

+ −

v1



i1 = 5e−t A

2

5H +

di1 dt + −

v2



i2 = 10e−t A

��������� 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 �t v2 � 5 �2 � 5 (15 e ) � 2 (5 e � t ) � �75 e � t � 10 e � t � �85 e � t V dt dt dt dt v1 � 3

��������������

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

���������



������������������������������������������������������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.44.

+

di3 dt

2

2H

i3

− +

v1



i1 = 10e−t A

2

4H

di1 dt − +

+



i2 = 10e−t A

��������� 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

��������������

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



��������� 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

��������� Applying KVL to Mesh 1, 30 sin t � 0.2

di1 di � 0.1 2 � 0 dt dt

�(i)

4.8����������������������

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 )

��������������

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. 2�

1�

j2 �



� j4 �

5�0��V

V2

j3 �

I1



I2

Loop 1



Loop 2

��������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.48. 2��

1�� j4��



j 3��



V2

5�0��V �

I1

j2�I2

� �

� �

���������

j2�I1

I2



������������������������������������������������������� Applying KVL to Loop 1, 5 � 0� � � 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 (5 � 0�)(1 � j 3) V2 � � 7.91 � � 18.43� V j2

��������������

Determine the ratio

V2 in the circuit of Fig. 4.49, if I1 = 0. V1

8�

2�

j2 �



� j8 �

V1

j2 �

V2

I1





I2

��������� Solution The equivalent circuit in terms of dependent sources is as shown in Fig. 4.50. 8��

2�� j8��



j 2��



V1

V2 �

I1

j2�I2

� �

� �

j 2�I1

I2



��������� Applying KVL to Mesh 1, V1 � 8I1 � j8I1 � j 2I 2 � 0 (8 � j8) I1 � j 2I 2 � V1

�(i)

4.8����������������������

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

�(ii)

�(iii)

Putting I1 � 0 in Eq (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

��������������

For the coupled circuit shown in Fig. 4.51, find input impedance at terminals A and B. A

j4��

3�

j3 �



j5��

V1

�j8��



B

��������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.52. A

j4��

3�

j 3�(I1 ��I2) ���� j5��



V1

�j8��

I1



� j 3�I I 1 2 �

B

��������� 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)

������������������������������������������������������� Substituting Eq (ii) in Eq (i), (3 � j15)I1 � j8 ( �2.67 I1 ) � V1 (3 � j 36.36) I1 � V1 Zi �

��������������

V1 � (3 � j 36.36) � � 36.48 � 85.28� � I1

Find equivalent impedance of the network shown in Fig. 4.53. j2 �

2�

j6 � j4 �

j4 �

j3 �

Zeq �j5 �

5�

��������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.54. j2��

2�

j4�(I1 ��I2) ���� j3��



� �

V1 �

I1

j 4 I1

5� � �

j4��

j 6 I2

I2

� j 6�(I1 ��I2) �

�j 5��

��������� 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����������������������

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 �

��������������

V1 (5 � j 5) (5 � j 5) � 7 � j3 � � 5.63 � � 47.15� � I1 5 � j4

Find the voltage across the 5 � resistor in Fig. 4.55 using mesh analysis. j5.66 � j5 �



j10 �

3�

50�0��V

I1



�j4 �

5�

I2

��������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.56. j5 �

j5.66 I2

j10 �

j5.66 I1

���� �

3�

50�0��V �

����

I1

�j4 �

5�

I2

��������� 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)

������������������������������������������������������� 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� � I2 � 3� j 1 �(3 � j 9.66) �(3 � j 9.66) 8 � j6 V5 � � 5 I 2 � 5 (3.82 � � 112.14�) � 19.1 � � 112.14� V

��������������

Find the voltage across the 5 � resistor in Fig. 4.57 using mesh analysis. k � 0.8 j5 �



j10 �

3�

50�0��V

I1



�j4 �

5�

I2

��������� Solution For a magnetically coupled circuit, X M � k X L1 X L2 � 0.8 (5) (10) � 5.66 � The equivalent circuit in terms of dependent sources is shown in Fig. 4.58. j5 �

j5.66 I2

j10 �

j5.66 I1

���� �

3�

50�0��V �

����

I1

�j4 �

5�

I2

��������� 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����������������������

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� � I2 � 3� j 1 �(3 � j 1.66) �(3 � j 1.66) 8 � j6 V5 � � 5 I 2 � 5 (8.62 � � 24.79�) � 43.1 � � 24.79� �

��������������

Find the current through the capacitor in Fig. 4.59 using mesh analysis. j4 �

3�

j3 �



j5 �

50�45��V �

I1

�j8 �

I2

��������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.60. 3�

j4 �

j3�(I1 ��I2) ���� j5 �

� 50�45��V �

I1

� �

�j8 � j 3 I1 I2

��������� 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 Networks—Analysis and Synthesis Applying KVL to Mesh 2, j 3 I1 � j 5 ( I 2 � I1 ) � j8 I 2 � 0 � j8 I1 � j 3 I 2 � 0

�(ii)

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 � 139.72� � I2 � 3 � j 15 � j8 � j8 � j3 IC � I 2 � 3.66 � 139.72��

��������������

Find the voltage across the 15 � resistor in Fig. 4.61 using mesh analysis. j10 �

20 �

j5 � j20 �

� 120�0��V �

15 �

I2

I1

��������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.62. j10 �

20 �

j 5�(I1 ��I2) ����

j20 �



15 �

120�0��V �

I1

� �

j 5�I2

I2

��������� 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����������������������

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�� I2 � 20 � j 20 � j15 � j15 15 � j 20 V15 � � 15 I 2 � 15 ( 2.53 � 10.12�) � 37.95 � 10.12�V

��������������

Find the current through the 6 � resistor in Fig. 4.63 using mesh analysis. 4� j2 � � j3 �

120�0��V �

j8 �

I1

I2

6�

��������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.64. 4� j3 �



j8 �

200�30��V �

I1

� �

j 2�I2

I2

� � j 2�(I1 ��I2)

6�

��������� 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)

������������������������������������������������������� 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�� I2 � 4� j 3 � j5 � j 5 6 � j 15

��������������

Determine the mesh current I3 in the network of Fig. 4.65. 12 �

j16��

j5 � 4�

7�

I3

�j8 �



6�

j4 �

200�30��V �

I1

I2 �j4 �

��������� Solution The equivalent circuit in terms of dependent sources is shown in Fig. 4.66. j16��

j 5�(I1 ��I2)

12 �

����

4�

7�

I3 j4 �



�j8 �

6�

200�30��V �

I1

� �

j 5�I3

���������

I2 �j4 �

4.8����������������������

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)

Applying KVL to Mesh 3, � 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)

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 � �� � � � � ( 4 � j 5 ) � ( 7 � j 13 ) 23 � j 8 I 0 � �� 3� � � By Cramer’s rule, � j4 4 � j4 200 � 30� � j4 13 � j8 0 �( 4 � j 5) �(7 � j13) 0 � 16.28 � 16.87�� I3 � � j4 �( 4 � j 5) 4 � j4 � j4 13 � j8 �(7 � j13) �( 4 � j 5) �(7 � j13) 23 � j8

��������������

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.

5� j5 �

5� j2 �

j5 �









10��0��V

10��90��V

�j10 �

��������� 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.

������������������������������������������������������� 5 � I1

f1

I2 5 � j2 �

j5 �

j5 �

f2

� 10��0�V

I1



� 10��90�V

I2



�j10 �

��������� From Fig. 4.68, it is seen that two fluxes �1 and �2 aid each other. Hence, dots are placed at the two coils as shown in Fig. 4.69. 5�

j2 �

j5 �

j5 �

5�



� �j10 �

10�0��V

10�90��V

I1





I2

��������� The equivalent circuit in terms of dependent sources is shown in Fig. 4.70. 5�

j5 �

j2 I2 ����



� �

10�0��V �

I1

j 2 I1

j5 �





����

5� �

� � 10�90��V

�j10 �

I2



��������� 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����������������������������������������������

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� �) � 10.08 � 24.03�V

����������������������������������������������� 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





������������������������� The equivalent circuit in terms of dependent sources is shown in Fig. 4.72. I1

+

I2 jwL1

jwL2

+

V1

V2 −

jwM I2

+ −

+ jwM I1 −



���������������������������� Applying KVL to Mesh 1, V1 � j� L1 I1 � j� M I 2 � j� L1 I1 � j� M I 2 �

�(4.11)

Applying KVL to Mesh 2, V2 � j� L2 I 2 � j� M I1 � 0 j� M I1 � j� L2 I 2 � V2

�(4.12)

������������������������������������������������������� Writing Eqs (4.11) and (4.12) in matrix form, � j� L1 j� M � � I1 � � V1 � �� j� M j� L2 �� �� I 2 �� � �� V2 ��

�(4.13)

Consider a T-network as shown in Fig. 4.73. Z1

I1

Z2

I2

+

+

Z3

V1

V2 −



������������������� 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 �

� (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 � L1 Z3 � j � M Z 2 � Z3 � j � L2 Solving these equations, Z1 � j � L1 � j � M � j � ( L1 � M ) Z 2 � j � L2 � j � M � j � ( L2 � M ) Z3 � j � 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 −

�������������������������������������������������

4.9����������������������������������������������

��������������

Find the conductively coupled equivalent circuit for the network shown in Fig. 4.75. �j4 �

j6 �

j2 � �

V1

j3 � �

j5 �

10 �

I1

I2

2�

��������� 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 � j� (L1 � M ) � j� L1 � j� M � j 3 � j 2 � j1 � Z� � j� ( L2 � M ) � j� L2 � j� M � j 5 � j 2 � j 3 � Z � � j� M � j 2 � The conductively coupled equivalent circuit is shown in Fig. 4.76. �j4 �

j1 �

j3 �

j6 �

� j2 �

V1

10 �

I2

I1



2�

���������

��������������

Draw the conductively coupled equivalent circuit of Fig. 4.77. j6 � j5 �



3�

V1 �

j10 �

I1

�j4 �

5�

I2

��������� Solution The current I1 enters from the dotted end and I2 leaves from the dotted end. Hence, the mutual inductance M is negative.

������������������������������������������������������� In the conductively coupled equivalent circuit, Z1 � j� ( L1 � M ) � j� L1 � j� M � j 5 � j 6 � � j1 � Z� � j� ( L2 � M ) � j� L2 � j� M � j10 � j 6 � j 4 � Z 3 � j� M � j 6 � The conductively coupled equivalent circuit is shown in Fig. 4.78. �j 1 �

j4 �

j6 �



V1

5�



(3 ��j 4) � I2

I1

���������

��������������

Find the conductively coupled equivalent circuit of the network in Fig. 4.79. j1�

4�

2�



V1

j2 � �

j4 �

I1

��������� Solution The currents I1 and I2 leave from the dotted terminals. Hence, mutual inductance is positive. In the conductively coupled equivalent circuit, Z1 � j� ( L1 � M ) � j� L1 � j� M � j 4 � j 2 � j 6 � Z� � j� (L2 � M ) � j� L2 � j� M � j 2 � j 2 � j 4 � Z � � � j� M � � j 2 � The conductively coupled equivalent circuit is shown in Fig. 4.80. j6 �

(4 ��j2) �

j4 �



V1

�j2 � �

I1

(2 ��j 4) �

I2

���������

��������������

Exercises 4.1

4.2

4.3

4.4

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]

4.6

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 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]

v(t )

Write mesh equations of the network shown in Fig. 4.83. L2

R1

R3

+

L1 i1



R2 i2

L3 i3

��������� 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

Find the equivalent inductance of the network shown in Fig. 4.81.

Find the input impedance at terminals AB of the coupled circuits shown in Fig. 4.84 to 4.85. (i) 3�

7H

j4 �

A

2H

3H

5H 4H

6H

j3 �

j5 �

�j8 �

��������� [10 H] 4.5

Find the effective inductance of the network shown in Fig. 4.82. 2H

B

��������� (ii) A

3H 2�

2�

4H

5H

j5 �

2H

j5 � j2 �

B

��������� [4.8 H]

���������

������������������������������������������������������� (iii)

j2 �

k ��0.5

j1 �

A �

�j 3 �

j5 �

�j 3 �

10�0��V 2�

j4 � j8 �

��������� [19.2� � 33.02�V ]

B 4�

4.11 Find the power dissipated in the 5 � resistor in the network of Fig. 4.90.

��������� �(a) �3 � j 36.3� � (b)��1 � j1.5� � � � � �(c)��6.22 � j 4.65� � � 4.8

5�



In the coupled circuit shown in Fig. 4.87, find V2 for which I1 = 0. What voltage appears at the 8 � inductive reactance under this condition? 5�

2�

j2 �

j3 �

j4 �

� 3�

100�0�V �

��������� [668.16 W]

2�

j2 �

5�

4.12 Find the current I in the circuit of Fig. 4.91. �

� j8 �

100�0�V

j2 �

V2



14 � �j15 �





10 �

j3 �





j10 �

100�20��V

���������

j4 �

j12 �

70��30��V �

I

[141.5� � 45�V, 100�0�V] 4.9

For the coupled circuit shown in Fig. 4.88, find the components of the current I2 resulting from each source V1 and V2. 2�

V1 � 10�0�V



j2 �

j4 �

� j3 �

I2





��������� [7.07�45� V, 1. 4.13 Obtain a conductively coupled circuit for the circuit shown in Fig. 4.92.

V2 � 10�0�V

��������� [0.77�112.6�������� ���������

2�

j5 �

j3 �

j4 �

� 100�0�V

3� �

4.10 Find the voltage across the 5 � resistor in the network shown in Fig. 4.89. ���������

�j 2 �

����������������������������� 2�

j2 �

�j 1 �



3�



j3 �

100�0�V

�j 2 �

���������

Objective-Type Questions 4.1

4.2

Two coils are wound on a common magnetic core. The sign of mutual inductance M for finding out effective inductance of each coil is positive if the (a) two coils are wound in the same sense. (b) fluxes produced by the two coils are equal (c) fluxes produced by the coils act in the same direction (d) fluxes produced by the two coils act in opposition When two coils having self-inductances of L1 and L2 are coupled through a mutual inductance M, the coefficient of coupling k is given by (a)

(c) 4.3

k�

k�

M 2 L1 L2 2M L1 L2

(b)

(d)

k�

k�

M L1 L2

L1 ��L2

(b) L1 � L2

1 1 ( L1 � L2 ) ( L1 � L2 ) (d) 4 2 Consider the following statements: The coefficient of coupling between two oils (c)

4.6

L1 L2 M

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)

4.4

4.5

4.7

4.8

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 (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. 4.94 is 4.9

j5 �

j2 �

j 10 �

j10 �

j2 �

��������� (a) j 29 � (c) j19 �

(b) j9 � (d) j39 �

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)

4.5 (d)

4.6

(c)

5 Graph Theory

������������������� The purpose of network analysis is to find voltage across and current through all the elements. When the network is complicated and has a large number of nodes and closed paths, network analysis can be done conveniently by using ‘Network Topology’. This theory does not make any distinction between different types of physical elements of the network but makes the study based on a geometric pattern of the network. The basic elements of this theory are nodes, branches, loops and meshes. Node Branch Loop Mesh

It is defined as a point at which two or more elements have a common connection. It is a line connecting a pair of nodes, the line representing a single element or series connected elements. Whenever there is more than one path between two nodes, there is a circuit or loop. It is a loop which does not contain any other loops within it.

������������������������� A linear graph is a collection of nodes and branches. The nodes are joined together by branches. The graph of a network is drawn by first marking the nodes and then joining these nodes by lines which correspond to the network elements of each branch. All the voltage and current sources are replaced by their internal impedances. The voltage sources are replaced by short circuits as their internal impedances are zero whereas current sources are replaced by open circuits as their internal impedances are infinite. Nodes and branches are numbered. Figure 5.1 shows a network and its associated graphs. Each branch of a graph may be given an orientation or a direction with the help of an arrow head which represents the assigned reference direction for current. Such a graph is then referred to as a directed or oriented graph. Branches whose ends fall on a node are said to be incident at that node. Degree of a node is defined as the number of branches incident to it. Branches 2, 3 and 4 are incident at Node 2 in Fig. 5.1(c). Hence, the degree of Node 2 is 3.

5.2�Circuit Theory and Networks—Analysis and Synthesis

(2)

3

2

1

3

(3)

(2)

3

(3)

(3)

(2) (1) V

2

1

2

1

(6)

(6)

(6)

(5)

(4)

(4)

(1)

(1)

(5)

(4)

(5)

+ − 4

4

4

(a) Network

(b) Undirected graph

(c) Directed or oriented graph

�������������������������������

�������������������������� 1. Planar Graph A graph drawn on a two-dimensional plane is said to be planar if two branches do not intersect or cross at a point which is other than a node. Figure 5.2 shows such graphs. (2) 2

1

(3)

(1) (5)

(8)

3

3

(4)

(1)

2

(2)

1

(6)

(3)

(6)

(7)

4

(9)

6 (4)

4

5

(5)

���������������������� 2. Non-planar Graph A graph drawn on a two-dimensional plane is said to be non-planar if there is intersection of two or more branches at another point which is not a node. Figure 5.3 shows non-planar graphs. 1

(2)

2

(2)

2

1 (5)

(3)

3

(6)

(1)

(4) 3

4 (3)

(1)

(7)

(8)

4 (5)

5

(6)

(4) 6

�������������������������� 3. Sub-graph It is a subset of branches and nodes of a graph. It is a proper sub-graph if it contains branches and nodes less than those on a graph. A sub-graph can be just a node or only one branch of the graph. Figure 5.4 shows a graph and its proper sub-graph.

5.3������������������������� (2)

(2)

2

1

(3) (1)

(4)

(5)

2

3

(5)

3 3

1 (6)

(4)

4

4 (a)

(b)

��������������������������������������� 4. Path It is an improper sub-graph having the following properties: 1. At two of its nodes called terminal nodes, there is incident only one branch of sub-graph. 2. At all remaining nodes called internal nodes, there are incident two branches of a graph. In Fig. 5.5, branches 2, 5 and 6 together with all the four nodes, constitute a path. (3) 2

1

(2)

3

(4)

(1)

(5)

(6)

4

������������� 5. Connected Graph A graph is said to be connected if there exists a path between any pair of nodes. Otherwise, the graph is disconnected. 6. Rank of a Graph

If there are n nodes in a graph, the rank of the graph is (n − 1).

7. Loop or Circuit A loop is a connected sub-graph of a connected graph at each node of which are incident exactly two branches. If two terminals of a path are made to coincide, it will result in a loop or circuit. Figure 5.6 shows two loops. (2) 1

2

(3) 3

(1)

(4)

(2) 1

4 Loops: {1, 2, 3, 4}

(1)

2

Loops: {1, 2}

�������������� Loops of a graph have the following properties: 1. 2. 3.

There are at least two branches in a loop. There are exactly two paths between any pair of nodes in a circuit. The maximum number of possible branches is equal to the number of nodes.

5.4�Circuit Theory and Networks—Analysis and Synthesis 8. Tree A tree is a set of branches with every node connected to every other node in such a way that removal of any branch changes this property. Alternately, a tree is defined as a connected sub-graph of a connected graph containing all the nodes of the graph but not containing any loops. Branches of a tree are called twigs. A tree contains (n � 1) twigs where n is the number of nodes in the graph. Figure 5.7 shows a graph and its trees. (6) 2

1

(1)

3

(2)

(3)

(4)

1

2

(4)

(5)

(3)

(5)

4 (a) Graph

2 (2)

1

3

(1)

3

(5)

4

4

(b)

(c)

Twigs: {1, 4, 5}

Twigs: {2, 3, 5}

����������������������������� Trees have the following properties: 1. 2. 3. 4. 5. 6.

There exists only one path between any pair of nodes in a tree. A tree contains all nodes of the graph. If n is the number of nodes of the graph, there are (n � 1) branches in the tree. Trees do not contain any loops. Every connected graph has at least one tree. The minimum terminal nodes in a tree are two.

9. Co-tree Branches which are not on a tree are called links or chords. All links of a tree together constitute the compliment of the corresponding tree and is called the co-tree. A co-tree contains b ��(n � 1) links where b is the number of branches of the graph. In Fig. 5.7 (b) and (c) the links are {2, 3, 6} and {1, 4, 6} respectively.

�������������

Draw directed graph of the networks shown in Fig. 5.8. R2

R1

C1

L

L1 L2

v C1

R1 R2 V

C2

+ −

C3 C2

(a)

R3 (b)

��������

5.3����������������������5.5

Solution For drawing the directed graph, 1. replace all resistors, inductors and capacitors by line segments, 2. replace the voltage source by a short-circuit, 3. assume directions of branch currents, and 4. number all the nodes and branches. The directed graph for the two networks are shown in Fig. 5.9. 1 (1)

1

(2)

4

(2)

(1)

2

5

(5)

(3)

(3)

(4)

(7) (8)

(6)

2

3

(a)

(b)

��������

�������������

Figure 5.10 shows a graph of the network. Show all the trees of this graph. (1) 1

(2)

2

3 (4) (5)

(3) 4

��������� Solution A graph has many trees. A tree is a connected sub-graph of a connected graph containing all the nodes of the graph but not containing any loops. Figure 5.11 shows various trees of the given graph. (2)

2

3

1

(1)

2

3

1 (1) 2

3

1 (4)

(3)

(4)

4 1

2

4

(3)

4 3

(4) (5)

(3)

(5)

1

2

(3)

(5)

1 (1) 2 (2) 3

(3)

4

4

���������

3

1 (1) 2 (2) 3

(4)

(5)

(5)

4 (2) 3

1 (1) 2 (2)

4

4

����Circuit Theory and Networks—Analysis and Synthesis

���������������������� A linear graph is made up of nodes and branches. When a graph is given, it is possible to tell which branches are incident at which nodes and what are its orientations relative to the nodes.

5.4.1 Complete Incidence Matrix (Aa) For a graph with n nodes and b branches, the complete incidence matrix is a rectangular matrix of order n � b.� Elements of this matrix have the following values: aij � 1, if branch j is incident at node i and is oriented away from node i. � �1, if branch j is incident at node i and is oriented towards node i. ��0, if branch j is not incident at node i. For the graph shown in Fig. 5.12, branch 1 is incident at nodes 1 and 4. It is oriented away from Node 1 and oriented towards Node 4. Hence, a11 � 1 and a41 � �1. Since branch 1 is not incident at nodes 2 and 3, a21 � 0 and a31 � 0. Similarly, other elements of the complete incidence matrix are written. (6) 2

1

(2)

3

(4)

(1)

(3)

(5)

Nodes �

Branches � 1

2

3

4

5

6

1

1

1

0

0

0

1

2

0 �1

1 �1

0

0

3

0

0

1 �1

4

�1

0

0 �1

1

0 �1

0

4

��������������� The complete incidence matrix is 1� � 1 1 0 0 0 � 0 �1 1 �1 0 0 � Aa � � � 1 1 �1� � 0 0 0 � �1 0 �1 0 �1 0 � It is seen from the matrix Aa that the sum of the elements in any column is zero. Hence, any one row of the complete incidence matrix can be obtained by the algebraic manipulation of other rows.

5.4.2 Reduced Incidence Matrix (A) The reduced incidence matrix A is obtained from the complete incidence matrix Aa by eliminating one of the rows. It is also called incidence matrix. It is of order (n � 1) ��b. Eliminating the third row of matrix Aa, � 1 1 0 0 0 1� A � � 0 �1 1 �1 0 0 � � � � �1 0 �1 0 �1 0 � When a tree is selected for the graph as shown in Fig. 5.13, the incidence matrix is obtained by arranging a column such that the first (n � 1) column corresponds to twigs of the tree and the last b � (n � 1) branches corresponds to the links of the selected tree.

5.4�Incidence Matrix���� 2

1

3

(2)

(4)

2

Twigs: {2, 3, 4} Links: {1, 5, 6}

(3)

Twigs 3 4

� 1 0 A � � �1 1 � 1 � 0 �1

4

1

Links 5 6

0 1 1 0 0 �1 1

0 1� 0 0� � 1 0�

�������������� The matrix A can be subdivided into submatrices At and Al. A [ At : Al ] Where At the is twig matrix and Al is the link matrix.

5.4.3 Number of Possible Trees of a Graph Let the transpose of the reduced incidence matrix A be AT. It can be shown that the number of possible trees of a graph will be given by Number of possible trees � |AAT| For the graphs shown in Fig. 5.12, the reduced incidence matrix is given by � 1 1 0 0 0 1� A � � 0 �1 1 �1 0 0 � � � � �1 0 �1 0 �1 0 � Then transpose of this matrix will be � 1 0 �1� � 1 �1 0 � � � 0 1 �1� AT � � �0 �1 0 � �0 0 �1� � � � 1 0 0� Hence, number of all possible trees of the graph � 1 0 �1� � 1 �1 0 � � 3 �1 �1� � 1 1 0 0 0 1� �0 1 �1� � � � �1 3 �1� AAT � � 0 �1 1 �1 0 0 � � � � � �0 �1 0 � � � �1 �1 3� � �1 0 �1 0 �1 0 � � 0 0 �1� � � � 1 0 0�

|

T

3 �1 �1 | � �1 3 �1 � 3(( �1 �1 3

Thus, 16 different trees can be drawn.

) � ( )(

) � 1((

) � 16

����Circuit Theory and Networks—Analysis and Synthesis

����������������������������������� When a graph is given, it is possible to tell which branches constitute which loop or circuit. Alternately, if a loop matrix or circuit matrix is given, we can reconstruct the graph. For a graph having n nodes and b branches, the loop matrix Ba is a rectangular matrix of order b columns and as many rows as there are loops. Its elements have the following values: bij � 1, if branch j is in loop i and their orientations coincide. � ��1, if branch j is in loop i and their orientations do not coincide. ��0, if branch j is not in loop i. A graph and its loops are shown in Fig. 5.14. (6) 2

1

3

(2)

(4) (3)

(1)

(5)

Loop 1: {1, 2, 3} Loop 2: {3, 4, 5} Loop 3: {2, 4, 6} Loop 4: {1, 2, 4, 5} Loop 5: {1, 5, 6} Loop 6: {2, 3, 5, 6} Loop 7: {1, 3, 4, 6}

4

��������������� All the loop currents are assumed to be flowing in a clockwise direction. Loops � 1 2 3 4 5 6 7

Branches � 1 2 3 4 �1 1 1 0 0 0 �1 �1 0 �1 0 1 �1 1 0 �1 �1 0 0 0 0 �1 �1 0 �1 0 1 1

� �1 1 1 0 � 0 0 �1 �1 � 1 � 0 �1 0 Ba � � �1 1 0 �1 � �1 0 0 0 � 0 �1 �1 0 � 1 1 � �1 0

5 0 1 0 1 1 1 0 0 1 0 1 1 1 0

6 0 0 1 0 1 1 1

0� 0� � 1� 0� 1� 1�� 1�

5.5.1 Fundamental Circuit (Tieset) and Fundamental Circuit Matrix When a graph is given, first select a tree and remove all the links. When a link is replaced, a closed loop or circuit is formed. Circuits formed in this way are called fundamental circuits or f-circuits or tiesets. Orientation of an f-circuit is given by the orientation of the connecting link. The number of f-circuits is same as the number of links for a graph. In a graph having b branches and n nodes, the number of f-circuits or tiesets will be (b � n � 1). Figure 5.15 shows a tree and f-circuits (tiesets) for the graph shown in Fig. 5.14.

5.5���������������������������������� (6) 1

2 (2)

3 (4) (3)

4 (a) Tree

(2) (1)

(4)

(2)

(4)

(3)

(3)

tieset 1

(2)

(4) (3)

(5)

tieset 5

tieset 1: {1, 2, 3} tieset 5: {5, 3, 4} tieset 6: {6, 2, 4}

tieset 6

(b) f-circuits (tiesets)

����������������������������� Here, b � 6 and n � 4. Number of tiesets � b � n � 1�� 6 � 4 � 1 ��3 f-circuits are shown in Fig. 5.15. The orientation of each f-circuit is given by the orientation of the corresponding connecting link. The branches 1, 2 and 3 are in the tieset 1. Orientation of tieset 1 is given by orientation of branch 1. Since the orientation of branch 1 coincides with orientation of tieset 1, b11 � 1. The orientations of branches 2 and 3 do not coincide with the orientation of tieset 1, Hence, b12� ��1 and b13 � ��1. The branches 4, 5 and 6 are not in tieset 1. Hence, b14 � 0, b15 � 0 and b16 � 0. Similarly, other elements of the tieset matrix are written. Then, the tieset schedule will be written as Branches �

Tiesets �

1

3

4 5 6

1

1 �1 �1

0 0 0

5

0

6

0 �1

2

0 �1 �1 1 0 0

1 0 1

Hence, an f-circuit matrix or tieset matrix will be given as � 1 �1 �1 0 0 0 � B � �0 0 �1 �1 1 0 � � � 1 0 1� �0 �1 0 Usually, the f-circuit matrix B is rearranged so that the first (n � 1) columns correspond to the twigs and b � (n � 1) columns to the links of the selected tree. 2

gs 3 4

Links 1 5 6

��11 B�� 0 � ��1

1 0 1 �1 0 1

1 0 0

0 1 0

0� 0� � 1�

The matrix B can be partitioned into two matrices Bt and Bl. B [ Bt : Bl ] [ Bt : U ] where Bt is the twig matrix, Bl is the link matrix and U is the unit matrix.

�����Circuit Theory and Networks—Analysis and Synthesis

5.5.2

Orthogonal Relationship between Matrix A and Matrix B

For a linear graph, if the columns of the two matrices Aa and Ba are arranged in the same order, it can be shown that Aa BaT � 0 Ba AaT � 0

or

The above equations describe the orthogonal relationship between the matrices Aa and Ba. If the reduced incidence matrix A and the f-circuit matrix B are written for the same tree, it can be shown that T �0 or B AT � 0 These two equations show the orthogonal relationship between matrices A and B.

��������������������

1

(2)

2

Consider a linear graph. By removing a set of branches without affecting the nodes, (5) (3) two connected sub-graphs are obtained and the original graph becomes unconnected. (1) The removal of this set of branches which results in cutting the graph into two parts (4) 4 3 are known as a cutset. The cutset separates the nodes of the graph into two groups, each being in one of the two groups. ��������������� Figure 5.16 shows a graph. Branches 1, 3 and 4 will form a cutset. This set of branches separates the graph into two parts. One having an isolated node 4 and other part having branches 2 and 5 and nodes 1, 2 and 3. Similarly, branches 1 and 2 will form a cutset. Each branch of the cutset has one of its terminals incident at a node in one part and its other end incident at other nodes in the other parts. The orientation of a cutset is made to coincide with orientation of defining branch. For a graph having n nodes and b branches, the cutset matrix Qa is a rectangular matrix of order b columns and as many rows as there are cutsets. Its elements have the following values: qij � 1, if the branch j is in the cutset i and the orientation coincide. � �� �1, if the branch j is in the cutset i and the orientations do not coincide. � �� 0, if the branch j is not in the cutset i. Figure 5.17 shows a directed graph and its cutsets. (6) 1

3

2 (2)

(4) (3)

(1)

(5)

Cutset 1: {1, 2, 6} Cutset 2: {2, 3, 4} Cutset 3: {3, 1, 5} Cutset 4: {4, 5, 6} Cutset 5: {5, 2, 3, 6} Cutset 6: {6, 1, 3, 4}

Cutsets � 1 2 3 4 5 6

Branches � 1 2 3 4 5 6 1 1 0 0 0 1 0 1 �1 1 0 0 1 0 1 0 1 0 0 0 0 1 1 �1 0 �1 1 0 1 �1 1 0 1 �1 0 1

4

������������������������ For the cutset 2, which cuts the branches 2, 3 and 4 and is shown by a dotted circle, the entry in the cutset schedule for the branch 2 is 1, since the orientation of this cutset is given by the orientation of the branch 2 and hence it coincides. The entry for branch 3 is �1 as orientation of branch 3 is opposite to that of cutset 2,

5.6�Cutset Matrix�����

i.e., branch 2 goes into cutset while branch 3 goes out of cutset. The entry for branch 4 is 1 as the branch 2 and the branch 4 go into the cutset. Thus their orientations coincide. Hence, the cutset matrix Qa is given as �1 1 0 0 �0 1 �1 1 � 1 0 1 0 Qa � � �0 0 0 1 �0 �1 1 0 � 1 0 1 �1 �

0 1� 0 0� � 1 0� 1 �1� 1 �1� 0 1��

5.6.1 Fundamental Cutset and Fundamental Cutset Matrix When a graph is given, first select a tree and note down its twigs. When a twig is removed from the tree, it separates a tree into two parts (one of the separated part may be an isolated node). Now, all the branches connecting one part of the disconnected tree to the other along with the twig removed, constitutes a cutset. This set of branches is called a fundamental cutset or f-cutset. A matrix formed by these f-cutsets is called an f-cutset matrix. The orientation of the f-cutset is made to coincide with the orientation of the defining branch, i.e., twig. The number of f-cutsets is the same as the number of twigs for a graph. Figure 5.18 shows a graph, selected tree and f-cutsets corresponding to the selected tree. (6) 1

(6) 3

2 (2)

(4)

(1)

(3)

(2) (5)

(2)

(4) (3)

(1)

(4) (3)

(5)

f-cutset 2: {2, 1, 6} f-cutset 3: {3, 1, 5} f-cutset 4: {4, 5, 6}

4 (a) Graph

(b) Tree

(c) f-cutsets

�������������������������������������������� The branches 2, 1, and 6 are in the f-cutset 2. Orientation of f-cutset 2 is given by orientation of the branch 2 which is moving away from the f-cutset 2. Since the orientations of branches 2, 1 and 6 coincide with the orientation of the f-cutset 2, q11 � 1, q12 � 1 and q16 � 0. The branches 3, 4 and 5 are not in the f-cutset 2. Hence, q13 � 0, q14 � 0 and q15 � 0. Similarly, other elements of the f-cutset matrix are written. The cutset schedule is f-cutsets � 2 3 4 Hence, the f-cutset matrix Q is given by � 1 Q�� 1 � � 0

1 1 1 0 1 0 0

Branches � 2 3 4 5 6 1 0 0 0 1 0 1 0 1 0 0 0 1 1 �1 0 1 0

0 0 1

0 1� 1 0� � 1 �1�

The f-cutset matrix Q is rearranged so that the first (n � 1) columns correspond to twigs and b � (n � 1) columns to links of the selected tree.

�����Circuit Theory and Networks—Analysis and Synthesis Twigs � Q�� � �

Links

2

3

4

1

5

1 0 0

0 1 0

0 0 1

1 1 0

0 1� 1 0� � 1 �1 1�

6

The matrix Q can be subdivided into matrices Qt and Ql. Q [Qt : Q U : Ql ] where Qt is the twig matrix, Ql is the link matrix and U is the unit matrix.

5.6.2

Orthogonal Relationship between Matrix B and Matrix Q

For a linear graph, if the columns of two matrices Ba and Qa are arranged in the same order, it can be shown that Q BaT � 0 or

B QaT � 0

If the f-circuit matrix B and the f-cutset matrix Q are written for the same selected tree, it can be shown that B QT � 0 or Q BT � 0 These two equations show the orthogonal relationship between matrices A and B.

�������������������������������������������������� Arranging the columns of matrices A, B and Q with twigs for a given tree first and then the links, we get the partitioned forms as A [ At : Al ] B Q

[ Bt : Bl ] [ Bt : U ] [Qt : Q U : Ql ]

From the orthogonal relation, ABT � 0, � Bt T � AB � [ At : Al ] � � � � � �� Bl T �� T

At Bt T � Al Bl T � 0 At Bt T � � Al Bl T Since At is non-singular, i.e., |A| ��0,�At�1 exists. Premultiplying with At�1, Bt T At�1 Al Bl T Since Bl is a unit matrix

Bt

Bl ( At 1 Al )T

Bt

( At�1. Al )T

5.7��������������������������������������������������

Hence, matrix B is written as B [ ( At�1 Al )T : U ]

�(5.1)

ABT � 0

We know that

Al Bl T � � At Bt T Postmultiplying with (

T �1

) , At Bt T ( Bl T ) �1 � � At Bt T (B ( Bl 1 )T

Al

At ( Bl 1 Bt )T

Hence matrix A can be written as A [ Al : At ( Bl�1 Bt )T ]

�(5.2)

� At [U : � ( Bl 1 Bt )T ] Similarly we can prove that Q

[U : � ( Bl 1 Bt )T ]

A

At Q

Q

At�1 A

�(5.3)

From Eqs (5.2) and (5.3), we can write

We have shown that

At�1[ At : Al ] � [U : At�11 Al ]

Bt

( At�1. Al )T

Bt T

( At�1 Al )

Q

[U : � Bt T ]

Hence, Q can be written as Bt T

Q

������������� For the circuit shown in Fig. 5.19, draw the oriented graph and write the (a) incidence matrix, (b) tieset matrix, and (c) f-cutset matrix. R5 R2 R1 V

L1

R4 R6

+ −

C

R3

I

(4)

��������� Solution For drawing the oriented graph, 1. replace all resistors, inductors and capacitors by line segments, 2. replace the voltage source by short circuit and the current source by an open circuit, 3. assume the directions of branch currents arbitrarily, and 4. number all the nodes and branches. The oriented graph is shown in Fig. 5.20.

1

2 (3) (1)

(2)

3

���������

�����Circuit Theory and Networks—Analysis and Synthesis (a)

Incidence Matrix (A) Nodes Branches � � �1 0 �1 1� � 1 2 3 4 Aa � � 0 1 1 �1� � � 1 �1 0 �1 1 1 � 1 0 0� � 2 0 1 1 �1 3 1 �1 0 0 Eliminating the third row from the matrix Aa, we get the incidence matrix A. � �1 0 �1 1� A� � � 0 1 1 �1��

(b)

Tieset Matrix (B) The oriented graph, selected tree and tiesets are shown in Fig. 5.21. (4) 1

2

Links: {3, 4} Tieset 3: {3, 1, 2} Tieset 4: {4, 1, 2}

(3) (1)

(2)

1

2

3 � �1 1 B� � 4� 1

3 4

1 1 0� 1 0 1��

3

(c)

��������� f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 5.22. (4) 1

2 (3)

(1)

Twigs: {1, 2} f-cutset 1: {1, 3, 4} f-cutset 2: {2, 3, 4}

1 2 3 1�1 0 1 Q� � 2 �0 1 1

(2)

4 1� 1��

3

���������

��������� ���� For the network shown in Fig. 5.23, draw the oriented graph and write the (a) incidence matrix, (b) tieset matrix, and (c) f-cutset matrix. 2� 2� 4� 2� V1 � �

10 � 1F

I 1H

2� � � V2

���������

4F

5.7��������������������������������������������������

Solution For drawing the oriented graph, 1. replace all resistors, inductors and capacitors by line segments, 2. replace all voltage sources by short circuits and current source by an open circuit, 3. assume directions of branch currents arbitrarily, and 4. number all the nodes and branches. The oriented graph is shown in Fig. 5.24. (a)

(6)

(7)

5 2

1

3

(1)

(2) (3)

(4)

(5)

4

Incidence Matrix (A)

���������� Branches � 2 3 4 5 6 7 0 0 1 0 1 0 �1 1 0 0 0 0 1 0 0 1 0 1 0 �1 �1 �1 0 0 0 0 0 0 �1 �1

Nodes � 1 1 1 2 �1 3 0 4 0 5 0

1 0 1 0� � 1 0 0 � �1 �1 1 0 0 0 0 � � � Aa � � 0 1 0 0 1 0 1� � 0 0 � 1 �1 �1 0 0 � �� 0 0 0 0 0 �1 �1��

Eliminating the last row from the matrix Aa, we get the incidence matrix A. 1 0 � 1 0 0 � �1 �1 1 0 0 A� � 1 0 0 1 � 0 � 0 0 �1 �1 �1 (b)

1 0 0 0

0� 0� � 1� 0�

Tieset Matrix (B) The oriented graph, selected tree and tiesets are shown in Fig. 5.25. (6)

(7)

5 2

1 (1)

3 Links: {2, 4, 7} Tieset 2: {2, 3, 5} Tieset 4: {4, 1, 3} Tieset 7: {7, 6, 1, 3, 5}

(2) (3)

(4)

(5)

1 2

3

4

5

1 0 �1 1 1 0 1 0 �1 1

2� 0 1 B � 4 � �1 0 � 7� 1 0

6 7 0 0� 0 0� � 1 1�

4

(c)

��������� f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 5.26. 5

(6)

(7) 2

1

3

(2)

(1) (3) (4)

(5)

4

���������

Twigs: {1, 3, 5, 6} f-cutset 1: {1, 4, 7} f-cutset 3: {3, 4, 7, 2} f-cutset 5: {5, 2, 7} f-cutset 6: {6, 7}

2

3

4 5

6

1�1 0 3 �0 �1 Q� � 5 �0 1 6 �0 0

1

0 1 0 0

1 1 0 0

0 �1� 0 �1� � 0 1� 1 1�

0 0 1 0

7

�����Circuit Theory and Networks—Analysis and Synthesis

������������� For the circuit shown in Fig. 5.27, draw the oriented graph and write (a) incidence matrix, (b) tieset matrix, and (c) cutset matrix. i L1

r2

r1

r5

r4

r6

r3

+

C1

V

L2



���������

1. replace all resistors, inductors and capacitors by line segments, 2. replace voltage source by short circuit and current source by an open circuit, 3. assume directions of branch currents arbitrarily, and 4. number the nodes and branches.

(1)

3

� �1 1 0 �1� Aa � � 0 0 1 1� � � � 1 �1 �1 0 �

Eliminating the third row from the matrix Aa, we get the incidence matrix A. � �1 1 0 �1� A� � � 0 0 1 1�� Tieset Matrix (B) The oriented graph, selected tree and tiesets are shown in Fig. 5.29.

(1)

(4) 2

(2)

(3)

Links: {1, 3} Tieset 1: {1, 2} Tieset 3: {3, 2, 4}

1 1�1 B� � 3 �0

3

��������� (c)

(3)

���������

Nodes Branches � � 1 2 3 4 1 �1 1 0 �1 2 0 0 1 1 3 1 �1 �1 0

1

2

(2)

The oriented graph is shown in Fig. 5.28. (a) Incidence Matrix (A)

(b)

(4)

1

Solution For drawing the oriented graph,

f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 5.30.

2

3

1 0 1 1

4 0� 1��

5.7�������������������������������������������������� (4)

1

(1)

2 Twigs: {2, 4} f-cutset 2: {2, 1, 3} f-cutset 4: {4, 3} (2)

1 2 3 4

(3)

2 � �1 1 1 0 � Q� � 4 � 0 0 1 1��

3

���������

������������� For the circuit shown in Fig. 5.31, (a) draw its graph, (b) draw its tree, and (c) write the fundamental cutset matrix. 2� 1F

1F 1�

1�

I

1� 1H (3)

��������� Solution (a) For drawing the oriented graph, 1. replace all resistors, inductors and capacitors by line segments, 2. replace the current source by an open circuit, 3. assume directions of branch currents, and 4. number all the nodes and branches. The oriented graph is shown in Fig. 5.32. (b) Tree The oriented graph and its selected tree are shown in Fig. 5.33. (3) 2

1

3

(2)

(4) (5)

(1)

(6)

Twigs: {2, 4, 5} f-cutset 2: {2, 1, 3} f-cutset 4: {4, 3, 6} f-cutset 5: {5, 1, 6}

4

��������� (c) Fundamental Cutset Matrix (Q) 1

2

3 4 5 6

2 � 1 1 1 0 0 0� Q � 4 �0 0 �1 1 0 1� � � 5 � 1 0 0 0 1 1�

2

1

3

(2)

(4) (5)

(1)

4

���������

(6)

�����Circuit Theory and Networks—Analysis and Synthesis

��������� ���� The graph of a network is shown in Fig. 5.34. Write the (a) incidence matrix, (b) tieset matrix, and (c) f-cutset matrix. (2) 1

3 (4)

(5) 2

(1)

(3)

(6)

4

��������� Solution (a) Incidence Matrix (A) 1 1 � �1 2� 0 Aa � � 3� 0 4� 1

2

3

4

5

6

1 0 1 0 0 0 �1 1 1 1 0 �1 0 �1 1 0 0

0� 1� � 0� 1�

The incidence matrix A is obtained by eliminating any row from the matrix Aa. 1 0 0� � �1 1 0 A � � 0 0 0 �1 1 1� � � � 0 �1 1 0 �1 0 � (b)

Tieset, Matrix (B) The oriented graph, selected tree and tiesets are shown in Fig. 5.35. (2) 1

3

(5)

(4) 2 (1)

(3)

(6)

Links: {1, 2, 3} Tieset 1: {1, 4, 6} Tieset 2: {2, 4, 5} Tieset 3: {3, 5, 6}

1 2 3 1�1 0 0 B � 2 �0 1 0 � 3 �0 0 1

4

5

1 0 1 �1 0 1

6 1� 0� � 1�

4

(c)

��������� f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 5.36. (2) 1

3

(5)

(4) 2 (1)

(6)

(3)

4

���������

Twigs: {4, 5, 6} f-cutset 4: {4, 1, 2} f-cutset 5: {5, 2, 3} f-cutset 6: {6, 1, 3}

1

2 3

4 5

6

4 � �1 1 0 1 0 0 � Q � 5 � 0 1 �1 0 1 0 � � � 6 � �1 0 1 0 0 1�

5.7��������������������������������������������������

�������������

For the graph shown in Fig. 5.37, write the incidence matrix, tieset matrix and f-cutset

matrix. (5) 2

1

3

(2)

4

(3)

(4) (6)

(1) (7) 5

��������� Solution (a) Incidence Matrix (A) 1

2

1� 1 2� 0 � Aa � 3 � 0 4� 0 5 �� �1

3

4

5

1 0 0 1 1 0 0 �1 1 0 0 �1 1 0 0 0

6

7 0� 0� � 0� 1� 1��

0 0 1 0 0 0 1 1 0 �1

The incidence matrix A is obtained by eliminating any row from the matrix Aa. �1 1 0 0 0 �0 �1 1 0 1 A� � 0 0 1 1 0 � � �0 0 0 �1 �1 (b)

(c)

Tieset Matrix (B) Links: {5, 6, 7} Tieset 5: {5, 3, 4} Tieset 6: {6, 1, 2, 3, 4} Tieset 7: {7, 1, 2, 3, 4}

1 5� 0 B � 6 � �1 � 7� 1

2

3

4

0 0� 0 0� � 0 0� 1 �1� 5

6 7

0 �1 �1 1 0 0 � 1 1 1 0 1 0� � 1 1 1 0 0 1�

f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 5.38. (5) 2

1 (2)

3 (4)

(3) (6)

(1)

(7) 5

���������

4

1 Twigs: {1, 2, 3, 4} f-cutset 1: {1, 6, 7} f-cutset 2: {2, 6, 7} f-cutset 3: {3, 5, 6, 7} f-cutset 4: {4, 5, 6, 7}

1 �1 2 �0 Q� � 3 �0 4 �0

2 3 4 5 0 1 0 0

0 0 1 0

0 0 0 1

6

0 1 0 �1 1 1 1 1

7 1� 1� � 1� 1�

�����Circuit Theory and Networks—Analysis and Synthesis

�������������

For the graph shown in Fig. 5.39, write the incidence matrix, tieset matrix and

f-cutset matrix. (1) 1

2 (5)

(6) (2)

(4) 5 (8)

(7) 3

4 (3)

��������� Solution (a) Incidence Matrix (A) 1

2

1� 1 2 � �1 � Aa � 3 � 0 4� 0 5 �� 0

3

4

5

6

7

8

1 1 0 0 0� 0 0 1 0 0� � 0 0 0 1 0� 1 0 0 0 1� 0 �1 �1 �1 �1��

0 0 1 0 1 1 0 �1 0 0

The incidence matrix is obtained by eliminating any one row. � 1 0 0 �1 � �1 1 0 0 A� � � 0 �1 1 0 � 0 0 �1 1 (b)

(c)

Tieset Matrix (B) Links: {2, 4, 6, 8} Tieset 2: {2, 7, 5, 1} Tieset 4: {4, 5, 7, 3} Tieset 6: {6, 5, 1} Tieset 8: {8, 7, 3}

1 2 3 4 2�1 4 �0 B� � 6 �1 8 �0

1 0 0 0

0 1 0 1

1 0 0 0

0 1 0 0

5 6

0 �1 1 1 0 �1 0 0

0� 0� � 0� 1�

0 0 1 0 7

0 1 0 �1 1 0 0 �1

8 0� 0� � 0� 1�

f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 5.40. (1) 1

2 (5)

Twigs: {1, 3, 5, 7} f-cutset 1: {1, 6, 2} f-cutset 3: {3, 4, 8} f-cutset 5: {5, 4, 6, 2} f-cutset 7: {7, 2, 8, 4}

(6) (2)

(4) 5 (8)

(7)) 3

4 (3)

���������

1 1�1 3 �0 Q� � 5 �0 7 �0

2 3

4

5

6

7

8

1 0 1 1

0 1 1 1

0 0 1 0

1 0 1 0

0 0 0 1

0� 1� � 0� 1�

0 1 0 0

5.7��������������������������������������������������

��������������

How many trees are possible for the graph of the network of Fig. 5.41. 2

3

+ −

1

���������� (2)

Solution To draw the graph, 1. replace all resistors, inductors and capacitors by line segments, 2. replace voltage source by short circuit and current source by an open circuit, 3. assume directions of branch currents arbitrarily, and 4. number all the nodes and branches. The oriented graph is shown in Fig. 5.42. The complete Incidence Matrix (Aa) is written as 1

2

3

2

3 (1)

(3) (4) 1

���������

4

1 � 1 0 �1 Aa � 2 � �1 1 0 � 3 � 0 �11 1

1� 0� � 1�

The reduced incidence matrix A is obtained by eliminating the last row from matrix Aa. � 1 0 �1 1� A� � � �1 1 0 0 �� � 1 �1� 1� � 3 �1� � 1 0 �1 1� � 0 AA � � � ��� � � 1 1 0 0 1 0 � � �� � � � �1 2� 1 0 � � 3 �1 T The number of possible trees � AA � � 6 � 1 � 5. �1 2 T

��������������

Draw the oriented graph from the complete incidence matrix given below;

� 1 2 3 4 5

(8)

Branches �

Nodes 1

2

3

4

5

6

7

8

1 0 0 0 1 0 0 1 0 1 0 0 �1 1 0 0 0 0 1 0 0 �1 1 �1 0 0 0 1 0 0 �1 0 �1 �1 �1 �1 0 0 0 0

1

(5)

(1)

2

(6)

(2)

(3)

3

(7)

4

(4) 5

���������

Solution First, note down the nodes 1, 2, 3, 4, 5 as shown in Fig. 5.43. From the complete incidence matrix, it is clear that the branch number 1 is between nodes 1 and 5 and it is going away from node 1 and towards node 5 as the entry against node 1 is 1 and that against 5 is �1. Hence, connect the nodes 1 and 5 by a line, point the arrow towards 5 and call it branch 1 as shown in Fig. 5.43. Similarly, draw the other oriented branches.

5.22�Circuit Theory and Networks—Analysis and Synthesis

��������� �����

The reduced incidence matrix of an oriented graph is given below. Draw the

graph. � 0 �1 1 1 0 � � � 0 0 �1 �1 �1� � � 1� � �1 0 0 0 Solution First, writing the complete incidence matrix from the matrix A such that the sum of all entries in each column of Aa will be zero, we have (4)

1

2

3

4

5

1 � 0 �1 1 2 � 0 0 �11 Aa � � 3 � �1 0 0 4� 1 1 0

1 1 0 0

0� 1� � 1� 0�

2

1 (3)

3 (5) (1)

(2)

4

��������� Now, the oriented graph can be drawn with matrix Aa as shown in Fig. 5.44.

��������������

The reduced incidence matrix of an oriented graph is � 0 �1 1 0 0 � A � � 0 0 �1 �1 �1� � � 1� � �1 0 0 0

(a) Draw the graph. (b) How many trees are possible for this graph? (c) Write the tieset and cutset matrices. Solution (a)

First, writing the complete incidence matrix Aa such that the sum of all the entries in each column of Aa is zero, we have 2

1

1 1� 0 2� 0 Aa � � 3 � �1 4� 1

2

3

4

5

1 1 0 �11 0 0 1 0

0 1 0 1

0� 1� � 1� 0�

(3)

3 (5)

(4)

(2)

(1)

4

��������� Now, the oriented graph can be drawn with the matrix Aa, as shown in Fig. 5.45. (b) The number of possible trees � |AAT| � 0 0 �1� � 0 �1 1 0 0 � � �1 0 0 � � 2 �1 0 � � � AAT � � 0 0 �1 �1 �1� � 1 �1 0 � � � �1 3 �1� � � � � 1� � 0 �1 0 � � 0 �1 2� � �1 0 0 0 �� 0 �1 1��

5.7��������������������������������������������������

2 �1 0 AAT � �1 3 �1 � 2(6 � 1) � 1( �2) � 8 0 �1 2 (c)

The number of possible trees � 8. Tieset Matrix (B) The oriented graph, selected tree and tiesets are shown in Fig. 5.46. 2

1

3 (3)

(5) (4)

(2)

1

(1)

2 3

4 5

1 �1 0 0 �1 1 � B� � 2 �0 1 1 �1 0 ��

Links: {1, 2} Tieset 1: {1, 4, 5} Tieset 2: {2, 3, 4}

��������� f-cutset Matrix (Q) The oriented graph, selected tree and f-cutsets are shown in Fig. 5.47 2

1

3 (3)

(5) (4)

(2)

(1)

Twigs: {3, 4, 5} f-cutset 3: {3, 2} f-cutset 4: {4, 2, 1} f-cutset 5: {5, 1}

1 3� 0 Q � 4� 1 � 5 � �1

2 3 4 5 1 1 0 0� 1 0 1 0� � 0 0 0 1�

4

���������

��������������

The fundamental cutset matrix of a network is given as follows; Twigs g a c e

Links b d f

1 0 0 0 1 0 0 0 1

1 0 1 0 1 1 1 1 1

Draw the oriented graph. Solution No. of links l � b � n � 1 No. of nodes n � b � l � 1 ��6 � 3 � 1 � 4 f-cutsets are written as, f-cutsets a: {a, b, f } f-cutsets c: {c, d, f } f-cutsets e: {e, b, d, f } The oriented graph is drawn as shown in Fig. 5.48.

(f)

(a) (b)

(d) (e)

����������

Twigs: {a, c, e} Links: {b, d, f} (c)

5.24�Circuit Theory and Networks—Analysis and Synthesis

��������������

Draw the oriented graph of a network with the f-cutset matrix as shown below: Twigs g 1 1 0 0 0

2 0 1 0 0

3 0 0 1 0

Links k 4 5 0 �1 0 1 0 0 1 0

Solution No. of links l � b � n � 1 No. of nodes n � b � l � 1 ��7 � 3 � 1 � 5 f-cutsets are written as f-cutset 1: {1, 5} f-cutset 2: {2, 5, 7} f-cutset 3: {3, 6, 7} f-cutset 4: {4, 6} Then oriented graph can be drawn as shown in Fig. 5.49.

6 0 0 1 1

7 0 1 1 0

(5)

(2)

(7)

(3)

(1) (4)

Twigs: {1, 2, 3, 4} Links: {5, 6, 7}

(6)

���������

������������������������������ KVL states that if vk is the voltage drop across the kth branch, then

� vk � 0 the sum being taken over all the branches in a given loop. If l is the number of loops or f-circuits, then there will be l number of KVL equations, one for each loop. The KVL equation for the f-circuit or loop ‘l’ can be written as b

� bik

k

�0

(i � 1, 2, �, l)

k �1

where bik is the elements of the tieset matrix B, b being the number of branches. The set of l KVL equations can be written in matrix form. BV Vb � 0

where

� v1 � �v � Vb � � 2 � is a column vector of branch voltages. ��� � vb �

and B is the fundamental circuit matrix.

������������������������������ KCL states that if ik is the current in the kth branch then at a given node

� ik � 0 the sum being taken over all the branches incident at a given node. If there are ‘n’ nodes, there will ‘n’ such equations, one for each node

����������������������������������������������b�����������������������������������������������������5.25 b



ik ik

�0

(i � 1, 2,� , n)

k �1

so that set of n equations can be written in matrix form. Aa I b � 0

…(5.4)

� i1 � �i � I b � � 2 � is a column vector of branch currents. where ��� �ib � and Aa is the complete incidence matrix. If one node is taken as reference node or datum node, we can write the Eq. (5.4) as, AI b � 0 where A is the incidence matrix of order (n � 1) ��b.

…(5.5)

A At Q We know that Equation (5.5) can be written as A QI QI b � 0 �1 Premultiplying with At , At 1 At Q I b � At�1. 0 I Q Ib � 0 Q Ib � 0 where Q is the f-cutset matrix.

5.10�

����� ��������������������������������������Vb�������� ���������������Vt�������������������������Vn BV Vb � 0

We know that

�Vt � : Bl ] ��� � 0 � � �Vl �

[

Bt Vt � Bl Vl � 0 Bl Vl � � Bt Vt Premultiplying with Bl�1 . Vl Now

Bl�11 Bt Vt � � ( Bl 1 Bt )Vt

�Vt � Vb � ��� � � �Vl � � � �� �� �



Vt �

� � � � U ��� � Vt �� �� �







� � � .Vt � �

�(5.6)

5.26�Circuit Theory and Networks—Analysis and Synthesis Also,

T

V Q

Vt

At�1 A AT ( AtT ) �1

QT

AT ( At 1 )T

Vb

AT ( AtT ) 1Vt � AT

Vn

( AtT ) �1Vt is node voltage matrix.

Hence Eq. (5.6) can be written as

where

�A

T t

Vt



AT Vn

����� ��������������������������������������Ib ���������� ������ ���������������Il A Ib � 0

We know that,

� It � [ A : At ] ��� � 0 � � � Il � At I t � Al I l � 0 At I t � � Al I l Premultiplying with At�1 , At�11 Al I l � � ( At 1 Al ) I l

It

1 1 � I t � � � ( At Al ) I l � � � ( At Al ) � � � � � � Ib � � � � � � .I l � � � � � � � Il �� �� �� U � I l � ��

Now

I b � BT I l

����������������������������������� 5.12.1 KVL Equation 1.

If there is a voltage source vsk in the branch k having impedance zk and carrying current ik, as shown in Fig. 5.50, vk

zk ik

vsk

(k � 1, 2,�,�b)

In matrix form,

vsk

Zk

+

−+

ik

Vb

Zb I b Vs

where Zb is the branch impedance matrix, Ib is the column vector of branch currents and Vs is the column vector of source voltages. Hence, KVL equation can be written as BV Vb � 0 B ( Zb I b Vs ) � 0 B Zb I b BV Vs

vk

�������������������������



5.12����������������������������������

Ib

Also,

BT I l

B Zb BT I l � BV Vs Z Il � E where E BV Vs and Z B Z b BT The matrix Z is called loop impedance matrix. 2.

If there is a voltage source in series with an impedance and a current source in parallel with the combination as shown in Fig. 5.51, �v � vsk ) ik � k � isk zk vk � zk ik zk isk vsk In matrix form, Vb Zb I b Zb I s Vs KVL equation is BV Vb � 0.

vk ik +

Zk

−+



vsk isk

�������������������������

BV Vb � B ( Zb I b Zb I s Vs ) � 0 B Zb I b BV Vs B Zb I s Now

Ib

BT I l

B Zb BT I l � BV Vs � B Zb I s Z I l � BV Vs � B Zb I s where Z

5.12.2 1.

B Zb BT is the loop impedance matrix. This is the generalised KVL equation.

KCL Equation

If the branch k contains an input current source isk and an admittance yk as shown in Fig. 5.52, i

yk vk

isk

Ib

Yb Vb

Is

ik +

vk yk



(k � 1, 2,�, b)

In the matrix form, where Yb is the branch admittance matrix. Hence KCL equation is given by, A Ib � 0 A (Yb Vb I s ) � 0 AY Yb Vb A I s Also

Vb

AT Vn

AY Yb AT Vn � A I s Y Vn � I where and

Y I

AY Yb AT A Is

isk

�������������������������

�����Circuit Theory and Networks—Analysis and Synthesis The matrix Y is called admittance matrix. This is the KCL equation in matrix form. In terms of f-cutset matrix, the KCL equation can be written as Q Ib � 0 Q (Yb Vb Vs ) � 0 QY Yb V Q I s Also

V

QT Vt

QY Yb QT Vt � Q I s Y Vt � I

2.

where Y QY Y QT and I Q Is This is the KCL equation in matrix form. If there is a voltage source in series with an impedance and a current source in parallel with the combination as shown in Fig. 5.53, 1 yk � zk i yk v � yk vsk isk In matrix form, I b Yb Vb Yb Vs I s

ik +

zk

vk

−+ vsk isk

�������������������������

KCL equation will be given by, A (Yb Vb

A Ib � 0 Yb Vs I s ) � 0 AY Yb Vb

Also

Vb

A Is

AY Yb Vs

AT Vn

AY Yb AT Vn � A I s � AY Yb Vs Y Vn � A I s � AV Vb Vs T

where Y AY Yb A is the node admittance matrix. This is a generalised KCL equation. In terms of f-cutset matrix, the KCL equation can be written as Q (Yb Vb

Q Ib � 0 Yb Vs I s ) � 0 QY Yb V

Also

V

QI

QY Yb Vs

QT Vt

QY Yb QT Vt � Q I s � QY Yb Vs Y Vt � Q I s � QY Yb Vs This is a generalised KCL equation.



5.12����������������������������������

Note (i) For a graph having b branches, the branch impedance matrix Zb is a square matrix of order b, having branch impedances as diagonal elements and the mutual impedances between the branches as nondiagonal elements. For a network having no mutual impedances, only diagonal elements will be present in the branch impedance matrix. (ii) For a graph having b branches, the branch admittance matrix Yb is a square matrix of order b, having branch admittances as diagonal elements and the mutual admittances between the branches as nondiagonal elements. For a network having no mutual admittances, only diagonal elements will be present in the branch admittance matrix. (iii) For a graph having b branches, the voltage source matrix or vector Vs is a rectangular matrix of order b � 1, having the value of the voltage source in the particular branch. The value will be positive if there is a voltage rise in the direction of current and will be negative if there is a voltage fall in the direction of current. (iv) For a graph having b branches, the current source matrix or vector Is is a rectangular matrix of order b � 1, having the value of the current source in the particular parallel branch. The value will be positive if the direction of the current source and the corresponding parallel branch current are not same. The value will be negative if the directions of the current source and corresponding parallel branch current are same.

�������������� Write the incidence matrix of the graph of Fig. 5.54 and express branch voltages in terms of node voltages. Write the tieset matrix and express branch currents in terms of loop currents. (8) (5)

(6)

2

1

3

(2)

(1)

5

(7)

(3)

(4)

��������� Solution (a) Incidence Matrix 1

2

3

4

5

6

7

8

1� 1 0 0 0 1 0 2� 0 1 0 0 �1 1 � Aa � 3 � 0 0 1 0 0 �1 1 4� 0 0 0 1 0 0 5 �� �1 �1 �1 �1 0 0

0 0 1 1 0

1� 0� � 1� 0� 0 ��

The incidence matrix is obtained by eliminating any one row. �1 �0 A� � �0 �0

0 1 0 0

0 0 1 0

0 1 0 0 1� 0 �1 1 0 0 � � 0 0 �1 1 �1� 1 0 0 �1 0 �

�����Circuit Theory and Networks—Analysis and Synthesis (b)

Branch voltages in terms of node voltages AT Vn

Vb

�V1 � � 1 �V2 � �0 � � � �V3 � �0 �V4 � � �0 �V5 � � 1 �V � �0 � 6� � �V7 � �0 ��V8 �� �� 1 (c)

0 0 1 0 0 1 0 0 1 0 1 �1 0 1 0 �1

0� 0� � 0 � �Vn1 � 1� �Vn2 � � � 0 � �Vn3 � � 0 � ��Vn4 �� 1� 0 ��

Tieset Matrix Selected tree and tiesets are shown in Fig. 5.55. (8) (5) 1

(6)

2

(2)

(1)

1

3

(7)

(3)

Links: {1, 4, 6, 8} Tieset 1: {1, 2, 5} Tieset 4: {4, 3, 7} Tieset 6: {6, 2, 3} Tieset 8: {8, 2, 3, 5}

1 �1 4 �0 B� � 6 �0 8 �0

2

3 4

1 0 0 �1 1 1 1 1

5 6 7 8

0 �1 1 0 0 0 0 �1

0 0 1 0

0 1 0 0

0� 0� � 0� 1�

4

5

(4)

��������� (d) Branch currents in terms of loop currents Ib

BT I l

� I1 � � 1 0 0 0 � � I 2 � � �1 0 �1 �1� � �I � � � � � I 3 � � 0 �1 1 1� � l1 � 1 0 0 � I l4 �I4 � � � 0 � I 5 � � �1 0 0 �1� �� I l6 �� �I � � 0 0 1 0 �� �� I l8 �� � 6� � 1 0 0� � I7 � � 0 �� I 8 �� �� 0 0 0 1��

��������������

Branch current and loop-current relationships are expressed in matrix form as

Draw the oriented graph.

� I1 � � 1 0 0 �I2 � � 0 1 0 � � � I 0 1 1 3 � � � 1 1 �I4 � � � 0 � I 5 � � 1 �1 0 � I � � 0 0 �1 � 6� � � I 7 � � �1 0 0 �� I 8 �� �� 0 0 0

1� 1� � 0 � � I l1 � 0 � � I l2 � � � 0 � � I l3 � � 0 � �� I l4 �� 0� 1��

5.12�Network Equilibrium Equation�5.31

Solution Writing the equation in matrix form, Ib

BT I l

� 1 0 0 �1� � 0 1 0 �1� � � 1 1 0� � 0 0 1 1 0� BT � �� 1 �1 0 0 � � 0 0 �1 0 � � � � �1 0 0 0 � �� 0 0 0 1��

1 � 1 � 0 �B � � � 0 � �11

2 3 4 0 1 0 1

0 1 1 0

5

0 1 1 �1 1 0 0 0

6

7

0 �1 0 0 1 0 0 0

8 0� 0� � 0� 1�

(8) (1)

1

2

(2)

3

(5)

(7)

From tieset matrix, No. of links l�4 No of branches b � 8 No of nodes n ��b � l � 1 ��8 � 4 � 1 ��5 The oriented graph is shown in Fig. 5.56.

(3)

4

5

(4)

(6)

���������

��������� ����� For the given graph shown in Fig. 5.57, write down the basic tieset matrix and taking a tree of branches 2, 4, 5, write down KVL equations from the matrix. (2) 1 (5) (1)

2

(3)

3

4 (4)

(6)

��������� Solution Selecting branches 2, 4, and 5 as the tree as shown in Fig. 5.58, (2) 1

2

1 2 (5) (1)

(3)

Links: {1, 3, 6} Tieset 1: {1, 5, 4} Tieset 3: {3, 2, 5} Tieset 6: {6, 2, 5, 4}

3

4 (4)

(6)

���������

3

1�1 0 0 B � 3 �0 1 1 � 6 �0 1 0

4 5 6 1 1 0� 0 1 0� � 1 1 1�

�����Circuit Theory and Networks—Analysis and Synthesis The KVL equation in matrix form is given by BV Vb � 0 �V1 � �V � � 1 0 0 �1 1 0 � � 2 � �0 1 1 0 1 0 � �V3 � � 0 � � �V4 � �0 1 0 �1 1 1� �V � 5 �V � � 6� V1 V4 � V5 � 0 V2 V3 � V5 � 0 V2 V4 � V5 V6 � 0

�������������� Obtain the f-cutset matrix for the graph shown in Fig. 5.59 taking 1, 2, 3, 4 as tree branches. Write down the network equations from the f-cutset matrix. 2

(5)

(6)

(3)

3 1 (2)

(7) (8)

(1)

(4)

4

��������� 1 2 3

Solution Twigs: {1, 2, 3, 4} f-cutset 1 : {1, 6, 7, 8} f-cutset 2 : {2, 5, 6, 7, 8} f-cutset 3 : {3, 5, 6} f-cutset 4 : {4, 6, 7}

1 �1 2 �0 Q� � 3 �0 4 �0

The KCL equation in matrix form is given by Q Ib � 0

�1 �0 � �0 �0

0 1 0 0

0 0 1 0

0 0 0 1

0 1 �1 1 �1 1 1 �1 0 0 1 �1

� I1 � �I2 � � � 1� � I 3 � �1� � I 4 � �0 � 0� � I 5 � � � 0� � I 6 � � I7 � �� I 8 ��

0 1 0 0

0 0 1 0

4 5

6

7

8

0 0 0 1

1 1 1 1

1 1 0 1

1� 1� � 0� 0�

0 1 1 0

5.12����������������������������������

I2

��������������

I1 I 6 � I 7 I 8 I 5 � I 6 I 7 � I8 I3 I5 � I6 I 4 I6 � I7

�0 �0 �0 �0

The reduced incidence matrix of a graph is given as

� 1 0 0 0 �1� A � � �1 �1 �1 0 0 � � � 1 �1 0 � � 0 0 Express branch voltages in terms of node voltages. Solution For the given graph, No of branches b � 5 No of nodes n � 3 Branch voltages can be expressed in terms of node voltages by Vb

AT Vn

�V1 � � 1 �1 0 � �V2 � � 0 �1 0 � �Vn1 � �� � � � � �V3 � � � 0 �1 1� �Vn2 � �V4 � � 0 0 �1� �Vn3 � ��V5 �� �� �1 0 0 �� V1 Vn1 � Vn2

��������������

V2

Vn2

V3

Vn2 � Vn3

V4

Vn3

V5

Vn1

The fundamental cutset matrix of a graph is given as

� �1 1 0 0 �1� Q � � 0 0 1 0 �1� � � �1 0 0 1 0� Express branch voltages in terms of twig voltages. Solution For the given graph, No. of branches b � 5 No. of twigs � 3 Branch voltages are expressed in terms of twig voltages by V

QT Vt

�V1 � � �1 0 �V2 � � 1 0 � � � 1 �V3 � � � 0 V 0 0 � 4� � ��V5 �� �� �1 �1

1� 0 � �Vt1 � � 0 � �Vt2 � � � 1� �Vt3 � 0 ��

�����Circuit Theory and Networks—Analysis and Synthesis Vt1 � Vt3

V1 V2

Vt1

V3

Vt2

V4

Vt3 Vt1 � Vt2

V5

�������������� For this network shown in Fig. 5.60, write down the tieset matrix and obtain the network equilibrium equation in matrix form using KVL. Calculate the loop currents and branch currents.

2�

1�

1�

2V

� �

2�

2� 1�

��������� Solution The oriented graph and one of its trees are shown in Fig. 5.61. 1

(1) (5)

3

(4) 2 (3)

1

(1)

(2)

(5)

(6)

4

3

(2)

(4)

Links: {1, 2, 3} Tieset 1: {1, 4, 5} Tieset 2: {2, 4, 6} Tieset 3: {3, 5, 6}

(6)

2

1 2 3 1 �1 0 0 B � 2 �0 1 0 � 3 �0 0 1

4

(3)

��������� The KVL equation in matrix form is given by B Zb BT I l � BV Vs � B Zb I s I s � 0,

Here, T

B Zb B I l � BV Vs

Zb

�1 �0 � �0 �0 �0 �0 �

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 2 0 0

0 0 0 0 2 0

0� � 2� � 1 0 0� �0 � �0 1 0� 0� � � � � � 1� 0 0� T �0 0 ;V Vs � � � ;B � 0� �0 � � 1 �1 0 � �0 � � 1 0 �1� 0� �0 � �0 �1 1� 2�� � � � �

4

5

1 1 1 0 0 �1

6 0� 1� � 1�

5.12����������������������������������

1 1 �1 0 0 B Z b � � 0 1 0 �1 0 � � 0 0 1 0 �1

2 2 �1 0 0 B Zb BT � �0 1 0 �2 0 � �0 0 1 0 �2

�1 �0 0� � 0 �1� � � 0 1� �� 0 �0 � �1 �0 0� � 0 �2� � � �1 2� � 1 �0 �

0 1 0 0 0 0

0 0 1 0 0 0

� 5 � �2 � 2 � �2

2 �2� � I l1 � � 2� 5 �2� � I l2 � � �0 � �� � � � 2 5� � I l3 � �0 �

Solving this matrix equation, 6 A 7 4 I l2 � A 7 4 I l3 � A 7 I l1 �

The branch currents are given by Ib

BT I l

� I1 � � 1 0 � I 2 � �0 1 � � � I 0 0 � 3� � � � I 4 � � 1 �1 � I5 � � 1 0 � I � �0 �1 � 6� �

�6� �7� �4� 0� � 6 � � � 0� � 7 � � 7 � � � � �4� 1� � 4 � � � � 0� � 7 � � 7 � 2 �1� � 4 � � � �7� � � � 1� � 7 � � � 2 � � �7� �0�

0 0 0 0 2 0

0� 0� � �1 0 0 0� � � 0 1 0 0� � 0 0 1 0� � � 2�

0 0� 1 0� � � 5 �2 �2� 0 1� � � �2 5 �2� �1 0 � � � � 2 � 2 5� 0 �1� � �1 1��

� 2� �0 � 1 1 0 � � � � 2� �1 0 0 0 BV Vs � �0 1 0 �1 0 �1� � � � �0 � � � �0 � � � �0 0 1 0 �1 1� �0 � �0 � �0 � � � The KVL equation in matrix form is given by

0 0 0 2 0 0

2 2 2 0 0 �2

0� 2� � 2�

�����Circuit Theory and Networks—Analysis and Synthesis

�������������� For the network shown in Fig. 5.62, write down the tieset matrix and obtain the network equilibrium equation in matrix form using KVL. Calculate loop currents. 2�

8V ��

4�

6�

Il 3

6�

2� I l1

12 V � �

4�

Il2

� � 6V

��������� Solution The oriented graph and its selected tree are shown in Fig. 5.63. (3)

(3)

2

1

(4)

3

(5)

(1)

(6)

2

1

(2)

(4)

3

(5)

(1)

(6)

4

Links: {1, 2, 3} Tieset 1: {1, 4, 6} Tieset 2: {2, 5, 6} Tieset 3: {3, 5, 4}

(2)

1 2

3

4

5

6

1�1 0 0 1 0 1� B � 2 �0 1 0 0 1 �1� � � 3 �0 0 1 �1 �1 0 �

4

��������� The KVL equation in matrix form is given by B Zb BT I l � BV Vs � B Zb I s I s � 0,

Here, T

B Zb B I l � BV Vs 0� �1 0 �0 1 0� � � 0� T �0 0 ;B � Zb 0� �1 0 �0 1 0� � � 1 �1 2� � �6 0 0 �0 4 0 1 0 1� � �1 0 0 0 0 2 B Zb � �0 1 0 0 1 �1� � � � �0 0 0 �0 0 1 �1 �1 0 � �0 0 0 �0 0 0 � �6 �0 � �0 �0 �0 �0 �

0 4 0 0 0 0

0 0 2 0 0 0

0 0 0 4 0 0

0 0 0 0 6 0

0� � 12� � �6 � 0� � � � 1� �8 ;V Vs � � � �1� � 0� � 0� �1� � � 0� 0� � � 0 0 0� 0 0 0� 4 0 � �6 0 0 0 0 0� � � 0 4 0 0 6 4 0 0� � � � 6 0 0 2 4 � 0 6 0� 0 0 2��

� 1 0 0� �0 1 0� 4 0 2� � �6 0 0 � � 12 �2 �4 � 1� � 0 0 BZb BT � �0 4 0 � �1 12 �6 � 0 6 �2 � � � � � 1 0 �1� � � � � �4 �6 12� 0 0 2 4 6 0 � � �0 1 �1� � � � � 1 �1 0 �

2� 2� � 0�

5.12����������������������������������

�1 0 0 BV Vs � � 0 1 0 � �0 0 1

� 12� � �6 � 0 1� � � � 12� �8 1 �1� � � � � �6 � � � 0� � � 1 0 � � � � �8� 0 � 0� � �

1 0 1

Hence, the KVL equation in matrix form is given by 4 � � I l1 � � 12 � 6 � � I l2 � � � �6 � �� � � � 12� � I l3 � � �8�

� 12 �22 � �22 12 � 6 � �44 Solving this matrix equation,

I l1 � 0 55A I l2 � �0.866 A I l3 � �0.916 A

��������� ����� For the network shown in Fig. 5.64, draw the oriented graph. Write the tieset schedule and hence obtain the equilibrium equation on loop basis. Calculate the values of branch currents and branch voltages. � 1V �

1A 1� 1�

1�

1�

1�

���������� Solution The oriented graph and one of its trees are shown in Fig. 5.65. 1

1 (4)

(1) (2)

2

(6)

3

(4) (1)

(3)

(5)

(2)

4

2

(6)

3

Links: {1, 2, 3} Tieset 1: {1, 4, 5, 6} Tieset 2: {2, 4, 5} Tieset 3: {3, 4}

(3)

(5)

4

��������� 1 2 3 1�1 0 0 B � 2 �0 1 0 � 3 �0 0 1

4

5

1 1 1 1 1 0

6 1� 0� � 0�

�����Circuit Theory and Networks—Analysis and Synthesis The KVL equation in matrix form is given by B Zb BT I l � BV Vs � B Zb I s 0� � 1 � 0 0� � � 0� T � 0 ;B � 0� � �1 � 1 0� � � �1 1� � �1 0 0 �0 1 0 � 1 0 0 �1 1 �1� � 0 0 1 B Zb � �0 1 0 �1 1 0 � � � � �0 0 0 �0 0 1 1 0 0 � �0 0 0 �0 0 0 � �1 �0 � 0 Zb � � �0 �0 � �0

0 1 0 0 0 0

0 0 1 0 0 0

�1 0 0 BZb BT � �0 1 0 � �0 0 1

�1 0 0 BV Vs � �0 1 0 � �0 0 1

�1 0 0 B Zb I s � �0 1 0 � �0 0 1

0 0 0 0 0 0

0 0 0 0 1 0

� 3 1 0 � � I l1 � � �1� � �1� � 0 � �1 2 0 � � I l � � � �1� � � 0 � � � �1� � �� 2� � � � � � � �0 0 1 � � I l3 � � 1� � 0 � � 1� Solving this matrix equation, � A 5 3 I l2 � � A 5 I l3 � 1 A

0 0 0 0 0 0

0� �0 � �0 � 0� � � � 0 1� ; Vs � � � ; I s 1� �1 � �0 � 0� � �0 � 0� � � 0 0� 0 0� � �1 0 0 0� � � 0 1 0 0� � 0 0 1 0� � 0 1 ��

� 1 0 0� � 0 1 0� 0 1 �1� � � �3 1 0� 0 0 1� � � 1 2 0� 0 1 0� � � � �1 �1 1� � � 0 0 0� � 0 0 1� 1 1 0� � � � � �1 0 0 � �0 � �0 � �1 1 �1� � � � �1� 0 �1 1 0 � � � � � �1� � �1 � � � 1 0 0 � � � � 1� 0 �0 � � � � �1� � 0� 0 1 �1� � � � �1� 0 0 1 0� � � � � 0� � � 0� � � 0 0 0� � � � 0� 0 � 0� � �

Hence, the KVL equation in matrix form is given by

I l1 �

0 1 0 �1 1 0

� �1� � 0� � � 0 �� � � 0� � 0� � 0� � � 0 0 1 0 0 1 1 0 0

1� 0� � 0�

5.12����������������������������������

The branch currents are given by Ib

BT I l

Is

� I1 � � 1 0 �I2 � � 0 1 � � � I 0 0 3 � ��� � I 4 � � �1 �1 � I5 � � 1 1 � � � � I 6 � � �1 0

� 4� �� 5 � � 3� � �1� � � � 0� � 5� � 1� 0� � � � 0� � 1 � � � � 5 1� � � � � 0 � � � � 3 � 7� 1� � � � � 0 � � � � 5� � � 5 � 0 � 1� 0 � � � 2� � � 0� � 0� � � � 5� � 1� �� � � 5�

The branch voltages are given by Vb

Zb I b Vs

�V1 � �1 �V2 � �0 � � � �V3 � � �0 �V4 � �0 �V5 � �0 � � � �V6 � �0

��������������

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 0 0 0

0 0 0 0 1 0

� 4� � 4� �� 5 � �� 5 � � 3� � 3� � � �0 � 0� � � 5� � � � � � � � 5� 0 � � 0 � 1 � � 0 � � 7 � � �0 � � � 1� � � 0 � � � �1 � � �1� 5 � � 0 � � 2 � �0 � � � 2 � � � � 1 � � � � �0 � � 5 � 5 � 1� � � �� � � � 1� � 5� � 5�

For the network shown in Fig. 5.66, obtain the loop equation in matrix form. R1 R2

I

V

R4 R3

��������� Solution The oriented graph and one of its trees are shown in Fig. 5.67. 1

1 (2)

(1)

2

(2) (4)

(1)

2

(3) 3

(3) 3

���������

Links: {1, 4} (4) Tieset 1: {1, 2, 3} Tieset 4: {4, 2, 3}

1 2 3 4 1 �1 1 1 0� B� � 4 �0 1 1 1 ��

�����Circuit Theory and Networks—Analysis and Synthesis The KVL equation in matrix form is given by B Zb BT I l � BV Vs � B Zb I s 0 0 0� � R1 � 0 R2 0 0� T Zb � � �; B 0 0 0� R 3 � 0 0 R4 � � 0 0 � R1 �1 1 1 0 � � 0 R2 B Zb � � � 0 �0 1 1 1 �� � 0 0 � 0

�0� �V � �1 0� �I � �0� �1 1� �� � ;Vs � � � ; I s � � � 1 1 0 �0� � � � � �0� �0� �0 1� 0 0� 0 0 � � R1 R2 R3 0� �� R3 0 � �� 0 R2 R3 R4 �� 0 R4 �

�1 0� 0 � �1 1 � � R1 � �� R4 �� �1 1 � �� �0 1� �V � �1 1 1 0 � � 0 � �V � BV Vs � � � �� �0 1 1 1 �� � 0 � �� 0 �� �0� �0 � 0 � � I � � R2 I � � R1 R2 R3 B Zb I s � � � �� � 0 R2 R3 R4 �� �0 � �� R2 I �� �0 � Hence, the KVL equation in matrix form is given by � R R2 B Zb B � � 1 � 0 R2 T

� R1 ��

R2 � R3 R2 R3

R3 R3

R2 � R3 � � I l1 � �V � � R2 � � R2 R3 � R4 �� �� I l4 �� �� 0 �� �� R2

R2 � R3 R2 R3

R2 � R3 � R2 R3 � R4 ��

I � �V R2 I � � I �� �� � R2 I ��

�������������� For the network shown in Fig. 5.68, write down the tieset matrix and obtain the network equilibrium equation in matrix form using KVL. 2A

5� j5 �

5�

2� �

10 V

�4� �j



���������

2�

5.12����������������������������������

Solution The oriented graph and its selected tree are shown in Fig. 5.69. (3)

(3)

2

1

(4)

3

(5)

(1)

(6)

2

1

(4)

(2)

(1)

4

3

(5) (6)

(2)

Links: {1, 2, 3} Tieset 1: {1, 4, 6} Tieset 2: {2, 5, 6} Tieset 3: {3, 5, 4}

1 2

3

4

5

1 �1 0 0 1 0 B � 2 �0 1 0 0 1 � 3 �0 0 1 �1 �1

4

��������� The KVL equation in matrix form is given by B Zb BT I l � BV Vs � B Zb I s �2 �0 � 0 Zb � � �0 �0 �0 �

0 2 0 0 0 0

0 0 5 0 0 0

0 0 0 5 0 0

1 �1 0 0 B Zb � �0 1 0 0 � � 0 0 1 �1

5 �2 0 0 B Z b BT � � 0 2 0 0 � �0 0 5 �5

0 0� �1 �0 0 0� � � 0 0� T �0 ;B � 0 0� �1 �0 j5 0� �1 0 � j 4 �� � �2 0 0 �0 2 0 0 1� � 0 0 5 1 �1� � � 0 0 0 �1 0 � �� 0 0 0 �0 0 0 � �1 0 �0 1 0 � j 4� � 0 0 j5 j 4� � � 1 0 � j5 0 � �� 0 1 � 1 �1 �

0 0� �10 � � 0� 1 0� � � � 0 1� 0 ;Vs � � � ; I s 0 �1� � 0� � 0� 1 �1� � 0� 1 0 �� � � 0 0 0� 0 0 0� � �2 0 0 0 0� � � 0 2 5 0 0� � 0 0 0 j5 0� � � 0 0 j 4�

5 0 0 j5 5 � j5

0 0 5

5 0 0 j5 5 � j5

0� 0� j4 �5� � �7 � j 4 1� � j 4 2 � j1 � � j 5� �1� � � �5 � j 5 10 � j 5� �1� � 0 ��

�10 � �0� 1 0 1� � � �10 0� �1 0 0 0 1 �1� � � � � 0 � BV Vs � �0 1 0 0 � �� 0 � � � �0 0 1 �1 �1 0 � � 0 � � 0 � �0� � � �2 0 0 B Zb I s � �0 2 0 � �0 0 5

� 0� � 0� � � �2 �� � � 0� � 0� � 0� � �

� 0� � 0� j 4� � � � 0� �2 j 4� � � � � 0� � � � 0� � 0 � � � � �10 � 0 � 0� � �

j 4� j 4� � 0�

6 1� 1� � 0�

�����Circuit Theory and Networks—Analysis and Synthesis Hence, the KCL equation in matrix form is given by j4 5� � I l1 � �10 � � 0 � �10 � �7 j 4 � j 4 2 � j1 j 5� � I l2 � � � 0 � � � 0 � � � 0 � � �� � � � � � � � 5 j 5 10 � j 5� � I l3 � � 0 � � �10 � �10 � � �5

�������������� For the network shown in Fig. 5.70, write down the tieset matrix and obtain the network equilibrium equation in matrix form using KVL. j 5 .66 �

j5 �

50 �0� V

j10 �

3�

� �

�4� �j

I1

5� I2

��������� Solution The branch currents are so chosen that they assume the direction out of the dotted terminals. Because of this choice of current direction, the mutual inductance is positive. The oriented graph and its selected tree are shown in Fig. 5.71. 1

1

(1)

(2)

(3)

2

(1)

(2)

(3)

Links: {1, 3} Tieset 1: {1, 2} Tieset 3: {3, 2}

1 �1 B�� �0

2

3

1 0� 1 1��

2

��������� The KVL equation in matrix form is given by

Here,

B Zb BT I l � BV Vs � B Zb I s I s � 0, B Zb BT I l � BV Vs j 5 66 � 0 � j5 � 1 0� �50�0�� Zb � � 0 3 � j4 0 � ; BT � 1 1� ; Vs � � 0 � � � � � � � 0 5 j10 � 1� � j 5 66 �0 � 0 � 0 j 5 66 � � j5 j 5.66 � 3 � j4 � j5 � 1 1 0� � B Zb � � 0 3 � j 4 0 ��� 3 � j 4 5 � j10 �� � � j 5 66 �0 �1 1�� � j 5 666 0 5 � j10 � � � 1 0� 3 j4 j 5.66 � � �3 � j 9.66 � � 3 � j1 1 �1� � � 8 � j 6 �� 3 j 4 5 � j10 �� � � � �3 � j 9 66 1� �0 �50�0�� � 1 1 0� � �50 �0�� BV Vs � � 0 �� � � 0 �1 1� � 0 � �� 0 �� � �

� j5 B Z b BT � � � j 5 66

5.12����������������������������������

Hence, the KVL equation in matrix form is given by �3 j 9.66 � � I l1 � �50 �0�� � 3 j1 � �� �33 j 9 66 8 j 6 �� �� I l2 �� �� 0 ��

�������������� For the network shown in Fig. 5.72, write down the tieset matrix and obtain the network equilibrium equation in matrix form using KVL. j4 �

3�

j3 �

� 50 �45� V

j5 � �

� j8 �

I2

I1

���������� Solution The branch currents are so chosen that they assume the direction out of the dotted terminals. Because of this choice of current direction, the mutual inductance is positive. The oriented graph and its selected tree are shown in Fig. 5.73. 1

1

(1)

(2)

(3)

(1)

(2)

(3)

Links: {1, 3} Tieset 1: {1, 2} Tieset 3: {3, 2}

1 �1 B�� �0

2 3 1 0� 1 1��

2

2

��������� The KVL equation in matrix form is given by B Zb BT I l � BV Vs � B Zb I s I s � 0,

Here, T

B Zb B I l � BV Vs j3 0 � �50 �45�� � 1 0� j 5 0 � ; BT � � 1 �1� ; Vs � � 0 � � � � � � 0 � j 8� 1� � 0 � �0 �3 � j 4 j 3 0 � 0 � �3 � j 7 j 8 � 1 1 0� � B Zb � � j3 j5 0 � � � � � � j 3 � j 5 � j8�� �0 �1 1�� � 0 0 � j 8� � �3 � j 4 Zb � � j 3 � � 0

� 1 0� j8 0� � �3 � j15 1 �1� � � j 5 � j8�� � � � � j8 1� �0 �50� 45�� �50� 45�� � 1 1 0� � BV Vs � � 0 ��� � � 0 1 1 � � 0 � � 0 �� � � �

�3 � j 7 B Z b BT � � � � j3

j 8� j 3��

�����Circuit Theory and Networks—Analysis and Synthesis Hence, the KVL equation in matrix form is given by, �3 j15 �� � j8

��������������

j8� � I l1 � �50 �45�� � j 3�� �� I l3 �� �� 0 ��

For the network shown in Fig. 5.74, obtain branch voltages using KCL equation on

node basis. 8�

3�

4�

4�

5� 12 V

24 V

��������� Solution The oriented graph is shown in Fig. 5.75. (3)

1

2

(2)

(4)

(1)

(5) 3

��������� The complete incidence matrix for the graph is 1

2

3

4 5

1 � 1 �1 1 0 Aa � 2 � 0 0 �1 1 � 3 � �1 1 0 �1 1

0� 1� � 1�

Eliminating the last row from the matrix Aa, we get the incidence matrix A. 1

2

3

4 5

1 � 1 �1 1 0 0 � A� � 2 �0 0 �1 1 1�� The KCL equation in matrix form is given by AY Yb AT Vn � AI s � AY Yb Vs Is � 0 ,

5.12����������������������������������

AY Yb AT Vn � AY Yb Vs

Here, �1 �4 0 0 � �0 1 0 � 3 � 1 Yb � � 0 0 8 � �0 0 0 � � �0 0 0 �

� 0� � � 0� � 1 0� 0 0� � � �1 0 � � 12� � T � � � � � � ; Vs � � 0 � 0 0� ; A � � � 1� � 0 � 24 � 1 �� 0 �� 0 �� 1�� 0� � 4 1� 0 � 5� 0

�1 �4 � �0 � � 1 �1 1 0 0 � � AY Yb � � �0 �0 0 �1 1 1�� � �0 � � �0 � 1 1 �1 � � 3 8 AY Yb AT � � 4 1 �0 0 � 8 � 1 1 �1 �4 � 3 8 AY Yb Vs � � 1 �0 0 � 8 � Hence, the KCL equation in matrix form is given by � 17 � 24 � 1 �� � 8

1� � � �V � n1 � �4 � � 4 � 8 � �� � � � � 23 �� ��Vn2 �� � 6 � � �6 � 40 �

Solving this matrix equation, Vn1 � 3 96 V Vn2 � �9 57 V

0

0

0

1 3

0

0

0

1 8

0

0

0

1 4

0

0

0

� 0� � 0� � �1 � � 0� � �4 � �0 0� � � 1� � 5�

� 1 0� � 17 � 0 � � �1 0 � � � � 1 �1� � � 24 1 1 �� � 1 1� � � � 0 4 5� � 0 8 � 1�� � �0� � 0 0 � �12 � � � � �4 � 0 � 1 1 �� � � �� 6 �� � 24 � 4 5� � 0 � � � 0

1� � � 8 23 �� 40 �



1 3

0

1 8 1 � 8

0 1 4

� 0� 1 �� 5�

5.46�Circuit Theory and Networks—Analysis and Synthesis Branch voltages are given by, Vb

AT Vn

� 3 96 � �V1 � � 1 0 � � �3 96 � �V2 � � �1 0 � � � 3 96 � � � � � � � � � V 1 1 � �� �9 57�� � 13.53� � 3� � 1� � �9 57� �V4 � � 0 �� �9 57�� ��V5 �� �� 0 1��

�������������� For the network shown in Fig. 5.76, write down the f-cutset matrix and obtain the network equilibrium equation in matrix form using KCL. 1�

1�

1� 1�

10 V

2A

��������� Solution The oriented graph and its selected tree are shown in Fig. 5.77. (3)

1

(1)

(2)

1

2

(3) 2

(1)

(4)

(2)

3

(4)

Twigs: {2, 4} f-cutset 2: {2, 1, 3} f-cutset 4: {4, 3}

1 2

2 � �1 1 1 0 � Q� � 4 � 0 0 �1 1��

3

��������� The KCL equation in matrix form is given by QY Yb QT Vt � Q I s � QY Yb Vs Yb

�1 �0 � �0 �0

0 1 0 0

� �1 1 QY Yb � � � 0 0

� �1 1 QY Yb QT � � � 0 0

0 0 1 0

3 4

0� � �1 0 � �10 � �0 � � 0� T � 1 0� �0 � 0� � � I � � �;V � � � ; Q � � 0 � s �0 � s � 0 � � 1 1� 1� 1� � 0 � 0� � 2� �1 0 0 0 � 1 0 � �0 1 0 0 � � �1 1 1 0 � � �� �1 1�� �0 0 1 0 � �� 0 0 �1 1�� �0 0 0 1 � � �1 0 � 1 0 � � 1 0 � � 3 1� � �� �1 1�� � 1 1� �� �1 2�� 1� � 0

5.12����������������������������������

�0 � � �1 1 1 0 � �0 � �0 � Q Is � � � �� � 0 0 �1 1�� �0 � �� 2�� � 2� �10 � � �1 1 1 0 � � 0 � � �10 � QY Yb Vs � � � �� � 0 0 �1 1�� � 0 � �� 0 �� �0� Hence, the KCL equation is given by 1� �Vt2 � � 0 � � �10 � �10 � � � � 2�� ��Vt4 �� �� 2�� �� 0 �� �� 2 ��

� 3 �� �1 Solving this matrix equation,

Vt2 � 4.4 V Vt4 � 3 2 V

��������������

Calculate the twig voltages using KCL equation for the network shown in Fig. 5.78. 2�

5� 5�

10 �

10 �

5�

� 910 V �

���������� Solution The oriented graph and one of the trees are shown in Fig. 5.79. (5)

(3) (1)

(5)

(4) (6)

(3) (2)

(1)

Twigs: {1, 2, 3} f-cutset 1: {1, 4, 5, 6} f-cutset 2: {2, 4, 5} f-cutset 3: {3, 4, 6}

(4) (6))

(2)

��������� The network equilibrium equation on node basis can be written as

Here,

QY Yb QT Vt � QI s � QY Yb Vs I s � 0, QY Yb QT Vt � QY Yb Vs

1 2

3

4

5

6

1 � 1 0 0 �1 �1 1� Q � 2 � 0 1 0 �1 � 1 0 � � � 3 �0 0 1 1 0 1�

5.48�Circuit Theory and Networks—Analysis and Synthesis 0 0 0 0 0� � 1 0 0� �0 2 �9110 � � 0 � 0 0.2 � 0� 0 0 0 0� 1 0� � � � � � � 0 0 0 2 0 0 0 0 0 1 T � ; Vs � � 0 � �; Q � � Yb � � 0 0 01 0 0� � �1 �1 �1� � 0 � 0� � �1 �1 0 � � 0 � 0� 0 0 0 0.5 0� � 1 0 � 0 � 0� 0 0 0 0 0 1�� 1�� � � � � 0 0 0 0 0� �0 2 � 0 0.2 0 0 0 0� 0 0 �0 1 �0 5 0 1� 1 �1 1� � �1 0 0 � �0 2 0 0 02 0 0 0� � QY Yb � �0 1 0 � 0 0.22 0 �0 1 �0 5 0� 1 �1 0 � � 0 0 01 0 0� � � � �� 0 0 0 0 2 � 0 1 0 0 . 1� 0 0 1 1 0 1 � �� 0 0 0 0 0.5 0� � � 0 0 0 0 0 0 1�� � � 1 0 0� � 0 1 0� 0 0 �0 1 �0 5 0 1� � �0 2 � �0 9 0 6 0 2� 0 1� � 0 T � 0 6 0 8 0 1� QY Yb Q � � 0 0.22 0 �0 1 �0 5 0� � � � � � �11 1 1� � 0 0 0 2 � 0 1 0 0 . 1 � � � �11 1 0 � � 0 2 0 1 0 4 � � 1 0 1�� � 1 � �910 � 0� 0 0 �0 1 �0 5 0 1� � �0 2 � �182� 0� � � � QY Yb Vs � � 0 0.22 0 �0 1 �0 5 0� � 0 � � � 0� � � 0 0 0 2 � 0 1 0 0 . 1 � � � 0� � 0� � 0� � � Hence, KCL equation can be written as, � 0 9 0 6 0 2� � vt1 � � �182� �0 6 0 8 0 1� � vt � � � 0 � � �� 2 � � � � 0 2 0 1 0 4 � � vt3 � � 0 � Solving this matrix equation, vt1 � �444.26 V vt2 � 315.28 V vt3 � 143.31 V For the network shown in Fig. 5.80, obtain equilibrium equation on node basis.



5

���������

5



10 A

10



5



��������������

5.12����������������������������������

Solution The oriented graph and its selected tree are shown in Fig. 5.81. (2)

(2)

2

1

1

(3)

(1)

(3)

(1)

(4) 3

2 (4)

Twigs: {1, 3} f-cutset 1: {1, 2} f-cutset 3: {3, 2, 4}

1

2

3 4

1 �1 �1 0 0 � Q� � 3 �0 �1 1 1 ��

3

��������� The KCL equation in matrix form is given by QY Yb QT Vt � Q I s � QY Yb Vs Vs � 0,

Here, T

QY Yb Q Vt � Q I s �5 �0 � �0 �0

0 0� 0 0� T Y � Q 5 0� 0 10 � �5 �1 �1 0 0 � �0 QY Yb � � � �0 �1 1 1 �� �0 �0 0 5 0 0

� 1 � �1 �� � 0 � 0 0 0 5 0 0 5 0 0

0� � �10 � � 0� �1� �; I � � � 1� s � 0 � 1� � 0� 0� 0� � 5 5 0 0� �� 0 � ��0 5 5 10 �� 1 � 10

� 1 0� 5 5 0 0 � � � �1 �1� �10 5� � QY Yb QT � � �� �� � 0 0 5 5 10 1� �� 5 20 �� � � � 1� � 0 � �10 � �1 �1 0 0 � � 0 � � �10 � Q Is � � � �� �0 �1 1 1 �� � 0 � �� 0 �� � 0� Hence, KCL equation will be written as �10 5� � vt1 � � �10 � �� 5 20 �� � vt � � �� 0 �� � 3� Solving this matrix equation, 8 vt1 � � V 7 2 vt3 � V 7

�������������� For the network shown in Fig. 5.82, write down the f-cutset matrix and obtain the network equilibrium equation in matrix form using KCL and calculate v.

�����Circuit Theory and Networks—Analysis and Synthesis 2v 2� 2� 2V

� v �

� �

2�

2�

��������� Solution The oriented graph and its selected tree are shown in Fig. 5.83 Since voltage v is to be determined, Branch 2 is chosen as twig, 1

(4)

(2)

2

1

(4)

(2)

(3)

(3)

(1)

(1) 3

2 Twigs: {2, 4} f-cutset 2: {2, 1, 3} f-cutset 3: {4, 3}

1

2

3 4

� �1 1 1 0 � Q�� � 0 0 �1 1��

3

���������� The KCL equation in matrix form is given by QY Yb QT Vt � Q I s � QY Yb Vs 0 0 0� � 2� � 0� � �1 0 � �0 5 �0� � 0� � 0 0.5 0 0� T � 1 0� Y � ; Vs � � � � ; Is � � � � Q �� 0 05 0� �0� � 0� � 1 �1� � 0 0 0 0.5� 1� �0� � �2v � � 0 � 0 0 0 0� �0 5 0� 0 0 � � �0 5 0 5 0.5 � �1 1 1 0 � � 0 0.5 QY Yb � � � ��� � 0 0 � 0 . 5 0 . 5�� 0 0 � 1 1 0 0 0 5 0 �� � � � 0 0 0 5� � 0 � �1 0 � 0 � � 1 0 � � 1 5 �0 5� � �0 5 0 5 0 5 T QY Yb Q � � � �� 1�� 0 �0 5 0 5�� � 1 �1� �� �0 5 � 0 1� � 0 0 � � � �1 1 1 0 � � 0 � � 0 � Q Is � � � �� � 0 0 �1 1�� � 0 � �� �2v �� � �2v � � 2� 0 � �0 � � �1� � �0 5 0 5 0 5 QY Yb Vs � � � �� 0 �0 5 0 5�� �0 � �� 0 �� � 0 �0 � 1 � � Q I s � QY Yb Vs � � � �2v ��

5.12����������������������������������

Hence, the KCL equation can be written as � 1 5 �0 5� � vt2 � � 1� � �� �0 5 1�� �� vt4 �� �� �2v �� From Fig. 5.82, vt2 v Solving this matrix equation, vt2 � 0 44 V vt4 � 0 66 V vt2 � 0 44 V

v

�������������� For the network shown in Fig. 5.84, write down the f-cutset matrix and obtain the network equilibrium equation in matrix form using KCL and calculate v. 4� 0.5 A

2� �

� v

1V � �

4�

2�

��������� Solution The voltage and current sources are converted into accompanied sources by source–shifting method as shown in Fig. 5.85. 4� 2� 1V

� �

� v � � 1V �

0.5 A 2�

4�

0.5 A

��������� The oriented graph and its selected tree are shown in Fig. 5.86. 2

(1) (2)

2 Twigs: {1, 2} f-cutset 1: {1, 4} f-cutset 2: {2, 3}

1

1 (4)

(1)

(3)

(2)

3

(4) (3)

3

���������

1

2

3

4

� 1 0 0 �1� Q�� �0 1 �1 0 ��

5.52�Circuit Theory and Networks—Analysis and Synthesis The KCL equation in the matrix form is given by QY Yb QT Vt � Q I s � QY Yb Vs 0 0� �0 25 0 � 0 0.5 0 0� T Yb � � �;Q 0 0 25 0 � � 0 0 0 0.5� � 0

�1 � � 0 � � 1 0� �1 � � 0 � � 0 1� � ; Vs � � � � � ; Is � � �0 � � 0.5 � � 0 �1� �0 � � �0.5� � �1 0 �

0 0� �0 25 0 0 0.5 0 0 � �0 25 0 �0.5� � 1 0 0 �1� � 0 QY Yb � � � ��� � 0 0 . 5 0 . 25 0 �� 0 1 1 0 0 0 0 25 0 � � �� � � � 0 0 0.5� � 0 � 1 0� 0 1� �0 75 0 � �0.5� � 0 �0 25 0 T QY Yb Q � � � �� � 0.5 �0.25 0 � � 0 �1� �� 0 0.75�� � 0 � �1 0 � � 0 � �1 0 0 �1� � 0 � � 0 5� Q Is � � � �� �0 1 �1 0 �� � 0 5 � �� �0 5�� � �0 5� �1 � 0 �0.5� �1 � �0 25� �0 25 0 Q Yb Vs � � � �� 0.5 �0.25 0 �� �0 � �� 0 5 �� � 0 �0 � �0 25� Q I s � Q Yb Vs � � � �1 �� Hence, the KCL equation can be written as 0 � � vt1 � �0 25� �0 75 � �� 0 0.75�� �� vt2 �� �� �1 �� Solving this matrix equation, vt1 � 0 33 V vt2 � �1 33 V From Fig. 5.85,

v 1 � vt2 � �0.33 V

��������������

Exercises 5.1

For the networks shown in Fig. 5.87–5.90, write the incidence matrix, tieset matrix and f-cutset matrix. (i)

C

R2

(4) 1

(2)

L2

R4

3

2 (1)

(6)

(5)

R1

4 (3) (7)

R3

L1

5

V + −

��������� ��������� The incidence matrix is given as follows:

5.3 (ii)

L C

Branches � 1 2 3 4 5 6 �1 �1 0 0 0 0 0 1 1 0 1 0 0 0 �1 �1 0 1 1 0 0 1 0 0

R1

I1

+ V 2 −

R2

���������

1�

1

2� 1� 2�

� 10 V �

2�

20 A

1�

5.5 (iv)

I R1

V

+ −

C1

2

i

L1 L2

��������� For the graph shown in Fig. 5.91, write the incidence matrix, tieset matrix and f-cutset matrix.

Branches � 4 5 6 7 1 1 0 0 0 0 0 �1 1

8

9

10

1 0 1 0 0 0 � 1 0 0 �1 0 0 �1 �1 1 1 1 0 0 0

Draw the oriented graph. For the network shown in Fig. 5.92, draw the oriented graph and obtain the tieset matrix. Use this matrix to calculate the current i. 1�

2� 2�

R2

5.2

3

0 0 1 0 �1 � 1 �1 1 0 1 0 0

���������

8 0 0 0 1

Draw oriented graph and write tieset matrix. The incidence matrix is given below:

5.4

(iii)

7 1 0 0 0

1�

1V

2V 3�

1�

��������� [0.91 A]

5.54�Circuit Theory and Networks—Analysis and Synthesis Using the principles of network topology, write the loop/node equation in matrix form for the network shown in Fig. 5.93.

5.6

R1 R2 Vg

ig

+ −

R4 R3

���������

Objective-Type Questions The number of independent loops for a network with n nodes and b branches is (a) n − 1 (b) b − n (c) b − n + 1 (d) independent of the number of nodes

5.1

A network has 7 nodes and 5 independent loops. The number of branched in the network is (a) 13 (b) 12 (c) 11 (d) 10

5.2

Identify which of the following is NOT a tree of the graph shown in Fig. 5.94.

5.3

5.5

(a) 3 (b) 4 (c) 6 (d) 7 Consider the network graph shown in Fig. 5.96.

��������� Which one of the following is NOT a tree of this graph? (a)

(b)

(c)

(d)

a 2 1

d

f

e

4

3

c

b

h

g

5

��������� (a) (c)

begh bdeg

(b) (d)

5.6

defg aegh

Figure 5.97 below shows a network and its graph is drawn aside. A proper tree chosen for analyzing the network will contain the edges.

The minimum number of equations required to analyze the circuit shown in Fig. 5.95 is

5.4

C

C

R

a

b

c

+ −

R

d V

R

C

���������

���������

R

(a) (c)

ab, bc, ad ab, bd, ca?

(b) (d)

ab, bc, ca ac, bd, ad

������������������������������������5.55

5.7

5.8

The graph of an electrical network has n nodes and b branches. The number of links with respect to the choice of a tree is given by (a) b � n � 1 (b) b ��n (c) n �� b � 1 (d) n �� 2b � 1.

(2)

2

1

(a) (c) 5.10

5

(5)

���������

8 7

(3)

(1)

In the graph shown in Fig. 5.98, one possible tree is formed by the branches 4, 5, 6, 7. Then one possible fundamental cutset is 6

(4)

4 5

(b) C (d) 8

Which one of the following is a cutset of the graph shown in Fig. 5.100? (3)

4 (2)

3

(4)

��������� (a) (c) 5.9

1, 2, 3, 8 1, 5, 6, 8

(b) 1, 2, 5, 6 (d) 1, 2, 3, 7, 8

(5)

(1)

Which one of the following represents the total number of trees in the graph given in Fig. 5.99?

���������� (a) (c)

1, 2, 3 and 4 1, 4, 5 and 6

Answers to Objective-Type Questions 5.1. (c) 5.6. (d)

5.2. (c) 5.7. (a)

5.3. (c) 5.8. (d)

(6)

5.4. (b) 5.9. (d)

5.5. (b) 5.10. (d)

(b) (d)

2, 3, 4 and 6 1, 2, 4 and 5

6

Time Domain Analysis of RLC Circuits

��6.1 ��������������� 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 ��������������������� 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 Networks—Analysis and Synthesis 2. Initial Conditions for the Inductor

For an inductor, current and voltage are related by, v(t ) � L

di dt

Voltage across the inductor is proportional to the rate of change of current. It is impossible to change the current through an inductor by a finite amount in zero time. This requires an infinite voltage across the inductor. An inductor does not allow an abrupt change in the current through it. The current through the inductor is given by, t

i( t ) �

1 v(t )dt � i(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



+−

���������������������������

6.2 ��������������������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

������������������������� 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.,

�������������

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 �

1H

100 V i(t)

�������� Solution At t � 0–, no current flows through the inductor. i(0 � ) � 0

6.4�Circuit Theory and Networks—Analysis and Synthesis At t � 0 � , the network is shown in Fig. 6.4.

10 �

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

10 �

di �0 dt

…(i)

di � 100 � 10i dt At t � 0�,

��������

1H

100 V i(t )

…(ii) ��������

di � (0 ) � 100 � 10i(0 � ) � 100 � 10(0) � 100 A / s dt

Differentiating Eq. (ii), d 2i dt 2

At t � 0�,

d i dt 2

2

� �10

(0 � ) � �10

di dt di � (0 ) � �10(100) � �1000 A / s 2 dt

�������������

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 � 1 �F

100 V i(t )

�������� Solution At t � 0�, the capacitor is uncharged. vC (0 � ) � 0 i( 0 � ) � 0 1000 �

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�)

��������

6.2 �����������������������

For t � 0, the network is shown in Fig. 6.8. Writing the KVL equation for t � 0, 100 � 1000i �

1

1000 �

t

� i dt � 0 1 � 10 �6

100 V

1 �F

…(i) i(t )

0

Differentiating Eq. (i),

��������

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

��

2

At t � 0�,

d i dt

(0 � ) � � 2

106 di 1000 dt 106 di � 106 (0 ) � � ( �100) � 105 A / s 2 1000 dt 1000

�������������

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 �

1H

10 �F

10 V i(t )

�������� Solution At t � 0–, i( 0 � ) � 0 vC (0 � ) � 0 10 �

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 V

i ( 0�)

���������

v C ( 0�)

6.6�Circuit Theory and Networks—Analysis and Synthesis For t � 0, the network is shown in Fig. 6.11. Writing the KVL equation for t � 0,

10 �

t

10 � 10i � 1

At t � 0�,

di 1 � i dt � 0 dt 10 � 10 �6 �0

10 � 10i(0 � ) �

…(i)

1H

10 �F

10 V

i(t )

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

�������������

(0 � ) � �10 � 10 � �100 A / s 2

In the network shown in Fig. 6.12, at t = 0, the switch is opened. Calculate v,

2

d v dt 2

dv and dt

at t = 0�. v(t )

1H

1A

100 �

��������� 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 �

v(0 � ) � 100 � 1 � 100 V ���������

6.2 �����������������������

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 �

Differentiating Eq. (i), ��������� 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

�������������

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,

2

d v dt 2

dv and dt

at t = 0�. v(t)

10 A

1 k�

1 �F

��������� 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 k�

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

���������

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 k�

���������

1 �F

����Circuit Theory and Networks—Analysis and Synthesis 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

������������� 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�.

2�

10 A

1H

0.5 �F v (t)

��������� 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� )

2� i L ( 0�)

iL (0 � ) � 0 v(0 � ) � 0

���������

For t � 0, the network is shown in Fig. 6.20. Writing the KCL equation for t � 0,

10 A

2�

t

v 1 dv � vdt � 0.5 � 10 �6 � 10 dt 2 1 �1 At t � 0�,

…(i) ���������

v(0 � ) dv � 0 � 0.5 � 10 �6 (0 � ) � 10 2 dt dv � (0 ) � 20 � 106 V / s dt

1H

0.5 �F v (t)

6.2 �����������������������

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

�������������

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 �

1 2

20 V

1H

20 � i(t )

���������

10 �

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

��������� 10 �

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 �

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, �20i � 10i � 1 At t � 0�,

�30i(0 � ) �

di �0 dt

��������� 10 �

…(i)

di � (0 ) � 0 dt di � (0 ) � �30 � 2 � �60 A / s dt

Differentiating Eq. (i), �30

di d 2 i � �0 dt dt 2

20 �

1H i( t )

���������

2A

�����Circuit Theory and Networks—Analysis and Synthesis �30

At t � 0�,

di � d 2i (0 ) � 2 (0 � ) � 0 dt dt d 2i dt 2

(0 � ) � 1800 A / s 2

�������������

In the network shown in Fig.6.25, the switch is changed from the position 1 to the d 2i di and 2 at position 2 at t � 0, steady condition having reached before switching. Find the values of i, dt dt t � 0�. 20 �

1 2

30 V

1 �F

10 � i(t)

��������� 20 �

Solution At t � 0–, the network attains steady-state condition. Hence, the capacitor acts as an open circuit. vC (0 ) � 30 V

i( 0�)

i( 0 � ) � 0

���������

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 �

vC (0 � ) � 30 V

10 �

30 i( 0 ) � � � �1 A 30 �

1 � 10 �6

1 �F

� i dt � 30 � 0

…(i)

10 �

0

Differentiating Eq. (i),

i(t )

��������� di �30 � 106 i � 0 dt

At t � 0�,

i ( 0�)

20 �

t

1

vC( 0�) 30 V

���������

For t � 0, the network is shown in Fig. 6.28. Writing the KVL equation for t � 0, �10i � 20i �

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 ������������������������

Differentiating Eq. (ii), �30 2

�30

At t � 0�,

d i dt

2

d 2i dt

2

� 106

di �0 dt

di � (0 ) � 0 dt

(0 � ) � 106

d 2i dt

(0 � ) � � 3

106 � 0.33 � 105 � �1.1 � 109 A / s 2 30

�������������

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 �

1 2

40 V

1 �F

1H i(t)

��������� 20 �

Solution At t � 0–, the network attains steady state. Hence, the capacitor acts as an open circuit.

v C ( 0� )

40 V



i ( 0�)

vC (0 ) � 40 V i( 0 � ) � 0

���������



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 � v C ( 0�) 40 V

vC (0 � ) � 40 V

i ( 0�)

i( 0 � ) � 0

���������

For t � 0, the network is shown in Fig. 6.32. Writing the KVL equation fo t � 0,

20 � 1 �F

t

�1 At t � 0�,

di 1 � 20i � � i dt � 40 � 0 dt 1 � 10 �6 0 �

di � (0 ) � 20i(0 � ) � 0 � 40 � 0 dt di � (0 ) � � 40 A / s dt

…(i)

1H i(t)

���������

40 V

�����Circuit Theory and Networks—Analysis and Synthesis Differentiating Eq. (i), � �

At t � 0�,

d 2i 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

��������������

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 k�

a b

100 V

1H

0.1 �F i(t )

��������� Solution At t � 0–, the network attains steady condition. Hence, the inductor acts as a short circuit. i( 0 � ) �

1 k�

100 V

100 � 0.1 A 1000

i ( 0� )

vC (0 � ) � 0

���������

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 k� v C ( 0�)

0.1 A



i ( 0�)

i(0 ) � 0.1 A vC (0 � ) � 0

���������

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 k�

0.1 �F

…(i)

1H i ( t)

0

���������

0.1 A

6.2 ������������������������

�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), 1



10 At t � 0�,

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

��������������

K

1H

vK 2V

1� 0.5 F

i(t )

��������� 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

1�

vC( 0�)

2 �2A 1

i ( 0�)

���������

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�)

1� i ( 0�)

���������

�����Circuit Theory and Networks—Analysis and Synthesis 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 � ,

��������� �����

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)

��������� 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. i1 (0 � ) �

R2

R1 vC( 0+)

V

V R1

i1( 0+)

i2 (0 � ) � 0

i2( 0+)

���������

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)

di 1 (i2 � i1 )dt � R2 i2 � L 2 � 0 C� dt

…(ii)

V � R1i1 �

and



R1

R2

L

C

V i1( t)

i2( t)

���������

6.2 ������������������������

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 (00 ) � � 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

��������� �����

In the network shown in Fig. 6.43, assuming all initial conditions as zero, find d 2 i2 di di d i i1 , i2 1 , 2 , 21 and at t � 0 � . dt dt dt dt 2 2

R2

C

V

L

R1 i2( t)

i1( t)

���������

�����Circuit Theory and Networks—Analysis and Synthesis 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+)

���������

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)

���������

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 � � R LC dt L � 1

d i2 2

...(iv)

6.2 ������������������������

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� � 2 � R1C dt R1 LC L R1C � L R1 C � R1 C R1 LC L2

�������������� 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 � 10 �

va(t )

20 �

vb(t )

5V

2H 10 �

��������� Solution Fig. 6.47.

At t � 0 � , the network is shown in

10 �

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 �

va(0�)

20 �

vb(0�)

5V iL(0�)

vb (0 � ) � 0

���������

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 � 10 �

va(0�)

10 �

���������

20 �

vb(0�) 2 A 3

�����Circuit Theory and Networks—Analysis and Synthesis 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

�������������� 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 � 10 �

va(t)

20 � vb(t)

5V

2F 10 �

��������� 10 �

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.

10 �

va(0�)

20 �

vb(0�)

5V

va ( 0 � ) � 5 V vb (0 � ) � 5 V

���������

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 � 10 �

vb (0 � ) � 5 V

va(0�)

20 �

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 �

5V

���������

va ( 0 � ) � 3 V

��������������

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 ������������������������ L

v1( t)

v2( t)

R2

R1

i (t )

C

��������� 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

���������

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), di 1 dv1 1 � ( v1 � v2 ) � R1 dt L dt

���������

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

v dv 1 ( 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

��������������

�����Circuit Theory and Networks—Analysis and Synthesis R1 i C (t ) V

L

+ v1(t )

R2

− + v2(t )

i L (t )

C



��������� 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 � ) �

���������

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 )

vC ( t )

Differentiating Eq. (i), dvC dv1 dv2 � � dt dt dt

i L (t )

…(ii) ���������

6.2 ������������������������ 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 v2 � R2 iL

…(vi)

dv2 di � R2 L dt dt

…(vii)

Also,

At t � 0�,

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 d 2 v2 At t � 0 , �

dt

2

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 Networks—Analysis and Synthesis

��������� ����� 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�. 2�

4�

i2(t )

0.5 F

6V 1H

i1(t )

1F

i 3( t )

��������� Solution At t � 0 � , the network is shown in Fig. 6.59. At t � 0–, the network attains steady-state condition. Hence, the inductor acts as a short circuit and the capacitors act as open circuits. i4 � (0 � ) � i1 (0 � ) �

2�

4�

v2(0�)

i1(0�) i3(0�)

i3 (0 � ) � 0 ���������

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 8 4 voltage source of V and V respectively. 3 3

2�

4� 6V

8 V 3 4 v2 ( 0 � ) � V 3

i1(0�)

v1 (0 � ) �

At t � 0�,

i2(0�)

8 V 3

i3(0�)

4 V 3

1A

���������

8 4 6 � 2i1 (0 � ) � � � 0 3 3 i1 (0 � ) � 1 A i1 (0 � ) � i3 (0 � ) � 1

Now,

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

��������������

In the network shown in Fig. 6.61, the switch K is closed at t � 0 connecting a voltdi di age V0 sin � 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

��������� �

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 Networks—Analysis and Synthesis 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 � t acts as a short circuit.

i2 (0+ )

i1 (0+ ) R

i1 (0 � ) � 0

R

vC (0+ )

i2 (0 � ) � 0 vC (0 � ) � 0 ���������

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 � t � R i2 � L 2 � 0 dt

V0 sin � t � R i1 � and

i 1 (t )

…(i) R

…(ii)

R

V0 sinwt C

Differentiating Eq. (i), V0 � cos � t � R

i2 (t )

di1 i1 � �0 dt C

L

���������

di1 V0� i � cos � t � 1 …(iii) dt R RC di1 � V� i (0 � ) V0� (0 ) � 0 cos � t � 1 � dt R RC R t � 0�

At t � 0 � , From Eq. (ii),

di2 V0 R � sin � t � i2 dt L L di2 � V R (0 ) � 0 sin � t � i2 (0 � ) � 0 dt L L t � 0�

At t � 0 � ,

�������������� 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

i1

���������

C1 i2

6.2 ������������������������

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

vC1(0−)

i1(0−)

i2 (0 � ) � 0

i3(0−) vC2(0−)

i2(0−)



i3 (0 ) � 0 vC3 (0 � ) � V

���������



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 i1

vC1(0+)

(0+)

vC2(0+)

i2(0+)

V i1 (0 ) � i2 (0 ) � � R1 � R2 � R3 �

i3(0+)



���������

i3 (0 � ) � 0

��������������

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 � iC

10 V

C

i2 10 �

vC

��������� 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 � i2(0�)

iC(0�) 10 V

vC(0�)

i1 (0 � ) � 0 i2 (0 � ) � 0 iC (0 � ) � 0 vC (0 � ) � 10 V

���������

10 �

6.26�Circuit Theory and Networks—Analysis and Synthesis 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 � iC(0�) 10 V

vC (0 � ) � 10 V

i2(0�) 10 V

10 �

vC(0�)

Writing the KVL equation at t � 0 � , ���������

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

��������������

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 � i2(t)

i1(t ) 20 �

20 �

1H

1 �F

100 V

��������� 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 � i2 (0�)

i1 (0�)

20 �

100 V 20 �

20 � 66.67 V 20 � 10

���������

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 � 3.33 A

���������

20 � 66.67 A

6.3�������������������������������

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 �

At t � 0 � ,

i2(t )

i1(t )

20 �

20 � 100 V

1 �F 1H

3.33 A 66.67 V

…(i) ��������� 1 10 �6

� i2 dt � 66.67 � 0

…(ii)

di1 � (0 ) � 100 � 20i1 (0 � ) � 100 � 20 (3.33) � 33.3 A/s dt

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

�������������������������������� 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. Applying KVL to the circuit for t � 0, di V � Ri � L � 0 dt

R

V

L i(t )

���������������������������

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

�����Circuit Theory and Networks—Analysis and Synthesis where ln denotes that the logarithm is of base e and k is an arbitrary constant. k can be evaluated from the initial condition. In the circuit, the switch is closed at t � 0, i.e., just before closing the switch, the current in the inductor is zero. Since the inductor does not allow sudden change in current, at t � 0�, just after the switch is closed, the current remains zero. Setting i � 0 at t � 0, � � �

L 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 V response or forced response or zero state response and transient R

for t � 0 i(t) V R

R

V � Lt e . t R O The natural response is a characteristic of the circuit. Its form may be found by considering the source-free circuit. The forced response ������������������� ��������� ��� ����������������� 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

response or natural response or zero input response

At one time constant, the current reaches 63.2 per cent of its final value 1

i(T ) �

V . R

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�������������������������������

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 � t� V� �1 � e L � R� � R

vR � Ri � R �

R � � t� 1� e L

�V � �

� �

v(t)

for t � 0

V

VL

Similarly, the voltage across the inductor is vL �

O

R � t� di V d� �L �1 � e L � dt R dt � �

� Ve

R � t L

VR

t

������������������������������� �����������������

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).

�������������� 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).

�����Circuit Theory and Networks—Analysis and Synthesis R1

1 2

V

L

R2 i ( t)

��������� 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− )

���������

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

���������

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�������������������������������

�������������� 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

��������� 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 � ) �

���������

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

���������

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 Networks—Analysis and Synthesis 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 L 1 � e � � R1 � R1 � R2 �

for t � 0

�������������� 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 �

20 V

20 �

K

10 V

1 H 2 i(t )

��������� 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 �

20 �

20 V i ( 0�)

10 V

Since the current through the inductor cannot change instantaneously,

���������

i(0 � ) � 0.6 A 20 �

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

1 H 2

10 V i(t )

���������

0.6 A

6.3���������������������������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

�������� ���� 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 �

1 2 50 V

20 mH

10 V i (t )

��������� 40 �

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 �

��������� Since current instantaneously,

through

the

inductor

cannot

change

i(0 � ) � 1.25 A 40 �

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 )

���������

1.25 A

6.34�Circuit Theory and Networks—Analysis and Synthesis 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

������������

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 5�

2 0.5 H

2�

20 V

i(t ) 40 V

��������� 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.

5� 20 V

20 i(0 � ) � �4A 5

i ( 0�)

���������

Since the current through the inductor cannot change instantaneously, i(0 � ) � 4 A For t � 0, the network is shown in Fig. 6.91. Writing the KVL equation for t � 0, di 40 � 2i � 0.5 � 0 dt di � 4i � 80 dt

2� 0.5 H 40 V

i(t )

���������

4A

6.3�������������������������������

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

������������ 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). 100 �

iL(t )

15 V

3000 �

90 mH

vL(t )

��������� 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 �

15 V

15 iL (0 ) � � 0.15 A 100

i L ( 0�)



Since current through the inductor cannot change instantaneously,

���������

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 �

90 mH iL(t)

���������

0.15 A

6.36�Circuit Theory and Networks—Analysis and Synthesis di � Pi � 0, dt

Comparing with the differential equation

P � 33.33 � 103 The solution of this differential equation is given by, iL (t ) � k e � Pt iL (t ) � k e �33.33�10

3

t

At t = 0, iL (0) � 0.15 A 0.15 � k iL (t ) � 0.15 e �33.33�10

3

t

for t � 0

di vL (t ) � L L dt

Also,

� 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

��������������

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 �

10 �

50 V

0.1 H 10 �

i(t )

���������

10 �

10 �



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

���������

Since current through the inductor cannot change instantaneously, i (0�) � 2.5 A 10 �

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 �

10 �

10 � i(t )

���������

0.1 H

2.5 A

6.3�������������������������������

For t � 0, Thevenin’s equivalent network is shown in Fig. 6.98. Writing the KVL equation for t � 0,

15 �

di 25 � 15i � 0.1 � 0 dt 25 V

di � 150i � 250 dt Comparing with the differential equation

0.1 H i(t )

di � Pi � Q, dt P � 150,

��������� 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

��������������

for t � 0

In Fig. 6.99, the switch is closed at t � 0. Find i (t) for t � 0. 2�

10 A

1�

2� i (t)

���������

Solution At t � 0–,

i(0 � ) � 0

Since current through inductor cannot change instantaneously, i(0 � ) � 0

1H

2.5 A

�����Circuit Theory and Networks—Analysis and Synthesis For t � 0, simplifying the network by source-transformation technique as shown in Fig. 6.100. 2�

1�

10 A

2�

2�

1H

0.67 �

10 A

i (t )

1H i (t)

(a)

(b)

Writing the KVL equation for t � 0, 6.67 � 2.67i � 1

2.67 �

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

����������

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 )

��������������

for t � 0

Find the current i (t) for t � 0. 60 �

25 A

20 �

140 �

0.3 H i (t)

����������

6.3�������������������������������

Solution At t � 0–, the inductor acts as a short circuit. Simplifying the network as shown in Fig. 6.102. 60 �

25 A

140 �

25 A

20 �

140 �

i (0�)

60 �

i (0�)

(a)

(b)

���������� 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 �

20 �

140 �

25 A

0.3 H

17.5 A

i (t)

���������� Simplifying the network by source transformation as shown in Fig. 6.104,

60 �

20 �

17.5 A

0.3 H

15 �

i (t)

0.3 H i (t)

(a)

(b)

���������� 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

�����Circuit Theory and Networks—Analysis and Synthesis The solution of this differential equation is given by, i(t ) � k e � Pt � k e �50 t At t � 0, i (0) � 17.5 A k � 17.5 i(t ) � 17.5 e �50 t

for t � 0

�������������� 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. 1�

a

b �

1V

1 � 2

1H

v2(t)

2H �

���������� 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 � 2

v2(t )

2H �

����������

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�������������������������������

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

�������������� 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 � � 3V

va(t)

5� 0.5 H



���������� Solution At t � 0–, the network has attained steady-state condition. Hence, the inductor acts as a short circuit. 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 � �

t

1 v v �3 va dt � a � a �0 0.5 �0 5 10

3V

5�



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

����������

6.42�Circuit Theory and Networks—Analysis and Synthesis Comparing with the differential equation

dv � Pv � 0, dt P�

20 3

The solution of this differential equation is given by, va (t ) � k e � Pt � k e



20 t 3

At t � 0, va (0) � 1 V 1� k va ( t ) � e

�������������� closed at t � 0.



20 t 3

for t � 0

In the network of Fig. 6.109, determine currents i1 (t) and i2 (t) when the switch is

10 � 5� 100 V

i2(t)

5�

0.01 H

i1(t )

���������� i1 (0 � ) � i2 (0 � ) � 0

Solution At t � 0–, At t � 0 , �

i1 (0 � ) � 0 100 i2 (0 � ) � � 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 �

…(i) …(ii)

5� 100 V

i2(t) i1(t )

0.01 H

From Eq. (ii), i2 �

100 � 10i1 15

Substituting in Eq. (i), di1 � 833i1 � 3333 dt

����������

5�

6.3���������������������������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

�������������� 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 �

10 V

� �

10 � i1(t ) i2(t )

���������� 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 Networks—Analysis and Synthesis For t � 0, the network is shown in Fig. 6.112. Writing the KVL equations for t � 0,

and

30 �

10 � 30(i1 � i2 ) � 10 i1 � 0

…(i)

di2 �0 dt

…(ii)

10 � 30(i1 � i2 ) � 0.2

10 V

� �

10 � i1(t )

i2(t)

0.2 H

From Eq. (i), ����������

10 � 30 i2 � 0.25 � 0.75 i2 …(iii) 40 Substituting Eq. (iii) into Eq. (ii), 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 di v2 (t ) � 0.2 2 dt d � 0.2 (0.067 � 0.067 e �37.5t ) dt � 0.5 e �37.5t

for t � 0

�������������� 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 �

60 �

1 2

500 V

� � 10i i(t )

����������

0.4 H

6.3�������������������������������

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 �

500 V

500 i(0 ) � �5A 40 � 60 through the inductor cannot �

Since current instantaneously,

60 �

i( 0�)

change

����������

i(0 � ) � 5 A 60 �

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,

����������

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

�������������� For the network shown in Fig. 6.116, find the current in the 20 � resistor when the switch is opened at t � 0. i 30 � 50 V

20 �

� � 10i i2(t)

i1(t )

2H

���������� 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.

i (0 � ) 30 � 50 V

i1(0 � )

� � � � 10i (0 ) i2(0 )

i(0�) = i2(0�) ����������

20 �

6.46�Circuit Theory and Networks—Analysis and Synthesis 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 �

Writing the KVL equation for t � 0, 10i2 � 30i2 � 20i2 � 2

di2 �0 dt

20 �

� � 10i2 i2(t)

di2 � 20i2 � 0 dt

2H

2.5 A

����������

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.

��������������

0.5 �

� 4e ��3t �

0.25 H i(t )

���������� Solution At t � 0 � ,

i(0 � ) � 0

Since current through the inductor cannot change instantaneously, i(0 � ) � 0

6.3�������������������������������

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

�������������� For the network shown in Fig. 6.120, a sinusoidal voltage source v � 150 sin(500t � � ) volts is applied at a time when � � 0. Find the expression for the current i(t). 50 �

150 sin(500t����q )

0.2 H i(t )

���������� Solution Writing the KVL equation for t � 0, 150 sin(500t � � ) � 50i � 0.2

di �0 dt

di � 250i � 750 sin(500t � � ) dt Comparing with the differential equation

di � Pi � Q, dt

P � 250, Q � 750 sin(500t � � )

�����Circuit Theory and Networks—Analysis and Synthesis 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 � � ) e 250 t � k e �250 t � � e 250 t � 750 e �250 t � �250 sin(500t � � ) � 500 cos(500t � � )�� � k e �250t 2 2 � ( 250) � (500) � � 0.6 sin(500t � � ) � 1.2 cos(500t � � ) � k e �250 t Acos � � 0.6

Let

A sin � � 1.2

and

A cos � � A sin 2 � � (0.6) 2 � (1.2) 2 � 1.8 A � 1.342 2

2

2

� 1.2 � � � tan �1 � � � 63.43� � 0.6 �

and

i(t ) � 1.342 cos(63.43�) sin(500t � � ) � 1.342 sin(63.43�) cos(500t � � ) � k e �250 t i(t ) � 1.342 sin(500t � � � 63.43�) � k e �250 t At t � 0, � � 0, i(0) � 0 0 � 1.342 sin( �63.43�) � k k � 1.2 i(t ) � 1.342 sin(500t � � � 63.43�) � 1.2 e �250 t

for t � 0

�������������� 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 �

150 cos(200t ��30�) 0.5 H

���������� �

Solution At t � 0 , the network is shown in Fig 6.122. At t � 0 � the network attains steady-state condition. V 150 �30� I� � � 0.67�3.43� A Z 200 � j � 200 � 0.5

200 � 150 cos(200t ��30�) i(t )

����������

0.5 H

6.4 ��������������������������������

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 �

di �0 dt

0.5 H

i(t )

di � 400 i � 0 dt

����������

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),

At t � 0,

0.67 cos( 200t � 3.43�) � k e �400 t 0.67 cos(3.43�) � k k � 6.67 i(t ) � 0.67 e �400 t

for t � 0

����������������������������������� 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

����������������������������

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

�����Circuit Theory and Networks—Analysis and Synthesis The constant k can be evaluated from initial condition. In the circuit shown, the switch is closed at t � 0. Since the capacitor never allows sudden change in voltage, it will act as short circuit at t � 0�. Hence, current in the V circuit at t � 0� is . R V Setting i � at t � 0, R ln

V �k R

1 V t � ln RC R V 1 ln i � ln � � t R RC ln i � �

1 i �� t RC �V � �� �� R

ln

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

vR � Ri � R � Ve



V � RC t e R

1 t RC

for t � 0

i(t ) V R

t

O

����������������������������������� series RC circuit

v(t )

Similarly, the voltage across the capacitor is V t

1 vC � � i dt C0 t



1 V C �0 R

� �Ve



VR

1 � t e RC

1 t RC

VC

�k

O

t

����������������������� ��������� ��� series RC circuit

6.4 ��������������������������������

At t � 0, vC (0) � 0

k�V 1 � � � t vC � V �1 � e RC � � �

Hence,

��������������

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 k�

1 2 100 V

50 V

� vC(t)

1 �F



���������� 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 k�

v C ( 0�)

100 V

vC (0 ) � 100 V �

Since the voltage instantaneously,

across

the capacitor cannot

change ����������

vC (0 ) � 100 V �

5 k�

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) �

����������

1 �F

�����Circuit Theory and Networks—Analysis and Synthesis At t � 0, vC (0) � 100 V

100 � –50 � k k � 150 vC (t ) � �50 � 150 e �200 t

for t � 0

�������������� 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 k�

10 V

4 k� iC(t )

1 k�

� vC(t) �

3 �F

���������� 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 k�

Since voltage across the capacitor cannot change instantaneously, vC (0 � ) � 0 At t � 0�,

10 V

� � 10 10 iT (0 � ) � � � � 1.02 mA � � 9 k � ( 4 k � 1 k ) � 9.8 k 1k iC (0 � ) � 1.02 m � � 0.204 mA 1k�4 k

4 k� iC (0 � )

v C ( 0�)

1 k�

����������

For t � 0, the network is shown in Fig. 6.131.

9 k�

4 k�

For t � 0, representing the network to the left of the capacitor by Thevenin’s equivalent network, 10 V

1k �1V Veq � 10 � 9 k �1 k Req � (9 k � 1 k) � 4 k � 4.9 k�

dvC v �1 � C �0 dt 4.9 � 103 dvC � 68.02 vC � 68.02 dt

3 �F

����������

For t � 0, Thevenin’s equivalent network is shown in Fig. 6.132. Writing the KCL equation for t � 0, 3 � 10 �6

� vC(t) �

1 k�

4.9 k�

1V

3 �F

����������

vC(t)

6.4 ��������������������������������

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

fo or t � 0

��������������

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 �

1200 V 300 �

� vC(t ) �

50 �F

����������

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 �

v C ( 0�)

1200 V

Since the voltage across the capacitor cannot change instantaneously, vC (0�) � 1200 V

����������

6.54�Circuit Theory and Networks—Analysis and Synthesis 100 �

For t � 0, the network is shown in Fig. 6.135. Writing the KCL equation for t � 0, 50 � 10 �6



dvC vC vC � 1200 � � �0 dt 300 100 dvC � 266.67 vC � 0.24 � 106 dt

300 �

1200 V

50 �F

vC (t) �

����������

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. 100 �

5V

2�

1F

� �

vC(t)

����������

Solution At t � 0�,

vC (0�) � 0 Since the voltage across the capacitor cannot change instantaneously, vC (0�) � 0 Since the resistor of 2 � is connected in parallel with the voltage source of 5 V, it becomes redundant. For t � 0, the network is 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 � 1F

5V

����������

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 �0.01t � 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. 1 F 4

a b

� v (t) �

5V 2�

2�

���������� 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.

2�

� v( 0�) �

2�

� v( 0�) �

5V

vC (0�) � 5 V v (0�) � 0

����������

At t � 0 , the network is shown in Fig. 6.140. 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

5V

2� i( 0�)

����������

�����Circuit Theory and Networks—Analysis and Synthesis

For t � 0, the network is shown in Fig. 6.141. Writing the KVL equation for t � 0,

1F 4

5V

t

�2i � 5 �

1 i dt � 2i � 0 1� 0 4

…(i) 2�

2�

i ( t)

Differentiating Eq. (i), �4

di � 4i � 0 dt di �i � 0 dt

����������

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

��������������

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 � � 6V

1 � 2

0.3 F

v2(t) �

���������� Solution At t � 0�, the switch is open. v2 (0�) � 0 Since voltage across capacitor cannot change instantaneously,

0.25 � � 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 � 2

���������

6.4 ��������������������������������

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.

�������������� The switch is moved from the position a to b at t � 0, having been in the position a for a long time before t � 0. The capacitor C2 is uncharged at t � 0. Find i (t) and v2 (t) for t � 0. R

a

R1

b

+ V0

C2

C1 i(t)

v2 (t) −

���������� 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

�����Circuit Theory and Networks—Analysis and Synthesis 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

Differentiating Eq. (i),





i di i � R1 � �0 C1 dt C2

����������

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 )

i(t)

� Pt

� ke



1 � C1 � C2 � t R1 �� C1 C2 ��

V0 R1 k�

V0 R1 1 � C1 � C2 � t C C ��

V � � i(t ) � 0 e R1 � R1

1

2

1



V0 � R1C t e R1

v2 ( t ) �

1 i dt C2 �0

where, C �

C1C2 C1 � C2

t

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 ��������������������������������

�������������� t � 0.

For the network shown in Fig.6.146, the switch is opened at t � 0. Find v (t) for � K 1�

10 A 1 � 2

1 F 2

v (t) �

���������� 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 � 2

v C (0 � ) v(0 � )

1�





Writing the KCL equation at t � 0 , �

���������� �

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 �

1�

� �

1 F 2

v (t)



����������

�����Circuit Theory and Networks—Analysis and Synthesis At t � 0, v(0) � 3.33 V 3.33 � 10 � k k � 6.67 v(t ) � 10 � 6.67 e �2t

��������������

For the network shown in Fig. 6.149, find the current i(t) when the switch is opened

at t � 0. 10 � i ( t) � �

100 V

5i 10 � 4 �F

���������� 10 �

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

����������

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. 25 �

1 4 � 10 �6

10 � v C ( 0� )

vC (0 � ) � 100 � 10i(0 � ) � 5i(0 � ) � 100 � 10(5) � 5(5) � 25 V



5i ( 0�)

5i ( 0�) 10 � 25 V

t

� i dt � 5i � 10i � 0

…(i)

0

����������

Differentiating Eq. (i), 0 � 0.25 � 106 i � 5

di di � 10 � 0 dt dt

di � 50000 i � 0 dt

� �

i(t) 5i 4 �F 25 V

di � Pi � 0, Comparing with the differential equation dt P � 50000

����������

10 �

6.4 ��������������������������������

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

��������������

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 � i (t ) 20 �

10 �

20i

2 �F

100 V � �

���������� 10 �

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 � � �

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 �

100 V � �

vc (0 �)

����������



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 � ) 20 i(0 � ) � 20 i2 (0 � ) � 10 i2 (0 � ) � 80 � 0 �



10 �

20 � � 20i(0 �) i 2 (0 �) �



20 i(0 ) � 20 i(0 ) � 10 i(0 ) � 80 � 0 i(0 � ) � 1.6 A

80 V

����������

vC (0 � ) � 80 V i(t)

For t � 0, the network is shown in Fig. 6.156.

20i � 20i2 � 10i2 �

10 �

20 �

From Fig. 6.156, i(t ) � �i2 (t ) Writing the KVL equation for t � 0, 1 2 � 10 �6

t

� i2 dt � 80 � 0 0

� 20i �

2 �F i 2 (t)

����������

80 V

6.62�Circuit Theory and Networks—Analysis and Synthesis 20i � 20i2 � 10i2 � 50i �

t

1 2 � 10 �6 1 2 � 10 �6

� i dt � 80 � 0 0

� i dt � 80 � 0

…(i)

Differentiating Eq. (i), 50

Comparing with the differential equation

di � 5 � 105 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.

��������������

0.2 �

4e �5t

� �

1F i(t)

���������� 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 �

4

� �

i(0 � )

v c (0 � )

����������

t

1 i dt � 0 1 �0

…(i)

6.4 ����������������������������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 source e � t to the network. At t � 0, vC (0) � 0.5 V. Determine v(t). 1� � e �t

1 � 0.5 V� 2 �

� �

1F

v (t) �

���������� 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 Networks—Analysis and Synthesis 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

�������������� In the network shown in Fig. 6.160, a sinusoidal voltage v � 100 sin(500t � � ) volts is applied to the circuit at a time corresponding to � � 45�. Obtain the expression for the current i(t). 15 �

100 sin (500t ��45� )

100 �F i(t)

���������� 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 ��������������������������������

� � 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 � � 3.2

Let

A cos � � 2.4

and

A sin � � A cos 2 � � (3.2) 2 � ( 2.4) 2 � 16 A�4 2

2

2

� 3.2 � � � 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 � � 1000 rad / s.

��������������

1 50 �

2 100 sin (w t ��30� )

50 � i(t)

20 �F

���������� 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 � 100 sin (wt ��30�) i(�)

20 �F

���������� …(i)

6.66�Circuit Theory and Networks—Analysis and Synthesis For t � 0, the network is as shown in Fig. 6.163. Writing the KVL equation for t � 0, �50i � 50i �

50 � 50 �

t

1 20 � 10

�6

� i dt � vC (0) � 0



i(t)

…(ii)

20 �F

vC (0 ) � �

0

Differentiating Eq. (ii),

����������

di di 1 i�0 �50 � 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

for t � 0

�������������������������������������������� 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

C

�����������������������������

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 �����������������������������������������

Let s1 and s2 be the roots of the equation. 2

s1 � �

1 R � R� � � � � � �� � � 2 � � 02 � �� � � � 2L � 2L LC

s2 � �

1 R � R� � � � � � �� � � 2 � � 02 � �� � � � 2L � 2L LC

2

��

where

�0 �

R 2L 1 LC

� � � 2 � � 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 � and �0, we have 3 cases of the response. Case I When � > �0 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–� t (k1 e� t � k2 e–� t) i � k1 e s1t � k2 e s2t

or

for t > 0

The current curve for an overdamped case is shown in Fig. 6.165.

i(t)

Case II When � � �0 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, i � e �� t ( k1 � k2 t ) for t � 0

t

����������������������� ������� ��������

The current curve for critically damped case is shown in Fig. 6.166. Case III

When � � �0 i.e.,

R � 2L

1 LC

The roots are complex conjugate and it gives an underdamped response.

�����Circuit Theory and Networks—Analysis and Synthesis In this case, the solution is given by, i(t ) � k1e s1t � k2 e s2t s1, 2 � �� � � 2 � � 02

where Let

� 2 � � 02 � �1 � 02 � � 2 � j� d j � �1

where and

� d � � 02 � � 2

Hence,

i(t ) � e �� t ( k1e j� d t � k2 e � j� d t ) �� � e j� d t � e � j � d t � � e j� d t � e � j� d t � �� � e �� t �( k1 � k2 ) � �� � � j ( k1 � k2 ) � 2 j2 �� � � � � � e �� t �( k1 � k2 ) cos � d t � j ( k1 � k2 ) sin � d t � for t � 0

The current curve for an underdamped case is shown in Fig. 6.167. i(t )

t

�������������������������������

�������������� current i(t ) for t � 0.

In the network of Fig. 6.168, the switch is closed at t � 0. Obtain the expression for 9�

1H

0.05 F

20 V i(t )

���������� 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 ����������������������������������������� 9�

For t � 0, the network is shown in Fig. 6.169.

1H

Writing the KVL equation for t � 0, t

20 � 9i � 1

di 1 � i dt � 0 dt 0.05 �0

20 V

…(i)

0.05 F i(t)

����������

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

�������������� 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 ).

�����Circuit Theory and Networks—Analysis and Synthesis 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 1� � 2 �� D � 5 D � �� i � 0 6 D1 � �0.03, D2 � �4.97 dt 2

�5

The solution of this differential equation is given by, i(t ) � k1 e �0.03t � k2 e �4.97t

…(ii)

Differentiating Eq. (ii),

At t � 0, i(0) � 0

di � �0.03k1 e �0.03t � 4.97k2 e �4.97t dt 0 � k1 � k2 di (0) � �0.03 k1 � 4.97 k2 dt

…(iii) …(iv) …(v)

6.5 �����������������������������������������

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

�������������� 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 � i2(t) 100 V

1H

10 �F

���������� 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 � i2 (0 �) vC (0 �)

100 V

100 � 10 A 10

����������



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)

10 A

t

di 1 i dt � 0 �1 2 � dt 10 � 10 �6 �0

1H

…(i) ����������

10 μF

�����Circuit Theory and Networks—Analysis and Synthesis 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. v (t)

� 10 V



1� 4

1F

���������� Solution At t � 0 � ,

iL (0 � ) � 0 v(0 � ) � vC (0 � ) � 10 V

1 H 2

6.5 �����������������������������������������

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



1� 4

1F

Differentiating Eq. (i),

1 H 2

���������� 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

�������������� In the network of Fig. 6.178, the switch is opened at t � 0 obtain the expression for v(t ). Assume zero initial conditions.

�����Circuit Theory and Networks—Analysis and Synthesis v (t)

0.5 �

2A

0.5 H

1F

���������� 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

1 v dv � vdt � 1 � 2 0.5 0.5 �0 dt

…(i)

2A

0.5 �

0.5 H

1F

Differentiating Eq. (i), 2

dv d 2v � 2v � 2 � 0 dt dt

d 2v dt 2

�2

����������

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 �����������������������������������������

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

�������������� The network shown in Fig. 6.180, a sinusoidal voltage v � 150 sin(200t � � ) is applied at � � 30�. Determine the current i(t). 10 �

0.5 H

200 �F

150 sin (200t ��f) i(t)

���������� 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�)

�����Circuit Theory and Networks—Analysis and Synthesis 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�) Asin � � �1.97

Let

A cos � � 0.26

and

A sin � � A cos 2 � � ( �1.97) 2 � (0.26) 2 � 3.95 A � 1.987 2

2

and

2

� �1.97 � � � 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 �����������������������������������������

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

�������������� (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

2�

(c) L � 5 H and C � 1 F

i(t ) �

4V

C vC (t)

2�



���������� 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 �� L �

1 H, C � 1 F 2

��

�0 �

R � 2L 1 LC

� � �0

2 2� �

1 2

�2 1

1 �1 2



1 0.5

� 1.414

�����Circuit Theory and Networks—Analysis and Synthesis This indicates an overdamped case. i(t ) � A1 e s1t � A2 e s2t where,

s1 � �� � � 2 � � 02 � �2 � 4 � 2 � �2 � 2 � �3.414

and

s2 � �� � � 2 � � 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 �, L � 1 H, C �1 F

��

R 2 2 � � �1 2L 2 � 1 2

�0 �

1



LC

1 1

�1

� � �0 This indicates a critically damped case. i(t) � e–� t (k1 � k2 t) � e–t (k1 � k2 t) At t � 0, i(0) � 0 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�����

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 �, L � 5 H, C � 1 F R 2 � � 0.2 2 L 10 1 1 �0 � � � 0.447 LC 5 � � �0

��

This indicates an underdamped case. i (t) � e–� t (B1 cos �d t � B2 sin �d t)

� d � � o2 � � 2 � (0.447) 2 � (0.2) 2 � 0.4

where,

s1, 2 � �� � j� 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 B2 � �1

and

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 i2(t ) b

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 1�

1 � i1(t )

a

10 V

6.2

1�

1H

i3(t )

1�

2�

2H

1F

V0

1� i1(t)

1F i(t)

���������� ���������� [1.66 A, 5 A, �3.33 A, �3.33 A/s, 2.22 A/s]

� 1 � �0, 2 V0 , � V0 � � �

�����Circuit Theory and Networks—Analysis and Synthesis 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 k� i(t )

K

R1

L1

V1

V3

10 k�

4A

4 mH

8 k�

R2 v (t)

+ −

C3

C1 C2

����������

V2

6

[i(t ) � 2.857 e �2 �10 t ]

R

6.7

���������� � 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. 6�

3� 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

30 �

2 M�

[i(t) � 25(1 � e�4t)] 6.8

� 1000 V �

10 �F

1 M� 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 k�

���������� � 17 2� � 400000 A / s � � �

5 k�

2 k�

vC (t) 1 k�

6.5

2H

����������

K

10 �F

240 V

In the network shown in Fig. 6.186, switch is closed at t � 0. Obtain the current i2 (t).

� �

5 �F

6V

����������

10 �

[vC(t) � 5e�20t] 50 V

10 � i1(t)

2 �F

i2(t )

���������� [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�����



6.13 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 k� i(t)

vC (t )

2 �F

5V

5 k�

12 �

8�





����������

2

20 V

1H

20 � i (t )

v( t) �

���������� ( t �100 ) � � � � v(t ) � 5 � 8e 160 � �� ��

10 �

1

2F

6F

5V

[(i) 5 V, 5 V, 1 mA, 0, (ii) 0.01 s, (iii) e�100 t 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.

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.

���������� [i (t) � 2 e�30t] 6.11 In the network shown in Fig. 6.192, the switch is closed at t � 0. Find v (t) for t � 0.

vR

10 �

vL

1H

10 V



���������� 1 � 2

3A

1 H 3

1�

[0.0693 s]

v (t )



���������� [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. 1�

1V

1

2

1H

� 0.5 �

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 k�

10 V

y

� vC (t) �

z

0.1 �F

5 k�

2 H v(t )

���������� �

���������� [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

�����Circuit Theory and Networks—Analysis and Synthesis 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 �

0.1 H

2�

1H

� v (t) � L 200 u(t )

� �

� v (t) � L �

10 �F u(t)

i(t )

� �

v(t) �

i(t)

����������

1 F 2

����������

[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 k�

� 100 V �

� vC1 �

2F 2H � �

6.3

40 k�

� vC 3 �

(b) eat � ebt (d) aeat � bebt

The differential equation for the current i(t) in the circuit of Fig. 6.201 is 2�

i(t)

vC 2 1F

2 k�

(a) eat � ebt (c) aedt � bebt

2H

3F

1F

sin t

����������

6.2

(a)

In the circuit of Fig. 6.200, the voltage v(t) is

(b)

2

d 2i dt 2

d 2i dt 2

�2

�2

di � i � t � � sin t dt

di � 2i � t � � cos t dt

1�

1�

eat

����������

(a) 80 V, 32 V, 48 V (b) 80 V, 48 V, 32 V (c) 20 V, 8 V, 12 V (d) 20 V, 12 V, 8 V

� v( t ) �

(c) 1H

ebt

(d) ����������

2

d 2i dt

d 2i dt

2

2

�2

�2

di � i � t � � cos t dt

di � 2i � t � � sin t dt

�������������������������������

6.4

At t � 0�, the current i1 in Fig. 6.202 is 1

C

1 0.63

2

V

R L

R i1(t )

(a) (c)

V 2R V � 4R �

t

0.5

C

i2(t )

(c)

����������

6.5

i(t)

(b)

i(t) 0.5 0.31

V R

(b)



(d)

zero

t

0.5

(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

���������� 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

���������� (a) sin (10 � 45°) (c) sin (103 t � 53) 3

6.6

L

(b)

sin (10 t � 45°)

(d)

sin (103 t � 53°)

+ v(t) −

C

For the RL circuit shown in Fig. 6.204, the input voltage vi(t) � u(t). The current i(t) is

2�

���������� i(t )

6.8

0.5 0.31

2

t

1 L 2 C

(b)

R�

L C

(d)

R�

(a)

R�

(c)

R�2

i(t )

(a)

v(t )

����������

1H

vi ( t )

R

3

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

�����Circuit Theory and Networks—Analysis and Synthesis the terminal pair, the time constant of the system will be

v1 (t ) 8 �F v2 (t)

1 M�

2 �F

i(t)

Network of linear resistors and independent sources

+



���������� (a) v2 (�) � v1 (�) � 0 (b) v2(�) � 2 V, v1 (�) � 8 V (c) v2 (�) � v1 (�) � 8 V (d) v2(�) � 8 V, v1 (�) � 2 V 6.9

A B

(a) i(t ) 4 mA

The time constant of the network shown in Fig. 6.208 is

8V

(0, 0)

v (t )

(b)

R

���������� 2R

10 V

C

���������� (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 �s (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

2�

2�

2V

2H 1V

���������� ���������� (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)

zero

(b)

0.5 A/s

(c)

1 A/s

(d)

2 A/s

6.13 A step function voltage is applied to an RLC series circuit having R � 2 �, 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 6.1 (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

Frequency Domain Analysis of RLC Circuits

��������������������� 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.

������������������������������� 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 � � � j� The function f (t) must satisfy the following condition to possess a Laplace transform, �

� | f (t ) | e 0

�� t

dt � �

7.2�Circuit Theory and Networks—Analysis and Synthesis where � is real and positive. The inverse Laplace transform L–1 �F � s �� is � � j�

f (t ) �

1 F ( s) e st ds 2� j � ��j�

������������������������������������������������������� 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

Putting st = x, dt �

dx 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 t�0 The Laplace transform of unit step function is �

L �u(t )� � � 1� e 0



� st

� e � st � 1 dt � � � � � s � �0 s

u (t)

1

4. Delayed or Shifted Unit-Step Function The delayed or shifted unit-step function (Fig 7.2) is defined by the equation u (t � a) � 1 � � � � � � ����� 0 The Laplace transform of u (t � a) is

t

0

Fig. 7.1������������������� u (t − a)

t�a t�a 0

a

t



� � e � st � e � as L �u(t � a)� � � 1� e � st dt � � � � � s � s �a a

�����������������������������������

7.3 ������������������������������������������������7.3

5. Unit-Ramp Function The unit-ramp function (Fig 7.3) is defined by the equation r(t) ��t t�0 � � � � ��� 0 t�0 The Laplace transform of the unit-ramp function is

t

0



L �r (t )� � � t e � st � 0

r (t)

1 ���������������������������

s2

r (t − a)

6. Delayed Unit-Ramp Function The delayed unit-ramp function (Fig 7.4) is defined by the equation r (t � a) � t t�a � � � � � � �����0 t�a The Laplace transform of r (t � a) is �

L �r (t � a)� � � t e � st dt � a

e � as

a

0

t

�����������������������������������

s2

7. Unit-Impulse Function The unit-impulse function (Fig 7.5) is defined by the equation � (t) � 0 t ��0

d (t)

1



and

� � (t ) dt � 1

t�0

t

0

��

The Laplace transform of the unit-impulse function is �

������������������������������ at

L �� (t )� � � � (t ) e � st dt � 1 0

8. Exponential Function (eat) The Laplace transform of the exponential function (Fig 7.6) is �

� � � e �( s � a)t � 1 L e at � � e at e � st dt � � e � ( s � a )t dt � � � � � a s � a s � � �0 0 0

� �

9. Sine Function

0

�����������������������������

1 � e j� t � e � j� t � . � 2j � The Laplace transform of the sine function is �1 � 1 1 � 1 1 � � � L{e j� t } � L{e � j� t }� � � L �sin � t � � L � (e j� t � e � j� t )� � � 2 � � � � 2 j � s � j� s � j� � s � � 2 �2 j � 2j 10. Cosine Function 1 We know that cos � t � ��e j� t � e � j� t �� . 2 The Laplace transform of the cosine function is We know that sin � t �

s 1� 1 1 � �1 � 1 � � L �cos � t � � L � (e j� t � e � j� t )� � �� L{e j� t } � L{e � j� t }�� � � 2 � s � j� s � j� �� s 2 � � 2 �2 � 2

t

����Circuit Theory and Networks—Analysis and Synthesis 11. Hyperbolic sine function









1 �t e � e �� t . 2 The Laplace transform of the hyperbolic sine function is 1� 1 1 � � �1 � 1 L �sinh � t � � L � (e� t � e �� t )� � �� L{e� t } � L{e �� t }�� � � � � 2 � s � � s � � �� s 2 � � 2 �2 � 2 12. Hyperbolic cosine function 1 We know that cosh � t � e� t � e �� t . 2 The Laplace transform of the hyperbolic cosine function is We know that sinh � t �

s 1� 1 1 � �1 � 1 L �cosh � t � � L � (e� t � e �� t )� � �� L{e� t } � L{e �� t }�� � � � � 2 � 2 � s � � s � � � s � �2 �2 � 2 13. Exponentially Damped Function Laplace transform of an exponentially damped function e�at f (t) is







L e � at f (t ) dt �



f (t )e � at e � st dt �

0�





f (t )e � ( s � a )t dt � F ( s � a)

0�

Thus, the transform of the function e�at f (t) is obtained by putting (s � a) in place of s in the transform of f (t). � � L e � at sin � t � L e � at sinh � t � 2 2 ( s � a) � � ( s � a) 2 � � 2









L e � at cos � t �

s�a ( s � a) � � 2









L e � at cosh � t �

2

s�a ( s � a) 2 � � 2

���������������������������������������� 7.4.1 Linearity If L{ f1 (t )} � F1 ( s) and L{f 2 (t )} � F2 ( s) then L{af1 (t ) � bf 2 (t )} � aF1 ( s) � bF2 ( s) where a and b are constants. �

Proof

L{ f (t )} �

� f (t ) e

� st

dt

0 �





0

0

0

L{af1 (t ) � bf 2 (t )} � � {af1 (t ) � bf 2 (t )}e � st dt � a � f1 (t )e � st dt � b � f 2 (t )e � st dt � aF1 � s � � bF2 � s �

7.4.2 Time Scaling If L{ f (t )} � F ( s) then L{f ( at )} �

1 � s� F� � a � a� �

Proof

L{ f (t )} �

� f (t ) e

� st

dt

0



L{ f ( at )} �

� f (at )e 0

� st

dt

7.4 ������������������������������������

Putting at � x, dt �

dx a �

L{( f ( at )} �



f ( x) e

� x� � s� � � a�

0

� s�



�� � x 1 � s� dx 1 � � f ( x ) e � a � dx � F � � a a0 a � a�

7.4.3 Frequency-Shifting Theorem If L{ f (t )} � F ( s) then L{e � at f (t )} � F ( s � a) �

Proof

L{ f (t )} �

� f ( t )t e

� st

d

0 �



0

0

L{e � at f (t )} � � e � at f (t )e � st dt � � f (t )e � ( s � a )t dt � F ( s � a)

7.4.4 Time-Shifting Theorem If L{ f (t )} � F ( s) then L{ f (t � a)} � e � as F ( s) �

Proof

� f ( t )e

L{ f (t )} �

� st

dt

0



� f (t � a) e

L{ f (t � a)} �

� st

dt

0

Putting When

t � a � x, dt � dx t � a, x � 0 t � �, x � � �



L{ f (t � a)} �

0





0

0

f ( x )e � s( a � x ) dx � e � as � f ( x )e � sx dx � e � as � f (t ) e � st dt � e � as F ( s)

7.4.5 Multiplication by t (Frequency-Differentiation Theorem) If L{ f (t )} � F ( s) then L{t f (t )} � �

d F ( s) ds �

Proof

L{ f (t )} � F ( s) �

� f ( t )e

� st

dt

0

Differentiating both the sides w.r.t s using DUIS, �



d d � F ( s) � � f (t )e � st dt � � f (t )e � st dt ds ds 0 s � 0 �



0

0

� � ( �t ) f (t )e � st dt � � {� t f ( t )}e � st dt � � L{t f (t )} d L{t f (t )} � ( �1) F ( s) ds

����Circuit Theory and Networks—Analysis and Synthesis

7.4.6 Division by t (Frequency-Integration Theorem) �

f (t ) � If L{ f (t )} � F ( s), then L �� � � � F ( s) ds � t � s Proof L � f (t )� � F ( s) �



� f (t ) e

� st

dt

0

Integrating both the sides w.r.t s from s to �, �

��

s

s 0

� F ( s) ds � � � f (t )e

� st

dt ds

Since s and t are independent variables, interchanging the order of integration, �



� �� � � � f (t ) � st �1 � st � st � F ( s ) ds � f ( t ) e ds dt f ( t ) e dt � � � � � � �� � �� �t � t e dt � � � s 0 �s 0 0 � s �

� f (t ) � L� � � � F ( s)ds � t � s

7.4.7 Time-Differentiation Theorem: Laplace Transform of Derivatives If L{ f (t )} � F ( s) then L{ f �(t )} � sF ( s) � f (0) L{ f ��(t )} � s 2 F ( s) � sf (0) � f �(0) In general,

L{ f ��(t )} � s ��F ( s) � s n �1 f (0) � s n � 2 f �(0) � s n �3 f ��(0)� � f ( n �1) (0)

Proof L � f �(t )� �



� f �( t ) e

� st

dt

0

Integrating by parts, �





0

0

L{ f �(t )} � �� f (t )e � st �� � � ( � s) f (t )e � st dt � � f (0) � s � f (t )e � st dt � � f (0) � sL � f (t )� 0

Similarly,

L{ f �� t} � � f � � sL{ f � t} � � f �(0) � s � � f (0) � sL{ f (t )}� � � f �(0) � sf (0) � s 2 L{ f (t )} In general,





L f n (t ) � s n F ( s) � s n �1 f (0) � s n � 2 f �(0) � s n � 3 f ��(0) � � f n �1 (0)

7.4.8 Time-Integration Theorem: Laplace Transform of Integral �� t �� F ( s) If L � f (t )� � F ( s) then L �� f (t )dt � � s �0 �

7.5 ���������������������������7.7

�� t �� � t �� L �� f (t )dt � � � � f (t ) dt �e � st dt �0 � 00 �

Proof Integrating by parts,



�� � �� t �� � t �� � e � st � � � e � st � � d t 1 1 F ( s) � � � f t dt dt f (t )e � st dt � L{ f (t )} � L �� f (t )dt � � � � f (t )dt � ( ) � � � � � � � � � � s s s � � s � �� 0 0 ��� � s � � dt 0 � �� �0 � �� 0 0

7.4.9 Initial Value Theorem If L{ f (t )} � F ( s) then lim f (t ) � lim sF ( s) t �0

s ��

Proof We know that, L{ f �(t )} � sF ( s) � f (0) �

sF ( s) � L{ f �(t )} � f (0) �

� f �( t ) e

� st

dt � f (0)

0





0

0

lim sF ( s) � lim � f �(t )e � st dt � f (0) �

s ��

s ��

f � ( t )e � slim[ ��

� st

]dt � f (0) � 0 � f (0) � f (0) � lim f (t ) t �0

7.4.10 Final Value Theorem If L{ f (t )} � F ( s) then lim f (t ) � lim sF ( s) t ��

s ��

Proof We know that L{ f �(t )} � sF ( s) � f (0) �

sF ( s) � L{ f �(t )} � f (0) �

� f �( t ) e

� st

dt � f (0)

0







0

0

0

lim sF ( s) � lim � f �(t ) e � st dt � f (0) � � lim �� f �(t )e � st �� dt � f (0) � s �� s�0 s�0 � f

� (t ) 0

� f �(t )dt � f (0)

� f (0) � lim f (t ) � f (0) � f (0) � lim f (t ) t ��

t ��

���������������������������������� If L{ f (t )} � F ( s) then f (t ) is called inverse Laplace transform of F ( s) and symbolically written as f (t ) � L�1{F ( s)} where L�1 is called the inverse Laplace transform operator. Inverse Laplace transform can be found by the following methods: (i) Standard results (ii) Partial fraction expansion

7.8�Circuit Theory and Networks—Analysis and Synthesis

7.5.1 Standard Results Inverse Laplace transforms of some simple functions can be found by standard results and properties of Laplace transform.

�������������

Find the inverse Laplace transform of

s 2 � 3s+ 4 .

s3 s � 3s � 4 1 3 4 � � 2� 3 F ( s) � s s s3 s 2

Solution

L�1{F ( s)} � 1 � 3t � 2t 2

�������������

Find the inverse Laplace transform of F ( s) �

Solution

3s � 4 2

s �9



3s 2

3s+ 4 . s2 + 9

4



2

s �9 s �9 4 L {F ( s)} � 3 cos 3t � sin 3t 3 4s+ 15 . Find the inverse Laplace transform of 16s 2 � 25 �1

�������������

F ( s) �

Solution

L�1{F ( s)} �

�������������

4 s � 15 16 s 2 � 25

Solution

4 s � 15 1 15 1 s � � � 2 25 � 4 s 2 � 25 16 s 2 � 25 16 � s � � � 16 16 16 �

1 5 3 5 cosh t � sinh t 4 4 4 4

Find the inverse Laplace transform of F ( s) �

Solution

�������������



2s � 2 s 2 � 2 s � 10



2s+ 2 2

.

s + 2s+ 10 2( s � 1)

( s � 1) 2 � 9

� s � �t L�1{F ( s)} � 2e � t L�1 � 2 � � 2e cos 3t � s � 9� 3s � 7 Find the inverse Laplace transform of 2 . s � 2s � 3 F ( s) �

3s � 7 2

s � 2s � 3



3( s � 1) � 10 2

( s � 1) � 4

�3

( s � 1) 2

( s � 1) � 4

� 10

1 ( s � 1) 2 � 4

� s � t �1 � 1 � t t L�1{F ( s)} � 3e t L�1 � 2 � � 10e L � 2 � � 3e cosh 2t � 5e sinh 2t � 4 � 4 s s � � � �

7.5.2 Partial Fraction Expansion P ( s) where P ( s) and Q( s) are polynomials in s. For performing Q ( s) partial fraction expansion, the degree of P ( s) must be less than the degree of Q( s). If not, P ( s) must be

Any function F ( s) can be written as

7.5 ���������������������������7.9

divided by Q( s), so that the degree of P ( s) becomes less than that of Q( s). Assuming that the degree of P ( s) is less than that Q( s), four possible cases arise depending upon the factors of Q( s). Case I

Factors are linear and distinct, F ( s) �

P ( s) ( s � a)( s � b)

By partial-fraction expansion, F ( s) �

A B � s�a s�b

Case II Factors are linear and repeated, P ( s) F ( s) � ( s � a)( s � b) n By partial-fraction expansion, F ( s) �

A B B2 Bn � 1 � � ... � s � a s � b ( s � b) 2 ( s � b) n

Case III Factors are quadratic and distinct, F ( s) �

P ( s) 2

( s � as � b)( s 2 � cs � d )

By partial-fraction expansion, F ( s) � Case IV

As � B



2

s � as � b

Cs � D 2

s � cs � d

Factors are quadratic are repeated, F ( s) �

P ( s) ( s 2 � as � b)( s 2 � cs � d ) n

By partial-fraction expansion, F ( s) �

�������������

As � B 2

s � as � b



C1s � D1 2

s � cs � d



C2 s � D2 2

( s � cs � d )

Find the inverse Laplace transform of

Solution

F ( s) �

s�2 s( s � 1)( s � 3)

F ( s) �

A B C � � s s �1 s � 3

2

� ... �

Cn s � Dn ( s � cs � d ) n

s�2 . s( s � 1)( s � 3)

By partial-fraction expansion,

A � sF ( s) s � 0 �

2

2 s�2 � ( s � 1)( s � 3) s � 0 3

7.10�Circuit Theory and Networks—Analysis and Synthesis B � ( s � 1) F ( s) s � �1 �

s�2 1 �� s( s � 3) s � �1 2

C � ( s � 3) F ( s) s � �3 �

1 s�2 �� 6 s( s � 1) s � �3

2 1 1 1 1 1 � � � � � 3 s 2 s �1 6 s � 3 �1� 1 � 1 � 1 �1 � 1 � 2 1 � t 1 �3t 2 L�1{F ( s)} � L�1 � � � L�1 � �� L � �� � e � e 3 6 �s� 2 � s � 1� 6 � s � 3� 3 2 F ( s) �

�������������

Find the inverse Laplace transform of F ( s) �

Solution

s�2

s�2 . s ( s � 3) 2

s 2 ( s � 3)

By partial-fraction expansion, F ( s) �

A B C � � s s2 s � 3

s � 2 � As( s � 3) � B( s � 3) � Cs 2 � As 2 � 3 As � Bs � 3 B � Cs 2 � ( A � C ) s 2 � (3 A � B ) s � 3 B Comparing coefficients of s 2 , s1 and s0 , A�C � 0 3A � B � 1 3B � 2 Solving these equations, A� F ( s) � L�1{F ( s)} �

�������������

1 2 1 , B� , C�� 9 3 9 1 1 2 1 1 1 � � � � � 9 s 3 s2 9 s � 3 1 �1 � 1 � 2 �1 � 1 � 1 �1 � 1 � 1 2 1 �3t L � �� L � 2�� L � �� � t� e 9 9 3 9 3 9 s � s � � �s� 3 � �

Find the inverse Laplace transform of

Solution

By partial-fraction expansion,

F ( s) �

s 2 � 15s � 11 .

(s+ 1)(s � 2)2

5s 2 � 15s � 11 ( s � 1)( s � 2) 2

7.5 ���������������������������7.11

F ( s) �

A B C � � s � 1 s � 2 ( s � 2) 2

5s 2 � 15s � 11 � A( s � 2) 2 � B( s � 1)( s � 2) � C ( s � 1) � A( s 2 � 4 s � 4) � B( s 2 � s � 2) � C ( s � 1) � As 2 � 4 As � 4 A � Bs 2 � Bs � 2 B � Cs � c � ( A � B)s2 � (4 A � B � C )s � (4 A � 2 B � C ) Comparing coefficients of s 2 , s1 and s0 , A� B � 5 4 A � B � C � 15 4 A � 2 B � C � �11 Solving these equations, A�1 B�4 C � �7 F ( s) �

1 4 7 � � s � 1 s � 2 ( s � 2) 2

1 � � 1 � �1 � 1 � �1 � L�1{F ( s)} � L�1 � � e � t � 4e 2t � 7te 2t � � 7L � � � 4L � 2� 1 2 s � s � � � � � � ( s � 2) �

�������������

Find the inverse Laplace transform of

Solution

F ( s) �

By partial-fraction expansion, F ( s) �

3s+ 1 (s+ 1)(s 2 + 2)

3s � 1 ( s � 1)( s 2 � 2) A Bs � C � s � 1 s2 � 2

3s � 1 � A( s 2 � 2) � ( Bs � C )( s � 1) � As 2 � 2 A � Bs 2 � Bs � Cs � C � ( A � B ) s 2 � ( B � C ) s � ( 2A � C ) 2 1 0 Comparing coefficients of s , s and s , A� B � 0 B�C � 3 2A � C � 1 Solving these equations,

2 2 7 A� � , B� , C � 3 3 3

.

7.12�Circuit Theory and Networks—Analysis and Synthesis s 2 1 2 7 1 � � 2 � � 2 F ( s) � � � 3 s �1 3 s � 2 3 s � 2 2 �1 � 1 � 2 �1 � s � 7 �1 � 1 � 2 �t 2 7 �1 L {F ( s)} � � L � sin 2t �� L � 2 �� L � 2 � � � e � cos 2t � 3 3 3 3 2 � s � 1� 3 � s � 2� 3 � s � 2�

�������������� Solution

Find the inverse Laplace transform of s

F ( s) � L�1{F ( s)} �

( s 2 � 1)( s 2 � 4)



s ( s 2 � 1)( s 2 � 4)

.

s � s2 � 4 � s2 � 1 � 1 � s s � � 2 � 2 �� � 2 2 3 � ( s � 1)( s � 4) � 3 � s � 1 s � 4 ��

1 � �1 � s � �1 � s � � 1 ��L � 2 � � � [cos t � cos 2t ] �L � 3 � � s 2 � 1� � s � 4 �� 3

������������������������������������������������������� Voltage–current relationships of network elements can also be represented in the frequency domain. 1. Resistor For the resistor, the v–i relationship in time domain is v (t) ��R i (t) The corresponding frequency–domain relation are given as V (s) ��RI (s) The transformed network is shown in Fig 7.7. I (s)

i (t) +

+ R

v (t)

R

V (s)





Fig. 7.7��������� For the inductor, the v–i relationships in time domain are di v(t ) � L dt t 1 i(t ) � � v(t ) dt � i(0) L0 The corresponding frequency-domain relation are given as V ( s) � Ls I ( s) � L i(0) 1 i( 0) I ( s) � V ( s) � Ls s The transformed network is shown in Fig 7.8.

2. Inductor

I (s)

I (s)

i (t ) +

+

+

i (0) v (t)

Ls V (s)

V (s) − +



Li (0) −



�����������������

Ls

i (0) s

7.7 ���������������������������7.13

3. Capacitor

For capacitor, the v�i relationships in time domain are t 1 v(t ) � � i(t ) dt � v(0) C0

dv dt The corresponding frequency–domain relations are given as 1 v ( 0) V ( s) � I ( s) � Cs s I ( s) � CsV ( s) � Cv(0) The transformed network is shown in Fig 7.9. i( t ) � C

i (t)

I (s)

I (s)

+

+

+ 1 Cs

v (t)

+ − −

1 Cs

V (s)

V (s)

C

Cv (0)

v (0) s −



������������������

���������������������������������� Consider a series RL circuit as shown in Fig. 7.10. The switch is closed at time t � 0. R

V

L i (t)

�������������������� For 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)

�����������������������������

� s�0

V R

�����Circuit Theory and Networks—Analysis and Synthesis 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

�������������� 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. 1�

1 2

10 V

1�

1H i (t)

��������� 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. 1�

Solution



10 � 10 A 1 Since the current through the inductor cannot change instantaneously, i(0 � ) �

10 V i (0�)

i (0 ) ��10 A �

���������

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

���������

7.7 �������������������������������

������������ 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. 2�

a

b � 1�

2V

1H

2H

v (t) �

���������

2�

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�)

��������� + s

1

V ( s) � 2 V ( s) V ( s) � � �0 2s 1 s 3� 1 � V ( s) � 1 � � � � � 2s � s

V (s)

2 −

���������

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 �

for t � 0

In the network of Fig. 7.18, the switch is opened at t � 0. Find i(t).

������������

10 � 3� 36 V

6� 0.1 H

i (t)

���������

iT (0�) 10 �

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 � ) �



3� 36 V iL (0�)

���������

6�

�����Circuit Theory and Networks—Analysis and Synthesis Since current through the inductor cannot change instantaneously, iL(0+) = 2 A For t > 0, the transformed network is shown in Fig. 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

������������ 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 short circuit. 36 36 iT (0 � ) � � �3A 10 � ( 4 � 4) 10 � 2 36 V 4 i( 0 � ) � 3 � � 1.5 A 4�4 Since current through the inductor cannot change instantaneously, i(0+) = 1.5 A For t > 0, the transformed network is shown in Fig. 7.23. Applying KVL to the mesh for t > 0,

4�

10 � iT (0� ) 4�

i (0� )

��������� 4

�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

�������������� 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.

7.7 ���������������������������7.17 2�

2H

2�

1V i1 (t)

3H i2 (t)

���������

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 �

2�

1V i1 (0�)

���������

1

2s

2

2 I1 (s)

3s I2 (s)

���������

1 s

2 1 ( s � 1) ( s � 1) �2 0 s �1 s �1 s 3 � � � I 2 ( s) � � 2 �2 2 � 2s 1� 1� ( 2 � 2 s)( 4 � 3s) � 4 s(3s � 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 Networks—Analysis and Synthesis 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

�������������� In the network of Fig. 7.27, the switch is closed at t � 0 with the network previously unenergised. Determine currents i1(t). 10 �

1H

1H 10 �

100 V i1 (t)

10 �

i2 (t)

��������� Solution For t > 0, the transformed network is shown in Fig. 7.28. 10

s

s 100 s

10 I1 (s)

I2 (s)

10

��������� 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) � � 2 2 s � 20 �10 s ( s � 10)( s � 30) ( s � 20) � 100 s( s � 40 s � 300) �10 s � 20

7.8 ����������������������������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 ) �

10 5 � 5e �10 t � e �30 t 3 3

for t � 0 R

����������������������������������� 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)

��������������������

7.20�Circuit Theory and Networks—Analysis and Synthesis 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

������������

1 Cs

�����������������������������

1

i( t ) �

I (s)

V � RC t e R

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).

1�

a

1�

b



10 V

6F

3F

vc (t)

i (t) �

��������� �

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. 1� v6 F (0�) � 10 V i (0�) ��0 v3 F (0�) ��0 v6F (0 � ) Since voltage across the capacitor cannot change 10 V � instantaneously, i (0 ) v6 F (0�) ��10 V v3 F (0�) ��0 ��������� For t � 0, the transformed network is shown in 7.33. Applying KVL to the mesh for t > 0, 1 10 1 1 � I ( s) � I ( s) � I ( s) � 0 s 6s 3s 1 1 1 10 I ( s) � I ( s) � I ( s) � 6s 1 V (s) 6s 3s s c 3s 10 60 10 I (s) 10 I ( s) � � � 1 1 � 6 s � 3 s � 0.5 s � s �1 � � � � 6 s 3s �

Solution

Taking the inverse Laplace transform, i(t) � 10e

�0.5t

��������� for t > 0

7.8 ����������������������������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 ) �

������������

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 �

10 � 2 F

10 V

vc (t)

���������

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 s � Vc ( s) � Vc ( s) � 0 1 10 10 2s

Vc ( s) �

1 2 sVc ( s) � 0.2Vc ( s) � s Vc ( s) �

1 0.5 � s( 2 s � 0.2) s( s � 0.1)

10 s

10

10

���������

1 2s

Vc (s)

7.22�Circuit Theory and Networks—Analysis and Synthesis 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

������������ 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

1� 1�

���������

Solution 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 v (0 � ) vc (0 � ) 1�

2A 1�

��������� 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

���������

7.8 ����������������������������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

for t � 0

In the network of Fig. 7.39, the switch is moved from a to b, at t � 0. Find v(t).

������������

4�

a b



6V

v (t) �

1F

2�

2�

���������

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.

4�

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

��������� 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

��������� 2

s�

2 3

Taking the inverse Laplace transform, v � t � � 2e

2 � t 3

2�

for t � 0

2

�����Circuit Theory and Networks—Analysis and Synthesis

������������ 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). 2� � 2�

4V

1F

1F

v (t)



���������

Solution At t � 0 , the network is shown in Fig 7.43. At �

2�

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

2 �2V 2�2

capacitor

2�

4V

cannot

change ���������

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

��������� 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 �����������������������������������������

�������������������������������������������� Consider a series RLC circuit shown in Fig. 7.45. The switch is closed at time t � 0. R

L V i (t)

C

��������������������� 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) � � � Cs s � �

�����������������������������

V s I ( s) � � LCs 2 � RCs � 1 s 2 � Cs

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� � � � � � �� � � 2 � � 02 � �� � � � � 2L 2L LC

s2 � �

1 R � R� � � � � � �� � � 2 � � 02 � �� � � � 2L � 2L LC

2

where

�� �0 �

and By partial-fraction expansion, of I(s),

1 Cs

R 2L 1 LC

� � � 2 � � 02

�����Circuit Theory and Networks—Analysis and Synthesis 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 � � �0 In this case, the solution is given by



i � t � � e �� t k1e �t � k2e � �t 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 � � �0 In this case, the solution is given by for t � 0 i(t) � e��t (k1 � k2 t) Case III When the roots are complex conjugate, it gives an underdamped response. R 1 � 2L LC � � �0 In this case, the solution is given by i(t ) � k1 e s1t � k2 e s2t Case II

where Let where and

s1,2 � �� � � 2 � � 02

� 2 � � 02 � �1 � 02 � � 2 � j� d j � �1

� d � � 02 � � 2

7.9 �������������������������������������7.27

Hence



i(t ) � e �� t k1 e j� d t � k2 e � j� d t � �� e � e �� t �( k1 � k2 ) � � �

j� d t



�� e j� d t � e � j� d t �� � � e � j� d t �� � � j ( k1 � k2 ) � �� 2 2j � � ��

� e �� t �( k1 � k2 ) cos � d t � j ( k1 � k2 )sin � d t �

������������

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 �

0.5 H

0.5 F

v (t) �

��������� 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 �



��������� 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



���������

4 ( s � 2)

2

7.28�Circuit Theory and Networks—Analysis and Synthesis Taking inverse Laplace transform, v � t � � 4t e �2t

for t � 0

�������� ���� 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 �

5V

0.5 H

200 �F

��������� 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 �

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�)

���������

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 ���������

I ( s) 0.5s I ( s) � 1 � 5000 �0 s I ( s) �

1 5000 0.55s � s



2s 2

s � 10000

Taking inverse Laplace transform, i � t � � 2 cos 100t

for t � 0

1

7.9 �������������������������������������7.29

������������ 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 � � s � 2 s � 2 ( s � 1) 2 � 1 ( s � 1) 2 � 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

7.30�Circuit Theory and Networks—Analysis and Synthesis 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)

��������� 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 �( s � 1) s�2 � s 2 � 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) � 3s � 5 s s2 � � ( 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���������������������������������������������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, s�2

i2 (t ) � 3e � t � 2te � t

for t � 0

��������������������������������������������������� 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)

�������������������� 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)

�����������������������������

7.32�Circuit Theory and Networks—Analysis and Synthesis � � 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 2

Comparing coefficients of s ,

B�

1 R

C�

L R2

7.10���������������������������������������������7.33

A�C � 0 A � �C � �

Taking inverse Laplace transform,

L

R2 L 1 1 1 L 1 I ( s) � � 2 � � 2 2 R s Rs R s� R L i( t ) � �

L R2

� R�



1 L �� � t t � 2 e � L� R R � R�

� (c)

�� � t 1 L t � 2 [1 � e � L � ] R R

for t � 0

When unit impulse signal is applied, v(t ) � � (t ) Taking Laplace transform,

V ( s) � 1 1 I ( s) � L

1 s�

R L

Taking inverse Laplace transform, � R�

i( t ) �

1 � �� L �� t e L

for t � 0

������������ 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). 1�

v (t)

v (t )

1

� �

1H i (t )

0

t

1

��������� 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 )

���������

�����Circuit Theory and Networks—Analysis and Synthesis 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

�������������� 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. 5�

2H

r (t � � � 3) � �

i(t)

��������� 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



2s I (s)

s2 e �3s 2

s ( 2 s � 5)



0.5 e �3s 2

s ( s � 2.5)

By partial-fraction expansion, s 2 ( s � 2.5)

+ −

A B C � � s s 2 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

���������

7.10�������������������������������������������������

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)

������������ Determine the expression for vL (t) in the network shown in Fig. 7.64. Find vL(t) when (i) vs(t) � � (t), and (ii) vs(t) � e�t u(t). 5� � 1 v (t ) 2H L

� �

Vs (t )



���������

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 ) � � (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)

���������

�����Circuit Theory and Networks—Analysis and Synthesis 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

������������� 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. 2�

2d (t � �� 3)

� �

1H i (t)

���������

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

������������

�2( t � 3)

2

2e−3s

+ −

s I (s)

��������� 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 �

50 sin 25 t

5H i (t)

���������

7.10���������������������������������������������7.37

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 2 s � 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

���������

250 � ( As � B )( s � 2) � C ( s 2 � 625) � ( A � C ) s 2 � ( 2 A � B ) s � ( 2 B � 625C ) Comparing coefficients,

Solving the equations,

A�C � 0 2A � B � 0 2 B � 625 C � 250 A � �0.397 B � 0.795 C � 0.397 �0.397 s � 0.795 0.397 0.397 s 0.795 0.397 I ( s) � � �� 2 � � s�2 s 2 � 625 s � 625 s 2 � 625 s � 2

Taking the inverse Laplace transform, i(t ) � �0.397 cos 25t � 0.032 sin 25t � 0.397e �2t

������������

for t � 0

Find impulse response of the current i(t) in the network shown in Fig. 7.70. 1�

i1 (t )

i (t ) d (t)

� �

1� 2H

��������� 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

���������

7.38�Circuit Theory and Networks—Analysis and Synthesis 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

�������� ���� The network shown in Fig. 7.72 is at rest for t < 0. If the voltage v(t ) � u(t ) cos t � A� (t ) is applied to the network, determine the value of A so that there is no transient term in the current response i(t). 1�

v (t )

2H i (t )

��������� v(t ) � u(t ) cos t � A� (t ) s V ( s) � 2 �A s �1

Solution For t � 0, the transformed network is shown in Fig. 7.73. 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� � s �1 2 � s � � ( s 2 � 1) s � � 2 2�

1

V (s )

2s I (s )

���������

The transient part of the response is given by the first term. Hence, for the transient term to vanish, K1 � 0. 1� � K1 � � s � � I ( s) s � � 1 � 2� 2 When

K1 � 0 5 1 A� 4 2 2 A � � 0.4 5

�1 � 5� � A� � � 4� 2 � � 5� 2� � � 4�

7.11���������������������������������������������7.39

��������������������������������������������������� Consider a series RC circuit as shown in Fig. 7.74. R

+ −

v (t )

C i (t )

�������������������� 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)

�����������������������������

When unit step signal is applied, v ( t ) � u( t )

V ( s) �

1 Cs

for t � 0

�����Circuit Theory and Networks—Analysis and Synthesis 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) � � (t) Taking Laplace transform, V ( s) � 1 1 1 1 � � � 1� � RC RC RC � � �1 � 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 � �� (t ) � � e R� RC �� �

for t � 0

�������� ���� 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)

���������

7.11�������������������������������������������������

v(t) � u(t) � u(t � T)

Solution

V ( s) �

� sT

1 e � s s



R � sT

1� e s

For t � 0, the transformed network is shown in Fig. 7.77. Applying KVL to the mesh for t � 0,

1 Cs

V (s ) + −

I (s)

��������� V ( s) � RI ( s) �

1 I ( s) � 0 Cs 1 � � s 1 � e � sT 1� 1 e � sT � V ( s) R � � � I ( s) � � 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)

��������� 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) �

By partial-fraction expansion,

2e �2 s 0.67e �2 s � 1 � s( s � 0.33) � s2 � 3 � � � s�

3

2e−2s + − s2

I(s)

���������

1 s

�����Circuit Theory and Networks—Analysis and Synthesis 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

�������������� 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. 5�

d (t )

� �

2F i (t )

��������� 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 ) � 0.2� (t ) � 0.02 e �0.1t u(t )

1

+ −

I(s)

���������

1 2s

7.11�������������������������������������������������

������������

For the network shown in Fig. 7.82, find the response v0 (t). 2� � vs (t ) =

1 cost u(t ) 2

� �

1 F 4

vo (t) �

���������

Solution For t � 0, the transformed network is shown in Fig. 7.83. 2 1 s 2 s2 � 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� ��������� 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) � � � � � s � 2 s2 � 1 s2 � 1 s � 2 s2 � 1 Taking the inverse Laplace transform, i(t ) � 0.4 cos t � 0.2 sin t � 0.4e �2t

for t � 0

������������ 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. 2�

v (t )

1H



� �



���������

1 F vc (t)

�����Circuit Theory and Networks—Analysis and Synthesis

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 −

+ −

���������

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�������������������������������������������������

������������ 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. 5�

5r (t � 1)

1H

� �

1 F 6

i (t)

���������

Solution For t � 0, the transformed network is shown in Fig. 7.87. Applying KVL to the mesh for t � 0, 5e � s s2

5

6 � 5 I ( s) � sI ( s) � I ( s) � 0 s 5 I ( s) � sI ( s) �

5e−s s2

6 5e � s I ( s) � 2 s s I ( s) �

5e � s 2

s( s � 5 s � 6 )



5e � s s( s � 3)( s � 2)

s

+ − I (s)

6 s

���������

By partial-fraction expansion, A B C 1 � � � 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

�������������� For the network shown in Fig. 7.88, the switch is closed at t � 0. Determine the current i(t) assuming zero initial conditions.

�����Circuit Theory and Networks—Analysis and Synthesis 2�

1H

sin t

0.5 F i(t )

��������� Solution For t � 0, the transformed network is shown in Fig. 7.89. Applying KVL to the mesh for t � 0, 1 2

s �1

� 2 I ( s) � s I ( s) �

2 I ( s) � 0 s

1 s2 + 1

2� 1 � �� 2 � s � �� I ( s) � 2 s s �1 I ( s) �

s 2

s

2

I (s)

���������

2

( s � 1)( s � 2 s � 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 0.2 s � 0.8 � 2 s2 � 1 s � 2s � 2 0.2 s 0.4 0.2 s � 0.2 � 0.6 � 2 � � s � 1 s 2 � 1 ( s � 1) 2 � (1) 2

I ( s) �



0.2 s � 0.4

0.2 s s2 � 1



0.4 0.2( s � 1) 0.6 � � s 2 � 1 ( s � 1) 2 � 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

2 s

7.11�������������������������������������������������

�������������� 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. 5�

6e�2t

1H

� �

0.25 F i(t )

��������� 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

s

5

6 s+2

+ −

I (s)

4 s

6s

I ( s) �

���������

( 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

������������ The network shown has zero initial conditions. A voltage vi(t) � ���(t) applied to two terminal network produces voltage vo(t) � [e�2 t � e�3 t] u(t). What should be vi(t) to give vo(t) � t e�2 t u(t)? + vi (t ) −

Network

���������

+ vo (t ) −

�����Circuit Theory and Networks—Analysis and Synthesis

Solution For vi(t) � ���(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) � � � H ( s) ( s � 2) 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

�������� ���� A unit impulse applied to two terminal black box produces a voltage vo(t)� 2e�t �e�3t. Determine the terminal voltage when a current pulse of 1 A height and a duration of 2 seconds is applied at the terminal. + is (t )

Black box

vo (t)



��������� Solution

vo (t ) � 2e � t � e �3t 2 1 Vo ( s) � � s �1 s � 3

When is (t ) � � (t ),

is (t)

1 0

I s ( s) � 1

2

���������

t

��������������

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 �1 � 2� � � � � � 2e �2 s � � � � 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.

2�� i (t)

100�� 2�� 2�� 50��

100 V

4H 24 V

i1 (t)

���������

��������� [i1(t) � 3 � e�25 t] 7.2

0.5 H

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.

�����Circuit Theory and Networks—Analysis and Synthesis � 0.33��

10 V

1 � 2

10 A

0.5 H

1 � 8

v (t)

1 F v (t)

1F



��������� [v(t) � 1 � 4 e�10t] 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

7.8

5��

���������� [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. 2��

1

2 �

v (t)

20 V

3��

3��

2F

2V

v (t )

0.5 F

1H

��������� [v(t) � 7.5 � 12.5 e�(4/15)t] 7.5

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 � seconds.



���������� 7.9

800��

[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

3��

1��

i (t)

12 V

7.6

400��

0.001 �F

200��

1�� i (t ) 2

i1 (t)

1 F 3

��������� 6 [i(t) � 0.00857 � 0.01143 e�8.75 � 10 t, 0.013 A]

���������� [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 � resistor.

i (t ) 10��

1F 50 V

20 V

5��

10��

0.5 F

1H

2H

5�� 20��

10 V

30��

���������� [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.

���������� [i(t) � 0.1818 � 0.265 e�13.14t � 0.083 e�41.86t ]

��������������

having been established previously. Find i(t) for t � 0.

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.

3�� i (t )

2��

2H

s1

s2

5��

1.8 H

0.1 F

3.5��

10 V

3H

10 V

1 �F

���������� [i(t) � 1.5124e�2.22t � 3.049e�2.5t]

���������� [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��

10��

i (t )

10��

0.2 F

50 V i1 (t )

40��

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. 7��

5��

20 �F i1 (t )

2.5 H

����������

����������

50 V

15��

5A

i2 (t )

5��

5 cos 2t

1H

� �

1 F v (t ) 2

0.1 H

���������� ���������� �100.5t

�9949.5t

� i1 (t ) � 0.101e � � 10.05e � � �9949.5t �100.5t � 0.05e �i2 (t ) � 5 � 5.05e � 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)] �t (ii) i(t) � 4e u(t) [4t e�t u(t)]

�����Circuit Theory and Networks—Analysis and Synthesis 10��

� 1��

i (t )

v (t)

1F

10 F vc (t)

r (t)



���������� 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 ��t is equal to (a)

s �� ( s � � )2 � � 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 1�H

1F

v (t) i (t)



���������� (a) (c)

( s � a) 2 s2

( s � � )2 none of the above



e � as

a

1

1��

7.6

e � as

s�� ( s � � )2 � � 2

s 2

s � s �1 s�2 s2 � s � 1

(b) (d)

s�2 2

s � s �1 s�2 s2 � 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) (c)

3V 4V

(b) �3 V (d) �4 V

���������������������������������������� 0.1 �F

3V

7.9

�j2 �

vi

vo

1�k�

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)

���������� 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

����������

1mV

(a) a constant voltage (b) a linearly increasing voltage (c) an ideal impulse (d) as exponential increasing voltage

����������

7.8

(a) 0.5 A (b) 2 A (c) 1 A (d) 0 A current impulse 5 ���(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

��������������������� 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�

I2

I1 � V1 �

(a)

Twoport network

2 � V2 � 2�

(b)

Fig. 8.1������������������������������������������ 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.

�������������������������������� 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��������������������������������������������������� 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,

��������������������������� 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 � (s).

�12 ( ) �

I2 ( ) I1 ( )

� 21 ( ) �

I1 ( ) I2 ( )

������������������������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).

���������������

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 Ls) s� R Ls 1 Cs L Z ( s) � � � 2 1 1 C 2 R LCs C R RCs Cs � 1 � (R ( R Ls) s � s� Cs L LC

������������

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 4 s � 8s3 � 2 s � 2s � � 2 s � �� � 2s � �� 4 � s2 � 1 2s � 2s 2s � 2s 2 s � 16 s �12 2s � � � s Z ( s) � 2 s � � 2s � 2 1 1 2 � 4s2 2 � 4s2 8s3 � 4 s 2 � 4s2 � 2s � 2s 2s 2s

8.4���������������������������������������������������

������������

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 ( s 2 ) s 1 s � s3 1 s4 � 3 2 1 Z ( s) � � � � � � � 1 s s 2s2 � 1 s 2s2 � 1 2 s3 � s s� �s s

������������

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 � 22 s 2 � 1 1 1 s 30 s 4 �15 � 5s 2 � 2 s 2 � 1 � 5s 2 30 s 4 � 22s � Z ( s ) � 3s � � � 3s � � 2 � 2 1 5s 5s 2 s � 1 5 ( 2 s 1) 5 ( 2 s 2 1) s� 2s 1 5s ( s 2 ) Y ( s) � � Z ( s) 30 s 4 � 22 s 2 � 1

������������

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

���������������������������������8.5

� 1� R� � � Cs � V2 ( s) � I1 ( s) 1 R� Cs V2 ( s) R � I1 ( s) RCs C �1 V ( s) 1 Z12 ( s) � 2 � 1 � I1 ( s) � C s� � � RC �

Solution

������������

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 1 � V1 ( s) � V1 ( s) RC V2 ( s) � V1 ( s) 1 1 RCs RC C �1 R� s� Cs RC 1 V2 ( s) RC � 1 V1 ( s) s� RC 1 Cs

Voltage transfer function

������������������������������������ 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��������������� 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 ���) 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��������������������������������������������������� 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.

������������

For the network shown in Fig. 8.10, determine transfer function 1�

I1

1�

I2 ��0



V2 . V1



V1

1F

V2

1F





Fig. 8.10 Solution The transformed network is shown in Fig. 8.11. Vb Ib Va I1

Hence,

V1 V2 V1

1

I

������������

Ib

I2 = 0

Vb

+ 1 s

For the network shown in Fig. 8.12, determine the voltage transfer function I1

s

s

Va

Ib

I2 = 0

Vb

+ V1

+ 1

V2

1





Fig. 8.12 Solution

1

V

a 1 V2 + V � 2 � sV V2 1 1 V1 s s − 1 I b � V2 � sV V2 � V2 ( s � 1)V2 Fig. 8.11 V V2 ( s 2 s)V2 � a � I b sV Va I b � s( s )V2 sV 1 s s)V2 � (s 1II1 � Va ( s 2 ( s )V2 � ( s 2 � 3s � 1)V2 1 � 2 s � 3s 3s 1

Vb � V2 V2 � V2 1 Va s I b V2 � sV V2 � V2 ( s � 1)V2 Va I1 � � I b ( s � 1)V2 V2 � ( s � 2)V2 1 V1 s I1 Va � s ( )V2 ( )V2 ( Ib �

2

)V2

V2 −

V2 . V1

���������������������������������8.7

V2 1 � V1 s 2 � 3s 3s 1

Hence,

������������

Find the network functions

V1 V2 V , and 2 for the network shown in Fig. 8.13. I1 V1 I1

1H

I1

1H

I2 = 0 +

+ V1

1F

V2

1F





Fig. 8.13 Solution The transformed network is shown in Fig. 8.14. Vb

V2 V I b � 2 � sV V2 1 s ( s 2 1)V2 Va s I b V2 � s ( sV V2 ) V2 � (s

I1

s

s

Va

I2 = 0

Vb

+ 1 s

1 s

V1

Fig. 8.14

((ss 4

3s 2 1)V2

V1 s 4 � 3s 3s 2 1 � I1 s3 � 2 s V2 1 � V1 s 4 � 3s 3s 2 1 V2 1 � 3 I1 s � 2 s

�������������

Find the network functions

I1

1 F 4

V1 V2 V , , and 2 for the network in Fig. 8.15. I 1 V1 I1 3H

I2 = 0

+ V1

+ 1 H 2



2F

V2 −

Fig. 8.15

V2 −



V I1 � a � I b sV Va I b � s ( s 2 � 1)V2 sV s V2 ( s3 2 s)V2 1 s ( s 2 � 1) ( s 4 2 s 2 � s 2 � 1) V1 s I1 Va � s ( s3 � 2 s) Hence,

Ib

+

8.8��������������������������������������������������� Solution The transformed network is shown in Fig. 8.16. Vb

V2 V I b � 2 � 2 sV V2 1 2s

Va

4 s

I1

3s

Va

Ib

I2 = 0

Vb

+

+

3s I b V2 � 3s ( 2 sV V2 ) V2 � (6 s

2

1 2s

s 2

V1 −

1)V2

� 14 s � 2 � 2 Va V2 � � � I b � (6 s 2 1)V2 � 2 sV � V2 s s s � � 2 � 6s4 4 4 � 14 s 2 � 2 � V1 � I1 Va � � V2 (6 s 2 1)V2 � � � s s� s � �

V2 −

2

I1 �

Hence,

Fig. 8.16 57s 57s 2 8 � � V2 s2 �

V1 6 s 4 57s 57 s 2 8 � I1 14 s3 � 2 s V2 s2 � 4 V1 6 s 57s 57 s 2 8 V2 s � I1 14 s 2 � 2

������������� For the ladder network of Fig. 8.17, find the driving point-impedance at the 1 – 1� terminal with 2 – 2� open. 1

I1

1�

1�

1H

1H

I2 � 0





V1 1�

1�

1H

1F

1F

1F



2

V2 �

2�

Fig. 8.17 Solution The transformed network is shown in Fig. 8.18. 1

1�

I1

1

s

1

a

s



Ia

Ib

V1

1 s

1 s



V2 V I b � 2 � s V2 1 s

s

Vc

I2 � 0 � 1 s

2

V2 �

Fig. 8.18 Vc

1

Vb

2�

���������������������������������8.9

V I �

V � 1 s

b

V

� s(

s

s

a

s s

V

I

� s

� 2s V

sV �

s



s � s

�1 V �

� 2s V

s � s �2

s2

s

V

s

s) V s

s

s

� Hence,

Z11



s

V

V

V I1

�������������

s

s

s

s

s

s s

Determine the voltage transfer function 1�

I1

V for the network shown in Fig. 8.19. V

2H

I �0





V

V

1�

1



Fig. 8.19 Solution The transformed network is shown in Fig. 8.20. I1

V

1

Ia

V

2s

V

+

V � Ic � � 1 1 s

s �1 V

V

1 s

I1

s

Fig. 8.20

a

�s

Hence,

s �

s 1

2s

�1

1V

�1

V V V

0 +

1

V −

�1 V

V 1 s

I

I s

V �

Vc Ib

s � 4s 2

s

s

4s

)V

8.10���������������������������������������������������

�������������

For the network shown in Fig. 8.21, determine the transfer function 3� 2

I1

1�

I2

2F 3



I2 . V1

I�0 �

V1

1� 6

2F

V2





Fig. 8.21 Solution The transformed network is shown in Fig. 8.22. 3 2 1

I1

3 2s

+

I2

Va

Ia

I=0

Vb

1 6

1 2s

V1

+

Ib

V2





Fig. 8.22 Vb

V2 V2 � � 2 sV V2 � 6 V2 ( 2 s � 6)V2 1 1 2s 6 �3 3 � � � ( 2 s 5)( s � 3) � (6 15) � 2 2s � � 9 � � 1� V2 V1 � � � 1� I1 V2 � � � 1 I1 V2 � ( 2 6)V2 V2 � � 3 3 � 6 6 �� 6 6 ( s � 1) � � � � � � 2 2s � � 2 s 2 � 6 s � 5s � 15 � s � 1� � 2 s 2 � 12 s � 16 � � ( 2 s 5)( s � 3) ( s 1) � � �� V V � 2 2 � � � � V2 � s �1 s �1 s �1 � � � � � � 6 s 8) 2 ( s 4)( s � 2) 2(ss 2 � 6s V2 � V2 � s �1 s �1 I2 ���Ib � �6 V2 I1

Also,

Va � V2 Ia � Ib �

I2 3 ( s 1) �� V1 ( s � 4)( s 2)

�������������

For the network shown in Fig. 8.23, compute � 12 (s) s � 1�

I2

I�0

� I1

V1

� 1F

1F

V2 �



Fig. 8.23

I2 V and Z12 (s) � 2 . I1 I1

���������������������������������8.11

Solution The transformed network is shown in Fig. 8.24.

1

Va

Hence, and

V2 + V2 I1 I2 � � sV V2 V1 1 s − V1 1 I 2 � Va sV V2 V2 � (s 1)V2 V I1 � 1 � I 2 sV V1 I 2 � s (s � 1)V � sV � s 2 � 2s)V2 1 s I2 1 � �12 ( ) � I1 s � 2 Z12 ( s) �

I2

Va

I=0 +

1 s

1 s

V2 −

Fig. 8.24

1 V2 � I1 s 2 � 2 s

�������� �����

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

3�

3H

I1

I�0

I2

1F 3





2�

V1

1�

1F

1F 2



V2 �

Fig. 8.25 Solution The transformed network is shown in Fig. 8.26. 3

I1

Va

3 s

+

3s

Ia 1 s

V1 −

I2

Vb

I=0

Vc

+

Ib 2 2 s

1

V2 −

Fig. 8.26 Vb � V2 V I 2 � 2 � V2 1

Vc

V s � 3s � 2 � � 2 � V V2 � � V2 2 1 2s � 2 2 � 2 s � 2 �� 2� s � 9 s 2 � 8s � 2 � 3s (3s � 2) Va � 3s I a � V2 � V2 � V2 � � � V2 2s � 2 � 2s � 2 � Ia

Ib � I2 �

V2

8.12��������������������������������������������������� I1 �

V



s 8s � 2 V 2s 2

s



3s 2 2s 2



� 9s

8 2 s 2s 2

V

V �

� 2� V 2

s 3� V 3� �

27

3� s

3 I � s �1





� 15s � � 25s � 8

2

2�

�5 2

� s � 2� (36 �V � �

s�



s

� 5s 2 2s 2

s

s � 5s 2

V

s �2 s�2

s

36



s � 25s � 8 s�2 V V I I , , and 2 . V I1 V I1

For the resistive two-port network of Fig. 8.27, find 2�

I1

2�





I �

V







V

1�



Fig. 8.27 Solution The network is redrawn as shown in Fig. 8.28. I1

V

� 25 � 8) V �2

s � 25s � 8 2s 2

V 36 � I1 s

+

2

s�2

s 36

�������������

s � �2

s �2

V V I I1 V I1

3 � s �1

s

36

Hence,



2

V I

2

V

V

I +

Ib 1

Fig. 8.28

1

V

1

���������������������������������8.13

V2 � �V2 1 3 I 2 3V2 Vb Vb 4 � � � Vb 4 V2 1 3 3 2 I b Vb 8 V2 � 3 V2 � 11 V2 Va � � I b � 11 V2 � 4 2 � 15 V2 1 � 2 a � a � 30 2 � 11 V2 � 441 1 V2 V1 � � I a � 41 V2 �15 15 5 V2 � 56 6 V2 1 1 � 41 1 � � 56 1 �� � 41 1 �� 56

I2 � � Vb Ib Va Ia V1 I1 V2 V1 V2 I1 I2 V1 I2 I1

Hence,

�������������

Find the network function Ia

1�

1�

V2 for the network shown in Fig. 8.29. V1 2V1

I2 � 0

� �

� 2Ia

V1



1�



V2 �

Fig. 8.29 Solution The network is redrawn as shown in Fig. 8.30. Ia

1

3Ia

1

2V1

+

V1

I2 = 0

+ −

3Ia 2Ia



1

+

V2 −

Fig. 8.30

8.14��������������������������������������������������� 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,

�������������

I2 for the network shown in Fig. 8.31. I1

Find the network function 2Ia

2�

I2

�� �



Ia I1

1�

I1 2

1�

1�

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

I2 +

2

I1 2

+

f

Ia I1 2

1

1

I2

1

V2 −

b

c

e

Fig. 8.32 From Fig. 8.32, I1 � I a

I �

3 I1 2

I2 � Ia � I2

I1 2 ...(i)

Applying KVL to the loop abcda, �1 � 1 I a � 2

a

�0

�I � 3

a

�0

I �3

a

�0

...(ii)

�������������������������������������8.15

Substituting Eq. (i) in Eq. (ii), 3 I1 2

Ia � I2 3 I1 2

3

a

0

I2 � 4

a

0

...(iii)

Applying KVL to the loop dcefd, 1

a

1I2

I � � 2 I2 � 1 � � 0 � 2� Ia

3

2

1

Ia

0 3 I 2 � I1

...(iv)

Substituting Eq. (iv) in Eq. (iii), 3 I1 2

I 2 � 4 (3 I 2

3 I1 2

I 2 � 12 I 2

I1 ) � 0 4

1

0

11 I1 133 I 2 � 0 2 13 I 2 � �

11 I1 2

I2 11 �� I1 26

Hence,

���������������������������������������� 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������������������� 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���������������������������������������������������

�������������

For the resistive bridged T network shown in Fig. 8.34, find

V2 V2 I 2 I , , and 2 . V1 I 1 V1 I1

2�

1�

I1

1�

I3



I2 �

V1

V2

0.5 �

I1



I2

1�



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, 0 �V1 � �1 � 0 � � �0 5 2 5 � � � 1 � 0 � � �1 V1 0 0 �1 � I1 � 15 � 05 �1

1� � I1 � 1� � I 2 � �� � 4 � � I3 � 0 5 �1 2.5 1 1 4 V (10 � 1) � 1 � V1 0 5 �1 9 25 1 1 4

1 5 V1 �1 05 0 1 �1 0 4 �2 �V ( 2 1) 1 � 1 � � V1 � I2 � 1 5 0 5 �1 � 9 3 05 25 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 V1 � � I1 3 3 3I 2 I1 I2 1 �� � V1 3 I2

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

...(iv)

...(iv)

...(v)

�������������������������������������8.17

1 � V1 1 I2 � 3 �� 3 I1 V1 �I2 1 V2 � � V1 �3 I 2 3 �I2 1 V2 � � � I1 �3 I 2 3

�������������

For the network of Fig. 8.35, find Z11, Z12 and G12. Za

a

I1

b

+

+

Zb

V1

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. ( Z a � Zb )

V1

I1

I1 2

Z11 �

V1 Z a Zb � 2 I1

V2

Zb

Z12 �

V2 Zb � Z a � 2 I1

+

�I � Z a � 1 � � ( Zb � 2�

I1 2

Za )

I1 2

V1

I1 2 Za

+

I1 2 Zb V2



Zb −

Fig. 8.36

By voltage-division rule, V2 � G12 �

�������������

Zb Za

Zb

V2 Zb � V1 Z a

Za

V1

Za

Zb

V1 �

Zb Za

Za V1 Zb

Za Zb

For the network shown in Fig. 8.37, determine Z11 (s), G12 (s) and Z12 (s).

Za

8.18��������������������������������������������������� 1F 1�

I1

1�

I2 � 0





V1

V2

1F





Fig. 8.37 Solution The transformed network is shown in Fig. 8.38. 1 s

1

1

I3

+

+ 1 s

V1 −

I2 = 0

I1

V2 −

Fig. 8.38 Applying KVL to Mesh 1, � 1� V1 � 1 � I1 � s�

I3

...(i)

Applying KVL to Mesh 2, V2 �

1 I1 s

I3

...(ii)

Applying KVL to Mesh 3, 1� � � I1 � � 2 � � I 3 � 0 � s� � s � I3 � � I1 � 2 s � 1��

...(iii)

Substituting Eq. (iii) in Eqs. (i) and (ii), � s 2 � 3s s � 3s 1 � � 1� � s � � s �1 V1 � 1 � I1 � � I1 � I1 � � � I1 � � s� � 2 s 1� � s 2 s � 1) � 2 s � 1� s (2 � � s 2 � 2 s � 1� s 1 V2 � I1 � I1 � � � I1 s 2s � 1 � s ( 2 s � 1) � Hence,

Z11 ( s) �

3s 1 V1 s 2 � 3s � I1 s ( 2 s 1)

Z12 ( s) �

2s 1 V2 s 2 � 2s � I1 s ( 2 s 1)

G12 ( s) �

2s 1 V2 s 2 � 2s � 2 V1 s � 3s 3s 1

�������������������������������������8.19

�������������

For the network shown in Fig.8.39, find the driving-point admittance Y11 and trans-

fer admittance Y12. 1� 1F

I1

V1

1F

� �

I2

1�

1�

Fig. 8.39 Solution The transformed network is shown in Fig. 8.40. 1 1 s

I1

I3

+ −

V1

1

I1

1 s

I2

I2

1

Fig. 8.40 Applying KVL to Mesh 1, V1 �

�1 � � 1 I1 �s �

1 I 2 � I3 s

...(i)

1� 1 � 2� I 2 � I3 � s� s

...(ii)

Applying KVL to Mesh 2, 0

1

Applying KVL to Mesh 3, 1 1 �2 � 0 � � I1 � I 2 � � � 1� I 3 �s � s s Writing these equations in matrix from, �1 1 � s �1 �V1 � � 1 �0� � � 1 2� � � � s �0� � 1 1 �� s � s �1 I1 � �

� � � � I1 � � �I2 � � �I � 2 � � 3� � 1� s � 1 s 1 s



...(iii)

8.20��������������������������������������������������� 1 �1 s ��

� �

2�

1 1 s

1 s

s2 � 5 s

1 s 1 1� � 2 � 1 � � � 2 � 1 � 1 � � 1� � 1� � 1� � � 1 � �� � � 1� �� 2 � � � � 1� � 2 � �1 1 (1) � 1� � 2 � � �(1) � � � � � � 2 � � � � s � �� s s� � s � s � � � s � s � s � � s� � s� � s� � 2 �1 s �

1 1 s

2

2

V1

1

�1 � 0

2�

0

1 s

I1 Hence,

1 s

2

2 �1 s

55ss � 1� � s � 2

� 2 s 2 5s � 1� V1 � 2 � 5s 2 � � s � 5s

I1 2 s 2 5s � 1 � V1 s 2 � 5s 5s 2 �2 I2 � �

Y11 �

1 � 1 V1 s �2 �

1 �

Hence,

1 s �2 �� 1 1� � 2 � 1 � � V1 � 2 � � � � 1� � 2 � � V1 � s s� � s � s � � ��



1 s

0 0

1 s � s 2 2 s � 1� 1 1� �2 � �V1 � � 1 � 2 � � �V1 � � �s s s � s2 � �



2 �1 s

I2

� s 2 � 2s 2s � 1 � V1 � 2 � � s � 5 2�

Y12 �

2s 1 I2 s 2 � 2s �� 2 V1 5s 2 s � 5s

��������������������������������������������� 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

���������������������������������������������������������������������8.21

F ( s) � H

( s z ))(s ( s z ) �( s zn ) ( s p ))(s ( s p ) �( s pm ) (s

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 � �. jw The total number of zeros is equal to the total number of poles. For any 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 s 0 infinite at poles, while the network function becomes zero at zeros. The 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 Fig. 8.41��������������� diagram is referred to as pole-zero plot.

where

������������������������������������������������������������� ������� �����������������������������N��������D�������������� (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.

���� ���������������������������������������������������� ������� �����������������������������N��������D�������������� (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��������������������������������������������������� (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.

������������� (a)

Test whether the following represent driving-point immittance functions.

5s4 � 3s 3 2 � 2s � 1 s

3

6s 6 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.

������������� (a) G21 =

Test whether the following represent transfer functions.

3s+ 2 3

2

5s + 4s + 1

(b) � 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.

�������������

Obtain the pole-zero plot of the following functions.

(a) F(s) =

s(s+ ( 2) (s+ 1)(s 1 + 3) 3

(b) F(s) =

(c) F(s) =

s(s+ ( 2) (s+ 1+ j1)(s+ 1 - j1)

(d) F(s) =

(e) F(s) =

s2 + 4 (s+ 22)(s 2 + 9) 9

s(s+ ( 1) (s+ 2 2)2 (s+ 3) (s+ 1 1)2 (s+ 5) (s+ 2)(s 2 + 3 � j2)(s+ 3 - j2)

���������������������������������������������������������������������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

−2

s

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 −j1

−j2

Fig. 8.45

s

8.24��������������������������������������������������� (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 −2

−1

s

0 −j 1

−j 2 −j 3

Fig. 8.46

�������������

Find poles and zeros of the impedance of the network shown in Fig. 8.47 and plot

them on the s-plane. 1F

1 H 2

2�

Z (s)

Fig. 8.47 1 s

Solution The transformed network is shown in Fig. 8.48. s �2 1 2 .5s � 2) 1 2s 2s 2 � s � 4 2(s 2(( 2 Z ( s) � � � � � � s s s s�4 s (s ) s (s ) �2 2 2 ( � 0.25 � j1.4)( � 0.25 � j1.4) � 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

−4

−3

−2

− 1 − 0.25

0

− j 1.4

Fig. 8.49

s

���������������������������������������������������������������������8.25

�������������

Determine the poles and zeros of the impedance function Z (s) in the network shown

in Fig. 8.50. 1 � 2

Z (s)

4F

1 � 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 � 8 1 4s 6 1 s � 2 0.. ( s � 2) Z ( s) � � � � � � � s �1 5 2 1 1 2 4 s � 6 2(( s ) 2 s � 3 � 4s 6 The function Z(s) has zero at s � �2 and pole at s � �1.5.

�������������

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)

4�

Fig. 8.52 Solution The transformed network is shown in Fig. 8.53. s

Z (s)

20 s

Fig. 8.53

4

8.26��������������������������������������������������� 20 �4 80 20 s ( s ) � 20 s 2 � 5s � 20 Z ( s) � s � s � s� � s� � � 20 4 s � 20 s�5 s�5 s�5 �4 s ( � 2.5 � j 3.71)( � 2.5 � j 3.71) � s�5 The function Z(s) has zeros at s � –2.5 � j3.71 and s � –2.5 –j3.71 and pole at s � –5. The pole-zero diagram is shown in Fig. 8.54. jw j 3 .71

−5

−4

−3

−2

s

0

−1

− j 3.71

Fig. 8.54

�������������

For the network shown in Fig. 8.55, plot poles and zeros of function

I0 . Ii

I0 4� Ii

0.5 F 2H

Fig. 8.55 Solution The transformed network is shown in Fig. 8.56. By current-division rule,

I0

� � � 4 2s � Ii � 2� � 4 2s � � � s�

I0 4 2 s

Ii

I0 s ( 4 2 s) s ( s � 2) s ( s � 2) � � � I i 4 s 2s 2 � 2 s 2 � 2s � 1 ( s � �1)( s � 1) The function has zeros at s � 0 and s � –2 and double poles at s � –1. The pole-zero diagram is shown in Fig. 8.57.

2s

Fig. 8.56

���������������������������������������������������������������������8.27 jw

−4 −3

−2

s

0

−1

Fig. 8.57

�������������

Draw the pole-zero diagram of I 2 for the network shown in Fig. 8.58. I1 I2

10 H 250 �F

I1

200 �

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 10 s � 200 250 � 10 �6 s 400 400 I2 � � 1 � j17 j17 32)( s � 10 � j17.32) I1 s 2 � 20 s � 400 ( s � 10 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

8.28���������������������������������������������������

������������� I1

For the network shown in Fig. 8.61, draw pole-zero plot of

Vc . V1

1�



V1

� �

� �

2V c

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 Vc � � Vc s 2 � 20 s � 2 � s 2 � Vc � � 2s � � Vc V1 5 I1 �

�0

�0 �0 � �5V1 ��

10 s 2

s � 20 s � 2

��

10 s ( s � 0.1)( s � 19.9)

jw

The function has zero at s � 0 and poles at s � ��0.1 and s � �19.9. The pole-zero diagram is shown in Fig. 8.63. −20

−2

− 1 − 0. 1

Fig. 8.63

0

s

���������������������������������������������������������������������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 10 I1 �

�2 � �1 �s �

2

0

� 2 � s� V2 � 0 10 I1 � � � s �� � s � 2� �� � V2 s �

10 I1

V2 � �

10 s I1 s�2

Substituting Eq. (ii) in Eq. (i), V1 � 0 1 I1 �

� 10 s � I � s � 2� 1 � 10 s � � 0 55V V1 � 0 05 I1 � s � 2� 2

0 5s � � I1 �1 � � 0 5V1 � s � 2 �� Hence,

Y11 �

I1 � V1

05 0.5 ( s � 2) 0 5s � 1 � � 0 5s s � 0 5s � 2 1 5s � 2 1� s�2

...(ii)

8.30��������������������������������������������������� The function has zero at s � �2 and pole at s � �1.33. The pole-zero diagram is shown in Fig 8.65. jw

− 3 − 2 −1.33 − 1

s

0

Fig. 8.65

�������������

For the network shown in Fig. 8.66, determine

V2 V . Plot the pole-zero diagram of 2 . Ig Ig I2 � 0

1H

� 1�

Ig

1F

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

Hence,

Vc

Vb � V2

Ib �

Vb Vc � � sV V2 � V2 1 1 s

Va

s I b Vb � s ( s � 1)V2 V2 � ( s 2 � s � 1)V2

Ig �

Va Va � � I b � ( s 2 � s � 1)V2 1 1 s 1

V2 � I g s3

2s2

( s � 1)V2

s ( s2

s 1)V2 � ( s �1 1)V2 � ( s3 � 2

3s � 2

The function has no zeros. It has poles at s � �1, s � �0.5 � j1.32 and s � �0.5 –j1.32, The pole-zero diagram is shown in Fig. 8.68.

2

� 3s � 2)V2

���������������������������������������������������������������������8.31 jw j1.32

−2

−1

s

0

− 0.5

−j1.32

Fig. 8.68

�������������

For the transfer function H(s) =

network shown in Fig. 8.69. Find L and C when R � 5 ��

V0 10 = , realise the function using the Vi s 2 + 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 RCs C �1 R� Cs R�

8.32��������������������������������������������������� R C �1 Vi RCs R Ls � RCs C �1

V0

1 V0 R LC � � 1 1 Vi RLC s 2 � Ls � R s2 � s� RC LC V0 10 � Vi s 2 � 3s 3s 10

But

...(i)

...(ii)

R�5�

and Comparing Eq. (ii) with Eq. (i),

1 �3 RC 1 � 10 LC Solving the above equations, L �1 5H 1 C� F 15

�������������

Obtain the impedance function Z(s) for which pole-zero diagram is shown in Fig. 8.72. jw Z( ) � 1

�3

�2

�1

0

s

Fig. 8.72 Solution The function Z(s) has poles at s � –1 and s � –3 and zeros at s � 0 and s � –2. � 2� s 2 �1 � � � s (s ) s� Z ( s) � H �H ( s )(s )( ) � 1� � 3� s 2 �1 � � �1 � � � s� � s� For s � �� Z( ) When

H

1 �H ( )( )

Z(�) � 1, H�1 Z ( s) �

s (s ) ( s )(s )( )

���������������������������������������������������������������������8.33

�������������

Obtain the admittance function Y(s) for which the pole-zero diagram is shown in

Fig. 8.73. jw

Y( ) � 1

j1

�2

s

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.

Y ( s) � H

s (s ) j )( s

(s

j)

�H

s (s (s

)2

� 2� s 2 �1 � � � s� �H 2 �H 2 2� � ( j )2 s � 2s � 2 s 2 �1 � � 2 � � s s � )

s (s

)

For s � ��, Y( ) When

H

() �H ()

Y (�) � 1, H�1 s (s ) Y ( s) � 2 s � 2s � 2

�������������

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 5

2 s

0

−j

√111 2

Fig. 8.74 1� R� 1 � s � �� Ls � R C� L Cs Z ( s) � � � 2 1 R 1 2 LCs C R RCs Cs � 1 s � s� ( Ls R) � Cs L LC ( Ls R)

Solution

...(i)

8.34��������������������������������������������������� From the pole-zero diagram, zero is at s � –3 and poles are at s

15

j

111 2

ds

15

j

111 2

s�3 111 � � 111 � j s 15 j � � 2 �� 2 �� s�3 s�3 �H 2 Z ( s) � H 2 s � 3 � 30 � 111 � � s � 1.5�2 � � j 2 � � � Z ( s) � H

When

� �s 1 5 �

Z (j0) � 1, � 3� 1� H � � � 30 � H � 10 10 ( s � 3) Z ( s) � 2 s � 3 30

...(ii)

Comparing Eq. (ii) with Eq. (i), R �3 L 1 � 10 C 1 � 30 LC Solving the above equations, 1 F 10 1 L� H 3 R �1�

C�

������������� 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

���������������������������������������������������������������������8.35

1� R� 1 � s � �� Ls � R C� L Cs Z ( s) � � � 2 1 1 R 2 LCs C R RCs Cs � 1 s � s� Ls � R � Cs L L LC ( Ls R)

Solution

...(i)

1 3 The poles are at � � j and zero is at –1. 2 2 Z ( s) � H

� �s �

1 2

s �1 3� � j s 2 �� ��

1 2

3� j 2 ��

s �1

�H � �� s

1� 2�

2

� �

j

3� 2 ��

2

�H

s �1 s2 � s � 1

Z (j0) � 1,

When

1� H

(1) (1)

H �1 Z ( s) �

s �1

...(ii)

s2 � 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�1�

�������������

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

−j4

Fig. 8.76 1� R� 1 � s � �� Ls � R C� L Cs Z ( s) � � � 2 1 R 1 2 LCs C R RCs Cs � 1 Ls � R � s � s� Cs L LC L ( 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��������������������������������������������������� Z ( s) � H

(s

s�2 j )( s

j )

�H

s�2 (s

)2

( j )2

�H

s�2 s 2 � 2 s � 17

� � 0, Z ( j0) � 2

At dc, i.e.,

2� H H � 17 Z ( s) � 17

2 17 s�2

...(ii)

2

s � 2 s 17

Comparing Eq. (ii) with Eq. (i), 1 � 17 C R �2 L 1 � 17 LC Solving the above equations, 1 F 17 L �1H R�2�

C�

�������������

The network shown in Fig. 8.77 has the driving-point admittance Y (s) of the form Y(s) ( �H

(s � s1 )(s s ) (s � s )

(a) Express s1, s2, s3 in terms of R, L and C. (b) When s1 � –10 � j104, s2 � –10 – j104 and Y (j0) � 10–1 mho, find the values of R, L and C and determine the value of s3.

Y (s)

1 Cs

Ls

R

Fig. 8.77 Solution 1 ( Ls R) Cs C � 1 LCs C 2 R RCs Cs � 1 Y ( s) � Cs � � � � Ls � R Ls � R Ls � R

(a)

But

Y ( s) �

H ( s s )(s )( s s ) (s s )

1 � R � C s2 � s � � � L LC � R s� L

...(i)

���������������������������������������������������������������������8.37 2

� s1 s2 �

where

R 4 � R� � � � � 2 � L� L LC R 1 � R� �� � � � � � � LC 2 2L 2L

R L s1 � �10 � j104 s2 � � 10 � j104

s3 � � (b) When

Y ( s) � H

(s

j

4

))(( s s s3

j

4

)

s 2 � 200 s � 108 s s3

�H

...(ii)

Comparing Eq. (ii) with Eq. (i), R � 20 L s3 � �20 Y ( s) � H

( s 2 � 20 0 s � 108 ) ( s � 20)

At s � j0, (

8

)

� 10 �1 20 H � 0.02 10 �6

Y( j ) � H

Y ( s) � 0.02 10 �6

( s2

s ( s � 20)

8

)

...(iii)

Comparing Eq. (iii) with Eq. (i), C � 0 02 0 � 10 �6 F � 0.02 �F 1 � 108 LC 1 L� H 2 R � 20 L R � 10 �

�������������

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

−j 1 −j 2 −j 3

8.38��������������������������������������������������� Solution It is easier to calculate Y(s) and then invert it to obtain Z(s). Y ( s) � G C Cs �

Z( ) �

C )( Lss R) � 1 LCs C 2 1 (G Cs � � Ls � R Ls � R

1 � Y ( ) LCs C 2

(

G RC ) s � 1 � GR (GL Ls � R 1� R� � s � �� Ls � R L C� � G R � 1 � GR � � � ) s � 1 � GR s 2 � s�� � � LC �� � C L ��

...(i)

From the pole-zero diagram, zero is at s � –2 and poles are at s � –3 ± j3. Z ( s) � H When

(s

(s ) j )( s

j )

�H

(s (s

)

2

) (j )

2

�H

s�2 2

s � 6 s � 18

Z (j0) � 1, 2 18

1� H H �9 Z( s) �

9( 2

(s � 6

2)

...(ii)

18)

Comparing Eq. (ii) with Eq. (i), 1 C R L G R � C L 1 � GR LC

�9 �2 �6 � 18

Solving the above equation, 1 F 9 9 L� H 10 4 G� � 9 9 R� � 5 C�

������������� 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.

���������������������������������������������������8.39 jw

j 25 Scale factor = 1 s

0

−1

−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

j

s ))(s (s

j

)

�H

s (s

)

2

(j

)

2

�H

s 2

s � 2 s � 626

H�1

Scale factor

Y ( s) �

s

...(i)

2

s � 2 s � 626

For a series RLC circuit, 1 LCs C 2 R RCs Cs � 1 C Z ( s) � R Ls � � � Cs Cs 1 s � Y ( s) � 1 � R Z ( s) � L � s2 � s � � � L LC �

1 � R � 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�2�

������������������������������������������������������ The time-domain behaviour of a system can be determined from the pole-zero plot. Consider a network function F ( s) � H

( s z ))(s ( s z ) � ( s zn ) ( s p ))(s ( s p )� ( s pm ) (s

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��������������������������������������������������� (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)

0

s

t

0

Fig. 8.80��������������� (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��������������������������������� (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���������������������������������� (iv)

For s � 0 �j�n, the response becomes f (t) � Ae� j�nt � A(cos �nt � j sin �nt).The exponential response e � j�nt may be interpreted as a rotating phasor of unit length. A positive sign of exponential e j�nt indicates counterclockwise rotation, while a negative sign of exponential e�j�nt indicates clockwise rotation. The variation of exponential function e j�nt 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������������ �����

t

���������������������������������������������������8.41

(v)

For s � �n � j�n, the response becomes f (t) � Aest ��Ae��n�j�nt� � Ae�nt ej�nt. The response e�nt is an exponentially increasing or decreasing function. The response ej�nt 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���������������������������������������������������������� �������������������������������������������������������� (vi)

The real part s of the pole is the displacement of the pole from the imaginary axis. Since � is also the damping factor, a greater value of � (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 0

t

s

f (t)

0

(a)

Fig. 8.85�����������������������������������������������������

t

8.42��������������������������������������������������� f (t)

jw t

0 s

0

f (t)

t

0

(b)

Fig.8.85������������

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 j� axis, the network is said to be marginally stable. When there are multiple poles on the j� axis, the network is said to be unstable. When the poles move away from j� axis towards the left half of the s-plane, the relative stability of the network improves.

������������������������������������������������������� Consider a network function, F ( s) � H

( s z ))(s ( s z ) � ( s zn ) ( s p ))(s ( s p )� ( s pm ) (s

By partial fraction expansion, F ( s) �

(s

K1 K2 � � p ) (s p )



(s

Km pm )

The residue Ki is given by Ki

( s � p ) F ( s ) �s � p

i

H

(p (p

z ))(( p p )( )( p

z )� ( p p )� ( p

zn ) pm )

Each term (pi – zi) represents a phasor drawn from zero zi to pole pi. Each term (pi – pk), i � k, represents a phasor drawn from other poles to the pole pi. Ki

H

Product of phasors (polar form) from each zero to pi a s (polar form) from other poles to pi Product of phasor

����������������������������������������������������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) s � . Plot the pole-zero pat(s � 1)(s � 2) tern in the s-plane and hence obtain i(t).

�������������

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� � � 2� � 2 ����� � �� K1 H � 1�0� �� a A Phasor from pole at B to pole at jw

B

A s

−2 −1

0

Fig. 8.87 From Fig. 8.88, K2

H

� 2 �180 80� � Phasor from zero at origin to pole at B � 2� �4 a B � 1 �0� �� Phasor from pole at A to pole at

8.44��������������������������������������������������� 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) � �

�������������

The voltage V(s) of a network is given by

V ( s) �

3s (s

)(s )( )( s 2

s

)

Plot its pole-zero diagram and hence obtain v (t). Solution

V ( s) �

3s (s

)(s )( )( s

2

s

)



(s

)(s )( )( s

3s j )(s )( s

j)

Poles are at s � –2 and s � –1 ± j1 and zero is at s � 0 as shown in Fig. 8.89. jw B

j1

A

−2

0

−1

s

−1 −j C

Fig. 8.89 By partial-fraction expansion, V ( s) �

K1 K2 K 2* � � s � 2 s 1 j1 s 1 j1

The coefficients K1, K2 and K2* can be evaluated from the pole-zero diagram. From Fig. 8.90, K1 �

3 (OA) ( BA) (CA)

� � 2�180 80� � 3� � � 3 180� � �3 2 � 135 35 2 135 35 ( ) ( ) � �

����������������������������������������������������8.45 jw B

j1 2

135�

A

180�

�2

s

0

�1

135�

2

�j 1 C

Fig. 8.90 From Fig. 8.91, � 3 � ( 2 135 35 ) � 3� �� ( AB ) (CB) � ( 2 45 ) ( 2 90 ) � 2 3 K 2* � 2 K2 �

3 (OB)

jw B

j1

2 A

2 135�

45�

�2

s

0

�1

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 �2tt

�������������

3� ( e 2�

j ))tt

� e(

j ))tt �



3e

2t

� e j1 � e � j1 � 3 � 2 � e �t � � � �3e 2 2 � �

2t

3e t cos t

Find the function v(t) using the pole-zero plot of following function: V(s) �

(s � 2)(s � 6) (s � 1)(s � 5)

Solution If the degree of the numerator is greater or equal to the degree of the denominator, we have to divide the numerator by the denominator such that the remainder can be expanded into partial fractions.

8.46��������������������������������������������������� V ( s) �

s 2 � 8s � 12 2

s � 6s � 5

� 1�

2s � 7 2

s � 6s � 5

� 1�

2(s ( .5) ( s ))(s ( )

By partial fraction expansion, V ( s) � 1 �

K1 K � 2 s �1 s � 5

K1 and K2 can be evaluated from the pole-zero diagram shown in Fig. 8.92 and Fig. 8.93. jw

jw

s

s 0

0

−5

−3.5

−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 80� � 3 � 1.5�180 K2 � 2 � � � 4 ����� �� 4 3 5 V ( s) � 1 � 4 � 4 s �1 s � 5 Taking inverse Laplace transform, 5 �tt 3 e � e 4 4

v (t ) � � (t (t )

�������� �����

5t

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

−1

0

Scale factor = 5 s

−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.

����������������������������������������������������8.47

Z ( s) � H

s j )( s

(s

j)

H�5

Scale factor

Z ( s) �

jw

5s j )( s

(s

j)

j1

A

By partial fraction expansion, Z ( s) �

K1 K1* � s 1 j1 s 1 j1

s

−j1

B

d K1* can be evaluated from the pole-zero

The coefficients K1

0

−1

Fig. 8.95

diagram. From Fig. 8.95, K1 �

5 (OA) ( BA)



5 ( 2 135 35 ) � 3 54 �45� 2�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) �

pole-zero plot at s � j2. F ( s) �

Solution

4s 2

s � 2s � 2



(s

4s j )( s

4s s

2

2s 2

from the

j)

The pole-zero plot is shown in Fig. 8.96. At s � j2, jw j2 j1 −1

s 0

−j1

Fig. 8.96 �

��

2 Product of phasor magnitudes from all zero to j 2 � � 0.447 h r magnitudes from all poles to j 2 ( 2 ) ( 10 ) Product of phaso

8.48���������������������������������������������������

� ( ) � tan �1

�������������

� 2� � 0�

� 3� � 1� tan �1 � � � tan �1 � 1� � 1�

90� 71.56� � 45 4 � � �26.56�

Using the pole-zero plot, find magnitude and phase of the function

Solution

F ( s) �

(s

)(s )(s ( ) at s s (s )

F ( s) �

(s

)(s )( ) s (s )

j 4.

The pole-zero plot is shown in Fig. 8.97. At s � j4, jw j4

s

−3

−2

−1

0

Fig. 8.97 �

� (� ) � tan �1

� 4� � 1�

�������������

j )��

) Product of phasor magnitudes from all zeros to j 4 ( ) ( � � 1.15 p r magnitudes from all poles to j 4 ( product of phaso )( )

� 4� � 4� tan �1 � � � tan1 � 3� � 0�

� 4� tan �1 � � � 75.96� � ������ � ��� � ������ � ������� � 2�

Plot amplitude and phase response for F(s) �

s s 10

Solution F( j ) � �F( j )��

j� j� � 10

� � � 100 2

����������������������������������������������������8.49



�F(j�)�

0

0

10

0.707

100

0.995

1000

1

1

The amplitude response is shown in Fig. 8.98.

� (� ) � tan �1

��� � 0�

F(jw)

0.7

w 0

10

100

1000

Fig. 8.98

��� ��� tan �1 � � � 90� � tan �1 � � � 10 � � 10 � f (w)



� ���

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

������������

Sketch amplitude and phase response for F(s) �

s 10 s 10

Solution F(jw)

j� � 10 F( j ) � j� � 10 �F f� )��

1

� 2 � 100 � 2 � 100

w 0

For all �, magnitude is unity. The amplitude response is shown in Fig. 8.100.

� (� ) � tan �1

��� � 10 �

� �� ��� tan �1 � � � � 2 tan �1 � � � 10 � � 10 �

The phase response is shown in Fig. 8.101. �

����

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���������������������������������������������������

Exercises Determine the driving-point impedance

8.1

transfer impedance ratio I1

V1 , I1

V2 and voltage transfer I1

V2 for the network shown in Fig. 8.102. V1 5�

I1

1

2H

V1 1�

1 F 2

2H

V1

2F

2F

V2 �



2�

Fig. 8.104 1 1 � � , G21 � 4 � Z 21 � 3 � 2 2 s 3s 4 s 7 s � 1� �

V2

For the two-port network shown in Fig. 8.105, determine Z11, Z21 and voltage transfer ratio G21(s).

8.4





2







I2 � 0



I2 � 0

2�

1H

Fig. 8.102 I1

�V1 7 s 7 s � 5 V2 2s ; � 2 ; � � I s 2 s 1 I s � s �1 1 � 1 V2 2s � � V1 7s 2 7s � 5 �� 2

2�

2H



� 1F

V1

1H



V2 �

Fig. 8.105 8.2

For the network shown in Fig. 8.103, V V determine 2 and 2 . V1 I1

� 2ss3 s 2 � 3s 2 s , Z 21 � 2 , � Z11 � 2 � � 2 2s 1 2s 1 2s s s s � s � G21 � 3 � 2 2ss s � 3s 2 �

2H I1 + V1

1 F 2

+ 1 F 2

8.5

Draw the pole-zero diagram of the following network functions:

V2 −



8.3

I2 = 0

1 F 2

(i) F ( s) �

Fig. 8.103

(ii) F ( s) �

�V2 s 2 � 1 V2 2s 2 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 5s � 12 2

s � 4 s � 13 s �1 (s

2

s ( s2 4

)2

s )

s � s2 � 1

����������8.51

s2

Z (s)

s

1H

s 2 � 3s � 2

Fig. 8.108

s 2 � 3s ( s2

(viii) F( s) �

( s2

)(s )( )( s )(s )( )( s 2

� 1.. s ( s 2 � 0.. ) � Z ( s) � 2 (s .707)( s 2 � 0. �

) s

)

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 and zeros of

I1

1 F 20

Z (s)

4�

Fig. 8.106 .5

j1..

2H

1�

)( )(ss s�5

.5

j1.

)� � �

10 �

10 F

1F

� V2 �

1�

Fig. 8.109 �V2 1 V 2( s3 � 2 s 2 � 2 s � 1) � , 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

� 5(s ( .01)( s � 0.. � Z ( s) � s ( s .03) �

V2 . I1 2H

1H

+

Fig. 8.107

8.8

1F



For the network shown in Fig. 8.107, draw the pole-zero plot of driving-point impedance function Z(s). 5�

V2 ( s) . V1 ( s)

� V1

� (s Z ( s) � � �

� � )�

Find network functions

8.9

2H

Z (s)

1H

s3 � s 2 � s � 2

(vii) F ( s) �

8.7

2F

s 4 � 5s3 � 6 s 2

(vi) F ( s) �

8.6

1F

s2 � s � 2

(v) F ( s) �

)� � �

Find the driving-point impedance of the network shown in Fig. 8.108. Also, find poles and zeros.

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��������������������������������������������������� 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 1H

V1

1H

V2



−2

0

−1

−j1



Fig. 8.111

Fig. 8.114 �V1 s 4 � 3s 3s 2 � � 3 2s � s � I1

1� � �

8.12 For the network shown in Fig. 8.112, plot the poles and zeros of transfer impedance function. I1

1H

1H



� V1

1�

1��

V2



2 (s (s ) � � � 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.



Ls

1 � �V2 � I � s � 2� � 1 �

Fig. 8.115

8.13 For the network shown in Fig. 8.113, determine V1 V and 2 . Plot the poles and zeros of transfer I1 V1 impedance function. 2H

4H +

+ V1

2F

2 � �1 �� ��� �� 17 �

� F� �

8.16 For the two-port network shown in Fig. 8.116, V 02 find R1, R2 and C. 2 � 2 V1 s � 3s 3s 2

1 F V2 R1

1H −

R

1 Cs

Z (s)

Fig. 8.112

I1

s



Fig. 8.113 �V1 16 s 4 � 100 s 2 � 1 V2 1 , � 3 , � � 3 I I 8 s 3 s 8 s 3s 1 1 � V2 1 � � V1 16 s 4 � 100 s 2 � 1��

+

+ V1

C



R2

V2 −

Fig. 8.116 1 �3 � � 5 �, 15 �, 0.5 F� � �

�������������������������8.53

8.17 For the given network function, draw the pole-zero diagram and hence obtain the time domain voltage. V ( s) �

(s

5 (s ( ) )(s )( ) [v(t) � 3e�2t � 2e�7t]

8.18 A

transfer

function is given by 10 s Y ( s) � . Find time(s j )(s )( s j ) domain response using graphical method.

�5 26 18 4� e �

((55

15) 15 5)

� 5 26 6

18 8 4� e

15) t �

(5



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

1� 1F

2�

−3

1F

s

0

−1

−j 1 N1 2H

Fig. 8.118 1�

(a)

3(

3)

s2 � 2

(b)

3

2(

3)

2

s � 2s 2

1F

2�

(c)

N2 1� 1�

1H 1F

8.3

3( s

2

3) 2s 2

1�

2( s

2

3

� 100 �F

vi (t) �

v0 �

1F

Fig. 8.119

N4

Fig. 8.117

2

10 mH

� 2H

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 k�

N3

(d)

(a)

1 06 s � 106 s 2 10

(b)

106 03 s � 106 s 2 10

8.54��������������������������������������������������� 103

(c)

s 2 1003 s � 106

8.7

106

(d)

s 2 1006 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 k�

� 4 �F

vi (t )

4 k�

1 �F

v0 (t)

Fig. 8.122





jw

Fig. 8.120 (a) 8 e–0.004 t (c) 8 u(t)

(b) 8 (1–e–0.004 t) (d) 8

A system is represented by the transfer 10 . The dc gain of this function ( 1 ))(( 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� RCs C

(d)

s 1� RCs C

of the network shown in Fig. 8.121. (c) 1� 1

2 1�

s

Fig. 8.123

(a) 1

Which one of the following is the ratio

8.6

t

0



1�

3

8.9

RCs C 1� RCs C

The driving-point admittance function of the network shown in Fig. 8.124 has a

4

R

L

1�

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

������������������������������������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) 1�

I2(s)

V1(s )

1H

1F

Fig. 8.125

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 (a) remains constant (b) becomes less oscillating (c) becomes more oscillating (d) 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 Two-Port Networks

���������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 known I1 I2 as output port as shown in Fig. 9.1. There are four variables + Two-port network V1, V2, I1 and I2 associated with a two-port network. Two V1 − of these variables can be expressed in terms of the other two variables. Thus, there will be two dependent 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 9.1�������������������� Parameter

Variables

Equation

Express

In terms of

Open-Circuit Impedance

V1, V2

I1, I2

V1 � Z11 I1 � Z12 I2 V2 � Z21 I1 � Z22 I2

Short-Circuit Admittance

I1, I2

V1, V2

I1 � Y11 V1 � Y12 V2 I2 � Y21 V1 � Y22 V2

Transmission

V1, I1

V2, I2

V1 � AV2 � BI2 I1 � CV2 � DI2

Inverse Transmission

V2, I2

V1, I1

V2 � A� V1 � B � I1 I 2 � C � V1 � D � I1

Hybrid

V1, I2

I1, V2

V1 � h11 I1 � h12 V2 I2 � h21 I1 � h22 V2

Inverse Hybrid

I1, V2

V1, I2

I1 � g11 V1 � g12 I2 V2 � g21 V1 � g22 I2

+ V2 −

9.2���������������������������������������������������

��������������������������������������������������������� 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. 9.2.

I1

+ Z11

Z22

V1

V2 Z12I2



+ −

+ Z I − 21 1



��������������������� �������� ��� ���� ��������� �������������������������������

9.2.1 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. 9.3, voltage Vs is applied at the input port with the output port short-circuited.

I2

+

I1 � Vs �

I2 Network

I2�

����������������������������������������� ���������������

���������������������������������������������������������

i.e.,

V1 � Vs V2 � 0 I2 � � I� � 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 I1 (b) As shown in Fig. 9.4, voltage Vs is applied at the I2 output port with input port short-circuited. � I1� Network V2 � Vs i.e., � Vs V1 � 0 I1 � � I1 � ��������������������������������������������� From the Z-parameter equations, ����������� 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 i.e.,

9.2.2 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,

������������������������������������������������������ Vs Vs � I1 I 2 Z11 � Z 22

i.e.,

��������� ���� Test results for a two-port network are (a) I1 = 0.1 � �� A, V1 = 5.2 � 5�� V, V2= 4.1 ��25� V with Port 2 open-circuited (b) I = 0.1 � �� A, V = 3.1 ��80� V, V =4.2 � 6�� V, with Port 1 2 1 2 open-circuited. Find Z parameters. Solution 5.2�50� V � � 52 �50� �, Z11 � 1 0.1�0� I1 I 2 � 0 Z 21

V � 2 I1

I2 �0

Z�� �

4.1� � �5� � � 41 � � �5� �� 0.1�0�

Hence, the Z-parameters are � Z11 �� Z 21

�������������

Z��

V � 2 I2



3.1� � 80� � 31 � � 80� � 0.1�0�



4.2�60� � 42 �60� � 0.1�0�

I1 � 0

I1 � 0

��� � 80�� Z12 � �52�50� � Z 22 �� �� 41� � �5� 42�60� ��

Find the Z parameters for the network shown in Fig. 9.5. I1

Z1

Z3

I2 +

+ Z2

V1

V2





�������� 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

���������������������������������������������������������

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. 9.6. 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

�������������

I2





Comparing Eqs (i) and (ii) with Z-parameter equations, � Z11 �� Z 21

V2

Z2

V1

��������

Z2 � Z 2 � Z3 ��

Find Z-parameter for the network shown in Fig. 9.7. I1

2�

1�

1H

I2





V1

2F

V2

1�





�������� Solution The transformed network is shown in Fig. 9.8. Z1 � 1 � 1� �� �� (1) 1 2s � Z2 � 1 2s � 1 �1 2s Z3 � s � 2

1

+ Z1

V1

1 2s

Z 21

V1 I1

V � 2 I1

� Z1 � Z 2 � 1 � I2� 0

I2 � 0

1 � Z2 � , 2s � 1

1 2s � 2 � , 2s � 1 2s � 1

1 Z 2



Z3

V2 −

��������

From definition of Z-parameters, Z11 �

s

2

+

Z12 � Z 22

V1 I2

V � 2 I2

� Z2 � I1 � 0

1 2s � 1

� Z 2 � Z3 � I1 � 0

1 2 s 2 � 5s � 3 �s�2 � 2s � 1 2s � 1

������������������������������������������������������

�������������

Find Z-parameters for the network shown in Fig. 9.9. I1

1�

1�

I2





V1

2�

V2

2�





�������� Solution The network is redrawn as shown in Fig. 9.10. Applying KVL to Mesh 1, V1 � 3I1 � 2 I 3 �(i) Applying KVL to Mesh 2, � V2 � 2 I 2 � 2 I 3 �(ii) Applying KVL to Mesh 3,

1�

1� �

2�

V1

�2 I1 � 2 I 2 � 5 I 3 � 0

I1

2 2 I 3 � I1 � I 2 5 5 Substituting Eq. (iii) in Eq. (i),

�(iii)

V2

2� I3

I2





���������

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 Z12 � � 5 � Z 22 �� �� 4 �5

�������������

4� 5� 6 �� 5�

Find the Z-parameters for the network shown in Fig. 9.11. I1

1�

1�

I2

� V1



1H



1H

V2 �

���������

���������������������������������������������������������

Solution The transformed network is shown in Fig. 9.12. Applying KVL to Mesh 1, V1 � ( s � 1) I1 � sI 3 Applying KVL to Mesh 2, V2 � sI 2 � sI 3 Applying KVL to Mesh 3,

�(i) �(ii)

1 + s

V1 I1

s I3



� sI1 � sI 2 � ( 2 s � 1) I 3 � 0 I3 �

1

+

s s I1 � I2 � 2s � 1 2s � 1 (iii)

V2 I2



���������

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 ��

������������� Find the open-circuit impedance parameters for the network shown in Fig. 9.13. Determine whether the network is symmetrical and reciprocal. 4�

I1

1�

3�

� V1

I2 �

2�



V2 �

���������

������������������������������������������������������ Solution The network is redrawn as shown in Fig. 9.14. 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 �(ii) � Applying KVL to Mesh 3, V1 �4 I 3 � 3 ( I 2 � I 3 ) � 1( I 3 � I1 ) � 0 � I1 � 3I 2 � 8 I 3 � 0 1 � I 3 � I1 � I 2 �(iii) 8 8 Substituting Eq. (iii) in Eq. (i),

4�

1�

I3

3� � V2

2� I1

I2



���������

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 ��Z��� the network is not symmetrical. Since Z12 ��Z��� the network is reciprocal. �

�(v)

����������������������������������������������������������� 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 ]

��������������������������������������������������������9.9

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. 9.15. I2

I1 + V1

+ Y11

Y12V2

Y21V1

Y22

2

V2





�����������������������������������������������������������������������������

9.3.1 Condition for Reciprocity (a) As shown in Fig. 9.16, voltage Vs is applied at input port with the output port short-circuited. V1 � Vs i.e, I1 I2 V2 � 0 I 2 � � I 2� � Network I2� From the Y-parameter equation, � Vs � I 2� � Y21 Vs I2 � �Y21 Vs

���������������������������������������������� �����������

������������������������������������������������������� (b) As shown in Fig. 9.17, voltage Vs is applied at output port with the input port short-circuited. i.e, V2 � Vs V1 � 0 I1 I2 I1 � � I1� From the Y-parameter equation, � I � Network

1

� I1� � Y12 Vs I1� � �Y12 Vs Hence, for the network to be reciprocal,



Vs

���������������������������������������������� �����������

I 2� I1� � Vs Vs Y12 � Y21

i.e,

9.3.2 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 1 Vs � I 2 Y22 Hence, for the network to be symmetrical, Vs Vs � I1 I 2 Y11 � Y22 i.e.,

�������������

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� �� � �

��������������������������������������������������������9.11

�������������

Find Y-parameters for the network shown in Fig. 9.18. I1

1�

3

3�

I2





V1

V2

2�





��������� Solution First Method Case 1 When the output port is short-circuited, i.e., V2 � 0 as shown in Fig. 9.19, 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

3�

V1

I2

2�



���������

1� 2 2 11 � 3� � � 1� 2 3 3

11 I2 3 3 I � 2 � � V2 V1 � 0 11

I1

1�

3�

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

1�

I1 �

When the input port is short-circuited, i.e., V1 � 0 as shown in Fig. 9.20, Req � 3 �

Now

2�3 6 11 � 1� � � 2�3 5 5

2�

V2 �

���������

Hence, the Y-parameters are 2� � 5 � � �Y11 Y12 � � 11 11 ��Y21 Y22 �� � � 2 3 �� �� � 11 11 � Second Method (Refer Fig. 9.18) V �V I1 � 1 3 1 � V1 � V3

�(i)

9.12��������������������������������������������������� 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 �

�������������

�(iv)

�(v)

�(vi)

Find Y-parameters of the network shown in Fig. 9.21. I1

V3

2�

2�



I2 �

1� V1

V2

3V2 �



��������� Solution From Fig. 9.21, V1 � V3 2 1 1 � V1 � V3 2 2

I1 �

�(i)

������������������������������������������������������������

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)

�������������� Determine Y-parameters for the network shown in Fig. 9.22. Determine whether the network is symmetrical and reciprocal. I1

3

1�

2

2�

� V1

I2 �

2�

4�



V2 �

��������� Solution From Fig. 9.22, V1 � V3 1 � V1 � V3

I1 �

�(i)

Applying KCL at Node 3, V3 V3 � V2 � 2 2 V2 � V3 � 2

I1 �

�(ii)

������������������������������������������������������� Applying KCL at Node 2, V2 V2 � V3 � 4 2 3 V � V2 � 3 4 2

I2 �

�(iii)

Substituting Eq. (i) in Eq. (ii), V2 2 V V V3 � 1 � 2 2 4

V1 � V3 � V3 �

�(iv)

Substituting Eq. (iv) in Eq. (ii), V1 V2 V2 � � 2 4 2 V1 V2 � � 2 4

I1 �

Substituting Eq. (iv) in Eq. (iii),

�(v)

3 1 �V V � 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)

��

��������������

Determine the short-circuit admittance parameters for the network shown in Fig. 9.23. I1

1�

1�

I2 �

� V1

V2

1F

1F





��������� Solution The transformed network is shown in Fig. 9.24. From Fig. 9.24, V1 � V3 1 � V1 � V3

I1

V1

I1 �

�(i)

1

3

1

2

I2

+

+ 1 s

1 s

V2 −



���������

������������������������������������������������������������

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)

��

��������������

Determine Y-parameters for the network shown in Fig. 9.25. I1

1 F 2

1 F 2

I2

+ V1

+

1H



1H

V2 −

���������

������������������������������������������������������� Solution The transformed network is shown in Fig. 9.26. From Fig. 9.26, 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( s 2 � 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 � V2 �� � � V1 � 2 4( s 2 � 1) � 2 4( s � 1) � �

s3 � 2 s 2

4( s � 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 � V1 � � � V2 2 4( s � 1) 4( s 2 � 1) � � 2s

s3 s4 � 6s2 � 4 V1 � V2 2 4( s � 1) 4 s( s 2 � 1) Comparing Eqs (iii) and (v) with Y-parameter equation, ��

�Y11 ��Y21

� s3 � 2 s � Y12 � � 4( s 2 � 1) � Y22 �� � s3 �� 2 � 4( s � 1)

� s3 � 4( s 2 � 1) � s4 � 6s2 � 4 � � 4 s( s 2 � 1) � �

�(v)

������������������������������������������������������������

��������������

Obtain Y-parameters of the network shown in Fig. 9.27. 1 s

I1

1

3

1

1

2

I2

+

+ 1 s

V1 −

V2 −

��������� 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 � 1)( s � 2) � 1� � s( s � 2) � 1� �� V1 � � � � V2 ( s � 2) � � � ( s � 2) � � s 2 � 3s � 1� � s 2 � 2 s � 1� �� V1 � � � � V2 � s�2 � � s�2 �

�(iv)

������������������������������������������������������� 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� � �� � V2 � V1 � � ( 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 � 2 s � 3s � 1 � �� s�2

�������������������������������������������������� 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.

�����������������������������������������������9.19

9.4.1 Condition for Reciprocity (a) As shown in Fig. 9.28, 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� ���������������������������������������������� ����������� Vs �B � I2 (b) As shown in Fig. 9.29, 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 � I� � � CVs � Vs B Vs B � I� � AD � BC Hence, for the network to be reciprocal, Vs Vs � I � � I� �

i.e.,

Network

� �

Vs

���������������������������������������������� �����������

From the transmission parameter equations,

i.e.,

I2

B AD � BC AD � BC � 1 B�

9.4.2 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

������������������������������������������������������� Hence, for network to be symmetrical, Vs Vs � I1 I 2 A� D

i.e.,

��������������

Find the transmission parameters for the network shown in Fig. 9.30. I1

1�

2�



I2 �

V1

5�



V2 �

��������� Solution First Method Case 1 When the output port is open-circuited, i.e., I2 � 0. V1 � 6 I1 V2 � 5 I1 and A�

V1 V2

I C� 1 V2 Case 2

Now and



6 I1 6 � 5 I1 5



1 � 5

I2 � 0

I2 � 0

When the output port is short-circuited, i.e., V2 � 0, as shown in Fig. 9.31, 1� 2� I1 5� 2 10 17 Req � 1 � � 1� � � � 5� 2 7 7 17 V1 � I1 V1 5� 7 5 5 � I 2 � ( � I1 ) � � I1 7 7 ��������� 17 I1 17 V1 B�� �� 7 � � 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

�����������������������������������������������9.21

Second Method (Refer Fig. 9.37) 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 � 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 �

��������������

Obtain ABCD parameters for the network shown in Fig. 9.32. I1

1�

1�

I2

� V1



2�

V2

2�





��������� Solution The network is redrawn as shown in Fig. 9.33. 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

1�

1� �

2� I1

V2

2� I3

I2



���������

�(iv)

9.22��������������������������������������������������� 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

��������������

�(vi)

5� 2� 3 �� 2�

Determine the transmission parameters for the network shown in Fig. 9.34. 1

I1

1

2

I2

+

+ s

V1

1 s

V2 −



��������� 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)

9.4 �������������������������������������������9.23

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 � �� s �� I 2 s � �

�(iii)

Comparing Eqs (ii) and (iii) with ABCD parameter equations, 1 � � s �1 �A B� � 2 � ��C D �� � � s � s � 1 s � 1� s s �� ��

��������������

Find transmission parameters for the two-port network shown in Fig. 9.35. 10 �

I1

1.5 V1



25 �

V1 I1

I2

���



20 �

I3



V2 I2



��������� Solution Applying KVL to Mesh 1, V1 � 10 I1 � 25 ( I1 � I 3 ) � 35I1 � 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)

Substituting Eq. (iii) in Eq. (i), 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)

Substituting Eq. (iii) in Eq. (ii),

9.24�Circuit Theory and Networks—Analysis and Synthesis From Eq. (v), I1 � 0.053 V2 � 0.81I 2

�(vi)

V1 � 11.5 (0.053 V2 � 0.81I 2 ) � 6 I 2 � 0.61 V2 � 3.32 I 2

�(vii)

Substituting Eq. (vi) in Eq. (iv),

Comparing Eqs (vi) and (vii) with ABCD parameter equations, � A B � � 0.61 3.32� ��C D �� � ��0.053 0.81��

����������������������������h������������ 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. Case 1 When the output port is short-circuited i.e., V2 = 0 h11 �

V1 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 h12 �

V1 V2

I1 � 0

where h12 is the open-circuit reverse voltage gain. h22 �

I2 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. 9.36.

������������������������������������������ h11

I1

I2

+

+

V1

h12V2

+ −

h21I1

h22







V2

�����������������������������������������������������������������������������

9.5.1 Condition for Reciprocity (a) As shown in Fig. 9.37, voltage Vs is applied at the input port and the output port is short-circuited. i.e.,

I1

V1 � Vs V2 � 0 I2 � � �I2

� Vs �

I2 I2�

Network

From the h-parameter equations, Vs � h11 I1 � I 2 � � h21 I1 Vs h � � 11 I2 � h21

������������������������������������������ ���������������

(b) As shown in Fig. 9.38, voltage Vs is applied at the output port with the input port short-circuited. V1 � 0 V2 � Vs I1 � � I1 �

i.e.,

I1 I1�

I2 Network

� �

Vs

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 � h21 � � h12

i.e.,

9.5.2 Condition for Symmetry The condition for symmetry is obtained from the Z-parameters. Z11 �

V1 I1

� I2 � 0

h11 I1 � h12V2 I1

� h11 � h12 I2 � 0

V2 I1

������������������������������������������������������� But with I2 � 0,

0 � h21 I1 � h22 V2 V2 h � � 21 I1 h22 Z11 � h11 �

h 12 h 21 h11h22 � h12 h21 �h � � h22 h22 h22

�h � h11h22 � h12 h21

where Similarly,

Z 22 � With I1 � 0,

V2 I2

I1 � 0

I 2 � h22 V2 Z 22 �

V2 I2

� I1 � 0

1 h22

Z11 � Z 22

For a symmetrical network,

�h 1 � h22 h22 �h � 1

i.e., i.e.,

h11 h22 � h12 h21 � 1

i.e.,

Table 9.2���������������������������������������� 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

h

h12 � � h21

h11 h22 � h12 h21 � 1

��������� �����

In the two-port network shown in Fig. 9.39, 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

���������

+ V2 −

������������������������������������������

Solution V1 I1 V

h11 �



2�

I2 I1

h21 �

0

� V2 � 0

25 � 25 �, 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 ��

��������������

Determine hybrid parameters for the network of Fig. 9.40. Determine whether the network is reciprocal. I1

1�

I2

2�





2�

V1 I1

V2

4� I3

I2





��������� Solution First Method Case 1 When Port 2 is short-circuited, i.e., V2 � 0 as shown in Fig. 9.41, Req � 1 � Now,

Also,

Case 2

2�2 �2� 2�2

V1 � 2I1 V h11 � 1 �2� I1 V2 � 0 2 I �� 1 2�2 2 1 I2 h21 � �� 2 I1 V2 � 0 I 2 � � I1 �

I1

1�

I2

2�



2�

V1

4�



���������

When Port 1 is open-circuited, i.e., I1 � 0 as shown in Fig. 9.42, ( 2 � 2) � 4 �2� 2�2�4 V1 � 2 I y

Req �

I Iy � 2 2 V2 � 4 I x I Ix � 2 2

I1 ��0

� V1

1�

I2

1� ly

2�



lx

4�

� V2 �

���������

�������������������������������������������������������

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. 9.40) 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.

� h12 � � 2 �� h22 �� � 1 � � 2

1� 2� 1 �� 2�

...(vi)

������������������������������������������������9.29

��������������

Find h-parameters for the network shown in Fig. 9.43. 1 F 2

I1

1 F 2

I2

+

+

1H

V1

V2

1H





��������� Solution As solved in Example 9.12, derive the equations for I1 and I2 in terms of V1 and V2. I1 �

s3 � 2 s s3 V V2 � 1 4( s 2 � 1) 4( s 2 � 1)

I2 � �

s3 2

4( s � 1)

V1 �

s4 � 6s2 � 4 4 s( s 2 � 1)

�(i)

V2

�(ii)

From Eq. (i), V1 �

s2 4( s 2 � 1) I V2 � 1 s( s 2 � 2) s2 � 2

�(iii)

Substituting Eq. (iii) in Eq (ii), 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

s �2

I1 �

2( s 2 � 1) s ( s 2 � 2)

V2

�(iv)

Comparing Eqs (iii) and (iv) with h-parameter equations,

� h11 � h21 �

� 4( s 2 � 1) � h12 � � s( s 2 � 2) � h22 �� � s2 �� 2 � s �2

� � s �2 � 2( s 2 � 1) � � s( s 2 � 2) � s2

2

��������������������������������������������������� 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.

9.6.1 Z-parameters in Terms of Other Parameters 1. Z-parameters in Terms of Y-parameters We know that I1 � Y11 V1 � Y12 V2 I 2 � Y21 V1 � Y22 V2

������������������������������������������������������� By Cramer’s rule, 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

where

�Y � Y11Y22 � Y12Y21

Comparing with

V1 � Z11I1 � Z12 , I 2 , Y Z11 � 22 �Y Y12 �Y

Z12 � �

Also, Comparing with

Y11 I1 Y21 I 2 Y Y V2 � � 11 I 2 � 21 I1 �Y �Y �Y V2 � Z 21 I1 � Z 22 I 2 , Z 22 �

Y11 �Y

Z 21 � �

Y21 �Y

2. Z-parameter in Terms of ABCD Parameters We know that V1 � AV2 � BI 2 I1 � CV2 � DI 2 Rewriting the second equation, V2 � Comparing with

D 1 I1 � I 2 C C

V2 � Z 21 I1 � Z 22 I 2 , 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 � � � � � � Z 21 �

Also, Comparing with

V1 � Z11 I1 � Z12 I 2 , Z11 �

A C

Z12 �

AD � BC C

����������������������������������������������������

3. Z-parameters in Terms of Hybrid Parameters We know that 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 h22 �� h22 h22 h22 h22 � h22 � � Comparing with V1 � Z11 I1 � Z12 I 2 , Also,

h11h22 � h12 h21 �h � h22 h22 h Z12 � 12 h22 Z11 �

9.6.2 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 � �

Z 21 �Z

������������������������������������������������������� Y22 �

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, Comparing with

1 B

A B

A � D � 1 � BC � AD � I1 � CV2 � D � � V1 � V2 � � V1 � � � V2 B � B B � B � � I1 � Y11 V1 � Y12 V2 , D B BC � AD AD � BC �T Y12 � �� �� B B B Y11 �

3. 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 h h h h11 � � � 11 � 11 11 I 2 � Y21 V1 � Y22 V2 , h21 h11 h11h22 � h12 h21 �h � � h11 h11

Y21 � Y22

����������������������������������������������������

9.6.3 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 22 D� 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

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 Y D � � 11 Y21

������������������������������������������������������� 3. 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, I1 � � Comparing with

h22 1 V2 � I2 h21 h21

I1 � CV2 � DI 2 , h C � � 22 h21 1 D�� h21

Also,

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 � �

Comparing with

V1 � AV2 � BI 2 , A�

h12 h21 � h11 h22 �h �� h21 h21

B��

h11 h21

9.6.4 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 Z12 h12 � Z 22 h11 �

����������������������������������������������������

2. Hybrid Parameters in terms of Y-parameters We know that I1 � Y11 V1 � Y12 V2 I 2 � Y21 V1 � Y22 V2 Rewriting the first equation, V1 � Comparing with

Y 1 I1 � 12 V2 Y11 Y11

V1 � h11 I1 � h12 V2 , h11 �

1 Y11

h12 � �

Y12 Y11

Also,

Y Y � 1 � �Y Y � Y Y � I 2 � Y21 � I1 � 12 V2 � � Y22 V2 � � 11 22 12 21 � V2 � 21 I1 Y11 � Y11 Y11 � Y11 � �

Comparing with

I 2 � h21 I1 � h22 V2 , h21 �

Y22 Y11

h22 �

Y11 Y22 � Y12 Y21 �Y � Y11 Y11

3. Hybrid Parameters in Terms of ABCD Parameters We know that V1 � AV2 � BI 2 I1 � CV2 � DI 2 Rewriting the second equation, I2 � � Comparing with

I 2 � h21 I1 � h22 V2 , h21 � � h22 �

Also, Comparing with

1 C I1 � V2 D D 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 �

������������������������������������������������������� ����������������������������������������������� �X � X 11 X 22 � X12 X 21 In terms of [Z]

[Y]

[T]

[h]

A C

�T C

Y11 �Y

1 C

D C

Y11 Y12

D B



1 B

A B

�h h22

h12 h22

h21 h22

1 h22

1 h11



h21 h11

�h h11

Z11

Z12

Z 21

Z 22

Z 22 �Z



Z 21 �Z

Z11 �Z

Z11 Z 21

�Z Z 21



Y22 Y21



1 Y21

A

B

1 Z 21

Z 22 Z 21



�Y Y21



Y11 Y21

C

D

�Z Z 22

Z12 Z 22

1 Y11



Y12 Y11

B D

�T D

h11 h12

Z 21 Z 22

1 Z 22

Y21 Y11

�Y Y11

1 D

C D

h21 h22

[Z] �

Z12 �Z

Y22 �Y



Y21 �Y

Y12 �Y

� �T B

[Y] �

Y21 Y22

[T]

[h] �





� �

�h h21

h22 h21

h12 h11

� �

h11 h21

1 h21

��������������� The Z parameters of a two-port network are Z11 = 20 �, Z22 = 30 �, Z12 = Z21 = 10 �. Find Y and ABCD parameters. 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 Y11 �

Y22

Hence, the Y-parameters are � 3 �Y11 Y12 � � 50 ��Y21 Y22 �� � � 1 �� � 50

10 1 Z12 �� �� � �Z 500 50 20 2 Z � 11 � � � �Z 500 50

Y12 � �

1� 50 � 2 �� 50 �



����������������������������������������������������

ABCD parameters �Z 500 � � 50 Z 21 10 30 Z �3 D � 22 � 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 ��

�������������� 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 � Y21 �

I1 � 0.5 �, V1 V2 � 0

Y12 �

I2 V1

Y22 �

� �0.2 �, V2 � 0

I1 V2

V1 � 0

I2 V2

V1 � 0

� �0.2 � � 1�

Hence, the Y-parameters are �Y11 Y12 � � 0.5 �0.2� ��Y21 Y22 �� � �� �0.2 1 �� Z-parameters �Y � Y11 Y22 � Y12 Y21 � (0.5)(1) � ( �0.2)( �0.2) � 0.46

� Z11 �� Z 21

1 Y22 � � 2.174 �, �Y 0.46 ( �0.2) Y � 0.434 �, � � 21 � � �Y 0.46

( �0.2) Y12 �� � 0.434 � �Y 0.46 Y 0.5 � 11 � � 1.087 � �Y 0.46

Z11 �

Z12 � �

Z 21

Z 22

Z12 � � 2.174 0.434 � � Z 22 �� �� 0.434 1.087 ��

ABCD parameters 1 Y22 �� � 5, Y21 �0.2 0.46 �Y C�� �� � 2.3, Y21 �0.2 A� �

1 1 �� �5 Y21 �0.2 0.5 Y D � � 11 � � � 2.5 Y21 �0.2 B��

������������������������������������������������������� Hence, the ABCD parameters are 5� �A B� � 5 ��C D �� � �� 2.3 2.5��

���������������

Using the relation Y � Z−1, show that |Z |�

1 � Z 22 Z11 � � . 2 �� Y11 Y22 ��

Solution We know that

�Y11 ��Y21

i.e.,

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 |�

���������������

1 � Z 22 Z11 � � 2 �� Y11 Y22 ��

For the network shown in Fig. 9.44, find Z and Y-parameters. 2�

I1

I2

� V1



1�

2�

V2

3I2





��������� Solution The network is redrawn by source transformation technique as shown in Fig. 9.45. Applying KVL to Mesh 1, V1 � I1 � I 3

�(i)

2� �

Applying KVL to Mesh 2, V2 � 2( I 2 � I 3 ) � 6 I 2 � �4 I 2 � 2 I 3 �(ii) Applying KVL to Mesh 3, �( I 3 � I1 ) � 2 I 3 � 2( I 2 � I 3 ) � 6 I 2 � 0 5 I 3 � I1 � 4 I 2 1 4 I 3 � I1 � I 2 5 5



2�

1�

V1 I1

I3



��������� �(iii)

� 6I2 �

V2 I2



����������������������������������������������������

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� 25 5 5 12 5 � 3 �, 8 2 � 5 2 � 1 Z 21 5 � � �, Y21 � � 8 �Z 4 � 5 Hence, the Y-parameters are 1� �3 � � �Y11 Y12 � � 2 2 � ��Y21 Y22 �� � � 1 1 � � � �4 2� Z Y11 � 22 � �Z

���������������



�(v)

4 5 � �1 � 8 2 � 5 4 1 Z11 � � 5 �� � 8 2 �Z � 5

Z Y12 � � 12 � �Z

Y22

�(iv)



Find Z and h-parameters for the network shown in Fig. 9.46. 4I1

2�

I1



2�

V1

2�

���

I1

I2



2� I3



V2 I2



��������� 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)

������������������������������������������������������� 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 � 5 I1 � I 2 Substituting Eq. (iii) in Eq. (ii), V2 � 4 I 2 � I1 � I 2 � � I1 � 3I 2 Comparing Eqs (iv) and (v) with Z-parameter equations, � Z11 �� Z 21

�(iii) �(iv) �(v)

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 � �, 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

��������������

h12 �

�16 h12 � � 3 � h22 �� �� 1 �3

1� 3� 1 �� 3�

Find Y and Z-parameters for the network shown in Fig. 9.47. I1

1� 2

3

1�

2

I2





2V1

V1 �

1� 2

V2 �

��������� 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)

����������������������������������������������������

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 �� � � �0 8 � 3 � � Z-parameters 16 � 8� �Y � Y11 Y22 � Y12 Y21 � ( 2) � � � 0 � � 3� 3 � 2� �� � �� 1 Y12 3 � � �� Z12 � � 16 �Y 8 3 Y 2 3 Z 22 � 11 � � � �Y 16 8 3

8 1 Y22 � 3 � �, Z11 � �Y 16 2 3 Y21 0 Z 21 � � �� � 0, 16 �Y 3 Hence, the Z-parameters are � Z11 �� Z 21

��������������

�1 Z12 � � 2 � Z 22 �� �� 0 �

1� 8� 3 �� 8�

For the network shown in Fig. 9.48, find Y and Z-parameters. I1

1

1�

� V1

3 V1

2

I2

���

2�



1�



V2 �

��������� 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)

������������������������������������������������������� 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 �, �Y �1

Z 21 � �

( �1) Y12 �� � �1 � �Y ( �1) 3 � 3 Y11 � � 2� � �Y �1 2

Z12 � �

2 Y21 � 2 �, �� �Y ( �1)

Z 22

Hence, the Z-parameters are � Z11 �� Z 21

��������������

� �2 �1� Z12 � � � 3� Z 22 �� � 2 � � 2�

Find Z-parameters for the network shown in Fig. 9.49. Hence, find Y and h-parameters. 0.9 I1 I1

1�

I2

� V1



10 �

V2

1�





��������� Solution The network is redrawn by source transformation technique as shown in Fig. 9.50. Applying KVL to Mesh 1, V1 � 2 I1 � I 2

�(i)

1�

�(ii)

Comparing Eqs (i) and (ii) with Z-parameter equations,

Z 22 11 � �, �Z 12 10 5 Z Y21 � � 21 � � � � �, �Z 12 6

9I1

���

V1

1�



I2 � 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 �

10 �



Applying KVL to Mesh 2, V2 � 9 I1 � 10 I 2 � 1( I1 � I 2 ) � 10 I1 � 11I 2

I1

1 Z12 �� � �Z 12 Z 2 1 � 11 � � � �Z 12 6

Y12 � � Y22

���������

����������������������������������������������������

Hence, the Y-parameters are � 11 �Y11 Y12 � � 12 ��Y21 Y22 �� � � 5 �� � 6



1� 12 � 1 �� 6 �

h-parameters �Z 12 � �, 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 � h11 h12 � � 11 11� �� h21 h22 �� � � 10 1 � � �� � 11 11�

��������������

Find Y and Z-parameters of the network shown in Fig. 9.51. 1

I1

1�



1�

V1

2V2

2V1

2

I2

���



2�

V2





��������� 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 �, �Y 3 ( �3) Y � �1 �, � � 21 � � �3 �Y

�(ii)

( �3) Y12 �� � �1 � �Y �3 Y 4 4 � 11 � � � � � �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� �

�������������������������������������������������������

��������������

Determine Y and Z-parameters for the network shown in Fig. 9.52. 1

I1

2

2�

I2





1�

V1

2�

V2

3I1





��������� 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. 9.53. I1

0.5 �

1

3

2

1�

� V1

0.5 Y12 � � 0.4 � �Y 1.25 Y 1.5 � 11 � � 1.2 � �Y 1.25

Z11 �

Hence, the Z-parameters are

��������������

0.5 Y22 �� � �0.4 �, �Y 1.25 4 Y � � 21 � � � �3.2 �, �Y 1.25

�(ii)

I2 �

1�

2V1



0.5 �

V2



���������

����������������������������������������������������

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 Y22 3 1 Z11 � � � �, �Y 8 3 2 Y12 1 Z 21 � � �� � 3� �Y 8 12

Z12

2 Y12 3 1 �� � � � �Y 8 12

Z 22 �

Y11 3 � � Y� 8

Hence, the Z-parameters are � Z11 �� Z 21

��������������

�1 1 � Z12 � � 3 12 � � 3 �� Z 22 �� �� 0 8� �

Determine Z and Y-parameters of the network shown in Fig. 9.54. I1

2�

4�

� V1

I2 �

0.05I2

� �

� 10V1 �

V2 �



���������

������������������������������������������������������� Solution Applying KVL to Mesh 1, V1 � 4I1 � 0.05 I2 � 0 V1 � 4I1 � 0.05I2 Applying KVL to Mesh 2, V2 � 2I2 � 10V1 � 0 V2 � 2I2 � 10V1 Substituting Eq. (i) in Eq. (ii), V2 � 2I2 � 40I1 � 0.5I2 � ������������������ �40I1 � 1.5I2 Comparing Eqs (i) and (iii) with Z-parameter equations, � Z11 �� Z 21

…(i)

...(ii)

...(iii)

Z12 � � 4 0.05� � Z 22 �� �� �40 1.5 ��

Y-parameters �Z � Z11Z 22 � Z12 Z 21 � ( 4)(1.5) � (0.05)( �40) � 8 Z 0.05 Z 1.5 � 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 �1.5 �Y11 Y12 � � 8 ��Y21 Y22 �� � � 40 � � 8

��������������



0.05 � 8 � 4 �� 8 �

Determine Z and Y-parameters of the network shown in Fig. 9.55. 2V3

1�

I1



3I2

I3

���

I2

� 2�

V1

2�

I1



� V3 �

V2 I2



��������� Solution Applying KVL to Mesh 1, V1 � 1I1 � 3I2 � 2(I1 � I2) � 0 V1 � 3I1 � 5I2

...(i)

����������������������������������������������

Applying KVL to Mesh 2, V2 � 2(I2 � I3) � 2(I1 � I2) � 0 V2 � 2I2 � 2I3 � 2I1 � 2I2 � 0 V2 � 2I1 � 4I2 � 2I3 …(ii) Writing equation for Mesh 3, ...(iii) I3 � 2V3 From Fig. 9.55, V3 � 2(I1 � I2) ...(iv) I3 � 2V3 � 4I1 � 4I2 Substituting Eq. (iv) in Eq. (ii), V2 � �6I1 � 4I2 ...(v) Comparing Eqs (i) and (v) with Z-parameter equations, 5� � Z11 Z12 � � 3 �� Z 21 Z 22 �� � �� �6 �4 �� Y-parameters �Z � Z�� Z�� � Z�� Z�� � (3)( �4) � (5)( �6) � 18 Z 4 2 5 Z Y11 � 22 � � � � �, Y12 � � 12 � � � �Z 18 9 �Z 18 ( �6) 1 Z 21 Z11 3 Y21 � � Y22 � � �� � �, � �Z 18 3 �Z 18 Hence, Y-parameters are 5� � 2 � � � �Y11 Y12 � � 9 18 ��Y21 Y22 �� � � 1 3 �� � 18 � � 3

��������������������������������������������� 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.

9.7.1 Cascade Connection Transmission Parameter Representation Figure 9.56 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� �

������������������������������������������������������� 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 � ...(9.1) �� I1 �� � ��C1 D1 �� �� � I 2 �� For the network N2,

�V1 � � � A2 �� I1 � �� � ��C2

B2 � � V2 � � D2 �� �� � I 2 � ��

...(9.2)

Since V1� � V2 and I 2� � � I 2 , we can write � V2 � � A2 �� � I 2 �� � ��C2

B2 � � V2 � � D2 �� �� � I 2 � ��

...(9.3)

Combining Eqs (9.1) and (9.3),

Hence,

�V1 � � A1 �� I1 �� � ��C1

B1 � � A2 D1 �� ��C2

B2 � � V2 � � � A B � � V2 � � � D2 �� �� � I 2 � �� ��C D �� �� � I 2 � ��

� A B � � A1 ��C D �� � ��C1

B1 � � A2 D1 �� ��C2

B2 � D2 ��

...(9.4)

Equation 9.4 shows that the resultant ABCD matrix of the cascade connection is the product of the individual ABCD matrices.

�������������� Two identical sections of the network shown in Fig. 9.57 are connected in cascade. Obtain the transmission parameters of the overall connection. 2�

I1

2�

I2





1�

V1

V2

2�





��������� Solution The network is redrawn as shown in Fig. 9.58. 2�

2�





1�

V1 I1

2� I3



V2 I2



���������

����������������������������������������������

Applying KVL to Mesh 1, V1 � 3I1 � I3

…(i)

Applying KVL to Mesh 2, V2 � 2I2 � 2I3 Applying KVL to Mesh 3, � �I1 � 2I2 � 5I3 � 0 I3 �

…(ii)

1 2 I1 � I 2 5 5

…(iii)

Substituting Eq. (iii) in Eq. (i), 2 � �1 V1 � 3I1 � � I1 � I 2 � �5 5 � �

14 2 I1 � I 2 5 5

…(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), �

…(v)

14 � 5 � 2 � V2 � 3I 2 �� � I 2 5 �2 5 � 7V2 � 8 I 2

V1 �

…(vi)

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��

��������������

Determine ABCD parameters for the ladder network shown in Fig. 9.59. I1

2�

2H

I2 �

� V1

2F



1F

1�

V2 �

���������

������������������������������������������������������� Solution The above network can be considered as a cascade connection of two networks N1 and N2. The network N1 is shown in Fig. 9.60. 2s

2

I1

I2 +

+

1 2s

V1

V2





��������� Applying KVL to Mesh 1, 1� 1 � V1 � � 2 � � I1 � I 2 � 2s � 2s

…(i)

Applying KVL to Mesh 2, V2 �

1 1� � I1 � � 2 s � � I 2 � 2s 2s �

…(ii)

From Eq. (ii), I1 � 2s V2 � (4s2 � 1) I2 Substituting Eq. (iii) in Eq. (i), 1� 1 � V1 � � 2 � � [2 s V2 � ( 4 s 2 � 1) I 2 ] � I 2 � 2s � 2s

...(iii)

� ( 4 s � 1)V2 � (8s 2 � 2 s � 2) I 2 Comparing Eqs (iv) and (iii) with ABCD parameter equations,

…(iv)

� A1 ��C1

B1 � � 4 s � 1 8s 2 � 2 s � 2� � � D1 �� �� 2 s 4s2 � 1 �

The network N2 is shown in Fig. 9.61. I2�

I1� �

1 s

V1�

1

V2�





I2�

I1� �



V1�

� Z(s) �

1 s ��1





(a)

V2�

(b)

��������� 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,

����������������������������������������������

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 �� � � 2 s 4s � 1 � � � � 4 s � 4 s 2 � 3s � 1 4s2 � 1 � �

9.7.2 Parallel Connection Figure 9.62 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

����������������������������� 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.

��������������

Determine Y-parameters for the network shown in Fig. 9.63.

������������������������������������������������������� 3�

I1

2�

2�

� V1

I2 �

2�



V2 �

��������� Solution The above network can be considered as a parallel connection of two networks, N1 and N2. The network N1 is shown in Fig. 9.64. Applying KCL at Node 3, 3 2� 2� I2 � I1� V3 I1 � � I 2 � � …(i) � � 2 From Fig. 9.64, V2 V1 2� V1 � V3 I1 � � …(ii) � � 2 V2 � V3 ��������� I2 � � …(iii) 2 Substituting Eqs (ii) and (iii) in Eq (i), V1 � V3 V2 � V3 V3 � � 2 2 2 3V3 � V1 � V2 V V V3 � 1 � 2 ...(iv) 3 3 Substituting Eq. (iv) in Eq. (ii), V 1 �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 � � 3� I2�� I1�� �Y11 � Y12 � � � 3 6 ��Y21 � Y22 � �� � � 1 1 � � � �� � � 6 3 � V1 V2 The network N2 is shown in Fig. 9.65. � � 1 1 V �V I1 � � � I 2 � � 1 2 � V1 � V2 3 3 3 ���������

����������������������������������������������

Hence, the Y-parameters are � 1 �Y11 � Y12 � � � 3 ��Y21 � Y22 � �� � � 1 �� � 3 The overall Y-parameters of the network are

1� � � 3 1 �� 3 �

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

��������������

1� � � 2 2 �� 3 �

Find Y-parameters for the network shown in Fig. 9.66. 2�

I1

I2





1�

V1

0.5 �

0.5 �

V2

0.5 � 2�





��������� Solution The above network can be considered as a parallel combination of two networks N1 and N2. The network N1 is shown in Fig. 9.67. 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 �

2�

2

I2�

� V1

Applying KCL at Node 2, I2 � �

1

I1 �



1�

0.5 �



V2 �

���������

������������������������������������������������������� The network N2 is shown in Fig. 9.68. Applying KCL at Node 3, V I1 � � I 2 � � 3 2 V1 � V3 � 2V1 � 2V3 I1 � � where 0.5 V �V I 2 � � 2 3 � 2V2 � 2V3 0.5 Substituting I1� and I2� in Eq (i), 2V1 � 2V3 � 2V2 � 2V3 � 0.5V3

I1��

…(i)

3

0.5 �

0.5 �



I2�� �

V1

V2

2�





���������

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 � � �Y11 � Y12 � � � 9 9 ��Y21 � Y22 � �� � � 8 10 � �� � � 9 9 � Hence, overall Y-parameters of the network are

…(iv)

� 3 10 � �Y11 Y12 � � Y11 � � Y11 � Y12 � � Y12 � � � 2 9 � � ��Y21 Y22 �� ��Y21 � � Y21 � Y22 � � Y22 � �� � 1 8 �� � � 2 9

��������������

1 8 � � 47 � � � � 2 9 � 18 5 10 �� �� 25 � � 2 9 � � 18

Find Y-parameters for the network shown in Fig. 9.69. 1F

I1

1F

2�

2�

� V1

I2 �

1�



2F

V2 �

���������

25 � 18 � 65 �� 18 �



����������������������������������������������

Solution The above network can be considered as a parallel connection of two networks, N1 and N2. The network N1 is shown in Fig. 9.70. I1 �

2�

2�

I2 �

� V1

2F



I1� �



V2

V1





2

3

2

I2� �

1 2s

V2 �

(b)

(a)

��������� Applying KCL at Node 3, I1 � � I 2 � � 2 s V3 From Fig. 9.70,

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 � �Y11 � Y12 � � � 8s � 4 8s � 4 � ��Y21 � Y22 � �� � � s � 1 �� 1 4 �� � 8s � 4 8s � 4 � I2 � �

…(vi)

������������������������������������������������������� The network N2 is shown in Fig. 9.71. I1��

1F

1F

I2��

� V1

1�



I1�� �



V2

V1





1 s

1 s

3

I2�� �

1

V2 �

(b)

(a)

��������� Applying KCL at Node 3, I1 � � I 2 � � V3

…(i)

From Fig. 9.71, 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 s � 1) �� �� 2 s � 1� I1 � �

� s2 � � s( s � 1) � V1 � � �� � V2 � � 2 s � 1� � 2s � 1 �

…(ii)

…(iii)

…(iv)

…(v)

Substituting Eq. (iv) in Eq. (iii), �� s � � s � � I 2 � � sV2 � s �� �� V1 � �� � V2 � 2 1 2 s � 1� �� s � � � s2 � � s( s � 1) � � �� � V1 � � 2 s � 1 � V2 s 2 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)

����������������������������������������������

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) � �

9.7.3 Series Connection Figure 9.72 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��







��������������������������� 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 Hence,

�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 ��V2 �� � ��V2 � � V2 � �� � �� Z 21 � � Z 21 � Z 22 � � Z 22 � �� �� I 2 �� � �� Z 21

Z12 � � I1 � Z 22 �� �� I 2 ��

Thus, the resultant Z-parameter matrix for the series-connected networks is the sum of Z matrices of each individual two-port network.

�������������� Two identical sections of the network shown in Fig. 9.73 are connected in series. Obtain Z-parameters of the overall connection. I1

2�

2�

� V1

I2 �

1�



V2 �

���������

������������������������������������������������������� 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

��������������

Z12 � �3 1� �3 1� �6 2� � � � Z 22 �� ��1 3�� ��1 3�� �� 2 6 ��

Determine Z-parameters for the network shown in Fig. 9.74. L

L

+

+

C

V2

V1 L −

C

C



��������� Solution The above network can be considered as a series connection of two networks, N1 and N2. The network N1 is shown in Fig. 9.75. Ls Ls Applying KVL to Mesh 1, I1 I2 1� � � 1� V1 � � � Ls � � I1 � � � I 2 � � Cs � Cs �

…(i)

…(ii)

…(i)

Applying KVL to Mesh 2, 1� � V2 � � ( Ls) I1 � � Ls � � I 2 � Cs �

1 Cs

V2 �





���������

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. 9.76. Applying KVL to Mesh 1, 1� � V1 � � � Ls � � I1 � ( Ls) I 2 � Cs �



V1�

Applying KVL to Mesh 2, 1� � 1� � V2 � � � � I1 � � Ls � � I 2 � Cs � � Cs �





� Ls

V1�� �

…(ii)

I2

I1

1 Cs

V2��

1 Cs

���������



����������������������������������������������

Comparing Eqs (i) and (ii) with Z-parameter equations, 1 � Ls � � Z11 � Z12 � � � Cs �� Z 21 � Z 22 � �� � � � Ls �

� � 1 �� Ls � Cs � Ls

Hence, the overall Z-parameters of the network are, � Z11 �� Z 21

2 � 2 Ls � Z12 � � Z11 � � Z11 � Z12 � � Z12 � � � Cs � � Z 22 �� �� Z 21 � � Z 21 � Z 22 � � Z 22 � �� �� 1 Ls � � Cs

1 � Cs � � � Ls � 1 � � 2 1 � � 2 � �� Cs � ��1 2�� 2 Ls � � Cs � Ls �

9.7.4 Series-Parallel Connection Figure 9.77 shows two networks connected in series-parallel. Here, the input ports of two networks are connected in series and the output ports are connected in parallel. I1 �

I2�

� V1� �

I2��

� V1��



N1

V1



I2

N2

V2





������������������������������������ 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.

��������������

Determine h-parameters for the network shown in Fig. 9.78.

������������������������������������������������������� I1

1�

1�



I2 �

2�

V1

1�

1�

1�

1�

V2

2� �



1�

1�

��������� 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. 9.79. Applying KVL to Mesh 1, 1� 1� I2 I1 V1 � 4I1 � 2I2 …(i) � � Applying KVL to Mesh 2, …(ii) V2 � 2I1 � 4I2 Rewriting Eq. (ii) V2 V1 2� 4 I 2 � �2 I1 � V2 1 1 …(iii) I 2 � � I1 � V2 2 4 � � Substituting Eq. (iii) in Eq. (i), 1� 1� 1 � � 1 V1 � 4 I1 � 2 � � I1 � V2 � ��������� � 2 4 � 1 � 3I1 � V2 …(iv) 2 Comparing Eqs (iii) and (iv) with h-parameters equations, 1� � 3 � h11 � h12 � � � 2� �� h21 � h22 � �� � � 1 1 � �� � � 2 4� For network N2, h-parameters will be same as the two networks are identical. 1� � 3 � h11 � h12 � � � 2� �� h21 � h22 � �� � � 1 1 � �� � � 2 4� Hence, the overall h-parameters of the network are 1� � 1� � 3 3 � 6 1� � � � h11 h12 � � h11 � h12 � � � h11 � h12 � � � 2 � 2� � � � � � 1� � � � � �� h21 h22 �� �� h21 � h22 � �� �� h21 � h22 � �� � � 1 1 � � � 1 1 � �� �1 2 �� � 2 4� � 2 4�

9.9���������������������

���������T�������� Any two-port network can be represented by an equivalent T network as shown in Fig. 9.80. 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

...(9.9)

Applying KVL to Mesh 2,

ZB

+

...(9.10)

V2

ZC

V1

V2 � Z B I 2 � ZC ( I 2 � I1 ) � ZC I1 � ( Z B � ZC ) I 2

I2

+





Comparing Eqs (9.9) and (9.10) with Z-parameter equations, Z11 � Z A � ZC Z12 � ZC 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

���������PI������������� Any two-port network can be represented by an equivalent pi (�) network as shown in Fig. 9.81. 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

...(9.11)

YB

1

2

I2 +

+

...(9.12) Comparing Eqs (9.11) and (9.12) with Y-parameter equations, Y11 � YA � YB Y12 � �YB Y21 � �YB Y22 � YB � YC

V1

YA

YA � Y11 � Y12 � Y11 � Y21 YB � �Y12 � �Y21 YC � Y22 � Y12 � Y22 � Y21

V2





Solving the above equations,

YC

�������������������

�������������������������������������������������������

�������������� The Z-parameters of a two-port network are: Z11 = 10 �, Z12 = Z21 = 5 �, Z22 = 20 �. Find the equivalent T-network. I1

Solution The T-network is shown in Fig. 9.82. Applying KVL to Mesh 1, V1 � (Z1 � Z2)I1 � Z2 I2 …(i) Applying KVL to Mesh 2, V2 � Z2I1 � (Z2 � Z3)I2 …(ii) Comparing Eqs (i) and (ii) with Z parameter equations, Z11 � Z1 � Z 2 � 10 Z12 � Z 2 � 5 Z 21 � Z 2 � 5 Z 22 � Z 2 � Z3 � 20 Solving the above equations, Z1 � 5 � Z2 � 5 � Z3 � 15 �

��������������

Z1

Z3

I2 +

+ Z2

V1

V2





���������

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. 9.83. Applying KCL at Node, 1, I1 �

V1 V1 � V2 � Ra Rb

1� 1 � 1 V1 � V2 �� � � Ra Rb �� Rb

Rb

I1

…(i)

Applying KCL at Node 2,

+ V1

V V �V I2 � 2 � 2 1 Rc Rb

+ Ra



1 1� � 1 V2 � � V1 � � � � 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 � � 0.0 07 Y22 � Rb Rc Y11 �

I2

Rc

V2



���������

9.9���������������������

Solving the above equations, Ra � 25 � Rb � 20 � Rc � 50 �

��������� ����� Find the parameters YA, YB and YC of the equivalent p network as shown in Fig. 9.84 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�

��������� 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)

�������������������������������������������������������

�������������� A network has two input terminals (a, b) and two output terminals (c, d) as shown in Fig. 9.85. 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

��������� 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) �

…(vi)

From Eqs (i) and (vi), ZA � (150 � j300) � From Eqs (iii) and (vi), ZB � (100 � j200) �

��������������

Find the equivalent �-network for the T-network shown in Fig. 9.86. I1

2�

2.5 �

� V1

I2 �

5�



V2 �

���������

9.9���������������������

Solution Figure 9.87 shows T-network and �-network. ZA

ZB

Z3

ZC

Z1

Z2

p-network (Delta network)

T-network (Star network)

��������� For converting a T-network (star network) into an equivalent �-network (delta network), we can use stardelta transformation technique. Z A ZC 2�5 � 2�5� � 11 � ZB 2.5 Z Z 2 � 2.5 Z3 � Z A � Z B � A B � 2 � 2.5 � � 5.5 � ZC 5 Z Z 2.5 � 5 Z 2 � Z B � ZC � B C � 2.5 � 5 � � 13.75 � ZA 2

5.5 �

Z1 � Z A � ZC �

11 �

The equivalent �-network is shown in Fig. 9.88.

��������������

13.75 �

���������

For the network shown in Fig. 9.89. Find the equivalent T-network. 8�

1�

I1

I3

4�

I2





2�

V1 �

I1

V2 I2



��������� 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)

������������������������������������������������������� 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. 9.90. 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),

I1

…(vi)

…(v)

ZA

ZB

+

+

…(vii) V 1

38 13 30 ZC � 13 62 Z B � ZC � 13

ZC



Z A � ZC �

I2

V2



���������

Solving the above equations, 8 � 13 32 ZB � � 13 30 ZC � � 13 ZA �

������������������������ A lattice network is one of the common two-port networks, shown in Fig. 9.91. 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

�������������������������

9.10����������������������

The lattice network can be redrawn as a bridge network as shown in Fig. 9.92. 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

V2



I2 ZB

ZA



Since the network is symmetric,

������������������������

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.

��������������

Find the lattice equivalent of a symmetrical T network shown in Fig. 9.93. I1

1�

1�



I2 �

2�

V1 �

V2 �

��������� 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 � Z12 � 2 �

…(ii)

Applying KVL to Mesh 2,

Z 21 � 2 � Z 22 � 3 �

������������������������������������������������������� 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 � Z B � Z11 � Z12 � 3 � 2 � 5 � The lattice network is shown in Fig. 9.94. 1�

I1

I2





5�

V1

5�

V2





1�

���������

��������������

Find the lattice equivalent of a symmetric �-network shown in Fig. 9.95. 5�

I1

I2





10 �

V1

10 �



V2 �

��������� Solution The network is redrawn as shown in Fig. 9.96. Applying KVL to Mesh 1, I1 V1 � 10 I1 � 10 I 3 …(i) � Applying KVL to Mesh 2, V2 � 10 I 2 � 10 I 3 …(ii) V1 I1 Applying KVL to Mesh 3, �10 I1 � 10 I 2 � 25 I 3 � 0 2 2 I3 � I � I 2 5 5 Substituting Eq (iii) in Eq (i),

5�



10 �

10 � I3



…(iii)

I2

V2 I2



���������

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)

9.11����������������������������������

Comparing the Eqs (iv) and (v) with Z-parameter equations, Z11 � 6 � Z12 � 4 � Z 21 � 4 � Z 22 � 6 �

2�

I1

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 � Z B � Z11 � Z12 � 6 � 4 � 10 �



10 �

V1

10 �



V2



2�

���������

The lattice network is shown in Fig. 9.97.

������������������������������������ 9.11.1 Driving-Point Impedance at Input Port A two-port network is shown in Fig. 9.98. 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



������������������������������������� 1. Input Impedance in Terms of Z-parameters We know that

From Fig. 9.98,

V1 � Z11 I1 � Z12 I 2 V2 � Z 21 I1 � Z 22 I 2 V2 � � Z L I 2 � I 2 Z L � Z 21 I1 � Z 22 I 2 Z 21 I2 � � I1 Z 22 � Z L Zin �

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

������������������������������������������������������� 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. 9.98, V2 � � Z L I 2 V I 2 � � 2 � �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. 9.98, 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

9.11����������������������������������

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 � 1 � h22 Z L h Z � � � � 22 L V1 ( h11h22 � h12 h21 ) Z L � h11 Zin � � 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

9.11.2 Driving-Point Impedance at Output Port A two-port network is shown in Fig. 9.99. 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 −

������������������������������������� 1. Output Impedance in terms of Z-parameters We know that V1 � Z11I1 � Z12 I 2 V2 � Z 21I1 � Z 22 I 2

������������������������������������������������������� From Fig. 9.99, 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. 9.99, 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

9.11����������������������������������

From Fig. 9.99, 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. 9.99, 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 �

��������������

h11 h11h22 � h12 h21

Measurements were made on a two-terminal network shown in Fig. 9.100. 1

I1

I2

� V1

2 �

Network

V2





1�

2�

����������

RL

������������������������������������������������������� (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-� 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 �

V1 I1

V Z 21 � 2 I1 Z 22 �

V2 I2

V Z12 � 1 I2



100 �0� � 10 � 10 �0�



25�0� � 2.5 � 10 �0�



100 �0� �5� 20 �0�



50 �0� � 2.5 � 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-� 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 100 � �1.74 A 57.5 Voltage across the resistor � �I2RL � �10(�1.74) � 17.4 V I2 � �

�������������� The Z-parameters of a two-port network shown in Fig. 9.101 are Z11 = Z22 = 10 �, Z21 = Z22 = 4 �. If the source voltage is 20 V, determine I1,I2,V2 and input impedance. I1

I2

� Vs

� �

V1

� Network



V2 �

����������

5�

9.11����������������������������������

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 V2 � 4 I1 � 10 I 2 �20 I 2 � 4 I1 � 10 I 2

…(i) …(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 � I1 2.1

��������� �����

The Z-parameters of a two-port network shown in Fig. 9.102 are, Z11 = 2 �, I V V Z12 = 1 �, Z21 = 2 �, Z22 = 5 �. Calculate the voltage ratio 2 , current ratio � 2 and input impedance 1 . I1 Vs I1 I1

I2

� Vs

� �

� Network

V1 �

V2

5�



���������� 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 � 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)

������������������������������������������������������� �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 1 I � 2 � 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 � � I1 5

�������������� The following equations give the voltages V1 and V2 at the two ports of a two-port network shown in Fig. 9.103. V1 � 5 I 1 � 2 I 2 V2 � 2 I 1 � I 2 A load resistor of 3 � is connected across port 2. Calculate the input impedance. I1 � V1

I2 Twoport network



� V2

3�



���������� Solution From Fig. 9.103, 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 � I1

...(i)

(ii)

9.11����������������������������������

��������� ����� The y-parameters for a two-port network shown in Fig. 9.104 are given as, Y11 = 4 , Y22 = 5 , Y12 = Y21 = 4 . If a resistor of 1 � is connected across port-1 of the network then find the output impedance. I2

I1 �



1�

V1

Network

V2





���������� Solution The two-port network can be represented in terms of Y-parameters. I1 � 4V1 � 4V2 I 2 � 4V1 � 5V2 When the 1-� 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 � � � I2 9

�������������� The following equation gives the voltage and current at the input port of a two-port network shown in Fig. 9.105. V1 � 5V2 � 3I 2 I1 � 6V2 � 2 I 2 A load resistance of 5 � is connected across the output port. Calculate the input impedance. I1 � V1

I2 Twoport network



� V2 �

���������� Solution From Fig. 9.105, 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 � � � I1 �32 I 2 8

5�

�������������������������������������������������������

��������� ����� The ABCD parameters of a two-port network shown in Fig. 9.106 are A = 2.5, B = 4 �, 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 � connected at Port 2? I1

I2





V1

Network



10 �

V2 �

���������� Solution The two-port network can be represented in terms of ABCD parameters. V1 � 2.5 V2 ���I2 I1 � V2 � 2I2 When the 10 � resistor is connected across Port 2, V2 � �10 I 2 � 10 I 2 � �1A V1 � 2.5(10) � 4( �1) � 29 V

…(i) …(ii) ...(iii)

�������������� The h-parameters of a two-port network shown in Fig. 9.107 are h11 = 4 �, h12 = 1, h21 = 1, h22 = 0.5 . Calculate the output voltage V2 when the output port is terminated in a 3 � resistance and a 1 V is applied at the input port. I1

I2



1V

� �

� Network

V1 �

V2

3�



���������� 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)

��������������

��������� ����� The h-parameters of a two-port network shown in Fig. 9.108 are h11 = 1 �, h12 = −h21 = 2, h22 = 1 . The power absorbed by a load resistance of 1 � connected across port-2 is 100W. The network is excited by a voltage source of generated voltage Vs and internal resistance 2 �. Calculate the value of Vs. 2�

I1

I2

� Vs

� �

� Network

V1

V2



2�



���������� Solution The two-port network can be represented in terms of h-parameters. V1 � I1 � 2 V2 I 2 � �2 I1 � V2 When the 1 � resister is connected across port-2, V22 1 V2 V2 I2

…(i) …(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 9.1

Determine Z-parameters for the network shown in Fig. 9.109. I1

1�

1�

I2

� V1

2�

9.2

0.5 �



Find Z-parameters for the network shown in Fig. 9.110. I1



+

V2

V1

1F

2F

I2 +

2H

2H





V2 −

����������

���������� �13 � Z��7 �2 �7

2� 7� 3 �� 7�

� 4s4 � 6s2 � 1 � 3 Z � � 4 s �3 s � 4s � � 4s2 � 1

4 s3 � � 4s2 � 1 � 3 4 s � 2s � � 4s2 � 1 �

������������������������������������������������������� Find Y-parameters of the network shown in Fig. 9.111.

9.3 I1

2�

2�

2�

1�

1�



8�

i

I2

� V1

2i





V2

V1







����������

� 0.36 �0.033� Y �� � �0.033 �0.36 �� Find Y-parameters for the network shown in Fig. 9.112. 3�

I1

1H

9.7



2�

2F

1F



I1

V2

+



V1

�10 s 2 � 13s � 2 � 5s � 6 Y �� 2 � �� � 5s � 6

� 1F 3

2H

1�



V2



� 7s � 6 � 12 Y �� � �s � 4



�1 � s 2 � � A B � � s2 ��C D �� � � 1 � � s

1 � 2s2 � � s3 � 2 1� s � � s2 �

Find ABCD parameters for the network shown in Fig. 9.116.

9.8 I1

4�

8�

6�



I2 �

V2 V1

1�

2�



����������

9.6

1H

����������

I2

� 2�

I2 +

� 2 � 5s � 6 � 2 5s � 6 s � 5 � 5s � 6 ��

1F 4

I1

1F



Find Y-parameters for the network shown in Fig. 9.113.

9.5

1F



����������

V1

1� � 3 � 20 � 20 � Y �� 1 �� �� 1 � 4 4 � Show the ABCD parameters of the network shown in Fig. 9.115.

I2

� V1

V2

2�

����������

9.4



4�



����������

s � � 4 � s2 � 4s � 2 � � 4s �

Find Y-parameters for the network shown in Fig. 9.114.

V2

� 27 206 � �A B� � � � ��C D �� �11 42 � �2 � 9.9

For the network shown in Fig. 9.117, determine parameter h21.

�������������� 2I1 1�

I1

I2





1F

V2

1�

V1 �



����������

I1

1�

V1

1�

V2

2I1



1� � 3 � 4 � 4� Y �� � �� 1 3 � � 4 4 � Two identical sections of the network shown in Fig. 9.121 are connected in parallel. Obtain Y-parameters of the connection. I1

1�

2�

�Y11 � 1 �, Y12 � �0.5 �, Y21 � 1.5 �, Y22 � 0.5 �



2�



���������� � � 1 Y �� �� 1 � 2

For the bridged T, R-C network shown in Fig. 9.119 determine Y-parameters using interconnections of two-port networks. 1 2F

9.14 1� 2

1�



I2



V2

I1



� V1 �

� s 2 � 8s � 8 � 2( s � 6) Y �� � ( s 2 � 6 s � 8) �� 2( s � 6) � 9.12

1� 2�



( s 2 � 6 s � 8) � � 2( s � 6) � s 2 � 10 s � 8 � � 2( s � 6) �

For the network of Fig. 9.120, find Y-parameters using interconnection of two-port networks.

I2 � V2 �

2� 1� 2

����������

1� � � 2 5 �� 4 �

Determine Y-parameters using interconnection of two-port networks for the network shown in Fig. 9.122. 1�



1 F 2

V2

4�



2 2 6 4 � Z11 � �, Z12 � �, Z 21 � � �, Z 22 � � � 5 5 5 5 �

V1

I2

� V1

����������

I1

V2

1�



9.13

I2



9.11

I2





2�

2�

����������

� V1

2�



�( 2 � s) � � � h21 � 1 � s � � �

2�

I1

2�



Determine Y and Z-parameters for the network shown in Fig. 9.118.

9.10

2�

1� 2

���������� � 3.1 �0.9� Y �� � �0.9 3.1 �� 9.15

Determine the transmission parameters of the network shown in Fig. 9.123 using the concept of interconnection of two two-port networks.

������������������������������������������������������� 1H

1H

Determine the h-parameters of the overall network.

1H

3� 1F

3�

1F

1�

1�

���������� �1 � 3s 2 � s 4 � 3 � 2s � s 9.16

3s � 4 s 3 � s 5 � � 1 � 3s 2 � s 4 �

���������� � 15 � 2 � 1 �� � 2

Two networks shown in Fig. 9.124 are connected in series. Obtain the Z-parameters of the resulting network. 1�

1�

The h-parameters of a two-port network shown in Fig. 9.126 are h11 � 2 �, h12 � 4, h21 = �5, h22 � 2 . Determine the supply voltage Vs if the power dissipated in the load resistor of 4 � is 25 W and Rs = 2 ��

9.18 2�

Rs ��2 �

(a) 10 �

20 �

Vs

� �

I1 � V1 �

I2 Network

� V2 �

4�

����������

5�

9.19 (b)

���������� �18 7 � �� 7 28�� 9.17

1� � � 2 5 �� 2 �

9.20

Two identical sections of the network shown in Fig. 9.125 are connected in series-parallel.

[58 V] The Z-parameters of a two-port network are Z11 � 2.1 �, Z12 � Z21 � 0.6 �, Z22 � 1.6 �. A resistor of 2 ��is connected across port 2. What voltage must be applied at port 1 to produce a current of 0.5 A in the 2 � resistor. [6 V] If a two-port network has Z11 � 25 �, Z12 � Z21 � 20 �, Z22 � 50 �, find the equivalent T-network. [10 �, 30 �, 20 �]

Objective-Type Questions 9.1

The open-circuit impedance matrix of the two-port network shown in Fig. 9.127 is 2�

1�

����������

3I1

9.2

� �2 1� (a) � � �8 3��

� �2 �8� (b) � 3 �� �1

�0 1 � (c) � �1 0 ��

� 2 �1� (d) � � �1 3 ��

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

�����������������������������

(a) (b) (c) (d) 9.3

z-parameter matrix h-parameter matrix y-parameter matrix ABCD parameter matrix

For a two-port network to be reciprocal (b) y21 � y12 (a) z11 � z22 (c) h21 � �h12

9.4

9.7

(a) �0.2 mho (b) 0.1 mho (c) �0.05 mho (d) 0.05 mho The Z-parameters Z11 and Z21 for the two-port network in Fig. 9.130 are, I1



4� V2

V1

� 10V1 � �



���������� 6 16 �, � 11 11

(b)

6 4 �, � 11 11

6 16 �, � � 11 11

(d)

4 4 �, � 11 11

(a) �

9.5

non-reciprocal and passive non-reciprocal and active reciprocal and passive reciprocal and active

(c) 9.8

A two-port network is shown in Fig. 9.128. The parameter h21 for this network can be given by I1

R

R

The impedance parameters Z11 and Z12 of a two-port network in Fig. 9.131. 2�

2�

I2

+

3�

1�

1�

+

V1

V2

R



���������� (a) 2.75 �, 0.25 � (c) 3 �, 0.25 �



���������� 9.9 1 (a) � 2 (c) � 9.6

I2

(d) AD � BC � 0

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)

2�



1 (b) 2

3 2

(d)

(b) 3 �, 0.5 � (d) 2.25 �, 0.5 �

The h parameters of the circuit shown in Fig. 9.132. 10 �

3 2 20 �

The admittance parameter Y12 in the two-port network in Fig. 9.129. 20 �

���������� 5�

����������

10 �

� 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��

������������������������������������������������������� 9.10

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

(c) 9.15

9.11

(a)

A � BRL C � DRL

(b)

ARL � C BRL � D

(c)

DRL � A BRL � C

(d)

B � ARL D � CRL

9.16

In the two-port network shown in Fig. 9.133, Z12 and Z21 are respectively

re

For the network shown in Fig. 9.134 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 YC

����������

9.12

9.13

9.14

YA

(d) re and �� r0 ���������� 9.17

port network will be

(d) 2, �1, �1, 1

two-port network has transmission �A B� . The input impedance parameters � �C D �� of the network at port 1 will be

A

A C

(b)

AD BC

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 two-

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

(a)

YB

(b) 0 and �� r0

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

(c) 1, 1, 1, 2

D C

Z11 Z22 � Z12 Z21 � 1 AD � BC � 1 h11 h22 � h12 h21 � 1 Y11 Y22 � Y12 Y21 � 1

ro

bI 1

(a) re and � r0 (c) 0 and � r0

(d)

A two-port network is symmetrical if (a) (b) (c) (d)

impedance seen at port 1 is given by

AB DC

9.18

�3 5 � (a) � � 2 25��

�18 7 � (b) � � 7 28��

�15 2� (c) � � 5 3��

(d) inderminate

If the � 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

���������������������������������������� 16 �

1 3F

1�

24 �

2H

24 �

���������� R1

2s � � 2s � 1 (a) � 3� 2s � � � 2s s� �

R2

R3

9.19

9.20

���������� 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 The impedance matrix for the network shown in Fig. 9.136 is

9.21

� 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� � 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 9.1. (a)

9.2. (d)

9.4. (b)

9.5. (a)

9.6. (c)

9.7. (c)

9.8. (a)

9.9. (d)

9.10. (d)

9.3. (b) , (c)

9.11. (b)

9.12. (d)

9.13. (b)

9.14. (a)

9.15. (c)

9.16. (c)

9.17. (b)

9.18. (a)

9.19. (c)

9.20. (a)

9.21. (b)

10

Synthesis of RLC Circuits

�������������������� 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 j� axis.

��������������������������� 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 j� -axis only. 3. If the polynomial P(s) is either even or odd, the roots of polynomial P(s) lie on the j� -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).

10.2�Circuit Theory and Networks—Analysis and Synthesis 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.

��������������

State for each case, whether the polynomial is Hurwitz or not. Give reasons in each case.

(a) s � 4s3 � 3s � 2 (b) s6 � 5s5 � 4s4 � 3s3 � 2 s 2 � s � 3 4

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.

��������������

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 � 4� �

�1 s� s �4 s 0

Since all the quotient terms are positive, P(s) is Hurwitz.

��������������

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

10.2�Hurwitz Polynomials�10.3

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.

��������������

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.

10.4�Circuit Theory and Networks—Analysis and Synthesis

��������������

Prove that polynomial P ( s) � s4 � s3 � 2 s 2 � 3 s � 2 is not Hurwitz.

Solution 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.

��������������

Prove that polynomial P ( s) � 2 s4 � 5 s3 � 6 s 2 � 3s � 1 is Hurwitz.

Solution 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

10.2�Hurwitz Polynomials�10.5

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.

��������������

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.

��������������

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

10.6�Circuit Theory and Networks—Analysis and Synthesis 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.

��������������

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

10.2�Hurwitz Polynomials�10.7

5s 4 �

20 2 s 3 7 2 8 � 18 �6 s � 2� s3 � s � s � 3 5 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.

���������������

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) � 5s 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

10.8�Circuit Theory and Networks—Analysis and Synthesis 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.

���������������

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)

10.2�Hurwitz Polynomials�10.9

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 � 2s� 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.

��������� ������

Test whether the polynomial P ( s) � s7 � 2 s6 � 2 s5 � s4 � 4 s3 � 8 s 2 � 8 s � 4 is

Hurwitz. 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)

10.10�Circuit Theory and Networks—Analysis and Synthesis 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

�3 �3 � 4� s 5 � 6 s � s �2 �2 3 5 s � 6s 2 0

s4

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.

���������������

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) ( 2s 2 � s � 1) Let

P1 ( s) � s 4 � 6 s 2 � 25

10.2�Hurwitz Polynomials�10.11

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 0 0 1 s 4 s3 � 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 � �

10.12�Circuit Theory and Networks—Analysis and Synthesis 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.

���������������

Test whether P ( s) � s4 � 7 s3 � 6 s 2 � 21s � 8 is Hurwitz.

Solution The Routh array is given by, s4 s3 s2

1 6 8 7 21 3 8 7 3 0 8

s1 s0

Since all the elements in the first column are positive, the polynomial P(s) is Hurwitz.

���������������

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

1 1 �1 5

s0

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.

���������������

Test whether P ( s) � s5 � 2 s4 � 4 s3 � 6 s 2 � 2 s � 5 is Hurwitz.

10.2�Hurwitz Polynomials�10.13

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.

���������������

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.

���������������

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.

���������������

Test whether P(s) = s5 + 12s4 + 45s3 + 60s2 + 44s + 48 is Hurwitz.

10.14�Circuit Theory and Networks—Analysis and Synthesis

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.

��������������� 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

10.2�Hurwitz Polynomials�10.15

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.

��������������� 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

��������������� 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

10.16�Circuit Theory and Networks—Analysis and Synthesis 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.

�������������������������������� 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

10.3.1 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 j�-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.

������� �������������������������������������������������������������� 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 j�-axis must be simple and the residues evaluated at these poles must be real and positive. 3. Re F (j�) � 0 for all �. 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 j�-axis are positive and real. Thus, if F(s) has a pair of poles at s � ± j�0, a partial-fraction expansion gives terms of the form

10.3�Positive Real Functions�10.17

K1 K1* � s � j� 0 s � j� 0 Since residues of complex conjugate poles are themselves conjugate, K1 � K1* and should be positive and real. Condition (3) requires that Re F(j�) must be positive and real for all �. Now, to compute Re F(j�) 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 � j� in the even polynomial gives the real part of F(s) and substituting s � j� into the odd polynomial gives imaginary part of F(s). Ev F ( s) s � j� � Re F ( j� ) Od F ( s) s � j� � j Im F ( j� ) We have to test Re F(j�) � 0 for all �. The denominator of Re F(j�) is always a positive quantity because m22 � n22

s � j�

�0

Hence, the condition that Ev F(j�) should be positive requires m1m2 � n1n2

s � j�

� A(� 2 )

should be positive and real for all � � 0.

���������������

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. 10.1. Thus, pole and zero are in the left half of the s-plane. (b) There is no pole on the j� axis. Hence, the residue test is not carried out.

s −3 −2 −1 0

���������

10.18�Circuit Theory and Networks—Analysis and Synthesis (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(� 2 ) � m1m2 � n1n2 |s � j� � (3)(1) � ( s)( s) |s � j� � 3 � s 2 |s � j� � 3 � � 2 A(�2) is positive for all � � 0. Since all the three conditions are satisfied, the function is positive real.

��������������� 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) � � D( s) s 2 � 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. 10.2. Thus, all the poles and zeros are in the left half of the s plane. (b) Since there is no pole on the j� axis, the residue test is not carried out.

−7 −6 −5 −4 −3 −2

−1

0

s

���������

(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(� 2 ) � m1m2 � n1n2 |s � j� � (s 2 � 5) (s 2 � 14) � �6s��9s� |s � j� � s 4 � 35s 2 � 70 |s � j� � � 4 � 35� 2 � 70 A(�2) is positive for all � � 0. Since all the three conditions are satisfied, the function is positive real.

��������������� 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

10.3�Positive Real Functions�10.19

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. 10.3. Thus, all the poles and zeros are in the left half of the s plane. (b) There is no pole on the j� axis, hence the residue test is not carried out.

jw

s −5

−4

−3

−2

−1

0

(c) Even part of N ( s) � m1 � 8s 2 Odd part of N ( s) � n1 � s3 � 15s

���������

Even part of D( s) � m2 � s � 4 Odd part of D( s) � n2 � 5s 2

A(� 2 ) � m1m2 � n1n2 |s � j� � (8s 2 )( s 2 � 4) � ( s3 � 15s)(5s) |s � j� � 3s 4 � 43s 2 |s � j� � 3� 4 � 43� 2 A(�2) is positive for all � � 0. Since all the three conditions are satisfied, the function is positive real.

��������������� 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. 10.4. Thus, all the poles and zeros are on the j� axis. (b) The poles on the j� axis are simple. Hence, residue test is carried out.

j1 0 −j1 −j2

F ( s) �

s2 � 1 s3 � 4 s



s2 � 1 s( s 2 � 4 )

���������

By partial-fraction expansion, F ( s) �

K1 K2 K 2* � � s s � j2 s � j2

The constants K1, K2 and K2* are called residues. K1 � s F ( s) |s � 0 �

s2 � 1 s �4 2

� s�0

K 2 � ( s � j 2) F ( s ) | s � � j 2 � 3 K 2* � K 2 � 8 Thus, residues are real and positive.

1 4

s2 � 1 s ( s � j 2)

� s� � j 2

3 �4 � 1 � ( � j 2)( � j 2 � j 2) 8

s

10.20�Circuit Theory and Networks—Analysis and Synthesis (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.

���������������

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 2 s3 � 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.

10.3�Positive Real Functions�10.21

� � s 2 � 1� 2 s3 � 2 s 2 � 3s � 2 � 2 s � 2 � � � 2s

2 s3

2s � s � 2 2

�2

2s2 s F ( s) � 2 s � 2 �

s s �1 2

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

10.22�Circuit Theory and Networks—Analysis and Synthesis 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 j� 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(� 2 ) � m1m2 � n1n2 |s � j� � (6s 2 � 3)(s 2 � 1) � ( s3 � 7s)( 2s) |s � j� � 4s 4 � 5s 2 � 3 |s � j� � 4� 4 � 5� 2 � 3 A(�2) is positive for all � � 0. Since all the three conditions are satisfied, the function is positive real.

���������������

Test whether F ( s) �

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 j� 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

10.3�Positive Real Functions�10.23

A(� 2 ) � m1m2 � n1n2 |s � j� � ( s 2 � 6)(s 2 � 1) � ( s)(s) |s � j� � s 4 � 6s 2 � 6 |s � j� � � 4 � 6� 2 � 6 For � � 2, A(�2) � 16 − 24 � 6 � −2 This condition is not satisfied. Hence, the function F(s) is not positve real.

���������������

Test whether F ( s) �

s2 � 4 is positive real function. s � 3s2 � 3s � 1 3

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 j� 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(� 2 ) � m1m2 � n1n2 |s � j� � ( s 2 � 4)(3s 2 � 1) � (0)( s3 � 3s) |s � j� � 3s 4 � 13s 2 � 4 |s � j� � 3� 4 � 13� 2 � 4 For � � 1, A(�)2 � 3 − 13 � 4 � −6 This condition is not satisfied. Hence, the function F(s) is not positive real.

���������������

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 j� axis are not simple. Hence, the function F (s) not positive real.

���������������

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.

10.24�Circuit Theory and Networks—Analysis and Synthesis 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.

�������������������������������������� 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)

Z(s)

Z2(s)

Here, Z1(s) is said to have been removed from Z(s) in forming the new function Z2(s) as shown in Y2(s) Y1(s) Y(s) Y(s) Fig. 10.5. If the removed network is associated with the pole or zero of the original network impedance then that pole or zero is also said to ���������������������������������������������������� have been removed. ������������������������ There are four important removal operations.

������� ���������������������������� 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 n � bn �1s n �1 � � � b1s � b0 a H � n �1 bn Z1 ( s) � Hs cn s n � cn �1s n �1 � � � c1s � c0

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. 10.6(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 ���������������������������������������������������� ������������������ capacitor of value C � H as shown in Fig. 10.6(b). and

Z 2 ( s) �

� Hs �

10.4�������������������������������10.25

10.4.2

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

Z2(s)

Y(s)

L

Z2(s)

Thus, removal of a pole from the impedance function Z(s) at the origin corresponds to the (a) (b) removal of a capacitor, and from admittance function Y(s) corresponds to removal of an ������������������������������������������������������ �������������� inductor as shown in Fig. 10.7.

10.4.3

Removal of Conjugate Imaginary Poles

If Z(s) contains poles on the imaginary axis, i.e., at s � ± j�1 then Z(s) will have factors (s � j�1) (s − j�1) � s2 � �12 in the denominator polynomial Z ( s) �

p( s) ( s � �12 ) q1 ( s) 2

By partial-fraction expansion, Z ( s) �

K1 K1* � � Z 2 ( s) s � j�1 s � j�1

For a positive real function, j� axis poles must themselves be conjugate and must have equal, positive and real residues. K1 � K1* 2K s Hence, Z ( s) � 2 1 2 � Z 2 ( s) s � �1 2K s 1 1 � Thus, Z1 ( s) � 2 1 2 � 2 Y � s � �1 s � a Yb � 1 2 K1 2 K1s where Ya � and Yb �

s 1 is the admittance of a capacitor of value C � 2 K1 2 K1

�12 2K is the admittance of an inductor of value L � 21 2 K1s �1

10.26�Circuit Theory and Networks—Analysis and Synthesis If the given function is an admittance function Y(s) then 2K s 1 � Y1 ( s) � 2 1 2 � s � �1 Z a � Zb where Z a �

1 s �2 � 1 2 K1 2 K1s

�2 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 �12 Thus, removal of conjugate imaginary poles from impedance function Z(s) corresponds to Z(s) C Z2(s) Y(s) Z2(s) L C the 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 ���������������������������������������������������� in Fig. 10.8. �������������������������

10.4.4

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) �

��������������� 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. Z(s) s The impedances are connected in the series branches whereas the admittances are connected in the parallel branches. The network is shown in Fig. 10.9. Y3 ( s) �

1H 1 F 2

���������

1H

10.4�������������������������������10.27

���������������

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 �. 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. 10.10. Y4 ( 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 Z ( s) � 3 �

5s � 4 2

2 s3 � 2 s

2

�4

� Z1 (ss) � Z 2 ( s)

Z1(s) � 3 represents the impedance of a resistor of value 3 �. Y2 ( s) �

1 2 s3 � 2 s � 2 Z 2 ( s) 5s � 4

1 H 5

1�

Z(s)

6 s3 � 5 s 2 � 6 s � 4 2 s3 � 2 s

.

10.28�Circuit Theory and Networks—Analysis and Synthesis 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. 10.11. Z5 ( s) �

���������������

3�

Z(s)

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) �

1 s3 � 2 s � 2 Z 2 ( s) 8s � 7

25 H 2

2 F 5

����������

1 F 10

10.4�������������������������������10.29

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 Z(s) 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. 10.12. Z5 ( s) �

���������������

Realise the network having admittance function Y ( s) �

Solution By long division of Y(s), � � 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)

64 H 9

1H

1 F 8

���������� 4 s2 � 6 s . s�1

9 F 56

10.30�Circuit Theory and Networks—Analysis and Synthesis 1 1 �� represents the impedance of a resistor of value 2 2 1 Z 4 ( s) � represents the impedance of a capacitor of value 2 F. Z(s) 2s The impedances are connected in the series branches, whereas the admittances are connected in the parallel branches. The network is shown in Fig. 10.13. Z3 ( s) �

���������������

Realise the admittance function Y ( s) �

1 � 2

2F

4F

����������

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 �. 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 �. 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. 10.14. Z 2 ( s) �

7 F 8

4 � 7

4 � 3

�����������

������������������������LC���������� 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 j�-axis. The poles and zeros interlace on the j�-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.

10.5�����������������������������10.31

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).

10.5.1 Foster Realisation Consider a general LC function F(s) given by F ( s) �

H ( s 2 � �12 )( s 2 � � 32 )... s( s 2 � � 22 )( s 2 � � 42 )...

where 0 � �12 � � 22 � � 32 � and H is positive. By partial-fraction expansion of F(s), F ( s) �

K0 K2 K2 � � � ... � K � s s s � j� 2 s � j� 2

Combining terms with conjugate poles, F ( s) �

K0 2K 2 s � ... � K � s s s � � 22

where K0, Ki and K� are the residues of F(s) at the origin, at j�i and at infinity respectively. These residues are given by K 0 � sF ( s)�s � 0 Ki � K� � Foster I Form

�s

2



� � i2 F ( s) 2s

s 2 � � � 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 � �2

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 � �i 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 � �i Ci s � s2 � Li s Li Ci 1

10.32�Circuit Theory and Networks—Analysis and Synthesis Ci �

1 2K and Li � 2i 2 Ki �i

����������������������������������������������������� 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 � s 2 � � i2 s 2 � 1 Li Ci

K� s � Ls

� i2

L� � K �

The network corresponding to Foster I form is shown in Fig. 10.15. C0

Z(s)

L1

Li

C1

Ci

L�

�������������������������������������� 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

2K2 s K0 � + �� K � s � Y1 � s � � Y2 � s � � �Yn � s � s s 2 � � 22

K0 1 represents the admittance of an inductor of henry. K0 s

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 � � i2 2

represent the admittance of a series combination of an inductor Li and

10.5�����������������������������10.33

For series combination of Li and Ci, � 1� �� L �� s 2K s i Y ( s) � � � 2 i 2 1 1 s � �i Li s � s2 � Ci s Li Ci 1

Li � ����������

1 2 Ki

and Ci �

2 Ki

� i2

�������������������������������������������

Admittance function

Element

K0 1 � s L0 s

� 1� �� L �� s i � s 2 � � i2 s 2 � 1 Li Ci 2 Ki s

L0 �

1 K0

Li �

1 2 Ki 2 Ki

Ci �

K� s � Cs

� i2

C� � K �

The network corresponding to the Foster II form is shown L0 L1 Li Y(s) C� in Fig. 10.16. 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� ��������������������������������������� is not present.

10.5.2 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.

10.34�Circuit Theory and Networks—Analysis and Synthesis 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. 10.17. L1

Z1

L3

Z(s)

C2

C4

Z3 Y2

Z(s)

(a)

Y4

(b)

������������������������������������� 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. 10.18. L2

Z(s)

Z2

L4

C1

C3

(a)

C5

Z(s)

Z4

Y1

Y3

Y5

(b)

������������������������������������� 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)

L2

L4

Thus, the final structure is a ladder network whose first element is a series capacitor and second element is a shunt inductor as �������������������������������������� shown in Fig. 10.19.

10.5�����������������������������10.35 C4 C2 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. 10.20. Z(s) L5 L1 L3 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) including any at infinity. ��������������������������������������

��������� ������

State whether the following functions are driving point immittance of LC

networks are not: (a) Z ( s) �

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

Solution (a) Z ( s) �

j1

5 s( s 2 � 4 ) ( s 2 � 1)( s 2 � 3)

−3 −2 −1

s

−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. 10.21. Since the poles and zeros do not interlace on the j� axis, the function Z(s) is not an LC impedance function. (b) Z ( s) �

0

−j √3 −j2 −j3

2( s 2 � 1)( s 2 � 9)

����������

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. 10.22. The poles and zeros are simple and lie on the j�-axis. The poles and zeros interlace on the j� 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

����������

10.36�Circuit Theory and Networks—Analysis and Synthesis

��������������� 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. 10.23. From the pole-zero diagram, it is clear that poles and zeros are simple and lie on the j� 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 � 1)( s � 9) 2

2

s( s � 4 ) 2

4 s � 40 s � 36 4



2

4 s � 16 s

−j2 −j3

s � 4s 3

s3 � 4 s) 4 s 4 � 40 s 2 � 36 ( 4 s 4

−j1

����������

2

24 s 2 � 36 24 s � 36 24 s 2 � 36 Z ( s) � 4 s � 3 � 4s � s � 4s s( s 2 � 4 ) 2

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

� s � �4 2

4( �4 � 1)( �4 � 9) 15 � 2( �4) 2

9 15s Z ( s) � 4 s � � 2 s s �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 s2 � LC By direct comparison, C�

1 F 15

10.5�����������������������������10.37

L�

15 H 4

1 F 9

4H

The network is shown in Fig. 10.24.

Foster II Form The Foster II form is obtained by partialfraction expansion of the admittance function Y(s). Y ( s) �

15 H 4

1 F 15

s( s 2 � 4 ) 4( s 2 � 1)( s 2 � 9) ����������

By partial-fraction expansion, Y ( s) � K1 �

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 ( s 2 � 1) Y ( s) 2s

( s 2 � 9) K2 � Y ( s) 2s Y ( s) �

� 3� �� �� s 32 s2 � 1





( �1 � 4) 3 � 8( �1 � 9) 64



( �9 � 4) 5 � 8( �9 � 1) 64

s � �1 2

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 2 s � 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. 10.25.

3 F 32

5 F 288

32 H 3

32 H 5

����������

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.

10.38�Circuit Theory and Networks—Analysis and Synthesis 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. 10.26.

48 H 5

4H

1 F 24

5 F 72

����������

Cauer II Form The Cauer II form is obtained from continued fraction expansion about pole at origin. Z ( s) �

4( s 2 � 1)( s 2 � 9) s( s � 4 ) 2



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 � � � � �s 36 � 9 s 2 � 4 � �Y 31s 2 � 4 s 4 � 4 s � s3 � � � 31s 4s �

16 3 s 31 15 3 � � 961 �Z s � 31s 2 � 4 s 4 � � 15s 31 � 31s 2

10.5�����������������������������10.39

� 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. 10.27.

����������������

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) s( s 2 � 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 � 3 2

s3 � 2 s

� 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 ( s 2 � 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

10.40�Circuit Theory and Networks—Analysis and Synthesis For a parallel LC network,

By direct comparison,

� 1� �� �� s C Z LC ( s) � 1 2 s � LC

1H

C � 2F 1 L= H 4 The network is shown in Fig. 10.28.

2 F 3

1 H 4

2F

����������

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 � 1)( s 2 � 3) 2

By partial-fraction expansion, Y ( s) � where

K1 � K2 �

Y ( s) �

2 K1s 2 K 2 s K1 K* K2 K 2* � 1 � � � � s � j1 s � j1 s � j 3 s � j 3 s 2 � 1 s 2 � 3 ( s 2 � 1) Y ( s) 2s ( s � 3) Y ( s) 2s

( �1 � 2) 1 � 2( �1 � 3) 4



1 �3 � 2 � 2( �3 � 1) 4

s � �1

2

� 1� �� �� s 2

� 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 The network is shown in Fig. 10.29. L1 � 2H, C1 �

1 F 2

1 F 6

2H

2H

����������

��������������� 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).

10.5�����������������������������10.41

By partial-fraction expansion, Z ( s) �

2K s 2K s K0 K1 K1* K2 K 2* K � � � � � 0� 2 1 � 2 2 s s s �2 s �4 s � j 2 s � j 2 s � j2 s � j2

K 0 � sZ ( s) s � 0 �

where

K1 � K2 �

(1)(3) 3 � ( 2)( 4) 8

( s 2 � 2) 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

( s 2 � 4) Z ( s) 2s � 1� �� �� s 4



s � �4 2

� 3� �� �� s 8

3 8 Z ( s) � � 2 � s s � 2 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 2 s � 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 ���������� The network is shown in Fig. 10.30. 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 2 � 1)( s 2 � 3)

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

10.42�Circuit Theory and Networks—Analysis and Synthesis 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 2 s � 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 ���������� The network is shown in Fig. 10.31.

��������������� 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 � 2s 2 � 1� 2 s � Z 2s2 1� s � s � Y s 0

10.5�����������������������������10.43

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. 10.32. 5H

2H

1F

1F

���������� Cauer II Form origin.

The Cauer II form is obtained from continued fraction expansion about the pole at the Z ( s) �

10 s 4 � 12 s 2 � 1

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 � � � � � 2s ��

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 � 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. 10.33.

2F

2 F 121

11 H 2

����������

110 H 2

10.44�Circuit Theory and Networks—Analysis and Synthesis

��������������� Realise the following network function in Cauer I form: Z ( s) �

6 s4 � 42 s 2 � 48

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. 10.34. 6 H 11

1F 6

49284 H 33792

121 F 222

����������

128 F 444

10.5�����������������������������10.45

��������������

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 � 3s3 � s5 � � 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. 10.35.

���������������

1H

1H

����������

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.

10.46�Circuit Theory and Networks—Analysis and Synthesis � � 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. 10.36. 1H

1H 2

1H

1F 2

1F

����������

��������������� Synthesize the following LC impedance function in Cauer II form: Z ( s) �

s3 � 2 s s � 4 s2 � 3 4

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

10.6�����������������������������10.47

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. 10.37.

5F 4

2H 3

2 H 25

5F

����������

������������������������RC���������� 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 d� 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 � �1 )( s � � 3 )... s( s � � 2 )...

10.6.1 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 � �1 s � � 2

10.48�Circuit Theory and Networks—Analysis and Synthesis where K0, K1, K2, … K� are residues of Z(s). K o � sZ ( s) s � 0 K i � ( s � � i ) Z ( s) s � �� 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 � �i capacitor Ci. For parallel combination of Ri and Ci, � 1 � Ri � � Ci s �� Ki Z ( s) � � 1 s � �i Ri � Ci s Ri �

Ki 1 and Ci � �i Ki

Table 10.3������������������������������������������� Impedance function

K0 1 � s C0 s

Ki � s � �i

Element

1 K0 Ki Ri � �i

C0 �

1 Ci s�

1 Ri Ci

K � � R�

Ci �

R� = K�

The network corresponding to the Foster-I form is shown in Fig. 10.38. 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) �������������������������������������� . pole, Foster II form is obtained by expanding s Y ( s) K o n Ki � �� � K� s s i �1 ( s � � i )

10.6�����������������������������10.49

Multiplying this equation by s, n

Ki s � K� s ( s � �i ) i �1

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 � �i Ci with Ri �

1 K ohms and Ci � i farads. �i Ki ������������������������������������������������������ Admittance function

Element

1 K0 � R0

Ro �

1 Ko

1 Ki K Ci � i �i

� 1� �� R �� s Ki i � 1 s � �i s� Ri Ci

Ri �

C� � K �

K� s � C� s The network corresponding to the Foster II form is shown in Fig. 10.39. Z(s)

R0

R1

Ri

C1

C�

Ci

���������������������������������������

10.6.2 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. 10.40.

1 C2 s �

1 R3 �

1 C4 s � ...

10.50�Circuit Theory and Networks—Analysis and Synthesis In the network shown in Fig. 10.40, 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

Rn − 1

C2

Z(s)

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 ������������������������������������� 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 1 C3 s � ... R4 The network is shown in Fig. 10.41. In the network shown in Fig. 10.41, if Z (s) has a pole at s � 0, �������������������������������������� the 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) �

��������������� (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. 10.42. 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

����������

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. 10.43. 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

����������

0

10.6�����������������������������10.51

���������������

Realise the Foster and Cauer forms of the impedance function Z ( s) �

( s � 1)( s � 3) s( s � 2 )

jw

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. 10.44. 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 � � 2 s � 2 s� s 2 � 4 s � 3 � 1 � �

s −3

−2

−1

0

����������

s2 � 2s 2s � 3 Z ( s) � 1 �

2s � 3 s2 � 2s

� 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 �. 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 1� Ci Z RC ( s) � 1 s� Ri Ci 1 � By direct comparison, 2F 4 1 R� � 4 C�2F The network is shown in Fig. 10.45. ���������� Z ( s) � 1 �

10.52�Circuit Theory and Networks—Analysis and Synthesis Foster II Form Y ( s) . s

The Foster II form is obtained by the partial-fraction expansion of admittance function 1 s( s � 2) � Z ( s) ( s � 1)( s � 3) Y ( s) s�2 � s ( s � 1)( s � 3) Y ( s) �

By partial-fraction expansion, Y ( s) K K � 1 � 2 s s �1 s � 3 Y ( s) ( �1 � 2) 1 K1 � ( s � 1) � � s s � �1 ( �1 � 3) 2 Y ( s) ( �3 � 2) 1 � K 2 � ( s � 3) � �3 � 1) 2 s s � �3 (�

where

1 1 Y ( s) � 2 � 2 s s �1 s � 3 1 1 s s Y ( s) � 2 � 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 �� C2 � F 6 R1 � 2 ��

C1 �

The network is shown in Fig. 10.46.

2�

2�

1F 2

1F 6

����������

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

10.6�����������������������������10.53

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. 10.47.

1�

4� 1F 2

1F 6

����������

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 � s2 � � Y � �5 2 4 2s � s2 5 1 2� 5 � 25 �Z s � s � s2 � � 2s 5 �2 5 s 2 � 1 �1 s2 � s2 � � Y � 5 �5 1 2 s 5 0

2F 3

2 25 F

5 � 4

The impedances are connected in the series branches whereas admittances are connected in the parallel branches. The network is shown in Fig. 10.48.

����������

5�

10.54�Circuit Theory and Networks—Analysis and Synthesis

��������������� 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 ( �2 � 1)( �2 � 3) ( �1)(1) 1 � � ( �2)( �2 � 4) ( �2)( 2) 4 ( �4 � 1)( �4 � 3) ( �3)( �1) 3 � � � ( �4)( �4 � 2) ( �4)( �2) 8

K1 � ( s � 2) Z ( s) s � �2 � K 2 � ( s � 4) Z ( s) s � �4

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 �� C1 � 4 F 8 3 8 �� C2 � F R2 � 32 3 R1 �

1� 8

4F

3 � 32

8F 3

����������

The network is shown in Fig. 10.49.

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

10.6�����������������������������10.55

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 �, C1 � F 3 2 1 R2 � 2 �� C2 � F 6 R1 �

1F

2� 3

2�

3F 2

1F 6

����������

The network is shown in Fig. 10.50.

��������������� 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.

10.56�Circuit Theory and Networks—Analysis and Synthesis 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 s � 4s � 3 2

� 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 �. The remaining terms represent the impedance of a parallel RC circuit for which 1 Ci Z RC ( s) � 1� 1 3 3� s� 2� Ri Ci By direct comparison, 1 1F 1F R1 � 3 �� C1 � F 3 3 1 R2 � �� C2 � 1 F 3 ���������� The network is shown in Fig. 10.51. 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

10.6�����������������������������10.57

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 �. 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 � �� C2 � F 3 64 The network is shown in Fig. 10.52. R1 � 8 ��

C1 �

16 � 3

8� 1 F 16

16 � 3

3 F 64

����������

��������������� 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

10.58�Circuit Theory and Networks—Analysis and Synthesis 15 � 7 � 49 � s � 6 �� s � Y 7�2 30 7 s 2 � 15 � 5 6� � � Z � 7 � 14 15 7 0

1 � 2

The impedances are connected in the series branches whereas the admittances are connected in the parallel branches. The network is shown in Fig. 10.53.

8 � 7

1 F 2

5 � 14

49 F 30

����������

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 2 6 � 4s � s 2 3 2� �3 4 s � s � 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

10.6�����������������������������10.59

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. 10.54.

1 F 3

5 F 49

14 � 5

7 � 8

2�

����������

��������������� Realise the RC impedance in Cauer I and Foster I forms Z ( s) �

s�4 ( 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



1

12 4 � �1 12� 1 � � Z � � 12 1 1F 4F 0 The impedances are connected in series branches, whereas the admittances are connected in parallel branches. The network is shown ���������� in Fig. 10.55. 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



10.60�Circuit Theory and Networks—Analysis and Synthesis ( �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 2 � 2 Z ( s) � 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

1 � 12

1� 4

By direct comparison, 1 �, C1 � 2 F 4 1 �, C2 � 2 F R2 � 12 R1 �

2F

2F

����������

The network is shown in Fig. 10.56.

The RC driving-point impedance function is given as Z ( s) � H Realise the impedance function in the ladder form, given Z( �2) � 1.

��������� ������

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) Z ( s) � 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 � � s 2 � 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

10.6�����������������������������10.61

The impedances are connected in the series branches whereas admittances are connected in the parallel branches. The network is shown in Fig. 10.57.

1�

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) �

2�

1 F 4

����������

4 � 5s � s 2 3s � s 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 � 9

The impedances are connected in the series branches whereas the admittances are connected in the parallel branches. The network is shown in Fig. 10.58.

���������������

11 � 2

����������

An impedance function has the pole-zero diagram as shown in Fig. 10.59. 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

����������

10.62�Circuit Theory and Networks—Analysis and Synthesis 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. 10.60. Foster II Form The Foster II form is obtained by partial fraction Y ( s) . expansion of s

2�

8� 1 F 4

����������

1 F 12

10.7�����������������������������10.63

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 where

K1 � ( s � 1)

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 4 Y ( s) � � 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 4�

By direct comparison, 1 R1 � 4 �, C1 � F 4 R2 � 4 �, C2 �

1 F 12

1 F 4

4� 1 F 12

����������

The network is shown in Fig. 10.61.

������������������������RL���������� RL 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 is a zero which may be at s � 0. 4. The highest critical frequency is a pole which may be at infinity. Z ( s) 5. 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 d� 7. Z RL � 0 � � Z RL (�) .

10.64�Circuit Theory and Networks—Analysis and Synthesis 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.

��������������� 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) �

s( s � 4)( s � 8) ( s � 1)( s � 6)

(b) Z ( s) �

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.

���������������

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

10.7�����������������������������10.65

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 where

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

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 � . 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 � �, L1 � H 4 8 5 5 H R2 � �, L2 � 4 24

1 � 2

1� 4

5 � 4

1 H 8

5 H 24

����������

The network is shown in Fig. 10.62. 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

10.66�Circuit Theory and Networks—Analysis and Synthesis �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 �. 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 4� By direct comparison, 4� 5 4 4 2� R1 � �, L1 � H 4 5 5 4 H H 5 3 4 R2 � 4 �, L2 � H 3 ���������� The network is shown in Fig. 10.63. Y1 ( s) �

��������������� 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

10.7�����������������������������10.67

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 Z ( s) � � 3 � 5 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 � �, L1 � H 3 9 2 2 H R2 � �, L2 � 5 25 The network is shown in Fig. 10.64.

4 � . The other two terms represent the 15

4 � 15

1� 3

2 � 5

1 H 9

2 H 25

����������

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 3 � 3 Y ( s) � 1 � s �1 s � 4

10.68�Circuit Theory and Networks—Analysis and Synthesis The first term represents the admittance of a resistor of 1 �. 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 3� By direct comparison, 12 � 1�

3 3 �, L1 � H 8 8 R2 � 12 �, L2 � 3 H R1 �

3 H 8

3H

����������

The network is shown in Fig. 10.65.

���������������

8

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 2s 2 �

9 s 2 7 � �8 s � 6� 4 s � 9 � � Y � �7 2 48 4s � 7

10.7�����������������������������10.69

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

2�

The impedances are connected in the series branches whereas the admittances are connected in the parallel branches. The network is shown in Fig. 10.66.

49 H 30

7� 8

42 � 15

����������

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

10.70�Circuit Theory and Networks—Analysis and Synthesis The impedances are connected in the series branches whereas the admittances are connected in the parallel branches. The network is shown in Fig. 10.67. 1� 2

5 � 14

8� 7

1 H 3

10 H 98

����������

��������������� 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) 5 � 1� � 5 s s �1 s � 6 21 4 s s 5 � 5 Z ( s) � s � s �1 s � 6

10.7�����������������������������10.71

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 � 4 � 5 5 Ri s Z RL ( s) � 1H R s� i Li 21 4 H H By direct comparison, 5 30 21 21 �, L1 � H R1 � 5 5 4 4 ���������� H R2 � �, L2 � 5 30 The network is shown in Fig. 10.68. 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 � � 140 0 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. 10.69.

10.72�Circuit Theory and Networks—Analysis and Synthesis 25 H 9

1H

28 H 18

140 � 27

5�

����������

Exercises 10.1 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 10.2 Determine whether the following functions are positive real: (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 10.3 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)

����������10.73

(iii) Z ( s) �

( s � 1)( s � 8) ( s � 2)( s � 4) Ks( s � 4) 2

(iv)

Z ( s) �

(v)

Z ( s) �

(vi)

Z ( s) �

(vii)

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)

Z ( s) � Z ( s) �

( s 2 � 1)( s 2 � 3) 2( s 2 � 1)( s 2 � 9) s ( s 2 � 2) 4( s � 1)( s � 3) s( s � 2) ( s 2 � 1)( s 2 � 3) s( s 2 � 2)

Z ( s) �

Z ( s) �

(ii)

Z ( s) �

(iii)

Z ( s) �

(iv) Y ( 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) �

s( s 2 � 4)( s 2 � 16) ( s 2 � 1)( s 2 � 8) s( s 2 � 4 )

3( s � 2)( s � 4) s( s � 3) 2( s 2 � 1)( s 2 � 9)

(iv) Z ( s) �

4( s � 1)( s � 3) ( s � 2)( s � 6) 4( s 2 � 4)( s 2 � 25) s( s 2 � 16)

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 ) ( s � 1)( s � 3) s( s � 2)

(vi) Z ( s) �

s�4 ( s � 2)( s � 6)

Z ( s) �

(vii)

Z ( s) �

(iii)

4( s � 1)( s � 3) F ( s) � ( s � 2)( s � 6)

(iv)

s�4 Z ( s) � ( s � 2)( s � 6)

(viii)

Z ( s) �

(v)

( s � 1)( s � 4) Z ( s) � s ( s � 2)

s( s 2 � 4 )

( s � 2)( s � 5) s( s � 4)( s � 6)

s( s 2 � 4 )

F ( s) �

(v)

(ii)

(vi) Y ( s) �

2( s 2 � 1)( s 2 � 9)

( s � 2)( s � 5) s( s � 4)( s � 6) 10.6 Realise the following functions in Cauer I form: Z ( s) �

( s 2 � 9)( s 2 � 25)

s2 � 2s � 2

s2 � s � 1 10.5 Realise the following functions in Foster II form: 3( s � 2)( s � 4) (i) Z ( s) � s( s � 3)

(v)

10.4 Realise the following functions in Foster I form: (i)

(vii)

(ix) Z ( s) � (x)

Z ( s) �

6( s � 2)( s � 4) s( s � 3) 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

10.74�Circuit Theory and Networks—Analysis and Synthesis 10.7 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)

(viii)

Z ( s) � Z ( s) �

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) 2( s 2 � 1)( s 2 � 9) s( s 2 � 4 )

���������� 10.9 An impedance function has the pole-zero diagram as shown in Fig. 10.71. 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)

���������� 10.10 For the realisation of a given function F(s).

s( s 2 � 2)( s 2 � 5) ( s 2 � 1)( s 2 � 3) s2 � 2s � 2 s2 � s � 1

10.8 An impedance function has the pole-zero diagram as shown in Fig. 10.70 below. If Z (−2) � 3, synthesise the impedance function in Foster and Cauer forms.

F ( s) �

n K0 sK � � 2 i 2 � sK � s i �1 ( s � � 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, �1 � 4, K2 � 10, �2 � 8, K� � 5, find the component values of realised network for F(s) � Z(s) and F(s) � Y(s). Draw neat diagrams.

Objective-Type Questions 10.1 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 10.2 The number of roots of s3 � 5s2 � 7s � 3 � 0 in the left half of s-plane is

(a) zero (c) two

(b) one (d) three

10.3 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) (b) (c) (d)

RL network only RC network only LC network only RC as well as RL network

�������������������������10.75

10.4 The pole-zero pattern of a particular network is shown in Fig. 10.72. It is that of an jw j2 j1 s −j1 −j2

���������� (a) LC network (c) RL network

(b) RC network (d) none of these

10.5 The first critical frequency nearest to the origin of the complex frequency plane for an RL driving-point impedance function will be (a) (b) (c) (d)

a zero in the left-half plane a zero in the right-half plane a pole in the left-half plane a pole in the right-half plane

10.6 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? (b) P2 (a) P1 (c) P3 (d) P4 10.7 For very high frequencies, the driving-point 4( s � 1)( s � 3) admittance function, Y ( s) � s ( s � 2)( s � 4) behaves as 3 � 2 (b) a capacitance of 4 F 1 (c) an inductance of H 4 (d) an inductance of 4 H (a) a resistance of

s�3 10.8 The driving-point impedance Z ( s) � s �4 behaves as

(a) (b) (c) (d)

a resistance of 0.75 � at low frequencies a resistance of 1 � at high frequencies both (a) and (b) above none of the above

10.9 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 (b) s � −1, s � −3 (c) s � 0, s � −1 (d) s � −3, s � −4 10.10 Consider the following from the point of view of possible realisation as driving-point impedances using passive elements: 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 10.11 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 10.12 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)

10.13 The circuit shown in Fig. 10.73 is

���������� (a) Cauer I form (b) Foster I form (c) Cauer II form (d) Foster II form

10.76�Circuit Theory and Networks—Analysis and Synthesis 10.14 For an RC driving-point impedance function, the poles and zeros (a) should alternate on the real axis (b) should alternate only on the negative real axis

(c) should alternate on the imaginary axis (d) can lie anywhere on the left half-plane

Answers to Objective-Type Questions 10.1 (a)

10.2 (a)

10.3 (b)

10.4 (a)

10.5 (a)

10.6 (b)

10.7 (c)

10.8 (c)

10.9 (b)

10.10 (a)

10.11 (b)

10.12 (c)

10.13 (b)

10.14 (b)

11 Filters

�������������������� 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.

��������������������������������� 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

����������������� Figure 11.1 shows a T network.

1�

2�

Fig. 11.1�����������

11.2���������������������������������������������������

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 � � 2 � Z1 Zin � � Z1 2 � Z 2 � Z0 2 But

Zin � Z 0 �Z � 2Z 2 � 1 � Z 0 � � 2 � Z1 Z0 � � 2 Z1 � 2Z 2 � 2Z 0 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 � Z1Z 2 4 Z12 � Z1Z 2 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 � Z 02T � � � � � � 2Z1 � 4 Z 2 � 2 4

Z0T � Zoc Z sc

11.3.2 Propagation Constant The propagation constant � of the network in Fig. 11.2 is given by,

11.3�����������11.3

� � log e

I1 I2

Z1

Z1

2

2

1

Applying KVL to Mesh 1,

2

� Vs

Z Vs � 1 I1 � Z 2 ( I1 � I 2 ) � 0 2 �Z � Vs � � 1 � Z 2 � I1 � Z 2 I 2 � 2 �

Z2 �

I1

Z0 I2

1�

2�

Fig. 11.2���������� 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 � e� I2 Z2 Z1 � Z 2 � Z 0 � Z 2 e� 2 Z0 � Z 2 (e� � 1) �

Z1 2

The characteristic impedance of the T-network is given by, Z12 � Z1Z 2 4

Z0 �

Z12 Z � Z1Z 2 � Z 2 (e� � 1) � 1 4 2 Z12 Z2 � Z1Z 2 � Z 22 (e� � 1) 2 � 1 � Z1Z 2 (e� � 1) 4 4 Z 22 (e� � 1) 2 � Z1Z 2 (1 � e� � 1) � 0 Z 22 (e� � 1) 2 � Z1Z 2 e� � 0 Z 2 (e� � 1) 2 � Z1e� � 0 (e� � 1) 2 � e 2� � 2e� � 1 � e �� (e 2� � 2e� � 1) �

Z1 � e Z2 Z1 Z 2 e �� Z1 Z2

11.4��������������������������������������������������� e� � e �� � 2 � Dividing both the sides by 2, Z e� � e �� � 1� 1 2 2Z 2 Z cosh � � 1 � 1 2Z 2

Z1 Z2

2

� Z Z2 Z � sinh � � cosh 2 � � 1 � �1 � 1 � � 1 � 1 � 1 � 12 � 1 � Z 2 4Z 2 � 2Z 2 �

Z1 � Z1 � � Z 2 �� 2Z 2 ��

2

Z2 Z 1 Z1Z 2 � 1 � 0T Z2 4 Z2 Z 0T Z0T sinh � tan nh � � � � cosh � � Z1 � Z � Z1 Z 2 �1 � 2 2 � 2Z �� �

2

But

Z0T � Zoc Z sc

and

Zoc �

Z1 � Z2 2

tanh � �

Z sc Zoc

Also,

sinh

� � 2

1 (cosh � � 1) � 2

� Z 1� 1 � 1 � 1� � � 2 � 2Z 2 �

Z1 4Z 2

����������������� Figure 11.3 shows a �-network. Z1 1

2

2 Z2

2 Z2

1�

Z0

2�

Fig. 11.3��-�������

11.4.1

Characteristic Impedance

For a � network, the value of input impedance when it is terminated by impedance Z0, is given by, � �Z 2 Z0 � 2Z 2 � Z1 � � Z 2 � Z0 �� � Zin � �Z 2 Z0 Z1 � � �Z2 �Z 2 � Z0

11.4�����������11.5

Zin � Z 0

But

�Z 2 Z0 � � 2Z 2 � Z1 � � � Z 2 � Z 0 �� Z0 � �Z 2 Z0 � �Z 2 Z1 � �Z 2 � Z0 Z 0 Z1 �

� Z 2 Z 02 � Z 2 ( 2Z1Z 2 � Z0 Z1 � 2Z0 Z 2 ) � 2Z 0 Z 2 � �Z 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 � network is given by Z0� �

Z1Z 2 � Z 1� 1 4Z 2

Z1Z 2 Z1Z 2 �

Z12 4

Z12 � Z1Z 2 4 ZZ � 1 2 Z0T

Z0T �

But

Z0�

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 � 4Z 2 1� 1 4Z 2

Z0� � Zoc Z sc

11.6���������������������������������������������������

11.4.2

Propagation Constant

The propagation constant of a symmetrical � network is same as that of a symmetrical T network.

��������������������������������� A filter transmits or passes desired range of frequencies without loss and attenuates all undesired frequencies. The propagation constant � � � � j� where � is attenuation constant and � is the phase constant. We know that sinh

� � 2

Z1 4Z 2

� � � � � � � j� � sinh � � sinh cos � j cosh sin � � 2 �� 2 2 2 2

Z1 4Z 2

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

� � sin 2 2 � sin 2 � cos 2 � sinh 2

Z1 � and sinh are real. 4Z 2 2

�0

� � � if � � 0 or n� where n � 0,1, 2, � �� cosh cannot be zero � 2 � �

�0 �1 �

Z1 4Z 2

� � 2 sinh �1

Z1 4Z 2

Z1 Z1 � 0 and Case (ii) If Z1 and Z2 are opposite types of reactances then is negative, i.e. 4Z 2 4Z 2 is purely imaginary. sinh j cosh cosh

Z1 4Z 2

� � cos � 0 2 2 Z1 � � sin � j 2 2 4Z 2 � � 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 � and �. Two conditions may arise

11.6����������������������������11.7

(a)

� � 0 i.e. � � 0 when � � 0 2 Z � Since 1 is negative and sin is real, 4Z 2 2 sinh

sin

� � 2

Z1 4Z 2

as cosh

� �1 2

This signifies the region of zero attenuation or pass band which is limited by the upper limit of sine terms. � sin � 1 2 Z �1 � 1 � 0 4Z 2 The phase angle in the pass band is given by, � � sin �1 2

Z1 4Z 2

� � 2 sin �1 (b)

Z1 4Z 2

� �0 2 � sin � �1 2 Z � Since 1 is negative and cosh is real, 4Z 2 2 Z1 � cosh � 2 Z2 cos

This signifies a stop band since � � 0. � � cosh �1 2

Z1 4Z 2

� � 2cosh �1 and

Z1 4Z 2

Z1 � �1 4Z 2

�����������������k���������������� A T or � network is said to be of the constant k type if Z1 and Z2 are opposite types of reactances satisfying the relation Z1Z 2 � k 2

11.8��������������������������������������������������� 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 �-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 �-section filters. Z1

Z1

2

2

Z1

2Z2

Z2

2Z2

(b) p-section

(a) T-section

Fig. 11.4������������������ In constant-k low pass filter,

Z1 � j� L Z2 � � j

1 1 � � c j� c

1. Nominal Impedance � 1 � k � Z1Z 2 � ( j� L) � � � j� c �� 2. Cut-off Frequency The cut-off frequencies are obtained when (i)

(ii)

When

When

Z1 Z � 0 and 1 � �1 4Z 2 4Z 2

Z1 �0 4Z 2 Z1 � 0 j� L � 0 ��0 f �0 Z1 � �1 4Z 2 Z1 � �4Z 2 4j �C � 2 LC � 4 4 �2 � LC j� L �

� � �c � f � fc �

2 LC 1

� LC

L C

11.6����������������������������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, � � 0 In stop band,

� � 2 cosh �1

Z1 � 2 LC �2 � 2 cosh �1 � 2 cosh �1 4Z 2 4 �c2 a

� f � ��� � 2 cosh �1 � � � 2 cosh �1 � � � �c � � fc � The attenuation constant � is zero throughout the pass band but increases gradually from the cut-off frequency. The variation of � is plotted in Fig. 11.5.

4. Phase Constant In pass band,

Fig. 11.5������������������������������

� � 2 sin �1

b

� f � Z1 � 2 sin �1 � � 4Z 2 � fc �

p

In stop band, � � � . The phase constant � is zero at zero frequency and increases gradually through the pass band, reaches � at cutoff frequency fc and remains at � for all frequency beyond fc. The variation of � is plotted in Fig. 11.6.

Z12 Z � � � Z1Z 2 � Z1Z 2 �1 � 1 � � 4 4Z 2 �

Z0

2



��� � f � L � � 2 LC � 1� � k 1� � � � k 1� � � C �� 4 �� � �c � � fc � Z0�

ZZ � 1 2 � Z0T

f

fc

0

Fig. 11.6������������������������������

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 ������������������������������ ������������������������ shown in Fig. 11.7.

11.10��������������������������������������������������� 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. Z0� is real when f < fc. At f � fc, Z0� is infinite and for f > fc, Z0� is imaginary. 6. Design of Filter L C 1

k� fc �

� LC

Solving these two equations, L�

k � fC

C�

1 � fC k

The constant-k, T and � section filters are shown in Fig. 11.8. L 2

L 2

L

C 2

C

C 2

(b) p-section

(a) T-section

Fig. 11.8���������������������������

��������������

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 �F 2 L � 50 mH

11.6����������������������������11.11

(a)

Nominal Impedance

(b)

0.2 � 10 �6

� 500 �

Cut-off frequency fc �

(c)

50 � 10 �3

L � C

k�

1



� LC

1

� 50 � 10

�3

� 0.2 � 10 �6

� 3.18 kHz

Pass band The pass band is from zero to 3.18 kHz.

�������������� A low-pass filter is composed of a symmetrical � 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 �F

Cut-off frequency

fc � (b)

1



� LC

1

� 1125.4 Hz

� 0.02 � 4 � 10 �6

Nominal impedance k�

(c)

C � 2 �F 2

L � 0.02 H,

Solution

0.02

L � C

4 � 10 �6

� 70.71 �

Characteristic impedance at 200 Hz Z0 �

k � f � 1� � � � fc �

2



70.71 � 200 � 1� � � 1125.4 ��

2

� 71.85 �

Characteristic impedance at 2000 Hz 70.71 k Z0 � � j 48.13 � � 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. ��0 At 2000 Hz, � f � � 2000 � � � 2 cosh �1 � � � 2 cosh �1 � � 2.35 � 1125.4 �� � f � c

(e)

Phase shift constant at 200 Hz and 2000 Hz At 200 Hz, � f � � 200 � � � 2 sin �1 � � � 2 sin �1 � � 20.47� � 1125.4 �� � fc �

11.12��������������������������������������������������� 2000 Hz frequency lies in the attenuation band. � � 180�

��������� ����� A � 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 �F 2

C � 0.32 �F (a)

Cut-off frequency fc �

(b)

1



� LC

1

� 20 � 10

�3

� 0.32 � 10 �6

� 3.98 kHz

Attenuation at 15 kHz � 15 � 103 � � f � � � 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.

� � 180� (d)

Nominal impedance k�

20 � 10 �3

L � C

0.32 � 10 �6

� 250 �

�������������� 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 �F

Cut-off frequency fc �

(b)

1

� LC

1

� 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 �

Characteristic impedance at 1 kHz 2

2

� 1 � 103 � � f � Z0 � k 1 � � � � 547.72 1 � � � � 514.36 � � fc � � 2.91 � 103 �

11.6����������������������������11.13

Characteristic impedance at 5 kHz 2

2

� 5 � 103 � � f � Z0 � k 1 � � � � 547.72 1 � � � � j 765.29 � � fc � � 2.91 � 103 �

��������� ����� Design a constant-k low-pass T and � section filters having cut-off frequency of 4 kHz and nominal characteristic impedance of 500 �. f c � 4 kHz, k � 500 �

Solution

500 k � � 39.79 mH � f c � � 4 � 103 1 1 C� � � 0.16 �F � f c k � � 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 �-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

�������������� Design a constant-k, T section low-pass filter is having 2.5 kHz cut-off frequency and nominal impedance of 700 �. 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 � 700 k � � 89.13 mH � f C � � 2.5 � 103 1 1 C� � � 0.18 �F � f C k � � 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���������������������������������������������������

� � 19.1 dB = 2.197 nepers

44.57 mH

44.57 mH

� f � � � 2 cosh �1 � � � fC � �1 �

0.18 μF

� 2.197 � 2 cosh � � 2.5 � 103 �� f � 4.17 kHz � is imaginary in pass band. In attenuation band,

f

Fig. 11.11

� � 180�

�����������������k����������������� 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 � section, high-pass filter. 2C

C

2C

L

2L

(b) p-section

(a) T-section

Fig. 11.12���������������������������� In a constant-k high-pass filter 1 1 � �C j� C Z 2 � j� L Z1 � � j

1. Nominal Impedance � 1 � L k � Z1Z 2 � � � j� L � � � j�C �� C 2. Cut-off Frequency The cut-off frequencies are obtained when (i)

When

Z1 �0 4Z 2 Z1 � 0 1 �0 j� C ��� f ��

(ii)

When

Z1 � �1 4Z 2

Z1 Z � 0 and 1 � �1 4Z 2 4Z 2

2L

11.7�����������������������������11.15

Z1 � �4Z 2 �j

1 � 4 j� L �c

� 2 LC � �2 �

1 4 1 4 LC

� � �c � f � fc �

1 a

2 LC 1

4� 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, � � 0 � f � Z1 � 2 cosh �1 � c � 4Z 2 � f �

In stop band, � � 2 cosh �1

fc

Fig. 11.13������������������������������ b

The attenuation constant � decreases gradually to zero at the cut-off frequency and remains at zero through the pass band. The variation of � is plotted in Fig. 11.13.

p

4. Phase Constant

fc

0

In pass band, � � 2 sin �1

� f � Z1 � 2 sin �1 � c � 4Z 2 � f �

In stop band, � � � . The phase constant � remains at constant value � in the stop band and decreases to –� in at fc and reaches zero value gradually as f increases in the pass band. The variation of � is plotted in Fig. 11.14.

f

f

−p

Fig. 11.14������������������������������� Z0

5. Characteristic Impedance Z 0T �

Z0p

Z12 Z � � � Z1Z 2 � Z1Z 2 �1 � 1 � � 4Z 2 � 4 k

� Z 0�

� f � 1 � L� 1� � k 1� � c � C �� 4� 2 LC �� � f �

ZZ � 1 2 � Z0T

2

Z0T 0

fc

k � f � 1� � c � � f �

2

Fig. 11.15������������������������������ ������������������������

The variation of characteristic impedance is plotted in Fig. 11.15.

f

11.16��������������������������������������������������� 6. Design of Filter k� fc �

L C 1 4� LC

Solving these two equations, k 4� f c 1 C� 4� f c k L�

��������������

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 �F, L � 50 mH C � 0.2 �F (a)

(b)

(c)

Characteristic impedance k�

L � C

fc �

1

50 � 10 �3 0.2 � 10 �6

� 500 �

Cut-off frequency 4� LC



1 4� 50 � 10

�3

� 0.2 � 10 �6

� 795.77 Hz

Pass band The pass band is from 795.77 Hz to infinite frequency.

�������������� 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 �F C � 0.5 �F

11.7�����������������������������11.17

(a)

Cut-off frequency fc �

(b)



1 4� 15 � 10

�3

� 0.5 � 10 �6

� 918.88 Hz

Infinite-frequency characteristic impedance At f � �,

(c)

1 4� LC

Z0 � k �

1.5 � 10�3 L � � 173.21 � 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 � � 2000 �� � f � Characteristic impedance at 2000 Hz � f � � 918.88 � Z0T � k 1 � � c � � 173.21 1 � � � 153.85 � � 2000 �� � f � (d)

Attenuation at 200 Hz � f � � 918.88 � � � 2 cosh �1 � c � � 2 cosh �1 � � 4.41 � 200 �� � f �

The frequency of 2000 Hz lies in the pass band. ��0 (e) Phase constant The frequency 200 Hz lies in the attenuation band. � � � radians At 2000 Hz, � f � � 918.88 � � � 2 sin �1 � C � � 2 sin �1 � � 54.7� � 2000 �� � f �

�������������� Design a constant-k high-pass T and � sections filters having a cut-off frequency of 2000 Hz and infinite frequency characteristic impedance of 300 �. Solution

f c � 2000 Hz, k � 300 � k 300 � � 11.94 mH � � 4 f c 4 � 2000 1 1 C� � � 0.13 �F 4� f c k 4� � 2000 � 300 L�

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 �-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��������������������������������������������������� 0.26 μF

0.26 μF

0.13 μF

11.94 mH

23.88 mH

(a)

23.88 mH

(b)

Fig. 11.17

��������������� Design a T-section constant-k high-pass filter having a cut-off frequency of 10 kHz and a design impedance of 600 �. Find its characteristic impedance and phase constant at 25 kHz. Solution

f c � 10 kHz, k � 600 � k 600 � � 4.77 mH 4� f c 4� � 10 � 103 1 1 C� � � 0.013 �F 4� f c k 4� � 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 � � f � � 25 � 103 � (b)

Phase constant at 25 kHz � 10 � 103 � � f � � � 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, � � 180� � � radians

������������������������ 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������������������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����������������� For series arm, L1 1 � 2 2� 0C1 1 � 02 � L1C1 1 � � 0 L2 � 0C 2 1 � 02 � L2C2 L1C1 � L2C2

�0

For shunt arm,

For series arm, Z1 � j� L1 �

� � 2 L1C1 � 1� j � j� � � c1 � c1 � �

For shunt arm, 1 j� L2 j� c2 Z2 � � 2 1 1 � L2 C2 � j� L2 � j� c2 j� L2

� � 2 L1C1 � 1� � j� L2 � L2 � � 2 L1C1 � 1 � L2 L Z1Z 2 � j � � � � � 1 � k2 � � � � � 2 2 C1 � 1 � � L2 C2 � C1 C2 � c1 � � � 1 � � L2 C2 � where k is constant. For constant k filter, at cut-off frequency, Z1 � �4Z 2 Z12 � �4Z1Z 2 � �4k 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 � j�1 L1 � � � � j� 2 L1 � j�1C1 � j� 2C1 � 1 � �12 L1C1 �

�1 2 � 2 L1C1 � 1 �2





11.20���������������������������������������������������

� 02 �

But 1�

1 L1C1

�12 �1 � � 02 � 2

� � 22 � � 2 � 1� � �0 �

(� 02 � �12 )� 2 � �1 (� 22 � � 02 )

� 02� 2 � �12� 2 � �1� 22 � �1� 02 � 02 (�1 � � 2 ) � �1� 2 (� 2 � �1 ) � 02 � �1� 2 � 0 � �1� 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������������������������������������������������������������������������������������������ ����������������������������������������� Design of Filter If the filter is terminated in a load resistance R � k then at lower cut-off frequency, Z1 � �2 jk 1 � j�1 L1 � �2 jk j�1C1 1 � �1 L1 � 2k �1C1 1 � �12 L1C1 � 2k�1C1 1�

�12 � 2k�1C1 � 02 2

� f � 1 � � 1 � � 4� kf1C1 � f0 �

1 � � 2 ��� � 0 � L C �� 1 1

11.8������������������11.21

f12 � 4� kf1C1 � f 0 � f1 f 2 f1 f 2 f 2 � f1 � 4� kf1 f 2C1 f �f C1 � 2 1 4� kf1 f 2 1 4� kf1 f 2 4� kf1 f 2 k � L1 � 2 � 2 � 2 2 � ( f � 0 C1 � 0 ( f 2 � f1 ) 4� f 0 � f 2 � f1 � 2 � f1 )





1�

For shunt arm, Z1Z 2 �

L2 L � 1 � k2 C1 C2

L2 � C1k 2 � C2 �

L1 k2



( f 2 � f1 )k 4� f1 f 2

1 � ( f 2 � f1 )k

��������������� 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 � � k ( f 2 � f1 ) f �f C1 � 2 1 4� kf1 f 2 1 4� kf1 f 2 C2 � C1 � k ( f 2 � f1 ) f 2 � f1 4 f1 f 2 100 � ( f 2 � f1 ) 2 100 � f 2 � f1 �

4 f 02 ( f 2 � f1 ) 2 2 f0



2 � 1000

100 100 BW � f 2 � f1 � 200 Hz

� 200

��������������� Design a band pass constant-k filter with cut-off frequency of 4 kHz and 10 kHz and nominal characteristic impedance of 500 � . Solution

f1 � 4 kHz, f 2 � 10 kHz, k � 500 � 500 k � � 26.53 mH L1 � � ( f2 � f1 ) � (10 � 103 � 4 � 103 ) C1 �

f2 � f1 10 � 103 � 4 � 103 � 0.024 �F � 4� kf1 f2 4� � 500 � 4 � 103 � 10 � 103

11.22��������������������������������������������������� ( f2 � f1 )k (10 � 103 � 4 � 103 ) 500 � � 5.97 mH 4� f1 f2 4� � 4 � 103 � 10 � 103 1 1 � � 0.11 �F C2 � � ( f2 � f1 )k � � (10 � 103 � 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 of The T-section filter consists of series combination of an inductor 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 �-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

������������������������ 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 �0. For series arm, � 0 L1 1 � 2 2� 0C1 1 � 02 � L1C1

11.9������������������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����������������� For shunt arm, 1 � 0 C2 1 � 02 � L2C2 L1C1 � L2C2 L L Z1Z 2 � 1 � 2 � k 2 C2 C1

� 0 L2 �

Similarly,

f 0 � f1 f 2 Z1 � �4 Z 2

and At cut-off frequencies,

Z1Z 2 � �4 Z 22 � 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

f1

f0

f2

f

k

Z0T

f

0

−p (b)

f

f1 f0 f2 (c)

Fig 11.24������������������������������������������������������������������������������������������ ������������������������������������������ Design of Filter If the load is terminated in load resistance R � k then at lower cut-off frequency,

11.24��������������������������������������������������� � 1 � k � �1 L2 � � j Z2 � j � � C 2 � 1 2 � 1 k � �1 L2 � �1C2 2 1 � �12 L2C2 � �1C2 1�

k 2

�12 k � �1C2 � 02 2

1 � � 2 ��� � 0 � L C �� 2 2

2

� f � 1 � � 1 � � k� f1C2 � f0 � C2 � L2 �

2 1 � � f1 � � 1 � 1 1 � 1 � f 2 � f1 � f �f �1 � � � � � � � � 2 1 k� f1 � � f 0 � � k� �� f1 f 2 �� k� �� f1 f 2 �� � k f1 f 2 � �

1

� 02C2



� kf1 f 2 2 � 0 ( f 2 � f1 )



k � kf1 f 2 � 2 4� f 0 � f 2 � f1 � 4� ( f 2 � f1 ) 2

L1 L2 � C2 C1 k ( f 2 � f1 ) L1 � k 2C2 � � f1 f 2 1 L2 C1 � 2 � 4� k ( f 2 � f1 ) k

k2 �

��������������� Design a constant-k band-stop filter having cut-off frequencies at 2000 Hz and 5000 Hz and characteristic resistance of 600 �. Solution

f1 � 2000 Hz,

f 2 � 5000 Hz, k � 600 �

k ( f 2 � f1 ) 600(5000 � 2000) � � 57.3 mH L1 � � f1 f 2 � � 2000 � 5000) 1 1 � � 0.044 �F C1 � 4� k ( f 2 � f1 ) 4� � 600 � (5000 � 2000) 600 k � � 15.92 mH 4� ( f 2 � f1 ) 4� � (5000 � 2000) 5000 � 2000 f �f � 0.16 �F C2 � 2 1 � � kf1 f 2 � � 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���������������������������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 �-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.

���������������������������������� A filter is composed of a number of sections. As the characteristic impedance of an equivalent T or � section does not match with each other, a half section is used for impedance matching between T and � 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 � section. Hence, matching can be obtained if a half section is connected in between a T and � section.

Constant-k Half Sections Figure 11.27(a) shows a constant-k T-section filter. If this section is bisected longitudinally, a half section is obtained as shown in Fig. 11.27(b). Z1

Z1

Z1

Z1

2

2

2

2

Z2

(a)

2Z 2

2Z 2

(b)

Fig. 11.27������������������������������������������������������� Similarly, a constant-k � section can be bisected into two half sections as shown in Fig. 11.28.

11.26��������������������������������������������������� Z1

2Z 2

2Z 2

Z1

Z1

2

2

2Z 2

2Z 2

(a)

(b)

Fig. 11.28������������������������������������������������������� Figure 11.29 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.

Z12 � Z oc Z sc �

Z � � ( 2Z 2 ) � 2Z 2 1 � � 2� � Z1 2Z 2 � 2

Z1

4Z1 Z 22 Z1 � 4Z 2

2

� Z 0�

�Z ��Z � Z34 � Zoc Z sc � � 1 � 2Z 2 � � 1 � � 2 �� 2 � Z � � � Z1 Z 2 �1 � 1 � � Z 0T � 4Z 2 �

1

3 2Z 2 4

2

Fig. 11.29������������������������

Thus, the half section has the impedance characteristics of a � 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 � 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 � section.

��������������� 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 � and the filter is cut off below 20000 Hz. Solution

k � 600 �,

f c � 20000 Hz

k 600 � � 2.39 mH 4� f c 4� � 20000 1 1 C� � � 0.0066 �F F 4� f c k 4� � 20000 � 600 L�

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. 11.30.

0.0133 μF

4.78 mH

Fig. 11.30

�������������������������11.27

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 �-section. � f c � 7.587 kHz, k � 1.907 k�,L � 80 mH,� � � C � � 0.011 �F� 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.

11.4 Design a �-section constant-k high-pass filter having a cut-off frequency f c � 8 kHz and nominal characteristic impedance k � 600 �. 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 �F,� � � Z � 805 �, � � 83.6�, � � 5 nepers� � � O� 11.5 Design a band-pass with f1 � 2 kHz and k � 500 �.

[ f � 1170 Hz] 11.3 Design a low-pass constant-k (a) T-section, and (b) �-section filter with f c � 6 kHz and RO � 500 �. Calculate � and � for the filters for f � 10 kHz. Also determine the frequency at which the attenuation is 10 dB. � � � 2.2 neper, � � 1.46 radians,� � � �� � � at 10 kHzT � L � 13.25 mH,� 2 � � � C � 0.106 �F� � L � 26.5 mH,� � � C � � � 0 053 � F . � � 2 � � f � 7 kHz � �

constant-k filter f 2 � 3 kHz and

� L1 � 159 mH, C1 � 0.026 �F, � � L � 20.8 mH, C � 0.637 �F� 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 �. � L1 � 6.6 mH, C1 � 0.0398 �F, � � L � 9.9 mH, C � 0.0265 �F� 2 � � 2 11.7 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 ]

Objective–Type Questions 11.1 For the design of low-pass prototype filter Fig. 11.31 of with load resistance RL � 1 � and angular frequency � � 1 rad/s the values of L and C would be L 2

L 2

(a) 1 H and I F (b) 1 H and 2 F (c) 2 H and I F (d) 2 H and 2 F 11.2 The passband of a typical filter network with Z1 and Z2 as the series and shunt-arm impedances is characterised by (a)

�1 �

(c)

0�

C

Fig. 11.31

Z1 �0 4Z2

Z1 �1 4Z2

Z1 �1 4Z2

(b)

�1 �

(d)

none of the above

11.3 The application of the bisection theorem finds out an equivalent lattice network if the

11.28��������������������������������������������������� original network is (a) symmetrical and balanced only (b) unsymmetrical and balanced only (c) symmetrical, balanced and unbalanced only (d) symmetrical and unbalanced only 11.4 Z1 and Z2 are the total series and shunt impedances of a T or �-filter. Consider the following zones of operation of filters and the conditions on the impedances

11.5

11.6

(a) Pass band

1.

(b)

Stop band

2.

(c)

Transition band

3.

Z1 � �1 4Z2 Z1 � �1 4Z2 Z �1 � 1 � 0 4Z2

Tick the correct combination: A B C (a) 1 2 3 (b) 3 1 2 (c) 3 2 1 (d) 2 3 1 Tick out the correct statement in case of a filter. (a) Characteristic impedance is resistive in stop band (b) Characteristic impedance is reactive in pass band (c) Characteristic impedance is resistive in pass band (d) None of the above If L is the total series inductance and C the total shunt capacitacne of a T or �-type lowpage filter, the pass band frequency range of the filter is given as

(a) 0 to

11.7

(b)

0 to

(c)

0 to

(d)

0 to

11.9

� LC 2

� LC 1 4� LC

Hz Hz Hz Hz

If C is the total series capatance, L is the total shunt inductance of a T or �-type high-pass filter, the frequency range for the stop band of the filter is (a) 0 to

11.8

1 2� LC 1

(b)

0 to

(c)

0 to

(d)

0 to

1 2� LC 1

� LC 2

� LC 2 4� LC

Hz Hz Hz Hz

If f1 and f2 are the lower and uppper cut-off frequencies of the band pass filter, the series impedance Z1 is (a) capacitive at f1 (b) inductive at f1 (c) resistive at f2 (d) none of the above The phase constant � of a filter during stop band is (a) Zero radian (b) 2 (c) � (d) 2�

Answers to Objective-Type Questions 11.1 (d)

11.2 (a)

11.8 (a)

11.9 (c)

11.3 (c)

11.4 (b)

11.5 (c)

11.6 (b)

11.7 (d)

APPENDIX Additional Solved Mumbai University Examination Questions

Circuit Theory Network (May 2018)

1. (a) Find the voltage drop across 5 � resistor in the circuit given below.

Fig. 1 Ans. Assign clockwise current in three meshes as shown in Fig. 2. From Fig. 2, i = I1 – I2 For Mesh 1, I1 = 5 (i) Applying KVL to Mesh 2, –1(I2 – I1) + 2i – 2I2 = 0 –1(I2 – I1) + 2(I1 – I2) – 2I2 = 0 (ii) 3I1 – 5I2 = 0 Applying KVL to Mesh 3, – 5I3 – 2i – 3(I3 – I1) = 0 – 5I3 – 2(I1 – I2) – 3(I3 – I1) = 0 I1 + 2I2 – 8I3 = 0 Solving Eqs (i), (ii) and (iii) I1 = 5 A I2 = 3 A I3 = 1.375 A V5� = 5I3 = 5(1.375) = 6.875 V

Fig. 2

1. (b) For the graph given below obtain the incidence matrix and find the number of possible trees.

Fig. 3

(iii)

A.2�Additional Solved Mumbai University Examination Questions—CTN (May 2018) Ans. Branches �

Nodes �

1

2

3

4

5

a

1

0

0

–1

0

b

–1

–1

1

0

0

c

0

1

0

0

–1

d

0

0

–1

1

1

� 1 0 0 �1 0 � � � �1 �1 1 0 0 � Aa = � � 0 1 0 0 �1� � � � 0 0 �1 1 1�

Eliminating fourth row from the matrix Aa, we get the incidence matrix A.

� 1 0 0 �1 0 � � � A = ��1 �1 1 0 0 � �� 0 1 0 0 �1�� � 1 �1 0 � � � � 1 0 0 �1 0 � � 0 �1 1� � � AAT = ��1 �1 1 0 0 � � 0 1 0� � � �� 0 1 0 0 �1�� ��1 0 0 � �� 0 0 �1��

� 2 �1 0 � � � = ��1 3 �1� �� 0 �1 2 �� � 2 �1 0 � � � Number of possible trees = |AAT| = ��1 3 �1� � 8 �� 0 �1 2 �� 1. (c) Find Y parameters for the two-port network shown in Fig. 4.

Fig. 4 Ans. Applying KCL at Node 1,

V1 V1 – V2 � 2 5 7 1 V � V = 10 1 5 2

I1 =

(i)

�������������������������������������������������������������������������A.3

Applying KCL at Node 2,

V2 V2 – V1 � 2 5 1 7 = � V1 � V2 5 10

I2 =

(ii)

Comparing Eqs (i) and (ii) with Y-parameter equations, � 7 �Y11 Y12 � � 10 � ��� �Y21 Y22 � �� 1 �� 5

1� � � 5 � 7 � 10 ��

1. (d) Check whether the following polynomials are Hurwitz. (i) P(s) = s4 + 7s3 + 6s2 + 21s + 8 (ii) P(s) = s5 + 2s3 + s Ans. (i) Refer Example 10.14 on page 10.12. (ii) The given polynomial contains odd functions only. P�(s) = 5s4 + 6s2 + 1 The Routh array is given by, s5

1

2

1

s4

5

6

1

s3

0.8

0.8

0

s2

1

1

s1

02

00

s0

1

Auxiliary equation, A(s) = s2 + 1 A�(s) = 2s

Since there is no sign change in the first column of the array, the polynomial P(s) is Hurwitz. 2. (a) Find the Thevenin’s equivalent across AB and find the power dissipated in a 25 � load.

Fig. 5 Ans. Step I: Calculation of VTh (Fig. 6) From Fig. 6, V1 = 5I1 Applying KVL to Mesh, 20 – 5I1 – 6I1 – 5V1 = 0

Fig. 6

A.4�Additional Solved Mumbai University Examination Questions—CTN (May 2018) 20 – 5I1 – 6I1 –5(5I1) = 0 I1 =

20 5 � A 36 9

Writing the VTh equation, 20 – 5I1 – VTh = 0 � 5� 20 – 5 � � – VTh = 0 � 9�

VTh = 17.22 V Step II: Calculation of IN From Fig. 7, V1 = 5I1 Applying KVL to Mesh 1, 20 – 5I1 – 6(I1 – I2) – 5V1 = 0 20 – 5I1 – 6(I1 – I2) –5(I1) = 0 36I1 – 6I2 = 20 Applying KVL to Mesh 2, 5V1 – 6(I2 – I1) – 8I2 = 0 5(5I1) – 6(I2 – I1) – 8I2 = 0 31I1 – 14I2 = 0 Solving Eqs (i) and (ii), I1 = 0.88 A I2 = 1.95 A IN = I2 = 1.95 A Step III: Calculation of RTh RTh =

(i) Fig. 7

(ii)

VTh 17.22 = 8.33 � � IN 1.95

Step IV: Power dissipated in 25 � load IL =

17.22 = 0.51 A 8.83 � 25

P25� = IL2RL = (0.51)2 � 25 = 6.5 W 2. (b) Find h parameters for the following two-port network.

Fig. 9

Fig. 8

�������������������������������������������������������������������������A.5

Ans. The network is redrawn as shown in Fig. 10. Applying KVL to Mesh 1, V1 = 3I1 – 2I3 Applying KVL to Mesh 2, V2 = 2I2 + 2I3 Applying KVL to Mesh 3, –2(I3 – I1) – 1I3 – 2(I3 + I2) = 0

(i) (ii) Fig. 10

–2I1 + 2I2 + 5I3 = 0 I3 =

2 2 I – I 5 1 5 2

(iii)

Substituting Eq. (iii) in Eq. (i), V1 = 3I1 – =

4 4 I1 + I2 5 5

11 4 I + I 5 1 5 2

(iv)

Substituting Eq. (iii) in Eq. (ii), V2 = 2I2 + =

4 4 I1 – I2 5 5

4 6 I1 + I2 5 5

(v)

Rewriting Eq. (v),

6 4 I2 = – I1 + V2 5 5 I2 = –

2 5 I1 � V2 3 6

(vi)

Substituting Eq. (vi) in Eq. (iv), V1 = =

11 4� 2 5 � I + � I � V 5 1 5 �� 3 1 6 2 �� 5 2 I + V 3 1 3 2

(vii)

Comparing Eqs (vi) and (vii) with h-parameter equations, � h11 � �h21

� 5 h12 � � 3 ��� h22 � � 2 ��� 3

2� 3� � 5� 6 ��

2. (c) In the network shown below the switch is closed al t = 0. Assuming all initial conditions to be zero, di d 2 i find i, for t = 0+. , dt dt 2

A.6�Additional Solved Mumbai University Examination Questions—CTN (May 2018)

Fig. 11 Ans.

At t = 0–

i(0–) = 0 vc (0–) = 0 + At t = 0 , the network is shown in Fig. 12. i(0+) = 0 vc (0+) = 0 For t > 0, the network is shown in Fig. 13. Writing the KVL equation for t > 0, 10 – 10i – At t = 0+,

1 2 � 10

t

�6

�0 idt � 5 � 10

10 – 10i(0+) – 0 – 5 � 10–3

�3

Fig. 12 di =0 dt

(i)

di + (0 ) = 0 dt

Fig. 13

10 � 10(0) di + (0 ) = = 2 � 103 A/s dt 5 � 10 �3 Differentiating Eq. (i), 0 – 10 At t = 0+,

0 – 10

d2i di 1 – i – 5 � 10–3 2 = 0 dt dt 2 � 10 �6

2 di � 1 � �3 d i (0 ) � i (0 ) 5 10 (0 � ) = 0 � � dt 2 � 10 �6 dt 2

d2i dt

2

(0+) =

�10(2 � 103 ) 5 � 10 �3

3. (a) Find the tie-set and f-cutset matrix for the oriented graph shown below.

Fig. 14

= –4 � 106 A/s2

�������������������������������������������������������������������������A.7

Ans. Links: {6, 7, 8, 9} Tieset 6: {6, 1, 2, 3, 4, 5}

1 2 3 4 5 � 1 1 1 �1 �1 � �0 �1 �1 1 0 �1 1 0 0 0 � 9 �0 0 0 1 1

6 7 B� 8

Tieset 7: {7, 2, 3, 4} Tieset 8: {8, 2, 1} Tieset 9: {9, 4, 5} Twigs: {1, 2, 3, 4, 5} f-cutset 1: {1, 6, 8} f-cutset 2: {2, 6, 7, 8} f-cutset 3: {3, 6, 7} f-cutset 4: {4, 6, 7, 9} f-cutset 5: {5, 6, 9}

1 1 �1 � 2 �0 Q � 3 �0 � 4 �0 5 ��0

2 0 1 0

6 1 0 0

7 0 1 0

8 0 0 1

9 0� � 0� 0� � 0 0 0 1�

3 0 0 1

4 5 6 7 8 9 0 0 �1 0 �1 0 � � 0 0 �1 1 �1 0 � 0 0 �1 1 0 0 � � 0 0 1 0 1 �1 0 �1� 0 0 0 1 0 0 0 �1��

3. (b) Realize the following function in Foster-I and Foster-II forms. Z(s) =

(s � 1)(s � 4) s( s � 2 )

Ans. 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. s 2 � 5s � 4 Z(s) = s2 � 2s s2 � 2 s )s2 � 5s � 4( 1 s2 � 2s 3s � 4 Z(s) = 1 +

3s � 4 2

s � 2s

� 1�

3s � 4 s(s � 2)

By partial fraction expansion, Z(s) = 1 + where

K0 K � 1 s s�2

K0 = sZ (s) s � 0 =

(1)(4) =2 2

K1 = (s � 2)Z (s) � S ��2 Z(s) = 1 +

2 1 � s s�2

(�2 � 1)(�2 � 4) �1 �2

A.8�Additional Solved Mumbai University Examination Questions—CTN (May 2018) The first term represents the impedance of a resistor of 1 �. The second term represents the impedance 1 of a capacitor of F. The third term represents the impedance of a parallel RC circuit. 2 1 Ci ZRC(s) = 1 s� Ri Ci By direct comparison, R1 =

1 �, 2

C1 = 1 F Fig. 15

The network is shown in Fig. 15. Foster II form: The Foster II form is obtained by partial fraction expansion of admittance Y ( s) function . s s(s � 2) Y(s) = (s � 1)(s � 4)

Y ( s) s�2 = s (s � 1)(s � 4) By partial fraction expansion, K K Y ( s) � 1 � 2 s s �1 s � 4 where

�1 � 2 1 � K1 = (s � 1)Z (s) s ��1 = �1 � 4 3 K2 = (s � 4)Z (s) s ��4 =

�4 � 2 2 � �4 � 1 3

1 2 Y ( s) 3 � � 3 s s �1 s � 4

1 2 s s 3 Y(s) = � 3 s �1 s � 4 These two terms represent the admittance of a series RC circuit. � 1� �� R �� s i YRC(s) = 1 s� Ri Ci

�������������������������������������������������������������������������A.9

By direct comparison, R1 = 3 �, R2 =

3 �, 2

C1 =

1 F 3

C2 =

1 F 6

Fig. 16

The network is shown in Fig. 16. 4. (a) A switch is in position A for a long time and then thrown to position B at t = 0. Find i(t) for t > 0. At what value of ‘t’ the current i(t) will become half of current at t = 0.

Fig. 17 Ans. At t = 0–, the network has attained steady-state condition. Hence, the capacitor of 1 F acts as an open circuit and will charge to 12 V. vc1 (0 � ) = 12 V vc2 (0 � ) = 0 Since the voltage across the capacitor cannot change instantaneously, vc1 (0 � ) = 12 V vc2 (0 � ) = 0 V

i(0+) =

12 =6A 2

For t > 0, the network is shown in Fig. 18. Writing the KVL equation for t > 0, 1 t 1 t idt � � idt � 2i = 0 1 �0 2 0 Differentiating Eq. (i), di 0 – i – 0.5i – 2 =0 dt 12 –

di + 0.75i = 0 dt di + Pi = 0 dt P = 0.75 The solution of this differential equation is given by, i(t) = ke–0.75t

Comparing with differential equation

Fig. 18 (i)

A.10�Additional Solved Mumbai University Examination Questions—CTN (May 2018) At t = 0,

when i =

i(0) = 6 A 6=k i(t) = 6e–0.75t

i(0) =3A 2 3 = 6e–0.75t t = 0.924 s

4. (b) For the following network, find the driving point impedance function.

Fig. 19 Ans. The transformed network is shown in Fig. 20. �1 � 2 S � � 5� �s � Z(s) = 5+ 1 2s � � 5 s =5+ =

=

s(2 � 10 s) 2 s2 � 5s � 1

10 s2 � 25s � 5 � 2 s � 10 s2

Fig. 20

2 s 2 � 5s � 1 20 s2 � 27s � 5 2 s2 � 5s � 1

4. (c) Find the condition for symmetry and reciprocity for a two port network using any one parameter. Ans. Refer Section 9.2.1 and 9.2.2 on pages 9.2 and 9.3 respectively. 5. (a) Obtain the ABCD parameters of the following network. If two such networks are cascaded find the overall ABCD parameter.

Fig. 21 Ans. The network is redrawn as shown in Fig. 22.

�������������������������������������������������������������������������A.11

Applying KVL to Mesh 1, V1 = 7I1 – 5I3

(i)

V2 = 3I2 + 3I3

(ii)

Applying KVL to Mesh 2, Applying KVL to Mesh 3, – 5(I3 – I1) – 2I1 – 3(I3 + I2) = 0 – 8I3 + 3I1 – 3I2 = 0 I3 =

Fig. 22

3 3 I1 – I2 8 8

(iii)

Substituting Eq. (iii) in Eq. (i), 3 � �3 V1 = 7I1 – 5 � I1 � I 2 � �8 8 �

=

41 15 I1 � I 2 8 8

(iv)

Substituting Eq. (iii) in Eq. (ii), 3 � �3 V2 = 3I2 + 3 � I1 � I 2 � �8 8 �

=

9 15 I1 � I 2 8 8

(v)

Rewriting Eq. (v),

9 15 I = V2 – I 8 1 8 2 I1 =

8 5 V – I 9 2 3 2

V1 =

41 � 8 5 � 15 V � I � I 8 �� 9 2 3 2 �� 8 2

(vi)

Substituting Eq. (vi) in Eq. (iv),

=

41 20 V2 � I2 9 3

(vii)

Comparing Eq. (vi) and (vii) with ABCD parameter equations, � 41 � A� B � � � 9 � ��� �C � D � � � 8 �� 9

20 � 3� � 5� 3 ��

Hence, overall ABCD parameters of the cascaded network are � 41 � � A B � � A � B � � � A �� B �� � = � 9 � ��� ��� � �8 �C D � �C � D � � �C �� D �� � �� 9

20 � � 41 3 �� 9 �� 5 �� 8 3 �� �� 9

20 � 3� � 5� 3 ��

A.12�Additional Solved Mumbai University Examination Questions—CTN (May 2018) � 2161 1120 � � 27 � �26.68 41.48� = � 81 ��� � 235 � � 5.53 8.7 � � 448 �� 81 27 ��

5. (b) Check whether the following function is positive real or not. F(s) = Ans.

(s2 � 6 s � 5) (s2 � 9s � 14)

Refer Example 10.24 on page 10.18.

5. (c) Find the oriented graph if the incidence matrix of the network is as given below. � 1 0 � 0 �1 A= � ��1 1 � � 0 0

0 1 1 1 0 �1 0 0 0 0 �1 0

0 0� � 0 0� 0 �1� � 1 0�

Ans. Writing the complete incidence matrix from the matrix A such that the sum of all entries in each column of Aa will be zero. 1 1 0 0� � 1 0 0 � � � 0 �1 1 0 �1 0 0 � Aa = ��1 1 0 0 0 0 �1� � � 1 0� � 0 0 0 �1 0 �� 0 0 �1 0 0 �1 1�� The oriented graph is shown in Fig. 23.

Fig. 23 6. (a) Find the mesh currents if the coupling factor k = 0.6.

Fig. 24

�������������������������������������������������������������������������A.13

Ans.

For a magnetically coupled circuit, XM = k X L1 X L2 = 0.6 (5)(10) = 4.24 �

The equivalent circuit in terms of dependent sources is shown in Fig. 25.

Fig. 25 Applying KVL to Mesh 1, 100�0° – j5I1 – j4.24I2 –(5 – j3)(I1 – I2) = 0 (5 + j2)I1 + (–5 + j7.24)I2 = 100

(i)

Applying KVL to Mesh 2, –(5 – j3)(I2 – I1) – j10I2 – j4.24I1 –10I2 = 0 (–5 + j7.24)I1 + (15 + j7)I2 = 0

(ii)

Writing Eqs (i) and (ii) in matrix form, �5 � j 7.24 � � I1 � �100 � � 5 � j2 � �� � � � � 15 � j 7 � �I2 � � 0 � ��5 � j 7.24 By Cramer’s rule,

I1 =

I2 =

100 �5 � j 7.24 0 15 � j 7 5 � j2 �5 � j 7.24 15 � j 7 �5 � j 7.24 5 � j2 100 �5 � j 7.24 0 5 � j2 �5 � j 7.24 15 � j 7 �5 � j 7.24

6. (b) Find i2(t) using Laplace transform.

Fig. 26

= 10.13�–32.22° A

= 5.39�–112.61° A

A.14�Additional Solved Mumbai University Examination Questions—CTN (May 2018) Ans.

For t > 0, the transformed network is shown in Fig. 27.

Writing KVL equations in matrix form, � 1 � � s � 12 �(s � 2)� � I1 � � � s � 5� � �� � ��(s � 2) s � 12 � � I 2 � � 0 � � � By Cramer’s rule, s � 12

I2(s) =

�(s � 2) s � 12 �(s � 2)

Fig. 27 1 s�5 0

�(s � 2) s � 12

s�2 s�5 = (s � 12)2 � (s � 2)2 s�2 s�5 = 2 s � 24 s � 144 � s2 � 4 s � 4 s�2 s�5 = 20 s � 140 =

s�2 20(s � 5)(s � 7)

By partial fraction expansion, I2(s) =

A B � s�5 s�7

A = (s � 5)I 2 (s) s ��5 =

(�5 � 2) 3 �� 20(�5 � 7) 40

B = (s � 7)I 2 (s) s ��7 =

�7 � 2 5 � 20(�7 � 5) 40

I2(s) = �

3 � 1 � 5 � 1 � � 40 �� s � 5 �� 40 �� s � 7 ��

Taking the inverse Laplace transform, i2(t) = �

3 �5t 5 �7t e � e 40 40

for t > 0

Circuit Theory Network (December 2017)

1. (a) Find voltages V1 and V2 by nodal analysis for the circuit given below.

Fig. 1 Ans.

From Fig. 1, I1 =

V1 � V2 2

Assume that the currents are moving away from the nodes. Applying KCL at Node 1, V1 V1 � V2 =5 � 1 2 1 � 1� �� 1 � 2 �� V1 � 2 V2 = 5

1.5V1 – 0.5V2 = 5 Applying KCL at Node 2,

(i)

V2 V2 � V1 = 2I1 � 1 2

V2 +

V2 � V1 � V �V � = 2� 1 2 � � 2 � 2

3 5 V � V =0 2 1 2 2

3V1 – 5V2 = 0 Solving Eqs (i) and (ii) V1 = 4.17 V V2 = 2.5 V

(ii)

1. (b) Find Z parameters of the following two port network.

Fig. 2

A.16�Additional Solved Mumbai University Examination Questions—CTN (December 2017) Ans. The network is redrawn as shown in Fig. 3. Applying KVL to Mesh 1, V1 = 5I1 + 10(I1 + I2) = 15I1 + 10I2 Applying KVL to Mesh 2, V2 = 5I2 + 10(I2 + I1) = 10I1 + 15I2 Comparing Eqs (i) and (ii) with Z-parameter equations, � Z1 � � Z3

(i)

(ii)

Fig. 3

Z 2 � �15 10 � �� Z 4 � ��10 15��

1. (c) Synthesize in Cauer I, Cauer II, Foster I and Foster II forms.

Ans.

Z(s) =

s s�2

Z(s) =

s � s�2

Foster I Form: 1 1�

2 s

Fig. 4

Foster II Form: Y(s) =

s�2 2 � 1� s s

Fig. 5

Cauer I Form: s�2 Y(s) = s s)s � 2(1 � Y s 2) s ( 2s � Z s 0

Fig. 6

Cauer II Form: Y(s) =

2�s s

s)2 � s( 2s � Y 2 s) s ( 1 � Z s 0

Fig. 7

������������������������������������������������������������������������������A.17

1. (d) For the network shown find

VC . Also draw pole-zero plot. V

Fig. 8 Ans. The transformed network is shown in Fig. 9.

Fig. 9 2 s � 16 Z2 = 2 8s � 2 8� s 8�

16 Z2 Vc 16 2 � 8s � 2 = = � 2 2 16 s � Z 4 V 32 s � 8s � 16 4 s � s � 2 2 4s � 8s � 2 �

0.5 s 2 � 0.25s � 0.5



0.5 ( s � 0.175 � j 0.7)( s � 0.175 � j 0.7)

The pole-zero plot is shown in Fig. 10.

Fig. 10

A.18�Additional Solved Mumbai University Examination Questions—CTN (December 2017) 2. (a) Find the current through 5 � resistor using superposition theorem.

Fig. 11 Ans. Step I: When the 18 V source is acting alone (Fig. 12) From Fig. 12, Vx = 1I� = I� Applying KVL to the Mesh, 18 – 1I� + 2Vx – 5I� = 0 18 – I� + 2I� – 5I� = 0 I� = 4.5 A (�) Step II: When the 3 A source is acting alone (Fig. 13) From Fig. 13, Vx = 1I1 = I1 (i) Meshes 1 and 2 will form a supermesh. Writing current equation for the supermesh, I2 – I1 = 3 (ii) Applying KVL to outer path of the supermesh, –1I1 + 2Vx – 5I2 = 0 –I1 + 2(I1) – 5I2 = 0 I1 – 5I2 = 0 Solving Eqs (ii) and (iii), I1 = –3.75 A I2 = –0.75 A I�� = I2 = –0.75 A (�) Step III: By superposition theorem, I5� = I� + I�� = 4.5 – 0.75 = 3.75 A (�)

Fig. 12

Fig. 13

2. (b) Draw the oriented graph for the following circuit and obtain its incidence matrix.

Fig. 14

(iii)

������������������������������������������������������������������������������A.19

Ans.

For drawing the oriented graph, 1. replace all resistors, inductors and capacitors by line segments. 2. replace all voltage sources by short circuits and current source by an open circuit. 3. number all the nodes and branches.

The oriented graph is shown in Fig. 15. Nodes �

Fig. 15

Branches �

1

1

0

1

0

0

1

2

–1

1

0

–1

0

0

3

0

–1

0

0

1

0

4

0

0

–1

1

–1

–1

1 0 0 1� � 1 0 � � � � 1 1 0 1 0 0 � Aa � � � 0 �1 0 0 1 0� � � � 0 0 �1 1 �1 �1�

Eliminating the last row from the matrix Aa, we get the incidence matrix A. � 1 0 1 0 0 1� � � A = ��1 1 0 �1 0 0 � �� 0 �1 0 0 1 0 �� 2. (c) Find the condition for symmetry and reciprocity in terms of Z parameter. Ans. Refer Section 9.2.1 and 9.2.2 on pages 9.2 and 9.3 respectively.

3. (a) Realise Z(s) in Foster I and Foster II form. Z(s) =

s( s 2 � 4 ) (s2 � 1)(s2 � 9)

Ans. The function Z(s) has poles at s = �j1 and s = �j3 and zeros at s = 0 and s = �j2 as shown in Fig. 16. From the pole-zero diagram, it is clear that poles and zeros are simple and lie on the j�-axis. Poles and zero are interlaced. Hence, the given function is an LC function. Foster I Form: The Foster I form is obtained by partial fraction expansion of the impedance function Z(s). K1 K* K2 K 2* 2 K1s 2 K 2 s � 1 + � Z(s) = = � s � j1 s � j1 s � j3 s � j3 s2 � 1 s2 � 9 Fig. 16 s2 � 1 �1 � 4 3 Z ( s) � � where K1 = 2s 2(�1 � 9) 16 2 S ��1

K2 =

Z(s) =

s2 � 9 Z ( s) 2s � 3� �� 8 �� s s2 � 1



� S 2 ��9

� 5� �� 8 �� s s2 � 9

�9 � 4 5 � 2(�9 � 1) 16

A.20�Additional Solved Mumbai University Examination Questions—CTN (December 2017) These two terms represent the impedance of a parallel LC network. For a parallel LC network, � 1� �� C �� s ZLC(s) = 1 s2 � LC By direct comparison, C1 =

8 F, 3

L1 =

3 H 8

C2 =

8 F, 5

L2 =

5 H 72

The network is shown in Fig. 17. Foster II Form: The Foster II form is obtained by partial fraction expansion of the admittance function Y(s). Y(s) =

(s2 � 1)(s2 � 9) s(s2 � 4)



Fig. 17

s 4 � 10 s2 � 9 s3 � 4 s

Since the degree of the numerator is greater than the degree of the denominator, division is first carried out. s3 � 4 s) s 4 � 10 s2 � 9(s s 4 � 4s2 6s2 � 9 Y(s) = s +

6s2 � 9 s(s2 � 4)

By partial-fraction expansion, Y(s) = s +

where

K0 K K1 K* 2K s � � 1 � s� 0 � 2 1 s s � j2 s � j2 s s �4

K0 = sY (s) s � 0 �

K1 =

s2 � 4 Y ( s) 2s

(1)(9) 9 � 4 4 � s2 ��4

(�4 � 1)(�4 � 9) 15 � 2(�4) 8

9 � 15 � s �� 4 �� Y(s) = s + 4 � 2 s s �4 The first term represents the admittance of capacitor of 1 F. The second term represents the admittance 4 of inductor of H. The third term represents the admittance of a series LC network. For a series LC 9 network,

������������������������������������������������������������������������������A.21

� 1� �� L �� s YLC(s) = 1 s2 � LC By direct comparison, L2 =

4 H, 15

C2 =

15 F 16

The network is shown in Fig. 18.

Fig. 18 3. (b) In the following the series RC circuit switch is closed at t = 0. Find i(t) and vC(t) for t > 0.

Fig. 19 At t = 0–, the capacitor is uncharged. vC (0–) = 0 i (0–) = 0 At t = 0+, the network is shown in Fig. 20. At t = 0+, the capacitor acts as a short circuit. vC (0+) = 0

Ans.

Fig. 20

10 =2A i (0+) = 5

For t > 0, the network is shown in Fig. 21. 10 – 5i –

1 t idt � 0 0.1 �0

Differentiating Eq. (i),

(i) Fig. 21

di – 10i = 0 0–5 dt di + 2i = 0 dt

A.22�Additional Solved Mumbai University Examination Questions—CTN (December 2017) di + Pi = 0 dt The solution of this differential equation is given by,

Comparing with the differential equation

i(t) = ke–Pt = ke–2t At t = 0,

i(0+) = 2 A 2=k i(t) = 2e–2t vC(t) =

for t > 0

1 t i dt 0.1 �0 t

= 10 � 2e �2 t dt 0

t

� e �2 t � � = 20 � � �2 � 0 = –10[e–2t– 1] = 10 – 10 e–2t for t > 0 3. (c) Test whether the given polynomial is Hurwitz. (i) s4 + 7s3 + 6s2 + 21s + 8 (ii) s5 + s3 + s Ans. (i) Refer Example 10.14 on page 10.12. (ii) Refer Example 10.17 on page 10.13. 4. (a) Find ABCD parameters of the following network.

Fig. 22 Ans. The network is redrawn as shown in Fig. 23.

Fig. 23

������������������������������������������������������������������������������A.23

Applying KVL to Mesh 1, V1 = 3I1 – I3

(i)

Applying KVL to Mesh 2, V2 = 3I2 + I3 Applying KVL to Mesh 3, –1(I3 – I1) – 2I3 – 2V1 – 1(I3 + I2) = 0

(ii)

I3 = 0.25I1 – 0.25I2 – 0.5V1 Substituting Eq. (iii) in Eq. (i), V1 = 3I1 – (0.25I1 – 0.25I2 – 0.5V1) 0.5V1 = 2.75I1 + 0.25I2

(iii)

V1 = 5.5I1 + 0.5I2 Substituting Eq. (iii) in Eq. (ii), V2 = 3I2 + (0.25I1 – 0.25I2 – 0.5V1) = 3I2 + 0.25I1 – 0.25I2 – 0.5(5.5I1 + 0.5I2) = 3I2 + 0.25I1 – 0.25I2 – 2.75I1 – 0.25I2 = –2.5I1 + 2.5I2 From Eq. (v), I1 = – 0.4V2 + I2

(iv)

(v) (vi)

Substituting Eq. (vi) in Eq. (iv), V1 = 5.5(– 0.4V2 + I2) + 0.5I2 = – 2.2V2 + 6I2 Comparing Eqs (vi) and (vii) with ABCD parameter equations, � A B � ��2.2 �6 � � ��� � �C D � ��0.4 �1� 4. (b) Test for positive real function. F ( s) � Ans.

s2 � 4 3

(s � 3s2 � 3s � 1)

Refer Example 10.30 on page 10.23.

4. (c) Find I2 using Mesh Analysis.

Fig. 24

(vii)

A.24�Additional Solved Mumbai University Examination Questions—CTN (December 2017) Ans. Applying KVL to Mesh 1, 100�0° – j7I1 – (4 – j3)(I1 – I2) = 0 (4 + j4)I1 – (4 – j3)I2 = 100

(i)

Applying KVL to Mesh 2, –(4 – j3)(I2 – I1) – j12I2 – 5I2 = 0 –(4 – j3)I1 + (9 + j9)I2 = 0

(ii)

Writing Eqs (i) and (ii) in matrix form, �(4 � j 3)� � I1 � �100 � � 4 � j4 � �� � � � � � (4 � j 3) 9 � j 9 � �I2 � � 0 � � By Cramer’s rule,

I2 =

4 � j 4 100 �(4 � j 3) 0 4 � j4 �(4 � j 3) �(4 � j 3) 9 � j9

= 5.19�–131.04° A

5. (a) Obtain equilibrium equation using KVL in matrix form. Hence, find link currents.

Fig. 25 Ans.

Refer Example 5.22 on page 5.34.

5. (b) In the network given below the switch is closed for a long time and opened at t = 0. Find dv dv 2 v(0 � ), (0 � ) and 2 (0 � ). dt dt

Fig. 26 Ans. At t = 0 –, switch is closed. Hence, the voltage across the capacitor and the current through inductor are zero. v(0–) = vC(0–) = 0

������������������������������������������������������������������������������A.25

iL(0–) = 0 At t = 0+, the network is shown in Fig. 27. At t = 0+, the capacitor acts as short circuit and the inductor acts as an open circuit. v(0+) = 0 For t > 0, the network is shown in Fig. 28. Writing the KCL equation for t > 0, v 1 t 1 dv = 10 � vdt � 5 2 �0 2 dt

(i)

v(0 � ) 1 dv � (0 ) = 10 �0� 5 2 dt

At t = 0+,

Fig. 27

Fig. 28

dv � (0 ) = 20 V/s dt

Differentiating Eq. (i), 1 dv 1 1 d2 v =0 � v� 5 dt 2 2 dt 2

At t = 0+,

1 dv � 1 1 d2 v � (0 ) � v(0 � ) � (0 ) = 0 5 dt 2 2 dt 2 d2 v dt 2

(0 � ) = –

2 dv � (0 ) � v(0 � ) 5 dt

2 (20) � 0 5 = – 8 V/s2

=–

5. (c) The switch is closed at t = 0. Determine current i(t), assuming zero initial conditions, using Laplace transform.

Fig. 29 Ans.

Refer Example 7.40 on page 7.47.

6. (a) For the ladder Network shown below obtain

V1 V2 , . V2 I1

A.26�Additional Solved Mumbai University Examination Questions—CTN (December 2017)

Fig. 30 Ans. The transformed network is shown in Fig. 31.

Fig. 31 Vc = V2 Ib =

V2 s

Vb= 1Ib + V2 =

V2 � s � 1� V + V2 = � � s �� 2 s

� s2 � s � 1� Vb V + Ib = sVb + Ib = (s + 1)V2 + 2 = � � V2 1 s s � � s � s2 � s � 1� � 2 s 2 � 3s � 3 � � s � 1� V2 � � V2 � � Va = 2Ia + Vb = 2 � � � V2 � � s � s s � � � � Ia =

I1 =

� 2 s 2 � 3s � 3 � � s2 � s � 1� � s3 � 3s2 � 4 s � 3 � Va � Ia � � V � V � 2 2 � � � � � V2 s s s2 s2 � � � � � �

� s3 � 3s2 � 4 s � 3 � � 2 s 2 � 3s � 3 � V � V1 = 1I1 + Va = � 2 � � � V2 s s2 � � � � � 3s3 � 6 s2 � 7s � 3 � V1 = � � V2 s2 � � Hence,

V1 3s3 � 6 s2 � 7s � 3 = V2 s2 V2 s2 = 3 V1 s � 3s2 � 4 s � 3

������������������������������������������������������������������������������A.27

6. (b) Find Z parameters in terms of Y parameters. Ans.

Refer Section 9.6.1 (1) on page 9.29.

6. (c) Obtain Tieset and f-cutset matrix for the following graph.

Fig. 32 Ans.

Tieset matrix

Links: {4, 5} Tieset 4: {4, 1, 3} Tieset 5: {5, 2, 3}

1 2 3 4 5 4 �1 0 1 1 0 � B� � � 5 �0 1 1 0 1 �

f-cutset matrix Twigs: {1, 2, 3} f-cutset 1: {1, 4} f-cutset 2: {2, 5} f-cutset 3: {3, 4, 5}

1 1 �1 � Q � 2 �0 3 ��0

2 3 4 5 0 0 �1 0 � � 1 0 0 �1� 0 1 �1 �1��

Electrical Network Analysis & Synthesis (May 2018)

1. (a) Check for Hurwitz polynomial Q(s) = s5 + s3 + s1 Q(s) = s4 + 6s3 + 8s2 + 10 Ans. (i) Refer Example 10.17 on page 10.13. (ii) Q(s) = s4 + 6s3 + 8s2 + 10 In the given polynomial, the term s is missing and it is neither an even nor an odd polynomial. Hence it is not Hurwitz. 1. (b) Obtain s-domain (Laplace Transform) equivalent circuit diagram of an inductor and capacitor with initial conditions. Ans. Refer Section 7.6 (2) and (3) on page 7.12. 1. (c) What are conditions for rational function F(s) with real coefficient to be p.r.f? Ans. Refer Section 10.32 on page 10.16. 1. (d) Explain Y-parameter in terms of Z-parameter. Ans. Refer Section 9.6.2(1) on page 9.31. 2. (a) Realise the following function in Foster-I and Foster-II forms. Z(s) =

3(s � 2)(s � 4) s(s � 3)

Ans. The function Z(s) has poles at s = 0 and s = –3 and zeros at s = –2 and s = –4. The poles and zeros are simple and lie 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. Foster I Form: The Foster I form is obtained by the partial fraction expansion of Z(s). Since the degree of the numerator is equal to the degree of the denominator, division is first carried out. Z(s) =

3(s � 2)(s � 4) 3s2 � 18s � 24 � s(s � 3) s2 � 3s

s2 � 3s) 3s2 � 18s � 24 (3 3s 2 � 9 s 9s � 24 Z(s) = 3 �

9s � 24 s2 � 3s

� 3�

By partial-fraction expansion, K K Z(s) = 1 + 1 � 2 s s�3

9s � 24 s(s � 3)

Fig. 1

��������������������������������������������������������������������������A.29

where

3(2)(4) =8 3 3(�3 � 2)(�3 � 4) K2 = (s � 3)Z (s) s ��3 � =1 �3 8 1 Z(s) = 3 + � s s�3 K1 = sZ (s) s � 0 �

The first-term represents the impedance of a resistor of 3 �. The second term represents the impedance 1 of a capacitor of F. The third term represents the impedance of parallel RC circuit for which 8

ZRC(s) =

1 Ci s�

1 Ri Ci

By direct comparison,

1 �, C=1F 3 The network is shown in Fig. 2. R=

Fig. 2

Foster II Form: The Foster II form is obtained by the partial-fraction expansion of admittance function Y ( s) . s s(s � 3) s�3 Y ( s) � = 3 s ( s � 2)( s � 4) 3( s � 2)(s � 4) s By partial-fraction expansion, K1 K Y ( s) � 2 = s�2 s�4 s where

K1 = (s � 2)

Y ( s) �2 � 3 1 � � s s ��2 3(�2 � 4) 6

K2 = (s � 4)

Y ( s) �4�3 1 � � s s ��4 3(� 4 � 2) 6

1 1 Y ( s) = 6 � 6 s�2 s�4 s 1 1 s s Y(s) = 6 � 6 s�2 s�4 These two terms represents the admittance of a series RC circuit. For a series RC circuit

A.30�Additional Solved Mumbai University Examination Questions—ENAS (May 2018) 1 s Ri YRC(s) = 1 s� Ri Ci By direct comparison, R1 = 6 �,

C1 =

1 F 12

R2 = 6 �,

C2=

1 F 24

Fig. 3

The network is shown in Fig. 3. 2. (b) Realise the following function in Cauer-I and Cauer-II forms. Z(s) = Ans.

(s � 1)(s � 3) s2 � 2s

Refer Example 10.49 on page 10.51.

3. (a) Obtain hybrid parameter of the inter connected network.

Fig. 4 Ans. The above network can be considered as a series-parallel connection of two networks N1 and N2. The network of N1 is shown in Fig. 5. From Fig. 5,

Vx = 10I1

Applying KVL to Mesh 1, V1 – 10I1 – 4(I1 + I2) – 3Vx = 0 V1 – 10I1 – 4(I1 + I2) – 3(10I1) = 0 V1 = 44I1 + 4I2 Applying KVL to Mesh 2, V2 – 8I2 – 4(I2 + I1) – 3Vx = 0 V2 – 8I2 – 4(I2 + I1) – 3(10I1) = 0

(i)

Fig. 5

��������������������������������������������������������������������������A.31

V2 = 34I1 + 12I2

(ii)

Rewriting Eq. (ii), 12I2 = –34I1 + V2 I2 = –

1 34 I1 + V 12 2 12

= –2.83I1 + 0.0833V2

(iii)

Substituting Eq. (iii) in Eq. (i), V1 = 44I1 + 4(–2.83I1 + 0.0833V2) = 32.68I1 + 0.33V2

(iv)

Comparing Eqs (iii) and (iv) with h-parameter equations,

� h12 � � �32.68 0.33 � � h11 � ��� � h h � � � 21 22 � ��2.83 0.0833� The network N2 is shown in Fig. 6. Applying KVL to Mesh 1, V1 = 4I1 – 4I3

(i)

V2 = 6I2 + 6I3

(ii)

Applying KVL to Mesh 2, Fig. 6

Applying KVL to Mesh 3, –4(I3 – I1) – 10I3 – 6(I3 + I2) = 0 I3 =

1 3 I � I 5 1 10 2

(iii)

Substituting Eq. (iii) in Eq. (i), 3 � �1 V1 = 4I1 – 4 � I1 � I 2 � �5 10 �

= Substituting Eq. (iii) in Eq. (ii),

16 6 I � I 5 1 5 2

(iv)

3 � �1 V2 = 6I2 + 6 � I1 � I 2 � �5 10 �

=

6 21 I � I 5 1 5 2

(v)

Rewriting Eq. (v),

21 6 I =V – I 5 2 2 5 1 2 5 I2 = � I1 � V2 7 21

(vi)

A.32�Additional Solved Mumbai University Examination Questions—ENAS (May 2018) Substituting Eq. (vi) in Eq. (iv), V1 = =

16 6� 2 5 � I1 � � � I1 � V2 � 5 5� 7 21 �

20 2 I1 � V2 7 7

(vii)

Comparing Eqs (vi) and (vii) with h-parameter equations, �� � h11 � h �� � 21

� 20 h12 �� � � 7 ��� h22 �� � � 2 ��� 7

2� 7� � 5� 21 ��

Hence, the overall h-parameters of the network are

� h12 � � � h11 �� h12 �� � � h11 h12 � � h11 � ��� ��� � � h22 � � �h21 �� h22 �� � �h21 h22 � �h21 � 20 �32.68 0.33 � � 7 �� ��� ��2.83 0.0833� �� 2 �� 7 3. (b) The switch is closed at t = 0, find values of i, inductor to be zero for circuit.

Fig. 7 Ans.

Refer Example 6.1 on page 6.3.

4. (a) Find the Thevenin’s equivalent network.

Fig. 8 Ans.

Refer Example. 2.75 on page 2.75.

2� 7 � � 35.53 0.616 � ��� � 5 � ��3.116 0.3212 � 21 ��

di d 2 i , at t = 0+. Assume all initial current of dt dt 2

��������������������������������������������������������������������������A.33

4. (b) Obtain i(t) for t > 0.

Fig. 9 Ans. For t > 0, the transformed network is shown in Fig. 10.

Fig. 10 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 2

s(s � 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 s 3(s � 3) 2(s � 2) � 1 e �2 s 2 e �2 s e �2 s � � � 3 s 3 s�3 s�2 Taking inverse Laplace transform, �

i (t ) �

1 2 u(t � 2) � e �3(t � 2)u(t � 2) � e �2(t � 2)u(t � 2) for t � 0 3 3

A.34�Additional Solved Mumbai University Examination Questions—ENAS (May 2018) 5. (a) The poles and zeros of the network shown below are as follows: Poles at –1 + j 5 , –1 –j 5 , zeros at –1, –3 and the scale factor is K. If Z(0) = 1. Find the values of R, R1, L and C.

Fig. 11

Ans.

1� � R L ( R1 � Ls) � R � � R1 R � 1 � RLs � � Cs � Cs C Z(s) = � 1 R1Cs � LCs2 � RCs � 1 R1 � Ls � R � Cs Cs =

R1 RCs � R1 � RLCs2 � Ls LCs2 � RCs � R1Cs � 1

� R � 1 � �R � 1 � R �s2 � s � 1 � � � L RC � RLC � RLCs � s( R1 RC � L ) � R1 = = � LCs2 � s( RC � R1C ) � 1 1 �R R � s2 � s � � 1 � � � L L � LC 2

Z(s) = K

=K

(s � 1)(s � 3) (s � 1 � j 5)(s � 1 � j 5) s2 � 4s � 3 s2 � 2s � 1 � 5

=K

�K

(i)

(s � 1)(s � 3) (s � 1)2 � ( j 5)2

s2 � 4s � 3 s2 � 2s � 6

When Z(0) = 1 � 3� K� � = 1 � 6� K=2

Z(s) = 2 Comparing Eqs (ii) with (i), R=2

R1 1 =4 � L RC R1 =3 RLC

s2 � 4s � 3 s2 � 2s � 6

(ii)

��������������������������������������������������������������������������A.35

R R1 =2 � L L 1 =6 LC Solving the above equations, R=2� R1 = 1 � L=

3 H 2

C=

3 F 20

5. (b) Find the current Ix using superposition theorem.

Fig. 12 Ans.

Refer Example 2.50 on page 2.49.

6. (a) Check whether the following functions are prf or not. F(s) =

2 s 4 � 7s3 � 11s2 � 12 s � 4 s 4 � 5s3 � 9s2 � 11s � 6

Ans. Refer Example 2(c) of Additional Solved Mumbai University Questions ENAS December 2017. 6. (b) Find Voltages across 5 � resistor using mesh analysis.

Fig. 13 Ans.

Refer Example 4.28 on page 4.30.

Electrical Network Analysis & Synthesis (December 2017)

1. (a) Obtain transmission parameters in terms of ‘Z’ parameters. Ans. Refer Section 9.6.3 (1) on page 9.33. 1. (b) If i1 = 2 A, find V.

Fig. 1 Ans. i1 = 2 A The current due to dependent source 3i1 = 3 � 2 = 6 A Current through 10 � resistor = 6 – 4 = 2 A Hence, V = 10i10� = 10(2) = 20 V 1. (c) Obtain s-domain (Laplace Transform) equivalent circuit diagram of an inductor and capacitor with initial conditions. Ans. Refer Section 7.6(1) and 7.6(2) on page 7.12. 1. (d) Check whether the polynomial is Hurwitz or not by continued fraction method. F(s) = s4 + s3+ 4s2 + 2s + 3 Ans.

Odd part of F(s) = n(s) = s3 + 2s

Even part of F(s) = m(s) = s4 + 4s2 + 3 m( s ) Q(s) = n(s) By continued fraction expansion,

Since all the quotient terms are positive, the polynomial F(s) is Hurwitz.

�������������������������������������������������������������������������������A.37

1. (e) List the type of damping in a series R-L-C circuit and mention the condition for each damping. Ans.

Types of Damping

Condition for Damping

(i) Overdamped

R � 2L

(ii) Critically damped

R � 2L

(iii) Under damped

R � 2L

1 LC 1 LC 1 LC

2. (a) Obtain hybrid parameter of the interconnected 2-port network.

Fig. 2 Ans. The above network can be considered as a series-parallel combination of two networks N1 and N2. The network N1 is shown in Fig. 3.

Fig. 3 Applying KVL to Mesh 1, V1 = 15I1– 5I3

(i)

V2 = 6I2 + 2I3

(ii)

Applying KVL to Mesh 2, Applying KVL to Mesh 3, –5(I3 – I1) – 3I3 – 2(I3 + I2) = 0 10I3 + 2I2 – 5I1 = 0 I3 =

1 1 I – I 2 1 5 2

(iii)

A.38�Additional Solved Mumbai University Examination Questions—ENAS (December 2017) Substituting Eq. (iii) in Eq. (i), 1 � �1 V1 = 15I1 – 5 � I1 � I 2 � �2 5 �

= 12.5I1+ I2

(iv)

Substituting Eq. (iii) in Eq. (ii), 1 � �1 V2 = 6I2 + 2 � I1 � I 2 � �2 5 �

= I1 + 5.6I2

(v)

Rewriting Eq. (v), 5.6I2 = V2 – I1 I2 = –0.1785I1 + 0.1785V2

(vi)

Substituting Eq. (vi) in Eq. (iv), V1 = 12.5I1 + 0.1785V2 – 0.1785I1 = 12.3215I1 + 0.1785V2 Comparing Eqs (vi) and (vii) with h-parameters equations, � � h11 � � �h21

(vii)

h12 � � �12.3215 0.1785� �� h22 � � ���0.1785 0.1785��

The network N2 is shown in Fig. 4. From Fig. 4, Vx = 10I1 Applying KVL to Mesh 1, V1 – 10I1 – 4(I1 + I2) – 3Vx = 0 V1 – 10I1 – 4(I1 + I2) – 3(10I1) = 0 V1 = 44I1 + 4I2 Applying KVL to Mesh 2, V2 – 8I2 – 4(I2 + I1) – 3Vx = 0 V2 = 8I2 – 4(I2 + I1) – 3(10I1) = 0 V2 = 34I1 + 12I2 Rewriting Eq. (ii), 12I2 = –34I1 + V2 I2 = –2.83I1 + 0.0833V2 Substituting Eq. (iii) in Eq. (i), V1 = 44I1 + 4(–2.83I1 + 0.0833V2) = 32.68I1 + 0.33V2 Comparing Eqs (iii) and (iv) with h-parameters equations, �� � h11 � h �� � 21

h12 �� � �32.68 0.33 � �� h22 �� � ���2.83 0.0833��

Fig. 4

(i)

(ii)

(iii)

(iv)

�������������������������������������������������������������������������������A.39

Hence, the overall h-parameters of the network are � h11 � �h21

h12 � � h11 � ��� h22 � �h21 �

h12 � � � h11 �� ��� h22 � � �h21 ��

h12 �� � � 12.32 0.1785� �32.68 0.33 � � � = h22 �� � ���0.1785 0.1785�� ���2.83 0.0833��

0.5085� � 45 = � � � 3.0085 0.2618 � �

2. (b) Find current through 15 � resistor.

Fig. 5 Ans. The equivalent circuit in terms of dependent sources is shown in Fig. 6.

Fig. 6 Applying KVL to Mesh 1, 120 – 5 I1 � j 20(I1 � I2 ) � j15I2 = 0 (5 + j20)I1 – j5 I2 = 120

(i)

Applying KVL to Mesh 2, j15I2 – j20(I2 – I1) – j10I2 – j15(I1 – I2) – 15I2 = 0 –j5I1 + 15I2 = 0 Writing Eqs (i) and (ii) in matrix form, �5 � j 20 � j 5� � I1 � �120 � � �� � � � � 15 � �I2 � � 0 � � � j5 By Cramer’s rule,

I2 �

5 � j 20 120 � j5 0 5 � j 20 � j 5 � j5 15

= 1.9�18.43° A

(ii)

A.40�Additional Solved Mumbai University Examination Questions—ENAS (December 2017) 2 s 4 � 7s3 � 11s2 � 12 s � 4

2. (c) Test whether F(s) =

Ans.

s 4 � 5s3 � 9s2 � 11s � 6

is a positive real function.

N (s) 2 s 4 � 7s3 � 11s2 � 12 s � 4 � 4 D( s ) s � 5s3 � 9s2 � 11s � 6

F(s) =

(i) By Routh array, N(s) = 2s4 + 7s3 + 11s2 +12s + 4 s4

D(s) = s4 + 5s3 + 9s2 + 11s + 6

4

s4

1

9

6

12

0

s3

5

11

0

53 7

4

0

s2

34 5

6

0

s1

440 53

0

s1

112 17

0

s0

4

s0

6

2

s3

7

s2

11

Since there is no sign change in first columns of the arrays, N(s) and D(s) are Hurwitz. (ii)

N(s) = 2s4 + 7s3 + 11s2 + 12s + 4 = (s + 2)(s + 0.5)(s + 0.5 + j1.323)(s + 0.5 – j1.323) D(s) = s4 + 5s3 + 9s2 + 11s + 6 = (s + 3)(s + 1)(s + 0.5 + j1.323)(s + 0.5 – j1.323) There are no poles on the j��axis. Hence, residue test is not carried out.

(iii) Even part of N(s) = m1 = 2s4 + 11s2 + 4 Odd part of N(s) = n1 = 7s3 + 12s Even part of D(s) = m2 = s4 + 9s2 + 6 Odd part of D(s) = n2 = 5s3 + 11s A(�)2 = m1m2 – n1n2 =

(2s4

=

2s8

+

+

11s2

18s6

3 3 + 4)(s4 + 9s2 + 6) – (7s � 12 s)(5s � 11s)

+ 6

s � j�

12s4

+

11s6

+

4

4

4

2

99s4

+

6

= 2 s � 6 s � 22 s � 30 s � 24

+

4s4

+

36s2

2

– (35s � 77s � 60 s � 23s ) 8

66s2 s � j�

s � j�

= 2(j�)8 – 6(j�)6 – 22(j�)4 – 30(j�)2 + 24 = 2�8 + 6�6 – 22�4 + 30�2 + 24 A(�)2

� 0 for all ��� 0 Since all the three conditions are satisfied, the function is positive real.

s � j�

+ 24

�������������������������������������������������������������������������������A.41

3. (a) Obtain Thevenin’s equivalent circuit.

Fig. 7 Hence find current flowing through 10 � load. Ans. Step I: Calculation of VTh (Fig. 8)

Fig. 8 From Fig. 8, Vx = –2I1 Ix = I2 Meshes 1 and 2 will form a supermesh. Writing the current equation for the supermesh, I2 – I1 = 4Vx = 4(–2I1) = –8I1 7I1 + I2 = 0

(i)

Applying KVL to the outer path of the supermesh, 4 – 2I1 – 5I1 – 4I2 – 7I2 – 8 – 3I2 – 1I1 = 0 8I1 + 14I2 = –4 Solving Eqs (i) and (ii), I1 = 0.044 A I2 = –0.31 A Writing the VTh equation, 6 + 8 + 7I2 – 3Ix –VTh = 0 14 + 7(–0.31) – 3(–0.31) – VTh = 0 VTh = 12.76 V

(ii)

A.42�Additional Solved Mumbai University Examination Questions—ENAS (December 2017) Step II: Calculation of IN

Fig. 9 From Fig. 9, Vx = –2I1 Ix = I2 – I3 Meshes 1 and 2 will form a supermesh Writing the current equation for the supermesh. I2 – I1 = 4Vx = 4(–2I1) = –8I1 7I1 + I2 = 0

(i)

Applying KVL to the outer path of the supermesh, 4 – 2I1 – 5I1 – 4I2 – 7(I2 – I3) – 8 – 3I2 – 1I1 = 0 –8I1 – 14I2 + 7I3 = 4

(ii)

Applying KVL to Mesh 3, 6 – 6I3 + 8 – 7(I3 – I2) – 8I3 – 3Ix = 0 6 – 6I3 + 8 – 7(I3 – I2) – 8I3 – 3(I2 – I3) =0 4I2 – 18I3 = –14

(iii)

Solving Eqs (i), (ii) and (iii), I1 = – 0.018 A I2 = 0.128 A I3 = 0.806 A IN = I3 = 0.806 A Step III: Calculation of RTh RTh =

VTh 12.76 � = 15.83 � IN 0.806

Step IV: Calculation of IL IL =

12.76 = 0.49 A 15.83 � 10

Fig. 10

�������������������������������������������������������������������������������A.43

3. (b) For the network shown in Fig. 11, find the voltage across the capacitor.

Fig. 11 Ans.

Refer Example 3.6 on page 3.5.

4. (a) Find IC and VC for t > 0.

Fig. 12 Ans. At t = 0–, there is no current through inductor and no voltage across capacitor. iL(0–) = 0 vC(0–) = 0 + At t = 0 , iL (0+) = 0 vC(0+) = 0 For t > 0, the transformed network is shown in Fig. 13. Fig. 13 Writing KVL equation in matrix form,

�20 � 2 � 10 �3 s �2 � 10 �3 s � � � � I1 � 1 �3 ��20 � 10 �3 s � �I � = 60 � 2 � 10 s � �6 � � 2 � �� 5 � 10 s � By Cramer’s rule, 480 20 � 2 � 10 �3 s I2 =

�2 � 10 �3 s �3

� 480 � � s � � � � 0 �

0 �3

20 � 2 � 10 s �2 � 10 s �2 � 10 �3 s

60 � 2 � 10 �3 s �

1 � 5 10 �6 s

A.44�Additional Solved Mumbai University Examination Questions—ENAS (December 2017)

=

=

= = =

480 (2 � 10 �3 s) s 1 � � �3 2 (20 � 2 � 10 �3 s) � 60 � 2 � 10 �3 s � � � (2 � 10 s) � 5 � 10 �6 s � 0.96 4 � 106 1200 � 0.04 s � � 0.12 s � 4 � 10 �6 s2 � 400 � 4 � 10 �6 s2 s 0.96 s 1200 s � 0.04 s2 � 4 � 106 � 0.12 s2 � 400 s 0.96 s 2

0.16 s � 1600 s � 4 � 106 6s 2

s � 10000 s � 25 � 106 6s

=

(s � 5000)2 By partial fraction expansion, I2 =

A B � s � 5000 (s � 5000)2

6s = A(s + 5000) + B Putting s = –5000, B = –30000 Putting s = 0,

A=6 I2 =

6 30000 � s � 5000 (s � 5000)2

Taking inverse Laplace transform, iC(t) = i2(t) = 6e–5000t – 30000te–5000t VC(s) =

1 5 � 10 �6 s

I 2 ( s) �

1

5 � 10 �6 s (s � 5000)2

By partial fraction expansion, VC =

A B � s � 5000 (s � 5000)2

12 � 105 = A(s + 5000) + B Putting s = –5000, B = 12 � 105 Putting s = 0,

A=0 VC(s) =

12 � 105 (S � 5000)2

6s



12 � 105 (s � 5000)2

�������������������������������������������������������������������������������A.45

Taking inverse Laplace transform, vC(t) = 12 � 105te–5000t 4. (b) Realise the following function in Foster-I and Foster-II form. Z ( s) �

(s � 1)(s � 3) (s � 2)(s � 4)

Ans. The function Z(s) has poles at s = –2 and s = –4, and zeros at s = –1 and –3. From the pole-zero diagram, it is clear that poles and zero are simple and lie on the negative real axis. Poles and zeros are interlaced and the critical frequency nearest to the origin is a zero. Hence, the given function is an RL function. Foster I Form: The Foster I form is obtained by the partial fraction Z ( s) . expansion of s (s � 1)(s � 3) Z ( s) = s(s � 2)(s � 4) s By partial fraction expansion, K K K Z ( s) = 0 � 1 � 2 s s�2 s�4 s where

K0 = s

Fig. 14

Z ( s) (1)(3) 3 � � s s � 0 (2)(4) 8

K1 = (s � 2)

1 Z ( s) (�2 � 1)(�2 � 3) = � 4 s s ��2 (�2)(�2 � 4)

K2 = (s � 4)

Z ( s) (�4 � 1)(�4 � 3) 3 � � s s ��4 (�4)(�4 � 2) 8

3 3 1 Z ( s) 8 8 4 = � � s s s�2 s�4 3 1 s s 3 4 8 Z(s) = � � 8 s�2 s�4 3 The first term represents the impedance of a resistor of �. The other two terms represent the impedance 8 of the parallel RL circuit for which

ZRL(s) =

Ri s R s� i Li

A.46�Additional Solved Mumbai University Examination Questions—ENAS (December 2017) By direct comparison, 1 R1 = �, 4

L1 =

1 H 8

3 3 �, L2 = H 8 32 The network is shown in Fig. 15. Foster II Form: The Foster II form is obtained by the Fig. 15 partial fraction expansion of Y(s). Since the degree of numerator is equal to the degree of denominator, division is carried out first.

R2 =

Y(s) =

(s � 2)(s � 4) s2 � 6s � 8 � (s � 1)(s � 3) s2 � 4 s � 3

s2 � 4 s � 3) s2 � 6 s � 8 (1 s2 � 4s � 3 2s � 5 Y(s) = 1 +

2s � 5 2

s � 4s � 3

� 1�

2s � 5 (s � 1)(s � 3)

By partial fraction expansion, Y(s) = 1 + where

K1 K � 2 s �1 s � 3

K1 = (s � 1)Y (s) s ��1 �

(�1 � 2)(�1 � 4) 3 � (�1 � 3) 2

K2 = (s � 3)Y (s) s ��3 �

(�3 � 2)(�3 � 4) 1 � (�3 � 1) 2

3 1 Y(s) = 1 + 2 � 2 s �1 s � 3 The first term represents the admittance of a resistor of 1 �. 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

By direct comparison, 2 R1 = �, 3 R2 = 6 �,

L1 =

2 H 3

L2 = 2 H

The network is shown in Fig. 16.

Fig. 16

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5. (a) Find driving point impedance

V1 for the network shown in Fig. 17. I1

Fig. 17 Ans. The transformed network is shown in Fig. 18.

Fig. 18

Fig. 19 From Fig. 19, e = V1 – 1I1 = V1 – I1 Applying KVL to the mesh, � 4 � I – 2e = 0 V1 – 1I1 – � � s � 3 �� 1 � 4 � I – 2(V1 – I1) = 0 V1 – I1 – � � s � 3 �� 1 4 � � –V1 = � 1 � �2 I � s � 3 �� 1

A.48�Additional Solved Mumbai University Examination Questions—ENAS (December 2017) � s � 3 � 4 � 2(s � 3) � V1 = – � � I1 s�3 � � V1 �(s � 3 � 4 � 2 s � 6) �(� s � 1) s � 1 � � � I1 s�3 s�3 s�3 5. (b) Find i1(t), i2(t) and i3(t) at t = 0+.

Fig. 20 Ans. At t =

0 –,

the network is shown in Fig. 21.

Fig. 21 At t = 0–, the network attains steady-state condition. Hence, the inductor acts as a short circuit and the capacitor acts as open circuit. Applying KVL to the loop, 10 – 2i1(0–) – 8i1(0–) – 5 = 0

�������������������������������������������������������������������������������A.49

i1(0–) = 0.5 A (0–)

v1

+ v2(0–) = 5 + 8(0.5) = 9 V

Since the charges on capacitor are equal when connected in series, Q1 = Q2, C1v1 = C2v2 v1 (0 � )



C2 3 � C1 4

v1(0–) =

3 v (0–) 4 2



v2 (0 )

3 v (0–) + v2(0–) = 9 4 2 36 v2(0–) = V 7

v1(0–) =

27 V 7

At t = 0+, the network is shown in Fig. 22.

Fig. 22 At t = 0+, the inductor is replaced by a current source of 0.5 A and the capacitors are replaced by voltage 27 36 V and V respectively. sources of 7 7 27 v1(0+) = V 7 v2(0+) = At t = 0+,

Also,

10 – 2i1(0+) –

36 V 7

27 36 – =0 7 7 i1(0+) = 0.5 A

i1(0+) – i3(0+) = 0.5

A.50�Additional Solved Mumbai University Examination Questions—ENAS (December 2017) I3(0+) = 0 Applying KVL to the Mesh 2, 27 – 4[i2(0+) – i3(0+)] = 0 7 27 5 – 8i2(0+) + 8(0.5) – – 4i2(0+) + 4(0) = 0 7 i2(0+) = 0.429 A

5 – 8[i2(0+) – i1(0+)] –

6. (a) Find the characteristic impedance, cut-off frequency and pass band for the network shown in Fig. 23.

Fig. 23 Ans.

Refer Example 11.7 on page 11.16.

6. (b) For given circuit, the switch is closed at t = 0. Find vC(t) for t > 0.

Fig. 24 Ans.

At t = 0–,

vC(0–) = 0

Since the voltage across the capacitor cannot change instantaneously, vC(0+) = 0 Since the resistor of 3 � is connected in parallel with the voltage source of 5 V, it becomes redundant. For t > 0, the network is shown in Fig. 25. Writing KCL equation for t > 0, vC � 5 dv �1 C = 0 100 dt 100

dvC + vC = 5 dt

dvC + 0.01vC = 0.05 dt

Fig. 25

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Comparing with the differential equation

dv � Pv � Q, dt

P = 0.01, Q = 0.05 The solution of this differential equation is given by, vC(t) = e–Pt � Qe � Pt dt � ke � Pt = e–0.01t � 0.05e0.01t dt � ke �0.01t 0.05 � ke �0.01t 0.01 = 5 + ke–0.01t =

At t = 0,

vC(0) = 0 0=5+k k = –5 vC(t) = 5 – 5e–0.01t = 5(1 – e–0.01t)

for t > 0

6. (c) The network shown in Fig. 26 reaches a steady state switch at position 1. At t = 0, the switch is di d 2 i , at t = 0+. changed from the position 1 to position 2. Find the value of i, dt dt 2

Fig. 26 Ans.

At t = 0 –, the network attains steady-state condition.

Hence, the inductor acts as a short circuit. 30 = 1.5 A 20 At t = 0+, the network is shown in Fig. 27.

i(0–) =

At t =

0+,

Fig. 27

the inductor is acts as a current source of 1.5 A. i(0+) = 1.5 A

For t > 0, the network is shown in Fig. 28. Writing the KVL equation for t > 0, –30i – 20i – 1

di =0 dt

(i) Fig. 28

A.52�Additional Solved Mumbai University Examination Questions—ENAS (December 2017) At t = 0+,

–50 i(0+) –

di + (0 ) = 0 dt di + (0 ) = –50 � 1.5 = 75 A/s dt

Differentiating Eq. (i), –50 At t = 0+,

–50

di d2i – =0 dt dt 2

di + d2i (0 ) – 2 (0+) = 0 dt dt d2i dt 2

(0+) = (–50)(–70) = 3500 A/s2

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 Attenuation Band 11.1

B Band-Pass Filter 11.1, 11.18 Band-Stop Filter 11.1, 11.22 Branch 2.1 Branch Admittance Matrix. 5.27

C Capacitance 1.3 Cascade Connection 9.47 Cauer I Form 10.33 Cauer II Form 10.34 Cauer Realisation or Ladder Realisation 10.33 Characteristic Impedance 11.2 Characteristic of Filters 11.6 Co-Tree 5.4 Coefficient of Coupling (k) 4.2 Complete Incidence Matrix (Aa) 5.6 Condition for Reciprocity 9.2 Condition for Symmetry 9.3 Conductively Coupled Equivalent Circuits 4.37 Connected Graph 5.3 Constant-k Half Sections 11.25 Constant-k High-pass filter 11.14 Constant-k Low Pass Filter 11.7

Cosine Function 7.3 Coupled Circuits 4.15 Critically Damped 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 Cutset Matrix 5.10

D Degree of a Node 5.1 Delayed Unit-Ramp Function 7.3 Delayed or Shifted Unit-Step Function 7.2 Dependent Sources 1.5 Differential Coupling 4.4 Directed or Oriented Graph 5.1 Dot Convention 4.9 Driving Point and Transfer Function 8.1 Driving-Point Admittance Function 8.2 Driving-Point Functions 8.1 Driving-Point Impedance at Input Port 9.69 Driving-Point Impedance at Output Port 9.71 Driving-point Impedance Function 8.2

E Elementary Synthesis Concepts 10.24 Energy 1.1 Exponential Function (eat) 7.3 Exponentially Damped Function 7.4

I.2�Index

F Farad 1.3 Final Value Theorem 7.7 Forced Response 6.28 Foster I Form 10.31 Foster II Form 10.32 Foster Realisation 10.31 Fundamental Circuit (Tieset) 5.8 Fundamental Circuit Matrix 5.8 Fundamental Cutset 5.11 Fundamental Cutset Matrix 5.11

G Graphical Method for Determination of Residue 8.42

H Henry 1.2 High-Pass Filter 11.1 Hurwitz Polynomials 10.1 Hybrid Parameters (h Parameters) 9.24 Hyperbolic Cosine Function 7.4 Hyperbolic Sine Function 7.4

I Incidence Matrix 5.6 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 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 Inter-relationships between the Parameters 9.29 Interconnection of Two-Port Networks 9.47 Inverse Laplace Transform 7.7

K Kirchhoff’s Current Law 5.24

Kirchhoff’s Current Law (KCL) 2.1 Kirchhoff’s Laws 2.1 Kirchhoff’s Voltage Law 5.24 Kirchhoff’s Voltage Law (KVL) 2.2

L Laplace Transformation 7.1 Laplace Transforms of Some Important Functions 7.2 Lattice Networks 9.66 Linear Graph 5.1 Linear and Non-linear Elements 1.6 Loop 2.1 Loop Impedance Matrix 5.27 Loop Matrix or Circuit Matrix 5.8 Loop or Circuit 5.3 Low-Pass Filter 11.1 Lumped and Distributed Elements 1.7

M Maximum Power Transfer Theorem 3.51, 2.106 Mesh 2.1 Mesh Analysis 2.2, 3.1 Millman’s Theorem 2.122, 3.68 Mutual Inductance 4.2 Mutually Induced Emf 4.2

N Natural Response 6.28 Network Topology 5.1 Network and Circuit 1.6 Node 2.1 Node Analysis 2.23 Node Analysis 3.9 Non-planar Graph 5.2 Norton’s Theorem 2.82, 3.41 Number of Possible Trees of a Graph 5.7

O Ohm 1.1 One-Port Network 8.1 Open-Circuit Impedance Parameters (Z Parameters) 9.2 Output Port 8.1 Overdamped 6.67

������I.3

P � Network 9.61, 11.4 Parallel Connection 9.51 Partial Fraction Expansion 7.8 Pass Band 11.1 Path 5.3 Planar Graph 5.2 Poles and Zeros of Network Functions 8.20 Positive Real Functions 10.16 Potential Difference 1.1 Propagation Constant 11.2 Properties of Hurwitz Polynomials 10.1 Properties of Laplace Transform 7.4 Properties of Positive Real Functions 10.16

R Rank of a Graph 5.3 Realisation of LC Functions 10.30 Realisation of RC Functions 10.47 Realisation of RL Functions 10.63 Reciprocity Theorem 2.118, 3.64 Reduced Incidence Matrix (A) 5.6 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, 6.67 Response of RC Circuit to Various Functions 7.31, 7.39

S Self-Induced emf 4.1 Self-Inductance 4.1 Series Connection 9.57 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-Parallel Connection 9.59 Short-Circuit Admittance Parameters (Y Parameters) 9.8 Sine Function 7.3 Source Shifting 1.19 Source Transformation 1.13 Sources 1.4 Stability of the Network 8.42

Star-Delta Transformation 1.10 Steady-State 6.28 Steady-State Response 6.1 Stop Band 11.1 Sub-graph 5.2 Supermesh Analysis 2.15 Supernode Analysis 2.36 Superposition Theorem 2.42, 3.14

T T-Network 9.61, 11.1 Terminal Pair or Port 8.1 Terminated Two-Port Networks 9.69 Terminating Half sections 11.25 Thevenin’s Theorem 2.62, 3.27 Time-Domain Behaviour from the Pole-Zero Plot 8.39 Time-invariant and Time-variant Networks 1.7 Transfer Admittance Function 8.3 Transfer Functions 8.2 Transfer Impedance Function 8.3 Transformed Circuit 7.12 Transient Period 6.1 Transient Response 6.1, 6.28 Transients 6.1 Transmission Parameters (ABCD Parameters) 9.18 Tree 5.4

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 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