Network Analysis
and
Synthesis
About the Author
Ravish R. Singh is presently Vice-Principal at Thakur College of Engineering and
Technology, Mumbai. He obtained a BE degree from University of Mumbai in 1991
and an MTech degree from IIT Bombay in 2001. He is currently involved in doctoral
research at the Faculty of Technology, University of Mumbai. He has published three
books, namely Electrical Networks, Engineering Mathematics and Basic Electrical
and Electronics Engineering with McGraw Hill Education (India). He is a member
of IEEE, ISTE, IETE, CSI and has published several acclaimed research papers in
national and international journals. His fields of interest include Circuits, Signals and
Systems and Engineering Mathematics.
Network Analysis
and
Synthesis
Ravish R Singh
Thakur College of Engineering and Technology
Mumbai
McGraw Hill Education (India) Private Limited
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Network Analysis and Synthesis
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Dedicated to
My Father
Late Shri Ramsagar Singh
and
My Mother
Late Shrimati Premsheela Singh
Contents
Preface
xiii
1. BASIC NETWORK CONCEPTS
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.1
Introduction 1.1
Resistance 1.1
Inductance 1.2
Capacitance 1.7
Sources 1.14
Some Definitions 1.15
Series and Parallel Combination of Resistors 1.16
Series and Parallel Combination of Inductors 1.26
Series and Parallel Combination of Capacitors 1.28
Star-Delta Transformation 1.32
Source Transformation 1.51
Source Shifting 1.58
Exercises 1.60
Objective-Type Questions 1.64
Answers to Objective-Type Questions 1.65
2. ELEMENTARY NETWORK THEOREMS
2.1
2.2
2.3
2.4
2.5
2.6
Introduction 2.1
Kirchhoff’s Laws 2.1
Mesh Analysis 2.20
Supermesh Analysis 2.35
Node Analysis 2.43
Supernode Analysis 2.63
Exercises 2.69
Objective-Type Questions 2.73
Answers to Objective-Type Questions
2.75
3. NETWORK THEOREMS (APPLICATION TO dc NETWORKS)
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
2.1
Introduction 3.1
Superposition Theorem 3.1
Thevenin’s Theorem 3.30
Norton’s Theorem 3.64
Maximum Power Transfer Theorem 3.91
Reciprocity Theorem 3.112
Millman’s Theorem 3.116
Tellegen’s Theorem 3.121
Substitution Theorem 3.124
3.1
viii Contents
3.10 Compensation Theorem 3.126
Exercises 3.128
Objective-Type Questions 3.133
Answers to Objective-Type Questions 3.134
4. SINGLE-PHASE ac CIRCUITS
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
4.13
4.14
4.15
Introduction 4.1
Generation of Alternating Voltages 4.1
Terms Related to Alternating Quantities 4.2
Root Mean Square (rms) or Effective Value 4.3
Average Value 4.4
Phasor Representations of Alternating Quantities 4.23
Mathematical Representations of Phasors 4.27
Behaviour of a Pure Resistor in an ac Circuit 4.34
Behaviour of a Pure Inductor in an ac Circuit 4.36
Behaviour of a Pure Capacitor in an ac Circuit 4.38
Series RL Circuit 4.41
Series RC Circuit 4.61
Series RLC Circuit 4.68
Parallel ac Circuits 4.81
Locus Diagram 4.100
Exercises 4.103
Objective-Type Questions 4.111
Answers to Objective-Type Questions 4.114
5. RESONANCE
5.1
5.2
5.3
5.4
5.1
Introduction 5.1
Series Resonance 5.1
Parallel Resonance 5.18
Comparison of Series and Parallel Resonant Circuits 5.21
Exercises 5.38
Objective-Type Questions 5.39
Answers to Objective-Type Questions 5.40
6. NETWORK THEOREMS
(APPLICATION TO ac CIRCUITS)
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
4.1
Introduction 6.1
Mesh analysis 6.1
Node Analysis 6.9
Superposition Theorem 6.14
Thevenin’s Theorem 6.27
Norton’s Theorem 6.41
Maximum Power Transfer Theorem 6.51
Reciprocity Theorem 6.64
Millman’s Theorem 6.68
6.1
Contents ix
6.10 Tellegen’s Theorem 6.72
6.11 Substitution Theorem 6.73
6.12 Compensation Theorem 6.76
Exercises 6.78
Objective-Type Questions 6.80
Answers to Objective-Type Questions
6.81
7. COUPLED CIRCUITS
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
7.10
Introduction 7.1
Self-Inductance 7.1
Mutual Inductance 7.2
Coefficient of Coupling (k) 7.2
Inductances in Series 7.3
Inductances in Parallel 7.4
Dot Convention 7.9
Coupled Circuits 7.15
Conductively Coupled Equivalent Circuits 7.41
Tuned Circuits 7.44
Exercises 7.51
Objective-Type Questions 7.54
Answers to Objective-Type Questions 7.55
8. THREE-PHASE CIRCUITS
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
8.10
8.11
8.12
8.13
7.1
8.1
Introduction 8.1
Advantages of a Three-Phase System 8.2
Some Definitions 8.2
Interconnection of Three Phases 8.2
Star, or WYE, Connection 8.3
Delta, or Mesh, Connection 8.4
Voltage, Current and Power Relations in a Balanced Star-Connected Load 8.4
Voltage, Current and Power Relations in a Balanced Delta-Connected Load 8.6
Balanced Y/Δ And Δ/Y Conversions 8.8
Relation between Power in Delta and Star Systems 8.8
Comparison between Star and Delta Connections 8.9
Three-Phase Unbalanced Circuits 8.28
Measurement of Three-Phase Power 8.38
Exercises 8.50
Objective-Type Questions 8.53
Answers to Objective-Type Questions 8.55
9. NETWORK TOPOLOGY
9.1 Introduction 9.1
9.2 Graph of a Network 9.1
9.3 Definitions Associated with a Graph 9.2
9.1
x Contents
9.4
9.5
9.6
9.7
9.8
9.9
9.10
Incidence Matrix 9.6
Loop Matrix or Circuit Matrix 9.8
Cutset Matrix 9.10
Relationship Among Submatrices of A, B and Q 9.12
Kirchhoff’s Voltage Law 9.24
Kirchhoff’s Current Law 9.24
Relation Between Branch Voltage Matrix Vb, Twig Voltage Matrix Vt and Node
Voltage Matrix Vn 9.25
9.11 Relation Between Branch Current Matrix Ib and Loop Current Matrix Il 9.26
9.12 Network Equilibrium Equation 9.26
9.13 Duality 9.52
Exercises 9.58
Objective-Type Questions 9.60
Answers to Objective-Type Questions 9.61
10. TRANSIENT ANALYSIS
10.1
10.2
10.3
10.4
10.5
Introduction 10.1
Initial Conditions 10.1
Resistor–Inductor Circuit 10.27
Resistor–Capacitor Circuit 10.49
Resistor–Inductor–Capacitor Circuit 10.66
Exercises 10.79
Objective-Type Questions 10.82
Answers to Objective-Type Questions 10.85
11. LAPLACE TRANSFORM AND ITS APPLICATION
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
11.10
11.11
11.12
11.13
11.14
11.15
10.1
Introduction 11.1
Laplace Transformation 11.1
Laplace Transforms of Some Important Functions 11.2
Properties of Laplace Transform 11.4
Laplace Transform of Periodic Functions 11.15
Waveform Synthesis 11.21
Inverse Laplace Transform 11.30
Solution of Differential Equations with Constant Coefficients 11.36
Solution of a System of Simultaneous Differential Equations 11.39
The Transformed Circuit 11.42
Resistor–Inductor Circuit 11.43
Resistor–Capacitor Circuit 11.49
Resistor–Inductor–Capacitor Circuit 11.54
Response of RL Circuit to Various Functions 11.60
Response of RC Circuit to Various Functions 11.68
Exercises 11.79
Objective-Type Questions 11.81
Answers to Objective-Type Questions 11.83
11.1
Contents xi
12. NETWORK FUNCTIONS
12.1
12.1
12.2
12.3
12.4
12.5
12.6
12.7
Introduction 12.1
Driving-Point Functions 12.1
Transfer Functions 12.2
Analysis of Ladder Networks 12.5
Analysis of Non-Ladder Networks 12.15
Poles and Zeros of Network Functions 12.20
Restrictions on Pole and Zero Locations for Driving-Point Functions [Common Factors in N(s)
and D(s) Cancelled] 12.21
12.8 Restrictions on Pole and Zero Locations for Transfer Functions [Common Factors in N(s) and
D(s) Cancelled] 12.21
12.9 Time-Domain Behaviour from the Pole-Zero Plot 12.39
12.10 Graphical Method for Determination of Residue 12.42
Exercises 12.50
Objective-Type Questions 12.53
Answers to Objective-Type Questions 12.55
13. TWO-PORT NETWORKS
13.1
13.2
13.3
13.4
13.5
13.6
13.7
13.8
13.9
13.10
13.11
13.12
13.13
13.14
Introduction 13.1
Open-Circuit Impedance Parameters (Z Parameters) 13.2
Short-Circuit Admittance Parameters (Y Parameters) 13.8
Transmission Parameters (ABCD Parameters) 13.18
Inverse Transmission Parameters (A′B′C′D′ Parameters) 13.24
Hybrid Parameters (h Parameters) 13.28
Inverse Hybrid Parameters (g Parameters) 13.33
Inter-relationships Between the Parameters 13.37
Interconnection of Two-Port Networks 13.63
T-Network 13.79
Pi (p )-Network 13.79
Lattice Networks 13.84
Terminated Two-Port Networks 13.87
Image Parameters 13.97
Exercises 13.101
Objective-Type Questions 13.104
Answers to Objective-Type Questions 13.107
14. FOURIER ANALYSIS
14.1
14.2
14.3
14.4
14.5
14.6
14.7
13.1
Introduction 14.1
Trigonometric Fourier Series 14.1
Waveform Symmetry 14.9
Exponential Fourier Series 14.25
Average Value of a Periodic Complex Wave 14.33
RMS Value of Periodic Complex Wave 14.33
Power Supplied by Complex Wave 14.34
14.1
14.8
14.9
14.10
14.11
14.12
Fourier Transform 14.37
Fourier Transforms of Some Useful Functions 14.38
Fourier Transform of Periodic Function 14.47
Properties of Fourier Transform 14.50
Energy Density Spectrum 14.58
Exercises 14.60
Objective-Type Questions 14.62
Answers to Objective-Type Questions 14.64
15. FILTERS AND ATTENUATORS
15.1
15.2
15.3
15.4
15.5
15.6
15.7
15.8
15.9
15.10
15.11
15.12
15.13
15.14
15.15
15.16
15.17
15.18
15.19
Introduction 15.1
Classification of Filters 15.1
T-Network 15.1
p Network 15.4
Characteristic of Filters 15.6
Constant-k Low Pass Filter 15.7
Constant-k High-pass Filter 15.14
Band-pass Filter 15.18
Band-stop Filter 15.22
m-Derived Filters 15.25
m-Derived Low-Pass Filter 15.28
m-Derived High-Pass Filter 15.31
Terminating Half Sections 15.34
Composite Filter 15.37
Attenuator 15.40
Lattice Attenuator 15.41
T-Type Attenuator 15.42
p -Type Attenuator 15.45
Ladder-Type Attenuator 15.47
Exercises 15.48
Objective-Type Questions 15.49
Answers to Objective-Type Questions
15.1
15.50
16. NETWORK SYNTHESIS
16.1
16.2
16.3
16.4
16.5
16.6
16.7
Index I.1
Introduction 16.1
Hurwitz Polynomials 16.1
Positive Real Functions 16.16
Elementary Synthesis Concepts 16.24
Realisation of LC Functions 16.30
Realisation of RC Functions 16.47
Realisation of RL Functions 16.63
Exercises 16.72
Objective-Type Questions 16.74
Answers to Objective-Type Questions 16.76
16.1
Preface
Overview
Network Analysis and Synthesis is one of the core subjects for students of Electrical Engineering, Electronics Engineering, Electronics and Telecommunication Engineering, Instrumentation Engineering, Biomedical
Engineering and related branches of engineering in the third/fourth semester course of almost all universities
in India. This subject is also one of the important topics of competitive examinations such as IAS, IES, etc.
and examinations conducted by various public-sector undertakings in this field.
About the Book
Network Analysis and Synthesis is a complete solution for the undergraduate student who wishes to prepare
for university and competitive examinations. This highly illustrative text makes the subject matter interesting
and easy for students.
As per the course requirement, five major topics that are mandatory to be covered by any book on the
subject are dc and ac circuits, transient analysis, network function, two-port networks and network synthesis.
Generally, numerical problems are expected in university examinations in this subject. The weightage given
to problems is more than 70-80% in examinations. This book attempts to cover almost all the topics and
solved problems on these important topics. The simplistic presentation style sequentially takes the readers
from basic to advance concepts and the crisp theory helps in ready understanding. Numerous solved examples
and graded problems have been provided in this book, especially in important chapters from the examination
viewpoint. Objective-type questions from various papers of competitive examinations are included in each
chapter. This will help students in sharpening their knowledge about core concepts. Covering both analysis
and synthesis of networks, the text provides a simple explanation about the concepts of electrical networks
with brief theory and a large number of problems. Numerous examples and exercise problems have been
included to help the reader develop an intuitive grasp of the contents.
Salient Features
●
●
●
●
●
Span of coverage meets curriculum requirements
Simplistic presentation and logical sequencing of topics
Graded problem-solving approach
Numerous solved examples from frequently asked examination questions
Rich pedagogy
*
*
*
Illustrations: 1320
Solved Examples: 1065
Practice Problems: 340
Chapter Organisation
The entire text is organised into 16 chapters, primarily based on the electrical and electronics engineering
syllabi in BE/B.Tech undergraduate courses of universities around the country. The summary of the book can
be organised in the following major parts:
●
●
●
Circuit Elements and Network Analysis
Network Theorems for dc and ac Circuits
Resonance and Coupled Circuits
xiv Preface
●
●
●
●
Network Topology/Graph Theory
Transient Analysis and Laplace Transforms
Two-Port Networks
Network Functions
Within this structure, a more in-depth and detailed breakdown can be obtained from the table of contents.
The fluid flow of text dealing with different topics as well as illustrations, examples, questions and numerical
problems are extremely relevant and appropriate for students as well as instructors. The main strength of the
book thus is its strong focus towards students’ readability and understanding with the scope for independent
study and problem-solving skill development. I am confident that the depth of this book has been judiciously
developed so that students not only treat this as a textbook, but, in addition, can also gather enough practical
and theoretical knowledge to appear in national-level competitive examinations and interviews.
Web Supplements
The text is supplemented with an exhaustive Online Learning Center, which can be accessed at
https://www.mhhe.com/singh/nas1
Acknowledgements
I would like to thank all the staff at McGraw Hill Education (India), especially Shalini Jha, Koyel Ghosh,
Sagar Divekar, Satinder Singh Baveja, Sohini Mukherjee and Anuj Kr. Shrivastava for coordinating with
me during the editorial, copyediting and production stages of this book. My acknowledgements would be
incomplete without a mention of the contribution of my family members. I feel indebted to my father and
mother for their lifelong inspiration. I also send a heartfelt thanks to my wife, Nitu; son, Aman; and daughter,
Aditri, for always motivating and supporting me during the preparation of the project. Finally, I would like
to appreciate the role of all those reviewers who, time and again have provided valuable inputs on the subject
of Network Analysis and Synthesis which have significantly enriched this text and provide their constructive
comments and inputs. Their names are given below:
Gagan Deep Yadav
Yamuna Institute of Engineering and Technology, Yamuna Nagar, Haryana
Mukesh Kumawat
Star Academy Technology and Management, Indore, Madhya Pradesh
Kshipra Kshingwekar
Jai Narain College of Technology (JNCT), Bhopal, Madhya Pradesh
Santosh Sharma
Rajasthan Technical University, Kota, Rajasthan
Nikhil Gupta
Malaviya National Institute of Technology, Jaipur, Rajasthan
Prithviraj Purkait
Haldia Institute of Technology, Haldia, West Bengal
K V V S R Chowdary
Kalinga Institute of Industrial Technology (KIIT) University, Bhubaneswar,
Odisha
Dhananjay Rao
Kalinga Institute of Industrial Technology (KIIT) University, Bhubaneswar,
Odisha
Pranita Joshi
Atharva College of Engineering, Mumbai, Maharashtra
Shamala Rajaram Mahadik Dr. J J Magdum College of Engineering, Kolhapur, Maharshtra
Preface xv
Ch. Venkata Ramesh
NITTE Meenakshi Institute of Technology, Bangalore, Karnataka
Rupesh K C
Siddaganga Institute of Technology, Tumkur, Karnataka
G V Chalapathi Rao
Maturi Venkata Subba Rao (MVSR) Engineering College, Hyderabad,
Andhra Pradesh
P Sravani
Mathrusri Engineering College, Hyderabad, Andhra Pradesh
R Dhanasekaran
Syed Ammal Engineering College, Ramnad, Tamil Nadu
A Pon Raj
Easwari Engineering College, Chennai, Tamil Nadu
R Arun Chithra
KLN Engineering College, Madurai, Tamil Nadu
R Manimekalai
Selvam College of Technology, Namakkal, Tamil Nadu
S Ramesh
KSR College of Engineering, Nammakal, Tamil Nadu
Suggestions for improvements will always be welcome.
Ravish R. Singh
Publisher’s Note
Do you have any further request or a suggestion? We are always open to new ideas (the best ones come from
you!). You may send your comments to tmh.elefeedback@gmail.com
Piracy-related issues may also be reported!
1
1.1
Basic Network
Concepts
INTRODUCTION
We know that like charges repel each other whereas unlike charges attract each other. To overcome this force
of attraction, a certain amount of work or energy is required. When the charges are separated, it is said that a
potential difference exists and the work or energy per unit charge utilised in this process is known as voltage
or potential difference.
The phenomenon of transfer of charge from one point to another is termed current. Current (I) is defined as
the rate of flow of electrons in a conductor. It is measured by the number of electrons that flow in unit time.
Energy is the total work done in the electric circuit. The rate at which the work is done in an electric circuit
is called electric power. Energy is measured in joules (J) and power in watts (W).
1.2
RESISTANCE
Resistance is the property of a material due to which it opposes the flow of electric current through it.
Certain materials offer very little opposition to the flow of electric current and are called conductors, e.g.,
metals, acids and salt solutions. Certain materials offer very high resistance to the flow of electric current and
are called insulators, e.g., mica, glass, rubber, Bakelite, etc.
The practical unit of resistance is ohm and is represented by the symbol Ω. A conductor is said to have
resistance of one ohm if a potential difference of one volt across its terminals causes a current of one ampere
to flow through it.
The resistance of a conductor depends on the following factors.
(i)
(ii)
(iii)
(iv)
It is directly proportional to its length.
It is inversely proportional to the area of cross section of the conductor.
It depends on the nature of the material.
It also depends on the temperature of the conductor.
Hence,
R∝
l
A
R=ρ
l
A
where l is length of the conductor, A is the cross-sectional area and r is a constant known as specific resistance
or resistivity of the material.
1.2 Network Analysis and Synthesis
1. Power Dissipated in a Resistor We know that v = R i
When current flows through any resistor, power is absorbed by the resistor which is given by
p=vi
The power dissipated in the resistor is converted to heat which is given by
t
t
∫ v i dt
∫ R i i dt = i
0
Example 1.1
2
Rt
0
A 25 W resistor has a voltage of 150 sin 377 t. Find the corresponding current i and
power p.
R = 25 Ω
v 150 si 377 t
v 150 sin 377 t
i= =
= 6 sin 377 t
R
25
Solution
p
Example 1.2
vi
sin 2 337 t
si
(150 si 377 t ) (6 sin
i 377 t))
A current waveform shown in Fig. 1.1 is applied to a 2 W resistor. Draw the voltage
waveform.
i (A )
5
0
1
2
3
4
t(s )
5
Fig. 1.1
v ( V)
Solution
For the resistor,
(a) For 0
(b) For 1
(c) For 3
1,
3,
4,
v
Ri
v
v
v
R i = 2 × 5 = 10 V
Ri = 2 × 0 = 0
R i = 2 × 5 = 10 V
The voltage waveform is shown in Fig. 1.2.
1.3
10
0
1
2
3
4
t( s)
Fig. 1.2
INDUCTANCE
Inductance is the property of a coil that opposes any change in the amount of current flowing through it. If
the current in the coil is increasing, the self-induced emf is set up in such a direction so as to oppose the rise
of current. Similarly, if the current in the coil is decreasing, the self-induced emf will be in the same direction
as the applied voltage.
Inductance is defined as the ratio of flux linkage to the current flowing through the coil. The practical unit
of inductance is henry and is represented by the symbol H. A coil is said to have an inductance of one henry
if a current of one ampere when flowing through it produces flux linkages of one weber-turn in it.
1.3 Inductance 1.3
The inductance of an inductor depends on the following factors.
(i)
(ii)
(iii)
(iv)
It is directly proportional to the square of the number of turns.
It is directly proportional to the area of cross section.
It is inversely proportional to the length.
It depends on the absolute permeability of the magnetic material.
Hence,
L∝
N 2A
l
L=μ
N 2A
l
where l is the mean length, A is the cross-sectional area and m is the absolute permeability of the magnetic
material.
1. Current–Voltage Relationships in an Inductor We know that
v
L
di
dt
Expressing inductor current as a function of voltage,
di =
1
v dt
L
di =
l
v dt
L ∫0
Integrating both the sides,
i(t )
∫
t
i(0)
t
i( t ) =
1
v dt i(0)
L ∫0
The quantity i(0) denotes the initial current through the inductor. When there is no initial current
through the inductor,
t
i( t ) =
1
v dt
L ∫0
2. Energy Stored in an Inductor Consider a coil of inductance L carrying a changing current I. When
the current is changed from zero to a maximum value I, every change is opposed by the self-induced emf
produced. To overcome this opposition, some energy is needed and this energy is stored in the magnetic
field. The voltage v is given by
v
L
di
dt
Energy supplied to the inductor during interval dt is given by
dE = v i dt = L
di
i dt
dt
L i dt
1.4 Network Analysis and Synthesis
Hence, total energy supplied to the inductor when current is increased from 0 to I amperes is
I
I
∫ ddEE
E
∫ L i di
0
0
5(1 − e
Example 1.3
An inductance of 3 mH has a current i
age and the maximum stored energy.
Solution
1 2
LI
2
5000t
). Find the corresponding volt-
L = 3 mH
i
v
I max
5(1 e −5000 t )
di
d
5000 t ⎤
5000
5000t
= 3 × 10 −3 ⎡⎣5(1 − e −5000
) ⎦ = 3 × 10 −3 (5 × 5000
0 e −5000t
)
dt
dt
i(∞) = 5(( − e −∞∞ ) = 5 A
L
75 e
5000 t
1
1
2
L I max
= × 3 × 10 −3 × ( ) 2 = 37.5 mJ
2
2
Emax
Example 1.4
A current waveform is applied to a 2 H inductor. Draw the voltage waveform for
Fig. 1.3.
i (A)
6
0
2
4
6
8
t( s)
Fig. 1.3
Solution
(a) For 0
v
L
di
dt
2,
v = 2×
(6 − 0 )
=6V
( 2 − 0)
6
(b) For 2 < t < 6,
v = 2×
(c) For 6
v( V)
(6 − 6 )
=0
( 6 − 2)
0
8,
v = 2×
( 0 − 6)
= −6 V
(8 − 6)
The voltage waveform is shown in Fig 1.4.
2
4
6
8
−6
Fig. 1.4
t ( s)
1.3 Inductance 1.5
Example 1.5
A voltage waveform shown in Fig. 1.5 is applied across a 1 H inductor. Draw the
current waveform.
v ( V)
1
0
1
2
t ( s)
3
Fig. 1.5
t
i=
Solution
(a) For 0
1
v dt i(0)
L ∫0
1,
v =1
t
i
1
1dt
d i(0) = [t ]t0
dt
1 ∫0
0
t
i(1) = 1 A
(b) For 1 t
2,
v=0
t
i
1
0 ddt i(1) = 0 + 1 = 1 A
1 ∫1
i ( A)
i( 2) = 1 A
(c) For 2
3,
v =1
2
t
i
1
dt i( 2) = [t ]t2 1 t − 2 + 1 = t − 1
1 ∫2
1
t ( s)
i(3) = 3 − 1 = 2 A
0
1
The current waveform is shown in Fig. 1.6.
Example 1.6
2
3
4
Fig. 1.6
A voltage waveform shown in Fig. 1.7 is applied across a 2 H inductor. Draw the
current waveform.
v ( V)
3
2
0
−1
2
4
Fig. 1.7
6
8
t ( s)
1.6 Network Analysis and Synthesis
t
1
v ( t ) + i( 0)
L ∫0
i=
Solution
(a) For 0
2,
v=2
i ( A)
t
i
1
2 ddt i(0) = 0.5[2tt ]t0
2 ∫0
0
t
3.5
i( 2) = 2 A
(b) For 2
8,
1.5
3
t + 3 = − 0.5t + 3
6
v
0
2
4
6
t ( s)
8
t
i
1
( 0.5t + 3) dt
dt + i( 2)
2 ∫2
Fig. 1.8
t
⎡
⎤
t2
= 0.5 ⎢ − 0.5 + 3 ⎥ + 2 = 0.5
5[[ −0
0.25
5 2 + 3t + 0.5 × 2 − 3 × 2] + 2 = 0.5[ −0.25
2
⎣
⎦2
i(8) 0.5
5[[ 0.25(64) 3(8) 5] 2 = 3 5 A
2
3
5] 2
The current waveform is shown in Fig. 1.8.
Example 1.7
R
2 Ω and L
The following current waveform i(t) is passed through a series RL circuit with
2 mH . Find the voltage across each element and sketch the same.
i ( A)
5
0
1
2
3
4
5
6
7
8
t (ms)
−5
Fig. 1.9
Solution
(a) Voltage across the resistor of 2 Ω
vR Ri
Voltage across the resistor is a trapezoidal waveform with a peak value of 10 V.
(b) Voltage across the inductor of 2 mH
vL
L
di
dt
1.4 Capacitance 1.7
For 0
i ( A)
1 ms,
⎛ 5−0 ⎞
vL = 2 × 10 −3 ⎜
= 10 V
⎝ 1 × 10 −3 − 0 ⎟⎠
For 1 s
5
3 ms,
0
−5 − 5
⎛
⎞
vL = 2 × 10 −3 ⎜
=0
−
3
−
3
⎝ 3 × 10 − 1 × 10 ⎟⎠
For 3
s
2
3
1
2
2
4
5
6
7
8
3
5
6
7
8
3
5
6
t (ms)
−5
vR ( V)
10
5 ms,
5−5
⎛
⎞
vL = 2 × 10 −3 ⎜
= −10 V
⎝ 5 × 10 −3 − 3 × 10 −3 ⎟⎠
For 5 ms < t < 7 ms,
−5 + 5
⎛
⎞
vL = 2 × 10 −3 ⎜
=0
⎝ 7 × 10 −3 − 5 × 10 −3 ⎟⎠
For 7 ms < t < 8 ms,
0
The voltage waveforms are shown in Fig 1.10.
t (ms)
−10
vL ( V)
10
0
0+5
⎛
⎞
vL = 2 × 10 −3 ⎜
= 10 V
⎝ 8 × 10 −3 − 7 × 10 −3 ⎟⎠
1.4
1
1
7
8
t (ms)
−10
Fig. 1.10
CAPACITANCE
Capacitance is the property of a capacitor to store an electric charge when its plates are at different potentials.
If Q coulombs of charge is given to one of the plates of a capacitor and if a potential difference of V volts is
applied between the two plates then its capacitance is given by
C=
Q
V
The practical unit of capacitance is farad and is represented by the symbol F. A capacitor is said to have
capacitance of one farad if a charge of one coulomb is required to establish a potential difference of one volt
between its plates.
The capacitance of a capacitor depends on the following factors.
(i) It is directly proportional to the area of the plates.
(ii) It is inversely proportional to the distance between two plates.
(iii) It depends on the absolute permittivity of the medium between the plates.
Hence,
A
d
A
C=ε
d
where d is the distance between two plates, A is the cross-sectional area of the plates and e is absolute
permittivity of the medium between the plates.
C∝
1.8 Network Analysis and Synthesis
1. Current–Voltage Relationships in a Capacitor The charge on a capacitor is given by
q = Cv
where q denotes the charge and v is the potential difference across the plates at any instant.
We know that
dq d
dv
i=
= Cv = C
dt dt
dt
Expressing capacitor voltage as a function of current,
dv =
1
i dt
C
dv
1
i dt
C ∫0
Integrating both the sides,
v(t )
∫
t
v( )
t
v(t ) =
1
i dt v( )
C ∫0
The quantity v (0) denotes the initial voltage across the capacitor. When there is no initial voltage on
the capacitor,
t
v(t ) =
1
i dt
C ∫0
2. Energy Stored in a Capacitor Let a capacitor of capacitance C farads be charged from a source of
V volts. Then current i is given by
i
C
dv
dt
Energy supplied to the capacitor during interval dt is given by
dE = v i dt = v C
dv
dt
dt
Hence, total energy supplied to the capacitor when potential difference is increased from 0 to V volts is
E
Example 1.8
V
V
0
0
∫ ddEE ∫ C v dv
A voltage is defined by
⎧0
⎪
v(t ) = ⎨2t
⎪4 e −(t( t
⎩
and is applied to the 10 μ F capacitor. Find i(t).
Solution
1
CV 2
2
i
C
dv
dt
)
t<0
0 t < 2s
t > 2s
1.4 Capacitance 1.9
For t < 0,
i=0
For 0
2 s,
v
2t
i = 10 × 10 −6
For t
d
( 2tt ) = 20 μ A
dt
2 s,
v
4 e − ( t − 2)
i = 10 × 10 −6
Example 1.9
d
[4ee − ( t
dt
2)
]
0
0 6[
e
( t − 2)
] = −40 e − ( t −2) μ A
A voltage waveform shown in Fig. 1.11 is applied to the capacitor. Draw the current
waveform.
v (V )
C=2F
4
2
0
1
2
3
4
t (s)
−2
Fig. 1.11
Solution
(a) 0
At t
At t
i
dv
C
dt
i
C
dv
⎛ 2 − 2⎞
= 2⎜
=0
⎝ 1 − 0 ⎟⎠
dt
i
C
dv
⎛ 4 − 2⎞
= 2⎜
=4A
⎝ 2 − 1 ⎟⎠
dt
1
0, v = 2 V
1, v = 2 V
(b) 1 t 2
At t 1, v = 2 V
At t 2, v = 4 V
i (A )
4
(c) 2 t 4
At t 2, v = 4 V
At t 4, v = −2 V
0
i
C
dv
⎛ −2 − 4 ⎞
⎛ −6 ⎞
= 2⎜
⎟⎠ = 2 ⎜⎝ ⎟⎠ = −6 A
⎝
dt
4−2
2
The current waveform is shown in Fig. 1.12.
1
2
3
−6
Fig. 1.12
4
t (s)
1.10 Network Analysis and Synthesis
Example 1.10
A voltage waveform shown in Fig. 1.13 is applied to the capacitor. Draw the current
waveform.
v (V)
C = 1000 μF
50
≈
2
0
≈
4 5
t (s)
6
−50
Fig. 1.13
Solution
(a) 0
At t
At t
i
(b) 2
At t
At t
2
0, v = 0 V
2, v = 50 V
i (A)
25 × 10−3
dv
⎛ 50 − 0 ⎞
⎛ 50 ⎞
C
= 10 −3 ⎜
= 10 −3 ⎜ ⎟ = 25 × 10 −3 A
⎝ 2 − 0 ⎟⎠
⎝ 2⎠
dt
0
5
2, v = 50 V
−33.33 × 10−3
5, v = −50 V
2
dv
⎛ −50 − 50 ⎞
⎛ −100 ⎞
= 10 −3 ⎜
= 10 −3 ⎜
= −33.33 × 10 −3 A
⎝ 5 − 2 ⎟⎠
⎝ 3 ⎟⎠
dt
(c) 5
6
At t 5, v = −50 V
At t 6, v = −50 V
i
C
i
C
4
5
6
t (s)
Fig. 1.14
dv
⎛ −50 + 50 ⎞
= 10 −3 ⎜
=0
⎝ 6 − 5 ⎟⎠
dt
The current waveform is shown in Fig. 1.14.
Example 1.11
A current waveform shown in Fig. 1.15 is applied to a 0.1 F capacitor. Draw the
voltage waveform.
i (A)
2
1
0
1
2
3
−2
Fig. 1.15
4
t (s)
1.4 Capacitance 1.11
t
v=
Solution
(a) For 0
i=2
1
i dt v(0)
C ∫0
1,
t
v
1
2 dt
d v(0) = 10[2tt ]t0
0 1 ∫0
0
20 t
v(1) = 20 V
(b) For 1 t
i =1
2,
vC (V)
t
v
1
dt v(1) = 10[t ]1t
0 1 ∫1
20 10(t − 1) + 20
30
v( 2) = 10( 2 − 1) + 20 = 30
3 V
(c) For 2 t 4,
i = −2
20
t
v
1
0 1 ∫2
dt v( 2) = 10[ −2tt ]t2
30 10( 2t + 4) + 30
1
0
2
3
4
t (s)
−10
v( 4) = 10( −2 × 4 + 4) + 30 = −10 V
Fig. 1.16
The voltage waveform is shown in Fig. 1.16.
Example 1.12
circuit with R = 1 W, L
A current with periodic waveform shown in Fig. 1.17 is applied to a series RLC
1 mH and C
μ F.
F Sketch the voltage across each element.
i (A)
2
0
1
2
3
4
t (ms )
Fig. 1.17
Solution
(a) Voltage across the resistor of 1 Ω
vR R i
Voltage across the resistor is a triangular waveform with peak value of 2 A and slope 2 103.
(b) Voltage across inductor of 1 mH
di
vL L
dt
For 0
1 ms,
⎛ 2−0 ⎞
vL = 1 × 10 −3 ⎜
=2V
⎝ 1 × 10 −3 − 0 ⎟⎠
For 1 s
2 ms,
0−2
⎛
⎞
vL = 1 × 10 −3 ⎜
= −2 V
⎝ 2 × 10 −3 − 1 × 10 −3 ⎟⎠
1.12 Network Analysis and Synthesis
Voltage across the inductor is a square waveform.
(c) Voltage across capacitor of 100 μF
t
vC =
For 0
1
i dt v(0)
C ∫0
1 ms,
2
i
100 × 10 −6
7
vC (
2⎞
⎛
4 2000t
7 2
2000
t
dt
+
0
=
1
0
⎜ 2 ⎟ = 10 t
∫
⎝
⎠
0
t
1
vC
For 1 s
t = 2000t
10 −3
((1
10 3 ) 2
10 V
2 ms,
2
i
10 −3
t + 4 = −2000t + 4
1
vC
t
∫
100 × 10 −6 1 ms
= 10 4 ( −1000
1000
2
( 2000
000t + 4)dt + v(1 ms) = 10 4 ( −1000
2
+ 4 − 3 × 10 −3 ) + 10
+ 4 ) − 20
⎡ −1000( 2 × 10 −3 ) 2 + 4( 2 × 10
1 −3 ) ⎤⎦ − 20 = 20 V
⎣
Voltage across the capacitor keeps rising continuously. The voltage waveforms are shown in Fig. 1.18.
vC ( 2
=
4
i (A)
2
0
v R (V)
2
1
2
3
4
0
v i (V)
2
1
2
3
4
0
1
2
3
4
t(ms)
t(ms)
t(ms)
−2
v c (V)
20
10
0
1
2
3
Fig. 1.18
4
t(ms )
1.4 Capacitance 1.13
Example 1.13 Draw the waveform for iR iL , iC for the network shown in Fig. 1.19 (a) when it is
excited by a voltage source having a waveform shown in Fig. 1.19 (b).
v (V)
iR
v (t)
iC
iL
10 Ω
2F
5H
10
0
2
4
t (s)
6
−10
(a)
Fig. 1.19
Solution
(a) Current through the resistor of 10 Ω
v
R
Current through the resistor is a rectangular waveform with a peak value of 1 A.
(b) Current through the inductor of 5 H
iR =
t
iL =
For 0
1
v dt i(0)
L ∫0
2,
v = 10
v (V)
t
iL
1
0 dt i(0) = 0.2[10tt ]t0
5 ∫0
0
2t
0
iL ( 2) = 2( 2) = 4 A
For 2 t 4,
v = −10
iL
0 dt i( 2) =
6
2
4
6
2
4
6
t (s)
1
0.2[ −10t ]t2
+4
⎛ 10 − 10 ⎞
iC = 2 ⎜
=0
⎝ 2 − 0 ⎟⎠
t
4
i R (A)
= 0.2( −10t 20) 4
2t + 8
iL ( 4)
2( 4) 8 = 0
(c) Current through the capacitor of 2 F
dv
iC C
dt
For 0
2,
For 2
2
−10
t
1
5 ∫2
10
4,
⎛ −10 + 10 ⎞
iC = 2 ⎜
=0
⎝ 4 − 2 ⎟⎠
The current waveforms are shown in Fig. 1.20.
0
t (s)
−1
i L (A)
4
0
t (s)
i C (A)
t (s)
0
Fig. 1.20
1.14 Network Analysis and Synthesis
1.5
SOURCES
Source is a basic network element which supplies energy to the networks. There are two classes of sources,
namely,
1. Independent sources
2. Dependent sources
1.5.1 Independent Sources
Output characteristics of an independent source are not dependent on any network variable such as a current
or voltage. Its characteristics, however, may be time-varying. There are two types of independent sources:
1. Independent voltage source
2. Independent current source
1. Independent Voltage Source An independent voltage
source is a two-terminal network element that establishes a
specified voltage across its terminals. The value of this voltage
at any instant is independent of the value or direction of the
v (t)
V
current that flows through it. The symbols for such voltage
sources are shown in Fig. 1.21.
The terminal voltage may be a constant, or it may be some
(a)
(b)
specified function of time.
2. Independent Current Source An independent current Fig. 1.21 Symbols for independent
source is a two-terminal network element which produces a
voltage source
specified current. The value and direction of this current at any
instant is independent of the value or direction of the voltage that
appears across the terminals of the source. The symbols for such
current sources are shown in Fig. 1.22.
i (t)
I
The output current may be a constant or it may be a function
of time.
1.5.2 Dependent Sources
(a)
(b)
If the voltage or current of a source depends in turn upon some other
voltage or current, it is called as dependent or controlled source. The Fig. 1.22 Symbols for independent
dependent sources are of four kinds, depending on whether the
current source
control variable is voltage or current and the controlled source is a
voltage source or current source.
1. Voltage-Controlled Voltage Source (VCVS) A
voltage-controlled voltage source is a four-terminal
network component that establishes a voltage vcd between
c
two points c and d in the circuit that is proportional to a a +
+
+
voltage vab between two points a and b.
mvab
vcd
vab
−
The symbol for such a source is shown in −
−
d
b
Fig. 1.23.
The (+) and (−) sign inside the diamond of the
component symbol identifies the component as a
Fig. 1.23 Symbol for VCVS
voltage source.
vcd = m vab
The voltage vcd depends upon the control voltage vab and the constant m, a dimensionless constant
called voltage gain.
1.6 Some Definitions 1.15
2. Voltage-Controlled
Current
Source
(VCCS) A voltage-controlled current source is a
four-terminal network component that establishes a
current icd in a branch of the circuit that is proportional
to the voltage vab between two points a and b.
The symbol for such a source is shown in
Fig. 1.24.
The arrow inside the diamond of the component
symbol identifies the component as a current source.
icd = gm vab
icd
a
gmvab
vab
b
c
+
+
vcd
−
−
d
Fig. 1.24 Symbol for VCCS
The current icd depends only on the control voltage vab and the constant gm, called the transconductance
or mutual conductance. The constant gm has dimension of ampere per volt or siemens (S).
3. Current-Controlled Voltage Source (CCVS) A
current-controlled voltage source is a four-terminal
network component that establishes a voltage vcd between
two points c and d in the circuit that is proportional to
the current iab in some branch of the circuit.
The symbol for such a source is shown in
Fig. 1.25.
vcd = r iab
a
b
iab
+
+
+
r iab
−
−
c
vcd
−
d
Fig. 1.25 Symbol for CCVS
The voltage vcd depends only on the control current
iab and the constant r called the transresistance or
mutual resistance. The constant r has dimension of
volt per ampere or ohm (Ω).
a
4. Current-Controlled
Current
Source
(CCCS) A current-controlled current source is a
four-terminal network component that establishes a
current icd in one branch of a circuit that is proportional
to the current iab in some branch of the network.
The symbol for such a source is shown in
Fig. 1.26.
icd = b iab
icd
iab
+
+
c
b iab
b
−
−
Fig. 1.26
d
Symbol for CCCS
The current icd depends only on the control current iab and the dimensionless constant b, called the current gain.
1.6
SOME DEFINITIONS
1. Network and Circuit The interconnection of two or more circuit elements (viz.,
voltage sources, resistors, inductors and
capacitors) is called an electric network. If the
network contains at least one closed path, it is
called an electric circuit. Every circuit is a network, but all networks are not circuits. Figure
1.27(a) shows a network which is not a circuit
and Fig. 1.27(b) shows a network which is a
circuit.
L
R
C
(a)
Fig. 1.27
L
R
V
C
(b)
(a) Network which is not a circuit
(b) Network which is a circuit
1.16 Network Analysis and Synthesis
2. Linear and Non-linear Elements If the resistance, inductance or capacitance offered by an element
does not change linearly with the change in applied voltage or circuit current, the element is termed as linear
element. Such an element shows a linear relation between
I
voltage and current as shown in Fig. 1.28. Ordinary resistors,
capacitors and inductors are examples of linear elements.
A non-linear circuit element is one in which the
current does not change linearly with the change in
t
applied voltage. A semiconductor diode operating in the
en
t
m
curved region of characteristics as shown in Fig. 1.28 is
en
le
m
E
e
l
common example of non-linear element.
ar
rE
ne
ea
Li
Other examples of non-linear elements are voltagen
Li
ndependent resistor (VDR), voltage-dependent capacitor
No
(varactor), temperature-dependent resistor (thermistor), lightV
0
dependent resistor (LDR), etc. Linear elements obey Ohm’s
law whereas non-linear elements do not obey Ohm’s law.
3. Active and Passive Elements An element Fig. 1.28 V-I characteristics of linear
and non-linear elements
which is a source of electrical signal or which is capable
of increasing the level of signal energy is termed as active element. Batteries, BJTs, FETs or OP-AMPs
are treated as active elements because these can be used for the amplification or generation of signals. All
other circuit elements, such as resistors, capacitors, inductors, VDR, LDR, thermistors, etc., are termed
passive elements. The behaviour of active elements cannot be described by Ohm’s law.
4. Unilateral and Bilateral Elements If the magnitude of current flowing through a circuit element
is affected when the polarity of the applied voltage is changed, the element is termed unilateral element.
Consider the example of a semiconductor diode. Current flows through the diode only in one direction.
Hence, it is called an unilateral element. Next, consider the example of a resistor. When the voltage is
applied, current starts to flow. If we change the polarity of the applied voltage, the direction of the current
is changed but its magnitude is not affected. Such an element is called a bilateral element.
5. Lumped and Distributed Elements A lumped element is the element which is separated physically,
like resistors, inductors and capacitors. Distributed elements are those which are not separable for analysis
purposes. Examples of distributed elements are transmission lines in which the resistance, inductance and
capacitance are distributed along its length.
6. Active and Passive Networks A network which contains at least one active element such as an
independent voltage or current source is an active network. A network which does not contain any active
element is a passive network.
7. Time-invariant and Time-variant Networks A network is said to be time-invariant or fixed if its
input–output relationship does not change with time. In other words, a network is said to time-invariant,
if for any time shift in input, an identical time-shift occurs for output. In time-variant networks, the
input–output relationship changes with time.
1.7
SERIES AND PARALLEL COMBINATIONS OF RESISTORS
Let R1 R2 and R3 be the resistances of three resistors
connected in series across a dc voltage source V as shown in
Fig. 1.29. Let V1 V2 and V3 be the voltages across resistances
R1 R2 and R3 respectively.
In series combination, the same current flows through
each resistor but voltage across each resistor is different.
I
R1
R2
R3
V1
V2
V3
V
Fig. 1.29 Series combination of resistors
1.7 Series and Parallel Combinations of Resistors 1.17
V V1 + V2 V3
RT I = R1 I + R2 I + R3 I
R1 + R2
RT
R3
Hence, when a number of resistors are connected in series, the equivalent resistance is the sum of all the
individual resistance.
1. Voltage Division and Power in a Series Circuit
I=
V1
V2
V3
Total power
PT
V
R1 + R2 + R3
R1
R1 I =
V
R1 + R2 + R3
R2
R2 I =
V
R1 + R2 + R3
R3
R3 I =
V
R1 R2 + R3
P1 + P2 P3
= I 2 R1 + I 2 R2 + I 2 R3
=
I
V12 V22 V32
+
+
R1 R2 R3
Figure 1.30 shows three resistors connected in parallel across
a dc voltage source V. Let I1 I 2 and I 3 be the current flowing
through resistors R1 R2 and R3 respectively.
In parallel combination, the voltage across each resistor is
same but current through each resistor is different.
I1
R1
I2
R2
I3
R3
V
Fig. 1.30
Parallel combination
of resistors
I = I1 + I 2 I 3
V
V V
V
=
+
+
RT R1 R2 R3
1
1
1
1
=
+
+
RT R1 R2 R3
R1 R2 R3
RT =
R2 R3 + R3 R1 + R1 R2
Hence, when a number of resistors are connected in parallel, the reciprocal of the equivalent resistance
is equal to the sum of reciprocals of individual resistances.
2. Current Division and Power in a Parallel Circuit
V
RT I
R1 I1
R2 I 2
R3 I 3
V
R I
R2 R3
= T =
I
R1
R1
R1 R2 + R2 R3 + R3 R1
V
R I
R1 R3
I2 =
= T =
I
R2
R2
R1 R2 + R2 R3 + R3 R1
V
R I
R1 R2
I3 =
= T =
I
R3
R3
R1 R2 + R2 R3 + R3 R1
I1 =
1.18 Network Analysis and Synthesis
Total power
P1 + P2
PT
=
=
+
P3
I 22 R2 + I 32 R3
2
3
V
V
V
+
+
R1 R2 R3
R1 R2
R1 R2
V RT I R1 I1 R2 I 2
V
R I
R2
I1 =
= T =
I
R1
R1
R1 R2
V
R I
R1
I2 =
= T =
I
R2
R2
R1 R2
RT =
Note: For two branch circuits,
Example 1.14
I12 R1
2
Find an equivalent resistance between A and B in the network of Fig. 1.31.
B
A
10 Ω
10 Ω
10 Ω
Fig. 1.31
Solution Marking all the junctions and redrawing the network (Fig. 1.32),
A
C
10 Ω
D
10 Ω
B
10 Ω
(a)
D
A
10 Ω
10 Ω
10 Ω
B
(b)
C
Fig. 1.32
RAB = 10 � 10 � 10 = 3.33 Ω
Example 1.15
Find an equivalent resistance between A and B in the network of Fig. 1.33.
6Ω
A
8Ω
10 Ω
B
12 Ω
Fig. 1.33
1.7 Series and Parallel Combinations of Resistors 1.19
Solution Marking all the junctions and redrawing the network (Fig. 1.34),
6Ω
A
8Ω
D
A
C
8Ω
10 Ω
B
D
6Ω
10 Ω
12 Ω
B
12 Ω
C
(b)
(a)
Fig. 1.34
RAB = 8 � 6 � 10 � 12 = 2.11 Ω
Example 1.16
Find the equivalent resistance between A and B in the network of Fig. 1.35.
2Ω
15 Ω
B
A
10 Ω
8Ω
10 Ω
20 Ω
30 Ω
40 Ω
Fig. 1.35
Solution Marking all the junctions and redrawing the network (Fig. 1.36),
2Ω
B
C
15 Ω
2Ω
A
10 Ω B
8Ω
25 Ω
D
F
10 Ω
20 Ω
30 Ω
E
40 Ω
A
8Ω
2Ω
C, E
D, F
(b)
B
50 Ω
12.5 Ω
50 Ω
A
8Ω
(c)
(a)
Fig. 1.36
RAB = 22.5 Ω
Example 1.17
What is the resistance between the terminals A and B in network of Fig. 1.37 when
the potential difference between C and D is zero?
1.20 Network Analysis and Synthesis
C
R
R
R
A
R
B
D
R
Fig. 1.37
Solution When the potential difference between C and D is zero, the resistor connected between C and D
is shorted. Hence, the points C and D are at same potential.
Redrawing the network (Fig. 1.38),
Simplifying the network (Fig. 1.39),
R
C, D
A
R
2
A
R
R
R
R
2
B
B
Fig. 1.38
Fig. 1.39
RAB =
Example 1.18
R R
+ =RΩ
2 2
Determine the current delivered by the source in the network of Fig. 1.40.
2Ω
2Ω
4Ω
2Ω
30 V
2Ω
1Ω
Fig. 1.40
3Ω
1Ω
2Ω
1.7 Series and Parallel Combinations of Resistors 1.21
Solution The network can be simplified by series–parallel reduction technique (Fig. 1.41).
4Ω
2Ω
30 V
4Ω
2Ω
2Ω
3Ω
2Ω
30 V
2Ω
1.5 Ω
3Ω
1Ω
1Ω
(b)
(a)
3.5 Ω
30 V
2Ω
2Ω
2Ω
30 V
1.27 Ω
1Ω
1Ω
(d)
(c)
2.27 Ω
2Ω
30 V
1.06 Ω
30 V
I
(f)
(e)
Fig. 1.41
I=
Example 1.19
30
= 28.3 A
1 06
Find the current delivered by the source in the network of Fig. 1.42.
2Ω
2Ω
2Ω
4Ω
2Ω
4Ω
4Ω
50 V
5Ω
2Ω
2Ω
1Ω
Fig. 1.42
3Ω
1Ω
2Ω
1.22 Network Analysis and Synthesis
Solution The network can be simplified by series–parallel reduction technique (Fig. 1.43).
4Ω
4Ω
5Ω
50 V
4Ω
2Ω
2Ω
4Ω
4Ω
2Ω
2Ω
5Ω
50 V
3Ω
2Ω
3Ω
4Ω
2Ω
1.5 Ω
1Ω
1Ω
(a)
(b)
5Ω
50 V
2Ω
2Ω
2Ω
1.5 Ω
1Ω
(c)
3.5 Ω
50 V
2Ω
5Ω
1.27 Ω
5Ω
50 V
2Ω
2Ω
1Ω
1Ω
(d)
50 V
5Ω
(e)
0.88 Ω
2.27 Ω 50 V
2Ω
I
(f)
(g)
Fig. 1.43
I=
50
= 56.82
82 A
0 88
Example 1.20 Three equal resistors of 30 W each are connected in parallel across a 120 V dc supply as shown in Fig. 1.44. What is the current through each of them (a) if one of the resistors burns out, or
(b) if one of the resistors gets shorted?
+
120 V
I1
I2
30 Ω
−
Fig. 1.44
I3
30 Ω
30 Ω
1.7 Series and Parallel Combinations of Resistors 1.23
Solution
(a) If one of the resistors burns out, it will act as an open circuit (Fig. 1.45).
+
I3 = 0
I1
I2 =
120
=4A
30
I1
I2
30 Ω
120 V
I3
30 Ω
−
Fig. 1.45
(b) If one of the resistors gets shorted, the effective resistance becomes zero (Fig. 1.46).
+
I1
I2 = 0
I1
I2
30 Ω
120 V
I3
30 Ω
−
Fig. 1.46
Example 1.21
A lamp rated at 100 V, 75 W is to be connected across a 230 V supply. Find the value
of resistance to be connected in series with the lamp. Also find the power loss occuring in the resistor.
Solution The network is shown in Fig. 1.47.
V1
P1 = 75 W, V = 230 V
100
I
R
L
100 V
230 V
Fig. 1.47
(a) Value of resistance
Rated current of the lamp
I=
P1 75
=
= 0.75
75 A
V1 100
Lamp will operate normally on 230 V supply if the current flowing through the lamp remains the rated
current, i.e., 0.75 A.
Voltage across the resistor R
V2 = 230 − 100 = 130 V
R=
130
= 173.33 Ω
0 75
P2 =
V22 (130) 2
=
= 97.55 W
R 173.33
(b) Power loss occurring in the resistor
1.24 Network Analysis and Synthesis
Example 1.22 A 100 V, 60 W lamp is connected in series with a 100 V, 100 W lamp and the combination is connected across 200 V mains. Find the value of the resistance that should be connected in parallel
with the first lamp so that each lamp may get the rated current at rated voltage.
Solution The network is shown in Fig. 1.48.
V1 = 100 V
R
P1 = 60 W
L1
V2 = 100 V
P2 = 100 W
L2
100 V
100 V
V = 200 V
Rated current of the lamp L1
200 V
P
60
I1 = 1 =
= 0.6
6A
V1 100
Fig. 1.48
Rated current of the lamp L2
I2 =
P2 100
=
=1A
V2 100
Let R be the value of resistance that should be connected in parallel with the lamp L1 so that rated current
flows through the lamp L1.
Current through resistor R
I = 1 − 0.6 = 0.4 A
R=
V1 100
=
= 250 Ω
I
04
Example 1.23 A 100 V, 60 W lamp is connected in series with a 100 V, 100 W lamp across 200 V
supply. What will be the current drawn by the lamps? What will be the power consumed by each lamp and
will such a combination work?
Solution The network is shown in Fig. 1.49.
V1
V2
L1
P1 = 60 W
P2 = 100 W
100
100
V = 200 V
L2
100 V
100 V
200 V
(a) Current drawn by the lamps
Resistance of the lamp L1
Fig. 1.49
R1 =
V12 (100) 2
=
= 166.67
67 Ω
P1
60
R2 =
V22 (100) 2
=
= 100 Ω
P2
100
Resistance of the lamp L2
Current drawn by the lamps
I=
V
R1
R2
=
200
= 0.75 A
166.67 + 100
1.7 Series and Parallel Combinations of Resistors 1.25
(b) Power consumed by the lamps
Power consumed by the lamp
L1
Power consumed by the lamp L2
I 2 R1 = (0.75) 2 ×166
166.67 = 93.75 W
I 2 R2 = (0.75) 2 × 100 = 56.25 W
If a 100 V, 60 W lamp draws a power of 93.75 W, its filament will be overheated and will burn out.
Hence, such a combination will not work.
Example 1.24
A resistor R is connected in series with two lamps of 12 V, 9 W across a 300 V
supply. Find the value of R so that both the lamps operate at rated conditions. If one of the lamps is short
circuited, find the current through the circuit and the power dissipited in each lamp.
Solution The network is shown in Fig. 1.50.
L1
V1
V2 = 12 V
P1
P2 = 9 W
L2
12 V
12 V
V = 300 V
R
v3
300 V
(a) Value of R
Rated current of lamps L1 and L2
Fig. 1.50
I=
P1 P2
9
=
=
= 0.75 A
V1 V2 12
Resistance of the lamps L1 and L2
R1
R2 =
V12 V22 (12) 2
=
=
= 16 Ω
P1
P2
9
Voltage across the resistor R
V3 = 300 − 12 − 12 = 276 V
R=
V3 276
=
= 368 Ω
I
0 75
(b) Current through the circuit if the lamp L2 is short circuited
If the lamp L2 is short circuited, resistance across terminals of L2 is zero. If the new current is I′ then
= I ′(16 + 368)
I ′ = 0.78 A
(c) Power dissipated in each lamp if the lamp L2 is short circuited
( I ′ ) 2 R1 = (0.78) 2 16
Power dissipated in the lamp
L1
Power dissipated in the lamp
L2 = 0
9.73 W
30
. It is connected in
I
series with a resistor across a 100 V supply. If the voltage across the arc and the resistor are equal, find the
ohmic value of the resistor.
Example 1.25
A dc arc has voltage/current relation expressed by V
44 +
1.26 Network Analysis and Synthesis
Solution The network is shown in Fig 1.51.
R
I
V = 100 V
V1 V2
V V1 + V2 = 100
V1 V2 = 50
v1
v2
100 V
For a dc arc,
Fig. 1.51
30
V1 = 44 +
I
30
50 = 44 +
I
I =5A
Ohmic value of the resistor
R=
Example 1.26
V2 50
=
= 10 Ω
I
5
Find the voltage VAB in the network as shown in Fig. 1.52.
10 Ω
10 V
30 Ω
A
B
20 Ω
40 Ω
Fig. 1.52
Solution The resistors of 10 Ω and 20 Ω are in series. Similarly, the resistors of 30 Ω and 40 Ω are in series.
By voltage-division rule,
20
VA = 10 ×
= 6 67 V
10 + 20
40
VB = 10 ×
= 5 71 V
30 + 40
VAB VA VB = 6.67 − 5.71 0.96 V
1.8
SERIES AND PARALLEL COMBINATION OF INDUCTORS
Let L1 L2 and L3 be the inductances of three inductors connected
in series across an ac voltage source v as shown in Fig. 1.53. Let
v1 v2 and v3 be the voltages across inductances L1 L2 and L3
respectively.
In series combination, the same current flows through each
inductor but the voltage across each inductor is different.
v = v1 + v1 + v3
di
di
di
di
LT
= L1 + L2 + L3
dt
dt
d
dt
dt
LT L1 + L2 L3
L1
i
v1
L2
v2
L3
v3
v
Fig. 1.53
Series connection of
inductors
1.8 Series and Parallel Combination of Inductors 1.27
Hence, when a number of inductors are connected in series, the equivalent inductance is the sum of all the
individual inductances.
L1
i1
Figure 1.54 shows three inductors connected in parallel
across an ac voltage source v. Let i1 i2 and i3 be the current
L2
i2
i
through each inductance L1 L2 and L3 respectively.
In parallel combination, the voltage across each inductor
L3
i3
is same but the current through each inductor is different.
i
1
LT
i1 + i2
1
i3
1
1
∫ v ddt = L1 ∫ v dt + L2 ∫ v ddt + L3 ∫ v dt
v
LT =
Parallel connection of inductors
Fig. 1.54
1
1
1
1
= +
+
LT L1 L2 L3
L1 L2 L3
L1 L2 + L2 L3 + L3 L1
Hence, when a number of inductors are connected in parallel, the reciprocal of the equivalent inductance
is equal to the sum of reciprocals of individual inductances.
Example 1.27
Find an equivalent inductance between terminals A and B in the network of Fig. 1.55.
8H
2H
1H
A
B
8H
Fig. 1.55
Solution The network can be simplified by series–parallel reduction technique (Fig. 1.56).
2H
4H
1H
B
A
(a)
7H
B
A
(b)
Fig. 1.56
LAB = 7 H
Example 1.28
Find an equivalent inductance of the network shown in Fig 1.57.
4H
5H
4H
2H
3H
3H
B
A
5H
4H
Fig. 1.57
1.28 Network Analysis and Synthesis
Solution The network can be simplified by series–parallel reduction technique (Fig. 1.58).
5H
1.33 H
2H
3H
3H
A
B
5H
(a)
5H
5H
1.33 H
3H
A
B
5H
(b)
5H
2.5 H
1.33 H
3H
B
A
(c)
11.83 H
B
A
(d)
Fig. 1.58
LAB = 11.83 H
1.9
SERIES AND PARALLEL COMBINATION OF CAPACITORS
Let C1 C2 and C3 be the capacitances of three capacitors connected in series across an ac voltage source v as
shown in Fig 1.59. Let v1 v2 and v3 be the voltages across capacitances
C1
C2
C3
C1 C2 and C3 respectively.
In series combination, the charge on each capacitor is same but
v1
v2
v3
voltage across each capacitor is different.
v = v1 + v2 + v3
1
1
1
1
v
id
dt =
i dt +
i ddt +
i dt
∫
∫
∫
CT
C1
C2
C3 ∫
Fig. 1.59 Series combination of
1
1
1
1
capacitors
=
+
+
CT C1 C2 C3
C1C2C3
CT =
C1C2 + C2C3 + C3C1
Hence, when a number of capacitors are connected in series, the reciprocal of the equivalent capacitance
is equal to the sum of reciprocals of individual capacitances.
1. Voltage Division in a Series Circuit
Q
CT V C1V1 C2V2 C3V3
Q CT V
C2 C3
V1 =
=
=
V
C1
C1
C1C2 + C2C3 + C3C1
Q CT V
C1 C3
V2 =
=
=
V
C2
C2
C1C2 + C2C3 + C3C1
Q CT V
C1 C2
V3 =
=
=
V
C3
C3
C1C2 + C2C3 + C3C1
1.9 Series and Parallel Combination of Capacitors 1.29
Figure 1.60 shows three capacitors connected in parallel across
an ac voltage source v. Let i1 i2 and i3 be the current through each
capacitance C1 C2 and C3 respectively.
In parallel combination, the voltage across each capacitor is same
but current through each capacitor is different.
i i1 + i2 i3
dv
dv
dv
dv
CT
= C1
+ C2
+ C3
dt
dt
d
dt
dt
CT C1 + C2 C3
Hence, when a number of capacitors are connected in parallel, the
equivalent capacitance is the sum of all the individual capacitance.
Example 1.29
i
i1
C1
i2
C2
i3
C3
v
Fig. 1.60
Parallel combination of
capacitors
What is the equivalent capacitance between terminals A and B in the network of
Fig. 1.61.
2 μF
1 μF
2 μF 2 μF
A
B
Fig. 1.61
Solution The network can be simplified by series–parallel reduction technique (Fig. 1.62).
2 μF
1 μF
3 μF
1 μF
A
1 μF
A
B
B
(b)
(a)
0.75 μF
B
A
(c)
Fig. 1.62
C AB = 0.75 μF
Example 1.30
Find the equivalent capacitance for the network of Fig. 1.63.
4 μF
2 μF
3 μF
A
3 μF
5 μF
B
2 μF
Fig. 1.63
1.30 Network Analysis and Synthesis
Solution The network can be simplified by series–parallel reduction technique (Fig. 1.64).
4 μF
2 μF
1.5 μF
5 μF
2 μF
A
B
5.5 μF
5 μF
A
B
2 μF
2 μF
(a)
(b)
1.47 μF
5 μF
3.47 μF
A
5 μF
A
B
B
2 μF
(c)
(d)
2.05 μF
A
B
(e)
Fig. 1.64
C AB = 2.05
05 μF
Example 1.31
What is equivalent capacitance between terminals A and B in the network of
Fig. 1.65?
6 μF
4 μF
A
4 μF
2 μF
B
6 μF
Fig. 1.65
Solution The network can be simplified by series–parallel reduction technique (Fig. 1.66).
6 μF
8 μF
A
6 μF
6 μF
A
2 μF
A
2 μF
3.429 μF
5.429 μF
6 μF
B
B
(a)
B
(b)
Fig. 1.66
(c)
1.9 Series and Parallel Combination of Inductors 1.31
CAB = 2.85 μF
Example 1.32
What is equivalent capacitance between terminals A and B in the network of Fig. 1.67?
3 μF
A
6 μF
3 μF
3 μF
6 μF
B
Fig. 1.67
Solution The network can be simplified by series–parallel reduction technique (Fig. 1.67).
3 μF
A
A
9 μF
3 μF
2.25 μF
3 μF
6 μF
6 μF
B
B
(a)
(b)
A
8.25 μF
3 μF
B
(c)
Fig. 1.68
CAB = 11.25 μF
Example 1.33 A combination of four capacitors is shown in Fig. 1.69. Find the value of C to obtain
an equivalent capacitance of 0.5 mF.
C
0.2 μF
0.6 μF
0.8 μF
Fig. 1.69
1.32 Network Analysis and Synthesis
Solution
(C + .6)(0.2 + 0.8)
= 0.5
C + 0.6 + 0.2 + 0.8
C
C + 0.8
0.5 C = 0.22
Ceq =
C = 0.4 μF
F
1.10
STAR-DELTA TRANSFORMATION
When a circuit cannot be simplified by normal series–parallel reduction technique, the star-delta transformation
can be used.
Figure 1.70 (a) shows three resistors RA, RB and RC connected in delta.
Figure 1.70 (b) shows three resistors R1, R2 and R3 connected in star.
1
1
R1
RC
RB
R3
R2
2
2
3
3
RA
(a)
Fig. 1.70
(b)
(a) Delta networks
(b) Star Network
These two networks will be electrically equivalent if the resistance as measured between any pair of
terminals is the same in both the arrangements.
1.10.1
Delta to Star Transformation
Referring to delta network shown in Fig. 1.70 (a), the resistance between terminals 1 and 2
RC ( RA RB )
= RC � ( RA RB ) =
RA RB + RC
Referring to the star network shown in Fig. 1.70 (b), the resistance between terminals 1 and 2 = R1
Since the two networks are electrically equivalent,
R1
Similarly,
R2
and
R3
RC ( RA + RB )
RA RB + RC
R ( R + RC )
R3 = A B
RA RB + RC
R ( R + RC )
R1 = B A
RA RB + RC
R2 =
R2 .
...(1.1)
...(1.2)
...(1.3)
Subtracting Eq. (1.2) from Eq. (1.1),
R1
R3 =
RB RC − RA RB
RA RB + RC
...(1.4)
1.10 Star-delta Transformation 1.33
Adding Eq. (1.4) and Eq. (1.3),
R1 =
R2 =
Similarly,
R3 =
RA
RB RC
RB + RC
RA
RA RC
RB + RC
RA RB
RA RB + RC
Thus, star resistor connected to a terminal is equal to the product of the two delta resistors connected to
the same terminal divided by the sum of the delta resistors.
1.10.2
Star to Delta Transformation
Multiplying the above equations,
R1 R2 =
R2 R3 =
R3 R1 =
RA RB RC2
...(1.5)
RB + RC ) 2
( RA
RA2 RB RC
( RA
...(1.6)
RB + RC ) 2
RA RB2 RC
( RA
...(1.7)
RB + RC ) 2
Adding Eqs (1.5), (1.6) and (1.7),
R1 R2 + R2 R3 + R3 R1 =
RA RB RC ( RA + RB
( RA
RB + RC )
RC )
2
=
RA RB RC
RA RB + RC
= RA R1 = RB R2 = RC R3
Hence,
RA =
R1 R2 + R2 R3 + R3 R1
= R2
R1
R3 +
R2 R3
R1
RB =
R1 R2 + R2 R3 + R3 R1
= R1
R2
R3 +
R3 R1
R2
RC =
R1 R2 + R2 R3 + R3 R1
= R1
R3
R2 +
R1 R2
R3
Thus, delta resistor connected between the two terminals is the sum of two star resistors connected to the
same terminals plus the product of the two resistors divided by the remaining third star resistor.
Note: (1) When three equal resistors are connected in delta (Fig. 1.71), the equivalent star resistance is given by
RY =
RΔ RΔ
R
= Δ
RΔ RΔ RΔ
3
RΔ
3RY
1.34 Network Analysis and Synthesis
A
A
RY
RΔ
RΔ
RY
RY
B
C
B
C
RΔ
Fig. 1.71 Equivalent star resistance for three equal delta resistors
(2) Star-delta transformation can also be applied to network containing inductors and capacitors.
Example 1.34
Find an equivalent resistance between A and B in the network of Fig. 1.72.
4Ω
4.5 Ω
4.5 Ω
3Ω
3Ω
A
7.5 Ω
B
7.5 Ω
3Ω
Fig. 1.72
Solution Converting the two delta networks formed by resistors of 4.5 Ω, 3 Ω and 7.5 Ω into equivalent
star networks (Fig. 1.73 and Fig. 1.74),
4Ω
R4
R3
R6
R1
B
A
R5
R2
3Ω
Fig. 1.73
4Ω
R1
R2
R3
4.5 × 7.5
= 2.25
5Ω
4.5 + 7.5 + 3
7 5×3
R5 =
=1 5Ω
4.5 + 7.5 + 3
4 5×3
R4 =
=09Ω
4.5 + 7.5 + 3
0.9 Ω
0.9 Ω
R6 =
2.25 Ω
2.25 Ω
B
A
1.5 Ω
1.5 Ω
3Ω
Fig. 1.74
1.10 Star-delta Transformation 1.35
Simplifying the network (Fig. 1.75),
5.8 Ω
A
B
2.25 Ω
2.25 Ω
6Ω
(a)
2.95 Ω
2.25 Ω
2.25 Ω
A
B
(b)
7.45 Ω
B
A
(c)
Fig. 1.75
RAB = 7 45 Ω
Example 1.35
Find an equivalent resistance between A and B in the network of Fig. 1.76.
10 Ω
A
10 Ω
C
10 Ω
D
B
10 Ω
10 Ω
Fig. 1.76
Solution Redrawing the network (Fig. 1.77),
C
10 Ω
10 Ω
10 Ω
A
10 Ω
B
10 Ω
D
Fig. 1.77
Converting the delta network formed by three resistors of 10 Ω into an equivalent star network (Fig. 1.78
and Fig. 1.79),
1.36 Network Analysis and Synthesis
C
C
R2
10
Ω
3
10 Ω
10 Ω
10
Ω
3
R1
A
A
B
B
10
Ω
3
10 Ω
R3
10 Ω
D
D
Fig. 1.79
Fig. 1.78
R2 = R3 =
R1
10 × 10
10
=
Ω
10 + 10 + 10 3
Simplifying the network (Fig. 1.80),
40
Ω
3
10
Ω
3
10
Ω
3
A
B
20
Ω
3
A
B
(b)
40
Ω
3
(a)
10 Ω
A
B
(c)
Fig. 1.80
RAB = 0 Ω
Example 1.36
Find an equivalent resistance between A and B in the network of Fig. 1.81.
A
6Ω
9Ω
1.5 Ω
3Ω
4Ω
C
B
1Ω
Fig. 1.81
1.10 Star-delta Transformation 1.37
Solution Converting the star network formed by resistors of 3 Ω, 4 Ω and 6 Ω into an equivalent delta
network (Fig. 1.82 and Fig. 1.83),
A
A
1.5 Ω
9Ω
1.5 Ω
9Ω
13
R2
18 Ω
.5
9Ω
R3
B
C
Ω
R1
B
C
1Ω
1Ω
Fig. 1.82
Fig. 1.83
6 4
= 188 Ω
3
6 3
R2 = 6 3 +
= 133.5 Ω
4
4 3
R3 = 4 3 +
=9Ω
6
R1 = 6 4 +
Simplifying the network (Fig. 1.84),
A
1.35 Ω
6Ω
RAB = 6 � (1.35 + 0.9)
= 6 � 2.25
B
= 6 Ω
C
0.9 Ω
Fig. 1.84
Example 1.37
Find an equivalent resistance between A and N by solving outer delta ABC in Fig. 1.85.
A
2Ω
12 Ω
12 Ω
N
2Ω
2Ω
C
B
12 Ω
Fig. 1.85
1.38 Network Analysis and Synthesis
Solution Converting outer delta ABC into a star network (Fig. 1.86),
A
A
4Ω
RY =
12 × 12
=
12 + 12 + 12
4Ω
Ω
2Ω
4Ω
N
2Ω
B
C
2Ω
M
B
C
Fig. 1.86
Simplifying the network (Fig. 1.87),
2Ω
M
2Ω
A
M
A
2Ω
2Ω
4Ω
4Ω
4Ω
2Ω
6Ω
6Ω
N
N
(a)
(b)
2Ω
M
A
A
A
4Ω
3Ω
N
4Ω
5Ω
2.22 Ω
N
N
(c)
(d)
(e)
Fig. 1.87
RAN = .
Example 1.38
Ω
Find an equivalent resistance between A and B in the network of Fig. 1.88.
4Ω
2Ω
41 Ω
6Ω
6Ω
17 Ω
A
15 Ω
15 Ω
4Ω
11 Ω
B
Fig. 1.88
1.10 Star-delta Transformation 1.39
Solution The resistors of 2 Ω and 4 Ω and the resistors of 4 Ω and 11 Ω are connected in series (Fig. 1.89).
15 Ω
41 Ω
6Ω
15 Ω
6Ω
6Ω
15 Ω
17 Ω
A
B
Fig. 1.89
Converting the two outer delta networks into equivalent star networks (Fig. 1.90),
41 Ω
5Ω
2Ω
6 6
=2Ω
6 6 6
15 × 15
RY2 =
=5Ω
15 + 15 + 15
RY1 =
2Ω
5Ω
A
B
17 Ω
2Ω
5Ω
Fig. 1.90
Simplifying the network (Fig. 1.91),
48 Ω
A
B
2Ω
5Ω
24 Ω
(a)
2Ω
5Ω
16 Ω
B
A
(b)
23 Ω
A
B
(c)
Fig. 1.91
RAB = 3 Ω
Example 1.39
Find an equivalent resistance between A and B in the network of Fig. 1.92.
15 Ω
20 Ω
A
25 Ω
30 Ω
45 Ω
35 Ω
B
40 Ω
Fig. 1.92
1.40 Network Analysis and Synthesis
Solution Drawing the resistor of 30 Ω from outside (Fig. 1.93),
15 Ω
20 Ω
A
25 Ω
30 Ω
35 Ω
45 Ω
B
40 Ω
Fig. 1.93
Converting the delta network formed by resistors of 20 Ω, 25 Ω and 35 Ω into an equivalent star network
(Fig. 1.94),
15 Ω
A
R1 =
20 × 35
= 8 75 Ω
20 + 35 + 25
R2 =
20 × 25
= 6 25
5Ω
20 + 35 + 25
R3 =
35 × 25
= 100.94
9 Ω
20 + 35 + 25
2
R2
R1
45 Ω
30 Ω
R3
B
40 Ω
Fig. 1.94
Redrawing the network (Fig. 1.95),
15 Ω
6.25 Ω
8.75 Ω
A
10.94 Ω
45 Ω
30 Ω
40 Ω
B
Fig. 1.95
Simplifying the network (Fig. 1.96),
15 Ω
6.25 Ω
A
50.94 Ω
45 Ω
38.75 Ω
B
(a)
15 Ω
6.25 Ω
15 Ω
A
A
45 Ω
22.01 Ω
45 Ω
B
B
(c)
(b)
Fig. 1.96 Continued
28.26 Ω
1.10 Star-delta Transformation 1.41
15 Ω
A
A
32.36 Ω
17.36 Ω
B
B
(e)
(d)
Fig. 1.96
RAB = 32.36 Ω
Example 1.40
Find an equivalent resistance between A and B in the network of Fig. 1.97.
10 Ω
A
20 Ω
5Ω
5Ω
15 Ω
25 Ω
10 Ω
2Ω
B
5Ω
30 Ω
Fig. 1.97
Solution The resistors of 5 Ω and 25 Ω and the resistors of 10 Ω and 5 Ω are connected in series (Fig. 1.98).
10 Ω
20 Ω
A
5Ω
15 Ω
30 Ω
15 Ω
2Ω
B
30 Ω
Fig. 1.98
Converting the delta network formed by the resistors of 20 Ω, 5 Ω and 15 Ω and 15 Ω into an equivalent
star network (Fig. 1.99),
10 Ω
A
20 × 5
= 25Ω
20 + 5 15
20 × 15
R2 =
=75Ω
20 + 5 15
5 15
R3 =
= 1.8875
5Ω
20 + 5 15
R1 =
R1
R2
30 Ω
15 Ω
R3
2Ω
B
30 Ω
Fig. 1.99
1.42 Network Analysis and Synthesis
Redrawing the network (Fig. 1.100),
12.5 Ω
2.5 Ω
10 Ω
A
A
7.5 Ω
1.875 Ω
30 Ω
15 Ω
15 Ω
37.5 Ω
3.875 Ω
2Ω
B
B
30 Ω
30 Ω
(a)
(b)
Fig. 1.100
Converting the delta network formed by the resistors of 3.875 Ω, 37.5 Ω and 30 Ω into an equivalent star
network (Fig. 1.101),
12.5 Ω
A
3.875 × 37.5
R4 =
= 2 004 Ω
3.875 + 37.5 30
R5 =
R6 =
R4
3.875 × 30
= 1.63 Ω
3.875 + 37.5 30
B
15 Ω
R6
37.5 30
= 15.76 Ω
3.875 + 37.5 30
R5
Fig. 1.101
Simplifying the network (Fig. 1.102),
12.5 Ω
2.04 Ω
A
14.54 Ω
A
15.76 Ω
15.76 Ω
B
B
16.63 Ω
1.63 Ω
15 Ω
(a)
(b)
7.76 Ω
23.52 Ω
15.76 Ω
B
A
A
(c)
B
(d)
Fig. 1.102
RAB = 23.52 Ω
1.10 Star-delta Transformation 1.43
Example 1.41
Find an equivalent resistance between A and B in the network of Fig. 1.103.
A
6Ω
4Ω
3Ω
5Ω
8Ω
5Ω
4Ω
B
Fig. 1.103
Solution Converting the star network formed by the resistors of 3 Ω, 5 Ω and 8 Ω into an equivalent delta
network (Fig. 1.104 and Fig. 1.105),
A
6Ω
4Ω
A
6Ω
9.875 Ω
R1
R2
4Ω
15.8 Ω
R3
26.33 Ω
5Ω
4Ω
5Ω
4Ω
B
B
Fig. 1.104
Fig. 1.105
3 5
= 9.875 Ω
8
3 8
R2 = 3 8 +
= 15.8 Ω
5
5 8
R3 = 5 8 +
= 26.33 Ω
3
The resistors of 15.8 Ω and 5 Ω and the resistors of 26.33 Ω and 4 Ω are connected in parallel (Fig. 1.106),
R1 = 3 5 +
6Ω
A
4Ω
9.875 Ω
3.47 Ω
3.8 Ω
B
Fig. 1.106
1.44 Network Analysis and Synthesis
Converting the delta network into a star network (Fig. 1.107),
A
6Ω
A
4Ω
6Ω
4Ω
2.19 Ω
R4
2Ω
R6
0.77 Ω
R5
B
B
Fig. 1.107
R4 =
3.8 × 9.875
= 2.199 Ω
3.8 + 9.875 + 3 47
R5 =
3.8 × 3.47
= 0 77 Ω
3.8 + 9.875 + 3 47
R6 =
3.47 9.875
=2Ω
3.8 + 9.875 + 3 47
Simplifying the network (Fig. 1.108),
4Ω
2Ω
0.77 Ω
A
B
6Ω
2.19 Ω
(a)
6Ω
0.77 Ω
A
B
8.19 Ω
(b)
3.46 Ω
0.77 Ω
A
B
(c)
4.23 Ω
A
B
(d)
Fig. 1.108
RAB = . 3 Ω
1.10 Star-delta Transformation 1.45
Example 1.42
Find an equivalent resistance between A and B in the network of Fig. 1.109.
6Ω
4Ω
A
3Ω
5Ω
4Ω
8Ω
B
Fig. 1.109
Solution Converting the star network formed by the resistors of 3 Ω, 4 Ω and 5 Ω into an equivalent delta
network (Fig. 1.110),
5Ω
3Ω
R3
5 4
R1 = 5 4 +
= 15
5.67 Ω
3
3 4
R2 = 3 4 +
=94Ω
5
5 3
R3 = 5 3 +
= 11.75 Ω
4
4Ω
⇒
R2
R1
(a)
(b)
Fig. 1.110
Similarly, converting the star network formed by the resistors of 4 Ω, 6 Ω and 8 Ω into an equivalent delta
network (Fig. 1.111),
6Ω
4Ω
R6
6 8
= 26
6Ω
4
4 8
R5 = 4 8 +
= 17
7.33 Ω
6
6 4
R6 = 6 4 +
= 13 Ω
8
R4 = 6 8 +
8Ω
⇒
R5
R4
(a)
(b)
Fig. 1.111
These two delta networks are connected in parallel between points A and B (Fig. 1.112).
13 Ω
A
11.75 Ω
26 Ω
15.67 Ω
9.4 Ω
17.33 Ω
B
A
6.17 Ω
Fig. 1.112
The resistors of 9.4 Ω and 17.33 Ω are in parallel with a short. Hence, the
equivalent resistance of this combination becomes zero.
9.78
. Ω
B
Fig. 1.113
1.46 Network Analysis and Synthesis
Simplifying the parallel networks (Fig. 1.113),
RAB = 6.17
17 � 9.78 = 3 78 Ω
Example 1.43
Determine the current supplied by the battery in the network of Fig. 1.114.
4Ω
6Ω
6Ω
4Ω
6Ω
50 V
2Ω
Fig. 1.114
Solution Converting the delta network formed by resistors of 6 Ω, 6 Ω and 6 Ω into an equivalent star
network (Fig. 1.115),
4Ω
2Ω
2Ω
4Ω
50 V
2Ω
2Ω
Fig. 1.115
Simplifying the network (Fig. 1.116),
6Ω
2Ω
3Ω
2Ω
5Ω
I
6Ω
2Ω
50 V
(a)
2Ω
50 V
(b)
Fig. 1.116
50 V
2Ω
(c)
1.10 Star-delta Transformation 1.47
50
= 7.14 A
5+ 2
I=
Example 1.44
Calculate the current flowing the 10 W resistor in Fig. 1.117.
180 V
F
8Ω
30 Ω
4Ω
17 Ω
34 Ω
A
D
B
12 Ω
30 Ω
12 Ω
C
13 Ω
E
10 Ω
Fig. 1.117
Solution Between terminals A and B resistors of 8 Ω and 4 Ω are connected in series. Similarly, between
terminals F and E, resistors of 17 Ω and 13 Ω are connected in series (Fig. 1.118),
180 V
F
30 Ω
12 Ω
A
34 Ω
B
12 Ω
12 Ω
30 Ω
D
30 Ω
C
E
10 Ω
Fig. 1.118
Converting delta ABC and DEF into an equivalent star network (Fig. 1.119),
180 V
34 Ω
10 Ω
4Ω
4Ω
10 Ω
10 Ω
4Ω
10 Ω
Fig. 1.119
1.48 Network Analysis and Synthesis
Simplifying the network (Fig. 1.120),
180 V
180 V
48 Ω
I
4Ω
I′
I
10 Ω
4Ω
16 Ω
10 Ω
24 Ω
(a)
(b)
Fig. 1.120
I=
180
=6A
4 + 16 + 10
By current division rule,
I′
Example 1.45
I 24 Ω = I
0Ω
48
=4A
24 + 48
= 6×
Find the current supplied by the battery in the network of Fig. 1.121.
40 Ω
20 Ω
50 Ω
10 Ω
30 Ω
5Ω
15 V
Fig. 1.121
Solution Converting the star network formed by resistors of 40 Ω, 20 Ω and 50 Ω into an equivalent delta
network (Fig. 1.122 and Fig. 1.123),
R1 = 40 + 20 +
40 × 20
= 76 Ω
50
20 × 50
= 95
9 Ω
40
190 Ω
R2
40 × 50
R2 = 40 + 50 +
= 190
90 Ω
20
R3 = 20 + 50 +
76 Ω
R1
95 Ω
R3
10 Ω
15 V
30 Ω
5Ω
Fig. 1.122
10 Ω
15 V
30 Ω
5Ω
Fig. 1.123
1.10 Star-delta Transformation 1.49
The resistors of 190 Ω and 10 Ω and the resistors of 95 Ω and 30 Ω are connected in parallel (Fig. 1.124).
76 Ω
9.5 Ω
22.8 Ω
5Ω
15 V
Fig. 1.124
Simplifying the network (Fig. 1.125),
I=
76 Ω
15
= 0.542 A
22.67 + 5
22.67 Ω
I
32.3 Ω
5Ω
15 V
5Ω
15 V
(a)
(b)
Fig. 1.125
Example 1.46
Find an equivalent inductance between terminals A and B in the network of Fig. 1.126.
6H
A
2H
2H
6H
6H
2H
B
Fig. 1.126
Solution Converting the star network formed by inductors of 2 H, 2 H and 2 H into an equivalent delta
network (Fig. 1.127),
6H
A
6H
6H
6H
6H
6H
B
Fig. 1.127
Simplifying the network (Fig. 1.128),
3H
A
A
3H
A
3H
B
3H
B
(a)
6H
2H
B
(b)
Fig. 1.128
(c)
1.50 Network Analysis and Synthesis
LAB = 2 H
Example 1.47
Determine the capacitance between terminals A and B in the network of Fig. 1.129.
2 μF
4 μF
3 μF
A
B
5 μF
1 μF
Fig. 1.129
Solution Converting delta network formed by capacitors of 2 μF, 1 μF and 3 μF into an equivalent
star network [Fig. 1.130 (a)],
C1 = 1 + 2 +
2 μF
1 2
= 3.667 μ F
3
C1 C3
1 3
C2 = 1 3 +
= 5.5 μ F
2
A
C2
3 μF
B
5 μF
1 μF
2 3
= 11 μ F
1
C3 = 2 3 +
4 μF
(a)
Simplifying the network (Fig. 1.130),
11 μF
4 μF
3.67 μF
2.93 μF
B
A
3.67 μF
B
A
5.5 μF
5 μF
2.62 μF
(c)
(b)
3.67 μF
2.21 μF
5.55 μF
A
B
A
(d)
B
(e)
Fig. 1.130
C AB = 2.21 μF
Example 1.48
network of Fig. 1.31.
If the combined capacitance of network shown is 5 μF, find capacitance C in the
1.11 Source Transformation 1.51
1 μF
2 μF
3 μF
3 μF
2 μF
C
4 μF
1 μF
1 μF
Fig. 1.131
Solution The capacitors of 1 μF and 1 μF and the capacitors of 1 μF and 3 μF are in parallel (Fig. 1.132),
4 μF
2 μF
4 μF
3 μF
2 μF
C
2 μF
Fig. 1.132
Replacing star network formed by capacitors of 4 μF, 4 μF
and 2 μF into an equivalent delta network (Fig. 1.133),
4 4
C1 =
= 1 6 μF
4 4 2
4 2
C2 C3 =
= 0 8 μF
4 4 2
1.6 μF
2 μF
C1
2 μF
3 μF
C3
0.8 μF
C
0.8 μF
C2
Fig. 1.133
Simplifying the network (Fig. 1.134),
3 μF
3.6 μF
0.6 μF
2.8 μF
1.575 μF
C
(a)
0.8 μF
3C
3+C
(b)
Fig. 1.134
1.575 + 0.8 +
1.11
3C
=5
3+C
3C
= 2.625
3+C
C = 21 μF
SOURCE TRANSFORMATION
A voltage source with a series resistor can be converted into a equivalent current source with a parallel
resistor. Conversely, a current source with a parallel resistor can be converted into a voltage source with a
series resistor as shown in Fig. 1.135.
1.52 Network Analysis and Synthesis
R
⇔I=
V
V
R
R
(a)
(b)
Fig. 1.135 Source transformation
Source transformation can be applied to dependent sources as well. The controlling variable, however
must not be tampered with any way since the operation of the controlled sources depends on it.
Example 1.49
Replace the given network of Fig. 1.136 with a single current source and a resistor.
A
10 A
6Ω
20 V
5Ω
B
Fig. 1.136
Solution Since the resistor of 5 Ω is connected in parallel with the voltage
source of 20 V it becomes redundant. Converting parallel combination
of current source and resistor into equivalent voltage source and resistor
(Fig. 1.137),
By source transformation (Fig. 1.138),
A
A
60 V
A
80 V
6Ω
6Ω
20 V
B
6Ω
13.33 A
B
Fig. 1.137
B
Fig. 1.138
Example 1.50
Reduce the network shown in Fig. 1.139 into a single source and a single resistor
between terminals A and B.
A
2Ω
1A
2Ω
4V
3Ω
1Ω
3V
6V
B
Fig. 1.139
1.11 Source Transformation 1.53
Solution Converting all voltage sources into equivalent current sources (Fig. 1.140),
A
1A
2Ω
2Ω
2A
2A
3Ω
1Ω
3A
B
Fig. 1.140
Adding the current sources and simplifying the network (Fig. 1.141),
A
1Ω
3A
0.75
. Ω
1A
B
Fig. 1.141
Converting the current sources into equivalent voltage sources (Fig. 1.142),
A
A
3V
3.75 V
1Ω
1.75 Ω
0.75 Ω
0.75 V
B
B
Fig. 1.142
Example 1.51
Replace the circuit between A and B in Fig. 1.143 with a voltage source in series
with a single resistor.
A
3A
5Ω
30 Ω
50 Ω
6Ω
20 V
B
Fig. 1.143
1.54 Network Analysis and Synthesis
Solution Converting the series combination of voltage source of 20 V and a resistor of 5 Ω into equivalent
parallel combination of current source and resistor (Fig. 1.144),
A
50 Ω
30 Ω
3A
4A
5Ω
6Ω
B
Fig. 1.144
Adding the two current sources and simplifying the circuit (Fig. 1.145),
A
30 || 50 || 5 || 6 = 2.38 Ω
7A
B
Fig. 1.145
By source transformation (Fig. 1.146),
2.38 Ω
A
16.67 V
B
Fig. 1.146
Example 1.52
Find the power delivered by the 50 V source in the network of Fig. 1.147.
3Ω
5Ω
50 V
2Ω
10 A
10 V
Fig. 1.147
Solution Converting the series combination of voltage source of 10 V and resistor of 3 Ω into equivalent
current source and resistor (Fig. 1.148),
1.11 Source Transformation 1.55
5Ω
2Ω
10 A
3Ω
3.33 A
50 V
Fig. 1.148
Adding the two current sources and simplifying the network (Fig. 1.149),
5Ω
1.2 Ω
13.33 A
50 V
Fig. 1.149
By source transformation (Fig. 1.150),
1.2 Ω
5Ω
I
50 V
16 V
Fig. 1.150
I=
50 − 16
= 5.48 A
5 + 1.2
Power delivered by the 50 V source = 50 × 5.48 = 274 W
Example 1.53
Find the current in the 4 W resistor shown in network of Fig. 1.151.
6V
5A
2Ω
2A
4Ω
Fig. 1.151
Solution Converting the parallel combination of the current source of 5 A and the resistor of 2 Ω into an
equivalent series combination of voltage source and resistor (Fig. 1.152),
1.56 Network Analysis and Synthesis
2Ω
6V
10 V
2A
4Ω
2A
4Ω
2A
4Ω
Fig. 1.152
Adding two voltage sources (Fig. 1.153),
2Ω
4V
Fig. 1.153
Again by source transformation (Fig. 1.154),
2Ω
2A
Fig. 1.154
Adding two current sources (Fig. 1.155),
2Ω
4A
4Ω
Fig. 1.155
By current-division rule,
I4 Ω = 4 ×
Example 1.54
2
= 1.33 A
2 4
Find the voltage across the 4 W resistor shown in network of Fig. 1.156.
3Ω
6V
2Ω
6Ω
Fig. 1.156
1Ω
3A
4Ω
1.11 Source Transformation 1.57
Solution Converting the series combination of the voltage source of 6 V and the resistor of 3 Ω into
equivalent current source and resistor (Fig. 1.157),
2Ω
3Ω
2A
1Ω
6Ω
4Ω
3A
Fig. 1.157
By series–parallel reduction technique (Fig. 1.158),
2Ω
1Ω
2Ω
2A
3A
4Ω
Fig. 1.158
By source transformation (Fig. 1.159),
2Ω
1Ω
1Ω
2Ω
4Ω
4Ω
3A
4Ω
3A
4V
4V
(a)
(b)
Fig. 1.159 (continued)
1Ω
1Ω
4Ω
1Ω
I
1A
4Ω
3A
4Ω
4Ω 4A
(c)
(d)
Fig. 1.159
I=
16
= 1.78 A
4 +1+ 4
Voltage across the 4 Ω resistor = 4 = 4 × 1.78 = 7.12 V
4Ω
4 Ω 16 V
(e)
1.58 Network Analysis and Synthesis
Example 1.55
Find the voltage at Node 2 of the network shown in Fig. 1.160.
50 Ω
1
2
I
100 Ω
15 V
100 Ω
+ 10 I
−
Fig. 1.160
Solution We cannot change the network between nodes 1 and 2 since the controlling current I, for the
controlled source, is in the resistor between these nodes. Applying source transformation to series combination
of controlled source and the 100 Ω resistor (Fig. 1.161),
1
I
50 Ω
1 I
2
100 Ω
0.1 I
15 V
1
I
100 Ω
50 Ω
50 Ω
2
0.1 I
15 V
2
50 Ω
50 Ω
+
−
15 V
5I
Fig. 1.161
Applying KVL to the mesh,
15 − 50 I − 50 I − 5
0
15
I=
= 0.143 A
105
Voltage at Node 2 = 15 − 50 I = 15 − 50 × 0.143 = 7.86 V
1.12
SOURCE SHIFTING
Source shifting is the simplification technique used when there is no resistor in series with a voltage source
or a resistor in parallel with a current source.
Example 1.56 Calculate the voltage across the 6 Ω resistor in the network of Fig. 1.162 using
source-shifting technique.
1.12 Source Shifting 1.59
3Ω
2
4Ω
1Ω
3
4
1
2Ω
18 V
+
Va
6Ω
−
Fig. 1.162
Solution Adding a voltage source of 18 V to the network and connecting to Node 2 (Fig. 1.163), we have
2
3Ω
4Ω
1Ω
3
1
4
18 V
2Ω
6Ω
18 V
+
Va
−
Fig. 1.163
Since nodes 1 and 2 are maintained at the same voltage by the sources, the connection between nodes 1
and 2 is removed. Now the two voltage sources have resistors in series and source transformation can be
applied (Fig. 1.164).
18 V
18 V
3Ω
4Ω
1Ω
+
2Ω
6Ω
Va
−
Fig. 1.164
Simplifying the network (Fig. 1.165),
1.60 Network Analysis and Synthesis
18 V
18 V
3Ω
3Ω
4.5 A
4.5 A
1Ω
1.33 Ω
+
4Ω
2Ω
6Ω
1Ω
Va
−
6Ω
−
(a)
18 V
1.33 Ω
+
Va
3Ω
5.985 V
Va
(b)
3Ω
1Ω
2.33 Ω
18 V
6Ω
5.985 V
6Ω
−
+
Va
(c)
(d)
Fig. 1.165
Applying KCL at the node,
Va 18 Va − 5.985 Va
+
+
=0
3
2 33
6
Va = 9.23 V
Exercises
1.1
Find the resistance between terminals A and B
in the network of Fig. 1.166.
1.2
Find the resistance between terminals A and B
in the network of Fig. 1.167.
24 Ω
4Ω
4Ω
4Ω
8Ω
A
B
8Ω
A
Fig. 1.166
[4 Ω]
4Ω
8Ω
4Ω
8Ω
8Ω
4Ω
B
Fig. 1.167
[2.67 Ω]
Exercises 1.61
Find the equivalent resistance between
terminals A and B in the network of
Fig. 1.168.
1.3
1Ω
A
2Ω
2Ω
3Ω
12 Ω
40 Ω
A
60 Ω
B
2Ω
3Ω
B
2Ω
2Ω
Fig. 1.172
Fig. 1.168
[3 Ω]
[8 Ω]
What is the equivalent resistance between
terminals A and B of the networks shown in
Fig. 1.169?
1.4
1.8
6Ω
A
1Ω
A
A
4Ω
B
6Ω
6Ω
4Ω
B
(a)
(b)
[(a) 0 (b) 0]
Find the equivalent resistance between terminals
A and B in the network of Fig. 1.170.
4Ω
6Ω
5Ω
B
3Ω
Fig. 1.173
1.9
[5 Ω]
Find the equivalent resistance between A and
B in the network of Fig. 1.174.
40 Ω
4Ω
A
3Ω
3Ω
Fig. 1.169
1.5
Find the equivalent resistance between A and
B in the network of Fig. 1.173.
12 Ω
60 Ω
B
4Ω
4Ω
20 Ω
30 Ω
A
B
100 Ω
Fig. 1.174
Fig. 1.170
1.6
[4 Ω]
Find the equivalent resistance between terminals
A and B in the network of Fig. 1.171.
A
[25 Ω]
1.10 Find the equivalent resistance between A and
B in the network of Fig. 1.175.
A
4Ω
4Ω
2R
R
R
4Ω
2R
2R
B
4Ω
Fig. 1.171
1.7
[4 Ω]
Find the equivalent resistance between terminals
A and B in the network of Fig. 1.172.
B
R
Fig. 1.175
⎡4 ⎤
⎢ 7 R⎥
⎣
⎦
1.62 Network Analysis and Synthesis
1.11 Find the equivalent resistance between A and
B in the network of Fig. 1.176.
A
3Ω
6Ω
20 Ω
3Ω
10 Ω
45 Ω
20 Ω
A
B
3.25 Ω
20 Ω
2.5 Ω
B
Fig. 1.176
Fig. 1.179
[17 Ω]
1.12 Find RAB by solving the outer delta (X-B-Y)
only in the network of Fig. 1.177.
Y
5Ω
1Ω
[3.5 Ω]
1.15 Find the equivalent resistance between terminals
A and B in the network of Fig. 1.180.
1Ω
7Ω
A
5Ω
5Ω
3Ω
2Ω
3Ω
X
3Ω
B
9Ω
3Ω
Fig. 1.177
1Ω
1Ω
[1.41 Ω]
5Ω
1.13 Find the equivalent resistance between A and
B in the network of Fig. 1.178.
Fig. 1.180
[1.82 Ω]
3Ω
2Ω
2Ω
2Ω
8Ω
A
1.16 Find the equivalent resistance between
the terminals A and B in the network of
Fig. 1.181.
3Ω
6Ω
3Ω
B
8Ω
6Ω
9Ω
9Ω
B
5Ω
A
Fig. 1.178
6Ω
[2.625 Ω]
1.14 Find the equivalent resistance between
the terminals A and B in the network of
Fig. 1.179.
4Ω
4Ω
Fig. 1.181
[10.32 Ω]
1.17 Find the equivalent resistance between A and
B in the network of Fig. 1.182.
Exercises 1.63
6Ω
3Ω
A
B
6Ω
1.21 Determine the power supplied to the network
in the network of Fig. 1.186.
6Ω
6Ω
6Ω
9Ω
6Ω
3Ω
6Ω
100 V
Fig. 1.182
[4.59 Ω]
1.18 Find the equivalent resistance between A and
B in the network of Fig. 1.183.
5Ω
3Ω
6Ω
9Ω
2Ω
A
Fig. 1.186
4Ω
4Ω
2Ω
4Ω
[4705.88 Ω]
1.22 Replace the given network with a single
voltage source and a resistor.
2Ω
4Ω
2Ω
B
2Ω
5A
2Ω
Fig. 1.183
[6.24 Ω]
1.19 Determine the current I in the network of
Fig. 1.184.
4Ω
I
6Ω
50 V
3Ω
10 A
2Ω
2Ω
10 V
3Ω
5Ω
4Ω
2Ω
Fig. 1.187
[8.6 V, 0.43 Ω]
1.23 Use source transformation to simplify the
network until two elements remain to the left
of terminals A and B.
3.5 kΩ
6 kΩ
8Ω
A
2Ω
2 kΩ
3 kΩ
20 mA
12 kΩ
Fig. 1.184
[8.59 A]
1.20 Find the voltage between terminals A and B in
the network of Fig. 1.185.
B
Fig. 1.188
[88.42 V, 7.92 k Ω]
1.24 Determine the voltage Vx in the network of
Fig. 1.189 by source-shifting technique.
A
3Ω
1Ω
1Ω
2Ω
1A
2Ω
1Ω
Vx
1Ω
2Ω
2V
B
2Ω
5Ω
1Ω
Fig. 1.185
[0.56 V]
Fig. 1.189
[1.129 V]
1.64 Network Analysis and Synthesis
Objective-Type Questions
1.1
1.2
A network contains linear resistors and ideal
voltage sources. If values of all the resistors
are doubled then the voltage across each
resistor is
(a) halved
(b) doubled
(c) increased by four times
(d) not changed
Four resistances 80 Ω, 50 Ω, 25 Ω, and R are
connected in parallel. Current through 25 Ω
resistor is 4 A. Total current of the supply is
10 A. The value of R will be
(a) 66.66 Ω
(b) 40.25 Ω
(c)
1.3
36.36 Ω
A
A
(a) 5 V source in series with a 10 Ω resistor
(b) 1 V source in series with a 2.4 Ω resistor
(c) 15 V source in series with a 2.4 Ω resistor
(d) 1 V source in series with a 10 Ω resistor
Consider the star network shown in Fig. 192.
The resistance between terminals A and B
with C open is 6 Ω, between terminals B and
C with A open is 11 Ω and between terminals
C and A with B open is 9 Ω, then resistacne
RA, RB and RC will be
RA
R2
B
C
B
R3
C
RB
15 Ω
1.4
(b)
(d)
3, 9, 1.5
3, 1.5, 9
C
Fig. 1.192
If each branch of a delta network has resistance
3 R, then each branch of the equivalent wye
network has resistance
(a)
R
3
3 3R
(a) RΑ = 4 Ω,
(b) RΑ = 2 Ω,
(c) RΑ = 3 Ω,
(d) RΑ = 5 Ω,
(b) 3 R
R
3
Viewed from the terminal AB, the network
of Fig.1.191 can be reduced to an equivalent
network of a single voltage source in series
with a single resistor with the following
parameters
(c)
1.5
1.5, 3, 9
9, 3, 1.5
RC
B
Fig. 1.190
(a)
(c)
4Ω
A
R1
30 Ω
5Ω
10 Ω
Fig. 1.191
A delta connected network with its wyeequivalent is shown in Fig.190. The resistances
R1, R2 and R3 (in ohms) are respectively.
A
5V
B
1.6
(d) 76.56 Ω
10 V
(d)
1.7
RΒ = 2 Ω, RC = 5 Ω
RΒ = 4 Ω, RC = 7 Ω
RΒ = 3 Ω, RC = 4 Ω
RΒ = 1 Ω, RC = 10 Ω
A 10 V battery with an internal resistance of 1
Ω is connected across a nonlinear load whose
V-I characteristic is given by 7 V 2 2V .
The current delivered by the battery is
(a) 0
(b) 10 A
(c) 5 A
(d) 8 A
Answers to Objective-Type Questions 1.65
1.8
1.9
If the length of a wire of resistance R is
uniformly stretched to n times its original
value, its new resistance is
(a) nR
(b)
(c) n2R
(d)
R
n
R
4
8
A
(d)
A
15
15
1.10 The current waveform in a pure resistor at 10
Ω is shown in Fig. 1.194. Power dissipated in
the resistor is
(c)
i
n2
All the resistances in Fig. 1.193 are 1 Ω each.
The value of I will be
9
0
I
3
t
6
Fig. 1.194
(a)
(c)
1V
Fig. 1.193
(a)
1
A
15
(b)
2
A
15
7.29 W
135 W
(b)
(d)
52.4 W
270 W
1.11 Two wires A and B of the same material and
length L and 2L have radius r and 2r respectively.
The ratio of their specific resistance will be
(a) 1 : 1
(b) 1 : 2
(c)
1:4
(d)
1:8
Answers to Objective-Type Questions
1.1. (d)
1.2. (c)
1.3. (d)
1.4. (a)
1.8. (c)
1.9. (d)
1.10. (d)
1.11. (b)
1.5. (b)
1.6. (b)
1.7. (c)
2
2.1
Elementary Network
Theorems
INTRODUCTION
In Chapter 1, we have studied basic network concepts. In network analysis, we have to find currents and
voltages in various parts of networks. In this chapter, we will study elementary network theorems like
Kirchhoff’s laws, mesh analysis and node analysis. These methods are applicable to all types of networks.
The first step in analyzing networks is to apply Ohm’s law and Kirchhoff’s laws. The second step is the
solving of these equations by mathematical tools.
2.2
KIRCHHOFF’S LAWS
The entire study of electric network analysis is based mainly on Kirchhoff’s laws. But before discussing this,
it is essential to familiarise ourselves with the following terms:
Node
A node is a junction where two or more network elements are connected together.
Branch An element or number of elements connected between two nodes constitute a branch.
Loop
A loop is any closed part of the circuit.
Mesh
A mesh is the most elementary form of a loop and cannot be further divided into other loops.
All meshes are loops but all loops are not meshes.
1. Kirchhoff’s Current Law (KCL) The algebraic sum of currents
meeting at a junction or node in an electric circuit is zero.
Consider five conductors, carrying currents I1, I2, I3, I4 and I5
meeting at a point O as shown in Fig. 2.1. Assuming the incoming
currents to be positive and outgoing currents negative, we have
I1 ( I 2 ) + I 3 ( I 4 ) + I 5 = 0
I1 − I 2 + I 3 I 4 + I 5 = 0
I1 I 3 + I 5 I 2 + I 4
I1
I2
O
I3
I5
Fig. 2.1
I4
Kirchhoff’s current law
Thus, the above law can also be stated as the sum of currents flowing towards any junction in an
electric circuit is equal to the sum of the currents flowing away from that junction.
2. Kirchhoff’s Voltage Law (KVL) The algebraic sum of all the voltages in any closed circuit or mesh
or loop is zero.
If we start from any point in a closed circuit and go back to that point, after going round the circuit,
there is no increase or decrease in potential at that point. This means that the sum of emfs and the sum of
voltage drops or rises meeting on the way is zero.
2.2 Network Analysis and Synthesis
3. Determination of Sign A rise in potential can be assumed to be positive while a fall in potential can
be considered negative. The reverse is also possible and both conventions will give the same result.
(i) If we go from the positive terminal of the battery or source to the negative terminal, there is a fall in
potential and so the emf should be assigned a negative sign (Fig. 2.2a). If we go from the negative
terminal of the battery or source to the positive terminal, there is a rise in potential and so the emf
should be given a positive sign (Fig. 2.2b).
(a) Fall in potential
(b) Rise in potential
Fig. 2.2 Sign convention
(ii) When current flows through a resistor, there is a voltage drop across it. If we go through the resistor
in the same direction as the current, there is a fall in the potential and so the sign of this voltage
drop is negative (Fig. 2.3a). If we go opposite to the direction of the current flow, there is a rise in
potential and hence, this voltage drop should be given a positive sign (Fig. 2.3b).
I
+
+
I
(b) Rise in potential
(a) Fall in potential
Fig. 2.3 Sign convention
Example 2.1 In Fig. 2.4, the voltage drop across the 15 W resistor is 30 V, having the polarity
indicated. Find the value of R.
2A
5Ω
5A
+
−
15 Ω
−
30 V
I
+
Fig. 2.4
Solution Current through the 15 Ω resistor
I=
Current through the 5 Ω resistor
resisto
i
30
=2A
15
5 2 7A
Applying KVL to the closed path,
−5 (7) − R ( 2) + 100 − 30 = 0
−35 − 2 R + 100 − 30 = 0
R = 17.5 Ω
R
−
100 V
3A
+
2.2 Kirchhoff’s Laws 2.3
Example 2.2
Determine the currents I1, I2 and I3 in Fig. 2.5.
I2
I1
12 Ω
I3
9A
8Ω
4A
16 Ω
Fig. 2.5
Solution Assigning currents to all the branches (Fig. 2.6),
I2
I1
12 Ω
I3
9A
8Ω
4A
16 Ω
(I1 − I2)
(I1 − I2 + 9 + I3 + 4)
(I3 + 4)
Fig. 2.6
From Fig. 2.6,
Also,
I1 I1 − I 2
I 2 I 3 = 13
−12 I1 − 8 ( 1 − 2 ) = 0
−20 I1 + 8 2 = 0
9 I3
4
…(i)
…(ii)
−12 I1 −16
16 I 3 = 0
and
Solving Eqs (i), (ii) and (iii),
…(iii)
I1 = 4 A
I 2 = 10 A
I 3 = −3 A
Example 2.3
Find currents in all the branches of the network shown in Fig. 2.7.
80 A
0.02 Ω
0.02 Ω
60 A
30 A
0.01 Ω
0.01 Ω
60 A
70 A
0.01 Ω
0.03 Ω
120 A
Fig. 2.7
2.4 Network Analysis and Synthesis
Solution Let
Then
I AF
x
I FE
x − 30
I ED
x + 40
I DC
x − 80
I CB
x − 20
I BA
x − 80
80 A
(x − 80) 0.02 Ω
0.02 Ω
30 A
A
60 A
B (x − 20)
x
F
0.01 Ω
0.01 Ω
C
60 A
(x − 80)
(x − 30) E
70 A
0.01 Ω
Applying KVL to the closed path AFEDCBA
(Fig. 2.8),
D
(x + 40)
0.03 Ω
120 A
Fig. 2.8
−0.02 − 0.01( − 30) − 0.01( x + 40) − 0.03 ( − 80) − 0.01( x − 2200) − 0.02 ( x − 80
8 )=0
x = 41 A
I AF = 41 A
I FE = 41 − 30 = 11 A
I ED = 41 + 40 = 81 A
I DC = 41 − 80 = −39 A
I CD = 39 A
I CB = 41 − 20 = 21 A
I BA = 41 − 80 = −39 A
I AB = 39 A
Example 2.4
Find currents in all the branches of the network shown in Fig. 2.9.
B
4Ω
C
3Ω
5Ω
2Ω
1A O
1A
A
1Ω
Fig. 2.9
Solution Assigning currents to all the branches (Fig. 2.10),
Applying KVL to the closed path OBAO,
−2(1 − x ) − 3 y + 1( ) = 0
3x − 3 y = 2
y ) 5( x + y ) = 0
9 x 12
1 y 4
4Ω
C
1A
(x + y)
y
3Ω
…(i)
Applying KVL to the closed path ABCA,
3 y 4(1 x
(1 −x − y)
B
5Ω
2Ω
(1 −x )
1A O
…(ii)
1Ω
x
Fig. 2.10
A
2.2 Kirchhoff’s Laws 2.5
Solving Eqs (i) and (ii),
= 0 57 A
y = −0.095 A
I OA = 0 57 A
I OB = 1 − 0.57 = 0.43 A
I AB = 0.095 A
I AC = 0.57 − 0.095 = 0.475 A
I BC = 1 − 0.57 + 0.095 = 0.525 A
Example 2.5
What is the potential difference between points x and y in the network shown in Fig. 2.11?
2Ω
−
4V
x
+
+
4V
−
I1
−
3Ω
−
3Ω
2V
+
5Ω
I2
y
Fig. 2.11
2
=04A
2 3
4
I2 =
=05A
3 5
Potential difference between points x and y = Vxy V x − Vy
I1 =
Solution
Writing KVL equation for the path x to y,
Vx
Vx
3I + 4 − 3I
3I
Vy = 0
3 (0 4) 4 3 (0 5) Vy = 0
Vx V y = −3 7
Vxy = −3.7 V
Example 2.6
Find the voltage between points A and B in Fig. 2.12.
B
10 Ω
A
I1
5Ω
20 V
5 V 12 Ω
Fig. 2.12
6Ω
4Ω
I2
15 V
+
2.6 Network Analysis and Synthesis
20
= 1 33 A
10 + 5
15
I2 =
=1 5A
4 6
I1 =
Solution
From Fig. 2.13,
Voltage between points A and B = VAB
VA VB
B
Writing KVL equation for the path A to B,
VA
VA
I
(1.33)
10 Ω
−
I1
−
+
VA VB = 17.65
VAB = 17.65 V
Example 2.7
−
+
5 15 + 6 I VB = 0
1 6 (1. ) VB = 0
A
5Ω
20 V
+
+
5 V 12 Ω
−
15 V
Fig. 2.13
Determine the potential difference VAB for the given network in Fig. 2.14.
2A
3Ω
A
10 Ω
2Ω
4Ω
5Ω
8V
5V
B
Fig. 2.14
Solution The resistor of 3 Ω is connected across a short circuit. Hence, it gets shorted (Fig. 2.15).
2A
A
10 Ω
+
2Ω
5V
I1
+
5Ω
−
8V
−
I2
B
Fig. 2.15
5
= 2.5 A
2
I2 = 2 A
VAB VA VB
I1 =
Potential difference
4Ω
Writing KVL equation for the path A to B,
VA 2 I + 8 5 I VB = 0
VA 2 ( 2.5) 8 5 ( 2) VB = 0
6Ω
I2
4Ω
2.2 Kirchhoff’s Laws 2.7
VA VB = 7
VAB = 7 V
Example 2.8
Find the voltage of the point A w.r.t. B in Fig. 2.16.
5A
5Ω
A
+
−
8V
3Ω
3Ω
10 V
+
−
I1
4Ω
I2
B
Fig. 2.16
10
= 1.25 A
5 3
I2 = 5 A
Applying KVL to the path from A to B,
VA 3 I 8 + 3 I VB = 0
VA 3 (1.25) 8 3 (5) VB = 0
I1 =
Solution
VA VB = −3 25
VAB = −3.225 V
Example 2.9
In Fig. 2.17, what values must R1 and R2 have
(a) when I1 = 4 A and I2 = 6 A both charging?
(b) when I1 = 2 A discharging and I2 = 20 A charging?
(c) when I1 = 0?
b (I1 + I2)
2Ω
c
d
I1
I2
R1
R2
110 V
80 V
a
f
50 V
e
Fig. 2.17
Solution Applying KVL to the closed path abcfa,
110 − 2 ( 1 2 ) 1 1 80 0
110 − 2
1
2 I 2 − R1 I1 − 80 = 0
( 2 1 ) 1 2 I 2 = 30
Applying KVL to the closed path fcdef,
80 + R1 I1 − R2 I 2 − 50 = 0
R1 I1 − R2 I 2 = −30
…(i)
…(ii)
2.8 Network Analysis and Synthesis
Case (a) I1 = 4 A and I2 = 6 A both charging
i.e.
I1 = 4 A and I2 = 6 A
Substituting I1 and I2 in Eq. (i),
(2 + R1) 4 + 2 (6) = 30
R1 = 2.5 Ω
Substituting R1, I1 and I2 in Eq. (ii),
2.5 (4)−R2 (6) = −30
R2 = 6.67 Ω
Case (b) I1 = 2 A discharging and I2 = 20 A charging
i.e.
I1 = −2 A and I2 = 20 A
Substituting I1 and I2 in Eq. (i),
(2 + R1) (−2) + 2 (20) = 30
R1 = 3 Ω
Substituting R1, I1 and I2 in Eq. (ii),
3 (−2) − R2 (20) = −30
R2 = 1.2 Ω
Case (c) I1 = 0
Substituting in Eq. (i)
(2 + R1) (0) + 2 I2 = 30
I2 = 15 A
Substituting I1 and I2 in Eq. (ii),
0 − 15 R2 = −30
R2 = 2 Ω
Example 2.10
In Fig. 2.18, find I1 and I2 when (a) R = 2.3 W, (b) R = 0.5 W, and (c) for what values
of R is I1 = 0?
b (I1 + I2)
0.2 Ω
c
e
I1
I2
0.2 Ω
130 V
R
110 V
a
d
f
Fig. 2.18
Solution Applying KVL to closed path abcda,
130 − 0.2 ( I1 + I 2 ) − 0.2 I1 − 110 = 0
0.44 I1 + 0 2 I 2 = 20
…(i)
Applying KVL to the closed path dcefd,
110 + 0 2 I1 − R I 2 = 0
0 2 I1 R I 2 = −110
…(ii)
2.2 Kirchhoff’s Laws 2.9
Case (a) R = 2.3 Ω
Substituting R in Eq. (ii),
0.22 I1 2 3 I 2 = −110
…(iii)
Solving Eqs (i) and (iii),
I1 = 25 A
I 2 = 50 A
Case (b) R = 0.5 Ω
Substituting R in Eq. (i),
0.22 I1 0 5 I 2 = −110
…(iv)
Solving Eqs (i) and (iv),
I1 = −50 A
I 2 = 200 A
Case (c)
I1 = 0
Substituting I1 in Eq. (i),
0 2 I 2 = 20
I 2 = 100 A
Substituting I1 and I2 in Eq. (ii),
0.2 (0)
(100) = −110
R = 1.1 Ω
Example 2.11
In Fig. 2.19, find the value of R.
10 Ω
3A
14 Ω
80 V
R
Fig. 2.19
Solution Assigning currents to all the branches (Fig. 2.20),
Applying KVL to the closed path abcda,
b
I
10 Ω
c
80 −10
10 I − 14 ( I − 3) 0
I = 5.08 A
Applying KVL to the closed path dcefd,
14 ( 3) 3R = 0
14 (5.08 3)
3) − 3
0
R = 9 71
7 Ω
3A
e
(I − 3)
14 Ω
80 V
a
d
Fig. 2.20
R
f
2.10 Network Analysis and Synthesis
Example 2.12
Determine current drawn by the ammeter shown in Fig. 2.21.
10 Ω
5Ω
9V
30 Ω
A
5Ω
Fig. 2.21
Solution Assigning currents to all the branches (Fig. 2.22),
Applying KVL to the closed path abcda,
(I1 + I2)
b
10 Ω
−5 ( 1 + 2 ) + 9 − 10 ( 1 + 2 ) − 30 I1 = 0
45 I1 +
+15
155 I 2 = 9 …(i)
I2
30 Ω
0 …(ii)
2
e
I1
9V
Applying KVL to the closed path dcefd,
30 I1 5
5Ω
c
A
5Ω
Solving Eqs (i) and (ii),
I 2 = 0.4 A
Current drawn by ammeter = 0.4 A
a
d
f
Fig. 2.22
Example 2.13
Find branch currents in the various branches of Fig. 2.23.
10 Ω
20 Ω
0.1 Ω
0.2 Ω
5Ω
2V
4V
Fig. 2.23
Solution Assigning currents to various branches (Fig. 2.24),
Applying KVL to the closed path abcda,
2 0.1
1
10
0
5 ( I1 I 2 ) = 0
15.1 1 5 I 2 = 2
b
1
1
2)
20
02 2 4=0
5 1 25
25 2 I 2 = 4
10 Ω
c
…(i)
Applying KVL to the closed path dcefd,
5(
I1
I2
20 Ω
e
(I1 − I2)
0.1 Ω
0.2 Ω
5Ω
2
Solving Eqs (i) and (ii),
I1 = 0.086 A
I 2 = −0.142 A
…(ii)
2V
4V
a
d
Fig. 2.24
f
2.2 Kirchhoff’s Laws 2.11
Example 2.14
In Fig. 2.25, find the value of R and current flowing through it when the current is
zero in the branch OA.
A
480 Ω
1.5 Ω
1Ω
O
4Ω
R
C
B
2Ω
10 V
Fig. 2.25
Solution Assigning currents to all branches (Fig. 2.26),
Applying KVL to the closed path OACO,
480 I 3 1 5 ( I1 − I 3 ) R ( I 2
I3 ) = 0
But current in the branch OA is zero,
i.e.
I3 = 0
−1 5 I1
A
I3
R I2 = 0
(I1 − I3)
480 Ω
…(i)
1.5 Ω
1Ω
Applying KVL to the closed path BOCB,
−4
2
− (
2
+
3 ) − 2 ( I1
+ I 2 ) + 10 = 0
B
−4
4 I2
Substituting I1 in Eq. (i) and (ii),
10 V
2Ω
(I1 + I2)
C
Fig. 2.26
I1
I1
0
4 I2
−6 2 + 2 = 0
−14 I 2 − R I 2 = −10
From Eq. (iv) and (v),
I 2 = 0.5 A
Current in branch OC = I 2
(I2 + I3)
I2
I3 = 0
But
Substituting I2 in Eq. (iv),
R
I1
I3 = 0
−2
2 I1 (6 R) I 2 = −10
...(ii)
Applying KVL to the closed path BOAB,
−4 I 2 + 480 I 3 + I1 = 0
But
and
O
4Ω
−6 (0.5) + (0.5) = 0
R=6Ω
I 3 = 0.5 A
…(iii)
…(iv)
…(v)
2.12 Network Analysis and Synthesis
Example 2.15
In Fig. 2.27, find the current supplied by the battery.
2Ω
6Ω
2Ω
50 V
5Ω
4Ω
3Ω
Fig. 2.27
Solution Assigning currents to all the branches (Fig. 2.28),
Applying KVL to the closed path OABEDO,
(I1 + I2)
50 − 2 ( 1 2 ) 6 I1 4 ( 1 3 ) 0
A
12 I1 + 2 2 4 I 3 = 50
…(i)
Applying KVL to the closed path BCEB,
−2 2 + 5 I 3 + 6 1 = 0
6 1 − 2 I2 + 5 3 = 0
…(ii)
2Ω
I1
3
− 3( I 2 + I3 ) + 4 ( 1 − 3 ) = 0
4 1 − 3 I 2 − 12 I 3 = 0
50 V
E
(I1 − I3)
I3
2Ω
3Ω
…(iii)
O
D
Fig. 2.28
I1 = 2.817 A
I 2 = 6.647 A
I 3 = −0.723 A
Current supplied by the battery = I1 + I 2 = 2.8
817
7 + 6.647 = 9.464 A
In Fig.2.29, find the current flowing through the 2 W resistor.
16 Ω
16 Ω
32 Ω
20 V
20 V
2Ω
Fig. 2.29
C
(I2 + I3)
5Ω
4Ω
Solving Eqs (i), (ii) and (iii),
Example 2.16
I2
6Ω
Applying KVL to the closed path ECDE,
−5
B
2.2 Kirchhoff’s Laws 2.13
Solution Assigning currents to all the branches (Fig. 2.30),
Applying KVL to the closed path OABFGO,
−
I
16 I
A
(I1 + I )
B
I
…(i)
0
32 Ω
16 Ω
Applying KVL to the closed path BCFB,
20 V
0
1−
1
48 −
+ 32 I = 0 …(ii)
(I1 − I )
= 20
20 V
I3
F
Applying KVL to the closed path GFCDEG,
−
(I1 + I − I ) D
C
2Ω
O
…(iii)
G
E
Fig. 2.30
Solving Eqs (i), (ii) and (iii),
I1 = 1 05 A
I 2 = 6 32 A
I 3 = 1 58 A
Current through th
resistor
Example 2.17
A
In Fig. 2.31, find the current through the 4 W resistor.
2Ω
2Ω
12 Ω
12 V
10 V
3Ω
1Ω
4Ω
24 V
Fig. 2.31
Solution Assigning currents to all the branches (Fig. 2.32),
Applying KVL to the closed path ABEDA,
1
+ 12 = 0
+ 12 I 3 = −12 …(i)
Applying KVL to the closed path BCFEB,
−
−I
+
− I3
−
0
= 0 … (ii)
)−
3(
−
+ 4 0
+
= −24 …(iii)
C
(I2 I3)
12 Ω
12 V
(I1 − I )
D
Applying KVL to the closed path DEFHGD,
1
2Ω
B I
A
10 V
I
(I1 I3)
3Ω
1Ω
F
I
G
24 V
4Ω
Fig. 2.32
H
2.14 Network Analysis and Synthesis
Solving Eqs (i), (ii) and (iii),
I1 = 4.11 A
I1 = 2 72 A
I 3 = 2 06 A
Current through the 4 Ω resisto
resistor
i
Example 2.18
.
1
A
In Fig. 2.33, find the current through the 10 W resistor.
5Ω
10 Ω
12 Ω
15 Ω
4V
8Ω
6V
Fig. 2.33
Solution Assigning currents to all the branches (Fig. 2.34),
Applying KVL to the closed path ABGHA,
−5 ( 1 +
2 ) − 15 1
(I1 + I2) 5 Ω B
A
+4=0
−20 I1 − 5
2
= −4
…(i)
I2
I1
I3
15 Ω
8Ω
4V
Applying KVL to the closed path BCFGB,
−10 I 2 − 8
3
+ 15
15 I1 −10
10 I 2 − 8
1
=0
3
=0
H
10 Ω C 12 Ω (I2 − I3)
G
F
Fig. 2.34
…(ii)
Applying KVL to the closed path CDEFC,
−12 (
2
−
3 ) − 6 + 8 I3
=0
−12 I 2 + 20
0 I3 = 6
…(iii)
Solving Eqs (i), (ii) and (iii),
I1 = 0 19 A
I 2 = 0.032 A
I 3 = 0 32 A
Current through the 10 Ω resistor
resisto =
Example 2.19
2
0. 03 A
In Fig. 2.35, determine the current supplied by each battery.
20 Ω
1Ω
2Ω
40 Ω
10 Ω
8V
12V
Fig. 2.35
D
6V
E
2.2 Kirchhoff’s Laws 2.15
Solution Assigning currents to all the branches (Fig. 2.36),
Applying KVL to the closed path ADEA,
40 I 2 + 8 − 1I1 = 0
I1 + 40 I 2 = 8
Applying KVL to the closed path ABDA,
−20 (
1
−
2 ) − 10 ( 1
−
2
+
0 I2
3 ) + 40
…(i)
B
(I1 − I2 + I3)
I2
I1
10 Ω
−30 I1 + 70 I 2 − 10 I 3 = 0
Applying KVL to the closed path BCDB,
2 3 12 10
0( 1 2 3 ) 0
10 I1 100 I 2 + 12 I 3 = 12
…(ii)
2Ω
40 Ω
1Ω
=0
20 Ω
(I1 − I2)
A
I3
8V
12 V
C
D
E
Fig. 2.36
…(iii)
Solving Eqs (i), (ii), and (iii),
I1 = 0.1005 A
I 2 = 0.197 A
I 3 = 1.081 A
Current supplied by the 8 V battery 1 0.1005 A
Current supplied by th
t e 12 V battery = I 3 = 1.081 A
Example 2.20
In Fig. 2.37, find the value of the unknown resistance R such that 2 A current flows
through it.
R
2A
2Ω
10 V
4Ω
3Ω
5Ω
Fig. 2.37
Solution Assigning currents to all the branches (Fig. 2.38),
Applying KVL to the closed path ABCDEA,
−2 R + 4 ( I1 − I 2 − 2) + 2 ( I1 − 2) = 0
6 1 − 4 I 2 − 2 = 12
Applying KVL to the closed path HEDGH,
10 − 2 ( 1 2) 3 2 0
2 1 3 I 2 = 14
A
…(i)
E
2) 5 ( I1 − I 2 ) = 0
9 1 12 2 8
Solving Eqs (i), (ii) and (iii),
2
2Ω
…(ii)
H
4 Ω (I1 − I2 − 2)
D
I2
3Ω
10 V
G
Fig. 2.38
4 ( I1 − I 2
…(iii)
I1 = 3 76 A
I 2 = 2.16 A
R = 0 98 Ω
Unknown resistance R = 0 98 Ω
(I1 − 2)
B
I1
Applying KVL to the closed path GDCFG,
3
R
2A
C
(I1 − I2)
5Ω
F
2.16 Network Analysis and Synthesis
Example 2.21
In Fig. 2.39, find the current delivered by the 12 V battery.
3Ω
5Ω
2Ω
4Ω
4Ω
12 V
4Ω
Fig. 2.39
Solution Assigning currents to all the branches (Fig. 2.40),
Applying KVL to the closed path ABCDEA,
3(
1
2)
2 I2 5( 2 3 ) 0
3 1 10
0 2 5 I3 = 0
…(i)
E
Applying KVL to the closed path HEDGH,
4(
1
…(ii)
3
2 I 2 − 12 + 4
4 1 2 I2 + 4
1
3
0
12
5Ω
G
…(iii)
I1
Fig. 2.40
Current delivered by the 12 V battery = I1 = 1 66 A
EXAMPLES WITH DEPENDENT SOURCES
In the network of Fig. 2.41, find I1, I2 and V.
V
I2
I1
Fig. 2.41
I2
C
I1
12 V
H
3Ω
B
4Ω
I1 = 1 66 A
I 2 = 0 93 A
I 3 = 0 87 A
2A
2Ω
I3
Solving Eqs (i), (ii), and (iii),
Example 2.22
(I1 − I2)
D
4Ω
Applying KVL to the closed path GDCFG,
4
(I2 − I3)
(I1 − I3)
5( 2 3 ) − 4 3 0
4 1 5 I 2 − 13 I 3 = 0
3)
3Ω
A
4V
6Ω
F
4Ω
2.2 Kirchhoff’s Laws 2.17
Solution Applying KCL at the node,
2+ 4V =
V V
+
3 6
V V
+ − 4V 2
3 6
7
− V =2
2
4
V
7
4
=− A
21
2
=− A
21
V =−
V
3
V
I2 =
6
I1 =
Example 2.23
Find voltages V1 and V2 in the network of Fig. 2.42.
2Ω
I
4I
− +
+ V1 −
6V
+
4 Ω V2
−
Fig. 2.42
Solution Applying KVL to the loop,
6 2I + 4 I
4I = 0
I =3A
From Fig. 2.42,
V1 = 2 I = 2 (3) = 6 V
V2 = 4 I = 4 (3) = 12 V
Example 2.24
Find the power delivered by the dependent source in the network of Fig. 2.43.
I
1Ω
3 Vx
− +
+
Vx
−
30 V
3Ω
Fig. 2.43
From Fig. 2.43,
Vx
0 5I
Applying KVL to the loop,
30 −1
1 I 3 Vx − 0 5 I − 3
0
30 − I + 3 (0
(0.5 ) 0.5 3 I = 0
0.5 Ω
2.18 Network Analysis and Synthesis
30 − 3
0
I = 10 A
Vx = 0.5 (10) = 5 V
Power supplied by the dependent source = (3Vx) (I) = 3 × 5 × 10 = 150 W
Example 2.25
Find the current I2 in the network of Fig. 2.44.
8V
+
4V
V1
10 V1
I2
16 V
−
Fig. 2.44
Solution Applying KVL to the left loop,
−4 + 8 − V1 = 0
V1 = 4 V
Applying KCL to the right part,
10 V1 I 2 = 0
10 ( 4) 2 0
I 2 = −40 A
Example 2.26
Find the voltage Vx in the network of Fig. 2.45.
+
2A
−
6Ω
Vx
+
Vy
−
3Ω
1V
6 y
Fig. 2.45
Solution Applying KCL at the node,
1
V
2 + Vy = x
6
9
…(i)
⎛V ⎞ V
Vy = 3 ⎜ x ⎟ = x
⎝ 9⎠
3
…(ii)
From Fig. 2.45,
Substituting Eq. (ii) in Eq. (i),
1 ⎛V ⎞ V
2+ ⎜ x ⎟ = x
6⎝ 3 ⎠
9
V V
2+ x − x = 0
18 9
Vx = 36 V
2.2 Kirchhoff’s Laws 2.19
Example 2.27
In the network of Fig. 2.46, find the current I1 and power dissipated in the 500 W
resistor.
I1 300 Ω
50 V
0.4 I1
500 Ω
Fig. 2.46
Solution Assigning currents to all the branches as shown in Fig. 2.47.
b
I1 300 Ω
e
c
0.6 I1
50 V
0.4 I1
500 Ω
a
d
f
Fig. 2.47
Applying KVL to the closed loop abcda,
50 − 300 I1 500 (0
0 6 I1 ) = 0
I1 = 0.083 A
Power dissipated in the 500 Ω resistor = 500 (0.6 I1)2 = 500 (0.6 × 0.083)2 = 1.24 W
Example 2.28
Find the current I in the network shown in Fig. 2.48.
2A
3Ω
5Ω
4V
+
Vx
−
2Ω
I
+
−
3 Vx
Fig. 2.48
Solution Assigning currents in all the branches as shown in Fig. 2.49.
2A
a
5Ω
c
(I − 2)
4V
3Ω
+
2Ω
Vx
−
b
d
e
I
+
−
3 Vx
f
Fig. 2.49
From Fig. 2.49,
Vx
Applying KVL to the closed loop fecdf,
3Vx 5 I − 4 2 ( I − 2)
3 [2 ( I 2)] 5 I − 4 2 ( I − 2)
2( I − 2)
0
0
…(i)
2.20 Network Analysis and Synthesis
6
12 5 I − 4 2 I + 4
0
− I = 12
I = −12 A
2.3
MESH ANALYSIS
A mesh is defined as a loop which does not contain any other loops within it. Mesh analysis is applicable only
for planar networks. A network is said to be planar if it can be drawn on a plane surface without crossovers.
In this method, the currents in different meshes are assigned continuous paths so that they do not split at a
junction into branch currents. If a network has a large number of voltage sources, it is useful to use mesh
analysis. Basically, this analysis consists of writing mesh equations by Kirchhoff’s voltage law in terms of
unknown mesh currents.
Steps to be Followed in Mesh Analysis
1.
2.
3.
Identify the mesh, assign a direction to it and assign an unknown current in each mesh.
Assign the polarities for voltage across the branches.
V2
Apply KVL around the mesh and use Ohm’s law to
express the branch voltages in terms of unknown mesh
currents and the resistance.
R1
V
4. Solve the simultaneous equations for unknown mesh 1
I1
I2
currents.
Consider the network shown in Fig. 2.50 which has three meshes.
R2
R4
I3
Let the mesh currents for the three meshes be I1, I2, and I3 and all
the three mesh currents may be assumed to flow in the clockwise
R5
V3
direction. The choice of direction for any mesh current is arbitrary.
Fig. 2.50
Applying KVL to Mesh 1,
V1
R3
I 2 ) − R2 ( I1 − I 3 ) = 0
R1 ( I1
V1
…(i)
R3 I 2 R4 ( I 2 I 3 ) − R1 ( I 2 − I1 ) = 0
− R1 I1 + ( R1 R3 + R4 ) I 2 − R4 I 3 = V2
…(ii)
( R1
R2 ) I1
R1 I 2
R2 I 3
Applying KVL to Mesh 2,
V2
Applying KVL to Mesh 3,
− R2 ( I 3 − I1 ) − R4 ( I 3 − I 2 ) − R5 I 3 + V3 = 0
− R2 I1 − R4 I 2 + ( R2 + R4 + R5 ) I 3 = V3
Writing Eqs (i), (ii), and (iii) in matrix form,
⎡ R1 R2
⎢ − R1
⎢
R2
⎣ −R
− R1
R1 + R3 R4
R4
R2
R2
⎤ ⎡ I1 ⎤ ⎡V1 ⎤
⎥ ⎢ I 2 ⎥ = ⎢V2 ⎥
− R4
⎥⎢ ⎥ ⎢ ⎥
R4 + R5 ⎦ ⎣ I 3 ⎦ ⎣V3 ⎦
In general,
⎡ R11
⎢ R21
⎢
⎣ R31
R12
R22
R32
R13 ⎤ ⎡ I1 ⎤ ⎡V1 ⎤
R23 ⎥ ⎢ I 2 ⎥ = ⎢V2 ⎥
⎥⎢ ⎥ ⎢ ⎥
R33 ⎦ ⎣ I 3 ⎦ ⎣V3 ⎦
…(iii)
2.3 Mesh Analysis 2.21
where, R11 = Self-resistance or sum of all the resistance of mesh 1
R12 = R21 = Mutual resistance or sum of all the resistances common to meshes 1 and 2
R13 = R31 = Mutual resistance or sum of all the resistances common to meshes 1 and 3
R22 = Self-resistance or sum of all the resistance of mesh 2
R23 = R32 = Mutual resistance or sum of all the resistances common to meshes 2 and 3
R33 = Self-resistance or sum of all the resistance of mesh 3
If the directions of the currents passing through the common resistance are the same, the mutual resistance
will have a positive sign, and if the direction of the currents passing through common resistance are opposite
then the mutual resistance will have a negative sign. If each mesh current is assumed to flow in the clockwise
direction then all self-resistances will always be positive and all mutual resistances will always be negative.
The voltages V1, V2 and V3 represent the algebraic sum of all the voltages in meshes 1, 2 and 3 respectively.
While going along the current, if we go from negative terminal of the battery to the positive terminal then its
emf is taken as positive. Otherwise, it is taken as negative.
Example 2.29
Find the current through the 5 W resistor is shown in Fig. 2.51.
1Ω
2Ω
3Ω
5Ω
10 V
5V
6Ω
4Ω
20 V
Fig. 2.51
Solution Assigning clockwise currents in three meshes as shown in Fig. 2.52.
Applying KVL to Mesh 1,
10 −1
1 1 3 ( I1
I2 ) − 6 ( 1 3 ) 0
10 I1 3 2 6 I 3 = 10
1Ω
2Ω
…(i)
−3 (
2
− 1) − 2 I2 − 5 2 − 5 = 0
−3 1 + 1
10
0 2 = −5
I2
…(ii)
3
− 1 ) + 5 − 4 3 + 20 = 0
−6 1 + 110
0 3 = 25
Writing Eqs (i), (ii) and (iii) in matrix form,
3 6 ⎤ ⎡ I1 ⎤ ⎡10 ⎤
⎡10 −3
⎢ −3 10 0 ⎥ ⎢ I 2 ⎥ = ⎢ −5⎥
⎢
⎥⎢ ⎥ ⎢ ⎥
⎣ −6 0 10 ⎦ ⎣ I 3 ⎦ ⎣ 25 ⎦
We can write matrix equation directly from Fig. 2.52,
⎡ R11
⎢ R21
⎢
⎣ R31
10 V
I1
Applying KVL to Mesh 3,
−6 (
R12
R22
R32
R13 ⎤ ⎡ I1 ⎤ ⎡V1 ⎤
R23 ⎥ ⎢ I 2 ⎥ = ⎢V2 ⎥
⎥⎢ ⎥ ⎢ ⎥
R33 ⎦ ⎣ I 3 ⎦ ⎣V3 ⎦
5Ω
3Ω
Applying KVL to Mesh 2,
5V
6Ω
4Ω
I3
…(iii)
20 V
Fig. 2.52
2.22 Network Analysis and Synthesis
R11 = Self-resistance of Mesh 1 = 1 + 3 + 6 = 10 Ω
R12 = Mutual resistance common to meshes 1 and 2 = −3 Ω
Here, negative sign indicates that the current through common resistance are in opposite direction.
R13 = Mutual resistance common to meshes 1 and 3 = −6 Ω
Similarly,
R21 = −3 Ω
R22 = 3 2 5 = 10 Ω
R23 = 0
where
R31 = −6 Ω
R32 = 0
R33 = 6 4 = 100 Ω
For voltage matrix,
V1 = 10 V
V2 = −5 V
V3 = algebraic sum of all the voltages in mesh 3 = 5 + 20 = 25 V
Solving Eqs (i), (ii) and (iii),
I1 = 4.27 A
I 2 = 0 78 A
I 3 = 5 06 A
I5
I 2 = 0 78 A
Example 2.30
Find the current through the 2 W resistor of the network shown in Fig. 2.53.
6Ω
2Ω
1Ω
10 V
3Ω
10 Ω
20 V
Fig. 2.53
Solution Assigning clockwise currents in three meshes as shown in Fig. 2.54,
Applying KVL to Mesh 1,
6Ω
10 − 6 1 1( I1 − I 2 ) = 0
…(i)
7 1 2 10
Applying KVL to Mesh 2,
−1( 2 − 1 ) − 2 I 2 − 3 ( 2 − 3 ) = 0
− I1 + 6 2 − 3 I 3 = 0
2Ω
1Ω
10 V
I1
3Ω
I2
20 V
…(ii)
Applying KVL to Mesh 3,
−3 ( 3 − 2 ) − 10 3 − 20 = 0
−3 2 + 113
3 3 = −20
Solving Eqs (i), (ii) and (iii),
I1 = 1 34 A
I 2 = −0 62 A
10 Ω
I3
Fig. 2.54
…(iii)
2.3 Mesh Analysis 2.23
I 3 = −1 68 A
I 2 = −0 62 A
I2
Example 2.31
Determine the current through the 5 W resistor of the network shown in Fig. 2.55.
10 V
1Ω
8V
2Ω
4Ω
3Ω
5Ω
12 V
Fig. 2.55
Solution Assigning clockwise currents in three meshes as shown in Fig. 2.56.
Applying KVL to Mesh 1,
8 1(
2 ) − 2( 1
1
3
1
2
3)
0
2 I3 = 8
…(i)
2
2Ω
3
− 1) − 3(
3Ω
I3
…(ii)
12 V
Applying KVL to Mesh 3,
−2 (
I2
I1
3 ( I 2 − I 3 ) 1( 2 1 ) 0
− I1 + 8 2 3 I 3 = 10
4Ω
1Ω
8V
Applying KVL to Mesh 2,
10 − 4
10 V
5Ω
Fig. 2.56
− 2 ) − 5 3 + 12 = 0
−2 1 − 3 I 2 + 10 I 3 = 12
3
…(iii)
Solving Eqs (i), (ii), and (iii),
I1 = 6 01 A
I 2 = 3 27 A
I 3 = 3 38 A
I5
I 3 = 3 38 A
Example 2.32
Find the current supplied by the battery of the network shown in Fig. 2.57.
3Ω
1Ω
I2
2Ω
4V
I1
4Ω
Fig. 2.57
5Ω
I3
6Ω
2.24 Network Analysis and Synthesis
Solution
Applying KVL to Mesh 1,
4 3 I1 1( 1
2 ) 4 ( I1 − I 3 ) = 0
8 1 2 4 I3 = 4
…(i)
Applying KVL to Mesh 2,
−2 2 − 5 ( I 2 − I 3 ) −1
1( 2 − 1 ) = 0
− I1 + 8 2 − 5 I 3 = 0
…(ii)
Applying KVL to Mesh 3,
−6 3 − 4 ( I 3 − I1 ) − 5 ( 3 − 2 ) = 0
−4 1 − 5 I 2 + 15 I 3 = 0
…(iii)
Solving Eqs (i), (ii) and (iii),
I1 = 0 66 A
I 2 = 0 24 A
I 3 = 0 26 A
Current supplied by the battery = I1 = 0.66 A.
Example 2.33
Find the current through the 4 W resistor in the network of Fig. 2.58.
2Ω
2Ω
2Ω
6V
8V
I2
4Ω
I1
2Ω
2Ω
I3
Fig. 2.58
Solution Applying KVL to Mesh 1,
8 2 I1 2 (
2)
1
6
1
2 ( I1 − I 3 ) = 0
2 I2 − 2 3 8
…(i)
Applying KVL to Mesh 2,
−2 (
2
− 1) − 2 I2 − 4 ( 2 − 3 ) − 6 = 0
−2 1 + 8 I 2 − 4 3 = −6
…(ii)
Applying KVL to Mesh 3,
−2 (
3
− 1 ) + 6 − 4 ( 3 − 2 ) − 2 I3 = 0
−2 1 − 4 I 2 + 8 3 = 6
Solving Eqs (i), (ii) and (iii),
I1 = 2 A
I2 = 0 5 A
I3 = 1 5 A
I4
I 3 − I 2 = 1.5 − 0.5 = 1 A
…(iii)
2.3 Mesh Analysis 2.25
Example 2.34
Determine the voltage V which causes the current I1 to be zero in the network
of Fig. 2.59.
6Ω
2Ω
20 V
3Ω
I2
1Ω
5Ω
I1
V
4Ω
I3
Fig. 2.59
Solution Applying KVL to Mesh 1,
20 − 6
1
2 ( I1 − I 2 ) 5 (
3)
1
V +13
13 I1 2
2
V
0
5 I 3 = 20
…(i)
Applying KVL to Mesh 2,
−2 (
2
− 1 ) − 3 I 2 −1
1(
2
−
3)
=0
− 6 I 2 + I3 = 0
…(ii)
+ V − 5( 3 − 1) = 0
V + 5 I1 + 2 − 10 3 = 0
…(iii)
2
1
Applying KVL to Mesh 3,
−1(
3
−
2 ) − 4 I3
Putting I1 = 0 in Eqs (i), (ii) and (iii),
V
2 I 2 − 5 I 3 = 20
−6 I 2 I 3 = 0
V
I 2 −10
− 10 I 3 = 0
…(iv)
Solving Eqs (i), (ii) and (iii),
V = 43.7 V
Example 2.35
Find the current through the 2 W resistor in the network of Fig. 2.60.
3Ω
12 Ω
6A
I1
36 V
2Ω
6Ω
I2
I3
9V
Fig. 2.60
Solution Mesh 1 contains a current source of 6 A. Hence, we can write current equation for mesh 1. Since
direction of current source and mesh current I1 are same,
I1 = 6
…(i)
2.26 Network Analysis and Synthesis
Applying KVL to Mesh 2,
36 −12
12 ( 2
36 −12
12 (
2
1) − 6 ( 2
3)
0
3
0
6) 6 I 2 + 6
6 I 3 = 108
18 I 2
Applying KVL to Mesh 3,
−6 ( 3 − 2 ) − 3 I 3 − 2
6
2
3
…(ii)
−9 = 0
− 11
11
3
=9
…(iii)
Solving Eqs (ii) and (iii),
I3 = 3 A
I2
I3 = 3 A
Example 2.36
Determine the mesh currents I1, I2 and I3 in the network of Fig. 2.61.
30 V
I1
I2
I3
1A
6Ω
15 Ω
50 V
10 Ω 5 Ω
Fig. 2.61
Solution Applying KVL to Mesh 1,
−30 − 6 1 − 15
15 ( 1 − 2 ) = 0
21 I1 − 15 I 2 = −30
…(i)
Applying KVL to Mesh 2,
−10 ( 2 − 3 ) − 15 ( 2 − 1 ) + 50 − 5 2 = 0
−15 I1 + 30 I 2 − 10 I 3 = 50
…(ii)
For Mesh 3,
I3 = 1
…(iii)
Solving Eqs (i), (ii) and (iii),
I1 = 0
I2 = 2 A
I3 = 1 A
Example 2.37
Find the current through the 5 W resistor in the network of Fig. 2.62.
3A
2Ω I
2
4A
I1
5Ω
I3
Fig. 2.62
2Ω
2V
I4
3A
2.3 Mesh Analysis 2.27
Solution Writing current equations for Meshes 1, 2 and 4,
I1 = 4
(i)
I2 = 3
… (ii)
I 4 = −3
…(iii)
Applying KVL to Mesh 3,
−5 (
3
− 1) − 2 (
3
−
2 ) − 2( 3
−
4) − 2
=0
…(iv)
Substituting Eqs (i), (ii) and (iii) in Eq. (iv),
−5 (
3
− 4) − 2 (
3
− 3) − 2 (
3
+ 3) − 2 = 0
I3 = 2 A
I 5 Ω = I1 − I 3 = 4 − 2 = 2 A
EXAMPLES WITH DEPENDENT SOURCES
Example 2.38
Obtain the branch currents in the network shown in Fig. 2.63.
IA
5Ω
10IB
+ −
5Ω
IB
10 Ω
5V
10 V
+
−
5IA
Fig. 2.63
Solution Assigning clockwise currents in two meshes as shown in Fig. 2.64,
From Fig. 2.64,
10IB
5Ω
IA
…(i)
I A I1
+ −
I B I 2 …(ii)
5Ω
Applying KVL to Mesh 1,
IB
10 Ω
5 5 I1 10 I 10 ( I1 − I 2 ) − 5
5 5 I1 10
0 I 2 −10
10 I1 + 10
0 I2 − 5
A
1
0
0
−20 I1 = −5
1
I1 = = 0.25 A
4
Applying KVL to Mesh 2,
5 A 10 ( 2 1 ) 5 I 2 − 10 = 0
5 1 10
0 2 10 1 5 I 2 = 10
15 I1 15
1 I 2 = 10
5V
10 V
I1
+
−
5IA
I2
Fig. 2.64
…(iii)
…(iv)
2.28 Network Analysis and Synthesis
Putting I1 = 0.25 A in Eq. (iv),
15 (0.25) 15
Example 2.39
10
I 2 = −0.416 A
2
Find the mesh currents in the network shown in Fig. 2.65.
2V2
4Ω
+ −
5Ω
2Ω
+ V2 −
+
V1
−
1Ω
10 V
−
2V1
+
5V
Fig. 2.65
Solution Assigning clockwise currents in the
two meshes as shown in Fig. 2.66,
From Fig. 2.66,
…(i)
V1
5 I1
V2 2 I 2
…(ii)
Applying KVL to Mesh 1,
2V2
+ −
5Ω
4Ω
2Ω
+ V2 −
+
V1
−
1Ω
10 V
5V
−
2V1
+
I1
−5 − 5 I1 − 2 2 − 4 I1 −1
1( 1 − 2 ) + 2V1 = 0
−5 − 5 I1 − 2 ( 2 I 2 ) − 4 1 − 1 + 2 + 2 ( −5 I1 ) = 0
20 I1 + 3 2
5
I2
Fig. 2.66
…(iii)
Applying KVL to Mesh 2,
−2V1 − 1( I 2 − I1 ) − 2
2
− 10 = 0
−2 ( −5 I1 ) − I 2 + I1 − 2
11 I1 − 3
2
2
= 10
= 10
…(iv)
Solving Eqs (iii) and (iv),
I1 = 0.161 A
I 2 = −2.742 A
Example 2.40
Find currents Ix and Iy of the network shown in the Fig. 2.67.
2Ix
+ −
5Ω
4 Ω Iy
2Ω
1Ω
10 V
5V
−
2Iy
+
Ix
Fig. 2.67
2.3 Mesh Analysis 2.29
Solution Assigning clockwise currents in the two Meshes as shown in Fig. 2.68.
From Fig. 2.68,
2Ix
4 Ω Iy
I y I1
+ −
I1 − I 2
Ix
5Ω
1Ω
Applying KVL to Mesh 1,
−5 − 5 I1 − 2
−5 − 5 I1 − 2 (
1
x
− 4 I1 −1
1(
−
2 ) − 4 I1
−5 − 5 I1 − 2 I1 + 2 I 2 − 4
1
1
−
2) + 2Iy
=0
−
2Iy
+
Ix
I1
5V
− I1 + I 2 + 2 I1 = 0
−
2Ω
10 V
I2
Fig. 2.68
+ 2 + 2 I1 = 0
−10 I1 + 3 2 = 5
1
…(iii)
Applying KVL to Mesh 2,
−2
y
− 1( I 2 − I1 ) − 2 I 2 − 10 = 0
−2
1
−
2
+ 1 − 2 I 2 = 10
− I1 − 3 2 = 10
...(iv)
Solving Eqs (iii) and (iv),
15
= −1.364 A
11
I 2 = −2.878 A
I1 = −
I y = −1.364 A
Ix
Example 2.41
I1 − I 2 = −1.364 + 2.878 = 1.514 A
Find the currents in the three meshes of the network shown in Fig. 2.69.
Ix
1Ω
1Ω
+ −
1Ω
Ix
+
I
− y
1Ω
5V
1A
1Ω
Iy
Fig. 2.69
Solution Assigning clockwise currents in the three meshes is shown in Fig. 2.70.
Ix
1Ω
1Ω
+ −
1Ω
Ix
+
I
− y
1Ω
5V
I1
1Ω
I2
Iy
Fig. 2.70
1A
I3
2.30 Network Analysis and Synthesis
From Fig. 2.70,
Ix = I1
I y I 2 − I3
...(i)
…(ii)
Applying KVL to Mesh 1,
5 1 I1 − I y − 1(
5
1
(
3)
2
(
−22
1
2)
0
1
2)
0
1
5
3
…(iii)
Applying KVL to Mesh 2,
−1(
−(
2
2
− 1 ) + I y − 1 I 2 − I x −11(
− 1) + (
2
−
3 ) − I2
2
−
3)
=0
− I1 − ( I 2 − I 3 ) = 0
−22
2
…(iv)
0
For Mesh 3,
I 3 = −1
…(v)
Solving Eqs (iii), (iv) and (v),
I1 = 2 A
I2 = 0
I 3 = −1 A
Example 2.42
For the network shown in Fig. 2.71, find the power supplied by the dependent voltage
source.
50 Ω
20 Ω
5A
+
V1
−
30 Ω
+
−
0.4 V1
0.01 V1
Fig. 2.71
Solution Assigning clockwise currents in three Meshes as shown in Fig. 2.72.
50 Ω
20 Ω
5A
+
V1
−
I1
30 Ω
I3
+
−
0.4 V1
Fig. 2.72
I2
0.01 V1
2.3 Mesh Analysis 2.31
From Fig. 2.72,
V1 20 ( I1 − I 3 ) 0 4 V1 = 0
20 I1 − 20 I 3
33 33 I1 − 33.33 I 3
0 6 V1
V1
…(i)
For Mesh 1,
I1 = 5
…(ii)
For Mesh 2,
0.33 I1
I2
I2
0 01 V1
0.33 I 3 = 0
0.01(33.33 I1 33.33 I 3 )
…(iii)
Applying KVL to Mesh 3,
−50 I 3 − 30 ( I 3 − I 2 ) − 20 ( I 3 − I1 ) = 0
−20 I1 − 30 I 2 + 100 I 3 = 0
…(iv)
Solving Eqs (ii), (iii) and (iv),
I1 = 5 A
I 2 = −1.47 A
I 3 = 0 56 A
V1
33 33 I1 − 33.33 I 3 = 33.33(
33 (5) 33.33 (0.56) = 148 V
Power supplied by the dependent voltage source = 0.4 V1 (I1 − I2) = 0.4 (148) (5 + 1.47) = 383.02 W
Example 2.43
Find the voltage Vx in the network shown in Fig. 2.73.
0.45 A
16.67 Ω
33.33 Ω
+ Vx −
25 Ω
30 V
−
+
2 Vx
10 V
`
Fig. 2.73
Solution Assigning clockwise currents in the three meshes as shown in Fig. 2.74.
0.45 A
16.67 Ω
I3
33.33 Ω
+ Vx −
25 Ω
−
+
30 V
I1
Fig. 2.74
10 V
I2
2 Vx
2.32 Network Analysis and Synthesis
From Fig. 2.74,
16 67 I1
Vx
Applying KVL to Mesh 1,
−30 − 16.67
67 I1 − 33 33 (
1
−
25 ( 1
3 ) − 25
−
2 ) − 10
…(i)
=0
−30 − 16.67
67 I1 − 33 33 I1 + 33.33 I 3 − 25
5 I1 + 25 I 2 − 10 = 0
−75 I1 + 25 I 2 + 33.33 I 3 = 40
…(ii)
Applying KVL to Mesh 2,
10 − 25 ( 2 1 ) + 2 Vx
10 − 25 ( 2 1 ) + 2 (16.67 1 )
0
0
10 − 25 I 2 + 25
25 I1 + 33.34 I1 = 0
58.34 I1 255 I 2 = −10
…(iii)
I 3 = 0 45
…(iv)
For Mesh 3,
Solving Eqs (ii), (iii) and (iv),
I1 = −0 9 A
I 2 = −1 7 A
I 3 = 0 45 A
Vx
Example 2.44
16 67 I1 = 16.67 ( −0.9) = −15 V
…(v)
For the network shown in Fig. 2.75, find the mesh currents I1, I2 and I3.
15 A
2Ω
I3
0.111 Vx
1Ω
+
Vx
−
I1
3Ω
1Ω
I2
2Ω
Fig. 2.75
Solution From Fig. 2.75,
Vx
3( I1 − I 2 )
…(i)
I 3 = 15
…(ii)
Writing current equation for the two current sources,
and
0.111 Vx I1
0.111 [3 ( 1 2 ) = I1
0.333
333 I1 0 333 I 2 − I1 I 3 = 0
−0
0.667
667 I1 0 333 I 2 + I 3 = 0
I3
I3
…(iii)
2.3 Mesh Analysis 2.33
Applying KVL to Mesh 2,
−3 ( 2 − 1 ) − 1( 2 − 3 ) − 2 2 = 0
−3 1 + 6 I 2 − I 3 = 0
…(iv)
Solving Eqs (ii), (iii) and (iv),
I1 = 17 A
I 2 = 11 A
I 3 = 15 A
Example 2.45
For the network shown in Fig. 2.76, find the magnitude of V0 and the current supplied
by it, given that power loss in RL = 2 W resistor is 18 W.
10 Ω
5Ω
5Ω
+ Vx −
V0
2Ω
4Ω
RL = 2 Ω
2Ω
2Vx
Fig. 2.76
Solution Assigning clockwise currents in meshes is shown in Fig. 2.77.
10 Ω
5Ω
5Ω
+ Vx −
V0
2Ω
I1
4Ω
2Ω
2Vx
I2
I3
RL = 2 Ω
I4
Fig. 2.77
From Fig. 2.77,
Vx
…(i)
I1
Also,
I 4 2 RL = 18
I 4 2 ( 2) 18
I4 = 3 A
…(ii)
Applying KVL to Mesh 1,
V0
5 I1 2 ( I1 I 2 ) = 0
7 I1 2 I 2 = V0
…(iii)
Applying KVL to Mesh 2,
−2 (
2
− 1) − 4 I2 = 0
−2 1 + 6 I 2 = 0
…(iv)
For Mesh 3,
10 I1
I 3 2Vx
I3 = 0
2 (5 I1 ) 10 I1
…(v)
2.34 Network Analysis and Synthesis
Applying KVL to Mesh 4,
−2 (
4
−
3 ) − 5 I4
−2
4
=0
−2 3 + 9 I 4 = 0
−2 3 + 9 (3) = 0
I 3 = 13.5 A
From Eq. (v),
I1 =
I 3 13.5
=
= 1.35 A
10 10
From Eq. (iv),
−2 (1.35) + 6
=0
I 2 = 0 45 A
2
From Eq. (iii),
7 (1.35) 2 (0.45
45)) = V0
V0 = 8 55 V
Current supplied by voltage source V0 = I1 = 1.35 A
Example 2.46
In the network shown in Fig. 2.78, find voltage V2 such that Vx = 0.
+
− 0 1 Vx
2A
10 Ω
24 V
3A
5Ω
+
Vx
−
20 Ω
V2
Fig. 2.78
Solution Assigning clockwise currents in four meshes as shown in Fig. 2.79.
From Fig. 2.79,
…(i)
Vx 20 ( I 3 − I 4 )
2A
Writing current equations for Meshes 1 and 2,
I1 = 2
I2 = 3
Applying KVL to Mesh 3,
24 −10
10 ( 3 1 ) − 20 ( 3 4 ) = 0
24 −10
10 ( 3 2) 20
0( 3 4) 0
−30
30 I 3 + 200 I 4 = −44
Applying KVL to Mesh 4,
−20 ( 4 − 3 ) − 5 ( 4 − 2 ) + V2 = 0
−20 ( 4 − 3 ) − 5 ( 4 − 3) + V2 = 0
20 I 3 − 25 I 4 = −V2 − 15
But
Vx = 0
20 ( I 3 I 4 ) = 0
I3 I 4
I2
+
− 0.1 Vx
I1
10 Ω
…(ii)
…(iii)
24 V
I3
3A
5Ω
+
Vx
−
V2
20 Ω
I4
Fig. 2.79
…(iv)
…(v)
2.4 Supermesh Analysis 2.35
From Eq. (iv),
30 I 3 + 20 I 3 = −44
I 3 = 4.4 A
I 4 = 4.4 A
From Eq. (v),
20 ( 4.4) 25 ( 4.4
4)) = − V2 − 15
V2 = 7 V
2.4
SUPERMESH ANALYSIS
Meshes that share a current source with other meshes, none of which contains a current source in the outer
loop, form a supermesh. A path around a supermesh doesn’t pass through a current source. A path around each
mesh contained within a supermesh passes through a current source. The total number of equations required
for a supermesh is equal to the number of meshes contained in the supermesh. A supermesh requires one
mesh current equation, that is, a KVL equation. The remaining mesh current equations are KCL equations.
Example 2.47
Find the current through the 10 W resistor of the network shown in Fig. 2.80.
5Ω
1Ω
2V
10 Ω
I1
15 Ω
4A
I2
I3
Fig. 2.80
Solution Applying KVL to Mesh 1,
2 1 I1 10 ( I1 − I 2 ) = 0
11 I1 10 I 2 = 2
…(i)
Since meshes 2 and 3 contain a current source of 4 A, these two meshes will form a supermesh. A
supermesh is formed by two adjacent meshes that have a common current source. The direction of the current
source of 4 A and current (I3 − I2) are same, i.e., in the upward direction.
Writing current equation to the supermesh,1
I2 = 4
…(ii)
Applying KVL to the outer path of the supermesh,
−10 ( 2 − 1 ) − 5 2 − 15
15 3 = 0
10 I1 −15
15 I 2 − 15 I 3 = 0
15
…(iii)
I3
Solving Eqs (i), (ii) and (iii),
I1 = −2 35 A
I 2 = −2 78 A
I 3 = 1.22 A
Current through the 10 Ω resistor = I1 − I2 = −(2.35) − (−2.78) = 0.43 A
…(iv)
2.36 Network Analysis and Synthesis
Example 2.48
Find the current in the 3 W resistor of the network shown in Fig. 2.81.
2Ω
1Ω
I2
7V
I1
7A
3Ω
1Ω
2Ω
I3
Fig. 2.81
Solution Meshes 1 and 3 will form a supermesh.
Writing current equation for the supermesh,
I3 = 7
I1
…(i)
Applying KVL to the outer path of the supermesh,
7 1(
2 ) − 3( 3
1
− I1 + 4
1I 3 = 0
2)
2
− 4 I 3 = −7
…(ii)
Applying KVL to Mesh 2,
−1(
2
− 1) − 2 I2 − 3(
I1 − 6
2
−
3)
=0
2 + 3 I3 = 0
…(iii)
Solving Eqs (i), (ii) and (iii),
I1 = 9 A
I2 = 2 5 A
I3 = 2 A
Current through the 3 Ω resistor = I2− I3 = 2.5 − 2 = 0.5 A
Example 2.49
Find the current in the 5 W resistor of the network shown in Fig. 2.82.
10 Ω
50 V
I1
2Ω
I2
3Ω
2A
1Ω
5Ω
Fig. 2.82
I3
2.4 Supermesh Analysis 2.37
Solution Applying KVL to Mesh 1,
50 −10
10 ( 1 2 ) − 5 ( 1 3 ) 0
15 I1 10 I 2 − 5 3 50
Meshes 2 and 3 will form a supermesh as these two meshes share a common current source of 2 A.
Writing current equation for the supermesh,
I 2 I3 = 2
…(i)
…(ii)
Applying KVL to the outer path of the supermesh,
−10 (
2
− 1) − 2
2
− 1I 3 − 5 (
3
− 1) = 0
−15 I1 +12
12 I 2 + 6
3
=0
…(iii)
Solving Eqs (i), (ii) and (iii),
I1 = 20 A
I 2 = 17.33 A
I 3 = 15.33 A
Current through the 5 Ω resistor = I1 − I3 = 20 − 15.33 = 4.67 A
Example 2.50
Determine the power delivered by the voltage source and the current in the 10 W
resistor of the network shown in Fig. 2.83.
5Ω
3A
5Ω
I2
10 Ω
50 V
I1
1Ω
10 A
I3
Fig. 2.83
Solution Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I1 I 2 = 3
…(i)
Applying KVL to the outer path of the supermesh,
50 − 5 1 5 I 2 −10
10 ( I 2 I 3 ) −1
1( 1 3 ) 0
−6
6 1 15
5 2 11 3
50
…(ii)
For Mesh 3,
I 3 = 10
Solving Eqs (i), (ii) and (iii),
I1 = 9 76 A
I 2 = 6 76 A
I 3 = 10 A
Power delivered by the voltage source = 50 I1 = 50 × 9.76 = 488 W
I10 Ω I 3 I 2 = 10 − 6.76 3.24 A
…(iii)
2.38 Network Analysis and Synthesis
Example 2.51
For the network shown in Fig. 2.84, find current through the 8 W resistor.
6Ω
I2
2Ω
3A
4Ω
5V
I1
11 Ω
8Ω
I4
7A
12 A
I3
10 V
Fig. 2.84
Writing current equations for Meshes 1 and 4,
I1 = −3
...(i)
I 4 = −12
...(ii)
Meshes 2 and 3 will form a supermesh.
Writing current equation for the supermesh,
I2 = 7
I3
…(iii)
Applying KVL to the outer path of the supermesh,
5 4 ( 2 1 ) − 6 2 8 ( 3 − 4 ) + 10 = 0
5 4 ( 2 + 3) 6 I 2 − 8 ( 3 12) 10 0
−10 I 2 8 I 3 93
…(iv)
Solving Eqs (iii) and (iv),
I 2 = −8 28 A
I 3 = −1.28 A
I 3 − I 4 = −1.28 12
1 10.72 A
I8
EXAMPLES WITH DEPENDENT SOURCES
Example 2.52
In the network of Fig. 2.85, find currents I1 and I2.
8Ω
3V0
V0
+ −
2Ω
−10 V
10 Ω
−3 A I
2
I1
Fig. 2.85
Solution From Fig. 2.85,
−10 − 8
1
−
0 =0
V0 = −10 − 8 I1
…(i)
Meshes 1 and 2 will form a supermesh.
Writing current equations for the supermesh,
I2
I1 = −3
…(ii)
2.4 Supermesh Analysis 2.39
Applying KVL to the outer path of the supermesh,
10 8 I1 − 3V0 − 10 I 2 = 0
−
10 − 8 1 ) − 10 2 = 0
1 3 ( −10
−10 − 8
16 I1 −10
10 I 2 = −20
…(iii)
Solving Eqs (ii) and (iii),
I1 = −8 33 A
I 2 = −11.33 A
Example 2.53
In the network of Fig. 2.86, find the current through the 3 W resistor.
−4 V
3Ω
+
Vx
−
2Ω
1Ω
+
−
2A
5 Vx
Fig. 2.86
Solution Assigning clockwise currents in two meshes as shown in Fig. 2.87.
From Fig. 2.87,
−4 V
Vx
2 I1
…(i)
Meshes 1 and 2 will form a supermesh.
Writing current equations for the supermesh,
2Ω
I1 = 2
I2
…(ii)
3Ω
+
Vx
−
1Ω
2A
I1
Applying KVL to the outer path of the supermesh,
1
− 3 I2 = 4
…(iii)
Solving Eqs (ii) and (iii),
I1 = 2 A
I2 = 4 A
I3
Example 2.54
I2 = 4 A
Find the currents I1 and I2 at the network shown in Fig. 2.88.
10 Ω
14 Ω
4Ω
I3
110 V
I2
Fig. 2.87
−2 1 − 4 − 3 2 − 5Vx = 0
−2 1 − 4 − 3 2 − 5 ( −2 1 ) = 0
8
+
−
2Ω
+
V1
−
I1
6Ω
0.5 V1
I2
Fig. 2.88
5 Vx
2.40 Network Analysis and Synthesis
Solution From Fig. 2.88,
V1 = 2 (I1 − I2)
Meshes 2 and 3 will form a supermesh.
Writing current equation for the supermesh,
I3
I 2 = 0.5 V1 = 0 5 × 2 ( I1
I3
I 2 ) = I1
I2
I1
Applying KVL to outer path of the supermesh,
−2 ( 2 − 1 ) − 10 3 − 6 I 2 = 0
−2 2 + 2 I1 − 10 I1 − 6 2 = 0
I1 = − I 2
Applying KVL to Mesh 1,
110 −14
14
1
4 1 2 ( I1 I 2 ) = 0
110 − 20 I1 + 2 2 0
110 + 20 I1 + 2
0
2
I 2 = −5 A
I1 = I 2 = 5 A
Example 2.55
For the network of Fig. 2.89, find current through the 8 W resistor.
5 Iy
+ −
10 Ω
6Ω
Iy
8Ω
50 V
Ix
52 V
0.5 Ix
Fig. 2.89
Solution Assigning clockwise currents to the three meshes as shown in Fig. 2.90.
From Fig. 2.90,
Ix
Iy
I1
I 2 − I3
…(i)
10 Ω
…(ii)
I3
5 Iy
6Ω
Iy
8Ω
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I 2 I1 = 0.5 I x
−0
0 5 I1 I 2 = 0
+ −
50 V
Ix
0.5 ( I1 )
…(iii)
I1
0.5 Ix
52 V
I2
Fig. 2.90
Applying KVL to the outer path of the supermesh,
50 −10
10 (
1
3) − 6( 2
−10
10 I1
8 I 2 − 52 = 0
14 I 2 +16
16 I 3 = 2
3)
…(iv)
2.4 Supermesh Analysis 2.41
Applying KVL to Mesh 3,
5 y − 6 ( I 3 − I 2 ) −10
10 ( I 3 − I1 ) = 0
−5 (
2
−
3) − 6
( I 3 − I 2 ) − 10 ( I 3 − I1 ) = 0
10 I1 + I 2 − 11 I 3 = 0
…(v)
Solving Eqs (iii), (iv) and (v),
I1 = −1 56 A
I 2 = −0 58 A
I 3 = −1.11 A
I8
I 2 = −0.58 A
Example 2.56
For the network shown in Fig. 2.91, find the current through the 10 W resistor.
10 Ω
20 V 5 Ω
15 V
4Ω
2A
I1
2 I1
I2
40 V
I3
Fig. 2.91
Solution Meshes 1, 2 and 3 will form a supermesh.
Writing current equations for the supermesh,
I1 I 2 = 2
…(i)
I 2 = 2 I1
I3 0
I3
I2
…(ii)
2 I1
and
Applying KVL to the outer path of the supermesh,
15 −10
10 I1 20
0 5 2 4 I 3 + 40 = 0
10 I1 + 5 2 4 I 3 = 35
…(iii)
Solving Eqs (i), (ii) and (iii),
I1 = 1 96 A
I 2 = −0 04 A
I 3 = 3 89 A
I10 Ω
I1 = 1.96 A
Example 2.57
In the network shown in Fig. 2.92, find the power delivered by the 4 V source and
voltage across the 2 W resistor.
2Ω
+ V2 −
5A
5Ω
4V
6Ω
4Ω
I1
1Ω
I2
I3
Fig. 2.92
V2
2
2.42 Network Analysis and Synthesis
Solution From Fig. 2.92,
V2
…(i)
2 I1
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I2
I1 = 5
…(ii)
Applying KVL to the outer path of the supermesh,
4 5 I 2 2 1 6 I1 4 ( 1 3 ) 1( I 2 − I 3 ) = 0
−12
12 I1 6 2 5 I 3 = −4
…(iii)
For Mesh 3,
V2 2 I1
=
= I1
2
2
I3 = 0
I3 =
I1
…(iv)
Solving Eqs (ii), (iii) and (iv),
I1 = −2 A
I2 = 3 A
I 3 = −2 A
Power delivered by the 4 V source = 4 I2 = 4 (3) = 12 W
V2
2 I1 2 ( −2
2)
4V
Example 2.58
Find currents I1, I2, I3, I4 of the network shown in Fig. 2.93.
1
Ω
10
1
Ω
5
6V
1
Ω
20
−
Vx
+
1
Ω
15
5 Vx
I2
I3
1
Ω
2
40 A
1
Ω
6
I1
I4
Fig. 2.93
From Fig. 2.93,
Vx
1
( I 2 − I1 )
5
…(i)
For Mesh 4,
I 4 = 40
Applying KVL to Mesh 1,
1
1
1
−6 − I1 − ( I1 − I 2 ) − ( I1 − I 4 ) = 0
10
5
6
1
1
1
1
1
−6 − I1 − I1 + I 2 − I1 + (40) 0
10
5
5
6
6
7
1
2
− I1 + I 2 = −
15
5
3
…(ii)
…(iii)
2.5 Node Analysis 2.43
Mesh 2 and 3 will form a supermesh.
Writing current equation for the supermesh,
I3
I1 2 I 2
I 2 = 5 Vx
⎡1
⎤
( I 2 − I1 )
5
⎣
⎦
I 2 − I1
…(iv)
0
I3
Applying KVL to the outer path of the supermesh,
1
1
1
1
− ( 2 − 1) −
I 2 − I3 − ( 3 − 4 ) = 0
5
20
15
2
1
1
1
1
1
1
− I 2 + I1 −
I 2 − I 3 − I 3 + ( 40) = 0
5
5
20
15
2
2
1
1
17
I1
I 2 − I 3 = −20
5
4
30
…(v)
Solving Eqs (iii), (iv) and (v),
I1 = 10 A
I 2 = 20 A
I 3 = 30 A
I 4 = 40 A
2.5
NODE ANALYSIS
Node analysis is based on Kirchhoff’s current law which states that the algebraic sum of currents meeting
at a point is zero. Every junction where two or more branches meet is regarded as a node. One of the nodes
in the network is taken as reference node or datum node. If there are n nodes in any network, the number of
simultaneous equations to be solved will be (n − 1).
Steps to be followed in Node Analysis
1.
2.
3.
4.
Assuming that a network has n nodes, assign a reference node and the reference directions, and
assign a current and a voltage name for each branch and node respectively.
Apply KCL at each node except for the reference node and apply Ohm’s law to the branch currents.
Solve the simultaneous equations for the unknown node voltages.
Using these voltages, find any branch currents required.
Example 2.59
Calculate the current through 2 W resistor for the network shown in Fig. 2.94.
VA
1Ω
0.5 Ω
1Ω
20 V
VB
2Ω
1Ω
20 V
Fig. 2.94
Solution Assume that the currents are moving away from the nodes.
2.44 Network Analysis and Synthesis
Applying KCL at Node A,
20 VA VA VB
+
+
=0
1
1
05
1
20
⎛1 1 1 ⎞
VB =
⎜⎝ + +
⎟⎠ VA
1 1 0.5
0.5
1
4 VA 2 VB = 20
2
…(i)
VB VA VB VB − 20
+
+
=0
05
2
1
1
20
⎛ 1 1 1⎞
−
VA
VB =
0.5 ⎝ 0.5 2 1⎠
1
−2 VA + 3.5 VB = 20
…(ii)
VA
Applying KCL at Node B,
Solving Eqs (i) and (ii),
VA = 11 V
VB = 12 V
Current through the 2 Ω resisto
resistor =
Example 2.60
VB
2
12
2
6A
Find the voltage at nodes 1 and 2 for the network shown in Fig. 2.95.
1Ω
V1
2Ω
1
1A
V2
2
2Ω
1Ω
2A
Fig. 2.95
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V V −V
1= 1 + 1 2
2
2
1
⎛ 1 1⎞
V2 = 1
⎜⎝ + ⎟⎠ V1
2 2
2
V1 0 5 V2 = 1
…(i)
Applying KCL at Node 2,
2=
V2 V2 − V1
+
1
2
1
⎛ 1⎞
− V1 + 1 + V2 = 2
⎝ 2⎠
2
−0
0.5
5V1 + 1 5 V2 = 4
…(ii)
2.5 Node Analysis 2.45
Solving Eqs (i) and (ii),
V1 = 2 V
V2 = 2 V
Example 2.61
Find the current in the 100 W resistor for the network shown in Fig. 2.96.
20 Ω
30 Ω
V1
50 Ω
1A
60 V
V2
100 Ω
40 V
Fig. 2.96
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V1 60 V1 − V2
+
=1
20
30
1⎞
1
60
⎛ 1
V2 =
+1
⎜⎝ + ⎟⎠ V1
20 30
30
20
…(i)
0.083 V1 − 0.033 V2 = 4
Applying KCL at Node 2,
V2 V1 V 2 − 40 V2
+
+
=0
30
50
100
1
1
1 ⎞
40
⎛ 1
− V1
V2 =
⎝
⎠
30
30 50 100
50
−0.0
033 V1 0 063 V2 = 0.8
…(ii)
Solving Eqs (i) and (ii),
V1 = 67.25 V
V2 = 48 V
Current through the 100 Ω resistor =
Example 2.62
V2
48
=
= 0 48 A
100 100
Find VA and VB for the network shown in Fig. 2.97.
1Ω
VA
2A
2Ω
2Ω
1V
Fig. 2.97
VB
2Ω
1A
2V
2.46 Network Analysis and Synthesis
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node A,
2=
⎛1 1 ⎞
⎜⎝ + + 1⎠ VA VB
2 2
2VA VB
VA VA − 1 VA − VB
+
+
2
2
1
1
2+
2
…(i)
2.5
Applying KCL at Node B,
VB VA VB − 2
+
=1
1
2
2
⎛ 1⎞
−V
VA 1
VB = 1 +
⎝ 2⎠
2
−V
VA 1 VB = 2
…(ii)
Solving Eqs (i) and (ii),
VA = 2.875 V
V B = 3 25 V
Example 2.63
Find currents I1, I2 and I3 for the network shown in Fig. 2.98.
10 Ω
V1
2Ω
V2
5Ω
2Ω
50 V
4Ω
I1
25 V
I2
I3
Fig. 2.98
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V1 V1 − 25 V1 V2
+
+
=0
2
5
10
1
25
⎛1 1 1 ⎞
V2 =
⎜⎝ + + ⎟⎠ V1
2 5 10
10
5
0.8 V1 0 1 V2 = 5
…(i)
V2 V1 V2 V2 − ( −50)
+ +
=0
10
4
2
1
50
⎛ 1 1 1⎞
− V1
V2 = −
⎝ 10 4 2 ⎠
10
2
−0 1 V1 + 0 85 V2 = −25
…(ii)
Applying KCL at Node 2,
2.5 Node Analysis 2.47
Solving Eqs (i) and (ii),
V1 = 2 61 V
V2 = −29.1 V
V
2 61
I1 = − 1 = −
= −1 31 A
2
2
V V2 2.61 ( 2.91)
I2 = 1
=
= 3 17 A
10
10
V + 50
5
−29
29.1 50
I3 = 2
=
= 10.45 A
2
2
Example 2.64
Find currents I1, I2 and I3 and voltages Va and Vb for the network shown in Fig. 2.99.
I1
0.2 Ω
I2 0.3 Ω
+ a
Va
120 V
−
I3 0.1 Ω
+ b
Vb
−
30 A
110 V
20 A
Fig. 2.99
Solution Applying KCL at Node a,
I1 30 I 2
120 − Va
V − Vb
= 30 + a
0.2
0.3
36 − 0.3
3 Va = 1.8 + 0.22Va − 0.2 Vb
0.55 Va − 0.2 Vb = 34.2
…(i)
Applying KCL at Node b,
0.1
1 Va
I 2 I 3 = 20
Va Vb 110 − Vb
+
= 20
0.3
0.1
0.1 Vb + 33 − 0 3 Vb
= 20
0 03
0.11 Va 0.4 Vb = −32.4
Solving Eqs (i) and (ii),
Va = 112 V
Vb = 109 V
120 − Va 120 − 112
=
= 40 A
0.2
0.2
V Vb 112 − 109
I2 = a
=
= 10 A
0.3
0.3
110 − Vb 110 − 109
I3 =
=
= 10 A
0.1
0.1
I1 =
…(ii)
2.48 Network Analysis and Synthesis
Example 2.65
Calculate the current through the 5 W resistor for the network shown in Fig. 2.100.
3Ω
V1
4A
5Ω
V2
V3
2Ω
4Ω
2Ω
8A
20 V
Fig. 2.100
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V1 V1 − V2
+
=0
2
3
1
⎛ 1 1⎞
V2 = 4
⎜⎝ + ⎟⎠ V1
2 3
3
4+
0.83
83 V1 0 33 V2 = −4
…(i)
Applying KCL at Node 2,
V1 V2 − ( −20) V2 V3
+
+
=0
3
2
5
1
1
20
⎛ 1 1 1⎞
− V1
V − V =−
⎝ 3 2 5⎠ 2 5 3
3
2
−0 333
3 V1 1 03 V2 − 0 2 V3 = −10
V2
…(ii)
Applying KCL at Node 3,
V3 V2 V3
+
=8
5
4
1
⎛ 1 1⎞
− V2
V3 = 8
⎝ 5 4⎠
5
−0.2 V2 0 45 V3 8
…(iii)
Solving Eqs (i), (ii) and (iii),
V1 = −8 76 V
V2 = −9 92 V
V3 = 13.37 V
Current through the 5 Ω resistor
resisto =
V3 V2
5
13.37 − ( −9.92)
5
4 66 A
2.5 Node Analysis 2.49
Example 2.66
Find the voltage across the 5 W resistor for the network shown in Fig. 2.101.
9A
4Ω
2Ω
V1
5Ω
V2
4Ω
V3
20 Ω
100 Ω
12 V
Fig. 2.101
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V1 12 V1 − V2 V1 − V3
+
+
+9
4
2
4
⎛ 1 1 1⎞
⎜⎝ + + ⎟⎠ V1
4 2 4
0
1
1
12
V2 − V3 = −9 +
2
4
4
V1 0 5 V2 − 0.25 V3
6
…(i)
Applying KCL at Node 2,
V2 V1 V2 V2 − V3
+
+
=0
2
100
5
1
1
⎛1
− V1
⎝ 2 100
2
1⎞
1
V2 − V3 = 0
5⎠
5
−0 5 V1 + 0.7
71 V2 − 0 2
3
=0
…(ii)
Applying KCL at Node 3,
V3 V2 V3 V3 − V1
+
+
=9
5
20
4
1
− V1
4
1
⎛ 1 1 1⎞
V2 + +
+
V3 = 9
⎝ 5 20 4 ⎟⎠
5
−0.
0 25
2 V1 − 0.22 V2
0 5 V3 = 9
Solving Eqs (i), (ii) and (iii),
V1 = 6 35 V
V2 = 11.76 V
V3 = 25.88 V
Voltages across the 5 Ω resistor
esisto V3
2
25.88 11.76 14.12 V
…(iii)
2.50 Network Analysis and Synthesis
Example 2.67
Determine the current through the 5 W resistor for the network shown in Fig. 2.102.
4Ω
36 V
2Ω
V1
4Ω
3A
5Ω
V2
100 Ω
V3
20 Ω
Fig. 2.102
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V1 V1 − V2 V1 − 36 − V3
+
+
=3
4
2
4
1
1
36
⎛ 1 1 1⎞
V2 − V3 = 3 +
⎜⎝ + + ⎟⎠ V1
4 2 4
2
4
4
V1 − 0.5 V2 0.25
2 V3 = 12
…(i)
Applying KCL at Node 2,
V2 V1 V2 V2 − V3
+
+
=0
2
100
5
1
1
1⎞
1
⎛1
− V1
V2 − V3 = 0
⎝
⎠
2
2 100 5
5
−0 5 V1 + 0.771 V2 − 0 2 3 = 0
…(ii)
Applying KCL at Node 3,
V3 V2 V3 V3 − ( −36) − V1
+
+
=0
5
20
4
1
1
⎛ 1 1 1⎞
− V1
V2 + +
+
V3 = −9
⎝ 5 20 4 ⎟⎠
4
5
−0.25
25 V1 0 2 2 0.5 V3
9
…(iii)
Solving Eqs (i), (ii) and (iii),
V1 = 13.41 V
V2 = 7 06 V
V3 = −8 47 V
Current through the 5 Ω resistor
resisto =
V2 V3
5
7.06 ( 8.47)
5
3
A
2.5 Node Analysis 2.51
Example 2.68
Find the voltage drop across the 5 W resistor in the network shown in Fig. 2.103.
4Ω
5Ω
2A
V2
V1
1Ω
1A
V3
2Ω
Fig 2.103
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
1+
V1 V2 V1 V3
+
=0
5
4
⎛ 1 1⎞
⎜⎝ + ⎟⎠ V1
5 4
0.45
45 V1 0 2
1
1
V2 − V3 = −1
5
4
2
0.2
25 V3 = −1
…(i)
Applying KCL at Node 2,
V2 V1 V2
+
=2
5
1
1
⎛1 ⎞
− V1
1 V2
⎝5 ⎠
5
2
−0.2 V1 1 2 V2 = 2
…(ii)
Applying KCL at Node 3,
V3 V3 − V1
+
+2
2
4
1
⎛1
− V1
⎝2
4
−0
0.25
5 V1 0
0
1⎞
V3 = −2
4⎠
V3 = −2 .
Solving Eqs (i), (ii) and (iii),
V1 = −4 V
V2 = 1 V
V3 = −4 V
V5
V2 − V1 = 1 ( 4) = 5 V
…(iii)
2.52 Network Analysis and Synthesis
Example 2.69
Find the power dissipated in the 6 W resistor for the network shown in Fig. 2.104.
V1
3Ω
20 V
1Ω
6Ω
V2
2Ω
V3
5A
5Ω
Fig. 2.104
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V1 20 V1 − V2 V1 − V3
+
+
=0
3
1
2
1⎞
1
20
⎛1
⎜⎝ + 1 + ⎟⎠ V1 V2 − V3 =
3
2
2
3
1 83 V1 V2 − 0.5
5 V3 = 6.67
…(i)
Applying KCL at Node 2,
V2 V1 V2 V3
+
=5
1
6
1
⎛ 1⎞
1
V2 − V3 = 5
⎝ 6⎠
6
1 17 V2 0.17 V3 = 5
…(ii)
V3 V1 V3 V3 − V2
+ +
=0
2
5
6
1
1
⎛ 1 1 1⎞
− V1
V2 +
+ + V3 = 0
⎝ 2 5 6 ⎟⎠
2
6
−0
0.5 V1 − 0.17 V2 0 87
8 V3 = 0
…(iii)
− V1
−V
V1
Applying KCL at Node 3,
Solving Eqs (i), (ii) and (iii),
V1 = 23.82 V
V2 = 27.4 V
V3 = 19.04 V
I6 Ω =
V2 V3 27.4 19.04
=
= 1.39 A
6
6
Power dissipated in the 6 Ω resistor = (1.39)2 × 6 = 11.59 W
Example 2.70
the 10 W resistor zero.
Find the voltage V in the network shown in Fig. 2.105 which makes the current in
2.5 Node Analysis 2.53
3Ω
10 Ω
V1
2Ω
V
7Ω
V2
5Ω
50 V
Fig. 2.105
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V1 V V1 V1 − V2
+ +
=0
3
2
10
1
1
⎛1 1 1 ⎞
V2 − V = 0
⎜⎝ + + ⎟⎠ V1
3 2 10
10
3
0.93 V1 0.1 V2 0 33 V 0
…(i)
Applying KCL at Node 2,
V2 V1 V2 V2 − 50
+ +
=0
10
5
7
1
50
⎛ 1 1 1⎞
− V1
V2 =
⎝ 10 5 7 ⎠
10
7
−0.1 V1 + 0.44 V2 7.14
…(ii)
V1 V2
=0
10
V2 = 0
I10 Ω =
V1
Solving Eqs (i), (ii) and (iii),
Example 2.71
V = 52.82 V
Find V1 and V2 for the network shown in Fig. 2.106.
50 Ω
Va
2A
10 Ω
80 V
20 Ω
Vb
+ V −
1
+
V2
50 Ω
−
Fig. 2.106
Solution
…(iii)
Assume that the currents are moving away from the nodes.
Vc
20 V
2.54 Network Analysis and Synthesis
Applying KCL at Node a,
80 Va − Vb
+
+2=0
50
10
1⎞
1
80
⎛ 1
Vb =
−2
⎜⎝ + ⎟⎠ Va
50 10
10
50
0.12
12 Va 0 1 Vb = −0 4
Va
…(i)
Applying KCL at Node b,
−
1
Va
10
Vb Va Vb Vb − Vc
+
+
=0
10
50
20
1
1⎞
1
⎛ 1
Vb − Vc = 0
⎝ 10 50 20 ⎠
20
−0..1 Va 0..17
17 Vb − 0.05 Vc 0
…(ii)
Node c is directly connected to a voltage source of 20 V. Hence, we can write voltage equation at Node c.
Vc = 20
…(iii)
Solving Eqs (i), (ii), and (iii),
Va = 3 08 V
Vb = 7 69 V
V1 Va Vb = 3.08 − 7.69 = −4 61 V
V2 Vb Vc = 7.69 20
12.31 V
Example 2.72
Find the voltage across the 100 W resistor for the network shown in Fig. 2.107.
50 Ω
VA
20 Ω
VB
20 Ω
12 V
60 V
0.6 A
50 Ω
20 Ω
50 Ω
VC
100 Ω
Fig. 2.107
Solution
Node A is directly connected to a voltage source of 20 V. Hence, we can write voltage equation at Node A.
…(i)
VA = 60
Assume that the currents are moving away from the nodes.
Applying KCL at Node B,
−
1
VA
20
VB VA VB VC VB
+
+
=06
20
20
20
1
1⎞
1
⎛ 1
VB − VC = 0..6
⎝ 20 20 20 ⎠
20
−0
0.0
05
5 VA 0.15
1 VB − 0.05
05 VC 0.6
…(ii)
2.5 Node Analysis 2.55
Applying KCL at Node C,
−
1
VA
50
VC VA VC VB VC 12 VC
+
+
+
=0
50
20
50
100
1
1
1
1 ⎞
12
⎛ 1
VB +
+
+
+
⎟ VC =
⎝ 50 20 50 100
20
0 ⎠
50
−0
0.02
02 VA 0 05
0 VB + 0.1V
1 VC = 0.24
…(iii)
Solving Eqs (i), (ii), and (iii),
VC = 31.68 V
Voltages across the 100 Ω resistor = 31.68 V
EXAMPLES WITH DEPENDENT SOURCES
Example 2.73
Find the voltage across the 5 W resistor in the network shown in Fig. 2.108.
I1
V1
20 Ω
10 Ω
10 Ω
5Ω
2A
−
+
30I1
50 V
Fig. 2.108
Solution From Fig. 2.108,
I1 =
V1 50 V1 − 50
=
20 + 10
30
…(i)
Assume that the currents are moving away from the node.
Applying KCL at Node 1,
2=
V1 V1 + 30 I1 V1 − 50
+
+
5
10
30
V1
5
V
2= 1
5
2=
⎛ V − 50 ⎞
V1 + 30 ⎜ 1
⎝ 30 ⎟⎠ V1 − 50
+
+
10
30
2V1 − 50 V1 − 50
+
+
10
30
Solving Eq. (ii),
V1 = 20 V
Voltage across the 5 Ω resistor = 20 V
…(ii)
2.56 Network Analysis and Synthesis
Example 2.74
For the network shown in Fig. 2.109, find the voltage Vx.
40 Ω
Vx
Iy
100 Ω
0.6 A
+
−
50 Ω
25 Iy
0.2 Vx
Fig. 2.109
Solution From Fig. 2.109,
Iy =
Vx
100
…(i)
Assume that the currents are moving away from the node.
Applying KCL at Node x,
Vx Vx Vx 0 2 Vx
+
+
100 50
40
Vx Vx 0 8 Vx
⎛ Vx ⎞
25 ⎜
+0 6 =
+
+
⎝ 100 ⎟⎠
100 50
40
25 I y + 0 6 =
1
1 0 8⎞
⎛1
−
−
⎜⎝ −
⎟ Vx = −0 6
4 100 50 40 ⎠
0.2
2 Vx = −0.6
Vx = −3 V
Example 2.75
For the network shown in Fig. 2.110, find voltages V1 and V2.
V1
5Ω
0.5 V1
+−
V2
10 Ω
20 Ω
2A
4A
2Ω
20 V
40 V
Fig. 2.110
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V 20 V1 − 0 5 V1 V2
2= 1
+
20
5
1
⎛ 1 1 0 5⎞
V2 = 3
⎜⎝ + −
⎟ V1
20 5 5 ⎠
5
0.15
15 V1 0.2 V2 = 3
…(i)
2.5 Node Analysis 2.57
Applying KCL at Node 2,
0 V1 − V1 V2 V2 − 40
+ +
=4
5
2
10
40
⎛ 0 5 1⎞
⎛1 1 1 ⎞
− ⎟ V1
V2 = 4 +
⎜⎝
⎠
⎝
⎠
5 5
5 2 10
10
−0
0.1V
1 V1 + 0 8 V2 = 8
V2
…(ii)
Solving Eqs (i) and (ii),
V1 = 40 V
V2 = 15 V
Example 2.76
Determine the voltages V1 and V2 in the network of Fig. 2.111.
0.5 Ω
2V
V1
1Ω
0.25 Ω
V2
V1
1Ω
Fig. 2.111
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V1 − 2 V1 V1 V2
+
+
=0
0.5
0.25
1
1
2
⎛ 1
⎞
+
+ 1⎟ V1 V2 =
⎜⎝
⎠
0.5 0.25
05
7 V7 V2 = 4
…(i)
Applying KCL at Node 2,
V2 V1 V 2
+
+ V1 = 0
1
1
2 V2 0
V2 = 0
From Eq. (i),
V1 =
4
V
7
…(ii)
2.58 Network Analysis and Synthesis
Example 2.77
In the network of Fig. 2.112, find the node voltages V1, V2 and V3.
1Ω
V1
1Ω
V2
V3
2V
− Vx +
2A
2Ω
2Ω
2Ω
+
−
4 Vx
Fig. 2.112
Solution From Fig. 2.112,
V2 − V1
Vx
…(i)
Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
⎛1 ⎞
⎜⎝ + 1⎠ V1
2
2=
V1 V1 − V2
+
2
1
2
2
1 5 V1 V2 = 2
…(ii)
Applying KCL at Node 2,
V2 V1 V2 V2 − V3
+ +
=0
1
2
1
−V
V1
⎛ 1 ⎞
1
1 V2 V3
⎝ 2 ⎠
0
−V
V1 2 5 V2 − V3 = 0
…(iii)
At Node 3,
4 Vx
2
4(V2 V1 )
2
V3
V3
4 V1 4 V2 + V3 = 2
Solving Eqs (ii), (iii) and (iv),
V1 = −1 33 V
V2 = −4 V
V3 = −8 67 V
…(iv)
2.5 Node Analysis 2.59
Example 2.78
For the network shown in Fig. 2.113, find the node voltages V1 and V2.
3A
0.5 Ω
V1
3V2
1A
Ix
1Ω
V2
0.25 Ω
2A
2 Ix
Fig. 2.113
Solution From Fig. 2.113,
Ix =
V1 V2
05
…(i)
Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
3 V2 1 =
V1 V1 − V2
+
+3
1
0.5
1 ⎞
1 ⎞
⎛
⎛
V2 = −2
⎜⎝1 +
⎟⎠ V1 ⎝ 3
0.5
5
0.5 ⎠
3 V1 5 V2 = −2
…(ii)
Applying KCL at Node 2,
3 2=
5=
V2 V1 V2
+
+ 2 Ix
0.5
0.25
V2 V1 V2
⎛ V −V ⎞
+
+ 2⎜ 1 2 ⎟
⎝ 05 ⎠
0.5
0.25
2 ⎞
1
2 ⎞
⎛ 1
⎛ 1
+
+
−
⎜⎝ −
⎟⎠ V1 + ⎝
⎟ V2 = 5
0.5 0.5
0.5 0.25 0 5 ⎠
2 V1 + 2 V2 = 5
Solving Eqs (ii) and (iii),
V1 = 1 31 V
V2 = 1.19 V
…(iii)
2.60 Network Analysis and Synthesis
Example 2.79
Find voltages V1 and V2 in the network shown in Fig. 2.114.
V1
V1
I1
2Ω
V2
1Ω
5A
1Ω
2I1
Fig. 2.114
Solution From Fig. 2.114,
V1 V2
2
Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V V −V
5 = 1 + 1 2 + V1
1
2
1
⎛ 1 ⎞
V2 = 5
⎜⎝1 + + 1⎟⎠ V1
2
2
2 5 V1 0 5 V2 = 5
Applying KCL at Node 2,
V2 V1 V2
+
= 2 I1 V1
1
1
⎛V V ⎞
V 2 V1 + V2 = 2 ⎜ 1 2 ⎟ + V1
⎝ 2 ⎠
3 V1 3 V2
V1 V2
I1 =
…(i)
…(ii)
…(iii)
Solving Eqs (ii) and (iii),
V1 = 2 5 V
V2 = 2 5 V
Example 2.80
Find the power supplied by the 10 V source in the network shown in Fig. 2.115.
2Ω
V1
10 V
4Ω
V2
+ V3 −
10 A
2Ω
Fig. 2.115
4V3
1Ω
2.5 Node Analysis 2.61
Solution Assume that the currents are moving away from the nodes.
From Fig. 2.115,
V3 V1 + 10 − V2
…(i)
Applying KCL at Node 1,
10 +
V1 V1 + 10 − V2 V1 − V2
+
+
=0
2
4
2
10
⎛ 1 1 1⎞
⎛ 1 1⎞
⎜⎝ + + ⎟⎠ V1 ⎝ + ⎠ V2 = −10 −
2 4 2
4 2
4
1 25
…(ii)
0 75 V2 = −12 5
Applying KCL at Node 2,
V2 10 V1 V2 V1 V2
+
+
= 4 V3
4
2
1
V2 10 V1 V2 V1 V2
+
+
= 4 (V1 10 V2 )
4
2
1
10
⎛ 1 1
⎞
⎛1 1
⎞
+ 40
⎜⎝ − − − 4⎟⎠ V1 + ⎝ + + 1 + 4⎟⎠ V2 =
4 2
4 2
4
−4.75
75 V1 + 5 75 V2 = 42.5
…(iii)
Solving Eqs (ii) and (iii),
V1 = −11.03 V
V2 = −1 72 V
V 10 V2 −11
11.03 + 10 − ( −1.72)
I10 V = 1
=
= 0.173 A
4
4
Power supplied by the 10 V source = 10 × 0.173 = 1.73 W
Example 2.81
For the network shown in Fig. 2.116, find voltages V1 and V2.
0.03 Vx
V1
20 Ω
+ Vx −
0.4 A
40 Ω
V2
100 Ω
V3
Iy
40 Ω
+
−
80 Iy
Fig. 2.116
Solution From Fig. 2.116,
Vx
V1 − V 2
V
Iy = 2
40
…(i)
…(ii)
2.62 Network Analysis and Synthesis
Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V
V −V
0.4 = 1 + 1 2 + 0.03 Vx
100
20
V1 V1 − V2
0.4 =
+
+ 0.03 ( 1
100
20
1
⎛ 1
⎞
⎛ 1
⎞
+
+ 0.03
03⎟ V1
+ 0 03 2 0.4
⎜⎝
⎠
⎝
⎠
100 20
20
0.09
09 V1 0 08 V2 = 0.4
2)
…(iii)
Applying KCL at Node 2,
V2 V1 V2 V2 − V3
+
+
=0
20
40
40
1
1
1⎞
1
⎛ 1
− V1
V2 − V3 = 0
⎝ 20 40 40 ⎠
20
40
− 0.05 V1 0
0..1 V2 − 0.025 V3 = 0
…(iv)
For Node 3,
V3
2 V2 V3
⎛V ⎞
80 I y = 80 ⎜ 2 ⎟ = 2 V2
⎝ 40 ⎠
…(v)
0
Solving Eqs (iii), (iv) and (v),
V1 = 40 V
V2 = 40 V
V3 = 80 V
Example 2.82
Find voltages Va , Vb and Vc in the network shown in Fig. 2.117.
Va
4A
2V
2Ω
2Ω
I1
Vb
3Ω
1Ω
Vc
2I1
Fig. 2.117
Solution From Fig. 2.117,
Va Vc
2
Assume that the currents are moving away from the nodes.
I1 =
5Ω
2.6 Supernode Analysis 2.63
Applying KCL at Node a,
4=
Va Va − Vc Va − 2 − Vb
+
+
1
2
2
1
1
⎛ 1 1⎞
⎜⎝1 + + ⎟⎠ Va − Vb − Vc = 5
2 2
2
2
2 Va
0.5 Vb − 0 5 Vc = 5
…(i)
Applying KCL at Node b,
Vb
Vb
2 Va Vb Vc
+
= 2 I1
2
3
2 Va Vb Vc
⎛ V Vc ⎞
+
= 2⎜ a
⎝ 2 ⎟⎠
2
3
2 Va Vb Vc
+
= Va Vc
2
3
⎛ 1 ⎞
⎛ 1 1⎞
⎛ 1⎞
Vb + 1 − ⎟ Vc = −1
⎜⎝ − − 1⎟⎠ Va ⎝
⎠
⎝ 3⎠
2
2 3
−1.5 Va + 0 83 Vb + 0.67 Vc = −1
Vb
…(ii)
Applying KCL at Node c,
Vc Vb Vc
+
= I1
3
5
Vc Vb Vc Va − Vc
+
=
3
5
2
1
1
⎛ 1 1 1⎞
− Va − Vb + + + ⎟ Vc = 0
⎝ 3 5 2⎠
2
3
−0.55 Va − 0 33 Vb + 1.033 Vc = 0
…(iii)
Solving Eqs (i), (ii), and (iii),
Va = 4.303 V
V b = 3 88 V
V c = 3 33 V
2.6
SUPERNODE ANALYSIS
Nodes that are connected to each other by voltage sources, but not to the reference node by a path of voltage
sources, form a supernode. A supernode requires one node voltage equation, that is, a KCL equation. The
remaining node voltage equations are KVL equations.
2.64 Network Analysis and Synthesis
Example 2.83
Determine the current in the 5 W resistor for the network shown in Fig. 2.118.
2Ω
V1
V3
5Ω
3Ω
10 A
20 V
V2
1Ω
2Ω
10 V
Fig. 2.118
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
10 =
V1 V1 − V2
+
3
2
1
⎛ 1 1⎞
V2 = 10
⎜⎝ + ⎟⎠ V1
3 2
2
0.83
83 V1 0 5 2 10
…(i)
Nodes 2 and 3 will form a supernode.
Writing voltage equation for the supernode,
V2
V3 = 20
…(ii)
Applying KCL at the supernode,
V2 V1 V2 V3 − 10 V3
+ +
+
=0
2
1
5
2
1
⎛1 ⎞
⎛ 1 1⎞
− V1
1 V2 + + ⎟ V3 = 2
⎝2 ⎠
⎝ 5 2⎠
2
−0
0.5
5 V1 1 5 V2 + 0 7 V3 = 2
…(iii)
Solving Eqs (i), (ii) and (iii),
V1 = 19.04 V
V2 = 11.6 V
V3 = −8 4 V
I5 Ω =
V3 − 10 −8 4 − 10
=
= −3.68 A
5
5
2.6 Supernode Analysis 2.65
Example 2.84
Find the power delivered by the 5 A current source in the network shown in Fig. 2.119.
10 V
V1
V2
3Ω
1Ω
2A
5A
V3
5Ω
2Ω
Fig. 2.119
Solution Assume that the currents are moving away from the nodes.
Nodes 1 and 2 will form a supernode.
Writing voltage equation for the supernode,
V1 V2 = 10
…(i)
Applying KCL at the supernode,
V1 V3 V2 V2 − V3
+ +
=5
3
5
1
1
⎛1 ⎞
⎛1 ⎞
V1 + + 1 V2 − + 1 V3 3
⎝5 ⎠
⎝3 ⎠
3
2+
0 33 V1 +1
1 2 V2 − 1.33 V3
3
…(ii)
V3 V1 V3 V2 V3
+
+ =0
3
1
2
1
1⎞
⎛1
− V1 V2 + + 1 + ⎟ V3 = 0
⎝3
3
2⎠
−0
0.33V
33 V1 V2 1.83
8 V3 = 0
…(iii)
Applying KCL at Node 3,
Solving Eqs (i), (ii) and (iii),
V1 = 13.72 V
V2 = 3 72 V
V3 = 4 51 V
Power delivered by the 5 A source = 5 V2 = 5 × 3.72 = 18.6 W
2.66 Network Analysis and Synthesis
Example 2.85
In the network of Fig. 2.120, find the node voltages V1, V2 and V3.
0.5 Ω
0.33 Ω
V1
V2
5V
V3
0.2 Ω
4A
1Ω
Fig. 2.120
Solution Assume that the currents are moving away from the nodes.
Applying KCL at Node 1,
V V V V
4= 1 2 + 1 3
0.33
0. 5
1 ⎞
1
1
⎛ 1
+
V2 −
V3 = 4
⎜⎝
⎟ V1
0.33 0.5 ⎠
0.33
0.5
5.003
3 V1 3 03 V2 − 2 V3 4
…(i)
Nodes 2 and 3 will form a supernode.
Writing voltage equation for the supernode,
V3 V2 = 5
…(ii)
V2 V1 V2 V3 V3 V1
+
+ +
=0
0.33 0.2 1
05
1
1 ⎞
1 ⎞
1 ⎞
⎛
⎛ 1
⎛
−
+
V3 = 0
⎜⎝ −
⎟ V1 + ⎝
⎟ V2 + ⎝1 +
0.33 0.5 ⎠
0 33 0.2 ⎠
0.5 ⎠
−5
5.03 V1 8.03 V2 + 3 V3 0
…(iii)
Applying KCL at the supernode,
Solving Eqs (i), (ii) and (iii),
V1 = 2 62 V
V2 = −0 17 V
V3 = 4 83 V
EXAMPLES WITH DEPENDENT SOURCES
Example 2.86
For the network shown in Fig. 2.121, determine the voltage Vx.
7Ω
6V
V1
8V
11 Ω
V2
+
3Ω
4Ω
10 Ω
2A
5V
+
2 Vx
−
Vx −
V3
12 A
Fig. 2.121
2.6 Supernode Analysis 2.67
Solution From Fig. 2.121,
V2 − V3
Vx
Assume that the currents are moving away from the nodes.
Node 1 and 2 will form a supernode.
Writing voltage equations for the supernode,
V1 V2 = 6
…(i)
Applying KCL at the supernode,
V1 + 5 V2 − 2Vx V2 8 V3 V2 V3
+
+
+
4
10
7
11
V1 + 5 V2 − 2(V2 V3 ) V2 − 8 − V3 V2 − V3
2=
+
+
+
4
10
7
11
2=
1
⎛ 1
V1
⎝ 10
4
1 1⎞
5 8
⎛1 1 1 ⎞
V2 +
− − ⎟ V3 = 2 − +
⎝ 5 7 11⎠
7 11⎠
4 7
0 25 V1 + 0.133V
133 2 − 0 033 V3 = 1 89
1
5
…(ii)
Applying KCL at Node 3,
V3 V2 V3 8 V2
+
+ 12 = 0
11
7
8
⎛ 1 1⎞
⎛ 1 1⎞
V3 = −12 −
⎜⎝ − − ⎟⎠ V2 ⎝
⎠
11 7
11 7
7
−0.233
33 V2 + 0 233 V3 = −13.14
…(iii)
Solving Eqs (i), (ii) and (iii),
V1 = 1 8 V
V2 = −4.2 V
V3 = −60.6 V
Example 2.87
Find the node voltages in the network shown in Fig. 2.122.
6Ω
V1
6A
5Ω
ix
V2 5ix
10 Ω
−+
20 Ω
V3
15 Ω
2Ω
V4
40 V
Fig. 2.122
Solution From Fig. 2.122,
Ix =
V2 V1
5
…(i)
For Node 4,
V4 = 40
…(ii)
2.68 Network Analysis and Synthesis
Applying KCL at Node 1,
V1 V1 − V2 V1 − V4
+
+
=0
10
5
6
V V − V V − 40
6+ 1 + 1 2 + 1
=0
10
5
6
1
40
⎛ 1 1 1⎞
V2 =
−6
⎜⎝ + + ⎟⎠ V1
10 5 6
5
6
7
1
2
V1
V2 =
15
5
3
Nodes 2 and 3 will form a supernode,
Writing voltage equation for the supernode,
6+
V3 V2 = 5 I x
V1 2V2 V3
…(iii)
⎛V V ⎞
5 ⎜ 2 1 ⎟ = V2 V1
⎝ 5 ⎠
…(iv)
0
Applying KCL to the supernode,
V2 V1 V2 V3 V3 V4
+
+ +
5
20 15
2
V2 V1 V2 V3 V3 − 40
+
+ +
5
20 15
2
1
1
1
1
1
⎛
⎞
⎛
⎞
− V1 + ⎜ + ⎟ V2 +
+ V3
⎝ 5 20 ⎠
⎝ 15 2 ⎟⎠
5
1
1
17
− V1 + V2 + V3
5
4
30
Solving Eqs (iii), (iv) and (v),
V1
V2
=0
=0
= 20
= 20
…(v)
= 10 V
= 20 V
V3 = 30 V
V4 = 40 V
Example 2.88
Find the node voltages in the network shown in Fig. 2.123.
V2
Vx
−
V1
+
14 A
0.5 Ω
2Ω
0.5 Vx
12 V
1Ω
V4
Fig. 2.123
+
− 2.5 Ω
Vy
+
−
0.5 Vy
V3
Exercises 2.69
Solution Selecting the central node as reference node,
V1 = −12 V
…(i)
Applying KCL at Node 2,
−
V2 V1 V2 V3
+
= 14
05
2
1
⎛ 1 1⎞
V2 − V3 = 14
⎝ 0.5 2 ⎠
2
1
V1
0.55
−2
2 V1 2 V2
0 5 V3 = 14
…(ii)
Nodes 3 and 4 will form a supernode,
Writing voltage equation for the supernode,
V3 V4 = 0.2Vy
0 2 (V4 V1 )
0.2V1 V3 − 1.2V4 = 0
…(iii)
Applying KCL to the supernode,
V3 V2
V V −V
− 0 5Vx + 4 + 4 1 = 0
2
1
2.5
V3 V2
V4 V1
− 0.5(
5 (V2 V1 ) + V4 +
=0
2
25
1 ⎞
1
1 ⎞
⎛
⎛1
⎞
⎛
1
V4 = 0
⎜⎝ 0 5 −
⎟ V1 ⎝ + 0 5⎠ V2 + V3 + ⎝1+
2 5⎠
2
2
2 5⎠
0 1 V1 V2 + 0 5 V3 +1
1 4 V3 = 0
…(iv)
Solving Eqs (i), (ii), (iii) and (iv),
V1 = −12 V
V2 = −4 V
V3 = 0
V4 = −2 V
Exercises
KIRCHHOFF’S LAWS
2.1
Find Ix and Vx in the network shown in Fig. 2.124.
4A
10 V
a
2.2
Find V1 and V2 in the network shown in
Fig. 2.125.
2A
b
9V
+
1A
−
5Ω
Vx
−
V2
+
12 V
c
d
3A
18 V
V1
10 V
3V
Fig. 2.125
Ix
[
Fig. 2.124
[
, −15 V ]
,,5 V ]
2.70 Network Analysis and Synthesis
2.3
Find the values of unknown currents in the
network shown in Fig. 2.126.
2.7
In the network shown in Fig. 2.130, find the
voltage between points A and B.
5Ω
5A
7A
I1
10 Ω
1 A I2
A
15 Ω
5V 4Ω
2.4
2 A, I 3
4 A, I 4
[
40 A
Find the current through the 10 Ω resistor in
the network shown in Fig. 2.131.
6Ω
0.1 Ω
2 V 10 Ω
Y
0.1 Ω
4Ω
70 A
20 V
[ .6
. 8 A]
[
]
for the network as shown in
Find I and VAB
Fig. 2.128.
15 Ω
Fig. 2.131
Fig. 2.127
2.5
5Ω
0.05 Ω
5V
50 A
]
MESH ANALYSIS
2.8
0.05 Ω
0.05 Ω
20 V
10 A ]
Find the current in the branch XY as shown in
Fig. 2.127.
160 A X
10 Ω
Fig. 2.130
Fig. 2.126
, I2
6Ω
30 V
I4
6A
8A
[
B
12 A
I3
10 A
4Ω
2.9
Find the current through the 20 Ω resistor in
the network shown in Fig. 2.132.
15 Ω
60 V
A
6A
1A
2Ω
5Ω
I
3Ω
12 V
B
+
− 4Ω
20 Ω
20 V
15 Ω
10 Ω
5Ω
40 V
5Ω
1A
Fig. 2.128
2.6
[ , 19 V ]
In the network shown in Fig. 2.129, find the
voltage between points A and B.
5Ω
3Ω
A
Fig. 2.132
[ .4
. 6 A]
2.10 Find the current through the 10 Ω resistor in
the network shown in Fig. 2.133.
10 Ω
2Ω
2A
5Ω
6Ω
30 V
6V
3Ω
B
30 Ω
2Ω
10 A
B
5Ω
20 Ω
100 V
Fig. 2.129
[
]
Fig. 2.133
[ .3
. 7 A]
Exercises 2.71
2.11 Find the current through the 1 Ω resistor in the
network shown in Fig. 2.134.
10 Ω
i1
20 Ω
1Ω
30 Ω
A
2Ω
3Ω
2Ω
1A
80 V
10 V
+
40 Ω
30 Ω
Fig. 2.134
[ .9
. 5 A]
[ .4 V,
V,
V,
V, 9977.3399 V]
2.15 Find currents I1, I2, and I3 in the network
shown in Fig. 2.138.
6V
2Ω
1Ω
4Ω
2A
3Ω
I1
Fig. 2.135
2Ω
Fig. 2.138
2.13 Find currents Ix and Iy in the network shown
in Fig. 2.136.
[
I3
2Ω
Ix
+−
20 Ω
4Ix
I1
+
− 10IY
2A
I2
25 Ω
Iy
2Ω
5V
, 11 A, 17 A]
2.16 Find currents Ix in the network shown in Fig.
2.139.
2IY
2Ix
1Ω
I3
[ .3
. 3 A]
+−
2Ω
I2
+ Vx −
1
V
9 x
15 A
5Ω
V3
Fig. 2.137
2.12 Find the current through the 4 Ω resistor in the
network shown in Fig. 2.135.
5A
−
2Ω
10 Ω
1.5 Ix
5Ω
5A
Ix
Fig. 2.139
Fig. 2.136
[ .3
. 3 A]
[ .5 A, 0.1 A]
2.14 In the network shown in Fig. 2.137, find V3 if
element A is a
(i) short circuit
(ii) 5 Ω resistor
(iii) 20 V independent voltage source, positive
reference on the right
(iv) dependent voltage source of 1.5 i1, with
positive reference on the right
(v) dependent current source 5 i1, arrow
directed to the right
2.17 Find currents I1 in the network shown in Fig.
2.140.
5 Ω I1
30 V
19 V
2Ω
4A
4Ω
1.5I1
25 V
Fig. 2.140
[−12 A]
2.72 Network Analysis and Synthesis
NODE ANALYSIS
[ .62 A ]
2.18 Find the current Ix in the network shown in
Fig. 2.141.
Ix
5Ω
2.22 Find the current through the branch ab in the
network shown in Fig. 2.145.
20 Ω
10 Ω
1Ω
a
2Ω
24 V
20 Ω
2A
36 V
10 Ω
10 V
2Ω
b
Fig. 2.141
1Ω
[ .93 A ]
2.19 Find VA and VB in the network shown in
Fig. 2.142.
1Ω
VA
[ .038 A ]
2.23 Find the current through the 4 Ω resistor in the
network shown in Fig. 2.146.
2Ω
2Ω
2A
2Ω
VB
Fig. 2.145
4Ω
1A
10 V
2V
1V
2Ω
Fig. 2.142
[ .88 V, 3.25 V ]
2.20 Find the current through the 6 Ω resistor in the
network shown in Fig. 2.143.
2Ω
2Ω
2Ω
2Ω
5A
6V
10 V 2 Ω
Fig. 2.146
10 Ω
4A
5Ω
3A
[ .34 A ]
6Ω
2.24 Find the current through the 4 Ω resistor in the
network shown in Fig. 2.147.
Fig. 2.143
2A
[ . 04 A ]
2.21 Calculate the current through the 10 Ω resistor
in the network shown in Fig. 2.144.
2Ω
2Ω
2V
6V
10 Ω
2Ω
4Ω
25 V
+
−
2Ω
4Ω
3A
−
+
7Ω
3Ω
12 V
Fig. 2.147
[1 A]
2Ω
3Ω
Fig. 2.144
2.25 Find the voltage Vx in the network shown in
Fig. 2.148.
Objective–Type Questions 2.73
1Ω
1Ω
4I1
+
Vx
−
2Ω
2.28 Determine V1 in the network shown in
Fig. 2.151.
50 Ω
I1
20 Ω
8V
4V
Fig. 2.148
5A
[ ..31 V]
2.26 Find the currents Vx in the network shown in
Fig. 2.149.
20 Ω
+
V1
−
+
0.4V1
−
0.01V1
Fig. 2.151
3V1
[140 V]
2.29 Find the voltage Vy in the network shown in
Fig. 2.152.
5Ω
+ Vx −
0.5Vy
4Vx
7A
+−
6Ω
4Ω
15 Ω
+
3Ω
4A
20 Ω Vy
6A
Fig. 2.149
−
[ .0
. 9 V]
2.27 Find the voltage Vx in the network shown in
Fig. 2.150.
1
Ω
3
1Ω
Fig. 2.152
[−10 V]
2.30 Find the voltage V2 in the network shown in
Fig. 2.153.
0.8V2
+−
0.5 Ω
Vx
2Ω
1V
3I
0.5 Ω
5A
5Ω
+
V2
−
8A
2.5 Ω
Fig. 2.150
Fig. 2.153
[ .2
. V]
[
Objective–Type Questions
2.1
Two electrical sub-networks N1 and N2 are
connected through three resistors as shown
in Fig. 2.154. The voltages across the 5 Ω
resistor and 1 Ω resistor are given to be 10 V
and 5 V respectively. Then the voltage across
the 15 Ω resistor is
(a) −105 V
(b) 105 V
(c) −15 V
(d) 15 V
+
5Ω −
10 V
N1
N2
15 Ω
1Ω
+
5V
−
Fig. 2.154
.9 V]
2.74 Network Analysis and Synthesis
2.2
2.3
1Ω
The nodal method of circuit analysis is based
on
(a) KVL and Ohm’s law
(b) KCL and Ohm’s law
(c) KCL and KVL
(d) KCL, KVL and Ohm’s law
The voltage across terminals a and b in
Fig. 2.155 is
2Ω
a
4Ω
(a) 6 A
(c) 12 A
1Ω
2Ω
+ Vx −
b
+
8V
−
(a) 0.5 V
(b) 3 V
(c) 3.5 V
(d) 4 V
The voltage V0 in Fig. 2.156 is
+ Vy −
+
2V
−
+ Vz −
+
1V
−
(a) −6, 3, −3
(c) 6, 3, 3
16 V
12 Ω
10 Ω
2.5 A
none of these
+
2V
−
Fig. 2.159
2Ω
8A
(b)
(d)
The value of Vx,Vy and Vz in Fig. 2.159
shown are
3A
Fig. 2.155
2.4
2Ω V
Fig. 2.158
2.7
+
1V
−
2Ω
IS
6Ω
+
V0
−
2.8
(b)
(d)
−6, −3, 1
6, 1, 3
The circuit shown in Fig. 2.160 is equivalent
to a load of
2Ω
I
Fig. 2.156
2.5
5Ω
V1
5
2.9
Fig. 2.157
(a)
(b)
(c)
(d)
2.6
delivers 80 W
absorbs 80 W
delivers 40 W
absorbs 40 W
If V = 4 in Fig. 2.158, the value of Is is given
by
2I
Fig. 2.160
5Ω
V1 = 20 V
+
−
4Ω
(a) 48 V
(b) 24 V
(c) 36 V
(d) 28 V
The dependent current source shown in
Fig. 2.157.
(a)
4
Ω
3
(b)
8
Ω
3
(b)
4Ω
(d)
2Ω
In the network shown in Fig. 2.161, the effective
resistance faced by the voltage source is
i
4
i
4Ω
V
Fig. 2.161
Answers to Objective-Type Questions 2.75
2.10
(a)
4Ω
(b)
3Ω
(c)
2Ω
(d)
1Ω
A network contains only an independent
current source and resistors. If the values of
all resistors are doubled, the value of the node
voltages will
(a) become half
(b) remain unchanged
(c) become double
(d) none of these
Answers to Objective-Type Questions
2.1. (a)
2.2. (b)
2.3. (c)
2.4. (d)
2.5. (a)
2.6. (d)
2.7. (a)
2.8. (a)
2.9. (d)
2.10. (b)
2.11. (b)
2.12. (b)
2.13. (c)
2.14. (b)
3
3.1
Network Theorems
(Application to dc
Networks)
INTRODUCTION
In Chapter 2, we have studied elementary network theorems like Kirchhoff’s laws, mesh analysis and node
analysis. There are some other methods also to analyse circuits. In this chapter, we will study superposition
theorem, Thevenin’s theorem, Norton’s theorem, maximum power transfer theorem, reciprocity theorem,
Millman’s theorem, Tellegen’s theorem, substitution theorem and compensation theorem. We can find
currents and voltages in various parts of the circuits with these methods.
3.2
SUPERPOSITION THEOREM
It states that ‘in a linear network containing more than one independent source and dependent source,
the resultant current in any element is the algebraic sum of the currents that would be produced by each
independent source acting alone, all the other independent sources being represented meanwhile by their
respective internal resistances.’
The independent voltage sources are represented by their internal resistances if given or simply with zero
resistances, i.e., short circuits if internal resistances are not mentioned. The independent current sources are
represented by infinite resistances, i.e., open circuits.
The dependent sources are not sources but dissipative components—hence they are active at all times. A
dependent source has zero value only when its control voltage or current is zero.
A linear network is one whose parameters are constant, i.e., they do not change with voltage and current.
Explanation
resistor R4.
Consider the network shown in Fig. 3.1. Suppose we have to find current I4 through
R1
V
R3
R2
R4
I
Fig. 3.1 Network to illustrate superposition theorem
3.2 Network Analysis and Synthesis
R1
R3
The current flowing through resistor R4 due to
constant voltage source V is found to be say IÄ4 (with
I4′
proper direction), representing constant current source
R4
R2
V
with infinite resistance, i.e., open circuit.
The current flowing through resistor R4 due to
constant current source I is found to be say I4″ (with
proper direction), representing the constant voltage Fig. 3.2 When voltage source V is acting alone
source with zero resistance or short circuit.
R1
R3
The resultant current I4 through resistor R4 is found
by superposition theorem.
I4′′
I 4 I 4′ + I 4″
R2
I
R4
Steps to be followed in Superposition Theorem
1. Find the current through the resistance when only
one independent source is acting, replacing all Fig. 3.3 When current source I is acting alone
other independent sources by respective internal
resistances.
2. Find the current through the resistance for each of the independent sources.
3. Find the resultant current through the resistance by the superposition theorem considering magnitude and
direction of each current.
Example 3.1
Find the current through the 2 W resistor in Fig. 3.4.
20 V
5Ω
2Ω
10 Ω
40 V
10 V
Fig. 3.4
Solution
Step I
When the 40 V source is acting alone (Fig. 3.5)
I
5Ω
I′
2Ω
10 Ω
40 V
Fig. 3.5
By series parallel reduction technique (Fig. 3.6),
40
I=
=6A
5 + 1.67
From Fig. 3.5, by current-division rule,
10
I′ = 6×
=5A(→)
10 + 2
I
5Ω
1.67 Ω
40 V
Fig. 3.6
3.2 Superposition Theorem 3.3
Step II
When the 20 V source is acting alone (Fig. 3.7)
I
20 V
5Ω
2Ω
I ′′
10 Ω
Fig. 3.7
By series–parallel reduction technique (Fig. 3.8),
I=
5Ω
I
20 V
20
=3A
5 + 1.67
1.67 Ω
From Fig. 3.7, by current-division rule,
I ′′ = 3 ×
Step III
10
= 2 5 A( ← ) = − 2 5 A ( → )
10 + 2
Fig. 3.8
When the 10 V source is acting alone (Fig. 3.9)
5Ω
2Ω
10 Ω
10 V
Fig. 3.9
I ′′′
By series–parallel reduction technique (Fig. 3.10),
I ′′′ =
Step IV
10
= 1 88 A( → )
3 33 + 2
2Ω
10 V
3.33 Ω
I ′′′
By superposition theorem,
I
Example 3.2
I ′ + I ′′ I ′′′ = 5 − 2.5 + 1.88 = 4.38 A ( → )
Fig. 3.10
Find the current through the 1 W resistor in Fig. 3.11.
4Ω
100 V
50 V
40 V
1Ω
2Ω
Fig. 3.11
3.4 Network Analysis and Synthesis
Solution
Step I When the 100 V source is acting alone (Fig. 3.12)
I
4Ω
I′
1Ω
100 V
2Ω
Fig. 3.12
I
By series–parallel reduction technique (Fig. 3.13),
I=
100
= 21.41 A
4 + 0.67
4Ω
0.67 Ω
100 V
From Fig. 3.12, by current-division rule,
I ′ = 21.41 ×
Step II
2
=
2 +1
Fig. 3.13
(↓)
When the 50 V source is acting alone (Fig. 3.14)
4Ω
50 V
2Ω
1Ω
Fig. 3.14
By series–parallel reduction technique (Fig. 3.15),
I ′′ =
Step III
50
=
1 + 1.33
I ′′
.46 A( ↑ )= − 21.46 A( ↓ )
50 V
1.33 Ω
1Ω
When the 40 V source is acting alone (Fig. 3.16)
Fig. 3.15
4Ω
I
I ′′′
1Ω
Fig. 3.16
40 V
2Ω
3.2 Superposition Theorem 3.5
By series–parallel reduction technique (Fig. 3.17),
40
I=
= 14.29 A
0 8+ 2
From Fig. 3.16, by current-division rule,
4
I ′′′ = 14.29 ×
=
4 +1
Step IV By superposition theorem,
I
I
40 V
0.8 Ω
2Ω
(↓ )
Fig. 3.17
I ′ + I ′′ I ′′′ = 14.27 − 21.46 + 11.43 = 4.24 A(↓)
Example 3.3
Find the current through the 8 W resistor in Fig. 3.18.
5Ω
10 Ω
15 Ω
4V
12 Ω
8Ω
6V
Fig. 3.18
Solution
Step I When the 4 V source is acting alone (Fig. 3.19)
I
5Ω
I1
10 Ω
12 Ω
I′
15 Ω
4V
8Ω
Fig. 3.19
By series–parallel reduction technique (Fig. 3.20),
I
5Ω
I1
10 Ω
15 Ω
4V
I
4.8 Ω
5Ω
I1
15 Ω
4V
14.8 Ω
(b)
(a)
I=
I
4
= 0.32 A
5 + 7.45
From Fig. 3.20(b), by current-division rule,
15
I1 = 0 32 ×
= 0.16 A
15 + 14.8
From Fig. 3.19, by current-division rule,
12
I ′ = 0.16 ×
=
(↓)
12 + 8
5Ω
7.45 Ω
4V
(c)
Fig. 3.20
3.6 Network Analysis and Synthesis
When the 6 V source is acting alone (Fig. 3.21)
Step II
5Ω
10 Ω
12 Ω
15 Ω
8Ω
6V
Fig. 3.21
By series–parallel reduction technique (Fig. 3.22),
10 Ω
12 Ω
12 Ω
I
I ′′
3.75 Ω
8Ω
13.75 Ω
6V
8Ω
(a)
(b)
I=
12 Ω
6
= 0.35 A
12 + 5 06
From Fig. 3.22(b), by current division rule,
5.06 Ω
13.75
I ′′ = 0 35 ×
=
13.75 + 8
By superposition theorem,
Step III
I
Example 3.4
I ′ + I ′′ = 0.096 + 0.22 =
(c)
(↓)
Fig. 3.22
4Ω
5Ω
40 V
8A
3Ω
Fig. 3.23
Solution
When the 40 V source is acting alone (Fig. 3.24)
I
40 V
12 Ω
I′
5Ω
Fig. 3.24
4Ω
3Ω
I
6V
(↓)
Find the current through the 4 W resistor in Fig. 3.23.
12 Ω
Step I
6V
3.2 Superposition Theorem 3.7
By series–parallel reduction technique (Fig. 3.25),
I
12 Ω
I′
5Ω
40 V
12 Ω
I
7Ω
2.92 Ω
40 V
(b)
(a)
Fig. 3.25
I=
40
= 2.68 A
12 + 2 92
From Fig. 3.25(a), by current-division rule,
I ′ = 2.68 ×
5
= 1 12 A( → )
5+ 7
− 1 12 A ( ← )
−1
Step II When the 8 A source is acting alone (Fig. 3.26)
12 Ω
4Ω
5Ω
8A
3Ω
Fig. 3.26
By series–parallel reduction technique (Fig. 3.27),
4Ω
I ′′
I ′′
3.53 Ω
3Ω
8A
7.53 Ω
(a)
(b)
Fig. 3.27
I″ = 8×
Step III
3
= 2 28 A( ← )
7 53 + 3
By superposition theorem,
I
3Ω
I ′ + I ′′ = −1.12 + 2.28 = 1.16 A ( ← )
8A
3.8 Network Analysis and Synthesis
Example 3.5
Find the current in the 10 W resistor in Fig. 3.28.
2Ω
1Ω
10 Ω
5Ω
4A
10 V
Fig. 3.28
Solution
Step I
When the 10 V source is acting alone (Fig. 3.29)
2Ω
1Ω
5Ω
10 Ω
10 V
Fig. 3.29
By series–parallel reduction technique (Fig. 3.30),
I
I
I ′′
1Ω
1Ω
7Ω
10 Ω
10 V
4.12 Ω
10 V
(a)
(b)
Fig. 3.30
I=
10
= 1.95 A
1 + 4.12
From Fig. 3.30 (a), by current-division rule,
I ′ = 1.95 ×
7
=
7 + 10
(↓ )
3.2 Superposition Theorem 3.9
Step II
When the 4 A source is acting alone (Fig. 3.31)
2Ω
I
I ′′
1Ω
10 Ω
5Ω
4A
Fig. 3.31
By series–parallel reduction technique (Fig. 3.32),
2Ω
I
I
0.91 Ω
4A
5Ω
2.91 Ω
4A
(a)
(b)
Fig. 3.32
I = 4×
5
= 2.53 A
2 91 + 5
From Fig. 3.31, by current-division rule,
I ′′ = 2.53 ×
Step III
1
−
1 + 10
(↓)
By superposition theorem,
I
Example 3.6
I ′ + I ′′ = 0.8 + 0.23 =
(↓)
Find the current through the 8 W resistor in Fig. 3.33.
8Ω
5A
12 Ω
Fig. 3.33
30 Ω
25 A
5Ω
3.10 Network Analysis and Synthesis
Solution
Step I
8Ω
I′
When the 5 A source is acting alone (Fig. 3.34)
I′ = 5×
12
= 1.2
12 + 8 + 30
(→)
30 Ω
12 Ω
5A
When the 25 A source is acting alone (Fig. 3.35)
Step II
I ′′ = 25 ×
Fig. 3.34
30
= 15 A (→)
30 + 12 + 8
I ′′
8Ω
By superposition theorem,
Step III
I
12 Ω
I ′ + I ′′ = 1.2 +15
+ 15 = 16 2 A(→)
30 Ω
Fig. 3.35
Example 3.7
Find the current through the 4 W resistor in Fig. 3.36.
2Ω
5A
6Ω
5Ω
4Ω
6Ω
20 V
Fig. 3.36
Solution
Step I
When the 5 A source is acting alone (Fig. 3.37)
2Ω
5A
6Ω
5Ω
4Ω
6Ω
Fig. 3.37
By series–parallel reduction technique (Fig. 3.38),
25 A
3.2 Superposition Theorem 3.11
2Ω
2Ω
I′
6Ω
4Ω
5A
8.73 Ω
5A
2.73 Ω
(a)
4Ω
(b)
Fig. 3.38
I′ = 5×
Step II
8 73
=
8 73 + 4
(↓ )
When the 20 V source is acting alone (Fig. 3.39)
2Ω
6Ω
5Ω
I
I ′′
4Ω
6Ω
20 V
Fig. 3.39
By series–parallel reduction technique (Fig. 3.40),
I
5Ω
I ′′
I
6Ω
20 V
10 Ω
3.75 Ω
20 V
(b)
(a)
Fig. 3.40
I=
20
= 2.29 A
5 + 3.75
From Fig. 3.40 (a), by current-division rule,
I ′′ = 2.29 ×
Step III
6
=
6 + 10
(↓ )
By superposition theorem
I
5Ω
I ′ + I ′′ = 3.43
3 + 0.86 =
(↓ )
3.12 Network Analysis and Synthesis
Example 3.8
Find the current through the 3 W resistor in Fig. 3.41.
2Ω
5Ω
3Ω
10 Ω
4Ω
5A
20 V
Fig. 3.41
Solution
Step I
When the 5 A source is acting alone (Fig. 3.42)
2Ω
5Ω
3Ω
10 Ω
4Ω
5A
Fig. 3.42
I′
By series–parallel reduction technique (Fig. 3.43),
I′ = 5×
Step II
15
=
15 + 2 + 3
(↓ )
2Ω
3Ω
15 Ω
5A
When the 20 V source is acting alone (Fig. 3.44)
2Ω
Fig. 3.43
3Ω
5Ω
I
I ′′
10 Ω
4Ω
20 V
Fig. 3.44
By series–parallel reduction technique (Fig. 3.45),
I ′′
I
20 Ω
4Ω
I
20 V
(a)
3.33 Ω
20 V
(b)
Fig. 3.45
3.2 Superposition Theorem 3.13
I=
20
=6A
3 33
From Fig. 3.45(a), by current-division rule,
I ′′ = 6 ×
4
=
20 + 4
(↑) = −1 A(↓)
By superposition theorem,
Step III
I
Example 3.9
I ′ + I ′′ = 3.75
75 − 1 =
(↓)
Find the current in the 1 W resistors in Fig. 3.46.
1A
2Ω
3Ω
1Ω
4V
3A
Fig. 3.46
2Ω
Solution
Step I
3Ω
I′
When the 4 V source is acting alone (Fig. 3.47)
1Ω
4V
4
I′ =
=
2 +1
(↓)
Fig. 3.47
Step II When the 3 A source is acting alone (Fig. 3.48)
By current-division rule,
2
I ′′ = 3 ×
=
1+ 2
Step III
2Ω
(↓ )
3Ω
I ′′
1Ω
When the 1 A source is acting alone (Fig. 3.49)
1A
2Ω
Fig. 3.48
3Ω
1Ω
Fig. 3.49
3A
3.14 Network Analysis and Synthesis
Redrawing the network (Fig. 3.50),
By current-division rule,
I ′′′
3Ω
2
I ′′′ = 1 ×
=
2 +1
2Ω
(↓)
1Ω
1A
Step IV By superposition theorem,
I
I ′ + I ′′ I ′′′ = 1.33 + 2 + 0.66 = 4 A (↓)
Example 3.10
Fig. 3.50
Find the voltage VAB in Fig. 3.51.
+
A
5Ω
6V
VAB
10 V
5A
−
B
Fig. 3.51
Solution
Step I When the 6 V source is acting alone (Fig. 3.52)
VABB′ = 6 V
+
A
5Ω
6V
VAB′
−
B
+
Fig. 3.52
A
5Ω
VAB′′
Step II When the 10 V source is acting alone (Fig. 3.53)
Since the resistor of 5 Ω is shorted, the voltage across it is zero.
10 V
VAB
″ = 10 V
−
B
Fig. 3.53
Step III When the 5 A source is acting alone (Fig. 3.54)
Due to short circuit in both the parts,
+
VAB
″′ = 0
5Ω
VAB′′′
Step IV By superposition theorem,
5A
VAB
VAB′ + VAB
″
A
−
VAB
″′ = 6 + 10 + 0 = 16 V
Fig. 3.54
B
3.2 Superposition Theorem 3.15
Example 3.11
Find the current through the 5 W resistor in Fig. 3.55.
5Ω
10 Ω
20 Ω
2A
24 V
36 V
Fig. 3.55
Solution
Step I
When the 24 V source is acting alone (Fig. 3.56)
5Ω
10 Ω
20 Ω
24 V
Fig. 3.56
I′
5Ω
By series–parallel reduction technique (Fig. 3.57),
I′ =
24
= 2.06
5 + 6.67
(→) = −2.06 A(←)
6.67 Ω
24 V
Step II When the 2 A source is acting alone
By series–parallel reduction technique (Fig. 3.58),
5Ω
Fig. 3.57
5Ω
10 Ω
I ′′
20 Ω
2A
6.67 Ω
2A
(a)
(b)
Fig. 3.58
By current-division rule,
I″ = 2×
6 67
= 1.14 (←)
5 + 6.67
Step III When the 36 V source is acting alone (Fig. 3.59)
By series–parallel reduction technique,
5Ω
10 Ω
I ′′′
20 Ω
10 Ω
I
36 V
(a)
4Ω
36 V
(b)
Fig. 3.59
I
3.16 Network Analysis and Synthesis
I=
36
= 2.57 A
10 + 4
By current-division rule,
I ′′′ = 2.57 ×
20
= 2 06 A(←)
20 + 5
Step IV By superposition theorem,
I ′ + I ′′ I ′′′ = −2.06 + 1.14 + 2.06
06 = 1 14 A (←)
I
Example 13.12
Find the current through the 4 W resistor in Fig. 3.60.
6V
2Ω
5A
4Ω
2A
Fig. 3.60
I′
Solution
I′ = 5×
2
=
2+4
2Ω
5A
Step I When the 5 A source is acting alone (Fig. 3.61)
By current-division rule,
(↓ )
Fig. 3.61
I ′′
Step II When the 2 A source is acting alone (Fig. 3.62)
By current-division rule,
I″ = 2×
2
=
2+4
4Ω
2Ω
2A
4Ω
(↓ )
Step III When the 6 V source is acting alone (Fig. 3.63)
Applying KVL to the mesh,
Fig. 3.62
−2l ′′′ − 6 − 4l ′′′ = 0
6V
I ′′′ = −1 A (↓)
2Ω
Step IV By superposition theorem,
4Ω
I ′′′
I
I ′ + I ′′ I ′′′ = 1.67 + 0.67 − 1 = 1.34
(↓ )
Fig. 3.63
3.2 Superposition Theorem 3.17
EXAMPLES WITH DEPENDENT SOURCES
Example 3.13
Find the current through the 6 W resistor in Fig. 3.64.
6Ω
I
8Ω
3I
15 V
10 V
Fig. 3.64
Solution
Step I When the 15 V source is acting alone (Fig 3.65)
From Fig. 3.65,
I ′ = I1
I′
…(i)
4
1
I1 = 3I ′
2
8Ω
3I ′
15 V
I1
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I2
6Ω
I2
Fig. 3.65
3I1
0
…(ii)
Applying KVL to the outer path of the supermesh,
15 − 6
1
8I 2 = 0
6
1
88II 2 = 15
…(iii)
Solving Eqs (ii) and (iii),
I1 = 0.39A
I 2 = 1.59A
I ′ = I1 = 0.39
(→)
Step II When the 10 V source is acting alone (Fig 3.66)
From Fig. 3.66,
I ′′ = I1
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I 2 I1 = 3I ′′ 3I1
41 2 0
Applying KVL to the outer path of the supermesh,
−6 1 − 88I 2 + 10 = 0
6 1 88II 2 = 10
I ′′ 6 Ω
8Ω
...(i)
3I ′′
I1
10 V
I2
…(ii)
Fig. 3.66
…..(iii)
3.18 Network Analysis and Synthesis
Solving Eqs (ii) and (iii),
I1 = 0 26 A
Step III
I 2 = 1 05 A
I ′′ I1 = 0.26
By superposition theorem,
(→)
I = I ′ + I ″ = 0.39 + 0.26 = 0.65
Example 3.14
(→)
Find the current Ix in Fig. 3.67.
Ix 5 Ω
1Ω
+
−
30 A
20 V
4Ix
Fig. 3.67
Solution
Step I When the 30 A source is acting alone (Fig. 3.68)
From Fig. 3.68,
I ′x
I1
Ix′ 5 Ω
…(i)
30 A
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I1
1Ω
I1
I 2 = 30
I2
Fig. 3.68
−5 1 − 1I 2 − 4 ′x = 0
−5 1 − 2 − 4 I1 = 0
91 2 0
Solving Eqs (ii) and (iii),
…(iii)
I1 = 3 A
I 2 = −27 A
I ′x I1 = 3 A (→)
Step II When the 20 V source is acting alone (Fig. 3.69)
Applying KVL to the mesh,
Step III
4Ix′
…(ii)
Applying KVL to the outer path of the supermesh,
20 − 5 x″
+
−
1I x″ − 4 ″x 0
1I
I x″ = 2 A( → )
Ix′′ 5 Ω
1Ω
+
−
20 V
By superposition theorem,
Ix
I x′ + I ″x
3+ 2
5 Α(→)
Fig. 3.69
4Ix′′
3.2 Superposition Theorem 3.19
Example 3.15
Find the current I1. in Fig. 3.70.
Vx
+ −
4Vx
I1
10 Ω
2A
2Ω
5V
Fig. 3.70
Solution
Step I When the 5 V source is acting alone (Fig 3.71)
From the figure,
Vx = 5 – 10I1′
Applying KVL to the mesh,
Vx
2Ω
5V
5 – 10I1′ – 4 (5 – 10I1′) – 2I1′ = 0
Fig. 3.71
5 – 10I1′ – 20 + 40I1′ – 2I1′ = 0
15
= 0.
0 54
28
(↑ )
Vx
Step II When the 2 A source is acting alone (Fig 3.72)
From Fig . 3.72,
10 I1′
Vx
I1′ = 2
+−
4Vx
I1′
…(i)
10 Ω
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I2
4Vx
10 Ω
5 – 10I1′ – 4Vx – 2I1′ = 0
I1′ =
+ −
I1′
2A
I1′
…(ii)
2Ω
I2
Fig. 3.72
Applying KVL to the outer path of the supermesh,
−10 I1′ − 4 x − 2 I 2 = 0
−10 I1′ − 4( −10
10 1′ ) − 2 I 2 = 0
30 I1′
2
2
0
…(iii)
Solving Eqs (ii) and (iii),
I1 = 0.14 (↑)
I 2 = 2.14 A
Step III
By superposition theorem,
I1
I1′ + I1″ = 0.54 0.14 = 0.68
(↑ )
3.20 Network Analysis and Synthesis
Example 3.16
Determine the current through the 10 W resistor in Fig. 3.73.
10 Ω
− +
10Vx
+
5Ω
100 V
Vx
10 A
−
Fig. 3.73
Solution
Step I When the 100 V source is acting alone (Fig. 3.74)
From the figure,
Vx = 5I′
Applying KVL to the mesh,
10 Ω
− +
I′
+
5Ω
100 V
100 – 10I′ + 10Vx – 5I′ = 0
100 – 10I′ + 10(5I′) – 5I′ = 0
10Vx
Vx
−
I ′ = −2.86
(→)
Fig. 3.74
Step II When the 10 A source is acting alone (Fig. 3.75)
From Fig. 3.75,
Vx
( I1 − I 2 )
…(i)
Applying KVL to Mesh 1,
−10 I1 +10
10Vx − 5( 1 − 2 ) = 0
−10 I1 +10
10
0{5( 1 − 2 )} − 5( I1 − I 2 ) = 0
35 I1 45 I 2 = 0
10 Ω
I ′′
− +
10Vx
+
5Ω
Vx
I1
10 A
I2
−
…(ii)
Fig. 3.75
For Mesh 2,
I 2 = −10
…(iii)
Solving Eqs (ii) and (iii),
I1 = −12.86 A
I 2 = −10 A
I ′′ = I1 = −12.86
(→)
Step III By superposition theorem,
I I ′ + I ′′ = −2.86 − 12
12.86 = −15.72 (→)
Example 3.17
Find the current I in the network of Fig. 3.76.
3Ω
17 V
I
2Ω
+
Vx
−
4Ω
1A
Fig. 3.76
+
−
5Vx
3.2 Superposition Theorem 3.21
Solution
Step I When the 17 V source is acting alone (Fig. 3.77)
From the figure,
Vx = –2I′
Applying KVL to the mesh,
–2I′ – 17 – 3I′ – 5Vx = 0
–2I′ – 17 – 3I′ – 5(–2I′) = 0
I′ =3.4 A (→)
Step II When the 1 A source is acting alone (Fig. 3.78)
From Fig. 3.78,
Vx
2 I1
…(i)
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I 2 I1 = 1
…(ii)
Applying KVL to the outer path of the supermesh,
I′
2Ω
+
Vx
−
2Ω
4Ω
I1
1A
I2
Fig. 3.78
Find the voltage V1 in Fig. 3.79.
4I
I
4Ω
+
+
−
V1
5A
20 V
−
Fig. 3.79
Solution
Step I When the 5 A source is acting alone (Fig. 3.80)
From Fig. 3.80,
I=
V1′
4
Applying KCL at Node 1,
V1′ − 4 I V1′
+ =5
1
4
+
−
5Vx
I ′′
+
Vx
−
…(iii)
1Ω
5Vx
3Ω
I ′′ I 2 = 1.6 (→)
By superposition theorem,
I = I′ + I′′ = 3.4 + 1.6 = 5 A (→)
Example 3.18
+
−
Fig. 3.77
−2 1 − 3I 2 − 5Vx = 0
−2 1 − 3I 2 − 5( −2 I1 ) = 0
81 3
3II 2 = 0
Solving Eqs (ii) and (iii),
I1 = 0 6 A
I2 = 1 6 A
Step III
3Ω
17 V
1Ω
4I
V1′
+
−
I
5A
Fig. 3.80
4Ω
3.22 Network Analysis and Synthesis
⎛ V ′⎞ V ′
V1′− 4 ⎜ 1 ⎟ + 1 = 5
⎝ 4⎠ 4
V1′ = 20 V
1Ω
Step II When the 20 V source is acting alone (Fig. 3.81)
Applying KVL to the mesh,
4I – I – 4I – 20 = 0
I = –20 A
V1′′ = 4I – 1(I) = 3I = 3 (–20) = –60 V
4I
4Ω
I
+
−
20 V
Step III By superposition theorem,
V1 = V1′ + V1′′ = 20 – 60 = –40 V
Example 3.19
V1′′
Fig. 3.81
Find the current in the 6 W resistor in Fig. 3.82.
1Ω
− +
2Vx
I
− Vx +
6Ω
3A
18 V
Fig. 3.82
1Ω
Solution
Step I When the 18 V source is acting alone (Fig. 3.83)
From Fig. 3.83,
Vx = –I′
Applying KVL to the mesh,
18 – I′ + 2Vx – 6I′ = 0
18 – I′ – 2I′ – 6I′ = 0
I′ = 2 A (↓)
Step II When the 3 A source is acting alone (Fig. 3.84)
From Fig. 3.84,
Vx
1 I1 = − I1
…(i)
I1 = 3
I′
6Ω
18 V
Fig. 3.83
1Ω
− +
− Vx +
I1
…(ii)
2Vx
− Vx +
2Vx
I ′′
6Ω
3A
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I2
− +
I2
Fig. 3.84
Applying KVL to the outerpath of the supermesh,
−1I1 + 2Vx − 6 2 = 0
− I1 + 2( − 1 ) − 6 I 2 = 0
3 1 66II 2 = 0
…(iii)
3.2 Superposition Theorem 3.23
Solving Eqs (ii) and (iii),
I1 = −2 A
I2 = 1 A
I ′′ = I 2 = 1 A (↓)
Step III
By superposition theorem,
I6 Ω = I′ + I′′ = 2 + 1 = 3 A (↓)
Example 3.20
Find the current Iy in Fig. 3.85.
Iy
4Ω
+ −
120 V
10Iy 8 Ω
12 A
40 V
Fig. 3.85
I y′
Solution
Step I When the 120 V source is acting alone (Fig. 3.86)
Applying KVL to the mesh,
4Ω
+ −
10I y′
8Ω
120 V
120 – 4Iy′ – 10Iy′ – 8Iy′ = 0
Iy′ = 5.45 A (→)
Fig. 3.86
Step II When the 12 A source is acting alone (Fig. 3.87)
From Fig. 3.87,
I y ″ I1
…(i)
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I 2 I1 = 12
Applying KVL to the outer path of the supermesh,
−4 1 − 1
10
0 y ″ − 8I 2 = 0
′′
10Iy′′
+ −
Iy
4Ω
8Ω
12 A
I1
I2
...(ii)
Fig. 3.87
−4 1 − 10
10 1 − 8 I 2 = 0
14 I1 8 2 0
…(iii)
Solving Eqs (ii) and (iii),
I1 = −4 36 A
I 2 = 7 64 A
Iy ″ = I1 = −4.36 A ( → )
Step III When the 40 V source is acting alone (Fig. 3.88)
Applying KVL to the mesh,
–4 Iy′′′ – 10Iy′′′ – 8Iy′′′ – 40 = 0
I y ″′ = −
40
= –1.82 A (→)
22
Iy′′′
10Iy′′′
+ −
4Ω
8Ω
40 V
Fig. 3.88
3.24 Network Analysis and Synthesis
Step IV By superposition theorem,
Iy = Iy′ + Iy′′ + Iy′′′ = 5.45 – 4.36 – 1.82 = –0.73 A (→)
Example 3.21
Find the voltage Vx in Fig. 3.89.
3Ω
+ Vx
6Ω
−
+ −
3Vx
24 Ω
18 V
36 V
5A
Fig. 3.89
Solution
Step I When the 18 V source is acting alone (Fig. 3.90)
From Fig. 3.90,
Vx′ = 3I
Applying KVL to the mesh,
18 – 3I – 6I – 3Vx′ = 0
18 – 3I – 6I – 3 (3I) = 0
I=1A
Vx′ = 3 V
3Ω
+ Vx
6Ω
+ −
3Vx′
−
18 V
Step II When the 5 A source is acting alone (Fig. 3.91)
From Fig. 3.91,
Vx′′ = –3I1
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I 2 I1 = 5
Applying KVL to the outer path of the supermesh,
I
Fig. 3.90
3Ω
+ Vx
…(i)
6Ω
−
24 Ω
5A
I1
+
−
I2
...(ii)
Fig. 3.91
−3 1 − 6 I 2 − 3V″x = 0
−3 1 − 6 I 2 − 3(3I1 ) = 0
12 I1 6 2 0
Solving Eqs (ii) and (iii),
3Vx ′
…(iii)
I1 = −1 67A
I 2 = 3 33 A
Vx″
3I1 = 3( −1.67) = −5 V
Step III When the 36 V source is acting alone (Fig. 3.92)
From the figure,
Vx′′′ = –3I
Applying KVL to the mesh,
3Ω
6Ω
+ −
3Vx′′′
+ Vx′′′ −
36 V
I
36 + 3Vx′′′ – 6I – 3I = 0
Fig. 3.92
3.2 Superposition Theorem 3.25
⎛ −V ′′′⎞
⎛ −V ′′′⎞
36 + 3Vx′′′ 6 ⎜ x ⎟ − 3 ⎜ x ⎟ = 0
⎝ 3 ⎠
⎝ 3 ⎠
36 + 3Vx′′′ 2Vx′′′ + Vx″′ = 0
Vx′′′ = −6 V
Step IV By superposition theorem,
Vx = Vx′ + Vx′′ + Vx′′′ = 3 – 5 – 6 = –8 V
Example 3.22
Find the voltage V in the network of Fig. 3.93.
8Ω
− V
15 Ω
5Ω
+
+
−
12 Ω
−5 A
10 V
8V
Fig. 3.93
Solution
Step I When the 10 V source is acting alone
(Fig. 3.94)
From Fig. 3.94,
V′
8 I1
...(i)
Applying KVL to Mesh 1,
−10 − 8 1 − 15
15
1
− 12 (
1
−
2)
8Ω
15 Ω
5Ω
− V′ +
+
−
12 Ω
I2
10 V
I1
8V′
=0
35 I1 12 I 2 = −10
...(ii)
Fig. 3.94
Applying KVL to Mesh 2,
−12( 2 − 1 ) − 5 I 2 − 8V ′ = 0
−12 I 2 +12
12 I1 − 5 2 − 8( −8
8 1) = 0
76 I1 17 I 2 = 0
...(iii)
Solving Eqs (ii) and (iii),
I1 = 0 54 A
I 2 = 2.4 A
V′
8 I1
8(0.54)
Step II When the –5 A source is acting alone
(Fig. 3.95)
From Fig. 3.95,
V″
8 I1
...(i)
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I1
I 2 = −5
8Ω
15 Ω
5Ω
− V ′′ +
I1
...(ii)
4.32 V
−5 A
I2
Fig. 3.95
12 Ω
I3
+
−
8 V ′′
3.26 Network Analysis and Synthesis
Applying KVL to the outer path of the supermesh,
−8 1 − 15
15 2 − 12( 2 − 3 ) = 0
−8 1 − 27
27 2 + 12 3 = 0
Applying KVL to Mesh 3,
...(iii)
−12( 3 − 2 ) − 5 I 3 − 8V ″ = 0
−12 I 3 +12
12 I 2 − 5 3 − 8( −8
8 1) = 0
64 I1 12 I 2 − 17 I 3 = 0
...(iv)
Solving Eqs (ii), (iii) and (iv),
I1 = 4 97 A
I 2 = 9 97 A
I 3 = 25.74 A
V″
8 I1
8( −4
4.97)
Step III
39.76 V
By superposition theorem,
V V ′ + V ″ = −4.32 − 39
39.76 = −44.08 V
Example 3.23
For the network shown in Fig. 3.96, find the voltage V0.
50 Ω
200 Ω
+
40 Ω
+
1A
V0
−
V1
25 V
+
−
0.5V1
−
Fig. 3.96
Solution
Step I When the 1 A source is acting alone (Fig. 3.97)
From Fig. 3.97,
V1 200 I 2
...(i)
For Mesh 1,
I1 = 1
...(ii)
Applying KVL to Mesh 2,
0.5 1 40 ( I 2 − I1 ) 200
00 I 2 = 0
0.5 ( 200 I 2 ) 40 I 2 + 40 I1 200 I 2 = 0
40 I1 140 I 2 = 0
Solving Eqs (ii) and (iii),
I1 = 1 A
I 2 = 0 29 A
V0 ′ 50 I1 − 200 I 2 = 0
V0 ′ − 50(1) − 200(0.29) 0
V0 ′ = 108 V
50 Ω
200 Ω
+
1A
+
V0′
−
40 Ω
V1
I1
+
−
0.5V1
I2
−
Fig. 3.97
...(iii)
3.2 Superposition Theorem 3.27
50 Ω
Step II When the 25 V source is acting alone (Fig. 3.98)
From Fig. 3.98,
V1 200 I − 25 = 0
200 I + 25
V1
...(i)
200 Ω
+
40 Ω
V0′′
25 V
V1
+
−
Applying KVL to Mesh 1,
0.5V1
I
−
0 5V1 40 I − 200 I − 25 = 0
0.5( 200 I 25)
25) 40 I − 200 I − 25 = 0
Fig. 3.98
I = −0 09 A
V0″ V1 = 200 I + 25 = 200(
200 −0.09) 25
7V
By superposition theorem,
Step III
V0 = V0′ + V0″ = 108 + 7 = 115 V
Example 3.24
For the network shown in Fig. 3.99, find the voltage Vx.
10 A
2Ω
4Ω
+
Vx
20 V
Vx
2
6Ω
−
Fig. 3.99
2Ω
Solution
Step I When the 20 V source is acting alone (Fig.
3.100)
From Fig. 3.100,
Vx′
6( I1 − I 2 )
...(i)
Applying KVL to Mesh 1,
20 − 2
6( I1 − I 2 ) = 0
81 6
6II 2 = 20
1
4Ω
+
20 V
Vx′
6Ω
Vx′
I1
2
I2
−
Fig. 3.100
...(ii)
For Mesh 2,
Vx′
6( I1 I 2 )
=
= 3I
3I1 3I 2
2
2
4II 2 = 0
4
I2 =
3
1
Solving Eqs (ii) and (iii),
I1 = 5 71 A
I 2 = 4.29 A
Vx′
6( I1
I2 )
6(5
(5.71 4.29) = 8 52 V
...(iii)
3.28 Network Analysis and Synthesis
Step II When the 10 A source is acting alone (Fig. 3.101)
From Fig. 3.101,
Vx″ = 6(II1 I 2 )
Applying KVL to Mesh 1,
−2( 1 − 3 ) − 6( 1 − 2 ) = 0
8 1 6I2 − 2 3 0
For Mesh 2,
V ″ 6( I1 I 2 )
I2 = x =
= 3I
3I1 3I 2
2
2
31 4
4II 2 = 0
For Mesh 3,
I 3 = −10
Solving Eqs (ii), (iii) and (iv),
I1 = −5 71 A
I 2 = −4.29 A
I 3 = −10 A
Vx ″
Step III
10 A
...(i)
2Ω
I3
4Ω
+
...(ii)
Vx′′
2
6Ω
Vx′′
I1
I2
−
Fig. 3.101
...(iii)
...(iv)
6( I1 − I 2 ) = 6( 5.71 + 4.29)
8.52 V
By superposition theorem,
Vx
Example 3.25
Vx ′ + Vx ″ = 8.52 − 8.52 = 0
Calculate the current I in the network shown in Fig. 3.102.
4Ω
2I1
20 Ω
70 V
− +
2Ω
50 V
I
10 Ω
Fig. 3.102
Solution
Step I When the 70 V source is acting alone (Fig. 3.103)
From Fig. 3.103,
I ′ I3
...(i)
Applying KVL to Mesh 1,
−4
1
− 2 I1 − 20 ( I1 − I 2 ) = 0
26 I1 20 I 2 = 0
Applying KVL to Mesh 2,
70 − 20( 2 1 ) − 2( 2 3 ) 0
−20 I1 + 22 I 2 − 2 3 = 70
...(ii)
4Ω
I1
20 Ω
2I1
I1
− +
2Ω
70 V
I2
I3
10 Ω
...(iii)
Fig. 3.103
I′
3.2 Superposition Theorem 3.29
Applying KVL to Mesh 3,
−2(
3
−
2
2 ) + 2 I1
−10
10 I 3 = 0
2 I 2 − 12 I 3 = 0
1
...(iv)
Solving Eqs (ii), (iii) and (iv),
I1 = 8 94 A
I 2 = 11.62 A
I 3 = 3 43 A
I′
I 3 = 3.43 (←)
Step II When the 50 V source is acting alone (Fig. 3.104)
From Fig. 3.104,
4Ω
I1
I ′′ = I 3 …(i)
20 Ω
2I1
I1
− +
Applying KVL to Mesh 1,
2Ω
−4 1 − 2 I1 − 20( I1 − I 2 ) = 0
26 I1 20
0 I 2 = 0 …(ii)
I2
50 V
I3
10 Ω
Applying KVL to Mesh 2,
−20(
2
− 1 ) − 2(
2
−
−20 I1 + 22 I 2 − 2
3)
=0
3
=0
I′′
Fig. 3.104
…(iii)
Applying KVL to Mesh 3,
−2(
3
−
2 ) + 2 I1
2
1
+ 50 − 10
0I3 = 0
2 I 2 − 12 I 3 = −50
…(iv)
Solving Eqs (ii), (iii), and (iv),
I1 = 1 06 A
I 2 = 1 38 A
I 3 = 4 57 A
I ′′
Step III
I 3 = 4.57
(←)
By superposition theorem,
I
Example 3.26
I ′ + I ′′ = 3.43 + 4.57 = 8 A (←)
Find the voltage V0 in the network of Fig. 3.105.
4V
5Ω
+
10 V
1A
V0
−
Fig. 3.105
2Ω
1Ω
+
−
V0
2
3.30 Network Analysis and Synthesis
Solution
5Ω
Step I When the 10 V source is acting alone
(Fig. 3.106)
Applying KCL at the node,
1Ω
+
V0′
10 V
V′
V0′ − 0
V0′ − 10 V0′
2 =0
+ +
5
2
1
⎛ 1 1 1⎞
⎜⎝ + + ⎟⎠ V0′ = 2
5 2 2
V0′ = 1.667 V
Step II When the 1A current source is acting
alone (Fig. 3.107)
Applying KCL at the node,
−
Fig. 3.106
5Ω
1Ω
+
1A
V ′′
V ′′− 0
V0′′
V0′′ 0 2
+ + +
5
2
1
4V
5Ω
+
V0″′
2
1Ω
+ V0″′
2
−
−
Fig. 3.108
By superposition theorem,
V0 = V0′ V0′′ V0′′′ = 1.67 0.83 + 3.33
3.3
+ V0″
− 2
2Ω
Fig. 3.107
V ′′′
V ′′′− 4 − 0
V0′′′ V0′′′ 0
2 =0
+ +
5
2
1
⎛ 1 1 1⎞
⎜⎝ + + ⎟⎠ V0′′′ = 4
5 2 2
V0′′′ = 3.33
33 V
Step IV
V0″
−
⎛ 1 1 1⎞
⎜⎝ + + ⎟⎠ V0′′= −1
5 2 2
V0′′= −0.83 V
Step III When the 4 V source is acting alone
(Fig. 3.108)
Applying KCL at the node,
+ V0′
− 2
2Ω
4.17 V
THEVENIN’S THEOREM
It states that ‘any two terminals of a network can be replaced by an equivalent voltage source and an
equivalent series resistance. The voltage source is the voltage across the two terminals with load, if any,
removed. The series resistance is the resistance of the network measured between two terminals with load
removed and constant voltage source being replaced by its internal resistance (or if it is not given with
zero resistance, i.e., short circuit) and constant current source replaced by infinite resistance, i.e., open
circuit.’
3.3 Thevenin’s Theorem 3.31
RTh
IL
Network
RL
IL
VTh
RL
(a)
(b)
Fig. 3.109 Network illustrating Thevenin’s theorem
Explanation
Consider a simple network as shown in Fig. 3.110.
R1
R3
A
R2
V
RL
B
Fig. 3.110
Network
R1
For finding load current through RL, first remove the load
resistor RL from the network and calculate open circuit voltage
VTh across points A and B as shown in Fig. 3.111.
VTh =
R2
R1
R2
R3 +
+
−
Fig. 3.111 Calculation of VTh
R1
R3
A
R1 R2
R1 R2
R2
R Th
Thevenin’s equivalent network is shown in Fig. 3.113.
IL =
VTh
RTh
RL
B
Fig. 3.112 Calculation of RTh
RTh
A
If the network contains both independent and dependent
sources, Thevenin’s resistance RTh is calculated as,
RTh
V
= Th
IN
where I N is the short-circuit current which would flow in a
short circuit placed across the terminals A and B. Dependent
sources are active at all times. They have zero values only when
the control voltage or current is zero. RTh may be negative in
VTh
A
VTh
R2
V
V
For finding series resistance RTh , replace the voltage source
by a short circuit and calculate resistance between points A and
B as shown in Fig. 3.112.
RTh
R3
RL
IL
B
Fig 3.113 Thevenin’s equivalent
network
B
3.32 Network Analysis and Synthesis
some cases which indicates negative resistance region of the device, i.e., as voltage increases, current
decreases in the region and vice-versa.
If the network contains only dependent sources then
VTh = 0
IN = 0
For finding RTh in such a network, a known voltage V is applied across the terminals A and B and current
is calculated through the path AB.
V
RTh =
I
RTh
or a known current source I is connected across the
A
terminals A and B and voltage is calculated across the
terminals A and B.
V
RTh =
I
B
Thevenin’s equivalent network for such a network is
shown in Fig. 3.114.
Fig. 3.114 Thevenin’s equivalent network
Steps to be Followed in Thevenin’s Theorem
1.
2.
3.
4.
5.
Remove the load resistance RL.
Find the open circuit voltage VTh across points A and B.
Find the resistance RTh as seen from points A and B.
Replace the network by a voltage source VTh in series with resistance RTh.
Find the current through RL using Ohm’s law.
IL =
Example 3.27
VTh
RTh
RL
Find the current through the 2 W resistor in Fig. 3.115.
5Ω
40 V
20 V
2Ω
10 Ω
10 V
Fig. 3.115
Solution
20 V
5Ω
Step I Calculation of VTh (Fig. 3.116)
Applying KVL to the mesh,
40 − 5 20 10
0
15 I = 20
I = 1.33 A
+
−
A
+
+
10 Ω
40 V
I
−
Fig. 3.116
B
VTh
−
10 V
3.3 Thevenin’s Theorem 3.33
Writing the VTh equation,
5Ω
A
B
RTh
10 I VTh + 10 = 0
VTh
Step II
10 I + 10 = 10(1
(1.33) 10
10 Ω
23.33 V
Calculation of RTh (Fig. 3.117)
Fig. 3.117
RTh = 5 10 = 3.33 Ω
3.33 Ω
A
Step III
Calculation of IL (Fig. 3.118)
2Ω
23.33 V
IL =
IL
23.33
= 4.38 A
3 33 + 2
B
Fig. 3.118
Example 3.28
Find the current through the 8 W resistor in Fig. 3.119.
5Ω
10 Ω
5Ω
250 V
8Ω
75 V
Fig. 3.119
5Ω
I
+
Solution
−
+
5Ω
250 V
VTh
−
250
= 25 A
5+5
−
75 V
Writing the VTh equation,
250 − 5
VTh
75
Fig. 3.120
5Ω
0
VTh = 175
1 − 5 I = 175 − 5( 25)
Step II
Calculation of RTh (Fig. 3.121)
RTh = (
A
+
Step I Calculation of VTh (Fig. 3.120)
I=
10 Ω
10 Ω
A
50 V
5Ω
) + 10 = 12.5 Ω
RTh
B
Fig. 3.121
B
3.34 Network Analysis and Synthesis
12.5 Ω
Calculation of IL (Fig. 3.122)
Step III
IL =
A
8Ω
50 V
50
= 2.44 A
12.5 + 8
IL
B
Fig. 3.122
Example 3.29
Find the current through the 2 W resistor connected between terminals A and B in
the Fig. 3.123.
2Ω
1Ω
3Ω
A
12 Ω
2V
2Ω
4V
2Ω
1Ω
B
Fig. 3.123
Solution
Step I Calculation of VTh (Fig. 3.124)
Applying KVL to Mesh 1,
+
2 2 I1 12( I1 − I 2 ) = 0
14 I1 12 I 2 = 2 ...(i)
−
+ −
+
−
12 Ω
2V
− +
I1
3Ω
A
+
+
−
V
I2
B
4V
−
Applying KVL to Mesh 2,
−12(
2
− 1 ) − 1 2 − 3I 2 − 4 = 0
−12 I1 +16
16 I 2 = −4
Fig. 3.124
...(ii)
Solving Eqs (i) and (ii),
I 2 = −0.4 A
2Ω
1Ω
A
Writing the VTh equation,
VTh
Step II
B
Fig. 3.125
Calculation of RTh (Fig. 3.125)
Calculation of IL (Fig. 3.126)
IL =
28
= 0.82 A
1.43 + 2
RTh
12 Ω
3I 2 − 4 = 0
VTh 4 3I 2 = 4 + 3( −0.4) = 2.8 V
1.43 Ω
RTh = [( 2 12)) + ] 3 = 1. 3 Ω
Step III
3Ω
A
2Ω
2.8 V
IL
B
Fig. 3.126
3.3 Thevenin’s Theorem 3.35
Example 3.30
Find the current through the 10 W resistor in Fig. 3.127.
6Ω
2Ω
1Ω
10 V
3Ω
10 Ω
20 V
Fig. 3.127
6Ω
Solution
Step I Calculation of VTh (Fig. 3.128)
Applying KVL to Mesh 1,
10 − 6
1
1( I1 − I 2 ) = 0
7
1
2
10
+
2Ω
−
+
−
+ −
I1
3Ω
I2
− +
2
VTh
−
−
− 1 ) − 2I 2 − 3
− I1 + 6
2
2
=0
=0
Fig. 3.128
…(ii)
6Ω
2Ω
A
Solving Eqs (i) and (ii),
I 2 = 0.24 A
…(ii)
1Ω
Writing the VTh equation,
3
2
VTh
20
VTh
Step II
Step III
3Ω
R Th
0
B
3I 2 20 3(0
(0.24) 20
19.28 V
=19.28 V (terminal B is positive w.r.t A)
Fig. 3.129
1.47 Ω
Calculation of RTh (Fig. 3.129)
RTh = [(6 1) + ] 3 = . 7 Ω
A
IL
Calculation of IL (Fig. 3.130)
IL =
10 Ω
19.28 V
19.28
= .68 ( ↑ )
1.47 + 10
B
Fig. 3.130
Example 3.31
B
20 V
Applying KVL to Mesh 2,
−1(
A
+
1Ω
10 V
…..(i)
+
Find the current through the 10 W resistor in Fig. 3.131.
10 Ω
10 A
5Ω
Fig. 3.131
30 Ω
20 Ω
100 V
3.36 Network Analysis and Synthesis
Solution
A
Step I Calculation of VTh (Fig. 3.132)
For Mesh 1,
1
+
+
5Ω
10 A
= 10
Th
I1
30 Ω
B
−
+
20 Ω
100 V
−
I2
−
Applying KVL to Mesh 2,
100 − 30 I 2
Fig. 3.132
20 I 2 = 0
R Th
I2 = 2 A
A
30 Ω
B
Writing the VTh equation,
5
1
20
VTh
Step II
5Ω
0
2
5 I1 20
2
20 Ω
5(10) − 20( 2) 10 V
Fig. 3.133
Calculation of RTh (Fig. 3.133)
17 Ω
RTh = 5 + ( 20 30) = 17 Ω
Step III
A
Calculation of IL (Fig. 3.134)
10 Ω
10 V
IL
10
= 0.37 A
17 + 10
IL =
B
Fig. 3.134
Example 3.32
Find the current through the 40 W resistor in Fig. 3.135.
50 Ω
25 V
20 Ω
10 Ω
40 Ω
30 Ω
10 V
Fig. 3.135
Solution
Step I Calculation of VTh (Fig.
3.136)
Since the 20 Ω resistor is connected
across the 25 V source, the resistor
becomes redundant.
50 Ω
+
25 V
20 Ω
10 Ω
−
+
+ A
VTh
I − B
V20 Ω = 25 V
Fig. 3.136
−
10 V
30 Ω
3.3 Thevenin’s Theorem 3.37
Applying KVL to the mesh,
25 − 50 I − 10 I + 10 = 0
I = 0.58 A
Writing the VTh equation,
VTh − 10I + 10 = 0
VTh
Step II
10( I ) − 10 = 10( .58)
.5 )
.2 V
= 4.2 V(terminal B is positive w . r .t )
Calculation of RTh (Fig. 3.137)
50 Ω
10 Ω
A
RTh
B
20 Ω
30 Ω
A
RTh
B
50 Ω
(a)
10 Ω
(b)
Fig. 3.137
8.33 Ω
A
RTh = 50 10 = 8 33 Ω
IL
40 Ω
4.2 V
Step III
Calculation of IL (Fig. 3.138)
IL =
Example 3.33
B
4.2
= 0.09 (↑)
8 33 + 40
Fig. 3.138
Find the current through the 10 W resistor in Fig. 3.139.
2V
6Ω
10 Ω
5Ω
15 Ω
4Ω
50 V
20 V
Fig. 3.139
Solution
Step I Calculation of VTh (Fig.
3.140)
50
=5A
6 4
20
I2 =
=1A
5 15
I1 =
6Ω
+
2V
−
A
+
+
V Th
5Ω
B
−
−
4Ω
50 V
I1
+
+
15 Ω
−
−
Fig. 3.140
20 V
I2
3.38 Network Analysis and Synthesis
Writing the VTh equation,
4 I1 2 V
h
− 15I
15 I 2 = 0
VTh 4 I1 2 − 15I
15 I 2 = 4(5) + 2 − 15(1) = 7 V
Calculation of RTh (Fig. 3.141)
Step II
6Ω
RTh = (
)+(
A
) = 6.15 Ω
R Th
5Ω
B
4Ω
15 Ω
Fig. 3.141
Step III
Calculation of IL (Fig. 3.142)
6.15 Ω
A
IL =
7
= 0.43 A
6 15 + 10
10 Ω
7V
IL
B
Fig. 3.142
Example 3.34
Determine the current through the 24 W resistor in Fig. 3.143.
30 Ω
20 Ω
220 V
24 Ω
50 Ω
5Ω
Fig. 3.143
Solution
Step I Calculation of VTh (Fig. 3.144)
220
I2 =
=88A
20 + 5
Writing the VTh equation,
VTh
30 I1 − 20 I 2 = 0
VTh 20 I 2 − 30 I1 = 20(8.8) − 30( 2.75) = 93 5 V
+
30 Ω +
220
I1 =
= 2 75 A
30 + 50
I1 −
+
220 V
A
50 Ω
Fig. 3.144
V Th
−
20 Ω
−
I2
B
5Ω
3.3 Thevenin’s Theorem 3.39
Step II
Calculation of RTh (Fig. 3.145)
30 Ω
20 Ω
A
R Th
B
50 Ω
5Ω
Fig. 3.145
Redrawing the circuit (Fig. 3.146),
RTh = (
)+(
30 Ω
) = 2 .7 5 Ω
20 Ω
A
B
50 Ω
5Ω
Fig. 3.146
Step III
Calculation of IL (Fig. 3.147)
22.75 Ω
A
IL =
24 Ω
93.5 V
93.5
=2A
22.75 + 24
IL
B
Fig. 3.147
Example 3.35
Find the current through the 3 W resistor in Fig. 3.148.
4Ω
5Ω
3Ω
8Ω
1Ω
2Ω
50 V
Fig. 3.148
3.40 Network Analysis and Synthesis
Solution
Step I Calculation of VTh (Fig. 3.149)
Applying KVL to Mesh 1,
50 − 2 1 1( I1 − I 2 ) 8( 1 2 )
11I1 9 2
4Ω
+
0
50
…(i)
+
Applying KVL to Mesh 2,
−4
2
− 5 I 2 − 8(
2
−
− 1 ) − 1( I 2 − I1 ) = 0
−9 1 + 1
18
8 2 =0
2Ω
...(ii)
−
I2
Step II
+
−
+
I1
50 V
Writing the VTh equation,
5 I 2 − 8(I
(I
−
B −
− 1 Ω −+
I1 = 7 69 A
I 2 = 3 85 A
I )=0
VTh 5 I 2 + 8(I
( I I ) = 5(( .85) ((33.85 − 7.69) = −11.47
4 V
= 11.47 V(the terminal B is positive w . r .t . )
Calculation of RTh (Fig. 3.150)
4Ω
5Ω
A
R Th
B
8Ω
1Ω
2Ω
Fig. 3.150
Redrawing the network (Fig. 3.151),
A
4Ω
5Ω
2Ω
1Ω
8Ω
B
Fig. 3.151
5Ω
A +
V Th
Solving Eqs (i) and (ii),
VTh
+
Fig. 3.149
+
−
8Ω
3.3 Thevenin’s Theorem 3.41
Converting the upper delta into equivalent star network (Fig. 3.152),
R1 =
4 2
= 0 73 Ω
4+2+5
R2 =
4 5
= 1 82 Ω
4+2+5
R3 =
5 2
= 0 91 Ω
4+2+5
A
A
1.82 Ω
R2
R1
1Ω
R3
0.73 Ω
8Ω
1Ω
0.91 Ω
8Ω
B
B
Fig. 3.152
Fig. 3.153
Simplifying the network (Fig. 3.153),
RTh = 1.82 + (1.73
73 8.91) = 3.27 Ω
A
1.82 Ω
1.73 Ω
8.91 Ω
B
Fig. 3.154
Step III
Calculation of IL (Fig. 3.155)
IL =
3.27 Ω
11.47
= .83 ( ↑ )
3 27 + 3
A
3Ω
11.47 V
IL
B
Fig. 3.155
3.42 Network Analysis and Synthesis
Example 3.36
Find the current through the 20 W resistor in Fig. 3.156.
120 V
45 V
20 Ω
10 Ω
15 Ω
5Ω
5Ω
20 V
Fig. 3.156
120 V
Solution
Step I Calculation of VTh (Fig. 3.157)
Applying KVL to Mesh 1,
45 − 120 − 15 I1 5(
2)
1
45 V
10
0( 1 2 ) 0
30 I1 15 I 2 = −75
...(i)
Applying KVL to Mesh 2,
20 − 5
2
10(
2
1)
5( I 2 − I1 ) = 0
I1
−
+
10 Ω
+
− B
−
+
−
−
Solving Eqs (i) and (ii),
−
+
5Ω
Writing the VTh equation,
45 − VTh − 10( I1 − I 2 ) = 0
Step II
I 2 ) = 45 −10
10[ −33.2 − ( −11.4)] = 63 V
Calculation of RTh (Fig. 3.158)
A
R Th
15 Ω
B
10 Ω
5Ω
5Ω
Fig. 3.158
5Ω
+
Fig. 3.157
I1 = −3 2 A
I 2 = −1.4 A
VTh = 45 − 10( I1
15 Ω
I2
...(ii)
−15
15 I1 + 20 I 2 = 20
+ A
V Th
20 V
+
−
3.3 Thevenin’s Theorem 3.43
Converting the delta formed by resistors of
10 Ω, 5 Ω and 5 Ω into an equivalent star
network (Fig. 3.159),
R1 =
10 × 5
= 2 5Ω
20
R2 =
10 × 5
= 2 5Ω
20
R3 =
5 5
= 1.25 Ω
20
Simplifying the network (Fig. 3.160),
A
RTh
B
A
B
R1
R2
2.5 Ω
15 Ω
R3
15 Ω
1.25 Ω
2.5 Ω
Fig. 3.159
Fig. 3.160
RTh = ( .25 | | 2.. ) + 2.5 = 4.67 Ω
A
RTh
B
2.5 Ω
16.25 Ω
2.5 Ω
Fig. 3.161
Step III
Calculation of IL
IL =
4.67 Ω
63
= 2.55 A
4 67 + 20
A
20 Ω
63 V
IL
B
Fig. 3.162
Example 3.37
Find the current through the 3 W resistor in Fig. 3.163.
12 Ω
3Ω
6Ω
6A
42 V
Fig. 3.163
Solution
Step I Calculation of VTh (Fig. 3.164)
For Mesh 1,
I1 = 6
Applying KVL to Mesh 2,
42 −12
12(
2
1) − 6 2
+
+ −
+
12 Ω
…(i)
− +
6A
I1
6Ω
42 V
I2
V Th
−
−
0
−12
12 I1 + 18 I 2 = 42
A
...(ii)
Fig. 3.164
B
3.44 Network Analysis and Synthesis
Solving Eqs (i) and (ii),
A
I 2 = 6.33 A
6Ω
12 Ω
R Th
Writing the VTh equation,
VTh
Step II
6 I 2 = 38 V
B
Calculation of RTh (Fig. 3.165)
Fig. 3.165
4Ω
RTh = 6 12 = 4 Ω
Step III
A
Calculation of IL (Fig. 3.166)
IL =
3Ω
38 V
IL
38
= 5.43 A
4+3
B
Fig. 3.166
Example 3.38
Find the current through the 30 W resistor in Fig. 3.167.
15 Ω
60 Ω
30 Ω
40 Ω
13 A
150 V
50 V
Fig. 3.167
Solution
Step I Calculation of VTh (Fig. 3.168)
Meshes 1 and 2 form a supermesh.
Writing current equation for supermesh,
I 2 I1 = 13
...(i)
Writing voltage equation for supermesh,
15 Ω
+
60 Ω
−
+
A
−
+
I1
V Th
B
−
40 Ω
13 A
150 V
+
I2
50 V
−
150 −15
15 I1 60 I 2 − 40 I 2 = 0
Fig. 3.168
15 I1 +100
100 I 2 = 150 ...(ii)
Solving Eqs (i) and (ii),
I1 = −10 A
I2 = 3 A
15 Ω
60 Ω
Writing the VTh equation,
40 I 2 VTh − 50 = 0
VTh 40 I 2 − 50 = 40(3) 50
Step II
40 Ω
70 V
Calculation of RTh (Fig. 3.169)
RTh = 75 40 = 26.09 Ω
A
Fig. 3.169
R Th
B
3.3 Thevenin’s Theorem 3.45
Step III
26.09 Ω
Calculation of IL (Fig. 3.170)
A
70
= 1.25 A
26.09 + 30
IL =
30 Ω
70 V
IL
B
Fig. 3.170
Example 3.39
Find the current through the 20 W resistor in Fig. 3.171.
10 Ω
5Ω
5A
20 Ω
100 V
Fig. 3.171
Solution
Step I Calculation of VTh (Fig. 3.172)
10 Ω
VTh = 100 V
+ A
V Th
− B
5Ω
5A
100 V
Fig. 3.172
Step II
Calculation of RTh (Fig.3.173)
10 Ω
A
R Th
B
5Ω
A
R Th
B
5Ω
(a)
(b)
Fig. 3.173
RTh = 0
Step III
Calculation of I L (Fig. 3.174)
IL =
100
= 5A
20
A
20 Ω
100 V
IL
B
Fig. 3.174
3.46 Network Analysis and Synthesis
Example 3.40
Find the current through the 20 W resistor in Fig. 3.175.
10 Ω
20 Ω
5Ω
4Ω
10 V
8Ω
2A
Fig. 3.175
Solution
Step I
10 Ω
Calculation of VTh (Fig. 3.176)
+
10
= 0.71 A
10 + 4
I2 = 2 A
A
−
I1 =
+
+
V Th
5Ω
B
+
−
4Ω
10 V
I1
−
+
8Ω
−
−
2A
I2
Writing the VTh equation,
4
1
Step II
8I 2 = 0
VTh = 4(0
(0.71) + 8( 2) = 18.84 V
Fig. 3.176
Calculation of RTh (Fig. 3.177).
10 Ω
R Th
A
4Ω
R Th
5Ω
B
A
8Ω
B
2.86 Ω
(a)
8Ω
(b)
Fig. 3.177
RTh = 10.86 Ω
Step III
Calculation of I L (Fig. 3.178)
10.86 Ω
A
IL =
18.84
= 0.61 A
10.86 + 20
20 Ω
18.84 V
IL
B
Fig. 3.178
3.3 Thevenin’s Theorem 3.47
Example 3.41
Find the current through the 5 W resistor in Fig. 3.179.
10 Ω
2Ω
50 V
50 V
5Ω
100 V
2Ω
3Ω
Fig. 3.179
Solution
Step I Calculation of VTh (Fig. 3.180)
Applying KVL to Mesh 1,
100 −10
10 I1 + 500 2 1 2( I1 I 2 ) = 0
14 I 2 2 I 2 = 150
Applying KVL to Mesh 2,
−2( 2 − 1 ) + 50 − 3I 2 = 0
…(ii)
−2 1 + 5 I 2 = 50
Solving Eqs (i) and (ii),
I1 = 12.88 A
I 2 = 15.15 A
...(i)
10 Ω
+
2Ω
50 V
−
+
50 V
−
A +
100 V
I1
2Ω
V Th
B −
Writing the VTh equation,
+
+ −
− +
3Ω
I2
Fig. 3.180
100 −10
10 I1 − VTh = 0
VTh = 100 − 10(12.88) = −28.8 V
= 28.8 V( terminal B is positiv
s e w.r.t. A)
Step II
Calculation of RTh (Fig. 3.181)
10 Ω
10 Ω
2Ω
A
A
2Ω
R Th
3Ω
R Th
B
B
(b)
(a)
10 Ω
RTh = 10 3.2 = 2.42 Ω
2Ω
3.2 Ω
A
R Th
B
(c)
Fig. 3.181
1.2 Ω
−
3.48 Network Analysis and Synthesis
Step III
2.42 Ω
Calculation of I L (Fig. 3.182)
IL =
28.8
= 3.88
2.42 + 5
A
(↑ )
IL
28.8 V
5Ω
B
Fig. 3.182
Example 3.42
Find the current through the 10 W resister in Fig. 3.183.
10 Ω
2Ω
2Ω
1Ω
15 V
1Ω
10 V
1Ω
Fig. 3.183
A
Solution
Step I Calculation of VTh (Fig. 3.183)
Applying KVL to Mesh 1,
+
2
= −25
+
...(i)
2
2Ω
−
2
0
− I1 + 4
2
10
...(ii)
+
−
1Ω
− +
I1
1 ) 2I 2 1
+ −
15 V
Applying KVL to Mesh 2,
10 −1
1(
−
2Ω
−15 − 2 1 − 1( I1 − I 2 ) − 100 − 1 1 = 0
4 1−
B
VTh
−
10 V
+
1Ω
Fig. 3.184
Solving Eqs (i) and (ii),
I1 = −6 A
I 2 = 1A
Writing the VTh equation,
−VTh + 2 I 2 + 2 I1 = 0
VTh = 2 I1 + 2 I 2 = 2( −6) + 2(1) = −10 V
= 10 V( the terminal B is positive w.r.t. A)
+
1Ω
I2
−
3.3 Thevenin’s Theorem 3.49
Step II
Calculation of RTh (Fig. 3.185)
2Ω
A
RTh
B
2Ω
1Ω
1Ω
1Ω
Fig. 3.185
Converting the star network formed by resistors of 2 Ω, 2 Ω and 1 Ω into an equivalent delta network
(Fig. 3.186),
A
2Ω
2 2
R1 = 2 + 2 +
=8Ω
1
2 1
R2 = 2 + 1 +
=4Ω
2
2 1
R3 = 2 + 1 +
=4Ω
2
RTh
B
2Ω
1Ω
1Ω
Fig. 3.186
Simplifying the network (Fig. 3.187),
RTh
A
RTh
A
B
B
8Ω
8Ω
4Ω
4Ω
0.8 Ω
0.8 Ω
(b)
RTh
A
1Ω
(a)
1Ω
B
1.33 Ω
(c)
Fig. 3.187
RTh = 1 33 Ω
1Ω
3.50 Network Analysis and Synthesis
Step III
Calculation of I L (Fig. 3.188)
1.33 Ω
A
IL =
10
= 0.88
1 33 + 10
(↑ )
IL
10 V
10 Ω
B
Fig. 3.188
Example 3.43
Find the current through the 1 W resistor in Fig. 3.189.
1A
2Ω
3Ω
4V
1Ω
3A
Fig. 3.189
Solution
Step I Calculation of VTh (Fig. 3.190)
1A
−
+
2Ω
+ I2
−
A +
4V
−
+
3Ω
+
−
V Th
I1
3A
B −
2Ω
Fig. 3.190
A
Writing the current equations for Meshes 1 and 2,
Writing the VTh equation,
4 2(
1
VTh
Step II
Step III
R Th
I1 = −3
I2 = 1
2 ) − VTh
=0
= 4 2(
3Ω
B
Fig. 3.191
1
−
Calculation of RTh (Fig. 3.191)
RTh = 2 Ω
Calculation of I L (Fig. 3.192)
12
IL =
= 4A
2 +1
2)
2Ω
= 4 2( −4) 12 V
A
1Ω
12 V
IL
B
Fig. 3.192
3.3 Thevenin’s Theorem 3.51
Example 3.44
Find the current through the 3 W resistor in Fig. 3.193.
2Ω
1Ω
2Ω
10 V
3Ω
10 A
Fig. 3.193
Solution
Step I Calculation of VTh (Fig. 3.194)
2Ω
1Ω
+
2Ω
10 V
10 A
A
V Th
−
B
Fig. 3.194
By source transformation (Fig. 3.195),
2Ω
+
1Ω
−
2Ω
−
10 V
I
A
+
+
20 V
Fig. 3.195
V Th
−
B
2Ω
1Ω
A
Applying KVL to the mesh,
10 − 2
2
20
4
0
2Ω
10
R Th
I = −2.5 A
B
Writing the VTh equation,
10 − 2
Step II
VTh 0
VTh = 10 − 2 I
2Ω
10 2( −2
2.5) = 15 V
A
Calculation of RTh (Fig. 3.196)
RTh = (
Step III
Fig. 3.196
) +1 = 1+1 = 2 Ω
Calculation of IL (Fig. 3.197)
IL =
15
= 3A
2+3
3Ω
15 V
IL
B
Fig. 3.197
3.52 Network Analysis and Synthesis
EXAMPLES WITH DEPENDENT SOURCES
Example 3.45
Obtain the Thevenin equivalent network for the given network of Fig. 3.198 at
terminals A and B.
A
I1
4Ω
2I1
8V
B
Fig. 3.198
Solution
Step I Calculation of VTh (Fig. 3.199)
From Fig. 3.199,
I1
+
I1
4Ω
2 I1
3I1 0
I1 = 0
2I1
V Th
8V
−
Writing the VTh equation,
A
B
Fig. 3.199
8 0 − VTh = 0
VTh = 8 V
Step II Calculation of IN (Fig. 3.200),
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I2
3
1
4Ω
I1 = 2 I1
2
0
A
I1
2I1
I1
...(i)
IN
I2
8V
B
Applying KVL to the outer path of the supermesh,
Fig. 3.200
8 4I
4 I1 = 0
I1 = 2
...(ii)
Solving Eqs (i) and (ii),
I2 = 6 A
IN I2 = 6 A
Step III
Calculation of RTh
1.33 Ω
A
8V
RTh =
VTh 8
= = 1 33 Ω
IN
6
Step IV Thevenin’s Equivalent Network (Fig. 3.201)
B
Fig. 3.201
3.3 Thevenin’s Theorem 3.53
Example 3.46
Find Thevenin’s equivalent network of Fig. 3.202.
2Ω
3Ω
+
4V
0.1 Vx
A
Vx
−
B
Fig. 3.202
2Ω
Solution
Step I Calculation of VTh (Fig. 3.203)
Vx
+
VTh
I1
3Ω
−
4V
+
0.1 Vx
A
Vx = VTh
I1
0 1 Vx
−
Writing the VTh equation,
B
Fig. 3.203
4 2 I1 − Vx = 0
2Ω
3Ω
4 2( −0.1Vx ) − Vx = 0
4 0.8Vx
+
0
4V
0.1 Vx
A
Vx IN
Vx = 5 V
−
Vx = VTh = 5 V
B
Fig. 3.204
Step II Calculation of IN (Fig. 3.204)
From Fig. 3.204,
2Ω
3Ω
A
Vx = 0
The dependent source 0.1 Vx depends on the controlling
variable Vx. When Vx = 0, the dependent source vanishes, i.e.,
0.1Vx = 0 as shown in Fig. 3.205.
IN =
Step III
4V
IN
B
4
= 0.8 A
2+3
Fig. 3.205
6.25 Ω
A
Calculation of RTh
RTh =
VTh
5
=
= 6 25 Ω
IN
08
5V
B
Step IV Thevenin’s Equivalent Network (Fig. 3.206)
Fig. 3.206
3.54 Network Analysis and Synthesis
Example 3.47
Obtain the Thevenin equivalent network of Fig. 3.207 for the terminals A and B.
1Ω
Vx
2Ω
4Vx
− +
A
2A
2V
B
Fig. 3.207
Solution
Step I Calculation of VTh (Fig. 3.208)
From Fig. 3.208,
2 2 I1 − Vx = 0
Vx = 2 2 I1
For Mesh 1,
I1 = −2 A
Vx = 2 2( −2
2)
Writing the VTh equation,
2 2 I1 − 0 4Vx − VTh = 0
2 2( −2) 0 + 4(6) − VTh = 0
VTh = 30 V
1Ω
Vx
4Vx
− +
−
2Ω
2V
−
6V
Step II Calculation of IN (Fig. 3.209)
From Fig. 3.209,
Vx 2 2 I1
Meshes 1 and 2 will form a supermesh,
Writing current equation for the supermesh
I 2 I1 = 2
Applying KVL to the outer path of the supermesh,
Vx
…(i)
1Ω
4Vx
− +
IN
2A
2V
I1
I2
…(ii)
B
Fig. 3.209
...(iii)
I1 = 0 73 A
I 2 = 2 73 A
I N I 2 = 2 73 A
10.98 Ω
A
30 V
Calculation of RTh
VTh
30
=
= 10.98 Ω
IN
2 73
Step IV Thevenin’s Equivalent Network (Fig. 3.210)
A
2Ω
Solving Eqs (ii) and (iii),
RTh =
B
Fig. 3.208
2 2 I1 1 2 + 4Vx = 0
2 2 I1 I 2 + 4( 2 − 2 1 ) 0
10 I1 + I 2 = 10
Step III
A
VTh
2A
+ I
1
+
B
Fig. 3.210
3.3 Thevenin’s Theorem 3.55
Example 3.48
Find the Thevenin equivalent network of Fig. 3.211 for the terminals A and B.
8I1
− +
1Ω
A
I1
10 Ω
10 Ω
5V
B
Fig. 3.211
Solution
8I1
− +
Step I Calculation of VTh (Fig. 3.212)
Applying KVL to the mesh,
5 10
1
10
I1 =
10 Ω
5
= 0.25 A
20
10 Ω
8
1
1
1
2
0 − VTh = 0
VTh = 5 2 I1 = 5 2(0.25)
4.5 V
8I1
− +
10
0( 1 2 ) 0
20 I1 10 I 2 = 5
...(i)
− 1 ) + 8 1 −1
− 1I 2 = 0
1Ω
A
IN
10 Ω
5V
18 I1 −11
11I 2 = 0
10 Ω
I1
I2
B
...(ii)
Fig. 3.213
Solving Eqs (i) and (ii),
I1 = 1.375 A
I 2 = 2.25 A
I N I 2 = 2.25 A
Step III
B
Fig. 3.212
Applying KVL to Mesh 2,
−10(
VTh
−
Step II Calculation of IN (Fig. 3.213)
Applying KVL to Mesh 1,
5 10
A
5V
Writing the VTh equation,
5 10
+
I1
0
1
1Ω
2Ω
A
Calculation of RTh
4.5 V
RTh
V
45
= Th =
=
IN
2.25
Ω
Step IV Thevenin’s Equivalent Network (Fig. 3.214)
B
Fig. 3.214
3.56 Network Analysis and Synthesis
Example 3.49
Find VTh and RTh between terminals A and B of the network shown in Fig. 3.215 .
1Ω
2Ω
12 V
Ix
A
1Ω
2Ix
B
Fig. 3.215
1Ω
Solution
Step I
2Ω
+
A
Calculation of VTh (Fig. 3.216)
12 V
Ix = 0
VTh
1Ω
The dependent source 2Ix depends on the controlling variable
Ix. When Ix = 0, the dependent source vanishes, i.e., 2 x 0 as
shown in Fig. 3.216.
−
B
Fig. 3.216
Writing the VTh equation,
VTh = 12 ×
1
=6V
1+1
1Ω
Step II Calculation of IN (Fig. 3.217)
From Fig. 3.217,
Ix =
V1
2
12 V
2Ω
V1
2Ix
1Ω
Ix
A
IN
Applying KCL at Node 1,
B
V1 12 V1 V1
+ + = 2I x
1
1 2
V1
⎛V ⎞
V1 V1 + − 12 = 2 ⎜ 1 ⎟
⎝ 2⎠
2
Fig. 3.217
V1 = 8 V
IN =
Step III
V1 8
= =4A
2 2
Calculation of RTh
RTh =
Example 3.50
terminals a and b .
VTh 6
= = 5Ω
IN
4
Obtain the Thevenin equivalent network of Fig. 3.218 for the given network at
3.3 Thevenin’s Theorem 3.57
3Ω
4Ω
Vx
5Vx
+ −
a
2Ω
2A
b
Fig. 3.218
3Ω
Solution
Step I Calculation of VTh (Fig. 3.219)
Applying KCL at Node x,
Vx
+
Vx − 5 Vx
a
VTh
−
Writing the VTh equation,
= −16 V
5Vx
+ −
2Ω
2A
V
2= x
2
Vx = 4 V
VTh
4Ω
b
Fig. 3.219
4 Vx
( the terminal a s
3Ω
ega ve w. . . b )
Step II Calculation of IN (Fig. 3.220)
Applying KCL at Node x,
4Ω
Vx
5Vx
+ −
a
IN
2Ω
2A
Vx Vx − 5 Vx
+
2
4
V
V
2 = x − Vx = − x
2
2
2=
b
Fig. 3.220
Vx = −4 V
V
Vx
IN = x
= −Vx = 4 A
4
Step III
−4 Ω
a
−16 V
Calculation of RTh
RTh =
VTh −16
=
= −4 Ω
IN
4
b
Step IV Thevenin’s Equivalent Network (Fig. 3.221)
Example 3.51
Fig. 3.221
Obtain the Thevenin equivalent network of Fig. 3.222 for the given network.
150 V 10 Ω
30 Ω
5A
+
Vx
15 Ω
−
Fig. 3.222
+
−
1
V
3 x
3.58 Network Analysis and Synthesis
Solution
Step I Calculation of VTh (Fig. 3.223)
From Fig. 3.223
Vx VTh
Applying KCL at the node,
1
Vx 1 0
Vx
3 + Vx + 5 = 0
10
15
150 V 10 Ω
30 Ω
A
+
VTh
+
Vx
5A
−
B
30 Ω
Step IV
IN
1
V
3 x
15 Ω
5A
B
Fig. 3.224
37.5 Ω
A
V
60
= x =
=2A
30 30
75 V
Calculation of RTh
VTh 75
=
= 37.5 Ω
IN
2
B
Thevenin’s Equivalent Network (Fig. 3.225)
Example 3.52
+
−
150 V 10 Ω
Vx
A
Step II Calculation of IN (Fig. 3.224)
Applying KCL at Node x,
1
Vx 1 0
Vx
Vx
Vx
3
+5+ +
=0
30
15
10
Vx Vx Vx Vx
+ +
−
= 15 − 5
30 15 10 30
Vx = 60 V
RTh =
1
V
3 x
Fig. 3.223
VTh = 75 V
Step III
+
−
−
Vx = 75 V
IN
15 Ω
Fig. 3.225
Find the Thevenin’s equivalent network of the network to the left of A-B in the
Fig. 3.226.
10 Ix
10 V
+ −
A
Ix
1A
5Ω
10 Ω
B
Fig. 3.226
10 Ix
Solution
Step I Calculation of VTh (Fig. 3.227)
From Fig. 3.227,
I x I1 − I 2
For Mesh 1,
I1 = 1
10 V
+ −
+
Ix
…(i)
+
5Ω
1A
I1
I2
10 Ω
VTh
−
−
…(ii)
Fig. 3.227
A
B
3.3 Thevenin’s Theorem 3.59
Applying KVL to Mesh 2,
−5(
−5(
2
2
− 1 ) − 10
− 1 ) − 10(
1
−
2
=0
2 ) − 10 2
=0
x
5
−1
10
0
5II 2 = 0
5
1
…(iii)
Solving Eqs (ii) and (iii),
I1 = 1 A
I 2 = −1 A
I x I1 I 2 = 1 ( 1) = 2 A
Writing the VTh equation,
10 I 2 10 VTh = 0
10( −11) 10 − VTh = 0
VTh = −20 V
Step II Calculation of IN (Fig. 3.228)
From Fig. 3.228,
I x I1 − I 2
For Mesh 1,
I1 = 1
Applying KVL to Mesh 2,
−5(
−5(
2
2
− 1 ) − 10
− 1 ) − 10(
1
−
2
−
3)
=0
2 ) − 10( 2
−
3)
=0
x
−1
10
0(
10 Ix
10 V
+ −
…(i)
…(ii)
A
Ix
5Ω
1A
I1
10 Ω
I2
IN
I3
B
Fig. 3.228
−5 1 − 5 I 2 + 10 I 3 = 0
…(iii)
Applying KVL to Mesh 3,
−10(
3
−
2 ) − 10
=0
10 I 2 10
0 I 3 = 10
…(iv)
Solving Eqs (ii), (iii) and (iv),
I1 = 1 A
I2 = 3 A
I3 = 2 A
I N I3 = 2 A
Step III
−20 Ω
A
−20 V
Calculation of RTh
VTh −20
=
= −10 Ω
IN
2
Step IV Thevenin’s Equivalent Network (Fig. 3.229)
RTh =
Example 3.53
Fig. 3.230.
B
Fig. 3.229
Find Thevenin’s equivalent network at terminals A and B in the network of
3.60 Network Analysis and Synthesis
2Ω
4Ω
A
4Vx
+
Vx
−
−
+
5Ω
B
Fig. 3.230
Solution
Since the network does not contain any independent source,
VTh = 0
IN = 0
But the RTh can be calculated by applying a known
voltage source of 1 V at the terminals A and B as shown
in Fig. 3.231.
V 1
RTh = =
I I
From Fig. 3.231,
Vx
( I1 − I 2 )
Applying KVL to Mesh 1,
−4Vx − 2 I1 − 5( 1 − 2 ) = 0
−4 [
−
] − 2 1 − 5I1 + 5
2
2Ω
4Ω
A
4Vx
−
+
I1
+
Vx
−
5Ω
1V
I2
B
Fig. 3.231
...(i)
=0
−27 I1 + 25 I 2 = 0
…(ii)
Applying KVL to Mesh 2,
−5(
2
− 1) − 4I2 − 1 = 0
5
1
9II 2 = 1
9
…(iii)
Solving Eqs (ii) and (iii),
A
I1 = −0 21 A
I 2 = −0 23 A
4.35 Ω
Hence, current supplied by voltage source of 1 V is 0.23 A.
1
RTh =
= 4 35 Ω
0 23
Hence, Thevenin’s equivalent network is shown in Fig. 3.232.
Example 3.54
B
Fig. 3.232
Find the current in the 9 W resistor in Fig. 3.233.
− +
6Ix
Ix
4Ω
6Ω
20 V
Fig. 3.233
9Ω
3.3 Thevenin’s Theorem 3.61
Solution
Step I Calculation of VTh (Fig. 3.234)
Applying KVL to the mesh,
−4 x 6
− +
Ix
4Ω
6I x = 0
x
Ix = 5 A
x
VTh
+
6Ω
−
20 V
Writing the VTh equation,
6
6Ix
−
− +
6Ix
Ix
Ix = 0
A
4Ω
6Ω
The dependent source 6Ix depends on the controlling
variable Ix. When I x = 0, the dependent source vanishes,
i.e., 6 x 0 as shown in Fig. 3.236.
20
IN =
=5A
4
IN
20 V
B
Fig. 3.235
6Ix
A
4Ω
IN
20 V
A
⇒
4Ω
IN
20 V
B
B
(a)
(b)
Fig. 3.236
6Ω
A
Calculation of RTh
RTh =
VTh 30
=
=6Ω
IN
5
9Ω
30 V
IL
Step IV Calculation of IL (Fig. 3.237)
B
30
IL =
=2A
6+9
Example 3.55
B
Fig. 3.234
Step II Calculation of IN (Fig. 3.235).
From Fig. 3.235,
− +
A
VTh
0
6(5) − VTh = 0
VTh = 30 V
Step III
+
Fig. 3.237
Determine the current in the 16 W resistor in Fig. 3.238.
10 Ω
6Ω
Ix
40 V
0.8Ix
Fig 3.238
16 Ω
3.62 Network Analysis and Synthesis
10 Ω
Solution
Step I Calculation of VTh (Fig. 3.239)
From Fig. 3.239,
Ix = 0
The dependent source 0.8Ix depends on the controlling
variable Ix. When Ix = 0, the dependent source vanishes, as
shown in Fig. 3.240.
i.e.,
0 8I x = 0
VTh = 40 V
I1
+
40 V
−
2
2
−
…(ii)
10 Ω
6Ω
…(iii)
40 V
0.8Ix
I1 = 3 A
5
I2 = A
3
IN
I2
B
Fig. 3.241
5
A
3
24 Ω
Calculation of RTh
A
RTh =
VTh 40
=
=
5
IN
3
Ω
16 Ω
40 V
IL
Step IV Calculation of IL (Fig. 3.242)
B
40
IL =
=1A
24 + 16
Example 3.56
Ix
A
0
40
I2 =
Fig. 3.242
Find the current in the 6 W resistor in Fig. 3.243.
1Ω
−
18 V
B
Fig. 3.240
I1
IN
A
VTh
Solving Eqs (ii) and (iii),
Step III
6Ω
40 V
Applying KVL to the outer path of the supermesh,
40 −10
10 I1 6
10 I1 + 6
B
Fig. 3.239
10 Ω
…(i)
0.8 I 2
A
VTh
0.8Ix
+
Step II Calculation of IN (Fig. 3.241)
From Fig. 3.241,
I x I2
Meshes 1 and 2 will form a supermesh,
Writing current equation for the supermesh,
I1 I 2 = 0.8 I x
1 8 I2 = 0
6Ω
Vx
− +
+
3A
Fig. 3.243
2Vx
6Ω
3.3 Thevenin’s Theorem 3.63
Solution
Step I Calculation of VTh (Fig. 3.244)
From Fig. 3.244,
Vx
1 1 = − I1
For Mesh 1,
I1 = −3 A
Vx = 3 V
1Ω
−
…(i)
Vx
− +
+
18 V
2Vx
+
3A
…(ii)
A
VTh
I1
−
Writing the VTh equation,
B
Fig. 3.244
18 −1
1 I1 2 Vx − VTh = 0
18 + 3 2(3) VTh 0
VTh = 27 V
Step II Calculation of IN (Fig. 3.245)
From Fig. 3.245,
Vx
I1
Meshes 1 and 2 will form a supermesh,
Writing current equation for supermesh,
I 2 I1 = 3
Applying KVL to the outer path of the supermesh,
18 −1
1I1 2 Vx = 0
18 − I1 + 2( 1 ) 0
1Ω
…(i)
−
Vx
− +
+
2Vx
IN
18 V
3A
…(ii)
I1
I2
B
Fig. 3.245
…(iii)
I1 = 6 A
Solving Eqs (ii) and (iii),
I2 = 9 A
Step III
IN
I2 = 9 A
RTh =
VTh 27
=
= 3Ω
IN
9
3Ω
A
Calculation of RTh
6Ω
27 V
IL
Step IV Calculation of IL (Fig. 3.246)
B
27
IL =
=3A
3+ 6
Example 3.57
Fig. 3.246
Find the current in the 10 W resistor.
10 Ω
100 V
A
− +
10Vx
+
Vx
−
Fig. 3.247
5Ω
10 A
3.64 Network Analysis and Synthesis
Solution
Step I Calculation of VTh (Fig. 3.248)
From the figure,
Vx = 10 × 5 = 50 V
+ VTh −
A
− +
10Vx
B
+
5Ω
−
100 V
Writing the VTh equation,
Vx
10 A
5Ω
10 A
100 − VTh + 10Vx − Vx = 0
100 − VTh + 9Vx
100 − VTh + 9(50)
0
0
Fig. 3.248
VTh = 550 V
Step II Calculation of IN (Fig. 3.249)
From Fig. 3.249,
Vx 5( I N + 10)
Applying KVL to Mesh 1,
A
B
− +
10Vx
IN
+
Vx
−
100 V
100 +10
10Vx − Vx = 0
Vx = −
100
−
=5
9
IN = −
Step III
100
9
N
Fig. 3.249
+ 50
550
A
45
−45 Ω
Calculation of RTh
A
550
RTh =
= −45 Ω
550
−
45
Step IV Calculation of IL (Fig. 3.250)
IL =
3.4
10 Ω
550 V
IL
B
550
110
=−
A
−45 + 10
7
Fig. 3.250
NORTON’S THEOREM
It states that ‘any two terminals of a network can be replaced by an equivalent current source and an equivalent
parallel resistance.’ The constant current is equal to the current which would flow in a short circuit placed across
the terminals. The parallel resistance is the resistance of the network when viewed from these open-circuited
terminals after all voltage and current sources have been removed and replaced by internal resistances.
IL
Network
(a)
RL
IL
IN
or
IN
RN
(b)
Fig. 3.251 Network illustrating Norton’s theorem
RL
3.4 Norton’s Theorem 3.65
Explanation
Consider a simple network as shown in Fig.3.252
R1
R3
A
V
R2
RL
B
Fig. 3.252
Network
For finding load current through RL , first remove the load
resistor RL from the network and calculate short circuit
current ISC or I N which would flow in a short circuit placed
across terminals A and B as shown in Fig. 3.253.
For finding parallel resistance RN , replace the voltage
source by a short circuit and calculate resistance between
points A and B as shown in Fig. 3.254.
RN
R3 +
R1
R3
A
V
R2
IN
B
R1 R2
R1 R2
Fig. 3.253
Calculation of IN
R1
R3
A
Norton’s equivalent network is shown in Fig.3.255.
IL
IN
RN
RN RL
R2
RN
If the network contains both independent and dependent
sources, Norton’s resistances RN is calculated as
B
Fig. 3.254
V
RN = Th
IN
Calculation of RN
IL
where VTh is the open-circuit voltage across terminals A and
B. If the network contains only dependent sources, then
IN
RN
RL
VTh = 0
IN = 0
Fig. 3.255
To find RTh in such network, a known voltage V or current
I is applied across the terminals A and B, and the current I or
the voltage V is calculated respectively.
RN =
V
I
Norton’s equivalent network for such a network is shown in
Fig. 3.256.
Norton’s equivalent network
A
RN
B
Fig. 3.256
Norton’s equivalent network
3.66 Network Analysis and Synthesis
Steps to be followed in Norton’s Theorem
1.
2.
3.
4.
5.
Remove the load resistance RL and put a short circuit across the terminals.
Find the short-circuit current ISC or IN.
Find the resistance RN as seen from points A and B.
Replace the network by a current source IN in parallel with resistance RN.
Find current through RL by current–division rule.
IL =
Example 3.58
I N RN
RN RL
Find the current through the 10 W resistor in Fig. 3.257.
5Ω
1Ω
10 Ω
15 Ω
4A
2V
Fig. 3.257
2 1I1 = 0
I1 = 2
1Ω
IN
...(i)
2V
Meshes 2 and 3 will form a supermesh.
Writing the current equation for the supermesh,
I2 = 4
I3
I1
2
− 15
15
3
B
Fig. 3.258
...(ii)
=0
5Ω
...(iii)
Solving Eqs (i), (ii) and (iii),
A
1Ω
I1 = 2 A
Fig. 3.259
A
Calculation of RN (Fig. 3.259)
RN = 1 (5 + 15) = 0 95 Ω
Step III
15 Ω
RN
B
I 2 = −3 A
I3 = 1A
I N I1 I 2 = 2 ( 3) = 5 A
Step II
15 Ω
4A
I3
I2
Applying KVL to the supermesh,
−5
5Ω
A
Solution
Step I Calculation of IN (Fig. 3.258)
Applying KVL to Mesh 1,
IL
5A
0.95 Ω
10 Ω
Calculation of IL (Fig.3.260)
IL = 5 ×
0 95
= 0.43 A
0 95 + 10
B
Fig. 3.260
3.4 Norton’s Theorem 3.67
Example 3.59
Find the current through the 10 W resistor in Fig. 3.258.
8Ω
12 V
V
2Ω
5Ω
10 Ω
Fig. 3.261
12 V
8Ω
A
Solution
Step I Calculation of IN (Fig. 3.262)
Applying KVL to Mesh 1,
20 V
5Ω
−5 1 + 2
20
0 − 2( I1 − I 2 ) = 0
7 1 − 2 I 2 = 20
...(i)
2Ω
I1
B
Applying KVL to Mesh 2,
−2(
2
IN
I2
Fig. 3.262
− 1 ) − 8 I 2 − 12 = 0
− 2 1 +1
10
0 2 = −12
8Ω
Solving Eqs (i) and (ii),
5Ω
2Ω
...(ii)
RN
I 2 = −0 67 A
IN
Step II
Fig. 3.263
Calculation of RN (Fig. 3.263)
RN = (
Step III
I 2 = −0 67 A
A
) + 8 = 9.43 Ω
I L = 0 67 ×
9.43 Ω
0.67 A
Calculation of IL (Fig. 3.264)
9 43
= 0.33
9 43 + 10
IL
(↑ )
B
Fig. 3.264
Example 3.60
Find the current in the 10 W resistor in Fig. 3.265.
50 Ω
50 V
40 Ω
20 Ω
10 V
Fig. 3.265
10 Ω
10 Ω
3.68 Network Analysis and Synthesis
50 Ω
Solution
Step I Calculation of IN (Fig. 3.266)
The resistance of
Ω becomes redundant as it is
connected across the 50 V source (Fig. 3.267).
Applying KVL to Mesh 1,
50 − 50 I1 20( I1 − I 2 ) − 10 = 0
70 I1 20 I 2 = 40
40 Ω
50 V
20 Ω
IN
10 V
...(i)
Fig. 3.266
Applying KVL to Mesh 2,
10 − 20(
2
1)
=0
−20
20 I1 + 20 I 2 = 0
Solving Eqs (i) and (ii),
Step II
50 Ω
...(ii)
40 Ω
20 Ω
50 V
I1
I1 = 1 A
I2 = 1 5 A
IN I2 = 1 5 A
10 V
IN
I2
Fig. 3.267
Calculation of RN (Fig. 3.268)
50 Ω
A
RN
40 Ω
RN = 50 20 = 14.28 Ω
20 Ω
B
Fig. 3.268
Step III
Calculation of IL (Fig. 3.269)
A
IL
IL = 1 5 ×
14.28 Ω
1.5 A
14.28
= 0 88 A
14.28 + 10
10 Ω
B
Fig. 3.269
Example 3.61
Find the current through the 10 W resistor in Fig. 3.270.
6Ω
10 V
2Ω
1Ω
3Ω
20 V
Fig. 3.270
10 Ω
3.4 Norton’s Theorem 3.69
6Ω
Solution
Step I Calculation of IN (Fig. 3.271)
Applying KVL to Mesh 1,
−6
7
1
2
10
A
1Ω
10 V
1( I1 − I 2 ) = 0
1
2Ω
I1
...(i)
3Ω
I2
B
Applying KVL to Mesh 2,
−1(
2
20 V
− 1 ) − 2 I 2 − 3(
2
−
− I1 + 6
2
− 3I 3 = 0
3)
=0
Fig. 3.271
...(ii)
6Ω
Applying KVL to Mesh 3,
−3(
3
IN
I3
2Ω
A
− 2 ) − 20 = 0
3 2 − 3I 3 = 20
1Ω
...(iii)
3Ω
RN
Solving Eqs (i), (ii) and (iii),
B
I 3 = −13.17 A
IN
Step II
Fig. 3.272
I 3 = −13.17 A
2Ω
Calculation of RN (Fig. 3.272)
A
IL
RN = [(6 1) + ] 3 = 1.46 Ω
Step III
1.46 Ω
13.17 A
10 Ω
Calculation of IL (Fig. 3.273)
I L = 13.17 ×
1.46
= .68 (↑)
1.46 + 10
Example 3.62
B
Fig. 3.273
Find the current through the 10 W resistor in Fig. 3.274.
10 Ω
20 Ω
30 Ω
20 Ω
20 Ω
50 V
100 V
40 V
Fig. 3.274
Solution
Step I Calculation of IN (Fig. 3.274)
Applying KVL to Mesh 1,
50 − 20( 1 2 ) − 40 = 0
20 I1 20 I 2 = 10
A
...(i)
Applying KVL to Mesh 2,
40 − 20(
2
1 ) − 20 I 2
20( 2 − 3 ) = 0
−20
20 I1 + 60 I 2 − 20 I 3 = 40 ...(ii)
IN
B
20 Ω
20 Ω
20 Ω
50 V
I1
30 Ω
I2
40 V
Fig. 3.275
100 V
I3
3.70 Network Analysis and Synthesis
Applying KVL to Mesh 3,
−20( 3 −
2 ) − 30 I 3
− 100 = 0
...(iii)
−20 I 2 + 50 I 3 = −100
Solving Eqs (i), (ii) and (iii),
I1 = 0 81 A
IN
Step II
I1 = 0 81 A
Calculation of RN (Fig. 3.276)
20 Ω
RN
A
RN = [( 20 30) +
30 Ω
B
20 Ω
] 20 = 12.3 Ω
20 Ω
Fig. 3.276
Step III
Calculation of IL (Fig. 3.277)
A
IL
I L = 0 81 ×
12.3 Ω
0.81 A
12.3
= 0 45 A
12.3 + 10
10 Ω
B
Fig. 3.277
Example 3.63
Obtain Norton’s equivalent network as seen by RL in Fig. 3.278.
30 Ω
40 V
10 Ω
30 Ω
60 Ω
120 V
RL
10 V
Fig. 3.278
30 Ω
Solution
Step I Calculation of IN (Fig. 3.279)
Applying KVL to Mesh 1,
90 I1 60 I 2 = 120
10 Ω
I1
A
IN
30 Ω
60 Ω
120 V
120 − 30 I1 60( I1 − I 2 ) = 0
40 V
I2
...(i)
Fig. 3.279
B
10 V
I3
3.4 Norton’s Theorem 3.71
Applying KVL to Mesh 2,
−60(
2
10 Ω
30 Ω
− 1 ) + 40 − 10 I 2 − 30(
30(
2
−
3)
A
RN
B
=0
−60 I1 +100
100 I 2 − 30 I 3 = 40
...(ii)
30 Ω
60 Ω
Applying KVL to Mesh 3,
−30(
3
−
2 ) + 10
=0
30 I 2 − 30 I 3 = −10
...(iii)
Fig. 3.280
Solving Eqs (i), (ii) and (iii),
A
I 3 = 4 67 A
IN
Step II
I 3 = 4 67 A
RN = [(30 60) +
Step III
15 Ω
4.67 A
Calculation of RN (Fig. 3.280)
] 30 = 15 Ω
RL
B
Norton’s equivalent network (Fig. 3.281)
Fig. 3.281
Example 3.64
Find the current through the 8 W resistor in Fig. 3.282.
5V
5A
12 Ω
2A
4Ω
8Ω
Fig. 3.282
Solution
Step I Calculation of IN (Fig. 3.283)
5V
A
4Ω
12 Ω
5A
2A
IN
B
Fig. 3.283
The resistor of the 4 Ω gets shorted as it is in parallel with the short circuit. Simplifying the network by
source transformation (Fig. 3.284),
12 Ω
5V
A
60 V
2A
I1
I2
IN
B
Fig. 3.284
3.72 Network Analysis and Synthesis
Meshes 1 and 2 will form a supermesh.
Writing the current equation for the supermesh,
I 2 I1 = 2
Applying KVL to the supermesh,
A
...(i)
60 −12
12 I1 − 5 0
12 I1 = 55
12 Ω
4Ω
B
...(ii)
Fig. 3.285
Solving Eqs (i) and (ii),
I1 = 4 58 A
I 2 = 6 58 A
I N I 2 = 6 58 A
Step II
A
IL
3Ω
6.58 A
8Ω
Calculation of RN (Fig. 3.285)
RN = 12 4 = 3 Ω
Step III
RN
B
Calculation of IL (Fig. 3.286)
I L = 6.58 ×
Example 3.65
Fig. 3.286
3
= 1.79 A
3+8
Find the current through the 1 W resistor in Fig. 3.287.
2Ω
1Ω
1A
3Ω
1V
2Ω
2Ω
Fig. 3.287
Solution
Step I Calculation of IN (Fig. 3.288)
A
2Ω
IN
1A
3Ω
1V
2Ω
B
2Ω
Fig. 3.288
3.4 Norton’s Theorem 3.73
By source transformation (Fig. 3.289),
A
2Ω
I3
3Ω
3V
IN
I1
1V
2Ω
I2
B
2Ω
Fig. 3.289
A
Applying KVL to Mesh 1,
−3 − 3I1 − 2(
− 3) +1 = 0
5 1 − 2 I 3 = −2
1
2Ω
...(i)
RN
3Ω
2Ω
Applying KVL to Mesh 2,
−1 − 2(
2
−
3) − 2 2
4
2
=0
− 2 I 3 = −1
B
2Ω
...(ii)
(a)
2Ω
Applying KVL to Mesh 3,
−2(
3
− 1 ) − 2( 3 − 2 ) = 0
−2 1 − 2 I 2 + 4 3 = 0
...(iii)
2Ω
A
B
Solving Eqs (i), (ii) and (iii),
3Ω
I1 = −0 64 A
I 2 = −0 55 A
I 3 = −0 59 A
I N I 3 = −0 59 A
Step II
2Ω
(b)
1.2 Ω
1Ω
A
B
(c)
Calculation of RN (Fig. 3.290)
Fig. 3.290
RN = 2.2 Ω
Step III
Calculation of IL (Fig. 3.291)
A
I L = 0 59 ×
2.2
= 0 41 A
2.2 + 1
0.59 A
2.2 Ω
1Ω
IL
B
Fig. 3.291
3.74 Network Analysis and Synthesis
EXAMPLES WITH DEPENDENT SOURCES
Example 3.66
Find Norton’s equivalent network across terminals A and B of Fig. 3.292.
3I1
I2
I1
10 Ω
5Ω
5V
+
10I2
−
10 Ω
− +
A
B
Fig. 3.292
Solution
Step I Calculation of VTh (Fig. 3.293)
From Fig. 3.293,
I2 I x
I1
3I1
I2
I1
10 Ω
Ix
5 10
Ix
5V
5 I x − 10 I 2 = 0
x
A
VTh
+
10I2
−
−
5 I x − 10 I x = 0
Ix = 0 2 A
x
+
5Ω
Applying KVL to the mesh,
5 10
10 Ω
− +
B
Fig. 3.293
I1 = −0 2 A
Writing the VTh equation,
5 10
3 I1 − VTh = 0
x
5 10 (0.2) 3 ( −0.2) − VTh = 0
VTh = 2.4 V
Step II Calculation of IN (Fig. 3.294)
From Fig. 3.294,
I2 I x
I1 I y − I x
Applying KVL to Mesh 1,
5 10 x 5( I x − I y ) − 10 I 2 = 0
5 10
5 Ix + 5
x
y
25 I x
10
x
0
5
y
5
3I1
…(i)
…(ii)
10
x
5
y
5(
5I x
x)
y
3(
y
10
y
0
12 I x 12 I y = 0
A
5Ω
Ix
+
10I2
−
IN
Iy
B
…(iii)
3I1 − 10 I y = 0
x)
I1
10 Ω
5V
Applying KVL to Mesh 2,
10 I 2
I2
10 Ω
− +
…(iv)
Fig. 3.294
3.4 Norton’s Theorem 3.75
Solving Eqs (iii) and (iv),
I x = 0 25 A
A
I y = 0 25 A
Step III
IN
I y = 0.25 A
RN =
VTh
2.4
=
=96Ω
IN
0 25
9.6 Ω
0.25 A
Calculation of RN
B
Step IV Norton’s Equivalent Network (Fig. 3.295)
Example 3.67
Fig. 3.295
For the network shown in Fig. 3.296, find Norton’s equivalent network.
20 Ω
A
5Ω
3Vx
15 Ω
+
Vx
2A
2Ω
−
B
Fig. 3.296
20 Ω
Solution
Step I Calculation of VTh (Fig. 3.297)
From Fig. 3.297,
Vx
+
−
5Ω
2I 2
+
…(i)
−
I1
15 Ω
For Mesh 1,
I1
3Vx
3( 2II 2 )
6I2
…(ii)
For Mesh 2,
I1
6I2
6( 2)
+
Vx
I2
I2 = 2
3Vx
VTh
+
2A
A
2Ω
−
−
…(iii)
B
Fig. 3.297
12 A
20 Ω
Writing the VTh equation,
A
VTh 0 I1 +15
+ 15( I1 I 2 ) − 2I
2I 2 = 0
VTh + 5( −12) + 15( −12 − 2) − 2( 2) = 0
VTh = 274 V
5Ω
IN
+
Step II Calculation of IN (Fig. 3.298)
From Fig. 3.298,
Vx
3Vx
I1
15 Ω
Vx
2A
2( I 2 − I 3 )
…(i)
For Mesh 2,
I2 = 2
…(ii)
I2
I3
2Ω
−
B
Fig. 3.298
3.76 Network Analysis and Synthesis
Meshes 1 and 3 will form a supermesh.
Writing the current equation for the supermesh,
I 3 I1 = 3Vx 3 [ 2 I 2
]
I3
6I 2 − 6I3
I1 6 I 2 7 I 3 = 0
Applying KVL to the outer path of the supermesh,
−5 1 − 20
20 3 − 2( I 3 − I 2 ) − 15( I1 − I 2 ) = 0
…(iii)
−20 I1 +17
17 I 2 − 22 I 3 = 0
…(iv)
Solving Eqs (ii), (iii) and (iv),
I1 = −0 16 A
I2 = 2 A
I 3 = 1 69 A
I N I 3 = 1 69 A
Step III
A
Calculation of RN
162.13 Ω
1.69 A
V
274
RN = Th =
= 162.13 Ω
I N 1 69
B
Step IV Norton’s Equivalent Network (Fig. 3.299)
Example 3.68
Fig. 3.299
Obtain Norton’s equivalent network across A-B in the network of Fig. 3.300.
2Ω
5Ω
A
I2
+
+
10I2
−
8Ω
V1
15 V
−
0.6V1
15 Ω
B
Fig. 3.300
Step I Calculation of VTh
(Fig. 3.301)
From Fig. 3.301,
V1
2Ω
5Ω
Solution
8( I x − I y )
+
V1
15 V
+
−
Ix
+
10I2
−
8Ω
Iy
I2
15 Ω
0.6V1
I2
+
VTh
−
−
…(i)
Applying KVL to Mesh 1,
A
B
Fig. 3.301
15 − 5
x
8( I x − I y ) = 0
13I x
8
y
15
…(ii)
Applying KVL to Mesh 2,
−8(
y
−
8
) − 2 I y −10
10 I 2 = 0
x
10
y
10
2
0
…(iii)
3.4 Norton’s Theorem 3.77
For Mesh 3,
I2
0 6V1 = 0.6 ⎡⎣8( I x
I y ) ⎤⎦
4.8 I y − I 2 = 0
4.8I
8I x
…(iv)
Solving Eqs (ii), (iii) and (iv),
I x = 3 28 A
2Ω
5Ω
I y = 3 45 A
A
I 2 = −0 83 A
+
V1
15 V
Writing the VTh equation,
+
10I2
−
8Ω
−
15 I 2 VTh = 0
15( −0
0.83) VTh 0
VTh = −12.45 V
0.6V1
15 Ω
IN
B
Fig. 3.302
Step II Calculation of IN (Fig. 3.302)
From Fig. 3.302,
The dependent source of 10 I2 depends
on the controlling variable I2. When
I 2 = 0, the dependent source vanishes,
i.e. 10 I 2 = 0 as shown in Fig. 3.303.
From Fig. 3.303,
2Ω
5Ω
A
I2 = 0
V1
I2
+
V1
15 V
Ix
−
8Ω
0.6V1
IN
Iy
B
Fig. 3.303
8( I x − I y )
…(i)
Applying KVL to Mesh 1,
15 − 5
8( I x − I y ) = 0
x
13I x
8
y
…(ii)
15
Applying KVL to Mesh 2,
−8(
y
−
−8
) − 2I y = 0
x
+1
10
0
y
=0
…(iii)
Solving Eqs (ii) and (iii),
I x = 2.27 A
I y = 1 82 A
Iy)
8( 2.27 1.82) = 3.66 V
V1
8( I x
IN
0 6 V1 = 0.6(3.6) = 2.16 A
A
For Mesh 3,
Step III
Calculation of RN
V
−12.45
RN = Th =
= −5 76 Ω
IN
2.16
Step IV Norton’s Equivalent Network (Fig. 3.304)
2.16 A
−5.76 Ω
B
Fig. 3.304
3.78 Network Analysis and Synthesis
Example 3.69
Find Norton’s equivalent network of Fig. 3.305.
A
I1
0.5I1 +
−
1Ω
2Ω
2V
B
Fig. 3.305
0.5I1 +
−
Solution
Step I Calculation of VTh (Fig. 3.306)
Applying KVL to the mesh,
2 − 2 I1 + 0 5 I1 −1
11
2 2.5 1
I1
0
0
VTh
−
2V
I1 = 0.8
8A
A
+
1Ω
2Ω
+
−
B
Fig. 3.306
Writing the VTh equation,
A
1 1 VTh
1(0.8) VTh
0
0
0.5I1 +
−
I1
1Ω
VTh = 0 8 V
Step II Calculation of IN (Fig. 3.307)
When a short circuit is placed across the 1 Ω resistor, it gets shorted.
I1 = 0
2V
B
The dependent source of 0 5 I1 depends on the controlling variable
I1. When I1 = 0, the dependent source vanishes, i.e. 0.5 I1 = 0 as shown
in Fig. 3.308.
2
IN = = 1 A
2
Step III Calculation of RN
Fig. 3.307
A
2Ω
IN
2V
V
08
RN = Th =
=08Ω
IN
1
B
Fig. 3.308
Step IV Norton’s Equivalent Network (Fig. 3.309)
A
1A
0.8 Ω
B
Fig. 3.309
IN
2Ω
3.4 Norton’s Theorem 3.79
Example 3.70
Find Norton’s equivalent network at the terminals A and B of Fig. 3.310.
6Ix
3Ω
2Ω
A
Ix
9V
6Ω
B
Fig. 3.310
6Ix
Solution
Step I Calculation of VTh (Fig. 3.311)
From Fig. 3.311,
I x I1
Applying KVL to Mesh 1,
9 3(
3Ω
+
…(i)
9
−
+
6Ω
+
A
VTh
I1
0
33II 2 = 9
1
Ix
9V
2) −6 1
1
2Ω
I2
−
−
…(ii)
B
Fig. 3.311
For Mesh 2,
I2
6
1
2
6I x
0
6 I1
…(iii)
Solving Eqs (ii) and (iii),
I1 = −1 A
I 2 = −6 A
Writing the VTh equation,
9 3( 1 2 ) + 2 2 VTh 0
9 3( −1 6) + 2( 6) − VTh = 0
VTh = −18 V
Step II Calculation of IN (Fig. 3.312)
From Fig. 3.312,
6Ix
Ix
I1 − I 3
…(i)
3Ω
Applying KVL to Mesh 1,
9 3(
1
2 ) − 6( 1
9
3)
3I 2 − 6
1
3
2Ω
I2
A
Ix
0
9
…(ii)
9V
6Ω
I1
IN
I3
For Mesh 2,
B
6
1
2
I2 6I x
6I3 = 0
Applying KVL to Mesh 3,
−6( 3 − 1 ) − 2( 3 − 2 ) = 0
−6 1 − 2 I 2 + 8 3 = 0
6( I1
I3 )
…(iii)
Fig. 3.312
…(iv)
3.80 Network Analysis and Synthesis
Solving Eqs (ii), (iii) and (iv),
A
I1 = 5 A
I2 = 3 A
I3 = 4 5 A
I N I3 = 4 5 A
Step III
−4 Ω
4.5 A
B
Calculation of RN
Fig. 3.313
V
−18
RN = Th =
= −4 Ω
IN
45
Step IV Norton’s Equivalent Network (Fig. 3.313)
Example 3.71
Find Norton’s equivalent network to the left of terminal A-B in Fig. 3.314.
A
6Ω
0.5I
4Ω
I
B
Fig. 3.314
V
Solution Since the network does not contain any
independent source,
VTh = 0
IN = 0
But RN can be calculated by applying a known current
source of 1 A at the terminals A and B as shown in Fig.
3.315.
From Fig. 3.315,
I=
A
6Ω
0.5I
4Ω
1A
I
B
Fig. 3.315
V
6
Applying KCL at the node,
V
V
+ 0 5I + = 1
6
4
V
⎛V ⎞ V
+ 0 5⎜ ⎟ + = 1
⎝ 6⎠ 4
6
⎛ 1 0 5 1⎞
+ ⎟V =1
⎜⎝ +
6 6 4⎠
V =2
V 2
RN = = = 2 Ω
1 1
Hence, Norton’s equivalent network is shown in Fig. 3.316.
A
2Ω
B
Fig. 3.316
3.4 Norton’s Theorem 3.81
Example 3.72
Find the current through the 2 W resistor in the network shown in Fig. 3.317.
2Ω
−10 V
−2Ix
Ix
+
−
4Ω
2A
10 Ω
Fig. 3.317
Solution
Step I Calculation of VTh (Fig. 3.318)
From Fig. 3.318,
Ix = 0
−2Ix
A
B Ix
V
+ Th −
+
−
−10 V
The dependent source of −2 Ix depends on
the controlling variable Ix. When I x = 0, the
dependent source vanishes, i.e. −2 x = 0 as
shown in Fig. 3.319.
I1 = 2
2A
10 Ω
Fig. 3.318
A
B Ix
VTh
+
−
Writing the VTh equation,
−10 − VTh − 4 I1 = 0
−10 V
−10 − VTh − 4( 2) = 0
+
−
10 Ω
I1
Step II Calculation of IN (Fig. 3.320)
From Fig. 3.320,
Ix
4Ω
2A
VTh = −18 V
Fig. 3.319
I1 = 2
−2Ix
A IN B Ix
…(i)
I1
Mesh 1 and 2 will form a supermesh.
Writing the current equation for
supermesh,
I2
4Ω
the
−10 V
+
−
…(ii)
4Ω
2A
I1
I2
Applying KVL to the outer path of the
supermesh,
10 Ω
I3
Fig. 3.320
−10 − 4(
2
−
−4
2
+ 4 I 3 = 10
3)
=0
…(iii)
For Mesh 3,
I3
2
1
3
( 2II x )
0
2I x
2 I1
…(iv)
3.82 Network Analysis and Synthesis
Solving Eqs (ii), (iii) and (iv),
I1 = 4 5 A
I2 = 6 5 A
I3 = 9 A
I N I1 = 4 5 A
Step III
A
IL
Calculation of RN
RN =
VTh −18
=
= −4 Ω
IN
45
−4 Ω
4.5 A
2Ω
Step IV Calculation of IL (Fig. 3.321)
B
−4
I L = 4.5 ×
=9A
−4 + 2
Example 3.73
Fig. 3.321
Find the current through the 2W resistor in the network of Fig. 3.322.
−
Vi
+
1Ω
2Ω
5V
−
+
4Vi
Fig. 3.322
−
Solution
Step I Calculation of VTh (Fig. 3.322)
From Fig. 3.323,
5 Vi 4Vi = 0
Vi = −1 V
Vi
+
+
1Ω
5V
−
+
Writing the VTh equation,
VTh
4Vi
−
−4Vi − VTh = 0
VTh = −4
4Vi = −4( −1) = 4 V
Applying KVL to the mesh,
−4Vi − 1I N = 0
I N = −4
4Vi = −4( −5) = 20 A
B
Fig. 3.323
−
Step II Calculation of IN (Fig. 3.324)
From Fig. 3.324,
5 Vi 0
Vi = −5 V
A
Vi
+
A
1Ω
5V
IN
−
+
4Vi
B
Fig. 3.324
3.4 Norton’s Theorem 3.83
Step III
Calculation of RN
A
RN =
VTh
4
=
=02Ω
IN
20
2Ω
0.2 Ω
20 A
Step IV Calculation of IL (Fig. 3.325)
B
02
I L = 20 ×
= 1.82
82 A
0 2+2
Example 3.74
Fig. 3.325
Find the current in the 2 W resistor in the network of Fig. 3.326.
1Ω
Ix
10 V
2Ix
3Ω
1A
2Ω
Fig. 3.326
Solution
Step I Calculation of VTh (Fig. 3.327)
Meshes 1 and 2 will form a supermesh.
Writing current equation for the supermesh,
I 2 I1 = 1
…(i)
Applying KVL to the outer path of the
supermesh,
10 −1
1 1 3I 2 = 0
I1 3I 2 10
…(ii)
Solving Eqs (i) and (ii),
1Ω
Ix
2Ix
+
+
10 V
3Ω
1A
I1
VTh
−
I2
A
−
B
Fig. 3.327
I1 = 1 75 A
I 2 = 2 75 A
Writing the VTh equation,
3 2 VTh 0
3( 2.75) − VTh = 0
VTh = 8 25 V
Step II Calculation of IN (Fig. 3.328)
From Fig. 3.328,
I x I1
…(i)
Meshes 1 and 2 will form a supermesh.
Writing the current equation for the supermesh,
I 2 I1 = 1
…(ii)
Applying KVL to the outer path of the supermesh,
10 −1
1 1 3( I 2 − I 3 ) = 0
1Ω
Ix
2Ix
1A
10 V
I1
I2
Fig. 3.328
3Ω
IN
I3
3.84 Network Analysis and Synthesis
3I 3 = 10
I1 3I 2
…(iii)
For Mesh 3,
I3
2
1
3
2I x
0
2 I1
…(iv)
Solving Eqs (ii), (iii) and (iv),
I1 = −3 5 A
I 2 = −2 5 A
I 3 = −7 A
IN
I 3 = −7 A
Calculation of RN
Step III
RN =
A
IL
VTh 8 25
=
= −1.18 Ω
IN
−7
−1.18 Ω
−7 A
Step IV Calculation of IL (Fig. 3.329)
B
−1.18
I L = −7 ×
= 10.07
07 A
−1.18 + 2
Example 3.75
2Ω
Fig. 3.329
Find the current through the 10 W resistor for the network of Fig. 3.330.
3Ix
− +
2Ω
Ix
5Ω
10 Ω
10 V
Fig. 3.330
3Ix
Solution
− +
Step I Calculation of VTh (Fig. 3.331)
Applying KVL to the mesh,
10 − 2
x
3I x − 5
3I
x
0
+
2Ω
+
5Ω
VTh
−
10 V
−
I x = 2.55 A
Writing the VTh equation,
5
0
3Ix
5( 2.5) VTh
0
− +
VTh = 12.5 V
Step II Calculation of IN (Fig. 3.332)
From Fig. 3.332,
Ix = 0
B
Fig. 3.331
VTh
x
A
Ix
2Ω
Ix
5Ω
10 V
Fig. 3.332
IN
3.4 Norton’s Theorem 3.85
The dependent source of 3 Ix depends on the controlling variable Ix.
When I x = 0, the dependent source 3 Ix vanishes, i.e. 3 Ix = 0 as
shown in Fig. 3.333.
N
Step III
A
2Ω
10
=
=5A
2
IN
10 V
B
Calculation of RN
Fig. 3.333
V
12.5
RN = Th =
=25Ω
IN
5
A
IL
Step IV Calculation of IL (Fig. 3.334)
IL = 5 ×
10 Ω
2.5 Ω
5A
25
=1A
2 5 + 10
B
Fig. 3.334
Example 3.76
Find the current through the 5 W resistor in the network of Fig. 3.335.
2Ω
12 V
4Ω
Ix
+
−
5Ω
4Ix
Fig. 3.335
Solution
Step I Calculation of VTh (Fig. 3.336)
Applying KVL to the mesh,
12 − 2 I x 4 I x − 4 x 0
12 V
12 − 10 I x = 0
I x = 1.2
2A
Writing the VTh equation,
12 − 2 x VTh 0
12 − 2(1.2) − VTh = 0
VTh = 9 6 V
Step II Calculation of IN (Fig. 3.337)
From Fig. 3.337,
I x I1
…(i)
Applying KVL to Mesh 1,
12 − 2 1 0
12 V
I1 = 6 A …(ii)
Applying KVL to Mesh 2,
−4 2 − 4 I x = 0
−4 2 − 4 I1 = 0
…(iii)
Solving Eqs (ii) and (iii),
I 2 = −6 A
I N I1 − I 2 = 6 − ( −6) = 12 A
2Ω
+
4Ω
Ix
−
+ A
V Th
− B
+
−
4Ix
+
−
4Ix
Fig. 3.336
2Ω
4Ω
Ix
IN
I1
I2
Fig. 3.337
3.86 Network Analysis and Synthesis
Step III
Calculation of RN
IL
VTh 9 6
=
=08Ω
IN
12
RN =
12 A
5Ω
0.8 Ω
Step IV Calculation of IL (Fig. 3.338)
I L = 12 ×
Example 3.77
08
= 1.66
66 A
0 8+5
Fig. 3.338
Find the current through the 10 W resistor for the network of Fig. 3.339.
5Ω
+
4Ω
0.5Vx
Vx
10 Ω
−
5V
Fig. 3.339
Solution
Step I Calculation of VTh (Fig. 3.340)
For the mesh,
0 5Vx = −0.5VTh
I
5Ω
+
4Ω
5V
0.5Vx
VTh = Vx
I
−
Writing the VTh equation,
5 4 I − 0 VTh
5Ω
5 4( −0.5VTh ) − VTh = 0
Step II Calculation of IN (Fig. 3.341)
From Fig. 3.341,
Vx = 0
+
4Ω
IN
Step III
0.5Vx
5V
Vx
−
A
IN
B
Fig. 3.341
The dependent source of 0 5V
Vx depends on the controlling
variable Vx. When Vx = 0, the dependent source vanishes, i.e. 0.5
Vx = 0 as shown in Fig. 3.342.
5
5
=
= A
4+5 9
B
Fig. 3.340
0
VTh = −5 V
A
5Ω
A
4Ω
IN
5V
B
Calculation of RN
RN =
VTh −5
=
= −9 Ω
5
IN
9
Step IV Calculation of IL (Fig. 3.343)
IL =
5
−9
×
= −5 A
9 −9 + 10
Fig. 3.342
A
IL
5A
9
−9 Ω
10 Ω
B
Fig. 3.343
3.4 Norton’s Theorem 3.87
Example 3.78
Find the current through the 10 W resistor in the network shown in Fig. 3.344.
1000 Ω
I1
+
+
2Vx
−
12 V
25 Ω
Vx
5I1
10 Ω
−
Fig. 3.344
Solution
Step I Calculation of VTh (Fig. 3.345)
From Fig. 3.345,
Vx
2 ( I1 ) = −125
125 I1 …(i)
1000 Ω
I1
+
−
12 V
+
2Vx
5I1
Vx
A
+
25 Ω
VTh
−
Applying KVL to Mesh 1,
−
12 −1000I
1000 1 − 2Vx 0
12 −1000
1000 I1 2( 125 I1 ) = 0
B
Fig. 3.345
…(ii)
I1 = 0.016 A
Vx
12 I1 = −125(0.016) = −2 V
Writing the VTh equation,
VTh
1000 Ω
I1
A
Vx = −2 V
Step II Calculation of IN (Fig. 3.346)
From Fig. 3.346,
+
−
12 V
+
2Vx
5I1
Vx
25 Ω
IN
−
Vx = 0
B
The dependent source of 2Vx depends
on the controlling variable Vx. When
Vx = 0, the dependent source vanishes,
i.e. 2 Vx = 0 as shown in Fig. 3.347.
I1 =
IN
Step III
12
= 0.012 A
1000
5 I1
5(0.012) = −0.06 A
Fig. 3.346
1000 Ω
A
12 V
5I1
IN
B
Fig. 3.347
Calculation of RN
RN =
I1
VTh
−2
=
= 33.33 Ω
IN
−0 06
A
IL
−0.06 A
33.33 Ω
10 Ω
Step IV Calculation of IL (Fig. 3.348)
I L = −0 06 ×
33.33
= −0.046
046 A
33.33 + 10
B
Fig. 3.348
3.88 Network Analysis and Synthesis
Example 3.79
Find the current through the 5 W resistor for the network of Fig. 3.349.
4V
1Ω
2Ω
+
1Ω
Vx
−
+
−
5Ω
4Vx
2A
Fig. 3.349
Solution
Step I Calculation of VTh (Fig. 3.350)
From Fig. 3.350,
Vx 2 I
For the mesh,
I =2
Vx = 2( 2) = 4 V
Writing the VTh equation,
4Vx
2 I + 1I
I 2 )] − 2
2
2 I1 − I 2
I1 + 4
Step III
…(i)
2Ω
+
Vx
1Ω
−
+
−
…(ii)
I
2A
4Vx
+ A
VTh
− B
Fig. 3.350
4V
…(i)
…(ii)
1Ω
I2
2Ω
+
1Ω
Vx
−
A
+
−
I1
2A
0
11I1 11I 2 = −4
Solving Eqs (ii) and (iii),
1Ω
4 VTh = 0
4( 4) + 2( 2) + 2 + 4 − VTh = 0
VTh = 26 V
Step II Calculation of IN (Fig. 3.351)
From Fig. 3.351,
Vx 2( I1 − I 2 )
For Mesh 1,
I1 = 2
Applying KVL to Mesh 2,
4Vx 2( I 2 I1 ) 1( 2 1 ) + 4 = 0
4[2( I1
4V
IN
4Vx B
…(iii)
Fig. 3.351
I1 = 2 A
I 2 = 2 36 A
I N I 2 = 2 36 A
Calculation of RN
RN =
VTh
26
=
= 11.02 Ω
IN
2 36
A
IL
2.36 A
11.02 Ω
Step IV Calculation of IL (Fig. 3.352)
11.02
I L = 2 36 ×
= 1.62 A
11.02 + 5
5Ω
B
Fig. 3.352
3.4 Norton’s Theorem 3.89
Example 3.80
Find the current through the 1 W resistor in the network of Fig. 3.353.
6Ω
Ix
3Ω
3Ix
12 V
1Ω
Fig. 3.353
Solution
Step I Calculation of VTh (Fig. 3.354)
From Fig. 3.354,
6Ω
Ix
I1
1
2
0
+
+
…(i)
3Ω
3Ix
12 V
Meshes 1 and 2 will form a supermesh.
Writing the current equation for the supermesh,
I 2 I1 = 3I x 3I1
4
Ix
I1
I2
−
VTh
−
Fig. 3.354
…(ii)
Applying KVL to the outer path of the supermesh,
12 − 6 1 3I 2 = 0
6
33II 2 = 12
1
…(iii)
Solving Eqs (ii) and (iii),
I1 = 0 67 A
I 2 = 2 67 A
6Ω
Ix
A
Writing the VTh equation,
3
VTh
2
0
I = 3I x
3I1
4
0
1
2
IN
B
Step II Calculation of IN (Fig. 3.355)
When a short circuit is placed across a 3 Ω resistor, it
gets shorted as shown in Fig. 3.356.
From Fig. 3.356,
I x I1 …(i)
Meshes 1 and 2 will form a supermesh.
Writing the current equation for the supermesh,
I2
3Ω
3Ix
12 V
3( 2.67) − VTh = 0
VTh = 8 V
Fig. 3.355
6Ω
Ix
A
3Ix
12 V
I1
IN
I2
B
…(ii)
Fig. 3.356
Applying KVL to the outer path of the supermesh,
12 − 6
1
0
I1 = 2
…(iii)
3.90 Network Analysis and Synthesis
Solving Eqs (ii) and (iii),
I1 = 2 A
I2 = 8 A
IN
Step III
A
IL
I2 = 8 A
Calculation of RN
1Ω
1Ω
8A
V
8
RN = Th = = 1 Ω
IN
8
B
Step IV Calculation of IL (Fig. 3.357)
Fig. 3.357
1
IL = 8 ×
=4A
1+1
Example 3.81
Find the current through the 1.6 W resistor in the network of Fig. 3.358.
3Ix
− +
Ix
1Ω
10 A
1.6 Ω
6Ω
Fig. 3.358
Solution
Step I Calculation of VTh (Fig. 3.359)
From Fig. 3.359,
3Ix
− +
Ix
Ix
I1 − I 2 …(i)
I1
For Mesh 1,
I1 = 10
2
6Ω
I2
VTh
−
−
…(ii)
A
B
Fig. 3.359
Applying KVL to Mesh 2,
−1(
+
1Ω
10 A
+
− 1 ) + 3I x − 6
− I 2 + I1 + 3(
2
=0
1 − 2 ) − 6I2 = 0
4 1 10
0 2 0
Solving Eqs (ii) and (iii),
I1 = 10 A
I2 = 4 A
Writing the VTh equation,
6 2 VTh 0
6( 4) − VTh = 0
VTh = 24 V
…(iii)
3.5 Maximum Power Transfer Theorem 3.91
Step II Calculation of IN (Fig. 3.360)
When a short circuit is placed across the 3 Ω resistor,
it gets shorted as shown in Fig. 3.361.
From Fig. 3.361,
I1 − I 2
Ix
…(i)
3Ix
− +
A
Ix
1Ω
10 A
6Ω
IN
For Mesh 1,
I1 = 10
B
…(ii)
Fig. 3.360
Applying KVL to Mesh 2,
3Ix
−1( 2 − 1 ) + 3I x = 0
− I 2 + I1 + 3( 1 − 2 ) = 0
4 1 4I2 = 0
Solving Eqs (ii) and (iii),
I1 = 10 A
− +
…(iii)
1Ω
10 A
IN
I1
I2
B
I 2 = 10 A
I N I 2 = 10 A
Step III
A
Ix
Fig. 3.361
Calculation of RN
A
IL
V
24
RN = Th =
= 2.4 Ω
IN
10
2.4 Ω
10 A
1.6 Ω
Step IV Calculation of IL (Fig. 3.362)
I L = 10 ×
3.5
2.4
=6A
2.4 + 1.6
B
Fig. 3.362
MAXIMUM POWER TRANSFER THEOREM
It states that ‘the maximum power is delivered from a source to a load when the load resistance is equal to
the source resistance.’
Proof From Fig. 3.363,
RS
I=
Power delivered to the load RL
V
Rs
RL
2
P = I RL =
V
V 2 RL
( Rs + RL ) 2
To determine the value of RL for maximum power to be transferred
Fig. 3.363
to the load,
dP
=0
dR
RL
dP
d
V2
=
RL
dR
RL dRL ( Rs RL ) 2
=
V 2 [( Rs + RL ) 2
( 2 RL )( Rs
( Rs + RL )
4
RL
I
RL )]
Network illustrating
maximum power transfer
theorem
3.92 Network Analysis and Synthesis
) 2 − 2 RL (
(
Rs 2
)=0
RL 2 + 2 Rs RL − 2 RL Rs − 2 RL2 = 0
Rs
RL
Hence, the maximum power will be transferred to the load when load resistance is equal to the source
resistance.
Steps to be followed in Maximum Power Transfer
Theorem
1.
2.
3.
4.
RTh
A
Remove the variable load resistor RL.
Find the open circuit voltage VTh across points A and B.
Find the resistance RTh as seen from points A and B.
Find the resistance RL for maximum power transfer.
RL
RL = RTh
VTh
IL
B
RTh
Thevenin’s equivalent network
Fig. 3.364
5. Find the maximum power (Fig. 3.364).
IL =
VTh
V
= Th
RTh RL 2 RTh
I L2 RL =
Pmax
2
VTh
2
4 RTh
× RTh =
2
VTh
4 RTh
Example 3.82 Find the value of resistance RL in Fig. 3.365 for maximum power transfer and
calculate maximum power.
RL
2Ω
2Ω
3V
10 V
6V
2Ω
Fig. 3.365
Solution
Step I Calculation of VTh (Fig. 3.366)
Applying KVL to the mesh,
3 2I − 2
2Ω
+
6=0
I = −0.75 A
+
+
VTh
2Ω
3V
I
Writing the VTh equation,
2Ω
5.5 V
= 5.5 V( terminal B is positive w.r.t A)
B
−
10 V
−
6V
6 2 I − VTh − 10 = 0
VTh = 6 2 I − 10 = 6 2( −0.75) 10
A
−
Fig. 3.366
3.5 Maximum Power Transfer Theorem 3.93
Step II
2Ω
Calculation of RTh (Fig. 3.367)
RTh = (
A
RTh
B
) + 2 = 3Ω
2Ω
Step III Calculation of RL
For maximum power transfer,
RL
RTh = 3 Ω
2Ω
Step IV Calculation of Pmax (Fig. 3.368)
Fig. 3.367
3Ω
A
Pmax =
2
VTh
4 RTh
2
=
( .5)
= 2.52 W
4 3
3Ω
5.5 V
B
Fig. 3.368
Example 3.83 Find the value of resistance RL in Fig. 3.369 for maximum power transfer and
calculate maximum power.
5Ω
1Ω
RL
5Ω
4A
10 V
8V
Fig. 3.369
Solution
Step I Calculation of VTh (Fig. 3.370)
I 2 I1 = 4
Applying KVL to the outer path,
8 1I1 5 1 5 I 2 − 10 = 0
−6
6 1 5I 2 = 2
5Ω
−
...(i)
1Ω
+
8V
+
+ A
V Th
− B I1
4A
I2
5Ω
−
10 V
Fig. 3.370
I1 = −2 A
I2 = 2 A
5Ω
...(ii)
Writing the VTh equation,
Step II
−
...(ii)
Solving Eqs (i) and (ii),
8 1I1 − VTh = 0
VTh = 8 I1
+
1Ω
8 − ( −2
2) 10 V
A
R Th
B
Calculation of RTh (Fig. 3.371)
RTh = 10 1 = 0.91 Ω
Fig. 3.371
5Ω
3.94 Network Analysis and Synthesis
0.91 Ω
Step III Calculation of RL
For maximum power transfer,
A
RTh = 0 91 Ω
RL
0.91 Ω
10 V
Step IV Calculation of Pmax
Pmax =
2
VTh
( )2
=
= 27.47 W
4 RTh 4 0.91
B
Fig. 3.372
Example 3.84 Find the value of the resistance RL in Fig. 3.373 for maximum power transfer and
calculate the maximum power.
10 Ω
50 A
2Ω
5Ω
RL
3Ω
Fig. 3.373
10 Ω
Solution
+
Step I Calculation of VTh (Fig. 3.374)
For Mesh 1,
I1 = 50
2Ω
−
+
−
+ −
5Ω
50 A
I1
+
+
3Ω
I2
− +
VTh
−
−
Applying KVL to Mesh 2,
−5(
2
− 1 ) − 2I 2 − 3
5 1 − 10
10
2
Fig. 3.374
=0
2 =0
I1 = 2 I 2
I 2 = 25 A
10 Ω
Step II
2Ω
A
5Ω
VTh = 3I 2 = 3( 25) = 75 V
3Ω
RTh = ( + ) 3 = 2.1 Ω
Fig. 3.375
2.1 Ω
Step III Calculation of RL
For maximum power transfer,
RL
RTh = 2.1 Ω
A
2.1 Ω
75 V
Step IV Calculation of Pmax (Fig. 3.376)
Pmax =
2
VTh
( )2
=
= 669.64 W
4 RTh 4 2.1
RTh
B
Calculation of RTh (Fig. 3.375)
B
Fig. 3.376
A
B
3.5 Maximum Power Transfer Theorem 3.95
Example 3.85 Find the value of resistance RL in Fig. 3.377 for maximum power transfer and
calculate maximum power.
3Ω
5Ω
RL
2Ω
6A
10 V
4Ω
Fig. 3.377
3Ω
Solution
−
Step I Calculation of VTh (Fig. 3.378)
Writing the current equation for the supermesh,
I2
I1 = 6
5Ω
+
...(i)
10 V
+
+
I1
6A
I2
2Ω
1
2I 2 = 0
5
1
2 I 2 = 10
VTh
−
−
Applying KVL to the supermesh,
10 − 5
A
B
4Ω
...(ii)
Fig. 3.378
3Ω
Solving Eqs (i) and (ii),
I1 = −0 29 A
I 2 = 5 71 A
A
5Ω
2Ω
RTh
Writing the VTh equation,
2 I 2 = 11.42 V
VTh
Step II
B
4Ω
Calculation of RTh (Fig. 3.379)
RTh = (
Fig. 3.379
) + 3 + 4 = 8.43 Ω
Step III Calculation of RL
For maximum power transfer,
RL
8.43 Ω
A
RTh = 8 43 Ω
8.43 Ω
11.42 V
Step IV
Calculation of Pmax (Fig. 3.380)
Pmax =
2
VTh
( .42) 2
=
= 3.87 W
4 RTh 4 8.43
B
Fig. 3.380
Example 3.86 Find the value of resistance RL in Fig. 3.381 for maximum power transfer and
calculate the maximum power.
3.96 Network Analysis and Synthesis
10 Ω
RL
120 V
6A
5Ω
Fig. 3.381
10 Ω
Solution
Step I Calculation of VTh (Fig. 3.382)
Applying KVL to Mesh 1,
120 −10
10 I1 5(
1
2)
+
−
120 V
0
I1
15 I1 5 I 2 = 120
...(i)
Writing current equation for Mesh 2,
I 2 = −6
Solving Eqs (i) and (ii),
5Ω
− + I2
B −
6A
10 Ω
Writing the VTh equation,
A
R Th
B
120 −10
10 I1 − VTh = 0
Step II
+ −
V Th
Fig. 3.382
...(ii)
I1 = 6 A
VTh = 120 − 10(6)
A +
5Ω
60 V
Calculation of RTh (Fig. 3.383)
Fig. 3.383
RTh = 10 5 = 3.33 Ω
3.33 Ω
Step III Calculation of RL
For maximum power transfer,
RL
A
RTh = 3 33 Ω
3.33 Ω
60 V
Step IV Calculation of Pmax (Fig. 3.384)
Pmax =
2
VTh
( )2
=
= 270.27 W
4 RTh 4 3.33
B
Fig. 3.384
Example 3.87 Find the value of resistance RL in Fig. 3.385 for maximum power transfer and
calculate the maximum power.
10 Ω
RL
3A
20 V
Fig. 3.385
25 Ω
6Ω
3.5 Maximum Power Transfer Theorem 3.97
10 Ω
Solution
Step I
+
Calculation of VTh (Fig. 3.386)
I1 = 3
A +
V Th
B −
…(i)
Applying KVL to Mesh 2,
−25( 2 − 1 ) − 10 I 2 − 6 2 = 0
−25 I1 + 41I 2 = 0
−
+ −
25 Ω
3A
I1
20 V
…(ii)
+
− +
6Ω
−
I2
Fig. 3.386
Solving Eqs (i) and (ii),
10 Ω
I 2 = 1.83 A
Writing the VTh equation,
20 + VTh − 10 I 2
6 2 0
VTh = −20 +10
10 (1.83) 6 (1.83)
R Th
A
25 Ω
9.28 V
Calculation of RTh (Fig. 3.387)
Step II
6Ω
B
Fig. 3.387
RTh = 25 || (10 + 6) = 9.76 Ω
9.76 Ω
Step III Calculation of RL
For maximum power transfer,
RL RTh = 9 76 Ω
A
9.76 Ω
9.28 V
Step IV Calculation of Pmax (Fig. 3.388)
Pmax =
2
VTh
( .28) 2
=
= 2.21 W
4 RTh 4 × 9.76
B
Fig. 3.388
Example 3.88 Find the value of resistance RL in Fig. 3.389 for maximum power transfer and
calculate maximum power.
1Ω
2Ω
1A
5V
5Ω
RL
3Ω
10 Ω
Fig. 3.389
Solution
Step I
Calculation of VTh (Fig. 3.390)
1Ω
+
2Ω
−
5V
I1
+ −
1A
I2
+
5Ω
−
3Ω
10 Ω
− +
Fig. 3.390
+
+
I3
−
A
VTh
−
B
3.98 Network Analysis and Synthesis
Meshes 1 and 2 will form a supermesh.
Writing the current equation for the supermesh,
I2
I1 = 1
…(i)
Writing the voltage equation for the supermesh,
5 1I1 10
0( I 2 − I 3 ) = 0
…(ii)
I1 +10
10 I 2 − 10 I 3 = 5
Applying KVL to Mesh 3,
−10(
3
−
2) − 2 3
− 3I 3 = 0
−10 I 2 +15
15 I 3 = 0
…(iii)
Solving Eqs (i), (ii) and (iii),
I1 = 0 38 A
I 2 = 1 38 A
I 3 = 0 92 A
Writing the VTh equation,
VTh
Step II
3I 3 = 2.76 V
Calculation of RTh (Fig. 3.391)
1Ω
2Ω
5Ω
A
3Ω
10 Ω
RTh
B
(a)
2Ω
5Ω
5Ω
A
3Ω
0.91 Ω
RTh
A
1.48 Ω
RTh
B
B
(b)
(c)
Fig. 3.391
RTh = 6 8 Ω
6.48 Ω
A
Step III Calculation of RL
For maximum power transfer,
RL
RTh = 6 48 Ω
6.48 Ω
2.76 V
Step IV Calculation of Pmax (Fig. 3.392)
Pmax =
2
VTh
( .76) 2
=
= 0.29 W
4 RTh 4 × 6.48
B
Fig. 3.392
3.5 Maximum Power Transfer Theorem 3.99
Example 3.89 For the network shown in Fig. 3.393, find the value of the resistance RL for maximum power transfer and calculate the maximum power.
2Ω
RL
1Ω
2Ω
2A
8V
3A
6V
Fig. 3.393
Solution
Step I Calculation of VTh (Fig. 3.394)
2Ω
+
1Ω
−
+
A
−
+
VTh
B
−
2Ω
8V
I1
2A
I2
3A
6V
Fig. 3.394
I1 = 2
I 2 = −3 A
I2
Solving Eqs (i), and (ii),
I1 = −5 A
…(i)
…(ii)
2Ω
1Ω
RTh
A
Writing the VTh equation,
8 2 I1 1
Step II
2
VTh
B
2Ω
6=0
VTh = 8 2( −5
5) ( 3) − 6 15 V
Fig. 3.395
Calculation of RTh (Fig. 3.395)
RTh = 5 Ω
Step III Calculation of RL
For maximum power transfer,
RL
5Ω
A
RTh = 5 Ω
5Ω
15 V
Step IV Calculation of Pmax (Fig. 3.396)
Pmax =
2
VTh
( )2
=
= 11.25 W
4 RTh 4 × 5
B
Fig. 3.396
3.100 Network Analysis and Synthesis
Example 3.90 For the value of resistance RL in Fig. 3.397 for maximum power transfer and calculate the maximum power.
RL
15 Ω
5Ω
18 Ω
15 Ω
10 Ω
27 Ω
20 Ω
9Ω
27 Ω
100 V
Fig. 3.397
Solution
Step I Calculation of VTh (Fig. 3.398)
15 Ω
A
+
5Ω
18 Ω
B
VTh
−
15 Ω
10 Ω
27 Ω
20 Ω
9Ω
27 Ω
100 V
Fig. 3.398
By star-delta transformation (Fig. 3.399),
I=
A
100
= 2.08 A
5 + 5 + 20 + 9 + 9
VTh
= 100 − 14( 2.08)
= 70.88 V
B
−
9Ω
+
−
5Ω
9I = 0
VTh = 100 − 14 I
VTh
5Ω
+
Writing the VTh equation,
100 − 5
5Ω
+
9Ω
20 Ω
I
100 V
Fig. 3.399
9Ω
−
3.5 Maximum Power Transfer Theorem 3.101
Step II
Calculation of RTh (Fig. 3.400)
A
RTh
B
9Ω
5Ω
9Ω
5Ω
5Ω
9Ω
20 Ω
(a)
5Ω
A
RTh
B
5Ω
5Ω
20 Ω
5Ω
9Ω
A
RTh
9Ω
B
9Ω
14 Ω
9Ω
34 Ω
5Ω
A
RTh
B
9Ω
9.92 Ω
(c)
(b)
(d)
Fig. 3.400
RTh = 23.92 Ω
23.92 Ω
Step III Calculation of RL
For maximum power transfer,
RL
A
RTh = 23.92 Ω
23.92 Ω
70.88 V
Step IV Calculation of Pmax (Fig. 3.401)
Pmax =
2
VTh
( .88) 2
=
= 52.51 W
4 RTh 4 × 23.92
B
Fig. 3.401
Example 3.91 For the value of resistance RL in Fig. 3.402 for maximum power transfer and calculate the maximum power.
2A
5Ω
10 Ω
20 Ω
80 V
20 V
RL
Fig. 3.402
3.102 Network Analysis and Synthesis
Solution
Step I Calculation of VTh (Fig. 3.403)
Applying KVL to Mesh 1,
80 − 5
1
10
0(
1
2)
20
0(
1
2)
2A
5Ω
+
−
20 0
−
+
35 I1 30 I 2 = 60 …(i)
10 Ω
+
−
80 V
I1
Writing the current equation for Mesh 2,
I2 = 2
I2
A +
−
+
20 Ω
+
−
20 V
V Th
B −
…(ii)
Solving Eqs (i) and (ii),
Fig. 3.403
I1 = 3.43 A
5Ω
Writing the VTh equation,
VTh
10 Ω
20 ( I1 − I 2 ) − 20 = 0
VTh = 20(3.43 − 2) + 20 = 48.6 V
Step II
20 Ω
A
R Th
Calculation of RTh (Fig. 3.404)
B
RTh = 15 || 20 = 8 57 Ω
Fig. 3.404
Step III Calculation of RL
For maximum power transfer,
RL
8.57 Ω
A
RTh = 8 57 Ω
Pmax =
8.57 Ω
48.6 V
Step IV Calculation of Pmax (Fig. 3.405)
2
VTh
( .6) 2
=
= 68.9 W
4 RTh 4 × 8.57
B
Fig. 3.405
Example 3.92 For the value of resistance RL in Fig. 3.406 for maximum power transfer and calculate the maximum power.
10 Ω
20 Ω
100 V
RL
30 Ω
Fig. 3.406
40 Ω
3.5 Maximum Power Transfer Theorem 3.103
I2
I1
10 Ω +
−
Solution
Step I Calculation of VTh (Fig. 3.407)
100
=25A
10 + 30
100
I2 =
= 1 66 A
20 + 40
I1 =
+
100 V
A
+
30 Ω
Writing the VTh equation,
V Th
+
−
B
+
−
−
20 Ω
−
40 Ω
Fig. 3.407
VTh 10 I1 − 20 I 2 = 0
VTh 20 I 2 −10
− 10 I1 = 20(1.66) − 10( 2.5) = 8.2 V
Step II
Calculation of RTh (Fig. 3.408)
10 Ω
20 Ω
A
R Th
30 Ω
B
40 Ω
Fig. 3.408
Redrawing the network (Fig. 3.409),
10 Ω
20 Ω
A
RTh = (
||
)+(
||
B
) = 0.83 Ω
30 Ω
40 Ω
Fig. 3.409
Step III Value of RL
For maximum power transfer,
RL
20.83 Ω
A
RTh = 20.83 Ω
Step IV Calculation of Pmax (Fig. 3.410)
Pmax =
20.83 Ω
8.2 V
2
VTh
( .2) 2
=
= 0.81 W
4 RTh 4 × 20.83
B
Fig. 3.410
3.104 Network Analysis and Synthesis
Example 3.93 For the value of resistance RL in Fig. 3.411 for maximum power transfer and calculate the maximum power.
6Ω
RL
72 V
2Ω
3Ω
4Ω
Fig. 3.411
Solution
Step I Calculation of VTh (Fig. 3.412)
Applying KVL to Mesh 1,
72 − 6
1
3( I1 − I 2 ) = 0
9
1
72 V
3I 2 = 72 …(i)
+
I1
+
Applying KVL to Mesh 2,
−3(
2
A
6Ω +
−
3Ω
− 1 ) − 2I 2 − 4
2
=0
−3 1 + 9 I 2 = 0 …(ii)
2Ω
−
I2
−
Fig. 3.412
Solving Eqs (i) and (ii),
I1 = 9 A
I2 = 3 A
Writing the VTh equation,
VTh
Step II
6 I1 − 2 I 2 = 0
VTh 6 I1 + 2 I 2 = 6(9) + 2(3) = 60 V
Calculation of RTh (Fig. 3.413)
A
6Ω
2Ω
R Th
B
6Ω
A
3Ω
4Ω
2Ω
3Ω
R Th
4Ω
B
Fig. 3.413
RTh = [(6 || 3) + ] || 4 = 2 Ω
−
+
V Th
−
B
+
4Ω
3.5 Maximum Power Transfer Theorem 3.105
2Ω
Step III Calculation of RL
For maximum power transfer,
RL
A
RTh = 2 Ω
2Ω
60 V
Step IV Calculation of Pmax (Fig. 3.414)
Pmax =
2
VTh
( )2
=
= 450 W
4 RTh 4 × 2
B
Fig. 3.414
Example 3.94 For the network shown in Fig. 3.415 find the value of the resistance RL for maximum
power transfer and calculate maximum power.
10 Ω
5Ω
25 A
2Ω
10 A
RL
10 Ω
30 V
10 Ω
30 V
Fig. 3.415
Solution
Step I Calculation of VTh (Fig. 3.416)
10 Ω
5Ω
25 A
2Ω
+ A
V Th
10 A
− B
Fig. 3.416
By source transformation, the current source of 25 A and the 5 Ω resistor is converted into an equivalent
voltage source of 125 V and a series resistor of 5 Ω. Also the voltage source of 30 V is connected across the
10 Ω resistor. Hence, the 10 Ω resistor becomes redundant (Fig. 3.417).
2Ω
10 Ω
+
5Ω
10 A
A
V Th
−
B
125 V
Fig. 3.417
30 V
10 Ω
3.106 Network Analysis and Synthesis
10 Ω
Applying KCL at the node,
VTh 12
V − 30
− 10 + Th
=0
15
2
2Ω
A
5Ω
B
VTh = 58.81 V
Step II
10 Ω
R Th
Calculation of RTh (Fig. 3.418)
Fig. 3.418
RTh = 15 || 2 = 1.76 Ω
Step III Value of RL
For maximum power transfer
RL
A
R Th
15 Ω
2Ω
B
RTh = 1 76 Ω
Step IV Calculation of Pmax (Fig. 3.420)
Fig. 3.419
1.76 Ω
Pmax
A
1.76 Ω
58.81 V
V2
( .81) 2
= Th =
= 491.28 W
4 RTh 4 × 1.76
B
Fig. 3.420
Example 3.95 For the network shown in Fig. 3.421, find the value of the resistance RL for maximum power transfer and calculate maximum power.
12 V
2Ω
4Ω
10 V
2Ω
6V
5Ω
10 Ω
RL
4A
8V
Fig. 3.421
Solution
Step I Calculation of VTh (Fig. 3.422)
12 V
2Ω
+
2Ω
−
4Ω
6V
+
A
+
5Ω
10 Ω
4 A 10 V
I
V Th
−
8V
−
Fig. 3.422
B
3.5 Maximum Power Transfer Theorem 3.107
Applying KVL to the outer path,
10 − 2 I 12 5 I − 8
0
I=−
10
= −1.43 A
7
Writing the VTh equation,
8 5 I + 6 VTh 0
VTh = 8 6 + 5 I
Step II
8 + 6 5( −1
1.43)
6.85 V
Calculation of RTh (Fig. 3.423)
2Ω
2Ω
A
4Ω
10 Ω
RTh = ( || ) + 2
= 3.43 Ω
5Ω
R Th
B
Fig. 3.423
Step III Value of RL
For maximum power transfer,
3.43 Ω
RL
RTh = 3 43 Ω
Step IV Calculation of Pmax (Fig. 3.424)
Pmax =
A
3.43 Ω
6.85 V
2
VTh
( .85) 2
=
= 3.42 W
4 RTh 4 × 3.43
B
Fig. 3.424
EXAMPLES WITH DEPENDENT SOURCES
Example 3.96 For the network shown in Fig. 3.425, find the value of RL for maximum power transfer. Also, calculate maximum power.
I
20 Ω
10 I
40 Ω
+
−
RL
50 V
Fig. 3.425
3.108 Network Analysis and Synthesis
Solution
Step I Calculation of VTh (Fig. 3.426)
Applying KVL to the mesh,
10 I
20 I
I
20 Ω
40 I − 50 = 0
I = −1 A
VTh
+
−
10 I
Writing the VTh equation,
40 Ω
A +
V Th
B
−
50 V
40 I − 50 = 0
VTh − 40( −1) − 50 = 0
Fig. 3.426
VTh = 10 V
I
Step II Calculation of IN (Fig. 3.427)
From Fig. 3.427,
I
Applying KVL to Mesh 1,
10 I
10 I 2
I2
20 I1 = 0
20 I1 = 0
20 Ω
…(i)
+
−
10I
IN
B
I1
−40 I 2 − 50 = 0
I 2 = −1.25
25 A
Solving Eqs (i), (ii) and (iii),
I1 = −0.625 A
I N I1 − I 2 = −0.625 + 1.25 = 0.625 A
Step III Calculation of RN
V
10
RTh = Th =
= 16 Ω
IN
0.625
Step IV Calculation of RL
For maximum power transfer,
RL
50 V
Fig. 3.427
…(iii)
16 Ω
A
RTh = 16 Ω
16 Ω
10 V
Calculation of Pmax (Fig. 3.428)
Pmax =
I2
…(ii)
Applying KVL to Mesh 2,
Step V
40 Ω
A
VTh
( )2
=
= 1.56 W
4 RTh 4 × 16
B
Fig. 3.428
Example 3.97 For the network shown in Fig. 3.429, calculate the maximum power that may be
dissipated in the load resistor RL.
− +
3Ω
2Ix
Ix
10 A
4Ω
Fig. 3.429
6Ω
RL
3.5 Maximum Power Transfer Theorem 3.109
2Ix
Solution
Step I Calculation of VTh (Fig. 3.430)
From Fig. 3.430,
Ix
3Ω
− +
+
…(i)
I2
6Ω
4Ω
10 A
I1
+
Ix
I2
For Mesh 1,
V Th
−
−
I1 = 10
…(ii)
A
B
Fig. 3.430
Applying KVL to Mesh 2,
−4( 2 − 1 ) + 2 I x − 6 2 = 0
−4 2 + 4 I1 + 2 2 − 6 I 2 = 0
4
8II 2 = 0
8
1
…(iii)
Solving Eqs (ii) and (iii),
I1 = 10 A
I2 = 5 A
Writing the VTh equation,
0 VTh = 0
6I2
VTh
6I2
6(5)
30 V
Step II Calculation of IN (Fig. 3.431)
From Fig. 3.431,
I x I 2 − I3
For Mesh 1,
I1 = 10
2Ix
6Ω
4Ω
10 A
…(ii)
−4
2
2
I1
I2
2
−
−
3)
=0
+6
3
=0
8I 2 + 4
3
0
2 ) − 3I 3
=0
2
3 ) − 6I2
4
IN
I3
B
− ) + 2 I x − 6(
+ 4 I + 2(
A
Ix
…(i)
Applying KVL to Mesh 2,
−4(
3Ω
− +
1
Fig. 3.431
…(iii)
Applying KVL to Mesh 3,
−6(
3
−
6
2
9II 3 = 0
9
Solving Eqs (ii), (iii) and (iv),
I1 = 10 A
I2 = 7 5 A
I3 = 5 A
IN
I3 = 5 A
…(iv)
3.110 Network Analysis and Synthesis
Step III
6Ω
Calculation of RTh
A
RTh
V
30
= Th =
=6Ω
IN
5
6Ω
30 V
Step IV Calculation of RL
For maximum power transfer,
B
RTh = 6 Ω
RL
Step V
Fig. 3.432
Calculation of Pmax
Pmax =
2
VTh
( )2
=
= 37.5 W
4 RTh 4 × 16
Example 3.98
For the network shown in Fig. 3.433, find the value of RL for maximum power
transfer. Also, find maximum power.
1Ω
−
2Vx
Vx
2V
+
+
−
1Ω
RL
1A
Fig. 3.433
Solution
Step I Calculation of VTh (Fig. 3.434)
From Fig. 3.434,
Vx
1I = − I
For Mesh 1,
I = −1
Vx = 1 V
Writing the VTh equation,
2Vx 1I + 2 VTh
1Ω
−
…(i)
2Vx
Vx
2V
+
+
A
1Ω
+
−
V Th
1A
I
…(ii)
−
B
Fig. 3.434
0
2(1) − ( −1) + 2 − VTh = 0
VTh = 5 V
Step II Calculation of IN (Fig. 3.435)
From Fig. 3.435,
Vx
1I1 = − I1
Meshes 1 and 2 will form a supermesh.
Writing the current equation for the supermesh,
I 2 I1 = 1
Applying KVL to the outer path of the supermesh,
2Vx 1I1 + 2 0
2( 1 ) 1 + 2 = 0
31 0
1Ω
−
…(i)
2Vx
…(ii)
+
−
Vx
2V
A
+
1Ω
IN
I1
1A
I2
B
Fig. 3.435
…(iii)
3.5 Maximum Power Transfer Theorem 3.111
Solving Eqs (ii) and (iii),
I1 = 0 67 A
I 2 = 1 67 A
I N I 2 = 1 67 A
Step III
Calculation of RTh
RTh =
VTh
5
=
=3Ω
I N 1 67
Step IV Calculation of RL
For maximum power transfer,
3Ω
A
RL
Step V
RTh = 3 Ω
Pmax =
Example 3.99
3Ω
5V
Calculation of Pmax ( Fig. 3.436)
2
VTh
( )2
=
= 2.08 W
4 RTh 4 × 3
B
Fig. 3.436
What will be the value of RL in Fig. 3.437 to get maximum power delivered to it?
What is the value of this power?
0.5 V
− +
4Ω
3A
+
4Ω
RL
V
−
Fig. 3.437
Solution
Step I Calculation of VTh (Fig. 3.438)
By source transformation,
From Fig. 3.438,
VTh 4 I
0.5 VTh
− +
4Ω
3A
+
4Ω
VTh
−
Applying KVL to the mesh,
12 − 4 I
0 5 VTh
B
Fig. 3.438
4I = 0
0.5 VTh
12 − VTh + 0 5 VTh − VTh = 0
− +
VTh = 8 V
Step II Calculation of IN (Fig. 3.439)
If two terminals A and B are shorted, the 4 Ω resistor gets
shorted.
=0
A
4Ω
12 V
+
4Ω
A
VTh
I
−
Fig. 3.439
B
3.112 Network Analysis and Synthesis
Dependent source 0.5 V depends on the controlling variable V . When V = 0, the dependent source vanishes,
ie. 0.5 V = 0 as shown in Fig. 3.441.
12
IN =
=3A
4
4Ω
0.5 V
− +
4Ω
4Ω
A
+
IN
V
B
−
A
12 V
IN
12 V
B
Fig. 3.440
Step III
Fig. 3.441
2.67 Ω
Calculation of RTh
RTh
V
8
= Th = = 2 67 Ω
IN
3
A
2.67 Ω
8V
IL
Step IV Calculation of RL
For maximum power transfer,
RL RTh = 2 67 Ω
B
Fig. 3.442
Step V Calculation of Pmax (Fig. 3.442)
Pmax =
3.6
2
VTh
( )2
=
=6W
4 RTh 4 × 2.67
RECIPROCITY THEOREM
It states that ‘in a linear, bilateral, active, single source network, the ratio of excitation to response remains
same when the positions of excitation and response are interchanged.’
In other words, it may be stated as ‘if a single voltage
source Va in the branch ‘a’ produces a current Ib in the branch
+
N t
k
I
‘b’ then if the voltage source Va is removed and inserted in V −
the branch ‘b’, it will produce a current Ib in branch ‘a’’.
Explanation Consider a network shown in Fig. 3.443.
Fig. 3.443 Network
When the voltage source V is applied at the port 1, it
produces a current I at the port 2. If the positions of the
excitation (source) and response are interchanged, i.e., if
+
I
Network
− V
the voltage source is applied at the port 2 then it produces a
current I at the port 1.
The limitation of this theorem is that it is applicable only
Fig. 3.444 Network when excitation
to a single-source network. This theorem is not applicable in
and response are
the network which has a dependent source. This is applicable
interchanged
only in linear and bilateral networks. In the reciprocity
theorem, position of any passive element (R, L, C) do not change. Only the excitation and response are
interchanged.
3.6 Reciprocity Theorem 3.113
Steps to be followed in Reciprocity Theorem
1. Identify the branches between which reciprocity is to be established.
2. Find the current in the branch when excitation and response are not interchanged.
3. Find the current in the branch when excitation and response are interchanged.
Example 3.100
Calculate current I and verify the reciprocity theorem for the network shown in
Fig. 3.445.
5Ω
4Ω
I
10 Ω
20 V
6Ω
Fig. 3.445
Solution
Case I Calculation of current I when excitation and
response are not interchanged (Fig. 3.446)
Applying KVL to Mesh 1,
20 − 5
10 1 2 ) 0
10(
15 I1 10 I 2 = 20
Applying KVL to Mesh 2,
−10(
5Ω
4Ω
I
10 Ω
20 V
I1
6Ω
I2
1
…(i)
Fig. 3.446
− 1) − 4 2 − 6I2 = 0
−10 I1 + 20 I 2 = 0
Solving. Eqs (i) and (ii),
I1 = 2 A
I2 = 1 A
I I2 = 1 A
2
…(ii)
Case II Calculation of current I when excitation and
response are interchanged (Fig. 3.447).
Applying KVL to Mesh 1,
−5 1 − 10
10(
0 1 − 2) = 0
15 I1 10 I 2 = 0
5Ω
4Ω
20 V
I
10 Ω
…(i)
I1
I2
6Ω
Applying KVL to Mesh 2,
−10(
2
− 1) − 4 2 − 2
20
0 − 6I2 = 0
−10 I1 + 200 I 2 = −20
Fig. 3.447
Solving Eqs (i) and (ii),
I1 = −1 A
I 2 = −1 5 A
I
I1 = 1 A
Since the current I remains the same in both the cases, reciprocity theorem is verified.
…(ii)
3.114 Network Analysis and Synthesis
Example 3.101
Find the current I and verify reciprocity theorem for the network shown in Fig. 3.446.
2Ω
2Ω
3Ω
5V
3Ω
4Ω
I
4Ω
Fig. 3.448
Solution
Case I Calculation of the current I when excitation and
response are not interchanged (Fig. 3.449)
Applying KVL to Mesh 1,
5V
4( I1 − I 3 ) = 0
9 1 3I 2 − 4 3 5
Applying KVL to Mesh 2,
5 2 I1 3(
−3(
2)
1
2
− 1 ) − 2I 2 − 3
2
2Ω
…(i)
2Ω
3Ω
I1
4Ω
=0
I3
−3 1 + 8
8I 2 = 0
Applying KVL to Mesh 3,
−4( 3 − 1 ) − 4 I 3 = 0
…(ii)
3Ω
I2
I
4Ω
Fig. 3.449
−4 1 + 8
8I 3 = 0
…(iii)
Solving Eqs (i), (ii) and (iii),
I1 = 0 85 A
I 2 = 0 32 A
I 3 = 0 43 A
I
I 3 = 0 43 A
Case II Calculation of current I when excitation and
response are interchanged (Fig. 3.450).
Applying KVL to Mesh 1,
−2 1 − 3( I1 − I 2 ) − 4(
9
1
−
3I 2 − 4
1
3)
=0
3
0
2Ω
I
2
− 1 ) − 2I 2 − 3
3Ω
I1
4Ω
I2
…(i)
Applying KVL to Mesh 2,
−3(
2Ω
I3
2
=0
−3 1 + 8
8I 2 = 0
…(ii)
4Ω
5V
Fig. 3.450
3Ω
3.6 Reciprocity Theorem 3.115
Applying KVL to Mesh 3,
−4(
3
− 1) + 5 − 4
3
=0
−4 1 + 88I 3 = 5
…(iii)
Solving Eqs (i), (ii) and (iii),
I1 = 0 43 A
I 2 = 0 16 A
I 3 = 0 84 A
I
I1 = 0 43 A
Since the current I remains the same in both the cases, reciprocity theorem is verified.
Example 3.102
Find the voltage V and verify reciprocity theorem for the network shown in
Fig. 3.451.
4Ω
2Ω
5Ω
−
10 A
V
+
6Ω
8Ω
Fig. 3.451
Solution
Case I Calculation of the voltage V when excitation and
response are not interchanged (Fig. 3.451)
For Mesh 1,
I1 = 10
…(i)
Applying KVL to Mesh 2,
−4(
2
− 1 ) − 2 I 2 − 5(
2
−
−4 1 + 1
11
1
2
− 5I 3 = 0
3)
4Ω
5Ω
−
10 A
I1
6Ω
=0
…(ii)
2Ω
I2
V
I3
+
8Ω
Fig. 3.452
Applying KVL to Mesh 3,
−6(
3
− 1 ) − 5(
3
−
2) −8 3
=0
−6 1 − 5 I 2 + 19 I 3 = 0
…(iii)
Solving Eqs (i), (ii) and (iii),
I1 = 10 A
I 2 = 5 76 A
I 3 = 4 67 A
V
( I 2 I3 )
5(5.76 4.67) = 5 45 V
3.116 Network Analysis and Synthesis
Case II Calculation of voltage V when excitation and response are interchanged (Fig. 3.453).
+
4Ω
2Ω
5Ω
V
6Ω
10 A
8Ω
−
Fig. 3.453
By source transformation (Fig. 3.454),
Applying KVL to Mesh 1,
+
−4 1 − 2 I1 − 50 − 5( 1 − 2 ) = 0
11I1 5 2
50 …(i)
Applying KVL to Mesh 2,
−6
2
− 5( I 2 − I1 ) + 50 − 8
−5 1 + 1
19
9
2
2
4Ω
6Ω
…(ii)
50 V
V
=0
= 50
2Ω
I1
5Ω
I2
8Ω
−
Solving Eqs (i) and (ii),
Fig. 3.454
I1 = −3 8 A
I 2 = 1 63 A
From Fig. 3.454,
V
4 I1 + 6 I 2 = 0
V + 4( −3.8) + 6(1.63) = 0
V = 5.42
42 V
Since the voltage V is same in both the cases, the reciprocity theorem is verified.
3.7
MILLMAN’S THEOREM
It states that ‘if there are n voltage sources V1 V2 ,… ,Vn with internal resistances R1 R2 ,… , Rn respectively
connected in parallel then these voltage sources can be replaced by a single voltage source Vm and a single
series resistance Rm, ’(Fig. 3.455).
A
R1
V1
R2
Rn
V2
Vn
A
⇒
Rm
Vm
B
Fig. 3.455 Millman’s network
where
Vm =
V1G1 + V2G2 + … + VnGn
G1 G2 + … + Gn
B
3.7 Millman’s Theorem 3.117
Rm =
and
1
1
=
Gm G1 + G
Gn
Explanation By source transformation, each voltage source in series with a resistance can be converted
to a current source in parallel with a resistance as shown in Fig. 3.456.
A
I1
R1
I2
R2
In
Rn
B
Fig. 3.456 Equivalent network
Let Im be the resultant current of the parallel current sources and Rm
be the equivalent resistance as shown in Fig. 3.457.
Im
Im
Rm
V V
V
I n = 1 + 2 + … + n = V1 G1 + V2 G2 + …Vn Gn
R1 R2
Rn
I +I
1
1
1
1
=
+
+…+
Rm R1 R2
Rn
Gm G1 + G2
Gn
Fig. 3.457 Equivalent network
A
Rm
By source transformation, the parallel circuit can be converted
into a series circuit as shown in Fig. 3.458.
I
V G + V G + … + Vn Gn
Vm I m Rm = m = 1 1 2 2
Gm
G1 G2 + … + Gn
Vm
B
Fig. 3.458 Millman’s equivalent
network
Dual of Millman’s Theorem
It states that ‘if there are n current sources I 1 I 2 ,… , I n with internal resistances R1 R2 ,… , Rn respectively,
connected in series then these current sources can be replaced by a single current source Im and a single
parallel resistance Rm ’ (Fig. 3.459).
I1
I2
In
Im
⇒
R1
R2
Rn
A
Rm
B
Fig. 3.459 Millman’s network
where
Im =
Rm
I1 R1 + I 2 R2 + … + I n Rn
R1 + R2 + … + Rn
R +R
Rn
A
B
3.118 Network Analysis and Synthesis
Steps to be followed in Millman’s Theorem
1. Remove the load resistance RL.
2. Find Millman’s voltage across points A and B.
Vm =
V1 G1 + V2 G2 + … + Vn Gn
G1 G2 + … + Gn
3. Find the resistance Rm between points A and B.
Rm =
G1
1
G2 + … + Gn
4. Replace the network by a voltage source Vm in series with the resistance Rm.
5. Find the current through RL using ohm’s law.
IL =
Example 3.103
Vm
Rm
RL
Find Millman’s equivalent for the left of the terminals A-B in Fig. 3.460.
A
4Ω
2Ω
1A
6V
B
Fig. 3.460
A
Solution By source transformation, the network is redrawn
as shown in Fig. 3.461.
Step I Calculation of Vm
⎛ 1⎞
⎛ 1⎞
6⎜ ⎟ + 2⎜ ⎟
⎝ 4⎠
⎝ 2⎠
V G +V G
Vm = 1 1 2 2 =
= 3.33
33 V
1 1
G1 G2
+
4 2
Step II
4Ω
2Ω
6V
2V
B
Fig. 3.461
A
Calculation of Rm
Rm =
1
1
1
=
=
= 1.33 Ω
1
1
Gm G1 + G2
+
4 2
Step III Millman’s Equivalent Network (Fig. 3.462)
1.33 Ω
3.33 V
B
Fig. 3.462
3.7 Millman’s Theorem 3.119
Example 3.104
Find the current through the 10 W resistor in the network of Fig. 3.463.
2Ω
4Ω
6Ω
5V
10 V
15 V
10 Ω
Fig. 3.463
Solution
Step I
Calculation of Vm
⎛ 1⎞
⎛ 1⎞
⎛ 1⎞
5 ⎜ ⎟ − 10 ⎜ ⎟ + 15 ⎜ ⎟
⎝
⎠
⎝
⎠
⎝ 6⎠
V G +V G +V G
2
4
Vm = 1 1 2 2 3 3 =
= 2.73
73 V
1 1 1
G1 G2 + G3
+ +
2 4 6
Step II
Calculation of Rm
1
1
=
= 1.09 Ω
Gm 1 1 1
+ +
2 4 6
Calculation of IL (Fig. 3.464)
Rm =
Step III
IL =
Example 3.105
1.09 Ω
2.73 V
2 73
= 0.25
25 A
1 09 + 10
10 Ω
IL
Fig. 3.464
Find the current through the 10 W resistor in the network of Fig. 3.465.
50 Ω
2Ω
50 V
40 Ω
10 Ω
100 V
Fig. 3.465
Solution
Since the 40 Ω resistor is connected in parallel with the 50 V
source, it becomes redundant. The network can be redrawn
as shown in Fig. 3.466.
Step I Calculation of Vm
V G +V G
Vm = 1 1 2 2 =
G1 G2
⎛ 1⎞
⎛ 1⎞
50 ⎜ ⎟ − 100 ⎜ ⎟
⎝ 50 ⎠
⎝ 20 ⎠
= −57.15
15 V
1
1
+
50 20
50 Ω
20 Ω
10 Ω
50 V
40 Ω
100 V
Fig. 3.466
3.120 Network Analysis and Synthesis
Step II
Calculation of Rm
1
1
1
=
=
= 14.29 Ω
1
1
Gm G1 + G2
+
50 20
Calculation of IL (Fig. 3.467)
Rm =
Step III
IL =
Example 3.106
IL
14.29 Ω
57.15 V
57.15
= 2.35
35 A
14.29 + 10
Fig. 3.467
Draw Millman’s equivalent network across terminals AB in the network of
Fig. 3.468.
C
2Ω
4A
8V
6V
2Ω
1Ω
6V
2Ω
3V
1Ω
A
B
D
5V
5Ω
Fig. 3.468
Step I
By source transformation, the network is redrawn as shown in Fig. 3.469.
C
8V
8V
6V
2Ω
2Ω
1Ω
6V
2Ω
3V
1Ω
A
B
D
5V
5Ω
Fig. 3.469
Step II
10 Ω
Applying Millman’s theorem at terminals CD,
Vm1
V G +V G +V G
= 1 1 2 2 3 3 =
G1 G2 + G3
⎛ 1⎞
⎛ 1⎞
−8 ⎜ ⎟ + 8 ⎜ ⎟ + 6(1)
⎝ 2⎠
⎝ 2⎠
=3V
1 1
+ +1
2 2
3.8 Tellegen’s Theorem 3.121
1
1
1
=
=
= 0.5 Ω
Gm1 G1 + G2 G3 1 1
+ +1
2 2
Applying Millman’s theorem at terminals CA,
4V
Rm1 =
Step III
Vm2
Rm2
A
3V
0.5 Ω
⎛ 1⎞
6 ⎜ ⎟ + 3(1)
⎝ 2⎠
V4 G4 + V5 G5
=
=
=4V
1
G4 G5
+1
2
1
1
1
=
=
=
= 0.67 Ω
Gm2 G4 + G4 1
+1
2
B
5V
5Ω
Fig. 3.470
A
6.17 Ω
Step IV Millman’s Equivalent Network (Fig. 3.470)
Simplifying Fig. 3.470 further, the Millman’s equivalent
network is shown in fig. 3.471.
3.8
0.67 Ω
2V
B
Fig. 3.471
TELLEGEN’S THEOREM
It states that ‘the algebraic sum of the powers in all branches of the network at any instant is zero’.
n
∑ Vk
Ik = 0
k =1
This condition is valid for the network which obeys kirchhoff’s voltage and current laws.
Explanation
Consider a network shown in Fig. 3.472.
I3
+
V3
−
I2
I1
V1
I5
+
+
−
V2
−
I4 + V −
5
+
V4
−
I6
+
V6
−
Fig. 3.472 Network illustrating Tellegen’s theorem
According to Tellegen’s theorem,
6
∑ Vk
I k = V1 I1 + V2 I 2 + V3 I 3 + V4 I 4 + V5 I 5 + V6 I 6
k =1
= V1 I1 + (V1 V4 ) I 2 (V1 − V6 ) I 3 + V4 I 4 + (V4 − V6 ) I 5 + V6 I 6
= V1 ( I1 + I 2 + I 3 ) + V4 ( − I 2 + I 4 + I 5 ) + V6 (−
− I3 − I5 + I6 )
Applying KCL at Node 1,
I1 I 2 + I 3 = 0
Applying KCL at Node 4,
I 2 I 4 + I5
3.122 Network Analysis and Synthesis
Applying KCL at Node 6,
I3 = I6
I5
6
∑ Vk
I k = V1 (0) V4 (0) +V
V6 (0) = 0
k =1
Hence, Tellegen’s theorem is verified.
Example 3.107
Determine the power supplied by the source in the network of Fig. 3.473.
I5
+
V5
I1
−
I3
+
V1
−
V
I2 + V −
3
+
I4
+
V2
V4
−
−
V1
V2
5
10
I1 2 A
I2 = 0 8 A
V3
V4
3
7
I3
I4
6A
3A
5
4A
V5 = 8 V,
Fig. 3.473
Solution By Tellegen’s theorem,
Power supplied by the source = V1 I1 + V2 I 2 + V3 I 3 + V4 I 4 + V5 I 5
= 5 × 2 + 10 × 0.8 + 3 × 6 + 7 × 3 + 8 × 4 = 89 W
Example 3.108
Prove the Tellegen’s theorem for the network shown in Fig. 3.474.
I6
+
+
I1
−
III
2 I
4
I2
1
V6
V2
−
+
I3
+
V1
V4
−
V3
−
I
II
3
I5
+
V5
−
Fig. 3.474
V1
I1
8
4A
V2
I2
4 V,
2 A,
Solution Applying KVL to Mesh 1,
V1 V2 − V3 = 0
V3 V1 − V2 = 8 4 = 4 V
Applying KVL to Mesh 2,
V3 V4 − V5 = 0
V5 V3 − V4 = 4 2 = 2 V
V4 = 2 V
I3 = 1 A
3.8 Tellegen’s Theorem 3.123
Applying KVL to Mesh 3,
−V6 + V4 + V2 = 0
V6 = V2 + V4 = 4 + 2 = 6 V
Applying KCL at Node 1,
I1
I 2 + I6 = 0
I6
I1 − I 2 = −44 2 = −6 A
Applying KCL at Node 2,
I2
I4
I3 + I 4
I 2 − I3 = 2 1 = 1 A
Applying KCL at Node 3,
I5
I 4 + I 6 = 1 ( 6) = −5 A
6
∑ Vb I b = V1 I1 + V2 I 2 + V3 I 3 + V4 I 4 + V5 I 5 + V6 I 6 = 8 × 4 + 4 × 2 + 4 × 1 + 2 × 1 + 2 × (
)+6×(
)=0
b=1
Hence, Tellegen’s theorem is verified.
Example 3.109
Find the value of source V2 in Fig. 3.475, using Tellegen’ theorem if the power
absorbed by V2 is 20 W.
10 Ω
5Ω
10 Ω
100 V
5Ω
V2
Fig. 3.475
Solution Finding Thevenin’s equivalent network across voltage source V2 (Fig. 3.476).
10 Ω
10 Ω
5Ω
5Ω
5Ω
10 Ω
100 V
10 Ω
+ A
VTh
− B
I1
Fig. 3.477
Fig. 3.476
I1 =
VTh
100
=5A
10 + 10
10 I1 = 10(5)
5Ω
RTh = (
50 V
) + 5 + 5 = 15 Ω
A
RTh
B
3.124 Network Analysis and Synthesis
Thevenin’s equivalent network is shown in Fig. 3.478.
I
15 Ω
A
V2 I = 20
Applying Tellegen’s theorem,
50 V
V2
2
−50 I +15
15 I + V2 I = 0
B
15 I 2 − 50 I + 20 = 0
I = 2 87 or 0.46 A
20
20
V2 =
=
= 6 97 V
I
2 87
20
V2 =
= 43.48 V
0 46
or
3.9
Fig. 3.478
SUBSTITUTION THEOREM
It states that ‘any branch in a network can be substituted by a different branch without disturbing the voltages
and currents in the entire network provided the new branch has the same set of terminal voltage and current
as the original network.’
Explanation Any branch can be replaced by any combination of elements that will have the same voltage
across it and same current through it as the original branch (Fig. 3.479).
A
A
A
+
VAB
⇒
VAB
⇒
IAB
−
IAB
B
B
(a) Branch of a
(b) Branch replaced
by a voltage
network whose VAB
and IAB are
source VAB
known
B
(c) Branch replaced
by a current
source IAB
Fig. 3.479 Substitution theorem
Steps to be followed in Substitution Theorem
1. Find the branch voltage and branch current in the network.
2. Replace the branch by an independent voltage source and find the voltage and current in the network.
3. Replace the branch by an independent current source and find the voltage and current in the
network.
Example 3.110 In the network of Fig. 3.480 verify substitution theorem by replacing the 6 W resistor by (a) a voltage source, and (b) a current source.
3.9 Substitution Theorem 3.125
2Ω
4Ω
6Ω
8Ω
100 V
Fig. 3.480
Solution
Step I Calculation of current through 6 Ω resistor (Fig.
3.481)
Applying KVL to Mesh 1,
100 − 2 1 8( I1 − I 2 ) = 0
10 I1 8 I 2 = 100
…(i)
Applying KVL to Mesh 2,
−8(
2
− 1) − 4I2 − 6
2
=0
−8 1 + 1
18
8
2
=0
2Ω
4Ω
6Ω
8Ω
100 V
I1
I2
Fig. 3.481
…(ii)
Solving Eqs (i) and (ii),
I1 = 15.52 A
I2 = 6 9 A
I6
V6
I2 = 6 9 A
6 I 2 6(6.9) = 41.4 V
Step II When the 6 Ω resistor is replaced by the voltage
source of 41.4 V (Fig. 3.482)
Applying KVL to Mesh 1,
100 − 2 1 8( I1 − I 2 ) = 0
10 I1 8 I 2 = 100
…(i)
Applying KVL to Mesh 2,
−8(
2
2Ω
8Ω
100 V
I1
− 1 ) − 4 I 2 − 41.4 = 0
−8 1 + 1
12
2
2
4Ω
41.4 V
I2
Fig. 3.482
= −41.4 …(ii)
Solving Eqs (i) and (ii),
I1 = 51.51 A
I2 = 6 9 A
By replacing the 6 Ω resistor by a voltage source has not affected the currents I1 and I2. Hence, the
substitution theorem is verified.
2Ω
4Ω
Step III When the 6 Ω resistor is replaced by current
source of 6.9 A (Fig. 3.483)
Applying KVL to Mesh 1,
100 − 2 1 8( I1 − I 2 ) = 0
10 I1 8 I 2 = 100
8Ω
100 V
I1
6.9 A
I2
…(i)
Fig. 3.483
3.126 Network Analysis and Synthesis
For Mesh 2,
I2 = 6 9
…(ii)
Solving Eqs (i) and (ii),
I1 = 15.52 A
I2 = 6 9 A
By replacing the 6 Ω resistor by a current source has not affected the currents I1 and I2. Hence, the
substitution theorem is verified.
3.10
COMPENSATION THEOREM
It states that ‘in any linear bilateral active network, if any branch carrying a current I has its resistance R changed
by an amount d R, the resulting changes that occur in the other branches are the same as those which would have
been produced by an opposing voltage source of value Vc (Id R) introduced into the modified branch.’
Explanation
Consider a network shown in Fig. 3.484 (a), having load resistance R.
RTh
RTh
R
VTh
R
VTh
I
(a)
I′
dR
(b)
Fig. 3.484 Network illustrating compensation theorem
I=
VTh
RTh
R
If the load resistance R be changed to R δ R as shown in Fig. 3.484(b) then the current flowing in the circuit
is
I′ =
RTh
VTh
+ R+δR
The change in the current is
δI
VTh
VTh
−
RTh + R + δ R RTh R
VTh δ R
=−
( RTh + R + δ R) ( RTh + R)
I′ − I =
VTh
δR
RTh + R RTh + R + δ R
I δR
=−
RTh + R + δ R
Vc
=−
RTh + R + δ R
=−
where Vc
I δ R and is called the compensation voltage.
3.10 Compensation Theorem 3.127
d I has the same direction as I. This shows that the change in
current d I due to a change in any branch in a linear network can be
calculated by determining the current in that branch in a network
obtained from the original network by removing all the independent
sources and placing a voltage source called compensation source
in series with the branch whose value is Vc I δ R, whose I is the
current through the branch before its resistance is changed and d R
is the change in resistance. (Fig. 3.485).
Example 3.111
RTh
R +dR
δI
Vc = Id R
Fig. 3.485 Compensation network
In the network of Fig. 3.486, the resistance of 4 W is changed to 2 W . Verify the
compensation theorem.
1Ω
4Ω
1V
8Ω
Fig. 3.486
Solution
Step I Calculation of change in current
Finding Thevenin’s equivalent network across the 4 Ω
resistor,
8
VTh = 1 ×
= 0 89 V
8 +1
1× 8
RTh =
= 0 89 Ω
1+ 8
Thevenin’s equivalent network is shown in Fig. 3.487.
0 89
IL =
= 0 18 A
0 89 + 4
When resistance of 4 Ω is changed to 2 Ω,
0 89
I L′ =
= 0 31 A
0 89 + 2
δ I L I L′ − I L = 0.31 − 0.18 = 0.13 A
Step II Calculation of change of current by
compensation theorem
δ R = 2 − 4 = −2 Ω
R δR = 4 − 2 = 2 Ω
Vc I L δ R = 0.18 × ( −2) = −0.36 V
The compensating network is shown in Fig. 3.489.
0 36
δ IL =
= 0.12
12 A
0 89 + 2
Since change in current is same in both the steps, the compensation
theorem is verified.
0.89 Ω
4Ω
0.89 V
IL
Fig. 3.487
0.89 Ω
0.89 V
IL′
2Ω
Fig. 3.488
0.89 Ω
2Ω
δIL
Fig. 3.489
0.36 V
3.128 Network Analysis and Synthesis
Exercises
3.4
Superposition Theorem
3.1
Find the current through the 10 Ω resistor in
Fig. 3.490.
10 Ω
10 Ω
30 Ω
5Ω
10 A
Calculate the current through the 10 Ω
resistor in Fig. 3.493.
4Ω
20 Ω
100 V
25 V
2Ω
[0.37 A ]
Find the current through the 8 Ω resistor in
Fig. 3.491.
12 Ω
30 Ω
3Ω
[1.62 A ]
3.5
Find the current through the 1Ω resistor in
Fig. 3.494.
2Ω
25 A
1Ω
3Ω
1A
2Ω
1V
Fig. 3.491
[16.2 A ]
3.3
2Ω
Find the potential across the 3 Ω resistor in
Fig. 3.492.
Fig. 3.494
[0.41 A ]
3.6
9Ω
Find the current through the 4 Ω resistor in
Fig. 3.495.
15 A
7Ω
2Ω
5Ω
5A
12 V
Fig. 3.493
8Ω
5A
−
+
7Ω
Fig. 3.490
3.2
2Ω
+
−
3Ω
2Ω
4Ω
10 V
2Ω
2Ω
2Ω
2Ω
5A
6V
4V
Fig. 3.495
Fig. 3.492
[3.3 V ]
[1.33 A ]
Exercises 3.129
3.7
Find the current Ix in Fig. 3.496.
Ix
5Ω
50 V
10 Ω
20 Ω
5Ω
100 V
20 Ω
2A
24 V
50 V
2Ω
3Ω
2Ω
36 V
Fig. 3.500
Fig. 3.496
[−
3.8
]
[3.87 A ]
3.12 Find the current through the 6 Ω resistor in
Fig. 3.501.
Find the voltage Vx in Fig. 3.497.
2Ω
4V
+
50 V
Vx
−
3A
3Ω
]
Determine the voltages V1 and V2 in Fig.
3.498.
10 A
V1
4Ω
+
−
0.5V2
V2
6Ω
Fig. 3.501
[−
4Ω
5Ω
2A
100 V
0.1Vx
Fig. 3.497
3.9
4Ω
4Ω
[1.26 A ]
3.13 Find the current through the 6 Ω resistor in
Fig. 3.502.
10 V 2 Ω
I1 4 Ω
0.25I1
20 V
4A
10 Ω
5Ω
6Ω
3A
Fig. 3.498
[6 V,12 V ]
3.10
Find the voltage Vx in Fig. 3.499.
10 Ω
4A
20 Ω
30 Ω
+
Vx
−
[ 2.04 A ]
3.14 Find the current through the 2 Ω resistor
connected between terminals A and B in Fig.
3.503.
60 V
I1
Fig. 3.502
2 Ω 12 V
0.4I1
10 V
Fig. 3.499
10 Ω
4Ω
5Ω
4A
8V
[7.5 V ]
Thevenin’s Theorem
3.11 Find the current through the 5 Ω resistor in
Fig. 3.500.
2Ω 6V A
2Ω
B
Fig. 3.503
[1.26 A ]
3.130 Network Analysis and Synthesis
3.15 Find the current through the 5 Ω resistor in
Fig. 3.504.
(i)
4Vx
10 V
5Ω
+ −
4Ω
40 V
100 V
20 Ω
20 V
6Ω
30 V
50 V
7Ω
6Ω
+
Vx
−
4A
4Ω
Fig. 3.507
20 V
10 Ω
[ 4.67 A ]
3.16 Find the current through the 20 Ω resistor in
Fig. 3.505.
8Ω
40 Ω
10 V
36 Ω
9I1
Vx
21 Ω
15 Ω
Ω]
I1 100 Ω
Fig. 3.504
15 Ω
[−
(ii)
15 Ω
100 Ω
Fig. 3.508
[9.09
7Ω
20 Ω
36 Ω
, 9.09 Ω ]
(iii)
0.5Vx
− +
100 V
+
Fig. 3.505
[1.54 A ]
3A
4Ω
4Ω
3.17 Calculate the current through the 10 Ω
resistor in Fig. 3.506.
Vx
−
Fig. 3.509
10 Ω
[8 V, 2.66 Ω]
(iv)
4Ω
25 V
+
−
3Ω
2Ω
−
+
7Ω
12 V
2Ω
2Ω
3Ω
+
Vx
−
5Ω
10 A
Vx
4
Fig. 3.506
[1.62 A ]
3.18 Determine Thevenin’s equivalent network
for figures 3.507 to 3.510 shown below.
Fig. 3.510
[150 V, 20 Ω]
Exercises 3.131
3.23 Find the current through the 2 Ω resistor in
Fig. 3.515.
3.19 Find the current Ix in Fig. 3.511.
Ix 5 Ω
2Ω
10 V
10 Ω
10Ix
+
−
5Ω
4A
1A
10 V
5Ω
5A
3Ω
2Ω
Fig. 3.511
[4 A ]
20 V
3.20 Find the current in the 24 Ω resistor in Fig.
3.512.
Fig. 3.515
[5 A ]
Ix 1000 Ω
48 V
3Vx
+
−
10Ix
+
Vx
−
13 Ω
24 Ω
3.24 Find the current through the 5 Ω resistor in
Fig. 3.516.
3Ω
6Ω
5Ω
6A
Fig. 3.512
10 V
Norton’s Theorem
4Ω
5V
5Ω
15 Ω
20 V
4Ω
2Ω
Fig. 3.516
3.21 Find the current through the 10 Ω resistor in
Fig. 3.513.
2 V 10 Ω
3Ω
10 Ω 2 A
[0.225 A ]
6Ω
6Ω
[ 4.13 A ]
3.25 Find Norton’s equivalent circuit for the
portion of network shown in Fig. 3.517 to the
left of ab. Hence obtain the current in the 10
Ω resistor.
a
Fig. 3.513
[0.68 A ]
3.22 Find the current through the 20 Ω resistor in
Fig. 3.514.
10 Ω
10 V
20 Ω
4Ω
4Ω
4Ω
7V
2Ω
10 Ω
12 V
8A
5Ω
8Ω
9Ω
6Ω
b
2A
Fig. 3.514
[0.61 A ]
Fig. 3.517
[0.053 A ]
3.26 Find Norton’s equivalent network and hence
find the current in the 10 Ω resistor in Fig.
3.518.
3.132 Network Analysis and Synthesis
2V
3.30 Find the value of the resistance RL in Fig. 3.522
for maximum power transfer and calculate the
maximum power.
3Ω
2Ω
I1
3Ω
10 Ω
2I1
8A
2A
Fig. 3.518
[0.25 A ]
4Ω
3.27 Find Norton’s equivalent network In Fig.
3.519.
2Ω
12 V
4 V/2 Ω
2Ω
8Ω
2 V/1 Ω
I2
+
V1
−
8I2
+
−
0.1 V1
2Ω
2Ω
RL
20 Ω
Fig. 3.522
[ 2.18 Ω, 29 35 ]
Fig. 3.519
[0.533 A, 31 Ω]
Maximum Power Transfer Theorem
3.31 Find the value of the resistance RL in Fig. 3.523
for maximum power transfer and calculate the
maximum power.
3.28 Find the value of the resistance RL in Fig. 3.520
for maximum power transfer and calculate the
maximum power.
3Ω
3Ω
2Ω
1Ω
6A
2Ω
3V
2Ω
6V
1Ω
2A
RL
RL
2Ω
Fig. 3.523
[3 Ω 2 52 ]
Fig. 3.520
[1.75 Ω,1.29 ]
3.29 Find the value of the resistance RL in Fig. 3.521
for maximum power transfer and calculate the
maximum power.
25 A
9Ω
RL
10 Ω
2Ω
6Ω
3Ω
110 V
3.32 Find the value of the resistance RL in Fig. 3.524
for maximum power transfer and calculate the
maximum power.
3Ω
5Ω
10 A
RL
10 Ω
30 A
Fig. 3.524
[1.76 Ω, 490.187 ]
Fig. 3.521
[ 2.36 Ω, 940 ]
Objective-Type Questions 3.133
Objective-Type Questions
3.1
The value of the resistance R connected across
the terminals A and B in Fig. 3.525, which
will absorb the maximum power is
3.5
The maximum power that can be transferred
to the load RL from the voltage source in Fig.
3.528 is
100 Ω
3 kΩ
4 kΩ
R
V
A
B
6 kΩ
RL
10 V
4 kΩ
Fig. 3.528
Fig. 3.525
(a)
(c)
3.2
4 kΩ
8 kΩ
(b)
(d)
(a)
(c)
4.11 kΩ
9kΩ
Superposition theorem is not applicable to
networks containing
3.6
(a) nonlinear elements
(b) dependent voltage source
(c) dependent current source
(d) transformers
3.3
(b)
(d)
10 W
0.5 W
For the circuit shown in Fig. 3.529, Thevenin’s
voltage and Thevenin’s equivalent resistance
at terminals a-b is
1A
0.5I1
The value of R required for maximum power
transfer in the network shown in Fig. 3.526 is
5Ω
1W
0.25 W
+
−
a I1
5Ω
5Ω
4Ω
10 V
b
Fig. 3.529
20 Ω
25 V
3A
R
(a) 5 V and 2 Ω
(c) 4 V and 2 Ω
3.7
Fig. 3.526
3.4
The value of RL in Fig. 3.530 for maximum
power transfer is
(a) 2 Ω (b) 4 Ω (c) 8 Ω (d) 16 Ω
In the network of Fig. 3.527, the maximum
power is delivered to RL if its value is
9Ω
6Ω
I1
20 Ω
0.5I1
40 Ω
RL
(b) 7.5 V and 2.5 Ω
(d) 3 V and 2.5 Ω
2V
6Ω
+
−
6Ω
1A
9Ω
RL
V
Fig. 3.530
Fig. 3.527
(a)
16 Ω
(c)
60 Ω
40
Ω
3
(d) 20 Ω
(b)
(a)
(c)
3Ω
4.1785 Ω
(b)
(d)
1.125 Ω
none of these
3.134 Network Analysis and Synthesis
Answers to Objective-Type Questions
3.1 (a)
3.2 (a)
3.3 (c)
3.4 (a)
3.5 (c)
3.6 (b)
3.7 (a)
4
4.1
Single-Phase ac
Circuits
INTRODUCTION
An alternating waveform changes its magnitude and direction periodically. Many times alternating voltages
and currents are represented by a sinusoidal waveform. A sinusoidal waveform is more useful than other
waveforms, like square, triangular, sawtooth etc. Any waveform can be written in terms of sinusoidal function
with the help of Fourier series. The derivative and integration of sinusoidal waveform result in sinusoidal
waveform only. The sinusoidal waveform can be easily analyzed and generated.
4.2
GENERATION OF ALTERNATING VOLTAGES
An alternating voltage can be generated either by rotating a coil in a stationary magnetic field or by a rotating
a magnetic field within a stationary coil. In both the cases, the magnetic field is cut by the conductors or coils
and an emf is induced in the coil according to Faraday’s laws of electromagnetic induction. The magnitude
of the induced emf depends upon the number of turns of the coil, the strength of the magnetic field and the
speed at which the coil or magnetic field rotates.
Consider a rectangular coil of N turns of area A m2 and rotating in anti-clockwise direction with angular
velocity of w radians per second in a uniform magnetic field as shown in Fig. 4.1.
Let fm be the maximum flux cutting the coil when its axis coincides with XX ′ axis (reference position of
the coil). Thus when the coil is along XX ′, the flux linking with it is maximum, i.e., fm. When the coil is along
YY ′, i.e., parallel to the lines of flux, the flux linking with it is zero.
The coil rotates through an angle θ ω t at any instant t. At this instant, the flux linking with the coil is
φ
φm cos ω t
w rad/s
N
S
(a)
4.2 Network Analysis and Synthesis
N
Y
X′
w rad/s
Coil
q = wt
0
X
fm cosw
wt
Y′
S
(b)
Fig. 4.1 Generation of alternating voltage
According to Faraday’s laws of electromagnetic induction,
dφ
d
e
N
= − N (φm c ω t ) N φm ω sin ω t Em i ω t
dt
dt
e
E m = Nφ m ω
where
= maximum value of induced emf
Em
When
ω t 0, i ω t = 0, e = 0
π
π
q = wt
When
ωt
, si
1 e = Em
p
p
3p
0
2p
2
2
2
2
If the induced emf is plotted against time, a sinusoidal −E
m
waveform is obtained as shown in Fig. 4.2.
Fig. 4.2 Sinusoidal wave form
4.3
TERMS RELATED TO ALTERNATING QUANTITIES
1. Waveform A waveform is a graph in which the instantaneous
value of any quantity is plotted against time. Figure 4.3 shows a
few waveforms.
2. Cycle One complete set of positive and negative values of an
alternating quantity is termed a cycle.
3. Frequency The number of cycles per second of an alternating
quantity is known as its frequency. It is denoted by f and is
measured in hertz (Hz) or cycles per second (c/s).
4. Time Period The time taken by an alternating quantity to
complete one cycle is called its time period. It is denoted by T
and is measured in seconds.
1
T=
f
v
v
t
0
t
0
v
0
Fig. 4.3
t
Alternating waveforms
5. Amplitude The maximum positive or negative value of an alternating quantity is called the amplitude.
6. Phase The phase of an alternating quantity is the time that has elapsed since the quantity has last
passed through zero point of reference.
7. Phase Difference This term is used to compare the phases of two alternating quantities. Two
alternating quantities are said to be in phase when they reach their maximum and zero values at the same
time. Their maximum value may be different in magnitude.
4.4 Root Mean Square (rms) or Effective Value 4.3
A leading alternating quantity is one which reaches its maximum or zero value earlier compared to the
other quantity.
A lagging alternating quantity is one which attains its maximum or zero value later than the other quantity.
A plus (+) sign, when used in connection with the phase
VB
VA
difference, denotes ‘lead’ whereas a minus (−) sign denotes
‘lag’.
Vm sin ω t
A
vB
Vm sin (ω t + φ )
Here, the quantity B leads A by a phase angle f as shown in
Fig. 4.4.
4.4
t
f
Fig. 4.4 Phase difference
ROOT MEAN SQUARE (rms) OR EFFECTIVE VALUE
i
Normally, the current is measured by the amount of work it will do
or the amount of heat it will produce. Hence, rms or effective value
i2
of alternating current is defined as that value of steady current(direct
current) which will do the same amount of work in the same time or
i1
would produce the same heating effect as when the alternating current
is applied for the same time.
Figure 4.5 shows the positive half cycle of a non-sinusoidal 0 t1 t2
alternating current waveform. The waveform is divided in m
Fig. 4.5
equal intervals of time each of duration t . Let the average
m
values of instantaneous currents during these intervals be respectively i1, i2,
i3
im
t3
t
m
t
Mid-ordinate method
…, im. This waveform
is applied to a circuit consisting of a resistance of R ohms. The work done in different intervals will be
t⎞ ⎛ 2
t⎞
t⎞
⎛ 2
⎛ 2
⎜⎝ i1 R × ⎟⎠ , ⎜⎝ i2 R × ⎟⎠ , …, ⎝ im R × ⎟⎠ joules.
m
m
m
Thus, the total work done in t seconds on applying an alternating current waveform to a resistance
i 2 i 2 + …+
+ im2
R= 1 2
× Rt joules
m
Let I be the value of the direct current that while flowing through the same resistance does the same amount
of work in the same time t.
i 2 i 2 + …im2
I 2 Rt = 1 2
× Rt
m
i 2 i 2 + …+ im2
I2 = 1 2
m
Hence, rms value of the alternating current is given by
I rms =
i12
i22 + …+
+ im2
= Mean value of (i ) 2
m
The rms value of any current i(t) over the specified interval t1 to t2 is expressed mathematically as
I rms =
1
t2
t2
t1 ∫t
1
i 2 (t )dt
4.4 Network Analysis and Synthesis
The rms value of an alternating current is of considerable importance in practice because the ammeters
and voltmeters record the rms value of alternating current and voltage respectively.
The rms Value of a Sinusoidal Waveform
v
Figure 4.6 shows a sinusoidal waveform.
Vm si θ
v
1
2π
Vrms
=
=
=
1
2π
Vm2
2π
Vm
< θ < 2π
2π
∫v
2
( ) dθ
p
0
2p
q
0
2π
∫ Vm
2
2
Fig. 4.6 Sinusoidal waveform
θ dθ
0
2π
∫ sin
2
θ dθ =
0
Vm2
2π
2π
2π
Vm2 ⎡ θ
i θ⎤
⎛ 1 − cos2θ ⎞
−
⎟⎠ dθ =
2
2π ⎢⎣ 2
4 ⎥⎦ 0
∫ ⎜⎝
0
Vm2 ⎡ 2
⎤ V
− 0 − 0 + 0 ⎥ = m = 0.707 Vm
⎢
2π ⎣ 2
2
⎦
Crest or Peak or Amplitude Factor It is defined as the ratio of maximum value to rms value of the
given quantity.
Maximum
u value
Peak factor ( p ) =
rms value
4.5
AVERAGE VALUE
The average value of an alternating quantity is defined as the arithmetic mean of all the values over one
complete cycle.
In case of a symmetrical alternating waveform (whether sinusoidal or non-sinusoidal), the average value
over a complete cycle is zero. Hence, in such a case, the average value is obtained over half the cycle only.
Referring to Fig. 4.5, the average value of the current is given by
I avg =
i1 i2 + …+ im
m
The average value of any current i(t) over the specified interval t1
to t2 is expressed mathematically as
I avg =
t2
1
t2
t1
∫ i(t ) dt
v
Vm
t1
p
0
2p
Average Value of a Sinusoidal Waveform
Figure 4.7 shows a sinusoidal waveform.
v
Vm si θ
Fig. 4.7
Sinusoidal waveform
< θ < 2π
Since this is a symmetrical waveform, the average value is calculated over half the cycle.
q
4.5 Average Value 4.5
π
Vavg
=
π
1
v
π ∫0
d
Vm
[+
π
]=
π
1
V
V
Vm sin θ dθ = m ∫ sin θ dθ = m [ −
∫
π0
π 0
π
]π0
2Vm
= 0.637 Vm
π
Form Factor It is defined as the ratio of rms value to the average value of the given quantity.
Form factor (
f
)=
rms value
Average value
Example 4.1
An alternating current takes 3.375 ms to reach 15 A for the first time after becoming
instantaneously zero. The frequency of the current is 40 Hz. Find the maximum value of the alternating
current.
i = 15 A, t = 3.375 ms, f = 40 Hz
i = Im sin 2p ft
15 = Im sin (2 × 180 × 40 × 3.375 × 10−3)
Im = 20 A
Solution
(angle in degrees)
Example 4.2
An alternating current of 50 c/s frequency has a maximum value of 100 A. (a) Calculate
1
its value
second after the instant the current is zero. (b) In how many seconds after the zero value will
600
the current attain the value of 86.6 A?
f = 50 c/s, Im = 100 A
1
(a) Value of current
second after the instant the current is zero
600
i = Im sin 2p ft
Solution
1 ⎞
⎛
= 100 sin 2 × 180 × 50 ×
⎟
⎝
600 ⎠
= 50 A
(b) Time at which current will attain the value of 86.6 A after the zero value
i = Im sin 2pft
86.6 = 100 sin (2 × 180 × 50 × t)
1
t=
second
300
Example 4.3
(angle in degrees)
(angle in degrees)
An alternating current varying sinusoidally with a frequency of 50 c/s has an rms
value of 20 A. Write down the equation for the instantaneous value and find this value at (a) 0.0025 s, and
(b) 0.0125 s after passing through zero and increasing positively. (c) At what time, measured from zero, will
the value of the instantaneous current be 14.14 A?
Solution
f = 50 c/s, Irms = 20 A
I m I rms × 2 = 20 2 = 28.28 A
4.6 Network Analysis and Synthesis
Equation of current, i = Im sin 2p ft
= 28.28 sin (100p × t)
= 28.28 sin (100 × 180 × t)
(a) Value of current at t = 0.0025 second
i = 28.28 sin (100 × 180 × 0.0025)
= 20 A
(b) Value of current at t = 0.0125 second
(c)
(angle in degrees)
(angle in degrees)
i = 28.28 sin (100 × 180 × 0.0125)
= − 20 A
Time at which value of instantaneous current will be 14.14 A
i = 28.28 sin (100 × 180 × t)
14.14 = 28.28 sin 18000 t
t = 1.66 ms
(angle in degrees)
(angle in degrees)
Example 4.4
An alternating current of 60 Hz frequency has a maximum value of 110 A. Calculate
(a) time required to reach 90 A after the instant current is zero and increasing positively, and (b) its value
1
second after the instant current is zero and its value decreasing thereafter.
600
i
Solution
f = 60 Hz, Im = 110 A
(a) Time required to reach 90 A after the instant current is zero
110 A
and increasing positively
i = Im sin 2p ft
90 = 110 sin (2 × 180 × 60 × t)
(angle in degrees)
0
t = 2.54 ms
1
(b) Value of current
second after the instant current is
600
zero and decreasing thereafter
From Fig. 4.8,
T
1
1
1
t= +
=
+
2 600 2 f 600
=
T
T
2
t
Fig. 4.8
1
1
+
= 0.01 s
2 × 60 600
i = Im sin 2p ft
= 110 sin (2 × 180 × 60 × 0.01)
= − 64.66 A
(angle in degrees)
Example 4.5 A sinusoidal wave of 50 Hz frequency has its maximum value of 9.2 A. What will be
its value at (a) 0.002 s after the wave passes through zero in the positive direction, and (b) 0.0045 s after
the wave passes through the positive maximum.
Solution
f = 50 Hz, Im = 9.2 A
(a) Value of current at 0.002 s after the wave passes through zero in the positive direction
4.5 Average Value 4.7
i = Im sin 2p ft
= 9.2 sin (2 × 180 × 50 × 0.002)
= 5.41 A
(b)
(angle in degrees)
i
Value of current at 0.0045 s after the wave passes through the
positive maximum
From Fig. 4.9,
9.2 A
t=
=
T
1
+ 0.0045 =
+ 0.0045
4
4f
0
T
4
1
+ 0.0045 = 9.5 ms
4 × 50
t
T
T
2
Fig. 4.9
i = Im sin 2p ft
= 9.2 sin (2 × 180 × 50 × 9.5 × 10−3)
= 1.44 A
(angle in degrees)
Example 4.6 An alternating current varying sinusoidally with a frequency of 50 Hz has an rms value
of current of 20 A. At what time measured from negative maximum value will the instantaneous current be
10 2 A?
f = 50 Hz, Irms = 20 A
Solution
Time at which instantaneous current will be 10 2 A
i = 10 2 A = 14.14 A
Im = Irms × 2 = 20 2 = 28.28 A
i = Im sin 2π ft
14.14 = 28.28 sin (2 × 180 × 50 × t)
t = 1.67 ms
i
(b) Time, measured from negative maximum value, at
which instantaneous current will be 10 2 A
28.28 A
From Fig. 4.10,
14.14 A
T
1
t = + 1.67 × 10 −3 =
+ 1.67 × 10 −3
0
4
4f
(a)
=
1
+ 1.67 × 10 −3 = 6.67 ms
4 × 50
(angle in degrees)
T
4
T
2
3T
4
T
t
Fig. 4.10
Example 4.7 An alternating current is given by i = 14.14 sin 377 t. Find (a) rms value of the
current, (b) frequency, (c) instantaneous value of the current when t = 3 ms, and (d) time taken by the
current to reach 10 A for first time after passing through zero.
i = 14.14 sin 377 t
Solution
(a)
The rms value of the current
I rms =
Im
2
=
14.14
2
= 10 A
4.8 Network Analysis and Synthesis
(b)
Frequency
2π f = 377
f =
(c)
377
= 60 Hz
2π
Instantaneous value of the current when t = 3 ms
i = 14.14 sin (377 × 3 × 10−3)
= 12.79 A
(angle in radians)
(d) Time taken by the current to reach 10 A for the first time after passing through zero
i = 14.14 sin 377 t
(angle in radians)
10 = 14.14 sin 377 t
t = 2.084 ms
Example 4.8 An alternating current varying sinusoidally at 50 Hz has its rms value of 10 A. Write
down an equation for the instantaneous value of the current. Find the value of the current at (a) 0.0025
second after passing through the positive maximum value, and (b) 0.0075 second after passing through zero
value and increasing negatively.
f = 50 Hz,
Solution
(a)
Irms = 10 A
Equation for instantaneous value of the current
Im
I rms × 2 = 10 2 = 14.14 A
i = Im sin 2p ft
= 14.14 sin (2 × 180 × 50 × t)
= 14.14 sin (18000 t)
(b)
(angle in degrees)
i
Value of the current at 0.0025 s after passing through
14.14 A
the positive maximum value
From Fig. 4.11,
t=
T
1
+ 0.0025 =
+ 0.0025
4
4f
0
T
4
1
+ 0.0025 = 7.5 ms
4 × 50
i = 14.14 sin(18000 × 7.5 × 10−3)
= 10 A
=
T
1
+ 0.0075 =
+ 0.0075
2
2f
1
+ 0.0075 = 17.5 ms
2 × 50
i = 14.14 sin (18000 × 17.5 × 10−3)
= −10 A
=
T
t
Fig. 4.11
(angle in degrees)
(c) Value of the current 0.0075 s after passing through
zero value and increasing negatively.
14.14 A
From Fig. 4.12,
t=
T
2
0
i
T
2
T
t
Fig. 4.12
(angle in degrees)
4.5 Average Value 4.9
Example 4.9 Draw a neat sketch in each case of the waveform and write expressions of instantaneous value for the following:
π
(a)) Sinusoidal current of amplitude 10 A, 50 Hz passing through its zero value at ω t = and increasing
3
positively
π
(b) Sinusoidal current of amplitude 8 A, 50 Hz passing through its zero value at ω t = − and increasing
6
positively.
Solution
i
(a) The current waveform is lagging in nature.
From Fig. 4.13,
I m sin( 2π ft
f − φ) =
i
π⎞
⎛
si
sin 2π × 50 × t − ⎟
⎝
3⎠
10 A
π⎞
⎛
= 10 sin 100π t − ⎟
⎝
3⎠
(b)
3
The current waveform is leading in nature.
From Fig. 4.14,
i
i
π⎞
⎛
I m sin ( 2π ft
ft + φ ) = sin 2π × 50 × t + ⎟
⎝
6⎠
= 8 si
q = wt
p
0
Fig. 4.13
8A
π⎞
⎛
100π t + ⎟
⎝
6⎠
−
q = wt
p 0
6
Fig. 4.14
Example 4.10
π⎞
⎛
The instantaneous current is given by l = 7.071 sin ⎜ 157.08t − ⎟ . Find its effective
⎝
4⎠
value, periodic time and the instant at which it reaches its positive maximum value. Sketch the waveform from
t = 0 over one complete cycle.
Solution
(a)
i
Effective value
I eff
(b)
π⎞
⎛
157.08t − ⎟
⎝
4⎠
7.707
7071si
I rms =
Im
2
=
7.071
2
= 5A
Periodic time
2p f = 157.08
f = 25 Hz
T=
(c)
1
1
=
= 0.04 s
f 25
Instant at which the current reaches its positive maximum value i.e., i = 7.071 A
4.10 Network Analysis and Synthesis
From Fig. 4.14,
π⎞
⎛
7.071 = 7.071sin
071sin 157.08t − ⎟
⎝
4⎠
1 = sin (157.08t − 0.785)
i
7.071 A
t = 0.015 s
(d)
0
The waveform is shown in Fig. 4.15.
p
4
5p
4
9p
4
wt
Fig. 4.15
Example 4.11
A 60 Hz sinusoidal current has an instantaneous value of 7.07 A at t = 0 and rms
value of 10 2 A. Assuming the current wave to enter positive half at t = 0, determine (a) expression for instantaneous current, (b) magnitude of the current at t = 0.0125 second, and (c) magnitude of the current at t
= 0.025 second after t = 0.
Solution
(a)
i(0) = 7.07 A,
f
I rms = 10 2 A
Expression for instantaneous current
Im
I rms × 2 = 10 2 × 2 = 20 A
Since i = 7.07 A at t = 0, the current is leading in nature.
At t = 0
I m sin ( 2π ft
f + φ)
7.07 = 20 sin
i ( × × + φ)
φ = 20.7°
i 20 sin (120
( π t + 20.7 )
i
20 A
Magnitude of current at t = 0.0125 second
i = 20 sin (120 × 180 × 0.0125 + 20.7°)
0
(angle in degrees)
= −18.7 A
(c) Magnitude of current at t = 0.025 second after t = 0
Fig. 4.16
From Fig. 4.16,
20.7°
20.7° 1 20 7° 1
Time corresponding to a phase shift of 20.7° =
×T =
× =
×
= 0.958 ms
360°
360° f
360° 60
Time 0.025 second after t = 0
(b)
t = 0.958 × 10−3 + 0.025 = 25.958 ms
i = 20 sin (120 × 180 × 25.958 × 10–3 + 20.7°)
= −13.22 A
Example 4.12
(
t
(angle in degrees)
A 50 Hz sinusoidal voltage applied to a single-phase circuit has an rms value of 200 V.
)
Its value at t = 0 is 2 × 200 V positive. The current drawn by the circuit is 5 A (rms) and lags behind the
voltage by one-sixth of a cycle. Write the expressions for the instantaneous values of voltage and current.
Sketch their waveforms and find their values at t = 0.0125 second.
4.5 Average Value 4.11
Solution
Vrms = 200 V,
f
I rms = 5 A
v (0) = 2 × 200 = 282.84 V
(a)
Instantaneous value of voltage
Vm
v
Vrms × 2 = 200 2 = 282.84 V
Vm sin ( 2π ft
f + φ)
At t = 0,
282.84 = 282
8 .84 sin (0 + φ )
sin φ = 1
φ = 90°
v = 282.8
84
4 si ( 2π × 50 × + 90°)
(b)
Instantaneous value of current
The current lags behind the voltage by one-sixth of a cycle.
φ=
1
× 360° = 60°
6
I m = I rms × 2 = 5 2 = 7.07 A
i = I m sin ( 2π ft + 90° − 60°) = 7.07 sin
s (2π 50
30 )
(c)
Voltage and current waveforms are shown in Fig. 4.17.
(d)
Value of voltage at t = 0.0125 s
v = 282.84 sin (100 × 180 × 0.0125 + 90°)
(angle in degrees)
= 200 V
(e)
282 84 i (100π t + 90°)
7 07 sin
7.07
si (100
( 00π t + 30 )
v
i
p
−
2
Value of current at t = 0.0125 s
i = 7.07 sin (100 × 180 × 0.0125 + 30°)
(angle in degrees)
= −6.83 A
0
Fig. 4.17
At the instant t = 0, the instantaneous value of a 50 Hz sinusoidal current is 5 A and
increases in magnitude further. Its rms value is 10 A. (a) Write the expression for its instantaneous value.
(b) Find the current at t = 0.01 s and t = 0.015 s. (c) Sketch the waveform indicating these values.
Example 4.13
Solution
(a)
i(0) = 5 A,
f
I rms = 10 A
Expression for instantaneous value of current
Im
Since
i
I rms × 2 = 10 2 = 14.14 A
5 A a t = 0, the current is leading in nature.
4.12 Network Analysis and Synthesis
i
I m sin ( 2π ft
f + φ)
At t = 0,
5 14.14 si ( 2 × 180 × 50 × 0 + φ )
5 14.14 sin φ
φ = 20.7°
i = 14.14 ssii (100
(100
(b)
(angle in degrees)
+ 20.7°)
Current at t = 0 01 s
i = 14.14 sin (100 × 180 × 0.01 + 20.7°)
= −5 A
(angle in degrees)
(c) Current at t = 0.015 s
i = 14.14 sin (100 × 180 × 0.015 + 20.7°)
= −13.23 A
i
(angle in degrees)
(d) Waveform
0.015
0
−5 A
0.01
14.14 A
The waveform is shown in Fig. 4.18.
t(s)
−13.23 A
Fig. 4.18
Example 4.14
In a certain circuit supplied from 50 Hz mains, the potential difference has a maximum
value of 500 V and the current has a maximum value of 10 A. At the instant t = 0, the instantaneous values of
potential difference and current are 400 V and 4 A respectively, both increasing in the positive direction. State
expressions for instantaneous values of potential difference and current at time t. Calculate the instantaneous
values at time t = 0.015 second. Find the phase angle between potential difference and current.
Solution
(a)
f
Vm = 500 V
Im
10 A v(0) = 400 V,
i( 0) = 4 A
Expression for instantaneous value of potential difference and current
Since v = 400 V and i = 4 A at t = 0, the voltage and current waveforms are leading in nature.
)
At t = 0
v
Vm sin ( 2π ft
f + φ1 )
400 = 500 sin ( 2π 50 0 + φ1 )
φ1 = 53.13°
v = 500 sin(100π t + 53.13°)
(ii)
At t = 0
i = Im sin (2p ft + f2)
4 10 in ( 2π × 500 × 0 + φ2 )
φ2 = 23.58°
i = 10 sin (100π t + 23.58°)
4.5 Average Value 4.13
(b)
Instantaneous values of potential difference and current at t = 0.015 second
(i)
v = 500 sin (100 × 180 × 0.015 + 53.13°)
= −300 V
(angle in degrees)
i = 10 sin (100 × 180 × 0.00155 + 23.58°)
(ii)
= −9.17 A
(angle in degrees)
(c) Phase angle between potential difference and current
φ φ1 − φ2 = 53.13° − 23.58° = 29.55°
Example 4.15 Find the following parameters of a voltage v = 200 sin 314 t: (a) frequency (b) form
factor, and (c) crest factor.
Solution
(a)
v
200 si 314 t
v
Vm sin 2π ft
Frequency
2π f = 314
314
f =
= 50 Hz
2π
For a sinusoidal waveform,
Vavg =
Vrms =
(b)
2Vm
π
Vm
2
Form factor
Vm
Vrms
kf =
= 2 = 1.11
Vavg 2Vm
π
(c)
Crest factor
kp =
Vm
V
= m = 1.414
Vrms Vm
2
Example 4.16 A non-sinusoidal voltage has a form factor of 1.2 and peak factor of 1.5. If the average value of the voltage is 10 V, calculate (a) rms value, and (b) maximum value.
Solution
kf
k p = 1.5, Vavg = 10 V
4.14 Network Analysis and Synthesis
(a)
rms value
kf =
Vrms
Vavg
Vrms
10
= 12 V
1.2 =
Vrms
(b)
Maximum value
Vm
Vrms
V
15= m
12
Vm = 18 V
kp =
Example 4.17 The waveform of a voltage has a form factor of 1.15 and a peak factor of 1.5. If the
maximum value of the voltage is 4500 V, calculate the average value and rms value of the voltage.
Solution
(a)
k p = 1.5, Vm = 4500 V
kf
rms value of the voltage
Vm
Vrms
4500
15=
Vrms
Vrms = 3000 V
kp =
(b)
Average value of the voltage
kf =
1.15 =
Vrms
Vavg
3000
Vavg
Vavg = 2608.7 V
Example 4.18
A 50 Hz sinusoidal current has a peak factor of 1.4 and a form factor of 1.1. Its average value is 20 A. The instantaneous value of the current is 15 A at t = 0. Write the equation of the current
and draw its waveform.
Solution
(a)
kp = 1 4
f
Equation of current
kf =
I rms
I avg
1.1 =
I rms
20
kf
. , I avg = 20 A, i(0) = 15
15 A
4.5 Average Value 4.15
I rms = 22 A
Im
I rms
I
1.4 = m
22
I m = 30.8 A
kp =
Since i = 15 Α at t = 0, the current is leading in nature as shown in Fig. 4.19.
π fftt + φ )
i
At t = 0
15 = 30.8
( 2π × 50
0 × 0 + φ)
φ = 29.14°
i = 30.8
8 sin (100π
(b)
29.14°)
The waveform is shown in Fig. 4.19.
i
30.8 A
wt
−29.14°
Fig. 4.19
Example 4.19
Find the average value and rms value of the waveform shown in Fig. 4.20.
v
Vm
q
p
0
3p
2p
Fig. 4.20
Solution
v = Vm sin q
(a) Average value of the waveform
0<q<p
π
1
v
π ∫0
Vavg
=
(b)
π
d
1
V
Vm sin θ dθ = m [ −
∫
π0
π
Vm
[ + ] = 0.637 Vm
π
rms value of the waveform
π
Vrms
1
v
π ∫0
π
d
π
1
Vm2
Vm2 sin 2 θ dθ =
sin 2 θ dθ
∫
π0
π ∫0
]π0
4.16 Network Analysis and Synthesis
V2
2
2
0
Example 4.20
2
−
4
sin 0 ⎤
= 0.707 V
4
sin 2
4
2
Find the average and rms values of the waveform shown in Fig. 4.21.
v
Vm
q
p
0
3p
2p
Fig. 4.21
v = Vm sin q
=0
Solution
(a)
Average value of the waveform
1
2
V
2
2
V sin
0
d
0
1
2
(b)
0<q <p
p < q < 2p
=
0
2
+
2
0 318 V
rms value of the waveform
V
2
1
2
0
d
d
0
1
=
V2
2
s
0
V
2
Example 4.21
⎤
2
V
2
−
4
2
π
2
− +
4
∫
1
0
sin 0 ⎤
4
2
d
2
0.5 V
Find the average value and rms value of the waveform shown in Fig. 4.22.
v
V
0
p
4
p
2p 9p
4
Fig. 4.22
3p
q
4.5 Average Value 4.17
v=0
Solution
0<q<
= Vm sin q
<q<p
4
p < q < 2p
=0
(a)
4
Average value of the waveform
1
2
V
2
V sin d
0
4
−
=
+ 707 = 0.272 Vm
4
(b)
rms value of the waveform
1
2
V
2
V2
V2
2
2
0
4
V
2
=
4
2
V2
∫
4
2
2
=
π
−
2
8
2
2
d
4
sin
4
1
⎤
2
0.476 V
4
4
Example 4.22 A full-wave rectified wave is clipped at 70.7% of its maximum value as shown in
Fig. 4.23. Find its average and rms values.
v
Vm
0.707 Vm
0
p
4
3p
4
2p
p
q
Fig. 4.23
v = Vm sin q
Solution
0<q<
4
3
<q<
4
4
3
<q<p
4
= 0.707 Vm
= Vm sin q
(a)
Average value of the waveform
V
V
0
⎤
3
4
4
d
0
4
d
3
4
4.18 Network Analysis and Synthesis
=
3
V
−
+ −
=
4
(b)
+
(0 2
293)
0.54 V
4
rms value of the waveform
⎡
1
1
0
=
Example 4.23
V
⎣
V2 ⎡
2
d
3
4
0
4
2 ⎤4
+ 0 499
4
0
−
⎤
3
4
4
3
4
+
⎡
4
2
2 ⎤
3
4
−
⎦
= 0.584 V
4
Find the rms value of the waveform shown in Fig. 4.24.
v
Vm
0.866 Vm
q
0
0.866 Vm
Fig. 4.24
Solution The equation of the waveform is given by v = Vm sin (q + f) where f is the phase difference.
When q = 0, v = 0.866 Vm.
0.866 Vm = Vm sin (q + f)
0.866 = sin (0 + f)
=
−
=
=
+
3
3
The time period of a complete sine wave is always 2p. Since some part of the waveform is chopped from
both the sides,
Time period =
V
−
1
4
3
π
3
4π
3
π
3
4
3
π
3
+
0
3
d
4.5 Average Value 4.19
4π
2 3
3Vm
=
Example 4.24
∫
4π
0
⎡
π⎞ ⎤
⎛
⎢1 − cos 2 ⎝ θ + 3 ⎟⎠ ⎥
⎢
⎥ dθ =
2
⎢
⎥
⎢
⎥
⎣
⎦
⎡
π⎞
⎛
sin 2 ⎜ θ + ⎟
2 ⎢
⎝
3Vm θ
3⎠
⎢ −
4π ⎢ 2
4
⎢
⎣
4π
⎤3
⎥
⎥ = 0.776 Vm
⎥
⎥
⎦0
Find the average and rms values of the waveform shown in Fig. 4.25.
v
Vm
0
T
2
t
T
Fig. 4.25
Solution
v = Vm
0<t<
T
<t<T
2
=0
(a) Average value of the waveform
⎡T
1
1 ⎢2
= ∫v t d
dt = ⎢ ∫ Vm dt
T0
T ⎢0
⎢⎣
T
Vavg
T
2
T
∫
T
2
T
⎤
⎥ 1 2
dt ⎥ = ∫ Vm dt
⎥ T 0
⎥⎦
Vm T2 Vm T
[t ] 0 =
. = 0.5 Vm
T
T 2
rms value of the waveform
=
(b)
T
2
T
Vrms =
=
Example 4.25
1
v
T ∫0
t d
dt =
1
Vm2 T2
Vm2 dt =
[t ] 0
∫
T 0
T
Vm2 T
. = 0.707 Vm
T 2
Find the average and rms values of the waveform shown in Fig. 4.26.
v
Vm
t
0
T
2T
Fig. 4.26
Solution
v=
Vm
t
T
0
t <T
3T
4.20 Network Analysis and Synthesis
(a)
Average value of the waveform
T
Vavg
(b)
T
T
1
1 V
V ⎡t2 ⎤
V T2
= ∫v t d
dt = ∫ m t dt = m2 ⎢ ⎥ = m2 .
= 0.5 Vm
T0
T0 T
2
T ⎣ 2 ⎦0 T
rms value of the waveform
T
T
1
1 Vm2 2
v
t
dt
d
=
⋅ t dt
T ∫0
T ∫0 T 2
Vrms =
T
Vm2 ⎡ t 3 ⎤
Vm2 ⎡ T 3 ⎤
=
⎢
⎥
⎢ ⎥ = 0.577 Vm
T 3 ⎣ 3 ⎦0
T3 ⎣ 3 ⎦
=
Example 4.26
Find the average and rms values of the waveform shown in Fig. 4.27.
v
1
1
0
2
3
t
4
Fig. 4.27
v=t
=1
Solution
0< t<1
1< t<2
(a) Average value of the waveform
Vavg =
=
(b)
T
1
1
1⎡
v
t
dt
d
=
t dt
⎢
T ∫0
2 ⎣⎢
2
∫
1
⎤
dt ⎥
⎥⎦
⎫ 1 1
1 ⎧⎪ ⎡ t
⎡
⎤
2⎪
⎨ ⎢ ⎥ + [t ]1 ⎬ = ⎢ − 0 + 2 − 1⎥ = 0.75 V
2 ⎪⎣ 2 ⎦
⎦
⎪⎭ 2 ⎣ 2
0
⎩
2 ⎤1
rms value of the waveform
T
Vrms =
=
1
v t ddt =
T ∫0
1⎡
⎢ t dt
2 ⎢⎣ 0
1
⎫
1 ⎧⎪ ⎡ t 3 ⎤
2⎪
⎨ ⎢ ⎥ + [t ]1 ⎬ =
2 ⎪⎣ 3 ⎦
0
⎩
⎭⎪
1
2
∫( )
1
2
⎤
dt ⎥
⎥⎦
1 ⎡1
⎤
− 0 + 2 − 1⎥ = 0.816 V
⎢
2 ⎣3
⎦
4.5 Average Value 4.21
Example 4.27
Find the average and rms values of the waveform shown in Fig. 4.28.
v
100
80
60
40
0
1
2
3
4
5
6
7
t
8
Fig. 4.28
Solution
(a) Average value of the waveform
Vavg =
(b)
0 + 40 + 60 + 80 + 100 + 80 + 60 + 40
= 57.5 V
8
rms value of the waveform
Vrms =
Example 4.28
0 2 + ( 40) 2 + (60) 2 + (80) 2 + (100) 2 + (80) 2 + (60) 2 + ( 40) 2
= 64.42 V
8
Find the average and rms values of the waveform shown in Fig. 4.29.
v
20
10
0
1
2
3
4
5
6
t
Fig. 4.29
Solution
(a) Average value of the waveform
Vavg =
(b)
0 + 10 + 20
= 10 V
3
rms value of the waveform
Vrms =
0 2 + (10) 2 + ( 20) 2
= 12.9 V
3
Example 4.29 Find the effective value of the resultant current which carries simultaneously a direct
current of 10 A and a sinusoidally alternating current with a peak value of 10 A.
4.22 Network Analysis and Synthesis
Solution
Figure 4.30 shows the current waveforms.
i
10 A
q
0
i
10 A
20 A
0
10 A
q
2p
p
q
0
Fig. 4.30
i = i1 + i2 = 10 + 10 sin q
1
2
=
2
+ 10 in
d
0
d
0
2π
1
2
+ 200 in +
=
100
2
2
⎡
+ sin
1
+
2
0
100
2
−
⎤
2
+
2π
−
2
π
4
+
2
0
100
2
d
− + co
θ d
+
0
−2
− +
+
2
⎤
2
4
−
sin 0 ⎤
100 ⎡
=
4
2
2
0
−
2
2
2
⎤
= 12.25 A
Example 4.30
Find the effective value of a resultant current in a wire which carries simultaneously a
direct current of i1 = 10 A and alternating current given by i = 12 si
Solution
t
n
t
6
+ 4 sin 5 t +
3
i1 = 10 A
ω
t+
6
t+
i
3 t
3
t+
6
3
= 10 + 12 sinq + 6 sin (3q − 30°) + 4 sin (5q + 60°)
1
2
2
2
0
1
2
2
∫
0
10 +
+
3
+
2
+ 60°) d = 14.07 A
.
4.6 Phasor Representations of Alternating Quantities 4.23
Example 4.31 Find the relative heating effects of two current waves of equal peak value, one sinusoidal and the other, rectangular in shape.
i
Solution Figure 4.31 shows the two current waves.
Im
rms value of the rectangular wave = Im
rms value of the sinusoidal current wave =
Im
0
2
Heating effect due to the rectangular current wave = (Im)2 RT
p
2p
q
Fig. 4.31
( I m ) RT
⎛I ⎞
Heating effect due to the sinusoidal current wave = ⎜ m ⎟ RT =
⎝ 2⎠
2
2
2
Relative heating effects =
4.6
( I m )2 RT : ( I
2
m)
2
RT =
1
:1 = 1: 2
2
PHASOR REPRESENTATIONS OF ALTERNATING QUANTITIES
The alternating quantities are represented by phasors. A phasor is a line of definite length rotating in an
anticlockwise direction at a constant angular velocity w. The length of a phasor is equal to the maximum
value of the alternating quantity, and the angular velocity is equal to the angular velocity of alternating
quantity.
As shown in Fig. 4.32, consider a phasor OP = Im, where Im is the maximum value of the alternating
current. Let this phasor rotate in an anticlockwise direction at a uniform angular velocity of w radians/
second. The projection of the phasor OP on the Y-axis at any instant gives the instantaneous value of that
alternating current.
OM = OP sin w t = Im sin w t = i
i
w rad/s
P
M
q
0
0
p
2p
q = wt
Fig. 4.32 Representation of alternating quantities in terms of phasors
Thus, if we plot the projections of the phasor on the Y-axis versus its angular position point by point, a
sinusoidal alternating current waveform is obtained.
Phasor Diagram Using rms Values Sinusoidal alternating currents and voltages can be represented by
phasors. Electrical measuring instruments like ammeters and voltmeters are calibrated to read the rms values
of ac quantities. Hence, instead of using maximum values, it is more convenient to draw phasor diagrams
using rms values of alternating quantities. However, such a phasor diagram will not generate a sine wave of
proper amplitude unless the length of the phasor is multiplied by 2.
4.24 Network Analysis and Synthesis
Example 4.32
Two currents i1 and i2 are given by the expressions
i1 = 10 sin
π⎞
⎛
t+
and
⎝
4⎠
π⎞
⎛
d i2 = 8 sin ω t −
. Find (a) i1 + i2, and (b) i1 − i2. Express the answers in the form i = I m sin( ω t φ )).
⎝
3⎠
Solution
i1
10 i
i2
8 i
π⎞
⎛
ωt +
⎝
4⎠
π⎞
⎛
ωt −
⎝
3⎠
y
(a)
Let phasors I1 and I 2 represent the alternating
currents i1 and i2 respectively in terms of their
maximum values.
I1 = 10∠45°
(i) Analytical method:
From Fig. 4.33,
Resolving
1
and I 2 into x- and y- components,
45°
∑ = 10 cos (45°) + 8 c
( −60°) = 11 07
∑y = 10 sin (45°) + 8 s
( −60°) = 0.14
14
Magnitude of (I1
x
0
60°
I 2 ) = (∑ x)2 + (∑ y)2
= (11.07)2 + (0.14)2
I2 = 8∠−60°
= 11.07
∑
y
⎛
⎞
⎛ 0 14 ⎞
Phase angle
e φ tan −1 ⎜
= tan −1
= 0.72°
⎟
⎝ 11.07 ⎠
⎝ ∑ x⎠
Fig. 4.33
i = i1 + i2 = 11.07 (w t + 0.72°)
(ii) Graphical method: The phasor sum I1 + I 2 is obtained by adding phasors I1 and I 2 by the
parallelogram law as shown in Fig. 4.34.
y
Scale: 1 cm = 2 A
I1
45°
0
I
60°
I2
Fig. 4.34
x
4.6 Phasor Representations of Alternating Quantities 4.25
−I2
(b) (i) Analytical method:
From Fig. 4.35,
Resolving I1 and −I 2 into x- and y-components,
y
I1 = 10∠45°
∑ = 10 cos ( 45°) − 8 c ( −60°°) = 3.07
∑ y = 10 sin ( 45°) − 8 si ( −60°°) = 14
Magnitude of (
2
) = (∑ x) + (
= ( ..07
07))
2
y)
60°
45°
0
2
x
60°
(14) 2
= 14.33 A
I2 = 8∠−60°
∑ y⎞
Phase angle φ = tan ⎜
⎝ ∑ x ⎟⎠
−1 ⎛
Fig. 4.35
⎛ 14 ⎞
= tan −1 ⎜
= 77.63
⎝ 3.07 ⎟⎠
i = i1 i2 = 14.33 sin (ω t + 77.63 )
(ii) Graphical method: The phasor sum I1 I 2 is obtained by adding phasors I1 and −I 2 by the
parallelogram law as shown in Fig. 4.36.
y
I
Scale: 1 cm = 2 A
I1
−I2
77.63°
45°
60°
0
x
60°
I2
Fig. 4.36
Example 4.33
π⎞
⎛
v3 = 30 sin ω t + ⎟ .
⎝
4⎠
Solution
π⎞
⎛
Three voltages are represented by v1 = 10 sin ω t,
t, v2 = 20 sin
in ω t − ⎟ and
⎝
6⎠
Find the magnitude and phase angle of the resultant voltage.
v1
v2
v3
10 si ω t
π⎞
⎛
20
0 si ω t − ⎟
⎝
6⎠
30 si
π⎞
⎛
ωt + ⎟
⎝
4⎠
4.26 Network Analysis and Synthesis
Let phasors V1 , V2 and V3 represent the alternating voltages v1, v2 and v3 respectively in terms of their
maximum values.
(a) Analytical method:
From Fig. 4.37,
Resolving V1 , V2 and V3 into x- and y-components,
y
V3 = 30∠45°
∑ = 10 + 20 cos ( −30°) + 30 cos ( 45°) = 48.53
∑ y = 20 sin ( −30°) + 30 si ( 45°)
.21
45°
Magnitude of (V1 V2 + V3 ) = ( ∑ x ) 2 + ( ∑ y ) 2
0
30°
= (48
( 48
8.53) 2 + (11.21) 2 = 49.81 V
⎛ ∑ y⎞
Phase angle φ = tan 1
⎝ ∑ x⎠
⎛ 11.21 ⎞
ta 1
⎝ 48.53 ⎠
V1 = 10∠0°
x
V2 = 20∠−30°
13°
Fig. 4.37
v = 49.81 sin(wt + 13°)
(b)
Graphical method: The phasor sum V1 + V2 + V3 is obtained by first adding phasors V1 and V2 by
the parallelogram law and then adding V3 to the resultant of V1 and V2 by the parallelogram law as
shown in Fig. 4.38.
y
V3
V
V1
45°
13°
30°
0
V2
x
V12
Fig. 4.38
Example 4.34
The instantaneous voltages across each of the four coils connected in series are given
π⎞
π⎞
⎛
⎛
by v1 = 100 sinw t, v2 = 250 cos w t, v3 = 150 sin ⎜ ω t + ⎟ , v4 = 200 sin ω t − ⎟ . Determine the resultant
⎝
⎠
⎝
6
4⎠
voltage by analytical method.
Solution
v1 = 100 sin w t,
v3
v2 = 250 cos w t = 250 sin(w t + 90°)
π⎞
⎛
150 i
t+
⎝
6⎠
v4
200 si
π⎞
⎛
ωt − ⎟
⎝
4⎠
Let phasors V1 , V2 , V3 and V4 represent the instantaneous voltages v1, v2, v3, and v4 respectively in terms
of their maximum values as shown in Fig. 4.39.
4.7 Mathematical Representations of Phasors 4.27
y
V2
V3
30°
x
0
45°
V1
V4
Fig. 4.39
Resolving V1 , V2 V3 and V4 into x- and y-components,
∑ = 100 + 50 cos (90°) + 50 cos (30°) + 200 cos ( −45°) = 371.33
∑ y = 250 sinn (90°))
sin ( ) + 200 sin ( −45°)
.58
2
∑
Magnitude of (
2
2
2
4414.23 V
⎛ ∑ y⎞
⎛ 183.58 ⎞
Phase angle φ = tan −1 ⎜
= tan −1 ⎜
= 26.31
⎟
⎝ ∑ x⎠
⎝ 371.33 ⎟⎠
v = v1 + v2 + v3 + v4 = 414.23 sin (w t + 26.31°)
4.7
MATHEMATICAL REPRESENTATIONS OF PHASORS
A phasor can be represented in four forms.
1. Rectangular Form
V = X ± jY
Magnitude of phasor, V
X 2 +Y 2
⎛Y ⎞
Phase angle φ = tan −1 ⎜ ⎟
⎝X⎠
2. Trigonometric Form
φ ± j sin
in φ )
V =V (
3. Exponential Form
V = Ve ±
4. Polar Form
jφ
V =V ∠± φ
Significance of Operator j The operator j is used in rectangular form. It is used to indicate anticlockwise
rotation of a phasor through 90°. Mathematically,
j = −1
4.28 Network Analysis and Synthesis
Whenever a phasor is multiplied by j, the phasor is rotated once in the anticlockwise direction through 90°.
The power of j represents the number of times the phasor should be rotated through 90° in the anticlockwise
directon.
Example 4.35 Two sinusoidal currents are given as
Find the expression for the sum of these currents.
Solution
+ 60 ).
t + 60°)
i
Writing currents i1 and i2 in the phasor form,
I1 =
10 2
I2 =
20 2
2
2
∠0
1
0
∠60° = 20 ∠60°
0° + 2 ∠ ° =
t+
i
Example 4.36
∠
=
9
t + 40.89 )
The following three sinusoidal currents flow into the junction
ω t + 30° ) and i = 6 2 sin( t
leaves the junction.
). Find the expression for the resultant current which
+ 30°),
Solution
Writing currents i1, i2 and i3 in the phasor form,
I1
I2
3 2
2
5 2
2
0
− 120°)
3∠0°
30
30
6 2
6
2
The resultant current which leaves the junction is given by
I
∠ + ∠ 0° + 6 ∠− 120° =
I3
i
Example 4.37
tant current.
Solution
−
=
1
t−
∠−
9°
°
In a circuit, four currents as indicated below, are meeting at a point. Find the resuli1 = 5 sin w t,
i2 = 10 sin (w t − 30°)
i3 = 5 cos (w t − 30°)
i4 = −10 sin (w t + 45°)
i1 = 5 sin w t, i2 = 10 sin (w t − 30°),
i3 = 5 cos (w t − 30°) = 5 sin (w t + 60°)
i4 = −10 sin (w t + 45°) = 10 sin (w t + 225°)
4.7 Mathematical Representations of Phasors 4.29
Writing currents i1, i2, i3, and i4 in the phasor form,
5
I1
0 3 54 ∠0°
2
10
I2
= 7 07
2
5
I3 =
∠60° = 3 5 60
2
10
I4 =
∠225 7 07∠225°
2
Re ultant current
=
° + 7∠− 30 3
∠
°+7
∠225°
= 8.44 ∠−40.36°
−
i
Example 4.38
=
sin
t
°)
.
Find the resultant voltage and its equation for the given voltages.
e1 = 20 sin w t,
= 40 cos
4
t+
6
Solution
t
4
=
os
E3 =
40
t+
6
=
+ 120°)
Writing voltages e1, e2, and e3 in the phasor form,
20
=
2
Resultant voltage
30
2
∠
21
°
14
e
Example 4.39
∠−
=
t
sin
∠120° = 28 ∠120°
2
∠120° = 75
33
°+
t+
°
Obtain the sum of the three voltages.
t
Solution
t+
t
v
=
=
v
t +188 1°
+
,
=
)
sin
+
°
Writing voltages v1, v2 and v3 in the phasor form,
V1 =
V =
V =
Re
147 3
2
294 6
2
8
2
∠188.1° = 104.16 ∠188 1°
∠45° = 208 31 ∠45°
∠135° = 62 51 ∠135°
= 104
tant voltage
v
t
∠
=
∠
.
s
+
51 ∠
°
° = 176 82 ∠90°
4.30 Network Analysis and Synthesis
Example 4.40
Find vectorially the resultant of the following four voltages.
6
sin
Solution
t
,
6
=
6
4
°
=
)
6
Writing voltages e1, e2, e3 and e4 in the phasor form,
25
E1
2
30
E2
2
E3 =
E4 =
30
2
20
2
17 68∠
30° = 21 21∠30°
∠90°
°
∠− 3
0°
21 21 ∠ ° +
Resultant voltage
∠90°
30
27 13°
t+
e
Example 4.41
=
t+
°
Two currents are represented by i = 15 sin
t+
and i = 25 sin t +
3
4
These currents are fed into a common conductor. Find the total current in the form i = Im sin(w t + f). If the
conductor has a resistance of 10 W, what will be the energy loss in 24 hours?
t+
3
4
R = 10 Ω, t = 24 hours = 86400 seconds
Solution
Writing currents i1 and i2 in the phasor form,
I1 =
I2 =
15
2
25
2
∠60° = 10.61∠60°
∠45° = 17.68∠45°
∠ °+
Total current
i
Energy loss in 24 hours, E
E
t
2
I Rt
68 ∠
°
=
68
∠
t+
62°
°
where I is the rms value of the current
×
= . ×
J
4.7 Mathematical Representations of Phasors 4.31
Example 4.42
The voltage drops across four series-connected impedances are given:
−
6
v3
t+
,
4
5
6
4)
Solution
Calculate the values of V4 m and f4 if the voltage applied across the series circuit is 140 sin
=
6
t
4
=
sn
=
t
t
5
6
t+
t+
3
5
75 s (
3
5
=
.
)
sin ( t + 108 °)
Writing voltages v1, v2, v3 and v in the phasor form,
60
V1
30° = 42.43∠30°
2
75
V2
−
2
V =
V=
100
2
140
2
° = 53 03∠−
°
∠135° = 70.71∠135°
∠108° = 98.99∠108°
For series–connected impedances,
+ V4
−
−
v
V
°−
t
m
= 80.79 V
4
=
.
43∠ °
6
t
150
70 71∠
°=
∠59.96
59 96°
°
Example 4.43 A circuit consists of three parallel branches. The branch currents are given as
i1 = 10 sinw t, i2 = 20 sin(w t + 60°) and i3 = 7.5 sin(w t − 30°). Find the resultant current and express it in the form
i = Im sin(w t + f). If the supply frequency is 50 Hz, calculate the resultant current when (a) t = 0 (b) t = 0.001 s.
Solution
i1 = 10 sin w t,
Writing currents i1, i2, i3 in the phasor form,
I1 =
I2 =
Resultant current I3 =
10
2
20
2
75
2
∠0
i2 = 20 sin (w t + 60°),
7 07∠0°
∠60
14.14 ∠60°
∠
= 5 3∠−
°
i3 = 7.5 sin (w t − 30°)
4.32 Network Analysis and Synthesis
1
I1 + I 2
I3
= 7.07 0° + 14.14 ∠60° + 5.3∠− 30° = 21.04 ∠27.13°
i
(a)
21.0
04 2 si ( t + 27.13°°) = 2299.76 in (ω t + 27.13°)
Resultant current at t = 0
i = 29.76 si (0 + 27.13°°) = 13.57 A
(b)
Resultant current at t = 0.001 s
i
( ft
f + 27.13°)
= 29.76 sin ( 2 × 180 × 50 × 0.001 + 7. 3°)
= 21.09
. A
Example 4.44
Estimate (a) I1
I 2 (b) I1 I 2
I1
Solution
(a)
(b)
(c)
10 ∠ 0°A
Two currents, I1
(c)
10 ∠50°
(angle in degrees)
d I 2 = 5 ∠− 100° A, flow in a single-phase ac circuit.
I1
.
I2
I 2 = 5∠− 100°A
I1 I 2 = 10 ∠50° + 5∠− 100° = 6.2 ∠ 26.21° A
I1 I 2 = (10
(10 50 ) (5
(5
100 ) = 50 ∠− 50° A
I1
10 ∠50°
=
= 2∠150° A
I 2 5∠− 100°
Example 4.45
Two voltages having rms values of 50 V and 75 V have a phase difference of 60°.
Find the resultant sum of these two voltages.
V2 = 75 V, φ = 60°
Solution
V1
Let
V1 = 50 ∠ 0° V
50
V2 = 75∠− 60° V
Resultant voltage V
V1 + V2 = 50 ∠0° + 75∠− 60° = 108.97 ∠
∠− 36.58° V
Example 4.46
Two single-phase alternators supply 300 A and 400 A respectively at a phase
difference of 20° to a common load. Find the resultant current and its phase relation to its component.
Solution
Let
I1
300 A
I 2 = 400 A, φ = 20°
I1 = 300 ∠0° A
I 2 = 400 ∠− 20° A
Resultant current I = I1 + I 2 = 300 ∠0° + 400 ∠− 20° = 689.59∠
∠− 11.44° A
Example 4.47 Two voltage sources have equal emfs and a phase difference α. When they are connected in series, the voltage is 200 V. When one source is reversed, the voltage is 15 V. Find their emfs and
phase angle a.
4.7 Mathematical Representations of Phasors 4.33
∠0°
E1
Solution
E2 = E∠ °,
E1 = E2 = E
E2
E2
180°−a
a
E1
a
E1
(a)
(b)
Fig. 4.40
When two sources are connected in series [Fig. 4.40 (a)],
E12 + E22 + 2 E1 E2 cos α = 200
E 2 + E 2 + 2 E 2 cos α = 200
2E 2 + E 2 cos α = 40000
…(i)
When one source is reversed [Fig. 4.40 (b)],
E12 + E22 − 2 E1 E2 cos α
15
E + E − 2 E cos α
15
2
2
2E
2
2
2 E cos α = 225
2
…(ii)
Adding Eqs (i) and (ii),
4 E 2 = 40225
E 2 = 10056.25
E = 100.28 V
From Eq. (i),
2
2
+ 2 E 2 cos α = 40000
20112.5 + 20112.5 cos α = 40000
cos α 0.988
α = 8.58°
Example 4.48 Two sinusoidal sources of emf have rms values of E1 and E2 and a phase difference
of a . When connected in series, the resultant voltage is 41.1 V. When one of the sources is reversed, the
resultant emf is 17.52 V. When phase displacement is made zero, the resultant emf is 42.5 V. Calculate E1,
E2 and a.
Solution
1∠0°,
E1
E2 = E2 ∠α °
E2
E2
180°−a
a
(a)
E1
a
E1
(b)
Fig. 4.41
E1
(c)
E2
4.34 Network Analysis and Synthesis
When two sources are connected in series [Fig. 4.41 (a)],
E12 + E22 + 2 E1 E2 cos α
E12
+
E22
41.1
41
…(i)
+ 2 E1 E2 cos α = 1689.21
When one of the sources is reversed [Fig. 4.41 (b)],
E12 + E22 − 2 E1 E2 cos α
E12
+
E22
17.52
17
− 2 E1 E2 cos α = 306.95
…(ii)
When phase displacement is made zero [Fig. 4.41 (c)],
E12 + E22 + 2 E1 E2
s 0° = 42.5
E2 = 42.5
E1
E1
42
…(iii)
E2
Adding Eqs (i) and (ii),
2
1
2
2)
= 1996.16
E12
E22
= 998.08
.5 − E2 ) 2
2
.08
E22
E22
= 998.08
2(
+
From Eq. (iii),
(
1806.25 − 85 E2 +
E22
+
− 42.5 E2 + 404.09 = 0
…(iv)
Solving Eq. (iv),
E2
28 14
E2 = 14.36 V
E1
14 36
E1 = 28.14 V
Subtracting Eq. (ii) from Eq. (i),
4 E1 E2 cos α 1382.26
4 14.36 28.14 cos α 1382.26
cos α 0.855
α = 31.24°
4.8
BEHAVIOUR OF A PURE RESISTOR IN AN ac CIRCUIT
Consider a pure resistor R connected across an alternating voltage source v as shown in Fig. 4.42. Let the
alternating voltage be v = Vm sin w t.
R
vR
i
v = Vm sin w t
Fig. 4.42
Purely resistive circuit
4.8 Behaviour of a Pure Resistor in an ac Circuit 4.35
1. Current The alternating current i is given by
i=
v Vm
=
sin ω t
R
R
V ⎞
⎛
I = m⎟
⎝ m
R⎠
I m sin
i ωt
where Im is the maximum value of the alternating current. From the voltage and current equation, it is
clear that the current is in phase with the voltage in a purely resistive circuit.
2. Waveforms
Figure 4.43 shows the voltage and current waveforms.
v
i
q
0
p
Fig. 4.43
2p
Waveforms
3. Phasor Diagram Figure 4.44 shows the phasor diagram.
I
V
Fig. 4.44 Phasor diagram
4. Impedance
circuit,
It is the resistance offered to the flow of current in an ac circuit. In a purely resistive
Z=
V Vm Vm
=
=
=R
I
I m Vm
R
5. Phase Difference Since the voltage and current are in phase with each other, the phase difference is
zero.
f = 0°
6. Power Factor It is defined as the cosine of the angle between the voltage and current phasors.
Power factor = cos f = cos (0°) = 1
7. Power Instantaneous power p is given by
p vi
= Vm i
t I m si ω t
= Vm I m sin 2 ω t
V I
= m m ( − ccos
os 2ω t )
2
V I
V I
= m m − m m co
c s 2ω t
2
2
V I
The power waveform is shown in Fig. 4.45. The power consists of a constant part m m and a fluctuating
2
V I
part m m cos 2w t. The frequency of the fluctuating power is twice the applied voltage frequency and its
2
average value over one complete cycle is zero.
4.36 Network Analysis and Synthesis
Average power P =
Vm I m Vm I m
=
= VI
2
2 2
p
0
q
2p
p
Fig. 4.45 Power waveform
Thus, power in a purely resistive circuit is equal to the product of rms values of voltage and current.
4.9
BEHAVIOUR OF A PURE INDUCTOR IN AN ac CIRCUIT
Consider a pure inductor L connected across an alternating voltage v as shown in Fig. 4.46. Let the
alternating voltage be v = Vm sin w t.
L
vL
i
v = Vm sin w t
Fig. 4.46
Purely inductive circuit
1. Current The alternating current i is given by
1
v dt
L∫
1
= ∫ Vm i ω t dt
L
Vm
=
( − cos ω t )
ωL
V
π⎞
⎛
= m sin ω t −
⎝
ωL
2⎠
i=
= Im i
π⎞
⎛
t−
⎝
2⎠
V ⎞
⎛
…⎜ Im = m ⎟
⎝
ωL⎠
where Im is the maximum value of the alternating current. From the voltage and current equation, it is
clear that the current lags behind the voltage by 90° in a purely inductive circuit.
4.9 Behaviour of a Pure Inductor in an ac Circuit 4.37
2. Waveforms Figure 4.47 shows the voltage and current waveforms.
v
i
0
q
2p
p
p
2
Fig. 4.47
Waveforms
3. Phasor Diagram Figure 4.48 shows the phasor diagram.
V
I
Fig. 4.48 Phasor diagram
4. Impedance
In a purely inductive circuit,
Z=
V Vm Vm
=
=
= ωL
Vm
I
Im
ωL
The quantity w L is called inductive reactance, is denoted by XL and is measured in ohms.
For a dc supply,
f=0
∴ XL = 0
Thus, an inductor acts as a short circuit for a dc supply.
5. Phase Difference It is the angle between the voltage and current phasors.
f = 90°
6. Power Factor It is defined as the cosine of the angle between the voltage and current phasors.
pf = cos f = cos (90°) = 0
7. Power Instantaneous powers p is given by
p
p
vi
= Vm i
t Im i
π⎞
⎛
ωt −
⎝
2⎠
= −Vm I m sin ω t cos ω t
=−
0
Vm I m
sin 2 ω t
2
The power waveform is shown in Fig. 4.49
The average power for one complete cycle, P = 0.
Hence, power consumed by a purely inductive circuit is zero.
p
2
p
3p
2
2p
Fig. 4.49 Power waveform
q
4.38 Network Analysis and Synthesis
4.10
BEHAVIOUR OF A PURE CAPACITOR IN AN ac CIRCUIT
Consider a pure capacitor C connected across an alternating voltage v as shown in Fig. 4.50.
Let the alternating voltage be v = Vm sin w t.
C
vC
i
v = Vm sin wt
Fig. 4.50
Purely capacitive circuit
1. Current The alternating current i is given by
dv
i C
dt
d
= C (Vm si ω t )
dt
= ωC Vm
ωt
= ωC Vm in (ω t + 90°)
= I m i (ω t + 90°)
( I m = ωC Vm )
where Im is the maximum value of the alternating current. From the voltage and current equation, it is
clear that the current leads the voltage by 90° in a purely capacitive circuit.
2. Waveforms Figure 4.51 shows the voltage and current waveforms.
v
i
−
p
2
p
0
Fig. 4.51
2p
Waveforms
3. Phasor Diagram Figure 4.52 shows the phasor diagram.
I
V
Fig. 4.52 Phasor diagram
4. Impedance
In a purely capacitive circuit,
q
4.10 Behaviour of a Pure Capacitor in an ac Circuit 4.39
Z=
Vm
V Vm
1
=
=
=
I
I m ωC Vm ωC
1
is called capacitive reactance, is denoted by XC and is measured in ohms.
ωC
For a dc supply, f = 0 ∴ XC = ∞
Thus, the capacitor acts as an open circuit for a dc supply.
The quantity
5. Phase Difference
It is the angle between the voltage and current phasors.
f = 90°
6. Power Factor It is defined as the cosine of the angle between the voltage and current phasors.
pf = cos f = cos(90°) = 0
p
7. Power Instantaneous power p is given by
p
vi
= Vm i
t Im i ( t +
= Vm I m sin ω t c
°)
ωt
p
2
0
p
3p
2
Vm I m
sin 2ω t
2
The power waveform is shown in Fig. 4.53.
The average power for one complete cycle, P = 0.
Fig. 4.53 Power waveform
Hence, power consumed by a purely capacitive circuit is zero.
=
2p
q
Example 4.49 An ac circuit consists of a pure resistance of 10 ohms and is connected across an ac
supply of 230 V, 50 Hz. Calculate (a) current, (b) power consumed, (c) power factor, and (d) write down the
equations for voltage and current.
Solution
(a) Current
R = 10 Ω,
I=
(b)
V = 230 V,
f = 50 Hz
V 230
=
= 23 A
R 10
Power consumed
P = VI = 230 × 23 = 5290 W
Power factor
Since the voltage and current are in phase with each other, f = 0°
pf = cos f = cos (0°) = 1
(d) Voltage and current equations
Vm
2 V = 2 × 230 = 325.27 V
(c)
Im
2 I = 2 × 23 = 32.53 A
w = 2πf = 2p × 50 = 314.16 rad/s
v = Vm sinw t = 325.27 sin 314.16 t
i = Im sinw t = 32.53 sin 314.16 t
4.40 Network Analysis and Synthesis
Example 4.50 An inductive coil having negligible resistance and 0.1 henry inductance is connected
across a 200 V, 50 Hz supply. Find (a) inductive reactance, (b) rms value of current, (c) power, (d) power
factor, and (e) equations for voltage and current.
L = 0.1 H,
Solution
V = 200 V,
f = 50 Hz
(a) Inductive reactance
XL = 2p f L = 2p × 50 × 0.1 = 31.42 Ω
(b)
rms value of current
I=
V
200
=
= 6 37 A
X L 31.42
(c)
Power
Since the current lags behind the voltage by 90° in purely inductive circuit, f = 90°
P = VI cos f = 200 × 6.37 × cos (90°) = 0
(d) Power factor
pf = cos f = cos (90°) = 0
(e) Equations for voltage and current
Vm
2 V = 2 × 200 = 282.84 V
Im
2 I = 2 × 6 37 = 9 A
w = 2p f = 2p × 50 = 314.16 rad/s
v = Vm sin w t = 282.84 sin 314.16 t
π⎞
π⎞
⎛
⎛
i I m sin ω t
9 i 314 16 t −
⎝
⎠
⎝
2
2⎠
Example 4.51
The voltage and current through circuit elements are
v = 100 sin (314 t + 45°) volts
i = 10 sin (314 t + 315°) amperes
(a) Identify the circuit elements. (b) Find the value of the elements. (c) Obtain an expression for power.
Solution
v = 100 sin (314 t + 45°)
i = 10 sin (314 t + 315°)
= 10 sin (314 t + 315° − 360°)
= 10 sin (314 t − 45°)
(a) Identification of elements
From voltage and current equations, it is clear that the current i lags behind the voltage by 90°. Hence,
the circuit element is an inductor.
(b) Value of elements
XL =
V Vm 100
=
=
= 10 Ω
I
Im
10
XL
ωL
10 = 314 L
L = 31.8 mH
4.11 Series RL Circuit 4.41
(c)
Expression for power
p= −
Vm I m
sin 2ω t
2
100 × 10
si ( 2
2
314 t ) = −500 sin 628 t
Example 4.52 A capacitor has a capacitance of 30 microfarads which is connected across a 230 V,
50 Hz supply. Find (a) capacitive reactance, (b) rms value of current, (c) power, (d) power factor, and (v)
equations for voltage and current.
C = 30 μF,
Solution
(a)
f = 50 Hz
Capacitive reactance
XC =
(b)
V = 230 V,
1
1
=
= 106.1 Ω
2π fC 2π × 50 × 30 × 10 −6
rms value of current
I=
V
230
=
= 2.17 A
XC 106.1
(c)
Power
Since the current leads the voltage by 90° in purely capacitive circuit, f = 90°
(d)
Power factor
P = VI cos f = 230 × 2.17 × cos (90°) = 0
pf = cos f = cos (90°) = 0
(e)
Equations for voltage and current
Vm
2 V = 2 × 230 = 325.27 V
Im
2 I = 2 × 2.17 = 3.07 A
w = 2p f = 2p × 50 = 314.16 rad/s
v = Vm sin w t = 325.27 sin 314.16 t
π⎞
π⎞
⎛
⎛
i I m sin ω t
3 07 i 314 16 t +
⎝
⎠
⎝
2
2⎠
4.11
SERIES RL CIRCUIT
Figure 4.54 shows a pure resistor R connected in series with a pure
inductor L across an alternating voltage v = Vm sin w t
Let V and I be the rms values of applied voltage and current.
Potential difference across the resistor = VR = R I
Potential difference across the inductor = VL = XLI
The voltage VR is in phase with the current I whereas
the voltage VL leads the current I by 90°.
1. Phase Diagram
Figure 4.55 shows the phasor diagram of series RL circuit.
R
L
vR
vL
i
v = Vm sin w t
Fig. 4.54 RL circuit
4.42 Network Analysis and Synthesis
V
V
VL
VL
f
cos f =
f
VR
VR
V
VR
I
(a)
(b)
Fig. 4.55 (a) Phasor Diagram, (b) Voltage triangle
2. Impedance
V
VR + VL
RI + jX L I
(
j
L)
I
V
= R + jX L = Z
I
Z = ∠φ
R 2 + ω 2 L2
R 2 + X L2
Z
⎛X ⎞
⎛ ωL⎞
φ = tan −1 ⎜ L ⎟ = tan −1 ⎜
⎝ R ⎠
⎝ R ⎟⎠
The quantity Z is called the complex impedance of the series RL circuit.
3. Impedance Triangle
Figure 4.56 shows the impedance triangle of series RL circuit.
Z
XL
cos f =
f
R
Z
R
Fig. 4.56
Impedance triangle
4. Current From the phasor diagram, it is clear that the current I lags behind the voltage V by an angle f.
If the applied voltage is given by v = Vm sin w t then the current equation will be
i = Im sin(w t − f)
Vm
Z
where
Im =
and
⎛ ωL⎞
φ = tan −1 ⎜
⎝ R ⎟⎠
5. Waveforms Figure 4.57 shows the voltage and current waveforms.
v
i
q
f
Fig. 4.57
Waveforms
4.11 Series RL Circuit 4.43
6. Power Instantaneous power p is given by
p = vi
= Vm sin w t Im sin (w t − f)
= Vm Im sin w t sin (w t − f)
⎡ cos φ − cos ( ω t − φ ) ⎤
= Vm I m ⎢
⎥
20
⎣
⎦
Vm I m
Vm I m
=
cos φ −
cos ( ω t − φ )
2
2
Vm I m
Vm I m
cos ( 2ω t φ )). The
cos φ and a fluctuating part
2
2
frequency of the fluctuating part is twice the applied voltage frequency and its average value over one
complete cyce is zero.
Thus, power consists of a constant part
Average powerr P =
Vm I m
cos φ
2
Vm I m
2
2
cos φ = VI cos φ
Thus, power is dependent upon the in-phase component of the current. The average power is also called
active power and is measured in watts.
We know that a pure inductor and capacitor consume no power because all the power received from the
source in a half cycle is returned to the source in the next half cycle. This circulating power is called reactive
power. It is a product of the voltage and reactive component of the current, i.e., I sin f and is measured in
VAR (volt-ampere-reactive).
Reactive power Q = VI sin f.
The product of voltage and current is known as apparent power (S) and is measured in volt-ampere (VA).
S
VI
V
P2 + Q2
7. Power Triangle In terms of circuit components,
R
cos φ =
Z
and
V = ZI
R
P VI
V cos φ ZII = I 2 R
Z
X
Q VI
V si φ ZII L = I 2 X L
Z
2
S = VI = Z I I = I Z
Figure 4.58 shows the power triangle of series RL circuit.
S
Q
cos f =
f
P
Fig. 4.58 Power triangle
8. Power Factor It is defined as the cosine of the angle between the voltage and current phasors.
pf = cos f
VR
V
R
From impedance triangle, pf =
Z
From voltage triangle,
pf =
P
S
4.44 Network Analysis and Synthesis
P
S
In case of series RL circuit, the power factor is lagging in nature.
From power triangle,
pf =
Example 4.53 An alternating voltage of 80 + j60 V is applied to a circuit and the current flowing is
4 − j2 A. Find the (a) impedance, (b) phase angle, (c) power factor, and (d) power consumed.
Solution
(a)
80 j 60
I = 4− j 2A
Impedance
V 80 + j 60
100 ∠36.87°
=
=
= 22.37
3 ∠63.43° Ω
I
4 − j2
4.47∠
∠− 26.56°
Z = 22.37 Ω
Z=
(b)
Phase angle
(c)
Power factor
f = 63.43°
pf = cos f = cos (63.43°) = 0.447 (lagging)
(d)
Power consumed
P = VI cos f = 100 × 4.47 × 0.447 = 199.81 W
The voltage and current in a circuit are given by V 1 0 ∠30°
d I = 2 ∠− 15° A.
If the circuit works on a 50 Hz supply, determine impedance, resistance, reactance, power factor and power
loss considering the circuit as a simple series circuit.
Example 4.54
Solution
(a)
V
1 0 ∠30° , I = 2 ∠− 15° A,
f = 50 Hz
Impedance
V 150 ∠30°
=
= 75 ∠45° Ω = 53.03
03 + j 53.03 Ω
I
2∠− 15°
Z = 75 Ω
Z=
(b)
Resistance
R = 53.03 Ω
(c)
Reactance
X = 53.03 Ω
(d)
Power factor
f = 45°
pf = cos f = cos (45°) = 0.707 (lagging)
(e)
Power loss
P = VI cos f = 150 × 2 × 0.707 = 212.1 W
Example 4.55 An rms voltage of 100 ∠0° V is applied to a series combination of Z1 and Z2 when
Z1 = 20 ∠ 30° W. The effective voltage drop across Z1 is known to be 40 ∠− 30° V. Find the reactive component
of Z2.
4.11 Series RL Circuit 4.45
Solution
V
100 ∠0°
, Z1 = 20
20 ∠30°Ω
Ω
V1 = 40 ∠−
∠ 30° V
I=
V1 40 ∠− 30°
=
= 2 ∠−
− 60°A
Z1
20 ∠ 30°
Z=
V1 100 ∠0°
=
= 50 ∠60° = 25 + j43
4 .3 Ω
I
2 ∠− 60°
Z1 = 20 ∠30° = 17.32 + j10 Ω
Z = Z1 + Z 2
Z2
Z Z1 = 25 + j 43.3 17.32
j10
7.68 + j33.3 Ω
Reactive component of Z 2 = 33.3 Ω
A voltage v(t) = 177 sin (314 t + 10°) is applied to a circuit. It causes a steady-state
current to flow, which is described by i(t) = 14.14 sin (314 t − 20°). Determine the power factor and average
power delivered to the circuit.
Example 4.56
Solution
v(t) = 177 sin (314 t + 10°),
(a) Power factor
Current i(t) lags behind voltage v(t) by 30°.
i(t) = 14.14 sin (314 t − 20°)
f = 30°
pf = cos f = cos(30°) = 0.866 (lagging)
(b) Average power
P
VI
VI cos φ =
177 14.14
×
× 0.866 = 1083.7 W
2
2
Example 4.57 When a sinusoidal voltage of 120 V(rms) is applied to a series RL circuit, it is found
that there occurs a power dissipation of 1200 W and a current flow given by i(t) = 28.3 sin (314t − f). Find
the circuit resistance and inductance.
Solution
(a)
V = 120 V,
P = 1200 W,
Resistance
I=
28.3
= 20.01 A
2
P = V I cos f
1200 = 120 × 20.01 × cos f
cos f = 0.499
f = 60.02°
V
120
Z= =
=6Ω
I 20.01
Z = Z ∠φ 6 ∠60 02° 3
R = 3Ω
(b)
i(t) = 28.3 sin (314t – f)
Inductance
XL = 5.2 Ω
XL = w L
5.2 = 314 × L
L = 0.0165 Η
j 5.2 Ω
j5
4.46 Network Analysis and Synthesis
Example 4.58
In a series circuit containing resistance and inductance, the current and voltage
2π ⎞
5π ⎞
⎛
⎛
are expressed as i(t) = 5 sin ⎜ 314t + ⎟ and v(t) = 20 sin ⎜ 314t + ⎟ (a) What is the impedance of the
⎝
⎝
3⎠
6 ⎠
circuit? (b) What are the values of resistance, inductance, and power factor? (c) What is the average power
drawn by the circuit?
⎛
i(t ) = 5 sin
sin 314t
⎝
Solution
(a)
200 sii
5π ⎞
⎛
314t + ⎟
314
⎝
6⎠
Impedance
Z=
(b)
2π ⎞
, v(t )
3⎠
V Vm 20
=
=
=4Ω
I Im
5
Power factor, resistance and inductance
Current i(t) lags behind voltage v(t) by an angle f = 150° − 120° = 30°
pf = cos f = cos(30°) = 0.866 (lagging)
Z = 4 ∠30° = 3.464 + j Ω
R = 3.464 Ω
XL = 2 Ω
XL = ωL
2 = 314 × L
L = 6.37 mH
(c)
Average power
P
VII cos φ =
V
20
2
×
5
2
× 0.866 = 43.3 W
Example 4.59 A series circuit consists of a non-inductive resistance of 6 W and an inductive reactance of 10 W. When connected to a single-phase ac supply, it draws a current i(t) = 27.89 sin (628t − 45°).
Calculate (a) the voltage applied to the series circuit in the form Vm in( ω t φ )), (b) inductance, and (c)
power drawn by the circuit.
Solution
(a)
(b)
R
6 , X L = 10 Ω, i(t ) = 27.89 sin (628t − 45 )
Voltage applied to the series circuit
Z = R + jX L = 6 + j 10 = 11 66 ∠59 04° Ω
27.89
I=
∠− 45° = 19.72 ∠
∠− 45° A
2
V Z I = ( .66 ∠59. ) ( .72
.72 ∠− 45°)
.95∠ 14.04° V
v = 229.95 2 sin
i (ω t + 14.04 ) = 325.2 si (
+ 1144.04°)
Inductance
X L = 10 Ω
XL ωL
10 = 628 × L
L = 15.9
9 mH
4.11 Series RL Circuit 4.47
(c)
Power drawn by the circuit
P
VII cos φ = 229.95 × 19.72 × cos(
V
.04°) =
.78 W
Example 4.60
When an inductive coil is connected to a dc supply at 240 V, the current in it is 16 A.
When the same coil is connected to an ac supply at 240 V, 50 Hz, the current is 12.27 A. Calculate (a) resistance, (b) impedance, (c) reactance, and (d) inductance of the coil.
Solution
(a)
For dc : V
240 V,
For ac : V = 240 V,
Resistance
When an inductive coil is connected to a dc supply,
f =0
XL
f =0
fL
I = 16 A
I = 12.27 A
The coil behaves like a pure resistor.
R=
(b)
Impedance
When the coil is connected to an ac supply,
Z=
(c)
V
240
=
= 19.56 Ω
I 12.27
Reactance
XL
(d)
V 240
=
= 15 Ω
I
16
Z 2 − R 2 = (19.56)2 − (15)2 = 12.55 Ω
Inductance
XL
fL
12.55 = 2π × 50 × L
L = 0 04 H
Example 4.61 An inductive coil draws 10 A current and consumes 1 kW power from a 200 V, 50
Hz ac supply. Determine (a) impedance in Cartesian and polar forms, (b) power factor, and (c) reactive and
apparent power.
Solution
(a)
I
10 A, P = 1 kW, V
Impedance in Cartesian and polar forms
V 200
=
= 20 Ω
I
10
P VI
V cos φ
1000 = 200 × 10 × cos φ
cos φ = 0.5
φ = 60°
Z=
Expressing Z in polar form,
f = 50 Hz
4.48 Network Analysis and Synthesis
Z = Z ∠ φ ° = 20 ∠60° Ω
Expressing Z in Cartesian form,
Z= 0
(b)
j 7.3 Ω
Power factor
pf = cos φ = cos(
(c)
) = 0.5 (lagging)
Reactive and apparent power
Q VI
V sin φ = 200 × 10 × sin (
) = 1.73 kVAR
S = VI = 200 × 10 = 2 kVA
Example 4.62
In the circuit of Fig. 4.59, a coil connected across a 250 V, 50 Hz supply takes a
current of 10 A at 0.8 lagging power factor. What will be the power taken by the choke coil when connected
across a 200 V, 25 Hz supply? Also calculate resistance and inductance of the coil.
XL
R
10 A
250 V, 50Hz
Fig. 4.59
Solution
(a)
V
f1 = 50 Hz, I1 = 10 A, pf = 0.8 (lagging)
V
f 2 = 25 Hz
Resistance and inductance of the coil
V
250
Z1 = 1 =
= 255 Ω
I1
10
φ1 = cos −1 (0.8) = 36.87°
Z1 = Z1∠φ1 = 25
25∠36.87
8 ° = 20 + j15 Ω
R = 20 Ω
X L 1 = 15 Ω
X
f1 L
1
15 = 2π 50 × L
L = 0.0477 H
(b)
Power taken by the choke coil when connected across a 200 V, 50 Hz supply.
X
2
2π × 25 × 0.0477 = 7.49 Ω
f2 L
Z 2 = R + jX
2
= 20 + j 7.49
21.36 ∠20.53° Ω
Z 2 = 21.36 Ω
φ2 = 20.53ο
V
200
I2 = 2 =
= 9 36 A
Z 2 21.36
P V2 I 2 cos φ2 = 200
00 × 9.36
s ( 20.53°) = 1.753 kW
4.11 Series RL Circuit 4.49
Example 4.63 A load of 22 kW operates at 0.8 lagging power factor when connected to a 420 V,
single-phase, 50 Hz source. Find (a) current in the load, (b) power factor angle, (c) impedance, (d) resistance
of load, (e) reactance of load, (f) voltage and current equations.
Solution
(a)
22 kW, pf
= 420 V,
0.8 (lagging),
f
z
Current in the load
P
V cos φ
VI
22 × 103 = 420 × I × 0.8
I = 65.48 A
(b)
Power-factor angle
φ = cos − (0.. ) = 36.87°
(c)
(d)
(e)
Impedance
Z=
V
420
=
= 6.41
41 Ω
I 65.48
R
Z cos φ = 6.41 × 0.8 = 5.13 Ω
Resistance of load
Reactance of load
XL
Z sin φ = 6.41 × sin
in (36..
°) = 3.85 Ω
(f) Voltage and current equations
v Vm sin 2π ft = 420 2 sin
si (
×
)t
593.97 si 100 π t
The current lags behind the voltage by 36.87°.
i
I m sin ( 2π ft
ft − φ ) = .48 2 sin ( 2π × 50t − 36..
= 92.6 sin (
t − .87
8 °)
Example 4.64
)
A choke coil is connected in series with fixed resistor as shown in Fig. 4.60. A 240 V,
50 Hz supply is applied and a current of 2.5 A flows. If the voltage drops across the coil and fixed resistor
are 140 V and 160 V respectively, calculate the value of the fixed resistance, the resistance and inductance
of the coil, and power drawn by the coil.
Choke coil
XL
r
2.5 A
140 V
R
160 V
240 V, 50 Hz
Fig. 4.60
4.50 Network Analysis and Synthesis
Solution
(a)
f = 50 Hz,
V
2 5 A,
Vcoil = 140 V,
VR = 160 V
Resistance of fixed resistor
R=
(b)
I
VR 160
=
=6 Ω
I
25
Resistance and inductance of the coil
Zcoil =
Vcoil 140
=
= 56 Ω
I
25
r 2 + X L2 = 56
Zcoil
,
r2
X L2 = 3136
…(i)
V 240
=
= 96 Ω
I
25
Z = ( R + r ) + jX L
Z=
Z
( R + r )2
X L2
96 = (64 + ) 2 + X L2
) 2 + X L2 = 9216
(
…(ii)
Subtracting Eq. (i) from Eq. (ii),
(
) 2 − r 2 = 6080
4096 + 128r + r 2 − r 2 = 6080
128r = 1984
r = 15.5 Ω
Substituting the value of r in Eq. (i),
( .5) 2 +
2
L
X L2
= 3136
= 2895.75
X L = 53.81 Ω
X = 2 fL
53.81 = 2π 50 × L
L = 0.17
1 H
(c)
Power drawn by the coil
Pcoil
I 2 r = ( 2.5) 2 × 15.5 = 96.875 W
Example 4.65 A 100 W resistor is connected in series with a choke coil as shown in Fig. 4.61. When
a 400 V, 50 Hz supply is applied to this combination, the voltages across the resistance and the choke coil are
200 V and 300 V respectively. Find the power consumed by the choke coil. Also, calculate the power factor
of the choke coil and the power factor of the circuit.
4.11 Series RL Circuit 4.51
Choke coil
XL
r
R = 100 Ω
300 V
200 V
400 V, 50 Hz
Fig. 4.61
Solution
R = 100 Ω,
V = 400 V,
f = 50 Hz,
VR = 200 V, Vcoil = 300 V
(a) Power consumed by choke coil
VR 200
=
=2A
R 100
300
V
= coil =
= 150 Ω
I
2
I=
Zcoil
r2
X L2 = 150
r2
X L2 = 22500
V 400
Z= =
= 200 Ω
I
2
Z = ( R + r ) + jX L
( R + r )2
Z
X L2 = 200
) 2 + X L2 = 40000
(
Subtracting Eq. (i) from Eq. (ii),
) 2 − r 2 = 17500
(
10000 + 200 r + r 2 − r 2 = 17500
200 r = 7500
r = 37.5 Ω
Substituting the value of r in Eq. (i),
(
.5) 2 +
2
L
= 22500
X L2
= 21093.75
X L = 145.24 Ω
Pcoil = I 2 r = ( ) 2 37.5 = 50 W
(b)
Power factor of the choke coil
pf coil =
(c)
r
Zccoil
=
37.5
= 0.25 (lagging)
150
Power factor of the circuit
pf circuit =
…(i)
R + r 100 + 37.5
=
= 0.6875 (lagging)
Z
200
…(ii)
4.52 Network Analysis and Synthesis
Example 4.66 A resistor of 25 W is connected in series with a choke coil as shown in Fig. 4.62. The
series combination when connected across a 250 V, 50 Hz supply, draws a current of 4 A which lags behind
the voltage by 65°. Calculate (a) resistance and inductance of the coil, (b) total power, (c) power consumed
by resistance, and (d) power consumed by choke coil.
Coil
XL
r
R = 25 Ω
4A
250 V, 50 Hz
f = 65°
Fig. 4.62
Solution
R = 25 Ω,
V = 250 V,
f = 50 Hz,
I = 4 A,
f = 65°
(a) Resistance and inductance of the coil
V 250
=
= 62.5 Ω
I
4
Z = Z ∠φ 62 ∠6 ° 26 41
Z=
j 56 64 Ω
Z = ( + ) + jX L
But
X L = 56.64 Ω
R + r = 26.41
r = 26.41 − 25 = 1.41 Ω
X
ffL
56.64 = 2π × 50 × L
L = 0.18
1 H
(b)
Total power
P = I 2 ( R + r ) = ( 4) 2 × 26.41 = 422.56 W
(c)
Power consumed by resistance
PR
(d)
I 2 R = ( 4) 2 × 25 = 400 W
Power consumed by choke coil
Pcoil
Example 4.67
I 2 r = ( 4) 2 × 1.41 = 22.56 W
When a resistor and a coil in series are connected to a 240 V supply as shown in
Fig. 4.63, a current of 3 A flows, lagging 37° behind the supply voltage. The voltage across the coil is 171
volts. Find the resistance and reactance of the coil, and the resistance of the resistor.
4.11 Series RL Circuit 4.53
Coil
XL
r
R
171 V
3A
240 V
f = 37°
Fig. 4.63
Solution
V = 240 V,
I = 3 A, f = 37°,
Vcoil = 171 V
(a) Resistance and reactance of the coil
V
171
Zcoil = coil =
= 57 Ω
I
3
r2
X L2 = 57
r2
X L2 = 3249
V 240
Z= =
= 80 Ω
I
3
Z = Z∠φ 80 ∠3 ° 63 89
jj48
48.15 Ω
Z = ( + ) + jX L
But
X L = 48.15 Ω
r
2
X L2 = 3249
r 2 + ( 48
8.155) 2 = 3249
r 2 = 931.04
r = 30.51 Ω
(b)
Resistance of the resistor
R + r = 63.89
R + 30.5
51 = 63.89
R = 33.38 Ω
Example 4.68 A choke coil and a resistor are connected in series across a 230 V, 50 Hz ac supply
as shown in Fig. 4.64. The circuit draws a current of 2 A at 0.866 lagging pf. The voltage drop across the
resistor is 100 V. Calculate the power factor of the choke coil.
Coil
r
XL
R
100 V
2A
230 V, 50 Hz
pf = 0.866 (lagging)
Fig. 4.64
4.54 Network Analysis and Synthesis
Solution
V = 230 V,
f = 50 Hz, I = 2 A,
R=
pf = 0.866 (lagging),
VR = 100 V
VR 100
=
= 50 Ω
I
2
V 230
=
= 115 Ω
I
2
pf = 0.866 (lagging)
Z=
φ = cos −11 (0.866) = 30°
Z = Z∠φ = 115∠30° = 99.59 + j 57.5 Ω
R + r = 99.59
50 + = 99.59
r = 49.59 Ω
X L = 57.5 Ω
Zcoil
r 2 + X L2 = ( 49.59) 2 + (57.5) 2 = 75 93 Ω
pf ccoil =
r
=
Zcoil
49.59
= 0.653 (lagging)
75.93
9
Example 4.69 A circuit consists of a pure resistor and coil in series as shown in Fig. 4.65. Power
dissipated in the resistor and in the coil are 1000 W and 250 W respectively. The voltage drops across the resistor and the coil are 200 V and 300 V respectively. Determine (a) value of pure resistance, (b) resistance and
reactance of the coil, (c) combined resistance of the circuit, (d) combined impedance, and (e) supply voltage.
Coil
r
XL
R
300 V
200 V
I
V
Fig. 4.65
Solution
PR = 1000 W,
(a) Value of pure resistance
VR2
R
( 200) 2
1000 =
R
R = 40 Ω
PR =
Pcoil = 250 W,
VR = 200 V,
Vcoil = 300 V
4.11 Series RL Circuit 4.55
(b)
Resistance and reactance of the coil
VR RI
200 = 40 I
I = 5A
Pcoil
I 2r
250 = (5) 2 × r
r = 10 Ω
V
300
Zcoil = coil =
= 60 Ω
I
5
2
X L = Zcoil
− r 2 = (60) 2 − (10) 2
(c)
. Ω
.2
Combined resistance of the circuit
RT = R + r = 40 + 10 = 50 Ω
(d)
Combined impedance
( R + r )2
Z
(e)
X L2 = (
)2 + (
. )2 =
.5 Ω
Supply voltage
V
ZI
ZI = 77.5 × 5 = 387.5 V
Example 4.70 In the circuits of Fig. 4.66, a coil A takes 2 A at a power factor of 0.8 lagging with
an applied p.d. of 10 V. A second coil B takes 2 A with a power of 0.7 lagging with an applied voltage of 5 V.
What voltage will be required to produce a total current of 2 A with coils A and B in series? Find the power
factor in this case.
Coil A
rA
Coil B
XA
rB
2A
XB
2A
5V
pf = 0.7 (lagging)
10 V
pf = 0.8 (lagging)
Fig. 4.66
Solution
For Coil A,
Coil A:
Coil B:
IA = 2 A,
IB = 2 A,
pfA = 0.8 (lagging),
pfB = 0.7 (lagging),
φ A = cos −1 (0.. ) = 36.87°
VA 10
=
=5Ω
IA
2
Z A = Z ∠φ
5∠36 87°
rA = 4 Ω
XA = 3Ω
ZA =
4
j3 Ω
VA = 10 V
VB = 5 V
4.56 Network Analysis and Synthesis
φ B = cos −1 (0.. ) = 45.57°
For Coil B,
VB 5
= = 25Ω
IB 2
Z B = Z ∠φ
2 ∠4
ZB =
° 1 5
j1.78 Ω
j1
rB = 1 75 Ω
X B = 1.78
. Ω
When coils A and B are connected in series as shown in Fig. 4.67,
Coil A
rA
Coil B
XA
rB
XB
2A
V
Fig. 4.67
Z = r + jX A + rB + jX B = 4 + j 3 + 1 75 + j1
j1.78 = 5.75 + j 4.78 = 7.48∠39.74° Ω
Z = 7.48 Ω
φ = 39.74°
V = Z I = 7.48 2 = 14.96 V
pf = cos φ cos ( .74°))
.77 (lagging
g
)
Example 4.71
In the circuits of Fig. 4.68, a voltage of 100 V is applied to a coil A, the current taken
is 8 A and the power is 120 W. When applied to a coil B, the current is 10 A and the power is 500 W. What
current and power will be taken when 100 V is applied to the two coils connected in series?
Coil A
rA
Coil B
XA
8A
rB
XB
10 A
100 V
100 V
PA = 120 W
PB = 500 W
Fig. 4.68
Solution
For Coil A,
Coil A:
Coil B:
VA = 100 V,
VB = 100 V,
ZA =
VA 100
=
= 12.5 Ω
IA
8
IA = 8 A,
IB = 10 A,
PA = 120 W
PB = 500 W
4.11 Series RL Circuit 4.57
I A2 rA
PA
120 = (8) 2 × rA
rA = 1.875 Ω
Z A2 − rA2 = (12.5) 2 − (1.875) 2 = 12.36
. Ω
XA
For Coil B,
ZB =
VB 100
=
= 10 Ω
IB
10
PB
I B2 rB
500 = (10) 2 × rB
rB = 5 Ω
XB
Z B2 − rB2 = (10) 2 − (5) 2 = 8.66 Ω
When coils A and B are connected in series as shown in Fig. 4.69,
Coil A
rA
Coil B
XA
rB
XB
I
100 V
Fig. 4.69
Z = r + jX A + rB + jX B = 1.875 + j12 36 + 5 + j8.66 = 6.875 + j 21.02 = 22.11 ∠71.89° Ω
Z = 22.11 Ω
φ = 71.89°
V
100
I= =
= 4.52 A
Z 22.11
P = I 2(
) = ( .52) 2 (6.
) = 140.64 W
Example 4.72 In a particular circuit, a voltage of 10 V at 25 Hz produces 100 mA, while the same
voltage at 75 Hz produces 60 mA. Find the values of components of the circuit.
V1 = 10 V,
V2 = 10 V,
Solution
Case (a)
Case (b)
f1 = 25 Hz,
f2 = 75 Hz,
I1 = 100 mA
I2 = 60 mA
V1
100
=
= 100 Ω
I1 100 × 10 −3
V
10
Z2 = 2 =
= 166.67 Ω
I 2 60 × 10 −3
Z1 =
As frequency increases, impedance of the circuit increases. In a series RL circuit, inductive reactance XL increases
with frequency. Hence, impedance increases. The circuit consists of a resistance R and an inductance L.
4.58 Network Analysis and Synthesis
R 2 + X L21
Z1
R2
25 × L) 2 = 100 Ω
(50 L) 2 = 10000
…(i)
R 2 + X L22
Z2
R2
R2 + (2
R2 + (2
75 × L) 2 = 166.67 Ω
(150 L) 2 = 27778.89
…(ii)
Solving Eqs (i) and (ii),
R = 88.1 Ω
L = 0.3 H
Example 4.73 When 1 A is passed through three coils A, B and C in series, the voltages across them
are 6 V, 3 V and 8 V respectively on a dc supply and 7 V, 5 V and 10 V respectively on an ac supply. Find the
power factor and the power dissipated in each coil and the power factor of the whole circuit.
I=1A
VA = 6 V,
VA = 7 V,
Solution
On dc supply,
On dc supply,
VB = 3 V,
VB = 5 V,
For dc supply, f = 0
f =0
fL
X
The coils behave like pure resistors.
VA 6
= =6Ω
I
1
VB 3
RB =
= = 3Ω
I
1
V
8
RC = C = = 8 Ω
I
1
V
7
ZA = A = = 7 Ω
I
1
RA =
For ac supply,
VB 5
= =5Ω
I
1
VC 10
ZC =
=
= 10 Ω
I
1
ZB =
(a)
XA
Z A2 − RA2 = (7) 2 − (6) 2 = 3 6 Ω
XB
Z B2 − RB2 = (5
(5) 2
XC
ZC2 − RC2 = (10) 2 − (8) 2
(3) 2 = 4 Ω
Power factor of Coil A
pf A =
RA 6
= = 0.857 (lagging)
ZA 7
6Ω
VC = 8 V
VC = 10 V
4.11 Series RL Circuit 4.59
(b)
Power factor of Coil B
pf B =
(c)
Power factor of Coil C
pfC =
(d)
RC
8
=
= 0.8 (lagging)
ZC 10
Power dissipated in Coil A
PA
(e)
RB 3
= = 0.6 (lagging)
ZB 5
I 2 RA = (1) 2 × 6 = 6 W
Power dissipated in Coil B
PB
I 2 RB = (1) 2 × 3 = 3 W
(f) Power dissipated in Coil C
PC
(g)
I 2 RC = (1) 2 × 8 = 8 W
Power factor of the whole circuit
Z = R + jX A + RB + jX B + R + jX = 6 + j 3.6 + 3 + j 4 + 8 + j 6 = 17
17 + j13.6 = 21.77∠38.68
6 °Ω
pfT = cos (38.68°
68 ) = 0.78 (lagging)
Example 4.74
An air-cored coil takes 5 A of current and consumes 600 W of power when connected
across a 200 V, 50 Hz ac supply. Calculate the value of the current drawn by the coil if the supply frequency
increases to 60 Hz.
Solution
For f = 50 Hz,
I = 5 A,
P = 600 W, V = 200 V
Z=
V 200
=
= 40 Ω
I
5
P
I 2r
600 = (5) 2 × r
r = 24 Ω
XL
Z 2 − r 2 = ( 40) 2 − ( 24) 2 = 32 Ω
X
fL
f
32 = 2π × 50 × L
L = 0.1019 H
For f = 60 Hz,
X L = 2π × 60 × 0.1019 = 38.4 Ω
r = 24 Ω
Z
I=
r 2 + X L2 = ( 24) 2 + (38.4) 2 = 45.28 Ω
V
200
=
= 4.417 A
Z 45.28
4.60 Network Analysis and Synthesis
Example 4.75 When an iron-cored choking coil is connected to a 12 V dc supply, it draws a current
of 2.5 A and when it is connected to a 230 V, 50 Hz supply, it draws a 2 A current and consumes 50 W of
power. Determine for this value of current (a) power loss in the iron core, (b) inductance of the coil, (c)
power factor, and (d) value of the series resistance which is equivalent to the effect of iron loss.
Solution
V = 12 V,
V = 230 V,
For dc
For ac
I = 2.5 A
I = 2 A,
P = 50 W
In an iron-cored coil, there are two types of losses.
(a) Losses in core known as core or iron loss
(b) Losses in winding known as copper loss
P = I 2 R + Pi
P
I
2
= R+
RT = R +
Pi
I2
Pi
I2
where R is the resistance of the coil and
For dc supply,
For ac supply,
(a)
I2
is the resistance which is equivalent to the effect of iron loss.
f=0
XL = 0
12
R=
=48Ω
25
230
Z=
= 115 Ω
2
Iron loss
Pi
RT =
XL
(b)
Pi
P − I 2 R = 50 − (2)2 × 4.88 = 30
30 8 W
P
I
2
=
50
(2)2
= 12.5 Ω
Z 2 − RT2 = (115)2
(1122.55))2 = 114.3 Ω
Inductance
XL
fL
114.3 = 2π × 50 × L
L = 0.363 H
(c)
Power factor
(d)
RT 12.5
=
= 0.108 (lagging)
Z
115
The series resistance equivalent to the effect of iron loss
P 30.8
Ri = 2i =
=77Ω
I
(2)2
pf =
4.12 Series RC Circuit 4.61
Example 4.76 An iron-cored coil takes 4 A at a power factor of 0.5 when connected to a 200 V, 50
Hz supply. When the iron core is removed and the voltage is reduced to 40 V, the current rises to 5 A at a pf
of 0.8. Find the iron loss in the core and inductance in each case.
With iron core
I = 4 A,
Without iron core
I = 5 A,
(a) Inductance of the coil
(i) When the iron core is removed,
V 40
Z= =
=8Ω
I
5
R
pf =
Z
R
08=
8
R=64Ω
Solution
XL
pf = 0.5,
pf = 0.8,
V = 200 V
V = 40 V
Z 2 − R 2 = (8)2 − (6.4)2 = 4.8 Ω
XL
fL
4 8 = 2π × 50 × L
L = 0.0153 H
(ii)
With iron core,
V 200
=
= 50 Ω
I
4
R
pf = T
Z
RT
05=
50
RT = 25 Ω
Z=
XL
Z 2 − RT2 = (50)2 − (25)2 = 43.3 Ω
XL
fL
43.3 = 2π × 50 × L
L = 0.1378 H
(b)
Iron loss
Pi P − I 2 R = VI cos φ − I 2 R = 200 × 4 × 0.5 − (4)2 × 6 4 = 297.6 W
4.12
SERIES RC CIRCUIT
Figure 4.70 shows a pure resistor R connected in series with a pure capacitor
C across an alternating voltage v = Vm sin w t.
Let V and I be the rms values of applied voltage and current.
Potential difference across the resistor = VR = R I
Potential difference across the capacitor = VC = XC I
The voltage VR is in phase with the current I whereas voltage
Vc lags behind the current I by 90°.
V VR + VC
R
C
vR
vC
i
v = Vm sin w t
Fig. 4.70 Series RC circuit
4.62 Network Analysis and Synthesis
1. Phasor Diagram Since the same current flows through R and C, the current I is taken as a reference
phasor. Figure 4.71 shows the phasor diagram of series RC circuit.
VR
VR
I
f
f
VC
VC
V
V
(a) Phasor diagram
cos f =
VR
V
(b) Voltage triangle
Fig. 4.71
2. Impedance
V
VR + VC
RI − jX C I
(
j
C )I
V
= R − jX C = Z
I
Z = ∠ −φ
1
ω 2C 2
⎛X ⎞
⎛ 1 ⎞
φ = tan −1 ⎜ C ⎟ = tan −1 ⎜
⎝ R ⎠
⎝ ω RC ⎟⎠
Z = R 2 + X C2 = R 2 +
The quantity Z is called the complex impedance of the series RC circuit.
3. Impedance Triangle
Figure 4.72 shows the impedance triangle of series RC circuit.
R
f
XC
Z
Fig. 4.72
cos f =
R
Z
Impedance triangle
4. Current From the phasor diagram, it is clear that the current I leads the voltage V by an angle f. If the
applied voltage is given by v = Vm sin w t then the current equation will be
i
where
and
Im =
I m sin (ω t + φ )
Vm
Z
⎛X ⎞
⎛ 1 ⎞
φ = tan −1 ⎜ C ⎟ = tan −1 ⎜
⎝ R ⎠
⎝ ω RC ⎟⎠
4.12 Series RC Circuit 4.63
5. Waveforms Figure 4.73 shows the voltage and current waveforms.
v
i
q
f
Fig. 4.73
Waveforms
6. Power
Active power
P
V cos φ
VI
I 2R
Reactive power
Q
V si φ
VI
I 2 XC
Apparent power
S
V
VI
I 2Z
7. Power Triangle Figure 4.74 shows the power triangle of series RC circuit.
P
f
S
Q
cos f =
P
S
Fig. 4.74 Power triangle
8. Power Factor It is defined as the cosine of the angle between voltage and current phasors.
pf = cos φ
From voltage triangle,
From impedance triangle
From power triangle,
VR
V
R
pf =
Z
P
pf =
S
pf =
In case of series RC circuit, the power factor is leading in nature.
Example 4.77 The voltage applied to a circuit is e = 100 sin ( t + 300 ) and the current flowing in
the circuit is i = 15 sin ( t + 60
0 ). Determine impedance, resistance, reactance, power factor and power.
Solution
(a)
Impedance
e 100 sin (ω t + 30°), i 155 sin (ω t + 60°)
E=
I=
100
2
15
2
∠30° V
∠60° A
4.64 Network Analysis and Synthesis
100
∠30°
E
2
I= =
= 6.67∠
∠−
− 30° = 5.77 − j33 33 = R − jX C
15
I
∠60°
2
Z = 6 67 Ω
(b)
Resistance
(c)
Reactance
(d)
Power factor
(e)
Power
R = 5 77 Ω
X C = 3 33 Ω
pf = cos φ = cos (
P
EII cos φ =
E
)° = 0.
100
2
×
15
2
(leading)
× 0.866 = 649.5 W
Example 4.78 A series circuit consumes 2000 W at 0.5 leading power factor, when connected to
230 V, 50 Hz ac supply. Calculate (a) current, (b) kVA, and (c) kVAR.
P = 2000 W,
Solution
(a)
Current
pf = 0.5 (leading), V = 230 V
P VI
V cos φ
2000 = 230 × I × 0.5
I = 17.39 A
(b)
Apparent power
S
(c)
VI
V =
P
2000
=
= 4kVA
cos φ
0.5
Reactive power
φ = cos −1 (0.. ) = 60°
Q = VI sin φ = 230 × 17.39 × sin (60°) = .464 kVAR
Example 4.79
A resistor R in series with a capacitor C is connected to a 240 V, 50 Hz ac supply.
Find the value of C so that R absorbs 300 W at 100 V. Find also the maximum charge and maximum stored
energy in C.
Solution
(a)
V = 240 V, f = 50 Hz,
P = 300 W,
Value of C
VR2
R
(100) 2
300 =
R
R = 33.33 Ω
P=
P
I 2R
300 = I 2 × 33.33
I = 3A
VR = 100 V
4.12 Series RC Circuit 4.65
Z=
V 240
=
= 80 Ω
I
3
Z 2 − R 2 = (80) 2 − (33.33) 2 = 72.72 Ω
XC
1
2π fC
1
72.72 =
2π 50
50 × C
C = 43.77 μ F
XC =
(b)
Maximum charge
VC
V 2 VR2 = ( 240) 2 − (100) 2 = 218.17 V
VC max = 218.17 × 2 = 308.54 V
Qmax = CV
CVCmax = 43.77 × 10 −6 × 308.54 = 0.0135 C
(c)
Maximum stored energy
Emax
1
1
C (VCmaax ) 2 = × 43.
2
2
6
(308.
) 2 = 2.08 J
Example 4.80 A capacitor of 35 mF is connected in series with a variable resistor. The circuit is
connected across 50 Hz mains. Find the value of the resistor for a condition when the voltage across the
capacitor is half the supply voltage.
C = 35 μF
F,
Solution
f = 50 Hz
VC
1
V
2
1
1
=
= 90.946 Ω
2π fC 2π × 50 × 35 × 10 −6
1
VC = V
2
1
X C I = ZI
2
XC =
XC
Z
Z
(2 X C )2
1
Z
2
2 XC
R 2 + X C2
R 2 + X C2
R 2 3X
X C2 = 3 × (90.946) 2 = 24813.35
3
R = 157.5 Ω
Example 4.81 A voltage of 125 V at 50 Hz is applied across a non-inductive resistor connected in
series with a capacitor. The current is 2.2 A. The power loss in the resistor is 96.8 W. Calculate the resistance
and capacitance.
4.66 Network Analysis and Synthesis
V = 125 V,
Solution
(a) Resistance
Z=
P
f = 50 Hz,
I = 2.2 A,
P = 96.8 W
V 125
=
= 56.82 A
I
2.2
I 2R
96.8 = ( 2.2) 2 × R
R = 20 Ω
(b)
Capacitance
Z 2 − R 2 = (56.82) 2 − ( 20) 2 = 53.18 Ω
1
XC =
2π fC
1
53.18 =
2π × 50 × C
C = 59.85 μF
XC
Example 4.82 A resistor and a capacitor are connected across a 250 V supply. When the supply
frequency is 50 Hz, the current drawn is 5 A. When the frequency is increased to 60 Hz, it draws 5.8 A. Find
the values of R and C and the power drawn in the second case.
Solution
V = 250 V,
f1 = 50 Hz,
I1 = 5 A,
f2 = 60 Hz,
I2 = 5.8 A
(a) Values of R and C
For f1 = 50 Hz,
Z1 =
V 250
=
= 50 Ω
I1
5
2
Z1
⎛ 1 ⎞
⎛ 1 ⎞
R2 + ⎜
= R2 + ⎜
⎝ 100π C ⎟⎠
⎝ 2π f C ⎟⎠
2
2
⎛ 1 ⎞
R2 + ⎜
= 2500
⎝ 100π C ⎟⎠
…(i)
For f2 = 60 Hz,
Z2 =
V
250
=
= 43.1 Ω
I 22 5 8
2
Z2
⎛ 1 ⎞
⎛ 1 ⎞
R2 + ⎜
= R2 + ⎜
⎝ 120π C ⎟⎠
⎝ 2π f C ⎟⎠
2
2
⎛ 1 ⎞
R2 + ⎜
= 1857.9 Ω
⎝ 120π C ⎟⎠
Solving Eqs (i) and (ii),
(b)
R = 19.96 Ω
C = 69.4 μF
Power drawn in the second case
P2 I 22 R = (5.8) 2 ×19
19.96 = 671.45 W
…(ii)
4.12 Series RC Circuit 4.67
Table showing various relationships for R, L, and C
R
L
C
Voltage
Vm i ω t
Vm i ω t
Vm i ω t
Current
I m si ω t
I m sin(ω t − 90°)
I m sin(ω t + 90°)
v
v
v
i
Waveform
0
i
i
p
2p
q
0
p
2
q
2p
p
V
Phasor
Diagram
p
2
p
0
I
V
I
I
Impedance
Phase
Difference
Power
Factor
Power
2p
V
R
j L
1
1
=−j
j C
ωC
0°
90°
90°
1
0
0
VI
0
0
Table showing various relationships for series RL and series RC circuits
Series RL Circuit
Voltage
Vm i ω t
Current
I m sin(ω t φ )
Series RC Circuit
Vm i ω t
I m sin(ω t φ )
v
v
i
i
Waveform
q
f
q
f
VR
V
Phasor Diagram
VL
f
I
V
VR
Impedance
R + jjX L , Z ∠
V
f
R − jjX C , Z∠− φ
VC
4.68 Network Analysis and Synthesis
Phase Difference
0° φ < 90°
0° φ < 90°
Power Factor
VR R P
= =
V
Z S
VR R P
= =
V
Z S
Power
P
V cos φ
VI
I 2R
P
VI
V cos φ
I 2R
Q
V sii φ
VI
I 2 XL
Q
V sii φ
VI
I 2 XC
S
V
VI
S
VI
V
4.13
I 2Z
I 2Z
SERIES RLC CIRCUIT
Figure 4.75 shows a pure resistor R, pure inductor L and pure capacitor C connected in series across an
alternating voltage v = Vm sin w t.
R
L
vR
vL
C
vC
i
v = Vm sin w t
Fig. 4.75 Series RLC circuit
Let V and I be the rms values of the applied voltage and current.
Potential difference across the resistor = VR = R I
Potential difference across the inductor = VL = XL I
Potential difference across the capacitor = VC = XC I
The voltage VR is in phase with the current I, the voltage VL leads the current I by 90° and the voltage
VC lags behind the current I by 90°.
V VR + VL VC
1. Phasor Diagram Since the same current flows through R, L and C, the current I is taken as a reference
phasor.
Case (i) XL > XC (Fig. 4.76)
VL
V
f
VL − VC
I
VR
VC
Fig. 4.76 Phasor diagram
The reactance X will be inductive in nature and the circuit will behave like a series RL circuit.
4.13 Series RLC Circuit 4.69
Case (ii) XC > XL (Fig. 4.77)
VL
VR
I
f
V
VC − VL
VC
Fig. 4.77 Phasor diagram
The reactance X will be capacitive in nature and the circuit will behave like a series RC circuit.
2. Impedance
V
VR + VL
VC = RI
j
jX C I = [ R + j ( X L − X C )] I
L
V
= R + j( X L − X C ) = Z
I
Z = ∠φ
Ζ = R2 + ( X L − X C )2
⎛ X − XC ⎞
φ = tan −1 ⎜ L
⎟⎠
⎝
R
3. Impedance Triangles Figure 4.78 shows the impedance triangles of series RLC circuit for two cases.
Case (i) XL > XC
XC > XL
Case (ii)
R
Z
f
XL − XC
XC − XL
Z
f
R
Fig. 4.78
Impedance triangles
4. Current Equation If the applied voltage is given by v = Vm sin w t then current equation will be
i = Im sin (w t ± f)
‘−’ sign is used when XL > XC.
‘+’ sign is used when XC > XL.
5. Waveforms Figure 4.79 shows the voltage and current waveforms.
Case (i) XL > XC
Case (ii)
XC > XL
4.70 Network Analysis and Synthesis
v
v
i
i
q
f
Fig. 4.79
6. Power
Average power
Reactive power
Apparent power
q
f
Waveforms
P = VI cos f = I2R
Q = VI sin f = I2X
S = VI = I2Z
7. Power Triangles Figure 4.80 shows the power triangles of series RLC circuit.
Case (i) XL > XC
Case (ii)
X C > XL
P
S
f
Q
S
f
Q
P
Fig. 4.80 Power triangles
8. Power Factor It is defined as the cosine of the angle between voltage and current phasors.
pf = cos f
pf =
VR R P
= =
V
Z S
Example 4.83
A resistor of 20 W, inductor of 0.05 H and a capacitor of 50 mF are connected in
series as shown in Fig. 4.81. A supply voltage 230 V, 50 Hz is connected across the series combination. Calculate the following: (a) impedance, (b) current drawn by the circuit, (c) phase difference and power factor,
and (d) active and reactive power consumed by the circuit.
R
L
C
230 V, 50 Hz
Fig. 4.81
Solution
(a) Impedance
R = 20 Ω,
L = 0.05 H,
C = 50 μF,
XL = 2p fL = 2p × 50 × 0.05 = 15.71 Ω
V = 230 V,
f = 50 Hz
4.13 Series RLC Circuit 4.71
1
1
=
= 63.66 Ω
2π fC 2π × 50 × 50 × 10 −6
Z = R jjX
X
jX
jX
20 j15.71 − j 63.66 = 51.95
9 ∠− 67.36° Ω
XC =
Z = 51.95 Ω
(b)
Phase difference
(c)
Current
f = 67.36°
I=
(d)
Power factor
(e)
Active power
V
230
=
= 4.43 A
Z 51.95
pf = cos f = cos (67.36°) = 0.385 (leading)
P = VI cos f = 230 × 4.43 × 0.385 = 392.28 W
(f) Reactive power
Q = VI sin f = 230 × 4.43 × sin (67.36°) = 940.39 VAR
Example 4.84 A circuit consists of a pure resistor, a pure inductor, and a capacitor connected in series as shown in Fig. 4.82. When the circuit is supplied with 100 V, 50 Hz supply, the voltages across inductor
and resistor are 240 V and 90 V respectively. If the circuit takes a 10 A leading current, calculate (a) value of
inductance, resistance and capacitance, (b) power factor of the circuit, and (c) voltage across the capacitor.
R
L
10 A
90 V
240 V
C
VC
100 V, 50 Hz
Fig. 4.82
Solution
V = 100 V,
f = 50 Hz,
VL = 240 V,
VR = 90 V, I = 10 A
(a) Value of inductance, resistance and capacitance
VR 90
=
=9 Ω
I
10
V
240
XL = L =
= 24 Ω
I
10
V 100
Z= =
= 10 Ω
I
10
Z = R jjX
X
jjX
X C R − j(
R=
Z
R2 + ( X C
X L )2
10 = (9) 2 + ( X C − 24) 2
X C = 28.36 Ω
X
fL
f
)
4.72 Network Analysis and Synthesis
24 = 2π × 50 × L
L = 0.076 H
1
2π fC
1
28.36 =
2π 50 × C
C = 112.24 μF
XC =
(b)
Power factor of the circuit
pf =
(c)
R 9
=
= 0.9 (leading)
Z 10
Voltage across the capacitor
VC = XC I = 28.36 × 10 = 283.6 V
Two impedances Z1 = 40∠30° W and Z2 = 30∠60° W are connected in series across
a single-phase 230 V, 50 Hz supply as shown in Fig. 4.83. Calculate the (a) current drawn, (b) pf, and
(c) power consumed by the circuit.
Example 4.85
Z1
Z2
230 V, 50 Hz
Fig. 4.83
Solution
(a) Current drawn
Z1
40∠30° Ω, Z 2 = 30∠60° Ω, V = 230 V
Z = Z1 + Z 2 = 40 ∠30° + 30 ∠60° = 67.66 ∠42.81° Ω
V
230
I= =
=34A
Z 67.66
(b)
Power factor
(c)
Power consumed
pf = cos f = cos (42.81°) = 0.734 (lagging)
P = VI cos f = 230 × 3.4 × 0.734 = 573.99 W
Example 4.86 A circuit takes a current of 3 A at a power factor of 0.6 lagging when connected to
115 V, 50 Hz supply. Another circuit takes a current of 5 A at a power factor of 0.707 leading when connected
to the same supply. If the two circuits are connected in series across a 230 V, 50 Hz supply, calculate
(a) current, (b) power consumed, and (c) power factor.
Solution
Circuit 1:
Circuit 2:
I1 = 3 A,
I2 = 5 A,
pf1 = 0.6 (lagging), V1 = 115 V
pf2 = 0.707 (leading), V2 = 115 V
V = 230 V
4.13 Series RLC Circuit 4.73
f1 = cos−1 (0.6) = 53.13°
For Circuit 1,
V1 115
=
= 38.33 Ω
I1
3
Z1 = Z1∠φ1 = 38.33∠53.13° Ω
Z1 =
f2 = cos−1 (0.707) = 45°
For Circuit 2,
V2 115
=
= 23 Ω
I2
5
Z 2 = Z 2 ∠φ 2 = 23∠− 45° Ω
Z2 =
When the two circuits are connected in series,
Z = Z1 + Z 2 = 38.33∠53.13° + 23∠
∠− 45° = 41.82∠20.14° Ω
(a)
Current
I=
V
230
=
= 5.5 A
Z 41.82
(b)
Power consumed
(c)
P = VI cos f = 230 × 5.5 × cos (20.14°) = 1.187 kW
Power factor
pf = cos f = cos (20.14°) = 0.939 (lagging)
Example 4.87
Two impedances Z1 and Z2, having the same numerical value, are connected
in series. If Z1 has a pf of 0.866 lagging and Z2 has a pf of 0.8 leading, calculate the pf of the series
combination.
Solution
pf1 = 0.866 (lagging),
pf2 = 0.8 (leading),
f1 = cos (0.866) = 30°
f2 = cos−1 (0.8) = 36.87°
Z1 = Z2 = Z
−1
Z1 = ∠φ1 = Z ∠30° = 0.866
866 Z + j0
j 0.5 Z Ω
Z 2 = ∠ φ 2 = Z ∠− 36.87° = 0.8 Z − j 0 6 Z Ω
For a series combination,
Z = Z1 + Z 2 = 0.866
866 Z + j 0.5 Z + 0.88 Z − j 0.6 Z = 1.66
66
− j 0. 1
= (1.666 − j 0.1) = 1.668 Z∠− 3.43° Ω
pf = cos(3.43°) = 0.9982
Example 4.88 A coil of 3 W resistance and an inductance of 0.22 H is connected in series with an
imperfect capacitor as shown in Fig. 4.84. When such a series circuit is connected across a 200 V, 50 Hz
supply, it has been observed that their combined impedance is (3.8 + j6.4) W . Calculate the resistance and
capacitance of the imperfect capacitor.
4.74 Network Analysis and Synthesis
Imperfect capacitor
Coil
XL
r
XC
R
200 V, 50 Hz
Fig. 4.84
Solution
r = 3 Ω,
L = 0.22 H,
V = 200 V,
f = 50 Hz,
Z = 3.8 + j6.4 Ω
(a) Resistance of the imperfect capacitor
Z = 3.8 + j 6.4 Ω
X = 2 f = 2π × 50 × 0.22 = 69
69.12 Ω
Z = 3 + j 69.12
69.12 + R −
3 + R = 3.8
R = 0.8 Ω
(b)
= (3 + R) + j (
12 − X C )
Capacitance of the imperfect capacitor
69.12 − X C = 6.4
X C = 62.72 Ω
1
2π fC
1
62.72 =
2π 50 × C
C = 50.75 μF
XC =
Example 4.89 An RLC series circuit has a current which lags the applied voltage by 45°. The
voltage across the inductance has a maximum value equal to twice the maximum value of voltage across the
capacitor. Voltage across the inductance is 300 sin (1000t) and R = 20 W. Find the value of inductance and
capacitance.
Solution
(a) Value of inductance
f = 45°,
vL = 300 sin (1000t),
VL(max) = 2VC(max)
VL = 2VC
IXL = 2I XC
XL = 2XC
R
cos φ =
Z
20
cos( 45°) =
Z
Z = 28.28 Ω
R = 20 Ω
4.13 Series RLC Circuit 4.75
For a series RLC circuit,
R2 + ( X L
Z
X C )2
28.28 = ( 20) 2 + ( 2 X C
X C ) 2 = 400 + X C 2
X C = 20 Ω
X L = 2 C = 40 Ω
XL = ωL
40 = 1000 × L
L = 0.04 H
(b) Value of capacitance
XC =
1
ωC
1
1000 × C
C = 50 μF
20 =
Example 4.90 A coil having a power factor of 0.5 is in series with a 79.57 mF capacitor and when
connected across a 50 Hz supply, the p.d. across the coil is equal to the p.d. across the capacitor. Find the
resistance and inductance of the coil.
Solution
(a) Resistance of the coil
pfcoil = 0.5,
C = 79.57 μF,
f = 50 Hz, Vcoil = VC
1
1
=
= 40 Ω
2π fC 2π 50 79 57 10 −6
VC
= IX C
XC =
Vcoil
IZcoil
X C = 40 Ω
R
pf coil = cos φ =
Zccoil
R
05=
40
R = 20 Ω
Zcoil
(b) Inductance of the coil
XL
2
Zcoil
− R 2 = ( 40) 2 − ( 20) 2 = 34.64 Ω
X
fL
f
34.64 = 2π 50 × L
L = 0 11 H
Example 4.91
A 250 V, 50 Hz voltage is applied to a coil having a resistance of 5 W and an
inductance of 9.55 H in series with a capacitor C. If the voltage across the coil is 300 V, find the value of C.
4.76 Network Analysis and Synthesis
V = 250 V,
Solution
f = 50 Hz,
R = 5 Ω, L = 9.55 H,
Vcoil = 300 V
f = 2π × 50 × 9 55 = 3000 Ω
fL
X
R 2 + X L2 = (5) 2 + (3000) 2 = 3000 Ω
Zcoil
Vccoil
300
=
= 0 1A
Zccoil 3000
V 250
Z= =
= 2500 Ω
I
01
I=
When XL > XC ,
R2 + ( X L
Z
X C )2
2500 = (5) 2 + (3000 −
)2
X C = 500 Ω
1
XC =
2π fC
2
500 =
2π 50 × C
C = 6 37 μF
When XC > XL ,
R2 + ( X C
Z
2500 = (5) 2 + (
X L )2
− 300) 2
X C = 5500 Ω
1
XC =
2π fC
1
2π 50 × C
C = 0 58 μF
5500 =
Example 4.92 Draw the phasor diagram for the series circuit shown in Fig. 4.85 when the current
in the circuit is 2 A. Find the values of V1 and V2 and show these voltages on the phasor diagram.
5Ω
2A
3Ω
6Ω
8Ω
4Ω
V1
V2
Fig. 4.85
Solution
Z1 = j 3 6 − j8 6 − j 5 7.81 ∠
∠− 39.8° Ω
Z 2 = 5 j 3 + 6 j8 + 4 = 15 j 5 = 15.81 ∠− 18.43
. °Ω
I =2A
4.13 Series RLC Circuit 4.77
(a)
Values of V1 and V2
I = 2∠0° Α
Let
(b)
V1
Z1I = (7.81
0 ) = 15.62 ∠
∠− 39.8° V
V2
Z 2 I = (15.81 ∠− 18.43
4 °) ( 2
39.8 ) (2
(2
0° )
0°
31.62 ∠− 18.43° V
Phasor diagram
The Phasor diagram is shown in Fig. 4.86.
I
3918.43°
.8
°
V2
V1
Fig. 4.86
Example 4.93
Draw a vector diagram for the circuit shown in Fig. 4.87 indicating terminal voltages V1 and V2 and the current. Find the value of (a) current, (b) V1 and V2 , and (c) power factor.
10 Ω
20 Ω
0.05 H
V1
0.1 H
50 μF
V2
200 V, 50 Hz
Fig. 4.87
Solution
(a) Current
X
1
ffL
L1
2π × 50 × 0.05 15.71 Ω
X
2
fL2
fL
2π × 50 × 0.1 = 31.42 Ω
XC =
1
1
=
= 63.66 Ω
2π fC 2π 50
50 × 50 10 −6
Z = 10 + j15 71 20
I=
(b)
V1 and V2
Let
jj31
31.42 − j 63.66 = 30 − j16.53 = 34.25∠− 28.85° Ω
V
200
=
= 5 84 A
Z 34.25
I = 5 84 ∠0° A
Z = 10 + j15.71 = 18.62 ∠57.52 Ω
V1 Z1I = (18.62∠57.52°)(5.84 ∠0°) = 108..74 ∠57.52° V
4.78 Network Analysis and Synthesis
Z 2 = 20 + j 31 42 − j 63.66 = 20 − j 32.24 = 37.94 ∠− 58.19° Ω
V2 = Z 2 I = (37.94 ∠− 58.19 ) (5
(5.84 ∠0°) = 221.57∠− 58.19° V
(c)
Power factor
(d)
pf = cos f = cos (28.85°) = 0.875 (leading)
Vector diagram
The vector diagram is shown in Fig. 4.88.
V1
57.52°
58.19°
I
V2
Fig. 4.88
Example 4.94
Find the values of R and C so that Vx = 3Vy , Vx and Vy are in quadrature in the
circuit of Fig. 4.89.
0.0255 H
6Ω
R
C
Vy
Vx
240 V, 50 Hz
Fig. 4.89
Solution
XL = 2p fL = 2p × 50 × 0.0255 = 8 Ω
Z x = 6 + j8 = 10 ∠53.13° Ω
Vx 3Vy
IZ x = 3IZ y
Zx
3Z y
Vx and Vy are in quadrature, i.e., phase angle between Vx and Vy is 90°. Hence, the angle between Zx and Zy
will be 90°. The impedance Zy is capcitive in nature.
Z y = Z y ∠− φ
10
∠(53.13 − 90)°
3
R = 2 66 Ω
XC = 2 Ω
1
XC =
2π fC
1
2=
2π 50 × C
C = 1.59 mF
Zy =
3.33∠− 36.87° = 2.66 − j 2 Ω
4.13 Series RLC Circuit 4.79
Example 4.95 In the circuit of Fig. 4.90, a pure resistor R, a choke coil and a pure capacitor of
15.91 mF are connected in series across a supply of V volts and carries a current of 0.25 A. The voltage
across the choke coil is 40 V, the voltage across the capacitor is 50 V and the voltage across the resistor
is 20 V. The voltage across the combination of R and the choke coil is 45 V. Calculate (a) supply voltage,
(b) frequency, and (c) power loss in the choke coil.
Choke Coil
R
XL
r
C = 15.91 μF
0.25 A
20 V
40 V
50 V
45 V
Fig. 4.90
Solution
C = 15.91 μF,
I = 0.25 A,
Vcoil = 40 V,
VC = 50 V,
VR = 20 V, VR-coil = 45 V
(a) Supply voltage
VR
20
=
= 80 Ω
I
0 25
V
40
Zcoil = coil =
= 160 Ω
I
0 25
V
50
XC = C =
= 200 Ω
I
0 25
V
45
Z R-coil = R-coil =
= 180 Ω
I
0 25
R=
Zcoil
r2
r 2 + X L2 = 160
X L2 = 25600
…(i)
Z R-coil = ( R r ) + j X L
Z R co
( R + r )2
180 = (
(
+ r )2
X L2
X L2
+ r ) 2 + X L2 = 32400
Subtracting Eq. (i) from Eq. (ii),
(80 + r)2 − r2 = 6800
6400 + 160 r + r2 − r2 = 6800
160 r = 400
r = 2.5 Ω
Substituting the value of r in Eq. (i),
( .5) 2 +
2
L
X L2
= 25600
= 25593.75
X L = 159.98 Ω
…(ii)
4.80 Network Analysis and Synthesis
Z = R + r + jX − jX C = 80 + 2.5 + j 159 98 − j200
j 200 = 82.5 − j 40.02 = 91
91.69∠− 25 88° Ω
V = ZI = 91.69 × 0.25 = 22.92
92 V
(b)
Frequency
XC =
200 =
1
2π fC
1
15
15 91 10 −6
2π f
f
(c)
H
Hz
Power loss in the choke coil
Pcoil = I 2 r = (0.25)2 × 2.5 = 0.156 W
Example 4.96
Two impedances, one inductive and the other capacitive, are connected in series
across the voltage of 120∠30° V and a frequency of 50 Hz as shown in Fig. 4.91. The current flowing in the
circuit is 3∠−15°A. If one of the impedances is (10 + j48.3) W, find the other. Also calculate the values of L
and C in the impedances.
Z1
Z2
3∠−15° A
120∠30° V
Fig. 4.91
Solution
(a)
V = 120∠30° V,
f = 50 Hz,
I = 3∠−15° A,
Z1 = 10 + j 48.3 Ω
Impedance Z2
Z=
V 120 ∠30°
=
= 40 ∠45° = 28.28 + j 28.28 Ω
I
3∠− 15°
Z = Z1 + Z 2
Z2
(b)
Z − Z1 = 28.28
8
j 28
28 288 10
Value of L
Z1 = 10 + j 48 3 = R1 + jX L
X L = 48.3 Ω
X = 2 fL
48.3
2π × 50 × L
L = 0.1537 H
j 48.3 = 18.28
28 − j 20.02 Ω
4.14 Parallel ac Circuits 4.81
(c)
Value of C
Z 2 = 18 28
j 20 02 = R2 − j X C
X C = 20.02 Ω
1
2π fC
1
20.02 =
2π 50 × C
C = 159 μF
XC =
4.14
PARALLEL ac CIRCUITS
In parallel circuits, resistor, inductor and capacitor or any combination of these elements are connected
across same supply. Hence, the voltage is same across each branch of the parallel ac circuit. The total current
supplied to the circuit is equal to the phasor sum of the branch currents.
Z1
i1
Z2
i2
i
v = Vm sin w t
Fig. 4.92 Parallel ac circuit
For the parallel ac circuit shown in Fig. 4.92,
1
1
1
=
+
Z Z1 Z 2
Y Y1 + Y2
where Y represents the admittance of the circuit and is defined as the reciprocal of impedance. The real part of
admittance is called conductance (G) and the imaginary part is called susceptance (B) and these are measured
in mhos ( ) or siemens (S).
If Z1
jjX
X
d Z 2 = − jX C ,
1
1
1
=
+
Z R + jX
− jX C
R − jX L
1
= 2
+j
2
XC
R + XL
=
R
2
R +
X L2
= G + jB
+j
⎛ 1
⎞
X
− 2 L 2⎟
X
⎝ C R + XL ⎠
4.82 Network Analysis and Synthesis
R
G=
where,
2
R
X L2
1
X
B=
− 2 L 2
XC R
XL
The current in the parallel ac circuit can be found as the phasor sum of the branch currents,
I = I1 + I 2
i.e.,
Note: 1. For a series RL circuit,
Z = R + jX L
G
1
1
R − jX L
R
X
Y=
=
= 2
= 2
− j 2 L 2 = G − jBL
2
2
Z R + jX L R + X L R + X L
R + XL
G=
where
R
R2
X2
and BL =
Y
XL
R2
Fig. 4.93
Admittance triangle
For a series RC circuit,
Z = R − jX C
1
1
R + jX C
R
X
Y= =
= 2
= 2
+ j 2 C 2 = G + jBC
2
2
Z R − jX C R + X C R + X C
R + XC
where G =
R
R
2
X C2
and BC =
XC
R
2
BL
X L2
The admittance triangle for series RL circuit is shown in Fig. 4.93.
2.
f
Y
BC
f
X C2
G
The admittance triangle for series RC circuit is shown in Fig. 4.94.
Fig. 4.94
Admittance triangle
Example 4.97 A coil having a resistance of 50 W and an inductance of 0.02 H is connected in
parallel with a capacitor of 25 mF across a single-phase 200 V, 50 Hz supply as shown in Fig. 4.95. Calculate
the current in coil and capacitance. Calculate also the total current drawn, total pf and total power consumed by the circuit.
I1
R
I2
C
L
I
200 V, 50 Hz
Fig. 4.95
R = 50 Ω,
Solution
L = 0.02 H,
C = 25 μF,
V = 200 V, f = 50 Hz
(a) Current in coil and capacitance
ffL = 2π × 50 × 0.02 = 6.28 Ω
1
1
XC =
=
= 7.3 Ω
2π fC 2π × 50 × 25 × 10 −6
Z 2 = R jX
0 j 6.288 = 50
50.39∠ 7. 6° Ω
X
4.14 Parallel ac Circuits 4.83
Z 2 = − jX C = − j127.32 = 127
12 .32∠ − 90° Ω
V = 200 ∠ 0° = 200 V
V
200
=
= 3.97
9 ∠− 7.16°A
Z1 50.39∠7
∠7.16°
V
200
I2 =
=
= 1.57
5 ∠90°A
Z 2 127.32∠− 90°
I1 =
(b)
Total current
I = I1 + I 2 = 3.997∠
∠− 7.16° + 1.57∠90° = 4.08∠ 5. 7°A
(c)
Total pf
(d)
pf = cos f = cos (15.27°) = 0.965 (lagging)
Total power consumed
P = VI cos f = 200 × 4.08 × 0.965 = 787.44 W
Two impedances Z1 30 ∠ ° Ω d Z 2 = 45 ∠30° Ω are connected in parallel
across a single-phase 230 V, 50 Hz supply. Calculate (a) current drawn by each branch, (b) total current,
and (c) overall power factor. Also draw the phasor diagram indicating the current drawn by each branch and
the total current, taking the supply voltage as reference.
Example 4.98
Solution
(a)
Z1
30 ∠ 4 ° Ω
Z 2 = 45
4 ∠ 30° Ω, V = 230 V
Current drawn by each branch
Let V
= 230 ∠ 0° = 230 V
V
230
=
= 7 67 − 45° A
Z1 30 ∠ 455°
V
230
I2 =
=
= 5 11 − 30° A
Z 2 45∠ 30°
I1 =
(b)
45°
V
30°
39.01
Total current
I = I1 + I 2 = 7.67∠
∠− 45° + 5.11∠
∠− 30° = 12.67
6 ∠
∠− 39.01° A
Overall power factor
pf = cos f = cos (39.01°) = 0.777 (lagging)
(d) Phasor diagram
The phasor diagram is shown in Fig. 4.96.
I2
(c)
I1
I
Fig. 4.96
Example 4.99
Two circuits, the impedances of which are given by Z1 (10 j15) h
d
are
connected
in
parallel
across
an
ac
supply
as
shown
in
Fig.
4.97.
If
the
total
current
Z 2 = (6 j8 ) ohms,
supplied is 15 A, what is the power taken by each branch?
Z1
Z2
15 A
Fig. 4.97
4.84 Network Analysis and Synthesis
Solution
Z1
Z=
j15 Ω
10
Z 2 = 6 − j8 Ω, I = 15 A
Z1Z 2
(10 + j 15)(6
(6 − j8)
=
= 10.32∠
∠− 20.45° Ω
Z1 Z 2 10 + j 15 + 6 − j8
V = ZI = 10.332
2 15 = 154.8 V
Let
V = 154.8∠ 0° = 154.8 V
I1 =
V
154.8
=
= 8.59∠
∠−
− 56.31° A
Z1 10 + j 15
I2 =
V 154.8
=
= 15.48 ∠ 53.13°A
Z 2 6 j8
P1 = I12 R1 = (8.59) 2 10
10 = 737.88 W
P2 = I 22 R2 = (15.48) 2 × 6 = 1437.78 W
Example 4.100 A circuit consists of a 25 W resistor, 64 mH inductor and 80 mF capacitor connected in parallel across a 110 V, 50 Hz single-phase supply as shown in Fig. 4.98. Calculate the individual
currents drawn by each element, the total current drawn from the supply and the overall power factor of the
circuit. Draw the phasor diagram.
R
L
I
C
110 V, 50 Hz
Fig. 4.98
Solution
R = 25 Ω,
L = 64 mH,
C = 80 μF,
V = 110 V, f = 50 Hz
(a) Individual currents
X
XC =
f = 2π × 50 × 64 × 10 −3 = 0.
fL
Ω
1
1
=
= 39.79 Ω
2π fC 2π × 50 × 80 × 10 −6
Z1 = R = 25 Ω
Z 2 = jX L = j 20.11 Ω
Z3 = − jX C = − j 39.79 Ω
4.14 Parallel ac Circuits 4.85
Let
V = 110 ∠ 0° = 110 V
V 110
I1 =
=
= 4.4 ∠ 0°A
Z1
25
I3
V
110
=
= 5 47 − 90° A
Z2
j 20.11
110
V
I3 =
=
= 2 76 90° A
Z3 − j 39.79
I2 =
I1
V
31.63°
(b) Total current
I = I1 + I 2 I3
= 4.4 ∠ 0° + 5.47∠− 90° + 2.76
76∠
∠ 90° = 5.17∠− 31.63° A
Overall power factor
pf = cos f = cos (31.63°) = 0.851 (lagging)
(d) Phasor diagram
The phasor diagram is shown in Fig. 4.99.
I
(c)
I2
Fig. 4.99
Example 4.101 An ac circuit connected across a 200 V, 50 Hz, supply has two branches A and B.
Branch A draws a current of 4 A at 0.8 lagging power factor, while the total current drawn by the parallel
combination is 5 A at unity power factor as shown in Fig. 4.100, Find (a) current and power factor of branch
B, and (b) admittances of branch A and B, and their parallel combination both in polar and rectangular
forms.
IA
IB
ZA
ZB
I
200 V, 50 Hz
Fig. 4.100
V = 200 V, f = 50 Hz, I A = 4 A
Solution
(a) Current and power factor of Branch B
φ A = cos −1 (0.. ) = 36.87°
pf A
pf
0.8 (lagging), I = 5 A, pf = 1
φ = cos −1 (1)) = 0°
I A = I A ∠− φ A = 4 ∠− 36.87° A
I = I ∠φ = 5 ∠ 0° A
I = IA + IB
I B I − I A = 5 ∠0° − 4 ∠− 36.87° = 3 ∠53. 3° A
pf B = cos (53.
°) = 0..
(leading)
4.86 Network Analysis and Synthesis
(b)
Admittances of branches A and B and their parallel combination
V = 200 ∠ 0° = 200 V
I
4 ∠− 36 87°
YA = A =
= 0.02 ∠
∠−
− 36.87°
V
200 ∠ 0°
Let
YB =
I B 3 ∠53 13°
=
= 0.0015 ∠53.13°
V
200 ∠0°
YA + YB = 0.02 ∠
∠−
− 36.87° + 0.015
0 ∠53.13°
Y
= 0.025
02 ∠0°
Example 4.102
Two circuits A and B are connected in parallel to a 115 V, 50 Hz supply as shown in
Fig. 4.101. The total current taken by the combination is 10 A at unity power factor. Circuit A consists of a 10 Ω
resistor and 200 mF capacitor connected in series. Circuit B consists of a resistor and an inductor in series.
Determine (a) current, (b) power factor, (c) impedance, (d) resistance, and (e) reactance of the circuit B.
ZA
IA
RA
C
ZB
RB
10 A
XB
IB
115 V, 50 Hz
Fig. 4.101
Solution
(a)
V = 115 V
V,
Current IB
f = 50 Hz, I = 10 A
A, pf
1,
RA = 0 Ω, C = 200 μF
1
1
XC =
=
= 5.9 Ω
2π fC 2π × 50 × 200 × 10 −6
Z A = 10 − j 15.92 = 18.8 ∠−
− 57 87° Ω
V
115
=
= 6.12 ∠57.87° A
Z A 18.8 ∠
∠− 57.87°
I = 10 ∠0° A
I B I I A = 10 ∠0° − 6.12 ∠57.87° = 8.5 ∠− 37.54° A
IA =
(b)
(c)
Power factor of Circuit B
pf B = cos(37.
°) = 0..
(lagging)
Impedance of Circuit B
Z
(d)
Resistance of Circuit B
(e)
Reactance of Circuit B
V
IB
115
8.5 ∠
∠− 37.54°
RB = 0.73 Ω
XB = 8
Ω
13 3 ∠3
4° Ω 10 3
j 8.24 Ω
4.14 Parallel ac Circuits 4.87
Example 4.103 Two circuits, the impedances of which are given by Z
= −
and are connected in parallel as shown in Fig. 4.102. If the applied voltage to the combination is 100 V, find
(a) current and pf of each branch, (b) overall current and pf of the combination, and (c) power consumed by
each impedance.
I1
Z
I2
Z
I
100 V
Fig. 4.102
Solution
(a)
Ω, V
=
Z
100 V
Current and pf of each branch
0
V
I1
I2
V
Z1
V
Z2
100 V
100
6+ 8
100
=
8 6
A
∠3
A
cos f1 = cos (53.13°) = 0.6 (lagging)
cos f2 = cos (36.9°) = 0.8 (leading)
(b)
Overall current and pf of the combination
+
.
(c)
∠ .9
1
∠− 13° A
pf = cos f = cos (8.13°) = 0.989 (lagging)
Power consumed by each impedance
Example 4.104
2
= (10) 2 ×
= 600 W
2
(10) ×
= 800 W
2
Two impedances
are connected in parallel across a supply
π
voltage v = 100 2 sin 314 t. The current flowing through two impedances are
4
π
respectively. Find the equation for instantaneous value of total current drawn
10 2 sin 314 t
4
from the supply. Also find values of R1, R2, X
1
and
2
Solution
v 100
i =
+
4
t
−
4
4.88 Network Analysis and Synthesis
(a)
Instantaneous value of total current
Writing v, i1 and i2 in polar form,
V = 100 ∠0° V
I1 = 10 ∠45°A
I 2 = 10 ∠− 45°A
I = I1 + I 2 = 10 ∠45° + 10 ∠ 45° = 14.14 ∠0° A
i = I m sin 2π ft = 1144.14 2 sin 314 t = 20
2 sin 314 t
(b)
Values of R1, R2, X C1
d X L2 .
V 100 ∠0°
=
= 10 ∠− 45° = 7.07
07
I1 10 ∠45°
R1 = 7 07 Ω
X C1 = 7 07 Ω
Z1 =
j 7.07 Ω
V
100 ∠0°
=
= 10 ∠45° = 7.07
07 + j 7.07 Ω
I 2 10 ∠− 45°
R2 = 7 07 Ω
Z2 =
X L2 = 7 07 Ω
An impedance of (7 + j5) W is connected in parallel with another impedance of
(10 − j8) W across a 230 V, 50 Hz supply as shown in Fig. 4.103. Calculate (a) admittance, conductance and
susceptance of the combined circuit, and (b) total current and power factor.
Example 4.105
Z1
Z2
I
230 V
Fig. 4.103
Solution
(a)
j Ω
Z1
Z 2 = 10 − j8 Ω, V = 230 V
Admittance, conductance and susceptance of the combined circuit
1
1
=
= 0.12 − 35.54°
Z1 7 + j5
1
1
Y2 =
=
= 0.08 38.66°
Z 2 10 − j8
Y1 =
Y
Y1 + Y2 = 0.122 ∠
∠−
− 35.54° + 0.08 ∠ 38.66°
= 0.16
6∠
∠−
− 7.04°
= 0.166 − j 0.02
0
4.14 Parallel ac Circuits 4.89
Y = 0 16
G = 0 16
B = 0 02
(b)
Total current and power factor
I VY=(
I = 36.8 A
∠ °) ( .16 ∠− 7..
°) = 36.8 ∠
∠− 7.04° A
pf = cos f = cos (7.04°) = 0.99 (lagging)
Example 4.106
Two impedances Z1 and Z 2 are connected in parallel across a 200 V, 50 Hz ac
supply. The current drawn by the impedance Z1 is 4 A at 0.8 lagging pf. The total current drawn from the
supply is 5 A at unity pf. Calculate the impedance Z 2 .
Solution
V
200 ,
I1 = 4 A
I1 = 4
I=5
0.8 lagging pf, I = 5 A at unity pf
s 1 (0.8)
8)
4 ∠− 36.87° A
1
0° A
s (1) = 5
I = I1 + I 2
I 2 I I1 = 5 0°
0 − 4 36.87° 3∠ 53.133° A
V
200 ∠0°
Z2 =
=
= 66.67∠ 53 13° Ω
I 2 3∠ 53 133°
Example 4.107
Compute Zeq and Yeq for the circuit as shown in Fig. 4.104.
5Ω
j5 Ω
− 10 Ω
−j
15 Ω
j 8.66 Ω
Fig. 4.104
Solution
j Ω
Z1
Z2
5 + j8.66 Ω, Z3
Y1 + Y2 + Y3 + Y4 =
Yeq
=
Zeq =
15 Ω
Z4
− j10
10 Ω
1
1
1
1
+
+
+
Z1 Z 2 Z3 Z 4
1
1
1
1
+
+ +
= 0.222 ∠
∠−
− 57.99°
j 5 5 + j 8.66 15 − j 10
1
1
=
= 4.54 ∠ 57.99° Ω
Yeq 0.22 ∠
∠− 57.99°
4.90 Network Analysis and Synthesis
Example 4.108
Find currents I1
d I 2 in Fig. 4.105.
25 ∠ 90° A
I1
I2
3Ω
10 Ω
−4Ω
−j
Fig. 4.105
Solution
Z1
3
j4 Ω
Z 2 = 10 Ω, I = 25 ∠ 90°A
By current-division rule,
I1
I2
Z2
10
= ( 25 ∠ 90°)
= 18.38 ∠ 107.1° A
Z1 Z 2
3 j 4 + 10
I I1 = 25∠ 90° − 18
1 .38 ∠ 107.1° 9.19 ∠ 54° A
I
Three impedances of 25 ∠ 53.1°Ω , 5 ∠− 53
53.
53.1°
3 1 Ω and 10 ∠ 36.9°Ω are connected
in parallel. The combination is in series with another impedance of 14.14 ∠ 45°Ω . Calculate the equivalent
impedance of the circuit.
Example 4.109
Solution
2 ∠ 53 1° Ω
25
Z1
Y=
Z2
5 ∠− 53.1° Ω , Z3
10 ∠36 9° Ω
Z 4 = 14.14 ∠45° Ω
1
1
1
1
1
1
1
=
+
+
=
+
+
= 0.23 ∠ 16.86 °
Z Z1 Z 2 Z3 25 ∠ 53. ° 5 ∠
∠− 53.1° 10 ∠ 36.9°
1
= 4.27 ∠
∠− 16.86 ° Ω
Y
= Z + Z 4 = 4.27
7∠
∠−
− 16.86° + 14.144 ∠ 45° = 16
16.58 ∠ 31.87
. °Ω
Z=
Zeq
A voltage of 200 ∠ 255° V is applied to a circuit composed of two parallel branches.
If the branch currents are 10 ∠ 40° A and 20 ∠ 30° A, determine the kVA, kVAR and kW in each branch.
Also, calculate the pf of the combined load.
Example 4.110
Solution
V
200 ∠ 2 ° , I1 = 10 ∠ 0° A,
A I 2 = 20 ∠− 30°A
Phase difference between V and I1
f1 = 40° − 25° = 15°
Phase difference between V and I2
f2 = 25° − (−30°) = 55°
cos f1 = cos (15°) = 0.97 (leading)
cos f2 = cos (55°) = 0.57 (lagging)
4.14 Parallel ac Circuits 4.91
(a)
(b)
(c)
kVA, kVAR and kW for the branch current of 10 ∠ 40° A
P1
Q
VII1 cos φ1 = 200 × 10 × 0.97 1.94 kW
V
VII1 sin φ1 = 200 × 10 × sin (15°) = 0.52 kVAR
V
S1
VII1 = 200 × 10 = 2 kVA
V
kVA, kVAR and kW for the branch current of 20 ∠− 30° A
P2
VII 2 cos φ2 = 200 × 20 × 0.57
V
Q
S2
VII 2 sin φ2 = 200 × 20 × sin (55°) = 3.28 kVAR
V
VI 2 = 200 × 20 = 4 kVA
VI
2.28 kW
Power factor of the combined load
I1 = 10 ∠ 40° A
I 2 = 20 ∠− 30° A
I = I1 + I 2 = 10 ∠ 40° + 20 ∠− 30° = 25.24 ∠
∠− 8.14°A
φ = 25° − ( −8.14°°) = 33.14°
pf = cos (33.14) = 0.84 (lagging)
Example 4.111 The load taken from a supply consists of a (a) lamp load of 10 kW at unity power
factor, (b) motor load of 80 kVA at 0.8 power factor lagging, and (c) motor load of 40 kVA at 0.7 power
factor lagging. Calculate the total load taken from the supply in kW and in kVA and the power factor of the
combined load.
Solution
Lamp load: P1 = 10 kW, pf1 = 1
Motor load: S2 = 80 kVA, pf2 = 0.8 (lagging)
Motor load: S3 = 40 kVA, pf3 = 0.7 (lagging)
(a) Total load in kW
For motor loads,
(b)
P2
S2 × pf 2 = 80 × 0 8 = 64 kW
P3
S3 × pf3 = 40 × 0 7 = 28 kW
P
P1 + P2
S1 =
P1 10
=
= 10 kVA
pf1 1
S1 + S2 S3 = 10 + 80 + 40 = 130 kVA
Total load in kVA
For lamp load,
S
(c)
P3 = 10 + 64 + 28 = 102 kW
Power factor of the combined load
pf =
P 102
=
= 0.785 (lagging)
S 130
4.92 Network Analysis and Synthesis
Example 4.112 Two circuits have the same numerical value of impedances. The pf of one is 0.8
lagging and that of the other is 0.6 lagging. What is the pf of combination if they are connected in parallel?
Solution
pf1 = 0.8 (lagging),
pf2 = 0.6 (lagging)
−1
Z1 = Z ∠ cos (0.8)
8) = Z ∠ 36.87° Ω
Z 2 = Z ∠ cos −1 (0.6)
6) = Z ∠ 53.13° Ω
For parallel combination,
Z=
Z1Z 2
( ∠ 36.87°)( Z ∠ 53.13°)
=
Z1 Z 2
Z ∠ 36.87° + ∠ 53.13°
=
Z 2 ∠90°
Z 2 ∠90°
=
= 0.505
Z (1 4 + j1.4) 1 98Z ∠45°
∠45° Ω
pf = cos(45°) = 0.707 (lagging)
Example 4.113 When a 240 V, 50 Hz supply is fed to a 15 W resistor in parallel with an inductor,
the total current is 22.1 A as shown in Fig. 4.106. What value must the frequency have for the total current
to be 34 A?
I1
I2
15 Ω
L
22.1 A
240 V, 50 Hz
Fig. 4.106
240 , R = 15 Ω, I = 22.1 A
Solution
V
Let
V = 240 ∠ 0° V
I1 =
V 240 ∠ 0°
=
= 16 ∠ 0° = 16 A
R
15∠ 0°
V
240 ∠ 0°
240
240
=
=
∠− 90° = − j
A
jX L
X L ∠90°
XL
XL
240
I = 16 − j
XL
I2 =
2
⎛ 240 ⎞
( )2 + ⎜
= 22.1
⎝ X L ⎟⎠
256 +
57600
X L2
= 488.41
X L = 5.7 Ω
X
ffL
4.14 Parallel ac Circuits 4.93
15.74 = 2π 50 × L
L = 0 05 H
Let the new frequency be f for the total current of 34 A.
2
⎛
⎞
240
( )2 + ⎜
= 34
⎝ 2π f 0.05 ⎟⎠
256 +
57600
0.0987 f 2
= 1156
f = 25.47 Hz
Example 4.114
Determine the current in the circuit of Fig. 4.107. Also, find the power consumed
as well as pf.
6Ω
4Ω
0.02 H
2Ω
200 μF
0.01 H
100 V, 50 Hz
Fig. 4.107
Solution
X
X
1
fL1
fL
2π × 50 × 0.01 = 3.14 Ω
2
fL2
fL
2π × 50 × 0.02 = 6.288 Ω
XC =
1
1
=
= 15.92 Ω
2π fC
f
2π × 50 × 200 × 10 −6
Z1 = 6
Z2 = 4
j 3.14 Ω
j 6.28 Ω
Z3 = 2
j15 92 Ω
Z 2 Z3
( 4 + j 6.28)( 2 − j15.92)
= (6 + 3.14 ) +
= 17.27
.27
27 ∠30.75° Ω
Z 2 Z3
( 4 + j 6.28) + ( 2 − j15
1 .92)
V
100 ∠0°
I= =
= 5.79
9 − 30.75° A
Z 17.27∠30.75°
Z = Z1 +
P = VI cos f = 100 × 5.79 × cos (30.75°) = 497.94 W
pf = cos f = cos(30.75°) = 0.86 (lagging)
Example 4.115
Two impedances Z A = (
j3) Ω
d Z B = (10 - j7) Ω are connected in parallel
and impedance ZC = (6 + j5)
5 Ω is connected in series with parallel combination of Z A
d Z B as shown
4.94 Network Analysis and Synthesis
in Fig. 4.108. If the voltage applied across the circuit is 200 V at 59 Hz, calculate (a) currents flowing in
,
and Z , and (b) total power factor of the circuit.
ZA
IA
ZC
IC
IC
ZB
IB
200 V, 59 Hz
Fig. 4.108
Solution
(a)
Z = +
Z
Currents flowing in
B
V = 200 V,
f
59 Hz
and ZC
Z A = + = ∠ 6 87° Ω
Z B = − j 7 = 12 2 ∠
Ω
ZC
+
Ω=
+
Z
IC =
1∠
=
Ω
∠ 6 87
3
1
V
200 ∠ °
=
Z 11 8 ∠
ZB
)
+
9 81
1
∠
17 Ω
∠
A
A
=
∠
°
−IA
(b)
∠−
1
( .21 ∠− 35° )
∠
°+
∠
∠
=
5
4° A
Total power factor of the circuit
pf = cos (32.17°) = 0.85 (lagging)
Example 4.116
In a series–parallel circuit, the parallel branches A and B are in series with Branch
16
C as shown in Fig. 4.109. The impedances are Z A
−
and ZC =
8 ) If
3
the current Ic = (25 + j0) A, determine the branch currents, voltages and the total voltage. Hence, calculate
active and reactive powers for each branch and the whole circuit.
IA
ZA
IC
IB
Z
Z
VA = V
V
V
Fig. 4.109
4.14 Parallel ac Circuits 4.95
Solution
(a)
16
3
Z
=
, Z
4
=
j
+j
= 25
IB
°−
16
3
16
3
°= ∠
+
∠
−
. °A
3 A
Branch voltages and total voltage
ZA
(
A
=( +
=
(c)
A
Branch currents
By current-division rule,
ZB
(b)
∠
) ( 25
∠−
∠−
.87° =
=
∠− .02 V
∠ 75 96° V
75 ° = 249 7
°
°
Active and reactive powers for each branch and the whole circuit
2
=
2
×
= 1600 W
2
=
2
×
= 900 W
2
=
2
2
=
2
× = 1200 VAR
2
=
2
×
2
Example 4.117
=
× = 1250 W
= 1600 +
+ 1250 = 3750 W
2
16
= 1200 VAR
3
× = 5000 VAR
= 1200 +
+ 5000
7400 VAR
Find the applied voltage VAB so that a 10 A current may flow through the capacitor
in Fig. 4.110.
I1
2Ω
0.0191
8 Ω 0.0318 H
A I
I
398 μ
7Ω
Fig. 4.110
Solution
π×
×
× 0 0191 6
× 0 0318 10
B
4.96 Network Analysis and Synthesis
XC =
1
1
=
=8Ω
2π fC 2π × 50 × 398 × 10 −6
Z1 = 2 + j 6 = 6.32 ∠71.56° Ω
Z2 = 7
j8 = 10.63 ∠− 48.8° Ω
Z3 = 8
j10 12.8 ∠51.34° Ω
Z=
Z1Z 2
(2
2 j6
6)(7
+ Z3 =
Z1 Z 2
(2 j6 + 7
j8)
+(8
j8)
j10) 19.91
9 ∠45.53° Ω
I 2 = 10 ∠ 0°A
V2 Z 2 I 2 = (10.63 ∠− 48.8°) (10 ∠ 0°) = 106.3
Let
V1
V2 = 106.3
I1 =
V1 106.3
48.8°
=
= 16.82 ∠− 120
1 .36° A
Z1
6.32 ∠ 71.56°
48.8° V
48.8° V
I = I1 + I 2 = 16.82 ∠
∠− 120.36° + 10 ∠ 0° = 14.58 ∠
∠− 84.09° A
VAB Z I = (19.91 ∠ 45.53 ) (14.58 ∠− 84.09 ) = 290.28
.288 ∠− 38.56° V
Example 4.118
If a voltage of 150 V applied between terminals A and B produces a current of 32 A
for the circuit shown in Fig. 4.111, calculate the value of the resistance R and pf of the circuit.
5Ω
A
R
B
j4 Ω
Fig. 4.111
Solution
V
1 0 V, I = 32 A
Z=
150
= 4.687 Ω
32
Z=
(5)( j4
j )
(5)( 4 ∠ °)
+R=
+ R = 3.125 ∠ 51
5 .34
34° + R = 1.95
5 + j4
64 ∠ 38.66°
( .95
)2
( 2.
) 2 = 4.687
( .95
)2
( 2.
) 2 = ( .687) 2
( .95
1.95
2
)
( 4.
4
)2 − (
)2
j 2.44 + R
4.14 Parallel ac Circuits 4.97
R=
05 Ω
Total resistance 1.95 + 2.05
pf =
=
= 0 853
5 (
Total impedance
4.687
)
Example 4.119 An impedance of R + jX ohms is connected in parallel with another impedance of
−j5 ohms as shown in Fig. 4.112. The combination is then connected in series with a pure resistance of 2 W.
When connected across a 100 V, 50 Hz ac supply, the total current drawn by the circuit is 20 A and the total
power consumed by the circuit is 2 kW. Calculate (a) currents through parallel branches, and (b) R and L.
IX
R
X
2Ω
− 5
−j
IC
20 A
VP
VR
100 V, 50 Hz
Fig. 4.112
Solution
(a)
P
2 k , V = 100 V, I = 20 A
Currents through parallel branches
P
VI
V cos φ
2000 = 100 × 20 × cos φ
cos φ = 1
φ = 0°
V = 100 ∠ 0° V
I = 20 ∠ 0° A
VR = 2 × 20° ∠ 0° = 40 ∠ 0° V
VP V − VR = 100 ∠ 0° − 40 ∠ 0° = 60 ∠ 0° V
IC =
IX
(b)
VP
60 ∠ 0°
=
= 12 ∠ 90° A
ZC 5 ∠− 90°
I IC = 20 ∠ 0° − 12 ∠ 90° = 23.32 ∠ − 30.96
9 °A
Resistance and inductance
60 ∠ 0°
= 2.57 ∠30.96° = 2.2 + j .3 Ω
23.32 ∠
∠− 30.96°
R= . Ω
XL = 3 Ω
ZX =
4.98 Network Analysis and Synthesis
X
ffL
1 32 = 2π × 50 × L
L = 4.2 mH
Example 4.120 The circuit of Fig. 4.113 takes 12 A at a lagging power factor and dissipates 1800
W. The reading of the voltmeter is 200 V. Find R1, X1 and X2.
X1
R1
10 Ω
20 Ω
200 V
V
X2
Fig. 4.113
12 A, P = 1800 W
Solution
I
Let
I = 12 ∠0° A
Z2 =
200
= 16.67 Ω
12
Z 2 = R22 + X 22
16.67
(10) 2 + X 22
277.88 = 100 + X 22
X 2 = 13.33 Ω
V2 = (12 ∠0°) (10 + j13.33)
P
(12
0°) (16.67 ∠53.13°) = 200 ∠53 13° V
V
VI cos φ
1800 = 200 × 12 × cos φ
cos φ 0.75
φ = 41.41°
Applied voltage
req
200 ∠41 41° V
Voltage across parallel branches = 200 ∠41.41° − 200 ∠533.13° = 40.84 ∠− 42.73° V
Current through capacitor =
Current through R1 and
a d
40.84 ∠
∠− 42.73°
= 2.04 ∠47.27°A
20 ∠− 90°
A
1
40.84 ∠
∠− 42.73°
= 3.81
10.72 ∠
∠− 8.03°
R1 = 3 13 Ω
X1 = . 7 Ω
Z1 =
34.7° = 3.13
j 2.17
4.14 Parallel ac Circuits 4.99
Example 4.121 For the circuit shown in Fig. 4.114, calculate (a) total admittance, total conductance and total susceptance, (b) total current and total pf, and (c) value of pure capacitance to be connected
in parallel with the above combination to make the total pf unity.
6Ω
8Ω
8Ω
6Ω
200 V, 50 Hz
Fig. 4.114
Solution
(a) Total admittance, total conductance and total susceptance
Z1 = 6
Z2 = 8
j8 = 10 ∠ 53.13° Ω
j 6 = 10 ∠− 36.87° Ω
1
1
=
= 0.1 ∠
∠− 53.13°
Z1 10 ∠ 53.13°
1
1
Y2 =
=
= 0.1 ∠ 36.87°
Z 2 10 ∠− 36.87°
Y1 =
Y
Y1 + Y2 = 0.1 ∠−
∠ 53.13° + 0.1 ∠ 36.87° = 0.14
14 − j 0.02
= 0.14 ∠
∠− 8.133°
Y = 0 14
G = 0 14
B = 0 02
(b)
Total current and total pf
Let
V = 200 ∠ 0° V
I V Y = ( 200 ∠ 0°°) (0.14 ∠− 8.13°°) = 28 ∠− 8.13° A
pf = cos (8.13°) = 0.989 (lagging)
(c) Value of pure capacitance to make total pf unity
Since the current lags behind voltage, the circuit is inductive in nature. In order to make the total pf
unity, a pure capacitor is connected in parallel so that pf becomes unity and imaginary part of Yreq
becomes zero.
Yreq Y1 + Y2 Y3
Yreq = 0.14
14 − j 0.02 + j 0.02 = 0.14
Y3 = YC =
1
= 0 02
XC
X C = 50 Ω
C=
1
= 63.66 μF
2π × 50 × 50
4.100 Network Analysis and Synthesis
4.15
LOCUS DIAGRAM
Locus diagrams are useful in analysing the behaviour of an RLC circuit when one of its parameters, i.e, R, L
or C is varied keeping frequency and voltage constant. The magnitude and phase of a current phasor in the
circuit depends upon the values of R, L and C and frequency of the supply.
The path traced by the terminus of the current phasor when any one parameter R, L or C is varied keeping
f and V constant is called current locus.
Locus diagrams can also be drawn for impedance and admittance when frequency is varied.
4.15.1 Locus Diagram of a Series RL Circuit
Figure 4.115 shows a series RL circuit with variable resistance and constant reactance.
XL
R
I
V
Fig. 4.115
I=
V
=
Z
V
R
2
X L2
In a series RL circuit, current I lags behind voltage V by an angle f. Figure 4.116 shows the phasor diagram
and impedance triangle for a series RL circuit.
V
f
I
Z
XL
f
R
Fig. 4.116 Phasor diagram and impedance
triangle of a series RL circuit
R
Z
X
sin φ = L
Z
cos φ =
…(4.1)
The current I has two components, Ix and Iy,
Ix
Iy
I cos φ =
V R VR
VR
= 2 = 2
ZZ Z
R
X L2
I sin φ = −
V XL
VX
VX
.
= − 2L = − 2 L 2
Z Z
Z
R + XL
…(4.2)
…(4.3)
4.15 Locus Diagram 4.101
Squaring and adding Eqs (4.2) and (4.3),
I x2
I y2 =
V 2 R2
V 2 X L2
V 2 ( R 2 + X L2 )
V2
+
,
=
,
=
( R2 X L 2 )2 ( R2 X L )2
( R2 X L 2 )2
R 2 + X L2
From Eq. (4.3),
I x2
Adding ⎛⎜ V ⎞⎟
⎝ 2XL ⎠
R2
X L2 = −
VX L
Iy
I x2
I y2 = −
VI y
I y2 +
VI y
XL
XL
=0
2
to both the sides,
I x2
Iy +
V 2 ⎛ V ⎞
) =⎜
⎝ 2 X L ⎟⎠
2XL
Equation (4.4) represent a circle with radius
2
…(4.4)
V
V ⎞
⎛
and centre ⎜ 0, −
.
2XL
⎝
2 X L ⎟⎠
V
Thus, the locus of the extremity of a current phasor is a circle.
But the maximum possible value for f is 90°. When R = 0, the
I1
locus will be a semicircle Hence, the locus diagram for a series RL
Center
V
V
circuit with variable resistance is a semicircle of radius
and
0, −
I2
2XL
2XL
V
V ⎞
⎛
2XL
centre ⎜ 0, −
. The locus diagram is shown in Fig. 4.117.
⎝
2 X L ⎟⎠
Similarly, the locus diagram of a series RL circuit with variable
reactance and constant resistance can be drawn. From Eq. (4.3),
VR
R 2 X L2 =
Ix
I
V
Fig. 4.117 Locus diagram of a series
I x2 I y2 = I x
R
RL circuit for variable R
V
I x2 I y2 − I x = 0
R
2
⎛V ⎞
Adding ⎜ ⎟ on both the sides,
⎝ 2R ⎠
2
V ⎞
⎛
⎛V ⎞
2
⎜⎝ I x −
⎟ + I y = ⎜⎝ ⎟⎠
2R
R⎠
2R
2
…(4.5)
V
⎛V ⎞
and centre ⎜
, 0 . Thus, the locus of the extremity of
⎝ 2 R ⎟⎠
2R
V
⎛V ⎞
and centre ⎜
, 0 . The locus diagram is shown in Fig. 4.118
a current phasor is a semicircle of radius
⎝ 2 R ⎟⎠
2R
Equation (4.5) represents a circle with radius
4.102 Network Analysis and Synthesis
V
,0
2R
V
V
2R
I2
I1
I
Fig. 4.118 Locus diagram of a series RL circuit for variable XL
4.15.2 Locus Diagram of a Series RC Circuit
Figure 4.119 shows a series RC circuit with variable resistance and constant reactance.
XC
R
I
I=
V
=
Z
V
R
2
X C2
V
Fig. 4.119
Series RC circuit
In a series RC circuit, current I leads voltage V by an angle f. Figure 4.120 shows the phasor diagram and
impedance triangle for series RC circuit.
R
Z
X
sin φ = C
Z
R
cos φ =
I
f
f
XC
Z
V
Fig. 4.120 Phasor diagram and impedance triangle of a
series RC circuit
I
It can be similarly proved that the locus of the extremity of a
V
current phasor is a semicircle with radius
and centre
2 XC
V ⎞
⎛
⎜⎝ 0, 2 X ⎟⎠ . The locus diagram is shown in Fig. 4.121.
C
For a series RC circuit with variable reactance and constant
0,
V
2XC
resistance, the locus of the extremity of a current phasor is a
V
⎛V ⎞
semicircle with radius
and centre ⎜
, 0 . The locus diagram
0
⎝ 2 R ⎟⎠
2R
is shown in Figure 4.122.
Fig. 4.121
V
2XC
I2
I1
V
Locus diagram of a series
RC circuit with variable R
Exercises 4.103
I
I2
I1
V
2R
V
V
,0
2R
Fig. 4.122
Locus diagram of a series RC circuit with variable XC.
Exercises
4.1
4.2
An alternating current varying sinusoidally with
a frequency of 50 Hz has an rms value of 20 A.
Write down the equation for the instantaneous
value and find this value (i) 0.0025 second,
and (ii) 0.0125 second after passing through
a positive maximum value. At what time,
measured from a positive maximum value, will
the instantaneous current be 14.14 A?
1 ⎤
⎡
t 20 A, − 20 A,
s
⎢⎣i 28 28 i 100 π t,
300 ⎥⎦
A certain waveform has a form factor of 1.2
and a peak factor of 1.5. If the maximum value
is 100, find the rms value and average value.
[66.67 V, 55.67 V]
3
I.
2
square value of its resultant is
4.6
Find the average values and rms values
of the waveforms shown in Fig. 4.123 to
Fig. 4.126.
v
Vm
0
p
4
2p
p
Fig. 4.123
4.3
4.4
4.5
Find the root mean square value, the average
value and the form factor of the resultant
current in a wire that carries simultaneously
a direct current of 5 A and a sinusoidal
alternating current with an amplitude of 5 A.
[6.12 A, 5 A, 1.224]
Find the relative heating effects of two inphase current waveform of equal peak values
and time periods, but one sinusoidal and the
other triangular.
[3:2]
Prove that if a dc current of I amperes is
superimposed in a conductor by an ac current
of maximum value I amperes, the root mean
[0.543 Vm, 0.674 Vm]
v
Vm
0.866 Vm
0
p
2
Fig. 4.124
[0.622 Vm, 0.687 Vm]
p
q
q
4.104 Network Analysis and Synthesis
v
v
20 V
15 V
0
2
4
6
8
t
10
0
15 V
3
6
9
12
t( s)
Fig. 4. 128
[10 V, 12.91 V, 16.67 W]
Fig. 4.125
[7.5 V, 8.666 V]
Find the resultants of the following voltages:
(i) v1 = 60 cos ,
4.9
v
v2 =
v3
Vm
−
3
,
15 sin
sin
.92°)]
t + 120°)
− 30 )
[107. sin
.35°)]
[43.
p
3
0
(ii)
t
p
2p
3
t
Fig. 4.126
[0.67 Vm, 0.745 Vm]
t + 135°),
(iii)
Find the rms value of the periodic waveform
with time period T given in Fig. 4.127.
4.7
t + 60°),
v
− 150 )
[
sin (
v
(iv)
10 V
)]
t
t
,
6
t
0
T
2T
T
4T
5T
2T
−
t
4
Fig. 4.127
[7.45 V]
4.8
A voltage wave has the variation shown in
Fig. 4.128.
(i) Find the average and effective values of
the voltage.
(ii) If this voltage is applied to a 10 Ω
resistance, find the dissipated power.
14 sin ( t + 23
]
[ .14 sin ( t −
°)]
(v )
4.10 Three
and
alternating
4
t+
currents
t+
6
2
are fed into a
Exercises 4.105
common conductor. Find graphically or
otherwise the equation of the resultant current
and its rms value.
[ . sin
45
54]
4.11 Four wires, p, q, r, s are connected to a
common point. The currents in lines p, q and
t
3 cos
t+
3
t
3
resp
3
and
ind the current
in the wire s.
.9 si
.
]
4.12 A sinusoidal voltage V
t is applied
to three parallel branches yielding branch
currents as follows:
i
− 45°
i
60°
i
t 60°
Find the complete expression for the source
current. Draw the phasor diagram in terms of
effective values. Use the voltage as reference.
4 n
15 ]
4.13 A sinusoidal voltage V
t is applied to
three parallel branches. Two of the branch
currents are given by
i
− 37°
i
− 143°
The source current is found to be
i
t + 12 8 ).
(i) Find the effective value of the current in
the third branch.
(ii) Write the complete time expression for
the instantaneous value of the current in
part (i).
(iii) Draw the phasor diagram showing the
source current and the three branch
currents. Use voltage as the reference
phasor.
95 A, 56.51 sin ( t + 53.
]
4.14 A voltage wave e (t) = 170 sin 120 t produces
a net current of 14.14 sin120 t + 7.07 cos (120
t + 30°).
(i)
Express the effective value of the
current as a single phasor quantity.
(ii) Draw the phasor diagram. Show the
components of the current as well as
the resultant.
(iii) Determine the power delivered by the
source.
[8.66 ∠30° A, 901.7 W]
4.15 The instantaneous values of two alternating
voltages are represented by v1 = 60 sin
and v2
− / 3)
Derive
an
expression for the instantaneous value of
(i) the sum, and
(ii) the difference of these voltages.
[ .17 sin ( − . ),
52. sin
89°)]
4.16 A 250 V, 50 Hz voltage is applied across a
circuit consisting of a pure resistance of 20 Ω.
Determine
(i) the current flowing through the circuit,
and (ii) power absorbed by the circuit.
Give the expressions for the voltage
and current.
[(i) 12.5 A, (ii) 3.125 kW,
(iii) v = 353.6 sin 314 t,
i = 17.68 sin 314 t]
4.17 A purely inductive circuit allows a current 20
A to flow through a 230 V, 50 Hz supply. Find
(i) inductive reactance, (ii) inductance of
the coil, (iii) power absorbed, and (iv)
equations for voltage and current.
⎤
(i) 11.5 Ω,
(ii) 36.62 mH
(iii) 0
(iv) v =
.
sin
i = 28.28 sin
t,
t−
2
4.18 A capacitor connected to a 230 V, 50 Hz
supply draws a 15 A current. What current will
it draw when the capacitance and frequency
are both reduced to half?
[3.75 A]
4.19 Calculate the impedance of the circuit shown
in Fig. 4.129.
4.106 Network Analysis and Synthesis
R
L
4.27 Voltage and current in an ac circuit are given
by
v = 200 sin 377t
5A
16 V
12 V
Fig. 4.129
4.28
4.20
4.21
4.22
4.23
[4 Ω]
An ac circuit has the following voltage and
current: v = 325 sin 314 t, i = 65 sin (314
t − 1.57). Find (i) frequency, (ii) rms value of
voltage and current, (iii) impedance, and (iv)
power factor.
[50 Hz, 229.81 V, 45.96 A, 5 Ω 0.99 (lagging)]
Two sources of an electromotive force
represented respectively by 200 sin ω t and
200 sin(ω t + π / ) are in series. Express the
resultant in vector notation with reference
to 200 sin ω t and calculate the rms current
and power supplied to a circuit of 8 + j6 Ω
impedance.
[273.2 ∠15° V, 27.32 ∠−21.87° A, 5971 W]
The voltage applied to a series circuit
consisting of two pure elements is given by v
= 180 sin ω t and the resulting current is given
by i = 2.5 sin (
). Find the average
power taken by the circuit and the values of
the elements.
[159 W, 50.91 Ω, 50.91 Ω]
Find an expression for the current and calculate
the power when a voltage represented by v =
283
100 π t is applied to a coil having R =
50 Ω and L = 0.159 H.
[4 sin (100π t π /4), 400 W]
4.24 A voltage V = (150 + j180) V is applied across
an impedance and the current is found to be
I = (5 − j4) A. Determine (i) scalar impedance,
(ii) reactance, and (iii) power consumed.
[36.6 Ω, 36.6 Ω, 30 W]
4.25 The current in a series circuit of R = 5 Ω
and L = 30 mH lags the applied voltage by
72°. Determine the source frequency and the
impedance Z.
[81.63 Hz, 16.18 Ω]
4.26 Current flowing through an inductive circuit
is 15 sin (wt + p /4). When the voltage applied
across it is 30 cos w t, find the pf of the circuit.
[0.707 (lagging)]
4.29
4.30
4.31
i = 8 sin (377t − p /6)
Determine true power, reactive power and
apparent power drawn by the circuit.
[692.82 W, 400 VAR, 800 VA]
A current of 5 A flows through a noninductive resistor in series with a choke coil
when supplied at 250 V, 50 Hz. If the voltage
drops across the coil and non-inductive
resistor are 200 V and 125 V respectively,
calculate the resistance and inductance of
the impedance coil, value of non-inductive
resistor and power drawn by the coil. Draw
the vector diagram.
[5.5 Ω, 0.126 H, 25 Ω, 137.5 W]
Two coils A and B are connected in series
across a 200 V, 50 Hz ac supply. The power
input to the circuit is 2 kW and 1.15 kVAR. If
the resistance and reactance of the coil A are
5 Ω and 8 Ω respectively, calculate resistance
and reactance of the coil B. Also, calculate
the active power consumed by coils A and B.
[10.3 Ω, 0.642 Ω, 665.3 W, 1334.7 W]
A choking coil and a pure resistor are
connected in series across a supply of 230 V,
50 Hz. The voltage drop across the resistor is
100 V and that across the chocking coil is 150
V. Find the voltage drop across the inductance
and resistance of the choking coil. Hence, find
their values if the current is 1 A.
[109.98 V, 102 V, 0.366 H, 102 Ω]
For Fig. 4.130, find R and L.
10 Ω
20 V
R
L
22.4 V
36 V/60 Hz
Fig. 4.130
[4.928 Ω, 0.0266 H]
4.32 Voltage and current for a circuit with two
pure elements in series are expressed as
follows:
Exercises 4.107
π⎞
⎛
v(t ) = 170 sin
si 6280t +
volts
⎝
3⎠
π⎞
⎛
i(t ) = 8.5 sin
si 6280t +
amperes
e
⎝
2⎠
4.33
4.34
4.35
4.36
Sketch the two waveforms. Determine
(i) frequency, (ii) power stating its nature,
and (iii) values of the elements.
[(ii) 0.866 (leading) (iii) 17.32 Ω, 16 μF]
A load consisting of a capacitor in series with
a resistor has an impedance of 50 Ω and a pf
of 0.707 leading. The load is connected in
series with a 40 Ω resistor across an ac supply
and the resulting current is of 3 A. Determine
the supply voltage and overall phase angle.
[249.69 V, 25.135°]
A capacitive load takes 10 kVA and 5 kVAR,
when connected to a 200 V, 50 Hz ac supply.
Calculate (i) resistance, (ii) capacitance, (iii)
active power, and (iv) pf.
[3.464 Ω, 1.59 × 10−3 F,
8.66 kW, 0.866 (leading)]
A resistor of 100 Ω is connected in series with
a 50 μF capacitor to a 50 Hz, 200 V supply.
Find:
(i) impedance
(ii) current
(iii) power factor
(iv) phase angle
(v) voltage across the resistor and across
the capacitor.
[118.6 Ω, 1.69 A,0.845 (leading),
32.48°, 168.712 V, 107.42 V]
In an RLC series circuit, the voltage across the
resistor, inductor and capacitor are 10 V, 15 V and
10 V respectively. What is the supply voltage?
[11.18 V]
4.37 A resistor R, inductor L = 0.01 H and capacitor
C are connected in series. When a voltage of
400 sin (3000 t − 10°) is applied to a series
combination, the current flowing is 10 2 cos
(3000 t − 55°). Find R and C.
[20 Ω, 6.66 μF]
4.38 A resistor of 20 Ω, an inductor of 0.2 H and
a capacitor of 100 μF are connected in series
across a 220 V, 50 Hz supply. Determine
(i) impedance
(ii) current flowing through the circuit
(iii) voltage drop across each element
(iv) power consumed
[36.9 Ω, 5.96 A, 119.2 V,
374.52 V, 189.64 V, 710.43 W]
4.39 A series RLC circuit having a resistance of
8 Ω, an inductance of 80 mH and a capacitance
of 100 μF is connected across a 150 V, 50 Hz
supply. Calculate (i) current, (ii) power factor,
and (iii) voltage across the inductance and
capacitance.
4.40
4.41
4.42
4.43
[(i) 14.35 A (ii) 0.766 (leading)
(iii) 360.47 V, 457.04 V]
An emf of 100 V, 50 Hz is applied to an
impedance Z1 = 8 + j6 Ω. An impedance
Z2 is added in series with Z1. The current
becomes half of the original and leads it by
20°. Determine Z2.
[(11.14 – j0.2) Ω]
In a series RLC circuit, the inductor is
imperfect and the capacitor has no leakage
but has a lead resistance. When the circuit
is connected to a 50 Hz supply, the voltage
drop across the resistor and the inductor is
200 V. The voltage drop across the resistor
and the capacitor is 250 V. The voltage
drop across the resistor alone is 150 V.
The voltage drop across the capacitor and
inductor are 125 V and 100 V respectively.
If the current flowing through the circuit is 1
A, find graphically
(i) capacitance
(ii) lead resistance of the capacitor
(iii) inductance
(iv) resistance of the inductor
(v) applied voltage
[33.5 μF, 81.25 Ω, 0.3 H, 25 Ω, 256.26 V]
Two impedances of 10 ∠ 30° Ω and 20 ∠−45°
Ω are connected in series. Calculate the power
factor of the series combination.
[0.9281 (lagging)]
Two impedances Z1 and Z2 are connected in
series across a 230 V, 50 Hz ac supply. The
total current drawn by the series combination
in 2.3 A. The pf of Z1 is 0.8 lagging. The
voltage drop across Z1 is twice the voltage
drop across Z2 and it is 90° out of phase with
it. Determine the value of Z2.
[44.719 Ω]
4.108 Network Analysis and Synthesis
4.44 In the arrangement shown in Fig. 4.131, C = 20
microfarads and the current flowing through
the circuit is 0.345 A. If the voltages are as
indicated, find the applied voltage, frequency
and loss in the iron-cored inductor. Draw the
phasor diagram.
L
R
25 V
40 V
55 V
Fig. 4.131
[34.2 V, 50 Hz, 1.89 W]
4.45 Two impedances of 14 + j5 Ω and 18 + j10 Ω
are connected in parallel across a 200 V, 50
Hz supply, Determine (i) admittance of each
branch and the entire circuit, (ii) current in
each branch and total current, (iii) power and
power factor of each branch, (iv) total power
factor, and (v) draw phasor diagram.
[0.067∠ 19
19.6
6° , 0.048
0 048 ∠− 29
29.05
05° ,
0.115∠− 23.75° ,13.45∠− 19.65°A,
9.6 ∠ 29
29.05
05°A
A, .9 ∠− 23.75° A,
2532.64 W,1658.8 W, 0.94 (lagging),
0.874 (lagging), 0.916 (lagging
g)]
4.46 In a parallel RC circuit shown in Fig. 4.132,
iR = 15 cos (5000 t −30°) amperes. Obtain the
current in capacitance.
iC
100 μF
10 Ω
2 mH
30 μF
Fig. 4.133
C
50 V
iR
25 Ω
V
[7.07 A]
4.48 A circuit comprises of a conductance G in parallel
with a susceptance B. Calculate the admittance
G + jB if the impedance is 10 + j5 Ω.
[0.08 S, 0.04 S]
4.49 If an admittance of a circuit is (8 + j6) and
the circuit current is 2 ∠ 30°A, find the pf of
the circuit. Also, calculate the apparent power
in the circuit.
[0.8 (leading), 0.2 VA]
4.50 A resistor of 50 Ω, and inductor of 0.15 H and
a capacitor of 100 μF are connected in parallel
across a 100 V, 50 Hz ac supply. Calculate (i)
current in each circuit, (ii) resultant current.
Draw individual phasor diagrams and the
overall phasor diagram.
[2 ∠0° A
A,, 2.12
2.12 ∠− 90°A ,
3.14 ∠90° A
A,, 2.25
2.25
25 ∠27
27.02
02° A]
4.51 The power dissipated in the coil A is 300 W and
in the coil B is 400 W. Each coil takes a current of
5 A when connected to a 110 V, 50 Hz supply.
Find the current drawn when the coils are
connected in parallel.
[9.93 ∠−50.11°A]
4.52 Find the total impedance, supply current and
pf of the entire circuit.
3Ω
4Ω
4Ω
3Ω
4.57 Ω
5.51 Ω
Fig. 4.132
[74.95 sin (5000 t + 150°)]
4.47 A voltage v = 100 sin (5000 t + p /4) is applied
across a three-branch circuit as shown in Fig.
4.133. Find the rms value of the current in the
inductor.
200 V 50 Hz
Fig. 4.134
[10.06 ∠36.68° Ω, 19.88 ∠−36.68° A,0.8 (lagging)]
Exercises 4.109
4.53 Determine kVA, kVAR and kW consumed by
the two impedances Z1 = ( 20 + 37.7) Ω and
Z 2 = (50 + 0) Ω, when connected in parallel
across a 230 V, 50 Hz supply.
[1.971 kVA, 1.095 kVAR, 1.64 kW]
4.54 In the parallel circuit shown in Fig. 4.135,
the power in the 3 Ω resistor is 666 W
and the total volt-amperes taken by the
circuit is 3370 VA. The power factor
of the whole circuit is 0.937 leading.
Find Z.
3Ω
Z
j6Ω
Fig. 4.135
[(2 − j2) Ω]
4.55 A coil is connected across a non-inductive
resistor of 120 Ω. When a 240 V, 50 Hz
supply is applied to this circuit, the coil draws
a current of 5 A and the total current is 6 A.
Determine the power and the power factor of
(i) the coil, and (ii) the whole circuit.
[420 W, 0.35 (lagging), 900 W, 0.625 (lagging)]
4.56 A coil takes a current of 10 A and dissipates
1410 W when connected to 220 V, 50 Hz
supply. If another coil is connected in parallel
with it, the total current taken from the supply
is 20 A at a power factor of 0.868. Determine
the current and the overall power factor when
the coils are connected in series across the
same supply.
[5.55 A, 0.8499 (lagging)]
4.57 Draw an impedance triangle between
terminals AB in Fig. 4.136, labelling its sides
with appropriate values and units.
− 8
−j
j4
B
A
6Ω
Fig. 4.136
[R = 3.84 Ω, X = 1.12 Ω (inductive), Z = 4 Ω]
4.58 Draw the impedance triangle between
terminals AB in Fig. 4.137, for the following
conditions:
(i) XC = 8 Ω, R = 6 Ω, XL = 4 Ω
(ii) XC = 4 Ω, R = 6 Ω, XL = 6 Ω
Label its side with appropriate values and units.
R
XC
B
A
XL
Fig. 4.137
[(i) R = 1.8461 Ω, X = 5.2308 Ω,
(capacitive), Z = 5.547 Ω
(ii) R = 3 Ω, X = 1 Ω,
(capacitive), Z = 3.1623 Ω]
4.59 Draw an admittance triangle between
terminals AB in Fig. 4.138, labelling its
sides with appropriate values and units in
case of
(i) XL = 4 Ω, and XC = 8 Ω
(ii) XL = 10 Ω, and XC = 5 Ω
jXL
1Ω
B
A
− C
−jX
Fig. 4.138
[(i) G = 0.015 , BL = 0.123 , Y = 0.124 , (ii) G
= 0.0098 , BC = 0.099 , Y = 0.1 ]
4.60 An inductive impedance BC is connected
in series with a parallel combination AB
consisting of a capacitor and a non-inductive
resistor. The circuit constants are so adjusted
that the current in the parallel branches AB are
equal, and that the voltage across AB is equal
to and in quadrature with the voltage across
BC. When a voltage of 200 V is applied to
AC, the total power absorbed is 1200 W.
Calculate the circuit constants and draw a
vector diagram.
[AB = R = 33.32 Ω in parallel with
XC = 33.32 Ω, BC = (16.66 + j16.66) Ω]
4.61 A 100 Ω resistor, shunted by a 0.4 H inductor is
in series with a capacitor C. A voltage of 250 V at
4.110 Network Analysis and Synthesis
4.62
4.63
4.64
4.65
50 Hz is applied to the circuit. Find
(i) the value of C to give unity power
factor
(ii) the total current
(iii) current in the inductive branch
[65.4 μF, 4.08 A, 2.55 A]
A non-inductive 10 Ω resistor is in series with a
coil of 1.3 Ω resistance and 0.018 H inductance.
If a voltage of maximum value of 100 V at a
frequency of 100 Hz is applied to this circuit,
what will be the voltage across the resistor?
[62.54 V]
Two impedances Z1 = 10 – j15 Ω and
Z 2 = 4 + j8 Ω are connected in parallel. The
supply voltage is 100 V, 25 Hz. Calculate (i)
the admittance, conductance and susceptance
of the combined circuit, and (ii) total current
drawn and pf.
[0.097 , 0.081 , 0.054 , 9.7 A,
0.83 (lagging)]
A voltage of 200 Æ 53.13° V is applied
across two impedances in parallel. The
values of the impedances are (12 + j16) Ω
and (10 – j20) Ω. Determine kVA, kVAR
and kW in each branch and the pf of the
whole circuit.
[2 kVA, 1.2 kW, 1.6 kVAR, 1.788 kVA,
0.8 kW, 1.6 kVAR, unity pf]
For the circuit shown in Fig. 4.139, evaluate
the current through and voltage across each
element.
l
10 Ω
l1
173 ∠0° V
20 Ω
l2
− 11.55 Ω
−j
Fig. 4.139
⎡ I1 5 ∠ 30o A, I 2 = 8.66 ∠60o A,
⎢
⎢⎣ V1 100∠ 30o , V2 = 100 ∠− 30o V,
⎡ V10 Ω = 100∠− 30o V ⎤
⎣
⎦
4.66 Two impedances Z1 and Z2 are connected
in parallel. The first branch takes a leading
current of 16 A and has a resistance of
5 Ω, while the second branch takes a lagging
current at a pf of 0.8. The total power supplied
is 5 kW, the applied voltage being (100 +
j200) V. Determine branch currents and total
current.
[16∠132.46° A, 20.8 ∠26.57° A,
22.49 A]
4.67 For the parallel branch shown in Fig. 4.140, find
the value of R2 when the overall power factor is
0.92 lag.
j2
1Ω
R2
Fig. 4.140
[1.35 Ω]
4.68 In a series−parallel circuit, two parallel branches
A and B are in series with C. The impedances are
ZA = (10 + j8) Ω, and ZB = (9 – j6) Ω and ZC
(3 + j2) Ω. If the voltage across ZC is 100 ∠0°
V, determine the values IA and IB.
[15.7∠–73.39° A, 18.59 ∠–1.04° A]
4.69 A capacitor is placed in parallel with two
inductive loads. The current through the first
inductor is 20 A at 30° lag and the current
through the second is 40 A at 60° lag. What
must be the current in the capacitor so that the
current in the external circuit is of unity power
factor?
[44.64 ∠90° A]
4.70 A circuit of 15 Ω resistance and 12 Ω inductive
reactance is connected in parallel with another
circuit consisting of a resistor of 25 Ω in
series with a capacitive reactance of 17 Ω.
This combination is energized from a 200 V,
40 Hz mains. Find the branch currents, total
current and power factor of the circuit. It is
desired to raise the power factor of this circuit
to unity by connecting a capacitor in parallel.
Determine the value of the capacitance of the
capacitor.
Objective-Type Questions 4.111
[10.42 ∠–38.56° A, 6.61 ∠34.21° A, 13.95
∠–11.56° A, 0.98 (lagging), 54.9 μF]
4.71 A resistor of 30 Ω and a capacitor of unknown
value are connected in parallel across a 110
V, 50 Hz supply. The combination draws a
current of 5 A from the supply. Find the value
of the unknown capacitance of the capacitor.
This combination is again connected across
a 110 V supply of unknown frequency. It is
observed that the total current drawn from the
mains falls to 4 A. Determine the frequency
of the supply.
[98.58 μF, 23.68 Hz]
4.72 Two reactive circuits have an impedance of
20 Ω each. One of them has a lagging power
factor of 0.8 and the other has a leading power
factor of 0.6. Find (i) voltage necessary to send
a current of 10 A through the two in series,
and (ii) current drawn from 200 V supply
if the two are connected in parallel. Draw a
phasor diagram in each case.
[282.8 V, 14.14 A]
4.73 Inductor loads of 0.8 kW and 1.2 kW
at lagging power factors of 0.8 and 0.6
respectively are connected across a 200 V,
50 Hz supply. Find the total current, power
factor and the value of the capacitor to be put
in parallel to both to raise the overall power
factor to 0.9 lagging.
[14.87 A, 0.673 (lagging), 98 μF]
Objective-Type Questions
Choose the correct alternative in the following
questions:
4.1 A current is said to be alternating when it
changes in
(a) magnitude only
(b) direction only
(c) both magnitude and direction
(d) neither magnitude nor direction
4.2 An alternating current of 50 Hz frequency
and 100 A maximum value is given as
(a)
200 si 628
(b) i 100 si 314 t
4.3
4.4
(c)
i 100 2 si 314 t
(d)
i 100 2 si 157 t
An alternating current of 50 Hz frequency
has a maximum value of 100 A. Its value
1
second after the instant the current is
600
zero will be
(a) 25 A
(b) 12.5 A
(c) 50 A
(d) 75 A
A sinusoidal voltage varies from zero to
a maximum of 250 V. The voltage at the
instant of 60° of the cycle will be
(a) 150 V
(b) 216.5 V
(c) 125 V
(d) 108.25 V
4.5
4.6
An alternating current is given by the
π⎞
⎛
expression i 200 si 314t + ⎟ amperes.
⎝
3⎠
The maximum value and frequency of the
current are
(a) 200 A, 50 Hz (b) 100 2 , 50 Hz
(c) 200 A,100 Hz (d) 200 A, 25 Hz
The average value of the current i = 200 sin t
π
from t 0 to t = is
2
400
π
1
π
(c)
(d)
400
400
The rms value of half-wave rectified
current is 50 A. Its rms value for full wave
rectification would be
(a) 100 A
(b) 70.7 A
(a)
4.7
(b)
50
100
A
(d)
A
π
π
When the two quantities are in quadrature,
the phase angle between them will be
(a) 45°
(b) 90°
(c) 135°
(d) 60°
(c)
4.8
400 π
4.112 Network Analysis and Synthesis
The phase difference between the two
waveforms can be compared when they
(a) have the same frequency
(b) have the same peak value
(c) have the same effective value
(d) are sinusoidal.
If two sinusoids of the same frequency but of
different amplitude and phase difference are
added, the resultant is a
(a) sinusoid of the same frequency
(b) sinusoid of double the original frequency
(c) sinusoid of half the original frequency
(d) non-sinusoid
Two alternating currents represented as i1 = 4
π⎞
⎛
sin w t and i2 10 i ω t +
are fed into
⎝
3⎠
a common conductor. The rms value of the
resultant current is
(a) 9.62 A
(b) 8.83 A
(c) 12.48 A
(d) 13.6 A
A constant current of 2.8 A exists in a
resistor. The rms value of the current is
(a) 2.8 A
(b) 2 A
(c) 1.4 A
(d) undefined.
The current through a resistor has a waveform
as shown in Fig. 4.141. The reading shown by
a moving-coil ammeter will be
4.9
4.10
4.11
4.12
4.13
4.15
4.16
4.17
4.18
i
4.19
5A
q
p
0
2p
3p
Fig. 4.141
5
A (d) 0
π
2
2
Which of the following statements about the
two alternating currents shown in Fig. 4.142
is correct?
(a)
4.14
5
i2
A
(b)
25
A
(c)
4.20
i1
q
0
4.21
Fig. 4.142
(a) The peak values of i1 and i2 are different.
(b) The rms values of i1 and i2 are different.
(c) The time period of current i1 is more
than that of the current i2.
(d) The frequency of current i1 is more than
that of the current i2.
The alternating voltage e = 200 sin 314 t is
applied to a device which offers an ohmic
resistance of 20 Ω to the flow of current in
one direction while entirely preventing the
flow in the opposite direction. The overage
value of the current will be
(a) 5 A
(b) 3.18 A
(c) 1.57 A
(d) 1.10 A
A 10 mH inductor carries a sinusoidal
current of 1 A rms at a frequency of 50 Hz.
The average power dissipated by the inductor
is
(a) 0
(b) 0.25 W
(c) 0.5 W
(d) 1 W
The power factor of an ordinary bulb is
(a) zero
(b) unity
(c) slightly more than unity
(d) slightly less than unity
When a sinusoidal voltage is applied across
RL series circuit having R = XL, the phase
angle will be
(a) 90°
(b) 45°lag
(c) 45° lead
(d) 90° lead
An ac source having voltage
π⎞
⎛
e 110 si ω t + ⎟ is connected in an ac
⎝
3⎠
circuit. If the current drawn from the circuit
varies as i 5 sin ⎛ t − π ⎞⎟ , the impedance
⎝
3⎠
of the circuit will be
(a) 22 Ω
(b) 16 Ω
(c) 30.8 Ω
(d) none of these
An ac source of 200 V rms supplies active
power of 600 W and reactive power of
800 VAR. The rms current drawn from the
source is
(a) 10 A
(b) 5 A
(c) 3.75 A
(d) 2.5 A
The voltage phasor of a circuit is 10 ∠ 15° V
45° A. The
and the current phasor is 2
active and reactive powers in the circuit are
Objective-Type Questions 4.113
4.22
4.23
4.24
4.25
(a) 10 W and 17.32 VAR
(b) 5 W and 8.66 VAR
(c) 20 W and 60 VAR
(d) 20 2 and 10 2 VAR
In a two element series circuit, the applied
voltage and resultant current are respectively
v(t) = 50 + 50 sin (5 × 103t) and i(t) = 11.2 sin
(5 × 103t + 63.4°). The nature of the elements
would be
(a) RL
(b) RC
(c) LC
(d) neither R, nor L, nor C
A boiler at home is switched on to the ac
mains supplying power at 230 V, 50 Hz. The
frequency of instantaneous power consumed
is
(a) 0
(b) 50 Hz
(c) 100 Hz
(d) 150 Hz
In a series RLC circuit, VR = 3 V, VL = 14 V,
VC = 10 V. The input voltage to the circuit is
(a) 10 V
(b) 5 V
(c) 27 V
(d) 24 V
A1, A2 and A3 are ideal ammeters, shown in
Fig. 4.143. If A1 reads 5 A, A2 reads 12 A then
A3 should read
R
4.27 In Fig. 4.144, A1, A2, and A3, are ideal
ammeters. If A1 and A3 read 5 A and 13 A
respectively, the reading of A2 will be
C
A2
A3
100 sin t
Fig. 4.144
(a) 8 A
(b) 12 A
(c) 18 A
(d) 10 A
4.28 In Fig. 4.145, i(t) under steady state is
1Ω
A3
A2
C
100 sin wt
1F
10 sin t
Fig. 4.145
zero
5
7.07 sin t
7.07 sin(t − 45°)
1
1
= Ω, L = H, C
4
3
= 3 F in Fig. 4.146 has input voltage v(t) = sin
2t. The resulting current i(t) is
4.29 The circuit shown with
i(t)
Fig. 4. 143
(a) 7 A
(b) 12 A
(c) 13 A
(d) 17 A
4.26 A series RLC circuit consisting of R = 10 Ω,
XL = 20 Ω and XC = 20 Ω is connected across
an ac supply of 200 V rms. The rms voltage
across the capacitor is
(a) 200 ∠− 90° V
(b) 200 ∠ 90° V
(c) 400 ∠− 90° V
(d) 400 ∠ 90° V
2H
5V
(a)
(b)
(c)
(d)
A1
R
A1
v(t )
R
L
Fig. 4.146
(a)
(b)
(c)
(d)
5 sin (2t + 53.1°)
5 sin (2t − 53.1°)
25 sin (2t + 53.1°)
25 sin (2t − 53.1°)
C
4.114 Network Analysis and Synthesis
Answers to Objective-Type Questions
4.1
4.9
4.17
4.25
(c)
(a)
(d)
(c)
4.2
4.10
4.18
4.26
(b)
(a)
(b)
(c)
4.3
4.11
4.19
4.27
(c)
(c)
(a)
(b)
4.4
4.12
4.20
4.28
(b)
(a)
(b)
(d)
4.5
4.13
4.21
4.29
(a)
(c)
(a)
(a)
4.6 (b)
4.14 (c)
4.22 (b)
4.7 (b)
4.15 (b)
4.23 (c)
4.8 (b)
4.16 (a)
4.24 (b)
5
Resonance
5.1
INTRODUCTION
Resonance is a very important phenomenon in many electrical applications. The study of resonance is very
useful in the telecommunication field. A circuit containing reactance is said to be in resonance if the voltage
across the circuit is in phase with the current through it. At resonance, the circuit thus behaves as a pure
resistor and the net reactance is zero.
5.2
SERIES RESONANCE
R
Consider the series RLC circuit as shown in Fig. 5.1. The
impedance of the circuit is
L
i
Z = R + jX − jX C = R + j ( X L − X C )
At resonance, the circuit is purely resistive.
XL
v = Vm sin wt
XC = 0
Fig. 5.1 Series circuit
XL
XC
ω0 L =
1
ω 0C
ω 02 =
1
LC
ω0 =
f0 =
1
LC
1
2π LC
where f 0 is called the resonant frequency of the circuit.
C
5.2 Network Analysis and Synthesis
1. Power Factor
Power factor = cos φ =
At resonance Z = R
R
Z
R
=1
R
Power factor =
2. Current Since impedance is minimum, the current is maximum at resonance. Thus, the circuit accepts
more current and as such, an RLC circuit under resonance is called an accepator circuit.
V V
=
Z R
I0 =
3. Voltage At resonance,
1
ω 0C
ω0 L =
ω 0 L I0 =
VL0
1
I0
ω 0C
VC0
Thus, potential difference across inductor equal to potential difference across capacitor being equal and
opposite cancel each other. Also, since I 0 is maximum, VL 0 and VC 0 will also be maximum. Thus, voltage
magnification takes place during resonance. Hence, it is also referred to as voltage magnification circuit.
4. Phasor Diagram Figure 5.2 shows the phasor diagram of a series resonant circuit.
VL
I
VR
VC
I
VR
Fig. 5.2 Phasor diagram
5. Behaviour of R, L and C with Change in
Frequency Resistance remains constant with the
change in frequencies. Inductive reactance X L is directly
proportional to frequency f. It can be drawn as a straight
line passing through the origin. Capacitive reactance X C
is inversely proportional to the frequency f. It can be
drawn as a rectangular hyperbola in the fourth quadrant.
Figure 5.3 shows the behaviour of R, L and C with change
in frequency.
Total impedance Z = R + j ( X L − X C )
XL
Z
R
0
XC
Fig. 5.3
Behaviour of R, L and C with
change in frequency
5.2 Series Resonance 5.3
(a) When f f 0 , impedance is capacitive and decreases up to f0. The power factor is leading in
nature.
(b) At f = f0, impedance is resistive. The power factor is unity.
(c) When f > f0, impedance is inductive and goes on increasing beyond f0. The power factor is lagging
in nature. Figure 5.4 shows the impedance curve of series resonant circuit.
Z
Capacitive
0
Inductive
f
f0
Fig. 5.4 Impedance curve
6. Bandwidth For the series RLC circuit, bandwidth
is defined as the range of frequencies for which the
P
power delivered to R is greater than or equal to 0
2
where P0 is the power delivered to R at resonance.
I
V
R
0.707 I0
I0 =
From the shape of the resonance curve shown in
Fig. 5.5, it is clear that there are two frequencies for
which the power delivered to R is half the power
at resonance. For this reason, these frequencies are
referred as those corresponding to the half-power
points. The magnitude of the current at each halfpower point is the same.
Bandwidth = w 2 − w 1
0
w1 w0 w2
w
Fig. 5.5 Resonance curve
1 2
I 0 R = I 22 R
2
where the subscript 1 denotes the lower half point and the subscript 2, the higher half point. It follows
then that
I12 R =
Hence,
I1
I2 =
I0
= 0.707 I 0
2
Accordingly, the bandwidth may be identified on the resonance curve as the range of frequencies over
which the magnitude of the current is equal to or greater than 0.707 of the current at resonance. In Fig. 5.5,
the bandwidth is w 2 −w1.
Expression for Bandwidth Generally, at any frequency w,
I=
V
=
Z
V
R
2
( X L − XC )
2
=
V
1 ⎞
⎛
R2 + ω L −
⎟
⎝
ωC ⎠
…(i)
2
5.4 Network Analysis and Synthesis
At half-power points,
But
I=
I0
I0 =
V
R
V
I=
2
…(ii)
2R
From Eqs (i) and (ii),
V
2
V
=
2R
R2
1 ⎞
⎛
ωL −
⎟
⎝
ωC ⎠
R2
1 ⎞
⎛
ωL −
⎟ = 2R
⎝
ωC ⎠
2
Squaring both the sides,
2
R
2
1 ⎞
⎛
2
ωL −
⎟ = 2R
⎝
ωC ⎠
2
1 ⎞
⎛
2
⎜⎝ ω L −
⎟ =R
ωC ⎠
1
±R=0
ωC
R
1
ω2 ± ω −
=0
L
LC
ωL −
ω=±
R
R2
1
±
+
2
2L
LC
4L
⎛ R2 ⎞
1
For low values of R, the term ⎜
can be neglected in comparison with the term
.
2⎟
LC
⎝ 4L ⎠
Then w is given by,
ω=±
R
1
R
1
±
=±
±
2L
LC
2L
LC
The resonant frequency for this circuit is given by
ω0 =
ω
ω1
1
LC
R
±
+ ω0
2L
R
ω0 −
2L
(considering only posititve sign of w0)
5.2 Series Resonance 5.5
R
2L
R
f0 −
4π L
R
f0 +
4π L
ω2 = ω0 +
and
f
f
and
R
L
R
Bandwidth = f − f1 =
2π L
Bandwidth = ω 2 − ω1 =
or
7. Quality Factor It is measure of voltage magnification in the series resonant circuit. It is also a measure
of selectivity or sharpness of the series resonant circuit.
Q0 =
Substituting values of VL0 and V,
Q0 =
Voltage across inductor or capacitor VL0 VC0
=
=
Voltage at resonance
V
V
I 0 X L0
I0 R
=
X L0
R
=
ω0 L
1
=
R
ω 0CR
Substituting values of w0,
⎛ 1 ⎞
⎜⎝
⎟L 1 L
LC ⎠
Q0 =
=
R
R C
Example 5.1 A series RLC circuit has the following parameter values: R = 10 W , L = 0.01 H,
C = 100 mF. Compute the resonant frequency, bandwidth, and lower and upper frequencies of the bandwidth.
Solution
R = 10 Ω,
L = 0.01 H,
C = 100 μF
(a) Resonant frequency
f0 =
1
2π LC
=
1
2π 0.01 × 100 × 10 −6
= 159.15 Hz
(b) Bandwidth
R
10
=
= 159.15 Hz
2π L 2π 0.01
(c) Lower frequency of bandwidth
BW =
f1
f0 −
BW
159.15
= 159.15 −
= 79.58 Hz
2
2
(d) Upper frequency of bandwidth
f2
f0 +
BW
159.15
= 159.15 +
= 238.73 Hz
2
2
5.6 Network Analysis and Synthesis
Example 5.2 An RLC series circuit with a resistance of 10 W, inductance of 0.2 H and a capacitance
of 40 μF is supplied with a 100 V supply at variable frequency. Find the following w.r.t. the series resonant
circuit:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
frequency of which resonance takes place
current
power
power factor
voltage across RLC at that time
quality factor
half-power points
resonance and phasor diagrams
R = 10 Ω,
Solution
L = 0.2 H,
C = 40 μF,
=
1
V = 100 V
(a) Resonant frequency
f0 =
1
2π LC
2π 0.2
40 10 −6
= 56.3 Hz
(b) Current
I0 =
V 100
=
= 10 A
R 10
P0
I 02 R = (10)2 × 10 = 1000 W
(c) Power
(d) Power factor
pf = 1
(e) Voltage across R, L, C
VR0
R I0 = 10 × 10 = 100 V
VL0
X L0 I0
VC0 = XC0 I 0 =
(f)
f0 LI 0
2π × 56.3 0.2
1
I0 =
2π f0C
2π
1
56 3 × 40 × 10 −6
Quality factor
Q=
1 L
1
02
=
= 7 07
R C 10 40 × 10 −6
(g) Half-power points
f1
f2
R
10
= 56.3 −
= 52.32 Hz
4π L
4π 0.2
R
10
f0 +
= 56.3 +
= 60 3 Hz
4π L
4π 0.2
f0 −
10 = 707.5 V
× 10 = 707.5 V
5.2 Series Resonance 5.7
(h) Resonance and phasor diagram are shown in Fig. 5.6.
I
I0
I0
√2
0
f1
f0
f
f2
V
I
(b)
(a)
Fig. 5.6
Example 5.3 A series RLC circuit is connected to a 200 V ac supply. The current drawn by the
circuit at resonance is 20 A. The voltage drop across the capacitor is 5000 V at series resonance. Calculate
resistance and inductance if capacitance is 4 μF. Also, calculate the resonant frequency.
Solution
V
200
, I 0 = 20 A, VC0
5
(a) Resistance
R=
V 200
=
= 10 Ω
I0
20
(b) Resonant frequency
X C0 =
X C0 =
250 =
VC0
I0
=
5000
= 250 Ω
20
1
2π f0C
1
2π
f0
4 × 10 −6
f0 = 159.15 Hz
(c) Inductance
At resonance
X C0
X L0
X L0 = 250 Ω
f0 L
250 = 2π × 159.15 × L
L = 0 25 H
, C = 4 μF
5.8 Network Analysis and Synthesis
Example 5.4 A resistor and a capacitor are connected in series with a variable inductor. When the
circuit is connected to a 230 V, 50 Hz supply, the maximum current obtained by varying the inductance is 2 A.
The voltage across the capacitor is 500 V. Calculate the resistance, inductance and capacitance of the circuit.
Solution
f0 = 50 Hz, I 0
V
2 A, VC0 = 500 V
I0 = 2 A
VC0 = 500 V
(a) Resistance
R=
V 230
=
= 115 Ω
I0
2
(b) Capacitance
X C0 =
X C0 =
250 =
VC0
I0
=
500
= 250 Ω
2
1
2π f0C
1
50 × C
2π
C = 12.73 μF
(c) Inductance
At resonance
X L0 = 250 Ω
X C0
X L0
f0 L
250 = 2π
50 × L
L = 0.795 H
Example 5.5
A coil of 2 W resistance and 0.01 H inductance is connected in series with a capacitor
across 200 V mains. What must be the capacitance in order that maximum current occurs at a frequency of
50 Hz? Find also the current and the voltage across the capacitor.
Solution
R
2 , L = 0.01 H, V
(a) Capacitance
f0 =
50 =
1
2π LC
1
2π 0.01 × C
C = 1013.2 μF
(b) Current
I0 =
V 200
=
= 100 A
R
2
f0 = 50 Hz
5.2 Series Resonance 5.9
(c) Voltage across capacitor
VC0
I X C0
I0
1
1
= 100 ×
= 314.16 V
2π f C
2π 50 × 1013.2 10 −6
Example 5.6
A series RLC circuit has a quality factor of 5 at 50 rad/s. The current flowing through
the circuit at resonance is 10 A and the supply voltage is 100 V. Find the circuit constants.
Solution
Q
/
I0
10 A, V = 100 V
10
V
R
100
10 =
R
R = 10 Ω
I0 =
ω0 L
R
50 × L
5=
10
L =1H
Q0 =
Q0 =
1 L
R C
1 1
10 C
C = 400 μF
5=
Example 5.7
A series RLC circuit has R = 2.5 W, C = 100 μF and a variable inductance. The applied
voltage is 50 V at 800 rad/s. The inductance is varied till the voltage across the resistance is maximum.
Find (a) value of inductance under this condition, (b) Q, (c) current, and (d) voltage across resistance,
capacitance and inductance.
Solution
= 2.5 Ω, C = 100 μF, V = 50 V, ω = 800 rad/s
(a) Value of inductance
ω0 =
800 =
1
LC
1
L × 100 × 10 −6
L = 0.0156 H
(b) Quality factor
Q0 =
ω 0 L 800 × 0.0156
=
=5
R
25
5.10 Network Analysis and Synthesis
(c) Current
I0 =
V 50
=
= 20 A
R 25
(d) Voltage across resistance, capacitance and inductance
VR0
I 0 R = 20 × 2.5 = 50 V
V
0
Q0 V = 5 50 = 250 V
V
0
Q0 V = 5 50 = 250 V
Example 5.8 A series RLC circuit which resonates at 500 kHz has R = 25 W, L = 100 μH and
C = 1000 pF. Determine the quality factor, new value of C required to resonate at 500 kHz when the value of
L is doubled, and the new quality factor.
Solution
R = 25 Ω, L =
f
μ , C = 1000 pF
(a) Quality factor
Q0 =
1 L
1
100 × 10 −6
=
= 12.65
R C 25 1000 × 10 −12
(b) New value of C when L = 200 μH
f0 =
1
2π LC
1
500 ×10
103 =
2π 200 ×10
10 −6 × C
C = 506.61 pF
(c) New Q factor
Q0 =
1 L
1
200 × 10 −6
=
= 25.13
R C 25 506.61 × 10 −12
Example 5.9
Find the values of R,L and C in a series RLC circuit that resonates at 1.5 kHz and
consumes 50 W from a 50 V ac source operating at the resonant frequency. The bandwidth is 0.75 kHz.
Solution
P = 50 W, V
f
2
P=
V
R
(50) 2
R
R = 50 Ω
50 =
0 , B W = 0.75 kHz
5.2 Series Resonance 5.11
R
2π L
50
0 75 × 103 =
2π × L
L = 0.011 H
BW =
f0 =
1 5 ×10
103 =
1
2π LC
1
2π 0.011 × C
C = 1.023 μF
Example 5.10 A 50 W resistor is connected in series with an inductor having internal resistance, a
capacitor and 100 V variable frequency supply as shown in Fig. 5.7. At a frequency of 200 Hz, a maximum
current of 0.7 A flows through the circuit and voltage across the capacitor is 200 V. Determine the circuit
constants.
50 Ω
r
L
C
100 V
Fig. 5.7
Solution
V
, R = 50 Ω
100
VC0 = 200 V
VC0
I XC 0
200 = 0 7 X C 0
X C 0 = 285.71 Ω
X C0 =
1
2π f C
1
2π × 200 × C
C = 2 79 μF
285.71 =
f0 =
200 =
1
2π LC
1
2π L 2.79 × 10 −6
L = 0.23 H
f
I 0 = 0.7 A
5.12 Network Analysis and Synthesis
At resonance, total resistance is
V 100
=
=
I
07
RT = R + r
RT =
.86 Ω
142.86 = 50 + r
r = 92.86 Ω
Example 5.11 In the circuit shown in Fig. 5.8, a maximum current of 0.1 A flows through the circuit
when the capacitor is at 5 μF with a fixed frequency and a voltage of 5 V. Determine the frequency at which
the circuit resonates, the bandwidth, quality factor Q and the value of resistance at resonant frequency.
R
5 μF
0.1 H
0.1 A
5V
Fig. 5.8
Solution
I0
R=
f0 =
A, C
μF,
F, V
V, L
0 1H
V
5
=
= 50 Ω
I0 0 1
1
2π LC
=
1
2π 0.1 5 × 10 −6
= 225.08 Hz
1 L
1
01
=
= 2 83
R C 50 5 10 −6
R
50
BW =
=
= 79.588 Hz
2π L 2π × 0.1
Q0 =
Example 5.12 A series RLC circuit has R = 10 W and L = 60 mH. At a frequency of 25 Hz, the pf of
the circuit is 45° lead. At what frequency will the circuit be resonant?
Solution
R = 10 Ω, L = 60 mH
XL = 2πf L = 2π × 25 × 60 × 10−3 = 9.42 Ω
For a series RLC circuit,
XC − X L
R
X C − 9.42
tan 45° =
10
tan φ =
5.2 Series Resonance 5.13
X C − 9 42
10
X C = 19.42 Ω
1=
XC =
1
2π fC
1
2π 25 × C
C = 327.82 μF
19.42 =
f0 =
1
=
2π LC
1
2π 60 10
−33
× 327.82 × 10
6
= 35.89
. Hz
A voltage v(t ) = 10 sin ω t is applied to a series RLC circuit. At the resonant frequency
of the circuit, the voltage across the capacitor is found to be 500 V. The bandwidth of the circuit is known to
be 400 rad/s and the impedance of the circuit at resonance is 100 W . Determine inductance and capacitance,
resonant frequency, upper and lower cut-off frequencies.
Example 5.13
Solution
v(t ) = 10 sin
n ω t , VC0 = 500 V, BW = 400 rad / s, R = 100 Ω
(a) Inductance and capacitance
V=
Vm
10
= 7 07 V
2
V 7 07
I0 = =
= 0.0707 A
R 100
R
BW =
L
100
400 =
L
2
=
L = 0.25 H
VC
500
Q0 = 0 =
= 70.72
V
7 07
Q0 =
1 L
R C
1 0 25
100 C
C = 4 99 nF
70.72 =
(b) Resonant frequency
f0 =
1
2π LC
=
1
2π
0.25 × 4.99 10 −9
= 4506.09 Hz
5.14 Network Analysis and Synthesis
(c) Lower cut-off frequency
f
f0 −
R
100
= 4506.09 −
= 4474.26 Hz
4π L
4π 0.25
f
f0 +
R
100
= 4506.09 +
= 4537.92 Hz
4π L
4π 0.25
(d) Upper cut-off frequency
Example 5.14 A series resonant circuit has an impedance of 500 W at resonant frequency. Cut-off
frequencies are 10 kHz and 100 Hz. Determine (a) resonant frequency, (b) value of L, C, and (c) quality
factor at resonant frequency.
Solution
f1 = 100 Hz
R
f 2 = 10 kHz
(a) Resonant frequency
BW = f
f
f
f1 = 10000 − 100 = 9900 Hz
R
4π L
R
f0 +
4π L
f0 −
Adding Eqs (i) and (ii),
f 2 = 2 f0
f f 2 100 + 10000
f0 =
=
= 5050 Hz
2
2
f
(b) Values of L and C
R
2π L
500
9900 =
2π L
L = 8.038 mH
BW =
X
At resonance
0
f 0 L = 2π × 5050 × 8.038 × 10 −3 = 255.05 Ω
X L0
X C0 = 255.05 Ω
X C0 =
1
2π f C
1
2π × 5050 × C
C = 0 12 μF
255.05 =
(c) Quality factor
Q0 =
1 L
1 8.038 × 10 −3
=
= 0.5176
R C 500 0 12 10 −6
…(i)
…(ii)
5.2 Series Resonance 5.15
Example 5.15 Impedance of a circuit is observed to be capacitive and decreasing from 1 Hz to
100 Hz. Beyond 100 Hz, the impedance starts increasing. Find the values of circuit elements if the power
drawn by this circuit is 100 W at 100 Hz, when the current is 1 A. The power factor of the circuit at 70 Hz
is 0.707.
Solution
P0 = 100 W, I 0 = 1 A, (pf
( )70 Hz
f
0.707
The impedance of the circuit is capacitive and decreasing from 1 Hz to 100 Hz. Beyond 100 Hz, the
impedance starts increasing.
f 0 = 100 Hz
P0
I 02 R
100 = (1) 2 × R
R = 100
00 Ω
1
f0 =
2π LC
100 =
1
2π LC
LC = 2 53 10 −6
…(i)
Power factor at 70 Hz is 0.707.
R
= 0.707
Z
100
= 0.707
Z
Z=
. Ω
Impedance at 70 Hz = Z70 = R 2 + ( X C − X L ) 2
1
⎛
⎞
141.44 = (100) 2 + ⎜
− 2π 70 × L⎟
⎝ 2π 70 × C
⎠
2.27 10 −3
− 439.82 L = 100.02
C
2
…(ii)
Solving Eqs (i) and (ii),
L = 0.2187 H
C = 11.58 μF
Example 5.16 A constant voltage at a frequency of 1 MHz is applied to an inductor in series with a
variable capacitor. When the capacitor is set to 500 pF, the current has its maximum value while it is reduced
to one-half when the capacitance is 600 pF. Find resistance, inductance and Q-factor of inductor.
5.16 Network Analysis and Synthesis
Solution
C1 = 500 pF, C2 = 600 pF
f
(a) Resistance and inductance of inductor
C = 500 pF
At resonance
f0 =
106 =
1
2π LC
1
2π L × 500 × 10 −12
L = 0 05 mH
f 0 L = 2π × 106 × 0.05 × 10 −3 = 314.16 Ω
X
When capacitance is 600 pF, the current reduces to one-half of the current at resonance.
1
1
=
= 65. 6 Ω
2π fC 2π × 106 × 600 × 10 −12
1
I0
2
1V
=
2R
2R
XC =
I
V
Z
Z
R2
R2
(XL
X C )2
(314.16 265.26) 2
2R
4 R2
3R 2 = 2391.21
R = 28.23 Ω
(b) Quality factor
Q0 =
1 L
1
0 05 10 −3
=
= 11.2
R C 28.23 500 × 10 −12
Example 5.17 In the circuit shown in Fig. 5.9, determine the circuit constants when the circuit
draws a maximum current at 10 mF with a 10 V, 100 Hz supply. When the capacitance is changed to 12 mF,
the current that flows through the circuit becomes 0.707 times its maximum value. Determine Q of the coil at
900 rad/s. Also find the maximum current that flows through the circuit.
R
L
10 V, 100 Hz
Fig. 5.9
C
5.2 Series Resonance 5.17
Solution
C
f0 =
μ , V = 10 V,
f 0 = 100 Hz
1
2π LC
1
100 =
2π L 10
10 10 −6
L = 0.225 H
When C = 12 μF, w = 900 rad/s
I
R2
(XL
0.707 I 0
V
V
= 0.707
Z
R
R
Z=
0.707
R
X C )2 =
0.707
2
1
R
⎛
⎞
R 2 + 900 × 0 25 −
=
−6 ⎟
⎝
⎠
0.707
900 × 12 × 10
R = 132.37 Ω
ω L 900 × 0 25
=
=1 7
R
132.37
V
10
I0 = =
= 0.076 A
R 132.37
Q0 =
Example 5.18
A series RLC circuit is excited from a constant voltage variable frequency source.
600
The current in the circuit becomes maximum at a frequency of
Hz and falls to half the maximum value
2π
400
at
Hz. If the resistance in the circuit is 3W, find L and C.
2π
Solution
f
f0 =
600
2π
1
R = 3Ω
2π LC
1
600
=
2π
2π LC
LC = 2 78 10 −6
At f =
400
Hz,
2π
I=
I0
2
…(i)
5.18 Network Analysis and Synthesis
V
V
=
Z 2R
Z 2R
R2
(XL
X C )2
2R
R2
(XL
X C )2
4 R2
(XL
X C ) 2 = 3R 2
3(3) 2
27
2
⎛
⎞
400
1
⎜
⎟
⎜ 2π × 2π × L −
⎟ = 27
400
⎜
2π ×
×C⎟
⎝
⎠
2π
2
1 ⎞
⎛
⎜⎝ 400 L −
⎟ = 27
400C ⎠
⎛
2 78 10
⎜ 400 ×
C
⎝
6
…(ii)
2
1 ⎞
−
= 27
400 C ⎟⎠
C = 267.12 μF
L=
5.3
2.78 10
C
6
=
2.78 × 10
1 −6
= 10.41 mH
267.12 × 10 −6
PARALLEL RESONANCE
Consider a parallel circuit consisting of a coil and a capacitor as shown in
Fig. 5.10. The impedances of two branches are
Z1 = R + jX L
Z 2 = − jX C
XC
i
1
1
R − jX L
=
=
Z1 R + jX L R 2 + X L2
1
1
j
Y2 =
=
=
Z 2 − jX C
XC
Y1 =
Admittance of the circuit
Y1 + Y2 =
Y
R − jX L
2
R +
X L2
+
v = Vm sin wt
Fig. 5.10
XL
R
−
X L2
1
=0
XC
XL
R
2
X L2
=
1
XC
X L X C = R 2 + X L2
ω0 L
1
= R 2 + ω 02 L2
ω0c
Parallel circuit
⎛ X
j
R
1 ⎞
= 2
− j⎜ 2 L 2 −
2
XC R + X L
⎝ R + X L X C ⎟⎠
At resonance, the circuit is purely resistive. Hence, the condition for resonance is
2
XL
R
Parallel Resonance 5.19
5.3
L
= R 2 + ω 02 L2
C
ω 02 L2 =
ω02 =
L
− R2
C
1
R2
− 2
LC L
1
R2
− 2
LC L
ω0 =
f0 =
1
2π
1
R2
− 2
LC L
where f0 is called the resonant frequency of the circuit.
If R is very small as compared to L then
1
ω0 =
f0 =
LC
1
2π LC
1. Dynamic Impedance of a Parallel Circuit At resonance, the circuit is purely resistive. The real
R
part of admittance is 2
. Hence, the dynamic impedance at resonance is given by
R
X L2
R 2 X L2
ZD =
R
At resonance,
R2
X L2 = X L X C =
ZD =
L
C
L
CR
2. Current Since impedance is maximum at resonance, the current is minimum at resonance.
I0 =
V
V
VCR
C
=
=
L
ZD
L
CR
3. Phasor Diagram At resonance, power factor of the
circuit is unity and the total current drawn by the circuit
is in phase with the voltage. This will happen only when
the current I C is equal to the reactive component of the
current in the inductive branch, i.e., I C I L sin φ
IC
IL cos f
f
V
I
IL sin f
IL
Fig. 5.11 Phasor diagram
V
5.20 Network Analysis and Synthesis
Hence, at resonance
and
IC
I L sin φ
I
I L cos φ
The phasor diagram of parallel resonant circuit is shown in Fig. 5.11.
4. Behaviour of Conductance G, Inductive Susceptance BL and Capacitive Susceptance
with Change in Frequency Conductance remains constant with the change in frequencies.
Inductive susceptance BL is
BL =
1
1
1
=−j
=−j
jX L
XL
2π fL
It is inversely proportional to the frequency. Thus, it decreases with the increase in the frequency.
Hence, it can be drawn as a rectangular hyperbola in the fourth quadrant.
Capacitive susceptance BC is
BC
Y
1
1
BC =
=−j
= j
− jX C
XC
fC
It is directly proportional to the frequency. It can be
drawn as a straight line passing through the origin.
Figure 5.12 shows the behaviour of G, BL and BC
with change in frequency.
G
0
BL
Fig. 5.12 Behaviour of G, BL and Bc with
change in frequency
(a) When f f 0 , inductive susceptance predominates.
Hence, the current lags behind the voltage and the
power factor is lagging in nature.
(b) When f f 0 , net susceptance is zero. Hence, the
admittance is minimum and impedance is maximum.
At f 0 , the current is in phase with the voltage and the
power factor is unity.
Z
ZD =
L
CR
0
f0
(c) When f f 0 , capacitive susceptance predominates.
Hence, the current leads the voltage and power factor
Fig. 5.13 Impedance curve
is leading in nature.
Figure 5.13 shows the impedance curve of a parallel resonant circuit.
f
I
5. Bandwidth The bandwidth of a parallel resonant circuit is
defined in the same way as that for a series resonant circuit.
Figure 5.14 shows the resonance curve of a parallel resonant
circuit.
6. Quality Factor It is a measure of current magnification in a
parallel resonant circuit.
Current through inductor or capacitor I C0
Q0 =
=
Current at resonance
I0
I0 =
V
ZD
0
f0
Fig. 5.14 Resonance curve
f
5.4 Comparison of Series and Parallel Resonant Circuits 5.21
Substituting values of I C0 and I 0 ,
V
1
X C0
X C0 ω 0 C ω 0 L
Q0 =
=
=
=
VCR
C
CR
CR
R
L
L
L
Neglecting the resistance R, the resonant frequency ω 0 is given by
1
ω0 =
LC
⎛ 1 ⎞
⎜⎝
⎟L 1 L
LC ⎠
Q0 =
=
R
R C
5.4
COMPARISON OF SERIES AND PARALLEL RESONANT CIRCUITS
Parameter
Series Circuit
Current at resonance
I=
Impedance at resonance
Power factor at resonance
Parallel Circuit
V
and is maximum
R
I=
VCR
C
and is minimum
L
Z = R and is minimum
Z=
L
and is maximum
CR
Unity
Unity
1
Resonant frequency
f0 =
Q-factor
Q=
It magnifies
Voltage across L and C
2π LC
2π f L
R
1
R2
− 2
LC L
f0 =
1
2π
Q=
2π f L
R
Current through L and C
Example 5.19
A coil having an inductance of L henries and a resistance of 12 W is connected in
parallel with a variable capacitor. At ω = 2.3 × 10 6 radd /s, resonance is achieved and at this instant, capacitance C = 0.021 μ F.
F Find the inductance of the coil.
Solution
R
Ω,
ω0 =
1
R2
− 2
LC L
0
6
d/ , C
0 021 μF
5.22 Network Analysis and Synthesis
1
2 3 × 106 =
L × 0.021 × 10
L = 89.7 μΗ
−6
−
(12) 2
L2
Example 5.20 A coil of 20 W resistance has an inductance of 0.2 H and is connected in parallel
with a condenser of 100 μF capacitance. Calculate the frequency at which this circuit will be have as a noninductive resistance. Find also the value of dynamic resistance.
Solution
R
20 Ω, L = 0.2 H, C =100 μF
f0 =
1
2π
(a) Resonant frequency
1
R2
1
− 2 =
LC L
2π
2
1
0 2 × 100 × 10
6
⎛ 20 ⎞
−⎜
= 31.83 Hz
⎝ 0 2 ⎟⎠
(b) Dynamic resistance
ZD =
L
02
=
= 00 Ω
CR 100 × 10 −6 × 20
Example 5.21 In the circuit shown in Fig. 5.15, an inductance at 0.1 H having a Q0 of 5 is in
parallel with a capacitor. Determine the value of capacitance and coil resistance at a resonant frequency of
500 rad/s.
R
L
C
Fig. 5.15
Solution
Q0 = 5
L
ω0 L
R
500 × 0 1
5=
R
R = 10 Ω
Q0 =
ω0 =
1
R2
− 2
LC L
500 =
1
(10) 2
−
0 1 × C (0.1) 2
C = 38.46 μF
ω 0 = 500 rad/s
5.4 Comparison of Series and Parallel Resonant Circuits 5.23
Example 5.22 A coil of 10 W resistance and 0.2 H inductance is connected in parallel with a
variable capacitor across a 220 V, 50 Hz supply. Calculate (a) the capacitance of the capacitor for resonance, (b) the dynamic impedance of the circuit, and (c) supply current.
Solution
R
0 Ω, L = 0.2 H, V
f0 =
1
2π
1
R2
− 2
LC L
50 =
1
2π
1
(10) 2
−
0 2 × C (0.2) 2
f 0 = 50 Hz
(a) Capacitance of the capacitor
C = 49.41 μF
(b) Dynamic impedance of the circuit
ZD =
L
02
=
= 404.78 Ω
CR 49.41 × 10 −6 × 10
(c) Supply current
I=
V
220
=
= 0.543 A
Z D 404.78
Example 5.23 Find the value of the dynamic impedance of the circuit shown in Fig. 5.16 at a
frequency of 500 kHz and for bandwidth of operation equal to 20 kHz. The resistance of the coil is 5 W..
5Ω
L
C
Fig. 5.16
Solution
f
BW
20 kHz, R = 5 Ω
R
2π L
5
20 × 103 =
2π L
L = 39.79 μΗ
BW =
f0 =
1
2π
1
R2
− 2
LC L
500 ×10
103 =
1
2π
1
39.79 ×10
10 −6 × C
−
(5) 2
(39.79 × 10 −6 ) 2
(39
5.24 Network Analysis and Synthesis
C = 2 54 nF
ZD =
L
39.79 × 10 −6
=
= 3 13 k Ω
CR 2 54 × 10 −9 × 5
Example 5.24 A coil having a resistance of 20 W and an inductance of 200 μH is connected in
parallel with a variable capacitor. This parallel combination is connected in series with a resistance of
8000 W. A voltage of 230 V at a frequency of 106 Hz is applied across the circuit as shown in Fig. 5.17.
Calculate (a) the value of capacitance at resonance, (b) Q factor of the circuit, (c) dynamic impedance of the
circuit, and (d) total circuit current.
200 μH
20 Ω
8000 Ω
C
230 V, 106 Hz
Fig. 5.17
Solution
R
20 Ω, L = 200 μH,
6
f
V = 230 V, RS = 8000 Ω
(a) Value of capacitance
f0 =
1
2π
106 =
1
2π
1
R2
− 2
LC L
1
200 × 10 −66 × C
−
( 20) 2
( 200 × 10 −6 ) 2
C = 126.65 pF
(b) Quality factor
Q0 =
2π f L 2π × 106 × 200 × 10
=
R
20
ZD =
L
200 × 10 −6
=
= 78958 Ω
CR 126.65 × 10 −12 × 20
6
= 62.83
(c) Dynamic impedance
(d) Current
Total impedance of the circuit at resonance = Z D + RS = 78958 + 8000 = 86958
5 Ω
230
I=
= 2.65 mA
86958
5.4 Comparison of Series and Parallel Resonant Circuits 5.25
Example 5.25 Derive the expression for resonant frequency for the parallel circuit shown in
Fig. 5.18. Also calculate the impedance and current at resonance.
R
L
C
v = Vm sin wt
Fig. 5.18
Solution
Z1
R
Z 2 = jjX
X
Z 3 = − jX C
(a) Resonant frequency of the circuit
For a parallel circuit,
1
1
1
1
=
+
+
Z Z1 Z 2 Z3
Y=
1
1
1
1
1
1
1
+
+
= −j
+j
= −
R jX
jX
− jX C R
XL
XC R
1 ⎞
⎛ 1
j⎜
−
⎝ X L X C ⎟⎠
At resonance, the circuit is purely resistive. Hence, the condition for resonance is
1
1
−
=0
X L XC
X L XC
1
ω 0C
1
ω 02 =
LC
1
ω0 =
LC
ω0 L =
f0 =
1
2π LC
where f 0 is called the resonant frequency of the circuit.
(b) Impedance at resonance.
At resonance, the circuit is purely resistive. Hence, the imaginary part of Y is zero.
YD =
ZD
1
R
R
5.26 Network Analysis and Synthesis
(c) Current at resonance
V
V
=
ZD R
I0 =
Example 5.26
Derive the expression for the resonant frequency of the parallel circuit as shown in
Fig. 5.19.
R1
L
R2
C
v = Vm sin wt
Fig. 5.19
Solution
Z1 = R1 + jX L
Z 2 = R2 − jX C
For a parallel circuit,
1
1
1
=
+
Z Z1 Z 2
Y=
⎛ X
⎞
1
1
R − jX L R2 + jX C
R
R
X
+
= 12
+ 2
= 2 1 2 + 2 2 2 − j⎜ 2 L 2 − 2 C 2 ⎟
2
2
R1 + jX L R2 − jX
XC
⎝ R1 X L R2 X C ⎠
R1 + X L R2 + X C
R1 X L R
XC
At resonance, the circuit is purely resistive. Hence, the condition for resonance is
XL
R12
X L2
−
XC
R22
X C2
XL
R12
X L2
=0
=
XC
R22
X C2
1
ω0 L
ω 0C
=
2
2 2
1
2
R1 ω 0 L
R2 + 2 2
ω0 C
ω 02 LC
ω 02C 2
=
R12 ω 02 L2 R22ω 02C 2 + 1
LC ( R 2
2
C2
ω 02 R22 LC 3
)
LC
C 2 (R
( R2
2 2
L)
C 2 R12 ω 02 L2C 2
ω 02 R22 LC 3 − ω 02 L2C 2 = C 2 R12 − LC
5.4 Comparison of Series and Parallel Resonant Circuits 5.27
ω 02 LC 2 (CR22
ω 02 LC (CR22
L)
C (CR12
L)
C 12
CR
ω 02 =
ω0 =
f0 =
L)
L
CR12 − L
LC (CR22 − L)
CR12 − L
LC (CR22 − L)
=
1
CR12 − L
2π LC
CR22 − L
1
CR12 − L
LC
CR22 − L
where f 0 is called the resonant frequency of the circuit.
Example 5.27
Derive the expression for resonant frequency for the parallel circuit shown in
Fig. 5.20.
r
L
R
C
v = Vm sin wt
Fig. 5.20
Solution
Z1
j
L
Z2 = R
Z3 = − jX
XC
For a parallel circuit,
1
1
1
1
=
+
+
Z Z1 Z 2 Z3
Y=
⎛
⎛ 1
⎞
1
1
1
j L 1
1
r
1⎞
X
+ +
= 2
+ +j
=⎜ 2
+ ⎟+ j
− 2 L 2⎟
2
2
r + jX L R − jX
XC r + X L
R
XC ⎝ r + X L
R⎠
⎝ XC r + X L ⎠
At resonance, the circuit is purely resistive, i.e., the imaginary part of Y is equal to zero.
1
XL
−
=0
X C r 2 X L2
1
X
= 2 L 2
XC r
XL
X L X C = r 2 + X L2
5.28 Network Analysis and Synthesis
1
= r 2 + ω 0 2 L2
ω 0C
L
= r 2 + ω 0 2 L2
C
L
ω 0 2 L2 = − r 2
C
ω0 L
ω02 =
ω0 =
f0 =
1
r2
− 2
LC L
1
r2
− 2
LC L
1
r2
− 2
LC L
1
2π
where f0 is called the resonant frequency of the circuit.
Example 5.28
Find the value of R in the circuit shown in Fig. 5.21 to achieve resonance.
R
−2Ω
−j
10 Ω
j10 Ω
Fig. 5.21
Solution
Z1
10
j10 Ω
Z2 = R − j 2 Ω
For a parallel circuit,
1
1
1
=
+
Z Z1 Z 2
Y=
1
1
10 − j10 R + j 2 ⎛ 10
R ⎞
2 ⎞
⎛ 10
+
=
+
=
+
−
⎟ − j ⎜⎝
⎟
10 + j10 R − j 2 100 + 100 R 2 + 4 ⎝ 200 R 2 + 4 ⎠
200 R 2 + 4 ⎠
At resonance, the circuit is purely resistive, i.e., the imaginary part of Y is equal to zero.
10
2
−
=0
200 R 2 + 4
2
1
=
2
20
R +4
R 2 + 4 = 40
R 2 = 36
R=6Ω
5.4 Comparison of Series and Parallel Resonant Circuits 5.29
Example 5.29
Find the value of R1 such that the circuit given in Fig. 5.22 is in resonance.
R1
j6 Ω
10 Ω
−4Ω
−j
Fig. 5.22
Solution
j 6 Ω, Z 2
R1
Z1
10 − j 4 Ω
For a parallel circuit,
1
1
1
=
+
Z Z1 Z 2
Y=
⎛ 6
1
1
R − j 6 10 + j44 ⎛ R1
10 ⎞
4 ⎞
+
= 12
+
=⎜ 2
+
− j⎜ 2
−
⎟
R1 + j 6 10 − j 4 R1 + 36 100 + 16 ⎝ R1 + 36 116 ⎠
⎝ R1 + 36 116 ⎟⎠
At resonance, the circuit is purely resistive, i.e., the imaginary part of Y is equal to zero.
6
4
−
=0
R12 + 36 116
6
4
=
2
R1 + 36 116
116 × 6
= 174
4
R12 = 138
R12 + 36 =
R1 = 11.75 Ω
Example 5.30
Find the value of C in the circuit shown in Fig. 5.23 to give resonance.
20 Ω
10 Ω
j 37.7 Ω
C
230 V, 50 Hz
Fig. 5.23
Solution
Z1
20
j3 7 Ω
Z2
1100 − jX C
5.30 Network Analysis and Synthesis
For a parallel circuit,
1
1
1
=
+
Z Z1 Z 2
1
1
20 − j 37 7
10 + jX C
Y=
+
=
+
20 + j 37 7 10 − jX C 400 + 1421.29 100 + X C2
⎛ 20
⎞
⎛ 37.7
10
XC ⎞
=⎜
+
− j⎜
−
⎟
2
2 .29 100 + X C ⎠
⎝ 1821
⎝ 1821.29 100 + X C2 ⎟⎠
At resonance, the circuit is purely resistive, i.e., the imaginary part of Y is equal to zero.
37.7
XC
−
=0
1821.29 100 + X C2
XC
100 + X C2
= 0.021
0.021 X C2 − X C + 2.1 = 0
Solving the quadratic equation,
Ω
XC
1
= 45.41
2π × 50 × C
C
μ
XC = . Ω
1
= 2.2
π
C
C = 1.44 mF
or
Example 5.31
Find the value of L at which the circuit resonates at a frequency of 1000 rad/s in the
circuit shown in Fig. 5.24.
5Ω
10 Ω
L
− 12 Ω
−j
Fig. 5.24
Solution
Z1
5
j
Ω
Z2
10 − j12
1 Ω
For a parallel circuit,
1
1
1
=
+
Z Z1 Z 2
Y=
⎛ XL
1
1
5 − jX L 10 + j12 ⎛
5
10 ⎞
12 ⎞
+
=
+
=
+
− j⎜
−
⎟⎠
2
5 + jX L 10 − j12 25 + X L2 100 + 144 ⎜⎝ 25 + X L2 244 ⎟⎠
244
⎝ 25 + X L
At resonance, the circuit is purely resistive, i.e., the imaginary part of Y is equal to zero.
5.4 Comparison of Series and Parallel Resonant Circuits 5.31
XL
25 + X L2
−
12
=0
244
XL
12
244
X L2
25 +
+ 300 = 0
12
Solving the quadratic equation,
=
×
1Ω
×
=
=
−
are connected in parallel
Find the value of X which will produce
Example 5.32
Two impedances
and this combination is connected in series with Z = (3
resonance.
Solution
+
Z
=
Z
20 +
20 +
=
− 30
+
+10 30
X)
−
+
1
H
jX Ω
Z3
19 23 −
=
+
− 3 85)
At resonance, the circuit is purely resistive, i.e., imaginary part of Z is equal to zero.
=
X
X
Example 5.33
Impedances Z 2 and Z in parallel are in series with an impedance Z1 across a
100 V, 50 Hz ac supply.
.
Determine the value of capacitance of XC such that the total current of the circuit will be in phase with the total voltage.
Solution
=
Z
Z=
+ Z1
55
+
+ − jX C
25 j 5 X C
+
10 jX
250 + 5 X C 2
100 + X
2
25
25 X C
−
C
2
+
Z
25
25 j 5 X C 10 + jX
×
+
10 jX C 10 + jX
+
5−
25 =
⎛
+
+ 1 25
250 + 5 X 2 ⎞
100 +
25 X
X C2
At resonance, the circuit is purely resistive, i.e., the imaginary part of Z is equal to zero.
25 X C
−
100 + X C 2
25 X C
100 + X
12
2
= 1 25
X + 125 = 0
2
⎞
− 1 25
5.32 Network Analysis and Synthesis
XC2
20 X C + 100 = 0
X C = 10
1
= 10
2π 50 × C
C = 318.31 μF
Example 5.34 Find the frequency at which the circuit shown in Fig. 5.25 will be at resonance. If the
capacitor and inductors are interchanged, what would be the value of resonant frequency?
−
j
Ω
2w
2
jw Ω
1Ω
Fig. 5.25
Solution
Z1
j
Ω
Z2 = −
j
Ω, Z3 = 1 Ω
2ω
⎛ j ⎞
⎜⎝ −
⎟ (1)
Z Z
j
j(2 + j )
1
2ω ⎞
2ω ⎠
⎛
Z = Z1 + 2 3 = jω +
= jω −
= jω −
=
+ j ω−
⎟
2
2
j
⎝
Z 2 Z3
2ω − j
4ω
ω 1
4ω 2 1⎠
−
+1
2ω
At resonance, the circuit is purely resistive, i.e., imaginary part of Z is equal to zero.
ω−
2ω
4ω 2 + 1
=0
ω=
2ω
4ω 2 + 1
1
1
=
2 4ω 2 + 1
4ω 2 + 1 = 2
4ω 2 1 = 0
ω = ± 0.5
Considering positive values of w,
d/s
ω = 0.5 d/s
If the capacitor and inductor are interchanged,
Z=−
j (1)( jω )
j
jω (1 − j )
j (ω 2 + jω )
ω2
ω ⎞
⎛ 1
+
=−
+
=
−
+
=
− j⎜
−
⎟
2
2
2
⎝
2ω 1 + jω
2ω
2ω
2ω 1 + ω 2 ⎠
1+ ω
1+ ω
1+ ω
5.4 Comparison of Series and Parallel Resonant Circuits 5.33
At resonance, the circuit is purely resistive, i.e., the imaginary part of Z is equal to zero.
1
ω
−
=0
2ω 1 + ω 2
1
ω
=
2ω 1 + ω 2
2ω 2 ω 2 1 = 0
ω 2 −11 0
ω = ± 1 rad s
Considering positive values of w,
ω = 1 rad / s
Example 5.35 A 400 V, 50 Hz supply feeds energy to a parallel circuit consisting of 10 ∠30° Ω
branch and 10 ∠− 60° Ω branch. Determine the impedance of a circuit element such that if connected in
series with the source, the system comes at resonance.
Zp =
Solution
( ∠ °) ( ∠− °)
= 7.07 ∠−
∠− 15° = 6.83 − j1.83 Ω
10 ∠30° + 10 ∠− 60°
Zeq = Z + Z p = Z + 6.83 − j 1.83
At resonance, the circuit is purely resistive, i.e., the imaginary part of Z is equal to zero. Hence Z must
be equal to j1.83 Ω.
Z= j
Ω = jX L
X L = 1 83
2π 50
1.83
L = 5.82 mH
Example 5.36
A coil of 400 W resistance and 318 μH inductance is connected in parallel with a
capacitor and the circuit resonates at 1 MHz. If a second capacitor of 23.5 pF capacitance is connected in
parallel with the first capacitor, find the frequency at which the circuit resonates.
Solution
R
00 Ω, L = 318 μH,
f0 =
1
2π
106 =
1
2π
C2 = 23
23.5 pF
f
1
R2
− 2
LC1 L
1
318 ×10
10 −6 × C1
−
( 400) 2
(318 × 10 −6 ) 2
C1 = 76.59
5 pF
When capacitor of 23.5 pF is connected in parallel with C1 = 76.59 pF ,
CT
C1 + C2 = 76.59 + 23.5 = 100.09 pF
5.34 Network Analysis and Synthesis
The new resonant frequency f 0′ =
1
2π
1
R2
1
− 2 =
LCT L
2π
1
318 × 10
−6
× 100.09 10
−12
12
−
( 400) 2
(318 × 10 −6 ) 2
= 869.34 kHz
Example 5.37 A series circuit consisting of a 12 W resistor, 0.3 H inductor and a variable capacitor
is connected across a 100 V, 50 Hz ac supply as shown in Fig. 5.26. The capacitance value is adjusted to
obtain maximum current. Find the capacitance value. Now, the supply frequency is raised to 60 Hz, the voltage remaining same at 100 V. Find the value of the capacitance C2 to be connected across the above series
circuit so that current drawn from the supply is minimum.
R
C1
L
100 V, 50 Hz
Fig. 5.26
Solution
Ω, L = 0.3 H, V
R
(a) Value of capacitance C1
The resonance occurs at
f = 50 Hz
f = 50 Hz
f0 =
50 =
1
2π LC1
1
2π 0.3 × C1
C1 = 33.77 μF
(b) Value of capacitance C2
The resonance occurs at 60 Hz for the circuit shown in Fig. 5.27.
R
C1
L
C2
100 V, 60 Hz
Fig. 5.27
f 0 L = 2π × 60 × 0.3 = 3. Ω
1
1
X C1 =
=
= 78.555 Ω
2π f C1 2π × 60 × 33 77 × 10 −6
Z1 = 12 + j113 1 j 78.55 36.57 ∠70.85° Ω
X
Y1 =
Yreq
1
1
=
= 0.027
02 ∠70.85° = 8.86
86 10
Z1 36.57 ∠70
∠70.85°
Y1 + Y2 = 8.86
86 10
3
j 0.0255 + Y2
3
j 0.0255 �
5.4 Comparison of Series and Parallel Resonant Circuits 5.35
At resonance, the imaginary part of Yreq becomes zero.
Y2 = j 0.0255 �
Y2 =
1
= 0.0255 �
X C2
X C2 = 39.22 Ω
X C2 =
1
2π f C2
1
2π 60 × C2
C2 = 67.63 μF
39.22 =
Example 5.38
Find active and reactive components of the current taken by a series circuit consisting
of a coil of inductance 0.1 H and a resistance of 8 W and a capacitor of 120 μF connected to a 240 V, 50 Hz
supply. Find the value of the capacitor that has to be connected in parallel with the above series circuit so
that the pf of the entire circuit is unity.
Solution
L
0.1
1
R = 8 Ω, C =
μ , V = 240 V,
f = 50 Hz
ffL = 2π × 50 × 0.1 = 3 . Ω
1
1
XC =
=
= 26.53 Ω
2π fC 2π × 50 × 120 × 10 −6
X
Z1 = R + jX − jX = 8
I=
j 31.42
j 26.53
9.38 ∠31.44° Ω
V
240
=
= 25.445 A
Z1 9 43
Active component of current = I cos φ = 25.45 cos(31. °) = 21.71 A
Reactiv
t e component of current = I sin φ = 25.45 sin(31. °) = 13.27 A
When a capacitor is connected in parallel with the above series circuit, to make pf of the entire circuit
unity,
Z1 = R + jX − jX = 8 j 31.42
42 j26
j 26 53 8 + j 4.89 Ω
Z 2 = − jX C
For a parallel circuit,
1
1
1
=
+
Z Z1 Z 2
Y=
8
1
1
8 − j 4 89
1
8
4 89 ⎞
⎛ 1
+
=
+j
=
+ j⎜
−
⎝ X C 87.91⎟⎠
j 4 89 − jX C 64 + 23 91
X C 87.91
At resonance, pf is unity and the imaginary part of Y is equal to zero.
1
4.89
−
=0
X C 87.91
1
= 0.056
XC
5.36 Network Analysis and Synthesis
X C = 17.86
1
= 17.86
2π × 50 × C
C = 178.23 μF
Example 5.39
A coil of 10 W resistance and 0.5 H inductance is connected in series with a capacitor.
On applying a sinusoidal voltage the current is maximum when the frequency is 50 Hz. A second capacitor
is connected in parallel with the circuit. What capacitance should it have so that the combination acts as a
non-inductive resistor at 100 Hz?
R 10 Ω, L = 0.5 H,
Solution
f0 =
50 =
f 0 = 50 Hz
1
2π LC
1
2π 0.5 × C
C = 20.26 μF
When a second capacitor is connected in parallel with the circuit so that the combination acts as a non-inductive
resistor at 100 Hz,
X L = 2π × 100 × 0.5 = 314.16 Ω
1
XC =
= 78.56 Ω
2π × 100 × 20.26 × 10 −6
Z1 = 10 + j 314 16 − j 78.56 = 10 + j 235.6 Ω
Z 2 = − jX C
For a parallel circuit,
1
1
1
=
+
Z Z1 Z 2
Y=
1
1
10 − j 235.6
1
10
235.6 ⎞
⎛ 1
+
=
+j
=
+j
−
⎝ X C 55607.36 ⎟⎠
10 + j 235 6 − jX C 100 + 55507 36
X C 555607.36
At resonance, the circuit acts as a non-inductive resistor, i.e., the imaginary part of Y is equal to zero.
1
235.6
−
=0
X C 55607.36
X C = 236.02
1
= 236.02
2π × 100 × C
C = 6 7 μF
Example 5.40 A coil having a resistance and inductance of 15 W and 8 mH respectively is connected
in parallel with another coil having a resistance and inductance of 4 W and 18 mH as shown in Fig. 5.28.
If this parallel combination is to be replaced by a single coil, calculate the value of resistance and inductance
5.4 Comparison of Series and Parallel Resonant Circuits 5.37
of that coil. What value of capacitance should be connected in parallel with this coil in order to get unity
power factor? Assume operating frequency to be 50 Hz.
R1
L1
R2
L2
Fig. 5.28
Solution
1 Ω
R1
L1 = 8
R2
4Ω
L2 = 18 mH,
f = 50 Hz
(a) Value of resistance and inductance of the coil
X
1
fL1
fL
2π × 50 × 8 10
X
2
ffL2
2π × 50 × 18 × 10 −3 = 5 65 Ω
Z1 = R1 + jX
Z 2 = R2 + jX
1
2
3
2.51 Ω
= 15 + j 2.51 155.21 9.5° Ω
=4
j 5.65 = 6.92
54.77° Ω
(15.21 ∠9.5°)(6.92 ∠5 .7° )
=
= 5.09
09∠40.96° Ω
Z1 Z 2
15.21 ∠9.5° + 6.92 ∠54.7°
R=38 Ω
XL = 3 3 Ω
X
ffL
3 34 2π × 50 × L
L = 10.63 mH
Z=
Z 1Z 2
3.84 + j 3.334 Ω
(b) Value of capacitance
When a capacitance C is connected in parallel with this coil as shown in Fig. 5.29, power factor becomes
unity. Hence, the circuit behaves like a parallel resonant circuit at 50 Hz.
R
L
C
Fig. 5.29
1
R2
− 2
LC L
f0 =
1
2π
50 =
1
1
(3.84) 2
−
2π 10.63 ×10
10 −3 × C ((10
10.63 × 10 −3 ) 2
C = 410 46 μF
5.38 Network Analysis and Synthesis
Exercises
5.1
5.2
5.3
A series RLC circuit has the following
parameter values:
R = 10 Ω, L = 0.014 H, C = 100 μF. Compute
the following.
(i) resonant frequency in rad/s
(ii) quality factor of the circuit
(iii) bandwidth
(iv) lower and upper frequency points of
the bandwidth
[ .2 rad/s, 1.185, 714.3 rad/s
487.9 rad/s, 1202.1 rad/s]
A series RLC circuit as R = 500 Ω, L = 60 mH
and C = 0.053 μF. Estimate
(i) circuit impedance at 50 Hz
(ii) resonance frequency in rad/s
(iii) quality factor of the circuit
(iv) bandwidth
(v) current at resonance frequency with
an applied voltage of 30 V
[ .0041 Ω, 17733 rad/s 2.128,
8333.
d/
]
A choking coil of 10 Ω resistance and 0.1
H inductance is connected in series with a
capacitor of 200 μF capacitance across a
supply voltage of 230 V, 50 Hz ac. Find the
following w.r.t. the resonant circuit.
(i) Neat circuit diagram indicating all the
relevant parameters.
(ii) At what frequency will the circuit
resonate.
(iii) Value of reactances at resonance
condition
(iv) Impedance of the coil at resonance
condition
(v) Voltage across the coil and voltage
across the capacitance
(vi) Half-power frequencies f1 and f2
(vii) Q factor of the circuit
(vii) Power and power factor
[ .59 Hz, 22.36 Ω,10
, Ω,
744.28 V, 514
14.28 V, 27.63 Hz,
43.55 Hz, 2.24, 5.29 kW, ]
5.4
5.5
5.6
5.7
5.8
A coil of 40 Ω resistance and 0.75 H
inductance forms part of a series circuit for
which the resonant frequency is 55 Hz. If the
supply is 250 V, 50 Hz, find
(i) the line current
(ii) the power factor
(iii) the voltage across the coil
[ .85 A, 0.616 (leading), 920.07 V ]
An RLC series circuit of 10 Ω resistance
should be designed to have a bandwidth of
100 Hz. Determine the values of L and C so
that the circuit resonates at 250 Hz.
[ .0159 H,, μ ]
A resistor and capacitor are connected in series
with a variable inductor. When the circuit is
connected to a 220 V, 50 Hz supply, the
maximum current obtained by varying the
inductance is 0.314 A. The voltage across the
capacitor is 800 V. Calculate the resistance,
inductance and capacitance of the circuit.
[
.63 Ω, 1.25 μ , 8.10 H]
A coil of 2 Ω resistance and 0.01 H inductance
is connected in series with a capacitor across
230 V mains. What must be the capacitance,
in order that maximum current occurs at a
frequency of 50 Hz? Find also the current and
the voltage across the capacitor.
[
, 115 A, 361.28 V ]
A choking coil of 10-ohm resistance and 0.1
H inductance is connected in series with a
capacitor of 200 μF capacitance. Calculate
the current, the coil voltage and the capacitor
voltage. The supply voltage is 230 V at 50 Hz.
At what frequency will the circuit resonate?
Calculate the voltages at resonant frequency
across the coil and capacitor. For this, assume
that supply voltage is 230 V of variable
frequency.
[ .47 A, 411.17, 198.47 V, 35.6 Hz,
563.3 V,, 514.
]
5.9
An inductive coil having an inductance of 0.04
H and a resistance of 25 Ω has been connected
in series with another inductive coil of 0.2 H
Objective-Type Questions 5.39
inductance and 15 Ω resistance. The whole
circuit has been energized from 230 V, 50 Hz
mains. Calculate power dissipation in each
coil and power factor of the whole circuit.
Draw the phasor diagram. Suggest a suitable
capacitor for the above circuit to resonate at
50 Hz.
[
.57 W,108.95 W, 0.468 (
42.
),
μ ]
5.10 A voltage v (t ) = 10 2 sin ω t is applied to
a series RLC circuit. At resonant frequency,
voltage across the capacitor is found to be 500
V. The bandwidth of the circuit is known to
be 400 rad/s and impedance at resonance, is
10 Ω. Determine resonant frequency, upper
and lower cut-off frequencies, L and C.
183 kHz, 3.151 kHz, 3.214 kHz,
0.025 H,, 0.. μ ]
5.11 An inductive coil having a resistance of 20
Ω and an inductance of 0.2 H is connected in
parallel with a 20 μF capacitor with variable
frequency, 230 V supply. Find the resonant
frequency at which the total current taken
from the supply is in phase with the supply
voltage. Also, find the value of this current.
[ .49 Hz, 4.6 A ]
Objective-Type Questions
5.1
5.2
5.3
5.4
5.5
For a series RLC resonant circuit, the total
reactance at the lower half power frequency
is
(a)
2 ∠45
5°
(b)
2
45°
(c) R
(d) −R
In a series RLC high Q circuit, the current
peaks at a frequency
(a) equal to the resonant frequency
(b) greater than the resonant frequency
(c) less than the resonant frequency
(d) none of the above.
In a series RLC circuit, R
k Ω, L = 1 H,
1
C=
μF. The resonant frequency is
400
1
(a) 2 10 4 Hz
(b)
× 10 4 Hz
π
(c) 10 4 Hz
(d) 2π 10 4 Hz
For a series resonant circuit at low frequency,
circuit impedance is ______ and at high
frequency circuit impedance is ______.
(a) capacitive, inductive
(b) inductive, capacitive
(c) resistive, inductive
(d) capacitive, resistive
A circuit with a resistor, inductor and
capacitor in series in resonant of f 0 Hz. If all
the component values are now doubled, the
new resonant frequency is
(a)
2 f0
(b)
f0
f0
(d) f 0
4
2
In a series RLC circuit at resonance, the
magnitude of the voltage developed across
the capacitor
(a) is always zero
(b) can never be greater than the input
voltage
(c) can be greater than the input voltage,
however it is 90° out of phase with the
input voltage
(d) can be greater than the input voltage and
is in phase with the input voltage
The power in a series RLC circuit will be half
of that at resonance when the magnitude of
the current is equal to
V
V
(a)
(b)
2R
3R
(c)
5.6
5.7
2V
R
2R
A series RLC circuit has a resonance frequency
of 1 kHz and a quality factor Q of 100. If each
of R, L and C is doubled from its original
value, the new Q of the circuit is
(c)
5.8
V
(d)
5.40 Network Analysis and Synthesis
(a) 25
(b) 50
(c) 100
(d) 200
5.9 A series RLC circuit has a Q of 100 and an
impedance of (
) Ω at its resonant
angular frequency of 107 rad/s. The values of
R and L are
(a) 100 Ω, 103 H
(b) 100 Ω, 10 −3 H
(c) 10 Ω, 10 H
(d) 10 Ω, 0.1 H
5.10 The following circuit in Fig. 5.30 resonates at
4H
(a) all frequencies (b) 0.5 rad/s
(c) 5 rad/s
(d) 1 rad/s
5.11 At resonance, the parallel circuit shown in
Fig. 5.31 behaves like
R
C
L
1F
10 Ω
Fig. 5.31
(a) an open circuit
(b) a short circuit
(c) a pure resistor of value R
(d) a pure resistor of value much higher
than R
1F
Fig. 5.30
Answers to Objective-Type Questions
5.1 (d)
5.7 (c)
5.2 (a)
5.8 (b)
5.3 (b)
5.9 (b)
5.4 (a)
5.10 (b)
5.5 (d)
5.11 (d)
5.6 (c)
6
6.1
Network Theorems
(Application to ac
Circuits)
INTRODUCTION
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.
6.2
MESH ANALYSIS
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,
Example 6.1
Find mesh currents I1 and I2 in the network of Fig. 6.1.
3Ω
j4 Ω
+
j 10 Ω
100∠45° V
−
I1
− 10 Ω
−j
I2
Fig. 6.1
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),
− j10I 2 = 100 ∠45°
100 ∠ 5°
I2 =
= 10 ∠135° A
− j10
…(i)
…(ii)
6.2 Network Analysis and Synthesis
Example 6.2
Find mesh current I1, I2 and I3 in the network of Fig. 6.2.
5Ω
−2Ω
−j
j5 Ω
+
3Ω
10∠30° V
−
I1
I2
5Ω
2Ω
I3
−2Ω
−j
Fig. 6.2
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),
−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,
10 ∠30°
−3
0
0
8 + j5
5
0
5
7 j2
I1 =
= 1.433 38.7° A
8 j2
−3
0
−3
8 + j5
5
0
5
7 j2
j 2 10 ∠30°
0
−3
0
−5
0
0
7 j2
I2 =
= 0.693∠ − 2.2° A
8 j2
−3
0
−3
8 + j5
5
0
5
7 j2
8
j2
−3 10 ∠30°
−3
8 + j5
0
0
5
0
I3 =
= 0.476
4 6 ∠13.8° A
8 j2
−3
0
−3
8 + j5
5
0
5
7 j2
8
… (i)
… (ii)
… (iii)
6.2 Mesh Analysis 6.3
Example 6.3
In the network of Fig. 6.3, find the value of V2 so that the current through (2 + j3) ohm
impedance is zero.
5Ω
j3 Ω
2Ω
+
4Ω
+
j5 Ω
30∠0° V
−
6Ω
I2
I1
I3
−
V2
Fig. 6.3
Solution 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 ⎥ ⎢ I2 ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ ⎥ ⎢
⎥
6
10 ⎦ ⎣ I3 ⎦ ⎣ V2 ⎦
⎣ 0
By Cramer’s rule,
5 + j 5 30 ∠0° 0
− j5
0
−6
0
V2
10
I2 =
=0
5 j5 − j5
0
− j 5 8 + j8 6
0
6
10
(
j )(
) (
)( j
)
V2 =
Example 6.4
0
j1500
= 35.36 ∠45° V
30 + j 30
Find the value of the current I3 in the network shown in Fig. 6.4.
10∠30° V
−4Ω
−j
4Ω
+
+
20∠0° V
−
−
j10 Ω
I1
20 Ω
I2
10 Ω
4Ω
I3
20 Ω
Fig. 6.4
−4Ω
−j
…(iii)
6.4 Network Analysis and Synthesis
Solution 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
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,
14 + j 6
j10
20 ∠0°
− j10
24 + j 6
10 30°
−10
−( 4 j 4)
0
I3 =
= 0.44 ∠ − 14° A
14 + j 6
10
10
− j10
24 j6
6 −( 4 j 4)
−10
−( 4 j4
4) 34 − j 4
Example 6.5
Find the voltage VAB in the network of Fig. 6.5.
1Ω
− 50 Ω
−j
100 Ω
A
B
I2
4Ω
100 Ω
j200 Ω
96 Ω
I1
+
−
10∠0° V
Fig. 6.5
Solution 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,
200
−(104 + j 200) ⎤ ⎡ I1 ⎤ ⎡10 ∠0°⎤
⎡ 200 + j200
=
⎢⎣ −(104 + j200
200)
205 + j150 ⎥⎦ ⎢⎣I2 ⎥⎦ ⎢⎣ 0 ⎥⎦
…(ii)
6.2 Mesh Analysis 6.5
By Cramer’s rule,
10 ∠0° −
+ 200
0
205 + j150
200 +
− 104 + 200
104 + 200
+ j150
I1
10
200 + 200 10 ∠0
−
+ j 200
0
200 +
− 104 + j 200
− 104 + j 200
+ j150
−
2
I2
=
Example 6.6
0
A
26 34° A
−
∠26 34° − ( 4 + j 200) ( 0.051
V
− 045∠
°
For the network shown in Fig. 6.6, find the voltage across the capacitor.
1Ω
2Ω
j2 Ω
1Ω
3Ω
I2
j2 Ω
1Ω
I1
+
I3
j3 Ω
5∠0° V
j1 Ω
Fig. 6.6
Solution Applying KVL to Mesh 1,
0 −
2)
0
5∠0°
…(i)
Applying KVL to Mesh 2,
−
2
−
Applying KVL to Mesh 3,
+ )
− + 2
− + )
=
+ 2I 3 = 0
+
+( +
…(ii)
=0
=0
…(iii)
Writing Eqs (i), (ii) and (iii) in matrix form,
−2
⎡
2
−
5
3
2
j2
1 + j 3 ⎡ I1 ⎤
I2
2 + j2
I3
⎤
5
0
0
6.6 Network Analysis and Synthesis
By Cramer’s rule,
5
5
2
−1
I2
1
0
0
−2
j2
2 + j2
1 + j3
130 51° A
2
+ j2
(
−
I3
3
−1
−2
5∠0°
j2
0
j2
0
−2
+ j3
2
2
−
V
Example 6.7
09
A
∠130 51° = 3 0
2 0 91
V
Find the voltage across the 2 W resistor in the network of Fig. 6.7.
j1 Ω
3Ω
2Ω
+
8∠45° V
2∠30° A
I1
Ω
I2
Fig. 6.7
Solution For Mesh 1,
I1 = 2
0°
…(i)
Applying KVL to Mesh 2,
−
−
)(
)−
− ∠45
0
=
…(ii)
45
Substituting I1 in Eq. (i),
8∠45°
I2 =
V2
2 1
2 2 0
9
3 19∠ − 65°
65 A
78
84 37
3 V
6.2 Mesh Analysis 6.7
Example 6.8
Find the current through 3 W resistor in the network of Fig. 6.8.
3Ω
5Ω
j2 Ω
+
1Ω
I2
I1
10∠0° V
1∠0° A
I3
j1 Ω
−
Fig. 6.8
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,
I 3 I 2 = 1∠0°
Applying KVL to the outer path of the supermesh,
−1( 2 − 1 ) − 5I 3 − j1 3 = 0
I1 I 2 − (5
…(ii)
1) I 3 = 0
…(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 −(5 j1) ⎦ ⎣ I 3 ⎦ ⎣ 0 ⎦
⎣ 1
By Cramer’s rule,
10 ∠0°
1
0
1∠0° −1
1
0
1 −(5 1)
I1 =
= 2.11∠ 28.01° A
4 j 2 −1
0
0
1
1
1
1 −(5 1)
I 3 Ω = I1 = 2.11
Example 6.9
28.01° A
Find the currents I1 and I2 in the network of Fig. 6.9.
2Vx
6Ω
−+
+
9∠0° V
−
Vx
I1
+
−
3Ω
−3 Ω
−j
Fig. 6.9
I2
6.8 Network Analysis and Synthesis
Solution From Fig. 6.9,
Vx = − j 3( 1 −
…(i)
2)
Applying KVL to Mesh 1,
9∠0° − 6
1
j 3 ( I1 − I 2 ) = 0
j ) I1 + j 3
(
2
9 ∠ 0°
…(ii)
Applying KVL to Mesh 2,
j3
j 3 ( 2 1 ) 2Vx − 3
j 3I1 + 2 [ jj3
j3
3 ( I1 I 2 )] − 3
j9
(3 j9
j 9)
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,
9∠0°
j3
0
3 − j9
I1 =
= 1.3∠2.49° A
6 j3
j3
j9
3 − j9
j 3 9 0°
j9
0
I2 =
= 1.24 ∠ − 15.95° A
6 j3
j3
j9
3 − j9
6
Example 6.10
Find the voltage across the 4 W resistor in the network of Fig. 6.10.
6∠30° V
−
+
j2 Ω
Ix
−1Ω
−j
2Ω
+
−
I1
2 Ix
4Ω
I2
Fig. 6.10
Solution From Fig. 6.10,
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
(
j ) I1 + j1 2 6 ∠30°
…(ii)
6.3 Node Analysis 6.9
Applying KVL to Mesh 2,
=0
4
0
=0
…(iii)
Writing Eqs (ii) and (iii) in matrix form,
⎡
1
⎤ I1
⎡
=
I2
1
−4
30 ⎤
0
By Cramer’s rule,
2
I2
1 6 0°
1
0
=
1
1
−4
A
(0 74 ∠ − 2.
V
6.3
4
2
∠−
1° V
NODE ANALYSIS
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.
Example 6.11
In the network shown in Fig. 6.11, determine Va and Vb.
j6 Ω
Va
3Ω
Vb
j5 Ω
+
j4 Ω
j6 Ω
10∠0° V
−
j4 Ω
Fig. 6.11
Solution Applying KCL at Node a,
+
1
j
1
j6
3
=0
1
10 ∠ °
V =
3
j6
1
3
= 1 67
90°
…(i)
Applying KCL at Node b,
V
−
−
1
3
V
1
3
+
1
j
−
+
1
1
j1
=0
0
b
0
…(ii)
6.10 Network 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∠75.17°
Va =
= 5.244
0 33
0.33
33Va
Example 6.12
75.17
1 °V
For the network shown in Fig. 6.12, find the voltages V1 and V2.
5Ω
V1
+
50∠0° V
4Ω
V2
2Ω
+
j2 Ω
−2Ω
−j
−
50∠90° V
−
Fig. 6.12
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 ( .7
.7
75
5
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
=
⎢⎣ −0.25
0.75 + j 0 5⎥⎦ ⎢⎣ V2 ⎥⎦ ⎢⎣ 25∠90°⎥⎦
By Cramer’s rule,
10 ∠0°
0.25
j 25 0.75
75 j 0.5
V1 =
= 24.7∠7 . 5° V
0.45
45 j 0.5
0.25
−0.25
0.75 + j 0 5
0.45
45 j 0.5 10 0°
−0
0 25
25 90°
V2 =
= 34.34
34 ∠552.82
8 °V
0.45
45 j 0.5
0.25
−0.25
0.75 + j 0 5
…(ii)
6.3 Node Analysis 6.11
Example 6.13
Find the voltage VAB in the network of Fig. 6.13.
j5 Ω
2Ω
I = 10∠0° A
2
j 10 Ω
A
I
1
B
3Ω
j4 Ω
Fig. 6.13
Solution 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 ⎠
−0
0.5V1 + (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
5 j 0.3
V1 =
= 21.8 56.42° V
0.62
62 j 0.16
−0 5
−0.5
0.5 − j 0 3
0.62
62 j 0.16 10 ∠0°
−0 5
0
V2 =
= 18.7 87.42
4 °V
0.62
62 j 0.16
−0 5
−0.5
0.5 − j 0 3
VA = V2 = 18.7∠87.442° V
V1
21.88∠
∠56. °
VB =
( j4) =
( j4) = 17.45∠93.32° V
3 + j4
3 + j4
VAB VA VB = ( .7∠87.. °) − ( .45∠93.. °) = 2.23∠34.1° V
…(ii)
6.12 Network Analysis and Synthesis
Example 6.14
Find the node voltages V1 and V2 in the network of Fig. 6.14.
−1 Ω
−j
V1
2Ω
V2
j2 Ω
2V2
2∠30° A
Fig. 6.14
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
…(i)
0
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 j1
⎢⎣ − 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,
0
( 2 + 1)
2∠30°
j 0.5
V1 =
= 2.466 ∠130.62° V
0.5 + j1 −( 2 + j1)
− j1
j 0.5
0.5 j1
0
− j1
2∠30°
V2 =
= 1.2233∠167.49° V
0..
j
( 2 + j1)
(2
− j1
j 0.5
Example 6.15
4 W resistor.
In the network of Fig 6.15, find the voltage V2 which results in zero current through
5Ω
V1
4Ω
V3
2Ω
+
+
j2 Ω
50∠0° V
−
−2Ω
−j
−
Fig. 6.15
V2
6.3 Node Analysis 6.13
Solution Applying KCL at Node 1,
+
5
+
2
1 1 1
+
+
5 j2 4
. 45
0
4
1
4
= 10 ∠0°
= 10 ∠0°
0
…(i)
Applying KCL at Node 3,
+
4
1
4
1
4
V2
+
2
2
1
−j
1
0
0 5V
= 0 5V
…(ii)
Writing Eqs (i) and (ii) in matrix form,
5
0.25
+j
−
⎡ V1 ⎤
10 ∠0°
By Cramer’s rule,
10 0
−
25
0.5
0 25
+ j0 5
5
−
0 5 10
.5V
05
0.25
. + j0 5
0 75
V1
V
I4 Ω
10 0 7
0.
5
0 125
1 5°
−
+
Example 6.16
5
4
2
0 55
10 .7
.
0 125 V
5°
.
05
0
0 55
1
25
5°
0
.
25
15 95
0 25V2
=V 01 0 )
5 5
V =
= 26
5
2 V
Find the voltage across the capacitor in the network of Fig. 6.16.
V1
2∠60 A
12∠30 V
V2
+
j1 Ω
Fig. 6.16
2Ω
j2 Ω
6.14 Network 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,
( j)
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°
V2 =
= 18.55∠157.42° V
1
1
− j1
j1 0 5 j 0.5
V2 = 18.55∠157
57.42
42° V
Vc
6.4
SUPERPOSITION THEOREM
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.
Example 6.17
Find the current through the 3 + j4 ohm impedance.
5Ω
50∠90° V
j5 Ω
+
3Ω
−
j4 Ω
−
+
50∠0° V
Fig. 6.17
Solution
Step I
j5 Ω
5Ω
When the 50 ∠90° V source is acting alone (Fig. 6.18)
(3 + j 4)( j5)
= 6.35∠23.2° Ω
3 + j9
50 ∠90°
IT =
= 7.87∠66.8° A
6.35
35∠
∠23.2°
ZT = 5 +
I′
+
3Ω
−
j4 Ω
50∠90° V
Fig. 6.18
6.4 Superposition Theorem 6.15
By current division rule,
⎛ j5 ⎞
I ′ = ( .87∠66.. °) ⎜
= 4.15∠85.3° (↓)
⎝ 3 + j 9 ⎟⎠
j5 Ω
5Ω
When the 50∠0° V source is acting alone (Fig. 6.19)
Step II
I″
5(3 + j 4)
ZT = j 5 +
= 6.74 ∠68.2° Ω
8 + j4
50 ∠0°
IT =
= 7.42∠
∠−
− 68.2° A
6.74∠
74 ∠68.2°
3Ω
−
j4 Ω
+
50∠0° V
Fig. 6.19
By current division rule,
⎛ 5 ⎞
I″ = ( .42∠− 68.. °) ⎜
= 4.15∠−
∠− 94.77° A ((↑
↑) = 4.15∠ .3° A (↓)
⎝ 8 + j 4 ⎟⎠
By superposition theorem,
I = I′ + I′′ = 4.15 ∠85.3° + 4.15 ∠85.3° = 8.31 ∠85.3°A (↓)
Step III
Example 6.18
Determine the voltage across the (2 + j5) ohm impedance for the network shown in
Fig.6.20.
j4 Ω
−3Ω
−j
+
2Ω
−
j5 Ω
50∠0° V
20∠30° A
Fig. 6.20
Solution
Step I
j4 Ω
When the 50∠0° V source is acting alone (Fig. 6.21)
I=
50 ∠0°
= 5.42∠ − 77.47° A
2 + j 4 + j5
Voltage cross (2 + j5) Ω impedance
2Ω
+
50∠0° V
−
I=(
⎛ j4 ⎞
∠ °) ⎜
= 8.68∠42.53° A
⎝ 2 + j 9 ⎟⎠
Voltage across (2 + j5) Ω impedance
V′′ = (2 + j5) (8.68 ∠42.53°) = 46.69 ∠110.72° V
j5 Ω
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. 6.22)
By current division rule,
−3Ω
−j
Fig. 6.21
−3Ω
−j
j4 Ω
I
2Ω
20∠30° A
j5 Ω
Fig. 6.22
6.16 Network Analysis and Synthesis
By superposition theorem,
Step III
V = V′ + V′′ = 29.16 ∠−9.28° + 46.69 ∠110.72° = 40.85 ∠72.53° V
Example 6.19
Determine the voltage VAB for the network shown in Fig. 6.23.
j5 Ω
−2Ω
−j
A
4∠0° A
+
50∠0° V
5Ω
−
B
Fig. 6.23
Solution
Step I
When the 50∠0° V source is acting alone (Fig. 6.24)
j5 Ω
A
−2Ω
−j
+
50∠0° V
5Ω
−
B
Fig. 6.24
′
VAB
= 50 ∠0° V
Step II
When the 4∠0° A source is acting alone (Fig. 6.25)
j5 Ω
−2Ω
−j
A
4∠0° A
5Ω
B
Fig. 6.25
″
VAB
=0
Step III
By superposition theorem,
VAB
′
″
VAB
+ VAB
= 50 ∠0° + 0 = 50 ∠0° V
6.4 Superposition Theorem 6.17
Example 6.20
Find the current I in the network shown in Fig. 6.26.
−5Ω
−j
j3 Ω
I 4Ω
2Ω
+
13 ∠25° V
+
3 ∠50° A
−
−
20 ∠−30° V
Fig. 6.26
Solution
Step I
When the 13∠25° V source is acting alone (Fig. 6.27)
j3 Ω
4Ω
−5Ω
−j
2Ω
+
13 ∠25° V
−
I′
Fig. 6.27
I′ =
13∠25°
= 2.057
0 ∠43.43°
6 − j2
(→)
Step II When the 20∠−30° V source is acting alone (Fig. 6.28)
4Ω
j3 Ω
−5Ω
−j
2Ω
+
20 ∠−30° V
−
I″
Fig. 6.28
I ′′ =
20 ∠ − 30° V
= 3.16
6 ∠ − 11.57°° A(←) = 3.16 ∠ 68. 3° A( → )
6 − j2
Step III When the 3∠50° A source is acting alone (Fig. 6.29)
4Ω
I′′′ j 3 Ω
−5Ω
−j
3 ∠50° A
Fig. 6.29
2Ω
6.18 Network Analysis and Synthesis
By current division rule,
2 − j5
= 2.56 ∠0.23°
6 − j2
Step IV By superposition theorem,
I ′′′ = 3∠50° ×
(←) = 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 (→)
Example 6.21
Find the current through the j3 W reactance in the network of Fig 6.30.
−5 Ω
−j
+
j5 Ω
−2 Ω
−j
5∠30° V
−
+
10∠60° V
−
j3 Ω
Fig. 6.30
Solution
Step I
When the 5∠30° V source is acting alone (Fig. 6.31)
−5 Ω
−j
+
j5 Ω
−2 Ω
−j
5∠30° V
−
j3 Ω
Fig. 6.31
When a short circuit is placed across j15 Ω reactance, it gets shorted as shown in Fig 6.32.
I′
−5 Ω
−j
+
−2 Ω
−j
5∠30° V
−
j3 Ω
Fig. 6.32
I′ =
5∠30°
= 2.5∠120° ( ← )
− j5 + j3
6.4 Superposition Theorem 6.19
When the 10∠60° V source is acting alone (Fig. 6.33)
Step II
−5 Ω
−j
+
j5 Ω
−2 Ω
−j
10∠60° V
−
j3 Ω
Fig. 6.33
When a short circuit is placed across the −j2 Ω reactance, it gets shorted as shown in Fig. 6.34
−5 Ω
−j
I′′
j5 Ω
+
10∠60° V
−
j3 Ω
Fig. 6.34
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°
Example 6.22
(←)
Find the current I0 in the network of Fig. 6.35.
2∠0° A
−2Ω
−j
6Ω
I0
8Ω
+
j4 Ω
10∠30° A
−
Fig. 6.35
−2Ω
−j
Solution
Step I
When the 10∠30° V source is acting alone (Fig. 6.36)
j 4(8 j 2)
ZT = 6 +
= 8.64 ∠24.12° Ω
j4 8 j2
10 ∠30°
IT =
= 1.16 ∠5.88° A
8.64∠
64 ∠2 . °
6Ω
IT
′
I0
8Ω
+
j4 Ω
10∠30° V
−
Fig. 6.36
6.20 Network Analysis and Synthesis
By current division rule,
I ′0 = 1.16
6 5.88° ×
Step II
j4
= 0.56 81.84°
j2 + j4
8
(↓)
When the 2∠0° A source is acting alone (Fig. 6.37)
2∠0° A
6Ω
−2Ω
−j
I0′′
j4 Ω
8Ω
Fig. 6.37
The network can be redrawn as shown in Fig. 6.38.
−2Ω
−j
−2Ω
−j
I0′′
I0′′
j4 Ω
8Ω
6Ω
2∠0° A
8Ω
2∠0° A
(1.85 + j2.77) Ω
(a)
(b)
Fig. 6.38
By current division rule,
I 0″ = 2 0° ×
Step III
1.85
85 j 2.77
= 0.67
6 ∠51.83°
1.85
85 j 2.77 + 8 j 2
(↓)
By superposition theorem,
I0
Example 6.23
I ′0 + I 0″ = 0.56 81.84° 0.67∠51
∠551.83° = 1.199 65.46°
(↓)
Find the current through the j5 W branch for the network shown in Fig. 6.39.
j5 Ω
−4Ω
−j
3Ω
+
+
+
15 ∠90° V
10 ∠0° V
−
−
Fig. 6.39
20 ∠0° V
−
6.4 Superposition Theorem 6.21
Solution
Step I
When the 10∠0° V source is acting alone (Fig. 6.40)
I′
j5 Ω
−4Ω
−j
3Ω
+
10 ∠0° V
−
Fig. 6.40
3( − j4
j )
= 4.04 ∠61.66° Ω
3 − j4
10 ∠0°
I′ =
= 2.488∠ − 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. 6.41)
( j5)( jj4)
ZT = 3 +
= 20.22∠
∠− 81.47° Ω
j5 j 4
IT =
IT
j5 Ω
3Ω
+
15∠90°
= 0.74 ∠171.47° A
20.22∠ − 81.47°
15 ∠90° V
−
By current division rule,
Fig. 6.41
− j4
I″ = 0.7 ∠171.47 ×
= .96 ∠ 8.53 A ( ← ) = 2 96∠171.47°
(→)
− j 4 + j5
Step III
When the 20 ∠0° V source is acting along (Fig. 6.42)
IT
I′′′
j5 Ω
−4Ω
−j
3Ω
+
20 ∠0° V
−
Fig. 6.42
3( j5
j )
= 3.47∠−
∠ 50.51° Ω
3 + j5
20 ∠0°
IT =
= 5.766 ∠50.51
0. ° A
3.47∠
∠− 50.51°
ZT = − j 4 +
By current division rule,
I″′ = 5.76
6 ∠50.51° ×
3
= 2.96 ∠
∠−
− 8.53° A ( ← ) = 2 96∠171.47°
3 + j5
(→ )
−4Ω
−j
6.22 Network Analysis and Synthesis
Step IV By superposition theorem,
I = I ′ + I ″ + I ″′ = 2.48∠
∠− 61.66° + 2.96 ∠ 7 . 7° + 2.96 ∠171.47° = 4.86
8 ∠− 164
6 .41° A
Example 6.24
Find the voltage drop across the capacitor for the network shown in Fig. 6.43.
20 ∠45° V
+
2Ω
j5 Ω
−
+
2Ω
−
10 ∠0° V
−2Ω
−j
4Ω
Fig. 6.43
Solution
Step I
When the 10∠0° V source is acting alone
(Fig. 6.44)
( 2 + j 5)( 2 j 2)
ZT = 4 +
2 + j5 2 j 2
= 7∠ − 5.91° Ω
10 ∠0°
IT =
= 1.43
3∠5 91° A
7∠ − 5.91°
I′
2Ω
+
2Ω
−
10 ∠0° V
4Ω
j5 Ω
Fig. 6.44
By current division rule,
I′ = ( .43∠5.
Step II
⎛
⎞
2 + j5
°) ⎜
= 1.54 ∠37.24° A ( → )
⎝ 2 + j 5 + 2 − j 2 ⎟⎠
When the 20∠45° V source is acting alone (Fig. 6.45)
20 ∠45° V
+
2Ω
I′′
−
2Ω
4Ω
−2Ω
−j
j5 Ω
Fig. 6.45
4(( + j )
= 4.48∠ − 8.84° Ω
4 + 2 + j5
20 ∠ 5°
I ′′ =
= 4.446
6 ∠53.84
8 °A (←)
4.48∠ − 8.84°
ZT = ( − j ) +
−4.46
.46 ∠53.84° A ( → )
−2Ω
−j
6.4 Superposition Theorem 6.23
By superposition theorem,
Step III
I = I ′ + I ″ = 1.54 ∠37.24 − 4.46 ∠53.84° = 3.01∠ − 117.78° A
Vc
Example 6.25
( j 2) I = ( − j 2) (3.01∠ − 117.78°))
6.02∠152
5 .22° V
Find the node voltage V2 in the network of Fig. 6.46.
5 ∠30° V
V1
V2
5Ω
10 ∠0° A
2Ω
j10 Ω
5 ∠0° V
Fig. 6.46
Solution
Step I
When the 10 ∠0° A source is acting alone (Fig. 6.47)
5 ∠30° V
V1′
5Ω
10 ∠0° A
V2′
2Ω
j10 Ω
Fig. 6.47
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
j 00.. )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 − j 0.
0. )V2′ = 0
…(ii)
6.24 Network Analysis and Synthesis
Writing Eqs (i) and (ii) in matrix form,
37 j 0.1
⎡ 0.37
⎢⎣ −(0
(0.17 j 0.1)
(0
(0.17 − j 0.1) ⎤ ⎡ V1′ ⎤ ⎡10 ∠0°⎤
(0
=
0.67
67 − j 0.2 ⎥⎦ ⎢⎣ V2′ ⎥⎦ ⎢⎣ 0 ⎥⎦
By Cramer’s rule,
0.37 j 0.1 10 0°
−
(
(0
0.17 j 0.1)
0
V2′ =
= 8.57∠
∠−
− 3.36° V
0.37
37 j 0.1
((00.17 − j 0.1)
−(0.17 − j 0 1)
0 67 − j0
j 0.2
When the 5∠0° A source is acting alone (Fig. 6.48)
Step II
5 ∠30° V
V2′′
2Ω
5Ω
j10 Ω
5 ∠0° A
Fig. 6.48
V2″
V ′′ V ″
+ 2 + 2 = 5 0°
5∠30° 5 2
j10
(0.61
11.93°)V2″
V2″
5∠0°
= 8.2∠
2 11.93° V
By superposition theorem,
V2 V2′ + V2″ = 8.57 − 3.36° + 8.2∠11.93° = 16.62∠4
62∠4.12° V
Step III
Example 6.26
Find current through inductor in the network of Fig. 6.49.
8∠135° V
−
j2 Ω
2∠0° A
+
−1 Ω
−j
2Ω
2∠90° A
8∠135° V
Fig. 6.49
Solution
Step I
j2 Ω
When the 8∠135° V source is acting alone (Fig. 6.50)
Applying KVL to the mesh,
8∠135
35° − ( − j1) ′ j 2I ′ = 0
8∠135°
I′ =
= 8 45°
j1
−
(←)
8∠ − 135° A (→)
+
I′
−1 Ω
−j
2Ω
Fig. 6.50
6.4 Superposition Theorem 6.25
When the 2∠0° A source is acting alone (Fig. 6.51)
Step II
j2 Ω
−1 Ω
−j
2Ω
2∠0° A
−1 Ω
−j
Fig. 6.51
The network can be redrawn as shown in Fig. 6.52.
By current division rule,
j2 Ω
I′′
⎛ − j1 ⎞
⎛ − j1⎞
I ′′ = 2∠0° ⎜
= 2∠0° ⎜
= 2∠180° A(→)
⎟
⎝ − j1+
1 j2⎠
⎝ j1 ⎟⎠
Step III
2Ω
2∠0° A
When the 2∠90° A source is acting alone (Fig. 6.53)
Fig. 6.52
j2 Ω
−1 Ω
−j
2Ω
2∠90° A
Fig. 6.53
j2 Ω
The network can be redrawn as shown in Fig. 6.54.
By current division rule,
−1 Ω
−j
⎛ − j1 ⎞
I ′′′ = 2∠90° ⎜
= 2∠ − 90°° A (←) = 2∠90° A (→)
⎝ − j1+
1 j 2 ⎟⎠
Step III
I
I′′′
2Ω
2∠90° A
By superposition theorem,
I ′ + I ′′ I ′′′ = 8 ∠− 135° + 2∠180
80° + 2∠90° = 8.49∠
∠− 154.47°A
Example 6.27
Fig. 6.54
Determine the source voltage VS so that the current through 2 W resistor is zero in
the network of Fig. 6.55.
3Ω
+
Vs
2Ω
j3 Ω
4Ω
+
−3Ω
−j
−
20∠90° V
−
Fig. 6.55
6.26 Network Analysis and Synthesis
Solution
Step I
When the voltage source Vs is acting alone (Fig. 6.56)
3Ω
2Ω
+
Vs
4Ω
j3 Ω
−
′
−3Ω
−j
I2′
I1
I 3′
Fig. 6.56
Appling KVL to Mesh 1,
3I1′ − j3
j 3(I1′ I′2 ) = 0
Vs
j )I1′ − j 3I ′2 = Vs
(
Appling KVL to Mesh 2,
− j 3( ′ − ′ ) − 2I′ + jj3
3(
′
2
−
− j 3 ′ + 2I ′ + j 3
Appling KVL to Mesh 3,
− j3(
′
3
−
′
3)
=0
′
3
=0
′
′
2 ) − 4 I3
=0
′
…(i)
…(ii)
′
3
…(iii)
j3
( 4 j3
j 3)
0
Writing Eqs (i), (ii) and (iii) in matrix form,
⎡ ′⎤
0 ⎤ ⎢ I1 ⎥ ⎡ Vs ⎤
⎡3 j 3 − j 3
⎢ − j3
2
j 3 ⎥ ⎢ I 2′ ⎥ = ⎢ 0 ⎥
⎢
⎥
⎢ ⎥
0
j
3
4
j 3⎦ ⎢ I ′ ⎥ ⎣ 0 ⎦
⎣
⎢⎣ 3 ⎥⎦
By Cramer’s rule,
3 j 3 Vs
0
− j3
0
j3
0
0 4 j3
(9
I 2′ =
=
3 j3 − j3
0
− j3
2
j3
0
j3 4 j3
Step II
j12) Vs
Δ
When the 20 ∠90° V source is acting alone (Fig. 6.57)
3Ω
2Ω
j3 Ω
I1′′
4Ω
+
−3Ω
−j
I2′′
20∠90° V
I3′′
−
Fig. 6.57
Applying KVL to Mesh 1,
−3
(
″
1
− j 3 ( I1″ − I ″2 ) = 0
j ) I1″ − j 3
″
2
0
…(i)
6.5 Thevenin’s Theorem 6.27
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
j 3)
j3
″
3
…(iii)
20 ∠90°
Writing Eqs (i), (ii) and (iii) in matrix form,
″
0 ⎤ ⎡ I1 ⎤ ⎡
0
⎡3 j 3 − j 3
⎤
⎢ ⎥
⎢ − j3
⎥
2
j 3 ⎥ ⎢ I ″2 ⎥ = ⎢
0
⎢
⎥ ″
⎢
⎥
⎢
⎥
0
j
3
4
j
3
−
20
∠
90
°
⎣
⎦ ⎣ I3 ⎦ ⎣
⎦
By Cramer’s rule,
3 j3
0
0
− j3
0
j3
0
20 90° 4 − j 3 −180 − j180
I ″2 =
=
3 j3 − j3
0
Δ
− j3
2
j3
0
j3 4 j3
Step III
By superposition theorem,
I2
(9
(9
j12
1 )V + ( −180 − j180)
=0
Δ
j12
1 )V + ( −180 − j180) = 0
(
6.5
I′2 + I 2″ =
j )V = 180 + j180
Vs = 16.97
9 ∠ − 8.13
13° V
THEVENIN’S THEOREM
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.
Example 6.28
Obtain Thevenin’s equivalent network for the terminals A and B in Fig. 6.58.
3Ω
−4Ω
−j
j5
−j
−4Ω
A
50∠0° V
+
4Ω
−
j6 Ω
B
Fig. 6.58
6.28 Network Analysis and Synthesis
Solution
Step I
Calculation of VTh (Fig. 6.59)
−4Ω
−j
3Ω
+
4Ω
+
50∠0° V
−j
−4Ω
j5
A
VTh
−
j6 Ω
I
−
B
Fig. 6.59
Applying KVL to the mesh,
50 ∠0° − (3 − j4) I − (4 + j6) I = 0
I=
50 ∠0°
= 6.87∠ − 15.95° A
(3 − j 4) + (4
( 4 + j 6)
VTh = (4 + j6) I = (4 + j6) (6.87 ∠−15.95°) = 49.5 ∠40.35° V
Step II
Calculation of ZTh (Fig. 6.60)
ZTh = (
−
3Ω
)+
( − j )( + j )
= 4.83∠ − 1.13° Ω
( − j )+( + j )
−4Ω
−j
−j
−4Ω
j5
A
4Ω
ZTh
j6 Ω
B
Fig. 6.60
Step III
Thevenin’s Equivalent Network (Fig. 6.61)
4.83 ∠−1.13° Ω
A
+
49.5 ∠40.35° V
−
B
Fig. 6.61
Example 6.29
Find Thevenin’s equivalent network for Fig. 6.62.
5Ω
j5 Ω
−2Ω
−j
A
+
10 ∠30° V
3Ω
5Ω
−
B
Fig. 6.62
6.5 Thevenin’s Theorem 6.29
Solution
Step I
Calculation of VTh (Fig. 6.63)
+
+
10 ∠30° V
j5 Ω
−2Ω
−j
5Ω
3Ω
−
I1
5Ω
I2
A
VTh
−
B
Fig. 6.63
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)
−3 ⎤ ⎡ I1 ⎤ ⎡10 ∠30°⎤
⎡8 j 2
=
⎢⎣ −3
8 + j 5⎥⎦ ⎢⎣I2 ⎥⎦ ⎢⎣ 0 ⎥⎦
By Cramer’s rule,
j 2 10 ∠30°
−3
0
I2 =
= 0.433
33∠9.7° A
8 j2
−3
−3
8 + j5
8
VTh
Step II
5(0.433∠9.7°))
Calculation of ZTh (Fig. 6.64)
ZTh
Step III
I2
−2Ω
−j
⎡ ⎧ ( − )3 ⎫
⎤
= ⎢⎨
⎬ + 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° Ω
6.94
94 j 4.735
5Ω
3Ω
5Ω
ZTh
B
Fig. 6.64
3.04 ∠33.4° Ω
A
+
−
B
Fig. 6.65
j5 Ω
A
Thevenin’s equivalent Network (Fig. 6.65)
2.16 ∠9.7° V
.16 ∠9.7° V
6.30 Network Analysis and Synthesis
Example 6.30
Obtain Thevenin’s equivalent network for Fig. 6.66.
4Ω
+
+
2Ω
−
j6 Ω
10 ∠0° V
A
−
5 ∠90° V
−4Ω
−j
B
Fig. 6.66
Solution
Step I
Calculation of VTh
4Ω
j6 Ω
+
+
2Ω
−
+
I
−
10 ∠0° V
−
+
A
5 ∠90° V
VTh
−4Ω
−j
−
B
Fig. 6.67
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. 6.67)
4Ω
A
2Ω
−4Ω
−j
j6 Ω
ZTh
B
Fig. 6.68
ZTh = 4 +
Step III
( 2 + j 6)(−
( j 4)
= 11.3∠
∠−
− 44.93° Ω
2 + j2
Thevenin’s Equivalent Network
11.3 ∠−44.93° Ω
A
+
18 ∠146.31° V
−
B
Fig. 6.69
6.5 Thevenin’s Theorem 6.31
Example 6.31
Obtain Thevenin’s equivalent network for Fig. 6.70.
j15 Ω
10 ∠0° A
2Ω
A
−5Ω
−j
3Ω
B
Fig. 6.70
I
Solution
j15 Ω
10 ∠0° A
Calculation of VTh (Fig. 6.71)
By current division rule,
Step I
2Ω
+
I=
Fig. 6.71
Calculation of ZTh (Fig. 6.72)
ZTh =
Step III
VTh
−
VTh = (−j5) I
= (5 ∠−90°) (13.42 ∠26.57°) = 67.08 ∠−63.43° V
Step II
−5Ω
−j
3Ω
( ∠ °)( )
= 13.42∠26.57° A
5 − j 5 + j15
j 15 Ω
( − j )(
)( + j )
= 7.07∠ − 81.86° Ω
− j 5 + 5 + j15
2Ω
A
Thevenin’s Equivalent Network
3Ω
7.07 ∠−81.86° Ω
−5Ω
−j
A
Fig. 6.72
−
B
Fig. 6.73
Example 6.32
Obtain Thevenin’s equivalent network for Fig. 6.74.
50 Ω
21 Ω
+
20 ∠0° V
A
−
12 Ω
j 24 Ω
Fig. 6.74
ZTh
B
+
67.08 ∠−63.43° V
B
30 Ω
j 60 Ω
A
B
6.32 Network Analysis and Synthesis
Solution
Step I
Calculation of VTh (Fig 6.75)
I2
I1
50 Ω
21 Ω
+
20 ∠0° V
A
−
+
VTh −
B
12 Ω
30 Ω
j 24 Ω
j 60 Ω
Fig. 6.75
20 ∠0°
= 0.49
9 − 36.02° A
21 + 12 + j 24
20 ∠0°
I2 =
= 0.2∠ − 36.86° A
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. 6.76)
21 Ω
50 Ω
A
B
12 Ω
j 24 Ω
30 Ω
j60 Ω
Fig. 6.76
ZTh =
Step III
21(12 + 24) 50(30 + 60)
+
= 47.4 ∠26.8° Ω
33 + j 24
80 + j 60
Thevenin’s Equivalent Network
47.4 ∠26.8° Ω
A
+
0.33 ∠171.12° V
−
B
Fig. 6.77
6.5 Thevenin’s Theorem 6.33
Example 6.33
Find Thevenin’s equivalent network across terminals A and B for Fig. 6.78.
A
1Ω
5Ω
2 ∠45° A
+
j2 Ω
10 ∠90° V
−
B
Fig. 6.78
Solution
Step I
Calculation of VTh (Fig. 6.79)
+
1Ω
5Ω
2 ∠45° A
VTh
+
j2 Ω
10 ∠90° V
−
−
Fig. 6.79
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. 6.80)
A
1Ω
5Ω
ZTh
j2 Ω
B
Fig. 6.80
ZTh =
A
5(1 + 2)
= 1 77∠45° Ω
5 + 1+ j2
B
6.34 Network Analysis and Synthesis
Thevenin’s Equivalent Network (Fig. 6.81)
Step III
1.77∠45° Ω
A
+
6.49 ∠112.5° V
−
B
Fig. 6.81
Example 6.34
Find the current through the (
) Ω impedance in the network of Fig. 6.82.
5Ω
+
3Ω
2Ω
20 ∠0° V
−
j2 Ω
5Ω
20 ∠0° A
−2Ω
−j
Fig. 6.82
Solution
Step I
Calculation of VTh (Fig. 6.83)
5Ω
V1
3Ω
+
+
VTh
B −
2Ω
A
20 ∠0° V
−
20 ∠0° A
−2Ω
−j
I2
Fig. 6.83
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∠29
9.05° V1 24 0°
V1 = 47.06
.06 ∠− 29.05° V
VTh = V1 = 47.06 ∠− 29.05° V
6.5 Thevenin’s Theorem 6.35
Calculation of ZTh (Fig. 6.84)
Step II
5Ω
3Ω
ZTh
2Ω
A
−2Ω
−j
B
Fig. 6.84
ZTh
5 ( 2 − 2)
= 3+
= 4.79∠
∠−
− 11.35° Ω
5 + 2 − j2
Calculation of IL (Fig. 6.85)
Step III
4.79∠−11.35° Ω
A
5Ω
+
47.06 ∠−29.05° V
−
IL
j2 Ω
B
Fig. 6.85
IL =
Example 6.35
47.06 ∠
∠− 29.05°
= 4.73∠
∠−
− 39.96° A
4.79∠
∠− 11.35° + 5 + j 2
Find the current through the 5 W resistor in the network of Fig. 6.86.
j5 Ω
6 ∠0° A
5Ω
4Ω
−
−j 2 Ω
4 ∠0° A
Fig. 6.86
Solution
Step I
Calculation of VTh (Fig. 6.87)
j5 Ω
V1
V2
+
6 ∠0° A
A
Vth
− B
4Ω
Fig. 6.87
−2Ω
−j
4 ∠0° A
6.36 Network Analysis and Synthesis
Applying KCL at Node 1,
V1 V1 − V2
+
+ 6 0° = 0
4
j5
⎛1 1 ⎞
V1
⎜⎝ 4 + j5⎠
5 ⎟⎠
( .25
j 0. )
j 0. 2
1
1
V2 = −66 0°
j5
6 ∠0°
2
…(i)
Applying KCL at Node 2,
V2
V1
j5
⎛ 1⎞
⎛ 1
V1
⎜⎝ − j 5⎠
⎟
5⎠
⎝ j5
+
V2
= 4 0°
j2
1⎞
V2 = 4 0°
j2⎠
j 0 2 V jj0
0 3 V2 = 4 ∠0°
Writing Eqs (i) and (ii) in matrix form,
25 j 0.2
⎡0.25
⎢⎣
j 0.22
…(ii)
j 0.2⎤ ⎡ V1 ⎤ ⎡ −66 0°⎤
=
j 0.3⎥⎦ ⎢⎣ V2 ⎥⎦ ⎢⎣ 4 ∠0° ⎥⎦
By Cramer’s rule,
−6
6 0° j 0 2
4 ∠0° j 0 3
V1 =
= 20.8∠− 126.87° V
0.25
25 j0
j 0 2 j 0 .2
j0 2
j 0.3
VThh = V1 = 20.8∠
∠− 126.87° V
Step II
Calculation of ZTh (Fig. 6.88)
j5 Ω
A
ZTh
4Ω
−2Ω
−j
B
Fig. 6.88
ZTh =
Step III
4( − 2 + 5)
= 2.4 ∠53
∠53.13° Ω
4 − j 2 + j 5)
Calculation of IL (Fig. 6.89)
2.4 ∠53.13° Ω
A
IL =
20.8∠
∠− 126.87°
= 3.1∠
∠− 143.47° Α
2.4 ∠53.13° + 5
+
20.8 ∠−126.87° V
5Ω
−
IL
B
Fig. 6.89
6.5 Thevenin’s Theorem 6.37
Example 6.36
In the network of Fig. 6.90, find the current through the 10 W resistor.
5 ∠30° V
+
+
2Ω
−
10 ∠0° V
1Ω
−
−2Ω
−j
10 Ω
5Ω
Fig. 6.90
Solution
Step I
5 ∠30° V
Calculation of VTh (Fig. 6.91)
Applying KVL to the mesh,
+
j2I 1I − 10 ∠0° 5
5II = 0
( j2 6)I = 10 ∠0°
2Ω
−
+
−
−2Ω
−j
I
VTh
5Ω
−
5I 10 0° 5∠30° 0 VTh = 0
5(1.58∠ − 161.57°) − 10 ∠0° 5∠30° − VTh = 0
VTh = 5.332∠ − 110.06° V
Step II
A
10 ∠0° V
1Ω
I = 1.58∠ − 161.57° A
Writing VTh equation,
+
B
Fig. 6.91
Calculation of ZTh (Fig. 6.92)
2Ω
A
ZTh = 2 +
5(1 − 2)
= 3.48
8∠ − 21.04° Ω
5 + 1 − j2
1Ω
5Ω
ZTh
−2Ω
−j
Step III
Calculation of IL (Fig. 6.93)
B
3.48 ∠−21.04° Ω
A
+
5.32 ∠−110.06° V
10 Ω
−
IL
B
Fig. 6.93
IL =
5.32∠
∠− 110.06°
= 0.4 ∠
∠− 104.67° A
3.48∠
∠− 21.04° + 10
Fig. 6.92
6.38 Network Analysis and Synthesis
Example 6.37
) Ω impedance in the network of Fig. 6.94.
Find the current through (
2Ω
j5 Ω
3Ω
4Ω
+
100 ∠0° V
−5Ω
−j
−
+
−
j6 Ω
50 ∠90° V
Fig. 6.94
Solution
Step I
Calculation of VTh (Fig. 6.95)
2Ω
j5 Ω
+
100 ∠0° V
−
I
3Ω
−5Ω
−j
+ A
VTh
−
B
+
−
50 ∠90° V
Fig. 6.95
Applying KVL to the mesh,
100 ∠0° 2I − j 5
3I + j 5 − 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 = 80.61
61
Step II
82.88° V
Calculation of ZTh (Fig. 6.96)
2Ω
j5 Ω
3Ω
−5Ω
−j
A
ZTh
B
Fig. 6.96
ZTh =
( + j )( − j )
= 6.28∠9.16° Ω
2 + j5 + 3 − j5
6.5 Thevenin’s Theorem 6.39
Calculation of IL (Fig. 6.97)
Step III
6.28∠9.16° Ω
A
4Ω
+
80.61 ∠−82.88° V
−
IL
j6 Ω
B
Fig. 6.97
IL =
Example 6.38
80.61∠ − 82.88°
= 6.552∠ − 117.34° A
6.28∠
28∠9.16° + 4 + j 6
Obtain Thevenin’s equivalent network across terminals A and B in Fig. 6.98.
4Ω
I
j2 Ω
A
− j1 Ω
+
10 ∠0° V
+
−
−
2I
B
Fig. 6.98
Solution
Step I
I
Calculation of VTh (Fig. 6.99)
Applying KVL to the mesh,
4Ω
10 ∠0° V
+
− j1 Ω
+
10 ∠0° 4I + j1I 2I = 0
I = 1.64 9.46° A
+
−
−
A
VTh
2I
−
Writing VTh equation,
10 ∠0° 4I − 0 VTh
j2 Ω
B
Fig. 6.99
0
10 ∠0° 4(1
(1.64 9.46°) − VTh = 0
VTh = 3 69 − 17° V
Step II
Calculation of IN (Fig. 6.100)
From Fig. 6.100,
I I1
Applying KVL to Mesh 1,
10 ∠0° 4I1 + j1( 1 2 ) 2I = 0
10 ∠0° 4I1 + j1 I1 j1 I2 2 1 0
(
j ) I1 + j1 2 = 10 0° …(i)
Applying KVL to Mesh 2,
2I j1(I − I ) j 2I 2 = 0
2I1 j1I − j1I1 j2 I 2 = 0
I
4Ω
j2 Ω
− j1 Ω
+
100 ∠0°V
−
A
I1
+
−
2I
IN
I2
B
Fig. 6.100
6.40 Network Analysis and Synthesis
…(ii)
(
j )I
) 1 j1 2 0
Writing Eqs (i) and (ii) in matrix form,
⎡6 j1 j1 ⎤ ⎡ I1 ⎤ ⎡10 ∠0°⎤
⎢⎣ 2 j1 − j1⎥⎦ ⎢⎣I2 ⎥⎦ = ⎢⎣ 0 ⎥⎦
By Cramer’s rule,
6 j1 10 ∠0°
2 j1
0
I2 =
= 2.71
7 ∠− 102.53° A
6 j1 j1
2 j1 − j1
I 2 = 2 71∠− 102
1 .53° A
IN
Calculation of ZTh
Step III
ZTh =
VTh
3 69∠ − 17°
=
= 1.36 ∠85.53° Ω
IN
2.71∠ − 102.53°
Step IV Thevenin’s Equivalent Network (Fig. 6.101)
1.36 ∠85.53° Ω
A
+
3.69 ∠−17° V
−
B
Fig. 6.101
Example 6.39
Find Thevenin’s equivalent network across terminals A and B for Fig. 6.102.
2Ω
j4 Ω
+
+
A
5 ∠0° V
−
0.2 Vx
Vx
1Ω
−
B
Fig. 6.102
Solution
Step I
2Ω
Calculation of VTh (Fig. 6.103)
j4 Ω
+
+
A
5 ∠0 ° V
From Fig. 6.103,
I = −0 2 Vx
−
… (i)
1Ω
0.2 Vx
Writing VTh equation,
−I + ∠ ° − 0 − Vx = 0
Vx = VTh
I
−
Fig. 6.103
B
6.6 Norton’s Theorem 6.41
0 2Vx
5 0° Vx = 0
Vx = 6 25 0° V
VTh
Step II
Vx = 6 25∠0° V
Calculation of IN (Fig. 6.104)
2Ω
j4 Ω
+
+
A
5 ∠0° V
−
Vx
0.2 Vx
IN
1Ω
−
B
Fig. 6.104
2Ω
From Fig. 6.104,
Vx = 0
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. 6.105.
IN =
Step III
+
5 ∠0° V
−
IN
1Ω
5∠ 0°
= 1∠−
− 53.13
13° A
1+ 2 + j4
B
Fig. 6.105
Calculation of ZTh
ZTh =
Step IV
j4 Ω
VTh
6 25∠0°
=
= 6.25
2 ∠53.13° Ω
IN
1∠− 53 13°
Thevenin’s Equivalent Network (Fig. 6.106)
6.25 ∠53.13° Ω
A
+
6.25 ∠0° V
−
B
Fig. 6.106
6.6
NORTON’S THEOREM
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.
6.42 Network Analysis and Synthesis
Example 6.40
Obtain Norton’s equivalent network between terminals A and B as shown in Fig. 6.107.
3Ω
j4 Ω
A
+
25 ∠0° V
4Ω
−
−j5 Ω
B
Fig. 6.107
Solution
Step I
Calculation of IN (Fig. 6.108)
When a short circuit is placed across (4 − j4) Ω impedance,
it gets shorted as shown in Fig. 6.109.
3Ω
j4 Ω
3Ω
A
+
25 ∠0° V
A
+
j4 Ω
4Ω
IN
−
−j5 Ω
25 ∠0° V
B
−
IN
Fig. 6.108
B
3Ω
Fig. 6.109
IN
Step II
A
25∠0°
=
= 5∠− 53.13
13° A
3 + j4
4Ω
ZN
Calculation of ZN (Fig. 6.110)
( + j )(
ZN =
3 + j4 4
Step III
j4 Ω
−j5 Ω
j )
= 4.53∠9.92° Ω
j5
B
Fig. 6.110
Norton’s Equivalent Network
A
5 ∠−53.13° A
4.53 ∠9.92° Ω
B
Fig. 6.111
Example 6.41
Obtain Norton’s equivalent network at the terminals A and B in Fig. 6.112.
5Ω
A
1Ω
4Ω
j2 Ω
j4 Ω
10 ∠30° A
B
Fig. 6.112
6.6 Norton’s Theorem 6.43
Solution
Step I
Calculation of IN (Fig. 6.113)
5Ω
1Ω
A
4Ω
10 ∠30° A
IN
j2 Ω
j4 Ω
B
Fig. 6.113
By series-parallel reduction technique (Fig. 6.114)
5Ω
10 ∠30° A
A
1.62 ∠58.24° Ω
IN
B
Fig. 6.114
62∠
∠58. ° ⎞
⎛ 1.62
I N = (10 ∠30°) ⎜
69∠75° A
⎟ = 2.69
⎝ 1.62∠
62∠58.24° + 5 ⎠
Step II
Calculation of ZN (Fig. 6.115)
5Ω
A
1Ω
4Ω
ZN
j2 Ω
j4 Ω
B
Fig. 6.115
ZN = 5 +
(1 + j 2) ( 4
1+ j2 4
j 4)
= 6.01∠13.24° Ω
j4
6.44 Network Analysis and Synthesis
Norton’s Equivalent Network (Fig. 6.116)
Step III
A
2.69 ∠75° A
6.01 ∠13.24° Ω
B
Fig. 6.116
Example 6.42
Find Norton’s equivalent network across terminals A and B in Fig. 6.117.
A
j4 Ω
10 Ω
4 ∠45° A
+
3Ω
25 ∠90° V
−
B
Fig. 6.117
Solution
Step I
Calculation of IN (Fig. 6.118)
A
10 Ω
j4 Ω
4 ∠45° A
IN
+
3Ω
25 ∠90° V
−
B
Fig. 6.118
When a short circuit is placed across the (
) Ω impedance, it gets shorted as shown in Fig. 6.119.
A
10 Ω
4 ∠45° A
IN
+
−
25 ∠90° V
B
Fig. 6.119
6.6 Norton’s Theorem 6.45
By source transformation, the network is redrawn as shown in Fig. 6.120.
A
4 ∠45° A
2.5 ∠90° A
10 Ω
A
4 ∠45° A
IN
2.5 ∠90° A
B
B
(a)
(b)
Fig. 6.120
I N = 4 ∠45° + 2.5∠90°
Step II
6.03∠62.04° A
Calculation of ZN (Fig. 6.121)
A
j4 Ω
10 Ω
ZN
3Ω
B
Fig. 6.121
ZN =
Step III
10 (3 + j 4)
= 3.68∠36.03° Ω
10 + 3 + j 4
Norton’s Equivalent Network (Fig. 6.122)
A
6.03 ∠62.04° A
3.68 ∠36.03° Ω
B
Fig. 6.122
Example 6.43
Obtain the Norton’s equivalent network for Fig. 6.123.
10 ∠0° A
IN
5Ω
j3 Ω
j5 Ω
A
j2 Ω
−5Ω
−j
B
Fig. 6.123
6.46 Network Analysis and Synthesis
Solution
Step I
Calculation of IN (Fig. 6.124)
5Ω
10 ∠0° A
j3 Ω
j5 Ω
A
−5Ω
−j
j2 Ω
IN
B
Fig. 6.124
By source transformation, the network can be redrawn as shown in Fig. 6.125.
Writing KVL equations in matrix form,
⎡5
⎢⎣ j 5
+
j 5⎤ ⎡ I1 ⎤ ⎡50 ∠0°⎤
=
0 ⎥⎦ ⎢⎣I2 ⎥⎦ ⎢⎣ 0 ⎥⎦
50 ∠0° V
A
I1
5 50 ∠0°
j5
0
I2 =
= 10 ∠− 90° A
5 j5
j5 0
Step II
j5 Ω
5Ω
By Cramer’s rule,
IN
j3 Ω
−
−5Ω
−j
j2 Ω
B
I 2 = 10 ∠− 90° A
Fig. 6.125
Calculation of ZN (Fig. 6.126)
5Ω
j3 Ω
j5 Ω
j2 Ω
−5Ω
−j
ZN
Fig. 6.126
Z N = j5 +
Step III
IN
I2
(5 + j 5) ( j5)
=5Ω
5 + j5 j5
Norton’s Equivalent Network (Fig. 6.127)
A
10 ∠−90° A
5Ω
B
Fig. 6.127
6.6 Norton’s Theorem 6.47
Example 6.44
Obtain the Norton’s equivalent network for Fig. 6.128.
5Ω
10 Ω
10 ∠45° V
+
−2Ω
−j
−
A
5Ω
B
10 Ω
j2 Ω
Fig. 6.128
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
I2 =
= 1∠41.28° A
15 − j 2
10 j 2
−5
−10 + j 2 15 j 2
0
−5
0
15 + j 2
Step II
IN
A
5 Ω I3
I1
B
10 Ω
j2 Ω
By Cramer’s rule,
IN
− 2 Ω I2
−j
+
10 ∠45° V
10 j 2
−5 ⎤ ⎡ I1 ⎤ ⎡10 ∠45°⎤
15 j 2
0 ⎥ ⎢I2 ⎥ = ⎢ 0 ⎥
⎥⎢ ⎥ ⎢
⎥
0
15 + j 2⎦ ⎣ I3 ⎦ ⎣ 0 ⎦
15 − j 2
−10 + j 2
−5
I3 =
15 − j 2
−10 + j 2
−5
5Ω
10 Ω
Calculation of IN (Fig. 6.129)
Writing KVL equations in matrix form,
Fig. 6.129
10 j 2 10 ∠45°
15 j 2
0
0
0
= 0.49 37.41° A
10 j 2
−5
15 − j 2
0
0
15 + j 2
I 3 − I 2 = 0.49∠37.41 − 1∠41.28° = 0.51∠− 135° A
Calculation of ZN (Fig. 6.130)
5Ω
10 Ω
−2Ω
−j
ZN
A
5Ω
B
j2 Ω
−j2 Ω
10 Ω
5Ω
A
B
10 Ω
5Ω
(a)
10 Ω
(b)
Fig. 6.130
j2 Ω
6.48 Network Analysis and Synthesis
ZN =
5(10 − j 2) 5(10 + j 2)
+
= 6.72 Ω
5 + 10 − j 2 5 + 10 + j 2
Norton’s Equivalent Network (Fig. 6.131)
Step III
A
0.51 ∠−135° A
6.72 Ω
B
Fig. 6.131
Example 6.45
Find the current through the 8 W resistor in the Network of Fig. 6.132.
5Ω
10 Ω
8Ω
+
20 ∠0° V
5 ∠0° A
−
j4 Ω
Fig. 6.132
Solution
Step I
Calculation of IN (Fig. 6.133)
5Ω
A
10 Ω
+
20 ∠0° V
IN
5 ∠0° A
−
j4 Ω
B
Fig. 6.133
When a short circuit is placed across the (
) Ω impedance, it gets shorted as shown in fig. 6.134.
5Ω
A
+
20 ∠0° V
IN
−
B
Fig. 6.134
5 ∠0° A
6.6 Norton’s Theorem 6.49
By source transformation, the network is redrawn as shown in Fig. 6.135.
A
5Ω
4 ∠0° A
IN
5 ∠0° A
B
Fig. 6.135
I N = 4 ∠0° + 5∠0° = 9∠0° A
Step II
Calculation of ZN (Fig. 6.136)
5Ω
A
10 Ω
ZN
j4 Ω
B
Fig. 6.136
ZN =
5(10 + j 4)
= 3.47∠6.87° Ω
5 + 10 + j 4
A
9 ∠0° A
Step III
3.47 ∠6.87 Ω
Calculation of IL (Fig. 6.137)
8Ω
B
9∠0°
IL =
= 0.79∠
∠−
− 2.08° A
3.47
47∠
∠6.87° + 8
Example 6.46
IL
Fig. 6.137
Obtain Norton’s equivalent network across the terminals A and B in Fig. 6.138.
5I
I
100 Ω
− 5Ω
−j
A
+
10 ∠0° V
−
j 10 Ω
B
Fig. 6.138
6.50 Network Analysis and Synthesis
Solution
Step I
Calculation of VTh (Fig. 6.139)
5I
100 Ω
I
+
−5Ω
−j
+
−
−
+
A
+
10 ∠0° V
j10 Ω
−
VTh
−
B
Fig. 6.139
I=
10 ∠0°
= 0.1∠
∠− 5.71° A
100 + j10
Writing VTh equation,
10 ∠0
0° 00
( j )(5 ) VTh 0
10 ∠0° − 100 (0
(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. 6.140)
5I
100 Ω
I
−5Ω
−j
+
A
−
+
10 ∠0° V
j10 Ω
−
IN
B
Fig. 6.140
By source transformation, the network is redrawn as shown in Fig. 6.141.
−5Ω
−j
100 Ω
I
− 25 I
−j
+ −
A
+
10 ∠0° V
j10 Ω
−
I1
IN
I2
B
Fig. 6.141
From Fig. 6.141,
I = I1
…(i)
6.7 Maximum Power Transfer Theorem 6.51
Applying KVL to Mesh 1,
10 ∠0° − 100 I1 j10
0 ( I1 − I 2 ) = 0
(100 + j10
10) I1 − j10 I 2 = 10 ∠0°
…(ii)
Applying KVL to Mesh 2,
− j10 (
− ) + j5
2
+ j 25 = 0
− j10 I + j100 I1 + j 5 + j 2255 1 = 0
j 35I − j5
j 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
I2 =
= 0.6 ∠30.96° A
100 + j10 − j10
j 35
− j5
I 2 = 0.6 ∠30.96° Α
IN
Step III
Calculation of ZN
ZN =
VTh
3.5∠85.1°
=
= 5.83∠54.14° Ω
IN
0.6 ∠30.96°
Step IV Norton’s Equivalent Network (Fig. 6.142)
A
0.6 ∠30.96° A
5.83∠54.14° Ω
B
Fig. 6.142
6.7
Zs
MAXIMUM POWER TRANSFER THEOREM
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. 6.143.
There are three possible cases for load impedance ZL.
+
Vs
ZL
−
Fig. 6.143
6.52 Network Analysis and Synthesis
Case (i)
When the load impedance is variable resistance (Fig. 6.144)
V
IL
V
s
S
L
Vs
IL
RS + jX S
+
ZL = RL
Vs
s
IL
The power delivered to the load is
Vs
2
L
Fig. 6.144
Purely resistive load
s
For power to be maximum,
dP
=0
L
V
2
⎡ {(
⎢
[(
s
2
R
0
− RL2
R
0
+ X s2
R
Hence, load resistance RL should be equal to the magnitude of the source impedance for maximum
power transfer.
S=
Case (ii) When the load impedance is a complex impedance with
variable resistance and variable reactance (Fig. 6.145)
IL
IL
RS + jX S
+
Vs
Vs
ZL = RL + jX L
Vs
Fig. 6.145 Complex impedance load
The power delivered to the load is
L
VS
=
L
2
L
For maximum value of PL, denominator of the equation should be small, ie.
PL =
VS
(
2
L
)2
.
6.7 Maximum Power Transfer Theorem 6.53
Differentiating the above equation w.r.t. RL and equating to zero,
−2 R
dPL
2
= Vs ⎢
dRL
⎣
s
+ RL
)2
(
− 2 RL (
=
−2
R
0
⎦
− 2R
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. 6.146)
S = RS + jX S
IL
V
+
Vs
Vs
IL
2
L
For maximum power,
dP
0
L
Vs
−2 R
2
s
+ RL
0
{(
−2
R
+
+
L
0
=0
L
2
R
0
)2
+(
R
=
L
=
XL
=
+
L
RL + jX L
Complex impedance load
Fig. 6.146
The power delivered to the load is
Vs
L
6.54 Network 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.
Example 6.47 For maximum power transfer, find the value of ZL in the network of Fig. 6.147 if
(i) ZL is an impedance, and (ii) ZL is pure resistance.
6Ω
−j 8 Ω
+
Vs
ZL
−
Fig. 6.147
Zs = ( − j ) Ω
Solution
(i) If ZL is an impedance
For maximum power transfer, Z L
(ii) If ZL is a resistance
For maximum power transfer, Z L
Example 6.48
Z*s = ( + j ) Ω
Z S = 6 + j8 = 10 Ω
For the maximum power transfer, find the value of ZL in the network of Fig. 6.148
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 W.
A
3Ω
2Ω
10 A
5 ∠0° V
j5 Ω
B
Fig. 6.148
Solution Thevenin’s impedance can be calculated by replacing voltage source by a short circuit and current
source by an open circuit.
A
2Ω
3Ω
ZTh
j5 Ω
B
Fig. 6.149
6.7 Maximum Power Transfer Theorem 6.55
ZTh =
3 ( 2 + 5)
= ( 2.1 + j 0.9) Ω
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 Ω
ZL is complex impedance with variable resistance and variable reactance
ZL
(iii)
Z*Th = ( .1 − j 0.. ) Ω
ZL is complex impedance with variable resistance and fixed reactance of j5 Ω
ZL
ZTh + j 5
j 0.9 + j 5 = 6 26 Ω
21
Example 6.49 Find the impedance ZL so that maximum power can be transferred to it in the network
of Fig. 6.150. Find maximum power.
3Ω
3Ω
+
5 ∠0° V
− j3 Ω
j3 Ω
−
ZL
Fig. 6.150
Solution
Step I
Calculation of VTh (Fig. 6.151)
IT 3 Ω
3Ω
+
5 ∠0° V
+
I
j3 Ω
−
− j3 Ω
A
VTh
−
Fig. 6.151
j 3(3 j 3)
= 6.71∠26.57° Ω
3 + j3 j3
5∠ 0°
IT =
= 0 75∠− 26.57° A
6.71∠
71∠26.57°
ZT = 3 +
By current division rule,
j3
= 0.75∠63.43° A
3 + j3 − j3
= ( − 3)(0.775
5∠63.433°) 2.24 ∠− 26.57° V
I = 0.75∠
∠−
− 26.57° ×
VTTh
6.56 Network Analysis and Synthesis
3Ω
Calculation of ZTh (Fig. 6.152)
ZTh = [(3 || j3) + 3] || (−j3)
= 3 ∠−53.12° Ω
= (1.8 − j2.4) Ω
Step II
3Ω
A
− j3 Ω
j3 Ω
B
Step III Calculation of ZL
For maximum power transfer, the load impedance
should be a complex conjugate of the source impedance.
Fig. 6.152
ZL = (1.8 + j2.4) Ω
Calculation of Pmax (Fig. 6.153)
Step IV
(1.8 − j 2 .4) Ω
A
+
2.24 ∠−26.57° V
(1.8 + j 2.4) Ω
−
B
Fig. 6.153
Pmax =
Example 6.50
|
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 maxi-
mum power.
j10 Ω
5Ω
100 ∠0° V
+
7Ω
ZL
−
− 20 Ω
−j
Fig. 6.154
Solution
Step I
I1
Calculation of VTh (Fig. 6.155)
100 ∠0°
= 8.94 − 63.43° A
5 j10
100 ∠0°
I2 =
= 4.72 70.7° A
7 j 20
I2
5Ω
I1 =
7Ω
+
100 ∠0° V
A
−
+ VTh −
B
j10 Ω
Fig. 6.155
− 20 Ω
−j
6.7 Maximum Power Transfer Theorem 6.57
VTh
Step II
VA − VB = ( .94 ∠ − 63.. °) ( j ) − (
2∠70 °) ( − j
) = 71.76 ∠97.33° V
Calculation of ZTh (Fig. 6.156)
5Ω
7Ω
A
B
j10 Ω
− 20 Ω
−j
Fig. 6.156
ZTh =
5( 10) 7( − 20)
50 ∠90°
140 ∠− 90°
+
=
+
= ( .23 − j 0.. ) Ω
5 + j10 7 − j 20 11.18∠63.43° 21.19∠− 70.7°
Step III For maximum power transfer, the load impedance should be complex conjugate of the source
impedance.
ZL = (10.23 + j0.18) Ω
Step IV Calculation of Pmax (Fig. 6.157)
(10.23 − j0 .18) Ω
A
+
71.76 ∠97.3° V
(10.23 + j0 .18) Ω
−
B
Fig. 6.157
Pmax =
|
Th
T
4 RL
|2
=
| .76 |2
= 125.84 W
4 × 10.23
Example 6.51 Find the value of load impedance ZL so that maximum power can be transferred to it in
the network of Fig. 6.158. Find maximum power.
3Ω
2Ω
+
50 ∠45° V
−
ZL
j10 Ω
Fig. 6.158
6.58 Network Analysis and Synthesis
Solution
Step I
Calculation of VTh (Fig. 6.159)
3Ω
2Ω
+
50 ∠45° V
+ A
VTh
T
− B
−
j10 Ω
I
Fig. 6.159
I=
VTh
50 ∠ 5°
= 4.47
7∠
∠−
− 18.43° A
3 + 2 + j10
( 2 j10) I = ( 2 + j10) ( 4.47∠− 18
1 .43 ) = 45.6 ∠60.266° V
Step II Calculation of ZTh (Fig. 6.160)
3Ω
2Ω
A
B
ZTh
T
j10 Ω
Fig. 6.160
ZTh =
Step III
3( 2 + 10)
= ( 2.64 + j 0.72) Ω
3 + 2 + j10
Calculation of ZL
For maximum power transfer, the load impedance should be complex conjugate of the source
impedance.
ZL = (2.64 − j0.72) Ω
Step IV Calculation of Pmax (Fig. 6.161)
(2.64 + j 0.72) Ω
A
+
45.6 ∠60.26° V
(2.64 − j 0.72) Ω
−
B
Fig. 6.161
Pmax =
|
Th
T
4 RL
|2
=
| .6 |2
= 196.91 W
4 × 2.64
6.7 Maximum Power Transfer Theorem 6.59
Example 6.52 Determine the load ZL required to be connected in the network of Fig. 6.162 for
maximum power transfer. Determine the maximum power drawn.
j1 Ω
2Ω
4 ∠0° A
4Ω
ZL
Fig. 6.162
Solution
Step I
Calculation of VTh (Fig. 6.163)
j1 Ω
I1
+
I2
2Ω
4 ∠0° A
4Ω
A
VTh
−
B
Fig. 6.163
I 2 = 4 0° ×
VTh
2
6
j1
= 1.315
31 ∠− 9.46° A
4 (1.315∠− 9.46°)
4I2
5.26 ∠− 9 46° V
j1 Ω
Step II
ZTh =
Step III
A
Calculation of ZTh (Fig. 6.164)
2Ω
4( 2 + 1)
= 1.47
7∠17.1° = (1.41 + j 0.43) Ω
4 + 2 + j1
ZL = (1.41 − j0.43) Ω
Step IV Calculation of Pmax (Fig. 6.165)
(1.41 + j 0.43) Ω
A
+
(1.41 − j 0.43) Ω
−
B
Fig. 6.165
Pmax =
|
Th
T
4 RL
|2
=
ZTh
B
Calculation of ZL
For maximum power transfer, the load impedance
should be the complex conjugate of the source impedance.
5.26 ∠−9.46° V
4Ω
| .26 |2
= 4.91 W
4 × 1.41
Fig. 6.164
6.60 Network Analysis and Synthesis
Example 6.53 In the network shown in Fig. 6.166, find the value of ZL for which the power transferred
will be maximum. Also find maximum power.
5 ∠60° Ω
10 ∠−30° Ω
+
10 ∠0° V
+
ZL
−
−
5 ∠90° V
Fig. 6.166
Solution
Step I
Calculation of VTh (Fig. 6.167)
5 ∠60° Ω
+
10 ∠0° V
10 ∠−30° Ω
+
A +
VTh
B −
−
5 ∠90° V
−
I
Fig. 6.167
Applying KVL to the mesh,
10 ∠0
0° (5 60°) ( 0 30°) 5 90° = 0
11.18∠− 26.57° − (11.18∠− 3.43°)
0
I = 1∠− 23 14° A
Writing VTh equation,
10 ∠0
0° (5 60°) VTh 0
10 ∠0
0° (5 60°) (1∠ − 23.14°) − VTh = 0
VTh = 6 71∠ −26.56° V
Step II
Calculation of ZTh (Fig. 6.168)
5 ∠60° Ω
10 ∠−30° Ω
A
ZTh
B
Fig. 6.168
ZTh =
(5∠60°)(10 ∠−30°)
= 4.47∠33.43° Ω = ( .73 + j 2.46) Ω
5∠60° 0 ∠−30°
6.7 Maximum Power Transfer Theorem 6.61
Calculation of ZL
For maximum power transfer, the load impedance should be the complex conjugate of the source
impedance.
Step III
Z*Th = ( .73 − j 2.
L
)Ω
Calculation of Pmax (Fig. 6.169)
Step IV
(3.73 + j2.46) Ω
A
+
6.71 ∠−26.56° V
(3.73 − j2.46) Ω
−
B
Fig. 6.169
Pmax =
VTh
4 RL
2
=
( .71) 2
= 3.02 W
4 × 3.73
Example 6.54 In the network shown in Fig. 6.170, find the value of ZL so that power transfer from
the source is maximum. Also find maximum power.
j9 Ω
j9 Ω
+
j9 Ω
10 ∠0° V
−
8Ω
ZL
Fig. 6.170
Solution
Step I
Calculation of VTh (Fig. 6.171)
Applying Star-delta transformation (Fig. 6.172)
j9 Ω
j9 Ω
Z1
Z 2 = Z3 =
( j 9)( j 9)
= j3 Ω
j9 j9 j9
VTh = Voltage drop across (8 + j3)Ω impedence
=( +
10 ∠0°
)
= 8.54 ∠−
∠ 16.31° V
8 + j3 + j3
j9 Ω
10 ∠0° V
A
+
VTh −
B
Fig. 6.171
8Ω
6.62 Network Analysis and Synthesis
Z1
Z2
Z3
10 ∠0° V
A
+
VTh −
B
8Ω
Fig. 6.172
Step II
Calculation of ZTh (Fig. 6.173)
j3 Ω
j3 Ω
j3 Ω
ZTh
8Ω
Fig. 6.173
ZTh = j 3 +
j 3(8 + j 3)
= 5.511∠82.49° Ω = (0.72 + j 5.46) Ω
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. ) Ω
Step IV Calculation of Pmax (Fig. 6.174)
(0.72 + j5.46) Ω
A
+
8.54 ∠−16.31° V
(0.72 − j5.46) Ω
−
B
Fig. 6.174
Pmax =
VTh
4 RL
2
=
( .54) 2
= 25.32 W
4 × 0.72
6.7 Maximum Power Transfer Theorem 6.63
Example 6.55 For the network shown in Fig. 6.175, find the value of ZL that will transfer maximum
power from the source. Also find maximum power.
4Ω
−
Vx
+
j 10 Ω
ZL
+
+
−
100 ∠0° V
−
5Vx
Fig. 6.175
−
Solution
Step I
4Ω
Calculation of VTh (Fig. 6.176)
From Fig 6.176,
Vx 4 I
Applying KVL to the mesh,
Vx
+
+
100 ∠0° V
100 ∠0°
= 3.85
85
24 + j10
VTh
B
−
−
I
+
−
5Vx
Fig. 6.176
100 ∠00° 4 I − j 10 I − 5Vx 0
100 ∠0
0° ( 4 + j10) I − 5 ( 4I
4 )=0
I=
j 10 Ω
+
A
22.62° Α
Writing VTh equation,
100 ∠ 0° 4 I − VTh = 0
100 ∠0
0° 4(3
(3.85
Step II
22.62°
62 ) VTh 0
VTh = 86 ∠3 95° V
Calculation of IN (Fig. 6.177)
From Fig. 6.177,
Vx
4Ω
4 I1
Applying KVL to Mesh 1,
100 ∠00° 4
−
Vx
+
−
0
1
I1 = 25 A
IN
+
100 ∠0° V
Applying KVL to Mesh 2,
− j10 I 2 − 5Vx = 0
− j10 I 2 − 5( 4I
4 1) = 0
− j10 I 2 − 5(100) = 0
I 2 = 50 ∠90°A
I N = I1 − I 2 = 25 − 50 ∠90°
B
I1
VTh
86 ∠3 95°
=
= 1.54 ∠67.38° Ω = (0.59 + j .
IN
55.9∠
∠− 63.43°
I2
Fig. 6.177
55.9∠− 63.43°Α
Step III Calculation of ZTh
ZTh =
j 10 Ω
A
)Ω
+
−
5Vx
6.64 Network Analysis and Synthesis
Step IV Calculation of ZL
For maximum power transfer,
Step V
Z∗Th = ( .59 − j1.
L
)Ω
Calculation of Pmax (Fig. 6.178)
(0.59 + j1.42) Ω
A
+
86 ∠3.95° V
(0.59 − j1.42) Ω
−
B
Fig. 6.178
Pmax =
6.8
VTh
2
=
4 RL
( )2
= 3133.9 W
4 × 0.59
RECIPROCITY THEOREM
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.’
Example 6.56
Find the current through the 6 W resistor and verify the reciprocity theorem.
−1Ω
−j
1Ω
I
+
5 ∠0° V
2Ω
j1 Ω
−
Fig. 6.179
Solution
Case I
Calculation of current I when excitation and response are not interchanged (Fig. 6.180)
−1Ω
−j
1Ω
I
+
5 ∠0° V
−
j1 Ω
I1
2Ω
I2
Fig. 6.180
Applying KVL to Mesh 1,
5∠0° −1
1 1 j1( I1 − I 2 ) = 0
(1 j1
1) I1 − j1 2 5∠0°
…(i)
6.8 Reciprocity Theorem 6.65
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
I2 =
= 1.39 56.31° Α
1 j1 − j1
− j1
2
I 2 = 1.39 56.31° Α
I
Case II Calculation of current I when excitation and response are interchanged (Fig. 6.181)
−1Ω
−j
1Ω
I
2Ω
j1 Ω
I1
+
I2
5 ∠0° V
−
Fig. 6.181
Applying KVL to Mesh 1,
−1I1 j1( I1 I 2 )
(1 j1
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 ⎥⎦ ⎢⎣ −55 0°⎥⎦
By Cramer’s rule,
0
j1
−5
5 0° 2
I1 =
= 1.38∠− 123.69° Α
1 j1 − j1
− j1
2
I
I1 = 1.39 56.31° Α
Since the current I is same in both the cases, the reciprocity theorem is verified.
…(ii)
6.66 Network Analysis and Synthesis
Example 6.57
In the network of Fig. 6.182, find the voltage Vx and verify the reciprocity theorem.
10 Ω
j5 Ω
j5 Ω
−2Ω
−j
20 ∠90° A
+
Vx
−
Fig. 6.182
Solution
Case I
Calculation of voltage Vx when excitation and response are interchanged. (Fig. 6.183)
Ix
10 Ω
j5 Ω
j5 Ω
−2Ω
−j
20 ∠90° A
+
Vx
−
Fig. 6.183
By current division rule,
( +j )
= 117.46 ∠77.91
91° Α
( + j ) (j − j )
( j )I x = ( − ) ( .46 ∠77.91°) 34.92 ∠−
∠ 12.09° V
Ix = (
Vx
∠
)
Case II Calculation of voltage Vx when excitation and response are interchanged (Fig. 6.184)
+
Ix
10 Ω
j5 Ω
Vx
j5 Ω
−2Ω
−j
20 ∠90° A
−
Fig. 6.184
(− j )
= 3.12∠−
∠ 38.66°Α
(− 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.
6.8 Reciprocity Theorem 6.67
Example 6.58
Find I and verify the reciprocity theorem for the network shown in Fig. 6.185.
3Ω
2Ω
4Ω
1Ω
2Ω
+
j3 Ω
4Ω
10 ∠45° V
2Ω
Fig. 6.185
Solution
Case I
Calculation of I when excitation and response are not interchanged (Fig. 6.186)
3Ω
j2 Ω
j4 Ω
1Ω
2Ω
I
+
j3 Ω
4Ω
10 ∠45 V
I2
I1
j2 Ω
3
Fig. 6.186
Applying KVL to Mesh 1,
10 ∠45 −
10
…(i)
5
Applying KVL to Mesh 2,
− 2
=0
− 3I 3 = 0
2
−
3
…(ii)
Applying KVL to Mesh 3,
−
−
−
=
=0
…(iii)
Writing Eqs (i), (ii) and (iii) in matrix form,
−4
+ 1
−
0 ⎤ I1
I2
I3
⎡10 ∠45°
0
0
By Cramer’s rule,
4
I3 =
0
4
0
−4 10 45
j1
0
3
0
=
−4
0
1
∠
72 Α
704 ∠
72° Α
6.68 Network Analysis and Synthesis
Case II Calculation of I when excitation and response are interchanged (Fig. 6.187)
j2 Ω
j4 Ω
3Ω
1Ω
2Ω
I
j2 Ω
j3 Ω
4Ω
I2
I3
+
10 ∠45° V
Fig. 6.187
Applying KVL to Mesh 1,
− +
I
−
+
−
=
…(i)
Applying KVL to Mesh 2,
2
0
− 3I 3 = 0
…(ii)
Applying KVL to Mesh 3,
+
∠ 5
0
= 10 ∠45°
−
…(iii)
Writing Eqs (i), (ii) and (iii) in matrix form,
−4
+ 1
−
0 ⎤ ⎡ I1 ⎤
I2
I3
⎡
0 ⎤
0
10 ∠45
By Cramer’s rule,
I1 =
=
0
0
10 ∠45
−
4
5+
− j3
−4
+ 1
=0
∠
0
2 + j5
0
0 704 30 72°Α
72°Α
Since the current I is same in both the cases, the reciprocity theorem, is verified.
6.9
MILLMAN’S THEOREM
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
+ +
+ … + Yn
1
6.9 Millman’s Theorem 6.69
Example 6.59
Find the current through the 40 W resistor for the network shown in Fig. 6.188.
10 Ω
30 Ω
− 20 Ω
−j
j 20 Ω
+
+
−
−
10 ∠0° V
40 Ω
20 ∠0° V
Fig. 6.188
Solution
Step I
Calculation of Vm
⎛
⎞
⎛
⎞
1
1
(10 ∠0 ) ⎜
+ ( 20 ∠0 ) ⎜
⎟
⎟⎠
⎝
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° Ω
1
1
+
30 − j 20 10 j 20
Step III Calculation of IL (Fig. 6.189)
20.15 ∠29.74° Ω
40 Ω
+
18.2 ∠−15.95° V
IL
−
Fig. 6.189
IL =
Example 6.60
18.2∠
∠− 15.95°
= 0.31∠
∠− 25.81° Α
20.15∠29.74° + 40
Find the current I in the network shown in Fig. 6.190.
− 20 Ω
−j
j20 Ω
I
10 Ω
30 Ω
+
10 ∠0° V
40 Ω
+
−
−
Fig. 6.190
20 ∠0° V
6.70 Network Analysis and Synthesis
Solution The network is redrawn as shown in Fig. 6.191.
I
− 20 Ω
−j
j20 Ω
40 Ω
30 Ω
10 Ω
+
+
10 ∠0° V
0V
−
−
20 ∠0° V
Fig. 6.191
Step I
Calculation of Vm
⎛
⎞
1
⎛ 1⎞
0 ⎜ ⎟ + ( 20 ∠0 ) ⎜
⎝
⎠
40
⎝ 10 + j 20 ⎟⎠
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 =
Step III
1
1
1
=
=
= 16.61∠41.63° Ω
1
1
Ym Y1 + Y2
+
40 10 + j 20
Calculation of I (Fig. 6.192)
I
− 20 Ω
−j
16.61 ∠41.63° Ω
30 Ω
10 ∠0° V
+
+
−
−
14.86 ∠−21.8° V
Fig. 6.192
I=
Example 6.61
10 − 14.86 ∠− 21.8°
= 0.15
15∠136.47° Α
30 − j 20 + 16.61∠41.63°
Apply the dual Millman theorem and find the power loss in the (2 + j2) W impedance.
10 A
20 A
4Ω
5Ω
2Ω
j2 Ω
Fig. 6.193
6.9 Millman’s Theorem 6.71
Solution
Step I
Calculation of Im
Im =
I1Z1 + I 2 Z 2 (10) ( 4) + ( 20) (5)
=
= 15.56 Α
Z1 Z 2
4+5
Calculation of Zm
Step II
Zm = Z1 + Z2 = 4 + 5 = 9 Ω
Calculation of P (Fig. 6.194)
Step III
IL
2Ω
15.56 A
9Ω
j2 Ω
Fig. 6.194
By current division rule,
I L = 15.56 ×
9
= 12.53∠ − 10.3° Α
9 + 2 + j2
I 2L RL = (12.53) 2 × 2 = 314 W
P
Example 6.62 In the network shown in Fig. 6.195, what load ZL will receive the maximum power.
Also find maximum power.
j5 Ω
5Ω
3Ω
−4Ω
−j
+
50 ∠0° V
+
ZL
−
−
25 ∠90° V
Fig. 6.195
Solution
Step I
Calculation of Vm
V Y + V2 Y2
Vm = 1 1
=
Y1 Y2
Step II
⎛ 1 ⎞
(50 ∠0 ) ⎜
+ ( 225∠90
⎝ 5 + j 5 ⎟⎠
1
1
+
5 + j5 3 − j 4
⎛ 1 ⎞
)⎜
⎝ 3 j 4 ⎟⎠
= 9.81∠
∠− 78.69°V
Calculation of Zm
Zm =
1
1
=
=
Ym Y1 Y2
1
1
1
+
5 + j5 3 j 4
= 4.39
39∠
∠ − 15. 6 Ω = ( 4.23 − j1.15) Ω
6.72 Network Analysis and Synthesis
Step III
Calculation of ZL
For maximum power transfer
Z*Th = ( .23 + j1. ) Ω
L
Step IV Calculation of Pmax (Fig. 6.196)
A
(4.23 − j1.15) Ω
(4.23 + j1.15) Ω
+
9.81 ∠−78.69° V
−
B
Fig. 6.196
Pmax =
6.10
Vm
2
4 RL
=
( .81) 2
= 5.69 W
4 × 4.23
TELLEGEN’S THEOREM
Tellegen’s theorem states that ‘The algebraic sum of the powers in all branches of the network at any instant
is zero’.
n
∑ Vk I k = 0
k =1
This condition is valid for the network which obeys Kirchhoff’s voltage and current laws.
Example 6.63
Verify Tellegen’s theorem for the network shown in Fig. 6.197.
I4
+
I1
I0
+ V −
1
+
V0
−
V4
−
I2
I3 + V −
2
+
V3
−
I5
+
Vs
−
V0
20 V
V1
V2
2V
V3 = 4 V
V4
14 V
V5
6V
I1
12 A
V0
16 A
16 V
I2
2A
I 3 = 14
1 A
I4
4A
I5 = 2 A
Fig. 6.197
Solution
5
∑ Vb Ib = V0 I0 + V1I1 + V2 I2 + V3I3 + V4 I4 + V5 I5
b=0
= 20( −16) + 16 ×12 + 2 × 2 + 4 × 14 + 14 × 4 6 × 2
Hence, Tellegen’s theorem is verified.
0
6.11 Substitution Theorem 6.73
Example 6.64
In the network shown in Fig. 6.198, check the validity of Tellegen’s theorem.
V6
I6
+
I2
I1
+
V1
−
+
V2
I
−
III
I4
−
+
+
V3
−
V4
−
II
I3
V1
I1
I5
+
V5
−
V2 = 4 V , V4 = 6 V
I 2 = 2 A, I3 = 4 A
10
2 A,
Fig. 6.198
Solution
Applying KVL to Mesh 1,
V1 V2 − V3 = 0
V3 V1 − V2 = 10 − 4
6V
Applying KVL to Mesh 2,
V3
V4 − V5 = 0
V5
V3 − V4 = 6 6 = 0
Applying KVL to Mesh 3,
− V6 + V4 + V2 = 0
V6 = V2 + V4 = 4 + 6 = 10 V
Applying KCL at Node 1,
I1 I 2 + I6 = 0
I6
I1 − I 2 = −22 2 = −4 A
Applying KCL at Node 2,
I2
I4
I3 + I 4
I 2 − I3 = 2 4 = −2 A
I5
I 4 + I6 = −2 ( 4) = −6 A
Applying KCL at Node 3,
6
∑ Vb Ib = V1I1 + V2 I2 + V3I3 + V4 I4 + V5 I5 + V6 I6
b =1
= 10 × 2 + 4 × 2 6 × 4 6 × ( −22) 0 × ( −66) 10 × ( 4)
=0
Hence, Tellegen’s theorem is verified.
6.11
SUBSTITUTION THEOREM
The substitution theorem states that ‘Any branch in a network can be substituted by a different branch
without disturbing the voltages and currents in the entire network, provided the new branch has the same set
of terminal voltage and current as the original network.’
6.74 Network Analysis and Synthesis
Example 6.65 In the network of Fig. 6.199, verify the substitution theorem by replacing the 6 W
resistor by (a) a voltage source (b) a current source.
20 Ω
6Ω
+
100 ∠0°
j10 Ω
−
j8 Ω
Fig. 6.199
Solution:
Step I
Calculation of current through and voltage across the 6 Ω resistor (Fig. 6.200)
20 Ω
6Ω
+
100 ∠0° V
−
j10 Ω
I1
I2
j8 Ω
Fig. 6.200
Applying KVL to Mesh 1,
100 ∠0
0° 20 1 j10( 1 2 ) 0
( 20 + j10
10) I1 − j100 I 2 = 100 ∠0°
…(i)
Applying KVL to Mesh 2,
− j10( − ) − (6 + j8) I 2 = 0
− j10 I + ((6
6 + j1188) 2 = 0
Writing Eqs (i) and (ii) in matrix form,
⎡ 20 + j10 − j10 ⎤ ⎡ I1 ⎤ ⎡100 ∠0°⎤
=
⎢⎣ − j10
6 j18⎥⎦ ⎢⎣ I 2 ⎥⎦ ⎢⎣ 0 ⎥⎦
By Cramer’s rule,
100 ∠0°
j10
0
6 + j18
I1 =
= 4.5∠− 13° A
20 + j10 − j10
− j10
6 j18
20 + j10 100 ∠0°
− j10
0
I2 =
= 2.37
3 ∠5.44° A
20 + j10 − j10
− j10
6 j18
…(ii)
6.11 Substitution Theorem 6.75
Step II
I6
I 2 = 2.37∠5.44° A
V6
6I2
6 ( 2.37∠55.44°) = 14.22∠5.44° V
When 6 Ω resistor is replaced by voltage source of 14.22∠5.44° V (Fig. 6.201)
20 Ω
+
+
100 ∠0° V
−
j10 Ω
−
I1
I2
14.22 ∠5.44° V
j8 Ω
Fig. 6.201
Applying KVL to Mesh 1,
100 ∠0° 20 1 j10( 1 2 ) 0
( 20 + j10
10) I1 − j10 I 2 = 100 ∠0°
…(i)
Applying KVL to Mesh 2,
− j10 ( − ) − 14 22∠5 44° − j8I 2 = 0
− j10 I + jj18
18I 2 = −14.22∠5.4 °
…(ii)
Writing Eqs (i) and (ii) in matrix form,
⎡ 20 + j10 − j10 ⎤ ⎡ I1 ⎤ ⎡ 100 ∠0° ⎤
=
⎢⎣ − j10
j18 ⎥⎦ ⎢⎣ I 2 ⎥⎦ ⎢⎣ −14.22∠5.44°⎥⎦
By Cramer’s rule,
100 ∠0°
j10
−14.22∠5.44° j18
I1 =
= 4.5∠− 13° A
20 + j10 − j10
− j10
j18
20 + j10
100 ∠0°
− j10
−14.22∠5
∠5.44°
I2 =
= 2.37
3 ∠5.44° A
20 + j10 − j10
− j10
j18
Since currents I1 and I2 remain the same, substitution theorem is verified.
Step III
When the 6 Ω resistor is replaced by a current source of 2.37∠5.44° A (Fig. 6.202)
Ω
2.37 ∠5.44° A
+
100 ∠0° V
−
j10 Ω
I1
I2
Fig. 6.202
Applying KVL to Mesh 1,
100 ∠0° 20
1
j10
0(
1
2)
0
j8 Ω
6.76 Network Analysis and Synthesis
(
j )I1 − j10 I 2 = 100 ∠0°
…(i)
For Mesh 2,
I 2 = 2.37
3 ∠5.44° A
…(ii)
Solving Eqs (i) and (ii),
I1 = 4.5∠ − 13° A
Since currents I1 and I2 remain the same, substitution theorem is verified.
6.12
COMPENSATION THEOREM
The compensation theorem states that ‘In any linear bilateral active network, if any branch carrying a current
I has its impedance Z changed by an amount δZ, the resulting changes that occur in the other branches
are the same as those which would have been provided by an opposing voltage source of value VC (I δZ)
introduced into the modified branch.’
Example 6.66 In the network of Fig. 6.203, the impedance (3+j2) W is changed to (5+j3) W.
Determine the change in current drawn from the source by direct calculation and then verify the result by
compensation theorem.
5Ω
3Ω
+
10 ∠0° V
−
I
j2 Ω
Fig. 6.203
Solution
Step I
Calculation of change in current
I=
10 ∠0°
= 1.211∠
∠−
− 14.04° A
5 + 3 + j2
When impedance (3 + j2) Ω is changed to (5+j3) Ω,
I′ =
δI
Step II
10 ∠0°
= 0.96 ∠
∠−
− 16.7° A
5 + 5 + j3
I′ − I = 0.96 ∠ − 16.7° − 1.21∠
∠− 14.04° = 0.25∠ 76.0 ° A
Calculation of change in current by compensation theorem
δZ = ( + j ) − ( + j ) = ( + ) Ω
Vc δ ZI = ( + )( .21∠ − 14.. °) = 2.71∠122 53° V
The compensating network is as shown in Fig. 6.204.
6.12 Compensation Theorem 6.77
5Ω
3Ω
j2 Ω
2.71
71∠
∠12.53°
5 + 3 + j 2 + 2 + j1
= 0.26 ∠175.83° A
δI = −
dI
(2 + j1) Ω
+
2.71 ∠12.53° V
−
Fig. 6.204
Example 6.67 In the network of Fig. 6.205, the impedance (5 + j2) W is changed to (1 + j1) W.
Find the change in current drawn from the supply by direct calculation and then verify by the compensation
theorem.
−j3 Ω
5Ω
5Ω
+
20 ∠0°V
6Ω
−
j2 Ω
Fig. 6.205
Solution
Step I
−3Ω
−j
5Ω
Calculation of change in current
Finding Thevenin’s equivalent circuit across impedance
of (5 + j2) Ω (Fig.6.206)
5(6 − 3)
ZTh =
= 2.94 ∠
∠−
− 11.31° Ω
5 + 6 − j3
I=
VTh
20 ∠0°
= 1.75∠15.26° A
5 + 6 − j3
(
j ) I = ( − j ) ( .75∠15..
20 ∠0° V
δ IL
6Ω
B
I
Fig. 6.206
°) = 11.74 ∠−
∠− 11.31° V
2.94 ∠−11.31° Ω
11.74 ∠
∠− 11.31°
= 1.47
7∠
∠−
− 21.55° A
2.94 ∠
∠− 11.31° + 5 + j 2
When impedance (5 + j2) Ω is changed to (1 + j1) Ω,
I′L =
VTh
−
Thevenin’s equivalent network is shown in Fig. 6. 207.
IL =
A
+
5Ω
+
11.74 ∠−11.31° V
11.74 ∠
∠− 11.31°
= 3∠−
− 17.53° A
2.94 ∠
∠− 11.31° + 1 + j1
I′L − I L = 3∠− 17.53° − 1.44 ∠−
− 25° = 1.533∠
∠− 13.69° A
−
IL
Fig. 6.207
.
j2 Ω
6.78 Network Analysis and Synthesis
Step II
Calculation of change in current by compensation theorem
δ Z = ( + j ) − ( + j ) = (− − ) Ω
Vc I L δ Z = ( .44 ∠ − 25°)
°) ( −4 − j1)) = .94 ∠169
6 .04° V
2.94 ∠−11.31° Ω
5Ω
The compensating network is shown in Fig. 6.208.
5.94 ∠ 69.0 °
δ IL = −
2.94 ∠
∠− 11.31° + 5 + j 2 4
j2 Ω
j1
d IL
(−4 − j1) Ω
= 1.52∠
∠−
− 17.18° A
+
−
Exercises
5.94 ∠169.04° V
Fig. 6.208
MESH ANALYSIS
6.1
Find the current through the 3 + j4 Ω
impedance in Fig. 6.209.
I1 j2 Ω
−4Ω
−j
I3
+
3Ω
20 Ω
j4 Ω
10 Ω
10 ∠0° V
+
50 ∠45° V
j5 Ω
5Ω
5Ω
−
j2 Ω
−
j2.5 Ω
Fig. 6.211
[11.6 ∠113.2° A]
Fig. 6.209
6.4
[0]
6.2
In the network of Fig.6.212, find V2 which
results in zero current through the 4 Ω
resistor.
In the network of Fig. 6.210, find V0.
2Ω
2Ω
5Ω
+
+
10 ∠0° V
−
j2 Ω
4Ω
2Ω
2Ω
j2 Ω
j2 Ω V0
+
50 ∠0° V
j2 Ω
−
−2Ω
−j
+
−
V2
−
Fig. 6.212
[26.3 ∠113.2° V]
Fig. 6.210
[1.56 ∠128.7° V]
6.3
Find the current I3 in the network of Fig.
6.211.
NODE ANALYSIS
6.5
For the network shown in Fig. 6.213, find the
voltage VAB.
Exercises 6.79
5Ω
100 ∠45° V
6.9
A
+
5Ω
−
j20 Ω
In the network of Fig. 6.217, find the current
through capacitance.
20 Ω
j1 Ω
5Ω
+
B
20 ∠0° V
Fig. 6.213
Find the voltages at nodes 1 and 2 in the
network of Fig. 6.214.
10 Ω
2Ω
1
3Ω
[4.86 ∠80.8° A]
THEVENIN’S THEOREM
−j10 Ω
j5 Ω
−
Fig. 6.217
2
+
50 ∠0° V
10 ∠0° V
−
[75.4 ∠55.2° V]
6.6
−j5 Ω
+
−
j4 Ω
6.10 Obtain Thevenin’s equivalent network for the
network shown in Fig. 6.218.
j20 Ω
Fig. 6.214
[15.95 ∠49.94° V, 12.9 ∠55.5° V]
6.7
In the network of Fig. 6.215, find the current
in the 10 ∠30° V source.
5Ω
−2Ω
−j
+
50 ∠0° V
50 Ω
40 Ω
A
−
B
1000 Ω
1000 Ω
− 400 Ω
−j
j5 Ω
Fig. 6.218
2Ω
+
10 ∠30° V
3Ω
−
5Ω
−2Ω
−j
[0.192 ∠−43.4° V, 88.7 ∠11.55° Ω]
6.11 Obtain Thevenin’s equivalent network for
Fig. 6.219.
10 Ω
Fig. 6.215
[1.44 ∠38.8° A]
10 ∠0° V
SUPERPOSITION THEOREM
6.8
For the network shown in Fig. 6.216, find the
current in the 10 Ω resistor.
10 Ω
+
100 ∠0° V
−
j5 Ω
10 Ω
+
−
−
j4 Ω
+
10 ∠90° V
Fig. 6.219
5Ω
−5Ω
−j
3Ω
+
j4 Ω
j16 Ω
−
50 ∠30° V
[11.17 ∠−63.4° V, 10.6 ∠45° Ω]
NORTON’S THEOREM
6.12 Find Norton’s equivalent network for Fig.
6.220.
Fig. 6.216
[73.4 ∠−21.84° A]
6.80 Network Analysis and Synthesis
50 ∠0°V
+
−
j5 Ω
j5 Ω
5Ω
j10 Ω
A
MAXIMUM POWER TRANSFER
THEOREM
B
6.14 Determine the maximum power delivered to
the load in the network shown in Fig. 6.222.
5Ω
10 Ω
j15 Ω
5 Ω −j6 Ω
+
−
50 ∠90°V
3Ω
−j10 Ω
50 ∠0° A
Fig. 6.220
j4 Ω
[2.77 ∠–33.7° A, 2.5 + j12.5 Ω]
6.13 Find the current through the (3 + j4) Ω
impedance in the network of Fig. 6.221.
j5 Ω
5Ω
3Ω
+
50 ∠90°V
−
j4 Ω
+
ZL
Fig. 6.222
[1032.35 W]
6.15 For the network shown in Fig. 6.223, find the
value of ZL that will receive the maximum
power. Determine also this power.
50 ∠0°V
−
j5 Ω
2Ω
50 ∠0°V
Fig. 6.221
ZL
+
−
4Ω
−j3 Ω
[8.3 ∠85.2° A]
Fig. 6.223
[3.82 − j1.03 Ω, 54.5 W]
Objective-Type Questions
6.1
In Fig. 6.224, the equivalent impedance seen
across terminals a, b, is
2Ω
6.2
The Thevenin equivalent voltage VTh
appearing between the terminals A and B of
the network shown in Fig. 6.225 is given by
4Ω
3Ω
j3 Ω
+
Zeq
2Ω
−j4 Ω
100 ∠0° V
16
Ω
3
⎛8
⎞
(C) ⎜ +j12 Ω
⎝3
⎠
−j6
j4
VTh
4Ω
−
Fig. 6.224
(a)
j2
A
B
Fig. 6.225
(b)
8
Ω
3
(d)
none of the above
6.3
(a) 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 Ω resistance
Answers to Objective-Type Questions 6.81
6.4
in series with a 1 H inductance. The load that
will obtain the maximum power transfer is
(a) 1 Ω resistance
(b) 1 Ω resistance in parallel with 1 H
inductance
(c) 1 Ω resistance in series with 1 F
capacitance
(d) 1 Ω resistance in parallel with 1 F
capacitance
For the network shown in Fig. 6.226, the
instantaneous current I1is
(a) 28 A
(b) 4 A
(c) 20 A
(d) cannot be determined
6.6
In the network of Fig. 5.228, the magnitudes
of VL and VC are twice that of VR. The
inductance of the coil is
VR
5Ω
C
5 ∠0°
L
−j2 Ω
j2 Ω
I1
VC
VL
Fig. 6.228
10 ∠60° A
3Ω
5 ∠0° A
(a)
(c)
6.7
Fig. 6.226
(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. 6.229.
10 3
∠− 90° A
2
I
V1 = 100 (1 + j )
10 Ω
V2 = 100 (1 − j )
6.5
In the network shown in Fig. 6.227, the current
supplied by the sinusoidal current source I is
Fig. 6.229
(a)
(c)
I
12 A
0°
−45°
16 A
Fig. 6.227
Answers to Objective-Type Questions
6.1 (b)
6.2 (d)
6.3 (c)
6.4 (a)
6.5 (c)
6.6 (c)
6.7 (d)
(b)
(d)
45°
−90°
j10 Ω
7
Coupled Circuits
7.1
INTRODUCTION
Two circuits are said to be coupled circuits when energy transfer takes place from one circuit to the other
without having any electrical connection between them. Such coupled circuits are frequently used in network
analysis and synthesis. Common examples of coupled circuits are transformer, gyrator, etc. In this chapter, we
will discuss self and mutual inductance, magnetically coupled circuits, dot conventions and tuned circuits.
7.2
SELF-INDUCTANCE
Consider a coil of N turns carrying a current i as shown in Fig. 7.1.
When current flows through the coil, a flux f is produced in the coil.
The flux produced by the coil links with the coil itself. If the current
flowing through the coil changes, the flux linking the coil also changes.
Hence, an emf is induced in the coil. This is known as self-induced emf.
The direction of this emf is given by Lenz’s law.
We know that
φ ∝i
φ
= k , a constant
i
φ=ki
i
v
Fig. 7.1 Coil carrying current
Hence, rate of change of flux = k × rate of change of current
dφ
di
=k
dt
dt
According to Faraday’s laws of electromagnetic induction, a self-induced emf can be expressed as
v
N
dφ
d
di
φ di
di
= − Nk = − N
= −L
dt
dt
i dt
dt
Nφ
and is called coefficient of self-inductance.
i
The property of a coil that opposes any change in the current flowing through it is called self-inductance or
inductance of the coil. If the current in the coil is increasing, the self-induced emf is set up in such a direction so
where L =
7.2 Network Analysis and Synthesis
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.
7.3
MUTUAL INDUCTANCE
If the flux produced by one coil links with the other coil, placed closed to the first coil, an emf is induced in
the second coil due to change in the flux produced by the first coil. This is known as mutually induced emf.
Consider two coils 1 and 2 placed adjacent to each other as shown in Fig. 7.2. Let Coil 1 has N1 turns while
Coil 2 has N2 turns.
Mutual flux
i1
+
+
L1
v1
−
L2
v2
Coil 2 −
Coil 1
Fig. 7.2 Two adjacent coils
If a current i1 flows in Coil 1, flux is produced and a part of this flux links Coil 2. The emf induced in Coil 2
is called mutually induced emf.
We know that
φ2 ∝ i1
φ2
= k , a constant
i1
φ2 = k i1
Hence, rate of change of flux = k × rate of change of current i1
dφ 2
dii
=k 1
dt
dt
According to Faraday’s law of electromagnetic induction, the induced emf is expressed as
v2
where M =
7.4
N2
dφ2
di
di
φ dii
dii
= −N2 k 1 = −N2 2 1 = −M 1
dt
dt
i1 dt
dt
N 2 φ2
and is called coefficient of mutual inductance.
i1
COEFFICIENT OF COUPLING (k)
The coefficient of coupling (k) between coils is defined as fraction of magnetic flux produced by the current
in one coil that links the other.
Consider two coils having number of turns N1 and N2 respectively. When a current i1 is flowing in Coil 1
and is changing, an emf is induced in Coil 2.
N φ
M= 2 2
i1
7.5 Inductances in Series 7.3
φ2
φ1
φ
k φ1
N k φ
M= 2 1 1
i1
k1 =
Let
…(7.1)
If the current i2 is flowing in Coil 2 and is changing, an emf is induced in Coil 1,
M=
N1 φ1
i2
φ1
φ2
φ k φ2
N k φ
M= 1 2 2
i2
k2 =
Let
…(7.2)
Multiplying Eqs (7.1) and (7.2),
N1φ1 N 2φ2
×
= k 2 L1 L2
i1
i2
M2
k1k2 ×
M
k L1 L2
where
k
k1k2
7.5
INDUCTANCES IN SERIES
1. Cumulative Coupling Figure 7.3 shows two coils 1 and 2 connected
in series, so that currents through the two coils are in the same direction in
order to produce flux in the same direction. Such a connection of two coils
is known as cumulative coupling.
Let
L1 = coefficient of self-inductance of Coil 1
L2 = coefficient of self-inductance of Coil 2
M = coefficient of mutual inductance
If the current in the coil increases by di amperes in dt seconds then
Coil 1
Coil 2
i
Fig. 7.3 Cu m u l a t i v e
coupling
di
dt
di
Self-induced emf in Coil 2 = −L2
dt
Self-induced emf in Coil 1 = −L
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
(L + L
M)
di
dt
…(7.3)
7.4 Network Analysis and Synthesis
If L is the equivalent inductance then total induced emf in that single coil would have been
v
L
di
dt
…(7.4)
Equating Eqs (7.3) and (7.4),
L
L1 + L2
2M
2. Differential Coupling Figure 7.4 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
Coil 1
Coil 2
i
Fig. 7.4 Differential
coupling
di
di
di
di
− L2 + 2M = −( L1 + L2 − 2 M )
dt
d
dt
dt
dt
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
7.6
L
di
dt
di
di
= −( L + L − M )
dt
dt
L = L1 + L2 − 2 M
INDUCTANCES IN PARALLEL
1. Cumulative Coupling Figure 7.5 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
dii1
dt
dii
Self-induced emf in Coil 2 = −L2 2
dt
L1
Self-induced emf in Coil 1 = −L
i2
Fig. 7.5
dii2
dt
dii1
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
dii1
di
−M 2
dt
dt
Coil 2
Cumulative coupling
7.6 Inductances in Parallel 7.5
Total induced emf in Coil 2 = − L2
dii2
dii
−M 1
dt
dt
As both the coils are connected in parallel, the emf induced in both the coils must be equal.
dii1
di
dii
dii
− M 2 = − L2 2 − M 1
dt
dt
dt
dt
dii1
dii1
dii2
di2
L1
−M
= L2
−M
dt
dt
dt
d
dt
dii1
dii2
(L M )
= (L M )
dt
dt
− L1
dii1 ⎛ L2
=
dt ⎜⎝ L1
Now,
i
M ⎞ dii2
M ⎟⎠ dt
...(7.5)
i1 + i2
di dii1 di2
=
+
dt dt dt
⎛ L M ⎞ dii2 di2
=⎜ 2
+
⎝ L1 M ⎟⎠ dt
dt
⎛ L M ⎞ dii2
=⎜ 2
+ 1⎟
⎝ L1 M
⎠ dt
L
+
L
−
⎛
2 2 M ⎞ dii2
=⎜ 1
⎝ L1 − M ⎟⎠ dt
...(7.6)
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
dii
di
L = L1 1 + M 2
dt
dt
dt
di 1 ⎛ dii1
di ⎞
=
L1
+ M 2⎟
dt L ⎝ dt
dt ⎠
v
L
1 ⎡ ⎛ L2 M ⎞ dii2
di ⎤
+M 2⎥
⎢ L1 ⎜⎝
⎟
L⎣
L1 − M ⎠ dt
dt ⎦
⎤ dii
1 ⎡ ⎛L −M⎞
= ⎢ L1 ⎜ 2
+ M⎥ 2
L ⎣ ⎝ L1 − M ⎟⎠
⎦ dt
=
Substituting Eq. (7.6) in Eq. (7.7),
⎤ dii2
⎛ L1 L2 − 2 M ⎞ dii2 1 ⎡ ⎛ L2 M ⎞
⎜⎝ L M ⎟⎠ dt = L ⎢ L1 ⎜⎝ L M ⎟⎠ + M ⎥ dt
1
1
⎣
⎦
⎛L M⎞
L1 ⎜ 2
+M
⎝ L1 M ⎟⎠
L=
L1 L2 − 2 M
L1 M
...(7.7)
7.6 Network Analysis and Synthesis
=
L1 L2 − L1 M + L1 M − M 2
L1 + L2 − 2 M
=
L1 L2 − M 2
L1 + L2 − 2 M
2. Differential Coupling Figure 7.6 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.
dii1
dt
dii
Self-induced emf in Coil 2 = −L2 2
dt
i1
Coil 1
Coil 2
i
i2
L1
Self-induced emf in Coil 1 = −L
Fig. 7.6
Differential coupling
dii2
dt
dii1
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
dii1
di
+M 2
dt
dt
dii
dii
Total induced emf in Coil 2 = − L2 2 + M 1
dt
dt
Total induced emf in Coil 1 = − L1
As both the coils are connected in parallel, the emf induced in the coils must be equal.
dii1
di
dii
dii
+ M 2 = − L2 2 + M 1
dt
dt
dt
dt
dii1
dii1
dii2
di2
L1
+M
= L2
+M
dt
dt
dt
d
dt
dii
dii
(L M ) 1 = (L M ) 2
dt
dt
− L1
dii1 ⎛ L2
=
dt ⎜⎝ L1
Now,
i
M ⎞ dii2
M ⎟⎠ dt
...(7.8)
i1 + i2
di dii1 di2
=
+
dt dt dt
⎛ L M ⎞ dii2 di2
=⎜ 2
+
⎝ L1 M ⎟⎠ dt
dt
⎛L
=⎜ 2
⎝ L1
M ⎞ dii2
+ 1⎟
⎠ dt
M
⎛ L + L2 + 2 M ⎞ dii2
=⎜ 1
⎝ L1 + M ⎟⎠ dt
...(7.9)
7.6 Inductances in Parallel 7.7
If L is the equivalent inductance of the parallel combination then the induced emf is given by
v
L
di
dt
Since induced emf in parallel combination is same as induced emf in any one coil,
L
di
dii
di
= L1 1 − M 2
dt
dt
dt
di 1 ⎛ dii1
di ⎞
=
L1
−M 2⎟
dt L ⎝ dt
dt ⎠
=
1 ⎡ ⎛ L2 M ⎞ dii2
di ⎤
−M 2⎥
⎢ L1 ⎜⎝
⎟
L⎣
L1 + M ⎠ dt
dt ⎦
=
⎤ di
1 ⎡ ⎛ L2 + M ⎞
− M⎥ 2
⎢ L1 ⎜⎝
⎟
L⎣
L1 + M ⎠
⎦ dt
...(7.10)
Substituting Eq. (7.9) in Eq. (7.10),
⎛ L1 L2 + 2 M ⎞ dii2 1 ⎡ ⎛ L2
⎜⎝ L M ⎟⎠ dt = L ⎢ L1 ⎜⎝ L
1
1
⎣
⎤ dii
M⎞
− M⎥ 2
⎟
M⎠
⎦ dt
⎛L M⎞
L1 ⎜ 2
−M
⎝ L1 M ⎟⎠
L=
L1 + L2 + 2 M
L1 M
=
L1 L2 + L1 M − L1 M − M 2
L1 + L2 + 2 M
=
L1 L2 M 2
L1 L2 + 2 M
Example 7.1 The combined inductance of two coils connected in series is 0.6 H or 0.1 H depending on relative directions of currents in the two coils. If one of the coils has a self-inductance of 0.2 H, find
(a) mutual inductance, and (b) coefficient of coupling.
Solution
L1
0.2 , Ldiff = 0.1 H, Lcum = 0 6 H
(a) Mutual inductance
Lcum
Ldiff
L1 + L2
L1 + L2
2 M = 0 .6
2 M = 0 .1
Adding Eqs (i) and (ii),
2(
1
L1
2)
0.7
L2 = 0 35
L2 = 0.35 0.2
0.15 H
...(i)
...(ii)
7.8 Network Analysis and Synthesis
Subtracting Eqs (ii) from Eqs (i),
4
0.5
M = 0.125 H
(b) Coefficient of coupling
M
k=
L1 L2
0.125
=
0.2 × 0.15
= 0 72
Example 7.2 Two coils with a coefficient of coupling of 0.6 between them are connected in series so
as to magnetise in (a) same direction, and (b) opposite direction. The total inductance in the same direction
is 1.5 H and in the opposite direction is 0.5 H. Find the self-inductance of the coils.
Solution
0.6, Ldiff = 0.55 H, Lcum = 1.5 H
k
L1 + L2
L1 + L2
Ldiff
Lcum
2 M = 0 .5
2 M = 1.5
...(i)
...(ii)
Subtracting Eq. (i) from Eq. (ii),
4
1
M = 0.225 H
Adding Eq. (i) and (ii),
2(
1
2)
2
L1
L2 = 1
...(iii)
M
k=
06=
L1 L2
0 25
L1 L2
L1 L2 = 0.1736
...(iv)
Solving Eqs (iii) and (iv),
L1 = 0 22 H
L2 = 0 78 H
Example 7.3 Two coils having self-inductances of 4 mH and 7 mH respectively are connected in
parallel. If the mutual inductance between them is 5 mH, find the equivalent inductance.
L1
Solution
For cumulative coupling,
4
L=
L2
7 mH,
M = 5 mH
L1 L2 − M 2
4 × 7 − (5) 2
=
= 3 mH
L1 + L2 − 2 M 4 + 7 − 2(5)
For differential coupling,
L=
L1 L2 − M 2
4 × 7 − (5) 2
=
= 0.143
143 mH
L1 + L2 + 2 M 4 + 7 + 2(5)
7.7 Dot Convention 7.9
Example 7.4 Two inductors are connected in parallel. Their equivalent inductance when the
mutual inductance aids the self-inductance is 6 mH and it is 2 mH when the mutual inductance opposes
the self-inductance. If the ratio of the self- inductances is 1:3 and the mutual inductance between the coils
is 4 mH, find the self-inductances.
Solution
Lcum
6
Lddiff
iff
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(
8) = 6(
1
2
L1
L2 + 8
L1
1
2
8)
3L
3L1 3
2
24
L2 = 16
L1
=1 3
L2
But
1 3 L2
L2 = 16
2 3 L2 = 16
L2 = 6 95 mH
L1
7.7
1 3 L2 = 9.035 mH
DOT CONVENTION
Consider two coils of inductances L1 and L2 respectively connected in series as shown in Fig. 7.7. 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
7.10 Network Analysis and Synthesis
shown in Fig 7.7 (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. 7.7 (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.
L1
L2
L1
L2
I
I
(a)
I
L1
(b)
L2
I
L1
(c)
L2
(d)
Fig. 7.7 Dot convention
Figure 7.7 shows the schematic connection of the two coils. It is not possible to state from Fig. 7.7(a) and
Fig. 7.7(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. 7.7(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. 7.7(d).
When two mutual fluxes aid each other, the mutual inductance is positive and polarity of the mutually
induced emf is same as that of the self-induced emf. When two mutual fluxes oppose each other, the
mutual inductance is negative and polarity of the mutually induced emf is opposite to that of the selfinduced emf.
Example 7.5
Obtain the dotted equivalent circuit for Fig. 7.8 shown below.
R
L1
+
C
v (t)
−
i(t)
L2
Fig. 7.8
Solution The current in the two coils is shown in Fig. 7.9. The corresponding flux due to current in each
coil is also drawn with the help of right-hand thumb rule.
7.7 Dot Convention 7.11
f1
R
L1
+
v (t)
C
−
L2
f2
Fig. 7.9
From Fig. 7.9, it is seen that, the flux f1 is in upward direction in
Coil 1, and flux f2 is in downward direction in Coil 2. Hence, fluxes
are opposing each other. The mutual inductances are negative and
mutually induced emfs have opposite polarities as that of self-induced
emf. The dots are placed in two coils to illustrate these conditions.
Hence, current i(t) enters from the dotted end in Coil 1 and leaves
from the dotted end in Coil 2.
The dotted equivalent circuit is shown in Fig. 7.10.
R
L1
+
v (t )
C
−
i(t)
L2
Fig. 7.10
Example 7.6
Obtain the dotted equivalent circuit for the circuit of Fig. 7.11.
j4 Ω
j2 Ω
j3 Ω
j3 Ω
j5 Ω
j6 Ω
Fig. 7.11
Solution The current in the three coils is shown in Fig. 7.12.
The corresponding flux due to current in each coil is also
drawn with the help of right-hand thumb rule.
From Fig. 7.12, it is seen that the flux is towards the left
in Coil 1, towards the right in Coil 2 and towards the left in
Coil 3. Hence, fluxes f1 and f2 oppose each other in coils
1 and 2, fluxes f2 and f3 oppose each other in coils 2 and
3, and fluxes f1 and f3 aid each other in coils 1 and 3. The
dots are placed in three coils to illustrate these conditions.
Hence, current enters from the dotted end in Coil 1, leaves
from the dotted end in Coil 2 and enters from the dotted end
in Coil 3.
The dotted equivalent circuit is shown in Fig. 7.13.
j4 Ω
j2 Ω
j3 Ω
f1
f 2f 3
i
j3 Ω
j5 Ω
j6 Ω
Fig. 7.12
j4 Ω
j3 Ω
j2 Ω
j5 Ω
j3 Ω
Fig. 7.13
j6 Ω
7.12 Network Analysis and Synthesis
Example 7.7
Obtain the dotted equivalent circuit for the circuit shown in Fig. 7.14.
j6 Ω
A
j5Ω
j4Ω
B
j3 Ω
j1 Ω
j2Ω
Fig. 7.14
Solution The current in the three coils is shown in
Fig. 7.15. The corresponding flux due to current in each coil
i
A
is also drawn with the help of right-hand thumb rule.
From Fig. 7.15, it is seen that all the three fluxes f1, f2, f3
j5 Ω
aid each other. Hence, all the mutual reactances are positive
and mutually induced emfs have same polarities as that of
self-induced emfs. The dots are placed in three coils to
illustrate these conditions. Hence, currents enter from the
dotted end in each of the three coils. The dotted equivalent
circuit is shown in Fig. 7.16.
f1
j6 Ω
j4 Ω
B
j3 Ω
j1 Ω
f3
f2
j2 Ω
Fig. 7.15
j6 Ω
j5 Ω
j4 Ω
j2 Ω
j1 Ω
j3 Ω
A
B
Fig. 7.16
Example 7.8
Obtain the dotted equivalent circuit for the coupled circuit of Fig. 7.17.
j3 Ω
j5 Ω
−j 8 Ω
10 Ω
+
−
50∠ 0°V
Fig. 7.17
Solution The current in the two coils is shown in Fig. 7.18 . The corresponding flux due to current in each
coil is also drawn with the help of right-hand thumb rule.
7.7 Dot Convention 7.13
f2
f1
j3 Ω
j5 Ω
−j 8 Ω
10 Ω
+
−
50∠ 0°V
Fig. 7.18
j5 Ω
From Fig. 7.18, it is seen that the flux f1 is in clockwise
direction in Coil 1 and in anti-clockwise direction in Coil 2.
Hence, fluxes are opposing each other. The dots are placed
in two coils to illustrate these conditions. Hence, current
enters from the dotted end in Coil 1 and leaves from the
dotted end in Coil 2. The dotted equivalent circuit is shown
in Fig. 7.19.
Example 7.9
10 Ω
j3 Ω
M12
+
−
50∠ 0°V
Fig. 7.19
Find the equivalent inductance of the network shown in Fig. 7.20.
1H
1H
0.5 H
i
1H
5H
2H
Fig. 7.20
Solution
L
Example 7.10
( L + M + M ) + (L
(L M + M ) (L + M
= ( + .5 + 1) + ( 2 + 1 + 0.5
5) (5 + 1 1)
= 13 H
M )
Find the equivalent inductance of the network shown in Fig. 7.21.
1H
2H
i
10 H
1H
6H
5H
Fig. 7.21
Solution
L
( L + M − M ) + (L
(L M
= ( + − ) + ( − + 2) + (
+ M ) (L − M
) 21 H
M )
−8Ω
−j
7.14 Network Analysis and Synthesis
Example 7.11
Find the equivalent inductance of the network shown in Fig. 7.22.
k3 = 0.65
i
12 H
k1 = 0.33
k2 = 0.37
14 H
14 H
Fig. 7.22
Solution
M12
M 21 = k1 L1 L2 = 0.33 (12)(14)
4.28 H
M 23
M 32 = k2 L2 L3 = 0.37 (14)(14)
5.18 H
M 31 = M113 = k3 L3 L1 = 0.65 (12)(14)
8.42 H
L
( L − M + M ) + (L
(L M − M ) (L + M
M )
= ( − .28 + 8. ) + (14
4 5.18 − 4.28) (14 8.42 − 5.18)
= 37.92 H
Example 7.12
Find the equivalent inductance of the network shown in Fig. 7.23.
8H
16 H
15 H
A
B
Fig. 7.23
Solution For Coil A,
LA
For Coil B,
LB
L1 − M12 = 15 − 8 = 7 H
L2 − M12 = 16 − 8 = 8 H
1
1
1
1 1 15
=
+
= + =
L LA LB 7 8 56
56
L=
= 3.73 H
15
Example 7.13
Find the equivalent inductance of the network shown in Fig. 7.24.
10 H
10 H
15 H
25 H
A
30 H
B
Fig. 7.24
35 H
C
7.8 Coupled Circuits 7.15
Solution For Coil A,
LA
L1 + M12
M13 = 25 + 10 − 10 = 25 H
LB
L2 − M 23 + M 21 = 35 − 15 + 10 = 25 H
LC
L3 − M 32
For Coil B,
For Coil C,
M 31 = 35 − 15 − 10 = 10 H
1
1
1
1
1
1
1
9
=
+
+
=
+
+
=
L LA LB LC 25 25 10 50
5
50
L=
= 5.555 H
9
Example 7.14
Find the equivalent impedance across the terminals A and B in Fig. 7.25.
5Ω
A
2Ω
3Ω
j2 Ω
j3 Ω
j4 Ω
B
Fig. 7.25
Solution
Z1
5 Ω, Z 2
Z = Z1 +
7.8
( 2 + j 4) Ω, Z3
(3
(3
j 3) Ω, Z M = j 2 Ω
Z 2 Z3 − Z 2M
( 2 + j44) (3 + j 3) − ( j 2) 2
= 5+
= 6.9 ∠ 24.16° Ω
Z 2 + Z 3 − 2Z M
2 + j 4 + 3 + j 3 − 2( j 2)
COUPLED CIRCUITS
Consider two coils located physically close to one another as shown
in Fig. 7.26.
When current i1 flows in the first coil and i2 = 0 in the second coil,
flux f1 is produced in the coil. A fraction of this flux also links the
second coil and induces a voltage in this coil. The voltage v1 induced
in the first coil is
dii
v1 L1 1
dt i2 = 0
The voltage v2 induced in the second coil is
dii
v2 M 1
dt
i2 = 0
i1
i2
M
+
v1
L1
+
L2
v2
−
−
Fig. 7.26
Coupled circuit
7.16 Network Analysis and Synthesis
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
dii
v2 L2 2
dt i1 = 0
and
v1
M
dii2
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. 7.26. 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
v1
v2
dii1
di
+M 2
dt
dt
dii
dii
M 1 + L2 2
dt
dt
L1
This can be represented in terms of dependent sources, as shown in Fig. 7.27.
i1
i2
+
+
L1
L2
v1
v2
+
−
Mdi2
dt
−
+
−
Mdi1
dt
−
Fig. 7.27 Equivalent circuit
Now consider the case when the dots are placed at the opposite ends in the two coils, as shown in Fig. 7.28.
i1
i2
M
+
v1
L1
+
L2
v2
−
−
Fig. 7.28 Coupled circuit
Due to i1, with i2 = 0, the dotted end in Coil 1 is positive, so the induced voltage in Coil 2 is positive at the
dot, which is the reverse of the designated polarity for v2. Similarly, due to i2, with i1 = 0, the dotted ends have
negative polarities for the induced voltages. The mutually induced voltages in both cases have polarities that
are the reverse of terminal voltages and the equations are
7.8 Coupled Circuits 7.17
di1
2
−
dt
dt
di
di
M 1 + L2 2
dt
dt
v
This can be repressed in terms of dependent sources as shown in Fig. 7.29.
i
i
+
+
L1
L2
v
v
Mdi
dr
+
Mdi
dt
+
Fig. 7.29 Equivalent circuit
The various cases are summarized in the table shown in Fig. 7.30.
Coupled circuit
Time-domain equivalent circuit
i
i
M
+
L1
v1
i
+
L2
v
i
+
L1
v
+
i
M
+
L1
v1
L2
v
+
+
L1
v1
v
+
i
M
+
v1
L1
+
Mdi1
dt
i
+
L2
+
v
−
+
L2
v
+
+
Mdi1
dt
j
+
2
v
j wM i1
+
i
j L1
j
+
2
v
v
j wM i
−
+
+
j wM i1
i
+
v
Mdi
dt
j L1
v
i
L1
i
i1
v
Mdi
dt
j wM i1
j wM i
+
v
−
i
v
L2
+
2
v
+
i
L1
j
i1
Mdi1
dt
i
+
L2
+
j L1
j wM i
+
L2
Mdi
dt
i
M
L1
i
v
i
v
−
i
v
+
Mdi1
dt
i
+
i
+
L2
Mdi
dt
i
Frequency domain equivalent
circuit
+
i
j L1
j
+
2
v
v
j wM i
Fig. 7.30 Coupled circuits for various cases
+
+
j wM i1
7.18 Network Analysis and Synthesis
Example 7.15
Write mesh equations for the network shown in Fig. 7.31.
R1
L1
L3
R2
+
v1(t)
M12
−
M23
L2
i1
R3
i2
Fig. 7.31
Solution Coil 1 is magnetically coupled to Coil 2. Similarly, Coil 2 is magnetically coupled with Coil 1
and Coil 3. By applying dot convention, the equivalent circuit is drawn with the dependent sources.
The equivalent circuit in terms of dependent sources is shown in Fig. 7.32.
L1
R1
M12 d (i1 − i2)
dt
+ −
L3
M23
d
(i − i )
dt 1 2
− +
R2
L2
+
v1(t)
−
−
di
M23 2 i
+
dt 2
i1
R3
+
di
M12 1
−
dt
Fig. 7.32
(a) In Coil 1, there is a mutually induced emf due to current (i1 − i2) in Coil 2. The polarity of the mutually
induced emf is same as that of self-induced emf because currents i1 and (i1 − i2) enter in respective coils
from the dotted ends.
(b) In Coil 2, there are two mutually induced emfs, one due to current i1 in Coil 1 and the other due to current
i2 in Coil 3. The polarity of the mutually induced emf in Coil 2 due to the current i1 is same as that of the
self-induced emf because currents i1 and (i1 − i2) enter in respective coils from dotted ends. The polarity
of the mutually induced emf in Coil 2 due to the current i2 is opposite to that of the self-induced emf
because current (i1 − i2) leaves from the dotted end in Coil 2 and the current i2 enters from the dotted end
in Coil 3.
(c) In Coil 3, there is a mutually induced emf due to the current (i1 − i2) in Coil 2. The polarity of the
mutually induced emf is opposite to that of self-induced emf because the current (i1 − i2) leaves from the
dotted end in Coil 2 and the current i2 enters from the dotted end in Coil 3.
Applying KVL to Mesh 1,
v1 (t ) R1 i1
L1
dii1
d
d
di
di
− M12 (i
(i1 i2 ) − R2 (i1 i2 ) L2 (i
(i1 i2 ) + M 23 2 − M112 1 = 0
dt
dt
dt
dt
dt
7.8 Coupled Circuits 7.19
(
) i1 (
)
dii1
− R2 i2 − (
dt
)
dii2
= v1 ( )
dt
…(i)
Applying KVL to Mesh 2,
M12
dii1
dii
d
− M 23 2 − L2 (i2
dt
dt
dt
d
Example 7.16
dii2
d
+ M 23 (i1 − i2 ) − R3i2 = 0
dt
dt
dii1
dii2
− R2 i1 − ( L2 + M12 + M 23 )
+ ( R2 + R3 ) i2 + ( L2 + L3 + 2M 23
=0
23 )
dt
dt
i1 ) R2 (i2
i1 ) L3
…(ii)
Write KVL equations for the circuit shown in Fig. 7.33.
R2
R1
+
v1(t )
−
R3
i3
C
M
i1
L1
+
v2(t )
−
L2
i2
Fig. 7.33
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.34.
R2
R3
R1
L1
+
v1(t )
−
C
i3
L2
i1
Mdi2
dt
+
−
+
−
Mdi1
dt
i2
+
v2
−
Fig. 7.34
Applying KVL to Mesh 1,
dii1
di
−M 2 =0
dt
dt
dii1
di2
R1 i1 + L1
+M
= v1 (t )
dt
dt
v1 (t ) R1 i1
L1
Applying KVL to Mesh 2,
t
M
dii1
dii
1
− L2 2 − R3 (i2 − i3 )
(i2
dt
dt
C ∫0
i3 ) dt v2 (t ) = 0
…(i)
7.20 Network Analysis and Synthesis
t
M
dii1
dii
1
− L2 2 − R3 (i2 − i3 )
(i2
dt
dt
C ∫0
i3 ) dt = v2 (t )
…(ii)
Applying KVL to Mesh 3,
t
1
− R2 i3 − ∫ (i3 − i2 )dt − R3 (i3 − i2 ) = 0
C0
Example 7.17
…(iii)
Write down the mesh equations for the network shown in Fig. 7.35.
Z1
M
V1
L2
L1
+
ZL
−
I1
Z2 I2
Fig. 7.35
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.36.
Z1
jwL1
ZL
+
V1
jwL2
j wM I2
−
I1
+
−
+
−
Z2
jwM I1
I2
Fig. 7.36
Applying KVL to Mesh 1,
V1 Z1I1 j
( Z1 j
j M
MII 2 Z 2 ( I1 I 2 ) = 0
Z 2 ) I1 ( Z 2 j M ) I 2 = V1
1I1
1
…(i)
Applying KVL to Mesh 2,
−Z 2 ( I 2
−(Z
(Z2
I1 )
j
I1 − j L2 I 2 Z L I 2 = 0
) I1 + (Z
(Z2
2 Z ) I2 = 0
…(ii)
7.8 Coupled Circuits 7.21
Example 7.18
Write mesh equations for the network shown in Fig. 7.37.
R
+
L3
L1
v(t )
L2
i
i
i
C
Fig. 7.37
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.38.
R
L
+
+
v1
i1
−
+
L2
+
M12 d (i − i )
dt
i
M13
+
−
di
dt
L3
d
M
(i1 − i )
i
di
dt
−
+
M
d
(i
dt 1
+
−
M
d
(i
dt
)
i )
C
Fig. 7.38
Applying KVL to Mesh 1,
vt
1
Ri1
d
(
dt
d
(i1
1
dt
) M12
1
12
d
dt
d
dt
+ M13
−
13
di3
0
dt
di
= v t)
dt
…(i)
Applying KVL to Mesh 2,
M13
di
+ M12
dt
di3
M12
dt
d
(
dt
d
(
dt
) − L1
)+
d
dt
d
dt
−
1
−
1
2
d
d
( − ) − M 21 (i
dt
dt
d
d
+ M 21 (i1
dt
dt
− M 23
3
di3
0
dt
di3
=0
dt
…(ii)
Applying KVL to Mesh 3,
di
d
+M
dt
dt
di
d
M 23
M 21
dt
dt
M 23
d
i
dt
d
+ L2
dt
L2
−
2
di3
dt
di
3
dt
M
M
d
(
dt
d
(
dt
)
)
32
d
1
(
)
i dt = 0
dt
C∫
d
1
− +
= 0 …(iii)
dt
C
7.22 Network Analysis and Synthesis
Example 7.19
Write KVL equations for the network shown in Fig. 7.39.
R
+
L3
L1
v(t )
i1
−
L2
i2
i3
C
Fig. 7.39
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.40.
R
L1
+
v1
−
i1
L2
+
−
M12 di2
dt
−
+
di3
dt
L3
+
−
M21 di1
dt
+
−
di3
dt
i2
M13
i3
M23
−
+
M31 di1
dt
+
−
M32 di2
dt
C
Fig. 7.40
Applying KVL to Loop 1,
v(t ) − R(i
i
R(i
i
dii1
− M12
dt
di
di
i ) + L1 1 + M12
dt
dii2
+ M13
dt
dii2
− M13
dt
dii3
=0
dt
dii3
= v(t )
dt
dii2
dii
− M 21 1 − M 23
dt
dt
di2
di
dii1
i ) + L2
+ M 21
+ M 23
dt
dt
dii3
=0
dt
dii3
= v(t )
dt
i ) − L1
…(i)
Applying KVL to Loop 2,
v(t ) − R(i
i
R(i
i
i ) − L2
…(ii)
Applying KVL to Loop 3,
dii3
dii
+ M 31 1 − M 32
dt
dt
di3
di1
R(i1 + i2 + i3 ) + L3
− M 331
+ M 332
dt
dt
v(t ) − R(i
i
i ) − L3
dii2 1
−
i3 dt = 0
dt C ∫
di2 1
+
i3 dt = v(t )
dt C ∫
…(iii)
7.8 Coupled Circuits 7.23
Example 7.20
In the network shown in Fig. 7.41, find the voltages V1 and V2.
M=2H
3H
+
5H
i3
−
v1
+
v2
−
i2 = 10e−t A
i1 = 5e−t A
Fig. 7.41
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.42.
+
di3
dt
2
3H
i3
+ −
+
v1
i1 = 5e−t A
2
5H
di1
dt
+ −
v2
−
i2 = 10e−t A
Fig. 7.42
From Fig. 7.42,
i3
i1 + i2
5e
t
10 e
dii1
dii
d
d
2 3 = 3 (5 e −tt ) + 2 (15 e t )
dt
dt
dt
dt
dii
dii
d
d
v2 = 5 3 2 1 = 5 (15
5 e t ) 2 (5 e t )
dt
dt
dt
dt
v1 = 3
Example 7.21
t
15 e − t A
15 e
t
30
30 e
t
45 e − t V
75 e
t
10 e
10
t
85 e − t V
In the network shown in Fig. 7.43, find the voltages V1 and V2.
M=2H
2H
+
v1
4H
i3
−
+
v2
i2 = 10e−t A
i1 = 10e−t A
Fig. 7.43
−
7.24 Network Analysis and Synthesis
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.44.
+
di3
dt
2
2H
i3
− +
− +
+
v1
i1 = 10e−t A
di1
dt
2
4H
−
i2 = 10e−t A
Fig. 7.44
From Fig. 7.44,
i3
dii1
dt
2
dii3
dt
2
v1 = 2
v2 = 4
Example 7.22
i1 + i2
10
0 e −tt + 100 e
= 20 e − t A
t
dii3
d
d
= 2 (10 e t ) 2 ( 20 e t )
dt
dt
dt
dii1
d
d
= 4 ( 20 e t ) 2 (10 e t )
dt
dt
dt
20 e −tt + 400 e
80 e −tt + 20 e
t
t
= 20 e − t A
= 60 e − t A
Calculate the current i2(t) in the coupled circuit of Fig. 7.45.
i1(t)
i2(t)
0.1 H
+
0.2 H
30 sint
0.2 H
−
Fig. 7.45
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.46.
i1(t)
+
i2(t)
0.2 H
0.2 H
30 sint
−
0.1
di2
dt
−
+
−
+
0.1
di1
dt
Fig. 7.46
Applying KVL to Mesh 1,
30 sin t − 0.2
di1
di
+0 1 2 = 0
dt
dt
…(i)
7.8 Coupled Circuits 7.25
Applying KVL to Mesh 2,
− 0.2
di2
di
+ 0.1 1 = 0
dt
dt
di1
di
=2 2
dt
dt
…(ii)
Substituting Eq. (ii) in Eq. (i),
⎛ di ⎞
30 sin t 0.2 ⎜ 2 2
⎝ dt ⎠
0.1
di2
dt
0
0.3
di2
dt
30 sin t
dii2
= 100 sin t
dt
dii2 = 100 sin t dt
Integrating both the sides,
t
i2 (t ) = 100∫ sin t dt
0
= 100 [ − cos t ]0
t
= 100 (1
(1 cos t )
Example 7.23
Find the voltage V2 in the circuit shown in Fig. 7.47 such that the current in the
left-hand loop (Loop 1) is zero.
2Ω
1Ω
j2 Ω
+
+
j4 Ω
5∠0° V
j3 Ω
V2
I1
−
I2
Loop 1
−
Loop 2
Fig. 7.47
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.48.
2Ω
1Ω
j4 Ω
+
j3 Ω
+
V2
5∠0° V
−
I1
j2 I2
−
+
−
+
Fig. 7.48
j2 I1
I2
−
7.26 Network Analysis and Synthesis
Applying KVL to Loop 1,
5
0° 2
1
− j4
j 2I 2 = 0
1
j4
4) I1 − j 2
(2
5 ∠ 0°
2
…(i)
Applying KVL to Loop 2,
− j 2 − j3
j 3I 2 − 1
2
−
2
=0
…(ii)
− j 2 − (1 + j 3) I 2 = V2
Writing Eqs (i) and (ii) in matrix form,
− j 2 ⎤ ⎡ I1 ⎤ ⎡5 0°⎤
⎡2 j 4
=
⎢⎣ − j2
2
(1+
1 j 3) ⎥⎦ ⎢⎣ I 2 ⎥⎦ ⎢⎣ V2 ⎥⎦
By Cramer’s rule,
0°
− j2
V2
−(1 j 3)
I1 =
2 j4
− j2
− j2
2
(1+
1 j 3)
5
But I1 = 0.
∴ −( ∠
)( + j ) + j 2 V2 = 0
V2 =
Example 7.24
Determine the ratio
( ∠
)( +
j2
)
= 7.91 ∠ − 18.43° V
V2
in the circuit of Fig. 7.49, if I1 = 0.
V1
8Ω
2Ω
j2 Ω
+
+
j8 Ω
V1
j2 Ω
V2
I1
−
−
I2
Fig. 7.49
Solution The equivalent circuit in terms of dependent sources is as shown in Fig. 7.50.
8Ω
2Ω
j8 Ω
+
j2 Ω
+
V1
V2
−
I1
j2 I2
+
−
+
−
j 2 I1
I2
−
Fig. 7.50
Applying KVL to Mesh 1,
V1 8I1 j8I1
(8 j8
8) I1
j 2I 2 0
j 2I 2 = V1
…(i)
7.8 Coupled Circuits 7.27
Putting I1 = 0 in Eq (i),
V1
…(ii)
2I 2 j 2I 2 j 2I1 0
j 2I1 ( 2 j 2) I 2 = V2
…(iii)
) I 2 = V2
…(iv)
2 I2
Applying KVL to Mesh 2,
V2
Putting I1 = 0 in Eq (iii),
(
From Eqs (ii) and (iv),
V2 ( 2 2)I 2 2 j 2
=
=
= 1.41 ∠
V1
j 2I 2
j2
Example 7.25
45° V
For the coupled circuit shown in Fig. 7.51, find input impedance at terminals A and B.
A
3Ω
j4 Ω
j3 Ω
+
V1
j5 Ω
−8Ω
−j
−
B
Fig. 7.51
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.52.
3Ω
A
j4 Ω
j 3 (I1 − I2)
+ −
j5 Ω
+
V1
−8Ω
−j
−
+ j3 I I
1 2
−
I1
B
Fig. 7.52
Applying KVL to Mesh 1,
V1 3I1
j 4 I1
j 3 ( I1 I 2 )
j ( I1 − I 2 ) j 3I1 0
(3 j15
15) I1 j8 I 2 = V1
…(i)
Applying KVL to Mesh 2,
j3
jj5
5 ( I 2 − I1 )
j8
j8 2 0
j8
j3 I2 = 0
j3
I2
jj88
I1 = −2 67 I1
j3
j3
…(ii)
7.28 Network Analysis and Synthesis
Substituting Eq (ii) in Eq (i),
(
j )
.67
67 I1 ) = V1
j8 (
1
.36) I1 = V1
(
Zi =
Example 7.26
V1
=(
I1
.36) Ω = 36.48 ∠ 85.28° Ω
Find equivalent impedance of the network shown in Fig. 7.53.
j2 Ω
2Ω
j6 Ω
j4 Ω
j4 Ω
j3 Ω
Zeq
−5Ω
−j
5Ω
Fig. 7.53
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.54.
2Ω
j2 Ω
j 4 (I1 − I2)
− +
j3 Ω
+
V1
−
+
−
I1
j 4 I1
5Ω
−
+
j4 Ω
I2
−
j 6 (I1 − I2)
+
−5Ω
−j
j 6 I2
Fig. 7.54
Applying KVL to Mesh 1,
V1 2I1
j 2I
2I1
j 4 ( I1 − I 2 )
j 3 ( I1 I 2 )
j 4 I1 − 5( I1 I 2 )
j6I2 = 0
(7
) I 2 = V1
j 3) I1 (
…(i)
7.8 Coupled Circuits 7.29
Applying KVL to Mesh 2,
+
−
+ 5
(5 + 5)I
5
5
5
4
…(ii)
Substituting Eq. (ii) in Eq. (i),
5
5
Example 7.27
5
4
1
V1
Zi
V1
I1
7
3−
5
=
4
3
5
Find the voltage across the 5 W resistor in Fig. 7.55 using mesh analysis.
j .66 Ω
j5 Ω
j10 Ω
+
3Ω
50∠0° V
1
j4 Ω
5Ω
I2
Fig. 7.55
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.56.
j5 Ω
j5.66 I2
j5
j10 Ω
+ −
+
I1
+ −
3Ω
50∠0 V
−
I
5Ω
I2
j4 Ω
Fig. 7.56
Applying KVL to Mesh 1,
50 ∠ 0
5
5 66
3
j4
0
50 ∠ °
…(i)
Applying KVL to Mesh 2,
−
−
)(
) − 10 I −
− − .66
− I2 = 0
2 =0
…(ii)
7.30 Network Analysis and Synthesis
Writing Eqs (i) and (ii) in matrix form,
−(3 j 9.66) ⎤ ⎡ I1 ⎤ ⎡50 ∠ 0°⎤
⎡ 3 j1
=
⎢⎣ −(3 j 9.66)
8 j 6 ⎥⎦ ⎢⎣ I 2 ⎥⎦ ⎢⎣ 0 ⎥⎦
By Cramer’s rule,
3 j1
50 ∠ 0°
−(3 j 9.66)
0
I2 =
= 3.82
8 ∠ −112
1 .14° Α
3 j1
−(3 j 9.66)
−(3 j 9.66)
8 j6
V5
Example 7.28
I2
5 (3.82 ∠ − 112.14°) = 19.1 ∠ − 112.14
14° V
Find the voltage across the 5 W resistor in Fig. 7.57 using mesh analysis.
j5 Ω
k = 0.8
+
j10 Ω
3Ω
50∠0° V
−
I1
−4Ω
−j
5Ω
I2
Fig. 7.57
Solution For a magnetically coupled circuit,
XM
k X L1 X L2
= 0.8 (5) (10)
= 5 66 Ω
The equivalent circuit in terms of dependent sources is shown in Fig. 7.58.
j5 Ω
j5.66 I2
− +
+
j5.66 I1
− +
3Ω
50∠0° V
−
j10 Ω
I1
−4Ω
−j
5Ω
I2
Fig. 7.58
Applying KVL to Mesh 1,
50 ∠ 0°
j 5 I1 + j 5.66 2 (3 − j 4) ( 1 2 ) 0
(3 j 1) I1 − (3
(3 j 1 66) I 2 = 50 ∠ 0°
…(i)
7.8 Coupled Circuits 7.31
Applying KVL to Mesh 2,
−( − j ) (
− ) − j 10 I 2 + j 5.66
66
−( + j .66)
− 5I 2 = 0
1
+ ((88 + j 6)
6) I 2 = 0
…(ii)
Writing Eqs (i) and (ii) in matrix form,
−(3 j 1.66) ⎤ ⎡ I1 ⎤ ⎡50 ∠ 0°⎤
⎡ 3 j1
=
⎢⎣ −(3 j 1.66)
8 j 6 ⎥⎦ ⎢⎣ I 2 ⎥⎦ ⎢⎣ 0 ⎥⎦
By Cramer’s rule,
3 j1
50 ∠ 0°
−(3 j 1.66)
0
I2 =
= 8.62
6 ∠ − 24.79° Α
3 j1
−(3 j 1.66)
−(3 j 1.66)
8 j6
V5
Example 7.29
5 (8.62
I2
24.79°)
43.1 ∠ − 24.79
79° Α
Find the current through the capacitor in Fig. 7.59 using mesh analysis.
j4 Ω
3Ω
+
j3 Ω
j5 Ω
50∠45° V
−
I1
−8Ω
−j
I2
Fig. 7.59
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.60.
j4 Ω
3Ω
j 3 (I1 − I2)
+ −
j5 Ω
+
50∠45° V
+
−
I1
−
−8Ω
−j
j 3 I1
I2
Fig. 7.60
Applying KVL to Mesh 1,
50 ∠ 45° − (3
j 4) I1 − j 3(
1
2)
j5 ( 1 −
(3 j 155)
2)
1
j3 1 0
j 8 I 2 = 50 ∠ 45
45°
…(i)
7.32 Network Analysis and Synthesis
Applying KVL to Mesh 2,
jj5
5 ( I 2 − I1 )
j3
j8
j8
− j8
2
0
j3 I2 = 0
j3
…(ii)
Writing Eqs (i) and (ii) in matrix form,
j8⎤ ⎡ I1 ⎤ ⎡50 ∠ 45°⎤
=
⎥⎦
j 3⎥⎦ ⎢⎣ I 2 ⎥⎦ ⎢⎣
0
⎡3 j 15
⎢⎣ − j8
By Cramer’s rule,
3
I2 =
IC
j 15 50 45°
− j8
0
= 3.66 ∠ − 310.33° Α
3 j 15
j8
− j8
j3
I 2 = 3.66 ∠ − 310.33°Α
Find the voltage across the 15 Ω resistor in Fig. 7.61 using mesh analysis.
Example 7.30
j10 Ω
20 Ω
j15 Ω
j20 Ω
+
120∠0° V
−
15 Ω
I2
I1
Fig. 7.61
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.62.
j10 Ω
20 Ω
j5 (I1 − I2)
+ −
j20 Ω
+
15 Ω
120∠0° V
−
I1
+
−
j5 I2
I2
Fig. 7.62
Applying KVL to Mesh 1,
120 ∠ 0° 20
j 20
0( 1 2 ) j 5 I 2 = 0
( 20 + j 20) I1 − j 15 I 2 = 120 ∠ 0°
1
…(i)
7.8 Coupled Circuits 7.33
Applying KVL to Mesh 2,
j5
j 20
20 (
2
1)
j 5 ( I1 I 2 ) − 15 I 2 = 0
j1
10
0
− j15 I
…(ii)
j 20
j20
0) I 2 = 0
(15
1
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
I2 =
= 2.53
20 + j 20
− j15
− j15
15 + j 20
V15 Ω
Example 7.31
1 I 2 = 15 ( 2.53
10.12 )
10.12°Α
37.95
10.12°V
Find the current through the 6 W resistor in Fig. 7.63 using mesh analysis.
4Ω
j2 Ω
+
j3 Ω
120∠0° V
−
j8 Ω
I1
I2
6Ω
Fig. 7.63
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.64.
4Ω
j3 Ω
+
j8 Ω
200∠30° V
−
I1
−
+
j2 I2
I2
−
+ j2 (I1 − I2)
6Ω
Fig. 7.64
Applying KVL to Mesh 1,
120 ∠ 0° 4 I1
j3
3( 1 2 ) j 2 I 2 = 0
( 4 j 3) I1 − j 5 I 2 = 120 ∠ 0°
…(i)
7.34 Network Analysis and Synthesis
Applying KVL to Mesh 2,
−j2
− j 3 ( I 2 − I1 ) − jj8
8
+ j 2 ( I1 − I 2 ) − 6 2 = 0
− j 5 + (6 + j15
15) 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,
j 3 120 ∠ 0°
− j5
0
I2 =
= 7.68
4 j3
− j5
− j 5 6 + j 15
4
Example 7.32
2.94°Α
Determine the mesh current I3 in the network of Fig. 7.65.
12 Ω
j16 Ω
j5 Ω
4Ω
7Ω
I3
+
6Ω
j4 Ω
200∠30° V
−
−8Ω
−j
I1
I2
−4Ω
−j
Fig. 7.65
Solution The equivalent circuit in terms of dependent sources is shown in Fig. 7.66.
j16 Ω
j5 (I1 − I2)
12 Ω
− +
4Ω
7Ω
I3
j4 Ω
+
−8Ω
−j
6Ω
200∠30° V
−
I1
−
+
j5 I3
Fig. 7.66
I2
−4Ω
−j
7.8 Coupled Circuits 7.35
Applying KVL to Mesh 1,
200 ∠ 30 −
+ 5
4
0
+ 5
20
…(i)
30
Applying KVL to Mesh 2,
−
−
2
I −
−
−
−
−
−
2
− (7
0
0
…(ii)
=0
=
…(iii)
Applying KVL to Mesh 3,
−
−
− j8
I1 − (7 −
+
−4
−
Writing Eqs. (i), (ii) and (iii) in matrix form,
−
− 4
3
7
3
5
−( 7
23
+ j5 ⎤ 1
13) I 2
8
I3
⎡ 200 ∠ 30°
⎢
0
0
200 ∠ 30°
0
0
+ j5
7
3
23 8
16 28 ∠ 16 87°Α
By Cramer’s rule,
−
I3
−
5
− 4
3
−7
3
− 4
j
−
5
−7
3
Example 7.33
Obtain the dotted equivalent circuit for the coupled circuit shown in Fig. 7.67 and
find mesh currents. Also find the voltage across the capacitor.
5Ω
j5 Ω
5Ω
j2 Ω
j5 Ω
+
+
10∠ 0 V
−
10∠ 90° V
j10 Ω
Fig. 7.67
Solution The currents in the coils are as shown in Fig. 7.68. The corresponding flux due to current in each
coil is also drawn with the help of right-hand thumb rule.
7.36 Network Analysis and Synthesis
f1
5 Ω I1
I2 5 Ω
j2 Ω
j5 Ω
j5 Ω
+
f2
+
10∠ 0°V
I1
−
I2
−
10∠ 90°V
− 10 Ω
−j
Fig. 7.68
From Fig. 7.68, it is seen that two fluxes f1 and f2 aid each other. Hence, dots are placed at the two coils as
shown in Fig. 7.69.
5Ω
j2 Ω
j5 Ω
+
j5 Ω
5Ω
+
−j 10 Ω
−j
10∠0° V
−
10∠90° V
I1
−
I2
Fig. 7.69
The equivalent circuit in terms of dependent sources is shown in Fig. 7.70.
5Ω
j5 Ω
j 2 I2
+ −
−
+
+
10∠0° V
−
−
I1
j 2 I1
j5 Ω
+
− +
5Ω
−
+
+
− 10 Ω
−j
10∠90° V
I2
−
Fig. 7.70
Applying KVL to Mesh 1,
10 ∠ 0°° − (5 + j 5)
− j 2 I 2 + j10 ( 1 + 2 ) = 0
(5 − j 5) I1 − j 8 2 = 10 ∠ 0°
…(i)
− jj2
2 I1 + 5 2 − 10 ∠ 90° = 0
− j 8 + (5 − j 5) 2 = 10 ∠ 90°
…(ii)
1
Applying KVL to Mesh 2,
− j10 (
+ ) + j5
2
7.8 Coupled Circuits 7.37
Writing Eqs. (i) and (ii) in matrix form,
⎡5 j 5 − j8 ⎤ ⎡ I1 ⎤ ⎡ 10 ∠ 0° ⎤
⎢⎣ − j8 5 − j 5⎥⎦ ⎢⎣ I 2 ⎥⎦ = ⎢⎣10 ∠ 90°⎥⎦
By Cramer’s rule,
10 ∠ 0°
j8
10 ∠ 90° 5 − j 5
I1 =
= 0.72
5 j 5 − j8
− j8 5 − j 5
82.97°A
5 − j 5 10 ∠ 0°
− j8 10 90°
I2 =
= 1.71
7 ∠ 106.96°A
5 j 5 − j8
− j8 5 − j 5
j 10 ( I
VC
Example 7.34
I )
( − j 10) (0.72 ∠ − 82.97° + 1.71 ∠ 106.96
= 10.08 ∠ 24.03
. °V
)
Find Thevenin’s and Norton’s equivalent network at terminals A-B for Fig. 7.71.
j 10 Ω
4Ω
3Ω
A
+
j5 Ω
j6 Ω
10∠0° V
−
4Ω
B
Fig. 7.71
Solution
Step I Calculation of VTh
The equivalent circuit in terms of dependent sources is shown in Fig. 7.72.
4Ω
j10 Ω
j6 I
− +
3Ω
+
+
A
−
+
− j6 I
+
10∠0° V
−
j5 Ω
I
VTh
+
4Ω
−
Fig. 7.72
−
B
7.38 Network Analysis and Synthesis
Applying KVL to Mesh,
10 ∠ 0
− 10
5I+
6
10 ∠ 0
0
8+
0
10 ∠ 0°
= 1 17 ∠ − 20 56°A
8 3
I=
Writing the VTh equation,
+
=0
0
0
=
V
−
17 ∠ − 20.
∠−
. V
Step II Calculation of IN
The equivalent circuit in terms of dependent sources is shown in Fig. 7.73.
j 10 Ω
4Ω
j 6( 1
2)
3Ω
− +
A
j5 Ω
+
−
j 6 I1
+
10∠0 V
I1
−
N
I2
4Ω
B
Fig. 7.73
Applying KVL to Mesh 1,
10 ∠ 0
1
+
5I
2
+ 6
(8
=0
= 10 ∠ 0°
3) I
…(i)
Applying KVL to Mesh 2,
1) I
2
0
=0
…(ii)
Writing Eqs. (i) and (ii) in matrix form,
⎡
1
−4
7
⎤ I1
I2
3
10 ∠ 0°
0
=
−4
7
⎡10 ∠ 0 ⎤
0
By Cramer’s rule,
I2 =
−(
−
)
1
6
9°A
7.8 Coupled Circuits 7.39
I 2 = 0.56 ∠ − 83.39°A
IN
Calculation of ZTh
Step III
ZTh =
VTh
4.82 ∠ − 34.6°
=
= 8.61 ∠ 48.79° Ω
IN
0.56 ∠ − 83.39°
Step IV Thevenin’s Equivalent Network
8.61∠48.79° Ω
A
+
4.82∠−34.6° V
−
B
Fig. 7.74
Step V
Norton’s Equivalent Network
A
0.56∠−83.39° A
8.61∠48.79° Ω
B
Fig. 7.75
Example 7.35 Find the value of ZL connected across terminals A and B in Fig. 7. 76 to receive the
maximum power when (a) ZL is complex impedance, and (b) ZL is pure resistance.
5Ω
j8 Ω
A
+
50∠0°V
−
j4 Ω
j 10 Ω
ZL
B
Fig. 7.76
Solution
Step I Calculation of VTh
The equivalent circuit in terms of dependent sources is shown in Fig. 7.77.
7.40 Network Analysis and Synthesis
5Ω
j 8Ω
j4 I
+ −
+
A
j 10Ω
+
50∠0°V
+
−
−
+
−
I
VTh
j4 I
−
B
Fig. 7.77
Applying KVL to the mesh,
50 ∠ 0° 5I − j 8
j 4 I − j 10
0I− j 4
0
50 ∠ 0° (5 + j 26) I = 0
50 ∠ 0°
I=
= 1.889 ∠ − 79.11° A
5 j 26
Writing the VTh equation,
j4
j 10
VTh 0
j 14 I VTh = 0
VTh = ( j 14) (1
(1.89
79.11111°)
26.46
10.89° V
Step II Calculation of IN
The equivalent circuit in terms of dependent sources is shown in Fig. 7.78.
4Ω
j8 Ω
j 4(I1 − I2)
+ −
j 10 Ω
+
50∠0° V
−
I1
+ j4
I1
−
IN
I2
Fig. 7.78
Applying KVL to Mesh 1,
50 ∠ 0° 5I1
j 4 ( I1 I 2 ) − j 10 ( I1 I 2 ) − j 4 1 0
(5 j 26
6) 1 j 14 I 2 = 50
5 ∠ 0°
Applying KVL to Mesh 2,
j4
j 10
10 ( 2 1 ) 0
j 14 I j 100 I 2 = 0
I1 0 71
7 I2
j8
1
Substituting Eq. (ii) in Eq. (i),
(
j
) ( .71
7 I 2 ) − j 14 I 2 = 50 ∠ 0°
(5.7 ∠ 51. ) 2 = 50 ∠ 0°
I 2 = 8.77
7 ∠ 51.48° A
I N I 2 = 8.77 ∠ 51.48° A
…(i)
…(ii)
7.9 Conductively Coupled Equivalent Circuits 7.41
Step III
Calculation of ZTh
ZTh =
VTh 26.46 ∠ 10.89
=
= 3.02 ∠ 62.37° Ω = (1.4 + j .67) Ω
IN
5.7 ∠
∠− 51.48
Step IV Calculation of ZL
(i) When ZL is complex impedance
ZL
For maximum power transfer,
(ii) When ZL is pure resistance
ZL
For maximum power transfer,
7.9
∗
ZTh
= ( .4 − j 2..
)Ω
ZTh = 1.4 + j 2.67 = 3.02 Ω
CONDUCTIVELY COUPLED EQUIVALENT CIRCUITS
For simplifying circuit analysis, it is desirable to replace a magnetically coupled circuit with an equivalent
circuit called conductively coupled circuit. In this circuit, no magnetic coupling is involved. The dot
convention is also not needed in the conductively coupled circuit.
Consider a coupled circuit as shown in Fig. 7.79.
I1
I2
jw M
+
+
jwL2
jwL1
V1
V2
−
−
Fig. 7.79 Coupled circuit
The equivalent circuit in terms of dependent sources is shown in Fig. 7.80.
I1
I2
jwL1
+
jwL2
+
V1
V2
−
j wM I2
+
−
+
jwM I1
−
−
Fig. 7.80 Equivalent circuit
Applying KVL to Mesh 1,
V1
j 1 I1
j L1 I1
j M I2 = 0
j M I 2 = V1
…(7.11)
j 2 I2
j M I1
j M I1 = 0
j 2 I 2 = V2
…(7.12)
Applying KVL to Mesh 2,
V2
Writing Eqs (7.11) and (7.12) in matrix form,
⎡ j L1 j M ⎤ ⎡ I1 ⎤ ⎡ V1 ⎤
⎢⎣ j M j L2 ⎥⎦ ⎢⎣ I 2 ⎥⎦ = ⎢⎣ V2 ⎥⎦
…(7.13)
7.42 Network Analysis and Synthesis
Consider a T-network as shown in Fig. 7.81.
Z1
I1
Z2
I2
+
+
Z3
V1
V2
−
−
Fig. 7.81 T-network
Applying KVL to Mesh 1,
V1 Z1 I1 Z3 ( I1 I 2 ) = 0
( Z1 Z3 ) I1 Z3 I 2 V1
… (7.14)
Z 2 I 2 Z3 ( I 2 I1 ) = 0
Z3 I1 + ( Z 2 + Z3 ) I 2 = V2
… (7.15)
Applying KVL to Mesh 2,
V2
Writing Eqs (7.14) and (7.15) in matrix form,
⎡ Z1 Z3
⎢⎣ Z3
Z3 ⎤ ⎡ I1 ⎤ ⎡ V1 ⎤
=
Z 2 + Z3 ⎥⎦ ⎢⎣ I 2 ⎥⎦ ⎢⎣ V2 ⎥⎦
Comparing matrix equations,
Z1 Z3 = j
Z3 = j
Z 2 Z3 = j
L1
M
L2
Solving these equations,
Z1 = j
Z2 = j
Z3 = j
L1 − j
L2 − j
M
M = j ( L1 − M )
M = j ( L2 − M )
Hence, the conductively coupled circuit of a magnetically coupled circuit is shown in Fig. 7.82.
I1
jw (L1 − M)
jw (L2 − M ) I2
+
+
j wM
V1
−
V2
−
Fig. 7.82 Conductively coupled equivalent circuit
7.9 Conductively Coupled Equivalent Circuits 7.43
Example 7.36
Find the conductively coupled equivalent circuit for the network shown in Fig. 7.83.
−4Ω
−j
j6 Ω
j2 Ω
+
V1
j3 Ω
−
j5 Ω
10 Ω
I1
I2
2Ω
Fig. 7.83
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 (
Z2 = j (
M ) = j L1 − j M = j 3
2
j 2 = j1 Ω
− M ) = j L2 − j M = j 5
j 2 = j3 Ω
Z3 = j M = j 2 Ω
The conductively coupled equivalent circuit is shown in Fig. 7.84.
−4Ω
−j
j1 Ω
+
j3 Ω
j6 Ω
j2 Ω
V1
−
10 Ω
I2
I1
2Ω
Fig. 7.84
Example 7.37
Draw the conductively coupled equivalent circuit of Fig. 7.85.
j6 Ω
j5 Ω
+
3Ω
V1
−
j10 Ω
I1
−4Ω
−j
5Ω
I2
Fig. 7.85
Solution The current I1 enters from the dotted end and I2 leaves from the dotted end. Hence, the mutual
inductance M is negative.
7.44 Network Analysis and Synthesis
−1Ω
−j
In the conductively coupled equivalent circuit,
Z1 = j ( L1 − M ) = jω L1 − jω M = j 5
j4 Ω
j 6 = − j1 Ω
Z2 = j ( L2 − M ) = jω L2 − jω M = j10 − j 6 = j 4 Ω
Z3 = j M = j 6 Ω
j6 Ω
+
V1
5Ω
−
(3 − j4) Ω I2
I1
The conductively coupled equivalent circuit is shown in
Fig. 7.86.
Fig. 7.86
Example 7.38
Find the conductively coupled equivalent circuit of the network in Fig. 7.87.
4Ω
2Ω
j1Ω
+
j2 Ω
V1
−
j4 Ω
I1
Fig. 7.87
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 ) = jω L1 + jω M = j 4 + j 2 = j 6 Ω
Z2 = j (L2 + M ) = jω L + jω M = j 2 + j 2 = j 4 Ω
Z3 = − j M = − j 2 Ω
The conductively coupled equivalent circuit is shown in Fig. 7.88.
(4 − j2) Ω
j6 Ω
j4 Ω
+
−2Ω
−j
V1
−
I1
(2 + j 4) Ω
I2
Fig. 7.88
7.10
TUNED CIRCUITS
At radio frequencies, the amplification of signals is done by tuned circuits. There are two types of tuned
circuits—single-tuned circuit and double-tuned circuit. In the single-tuned circuit, only the secondary is
tuned. In the double-tuned circuits, both primary and secondary of the coupled circuits are tuned. The tuned
circuits are used in TV receivers to couple the antenna to the RF stage. In RF circuit design, the tuned circuits
are generally employed for obtaining maximum power transfer to the load connected to secondary or for
obtaining maximum possible value of the secondary voltage.
7.10 Tuned Circuits 7.45
7.10.1 Single-Tuned Circuit
Consider a single-tuned circuit as shown in Fig. 7.89.
I1
I2
M
R2
+
R1
jwL1
+
jwL2
−
−j
V1
−
1
wC2
V2
−
Fig. 7.89
Single-tuned circuit
The conductively coupled equivalent circuit is shown in Fig. 7.90.
jw (L1 − M )
jw (L2 − M )
R2
+
R1
−
−j
jw M
+
I1
V1
−
I2
1
wC2
V2
−
Fig. 7.90 Conductively coupled equivalent circuit
Applying KVL to Mesh 1,
V1 − R1I1−jw(L1−M)I1− jwM(I1−I2) = 0
(R1 + jw L1)I1 − jw M I2 = V1
Applying KVL to Mesh 2,
1
− j M ( − ) − j ( L − M ) 2 − R2 I 2 + j
I2 = 0
ωC2
…(7.16)
⎡
1 ⎞⎤
⎛
− j M I1 + ⎢ R + j ω L2 −
⎥ I2 = 0
⎝
ωC2 ⎟⎠ ⎦
⎣
…(7.17)
From Eq. (7.17),
R2 + j
I1
Substituting Eq. (7.18) in Eq. (7.16),
(R
⎡
⎢R
jω L1 ) ⎢
⎢
⎢
⎣
⎧
⎡
⎪ ( R1 + j L1 ) ⎢ R
⎪
⎣
I2 ⎨
⎪
⎪⎩
j
1 ⎞⎤
⎛
L2 −
⎝
ωC2 ⎟⎠ ⎥⎥
I2
⎥
j M
⎥
⎦
j MI 2 = V1
⎫
1 ⎞⎤
⎛
j ω L2 −
+ ω2M 2 ⎪
⎝
ωC2 ⎟⎠ ⎥⎦
⎪
⎬ = V1
j M
⎪
⎪⎭
1 ⎞
⎛
L2 −
⎝
ωC2 ⎟⎠
I2
j M
…(7.18)
7.46 Network Analysis and Synthesis
j M V1
I2
(
1
C2
1
+
2
2
1. Output Voltage
1
V
1
I2
C2
I
MV
=
C
1
+
+
1
C
L2
+
2
2
2. Voltage Amplification
M
V2
V1
A
C
1
+
1
C2
L2
+
2
2
If R1 >> jwL1 and secondary circuit resonates at
1
0
then, at the resonant frequency w0,
I2 =
+
2
0
2
MV
V
C
1
M
A
C
1
Output current, output voltage and voltage amplification depends on the mutual inductance M at resonant
frequency. The maximum amplification or maximum output voltage can be obtained by
dA
dM
C
−
1R
C
0
=0
2
1
+
0
1 2
M=
0
where M =
is the critical value of mutual inductance.
0
7.10 Tuned Circuits 7.47
Hence, maximum value of I2, V2 and A is
I 2max =
V2max =
Amax =
V0
V1
k = k1
2 R1 R2
V1
2ω 0C2 R1 R2
k = k2
1
k = k3
2ω 0C2 R1 R2
k1 > k2 > k3
w
The variation of output voltage V0 with angular 0
w0
frequency w for different values of coefficient of
coupling k is shown in Fig. 7.91.
Fig. 7.91 Variation of V0 with w for
different values of k
7.10.2 Double-Tuned Circuit
Consider a double-tuned circuit with tuning capacitors placed both in primary and secondary circuits as
shown in Fig. 7.92.
−
−j
1
wC1
R2
M
R1
+
jwL1
+
jwL2
−
−j
V1
−
1
wC2
V2
−
Fig. 7.92
Double-tuned circuit
The conductively coupled equivalent circuit is shown in Fig. 7.93.
−
−j
1
wC1
jw (L1 − M )
jw (L2 − M )
R2
+
R1
−
−j
jw M
+
I1
V1
−
1
wC2
V2
−
Fig. 7.93 Conductively coupled equivalent circuit
Applying KVL to Mesh 1,
V1
R1I1 + j
1
I1
ω C2
j (
1
M )I1
jω M ( I1 I 2 ) = 0
1 ⎞
⎛
⎜⎝ R1 + j L1 − j ωC ⎟⎠ I1
1
j
I 2 = V1
...(7.19)
7.48 Network Analysis and Synthesis
Applying KVL to Mesh 2,
−
−
L −
−
1
I2
C2
0
1 ⎤
I2
C2
0
+j
+
−
...(7.20)
From Eq. (7.20),
1
C
−
2
I
j M
I2
Substituting Eq. (7.21) in Eq. (7.19),
⎡
⎢
1
C2
−
1
I2
1
C2
−
1
C2
−
2
j M
I2
j M
⎤
1
C
+
V1
j M
j MV1
I2 =
−
1
1
C
1
+
C2
−
2
2
2
1. Output Voltage
1
1
I2
C2
MV1
=
⎡
C
+
−
⎤⎡
1
C
R2 +
2
1 ⎤
+
C2
2
2
2. Voltage Amplification
A=
V
=
V1
M
R1 +
−
1
C2
2
−
If the primary and secondary circuits resonate at the same frequency w0,
1
0
1
1
C2
2
2
...(7.21)
7.10 Tuned Circuits 7.49
Then, at the resonant frequency w 0,
ω 0 MV
V1
R1 R2 + ω 02 M 2
MV
V1
V2 =
2
2
C2 ( R1 R2
0M )
M
A=
2
2
C2 ( R1 R2
0M )
I2 =
The maximum amplification or maximum output voltage can be obtained by,
dA
=0
dM
2
2
0 M ) − 2C2
C2 ( R1 R2
(R R
⎡C2 ( R1 R2
⎣
1 2
2
2
0M
+
)
2
2
0M
2
2 2
0 M )⎤
⎦
2ω 02 M 2
ω 02 M 2
=0
0
R1 R2
M=
R1 R2
ω0
V0
where M =
inductance.
R1 R2
ω0
is the critical value of mutual
k = k1
Hence, maximum value of I2, V2 and A is
I 2max =
V2max =
Am
max =
k1 = kc
k2 > k1
k3 < k1
k = k2
V1
2 R1 R2
V1
k = k3
w
w0
0
2ω 0C2 R1 R2
1
Fig. 7.94
2ω 0C2 R1 R2
Variation of V0 with w for
different values of k
The variation of the output voltage V0 with the angular frequency w for different values of coefficient of
coupling k is shown in Fig. 7.94.
Example 7.39 For the single-tuned circuit shown in Fig. 7.95, determine (a) the resonant frequency,
(b) Output voltage at resonance, and (c) maximum output voltage. Assume R1>>w0L1 and k = 0.8.
10 Ω
5Ω
M
+
+
60 μH
1 μF
50∠0°V
0.1 μF
V2
−
−
Fig. 7.95
7.50 Network Analysis and Synthesis
Solution
k L1 L2 = 0.8 (1 10 −66 )(660
0
M
0 6 ) = 6. 2 μ H
(a) Resonant frequency
1
ω0 =
f0 =
L2C2
1
=
−6
−6
(60 10 )(0
(0.1 × 10 )
= 0 41 106 rad/s
ω 0 0 41 106
=
= 0.
0 0653 MHz = 65.3 kHz
2π
2π
(b) Output voltage at resonance
V0 =
(
MV
V1
)
2
2
0M
C2 R1 R2
=
6 2 × 10 −6 × 50
(0 42
⎡
0.1
1 × 10 −6 (10 × 5)
⎣
106
) (6 2 ×10 ) ⎤⎥⎦
2
−6 2
= 54.6 V
(c) Maximum output voltage
V2max =
V1
2ω 0C2 R1 R2
=
50
6
2(0.42 × 10 )(0.1 10 6 ) 10 × 5
= 84.18 V
Example 7.40
The resonant frequency of the tuned circuit shown in Fig. 7.96 is 1000 rad/s. Calculate the self-inductances of two coils and critical value of mutual inductance.
1 μF
5Ω
3Ω
M
+
L1
Vs
L2
−
Fig. 7.96
Solution
ω0 =
L1 =
L2 =
Mc =
1
L1C1
1
ω 02C1
1
ω 02C2
=
=
=
1
L2C2
1
(1000)
2
(1
1
(1000)
2
(
10
6
)
=1H
)
=05H
R1 R2
5 3
=
= 3 87 mH
ω0
1000
2 μF
Exercises 7.51
Example 7.41
The tuned frequency of a double-tuned circuit shown in Fig. 7.97 is 104 rad/s. If the
source voltage is 2 V and has a resistance of 0.1 W, calculate the maximum output voltage.
C1
0.01 Ω
0.1 Ω
M
0.1 Ω
2 μH
25 μH
C2
2V
Fig. 7.97
Solution
MC =
ω0 =
10 4 =
RS ) R2
( R1
ω0
=
(0.01 + 0.1) (0.1)
10 4
= 10.49 μH
1
L2C2
1
(25
)C
2
C2 = 0.4 mF
V0 max =
V1
2ω 0C2
( RS + R1 ) R2
=
2
2
0
4
0.4 10
3
(0.1
0.01)( 0.1)
= 2 38 V
Exercises
7.1
Two coupled coils have inductances of 0.8 H
and 0.2 H. The coefficient of coupling is 0.90.
Find the mutual inductance and the turns ratio
N1
.
N2
[0.36 H, 2]
7.2
Two coils with coefficient of coupling 0.5 are
connected in such a way that they magnetise
(i) in the same direction, and (ii) in opposite
directions. The corresponding equivalent
inductances are 1.9 H and 0.7 H. Find selfinductances of the two coils and the mutual
inductance between them.
[0.4 H, 0.9 H, 0.3 H]
Two coils having 3000 and 2000 turns are
wound on a magnetic ring. 60% of the flux
7.3
produced in the first coil links with the
second coil. A current of 3 A produce a flux
of 0.5 mwb in the first coil and 0.3 mwb in the
second coil. Determine the mutual inductance
and coefficient of coupling.
[0.2 H, 0.63]
7.4
Find the equivalent inductance of the network
shown in Fig. 7.98.
7H
2H
3H
5H
4H
6H
Fig. 7.98
[10 H]
7.52 Network Analysis and Synthesis
Find the effective inductance of the network
shown in Fig. 7.99.
7.5
(ii)
A
2H
3H
2Ω
2Ω
4H
5H
j5 Ω
2H
B
Fig. 7.99
Fig. 7.102
[4.8 H]
(iii)
Write mesh equations of the network shown
in Fig. 7.100.
7.6
R1
L2
A
R3
2Ω
j4 Ω
L1
−
i1
R2
i2
−3Ω
−j
j5 Ω
+
v(t)
j5 Ω
j2 Ω
L3
j8 Ω
i3
B
Fig. 7.100
4Ω
Fig. 7.103
d
dii
dii ⎤
⎡
(i1 i2 ) + M12 2 + M13 3 ⎥
⎢ v i1 R1 L1 d
dt
dt
dt
⎢
⎥
di
i
d
⎢ R (i − i ) R i L 3 + M
(i1 i2 ) ⎥
33
3
13
⎢ 2 3 2
⎥
dt
dt
⎢
⎥
dii
⎢ −M
⎥
M 23 2 = 0
⎢⎣
⎥⎦
dt
7.7
⎡(a) (
⎢
2
⎢⎣(c) (6.22
7.8
Find the input impedance at terminals AB of
the coupled circuits shown in Fig. 7.101 to
7.103.
(i)
j 4.65) Ω
(b) (1
5Ω
j2 Ω
2Ω
+
+
j8 Ω
j2 Ω
−
j4 Ω
j1.5) Ω ⎤
⎥
⎥⎦
In the coupled circuit shown in Fig. 7.104,
find V2 for which I1 = 0. What voltage appears
at the 8 Ω inductive reactance under this
condition?
100∠0°V
3Ω
)
V2
−
A
Fig. 7.104
j3 Ω
j5 Ω
−8Ω
−j
[
7.9
B
Fig. 7.101
.5
45°V, 100∠0°V]
For the coupled circuit shown in Fig. 7.105,
find the components of the current I2 resulting
from each source V1 and V2.
Exercises 7.53
2Ω
V1 =
10∠0°V
+
j2 Ω
j4 Ω
+
j3 Ω
−
−
I2
V2 =
10∠0°V
7.13 For the circuit shown in Fig. 7.109, what
load connected to terminals A and B absorbs
maximum power and what is this power?
30 Ω
j 25 Ω
A
Fig. 7.105
+
[ .77∠112.6°Α,1.72
6
∠86.05°Α]
k = 0.5
j40 Ω
100∠15° V
−
7.10 Find the voltage across the 5 Ω resistor in the
network shown in Fig. 7.106.
j2 Ω
j10 Ω
B
Fig. 7.109
j1 Ω
56.3° Ω, 83.3 W]
[ .1
7.14 Find the current I in the circuit of Fig. 7.110.
+
−3Ω
−j
10∠0° V
5Ω
− 15 Ω
14 Ω −j
−
+
−
[ .2
33..
j2 Ω
j3 Ω
j4 Ω
5Ω
Fig. 7.110
[ .3
. 8∠39
39..4° A]
7.15 Obtain a conductively coupled circuit for the
circuit shown in Fig. 7.111.
2Ω
3Ω
−
j5 Ω
j3 Ω
j4 Ω
+
−
Fig. 7.107
[
.16 W]
Fig. 7.111
7.12 Obtain Thevenin’s and Norton’s equivalent
network at terminals AB of the coupled circuit
shown in Fig. 7.108.
4Ω
j8 Ω
j5 Ω
j8 Ω
−2Ω
−j
3Ω
100∠0°V
2Ω
4Ω
+
A
+
−
B
−
10∠0°V
70∠−30° V
−
° ]
+
100∠0°V
j 12 Ω
I
7.11 Find the power dissipated in the 5 Ω resistor
in the network of Fig. 7.107.
2Ω
j3 Ω
10 Ω
+
j 10 Ω
100∠20° V
Fig. 7.106
j4 Ω
j2 Ω
−1Ω
−j
+
3Ω
−
j3 Ω
−2Ω
−j
100∠0°V
10∠90°V
Fig. 7.108
[ .0
. 7 45° V, 1.04∠ − 27.9°A, 6.8∠72.9° Ω]
Fig. 7.112
7.54 Network Analysis and Synthesis
Objective-Type Questions
7.1
7.2
Two coils are wound on a common magnetic
core. The sign of mutual inductance M for
finding out effective inductance of each coil
is positive if the
(a) two coils are wound in the same sense.
(b) fluxes produced by the two coils are
equal
(c) fluxes produced by the coils act in the
same direction
(d) fluxes produced by the two coils act in
opposition
When two coils having self-inductances
of L1 and L2 are coupled through a mutual
inductance M, the coefficient of coupling k is
given by
k=
(a)
k=
(c)
7.3
k=
(b)
2M
L1 L2
L1 + L2
M
L1 L2
M
(b) L1 − L2
1
1
( 1
( 1
2)
2)
(d)
4
2
Consider the following statements:
The coefficient of coupling between two oils
depends upon
1. Orientation of the coils
2. Core material
3. Number of turns on the two coils
4. Self-inductance of the two coils
of these statements,
(a) 1, 2 and 3 are correct
(b) 1 and 2 are correct
(c) 3 and 4 are correct
(d) 1, 2 and 4 are correct
(c)
7.7
7.8
k=
(d)
7.6
L1 L2
The overall inductance of two coils connected
in series, with mutual inductance aiding selfinductance is L1; with mutual inductance
opposing self-inductance, the overall
inductance is L2. The mutual inductance M is
given by
(a)
7.4
M
2 L1 L2
7.5
Two coupled coils connected in series have
an equivalent inductance of 16 mH or 8 mH
depending on the inter connection.
Then the mutual inductance M between the
coils is
(a) 12 mH
(b) 8 2 mH
(c) 4 mH
(d) 2 mH
Two coupled coils with L1 = L2= 0.6 H have
a coupling coefficient of k = 0.8. The turns
N1
ratio
is
N2
(a) 4
(b) 2
(c) 1
(d) 0.5
The coupling between two magnetically
coupled coils is said to be ideal if the
coefficient of coupling is
(a) zero
(b) 0.5
(c) 1
(d) 2
The mutual inductance between two coupled
coils is 10 mH . If the turns in one coil are
doubled and that in the other are halved then
the mutual inductance will be
(a) 5 mH
(b) 10 mH
(c) 14 mH
(d) 20 mH
Two perfectly coupled coils each of 1 H selfinductance are connected in parallel so as
to aid each other. The overall inductance in
henrys is
(a) 2
(b) 1
1
(c)
(d) Zero
2
7.10 The impedance Z as shown in Fig.7.113 is
7.9
j5 Ω
j2 Ω
j 10 Ω
j10 Ω
j2 Ω
Fig. 7.113
(a)
(c)
j 29 Ω
j19 Ω
(b)
(d)
j9 Ω
j39 Ω
Answers to Objective-Type Questions 7.55
Answers to Objective-Type Questions
1. (c)
7. (c)
2.
8.
(b)
(b)
3.
9.
(c)
(b)
4. (d)
10. (b)
5.
(d)
6.
(c)
8
8.1
Three-Phase
Circuits
INTRODUCTION
A system which generates a single alternating voltage and current is termed a single-phase system. It utilises
only one winding. A polyphase system utilises more than one winding. It will produce as many induced
voltages as the number of windings.
A three-phase system consists of three separate but identical windings displaced by 120 electrical degrees
from each other as shown in Fig. 8.1. When these three windings are rotated in an anticlockwise direction
with constant angular velocity in a uniform magnetic field, the emfs induced in each winding have the same
magnitude and frequency but are displaced 120° from one another.
w rad/s
R
Y1
B1
S
N
Y
B
R1
Fig. 8.1 Three-phase system
The instantaneous values of generated voltage in windings RR1, YY1 and BB1 are given by
eR
eY
eB
Em sin θ
Em sin (θ − 120°)
Em sin (θ − 240°)
where Em is the maximum value of the induced voltage in each winding. The waveforms of these three
voltages are shown in Fig. 8.2.
e
eR
eY
eB
q
0
Fig. 8.2 Voltage waveforms
8.2 Network Analysis and Synthesis
Figure 8.3 shows the phasor diagram of these three induced voltages.
ER
120°
120°
120°
EB
EY
Fig. 8.3 Phasor diagram
8.2
ADVANTAGES OF A THREE-PHASE SYSTEM
1. In a single-phase system, the instantaneous power is fluctuating in nature. However, in a three-phase
system, it is constant at all times.
2. The output of a three-phase system is greater than that of a single-phase system.
3. Transmission and distribution of a three-phase system is cheaper than that of a single-phase
system.
4. Three-phase motors are more efficient and have higher power factors than single-phase motors of
the same frequency.
5. Three-phase motors are self-starting whereas single-phase motors are not self-starting.
8.3
SOME DEFINITIONS
1. Phase Sequence The sequence in which the voltages in the three phases reach the maximum positive
value is called the phase sequence or phase order. From the phasor diagram of a three-phase system, it is
clear that the voltage in the coil R attains maximum positive value first, next in the coil Y and then in the
coil B. Hence, the phase sequence is R-Y-B.
2. Phase Voltage The voltage induced in each winding is called the phase voltage.
3. Phase Current The current flowing through each winding is called the phase current.
4. Line Voltage The voltage available between any pair of terminals or lines is called the line voltage.
5. Line Current The current flowing through each line is called the line current.
6. Symmetrical or Balanced System A three-phase system is said to be balanced if the
(a) voltages in the three phases are equal in magnitude and differ in phase from one another by 120°, and
(b) currents in the three phases are equal in magnitude and differ in phase from one another by 120°.
7. Balanced Load The load is said to be balanced if loads connected across the three phases are identical,
i.e., all the loads have the same magnitude and power factor.
8.4
INTERCONNECTION OF THREE PHASES
In a three-phase system, there are three windings. Each winding has two terminals, viz., ‘start’ and ‘finish’. If
a separate load is connected across each winding as shown in Fig. 8.4, six conductors are required to transmit
and distribute power. This will make the system complicated and expensive.
8.5 Star, or WYE, Connection 8.3
F
Load
S
F
Load
S
F
Load
S
Fig. 8.4 Non-interlinked three-phase system
In order to reduce the number of conductors, the three windings are connected in the following two
ways:
1. Star, or Wye, connection
2. Delta, or Mesh, connection
8.5
STAR, OR WYE, CONNECTION
In this method, similar terminals (start or finish) of the three windings are joined together as shown in
Fig. 8.5. The common point is called star or neutral point.
Figure 8.6 shows a three-phase system in star connection.
R
IR
R
S
Zph
F
F
S
IR + IY + IB
O
N
F
Y
S
B
Fig. 8.5
N
Zph
Three-phase star connection
Fig. 8.6
IY
Y
IB
B
Zph
Three-phase, four-wire system
This system is called a three-phase, four-wire system. If three identical loads are connected to each phase,
the current flowing through the neutral wire is the sum of the three currents IR, IY and IB. Since the impedances
are identical, the three currents are equal in magnitude but differ in phase from one another by 120°.
iR
iY
iR
iB
iY + iB
I m sin θ
I m sin (θ − 120°)
I m sin(θ − 240°)
I m sin θ + I m i (
°) I m sin (θ − 240°) = 0
8.4 Network Analysis and Synthesis
Hence, the neutral wire can be removed without any way affecting the voltages or currents in the circuit
as shown in Fig. 8.7. This constitutes a three-phase, three-wire system. If the load is not balanced, the neutral
wire carries some current.
IR
R
Zph
N
O
Zph
IY
Y
IB
B
Zph
Fig. 8.7 Three-phase, three-wire system
8.6
DELTA, OR MESH, CONNECTION
In this method, dissimilar terminals of the three windings are
joined together, i.e., the ‘finish’ terminal of one winding is connected to the ‘start’ terminal of the other winding, and so on,
as shown in Fig. 8.8. This system is also called a three-phase,
three-wire system.
For a balanced system, the sum of the three-phase voltages
round the closed mesh is zero. The three emfs are equal in
magnitude but differ in phase from one another by 120°.
R
eY
eB
eR + eY + eB
8.7
8.7.1
R
F
S
F
S
Y
F
B
Three-phase delta connection
Fig. 8.8
Em sin θ
Em sin (θ − 120°)
Em sin (θ − 240°)
Em si θ + Em i (
°) Em sin (θ − 240°) = 0
VOLTAGE, CURRENT AND POWER RELATIONS IN A BALANCED
STAR-CONNECTED LOAD
Relation between Line Voltage and Phase Voltage
Figure 8.9 shows a balanced star-connected load.
Since the system is balanced, the three-phase
voltages VRN, VYN, and VBN are equal in magnitude and
differ in phase from one another by 120°.
VRN VYN = VBN V ph
where V ph indicates the rms value of phase voltage.
Let
IR
R
IY
VYB = VBR
IR
IB
Y
VBR
Zph
N
IY
VYB
IB
VYN = V pph ∠− 120°
VBN = V pph ∠− 240°
VRY
Zph
VRY
VRN = V pph ∠0°
Let
S
B
VL
Fig. 8.9 Star connection
Zph
8.7 Voltage, Current and Power Relations in a Balanced Star-Connected Load 8.5
where VL indicates the rms value of line voltage.
Applying Kirchhoff’s voltage law,
VRN + VNY
VRY
= VRN
VYN
= V ph ∠0° V pph ∠− 120°
=(
p
+
) − (−
( .5 V p − j 0.
p
)
= 1.5 V p + j 0.866 V ph
= 3 V ph ∠30°
Similarly,
VYB
VYN + VNB = 3 V pph ∠30°
VBR
VBN + VNR = 3 V pph ∠30°
Thus, in a star-connected, three-phase system, VL
voltages by 30°.
8.7.2
3 V ph and line voltages lead respective phase
Relation between Line Current and Phase Current
From Fig. 8.9, it is clear that line current is equal to the phase current.
IL = Iph
8.7.3
Phasor Diagram (Lagging Power Factor)
Figure 8.10 shows the phasor diagram of a balanced star-connected inductive load.
VRY
VRN
VNY
f
IB f
VBN
VNB
IR
30°
f
30°
VYB
30°
IY
VYN
VNR
VBR
Fig. 8.10 Phasor diagram
8.7.4
Power
The total power in a three-phase system is the sum of powers in the three phases. For a balanced load, the
power consumed in each load phase is the same.
8.6 Network Analysis and Synthesis
Total active power P = 3 × power in each phase = 3 Vph Iph cos f
In a star-connected, three-phase system,
V ph =
VL
I ph
IL
3
P = 3×
VL
3
φ
× IL
3 VL I L cos φ
where f is the phase difference between phase voltage and corresponding phase current.
Similarly, total reactive power Q = 3Vph Iph sin f = 3 VL I L sin φ
Total apparent power S
V phh I ph
3 VL I L
8.8
VOLTAGE, CURRENT AND POWER RELATIONS IN
A BALANCED DELTA-CONNECTED LOAD
8.8.1
Relation between Line Voltage and Phase Voltage
Figure 8.11 shows a balanced delta-connected load.
IR
R
R
IBR
VRY
Zph
Zph
Y
VYB
B
IRY
VBR
IB
Zph
B
IYB
Y
IY
Fig. 8.11 Delta connection
From Fig. 8.11, it is clear that line voltage is equal to phase voltage.
VL
8.8.2
V ph
Relation between Line Current and Phase Current
Since the system is balanced, the three-phase currents IRY, IYB, and IBR are equal in magnitude but differ in
phase from one another by 120°.
Let
I RY IYB = I BR I ph
where Iph indicates rms value of the phase current.
I RY = I ph ∠0°
IYB = I ph ∠− 120°
I BR = I ph ∠− 240°
Let
IR
IY = I B
IL
8.8 Voltage, Current and Power Relations in a Balanced Delta-Connected Load 8.7
where IL indicates rms value of the line current.
Applying Kirchhoff’s current law,
IR
I BBRR = I RY
I RY − I BR
IR
= I ph ∠0° I ph ∠− 240°
=(
p
+
) − (−
( .5 I p + j 0.
p
)
= 1.5 I p − j 0.866 I ph
= 3
Similarly,
ph
∠−
− 30°
IY
IYB
I RY = 3 I ph ∠− 30°
IB
I BR
IYB = 3 I ph
ph ∠− 30°
Thus, in a delta-connected, three-phase system, I L
phase currents by 30°.
8.8.3
3 I ph and line currents lag behind the respective
Phasor Diagram (Lagging Power Factor)
Figure 8.12 shows the phasor diagram of a balanced delta connected inductive load.
VRY
IR
IRY
−IYB
f
30°
IB
−IBR
30°
IBR f
−IRY
VBR
f
30°
IYB
IY
Fig. 8.12 Phasor diagram
8.8.4
Power
P 3V ph I ph cos φ
In a delta-connected, three-phase system,
V ph
I ph =
P
VL
IL
3
3 VL ×
IL
3
× cos φ = 3 VL I L cos φ
VYB
8.8 Network Analysis and Synthesis
Total reactive power Q
V phh I ph i φ
Total apparent
a
power S
V phh I ph
8.9
3 VL I L sin
i φ
3 VL I L
BALANCED Y/D AND D/Y CONVERSIONS
Any balanced star-connected system can be converted into the equivalent delta-connected system and vice versa.
For a balanced star-connected load,
Line voltage = VL
Line current = IL
Impedance/phase = ZY
V ph =
I ph
ZY =
VL
3
IL
V ph
I ph
=
VL
3 IL
For an equivalent delta-connected system, the line voltages and currents must have the same values as in
the star-connected system, i.e.,
Line voltage = VL
Line current = IL
Impedance/phase = ZΔ
V ph
I ph =
ZΔ =
ZY
VL
IL
3
V ph
I ph
=
VL
V
= 3 L = 3ZY
IL
IL
3
1
ZΔ
3
Thus, when three equal phase impedances are connected in delta, the equivalent star impedance is one-third
of the delta impedance.
8.10
RELATION BETWEEN POWER IN DELTA AND STAR SYSTEMS
Let a balanced load be connected in star having impedance per phase as Zph.
For a star-connected load,
V ph =
I ph =
IL
VL
3
V ph
Z ph
=
I ph =
VL
3 Z ph
VL
3 Z ph
8.11 Comparison between Star and Delta Connections 8.9
Now
PY
φ
3 VL I L
3 VL ×
VL
3Z ph
× cos φ =
VL2
cos φ
Z ph
For a delta-connected load,
Now
V ph
VL
I ph =
V ph
V
= L
Z ph Z ph
IL
3 I pphh = 3
VL
Z ph
PΔ
3 VL I L
φ
PY
3 VL × 3
VL
×c
Z ph
φ=3
VL2
cos φ = 3 PY
Z ph
1
PΔ
3
Thus, power consumed by a balanced star-connected load is one-third of that in the case of a deltaconnected load.
8.11
COMPARISON BETWEEN STAR AND DELTA CONNECTIONS
Star Connection
1. VL
2. I L
Delta Connection
1. VL
3 V ph
I ph
2. I L
V ph
3 I ph
3. Line voltage leads the respective phase 3. Line current lags behind the respective
voltage by 30°.
phase current by 30°
4. Power in star connection is one-third of 4. Power in delta connection is 3 times of
power in delta connection.
the power in star connection.
5. Three-phase, three-wire and three- 5. Only three-phase, three-wire system is
phase, four-wire systems are possible.
possible.
6. The phasor sum of all the phase currents 6. The phasor sum of all the phase voltages
is zero.
is zero.
Example 8.1 Three equal impedances, each of (8 + j10) ohms, are connected in star. This is further
connected to a 440 V, 50 Hz, three-phase supply. Calculate (a) phase voltage, (b) phase angle, (c) phase
current, (d) line current, (e) active power, and (f ) reactive power.
Solution
For a star-connected load,
(a) Phase voltage
Z ph = ( + j ) Ω
V ph =
VL
3
=
440
3
VL = 440 V,
= 254.03 V
f = 50 Hz
8.10 Network Analysis and Synthesis
(b) Phase angle
Z ph = 8 + j10 = 12.81 ∠51.34° Ω
Z ph =
.8 Ω
φ = 51.34°
(c) Phase current
I ph =
V ph
IL
I ph = 19.83 A
Z ph
=
254.03
= 19.83 A
12.81
(d) Line current
(e) Active power
P
(f)
3 VL I L
φ = 3 × 440 × 19.83 × cos (51.34°) = 9 44 kW
Reactive power
Q
VL I L i φ = 3 × 440 × 19.83 × sin (51.34°) = 11.81 kVAR
Example 8.2 A balanced delta-connected load of impedance (8 − j6) ohms per phase is connected to
a three-phase, 230 V, 50 Hz supply. Calculate (a) power factor, (b) line current, and (c) reactive power.
Solution
For a delta-connected load,
(a) Power factor
Z ph = ( − j ) Ω
VL = 230 V,
f = 50 Hz
Z ph = 8 − j 6 = 10 ∠ − 36.87° Ω
Z ph = 10 Ω
φ = 36.87°
pf = cos φ = cos (36.87°) = 0.8 (
)
(b) Line current
V ph
VL = 230 V
I ph =
V ph
IL
Z ph
=
230
= 23 A
10
3 I pphh = 3 × 23 = 39.84 A
(c) Reactive power
Q
VL I L i φ = 3 × 230 × 39.84 × sin (36.87°) = 9 52 kVAR
Example 8.3 Three coils, each having a resistance and an inductance of 8 W and 0.02 H respectively, are connected in star across a three-phase, 230 V, 50 Hz supply. Find the (a) power factor, (b) line
current, (c) power, (d) reactive volt-amperes, and (e) total volt-amperes.
Solution
R
8 Ω, L = 0.02 H, V
f = 50 Hz
8.11 Comparison between Star and Delta Connections 8.11
For a star-connected load,
(a) Power factor
=
×
×
Z ph =
2=
38.13° Ω
28
Z ph
.
= cos (38.13° = 0 786 (
pf
)
(b) Line current
V ph =
I ph
V
=
230
132 79 V
3
V ph 132 79
=
=
13.05 A
Z ph 10 17
= 13 A
3
(c) Power
×
×
.05 ×
×
×
05 ×
=
088 kW
(d) Reactive volt-amperes
=
° = 3 21 kVAR
(e) Total volt-amperes
= 3×
× 13
198 kVA
Example 8.4 Three coils, each having a resistance of 8 W and an inductance of 0.02 H, are connected
in delta to a three-phase, 400 V, 50 Hz supply. Calculate the (a) line current, and (b) power absorbed.
Solution
For a delta-connected load,
(a) Line current
R
50 Hz
= 400 V
= π×
R
Z ph
×
2=
38 13° Ω
28
L
Z ph
=
I ph =
I
°
V ph
Z ph
=
400
10.17
=
39.33 A
×
=
12 A
×
×
(b) Power absorbed
=
.12 ×
°
37 12 kW
8.12 Network Analysis and Synthesis
Example 8.5 The three equal impedances of each of 10 Æ 60°W, are connected in star across a
three-phase, 400 V, 50 Hz supply. Calculate the (a) line voltage and phase voltage, (b) power factor and
active power consumed. (c) If the same three impedances are connected in delta to the same source of supply,
what is the active power consumed?
Solution
Z ph = 10 ∠60° Ω, VL
f = 50 Hz
For a star-connected load,
(a) Line voltage and phase voltage
VL = 400 V
V
400
V ph = L =
= 230.94 V
3
3
(b) Power factor and active power consumed
φ = 60°
pf = cos φ = cos (60°) = 0.5 (lagging)
I ph =
IL
P
V ph
Z ph
230.94
= 23.094 A
10
=
I ph = 23.094 A
φ = 3 × 400 × 23.094 × 0 5 = 8 kW
3 VL I L
(c) Active power consumed for delta-connected load
VL = 400 V
Z ph = 10 Ω
V ph
VL = 400 V
I ph =
V ph
Z ph
400
= 40 A
10
=
IL
3 I pphh = 3 × 40 = 69.28 A
P
3 VL I L
φ = 3 × 400 × 69.28
28 c (60
(60 )
Example 8.6
24 kW
Three similar coils A,B, and C are available. Each coil has a 9 W resistance and a 12 W
reactance. They are connected in delta to a three-phase, 440 V, 50 Hz supply. Calculate for this load, the (a)
phase current, (b) line current, (c) power factor, (d) total kVA, (e) active power, and (f) reactive power. If
these coils are connected in star across the same supply, calculate all the above quantities.
Solution
R
9 Ω,
XL =
Ω
V
For a delta-connected load,
(a) Phase current
VL
V ph = 440 V
Z ph = R
jjX
Z ph = 15 Ω
9
j12 = 15 ∠53 13° Ω
f = 50 Hz
8.11 Comparison between Star and Delta Connections 8.13
φ = 53.13°
V ph 440
I ph =
=
= 29.33 A
Z ph
15
(b) Line current
IL
(c) Power factor
3 I pphh = 3 × 29.33 = 50.8 A
pf = cos φ = cos (
.13°) = .6 (lagging)
(d) Total kVA
VL I L = 3 × 440 × 50.8 = 38
38.71 kVA
S
(e) Active power
P
(f)
φ = 3 × 440 × 50.8 × 0.6 = 23
23.23 kW
3 VL I L
Reactive power
VL I L i φ = 3 × 440 × 50.8 × in (53.13°°) = 30 97 kVAR
Q
If these coils are connected in star across the same supply,
(a) Phase current
VL = 440 V
Z ph = 5 Ω
V ph =
VL
I ph =
V ph
=
3
Z ph
440
= 254.03 V
3
254.03
=
= 16.94 A
15
(b) Line current
IL
I ph = 16.94 A
(c) Power factor
pf = 0.6 (lagging)
(d) Total kVA
S
VL I L = 3 × 440 × 16.94 = 12.91 kVA
(e) Active power
P
(f)
3 VL I L
φ = 3 × 440 × 16.94 × 0.6 = 7.74 kW
Reactive power
Q
VL I L i φ = 3 × 440 × 16.94 × sin (53.13°) = 12.33 kVAR
Example 8.7
A 415 V, 50 Hz, three-phase voltage is applied to three star-connected identical
impedances. Each impedance consists of a resistance of 15 W , a capacitance of 177 μF and an inductance of
0.1 henry in series. Find the (a) power factor, (b) phase current, (c) line current, (d) active power, (e) reactive power, and (f) total VA. Draw a neat phasor diagram. If the same impedances are connected in delta,
find the (a) line current, and (b) power consumed.
Solution
V
f = 50 Hz, R
Ω
C = 177
1 μF, L = 0.1 H
8.14 Network Analysis and Synthesis
For a star-connected load,
(a) Power factor
ffL = 2π × 50 × 0.1 = 3 .
X
XC =
Ω
1
1
=
= 17.98 Ω
2π fC 2π × 50 × 177 × 10 −6
Z ph = R
jjX L
Z ph = 0.
jjX
XC
j 31.42 − j17 98 15
15
j13.44 = 20.14 ∠41.86° Ω
Ω
φ = 41.86°
pf = cos φ = cos ( 41.86°) = 0.744 (lagging)
(b) Phase current
V ph =
VL
I ph =
V ph
=
3
Z ph
415
= 239.6 V
3
239.6
=
= 11.9 A
20.14
(c) Line current
I ph = 11.9 A
IL
(d) Active power
P
φ = 3 × 415 × 11.9 × 0.744 = 6.36 kW
3 VL I L
(e) Reactive power
(f)
Q
VL I L i φ = 3 × 415 × 11.9 × in ( 41.86°°) = 5.71 kVAR
S
VL I L = 3 × 415 × 11.9 = 8.55 kVA
Total VA
The phasor diagram is shown in Fig. 8.13.
VRY
VRN
VNY
IR
30°
VNB
f
VYB
IB
30°
f
f
30°
IY
VBN
VYN
VNR
VBR
Fig. 8.13
f = 41.86°
8.11 Comparison between Star and Delta Connections 8.15
If the same impedances are connected in delta,
(a) Line current
VL V ph = 415 V
Z ph = 0. Ω
I ph =
IL
V ph
Z ph
=
415
= 20.61 A
20.14
3 I ph
ph = 3 × 20.61 = 35.69 A
(b) Power consumed
P
3 VL I L
φ = 3 × 415 × 35.69 × 0.774 = 19.09 kW
Example 8.8 Each phase of a delta-connected load consists of a 50 mH inductor in series with a parallel combination of a 50 W resistor and a 50 μF capacitor. The load is connected to a three-phase, 550 V,
800 rad/s ac supply. Find the (a) power factor, (b) phase current, (c) the line current, (d) power drawn, (e)
reactive power, and (f) kVA rating of the load.
Solution
L
50
0
, R = 500 Ω
C
μF, VL = 5500 V, ω = 800 rad/s
For a delta-connected load,
(a) Power factor
ω L = 800 × 50 × 10 −3 = 0 Ω
1
1
XC =
=
= 25 Ω
ωC 800 × 50 × 10 −6
R( jX
j C)
( − j 25)
Z ph = jX L +
= j 0+
= 10 + j 20 = 22.36 ∠63.43° Ω
R jX
j C
50 − j 25
Z ph = .36 Ω
φ = 63.43°
pf = cos φ = cos (63.43°) = 0.447 (
)
XL
(b) Phase current
VL
I ph
V ph = 550 V
V ph
550
=
=
= 24.6 A
Z ph 22.36
(c) Line current
IL
3 I pphh = 3 × 24.6 = 42
42.61 A
P
3 VL I L
(d) Power drawn
φ = 3 × 550 × 42.61 × 0.447 = 18.14 kW
(e) Reactive power
(f)
Q
VL I L i φ = 3 × 550 × 42.61 × sin (63.43°) = 36.3 kVAR
S
VL I L = 3 × 550 × 42.61 = 40.59 kVA
kVA rating of the load
8.16 Network Analysis and Synthesis
Example 8.9 A balanced star-connected load is supplied from a symmetrical three-phase 400 volts,
50 Hz system. The current in each phase is 30 A and lags 30° behind the phase voltage. Find the (a) phase
voltage, (b) resistance and reactance per phase, (c) load inductance per phase, and (d) total power consumed.
Solution
f = 50 Hz, I ph = 30 A, φ = 30°
VL
For a star-connected load,
(a) Phase voltage
V ph =
VL
3
=
400
3
= 230.94 V
(b) Resistance and reactance per phase
Z ph
V ph
230.94
=
=77Ω
I ph
30
= Z ph ∠φ = 7.7∠30° = (6.67 + j 3.85) Ω
Z ph =
R ph = 6 67 Ω
X ph = 3 85 Ω
(c) Load inductance per phase
Xp
ffL
L ph
3 85 = 2π × 50 × L ph
L ph = 0.01225 H
(d) Total power consumed
P
φ = 3 × 230.94 × 30 × cos (30°) = 18 kW
3V phh I phh
Example 8.10
A symmetrical three-phase 400 V system supplies a basic load of 0.8 lagging power
factor and is connected in star. If the line current is 34.64 A, find the (a) impedance, (b) resistance and reactance per phase, (c) total power, and (d) total reactive voltamperes.
Solution
VL
400
V ph =
VL
pf
0.8 (lagging), I L = 34.64 A
For a star-connected load,
(a) Impedance
I ph
Z ph
=
400
= 230.94 V
3
3
I L = 34.64 A
V ph 230.94
=
=
= 6.67
6 Ω
I ph
34.64
(b) Resistance and reactance per phase
pf = cos φ = 0.8 (lagging)
φ = cos −1 (0.. ) = 36.87°
Z ph = Z ph ∠φ = 6.67∠36.87° = (5.33
j 4) Ω
8.11 Comparison between Star and Delta Connections 8.17
R ph = 5 33 Ω
X ph = Ω
(c) Total power
P
φ = 3 × 400 × 34.64 × 0.8 = 19.19 kW
3 VL I L
(d) Total reactive volt-amperes
VL I L i φ = 3 × 400 × 34.64 × sin (36.87 ) = 14.4 kVAR
Q
Example 8.11 A balanced star-connected load is supplied by a 415 V, 50 Hz three-phase system.
Current in each phase is 20 A and lags 30° behind its phase voltage. Find the (a) phase voltage, (b) power,
and (c) circuit parameters. Also, find power consumed when the same load is connected in delta across the
same supply.
Solution
f = 50 Hz, I ph = 20 A, φ = 30°
VL
For a star-connected load,
(a) Phase voltage
V ph =
VL
=
3
415
3
= 239.6 V
(b) Power
IL
I ph = 20 A
P
3 VL I L
φ = 3 × 415 × 20 × cos (30 ) = 12.45 kW
(c) Circuit parameters
Z ph =
V ph
I ph
=
239.6
=
20
Z ph = Z p ∠φ
.98 Ω
11.98∠30
j 6) Ω
R ph = 0.37 Ω
X ph = 6 Ω
Since
Xp
6
fL ph
fL
2π × 50 × L ph
L ph = 19.1 mH
(d) Power consumed by same delta load across the same supply
PΔ
3PY = 3 × 12.45 × 103 = 37.35 kW
Example 8.12
Three identical coils connected in delta to a 440 V, three-phase supply take a total
power of 50 kW and a line current of 90 A. Find the (a) phase current, (b) power factor, and (c) apparent
power taken by the coils.
Solution
VL
440
P = 50 kW, I L = 90 A
8.18 Network Analysis and Synthesis
For a delta-connected load,
(a) Phase current
I ph =
IL
=
3
90
= 51.96 A
3
(b) Power factor
3 VL I L cos φ
P
50 × 103 = 3 × 440 × 90 × cos φ
pf = cos φ = 0.73 (
)
(c) Apparent power
VL I L = 3 × 440 × 90 = 68.59 kVA
S
Example 8.13 Three similar choke coils are connected in star to a three-phase supply. If the line
current is 15 A, the total power consumed is 11 kW and the volt-ampere input is 15 kVA, find the line and
phase voltages, the VAR input and the reactance and resistance of each coil. If these coils are now connected
in delta to the same supply, calculate phase and line currents, active and reactive power.
Solution
15 A, P = 11 kW, S = 15 kVA
IL
For a star-connected load,
(a) Line voltage
S
VL I L
3
15 × 10 = 3 × VL × 15
VL = 577.35 V
(b) Phase voltage
V ph =
VL
=
3
577.35
3
= 333.33 V
(c) VAR input
P 11 × 103
S 15 × 103
φ = 42.86°
cos φ
Q=
VL I L sin
i φ
0.733
3 × 577.35 × 15 × si
s n ( 42.86°) = 10.2 kVAR
(d) Reactance and resistance of coil
I ph
I L = 15 A
Z ph =
V ph
I ph
=
333.33
=
15
.
Ω
R = Z phh cos φ = 22.22 × 0.733 = 16.29 Ω
XL
If these coils are now connected in delta,
Z ph sin φ = 22.22 × sin ( 42.
) = 155 11 Ω
8.11 Comparison between Star and Delta Connections 8.19
(a) Phase current
V ph VL = 577.35 V
Z ph = . Ω
V ph 577.35
I ph =
=
= 25.98 A
Z ph
22.22
(b) Line current
IL
3 I pphh = 3 × 25.98 = 45 A
(c) Active power
P
3 VL I L
φ = 3 × 577.35 × 45 × 0.733 = 32.98 kW
(d) Reactive power
VL I L i φ = 3 × 577.35 × 45 × sin ( 42.86 ) = 30.61 kVAR
Q
Example 8.14
A three-phase, star-connected source feeds 1500 kW at 0.85 power factor lag to a
balanced mesh-connected load. Calculate the current, its active and reactive components in each phase of
the source and the load. The line voltage is 2.2 kV.
Solution
P
1 00 k , pf
P
3 VL I L cos φ
0 8 (lagging), VL = 2.2 kV
For a mesh or delta-connected load,
(a) Line current
3
1500 × 10 = 3 × 2.2 × 103 × I L × 0.85
I L = 463.12 A
(b) Active component of current in each phase of the load
I ph =
IL
=
463.12
= 267.38 A
3
3
I ph cos φ = 267.38 × 0.85 = 227.27 A
(c) Reactive component of current in each phase of the load
I ph sin φ = 267.38 × sin (cos −1 0.85)
.38 × 0.526 = 140.85 A
For a star-connected source, the phase current in the source will be the same as the line current drawn
by the load.
(d) Active component of this current in each phase of the source
I L cos φ = 463.12 × 0.85 = 393.65 A
(e) Reactive component of this current in each phase of the source
I L sin φ = 463.12 × 0.526 = 243.6 A
Example 8.15
A three-phase, 208-volt generator supplies a total of 1800 W at a line current of 10 A
when three identical impedances are arranged in a wye connection across the line terminals of the generator.
Compute the resistive and reactive components of each phase impedance.
Solution
VL
208
P = 1800 W
I L = 10 A
8.20 Network Analysis and Synthesis
For a wye-connected load,
VL
V ph =
=
208
3
3
I L = 10 A
I ph
= 120.09 V
Z ph =
V ph 120.09
=
=
I ph
10
P
3 VL I L cos φ
Ω
1800 = 3 × 208 × 10 × cos φ
cos φ = 0.5
φ = 60°
R ph = Z ph cos φ = 12 × 0.5 = 6 Ω
X ph = Z ph sin φ = 12 × sin (60°) = 10.39 Ω
Example 8.16
A balanced, three-phase, star-connected load of 100 kW takes a leading current of
80 A, when connected across a three-phase, 1100 V, 50 Hz supply. Find the circuit constants of the load per
phase.
Solution
P
100
00 kW, I L = 80 A
f = 500 Hz
V
For a star-connected load,
V ph =
I ph
Z ph =
P
VL
=
1100
=
635.08
=79 Ω
80
3
3
I L = 80 A
V ph
I ph
= 635.08 V
3 VL I L cos φ
100 × 10 = 3 × 1100 × 80 × cos φ
cos φ = 0.656 (leading)
φ = 49°
3
R ph = Z ph cos φ = 7.94 × 0.656 = 5.21 Ω
X ph = Z ph sin φ = 7.94 × sin ( 49°) = 6 Ω
This reactance will be capacitive in nature as the current is leading.
1
2π fC
1
6=
2π × 50 × C
C = 530.52 μF
XC =
8.11 Comparison between Star and Delta Connections 8.21
Example 8.17 Three identical impedances are connected in delta to a three-phase supply of 400 V.
The line current is 34.65 A, and the total power taken from the supply is 14.4 kW. Calculate the resistance
and reactance values of each impedance.
Solution
I L = 34.65
65 A, P = 14.4 kW
VL
400
VL
V ph = 400 V
For a delta-connected load,
I ph =
IL
Z ph =
V ph
I ph
P
3
=
34.65
= 20 A
3
400
=
= 20 Ω
20
3 VL I L cos φ
14.4 × 10 = 3 × 400 × 34.65 × cos φ
cos φ = 0.6
φ = 53.13°
R ph = Z ph cos φ = 20 × 0.6 = 12 Ω
3
X ph = Z ph sin φ = 20 × sin (53.13°) = 16 Ω
Example 8.18 A balanced, three-phase load connected in delta draws a power of 10.44 kW at 200 V
at a power factor of 0.5 lead. Find the values of the circuit elements and the reactive volt-amperes drawn.
Solution
P
10 44 k , VL = 200 V, pf = 0.5 (
)
For a delta-connected load,
(a) Values of circuit elements
VL
P
V ph = 200 V
3 VL I L cos φ
3
10.44 × 10 = 3 × 200 × I L × 0.5
I L = 60.28 A
I
60.28
I ph = L =
= 34.8 A
3
3
V ph 200
Z ph =
=
= 5 75 Ω
I ph 34.8
R ph
Z phh cos φ = 5.75 × 0.5 = 2.875 Ω
X ph
Z phh sin φ = 5.75 × sin
in (cos
( −1 0.5) = 5.747 × 0.866 = 4 98 Ω
(b) Reactive volt-amperes drawn
Q
VL I L i φ = 3 × 200 × 60.28 × 0.866 = 18.08 kVAR
8.22 Network Analysis and Synthesis
Example 8.19 Each leg of a balanced, delta-connected load consists of a 7 W resistance in series
with a 4 W inductive reactance. The line-to-line voltages are
2360 0 ° V , Ebc = 2360 ∠ 120° V , Eca = 2360 ∠120° V
Eab
Determine
(a) phase current Iab, Ibc, and Ica both magnitude and phase)
(b) each line current and its associated phase angle
(c) the load power factor
Solution
7 Ω,
R
XL =
Ω
VL = 2360 V
For a delta-connected load,
(a) Phase current
V ph VL = 2360 V
Z ph = 7 + j 4 = 8.06 ∠29.74° Ω
Eab
2360 ∠0°
=
= 292.8∠
∠− 29.74° A
Z ph 8.06 ∠29.74°
E
2360 ∠− 120°
= bc =
= 292.8∠− 149.71° A
Z ph 8.06 29.74°
E
2360 ∠120°
= ca =
= 292.8 90.26° A
Z ph 8 06 29 74°
Iab =
Ibc
Ica
(b) Line current
In a delta-connected, three-phase system, line currents lag behind respective phase currents by 30°.
IL
3 I pphh = 3 × 292.8 = 507.14 A
I La = 507.14 ∠ − 59.71°A
I Lb = 507.14 ∠−
∠ 179.71°A
I Lc = 507.14 ∠
∠60
60.26°A
(c) Load power factor
pf = cos ( 29.
) = 0.
(lagging)
Example 8.20
A three-phase, 200 kW, 50 Hz, delta-connected induction motor is supplied from a
three-phase, 440 V, 50 Hz supply system. The efficiency and power factor of the three-phase induction motor
are 91% and 0.86 respectively. Calculate (a) line currents, (b) currents in each phase of the motor, (c) active,
and (d) reactive components of phase current.
Solution
Po
200 kW
VL = 440 V,
For a delta-connected load (induction motor),
(a) Line current
η=
Output power Po
=
Input power
Pi
200 × 103
Pi
Pi = 219.78 kW
0 91 =
f = 50 Hz, η = 91%, pf = 0 86
8.11 Comparison between Star and Delta Connections 8.23
3 VL I L cos φ
Pi
3
219.78 × 10 = 3 × 440 × I L × 0.86
I L = 335.3 A
(b) Currents in each phase of motor
I ph =
IL
3
=
335.3
3
= 193.6 A
(c) Active component of phase current
I ph cos φ = 193.6 × 0.86 = 166.5 A
(d) Reactive component of phase current
I ph sin φ = 193.6 × sin
in (cos
( −1 0.86)
.6 × 0.51 = 98.7 A
Example 8.21 A three-phase, 400 V, star-connected alternator supplies a three-phase, 112 kW,
mesh-connected induction motor of efficiency and power factor 0.88 and 0.86 respectively. Find the (a) current in each motor phase, (b) current in each alternator phase, and (c) active and reactive components of
current in each motor phase, (d) active and reactive components of current in each alternator phase.
Solution
VL
Po = 112 kW, η = 0 88
400
pf = 0.86
For a mesh-connected load (induction motor),
(a) Current in each motor phase
V ph
η=
VL = 400 V
Output power Po
=
Input power
Pi
112 × 103
Pi
Pi = 127.27 kW
0 88 =
Pi
3 VL I L cos φ
3
127.27 × 10 = 3 × 400 × I L × 0.86
I L = 213.6 A
I
213.6
I ph = L =
= 123.32 A
3
3
Current in a star-connected alternator phase will be same as the line current drawn by the motor.
(b) Current in each alternator phase
I L = 213.6 A
(c) Active component of current in each phase of a motor
I ph cos φ = 123.32 × 0.86 = 105.06 A
Reactive component of current in each phase of the motor
I ph sin φ = 123.32 × sin (cos −1 0.86)
.32 × 0.51 = 62.89 A
8.24 Network Analysis and Synthesis
(d) Active component of current in each alternator phase
I L cos φ = 213.6 × 0.86 = 183.7 A
Reactive component of current in each alternator phase
I L sin φ = 213.6 × sin
in (cos
( −1 0.86)
.6 × 0.51 = 108.94 A
Example 8.22
Three similar resistors are connected in star across 400 V, three-phase lines. The
line current is 5 A. Calculate the value of each resistor. To what value should the line voltage be changed to
obtain the same line current with the resistors connected in delta?
Solution
VL
IL = 5 A
400
For a star-connected load,
V ph =
I ph
Z ph
VL
=
400
= 230.94 V
3
3
IL = 5 A
V ph 230.94
= R phh =
=
= 46.19 Ω
I ph
5
For a delta-connected load,
IL = 5 A
R ph = 6. 9 Ω
I
5
I ph = L =
A
3
3
5
V ph I phh R phh =
× 46.19 = 133.33 V
3
VL = 133.33 V
Voltage needed is one-third of the star value.
Example 8.23
Three 100 W, non-inductive resistors are connected in (a) star, and (b) delta across
a 400 V, 50 Hz, three-phase supply. Calculate the power taken from the supply in each case. If one of the
resistors is open-circuited, what would be the value of total power taken from the mains in each of the two
cases?
Solution
VL
For a star-connected load,
V ph =
Z ph = 100 Ω
400
VL
400
= 230.94 V
3
V ph 230.94
I ph =
=
= 2 31 A
Z ph
100
I L I ph = 2 31 A
3
=
cos φ = 1
P = 3 VL I L cos φ = 3 × 400 × 2.31 × 1 = 1600.41 W
( For pure resistor, pf = 1)
8.11 Comparison between Star and Delta Connections 8.25
For a delta-connected load,
V ph
VL = 400 V
I ph =
V ph 400
=
=4A
Z ph 100
IL
3 I pphh = 3 × 4 = 6.93 A
P
3 VL I L
φ = 3 × 400 × 6.93 × 1 = 4801.24 W
A
When One of the Resistors is Open-circuited
100 Ω
400 V
1. Star Connection The circuit consists of two 100 Ω resistors
in series across a 400 V supply as shown in Fig. 8.14.
100 Ω
C
400
Currents in lines A and
dC
A
200
Power taken from the mains 400
0 × 2=800 W
B
Fig. 8.14 Delta connection
Hence, when one of the resistors is open circuited, the power
consumption is reduced by half.
2. Delta Connection In this case, currents in A and C remains as
usual 120° out of phase with each other as shown in Fig. 8.15.
A
100 Ω
400 V
C
100 Ω
400
Currents in each phase =
=4A
100
Power taken from the mains 2 × 4 × 400 = 3200 W
B
Fig. 8.15 Delta connection
Hence, when one of the resistors is open-circuited, the power consumption is reduced by one-third.
Three identical impedances of 10 ∠30° Ω each are connected in star and another
set of three identical impedances of 18 ∠60° Ω are connected in delta. If both the sets of impedances are
connected across a balanced, three-phase 400 V, supply, find the line current, total volt-amperes, active
power and reactive power.
Example 8.24
Solution
ZY
10 ∠30° Ω
Z = 18∠
Ω
VL = 400 V
Three identical delta impedances can be converted into equivalent star impedances.
Z′Y =
Z Δ 18 ∠60°
=
= 6 ∠60° Ω
3
3
Now two star-connected impedances of 10 ∠30 Ω and 6 ∠60 Ω are connected in parallel across a threephase supply.
Zeq =
( ∠ )(6
)(6 ∠60°)
= 3.87∠48.83° Ω
10 ∠30° + 6 ∠60°
8.26 Network Analysis and Synthesis
For a star-connected load,
(a) Line current
V
=
I ph =
IL =
V
3
V ph
Z ph
ph
400
=
=
3
V ph
= 230 94 V
230 94
3 87
Z
59 67 A
59.67 A
(b) Total volt-amperes
3×
× 59.
41.34 kVA
(c) Active power
=
×
×
.67 ×
= 27 21 kW
=
×
×
.67 ×
= 31 12 kVAR
(d) Reactive power
Example 8.25
Three star-connected impedances ZY =
+ j37.7
per phase are connected in
parallel with three delta-connected impedances Z Δ
−
per phase. The line voltage is 398 V.
Find the line current, pf, active and reactive power taken by the combination.
Solution
V
Z
39
Three identical delta-connected impedances can be converted into equivalent star impedances.
Z′Y =
162 ∠− 9 3
3
−
Now two star-connected impedances of 42
across the three-phase supply.
( .68
Zeq =
42 68
Ω
and 54.03∠− 79.3 Ω are connected in parallel
62 05
=
3°
9 88
For a star-connected load,
(a) Line current
V ph
I ph
IL
(b) Power factor
pf =
V
398
3
V ph
Z ph
ph
3
V ph
Zeq
3 36
= 229.79 V
229 79
68.33
= os .88 )
3 36 A
99 (lagging)
(c) Active power
×
×
×
×
×
6×
=
9 kW
(d) Reactive power
=
= 397.43 VAR
8.11 Comparison between Star and Delta Connections 8.27
Example 8.26 Three coils, each having a resistance of 20 W and a reactance of 15 W, are connected
in star to a 400 V, three-phase, 50 Hz supply. Calculate(a) line current, (b) power supplied, and (c) power
factor. If three capacitors, each of same capacitance, are connected in delta to the same supply so as to
form a parallel circuit with the above coils, calculate the capacitance of each capacitor to obtain a resultant
power factor of 0.95 lagging.
Solution
R ph
20 Ω,
X ph = 15 Ω, VL = 400 V
For a star-connected load,
(a) Line current
Z ph = R ph
p
V pph =
VL
I ph =
V ph
IL
jX ph
p
=
3
Z ph
20
j15 = 25 ∠36.87° Ω
400
= 230.94 V
3
230.94
=
= 9 24 A
25
I ph = 9 24 A
(b) Power supplied
P1
3 VL I L
φ1
3 × 400 × 9.24
s (36.87 )
5.12 kW
(c) Power factor
pf = cos φ1 = cos (
.87°) = .8 (lagging)
(d) Value of capacitance of each capacitor
Q
VL I L i φ1
3 × 400 × 9.24
i (36.87 )
3.84 kVAR
When capacitors are connected in delta to the same supply,
pf = 0 95
φ2 = cos −11 (0.95) = 18 19°
tan φ2 = tan (18.19°) = 0.33
Since capacitors do not absorb any power, power remains the same even when capacitors are connected.
But reactive power changes.
P2 = 5.12 kW
Q
P2 tan φ2 = 5.12 0.33 = 1.69 kVAR
Difference in reactive power is supplied by three capacitors.
From Fig. 8.16,
Q
Q
2.15 103
Q1 − Q2 = 3.84 − 1.69 = 2.15 kVAR
VL I L sin φ
P1 = P2 = P
f2
f1
Q2
S2
S1
3 × 400 × I L × sin (90 )
IL = 3 1 A
Fig. 8.16
Q1
8.28 Network Analysis and Synthesis
I ph =
IL
I ph =
V ph
C=
8.12
3
= 1 79 A
= Vp
XC
fC
I ph
Vp
f
=
1 79
= 14
1 .24 μF
400 × 2π × 50
THREE-PHASE UNBALANCED CIRCUITS
If the loads connected across the three phases are not identical to each other, i.e., the loads have different
magnitude and power factors, the loads are said to be unbalanced. The phase currents in delta and phase or
line current in star connection differ in unbalanced loading giving rise to flow of current, in a neutral wire.
I R + IY
IN
IB
There may be three cases of unbalanced loads:
(i) Unbalanced delta-connected load
(ii) Unbalanced three-wire star-connected load
(iii) Unbalanced four-wire star-connected load
8.12.1 Unbalanced Delta-connected Load
Figure 8.17 shows an unbalanced delta-connected load connected to a balanced three-phase supply.
For a delta-connected load,
Let
VL
V ph
VRRY = V ph ∠0°
VYB = V ph ∠− 120°
IR
R
VBR = V ph ∠− 240°
IBR
Phase currents are given by
V
= RRY
Z RY
Y
IYB =
VYYB
ZYB
B
I BR =
VBBR
Z BR
I RY
IY
ZRY
ZBR
ZYB
IB
IYB
Fig. 8.17 Unbalanced delta-connected load
Line currents are given by
IR
I RY − I BR
IY
IYB
I RY
IB
I BR
IYB
IRY
8.12 Three-Phase Unbalanced Circuits 8.29
8.12.2
Unbalanced Four-wire Star-connected Load
Figure 8.18 shows an unbalanced star-connected load
connected to a balanced 3-phase, 4-wire supply. The
neutral point N of the load is connected to the neutral
point O of the supply. Hence, voltage across the three
load impedances is equal to the phase voltage of the
supply.
IR
R
ZR
IN
N
For a star-connected load,
N
IY
VL
VRRN = V ph ∠0°
Let
ZB
Y
3 V ph
ZY
IB
VYN = V pph ∠− 120°
B
VBN = V pph ∠− 240°
Fig. 8.18
Phase currents are given by
Unbalanced four-wire
star-connected load
VRN
ZR
V
IY = YN
ZY
VBN
IB =
ZB
IR =
For a star-connected load, phase currents are equal to the line currents.
The current in a neutral wire is given by
IN
I R + IY
IB
8.12.3 Unbalanced Three-wire Star-connected Load
Figure 8.19 shows an unbalanced three-wire star-connected load.
R
ZR
N
O
ZB
ZY
Y
B
Fig. 8.19
Unbalanced three-wire star-connected load
If the neutral point N of the load is not connected to the neutral point O of the supply, there exists a voltage
between the supply neutral point and the load neutral point. The load phase voltage is not the same as the
8.30 Network Analysis and Synthesis
supply phase voltage. There are many methods to solve such unbalanced star-connected loads. Two most
commonly used methods are
(i) Star-delta transformation
(ii) Millman’s theorem,
In star-delta transformation, the star-connected loads, are replaced by an equivalent delta-connected
load.
Z R ZY
ZB
ZY Z B
= ZY + Z B +
ZR
Z Z
= ZB + ZR + B R
ZY
Z RY = Z R + ZY +
ZYB
Z BR
The problem is then solved as an unbalanced delta-connected load. The line currents so calculated are
equal to the line currents of the original star-connected load.
An unbalanced three-wire star-connected load can also be solved by Millman’s theorem. Let
VRO , VYO and VBO be the phase voltages of the supply which are equal in magnitude but differ in phase from
one another by 120°. Let VRN , VYN and VBN be the load phase voltages which are unequal in magnitude
as well as differ in phase by unequal angles. The voltage between the neutral points, N and O, i.e, VNO , is
calculated using Millman’s theorem.
VNO =
VRO YR + VYO YY + VBO YB
YR YY + YB
where YR YY and YB are the admittances of the branches of the unbalanced star-connected load. The voltage
across phase impedance, i.e., load phase voltage can be calculated as,
VRN
VYN
VRO − VNO
VYO − VNO
VBN
VBO − VNO
The phase currents in the load are given by
VRN
ZR
V
IY = YN
ZY
V
I B = BR
ZB
IR =
The line currents are equal to the phase currents for a star-connected load.
Example 8.27
A three-phase supply with a line voltage of 250 V has an unbalanced delta-connected
load as shown in Fig. 8.20.
8.12 Three-Phase Unbalanced Circuits 8.31
IA
A
A
V
CA
25 V
ZCA
25 ∠90° Ω
IB
B
AB = 20
∠0° Ω
IAB
IC
C
BC
B
C
ZBC 15 ∠20° Ω
Fig. 8.20
Determine (a) phase currents, (b) line currents, (c)
power, if phase sequence is ABC.
=
Solution Let
Z
=1
Z
∠−
total active power and (d)
°
total reactive
° V,
5 90° Ω
20
(a) Phase currents
VAB 250 ∠0°
=
∠ °
Z AB
20 ∠0°
VBC 250 ∠− 120
.67∠− 140 A
Z BC
15 ∠20
V
250 ∠− 240°
=
=
= 10 ∠30°A
ZCA
25 ∠90°
I AB =
I BC
I
(b) Line currents
A
∠ °−
∠− 14
− .48°A
27.45∠− 157 02°
B
=
C
= 10 ∠30° −16.67∠− 140° = 2
0
∠36 25
(c) Total active power
=
×
=2
=
×
+
=
× co
13 kW
× os ( 20 )
× cos ( ) 0
+
3 92 kW
.05 kW
(d) Total reactive power
+
Example 8.28
=
×
× sin °) 0
= 250 × 16 67 ×
43 kVAR
=
× × sin ( ) 6.25 kVAR
= 0 + 1 43 +
.68 kVAR
In the circuit of Fig. 8.21, a 400 V, 50 Hz, 3-phase supply of phase sequence ABC is
supplied to a delta-connected load consisting of a 100 W resistor between lines A and B, a 378 mH inductor
between lines B and C, and a 37.8 μF capacitor between lines C and A. Determine phase and line currents.
8.32 Network Analysis and Synthesis
IA
A
ICA
VAB
B
VCA
B
ZCA
IB
ZAB
IAB
VBC
ZBC
IC
C
IBC
Fig. 8.21
VL = 400 V,
Solution
f = 50 Hz, R 100 Ω, L = 318 mH, C = 31.8 μF
VAB = 400 ∠0° V
VBC = 400 ∠ − 120° V
VCA = 400 ∠ − 240° V
Let
ffL = 2π × 50 × 318 × 10 −3 = 99.9 Ω
1
1
XC =
=
= 103. 35 Ω
2π fC 2π × 50 × 31 8 × 10 −6
X
Z AB = R = 100 ∠0 Ω
Z BC = jX L = j 99.9 = 99
99.9 ∠90° Ω
ZCA = − jX C = − j103.35 = 103.35
3 ∠− 90° Ω
(a) Phase currents
VAB 400 ∠0°
=
= 4 ∠0°A
Z AB 100 ∠0°
V
400 ∠− 120°
= BC =
= 4 ∠150°A
Z BC
99.9∠90°
V
400 ∠− 240°
= CA =
= 3 87∠− 150°A
ZCA 103.35∠− 90°
I AB =
I BC
ICA
(b) Line currents
I A I AB − ICA = 4 ∠0° − 3.87∠− 150° = 7.6 ∠14.75°°A
I B I BC − I AB = 4 ∠150° − 4 ∠0° = 7.73∠165°A
IC = ICA − I BC = 3.87
87
8 ∠−150°
1500° − 4 ∠150
150
50° = 3.94 ∠
∠− 88.36°A
Example 8.29
Z A (10
conductor.
A three-phase, 400 V, 4-wire system of Fig. 8.22 has a star-connected load with
j0 ) Ω , Z B = (15 + j10
j10 ) , ZC (0 + j5 ) Ω . Find the line currents and current through neutral
8.12 Three-Phase Unbalanced Circuits 8.33
IA
A
VL = 400
0V
ZA
IB
IN
B
N
ZC
ZB
IC
C
Fig. 8.22
Solution
VL = 400 V
Z A = (10 + j10) Ω = 10 ∠0 Ω
Z B = ( + j ) Ω = 18.03∠ .69 Ω
Z C = ( + j ) Ω = 5∠ 9 0 Ω
For a star-connected load,
V ph =
VL
3
=
400
3
= 230.94 V
The phase sequence is assumed as A-B-C.
VAN = 230.94 ∠0° V
VBN = 230.94 ∠− 120° V
VCN = 230.94 ∠− 240° V
Phase currents
VAN 230.94 ∠0°
=
= 23.09∠0°A
ZA
10 ∠0°
V
230.94 ∠− 120°
I B = BN =
= 12.81
8 ∠− 153.69°A
ZB
18.03 33 69°
V
230.94 ∠− 240°
IC = CN =
= 46.19∠30
3 °A
ZC
5 90°
IA =
Line currents are equal to the phase currents in a star-connected load.
I L1
I A = 23.09∠0°A
I L2
I B = 12.81∠ − 153.69°A
I L3
IC = 46.19∠30°A
Current through neutral conductor
IN
IA + IB
IC = 23.09∠0° + 12.81∠
∠− 153.69° + 46.19∠30° = 544.447∠18.65°A
8.34 Network Analysis and Synthesis
Example 8.30
A 3-phase, 4 wire, 208 V, CBA system as shown in Fig. 8.23 has a star-connected
load with Z A 5 0° Ω Z B = 5 30° Ω , ZC
current through neutral wire.
10
10
60° Ω . Obtain the phase currents, line currents and
IC
C
VL = 208
8V
ZC
IB
B
N
ZA
ZB
IA
A
Fig. 8.23
Solution
VL = 208 V
Z A = 5∠0 Ω, Z B = 5∠30° Ω
C
= 10 ∠− 60° Ω
For a star-connected load,
VL
V ph =
VCCN
VBN
VCN
=
208
= 120.09 V
3
3
= 120.09∠0°V
= 120.09∠− 120° V
= 120.09∠− 240° V
(a) Phase currents
VCN 120.09∠0°
=
= 12∠60° A
ZC
10 ∠− 60°
V
120.09∠− 120°
I B = BN =
= 24.02∠− 150° A
ZB
5∠30°
V
120.09∠− 240°
I A = AN =
= 24.02∠120° A
ZA
5∠0°
IC =
(b) Line currents are equal to the phase currents in a star-connected load.
I L1
IC = 12∠60°A
I L2
I B = 24.02∠− 150°A
I L3
I A = 24.02∠120°A
(c) Current through neutral wire
IN
IC + I B
I A = 12∠60° + 24.02∠
∠− 150° + 24.02∠120°
32.97∠144.42°° A
8.12 Three-Phase Unbalanced Circuits 8.35
Example 8.31 A symmetrical 440 V, 3-phase system supplied a star-connected load with the branch
j Ω , Z B = − j5 Ω as shown in Fig. 8.24. Calculate line currents and voltage
impedances Z R 10 Ω ZY = j5
across each phase impedance by Millman’s theorem. The phase sequence is RYB.
IR
R
ZR
VRY
O
Y
ZB
VBR
IY
N
ZY
VYB
IB
B
Fig. 8.24
Solution
V
10 Ω
440
j5 Ω
B
= − j5 Ω
For a star-connected load,
V ph =
VL
3
=
440
3
= 254.03 V
Let O be the neutral point of the supply system.
Let
VRO = 254.03∠0° V
VYO = 254.03∠ − 120° V
VBO = 254.03∠ − 240° V
YR =
1
1
1
=
=
= 0 1∠0° �
Z R 10 10 ∠0°
YY =
1
1
1
=
=
= 0 2∠− 90° �
ZY
j 5 5∠ 9 0°
YB =
1
1
1
=
=
= 0 2∠90° �
Z B − j 5 5∠− 90°
By Millman’s theorem,
VRO YR + VYO YY + YBO YB
YR YY + YB
(
.03∠0°)(0
)(0. ∠ ) + (
.03∠− 120°)(0.2∠
∠− 90
90°) + ( 254.03∠− 240°)(0.2∠90
90°)
=
0.1∠0° + 0.2∠
∠− 90° + 0.2 90°
= 625.96 ∠180° V
VNO =
8.36 Network Analysis and Synthesis
Voltage across phase impedance
− VNO
− VNO 254 0 ∠− 120
−
= 254.03∠− 24 °
0 V
=
29∠− 23 79 V
° 545 29 23 79
Phase currents/ line currents
IR
IY
IB
880 ∠0
= ∠0 A
10 ∠0°
545.29∠− .79
5∠90°
545. ∠23.79°
109
5
90°
VRN
ZR
VYN
ZY
VBN
ZB
−
79 A
113 79 A
Example 8.32
In the circuit of Fig. 8.25, a symmetrical 3-phase 100 V, three-wire supply feeds
an unbalanced star-connected load, with impedances of the load as Z
=
=4
. Find the line currents and voltage across the impedance using star-delta transformation
method.
IR
R
5Ω
N
−j4 Ω
IY
Y
j2 Ω
IB
C
Fig. 8.25
Solution
= ∠
V
∠− 90 Ω
The unbalanced star-connected load can be converted into equivalent delta-connected load by star-delta
transformation technique.
+
+
+
ZB
ZR
ZY
= ∠
+
= ∠
=4
−
9
The equivalent delta-connected load is shown in Fig. 8.26.
For a delta-connected load,
Let
VRY
= 100 V
0 V
∠
°
4 ∠− 90
∠
+
5∠0°
4 90
+
2 90°
38 66°
°
=
∠−
34° Ω
6.4 ∠− 141.34° Ω
8.12 Three-Phase Unbalanced Circuits 8.37
VYB =
VBR =
∠ 120° V
∠ 240° V
IR
R
IBR
Phase currents
I RY
IYB
I BR
100 ∠0°
∠38 66
100 ∠−
6 5 4
100 ∠− 240
6 4 ∠− 1 34°
VRY
Z RY
VYB
ZYB
VBR
Z BR
ZBR
IY
Y
38 66 A
ZRY
IRY
39 06 − 68 66° A
ZYB
IB
B
IYB
98 66° A
15
Fig. 8.26
Line currents
°
R
∠−
−
39.
−
−
°
.66
19 7
−
.
−
.
5
14° Α
36° Α
These line currents are equal to the line currents of the original star-connected load.
For a star-connected load,
I
I RN
IL
5 A
IYN
I BN
=
−121 14° A
128 36° A
= ∠
( 2∠9
= 4 90
− .
−
26 69∠
= 135.3∠− 8.65 V
39 4
1 14° V
36
.
V
Example 8.33
A three-phase three-wire unbalanced load is star-connected as shown in Fig. 8.27.
V =
V Calculate voltage
The phase voltage of two of the arms are
between star point of the load and the supply neutral.
R
O
N
Y
B
Solution
V
Let
VRO
VYO
=
V 0
V 120°
Fig. 8.27
°V
8.38 Network Analysis and Synthesis
From Fig. 8.27,
VRO
VNO
VRN + VNO
VRO − VRN = V∠0° − 100°∠− 10°
…(i)
VNO
VYO − VYN = V∠ − 120° − 11500 ∠100°
…(ii)
Similarly,
From Eqs. (i) and (ii),
V∠0° 100 ∠ 10° = V∠−120
V∠ 120° −150 ∠100°
V∠0° V∠
V∠− 120° = 100 ∠− 10° −150 ∠100°
3 V 30° 206.79 52 97°
V = 119.39∠− 82.97
VNO = 119.39∠
∠− 82.97 − 100 ∠− 10 = 131.88∠
∠− 129.67° V
8.13
MEASUREMENT OF THREE-PHASE POWER
In a three-phase system, total power is the sum of powers in three phases. The power is measured by wattmeter.
It consists of two coils: (i) Current coil, and (ii) Voltage coil. Current coil is connected in series with the load
and it senses current. Voltage coil is connected across supply terminals and it senses voltages.
There are three methods to measure three-phase power:
1. Three-wattmeter method
2. Two-wattmeter method
3. One-wattmeter method
8.13.1 Three-Wattmeter Method
This method is used for balanced as well as unbalanced loads. Three wattmeters are inserted in each of the
three phases of the load whether star connected or delta connected as shown in Fig. 8.28. Each wattmeter will
measure the power consumed in each phase.
Forr balanced load , W1 W2 = W3
Forr unbalanced load , W1 W2 ≠ W3
Total
t power P = W1 W2 + W3
W1
R
Zph
R
N
N
Zph
Zph
Zph
Y
W2
Y
W1
W3
Zph
Zph
B
W2
B
W3
Fig. 8.28
Three-wattmeter method
8.13 Measurement of Three-Phase Power 8.39
8.13.2 Two-Wattmeter Method
This method is used for balanced as well as unbalanced loads. The current coils of the two wattmeters are
inserted in any two lines and the voltage coil of each wattmeter is joined to a third line. The load may be
star or delta connected as shown in Fig. 8.29. The sum of the two wattmeter readings gives three-phase
power.
R
R
W1
Zph
Y
W2
Y
N
Zph
W1
Zph
W2
Zph
Zph
Zph
B
B
(b)
(a)
Two-wattmeter method
Fig. 8.29
Total power P
W1 + W2
8.13.3 One-Wattmeter Method
This method is used for balanced loads only. When the load is balanced, total power is given by
3 V ph I ph cos φ
P
Hence, one wattmeter is used to measure power in one phase. The wattmeter reading is then multiplied by
three to obtain three-phase power. The load may be star or delta connected as shown in Fig. 8.30.
R
W1
R
Y
Zph
N
N
Zph
Zph
Zph
Zph
Zph
B
W
B
Fig. 8.30
One-wattmeter method
8.13.4 Measurement of Power by Two-wattmeter Method
Figure 8.31 shows a balanced star-connected load and this load may be assumed to be inductive. Let VRN, VYN
and VBN be the three phase voltages and IR, IY and IB be the phase currents. The phase currents will lag behind
their respective phase voltages by angle f.
Current through current coil of W1 I R
Voltage across voltage coil of W1 VRB = VRN
VNB = VRN
VBN
8.40 Network Analysis and Synthesis
Figure 8.31 shows the phasor diagram of a balanced star-connected inductive load.
VRB
VRN
IR
VNB
°
30
f
VYB
30°
f
VBN
VYN
IY
Fig. 8.31 Phasor diagram
φ ).
)
From the phasor diagram, it is clear that the phase angle between VRB and IR is (
W1
Current through current coil of W2
Voltage across voltage coil of W2
VRB I R cos (30° − φ )
IY
VYB = VYN
VNB = VYN
VBN
From the phasor diagram, it is clear that phase angle between VYB and IY is (
But
W2
IR
VRB
W1
W2
φ )).
VYB IY cos (30° + φ )
IY = I L
VYB = VL
VL I L cos (30° − φ )
VL I L cos (30° + φ )
W1 W2 = VL I L [cos (30
φ ) + cos (30° φ ))] = VL I L ( 2 c 30°
Thus, the sum of two wattmeter readings gives three-phase power.
8.13.5 Measurement of Power Factor by Two-wattmeter Method
1. Lagging Power Factor
W1
W2
VL I L cos (30° − φ )
VL I L cos (30° + φ )
∴W
W1 W2
W1 W2 = 3 VL I L cos φ
W1 W2 = VL I L [cos (30 φ ) − cos (30° φ ) = VL I L sin φ
W1 − W2
V I sin φ
= L L
W1 + W2
3VL I L cos φ
W − W2
tan φ = 3 1
W1 + W2
s φ)
3 VL I L cos
o φ
8.13 Measurement of Three-Phase Power 8.41
⎛ W − W2 ⎞
φ = tan −1 ⎜ 3 1
⎝ W1 + W2 ⎟⎠
⎧
⎛ W − W2 ⎞ ⎫
pf = cos φ = cos ⎨tan −1 ⎜ 3 1
⎬
⎝ W1 + W2 ⎟⎠ ⎭
⎩
2. Leading Power Factor
Figure 8.32 shows the phasor diagram of a balanced
star-connected capacitive load.
VRB
VRN
W1 VL I L cos (30° + φ )
W2 VL I L cos (30° − φ )
∴W
W1 < W2
f
VBN
−
+
)⎞
) ⎟⎠
IB
30°
f
f
W1 W2 = 3 VL I L cos φ
W1 W2 = −VL I L sin φ
(
)
tan φ = − 3
(
)
⎛
(
φ = tan −1 − 3
⎝
(
VNB
IR
3 °
30
VYB
IY
VYN
Fig. 8.32 Phasor diagram
⎧
W − W2 ⎞ ⎫
⎛
pf = cos φ = cos ⎨tan −1 ⎜ − 3 1
⎬
⎝
W1 + W2 ⎟⎠ ⎭
⎩
Example 8.34
Two wattmeters are used to measure power in a three-phase balanced load. Find the
power factor if (a) two readings are equal and positive, (b) two readings are equal and opposite, and (c) one
wattmeter reads zero.
Solution
(a)
Power factor if two readings are equal and positive
W1
W2
tan φ = 3
W1 W2
W1 W2
3 ( 0)
0
φ = 0°
Power factor = cos φ = cos (0°) = 1
(b) Power factor if two readings are equal and opposite
W1
W2
W W2
tan φ = 3 1
=∞
W1 W2
φ = 90°
Power factor = cos φ = cos (90°) = 0
8.42 Network Analysis and Synthesis
(c) Power factor if one wattmeter reads zero
W2 = 0
W1 W2
W1 W2
φ = 60°
tan φ = 3
⎛W ⎞
3⎜ 1⎟ = 3
⎝ W1 ⎠
Power factor = cos φ = cos (60°) = 0 5
Example 8.35 What will be the relation between readings on the wattmeter connected to measure
power in a three-phase balanced circuit with (a) unity power factor, (b) zero power factor, and (c) power
factor = 0.5.
Solution
(a) Relation between wattmeter readings with power factor = 1
cos φ = 1
φ = 0°
tan φ tan ( ) = 0
W1 − W2
W1 + W2
W − W2
0= 3 1
W1 + W2
W1 = W2
tan φ = 3
(b) Relation between wattmeter readings with power factor = 0
cos φ = 0
φ = 90°
tan φ tan (
tan φ = 3
)=∞
W1 − W2
W1 + W2
W1 + W2 = 0
W1 = −W2
(c) Relation between wattmeter readings with power factor = 0.5
cos φ = 60°
tan φ tan (60°) = 1.732
W − W2
tan φ = 3 1
W1 + W2
W1 W2
W1 + W2
W2 = W1 + W2
W2 = 0
1.732 = 3
W1
8.13 Measurement of Three-Phase Power 8.43
Example 8.36
In a balanced three-phase system, the power is measured by two-wattmeter method
and the ratio of two-wattmeter readings is 4:1. The load is inductive. Determine the load power factor.
Solution
W1 4
=
W2 1
Load power factor
W1
4 W2
tan φ = 3
W1 W2
W1 W2
3
(3W2 )
⎛ 3⎞
= 3 ⎜ ⎟ = 1.039
⎝ 5⎠
(5W2 )
φ = 46.1°
pf = cos φ = cos ( 46.1°) = 0.0693 (
)
Example 8.37
Find the power and power factor of the balanced circuit in which the wattmeter
readings are 5 kW and 0.5 kW, the latter being obtained after the reversal of the current coil terminals of
the wattmeter.
Solution
W1 5 k , W2 0 5 kW
(a) Power
When the latter reading is obtained after the reversal of the current coil terminals of the wattmeter,
W1 = 5 kW
W2 = −0 5 kW
Power = W1 W2 = 5 ( 0.5) 4.5 kW
(b) Power factor
W1 − W2
(
= 3
W1 + W2
(
φ = 64.72°
.5)
= 2.12
.5)
tan φ = 3
Power factor = cos
o φ = cos (64.
) = 0.43
Example 8.38
Two wattmeters connected to measure the input to a balanced, three-phase circuit
indicate 2000 W and 500 W respectively. Find the power factor of the circuit (a) when both readings are positive and (b) when the latter is obtained after reversing the connection to the current coil of one instrument.
Solution
W1
2000
W2 = 500 W
(a) Power factor of the circuit when both readings are positive
W1 = 2000 W
W2 = 500 W
W − W2
(
tan φ = 3 1
= 3
W1 + W2
(
φ = 46.102°
Power factor = cos f = cos (46.102°) = 0.693
)
= 1.039
)
8.44 Network Analysis and Synthesis
(b) Power factor of the circuit when the latter reading is obtained after reversing the connection to the
current coil of one instrument
W1 = 2000 W
W2 = −500 W
W W2
tan φ = 3 1
W1 W2
φ = 70.89°
3
( 2000 + 500)
= 2.887
( 2000 − 500)
Power factor = cos f = cos (70.89°) = 0.33
Example 8.39 A three-phase, 10 kVA load has a power factor of 0.342. The power is measured by
the two-wattmeter method. Find the reading of each wattmeter when the (a) power factor is leading, and the
(b) power factor is lagging.
Solution
S = 10 kVA, pf = 0.342
S
VL I L
3
10 × 10 = 3 VL I L
VL I L = 5.77 kVA
cos φ = 0.342
φ = 72°
(a) Reading of each wattmeter when the power factor is leading
W1
VL I L cos (30° + φ ) = 5.77 × 103
W2
VL I L cos (30° − φ )
s (30° 70°)
3
..77
7 × 100 coss (30° 70°)
1 kW
.42 kW
(b) Reading of each wattmeter when the power factor is lagging
W1
W2
VL I L cos (30° φ ) = 4.42 kW
VL I L cos (30° + φ ) = −1 kW
Example 8.40
A three-phase, star-connected load draws a line current of 20 A. The load kVA
and kW are 20 and 11 respectively. Find the readings on each of the two wattmeters used to measure the
three-phase power.
Solution
IL
S
, S = 20 kVA, P = 11 kW
VL I L
3
20 × 10 = 3 VL I L
VL I L = 11.55 kVA
P
3 VL I L cos φ
11 × 10 = 20 × 103 × cos φ
cos φ = 0.55
φ = 56.63°
3
W1 = VL I L cos (30° φ ) = 11.55 × 103 × cos (30° 56.63 ) = 10.32 kW
W
W2
VL I L cos (30° + φ ) = 11.55 × 103 × cos (30° + 56
56.63
.63 ) = 0 68 kW
8.13 Measurement of Three-Phase Power 8.45
Example 8.41 Calculate the total power and readings of the two wattmeters connected to measure
power in three-phase balanced load, if the reactive power is 15 kVAR and load pf is 0.8 lagging.
Solution
Q = 15 kVAR, pf = 0.8 (lagging)
(a) Reading of the two wattmeters
cos φ 0.8
φ = 36.87°
Q = 3 VL I L sin φ
15 × 103 = 3 VL L in (36.87 )
VL I L = 14.43 kVA
W1
W2
VL I L cos (30° φ ) = 14.43 × 103 × cos (30° − 36.87
. ) 14 03 kW
VL I L co
c s (30° + φ )
.43 × 103 × cos (30
.43
(30° + 36.
36.. ) 5.67 kW
(b) Total power
P
W1 + W2 = 14.03 + 5.67 = 19.7 kW
Example 8.42 Two wattmeters are connected to measure power in a three-phase circuit. The reading of one of the wattmeters is 5 kW when the load power factor is unity. If the power factor of the load is
changed to 0.707 lagging without changing the total input power, calculate the readings of the two wattmeters.
Solution
W1 = 5 kW
(a) When the power factor is unity,
W1 W2 = 5 kW
Power = W1 W2 = 5 + 5 = 10 kW
…(i)
(b) When the power factor is changed to 0.707 lagging,
pf = cos φ = 0.707 (lagging)
φ = 45°
tan φ = tan ( 45°) = 1
W1 − W2
W1 + W2
W − W2
1= 3 1
10
10
W1 − W2 =
= 5.77 kW
3
tan φ = 3
…(ii)
Solving Eq. (i) and (ii),
W1 = 7 89 kW
W2 = 2.11 kW
Example 8.43
A three-phase RYB system has effective line voltage of 173.2 V. Wattmeters in line R
and Y read 301 W and 1327 W respectively. Find the impedance of the balanced star-connected load.
Solution
VL
173 2 , W1 = 301 W, W2 = 1327 W
8.46 Network Analysis and Synthesis
If the load is capacitive and pf is leading then W1 W2 .
W − W2
(
)
tan φ = − 3 1
=− 3
= −1.09
W1 + W2
(
)
φ = 47.47°
P = W1 + W2 = 301 + 1327 = 1628 W
3 VL I L cos φ
P
1628 = 3 × 173.2 × I L × cos ( 47.47°)
I L = 8 03 A
For a star-connected load,
V ph =
I ph
Z ph
VL
=
173.2
= 100 V
3
3
I L = 8 03 A
V ph 100
=
=
= .45 Ω
I ph 8 03
Example 8.44
Three coils each with a resistance of 10 W and reactance of 10 W are connected in
star across a three-phase, 50 Hz, 400 V supply. Calculate (a) line current, and (b) readings on the two wattmeters connected to measure the power.
Solution
For a star-connected load,
(a) Line current
R
V ph =
0 Ω,
VL
3
=
XL =
400
3
Ω
VL = 400 V
= 230.94 V
Z ph = R jjX L 10 j10 = 14.14 ∠45° Ω
Z ph = . Ω
φ = 45°
Power factor = cos φ = cos ( 45°) = 0.707 (lagging)
I ph =
IL
V ph
230.94
=
= 16.33 A
Z ph
14.14
I ph = 16.33 A
(b) Readings on the two wattmeters
Also,
P
3 VL I L
φ = 3 × 400 × 16.33 × 0.707 = 7998.83 W
W1 W2 = 7998.83 W
W − W2
tan φ = 3 1
W1 + W2
W − W2
tan 45° = 3 1
7998.83
W1 − W2 = 4618.13 W
Solving Eqs. (i) and (ii),
W1 = 6308.48 W
W2 = 1690.35 W
...(i)
...(ii)
8.13 Measurement of Three-Phase Power 8.47
Example 8.45 Three coils each having a resistance of 20 W and reactance of 15 W are connected
in (a) star, and (b) delta, across a three-phase, 400 V, 50 Hz supply. Calculate in each case, the readings on
two wattmeters connected to measure the power input.
Solution
0 Ω,
R
XL =
Ω
VL = 400 V
(a) Readings on two wattmeters for a star-connected load
V ph =
Z ph
VL
=
400
= 230.94 V
3
3
= 20 + j 15 = 25∠36.87° Ω
Z ph = 25 Ω
φ = 36.87°
I ph =
V ph
Z ph
=
230.94
= 9 24 A
25
IL
I ph = 9 24 A
W1
VL I L cos (30° φ ) = 400 × 9.24
W2
VL I L co
c s (30° + φ )
s (30° 36.87
. ) = 3669.46 W
.24 coss (30° 36
36.
) = 1451.86 W
(b) Readings on two wattmeters for a delta-connected load
V ph
VL = 400 V
Z ph = 25 Ω
φ = 36.87°
I ph =
IL
V ph
Z ph
=
400
= 16 A
25
3 I pphh = 3 × 16 = 27.72 A
W1
VL I L cos (30° φ ) = 400 × 27.72 × cos (30° − 36.87
. ) = 11008.39 W
W2
VL I L cos (30° + φ )
.72 × cos (30°
(30° + 36.
36..
36
) = 4355.57 W
Example 8.46 Two wattmeters connected to measure three-phase power for star-connected load
reads 3 kW and 1 kW. The line current is 10 A. Calculate (a) line and phase voltage (b) resistance and
reactance per phase.
Solution
W1
3 k , W2
1 kW, I L = 10 A
(a) Line and phase voltage
W1 − W2
(
= 3
W1 + W2
(
φ = 40.89°
tan φ = 3
)
= 0.866
)
Power factor = cos φ = cos ( 40.89°) = 0.756
P
W1 + W2 = 3 + 1 = 4 kW
8.48 Network Analysis and Synthesis
3 VL I L cos φ
P
3
4 × 10 = 3 × VL × 10 × 0.756
VL = 305.48 V
For a star-connected load,
V ph =
VL
3
=
305.48
3
= 176.37 V
(b) Resistance and reactance per phase
I ph
Z ph
I L = 10 A
V ph 176.37
=
=
= 7.637 Ω
I ph
10
R
XL
Z phh cos φ = 17.637 × 0.756 = 13.33 Ω
Z ph sin φ = 17.637 × sin ( 40. ) = .55 Ω
Example 8.47 The power input to a 2000 V, 50 Hz, three-phase motor running on full load at an
efficiency of 90% is measured by two wattmeters which indicate 300 kW and 100 kW respectively. Calculate
the (a) input power, (b) power factor, and (c) line current.
Solution
VL
2000 V, η
90%, W1 = 300 kW
W2 = 100 kW
(a) Input power
P
W1 + W2 = 300 + 100 = 400 kW
(b) Power factor
W1 − W2
(
= 3
W1 + W2
(
φ = 40.89°
pf = cos φ = co
c s ( 40. ) = 0..
tan φ = 3
)
= 0.866
)
(lagging)
(c) Line current
P
3 VL I L cos φ
3
400 × 10 = 3 × 2000 × I L × 0 76
I L = 151.93 A
Example 8.48 A three-phase, 400 V, 50 Hz induction motor has a full load output of 14.9 kW at
which the efficiency and power factor are 0.88 and 0.8 respectively. What is the line current? Find the readings on the two wattmeters connected to measure the power input to the motor.
Solution
V
(a) Line current
Po
Pi
14.9 × 103
0 88 =
Pi
Pi = 16.93 kW
η=
f = 50 Hz, Po = 14 9 kW, pf = 0.8, η = 0.88
8.13 Measurement of Three-Phase Power 8.49
pf = cos φ = 0 8
φ = 36.87°
Pi = 3 VL I L cos φ
16.93 × 103 3 × 400 × I L × 0.8
I L = 30.55 A
(b) Readings on the two wattmeters
W1
W2
VL I L cos (30° φ ) = 400 × 30.55 × cos (30° − 36.87
. ) 12.13 kW
VL I L co
c s (30° + φ )
.55 × cos (30°
(30° + 36.
36.. ) 4.8 kW
36
Example 8.49
A three-phase, 220 V, 50 Hz, 11.2 kW induction motor has a full load efficiency of
88 per cent and draws a line current of 38 A under full load, when connected to a three-phase, 220 V supply.
Determine power factor at which the motor is operating. Find the reading on two wattmeters connected in
the circuit to measure the input to the motor.
Solution
VL
220
Po = 11.2 kW,
k , η = 88
I L = 38 A
(a) Power factor at which the motor is operating
Po
Pi
11.2 × 103
0 88 =
Pi
Pi = 12.73 kW
η=
But
Pi
3 VL I L cos φ
12.73 × 10 = 3 × 220 × 38 × cos φ
pf = cos φ = 0.88 (lagging)
3
(b) Reading on two wattmeters
φ = 28.36°
W1 = VL I L cos (30° φ ) = 220 × 38 × cos (30
W2 = VL I L cos (30° + φ ) = 220 × 38 × cos (30
. 36° ) = 8356.58 W
. 36° ) = 4385.49 W
Example 8.50 Find the reading of a wattmeter when the network shown in Fig. 8.33 is connected to
a symmetrical 440 V, 3-phase supply. The phase sequence is RYB.
W
R
IR
−j 40 Ω
IRY
−j 53 Ω
50 Ω
IBR
B
Y
Fig. 8.33
8.50 Network Analysis and Synthesis
Solution
VL = 440 V
VRRY = 440 ∠0°V
Let
IRY
VYB = 440 ∠− 120°V
VBR
VBR = 440 ∠− 240°V
IBR
Z RY = − j 53 = 53∠− 90 Ω
Z BR = 50 + j 40 = 64.03∠38.66° Ω
V
440 ∠0°
I RY = RRY =
= 8 3∠90° A
Z RY 53∠− 90°
V
440 ∠− 240°
I BR = BBR =
= 6.87∠81.34° A
Z BR 64.03∠38.66°
I R = I RY − I BR = 8.3∠90° − 6.87∠81.34° = 1.83∠124
2 .44° A
IR
115.56°
VYB
The phasor diagram of the system is shown in Fig. 8.34.
From the phasor diagram, it is clear that angle between VYB and IR is 115.56°
W
VRY
Fig. 8.34
VYB I R cos (115.56°) = 440 × 1.83 × cos (115.56°) = 347.41 W
Exercises
8.1
Three coils, each of 5 Ω resistance, and 6 Ω
inductive reactance are connected in closed
delta and supplied from a 440 V, three-phase
system. Calculate the line and phase currents,
the power factor of the system and the intake
in watts.
[ .58 A, 56.33 A, 0.64 (
), 47.61 kW]
8.2
Three coils, each having a resistance of 10
Ω and inductance of 0.02 H are connected
(i) in star, (ii) in delta to a three-phase, 50
Hz supply, the line voltage being 500 volts.
Calculate for each case the line current and
the total power taken from the supply.
[(a))
: . A, 17.94 kW,
( ) delta : 73.39 A, .83 kW]
8.3
A balanced delta-connected load of (8 + j6)
ohms per phase is connected to a three-phase,
230 V supply with phase sequence R-Y-B.
Find the line current, power factor, power,
reactive volt-amperes and the total voltamperes. Draw the phasor diagram.
[ .85 A,, 0.8 (
),
12.74 kW, 9.52 k A , 15..
k
]
8.4
A balanced star-connected load with (6 + j8)
Ω per phase is connected to a three-phase,
440 V supply. Find the line current, power
factor, power, reactive volt-amperes and voltamperes total.
[ .404 A, . (lagging),
11.616 kW,15.488 kVAR , 19.
8.5
8.6
8.7
k
]
Calculate the active and reactive components
of the current in each phase of a star-connected
5000 V, three-phase, alternator supplying
3000 kW at a power factor of 0.8.
[
, 150
50 A ]
Three similar coils, connected in star, take
a total power of 1.5 kW at a power factor
of 0.2 lagging from a three-phase, 440 V,
50 Hz supply. Calculate the resistance and
inductance of each coil.
[ .16 Ω, 0.08 H]
A balanced three-phase load connected in
delta draws a power of 10 kW at 440 V at a
power factor of 0.6 lead. Find the values of the
circuit elements and the reactive volt-amperes
drawn.
[
.91 Ω, 27.88 Ω, 13.33 kVAR ]
Exercises 8.51
8.8
A balanced star-connected load, connected to
a 400 V, 50 Hz, three-phase ac supply draws a
phase current of 50 A at 0.6 power factor lagging.
Calculate (a) phase voltage, (b) total power, and
(c) parameters in the star-connected load.
.94 V, 20.84 kW,( 2.77 + j 3.7)
[
8.9
]
Three equal star-connected inductors,
consume 8 kW power at 0.8 power factor when
connected to 415 V, three-phase, three-wire,
50 Hz supply. Estimate the load parameters
per phase and determine the line currents.
[ .2 Ω, 0.0386 H,13.83 A ]
8.10 A balanced three-phase, star-connected load
of 150 kW takes a leading current of 100 A
with a line voltage of 1100 V, 50 Hz. Find the
circuit constant of the load per phase.
[ Ω, 813 μF]
8.11 Three pure elements connected in star draw x
kVAR. What will be the value of elements that
will draw the same kVAR, when connected in
delta across the same supply?
[
]
8.12 A balanced wye-connected load with (10 + j20)
ohms per phase is connected to a three-phase,
400 V supply. Determine the voltage across,
current through and power dissipated in each
resistor. Also, determine the total power.
[
.2 V,10
10.3
32
2 V, 1067 W,3201
, 3201 W ]
8.13
A delta-connected three-phase load is supplied
from a 3- phase, 400 volts balanced supply
system. The line current is 20 A and power taken
by the load is 10 kW. Find (a) impedance in each
branch, (b) line current, power factor and power
consumed if the same load is connected in star.
[( 24.95 + j 24.05) ,
6.66 A, 0. (lagging), 3323.
]
8.14
A balanced star-connected load is supplied from
a symmetrical three-phase, 400 V system. The
current in each phase is 30 A and lags 30°
behind the phase voltage. Find (a) phase voltage,
(b) the circuit elements, and (c) draw the vector
diagram showing the currents and the voltages.
[
.94 V, 6.67 Ω, 3.849
8 9 Ω]
8.15 A three-phase, delta-connected load having a
(3 + j4) ohms per phase is connected across
a 230 V, three-phase source. Calculate the
magnitude of the line current.
[ .21 A ]
8.16 A 220 V, three-phase voltage is applied to a
balanced delta-connected load. The rms value
of the phase current is 20 ∠ −30°A. Determine
(i) magnitude and phase of the line current
(ii) total power received by the three-phase
load
(iii) value of the resistive portion of the
phase impedance
Also, draw the phasor diagram showing
clearly the line voltages, phase current and
line currents.
[ .65 ∠− 60°A, 11.43 kW, 9.53 Ω]
8.17 A three-phase, 37.3 kW, 440 V, 50 Hz
induction motor operates on full load with
an efficiency of 89% and at a power factor
of 0.85 lagging. Calculate total kVA rating
of capacitance required to raise power factor
at 0.95 lagging. What will be the value of
capacitance/phase if capacitors are (a) delta
connected (b) star connected?
[ .19 kVA, 66.8 μF
F, 200.4
200 4 μF]
8.18 Three coils each having impedance (4 +
j3) ohms are connected in star to a 440 V,
three-phase, 50 Hz supply. Calculate the line
current and active power. Now if three pure
capacitors, each of C farads, connected in
delta, are connected across the same supply,
it is found that the total power factor of the
circuit becomes 0.96 lag. Find the value of C.
Also, find the total line current.
[ .8 A, 30.976 kW, 77.75 μF, 42.34 A]
8.19 Three identical coils each having an resistance
of 8 Ω and inductance of 0.02 H are connected
in (i) star, and (ii) delta across a 3f, 400 V,
50 Hz supply. Draw a neat phasor diagram
and calculate the reading of two wattmeters
connected to measure power. Also, calculate
pf of the circuit.
.99 kW, 3.3 kW, 0.7866 (
),
26.98 kW,10.14 kW, 0.
(lagging
g)]
8.52 Network Analysis and Synthesis
8.20 Three identical coils each having a reactance
of 20 Ω and resistance of 10 Ω are connected
in (i) star, (ii) delta across a 440 V, three-phase
line. Calculate for each method of connection
the line current and readings of each of the
two wattmeters connected.
36 A, 4.17 kW, 299.07 W,
34.08 A, 12.52 kW,, 897
89 .
]
8.21 A 3-phase motor load has a pf of 0.397 lagging.
Two wattmeters connected to measure power
show the input as 30 kW. Find the reading on
each wattmeter.
[ k , − 5 kW ]
8.22 Each of the wattmeters connected to measure
the input to a 3-phase induction motor reads
10 kW. If the power factor of the motor be
changed to 0.866 lagging, determine the
readings of the two wattmeters, the total input
power remaining unchanged.
[ .67 kW, 13.33 kW ]
8.23 A 3f, star-connected load draws a line current
of 25 A. The load kVA and kW are 20 and
16 respectively. Find the readings on each of
the two wattmeters used to measure the 3f
power.
[ .46 kW, 4.54 kW ]
8.24 Three similar coils are star-connected to a 3f,
50 Hz supply. The line current taken is 25 A
and the two wattmeters connected to measure
the input indicate 5.185 kW and 10.37 kW
respectively. Calculate (a) the line and phase
voltages, and (b) the resistance and reactance
of each coil.
[
, 240 V, 5.31 Ω, 4.8 Ω]
8.25 A three-phase, 500 V motor load has a power
factor of 0.4. Two wattmeters connected to
measure power show the input to be 30 kW.
Find the reading on each instrument.
[ k , − 5 kW ]
8.26 The power in a three-phase circuit is measured
by two wattmeters. If the total power is 100
kW and power factor is 0.66 leading, what
will be the reading of each wattmeter?
[ .26 kW, 82.74 kW ]
8.27 Two wattmeters are connected to measure
the input to a 400 V, three-phase connected
motor outputting 24.4 kW at a power factor
of 0.4 lag and 80% efficiency. Calculate (i)
resistance and reactance of motor per phase,
(ii) reading of each wattmeter.
[ .55 Ω, 5.58 Ω, 34915 W, − 4850
48 0 W ]
8.28 In a balanced three-phase, 400 V circuit, the line
current is 115.5 A. When power is measured
by the two wattmeter method, one meter reads
40 kW and the other, zero. What is the power
factor of the load? If the power factor were
unity and the line current the same, what would
be the reading of each wattmeter?
[ .5, 40 kW, 40 kW ]
8.29 A 440 V, three-phase, delta-connected
induction motor has an output of 14.92 kW
at pf of 0.82 and efficiency of 85%. Calculate
the readings on each of the two wattmeters
connected to measure the input. If another starconnected load of 10 kW at 0.85 pf lagging is
added in parallel to the motor, what will be
the current drawn from the line and the power
taken from the line?
[ .35 kW, 5.26 kW, 43.56 A, 27.6 kW ]
8.30 Balanced delta-connected impedances, each
of 10 ∠30 Ω are connected across threephase 400 V mains. Determine the twowattmeter readings if the current coils of the
two wattmeters are connected in lines R and Y
and the pressure coils are connected between
R and B and Y and B lines respectively.
[ .7 kW, 13.86 kW ]
8.31 A balanced star-connected load, each phase
having a resistance of 10 Ω and the inductive
reactance of 30 Ω is connected to 400 V, 50
Hz supply. The phase rotation is R, Y and B.
Wattmeters connected to read total power
have their currents coils in the red and blue
Objective-Type Questions 8.53
voltage, current in each phase, load parameter,
power in each phase, total power, readings of
the wattmeters connected in the load circuit to
measure the total power. Draw a neat circuit
diagram and vector/phase diagram.
lines respectively. Calculate the reading on
each wattmeter.
[
, 583 W ]
8.32 A balanced star-connected load is supplied
from a symmetrical three-phase 400 V, 50 Hz
supply system. The current in each phase is 20
A and lags behind its phase voltage by an angle
of 2p/9 radians. Calculate line voltage, phase
, 254.034 V, 20 A,, 9.73 Ω, 0.026 H,
[
3.892 kW, 11.676 kW, 8.666 kW, 3.009 kW ]
Objective-Type Questions
Choose the correct alternative in the following
questions:
8.1 In a three-phase system, voltages differ in
phase by
(a) 30°
(b) 60°
(c) 90°
(d) 120°
8.2 The rated voltage of a three-phase power
system is given as
(a) rms phase voltage
(b) peak phase voltage
(c) rms line to line voltage
(d) peak line to line voltage
8.3
8.4
Total instantaneous power supplied by a threephase ac supply to a balanced R-C load is
(a) zero
(b) constant
(c) pulsating with zero average
(d) pulsating with non-zero average
Which of the following equations is valid for a
three-phase, balanced star-connected system?
(a)
(b)
(c)
(d)
IR
IR
R
Y
B
Fig. 8.35
(a)
(c)
8.7
8.8
IY + I B = 0
I B = IY
A three-phase load is balanced if all the three
phases have the same
(a) impedance
(b) power factor
(c) impedance and power factor
(d) none of the above
8.6
The phase sequence of a three-phase system
shown in Fig. 8.35 is
(b) RBY
(d) YBR
In a 3-phase system, VYN = 100∠−120° V and
VBN = 100∠120° V. Then VYB will be
∠90° V
(a)
(b)
173∠ − 90° V
(c)
(d)
200 ∠60° V
none of the above.
If a balanced delta load has an impedance of
(6 + j9) ohms per phase then the impedance of
each phase in the equivalent star load is
(a) (6 + j9) ohms
(b) (2 + j3) ohms
(c) (2 + j8) ohms
(d) (3 + j4.5) ohms
I R IY − I B = 0
VR VB + VY
= IN
Z
8.5
RYB
BRY
8.9
Three equal impedances are first connected
in delta across a three-phase balanced supply.
If the same impedances are connected in star
across the same supply
(a) phase current will be one-third
(b) line current will be one-third
(c) power consumed will be one-third
(d) none of the above.
8.54 Network Analysis and Synthesis
8.10
8.11
Three identical resistors connected in star
carry a line current of 12 A. If the same
resistors are connected in delta across the
same supply, the line current will be
(a) 12 A
(b) 4 A
(c) 8 A
(d) 36 A
If one of the resistors in Fig. 8.38 is opencircuited, power consumed in the circuit is
10 Ω
400 V
60 kW
40 kW
Fig. 8.38
The power consumed in the star-connected load
shown in Fig. 8.36 is 690 W. The line current is
(a)
(c)
8.15
R
R
400 V
Y
400 V
(b)
(d)
4000 W
none of the above
IR
R
R
R1
IB
R
B
R1
Fig. 8.36
(b) 1 A
(a)
2.5 A
(c)
1.725 A
(d)
none of the above
ZL
Fig. 8.39
ZL
90 ∠32.44°
(c)
80 ∠− 32.44° (d)
(b)
80 ∠32.44°
90 ∠− 32.44°
1:1: 3
(b)
1:1: 2
(c)
1:1: 0
(d)
1:1:
3
2
The phase sequence RYB denotes that
(a) emf of phase Y lags behind that of phase
R by 120°
(b) emf of phase Y leads that of phase R by
120°
(c) emf of phase Y and phase R are in phase
(d) none of the above
8.17
In the two wattmeter method of measurement,
if one of the wattmeters reads zero, then power
factor will be
(a) zero
(b) unity
(c) 0.5
(d) 0.866
ZL
(a)
(a)
8.16
400 V
Fig. 8.37
IY
Y
If the three-phase balanced source in Fig. 8.37
delivers 1500 W at a leading power factor
of 0.844 then the value of ZL (in ohms) is
approximately
3f
balanced
source
8000 W
16000 W
For the three-phase circuit shown in Fig. 8.39,
the ratio of the currents I R IY : I B is given by
B
8.13
10 Ω
400 V
(b) 20 kW
(d) 180 kW
400 V
10 Ω
400 V
Three delta-connected resistors absorb 60 kW
when connected to a 3-phase line. If the resistors
are connected in star, the power absorbed is
(a)
(c)
8.12
8.14
Answers to Objective-Type Questions 8.55
8.18 Two wattmeters, which are connected to
measure the total power on a three-phase
system, supplying a balanced load, read 10.5
kW and −2.5 kW, respectively. The total
power and the power factor, respectively
are
(a) 13 kW, 0.334
(b) 13 kW, 0.684
(c) 8 kW, 0.52
(d) 8 kW, 0.334
8.19
The minimum number of wattmeter(s)
required to measure three-phase, three-wire
balanced or unbalanced power is
(a) 1
(b) 2
(c) 3
(d) 4
8.20
One of the two wattmeters has read zero in the
two-wattmeter method of power measurement.
This indicated that the load phase angle is
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Answers to Objective-Type Questions
8.1 (d)
8.5 (c)
8.9 (c)
8.2 (c)
8.6 (b)
8.10 (d)
8.3 (b)
8.7 (b)
8.11 (b)
8.4 (a)
8.8 (b)
8.12 (b)
8.13 (d)
8.17 (c)
8.14 (a)
8.18 (d)
8.15 (a)
8.19 (b)
8.16 (a)
8.20 (c).
9
Network Topology
9.1
INTRODUCTION
The purpose of network analysis is to find voltage across and current through all the elements. When the
network is complicated and has a large number of nodes and closed paths, network analysis can be done
conveniently by using ‘Network Topology’. This theory does not make any distinction between different
types of physical elements of the network but makes the study based on a geometric pattern of the network.
The basic elements of this theory are nodes, branches, loops and meshes.
Node
Branch
Loop
Mesh
9.2
It is defined as a point at which two or more elements have a common connection.
It is a line connecting a pair of nodes, the line representing a single element or series connected
elements.
Whenever there is more than one path between two nodes, there is a circuit or loop.
It is a loop which does not contain any other loops within it.
GRAPH OF A NETWORK
A linear graph is a collection of nodes and branches. The nodes are joined together by branches.
The graph of a network is drawn by first marking the nodes and then joining these nodes by lines which
correspond to the network elements of each branch. All the voltage and current sources are replaced by their
internal impedances. The voltage sources are replaced by short circuits as their internal impedances are zero
whereas current sources are replaced by open circuits as their internal impedances are infinite. Nodes and
branches are numbered. Figure 9.1 shows a network and its associated graphs.
Each branch of a graph may be given an orientation or a direction with the help of an arrow head which
represents the assigned reference direction for current. Such a graph is then referred to as a directed or
oriented graph.
Branches whose ends fall on a node are said to be incident at that node. Degree of a node is defined as
the number of branches incident to it. Branches 2, 3 and 4 are incident at Node 2 in Fig. 9.1(c). Hence, the
degree of Node 2 is 3.
9.2 Network Analysis and Synthesis
(6)
(6)
1
(2)
3
3
(3)
2
1
(2)
3
(3)
(3)
(2)
(1)
V
2
1
2
(6)
(5)
(4)
(1)
(4)
(5)
(1)
(4)
(5)
+
−
4
4
4
(a) Network
(b) Undirected graph
(c) Directed or oriented graph
Fig. 9.1 Network and its graphs
9.3
DEFINITIONS ASSOCIATED WITH A GRAPH
1. Planar Graph A graph drawn on a two-dimensional plane is said to be planar if two branches do not
intersect or cross at a point which is other than a node. Figure 9.2 shows such graphs.
(2)
2
1
(3)
(1)
(5)
(8)
3
3
(4)
(1)
2
(2)
1
(6)
(3)
(6)
(7)
4
(9)
6
(4)
4
5
(5)
Fig. 9.2 Planar graphs
2. Non-planar Graph A graph drawn on a two-dimensional plane is said to be non-planar if there is
intersection of two or more branches at another point which is not a node. Figure 9.3 shows non-planar
graphs.
1
(2)
2
(2)
2
1
(5)
(3)
3
(6)
(1)
(4)
3
4
(3)
(1)
(7)
(8)
4
(5)
5
(6)
(4)
6
Fig. 9.3 Non-planar graphs
3. Sub-graph It is a subset of branches and nodes of a graph. It is a proper sub-graph if it contains
branches and nodes less than those on a graph. A sub-graph can be just a node or only one branch of the
graph. Figure 9.4 shows a graph and its proper sub-graph.
9.3 Definitions Associated with a Graph 9.3
(2)
(2)
2
1
(3)
(1)
(4)
(5)
2
3
(5)
3
3
1
(6)
(4)
4
4
(a)
(b)
Fig. 9.4 (a) Graph (b) Proper sub-graph
4. Path It is an improper sub-graph having the following properties:
1. At two of its nodes called terminal nodes, there is incident only one branch of sub-graph.
2. At all remaining nodes called internal nodes, there are incident two branches of a graph.
In Fig. 9.5, branches 2, 5 and 6 together with all the four nodes, constitute a path.
(3)
2
1
(2)
3
(4)
(1)
(5)
(6)
4
Fig. 9.5
Path
5. Connected Graph A graph is said to be connected if there exists a path between any pair of nodes.
Otherwise, the graph is disconnected.
6. Rank of a Graph
If there are n nodes in a graph, the rank of the graph is (n − 1).
7. Loop or Circuit A loop is a connected sub-graph of a connected graph at each node of which are incident
exactly two branches. If two terminals of a path are made to coincide, it will result in a loop or circuit.
Figure 9.6 shows two loops.
(2)
1
2
(3)
3
(1)
(4)
(2)
1
4
Loops: {1, 2, 3, 4}
Fig. 9.6
(1)
2
Loops: {1, 2}
Loops
Loops of a graph have the following properties:
1.
2.
3.
There are at least two branches in a loop.
There are exactly two paths between any pair of nodes in a circuit.
The maximum number of possible branches is equal to the number of nodes.
9.4 Network Analysis and Synthesis
8. Tree A tree is a set of branches with every node connected to every other node in such a way that
removal of any branch destroys this property.
Alternately, a tree is defined as a connected sub-graph of a connected graph containing all the nodes of
the graph but not containing any loops.
Branches of a tree are called twigs. A tree contains (n − 1) twigs where n is the number of nodes in the
graph. Figure 9.7 shows a graph and its trees.
(6)
2
1
(1)
3
(2)
(3)
(4)
1
2
(4)
(5)
2 (2)
1
3
(1)
(3)
(5)
(5)
4
4
4
(a) Graph
3
(b)
(c)
Twigs: {1, 4, 5}
Twigs: {2, 3, 5}
Fig. 9.7 Graph and its trees
Trees have the following properties:
1.
2.
3.
4.
5.
6.
There exists only one path between any pair of nodes in a tree.
A tree contains all nodes of the graph.
If n is the number of nodes of the graph, there are (n − 1) branches in the tree.
Trees do not contain any loops.
Every connected graph has at least one tree.
The minimum terminal nodes in a tree are two.
9. Co-tree Branches which are not on a tree are called links or chords. All links of a tree together
constitute the compliment of the corresponding tree and is called the co-tree.
A co-tree contains b − (n − 1) links where b is the number of branches of the graph.
In Fig. 9.7 (b) and (c) the links are {2, 3, 6} and {1, 4, 6} respectively.
Example 9.1
Draw directed graph of the networks shown in Fig. 9.8.
R2
R1
C1
L
L1
L2
v
C1
R1
R2
V
C2
+
−
C3
C2
(a)
R3
(b)
Fig. 9.8
9.3 Definitions Associated with a Graph 9.5
Solution For drawing the directed graph,
1. replace all resistors, inductors and capacitors by line segments,
2. replace the voltage source by a short-circuit,
3. assume directions of branch currents, and
4. number all the nodes and branches.
The directed graph for the two networks are shown in Fig. 9.9.
1
(1)
1
(2)
4
(2)
(1)
2
5
(5)
(3)
(3)
(4)
(7)
(8)
(6)
2
3
(a)
(b)
Fig. 9.9
Example 9.2
Figure 9.10 shows a graph of the network. Show all the trees of this graph.
(1)
1
(2)
2
3
(4)
(5)
(3)
4
Fig. 9.10
Solution A graph has many trees. A tree is a connected sub-graph of a connected graph containing all the
nodes of the graph but not containing any loops. Figure 9.11 shows various trees of the given graph.
(2)
2
3
1
(1)
2
3
1 (1) 2
3
1
(4)
(3)
(4)
4
1
2
4
(3)
4
3
(4) (5)
(3)
(5)
1
2
(3)
(5)
1 (1) 2 (2) 3
(3)
4
4
Fig. 9.11
3
1 (1) 2 (2) 3
(4)
(5)
(5)
4
(2) 3
1 (1) 2 (2)
4
4
9.6 Network Analysis and Synthesis
9.4
INCIDENCE MATRIX
A linear graph is made up of nodes and branches. When a graph is given, it is possible to tell which branches
are incident at which nodes and what are its orientations relative to the nodes.
9.4.1 Complete Incidence Matrix (Aa)
For a graph with n nodes and b branches, the complete incidence matrix is a rectangular matrix of order n × b.
Elements of this matrix have the following values:
aij = 1, if branch j is incident at node i and is oriented away from node i.
= −1, if branch j is incident at node i and is oriented towards node i.
= 0, if branch j is not incident at node i.
For the graph shown in Fig. 9.12, branch 1 is incident at nodes 1 and 4. It is oriented away from Node 1 and
oriented towards Node 4. Hence, a11 = 1 and a41 = −1. Since branch 1 is not incident at nodes 2 and 3, a21 =
0 and a31 = 0. Similarly, other elements of the complete incidence matrix are written.
(6)
2
1
(2)
3
(4)
(1)
(3)
(5)
Nodes
↓
Branches →
1
2
3
4
5
6
1
1
1
0
0
0
1
2
0 −1
1 −1
0
0
3
0
0
1 −1
4
−1
0
0 −1
1
0 −1
0
4
Fig. 9.12 Graph
The complete incidence matrix is
1⎤
⎡ 1 1 0 0 0
⎢ 0 −1 1 −1 0 0 ⎥
Aa = ⎢
⎥
1 1 −1⎥
⎢ 0 0 0
⎢⎣ −1 0 −1 0 −1 0 ⎥⎦
It is seen from the matrix Aa that the sum of the elements in any column is zero. Hence, any one row of the
complete incidence matrix can be obtained by the algebraic manipulation of other rows.
9.4.2 Reduced Incidence Matrix (A)
The reduced incidence matrix A is obtained from the complete incidence matrix Aa by eliminating one of the
rows. It is also called incidence matrix. It is of order (n − 1) × b.
Eliminating the third row of matrix Aa,
⎡ 1 1 0 0 0 1⎤
A = ⎢ 0 −1 1 −1 0 0 ⎥
⎢
⎥
⎣ −1 0 −1 0 −1 0 ⎦
When a tree is selected for the graph as shown in Fig. 9.13, the incidence matrix is obtained by arranging
a column such that the first (n − 1) column corresponds to twigs of the tree and the last b − (n − 1) branches
corresponds to the links of the selected tree.
9.4 Incidence Matrix 9.7
2
1
Twigs
3
(2)
(4)
2
Twigs: {2, 3, 4}
Links: {1, 5, 6}
(3)
3
⎡ 1 0
A = ⎢ −1 1
⎢
⎣ 0 −11
4
Links
4
1
0
1
1 0
0 −11
5 6
0 1⎤
0 0⎥
⎥
1 0⎦
Fig. 9.13 Tree
The matrix A can be subdivided into submatrices At and Al.
A [ At : Al ]
Where At the is twig matrix and Al is the link matrix.
9.4.3 Number of Possible Trees of a Graph
Let the transpose of the reduced incidence matrix A be AT. It can be shown that the number of possible trees
of a graph will be given by
Number of possible trees = |AAT|
For the graphs shown in Fig. 9.12, the reduced incidence matrix is given by
⎡ 1 1 0 0 0 1⎤
A = ⎢ 0 −1 1 −1 0 0 ⎥
⎢
⎥
⎣ −1 0 −1 0 −1 0 ⎦
Then transpose of this matrix will be
⎡ 1 0 −1⎤
⎢ 1 −1 0 ⎥
⎢
⎥
0
1 −1⎥
AT = ⎢
⎢0 −1 0 ⎥
⎢0 0 −1⎥
⎢
⎥
⎣ 1 0 0⎦
Hence, number of all possible trees of the graph
⎡ 1 0 −1⎤
⎢ 1 −1 0 ⎥
⎥
3 −1 −1⎤
⎡ 1 1 0 0 0 1⎤ ⎢0
1 −1⎥ ⎡
AAT = ⎢ 0 −1 1 −1 0 0 ⎥ ⎢
= ⎢ −1 3 −1⎥
⎢
⎥ ⎢0 −1 0 ⎥ ⎢
⎥
⎣ −1 0 −1 0 −1 0 ⎦ ⎢
⎣ −1 −1 3⎦
⎥
0 0 −1
⎢
⎥
⎣ 1 0 0⎦
|
T
3 −1 −1
| = −1 3 −1 = 3((
−1 −1 3
Thus, 16 different trees can be drawn.
) + ( )(
) − 1((
) = 16
9.8 Network Analysis and Synthesis
9.5
LOOP MATRIX OR CIRCUIT MATRIX
When a graph is given, it is possible to tell which branches constitute which loop or circuit. Alternately, if a
loop matrix or circuit matrix is given, we can reconstruct the graph.
For a graph having n nodes and b branches, the loop matrix Ba is a rectangular matrix of order b columns
and as many rows as there are loops.
Its elements have the following values:
bij = 1, if branch j is in loop i and their orientations coincide.
= − 1, if branch j is in loop i and their orientations do not coincide.
= 0, if branch j is not in loop i.
A graph and its loops are shown in Fig. 9.14.
(6)
2
1
3
(2)
(4)
(3)
(1)
(5)
Loop 1: {1, 2, 3}
Loop 2: {3, 4, 5}
Loop 3: {2, 4, 6}
Loop 4: {1, 2, 4, 5}
Loop 5: {1, 5, 6}
Loop 6: {2, 3, 5, 6}
Loop 7: {1, 3, 4, 6}
4
Fig. 9.14
Graph
All the loop currents are assumed to be flowing in a clockwise direction.
Loops
↓
1
2
3
4
5
6
7
Branches →
1 2 3 4
−1 1 1 0
0 0 −1 −1
0 −1 0 1
−1 1 0 −1
−1 0 0 0
0 −1 −1 0
−1 0 1 1
⎡ −1 1 1 0
⎢ 0 0 −1 −1
⎢
1
⎢ 0 −1 0
Ba = ⎢ −1 1 0 −1
⎢ −1 0 0 0
⎢ 0 −1 −1 0
⎢
1 1
⎣ −1 0
5
0
1
0
1
1
1
0
0
1
0
1
1
1
0
6
0
0
1
0
1
1
1
0⎤
0⎥
⎥
1⎥
0⎥
1⎥
1⎥⎥
1⎦
9.5.1 Fundamental Circuit (Tieset) and Fundamental Circuit Matrix
When a graph is given, first select a tree and remove all the links. When a link is replaced, a closed loop or
circuit is formed. Circuits formed in this way are called fundamental circuits or f-circuits or tiesets.
Orientation of an f-circuit is given by the orientation of the connecting link. The number of f-circuits is
same as the number of links for a graph. In a graph having b branches and n nodes, the number of f-circuits or
tiesets will be (b − n + 1). Figure 9.15 shows a tree and f-circuits (tiesets) for the graph shown in Fig. 9.14.
9.5 Loop Matrix or Circuit Matrix 9.9
(6)
1
2
(2)
3
(4)
(3)
4
(a) Tree
(2)
(1)
(4)
(2)
(4)
(3)
(3)
tieset 1
(2)
(4)
(3)
(5)
tieset 5
tieset 1: {1, 2, 3}
tieset 5: {5, 3, 4}
tieset 6: {6, 2, 4}
tieset 6
(b) f-circuits (tiesets)
Fig. 9.15
Tree and f-circuits
Here, b = 6 and n = 4.
Number of tiesets = b − n + 1 = 6 − 4 + 1 = 3
f-circuits are shown in Fig. 9.15. The orientation of each f-circuit is given by the orientation of the
corresponding connecting link.
The branches 1, 2 and 3 are in the tieset 1. Orientation of tieset 1 is given by orientation of branch 1. Since
the orientation of branch 1 coincides with orientation of tieset 1, b11 = 1. The orientations of branches 2 and 3
do not coincide with the orientation of tieset 1, Hence, b12= − 1 and b13 = − 1. The branches 4, 5 and 6 are not
in tieset 1. Hence, b14 = 0, b15 = 0 and b16 = 0. Similarly, other elements of the tieset matrix are written.
Then, the tieset schedule will be written as
Branches →
Tiesets
↓
1
3
4 5 6
1
1 −1 −1
0 0 0
5
0
6
0 −1
2
0 −1 −1 1 0
0
1 0 1
Hence, an f-circuit matrix or tieset matrix will be given as
⎡ 1 −1 −1 0 0 0 ⎤
B = ⎢0 0 −1 −1 1 0 ⎥
⎢
⎥
1 0 1⎦
⎣0 −1 0
Usually, the f-circuit matrix B is rearranged so that the first (n − 1) columns correspond to the twigs and
b − (n − 1) columns to the links of the selected tree.
Twigs
gs
2 3 4
1
⎡−1
B=⎢ 0
⎢
⎣−1
1 0
1 −1
0 1
Links
1 5 6
1
0
0
0
1
0
0⎤
0⎥
⎥
1⎦
The matrix B can be partitioned into two matrices Bt and Bl.
B [ Bt : Bl ] [ Bt : U ]
where Bt is the twig matrix, Bl is the link matrix and U is the unit matrix.
9.10 Network Analysis and Synthesis
9.5.2
Orthogonal Relationship between Matrix A and Matrix B
For a linear graph, if the columns of the two matrices Aa and Ba are arranged in the same order, it can be
shown that
Aa BaT = 0
Ba AaT = 0
The above equations describe the orthogonal relationship between the matrices Aa and Ba.
If the reduced incidence matrix A and the f-circuit matrix B are written for the same tree, it can be shown
that
A BT = 0
or
B AT = 0
or
These two equations show the orthogonal relationship between matrices A and B.
9.6
CUTSET MATRIX
1
(2)
2
Consider a linear graph. By removing a set of branches without affecting the nodes,
(5)
(3)
two connected sub-graphs are obtained and the original graph becomes unconnected. (1)
The removal of this set of branches which results in cutting the graph into two parts
(4)
4
3
are known as a cutset. The cutset separates the nodes of the graph into two groups,
each being in one of the two groups.
Fig. 9.16 Graph
Figure 9.16 shows a graph.
Branches 1, 3 and 4 will form a cutset. This set of branches separates the graph into two parts. One having
an isolated node 4 and other part having branches 2 and 5 and nodes 1, 2 and 3.
Similarly, branches 1 and 2 will form a cutset. Each branch of the cutset has one of its terminals incident
at a node in one part and its other end incident at other nodes in the other parts. The orientation of a cutset is
made to coincide with orientation of defining branch.
For a graph having n nodes and b branches, the cutset matrix Qa is a rectangular matrix of order b columns
and as many rows as there are cutsets. Its elements have the following values:
qij = 1, if the branch j is in the cutset i and the orientation coincide.
= −1, if the branch j is in the cutset i and the orientations do not coincide.
= 0, if the branch j is not in the cutset i.
Figure 9.17 shows a directed graph and its cutsets.
(6)
1
3
2
(2)
(4)
(3)
(1)
(5)
Cutset 1: {1, 2, 6}
Cutset 2: {2, 3, 4}
Cutset 3: {3, 1, 5}
Cutset 4: {4, 5, 6}
Cutset 5: {5, 2, 3, 6}
Cutset 6: {6, 1, 3, 4}
Cutsets
↓
1
2
3
4
5
6
Branches →
1 2 3 4 5 6
1 1 0 0 0 1
0 1 −1 1 0 0
1 0 1 0 1 0
0 0 0 1 1 −1
0 −1 1 0 1 −1
1 0 1 −1 0 1
4
Fig. 9.17 Directed graph
For the cutset 2, which cuts the branches 2, 3 and 4 and is shown by a dotted circle, the entry in the cutset
schedule for the branch 2 is 1, since the orientation of this cutset is given by the orientation of the branch 2
and hence it coincides. The entry for branch 3 is −1 as orientation of branch 3 is opposite to that of cutset 2,
9.6 Cutset Matrix 9.11
i.e., branch 2 goes into cutset while branch 3 goes out of cutset. The entry for branch 4 is 1 as the branch 2
and the branch 4 go into the cutset. Thus their orientations coincide.
Hence, the cutset matrix Qa is given as
⎡1 1 0 0
⎢0 1 −1 1
⎢
1 0 1 0
Qa = ⎢
0
0 0 1
⎢
⎢0 −1 1 0
⎢ 1 0 1 −1
⎣
0 1⎤
0 0⎥
⎥
1 0⎥
1 −1⎥
1 −1⎥
0 1⎥⎦
9.6.1 Fundamental Cutset and Fundamental Cutset Matrix
When a graph is given, first select a tree and note down its twigs. When a twig is removed from the tree, it
separates a tree into two parts (one of the separated part may be an isolated node). Now, all the branches
connecting one part of the disconnected tree to the other along with the twig removed, constitutes a cutset.
This set of branches is called a fundamental cutset or f-cutset. A matrix formed by these f-cutsets is called an
f-cutset matrix. The orientation of the f-cutset is made to coincide with the orientation of the defining branch,
i.e., twig. The number of f-cutsets is the same as the number of twigs for a graph.
Figure 9.18 shows a graph, selected tree and f-cutsets corresponding to the selected tree.
(6)
1
(6)
3
2
(2)
(2)
(4)
(1)
(3)
(5)
(2)
(4)
(3)
(1)
(4)
(3)
(5)
f-cutset 2: {2, 1, 6}
f-cutset 3: {3, 1, 5}
f-cutset 4: {4, 5, 6}
4
(a) Graph
(b) Tree
(c) f-cutsets
Fig. 9.18 Graph, selected tree and f-cutsets
The branches 2, 1, and 6 are in the f-cutset 2. Orientation of f-cutset 2 is given by orientation of the branch
2 which is moving away from the f-cutset 2. Since the orientations of branches 2, 1 and 6 coincide with the
orientation of the f-cutset 2, q11 = 1, q12 = 1 and q16 = 0. The branches 3, 4 and 5 are not in the f-cutset 2.
Hence, q13 = 0, q14 = 0 and q15 = 0.
Similarly, other elements of the f-cutset matrix are written.
The cutset schedule is
f-cutsets
↓
2
3
4
Hence, the f-cutset matrix Q is given by
⎡ 1
Q=⎢ 1
⎢
⎣ 0
1
1
1
0
Branches →
2 3 4 5 6
1 0 0 0 1
0 1 0 1 0
0 0 1 1 −1
0 1⎤
1 0⎥
⎥
1 −1⎦
The f-cutset matrix Q is rearranged so that the first (n − 1) columns correspond to twigs and b − (n − 1)
columns to links of the selected tree.
1
0
0
0
1
0
0
0
1
9.12 Network Analysis and Synthesis
Twigs
Links
2
3
4
1
5
6
⎡ 1
Q=⎢ 0
⎢
⎣ 0
0
1
0
0
0
1
1
1
0
0 1⎤
1 0⎥
⎥
1 −11⎦
The matrix Q can be subdivided into matrices Qt and Ql.
Q
Qt : Q
U : Ql ]
where Qt is the twig matrix, Ql is the link matrix and U is the unit matrix.
9.6.2
Orthogonal Relationship between Matrix B and Matrix Q
For a linear graph, if the columns of two matrices Ba and Qa are arranged in the same order, it can be shown
that
Q BaT = 0
B QaT = 0
or
If the f-circuit matrix B and the f-cutset matrix Q are written for the same selected tree, it can be shown that
B QT = 0
or
Q BT = 0
These two equations show the orthogonal relationship between matrices A and B.
9.7
RELATIONSHIP AMONG SUBMATRICES OF A, B AND Q
Arranging the columns of matrices A, B and Q with twigs for a given tree first and then the links, we get the
partitioned forms as
A [ At : Al ]
B [ Bt : Bl ] [ Bt : U ]
Q
Qt : Q
U : Ql ]
From the orthogonal relation, ABT = 0,
⎡ Bt T ⎤
AB = [ At : Al ] ⎢ … ⎥
⎢
⎥
⎢⎣ Bl T ⎥⎦
T
At Bt T + Al Bl T = 0
At Bt T = − Al Bl T
Since At is non-singular, i.e., |A| ≠ 0, At−1 exists.
Premultiplying with At−1,
Bt T
At−1 Al Bl T
Since Bl is a unit matrix
Bt
Bl ( At 1 Al )T
Bt
( At−1. Al )T
9.7 Relationship Among Submatrices of A, B and Q 9.13
Hence, matrix B is written as
B [ ( At−1 Al )T : U ]
…(9.1)
ABT = 0
We know that
Al Bl T = − At Bt T
Postmultiplying with (
T −1
) ,
At Bt T ( Bl T ) −1 = − At Bt T (B
( Bl 1 )T
Al
At ( Bl 1 Bt )T
Hence matrix A can be written as
A [ Al : At ( Bl−1 Bt )T ]
…(9.2)
= At [U : − ( Bl 1 Bt )T ]
Similarly we can prove that
Q
U : − ( Bl 1 Bt )T ]
A
At Q
Q
At−1 A
…(9.3)
From Eqs (9.2) and (9.3), we can write
We have shown that
At−1[ At : Al ] = [U : At−11 Al ]
Bt
( At−1. Al )T
Bt T
( At−1 Al )
Q
U : − Bt T ]
Hence, Q can be written as
Bt T
Q
Example 9.3 For the circuit shown in Fig. 9.19, draw the oriented graph and write the (a) incidence
matrix, (b) tieset matrix, and (c) f-cutset matrix.
R5
R2
R1
V
L1
R4
R6
+
−
C
R3
I
(4)
Fig. 9.19
Solution For drawing the oriented graph,
1. replace all resistors, inductors and capacitors by line segments,
2. replace the voltage source by short circuit and the current source by an open
circuit,
3. assume the directions of branch currents arbitrarily, and
4. number all the nodes and branches.
The oriented graph is shown in Fig. 9.20.
1
2
(3)
(1)
(2)
3
Fig. 9.20
9.14 Network Analysis and Synthesis
(a)
Incidence Matrix (A)
Nodes Branches →
⎡ −1 0 −1 1⎤
↓
1 2 3 4
Aa = ⎢ 0
1 1 −1⎥
⎢
⎥
1
−1 0 −1 1
⎣ 1 −1 0 0 ⎦
2
0 1 1 −1
3
1 −1 0 0
Eliminating the third row from the matrix Aa, we get the incidence matrix A.
⎡ −1 0 −1 1⎤
A= ⎢
⎣ 0 1 1 −1⎥⎦
(b) Tieset Matrix (B)
The oriented graph, selected tree and tiesets are shown in Fig. 9.21.
(4)
1
2
Links: {3, 4}
Tieset 3: {3, 1, 2}
Tieset 4: {4, 1, 2}
(3)
(1)
(2)
1
2
3 ⎡ −11
B= ⎢
4⎣ 1
3 4
1 1 0⎤
1 0 1⎥⎦
3
(c)
Fig. 9.21
f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.22.
(4)
1
2
(3)
(1)
Twigs:
{1, 2}
f-cutset 1: {1, 3, 4}
f-cutset 2: {2, 3, 4}
1 2 3
1⎡1 0 1
Q= ⎢
2 ⎣0 1 1
(2)
4
1⎤
1⎥⎦
3
Fig. 9.22
Example 9.4
For the network shown in Fig. 9.23, draw the oriented graph and write the
(a) incidence matrix, (b) tieset matrix, and (c) f-cutset matrix.
2Ω
2Ω
4Ω
1F
2Ω
V1
10 Ω
+
−
I
1H
2Ω
+
− V2
Fig. 9.23
4F
9.7 Relationship Among Submatrices of A, B and Q 9.15
Solution For drawing the oriented graph,
1. replace all resistors, inductors and capacitors by line segments,
2. replace all voltage sources by short circuits and current source
by an open circuit,
3. assume directions of branch currents arbitrarily, and
4. number all the nodes and branches.
The oriented graph is shown in Fig. 9.24.
(6)
(7)
5
2
1
3
(1)
(2)
(3)
(4)
(5)
4
(a) Incidence Matrix (A)
Fig. 9.24
Nodes
↓
1
1
1
2
−1
3
0
4
0
5
0
2
0
−1
1
0
0
Branches →
3 4 5 6 7
0 1 0 1 0
1 0 0 0 0
0 0 1 0 1
−1 −1 −1 0 0
0 0 0 −1 −1
1 0
1 0⎤
⎡ 1 0 0
⎢ −1 −1 1 0 0 0 0 ⎥
⎢
⎥
Aa = ⎢ 0
1 0 0
1 0
1⎥
⎢ 0 0 −1 −1 −1 0 0 ⎥
⎢⎣ 0 0 0 0 0 −1 −1⎥⎦
Eliminating the last row from the matrix Aa, we get the incidence matrix A.
1 0
⎡ 1 0 0
⎢ −1 −1 1 0 0
A= ⎢
1 0 0
1
⎢ 0
⎢⎣ 0 0 −1 −1 −1
(b)
1
0
0
0
0⎤
0⎥
⎥
1⎥
0 ⎥⎦
Tieset Matrix (B)
The oriented graph, selected tree and tiesets are shown in Fig. 9.25.
(6)
(7)
5
2
1
(1)
3 Links: {2, 4, 7}
Tieset 2: {2, 3, 5}
Tieset 4: {4, 1, 3}
Tieset 7: {7, 6, 1, 3, 5}
(2)
(3)
(4)
(5)
1 2
3
4
5
1 0 −1
1 1 0
1 0 −11
2⎡ 0 1
B = 4 ⎢ −1 0
⎢
7⎣ 1 0
6 7
0 0⎤
0 0⎥
⎥
1 1⎦
4
(c)
Fig. 9.25
f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.26.
5
(6)
(7)
2
1
3
(2)
(1)
(3)
(4)
(5)
4
Fig. 9.26
Twigs: {1, 3, 5, 6}
f-cutset 1: {1, 4, 7}
f-cutset 3: {3, 4, 7, 2}
f-cutset 5: {5, 2, 7}
f-cutset 6: {6, 7}
1
2
3
4 5
6
1⎡1 0
3 ⎢0 −1
Q= ⎢
5 ⎢0
1
6 ⎢⎣0 0
0
1
0
0
1
1
0
0
0 −1⎤
0 −1⎥
⎥
0
1⎥
1 1⎥⎦
0
0
1
0
7
9.16 Network Analysis and Synthesis
Example 9.5 For the circuit shown in Fig. 9.27, draw the oriented graph and write (a) incidence
matrix, (b) tieset matrix, and (c) cutset matrix.
i
L1
r2
r1
r5
r4
r6
r3
+
V
C1
−
L2
Fig. 9.27
1.
2.
3.
4.
replace all resistors, inductors and capacitors by line segments,
replace voltage source by short circuit and current source by an open
circuit,
assume directions of branch currents arbitrarily, and
number the nodes and branches.
(1)
3
⎡ −1 1 0 −1⎤
Aa = ⎢ 0 0
1 1⎥
⎢
⎥
⎣ 1 −1 −1 0 ⎦
Eliminating the third row from the matrix Aa, we get the incidence matrix A.
⎡ −1 1 0 −1⎤
A= ⎢
⎣ 0 0 1 1⎥⎦
Tieset Matrix (B)
The oriented graph, selected tree and tiesets are shown in Fig. 9.29.
(1)
(4)
2
(2)
(3)
Links: {1, 3}
Tieset 1: {1, 2}
Tieset 3: {3, 2, 4}
1
1⎡1
B= ⎢
3 ⎣0
3
Fig. 9.29
(c)
(3)
Fig. 9.28
Nodes Branches →
↓
1 2 3 4
1
−1 1 0 −1
2
0 0 1 1
3
1 −1 −1 0
1
2
(2)
The oriented graph is shown in Fig. 9.28.
(a) Incidence Matrix (A)
(b)
(4)
1
Solution For drawing the oriented graph,
f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.30.
2
3
1 0
1 1
4
0⎤
1⎥⎦
9.7 Relationship Among Submatrices of A, B and Q 9.17
(4)
1
(1)
2 Twigs:
{2, 4}
f-cutset 2: {2, 1, 3}
f-cutset 4: {4, 3}
(2)
1 2 3 4
(3)
2 ⎡ −1 1 1 0 ⎤
Q= ⎢
4 ⎣ 0 0 1 1⎥⎦
3
Fig. 9.30
Example 9.6 For the circuit shown in Fig. 9.31, (a) draw its graph, (b) draw its tree, and (c) write
the fundamental cutset matrix.
2Ω
1F
1F
1Ω
1Ω
I
1Ω
1H
(3)
Fig. 9.31
Solution
(a) For drawing the oriented graph,
1. replace all resistors, inductors and capacitors by line segments,
2. replace the current source by an open circuit,
3. assume directions of branch currents, and
4. number all the nodes and branches.
The oriented graph is shown in Fig. 9.32.
(b) Tree
The oriented graph and its selected tree are shown in Fig. 9.33.
(3)
2
1
3
(2)
(4)
(5)
(1)
(6)
Twigs:
{2, 4, 5}
f-cutset 2: {2, 1, 3}
f-cutset 4: {4, 3, 6}
f-cutset 5: {5, 1, 6}
4
Fig. 9.33
(c) Fundamental Cutset Matrix (Q)
1
2
3 4 5 6
2 ⎡ 1 1 1 0 0 0⎤
Q = 4 ⎢0 0 −1 1 0 1⎥
⎢
⎥
5 ⎣ 1 0 0 0 1 1⎦
2
1
3
(2)
(4)
(5)
(1)
4
Fig. 9.32
(6)
9.18 Network Analysis and Synthesis
Example 9.7 The graph of a network is shown in Fig. 9.34. Write the (a) incidence matrix,
(b) tieset matrix, and (c) f-cutset matrix.
(2)
1
3
(4)
(5)
2
(1)
(3)
(6)
4
Fig. 9.34
Solution
(a) Incidence Matrix (A)
1
1 ⎡ −1
2⎢ 0
Aa = ⎢
3⎢ 0
4 ⎢⎣ 1
2
3
4
5
6
1 0
1 0
0 0 −1 1
1 1 0 −1
0 −11 0 0
0⎤
1⎥
⎥
0⎥
1⎥⎦
The incidence matrix A is obtained by eliminating any row from the matrix Aa.
1 0 0⎤
⎡ −1 1 0
A = ⎢ 0 0 0 −1 1 1⎥
⎢
⎥
⎣ 0 −1 1 0 −1 0 ⎦
(b)
Tieset, Matrix (B)
The oriented graph, selected tree and tiesets are shown in Fig. 9.35.
(2)
1
3
(5)
(4)
2
(1)
(3)
(6)
Links: {1, 2, 3}
Tieset 1: {1, 4, 6}
Tieset 2: {2, 4, 5}
Tieset 3: {3, 5, 6}
1 2 3
4
1⎡1 0 0
B = 2 ⎢0 1 0
⎢
3 ⎣0 0 1
5
1 0
1 −1
0
1
6
1⎤
0⎥
⎥
1⎦
4
(c)
Fig. 9.35
f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.36.
(2)
1
3
(5)
(4)
2
(1)
(6)
(3)
4
Fig. 9.36
Twigs:
{4, 5, 6}
f-cutset 4: {4, 1, 2}
f-cutset 5: {5, 2, 3}
f-cutset 6: {6, 1, 3}
1
2
3
4
5
6
4 ⎡ −1 1 0 1 0 0 ⎤
Q = 5 ⎢ 0 1 −1 0 1 0 ⎥
⎢
⎥
6 ⎣ −1 0
1 0 0 1⎦
9.7 Relationship Among Submatrices of A, B and Q 9.19
Example 9.8
For the graph shown in Fig. 9.37, write the incidence matrix, tieset matrix and f-cutset
matrix.
(5)
2
1
4
3
(2)
(3)
(4)
(6)
(1)
(7)
5
Fig. 9.37
Solution
(a) Incidence Matrix (A)
1
2
1⎡ 1
2⎢ 0
⎢
Aa = 3 ⎢ 0
4⎢ 0
5 ⎢⎣ −1
3
4
5
1 0 0
1 1 0
0 −1 1
0 0 −11
0 0 0
6
7
0⎤
0⎥
⎥
0⎥
1⎥
1⎥⎦
0 0
1 0
0 0
1 1
0 −1
The incidence matrix A is obtained by eliminating any row from the matrix Aa.
⎡1 1 0 0 0
⎢0 −1 1 0
1
A= ⎢
⎢0 0 −1 1 0
⎢⎣0 0 0 −1 −1
(b)
(c)
Tieset Matrix (B)
Links: {5, 6, 7}
Tieset 5: {5, 3, 4}
Tieset 6: {6, 1, 2, 3, 4}
Tieset 7: {7, 1, 2, 3, 4}
1
5⎡ 0
B = 6 ⎢ −1
⎢
7⎣ 1
2
3
4
0 0⎤
0 0⎥
⎥
0 0⎥
1 −1⎥⎦
5
6 7
0 −1 −1 1 0 0 ⎤
1 1 1 0 1 0⎥
⎥
1 1 1 0 0 1⎦
f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.38.
(5)
2
1
(2)
3
(4)
(3)
(6)
(1)
(7)
5
Fig. 9.38
4
1
Twigs:
{1, 2, 3, 4}
f-cutset 1: {1, 6, 7}
f-cutset 2: {2, 6, 7}
f-cutset 3: {3, 5, 6, 7}
f-cutset 4: {4, 5, 6, 7}
1 ⎡1
2 ⎢0
Q= ⎢
3 ⎢0
4 ⎢⎣0
2 3 4 5
0
1
0
0
0
0
1
0
0
0
0
1
6
0
1
0 −1
1 1
1 1
7
1⎤
1⎥
⎥
1⎥
1⎥⎦
9.20 Network Analysis and Synthesis
Example 9.9
For the graph shown in Fig. 9.39, write the incidence matrix, tieset matrix and
f-cutset matrix.
(1)
1
2
(5)
(6)
(2)
(4)
5
(8)
(7)
3
4
(3)
Fig. 9.39
Solution
(a) Incidence Matrix (A)
1
2
1⎡ 1
2 ⎢ −1
⎢
Aa = 3 ⎢ 0
4⎢ 0
5 ⎢⎣ 0
3
4
5
6
7
8
1 1 0 0 0⎤
0 0
1 0 0⎥
⎥
0 0 0
1 0⎥
1 0 0 0
1⎥
0 −1 −1 −1 −1⎥⎦
0 0
1 0
1 1
0 −1
0 0
The incidence matrix is obtained by eliminating any one row.
⎡ 1 0 0 −1
⎢ −1 1 0 0
A= ⎢
⎢ 0 −1 1 0
⎢⎣ 0 0 −1 1
(b) Tieset Matrix (B)
Links: {2, 4, 6, 8}
Tieset 2: {2, 7, 5, 1}
Tieset 4: {4, 5, 7, 3}
Tieset 6: {6, 5, 1}
Tieset 8: {8, 7, 3}
(c)
1 2 3 4
2⎡1
4 ⎢0
B= ⎢
6 ⎢1
8 ⎢⎣0
1
0
0
0
0
1
0
1
1
0
0
0
0
1
0
0
5 6
0 −1
1 1
0 −1
0 0
0⎤
0⎥
⎥
0⎥
1⎥⎦
0
0
1
0
7
0
1
0 −1
1 0
0 −1
8
0⎤
0⎥
⎥
0⎥
1⎥⎦
f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.40.
(1)
1
2
(5)
Twigs:
{1, 3, 5, 7}
f-cutset 1: {1, 6, 2}
f-cutset 3: {3, 4, 8}
f-cutset 5: {5, 4, 6, 2}
f-cutset 7: {7, 2, 8, 4}
(6)
(2)
(4)
5
(8)
(7))
3
4
(3)
Fig. 9.40
1
1⎡1
3 ⎢0
Q= ⎢
5 ⎢0
7 ⎢⎣0
2 3
4
5
6
7
8
1
0
1
1
0
1
1
1
0
0
1
0
1
0
1
0
0
0
0
1
0⎤
1⎥
⎥
0⎥
1⎥⎦
0
1
0
0
9.7 Relationship Among Submatrices of A, B and Q 9.21
Example 9.10
How many trees are possible for the graph of the network of Fig. 9.41.
2
3
+
−
1
Fig. 9.41
(2)
Solution To draw the graph,
1. replace all resistors, inductors and capacitors by line segments,
2. replace voltage source by short circuit and current source by an open
circuit,
3. assume directions of branch currents arbitrarily, and
4. number all the nodes and branches.
The oriented graph is shown in Fig. 9.42.
The complete Incidence Matrix (Aa) is written as
3
2
3
(1)
(3)
(4)
1
Fig. 9.42
4
1 ⎡ 1 0 −1
Aa = 2 ⎢ −1 1 0
⎢
3 ⎣ 0 −1
1 1
1⎤
0⎥
⎥
1⎦
The reduced incidence matrix A is obtained by eliminating the last row from matrix Aa.
⎡ 1 0 −1 1⎤
A= ⎢
⎣ −1 1 0 0 ⎥⎦
⎡ 1 −1⎤
1
0
−
1
1
1⎥ ⎡ 3 −1⎤
⎡
⎤⎢ 0
AAT = ⎢
⎢
⎥=
⎥
⎣ −1 1 0 0 ⎦ ⎢ −1 0 ⎥ ⎢⎣ −1 2⎥⎦
⎢⎣ 1 0 ⎥⎦
3 −1
The number of possible trees = AAT =
= 6 − 1 = 5.
−1 2
Example 9.11
Draw the oriented graph from the complete incidence matrix given below;
↓
1
2
3
4
5
(8)
Branches →
Nodes
1
2
3
4
5
6
7
8
1 0 0 0 1 0 0 1
0 1 0 0 −1 1 0 0
0 0 1 0 0 −1 1 −1
0 0 0 1 0 0 −1 0
−1 −1 −1 −1 0 0 0 0
1
(5)
(1)
2
(6)
(2)
(3)
3
(7)
4
(4)
5
Fig. 9.43
Solution First, note down the nodes 1, 2, 3, 4, 5 as shown in Fig. 9.43. From the complete incidence matrix, it
is clear that the branch number 1 is between nodes 1 and 5 and it is going away from node 1 and towards node 5
as the entry against node 1 is 1 and that against 5 is −1. Hence, connect the nodes 1 and 5 by a line, point the arrow
towards 5 and call it branch 1 as shown in Fig. 9.43. Similarly, draw the other oriented branches.
9.22 Network Analysis and Synthesis
Example 9.12
The reduced incidence matrix of an oriented graph is given below. Draw the
graph.
⎡ 0 −1 1 1 0 ⎤
A = ⎢ 0 0 −1 −1 −1⎥
⎢
⎥
1⎦
⎣ −1 0 0 0
Solution First, writing the complete incidence matrix from the matrix A such that the sum of all entries in
each column of Aa will be zero, we have
(4)
1
2
3
4
5
1 ⎡ 0 −1 1
2 ⎢ 0 0 −1
1
Aa = ⎢
3 ⎢ −1 0 0
4 ⎢⎣ 1 1 0
1
1
0
0
0⎤
1⎥
⎥
1⎥
0 ⎥⎦
2
1
(3)
3
(5)
(1)
(2)
4
Fig. 9.44
Now, the oriented graph can be drawn with matrix Aa as shown in Fig. 9.44.
Example 9.13
The reduced incidence matrix of an oriented graph is
⎡ 0 −1 1 0 0 ⎤
A = ⎢ 0 0 −1 −1 −1⎥
⎢
⎥
1⎦
⎣ −1 0 0 0
(a) Draw the graph. (b) How many trees are possible for this graph? (c) Write the tieset and cutset matrices.
Solution
(a) First, writing the complete incidence matrix Aa such that the sum of all the entries in each column of Aa
is zero, we have
2
1
1
1⎡ 0
2⎢ 0
Aa = ⎢
3 ⎢ −1
4 ⎢⎣ 1
2
3
4
5
1 1
0 −1
1
0 0
1 0
0
1
0
1
0⎤
1⎥
⎥
1⎥
0 ⎥⎦
(3)
3
(5)
(4)
(2)
(1)
4
Fig. 9.45
Now, the oriented graph can be drawn with the matrix Aa, as shown in Fig. 9.45.
(b) The number of possible trees = |AAT|
⎡ 0 0 −1⎤
⎡ 0 −1 1 0 0 ⎤ ⎢ −1 0 0 ⎥ ⎡ 2 −1 0 ⎤
⎢
⎥
AAT = ⎢ 0 0 −1 −1 −1⎥ ⎢ 1 −1 0 ⎥ = ⎢ −1 3 −1⎥
⎢
⎥
⎢
⎥
1⎦ ⎢ 0 −1 0 ⎥ ⎣ 0 −1 2⎦
⎣ −1 0 0 0
⎢⎣ 0 −1 1⎥⎦
9.7 Relationship Among Submatrices of A, B and Q 9.23
2 −1 0
AAT = −1 3 −1 = 2(6 − 1) + 1( −2) = 8
0 −1 2
The number of possible trees = 8.
(c) Tieset Matrix (B)
The oriented graph, selected tree and tiesets are shown in Fig. 9.46.
2
1
3
(3)
(5)
(4)
(2)
1
(1)
2 3
4 5
1 ⎡1 0 0 −1 1 ⎤
B= ⎢
2 ⎣0 1 1 −1 0 ⎥⎦
Links: {1, 2}
Tieset 1: {1, 4, 5}
Tieset 2: {2, 3, 4}
Fig. 9.46
f-cutset Matrix (Q)
The oriented graph, selected tree and f-cutsets are shown in Fig. 9.47
2
1
3
(3)
(5)
(4)
(2)
(1)
Twigs:
{3, 4, 5}
f-cutset 3: {3, 2}
f-cutset 4: {4, 2, 1}
f-cutset 5: {5, 1}
1
3⎡ 0
Q = 4⎢ 1
⎢
5 ⎣ −1
2 3 4 5
1 1 0 0⎤
1 0 1 0⎥
⎥
0 0 0 1⎦
4
Fig. 9.47
Example 9.14
The fundamental cutset matrix of a network is given as follows;
Twigs
g
a c e
Links
b d f
1 0 0
0 1 0
0 0 1
1 0 1
0 1 1
1 1 1
Draw the oriented graph.
Solution No. of links l = b − n + 1
No. of nodes n = b − l + 1 = 6 − 3 + 1 = 4
f-cutsets are written as,
f-cutsets a: {a, b, f }
f-cutsets c: {c, d, f }
f-cutsets e: {e, b, d, f }
The oriented graph is drawn as shown in Fig. 9.48.
(f)
(a)
(b)
(d)
(e)
Fig. 9.48
Twigs: {a, c, e}
Links: {b, d, f}
(c)
9.24 Network Analysis and Synthesis
Example 9.15
Draw the oriented graph of a network with the f-cutset matrix as shown below:
Twigs
g
1
1
0
0
0
2
0
1
0
0
3
0
0
1
0
Links
k
4 5
0 −1
0
1
0 0
1 0
Solution No. of links l = b − n + 1
No. of nodes n = b − l + 1 = 7 − 3 + 1 = 5
f-cutsets are written as
f-cutset 1: {1, 5}
f-cutset 2: {2, 5, 7}
f-cutset 3: {3, 6, 7}
f-cutset 4: {4, 6}
Then oriented graph can be drawn as shown in
Fig. 9.49.
6
0
0
1
1
7
0
1
1
0
(5)
(2)
(7)
(3)
(1)
(4)
Twigs: {1, 2, 3, 4}
Links: {5, 6, 7}
(6)
Fig. 9.49
9.8
KIRCHHOFF’S VOLTAGE LAW
KVL states that if vk is the voltage drop across the kth branch, then
∑ vk = 0
the sum being taken over all the branches in a given loop. If l is the number of loops or f-circuits, then there
will be l number of KVL equations, one for each loop. The KVL equation for the f-circuit or loop ‘l’ can be
written as
b
∑ bik
k
=0
(k = 1, 2, …, l)
k =1
where bik is the elements of the tieset matrix B, b being the number of branches. The set of l KVL
equations can be written in matrix form.
BV
Vb = 0
⎡ v1 ⎤
⎢v ⎥
Vb = ⎢ 2 ⎥ is a column vector of branch voltages.
⎢ ⎥
⎢⎣ vb ⎥⎦
where
and B is the fundamental circuit matrix.
9.9
KIRCHHOFF’S CURRENT LAW
KCL states that if ik is the current in the kth branch then at a given node
∑ ik = 0
the sum being taken over all the branches incident at a given node. If there are ‘n’ nodes, there will ‘n’ such
equations, one for each node
9.10 Relation Between Branch Voltage Matrix Vb, Twig Voltage Matrix Vt And Node Voltage Matrix Vn
9.25
b
∑ aik ik = 0
(k = 1, 2,… , n)
k =1
so that set of n equations can be written in matrix form.
Aa I b = 0
…(9.4)
⎡ i1 ⎤
⎢i ⎥
where
I b = ⎢ 2 ⎥ is a column vector of branch currents.
⎢ ⎥
⎢⎣ib ⎥⎦
and Aa is the complete incidence matrix.
If one node is taken as reference node or datum node, we can write the Eq. (9.4) as,
AI b = 0
where A is the incidence matrix of order (n − 1) × b.
We know that
A
…(9.5)
At Q
Equation (9.5) can be written as
AQ
QII b = 0
Premultiplying with At−1,
At 1 At Q I b = At−1. 0
I Q Ib = 0
Q Ib = 0
where Q is the f-cutset matrix.
9.10
RELATION BETWEEN BRANCH VOLTAGE MATRIX Vb, TWIG
VOLTAGE MATRIX Vt AND NODE VOLTAGE MATRIX Vn
BV
Vb = 0
We know that
⎡Vt ⎤
: Bl ] ⎢ ⎥ = 0
⎢ ⎥
⎣Vl ⎦
[
Bt Vt + Bl Vl = 0
Bl Vl = − Bt Vt
Premultiplying with Bl−1 .
Vl
Now
Bl−11 Bt Vt = − ( Bl 1 Bt )Vt
⎡Vt ⎤
Vb = ⎢ ⎥
⎢ ⎥
⎣Vl ⎦
⎡
⎢
=⎢
⎢−
⎢⎣
(
Vt
…
⎤ ⎡
⎥ ⎢ U
⎥=⎢ …
Vt ⎥⎥ ⎢⎢ −
⎦ ⎣
)
(
)
⎤
⎥
⎥ .Vt
⎥
⎥⎦
…(9.6)
9.26 Network Analysis and Synthesis
Also,
V
QT Vt
Q
At−1 A
AT ( AtT ) −1
QT
AT ( At 1 )T
Vb
AT ( AtT ) 1Vt = AT
Vn
( AtT ) −1Vt is node voltage matrix.
Hence Eq. (9.6) can be written as
where
9.11
{A
T
t
Vt
}
AT Vn
RELATION BETWEEN BRANCH CURRENT MATRIX Ib AND LOOP
CURRENT MATRIX Il
We know that, A I b = 0
⎡ It ⎤
: At ] ⎢…⎥ = 0
⎢ ⎥
⎣ Il ⎦
[
At I t + Al I l = 0
At I t = − Al I l
Premultiplying with At−1 ,
At−11 Al I l = − ( At 1 Al ) I l
It
1
1
⎡ I t ⎤ ⎡ − ( At Al ) I l ⎤ ⎡ − ( At Al ) ⎤
⎢
⎥
⎢
I b = ⎢…⎥ =
…
=
… ⎥ .I l
⎥ ⎢
⎥
⎢ ⎥ ⎢
Il
⎥⎦ ⎢⎣ U
⎥⎦
⎣ I l ⎦ ⎢⎣
Now
I b = BT I l
9.12
9.12.1
1.
NETWORK EQUILIBRIUM EQUATION
KVL Equation
If there is a voltage source vsk in the branch k having impedance zk and carrying current ik, as shown in
Fig. 9.50,
vk
zk ik
vsk
Vb
Zb I b Vs
(k = 1, 2,…, b)
In matrix form,
vsk
Zk
+
−+
ik
where Zb is the branch impedance matrix, Ib is the column vector
of branch currents and Vs is the column vector of source voltages.
Hence, KVL equation can be written as
BV
Vb = 0
B ( Zb I b Vs ) = 0
B Zb I b BV
Vs
vk
Fig. 9.50 Circuit diagram
−
9.12 Network Equilibrium Equation 9.27
Ib
Also,
BT I l
B Zb BT I l = BV
Vs
Z Il = E
where
E BV
Vs
and
Z B Z b BT
The matrix Z is called loop impedance matrix.
2.
If there is a voltage source in series with an impedance and
a current source in parallel with the combination as shown in
Fig. 9.51,
(v + vsk )
ik = k
− isk
zk
vk = zk ik zk isk vsk
In matrix form,
Vb Zb I b Zb I s Vs
vk
ik +
Zk
−+
−
vsk
isk
Fig. 9.51 Circuit diagram
KVL equation is BV
Vb = 0.
BV
Vb = B ( Zb I b Zb I s Vs ) = 0
B Zb I b BV
Vs B Zb I s
Now
Ib
BT I l
B Zb BT I l = BV
Vs − B Zb I s
Z I l = BV
Vs − B Zb I s
where Z
9.12.2
1.
B Zb BT is the loop impedance matrix. This is the generalised KVL equation.
KCL Equation
If the branch k contains an input current source isk and an
admittance yk as shown in Fig. 9.52,
i
yk vk
isk
Ib
Yb Vb
Is
ik +
vk
yk
−
(k = 1, 2,…, b)
In the matrix form,
where Yb is the branch admittance matrix.
Hence KCL equation is given by,
A Ib = 0
A (Yb Vb I s ) = 0
AY
Yb Vb A I s
Also
Vb
AT Vn
AY
Yb AT Vn = A I s
Y Vn = I
where
and
Y
I
AY
Yb AT
A Is
isk
Fig. 9.52 Circuit diagram
9.28 Network Analysis and Synthesis
The matrix Y is called admittance matrix. This is the KCL equation in matrix form.
In terms of f-cutset matrix, the KCL equation can be written as
Q Ib = 0
Q (Yb Vb Vs ) = 0
QY
Yb V Q I s
Also
V
QT Vt
QY
Yb QT Vt = Q I s
Y Vt = I
2.
where
Y QY
Y QT
and
I Q Is
This is the KCL equation in matrix form.
If there is a voltage source in series with an impedance and a
current source in parallel with the combination as shown in
Fig. 9.53,
1
yk =
zk
i
yk v + yk vsk isk
In matrix form,
I b Yb Vb Yb Vs I s
ik +
zk
vk
−+
vsk
isk
Fig. 9.53 Circuit diagram
KCL equation will be given by,
A Ib = 0
A (Yb Vb Yb Vs I s ) = 0
AY
Yb Vb A I s
Also
Vb
AY
Yb Vs
AT Vn
AY
Yb AT Vn = A I s − AY
Yb Vs
Y Vn = A I s − AV
Vb Vs
where Y AY
Yb AT is the node admittance matrix. This is a generalised KCL equation.
In terms of f-cutset matrix, the KCL equation can be written as
Q Ib = 0
Q (Yb Vb Yb Vs I s ) = 0
QY
Yb V Q I
Also
V
QY
Yb Vs
QT Vt
QY
Yb QT Vt = Q I s − QY
Yb Vs
Y Vt = Q I s − QY
Yb Vs
This is a generalised KCL equation.
−
9.12 Network Equilibrium Equation 9.29
Note
(i) For a graph having b branches, the branch impedance matrix Zb is a square matrix of order b, having
branch impedances as diagonal elements and the mutual impedances between the branches as nondiagonal elements. For a network having no mutual impedances, only diagonal elements will be present
in the branch impedance matrix.
(ii) For a graph having b branches, the branch admittance matrix Yb is a square matrix of order b, having
branch admittances as diagonal elements and the mutual admittances between the branches as nondiagonal elements. For a network having no mutual admittances, only diagonal elements will be present
in the branch admittance matrix.
(iii) For a graph having b branches, the voltage source matrix or vector Vs is a rectangular matrix of order
b × 1, having the value of the voltage source in the particular branch. The value will be positive if there
is a voltage rise in the direction of current and will be negative if there is a voltage fall in the direction
of current.
(iv) For a graph having b branches, the current source matrix or vector Is is a rectangular matrix of order
b × 1, having the value of the current source in the particular parallel branch. The value will be positive if
the direction of the current source and the corresponding parallel branch current are not same. The value
will be negative if the directions of the current source and corresponding parallel branch current are same.
Example 9.16 Write the incidence matrix of the graph of Fig. 9.54 and express branch voltages in
terms of node voltages. Write the tieset matrix and express branch currents in terms of loop currents.
(8)
(5)
(6)
2
1
3
(2)
(1)
5
(7)
(3)
(4)
Fig. 9.54
Solution
(a) Incidence Matrix
1
2
3
4
5
6
7
8
1⎡ 1 0 0 0
1 0
2⎢ 0
1 0 0 −1 1
⎢
Aa = 3 ⎢ 0 0
1 0 0 −11
4⎢ 0 0 0
1 0 0
5 ⎢⎣ −1 −1 −1 −1 0 0
0
0
1
1
0
1⎤
0⎥
⎥
1⎥
0⎥
0 ⎥⎦
The incidence matrix is obtained by eliminating any one row.
⎡1
⎢0
A= ⎢
⎢0
⎢⎣0
0
1
0
0
0
0
1
0
0
1 0 0
1⎤
0 −1 1 0 0 ⎥
⎥
0 0 −1 1 −1⎥
1 0 0 −1 0 ⎥⎦
9.30 Network Analysis and Synthesis
(b)
Branch voltages in terms of node voltages
AT Vn
Vb
⎡V1 ⎤ ⎡ 1
⎢V2 ⎥ ⎢0
⎢ ⎥ ⎢
⎢V3 ⎥ ⎢0
⎢V4 ⎥ = ⎢0
⎢V5 ⎥ ⎢ 1
⎢V ⎥ ⎢0
⎢ 6⎥ ⎢
⎢V7 ⎥ ⎢0
⎢⎣V8 ⎥⎦ ⎢⎣ 1
(c)
0 0
1 0
0
1
0 0
1 0
1 −1
0
1
0 −1
0⎤
0⎥
⎥
0 ⎥ ⎡Vn1 ⎤
1⎥ ⎢Vn2 ⎥
⎢ ⎥
0 ⎥ ⎢Vn3 ⎥
0 ⎥⎥ ⎢⎣Vn4 ⎥⎦
1⎥
0 ⎥⎦
Tieset Matrix
Selected tree and tiesets are shown in Fig. 9.55.
(8)
(5)
1
(6)
2
(2)
(1)
1
3
(7)
(3)
Links: {1, 4, 6, 8}
Tieset 1: {1, 2, 5}
Tieset 4: {4, 3, 7}
Tieset 6: {6, 2, 3}
Tieset 8: {8, 2, 3, 5}
1 ⎡1
4 ⎢0
B= ⎢
6 ⎢0
8 ⎢⎣0
2
3 4
1 0
0 −1
1 1
1 1
5 6 7 8
0 −1
1 0
0 0
0 −1
0
0
1
0
0
1
0
0
0⎤
0⎥
⎥
0⎥
1⎥⎦
4
5
(4)
Fig. 9.55
(d) Branch currents in terms of loop currents
Ib
BT I l
⎡ I1 ⎤ ⎡ 1 0 0 0 ⎤
⎢ I 2 ⎥ ⎢ −1 0 −1 −1⎥
⎢ ⎥ ⎢
⎥ ⎡I ⎤
⎢ I 3 ⎥ ⎢ 0 −1 1 1⎥ ⎢ l1 ⎥
1 0 0 ⎥ I l4
⎢I4 ⎥ = ⎢ 0
⎢ I 5 ⎥ ⎢ −1 0 0 −1⎥ ⎢⎢ I l6 ⎥⎥
⎢I ⎥ ⎢ 0 0
1 0 ⎥⎥ ⎢⎣ I l8 ⎥⎦
⎢ 6⎥ ⎢
1 0 0⎥
⎢ I7 ⎥ ⎢ 0
⎢⎣ I 8 ⎥⎦ ⎢⎣ 0 0 0
1⎥⎦
Example 9.17
Branch current and loop-current relationships are expressed in matrix form as
Draw the oriented graph.
⎡ I1 ⎤ ⎡ 1 0 0
⎢I2 ⎥ ⎢ 0
1 0
⎢ ⎥ ⎢
I
0
1 1
3
⎢ ⎥ ⎢
1 1
⎢I4 ⎥ = ⎢ 0
⎢ I 5 ⎥ ⎢ 1 −1 0
⎢ I ⎥ ⎢ 0 0 −1
⎢ 6⎥ ⎢
⎢ I 7 ⎥ ⎢ −1 0 0
⎢⎣ I 8 ⎥⎦ ⎢⎣ 0 0 0
1⎤
1⎥
⎥
0 ⎥ ⎡ I l1 ⎤
0 ⎥ ⎢ I l2 ⎥
⎢ ⎥
0 ⎥ ⎢ I l3 ⎥
0 ⎥⎥ ⎢⎣ I l4 ⎥⎦
0⎥
1⎥⎦
9.12 Network Equilibrium Equation 9.31
Solution Writing the equation in matrix form,
Ib
BT I l
⎡ 1 0 0 −1⎤
⎢ 0
1 0 −1⎥
⎢
⎥
0
1 1 0⎥
⎢
0
1 1 0⎥
BT = ⎢⎢
1 −1 0 0 ⎥
⎢ 0 0 −1 0 ⎥
⎢
⎥
⎢ −1 0 0 0 ⎥
⎢⎣ 0 0 0
1⎥⎦
1
⎡ 1
⎢ 0
∴B = ⎢
⎢ 0
1
⎢⎣ −1
2 3 4
0
1
0
1
0
1
1
0
5
0
1
1 −1
1 0
0 0
6
7
0 −1
0 0
1 0
0 0
8
0⎤
0⎥
⎥
0⎥
1⎥⎦
(8)
(1)
1
2
(2)
3
(5)
(7)
From tieset matrix,
No. of links
l=4
No of branches b = 8
No of nodes
n= b − l + 1 = 8 − 4 + 1 = 1
The oriented graph is shown in Fig. 9.56.
(3)
4
5
(4)
(6)
Fig. 9.56
Example 9.18 For the given graph shown in Fig. 9.57, write down the basic tieset matrix and
taking a tree of branches 2, 4, 5, write down KVL equations from the matrix.
(2)
1
(5)
(1)
2
(3)
3
4
(4)
(6)
Fig. 9.57
Solution Selecting branches 2, 4, and 5 as the tree as shown in Fig. 9.58,
(2)
1
2
1 2
(5)
(1)
(3)
Links: {1, 3, 6}
Tieset 1: {1, 5, 4}
Tieset 3: {3, 2, 5}
Tieset 6: {6, 2, 5, 4}
3
4
(4)
(6)
Fig. 9.58
3
1⎡1 0 0
B = 3 ⎢0 1 1
⎢
6 ⎣0 1 0
4 5 6
1 1 0⎤
0 1 0⎥
⎥
1 1 1⎦
9.32 Network Analysis and Synthesis
The KVL equation in matrix form is given by
BV
Vb = 0
⎡V1 ⎤
⎢V ⎥
⎡ 1 0 0 −1 1 0 ⎤ ⎢ 2 ⎥
⎢0 1 1 0 1 0 ⎥ ⎢V3 ⎥ = 0
⎢
⎥ ⎢V4 ⎥
⎣0 1 0 −1 1 1⎦ ⎢V ⎥
5
⎢V ⎥
⎣ 6⎦
V1 V4 + V5 = 0
V2 V3 + V5 = 0
V4 + V5 V6 = 0
V2
Example 9.19 Obtain the f-cutset matrix for the graph shown in Fig. 9.59 taking 1, 2, 3, 4 as tree
branches. Write down the network equations from the f-cutset matrix.
2
(5)
(6)
(3)
3
1
(2)
(7)
(8)
(1)
(4)
4
Fig. 9.59
1 2 3
Solution Twigs:
{1, 2, 3, 4}
f-cutset 1 : {1, 6, 7, 8}
f-cutset 2 : {2, 5, 6, 7, 8}
f-cutset 3 : {3, 5, 6}
f-cutset 4 : {4, 6, 7}
1 ⎡1
2 ⎢0
Q= ⎢
3 ⎢0
4 ⎢⎣0
The KCL equation in matrix form is given by
Q Ib = 0
⎡1
⎢0
⎢
⎢0
⎢⎣0
0
1
0
0
0
0
1
0
0
0
0
1
0
1 −1
1 −1 1
1 −1 0
0
1 −1
⎡ I1 ⎤
⎢I2 ⎥
⎢ ⎥
1⎤ ⎢ I 3 ⎥
−1⎥ ⎢ I 4 ⎥
=0
⎥
0⎥ ⎢ I 5 ⎥
0 ⎥⎦ ⎢⎢ I 6 ⎥⎥
⎢ I7 ⎥
⎢⎣ I 8 ⎥⎦
0
1
0
0
0
0
1
0
4
5
6
7
8
0
0
0
1
0
1
1
0
1
1
1
1
1
1
0
1
1⎤
1⎥
⎥
0⎥
0 ⎥⎦
9.12 Network Equilibrium Equation 9.33
I1
I2
Example 9.20
I6 − I7
I5 − I6
I3
I4
I8 = 0
I 7 − I8 = 0
I5 − I6 = 0
I6 − I7 = 0
The reduced incidence matrix of a graph is given as
⎡ 1 0 0 0 −1⎤
A = ⎢ −1 −1 −1 0 0 ⎥
⎢
⎥
1 −1 0 ⎦
⎣ 0 0
Express branch voltages in terms of node voltages.
Solution For the given graph,
No of branches b = 5
No of nodes n = 3
Branch voltages can be expressed in terms of node voltages by
Vb
T
Vn
⎡V1 ⎤ ⎡ 1 −1 0 ⎤
⎢V2 ⎥ ⎢ 0 −1 0 ⎥ ⎡Vn1 ⎤
⎢ ⎥ ⎢
⎥⎢ ⎥
⎢V3 ⎥ = ⎢ 0 −1 1⎥ ⎢Vn2 ⎥
⎢V4 ⎥ ⎢ 0 0 −1⎥ ⎣Vn3 ⎦
⎢⎣V5 ⎥⎦ ⎢⎣ −1 0 0 ⎥⎦
V1 Vn1 − Vn2
Example 9.21
V2
Vn2
V3
Vn2 + Vn3
V4
Vn3
V5
Vn1
The fundamental cutset matrix of a graph is given as
⎡ −1 1 0 0 −1⎤
Q = ⎢ 0 0 1 0 −1⎥
⎢
⎥
⎣1 0 0 1 0⎦
Express branch voltages in terms of twig voltages.
Solution For the given graph,
No. of branches b = 5
No. of twigs = 3
Branch voltages are expressed in terms of twig voltages by
V
QT Vt
⎡V1 ⎤ ⎡ −1 0
⎢V2 ⎥ ⎢ 1 0
⎢ ⎥ ⎢
1
⎢V3 ⎥ = ⎢ 0
⎢V4 ⎥ ⎢ 0 0
⎢⎣V5 ⎥⎦ ⎢⎣ −1 −1
1⎤
0 ⎥ ⎡Vt1 ⎤
⎥
0 ⎥ ⎢Vt2 ⎥
⎢ ⎥
1⎥ ⎣Vt3 ⎦
0 ⎥⎦
9.34 Network Analysis and Synthesis
Vt1 + Vt3
V1
V2
Vt1
V3
Vt2
V4
Vt3
Vt1 − Vt2
V5
Example 9.22 For this network shown in Fig. 9.60, write down the tieset matrix and obtain the
network equilibrium equation in matrix form using KVL. Calculate the loop currents and branch currents.
2Ω
1Ω
1Ω
2V
+
−
2Ω
2Ω
1Ω
Fig. 9.60
Solution The oriented graph and one of its trees are shown in Fig. 9.61.
1
(1)
(5)
3
(4)
2
(3)
1
(1)
(2)
(5)
(6)
4
3
(2)
(4)
Links: {1, 2, 3}
Tieset 1: {1, 4, 5}
Tieset 2: {2, 4, 6}
Tieset 3: {3, 5, 6}
(6)
2
1 2 3
1⎡1 0 0
B = 2 ⎢0 1 0
⎢
3 ⎣0 0 1
4
(3)
Fig. 9.61
The KVL equation in matrix form is given by
Here,
B Zb BT I l = BV
Vs − B Zb I s
I s = 0,
B Zb BT I l = BV
Vs
Zb
⎡1
⎢0
⎢
⎢0
⎢0
⎢0
⎢0
⎣
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
2
0
0
0
0
0
0
2
0
0⎤
⎡ 1 0 0⎤
⎡ 2⎤
⎢0
⎢0 ⎥
0⎥
1 0⎥
⎥
⎢
⎥
⎢ ⎥
0⎥ T ⎢0 0
1⎥
0
;B =
; Vs = ⎢ ⎥
0⎥
1
−
1
0
⎢
⎥
⎢0 ⎥
⎢ 1 0 −1⎥
⎢0 ⎥
0⎥
⎥
⎢
⎥
⎢0 ⎥
2⎦
⎣0 −1 1⎦
⎣ ⎦
4
5
1 1
1 0
0 −1
6
0⎤
1⎥
⎥
1⎦
9.12 Network Equilibrium Equation 9.35
1 1
⎡1 0 0
B Zb = ⎢0 1 0 −1 0
⎢
⎣0 0 1 0 −1
2 2
⎡1 0 0
B Zb BT = ⎢0 1 0 −2 0
⎢
⎣0 0 1 0 −2
1 1
⎡1 0 0
BV
Vs = ⎢0 1 0 −1 0
⎢
⎣0 0 1 0 −1
⎡1 0 0 0 0 0⎤
⎢0 1 0 0 0 0⎥
0⎤ ⎢
⎥ ⎡1
0 0 1 0 0 0⎥ ⎢
−1⎥ ⎢
= 0
⎥ ⎢0 0 0 2 0 0 ⎥ ⎢
1⎦ ⎢
0
0 0 0 0 2 0⎥ ⎣
⎢ 0 0 0 0 0 2⎥
⎣
⎦
1
0
0
⎡
⎤
⎢0
1 0⎥
0⎤ ⎢
⎥ ⎡ 5 −2
0 0
1⎥ ⎢
−2⎥ ⎢
= −2
5
⎥ 1 −1 0 ⎥ ⎢
2⎦ ⎢⎢
−
2
−
2
1 0 −1⎥ ⎣
⎢0 −1 1⎥
⎣
⎦
⎡ 2⎤
⎢0 ⎥
0 ⎤ ⎢ ⎥ ⎡ 2⎤
0
−1⎥ ⎢ ⎥ = ⎢0 ⎥
⎥ 0
⎢ ⎥
1⎦ ⎢⎢ ⎥⎥ ⎣0 ⎦
0
⎢0 ⎥
⎣ ⎦
The KVL equation in matrix form is given by
⎡ 5
⎢ −2
⎢
2
⎣ −2
2 −2⎤ ⎡ I l1 ⎤ ⎡ 2⎤
5 −2⎥ ⎢ I l2 ⎥ = ⎢0 ⎥
⎥⎢ ⎥ ⎢ ⎥
2
5⎦ ⎣ I l3 ⎦ ⎣0 ⎦
Solving this matrix equation,
6
A
7
4
I l2 = A
7
4
I l3 = A
7
I l1 =
The branch currents are given by
Ib
BT I l
⎡ I1 ⎤ ⎡ 1 0
⎢ I 2 ⎥ ⎢0
1
⎢ ⎥ ⎢
I
0
0
⎢ 3⎥ = ⎢
⎢ I 4 ⎥ ⎢ 1 −1
⎢ I5 ⎥ ⎢ 1 0
⎢ I ⎥ ⎢0 −1
⎣ 6⎦ ⎣
⎡6⎤
⎢7⎥
⎢4⎥
0⎤ ⎡ 6 ⎤ ⎢ ⎥
0⎥ ⎢ 7 ⎥ ⎢ 7 ⎥
⎥ ⎢ ⎥ ⎢4⎥
1⎥ ⎢ 4 ⎥ ⎢ ⎥
=
0⎥ ⎢ 7 ⎥ ⎢ 7 ⎥
2
−1⎥ ⎢ 4 ⎥ ⎢ ⎥
⎢7⎥
⎢
⎥
⎥
1⎦ ⎣ 7 ⎦ ⎢ ⎥
2
⎢ ⎥
⎢7⎥
⎢⎣ 0 ⎥⎦
0 0
1 0
0 1
−2⎤
−2⎥
⎥
5⎦
2 2
2 0
0 −2
0⎤
2⎥
⎥
2⎦
9.36 Network Analysis and Synthesis
Example 9.23 For the network shown in Fig. 9.62, write down the tieset matrix and obtain the
network equilibrium equation in matrix form using KVL. Calculate loop currents.
2Ω
+−
4Ω
6Ω
Il 3
6Ω
8V
2Ω
12 V +
−
Il1
4Ω
+
− 6V
Il2
Fig. 9.62
Solution The oriented graph and its selected tree are shown in Fig. 9.63.
(3)
(3)
2
1
(4)
3
(5)
(6)
(1)
2
1
(2)
(4)
3
(5)
(6)
(1)
4
Links: {1, 2, 3}
Tieset 1: {1, 4, 6}
Tieset 2: {2, 5, 6}
Tieset 3: {3, 5, 4}
(2)
1 2
3
4
5
6
1⎡1 0 0
1 0
1⎤
B = 2 ⎢0 1 0 0
1 −1⎥
⎢
⎥
3 ⎣0 0 1 −1 −1 0 ⎦
4
Fig. 9.63
The KVL equation in matrix form is given by
B Zb BT I l = BV
Vs − B Zb I s
I s = 0,
Here,
T
B Zb B I l = BV
Vs
⎡6
⎢0
⎢
⎢0
⎢0
⎢0
⎢0
⎣
0⎤
⎡1 0
⎥
⎢0
0
1
⎥
⎢
0⎥ T ⎢0 0
Zb
;B =
0⎥
⎢1 0
⎢0
0⎥
1
⎢ 1 −1
2⎥⎦
⎣
⎡6 0 0
⎢0 4 0
1 0
1⎤ ⎢
⎡1 0 0
0 0 2
B Zb = ⎢0 1 0 0
1 −1⎥ ⎢
⎢
⎥ ⎢0 0 0
⎣0 0 1 −1 −1 0 ⎦ ⎢0 0 0
⎢0 0 0
⎣
0
4
0
0
0
0
0
0
2
0
0
0
0
0
0
4
0
0
0
0
0
0
6
0
⎡1 0
⎢0
1
4
0
2⎤ ⎢
⎡6 0 0
0
0
BZb BT = ⎢0 4 0
0
6 −2⎥ ⎢
⎢
⎥ ⎢1 0
⎣0 0 2 −4 −6 0 ⎦ ⎢0
1
⎢ 1 −1
⎣
0⎤
⎡ 12⎤
⎥
⎢ −6 ⎥
0
⎥
⎢ ⎥
1⎥
−8
; Vs = ⎢ ⎥
−1⎥
⎢ 0⎥
⎢ 0⎥
−1⎥
⎢ 0⎥
0 ⎥⎦
⎣ ⎦
0 0 0⎤
0 0 0⎥
4
0
⎥ ⎡6 0 0
0 0 0⎥ ⎢
= 0 4 0
0
6
4 0 0⎥ ⎢
0 0 2 −4 −6
0 6 0⎥ ⎣
0 0 2⎥⎦
0⎤
0⎥
⎥ ⎡ 12 −2 −4 ⎤
1⎥ ⎢
= −1 12 −6 ⎥
−1⎥ ⎢
⎥
−4 −6 12⎦
−1⎥ ⎣
0 ⎥⎦
2⎤
2⎥
⎥
0⎦
9.12 Network Equilibrium Equation 9.37
⎡1 0 0
BV
Vs = ⎢0 1 0
⎢
⎣0 0 1
⎡ 12⎤
⎢ −6 ⎥
0
1⎤ ⎢ ⎥ ⎡ 12⎤
−8
1 −1⎥ ⎢ ⎥ = ⎢ −6 ⎥
⎥ 0
⎢ ⎥
1 0 ⎦ ⎢⎢ ⎥⎥ ⎣ −8⎦
0
⎢ 0⎥
⎣ ⎦
1
0
1
Hence, the KVL equation in matrix form is given by
4 ⎤ ⎡ I l1 ⎤ ⎡12 ⎤
6 ⎥ ⎢ I l2 ⎥ = ⎢ −6 ⎥
⎥⎢ ⎥ ⎢ ⎥
12⎦ ⎣ I l3 ⎦ ⎣ −8⎦
2
⎡ 12 −2
⎢ −2
2 12
⎢
4
6
⎣ −4
Solving this matrix equation,
I l1 = 0 55A
I l2 = −0.866 A
I l3 = −0.916 A
Example 9.24 For the network shown in Fig. 9.64, draw the oriented graph. Write the tieset
schedule and hence obtain the equilibrium equation on loop basis. Calculate the values of branch currents
and branch voltages.
+ 1V
−
1A
1Ω
1Ω
1Ω
1Ω
1Ω
Fig. 9.64
Solution The oriented graph and one of its trees are shown in Fig. 9.65.
1
1
(4)
(1)
(2)
2
(6)
3
(4)
(1)
(3)
(5)
(2)
4
2
(6)
3
Links: {1, 2, 3}
Tieset 1: {1, 4, 5, 6}
Tieset 2: {2, 4, 5}
Tieset 3: {3, 4}
(3)
(5)
4
Fig. 9.65
1 2 3
1⎡1 0 0
B = 2 ⎢0 1 0
⎢
3 ⎣0 0 1
4
5
1 1
1 1
1 0
6
1⎤
0⎥
⎥
0⎦
9.38 Network Analysis and Synthesis
The KVL equation in matrix form is given by
B Zb BT I l = BV
Vs − B Zb I s
⎡1
⎢0
⎢
0
Zb = ⎢
⎢0
⎢0
⎢0
⎣
0⎤
⎡ 1
⎢ 0
0⎥
⎥
⎢
0⎥ T ⎢ 0
;B =
0⎥
⎢ −1
⎢ 1
0⎥
⎢ −1
1⎥⎦
⎣
⎡1 0 0
⎢0 1 0
⎡ 1 0 0 −1 1 −1⎤ ⎢
0 0 1
B Zb = ⎢0 1 0 −1 1 0 ⎥ ⎢
⎢
⎥ ⎢0 0 0
⎣0 0 1 1 0 0 ⎦ ⎢0 0 0
⎢0 0 0
⎣
0
1
0
0
0
0
0
0
1
0
0
0
⎡1 0 0
BZb BT = ⎢0 1 0
⎢
⎣0 0 1
⎡1 0 0
BV
Vs = ⎢0 1 0
⎢
⎣0 0 1
0
0
0
0
0
0
0
0
0
0
1
0
0
1
0
−1
1
0
⎡ 1 0 0⎤
⎢ 0
1 0⎥
0 1 −1⎤ ⎢
⎥ ⎡3 1 0⎤
0
0
1⎥ ⎢
0 1 0⎥ ⎢
= 1 2 0⎥
⎥ ⎢ −1 −1 1⎥ ⎢
⎥
0 0 0⎦ ⎢
0 0 1⎦
1 1 0⎥ ⎣
⎢ −1 0 0 ⎥
⎣
⎦
⎡0 ⎤
⎢0 ⎥
−1 1 −1⎤ ⎢ ⎥ ⎡ −1⎤
0
−1 1 0 ⎥ ⎢ ⎥ = ⎢ −1⎥
⎥ ⎢1 ⎥ ⎢ ⎥
1 0 0 ⎦ ⎢ ⎥ ⎣ 1⎦
0
⎢0 ⎥
⎣ ⎦
⎡ −1⎤
⎢ 0⎥
⎡ 1 0 0 0 1 −1⎤ ⎢ ⎥ ⎡ −1⎤
0
B Zb I s = ⎢0 1 0 0 1 0 ⎥ ⎢ ⎥ = ⎢ 0 ⎥
⎢
⎥ ⎢ 0⎥ ⎢ ⎥
⎣0 0 1 0 0 0 ⎦ ⎢ 0 ⎥ ⎣ 0 ⎦
⎢ 0⎥
⎣ ⎦
Hence, the KVL equation in matrix form is given by
⎡ 3 1 0 ⎤ ⎡ I l1 ⎤ ⎡ −1⎤ ⎡ −1⎤ ⎡ 0 ⎤
⎢1 2 0 ⎥ ⎢ I l ⎥ = ⎢ −1⎥ − ⎢ 0 ⎥ = ⎢ −1⎥
⎢
⎥⎢ 2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣0 0 1 ⎦ ⎣ I l3 ⎦ ⎣ 1⎦ ⎣ 0 ⎦ ⎣ 1⎦
Solving this matrix equation,
1
A
5
3
I l2 = − A
5
I l3 = 1 A
I l1 =
0
0
0
0
0
0
0⎤
⎡0 ⎤
⎢0 ⎥
0⎥
⎥
⎢ ⎥
1⎥
0
; Vs = ⎢ ⎥ ; I s
1⎥
⎢1 ⎥
⎢0 ⎥
0⎥
⎢0 ⎥
0 ⎥⎦
⎣ ⎦
0 0⎤
0 0⎥
⎥ ⎡1 0
0 0⎥ ⎢
= 0 1
0 0⎥ ⎢
0 0
1 0⎥ ⎣
0 1 ⎥⎦
⎡ −1⎤
⎢ 0⎥
⎢ ⎥
0
=⎢ ⎥
⎢ 0⎥
⎢ 0⎥
⎢ 0⎥
⎣ ⎦
0 0 1
0 0 1
1 0 0
1⎤
0⎥
⎥
0⎦
9.12 Network Equilibrium Equation 9.39
The branch currents are given by
I b = BT I l
Is
⎡ I1 ⎤ ⎡ 1 0
⎢I2 ⎥ ⎢ 0
1
⎢ ⎥ ⎢
⎢ I3 ⎥ = ⎢ 0 0
⎢ I 4 ⎥ ⎢ −1 −1
⎢ I5 ⎥ ⎢ 1 1
⎢ ⎥ ⎢
⎣ I 6 ⎦ ⎣ −1 0
⎡ 4⎤
⎢− 5 ⎥
⎢ 3⎥
⎡ −1⎤ ⎢ − ⎥
0⎤
⎢ 5⎥
⎡ 1⎤
0⎥ ⎢ ⎥ ⎢ 0⎥ ⎢ 1 ⎥
⎥
⎢ ⎥
1⎥ ⎢ 5 ⎥ + ⎢ 0 ⎥ = ⎢ ⎥
3
⎢ 7⎥
1⎥ ⎢ − ⎥ ⎢ 0 ⎥ ⎢ ⎥
⎢ 5⎥
0⎥ ⎣ 1 ⎦ ⎢ 0⎥ ⎢ 5 ⎥
⎥
⎢ ⎥
2
0⎦
⎣ 0⎦ ⎢ − ⎥
⎢ 5⎥
⎢ 1⎥
⎢− ⎥
⎣ 5⎦
The branch voltages are given by
Vb
Zb I b Vs
⎡V1 ⎤ ⎡1
⎢V2 ⎥ ⎢0
⎢ ⎥ ⎢
⎢V3 ⎥ = ⎢0
⎢V4 ⎥ ⎢0
⎢V5 ⎥ ⎢0
⎢ ⎥ ⎢
⎣V6 ⎦ ⎣0
Example 9.25
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
⎡ 4⎤
⎡ 4⎤
⎢− 5 ⎥
⎢− 5 ⎥
⎢ 3⎥
⎢ ⎥
0⎤ ⎢ − ⎥ ⎡0 ⎤ ⎢ − 3 ⎥
⎢ 5⎥ ⎢ ⎥
⎥
⎢ 5⎥
0 ⎢ ⎥ 0
⎥
1 ⎢ ⎥ ⎢ ⎥
0⎥ ⎢ 7 ⎥ − ⎢0 ⎥ =
1
⎢ ⎥
0 ⎥ ⎢ ⎥ ⎢1 ⎥ ⎢ −1⎥
⎢ 5⎥
0 ⎥ ⎢ 2 ⎥ ⎢0 ⎥ ⎢ − 2 ⎥
⎥ −
⎢ ⎥
1 ⎦ ⎢ ⎥ ⎣0 ⎦ ⎢ 5 ⎥
5
⎢ 1⎥
⎢ ⎥
⎢− ⎥
⎢ − 1⎥
⎣ 5⎦
⎢⎣ 5 ⎥⎦
For the network shown in Fig. 9.66, obtain the loop equation in matrix form.
R1
R2
I
V
R4
R3
Fig. 9.66
Solution The oriented graph and one of its trees are shown in Fig. 9.67.
1
1
(2)
(1)
2
(2)
(4)
(1)
2
(3)
3
(3)
3
Fig. 9.67
Links: {1, 4}
(4) Tieset 1: {1, 2, 3}
Tieset 4: {4, 2, 3}
1 2 3 4
1 ⎡1 1 1 0 ⎤
B= ⎢
4 ⎣0 1 1 1 ⎥⎦
9.40 Network Analysis and Synthesis
The KVL equation in matrix form is given by
B Zb BT I l = BV
Vs − B Zb I s
0 0
0⎤
⎡ R1
⎢ 0 R2
0
0⎥ T
Zb = ⎢
⎥; B
0
0
R
0⎥
3
⎢
0 0 R4 ⎦⎥
⎢⎣ 0
0
⎡ R1
⎡1 1 1 0 ⎤ ⎢ 0 R2
B Zb = ⎢
⎢
0
⎣0 1 1 1 ⎥⎦ ⎢ 0
0
⎢⎣ 0
⎡1 0⎤
⎡V ⎤
⎡0 ⎤
⎢1 1⎥
⎢0⎥
⎢I ⎥
=⎢
⎥ ; Vs = ⎢ ⎥ ; I s = ⎢ ⎥
1
1
0
⎢
⎥
⎢ ⎥
⎢0 ⎥
0
1
0
⎢⎣
⎥⎦
⎢⎣ ⎥⎦
⎢⎣0 ⎥⎦
0
0⎤
0
0 ⎥ ⎡ R1 R2 R3
0⎤
⎥=
R3
0 ⎥ ⎢⎣ 0 R2 R3 R4 ⎥⎦
0 R4 ⎥⎦
⎡1 0⎤
0 ⎤ ⎢1 1 ⎥ ⎡ R1 + R2 + R3
⎢
⎥=
R4 ⎥⎦ ⎢1 1 ⎥ ⎢⎣
R2 R3
0
1
⎢⎣
⎥⎦
⎡V ⎤
⎡1 1 1 0 ⎤ ⎢ 0 ⎥ ⎡V ⎤
BV
Vs = ⎢
⎢ ⎥=
⎣0 1 1 1 ⎥⎦ ⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦
⎢⎣ 0 ⎥⎦
⎡0 ⎤
0 ⎤ ⎢ I ⎥ ⎡ R2 I ⎤
⎡ R1 R2 R3
B Zb I s = ⎢
⎢ ⎥=
⎣ 0 R2 R3 R4 ⎥⎦ ⎢0 ⎥ ⎢⎣ R2 I ⎥⎦
⎢⎣0 ⎥⎦
Hence, the KVL equation in matrix form is given by
⎡ R R2
B Z b BT = ⎢ 1
⎣ 0 R2
⎡ R1
⎢⎣
R2 + R3
R2 R3
R3
R3
R2 + R3
⎤ ⎡ I l1 ⎤ ⎡V ⎤ ⎡ R2
=
−
R2 R3 + R4 ⎥⎦ ⎢⎣ I l4 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ R2
R2 + R3
⎤
R2 R3 + R4 ⎥⎦
I ⎤ ⎡V R2 I ⎤
=
I ⎥⎦ ⎢⎣ − R2 I ⎥⎦
Example 9.26 For the network shown in Fig. 9.68, write down the tieset matrix and obtain the
network equilibrium equation in matrix form using KVL.
2A
5Ω
j5 Ω
5Ω
2Ω
+
10 V
−4Ω
−j
−
Fig. 9.68
2Ω
9.12 Network Equilibrium Equation 9.41
Solution The oriented graph and its selected tree are shown in Fig. 9.69.
(3)
(3)
2
1
(4)
3
(5)
(1)
(6)
2
1
(4)
(2)
(1)
4
3
(5)
(6)
(2)
Links: {1, 2, 3}
Tieset 1: {1, 4, 6}
Tieset 2: {2, 5, 6}
Tieset 3: {3, 5, 4}
1 2
3
4
5
1⎡1 0 0
1 0
B = 2 ⎢0 1 0 0
1
⎢
3 ⎣0 0 1 −1 −1
4
Fig. 9.69
The KVL equation in matrix form is given by
B Zb BT I l = BV
Vs − B Zb I s
⎡2
⎢0
⎢
0
Zb = ⎢
⎢0
⎢0
⎢0
⎣
0
0⎤
⎡1
⎢0
0
0⎥
⎥
⎢
0
0 ⎥ T ⎢0
;B =
0
0⎥
⎢1
⎢0
j5
0⎥
⎢1
0 − j 4 ⎥⎦
⎣
⎡2 0 0
⎢0 2 0
1 0
1⎤ ⎢
⎡1 0 0
0 0 5
B Zb = ⎢0 1 0 0
1 −1⎥ ⎢
⎢
⎥ ⎢0 0 0
⎣0 0 1 −1 −1 0 ⎦ ⎢0 0 0
⎢0 0 0
⎣
0
2
0
0
0
0
0
0
5
0
0
0
0
0
0
5
0
0
⎡1 0
⎢0
1
5
0 − j 4⎤ ⎢
⎡2 0 0
0
0
B Z b BT = ⎢ 0 2 0 0
j5
j 4⎥ ⎢
⎢
⎥ 1 0
0 ⎦ ⎢⎢
⎣0 0 5 −5 − j 5
0
1
⎢ 1 −1
⎣
0 0⎤
⎡10 ⎤
⎢ 0⎥
1 0⎥
⎥
⎢ ⎥
0
1⎥
0
; Vs = ⎢ ⎥ ; I s
0 −1⎥
⎢ 0⎥
⎢ 0⎥
1 −1⎥
⎢ 0⎥
1 0 ⎥⎦
⎣ ⎦
0 0
0⎤
0 0
0⎥
⎥ ⎡2 0
0 0
0⎥ ⎢
= 0 2
5 0
0⎥ ⎢
0 0
0 j5
0⎥ ⎣
0 0
j 4 ⎥⎦
0⎤
0⎥
j4
⎥ ⎡7 − j 4
1⎥ ⎢
=
j 4 2 + j1
−1⎥ ⎢
−5
− j5
−1⎥ ⎣
⎥
0⎦
⎡10 ⎤
⎢0⎥
1 0
1⎤ ⎢ ⎥ ⎡100 ⎤
⎡1 0 0
0
BV
Vs = ⎢0 1 0 0
1 −1⎥ ⎢ ⎥ = ⎢ 0 ⎥
⎢
⎥⎢0 ⎥ ⎢ ⎥
⎣0 0 1 −1 −1 0 ⎦ ⎢ 0 ⎥ ⎣ 0 ⎦
⎢0⎥
⎣ ⎦
⎡2 0 0
B Zb I s = ⎢0 2 0
⎢
⎣0 0 5
5
0
0
j5
5 − j5
⎡ 0⎤
⎢ 0⎥
j 4⎤ ⎢ ⎥ ⎡ 0⎤
−2
j 4⎥ ⎢ ⎥ = ⎢ 0⎥
⎥ ⎢ 0⎥ ⎢
⎥
0 ⎦ ⎢ ⎥ ⎣ −10 ⎦
0
⎢ 0⎥
⎣ ⎦
⎡ 0⎤
⎢ 0⎥
⎢ ⎥
−2
=⎢ ⎥
⎢ 0⎥
⎢ 0⎥
⎢ 0⎥
⎣ ⎦
0
0
5
5
0
0
j5
5 − j5
−5⎤
− j 5⎥
⎥
10 + j 5⎦
j 4⎤
j 4⎥
⎥
0⎦
6
1⎤
1⎥
⎥
0⎦
9.42 Network Analysis and Synthesis
Hence, the KCL equation in matrix form is given by
j4
5⎤ ⎡ I l1 ⎤ ⎡10 ⎤ ⎡ 0 ⎤ ⎡10 ⎤
⎡7 j 4
⎢
j 4 2 + j1
j 5⎥ ⎢ I l2 ⎥ = ⎢ 0 ⎥ − ⎢ 0 ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ ⎥ ⎢ ⎥ ⎢
⎥ ⎢ ⎥
5
j 5 10 + j 5⎦ ⎣ I l3 ⎦ ⎣ 0 ⎦ ⎣ −10 ⎦ ⎣10 ⎦
⎣ −5
Example 9.27 For the network shown in Fig. 9.70, write down the tieset matrix and obtain the
network equilibrium equation in matrix form using KVL.
j5.66 Ω
j5 Ω
50 ∠0° V
j10 Ω
3Ω
+
−
−4Ω
−j
I1
5Ω
I2
Fig. 9.70
Solution The branch currents are so chosen that they assume the direction out of the dotted terminals.
Because of this choice of current direction, the mutual inductance is positive. The oriented graph and its
selected tree are shown in Fig. 9.71.
1
1
(1)
(2)
(3)
2
(1)
(2)
(3)
Links: {1, 3}
Tieset 1: {1, 2}
Tieset 3: {3, 2}
1
⎡1
B=⎢
⎣0
2
3
1 0⎤
1 1⎥⎦
2
Fig. 9.71
The KVL equation in matrix form is given by
Here,
B Zb BT I l = BV
Vs − B Zb I s
I s = 0,
B Zb BT I l = BV
Vs
0
j 5 66 ⎤
⎡ j5
⎡ 1 0⎤
⎡50 ∠0°⎤
Zb = ⎢ 0
3 − j4
0 ⎥ ; BT ⎢ 1 1⎥ ; Vs = ⎢ 0 ⎥
⎢
⎥
⎢
⎥
⎢
⎥
0
5 j10 ⎦
1⎦
⎣ j 5 66
⎣0
⎣ 0 ⎦
0
j 5 66 ⎤
⎡ j5
3 − j4
j 5.66 ⎤
⎡ 1 1 0⎤ ⎢
⎡ j5
B Zb = ⎢
0
3
−
j
4
0 ⎥=⎢
⎥
0
−
1
1
j
5
66
3
+
j
4
5
+ j10 ⎥⎦
⎢
⎥
⎣
⎦ j 5 666
0
5 + j10 ⎦ ⎣
⎣
⎡ 1 0⎤
3 j4
j 5.66 ⎤ ⎢
−3 + j 9.66 ⎤
⎡ 3 + j1
1 −1⎥ = ⎢
⎥
3 j 4 5 + j10 ⎦ ⎢
−3 + j 9 66
8 + j 6 ⎥⎦
⎥
1⎦ ⎣
⎣0
⎡50 ∠0°⎤
⎡ 1 1 0⎤ ⎢
⎡50 ∠0°⎤
BV
Vs = ⎢
0 ⎥=⎢
⎥
0
−
1
1
⎢
⎥
⎣
⎦
⎣ 0 ⎥⎦
⎣ 0 ⎦
⎡ j5
B Z b BT = ⎢
⎣ j 5 66
9.12 Network Equilibrium Equation 9.43
Hence, the KVL equation in matrix form is given by
−3 j 9.66 ⎤ ⎡ I l1 ⎤ ⎡50 ∠0°⎤
⎡ 3 j1
=
⎢⎣ −3
3 j 9 66
8 j 6 ⎥⎦ ⎢⎣ I l2 ⎥⎦ ⎢⎣ 0 ⎥⎦
Example 9.28 For the network shown in Fig. 9.72, write down the tieset matrix and obtain the
network equilibrium equation in matrix form using KVL.
j4 Ω
3Ω
j3 Ω
+
50 ∠45° V
j5 Ω
−
−j8 Ω
I2
I1
Fig. 9.72
Solution The branch currents are so chosen that they assume the direction out of the dotted terminals.
Because of this choice of current direction, the mutual inductance is positive. The oriented graph and its
selected tree are shown in Fig. 9.73.
1
1
(1)
(2)
(3)
(1)
(2)
(3)
Links: {1, 3}
Tieset 1: {1, 2}
Tieset 3: {3, 2}
1
⎡1
B=⎢
⎣0
2 3
1 0⎤
1 1⎥⎦
2
2
Fig. 9.73
The KVL equation in matrix form is given by
B Zb BT I l = BV
Vs − B Zb I s
Here,
I s = 0,
B Zb BT I l = BV
Vs
⎡3 + j 4
Zb = ⎢ j 3
⎢
⎣ 0
j3 0 ⎤
⎡ 1 0⎤
⎡50 ∠45°⎤
j 5 0 ⎥ ; BT = ⎢ 1 −1⎥ ; Vs = ⎢ 0 ⎥
⎥
⎢
⎥
⎢
⎥
0 − j 8⎦
1⎦
⎣0
⎣ 0 ⎦
⎡3 + j 4 j 3 0 ⎤
0 ⎤
⎡ 1 1 0⎤ ⎢
⎡3 + j 7 j 8
B Zb = ⎢
j3
j5 0 ⎥ = ⎢
⎥
0
−
1
1
−
j
3
−
j
5
−
j8⎥⎦
⎢
⎥
⎣
⎦
0 − j 8⎦ ⎣
⎣ 0
⎡ 1 0⎤
j8
0⎤ ⎢
⎡3 + j15
1 −1⎥ = ⎢
j 5 − j8⎥⎦ ⎢
⎥ ⎣ − j8
1⎦
⎣0
⎡50 ∠45°⎤
⎡ 1 1 0⎤ ⎢
⎡50 ∠45°⎤
BV
Vs = ⎢
0 ⎥=⎢
⎥
0
−
1
1
⎢
⎥
⎣
⎦
⎣ 0 ⎥⎦
⎣ 0 ⎦
⎡3 + j 7
B Z b BT = ⎢
⎣ − j3
j 8⎤
j 3⎥⎦
9.44 Network Analysis and Synthesis
Hence, the KVL equation in matrix form is given by,
⎡3 j15
⎢⎣ − j8
Example 9.29
j8⎤ ⎡ I l1 ⎤ ⎡50 ∠45°⎤
=
j 3⎥⎦ ⎢⎣ I l3 ⎥⎦ ⎢⎣ 0 ⎥⎦
For the network shown in Fig. 9.74, obtain branch voltages using KCL equation on
node basis.
8Ω
3Ω
4Ω
4Ω
5Ω
12 V
24 V
Fig. 9.74
Solution The oriented graph is shown in Fig. 9.75.
(3)
1
2
(2)
(4)
(1)
(5)
3
Fig. 9.75
The complete incidence matrix for the graph is
1
2
3
4 5
1 ⎡ 1 −1 1 0
Aa = 2 ⎢ 0 0 −1 1
⎢
3 ⎣ −1 1 0 −1
0⎤
1⎥
⎥
1⎦
Eliminating the last row from the matrix Aa, we get the incidence matrix A.
1
2
3
4 5
1 ⎡ 1 −1 1 0 0 ⎤
A= ⎢
2 ⎣0 0 −1 1 1⎥⎦
The KCL equation in matrix form is given by
AY
Yb AT Vn = AI s − AY
Yb Vs
Is = 0 ,
9.12 Network Equilibrium Equation 9.45
AY
Yb AT Vn = AY
Yb Vs
Here,
⎡1
⎢4 0 0
⎢
⎢0 1 0
⎢
3
⎢
1
Yb = ⎢ 0 0
8
⎢
⎢0 0 0
⎢
⎢
⎢0 0 0
⎣
⎤
0⎥
⎥
⎡ 1 0⎤
⎡ 0⎤
0 0⎥
⎥
⎢ −1 0 ⎥
⎢ 12⎥
⎥ T ⎢
⎥
⎢ ⎥
− ⎥ ; Vs = ⎢ 0 ⎥
0 0⎥ ; A = ⎢
⎥
1⎥
⎢ 0
⎢ 24 ⎥
1
⎢⎣ 0
⎢⎣ 0 ⎥⎦
1⎥⎦
0⎥
⎥
4
1⎥
0
⎥
5⎦
0
⎡1
⎢4
⎢
⎢0
⎢
⎡ 1 −1 1 0 0 ⎤ ⎢
AY
Yb = ⎢
⎢0
⎣0 0 −1 1 1⎥⎦ ⎢
⎢0
⎢
⎢
⎢0
⎣
1
1
⎡1
−
⎢
3
8
AY
Yb AT = ⎢ 4
1
⎢0
0 −
8
⎣
1
1
⎡1
⎢4 − 3
8
AY
Yb Vs = ⎢
1
⎢0
0 −
8
⎣
Hence, the KCL equation in matrix form is given by
⎡ 17
⎢ 24
⎢ 1
⎢−
⎣ 8
1⎤
− ⎥ ⎡V ⎤
n1
⎡ −4 ⎤ ⎡ 4 ⎤
8
= −⎢ ⎥ = ⎢ ⎥
23 ⎥⎥ ⎢⎣Vn2 ⎥⎦
⎣ 6 ⎦ ⎣ −6 ⎦
40 ⎦
Solving this matrix equation,
Vn1 = 3 96 V
Vn2 = −9 57 V
0
0
0
1
3
0
0
0
1
8
0
0
0
1
4
0
0
0
⎤
0⎥
⎥
0⎥
⎥ ⎡1
⎥ ⎢
0⎥ = ⎢ 4
⎥ ⎢0
0⎥ ⎣
⎥
1⎥
⎥
5⎦
⎡ 1 0⎤
⎤
⎡ 17
0 ⎥ ⎢ −1 0 ⎥ ⎢
⎢
⎥
1 −1⎥ = ⎢ 24
1 1 ⎥⎥ ⎢
1
1⎥ ⎢ −
⎢ 0
4 5⎦ ⎢ 0
8
⎣
1⎥⎦
⎣
⎡0⎤
⎤⎢ ⎥
0 0 ⎥ 12
⎢ ⎥ ⎡ −4 ⎤
0 =
1 1 ⎥⎥ ⎢ ⎥ ⎢⎣ 6 ⎥⎦
⎢ 24 ⎥
4 5⎦ ⎢ 0 ⎥
⎣ ⎦
0
1⎤
− ⎥
8
23 ⎥⎥
40 ⎦
−
1
3
0
1
8
1
−
8
0
1
4
⎤
0⎥
1 ⎥⎥
5⎦
9.46 Network Analysis and Synthesis
Branch voltages are given by,
Vb
AT Vn
⎡V1 ⎤ ⎡ 1 0 ⎤
⎡ 3 96 ⎤
⎢V2 ⎥ ⎢ −1 0 ⎥
⎢ −3 96 ⎥
⎢ ⎥ ⎢
⎥ ⎡ 3 96 ⎤ ⎢
⎥
⎢V3 ⎥ = ⎢ 1 −1⎥ ⎢⎣ −9 57⎥⎦ = ⎢ 13.53⎥
1⎥
⎢V4 ⎥ ⎢ 0
⎢ −9 57⎥
⎢⎣V5 ⎥⎦ ⎢⎣ 0
⎢⎣ −9 57⎥⎦
1⎥⎦
Example 9.30 For the network shown in Fig. 9.76, write down the f-cutset matrix and obtain the
network equilibrium equation in matrix form using KCL.
1Ω
1Ω
1Ω
1Ω
10 V
2A
Fig. 9.76
Solution The oriented graph and its selected tree are shown in Fig. 9.77.
(3)
1
(1)
(2)
1
2
(3)
2
(1)
(4)
(2)
3
Twigs:
{2, 4}
f-cutset 2: {2, 1, 3}
f-cutset 4: {4, 3}
(4)
1 2
2 ⎡ −1 1 1 0 ⎤
Q= ⎢
4 ⎣ 0 0 −1 1⎥⎦
3
Fig. 9.77
The KCL equation in matrix form is given by
QY
Yb QT Vt = Q I s − QY
Yb Vs
⎡1
⎢0
⎢
⎢0
⎢⎣0
0⎤
⎡0 ⎤
⎢0 ⎥
0⎥
Yb
⎥ I s = ⎢ ⎥ ; Vs
0⎥
⎢0 ⎥
1 ⎥⎦
⎢⎣ 2⎥⎦
⎡1 0 0
−
1
1
1
0
⎡
⎤ ⎢0 1 0
QY
Yb = ⎢
⎢
⎣ 0 0 −1 1⎥⎦ ⎢0 0 1
⎢⎣0 0 0
0
1
0
0
0
0
1
0
⎡ −1
⎡ −1 1 1 0 ⎤ ⎢ 1
QY
Yb Q = ⎢
⎢
⎣ 0 0 −1 1⎥⎦ ⎢ 1
⎢⎣ 0
T
⎡10 ⎤
⎡ −1 0 ⎤
⎢ 0⎥ T ⎢ 1 0⎥
= ⎢ ⎥;Q = ⎢
⎥
⎢ 0⎥
⎢ 1 1⎥
1⎥⎦
⎢⎣ 0 ⎥⎦
⎢⎣ 0
0⎤
0 ⎥ ⎡ −1 1 1 0 ⎤
⎥=
0 ⎥ ⎢⎣ 0 0 −1 1⎥⎦
1 ⎥⎦
0⎤
0⎥ ⎡ 3
⎥=
1⎥ ⎢⎣ −1
1⎥⎦
1⎤
2⎥⎦
3 4
9.12 Network Equilibrium Equation 9.47
⎡0 ⎤
⎡ −1 1 1 0 ⎤ ⎢0 ⎥ ⎡0 ⎤
Q Is = ⎢
⎢ ⎥=
⎣ 0 0 −1 1⎥⎦ ⎢0 ⎥ ⎢⎣ 2⎥⎦
⎢⎣ 2⎥⎦
⎡10 ⎤
−
1
1
1
0
⎡
⎤ ⎢ 0 ⎥ ⎡ −10 ⎤
QY
Yb Vs = ⎢
⎢ ⎥=
⎣ 0 0 −1 1⎥⎦ ⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦
⎢⎣ 0 ⎥⎦
Hence, the KCL equation is given by
1⎤ ⎡Vt2 ⎤ ⎡0 ⎤ ⎡ −10 ⎤ ⎡10 ⎤
=
−
=
2⎥⎦ ⎢⎣Vt4 ⎥⎦ ⎢⎣ 2⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 2 ⎥⎦
⎡ 3
⎢⎣ −1
Solving this matrix equation,
Vt2 = 4.4 V
Vt4 = 3 2 V
Example 9.31
Calculate the twig voltages using KCL equation for the network shown in Fig. 9.78.
2Ω
5Ω
5Ω
10 Ω
10 Ω
5Ω
+
910 V −
Fig. 9.78
Solution The oriented graph and one of the trees are shown in Fig. 9.79.
(5)
(3)
(1)
(5)
(4)
(6)
(3)
(2)
(1)
Twigs:
{1, 2, 3}
f-cutset 1: {1, 4, 5, 6}
f-cutset 2: {2, 4, 5}
f-cutset 3: {3, 4, 6}
(4)
(6))
(2)
Fig. 9.79
The network equilibrium equation on node basis can be written as
Here,
QY
Yb QT Vt = QI s − QY
Yb Vs
I s = 0,
QY
Yb QT Vt = QY
Yb Vs
1 2
3
4
5
6
1 ⎡ 1 0 0 −1 −1 1⎤
Q = 2 ⎢0 1 0 −1 −1 0 ⎥
⎢
⎥
3 ⎣0 0 1 1 0 1⎦
9.48 Network Analysis and Synthesis
0
0
0
0
0⎤
⎡0 2
⎡ 1
⎢ 0 0.2
⎢ 0
0
0
0
0⎥
⎢
⎥
⎢
0
0 02
0
0
0⎥ T ⎢ 0
Yb = ⎢
;Q =
0
0 01
0
0⎥
⎢ 0
⎢ −1
⎢ 0
⎢ −1
0
0
0 0.5
0⎥
⎢ 0
⎥
⎢ 1
0
0
0
0
0
1
⎣
⎦
⎣
0
0
0
⎡0 2
⎢ 0 0.2
0
0
1 −1 1⎤ ⎢
⎡1 0 0
0
0
0
2
0
QY
Yb = ⎢0 1 0
1 −1 0 ⎥ ⎢
0
0 01
⎢
⎥⎢ 0
⎣0 0 1 1 0 1⎦ ⎢ 0
0
0
0
⎢ 0
0
0
0
⎣
0 0⎤
⎡9110 ⎤
⎢ 0⎥
1 0⎥
⎥
⎢
⎥
0
1⎥
0⎥
; Vs = ⎢
−1 −1⎥
⎢ 0⎥
⎢ 0⎥
−1 0 ⎥
⎢ 0⎥
0
1⎥⎦
⎣
⎦
0
0⎤
0
0⎥
0
0 −0.1 −0.5 0.1⎤
⎥ ⎡0.2
0
0⎥ ⎢
= 0 0.2
0 −0.1 −0 5
0⎥
0
0⎥ ⎢
⎥
0
0 0.2 −0.1
0 0.1⎦
0.5
0⎥ ⎣
0 0 1⎥⎦
0
0 −0.1 −0.5
⎡0.2
QY
Yb QT = ⎢ 0 0.2
0 −0.1 −0 5
⎢
0 0.2 −0.1
0
⎣ 0
0
0 −0.1 −0.5
⎡0.2
QY
Yb Vs = ⎢ 0 0.2
0 −0.1 −0 5
⎢
0 0.2 −0.1
0
⎣ 0
⎡ 1 0 0⎤
⎢ 0
1 0⎥
0.1⎤ ⎢
⎥ ⎡ 0.9 0.6 0 2⎤
0
0
1⎥ ⎢
0⎥ ⎢
= 0.6 0.8 0 1⎥
⎥ ⎢ −11 1 1⎥ ⎢
⎥
0.1⎦ ⎢
0.2 0.1 0 3⎦
−11 1 0 ⎥ ⎣
⎢ 1 0
1⎥⎦
⎣
1 ⎤
⎡910
⎢ 0⎥
0.1⎤ ⎢
⎥ ⎡182⎤
0⎥ ⎢ ⎥
0⎥ ⎢
=
0
0⎥ ⎢ ⎥
⎥
0.1⎦ ⎢⎢
0
0⎥ ⎣ ⎦
⎢ 0⎥
⎣
⎦
Hence, KCL equation can be written as,
⎡ 0.9 0.6 0 2⎤ ⎡ vt1 ⎤ ⎡ −182⎤
⎢0.6 0.8 0 1⎥ ⎢ vt ⎥ = ⎢ 0 ⎥
⎢
⎥⎢ 2⎥ ⎢
⎥
⎣ 0.2 0.1 0 3⎦ ⎣ vt3 ⎦ ⎣ 0 ⎦
Solving this matrix equation,
vt1 = −460 V
vt2 = 320 V
vt3 = 200 V
For the network shown in Fig. 9.80, obtain equilibrium equation on node basis.
Ω
5
Fig. 9.80
5
Ω
10 A
10
Ω
5
Ω
Example 9.32
9.12 Network Equilibrium Equation 9.49
Solution The oriented graph and its selected tree are shown in Fig. 9.81.
(2)
(2)
2
1
1
(3)
(1)
(3)
(1)
(4)
3
2
(4)
Twigs:
{1, 3}
f-cutset 1: {1, 2}
f-cutset 3: {3, 2, 4}
1
2
3 4
1 ⎡1 −1 0 0 ⎤
Q= ⎢
3 ⎣0 −1 1 1 ⎥⎦
3
Fig. 9.81
The KCL equation in matrix form is given by
Here,
QY
Yb QT Vt = Q I s − QY
Yb Vs
Vs = 0,
QY
Yb QT Vt = Q I s
Y
⎡5
⎢0
⎢
⎢0
⎢⎣0
0
5
0
0
⎡1 −1
QY
Yb = ⎢
⎣0 −1
⎡ 5 −5
QY
Yb QT = ⎢
⎣0 −5
⎡1 −1
Q Is = ⎢
⎣0 −1
0 0⎤
⎡ 1 0⎤
⎡ −10 ⎤
⎥
⎢
⎥
⎢ 0⎥
0 0
−1 −1
T
⎥ Q =⎢
⎥ ; Is = ⎢
⎥
5 0⎥
1⎥
⎢ 0
⎢ 0⎥
0 10 ⎥⎦
1⎥⎦
⎢⎣ 0
⎢⎣ 0 ⎥⎦
⎡ 5 0 0 0⎤
0 0⎤ ⎢0 5 0 0 ⎥ ⎡ 5 5 0 0 ⎤
⎢
⎥=
1 1 ⎥⎦ ⎢0 0 5 0 ⎥ ⎢⎣0
5 5 10 ⎥⎦
1 ⎥⎦
⎢⎣0 0 0 10
⎡ 1 0⎤
0 0 ⎤ ⎢ −1 −1⎥ ⎡10 5⎤
⎢
⎥=
5 10 ⎥⎦ ⎢ 0
1⎥ ⎢⎣ 5 20 ⎥⎦
1⎥⎦
⎢⎣ 0
⎡ −10 ⎤
0 0 ⎤ ⎢ 0 ⎥ ⎡ −10 ⎤
⎢
⎥=
1 1 ⎥⎦ ⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦
⎢⎣ 0 ⎥⎦
Hence, KCL equation will be written as
⎡10 5⎤ ⎡ vt1 ⎤ ⎡ −10 ⎤
⎢⎣ 5 20 ⎥⎦ ⎢ vt ⎥ = ⎢⎣ 0 ⎥⎦
⎣ 3⎦
Solving this matrix equation,
8
vt1 = − V
7
2
vt3 = V
7
Example 9.33 For the network shown in Fig. 9.82, write down the f-cutset matrix and obtain the
network equilibrium equation in matrix form using KCL and calculate v.
9.50 Network Analysis and Synthesis
2v
2Ω
2Ω
2V
+
v
−
+
−
2Ω
2Ω
Fig. 9.82
Solution The oriented graph and its selected tree are shown in Fig. 9.83
Since voltage v is to be determined, Branch 2 is chosen as twig,
1
(4)
(2)
2
1
(2)
(3)
(1)
(4)
(3)
(1)
3
2
Twigs:
{2, 4}
f-cutset 2: {2, 1, 3}
f-cutset 3: {4, 3}
1
2
3 4
⎡ −1 1 1 0 ⎤
Q=⎢
⎣ 0 0 −1 1⎥⎦
3
Fig. 9.83
The KCL equation in matrix form is given by
QY
Yb QT Vt = Q I s − QY
Yb Vs
0
0
0⎤
⎡0 5
⎡ −1 0 ⎤
⎡ 0⎤
⎡ 2⎤
⎢ 0 0.5
⎢ 0⎥
⎢0 ⎥
0
0⎥ T ⎢ 1 0⎥
Y
⎢
⎥ Q =⎢
⎥ ; Is = ⎢
⎥ ; Vs = ⎢ ⎥
0 05
0⎥
⎢ 0
⎢ 1 −1⎥
⎢ 0⎥
⎢0 ⎥
0
0 0.5⎥⎦
1⎥⎦
⎢⎣ 0
⎢⎣ 0
⎢⎣ −2v ⎥⎦
⎢⎣0 ⎥⎦
0
0
0⎤
⎡0 5
⎢
0
0 ⎥ ⎡ −0.5 0.5 0.5
0⎤
⎡ −1 1 1 0 ⎤ 0 0.5
QY
Yb = ⎢
⎢
⎥=
⎥
0 05
0 ⎥ ⎢⎣ 0
0 −0.5 0.5⎥⎦
⎣ 0 0 −1 1⎦ ⎢ 0
0
0 0 5⎥⎦
⎢⎣ 0
⎡ −1 0 ⎤
0 ⎤ ⎢ 1 0 ⎥ ⎡ 1 5 −0 5⎤
⎡ −0.5 0.5 0 5
T
QY
Yb Q = ⎢
⎢
⎥=
0 −0.5 0.5⎥⎦ ⎢ 1 −1⎥ ⎢⎣ −0 5
1⎥⎦
⎣ 0
1⎥⎦
⎢⎣ 0
⎡ 0⎤
⎡ −1 1 1 0 ⎤ ⎢ 0 ⎥ ⎡ 0 ⎤
Q Is = ⎢
⎢
⎥=
⎣ 0 0 −1 1⎥⎦ ⎢ 0 ⎥ ⎢⎣ −2v ⎥⎦
⎢⎣ −2v ⎥⎦
⎡ 2⎤
0 ⎤ ⎢0 ⎥ ⎡ −1⎤
⎡ −0.5 0.5 0 5
QY
Yb Vs = ⎢
⎢ ⎥=
0 −0.5 0.5⎥⎦ ⎢0 ⎥ ⎢⎣ 0 ⎥⎦
⎣ 0
⎢⎣0 ⎥⎦
⎡ 1⎤
Q I s − QY
Yb Vs = ⎢
⎣ −2v ⎥⎦
9.12 Network Equilibrium Equation 9.51
Hence, the KCL equation can be written as
⎡ 1.5 −0.5⎤ ⎡ vt2 ⎤ ⎡ 1⎤
=
⎢⎣ −0 5
1⎥⎦ ⎢⎣ vt4 ⎥⎦ ⎢⎣ −2v ⎥⎦
From Fig. 9.82, vt2 v
Solving this matrix equation,
vt2 = 0 44 V
vt4 = 0 66 V
vt2 = 0 44 V
v
Example 9.34 For the network shown in Fig. 9.84, write down the f-cutset matrix and obtain the
network equilibrium equation in matrix form using KCL and calculate v.
4Ω
0.5 A
2Ω
−
+ v
1V +
−
4Ω
2Ω
Fig. 9.84
Solution The voltage and current sources are converted into accompanied sources by source–shifting
method as shown in Fig. 9.85.
4Ω
2Ω
1V
+
−
+ v −
+
1V −
0.5 A
2Ω
4Ω
0.5 A
Fig. 9.85
The oriented graph and its selected tree are shown in Fig. 9.86.
2
(1)
(2)
2
Twigs:
{1, 2}
f-cutset 1: {1, 4}
f-cutset 2: {2, 3}
1
1
(4)
(1)
(3)
(2)
3
(4)
(3)
3
Fig. 9.86
1
2
3
4
⎡ 1 0 0 −1⎤
Q=⎢
⎣0 1 −1 0 ⎥⎦
9.52 Network Analysis and Synthesis
The KCL equation in the matrix form is given by
QY
Yb QT Vt = Q I s − QY
Yb Vs
0
0⎤
⎡0 25 0
⎡ 1
⎢ 0
0.5
0
0⎥ T ⎢ 0
Yb = ⎢
⎥;Q
⎢
0 0 25 0 ⎥
⎢ 0
⎢ 0
0
0
0.5⎥⎦
⎢⎣ 0
⎢⎣ −1
0
⎡0 25 0
0.5
0
⎡ 1 0 0 −1⎤ ⎢ 0
QY
Yb = ⎢
⎢
0 0 25
⎣0 1 −1 0 ⎥⎦ ⎢ 0
0
0
⎢⎣ 0
0⎤
⎡ 0 ⎤
⎡1 ⎤
⎢ 0 ⎥
⎢1 ⎥
1⎥
⎥; I = ⎢
⎥; V = ⎢ ⎥
−1⎥ s ⎢ 0.5 ⎥ s ⎢0 ⎥
0 ⎥⎦
⎢⎣ −0.5⎥⎦
⎢⎣0 ⎥⎦
0⎤
0 ⎥ ⎡0.25 0
0
−0.5⎤
⎥=
0 ⎥ ⎢⎣ 0
0.5 −0.25
0 ⎥⎦
0.5⎥⎦
⎡ 1 0⎤
⎢ 0
0
.
25
0
0
−
0
.
5
1⎥ ⎡0 75
0 ⎤
⎡
⎤
QY
Yb QT = ⎢
⎢
⎥=⎢
⎥
0
0
.
5
−
0
.
25
0
0
−
1
0
0
.
75⎥⎦
⎣
⎦⎢
⎥ ⎣
−
1
0
⎢⎣
⎥⎦
⎡ 0 ⎤
⎡1 0 0 −1⎤ ⎢ 0 ⎥ ⎡ 0 5⎤
Q Is = ⎢
⎢
⎥=
⎣0 1 −1 0 ⎥⎦ ⎢ 0 5 ⎥ ⎢⎣ −0 5⎥⎦
⎢⎣ −0 5⎥⎦
⎡1 ⎤
0
−0.5⎤ ⎢1 ⎥ ⎡0 25⎤
⎡0.25 0
Q Yb Vs = ⎢
⎢ ⎥=
0.5 −0.25
0 ⎥⎦ ⎢0 ⎥ ⎢⎣ 0 5 ⎥⎦
⎣ 0
⎢⎣0 ⎥⎦
⎡0 25⎤
Q I s − Q Yb Vs = ⎢
⎣ −1 ⎥⎦
Hence, the KCL equation can be written as
0 ⎤ ⎡ vt1 ⎤ ⎡0 25⎤
⎡0 75
=
⎢⎣ 0
0.75⎥⎦ ⎢⎣ vt2 ⎥⎦ ⎢⎣ −1 ⎥⎦
Solving this matrix equation,
vt1 = 0 33 V
vt2 = −1 33 V
From Fig. 9.85, v 1 + vt2 = −0.33 V
9.13
DUALITY
Two networks are said to be the dual of each other when the mesh equations of one network are the same
as the node equations of the other. Kirchhoff ’s voltage law and current law are same, word for word, with
voltage substituted for current, independent loop for independent node pair, etc. Similarly, two graphs are
said to be dual of each other if the incidence matrix of any one of them is equal to the circuit matrix of the
other. Only planar networks have duals.
9.13 Duality 9.53
Table 9.1 Conversion for dual electrical circuits
Loop basis
Node basis
Current
Voltage
Resistance
Conductance
Inductance
Capacitance
Branch current
Branch voltage
Mesh
Node
Short circuit
Open circuit
Parallel path
Series path
The following steps are involved in constructing the dual of a network:
1. Place a node inside each mesh of the given network. These internal nodes correspond to the
independent nodes in the dual network.
2. Place a node outside the given network. The external node corresponds to the datum node in the dual
network.
3. Connect all internal nodes in the adjacent mesh by dashed lines crossing the common branches.
Elements which are the duals of the common branches will form the branches connecting the
corresponding independent node in the dual network.
4. Connect all internal nodes to the external node by dashed lines corresponding to all external branches.
Duals of these external branches will form the branches connecting independent nodes and the datum
node.
5. A clockwise current in a mesh corresponds to a positive polarity (with respect to the datum node) at
the dual independent node.
6. A voltage rise in the direction of a clockwise mesh current corresponds to a current flowing towards
the dual independent node.
Example 9.35
Draw the dual of the network shown in Fig. 9.87.
C
R1
V
+
−
L
R2
Fig. 9.87
Solution The following steps are involved in constructing the dual of the network as shown in Fig. 9.88.
1. Place a node inside each mesh.
2. Place a node outside the mesh which will correspond to the datum node.
3. Connect two internal nodes through a dashed line. The element which is dual of the common branch
(here capacitance) will form the branch connecting the corresponding independent node in the dual
network.
4. Connect all internal nodes to the external node by dashed lines crossing all the branches. The dual of
these branches will form the branches connecting the independent node and datum.
9.54 Network Analysis and Synthesis
R1
V
+
−
C
L
R2
1
2
0
Fig. 9.88
The dual network is shown in Fig. 9.89.
C
1
2
G1
I
G2
L
O
Fig. 9.89
Example 9.36
Draw the dual of the network of Fig. 9.90.
C2
C4
R1
I
L3
R5
Fig. 9.90
Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in Fig. 9.91.
C2
R1
I
1
C4
L3
2
3
0
Fig. 9.91
R5
9.13 Duality 9.55
The dual network is shown in Fig. 9.92.
G1
1
V
C3
2
+
−
3
L2
L4
G5
O
Fig. 9.92
Example 9.37
Draw the dual of the network shown in Fig. 9.93.
V
+−
R1
R2
R3
R4
C
R5
L
Fig. 9.93
Solution For drawing the dual network, proceed in the same way as in Example 9.35, as shown in Fig. 9.94.
V
+−
R1
R2
1
R3
R4
3
2
C
L
R5
0
Fig. 9.94
The dual network is shown in Fig. 9.95.
G4
G3
1
G1
G2
L
2
3
C
I
O
Fig. 9.95
G5
9.56 Network Analysis and Synthesis
Example 9.38
Draw the dual of the network shown in Fig. 9.96.
L2
V
C
+
−
R1
L1
R3
R2
I
Fig. 9.96
Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in Fig. 9.97.
C
1
V
L2
2
+
−
R1
L1
R3
R2
3
I
O
Fig. 9.97
The dual network is shown in Fig. 9.98.
C1
L
1
I
G1
G2
2
C2
O
Fig. 9.98
G3
3
−
+
V
9.13 Duality 9.57
Example 9.39
Draw the dual of the network shown in Fig. 9.99.
8F
3Ω
4H
10 sin w t
5Ω
Fig. 9.99
Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in Fig. 9.100.
8F
3Ω
1
2
4H
10 sin w t
5Ω
0
Fig. 9.100
The dual network is shown in Fig. 9.101.
4F
2
Ω
3
10 sin w t
8H
O
Fig. 9.101
Example 9.40
Draw the dual of the network shown in Fig. 9.102.
4F
2H
5V
+
−
1H
1A
2Ω
1F
Fig. 9.102
5
Ω
1
9.58 Network Analysis and Synthesis
Solution For drawing the dual network, proceed in the same way as in Example 9.35 as shown in
Fig. 9.103.
4F
2H
5V
1
+
−
2
1H
1A
2Ω
1F
0
Fig. 9.103
The dual network is shown in Fig. 9.104.
1V
5A
2F
2
−+
1H
1F
4H
2
Ω
1
O
Fig. 9.104
Exercises
9.1
For the networks shown in Fig. 9.105–9.108,
write the incidence matrix, tieset matrix and
f-cutset matrix.
(i)
C
R2
R4
(iii)
1Ω
2Ω
L2
1Ω
R1
2Ω
+
10 V −
R3
L1
V +
−
1Ω
Fig. 9.107
Fig. 9.105
(ii)
(iv)
I
R1
L
C
R1
V
I1
2Ω
20 A
R2
+ V
2
−
+
−
C1
L1
L2
R2
Fig. 9.108
Fig. 9.106
Exercises 9.59
For the graph shown in Fig. 9.109, write the
incidence matrix, tieset matrix and f-cutset
matrix.
9.2
Draw the dual networks for the circuits shown
in Fig. 9.111–9.115.
(i)
9.6
(4)
1
3
2
(2)
(1)
(6)
(5)
4
4A
(3)
4Ω
2F
3H
(7)
Fig. 9.111
(ii)
5
R1
R3
Fig. 9.109
+
The incidence matrix is given as follows:
9.3
1 2
−1 −1
0
1
0 0
1 0
Branches →
3 4 5 6
0 0 0 0
1 0 1 0
−1 −1 0 1
0
1 0 0
7
1
0
0
0
v (t )
8
0
0
0
1
Fig. 9.112
(iii)
R
Draw oriented graph and write tieset matrix.
9.4 The incidence matrix is given below:
1
2
3
0 0
1
0 −1 −1
−1 1 0
1 0 0
9.5
Branches →
4
5 6 7
1 1
0 0
0 0
0 −1
1
8
9
V
1Ω
L1
L2
R1
R2
R3
C
Fig. 9.114
(v)
1V
2V
3Ω
C
+
−
V
2Ω
2Ω
+
−
(iv)
1 0
1 0 0
0 −1 0 0 −1
0 0 −1 −1 1
1 1 0 0 0
1Ω
L
Fig. 9.113
10
Draw the oriented graph.
For the network shown in Fig. 9.110, draw the
oriented graph and obtain the tieset matrix.
Use this matrix to calculate the current i.
i
C
R2
−
1Ω
2H
V
3F
+
−
3Ω
3Ω
3Ω
4F
Fig. 9.110
1A
[0.91 A]
Fig. 9.115
9.60 Network Analysis and Synthesis
Using the principles of network topology,
write the loop/node equation in matrix form
for the network shown in Fig. 9.116.
9.7
R1
R2
Vg
ig
+
−
R4
R3
Fig. 9.116
Objective-Type Questions
The number of independent loops for a
network with n nodes and b branches is
(a) n − 1
(b) b − n
(c) b − n + 1
(d) independent of the number of nodes
9.1
A network has 7 nodes and 5 independent
loops. The number of branched in the network
is
(a) 13
(b) 12
(c) 11
(d) 10
9.2
Identify which of the following is NOT a tree
of the graph shown in Fig. 9.117.
9.3
9.5
(a) 3
(b) 4
(c) 6
(d) 7
Consider the network graph shown in Fig.
9.119.
Fig. 9.119
Which one of the following is NOT a tree of
this graph?
(a)
(b)
a
2
1
d
f
e
4
3
c
b
h
(d)
(c)
g
5
Fig. 9.117
(a) begh
(c) adfg
9.6
(b) defg
(d) aegh
Figure 9.120 below shows a network and its
graph is drawn aside. A proper tree chosen for
analyzing the network will contain the edges.
The minimum number of equations required
to analyze the circuit shown in Fig. 9.118 is
9.4
C
C
a
b
c
+
−
R
R
d
V
R
C
Fig. 9.118
R
Fig. 9.120
(a)
(c)
ab, bc, ad
ab, bd, ca?
(b)
(d)
ab, bc, ca
ac, bd, ad
Answers to Objective-Type Questions 9.61
9.7
9.8
The graph of an electrical network has n nodes
and b branches. The number of links with
respect to the choice of a tree is given by
(a) b − n + 1
(b) b + n
(c) n − b + 1
(d) n − 2b − 1.
(2)
2
1
(a)
(c)
9.10
5
(5)
Fig. 9.122
8
7
(3)
(1)
In the graph shown in Fig. 9.121, one
possible tree is formed by the branches 4, 5,
6, 7. Then one possible fundamental cutset is
6
(4)
4
5
(b)
(d)
C
8
Which one of the following is a cutset of the
graph shown in Fig. 9.123?
(3)
4
(2)
3
(4)
Fig. 9.121
(a)
(c)
9.9
1, 2, 3, 8
1, 5, 6, 8
(b)
(d)
1, 2, 5, 6
1, 2, 3, 7, 8
(5)
(1)
Which one of the following represents the
total number of trees in the graph given in
Fig. 9.122?
Fig. 9.123
(a)
(c)
1, 2, 3 and 4
1, 4, 5 and 6
Answers to Objective-Type Questions
9.1. (c)
9.6. (d)
9.2. (c)
9.7. (a)
9.3. (c)
9.8. (d)
(6)
9.4. (b)
9.9. (d)
9.5. (b)
9.10. (d)
(b)
(d)
2, 3, 4 and 6
1, 2, 4 and 5
10
Transient Analysis
10.1
INTRODUCTION
Whenever a network containing energy storage elements such as inductor or capacitor is switched from one
condition to another, either by change in applied source or change in network elements, the response current
and voltage change from one state to the other state. The time taken to change from an initial steady state
to the final steady state is known as the transient period. This response is known as transient response or
transients. The response of the network after it attains a final steady value is independent of time and is called
the steady-state response. The complete response of the network is determined with the help of a differential
equation.
10.2
INITIAL CONDITIONS
In solving the differential equations in the network, we get some arbitary constant. Initial conditions are
used to determine these arbitrary constants. It helps us to know the behaviour of elements at the instant of
switching.
To differentiate between the time immediately before and immediately after the switching, the signs ‘–’
and ‘+’ are used. The conditions existing just before switching are denoted as i (0–), v (0–), etc. Conditions
just after switching are denoted as i (0+), v (0+).
Sometimes conditions at t = ∞ are used in the evaluation of arbitrary constants. These are known as final
conditions.
In solving the problems for initial conditions in the network, we divide the time period in the following
ways:
1. Just before switching (from t = –∞ to t = 0–)
2. Just after switching (at t = 0+)
3. After switching (for t > 0)
If the network remains in one condition for a long time without any switching action, it is said to be under
steady-state condition.
1. Initial Conditions for the Resistor For a resistor, current and voltage are related by v(t) = Ri(t). The
current through a resistor will change instantaneously if the voltage changes instantaneously. Similarly, the
voltage will change instantaneously if the current changes instantaneously.
10.2 Network Analysis and Synthesis
2. Initial Conditions for the Inductor
For an inductor, current and voltage are related by,
v(t ) = L
di
dt
Voltage across the inductor is proportional to the rate of change of current. It is impossible to change
the current through an inductor by a finite amount in zero time. This requires an infinite voltage across the
inductor. An inductor does not allow an abrupt change in the current through it.
The current through the inductor is given by,
t
i( t ) =
1
v(t )dt + i( )
L ∫0
where i(0) is the initial current through the inductor.
If there is no current flowing through the inductor at t = 0–, the inductor will act as an open circuit at t = 0+.
If a current of value I0 flows through the inductor at t = 0–, the inductor can be regarded as a current source
of I0 ampere at t = 0+.
3. Initial Conditions for the Capacitor
For the capacitor, current and voltage are related by,
i( t ) = C
dv(t )
dt
Current through a capacitor is proportional to the rate of change of voltage. It is impossible to change
the voltage across a capacitor by a finite amount in zero time. This requires an infinite current through the
capacitor. A capacitor does not allow an abrupt change in voltage across it.
The voltage across the capacitor is given by,
t
v(t ) =
1
i(t )dt + v( )
C ∫0
where v(0) is the initial voltage across the capacitor.
If there is no voltage across the capacitor at t = 0–, the capacitor will act as a short circuit at t = 0+. If the
capacitor is charged to a voltage V0 at t = 0–, it can be regarded as a voltage source of V0 volt at t = 0+. These
conditions are summarized in Fig. 10.1.
Element with initial conditions
Equivalent circuit at t = 0+
R
R
L
OC
C
SC
I0
I0
+
V0
−
+−
Fig. 10.1
Initial conditions
V0
10.2 Initial Conditions 10.3
Similarly, we can draw the chart for final conditions as shown in Fig. 10.2
Equivalent circuit at t = ∞
Element with initial conditions
R
R
L
SC
C
OC
SC
I0
+
V0
V0
−
+−
I0
OC
Fig. 10.2 Final conditions
4. Procedure for Evaluating Initial Conditions
(a) Draw the equivalent network at t = 0–. Before switching action takes place, i.e., for t = –∞ to t = 0–, the
network is under steady-state conditions. Hence, find the current flowing through the inductors iL (0–)
and voltage across the capacitor vC(0–).
(b) Draw the equivalent network at t = 0+, i.e., immediately after switching. Replace all the inductors
with open circuits or with current sources iL(0+) and replace all capacitors by short circuits or voltage
sources vC (0+). Resistors are kept as it is in the network.
(c) Initial voltages or currents in the network are determined from the equivalent network at t = 0+.
di + dv + d 2 i + d 2 v +
( )), (0 ), 2 ( )), 2 (0 ) are determined by writing integrodt
dt
dt
ddt
differential equations for the network for t > 0, i.e., after the switching action by making use of initial
condition.
(d) Initial conditions, i.e.,
Example 10.1
In the given network of Fig. 10.3, the switch is closed at t = 0. With zero current in
di
d 2i
the inductor, find i,
and 2 at t = 0+.
dt
dt
10 Ω
1H
100 V
i(t)
Fig. 10.3
Solution
At t = 0–, no current flows through the inductor.
i(
)
0
10.4 Network Analysis and Synthesis
10 Ω
At t = 0 + , the network is shown in Fig. 10.4.
At t = 0 + , the inductor acts as an open circuit.
100 V
+
i(
)
i( 0+)
0
For t > 0, the network is shown in Fig. 10.5.
Writing the KVL equation for t > 0,
100 − 10i
10 − 1
di
dt
At t = 0 + ,
+
10 Ω
0
…(i)
) = 100 − 10i(
1H
100 V
i(t)
di
= 100 − 10i
dt
di
(
dt
Fig. 10.4
…(ii)
Fig. 10.5
+
) = 100 − 10( ) = 100 A / s
Differentiating Eq. (ii),
d 2i
dt
2
d i
At t = 0 + ,
dt
2
2
= −10
(0 + )
10
di
dt
di +
(0 ) = −10(100) = −1000 A / s 2
(0
dt
Example 10.2
In the network of Fig. 10.6, the switch is closed at t = 0. With the capacitor
di
d 2i
uncharged, find value for i,
and 2 at t = 0+.
dt
dt
1000 Ω
1 μF
100 V
i(t)
Fig. 10.6
Solution
At t = 0−, the capacitor is uncharged.
vC (
)
0
i(
)
0
1000 Ω
At t = 0 + , the network is shown in Fig. 10.7.
At t = 0+, the capacitor acts as a short circuit.
vC (
+
)
0
100
i( + ) =
= 0.1 A
1000
v C ( 0+)
100 V
i( 0+)
Fig. 10.7
10.2 Initial Conditions 10.5
For t > 0, the network is shown in Fig. 10.8.
Writing the KVL equation for t > 0,
100 −1000
1000i −
t
1
1 10
1000 Ω
−6
100 V
∫ i dt = 0
1 μF
…(i)
i(t)
0
Differentiating Eq. (i),
Fig. 10.8
di
0 − 1000 − 106 i = 0
dt
di
106
=−
i
dt
1000
+
At t = 0 ,
di
(
dt
+
)=−
…(ii)
106
i(
1000
+
)=−
106
( .1)
1000
/s
Differentiating Eq. (ii),
d 2i
dt 2
At t = 0 + ,
=−
d 2i
dt
(0 + ) = −
2
106 di
1000 dt
106 di +
106
(0 ) = −
(0
( 100) = 105 A / s 2
1000 dt
1000
Example 10.3
In the network shown in Fig. 10.9, the switch is closed. Assuming all initial condidi
d 2i
and 2 at t = 0+.
tions as zero, find i,
dt
dt
10 Ω
1H
10 μF
10 V
i(t)
Fig. 10.9
Solution
At t = 0–,
i(
)
0
vC (
)
0
At t = 0 + , the network is shown in Fig. 10.10.
At t = 0+, the inductor acts as an open circuit and
the capacitor acts as a short circuit.
i(
vC (
+
+
)
0
)
0
10 Ω
10 V
i( 0+)
Fig. 10.10
v C ( 0+)
10.6 Network Analysis and Synthesis
10 Ω
For t > 0, the network is shown in Fig. 10.11.
Writing the KVL equation for t > 0,
t
10 − 10i − 1
At t = 0 + ,
di
1
−
i dt = 0
dt 10 × 10 −6 ∫0
10 −10
10i(0 + )
…(i)
1H
10 μF
10 V
i(t)
Fig. 10.11
di +
(0 ) − 0 0
dt
di +
(0 ) 10 A / s
dt
Differentiating Eq. (i),
0 10
At t = 0+,
0 10
di d 2 i
1
−
−
dt dt 2 10 × 10
dt
and
d 2v
dt 2
i=0
di +
d 2i
1
(0 ) − 2 (0 + ) −
i( 0 + )
dt
dt
10 5
d 2i
Example 10.4
6
2
(0 + )
0
10 10 = −100 A / s 2
In the network shown in Fig. 10.12, at t = 0, the switch is opened. Calculate v,
dv
dt
at t = 0+.
v(t)
1H
1A
100 Ω
Fig. 10.12
Solution
At t = 0 , the switch is closed. Hence, no current flows through the inductor.
–
iL (
)
0
v( 0+)
At t = 0 + , the network is shown in Fig. 10.13.
At t = 0+, the inductor acts as an open circuit.
iL (
v(
+
+
)
0
i L ( 0+)
100 Ω
1A
) = 100 × 1 = 100 V
Fig. 10.13
10.2 Initial Conditions 10.7
For t > 0, the network is shown in Fig. 10.14.
Writing the KCL equation for t > 0,
v(t)
t
v 1
+ v dt = 1
100 1 ∫0
1H
1A
…(i)
100 Ω
Differentiating Eq. (i),
1 dv
+v =0
100 dt
dv +
( ) = −100 v (
dt
At t = 0+,
Fig. 10.14
…(ii)
+
) = −100 × 100 = −10000 V / s
Differentiating Eq. (ii),
1 d 2 v dv
+
=0
100 dt 2 dt
d 2v
At t = 0+,
dt 2
Example 10.5
and
d 2v
dt 2
(0 + ) = −100
dv +
(0 ) = −100 × ( 10 4 ) = 106 V / s 2
(0
dt
In the given network of Fig. 10.15, the switch is opened at t = 0. Solve for v,
dv
dt
at t = 0+.
v(t)
10 A
1 kΩ
1 μF
Fig. 10.15
Solution At t = 0–, switch is closed. Hence, the voltage
across the capacitor is zero.
v(
)
)=0
vC (
v( 0+)
10 A
1 kΩ
V C ( 0+)
+
At t = 0 , the network is shown in Fig. 10.16.
At t = 0+, the capacitor acts as a short circuit.
v(
+
)
vC (
+
Fig. 10.16
)=0
For t > 0, the network is shown in Fig. 10.17.
Writing the KCL equation for t > 0,
v
dv
+10
10 −6
= 10
1000
dt
At t = 0+,
v( + )
dv
+ 10 −66 (
1000
dt
+
) 10
v(t)
…(i)
10 A
1 kΩ
Fig. 10.17
1 μF
10.8 Network Analysis and Synthesis
dv
(
dt
+
)=
10
10 −6
= 10 × 106 V / s
Differentiating Eq. (i),
1 dv
d 2v
+ 10 −6 2 = 0
1000 dt
dt
1 dv +
d 2v
(0 ) 10 −6 2 (0 + ) = 0
1000 dt
dt
At t = 0+,
d 2v
dt 2
Example 10.6
and
d 2v
dt 2
(0 + ) = −
1
1000 × 10
6
× 10 × 106 = −10 × 109 V / s 2
For the network shown in Fig. 10.18, the switch is closed at t = 0, determine v,
dv
dt
at t = 0+.
2Ω
10 A
1H
0.5 μF v (t)
Fig. 10.18
Solution At t = 0–, no current flows through the inductor and there is no voltage across the capacitor.
iL (
)
0
v(
)
0
At t = 0 + , the network is shown in Fig. 10.19.
At t = 0+, the inductor acts as an open circuit
and the capacitor acts as a shot circuit.
iL (
v(
+
+
)
0
)
0
v( 0+)
2Ω
10 A
i L ( 0+)
Fig. 10.19
For t > 0, the network is shown in Fig. 10.20.
Writing the KCL equation for t > 0,
10 A
2Ω
1H
t
v 1
dv
+ ∫ vdt + 0 5 × 10 −6
= 10
2 11
dt
At t = 0+,
v(
+
2
)
+ 0 0.
…(i)
Fig. 10.20
dv +
(0 ) = 10
dt
dv
(
dt
+
)
20 106 V / s
20
0.5 μF v (t)
10.2 Initial Conditions 10.9
Differentiating Eq. (i),
1 dv
d 2v
+ v + 0 5 × 10 −6 2 = 0
2 dt
dt
1 dv +
(0 )
2 dt
At t = 0+,
((0
0 + ) + 0.5 10 −6
d 2v
ddt
2
d 2v
dt 2
(0 + )
(0 + )
0
20 1012 V /s
/ s2
Example 10.7
In the network shown in Fig. 10.21, the switch is changed from the position 1 to the
di
d 2i
position 2 at t = 0, steady condition having reached before switching. Find the values of i,
and 2 at
dt
dt
t = 0+.
10 Ω
1
2
20 V
1H
20 Ω
i(t)
Fig. 10.21
10 Ω
20 V
Solution At t = 0–, the network attains steady-state condition.
Hence, the inductor acts as a short circuit.
i(
)=
i( 0−)
20
=2A
10
Fig. 10.22
10 Ω
At t = 0 + , the network is shown in Fig. 10.23.
At t = 0+, the inductor acts as a current source of 2 A.
i(
+
)
20 Ω
2A
For t > 0, the network is shown in Fig. 10.24.
Writing the KVL equation for t > 0,
At t = 0 ,
+
di
−20i −10
10i − 1 = 0
dt
di
−30i(0 + ) − (0 + ) = 0
dt
di +
(0 ) = −30 × 2 = −60 A / s
dt
Differentiating Eq. (i),
−30
2A
i( 0+)
di d 2 i
−
=0
dt dt 2
Fig. 10.23
10 Ω
…(i)
20 Ω
1H
i(t )
Fig. 10.24
2A
10.10 Network Analysis and Synthesis
−30
At t = 0+,
di +
d 2i
(0 ) − 2 (0 + ) = 0
dt
dt
d 2i
dt
2
(0 + ) = 1800 A / s 2
Example 10.8
In the network shown in Fig.10.25, the switch is changed from the position 1 to the
di
d 2i
and 2 at
position 2 at t = 0, steady condition having reached before switching. Find the values of i,
dt
dt
t = 0+.
20 Ω
1
2
30 V
1 μF
10 Ω
i(t)
Fig. 10.25
20 Ω
Solution At t = 0–, the network attains steady-state condition.
Hence, the capacitor acts as an open circuit.
vC( 0−)
30 V
vC (
)
30 V
i(
)
0
i ( 0−)
Fig. 10.26
At t = 0 + , the network is shown in Fig. 10.27.
At t = 0+, the capacitor acts as a voltage source of 30 V.
vC (
i(
+
+
)
20 Ω
30 V
)=−
10 Ω
i( 0+)
30
= −1 A
30
Fig. 10.27
For t > 0, the network is shown in Fig. 10.28.
Writing the KVL equation for t > 0,
−10i − 20i −
20 Ω
t
1
1 × 10 −6
∫ i dt − 30 = 0
…(i)
1 μF
10 Ω
0
Differentiating Eq. (i),
i(t)
Fig. 10.28
di
−30 − 106 i = 0
dt
At t = 0+,
vC( 0+)
30 V
−30
…(ii)
di +
( 0 ) − 10 6 ( 0 + ) = 0
dt
di +
106 ( −1)
(0 ) =
= 0.33 × 105 A / s
dt
30
30 V
10.2 Initial Conditions 10.11
Differentiating Eq. (ii),
−30
2
−30
At t = 0+,
d i
dt
d 2i
dt
2
− 106
di +
(0 ) = 0
dt
( 0 + ) − 10 6
2
di
=0
dt
d 2i
dt
(0 + ) = −
3
106 × 0 33 × 105
= −1.1 × 109 A s 2
30
Example 10.9
In the network shown in Fig. 10.29, the switch is changed from the position 1 to the
d 2i
di
position 2 at t = 0, steady condition having reached before switching. Find the values of i,
and 2 at
dt
dt
t = 0 +.
20 Ω
1
2
40 V
1 μF
1H
i(t)
Fig. 10.29
20 Ω
Solution At t = 0–, the network attains steady state. Hence, the
capacitor acts as an open circuit.
vC (
)
40 V
i(
)
0
40 V
i( 0−)
Fig. 10.30
At t = 0 + , the network is shown in Fig. 10.31.
At t = 0+, the capacitor acts as a voltage source of 40 V and the
inductor acts as an open circuit.
vC (
i(
+
+
)
40 V
)
0
20 Ω
v C ( 0+)
40 V
i( 0+)
Fig. 10.31
For t > 0, the network is shown in Fig. 10.32.
Writing the KVL equation fo t > 0,
20 Ω
1 μF
t
di
1
−1 − 20i −
∫ i dt − 40 = 0
dt
1 × 10 −6 0
At t = 0+,
−
di
(
dt
+
) − 20i((
+
) − 0 − 40 = 0
di
(
dt
v C ( 0−)
+
) = − 40 A / s
…(i)
1H
i(t)
Fig. 10.32
40 V
10.12 Network Analysis and Synthesis
Differentiating Eq. (i),
−
2
−
At t = 0+,
d i
dt
2
d 2i
dt
2
(0 + ) − 20
− 20
di
− 106 i − 0 = 0
dt
di +
(0 ) − 106 i(0 + ) = 0
dt
d 2i
dt
2
(0 + ) = 800 A / s 2
Example 10.10
In the network of Fig. 10.33, the switch is changed from the position ‘a’ to ‘b’ at
di
d 2i
and 2 at t = 0+.
t = 0. Solve for i,
dt
dt
1 kΩ
a
b
100 V
0.1 μF
1H
i(t)
Fig. 10.33
Solution At t = 0–, the network attains steady condition. Hence,
the inductor acts as a short circuit.
i(
vC (
)=
100
= 0.1 A
1000
)
0
100 V
i ( 0−)
Fig. 10.34
At t = 0 + , the network is shown in Fig. 10.35.
At t = 0+, the inductor acts as a current source of 0.1 A and
the capacitor acts as a short circuit.
i(
vC (
+
+
)
0.1 A
)
0
1 kΩ
1 kΩ
v C ( 0+)
Fig. 10.35
For t > 0, the network is shown in Fig. 10.36.
Writing the KVL equation for t > 0,
−
t
1
0 1 × 10
−6
di
1000i − 1 = 0
∫ i dt −1000
dt
0.1 A
i( 0+)
1 kΩ
…(i)
0.1 μF
1H
i(t )
0
Fig. 10.36
0.1 A
10.2 Initial Conditions 10.13
−0 − 1000i(0 + ) −
At t = 0+,
di +
(0 ) = 0
dt
di +
(0 ) = −1000 i (0 + ) = −1000 × 0.1 = −100 A /s
/
dt
Differentiating Eq. (i),
−
At t = 0+,
1
10
i − 1000
−7
−107 i(0 + ) − 1000
di d 2 i
−
=0
dt dt 2
di +
d 2i
(0 ) − 2 (0 + ) = 0
dt
dt
d 2i
dt 2
(0 + ) = −107 (0.1) − 1000
0 ( 100) = −99 105 A / s 2
Example 10.11
The network of Fig. 10.37 attains steady-state with the switch closed. At t
dv
switch is opened. Find the voltage across the switch vK and K at t 0 + .
dt
K
0, the
1H
vK
1Ω
2V
0.5 F
i(t)
Fig. 10.37
Solution: At t = 0 − , the network is shown in Fig.
10.38. At t = 0 − , the network attains steady-state
condition. The capacitor acts as an open circuit and the
inductor acts as a short circuit.
i(
vC (
)=
2
=2A
1
)
0
At t = 0 + , the network is shown in Fig. 10.39.
At t = 0 + , the capacitor acts as a short circuit
and the inductor acts as a current source of 2 A.
2V
i(
vC (
vK (
+
+
+
)
2A
)
0
)
0
vK( 0−)
2V
1Ω
vC( 0−)
i( 0−)
Fig. 10.38
2A
vC( 0+)
1Ω
i( 0+)
Fig. 10.39
10.14 Network Analysis and Synthesis
1
i dt
C∫
dvK
i
=
dt
C
vK =
Also,
dvK
(
dt
At t = 0 + ,
+
)=
+
i(
C
)
=
2
= 4 A /s
0.5
Example 10.12
In the network shown in Fig. 10.40, assuming all initial conditions as zero, find
dii
di
d 2i
d 2 i2 +
i1 (0 + ) i (0 + ), 1 (0 + ), 2 (0 + ), 21 (0 + ) and
( ).
dt
dt
d
dt
dt 2
R2
R1
V
L
C
i1(t)
i2(t)
Fig. 10.40
Solution At t = 0–, all initial conditions are zero.
vC (
)
0
i1 (
)
0
i2 (
)
0
At t = 0 + , the network is shown in Fig. 10.41.
At t = 0+, the inductor acts as an open circuit and the
capacitor acts as a short circuit.
i1 (0 + ) =
V
R1
i2 (0 + )
0
+
vC (0 )
i1( 0+)
Fig. 10.41
t
and
−
i2( 0+)
0
1
(i1 i2 )dt = 0
C ∫0
…(i)
1
di
(i2 − i1 )dt − R2 i2 − L 2 = 0
C∫
dt
…(ii)
R1i1 −
vC( 0+)
V
For t > 0, the network is shown in Fig. 10.42.
Writing the KVL equations for two meshes for t > 0,
V
R2
R1
R1
R2
L
C
V
i1(t)
i2(t)
Fig. 10.42
10.2 Initial Conditions 10.15
From Eq. (ii), at t = 0 + ,
1
−
C
0+
∫ (i2 − i1 )dt − R2i2 (0
+
di2 +
(0 ) = 0
dt
)−
0
di2 +
(0 ) = 0
dt
Differentiating Eq. (i),
di1 1
− (i1 i2 ) = 0
dt C
di1 +
1
1
(0 ) − i1 (0 + ) + i2 (0 + ) 0
1
dt
C
C
di
1 V
R1 1 (0 + ) +
=0
dt
C R1
di1 +
V
(0
0 )=− 2
dt
R1 C
0 − R1
At t = 0 + ,
0
…(iii)
Differentiating Eq. (iii),
− R1
2
At t = 0+,
− R1
d i1
dt
2
(0 + ) −
d 2 i1
dt
2
−
1 dii1 1 di2
+
=0
C dt C dt
1 dii1 +
1 di2 +
(0 ) +
(0 ) = 0
C dt
C dt
d 2 i1
dt
2
(0 + ) =
V
R13C 2
Differentiating Eq. (ii),
−
1
di
d 2i
(i2 − i1 ) − R2 2 − L 22 = 0
C
dt
dt
d 2 i2
At t = 0+,
dt
2
R2 di2 +
1
V
((00 ) −
[i2 (0 + ) i1 (0 + )] =
[i
L dt
LC
R1 LC
(0 + )
Example 10.13
In the network shown in Fig. 10.43, assuming all initial conditions as zero, find
dii di d i
d 2 i2
i1 , i2 1 , 2 , 21 and
at t 0 + .
dt dt dt
dt 2
2
R2
C
V
L
R1
i2(t)
i1(t)
Fig. 10.43
10.16 Network Analysis and Synthesis
Solution At t = 0 − , all initial conditions are zero.
vC (
)
0
i1 (
)
0
i2 (
)
0
At t = 0 + , the network is shown in Fig. 10.44.
At t = 0 + , the capacitor acts as a short circuit and the
inductor acts as an open circuit.
i1 (0 + ) =
V
R1
i2 (0 + )
0
+
vC (0 )
vC( 0+)
R2
R1
V
i1( 0+)
i2( 0+)
Fig. 10.44
0
For t > 0, the network is shown in Fig. 10.45.
C
R2
Writing the KVL equation for t > 0,
V
t
1
V − ∫ i1 dt
C0
and
R1 (i1 i2 ) = 0
…(i)
dii2
=0
dt
…(ii)
− R1 (i2 − i1 ) − R2 i2 − L
L
R1
i1(t)
i2(t)
Fig. 10.45
From Eq. (ii),
At t = 0 + ,
dii2 1
= [ R1i1 − ( R1 R2 )i2 ]
dt
L
1⎡ V
R2 )i2 (0 + )] = ⎢ R1 − (R
( R1
L ⎣ R1
dii2 +
1
(0 ) = [ R1i1 (0 + ) ( R1
dt
L
…(iii)
⎤ V
R2 )0 ⎥ =
⎦ L
Differentiating Eq. (i),
0
At t = 0 + ,
i1
C
1
di1
dt
1
di2
0
dt
di1 di2
i
=
− 1
dt
dt R1C
...(iv)
di2 + i1 (0 + ) V
V
(0 ) −
(0
= − 2
dt
R1C
L R1 C
dii1
(0 )
dt
Differentiating Eq. (iii),
d 2 i2
dt
At t = 0 + ,
2
d i2
dt 2
2
=
1 ⎡ dii1
R1
− ( R1
L ⎢⎣ dt
R ⎞
⎛ 1
(0 + ) = −V ⎜
+ 2
⎝ R1 LC L2 ⎟⎠
R2 )
dii2 ⎤
dt ⎥⎦
10.2 Initial Conditions 10.17
Differentiating Eq. (iv),
d 2 i1
dt
2
d i1
At t = 0 + ,
dt
2
(0 + )
2
d i2
dt
2
((0
0+ ) −
2
=
d 2 i2
dt
−
2
1 dii1
R1C dt
1 dii1 +
V
VR
R
1 ⎛V
V ⎞
V
2V
VR
R
(0 ) = −
− 22 −
− 2 ⎟= 3 2−
− 2
⎜
R1C dt
R1 LC L
R1C ⎝ L R1 C ⎠ R1 C
R1 LC L2
Example 10.14 In the network shown in Fig. 10.46, a steady state is reached with the switch open.
At t = 0, the switch is closed. For the element values given, determine the value of va (0–), vb (0–), va (0+) and
vb (0+).
10 Ω
10 Ω
va(t)
20 Ω
vb(t)
5V
2H
10 Ω
Fig. 10.46
Solution At t = 0 − , the network is shown in
Fig. 10.47.
10 Ω
10 Ω
At t = 0–, the network attains steady-state condition.
Hence, the inductor acts as a short circuit.
5
)=
vb (
)
0
)
20
5×
= 3.33 V
30
va (
(
)
=
5
2
= A
7.5 3
iL (
+
) 5 va ( + ) va (
+
+
10
10
20 Ω
vb(0−)
5V
iL(0−)
Fig. 10.47
At t = 0 + , the network is shown in Fig. 10.48.
At t = 0+, the inductor acts as a current source
2
of A.
3
2
iL ( + ) = A
3
Writing the KCL equations at t = 0+,
va (
va(0−)
+
) vb (
20
+
)
=0
10 Ω
10 Ω
5V
va(0+)
10 Ω
Fig. 10.48
20 Ω
vb(0+)
2
A
3
10.18 Network Analysis and Synthesis
vb (
and
+
) va (
20
+
)
+
vb (
+
) 5 2
+ =0
10
3
Solving these two equations,
va (
vb (
+
+
) 1.9 V
)
0.477 V
Example 10.15 In the accompanying Fig. 10.49 is shown a network in which a steady state is
reached with switch open. At t = 0, switch is closed. Determine va (0–), va (0+), vb(0–) and vb (0+).
10 Ω
10 Ω
va(t)
20 Ω
vb(t)
5V
2F
10 Ω
Fig. 10.49
10 Ω
Solution At t = 0 − , the network is shown in
Fig. 10.50.
At t = 0–, the network attains steady-state
condition. Hence, the capacitor acts as an open
circuit.
10 Ω
va(0−)
20 Ω
vb(0−)
5V
va (
)
5V
vb (
)
5V
Fig. 10.50
At t = 0 + , the network is shown in Fig. 10.51.
At t = 0+, the capacitor acts as a voltage source
of 5 V.
vb (
+
)
10 Ω
10 Ω
5V
Writing the KCL equation at t = 0+,
va (
5V
+
) 5 va ( + ) va ( + ) 5
+
+
=0
10
10
20
0..
va ( 0 + )
va (
Example 10.16
+
)
.75
va(0+)
10 Ω
20 Ω
vb(0+)
5V
Fig. 10.51
3V
The network shown in Fig. 10.52 has two independent node pairs. If the switch is
dv
dv
opened at t = 0. Find v1 , v2 , 1 and 2 at t = 0+.
dt
dt
10.2 Initial Conditions 10.19
L
v1(t)
v2(t)
R2
R1
i(t)
C
Fig. 10.52
Solution At t = 0–, no current flows through the inductor and there is no voltage across the capacitor.
iL (
)
0
vC (
)
v2 (
)=0
At t = 0 + , the network is shown in Fig. 10.53.
At t = 0+, the inductor acts as an open circuit and the
capacitor acts as a short circuit.
iL (
v1 (
v2 (
+
+
+
iL(0+)
i( 0)
)
0
)
R1 ii((
)
0
+
v2(0+)
v1(0+)
R2
R1
)
Fig. 10.53
For t > 0, the network is shown in Fig. 10.54.
Writing the KCL equation at Node 1 for t > 0,
L
v1(t)
v2(t )
t
v1 1
+
( v1 − v2 )dt = i(t )
R1 L ∫0
…(i)
i(t)
R1
R2
C
Differentiating Eq. (i),
Fig. 10.54
1 dv1 1
di
+ ( v1 − v2 ) =
R1 dt L
dt
dv1 +
(0 )
dt
At t = 0 + ,
R1
⎡ di +
(0 )
(0
dt
⎣d
1
⎤
R1i(0 + ) ⎥
L
⎦
Writing the KCL equation at Node 2 for t > 0,
t
1
( v2
L ∫0
At t = 0+,
0+
v1 )dt +
v2
dv
+C 2 = 0
R2
dt
v2 ( 0 + )
dv
+ C 2 (0 + )
R2
dt
dv2 +
(0 )
dt
…(ii)
0
0
In the network shown in Fig. 10.55, the switch is closed at t = 0, with zero capacidv dv
d 2 v2
tor voltage and zero inductor current. Solve for v1 , v2 , 1 , 2 and
at t = 0+.
dt dt
dt 2
Example 10.17
10.20 Network Analysis and Synthesis
R1
iC (t )
+
v1(t )
iL (t )
L
V
−
+
v2(t )
C
R2
−
Fig. 10.55
Solution At t = 0–, no current flows through the inductor and there is no voltage across the capacitor.
vC (
)
0
v1 (
)
0
v2 (
)
0
iL (
)
0
iC (
)
0
At t = 0 + , the network is shown in Fig. 10.56.
At t = 0+, the inductor acts as an open circuit
and the capacitor acts as a short circuit.
vC (
v1 (
v2 (
iL (
iC (
+
+
+
+
+
)
0
)
0
)
0
)
0
R1
iC (0+)
iL (0+)
v1 (0+)
V
R2
v2 (0+)
Fig. 10.56
V
)=
R1
R1
For t > 0, the network is shown in Fig. 10.57.
iC (t )
Writing the KVL equation for t > 0,
vC (t ) = v1 (t ) + v2 (t )
…(i)
V
C
L
v1(t )
R
v2(t )
vC (t )
Differentiating Eq. (i),
dvC dv1 dv2
=
+
dt
dt
dt
iL (t )
…(ii)
Fig. 10.57
10.2 Initial Conditions 10.21
t
vC =
Now,
1
iC dt
C ∫0
…(iii)
dvC iC
=
dt
C
At t = 0+,
dvC
(
dt
)=
Also
v1
iC ( + )
V
=
V /s
C
R1C
L
diiL
dt
…(iv)
diiL v1
=
dt
L
At t = 0+,
diiL
(
dt
Also,
+
)=
v2
…(v)
v1 ( + )
=0
L
…(vi)
R2 iL
dv2
dii
= R2 L
dt
dt
At t = 0+,
…(vii)
dv2 +
(0 )
dt
R2
diiL +
((00 ) = 0
dt
dvC +
(0 )
dt
dv1
dv
(0 ) + 2 (0 + )
dt
dt
dv1 +
V
(0 ) =
dt
R1C
/s
Differentiating Eq. (vii),
d 2 v2
dt 2
At t = 0+,
d 2 v2
dt 2
= R2
(0 + )
R2
d 2 iL
dt 2
d 2 iL +
((00 )
dt
Differentiating Eq. (v),
d 2 iL
dt
At t = 0+,
d 2 iL
dt
2
d 2 v2
dt
2
2
=
1 dv1
L dt
(0 + ) =
1 dv1 +
1 V
(0 ) =
(0
L dt
L R1C
(0 + ) =
R2V
V / s2
R1 LC
10.22 Network Analysis and Synthesis
Example 10.18 In the network shown in Fig. 10.58, a steady state is reached with switch open. At
t = 0, switch is closed. Find the three loop currents at t = 0+.
2Ω
4Ω
i2(t )
0.5 F
6V
1H
i1(t )
1F
i3(t )
Fig. 10.58
Solution At t = 0 − , the network is shown in Fig.
10.59.
At t = 0–, the network attains steady-state condition.
Hence, the inductor act as a short circuit and the
capacitors act as open circuits.
i4 (0 )
i1 (0 − ) =
i2 (0 )
0
i3 (0 )
0
v1 (0 ) v2 (0 ) = 6 ×
2Ω
4Ω
v1(0−)
6V
6
=1A
6
v2(0−)
i1(0−)
(0−)
i3
Fig. 10.59
4
=4V
6
Since the charges on capacitors are equal when connected in series,
Q
Q2
C1v1 = C2 v2
v1 (0 − )
−
v2 ( 0 )
and
i2(0−)
=
C2
1
=
=2
C1 0 5
v1 (0 − ) =
8
V
3
v2 ( 0 ) =
4
V
3
10.2 Initial Conditions 10.23
At t = 0 + , the network is shown in Fig. 10.60.
At t = 0+, the inductor is replaced by a current
source of 1 A and the capacitors are replaced by a
8
4
voltage source of V and V respectively.
3
3
2Ω
4Ω
6V
8
V
3
4
v2 ( 0 + ) = V
3
i1(0+)
v1 (0 + ) =
At t = 0
+,
1A
i2(0+)
8 V
3
i3(0+)
4 V
3
Fig. 10.60
8 4
6 2i1 (0 ) − − = 0
3 3
+
i1 (0 + ) 1 A
Now,
i1 (0 + ) i3 (0
(0 + ) = 1
i3 (0 + )
0
Writing the KVL equation for Mesh 2,
8
=0
3
8
−4 2 (0 + ) + 4 − = 0
3
−4 [ 2 (0 + ) − 1 (0 + )] −
i2 (0 + ) =
1
A
3
Example 10.19
In the network shown in Fig. 10.61, the switch K is closed at t = 0 connecting a
dii
di
voltage V0 sin ω t to the parallel RL-RC circuit. Find (a) i ((00 + ) andd i2 (0 + ) (b) 1 (0 + ) andd 2 (0 + ).
dt
dt
K
i1 (t )
i2 (t )
R
R
C
L
V0 sinwt
Fig. 10.61
Solution At t = 0 − , no current flows in the inductor and there is no voltage across the capacitor.
vC (
)
0
i1 (
)
0
i2 (
)
0
10.24 Network Analysis and Synthesis
At t = 0 + , the network is shown in Fig. 10.62.
At t = 0 + , the inductor acts as an open circuit and the capacitor
acts as a short circuit. The voltage source V0 i ω t acts as a short
circuit.
i1 (0 + )
0
i2 (0 + )
0
vC (0 + )
0
R
Fig. 10.62
and
V0 i
1
i1 dt = 0
C∫
dii
R i2 − L 2 = 0
dt
R i1 −
t
t
i1 (t )
…(i)
…(ii)
ωt − R
i2 (t )
R
R
C
L
V0 sinwt
Differentiating Eq. (i),
V0 ω
R
vC (0+ )
For t > 0, the network is shown in Fig. 10.63.
Writing the KVL equation for t > 0,
V0 i
i2 (0+ )
i1 (0+ )
dii1 i1
− =0
dt C
Fig. 10.63
dii1 V0ω
i
=
cos ω t − 1 …(iii)
dt
R
RC
dii1
Vω
i (0 + ) V0ω
(0 ) = 0 cos ω t
− 1
=
dt
R
RC
R
t = 0+
At t = 0 + ,
From Eq. (ii),
dii2 V0
R
= sin ω t − i2
dt
L
L
dii2
(0 )
dt
At t = 0 + ,
V0
i
L
t
t = 0+
R
i2 (0 + ) = 0
L
Example 10.20 In the network of Fig. 10.64, the switch K is changed from ‘a’ to ‘b’ at t
(a steady state having been established at the position a). Find i1 , i2 and i3 at t = 0 + .
a
C3
R2
L2
K
V
R3
b
i3
C2
L1
R1
i1
Fig. 10.64
C1
i2
0
10.2 Initial Conditions 10.25
Solution At t = 0 − , the network is shown in Fig. 10.65.
At t = 0 − , the network attains steadystate condition. Hence, the capacitors act
as open circuits and inductors act as short
circuits.
V
i1 (0 )
0
i2 (0 )
0
i3 (0 )
0
vC3(0−)
R3
i1(0−)
i2(0−)
0
vC1 (0 )
0
V
At t = 0 , the network is shown in Fig. 10.66.
At t = 0 + , the capacitor C3 acts as a voltage
source of V volts and capacitors C1 and C2 act as
short circuts. The inductors act as open circuits.
vC3(0+)
R3
R1
i1 (0 )
i2 (0
(0 ) = −
i3 (0 + )
0
0. At t
vC2(0+)
i2(0+)
V
R2 + R3
Fig. 10.66
Example 10.21
to t
R1
vC2(0−)
i3(0+)
vC1(0+)
(0+)
i1
+
vC1(0−)
R2
+
+
i3(0−)
Fig. 10.65
vC3 (0 ) V
vC2 (0 )
R2
In the network of Fig. 10. 67, the switch K1 has been closed for a long time prior
0, the switch K2 is closed. Find v (0 + ) andd iC (0 + ).
K1
K2
i1 10 Ω
iC
C
10 V
i2
10 Ω
vC
Fig. 10.67
Solution At t = 0 − , the network is shown
in Fig. 10.68. At t = 0 − , the network attains
steady-state condition. Hence, the capacitor acts
as an open circuit.
i1 (0 )
0
i2 (0 )
0
iC (0 )
0
vC (0 ) 10 V
i1(0−)
10 Ω
i2(0−)
iC(0−)
10 V
vC(0−)
Fig. 10.68
10 Ω
10.26 Network Analysis and Synthesis
At t = 0 + , the network is shown in Fig. 10.69.
At t = 0 + , the capacitor acts as a voltage source
of voltage V.
vC (
+
i1(0+) 10 Ω
iC(0+)
10 V
vC(0+)
) 10 V
i2(0+)
10 Ω
10 V
Writing the KVL equation at t = 0 + ,
10 −10
10 i1 (0 + ) 10
0
10 −10
10 i2 (0 + )
0
i1 (0 + )
0
and
+
i2 (0 )
Fig. 10.69
1A
+
iC (0
(0 + ) + i2 (0 + )
i1 (0 )
iC (0 + ) 1 A
Example 10.22
In the network shown in fig. 10.70, a steady state is reached with the switch open.
dii
di
At t = 0, the switch is closed. Determine vC (0 ), i1 (0 + )), i2 (0
(0 + ), 1 (0 + ) andd 2 (0 + ).
dt
dt
10 Ω
i2(t)
i1(t)
20 Ω
20 Ω
1H
1 μF
100 V
Fig. 10.70
Solution At t = 0 − , the network is shown in
Fig. 10.71.
At t = 0 − , the network is in steady-state.
Hence, the inductor acts as a short circuit and the
capacitor acts as an open circuit.
vC (
) = 100 ×
10 Ω
i1 (
i2 (
)
20 Ω
Fig. 10.71
i1(0+)
0
At t = 0 + , the network is shown in Fig. 10.72.
At t = 0 + , the inductor acts as a current
source of 3.33 A and the capacitor acts as a
voltage source of 66.67 V.
20 Ω
100 V
20
= 66.67 V
20 + 10
66.67
)=
= 3.33 A
20
i2 (0−)
i1 (0−)
100 V
i2(0+)
20 Ω
3.33 A
Fig. 10.72
20 Ω
66.67 A
10.3 Resistor-Inductor Circuit 10.27
vC (
i1 (
+
+
)
66.67 V
3.33 A
100 − 66.67
i2 ( + ) =
= 1.67 A
20
For t > 0 − , the network is shown in
Fig. 10.73.
Writing the KVL equations for
t > 0,
100 − 20i1 − 1
and
di1
dt
0
)
20 Ω
20 Ω
100 V
1 μF
1H
3.33 A
66.67 V
…(i)
Fig. 10.73
100 − 20 i2 −
1
∫ i2 dt − 66.67 = 0
10 −6
…(ii)
dii1 +
(0 ) = 100 − 20i1 (0
(0 + ) = 100 − 20 (3.33)
dt
At t = 0 + ,
i2(t)
i1(t)
33.3 A/s
Differentiating Eq. (ii),
0 20
At t = 0 + ,
di2
dt
106
20
di2 +
(0 )
dt
2
0
106
(0
2 (0
+
)
di2 +
106
(0 ) = −
× 1 67 = −83500 A/s 2
dt
20
10.3
RESISTOR-INDUCTOR CIRCUIT
Consider a series RL circuit as shown in Fig. 10.74. The
switch is closed at time t = 0. The inductor in the circuit is
initially un-energised.
Applying KVL to the circuit for t > 0,
V
L
i(t)
di
Ri L = 0
dt
V
R
Fig. 10.74 Series RL circuit
This is a linear differential equation of first order. It can be solved if the variables can be separated.
(
) dt L di
L di
= dt
V Ri
Integrating both the sides,
−
L
ln (V − Ri ) = t + k
R
10.28 Network Analysis and Synthesis
where ln denotes that the logarithm is of base e and k is an arbitrary constant. k can be evaluated from the
initial condition. In the circuit, the switch is closed at t = 0, i.e., just before closing the switch, the current in
the inductor is zero. Since the inductor does not allow sudden change in current, at t = 0+, just after the switch
is closed, the current remains zero.
Setting i = 0 at t = 0,
−
−
−
L
[
R
L
ln V = k
R
L
L
ln (V − Ri ) = t − ln V
R
R
−
]= t
−
R
− t
V − Ri
=e L
V
V − Ri = Ve
R
− t
L
Ri = V − Ve
R
− t
t
R
i=
V V − Lt
− e
R R
The complete response is composed of two parts, the steady-state
V
response or forced response or zero state response
and transient
R
for t > 0
i(t)
V
R
R
V − Lt
e .
t
R
O
The natural response is a characteristic of the circuit. Its form may
be found by considering the source-free circuit. The forced response Fig. 10.75 Current response of
series RL circuit
has the characteristics of forcing function, i.e., applied voltage. Thus,
when the switch is closed, response reaches the steady-state value
after some time interval as shown in Fig. 10.75.
Here, the transient period is defined as the time taken for the current to reach its final or steady state value
from its initial value.
L
The term
is called time constant and is denoted by T.
R
L
T=
R
V
At one time constant, the current reaches 63.2 per cent of its final value .
R
response or natural response or zero input response
1
i(T ) =
V V − T t V V −1 V
V
V
− e
= − e = − 0.368 = 0.632
R R
R R
R
R
R
10.3 Resistor-Inductor Circuit 10.29
Similarly,
i( T )
V V
e
R R
V
V
V
0.135 = 0.865
R
R
R
i( T )
V V
e
R R
i(5T ) =
V V −5 V
V
V
− e = − 0.0067 = 0.993
R R
R
R
R
2
V
V
V
0.0498 = 0.950
R
R
R
3
After 5 time constants, the current reaches 99.33 per cent of its final value. The voltage across resistor is
vR
R×
Ri
⎛
= V ⎜1
⎝
R
− t⎞
V⎛
⎜1 − e L ⎟
R⎝
⎠
R
− t⎞
e L
⎟
⎠
f
v(t)
t>0
V
VL
Similarly, voltage across inductor is
vL =
O
R
− t⎞
di
V d⎛
=L
⎜1 − e L ⎟
dt
R dt ⎝
⎠
= Ve
R
− t
L
VR
Fig. 10.76
t
Voltage response of
series RL circuit
for t > 0
Note:
1. Consider a homogeneous equation,
di
+ Pi = 0
dt
where P is a constant.
The solution of this equation is given by,
i (t) = k e−Pt
The value of k is obtained by putting t = 0 in the equation for i (t).
2. Consider a non-homogeneous equation,
di
+ Pi = Q
dt
where P is a constant and Q may be a function of the independent variable t or a constant.
The solution of this equation is given by,
i(t) = e−Pt ∫Q ePt dt + k e−Pt
The value of k is obtained by putting t = 0 in the equation of i(t ).
Example 10.23 In the network of Fig. 10.77, the switch is initially at the position 1. On the steady
state having reached, the switch is changed to the position 2. Find current i(t).
10.30 Network Analysis and Synthesis
R1
1
2
V
L
R2
i (t)
Fig. 10.77
Solution At t = 0 − , the network is shown in Fig. 10.78.
At t = 0−, the network has attained steady-state condition. Hence,
the inductor acts as a short circuit.
R1
V
V
)=
R1
i(
i ( 0−)
Fig. 10.78
Since the inductor does not allow sudden change in current,
+
i(
)=
V
R1
R1
For t > 0, the network is shown in Fig. 10.79.
Writing the KVL equation for t > 0,
− R2 i − R1i − L
i(t )
di ( R1 + R2 )
+
i=0
dt
L
Comparing with the differential equation
Fig. 10.79
di
+ Pi = 0,
dt
P=
R1
R2
L
The solution of this differential equation is given by,
i(t ) = k e − Pt
i( t ) = k e
At t = 0, i( ) =
L
R2
di
=0
dt
⎛R R ⎞
−⎜ 1 2 ⎟ t
⎝ L ⎠
V
R1
V
= k e0 = k
R1
⎛ R1 R2 ⎞
⎟t
L ⎠
V −⎜
i( t ) = e ⎝
R1
for t > 0
V
R1
10.3 Resistor-Inductor Circuit 10.31
Example 10.24 In the network shown in Fig. 10.80, the switch is closed at t = 0, a steady state
having previously been attained. Find the current i (t).
R2
R1
V
i(t)
L
Fig. 10.80
Solution At t = 0 − , the network is shown in Fig. 10.81.
At t = 0−, the network has attained steady-state condition.
Hence, the inductor acts as a short circuit.
)=
i(
R2
R1
V
V
R1
i( 0−)
R2
Since the current through the inductor cannot change
instantaneously,
i(
+
)=
Fig. 10.81
V
R1
R2
For t > 0, the network is shown in Fig. 10.82.
Writing the KVL equation for t > 0,
V
R1 i L
R1
V
di
=0
dt
L
i (t)
di R1
V
+ i=
dt L
L
Fig. 10.82
Comparing with the differential equation
P=
di
+ Pi = Q,
dt
R1
V
, Q=
L
L
The solution of this differential equation is given by,
Pt
i( t ) = e
=e
−
∫ QQe
R1
t
L
Pt
R
R
− 1t
V − L1 t
L
e
dt
k
e
∫L
R
=
dt k e − Pt
− 1t
V
+ke L
R1
V
R1 + R2
10.32 Network Analysis and Synthesis
At t = 0, i( ) =
V
R1
R2
V
R1
=
R2
V
+k
R1
k=−
VR
R2
R1 ( R1 + R2 )
R
i( t ) =
− 1t
V
VR
R2
−
e L
R1 R1 ( R1 + R2 )
=
R
− 1t ⎞
V ⎛
R2
e L ⎟
⎜1 −
R1 ⎝ R1 R2
⎠
for t > 0
Example 10.25
t
In the network of Fig. 10.83, a steady state is reached with the switch K open. At
0, the switch K is closed. Find the current i(t ) for t 0.
30 Ω
20 Ω
20 V
K
10 V
1 H
2
i(t)
Fig. 10.83
Solution At t = 0 − , the network is shown in Fig. 10.84.
At t = 0 − , the network has attained steady-state condition.
Hence, the inductor acts as a short circuit
i(
)=
20 + 10
= 0.6 A
30 + 20
30 Ω
20 V
i( 0−)
10 V
Since the current through the inductor cannot change
instantaneously,
i(
+
)
1 di
=0
2 dt
di
+ 40i = 20
dt
Fig. 10.84
0.6 A
20 Ω
For t > 0, the network is shown in Fig. 10.85.
Writing the KVL equation for t > 0.
10 − 20i −
20 Ω
1 H
2
10 V
i(t)
Fig. 10.85
0.6 A
10.3 Resistor-Inductor Circuit 10.33
Comparing with the differential equation
di
+ Pi = Q,
dt
P
Q = 20
The solution of this differential equation is given by,
− Pt
∫ Q e ddt k e
= e −40 t ∫ 0 e 40 t ddt + k e −40 t
Pt
i( t ) = e
=
At t = 0, i( )
Pt
20
+ k e −40 t
40
= 0..5 + k e −40 t
0.6 A
0.6 = 0.5 + k
k=01
40
i(( ) = 0.5 + 0.1 e −40t
fo t > 0
Example 10.26 The network of Fig. 10.86 is under steady state with switch at the position 1. At
t = 0, switch is moved to position 2. Find i (t).
40 Ω
1
2
50 V
20 mH
10 V
i(t )
Fig. 10.86
40 Ω
Solution At t = 0 − , the network is shown in Fig. 10.87.
At t = 0−, the network has attained steady-state condition.
Hence, the inductor acts as a short circuit.
50 V
i( 0−)
50
)=
= 1.25 A
40
i(
Fig. 10.87
Since current
instantaneously,
through
i(
+
the
inductor
cannot
change
) 1.25 A
40 Ω
For t > 0, the network is shown in Fig. 10.88.
Writing the KVL equation for t > 0,
10 − 40i − 20 × 10 −3
di
=0
dt
di
+ 2000i = 500
dt
10 V
20 mH
i(t)
Fig. 10.88
1.25 A
10.34 Network Analysis and Synthesis
Comparing with the differential equation
di
+ Pi = Q,
dt
P
Q = 500
The solution of this differential equation is given by,
− Pt
∫ Q e ddt k e
= e −2000 t ∫ 500 e 2000 t + k e −2000 t
i( t ) = e
=
Pt
Pt
500
+ k e −2000 t
2000
= 0 25 + k e −2000 t
At t = 0, i( ) 1.25 A
1.25 0.25 + k
k =1
2000 t
i(( ) = 0 25
Example 10.27
f
0
In the network of Fig. 10.89, the switch is moved from 1 to 2 at t = 0. Determine i(t).
1
5Ω
2
2Ω
20 V
0.5 H
i(t)
40 V
Fig. 10.89
Solution At t = 0 − , the network is shown in Fig. 10.90.
At t = 0−, the network has attained steady-state condition. Hence,
the inductor acts as a short circuit.
i(
5Ω
20 V
20
)=
=4A
5
i( 0−)
Fig. 10.90
Since the current through the inductor cannot change instantaneously,
i(
+
)
4A
For t > 0, the network is shown in Fig. 10.91.
2Ω
Writing the KVL equation for t > 0,
di
40 − 2 0.5
dt
di
+4
dt
0.5 H
0
80
40 V
i(t)
Fig. 10.91
4A
10.3 Resistor-Inductor Circuit 10.35
Comparing with the differential equation
di
+ Pi = Q,
dt
P
Q = 80
The solution of this differential equation is given by,
− Pt
∫ Q e ddt k e
= e −4tt ∫ 80 e 4 t ddt + k e −4 t
i( t ) = e
Pt
Pt
80
+ k e −4 t
4
= 20 + k e −4 t
=
At t
0, i(0) = 4 A
4 = 20 + k
k = −16
4
i(t ) = 20 − 16 e −4t
fo t > 0
Example 10.28 For the network shown in Fig. 10.92, steady state is reached with the switch closed.
The switch is opened at t = 0. Obtain expressions for iL (t) and vL (t).
100 Ω
iL(t)
15 V
3000 Ω
90 mH
vL(t)
Fig. 10.92
Solution At t = 0 − , the network is shown in Fig. 10.93.
At t = 0−, the network has attained steady-state condition. Hence, the
inductor acts as a short circuit.
iL (
100 Ω
15 V
15
)=
= 0.15 A
100
i L ( 0−)
Since current through the inductor cannot change instantaneously,
Fig. 10.93
iL (0+) = 0.15 A
For t > 0, the network is shown in Fig. 10.94.
Writing the KVL equation for t > 0,
−3000iL − 90 × 10 −3
diiL
=0
dt
diiL
+ 33.33 × 103 iL = 0
dt
3000 Ω
90 mH
iL(t)
Fig. 10.94
0.15 A
10.36 Network Analysis and Synthesis
di
+ Pi = 0,
dt
Comparing with the differential equation
P = 33.33 × 103
The solution of this differential equation is given by,
iL (t ) = k e − Pt
iL (t ) = k e −33.33×10
At t = 0, iL ( )
3
t
0.15 A
0 15 = k
iL ( ) = 0 15
vL (t ) = L
Also,
33 33×103 t
f
0
diiL
dt
= 90 × 10 −3
3
d
(0.15 e −33 33×10 t )
dt
= −90 × 10 −3 × 0.15 × 33.33 × 103 × e −33.33×10
= −450 e −33.33×10
Example 10.29
3
t
3
t
for t > 0
In the network of Fig. 10.95, the switch is open for a long time and it closes at
t = 0. Find i (t).
10 Ω
10 Ω
50 V
0.1 H
10 Ω
i(t)
Fig. 10.95
10 Ω
10 Ω
−
Solution At t = 0 , the network is shown in Fig. 10.96.
At t = 0−, the network has attained steady-state condition.
Hence, the inductor acts as a short circuit.
i(
)=
50 V
i( 0−)
50
= 2.5 A
10 + 10
Fig. 10.96
Since current through the inductor cannot change instantaneously,
i (0+) = 2.5 A
10 Ω
For t > 0, the network is shown in Fig. 10.97.
For t > 0, representing the network to the left of
the inductor by Thevenin’s equivalent network,
50 V
10
Veq = 50 ×
= 25 V
10 + 10
Req = (10 10) + 10 = 15 Ω
10 Ω
10 Ω
i(t)
Fig. 10.97
0.1 H
2.5 A
10.3 Resistor-Inductor Circuit 10.37
For t > 0, Thevenin’s equivalent network is shown in Fig. 10.98.
Writing the KVL equation for t > 0,
25 − 15i − 0 1
15 Ω
di
=0
dt
25 V
di
+ 150i = 250
dt
Comparing with the differential equation
0.1 H
i(t)
di
+ Pi = Q,
dt
Fig. 10.98
Q = 250
P
The solution of this differential equation is given by,
i( t ) = e
Pt
∫Q e
Pt
ddt k e − Pt
= e −150 t ∫ 50 e150 t ddt + k e − Pt
=
250
+ k e −150 t
150
= 1.667 + k e −150 t
At t = 0, i( )
2 .5 A
2.5 = 1.667 + k
k = 0.833
150
i(( ) = 1.667
.667 + 0.833 e −150t
Example 10.30
for t > 0
In Fig. 10.99, the switch is closed at t = 0. Find i (t) for t > 0.
2Ω
1Ω
10 A
2Ω
i (t)
Fig. 10.99
Solution At t = 0–,
i(
)
0
Since current through inductor cannot change instantaneously,
i(
+
)
0
1H
2.5 A
10.38 Network Analysis and Synthesis
For t > 0, simplifying the network by source-transformation technique as shown in Fig. 10.100.
2Ω
1Ω
10 A
2Ω
2Ω
1H
0.67 Ω
10 A
i (t)
1H
i (t)
(a)
(b)
Writing the KVL equation for t > 0,
6.67 2.67i − 1
2.67 Ω
di
dt
0
6.67 V
di
+ 2.67
dt
1H
6.67
Comparing with the differential equation
i (t)
(c)
di
+ Pi = Q,
dt
Fig. 10.100
P = 2.67, Q = 6.67
The solution of this differential equation is given by,
− Pt
∫ Q e ddt k e
= e −2.67t ∫ 6.67 e 2.67t ddt + k e −2.67t
i( t ) = e
=
Pt
Pt
6 67
+ k e −2 67t
2 67
= 2 5 + k e −2 67t
At t = 0, i (0) = 0
0
2 .5 + k
k = −2 5
i(( ) = 2.5 − 2.5 e −2 67t
2 67 t
= 2.5(1
Example 10.31
)
f
0
Find the current i (t) for t > 0.
60 Ω
25 A
140 Ω
20 Ω
0.3 H
i (t)
Fig. 10.101
10.3 Resistor-Inductor Circuit 10.39
Solution At t = 0–, the inductor acts as a short circuit. Simplifying the network as shown in Fig. 10.102.
60 Ω
20 Ω
140 Ω
25 A
25 A
140 Ω
i (0−)
60 Ω
i (0−)
(a)
(b)
Fig. 10.102
i(
)
25 ×
25
140
= 17.5 A
140 + 60
Since current through the inductor cannot change instantaneously,
i(
+
) 17.5 A
For t > 0, the network is shown in Fig. 10.103.
60 Ω
20 Ω
140 Ω
25 A
0.3 H
17.5 A
i (t)
Fig. 10.103
Simplifying the network by source transformation as shown in Fig. 10.104,
60 Ω
20 Ω
17.5 A
0.3 H
15 Ω
i (t)
0.3 H
i (t)
(a)
(b)
Fig. 10.104
Writing the KVL equation for t > 0,
−15i − 0 3
di
=0
dt
di
+ 50i = 0
dt
Comparing with the differential equation
di
+ Pi = 0,
dt
P = 50
17.5 A
10.40 Network Analysis and Synthesis
The solution of this differential equation is given by,
i(t ) = k e − Pt = k e −50 t
At t = 0, i (0) = 17.5 A
k = 17.5
50 t
i(t ) = 17.5 e −50
f
t>0
Example 10.32 In the network of Fig. 10.105, the switch is in position ‘a’ for a long time. At t = 0,
the switch is moved from a to b. Find v2 (t). Assume that the initial current in the 2 H inductor is zero.
1Ω
a
b
+
1V
1 Ω
2
1H
v2(t)
2H
−
Fig. 10.105
Solution At t = 0 , the switch is in the position a. The network has attained steady-state condition. Hence,
the inductor acts as a short circuit.
Current through the 1 H inductor is given by
–
1
) = =1A
1
) 0
(
v2 (
Since current through the inductor cannot change instantaneously,
+
i(
+
v2 (
) 1A
1×
)
1
= −0.5 V
2
For t > 0, the network is shown in Fig. 10.106.
Writing the KCL equation for t > 0,
t
+
1H
t
1
v
1
v2 dt + 1 + 2 + ∫ v2 d
dtt = 0
∫
1
10
20
2
Differentiating Eq. (i),
v2 + 2
…(i)
1 Ω
2
dv
+ Pv = 0,
dt
P=
3
4
v2(t)
2H
−
Fig. 10.106
dv2 1
+ v2 = 0
dt 2
dv2 3
+ v2 = 0
dt 4
Comparing with the differential equation
1A
10.3 Resistor-Inductor Circuit 10.41
The solution of this differential equation is given by,
v2 (t ) = K e − Pt = k e
3
− t
4
At t = 0, v2 (0) = –0.5 V
−0 5 = k e°
k = −0 5
v2 ( ) = −0 5 e
3
− t
4
fo t > 0
Example 10.33 In the network shown in Fig. 10.107, a steady-state condition is achieved with
switch open. At t = 0 switch is closed. Find va (t).
10 Ω
+
5Ω
3V
va(t)
0.5 H
−
Fig. 10.107
Solution At t = 0–, the network has attained steady-state condition. Hence, the inductor acts as a short
circuit.
iL (
)
0
va (
)
3×
5
=1V
10 + 5
Since current through inductor cannot change instantaneously,
iL (
va (
+
+
)
0
) 1V
For t > 0, the network is shown in Fig. 10.108.
Writing the KCL equation for t > 0,
10 Ω
+
t
1
v
v −3
va dt + a + a
=0
0 5 ∫0
5
10
3V
5Ω
−
Differentiating Eq. (i),
2
a
0.5 H va(t)
0.2
dva
dv
0.1 a 0
dt
dt
dva 20
+
va = 0
dt
3
Fig. 10.108
10.42 Network Analysis and Synthesis
Comparing with the differential equation
dv
+ Pv = 0,
dt
P=
20
3
The solution of this differential equation is given by,
va (t ) = k e − Pt = k e
−
20
t
3
At t = 0, va (0) = 1 V
1= k
va ( t ) = e
Example 10.34
is closed at t = 0.
−
20
t
3
fo t > 0
In the network of Fig. 10.109, determine currents i1 (t) and i2 (t) when the switch
10 Ω
5Ω
100 V
i2(t)
5Ω
0.01 H
i1(t)
Fig. 10.109
Solution At t = 0–,
At t = 0+,
i1 (0 )
i2 (0 ) = 0
i1 (0 + )
0
i2 (0 + ) =
100
= 6 67 A
15
For t > 0, the network is shown in Fig. 10.110.
Writing the KVL equations for t > 0,
di1
=0
dt
100 −10
10( 1 + 2 ) − 5 2 0
100 −10
10( 1 + 2 ) − 5 1 0.01
and
10 Ω
…(i)
…(ii)
5Ω
100 V
i2(t)
i1(t)
0.01 H
From Eq. (ii),
i2 =
100 − 10i1
15
Substituting in Eq. (i),
dii1
+ 833i1 = 3333
dt
Fig. 10.110
5Ω
10.3 Resistor-Inductor Circuit 10.43
Comparing with the differential equation
di
+ Pi = Q,
dt
P
Q = 3333
The solution of this differential equation is given by,
− Pt
d ke
∫ Q e dt
= e −833t ∫ 3333 e833t ddt + k e −833t
Pt
i1 (t ) = e
=
Pt
3333
+ k e −833t
833
= 4 + k e −833t
At t = 0, i1 (0) = 0
4+k
0
k = −4
i1 (t ) = 4 4 e −833t
833t
= 4(1 − e −833t
)
100 − 10i1
i2 (t ) =
15
=
100 − 10( 4 − 4
15
fo t > 0
−833t
= 4 + 2.67 e −833t
)
fo t > 0
Example 10.35 The switch in the network shown in Fig. 10.111 is closed at t = 0. Find v (t) for
all t > 0. Assume zero initial current in the inductor.
30 Ω
10 V
+
−
10 Ω
i1(t)
i2(t)
Fig. 10.111
Solution At t = 0 − ,
i1 (0 )
0
i2 (0 )
0
Since current through the inductor cannot change instantaneously,
i2 (0 + )
0
i1 (0 + ) =
10
= 0 25 A
30 + 10
0.2 H
10.44 Network Analysis and Synthesis
For t > 0, the network is shown in Fig. 10.112.
Writing the KVL equations for t > 0,
and
30 Ω
10 − 30( 1 + 2 ) −10
10 i1 = 0
…(i)
di2
=0
dt
…(ii)
10 − 30( 1 + 2 ) − 0 2
10 V
+
−
10 Ω
i1(t)
i2(t)
0.2 H
From Eq. (i),
10 − 30 i2
= 0.25 0.75 i2 …(iii)
40
Substituting Eq. (iii) into Eq. (ii),
dii2
+ 37.5 i2 2.5
dt
di
Comparing with the differential equation
+ Pi = Q,
dt
P
Fig. 10.112
i1 =
Q = 2.5
The solution of this differential equation is given by,
− Pt
d ke
∫ Q e dt
37.5t
5t
37.5t
−37.5t
= e −37
∫ 5 e ddt + k e
i2 (t ) = e
=
At t
Pt
Pt
25
+ k e −37.5t
37.5
= 0.067 + k e −37.5t
0, i2 (0) = 0
0 0.067 + k
k = −0.067
i2 ( ) = 0.067 − 0.067 e −37.5t
di
v2 ( ) = 0 2 2
dt
d
= 0.2 (0.067 − 0.067 e
dt
= 0 5e
5 e −37 5t
37 5t
)
fo t > 0
Example 10.36 For the network shown in Fig. 10.113, find the current i(t) when the switch is
changed from the position 1 to 2 at t = 0.
40 Ω
60 Ω
1
2
500 V
+
− 10i i(t)
Fig. 10.113
0.4 H
10.3 Resistor-Inductor Circuit 10.45
Solution At t = 0 − , the network is shown in Fig. 10.114.
At t = 0 − , the network attains steady-state condition. Hence,
the inductor acts as a short circuit.
Since current
instantaneously,
i(
)
60 Ω
500 V
500
i( ) =
=5A
40 + 60
through the inductor cannot
+
40 Ω
i( 0−)
change
Fig. 10.114
5A
60 Ω
For t > 0, the network is shown in Fig. 10.115.
Writing the KVL equation for t > 0,
10i 60i − 0 4
di
=0
dt
+
−
10i
0.4 H
5A
i (t)
di
+ 125i
125i = 0
dt
di
Comparing with the differential equation
+ Pi = 0,
dt
P = 125
The solution of this differential equation is given by,
Fig. 10.115
i(t ) = k e − Pt = k e −125t
At t
0, i(0) = 5 A
5=k
i( t ) = 5
Example 10.37
the switch is opened at t
125t
f
0
For the network shown in Fig. 10.116, find the current in the 20 W resistor when
0.
i
30 Ω
50 V
20 Ω
+
− 10i i2(t)
i1(t)
2H
Fig. 10.116
Solution At t = 0 − , the network is shown in Fig.
10.117.
At t = 0 − , the network attains steady-state
condition. Hence, the inductor acts as a short circuit.
i (0 − )
30 Ω
50 V
i1(0 − )
+
−
−
− 10i (0 ) i2(0 )
i(0−) = i2(0−)
Fig. 10.117
20 Ω
10.46 Network Analysis and Synthesis
Writing the KVL equations at t = 0 − ,
50 − 30( 1
10i2
2 ) − 10i2
=0
30(( 2 − 1 ) 200i2 = 0
30
Solving these equations,
i1 (0 )
3.33 A
i2 (0 )
2.5 A
Since the current through the inductor cannot change instantaneously,
i2 (0 + )
2.5 A
For t > 0, the network is shown in Fig. 10.118.
i(t) = i2(t)
30 Ω
Writing the KVL equation for t > 0,
10i2
30i2 − 20i2 − 2
di2
dt
20 Ω
+
− 10i2 i2(t)
0
di2
+ 20i
20i2 = 0
dt
2H
2.5 A
Fig. 10.118
di
Comparing with the differential equation
+ Pi = 0,
dt
P = 20
The solution of this differential equation is given by,
i2 (t ) = k e − Pt = k e −20 t
At t
0, i2 (0) = 2.5 A
25=k
20t
i2 ( ) = 2 5 e −20
fo t > 0
In the network of Fig. 10.119, an exponential voltage v(t) = 4 e −3t is applied at
0. Find the expression for current i(t). Assume zero current through inductor at t 0.
Example 10.38
t
0.5 Ω
+
4e − 3t −
0.25 H
i(t)
Fig. 10.119
Solution At t = 0 − ,
i(
)
0
Since current through the inductor cannot change instantaneously,
i(
+
)
0
10.3 Resistor-Inductor Circuit 10.47
Writing the KVL equation for t > 0,
4
3t
0.5
0.25
di
=0
dt
di
+ 2i 16 e −3t
dt
Comparing with the differential equation
di
+ Pi = Q,
dt
Q = 16 e −3t
P
The solution of this differential equation is given by,
− Pt
∫ Q e ddt k e
= e −2tt ∫ 6 e 3t e 2t dt
k e 2t
= 16 e −2t ∫ e t dt k e −2t
i( t ) = e
Pt
= −16 e
At t
Pt
3t
ke
2t
0, i(0) = 0
0
16 + k
k = 16
i(t ) = −16 e −33 +16
166 e
2t
fo t > 0
Example 10.39 For the network shown in Fig. 10.120, a sinusoidal voltage source
v = 150 sin(500t + θ ) volts is applied at a time when θ = 0. Find the expression for the current i(t).
50 Ω
150 sin(500t + q )
0.2 H
i(t)
Fig. 10.120
Solution
Writing the KVL equation for t > 0,
150 sin(500t θ ) 50
0.2
di
=0
dt
di
+ 250i
250i
dt
Comparing with the differential equation
P
750 sin(500t + θ )
di
+ Pi = Q,
dt
Q = 750 sin(500t + )
10.48 Network Analysis and Synthesis
The solution of this differential equation is given by,
i t =e
e dt
e
Pt
=e
e
=
⎡
e 250 t
e −250 t ⎢
2
⎣ ( 250) +
=
+
2
Aco
.
and
A si
.
=
tan −1
=1 8
12
=
06
.
it =
i
At t
= .
+ k e −250 t
t+
+ k e −250
.
.6
A 1 342
and
{250 sin(500t
+
Let
sin
250 t
ke
°
°) sin
t
t
° +ke
sn
)+ke
250 t
250 t
i 0 =0
−63 43° + k
k =1 2
i
=
+
° +1 2
>0
Example 10.40 For the network shown in Fig. 10.121, find the transient current when the switch is
moved from the position 1 to 2 at t 0 The network is in steady state with the switch in the position 1. The
voltage applied to the network is v =
+ 30 )V
1
2
200 Ω
150 cos(200t + 30 )
0.5 H
Fig. 10.121
Solution At t = 0 − , the network is shown in Fig
10.122.
At t = 0 − the network attains steady-state condition.
V
150 ∠ °
I= =
=
Z 200 + × 200 × 0 5
200 Ω
150 cos(200t + 30 )
i t)
∠ 43 A
Fig. 10.122
0.5
10.4 Resistor–Capacitor Circuit 10.49
The steady-state current passing through the network when the switch is in the position 1 is
i
0.67 cos( 200t + 3.43°)
…(i)
For t > 0, the network is shown in Fig. 10.123.
Writing the KVL equation for t > 0,
−200 i − 0 5
200 Ω
di
=0
dt
0.5 H
i(t)
di
+ 400 i = 0
dt
Fig. 10.123
Comparing with the differential equation
di
+ Pi = 0,
dt
P = 400
The solution of this differential equation is given by,
p
i(t ) = k e − pt
= k e −400 t
...(ii)
From Eqs (i) and (ii),
0.67 cos( 200 3.43
3 ) k e −400 t
0.67 c (3.43°) = k
k = 6 67
At t = 0,
i(( ) = 0 67
10.4
400 t
f
0
RESISTOR–CAPACITOR CIRCUIT
Consider a series RC circuit as shown in Fig. 10.124. The switch
is closed at time t = 0. The capacitor is initially uncharged.
Applying KVL to the circuit for t > 0,
1
V
Ri −
R
V
C
i(t)
1
i dt = 0
C ∫0
Fig. 10.124
Series RC circuit
Differentiating the above equation,
0
di
dt
i
C
0
di
1
+
i=0
dt RC
This is a linear differential equation of first order. The variables may be separated to solve the equation.
di
dt
=−
i
RC
Integrating both the sides,
ln i = −
1
t k
RC
10.50 Network Analysis and Synthesis
The constant k can be evaluated from initial condition. In the circuit shown, the switch is closed at t = 0. Since
the capacitor never allows sudden change in voltage, it will act as short circuit at t = 0+. Hence, current in the
V
circuit at t = 0+ is .
R
V
Setting i = at t = 0,
R
ln
V
=k
R
1
V
t ln
RC
R
V
1
ln i − l n = −
t
R
RC
ln i = −
ln
i
1
=−
t
RC
⎛V ⎞
⎜⎝ ⎟⎠
R
1
−
t
i
= e RC
V
R
1
V − t
i = e RC fo t > 0
R
When the switch is closed, the response decays with time as shown
in Fig. 10.125(a).
The term RC is called time constant and is denoted by T.
T = RC
After 5 time constants, the current drops to 99 per cent of its initial
value.
The voltage across the resistor is
1
vR
V − t
R e RC
R
Ri
= Ve
−
1
t
RC
for t > 0
i(t)
V
R
t
O
Fig. 10.125(a) Current response
of series RC circuit
v(t)
Similarly, the voltage across the capacitor is
V
t
1
vC = ∫ i dt
C0
t
=
1 V
C ∫0 R
= −Ve
−
VR
1
−
t
e RC
1
t
RC
VC
+k
O
t
Fig. 10.125(b) Voltage response
of series RC circuit
10.4 Resistor–Capacitor Circuit 10.51
At t = 0, vC (0) = 0
k=V
Example 10.41
1 ⎞
⎛
−
t
V ⎜1 − e RC ⎟
⎝
⎠
vC
Hence,
f
t>0
The switch in the circuit of Fig. 10.126 is moved from the position 1 to 2 at t = 0. Find vC (t).
5 kΩ
1
2
100 V
50 V
+
vC(t)
1 μF
−
Fig. 10.126
Solution At t = 0−, the network is shown in Fig. 10.127.
At t = 0−, the network has attained steady-state condition.
Hence, the capacitor acts as an open circuit.
5 kΩ
v C ( 0−)
100 V
vC (0 ) = 100 V
−
Since the voltage across the capacitor cannot change
instantaneously,
Fig. 10.127
vC (0 ) = 100 V
+
5 kΩ
For t > 0, the network is shown in Fig. 10.128.
Writing the KCL equation for t > 0,
dvC vC + 50
1 × 10
+
=0
dt
5000
dvC
+ 200 vC = 10 4
dt
dv
Comparing with the differential equation
+ Pv = Q,
dt
−6
50 V
−
Fig. 10.128
Q = 10 4
P
Solution of this differential equation is given by,
− Pt
∫ Q e ddt k e
= e −200 t ∫ 0 4 e 200 t ddt + k e −200 t
vC (t ) = e
=
Pt
+
vC(t)
Pt
10 4
+ k e −200 t
200
= −50 + k e −200 t
1 μF
10.52 Network Analysis and Synthesis
At t = 0, vC (0) = 100 V
50
50 + k
1
k = 150
vC ( ) = −50 + 150 e −200 t
for t > 0
Example 10.42 In the network shown in Fig. 10.129, the switch closes at t = 0. The capacitor is
initially uncharged. Find vC (t) and iC (t).
9 kΩ
10 V
4 kΩ iC(t)
1 kΩ
+
vC(t)
−
3 μF
Fig. 10.129
Solution At t = 0−, the capacitor is uncharged. Hence, it acts as a short circuit.
vC (
) 0
iC (
) 0
At t = 0+, the network is shown in Fig. 10.130.
iT (0 + ) 9 kΩ
Since voltage across the capacitor cannot change
instantaneously,
vC (
At t = 0+,
iT (
iC (
+
+
+
10 V
) 0
⎡
10
)=⎢
⎣9 k (
) 1.02 m ×
⎤
10
=
= 1.02 mA
⎥
) ⎦ 9.8 k
4 kΩ iC (0 + )
v C ( 0+)
1 kΩ
Fig. 10.130
1k
= 0.204 mA
1k+4 k
For t > 0, the network is shown in Fig. 10.131.
9 kΩ
4 kΩ
For t > 0, representing the network to the left of the
capacitor by Thevenin’s equivalent network,
10 V
1k
=1V
9 k +1 k
Req = (9 k 1 k) + 4 k = 4.9 kΩ
+
vC(t)
−
1 kΩ
3 μF
Veq = 10 ×
Fig. 10.131
For t > 0, Thevenin’s equivalent network is shown in Fig. 10.132.
Writing the KCL equation for t > 0,
3 10 −6
dvC
v −1
+ C
=0
dt
4 9 × 103
dvC
+ 68.02 vC = 68.02
dt
4.9 kΩ
1V
3 μF
Fig. 10.132
vC(t)
10.4 Resistor–Capacitor Circuit 10.53
Comparing with the differential equation
dv
+ Pv = Q,
dt
Q = 68.02
P
The solution of this differential equation is given by,
vC (t ) = e
Pt
∫Q e
Pt
ddt k e − Pt
= e −68.02t ∫ 68.0 e68.02t ddt + k e −68.02t
= 1 + k e −68.02 t
At t
0, vC (0) = 0
0 1+ k
k = −1
vC (t ) = 1
iC (t ) = C
68 02 t
f
dvC
dt
= 3 10 −6
= 3 10
d
(
dt
6
e
68.02
02e
= 204.06 × 10 −66 e
Example 10.43
0
68 02 t
)
68.02 t
68.02 t
foor t > 0
For the network shown in Fig. 10.133, the switch is open for a long time and closes
at t = 0. Determine vC (t).
100 Ω
1200 V
300 Ω
+
vC(t)
−
50 μF
Fig. 10.133
Solution At t = 0−, the network is shown in Fig. 10.134.
100 Ω
At t = 0–, the network has attained steady-state condition.
Hence, the capacitor acts as an open circuit.
vC (0−) = 1200 V
v C ( 0−)
1200 V
Since the voltage across the capacitor cannot change
instantaneously,
vC (0+) = 1200 V
Fig. 10.134
10.54 Network Analysis and Synthesis
For t > 0, the network is shown in Fig. 10.135.
100 Ω
Writing the KCL equation for t > 0,
50 × 10 −6
+
dvC vC vC − 1200
+
+
=0
dt 300
100
dvC
+ 266.67
67 vC = 0.24 106
dt
Comparing with the differential equation
300 Ω
1200 V
vC (t)
−
50 μF
Fig. 10.135
dv
+ Pv = Q,
dt
Q = 0 24 × 106
P
The solution of this differential equation is given by,
d
∫ Q e dt
= e −266.67t ∫ 0.
vC (t ) = e
=
Pt
Pt
k e − Pt
06 e 266.67t ddt + k e −266 67t
0 24 × 106
+ k e −266.67 t
266.67
= 900 + k e −266.67 t
At t
0, vC (0) = 1200 V
1200 = 900 + k
k = 300
vC (t ) = 900 + 300 e −266 67 t
Example 10.44
for t > 0
In Fig. 10.136, the switch is closed at t = 0 Find vC (t) for t > 0.
100 Ω
5V
2Ω
1F
+
−
vC(t)
Fig. 10.136
Solution At t = 0−,
vC (0−) = 0
Since the voltage across the capacitor cannot change instantaneously,
vC (0+) = 0
Since the resistor of 2 Ω is connected in parallel with the
voltage source of 5 V, it becomes redundant.
For t > 0, the network is as shown in Fig. 10.137.
Writing KCL equation for t > 0,
vC
dv
+1 C = 0
100
dt
dvC
100
+ vC = 5
dt
100 Ω
1F
5V
Fig. 10.137
vC (t)
10.4 Resistor–Capacitor Circuit 10.55
dvC
+ 0.01vC = 0.05
dt
dv
+ Pv = Q,
dt
P
Q = 0.05
The solution of this differential equation is given by,
Comparing with the differential equation
− Pt
d ke
∫ Q e dt
= e −0.01 ∫ 0.05 e 0.01t ddt + k e 0.01t
Pt
vC (t ) = e
=
Pt
0 05
+ k e −0 01t
0 01
= 5 + k e −0 01t
At t
0, vC (0) = 0
0 5+ k
k = −5
vC (t ) = 5 5 e −0 01t
= 5(1 − e −0 01t )
Example 10.45
for t > 0
In the network shown, the switch is shifted to position b at t = 0. Find v (t) for t > 0.
1 F
4
a
b
5V
2Ω
+
v (t)
−
2Ω
Fig. 10.138
Solution At t = 0 , the network is shown in Fig. 10.139.
At t = 0−, the network has attained steady-state condition.
Hence, the capacitor acts as an open circuit.
vC( 0−)
−
5V
2Ω
vC (0−) = 5 V
v (0−) = 0
Fig. 10.139
At t = 0+, the network is shown in Fig. 10.140.
5V
At t = 0+, the capacitor acts as a voltage source of 5 V.
i(
v(
+
+
)=−
)
+
v( 0−)
−
5
= −1.25 A
4
1.25 × 2
2.5 V
2Ω
2Ω
i( 0+)
Fig. 10.140
+
v( 0+)
−
10.56 Network Analysis and Synthesis
For t > 0, the network is shown in Fig. 10.141.
1F
4
5V
Writing the KVL equation for t > 0,
t
−2 − 5 −
1
i dt − 2 = 0
1∫
0
4
…(i)
2Ω
2Ω
i(t )
Differentiating Eq. (i),
−4
di
− 4i = 0
dt
di
+i = 0
dt
Fig. 10.141
di
+ Pi = 0,
dt
P =1
The solution of this differential equation is given by,
Comparing with the differential equation
i(t ) = k e − Pt = k e − t
At t
0, i(0) = −1.25 A
k = −1.25
i(t ) = −1.25 e − t f
v(t ) = 2i(t )
= −2 5e − t
Example 10.46
t>0
fo t > 0
In the network of Fig. 10.142, the switch is open for a long time and at t = 0, it is
closed. Determine v2 (t).
0.25 Ω
+
6V
1 Ω
2
0.3 F
v2(t )
−
Fig. 10.142
Solution At t = 0−, the switch is open.
v2 (0−) = 0
Since voltage across capacitor cannot change instantaneously,
0.25 Ω
+
6V
0.3 F
v2 (0+) = 0
v2
dv
v −6
+0 3 2 + 2
=0
1
dt
0 25
2
v2(t )
−
For t > 0, the network is shown in Fig. 10.143.
Writing KCL equation for t > 0,
1 Ω
2
Fig 10.143
10.4 Resistor–Capacitor Circuit 10.57
dv2
+ 20 v2 = 80
dt
Comparing with the differential equation
dv
+ Pv = Q,
dt
Q = 80
P
The solution of this differential equation is given by,
− Pt
∫ Q e ddt k e
= e −20 t ∫ 80 e 20 t ddt + k e −20 t
v(t ) = e
=
Pt
Pt
80
+ k e −20 t
20
v2 t = 4 + k e −20 t
At t
0, v2 (0) = 0
0 4+k
k = −4
v2 (t ) = 4 4 e −20 t
20 t
= 4(1 − e −20t
)
for t > 0.
Example 10.47 The switch is moved from the position a to b at t = 0, having been in the position a
for a long time before t = 0. The capacitor C2 is uncharged at t = 0. Find i (t) and v2 (t) for t > 0.
R
a
R1
b
+
V0
C2
C1
i(t)
v2 (t)
−
Fig. 10.144
Solution At t = 0–, the network has attained steady-state condition. Hence, the capacitor C1 acts as an open
circuit and it will charge to V0 volt.
vC (0−) = V0
1
vC (0−) = 0
2
Since the voltage across the capacitor cannot change instantaneously,
vC1 (0 + ) V0
vC2 (0 + )
i( 0 + ) =
0
V0
R1
10.58 Network Analysis and Synthesis
For t > 0, the network is shown in Fig. 10.145.
Writing the KVL equation for t > 0,
t
1
V0 − ∫ i dt
d
C1 0
R1
+
t
1
R1i −
i dt = 0
C2 ∫0
C1
...(i)
C2
V0
i(t )
−
Differentiating Eq. (i),
−
i
di
i
− R1 −
=0
C1
dt C2
Fig. 10.145
di 1 ⎛ C1 + C2 ⎞
+
i=0
dt R1 ⎜⎝ C1 C2 ⎟⎠
Comparing with the differential equation
di
+ Pi = 0,
dt
P=
1 ⎛ C1 C2 ⎞
R1 ⎜⎝ C1 C2 ⎟⎠
The solution of this differential equation is given by,
i( t ) = k e
At t
0, i(0) =
v2(t)
− Pt
= ke
−
1 ⎛ C1 C2 ⎞
t
R1 ⎜⎝ C1 C2 ⎟⎠
V0
R1
k=
V0
R1
1 ⎛ C1 C2 ⎞
t
C1 C2 ⎠⎟
V − ⎜
i(t ) = 0 e R1 ⎝
R1
1
V − t
= 0 e RC
R1
where, C =
C1C2
C1 + C2
t
1
v2 ( t ) =
i dt
C2 ∫0
t
=
t
1 V0 − R1C
e
dt
C2 ∫0 R1
1 ⎞
⎛
−
t
V0
R1C
=
R1C ⎜1 − e
⎟
⎜⎝
⎟⎠
R1C2
=
V0C1
C1 + C2
t ⎞
⎛
−
t
⎜1 − e R1C ⎟
⎜⎝
⎟⎠
for t > 0
10.4 Resistor–Capacitor Circuit 10.59
Example 10.48
t > 0.
For the network shown in Fig.10.146, the switch is opened at t = 0. Find v (t) for
+
K
1
F
2
1Ω
10 A
1
Ω
2
v (t)
−
Fig. 10.146
Solution At t = 0 − , the network is shown in Fig.
10.147.
At t = 0 − , the network attains steady-state condition.
Hence, the capacitor acts as an open circuit.
vC(0−) = 0
+
1
Ω
2
10 A
v C (0 − ) v(0 − )
1Ω
−
−
Writing the KCL equation at t = 0 ,
v(
)
1
Fig. 10.147
)
= 10
1
2
3v(( ) = 10
+
v(
v(
)
3.33 V
Since the voltage across the capacitor cannot change instantaneously,
vC (
+
)
v(
v(
+
) = 3.33 V
For t > 0, the network is shown in Fig. 10.148.
Writing the KCL equation for t > 0,
+
1 dv v
+ = 10
2 dt 1
1Ω
10 A
+
−
1
F
2
v (t)
−
dv
+ 2v
dt
20
Fig. 10.148
dv
Comparing with the differential equation
+ Pv = Q,
dt
P
Q = 20
The solution of this differential equation is given by,
− Pt
d ke
∫ Q e dt
= e −2tt ∫ 0 e 2t dt + k e −2t
v(t ) = e
Pt
Pt
20
+ k e −2t
2
= 10 + k e −2t
=
10.60 Network Analysis and Synthesis
At t
0, v(0) = 3.33 V
3 33 10 + k
k = 6 67
v(( ) = 10 + 6 67 e −2t
Example 10.49
opened at t
For the network shown in Fig. 10.149, find the current i(t) when the switch is
0.
10 Ω
i(t)
+
−
100 V
5i
10 Ω
4 μF
Fig. 10.149
10 Ω
Solution At t = 0 − , the network is shown in Fig. 10.150.
At t = 0 − , the network attains steady-state condition.
Hence, the capacitor acts as open circuit.
i(
vC (
)=
+
−
100 V
100
=5A
10 + 10
) 100
00 10
0ii((
i( 0−)
) 5i(
i(
Fig. 10.150
At t = 0 + , the network is shown in Fig. 10.151.
+
25 + 5 (0
(0 ) −10
10
0i ( 0 )
+
i( 0 )
v c ( 0+) = 2 5 V
0
i( 0+)
+
−
5A
For t > 0, the network is shown in Fig. 10.152.
25 −
1
4 10 −6
10 Ω
v C ( 0−)
) = 100 − 10( ) 5( ) = 25 V
−
5i( 0−)
5i( 0+)
10 Ω
25 V
t
∫ i dt + 5
10
0
…(i)
0
Fig. 10.151
Differentiating Eq. (i),
0 0.25 × 106 i + 5
di
di
10
dt
dt
0
di
+ 50000 i = 0
dt
+
−
i(t )
5i
4 μF
25 V
di
Comparing with the differential equation
+ Pi = 0,
dt
P = 50000
Fig. 10.152
10 Ω
10.4 Resistor–Capacitor Circuit 10.61
The solution of this differential equation is given by,
i(t ) = k e − Pt = k e −50000 t
At t
0, i(0) = 5 A
5=k
i( t ) = 5
Example 10.50
opened at t
50000 t
f
0
For the network shown in Fig. 10.153, find the current i(t) when the switch is
0.
10 Ω
i(t)
20 Ω
10 Ω
20i
2 μF
100 V
+
−
Fig. 10.153
10 Ω
Solution At t = 0 − , the network is shown in Fig.
10.154. At t = 0 − , the network attains steady-state
condition. Hence, the capacitor acts as an open circuit.
Writing the KVL equation at t = 0 − ,
100 −10
10 i(0 ) 20 (0 − ) − 20 i(0 )
20 i(0 ) 20 (0 ) − 0
vC (
)
C (0
+
−
)=0
+
)
i2 (
+
10 Ω
20 Ω
+
20 i(0
(0 ) 20
0 (0
(0 ) + 10 i(0 ) 80
i(
vC (
+
+
vc (0 −)
i(0 +)
)
+
+
−
80 V
20 i(0 + ) 20
0 2 ((0
0 + ) 10 i2 (0 + ) 80
+
20i (0−)
Fig. 10.154
At t = 0 + , the network is shown in Fig. 10.155.
From Fig. 10.155, i(
10 Ω
2A
) = 40( )
40 i((
20 Ω
100 V
0
i( 0 )
Also,
i (0 −)
+
20i(0 +)
i 2 (0 +)
−
0
0
) 1.6 A
)
Fig. 10.155
80 V
i(t)
For t > 0, the network is shown in Fig. 10.156.
10 Ω
20 Ω
From Fig. 10.156, i(t ) = −i2 (t )
Writing the KVL equation for t > 0,
20i 20i2 10i2
80 V
1
2 10 −6
t
∫ i2 dt − 80 = 0
0
+
20i
−
2 μF
i 2 (t)
Fig. 10.156
80 V
10.62 Network Analysis and Synthesis
20i 20i2 + 10i2
50i
t
1
2 10 −6
1
2 10 −6
∫ i dt − 80 = 0
0
∫ i dt − 80 = 0
…(i)
Differentiating Eq. (i),
50
Comparing with the differential equation
di
+ 5 105
dt
di
+ 1 × 10 4
dt
0
0
di
+ Pi = 0,
dt
P = 1 × 10 4
The solution of this differential equation is given by,
i(t ) = k e − Pt = k e −1×10
At t
4
t
0, i(0) = 1.6 A
16=k
i(( ) = 1 6 e −11
10 4 t
fo t > 0
In the network of Fig. 10.157, an exponential voltage 4 e −5t is applied at time
0. Find the expression for current i(t). Assume zero voltage across the capacitor at t 0.
Example 10.51
t
0.2 Ω
+
−
4e −5t
1F
i(t)
Fig. 10.157
Solution At t = 0 − ,
vC (
)
0
i(
)
0
+
0.2 Ω
At t = 0 , the network is shown in Fig. 10.158.
Since voltage across the capacitor cannot change instantaneously,
vC (
+
)
0
4
i( + ) =
= 20 A
0.2
Writing the KVL equation for t > 0,
4
+
−
i(0 + )
v c (0 + )
Fig. 10.158
t
4
5t
0.2i −
1
i dt = 0
1 ∫0
…(i)
10.4 Resistor–Capacitor Circuit 10.63
Differentiating Eq. (i),
di
−i = 0
dt
di
+ 5i = −100
100 e −5t
dt
−20 e −5t − 0 2
Comparing with the differential equation
di
+ Pi = Q,
dt
Q = −100 e −5t
P
The solution of this differential equation is given by,
− Pt
∫ Q e ddt k e
= e −5tt ∫ 00 e 5t e5t dt
i( t ) = e
Pt
Pt
ke
5t
= −100t e −5t + k e −5t
At t
0, i(0) = 20 A
20 = k
i(t ) = −100 t e −55 + 200 e
55t
fo t > 0
Example 10.52
source e
−t
In the network shown in Fig. 10.159, the switch is closed at t
to the network. At t 0,
0 v (0) = 0.5 V. Determine v(t).
1Ω
+
e −t
1
Ω 0.5 V+
2
−
+
−
1F
v (t )
−
Fig. 10.159
Solution At t = 0 − ,
v( )
) = 0.5 V
vC (
Since voltage across the capacitor cannot change instantaneously,
v(
+
)
vC (
+
) = 0.5 V
Writing the KCL equation for t > 0,
v e−t v
dv
+ +1 = 0
1
1
dt
2
dv
+ 3v = e − t
dt
Comparing with the differential equation
dv
+ Pv = Q,
dt
P
Q = e−t
0 connecting a
10.64 Network Analysis and Synthesis
The solution of this differential equation is given by,
− Pt
∫ Q e ddt k e
= e −3tt ∫ e t e3t dt k e 3t
= e −3t ∫ e 2t dt + k e −3t
Pt
v(t ) = e
1 −t
e + k e −3t
2
=
At t
0,
Pt
v(0) = 0.5 V
1
+k
2
k=0
05=
v(( ) = 0 5 e − t
Example 10.53 In the network shown in Fig. 10.160, a sinusoidal voltage v = 100 sin(500t + θ )
volts is applied to the circuit at a time corresponding to θ = 45°. Obtain the expression for the current i(t).
15 Ω
100 sin (500t + 45°)
100 μF
i(t)
Fig. 10.160
Solution
Writing the KVL equation for t > 0,
100 sin(500t 45
5°) 15i −
t
1
100 × 10 −6
∫ i dt = 0
…(i)
0
Differentiating Eq. (i),
(
)(
) cos(500t + 45°) − 15
di
− 10 4 i = 0
dt
di
+ 666.67i
67i = 3333.33 cos
o(
dt
Comparing with the differential equation
)
di
+ Pi = Q,
dt
Q = 3333.33 cos(500t + 5°)
P
The solution of this differential equation is given by,
− Pt
∫ Q e ddt k e
= e −666 67t ∫ 3333.33 cos(
i( t ) = e
Pt
Pt
t
) e666.67t + k e
666.67 t
10.4 Resistor–Capacitor Circuit 10.65
⎡
⎤
e666.67t
= 3333.33 e −666.67t ⎢
{666.67 cos(500t + 45°) + 500 sin(500 + 455°)}⎥ + k e −666.67t
2
2
⎣ (666.67) + (500)
⎦
= 3.2 cos(500 + 45°) + 2.4 sin(500t 45
5 ) k e −666.67t
Let
and
2
A si
2
φ
Asin φ
3.2
A cos φ
2.4
A cos
cos φ = ( .2) 2
2
2
( . ) 2 = 16
(2
A=4
⎛ 3.2 ⎞
φ = tan −1 ⎜
= 53.13°
⎝ 2.4 ⎟⎠
and
i(t ) = 4 sin(
i (53..
) cos(500t
5°)
c (
cos(
. 3°) si (
) + k e −666.67t
t
= 4 sin(500t 98. 3 ) k e −666.67t
Putting t = 0, in Eq. (i),
100 sin( 45°)) 15 (0) 0 = 0
i(0) 4.71
4.71 = 4 sin(98
in((98.13°) + k
k = 0 75
i(t
(t )
si (
t
.13°))
.75 e −666 67t
f
t>0
Example 10.54 In the network of Fig. 10.161, the switch is moved from the position 1 to 2 at t 0.
The switch is in the position 1 for a long time. Initial charge on the capacitor is 7 10 −4 coulombs. Determine the current expression i(t), when ω = 1000 rad / s.
1
50 Ω
2
100 sin (wt + 30°)
50 Ω
i(t)
20 μF
Fig. 10.161
Solution At t = 0 − , the network is shown in Fig. 10.162.
At t = 0 − , the network attains steady-state condition.
I=
V
=
Z
100 ∠30°
= 1.41 ∠75° A
1
50 − j
1000 × 20 × 10 −6
The steady-state current passing through the network when
the switch is in the position 1 is
i 1.41si (1000t + 75°)
50 Ω
100 sin (wt + 30°)
i(∞)
20 μF
Fig. 10.162
…(i)
10.66 Network Analysis and Synthesis
For t > 0, the network is as shown in Fig. 10.163.
Writing the KVL equation for t > 0,
−50i − 50i −
50 Ω
50 Ω
t
1
∫ i dt − vC (0) = 0
20 × 10 −6
…(ii)
i(t)
0
Differentiating Eq. (ii),
+
vC (0 −) −
20 μF
Fig. 10.163
di
di
1
−50 − 50 −
i=0
dt
dt 20 × 10 −6
di
+ 500i = 0
dt
di
+ Pi = 0,
dt
Comparing with differential equation
P = 500
The solution of this differential equation is given by,
i(t ) = k e − Pt = k e −500 t
…(iii)
From Eqs (i) and (ii),
1.41 sin(1000
At t = 0,
75 )
k e −500 t
1.41si (75 ) = k
k = 1 36
i(( ) = 1 36
10.5
500 t
f
0
RESISTOR–INDUCTOR–CAPACITOR CIRCUIT
Consider a series RLC circuit as shown in Fig. 10.164. The
switch is closed at time t = 0. The capacitor and inductor are
initially uncharged.
Applying KVL to the circuit for t > 0,
Ri L
L
V
i(t)
t
V
R
di 1
−
i dt = 0
dt C ∫0
C
Fig. 10.164 Series RLC circuit
Differentiating the above equation,
0−R
di
d 2i 1
−L 2 − i=0
dt
C
dt
d 2i
dt 2
+
R di
1
+
i=0
L dt LC
This is a second-order differential equation. The auxiliary equation or characteristic equation will be given by,
⎛ R⎞
⎛ 1 ⎞
s2 + ⎜ ⎟ s + ⎜
=0
⎝ L⎠
⎝ LC ⎟⎠
10.5 Resistor–Inductor–Capacitor Circuit 10.67
Let s1 and s2 be the roots of the equation.
2
s1 = −
R
1
⎛ R⎞
+ ⎜ ⎟ −
= −α + α 2 − ω 02 = −α + β
⎝ 2L ⎠
2L
LC
s2 = −
R
1
⎛ R⎞
− ⎜ ⎟ −
= −α − α 2 − ω 02 = −α − β
⎝
⎠
2L
2L
LC
2
α=
where
ω0 =
β
and
R
2L
1
LC
α 2 − ω 02
The solution of the above second-order differential equation will be given by,
i(t ) = k1e s t + k2 e s2t
where k1 and k2 are constants to be determined and s1 and s2 are the roots of the equation.
Now depending upon the values of a and w0, we have 3 cases of the response.
Case I When a > w0
i.e.,
R
>
2L
1
LC
The roots are real and unequal and it gives an overdamped
response.
In this case, the solution is given by,
i(t)
t
i = e–a t (k1 eb t + k2 e–b t)
k1 e s t
i
or
k2 e s2t
for t > 0
The current curve for an overdamped case is shown in Fig. 10.165.
i(t )
Case II When a = w0
i.e.,
Fig. 10.165 Overdamped
response
R
=
2L
1
LC
The roots are real and equal and it gives a critically damped
response.
In this case the solution is given by,
e −α t ( k
i
t
Fig. 10.166
k t ) for t > 0
The current curve for critically damped case is shown in Fig. 10.166.
Case III When a < w0
i.e.,
R
<
2L
1
LC
The roots are complex conjugate and it gives an underdamped response.
Critically damped
response
10.68 Network Analysis and Synthesis
In this case, the solution is given by,
it
where
Let
2
0
s1 2
2
2
where
= −
j
= j
1
2
and
2
−
i t =e
Hence,
d
d
−
⎥+
2
−
−
⎤
)
2
−
for t > 0
The current curve for an underdamped case is shown in Fig. 10.167.
i t)
t
Fig. 10.167
Example 10.55
Underdamped response
In the network of Fig. 10.168, the switch is closed at t
0 Obtain the expression
for current i t for t >
9Ω
1H
0.05 F
20 V
t)
Fig. 10.168
Solution At t = 0 − , the switch is open.
i
)
0
vC
)
0
Since current through the inductor and voltage across the capacitor cannot change instantaneously,
i
)
0
vC
)
0
10.5 Resistor–Inductor–Capacitor Circuit 10.69
9Ω
For t > 0, the network is shown in Fig. 10.169.
1H
Writing the KVL equation for t > 0,
t
20 − 9
di
1
1 −
i dt = 0
dt 0 05 ∫0
…(i)
0.05 F
20 V
i(t )
Fig. 10.169
Differentiating Eq. (i),
0 9
di d 2 i
−
− 20i = 0
dt dt 2
d 2i
dt
(
2
+9
2
di
dt
20
)i
)
0
0
D1
4, D2
5
The solution of this differential equation is given by,
i(t ) = k1 e −4 t + k2 e
5t
…(ii)
Differentiating Eq. (ii),
di
= −4 k1 e −4 t − 5k2 e
dt
At t
5t
…(iii)
0, i(0) = 0
0 = k1 + k2
di
( )
dt
Putting t = 0 in Eq. (i),
20 − 9 (0 + ) −
…(iv)
…(v)
4 k1 5k2
di +
(0 ) 0 = 0
dt
di +
(0 )
dt
20 9i(0 + )
20 A / s
From Eq. (v),
20 = −4 k1 5k2
…(vi)
Solving Eqs (iv) and (vi),
k1 = 20
k2 = 20
4t
i(t ) = 20 e −4t
Example 10.56
at t
0e
5t
for t > 0
In the network shown in Fig. 10.170, the switch is moved from the position 1 to 2
0. The switch is in the position 1 for a long time. Determine the expression for the current i(t ).
10.70 Network Analysis and Synthesis
10 Ω
2H
1
2
20 V
3F
50 V
i(t )
Fig. 10.170
Solution At t = 0 − , the network is shown in
Fig. 10.171.
At t = 0 − , the network attains steady-state
condition. Hence, the inductor acts as a short circuit
and the capacitor acts as an open circuit.
vC (
)
20 V
i(
)
0
10 Ω
20 V
vC (0−)
i(0 −)
Fig. 10.171
Since the current through the inductor and the voltage across the capacitor cannot change instaneously,
vC (
i(
+
+
)
20 V
)
0
For t > 0, the network is shown in Fig. 10.172.
Writing the KVL equation for t > 0,
10 Ω
2H
t
20 − 10i − 2
di 1
− i dt − 20 = 0
dt 3 ∫0
…(i)
3F
20 V
20 V
i(t)
Differentiating Eq. (i),
0 10
di
dt
2
d 2i
dt 2
d 2i
1
− i−0
3
Fig. 10.172
0
di 1
+ i=0
dt 6
dt
1⎞
⎛ 2
⎜⎝ D + 5 D + ⎟⎠ i = 0
6
2
+5
D1
0 03
2
4.97
The solution of this differential equation is given by,
03t
i(t ) = k1 e −0.03
+ k2 e
4.97 t
…(ii)
Differentiating Eq. (ii),
At t
0, i(0) = 0
di
= −0.03k1 e −0 03t − 4.97k2 e
dt
0 = k1 + k2
di
( )
dt
0.03 k1
4.97 t
…(iii)
…(iv)
.97 k2
…(v)
10.5 Resistor–Inductor–Capacitor Circuit 10.71
Putting t = 0 in Eq. (i),
20 −10
10 i(0 + ) 2
di +
(0 ) 0 = 0
dt
di +
20 − 10 i(0 + )
(0 ) =
= 10 A / s
dt
2
From Eq. (v),
10 = −0.03
03 k1 4 97 k2
…(vi)
Solving Eqs (iv) and (vi),
k1 = 2 02
k2 = 2 02
i(t ) = 2.02 e
0 03t
03t
.0 e
4.97 t
for t > 0
Example 10.57
the network. At t
In the network of Fig. 10.173, the switch is closed and a steady state is reached in
0, the switch is opened. Find the expression for the current i (t) in the inductor.
10 Ω
i2(t)
100 V
10 μF
1H
Fig. 10.173
Solution At t = 0 − , the network is shown in
Fig. 10.174.
At t = 0 − , the network attains steady-state
condition. Hence, the inductor acts as a short circuit
and the capacitor acts as an open circuit.
i2 (0 ) =
vC (0 )
10 Ω
i2 (0 −)
vC (0 −)
100 V
100
= 10 A
10
Fig. 10.174
0
Since current through the inductor and voltage across capacitor cannot change instantaneously,
i2 (0 + ) 10 A
vC (0 + )
0
For t > 0, the network is shown in Fig. 10.175.
Writing the KVL equation for t > 0,
10 A
t
di
1
−1 2 −
i dt = 0
dt 10 × 10 −6 ∫0
i2(t)
1H
…(i)
Fig. 10.175
10 μF
10.72 Network Analysis and Synthesis
Differentiating Eq. (i),
d 2 i2
−
dt 2
d 2 i2
dt 2
− 105 i = 0
+ 105 i = 0
( D 2 + 05 )i = 0
D = j 316 D = − j 316
The solution of this differential equation is given by,
i2 (t ) = k1 cos 3 6t + k2 s 3 6t
Differentiating Eq. (ii),
dii2
= −316 k1
t + 316 k2 cos 3 6t
dt
At t 0, i2 (0) = 10 A
10 = k1
dii2
(0) = 316 k2
dt
Putting t = 0 in Eq. (i),
−
di
(
dt
+
…(ii)
…(iii)
…(iv)
…(v)
)−0 = 0
di
(
dt
+
)=0
From Eq. (v),
0 = 316 k2
k2 = 0
i2 (t ) = 10 cos 316
316t
fo t > 0
Example 10.58 In the network of Fig. 10.176, capacitor C has an initial voltage vc (0 ) of
10 V and at the same instant, current in the inductor L is zero. The switch is closed at time t 0.
Obtain the expression for the voltage v(t ) across the inductor L.
v (t)
+
10 V
−
1Ω
4
1F
Fig. 10.176
Solution At t = 0 − ,
iL (
)
0
v(
)
vC (
) = 10 V
1
H
2
10.5 Resistor–Inductor–Capacitor Circuit 10.73
Since current through the inductor and voltage across capacitor cannot change instantaneously,
+
iL (
+
v(
)
0
)
vC (
+
) = 10 V
For t > 0, the network is shown in Fig. 10.177.
Writing the KCL equation for t > 0,
v (t)
+
t
1
dv v 1
+ +
v dt = 0
dt 1 1 ∫0
4 2
…(i)
10 V
−
Differentiating Eq. (i),
1
H
2
Fig. 10.177
d 2v
dt
1Ω
4
1F
2
(D2
+4
dv
+ 2v = 0
dt
4 D 2)v = 0
4D
D1
1 D2
3
The solution of this differential equation is given by,
v(t ) = k1 e −tt + k2 e
3t
…(ii)
Differentiating Eq. (iii),
dv
= − k1 e −tt − 3 k2 e
dt
At t
3t
…(iii)
0, v(0) = 10 V
10 = k1 + k2
dv
( )
dt
…(iv)
…(v)
k1 3 k2
Putting t = 0 in Eq. (i),
dv
(
dt
+
) 4 v(
v( ) 0 = 0
dv
(
dt
+
)
40 V / s
From Eq. (v),
−40 = − k1 − 3 k2
…(vi)
Solving Eqs (iv) and (vi),
k1 = −5
k2 = 15
v(t ) = −5 e
t
15 e
3t
f
t
0
Example 10.59 In the network of Fig. 10.178, the switch is opened at t
for v(t ). Assume zero initial conditions.
0 obtain the expression
10.74 Network Analysis and Synthesis
v (t )
0.5 Ω
2A
0.5 H
1F
Fig. 10.178
Solution At t = 0 − ,
iL (
)
0
v(
)
vC (
)=0
Since current through the inductor and voltage across the capacitor can not change instantaneously,
iL (
v(
+
+
)
0
)
vC (
+
)=0
For t > 0, the network is shown in Fig. 10.179.
Writing the KCL equation for t > 0,
v (t)
t
v
1
dv
+
vdt + 1
∫
0.5 0.5 0
dt
2
…(i)
0.5 Ω
2A
0.5 H
1F
Differentiating Eq. (i),
2
dv
d 2v
+ 2v
2v + 2 = 0
dt
dt
d 2v
dt
2
(D2
+2
Fig. 10.179
dv
+ 2v = 0
dt
2 D 2) v = 0
D
1
j1
D
1
j1
The solution of this differential equation is given by,
v(t ) = e t ( k cos
cos t k2 sin t )
… (ii)
Differentiating Eq. (ii),
dv
= −e t ( k cos
cos t k2 sin t ) e t ( −kk s t + k2 cos t )
dt
= e − t [ − k (cos
(cos t + si t ) + k (cos t − sin t )]
…(iii)
At t = 0, v(0) = 0
0 = k1
…(iv)
dv
( ) = − k1 k2
dt
…(v)
Putting t = 0 in Eq. (i),
2 ( 0) + 0
dv
( 0) = 2
dt
10.5 Resistor–Inductor–Capacitor Circuit 10.75
dv
( )
dt
2 V /s
From Eq. (v),
2 = −k1 + k2
…(vi)
Solving Eq. (iv) and (vi),
k1 = 0
k2 = 2
v(t ) = 2 e
t
i t
f
t
0
Example 10.60 The network shown in Fig. 10.180, a sinusoidal voltage v = 150 sin(200t + φ ) is
applied at φ = 30°. Determine the current i(t).
10 Ω
0.5 H
200 μF
150 sin (200t + f)
i(t )
Fig. 10.180
Solution Writing the KVL equation for t > 0,
t
150 sin( 200t 30°) 10
di
1
−
i dt = 0
dt 200 × 10 −6 ∫0
0.5
…(i)
Differentiating Eq. (i),
di
dt
30000 cos( 200t + 30°) 10
d 2i
dt
(
2
0.5
+ 20
d 2i
dt 2
− 5000i
5000i = 0
di
+ 10000
0
i
dt
60000 cos( 200t + 30°)
)i
60000 cos( 200
00t + 30°)
2
…(ii)
The roots of the characteristic equation are
D
10
j 99 5
D
10
j 99 5
The complimentary function is
iC
e −10 t ( K1 c
iP
A cos( 200t 30°) B sin(
99 5t + K 2 i 99.5t )
Let the particular function be
t+
°)
iP′
00 A sin( 200t + 30°) + 200 B cos( 200t + 30°)
iP″
0000 A cos( 200t 30°)
B sin(200
i ( 200t + 30°)
10.76 Network Analysis and Synthesis
Substituting these values in Eq. (ii),
30° − 40000 B
+
(
200t +
+ 30 + B sin 200t + 30
° +
60000 cos( 200t +
−
B 4000 A
=
−200 A
200t + 0° + 200
cos 200 + 30°
°)
+
+
200t +
)
°)
+ 30°)
Equating the coefficients,
B 4000 A +
+ 4000 +
=0
= 60000
Solving these equations,
A
1 97
B = 0 26
t+
i
Let
Asin
and
A cos
° +
+ 30°
. 7
. 6
=
cos
= 3.95
A = 1.987
−1 97
= −82.48
0 26
= tan −1
and
t + 30° + 1.
i
. °) sin( 200 +
cos(
°)
t + 30° − 82 48 )
− 52 48°)
The solution of the differential equation is given by,
it
e
k
.
sin 9. t
t
…(iii)
48 )
Differentiating Eq. (iii),
di
dt
k
10
At t
99 5t
k1 os 99
n
. t
1
(
)
t
.
)
0
0
1
k1
in(
. °)
…(iv)
1 58
di
dt
−52.
=
)
1 5 ) 242.03
226.23
…(v)
10.5 Resistor–Inductor–Capacitor Circuit 10.77
Putting t = 0 in Eq. (i),
150 sin(30°)) 10(0) 0.5
di
( 0) − 0 0
dt
di
(0) = 150 A / s
dt
From Eq. (v),
150 = 99.5 k2 + 226.23
k2 = −0 77
i(( ) = e −10 t (1
(1.58
Example 10.61
(a) L =
s 99.5t − 0.77 si 99.5 )
.987 sin( 200t − 52.
.987
) fo t > 0
The switch in the network of Fig. 10.181 is opened at t = 0. Find i (t) for t > 0 if,
1
H and C = 1 F
2
(b) L = 1 H and C = 1 F
2Ω
L
4V
(c) L = 5 H and C = 1 F
i(t)
+
C vC (t)
2Ω
−
Fig. 10.181
Solution At t = 0–, the network has attained steady-state condition. Hence, the inductor acts as a short
circuit and the capacitor acts as an pen circuit.
vC (
)
4×
i(
)
0
2
=2V
2 2
Since current through the inductor and voltage across the capacitor cannot change instantaneously,
vC (
i(
Case I
When R = 2 Ω, L =
+
+
)
2V
)
0
1
H, C = 1 F
2
α=
ω0 =
R
=
2L
1
LC
α > ω0
2
1
2×
2
=
=2
1
1
×1
2
=
1
05
= 1.414
10.78 Network Analysis and Synthesis
This indicates an overdamped case.
i(t ) = A1 e s1t − A2 e s2t
where,
s1 = −α − α 2 − ω 02 = −2
4−2
and
s2 = −α + α 2 − ω 02 = −2
2 = −0.586
i(t ) = k1 e
−3 414 t
2− 2
3.414
0.586 t
+ k2 e
At t = 0, i (0) = 0
k1 k2 = 0
…(i)
Also vL(0+) + vC(0+) + vR(0+) = 0
vL (
vL (
di
(
dt
+
+
+
)
+
vR (
di
(
dt
) − vC (
+
+
)
2i(
)
L
)=
vL ( + )
2
=−
= −4 A / s
L
0.5
+
) vC (
+
) = −2 V
…(ii)
)
Differentiating the equation of i (t) and putting the condition at t = 0,
− 3.414 k1 − 0.586 k2 = −4
…(iii)
Solving Eqs (i) and (iii), we get
k1 = 1.414
(e −3.414 t
(t ) = 1.
Case II
k2 = − 1.414
and
e
0.586 t
)
When R = 2 Ω, L = 1 H, C =1 F
α=
R
2
2
=
= =1
2L 2 × 1 2
1
ω0 =
α
LC
=
1
1
=1
ω0
This indicates a critically damped case.
i(t) = e–a t (k1 + k2 t) = e–t (k1 + k2 t)
At t = 0, i(0) = 0
k1 = 0
Also,
di +
(0 )
dt
vL (0 + )
L
di
(
dt
vL ( + )
2
= − = −2 A / s
L
1
+
)=
for t > 0
Exercises 10.79
Differentiating the equation of i(t) and putting the condition at t = 0,
di
dt
= − k1 k2 = −2
t =0
k2 = −2
i(t ) = −2t e − t
Case III
f
t>0
When R = 2 Ω, L = 5 H, C = 1 F
R
2
=
=02
2 L 10
1
1
ω0 =
=
= 0.447
LC
5
α < ω0
α=
This indicates an underdamped case.
i (t) = e–a t (B1 cos wd t + B2 sin wd t)
ωd
where,
ω o2 − α 2 = (0.447) 2 − (0.2) 2 = 0 4
± jω d = −0.2 ± j 0.4
s,
i( t ) = e
0 2t
( B1 cos 0 4t
(B
B2 sin 0.4t )
Applying the initial condition,
i(0+) = 0
di
(
dt
and
+
vL ( + )
2
=−
L
5
i( 0) 0
)=−
B1
B2 = −1
i(t ) = −e −0 2t si 0.4t
for t > 0
Exercises
10.1 The switch in Fig. 10.182 is moved from the
position a to b at t = 0, the network having been
in steady state in the position a. Determine
dii
dii3 +
i1 (0 + ), i2 (0 + ), i3 (0 + ), 2 (0 + )
(0 ).
dt
dt
1Ω
1 Ω i1(t)
a
i2(t)
10 V
10.2 The switch K is closed at t = 0 in the
network shown in Fig. 10.183. Determine
di
d 2i +
i( + )), (0 + )
(0 ).
dt
ddt 2
b
1Ω
1H
i3(t )
1Ω
2Ω
2H
1F
V0
1Ω
i1(t)
1F
i(t)
Fig. 10.183
Fig. 10.182
[1.66 A, 5 A, −3.33 A, −3.33 A/s, 2.22 A/s]
⎡ 1
⎢0, 2 V0
⎣
⎤
V0 ⎥
⎦
10.80 Network Analysis and Synthesis
10.3 In the network of Fig. 10.184, the switch K
is closed at t = 0. At t = 0−, all capacitor
voltages and inductor currents are zero. Find
dv
dv
dv3
v1 , 1 , v2 , 2 , v3 and
at t = 0 + .
dt
d
dt
d
dt
10.6 The network shown in Fig. 10.187 is under
steady-state when the switch is closed. At t =
0, it is opened. Obtain an expression for i (t).
4 kΩ
i(t)
K
R1
L1
V1
V3
10 kΩ
4A
4 mH
8 kΩ
R2
v (t)
+
−
C3
C1
C2
Fig. 10.187
V2
R
.857 e −2 ×10 t ]
6
[ (t )
Fig. 10.184
⎡ 1 (0 + )
⎤
, 0, 0, 0, 0 ⎥
⎢0,
⎣ C R1
⎦
10.7 The switch in Fig. 10.188 is open for a long
time and closes at t = 0. Determine i (t) for
t > 0.
6Ω
3Ω
i(t)
10.4 In the network at Fig. 10.185, the capacitor C1
is charged to voltage 1000 V and the switch K
d 2 i2
is closed at t = 0. Find
at t = 0 + .
dt 2
2H
Fig. 10.188
K
10 μF
30 Ω
240 V
2 MΩ
+
1000 V
−
10 μF
1 MΩ
i1
[i(t) = 25(1 − e−4t)]
i2
10.8 In the network shown in Fig. 10.189, the steady
state is reached with the switch open. At t = 0,
the switch is closed Find vC (t) for t > 0.
3 kΩ
Fig. 10.185
⎡ 17
2⎤
⎢ 400000 A / s ⎥
⎣
⎦
5 kΩ
2 kΩ
vC (t)
1 kΩ
10.5 In the network shown in Fig. 10.186, switch is
closed at t = 0. Obtain the current i2 (t).
10 Ω
+
−
5 μF
6V
Fig. 10.189
[vC(t) = 5e−20t]
10 Ω
50 V
i1(t )
2 μF
i2(t)
Fig. 10.186
[i2(t) = 5 e–100000t]
10.9 The circuit shown in Fig. 10.190 has acquired
steady state before switching at t = 0.
(i) Obtain vC (0+), vC (0−), i (0+) and i (0−).
(ii) Obtain time constant for t > 0.
(iii) Find current i (t) for t > 0.
Exercises 10.81
+
vC (t)
10.13 In Fig. 10.194, the switch is open until time
t = 100 seconds and is closed for all times
thereafter. Find v (t) for all times greater than
100 if v (100) = −3 V.
10 kΩ
i(t )
2 μF
5V
5 kΩ
12 Ω
8Ω
+
−
Fig. 10.190
2
20 V
20 Ω
1H
i (t)
v (t )
−
Fig. 10.194
(t
⎡
−
⎢ v(t ) = 5 − 8e 160
⎢⎣
10 Ω
1
2F
6F
5V
[(i) 5 V, 5 V, 1 mA, 0, (ii) 0.01 s, (iii) e−100 t mA]
10.10 In the network shown in Fig. 10.191, the
switch is initially at the position 1 for a long
time. At t = 0, the switch is changed to the
position 2. Find current i (t) for t > 0.
)⎤
⎥
⎥⎦
10.14 A series RL circuit shown in Fig. 10.195 has
a constant voltage V applied at t = 0. At what
time does vR = vL.
Fig. 10.191
[i (t) = 2 e−30t]
10.11 In the network shown in Fig. 10.192, the
switch is closed at t = 0. Find v (t) for t > 0.
vR
10 Ω
vL
1H
10 V
+
Fig. 10.195
1 Ω
2
3A
1
H
3
1Ω
[0.0693 s]
v (t )
−
Fig. 10.192
[v(t) = e−t]
10.12 In the network shown in Fig. 10.193, the
switch is in the position 1 for a long time and
at t = 0, the switch is moved to the position .
Find v (t) for t > 0.
1Ω
1V
1
2
1H
+
0.5 Ω
10.15 In the circuit shown in Fig. 10.196, at time
t = 0, the voltage across the capacitor is zero
and the switch is moved to the position y. The
switch is kept at position y for 20 seconds
and then moved to position z and kept in that
position thereafter. Find the voltage across the
capacitor at t = 30 seconds.
10 kΩ
10 V
y
+
vC (t)
−
z
0.1 μF
5 kΩ
2 H v(t )
−
Fig. 10.193
[v (t) = −0.5
−3 t
e 4]
Fig. 10.196
[0]
10.16 Determine whether RLC series circuit shown
in Fig 10.197 is underdamped, overdamped or
10.82 Network Analysis and Synthesis
critically damped. Also, find vL (
and i (∞).
200 Ω
+
)),
di +
(0 )
dt
or
critically
damped.
Also
find
2
di
d v
vL ( + )), (0 + ), 2 ( + ) if v(t ) = u(t ).
dt
dt
0.1 H
2Ω
+ v (t) −
L
200 u(t )
+
−
1H
+ v (t) −
L
10 μF
u(t)
i(t)
+
−
+
v(t)
−
i(t)
Fig. 10.197
1 F
2
Fig. 10.198
[critically damped, 200 V, 2000 A/s, 0]
10.17 Determine whether RLC circuit of Fig
10.198 is underdamped, overdamped
[underdamped 1 V, 1 A/s, 2 V/s2]
Objective-Type Questions
10.1 The voltages vC1 vC2 and vC3 across the
capacitors in the circuit in Fig. 10.199 under
steady state are respectively
1H
10 kΩ
+
100 V
−
2F 2H
+ −
vC 2
+
vC1
−
1F
2 kΩ
(a)
eat − ebt
(b)
eat + ebt
(c)
aedt − bebt
(d)
aeat + bebt
10.3 The differential equation for the current i(t) in
the circuit of Fig. 10.201 is
40 kΩ
vC 3
−
2Ω
i(t)
+
2H
3F
1F
sin t
Fig. 10.199
Fig. 10.201
(a) 80 V, 32 V, 48 V
(b) 80 V, 48 V, 32 V
(a)
(c) 20 V, 8 V, 12 V
(d) 20 V, 12 V, 8 V
10.2 In the circuit of Fig. 10.200, the voltage v(t) is
1Ω
1Ω
eat
(b)
+
v ( t)
−
(c)
1H
ebt
(d)
Fig. 10.200
2
d 2i
dt
d 2i
dt 2
2
2
+2
d 2i
dt 2
d 2i
dt 2
+2
di
+ 2i ( t ) = cos t
dt
+2
+2
di
+ i ( t ) = sin t
dt
di
+ i ( t ) = cos t
dt
di
+ 2i ( t ) = sin t
dt
Objective-Type Questions 10.83
10.4 At t = 0+, the current i1 in Fig. 10.202 is
1
i(t )
(b)
C
1
0.63
2
V
R
L
R
i1(t)
i2(t )
(c)
Fig. 10.202
(a)
(c)
V
2R
V
−
4R
−
(b)
t
0.5
C
i(t)
0.5
0.31
−
V
R
t
0.5
(d) zero
(d)
10.5 For the circuit shown in Fig. 10.203, the time
constant RC = 1 ms. The input voltage is vi(t)
= 2 sin 103 t. The output voltage v0(t) is
equal to
i(t )
1
0.63
t
2
R
Fig. 10.205
vi (t )
C
vo (t)
10.7 The condition on R, L and C such that the step
response v(t) in Fig. 10.206 has no oscillations
is
Fig. 10.203
(a)
sin (103 − 45°)
(c)
sin (103 t − 53)
L
(b) sin (103 t + 45°)
(d) sin (103 t + 53°)
10.6 For the RL circuit shown in Fig. 10.204, the
input voltage vi(t) = u(t). The current i(t) is
+
v(t)
−
C
2Ω
(a)
1 L
2 C
(b)
R≥
L
C
L
C
(d)
R=
1
(a)
R≥
(c)
R≥2
i(t)
Fig. 10.204
v(t)
Fig. 10.206
1H
vi (t )
R
LC
10.8 The switch S in Fig. 10.207 closed at t = 0.
If v2 (0) = 10 V and vg(0) = 0 respectively,
the voltages across capacitors in steady state
will be
i(t )
0.5
0.31
2
t
10.84 Network Analysis and Synthesis
the terminal pair, the time constant of the
system will be
v1 (t )
8 μF
v2 (t )
1 MΩ
2 μF
i(t )
Network of
linear resistors
and independent
sources
Fig. 10.207
(a) v2 (∞) = v1 (∞) = 0
(b) v2(∞) = 2 V, v1 (∞) = 8 V
(c) v2 (∞) = v1 (∞) = 8 V
(d) v2(∞) = 8 V, v1 (∞) = 2 V
+
A
−
B
(a)
i(t )
4 mA
10.9 The time constant of the network shown in
Fig. 10.208 is
8V
(0, 0)
v( t)
(b)
R
Fig. 10.210
2R
10 V
C
Fig. 10.208
(a)
2 RC
(b)
3 RC
1
2
RC
(d)
RC
2
3
10.10 In the series RC circuit shown in Fig. 10.209,
the voltage across C starts increasing when the
dc source is switched on. The rate of increase
of voltage across C at the instant just after the
switch is closed i.e., at t = 0+ will be
(c)
C
(a) 3 μs
(b) 12 s
(c) 32 s
(d)
unknown, unless actual network is specified
10.12 In the network shown in Fig. 10.211, the circuit
was initially in the steady-state condition with
the switch K closed. At the instant when the
switch is opened, the rate of decay of current
through inductance will be
K
R
2Ω
2Ω
2V
2H
1V
Fig. 10.211
Fig. 10.209
(a)
zero
(b)
(c)
RC
(d)
infinity
1
RC
10.11 The v – i characteristic as seen from the
terminal pair (A – B) of the network of Fig.
10.210(a) is shown in Fig. 10.210(b). If an
inductance of value 6 mH is connected across
(a)
(c)
zero
1 A/s
(b)
(d)
0.5 A/s
2 A/s
10.13 A step function voltage is applied to an RLC
series circuit having R = 2 Ω, L = 1 H and C
= 1 F. The transient current response of the
circuit would be
(a) over damped (b) critically damped
(c) under damped (d) none of these
Answers to Objective-Type Questions 10.85
Answers to Objective-Type Questions
10.1 (b)
10.2 (d)
10.3 (c)
10.4 (d)
10.5 (a)
10.6 (b)
10.8 (d)
10.9 (d)
10.10 (d)
10.11 (a)
10.12 (d)
10.13 (b)
10.7 (c)
11
11.1
Laplace Transform
and Its Application
INTRODUCTION
Time-domain analysis is the conventional method of analysing a network. For a simple network with firstorder differential equation of network variable, this method is very useful. But as the order of network variable
equation increases, this method of analysis becomes very tedious. For such applications, frequency domain
analysis using Laplace transform is very convenient. Time-domain analysis, also known as classical method,
is difficult to apply to a differential equation with excitation functions which contain derivatives. Laplace
transform methods prove to be superior. The Laplace transform method has the following advantages:
(1)
(2)
(3)
Solution of differential equations is a systematic procedure.
Initial conditions are automatically incorporated.
It gives the complete solution, i.e., both complementary and particular solution in one step.
Laplace transform is the most widely used integral transform. It is a powerful mathematical technique which
enables us to solve linear differential equations by using algebraic methods. It can also be used to solve
systems of simultaneous differential equations, partial differential equations and integral equations. It is
applicable to continuous functions, piecewise continuous functions, periodic functions, step functions and
impulse functions. It has many important applications in mathematics, physics, optics, electrical engineering,
control engineering, signal processing and probability theory.
11.2
LAPLACE TRANSFORMATION
The Laplace transform of a function f (t) is defined as
F ( s) = L { f (t )}
∞
∫ f (t ) e
− st
dt
0
where s is the complex frequency variable.
jω
s
The function f (t) must satisfy the following condition to possess a Laplace transform,
∞
∫ | f (t ) | e
0
−σ t
dt < ∞
11.2 Network Analysis and Synthesis
where s is real and positive.
The inverse Laplace transform L–1 {F ( s )} is
σ + j∞
f (t ) =
11.3
1
F ( s) e st dds
2π j σ −∫j∞
LAPLACE TRANSFORMS OF SOME IMPORTANT FUNCTIONS
1. Constant Function k
The Laplace transform of a constant function is
∞
∞
L{k} = ∫ ke
− st
0
⎡ e − st ⎤
k
dt = k ⎢
⎥ =
s
⎣ −s ⎦
0
2. Function tn
The Laplace transform of f(t) is
∞
L{t n } = ∫ t n e − st dt
0
dx
Putting st = x, dt =
s
∞
1 ∞
n
dx
⎛ x⎞
L{t n } = ∫ ⎜ ⎟ e − x
= s n+
⎝ s⎠
s
0
n +1
n!
If n is a positive integer,
L{t n } =
∫x
n
e x ddx =
0
n +1
s n +1
, s > 0, n + 1 > 0
n!
s n +1
3. Unit-Step Function
The unit-step function (Fig 11.1) is defined by the
equation,
u(t) = 1
=0
t>0
t<0
u (t )
1
The Laplace transform of unit step function is
L {u(t )}
∞
∫1 e
0
∞
− st
⎡ e − st ⎤
1
dt = ⎢ −
⎥ =
⎣ s ⎦0 s
4. Delayed or Shifted Unit-Step Function
The delayed or shifted unit-step function (Fig 11.2) is defined
by the equation
u (t − a) = 1
=0
The Laplace transform of u (t − a) is
a
Fig. 11.1 Unit-step function
u (t − a)
t>a
t<a
0
a
t
∞
∞
L {u(t a)} = ∫ 1⋅ e
t
0
st
⎡ e − st ⎤
e − as
dt = ⎢ −
⎥ =
s
⎣ s ⎦a
Fig. 11.2
Shifted unit-step function
11.3 Laplace Transforms of Some Important Functions 11.3
5. Unit-Ramp Function
The unit-ramp function (Fig 11.3) is defined by the equation
r(t) = t
t>0
=0
t<0
The Laplace transform of the unit-ramp function is
∞
L rt
r (t )
− st
=
0
1
e
− st
s
a
r (t
as
∫
a)
Fig. 11.4 Delayed unit-ramp function
2
d (t )
1
=
) dt
t
0
∞
Fig. 11.5
The Laplace transform of the unit-impulse function is
L δ
t
a
0
7. Unit-Impulse Function
The unit-impulse function (Fig 11.5) is defined by the equation
d (t) = 0
t≠0
and
Unit-ramp function
Fig. 11.3
s2
6. Delayed Unit-Ramp Function
The delayed unit-ramp function (Fig 11.4) 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
0
Unit-impulse function
at
=1
0
8. Exponential Function (eat)
The Laplace transform of the exponential function (Fig 11.6) is
L e at =
at
− st
= e
0
dt = −
0
t
e
⎤
s a ⎦
0
=
1
t
0
s
Fig. 11.6 Exponential function
9. Sine Function
1
2j
The Laplace transform of the sine function is
⎫ 1
1
L
j
j
⎭
We know that i
}
1 ⎡
2
1
1
s2
s
2
10. Cosine Function
1
2
The Laplace transform of the cosine function is
We know that co
L
1
e
1
e
}
1⎡ 1
2 s
+
1
s+
s
s
2
2
11.4 Network Analysis and Synthesis
11. Hyperbolic sine function
1
−
2
The Laplace transform of the hyperbolic sine function is
1
1
1
1
=
2
2
2 s
We know that
t
=
s+
s2
2
12. Hyperbolic cosine function
1
We know that cosh t
e
2
The Laplace transform of the hyperbolic cosine function is
1
2
1
2
1
1
2 s
=
+
=
s
s
s
2
2
13. Exponentially Damped Function
Laplace transform of an exponentially damped function e−at f (t) is
L e−
− at
t dt =
− st
=
0
t e
d
s
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).
11.4
L e
t
L e − at co
t
L e
a
s a
t
a −
s a
L
+
−
PROPERTIES OF LAPLACE TRANSFORM
11.4.1 Linearity
If
f
where a and b are constants.
F s) then L f
t
( s)
F
∞
Proof
f t
f t e
st
dt
0
∞
L f
t
f
st
t e
d
a
0
Example 11.1
Solution
s
L t
Example 11.2
Solution
Lt
t e
Find the Laplace transform of
s
1
+
2
3
+
e
1
+
dt
( s)
a
s
36
.
3
2
+9
+
1
s−2
8
3
3
2
+9
+
3t .
t =Lt
e
2
t
t
4⋅
t
e
t e
0
Find the Laplace transform of 4t + si
t
∫
dt
0
−
e
+
2
L1
6t
1
s 2
11.4 Properties of Laplace Transform 11.5
Example 11.3
Solution
)2 .
Find the Laplace transform of (
L{(si 2t − cos t ) 2 }
L{sin
i 2 2t cos 2 2t 2 cos 2t si 2t}
L{1 sin
i t} = L{ } L{sin 4t}
1
4
= − 2
s s + 16
Example 11.4
Solution
Find the Laplace transform of cos ( t b) .
L{cos( t b)}
L{cos ω t cos b − sin ω t s b} cos
c bL{cos
{c ω t} s bL{sin ω t}
s
ω
= coss b 2
sin
sin b ⋅ 2
s + ω2
s + ω2
11.4.2 Time Scaling
If L{ f (t )}
F ( s) then L{
}=
1 ⎛ s⎞
F⎜ ⎟
a ⎝ a⎠
∞
Proof
∫ f (t ) e
L{ f (t
− st
dt
0
∞
L{ f ( at )} =
a )e
∫ f (at
− st
dt
0
Putting at = x, dt =
dx
a
∞
L
f ( at )}
∫ f ( x) e
0
Example 11.5
⎛ x⎞
− s⎜ ⎟
⎝ a⎠
dx
a
⎛ s⎞
∞
−⎜ ⎟ x
1
1 ⎛ s⎞
f ( x ) e ⎝ a ⎠ ddx = F ⎜ ⎟
∫
a0
a ⎝ a⎠
⎛ s 3⎞
If L{ f (t )} log ⎜
, find L{f(2t)}.
⎝ s 1 ⎟⎠
⎛ s + 3⎞
L{ f (t )} = log ⎜
⎝ s + 1 ⎟⎠
Solution
By time-scaling property,
⎛s
⎞
+3
1
⎜2
⎟
L{ f ( 2t )} = log ⎜
⎟
s
2
⎜ + 1⎟
⎝2 ⎠
Example 11.6
If L{ f (t )} =
Solution
By time-shifting property,
L{ f (t )} =
2
s
2
s
3
3
1
⎛ s + 6⎞
log ⎜
⎝ s + 2 ⎟⎠
2
e − s , find L{ f (3t )} .
e−s
s
s
s
1 2 − 3 1 54 − 3 18 − 3
L{ f (3t )} =
e =
e = 3e
3 ⎛ s⎞ 3
3 s3
s
⎜⎝ ⎟⎠
3
11.6 Network Analysis and Synthesis
11.4.3 Frequency-Shifting Theorem
If
( s a)
f
Proof
f t
f t te
d
0
− st
f )}
Le
=
0
Example 11.7
s
t4 .
Find the Laplace transform of e
Solution
t
0
4
Lt
s5
By frequency-shifting theorem,
4!
Le
Example 11.8
)5
(s
)2 et .
Find the Laplace transform of (
L t+
Solution
=
t
2
2
s
s
+
1
s
By frequency-shifting theorem,
L t+
Example 11.9
=
2
3
s
2
+
1
4
t
− sin 3t} =
1
s 1
in3t .
4t
Find the Laplace transform of
L
+
2
3
3
4s
s
Solution By frequency-shifting theorem,
3
t} =
Le
3
s
Example 11.10
Solution
3
4s
s
3
6
=
s2
s
2
− s
s 2 − s + 5)
Find the Laplace transform of cosh at cos at.
e at
cosh at cos at
L{ s }
L osh
cos at
e
2
at
⎞
co at
1
e
at
− at
os at )
s
2
a2
s
1
−
s
e
os at}
By frequency-shifting theorem,
L
at
1⎡
⎢
2⎣
at
=
1⎡
⎢
2⎣
s a
a
a s
a
+
s a
a
as +
⎤
a ⎦
1⎡
s a
2 s2 +
2 s
2
a s +
a s
2
2 s)
=
⎤
s a
as
s
s
4
3
a4
11.4 Properties of Laplace Transform 11.7
11.4.4 Time-Shifting Theorem
If L{ f (t )}
F ( s) then L{ f (t a)}
a)} = e
as
F ( s)
∞
Proof
∫ f ( t )e
L{ f (t
− st
dt
0
∞
L{ f (t aa)}
)} =
a) e − st dt
∫ f (t
0
t a = x, dt
Putting
When
d
dx
t a, x = 0
t→ , x→∞
∞
L{ f (t aa)}
)} =
∫
∞
s(a
( x)
f ( x )e
ddx
e
as
as
0
Example 11.11
∫
∞
f ( xx))e
sx
ddx
0
Find the Laplace transform of cos (t a) t > a.
L{ f (t )}
s
F
F(( s) =
2
s +1
By time-shifting theorem,
L{cos(t a)} e − as
Example 11.12
Solution
s
2
s +1
Find the Laplace transform of e t
a
t > a.
Let f (t ) = e t
L{ f (t )}
F(
F ( s) =
1
s −1
By time-shifting theorem,
L{e t a } = e − as
Example 11.13
Solution
By time-shifting theorem,
1
s −1
⎛ π⎞
Find the Laplace transform of sin t − ⎟
⎝ 4⎠
t>
Let f (t ) = sin t
L{ f (t )}
F(
F ( s) =
1
s2 + 1
πs
−
⎧ ⎛ π⎞⎫
1
L ⎨si t − ⎟ ⎬ = e 4 2
⎝
⎠
4
s +1
⎩
⎭
Example 11.14
Find the Laplace transform of (
as
∫ f (t ) e
0
Let f(t) = cos t
Solution
e
)3
t
1.
π
.
4
− st
as
dt = e −as
F ( s)
11.8 Network Analysis and Synthesis
Let f(t) = t3
Solution
3!
f
s4
By time-shifting theorem,
3
e−s
L t
s4
11.4.5 Multiplication by t (Frequency-Differentiation Theorem)
If
F ( s) then L t f )} = −
f t
d
F s
ds
∞
Proof
t e − st dt
f
0
Differentiating both the sides w.r.t s using DUIS,
∞
d
F s
ds
∞
d
ds 0
=
t
t e − st dt
∂
∂s
0
Example 11.15
at}
L
s
L t os t
t} = L
d
L
ds
L
ds
t}
t
4
3 d
1
4 ds s 2
−
s
1 d 1
s
+
2 ds s s
si t
t =−
3s s
2
2
}=
1 1
s
+ 2
2 s s +4
=−
1
−
1
s
+
⎤
− ⋅
s
s
=
Find the Laplace transform of t sin3 t .
L
L t in t
2as
d
a
=
2
ds s a 2
at
1 co 2t
1
=
2
2
t =−
Example 11.17
Solution
a2
Find the Laplace transform of t cos 2 t .
Example 11.16
L
at.
a
2
L t in at
Solution
d
F
ds
Find the Laplace transform of t s
Solution
( )}
0
−
( )}
dt
t e − st dt = − L
−
0
L
t e
−
1
3
4 s2
1
2
s +9
=
3s
⋅
2 (
1
2
s +9
=−
3
1
4 s2 + 1
3
−2 s
4 ( s2 + 1
+
+9
2s
2
s + 9) ⎦
s +5
s
1
2
=
3s
2
24 s s + 5
s
s
1
s
+
s2
2
4
11.4 Properties of Laplace Transform 11.9
Example 11.18
Solution
Find the Laplace transform of t sin 2t cosh t .
L{sin 2t cosh t}
⎧⎪
⎛ e t e t ⎞ ⎫⎪ 1
t
L ⎨sin
i 2t ⎜
i 2t e −tt sin t}
⎟ ⎬ = 2 L{e sin
2
⎝
⎠ ⎪⎭
⎩⎪
⎤
1⎡
2
2
1
1
+
+ 2
⎢
⎥= 2
2
2
2 ⎣ ( − 1) + 4 ( + 1) + 4 ⎦ s − 2 + 5 s + 2 + 5
=
d
d ⎛
L{sin 2t cosh t} = − ⎜ 2
ds
d
ds ⎝ s
L{{ sin 2t cosh t}
1
2s + 5
+
2s − 2
2s 2
⎞
+ 2
⎟⎠ = 2
2
s + 2s + 5
( s − 2 5)
( s + 2 5) 2
1
2
11.4.6 Division by t (Frequency-Integration Theorem)
∞
If L{ f (t )}
Proof
f (t ) ⎫
F ( s), then L ⎧⎨
⎬ = ∫ F ( s) ds
⎩ t ⎭ s
∞
L { f (t )}
∫ f (t ) e
F (s
F(s
( s)
− st
dt
0
Integrating both the sides w.r.t s from s to ∞,
∞
∞
∫ F ( s) ds
∫ f (t )e
st
dt dds
s 0
s
Since s and t are independent variables, interchanging the order of integration,
∞
∞
∞ ⎡∞
∫ F ( s)ds = ⎢⎢ ∫ f (t )e
0
s
⎣s
st
∞
∞
⎤
f (t ) − st
⎡1
⎤
ds
d ⎥ dt = ∫ ⎢ f (t )e − st ⎥ dt = ∫
e dt
−
t
t
⎦s
⎥⎦
0⎣
0
∞
⎧ f (t ) ⎫
L⎨
⎬ = ∫ F ( s)ds
⎩ t ⎭ s
Example 11.19
Solution
L{
Find the Laplace transform of
e t} =
1 e−t
.
t
1
1
−
s s +1
∞
∞
⎧1 − e − t ⎫
1 ⎞
⎛1
∞
−t
L⎨
g s + 1) s
⎬ = ∫ L 1 e ds = ∫ −
⎟⎠ ds = log s − log(
⎝
t
s
s
+
1
⎩
⎭ s
s
∞
∞
s ⎤
⎡
= ⎢log
s + 1⎦ s
⎣
⎡
⎤
⎢ 1 ⎥
log ⎢
⎥ = log
⎢1 + 1 ⎥
⎣ s ⎦s
⎛
⎞
s
s +1
⎜ 1 ⎟
log ⎜
= − log
= log
⎟
1
s +1
s
⎜1+ ⎟
⎝
s⎠
11.10 Network Analysis and Synthesis
Example 11.20
Solution
L
e
e
∞
− bt
−e
t
⎩
Example 11.21
s
s a
log
s+b
1
=
s
s
1
= log
2
Example 11.22
L
s b
1
ds =
s b
et
t
=
1
e
L
2
1
s +1
−
+
=
1⎡
log
s 1
s +1
∞
s
cos h2t sin 2t
.
t
e
2t
⎞
t
=∫
sin 2t
⎭
1
e2
L
2
t
+L
e
t
t
t
+4
n 2t
= L
t
s
L
s a
s+b
= log
s+b
s a
2
t} =
L
t
− log
1
s 1 1
s +1
− log
= log
2
s +1
s 1
Find the Laplace transform of
cosh t
t
1 1
2 s
1
log( s −
2
1⎞
s
− og
1
1+
s
1
s⎥ = 1
1
2
1+
s ⎦s
1
1+
∞
s
+
sinh t
.
t
1
ds
s +1
−
s
e
2
−
a
s
− og
b
1+
s
1+
t =L
L
Solution
s a
Find the Laplace transform of
Solution
sinh t
=
t
1
a⎤
s
= log
=
b
1+
s ⎦s
⎣
s
.
1
∞
⎡
bt
e
t
1
s a
∫
d
s
⎡
L
=
Le
−
e
Find the Laplace transform of
s
2
s +
ds = tan −1
s⎤
= − tan
2 s
2
2
cot
By first shifting theorem,
L
cosh t
t
t
2
L
e
t
+L
e
t
1
cot
2
1
2
cot
11.4.7 Time-Differentiation Theorem: Laplace Transform of Derivatives
If
f t
F ( s) then
− f( )
f
f
2
− sf
−
+2
2
2
11.4 Properties of Laplace Transform 11.11
In general,
f
−
s ′′F s
t
Proof
f ′ t e − st dt
L f′ t
0
Integrating by parts,
f t e − st
f t
L fn t
Example 11.23
n
t e
(0) + sL {
d
t
}
0
′ + sL{ f t}
f t
s∫
0
Similarly,
In general,
t e − st dt = − f
s
0
s −
+
f′
f t
f )}
−
−
Find L f (t ) and L f ′(t ) of
sin t
.
t
t
Solution
L f t
L
L
=
′
sin t
=
t
s
s
1
−
Example 11.24
∞
− lim
t →0
f
e
=
Example 11.25
Solution
−
∞
s
ta
2
−
−
s
s
s
− st
0
t
t
5
5.
∞
5
t dt =
t=
d
st 5
e
s
0
3
0 = ⋅ (1 − e −
s
+ =
0
−3
s
−
−
− s
− =
Find L f (t ) and L f ′(t ) of f(t) = e −5t sin t .
L f t
L f′ t
Example 11.26
s co
3
0
s
f t
s +1
an −1 s
ds
Find L f (t ) and L f (t ) of the following function:
t
Solution
si t
t
1
t}
Le
−
⋅
1
+1
s
1
s +
s
0
+ 26
2
Find L f (t ) and L f ′(t ) of the following function:
t
t
6
0
t
t
3
3
+
26
=
−5
s
11.12 Network Analysis and Synthesis
3
e − st ⎤
⋅ t⎥
⎣ −s ⎦0
1
3
Solution
L f t
e
0
=−
3
s
t dt
e
t dt
e
dt
3
−
−
+
1
s
3
e
s2 ⎦0
+6
e
st
s
∞
3
3
−
s s2
+
= 2
s
s
s
1
1
3s
f ( 0) = + e
3
s
s
=
L
st
11.4.8 Time-Integration Theorem: Laplace Transform of Integral
t
If L f t
F s
L
∫
F s
s
t dt
⎭
0
t
Proof
L
t
⎛ e − st
t
f t t
f t t
s
∞
− st
f t
⎩0
Integrating by parts,
L
t
f t t
⎡ e − st
∫⎢
t
d
dt ∫0
0
t dt
t
Example 11.27
dt
⎭
0
Find the Laplace transform of
∫e
−
⎞
dt =
1
s
0
=
1
L f
s
}=
F
s
t 3 dt .
0
Solution
t
L
Le
t
e−
t
3!
s
6
=
4
4
s
1
6
L e −2 t 3 =
s
s s+2
0
4
t
Example 11.28
Find the Laplace transform of
∫ t cosh t dt .
0
Solution L t cosh t
et
L t
t
t cosh t dt
L
0
e t⎞
1
=
2
⎭ 2
1
s
=
te
s
t cosh t
1
1
2 ( s 1)
⎤
1
(
)
1
ss
t
Example 11.29
Find the Laplace transform of the
∫ te
4t
sin 3t dt .
0
Solution
L t in 3t
L te
in t}
d
L
ds
6
[( s +
t =−
4
d
3
ds s
6
4
s
6s
s
2
1 2
1
s2 + 1
=
2
−
11.4 Properties of Laplace Transform 11.13
t
L
t e−
1
s
t dt
0
−
=
6 s+4
s s2 + s +
)2
t
Example 11.30
t sin 3t dt .
Find the Laplace transform of e
0
Solution
d
L
ds
L t in 3t
t
1
s
t sin 3t
L
0
2
s
s +
6
t
6s
6
t
L e −4 t ∫ t
d
3
ds s +
t
6
[s
0
11.4.9 Initial Value Theorem
If
f t
F ( s) then lim ( = lim
( s)
s→
Proof We know that,
− f( )
f
∞
L f t )} +
sF s
t e − dt +
0
∞
lim
( s)
s
s →∞
e − st dt +
f′
lim
0
∞
lim[
0
s →∞
dt +
t
t →0
11.4.10 Final Value Theorem
If
f t
F ( s) then lim (
lim
( s)
s →∞
t
Proof We know that
− f( )
f
∞
sF s
L f t )} +
t e − dt +
0
lim
s
( s)
f′
lim
s→ 0
t
e
st
dt +
0
lim f ′
e
st
d
0
0
+
t →∞
=
t dt +
0
+ f (0
im (
t →∞
(t )
11.14 Network Analysis and Synthesis
Example 11.31
Verify the initial and final value theorems for
f (t ) = e −tt (t ) 2 = e t (t
(t 2
2
2
F ( s) =
+
+
3
(s )
( s )2
2s
2s
sF ( s) =
+
+
3
( s 1)
( s 1) 2
lim f (t ) = 1
Solution
t
t + 1)2 .
t )
1
s +1
s
s +1
t →0
⎡
⎤
2
2
⎢
⎥
2
1 ⎥
s
lim sF ( s) = lim ⎢ s 3 +
+
=1
2
1⎥
s →∞
s →∞ ⎢
1⎞
1⎞
⎛
⎛
1
+
⎢ ⎜1 + ⎟
⎥
⎜⎝1 + ⎟⎠
s⎥
s
⎢⎣ ⎝ s ⎠
⎦
Hence, the initial value theorem is verified.
lim f (t ) = 0
t →∞
lim sF ( ) = 0
s→ 0
Hence, the final value theorem is verified.
Verify the initial and final value theorems for e t (t 2 + cos3t).
Example 11.32
f (t ) = e − t (t 2
Solution
F ( s) =
sF ( s) =
2
(s
)
3
2s
(s
lim f (t ) = 1
)3
cos 3t )
+
+
s +1
( s )2 + 9
s( s )
(s
)2 + 9
t →0
⎡
⎛ 1⎞ ⎤
2
⎢
⎜⎝1 + ⎟⎠ ⎥
2
s
⎥ =1
lim sF ( s) = lim ⎢ s 3 +
2
⎢
⎥
s →∞
s →∞
1
1
9
⎛
⎞
⎢ ⎛⎜1 + ⎞⎟
⎥
1
+
+
⎜⎝
⎟
s⎠
⎢⎣ ⎝ s ⎠
s 2 ⎥⎦
Hence, the initial value theorem is verified.
lim f (t ) = lim(
li ( 2 cos 3t )e − t = 0
t→
t →∞
⎡ 2s
s( s + ) ⎤
⎥=0
lim sF ( ) = lim ⎢
+
s→ 0
s→ 0 ⎢ (
)3 ( s + 1)2 + 9 ⎥⎦
⎣
Hence, the final value theorem is verified.
Example 11.33
F(s) =
2s+ 1
s3 + 6s 2 + 11s+ 6
Find the initial and final values of the function whose Laplace transform is
.
11.5 Laplace Transform of Periodic Functions 11.15
F ( s) =
Solution
2 1
+
s s2
sF ( s) = lim 3
=
lim
=0
1
6
s →∞
s →∞ s
6 s 2 11s 6 s→∞ 1 + 6 + 11
6s
+
s s 2 s3
2s2 + s
sF ( s) = lim 3
=0
s →0
s → 0 s + 6 s 2 + 11s + 6
f
Find the final value of the function whose Laplace transform is I(s)
s =
I ( s) =
Solution
s+6
s( s )
s→0
s+ 6 .
s(s+
( 3)
s+6
=2
s+3
li s ( s) = lim
I( )
11.5
s + 6s
6 s 2 + 11s + 6
2s2 + s
f
Example 11.34
2s + 1
3
s→0
LAPLACE TRANSFORM OF PERIODIC FUNCTIONS
A function f(t) is said to be periodic if there exists a constant T(T > 0) such that
values of t.
f (t T ) f (t T T ) f (t
(t T ) = f (t )
(t T )
f (t ), for all
In general, f (t nT ) = f (t ) for all t, where n is an integer (positive or negative) and T is the period of the
function.
If f (t) is a piecewise continuous periodic function with period T then
L{ f (t )} =
1
1 − e −Ts
∞
Proof
L{ f (t
∫ f (t(t ) e
T
∫ f ( t )e
dt
0
− st
0
In the second integral, putting t
When
st
∞
T
dt = ∫ f (t )e
st
∫ f ( t )e
ddt
0
T
x + T , dt dx
d
t T, x = 0
t→ , x→∞
T
L{ f (t
∫
0
∞
f (t
(t )e − st d
dt + ∫ f ( x T )e − s( x
0
∞
T
= ∫ f (tt))e
st
0
0
T
dt e −Ts
d
∫ f ( x )e
dx
∞
st
dt e −Ts ∫ f (t )e
0
T
= ∫ f (t )e − st dt + e
0
sx
0
T
= ∫ f ( t )e
T)
Ts
L{ f (t )}
st
dt
dx
− st
dt
11.16 Network Analysis and Synthesis
T
(
Ts
) L{ f (t )} = ∫ f (t )e − st dt
0
L{ f (t )} =
Example 11.35
T
1
1 − e −Ts
∫ f ( t )e
− st
dt
0
Find the Laplace transform of the waveform shown in Fig. 11.7.
f (t )
A
T
0
2T
t
3T
Fig. 11.7
Solution The function f (t) is a periodic function with period T.
At
f (t ) =
0<t <t
T
T
1
st
L{ f (t )} =
∫ f (t )e dt
1 − e −Ts 0
=
=
T
1
1− e
−Ts
At
∫T
e
st
dt
0
T
1
1 − e −Ts
A
te − st dt
T ∫0
T
⎡ e − st e − st ⎤
=
− 2 ⎥
⎢t
T (1 e −Ts ) ⎣ − s
s ⎦0
⎛
A
e −Ts e −Ts 1 ⎞
=
−T
− 2 + 2⎟
−Ts ⎜
s
T (1 − e ) ⎝
s
s ⎠
A
⎡ Te −Ts 1
⎤
+ 2 (1 − e −Ts ) ⎥
⎢−
s
T (1 − e ) ⎣
s
⎦
−Ts
A
Ae
= 2−
Ts
s(1 e −Ts )
A
=
Example 11.36
−Ts
Find the Laplace transform of the waveform shown in Fig. 11.8.
f (t )
1
0
1
2
Fig. 11.8
3
t
11.5 Laplace Transform of Periodic Functions 11.17
Solution The function f(t) is a periodic function with period 2.
(t ) = t 0 t < 1
= 0 1< t < 2
L{ f (t )} =
2
1
∫ f ( t )e
1 − e−2s
st
dt
0
=
=
=
=
Example 11.37
2
⎡ 1 −stst
te
dt
d
+
⎢∫
∫ 0⋅e
⎢⎣ 0
1
1
1− e
−2s
⎤
dt ⎥
⎥⎦
1
⎡ e − st
e − st ⎤
t− 2 ⎥
⎢
s ⎦0
⎣ −s
1
1 − e−2s
1
1 − e −2 s
⎛ e−s e−s 1 ⎞
⎜ −s − 2 + 2 ⎟
s
s ⎠
⎝
1
2
st
s (1 − e
−2 s
)
(1 − e − s − se − s )
Find the Laplace transform of the waveform shown in Fig. 11.9.
f (t )
1
0
2a
a
3a
4a
t
Fig. 11.9
Solution The function f(t) is a periodic function with period 2a.
t
a
1
= ( a t)
a
f (t ) =
L{ f (t )} =
1
1− e
− 2 as
0<t
a
t < 2a
a
2a
∫
f (t ) e
st
dt
0
2a
⎡ a t st
⎤
1
e
dt
+
( 2a t ) e st dt ⎥
⎢
∫
− 2 as ∫ a
a
1− e
⎢⎣ a
⎥⎦
a
a
2a
⎧ ⎡ e − st
1
e − st ⎤ ⎡ e − st
e st ⎤ ⎫⎪
⎪
=
t− 2 ⎥ +⎢
( a t) + 2 ⎥ ⎬
2 as ⎨ ⎢ − s
a((
) ⎪⎣
s ⎦0 ⎣ −s
s ⎦a ⎪
⎩
⎭
− as
−2 as
− as ⎞
⎛
1
e
1 e
e
=
− 2 + 2+ 2 − 2 ⎟
−2 as ⎜
a( e
s
)⎝ s
s
s ⎠
=
1
11.18 Network Analysis and Synthesis
=
=
=
=
=
Example 11.38
−2e − as + 1 + e −2 as
as 2 (1 − e −2 as )
( −
2
as ( −
1− e
− as 2
)
− as
)( +
)(
− as
)
− as
− as
2
as ( +
)
− as
−e 2
as
e2
⎛ as
as 2 ⎜ e 2
a
⎝
e
− as ⎞
2
⎟
⎠
⎛ as ⎞
tanh ⎜ ⎟
⎝ 2⎠
as 2
Find the Laplace transform of the waveform shown in Fig. 11.10.
f (t )
1
a
0
2a
3a
4a
t
−1
Fig. 11.10
Solution The function f(t) is periodic with period with period 2a.
f (t ) = 1
0<t a
= −1
a t < 2a
L{ f (t )} =
1
1− e
− 2 as
2a
∫
f ( t )e
st
dt
0
2aa
⎡ a st
⎤
e
dt
d
e − st ( ) dt ⎥
⎢
∫
− 2 as ∫
1− e
⎢⎣ 0
⎥⎦
a
a
2
a
⎧ ⎡ e − st ⎤ ⎡ e − st ⎤ ⎫
1
⎪
⎪
=
⎥ +⎢
⎥ ⎬
⎨⎢
1 − e − 2 as ⎪ ⎣ − s ⎦ 0 ⎣ s ⎦ a ⎪
⎩
⎭
− as
−2 as
⎛
1
e
1 e
e − as ⎞
=
−
+
+
−
⎜
s
s ⎟⎠
1 − e −2 as ⎝ − s s
=
=
1
(1 − e − as ) 2
s(1 + e − as )(1 − e − as )
11.5 Laplace Transform of Periodic Functions 11.19
1 − e − as
=
s(1 + e − as )
as
−
as
1 e 2 −e 2
= ⋅
as
s ⎛ as
− ⎞
2 +e 2
e
⎜
⎟
⎝
⎠
1
⎛ as ⎞
= tanh ⎜ ⎟
⎝ 2⎠
s
Example 11.39
Find the Laplace transform of the waveform shown in Fig. 11.11.
f (t )
a
p
w
O
2p
w
t
3p
w
Fig. 11.11
Solution The function f(t) is known as a half-sine wave rectifier function with period
f (t ) = a si ω t
=0
π
ω
π
2π
<t<
ω
ω
0<t<
The function f(t) is a periodic function.
L{ f (t )} =
1− e
=
2π
ω
1
∫
⎛ 2π ⎞
−⎜ ⎟ s 0
⎝ω⎠
1
1− e
−
2π s
ω
f ( t )e
st
dt
2π
⎛π
ω
⎜ω
− st
a
si
ω
t
e
d
dt
+
⎜∫
∫ 0⋅e
π
⎜0
⎝
ω
st
⎞
⎟
dt ⎟
⎟
⎠
π
=
a
1− e
=
−
2π s
ω
a
1− e
−
2π s
ω
⎡ 1
⎤ω
⋅ e − st ( s sinn ω t ω cos
cos ω t ) ⎥
⎢ 2
2
⎣s +ω
⎦0
⋅
⎡ −πs
⎤
ω (ω ) + ω ⎥
⎢
e
s 2 + ω 2 ⎢⎣
⎥⎦
1
2π
.
ω
11.20 Network Analysis and Synthesis
=
−π s ⎞
⎛
aω ⎜1 + e ω ⎟
⎝
⎠
−π s ⎞
⎛
1+ e ω
⎜
⎝
=
Example 11.40
⎟⎜
⎠⎝
⋅
⎟
⎠
1
s + ω2
2
aω
1
⋅ 2
−π s ⎞
⎛
s + ω2
ω
⎜1 − e ⎟
⎝
⎠
Find the Laplace transform of
f (t ) = t 2
f (t ) = f (t
if
−π s ⎞
⎛
1− e ω
t<2
0
).
Solution The function f(t) is a periodic function with period 2.
L{ f (t )} =
2
1
1− e
−2ss
∫e
st
f ((tt ) dt =
0
2
1
1− e
−2 s
∫e
− st
t 2 dt
0
2
=
=
=
Example 11.41
if
⎡ 2 ⎛ e − st ⎞
⎛ e − st ⎞
⎛ e − st ⎞ ⎤
−
2
t
+
2
⎢t ⋅ ⎜
⎟
⎜ 2 ⎟
⎜ 3 ⎟⎥
⎝ s ⎠
⎝ − s ⎠ ⎥⎦ 0
⎢⎣ ⎝ − s ⎠
1
1 − e −2 s
1
1 − e −2 s
(
⎛ e −2 s
e 2s
e −2 s 2 ⎞
−
4
−
4
−
2
+ 3⎟
⎜
s
s2
s3
s ⎠
⎝
1
− e−
s
)
s3
(2 − 2e
−2 s
Find the Laplace transform of
f (t ) = e t
f (t ) = f (t
t < 2π
0
).
Solution The function f(t) is a periodic function with period 2p.
L{ f (t )} =
=
=
=
1
1− e
− 2π s
1
1 − e − 2π s
1
1 − e − 2π s
1
1 − e − 2π s
2π
∫e
st
f (t ) dt
0
2π
∫e
− st t
e dt
0
2π
∫e
(
s)t
dt
0
2π
⎡ e( s )t ⎤
⎢
⎥
⎣ 1 − s ⎦0
− 4 se
2
4s2e
2s
)
11.6 Waveform Synthesis 11.21
=
=
Example 11.42
1
1 − e − 2π s
⎡ e(1−s ) 2π
1 ⎤
−
⎢
⎥
1− s⎦
⎣ 1− s
e(1−s ) 2π − 1
(1− e −2π s )(1 − s)
Find the Laplace transform of the function shown in Fig. 11.12.
f (t)
3
2
1
T
2T
3T
t
4T
Fig. 11.12
Solution The function f(t) can be represented in terms of Heaviside unit step function.
f (t ) = [u
[u(t − T ) u(t 2T )] + 2[u
[u(t − 2T ) u(t 3T )
L{ f (t )}
[u(t
(t
t )] + … ∞
T ) u(t
u( t
= u( t − T ) + u( t − T ) + u( t − T ) + … ∞
L
L{u
{u(t − T ) u(t 2T ) u(t − 3T )
}
1
1
1
= e −Ts + e −2Ts + e −3Ts + …
s
s
s
1
= ⎡⎣e −Ts + e −2 + e −3Ts + …⎤⎦
s
=
11.6
e −Ts
s(1 − e −Ts )
WAVEFORM SYNTHESIS
Any waveform can be constructed with unit step, unit ramp and unit impulse
functions, etc. We know the Laplace transforms of these special functions.
Hence, we can find the Laplace transform of any function in terms of Laplace
transform of these functions.
There is another way of synthesising the waveforms. Any function can
be expressed in terms of a gate function. The gate function is shown in
Fig. 11.13.
This function can be expressed in terms of unit-step functions.
f (t ) = Au(t
Au(t ) Au(t T )
f (t )
A
0
T
t
Fig. 11.13 Gate function
11.22 Network Analysis and Synthesis
Example 11.43
Find the Laplace transform of the unit-doublet function.
f (t )
Solution The unit-doublet function d ′(t) is shown in Fig. 11.14.
δ()
L{{δ ( )}
d
δ( )
dt
⎧d
⎨ δ(
⎩ dt
t
⎫
)⎬
⎭
0
sL{δ (t )} = s( )
sL
s
Fig. 11.14
Example 11.44
Find the Laplace transform of a rectangular pulse shown in Fig. 11.15.
f (t )
1
t
T
0
Fig. 11.15
Solution The rectangular pulse can be constructed from two functions as shown in Fig. 11.16. This function
is known as gate function.
f2 (t )
f1 (t )
1
u (t )
0
−1
t
0
T
−u (t − T )
(a)
(b)
Fig. 11.16
f ( t ) = f ( t ) + f ( t ) = u( t ) − u( t T )
1 1 T
1
F ( s) = L{u(t )} L{u(t − T )} = − e −Ts
= (1 − e −Ts )
s s
s
Example 11.45
Find the Laplace transform of a sawtooth waveform shown in Fig. 11.17.
f (t )
A
0
T
Fig. 11.17
t
11.6 Waveform Synthesis 11.23
Solution The sawtooth waveform can be constructed from three functions as shown in Fig. 11.18.
f1(t)
f2(t)
A r (t)
T
f3(t)
A
t
T
0
t
T
0
A
−A u(t − T )
− A r (t − T )
T
(a)
t
T
0
(b)
(c)
Fig. 11.18
A
A
r (t ) − r (t T ) Au(t T )
T
T
A
A
A A 1 −Ts A −Ts
F ( s) = L{r (t )} − L{r (t T )} − AL{u(t − T )} = 2 −
e − e
T
T
T s2
s
Ts
f (t ) = f (t ) + f (t ) + f (tt)) =
Example 11.46
Find the Laplace transform of a triangular waveform shown in Fig. 11.19.
f (t )
1
0
T
2
t
T
Fig. 11.19
Solution The triangular waveform can be constructed from three ramp functions as shown in Fig. 11.20.
f1(t )
f3(t )
f2(t )
2 r (t )
T
2 r (t −T )
T
t
0
0
t
T
2
t
T
0
−4 r t −T
2
T
(a)
(c)
(b)
Fig. 11.20
f (t ) = f (t ) + f (t ) + f (tt)) =
2
4 ⎛ T⎞ 2
r(t ) − r t − ⎟ + r(t T )
T
T ⎝
2⎠ T
Ts
2
4 ⎧ ⎛ T ⎞⎫ 2
2
4 −
2
F ( s) = L{r (t )} − L ⎨r ⎜ t − ⎟ ⎬ + L {r(t
(t T )} = 2 − 2 e 2 + 2 e −Ts
T
T ⎩ ⎝
2⎠⎭ T
Ts
T
Ts
Ts
11.24 Network Analysis and Synthesis
Example 11.47
Find the Laplace transform of a trapezoidal pulse shown in Fig. 11.21.
f(t )
1
t
0
a
2a
3a
4a
Fig. 11.21
Solution The trapezoidal waveform can be constructed from four ramp functions as shown in Fig.11.22.
f1(t )
f2(t )
1
a r (t )
t
a
0
t
0
− a1 r (t − a)
(a)
(b)
f3(t )
f4(t )
1
a r (t − 4a)
t
0
3a
− a1 r (t − 3a)
(c)
t
4a
0
(d)
Fig. 11.22
f (t ) = f (tt)) + f (t ) + f3 (t ) + f (t ) =
1
1
1
r ( t ) − r ( t a) − r ( t
a
a
a
a) +
1
r(
a
)
1
1
1
1
1 1 1 1 − as 1 1 −3as 1 1 −4 s
L{r (t )} − L{r (t aa)}
L{
)} − L{r (t 3a)}
a)} + L{r (t 4 a)}
a)} =
−
e −
e
+
e
a
a
a
a
a s2 a s2
a s2
a s2
1
= 2 (1 − e − as − e −3 + e −4 as )
as
F ( s) =
11.6 Waveform Synthesis 11.25
Example 11.48
Find the Laplace transform of a sinusoidal waveform shown in Fig. 11.23.
f (t )
A
t
T
2
0
Fig. 11.23
Solution The waveform can be constructed from two functions as shown in Fig. 11.24.
f2(t )
f1(t )
A
0
T
2
T
t
0
t
T
2
(a)
T
3T
2
(b)
Fig. 11.24
⎛ T⎞ ⎛ T⎞
f (t ) = f (t ) + f (t ) = A si ω t u ( t ) + A si ω t t − ⎟ u t − ⎟
⎝
2⎠ ⎝
2⎠
2π
where
ω=
T
Ts
Ts
−
− ⎞
⎧
Aω
Aω
Aω ⎛
⎛ T ⎞ ⎛ T ⎞⎫
2
F ( s) = A L {sin ω t u(t
u((t
(t )} AL ⎨si ω t ⎜ t − ⎟ u t − ⎟ ⎬ = 2
+
e
= 2
⎜1 + e 2 ⎟
⎝
2⎠ ⎝
2 ⎠ ⎭ s + ω 2 s2 + ω 2
s + ω2 ⎝
⎠
⎩
Example 11.49
Find the Laplace transform of the waveform shown in Fig. 11.25.
f (t )
2
1
0
1
2
3
t
Fig. 11.25
Solution The function f (t ) can be expressed as sum of four step functions.
f (t ) = u(t ) + u(t ) − u(t
( t ) − u( t )
u(t − 2)}
F ( s) = L{u(t )} L{u(t )} − L{u(
{ ( 3)} = 1s + 1s e − s − 1s e −2 − 1s e
3s
=
(
1
11+
+
s
−
2
−
3
)
11.26 Network Analysis and Synthesis
Example 11.50
Determine the Laplace transform of the waveform shown in Fig. 11.26.
f (t )
2
1
1
0
2
t
3
Fig. 11.26
Solution The function f (t ) can be expressed as sum of four functions as shown in Fig. 11.27.
f1(t )
f2(t )
r (t )
t
t
1
0
0
(a)
r (t − 1)
(b)
f3(t )
f4(t )
u (t − 2)
1
t
0
0
t
2
3
−2
−2u (t − 3)
(c)
(d)
Fig. 11.27
f (t ) = f (t ) + f (t ) + f3 (t ) + f (t ) = r (t ) + r (t
F ( s)
{ ( )}
Example 11.51
{r((
)}
{u(
)}
) + u(t
(t
{u((
) − 2u(t
(
)
1 1 − s 1 −2ss 2
)} = 2 + 2 e + e − e
s
s
s
s
Find the Laplace transform of the waveform shown in Fig. 11.28.
f (t )
2
1
2t 2
0
1
2(4 − t ) 2
t
2
3
4
Fig. 11.28
3s
11.6 Waveform Synthesis 11.27
Solution The given parabolic waveform can be constructed from three functions as shown in Fig.11.29.
f1(t )
2
0
f3(t )
f2(t )
2t 2u(t )
t
t
2
0
t
4
0
− 16 r ( t − 2 )
(a)
−2 ( t − 4)2 u(t − 4)
(b)
(c)
Fig. 11.29
F (t ) = f (t ) + f (t ) + f (t ) = 2tt 2 u(t
u( t )
6r(t
r(t
(
(t
F ( s) = 2 L{t
{t 2 u(t )} − 16L{{r (t 2)} 2L{(
L{(t 4) 2 u
L{(
u((t − 4)} = 2
Example 11.52
)
2
s3
− 16
(t
(t
) 2 u( t
1
e −2ss − 2
s2
)
2
s3
e
s
=
4
s3
(1 − e −4 s ) −
Find the Laplace transform of the waveform as shown in Fig. 11.30.
f (t )
3
2
1
0
1
t
2
Fig. 11.30
Solution The given waveform can be constructed from four functions as shown in Fig. 11.31.
f1(t )
f2(t )
3r ( t )
0
t
1
t
0
3r ( t −1)
(a)
(b)
Fig. 11.31
f4(t)
f3(t)
0
t
1
2
0
−1
−u (t − 1)
−2
(c)
−2u (t − 2)
(d)
Fig. 11.32
16
s2
e
2s
11.28 Network Analysis and Synthesis
f (t ) = f (t ) + f (t ) + f3 (t ) + f (t ) = 3rr((t ) − 3r (t
) − u(t
(t
) − 2u(t
u( t
3
F ( s) = 3L
L{r (t )} 3L{r (t − 1)}
)} − L{u(t 1)} 2L{
L{u(t − 2)} =
Example 11.53
s
2
−
)
3
1
2
e − s − e −ss − e
s
s
s
2s
2
Find the Laplace transform of the periodic waveform shown in Fig. 11.33.
f (t )
1
1
0
2
3
t
4
−1
Fig. 11.33
Solution The function f (t ) is a periodic function with period 2.
The function f (t ) can be constructed from three functions by waveform synthesis.
f (t ) = u(t ) − 2u
u(t 1) + u(t 2)
F1 ( s) = L{u(t )} 2 L{u(t − 1)}
1)} + L{u(t − 2)} =
The Laplace transform of periodic function f (t ) is
L{ f (t )} =
1
F1 ( s)
1 − e −2 s
=
1 ⎛ 1 2 −ss 1
⎜ − e + e
s
1 − e −2 s ⎝ s s
=
1 ⎛ 1 − 2e − s + e −22 s ⎞
⎜
⎟
s
1 − e −2 s ⎝
⎠
=
(1 − e − s ) 2
s(1− e − s )(1 + e − s )
=
1 1 − e−s
s 1 + e−s
s
1 e2
=
s s
e2
e
e
−
s
2
−
s
2
1
⎛ s⎞
= tanh ⎜ ⎟
⎝ 2⎠
s
2s ⎞
⎟⎠
1 2 − s 1 −2 s
− e + e
s s
s
11.6 Waveform Synthesis 11.29
Example 11.54
Find the Laplace transform of the waveform shown in Fig. 11.34.
f (t )
A
0
T
2T
3T
t
Fig. 11.34
Solution The function f (t ) is a periodic function with period T.
The function f (t ) can be constructed from three functions by waveform synthesis.
A
A
f (t ) = r (t ) − r (t T ) Au(t T )
T
T
A
A
A A 1 −Ts
A
F1 ( s) = L{r (t )} − L{r (t T )} − A L{u(t T )} = 2 −
e Ts − e − Ts
2
T
T
T s
s
Ts
The Laplace transform of the periodic function f (t ) is
1
1 ⎛ A A −Ts A −Ts ⎞
L { f (t )} =
F ( s) =
− e − e ⎟
⎜
−Ts 1
⎠
s
1− e
1 − e − Ts ⎝ Ts 2 T
Example 11.55
Find the Laplace transform of periodic waveform shown in Fig. 11.35.
f (t )
A
0
T
2
T
t
Fig. 11.35
Solution The function f (t ) is a periodic function with period T.
The function f (t ) can be constructed from two functions by waveform synthesis.
⎛ T⎞ ⎛ T⎞
f (t ) = A si t u(t ) A si ω t ⎜ t − ⎟ u t − ⎟
⎝
2⎠ ⎝
2⎠
Ts
Ts
−
− ⎞
⎧
ω
ω
Aω ⎛
⎛ T ⎞ ⎛ T ⎞⎫
2 =
2
F1 ( s) = AL{sin ω t u(t )} AL ⎨sin
i ωt t − ⎟ u t − ⎟⎬ = A 2
+
A
e
1
+
e
⎜
⎟
⎝
2⎠ ⎝
2⎠⎭
s + ω2
s2 + ω 2
s2 + ω 2 ⎝
⎩
⎠
The Laplace transform of the periodic function f (t ) is
1
L{ f (t )} =
F1 ( s)
1 − e −Ts
Ts
− ⎞
1
Aω ⎛
2
=
1
+
e
⎜
⎟
1 − e −Ts s 2 + ω 2 ⎝
⎠
−
Ts
Aω
1+ e 2
= 2
2
Ts
Ts
s +ω ⎛
− ⎞⎛
− ⎞
⎜1 − e 2 ⎟ ⎜ + e 2 ⎟
⎝
⎠⎝
⎠
Aω
1
= 2
Ts
−
s + ω2
1− e 2
11.30 Network Analysis and Synthesis
11.7
If L{ f (t )}
INVERSE LAPLACE TRANSFORM
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
(iii) Convolution theorem
11.7.1 Standard Results
Inverse Laplace transforms of some simple functions can be found by standard results and properties of
Laplace transform.
Example 11.56
F ( s) =
Solution
s2
3s+ 4 .
s3
1 3 4
= − 2+ 3
s s
s
3t + 2t 2
3s+ 4 .
Find the inverse Laplace transform of
F ( s) =
Solution
3s + 4
s3
L 1{F ( s)}
Example 11.57
s2
Find the inverse Laplace transform of
3s + 4
2
s +9
=
3s
s2 + 9
4
+
2
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
Example 11.58
F ( s) =
Solution
L 1{F ( s)}
Example 11.59
Solution
4 s + 15
2
16 s − 25
=
4 s + 15
1
s
15
1
=
+
2
⎛ 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) =
L 1{F ( s)}
2s + 2
2
s + 2 s + 10
=
2s+ 2
2
s + 2s+ 10
2(s
(s
(s
)
2
) +9
⎧ s ⎫
2e − t L−1 ⎨ 2
⎬=2
⎩ s + 9⎭
−t
3t
.
11.7 Inverse Laplace Transform 11.31
Example 11.60
Solution
3s + 7
Find the inverse Laplace transform of
F ( s) =
3s + 7
s
2
=
2s − 3
3(s
(
) + 10
(s
2
) −4
=3
s2
(s
(s
2s 3
)
)
2
4
+ 10
.
1
(s
⎧ s ⎫
t −1 ⎧ 1 ⎫
t
L−11{F ( s)} = 3e t L 1 ⎨ 2
⎬ + 10e L ⎨ 2
⎬ = 3e cosh 2
⎩s − 4⎭
⎩s − 4⎭
)2 − 4
5e t sinh 2t
11.7.2 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
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,
P ( s)
F ( s) =
( s a)(s
( s b)
By partial-fraction expansion,
A
B
F ( s) =
+
s+a s b
Case II Factors are linear and repeated,
P ( s)
F ( s) =
( s a)(s
( s b) n
By partial-fraction expansion,
A
B
B2
Bn
F ( s) =
+ 1 +
+ ... +
s + a s b ( s b) 2
( s b) n
Case III Factors are quadratic and distinct,
F ( s) =
P ( s)
( s2
as b)( s 2
cs d )
By partial-fraction expansion,
F ( s) =
As + B
+
s 2 + as b
Cs + D
s 2 + cs d
Case IV 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
s 2 + as b
+
C1s + D1
s 2 + cs d
+
C2 s + D2
( s2
cs d ) 2
+ ... +
Cn s + Dn
(
2
)n
11.32 Network Analysis and Synthesis
Example 11.61
s+2
.
s( s +1)( s + 3)
Find the inverse Laplace transform of
F ( s) =
Solution
s( s
s+2
)(s
)(s
(
)
By partial-fraction expansion,
F ( s) =
A
F ( s) =
L 1{F ( s)}
Example 11.62
A
B
C
+
+
s s +1 s + 3
s+2
2
=
sF ( s) s = 0 =
( s + )(s
)( s + ) s = 0 3
B
( s + ) F ( s) s = −1 =
s+2
1
=−
s( s + ) s = −1
2
C
( s + ) F ( s) s = −3 =
s+2
1
=−
s( s + ) s = −3
6
2 1 1 1
1 1
− ⋅
− ⋅
3 s 2 s +1 6 s + 3
2 −1 ⎧ 1 ⎫ 1 −1 ⎧ 1 ⎫ 1 −−11 ⎧ 1 ⎫ 2 1
1
L ⎨ ⎬− L ⎨
⎬− L ⎨
⎬= − e − e
3
6
⎩s⎭ 2
⎩ s + 1⎭ 6
⎩ s + 3⎭ 3 2
Find the inverse Laplace transform of
F ( s) =
Solution
s+2
s2 ( s
3t
s 2 .
s 2 ( s + 3)
)
By partial-fraction expansion,
F ( s) =
A B
C
+ 2+
s s
s+3
s 2
As(s
A(
) + B( s
) + Cs 2
= As 2 + 3 A
As + B
Bss + 3 B C
Cs 2
= ( A C)
C s2
2
1
(3A
A + B)
B) s + 3B
0
Comparing coefficients of s , s and s ,
A C=0
3A + B = 1
3B = 2
Solving these equations,
1
2
1
, B= , C=−
9
3
9
1 1 2 1 1 1
F ( s) =
+ ⋅ − ⋅
9 s 3 s2 9 s + 3
A
L 1{F ( s)}
1 −1 ⎧ 1 ⎫ 2 −1 ⎧ 1 ⎫ 1 −1 ⎧ 1 ⎫ 1 2
1 −3t
L ⎨ ⎬+ L ⎨ 2⎬− L ⎨
⎬= + t− e
9
s
3
9
s
+
3
9
3
9
⎭
⎩ ⎭
⎩
⎩s ⎭
11.7 Inverse Laplace Transform 11.33
Example 11.63
s 2 − 15s − 11 .
Find the inverse Laplace transform of
F( s) =
Solution
(s+ 1)(s
1 − 2)2
5 s 2 15 s − 11
(s
)( s
)(s
)(
)2
By partial-fraction expansion,
F ( s) =
5
2
15
A
B
C
+
+
s + 1 s − 2 ( s )2
A( s 2) 2 + B( s + 1)( s 2) C ( s + 1)
11
= A( s 2 − 4 s 4) B( s 2
2
s − 2) + C ( s + )
2
= As − 4 A
Ass + 4 A + Bs − Bs − 2 B + Cs + c
= ( A + B) s2 − ( A + B − C ) s + ( 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
1
4
7
F ( s) =
+
−
s + 1 s − 2 ( s )2
1 ⎫
⎧ 1 ⎫
−1 ⎧ 1 ⎫
−1 ⎧
L−1 ⎨
= e−t + 4
⎬ + 4L ⎨
⎬ − 7L ⎨
2⎬
(
s
−
2
)
⎩ s + 1⎭
⎩ s − 2⎭
⎩
⎭
L 1{F ( s)}
Example 11.64
Find the inverse Laplace transform of
Solution
F( s) =
By partial-fraction expansion,
3s+ 1
(s+ 1)(s
1 2 + 2)
2
3s + 1
)( s 2
)(s
)(
(s
)
F ( s) =
A
Bs + C
+ 2
s +1 s + 2
3s 1
A( s 2
2) ( Bs C )( s 1)
2
= As + 2 A + Bss 2
= ( A + B) s
2
Bs Cs
C C
( B + C )s + (
+ )
.
2
− 7te 2t
11.34 Network Analysis and Synthesis
Comparing coefficients of
2
Solving these equations,
1
0
2
0
=3
=1
2
3
F s =−
L
2 1
2
s
3 s + 1 3 s2
1
1
+ L
3
s +1
3
Example 11.65
Solution
2
3
7
3
7
1
s +2
s
7
+ L
2
3
s +2
2
1
2
= − e−
3
s +2
1
2
s
Find the inverse Laplace transform of
F s =
s
s
s
L
s
4
s
s
=
7
3 2
in 2t
.
1⎡ s
−
3 s2 + 1
1
3
s
1
L
s2 + 1
3
−s −
s s
3⎣ s
=
s
s
2
3
s
2
+4
− cos
11.7.3 Convolution Theorem
If L
t
1
F
t
then L
(t u )du
F
0
t
where
∫
t)
t
∞
Proof
F
e
f u u e − sv
0
∫
∞
∫e
0
Putting
When
v
t dv
−
∞
v dv
0
e−
f
v u dv
00
f v v du
0
dt
=u
t→∞
v
∞
F
∞
dt ⎥
s
0
dt du
0 0
u
u
The region of integration is bounded by the lines u 0 and u t.
To change the order of integration, draw a vertical strip which starts
from line u = 0 and terminates on the line u t. Hence, u varies
from 0 to t and t varies from 0 to ∞
∞
F
∫e
0
t
t
f
0
=L
f
0
(t u ) du
u
Fig. 11.36
t
11.7 Inverse Laplace Transform 11.35
t
Hence,
F
du
Convolution operation is commutative, i.e.,
Note
0
t
L
t
(t u )du =
f
u du
0
Example 11.66
⎭
0
Find the inverse Laplace transform of
1
F s
Solution
Let
s
s
1
s+2
F
1
.
)(s 1)
(s
1
s 1
s
t =
e
By convolution theorem,
t
L
t
s
e
0
Example 11.67
du
e
t
0
Find the inverse Laplace transform of
Solution
F s =
=
0
et
(1 e
3
1
s2 s
3t
)
.
2
1
F s
Let
t
e −3u
3
s2 s
1
(s
)
2
1
F s
2
s2
−t
t
By convolution theorem,
t
L
s
t
−
−u
u
0
e
−
du
−
t− u e
)
0
t
0
te −
Example 11.68
Find the inverse Laplace transform of
Solution
F s =
Let
F s =
t =
By convolution theorem,
t
L
−2 u
0
=
16
t −u
t
=
0
−
e 2
4
e −2
16
1
16
s
t
.
2
s
1
( s 2)
2
1
F s
2t
s 2
=e
2t
0
)2
(s 2)(s
1
⎡ ue 4
=e ⎢
⎣ 4
t
t
1
4t e
2
t
e u⎤
16
t
e 2t
0
⎛ −te
4
16
1⎞
16
11.36 Network Analysis and Synthesis
Example 11.69
Find the inverse Laplace transform of
F ( s) =
Solution
.
1
2
s ( s2
)
1
F1 ( s) =
Let
1
s 2 ( s 2 1)
F2 ( s) =
2
s +1
f (t ) = sin
nt
f 2 (t )
1
s2
t
By convolution theorem,
t
∫ sin u (t
L 1{F ( s)}
u ) du = [ t u −
u ]0 = t si t
t
u
0
Example 11.70
Find the inverse Laplace transform of
Solution
F( s) =
Let
F1 ( s) =
1
(s+ 1)(s
1 2 + 1)
1
.
1
)( s 2
)(s
)(
(s
1
2
s +1
f (t ) = sin
nt
)
F2 ( s) =
f 2 (t )
1
s +1
e−t
By convolution theorem,
t
∫
1
L {F ( s)}
0
=
11.8
t
t
ue
t u
du
d
∫e
u t
sin u du
1
⎡ eu
⎤
e−t t
e ⎢ (sin u cos u ) ⎥ =
[e (sin t − cos t ) +1]
2
⎣2
⎦0
−t
1
1
(sin t − cos t ) + e − t
2
2
SOLUTION OF DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
The Laplace transform is useful in solving linear differential equations with given initial conditions by using
algebraic methods. Initial conditions are included from the very beginning of the solution.
Linear
differential
equation
Solution of
differential
equation
Laplace
transform
Inverse
Laplace
transform
Algebraic
equation
Solution of
algebraic
equation
11.8 Solution of Differential Equations with Constant Coefficients 11.37
Example 11.71
Solve
dy
+ 22y = e -3t , y(0) = 1.
dt
Solution Taking Laplace transform of both the sides,
1
sY ( s) − y(
y( ) + 2Y ( s) =
s+3
1
sY ( s) 1 2Y ( s) =
s+3
1
s+4
( s )Y ( s) =
+1 =
s+3
s+3
s+4
Y ( s) =
( s + 2)( 3)
By partial-fraction expansion,
A
B
Y ( s) =
+
s+2 s+3
[ y(0))
]
s+4
=2
s + 3 s = −2
s+4
B ( s + ) Y ( s) s = −3 =
= −1
s + 2 s = −3
2
1
Y ( s) =
−
s+2 s+3
Taking inverse Laplace transform of both the sides.
y(t ) = 2e −2t e 3t
A
Example 11.72
( s + ) Y ( s) s = −2 =
Solve y ″ + y = t , y(0 )
1, y ′(0 ) = 0 .
Solution Taking Laplace transform of both the sides,
1
[ 2 ( s)) y(0) y ′(0)] + Y ( ) = 2
s
1
s 2Y ( ) − s + Y ( ) = 2
s
1
s3 + 1
( 2 )Y ( ) = 2 + s = 2
s
s
s3 + 1
s
1
Y( ) = 2 2
= 2
+ 2 2
=
s (
) s + 1 s ( s + 1)
y(
[∵ y(
s
2
1
+
s2 +
+11 − s 2
s
1
1
= 2
+ 2− 2
2 2
( s + 1) s 1 s
s +1
Taking inverse Laplace transform of both the sides,
y(t ) = cos t t sin t
Example 11.73
Solve y ″ + y = t 2
2t , y(0 )
4, y ′(0 ) = −2.
Solution Taking Laplace transform of both sides,
2 2
+
s3 s 2
2 2
s 2Y ( s) 4s
4 s 2 + sY ( s) − 4 = 3 + 2
s
s
⎡ s 2Y ( s) sy
s ( 0)
⎣
y ′(0) ⎤⎦ + [ sY ( s)
) 1, y ′(0) =
y ( 0) ] =
]
11.38 Network Analysis and Synthesis
Y
=
2
=
2
3
+
2
2
2
+ s
+
s
4
s
2
s
+
+ s
s3
2
s
2
2
s
s
4
4
2
2
2 2
+ −
= 4+
s s +1 s
s
s
2
s +1
Taking inverse Laplace transform of both the sides,
t3
3
yt
Example 11.74
Solve
2 + 2e − t
= 0 y′ 0
y
0
Solution Taking Laplace transform of both the sides,
sY s
0 +
s
s Y s
s
(
s
′
y0
0]
1
Y s
2
s +4
Taking inverse Laplace transform of both the sides,
1
y t = sin 2t
2
Example 11.75
Solution
Solve
0
0
0
Taking Laplace transform of both the sides,
0 +
sY s
sY s
s = e−s
0 +
s
=
sY s
(
)Y
e
Y
e
−s
2
s +3 +2
Taking inverse Laplace transform of both the sides,
−
yt
Example 11.76
−s
=
Solve
s
e−s
s+2
e−s
1
s
1
s+2
t
y 0
1
Solution Taking Laplace transform of both the sides,
0 +
s Y s
s =
Y ( s)
s Y s
2
e −2 s
s
e −2 s
s
−2 s
e
s
e 2
4
[ y(
1
1
1
s
−
s s2
s
s +4
Taking inverse Laplace transform of both the sides,
Y(
=
e
yt =
1
4
−
1
4
(t
1
s2 + 4
ut
1
t
′
1]
11.9 Solution of a System of Simultaneous Differential Equations 11.39
11.9
SOLUTION OF A SYSTEM OF SIMULTANEOUS DIFFERENTIAL EQUATIONS
The Laplace transform can also be used to solve two or more simultaneous differential equations. The Laplace
transform method transforms the differential equations into algebraic equations.
Example 11.77
Solve
dx
+ y = sin t .
dt
dy
+ x = cos t
dt
where x(0) = 0 and y(0) = 2.
Solution Taking Laplace transform of both the equations,
sX ( s) − x(
x ( ) + Y ( s) =
sX ( s) Y ( s) =
and
sY ( s) − y(
y ( ) + X ( s) =
sY ( s)
X ( s) =
sY ( s)
X ( s) =
1
2
s +1
1
s2 + 1
s
…(i)
s2 + 1
s
+2
s2 + 1
2s2 + s + 2
s2 + 1
…(ii)
Multiplying Eq. (i) by s,
s 2 X ( s) + sY ( s) =
s
2
s +1
…(iii)
Subtracting Eq. (iii) from Eq. (ii),
(
2
)X( )
2
X ( s) = −
2
2
s −1
…(iv)
Substituting X ( s) in Eq. (i),
1
s
+2 2
s2 + 1
s −1
Taking inverse Laplace transform of Eqs (iv) and (v),
Y ( s) =
x(t ) = −2 sinh t
y(t ) = sin t 2 cosh t
and
Example 11.78
Solve
where x(0) = 1 and y(0) = 0.
dx
− y et
dt
dy
+ x = sin t
dt
…(v)
11.40 Network Analysis and Synthesis
Solution Taking Laplace transform of both the equations,
1
s −1
1
s
sX ( s) Y ( s) =
+1 =
s −1
s −1
1
sY ( s) − yy(( ) + X ( s) = 2
s +1
1
sY ( s) X ( s) = 2
s +1
sX ( s) x(
x ( ) − Y ( s) =
and
…(i)
…(ii)
Multiplying Eq. (i) by s,
s 2 X ( s) sY ( s) =
s2
s −1
…(iii)
Adding Eqs (ii) and (iii),
(
2
)X( ) =
X( ) =
=
1
2
s +1
s2
s −1
+
1
(
2
(
2
)2
+
s2
(
)(
2
)
1⎛ 1
s
1 ⎞
+ ⎜
+ 2
+ 2 ⎟
2 ⎝ s − 1 s +1
+1)
+ 1 s + 1⎠
1
…(iv)
2
Substituting X(s) in Eq. (i),
Y ( s) = sX ( s) −
Y ( s) =
=
s
s
=
−
s − 1 ( s 2 ) (s
(s
s
( s2
)2
−
1⎛ 1
s
1 ⎞
− ⎜
− 2
+ 2 ⎟
2 ⎝ s − 1 s + 1 s + 1⎠
(s + )
s
2
Solve
dx
+ 55x - 2y = t
dt
dy
+ 2x
2 + y=0
dt
0 = 0.
where x(0) = 0 and y(0)
)
−
s
s −1
s
2
1
1 t
1
(sin t t cos t )
(e cos t s n t ) = (e t cos t + 2 sin t t cos t )
2
2
2
1
1 t
1
and y(t ) = t sin t
(e cos t sin t )
(t sin t e t + cos
cos t sin t )
2
2
2
Example 11.79
))(s
(s
2
( s 1)( s 2 + 1)
Taking the inverse Laplace transform of Eqs. (iv) and (v),
x(t ) =
s3
…(v)
11.9 Solution of a System of Simultaneous Differential Equations 11.41
Solution Taking Laplace transform of both the equations,
sX s
+5X s −
x
=
1
s2
1
Y( )
and
+ 2X s +
sY s
2
Multiplying Eq. (i) by
1
(
2
…(i)
s2
s =0
1
( s)
…(ii)
0
1),
1
2
−
1X
1 Y( ) =
s +1
…(iii)
2s2
Adding Eqs (ii) and (iii),
X s =
s +1
2
…(iv)
2
s s
Substituting X(s) in Eq. (ii),
2
Y s =−
Now,
s2 s
…(v)
2
s +1
X s
2
s s
2
+C s+
s
By partial-fraction expansion,
A
s
X s
B
s2
+
C
As s +
s+
+
Comparing coefficients of
3
2
+
s1
D
2
s
s+
=
=
+
s
+
s +
+
s0 ,
=0
3
A
B =1
9B 1
Solving these equations,
1
27
=
1
C
9
1
2
D=−
27
9
+
+ Ds 2
+
+ Bs 2 + 6 Bs
=0
6
+
…(vi)
s
+
s +
+
+
+ s2
s+ B
11.42 Network Analysis and Synthesis
1 1 1 1
1
1
2
1
⋅ + ⋅ − ⋅
− ⋅
27 s 9 s 2 27 s + 3 9 ( s ) 2
Taking inverse Laplace transform of both the sides,
1 1
1 −3t 2 3t
x(t ) =
+ t
e − te
27 9
27
9
Similarly,
−2
A B
C
D
4 1 2 1
4
1
2
1
Y ( s) = 2
= + +
+
=
⋅ − ⋅ − ⋅
− ⋅
s ( s ) 2 s s 2 s + 3 ( s ) 2 27 s 9 s 2 27 s + 3 9 ( + 3) 2
Taking inverse Laplace transform of both the sides,
4 2
4 −3t 2 3t
y(t ) =
− t
e − te
27 9
27
9
X ( s) =
11.10
THE TRANSFORMED CIRCUIT
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 11.37.
I (s)
i (t)
+
+
R
v (t)
R
V (s)
−
−
Fig. 11.37 Resistor
For the inductor, the v–i relationships in time domain are
di
v(t ) = L
dt
t
1
i(t ) = ∫ v(t ) dt + i( )
L0
The corresponding frequency-domain relation are given as
V ( s) = Ls I ( s) − L i( )
1
i( )
I ( s) = V ( ss)) +
Ls
s
The transformed network is shown in Fig 11.38.
2. Inductor
I (s)
I (s)
i (t )
+
+
+
i (0)
v (t)
Ls
V (s)
V (s)
−
+
−
Li (0)
−
Fig. 11.38 Inductor
−
Ls
i (0)
s
11.11 Resistor–Inductor Circuit 11.43
3. Capacitor
For capacitor, the v−i relationships in time domain are
t
1
v(t ) = ∫ i(t ) dt + v( )
C0
dv
dt
The corresponding frequency–domain relations are given as
1
v( )
V ( s) =
I ( s) +
Cs
s
I ( s) = CsV ( s) − Cv( )
The transformed network is shown in Fig 11.39.
i( t ) = C
i (t)
I (s)
I (s)
+
+
+
1
Cs
v (t)
+
−
Cv (0)
v (0)
s
−
−
−
1
Cs
V (s)
V (s)
C
Fig. 11.39 Capacitor
11.11
RESISTOR–INDUCTOR CIRCUIT
Consider a series RL circuit as shown in Fig. 11.40. The switch is closed at time t = 0.
R
V
L
i (t)
Fig. 11.40 RL circuit
For t > 0, the transformed network is shown in Fig. 11.41.
Applying KVL to the mesh,
V
− RI ( s) Ls I ( s) = 0
s
V
L
I ( s) =
R⎞
⎛
s s+ ⎟
⎝
L⎠
By partial-fraction expansion,
A
B
I ( s) = +
R
s
s+
L
V
L
A sI ( s) |s = 0 = s ×
R⎞
⎛
s s+ ⎟
⎝
L⎠
R
V
s
Ls
I (s)
Fig. 11.41 Transformed network
=
s=0
V
R
11.44 Network Analysis and Synthesis
B
V
R⎞
R⎞
⎛
⎛
L
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 ) =
f
t>0
Example 11.80 In the network of Fig. 11.42, the switch is moved from the position 1 to 2 at t = 0,
steady-state condition having been established in the position 1. Determine i (t) for t > 0.
1Ω
1
2
10 V
1Ω
1H
i (t)
Fig. 11.42
At t = 0−, the network is shown in Fig 11.43. At t = 0−, the network has attained steady-state
condition. Hence, the inductor acts as a short circuit.
1Ω
Solution
10
= 10 A
1
Since the current through the inductor cannot change instantaneously,
i(
)=
i (0+) = 10 A
For t > 0, the transformed network is shown in Fig. 11.44.
Applying KVL to the mesh for t > 0,
− I ( s) − I ( s) − sI ( s) + 10 = 0
I ( s ) ( s + ) = 10
10
I ( s) =
s+2
Taking inverse Laplace transform,
i (t) = 10e−2t for t > 0
10 V
i (0−)
Fig. 11.43
1
s
1
I (s)
10
Fig. 11.44
11.11 Resistor–Inductor Circuit 11.45
Example 11.81 The network of Fig. 11.45 was initially in the steady state with the switch in the
position a. At t = 0, the switch goes from a to b. Find an expression for voltage v (t) for t > 0.
2Ω
a
b
+
1Ω
2V
1H
2H
v (t)
−
Fig. 11.45
2Ω
At t = 0−, the network is shown in Fig 11.46. At
t = 0−, the network has attained steady-state condition. Hence, the
2V
inductor of 2H acts as a short circuit.
2
i( ) = = 1 A
2
Since current through the inductor cannot change instantaneously,
i (0+) = 1 A
For t > 0, the transformed network is shown in Fig. 11.47.
2s
Applying KCL at the node for t > 0,
Solution
Fig. 11.46
+
s
1
V ( s) + 2 V ( s) V ( s)
+
+
=0
2s
1
s
3⎞
1
⎛
V ( s) ⎜ 1 + ⎟ = −
⎝ 2s ⎠
s
2
V (s)
−
Fig. 11.47
1
−
s =− 2 =− 1
V ( s) =
2s + 3
2 3
s +1 5
2s
Taking the inverse Laplace transform,
v (t) = − e−1.5 t
Example 11.82
i (0−)
for t > 0
In the network of Fig. 11.48, the switch is opened at t = 0. Find i(t).
10 Ω
3Ω
36 V
6Ω
0.1 H
i (t)
Fig. 11.48
iT (0−) 10 Ω
Solution At t = 0 , the network is shown in Fig. 11.49. At t = 0 ,
the switch is closed and steady-state condition is reached. Hence,
the inductor acts as a short circuit.
36
36
iT ( ) =
=
=3A
10 + (
) 10 + 2
6
iL ( ) 3 ×
=2A
6 3
−
–
3Ω
36 V
−
iL (0 )
Fig. 11.49
6Ω
11.46 Network Analysis and Synthesis
Since current through the inductor cannot change instantaneously,
iL(0+) = 2 A
For t > 0, the transformed network is shown in Fig. 11.50.
Applying KVL to the mesh for t > 0,
−0. − 0.1s
3
I( ) =
6
0.1s
( ) − 3 ( ) − 6I ( s) = 0
0. ( ) + ( ) = 0.2
I(s)
0.2
02
2
=
0.1 9 s + 90
Taking inverse Laplace transform,
i(t) = 2e−90 t
Fig. 11.50
for t > 0
Example 11.83 The network shown in Fig. 11.51 has acquired steady-state with the switch closed
for t < 0. At t = 0, the switch is opened. Obtain i (t) for t > 0.
10 Ω
4Ω
4Ω
36 V
2H
i (t)
Fig. 11.51
Solution At t = 0−, the network is shown in Fig 11.52. At t = 0−, the switch is closed and the network has
acquired steady-state. Hence, the inductor acts as a
short circuit.
36
36
iT ( ) =
=
=3A
10 + (
) 10 + 2
36 V
4
i( ) 3 ×
= 1.5 A
4 4
Since current through the inductor cannot change
instantaneously,
i(0+) = 1.5 A
For t > 0, the transformed network is shown in Fig. 11.53.
Applying KVL to the mesh for t > 0,
4Ω
10 Ω
iT (0− )
4Ω
i (0− )
Fig. 11.52
4
−4I ( s ) − 4I (s ) − 2sI ( s ) + 3 = 0
8
2s
( ) + 2sI ( s) = 3
I ( s) =
Taking the inverse Laplace transform,
i(t) = 1. 5e−4 t
4
3
15
=
2s + 8 s + 4
I (s)
3
Fig. 11.53
for t > 0
Example 11.84 In the network shown in Fig. 11.54, the switch is closed at t = 0, the steady-state
being reached before t = 0. Determine current through inductor of 3 H.
11.11 Resistor–Inductor Circuit 11.47
2Ω
2H
2Ω
1V
i1 (t)
3H
i2 (t)
Fig. 11.54
Solution At t = 0−, the network is shown in Fig. 11.55. At t = 0−,
steady-state condition is reached. Hence, the inductor of 2 H acts
as a short circuit.
1
i1 (0 ) = A
2
i2 (0 ) 0
Since current through the inductor cannot change instantaneously,
1
i1 (0 + ) = A
2
i2 (0 + ) 0
For t > 0, the transformed network is shown in Fig. 11.56.
Applying KVL to Mesh 1,
1
1
s
− 2 s 1 ( ) 1 2[ 1 ( ) 2 ( )] 0
s
1
( 2 + 2s) I1 ( s) 2 2 ( s) = 1 +
s
Applying KVL to Mesh 2,
−2 [I2(s) − I1(s)] − 2I2(s) − 3s I2 (s) = 0
−2I1(s) + (4 + 3s) I2(s) = 0
By Cramer’s rule,
2 + 2s 1 +
2Ω
1V
i1 (0−)
Fig. 11.55
1
2s
2
2
I1 (s)
3s
I2 (s)
Fig. 11.56
1
s
2
1
( s 1)
( s + 1)
−2
0
s +1
s +1
s
3
I 2 ( s) =
=
=
=
=
2 + 2s
−2
( 2 2 s)(4
( 4 3s) − 4 s(3s 2 + 7s
7 s 2 ) 3s ⎛ s + 1 ⎞ ( s + 2 ) s ( s + 2 ) ⎛ s + 1 ⎞
⎟
⎟
−2
4 + 3s
⎝
⎝
3⎠
3⎠
By partial-fraction expansion,
I 2 ( s) =
A
B
A
B
C
+
+
1
s s+2
s+
3
1
( s +1)
3
s I 2 ( s) s = 0 =
1⎞
⎛
( s + 2) s + ⎟
⎝
3⎠
( s + ) I 2 ( s) S = −2
1
(s + )
= 3
1⎞
⎛
s s+ ⎟
⎝
3⎠
=
1
2
s=0
=−
s = −2
1
10
11.48 Network Analysis and Synthesis
C
1
( s +1)
1⎞
⎛
s + ⎟ I 2 ( s) s = − 1 = 3
⎝
3⎠
s ( s + 2)
3
=−
s= −
I 2 ( s) =
2
5
1
3
11 1 1
2 1
−
−
1
2 s 10 s + 2 5
s+
3
Taking inverse Laplace transform
1
i2 (t ) =
1 1 −2t 2 − 3 t
− e − e
for t > 0
2 10
5
Example 11.85 In the network of Fig. 11.57, the switch is closed at t = 0 with the network previously unenergised. Determine currents i1(t).
10 Ω
1H
1H
10 Ω
100 V
i1 (t)
10 Ω
i2 (t)
Fig. 11.57
Solution For t > 0, the transformed network is shown in Fig. 11.58.
10
s
s
100
s
10
I1 (s)
I2 (s)
10
Fig. 11.58
Applying KVL to Mesh 1,
100
−10I
10 I1 ( s) sI1 ( s) 10 [ I1 s
s
I2 s ] = 0
( s 20) I1 ( s) 100 I 2 ( s) =
Applying KVL to Mesh 2,
−10 [
−
100
0
s
10
0 I 2 ( s) = 0
] − s I 2 ( s) −10
−10 I1 ( ) + ( s + 20
0) I 2 ( s) = 0
By Cramer’s rule,
100
−10
100
s
( s 20)
0 s 20
100(
I1 ( s) =
= s
=
s + 20 −10
( s 20) 2 − 100 s(( 2
−10 s + 20
)
)
=
100(
s((
)
)(
)
11.12 Resistor–Capacitor Circuit 11.49
By partial-fraction expansion,
I1 ( s) =
A
A
B
C
+
+
s s + 10 s + 30
100( s 20)
2
20
s I1 ( s) |s 0 =
=
( s 10)(s
( s 30) s= 0 3
B
( s + 10) I1 ( s) |s
10 =
C
( s + 30) I1 ( ) |s=−30 =
100( s + 20)
= −5
s( s + 30) s=−10
100(
s((
)
)
=−
s =−30
5
3
20 1
5
5 1
I1 ( ) =
−
−
3 s s + 10 3 s + 30
Taking inverse Laplace transform,
20
5 30 t
i1 (t ) =
− 5e −10 t
e
3
3
Similarly,
100
s + 20
1000
s
−10
0
1000
1000
s
I 2 ( s) =
=
=
=
s + 20 −10
( s 20) 2 100 s( s 2 + 40 s + 300) s( s +10)( s + 30)
−10 s + 20
By partial-fraction expansion,
A
B
C
I 2 ( s) = +
+
s s + 10 s + 30
10
1000
A sI 2 ( s) |s 0 =
=
( s 10)(s
( s 30) s = 0 3
B=(
) I 2 ( ) |s=
s −10 =
1000
= −5
s((
) s = −10
(
) I 2 ( ) |s= −30 =
1
1000
5
=
s( s 10) s = −30 3
C
I 2 ( s) =
10 1
5
5 1
−
+
3 s s + 10 3 s + 30
Taking inverse Laplace transform,
i2 (t ) =
11.12
10
− 5e −10 t
3
5
e
3
30 t
fo t > 0
R
RESISTOR–CAPACITOR CIRCUIT
Consider a series RC circuit as shown in Fig. 11.59. The switch is
closed at time t = 0.
For t > 0, the transformed network is shown in Fig. 11.60.
Applying KVL to the mesh,
V
1
− RI ( s) −
I ( s) = 0
s
Cs
V
C
i (t)
Fig. 11.59
RC circuit
11.50 Network Analysis and Synthesis
1⎞
V
⎛
⎜⎝ R + ⎟⎠ I ( s) =
Cs
s
R
V
V
s
R
I ( s) =
=
=
1
RCs
C +1
1
R+
s+
Cs
Cs
RC
Taking the inverse Laplace transform,
V
s
V
s
Fig. 11.60
1
i( t ) =
Example 11.86
I (s)
V − RC t
e
R
1
Cs
Transformed network
fo t > 0
In the network of Fig. 11.61, the switch is moved from a to b at t = 0. Determine
i (t) and vc (t).
1Ω
10 V
a
6F
1Ω
b
+
3F
vc (t)
i (t)
−
Fig. 11.61
−
At t = 0 , the network is shown in Fig. 11.62. At t = 0–, the network has attained steady-state
condition. Hence, the capacitor of 6 F acts as an open circuit.
1Ω
v6 F (0−) = 10 V
i (0−) = 0
v3 F (0−) = 0
v6F (0 − )
Since voltage across the capacitor cannot change 10 V
instantaneously,
i (0 − )
v6 F (0+) = 10 V
v3 F (0+) = 0
Fig. 11.62
For t > 0, the transformed network is shown in 11.63.
Applying KVL to the mesh for t > 0,
1
10 1
1
− I ( s) − I ( s) − I ( s) = 0
s 6s
3s
1
1
1
10
I ( s) + I ( s) + I ( s) =
6s
1 V (s)
6s
3s
s
c
3s
10
60
10
I (s)
10
I ( s) =
=
=
1 1 ⎞ 6 s + 3 s + 0.5
s
⎛
s ⎜1 + + ⎟
⎝ 6 s 3s ⎠
Solution
Taking the inverse Laplace transform,
i(t) = 10e
for t > 0
Voltage across the 3 F capacitor is given by
1
10
Vc ( s) = I ( s) =
3s
3s(s
(
.5)
−0.5t
Fig. 11.63
11.12 Resistor–Capacitor Circuit 11.51
By partial-fraction expansion,
Vc ( s) =
A
B
+
s s + 0.5
10
A
sV
Vc ( s) s = 0 =
B
( s + .5)Vc ( s) s = −0.5 =
3(s
(
.5) s = 0
10
3s
=
20
3
=−
s = −0.5
20
3
20 1 20 1
Vc ( s) =
−
3 s 3 s + 0.5
Taking the inverse Laplace transform,
20 20 −0.5t
− e
3
3
20
=
( e 0 5t )
3
vc (t ) =
f
t>0
Example 11.87 The switch in the network shown in Fig. 11.64 is closed at t = 0. Determine the
voltage cross the capacitor.
10 Ω
10 Ω 2 F
10 V
vc (t)
Fig. 11.64
Solution At t = 0 , the capacitor is uncharged.
−
vc(0−) = 0
Since the voltage across the capacitor cannot change instantaneously,
vc(0+) = 0
For t > 0, the transformed network is shown in Fig. 11.65.
Applying KCL at the node for t > 0,
10
10
s
Vc ( s) −
V
(
s
)
V
(
s
)
s + c
+ c
=0
1
10
10
2s
1
2 Vc ( ) 0.2
2V
Vc ( s) =
s
1
05
Vc ( s) =
=
s( 2 0.2) s( s 0.1)
By partial-fraction expansion,
Vc ( s) =
A
A
B
+
s s + 0.1
sV
Vc ( s) s = 0 =
0.5
0.5
=
=5
s + 0.1 s = 0 0.1
10
10
Fig. 11.65
1
2s
Vc (s)
11.52 Network Analysis and Synthesis
B
Vc ( s) =
( s + .1)Vc ( s) s = −0.1 =
0.5
0.5
=−
= −5
s s = −0.1
0.1
5
5
−
s s + 0.1
Taking inverse Laplace transform,
vc ( t ) = 5 − 5e −0 1t
for t > 0
Example 11.88 In the network of Fig. 11.66, the switch is closed for a long time and at t = 0, the
switch is opened. Determine the current through the capacitor.
v (t)
i1 (t)
i2 (t)
0.5 F
2A
1Ω
1Ω
Fig. 11.66
At t = 0−, the network is shown in Fig. 11.67. At t = 0−, the switch is closed and steady-state
condition is reached. Hence, the capacitor acts as an open circuit.
vc (0−) = 0
Solution
v (0 − )
vc (0 − )
1Ω
2A
1Ω
Fig. 11.67
Since voltage across the capacitor cannot change instantaneously,
vc (0+) = 0
For t > 0, the transformed network is shown in Fig. 11.68.
Applying KVL to two parallel branches,
2
I1 ( s) + I1 ( s) = I 2 ( s)
s
Applying KCL at the node for t > 0,
2
= I1 ( s) + I 2 ( s)
s
2
2
I1 ( s) + I1 ( ss)) = − I1 ( s)
s
s
2
2
I1 ( s) + 2 I1 ( s) =
s
s
2
1
I1 ( s) = s =
2
s +1
+2
s
V (s)
I1(s)
2
s
2
s
I2(s)
1
1
Fig. 11.68
11.12 Resistor–Capacitor Circuit 11.53
Taking the inverse Laplace transform,
i1 ( t ) = e − t
for t > 0
In the network of Fig. 11.69, the switch is moved from a to b, at t = 0. Find v(t).
Example 11.89
4Ω
a
b
+
6V
v (t)
−
1F
2Ω
2Ω
Fig. 11.69
Solution At t = 0−, the network is shown in Fig 11.70. At t = 0−, steady-state condition is reached. Hence,
the capacitor acts as an open circuit.
v(
)
Since voltage
instantaneously,
6×
2
4 2
across
4Ω
=2V
the
capacitor
cannot
change
v (0+) = 2 V
For t > 0, the transformed network is shown in Fig. 11.71.
Applying KCL at the node for t > 0,
2
V ( s) −
V ( s)
s + V ( s) = 0
+
1
6
2
s
⎛2 ⎞
V ( s ) ⎜ + s⎟ = 2
⎝3 ⎠
V ( s) =
+
v (0−)
−
6V
2Ω
Fig. 11.70
4
V (s)
1
s
2
2
s
2
Fig. 11.71
2
s+
2
3
Taking the inverse Laplace transform,
v ( t ) = 2e
2
− t
3
fo t > 0
Example 11.90 The network shown in Fig. 11.72 has acquired steady-state at t < 0 with the switch
open. The switch is closed at t = 0. Determine v (t).
2Ω
+
4V
2Ω
1F
1F
v (t)
−
Fig. 11.72
11.54 Network Analysis and Synthesis
Solution At t = 0−, the network is shown in Fig 11.73. At
2Ω
t = 0–, steady-state condition is reached. Hence, the capacitor of
1 F acts as an open circuit.
v(
)
2
4×
2Ω
4V
=2V
2 2
Since voltage across the capacitor cannot change instantaneously,
v(0+) = 2 V
For t > 0, the transformed network is shown in Fig. 11.74.
Fig. 11.73
2
1
s
4
s
2
1 v (s)
s
2
s
Fig. 11.74
Applying KCL at the node for t > 0,
4
2
V ( s) −
V
(
s
)
s+
s + V ( s) = 0
+
1
1
2
2
s
s
2
2 sV ( s) V ( s) = + 2
s
2
+2
2s + 2
2
2
2
1
V ( s) = s
=
= −
= −
2 s + 1 s( s ) s 2 s + 1 s s + 0.5
V ( s) −
Taking the inverse Laplace transform,
v ( t ) = 2 − e −0 5t
11.13
fo t > 0
RESISTOR–INDUCTOR–CAPACITOR CIRCUIT
Consider a series RLC circuit shown in Fig. 11.75. The switch is closed at time t = 0.
R
L
V
i (t)
Fig. 11.75 RLC circuit
For t > 0, the transformed network is shown in Fig. 11.76.
C
v (0−)
11.13 Resistor–Inductor–Capacitor Circuit 11.55
Applying KVL to the mesh,
R
V
1
− RI ( s) Ls I ( s) −
I ( s) = 0
s
Cs
1⎞
V
⎛
⎜⎝ R + Ls + ⎟⎠ I ( s) =
Cs
s
Ls
V
s
I (s)
⎛ LCs
C 2 RCs
RCs + 1⎞
V
I ( s) =
⎜
⎟
Cs
s
⎝
⎠
Fig. 11.76 Transformed network
V
s
I ( s) =
=
LCs
C 2 + RCs + 1 s 2 +
Cs
V
V
L
L
=
R
1
( s − s )( s − s2 )
s+
L
LC
⎛ R⎞
⎛ 1 ⎞
where s1 and s2 are the roots of the equation s 2 + ⎜ ⎟ s + ⎜
= 0.
⎝ L⎠
⎝ LC ⎟⎠
2
s1 = −
R
1
⎛ R⎞
+ ⎜ ⎟ −
= −α + α 2 − ω 02 = −α + β
⎝ 2L ⎠
2L
LC
s2 = −
R
1
⎛ R⎞
− ⎜ ⎟ −
= −α − α 2 − ω 02 = −α − β
⎝ 2L ⎠
2L
LC
2
α=
where
ω0 =
and
β
By partial-fraction expansion, of I(s),
I ( s) =
A
B
I ( s) =
R
2L
1
LC
α 2 − ω 02
A
B
+
s s1 s s2
( s s ) I ( s) s
s1
( s − s ) I ( s) s
s2
=
=
V
L
s1
s2
V
L
s2
s1
=−
V
1 ⎤
⎡ 1
−
⎢
s ) ⎣ s − s1 s − s2 ⎥⎦
V
⎡e s t
s )⎣
L( s
V
L
s1
s2
Taking the inverse Laplace transform,
i( t ) =
1
Cs
L( s
e s2t ⎤⎦ = k1 e s t + k2 e s2t
where k1 and k2 are constants to be determined and s1 and s2 are the roots of the equation.
11.56 Network Analysis and Synthesis
Now depending upon the values of s1 and s2, we have 3 cases of the response.
Case I
When the roots are real and unequal, it gives an overdamped response.
R
1
>
2L
LC
> 0
In this case, the solution is given by
i t) = e
or
e
t
for t > 0
it
When the roots are real and equal, it gives a critically damped response.
R
1
2L
LC
= 0
In this case, the solution is given by
i(t) = e−at (k1 + k2 t)
for t > 0
Case III When the roots are complex conjugate, it gives an underdamped response.
R
1
<
2L
LC
< 0
In this case, the solution is given by
it
Case II
2
0
s1 2 = −
where
2
Let
2
where
2
=
= j
d
1
j
2
and
Hence
it
2
−
2
dt
e
e
⎡
2
−
Example 11.91
+
−
−
2j
−
for t > 0
The switch in Fig. 11.77 is opened at time t = 0. Determine the voltage v(t) for t > 0.
+
2A
0.5 Ω
Fig. 11.77
0.5 H
0.5 F
v()
11.13 Resistor–Inductor–Capacitor Circuit 11.57
At t = 0−, the network is shown in Fig. 11.78. At t = 0−, the network has attained steady-state
condition. Hence, the inductor acts as a short circuit and the capacitor acts as an open circuit.
Solution
+
iL (0 − )
2A
v (0 − )
0.5 Ω
−
Fig. 11.78
iL(0−) = 0
v(0−) = 0
Since current through the inductor and voltage across the capacitor cannot change instantaneously,
iL(0+) = 0
v(0+) = 0
For t > 0, the transformed network is shown in Fig. 11.79.
Applying KCL at the node for t > 0,
V ( s) V ( s) V ( s)
+
+
=
1
0.5 0.5s
0.5s
2
2V ( s) + V ( s) + 0.. sV ( s) =
s
2
s
V ( s) =
=
2
+ 0.5s + 2
s
+
2
s
2
s
0.5
0.5s
2
s
1
0.5s
V(s)
−
4
2
s + 4s + 4
=
Fig. 11.79
4
(s
)
2
Taking inverse Laplace transform,
v ( t ) = 4tt e −2t
for t > 0
Example 11.92 In the network of Fig. 11.80, the switch is closed and steady-state is attained. At
t = 0, switch is opened. Determine the current through the inductor.
2.5
. Ω
5V
200 μF
0.5 H
Fig. 11.80
Solution At t = 0−, the network is shown in Fig. 11.81. At t = 0–, the switch is closed and steady-state
condition is attained. Hence, the inductor acts as a short circuit and the capacitor acts as an open circuit.
Current through inductor is same as the current through the resistor.
11.58 Network Analysis and Synthesis
iL (
)=
2.5 Ω
5
=2A
2.5
Voltage across the capacitor is zero as it is connected in parallel
with a short.
vc (0−) = 0
vc (0−)
5V
iL (0−)
Fig. 11.81
Since voltage across the capacitor and current through the
inductor cannot change instantaneously,
iL (0+) = 2 A
vc (0+) = 0
−
1
200 × 10 −6 s
0.5s
1
200 × 10−6s
For t > 0, the transformed network is shown in Fig. 11.82.
Applying KVL to the mesh for t > 0,
I (s)
1
I ( s) − 0.5s I ( s) + 1 = 0
Fig. 11.82
I ( s)
0.5s I ( s) − 1 + 5000
=0
s
I ( s) =
1
5000
0.5
5s +
s
=
2s
2
s + 10000
Taking inverse Laplace transform,
i ( t ) = 2 cos 100t
fo t > 0
Example 11.93 In the network shown in Fig. 11.83, the switch is opened at t = 0. Steady-state condition is achieved before t = 0. Find i(t).
0.5 H
1V
1Ω
1F
i (t)
Fig. 11.83
Solution At t = 0−, the network is shown in Fig 11.84. At
t = 0−, the switch is closed and steady-state condition is achieved.
Hence, the capacitor acts as an open circuit and the inductor acts as
a short circuit.
vc (0−) = 1 V
i (0−) = 1 A
Since current through the inductor and voltage across the
capacitor cannot change instantaneously,
1Ω
1V
i (0−)
Fig. 11.84
11.13 Resistor–Inductor–Capacitor Circuit 11.59
vc(0+) = 1 V
i(0+) = 1 A
For t > 0, the transformed network is shown in Fig. 11.85.
Applying KVL to the mesh for t > 0,
0.5s
0.5
1
s
1
1
I(s)
s
1 1
− I ( s) − 0.5s I ( s) 0.5 I ( s) = 0
s s
Fig. 11.85
1 1
0.5 + = I ( s) + 0.5 s I ( s) I ( s)
s s
1
⎡ 1
⎤
I ( s) ⎢1 + + 0.5s ⎥ = 0.5 +
s
⎣ s
⎦
s+2
(s ) 1
s +1
1
I ( s) = 2
=
=
+
2
2
s + 2s
2 s 2 ( s ) 1 ( s +1) + 1 ( s + 1) 2 + 1
Taking the inverse Laplace transform,
i ( t ) = e −tt cos t + e t si t
for t > 0
Example 11.94 In the network shown in Fig. 11.86, the switch is closed at t = 0. Find the currents
i1(t) and i2(t) when initial current through the inductor is zero and initial voltage on the capacitor is 4 V.
1Ω
1Ω
10 V
i1 (t)
1Ω
+
4V
−
i2 (t)
1H
1F
Fig. 11.86
Solution For t > 0, the transformed network is shown in Fig. 11.87.
1
10
s
1
I1 (s)
s
1
1
s
4
s
I2 (s)
Fig. 11.87
Applying KVL to Mesh 1,
10
− I1 ( s) (1
s
)[
1
( s 2) I1 ( ) (
2
]
1) I 2 ( s) =
0
10
s
11.60 Network Analysis and Synthesis
Applying KVL to Mesh 2,
−( + ) [
1
4
) − I2( ) − = 0
s
s
1
4
⎛
⎞
−( + ) I1 ( ) + s + 2 + ⎟ I 2 ( s) = −
⎝
s⎠
s
−
] − I2(
By Cramer’s rule,
10
( s 1)
−(s
s
2
⎛ 10 ⎞ ⎛ s + 2 s 1⎞
⎛ 4⎞
10
4
4
1
⎜⎝ ⎟⎠ ⎜
(ss + 1) 2 ( s + 1)
−
s+ 2+
⎟ − ( s + 1) ⎜⎝ s ⎟⎠
2
s
s
⎝
⎠
s
s
s
s
I1 ( s) =
=
=
2
s+2
( s 1)
⎛ s 2 + 2s
⎞
( s +1)
2s 1
2
( s + 2)
− ( s + 1) 2
( s + 2) ⎜
⎟ − ( s + 1)
1
s
s
⎝
⎠
−(s
( s 1) s + 2 +
s
10
4
( s +1) −
2
3s + 5
s
s
=
=
( s +1)
s((
)
( s + 2)
−(
)
s
By partial-fraction expansion,
A
B
I1 ( s) = +
s s +1
3s 5
A sI1 ( s) |s 0 =
=5
s + 1 s=0
B
I1 ( s) =
Taking inverse Laplace transform,
Similarly,
( s 1) I1 ( s) |s
1=
3s + 5
= −2
s s = −1
5
2
−
s s +1
i1 ( t ) = 5 2e − t
fo t > 0
10
s
4
−(s
( s 1) −
3s 1 3s 3 − 2 3( + 1) − 2
3
2
s
I 2 ( s) =
=
=
=
=
−
2
2
2
s+2
( s 1)
s + 1 ( +1) 2
( s 1)
( +1)
( +1)
1
−(s
( s 1) s + 2 +
s
Taking inverse Laplace transform,
i2 (t ) = 3e
3e t 2te t
fo t > 0
s+2
11.14
RESPONSE OF RL CIRCUIT TO VARIOUS FUNCTIONS
Consider a series RL circuit shown in Fig. 11.88. When the switch is closed at t = 0, i(
)
i(
i(
+
) = 0.
11.14 Response of RL Circuit to Various Functions 11.61
R
v (t)
+
−
L
i(t)
Fig. 11.88 RL circuit
R
For t > 0, the transformed network is shown in Fig. 11.89.
Applying KVL to the mesh,
V ( s) − R I (s
( s) Ls I ( s) = 0
V ( s)
1 V ( s)
I ( s) =
=
R
R Ls L
s+
L
(a) When the unit step signal is applied,
v ( t ) = u( t )
Taking Laplace transform,
1
V ( s) =
s
1
1 s
I ( s) =
R
L
s+
L
1
1
=
R⎞
L ⎛
s s+ ⎟
⎝
L⎠
By partial-fraction expansion,
⎛
⎞
1⎜A
B ⎟
I ( s) = ⎜ +
R⎟
L s
s+ ⎟
⎜
⎝
L⎠
=
1
V (s) +
−
Fig. 11.89 Transformed network
=
L
R
A
s I ( s)
B
R⎞
1
L
⎛
s + ⎟ I ( s) s = − R =
=−
⎝
s s= − R
L⎠
R
L
s=0
R
s+
L
s=0
L
⎛
⎞
1⎜L1 L 1 ⎟
I ( s) = ⎜
−
R⎟
L Rs R
s+ ⎟
⎜
⎝
L⎠
⎛
⎞
1 ⎜1
1 ⎟
= ⎜ −
R⎟
R s
s+ ⎟
⎜
⎝
L⎠
Ls
I (s)
11.62 Network Analysis and Synthesis
Taking inverse Laplace transform,
i( t ) =
(b)
1
[
R
e
⎛ R⎞
−⎜ ⎟ t
⎝ L⎠
When unit ramp signal is applied,
v(t ) = r (t ) = t
Taking Laplace transform,
V ( s) =
]
f
t>0
for t > 0
1
s2
1
I ( s) =
L
1
R⎞
⎛
s2 s + ⎟
⎝
L⎠
By partial-faction expansion,
1
L
1
R⎞
⎛
s2 s + ⎟
⎝
L⎠
=
A B
C
+ 2+
R
s s
s+
L
1
R⎞
R⎞
⎛
⎛
= As s + ⎟ + B s + ⎟ + Cs 2
⎝
⎝
L
L⎠
L⎠
Putting s = 0,
Putting s = −
R
,
L
Comparing coefficients of s2,
B=
1
R
C=
L
R2
A C=0
A
C=−
L
R2
L 1 1 1
L 1
I ( s) = − 2 +
+ 2
2
R s Rs
R s+ R
L
Taking inverse Laplace transform,
i( t ) = −
L
R2
⎛ R⎞
+
1
L −⎜ ⎟ t
t + 2 e ⎝ L⎠
R
R
1
L
= t− 2[
R
R
(c) When unit impulse signal is applied,
v(t ) = δ (t )
Taking Laplace transform,
V ( s) = 1
1 1
I ( s) =
R
L
s+
L
e
⎛ R⎞
−⎜ ⎟ t
⎝ L⎠
]
f
t>0
11.14 Response of RL Circuit to Various Functions 11.63
Taking inverse Laplace transform,
⎛ R⎞
1 −⎜ ⎟ t
i( t ) = e ⎝ L ⎠
L
fo t > 0
Example 11.95 At t = 0, unit pulse voltage of unit width is applied to a series RL circuit as shown
in Fig. 11.90. Obtain an expression for i(t).
1Ω
v (t )
+
−
v (t)
1
1H
i (t)
0
t
1
Fig. 11.90
Solution
v(t ) = u ( t ) − u ( t − 1)
1 e −ss 1 − e s
−
=
s
s
s
For t > 0, the transformed network is shown in Fig. 11.91.
Applying KVL to the mesh,
V ( s) − I ( s) − sI ( s) = 0
V ( s)
I ( s) =
s +1
1 − e−s
=
s( s )
V ( s) =
1
V (s)
+
−
s
I (s)
Fig. 11.91
−s
=
1
e
−
s( s ) s( s + 1)
=
1
1
e−s e−s
−
−
+
s s +1 s
s +1
Taking inverse Laplace transform,
i(t ) = u(t ) − e −tt u(t ) u(t
u( t
=(
t
e )u(t
(t ) [
) e
e
( t −1)
(t
)
u(t
(t
)
] (t − 1)
f
>0
Example 11.96 For the network shown in Fig. 11.92, determine the current i(t) when the switch is
closed at t = 0. Assume that initial current in the inductor is zero.
5Ω
r (t − 3) +
−
2H
i(t)
11.64 Network Analysis and Synthesis
Fig. 11.92
5
Solution For t > 0, the transformed network is shown in Fig. 11.93.
Applying KVL to the mesh for t > 0,
e
−3 s
s2
− 5 I ( s) − 2 s I ( s) = 0
5 I ( s) + 2 s I ( s) =
I ( s) =
e−3s
s2
+
−
e −3s
2s
I (s)
Fig. 11.93
s2
e −3s
=
s 2 ( 2 s + 5)
0 5 e −3s
s2 (
2.5)
By partial-fraction expansion,
05
2
s (
2.5)
=
A B
C
+ 2+
s s
s+2 5
0.5
((ss 2.5
5)) B(
= As 2 + 2.5 A
=(
2
)s2
2.5) Cs 2
2.5 B C
Cs 2
( .5
.5 A + B))
.5 B
0
Comparing coefficients of s , s and s ,
A C=0
2 5A + B = 0
2.5 B = 0.5
Solving these equations,
A = −0 08
B=02
C = 0 08
⎛ 0.08 0.2 0.08 ⎞
I ( s) = e −3s −
+ 2 +
⎟
⎝
s
s + 2 5⎠
s
= −0.08
e −33
e 3s
e −3s
+ 0.2 2 + 0.08
s
s + 2.5
s
Taking inverse Laplace transform,
i(t ) = −0.. u(t − 3)
.2r (t
) + 0. e
2 ( t 3)
u(t − 3)
Example 11.97 Determine the expression for vL (t) in the network shown in Fig. 11.94. Find vL(t)
when (i) vs(t) = d (t), and (ii) vs(t) = e−t u(t).
5Ω
+
Vs (t )
+
−
1
v (t )
2H L
−
Fig. 11.94
11.14 Response of RL Circuit to Various Functions 11.65
Solution For t > 0, the transformed network is shown in Fig. 11.95.
By voltage-division rule,
VL ( s) = Vs ( s) ×
(a)
5
s
2
s
+5
2
=
s
Vs ( s)
s + 10
+
Vs (s )
For impulse input,
Vs ( s) = 1
+
−
s V (s )
2 L
−
s
s + 10 − 10
10
=
= 1−
s + 10
s + 10
s + 10
Taking inverse Laplace transform,
VL ( s) =
VL (t ) = δ (t
(t ) 10e −10 t u(t )
(b)
Fig. 11.95
for t > 0
For vs (t ) = e − t u(t ),
Vs ( s) =
VL ( s) =
1
s +1
(s
s
))(s
(s
)
By partial-fraction expansion,
VL ( s) =
A
B
+
s + 10 s + 1
A
(s
B
(s
)VL ( s) s = −10 =
)VL ( s) s = −1 =
s
10
=
s + 1 s = −10 9
s
s + 10
=−
s = −1
1
9
10 1
1 1
VL ( s) =
−
9 s + 10 9 s + 1
Taking inverse Laplace transform,
10 −10 t
1 t
e u(t
u( t )
e u( t )
9
9
⎛ 10 −10 t 1 t ⎞
=
e
e ⎟ u( t )
⎝ 9
9 ⎠
vL (t ) =
for t > 0
Example 11.98 For the network shown in Fig. 11.96, determine the current i (t) when the switch is
closed at t = 0. Assume that initial current in the inductor is zero.
2Ω
2d (t − 3)
+
−
1H
i (t)
Fig. 11.96
11.66 Network Analysis and Synthesis
Solution For t > 0, the transformed network is shown in Fig. 11.97.
2
Applying KVL to the mesh for t > 0,
2e
3s
2 I ( s) sI ( s) = 0
2 ( s) s ( s)
I ( s) =
2e
2e−3s
−3 s
Fig. 11.97
Taking inverse Laplace transform,
Example 11.99
s
I (s)
2e −3s
s+2
i(t ) = 2ee −2(t(t(
+
−
)
u(t
(
fo t > 0
)
Determine the current i(t) in the network shown in Fig. 11.98, when the switch is
closed at t = 0.
10 Ω
50 sin 25 t
5H
i (t)
Fig. 11.98
10
Solution For t > 0, the transformed network is shown in
Fig. 11.99.
Applying KVL to the mesh for t > 0,
1250
− 10I
10 I ( s) − 5 ( ) 0
2
s + 625
I ( s) =
1250
s 2 + 625
5s
I (s)
250
Fig. 11.99
2
( s + 625)( s + 2)
By partial-fraction expansion,
I ( s) =
As + B
2
s + 625
+
250 = ( As B)( s
= ( A C )s
Comparing coefficients,
2
C
s+2
) + C ( s2
( A + B) s ((2 B
)
C)
A C=0
2A + B = 0
2B
C = 250
Solving the equations,
A = −0.397
B = 0.795
C = 0.397
−0.397 s + 0.795 0.397
0.397 s
0.795
0.397
I ( s) =
+
=− 2
+
+
s+2
s 2 + 625
s + 625 s 2 + 625 s + 2
11.14 Response of RL Circuit to Various Functions 11.67
Taking the inverse Laplace transform,
i(t ) = −0.397 cos 25t 0.032 si 25t + 0.397e −2t
Example 11.100
fo t > 0
Find impulse response of the current i(t) in the network shown in Fig. 11.100.
1Ω
i1 (t)
i (t)
d (t)
+
−
1Ω
2H
Fig. 11.100
Solution The transformed network is shown in Fig. 11.101.
1(( s ) 2 s + 1
=
2s + 1 + 1 2s + 2
V ( s)
1
2s 2
I1 ( s) =
=
=
Z ( s) 2 s 1 2 s 1
2s 2
By current-division rule,
1
1
2s + 2
1
1 1
I ( s) = I1 ( s) ×
=
×
=
=
2 s + 2 2 s + 2 2 s + 1 2 s + 1 2 s + 0.5
Taking inverse Laplace transform,
1
i(t ) = e −0.5t u(t
(t )
fo t > 0
2
1
I1 (s)
I (s)
Z ( s) =
1
+
−
1
2s
Fig. 11.101
Example 11.101
The network shown in Fig. 11.102 is at rest for t < 0. If the voltage
v(t ) = u(t ) cos t Aδ (t ) is applied to the network, determine the value of A so that there is no transient term
in the current response i(t).
1Ω
v (t)
2H
i (t)
Fig. 11.102
v(t ) = u(t ) cos t Aδ (t )
s
V ( s) = 2
+A
s +1
11.68 Network Analysis and Synthesis
Solution For t > 0, the transformed network is shown in Fig. 11.103.
1
Applying KVL to the mesh for t > 0,
V ( s) = 2 sI (s
( s) I ( s) =
I ( s) =
s
s2 + 1
+A
V (s)
2s
I (s)
s A
A(s
( s2 )
K
K s + K3
= 1 + 22
1
1⎞
⎛
s +1
2 ⎜ s + ( s2 ) s +
⎝
2
2⎠
Fig. 11.103
The transient part of the response is given by the first term. Hence, for the transient term to vanish, K1 = 0.
K1
When
11.15
−1
⎛ 5⎞
+ A⎜ ⎟
⎝ 4⎠
1
2
⎛
⎞
s + ⎟ I ( s) s = − 1 =
⎝
2⎠
⎛ 5⎞
2
2⎜ ⎟
⎝ 4⎠
K1 = 0
5
1
A=
4
2
2
A= =04
5
RESPONSE OF RC CIRCUIT TO VARIOUS FUNCTIONS
Consider a series RC circuit as shown in Fig. 11.104.
R
v (t)
+
−
C
i (t)
Fig. 11.104 RC circuit
For t > 0, the transformed network is shown in Fig. 11.105.
Applying KVL to the mesh,
V ( s) − RI ( s) −
(a)
1
I ( s) = 0
Cs
V ( s)
sV ( s)
I ( s) =
=
1
1 ⎞
+ R R⎛s +
⎟
⎝
Cs
RC ⎠
When unit step signal is applied,
v ( t ) = u( t )
R
1
Cs
V (s) +
−
I (s)
Fig. 11.105 Transformed network
11.15 Response of RC Circuit to Various Functions 11.69
Taking Laplace transform,
1
s
V ( s) =
1
1
s
I ( s) =
=
1 ⎞
1 ⎞
⎛
⎛
R s+
⎟ R⎝s +
⎟
⎝
RC ⎠
RC ⎠
s×
Taking inverse Laplace transform,
1
1 − RC t
e
R
i( t ) =
(b)
fo t > 0
When unit ramp signal is applied,
v(t ) = r (t ) = t
Taking Laplace transform,
V ( s) =
1
s2
1
1
2
s
R
I ( s) =
=
1 ⎞
1 ⎞
⎛
⎛
R s+
⎟ s⎝s +
⎟
⎝
RC ⎠
RC ⎠
s×
By partial-fraction expansion,
I ( s) =
A
B
A
+
s
B
s+
s I ( s)
1
RC
s=0
1
R
=
s+
1
RC
=C
s=0
1
1 ⎞
⎛
R
s+
⎟ I ( s) s = − 1 =
⎝
RC ⎠
s
R
RC
= −C
s= −
I ( s) =
C
−
s
1
RC
C
s+
1
RC
Taking inverse Laplace transform,
i(t ) = C Ce
(c)
When unit inpulse signal is applied,
v(t) = d (t)
−
1
t
RC
fo t > 0
11.70 Network Analysis and Synthesis
Taking Laplace transform,
V ( s) = 1
1
1
1 ⎞
⎛
−
1
⎜
⎟
RC RC =
I ( s) =
=
1 − RC ⎟
⎜
1
1 ⎞
1 ⎞
R
⎛
⎛
⎜ s+
⎟
R s+
R s+
⎟
⎟
⎝
⎝
⎝
RC ⎠
RC ⎠
RC ⎠
s+
s
Taking inverse Laplace transform,
1⎡
1 − RC t ⎤
⎢δ (t ) −
⎥
e
R⎢
RC
⎥⎦
⎣
1
i( t ) =
fo t > 0
Example 11.102 A rectangular voltage pulse of unit height and T-seconds duration is applied to
a series RC network at t = 0. Obtain the expression for the current i(t). Assume the capacitor to be initially
uncharged.
v (t )
R
1
0
v (t)
T
+
−
C
i (t)
t
(a)
(b)
Fig. 11.106
v(t) = u(t) − u(t − T)
sT
1 e −sT
1 − e − sT
V ( s) = −
=
s
s
s
Solution
R
V (s) +
−
I (s)
For t > 0, the transformed network is shown in Fig. 11.107.
Applying KVL to the mesh for t > 0,
1
Fig. 11.107
V ( s) − RI ( s) −
I ( s) = 0
Cs
1
⎡
⎤
s
V ( s)
1 − e − sT
1⎢ 1
e − sT ⎥
R
I ( s) =
=
V ( s) =
= ⎢
−
⎥
1
1
1
1 ⎥
1 ⎞ R⎢
⎛
R+
s+
s
+
s
+
R s+
⎟
⎝
Cs
RC
RC
RC ⎦
⎣
R ⎠
RC
1
Cs
Taking inverse Laplace transform,
⎛ 1 ⎞
⎡ ⎛ 1⎞
−⎜
⎟ (t
1 ⎢ − ⎝⎜ RC ⎠⎟ t
i( t ) =
e
u(t ) e ⎝ RC ⎠
u(t
R⎢
⎣
T)
⎤
u( t T ) ⎥
⎥
⎦
fo t > 0
Example 11.103 For the network shown in Fig. 11.108, determine the current i(t) when the switch
is closed at t = 0 with zero initial conditions.
11.15 Response of RC Circuit to Various Functions 11.71
3Ω
+
−
2r (t − 2)
1F
i (t)
Fig. 11.108
Solution For t > 0, the transformed network is shown in Fig. 11.109.
Applying KVL to the mesh for t > 0,
2e −2 s
s2
3
2e−2s +
−
s2
1
− 3 I ( s) − I ( s) = 0
s
1⎞
2e −2 s
⎛
3
+
I
(
s
)
=
⎜⎝
⎟
s⎠
s2
I ( s) =
I(t)
1
s
Fig. 11.109
2e −2 s
0 67e −2 s
=
1 ⎞ s( s + 0.33)
⎛
s2 ⎜ 3 + ⎟
⎝
s⎠
By partial-fraction expansion,
0 67
A
= +
s(( 0.33)
B
0.33
A=
0 67
=2
s + 0 33 s = 0
B=
0 67
= −2
s s = −0 33
I ( s) = e −2 s
2 ⎞
e −2 s
e 2s
⎛2
−
=
2
−
2
⎝ s s + 0.33 ⎟⎠
s
s + 0.33
Taking inverse Laplace transform,
) − 2ee −0.
i(t ) = 2u(t
u( t
((tt − 2 )
u( t
u(t
fo t > 0
)
Example 11.104 For the network shown in Fig. 11.110, determine the current i(t) when the switch
is closed at t = 0 with zero initial conditions.
5Ω
d (t)
+
−
2F
i (t)
Fig. 11.110
Solution For t > 0, the transformed network is shown in Fig. 11.111.
11.72 Network Analysis and Synthesis
Applying KVL to the mesh for t > 0,
1
I ( s) = 0
2s
1⎞
⎛
⎜⎝ 5 + ⎟⎠ I ( s) = 1
2s
1 5 I ( s) −
5
1
I ( s) =
1
2s
2s
=
10 s + 1
0 2s
=
s+0 1
0.2( s + 0.1 − 0.1)
=
s+0 1
01 ⎞
⎛
= 0 2 1−
⎝ s + 0 1⎟⎠
0 02
= 0 2−
s + 0.1
5+
1
+
−
I(s)
1
2s
Fig. 11.111
Taking inverse Laplace transform,
.02 e −0.1t u(t )
i(t ) = 0.
0. δ (t )
Example 11.105
For the network shown in Fig. 11.112, find the response v0 (t).
2Ω
+
vs (t) =
1
cost u(t )
2
+
−
1
F
4
vo (t)
−
Fig. 11.112
Solution For t > 0, the transformed network is shown in Fig. 11.113.
1 s
2 s2 + 1
By voltage-division rule,
4
2V ( s)
Vo ( s) = Vs ( s) × s = s
= 2
4
s+2
(s
2+
s
By partial-fraction expansion,
2
Vs ( s) =
Vo ( s) =
s
s
+
Vs (s)
+
−
4
s Vo (s)
s
))(
)(s
(s
As + B
2
−
)
+
C
s+2
s +1
( As B)( s
2
Fig. 11.113
) + c(s
( s2
)
( A C ) s + ( 2 A B) s + (
)
11.15 Response of RC Circuit to Various Functions 11.73
Comparing coefficient of s2, s and s0,
A C=0
2A + B = 1
2B C = 0
Solving the equations,
A=04
B=02
C = −0 4
0.4 s + 0.2 0.4
0.4 s
02
04
Vo ( s) =
−
= 2
+ 2
−
2
s
+
2
s
+2
s +1
s +1 s +1
Taking the inverse Laplace transform,
i(t ) = 0.4 cos t 0.2 si t − 0.4e
2t
fo t > 0
Example 11.106 Find the impulse response of the voltage across the capacitor in the network
shown in Fig. 11.114. Also determine response vc (t) for step input.
2Ω
v (t)
1H
+
+
−
−
1 F vc (t)
Fig. 11.114
Solution For t > 0, the transformed network is shown in Fig. 11.115.
2
By voltage-division rule,
Vc ( s) = V ( s) ×
1
s
1
2+ s+
s
V ( s)
V ( s)
= 2
=
s + 2s + 1 ( s )2
(a)
For impulse input,
V (s)
1
)2
(s
Taking inverse Laplace transform,
vc (t ) = te − t u(t )
(b)
for t > 0
For step input,
V ( s) =
Vc ( s) =
+ 1
Vc (s )
s
−
+
−
Fig. 11.115
V ( s) = 1
Vc ( s) =
s
1
s
1
s( s
)2
11.74 Network Analysis and Synthesis
By partial-fraction expansion,
Vc ( s) =
A
B
C
+
+
s s + 1 ( s )2
) 2 + Bs(s
Bs( s
1 = A( s
) Cs
C
s
) B ( s 2 + s ) Cs
= ( A B) s2
( A B C )s A
= A( s
2
Comparing coefficient of s2, s1 and s0,
A=1
A+ B = 0
B
A = −1
2A + B + C = 0
A − B = −2 + 1 = −1
1
1
1
Vc ( s) = −
−
s s + 1 ( s )2
C
Taking inverse Laplace transform,
vc (t ) = u(t ) − e −tt u(t ) te t u(t )
=(
e
t
t t )utt
te
f
t>0
Example 11.107 For the network shown in Fig. 11.116, determine the current i(t) when the switch
is closed at t = 0 with zero initial conditions.
5Ω
5r (t − 1)
1H
+
−
1
F
6
i (t)
Fig. 11.116
Solution For t > 0, the transformed network is shown in Fig. 11.117.
Applying KVL to the mesh for t > 0,
5e − s
s2
5
6
− 5I
5 I ( s) − sI ( s) − I ( s) = 0
s
5 I ( s) + sI ( s) +
6
5e − s
I ( s) = 2
s
s
I ( s) =
5e−s
s2
5e
2
−ss
s( s + s + )
s
+
−
I (s)
s
=
5e
s( s + )(s
)( s + )
Fig. 11.117
6
s
11.15 Response of RC Circuit to Various Functions 11.75
By partial-fraction expansion,
1
s( s 3)(
A
B
C
+
+
2) s s + 3
2
1
1
A=
=
( s 3)( 2) s = 0 6
=
B=
1
1
=
s( s 2) s = −3 3
C=
1
1
=−
s( s + 3) s = −2
2
⎡1
1
1 ⎤ 5 e−s 5 e−s 5 e−s
I ( s) = 5e
5e − s ⎢ +
−
+
−
⎥=
3 s+3 2 s+2
⎣ 6 s 3( 3) 2( s 2) ⎦ 6 s
Taking inverse Laplace transform,
i( t ) =
5
u( t
6
5
) + e −3(t(t(
3
)
u( t
u(t
)
5
e
2
2(t
(t
)
u( t
)
f
t>0
Example 11.108 For the network shown in Fig. 11.118, the switch is closed at t = 0. Determine
the current i(t) assuming zero initial conditions.
2Ω
1H
sin t
0.5 F
i(t)
Fig. 11.118
Solution For t > 0, the transformed network is shown in Fig. 11.119.
Applying KVL to the mesh for t > 0,
1
2
s +1
− 2I
2 I ( s) − s I ( s) −
2
I ( s) = 0
s
1
s2 + 1
2⎞
1
⎛
⎜⎝ 2 + s + ⎟⎠ I ( s) = 2
s
s +1
I ( s) =
s
2
( s + 1)( s
s
2
I (s)
Fig. 11.119
2
2 s + 2)
2s
By partial-fraction expansion,
I ( s) =
s
As + B
2
s +1
+
Cs + D
2
s + 2s + 2
( As B)( s 2
= As3 + 2 As 2
s
) + (C
((Cs
Cs D )( s 2
C
)
2 As + Bs 2 + 2 Bs + 2 B + Cs3 + Cs + Ds 2 + D
= ( A + C ) s3 + ( 2 A
)s2
(2 A + 2 B + C )s + (2 B + D)
2
s
11.76 Network Analysis and Synthesis
Comparing coefficients of s3, s2, s1 and s0,
A C=0
2A + B + D = 0
2A 2B + C = 1
2B + D = 0
Solving these equations,
A = 0.2, B = 0.4, C = −0.2, D = −0.8
I ( s) =
0.2 s + 0.4
−
0.2 s + 0.8
s +1
+1
s2 + 2s + 2
0.2 s
0.4
0.2 s + 0.2 0.6
= 2
+
−
s + 1 s 2 + 1 ( +11) 2 (1) 2
=
2
0.2 s
2
s +1
+1
+
0.4
2
0.2( s + 1)
−
2
−
06
s +1
+ 1 ( + 1) + 1 ( + 1) 2 + 1
Taking inverse Laplace transform,
i(t ) = 0.
0.2 cos t 0.4 si t − 0.2 e −tt cos t − 0.6 e
t
si t
−t
= 0.2 cos t 0.4 si t − e ( .2 cos t + 0.6 sin t )
fo t > 0
Example 11.109
For the network shown in Fig. 11.120, the switch is closed at t = 0. Determine
the current i(t) assuming zero initial conditions in the network elements.
5Ω
6e−2t
1H
+
−
0.25 F
i(t)
Fig. 11.120
Solution For t > 0, the transformed network is shown in Fig. 11.121.
Applying KVL to the mesh for t > 0,
6
4
− 5I
5 I ( s) − s I ( s) − I ( s) = 0
s+2
s
4⎞
6
⎛
⎜⎝ 5 + s + ⎟⎠ I ( s) =
s
s+2
I ( s) =
=
s
5
6
s+2
+
−
I (s)
6s
( s + 2)(
2
)
6s
(
)(
)(
)(
)
Fig. 11.121
4
s
11.15 Response of RC Circuit to Various Functions 11.77
By partial-fraction expansion,
I ( s) =
A
C
B
+
+
s + 2 s +1 s + 4
6s
=6
( s )(
)( s ) s = −2
)(s
6s
B ( s + 1) ( s) |s = −1 =
= −2
( + 2)( + 4) s = −1
6s
C = ( + 4) I ( ) |s = −4 =
= −4
( + 2)( s + 1) s = −4
6
2
4
I ( s) =
−
−
s + 2 s +1 s + 4
Taking inverse Laplace transform,
A
(s
) I ( s) |s = −2 =
i(t ) = 6e −2t u(t
uu((t
(t ) 2e t u(t ) − 4 e −4 t u(t
(t )
f
t>0
Example 11.110 The network shown has zero initial conditions. A voltage vi(t) = d (t) applied to two
terminal network produces voltage vo(t) = [e−2 t + e−3 t] u(t). What should be vi(t) to give vo(t) = t e−2 t u(t)?
+
vi (t )
−
Network
+
vo (t )
−
Fig. 11.122
Solution For vi(t) = d (t),
Vi ( s) = 1
vo (t ) = [e
[e
2t
3t
]u(t )
1
1
Vo ( s) =
+
s+2 s+3
System function H ( s) =
e
Vo ( s)
Vi ( s)
=
1
1
2 +5
+
=
s + 2 s + 3 ( s + 2)( + 3)
For vo (t ) = te −2t u(t ),
Vo ( s) =
1
(s
)2
From Eq. (i),
Vi ( s) =
By partial-fraction expansion,
Vo ( s)
1
( s )(s
)(
)
(s )
=
×
=
2
H ( s) ( s )
2s + 5
2(s
( s .5)( s + 2)
A
B
+
s + 2 s + 2.5
A=1
B = −0.5
1
0.5
Vi ( s) =
−
s + 2 s + 2.5
Vi ( s) =
…(i)
11.78 Network Analysis and Synthesis
Taking inverse Laplace transform,
vi (t ) = e −22tt
0.5e
2.5t
for t > 0
Example 11.111 A unit impulse applied to two terminal black box produces a voltage
vo(t) = 2e−t −e−3t. Determine the terminal voltage when a current pulse of 1 A height and a duration of 2 seconds
is applied at the terminal.
+
is (t)
vo (t )
Black box
−
Fig. 11.123
Solution
vo (t ) = 2e −tt
e
3t
is (t )
2
1
Vo ( s) =
−
s +1 s + 3
When is (t ) = δ (t ),
1
I s ( s) = 1
0
Vo ( s) = Z ( s) I s ( s)
Z ( s) =
…(i)
t
2
Fig. 11.124
Vo ( s)
2
1
=
−
I s ( s) s + 1 s + 3
When is (t) is a pulse of 1 A height and a duration of 2 seconds then,
is (t ) = u(t ) − u(t
I s ( s) =
)
−2 s
1 e
−
s
s
From Eq. (i),
1 ⎤ ⎡ 1 e −2 s ⎤
⎡ 2
Vo ( s) = ⎢
−
⎥⎢ − s ⎥
⎣ s + 1 s + 3⎦ ⎣ s
⎦
=
2
1
2e −2 s
e −2 s
−
−
+
s( s ) s( s ) s(
) s((
)
1 ⎤ 1 ⎡1
1 ⎤
1 ⎤ e −2 s ⎡ 1
1 ⎤
⎡1
−2 s ⎡ 1
= 2⎢ −
−
−
−
2
e
−
+
−
⎥
⎢
⎥
⎢
⎥
3 ⎣ s s + 3 ⎥⎦
⎣ s s + 1⎦ 3 ⎣ s s + 3 ⎦
⎣ s s + 1⎦
Taking the inverse Laplace transform,
v(t ) = 2[u
[u(t ) e t u(t )]
1
[u
u((t ) e
3
3t
u(t )] − 2[[u
u ( t − 2) e
(t
(t
)
1
u(t − 2)] + [ (t − 2) −
3
−33( − 2 )
(t − 2)]
for t > 0
Exercises 11.79
Exercises
⎛ 1 − cos 2t ⎞
11.1 Find L{ f ′(t )} of f (t ) = ⎜⎝
⎟⎠
t
⎡
⎛ s2 + 4 ⎞ ⎤
⎢ s log ⎜
⎟⎥
⎜⎝
s ⎟⎠ ⎥
⎢
⎣
⎦
`
11.2 Find Laplace transform of the follwoing
function:
2
f (t ) = t + 1 0
t>2
=3
⎡1
⎢ s (1 e
⎣
1
Ω
8
v (t )
2s
⎤
)⎥
⎦
11.3 For the network shown in Fig 11.125, the
switch is closed at t = 0. Find the current i1(t)
for t > 0.
1
Ω
2
10 A
1F
Fig. 11.127
[v(t) = 1 + 4 e−10t]
11.6 The circuit of Fig. 11.128, has been in the
condition shown for a long time. At t = 0,
switch is closed. Find v(t) for t > 0.
5Ω
v (t )
20 V
3Ω
2F
100 Ω
50 Ω
100 V
Fig. 11.128
[v(t) = 7.5 + 12.5 e−(4/15)t]
4H
i1 (t )
Fig. 11.125
[i1(t) = 3 − e−25 t]
11.7 Figure 11.129 shows a circuit which is in the
steady-state with the switch open. At t = 0, the
switch is closed. Determine the current i (t).
Find its value at t = 0.114 μ seconds.
800 Ω
11.4 Determine the current i(t) in the network of
Fig. 11.126, when the switch is closed at t = 0.
The inductor is initially unenergized.
i (t )
12 V
2Ω
400 Ω
0.001 μF
200 Ω
i (t )
2Ω
24 V
2Ω
0.5 H
Fig. 11.129
6
[i(t) = 0.00857 + 0.01143 e−8.75 × 10 t, 0.013 A]
11.8 Find i(t) for the network shown in Fig.
11.130.
i (t ) 10 Ω
Fig. 11.126
[i(t) = 4(1 − e−6t)]
11.5 In the network of Fig. 11.127, after the switch
has been in the open position for a long time,
it is closed at t = 0. Find the voltage across the
capacitor.
1F
50 V
5Ω
Fig. 11.130
0.5 F
5Ω
11.80 Network Analysis and Synthesis
[i(t) = 0.125 e−0.308t + 3.875 e−0.052t]
11.9 Determine v(t) in the network of Fig. 11.131
where iL(0−) = 15 A and vc(0−) = 5 V.
10 Ω
2H
1H
20 Ω
10 V
0.5 H
30 Ω
+
0.33 Ω
10 V
1 F v (t )
Fig. 11.134
−
Fig. 11.131
[v(t) = 10 − 10e−t + 5e−2t]
11.10 The network shown in Fig. 11.132 has acquired
steady state with the switch at position 1 for
t < 0. At t = 0, the switch is thrown to the
position 2. Find v(t) for t > 0.
2Ω
1
[i(t) = 0.1818 − 0.265 e−13.14t + 0.083 e−41.86t ]
11.13 The network shown in Fig. 11.135 is in steady
state with s1 closed and s2 open. At t = 0, s1
is opened and s2 is closed. Find the current
through the capacitor.
2Ω
2H
s1
3H
10 V
2
s2
1 μF
+
3Ω
2V
v (t )
0.5 F
1H
−
Fig. 11.135
[i(t) = 5 cos (0.577 × 103 t)]
11.14 In the network shown in Fig. 11.136, find
currents i1(t) and i2(t) for t > 0.
10 Ω
Fig. 11.132
[v(t) = 4e−t − 2 e−2t]
11.11 In the network shown in Fig. 11.133, the
switch is closed at t = 0. Find current i1(t) for
t > 0.
3Ω
20 V
i1 (t )
1H
0.2 F
50 V
i1 (t )
40 Ω
i2 (t )
1Ω
1 Ω i (t )
2
1
F
3
Fig. 11.136
[i1(t) = 5 e−0.625t, i2(t) = 1 − e−0.625t ]
11.15 For the network shown in Fig. 11.137, find
currents i1(t) and i2(t) for t > 0.
5Ω
Fig. 11.133
[i1(t) = 5 + 5e−2t − 10e−3t]
11.12 In the network shown in Fig. 11.134, the
switch is closed at t = 0. Find the current
through the 30 Ω resistor.
20 μF
50 V
i1 (t )
Fig. 11.137
i2 (t )
5Ω
0.1 H
Objective-Type Questions 11.81
7Ω
99499.5t ⎤
⎡ i1 (t ) = 0.101
.101e
101e −100.5t 0.05e 99
⎢
⎥
⎢⎣i2 (t ) = 5 5.05e −100.5t + 0.05e −9949.5t ⎥⎦
11.16 In the network shown in Fig. 11.138, the
switch is opened at t = 0, the steady state
having been established previously. Find i(t)
for t > 0.
3Ω
i (t )
5Ω
10 V
0.1 F
1.8 H
3.5 Ω
5 cos 2t
1H
+
−
1 F v (t )
2
Fig. 11.140
6 −tt
⎡
⎢ v(t ) = − 5 e
⎣
9
e
10
6t
+
3
21
⎤
cos 2t
s i 2t ⎥
10
10
⎦
11.19 For the network shown in Fig. 11.141,
determine v(t) when the input is
(i) an impulse function
[e−t u(t)]
−t
(ii) i(t) = 4e u(t)
[4t e−t u(t)]
Fig. 11.138
[i(t) = 1.5124e−2.22t + 3.049e−2.5t]
11.17 Find the current i(t) in the network of Fig.
11.139, if the switch is closed at t = 0. Assume
initial conditions to be zero.
20 Ω
i (t )
10 Ω
15 Ω
+
1Ω
i (t )
v (t )
1F
−
Fig. 11.141
11.20 For a unit-ramp input shown in Fig. 11.142,
find the response vc(t) for t > 0.
5A
10 Ω
1H
2.5 H
10 F vc (t )
r (t )
Fig. 11.139
[i(t) = 3 + 0.57e−7.14 t]
11.18 In the network shown in Fig. 11.140, find the
voltage v(t) for t > 0.
Fig. 11.142
[vc(t) = −100 u(t) + 100e−0.01t u(t) + tu(t)]
Objective-Type Questions
11.1 If the Laplace transform of the voltage across
1
a capacitor of value F is
2
1
Vc ( s) = 2
s +1
the value of the current through the capacitor
at t = 0+ is
(a) 0
(b) 2 A
(c)
1
A
2
(d)
1A
11.2 The
response
of
an
initially
relaxed
linear
constant
parameter
network to a unit impulse applied at
t = 0 is 4 e−2t u(t). The response of this network
to a unit-step function will be
11.82 Network Analysis and Synthesis
2[1 − e −2t] u(t)
4[e−t − e–2t] u(t)
sin 2t
(1 − 4 e−4t) u(t)
11.3 The Laplace transform of a unit-ramp
function starting at t = a is
(a)
(b)
(c)
(d)
(a)
(c)
1
(b)
)2
(
e − as
3V
vi
(b)
( s + a) 2
11.7 A 2 mH inductor with some initial current can
be represented as shown in Fig. 11.145. The
value of the initial current is
a
s2
I (S)
−+
0.002s
1mV
Fig. 11.145
(a) 0.5 A
(b) 2 A
(c) 1 A
(d) 0
11.8 A current impulse 5 d (t) is forced through
a capacitor C. The voltage vc(t) across the
capacitor is given by
( s − α )2 + α 2
s+α
( s − α )2 + α 2
1
(
)2
(d) none of the above
11.5 The circuit shown in Fig. 11.143 has initial
current i(0−) = 1 A through the inductor and
an initial voltage vc(0−) = −1 V across the
capacitor. For input v(t) = u(t), the Laplace
transform of the current i(t) for t ≥ 0 is
1Ω
vo
Fig. 11.144
s −α
(c)
1 kΩ
t
2s
11.4 The Laplace transform of eat cos a t is equal
to
(a)
−2Ω
−j
e − as
(d)
s2
0.1 μF
(a)
5t
(b)
5 u(t) − C
(c)
5
t
C
(d)
5u(t )
C
11.9 In the circuit shown in Fig. 11.146, it is desired
to have a constant direct current i(t) through
the ideal inductor L. The nature of the voltage
source v(t) must be
1H
i (t )
+
v (t )
1F
v (t )
L
i (t )
−
Fig. 11.143
(a)
(c)
s
2
s + s +1
s−2
s2 + s + 1
(b)
(d)
Fig. 11.146
s+2
2
s + s +1
s−2
s2 + s + 1
11.6 A square pulse of 3 volts amplitude is applied
to an RC circuit shown in Fig. 11.144. The
capacitor is initially uncharged. The output
voltage v0 at time t = 2 seconds is
(a) 3 V
(b) −3 V
(c) 4 V
(d) −4 V
(a) a constant voltage
(b) a linearly increasing voltage
(c) an ideal impulse
(d) as exponential increasing voltage
11.10 When a unit-impulse voltage is applied to an
inductor of 1 H, the energy supplied by the
source is
(a) ∞
(b) 1 J
(c)
1
J
2
(d)
0
Answers to Objective-Type Questions 11.83
Answers to Objective-Type Questions
11.1 (c)
11.7 (a)
11.2 (a)
11.8 (d)
11.3 (c)
11.9 (c)
11.4 (a)
11.10 (c)
11.5 (b)
11.6 (b)
12
Network Functions
12.1
INTRODUCTION
A network function gives the relation between currents or voltages at different parts of the network. It is
broadly classified as driving point and transfer function. It is associated with terminals and ports.
Any network may be represented schematically by a rectangular box. Terminals are needed to connect any
network to any other network or for taking some measurements. Two such associated terminals are called
terminal pair or port. If there is only one pair of terminals in the network, it is called a one-port network.
If there are two pairs of terminals, it is called a two-port network. The port to which energy source is connected
is called the input port. The port to which load is connected is known as the output port. One such network
having only one pair of terminals (1 – 1′ ) is shown in Fig. 12.1 (a) and is called one-port network. Figure
12.1 (b) shows a two-port network with two pairs of terminals. The terminals 1 – 1′ together constitute a port.
Similarly, the terminals 2 – 2′ constitute another port.
1
1′
I
+
V
−
Oneport
network
1
1′
(a)
I2
I1
+
V1
−
Twoport
network
2
+
V2
−
2′
(b)
Fig. 12.1 (a) One-port network (b) Two-port network
A voltage and current are assigned to each of the two ports. V1 and I1 are assigned to the input port,
whereas V2 and I2 are assigned to the output port. It is also assumed that currents I1 and I2 are entering into
the network at the upper terminals 1 and 2 respectively.
12.2
DRIVING-POINT FUNCTIONS
If excitation and response are measured at the same ports, the network function is known as the driving-point
function. For a one-port network, only one voltage and current are specified and hence only one network
function (and its reciprocal) can be defined.
12.2 Network Analysis and Synthesis
1. Driving-point Impedance Function It is defined as the ratio of the voltage transform at one port
to the current transform at the same port. It is denoted by Z (s).
Z ( s) =
V ( s)
I ( s)
2. Driving-point Admittance Function It is defined as the ratio of the current transform at one port
to the voltage transform at the same port. It is denoted by Y (s).
Y ( s) =
I ( s)
V ( s)
For a two-port network, the driving-point impedance function and driving-point admittance function at
port 1 are
Z11 ( s) =
V1 ( s)
I1 ( s)
Y11 ( s) =
I1 ( s)
V1 ( s)
Z 22 ( s) =
V2 ( s)
I 2 ( s)
Y22 ( s) =
I 2 ( s)
V2 ( s)
Similarly, at port 2,
12.3
TRANSFER FUNCTIONS
The transfer function is used to describe networks which have at least two ports. It relates a voltage or current
at one port to the voltage or current at another port. These functions are also defined as the ratio of a response
transform to an excitation transform. Thus, there are four possible forms of transfer functions.
1. Voltage Transfer Function It is defined as the ratio of the voltage transform at one port to the
voltage transform at another port. It is denoted by G (s).
V2 ( s)
V1 ( s)
V ( s)
G21 ( s) = 1
V2 ( s)
G12 ( s) =
2. Current Transfer Function It is defined as the ratio of the current transform at one port to the current
transform at another port. It is denoted by a (s).
α12 ( ) =
I2 ( )
I1 ( )
α 21 ( ) =
I1 ( )
I2 ( )
12.3 Transfer Functions 12.3
3. Transfer Impedance Function It is defined as the ratio of the voltage transform at one port to the
current transform at another port. It is denoted by Z (s).
Z12 ( s) =
V2 ( s)
I1 ( s)
Z 21 ( s) =
V1 ( s)
I 2 ( s)
4. Transfer Admittance Function It is defined as the ratio of the current transform at one port to the
voltage transform at another port. It denoted by Y (s).
Example 12. 1
Y12 ( s) =
I 2 ( s)
V1 ( s)
Y21 ( s) =
I1 ( s)
V2 ( s)
Determine the driving-point impedance function of a one-port network shown in Fig. 12.2.
R
L
C
R
Fig. 12.2
Solution The transformed network is shown in Fig. 12.3.
1
R
( R Ls)
s+
R Ls
1
Cs
L
Z ( s) =
=
=
2
1
1
C 2 R
LCs
C
R
RCs
Cs
+
1
+ (R
( R Ls)
s + s+
Cs
L
LC
Example 12.2
1
Cs
Z(s)
Ls
Fig. 12.3
Determine the driving-point impedance of the network shown in Fig. 12.4.
2s
2s
1
2s
Z (s)
1
2s
Fig. 12.4
Solution
1 ⎛
1⎞
1 ⎛
1⎞
1
1
2s +
4 s + 8s3 + 2 s +
⎜⎝ 2 s + ⎟⎠
⎜⎝ 2 s + ⎟⎠
4
+ s2 + 1
2s
2s
2s
2s
2
s
2
s = 16 s +12
Z ( s) = 2 s +
= 2s +
=
2
s
+
=
1
1
2 + 4s2
2 + 4s2
2 + 4s2
8s3 + 4 s
+ 2s +
2s
2s
2s
12.4 Network Analysis and Synthesis
Example 12.3
Determine the driving-point impedance of the network shown in Fig. 12.5.
1
s
1
s
s
Z (s)
s
Fig. 12.5
Solution
⎛1 ⎞
s ⎜ + s⎟
⎝ s ⎠ 1 ( s 2 ) s 1 s + s3
1
s4 + 3 2 1
Z ( s) = +
= +
=
+
=
1
s
s 2s2 + 1
s 2s2 + 1
2 s3 + s
s+ +s
s
Example 12.4
Find the driving-point admittance function of the network shown in Fig. 12.6.
3s
s
1
5s
1
2s
Y (s)
Fig. 12.6
Solution
⎛ 1⎞
s⎜ ⎟
⎝ 2s ⎠
1
1
s
30 s 4 +15
+ 5s 2 + 2 s 2 + 1 + 5s 2 30 s 4 + 22s
22 s 2 + 1
Z ( s ) = 3s + +
= 3s + + 2
=
=
1
5s
5s 2 s + 1
5 ( 2 s 2 1)
5 ( 2 s 2 1)
s+
2s
1
5s ( s 2 )
Y ( s) =
=
Z ( s) 30 s 4 + 22 s 2 + 1
Example 12.5
Find the transfer impedance function Z12(s) for the network shown in Fig. 12.7.
I1 (s)
+
V1 (s)
+
R
1
Cs
−
V2 (s)
−
Fig. 12.7
12.4 Analysis of Ladder Networks 12.5
⎛ 1⎞
R⎜ ⎟
⎝ Cs ⎠
V2 ( s) = I1 ( s)
1
R+
Cs
V2 ( s)
R
=
I1 ( s) RCs
C +1
V2 ( s)
1
Z12 ( s) =
=
1 ⎞
I1 ( s)
⎛
C s+
⎟
⎝
RC ⎠
Solution
Example 12.6
Find voltage transfer function of the two-port network shown in Fig. 12.8.
I1(s)
R
I2(s)
+
V1(s)
+
1
Cs
−
V2(s)
−
Fig. 12.8
Solution By voltage division rule,
1
Cs
Voltage transfer function
12.4
1
1
V2 ( s) = V1 ( s)
= V1 ( s)
= V1 ( s) RC
1
1
RCs
RC
C +1
R+
s+
Cs
RC
1
V2 ( s)
= RC
1
V1 ( s)
s+
RC
ANALYSIS OF LADDER NETWORKS
Z1
Z3
I1
I2 = 0
Va
Ib
Vb
The network functions of a ladder network can be
obtained by a simple method. This method depends +
upon the relationships that exist between the branch
Y4
Y2
currents and node voltages of the ladder network. V1
Consider the network shown in Fig. 12.9 where all the
−
impedances are connected in series branches and all
the admittances are connected in parallel branches.
Fig. 12.9 Ladder network
Analysis is done by writing the set of equations.
In writing these equations, we begin at the port 2 of
the ladder and work towards the port 1.
Vb = V2
Ib = Y4 V2
Va = Z3 Ib + V2 = (Z3Y4 + 1) V2
I1 = Y2 Va + Ib = [Y2 (Z3 Y4 + 1) + Y4] V2
V1 = Z1 I1 + Va = [Z1 {Y2 (Z3 Y4 + 1) + Y4} + (Z3 Y4 + 1)] V2
+
V2
−
12.6 Network Analysis and Synthesis
Thus, each succeeding equation takes into account one new impedance or admittance. Except the first two
equations, each subsequent equation is obtained by multiplying the equation just preceding it by imittance
(either impedance or admittance) that is next down the line and then adding to this product the equation twice
preceding it. After writing these equations, we can obtain any network function.
Example 12.7
For the network shown in Fig. 12.10, determine transfer function
1Ω
I1
1Ω
I2 = 0
+
V
V
+
V
1F
V
1F
Fig. 12.10
Solution The transformed network is shown in Fig. 12.11.
V
Ib
1
I1
V
V
1
s
1
Ib
V
I
0
+
+
sV
+V
1
s
1
s
V
−
V
−
s +1 V
I1
Hence,
Fig. 12.11
V
1
s
V
+V
V
1
= 2
V
s
s
Example 12.8
s
b
s
s
sV
s
sV
s + )V
For the network shown in Fig. 12.12, determine the voltage transfer function
I1
s
s
Va
I
V
I
+
+
V1
1
V
1
Fig. 12.12
Solution
Vb = V2
Ib
V
1
I1
V
1
0
V
V
= s+2 V
I
s
V
)V
V
.
V
12.4 Analysis of Ladder Networks 12.7
V2
1
=
V1 s 2 + 3s
3s 1
Hence,
Example 12.9
Find the network functions
1H
I1
V1 V2
V
, and 2 for the network shown in Fig. 12.13.
I1 V1
I1
1H
I2 = 0
+
+
V1
1F
V2
1F
−
−
Fig. 12.13
Solution The transformed network is shown in Fig. 12.14.
I1
Vb
V2
V
I b = 2 = sV
V2
1
s
Va s I b V2 = s ( sV
V2 ) V2 = (s
( s 2 1)V2
I1 =
V1
Hence,
s
s
Va
Ib
I2 = 0
Vb
+
+
1
s
1
s
V1
−
−
Fig. 12.14
Va
+ I b sV
Va I b = s ( s 2 + 1)V2 sV
V2 ( s3 2 s)V2
1
s
s I1 Va = s ( s3 + 2 s)
( s 2 + 1)
( s 4 2 s 2 + s 2 + 1)
((ss 4
3s 2 1)V2
V1 s 4 + 3s
3s 2 1
=
I1
s3 + 2 s
V2
1
= 4
V1 s + 3s
3s 2 1
V2
1
= 3
V1 s + 2 s
Example 12.10
Find the network functions
I1
1
F
4
V1 V2
V
, , and 2 for the network in Fig. 12.15.
I 1 V1
I1
3H
I2 = 0
+
V1
+
1
H
2
−
2F
V2
−
Fig. 12.15
V2
12.8 Network Analysis and Synthesis
Solution The transformed network is shown in Fig. 12.16.
Vb
V2
V
I b = 2 = 2 sV
V2
1
2s
Va
4
s
I1
Ib
I2 = 0
Vb
+
3s I b V2 = 3s ( 2 sV
V2 ) V2 = (6 s
2
1
2s
s
2
V1
−
1)V2
⎛ 14 s 2 + 2 ⎞
Va
2
+ I b = (6 s 2 1)V2 + 2 sV
V2 = ⎜
⎟ V2
s
s
s
⎝
⎠
2
⎛ 6s4
4
4 ⎛ 14 s 2 + 2 ⎞
2
V1 = I1 Va = ⎜
V
(
6
s
1
)
V
=
2
⎟ 2
⎜
s
s⎝
s
⎠
⎝
V2
−
I1 =
Hence,
3s
Va
+
Fig. 12.16
557s
7s2 8 ⎞
⎟ V2
s2
⎠
V1 6 s 4 557s
7s2 8
=
I1
14 s3 + 2 s
V2
s2
= 4
V1 6 s 57s
57 s 2 8
V2
s
=
I1 14 s 2 + 2
Example 12.11 For the ladder network of Fig. 12.17, find the driving point-impedance at the 1 – 1Ä
terminal with 2 – 2Ä open.
1
I1
1Ω
1Ω
1H
1H
I2 = 0
+
+
V1
1′
1Ω
1H
1F
1F
1F
−
2
V2
−
2′
Fig. 12.17
Solution The transformed network is shown in Fig. 12.18.
1
1′
I1
1
s
1
a
s
1
Vb
+
Ia
Ib
V1
1
s
1
s
−
V2
V
I b = 2 = s V2
1
s
Vc
I2 = 0
+
1
s
2
V2
−
Fig. 12.18
Vc
s
2′
12.4 Analysis of Ladder Networks 12.9
V
Ia =
V
+
1
s
V
I1
V
1
s
s +2
s
s
a
=s s
s2
s
V
= s
s(
b
s
+ 2s V
sV +
s
+
s + s
s
+
+
+1 V
s + 2s V
s) V
s
s
+
s
V
V
Hence,
Z11
V
I1
Example 12.12
s
s
s
s
s
s
Determine the voltage transfer function
1Ω
I1
V
for the network shown in Fig. 12.19.
V
2
I
0
+
+
V
1F
V
1Ω
1F
−
Fig. 12.19
Solution The transformed network is shown in Fig. 12.20.
I1
=V
1
Ia
V
2s
V
+
+I
V
1
s
s
+1 V
s +1 V
1
Ib
V
1
s
s
V
=
I1 =
V
+
1
s
Fig. 12.20
a
s
V
Hence,
V
=
V
2s
s +
1V
+1
s
1
s + 4s 2
+
s
Vc
+1 V
s
4s
V
I
0
+
I
1
V
12.10 Network Analysis and Synthesis
Example 12.13
For the network shown in Fig. 12.21, determine the transfer function
3Ω
2
I1
1Ω
I2
2F
3
+
I2
.
V1
I=0
+
V1
1Ω
6
2F
V2
−
−
Fig. 12.21
Solution The transformed network is shown in Fig. 12.22.
3
2
1
I1
3
2s
+
I2
Va
Ia
I=0
Vb
1
6
1
2s
V1
+
Ib
V2
−
−
Fig. 12.22
Vb
Va = V2
V2 V2
+
= 2 sV
V2 + 6 V2 ( 2 s + 6)V2
1
1
2s 6
⎛3 3
⎞
×
⎡ ( 2 s 5)( s + 3) ⎤
(6 15)
⎜ 2 2s ⎟
⎛ 9
⎞
V1 = ⎜
+ 1⎟ I1 V2 = ⎜
+ 1⎟ I1 V2 =
( 2 6)V2 V2 = ⎢
+ 1⎥ V2
3 3
⎝6 6 ⎠
6 6
( s + 1)
⎣
⎦
⎜ +
⎟
⎝ 2 2s ⎠
⎛ 2 s 2 + 6 s + 5s + 15 + s + 1⎞
⎛ 2 s 2 + 12 s + 16 ⎞
⎡ ( 2 s 5)( s + 3) ( s 1) ⎤
=⎢
V2 = ⎜
V2 = ⎜
⎟
⎟ V2
⎥
s +1
s +1
s +1
⎝
⎠
⎝
⎠
⎣
⎦
2(ss 2 + 6s
6 s 8)
2 ( s 4)( s + 2)
=
V2 =
V2
s +1
s +1
I2 = −Ib = −6 V2
I1 = I a + I b =
Also,
I2
3 ( s 1)
=−
V1
( s + 4)( s 2)
Example 12.14
For the network shown in Fig. 12.23, compute α 12 (s)
s =
1Ω
I2
I=0
+
I1
V1
+
1F
1F
V2
−
−
Fig. 12.23
I2
V
and Z12 ((s)
s) = 2 .
I1
I1
12.4 Analysis of Ladder Networks 12.11
Solution The transformed network is shown in Fig. 12.24.
1
Va
Hence,
and
V2
+
V2
I1
I2 =
= sV
V2
V1
1
s
−
V1 1 I 2 + Va sV
V2 V2 = (s 1)V2
V
I1 = 1 + I 2 sV
V1 I 2 = s (s + 1)V
V + sV = ( s 2 + 2s)V2
1
s
I2
1
α12 ( ) =
=
I1 s + 2
Z12 ( s) =
I2
Va
I=0
+
1
s
1
s
V2
−
Fig. 12.24
V2
1
=
I1 s 2 + 2 s
Example 12.15
Determine the voltage ratio
V2
V2
I
and
, current ratio 2 , transfer impedance
I1
V1
I1
V1
for the network shown in Fig. 12.25.
I1
driving- point impedance
3Ω
3H
I1
+
I=0
I2
1F
3
+
2Ω
V1
1Ω
1F
1F
2
−
V2
−
Fig. 12.25
Solution The transformed network is shown in Fig. 12.26.
3
I1
Va
3
s
+
3s
Ia
1
s
V1
−
I2
Vb
I=0
Vc
+
Ib
2
2
s
1
V2
−
Fig. 12.26
Vb = V2
V
I 2 = 2 = V2
1
Vc
V
s
⎛ 3s + 2 ⎞
+ 2 =
V V2 = ⎜
V2
2 1 2s + 2 2
⎝ 2 s + 2 ⎟⎠
2+
s
⎛ 9 s 2 + 8s + 2 ⎞
3s (3s + 2)
Va = 3s I a + V2 =
V2 + V2 = ⎜
⎟ V2
2s + 2
⎝ 2s + 2 ⎠
Ia
Ib + I2 =
V2
12.12 Network Analysis and Synthesis
I1 =
V
a
s
3⎞
⎛
3×
s
V
3+
⎡ 27
⎣
⎛ 36
3
I +
s +1
=
V
=
V
36
I2
=
I1
s
V
I1
s
=
⎛ 9s
3s 2
s
3 ⎛
s +1
2
+5
2
⎞
8 2
s
2s 2
2
V +
⎛
2
+
+ 15s +
s +
+2
s
+ 25s + 8 ⎞
+
s+
s+2
(36
V
+
V
s + 25s + 8
2s 2
s + 5s 2
2s 2
+ 5s 2
Example 12.16
s +2
36
s + 25s + 8
s+2
s
s+2
For the resistive two-port network of Fig. 12.27, find
2Ω
I1
2Ω
2Ω
+
V V I2
I
, , and 2
V I1 V
I1
I
+
1Ω
V
1Ω
1Ω
1Ω
V
Fig. 12.27
Solution The network is redrawn as shown in Fig. 12.28.
I1
+2
V
2
+ 25 + 8)
V
+2
s+2
s
V
36
=
I1
s
V
s
+2
Hence,
s 8s + 2
V
2s 2
s
+
2
V
2
V
+
Ia
I
V
1
1
Fig. 12.28
V
I
+
1
V
1
12.4 Analysis of Ladder Networks 12.13
V2
= −V2
1
3 I 2 3V2
I2 = −
Vb
Vb Vb 4
+
= Vb 4 V2
1
3 3
2 I b Vb 8 V2 + 3 V2 = 11 V2
V
= a + I b = 11 V2 + 4 2 = 15 V2
1
= 2 a + a = 30 2 + 11 V2 = 4411 V2
V
= 1 + I a = 41 V2 +15
155 V2 = 566 V2
1
1
=
41
1
=
Ω
56
1
=−
41
1
=−
56
Ib =
Va
Ia
V1
I1
V2
V1
V2
I1
I2
V1
I2
I1
Hence,
Example 12.17
Find the network function
Ia
1Ω
1Ω
+
V2
for the network shown in Fig. 12.29.
V1
2V1
2Ia
V1
I2 = 0
+ −
+
1Ω
−
V2
−
Fig. 12.29
Solution The network is redrawn as shown in Fig. 12.30.
Ia
1
3Ia
1
+
V1
2V1
I2 = 0
+ −
3Ia
2Ia
−
1
+
V2
−
Fig. 12.30
12.14 Network Analysis and Synthesis
From Fig. 12.30,
V2 = 1 (3 Ia) = 3 Ia
Applying KVL to the outermost loop,
V1 − 1 (Ia) − 1 (3 Ia) − 2 V1 − 1 (3 Ia) = 0
V1 = −7 Ia
V2
3
=−
V1
7
Hence,
Example 12.18
I2
for the network shown in Fig. 12.31.
I1
Find the network function
2Ia
2Ω
I2
− +
+
Ia
I1
1Ω
I1
2
1Ω
1Ω
V2
−
Fig. 12.31
Solution The network is redrawn as shown in Fig. 12.32.
2 Ia
a
Ia + I2 +
− +
I
I1
I1
2
d
2
I2 +
I1
2
+
f
Ia
I1
2
1
1
I2
1
V2
−
b
c
e
Fig. 12.32
From Fig. 12.32,
I1 + I a + I 2 +
I
=
3
I1
2
Ia + I2
I1
2
...(i)
Applying KVL to the loop abcda,
−1 I − 1 I a − 2
−I − 3
I +3
a
a
a
=0
=0
=0
...(ii)
12.5 Analysis of Non-Ladder Networks
12.15
Substituting Eq. (i) in Eq. (ii),
3
I1
2
Ia + I2
3
I1
2
3
a
0
I2 + 4
a
0
...(iii)
Applying KVL to the loop dcefd,
1
a
1I2
I ⎞
⎛
2 I2 + 1 ⎟ = 0
⎝
2⎠
Ia
3
2
1
Ia
0
3 I 2 + I1
...(iv)
Substituting Eq. (iv) in Eq. (iii),
3
I1
2
I 2 + 4 (3 I 2
3
I1
2
I 2 + 12 I 2
I1 ) = 0
4
1
0
11
I1 13 I 2 = 0
2
13 I 2 = −
I2
11
=−
I1
26
Hence,
12.5
11
I1
2
ANALYSIS OF NON-LADDER NETWORKS
The above method is applicable for ladder networks. There are other network configurations to which the
technique described is not applicable. Figure 12.33 shows one such network.
Z4
I1
Z2
Z1
+
V1
I2
+
Z3
−
V2
−
Fig. 12.33 Non-ladder network
For such a type of network, it is necessary to express the network functions as a quotient of determinants,
formulated on KVL and KCL basis.
12.16 Network Analysis and Synthesis
Example 12.19
For the resistive bridged T network shown in Fig. 12.34, find
V2 V2 I 2
I
, , and 2 .
V1 I 1 V1
I1
2Ω
1Ω
I1
1Ω
I3
+
I2
+
0.5 Ω
V1
I1
−
V2
I2
1Ω
−
Fig. 12.34
Solution Applying KVL to Mesh 1,
V1 = 1.5 I1 + 0.5 I2 − I3
Applying KVL to Mesh 2,
0 = 0.5 I1 + 2.5 I2 + I3
Applying KVL to Mesh 3,
0 = −I1 + I2 + 4 I3
Writing these equations in matrix form,
0
⎡V1 ⎤ ⎡1
⎢ 0 ⎥ = ⎢0.5 2.5
⎢ ⎥ ⎢
1
⎣ 0 ⎦ ⎣ −1
V1
0
0
Δ1
I1 =
=
1.5
Δ
0.5
−1
1⎤ ⎡ I1 ⎤
1⎥ ⎢ I 2 ⎥
⎥⎢ ⎥
4 ⎦ ⎣ I3 ⎦
0 5 −1
2 .5 1
1 4
V (10 − 1)
= 1
= V1
0.5 −1
9
2.5 1
1 4
1 5 V1 −1
05 0
1
−1 0
4
Δ2
−V ( 2 + 1)
1
I2 =
=
= 1
= − V1
1.5 0.5 −1
Δ
9
3
0.5 2.5 1
−1
1 4
From Fig. 12.34,
From Eq. (v),
From Eqs. (iv) and (v),
Hence,
V2 = −1 (I2) = −I2
V1 = −3 I2
1
1
V1 = − I1
3
3
I1
3I 2
I2
1
=− �
V1
3
I2
...(i)
...(ii)
...(iii)
...(iv)
...(iv)
...(v)
12.5 Analysis of Non-Ladder Networks
12.17
1
− V1
I2
1
= 3 =−
I1
V1
3
V2
−I2
1
=
=
V1 −3 I 2 3
V2
−I2
1
=
= Ω
I1 −3 I 2 3
Example 12.20
For the network of Fig. 12.35, find Z11, Z12 and G12.
Za
a
I1
b
+
+
V1
Zb
V2
Zb
−
−
d
c
Za
Fig. 12.35
Solution The network can be redrawn as shown in Fig. 12.36. Since the network consists of two identical
impedances connected in parallel, the current in I1 divides equally in each branch.
V1
( Z a + Zb )
I1
I1
2
Z11 =
V1 Z a Zb
=
I1
2
V2
Zb
Z12 =
V2 Zb − Z a
=
I1
2
+
⎛I ⎞
Z a ⎜ 1 ⎟ = ( Zb
⎝ 2⎠
I1
2
Za )
I1
2
V1
I1
2
Za
+
I1
2
Zb
V2
−
Zb
−
Fig. 12.36
By voltage-division rule,
V2 =
G12 =
Example 12.21
Zb
Za
Zb
V2 Zb
=
V1 Z a
Za
V1
Za
Zb
V1 =
Zb
Za
Za
V1
Zb
Za
Zb
For the network shown in Fig. 12.37, determine Z11 (s), G12 (s) and Z12 (s).
Za
12.18 Network Analysis and Synthesis
1F
1Ω
I1
1Ω
I2 = 0
+
+
V1
V2
1F
−
−
Fig. 12.37
Solution The transformed network is shown in Fig. 12.38.
1
s
1
1
I3
+
+
1
s
V1
−
I2 = 0
I1
V2
−
Fig. 12.38
Applying KVL to Mesh 1,
⎛ 1⎞
V1 = 1 +
I1
⎝ s⎠
I3
...(i)
Applying KVL to Mesh 2,
V2 =
1
I1
s
I3
...(ii)
Applying KVL to Mesh 3,
1⎞
⎛
− I1 + ⎜ 2 + ⎟ I 3 = 0
⎝
s⎠
⎛ s ⎞
I3 = ⎜
I1
⎝ 2 s + 1⎟⎠
...(iii)
Substituting Eq. (iii) in Eqs. (i) and (ii),
⎡ s 2 + 3s
s ⎞
3s 1 ⎤
⎛ 1⎞
⎛ s ⎞
⎛ s +1
V1 = 1 + I1 − ⎜
I
=
I
=
⎢
⎥ I1
1
1
⎟
⎝ s⎠
⎝ 2 s 1⎠
⎝ s
2 s 1⎠
⎣ s (22 s + 1) ⎦
⎡ s 2 + 2 s + 1⎤
1
s
V2 = I1 +
I1 = ⎢
⎥ I1
s
2s + 1
⎣ s ( 2 s + 1) ⎦
Hence,
Z11 ( s) =
V1 s 2 + 3s
3s 1
=
I1
s ( 2 s 1)
Z12 ( s) =
V2 s 2 + 2s
2s 1
=
I1
s ( 2 s 1)
G12 ( s) =
V2 s 2 + 2s
2s 1
= 2
V1 s + 3s
3s 1
12.5 Analysis of Non-Ladder Networks
Example 12.22
12.19
For the network shown in Fig.12.39, find the driving-point admittance Y11 and
transfer admittance Y12.
1Ω
1F
I1
V1
1F
+
−
I2
1Ω
1Ω
Fig. 12.39
Solution The transformed network is shown in Fig. 12.40.
1
1
s
I1
+
−
V1
I3
1
I1
1
s
I2
I2
1
Fig. 12.40
Applying KVL to Mesh 1,
V1 =
1
⎛1 ⎞
+ 1 I1 + I 2 − I 3
⎝s ⎠
s
...(i)
1⎞
1
⎛
2+
I 2 + I3
⎝
s⎠
s
...(ii)
Applying KVL to Mesh 2,
0
1
Applying KVL to Mesh 3,
1
1
⎛2 ⎞
0 = − I1 + I 2 + ⎜ + 1⎟ I 3
⎝s ⎠
s
s
Writing these equations in matrix from,
⎡1
1
⎢ s +1
⎡V1 ⎤ ⎢
1
⎢0⎥ = ⎢ 1
2+
⎢ ⎥ ⎢
s
⎣0⎦ ⎢ 1
1
⎢−
s
⎣ s
Δ
I1 = 1
Δ
⎤
⎥
⎥ ⎡ I1 ⎤
⎥ ⎢I2 ⎥
⎥ ⎢I ⎥
2 ⎥ ⎣ 3⎦
+ 1⎥
s ⎦
1
s
1
s
−
...(iii)
12.20 Network Analysis and Synthesis
1
+1
s
Δ=
−
=
2+
1
1
s
1
s
s2 + 5
s
1
s
1
1⎞ ⎛ 2 ⎞ 1 ⎤ ⎡ ⎛ 2 ⎞ 1 ⎤ 1 ⎡ ⎛ 1⎞ ⎛ 1⎞ ⎛
1⎞ ⎤
⎛ 1 ⎞ ⎡⎛
=
+ 1⎟ ⎢⎜ 2 + ⎟ ⎜ + 1⎟ − 2 ⎥ −11 (1) + 1⎟ + 2 ⎥ − ⎢(1) ⎜ ⎟ + ⎜ ⎟ ⎜ 2 + ⎟ ⎥
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
s
s
s s
s
s
s
s⎠ ⎦
s ⎦ ⎣
s ⎦ s⎣
⎣
2
+1
s
−
1
1
s
2
2
V1
1
Δ1 = 0
2+
0
1
s
I1
Hence,
1
s
2
+1
s
5s + 1⎞
5s
⎟
s
⎠
2
I1 2 s 2 5s + 1
=
V1 s 2 + 5s
5s 2
Δ
I2 = 2
Δ
Y11 =
Δ2 =
1
−
12.6
2
⎛ 2 s 2 5s + 1⎞
V1 ⎜ 2
⎟
5s 2 ⎠
⎝ s + 5s
1
+ 1 V1
s
Hence,
1
s
⎛2
⎡⎛
1
1⎞ ⎛ 2 ⎞ 1 ⎤
= V1 ⎢ 2 + ⎟ ⎜ + 1⎟ − 2 ⎥ = V1 ⎜
⎝
⎠
⎝
⎠
s
s s
s ⎦
⎝
⎣
−
1
s
0
0
1
s
⎛ s 2 2 s + 1⎞
1
1⎞
⎛2
= −V1 ⎜ + 1 + 2 ⎟ = −V1 ⎜
⎟
⎝s
s
s ⎠
s2
⎝
⎠
−
2
+1
s
I2
⎛ s 2 + 2s
2s 1 ⎞
V1 ⎜ 2
⎟
⎝ s + 5 2⎠
Y12 =
I2
s 2 + 2s
2s 1
=− 2
V1
s + 5s
5s 2
POLES AND ZEROS OF NETWORK FUNCTIONS
The network function F(s) can be written as ratio of two polynomials.
F ( s) =
N ( s) an s n + an 1 s n −1 + … + a1 s + a0
=
D( s) bm s m + bm 1 s m −1 + … + b1 s + b0
where a0, a1, …, an and b0, b1, …, bm are the coefficients of the polynomials N(s) and D(s). These are real
and positive for networks of passive elements. Let N(s) = 0 have n roots as z1, z2, ……, zn and D(s) = 0
have m roots as p1, p2, ……, pm. Then F(s) can be written as
12.8 Restrictions on Pole and Zero Locations for Transfer Functions 12.21
F ( s) = H
( s z )(s
)( s z ) …( s zn )
( s p )(s
)( s p ) …( s pm )
an
is a constant called scale factor and z1, z2, …, zn, p1, p2, …, pm are complex frequencies. When
bm
the variable s has the values z1, z2, …, zn, the network function becomes zero; such complex frequencies
are known as the zeros of the network function. When the variable s has values p1, p2, …, pm, the network
function becomes infinite; such complex frequencies are known as the poles of the network function.
A network function is completely specified by its poles, zeros and the scale factor.
If the poles or zeros are not repeated, then the function is said to be having simple poles or simple zeros. If
the poles or zeros are repeated, then the function is said to be having multiple poles or multiple zeros. When
n > m, then (n – m) zeros are at s = ∞, and for m > n, (m − n) poles are at s = ∞.
The total number of zeros is equal to the total number of poles. For any
jw
network function, poles and zeros at zero and infinity are taken into account
in addition to finite poles and zeros.
Poles and zeros are critical frequencies. The network function becomes
infinite at poles, while the network function becomes zero at zeros. The
s
0
network function has a finite, non-zero value at other frequencies.
Poles and zeros provide a representation of a network function as shown
in Fig. 12.41. The zeros are shown by circles and the poles by crosses. This
diagram is referred to as pole-zero plot.
Fig. 12.41 Pole-zero plot
where
12.7
(1)
(2)
(3)
(4)
(5)
(6)
(7)
12.8
(1)
(2)
(3)
(4)
RESTRICTIONS ON POLE AND ZERO LOCATIONS FOR DRIVING-POINT
FUNCTIONS [COMMON FACTORS IN N(S) AND D(S) CANCELLED]
The coefficients in the polynomials N(s) and D(s) must be real and positive.
The poles and zeros, if complex or imaginary, must occur in conjugate pairs.
The real part of all poles and zeros, must be negative or zero, i.e., the poles and zeros must lie in
left half of s plane.
If the real part of pole or zero is zero, then that pole or zero must be simple.
The polynomials N(s) and D(s) may not have missing terms between those of highest and lowest
degree, unless all even or all odd terms are missing.
The degree of N(s) and D(s) may differ by either zero or one only. This condition prevents
multiple poles and zeros at s = ∞.
The terms of lowest degree in N(s) and D(s) may differ in degree by one at most. This condition
prevents multiple poles and zeros at s = 0.
RESTRICTIONS ON POLE AND ZERO LOCATIONS FOR TRANSFER
FUNCTIONS [COMMON FACTORS IN N(S) AND D(S) CANCELLED]
The coefficients in the polynomials N(s) and D(s) must be real, and those for D(s) must be positive.
The poles and zeros, if complex or imaginary, must occur in conjugate pairs.
The real part of poles must be negative or zero. If the real part is zero, then that pole must be
simple.
The polynomial D(s) may not have any missing terms between that of highest and lowest degree,
unless all even or all odd terms are missing.
12.22 Network Analysis and Synthesis
(5)
(6)
(7)
(8)
The polynomial N(s) may have terms missing between the terms of lowest and highest degree,
and some of the coefficients may be negative.
The degree of N(s) may be as small as zero, independent of the degree of D(s).
For voltage and current transfer functions, the maximum degree of N(s) is the degree of D(s).
For transfer impedance and admittance functions, the maximum degree of N(s) is the degree of
D(s) plus one.
Example 12.23
(a)
Test whether the following represent driving-point immittance functions.
5s4 + 3s
3 2 − 2s + 1
s3 6s
6 20
(b)
s3 + s 2 + 5s+ 2
s4 + 6s3 + 9s 2
(c)
s 2 + 3s+ 2
s 2 + 6s+ 2
Solution
(a) The numerator and denominator polynomials have a missing term between those of highest and lowest
degree and one of the coefficient is negative in numerator polynomial. Hence, the function does not
represent a driving-point immittance function.
(b) The term of lowest degree in numerator and denominator polynomials differ in degree by two. Hence, the
function does not represent a driving-point immittance function.
(c) The function satisfies all the necessary conditions. Hence, it represents a driving-point immittance function.
Example 12.24
(a) G21 =
Test whether the following represent transfer functions.
3s+ 2
3
2
5s + 4s + 1
(b) a 12 =
2s 2 + 5s+ 1
s+7
(c) Z 21 =
1
3
s + 2s
Solution
(a) The polynomial D(s) has a missing term between that of highest and lowest degree. Hence, the function
does not represent a transfer function.
(b) The degree of N(s) is greater than D(s). Hence the function does not represent a transfer function.
(c) The function satisfies all the necessary conditions. Hence, it represents a transfer function.
Example 12.25
Obtain the pole-zero plot of the following functions.
(a) F(s) =
s(s+
( 2)
(s+ 1)(s
1 + 3)
3
(b) F(s) =
(c) F(s) =
s(s+
( 2)
(s+ 1+ j1)(s+ 1 - j1)
(d) F(s) =
(e) F(s) =
s2 + 4
(s+ 22)(s 2 + 9)
9
s(s+
( 1)
(s+ 22)2 (s+ 3)
(s+ 1)
1 2 (s+ 5)
(s+ 2)(s
2 + 3 + j2)(s+ 3 - j2)
12.8 Restrictions on Pole and Zero Locations for Transfer Functions
12.23
Solution
(a) The function F(s) has zeros at s = 0 and s = − 2 and poles at s = − 1 and s = − 3.
The pole-zero plot is shown in Fig. 12.42.
jw
s
0
−3 −2 −1
Fig. 12.42
(b) The function F(s) has zeros at s = 0 and s = −1 and poles at s = −2, −2 and s = −3.
The pole-zero plot is shown in Fig. 12.43.
jw
s
0
−3 −2 −1
Fig. 12.43
(c) The function F(s) has zeros at s = 0 and s = −2 and poles at s = −1 −j1 and s = −1 + j1.
The pole-zero plot is shown in Fig. 12.44.
jw
j1
−2
s
0
−1
−j 1
Fig. 12.44
(d) The function F (s) has zeros at s = −1, −1 and s = −5 and poles at s = −2, s = −3 + j2 and s = −3 − j2. The
pole-zero plot is shown in Fig. 12.45.
jw
j2
j1
−5 −4 −3 −2
−1
0
−j 1
−j 2
Fig. 12.45
s
12.24 Network Analysis and Synthesis
(e) The function F (s) has zeros at s = j2 and s = −j2 and poles at s = −2, s = j3 and s = −j3. The pole-zero
plot is shown in Fig. 12.46.
jw
j3
j2
j1
−2
−1
s
0
−j 1
−j 2
−j 3
Fig. 12.46
Example 12.26
Find poles and zeros of the impedance of the network shown in Fig. 12.47 and plot
them on the s-plane.
1F
1
H
2
2Ω
Z (s)
Fig. 12.47
1
s
Solution The transformed network is shown in Fig. 12.48.
s
×2
1 2
1
2s
2 s 2 + s + 4 2(s
2(( 2
.5s + 2)
Z ( s) = +
= +
=
=
s s
s s+4
s (s )
s (s )
+2
2
2 ( s + 0.25 + j1.4)( s + 0.25 − j1.4)
=
s ( + 4)
Z (s)
2
s
2
Fig. 12.48
The function Z (s) has zeros at s = −0.25 + j1.4 and s = −0.25 − j1.4 and poles at s = 0 and s = −4 as shown
in Fig. 12.49.
jw
j 1.4
−4
−3
−2
− 1 − 0.25
0
− j 1 .4
Fig. 12.49
s
12.8 Restrictions on Pole and Zero Locations for Transfer Functions
Example 12.27
12.25
Determine the poles and zeros of the impedance function Z (s) in the network
shown in Fig. 12.50.
1
Ω
2
Z (s)
4F
1
Ω
6
Fig. 12.50
Solution The transformed network is shown in Fig. 12.51.
1
2
1
4s
Z (s)
1
6
Fig. 12.51
1 1
×
1 4s 6 1
1
4s + 8
s + 2 0.. ( s + 2)
Z ( s) = +
= +
=
=
=
2 1 1 2 4 s + 6 2(( s ) 2 s + 3
s +1 5
+
4s 6
The function Z(s) has zero at s = −2 and pole at s = −1.5.
Example 12.28
Determine Z(s) in the network shown in Fig. 12.52. Find poles and zeros of Z(s)
and plot them on s-plane.
1H
1
F
20
Z (s)
4Ω
Fig. 12.52
Solution The transformed network is shown in Fig. 12.53.
s
Z (s)
20
s
Fig. 12.53
4
12.26 Network Analysis and Synthesis
20
×4
80
20
s ( s ) + 20 s 2 + 5s + 20
Z ( s) = s + s
= s+
= s+
=
=
20
4 s + 20
s+5
s+5
s+5
+4
s
( s + 2.5 + j 3.71)( s + 2.5 − j 3.71)
=
s+5
The function Z(s) has zeros at s = –2.5 + j3.71 and s = –2.5 –j 3.71 and pole at s = –5.
The pole-zero diagram is shown in Fig. 12.54.
jw
j 3.71
−5
−4
−3
−2
s
0
−1
−j 3.71
Fig. 12.54
Example 12.29
For the network shown in Fig. 12.55, plot poles and zeros of function
I0
.
Ii
I0
4Ω
Ii
0.5 F
2H
Fig. 12.55
Solution The transformed network is shown in Fig. 12.56.
By current-division rule,
I0
⎛
⎞
⎜ 4 2s ⎟
Ii ⎜
2⎟
⎜ 4 + 2s + ⎟
⎝
s⎠
I0
4
2
s
Ii
I0
s ( 4 2 s)
s ( s + 2)
s ( s + 2)
=
=
=
I i 4 s 2s2 + 2 s2 + 2s + 1 ( s +
+1)( s + 1)
The function has zeros at s = 0 and s = –2 and double poles at s = –1.
The pole-zero diagram is shown in Fig. 12.57.
2s
Fig. 12.56
12.8 Restrictions on Pole and Zero Locations for Transfer Functions
jw
−4 −3
−2
s
0
−1
Fig. 12.57
Example 12.30
Draw the pole-zero diagram of I 2 for the network shown in Fig. 12.58.
I1
I2
10 H
250 μF
I1
200 Ω
Fig. 12.58
Solution The transformed network is shown in Fig. 12.59.
I2
10s
1
I1
250 × 10−6s
200
Fig. 12.59
By current-division rule,
1
250 × 10 −6 s
I 2 I1
1
+ 10 s + 200
250 × 10 −6 s
I2
400
400
= 2
=
I1 s + 20 s + 400 ( s + 10
1 − j17
17 32)( s + 10 + j17.32)
The function has no zero and poles at s = –10 + j17.32 and s = −10 −j17.32.
The pole-zero diagram is shown in Fig. 12.60.
jw
j17.32
−10
0
− 17.32
−j
Fig. 12.60
s
12.27
12.28 Network Analysis and Synthesis
Example 12.31
I1
For the network shown in Fig. 12.61, draw pole-zero plot of
Vc
.
V1
1Ω
+
V1
−
+
+
−
2Vc
1
F
2
1H
5I1
Vc
−
Fig. 12.61
Solution The transformed network is shown in Fig. 12.62.
I1
1
c
+
V1
−
+
+
−
2Vc
2
s
s
5I1
Vc
−
Fig. 12.62
Applying KVL to the left loop,
V1 − 1I1 + 2Vc = 0
I1 = V1 + 2Vc
Applying KCL at Node C,
Vc Vc
+
2
s
s
Vc s
5 (V1 2Vc ) + + Vc
s 2
Vc s
5V1 10 Vc + + Vc
s 2
⎛ 20 s + 2 + s 2 ⎞
Vc ⎜
⎟
2s
⎝
⎠
Vc
V1
5 I1 +
=0
=0
=0
= −5V1
=−
10 s
2
s + 20 s + 2
=−
10 s
( s + 0.1)( s + 19.9)
jw
The function has zero at s = 0 and poles at s = − 0.1 and s = −19.9.
The pole-zero diagram is shown in Fig. 12.63.
−20
−2
− 1 − 0.1
Fig. 12.63
0
s
12.8 Restrictions on Pole and Zero Locations for Transfer Functions
Example 12.32
12.29
Find the driving point admittance function and draw pole-zero plot for the network
shown in Fig. 12.64.
I1
2
2
+
V1
−
+
+
−
0.1V2
10I1
0.5s
1
V2
−
Fig. 12.64
Solution Applying KVL to the left loop,
V1 2 I1 0.1 V2
0
V 0 1 V2
I1 = 1
2
...(i)
Applying KCL at Node 2,
10 I1 +
V2 V2
+
=0
0 5s 1
2
10 I1 + V2 V2 = 0
s
10 I1 +
⎛2 ⎞
+1
⎝s ⎠
2
0
⎛ 2 + s⎞
10 I1 + ⎜
V2 = 0
⎝ s ⎟⎠
⎛ s + 2⎞
⎜⎝
⎟ V2
s ⎠
10 I1
V2 = −
10 s
I1
s+2
Substituting Eq. (ii) in Eq. (i),
V1 + 0 1
I1 =
⎛ 10 s ⎞
I
⎝ s + 2⎠ 1
⎛ 10 s ⎞
= 0.55V
V1 + 0 05
I1
⎝ s + 2⎠
2
0 5s ⎞
⎛
I1 ⎜1 +
= 0 5V1
⎝ s + 2 ⎟⎠
Hence,
Y11 =
I1
=
V1
05
0.5 ( s + 2) 0 5s + 1
=
=
0 5s s + 0 5s + 2 1 5s + 2
1+
s+2
...(ii)
12.30 Network Analysis and Synthesis
The function has zero at s = −2 and pole at s = −1.33. The pole-zero diagram is shown in Fig 12.65.
jw
− 3 − 2 −1.33 − 1
s
0
Fig. 12.65
Example 12.33
For the network shown in Fig. 12.66, determine
V2
V
. Plot the pole-zero diagram of 2 .
Ig
Ig
I2 = 0
1H
+
1Ω
Ig
1F
1F
V2
−
Fig. 12.66
Solution The transformed network is shown in Fig. 12.67.
s
Va
Ib Vb
I2 = 0
Vc
+
Ig
1
1
s
1
s
1
V2
−
Fig. 12.67
Hence,
Vc
Vb = V2
Ib =
Vb Vc
+
= sV
V2 + V2
1
1
s
Va
s I b Vb = s ( s + 1)V2 V2 = ( s 2 + s + 1)V2
Ig =
Va Va
+ + I b = ( s 2 + s + 1)V2
1
1
s
1
V2
=
I g s3
2s2
( s + 1)V2
s ( s2
s +11)V2 + ( s +11)V2 = ( s3 + 2
3s + 2
The function has no zeros. It has poles at s = −1, s = −0.5 + j1.32 and s = −0.5 –j1.32,
The pole-zero diagram is shown in Fig. 12.68.
2
+ 3s + 2)V2
12.8 Restrictions on Pole and Zero Locations for Transfer Functions
12.31
jw
j1.32
−2
−1
s
0
− 0.5
−j1.32
Fig. 12.68
Example 12.34
For the transfer function H(s) =
network shown in Fig. 12.69. Find L and C when R = 5 W.
V0
10
= 2
, realise the function using the
Vi s + 3s+ 10
L
+
Vi
+
−
C
R
V0
−
Fig. 12.69
Solution The transformed network is shown in Fig. 12.70.
Ls
+
Vi
+
−
1
Cs
R
V0
−
Fig. 12.70
Simplifying the network as shown in Fig. 12.71,
Ls
+
Vi
+
−
Z (s)
V0
−
Fig. 12.71
1
R
Cs
Z ( s) =
=
1
RCs
C +1
R+
Cs
R×
12.32 Network Analysis and Synthesis
R
C +1
Vi RCs
R
Ls +
RCs
C +1
V0
1
V0
R
LC
=
=
1
1
Vi RLC s 2 + Ls + R
s2 +
s+
RC
LC
V0
10
=
Vi s 2 + 3s
3s 10
But
...(i)
...(ii)
R=5Ω
and
Comparing Eq. (ii) with Eq. (i),
1
=3
RC
1
= 10
LC
Solving the above equations,
L = 1.5 H
1
C= F
15
Example 12.35
Obtain the impedance function Z(s) for which pole-zero diagram is shown in Fig. 12.72.
jw
Z (∞ ) = 1
−3
−2
−1
0
s
Fig. 12.72
Solution The function Z(s) has poles at s = –1 and s = –3 and zeros at s = 0 and s = –2.
s (s )
Z ( s) = H
=H
( s )(s
)(
)
For s = ∞,
Z( )
When
H
1
=H
( )( )
Z(∞) = 1,
H=1
Z ( s) =
s (s )
( s ))(s
(
(s
)
⎛ 2⎞
s 2 ⎜1 + ⎟
⎝
s⎠
⎛ 1⎞ ⎛ 3⎞
s 2 ⎜1 + ⎟ ⎜1 + ⎟
⎝ s⎠ ⎝ s⎠
12.8 Restrictions on Pole and Zero Locations for Transfer Functions
Example 12.36
12.33
Obtain the admittance function Y(s) for which the pole-zero diagram is shown in
Fig. 12.73.
jw
Y( ∞ ) = 1
j1
−2
s
0
−1
−j 1
Fig. 12.73
Solution The function Y(s) has poles at s = –1 + j1 and s = −1 −j1 and zeros at s = 0 and s = −2.
Y ( s) = H
s (s )
j )( s
(s
j)
=H
s (s
(s
)2
⎛ 2⎞
s 2 ⎜1 + ⎟
⎝
s⎠
=H 2
=H
2
⎛ 2 2⎞
(j )
s + 2s + 2
s 2 ⎜1 + + 2 ⎟
⎝
s s ⎠
)
s (s
)
For s = ∞,
Y( )
When
H
()
=H
()
Y (∞) = 1,
H=1
s (s )
Y ( s) = 2
s + 2s + 2
Example 12.37 A network and its pole-zero configuration are shown in Fig. 12.74. Determine the
values of R, L and C if Z (j0) = 1.
jw
j
√111
R
1
Cs
Z (s)
Ls
−3
− 1.5
5
2
s
0
−j
√111
2
Fig. 12.74
1⎛
R⎞
1
⎜⎝ s + ⎟⎠
Ls
+
R
C
L
Cs =
Z ( s) =
=
2
1
LCs
C
RCs
RCs + 1 s 2 + R s + 1
( Ls R) +
Cs
L
LC
( Ls R)
Solution
...(i)
12.34 Network Analysis and Synthesis
From the pole-zero diagram, zero is at s = –3 and poles are at s
15
j
111
2
ds
15
j
111
2
s+3
111 ⎞ ⎛
111 ⎞
j
s 15 j
⎟
⎜
2 ⎠⎝
2 ⎟⎠
s+3
s+3
Z ( s) = H
=H 2
2
s + 3 + 30
⎛ 111 ⎞
( s + 1.5)2 − ⎜ j 2 ⎟
⎝
⎠
Z ( s) = H
When
⎛
⎜s 1 5
⎝
Z (j0) = 1,
⎛ 3⎞
1= H ⎜ ⎟
⎝ 30 ⎠
H = 10
10 ( s + 3)
Z ( s) = 2
s + 3 30
...(ii)
Comparing Eq. (ii) with Eq. (i),
R
=3
L
1
= 10
C
1
= 30
LC
Solving the above equations,
1
F
10
1
L= H
3
R =1Ω
C=
Example 12.38
A network is shown in Fig. 12.75. The poles and zeros of the driving-point function
Z(s) of this network are at the following places:
1
3
Poles at − ± j
2
2
Zero at –1
If Z (j0) = 1, determine the values of R, L and C.
R
Z (s)
1
Cs
Ls
Fig. 12.75
12.35
12.8 Restrictions on Pole and Zero Locations for Transfer Functions
1⎛
R⎞
1
⎜⎝ s + ⎟⎠
Ls
+
R
C
L
Cs =
Z ( s) =
=
2
1
R
LCs
C
RCs
RCs + 1 s 2 + s + 1
Ls + R +
Cs
L
L
LC
( Ls R)
Solution
...(i)
1
3
The poles are at − ± j
and zero is at –1.
2
2
Z ( s) = H
⎛
1
⎜s 2
⎝
s +1
3⎞ ⎛
1
j
s
⎟
⎜
2 ⎠⎝
2
j
3⎞
2 ⎟⎠
s +1
=H
1⎞
⎛
⎜⎝ s
2⎠
2
⎛
⎝
j
3⎞
2 ⎟⎠
2
=H
s +1
2
s + s +1
Z (j0) = 1,
When
1= H
(1)
(1)
H =1
Z ( s) =
s +1
...(ii)
2
s + s +1
Comparing Eq. (ii) with Eq. (i),
C =1
R
=1
L
1
=1
LC
Solving the above equations,
C=1F
L=1H
R=1Ω
Example 12.39 The pole-zero diagram of the driving-point impedance function of the network of
Fig. 12.76 is shown below. At dc, the input impedance is resistive and equal to 2 W. Determine the values of
R, L and C.
jw
j4
Z (s)
1
Cs
R
Ls
−2
−1
0
s
−j 4
Fig. 12.76
1⎛
R⎞
1
⎜⎝ s + ⎟⎠
Ls
+
R
C
L
Cs =
Z ( s) =
=
2
1
R
LCs
C
RCs + 1 s 2 + s + 1
RCs
Ls + R +
Cs
L
L
LC
( Ls R)
Solution
From the pole-zero diagram, zero is at s = –2 and poles are at s = –1+ j4 and s = −1 − j4.
...(i)
12.36 Network Analysis and Synthesis
Z ( s) = H
(s
s+2
j )( s
j )
=H
s+2
(s
)2
( j )2
=H
s+2
s 2 + 2 s + 17
w = 0, Z ( j0) = 2
At dc, i.e.,
2= H
H = 17
Z ( s) = 17
2
17
s+2
...(ii)
s 2 + 2 s 17
Comparing Eq. (ii) with Eq. (i),
1
= 17
C
R
=2
L
1
= 17
LC
Solving the above equations,
1
F
17
L =1H
R=2Ω
C=
Example 12.40
The network shown in Fig. 12.77 has the driving-point admittance Y (s) of the form
Y(s)
( =H
(s − s1 )(s s )
(s − s )
(a) Express s1, s2, s3 in terms of R, L and C.
(b) When s1 = –10 + j104, s2 = –10 – j104 and Y (j0) = 10–1 mho, find the values of R, L and C and
determine the value of s3.
Y (s)
Ls
1
Cs
R
Fig. 12.77
Solution
R
1 ⎞
⎛
C s2 + s +
⎟
⎝
1
( Ls R) Cs
(Ls
C + 1 LCs
C
RCs + 1
RCs
L
LC ⎠
Y ( s) = Cs +
=
=
=
R
Ls + R
Ls + R
Ls + R
s+
L
H ( s s ))(s
(s s )
Y ( s) =
(s s )
2
(a)
But
...(i)
12.8 Restrictions on Pole and Zero Locations for Transfer Functions
12.37
2
−
s1 s2 =
where
R
4
⎛ R⎞
± ⎜ ⎟ −
2
⎝ L⎠
L
LC
R
1
⎛ R⎞
=−
± ⎜ ⎟ −
⎝ 2L ⎠
2
2L
LC
R
L
s1 = −10 + j104
s2 = − 10 − j104
s3 = −
(b) When
Y ( s) = H
(s
j
4
))(( s
s s3
j
4
)
s 2 + 200 s + 108
s s3
=H
...(ii)
Comparing Eq. (ii) with Eq. (i),
R
= 20
L
s3 = −20
Y ( s) = H
( s 2 + 200 s + 108 )
( s + 20)
At s = j0,
(
8
)
= 10 −1
20
H = 0.02 10 −6
Y( j ) = H
Y ( s) = 0.02 10 −6
( s2
s
( s + 20)
8
)
...(iii)
Comparing Eq. (iii) with Eq. (i),
C = 0.02
0 × 10 −6 = 0.02 μF
1
= 108
LC
1
L= H
2
R
= 20
L
R = 10 Ω
Example 12.41
A network and pole-zero diagram for driving-point impedance Z(s) are shown in
Fig. 12.78. Calculate the values of the parameters R, L, G and C if Z(j0) = 1.
jw
j3
j2
j1
R
Z (s)
G
1
Cs
Ls
Fig. 12.78
−3
−2
−1
0
s
− j1
− j2
− j3
12.38 Network Analysis and Synthesis
Solution It is easier to calculate Y(s) and then invert it to obtain Z(s).
Y ( s) = G Cs
C +
Z( ) =
1
(G Cs
C )( Lss R) + 1 LCs
C 2
=
=
Ls + R
Ls + R
1
=
Y ( ) LCs
C 2
(
(GL
G
RC ) s + 1 + GR
Ls + R
1⎛
R⎞
⎜ s + ⎟⎠
Ls + R
C⎝
L
=
G
R
) s + 1 + GR s 2 + ⎛ + ⎞ s + ⎛ 1 + GR ⎞
⎜⎝
⎟
⎝ C L ⎟⎠
LC ⎠
...(i)
From the pole-zero diagram, zero is at s = –2 and poles are at s = –3 ± j3.
Z ( s) = H
When
(s
(s )
j )( s
j )
=H
(s
(s
)
2
)
(j )
2
=H
s+2
2
s + 6 s + 18
Z (j0) = 1,
2
18
1= H
H =9
Z( s) =
9(
2
(s + 6
2)
...(ii)
18)
Comparing Eq. (ii) with Eq. (i),
1
C
R
L
G R
+
C L
1 + GR
LC
=9
=2
=6
= 18
Solving the above equation,
1
F
9
9
L=
H
10
4
G= �
9
9
R= Ω
5
C=
Example 12.42
A series R-L-C circuit has its driving-point admittance and pole-zero diagram is
shown in Fig. 12.79. Find the values of R, L and C.
12.9 Time-Domain Behaviour from the Pole-Zero Plot 12.39
jw
j 25
Scale factor = 1
s
0
−1
−j 25
Fig. 12.79
Solution The function Y (s) has poles at s = –1 + j25 and s = –1 – j25 and zero at s = 0.
Y ( s) = H
(s
j
s
))(s
(s
j
)
=H
s
(s
)
2
(j
)
2
=H
s
2
s + 2 s + 626
H=1
Scale factor
Y ( s) =
s
...(i)
s 2 + 2 s + 626
For a series RLC circuit,
R
1 ⎞
⎛
L s2 + s +
⎟
⎝
1
LCs
C
R
RCs
Cs + 1
C
L
LC ⎠
Z ( s) = R + Ls +
=
=
Cs
Cs
s
1
s
Y ( s) =
=
R
1 ⎞
Z ( s)
⎛
L ⎜ s2 + s +
⎟
⎝
L
LC ⎠
2
...(ii)
Comparing Eq. (i) with Eq. (ii),
L =1H
1
= 626
LC
1
C=
F
626
R
=2
L
R=2Ω
12.9
TIME-DOMAIN BEHAVIOUR FROM THE POLE-ZERO PLOT
The time-domain behaviour of a system can be determined from the pole-zero plot. Consider a network
function
F ( s) = H
( s z )(s
)( s z )… ( s zn )
( s p )(s
)( s p )… ( s pm )
The poles of this function determine the time-domain behaviour of f(t).The function f(t) can be determined
from the knowledge of the poles, the zeros and the scale factor H. Figure 12.80 shows some pole locations
and the corresponding time-domain response.
12.40 Network Analysis and Synthesis
(i) When pole is at origin, i.e., at s = 0, the function f(t) represents steady-state response of the
circuit i.e., dc value. (Fig. 12.80)
jw
f (t)
0
s
t
0
Fig. 12.80 Pole at origin
(ii)
When pole lies in the left half of the s-plane, the response decreases exponentially. (Fig. 12.81)
jw
0
f (t)
s
0
t
Fig. 12.81 Pole in left half of the s-plane
(iii)
When pole lies in the right half of the s-plane, the response increases exponentially. A pole in the
right-half plane gives rise to unbounded response and unstable system. (Fig. 12.82)
jw
0
f (t)
s
0
t
Fig. 12.82 Pole in right half of the s-plane
(iv)
For s = 0 +jwn, the response becomes f (t) = Ae± jwnt = A(cos wnt ± j sin wnt).The exponential
response e ± jwnt may be interpreted as a rotating phasor of unit length. A positive sign of
exponential e jwnt indicates counterclockwise rotation, while a negative sign of exponential
e−jwnt indicates clockwise rotation. The variation of exponential function e jwnt with time is thus
sinusoidal and hence constitutes the case of sinusoidal steady state. (Fig. 12.83)
f (t)
jw
0
s
0
Fig. 12.83 Poles on jw -axis
t
12.9 Time-Domain Behaviour from the Pole-Zero Plot 12.41
(v)
For s = sn + jwn, the response becomes f (t) = Aest = Ae(sn+jwnt) = Aesnt ejwnt. The response esnt is
an exponentially increasing or decreasing function. The response ejwnt is a sinusoidal function.
Hence, the response of the product of these responses will be over damped sinusoids or under
damped sinusoids (Fig. 12.84).
f (t)
jw
s
0
t
0
(a)
f (t)
jw
s
0
t
0
(b)
Fig. 12.84 (a) Complex conjugate poles in left half of the S-plane
(b) Complex conjugate poles in right half of the S-plane
(vi)
The real part s of the pole is the displacement of the pole from the imaginary axis. Since s is
also the damping factor, a greater value of s (i.e., a greater displacement of the pole from the
imaginary axis) means that response decays more rapidly with time. The poles with greater
displacement from the real axis correspond to higher frequency of oscillation (Fig. 12.85).
f (t)
jw
0
0
t
s
f (t)
0
t
(a)
Fig. 12.85 Nature of response with different positions of poles
12.42 Network Analysis and Synthesis
f (t)
jw
t
0
s
0
f (t)
t
0
(b)
Fig. 12.85 (Continued)
12.9.1 Stability of the Network
Stability of the network is directly related to the location of poles in the s-plane.
(i)
(ii)
(iii)
(iv)
(v)
When all the poles lie in the left half of the s-plane, the network is said to be stable.
When the poles lie in the right half of the s-plane, the network is said to be unstable.
When the poles lie on the jw axis, the network is said to be marginally stable.
When there are multiple poles on the jw axis, the network is said to be unstable.
When the poles move away from jw axis towards the left half of the s-plane, the relative stability of
the network improves.
12.10
GRAPHICAL METHOD FOR DETERMINATION OF RESIDUE
Consider a network function,
F ( s) = H
( s z )(s
)( s z )
( s p )(s
)( s p )
( s zn )
( s pm )
By partial fraction expansion,
F ( s) =
K1
K2
+
+
(s p ) (s p )
+
Km
( s pm )
( s − p ) F ( s ) |s → p
(p
(p
The residue Ki is given by
Ki
i
H
z )(
)( p
p )(
)( p
z )
p )
(p
(p
zn )
pm )
Each term (pi – zi) represents a phasor drawn from zero zi to pole pi.
Each term (pi – pk), i ≠ k, represents a phasor drawn from other poles to the pole pi.
Ki
H
Product of phasors (polar form) from each zero to pi
Product of phasor
a s (polar form) from other poles to pi
12.10 Graphical Method for Determination of Residue 12.43
The residues can be obtained by graphical method in the following way:
(1)
(2)
(3)
(4)
(5)
(6)
Draw the pole-zero diagram for the given network function.
Measure the distance from each of the other poles to a given pole.
Measure the distance from each of the other zeros to a given pole.
Measure the angle from each of the other poles to a given pole.
Measure the angle from each of the other zeros to a given pole.
Substitute these values in the required residue equation.
The graphical method can be used if poles are simple and complex. But it cannot be used when there are
multiple poles.
2s
The current I(s) in a network is given by I(s)
s =
. Plot the pole-zero
(s
+
1)(s
+ 2)
pattern in the s-plane and hence obtain i(t).
Example 12.43
Solution Poles are at s = –1 and s = –2 and zero is at s = 0. The pole-zero plot is shown in Fig. 12.86.
By partial-fraction expansion,
I ( s) =
K1
K
+ 2
s +1 s + 2
jw
s
0
−2 −1
Fig. 12.86
The coefficients K1 and K2, often referred as residues, can be evaluated from he pole-zero diagram. From
Fig. 12.87,
Phasor from zero at origin to pole at A
⎛ 1∠180° ⎞
K1 H
= 2⎜
= 2 ∠180° = −2
⎝ 1∠0° ⎟⎠
Phasor from pole at B to pole at
a A
jw
B
A
s
−2 −1
0
Fig. 12.87
From Fig. 12.88,
K2
H
⎛ 2 ∠180
Phasor from zero at origin to pole at B
80° ⎞
= 2⎜
=4
Phasor from pole at A to pole at
a B
⎝ 1 80° ⎟⎠
12.44 Network Analysis and Synthesis
jw
B
A
s
0
−2 −1
Fig. 12.88
2
4
+
s +1 s + 2
Taking inverse Laplace transform,
i (t) = –2e−t + 4e−2t
I ( s) = −
Example 12.44
The voltage V(s) of a network is given by
V ( s) =
3s
(s
)(s
)(
)( s 2
s
)
Plot its pole-zero diagram and hence obtain v (t).
Solution
V ( s) =
3s
(s
)( s
)(s
)(
2
s
)
=
(s
)(s
)(
)( s
3s
j ))(s
(s
j)
Poles are at s = –2 and s = –1 ± j1 and zero is at s = 0 as shown in Fig. 12.89.
jw
B
j1
A
−2
0
−1
C
s
−1
−j
Fig. 12.89
By partial-fraction expansion,
V ( s) =
K1
K2
K 2*
+
+
s + 2 s 1 j1 s 1 j1
The coefficients K1, K2 and K2* can be evaluated from the pole-zero diagram.
From Fig. 12.90,
K1 =
⎡
⎤
2∠180
80°
= 3⎢
⎥ = 3 180° = −3
( BA) (CA)
35 ) ( 2 135
35 ) ⎦
⎣ ( 2 − 135
3 (OA)
12.10 Graphical Method for Determination of Residue 12.45
jw
B
j1
√2
135°
A
180°
−2
s
0
−1
135°
√2
−j 1
C
Fig. 12.90
From Fig. 12.91,
⎡
( 2 135
35 )
= 3⎢
( AB ) (CB)
5 ) ( 2 90
⎣ ( 2 45
3
K 2* =
2
K2 =
3 (OB)
⎤ 3
⎥=
)⎦ 2
jw
B
j1
√2
√2
A
45°
−2
−1
135°
0
s
90°
−1
−j
C
Fig. 12.91
V ( s) = −
3
(s
)
+
3
2
(s
j)
+
3
2
(s
j)
Taking inverse Laplace transform,
v(t ) = −3e −2tt
Example 12.45
3 (
⎡e
2⎣
j )t
)t
+ e(
j )t ⎤
⎦
3e
2t
⎛ e j1 + e − j1 ⎞
3
+ 2 × e−t ⎜
⎟ = −3e
2
2
⎝
⎠
2t
3e t cos t
Find the function v(t) using the pole-zero plot of following function:
V(s) =
(s + 2)(s + 6)
(s + 1)(s + 5)
Solution If the degree of the numerator is greater or equal to the degree of the denominator, we have to
divide the numerator by the denominator such that the remainder can be expanded into partial fractions.
12.46 Network Analysis and Synthesis
V ( s) =
s 2 + 8 s + 12
2
s + 6s + 5
= 1+
2s + 7
2
s + 6s + 5
= 1+
2(s
(
.5)
( s )(s
)(
)
By partial fraction expansion,
V ( s) = 1 +
K1
K
+ 2
s +1 s + 5
K1 and K2 can be evaluated from the pole-zero diagram shown in Fig. 12.92 and Fig. 12.93.
jw
jw
s
s
0
−5
− 3.5
0
−1
−5
−3.5
Fig. 12.92
−1
Fig. 12.93
From Fig. 12.92
⎛ 2 5∠0° ⎞ 5
K1 = 2 ⎜
=
⎝ 4 ∠0° ⎟⎠ 4
From Fig. 12.93
80° ⎞ 3
⎛ 1 5∠180
K2 = 2 ⎜
=
⎝ 4 ∠180° ⎟⎠ 4
5
3
4
4
V ( s) = 1 +
+
s +1 s + 5
Taking inverse Laplace transform,
5 −tt 3
e + e
4
4
v (t ) = δ (t
(t )
Example 12.46
5t
The pole-zero plot of the driving-point impedance of a network is shown in
Fig. 12.94. Find the time-domain response.
jw
j1
−1
Scale factor = 5
0
s
−1
−j
Fig. 12.94
Solution The function Z(s) has poles at s = −1 + j1 and s = −1 −j1 and zero at s = 0.
12.10 Graphical Method for Determination of Residue 12.47
Z ( s) = H
s
j )( s
(s
j)
H=5
Scale factor
Z ( s) =
jw
5s
j )( s
(s
j)
j1
A
By partial fraction expansion,
Z ( s) =
K1
K1*
+
s 1 j1 s 1 j1
s
−1
−j
B
d K1* can be evaluated from the pole-zero
The coefficients K1
0
−1
Fig. 12.95
diagram. From Fig. 12.95,
K1 =
5 (OA)
( BA)
=
5 ( 2 135
35 )
= 3 54 ∠455°
2∠90°
K1*
= 3 54 ∠ 45
5°
3.54 ∠45
5° 3.54 ∠ 455°
Z ( s) =
+
s 1 j1
s 1 j1
Taking inverse Laplace transform,
z(t) = 3.54 ∠ 45° e(−1−j1)t + 3.54 ∠−45°e(−1+j1)t
Evaluate amplitude and phase of the network function F(s) =
Example 12.47
pole-zero plot at s = j2.
F ( s) =
Solution
4s
s2 + 2s + 2
=
(s
4s
j )( s
4s
s
2
2s 2
from the
j)
The pole-zero plot is shown in Fig. 12.96.
At s = j2,
jw
j2
j1
−1
s
0
−j 1
Fig. 12.96
|
|=
Product of phasor magnitudes from all zero to j 2
2
=
= 0.447
Product of phaso
h r magnitudes from all poles to j 2 ( 2 ) ( 10 )
12.48 Network Analysis and Synthesis
φ ( ) = tan −1
Example 12.48
⎛ 2⎞
⎝ 0⎠
⎛ 3⎞
⎛ 1⎞
tan −1 ⎜ ⎟ − tan −1
⎝ 1⎠
⎝ 1⎠
90° 71.56° − 45
4 ° = −26.56°
Using the pole-zero plot, find magnitude and phase of the function
Solution
F ( s) =
(s
)(
)(s
)
at s
s (s )
F ( s) =
(s
)(
)(s
)
s (s )
j 4.
The pole-zero plot is shown in Fig. 12.97.
At s = j4,
jw
j4
s
−3
−2
−1
0
Fig. 12.97
⎪ ( j )⎪=
φ (ω ) = tan −1
⎛ 4⎞
⎝ 1⎠
Example 12.49
Product of phasor magnitudes from all zeros to j 4 ( ) (
)
=
= 1.15
product of phaso
p
r magnitudes from all poles to j 4 (
)( )
⎛ 4⎞
⎛ 4⎞
tan −1 ⎜ ⎟ − tan1
⎝ 3⎠
⎝ 0⎠
⎛ 4⎞
tan −1 ⎜ ⎟ = 75.96° + 53.13° − 90° − 63.43° = −24.34°
⎝ 2⎠
Plot amplitude and phase response for
F(s) =
s
s 10
Solution
F( j ) =
⎪F(( j )⎪=
jω
jω + 10
ω
ω 2 + 100
12.10 Graphical Method for Determination of Residue 12.49
v
ÃF(jv)Ã
0
0
10
0.707
100
0.995
1000
1
1
The amplitude response is shown in Fig. 12.98.
φ (ω ) = tan −1
⎛ω⎞
⎝ 0⎠
F(jw)
0.7
w
0
10
100
1000
Fig. 12.98
⎛ω⎞
⎛ω⎞
tan −1 ⎜ ⎟ = 90° − tan −1 ⎜ ⎟
⎝ 10 ⎠
⎝ 10 ⎠
f (w)
v
e (v)
0
90°
10
45°
100
5.7°
1000
0°
90°
45°
w
10
The phase response is shown in Fig. 12.99
100
1000
Fig. 12.99
Example 12.50
Sketch amplitude and phase response for F(s) =
s 10
s 10
Solution
F(jw)
jω + 10
F( j ) =
jω − 10
⎪F ( fω )⎪=
1
ω 2 + 100
ω 2 + 100
w
0
For all w, magnitude is unity.
The amplitude response is shown in Fig. 12.100.
φ (ω ) = tan −1
⎛ω⎞
⎝ 10 ⎠
⎛ ω⎞
⎛ω⎞
tan −1 ⎜ − ⎟ = 2 tan −1 ⎜ ⎟
⎝ 10 ⎠
⎝ 10 ⎠
The phase response is shown in Fig. 12.101.
v
e(v)
0
0°
10
90°
100
168.6°
1000
178.9°
Fig. 12.100
f (w)
180°
90°
w
0
10
100
1000
Fig. 12.101
12.50 Network Analysis and Synthesis
Exercises
12.1 Determine the driving-point impedance
transfer impedance
ratio
I1
V1
,
I1
V2
and voltage transfer
I1
V2
for the network shown in Fig. 12.102.
V1
5Ω
1
I1
2H
V1
1′
1
F
2
2H
V1
2F
2F
V2
−
−
2′
Fig. 12.104
1
1
⎡
⎤
, G21 = 4
⎢ Z 21 = 3
⎥
2
2
s
3
s
4
s
7
s
+
1
⎣
⎦
V2
12.4 For the two-port network shown in Fig.
12.105, determine Z11, Z21 and voltage transfer
ratio G21(s).
−
−
2
+
+
+
I2 = 0
+
I2 = 0
2Ω
1H
Fig. 12.102
I1
⎡V1 7 s 7 s + 5 V2
2s
; = 2
;
⎢ =
s 2 s 1 I1 s + s + 1
⎣ I1
V2
2s
⎤
= 2
V1 7 s 7 s + 5 ⎥⎦
2
2Ω
2H
+
+
1F
V1
1H
−
V2
−
Fig. 12.105
12.2 For the network shown in Fig. 12.103,
V
V
determine 2 and 2 .
V1
I1
2H
I1
+
V1
I2 = 0
1 F
2
1
F
2
+
1
F
2
12.5 Draw the pole-zero diagram of the following
network functions:
V2
−
−
⎡
2ss3
s 2 + 3s 2
s
, Z 21 = 2
,
⎢ Z11 =
2
s + 2s
2s 1
s + 2s
2s 1
⎣
s
⎤
G21 = 3
⎥
2
2ss
s + 3s 2 ⎦
(i) F ( s) =
Fig. 12.103
(ii) F ( s) =
⎡V2
s 2 + 1 V2
2s2 2 ⎤
; =
⎢ = 2
⎥
2
⎣ V1 2 s 1 I1 s (3s 2) ⎦
(iii) F ( s) =
12.3 Find the open-circuit transfer impedance Z21
and open-circuit voltage ratio G21 for the
ladder network shown in Fig. 12.104.
(iv) F ( s) =
s2 + 4
s2 + 6s + 4
5 s − 12
2
s + 4 s + 13
s +1
(s
2
s ( s2
4
)2
s
)
2
s + s +1
Exercises 12.51
2
(v) F s
s2
(vi) F s =
Z (s
s
1
s
s+
2
Fig. 12.108
s
s
⎡
⎢Z s
⎣
s
s
s
s
12.6 For the network shown in Fig. 12.106, draw
the pole-zero plot of the impedance function
Z(s).
)
V s
V s
1Ω
I1
4Ω
s + 0.
s
V
V
and
for the
V
I1
network shown in Fig. 12.109 and plot poles
and zeros of
1 F
20
1. s s 2 + 0
12.9 Find network functions
2
Z (s)
H
s −s 2
s2
(viii) F s =
2F
+ s2
s
(vii) F s
1F
s 2
2
+
+
V
V
1
−
1Ω
Fig. 12.106
Z s
⎡ s
j1.
⎤
s
s+5
⎡V
12.7 For the network shown in Fig. 12.107, draw
the pole-zero plot of driving-point impedance
function Z(s).
5Ω
Z (s)
Fig. 12.109
10 Ω
1
⎣V
2s
2s
s + 2s 1 ⎤
2s
s + 2s 1 ⎦
12.10 For the network shown in Fig. 12.110,
V
V
determine
and
Plot the poles and
I1
I1
zeros of
10
I1
V
.
I1
2H
1H
+
Fig. 12.107
Z s =
V
s + 2 s 1 I1
5(s
s s
s + 0.
.03)
12.8 Find the driving-point impedance of the
network shown in Fig. 12.108. Also, find
poles and zeros.
V
+
1
1
2
V
Fig. 12.110
V
s
s +2 V
=
,
I1
I1
s
s
2
s
⎤
⎥
s⎦
12.52 Network Analysis and Synthesis
12.11 For the network shown in Fig. 12.111,
V
V
determine 1 and 2 . Plot the pole and zeros
I1
V1
V2
for
.
V1
I1
12.14 Obtain the impedance function for which the
pole-zero diagram is shown in Fig. 12.114.
jw
j1
2F
1F
Z ( j 0) = 1
+
+
1H
V1
1H
−2
0
−1
−j1
V2
−
−
Fig. 12.114
Fig. 12.111
⎡V1 s + 3s
3s 1⎤
⎢ =
⎥
3
2s + s ⎦
⎣ I1
4
1H
1H
+
+
1Ω
1Ω
V1
2 (s
(s ) ⎤
⎡
⎢ Z ( s) = 2
⎥
s + 2s + 2 ⎦
⎣
2
12.12 For the network shown in Fig. 12.112, plot the
poles and zeros of transfer impedance function.
I1
s
12.15 For the network shown in Fig. 12.115, poles
and zeros of driving point function Z(s)
are,
Poles: (–1 ± j4); zero: –2
If Z (j0) = 1, find the values of R, L and C.
V2
−
−
R
1
Cs
Z (s)
Ls
Fig. 12.112
1 ⎤
⎡V2
⎢ I = s + 2⎥
⎣ 1
⎦
12.13 For the network shown in Fig. 12.113,
V
V
determine 1 and 2 . Plot the poles and
I1
V1
zeros of transfer impedance function.
I1
2H
4H
2
⎡
⎢1 Ω, 0.5 Η, 17
⎣
R1
1H
2F
−
1 F V2
+
+
−
Fig. 12.113
⎡V1 16 s 4 + 100 s 2 + 1 V2
1
, = 3
,
⎢ =
3
I
I
8s 3s
8s 3s
1
⎣ 1
V2
1
⎤
=
⎥
4
2
V1 16 s +10
0 s + 1⎦
⎤
F⎥
⎦
12.16 For the two-port network shown in Fig.
V
02
12.116, find R1, R2 and C. 2 = 2
V1 s + 3s
3s 2
+
+
V1
Fig. 12.115
V1
C
−
R2
V2
−
Fig. 12.116
1
⎡3
⎤
⎢ 5 Ω, 15 Ω, 0.5 F⎥
⎣
⎦
Objective-Type Questions 12.53
12.17 For the given network function, draw the
pole-zero diagram and hence obtain the time
domain voltage.
V ( s) =
5 (s
(
)
( s )(s
)(s
(
)
[v(t) = 3e−2t + 2e−7t]
12.18 A
transfer
function is given by
10 s
Y ( s) =
. Find time(s
j )(s
)( s
j )
domain response using graphical method.
⎡5.26 18.4° e
⎣
(5
(5
15)
15
5)
+ 5.266
188.4° e
( 5 j 15) t ⎤
⎦
Objective-Type Questions
12.1 Of the four networks N1, N2, N3 and N4 of Fig.
12.117, the networks having identical drivingpoint functions are
(a) N1 and N2
(b) N2 and N4
(c) N1 and N3
(d) N1 and N4
12.2 The driving-point impedance Z(s) of a
network has the pole-zero locations as shown
in Fig. 12.118. If Z(0) = 3, then Z(s) is
jw
j1
1Ω
2H
1F
2Ω
−3
1F
s
0
−1
−j 1
N1
Fig. 12.118
1Ω
2H
(a)
2Ω
3(
3)
2
s +2
(b)
3
2(
3)
2
s + 2s
2s 2
1F
N2
1Ω
1Ω
1H
1F
(c)
3(
s
2
3)
2s 2
1Ω
2(
s
2
3
+
100 μF
vi (t)
−
v0
−
1F
Fig. 12.119
N4
Fig. 12.117
2
10 mH
+
2H
3)
12.3 For the circuit shown in Fig. 12.119, the
initial conditions are zero. Its transfer function
V ( s)
H ( s) = 0
is
V1 ( s)
10 kΩ
N3
(d)
(a)
1
s
2
6
100 s + 10
6
(b)
106
s
2
1003 s + 106
12.54 Network Analysis and Synthesis
103
(c)
s 2 10
03 s + 106
106
(d)
s 2 10
06 s + 106
12.4 In Fig. 12.120, assume that all the capacitors
are initially uncharged. If vi(t) = 10 u(t), then
v0(t) is given by
12.7 A network has response with time as shown
in Fig. 12.122. Which one of the following
diagrams represents the location of the poles
of this network?
x (t)
1 kΩ
+
4 μF
vi (t)
1 μF
4 kΩ
t
0
+
v0 (t)
Fig. 12.122
−
−
jw
Fig. 12.120
(a) 8 e–0.004 t
(c) 8 u(t)
(b) 8 (1–e–0.004 t)
(d) 8
12.5 A system is represented by the transfer
10
function
. The dc gain of this
(
1
1)
)(
)( 2)
system is
(a)
jw
s
0
(b)
0
jw
jw
(c)
s
s
0
(d)
0
s
(a) 1
(b) 2
Fig. 12.123
(c) 5
(d) 10
12.8 The transfer function of a low-pass RC
network is
12.6 Which one of the following is the ratio
V24
V13
(a) (RCs) (1 + RCs)
(b)
1
1+ RCs
C
(d)
s
1+ RCs
C
of the network shown in Fig. 12.121.
(c)
1Ω
1
2
1Ω
1Ω
3
RCs
C
1+ RCs
C
12.9 The driving-point admittance function of the
network shown in Fig. 12.124 has a
4
R
1Ω
L
Fig. 12.121
(a)
1
3
(b)
2
3
(c)
3
4
(d)
4
3
Fig. 12.124
(a)
(b)
(c)
(d)
pole at s = 0 and zero at s = ∞
pole at s = 0 and pole at s = ∞
pole at s = ∞ and zero at s = 0
pole at s = ∞ and zero at s = ∞
C
Answers to Objective-Type Questions 12.55
I 2 ( s)
for the
V1 ( s)
network shown in Fig. 12.125 is
12.10 The transfer function Y12 ( s) =
(a)
(c)
1Ω
I2(s)
V1(s)
1H
1F
Fig. 12.125
s2
(b)
2
s + s +1
1
s +1
(d)
s
s +1
s +1
s2 + 1
12.11 As the poles of a network shift away from
the x axis, the response
(a) remains constant
(b) becomes less oscillating
(c) becomes more oscillating
(d) none of these
Answers to Objective-Type Questions
1. (c)
2. (b)
3. (d)
4. (c)
5. (c)
7. (d)
8. (b)
9. (a)
10. (a)
11. (c)
6. (a)
13
Two-Port Networks
13.1
INTRODUCTION
A two-port network has two pairs of terminals, one pair at
the input known as input port and one pair at the output
I1
I2
known as output port as shown in Fig. 13.1. There are
+
Two-port network
four variables V1, V2, I1 and I2 associated with a two-port V1
−
network. Two of these variables can be expressed in terms
of the other two variables. Thus, there will be two dependent
Fig. 13.1 Two-port network
variables and two independent variables. The number of
possible combinations generated by four variables taken
two at a time is 4C2, i.e., six. There are six possible sets of equations describing a two-port network.
Table 13.1 Two–port parameters
Parameter
Variables
Equation
Express
In terms of
Open-Circuit Impedance
V1, V2
I1, I2
V1 = Z11 I1 + Z12 I2
V2 = Z21 I1 + Z22 I2
Short-Circuit Admittance
I1, I2
V1, V2
I1 = Y11 V1 + Y12 V2
I2 = Y21 V1 + Y22 V2
Transmission
V1, I1
V2, I2
V1 = AV2 − BI2
I1 = CV2 − DI2
Inverse Transmission
V2, I2
V1, I1
V2
I2
Hybrid
V1, I2
I1, V2
V1 = h11 I1 + h12 V2
I2 = h21 I1 + h22 V2
Inverse Hybrid
I1, V2
V1, I2
I1 = g11 V1 + g12 I2
V2 = g21 V1 + g22 I2
A′ V1
C ′ V1
B′ I1
D ′ I1
+
V2
−
13.2 Network Analysis and Synthesis
13.2
OPEN-CIRCUIT IMPEDANCE PARAMETERS (Z PARAMETERS)
The Z parameters of a two-port network may be defined by expressing two-port voltages V1 and V2 in terms
of two-port currents I1 and I2.
( , V2 ) f ( I1 , )
V1 Z11 I1 Z12 I 2
V2
Z 21 I1 Z 22 I 2
In matrix form, we can write
⎡V1 ⎤ ⎡ Z11
⎢⎣V2 ⎥⎦ = ⎢⎣ Z 21
[V ] [ Z ] [ I ]
Z12 ⎤ ⎡ I1 ⎤
Z 22 ⎥⎦ ⎢⎣ I 2 ⎥⎦
The individual Z parameters for a given network can be defined by setting each of the port currents equal
to zero.
Case 1 When the output port is open-circuited, i.e., I2 = 0
V
Z11 = 1
I1 I 2 = 0
where Z11 is the driving-point impedance with the output port open-circuited. It is also called open-circuit
input impedance.
Similarly,
V
Z 21 = 2
I1 I 2 = 0
where Z21 is the transfer impedance with the output port open-circuited. It is also called open-circuit forward
transfer impedance.
Case 2 When input port is open-circuited, i.e., I1 = 0
Z12 =
V1
I2
I1 = 0
where Z12 is the transfer impedance with the input port open-circuited. It is also called open-circuit reverse
transfer impedance.
Similarly,
V
Z 22 = 2
I 2 I1 = 0
where Z22 is the open-circuit driving-point impedance
with the input port open-circuited. It is also called open
circuit output impedance.
As these impedance parameters are measured with
either the input or output port open-circuited, these are
called open-circuit impedance parameters.
The equivalent circuit of the two-port network in
terms of Z parameters is shown in Fig. 13.2.
I1
+
+
Z11
V1
Z12I2
−
Fig. 13.2
13.2.1 Condition for Reciprocity
A network is said to be reciprocal if the ratio of excitation
at one port to response at the other port is same if excitation
and response are interchanged.
(a) As shown in Fig. 13.3, voltage Vs is applied at the
input port with the output port short-circuited.
I2
Z22
+
−
Fig. 13.3
−
Equivalent circuit of the two-port
network in terms of Z parameter
I1
+
Vs
−
V2
+ Z I
− 21 1
I2
Network
I2′
Network for deriving condition
for reciprocity
13.2 Open-Circuit Impedance Parameters (Z Parameters) 13.3
i.e.,
V1 Vs
V2 = 0
I2
I2 ′
From the Z-parameter equations,
Vs Z11 I1 Z12 I 2 ′
0 = Z 21 I1 − Z 22 I 2 ′
Z
I1 = 22 I 2 ′
Z 21
Z 22
Vs Z11
I 2 ′ Z12 I 2 ′
Z 21
Vs
Z Z − Z12 Z 21
= 11 22
I2 ′
Z 21
(b) As shown in Fig. 13.4, voltage Vs is applied at the
output port with input port short-circuited.
I1′
V2 Vs
i.e.,
V1 = 0
i.e.,
I1
I1
I1 ′
Fig. 13.4
From the Z-parameter equations,
0 = − Z11 I1 ′ + Z12 I 2
Vs = − Z 221 I1 ′ + Z 222 I 2
Z
I 2 = 111 I1 ′
Z112
Z
Vs
Z 21 I1 ′ Z 22 11 I1 ′
Z12
Vs Z11Z 22 − Z12 Z 21
=
I1 ′
Z12
Hence, for the network to be reciprocal,
Vs Vs
=
I1 ′ I 2 ′
Z12 Z 21
I2
Network
+
−
Vs
Network for deriving condition for
reciprocity
13.2.2 Condition for Symmetry
For a network to be symmetrical, the voltage-to-current ratio at one port should be the same as the voltageto-current ratio at the other port with one of the ports open-circuited.
(a) When the output port is open-circuited, i.e., I2 = 0
From the Z-parameter equation,
Vs Z11 I1
Vs
= Z11
I1
(b) When the input port is open-circuited, i.e., I1 = 0
From the Z-parameter equation,
Vs Z 22 I 2
Vs
= Z 22
I2
Hence, for the network to be symmetrical,
13.4 Network Analysis and Synthesis
Vs Vs
=
I1 I 2
Z11 Z 22
i.e.,
Example 13.1 Test results for a two-port network are (a) I1 = 0.1 Æ 0é A, V1 = 5.2 Æ 50é V,
V2= 4.1 Æ-25é V with Port 2 open-circuited (b) I = 0.1 Æ 0é A, V = 3.1 Æ−80é V, V =4.2 Æ 60é V, with Port 1
2
1
2
open-circuited. Find Z parameters.
Solution
V
5 2∠50°
Z11 = 1
=
= 52 ∠50
∠50° Ω,
I1 I 2 = 0
0 1∠0°
Z 221 =
V2
I1
=
I2 =0
Z112 =
4.1∠ − 25°
= 41 ∠ − 25° Ω,
0 1∠0°
Hence, the Z-parameters are
⎡ Z11
⎢⎣ Z 21
Example 13.2
Z222 =
V2
I2
=
3 1∠ − 80°
= 31 ∠ 80° Ω
0.1∠0°
=
4.2∠60°
= 42 ∠60° Ω
0 1∠0°
I1 = 0
I1 = 0
Z12 ⎤ ⎡52∠50°
31∠ − 80°⎤
=
Z 22 ⎥⎦ ⎢⎣ 41∠ − 25° 42∠60° ⎥⎦
Find the Z parameters for the network shown in Fig. 13.5.
I1
Z1
Z3
I2
+
+
V2
Z2
V1
−
−
Fig. 13.5
Solution
First Method
Case 1 When the output port is open-circuited, i.e., I2 = 0.
Applying KVL to Mesh 1,
V1 ( Z1 + Z 2 ) I1
Z11 =
Also
V1
I2
V2
Z 21 =
V1
I1
= Z1 Z 2
I2 = 0
Z 2 I1
V2
I1
= Z2
I2 = 0
Case 2 When the input port is open-circuited, i.e., I1 = 0.
Applying KVL to Mesh 2,
V2 ( Z 2 + Z3 ) I 2
Z 22 =
V2
I2
= Z2
I1 = 0
Z3
13.2 Open-Circuit Impedance Parameters (Z Parameters) 13.5
Also
V1
Z2 I 2
Z12 =
V1
I2
Hence, the Z-parameters are
⎡ Z11
⎢⎣ Z 21
= Z2
I1 = 0
Z12 ⎤ ⎡ Z1 Z 2
=
Z 22 ⎥⎦ ⎢⎣ Z 2
Z2 ⎤
Z 2 + Z3 ⎥⎦
Second Method
The network is redrawn as shown in Fig. 13.6.
Applying KVL to Mesh 1,
V1
Z1 I1
= ( Z1
Z 2 ( I1
I2 )
Z 2 ) I1 Z 2 I 2
Z1
…(i)
Z3
+
+
Applying KVL to Mesh 2,
V2
Z3 I 2
Z 2 ( I1
= Z 2 I1 + ( Z 2
Z3 ) I 2
I1
…(ii)
Z12 ⎤ ⎡ Z1 Z 2
=
Z 22 ⎥⎦ ⎢⎣ Z 2
Example 13.3
I2
−
−
Comparing Eqs (i) and (ii) with Z-parameter
equations,
⎡ Z11
⎢⎣ Z 21
V2
Z2
V1
I2 )
Fig. 13.6
Z2 ⎤
Z 2 + Z3 ⎥⎦
Find Z-parameter for the network shown in Fig. 13.7.
I1
2Ω
1Ω
1H
I2
+
+
V1
1Ω
2F
V2
−
−
Fig. 13.7
Solution The transformed network is shown in Fig. 13.8.
Z1 = 1
⎛ 1⎞
⎜⎝ ⎟⎠ (1)
1
2s
Z2 =
=
1
2s 1
+1
2s
Z3 s + 2
1
+
+
Z1
V1
Z 21
2 =
V1
I1
V2
I1
= Z1
Z2 = 1 +
I2 0
= Z2 =
I2 = 0
1
2s 1
1
2s 2
=
,
2s 1 2s 1
1
2s
1 Z
2
−
Z3
V2
−
Fig. 13.8
From definition of Z-parameters,
Z11 =
s
2
Z12 =
Z 222 =
V1
I2
V2
I2
= Z2 =
I1 = 0
1
2s 1
= Z 2 + Z3 =
I1 = 0
1
2 2 5s + 3
+s+2 =
2s 1
2s 1
13.6 Network Analysis and Synthesis
Example 13.4
Find Z-parameters for the network shown in Fig. 13.9.
I1
1Ω
1Ω
I2
+
+
2Ω
V1
V2
2Ω
−
−
Fig. 13.9
Solution The network is redrawn as shown in Fig. 13.10.
Applying KVL to Mesh 1,
…(i)
V1 3I1 2 I 3
Applying KVL to Mesh 2,
V2 2 I 2 2 I 3
Applying KVL to Mesh 3,
−2 1 + 2 I 2 + 5
3
…(ii)
1Ω
1Ω
+
+
2Ω
V1
=0
I1
2
2
I1 − I 2
5
5
Substituting Eq. (iii) in Eq. (i),
…(iii)
I3 =
2Ω
I3
V2
I2
−
−
Fig. 13.10
4
4
I1
I2
5
5
11
4
= I1
I2
5
5
V1 = 3I1
…(iv)
Substituting Eq. (iii) in Eq. (ii),
2I 2 +
V2
=
4
I1
5
4
4
I1
I2
5
5
6
I2
5
…(v)
Comparing Eqs (iv) and (v) with Z-parameter equations,
⎡ Z11
⎢⎣ Z 21
⎡11 4 ⎤
Z12 ⎤ ⎢ 5 5 ⎥
=
Z 22 ⎥⎦ ⎢⎢ 4 6 ⎥⎥
⎣ 5 5⎦
Example 13.5
Find the Z-parameters for the network shown in Fig. 13.11.
I1
1Ω
1Ω
I2
+
V1
+
1H
−
1H
V2
−
Fig. 13.11
13.2 Open-Circuit Impedance Parameters (Z Parameters) 13.7
Solution The transformed network is shown in Fig. 13.12.
Applying KVL to Mesh 1,
1
V1 ( s + 1) I1 sI 3
…(i) +
Applying KVL to Mesh 2,
V2 sI
sI 2 sI 3
…(ii) V1
I1
Applying KVL to Mesh 3,
+
s
s
I3
−
− sI1 + sI 2 + ( 2 s + 1) I 3 = 0
I3 =
1
s
s
I1 −
I2 …
2s 1
2s 1
(iii)
V2
I2
−
Fig. 13.12
Substituting Eq. (iii) in Eq. (i),
V1
( s + 1) I1
⎛ s
s⎜
I1
⎝ 2s 1
s
⎞
I2 ⎟
2s 1 ⎠
⎛ s 2 + 3s
⎛ s2 ⎞
3s 1⎞
=⎜
I
+
1
⎟
⎜ 2 s + 1⎟ I 2
⎝ 2s 1 ⎠
⎝
⎠
…(iv)
Substituting Eq. (iii) in Eq. (ii),
V2
s 2
sI
⎛ s
s⎜
I1
⎝ 2s 1
s
⎞
I2 ⎟
2s 1 ⎠
⎛ s2 ⎞
⎛ s2 + s ⎞
=⎜
I1 + ⎜
⎟
⎟ I2
⎝ 2 s 1⎠
⎝ 2s 1⎠
…(v)
Comparing Eqs (iv) and (v) with Z-parameter equations,
⎡ Z11
⎢⎣ Z 21
⎡ s 2 + 3s
3s 1
Z12 ⎤ ⎢ 2 s +1
1
=⎢
Z 22 ⎥⎦ ⎢ s 2
⎢⎣ 2 s 1
s2 ⎤
⎥
2s 1 ⎥
2s
s2 + s ⎥
2 s 1 ⎥⎦
Example 13.6 Find the open-circuit impedance parameters for the network shown in Fig. 13.13.
Determine whether the network is symmetrical and reciprocal.
4Ω
I1
1Ω
3Ω
+
V1
I2
+
2Ω
−
V2
−
Fig. 13.13
13.8 Network Analysis and Synthesis
Solution The network is redrawn as shown in Fig. 13.14.
Applying KVL to Mesh 1,
V
I
+ I2
0
V
…(i)
Applying KVL to Mesh 2,
V
I
+ I2 0
V
Applying KVL to Mesh 3,
−
−
−
=
2+
I −
+ I
0
1
8
Substituting Eq. (iii) in Eq. (i),
1Ω
I3
3Ω
…(ii) +
+
I1
3
− I2
8
…(iii)
1
8
V
2Ω
V
V
23
8
4Ω
I
Fig. 13.14
3
− I2
8
19
8
…(iv)
Substituting Eq. (iii) in Eq. (ii),
1
V
3
− I2
8
19
31
8
8
Comparing Eqs (iv) and (v) with Z-parameter equations,
23 19 ⎤
Z
Z12 ⎤
8
8
Z
Z 22
19 31
8
8⎦
Since Z11 ≠ Z22, the network is not symmetrical.
Since Z12 = Z21, the network is reciprocal.
13.3
…(v)
SHORT-CIRCUIT ADMITTANCE PARAMETERS (Y PARAMETERS)
The Y parameters of a two-port network may be defined by expressing the two-port currents I1 and I2 in terms
of the two-port voltages V1 and V2.
2
( ,
V
V
In matrix form, we can write
I1
I2
I
⎡Y
Y
Y V
Y ⎤ ⎡V ⎤
Y
V
13.3 Short-Circuit Admittance Parameters (Y Parameters) 13.9
The individual Y parameters for a given network can be defined by setting each of the port voltages equal to
zero.
Case 1 When the output port is short-circuited, i.e., V2 = 0
I
Y11 = 1
V1 V2 = 0
where Y11 is the driving-point admittance with the output port short-circuited. It is also called short-circuit
input admittance.
Similarly,
I
Y21 = 2
V1 V2 = 0
where Y21 is the transfer admittance with the output port short-circuited. It is also called short-circuit forward
transfer admittance.
Case 2 When the input port is short-circuited, i.e., V1 = 0
I
Y12 = 1
V2 V1 = 0
where Y12 is the transfer admittance with the input port short-circuited. It is also called short-circuit reverse
transfer admittance.
Similarly,
I
Y22 = 2
V2 V1 = 0
where Y22 is the short-circuit driving-point admittance with the input port short-circuited. It is also called the
short circuit output admittance.
As these admittance parameters are measured with either input or output port short-circuited, these are
called short-circuit admittance parameters.
The equivalent circuit of the two-port network in terms of Y parameters is shown in Fig. 13.15.
I2
I1
+
V1
+
Y11
Y12V2
Y22
V2
−
−
Fig. 13.15
Y21V1
2
Equivalent circuit of the two-port network in terms of Y-parameters
Condition for Reciprocity
(a) As shown in Fig. 13.16, voltage Vs is applied at input port with the output port short-circuited.
i.e,
V1 Vs
I1
I2
V2 = 0
I2
I 2′
+
Network
I2′
From the Y-parameter equation,
− Vs
− I 2′ Y21 Vs
I2
= −Y21
Vs
Fig. 13.16 Network for deriving condition
for reciprocity
13.10 Network Analysis and Synthesis
(b) As shown in Fig. 13.17, voltage Vs is applied at output port with the input port short-circuited.
i.e,
V2 Vs
V1 = 0
I1
I2
I1
I1′
From the Y-parameter equation,
+
I ′
Network
1
− I1′
−
Y12 Vs
I1′
= −Y12
Vs
Hence, for the network to be reciprocal,
Fig. 13.17
Vs
Network for deriving condition for
reciprocity
I 2′ I1′
=
Vs Vs
Y12 Y21
i.e,
13.3.2 Condition for Symmetry
(a) When the output port is short-circuited, i.e., V2 = 0.
From the Y-parameter equation,
I1 Y11 Vs
Vs
1
=
I1 Y11
(b) When the input port is short-circuited, i.e., V1 = 0.
From the Y-parameter equation,
I 2 Y22 Vs
Vs
1
=
I 2 Y22
Hence, for the network to be symmetrical,
Vs Vs
=
I1 I 2
Y11 Y22
i.e.,
Example 13.7
Test results for a two-port network are
(a) Port 2 short-circuited: V1
(b) Port 1 short-circuited: V2
Find Y-parameters.
50
0 ∠0° V I 1
50
0 ∠0° V I 2
2.1
3
30° A, I 2
15° A, I 1
1.
.
0° A
20° A.
Solution
Y11 =
I1
2.1∠ − 30°
=
= 0.042
0 ∠ − 30° ,
V1 V2 = 0
50 ∠0°
I
Y221 = 2
V1
V2 = 0
−1.1∠ − 20°
=
= −0.022∠ − 20
2 ° ,
50 ∠0°
Y12 =
I1
V2
2
22
I2
V2
V1 = 0
−1.1∠ − 20
2 °
= −0.022∠ − 20°
50 ∠0°
V1 = 0
3
15°
50 ∠0°
=
Hence, the Y-parameters are
0 ∠ − 20°⎤
⎡ Y11 Y12 ⎤ ⎡ 0.042∠ − 30° −0.022
⎢Y 21 Y22 ⎥ = ⎢⎣ −0.022
0
∠
−
20
0
°
0
.
06
15° ⎥⎦
⎣
⎦
0.06
155°
13.3 Short-Circuit Admittance Parameters (Y Parameters) 13.11
Example 13.8
Find Y-parameters for the network shown in Fig. 13.18.
I1
3
1Ω
3Ω
I2
+
+
V2
2Ω
V1
−
−
Fig. 13.18
Solution
First Method
Case 1 When the output port is short-circuited, i.e., V2 = 0 as shown in Fig. 13.19,
Req = 1 +
Now,
Also,
2×3
6 11
= 1+ = Ω
2+3
5 5
2
2 5
( I1 ) = − × V1
5
5 11
I2
2
Y21 =
=−
V1 V2 = 0
11
I2
1Ω
I1
11
I1
5
I
5
Y11 = 1
=
V1 V2 = 0 11
V1
3Ω
I2
+
2Ω
V1
2
V1
11
−
Fig. 13.19
Case 2 When the input port is short-circuited, i.e., V1 = 0 as shown in Fig. 13.20,
Req = 3 +
Now
V2
Y22
Also
1× 2
2 11
= 3+ = Ω
1+ 2
3 3
I1
2
2 3
( I 2 ) = − × V2
3
3 11
I1
2
Y12 =
=−
V2 V1 = 0
11
3Ω
I2
+
11
I2
3
I
3
= 2
=
V2 V1 = 0 11
I1
1Ω
2Ω
V2
−
2
V2
11
Fig. 13.20
Hence, the Y-parameters are
2⎤
⎡ 5
− ⎥
⎢
Y
Y
⎡ 11 12 ⎤
11
11
⎥
⎢⎣Y21 Y22 ⎥⎦ = ⎢ 2
3
⎢−
⎥
⎣ 11 11 ⎦
Second Method (Refer Fig. 13.18)
V V
I1 = 1 3
1
= V1 V3
…(i)
13.12 Network Analysis and Synthesis
V2 V3
3
V2 V3
=
−
3 3
I2 =
…(ii)
Applying KCL at Node 3,
I1
I2 =
V3
2
…(iii)
Substituting Eqs (i) and (ii) in Eq. (iii),
V V
V
V1 V3 + 2 − 3 = 3
3 3
2
V2 11
V1 +
= V3
3
6
6
2
V3
V1 + V 2
11
11
Substituting Eq. (iv) in Eq. (i),
6
2
I1 = V1
V1
V2
11
11
5
2
= V1
V2
11
11
Substituting Eq. (iv) in Eq. (ii),
V 1⎛ 6
2 ⎞
I 2 = 2 − ⎜ V1
V2 ⎟
⎝
3 3 11
11 ⎠
2
3
= − V1
V2
11
11
Comparing Eqs (v) and (vi) with Y-parameter equations,
2⎤
⎡ 5
− ⎥
⎡Y11 Y12 ⎤ ⎢ 11
11
⎢⎣Y21 Y22 ⎥⎦ = ⎢ 2
3 ⎥⎥
⎢−
⎣ 11 11 ⎦
Example 13.9
…(iv)
…(v)
…(vi)
Find Y-parameters of the network shown in Fig. 13.21.
I1
2Ω
2Ω
V3
+
I2
+
1Ω
V1
V2
3V2
−
−
Fig. 13.21
Solution From Fig. 13.21,
V1 V3
2
1
1
= V1
V3
2
2
I1 =
…(i)
13.3 Short-Circuit Admittance Parameters (Y Parameters) 13.13
V2 V3
2
1
1
= V2
V3
2
2
I2 =
…(ii)
Applying KCL at Node 3,
I1 I 2 + 3 V2
Substituting Eqs (i) and (ii) in Eq. (iii),
V1 V3 V2 V3
+
+ 3V
3 V2
2
2
2 V3
…(iii)
0
0
V1 7 V2
1
7
V1 + V2
2
2
V3
…(iv)
Substituting Eq. (iv) in Eq. (i),
1
1 ⎛1
V1 − ⎜ V1
2
2 ⎝2
1
7
= V1
V2
4
4
I1
7 ⎞
V2 ⎟
2 ⎠
…(v)
Substituting Eq. (iv) in Eq. (ii),
1
1 ⎛1
7 ⎞
V2 − ⎜ V1
V2 ⎟
2
2 ⎝2
2 ⎠
1
5
= − V1
V2
4
4
Comparing Eqs (v) and (vi) with Y-parameter equations,
7⎤
⎡ 1
− ⎥
⎡Y11 Y12 ⎤ ⎢ 4
4
⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1
5⎥
⎢−
− ⎥
⎣ 4
4⎦
I2
…(vi)
Example 13.10 Determine Y-parameters for the network shown in Fig. 13.22. Determine whether
the network is symmetrical and reciprocal.
I1
3
1Ω
2
2Ω
+
V1
I2
+
2Ω
4Ω
−
V2
−
Fig. 13.22
Solution From Fig. 13.31,
V1 V3
1
= V1 V3
I1 =
…(i)
Applying KCL at Node 3,
V3 V3 − V2
+
2
2
V2
= V3 −
2
I1 =
…(ii)
13.14 Network Analysis and Synthesis
Applying KCL at Node 2,
V2 V2 − V3
+
4
2
3
V
= V2 − 3
4
2
I2 =
…(iii)
Substituting Eq. (i) in Eq. (ii),
V2
2
V V
V3 = 1 + 2
2 4
V1 V3 = V3 −
…(iv)
Substituting Eq. (iv) in Eq. (ii),
V1 V2 V2
+ −
2 4
2
V1 V2
= −
2 4
I1 =
Substituting Eq. (iv) in Eq. (iii),
…(v)
3
1 ⎛V V ⎞
V2 − ⎜ 1 + 2 ⎟
4
2⎝ 2 4 ⎠
I2
V1
V
+ 2
4
8
Comparing Eqs (v) and (vi) with Y-parameter equations,
1⎤
⎡ 1
− ⎥
⎡Y11 Y12 ⎤ ⎢ 2
4
⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1 5 ⎥
⎢−
⎥
⎣ 4 8 ⎦
Since Y11 ≠ Y22, the network is not symmetrical.
Since Y12 = Y21, the network is reciprocal.
…(vi)
=−
Example 13.11
Determine the short-circuit admittance parameters for the network shown in Fig. 13.23.
I1
1Ω
1Ω
I2
+
+
V1
V2
1F
1F
−
−
Fig. 13.23
Solution The transformed network is shown in Fig. 13.24.
From Fig. 13.24,
V1 V3
1
= V1 V3
I1
…(i)
3
1
2
I2
+
V1
I1 =
1
+
1
s
1
s
V2
−
−
Fig. 13.24
13.3 Short-Circuit Admittance Parameters (Y Parameters) 13.15
Applying KCL at Node 3,
V3 (V3 V2 )
+
1
1
s
= ( s 1) V3 V2
I1 =
…(ii)
Applying KCL at Node 2,
I2 =
V2 (V2 V3 )
+
1
1
s
…(iii)
= ( s 1) V2 V3
Substituting Eq. (i) in Eq. (ii),
V1 V3 = ( s 1) V3 V2
( s 2) V3 = V1 V2
V3 =
1
1
V1 +
V2
s+2
s+2
I1
( s + 1)
…(iv)
Substituting Eq. (iv) in Eq. (ii),
=
1
⎛ 1
⎞
V1 +
V2 V2
⎝s+2
s+2 ⎠
s +1
V1
s+2
1
V2
s+2
…(v)
Substituting Eq. (iv) in Eq. (iii),
I2
1
⎛ 1
⎞
( s + 1) V2 − ⎜
V1 +
V2 ⎟
⎝s+2
s+2 ⎠
1
s 2 + 3s
3s 1
V1 +
V2
s+2
s+2
Comparing Eqs (v) and (vi) with Y-parameter equations,
1 ⎤
⎡ s +1
−
⎡Y11 Y12 ⎤ ⎢ s + 2
s+2 ⎥
⎥
⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1
s 2 + 3s
3s 1 ⎥
⎢−
⎢⎣ s + 2
s + 2 ⎥⎦
…(vi)
=−
Example 13.12
Determine Y-parameters for the network shown in Fig. 13.25.
I1
1
F
2
1
F
2
I2
+
V1
+
1H
−
1H
V2
−
Fig. 13.25
13.16 Network Analysis and Synthesis
Solution The transformed network as shown in Fig. 13.26.
From Fig. 13.26,
V V
I1 = 1 3
+
2
s
V1
s
s
…(i) −
= V1 − V3
2
2
Applying KCL at Node 3,
I1
2
s
2
s
3
2
I2
+
s
s
V2
−
Fig. 13.26
s
V
s
(V1 V3 ) = 3 + (V
(V3 V2 )
2
s 2
s
1
s
s
s
V3 + V3 + V3 = V1 + V2
2
s
2
2
2
V3 =
s2
2( s 2 1)
V1 +
s2
((ss 2 1)
V2
…(ii)
Substituting Eq. (ii) in Eq. (i),
⎤
s
s ⎡ s2
s2
I1 = V1 − ⎢ 2
V1 +
V2 ⎥
2
2
2 ⎣ 2( s 1)
2( s 1) ⎦
⎡s
s3 ⎤
s3
=⎢ −
V
−
V2
⎥
1
2
4( s 2 1)
⎣ 2 4( s 1) ⎦
=
s3 + 2 s
4( s 2 1)
V1 −
s3
4( s 2 1)
V2
…(iii)
Applying KCL at Node 2,
I2 =
=
V2 s
+ (V2 V3 )
s 2
s2 + 2
V2
2s
s
V3
2
…(iv)
Substituting Eq. (ii) in Eq. (iv),
I2 =
⎤
s2 + 2
s ⎡ s2
s2
V2 − ⎢ 2
V1 +
V2 ⎥
2
2s
2 ⎣ 2( s 1)
2( s 1) ⎦
=−
⎡ s2 + 2
s3
s3 ⎤
V
+
−
⎢
⎥ V2
1
4( s 2 1)
4( s 2 1) ⎦
⎣ 2s
s3
s4 + 6 2 4
V
+
V2
1
4( s 2 1)
4 s( s 2 1)
Comparing Eqs (iii) and (v) with Y-parameter equation,
=−
⎡Y11
⎢⎣Y21
⎡ s3 + 2 s
⎢
Y12 ⎤ ⎢ 4( s 2 1)
=
Y22 ⎥⎦ ⎢
s3
⎢−
2
⎢⎣ 4( s 1)
⎤
s3
⎥
2
4( s 1) ⎥
s4 + 6s2 + 4 ⎥
⎥
4 s( s 2 1) ⎥⎦
−
…(v)
13.3 Short-Circuit Admittance Parameters (Y Parameters) 13.17
Example 13.13
Obtain Y-parameters of the network shown in Fig. 13.27.
1
s
I1
1
3
1
2
1
+
I2
+
1
s
V1
V2
−
−
Fig. 13.27
Solution
Applying KCL at Node 1,
V1 V3 V1 V2
+
1
1
s
= ( s 1) V1 s V2 − V3
…(i)
V2 V3 V2 V1
+
1
1
s
= ( s 1) V2 s V1 − V3
…(ii)
I1 =
Applying KCL at Node 2,
I2 =
Applying KCL at Node 3,
V3 V3 V1 V3 V2
+
+
=0
1
1
1
s
( s 2) V3 V1 V2 0
V3 =
1
1
V1 +
V2
s+2
s+2
I1
( s + 1) V1
…(iii)
Substituting Eq. (iii) in Eq. (i),
sV
V2
1
⎛ 1
⎞
V1 +
V2 ⎟
⎝s+2
s+2 ⎠
⎡ ( s +11)( s 2) − 1⎤
⎡ s ( s + 2) 1 ⎤
=⎢
V1 ⎢
⎥
⎥ V2
( s + 2)
⎣
⎦
⎣ ( s + 2) ⎦
⎛ s 2 + 3s
⎛ s 2 + 2s
3s 1⎞
2 s 1⎞
=⎜
V1 − ⎜
⎟
⎟ V2
⎝ s+2 ⎠
⎝ s+2 ⎠
…(iv)
13.18 Network Analysis and Synthesis
Substituting Eq. (iii) in Eq. (ii),
I2
1
⎛ 1
⎞
V1 +
V2 ⎟
⎝s+2
s+2 ⎠
⎡ s ( s + 2) 1 ⎤
⎡(
)(
)(
) − 1⎤
= −⎢
V1 + ⎢
⎥
⎥ V2
(
)
⎣ ( s + 2) ⎦
⎣
⎦
( s + 1) V2
s V1
⎛ s 2 + 2 s + 1⎞
⎛ s 2 + 3s + 1⎞
= −⎜
V1 + ⎜
⎟
⎟ V2
⎝ s+2 ⎠
⎝ s+2 ⎠
Comparing Eqs (iv) and (v) with Y-parameter equations,
⎡Y11
⎢⎣Y21
13.4
⎡ s 2 + 3s
3s 1
Y12 ⎤ ⎢
s+2
=⎢
Y22 ⎥⎦ ⎢ ( s 2 + 2ss 1)
⎢⎣ −
s+2
−
…(v)
( s 2 + 2s
2 s 1) ⎤
⎥
s+2
⎥
s2 + 3 1 ⎥
⎥⎦
s+2
ΤRANSMISSION PARAMETERS (ABCD PARAMETERS)
The transmission parameters or chain parameters or ABCD parameters serve to relate the voltage and current
at the input port to voltage and current at the output port. In equation form,
( , I1 ) f (V2 ,
)
V1
AV
AV2
B 2
BI
I1 CV
V2 DI
D 2
Here, the negative sign is used with I2 and not for parameters B and D. The reason the current I2 carries a
negative sign is that in transmission field, the output current is assumed to be coming out of the output port
instead of going into the port.
In matrix form, we can write
⎡V1 ⎤ ⎡ A B ⎤ ⎡ V2 ⎤
⎢⎣ I1 ⎥⎦ = ⎢⎣C D ⎥⎦ ⎢⎣ − I 2 ⎥⎦
⎡A B⎤
is called transmission matrix.
where matrix ⎢
⎣C D ⎥⎦
For a given network, these parameters are determined as follows:
Case 1 When the output port is open-circuited, i.e., I2 = 0
V
A= 1
V2 I 2 = 0
where A is the reverse voltage gain with the output port open-circuited.
I
Similarly,
C= 1
V2 I 2 = 0
where C is the transfer admittance with the output port open-circuited.
Case 2 When output port is short-circuited, i.e., V2 = 0
V
B=− 1
I 2 V2 = 0
where B is the transfer impedance with the output port short-circuited.
I
Similarly,
D=− 1
I 2 V2 = 0
where D is the reverse current gain with the output port short-circuited.
13.4 Τransmission Parameters (ABCD Parameters) 13.19
13.4.1 Condition for Reciprocity
(a) As shown in Fig. 13.28, voltage Vs is applied at the input port with the output port short-circuited.
i.e.,
V V
1
s
I1
V2 = 0
I 2′
From the transmission parameter equations,
+
− Vs
I2
I2
I2′
Network
Vs B I 2′ Fig. 13.28 Network for deriving condition for
reciprocity
Vs
=B
′
I2
(b) As shown in Fig. 13.29, voltage Vs is applied at the output port with the input port short-circuited.
i.e.,
V2 Vs
V1 = 0
I1′
I1
I1′
I1
0 = AV
Vs − BI 2
− I1 ′ = CVs D
DI 2
A
I 2 = Vs
B
AD
−II1 ′ CV
Vs −
Vs
B
Vs
B
=
I1 ′ AD − BC
Hence, for the network to be reciprocal,
Vs Vs
=
I 2 ′ I1 ′
i.e.,
Network
+
−
Vs
Fig. 13.29 Network for deriving condition for
reciprocity
From the transmission parameter equations,
i.e.,
I2
B
AD − BC
AD − BC = 1
B=
13.4.2 Condition for Symmetry
(a) When the output port is open-circuited, i.e., I2 = 0.
From the transmission-parameter equations,
Vs AV
V2
I1 CV
V2
Vs A
=
I1 C
(b) When the input port is open-circuited, i.e., I1 = 0.
From the transmission parameter equation,
CV
Vs = DI 2
Vs D
=
I2 C
13.20 Network Analysis and Synthesis
Hence, for network to be symmetrical,
Vs Vs
=
I1 I 2
A D
i.e.,
Example 13.14
Find the transmission parameters for the network shown in Fig. 13.30.
I1
1Ω
2Ω
+
I2
+
5Ω
V1
−
V2
−
Fig. 13.30
Solution
First Method
Case 1 When the output port is open-circuited, i.e., I2 = 0.
V1 6 I1
V2
I1
and
A=
V1
V2
I
C= 1
V2
Case 2
Now
and
=
6 I1 6
=
5 I1 5
=
1
5
I2 = 0
I2 = 0
When the output port is short-circuited, i.e., V2 = 0, as shown in Fig. 13.31,
1Ω
2Ω
I1
5× 2
10 17
Req = 1 +
= 1+
=
Ω
+
5+ 2
7
7
17
V1
I1
V1
5Ω
7
5
5
−
I2
( I1 ) = − I1
7
7
Fig. 13.31
17
I1
V1
17
B=−
=− 7
=
Ω
5
I 2 V2 = 0
5
− I1
7
I1
7
D=−
=
I 2 V2 = 0 5
Hence, the transmission parameters are
⎡ 6 17 ⎤
⎡A B⎤ ⎢5 5 ⎥
⎢⎣C D ⎥⎦ = ⎢ 1 7 ⎥
⎢
⎥
⎣5 5 ⎦
I2
13.4 Τransmission Parameters (ABCD Parameters) 13.21
Second Method (Refer Fig. 13.37)
Applying KVL to Mesh 1,
V1 6 I1 5 I 2
Applying KVL to Mesh 2,
V2
I1 7 I 2
Hence,
51
2 7I 2
1
7
I1
V2 − I 2
5
5
Substituting Eq. (iii) in Eq. (i),
7 ⎞
⎛1
V1 6 V2 − I 2
I2
⎝5
5 ⎠
6
17
= V2
I2
5
5
Comparing Eqs (iii) and (iv) with transmission parameter equations,
…(i)
…(ii)
…(iii)
…(iv)
⎡ 6 17 ⎤
⎡A B⎤ ⎢5 5 ⎥
⎢⎣C D ⎥⎦ = ⎢ 1 7 ⎥
⎢
⎥
⎣5 5 ⎦
Example 13.15
Obtain ABCD parameters for the network shown in Fig 13.32.
I1
1Ω
1Ω
I2
+
V1
+
2Ω
V2
2Ω
−
−
Fig. 13.32
Solution The network is redrawn as shown in Fig 13.33.
Applying KVL to Mesh 1,
V1 3I1 2 I 3
…(i)
+
Applying KVL to Mesh 2,
V2 2 I 2 2 I 3 …(ii) V
1
Applying KVL to Mesh 3,
−2 ( 3 − 1 ) − 3 − 2 ( 3 + 2 ) = 0
−
5 3 = 2 I1 − 2 I 2
2
2
I 3 = I1 − I 2 …(iii)
5
5
Substituting Eq. (iii) in Eq. (i),
2 ⎞
⎛2
V1 3I1 2 ⎜ I1
I2 ⎟
⎝5
5 ⎠
11
4
=
I1
I2
5
5
1Ω
1Ω
+
2Ω
I1
2Ω
I3
V2
I2
−
Fig. 13.33
…(iv)
13.22 Network Analysis and Synthesis
Substituting Eq. (iii) in Eq. (ii),
⎛2
2 ⎜ I1
⎝5
4
6
= I1
I2
5
5
4
6
I1 V2 − I 2
5
5
5
3
I1
V2 − I 2
4
2
V2
2I 2
2 ⎞
I2 ⎟
5 ⎠
…(v)
Substituting Eq. (v) in Eq. (iv),
11 ⎛ 5
3 ⎞
V2 − I 2
5 ⎝4
2 ⎠
11
5
= V2
I2
4
2
Comparing Eqs (v) and (vi) with ABCD parameter equations,
V1
⎡11
⎡A B⎤ ⎢ 4
⎢⎣C D ⎥⎦ = ⎢ 5
⎢
⎣4
Example 13.16
4
I2
5
…(vi)
5⎤
2⎥
3 ⎥⎥
2⎦
Determine the transmission parameters for the network shown in Fig. 13.34.
1
I1
1
2
I2
+
+
s
V1
1
s
V2
−
−
Fig. 13.34
Solution
Applying KCL at Node 1,
V1
+ (V1 V2 )
s
s +1
=
V1 V2
s
I1 =
…(i)
Applying KCL at Node 2,
V2
+ (V2 V1 )
1
s
= ( s 1) V2 V1
V1 ( s + 1) V2 I 2
I2 =
…(ii)
13.4 Τransmission Parameters (ABCD Parameters) 13.23
Substituting Eq. (ii) in Eq. (i),
s +1
s
⎡ ( s + 1) 2
⎢
⎣ s
I1
2
=
−I
⎤
1 V
⎦
s +1
s
s 1
s +1
V −
s
s
…(iii)
2
Comparing Eqs (ii) and (iii) with ABCD parameter equations,
s+
−
⎡A B⎤
= 2 s 1 s +1
C D
s
s
Example 13.17
Find transmission parameters for the two-port network shown in Fig. 13.35.
1.5 V
10 Ω
I1
I
+
+
25 Ω
V
I1
20 Ω
V
I
I
−
Fig. 13.35
Solution Applying KVL to Mesh 1,
+ 25 (
)
=
…(i)
Applying KVL to Mesh 2,
+ I3
I
V
Applying KVL to Mesh 3,
− (
− )+
+
−
+
−
−
2
+
…(ii)
=0
) − 20 I 2 −
0
=0
20
− 20 I 2
82
I3
…(iii)
0 94
Substituting Eq. (iii) in Eq. (i),
− 25 0 9 I
I2
…(iv)
Substituting Eq. (iii) in Eq. (ii),
18
I + 20 0 9 I
I
I2
…(v)
From Eq. (v),
V − 0 81I 2
…(vi)
13.24 Network Analysis and Synthesis
Substituting Eq. (vi) in Eq. (iv),
V1
11. (0 0 3 V2 − 0.81I 2 ) 6 I 2
= 0.61V
61 V2 3 32 I 2
Comparing Eqs (vi) and (vii) with ABCD parameter equations,
⎡ A B ⎤ ⎡ 0.61 −3.32⎤
⎢⎣C D ⎥⎦ = ⎢⎣0.053 −0.81⎥⎦
13.5
…(vii)
INVERSE TRANSMISSION PARAMETERS (A�B�C�D� PARAMETERS)
The inverse transmission parameters serve to relate the voltage and current at the outport port to the voltage
and current at the input port. In equation form,
( , I 2 ) f (V1 , )
V2 A′ V1 B′ I1
I2
C ′ V1
D ′ I1
In matrix form, we can write
⎡V2 ⎤ ⎡ A′
⎢⎣ I 2 ⎥⎦ = ⎢⎣C ′
B′ ⎤ ⎡ V1 ⎤
D ′ ⎥⎦ ⎢⎣ − I1 ⎥⎦
⎡ A′ B′ ⎤
where matrix ⎢
is called the inverse transmission matrix.
⎣C ′ D ′ ⎥⎦
For a given network, these parameters are determined as follows:
Case 1 When the input port is open-circuited, i.e., I1 = 0
V
A′ = 2
V1 I1 = 0
where A′ is the forward voltage gain with the input port open-circuited.
I
Similarly,
C′ = 2
V1 I1 = 0
where C ′ is the transfer admittance with the input port open-circuited.
Case 2 When the input port is short-circuited, i.e., V1 = 0
V
B′ = − 2
I1 V1 = 0
where B′ is the transfer impedance with the input port short-circuited.
Similarly,
I
D′ = − 2
I1 V1 = 0
where D′ is the forward current gain with the input port short-circuited.
13.5.1 Condition for Reciprocity
(a) As shown in Fig 13.36, voltage Vs is applied at
input port with the output port short-circuited.
i.e.,
V1 Vs
V2 = 0
I2
I2 ′
I1
+
Vs
−
I2
Network
I2′
Fig. 13.36 Network for deriving condition
for reciprocity
13.5 Inverse Transmission Parameters (A′B′C′D′ Parameters) 13.25
From the inverse transmission parameter equations,
0 = A′ Vs − B′ I1
− I 2 ′ = C ′ Vs D ′ I1
I 2 ′ A′ D ′ − B ′C ′
=
Vs
B′
(b) As shown in Fig 13.37, voltage Vs is applied at the output port with the input port short-circuited.
V2 Vs
i.e.,
I1
I2
V1 = 0
+
I1′
I1
I1 ′
Network
− Vs
From the inverse transmission parameter equations,
Vs B′ I1 ′
Fig. 13.37 Network for deriving condition for
A′
I2 =
Vs
reciprocity
B′
I1 ′ 1
=
Vs B′
Hence, for the network to be reciprocal,
I 2 ′ I1 ′
=
Vs Vs
i.e.,
A′ D ′ − B′C ′ = 1
13.5.2 Condition for Symmetry
The condition for symmetry is obtained from the Z-parameters.
V
D′
Z11 = 1
=0=
I1 I 2 = 0
C′
Similarly,
V
A′
Z 22 = 2
=0=
I 2 I1 = 0
C′
For symmetrical network
Z11 Z 22
A′ D ′
Example 13.18
Find the inverse transmission parameters for the network shown in Fig. 13.38.
I1
1Ω
2Ω
+
V1
+
3Ω
−
V2
−
Fig. 13.38
Solution
First method
Case 1 When the input port is open-circuited, i.e., I1 = 0
V1 3I 2
V2
I2
13.26 Network Analysis and Synthesis
A′ =
V2
V1
I
C′ = 2
V1
=
5
3
=
1
3
I1 = 0
I1 = 0
Case 2 When the input port is short-circuited, i.e., V1 = 0 as shown in Fig. 13.39.
By current division rule,
1Ω
2Ω
I1
3
I1
I2
4
I
4
3Ω
D′ = − 2
=
I1 V1 = 0 3
3 1
3 11
Req = 2 +
= 2+ = Ω
3 1
4 4
Fig. 13.39
11
11 ⎛ 4 ⎞
11
V2
I 2 = ⎜ − ⎟ I1
I1
4
4 ⎝ 3⎠
3
V
11
B′ = − 2
= Ω
I1 V1 = 0 3
Second Method Refer Fig. 13.38.
Applying KVL to Mesh 1,
V1 = 4I1 + 3I2
Applying KVL to Mesh 2,
V2 = 3I1 + 5I2
Hence,
3I2 = V1 − 4I1
1
4
I2
V1 − I1
3
3
Substituting Eq. (iii) in Eq. (ii),
4 ⎞
⎛1
V2 3I1 5 ⎜ V1
I1 ⎟
⎝3
3 ⎠
5
11
= V1
I1
3
3
Comparing Eqs (iii) and (iv) with inverse transmission parameter equations,
⎡ 5 11⎤
⎡ A′ B′ ⎤ ⎢ 3 3 ⎥
⎢⎣C ′ D ′ ⎥⎦ = ⎢ 1 4 ⎥
⎢
⎥
⎣3 3 ⎦
Example 13.19
I2
+
V2
−
...(i)
...(ii)
…(iii)
…(iv)
Find the inverse transmission parameters for the network shown in Fig. 13.40
2Ω
I1
I2
+
V1
+
2Ω
2Ω
−
V2
−
Fig. 13.40
13.5 Inverse Transmission Parameters (A′B′C′D′ Parameters) 13.27
Solution The network is redrawn as shown in Fig 13.41.
Applying KVL to Mesh 1,
I1
…(i)
+
V1 2 I1 2 I 3
Applying KVL to Mesh 2,
V2
Applying KVL to Mesh 3,
…(ii)
V1
=0
1
1
I 3 = I1 − I 2 …(iii)
3
3
Substituting Eq. (iii) in Eq. (i),
−
−2 1 + 2 I 2 + 6
2I 2
2I3
2Ω
I2
+
2Ω
I1
2Ω
V2
I2
I3
−
3
Fig. 13.41
⎛1
2 I1 2 ⎜ I1
⎝3
4
2
= I1
I2
3
3
Substituting Eq. (iii) in Eq. (ii),
1 ⎞
I2 ⎟
3 ⎠
⎛1
2 ⎜ I1
⎝3
4
I2
3
1 ⎞
I2 ⎟
3 ⎠
V1
V2
2I 2
=
2
I1
3
...(iv)
...(v)
Rewriting Eq. (iv),
2
I2
3
V1 −
4
I1
3
3
V1 − 2 I1
2
Substituting Eq. (vi) in Eq. (v),
I2
V2
...(vi)
2
4⎛3
⎞
I1 + ⎜ V1 2 I1 ⎟
⎝
⎠
3
3 2
= 2V
2V1 2 I1
...(vii)
Comparing Eq. (vi) and (vii) with inverse transmission parameter equations,
⎡ A′
⎢⎣C ′
Example 13.20
⎡2
B′ ⎤ ⎢
=
3
D ′ ⎥⎦ ⎢
⎣2
2⎤
⎥
2⎥
⎦
Find the inverse transmission parameters for the network shown in Fig. 13.42.
I1
1Ω
1Ω
I2
+
V1
+
2Ω
−
1Ω
V2
−
Fig. 13.42
13.28 Network Analysis and Synthesis
Solution The network is redrawn as shown in Fig. 13.43.
Applying KVL to Mesh 1,
I1
1Ω
…(i)
V1 3I1 2 I 3
+
Applying KVL to Mesh 2,
V2 I 2 + I 3
…(ii)
V1
Applying KVL to Mesh 3,
I1
−2 1 + 2 + 4 I 3 = 0
−
1
1
I 3 = I1 − I 2 …(iii)
2
4
Substituting Eq. (iii) in Eq. (i),
1 ⎞
⎛1
V1 3I1 2 ⎜ I1
I2 ⎟
⎝2
4 ⎠
= 2 I1
1Ω
I2
+
2Ω
1Ω
I3
V2
I2
−
Fig. 13.43
1
I2
2
…(iv)
Substituting Eq. (iii) in Eq. (ii),
V2
Rewriting Eq. (iv),
1
I2
2
I2
1
1
I1
I2
2
4
1
3
= I1
I2
2
4
I2 +
…(v)
V1 − 2 I1
2
1
4 I1
...(vi)
Substituting the Eq. (vi) in Eq. (v),
1
3
I1 + ( 2V1 4 1 )
2
4
3
5
= V1
I1
2
2
Comparing Eqs (vi) and (vii) with inverse transmission parameter equations,
V2
⎡ A′
⎢⎣C ′
13.6
⎡3
B′ ⎤ ⎢
=
D ′ ⎥⎦ ⎢ 2
⎣2
...(vii)
5⎤
2⎥
4 ⎥⎦
HYBRID PARAMETERS (h PARAMETERS)
The hybrid parameters of a two-port network may be defined by expressing the voltage of input port V1 and
current of output port I2 in terms of current of input port I1 and voltage of output port V2.
( , I 2 ) f ( I1 , )
V1 h11 I1 h12 V2
I 2 h21 I1 h22 V2
In matrix form, we can write
⎡V1 ⎤ ⎡ h11 h12 ⎤ ⎡ I1 ⎤
⎢⎣ I 2 ⎥⎦ = ⎢⎣ h21 h22 ⎥⎦ ⎢⎣V2 ⎥⎦
The individual h parameters can be defined by setting I1 = 0 and V2 = 0.
13.6 Hybrid Parameters (h Parameters) 13.29
Case 1 When the output port is short-circuited i.e., V2 = 0
V
h11 = 1
I1 V2 = 0
where h11 is the short-circuit input impedance.
h21 =
I2
I1 V2 = 0
where h21 is the short-circuit forward current gain.
Case 2 When the input port is open-circuited, i.e., I1 = 0
V
h12 = 1
V2 I1 = 0
where h12 is the open-circuit reverse voltage gain.
I
h22 = 2
V2
I1 = 0
where h22 is the open-circuit output admittance.
Since h parameters represent dimensionally an impedance, an admittance, a voltage gain and a current
gain, these are called hybrid parameters.
The equivalent circuit of a two-port network in terms of hybrid parameters is shown in Fig 13.44.
I1
h11
I2
+
V1
+
h12V2
+
−
h21I1
h22
V2
−
−
Fig. 13.44 Equivalent circuit of the two-port network in terms of h-parameters
13.6.1 Condition for Reciprocity
(a) As shown in Fig. 13.45, voltage Vs is applied at the input port and the output port is short-circuited.
i.e.,
V1 Vs
V2 = 0
I2 ′
I2
From the h-parameter equations,
I1
+
Vs
−
Vs h11 I1
− I 2 ′ h21 I1
Fig. 13.45
Vs
h
= − 11
I2 ′
h21
I1
(b) As shown in Fig. 13.46, voltage Vs is applied
at the output port with the input port shortI1′
circuited.
i.e.,
V1 = 0
V2 Vs
Fig. 13.46
I1
I1 ′
I2
I2′
Network
Network for deriving condition
for reciprocity
I2
Network
+
−
Vs
Network for deriving condition for
reciprocity
13.30 Network Analysis and Synthesis
From the h-parameter equations,
0 = h11
1 I1 + h1122 Vs
h12 Vs = − h11
1 I1 = h11
11 I1 ′
Vs h111
=
I1 ′ h112
Hence, for the network to be reciprocal,
Vs Vs
=
I 2 ′ I1 ′
i.e.,
h21
h12
13.6.2 Condition for Symmetry
The condition for symmetry is obtained from the Z-parameters.
V
h I +h V
V
Z11 = 1
= 11 1 12 2
= h11 h12 2
I1 I 2 0
I1
I1
I2 = 0
But with I2 = 0,
0 = h21
2 I1 + h2222 V2
V2
h
= − 221
I1
h22
h h
h h −h h
Δh
Z11 h11 − 12 21 = 11 22 12 21 =
h22
h22
h22
where
Δh = h11h22 − h12 h21
Similarly,
V
Z 22 = 2
I 2 I1 = 0
With I1 = 0,
I 2 h22 V2
For a symmetrical network,
V2
I2
Z11
Z 22
=
I1 = 0
1
h22
Δh
1
=
h22 h22
Δh = 1
i.e.,
i.e.,
i.e.,
Z 22 =
h11 h22 − h12 h21 = 1
Example 13.21 In the two-port network shown in Fig. 13.47, compute h-parameters from the
following data:
(a) With the output port short-circuited:V
V1 2 V I 1 1 A, I 2
A
(b) With the input port open-circuited:V
V1 10 V V2 50 V , I 2
A
I1
+
V1
−
I2
Two-port network
Fig. 13.47
+
V2
−
13.6 Hybrid Parameters (h Parameters) 13.31
Solution
V1
I1 V
h11 =
=
2=
I2
I1
h21 =
0
=
V2 = 0
25
= 25
5 Ω,
1
h12 =
2
= 2,
1
h222 =
V1
V2
I2
V2
=
10
= 0.2
50
=
2
= 0.04
50
I1 = 0
I1 = 0
Hence, the h-parameters are
⎡ h11 h12 ⎤ ⎡ 25 0 2 ⎤
⎢⎣ h 21 h22 ⎥⎦ = ⎢⎣ 2 0.04 ⎥⎦
Example 13.22 Determine hybrid parameters for the network of Fig. 13.48.
Determine whether the network is reciprocal.
1Ω
I1
2Ω
I2
+
+
2Ω
V1
I1
4Ω
V2
I3
I2
−
−
Fig. 13.48
Solution
First Method
Case 1 When Port 2 is short-circuited, i.e., V2 = 0 as shown in Fig. 13.49,
2×2
=
2+2
Req = 1 +
Now,
Also,
I1 ×
2
2 2
V2 = 0
I2
2Ω
V1
=−
=−
2Ω
+
2I1
V
h11 = 1
=2Ω
I1 V2 = 0
I2
1Ω
I1
V1
I
h21 = 2
I1
Case 2
Ω
I1
2
4Ω
−
Fig. 13.49
1
2
When Port 1 is open-circuited, i.e., I1 = 0 as shown in Fig. 13.50,
Req =
V1
( + )×4
=
2+2+4
2I y
I
Iy = 2
2
V2 4 I x
I
Ix = 2
2
Ω
I1 = 0
+
V1
1Ω
1Ω
ly
2Ω
−
I2
lx
4Ω
+
V2
−
Fig. 13.50
13.32 Network Analysis and Synthesis
I1 = 0
I2
V2
I1 = 0
h22 =
I2
2 =1
=
=
I
4I x
2
4× 2
2
2I
1
= x =
4I x 2
2I y
V
h12 = 1
V2
2×
Hence, the h-parameters are
⎡
2
h12 ⎤ ⎢
=
⎢
h22 ⎥⎦ ⎢ 1
−
⎣ 2
⎡ h11
⎢⎣ h21
1⎤
2⎥
1 ⎥⎥
2⎦
Second Method (Refer Fig. 13.48)
Applying KVL to Mesh 1,
V1
3I1 2 I 3
…(i)
V2
4I2
…(ii)
Applying KVL to Mesh 2,
Applying KVL to Mesh 3,
−2 (
3
− 1 ) − 2I3 − 4 (
3
+
4I3
=0
= 2 I1 − 4 I 2
I
I
I3 = 1 − 2
4 2
2)
8
3
…(iii)
Substituting Eq. (iii) in Eq. (i),
V1
=
3I1
I ⎞
⎛I
2⎜ 1 − 2⎟
⎝4 2⎠
5
I1
2
I2
...(iv)
Substituting Eq. (iii) in Eq. (ii),
I2⎞
⎛I
4⎜ 1 − ⎟
⎝4 2⎠
= 4I
4 I 2 I1 2 I 2
= I1 + 2 I 2
1
1
I 2 = − I1 + V2
2
2
V2
4I2
...(v)
Substituting Eq. (v) in Eq. (iv),
5
1
I1 − I1
2
2
1
= 2 I1
V2
2
Comparing Eqs (v) and (vi) with h-parameter equations,
V1
⎡ h11
⎢ h21
⎣
Since h12 = − h21, the network is reciprocal.
⎡
2
h12 ⎤ ⎢
=⎢
⎥
h22 ⎦ ⎢ 1
−
⎣ 2
1⎤
2⎥
1 ⎥⎥
2⎦
1
V2
2
...(vi)
13.7 Inverse Hybrid Parameters (g Parameters) 13.33
Example 13.23
Find h-parameters for the network shown in Fig. 13.51.
1
F
2
I1
1
F
2
I2
+
+
1H
V1
1H
V2
−
−
Fig. 13.51
Solution As solved in Example 13.12, derive the equations for I1 and I2 in terms of V1 and V2.
I1 =
s3 + 2 s
s3
V
−
V2
1
4( s 2 1)
4( s 2 1)
I2 = −
s3
4( s
2
1)
s 4 + 6s
6s2
V1 +
4ss( s
2
…(i)
4
1)
V2
…(ii)
From Eq. (i),
V1 =
Substituting Eq. (iii) in Eq (ii),
4( s 2 1)
s2
I
+
V2
1
s ( s 2 + 2)
s2 2
I2 = −
=−
s3
4( s
2
s2
2
⎡ 4( s 2 1)
⎤ s 4 + 6s
s2
6s2 4
I1 + 2
V2 ⎥ +
V2
⎢ 2
1) ⎣ s( s + 2)
s 2 ⎦ 4 s( s 2 + 1)
I1 +
2
2(
2
+ 1)
s +2
s(( + 2)
Comparing Eqs (iii) and (iv) with h-parameter equations,
⎡ h11
⎢ h21
⎣
13.7
…(iii)
⎡ 4( s 2 1)
⎢
h12 ⎤ ⎢ s( s 2 + 2)
=
h22 ⎥⎦ ⎢
s2
⎢− 2
⎢⎣ s + 2
V2
…(iv)
⎤
⎥
s 2 ⎥
2( s 2 1) ⎥
⎥
s( s 2 + 2)
2 ⎥⎦
s2
2
INVERSE HYBRID PARAMETERS (g PARAMETERS)
The inverse hybrid parameters of a two-port network may be defined by expressing the current of the input
port I1 and voltage of the output port V2 in terms of the voltage of the input port V1 and the current of the
output port I2.
( , V2 ) f (V1 , )
I g11 V g12 I 2
V
g21 V g22 I 2
In matrix form, we can write
g12 ⎤ ⎡V1 ⎤
⎡ I1 ⎤ ⎡ g
⎢⎣V2 ⎥⎦ = ⎢⎣ g
g22 ⎥⎦ ⎢⎣ I 2 ⎥⎦
The individual g parameters can be defined by setting V1 = 0 and I2 = 0.
13.34 Network Analysis and Synthesis
Case 1
When the output port is open-circuited, i.e., I2 = 0
g11 =
I1
V1
I2 = 0
where g11 is the open-circuit input admittance.
V
g21 = 2
V1 I 2 = 0
where g21 is the open-circuit forward voltage gain.
Case 2 When the input port is short-circuited, i.e., V1 = 0
g12 =
I1
I 2 V1 = 0
where g12 is the short-circuit reverse current gain.
V
g22 = 2
I 2 V1 = 0
where g22 is the short-circuit output impedance.
The equivalent circuit of a two-port network in terms of inverse hybrid parameters is shown in
Fig. 13.52.
g22
I2
+
V1
I2
+
g11
g12I2
−
+
g V
− 21 1
V2
−
Fig. 13.52 Equivalent circuit of two-port network in terms of g-parameters
13.7.1 Condition for Reciprocity
(a) As shown in Fig. 13.53, voltage Vs is applied at the input port and the output port is short-circuited,
i.e.,
V1 Vs
I1
I2
V2 = 0
+
I2 ′
I2
Network
Vs
I2′
−
From the g-parameter equation,
g Vs = g I 2 ′
Fig. 13.53 Network for deriving condition for
Vs
g
= 22
reciprocity
I 2 ′ g21
(b) As shown in Fig. 13.54, voltage Vs is applied at the output port with the input port short-circuited,
i.e.,
V1 = 0
I1
V2 Vs
I2
I1
I1 ′
+
I1′
From the g-parameter equations,
Network
− Vs
− I ′ g12 I 2
V
g22 I 2
Fig. 13.54 Network for deriving condition for
Vs
g
reciprocity
= − 22
I1 ′
g12
13.7 Inverse Hybrid Parameters (g Parameters) 13.35
Hence, for the network to be reciprocal,
Vs Vs
=
I 2 ′ I1 ′
g
g12
i.e.,
13.7.2 Condition for Symmetry
The condition for symmetry is obtained from the Z-parameters.
V
1
Z11 = 1
=
I I 2 = 0 g11
Similarly,
Z 22 =
V2
I2
I1 = 0
0 = g V1 + g I 2
g
V1 = − 12 I 2
g11
⎛ g ⎞
V
g21 − 12 ⎟ I
⎝ g11 ⎠
Z 22 =
For a symmetrical network, Z11
V2
I2
=
I1 = 0
g22 I 2
g g22 − g g21
g11
Z 22 .
1
g g − g12 g21
= 11 22
g11
g11
g g22 − g g21 = 1
i.e.,
i.e.,
Table 13.2 Conditions for reciprocity and symmetry
Parameter
Condition for Reciprocity
Condition for Symmetry
Z
Z12
Z 21
Z11
Z 22
Y
Y12
Y21
Y11
Y22
A
D
T
AD − BC = 1
T′
h
A′ D ′ − B′C ′ = 1
h12
h21
g
Example 13.24
g
A′ D ′
h11 h22 − h12 h21 = 1
g g22 − g
g21
Find g-parameters for the network shown in Fig. 13.55.
I1
1Ω
2Ω
+
V1
I2
+
3Ω
−
V2
−
Fig. 13.55
g21 = 1
13.36 Network Analysis and Synthesis
Solution
First Method
Case 1 When the output port is open-circuited, i.e., I2 = 0
Applying KVL to Mesh 1,
Also
V1
4 I1
g11 =
I1
V1
V2
3 I1
g21 =
V2
V1
=
1
4
=
3 I1 3
=
4 I1 4
I2 = 0
I2 = 0
Case 2 When the input port is short-circuited, i.e., V1 = 0 as shown in Fig. 13.56.
By current division rule,
I1
g12 =
3
I2
4
I1
I1
3
=−
I 2 V1 = 0
4
1Ω
2Ω
I2
+
3Ω
3 ×1
3 11
Req = 2 +
= 2+ = Ω
3 +1
4 4
11
V2
I2
4
V
11
g22 = 2
= Ω
I 2 V1 = 0 4
V2
−
Fig. 13.56
Second Method (Refer Fig. 13.55).
Applying KVL to Mesh 1,
V1
4 I1 3 I 2
…(i)
V2
4 1
3 I1 5 I 2
1 3 I2
…(ii)
1
3
V1 − I 2
4
4
…(iii)
Applying KVL to Mesh 2,
Hence,
I1
Substituting Eq. (iii) in Eq. (ii),
3 ⎞
⎛1
V1 − I 2
⎝4
4 ⎠
3
11
= V1
I2
4
4
Comparing Eqs (iii) and (iv) with g-parameter equations,
V2
⎡g
⎢⎣ g
3
⎡1
g12 ⎤ ⎢ 4
=
g22 ⎥⎦ ⎢⎢ 3
⎣4
3⎤
− ⎥
4
11 ⎥⎥
4 ⎦
I2
…(iv)
13.8 Inter-relationships between the Parameters 13.37
Example 13.25
Find g-parameters for the network shown in Fig. 13.57.
5I1
I1
1Ω
I2
+
1Ω
V2
1Ω
V1
+
−
−
Fig. 13.57
Solution The network is redrawn using source transformation as shown in Fig. 13.58.
Applying KVL to Mesh 1,
1Ω
1Ω
I1
V1 1 I1 1 ( I1 I 2 ) = 0
V1 2 I1 + I 2 …(i) +
Applying KVL to Mesh 2,
V1
1Ω
V2 5 I1 1 I 2 1 ( I 2 I1 ) 0
−
V2 6 I1 2 I 2 …(ii)
2 I1 V1 − I 2
Hence,
I1
1
1
V1 − I 2
2
2
V2
6
Substituting Eq (iii) in Eq. (ii),
1 ⎞
⎛1
V1 − I 2
⎝2
2 ⎠
− +
I2
+
V2
−
Fig. 13.58
…(iii)
2 I2
= 3 V1 − I 2
Comparing Eqs (iii) and (iv) the g- parameter equations,
1⎤
⎡1
g12 ⎤ ⎢
− ⎥
⎡g
=
2
2
⎢⎣ g
g22 ⎥⎦ ⎢
⎥
3
1
⎣
⎦
13.8
5I1
…(iv)
INTER-RELATIONSHIPS BETWEEN THE PARAMETERS
When it is required to find out two or more parameters of a particular network then finding each parameter
will be tedious. But if we find a particular parameter then the other parameters can be found if the interrelationship between them is known.
13.8.1 Z-parameters in Terms of Other Parameters
1. Z-parameters in Terms of Y-parameters We know that
1
By Cramer’s rule,
I2
Y11
11 V1 Y12 V2
Y2211 V1 Y22 V2
I1
I2
V1 =
Y11
Y21
Y12
Y22
Y I Y I
Y
Y
= 22 1 12 2 = 22 I1 − 12 I 2
Y12 Y11Y22 − Y12Y21 ΔY
ΔY
Y22
13.38 Network Analysis and Synthesis
where
Comparing with
Also,
Comparing with
ΔY = Y11Y22 − Y12Y21
V1 Z11I1 Z12 , I 2 ,
Y
Z11 = 22
ΔY
Y
Z12 = − 12
ΔY
Y11 I1
Y21 I 2
Y
Y
V2 =
= 11 I 2 − 21 I1
ΔY
ΔY
ΔY
V2 Z 21 I1 Z 22 I 2 ,
Y
Z 22 = 11
ΔY
Y
Z 21 = − 21
ΔY
2. Z-parameter in Terms of ABCD Parameters We know that
V1 AV
AV2 BI
B 2
I1 CV
V2 D
DI 2
Rewriting the second equation,
V2 =
Comparing with
V2
1
D
I1 + I 2
C
C
Z 21 I1 Z 22 I 2 ,
1
C
D
Z 22 =
C
D ⎤
⎡1
V1 A
I1 + I 2
C ⎦
⎣C
V1 Z11 I1 Z12 I 2 ,
Z 21 =
Also,
Comparing with
B 2=
BI
A
⎡ AD
I1 +
C
⎣ C
A
⎤
⎡ AD − BC ⎤
B I 2 = I1 + ⎢
⎥ I2
C
C
⎦
⎣
⎦
A
C
AD − BC
Z12 =
C
Z11 =
3. Z-parameters in Terms of A�B�C�D� Parameters We know that
V2 A′ V1 B′ I1
I 2 C ′ V1 D ′ I1
Rewriting the second equation,
V1 =
Comparing with
V1
Z11 =
D′
1
I1 + I 2
C′
C′
Z11 I1 Z12 I 2 ,
D′
C′
13.8 Inter-relationships between the Parameters 13.39
Z12 =
Also,
V2
Comparing with
V2
1
C′
1 ⎤
⎡ D′
I1 +
I2
C
′
C
′ ⎦
⎣
Z 21 I1 Z 22 I 2 ,
A′
A′
⎡ A′ D ′ − B′C ′ ⎤
B′ I1 = ⎢
⎥ I1 + C ′ I 2
C
′
⎣
⎦
A′ D ′ − B′C ′ ΔT ′
=
C′
C′
A′
Z 22 =
C′
4. Z-parameters in Terms of Hybrid Parameters We know that
Z 21 =
V1
h11 I1 h12 V2
I 2 h21 I1 h22 V2
Rewriting the second equation,
h
1
V2 = − 21 I1 +
I2
h22
h22
Comparing with
V2 Z 21 I1 Z 22 I 2 ,
h21
h22
1
=
h22
Z 21 = −
Z 22
1
h
h h
h112
⎡ h
⎤
⎡ h1 h222 − h12
1 h221 ⎤
h11 I1 h12 ⎢ − 21 I1 +
I2
h11 I1 + 12 I 2 − 12 21 I1 = ⎢ 11
⎥ I1 + h I 2
h
h
h
h
h
⎣ 22
22
⎦
22
22
⎣
222
⎦
222
Comparing with
V1 Z11 I1 Z12 I 2 ,
Also,
V1
h11h22 − h12 h21 Δh
=
h22
h22
h12
Z12 =
h22
Z11 =
5. Z-parameters in Terms of Inverse Hybrid Parameters We know that
I
V
Rewriting the first equation,
g11 V
g 21 V
V1 =
Comparing with
V1
g12 I 2
g22 I 2
1
g
I1 − 12 I 2
g11
g11
Z11 I1 Z12 I 2 ,
1
g11
g
= − 12
g11
Z11 =
Z12
Also
V
⎡ 1
⎤
g
g21 ⎢
I1 − 12 I
g11 ⎦
⎣ g11
g 22 I 2 =
⎡ g g22 − g g21 ⎤
g21
I1 + ⎢
⎥ I2
g11
g11
⎣
⎦
13.40 Network Analysis and Synthesis
Comparing with
V2
Z 21 I1 Z 22 I 2 ,
g21
g11
g g22 − g
=
g11
Z 21 =
Z 22
g21
=
Δg
g11
13.8.2 Y-parameters in Terms of Other Parameters
1. Y-parameters in terms of Z-parameters We know that
By Cramer’s rule,
V1
V2
Z11 I1 Z12 I 2
Z 21 I1 Z 22 I 2
V1 Z12
V2 Z 22
Z V Z12 V2
Z
Z
I1 =
= 22 1
= 22 V1 − 12 V2
Z11 Z12
Z11 Z 22 − Z12 Z 21 ΔZ
ΔZ
Z 21 Z 22
where
Comparing with
ΔZ = Z11 Z 22 − Z12 Z 21
I1 Y1111 V1 Y12 V2 ,
Z 22
ΔZ
Z
Y12 = − 12
ΔZ
Z11 V1
Z 21 V2
Z V Z12V1
Z
Z
I2 =
= 11 2
= − 21 V1 + 11 V2
ΔZ
ΔZ
ΔZ
ΔZ
I 2 Y21
21 V1 Y22 V2 ,
Y11 =
Also,
Comparing with
Y21 = −
Y22 =
Z 21
ΔZ
Z11
ΔZ
2. Y-parameters in Terms of ABCD Parameters We know that
V1 A
AV
V2 B
BI 2
I1
CV
V2
D
DI 2
I2 =
1
A
V1 + V2
B
B
Y21
21 V1 Y22 V2 ,
Rewriting the first equation,
Comparing with
I2
Y21 = −
Y22 =
Also,
I1
1
B
A
B
CV
V2
A ⎤ D
⎡ 1
⎡ BC − AD ⎤
D ⎢ − V1 + V2 ⎥ = V1 + ⎢
⎥ V2
B
B
B
B
⎣
⎦
⎣
⎦
13.8 Inter-relationships between the Parameters 13.41
Comparing with
I1
Y1111 V1 Y12 V2 ,
D
B
BC − AD
AD − BC
ΔT
Y12 =
=−
=−
B
B
B
Y11 =
3. Y-parameters in Terms of A�B�C�D� Parameters We know that
V2 A′V1 B′ I1
C ′V1
I2
D ′ I1
Rewriting the first equation,
I1 =
Comparing with
I1
1
A′
V1 − V2
B′
B′
Y1111 V1 Y12 V2 ,
A′
B′
1
Y12 = −
B′
Y11 =
Also,
I2
C ′V1
Comparing with
I2
Y2211 V1
and
⎡ A′
D ′ ⎢ V1
⎣ B′
Y22 V2 ,
1 ⎤
D′
⎡ A′D ′ − B′C ′ ⎤
V2 = − ⎢
⎥ V1 + B′ V2
B′ ⎥⎦
B
′
⎣
⎦
ΔT ′
⎡ A′D ′ − B′C ′ ⎤
Y21 = − ⎢
=−
⎥
B′
B′
⎣
⎦
D′
Y22 =
B′
4. Y-parameters in Terms of Hybrid Parameters We know that
V1
h11 I1 h12 V2
I2
h21 I1 h22 V2
I1 =
1
h
V1 − 12 V2
h11
h11
Y1111 V1 Y12 V2 ,
Rewriting the first equation,
Comparing with
I1
1
h11
h
Y12 = − 12
h11
Y11 =
Also
I2
Comparing with
I2
h
h
⎡ 1
⎤
⎡h h − h h ⎤
h21 ⎢ V1 − 12 V2 h22V2 = 21 V1 + ⎢ 11 22 12 21 ⎥ V2
h
h
h
h11
⎣ 11
11
⎦
11
⎣
⎦
Y2211 V1 Y22 V2 ,
h21
h11
h h −h h
Δh
= 11 22 12 21 =
h11
h11
Y21 =
Y22
13.42 Network Analysis and Synthesis
5. Y-parameters in Terms of Inverse Hybrid Parameters We know that
I
g11 V
g12 I 2
V
g21 V g22 I 2
Rewriting the second equation,
g
1
I 2 = − 21 V1 +
V2
g22
g22
Comparing with
I 2 Y2211 V1 Y22 V2 ,
g21
g22
1
=
g22
Y21 = −
Y22
Also,
I
g11 V
Comparing with
I1
Y1111 V1
⎡ g
⎤ ⎡ g g22 − g g21 ⎤
1
g
g12 ⎢ − 21 V1 +
V2 ⎥ = ⎢
V1 + 12 V2
⎥
g22 ⎦ ⎣
g22
g22
⎣ g22
⎦
Y12 V2 ,
g g22 − g g21 Δg
=
g22
g22
g
Y12 = 12
g22
Y11 =
13.8.3 ABCD Parameters in Terms of Other Parameters
1. ABCD Parameters in Terms of Z-parameters We know that
V1 Z11 I1 Z12 I 2
V2 Z 21 I1 Z 22 I 2
Rewriting the second equation,
1
Z
I1 =
V2 − 22 I 2
Z 21
Z 21
Comparing with
I1 CV
V2 D
DI 2 ,
1
C=
Z 21
Z 22
D=
Z 21
Also,
V1
Comparing with
V1
Z
Z11
Z Z
⎡ 1
⎤
Z11 ⎢
V2 − 22 I 2
Z12
V2 − 22 11 I 2
12 I 2 =
Z
Z
Z
Z 21
⎣ 21
21
⎦
21
Z
⎡ Z Z − Z12 Z 21 ⎤
= 11 V2 − ⎢ 11 22
⎥ I2
Z 21
Z 21
⎣
⎦
AV
AV2 B
BI 2 ,
Z11
A=
Z 21
Z Z − Z12 Z 21 ΔZ
B = 11 22
=
Z 21
Z 21
Z12 I 2
13.8 Inter-relationships between the Parameters 13.43
2. ABCD Parameters in terms of Y-parameters We know that
I1
Y11
11 V1 Y12 V2
I2
Y2211 V1 Y22 V2
Rewriting the second equation,
Y22
1
V2 +
I2
Y21
Y21
V1 A
AV
V2 B
BI 2 ,
Y
A = − 22
Y21
1
B=−
Y21
V1 = −
Comparing with
Also,
Comparing with
1
Y
⎡ Y
⎤
⎡Y Y − Y Y ⎤
Y11 ⎢ − 22 V2 +
I 2 Y1122 V2 = ⎢ 12 21 11 22 ⎥ V2 + 111 I 2
Y21 ⎦
Y21
Y221
⎣ Y21
⎣
⎦
I1 CV
V2 D
DI 2 ,
Y12 Y21 − Y11 Y22
ΔY
C=
=−
Y21
Y21
Y
D = − 11
Y21
I1
3. ABCD Parameters in terms of A�B�C�D� Parameters We know that
V2 A′V1 B′ I1
I2
C ′V1
D ′ I1
By Cramer’s rule,
V2 B′
I 2 D′
D′
V1 =
=
V2
A′ B′
ΔT ′
C ′ D′
where
Comparing with
Also,
Comparing with
B′
I2
ΔT ′
ΔT ′ = A′ D ′ − B′C ′
V1 AV
AV2 BI
B 2,
D′
A=
ΔT ′
B′
B=
ΔT ′
A′ V2
C′ I2
C′
A′
− I1 =
=−
V2 +
I2
ΔT ′
ΔT ′
ΔT ′
C′
A′
I1 =
V2 −
I2
ΔT ′
ΔT
ΔT ′
I1 CV
V2 D
DI 2 ,
C′
C=
ΔT ′
A′
D=
ΔT ′
13.44 Network Analysis and Synthesis
4. ABCD Parameters in Terms of Hybrid Parameters We know that
V1 h11 I1 h12 V2
I 2 h21 I1 h22 V2
Rewriting the second equation,
h
1
I1 = − 22 V2 +
I2
h21
h21
I1 CV
V2 D
DI 2 ,
Comparing with
h
C = − 22
h21
1
D=−
h21
Also,
Comparing with
h
h
⎡ 1
⎤
⎡h h − h h ⎤
h11 ⎢
I 2 − 22 V2 h12 V2 = ⎢ 12 21 11 22 ⎥ V2 + 11 I 2
h21 ⎦
h21
h21
⎣ h21
⎣
⎦
V1 A
AV
V2 B
BI 2 ,
h12 h21 − h11 h22
Δh
A=
=−
h21
h21
h
B = − 11
h21
V1
5. ABCD Parameters in Terms of Inverse Hybrid Parameters We know that
I g11 V g12 I 2
V
g21 V g22 I 2
Rewriting the second equation,
1
g
V1 =
V2 − 22 I 2
g21
g21
Comparing with
V1 A
AV
V2 B
BI 2 ,
1
A=
g21
g
B = 22
g21
Also,
Comparing with
⎡ 1
g
g11 ⎢
V2 − 22 I
g
g21
⎣ 21
I1 CV
V2 D
DI 2 ,
g
C = 11
g21
g g22 − g g21
D=
=
g21
I
⎤
⎦
g12 I 2 =
⎡ g g22 − g g21 ⎤
g11
V2 − ⎢
⎥ I2
g21
g21
⎣
⎦
Δg
g21
13.8.4 A�B�C�D� Parameters in Terms of Other Parameters
1. A�B�C �D� Parameters in Terms of Z-parameters We know that
V1
V2
Z11 I1 Z12 I 2
Z 21 I1 Z 22 I 2
13.8 Inter-relationships between the Parameters 13.45
Rewriting the first equation,
Comparing with
I2 =
1
Z
V1 − 11 I1
Z12
Z12
I2
C ′V1
D ′ I1 ,
1
Z12
Z
D ′ = 11
Z12
C′ =
Also,
V2
Comparing with
V2
Z
Z
Z Z
⎡ 1
⎤
Z 22 ⎢
V1 − 11 I1
Z1122 I1 + 22 V1 − 22 11 I1
Z12 ⎦
Z12
Z12
⎣ Z12
Z
⎡ Z Z − Z12 Z 21 ⎤
= 22 V1 − ⎢ 11 22
⎥ I1
Z12
Z12
⎣
⎦
Z 21 I1
A′ V1
B′ I1 ,
Z 22
Z12
Z Z − Z12 Z 21 ΔZ
B′ = 11 22
=
Z12
Z12
A′ =
2. A�B�C �D� Parameters in Terms of Y-parameters We know that
I1
Y1111 V1 Y12 V2
I2
Y21
21 V1 Y22 V2
Rewriting the first equation,
Y11
1
V1 +
I1
Y12
Y12
V2 A′V1 B′ I1 ,
Y
A′ = − 11
Y12
1
B′ = −
Y12
V2 = −
Comparing with
Also,
Comparing with
1 ⎤ ⎡ Y12 Y21 − Y11 Y22 ⎤
Y
⎡ Y
Y2211 V1 Y22 ⎢ − 11 V1 +
I1 ⎥ = ⎢
V1 + 222 I1
⎥
Y12 ⎦ ⎣
Y12
Y112
⎣ Y12
⎦
I 2 C ′V1 D ′ I1 ,
Y Y −Y Y
ΔY
C ′ = 12 21 11 22 = −
Y12
Y12
Y
D ′ = − 22
Y12
I2
3. A� B� C � D� Parameters in Terms of ABCD Parameters We know that
V1
I1
AV
AV2
CV
V2
B
BI 2
DI
D 2
13.46 Network Analysis and Synthesis
By Cramer’s rule,
where
Comparing with
Also,
Comparing with
V1 B
I1 D
D
B
V2 =
=
V2
I2
A B
ΔT
ΔT
C D
ΔT = AD − BC
V2 A′V1 B′ I1 ,
D
A′ =
ΔT
B
B′ =
ΔT
A V1
C I1
C
A
−I2 =
=−
V1 +
I1
ΔT
ΔT
ΔT
C
A
I2 =
V1 −
I1
ΔT
ΔT
I 2 C ′ V1 D ′ I1 ,
C
ΔT
A
D′ =
ΔT
C′ =
4. A� B� C � D� Parameters in Terms of Hybrid Parameters We know that
V1
h11 I1 h12 V2
I2
h21 I1 h22 V2
V2 =
1
h
V1 − 11 I1
h12
h12
A′ V1 B′ I1 ,
Rewriting the first equation,
Comparing with
V2
1
h12
h
B′ = 11
h12
A′ =
Also,
I2
Comparing with
I2
C′ =
h
⎡ 1
⎤ h
⎡h h − h h ⎤
h21 I1 h22 ⎢ V1 − 11 I1 ⎥ = 22 V1 + ⎢ 11 22 12 21 ⎥ I1
h12 ⎦ h12
h12
⎣ h12
⎣
⎦
C ′ V1 D ′ I1 ,
h22
h12
⎡ h h − h h ⎤ Δh
D ′ = ⎢ 11 22 12 21 ⎥ =
h12
⎣
⎦ h12
5. A� B� C� D� parameters in Terms of Inverse Hybrid Parameters We know that
I
V
g11 V
g 21 V
g12 I 2
g22 I 2
13.8 Inter-relationships between the Parameters 13.47
Rewriting the first equation,
g11
1
V1 +
I1
g12
g12
C ′ V1 D ′ I1 ,
I2 = −
Comparing with
I2
g11
g12
1
D′ = −
g12
C=−
Also,
V
g21 V
Comparing with
⎡ g
⎡ g g22 − g g21 ⎤
1 ⎤
g22
g22 ⎢ − 11 V1 +
I1 ⎥ = − ⎢
⎥ V2 + g I1
g
g
g
12
12
12
⎣ 12
⎦
⎣
⎦
V2 A′ V1 B′ I1 ,
⎡ g g22 − g g21 ⎤
Δg
A′ = − ⎢
=−
⎥
g12
g12
⎣
⎦
g22
B′ =
g12
13.8.5 Hybrid Parameters in Terms of Other Parameters
1. Hybrid Parameters in terms of Z-parameters We know that
V1
V2
Rewriting the second equation,
Z11 I1 Z12 I 2
Z 21 I1 Z 22 I 2
Z 21
1
I1 +
V2
Z 22
Z 22
h21 I1 h22 V2 ,
I2 = −
Comparing with
I2
Z 21
Z 22
1
h22 =
Z 22
h21 = −
Also,
V1
Z11 I1
Comparing with
V1
h11 I1
1
Z112
⎡ Z
⎤ ⎡ Z Z − Z12 Z 21 ⎤
Z12 ⎢ − 21 I1 +
V2 ⎥ = ⎢ 11 22
⎥ I1 + Z V2
Z
Z
Z
⎣ 22
22
⎦ ⎣
22
⎦
222
h12 V2 ,
Z11 Z 22 − Z12 Z 21 ΔZ
=
Z 22
Z 22
Z12
h12 =
Z 22
h11 =
2. Hybrid Parameters in terms of Y-parameters We know that
I1 Y11
11 V1 Y12 V2
I 2 Y2211 V1 Y22 V2
Rewriting the first equation,
1
Y
V1 =
I1 − 12 V2
Y11
Y11
13.48 Network Analysis and Synthesis
Comparing with
V1
h11 I1 h12 V2 ,
1
Y11
Y
h12 = − 12
Y11
h11 =
Also,
I2
Comparing with
I2
Y
Y
⎡ 1
⎤
⎡Y Y − Y Y ⎤
Y21 ⎢
I1 − 12 V2 Y22 V2 = ⎢ 11 22 12 21 ⎥ V2 + 21 I1
Y11 ⎦
Y11
Y11
⎣ Y11
⎣
⎦
h21 I1 h22 V2 ,
Y22
Y11
Y Y −Y Y
ΔY
h22 = 11 22 12 21 =
Y11
Y11
h21 =
3. Hybrid Parameters in Terms of ABCD Parameters We know that
V1 AV
AV2 BI
B 2
I1 CV
V2 DI
D 2
Rewriting the second equation,
1
C
I2 =
I1 + V2
D
D
Comparing with
I 2 h21 I1 h22 V2 ,
h21 = −
1
D
h22 =
C
D
Also,
V1
AV2
AV
Comparing with
V1
h11 I1
C ⎤ B
⎡ 1
⎡ AD − BC ⎤
B ⎢ − I1 + V2 ⎥ = I1 + ⎢
⎥ V2
D
D
D
D
⎣
⎦
⎣
⎦
h12 V2 ,
B
D
AD − BC ΔT
h12 =
=
D
D
h11 =
4. Hybrid Parameters in Terms of A�B�C�D� Parameters We know that
V2 A′ V1 B′ I1
I 2 C ′ V1 D ′ I1
Rewriting the first equation,
B′
1
V1 =
I1 + V2
A′
A′
Comparing with
V1 h11 I1 h12 V2 ,
B′
A′
1
h12 =
A′
h11 =
13.8 Inter-relationships between the Parameters 13.49
Also,
I2
Comparing with
I2
1 ⎤
C′
⎡ B′
⎡ A′ D ′ − B′ C ′ ⎤
I1 + V2
D ′ I1 = − ⎢
⎥ I1 + A′ V2
A
′
A
′
A
′
⎣
⎦
⎣
⎦
h21 I1 h22 V2 ,
C′
ΔT ′
⎡ A′ D ′ − B′C ′ ⎤
h21 = − ⎢
=−
⎥
A′
A′
⎣
⎦
C′
h22 =
A′
5. Hybrid Parameters in Terms of Inverse Hybrid Parameters We know that
I
g11 V
g12 I 2
V
g 21 V
g22 I 2
By Cramer’s rule,
I
V
V1 =
g
g
where
Comparing with
g12
g 22
g
g
= 22 I1 − 12 V2
g12
Δg
Δg
g22
Δg = g11 g − g12 g21
V1 h11 I1 h12 V2 ,
g22
Δg
g
h12 = − 12
Δg
h11 =
Also,
I2 =
Comparing with
I2
g
g
g
g
= − 21 I1 + 11 V2
Δg
Δg
Δg
h21 I1 h22 V2 ,
h21 = −
h22 =
I1
V2
g21
Δg
g11
Δg
13.8.6 Inverse Hybrid Parameters in Terms of Other Parameters
1. Inverse Hybrid Parameters in Terms of Z-parameters We know that
V1
V2
Z11 I1 Z12 I 2
Z 21 I1 Z 22 I 2
I1 =
1
Z
V1 − 12 I 2
Z11
Z11
g11 V g12 I 2 ,
Rewriting the first equation,
Comparing with
I
g11 =
1
Z11
13.50 Network Analysis and Synthesis
g12 = −
Also,
V2
Comparing with
V
Z12
Z11
Z
Z 21
⎡ 1
⎤
⎡ Z Z − Z12 Z 21 ⎤
Z 21 ⎢
V1 − 12 I 2
Z 22
V1 + ⎢ 11 22
22 I 2 =
⎥ I2
Z11 ⎦
Z11
Z11
⎣ Z11
⎣
⎦
g21 V g22 I 2 ,
Z 21
Z11
Z Z − Z12 Z 21 ΔZ
= 11 22
=
Z11
Z11
g21 =
g22
2. Inverse Hybrid Parameters in Terms of Y-parameters We know that
I1 Y1111 V1 Y12 V2
I2
Rewriting the second equation,
Y21
21 V1 Y22 V2
Y21
1
V1 +
I2
Y22
Y22
g21 V g22 I 2 ,
V2 = −
Comparing with
V
Y21
Y22
1
=
Y22
g21 = −
g22
Also,
I1
Comparing with
I
1
Y
⎡ Y
⎤ ⎡Y Y − Y Y ⎤
Y1111 V1 Y12 ⎢ − 21 V1 +
I 2 ⎥ = ⎢ 11 22 12 21 ⎥ V1 + 112 I 2
Y
Y
Y
Y
⎣ 22
22
⎦ ⎣
22
⎦
222
g11 V g12 I 2 ,
Y11 Y22 − Y12 Y21 ΔY
=
Y22
Y22
Y
g12 = 12
Y22
g11 =
3. Inverse Hybrid Parameters in Terms of ABCD Parameters We know that
V1 A
AV
V2 B
BI 2
I1 CV
V2 DI
D 2
Rewriting the first equation,
1
B
V2 = V1 + I 2
A
A
Comparing with
V
g21 V g22 I 2 ,
1
A
B
=
A
g21 =
g22
Also,
I1
C
B ⎤
C
⎡1
V1 + I 2
D 2 = V1
DI
A
A
A
⎣
⎦
⎡ AD − BC ⎤
⎢
⎥ I2
A
⎣
⎦
13.8 Inter-relationships between the Parameters 13.51
Comparing with
I
g11 V
g12 I 2 ,
C
g11 =
A
ΔT
⎡ AD − BC ⎤
g12 = − ⎢
=−
⎥
A
A
⎣
⎦
4. Inverse Hybrid Parameters in Terms of A�B�C�D� Parameters We know that
V2
A′ V1
B′ I1
I2
C ′ V1
D ′ I1
Rewriting the second equation,
I1 =
Comparing with
I
C′
1
V1 −
I2
D′
D′
g11 V g12 I 2 ,
C′
D′
1
g12 = −
D′
g11 =
Also,
V2
Comparing with
V
⎡ C′
B′ ⎢ V1
⎣ D′
g21 V g22 I 2 ,
A′ V1
1 ⎤ ⎡ A′ D ′ − B′C ′ ⎤
B′
I2 =
⎥ V1 + D ′ I 2
D ′ ⎥⎦ ⎢⎣
D′
⎦
⎡ A′ D ′ − B′C ′ ⎤ ΔT ′
g21 = ⎢
⎥ = D′
D′
⎣
⎦
B′
g22 =
D′
5. Inverse Hybrid Parameters in Terms of Hybrid Parameters We know that
By Cramer’s rule,
V1
h11 I1 h12 V2
I2
h21 I1 h22 V2
V1 h12
I 2 h22
h
h
I1 =
= 22 V1 − 12 I 2
h11 h12
Δh
Δh
h21 h22
where
Comparing with
Δh = h11 h22 − h12 h21
I g11 V g12 I 2 ,
h22
Δh
h
g12 = − 12
Δh
h11 V1
h21 I 2
h
h
V2 =
= − 21 V1 + 11 I 2
Δh
Δh
Δh
V
g21 V g22 I 2 ,
g11 =
Also,
Comparing with
13.52 Network Analysis and Synthesis
g21 = −
g22 =
h21
Δh
h11
Δh
Table 13.3 Inter-relationship between parameters
ΔX = X11 X 22 − X12 X 21
In terms of
[Z]
[Z]
[Y]
−
[T]
[T ′]
[h]
−
[g]
[Y]
ΔT
C
D′
C′
Y11
ΔY
1
C
D
C
ΔT ′
C′
A′
C′
Y11 Y12
D
B
−
A′
B′
−
1
B
A
B
Z12
Z 21
Z 22
Z 22
ΔZ
−
Z 21
Δ
Z11
ΔZ
Z11
Z 21
ΔZ
Z 21
−
Y22
Y21
−
1
Y21
A
1
Z 221
Z 222
Z 221
−
ΔY
Y21
−
Y11
Y21
Z 22
Z12
ΔZ
Z12
−
Y11
Y12
−
1
Z112
Z111
Z112
−
ΔY
Y12
ΔZ
Z 22
Z12
Z 22
Z 21
Z 22
1
Z 22
1
Z111
−
Z 21
Z11
ΔZ
Z11
Z12
ΔZ
Y22
ΔY
−
Y21
Δ
Y12
ΔY
−
ΔT
B
.
1
C′
−
1
B′
[g]
Δh
h22
h12
h22
1
g11
−
h21
h22
1
h22
g21
g11
Δg
g11
1
h111
−
Δg
Δ
g22
g12
g22
h21
h11
Δh
h11
g21
g
1
g22
h112
h111
g12
g11
ΔT ′
B′
D′
B′
B
D′
ΔT ′
B′
ΔT ′
−
Δh
h21
−
h11
h21
1
g21
g22
g21
C
D
C′
ΔT ′
A′
ΔT ′
−
h22
h21
−
1
h21
g11
g21
Δg
g21
1
Y12
D
ΔT
B
ΔT
A′
B′
1
h112
h111
h112
−
Δg
g12
−
g22
g12
−
Y22
Y12
C
ΔT
A
ΔT
C′
D′
h22
h12
Δh
h12
−
g11
g
−
1
g12
1
Y111
−
Y112
Y111
B
D
ΔT
D
B′
A′
1
A′
h11 h12
g22
Δg
−
g12
Δg
Y21
Y11
ΔY
Y11
1
D
C
D
ΔT ′
A′
C′
A′
h21 h22
g21
Δg
g11
Δg
ΔY
Y22
Y12
Y22
C
A
−
C′
D′
−
g
g12
Y21
Y22
1
Y22
1
A
B
A
g
g22
Y21 Y22
Z112
Z111
[h]
A
C
Z11
−
[T �]
[T]
−
−
ΔT
A
−
ΔT ′
D′
1
D′
B′
D′
−
h22
Δh
−
h12
Δh
h21
Δh
h11
Δh
−
−
Example 13.26 The Z parameters of a two-port network are Z11 = 20 W, Z22 = 30 W, Z12 = Z21
= 10 W. Find Y and ABCD parameters.
13.8 Inter-relationships between the Parameters 13.53
Solution
ΔZ = Z11 Z 22 − Z12 Z 21 = ( 20)(30) − (10)(10) = 500
Y-parameters
Z 22
30
3
=
=
,
ΔZ 500 50
Z
10
1
Y21 = − 21 = −
=−
,
ΔZ
5
500
50
Hence, the Y-parameters are
1⎤
⎡ 3
− ⎥
⎡Y11 Y12 ⎤ ⎢ 50
50
⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1
2 ⎥⎥
⎢−
⎣ 50 50 ⎦
ABCD parameters
Y11 =
Z12
10
1
=−
=−
ΔZ
500
50
Z
20
2
= 111 =
=
ΔZ 500 50
Y12 = −
222
ΔZ 500
=
= 50
Z 21 10
Z
30
D = 22 =
=3
Z 21 10
1
Z11 20
=
= 2,
Z 21 10
1
1
C=
=
= 0.1,
Z 21 10
A=
B=
Hence, the ABCD parameters are
⎡ A B ⎤ ⎡ 2 50 ⎤
⎢⎣C D ⎥⎦ = ⎢⎣0 1 3 ⎥⎦
Example 13.27 Currents I1 and I2 entering at Port 1 and Port 2 respectively of a two-port network
are given by the following equations:
I 1 0 V1 − 0.2V2
I2
0 2V1 + V2
Find Y, Z and ABCD parameters for the network.
Solution
Y11 =
I1
V1 V2
= 0.5 ,
Y12 =
0
I2
= −0.2 ,
V1 V2 = 0
Hence, the Y-parameters are
⎡Y11 Y12 ⎤ ⎡ 0.5 −0.2⎤
⎢⎣Y21 Y22 ⎥⎦ = ⎢⎣ −0 2
1 ⎥⎦
Y21 =
Y22 =
I1
V2
V1 = 0
I2
V2
V1 = 0
= −0 2
=1
Z-parameters
ΔY = Y11 Y22 − Y12 Y21 = (0.5)(1) − ( −0.2)( −0.2) = 0.46
Y22
1
=
= 2.174 Ω,
ΔY 0 46
Y2
( −0.2)
Z 21 = − 21
=−
= 0.434 Ω,
ΔY
0 46
Z12 ⎤ ⎡ 2.174 0.434 ⎤
=
Z 22 ⎥⎦ ⎢⎣ 0.434 1.087 ⎥⎦
Z11 =
⎡ Z11
⎢⎣ Z 21
Y12
( −0.2)
=−
= 0.434
3 Ω
ΔY
0 46
Y
05
= 111 =
= 1.087 Ω
ΔY 0 46
Z12 = −
Z 222
13.54 Network Analysis and Synthesis
ABCD parameters
Y22
1
=−
= 5,
Y21
−0 2
ΔY
0 46
C=−
=−
= 2.3,
Y21
−0 2
Hence, the ABCD parameters are
5⎤
⎡A B⎤ ⎡ 5
⎢⎣C D ⎥⎦ = ⎢⎣ 2.3 2.5⎥⎦
1
1
=−
=5
Y21
−0 2
Y
05
D = − 11 = −
=25
Y21
−0 2
A= −
Example 13.28
B=−
Using the relation Y = Z−1, show that | |=
1 ⎛ Z 22
Z ⎞
2
+ 111 ⎟ .
2 ⎝ Y111 Y222 ⎠
Solution We know that
Z −1
Z ⎤
⎡ Z 22
− 12 ⎥
Y12 ⎤ ⎢ ΔZ
ΔZ
=
Y22 ⎥⎦ ⎢⎢ Z 21
Z11 ⎥⎥
−
⎣ ΔZ
ΔZ ⎦
| Z | Z11 Z 22 Z12 Z 21
Y
⎡Y11
⎢⎣Y21
i.e.,
⎛
⎞
1 ⎛ Z 222 Z111 ⎞ 1 ⎜ Z 222 Z111 ⎟ 1
+
=
+
= ( ΔZ + ΔZ
Z)
2 ⎜⎝ Y111 Y222 ⎟⎠ 2 ⎜ Z 222 Z111 ⎟ 2
⎜⎝
⎟
ΔZ ΔZ ⎠
|
Example 13.29
|=
1
( ΔZ ) = ΔZ
2
Z11 Z12
Z12 Z 21
1 ⎛ Z 22 Z11 ⎞
+
2 ⎝ Y11 Y22 ⎟⎠
For the network shown in Fig 13.59, find Z and Y-parameters.
2Ω
I1
I2
+
V1
+
1Ω
2Ω
−
3I2
V2
−
Fig. 13.59
Solution The network is redrawn by source transformation technique as shown in Fig 13.60.
Applying KVL to Mesh 1,
2Ω
V1 I1 − I 3
…(i)
Applying KVL to Mesh 2,
+
V2 2( I 2 I 3 ) 6 I 2
2Ω
= −4 I 2 2 I 3
…(ii) V
1Ω
1
Applying KVL to Mesh 3,
I1
I3
−
6I2
−( − ) − 2 I 3 − 2( + ) + 6 I 2 = 0
+
−
5 I 3 = I1 + 4 I 2
1
4
…(iii)
Fig. 13.60
I 3 = I1 + I 2
5
5
+
V2
I2
−
13.8 Inter-relationships between the Parameters 13.55
Substituting Eq. (iii) in Eq. (i),
V1
1
4
I1 − I1
I2
5
5
4
4
= I1
I2
5
5
…(iv)
Substituting Eq. (iii) in Eq. (ii),
4I2 + 2
V2
⎛1
I1
⎝5
4 ⎞
I2 ⎟
5 ⎠
2
12
I1
I2
5
5
Comparing Eqs (iv) and (v) with Z-parameter equations,
4⎤
⎡4
− ⎥
⎡ Z11 Z12 ⎤ ⎢ 5
5
⎢⎣ Z 21 Z 22 ⎥⎦ = ⎢ 2
12 ⎥⎥
⎢
−
5⎦
⎣5
Y-parameters
40
8
⎛ 4 ⎞ ⎛ 12 ⎞ ⎛ 4 ⎞ ⎛ 2 ⎞
ΔZ = Z11 Z 22 − Z12 Z 21 = ⎜ ⎟ ⎜ − ⎟ − ⎜ − ⎟ ⎜ ⎟ = −
=−
⎝ 5⎠ ⎝ 5 ⎠ ⎝ 5⎠ ⎝ 5⎠
255
5
12
4
−
−
Z 22
3
Z
1
Y11 =
= 5 =
,
Y12 = − 12 = 5 = −
8
8
ΔZ
2
ΔZ
2
−
−
5
5
2
4
−
Z 21
1
Z111
1
5
Y21 = −
=
=
,
= 5 =−
222 =
8 4
8
ΔZ
ΔZ
2
−
−
5
5
Hence, the Y-parameters are
1⎤
⎡3
− ⎥
⎡Y11 Y12 ⎤ ⎢ 2
2
⎥
⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1
1
⎢
− ⎥
⎣4
2⎦
=
Example 13.30
…(v)
Find Z and h-parameters for the network shown in Fig 13.61.
4I1
2Ω
I1
+
2Ω
V1
2Ω
+−
I1
I2
+
2Ω
I3
−
V2
I2
−
Fig. 13.61
Solution Applying KVL to Mesh 1,
V1 2 I1 2( I1 I 3 )
= 4I
4 I1 2 I 3
Applying KVL to Mesh 2,
V2 2 I 2 2( I 2 I 3 )
= 4I
4 I 2 2I3
…(i)
…(ii)
13.56 Network Analysis and Synthesis
Applying KVL to Mesh 3,
−2( 3 − 1 ) − 4 I1 − 2(
+ 2) = 0
I1 + I 2 = −2 I 3
3
…(iii)
Substituting Eq. (iii) in Eq. (i),
V1 = 4 I1
I1
= 5 I1
I2
Substituting Eq. (iii) in Eq. (ii),
V2
I2
4 I 2 − I1
…(iv)
I2
…(v)
= − I1 3I 2
Comparing Eqs (iv) and (v) with Z-parameter equations,
⎡ Z11
⎢⎣ Z 21
Z12 ⎤ ⎡ 5 1⎤
=
Z 22 ⎥⎦ ⎢⎣ −1 3⎥⎦
h-parameters
ΔZ = Z11 Z 22 − Z12 Z 21 = (5) (3) − (1) ( −1) = 15 + 1 = 16
ΔZ 16
=
Ω,
Z 22
3
Z
1
h21 = − 21 = ,
Z 22 3
Hence, the h-parameters are
Z12 1
=
Z 22 3
1
1
h22 =
=
Z 22 3
h11 =
⎡16
h12 ⎤ ⎢ 3
=
h22 ⎥⎦ ⎢⎢ 1
⎣3
⎡ h11
⎢⎣ h22
Example 13.31
h12 =
1⎤
3⎥
1 ⎥⎥
3⎦
Find Y and Z-parameters for the network shown in Fig 13.62.
1Ω
2
I1
3
1Ω
2
I2
+
+
1Ω
2
2V1
V1
−
V2
−
Fig. 13.62
Solution Applying KCL at Node 3,
2(
Now,
1
3)
2V1 (
V2
V3 =
3
I1
3
2)
2V1 + (V3 V2 )
V
= 2V1 + 2 − V2
3
2
= 2V1
V2
3
…(i)
…(ii)
13.8 Inter-relationships between the Parameters 13.57
I2
2V2 + (V2 V3 )
V
= 3V2 − 2
3
8
= V2
3
…(iii)
Comparing Eqs (ii) and (iii) with Y-parameter equations,
2⎤
⎡
2 − ⎥
⎡Y11 Y12 ⎤ ⎢
3
⎥
⎢⎣Y21 Y22 ⎥⎦ = ⎢
8
⎢0
⎥
3 ⎦
⎣
Z-parameters
16
⎛ 8⎞
ΔY = Y11 Y22 − Y12 Y21 = ( 2) ⎜ ⎟ − 0 =
⎝ 3⎠
3
⎛ 2⎞
⎜⎝ − ⎟⎠ 1
Y
3
Z12 = − 12 = −
= Ω
16
ΔY
8
3
Y
2 3
Z 222 = 111 =
= Ω
ΔY 16 8
3
8
Y22
1
Z11 =
= 3 = Ω,
ΔY 16 2
3
Y21
0
Z 21 = −
=−
= 0,
16
ΔY
3
Hence, the Z-parameters are
⎡ Z11
⎢⎣ Z 21
Example 13.32
⎡1
Z12 ⎤ ⎢ 2
=
Z 22 ⎥⎦ ⎢⎢
0
⎣
1⎤
8⎥
3 ⎥⎥
8⎦
For the network shown in Fig 13.63, find Y and Z-parameters.
I1
1
1Ω
+
V1
3 V1
2
I2
+−
2Ω
+
1Ω
−
V2
−
Fig. 13.63
Solution Applying KCL at Node 1,
V1 V1 − 3V1 V2
+
2
1
3
= − V1 V2
2
I1 =
…(i)
Applying KCL at Node 2,
V2 V2 + 3V1 V1
+
1
1
= 2V
2V1 2V2
I2 =
…(ii)
13.58 Network Analysis and Synthesis
Comparing Eqs (i) and (ii) with Y-parameter equations,
⎡ 3
⎡Y11 Y12 ⎤ ⎢ −
=
⎢⎣Y21 Y22 ⎥⎦ ⎢ 2
⎣ 2
⎤
−1⎥
2 ⎥⎦
Z-parameters
⎛ 3⎞
ΔY = Y11 Y22 − Y12 Y21 = ⎜ − ⎟ ( 2) − ( −1) ( 2) = −3 + 2 = −1
⎝ 2⎠
Z11 =
Y22
2
=
= −2 Ω,
ΔY −1
Z 21 = −
Y21
2
=−
=
ΔY
( −1)
Y12
( −1)
=−
= −1 Ω
ΔY
( −1)
3
−
Y111
3
=
= 2= Ω
ΔY
−1 2
Z12 = −
Ω,
Z 222
Hence, the Z-parameters are
⎡ Z11
⎢⎣ Z 21
Example 13.33
2
⎡ −2
Z12 ⎤ ⎢
=
Z 22 ⎥⎦ ⎢ 2
⎣
1⎤
3⎥
⎥
2⎦
Find Z-parameters for the network shown in Fig 13.64. Hence, find Y and h-parameters.
0.9 I1
I1
1Ω
I2
+
+
10 Ω
V2
1Ω
V1
−
−
Fig. 13.64
Solution The network is redrawn by source transformation technique as shown in Fig 13.65.
Applying KVL to Mesh 1,
9I1
1Ω
10 Ω
I1
V1 2 I1 + I 2
…(i)
+
Applying KVL to Mesh 2,
9 I1 10 I 2 1( I1 + I 2 )
V2
−+
= 10 I1 11I 2
…(ii)
Comparing Eqs (i) and (ii) with Z-parameter equations,
1Ω
V1
−
Z 22 11
=
,
ΔZ 12
Z
10
5
Y21 = − 21 = − = −
ΔZ
12
6
Y22
Fig. 13.65
Z12
1
=−
ΔZ
12
Z
2 1
= 11 =
=
ΔZ 12 6
Y12 = −
,
+
V2
−
⎡ Z11 Z12 ⎤ ⎡ 2 1 ⎤
⎢⎣ Z 21 Z 22 ⎥⎦ = ⎢⎣10 11⎥⎦
Y-parameters
ΔZ = Z11 Z 22 − Z12 Z 21 = ( 2) (11) − (1) (10) = 22 − 10 = 12
Y11 =
I2
13.8 Inter-relationships between the Parameters 13.59
Hence, the Y-parameters are
⎡ 11
⎡Y11 Y12 ⎤ ⎢ 12
⎢⎣Y21 Y22 ⎥⎦ = ⎢ 5
⎢−
⎣ 6
−
1⎤
12 ⎥
1 ⎥⎥
6 ⎦
h-parameters
ΔZ 12
=
Ω,
Z 22 11
Z
10
h21 = − 21 = − ,
Z 22
11
Z12
1
=
Z 22 11
1
1
h22 =
=
Z 22 11
1
h11 =
Hence, h-parameters are
h12 =
1⎤
⎡ 12
⎡ h11 h12 ⎤ ⎢ 11 11⎥
⎢⎣ h21 h22 ⎥⎦ = ⎢ 10 1 ⎥
⎢−
⎥
⎣ 11 11⎦
Example 13.34
Find Y and Z-parameters of the network shown in Fig 13.66.
1
I1
1Ω
+
1Ω
V1
2V2
2V1
2
I2
−+
+
2Ω
V2
−
−
Fig. 13.66
Solution Applying KCL at Node 1,
I1 2V2 =
I1
V1 3V1 − V2
+
1
1
4V1 3V2
…(i)
Applying KCL at Node 2,
V2 V2 − 2V1 V1
+
2
1
= −3V1 1 V2
Comparing Eqs (i) and (ii) with Y-parameter equations,
3⎤
⎡Y11 Y12 ⎤ ⎡ 4
⎢⎣Y21 Y22 ⎥⎦ = ⎢⎣ −3 1.5⎥⎦
Z-parameters
ΔY = Y11Y22 − Y12Y21 = ( 4) (1.5) − ( −3)( −3) = −3
I2 =
Y22
15
=−
= −0.5 Ω,
ΔY
3
Y
( −3)
= − 21 = −
= −1 Ω,
ΔY
−3
Z11 =
Z 21
Hence, the Z-parameters are
⎡ Z11
⎢⎣ Z 21
⎡ −0 5 −1 ⎤
Z12 ⎤ ⎢
=
4⎥
Z 22 ⎥⎦ ⎢ −1 − ⎥
3⎦
⎣
…(ii)
Y12
( −3)
=−
= −1 Ω
ΔY
−3
Y
4
4
= 111 = − = − Ω
ΔY
3
3
Z12 = −
222
13.60 Network Analysis and Synthesis
Example 13.35
Determine Y and Z-parameters for the network shown in Fig 13.67.
1
I1
2
2Ω
I2
+
+
1Ω
V1
2Ω
V2
3I1
−
−
Fig. 13.67
Solution Applying KCL at Node 1,
V1 V1 − V2
+
1
2
= 1.5V1 0 5V2
I1 =
…(i)
Applying KCL at Node 2,
V2
V V
+ 3I1 + 2 1
2
2
V2
V V
=
+ 3 (1.5V1 0. V2 ) + 2 1
2
2
= 0.5
5V2 + 4 5V1 − 1.55V2 + 0 5V2 − 0 5V1
= 4 V1 − 0 5V2
Comparing Eqs (i) and (ii) with the Y-parameter equation,
⎡Y11 Y12 ⎤ ⎡1.5 −0.5⎤
⎢⎣Y21 Y22 ⎥⎦ = ⎢⎣ 4
0.5⎥⎦
Z-parameters
ΔY = Y11Y22 − Y12Y21 = (1.5)( −0.5) − ( −0.5)( 4) = 1.25
I2 =
Y22
05
=−
= −0.4 Ω,
ΔY
1.25
Y
4
= − 21 = −
= −3.2 Ω,
ΔY
1.25
Z11 =
Z 21
Hence, the Z-parameters are
⎡ Z11
⎢⎣ Z 21
Example 13.36
Y12
05
=
=04Ω
ΔY 1.25
Y
15
= 111 =
= 1.2 Ω
ΔY 1.25
Z12 = −
22
Z12 ⎤ ⎡ −0.4 0.4 ⎤
=
Z 22 ⎥⎦ ⎢⎣ −3.2 1.2 ⎥⎦
Determine the Y and Z-parameters for the network shown in Fig 13.68.
I1
0.5 Ω
1
3
2
1Ω
+
V1
…(ii)
I2
+
1Ω
2V1
−
0.5 Ω
V2
−
Fig. 13.68
13.8 Inter-relationships between the Parameters 13.61
Solution Applying KCL at Node 1,
V1 V1 − V3
+
1
0.5
= 3V
3V1 2V3
…(i)
V2 V2 − V3
+
05
1
= 3V2 V3
…(ii)
I1 =
Applying KCL at Node 2,
I2 =
Applying KCL at Node 3,
V3 V1
V V
+ 2V1 + 3 2 = 0
05
1
1
V3
V2
3
Substituting Eq. (iii) in Eqs (i) and (ii),
…(iii)
2
3V1 − V2
3
8
I 2 0V1 + V2
3
Comparing Eqs (iv) and (v) with Y-parameter equations,
2⎤
⎡
3 − ⎥
⎡Y11 Y12 ⎤ ⎢
3
⎥
⎢⎣Y21 Y22 ⎥⎦ = ⎢
8
⎢0
⎥
3 ⎦
⎣
Z-parameters
I1
…(iv)
…(v)
⎛ 8⎞
Y11Y22 Y12Y21 = (3) ⎜ ⎟ − 0 = 8
⎝ 3⎠
8
Y
1
Z11 = 22 = 3 = Ω,
ΔY 8 3
2
Y12
1
Z 21 = −
= 3=
Ω,
ΔY 8 12
ΔY
Z12
2
Y12 3 1
=−
= =
Ω
ΔY 8 12
Z 22 =
Y11 3
= Ω
YΔ
Δ 8
Hence, the Z-parameters are
⎡ Z11
⎢⎣ Z 21
Example 13.37
⎡1 1 ⎤
Z12 ⎤ ⎢ 3 12 ⎥
=
Z 22 ⎥⎦ ⎢⎢
3 ⎥⎥
0
8⎦
⎣
Determine Z and Y-parameters of the network shown in Fig. 13.69.
I1
2Ω
4Ω
+
V1
I2
+
0.05I2
+
−
−
10V1
+
V2
−
−
Fig. 13.69
13.62 Network Analysis and Synthesis
Solution Applying KVL to Mesh 1,
V1 − 4I1 − 0.05 I2 = 0
V1 = 4I1 + 0.05I2
Applying KVL to Mesh 2,
V2 − 2I2 + 10V1 = 0
V2 = 2I2 − 10V1
Substituting Eq. (i) in Eq. (ii),
V2 = 2I2 − 40I1 − 0.5I2
= −40I1 + 1.5I2
Comparing Eqs (i) and (iii) with Z-parameter equations,
⎡ Z11 Z12 ⎤ ⎡ 4 0.05⎤
⎢⎣ Z 21 Z 22 ⎥⎦ = ⎢⎣ −40 1 5 ⎥⎦
Y-parameters
ΔZ Z11Z 22 Z12 Z 21 = ( 4)(1.5) − (0.05)( −40) = 8
Z 22 1 5
=
,
ΔZ
8
Z
( −40) 40
Y221 = − 221 = −
=
ΔZ
8
8
Y11 =
Hence, the Y-parameters are
⎡1.5
⎡Y11 Y12 ⎤ ⎢ 8
⎢⎣Y21 Y22 ⎥⎦ = ⎢ 40
⎢
⎣ 8
Example 13.38
−
…(i)
...(ii)
...(iii)
Z112
0 05
=−
ΔZ
8
Z111 4
=
=
ΔZ 8
Y12 = −
,
Y222
0.05 ⎤
8 ⎥
4 ⎥⎥
8 ⎦
Determine Z and Y-parameters of the network shown in Fig. 13.70.
2V3
1Ω
I1
+
3I2
I3
+−
I2
+
2Ω
V1
2Ω
I1
−
+
V3
−
V2
I2
−
Fig. 13.70
Solution Applying KVL to Mesh 1,
V1 − 1I1 − 3I2 − 2(I1 + I2) = 0
V1 = 3I1 + 5I2
Applying KVL to Mesh 2,
V2 − 2(I2 − I3) − 2(I1 + I2) = 0
V2 − 2I2 + 2I3 − 2I1 − 2I2 = 0
V2 = 2I1 + 4I2 − 2I3
...(i)
…(ii)
13.9 Interconnection of Two-Port Networks 13.63
Writing equation for Mesh 3,
I3 = 2V3
...(iii)
From Fig. 13.70,
V3 = 2(I1 + I2)
I3 = 2V3 = 4I1 + 4I2
...(iv)
Substituting Eq. (iv) in Eq. (ii),
V2 = −6I1 − 4I2
Comparing Eqs (i) and (v) with Z-parameter equations,
5⎤
⎡ Z11 Z12 ⎤ ⎡ 3
⎢⎣ Z 21 Z 22 ⎥⎦ = ⎢⎣ −6
6
4 ⎥⎦
Y-parameters
ΔZ = Z111 Z22 − Z112 Z221 = ( )(−
)( ) − ( )(−
)( ) = 18
Z 22
4
2
Z
5
Y11 =
=− =−
,
Y112 = − 12 = −
ΔZ
ΔZ
18
9
18
Z
( −66) 1
Z11 3
Y21 = − 21 = −
=
,
=
222 =
ΔZ
18
3
ΔZ 18
Hence, Y-parameters are
5⎤
⎡ 2
−
− ⎥
⎡Y11 Y12 ⎤ ⎢ 9
18
⎢⎣Y21 Y22 ⎥⎦ = ⎢ 1
3 ⎥⎥
⎢
18 ⎦
⎣ 3
13.9
...(v)
INTERCONNECTION OF TWO-PORT NETWORKS
We shall now discuss the various types of interconnections of two-port networks, namely, cascade, parallel,
series, series-parallel and parallel-series. We shall derive the relation between the input and output quantities
of the combined two-port networks.
13.9.1 Cascade Connection
1. Transmission Parameter Representation Figure 13.71 shows two-port networks connected in
cascade. In the cascade connection, the output port of the first network becomes the input port of the
second network. Since it is assumed that input and output currents are positive when they enter the
network, we have
I1 ′
I1
−
N1
I2′
I1′
I2
+
V1
I2
+
+
V2
V1′
−
−
+
N2
V2′
−
Fig. 13.71 Cascade Connection
Let A1,B1,C1,D1 be the transmission parameters of the network N1 and A2,B2,C2,D2 be the transmission
parameters of the network N2.
For the network N1,
⎡V1 ⎤ ⎡ A1 B1 ⎤ ⎡ V2 ⎤
...(13.1)
⎢⎣ I1 ⎥⎦ = ⎢⎣C1 D1 ⎥⎦ ⎢⎣ − I 2 ⎥⎦
13.64 Network Analysis and Synthesis
For the network N2,
Since V1′
V2 and I 2′
⎡V1 ′ ⎤ ⎡ A2
⎢⎣ I1 ′ ⎥⎦ = ⎢⎣C2
B2 ⎤ ⎡ V2 ′ ⎤
D2 ⎥⎦ ⎢⎣ − I 2 ′ ⎥⎦
...(13.2)
I 2 , we can write
⎡ V2 ⎤ ⎡ A2 B2 ⎤ ⎡ V2 ′ ⎤
⎢⎣ − I 2 ⎥⎦ = ⎢⎣C2 D2 ⎥⎦ ⎢⎣ − I 2 ′ ⎥⎦
Combining Eqs (13.1) and (13.3),
⎡V1 ⎤ ⎡ A1 B1 ⎤ ⎡ A2 B2 ⎤ ⎡ V2 ′ ⎤ ⎡ A B ⎤ ⎡ V2 ′ ⎤
⎢⎣ I1 ⎥⎦ = ⎢⎣C1 D1 ⎥⎦ ⎢⎣C2 D2 ⎥⎦ ⎢⎣ − I 2 ′ ⎥⎦ = ⎢⎣C D ⎥⎦ ⎢⎣ − I 2 ′ ⎥⎦
...(13.3)
⎡ A B ⎤ ⎡ A1 B1 ⎤ ⎡ A2 B2 ⎤
...(13.4)
⎢⎣C D ⎥⎦ = ⎢⎣C1 D1 ⎥⎦ ⎢⎣C2 D2 ⎥⎦
Equation 13.4 shows that the resultant ABCD matrix of the cascade connection is the product of the
individual ABCD matrices.
2. Inverse Transmission Parameter Representation Figure 13.72 shows two-port networks
connected in cascade. Since it is assumed that input and output currents are positive when they enter the
network, we have
Hence,
−I1′ = I2
I1
N1
V1
−
I2′
I1′
I2
+
+
+
V2
V1′
−
−
+
N2
V2′
−
Fig. 13.72 Cascade connection
A1′
B1′ , C1′
D1′
Let
be the transmission parameters of the network N1 and A2′ B2′ , C2′ D2′ be the
transmission parameters of the network N2.
For the network N1,
⎡V2 ⎤ ⎡ A1 ′ B1 ′ ⎤ ⎡ V1 ⎤
...(13.5)
⎢⎣ I 2 ⎥⎦ = ⎢⎣C1 ′ D1 ′ ⎥⎦ ⎢⎣ − I1 ⎥⎦
For the network N2,
⎡V2 ′ ⎤ ⎡ A2 ′ B2 ′ ⎤ ⎡ V2 ′ ⎤
...(13.6)
⎢⎣ I 2 ′ ⎥⎦ = ⎢⎣C2 ′ D2 ′ ⎥⎦ ⎢⎣ − I1 ′ ⎥⎦
Since V1′ V2 and − I1′ I 2 , we can write
⎡V2 ′ ⎤ ⎡ A2 ′ B2 ′ ⎤ ⎡V2 ⎤
⎢⎣ I 2 ′ ⎥⎦ = ⎢⎣C2 ′ D2 ′ ⎥⎦ ⎢⎣ I 2 ⎥⎦
Combining equations (13.5) and (13.7),
⎡V2 ′ ⎤ ⎡ A2 ′ B2 ′ ⎤ ⎡ A1 ′ B1 ′ ⎤ ⎡ V1 ⎤ ⎡ A′
⎢⎣ I 2 ′ ⎥⎦ = ⎢⎣C2 ′ D2 ′ ⎥⎦ ⎢⎣C1 ′ D1 ′ ⎥⎦ ⎢⎣ − I1 ⎥⎦ = ⎢⎣C ′
⎡ A′ B′ ⎤ ⎡ A2 ′ B2 ′ ⎤ ⎡ A1 ′ B1 ′ ⎤
Hence,
⎢⎣C ′ D ′ ⎥⎦ = ⎢⎣C2 ′ D2 ′ ⎥⎦ ⎢⎣C1 ′ D1 ′ ⎥⎦
...(13.7)
B′ ⎤ ⎡ V1 ⎤
D ′ ⎥⎦ ⎢⎣ − I1 ⎥⎦
...(13.8)
Equation 13.8 shows that the resultant A′B′C′D′ matrix of the cascade connection is the product of the
individual A′B′C′D′ matrices.
13.9 Interconnection of Two-Port Networks 13.65
Example 13.39 Two identical sections of the network shown in Fig. 13.73 are connected in cascade. Obtain the transmission parameters of the overall connection.
2Ω
I1
2Ω
I2
+
+
1Ω
V1
2Ω
V2
−
−
Fig. 13.73
Solution The network is redrawn as shown in Fig. 13.74.
Applying KVL to Mesh 1,
V1 = 3I1 − I3
…(i)
Applying KVL to Mesh 2,
+
V2 = 2I2 + 2I3
…(ii)
Applying KVL to Mesh 3,
V1
−I1 + 2I2 + 5I3 = 0
1
2
I3
I1 − I 2
…(iii) −
5
5
Substituting Eq. (iii) in Eq. (i),
2 ⎞
⎛1
V1 3I1 − ⎜ I1
I2 ⎟
⎝5
5 ⎠
=
14
I1
5
2
I2
5
2I 2
⎛1
2 ⎜ I1
⎝5
2Ω
2Ω
+
1Ω
I1
2Ω
I3
V2
I2
−
Fig. 13.74
…(iv)
Substituting Eq. (iii) in Eq. (ii),
V2
=
I1
2 ⎞
I2 ⎟
5 ⎠
2
6
I1
I2
5
5
5
V2 − 3I 2
2
…(v)
Substituting Eq. (v) in Eq. (iv),
V1
14 ⎛ 5
⎞
V2 − 3I 2
⎝
⎠
5 2
= 7V
7V2 8 I 2
2
I2
5
Comparing the Eqs (vi) and (v) with ABCD parameter equations,
⎡ A1 B1 ⎤ ⎡ 7 8⎤
⎢⎣C1 D1 ⎥⎦ = ⎢⎣ 2 5 3⎥⎦
Hence, transmission parameters of the overall cascaded network are
⎡ A B ⎤ ⎡ 7 8⎤ ⎡ 7 8⎤ ⎡69 80 ⎤
⎢⎣C D ⎥⎦ = ⎢⎣ 2.5 3⎥⎦ ⎢⎣ 2.5 3⎥⎦ = ⎢⎣ 25 29⎥⎦
…(vi)
13.66 Network Ana