KARNATAKA STATE OPEN UNIVERSITY
COURSE NAME:
DIPLOMA IN ELECTRICAL ENGINEERING
YEAR/SEMESTER:
3RD SEMESTER
PAPER NAME:
CIRCUIT THEORY
PAPER CODE:
DEE 301
NATIONAL COLLABORATIVE PARTNER
Syllabus as per Karanataka State Open University
BLOCK 1
DC Circuits
Unit 1: Electro Statics
Unit 2: Basic Laws
Unit 3: Series Circuits
Unit 4: Parallel Circuits
Unit 5: Problems in Series and Parallel Circuits
BLOCK 2
Network Theorems
Unit 1: Node voltage Analysis
Unit 2: Mesh Current Analysis
Unit 3: Star Delta Transformations
Unit 4: Thevenins and Norton Theorems
Unit 5: Superposition and Maximum Power Transform Theorem
BLOCK 3
Single Phase AC Circuits
Unit 1: Basic Definitions
Unit 2: Resistor, Capacitor and Inductor in AC Circuits
Unit 3: RL and RC Circuits
Unit 4: Series RLC Circuits
Unit 5: Parallel RLC Circuits
BLOCK 4
Three Phase AC Circuits
Unit 1: Star Delta Connections
Unit 2: Balanced Load and Unbalanced Load in Three Phase Circuits
Unit 3: Measurements of Three Phase Power
Unit 4: Effects of unbalanced Load
Unit 5: Problems in Three Phase Circuits
BLOCK 5
DC Transients
Unit 1: Basics of Transients
Unit 2: Transients in RL
Unit 3: Transients in RC
Unit 4: Transients in RLC
Unit 5: Problems
3
CONTENTS
Page No.
BLOCK 1 DC CIRCUITS
Unit 1
Unit 2
Unit 3
Unit 4
ELECTRO STATIC’S
1.0
Objectives
1.1
Introduction
1.2
Basic Definitions
1.3
Coulomb‟s Laws of Electrostatics
1.4
Capacitor
1.5
Capacitance of Parallel Plate Capacitor
1.6
Energy Stored in a Capacitor
1.7
Types of Capacitors
1.7.1 Fixed Capacitors
1.7.2 Variable Capacitor
1.8
Solved Problems
1.9
Key Words
1.10 Question for Discussions
1.11 Suggested Readings
BASIC LAWS
2.0
Objectives
2.1
Introduction
2.2
Basic definitions
2.2.1 Current
2.2.2 Potential difference
2.2.3 Resistance
2.3
Ohms Law
2.4
Law of Resistance
2.5
Kirchhoff‟s Law
2.6
Solved Problems
2.7
Key Words
2.8
Question for Discussions
2.9
Suggested Readings
SERIES CIRCUITS
3.0
Objectives
3.1
Introduction
3.2
Series Circuit
3.3
Simple series circuits
3.4
Key Words
3.5
Question for Discussions
3.6
Suggested Readings
PARALLEL CIRCUITS
4.0
Objectives
4.1
Introduction
4.2
Parallel Circuits
4.3
Simple Parallel circuit
4.4
Key Words
4.5
Question for Discussions
4.6
Suggested Readings
4
2
2
3
3
3
4
5
6
7
7
8
8
10
11
11
13
13
14
14
14
14
15
15
17
18
19
27
27
28
30
30
31
31
32
34
34
34
36
36
37
37
38
40
40
40
Unit 5
PROBLEMS IN SERIES AND PARALLEL CIRCUITS
5.0
Objectives
5.1
Introduction
5.2
Solved Problem
5.3
Question for Discussions
5.4
Suggested Readings
Unit 1
NODE VOLTAGE ANALYSIS
1.0
Objectives
1.1
Introduction
1.2
Node voltage method
1.3
Node voltage rules
1.4
Solved Problems
1.5
Key Words
1.6
Question for Discussions
1.7
Suggested Readings
MESH CURRENT ANALYSIS
2.0
Objectives
2.1
Introduction
2.2
Mesh current method
2.2.1 Mesh Current, conventional method
2.2.2 Mesh current by inspection
2.3
Mesh current rules
2.4
Solved Problems
2.5
Key Words
2.6
Question for Discussions
2.7
Suggested Readings
STAR DELTA TRANSFORMATIONS
3.0
Objectives
3.1
Introduction
3.2
Star delta Connections
3.3
Delta star transformation
3.4
Star Delta Transformation
3.5
Solved Problems
3.6
Key Words
3.7
Question for Discussions
3.8
Suggested Readings
THEVENIN’S AND NORTON THEOREM
4.0
Objectives
4.1
Introduction
4.2
Thevenin Theorem
4.3
Step to Solve the Thevenin‟s Theorem
4.4
Norton‟s Theorem
4.5
Step to Solve Norton‟s Theorem
4.6
Thevenin-Norton equivalencies
4.7
Simple Problem
4.8
Key Words
4.7
Question for Discussions
4.8
Suggested Readings
42
42
42
44
49
50
BLOCK 2 NETWORK THEOREMS
Unit 2
Unit 3
Unit 4
5
52
52
53
53
55
55
59
59
60
61
61
62
62
62
65
66
66
69
70
71
72
72
73
73
74
75
77
78
78
79
80
80
81
81
82
84
85
87
87
90
90
92
Unit 5
SUPERPOSITION & MAXIMUM POWER THEOREM
5.0
Objectives
5.1
Introduction
5.2
Superposition Theorem
5.3
Demerits of Superposition Theorem
5.4
Maximum Power Transfer Theorem
5.5
Demerits of Power Transfer Theorem
5.6
Simple Problem
5.7
Key Words
5.8
Question for Discussions
5.9
Suggested Readings
93
93
94
94
97
98
99
99
102
102
103
Block 3 SINGLE PHASE AC CIRCUITS
Unit 1
Unit 2
Unit 3
Unit 4
BASIC DEFINITIONS
1.0
Objectives
1.1
Introduction
1.2
Sinusoidal Voltage and Current
1.3
Basic Terms in AC circuits
1.4
Average and Effective Value
1.5
Form and Peak Factor
1.6
Phasor
1.7
Simple Problems
1.8
Key Words
1.9
Questions for Discussions
1.10 Suggested Readings
RESISTOR, CAPACITOR AND INDUCTOR IN AC
CIRCUITS
2.0
Objectives
2.1
Introduction
2.2
Reactance and Impedance– Inductive
2.2.1 AC resistor circuits
2.2.2 AC Inductor Circuits
2.2.3 AC capacitor circuits
2.3
Simple Problem
2.4
Key Words
2.5
Question for Discussions
2.6
Suggested Readings
RL AND RC CIRCUITS
3.0
Objectives
3.1
Introduction
3.2
Series RL circuits
3.3
Series RC circuits
3.4
Parallel RL circuits
3.5
R-C Parallel Circuits
3.6
Simple Problems
3.7
Key Words
3.8
Question for Discussions
3.9
Suggested Readings
SERIES RLC CIRCUITS
4.0
Objectives
6
105
105
106
106
107
108
110
111
111
113
113
114
115
115
116
116
116
117
120
122
123
124
124
125
125
126
126
127
128
128
129
134
134
135
136
136
4.1
4.2
4.3
4.4
Unit 5
R-L-C Series Circuit Constructions
Impedance for RLC Series Circuit
Power in RLC series Circuit
RLC series Resonance
4.4.1 Effects of Series Resonance
4.4.2 Quality Factor of Series Resonance Circuit
4.4.3 Resonance Cure
4.5
Solved Problem
4.6
Key Words
4.7
Question for Discussions
4.8
Suggested Readings
PARALLEL RLC CIRCUITS
5.0
Objectives
5.1
Introduction
5.2
RLC Parallel Resonance Circuit
5.3
Parallel Resonance Characteristics
5.4
Key Words
5.5
Question for Discussions
5.6
Suggested Readings
137
137
138
138
139
139
140
141
142
142
143
145
145
146
146
147
149
149
149
BLOCK 4 THREE PHASE AC CIRCUITS
Unit 1
Unit 2
STAR DELTA CONNECTIONS
1.0
Objectives
1.1
Introduction
1.2
Generation of three Phase System
1.3
Advantages of 3 Phase System
1.4
Phase Sequence
1.5
Interconnection of three phases
1.5.1 Star or Wye Connection
1.5.2 Delta or Mesh Connection
1.6
Three Phase Voltage
1.6.1 Line Voltage
1.6.2 Phase Voltage
1.7
Three Phase Current
1.7.1 Phase Current
1.7.2 Line Current
1.8
Key Words
1.9
Question for Discussions
1.10 Suggested Readings
BALANCED LOAD AND UNBALANCED LOAD IN THREE
PHASE CIRCUITS
2.0
Objectives
2.1
Introduction
2.2
Voltage in balanced Delta Connection
2.3
Line & Phase Current in balanced Delta Connection
2.4
Voltage in balanced Star Connection
2.5
Line & Phase Current in Balanced Star Connection
2.7
Key Words
2.8
Question for Discussions
2.9
Suggested Readings
7
151
151
152
152
153
154
154
154
155
155
155
155
156
156
156
156
156
157
158
158
159
159
160
160
162
162
162
162
Unit 3
Unit 4
Unit 5
MEASUREMENTS OF THREE PHASE POWER
3.0
Objectives
3.1
Introduction
3.2
Single Wattmeter Method
3.3
Two wattmeter Method
3.4
Three Wattmeter Method
3.5
Key Words
3.6
Question for Discussions
3.7
Suggested Readings
EFFECTS OF UNBALANCED LOAD
4.0
Objectives
4.1
Introduction
4.2
Unbalanced Delta connected Load
4.3
Unbalanced 3phase, 3 wire Star connected Load
4.4
Unbalanced 3 phase, 4 wire Star Connected Load
4.5
Key Words
4.6
Question for Discussions
4.7
Suggested Readings
PROBLEMS IN THREE PHASE CIRCUITS
5.0
Objectives
5.1
Introduction
5.2
Star – Delta Problems
5.3
Power Measurement Problems
5.4
Question for Discussions
5.5
Suggested Readings
163
163
164
164
166
168
169
169
169
170
170
171
171
172
173
174
174
174
175
175
176
176
181
185
186
BLOCK 5 DC TRANSIENTS
Unit 1
Unit 2
Unit 3
BASICS OF TRANSIENTS
1.0
Objectives
1.1
Circuit Transients
1.2
Laplace Transformation
1.3
Advantages of Laplace Transformation Technique
1.4
Key Words
1.5
Question for Discussions
1.6
Suggested Readings
TRANSIENTS IN RL
2.0
Objectives
2.1
DC Transient in RL Circuit
2.2
Voltage across Resistor in RL transient Circuit
2.3
Voltage across Inductor in RL transient circuit
2.4
RL Decay Transmit
2.5
Key Words
2.6
Question for Discussions
2.7
Suggested Readings
TRANSIENTS IN RC
3.0
Objectives
3.1
DC Transient in RC Circuit
3.2
Voltage across Resistor in RC transient
3.3
Voltage across Inductor in RC transient
3.4
Decaying Transient in RC Circuit
8
188
188
189
189
190
191
191
191
192
192
193
195
195
196
197
197
197
198
198
199
200
200
201
Unit 4
Unit 5
3.5
Key Words
3.6
Question for Discussions
3.7
Suggested Readings
TRANSIENTS IN RLC
4.0
Objectives
4.1
DC Transient in RLC Circuit
4.2
Varies result of Current in the RLC transient
4.3
Key Words
4.4
Question for Discussions
4.5
Suggested Readings
PROBLEMS
5.0
Objectives
5.1
Problem for RL Transient Circuit
5.2
Problem for RC Transient Circuit
5.3
Problem for RLC Transient Circuit
5.4
Question for Discussions
5.5
Suggested Readings
9
201
201
201
203
203
204
204
206
207
207
208
208
209
212
215
217
217
10
UNIT
1
ELECTRO STATIC
CONTENTS
1.0 Objectives
1.1 Introduction
1.2 Basic Definitions
1.3 Coulomb‟s Laws of Electrostatics
1.4 Capacitor
1.5 Capacitance of Parallel Plate Capacitor
1.6 Energy Stored in a Capacitor
1.7 Types of Capacitors
1.7.1 Fixed Capacitors
1.7.2 Variable Capacitor
1.8 Solved Problems
1.9 Keywords
1.10 Questions for Discussion
1.11 Suggested Readings
1.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to





Define the electrical field, flux, flux density, electrical potential and field intensity.
Understand the Coulomb‟s Law of Electrostatic.
Explain the construction details of Parallel Plate Capacitor.
Understand the various types of capacitor.
Solve the simple series and parallel connection of capacitors.
1.1 Introduction
11
Electrostatics is the branch of science which deals with the phenomena associated
with electricity at rest. Static electricity is generated due to friction between two surfaces. The
excess or deficit of electronics in body is referred. When a dry glass rod is rubbed with silk
cloth, the glass rod is positively charged and silk cloth is negatively charged. The electric
charge on the silk cloth is stationary and is known as static electricity.
1.2 Basic Definitions
Electric Field:
The space or field in which a change experiences a force is called an
electric field. The electric field around a charged body is represented by
imaginary lines, called electric lines of force.
Electric flux:
The total electric lines of force which flow outward from a positive
charge are called electric flux. It is measured in Coulombs. The symbol
of electric flux is Ф.
Electric flux
Density:
The flux density of an electric field is defined as the electric flux
crossing normally per unit area.
i.e. Electric flux Density (D) = Ф / A Coulombs/m2
Electric Field
Intensity:
The electric field intensity at any point is defined as the force on a unit
positive charge placed at that point.
i.e., Electric Field intensity (D) = F / Q
Where F – Force in Newtons acting on Q Coulombs.
Q – Charge in Coulombs placed at that point.
Electrical
Potential:
Electric potential at a point in an electric field may be defined as the
amount of work done in bringing positive charge of one coulomb from
infinity to that point against the electric field.
1.3 Coulomb’s Laws of Electrostatics
First Law:
Like charges of electricity repeal each other, whereas unlike charges attract each
other.
Second Law:
The force exerted between two point charges is directly proportional to the product of
their strengths and inversely proportional to the square of the distance between them.
Q1
d
Q2
12
Fig 1.1 Coulombs Law
Mathematically,
F α
F =K
Q1Q2
d2
Q1Q2
d2
Where, K is the constant of proportionality and it is given by
1
K =
4πε0εr
Where
ε0
- Absolute permittivity of vacuum or air
εr
- Relative permittivity of the medium in which the charges are placed.
The value of εr for vacuum or air is one(1). And the value of ε0 is 8.854 x 10-12 Farad / Meter
Q1Q2
F =
4πε0εrd2
1.4 Capacitor
A capacitor essentially consists of two conducting surfaces separated by an insulating
material called dielectric. It has the property to store electrical energy in the form of
electrostatic stress in the dielectric.
The ability of a capacitor to store capacitor is known as its capacitance. It has been
found experimentally that the charge Q stored in a capacitor is directly proportional to the
potential difference across it.
QαV
Q = CV
Where, C is a constant, called the capacitance of the capacitor. Hence the capacitance is
defined as the charge required per unit potential difference.
i.e.,
The unit of capacitance is Farad.
Farad is actually too large for practical purpose. Hence much smaller units like micro farad
(μF) and pico farad (ρF) are generally employed.
1 μF = 10-6 Farad
1 ρF = 10-12 Farad
1.5 Capacitance of Parallel Plate Capacitor
Consider a parallel plate capacitor consisting of two plates M and N each of area A m2
and separated by a uniform dielectric medium of thickness “d” meters and relative
permittivity εr as shown in the fig1.2.
13
Fig 1.2 Parallel Plate Capacitor
Let a potential difference of V volts applied between the plates.
Q – Charge on the plates
E – Electric field intensity
D – Flux density between the plates
Now, the flux density D = Q/A coulombs/m2
………………… (1.1)
Field intensity E = V/d Volt/meter
………………… (1.2)
But, D = ε E
………………… (1.3)
Equating (1.1) and (1.3) equation,
Q
= εE
A
Sub suite the E value in the above equation
Q
V
= ε
A
d
Q
εA
=
V
d
Q
ε εA
= 0 r
V
d
C =
ε0εrA
d
Where C = Q/V
1.6 Energy Stored in a Capacitor
Consider a capacitor of C farads being charged from a voltage of V volts as shown in fig 1.3.
Fig 1.3 Capacitor connected to Battery
14
Where V – The instantaneous voltage across the capacitor
i – The charging current through the capacitor
The instantaneous power P = V.i
Therefore, The energy stored in time dt is dW = Power x Time
dW = V i dt
= V.C
dV
dt
dt
dW = V C dV
1
CV2 Joules
2
W =
1.7 Types of Capacitors
Capacitors are mainly classified into two types. They are
 Fixed Capacitors
 Variable Capacitors
1.7.1 Fixed Capacitors
The fixed capacitors can be classified according to the nature of the dielectric used.
The main categories are mentioned below
i.
Paper Capacitors
In paper capacitors, the two electrodes are of metal foils interleaved with waxed or
oiled paper and rolled into a compact form.
The applications of paper capacitors are
 Used for decoupling of stages and circuits
 Used for timing circuits
 Used for smoothing the circuits
 Used for power factor correction
 Used for Motor starting
 Used for contact protection
ii.
Mica capacitor
These are similar to paper capacitors except the dielectric used. Both the metal foils are
separated by a flat mica sheet. Losses in these capacitors are extremely low and these are
used in high quality equipments.
15
The application of mica capacitors are
 Used in the Blocking circuits
 Used in the bypass circuits
 Used in the buffer circuits
 Used for filter circuits
 Used for coupling circuits
 Used in fine tuning circuits
iii.
Ceramic Capacitors
In these types of capacitors ceramic materials are used as dielectric between metallic
electrodes.
These types of capacitor are used for the following circuits
 By pass radio frequency
 Coupling the stages
 Filtering
 Temperature compensation
iv.
Electrolytic Capacitors
In this, two aluminum foils are used as electrodes. One of them is coated with oxide
film and the other is without coat. The two foils are separated by paper saturated, with
ammonium borated electrolyte. The oxide film acts as dielectric between the plates. As the
film is very thin, a very high capacitance may be obtained.
These types of capacitor are used for the following circuits
 Ripple filters
 Blocking DC
 AF transistor amplifier
 Used for single phase induction motor starting
 Used for photo flash
1.7.2 Variable Capacitor
A capacitor whose capacitance can be varied is called variable capacitor. This is done
by varying the thickness of the dielectric.
In this type there are two sets of plates, one is fixed and the other is movable.
Capacitance varies with the distance between the plates. Air is the dielectric medium of this
type of capacitor.
Applications
 are used in radar systems
1.8 Solved Problems
1. A charge of 10 micro coulomb is placed 100 cm away from a charge of 50 micro
coulomb in a medium of relative permittivity 4.5. Calculate the force between them.
16
Given data:
Q1 = 10 μC = 10 x 10-6C
Q2 = 50 μC = 50 x 10-6C
d = 100 cm = 100 x 10-2 m
εr = 4.5
Find data:
Force (F)
Solution:
F =
Q1Q2
4πε0εrd2
F =
10 x 10-6 x 50 x 10-6
4π x 8.854 x 10-12 x 4 x (100 x 10-2)2
F = 1N
Result:
Force is 1N
2. The repelling force between two equal charges is found to be 0.5 N. if they are placed
50 cms apart, find the charge on each.
Given Data:
Force = 0.5 N
d = 50 cm = 50 x 10-2 m
Q1 = Q2 (Repelling force)
Find data:
Q1 & Q2
Solution:
F =
Assume Q1 = Q2 = Q
0.5 =
Q1Q2
4πε0εrd2
Q
4π x 8.854 x 10 x 1 x (50 x 10-2)2
-12
Q = 3.729 x 10-6 Coulombs
Result:
Q1 = Q2 = Q = 3.729 x 10-6 C
3. A capacitor supplied from 250 V DC mains, takes 25 milli coulombs. Find its
capacitance. What will be its charge if the voltage is raised to 1000 V.
Given data:
V = 250 Volts
17
Q = 25 mC = 25 x 10-3 C
Find data:
C
Q at 1000V
Solution:
Q = CV
C = Q/V
C = 25 x 10-3/250
= 1 x 10-4 Farad = 100 μF
Charge if the voltage is raised to 1000V
Q = CV
= 1 x 10-4 x 1000
= 0.1 C
Result:
C = 1 x 10-4 F
Q = 0.1 C (At 1000V raised)
1.9 Keywords














Electrostatics
Electric Field
Electric flux
Electric flux Density
Electric charge
Electric Field Intensity
Electrical Potential
Coulomb
Capacitor
Capacitance
Potential difference
Farad
Dielectric medium
Permittivity
1.10 Questions for Discussion
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
What is static electricity?
Define electric field.
Define electric flux.
Define electric flux density.
Define electric Potential.
What is capacitor? State the unit of Capacitor.
Define the Farad.
Sate and explain Coulomb‟s Laws of electrostatics.
Derive the expression for energy stored in a capacitor.
Explain various types of capacitors and state their applications.
Derive the expression Capacitance of parallel plate capacitor.
18
The positive charges of 10μC and 20μC are placed in a medium of relative
permittivity 20. Find the force between them if the distance between the
charges is 10cms.
13. The distance between two charges of 20μC and 50μC is 200cms. The relative
permittivity of the medium is 5. Calculate the force between them.
12.
1.11 Suggested Readings
a. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
b. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
c. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
d. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
e. Tony Kuphaldt,Using the Spice Circuit Simulation Program, in“Lessons in
Electricity,
Reference”,
Volume
5,
Chapter
7,
at
http://www.ibiblio.org/obp/electricCircuits/Ref/
f. Electrical Circuit Theory, Gopalsamy, Veni Publication
19
UNIT
2
BASIC LAWS
CONTENTS
2.0 Objectives
2.1 Introduction
2.2 Basic definitions
2.2.1 Current
2.2.2 Potential difference
2.2.3 Resistance
2.3 Ohms Law
2.4 Law of Resistance
2.5 Kirchhoff‟s Law
2.6 Solved Problems
2.7 Keywords
2.8 Questions for Discussion
2.9 Suggested Readings
2.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to





Define the current, voltage, resistance, power, work and energy.
Understand the Ohms Law.
Understand the various factor to be affected the resistance value
Explain the Kirchhoff‟s Law.
Solve the simple circuit using ohms law and Kirchhoff‟s Law.
20
2.1 Introduction
Electrical energy is produced by batteries and generators. All the matter has electricity
in the form of electrons and protons. Electron motion in a closed path provides electric
current. Potential difference must exist for the current to flow in a closed path. The flow of
current can be felt only by its effect. Electrical motors, fans, lamps, electrical bell, heater,
soldering iron are some of the appliance which works on electricity.
Every atom in which the positive and negative charges are equal has no net positive or
negative charge. The structural arrangement of an atom consists of a central positive nucleus
around which a number of electrons circulate in various orbits like the planets round the sun,
except that the orbits are described in different planes.
The Direct Current is known as DC current. It is generated by a battery which consists
of an anode and cathode. The current flows between cathode and anode. The amplitude of the
DC current is constant and it is independent of time. That is, DC current is having fixed
polarity.
2.2 BASIC DEFINITIONS
Electricity is most essential for human life. Basic laws in electricity and definitions
are explained as follows.
2.2.1 Current
The flow of electrons in one direction along any path or around any circuit is called
electric current. It is also state that, the flow of free electrons in any conductor.
Its symbol is I and its unit is Ampere (A).
I =
dQ
dt
Ampere:
One ampere of current is said to flow through a wire if one coulomb of charge
per second flows through any section of wire.
The current in a circuit is measured by the instrument known as Ammeter. Ammeter is
always connected in series with circuit.
2.2.2 Potential difference
It is the difference of electric potential between the two points in an electric circuit. It
is also defined as work done in Joules in moving a Coulomb of charge between two points.
Its symbol is „V‟ or „E‟ and the unit is Volts.
V =
21
dW
dQ
Volts: The potential difference between two points is one volts if one joule of work is done
in transforming +1 coulomb of charge from a point of lower potential to a point of higher
potential.
The voltage in a circuit is measured by the instrument known as Voltmeter. Voltmeter is
always connected in parallel with circuit.
2.2.3 Resistance
The property of conductor which opposes the flow of current through it is known as
resistance.
Its symbol is „R‟, and its unit is ohm (Ω).
Ohm (Ω):
It is defined as that resistance offered by a conductor which will allow one
ampere of current to flow if one volt is applied across its terminals.
The instruments that measures resistance are known as ohmmeter. Good conductors have
small resistance and insulators have more resistance.
2.3 Ohms Law
Whenever an electric current flows through a conductor, the following three factors must be
present,
1. A potential difference „V‟ across the conductor making the current to flow
2. An opposing force to the flow of current, called the resistance „R‟ of the
conductor.
3. The current „I‟ which is retained in the conductor.
Ohm‟s law gives a definite relationship between three quantities. The law states that:
“At constant temperature, the current flow through a conductor is directly proportional
to the potential difference between the two ends of the conductor”
ie.
VαI
V = IR
V
R
I
Fig 2.1 Ohm’s law
Where R – Resistance of the given conductor between the two points. The various
relationships derived from the ohms law are given by
V = IR
=
V
R
R =
V
I
22
I
2.3.1 Work
Work is said to be done on a body. When, a force acts on it, the body moves through
some distance. The unit of work is joule or Newton-meter.
Work = Force x Distance
2.3.2 Power
The rate at which work is done in an electric circuit is called electric power. It is obvious
that the rate at which work is done in moving electrons in the circuit depends upon,
a. How many electrons are to be moved? (Voltage)
b. The speed at which electrons are to travel (Current)
Power = Voltage x Current = V x I
Its symbol is „P‟, and its unit is Watts. Thus the power (P) is measured by the product of
voltage and current.
Watt : The power consumed in a circuit is one watt if a potential difference of one volt
causes one ampere of current to flow through the circuit.
Watt is denoted by a letter „W‟ and is measured by an instrument known as wattmeter.
2.3.3 Energy
The total amount of work done in an electric circuit is called electric energy.
Electrical Energy
= Power x Time
=Pxt =VxIxt
The basic unit of electrical energy is joules or watt-sec.
Joules:It is energy used in a circuit, when one watt has been absorbed for one second.
Therefore, one joule is equal to 1 watt second.
The joule or watt-sec is very small unit of electrical energy. In practice, for the
measurement of electrical energy, bigger units namely watt-hour and Kilo watt hour (KWh)
are used.
2.4 Law of Resistance
The resistance offered by a conductor depends upon the following factors:
 The resistance R of the conductor is directly proportional to its length ℓ.
 The resistance of the conductor is inversely proportional to the area of the cross
section.
 The resistance of the conductor depends upon the nature of the material.
 The resistance of a conductor depends upon temperature
23
R α
ℓ
A
R = ρ
ℓ
A
ρ
ℓ
A
Where
Length of the conductor
Area of cross section of the Conductor
(Greek, letter “Rho”) Resistivity or Specific Resistance of the conductor
2.4.1 Specific Resistance
The specific resistance or resistivity of a material may be defined as the resistance
offered to a current if passed between the opposite faces of the unit cube of the material.
The SI unit of the specific resistance is ohm-meter (Ω-m).
2.4.2 Conductance
Conductance is the reciprocal of resistance.
i.e.
G =
1
R
R = ρ
G =
ℓ
A
1 A
ρ ℓ
G = σ
A
ℓ
Where σ – (Sigma) Conductivity or Conductance of a conductor.
The unit of conductivity is SIEMENS / METER (S/M), where as old unit is Mho-meter-1
2.5 Kirchhoff’s Law
Kirchhoff‟s laws are very useful in solving the circuit which cannot be easily solved by
ohm‟s Law.
2.5.1 First Law
24
The algebraic sum of all the current at any junction in an electric circuit is zero.
Consider four wires meeting at a point “O” and carrying current I1, I2, I3 and I4 as shown in
fig 2.2. Assume the sign of the currents which are flowing towards Junction “O” as positive
and the sign of the currents flowing away from Junction “O” as Negative.
Applying Kirchoff‟s first law to Junction “O”
Algebraic sum of currents at “O”
=0
i.e., (+I1) + (+I2) + (-I3) + (-I4)

I1 + I2 = I3 + I4

Incoming Currents
=0
= Outgoing currents
I1
I2
O
I3
I4
Fig 2.2 Circuits for KCL
Therefore, the Kirchhoff‟s first law can also be stated at any junction, as the sum of incoming
current is equal to the sum of the outgoing currents. This law is also known as Point law or
Kirchhoff’s Current law (KCL).
2.5.2 Second Law
In any closed circuit or mesh the algebraic sum of the product of current and resistances
(Voltage drop) plus the algebraic sum of all the e.m.f‟s in that circuit is zero.
That is, in any closed circuit or mesh,
Algebraic sum of e.m.f‟s + Algebraic sum of voltage drops = 0
While applying Kirchhoff‟s second law to a closed circuit or mesh, algebraic sums are taken,
therefore, proper sign must be given for e.m.f and voltage drops.
R1
25
R1
I
+
R2
E
R2
‘E’
-
R3
R2
Fig 2.3 Circuits for KVL
E = V1 + V2 + V3
V1 = IR1
V2 = IR2
V3 = IR3
E = IR1 + IR2 + IR3
2.6 Solved Problems:
4
Find the current in the resistor 100 Ω, when a supply of 50 Volt is connected across it.
Given Data:
Resistor (R) = 100 Ω
Voltage (V) = 50 V
Find data:
Current (I) = ?
Solution:
Formula used
I
=
V
R
I
=
50
100
I
=
0.5
Result:
Current in the Resistor is 0.5 amps
5. The resistance of a 230V incandescent lamp is 270 Ω. Calculate the current taken by the
lamp.
Given Data:
Resistor (R) = 270 Ω
Voltage (V) = 230 V
Find data:
Current (I) = ?
Solution:
Formula used
26
I
=
V
R
I
=
230
270
I
= 0.85
Result:
Current in the incandescent lamp is 0.85 amps
6. The resistance of the motor is 200 ohms and the maximum current that it can take is 2.2
A. Find the main supply voltage on which it works effectively.
Given Data:
Resistor (R)
Current (I)
= 200 Ω
= 2.2 A
Find data:
Voltage (V) = ?
Solution:
Formula used
V = IR
V = 2.2 x 200
= 440
Result:
The maximum supply voltage required for effective operation is 440 V
7. An electrical heater takes 8A when connected across a supply main of 220V. Find the
resistance of heating element used. What is its power rating?
Given Data:
Voltage (V) = 220 V
Current (I)
= 8A
Required Data:
Resistance(R)
Power (P)
Solution:
R =
V
I
R =
220
8
R = 27.5 Ω
P = VI
= 220 x 8
P = 1760 W
27
Result:
The resistance of the heating element is 27.5 Ohms
The power rating of the electric heater is 1760 watts.
8. What will be the current drawn by a lamp rated at 230V, 20W connected to 200V
supply.
Given data:
Rated Power (WL)
= 20 W
Rated Voltage(VL) = 230 V
Supply voltage (VS) = 200 V
Find Data:
Current (I)
Solution:
Step 1: Find the Lamp Resistance
VL2
RL =
PL
RL =
2302
20
RL = 2.645 KΩ
Step 2: Find the Lamp current
VS
I =
RL
R =
I
200
2645
= 0.076 A
Result:
The current drawn by the lamp is 0.076 Amps.
9. An electrical lamp consumes 100 watts of power. The supply voltage is 230 volts.
Determine (a) the current flowing through the filament (b) its resistance (c) the
energy consumed in 45 minutes.
Given data:
Power (P)
= 100 w
Supply voltage= 230 V
Time (t)
= 45 min
= 45 / 60 sec
Find Data:
a. Current (I)
b. Resistance (R)
c. Energy (E)
28
Solution:
(a)
Finding the Current
(b)
I
=
P
V
I
=
100
230
I
= 0.44 A
Finding the Resistance
R =
V
I
R =
230
0.44
R = 522.7 Ω
(c)
Finding the Energy
E =
Pxt
= 100 x 45
60
E = 0.075 KW
Result:
Current flowing through the filament
Resistance of the filament
Energy consumed at 45 minutes
=
=
=
0.44 A
522.7 ohms
0.075 kw
10. Find the current in each branch of the circuit shown in fig 2.4.
Fig 2.4 Branch Circuit
29
Solution:
Step 1: Marking the current direction
The current I1 and I2 are marked as shown in the diagram.
Step 2: Selecting any closed circuit
The closed circuits ABEFA and CBEDC are selected.
Step 3: Algebraic sum of EMF’s and Voltage drops
Applying Kirchhoff‟s second law to closed mesh.
In the closed circuit ABEFA
3 = 3I1 + 5(I1 + I2)
3 = 8 I1 + 5 I2
………….. (1)
In the closed circuit CBEDC
4 = 7I2 + 5(I1 + I2)
4 = 5 I1 + 12 I2
………….. (2)
Step 4: Solve Equation
(1) x 5 
15 = 40 I1 + 25 I2
(2) x 8 
32 = 40 I1 + 96 I2
_____________________________
Subtracting,
17 = 71 I2
I2 = (17 / 71)
I2 = 0.24 A
Substituting I2 in the equation (1)
3 = 8I1 + 5(0.24)
8I1 = 3 – 1.2
I1 = ( 1.8 / 8)
I1 = 0.225 A or 0.23A
Step 5: Result
The current through 3 Ω resistor (I1)
The current through 7 Ω resistor (I2)
The current through 5 Ω resistor (I1+I2)
………….. (3)
= 0.23 A
= 0.24 A
= (0.23 + 0.24) = 0.47A
11. Find the voltage V for the given circuit shown in fig 2.5
Fig 2.5
30
Solution:
Step 1: Marking the current direction
The current I1 and I2 are marked as shown in the diagram.
Step 2: Selecting any closed circuit
The closed circuits ABEFA and CBEDC are selected.
Step 3: Algebraic sum of EMF’s and Voltage drops
Applying Kirchhoff‟s second law to closed mesh.
In the closed circuit ABEFA
V = 50 I1 + 100(I1 - I2)
V = 150 I1 - 100 I2
………….. (1)
In the closed circuit BCDEB
0 = 20I2 + 50 I2 - 100(I1 - I2)
0 = - 100 I1 + 170 I2
100 I1 = 170 I2
………….. (2)
I1 = (170 / 100) I2
In the circuit diagram, I2 = 2A. Substituting the value of I1, I2 in the equation (1)
V = 150 x (17+0 / 100) x (2) – 100 (2)
= 510 - 200
= 310 Volts
Result:
V = 310 Volts
12. Find the resistance of a copper strip 600 meter long and having a rectangular crossing
section of 2.5 cm x 0.05 cm. Take ρ = 1.73 x 10-8 Ω-m
Given Data:
Length of the copper strip (l) = 600 meters
Cross sectional area (A)
= 2.5 cm x 0.05 cm
= 2.5x10-2 x 0.05x10-2
= 1.25 x 10-5 metes
The resistivity (ρ)
= 1.73 x 10-8 Ω-m
Find data:
Resistance(R)
Solution:
ℓ
R = ρ
A
R = 1.73 x 10-8 x
R = 0.83
600
1.25 x 10-5
Ohms
31
Result:
The resistance of the copper strip is 0.83 ohms
13 A current 1.5A is passed through a coil of iron wire which has a cross sectional area
of 0.015 cm2. If the resistivity of iron is 10 x 10-8 Ωm. The voltage across the ends of
the coil is 20 volts. (a) What is the length of the wire? (b) What is the conductivity of
the wire?
Given Data:
The current flowing through the coil (I)
Cross sectional area (A)
The resistivity of the wire (ρ)
The voltage across the ends of the coil (V)
Find Data:
Length of the wire (l)
Conductivity of the wire (σ)
Solution:
(a)
Resistance of the coil (R)
V
R =
I
R =
20
1.5
R =
13.3 Ω
R = ρ
(b)
ℓ
A
ℓ
=
AR
ρ
ℓ
=
0.015 x 10-4 x 13.3
10 x 10-8
ℓ
= 199.5 m
The conductivity (σ)
1
σ =
ρ
1
1.25 x 10-8
σ
=
σ
= 10 x 106 Mho-meter
ℓ
σ
= 199.5 m
= 10 x 106 Mho-meter
Result:
32
= 1.5 A
= 0.015 cm2 = 0.015 x 10-4 m2
= 10 x 10-8 Ωm
= 20 volts
2.6 Keywords

















Electrical energy
Atom
Negative charges
Positive charge
Direct Current
Cathode
Anode
Current
Conductor
Ampere
Instrument
Ammeter
Electric potential
Volts
Resistance
Ohm
Kirchoff‟s Law
2.8 Questions for Discussion
1.
2.
3.
4.
5.
6.
State the unit of Resistance & Current.
State the relationship between voltage & Power
Explain the Ohms law.
State the Kirchoff‟s Laws.
Explain the relationship between voltage, power and energy.
An electric kettle of 30 ohm resistance takes 5A current. Calculate the heat developed
in joules in one minutes. How much energy would be consumed in 6 hours?
7. Calculate the filament resistance in 100 Watts 230V lamp.
8. Calculate the power rating of a heater coil when used on 220V supply taking 5 amps.
9. Find the resistance of a copper wire of 0.75 km long and having a cross sectional area
of 0.01 cm2. Take ρ = 1.72 x 10-8.
10. A potential difference of 60V exists between the ends of a wire of length 110cms. If
its diameter is 0.14mm and the current through it is 2A. Calculate the specific
resistance.
11. Find the amount of heat produced in kilo calories in 30 minutes in an electric iron
taking a current of 4.2 amps and having a heating element resistance of 50 ohms.
12. The maximum resistance of a rheostat is 20 Ω and minimum value of it is 1 Ω. Find
for each condition the voltage across the rheostat when the current flowing through it
is 1 A.
13. If a resistor is to dissipate energy at the rate of 200 W. Find its resistance for a
terminal voltage of 110 V.
14. An electric heater having a resistance of 30 ohms is connected to 220 V mains supply
through a cable having a total resistance of 0.3 ohm. Calculate (a) The power
dissipated by heater (b) Power dissipated in cable and (c) Total energy consumed in 3
hrs.
33
15. Determine the current through the 5 KΩ resistors when the power dissipated by the
element is 20Watts.
2.9 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Tony Kuphaldt,Using the Spice Circuit Simulation Program, in“Lessons in
Electricity,
Reference”,
Volume
5,
Chapter
7,
at
http://www.ibiblio.org/obp/electricCircuits/Ref/
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
34
UNIT
3
SERIES CIRCUITS
CONTENTS
3.0 Objectives
3.1 Introduction
3.2 Series Circuit
3.3 Simple series circuits
3.4 Keywords
3.5 Questions for Discussion
3.6 Suggested Readings
3.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
 Derive the equivalent resistance for series circuit.
3.1 Introduction
Special components called resistors are made for the express purpose of creating a
precise quantity of resistance for insertion into a circuit. They are typically constructed of
metal wire or carbon, and engineered to maintain a stable resistance value over a wide range
of environmental conditions. Unlike lamps, they do not produce light, but they do produce
heat as electric power is dissipated by them in a working circuit. Typically, though, the
purpose of a resistor is not to produce usable heat, but simply to provide a precise quantity of
electrical resistance.
Because the relationship between voltage, current, and resistance in any circuit is so
regular, we can reliably control any variable in a circuit simply by controlling the other two.
Perhaps the easiest variable in any circuit to control is its resistance. This can be done by
changing the material, size, and shape of its conductive components.
The most common schematic symbol for a resistor is a zig-zag line:
35
Resistor values in ohms are usually shown as an adjacent number, and if several
resistors are present in a circuit, they will be labeled with a unique identifier number such as
R1, R2, R3, etc. As you can see, resistor symbols can be shown either horizontally or
vertically.
The resistance can be connected in three ways
1. Series combination
2. Parallel combination
3. Series – Parallel combination
3.2 Series Circuit
It is circuit in which the resistances are connected end to end so that there is only one
path. In a series circuits,
1. The same amount of current flows through all the resistances.
2. For each resistance, there will be a voltage drop according to Ohm‟s law.
3. The sum of the voltage drop will be equal to the applied voltage.
The fig 3.1 shows a series combination circuit. It consists of three resistances R1, R2 & R3.
All are connected in series across the voltage „V‟. The current „I‟ is flowing through the
circuit.
Fig 3.1 Series Connection
Let V1, V2 & V3 be the voltage drop across the resistance R1, R2 & R3 respectively. The total
voltage is given by,
V
= V1 + V2 + V3
= I1R1 + I2R2 + I3R3
[ V = IR]
Since I1 = I2 = I3 = I (Same current flow through all resistors)
V
= IR1 + IR2 + IR3
= I ( R 1 + R2 + R 3 )
V/I
= R1 + R2 + R3
R
= R1 + R2 + R3
Where, R is the total circuit resistance of the series combination.
Hence, when a number of resistances are connected in series, the total resistance is equal to
sum of individual resistances.
36
3.3 Simple series circuits
Let‟s start with a series circuit consisting of three resistors and a single battery:
Fig 3.2 Example Circuit
The first principle to understand about series circuits is that the amount of current is
the same through any component in the circuit. This is because there is only one path for
electrons to flow in a series circuit, and because free electrons flow through conductors like
marbles in a tube, the rate of flow (marble speed) at any point in the circuit (tube) at any
specific point in time must be equal.
From the way that the 9 volt battery is arranged, we can tell that the electrons in this
circuit will flow in a counter-clockwise direction, from point 4 to 3 to 2 to 1 and back to 4.
However, we have one source of voltage and three resistances. How do we use Ohm‟s Law
here? An important caveat to Ohm‟s Law is that all quantities (voltage, current, resistance,
and power) must relate to each other in terms of the same two points in a circuit. For instance,
with a single-battery, single-resistor circuit, we could easily calculate any quantity because
they all applied to the same two points in the circuit:
This brings us to the second principle of series circuits: the total resistance of any
series circuit is equal to the sum of the individual resistances. This should make intuitive
sense: the more resistors in series that the electrons must flow through, the more difficult it
will be for those electrons to flow. In the example problem, we had a 3 k, 10 k, and 5 k
resistor in series, giving us a total resistance of 18 k:
Fig 3.3 Result Circuit
In essence, we‟ve calculated the equivalent resistance of R1, R2, and R3 combined.
Knowing this, we could re-draw the circuit with a single equivalent resistor representing the
series combination of R1, R2, and R3:
37
Now we have all the necessary information to calculate circuit current, because we
have the voltage between points 1 and 4 (9 volts) and the resistance between points 1 and 4
(18 k):
Knowing that current is equal through all components of a series circuit (and we just
determined the current through the battery), we can go back to our original circuit schematic
and note the current through each component:
Now that we know the amount of current through each resistor, we can use Ohm‟s
Law to determine the voltage drop across each one (applying Ohm‟s Law in its proper
context):
In summary, a series circuit is defined as having only one path for electrons to flow.
From this definition, three rules of series circuits follow: all components share the same
current; resistances add to equal a larger, total resistance; and voltage drops add to equal a
larger, total voltage. All of these rules find root in the definition of a series circuit. If you
understand that definition fully, then the rules are nothing more than footnotes to the
definition.
3.4 Keywords
Components
Series
Parallel
Battery
Electrons
38
3.5 Questions for Discussion
1.
2.
3.
4.
What are the various connections of resistors?
What is series connection?
Derive the expression for series connected resistors?
What is the formula for series connected resistors?
3.6 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Tony Kuphaldt,Using the Spice Circuit Simulation Program, in“Lessons in
Electricity,
Reference”,
Volume
5,
Chapter
7,
at
http://www.ibiblio.org/obp/electricCircuits/Ref/
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
39
UNIT
4
PARALLEL CIRCUITS
CONTENTS
4.0 Objectives
4.1 Introduction
4.2 Parallel Circuits
4.3 Simple Parallel circuit
4.4 Keywords
4.5 Questions for Discussion
4.6 Suggested Readings
4.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
 Derive the equivalent resistance for parallel circuit.
4.1 Introduction
It is a circuit in which one end of each resistance is connected to a common point and the
other end of each resistance is connected to another common point so that there are as many
current paths as the number of resistance. In a parallel circuit,
a.
b.
c.
d.
The voltage across each resistance of the parallel circuit is the same.
The number of current paths equal to the number of braches.
The current in each branch is given by Ohm‟s law.
The total current is equal to the sum of branch currents.
4.2 Parallel Circuits
The figure 4.1 shows a parallel circuit. It consists of three resistances R1, R2 & R3.
They are connected in parallel across a voltage of „V‟ volts. The total circuit current flowing
is „I‟ amperes. Let I1,I2 and I3 be the current in resistance R1, R2 and R3 respectively.
40
Fig 4.1 Parallel Circuit
I =
I1
+
I2
+
I3
I =
V1
R1
+
V2
R2
+
V3
R3
Since V1 = V2 = V3 = V
I =
V
R1
I
=
V
1
=
R
1
R1
1
R1
+
+
+
V
R2
1
R2
1
R2
+
+
+
V
R3
1
R3
1
R3
Where R – The total resistance of the parallel circuit. Hence, when a number of
resistances are connected in parallel, the reciprocal of total resistance is equal to the sum of
reciprocal of individual resistance.
4.3 Simple Parallel Circuits
Let‟s start with a parallel circuit consisting of three resistors and a single battery:
Fig 4.2 Example Circuit
The first principle to understand about parallel circuits is that the voltage is equal
across all components in the circuit. This is because there are only two sets of electrically
common points in a parallel circuit, and voltage measured between sets of common points
must always be the same at any given time.
Therefore, in the above circuit, the voltage across R1 is equal to the voltage across R2
which is equal to the voltage across R3 which is equal to the voltage across the battery. This
equality of voltages can be represented in another table for our starting values:
41
Just as in the case of series circuits, the same caveat for Ohm‟s Law applies: values
for voltage, current, and resistance must be in the same context in order for the calculations to
work correctly.
However, in the above example circuit, we can immediately apply Ohm‟s Law to
each resistor to find its current because we know the voltage across each resistor (9 volts) and
the resistance of each resistor
The total circuit current is equal to the sum of the individual branch currents. Using
this principle, we can fill in the IT spot on our table with the sum of IR1, IR2, and IR3:
Finally, applying Ohm‟s Law to the rightmost (”Total”) column, we can calculate the
total circuit resistance:
42
The total circuit resistance is only 625 : less than any one of the individual resistors.
In the series circuit, where the total resistance was the sum of the individual resistances, the
total was bound to be greater than any one of the resistors individually. Here in the parallel
circuit, however, the opposite is true: we say that the individual resistances diminish rather
than add to make the total. This principle completes our triad of ”rules” for parallel circuits,
just as series circuits were found to have three rules for voltage, current, and resistance.
Mathematically, the relationship between total resistance and individual resistances in a
parallel circuit looks like this:
In summary, a parallel circuit is defined as one where all components are connected
between the same set of electrically common points. Another way of saying this is that all
components are connected across each other‟s terminals. From this definition, three rules of
parallel circuits follow:
 All components share the same voltage;
 Resistances diminish to equal a smaller, total resistance;
 Branch currents add to equal a larger, total current.
Just as in the case of series circuits, all of these rules find root in the definition of a parallel
circuit. If you understand that definition fully, then the rules are nothing more than footnotes
to the definition.
4.4 Keywords
 Current paths
 Branch
 Voltage drop
4.5 Questions for Discussion
a.
b.
c.
d.
e.
What is parallel circuit?
Draw the parallel connection of three resistors?
What are the properties of parallel connection?
What is the formula for two resistors connected in parallel?
Derive the expression for three resistor connected in parallel?
43
4.6 Suggested Readings
a. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
b. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
c. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
d. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
e. Tony Kuphaldt,Using the Spice Circuit Simulation Program, in“Lessons in
Electricity,
Reference”,
Volume
5,
Chapter
7,
at
http://www.ibiblio.org/obp/electricCircuits/Ref/
f. Electrical Circuit Theory, Gopalsamy, Veni Publication
44
UNIT
5
PROBLEMS IN SERIES AND PARALLEL CIRCUITS
CONTENTS
5.0 Objectives
5.1 Introduction
5.2 Solved Problem
5.3 Questions for Discussion
5.4 Suggested Readings
5.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
 Solve the simple series, parallel and combination of series & parallel circuit of
resistance connection.
 Find the current & voltage for the simple circuit.
5.1
Introduction
The circuit in which resistance are connected end to end as shown in the fig 5.1 is called a
series circuit.
Fig 5.1 Series Circuit
R = R1 + R2 + R3
Two or more resistances are connected in such a way that one end of them are joined to a
common point and the other ends are joined to another common point or connected across
one another as shown in fig 5.2 is called parallel circuits.
45
Fig 5.2 Parallel circuit
1
1
1
1
=
+
+
R
R1
R2
R3
Series Circuits
 Voltage drops add to equal total voltage.
 All components share the same (equal) current.
 Resistances add to equal total resistance.
Parallel Circuits
 All components share the same (equal) voltage.
 Branch currents add to equal total current.
 Resistances diminish to equal total resistance.
The circuit contains series and parallel combinations of resistances are known as series
parallel circuit. The fig 5.3
Fig 5.3 Series parallel Circuits
The fig 5.3 is simplified by the two parallel circuits like R1|| R2 and R3||R4. This combination
is connected in series. Fig 5.4 shows the simplified circuit.
46
Fig 5.4 simplified circuit
R = R1||R2 + R3||R4
R =
R1.R2
R3.R4
+
R1 + R2
R3 + R4
5.2 Solved Problems
14. Three resistance of 8 Ω, 9 Ω, and 16 Ω are connected in parallel. Find out the total
resistance.
Given Data:
R1 = 8 Ω
R2 = 9 Ω
R3 = 16 Ω
Find data:
RT
Solution:
Result:
RT
1
RT
=
1
R1
+
1
R2
+
1
R3
1
RT
=
1
8
+
1
9
+
1
16
1
RT
=
0.125
1
RT
=
0.335
RT
=
2.99
+
0.11
+
0.1
Ω
= 2.99 Ω
15 Three resistor 50 Ω, 60 Ω and 75 Ω are connected in series across 220V mains. Calculate
(a) Total resistance of the circuit (b) Current flowing through the circuit.
Given Data:
R1 = 50 Ω
R2 = 60 Ω
47
R3 = 75 Ω
Find data:
(a) Total Resistance (RT)
(b) Current (I)
Solution:
(a)
Total Resistance (RT)
RT
= R1 + R2 + R3
= 50 + 60 + 75
RT
= 185 Ω
(b)
Current (I)
I
=V/R
= 220 / 185
= 1.19 A
Result:
The total effective resistance is 185 Ω
Current flowing through the circuit is 1.19 A
16 The equivalent resistance of four resistors joined in parallel is 30 Ω. The current flowing
through them is 0.5, 0.4, 0.6 and 0.1 A. Find the value of each resistor.
Given Data:
RT = 30 Ω
I1 = 0.5 A
I2 = 0.4 A
I3 = 0.6 A
I4 = 0.1 A
Find data:
R1, R2, R3 R4
Solution:
I = I1 + I2 + I3
= 0.5 + 0.4 + 0.6 + 0.1
= 1.6 A
V = IRT
= 1.6 x 30
= 48 V
As per the Ohms law
R1 = V / I1
R1 = 48 / 0.5
= 96 Ω
R2 = V / I2
= 48 / 0.4
= 120 Ω
R3 = V / I3
= 48 / 0.6
= 80 Ω
R4 = V / I4
= 48 / 0.1
= 480 Ω
48
Result
R1
= 96 Ω
17 Three resistance 2 Ω, 4 Ω and 5 Ω are connected as shown in the fig 5.5 across 15 volts
battery. Calculate (a) The current supply by the battery and (b) The current through each
resistor.
Fig 5.5 Resistor Connection
Given Data:
Supply Voltage (V) = 15 V
R1
=2Ω
R2
=4Ω
R3
=5Ω
Find data:
(a)
Total Current (I)
(b)
Each resistance Current (I1, I2)
Solution:
1
RBC
=
1
R2
1
RBC
=
1
5
+
1
R3
+
1
4
RBC
= 2.22 Ω
The total Resistance (RT)= R1 + RBC
= 2 + 2.22
= 4.22 Ω
a.
Current supplied by the Battery
I=V/R
= 15 / 4.22
= 3.6 A
b.
Current through Each resistor
VAB = 3.6 x 2
= 7.2 V
VAB =15 – 7.2
= 7.8 V
Current flowing through 4 Ω = 7.8 / 4
Current flowing through 5 Ω = 7.8 / 5
Current flowing through 2 Ω = I
= 1.97 A
= 1.56 A
= 3.6 A
49
Result:
Current flowing through 4 Ω = 1.97 A
Current flowing through 5 Ω = 1.56 A
Total Current flowing through Ω = 3.6 A
18. Three resistors of 3,4,5 ohms respectively are connected in parallel. This combination
is put in series with a 2.5 ohm resistor. Determine the equivalent resistance of the
combination.
Given Data:
R1 = 3 Ω
R2 = 4 Ω
R4 = 2.5 Ω
Series Combination
Find data:
RT
Solution:
Step 1: Solve the parallel Circuit
R3 = 5 Ω Parallel Combination
1
RP
=
1
R1
+
1
R2
+
1
R3
1
RP
=
1
3
+
1
4
+
1
5
RP
=
1.28 Ω
Resistance RT = RP + R
= 1.28 + 2.5
= 3.78 Ω
Result:
RT = 3.78 ohms
5.2 Questions for Discussion
1. A parallel branch contains two resistnance each 2 ohms. This is connected to a 3 ohms
resistance in series. An other 4 ohms resistance is shunted across this series parallel
combination. Calculate the total resistance across the ends.
2. A circuit consist of three resistor of 3Ω, 3 Ω and 6 Ω in parallel and fourth resistor of
4 Ω in series. A batter emf of 12 V is connected across the circuit: (a) Draw a crcuit
arragement (b) Find the total resistance (c) What is the total current in the circuit.
3. A resistance of 10 Ω is connected in sereis with two resistors of 15 Ω each arranged in
parallel. What must be shunted across the parallel combination so that the total current
taken shall be 1.5 A with 20 V applied.
50
4. A bulb rated 110V, 60 Watts is connected with another bulb rated 110 V, 100 watts
across a 220 V mains. Calculate the resistance which should be joined in parallel with
the 60 watts bulb so that may take their rated power.
5. A circuit is made of 0.4 ohm wire, a 150 ohm bulb and a 120 ohm rehostat are
connected in series. Determine the total resistance of the circuit.
6. Three resistors R1, R2 and R3 are connected in series across a constant voltage of „V‟
volts. The voltage drop across R1 is 20 volts, the power loss in R2 is 25 watts and R3
has a resistance of 2 ohm. Find the voltgae „V‟ if the current is 5 amps.
7. What will be the effect value if three resistances having equal valve of „R‟ ohm are
connected in (i) Series and (ii) Parallel.
8. Two resistors connected in parallel across a 200 volts mains takes a total current of 10
amps. The power dissipated in one of the resistors is 1200 watts. Find the value of
each resistor.
9. Find resistors of 8 ohms, 10 ohms, 20 ohms and X ohms are connected in parallel.
The total current taken is 6.5 amps. Find (i) the value of X and (ii) the circuit
resistance.
10. The equivalent resistance of four resistors jointed in parallel is 20 Ω. The current
flowing through them are 0.6, 0.3, 0.2 and 0.1 Amps. Find the value of each resistor.
11. two resistors one of 30 Ω and another of unknown value are connected in parallel. The
total power dissipated in the circuit is 450 watts. When the applied voltage is 90 volts.
Find the value of unknown resistance.
12. Three resistors R1, R2 and R3 are connected in parallel across a supply voltage of 240
V. If R2 = 2R1, R3 = 3R1 and the total power taken by the circuit is 480 watts.
Calculate the power in each resistor.
13. A resistance of R ohm is connected in series with a parallel circuit comprising of two
resistance 12 Ω and 8 Ω respectively. The total power dissipated in the circuit is 70
watts when the applied voltage is 20 V. Calculate the value of R.
5.3 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Electrical Circuit Theory, Gopalsamy, Veni Publication
51
52
UNIT
1
NODE VOLTAGE ANALYSIS
CONTENTS
1.0 Objectives
1.1 Introduction
1.2 Node voltage method
1.3 Node voltage rules
1.4 Solved Problems
1.5 Keywords
1.6 Questions for Discussion
1.7 Suggested Question
1.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
 Understand the needs of Node voltage method.
 Simplify the circuit using Node voltage method.
 Solve the electrical circuit by using Node Voltage Analysis.
1.1 Introduction
Generally speaking, network analysis is any structured technique used to
mathematically analyze a circuit (a “network” of interconnected components). Quite often the
technician or engineer will encounter circuits containing multiple sources of power or
component configurations which defy simplification by series/parallel analysis techniques. In
those cases, he or she will be forced to use other means.
Some circuit configurations (“networks”) cannot be solved by reduction according to
series/ parallel circuit rules, due to multiple unknown values.
Mathematical techniques to solve for multiple unknowns (called “simultaneous
equations” or “systems”) can be applied to basic Laws of circuits to solve networks.
53
1.2 Node voltage method
The node voltage method of analysis solves for unknown voltages at circuit nodes in terms of
a system of KCL equations. This analysis looks strange because it involves replacing voltage
sources with equivalent current sources. Also, resistor values in ohms are replaced by
equivalent conductance‟s in siemens, G = 1/R. The siemens (S) is the unit of conductance,
having replaced the mho unit. In any event S = −1. And S = mho (obsolete).
We start with a circuit having conventional voltage sources. A common node E0 is chosen as
a reference point. The node voltages E1 and E2 are calculated with respect to this point.
Fig 1.1
A voltage source in series with a resistance must be replaced by an equivalent current source
in parallel with the resistance. We will write KCL equations for each node. The right hand
side of the equation is the value of the current source feeding the node.
Fig 1.2
Replacing voltage sources and associated series resistors with equivalent current sources and
parallel resistors yields the modified circuit. Substitute resistor conductance‟s in siemens for
resistance in ohms.
I1 = E1/R1
I2 = E2/R5
G1 = 1/R1
G2 = 1/R2
G3 = 1/R3
G4 = 1/R4
G5 = 1/R5
= 10/2 = 5 A
= 4/1 = 4 A
= 1/2 = 0.5 S
= 1/4 = 0.25 S
= 1/2.5 = 0.4 S
= 1/5 = 0.2 S
= 1/1 = 1.0 S
Fig 1.3
The Parallel conductance‟s (resistors) may be combined by addition of the conductance‟s.
Though, we will not redraw the circuit. The circuit is ready for application of the node
voltage method.
54
GA = G1 + G2 = 0.5 S + 0.25 S = 0.75 S
GB = G4 + G5 = 0.2 S + 1 S = 1.2 S
Deriving a general node voltage method, we write a pair of KCL equations in terms of
unknown node voltages V1 and V2 this one time. We do this to illustrate a pattern for writing
equations by inspection.
GAE1 + G3(E1 - E2) = I1 (1)
GBE2 - G3(E1 - E2) = I2 (2)
(GA + G3 )E1 -G3E2 = I1 (1)
-G3E1 + (GB + G3)E2 = I2 (2)
The coefficients of the last pair of equations above have been rearranged to show a pattern.
The sum of conductance‟s connected to the first node is the positive coefficient of the first
voltage in equation (1). The sum of conductances connected to the second node is the positive
coefficient of the second voltage in equation (2).
The other coefficients are negative, representing conductances between nodes. For both
equations, the right hand side is equal to the respective current source connected to the node.
This pattern allows us to quickly write the equations by inspection. This leads to a set of rules
for the node voltage method of analysis.
1.3 Node voltage rules
The following rule is followed to solve the electrical circuit by applying the Node
voltage method.
 Convert voltage sources in series with a resistor to an equivalent current source with
the resistor in parallel.
 Change resistor values to conductances.
 Select a reference node(E0)
 Assign unknown voltages (E1)(E2) ... (EN)to remaining nodes.
 Write a KCL equation for each node 1,2, ... N. The positive coefficient of the first
voltage in the first equation is the sum of conductance‟s connected to the node. The
coefficient for the second voltage in the second equation is the sum of conductance‟s
connected to that node. Repeat for coefficient of third voltage, third equation, and
other equations. These coefficients fall on a diagonal.
 All other coefficients for all equations are negative, representing conductances
between nodes. The first equation, second coefficient is the conductance from node 1
to node 2; the third coefficient is the conductance from node 1 to node 3. Fill in
negative coefficients for other equations.
 The right hand side of the equations is the current source connected to the respective
nodes.
 Solve system of equations for unknown node voltages.
1.4 Solved Problems
55
Problem 1
Solve the network in Fig 1.4 by nodal analysis and find the powers supplied by the sources.
Fig 1.4
Solution:No of Nodes =
3 (1,2,3)
Node 3 is the reference node.
Therefore, No of equations for 2 nodes in matrix form
(
1
1
1
+
+
R1
R2
R3
(
-1
R2
(
1
1
1
+
+
0.25
0.5
1
(
-1
0.5
)
-
(
1
R2
)
250
0.25
V1
=
)
(
1
1
1
+
+
R2
R3
R5
)
-
(
)
1
0.5
220
0.2
V2
)
250
0.25
V1
=
7
-2
∆ =
)
(
-2
8
V1
V2
7
-2
-2
8
∆V1 =
1000
1100
∆V2 =
7
-2
1
1
1
+
+
0.5
0.2
1
=
)
1000
1100
= 7 x 8 – (-2x-2) = 52
-2
8
1000
1100
= 1000 x 8 – (1100x-2) = 10200
= 7x 1100 – (1000x-2) = 9700
V1 =
∆V1
10200
=
= 196.15 V
∆
52
V2 =
∆V2
9700
=
= 186.54 V
∆
52
I1 =
V2
250 - V1
250 – 196.15
=
= 215.4 A
R1
0.25
56
220
0.2
I2 =
V1 – V2
196.15 – 186.54
=
= 19.22 A
R2
0.5
I3 =
220 – V2
220 – 186.54
=
= 167.3 A
R3
0.2
I4 =
V1
196.15
=
= 196.15 A
R4
1
I5 =
V2
186.54
=
= 186.54 A
R5
1
Power supplied by 250 V battery
Power supplied by 220 V battery
= V x I1
= 250 x 215.4
= 53850 Watts
= V x I3
= 220 x 167.3
= 36806 Watts.
Result:Power supplied by 250 V battery is 53850 Watts
Power supplied by 220 V battery is 36806 Watts
Problem 2
Write down the nodal equation in the matrix form for the network for the network shown in
fig 1.5 below taking 4 as reference node.
Fig 1.5
Solution:Node 4 is the reference node, the voltages of nodes 1,2 and 3 be V1, V2 and V3 with respect to
node 4.
General matrix form is
G11
G21
G31
G12
G22
G32
G13
G23
G33
V1
V2
V3
I1
I2
I3
=
(1/1+1/2.5)
-1/1
0
-1/5
(1/1+1/2+1/4)
-1/2
0
-1/2
(1/2+1/5)
V1
V2
V3
1.4
-1
-1
1.75
0
-0.5
V1
57
V2
=
4-3
3-3
3-4
=
1
0
0
-0.5
V3
0.7
-1
Problem 3
Write down the nodal equation in the matrix form for the network shown in fig 1.6 below,
taking node 4 as reference.
Fig 1.6
Solution:Node 4 is the reference node, the voltages of nodes 1,2 and 3 be V1, V2 and V3 with respect to
node 4.
General matrix form is
G11
G21
G31
G12
G22
G32
G13
G23
G33
V1
V2
V3
I1
I2
I3
=
V1
(1/1+1/0.5)
-1/0.5
0
-1/0.5
(1/0.5+1/0.25+1/
0.33)
-1/0.25
0
-1/0.25
(1/0.25+1/0.25)
3
-2
0
-2
9.03
-4
0
-4
8
V2
V1
V2
V3
1.5 Keywords








-2
Network
Configurations
Simultaneous equations
Node
Voltage sources
Current sources
Siemens
Conductance
1.6 Questions for Discussion
58
=
V3
=
9
2
-2
9
2
1.
2.
3.
4.
5.
What is meant by node?
What is meat by branch?
Sketch any one simple electrical network.
State the basis for node voltage analysis.
The circuit is having 7 branch and 3 nodes. How many equations are possible in this
circuit?
6. What are the rules to be followed while solving the circuit by using Node analysis
method?
7. Explain node analysis method with example.
8. Define network.
9. Give the current equivalent of voltage source
10. For the Network shown below mark (i) Branches (ii) Nodes (iii) Loops
1. Using node voltage technique finds the current in 4 Ω resistors, after doing source
transformation.
Suggested Question
1) Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2) Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3) Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication,
Pollachi Tamilnadu
4) A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5) Electrical Circuit Theory, Gopalsamy, Veni Publication
UNIT
2
59
MESH CURRENT ANALYSIS
CONTENTS
2.0 Objectives
2.1 Introduction
2.2 Mesh current method
2.2.1 Mesh Current, conventional method
2.2.2 Mesh current by inspection
2.3 Mesh current rules
2.4 Solved Problems
2.5 Keywords
2.6 Questions for Discussion
2.7 Suggested Readings
2.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
 Understand the needs of Mesh Current method.
 Simplify the circuit using Mesh current method.
 Solve the electrical circuit by using Mesh Current Analysis.
2.1 Introduction
The first and most straightforward network analysis technique is called the Branch
Current Method. In this method, we assume directions of currents in a network, then write
equations describing their relationships to each other through Kirchhoff ‟s and Ohm‟s Laws.
Once we have one equation for every unknown current, we can solve the simultaneous
equations and determine all currents, and therefore all voltage drops in the network.
2.2 Mesh current method
The Mesh Current Method, also known as the Loop Current Method, is quite similar
to the Branch Current method in that it uses simultaneous equations, Kirchhoff ‟s Voltage
Law, and Ohm‟s Law to determine unknown currents in a network. It differs from the Branch
Current method in that it does not use Kirchhoff ‟s Current Law, and it is usually able to
solve a circuit with less unknown variables and less simultaneous equations, which is
especially nice if you‟re forced to solve without a calculator.
60
2.2.1 Mesh Current, conventional method
Let‟s see how this method works on the same example problem:
Fig 2.1
The first step in the Mesh Current method is to identify “loops” within the circuit
encompassing all components. In our example circuit, the loop formed by B1, R1, and R2
will be the first while the loop formed by B2, R2, and R3 will be the second. The strangest
part of the Mesh Current method is envisioning circulating currents in each of the loops. In
fact, this method gets its name from the idea of these currents meshing together between
loops like sets of spinning gears:
Fig 2.2
The choice of each current‟s direction is entirely arbitrary, just as in the Branch
Current method, but the resulting equations are easier to solve if the currents are going the
same direction through intersecting components (note how currents I1 and I2 are both going
“up” through resistor R2, where they “mesh,” or intersect). If the assumed direction of a mesh
current is wrong, the answer for that current will have a negative value.
The next step is to label all voltage drop polarities across resistors according to the
assumed directions of the mesh currents. Remember that the “upstream” end of a resistor will
always be negative, and the “downstream” end of a resistor positive with respect to each
other, since electrons are negatively charged. The battery polarities, of course, are dictated by
their symbol orientations in the diagram, and may or may not “agree” with the resistor
polarities (assumed current directions):
Fig 2.3
Using Kirchhoff ‟s Voltage Law, we can now step around each of these loops,
generating equations representative of the component voltage drops and polarities. As with
the Branch Current method, we will denote a resistor‟s voltage drop as the product of the
61
resistance (in ohms) and its respective mesh current (that quantity being unknown at this
point). Where two currents mesh together, we will write that term in the equation with
resistor current being the sum of the two meshing currents.
Tracing the left loop of the circuit, starting from the upper-left corner and moving
counterclockwise (the choice of starting points and directions is ultimately irrelevant),
counting polarity as if we had a voltmeter in hand, red lead on the point ahead and black lead
on the point behind, we get this equation:
-28 + 2(I1 + I2) + 4I1 = 0
Notice that the middle term of the equation uses the sum of mesh currents I1 and I2 as
the current through resistor R2. This is because mesh currents I1 and I2 are going the same
direction through R2, and thus complement each other. Distributing the coefficient of 2 to the
I1 and I2 terms, and then combining I1 terms in the equation, we can simplify as such:
- 28 + 2(I1 + I2) + 4I1 = 0 Original form of equation
. . . distributing to terms within parentheses . . .
. . . combining like terms . . .
- 28 + 6I1 + 2I2 = 0
- 28 + 2I1 + 2I2 + 4I1 = 0
Simplified form of equation
At this time we have one equation with two unknowns. To be able to solve for two unknown
mesh currents, we must have two equations. If we trace the other loop of the circuit, we can
obtain another KVL equation and have enough data to solve for the two currents. Creature of
habit that I am, I‟ll start at the upper-left hand corner of the right loop and trace
counterclockwise:
- 2(I1 + I2) + 7 - 1I2 = 0
Simplifying the equation as before, we end up with:
- 2I1 - 3I2 + 7 = 0
Now, with two equations, we can use one of several methods to mathematically solve for the
unknown currents I1 and I2:
- 2I1 - 3I2 + 7 = 0
- 28 + 6I1 + 2I2 = 0
6I1 + 2I2 = 28
-2I1 - 3I2 = -7
. . . rearranging equations for easier solution . . .
Solutions:
I1 = 5 A
I2 = -1 A
Knowing that these solutions are values for mesh currents, not branch currents, we must go
back to our diagram to see how they fit together to give currents through all components:
62
Fig 2.4
The solution of -1 amp for I2 means that our initially assumed direction of current was
incorrect. In actuality, I2 is flowing in a counter-clockwise direction at a value of (positive) 1
amp:
Fig 2.5
This change of current direction from what was first assumed will alter the polarity of the
voltage drops across R2 and R3 due to current I2. From here, we can say that the current
through R1 is 5 amps, with the voltage drop across R1 being the product of current and
resistance (E=IR), 20 volts (positive on the left and negative on the right). Also, we can
safely say that the current through R3 is 1 amp, with a voltage drop of 1 volt (E=IR), positive
on the left and negative on the right. But what is happening at R2?
Mesh current I1 is going “up” through R2, while mesh current I2 is going “down” through
R2. To determine the actual current through R2, we must see how mesh currents I1 and I2
interact (in this case they‟re in opposition), and algebraically add them to arrive at a final
value. Since I1 is going “up” at 5 amps, and I2 is going “down” at 1 amp, the real current
through R2 must be a value of 4 amps, going “up:”
Fig 2.6
2.2.2 Mesh current by inspection
We take a second look at the “mesh current method” with all the currents running
counterclockwise (ccw). The motivation is to simplify the writing of mesh equations by
ignoring the resistor voltage drop polarity. Though, we must pay attention to the polarity of
voltage sources with respect to assumed current direction. The sign of the resistor voltage
drops will follow a fixed pattern.
63
If we write a set of conventional mesh current equations for the circuit below, where we do
pay attention to the signs of the voltage drop across the resistors, we may rearrange the
coefficients into a fixed pattern:
Fig 2.7
2.3 Mesh current rules
The following rule is followed to solve the electrical circuit by applying the Mesh
Current method.
 This method assumes electron flow (not conventional current flow) voltage sources.
Replace any current source in parallel with a resistor with an equivalent voltage
source in series with an equivalent resistance.
 Ignoring current direction or voltage polarity on resistors, draw counterclockwise
current loops traversing all components. Avoid nested loops.
 Write voltage-law equations in terms of unknown currents currents: I1, I2, and I3.
Equation 1 coefficient 1, equation 2, coefficient 2, and equation 3 coefficient 3 are the
positive sums of resistors around the respective loops.
 All other coefficients are negative, representative of the resistance common to a pair
of loops. Equation 1 coefficient 2 is the resistor common to loops 1 and 2, coefficient
3 the resistor common to loops 1 an 3. Repeat for other equations and coefficients.
+(sum of R‟s loop 1)I1 - (common R loop 1-2)I2 - (common R loop 1-3)I3 = E1
-(common R loop 1-2)I1 + (sum of R‟s loop 2)I2 - (common R loop 2-3)I3 = E2
-(common R loop 1-3)I1 - (common R loop 2-3)I2 + (sum of R‟s loop 3)I3 = E3
 The right hand side of the equations is equal to any electron current flow voltage
source. A voltage rise with respect to the counterclockwise assumed current is
positive, and 0 for no voltage source.
 Solve equations for mesh currents:I1, I2, and I3 . Solve for currents through
individual resistors with KCL. Solve for voltages with Ohms Law and KVL.
2.4 Solved Problems
Problem 4
64
In the circuit of Fig 1.4 obtain the load current and the power delivered to the load.
Fig 1.4
Solution:Consider I1, I2 and I3 are the loop currents as shown in fig (1.4). By inspection we can write
by matrix from
[
[
R11 R12 R13
R21 R22 R23
R31 R32 R33
(4+12)
-12
-12
(6+9+12)
0
∆ =
|
∆ = 16
I1
I2
V1
V2
][ ] [ ]
][ ] [ ]
=
I3
0
-9
V3
I1
I2
-9
(3+15+9)
16 -12 0
-12 27 -9
0
-9 27
=
I3
0
|
27 -9
-12 -9
-12 27
-(-12)
+0
-9 27
0
27
0
-9
= 16[27x27-(-9x-9)]+12[-12x27+0x9]+0
= 6480
∆3 =
|
∆3 = 16
16 -12 120
-12 27
0
0
-9
0
27
-9
0
0
-(-12)
|
-12
0
0
0
+120
-12
0
27
-9
= 16[27x0+9x0]+12[-12x0-0x0]+120[-12x-9 – 0x27]
= 12960
I3 =
∆3
12960
=
= 2 Amps
∆
6480
P = I2RL
= 22x 15
65
120
0
= 60 Watts
Result:Current through Load resistance is = 2 Amps
Power in Load (P)
= 60 Watts
Problem 5
A wheat stone bridge ABCD has resistance AB = CD = 4 Ω, BC = 14 Ω, DA = 8 Ω and BD =
12 Ω. Between A and C, a 100V battery is connected with A as positive. Using mesh current
method, find the current in 12 ohms resistor.
Given Data:-
Fig 1.5
Find Data:Current through BD arm
Solution:-
[
[
R11 R12 R13
R21 R22 R23
R31 R32 R33
(4+12+8)
-12
-12
(14+4+12)
-8
-4
[
∆ =
|
24 -12 -8
-12 30 -4
-8
-4
12
24 -12 -8
-12 30 -4
-8 -4 12
I1
I2
V1
V2
][ ] [ ]
][ ] [ ]
][ ] [ ]
=
I3
V3
-8
-4
I1
I2
(8+4)
I3
I1
I2
I3
=
0
0
=
100
0
0
100
|
= 24[30x12 – (-4 x-8)] –(-12)[-12x12 –(-8x-4)] -8[0x-4 – 10x30]
= 3840
∆1 =
|
0
-12 -8
|
66
0 30
100 -4
-4
12
= 0[30 x 12 – (-4x-4)] -0[-12x12 – (-4x-8)]+100[-12x-4 – (30x-8)]
= 28800
∆2 =
|
24
0 -8
-12 0 -4
-8 100 12
|
= 24[0 x 12 – (100x-4)] -0[-12x12 – (-4x-8)]-8[-12x100 – (0x-8)]
= 19200
I1 =
∆1
28800
=
= 7.5 Amps
∆
3840
∆2
19200
=
= 5 Amps
∆
3840
Current through arm BD is
= I1 – I2
= 7.5 – 5
= 2.5 A
Result:Current through BD arm is 2.5 A
I2 =
2.5 Keywords
Branch Current
Mesh current
Loop Current
Simultaneous equations
Polarities
Counterclockwise
Conventional
2.6 Questions for Discussion
1. What is mesh?
2. Explain how the impedance matrix can be written by inspection in loop current
method (Mesh equation method)
3. What are the rules to be followed while solving the circuit using mesh current
analysis?
4. Explain the mesh current conventional method with example circuit.
5. A wheat stone bridge has AB = 4 Ω, BC = 3 Ω, CD = 6 Ω and DA = 5 Ω. A
galvanometer of 110 Ω is connected across AC. A battery of 2 V is connected across
BD. Find the current in galvanometer.
6. Find the current in 6 Ω resistor using mesh current analysis.
67
7. For the network shown below write the loop equation and loop matrix
2.7 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Electrical Circuit Theory, Gopalsamy, Veni Publication
68
UNIT
3
STAR DELTA TRANSFORMATIONS
CONTENTS
3.0 Objectives
3.1 Introduction
3.2 Star delta Connections
3.3 Delta star transformation
3.3 Star Delta Transformation
3.4 Solved Problems
3.5 Keywords
3.6 Questions for Discussion
3.7 Suggested Readings
3.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
 Understand the needs of Mesh Current method.
 Simplify the circuit using Mesh current method.
 Solve the electrical circuit by using Mesh Current Analysis.
3.1 Introduction
A given voltage source with a series resistance can be converted into an equivalent
current source with a parallel resistance. Conversely a current source with a parallel
resistance can be converted an equivalent voltage source with a series resistance.
(a) Voltage Source
(b) Current Source
Fig 3.1 Source Conversion
Fig 3.1 (a) shows the voltage source B1 in series with resistance R1.
69
Voltage between the Ends (AB)
= VAB
= B1 – iR1 …………………. (1)
Fig 3.2 (b) shows the current source I1 in parallel with a resistance R1.
Voltage between the Ends (ab)
= Vab
= (I1 – i) R1 …………………. (2)
If the two circuits are identical
VAB = Vab
B1 – iR1= (I1 – i) R1
i.e.,
B1
= I1R1
I1
= B1 / R1
Hence a current source supplying this current I1 and having the same resistance R1,
connected in parallel with it represents the equivalent source of the voltage source.
Similarly a current source of I1 and a parallel resistance R1 can be converted into a
voltage source of B1= I1R1 and a resistance R1 series with it.
3.2 Star delta Connections
In fig 3.2 (a) the three resistors RA, RB and RC are connected in star between points A,
B and C. and in fig 3.2 (b) RAB, RBC and RCA are connected in delta.
In some networks, for simplification the star connected resistances are replaced by an
equivalent delta or sometimes a delta connected resistances are replaced by an equivalent
star.
(a) Star
(b) Delta
Fig 3.2 Star delta connection
For two systems to be equivalent, the currents in the other parts of the circuit should
not be affected. It is necessary that the resistances between any two pairs of terminals in star
delta should be same.
3.3 Delta star transformation
In the two systems are equivalent the resistance between points A and B in delta is
equal resistance between points A and B in star. Fig 3.3 shows conversion of delta to star.
70
Fig 3.3 Delta star transformation
Resistance between points A and B in delta
= RAB || (RBC + RCA)
= RAB(RBC + RCA)
(RAB + RBC + RCA)
Resistance between points A and B in star
= RA + R B
Equating (3) and (4)
RA + R B =
…………… (3)
…………… (4)
RAB(RBC + RCA)
RAB + RBC + RCA
…………… (5)
Similarly equating the resistance between points B and C
RB + RC =
RBC(RCA + RAB)
RAB + RBC + RCA
…………… (6)
And between point C and A
RC + R A =
RCA(RAB + RBC)
RAB + RBC + RCA
…………… (7)
Adding the three equations (5), (6) and (7)
2(RA + RB + RC) =
2(RABRBC + RBCRCA + RCARAB)
RAB + RBC + RCA
Therefore,
RA + R B + RC =
RABRBC + RBCRCA + RCARAB
RAB + RBC + RCA
…… (8)
(8) – (4)
RA
=
RCA RAB
RAB + RBC + RCA
71
…………… (9)
Similarly
RB
=
RAB RBC
RAB + RBC + RCA
RC
=
RBC RCA
RAB + RBC + RCA
…………… (10)
…………… (11)
RA, RB and RC are equivalent star resistances for the delta connected resistance RAB, RBC and
RCA.
3.4 Star Delta Transformation
Use the above equation (9), (10) and (11), we find the Delta values
(9) ÷ (10) gives
…………… (12)
RA
RCA
=
RB
RBC
Similarly (10) ÷ (11) gives
RB
RAB
=
RC
RCA
(11) ÷ (9)
RC
RBC
=
RA
RAB
…………… (13)
…………… (14)
Equating (9) dividing both numerator and denominator by RCA
RA
=
RCA RAB /RCA
(RAB + RBC + RCA) / RCA
RA
=
RAB
RAB/RCA + RBC/RCA + 1
…………… (15)
Substitutes the equation (13) & (14)
RA
=
RAB
RB/RC + RB/RA + 1
RA
=
RAB
(RB RA + RB RC +RA RC ) / RA RC
RA
=
RAB RARC
RB RA + RB RC +RA RC
RAB
=
RA RB + RB RC +RC RA
RC
Therefore RAB is
72
…………… (16)
Similarly,
RA RB + RB RC +RC RA
RA
RBC
=
RCA
RA RB + RB RC +RC RA
=
RB
…………… (17)
…………… (18)
3.5 Solved Problems
Problem 6:
Three resistors of 6 Ω, 2 Ω and 3 Ω are connected in delta. Determine the resistance for an
equivalent star connection.
Given Data:
3Ω
6Ω
2Ω
Solution:
RA
=
RCA RAB
RAB + RBC + RCA
RA
=
3x6
6+2+3
RA
= 1.64 Ω
RB
=
RAB RBC
RAB + RBC + RCA
RB
=
6x2
6+2+3
RB
= 1.09 Ω
RC
=
RBC RCA
RAB + RBC + RCA
RC
=
2x3
6+2+3
RC
= 0.545 Ω
73
Result
RA = 1.64 Ω
RB = 1.09 Ω
RC = 0.545 Ω
Problem 7
Three resistors of 20 Ω, 30 Ω and 60 Ω are connected in star. Determine the resistance for an
equivalent delta connection.
Given data:
RA = 20 Ω
Find data:
RAB
RBC
RB = 30 Ω
RC = 60 Ω
RCA
Solution:
RAB
=
RA RB + RB RC +RC RA
20x30+30x60+60x20
=
= 60 Ω
RC
60
RBC
=
RA RB + RB RC +RC RA
20x30+30x60+60x20
=
= 180 Ω
RA
20
RCA
=
RA RB + RB RC +RC RA
=
RB
Result:
RAB = 60 Ω
RBC
= 180 Ω
20x30+30x60+60x20
20
RCA
=
120 Ω
= 120 Ω
3.6 Keywords





Equivalent current source
Equivalent voltage source
Star
Delta
Transformation
3.7 Questions for Discussion
1.
2.
3.
4.
5.
6.
Write the expression for star to delta transformation
Write the expression for delta to star transformation
Calculate the delta equivalent of three star connected resistors 10, 15 and 20 ohms.
Obtain the star equivalent resistance from a given delta network of R12, R23 and R31.
Derive the equation for delta-star transformation.
Three resistors of 20 Ω, 30 Ω and 60 Ω are connected in delta. Determine the
resistance for an equivalent star connection.
7. Three resistors of 16 Ω, 22 Ω and 35 Ω are connected in delta. Determine the
resistance for an equivalent star connection.
8. Three resistors of 60 Ω, 25 Ω and 30 Ω are connected instar. Determine the resistance
for an equivalent delta connection.
74
Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Electrical Circuit Theory and Technology, Bird John
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
7. Circuit Theory, Gangadhar K.A., Kanna Publisher
75
UNIT
4
THEVENIN’S AND NORTON THEOREM
CONTENTS
4.0 Objectives
4.1 Introduction
4.2 Thevenin Theorem
4.3 Step to Solve the Thevenin‟s Theorem
4.4 Norton‟s Theorem
4.5 Step to Solve Norton‟s Theorem
4.6 Thevenin-Norton equivalencies
4.7 Simple Problem
4.8 Keywords
4.9 Questions for Discussion
4.10 Suggested Readings
4.0 OBJECTIVES
On completion of the following units of syllabus contents, the students must be able to
 Explain Thevenin‟s theorem and Norton‟s theorem
 Solve problems by using Thevenin & Norton‟s theorems
 Understand the Thevenin – Norton Equivalence
4.1 Introduction
Semiconductor (amplifier) circuits, where sometimes AC is often mixed (superimposed) with
DC. Because AC voltage and current equations (Ohm‟s Law) are linear just like DC, we can
use Superposition to analyze the circuit with just the DC power source, and then just the AC
power source, combining the results to tell what will happen with both AC and DC sources in
effect. For now, though, Superposition will suffice as a break from having to do simultaneous
equations to analyze a circuit.
76
4.2 Thevenin’s Theorem
Thevenin‟s Theorem states that it is possible to simplify any linear circuit, no matter how
complex, to an equivalent circuit with just a single voltage source and series resistance
connected to a load. The qualification of “linear” is identical to that found in the
Superposition Theorem, where all the underlying equations must be linear (no exponents or
roots). However, there are some components (especially certain gas-discharge and
semiconductor components) which are nonlinear: that is, their opposition to current changes
with voltage and/or current. As such, we would call circuits containing these types of
components, nonlinear circuits
Thevenin‟s Theorem is especially useful in analyzing power systems and other circuits where
one particular resistor in the circuit (called the “load” resistor) is subject to change, and
recalculation of the circuit is necessary with each trial value of load resistance, to determine
voltage across it and current through it. Let‟s take another look at our example circuit as
shown in the fig 4.1:
Fig 4.1 Simple Circuit for Thevenin Theorem
Let‟s suppose that we decide to designate R2 as the “load” resistor in this circuit. We already
have two methods of analysis at our disposal (Branch Current, Mesh Current,) to use in
determining voltage across R2 and current through R2, but each of these methods are timeconsuming. Imagine repeating any of these methods over and over again to find what would
happen if the load resistance changed (changing load resistance is very common in power
systems, as multiple loads get switched on and off as needed. the total resistance of their
parallel connections changing depending on how many are connected at a time). This could
potentially involve a lot of work!
Thevenin‟s Theorem makes this easy by temporarily removing the load resistance from the
original circuit and reducing what‟s left to an equivalent circuit composed of a single voltage
source and series resistance. The load resistance can then be re-connected to this “Thevenin
equivalent circuit” and calculations carried out as if the whole network were nothing but a
simple series circuit:
Fig 4.2 Example of Thevenin Circuit
77
. . . After Thevenin conversion . . .
Thevenin Equivalent Circuit
Fig 4.3 Thevenin Equivalent Circuit
The “Thevenin Equivalent Circuit” is the electrical equivalent of B1, R1, R3, and B2 as seen
from the two points where our load resistor (R2) connects.
The Thevenin equivalent circuit, if correctly derived, will behave exactly the same as the
original circuit formed by B1, R1, R3, and B2. In other words, the load resistor (R2) voltage
and current should be exactly the same for the same value of load resistance in the two
circuits. The load resistor R2 cannot “tell the difference” between the original network of B1,
R1, R3, and B2, and the Thevenin equivalent circuit of EThevenin, and RThevenin, provided that
the values for EThevenin and RThevenin have been calculated correctly.
The advantage in performing the “Thevenin conversion” to the simpler circuit, of course, is
that it makes load voltage and load current so much easier to solve than in the original
network. Calculating the equivalent Thevenin source voltage and series resistance is actually
quite easy.
4.3 Step to Solve the Thevenin Theorem
Step 1:
First, the chosen load resistor is removed from the original circuit, replaced with a break
(open circuit):
Fig 4.4 First step of Thevenin Theorem
Step 2:
Next, the voltage between the two points where the load resistor used to be attached is
determined. Use whatever analysis methods are at your disposal to do this. In this case, the
original circuit with the load resistor removed is nothing more than a simple series circuit
78
with opposing batteries, and so we can determine the voltage across the open load terminals
by applying the rules of series circuits, Ohm‟s Law, and Kirchhoff‟s Voltage Law:
Fig 4.5 Thevenin Theorem Second Step
The voltage between the two load connection points can be figured from the one of the
battery‟s voltage and one of the resistor‟s voltage drops, and comes out to 11.2 volts. This is
our “Thevenin voltage” (EThevenin) voltage.
Step 3:
To find the Thevenin series resistance for our equivalent circuit, we need to take the original
circuit (with the load resistor still removed), remove the power sources (in the same style as
we did with the Superposition Theorem: voltage sources replaced with wires and current
sources replaced with breaks), and figure the resistance from one load terminal to the other:
Fig 4.6 Thevenin Theorem third Step
With the removal of the two batteries, the total resistance measured at this location is equal to
R1 and R3 in parallel: 0.8. This is our “Thevenin resistance” (RThevenin) for the equivalent
circuit:
79
Step 4:
Fig 4.7 Thevenin Theorem fourth Step
Step 5:
Final we get voltage source. Use this source, we find the load current, load voltage or
load power by using Ohm‟s law or Kirchoff‟s Laws
4.4 Norton’s Theorem
`Norton‟s Theorem states that it is possible to simplify any linear circuit, no matter how
complex, to an equivalent circuit with just a single current source and parallel resistance
connected to a load. Just as with Thevenin‟s Theorem, the qualification of “linear” is
identical to that found in the Superposition Theorem: all underlying equations must be linear
(no exponents or roots).
Contrasting our original example circuit against the Norton equivalent: it looks something
like this:
Fig 4.8 Example circuit for Norton’s Theorem
Fig 4.9 Norton’s Equivalent Circuit
80
Remember that a current source is a component whose job is to provide a constant
amount of current, outputting as much or as little voltage necessary to maintain that constant
current. As with Thevenin‟s Theorem, everything in the original circuit except the load
resistance has been reduced to an equivalent circuit that is simpler to analyze. Also similar to
Thevenin‟s Theorem is the steps used in Norton‟s Theorem to calculate the Norton source
current (INorton) and Norton resistance (RNorton).
4.5 Step to Solve Norton’s Theorem
Step 1:
As before, the first step is to identify the load resistance and remove it from the original
circuit:
Fig 4.10 Remove the Load Resistor
Step 2:
Then, to find the Norton current (for the current source in the Norton equivalent
circuit), place a direct wire (short) connection between the load points and determine the
resultant current. Note that this step is exactly opposite the respective step in Thevenin‟s
Theorem, where we replaced the load resistor with a break (open circuit):
Fig 4.11 Short circuit condition
With zero voltage dropped between the load resistor connection points, the current
through R1 is strictly a function of B1‟s voltage and R1‟s resistance: 7 amps (I=E/R).
Likewise, the current through R3 is now strictly a function of B2‟s voltage and R3‟s
resistance: 7 amps (I=E/R). The total current through the short between the load connection
points is the sum of these two currents: 7 amps + 7 amps = 14 amps. This figure of 14 amps
becomes the Norton source current (INorton) in our equivalent circuit:
81
Step 3:
To calculate the Norton resistance (RNorton), we do the exact same thing as we did for
calculating Thevenin resistance (RThevenin): take the original circuit (with the load resistor still
removed), remove the power sources (in the same style as we did with the Superposition
Theorem: voltage sources replaced with wires and current sources replaced with breaks), and
figure 4.12 total resistance from one load connection point to the other:
Fig 4.12 Thevenin Equivalent Resistor
Step 4:
Now our Norton equivalent circuit looks like this:
Fig 4.13 Norton’s Equivalent
4.6 Thevenin-Norton equivalencies
1. Find the Norton source current by removing the load resistor from the original circuit
and calculating current through a short (wire) jumping across the open connection
points where the load resistor used to be.
2. Find the Norton resistance by removing all power sources in the original circuit
(voltage sources shorted and current sources open) and calculating total resistance
between the open connection points.
3. Draw the Norton equivalent circuit, with the Norton current source in parallel with the
Norton resistance. The load resistor re-attaches between the two open points of the
equivalent circuit.
4. Analyze voltage and current for the load resistor following the rules for parallel
circuits.
82
4.7 Simple Problem
Problem 8
Find the voltage drop in 10 Ω resistor is shown in fig 3.14 by using Thevenin theorem.
Fig 4.14 Circuit
Solution:
Step 1: Find the Open circuit Voltage
To find VO first RL is removed and voltage between AB is called VO
The voltage difference in the circuit = 9 – 6
The current flowing in the circuit
VO
=3V
= 3 / (5+6)
= 0.2727 A
= 6 + (6 x 0.2727)
= 7.6364 Volts
Step 2: Find the R Thevenin
To find the RTH voltage source is replaced with its internal resistance and resistance between
AB terminals is called RTH
RTH
= R1 + (R2. R3 / R2 + R3)
= 4 + [(5 x 6) / (5 +6)]
= 6.7273 Ω
Step 3: Draw the Thevenin’s Equivalent circuit
83
EThevenin
RThevenin
Load
Step 4: Find the load current
IL
VL
= VO = 7.6364 V
= RTH = 6.7373 Ω
= RL = 10 Ω
= VO / (RL + RTH)
= 0.4565 A
= IL . RL
= 0.4565 x 10
= 4.565 V
Result:
Voltage across 10 ohm resistor
= 4.565 Volts
Problem 9
Find the current in 20 ohms resistance in the fig 3.15 using Norton‟s theorem.
Fig 4.15 Simple Circuits
Solution:
Step 1: Find the Short circuit Current
Req
= 30 + [(30 x 45) / (30 + 45)]
= 48 Ω
I
= V / Req
= 120 / 48
= 2.5 A
ISC
= I x 30 / (30 + 45)
=1A
Step 2: Find the RTH
RTH = R1+ (R2 || R3)
= 45 + [(30 x 30) / (30 + 30)]
84
= 60 Ω
Step 3: Draw the Norton’s equivalent circuit
INorton = ISC = 1 A
RNorton = RTH = 60 Ω
Load = RL = 20 Ω
Step 4: Find IL
IL
= ISC.RTH / (RTH + RL)
= 0.75 A
Result:
The current through 20 ohm resistor is 0.75A
4.8 Keywords
Semiconductor
Thevenin‟s Theorem
Linear
Nonlinear
Power systems
Thevenin equivalent
Load resistor
Norton‟s Theorem
Norton resistance
4.9 Questions for Discussion
1.
2.
3.
4.
5.
6.
State the Thevenin theorem
State the Norton‟s Theorem
Explain the steps to solve the Thevenin theorem
Explain the steps to solve the Norton‟s theorem
Explain the Thevenin - Norton Equivalences.
Using Thevenin theorem find the current in the 20 ohm resistor in the fig given below.
7. Obtain the Thevenin equivalent circuit at terminal AB of the circuit given below
85
8.
Solve the Norton‟s Theorem
9. Apply Norton‟s theorem to calculate current flowing through 5 ohm resistor.
4.10 Suggested Readings
1) Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2) Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3) Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4) A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5) Tony Kuphaldt,Using the Spice Circuit Simulation Program, in“Lessons in
Electricity,
Reference”,
Volume
5,
Chapter
7,
at
http://www.ibiblio.org/obp/electricCircuits/Ref/
6) Electrical Circuit Theory, Gopalsamy, Veni Publication
86
UNIT
5
SUPERPOSITION & MAXIMUM POWER THEOREM
CONTENTS
5.0 Objectives
5.1 Introduction
5.2 Superposition Theorem
5.3 Demerits of Superposition Theorem
5.4 Maximum Power Transfer Theorem
5.5 Demerits of Power Transfer Theorem
5.6 Simple Problem
5.7 Keywords
5.8 Questions for Discussion
5.9 Suggested Readings
5.0 OBJECTIVES
On completion of the following units of syllabus contents, the students must be able to
 State and explain Superposition theorem and Maximum Power Transfer theorem
 Solve problems by using Superposition & Maximum Power Transfer theorems
5.1 Introduction
Anyone whose studied geometry should be familiar with the concept of a theorem: a
relatively simple rule used to solve a problem, derived from a more intensive analysis using
fundamental rules of mathematics. At least hypothetically, any problem in math can be
solved just by using the simple rules of arithmetic (in fact, this is how modern digital
computers carry out the most complex mathematical calculations: by repeating many cycles
of additions and subtractions!), but human beings aren‟t as consistent or as fast as a digital
computer. We need “shortcut” methods in order to avoid procedural errors.
In electric network analysis, the fundamental rules are Ohm‟s Law and Kirchhoff‟s
Laws. While these humble laws may be applied to analyze just about any circuit
configuration (even if we have to resort to complex algebra to handle multiple unknowns),
there are some “shortcut” methods of analysis to make the math easier for the average human.
87
As with any theorem of geometry or algebra, these network theorems are derived
from fundamental rules. In this chapter, I‟m not going to delve into the formal proofs of any
of these theorems. If you doubt their validity, you can always empirically test them by setting
up example circuits and calculating values using the “old” (simultaneous equation) methods
versus the “new” theorems, to see if the answers coincide. They always should!
5.2 Superposition Theorem
Superposition theorem is one of those strokes of genius that takes a complex subject
and simplifies it in a way that makes perfect sense. Superposition, on the other hand, is
obvious. The strategy used in the Superposition Theorem is to eliminate all but one source of
power within a network at a time, using series/parallel analysis to determine voltage drops
(and/or currents) within the modified network for each power source separately. Then, once
voltage drops and/or currents have been determined for each power source working
separately, the values are all superimposed” on top of each other (added algebraically) to find
the actual voltage drops/currents with all sources active.
Let‟s look at our example circuit and apply Superposition Theorem to it:
Fig 5.1 Simple Network
Since we have two sources of power in this circuit, we will have to calculate two sets
of values for voltage drops and/or currents, one for the circuit with only the 28 volt battery in
effect. . .
(a) 28 volt battery in effect
(b) 7 volt battery in effect
Fig 5.2 Remove sources voltage
88
When re-drawing the circuit for series/parallel analysis with one source, all other
voltage sources are replaced by wires (shorts), and all current sources with open circuits
(breaks). Since we only have voltage sources (batteries) in our example circuit, we will
replace every inactive source during analysis with a wire.
Analyzing the circuit with only the 28 volt battery, we obtain the following values for
voltage and current:
Analyzing the circuit with only the 7 volt battery, we obtain another set of values for
voltage and current:
When superimposing these values of voltage and current, we have to be very careful
to consider polarity (voltage drop) and direction (electron flow), as the values have to be
added algebraically.
Applying these superimposed voltage figures to the circuit; the end result looks something
like this:
89
Fig 5.3 Based on the Voltage Results
Currents add up algebraically as well, and can either be superimposed as done with
the resistor voltage drops, or simply calculated from the final voltage drops and respective
resistances (I=E/R). Either way, the answers will be the same. Here I will show the
superposition method applied to current:
Once again applying these superimposed figures to our circuit:
Fig 5.4 Based on the current results
90
5.3 Demerits of Super position Theorem
Theorem is only applicable for determining voltage and current, not power!!! Power
dissipations, being nonlinear functions; do not algebraically add to an accurate total when
only one source is considered at a time. The need for linearity also means this Theorem
cannot be applied in circuits where the resistance of a component changes with voltage or
current. Hence, networks containing components like lamps (incandescent or gas-discharge)
or varistors could not be analyzed.
Another prerequisite for Superposition Theorem is that all components must be
“bilateral,” meaning that they behave the same with electrons flows either direction through
them. Resistors have no polarity-specific behavior, and so the circuits we‟ve been studying so
far all meet this criterion.
5.4 Maximum Power Transfer Theorem
The Maximum Power Transfer Theorem is not so much a means of analysis as it is an
aid to system design. Simply stated, the maximum amount of power will be dissipated by a
load resistance when that load resistance is equal to the Thevenin / Norton resistance of the
network supplying the power. If the load resistance is lower or higher than the Thevenin /
Norton resistance of the source network, its dissipated power will be less than maximum.
This is essentially what is aimed for in radio transmitter design, where the antenna or
transmission line “impedance” is matched to final power amplifier “impedance” for
maximum radio frequency power output. Impedance, the overall opposition to AC and DC
current, is very similar to resistance, and must be equal between source and load for the
greatest amount of power to be transferred to the load. Load impedance that is too high will
result in low power output. A load impedance that is too low will not only result in low
power output, but possibly overheating of the amplifier due to the power dissipated in its
internal (Thevenin or Norton) impedance.
Taking our Thevenin equivalent example circuit, the Maximum Power Transfer
Theorem tells us that the load resistance resulting in greatest power dissipation is equal in
value to the Thevenin resistance (in this case, 0.8 ):
Fig 5.5 Condition for Maximum power Transfer
91
With this value of load resistance, the dissipated power will be 39.2 watts:
5.5 Demerits of Maximum Power Transfer Theorem
 Maximum power transfer does not coincide with maximum efficiency. Application of
The Maximum Power Transfer theorem to AC power distribution will not result in
maximum or even high efficiency. The goal of high efficiency is more important for
AC power distribution, which dictates relatively low generator impedance compared
to load impedance.
 Maximum power transfer does not coincide with the goal of lowest noise. For
example, the low-level radio frequency amplifier between the antenna and a radio
receiver is often designed for lowest possible noise. This often requires a mismatch of
the amplifier input impedance to the antenna as compared with that dictated by the
maximum power transfer theorem.
5.6 Simple Problem
Problem 10
Using Superposition theorem find the current through 4 ohm resistor in the circuit fig 5.6
shown.
Fig 5.6
Solution:
Step 1: Remove any One source (20V source only acting)
Step 2: Find the current through 4 ohm in the above circuit
Req
= 4 + [(4 x 2) / (4 +2)]
92
Req
= 5.33 Ω
The total current (I)
=V/R
= 20 / 5.33
= 3.75 A
The current Through 4 Ω is
I1
= 3.75 x 2 / (4+2)
= 1.25 A
Step 3: Remove another Source (10 V source only acting)
Step 4: Find the current in 4 Ω for the above circuit
Req
= 2 + [ (4 x 4) / (4 + 4)]
=4Ω
The total current is (I) = 10 / 4
= 2.5 A
The current through 4 Ω is
I2
= 2.5 x 4 / (4 + 4)
= 1.25 A
Step 5: Super impose the response of each source
The current through 4 Ω is I1 + I2
= 1.25 + 1.25
= 2.5 A
Result:
Current through 4 ohms is 2.5 A.
Problem 11
In the circuit shown (fig 5.7) below, find the resistance R to be connected between A & B. So
that the power dissipated in this is maximum. Find also the maximum power.
93
Fig 5.7
Solution:
Step 1: Remove resistor R find voltage between A and B.
The voltage at point A
= (20 x 7) / 10
VA
= 14
The voltage at point B
= (40 x 9) / (9+1)
VB
= 36 V
The voltage between A & B is
= VB – VA
= 36 – 14
= 22 V
Step 2: Find lock resistance from the terminals A & B
VAB
RTH
= [(3 x 7) / (3 + 7)] + [(9 x 1) / (9 + 1)]
=3Ω
Step 3: Draw the equivalent Circuit
By Maximum power transfer theorem RTH
IL
Maximum Power
= RL
= VTH / (RTH + RL)
= 3.67 A
= IL2.RL
= 40.40 Watts
Result :
Load Current is 3.67 A
Max. Power is 40.40W
5.7 Keywords
Superposition theorem
Complex
Superimposed
Maximum Power Transfer
Power dissipation
Efficiency
94
5.8 Questions for Discussion
1.
2.
3.
4.
State the super position theorem
State and explain the Maximum power transfer Theorem
What are the demerits of Maximum power transfer theorem
Find the current in 150 Ω load resistor and the power consumed in it by the principle
of super position.
5. Find the ohmic value of RL in the following circuit when its power is a maximum.
Find also i) The maximum load power. ii) The total power delivered by both the
batteries. iii) The Overall efficiency
5.9 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. http://www.ibiblio.org/obp/electricCircuits/Ref/
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
95
96
UNIT
1
BASIC DEFINITIONS
CONTENTS
1.0 Objectives
1.1 Introduction
1.2 Sinusoidal Voltage and Current
1.3 Basic Terms in AC circuits
1.4 Average and Effective Value
1.5 Form and Peak Factor
1.6 Phasor
1.7 Simple Problems
1.8 Key words
1.9 Questions for Discussions
1.10 Suggested Readings
1.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
 Understand the sinusoidal wave and voltage, current.
 Define the Cycle, instantaneous value, peak value and frequency.
 Define the Average value, form factor, peak factor.
1.1 Introduction
Most students of electricity begin their study with what is known as direct current
(DC), which is electricity flowing in a constant direction, and/or possessing a voltage with
constant polarity. DC is the kind of electricity made by a battery (with definite positive and
negative terminals), or the kind of charge generated by rubbing certain types of materials
against each other.
97
Whereas the familiar battery symbol is used as a generic symbol for any DC voltage
source, the circle with the wavy line inside is the generic symbol for any AC voltage source.
One might wonder why anyone would bother with such a thing as AC. It is true that in
some cases AC holds no practical advantage over DC. In applications where electricity is
used to dissipate energy in the form of heat, the polarity or direction of current is irrelevant,
so long as there is enough voltage and current to the load to produce the desired heat (power
dissipation). However, with AC it is possible to build electric generators, motors and power
distribution systems that are far more efficient than DC, and so we find AC used
predominately across the world in high power applications.
The benefits of AC over DC with regard to generator design are also reflected in
electric motors. While DC motors require the use of brushes to make electrical contact with
moving coils of wire, AC motors do not. In fact, AC and DC motor designs are very similar
to their generator counterparts (identical for the sake of this tutorial), the AC motor being
dependent upon the reversing magnetic field produced by alternating current through its
stationary coils of wire to rotate the rotating magnet around on its shaft, and the DC motor
being dependent on the brush contacts making and breaking connections to reverse current
through the rotating coil every 1/2 rotation (180 degrees).
1.2 Sinusoidal Voltage and Current
Commercial alternators produce sinusoidal alternating voltage i.e., alternating voltage
is a sine wave. A sinusoidal a.c. voltage can be produced by rotating a coil in a uniform
magnetic field. This alternating wave form follows sine law.
The sinusoidal alternating voltage can be expressed by the equation
e = Emsinωt
Where,
e – Instantaneous value of alternating voltage
Em – Maximum value of alternating voltage
ω – Angular Velocity of the coil
Sinusoidal voltage always produces sinusoidal currents. Therefore, a sinusoidal current can
be expressed in the same way as voltage.
i.e.,
i = Imsinωt
The shape of the curve obtained by plotting the instantaneous values of voltage or current Vs
time is called wave form.
Fig 1.1 AC Waveform (sine wave)
98
1.3 Basic Terms in AC circuits
1.3.1 Instantaneous Value
The value of an alternating quantity at any instant is called instantaneous value. The
instantaneous values of alternating voltage and current are represented by „e‟ and „i‟
respectively.
1.3.2 Cycle
One complete set of positive and negative values of an alternating quantity is called a cycle.
1.3.3 Time period
The time taken to complete one cycle of an alternating quantity is called its time period. It is
represented by „T‟
1.3.4 Frequency
The number of cycle made by an alternating quantity per second is called its frequency. The
unit of frequency is Herz (Hz). It is represented by „f‟.
1.3.5 Amplitude or Peak Value
The maximum +ive (Positive) or –ive (Negative) value of an alternating quantity is called
amplitude or peak value.
Fig 1.2 AC Peak Value
1.4 Average and Effective Value
As the current or voltage in an AC circuit is varying from instant to instant it becomes very
difficult to specify the quantity. Two normal ways used to express AC quantities: (1) the
average Value and (2) the Effective (RMS) value.
1.4.1 Average Value
This is the average of the instantaneous values of an alternating quantity over one complete
cycle of the wave.
99
Fig 1.3 Average value of a sine wave
Fig 1.2 shows a current wave for one complete cycle. To obtain the average value, divide the
period into „n‟ equal intervals. For these individual intervals the currents are i1, i2, i3, …. in
Then the average value of the current is
Iav =
i1+ i2 + i3 + ….. + in
n
The average value can also be obtained by finding the area under the curve and dividing it by
the base.
Fig 1.4
Let
i – The instantaneous value of the current and i = Imsinθ
Where, Im is the maximum value
Area under the curve
Base Length
Average value =
Iav =
0
∫
π
Imsinθ dθ
π
For symmetric waves take the half cycle only.
Iav
I
= m
π
π
- cosθ
0
100
Iav =
Im
π
Iav =
2Im
π
Iav =
0.637 Im
[ - ( cos π – cos 0)]
Similarly, the average voltage for sine wave is
Vav
=
0.637 Vm
1.4.2 Root Mean Square Value (RMS value)
When an alternating current flows through a resistance for a certain time, certain amount of
heat is produced. Now assume that a direct current is passed through the same resistance for
the same time such that the same amount of heat is developed. This value of steady current
which has caused the same heat as that of alternating current is known as Root Mean Square
Value or Effective Value.
The current passing through a resistor produces heat regardless of its direction. The heat
produced by an alternating current of say maximum value Im amps will not be equal to that
produced may by a direct current of I amps. The heating effect produced may be used to
specify alternating current or voltage. The effective value of an AC is defined as that value of
DC which on passing through a resistance R ohms for a given time T seconds, produce the
same heat as the AC passing through R for the same time T.
√
RMS value =
For Half cycle
IRMS =
√
IRMS =
Im
√2
0
∫
Area Squared curve
Base
π
(Imsinθ)2dθ
π
IRMS = 0.707 Im
Similarly,
VRMS = 0.707 Vm
101
1.5 Form and Peak Factor
1.5.1 Form Factor
The ratio of RMS value to the average value of an alternating quantity is called form factor.
Form Factor =
RMS Value
Average Value
Form Factor =
Im / √2
2Im / π
For sine wave
Form Factor = 1.11
1.5.2 Peak Factor
The ratio of maximum value to the RMS value of an alternating quantity is called peak factor.
Peak Factor =
Maximum Value
RMS Value
For sine wave
=
Im
Im / √2
= 1.414
1.6 Phasor
The fig 1.4 shows two sinusoidal waves. One wave (A) represents the voltage and the other
(B) current.
Fig 1.5 Out of Phase Waveforms
The maximum value of voltage is Vmax and that of current is Imax. The current reaches its
maximum value of 90 later than the voltage wave. Hence the current wave is written as,
i = Imsin(ωt – 90)
Therefore, the current wave is said to be lagging behind the voltage by an angle 90. It can
also be stated that the voltage leads the current by an angle 90.
102
Hence when two alternating quantities of the same frequency have different zero points, they
are said to have a phase difference.
1.7 Simple Problems
Problem 1
The equation of a sinusoidal current is 141.4sin314t. Find the RMS value and frequency of
the current.
Given Data:
i = 141.4sin314t
Im = 141.4
ω = 314
Find data:
RMS Value
Frequency
Solution:
IRMS =
Im
√2
IRMS =
141.4
√2
IRMS = 100 A
ω = 2πf
f =
ω
2π
f =
314
2π
f = 50 Hz
Result:
IRMS = 100 A
f
= 50 Hz
Problem 2
The alternating current passing through a circuit is being by I = 141.4sin314.2t. What are the
values of (a) Maximum value of current (b) The instantaneous value of the current when t =
0.02 sec.
103
Given Data:
i = 141.4sin314.2t
Solution:
i = Imsinωt
Im =141.4
i = 141.4 x sin(314.2 x 0.02)
= 141.4 x sin (360)
=0
Result:
i
=0A
Im
=141.4 A
Problem 3
A conductor carries a sinusoidal AC of peak value 10 amps superimposed on a direct current
of amplitude 10 Amps. Find the RMS value.
Given Data:
Peak Value = 10 Amps
Superimposed on direct current = 10 Amps
i = 10 + 10 sin θ
Solution:
IRMS =
IRMS =
√
√
0
0
∫
∫
π
(10 + 10 sinθ)2dθ
π
π
(100 + 100 sin2θ + 200 sinθ)dθ
π
IRMS = 12.25 A
Result:
IRMS
= 12.25 A
1.8 Key words









Sinusoidal
Instantaneous Value
Average Value
Cycle
Frequency
Form Factor
Peak Factor
RMS Value
Peak Value
104
1.9 Questions for Discussions
1. What is meant by Sinusoidal?
2. Define instantaneous value of AC quantity?
3. Define the cycle and frequency?
4. What is meant by RMS vale?
5. What is meant by average value?
6. What is meant by form factor and Peak factor?
7. What is the relationship between form factor and peak factor?
8. What is the value of form factor in the sine wave?
9. Define time period.
10. A conductor carries a sinusoidal AC of peak value 15 amps superimposed on a direct
current of amplitude 1 Amps. Find the RMS value.
11. The alternating current passing through a circuit is being by V = 100sin314.2t. What
are the values of (a) Maximum value of voltage (b) The instantaneous value of the
voltage when t = 2 msec.
12. The equation of a sinusoidal current is 200sin314t. Find the RMS value and frequency
of the current.
13. The peak value of current is 300 Amps and frequency is applied in the circuit is 60
Hz. Find the equation of sinusoidal current and instantaneous value of current when t
= 30 msec.
1.10
Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. http://www.ibiblio.org/obp/electricCircuits/Ref/
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
105
UNIT
2
RESISTOR, CAPACITOR AND INDUCTOR IN AC
CIRCUITS
CONTENTS
2.0 Objectives
2.1 Introduction
2.2 Reactance and Impedance– Inductive
2.2.1 AC resistor circuits
2.2.2 AC Inductor Circuits
2.2.3 AC capacitor circuits
2.3 Simple Problem
2.4 Key words
2.5 Questions for Discussion
2.6 Suggested Readings
2.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to




Understand the difference between DC and AC Circuits
Define the Reactance, Impedance
Explain the AC Resistance, Inductance and Capacitance circuit
Solve the simple problem
2.1 Introduction
Most students of electricity begin their study with what is known as direct current
(DC), which is electricity flowing in a constant direction, and/or possessing a voltage with
constant polarity. DC is the kind of electricity made by a battery (with definite positive and
negative terminals), or the kind of charge generated by rubbing certain types of materials
against each other. As useful and as easy to understand as DC, it is not the only “kind” of
106
electricity in use. Certain sources of electricity (most notably, rotary electro-mechanical
generators) naturally produce voltages alternating in polarity, reversing positive and negative
over time. Either as a voltage switching polarity or as a current switching direction back and
forth, this “kind” of electricity is known as Alternating Current (AC): Figure 2.1
Fig 2.1 Direct Vs alternating current
Whereas the familiar battery symbol is used as a generic symbol for any DC voltage
source, the circle with the wavy line inside is the generic symbol for any AC voltage source.
One might wonder why anyone would bother with such a thing as AC. It is true that in some
cases AC holds no practical advantage over DC. In applications where electricity is used to
dissipate energy in the form of heat, the polarity or direction of current is irrelevant, so long
as there is enough voltage and current to the load to produce the desired heat (power
dissipation). However, with AC it is possible to build electric generators, motors and power
distribution systems that are far more efficient than DC, and so we find AC used
predominately across the world in high power applications. To explain the details of why this
is so, a bit of background knowledge about AC is necessary
2.2 REACTANCE AND IMPEDANCE– INDUCTIVE
2.2.1 AC resistor circuits
If we were to plot the current and voltage for a very simple AC circuit consisting of a
source and a resistor (Figure 2.2), it would look something like this: (Figure 2.3)
Fig 2.2 Pure resistive AC circuit
Fig 2.3 Voltage and current “in phase” for resistive circuit
107
Because the resistor simply and directly resists the flow of electrons at all periods of
time, the waveform for the voltage drop across the resistor is exactly in phase with the
waveform for the current through it. We can look at any point in time along the horizontal
axis of the plot and compare those values of current and voltage with each other (any
“snapshot” look at the values of a wave are referred to as instantaneous values, meaning the
values at that instant in time). When the instantaneous value for current is zero, the
instantaneous voltage across the resistor is also zero. Likewise, at the moment in time where
the current through the resistor is at its positive peak, the voltage across the resistor is also at
its positive peak, and so on. At any given point in time along the waves, Ohm‟s Law holds
true for the instantaneous values of voltage and current.
Let „i‟ is the alternating current through the circuit.
ER – Voltage drop across the resistor.
As per the Ohms Law V = i x R
i=V/R
i = ER sinωt / R
i = (ER/R) sinωt
= Im sinωt
Where, Im = ER/R
Instantaneous Power = v i
= Emsinωt x Imsinωt
= EmImsin2ωt
Average Power
= VI watts
2.2.2 AC Inductor Circuits
Inductors do not behave the same as resistors. Whereas resistors simply oppose the
flow of electrons through them (by dropping a voltage directly proportional to the current),
inductors oppose changes in current through them, by dropping a voltage directly
proportional to the rate of change of current. In accordance with Lenz’s Law, this induced
voltage is always of such a polarity as to try to maintain current at its present value. That is, if
current is increasing in magnitude, the induced voltage will “push against” the electron flow;
if current is decreasing, the polarity will reverse and “push with” the electron flow to oppose
the decrease. This opposition to current change is called reactance, rather than resistance.
Expressed mathematically, the relationship between the voltage dropped across the inductor
and rate of current change through the inductor is as such:
The expression di/dt is one from calculus, meaning the rate of change of instantaneous
current (i) over time, in amps per second. The inductance (L) is in Henrys, and the
instantaneous voltage (e), of course, is in volts. Sometimes you will find the rate of
instantaneous voltage expressed as “v” instead of “e” (v = L di/dt), but it means the exact
same thing. To show what happens with alternating current, let‟s analyze a simple inductor
circuit: (Figure 2.4)
108
Fig 2.4 Pure inductive circuit
If we were to plot the current and voltage for this very simple circuit, it would look
something like this: (Figure 2.5)
Fig 2.5 Pure inductive circuit, waveforms
An inductor‟s opposition to change in current translates to an opposition to alternating
current in general, which is by definition always changing in instantaneous magnitude and
direction. This opposition to alternating current is similar to resistance, but different in that it
always results in a phase shift between current and voltage, and it dissipates zero power.
Because of the differences, it has a different name: reactance. Reactance to AC is expressed
in ohms, just like resistance is, except that its mathematical symbol is X instead of R. To be
specific, reactance associate with an inductor is usually symbolized by the capital letter X
with a letter L as a subscript, like this: XL.
Since inductors drop voltage in proportion to the rate of current change, they will drop
more voltage for faster-changing currents, and less voltage for slower-changing currents.
What this means is that reactance in ohms for any inductor is directly proportional to the
frequency of the alternating current. The exact formula for determining reactance is as
follows:
i.e.,
Applied Voltage
= Back emf
Applied Voltage (ET) = Vmsinωt
Back e.m.f induced in the coil = L.(di/dt)
Therefore,
ET
Vmsinωt
di
= L di/dt
= L di / dt
= Vm sinωt dt
t
Integrating on both sides, we get
∫di
= ∫(Vm / L)sinωt dt
109
i =
=
Vm
[-sin( 90 – ωt)]
ωL
Vm
ωL
= Im
sin(ωt – 90)
sin (ωt – 90)
Instantaneous Power = v x i
= Vm sinωt x Im sin(ωt-90)
= Vm Im sinωt sin(ωt-90)
Average Power = 0
Example
Fig 2.6 Inductance Reactance
2.2.3 AC capacitor circuits
Capacitors do not behave the same as resistors. Whereas resistors allow a flow of
electrons through them directly proportional to the voltage drop, capacitors oppose changes
in voltage by drawing or supplying current as they charge or discharge to the new voltage
level. The flow of electrons “through” a capacitor is directly proportional to the rate of
change of voltage across the capacitor. This opposition to voltage change is another form of
reactance, but one that is precisely opposite to the kind exhibited by inductors.
Expressed mathematically, the relationship between the current “through” the
capacitor and rate of voltage change across the capacitor is as such:
The expression de/dt is one from calculus, meaning the rate of change of
instantaneous voltage (e) over time, in volts per second. The capacitance (C) is in Farads, and
110
the instantaneous current (i), of course, is in amps. Sometimes you will find the rate of
instantaneous voltage change over time expressed as dv/dt instead of de/dt: using the lowercase letter “v” instead or “e” to represent voltage, but it means the exact same thing. To show
what happens with alternating current, let‟s analyze a simple capacitor circuit: (Figure 2.7)
Fig 2.7 Pure capacitive circuit: capacitor voltage lags capacitor current by 90o
If we were to plot the current and voltage for this very simple circuit, it would look
something like this: (Figure 2.8)
Fig 2.8 Pure capacitive circuit waveforms
A capacitor‟s opposition to change in voltage translates to an opposition to alternating
voltage in general, which is by definition always changing in instantaneous magnitude and
direction. For any given magnitude of AC voltage at a given frequency, a capacitor of given
size will “conduct” a certain magnitude of AC current. Just as the current through a resistor is
a function of the voltage across the resistor and the resistance offered by the resistor, the AC
current through a capacitor is a function of the AC voltage across it, and the reactance
offered by the capacitor. As with inductors, the reactance of a capacitor is expressed in ohms
and symbolized by the letter X (or XC to be more specific).
Since capacitors “conduct” current in proportion to the rate of voltage change, they
will pass more current for faster-changing voltages (as they charge and discharge to the same
voltage peaks in less time), and less current for slower-changing voltages. What this means is
that reactance in ohms for any capacitor is inversely proportional to the frequency of the
alternating current.
When the alternating voltage is given to a capacitor, the capacitor is charged in one direction
and then in the opposite direction.
Q
= CV
= CVmsinωt
111
Now the current in the circuit is
i
= dq / dt
= d(CVmsinωt)
dt
= CVmωcosωt
= Imcosωt
= Imsin(ωt + 90)
= Vm /XC
……………. (2.1)
Where, Im = CVmω
XC = 1 / ωC
It is clear that from the equations i and v by an angle 90
Power factor = cosθ
= cos 90
=0
Power in the pure capacitor = VIcosθ
= 0. Watts.
Example
Fig 1.9 Capacitive Reactance
2.3 Simple Problem
Problem 4
The Inductance of 20mH is connected to AC supply of 50Hz. Find the inductive reactance.
Given Data:
L = 20mH
= 20 x 10-3
f = 50 Hz
Find Data:
XL
Solution:
XL
= 2πfL
= 2 x (22/7) x 50 x 20 x 10-3
= 6.28 Ω
Result:
Inductive Reactance is 6.28 Ω
112
Problem 5
The capacitor of 20 μF is connected to AC supply of 60 Hz. Find the capacitive reactance.
Given Data:
C = 20 μF
= 20 x 10-6
f = 50 Hz
Find Data:
XC
Solution:
XC
= 1 / 2πfC
= 1 / (2 x π x 60 x 20 x 10-6)
= 159.15 Ω
Result:
Capacitive Reactance is 159.15 Ω
Problem 6
The inductive capacitance is 20 ohms. Find the Capacitor of the supply frequency is 60 Hz.
Given data:
XC = 20 Ω
f = 60 Hz
Find data:
C
Solution:
XC
C
= 1 / 2πfC
= 1 / 2 . π . f . XC
= 0.000132 F
Result:
The Capacitor is 0.000132 F.
2.4 Key words
Alternating Current
Inductive
Capacitive
Reactance
Impedance
2.5 Questions for Discussion
1.
2.
3.
4.
Define the inductive reactance
What is relationship between frequency and inductive reactance?
In the circuit, 40mH is connected in the 40Hz. Find the value of inductive reactance.
The capacitive reactance of the circuit is 3 ohms. This capacitor is connected to the
70Hz supply frequency and 200V. What is the value of the capacitor?
5. Define the capacitive reactance.
6. Draw the vector diagram of pure inductor and pure capacitor circuit.
113
2.6 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. http://www.ibiblio.org/obp/electricCircuits/Ref/
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
114
UNIT
3
RL AND RC CIRCUITS
CONTENTS
3.0 Objectives
3.1 Introduction
3.2 Series RL circuits
3.3 Series RC circuits
3.4 Parallel RL circuits
3.5 R-C Parallel Circuits
3.6 Simple Problems
3.7 Keywords
3.8 Questions for Discussions
3.9 Suggested Reading
3.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
Derive voltage and current in RL and RC Series Circuit
Derive the voltage, current and impedance of the RL and RC parallel circuits
Solve problems and draw vector diagram of RL and RC series and parallel circuits
3.1 Introduction
Figure 3.1: Single phase power system schematic diagram
115
Depicted above (Figure 3.1) is a very simple AC circuit. If the load resistor‟s power
dissipation were substantial, we might call this a “power circuit” or “power system” instead
of regarding it as just a regular circuit. The distinction between a “power circuit” and a
“regular circuit” may seem arbitrary, but the practical concerns are definitely not.
One such concern is the size and cost of wiring necessary to deliver power from the
AC source to the load. Normally, we do not give much thought to this type of concern if
we‟re merely analyzing a circuit for the sake of learning about the laws of electricity.
However, in the real world it can be a major concern. If we give the source in the above
circuit a voltage value and also give power dissipation values to the two load resistors, we
can determine the wiring needs for this particular circuit: (Figure 3.2)
Figure 3.2: As a practical matter, the wiring for the 20 kW loads at 120 Vac
3.2 Series RL circuits
Consider resistance of R ohms and inductance of L Henry are connected in series
across AC supply of EmSinωt. The voltage across resistance is ER and same that of inductance
is EL.
The total voltage is vector sum at ER and EL
Fig 3.3 Series resistor inductor circuit
ET = ER + EL
I is the RMS value of Current.
/ Voltage drop across R, ER = IR
Voltage drop across L, EL = IXL
/ The total applied voltage (ET) = ER + j EL
ET = IR + j IXL
2
2
|Z| = √ (R + XL )
Where, Z is called the impedance of the RL circuit. Z has a magnitude and phase angle.
|Z = θ = tan -1 (XL / R)
θ - The angle by which the total voltage E is out of the phase from the current I..
Power Consumed is VI
= VI Cosθ Watts
116
3.3 Series RC circuits
Consider resistance of „R‟ ohm and capacitance of „C‟ farad are connected in series
across AC supply of EmSinωt. The voltage across resistance is IR and same that at capacitor
is IXC.
According to the KVL total voltage is vector sum of ER and VC.
Fig 3.4 Series RC Circuit
ET = ER + EC
I is the RMS value of Current.
/ Voltage drop across R, ER = IR
Voltage drop across L, EC = IXC
/ The total applied voltage (ET) = ER - j EC
ET = IR - j IXL
= I(R - j XL)
|Z| = √ (R2 + XC2)
Where, Z is called the impedance of the RL circuit. Z has a magnitude and phase angle.
|Z = θ = tan -1 (XC / R)
The total voltage E lags behind the current I by θ
Power Consumed is VI
= VI Cosθ Watts
3.4 Parallel RL circuits
Consider of R ohms and inductances of L Hentry are connected in parallel across AC
supply at VmSinwt. The circuit is as shown in the fig 3.5.
Fig 3.5 Parallel R-L circuit
The total current is the vector sum of IR and IL .
I = IR + IL
= (V/R) + (V / XL)
= V ( 1/R + 1/jwL)
= V (G -jBL)
= V.Y
Where,
G = 1/R  Conductance
117
B = 1 / XL  Susceptance
In R-L parallel circuit current lags the voltage by Ө. The power consumed is VI cos Ө.
3.5 R-C Parallel Circuit
Consider the resistor of R ohm and capacitor of C farad are connected in parallel
across AC supply of VmSinwt. The circuit is as shown in fig 3.6.
Fig 3.6 Parallel RC Circuit
The total current is vector sum of IR and IC.
I
= IR + IC
= (V / R) + (V / XC)
= V ( 1 / R +j / XC)
= V (G + jBC)
= V.Y
Where, Y = admittance (G + jBC)
In this circuit current leads voltage by θ. The power is VIcosθ.
3.6 Simple Problems
Problem 7
A current of 20 amps flows in a circuit with 30 angle of lag when the applied voltage is
200V. Find the resistance, reactance and impedance.
Given data:
Current (I)
= 20 A
Angle (θ)
= 30
Voltage (V) = 200 V
Find data:
Resistance (R)
Reactance (X)
Impedance (Z)
Solution:
Z =
V
I
200
20
= 10 Ω
=
118
PF = Cosθ
= Cos30
= 0.866
PF =
R
Z
R = Z.PF
= 10 x 0.866
= 8.66 Ω
X2 = (Z2-R2 )
X
= √(102-8.662)
= 5.1 Ω
Results:
R = 8.66 Ω
X = 5.1 Ω
Z = 10 Ω
Problem 8
A 120 V, 60 W lamps is to be connected across a 220V 50Hz supply mains. Calculate the
value of inductance required in order that is not over run.
Given data:
Find data:
Inductance (L)
Solution:
Lamp resistance (R) = V2/P
= 1202/60
= 240 Ω
Current through the lamp (I) = P/V
= 60 / 120
= 0.5 A
Impedance (Z) = V / I = 220 / 0.5
= 440 Ω
Z
= √(R2 + X2)
X
= √(4402 – 2402)
= 368.78 Ω
L
= (XL / 2πf)
= 1.17 H
119
Result
Inductance required is 1.17 H
Problem 9
A coil of resistance 15 ohms and inductance 0.05 H is connected in parallel with a non
inductive resistor of 20 ohm. Find (a) the current in each branch circuit (b) The total current
(c) Phase angle of the combination when a voltage of 200 V of 50 Hz is applied.
Given data:
Solution:
Current through 20 Ω i1
= V/R
= 200 / 20
= 10 A
Current through the coil i2
Total current (i) = i1 + i2
Phase angle
= V/Z
= 200 / (15+j15.7)
= 6.3627 – j6.66
= 9.2106|-46.306
= 10 + 6.3627 – j6.66
= 16.3627 – j6.66
= 17.666|-22.147
= 22.147 Lag
Result:
i1
= 10 A
i2
= 9.2106 A
i
= 17.666 A
Phase angle = 22.147
Problem 10
A circuit consisting of a resistor in series with a capacitor takes 80 watts at a power factor of
0.4 from a 100 V, 50 Hz supply. Find the resistance and capacitance.
Given Data:
Power (P)
= 80 Watts
Power factor = 0.4
V
= 100 V
f
= 50 Hz
Find data:
Resistance (R)
Capacitance (C)
120
Solution:
R
Z
PF =
R
= PF x Z
Where, Z is
Z =
V
I
I =
P
VCosФ
Where, I is
=
80
100 x 0.4
= 2A
Therefore Z becomes
V
100
=
= 50 Ω
I
2
Z =
R = PF x Z
= 0.4 x 50
= 20 Ω
Finding C
XC = √ Z2 - R2
XC = √ 502
- 202
XC = 45.63 Ω
XC =
1
2πfC
C
=
1
2πf XC
C
=
1
2x π x 50 x 45.83
C
=
69.45 x 10-6 Farad or 69.45 mFarad
R
=
20 Ω
C
=
69.45 x 10-6 Farad or 69.45 mFarad
Result:
121
Problem 11
Two impedance Z1= (10 + j5) and Z2 = (8 + j6) are connected in parallel across a voltage of
200 V. Find the total current taken and also the power factor of the circuit.
Given Data:
Z1= (10 + j5)
Z2 = (8 + j6)
V = 200 V
Find data:
Total Current (I)
Power factor (PF)
Solution:
Z1= (10 + j5) = √(102 + 52) / θ=tan-1(5/10) = 11.18 /26.56 Ω
Z2= (8 + j6) = √(82 + 62) / θ=tan-1(6/8)
= 10 /36.87 Ω
I1 =
V
Z1
I1 =
200
11.18
I1 = 17.89 = 16 – j8
I2 =
V
Z2
I1 =
200
10
I1 =
20 -
= 16 – j12
I
=
=
=
=
I1 + I2
16 – j8 + 16 – j12
32 – j20
37.73
Amps
Power factor of the circuit
PF = Cos θ
= 0.84 (lagging)
Result:
I = 37.73 / 32 Amps
PF = 0.84
122
3.7 Keywords
Series Circuit
Impedance
Magnitude
Phase Angle
Parallel Circuit
Suceptance
Admittance
Henry
Power factor
3.8 Questions for Discussions
1. Explain the RL series circuit and derive the impedance for this circuit.
2. Explain the construction details of RC series circuit.
3. Derive the impedance and power of the RC series circuit.
4. Explain and derive the impedance of RL parallel circuit.
5. Explain the RC parallel circuit.
6. Prove the power of RL series circuit is VIcosФ
7. A coil is connected across a 200V, 50 Hz supply and takes a current of 8A. The power
loss in the cable is 560 Watts. Calculate (i) Resistance (ii) Reactance (iii) Impedance
(iv) Power factor and draw the vector diagram.
8. A current of 5A flows through a non inductive resistance in series with a choking coil
when supplied at 250V, 50 Hz. If the voltage across the resistance is 125 V and across
the coil 200V. Calculate (a) the impedance, reactance and resistance of coil. (b) the
power absorbed by the coil (c) total power.
9. A choke coil connected across a 250 V, 50 Hz supply takes a current by choke when it
is connected across a 230V, 25Hz.
10. A circuit consists of a resistance R ohm and inductance L Henry connected in series.
If it is connected to a 240V, 50Hz supply mains it consumes 300 watts and voltage
drop across „R‟ is 100 V. Calculate the value of inductance L.
11. A 100V, 100W lamp is to work at its rated values from a 200V, 50 Hz ac mains. Find
the capacitance of the capacitor required.
12. A voltage of V = 100sin314t is applied to a circuit consisting of a 20 ohms resistor
and a 100 mfd capacitor in series. Find (a) The power consumed (b) The p.d. across
the capacitor.
123
3.9 Suggested Reading
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Tony Kuphaldt,Using the Spice Circuit Simulation Program, in“Lessons in
Electricity,
Reference”,
Volume
5,
Chapter
7,
at
http://www.ibiblio.org/obp/electricCircuits/Ref/
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
124
UNIT
4
SERIES RLC CIRCUITS
CONTENTS
4.0 Objectives
4.1 R-L-C Series Circuit Constructions
4.2 Impedance for RLC Series Circuit
4.3 Power in RLC series Circuit
4.4 RLC series Resonance
4.4.1 Effects of Series Resonance
4.4.2 Quality Factor of Series Resonance Circuit
4.4.3 Resonance Cure
4.5 Solved Problem
4.6 Keywords
4.7 Questions for Discussions
4.8 Suggested Reading
4.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
Derive voltage and current RLC Series Circuit
Solve the RLC series circuits problems.
4.1 R-L-C Series Circuit Constructions
Let us consider resistor R ohms, inductance of L henry and capacitance of C farad is
connected in series across AC supply of VmSinwt.
The total voltage is vector sum of drop across each component. This circuit has two possible
cases.
125
Fig 4.1 RLC Series Circuit
4.2 Impedance for RLC Series Circuit
E – RMS value of applied voltage
I – RMS value of current.
Voltage drop across R, ER
= IR (In phase with I)
Voltage drop across L, EL
= IXL (Leading I by 90 )
Voltage drop across C, EC
= IXC (Lagging I by 90)
The total applied voltage E is the vector sum of three voltage drops.
E = ER + EL + EC
= IR + jIXL –jIXC
= I ( R + j (XL – XC))
= IZ
Where, Z is the Impedance of the circuit.
Z = √ (R2 + (XL - XC)2)
(XL -XC)
θ = tan-1
R
Let,
(
)
For the given circuit,
i)
If XL > XC then the circuit behaves like RL circuit.
ii)
If XL < XC then the circuit behaves like RC circuit.
Fig 4.2
diagram
E
E
Circuit
L
L
The phasor
are as
the fig 4.2.
EL EC
E
EC EL
E
R
C
C
4.3 Power in RLC series Circuit
Let,
V = Vmsinωt
I = Imsin(ωt – Ф)
= Imsin(ωt + Ф)
Average Power is P,
E
diagrams
shown in
R
E
E
Vector
of RLC
---------------------
126
(XL > XC)
(XL < XC)
E
2π
P =
Vmsinωt Imsin(ωt – Ф)
dωt
2π
∫
0
Note:
1.
2.
For, XL > XC write the expression for power in RL series circuit.
For, XL < XC write the expression for power in RC series circuit.
4.4 RLC series Resonance
An RLC series circuit is said to be in resonance when circuit power factor in unity,
i.e., at XL = XC, the frequency at which resonance occurs is called the resonance frequency
(fr). The resonance in an RLC series circuit can be achieved by changing the supply
frequency. At a certain frequency called the resonance occurs.
At resonance,
XL = XC
2πfrL
=
1
2πfrC
fr
=
1
2π√(LC)
Fig 4.3 RLC Series Resonance Circuit
A RLC series circuit is shown in fig 4.3. (a) and the phasor diagram for series resonance is
shown in fig 4.4 (b). In phasor diagram of resonance circuit
127
VL
VL = V C
VR
VC
Fig 4.4 Series Resonance – Phasor Diagram
Voltage drop across R
Voltage drop across L
Voltage drop across C
= VR = IR (In phase with I)
= VL = IXL (Leading I by 90 )
= VC = IXC (Lagging I by 90)
= VC in magnitude but opposite in direction.
VL
4.4.1 Effects of Series Resonance
The effects of RLC series resonance circuit are
When series resonance occurs, the inductive resonance occurs; the inductive reactance
XL and capacitive reactance XC are equal and opposite and cancel each other.
The impedance of the circuit is minimum and equal to the resistance of the circuit. (Zr
= R)
The current in the circuit is maximum, ( Ir = V / Zr)
The power factor of the circuit is unity.
The voltage drop across L and C is very large.
4.4.2 Quality Factor of Series Resonance Circuit
At series resonance the voltage across L and C is many times greater than the applied
voltage. This voltage magnification is called Q factor or Quality factor of the series resonant
circuit.
Q Factor =
Voltage across L or C
Applied Voltage
=
VL
V
=
IXL
IR
=
XL
R
=
IXL
IZr
=
ωr L
R
But resonance frequency fr is
128
fr
=
1
2π√(LC)
2πωr
=
1
2π√(LC)
ωr
=
1
√(LC)
Substituting the value of ωr in above equation, we get
Q factor
=
1
L
x
√(LC)
R
Q factor
=
1
R
√
L
C
4.4.3 Resonance Cure
Im= V/R
The curve drawn between current and frequency is known as resonance curve. Fig 4.5
shows the resonance curve of a typical RLC series circuit. At resonance frequency (f r) the
inductive reactance XL is equal to capacitive reactance XC. The impedance of the circuit is
only resistive and equal to R. Hence the current at resonance is the maximum and power
factor is unity.
0.707 Im
F1
Fr F2
Frequency
Fig 4.5 Series Resonance Curve
If the frequency of the circuit is less than resonance frequency XC becomes greater
than XL and the circuit behaves as a RC circuit. The current is decreased due to increased
impedance and the power factor is leading.
If the frequency of the circuit is greater than resonance frequency X L becomes greater
than XC and the circuit behaves as a RL circuit. The current is decreased due to increased
impedance and the power factor is lagging.
129
4.5 Solved Problem
Problem 10
A resistance of 20 ohms an inductance of 0.2 H and a capacitance 100 micro Farad connected
in series across a 220 V, 50 Hz. Determine (i) Impedance (ii) Current (iii) Voltage drop
across resistance, inductance and capacitance (iv) Power factor (v) Phase angle (vi) power in
watts.
Given data:
20 Ω
220 V
50 Hz
0.2 H
100 μF
Find data:
Impedance (Z)
Current (I)
Voltage drop VR, VL and VC Power factor(PF)
Phase angle (θ)
Power (P)
Solution:
Inductance Reactance (XL) = 2πfL
= 2π x 50 x 0.2
= 62.83 Ω
Capacitance Reactance (XC) = 1 / 2πfC
= 1/(2π x 50 x 100 x 10-6)
= 31.83 Ω
Impedance
(Z)
= √(R2 + (XL2 – XC2))
Z
= 36.89 Ω
Current (I)
=V/Z
= 220 / 36.89
= 5.96 A
Voltage drop across resistance (VR) = I.R
= 5.96 x 20
= 119.2 V
Voltage drop across inductance (VL) = I.XL
= 5.96 x 62.83
= 374.46 V
Voltage drop across capacitance (VC)= IXC
= 5.96 x 31.83
= 189.7 V
Power factor (PF) = Cosθ
= R/Z = 20/36.89
= 0.5421 lag
Phase angle (θ)
= Cos-1(PF)
= 57.17
Power = VICosθ
= 220 x 5.96 x 0.5421
= 710.87 Watts
Result:
Impedance = 36.87 Ω
130
Current
= 5.96 A
VR
= 119.2 V
VL
= 374.46 V
VC
= 189.7 V
Power factor = 0.5421
Phase angle = 57.17
Power
= 710.87 Watts
4.6 Keywords
RL Circuit
RC Circuit
Resonance
Q Factor
Bandwidth
Voltage drop
Power
4.7 Questions for Discussions
1.
2.
3.
4.
5.
Explain the construction of RLC series circuit.
Derive the impedance for RLC series circuit.
Define Q factor
What is resonant condition derive resonant frequency in RLC series circuit.
A series RLC circuit consists of 5 Ω, L = 1mH and C = 100μF. Calculate the
frequency at which resonance will take place, if the applied voltage is 230V at 50 Hz
calculate the current and voltage drop across RLC.
6. Sketch the response curve for RLC series circuit and mark (i) Band width (ii) Both cut
off frequencies (iii) Resonant frequency.
7. Prove that Q0 = (1/R)(√(L/C)
4.8 Suggested Reading
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Electrical Circuit Theory, Gopalsamy, Veni Publication
6. Circuit theory, Ganagadhar K.A., Kanna Publisher
7. Electrical Circuit Theory and Technology, Bird John
131
UNIT
5
PARALLEL RLC CIRCUITS
CONTENTS
5.0 Objectives
5.1 R-L-C Parallel Circuit Constructions
5.2 RLC Parallel Resonance Circuit
5.3 Parallel Resonance Characteristics
5.4 Keywords
5.5 Questions for Discussion
5.6 Suggested Readings
5.0 Objectives
On completion of the following units of syllabus contents, the students must be able to
Derive voltage and current RLC Parallel Circuit
Explain the parallel resonance circuit
Understand the characteristics of parallel resonance
5.1 RLC Parallel Circuit Constructions
Consider resistor R, inductor L and capacitor C are connected in parallel across AC supply of
VmSinwt. The circuit is as shown in fig 5.1.
Fig 5.1 RLC Parallel Circuit
The total current may lead or lag the applied voltage depends on XL > XC or XL < XC.
Applying the Kirchoff‟s Current Law,
132
I = IR + IL + IC
I =
E
E
E
+
+
Z1
Z2
Z3
I
1
1
1
=
+
+
E
Z1
Z2
Z3
1
1
1
1
=
+
+
ZEq
Z1
Z2
Z3
This relationship is identical to what we had for parallel connected resistor.
The following conditions to be applied for the RLC parallel circuits,
If XL > XC the circuit behaves like R-L like parallel circuit.
If XL < XC the circuit behaves like R-C like parallel circuit.
5.2 RLC Parallel Resonance Circuit
A resistor R, inductance L and a capacitance C are connected in parallel across a
sinusoidal voltage of variable frequency source. The circuit diagram is shown in fig 5.2.
Fig 5.2 RLC Parallel Resonance
The vector diagram is shown in fig 5.3. In such a circuit resonance occurs when the current IL
and IC are equal.
Fig 5.3 Vector diagram of RLC Parallel
ie., at resonance,
IL
=
IC
V
=
V
133
XL
XC
2πfL
fr
=
1
2πfC
=
1
2π√LC
5.3 Parallel Resonance Characteristics
5.3.1 Impedance Vs Frequency Curve
The impedance Vs Frequency curve of two branch RLC parallel circuit is shown in
fig 5.4. The impedance of the circuit is maximum at resonance. As the frequency changes
from resonance the circuit impedance decreases very rapidly.
Zm
fr
Fig 5.4
For frequencies below resonance, the capacitive reactance XC is higher. Therefore
more current will flow through coil. Thus the circuit becomes more inductive and the current
lag behind the applied voltage. For the frequencies above resonance, inductance X L is higher.
Therefore more current will flow through the capacitor. Thus the circuit behaves as capacitive
and the circuit current leads the applied voltage.
5.3.2 Current Vs Frequency Curve
IM = Current at resonance
fr = Frequency at resonance
Zr = Impedance at resonance
The fig 5.5 shows the current frequency curve of the parallel circuit. The current at
resonance is minimum. As the frequency changes from resonance, the circuit increases
rapidly.
134
I
Frequency
Fig 5.5 Current Vs Frequency
5.4 Keywords
Parallel Resonance
Sinusoidal Voltage
Vector
Impedance
Frequency
RL Parallel Circuit
RC Parallel Circuit
5.5 Questions for Discussion
1.
2.
3.
4.
Explain the construction and working of RLC Parallel resonance circuit.
Explain the RLC parallel resonance circuit
Explain the characteristics of RLC parallel resonance circuit
Draw the Vector diagram of RLC parallel circuit
5.6 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Electrical Circuit Theory, Gopalsamy, Veni Publication
135
136
UNIT
1
STAR DELTA CONNECTIONS
CONTENTS
1.0 Objectives
1.1 Introduction
1.2 Generation of three Phase System
1.3 Advantages of 3 Phase System
1.4 Phase Sequence
1.5 Interconnection of three phases
1.5.1 Star or Wye Connection
1.5.2 Delta or Mesh Connection
1.6 Three Phase Voltage
1.6.1 Line Voltage
1.6.2 Phase Voltage
1.7 Three Phase Current
1.7.1 Phase Current
1.7.2 Line Current
1.8 Keywords
1.9 Questions for Discussions
1.0 Objectives:
1.10 Suggested Readings
On completion of the following units of syllabus contents, the students must be able to
Understand 3 supply and advantages of three phase over single phase system
Understand the basis terms in the three phase system
Explain the interconnection of three phase system
1.1 Introduction
137
The electrical system using more one phase is called poly phase system. A poly phase
system should be balanced and symmetrical. When the emf‟s and phase angles of each phase
are equal the system is balanced. When the emf‟s are equal and are displaced from each other
by equal angles the system is symmetrical.
Large power generation, transmission and distribution is by poly phase system which
has several sources of equal emf‟s, phase angles of these emfs from each other being 2π / n,
where n is the number of phases. For 2- phase system the difference is π /2 and for 3 phase
system 2π / 3 is 120.
Most electrical systems are 3 phase systems as they have distinct economic and
operational advantages over single phase system. For same power transmission the cost of
transmission line is less. The torque produced in 3 phase motor is uniform and smooth; the
torque in single phase motor is pulsating. Three phase motor are self starting, single phase
motor require phase splitting. Power factor of 3 phase motor is high while in single phase
motors it is low.
Generation of Three Phase System:
In a three phase system, three insulated coils are wound on the armature. A phase difference
is 120 is kept between the coils. The fig 1.1 shows the three phase generation by using three
coils.
Fig 1.1 Three Phase generation of supply
The three starting ends of each coil and brought to the slip rings. The three remaining ends
are connected together at a point in the winding. The three coils have same number of turns.
All the coils rotate at the same speed in the magnetic filed. Therefore, the emf‟s generated
have the same peak value and frequency. The induced emfs in the three colis have a phase
difference is 120. The emfs induced in the three coils are given by
EX = EMAX.sinθ
EY = EMAX.sin(θ - 120 )
EZ = EMAX.sin(θ - 240 )
The fig 1.2 shows the vector diagram of three phase system.
138
Fig 1.2 Vector diagram of three phase supply.
1.3 Advantage of 3 phase system
Electrical power is generated, transmitted and distributed in the form of 3 phase power. Home
and small industries are used for single phase power. The three phase power is preferred over
single phase power for the following reasons.
1. Phase power has a constant magnitude but single phase power is pulsating one.
2. A 3 phase system can be set up a rotating magnetic field in stationary winding. This is
not possible in single phase.
3. For the same rating 3 phase machines are smaller in size and have better operating
characteristics than single phase machine.
4. Three phases has better power factor and efficiency.
5. There is saving of conductor material when the same power is to be transmitted over a
given distance by a 3 phase system compared to a single phase system.
6. Generation, transmission and utilization of power are more economical in three phase
system compared to single phase system.
1.4 Phase sequence
The order in which the voltage in the three phases reach their maximum values is
called phase sequence. This is determined by the direction of rotation of the alternator.
Referring the fig 1.3 is easy to say that voltage in coil A attains maximum positive value first,
next coil B and then coil C. Hence the phase sequence is ABC.
Fig 1.3 Three phase wave form
139
1.5 Interconnection of three phases
In a 3 phase alternator, there are three windings or phases. Each phase has two terminals ie,
start and finish. In practice, the three windings are interconnected in two methods.
I. Star or Wye connection (Y)
II. Mesh or Delta connection (Δ)
1.5.1 Star or Wye Connection
In this method, similar ends (Start or finish) of the three phases are joined together to form a
common junction N as shown in the fig 1.4. The junction N is called the star point or neutral
point. The three line conductors are connected from the three free ends of the three phases.
The three lines are named as R, Y & B. This constitutes a 3 phase, 3 wire star connected
system.
Fig 1.4 Star Connections
Sometimes a fourth wire called neutral wire is connected from the neutral point as shown in
figure. This system is called to 3 phase 4 wire star connected system.
1.5.2 Delta or mesh connection (Δ)
In this method of interconnection, the dissimilar ends of the three phase windings are
joined together ie, finishing end of one phase is connected to the starting end of the other
phase and so on to obtain mesh or delta connection as shown in fig 1.5. The three line
conductors are taken from the three junctions of the mesh and are named as R, Y, B. This is
called 3-phase, 3wire, delta connected system. Since no neutral exists in a delta connection,
only 3phase, 3wire system can be formed.
Fig 4.5 Delta Connection
1.6 Three Phase Voltages
140
In the three phase system consist of two voltages. This are given below
i. Line Voltage
ii. Phase Voltage
1.6.1 Line Voltage (VL)
In 3-phase system the voltage between any two lines or phase is called the line voltage.
1.6.2 Phase Voltage (Vph)
The voltage between any one line and neutral is called phase voltage.
1.7 Three Phase Currents:
In the three phase system consist of two voltages. This are given belo
i. Line Current
ii. Phase Current
1.7.1 Phase currant Iph
The currant flowing in a phase is called the phase current.
1.7.2 Line Current IL
The Current flowing in a line is called the line current.
1.8 Keywords
Poly-phase
Balanced Load
Symmetrical
Generation
Armature
Slip Ring
Magnetic field
Efficiency
Star
Delta
Mesh
Line
Phase
1.9 Questions for Discussions
1.
2.
3.
4.
5.
What is meant by three phase system?
What are the advantages of three phase system?
Explain how the three phases is generated?
What is line voltage?
What is Phase Voltage?
141
6. Define phase Current & Line current?
7. Explain the star connected system.
8. Explain the construction of delta connected system.
1.10 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. http://www.ibiblio.org/obp/electricCircuits/Ref/
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
142
UNIT
2
BALANCED LOAD AND UNBALANCED LOAD IN
THREE PHASE CIRCUITS
CONTENTS
2.0 Objectives
2.1 Introduction
2.2 Voltage in balanced Delta Connection
2.3 Line & Phase Current in balanced Delta
Connection
2.4 Voltage in balanced Star Connection
2.5 Line & Phase Current in Balanced Star
Connection
Keywords
2.02.6
Objectives:2.7 Questions for Discussion
On completion of the following units of syllabus contents, the students must be able to
2.8 Suggested Readings
Derive the relationship between Line current & Phase current and Line voltage &
Phase Voltage for Star and Delta connected system in Balanced condition.
2.1 Introduction
A poly phase system consists of two or more single phase systems. The voltage or current in
the d
. The three voltages
may be interred connected; they are of the same magnitude and frequencies but differ in
phase as shown in fig 2.1.
143
Fig 2.1
. In an n-phase
system, n voltages equal in magnitude and frequency, are displaced from each other by 360/n
electrical degrees.
2.2 Voltages in balanced Delta connection
Since the system is balanced the three phase voltages are equal in magnitude but displaced
120 from each another.
Fig 2.2 Balanced 3 phase Delta Connected System
From the connection diagram shown in fig 2.2 only one phase winding is connected between
any pair of lines. Hence in delta connections, the line voltage is equal to the phase voltage.
Ie. VL = VPH
2.3 Line Current & Phase Current in balanced Delta Connection
Since the system is balance, the three phase currents are equal in magnitude but
displaced 120 from one another as shown in the vector diagram in the fig 2.3.
Fig 2.3 Vector diagram of Delta connection
144
ie.
IRY = IYB = IBR = IPH
Similarly three line currents are equal.
ie.
IR = IY = IB = IL
At point 1 apply KCL,
IR = IBR - IRY
From the vector diagram
Similary
IB = IYB - IBR
Power
Total power,
IR = √(IBR2 + IRY2 + 2IBR. IRY. cos 60)
= √3 . IPH
IY = IRY - IYB
= √3 . IPH
= √3 . IPH
P = 3 x Power / Phase
= 3 x VPH.IPH.cosØ
= √3VL.IL. cosØ
2.4 Voltages in balanced Star Connection
The fig 2.4 shows a balanced 3 phase Y-connected system. The emf generated in the three
phases is ERN, EYN and EBN is equal in magnitude but displaced 120 from one another.
Fig 2.4 Balanced 3 Phase Y Connected System
ie.
ERN = EYN = EBN = VPH
VPH = Phase voltage ie. Voltage between phase and neutral
VRY = VYB = VBR = VL
VL = Line voltage ie. Voltage between two lines
VRY = ERN + ENY Phasor sum
VRY = ERN - EYN Phasor difference
Fig 2.5 Vector diagram of Y Connected System
145
From the vector diagram shown in fig 2.5.
VRY = √ (ERN2 + EYN2 + 2ERN. EYN. cos 60)
VRY = √3.VPH
Similarly
VYB = EYN – EBN
= √3.VPH
VBR = EBN – ERN
= √3.VPH
Hence in a balanced 3 phase Y connection
Line voltage VL = √3.VPH
2.5 Line and phase current in balanced Star Connection
In Y connection each line conductor is connected in a series to a separate phase as shown in
fig 2.5. Therefore line current is equal to the phase current
ie. IL = IPH
2.6 Keywords




Line Current
Line Voltage
Phase Current
Phase Voltage
2.7 Questions for Discussion
1. Prove that is star connected load VL = √3. VPH
2. What is the relationship between Phase current & Line current in the delta system?
And prove the relationship.
3. What is the relationship between Phase current & Line current in the star system?
4. Derive the relationship between line and phase voltage in delta connected circuit.
2.8 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Electrical Circuit Theory, Gopalsamy, Veni Publication
146
UNIT
3
MEASUREMENTS OF THREE PHASE POWER
CONTENTS
3.0 Objectives
3.1 Introduction
3.2 Single Wattmeter Method
3.3 Two wattmeter Method
3.4 Three Wattmeter Method
3.5 Keywords
3.6 Questions for Discussion
3.0 Objectives:
3.7 Suggested Readings
On completion of the following units of syllabus contents, the students must be able to
 Measure the power in the three phase system by using wattmeter.
 Derive the expression for power and power factor in a three phase circuit by using
single, two and three wattmeter for balanced load.
3.1 Introduction
Poly phase system is an ac system which is having a group of equal voltages arranged to have
equal phase difference between adjacent e.m.f‟s. A poly phase system is combination of
single phase voltage having same amplitude and frequency but different phase. The electrical
displacement depends on the number of phases and is given by
Electrical displacement
= 360 Electrical….
Number of Phases
The above relation is not applicable for two phase system where the voltages are displaced by
90. The most commonly used poly phase system.
Three phase power can be measured by the following method.
 Single wattmeter method
 Two wattmeter method
 Three wattmeter method
147
3.2 Single wattmeter Method
The total power in 3 phase balanced loads can be measured by the following methods
by using single wattmeter.
1.
2.
3.
4.
Potential Lead shift method
T method
Artificial Neutral method
Current Transformer method
If the load is balanced, the power in any phase can be measured by a single wattmeter.
The total power is given by multiplying the wattmeter reading by three.
Fig 3.1 Power measurement with Single wattmeter
Fig 3.1 shows single wattmeter method of measuring three phase power for a balanced star
connected load.
Total Power = 3 x Wattmeter Reading
Another method of power measurement is as shown in the fig 3.2. The current coli is
connected in one of the lines and one end of the pressure coil to the same line, other being
connected alternately to the other two lines. The phasor diagram is shown in fig 3.3.
Fig 3.2 Power measured by using single wattmeter with two way switch
148
Fig 3.3 Phasor diagram for single wattmeter method
When the switch is at position 3 means,
= V13I1Cos (30 - Ф)
= √3 V.I.Cos(30 – Ф)
When the switch is at position 2, Reading of the wattmeter is
W2
= V12I1Cos (30 + Ф)
= √3 V.I.Cos(30 + Ф)
Sum of the two wattmeter readings are
W1
………. (1)
………. (2)
= √3VI[Cos(30 – Ф) + Cos(30 + Ф)]
= √3 VI [cos30 cosФ – sin30 sinФ + cos30 cosФ + sin30 sinФ]
= √3 VLIL [2 cos30 cosФ]
= 3 VLIL CosФ
……… (3)
From the above the sum of the two wattmeter readings is the power consumed by the load.
Total Power (W)
= W1 + W2
……… (4)
W1 + W2
3.3 Two wattmeter method
A more convenient method is to use two wattmeter‟s and this method is valid whether
the load is balanced or not. Consider the two wattmeters are as shown in fig 3.4 for the star
connected load. In this method current coils of the two wattmeters are connected in any two
lines and pressure coil of each joined to the third line. The algebraic sum of the two readings
gives the total power of the circuit.
Fig 3.4 Power measurement by using two wattmeter’s
149
When a three phase star or delta connected load is balanced and power factor of the
load can be determined from the two wattmeter readings.
Assume the load is inductive and the power factor angle is Ф (lagging). Let VRN, VYN, VBN be
the three phase voltages. IR, IY, IB are the three line currents flowing through the load.
Wattmeter 1 (W1)
Current through coil of W1 = IR
Voltage across pressure coil of W1 is = VRY
= VRN + VNY
= VRN - VYN
From the phasor diagram shown in Fig 3.5
Fig 3.5 Phasor diagram of 2 wattmeter method
Angle between VRY and IR is = (30 + Ф)
Wattmeter Reading W1
= VRYIRcos(30 + Ф)
……… (3.1)
Wattmeter 2 (W2)
Current through coil of W2 = IB
Voltage across pressure coil of W2 is = VBY
= VBN + VNY
= VBN - VYN
From the phasor diagram shown in Fig 3.3
Angle between VBY and IB is = (30 - Ф)
Wattmeter Reading W2
= VBYIBcos(30 - Ф)
……… (3.2)
Since the load is balanced
VRY = VBY = VBR = VL (Line voltage)
IR
= IY = IB = IL (Line current)
W1
= VLILcos (30 + Ф)
………. (3.3)
W2
= VLILcos (30 – Ф)
……….. (3.4)
W1 + W2
= VLILcos (30 + Ф) + VLILcos(30 – Ф)
= VLIL [cos(30 + Ф) + cos(30 – Ф)]
= VLIL [cos30 cosФ – sin30 sinФ + cos30 cosФ + sin30 sinФ]
= VLIL [2 cos30 cosФ]
= VLIL [2 x √3/2 x cosФ]
= √3 VLILCosФ
……..... (3.5)
150
Hence the sum of the two wattmeter readings gives the total power of the three phase
load.
The power factor is
W 1 + W2
= √3 VLILCosФ
W1 – W2
= VLILSinФ
||ly
...……. (3.6)
(3.6) ÷ (3.5) We get
W2 – W1
W2 + W1
=
VLILSinФ
√3 VLILCosФ
SinФ
CosФ
= √3
W2 – W1
W2 + W1
tanФ
= √3
W2 – W1
W2 + W1
Ф
=
tan-1
√3
W2 – W1
W2 + W1
Thus the two wattmeter readings, we can find Ф and hence the load power factor CosФ.
3.4 Three wattmeter method
If all the phase leads are accessible, three wattmeters can be connected as shown in the fig
3.6. The total power is sum of the three wattmeter readings.
Fig 3.6 Power measurement with three wattmeter
Total Power = W1 + W2 + W3
Where,
W1
- Wattmeter 1 Reading
W2
- Wattmeter 2 Reading
W3
- Wattmeter 3 Reading
The above scheme is rarely used. The connections are difficult to make because usually the
delta may not be broken and also the star point may not be available.
151
3.5 Keywords








Wattmeter
Potential Lead
Artificial
Current Transformer
Pressure Coil
Current Coil
Lagging Power factor
Leading Power factor
3.6 Questions for Discussion
1.
2.
3.
4.
5.
6.
7.
8.
How to measure the power in three phase system?
What are the methods available for three phase power measurement?
Explain the single wattmeter method to measure the 3 phase power?
Explain the construction detail of two wattmeter method to measure the 3 phase
power.
Derive the expression for measuring power and power factor in a three phase circuit
by using two wattmeters for balanced load.
Explain the three wattmeter method to measure the 3 phase power.
Why the 3 wattmeter methods are rarely used?
What are the advantages of two wattmeters method?
Suggested Reading
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. http://www.ibiblio.org/obp/electricCircuits/
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
152
UNIT
4
EFFECTS OF UNBALANCED LOAD
CONTENTS
4.0 Objectives
4.1 Introduction
4.2 Unbalanced Delta connected Load
4.3 Unbalanced 3phase, 3 wire Star connected Load
4.4 Unbalanced 3 phase, 4 wire Star Connected Load
4.5 Keywords
4.6 Questions for Discussion
4.7 Suggested Readings
4.0 Objectives
On completion of the following units of syllabus contents, the students must be able to
 Define unbalanced load system
 Understand the needs of unbalanced load.
 Explain the unbalanced star and delta connected loads
4.1 Introduction
The unbalanced means satisfied the following conditions




The line or phase currents are different
Different power factor for all the lines
The load impedance in the three phase are not equal
Displaced from one another by unequal angles.
Most of the industries are unbalanced load conditions. Because, each phases are different
loads or different timing of operation. This will leads to change the load current and phase
angle.
153
4.2 Unbalanced Delta Connected Load
Fig 4.1 shows an unbalanced delta connected load. Assume a phase sequence of RYB.
The unbalanced delta connected load supplies from a balanced three phase supply do not
present any new problems because the voltage across the load phase is fixed. It is
independent of the nature of the load.
Fig 4.1 Unbalanced Delta connected Load
For delta connection line voltage is equal to phase voltage. Taking VRY as the reference
phasor voltage. Assuming the phase sequence is RYB.
Phase currents are
Line Currents are
154
4.3 Unbalanced 3-phase, 3-wire Star Connected Load
Fig 4.2 shows the unbalanced 3 phase 3 wire star connected load. Consider a 3-phase,
3 wire star connected; the following methods are used for analysis of such networks.
a. Equivalent delta method.
b. Mesh method.
Fig 4.2 Unbalanced 3-phases, 3-wire star connected load
a. Equivalent delta method
In this method, the unbalanced star load is replaced by an equivalent delta connected load and
then it can be solved as before. The equivalent delta load impedances can be obtained by star
delta transformation.
b. Mesh Method
In this method two mesh currents are chosen. Then the mesh equations are written as
ERY = I1(ZR + ZY) – I2ZY
EYB = -I1ZY + I2(ZY + ZB)
The above two equations are solved for I1 and I2 then find the value of IR, IY and IB from I1
and I2.
i.e.,
IR
IB
= I1
= - I2
4.4 Unbalanced 3-pahse, 4-wire Star Connected Load
Fig 4.3 shows an unbalanced star connected load to a balanced 3 phase 4 wire supply. In this
system the star point of the load is connected with the neutral of the supply. Hence the load
phase voltages are same as supply phase voltages. However the current in each phase will be
different.
Fig 4.3 Unbalanced 3-phase, 4-wire Star Connected load
155
The phase voltages are
The phase currents and the line currents are equal
The current through the neutral IN is the vector sum of the three line currents.
4.5 Keywords




Unbalanced
Phasor Voltage
Mesh Current
Vector
4.6 Questions for Discussion
1.
2.
3.
4.
What is meant by unbalanced load?
Define unbalanced load in 3 phase system.
Explain unbalanced delta connected load circuit.
Explain with neat circuit diagram of unbalanced 3 phase, 4 wire star connected load
system.
5. Explain the circuit of unbalanced 3 phase, 3 wire star connected system
4.7 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. http://www.ibiblio.org/obp/electricCircuits
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
156
UNIT
5
PROBLEMS IN THREE PHASE CIRCUITS
CONTENTS
5.0 Objectives
5.1 Introduction
5.2 Star – Delta Problems
5.3 Power Measurement Problems
5.4 Questions for Discussion
5.5 Suggested Readings
Objectives:
On completion of the following units of syllabus contents, the students must be able to
 Solve simple problems of Star and delta connection circuit
 Find the power and power factor of the simple circuit by using wattmeter methods.
5.1 Introduction
In this method, similar ends (Start or finish) of the three phases are joined together to
form a common junction N. The junction N is called the star point or neutral point. In star
connection the voltage and current is written by
VL = √3.VPH
IL = IPH
In this connection power is
P = √3VL.IL. cosØ
In this method of interconnection, the dissimilar ends of the three phase windings are
joined together ie, finishing end of one phase is connected to the starting end of the other
phase and so on to obtain mesh or delta connection. In this method, voltage and currents are
written by
IL = √3 . IPH
Power
Total power,
VL = VPH
P = √3VL.IL. cosØ
157
The above two method power can be measured by using wattmeters. Wattmeter can be one or
two or three is used to measure the total power.
Most commonly used method is two wattmeter methods. The total power of this method is
sum of the two wattmeter readings. The power factor of this method is
Ф
=
tan-1
√3
W2 – W1
W2 + W1
Total power (P) = W1 + W2
5.2 Star – Delta Problems
Problem 1
Three identical coils each having a resistance at 15 ohms and inductance of 0.5 H are
connected in delta to a 415V, 50 Hz, 3 phase supply. Determine (a) line current (b) power
factor (c) power
Given Data:
Find data:
Line Current (IL)
Power factor (PF)
Power (P)
Solution:
The phase inductive reactance (XL) = 2πfL
= 2π x 50 x 0.5
= 157.07 Ω
The Phase Impedance (ZPH)
= √(RPH2 + XPH2)
= √(152 + 157.072)
= 157.79 Ω
The phase Current (IPH)
= VPH/ZPH
= 415 / 157.79
= 2.63 A
In delta connection Line current (IL) = √3. IPH
= √3 x 2.63
= 4.555 A
Power factor (PF) = Cosθ = R/Z
= 15 / 157.79
= 0.095 lag
Power (P)
= √3.VL.IL.Cosθ
= √3 x 415 x 4.555 x 0.095
= 311 Watts
158
Result
IL
= 4.555 A
PF
= 0.095
P
= 311 Watts
Problem 2
Three identical coils each having a resistance of 10 ohms and an inductance of 0.4 H are
connected in star across 400 V, 3 phase 50 Hz supply. Calculate the power factor and total
power.
Given Data:
Find data:
Power factor (PF)
Total Power (P)
Solution:
Phase impedance (ZPH)
= √(RPH2+XLPH2)
= √(102 + 125.662)
= 126.06 Ω
Star Connection (VL) = √3. VPH
VPH = 400 / √3
= 231 V
IPH = IL = VPH / ZPH = 231 / 126.06
= 1.832 A
Power factor = Cosθ = R / Z = 10 / 126.06
= 0.079 lag
Power (P) = √3.VLILCosθ
= √3 x 400 x 1.832 x 0.079
= 100 Watts
Result:
PF = 0.079
P = 100 Watts
Problem 3
Three similar coils are connected in star taken at a total power of 1.5 KW at a power factor of
0.2 lagging from a 3 phase 400 V, 50 Hz supply. Calculate the resistance and inductance of
each phase.
Given data:
P = 1.5 KW
PF = CosФ
= 1.5 x 103 W = 1500 W
= 0.2
159
VL = 400 volts
f = 50 Hz
Find Data:
Resistance (RPh)
Inductance (LPh)
Solution:
In star system,
= √3 VPh and IL = IPh
= √3 VLILCosФ
= P / (√3 VLCosФ)
= 1500 / (√3 x 400 x 0.2)
= 10.83 A
IL
= IPh = 10.83 A
VPh
= VL / √3
= 400 / √3
= 230.94 Volts
ZPh
= VPh / IPh
= 230.94 / 10.83
= 21.32 Ω
CosФ = RPh / ZPh
RPh
= ZPh.CosФ
= 21.32 x 0.2
= 4.264 Ω
XL
= √(ZPh2 – RPh2)
= √(10.832 – 4.2642)
= 20.89 Ω
XL
= 2πfL
L
= XL / 2πf
= 20.89 / (2 x π x 50)
= 0.066 H
VL
P
IL
Result:
Resistance (RPh) = 4.246 Ω
Inductance (LPh) = 0.066 H
Problem 4:
The power input to 400 Volts, 50 Hz, 3 phase motor is measured by two wattmeter which
indicate 300 KW and 100 KW respectively. Calculate (a) the input power (b) Power factor (c)
line current.
Given Data:
VL
f
W1
W2
= 400 V
= 50 Hz
= 300 KW
= 100 KW
= 300 x 103
= 100 x 103
Find Data:
1. Input Power
2. Power Factor
3. Line current
Solution:
(a)
Input Power
160
P
(b)
= W1 + W2
= 300 + 10
= 400 KW = 400 x 103 W
Power Factor (CosФ)
Cos Ф
=
Cos
(
tan-1
√3
W2 – W 1
W2 + W 1
)
Cos Ф
=
Cos
(
tan-1
√3
100 – 300
100 + 300
)
Cos Ф
=
0.76
(c)
P
IL
Line Current (IL)
= √3 VLILCosФ
= P / (√3 VLCosФ)
= 400 x 103 / (√3 x 400 x 0.76)
= 760 A
Result:
(a) Input Power
= 400 KW
(b) Power Factor
= 0.76
(c) Line Current
= 760 A
5.3 Power Measurement Problems
Problem 5:
A three phase 440 Volt motor operates with a power factor of 0.4. two wattmeters are
connected to measure the input power, and the total power taken from the mains is 30 KW.
Find the readings of each Wattmeter.
Given Data:
PF
P
VL
f
= CosФ
= 0.4
= Total Power = 30 KW
= 440 V
= 50 Hz
= 30 x 103
Find Data:
W1 (First Wattmeter Reading)
W2 (Second Wattmeter Reading)
Solution:
P
Ф
= W1 + W2
= Cos-1(PF)
= Cos-1(0.4)
= 66.42
= 30 x 103 = 30,000
tan Ф
=
√3
W2 – W1
W2 + W1
tan 66.42
=
√3
W2 – W1
30 x 103
161
…………….. (1)
W2 – W1
…………….. (2)
= 39,665
(1) + (2) 
2W2 = 69,665
W2
= 34832.5 Watts or 34.8325 KW
From equation (1), Substitute the W2
We get,
W1
= - 4832.5 Watts or - 4.8325 KW
Result:
W1
W2
= - 4.8325 KW
= 34.8325 KW
Problem 6:
A 500 Volt, 3 phase motor has an output of 3.73 KW and operate at a power factor 0.85 with
an efficiency of 90% calculate the reading on each of the two wattmeter connected to
measure the input.
Given data:
V
= 500 volt
Output Power = 3.73 KW
= 3.73 x 103
PF
= 0.85
Efficiency (η) = 90 %
Find Data:
Wattmeter Reading W1 and W2
= 3730 W
Solution:
Efficiency
= Output
Input Power
= Output
Input
Efficiency
Ф
= 3730
0.9
Input Power = 4144.44 Watts
i.e., W1 + W2 = 4144.44
= Cos-1(PF)
= Cos-1(0.85)
= 31.78
(1) + (2) 
tan Ф
=
√3
W2 – W1
W2 + W1
tan 31.78
=
√3
W2 – W1
4144.44
W2 – W1
= 1483.53
2W2
W2
……………………. (1)
…………….. (2)
= 5627.97
= 2813.98 Watts
162
From equation (1), Substitute the W2
We get,
W1
= 1330.46 Watts
Result:
W1
W2
= 1330.46 W
= 2813.98 W
Problem 7:
Two wattmeter connected to measure the power of a 3 phase circuit indicated 2500 W and
500 W respectively. Find the power factor of the circuit when (a) Both the readings are
positive and (b) the later reading is obtained after reversing the connection of the current coli
of the wattmeter.
Given Data:
W1
W2
= 2500 W
= 500 W
Find data:
(a)
(b)
Solution:
Power factor (PF)
Both the readings are Positive
After reversing the connection of the current coil of the wattmeter
tan Ф
(a)
W1
W2
(b)
=
√3
W2 – W1
W2 + W1
When the two reading are positive
= 500 Watts
= 2500 Watts
tan Ф
=
√3
W2 – W1
W2 + W1
tan Ф
=
√3
2500 - 500
2500 + 500
tan Ф
= 1.156
Ф
=
Ф
=
PF
=
PF
= 0.65
tan-1(1.156)
CosФ
= Cos49.1
Second reading is negative
163
W1
W2
= -500 Watts
= 2500 Watts
tan Ф
=
√3
W2 – W1
W2 + W1
tan Ф
=
√3
2500 – (-500)
2500 + (-500)
tan Ф
=
2.598
Ф
=
tan-1(2.598)
Ф
=
CosФ
PF
=
PF
= 0.36
= Cos49.1
Result:
PF at both reading are positive
= 0.65
PF at reading is obtained after reversing the
connection of the current coil of the wattmeter
= 0.36
Problem 8:
The power input to a 3 phase induction motor is read by two wattmeters. The reading are 860
W and 240 W. What is the input power and power factor of the motor?
Given Data:
W1
W2
= 860 W
= 240 W
Find Data:
Power (P)
Power Factor (PF)
Solution:
P
= W1 + W2
= 860 + 240
= 1100 Watts
tan Ф
=
√3
W2 – W1
W2 + W1
tan Ф
=
√3
240 – 860
860 + 240
164
tan Ф
Result:
P
PF
= 0.976
Ф
=
tan-10.976
PF
=
Cos Ф
= 44.31
= Cos 44.31 = 0.72
= 1100 Watts
= 0.72
Questions for Discussion
1. Three identical impedances are connected in delta to a 3 phase 400 V supply. The line
current is 34.65A and the total power taken from the supply is 14.4 KW. Calculate the
resistance values of each impedance.
2. Three similar coils are connected in star taken at a total power of 1.5 KW at a PF of
0.2 lagging from a 3 phase 400V, 50 Hz supply. Calculate the resistance and
impedance of each phase.
3. A balanced three phase star connected load of 150 KW takes a leading current of
100A with a line voltage of 1100 V, 50 Hz. Find the circuit constant of the load per
phase.
4. Two wattmeters are connected to measure the power in a 3 phase balanced load.
Determine the total power factor if the two wattmeters read 1000 watts each (1) both
positive and (2) second reading is negative.
5. Three identical coils are each having a resistance of 10 ohms and reactance of 10 ohm
are connected in delta across 400V, 3 phase supply. Find the line current and the
readings on each of the two wattmeters connected to measure the power.
6. Three identical coils having a resistance of 10 W and inductance of 0.5 H are
connected in star across 400V, 3 phase 50 Hz circuit. Calculate the line current and
total power consumed.
7. A balanced three phase, delta connected load of 150 KW takes a leading current of
100A with a line voltage of 1100V, 50 Hz. Find the circuit constant of the load per
phase.
8. The power input to a 2KV, 50 Hz, 3 phase induction motor running on full load with
90 % of efficiency, is measured by two wattmeters which read 300KW and 100KW.
Find (i) the power input (ii) Power output, (iii) Power factor and (iv) the line current.
165
5.5 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. http://www.ibiblio.org/obp/electricCircuits/Ref/
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
166
167
UNIT
1
BASICS OF TRANSIENTS
CONTENTS
1.1 Objectives
1.2 Circuit Transients
1.3 Laplace Transformation
1.4 Advantages of Laplace Transformation
Technique
1.5 Keywords
1.6 Questions for Discussion
1.7 Suggested Readings
1.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
 Understand needs of Transients
 Known the Laplace formulas
 Understand the needs of Laplace Transform
1.1 Circuit transients
When, a circuit is switched from one state to another, the current in the circuit changes. The
change will depend on the properties of the circuit.
Example, if an e.m.f of 200 volts is applied to a 10 ohm pure resistor as in fig 1.1 (a) the
current rises abruptly, almost instantaneously to its steady value of 20 amps as in Fig 1.1(b).
Fig 1.1 Basic Resistor transient circuit
168
If the same resistor is coiled around a magnetic core, as shown in fig 1.2 (a) the 20 amps
current value is reached only after a long time as shown in fig 1.2 (b). This is because time is
required to store energy in the magnetic core.
Fig 1.2 Basic Inductor transient circuit
If this circuit is switched off, time is required to dissipate the stored magnetic energy. The
current of 20 A is falling down to zero only after a definite time interval. Depending upon the
circuit parameters the response from an initial steady state to final steady state in a time
period is known as transient period. The response is known as transients.
1.2 Laplace transformation
In circuit transients, we often come across integro-differential equations. For solving such
equations we shall adopt Laplace transformation technique. Table 5.1 shows the some of the
Laplace transformation functions.
Table 5.1 Laplace transformation of some importance functions
Sl.No
F(t)
1
u(t)
2
e-at
3
eat
4
t
5
te-at
6
f1(t) + f2(t)
7
sin ωt
8
cos ωt
9
tn
10
df(t)/dt
11
∫f(t) dt
F(s)
1
S
1
S+a
1
S-a
1
S2
1
(S+a)2
F1(S) + F2(S)
ω
S2+ω2
S
2
S +ω2
n!
Sn+1
SF(S) – f(0+)
F(S)
+
S
169
f-1(0+)
S
1.3 Advantages of Laplace Transformation Technique
The classical method of solving ordinary differential equations is rather involved. In the
classical method, the complementary function and particular integral have to be determined
and finally the arbitrary constants have to be obtained from known initial conditions. The
Laplace transform method is superior to the classical method due to the following reasons:
 It simplifies functions Laplace transformation transforms exponential and
trigonometric functions into simple algebraic functions.
 It simples operations. Laplace transformation transforms differentiation and
integration, respectively, into multiplication and divisions. It transforms integro
differential equations into algebraic equations which are easier to handle.
 The arbitrary constants need not be determined separately. In solving a differential
equation, by the Laplace transformation method, the solution comes out complete.
 It makes use of step and impulse response.
1.4 Keywords
 Transient
 Laplace Transform
 Intgro Differential
1.5 Questions for Discussion
1.
2.
3.
4.
5.
6.
7.
8.
What is transient?
State any one reason for transients in DC circuits.
What is transient analysis?
Explain the Resistance transient circuit?
Explain the inductor transient response?
What are the advantages of Laplace transform?
Write some of Laplace transforms formula used in the integral equations.
What is the use of Laplace transform?
1.6 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Tony Kuphaldt,Using the Spice Circuit Simulation Program, in“Lessons in
Electricity, Reference”,
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
170
UNIT
2
TRANSIENTS IN RL
CONTENTS
2.0 Objectives
2.1 DC Transient in RL Circuit
2.2 Voltage across Resistor in RL transient Circuit
2.3 Voltage across Inductor in RL transient circuit
2.4 RL Decay Transmit
2.5 Keywords
2.6 Questions for Discussion
2.7 Suggested Readings
2.0 Objectives:On completion of the following units of syllabus contents, the students must be able to
 Derive the expression of Voltage and current for RL Transient Circuit
 Explain the RL Decay transmit circuit.
2.1 DC transients in RL Circuit
An RL circuit is shown in fig 2.1. it is connected to a battery through a switch. The current
during the time interval before the steady conditions are reached is called transient current
Fig 2.1 RL Transient Circuit
171
Assume that the switch S is closed at time t = 0 and also assume that at the time of switching
the current is zero. (i.e., initial current = 0)
Applying the Kirchoff‟s Voltage Law for the loop, we get
V
V
= VR + V L
= iR + L. (di/dt)
…………………….. (2.1)
This is a differential equation and solve for i, we use Laplace transformation technique.
V/s
= RI(s) + L[sI(s) – i(0)]
……………….. (2.2)
`
We known that just before closing the switch, the current i(0) = 0. Therefore the equation
(2.2) becomes
V/s
= RI(s) + LsI(s)
I(s) (R+Ls) = V/s
I(s)
= V/s(R+sL)
……………… (2.3)
Divide both Nr and Dr by L
I(s)
= (V/L) / s(R/L + s) ……………… (2.4)
Using partial fraction technique
I(s)
=
=
V
L
=
s(
A
s
A(
V/L
R / L + s)
+
B
(R/L + s)
R/L + s )
Bs ……………… (2.5)
Put s = 0 on the equation (2.5)
V
L
=
AR
L
A
=
V
R
0
……………… (2.6)
Put s = 1 on the equation (2.5)
V
L
=
A(
V
L
=
AR
L
B
=
V
L
R/L + 1 )
B
+A+B
-
AR
L
172
-A
……………… (2.7)
Put value of A on the equation (2.7)
B
=
-A
B
=
-
V
R
……………… (2.8)
Put values of values of A and B on the equation (2.4)
I(s)
=
V/R
s
I(s)
=
V
R
-
[
1
s
V/R
(R/L + s)
-
1
(R/L + s)
]
Taking inverse Laplace transform
L-1I(s)
i(t)
=
V
R
=
V
R
[
[
1
1
s
L-1
-
e-(R/L)t
]
1
(R/L + s)
L-1
-
]
……………… (2.9)
Accord
ing to
the above equation the current rises exponentially as shown in the fig 2.2.
Fig 2.2 Transient response for RL circuit
The final value of current may be obtained by substituting t = α in the equation (2.9)
|
i(t)
i(t)
=
t=α
=
V
R
[
1
-
e-α
]
V
R
……………… (2.9)
[Since eα = 0]
2.2 Voltage across Resistor in RL transient Circuit
The transient voltage across the elements of R, L circuits are obtained from the current.
VR
………………….. (2.10)
=iR
Put i(t) value of the equation (2.9) on the above equation (2.10)
173
VR
=
V
R
[
1
-
e-(R/L)t
]
VR
=
V
[
1
-
e-(R/L)t
]
R
i.e.,
2.3 Voltage across Inductor in RL transient circuit
VL
………………….. (2.11)
= L. (di / dt)
Put i(t) value of the equation (2.9) on the above equation (2.11)
d
dt
V
R
[
VL
=
L
VL
=
VL
R
[
d
dt
=
VL
R
[
d
dt
=
VL
R
[
=
V
VL
0
(
(
1
1
(1)
-
-
R
L
d
dt
e-(R/L)t
e-(R/L)t
) ]
) ]
e-(R/L)t
-
-
e-(R/L)t
]
]
e-(R/L)t
2.4 RL Decay Transmit
Consider the circuit as shown in fig 2.3. The switch S is previously in position 1. The switch
has been in position (1) till steady state conditions are reached. The current then flowing is i =
V/R amps. At t = 0, the switch is tuned to position (2). Now the supply source is removed and
puts a short circuit across the RL circuit.
Fig 2.3 RL Decay Transmit Circuit
Apply Kirchoff‟s Voltage Law to the circuit at position (2).
VR + VL
=0
iR + L.(di / dt) = 0
…………………………. (2.12)
And this equation is to be solved with the initial condition i(0) = V / R (steady state current)
Taking Laplace transform on both sides in equation (2.12)
174
RI(s) + L [sI(s) – i(0)]
(
=0
)
RI(s) + L sI(s) – V/R
=0
I(s) [R + sL] – (VL / R)
=0
I(s) [R + sL] = VL/R
I(s)
VL
R
=
(
1
R+sL
)
Divide both Nr and Dr by L
I(s) =
V
R
1
R/L+s
(
)
To get i(t), taking inverse Laplace on both sides
L-1I(s) =
i(t)
V
R
(
=
V
R
L-1
1
R/L+s
)
e-(R/L)t
2.5 Keywords




RL Circuit
Switch
Transient response
Decay Transmit
2.6 Questions for Discussion
1. Explain the construction details of RL transient circuit.
2. Derive the current and voltage drop across Resistor and inductor expression for RL
transient circuit.
3. Explain RL Decay transient.
2.7 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Tony Kuphaldt,Using the Spice Circuit Simulation Program, in“Lessons in
Electricity, Reference, at http://www.ibiblio.org/obp/electricCircuits/Ref/
6. Electrical Circuit Theory, Gopalsamy, Veni Publication
175
UNIT
3
TRANSIENTS IN RC
CONTENTS
3.0 Objectives
3.1 DC Transient in RC Circuit
3.2 Voltage across Resistor in RC transient
3.3 Voltage across Inductor in RC transient
3.4 Decaying Transient in RC Circuit
3.5 Keywords
3.6 Questions for Discussion
3.7 Suggested Readings
3.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
 Derive the expression of Voltage and current for RC Transient Circuit
 Explain the Decaying transient RC circuit.
3.1 DC Transients in RC Circuit
Consider the RC series circuit with a DC voltage applied through a switch in fig 3.1. Let the
capacitance have an initial charge of Q0 coulombs.
Fig 3.1 RC Transient Circuit
176
Hence the initial voltage on the capacitor is V0 = Q0 / C
Apply Kirchoff‟s voltage law to the circuit, we get
VR + VC = V
iR +
1
C
∫
idt + V0 = V
iR +
1
C
∫
idt = V - V0 …………… (3.1)
Assume V0 = Q0 / C = 0. Hence the equation (3.1) becomes
iR +
1
C
∫
idt = V
This is the integral equation; to solve it we shall take Laplace transform on both sides.
1
V
I(s) =
Cs
s
I(s)R +
I(s)
[
1
Cs
R +
]
V
s
=
Therefore I(s) is
V
I(s) =
s
(
1
Cs
R +
)
Divide both Nr and Dr by R
V/R
I(s) =
(
s +
1
RC
)
Taking the inverse Laplace Transform on both sides.
V/R
L-1[I(s)] = L-1
(
i(t)
=
V
R
s +
e-(1/RC)t
3.2 Voltage across Resistor in the RC transient
177
1
RC
)
………………… (3.2)
Voltage across the resistor
………………… (3.3)
VR = iR
Put equation (3.2) in equation (3.3)
VR
=
V
R
e-(1/RC)t
VR
=
V
e-(1/RC)t
R
3.3 Voltage across the Capacitor in the RC transient
Voltage across the capacitor VC is
VC =
VC
VC
1
C
∫
i(t)dt
=
V
CR
=
V
CR
=
V
0∫
t
e-(1/RC)tdt
t
1
.
[
[
(-1/RC)
1 -
e-(1/RC)t
e-(1/RC)t
]
0
]
3.4 Decaying Transient in RC Circuit
Consider the circuit as shown in the fig 3.2. The switch is in position (1) for sufficient time
that is up to steady state condition and at t = 0 the switch is moved to position (2).
Fig 3.2 Decaying Transient RC Circuit
Before the switch is moved to position (2) the capacitor gets charged to the voltage V with
the polarity as shown in fig 3.2.
At position (2), the differential equation of the circuit is
178
iR +
1
C
∫
idt + V = 0
…………. (3.3)
Where, V is the voltage across the capacitor. Taking Laplace on both sides.
I(s)
[
R +
1
Cs
]
=-
V
s
The solution is as in the previous RC circuit, and it is in the equation (3.4)
i(t)
= -
V
R
e-(1/RC)t
………………… (3.4)
Keywords




Initial charge
Kirchoff‟s voltage law
Inverse Laplace Transform
Decaying Transient
Questions for Discussion
1. Explain the construction details of RC transient circuit?
2. Derive the current and voltage drop across resistor & Capacitor expression for RC
transient Circuit.
3. Explain RL Decay transient.
3.7 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Electrical Circuit Theory, Gopalsamy, Veni Publication
179
UNIT
4
TRANSIENTS IN RLC
CONTENTS
4.0 Objectives
4.1 DC Transient in RLC Circuit
4.2 Varies result of Current in the RLC transient
4.3 Keywords
4.4 Questions for Discussion
4.5 Suggested Readings
4.0 Objectives:On completion of the following units of syllabus contents, the students must be able to
 Explain the construction of RLC Transient Circuit.
 Derive the current equation for RLC transient circuit.
 Understand the effects of current in the RLC Transient circuit.
4.1 DC Transients in RLC circuit
A RLC circuit is shown in fig 4.1. Assume that the switch S is closed at time t = 0 and also
assume that at the time of switching the current i(0) = 0.
Fig 4.1 RLC Transient Circuit
180
By applying Kirchoff‟s Voltage Law to the circuit.
iR + L . (di/dt) + (1/C)∫idt
=V
[Since i(0) =0]
Assume no initial charge on the capacitor. i.e., Q(0) = 0 Taking the Laplace transform on the
both sides.
1
V
I(s) =
Cs
s
RI(s) + L[sI(s) – i(0)]
I(s)
[
R + sL +
1
Cs
]
=
V
S
V
I(s) =
s
[
R + sL +
1
Cs
]
Divide both Nr & Dr by L
V/L
I(s) =
[
s2 + s(R/L) +
1
LC
]
4.2 Varies result of Current in the RLC transient
The denominator s2 + (R/L)s + (1/LC) is the quadratic equation. Hence the roots of the
denominator are
=
√ (
-R
±
2L
R
2L
)2
)2
-
-
1
LC
= α ± β
Where,
α =
-R
2L
β =
√ (
R
2L
1
LC
Looking at the discriminate, it is obvious that there are three possibilities.
181
Fig 4.2 Response Curve for RLC transient
Case 1: When the discriminant is positive
R
( )
2
1
>
2L
LC
The roots are real and different values. Then the two roots are
(α + β) and (α – β)
I(s) =
A
B
+
s – (α + β)
s – (α - β)
The solution given by
i(t)
= A e(α + β)t + B e(α - β)t
= eαt[A.eβt + B e-βt]
The resulting current is said to be over damped as shown in fig 4.2.
Case 2: When the discriminant is zero
R
( )
2
1
=
2L
LC
The roots are equal
I(s) =
A
B
+
(s – α)2
(s – α)
The solution is given by
i(t)
= Ateαt + B eαt
= eαt [At + B]
The resulting current is critically damped as shown in fig 4.2.
Case 3: When the discriminant is negative
R
( )
2
182
<
1
2L
LC
The roots are complex and conjugate of one another. Then the two roots are
(α + jβ) and (α – jβ).
I(s) =
The solution is given by
i(t)
A
B
+
s – (α + jβ)
s – (α - jβ)
= eαt[Acosβt + Bsinβt]
The resulting current is oscillatory and at the same time decays in short time. This is shown in
fig 4.2.
4.3 Keywords
1.
2.
3.
4.
5.
6.
RLC circuit
Initial charge
Laplace transform
Damping
Critically damping
Roots
4.4 Questions for Discussion
1. Explain the construction details of RLC transient circuit?
2. Derive the current expression for RLC transient circuit.
3. Explain RL Decay transient.
4.5 Suggested Readings
1. Electric circuit theory, Dr. M. Arumugam & Dr. N. Premkumaran., Khanna
Publishers, New Delhi
2. Basic Electrical Electronics and computer Engineering, Dr.G.Nagarajan., AR
Publications, Sirkali – 609111
3. Electrical Circuit Theory., A.Balakrishnan & T.Vasantha, N.V. Publication, Pollachi
Tamilnadu
4. A.E. Fitzergerald, David E. Higginbotham, Arvin Grabel, Basic Electrical
Engineering, (McGraw-Hill, 1975).
5. Electrical Circuit Theory, Gopalsamy, Veni Publication
183
UNIT
5
PROBLEMS
CONTENTS
5.0 Objectives
5.1 Problem for RL Transient Circuit
5.2 Problem for RC Transient Circuit
5.3 Problem for RLC Transient Circuit
5.4 Questions for Discussion
5.5 Suggested Readings
5.0 Objectives:
On completion of the following units of syllabus contents, the students must be able to
Solve the simple problem for RL, RC and RLC transient circuits.
5.1 Problem for RL Transient Circuit
Problem 1:A series RL circuit with R = 100 ohms and L = 20 Henry has a DC voltage of 200 applied
through a switch at t = 0. Find (a) the equation for the current and voltage across the different
elements (b) Current at t = 0.5 second (c) the current at t = 1second (d) the time at which VR =
VL.
Given Data:
Fig 5.1
184
Solution:
(a) The equation for Current & Voltage
The differential equation for the circuit is
di
= V
dt
iR + L
di
= 200
dt
100i + 20
Initial condition current i(0) = 0
Solution for the current is using Laplace transform technique is
across R (VR)
i(t)
=
i(t)
=
i(t)
=
V
R
200
100
[
1
1
e-(100/20)t
-
]
]
e-5t
-
]
e-(R/L)t
-
[
[
2
1
……….. (5.1)
Volta
ge
= i(t) R
VR
=
200
[1
-
e-5t
]
……….. (5.2)
Volta
ge across L (VL)
e-(R/L)t
VL
=
V
VL
=
200
………… (5.3)
e-5t
(b)
The current at t = 0.5 second
Substitute the t value in the equation (5.1)
i(t)
[
=
2
1
=
1.836 A
=
2
=
1.987 A
e-5 x 0.5
-
]
(c) The current at t = 1 Second
Substitute the t value in the
equation (5.1)
i(t)
[
1
]
e-5 x 1
-
(d) The time at which VR = VL
Substitute the equation (5.2) and
(5.3), We get
[
200
1
2e-5t
-
1
e-5t
=
=
e-5t
e-5t
1
185
]
=
200
e-5t
e-5t
=
0.5
Taking log for both sides, we get
-5t
=
log 0.5
t
=
0.138 Sec
Result:
(a)
i(t)
=
VR
=
VL
=
[
2
200
1
-
[
1
]
e-5t
-
e-5t
]
e-5t
200
(b)
i(t)
=
1.836 A
(c)
i(t)
=
1.987 A
(d)
t
=
0.138 Sec
Problem 2:Find the transient current passing through the RL series circuit as shown in fig 5.2. When the
switch is closed at time t = 0.
Fig 5.2
Solution:
The differential equation for the circuit is
iR + L
di
= V
dt
di
= 50
dt
25 i + 0.01
Initial condition current i(0) = 0
Solution for the current is using Laplace transform technique is
i(t)
i(t)
=
=
V
R
50
[
[
1
1 -
e-(R/L)t
e
-(25 / 0.01)t
186
]
]
25
i(t)
=
2
[
1
]
e-2500t
-
Result:
i(t)
=
2
[
1
]
e-2500t
-
5.2 Problem for RC transient Circuit
Problem 3:The 20 μF capacitor in circuit of fig 5.3 has an intial charge Q0 = 0.001 coulombs as
shown. The switch is closed at t = 0. Find the transient current.
Fig 5.3
Solution:Q0
= 0.001 coulomb
The initial voltage on the capacitor (V0) = Q0 / C
The differential equation is
iR +
1
C
∫
idt - V0 = V
Since the capacitor is charged in reversed direction, hence put –V0.
iR +
1
C
∫
idt = V +
= 50 +
Q0
C
0.001
20x10-6
= 100
Solving the above equation by Laplace transform, we get
i(t)
=
V
R
e-(1/RC)t
-6
i(t)
=
100
-(1/ 100 x 20 x 10 )t
e
100
=
e-500t
Result:
i(t)
= e- 500t
Problem 4:In the circuit of fig 5.4 the switch is closed on position is closed 1 at t = 0 and after 1
time constant is moved to position 2. Find the current before and after moving to position 2.
Assume no initial charge on the capacitor.
187
Fig 5.4
Solution:
When the switch is on position 1, the current equation is
No initial charge on capacitor given
i(t)
V
R
=
e-(1/RC)t
-6
i(t)
100
=
-(1/ 500 x 20 x 10 )t
e
500
=
………. (1)
0.2 e-100t
This continues up to 1 times constant. One time constant means, t = RC.
t
= 500 x 20 x 10 -6
= 0.01 sec
Therefore, the current at t
= 0.01 sec.
Substitute „t‟ in the equation (1), we get
i(t)
=
=
=
0.2 e-100t
0.2 e-100 x 0.01
0.0736 Amp
The capacitor then
has a voltage of VC, becomes
= V1(1 – e-t/RC)
[Since t = RC]
-1
= V1(1 – e )
= 63.2 Volts
The switch is then closed on the position 2 at t1=0 or t = 0.01 sec.
Then the equation for the circuit is
VC
iR +
1
C
∫
idt + V0 = V2
At position (2) the initial voltage across the capacitor V0 is 63.2 volts.
1
C
500 i + 1
C
500 i +
∫
∫
idt1 +
idt1
63.2
= -100
= -100
-
63.2
Assume V3 = -163.2 Volts
Apply Laplace transform for solving the above equation, we get
188
i(t)
V
R
=
e-(1/RC)t
-6
i(t)
-(1/ 500 x 20 x 10 )t1
163.2
= -
e
500
Since t1 = t – 0.01
-100(t – 0.01)
i(t)
= -
0.3264
e
= -
0.3264 e-100(t – 0.01)
= = -
0.3264 e-100 (0.01 – 0.01)
0.3264 A
At t1 = 0, (t = 0.01 sec)
i(t)
Problem 5
The RC series circuit in the fig 5.5 has an initial charge Q0 = 2 x 10 -2 Coulomb as shown.
Find the transient current if the switch is closed at t = 0.
Fig 5.5
Solution:
Q0 = 2 x 10-2 coulomb
After closing the switch the equation for the circuit is,
iR +
1
C
∫
idt + V0 = V
Since V0 = Q0 / C
∫
∫
∫
1
C
1
10 i +
C
1
10 i +
C
10 i +
idt =
V
-
idt =
100
-
Q0
C
2 x 10-2
100x10-6
idt = - 100 Volts
Solving the above equation by Laplace transform, we get
i(t)
=
V
R
e-(1/RC)t
-6
i(t)
= -
-(1/ 10 x 100 x 10 )t
100
e
10
=-
10 e-1000t
189
5.3 Problem for RLC transient Circuit
Problem 6:A series RLC circuit with R = 1000 ohms L = 0.1 Henry and C = 100 μF has DC voltage of
200 volts applied to it at t = 0 through a switch. Find the expression for transient current.
Assume initially relaxed circuit condition.
Given Data:
Fig 5.6
Solution:
When the switch is closed, the integro differential equation for the circuit is
iR + L
∫
di
1
+
dt
C
100 i + 0.1
idt = V
di
1
+
dt
100x 10-6
∫
idt = 200
All the values for initial condition zero.
Taking Laplace Transforms on both sides, we get
1
200
I(s) =
100x 10-6
s
100 I(s) + 0.1 [sI(s) – i(0)] +
I(s)
(
I(s)
=
104
s
100 + 0.1s +
)
=
200
s
200
s
(
100 + 0.1s +
104
s
)
2000
I(s)
=
s2 + 1000s + 105
The solution of s2 + 1000s + 105 is root of (s + 112.7) and (s + 887.3)
As the RLC circuits
2000
I(s)
=
(s + 112.7) (s + 887.3)
Taking inverse Laplace, we get
i(t)
= 2.582 e-112.7t – 2.582 e-887.3t Amp
190
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