BASIC ELECTRICAL ENGINEERING
V.HimaBindu
V.V.S Madhuri
Chandrashekar.D
GOKARAJU RANGARAJU
INSTITUTE OF ENGINEERING AND TECHNOLOGY
(Autonomous)
Index:
1. Syllabus……………………………………………….……….. .1
2. Ohm’s Law………………………………………….…………..3
3. KVL,KCL…………………………………………….……….. .4
4. Nodes,Branches& Loops…………………….……….………. 5
5. Series elements & Voltage Division………..………….……….6
6. Parallel elements & Current Division……………….………...7
7. Star-Delta transformation…………………………….………..8
8. Independent Sources …………………………………..……….9
9. Dependent sources……………………………………………12
10. Source Transformation:…………………………………….…13
11. Review of Complex Number…………………………………..16
12. Phasor Representation:………………….…………………….19
13. Phasor Relationship with a pure resistance……………..……23
14. Phasor Relationship with a pure inductance………………....24
15. Phasor Relationship with a pure capacitance………..……….25
16. Series and Parallel combinations of Inductors………….……30
17. Series and parallel connection of capacitors……………...…..32
18. Mesh Analysis…………………………………………………..34
19. Nodal Analysis……………………………………………….…37
20. Average, RMS values……………….……………………….....43
21. R-L Series Circuit……………………………………………...47
22. R-C Series circuit……………………………………………....50
23. R-L-C Series circuit…………………………………………....53
24. Real, reactive & Apparent Power…………………………….56
25. Power triangle……………………………………………….....61
26. Series Resonance……………………………………………….66
27. Parallel Resonance……………………………………………..69
28. Thevenin’s Theorem…………………………………………...72
29. Norton’s Theorem……………………………………………...75
30. Superposition Theorem………………………………………..79
31. Maximum Power Transfer Theorem………………………....82
32. Reciprocity Theorem…………………………………………..84
33. Single Phase Transformer……………………………………..88
34. Single Phase Induction Motor…………………………………95
35. DC Motor……………………………………………………….97
GOKARAJU RANGARAJU INSTITUTE OF ENGG AND TECHNOLOGY
(AUTONOMOUS) ACADEMIC YEAR 2014-15, Semester-I/II
BASIC ELECTRICAL ENGINEERING (BEE)
Sub code: GR14 Regulation L T P C
3 2 0 4
PreRequities : Fundamental in Engineering Mathematics and Physics
Course Objectives: BEE (Basic Electric Engineering) is common to first year branches of
UG Engineering(expect BT). At the end of the course the student is expected to
1. Know the fundamental of Electrical Engineering and practical.
2. Practical implementation of fundamental theory concepts.
Course Outcomes :
1. Students will learn strong basics of Electrical Engineering and practical
implementation of Electrical fundamentals.
2. Students will learn different applications of commonly used electrical
machinery
UNIT 1: Basic Laws: Ohm’s law, Kirchhoff’s voltage and current laws, Nodes-Branches
and loops, Series elements and Voltage Division, Parallel elements and Current Division,
Star-Delta transformation, Independent sources and Dependent sources, source
transformation.
UNIT 2: AC Fundamentals-I: Reviews of Complex Algebra, Sinusoids, phasors, Phasor
Relations of circuit elements, Impedance and admittance, Impedance Combinations, Series
and Parallel combination of Inductors and capacitor, Mesh analysis and Nodal analysis .
UNIT 3: AC Fundamental-II: RMS and average values, Form factors, Steady state Analysis
of series, Parallel and Series Parallel combination of R,L,C with Sinusoidal excitation,
Instantaneous power, Real power, Reactive power and Apparent power, concept of Power
factor, Frequency.
UNIT 4: Network Theorems and Resonance: Superposition theorem, Thevinin’s theorem,
Nortorn’s theorem, Maximum Power Transfer Theorem, Reciprocity theorem, Resonance in
Electrical circuits: Analysis of series and parallel Resonance.
UNIT 5: Fundamentals of Electrical Machines: Construction, Principle, Operation and
Application of –(i) Single phase Transformer (ii) Single phase Induction motor (iii) DC
Motor.
TEXTBOOK:
1. Fundamentals of Electrical Circuits by Charles k.Alexander, Mattew N.O.Saidiku, Tata
McGraw Hill company.
Reference Book:
1.Circuit theory(Analysis and Synthesis) by A. Chakrabarti-Dhanpat Rai&Co.
2.Network Theory by Prof.B.N.Yoganarasimham.
3. Circuit Theory by Sudhakar and ShyamMohan.
4.Electrical Machines-I by B.I.Theraja.
UNIT 1: Basic Laws: Ohm’s law, Kirchhoff’s voltage and current laws, Nodes-Branches
and loops, Series elements and Voltage Division, Parallel elements and Current Division,
Star-Delta transformation, Independent sources and Dependent sources, source
transformation.
OHM’S LAW
At constant temperature, the current flowing through a conductor is directly
proportional to the potential difference(p.d) in volts across the two ends of the given
conductor and inversely proportional to the resistance (R) in ohms (Ω) between the ends of
the same conductor.
µ

µ
/
In all practical problems of electrical calculations, it is assumed that the temperature rise is
within limits, so that electrical properties such as insulation and conduction properties of the
    .  ℎ’   ℎ  
= /  =    = /
=

=
A
Solved Examples:
1. An electrical iron carrying 2A at 120V. Find resistance of the device?
Soln : = (/) = 120/2 = 60
2. The essential component of a toaster is an electrical element (a resistor) that converts
electrical energy to heat energy. How much current is drawn by a toaster with
resistance 12 at 110V?
Soln: 9.167 A
3. In the given circuit, calculate current I, the conductance G and the power P?
Soln:
i =
= 6mA
G =
= 0.2m mhos
P = VI = I
R = 180mW
4. A voltage source of 20Sinπt V is connected across a 5KΩ resistor. Find the current
through the resistor and power dissipated.
Soln: =
= 4
(

)
 , =  = 80
() .
5. A resistor absorbs an instantaneous power of 20Cos
2
t mW when connected to a
voltage source V=10Cost V. Find I and R.
Soln: I = 2Cos(t) mA R = 5KΩ.
KIRCHOFF’S CURRENT AND VOLTAGE LAWS ( KCL and KVL)
KIRCHOFF’S CURRENT LAW : KCL states that the total current entering a junction is
equal to the total current leaving the junction.
(or)
The algebraic sum of the currents at the junction (node) will be zero.
At node n, (2 + 3 + 4) = (1 + 6 + 5)
Or 2 + 3 + 4 − 1 − 6 − 5 = 0.
KIRCHOFF’S VOLTAGE LAW: KVL is based on the law of the law of conservation of
the energy, states that the algebraic sum of voltage drops in a closed loop is zero.
1 + 2 + 3 =
1 + 2 + 3 – = 0
{Flow of currents in loop is assumed +ve from higher
potential to lower potential in elements and +ve from lower to higher potential in Sources}
Problem:
1. Write KCL and KVL equations for the given circuits:
2. Find the current measured in galvanometer shown
in the wheatstone’s bridge.
Soln: 0.015625 A (C to A)
3. Determine the voltage across 1Ω resistor in the given
circuit.
Soln: 2.353 V
4.
5. From the given circuit, find the voltage across ab, cb,
and db.
Soln:

= 37.5 

= 0 

= −62.5
NODES, BRANCHES AND LOOPS:
Network is an interconnection of elements or devices, circuit is a network providing one or
more closed paths.
A BRANCH represents a single element such as voltage source or a resistor.
A NODE is the point of connection of two or more branches.
A LOOP is any closed path in a circuit.
A network with ‘b’ branches, ‘l’ loops and ‘n’ nodes should satisfy the theorem of n/w
topology.
= +  – 1
Solved examples:
1. Determine the number of branches and nodes in the
circuit shown. Identify which elements are in series and
which are in parallel.
Soln: Branches 4; Nodes 3: 10V source and 5Ω resistance are connected in series. The
combination is in parallel with 6Ω resistance and 2A
source.
2. A bulb rated 110V , 60W has to be operated from a 230V supply. Show the
arrangement.
Soln: =


= 0.5454;
= 120/ = 220.02Ω
3. A 50Ω and a 100Ω rheostats are each rated at 100W. What is the maximum voltage
that may be applied without causing overheating of either rheostat (i) when they are
connected in series and (ii) when they are in parallel.
Soln:
SERIES ELEMENTS AND VOLTAGE DIVISION
Two or more elements are in series if they are cascaded or connected sequentially and
consequently carry the same current.
The equivalent resistances of any number of resistors connected in series is the sum of the
individual resistances.
Three resistances R1,R2,R3 are connected in series. The current passing through all the
elements connected in series is same, I.
In the above figure1, according to KVL, sum of voltage drops in a closed loop is zero.
 –
–
–
= 0 (
,
,
are drop across
,
,
)
 =
+ 
+ 
= (
+ 
+
)……………………(1)
In figure2,
=

………………………….(2)
From equations (1) and (2)

=
+ 
+
Therefore voltage applied divides across the series connected elements.
Applications of Series Circuits :
1. Decorative serial sets, number of bulbs with low voltage ratings connected in series
across existing rated voltage.
2. Voltage distribution or tapping.
Solved Examples:
1. Find

for the given circuit.
Soln : 14.4Ω
2. Find
and
in the given circuit. Calculate the power
dissipated in 3Ω.
Soln :
= 3.33, 
= 4,
= 5.33.
3. Find
,
,
,
and power dissipated in 12Ω and
40Ω.
Soln :
= 5, 
= 10 ,
= 416.7 ,
= 250 ,

= 2.083 ,

= 2.5.
PARALLEL ELEMENTS AND CURRENT DIVISION
Two or more elements are in parallel if they are connected to the same two nodes and
consequently have the same voltage across them.
The parallel connected elements have the same voltage across them.
=
=
……………………………(1)
From KCL, at node A, =
+
…………….(2)
From (1) , = /
+ /
/ = 1/
+ 1/
or

= (
)/(
+ 
)
The equivalent resistance of the parallel resistors is equal to the product of their resistances
divided by their sum.

=
+
G : conductance
The equivalent conductance connected in parallel is the sum of their individual conductances.
From (2) ,
=
–
= (

−
)
(

)/
= (
)/(
(
+ 
)
Therefore
= (
)/(
+ 
) and 2 = (
)/(
+ 
)
Solved Examples:
1. Two batteries of 24V and 20V with internal resistances of 0.4Ω and 0.25Ω
respectively are connected in parallel across a load of 4Ω. Calculate (i) the current
supplied by each battery and (ii) voltage across the load.
Soln: 8.1482A and -2.963A , 20.741V.
2. Two coils are connected in parallel and a voltage of 200V is applied to the terminals.
The total current taken is 25A and the power dissipated in one of the coils is 1500W.
What is the resistance of each coil?
Soln: 26.67Ω and 11.43Ω
3. Three impedances
= (5 + 5)Ω ,
= −8Ω and
= 4Ω are connected in
series to an unknown voltage source V. find I and V, if the voltage drop across
is
63.2∟18.45°.
Soln:

=
+ 
+ 
= 9.487∟ − 18.434° Ω
I = 15.8∟18.45°A V= 150∟0°V.
STAR – DELTA TRANSFORMATION
Star connection and delta connection are the two different methods of connecting three basic
elements which cannot be further simplified into series or parallel.
The two ways of representation can have equivalent circuits in either form.
Assume some voltage source across the terminals AB.

=
+

=
(
+
)/(
+ 
+ 
)
Therefore
+
=
(
+
)/(
+ 
+ 
)…………(1)
Similarly
+
=
(
+
)/(
+ 
+ 
)…………(2)
+
=
(
+
)/(
+ 
+ 
)…………..(3)
Subtracting (2) from (1) and adding to (3) ,
=
/(
+ 
+ 
)…………….(4)
=
/(
+ 
+ 
)…………….(5)
=
/(
+ 
+ 
)…………….(6)
A delta connection of
can be replaced by an equvivalent star connection with the
values from equations (4),(5),(6).
Multiply (4)(5) ; (5)(6) ; (4)(6) and then adding the three we get,
+
+
=
/(
+ 
+ 
)
Dividing LHS by
gives
, by
gives
, by
gives
.
= (
+
+
) /
= (
+
+
) /
= (
+
+
) /
Solved Examples :
1. Obtain the star connected equivalent for the given delta circuit.
Soln :
2. For the bridge network shown, determine the total resistance seen
from the terminals AB using star-delta
transformation.
Soln: 1.182Ω
3. Calculate the voltage across AB in the network and
indicate the polarity of the voltage using star-delta conversion.
Soln:
4. Determine the equivalent resistance Req.
Soln: 4.93Ω
INDEPENDENT SOURCES AND DEPENDENT SOURCES
An ideal voltage source is one which maintains a constant voltage at its terminals,
irrespective of the current delivered to the network.
Practically a voltage sources has internal resistances and hence, when it delivers a current,
there is always an internal voltage drop which increases as the supplies more and more
current.
Thus its terminals voltage progressively falls.
Ideal Voltage Source Practical Voltage Source
An IDEAL CURRENT SOURCE is one which delivers a current of constant magnitude,
totally independent of the external network connected.
In practical, no current source can be ideal, practical current source is always shown with
resistance in parallel.
Ideal Current Source Practical Current Source
Solved Examples:
1. In the given circuit, determine all branch currents.
Soln :
= 0.222 and
= 0.888.
2. For the given circuit, find
,
,
,
,
.
Soln :
= 0.0196A ;
= 0.01225A ;
= 0.00735A ;
= 0.0196A and
= 0.098V.
3. In the given circuit, find voltage across 4Ω
using nodal analysis.
Soln:
= 0.01V.
DEPENDENT OR CONTROLLED SOURCES:
In some network, in which some of the voltage sources or current sources are controlled by
changing of current or voltage elsewhere in the circuit. Such sources are termed as
“Dependent or Controlled sources”.
There are four types of dependent sources.
Dependent Voltage Source(DVS) Dependent Current Source(DCS)
CDVS( Current DVS)
CDCS(Current DCS)
VDVS(Voltage DVS)
VDCS(Voltage DCS)
SOURCE TRANSFORMATION:
A practical voltage source can be replaced by a current source and vice versa.
This can be established if an equivalence between a voltage source and a current source.
=
–
=
–
since
=
Or
= (
–
)/
………..(1)
For Current source ,
=
+
/
or
=
−
/
Comparing (1) and (2) ,
=
/
.
A voltage source
with resistance
in series with its equivalent to a current source
=
/
, in parallel with resistance
.
Example : Obtain the equivalent voltage source for the current source as shown in figure.
= 10 and
= 20Ω ,
= 200.
EXERCISE PROBLEMS
1. Two batteries of 24V and 20V with internal resistances of 0.4Ω and 0.25Ω respectively are
connected in parallel across a load of 4Ω. Calculate (i) the current supplied by each battery
and (ii) voltage across the load.
(Soln: 8.1482A , -2.963A , 20.741V)
2. A resistance R is connected in series with a parallel circuit comprising of resistances of 4Ω
and 6Ω respectively. When the applied voltage is 15V, the power dissipated in 4Ω resistor is
36W, calculate R.
(Soln : 0.6Ω)
3. In the circuit shown, find current through R
L
using Kirchhoff’s
laws.
(Soln : 30.488A)
4. Find using KCL, the current through the 6Ω resistor of the circuit
shown.
(Soln : 0.641A flowing from A to B)
5. Find current through 10Ω resistor in the given circuit, use
nodal analysis.
6. Determine the current in 10Ω resistor in the
network shown , use star-delta conversion.
7. For the 100Ω potentiometer shown, find the resistance R for a
load current of 10A.
8. The resistance of two wires is 15Ω when connected in series and 7Ω when connected in
parallel, find the resistances of each one.
9. Find the equivalent resistance across the battery terminals with
(a) Switch S open
(b) Switch S closed.
10. In the given circuit, find the value of V2 that will
cause the voltage across 20Ω to be zero by using
mesh analysis.
Unit II: AC Fundamentals-I:
Review of Complex Algebra, Sinusoids, Phasors, Phasor Relations of Circuit elements, Impedance
and Admittance, Impedance Combination, Series and Parallel combination of Inductors and
Capacitors, Mesh analysis and Nodal Analysis.
Review of Complex Number:
In order to analyze AC circuit, it is necessary to represent multi-dimensional quantities. In
order to accomplish this task, scalar numbers were abandoned and complete numbers were
used to express the two dimensions of frequency and phase shift at one time
In mathematics, ‘i’ is used to represent imaginary numbers. In the study of electricity and
electronics, j is used to represent imaginary numbers so that there is no confusion with I,
which in electronics represents current (i). It is also customary for scientist to write the
complex number in the form of + .
1. A complex number may be written in Rectangular form as:
= + . Where =  ( ) is real and = ( ) is imaginary number.
= 

(/).
2. A second way of representing the complex number is by specifying the magnitude(r)
and angle (θ) in polar form
=
∠
;
=
√
+ 
; = 

(/).
3. The third way of representing the complex number is the Exponential form.
=

, = tan

(
), =
√
+ 
Mathematical Operations of Complex Numbers:
Assume
=
+ 
 
∠
1
and
=
+ 
 
∠
2
are two complex numbers,
: 
+ 
= (
+ 
) + (
+ 
)
:
− 
= (
− 
) + (
− 
)
:
=
∠(
1
+
2
)
:
/
=(
/
)∠(
1
−
2
)
: 1/
= (1/
)∠−
1
Square root:
=
∠
2
 : 
∗
=  −  =
√
∠
=

Explain Generation of Sinusoidal wave?
Generation of sinusoidal AC voltage
Consider a rectangular coil of N turns placed in a uniform magnetic field as shown in the
figure. The coil is rotating in the anticlockwise direction at an uniform angular velocity of ω
rad/sec.
When the coil is in the vertical position, the flux linking the coil is zero because the plane of
the coil is parallel to the direction of the magnetic field. Hence at this position, the emf induced
in the coil is zero. When the coil moves by some angle in the anticlockwise direction, there is a
rate of change of flux linking the coil and hence an emf is induced in the coil according to
Faradays Law. When the coil reaches the horizontal position, the flux linking the coil is
maximum, and hence the emf induced is also maximum. When the coil further moves in the
anticlockwise direction, the emf induced in the coil reduces. Next when the coil comes to the
vertical position, the emf induced becomes zero. After that the same cycle repeats and the emf
is induced in the opposite direction. When the coil completes one complete revolution, one
cycle of AC voltage is generated.
The generation of sinusoidal AC voltage can also be explained using mathematical equations.
Consider a rectangular coil of N turns placed in a uniform magnetic field in the position shown
in the figure. The maximum flux linking the coil is in the downward direction as shown in the
figure. This flux can be divided into two components, one component acting along the plane of
the coil Φ
max
sinωt and another component acting perpendicular to the plane of the coil
Φ
max
cosωt.
The component of flux acting along the plane of the coil does not induce any flux in the coil.
Only the component acting perpendicular to the plane of the coil ie Φ
max
cosωt induces an
emf in the coil.
∅ = ∅

cos ()
= −
∅

= −
(∅

cos ())

= ∅

∗ sin ()
=

sin ()
Hence the emf induced in the coil is a sinusoidal emf. This will induce a sinusoidal current
in the circuit given by
=
 ()
Where
  ,
 
  
Angular Frequency (ω)
Angular frequency is defined as the number of radians covered in one second(ie the angle
covered by the rotating coil). The unit of angular frequency is rad/sec.
=
2
Problem-An alternating current is given by = .  ()
Find i) The maximum value
ii) Frequency
iii) Time Period
iv) The instantaneous value when = 3 = 141.4 (314).
Solution:
=
 ()_____________________(1)
Compare given equation with eq-1,
  
= 141.4
= 314 /
= /2 =50
= 1/ = 0.02.
= 3 ,
= 141.4 ( 314 ∗ 3 ∗ 10

) = 114.35
Explain about phasor, and Lead and lagging ?
Phasor Representation:
An alternating quantity can be represented using
i) Waveform
ii) Equations
iii) Phasor
A sinusoidal alternating quantity can be represented by a rotating line called a Phasor. A
phasor is a line of definite length rotating in anticlockwise direction at a constant angular
velocity
The waveform and equation representation of an alternating current is as shown. This
sinusoidal quantity can also be represented using phasors.
In phasor form the above wave is written as
̅
=
∠0
Draw a line OP of length equal to I
m
. This line OP rotates in the anticlockwise direction with
a uniform angular velocity ω rad/sec and follows the circular trajectory shown in figure. At
any instant, the projection of OP on the y-axis is given by OM=OPsinθ = I
m
sinωt. Hence the
line OP is the phasor representation of the sinusoidal current.
w rad/sec
Phase
Phase is defined as the fractional part of time period or cycle through which the
quantity has advanced from the selected zero position of reference
Phase of +E
m
is π/2rad or T/4 sec
Phase of -E
m
is π/2rad or 3T/4 sec
Phase Difference
When two alternating quantities of the same frequency have different zero points, they are
said to have a phase difference. The angle between the zero points is the angle of phase
difference.
In Phase
Two waveforms are said to be in phase, when the phase difference between them is zero.
That is the zero points of both the waveforms are same. The waveform, phasor and equation
representation of two sinusoidal quantities which are in phase is as shown. The figure shows
that the voltage and current are in phase.
=
sin
(

)
=
sin
(

)
Lagging
In the figure shown, the zero point of the current waveform is after the zero point of the voltage
waveform. Hence the current is lagging behind the voltage. The waveform, phasor and equation
representation is as shown.
=
sin
(

)
=>
=
∠0
=
sin
(
 − 
)
=>
̅
=
∠ − 
Leading
In the figure shown, the zero point of the current waveform is before the zero point of the voltage
waveform. Hence the current is leading the voltage. The waveform, phasor and equation
representation is as shown.
=
sin
(

)
=>
=
∠0
=
sin
(
 + 
)
=>
̅
=
∠
Problem: if
= 
(
 + 
)
  
= 
(
 − 
)
, find their sum and draw
phasor diagram.
Q: Explain Phasor Relationship with Circuit Elements:
AC circuit with a pure resistance
Consider an AC circuit with a pure resistance R as shown in the figure. The alternating voltage v
is given by
=
sin
(

)
The current flowing in the circuit is i. The voltage across the resistor is given as V
R
which is
the same as v.
Using ohms law, we can write the following relations
=
=
sin ()
=
sin ()
Where
=
From equation (1) and (2) we conclude that in a pure resistive circuit, the voltage and current are
in phase. Hence the voltage and current waveforms and phasors can be drawn as below.
AC circuit with a pure inductance
Consider an AC circuit with a pure inductance L as shown in the figure. The alternating voltage
v is given by
=
sin
(

)
The current flowing in the circuit is i. The voltage across the inductor is given as V
L
which is
the same as v.
We can find the current through the inductor as follows
=


sin
(

)
=


 =
∗ sin
(

)
=
∗ sin
(

)

=

∗ −cos ()
=

∗ ( −
2
)
=
∗ (−
2
)
Where
=

From equation (1) and (2) we observe that in a pure inductive circuit, the current lags behind the
voltage by 90⁰. Hence the voltage and current waveforms and phasors can be drawn as below.
Inductive reactance
The inductive reactance X
L
is given as
= 2
=
 ℎ  
= 
̅
It is equivalent to resistance in a resistive circuit. The unit is ohms ( )
Problem:
The voltage = 
+ 
is applied to a 0.1H inductor. Find the steady-state current
through the inductor.
Solution: from equation
V
= 
̅
= 12∠45

= 60

̅
=
12∠45
 ∗60 ∗ 0.1
=
12∠45
60 ∗ 0.1∠90
= 2
= 2cos
(
60− 45
)
AC circuit with a pure capacitance
Consider an AC circuit with a pure capacitance C as shown in the figure. The alternating voltage
v is given by
=
sin
(

)
=>
=
∠0
____________(1)
The current flowing in the circuit is i. The voltage across the capacitor is given as V
C
which is
the same as .
We can find the current through the capacitor as follows
= 
q = 
sin
(

)


= 
cos ()
= 
cos ()
= 
sin( +
2
)____________(2)
=
sin( +
2
)
= 
=>
=
=
1

From equation (1) and (2) we observe that in a pure capacitive circuit, the current leads the
voltage by 90⁰. Hence the voltage and current waveforms and phasors can be drawn as below.
Capacitive reactance
The capacitive reactance X
C
is given as
=

=
It is equivalent to resistance in a resistive circuit. The unit is ohms ( )
=
∠0
= + 0
̅
=
∠90
= 0 + 
=
∠
∠
=
∠ − 
=
∠ − 
Problem:
The voltage =  ( + 
) is applied to a   capacitor, calculate the current
through the capacitor, and Draw phasor diagram?
Impedance:
relationship between Current and Voltage to different circuit elements are,
1. To Resistor: =
̅
2. To Capacitor:
=
̅
3. To inductor:
=
̅
This shows that a pure resistance within an AC circuit produces a relationship between its
voltage and current phasors in exactly the same way as it would relate the same resistors voltage
and current relationship within a DC circuit.
Reciprocal of impedance is called as Admittance and units are mhos.
Impedance Combinations:
Series:
Let us assume
,
,…,
are N impedances are connected in series than equivalent impedance
is

obtained by
Apply KVL in loop
= (
+ 
+,…….
)
=
+ 
+ ⋯

=
+ 
+ ⋯
Suppose N=2:

=
+ 
=

=

Parallel:
Let us assume
,
,…,
are N impedances are connected in parallel ,than equivalent is given
by
Apply current (I) between two nodes and assume voltage across nodes=V,
According to the KCL =
+ 
…

=
+
+ ⋯+

=
+
+ ⋯+
Suppose N=2; than

=
+
=

=

Problem:
Find equivalent impedance of below circuit.
Solution: -j10 and +j5 are in ||
el
after simplifying
Therefore Z
in
= (60 series in with j10) = 60+ j10
Problem:
Find Z in below figure
Problem:
Find impedance Z
in
of below circuit.
Solution:
Problem:
Calculate the value of
in below circuit.
Series and Parallel combination of Inductors:
Series: Consider N inductor are connected in series, and voltage drop across each inductor is
,
,…..
.
According to the KVL
=
+ 
…..+
But =


=


+ 


…..+


= (
+ 
…..+
)


=



So

=
+ 
…..+
The equivalent inductance of series connected inductors is the sum of the individual
inductors.
Parallel: Consider N inductor are connected in parallel,
According to KCL
The equivalent impedance of parallel inductors is reciprocal of the sum of the reciprocals
of the individual inductances.
Problem:
Find L
eq
between ‘ab’ terminals in below figure.
Problem:
Find L
eq
In below figure.
Problem:
Find L
eq
between ‘ab’ terminals in below figure?
Problem:
Find L
eq
in below figure.
Series and parallel connection of capacitors:
Parallel Connection of Capacitors:
Consider N capacitors are connected in parallel,
Apply KCL
The equivalent capacitance of N parallel connected capacitors is the sum of all individual
capacitance.
Series connection of capacitors: Consider N capacitors are connected in series.
Apply KVL above circuit (left)
The equivalent capacitance of series connected capacitors is the reciprocal of the sum of the
reciprocals of the individual capacitors.
Problem:
Find

in below figure.
Problem:
Find C
eq
in below figure, If the value of all capacitor 4mf
Mesh Analysis:
Mesh analysis provides another general procedure for analyzing circuits, using mesh current
as the circuit variables.
Definition: Mesh is a loop which does not contain any other loop within it.
Steps to Determine Mesh Currents:
Step 1: Determine the number of meshes n.
Step 2: Assign mesh current i
1
, i
2
, , i
n
, to the n meshes.
o The direction of the mesh current is arbitrary-(clockwise or counterclockwise)-
and does not affect the validity of the solution.
o For convenience, we de ne currents flow in the clockwise (CW) direction.
Step 3: From the current direction in each mesh, denote the voltage drop polarities.
Step 4: Apply KVL to each of the n meshes. Use Ohm's law to express the voltages in
terms of the mesh current.
Step 5: Solve the resulting n simultaneous equations to get the mesh current.
Problem: Find I
1
, I
2
and I
3
in the below circuit using mesh analysis.
Solution:
Step1: In given circuit, two node are present.
Step2: Current in to meshes is assumed as i
1
and i
2.
Step3:
. KVL equation to mesh 1, is
-15+5 i
1
+10(i
1
-i
2
)+10=0---------------(1)
KVL equation to mesh 2, is
+6i
2
+4i
2
-10+10(i
2
-i
1
)=0---------------(2)
By solving (1) and (2) we will get
i
1
=2.0 A and i
2
=1.50A
But comparing given circuit with circuit in Step3,
I
1
=i
1
=2.0A
I
2
=i
2
=1.5A
I
3
=i
1
-i
2
=0.5A
Supermesh: A supermesh results when two meshes have a (dependent or independent)
current source in common.
Problem:
Find I using mesh analysis.
Problem:
Determine current in each loop.
Problem:
Determine current in each loop using mesh analysis
Node Analysis:
Steps to determine Node Analysis:
Step1: Determine the number of nodes n.
Step2: Select a node as reference node (ground node). Assign voltages V
1
, V
2,
….…V
n-1
to the
remaining n-1 nodes. The voltages are referenced with respect to the reference node.
Ground node is assumed to have 0(zero) potential.
Step3: Apply KCL to each of the n-1 non-reference nodes. Use Ohm’s law to express the
branch current in terms of node voltages.
Step4: Solve the resulting simultaneous equations to obtain the un-known node voltages.
a) Current flows from a higher potential to lower potential in resistor.
b) If a voltage source is connected between the reference node and a non-reference node,
we simply set the voltage at the non-reference node equal to the voltage of the source.
c) Multiple methods to solve the simultaneous equations in Step4.
1) Method 1: Elimination technique (good for few variables)
2) Method 2: Write in terms of matrix and vectors (write Ax=b) then use Cramer’s
rule.
3) Method 3: Use computer or calculators.
Problem:
Find the value of V
4
in below circuit.
Step1: Total number of nodes are 4 they are n0, n1,n2 and n3.
Step2: node n0 is selected as reference node. V
n0
=0
and voltage at other nodes are V
n1
, V
n2
and V
n3.
Step3: Assume voltage at node n2 is more than all other
node voltages.
So current is going away from this node in all branches.
=



=


=



=


=



=

()
    2 
+ 
+ 
= 0


+


+

()
= 0
By solving above equation

= 2 
 
=

()
=
= 1.25
  ℎ
 
=
∗ 4 = 1.25 ∗ 4=5 
Supernode: A supernode is formed by enclosing a (dependent or independent voltage source
connected between two non-reference nodes and any elements connected parallel with it.
Problem:
Calculate the value of I in below figure using node analysis.
Problem:
Determine voltage at node in below figure.
Problem:
Find v and i in below figure using node analysis.
Problem:
Find
 
using nodal analysis.
Two marks questions:
1. What are the different type of alternative signals and draw them.
2. Explain about phase difference.
3. What is leading, and explain with example.
4. What is lagging, and explain with example.
5. Which laws are used to solve using mesh analysis.
6. Which laws are used to solve using mesh analysis.
7. What are the different methods to solve un-know parameter in network.
8. Find the equivalent capacitance of two capacitor when
a. Connected in parallel
b. Connected in series.
9. Find the equivalent inductance of two inductor when
a. Connected in series.
b. Connected in parallel.
10. = 40 ∗ sin
(
+ 30
)
and = 20 ∗ (+ 30
), find phase difference between two
waves.
11. Find X
L
and X
_C
in terms of frequency.
12. How to calculate period of a given signal.
13. Draw phasor diagram for =40 ∗ sin
(
+30
)
and =20 ∗ ( + 30
).
14. Draw the phasor diagram for a given AC Voltage applied to
a. Pure Resistor.
b. Pure Inductor.
c. Pure Capacitor.
Unit III: AC Fundamentals-II
RMS and Average values, Form factor, Steady State Analysis of Series, Parallel and
Series Parallel combinations of R, L,C with Sinusoidal excitation, Instantaneous
power, Average power, Real power, Reactive power and Apparent power, concept of
Power factor, Frequency.
Q: Define Average value, RMS value and Form Factor and Calculate for sinusoidal
wave.
Average Value
The arithmetic average of all the values of an alternating quantity over one cycle is called
its average value.
 =
   


=

∫
()

For Symmetrical waveforms, the average value calculated over one cycle becomes equal
to zero because the positive area cancels the negative area. Hence for symmetrical
waveforms, the average value is calculated for half cycle.
 =
   


=
1
()
Average value of a sinusoidal current:
=
sin ()

=
∫
()

=
∫
sin ()()

=

=0.637
Average value of a full wave rectifier output
=
sin ()

=
∫
()

=
∫
sin ()()

=

=0.637
Average value of a half wave rectifier output
=
sin ()

=

∫
()


=

∫
sin ()()


=


=0.318
RMS or Effective Value
The effective or RMS value of an alternating quantity is that steady current (dc) which
when flowing through a given resistance for a given time produces the same amount of
heat produced by the alternating current flowing through the same resistance for the same
time.
=
   


=
1
2
()

RMS value of a sinusoidal current:
=
sin ()

=
1
2
()


=
1

()()

=
(1 − cos (2)
2
()

=
√
2
=0.707 
RMS value of a full wave rectifier output:
=
sin ()

=
1
()

=
1

()()

=
(1 − cos (2)
2
()

=
√
2
=0.707
RMS value of a half wave rectifier output
=
sin
(

)
0≤≤180
=0 180
≤≤360

=
1
2
()


=
1
2

() ()

=
2
(1 − cos (2)
2
()

=
2
=0.5
Form Factor:
It is the ratio of RMS value to the average value of an alternating quantity is known as
Form Factor
=
 
 
Peak Factor or Crest Factor:
It is the ratio of maximum value to the RMS value of an alternating quantity is known as
the peak factor.
=
 
 
For a sinusoidal waveform and For full wave rectifier output:

=

=0.637

=
√
=0.707
=


=
.
.
=1.11
=


=
.
=1.414
For a Half Wave Rectifier Output
:

=
2
=0.5

=
2
2
=0.318
=


=
.
.
=1.57
=


=
.
=2
Problem: Find Form Factor of figure show below.
Solution:
From figure:
()=4 ∗  0≤≤1
=4 − 4(− 1) 1≤≤2
Average value:
1/
(
)
=1/4 4 +  4 − 4
(
− 1
)

/
¼4 ∗
+ 4
]
− 4 ∗
+ 4
]
 ∗ (4 ∗ [1/2] + 4 − 4[4/2 − 1/2] + 4)=0.25

=
∫
()
 
=/2

=
1
2
()


=
1
2
 16 ∗ ()
+ 16 ∗ (1 − 1
(
− 1
)
)


=
16
2

+ (−
(
 − 2
)
)


=
8 ∗ 
+ (
− 4 + 4) 

=
8 ∗ 
+ 
(
 − 2
)


=
8 ∗ 
3
+ (4)
−
4
2
+
3

=
8 ∗ 
1
3
−
0
3
+ 4
(
2 − 1
)
− 4 ∗
2
2
− 4 ∗
1
2
 +
2
3
−
1
3

=
14 ∗
8
3
Problem:
Calculate Average value, Rms value and Form factor of the sawtooth wave show in the
figure.
Steady State Analysis of Series, Parallel and Series Parallel combinations of R, L,C with
Sinusoidal excitation:
R-L Series Circuit:
Consider an AC circuit with a resistance R and an inductance L connected in series as shown
in the figure. The alternating voltage v is given by
=
sin ()
The current flowing in the circuit is i. The voltage across the resistor is V
R
and that across the
inductor is V
L
V
R
=IR is in phase with I
V
L
=IX
L
leads current by 90 degrees
With the above information, the phasor diagram can be drawn as shown.
The current I is taken as the reference phasor. The voltage V
R
is in phase with I and the
voltage V
L
leads the current by 90⁰. The resultant voltage V can be drawn as shown in the
figure. From the phasor diagram we observe that the voltage leads the current by an angle Φ or
in other words the current lags behind the voltage by an angle Φ.
The waveform and equations for an RL series circuit can be drawn as below.
=
sin ()
=
sin (− ∅)
From the phasor diagram, the expressions for the resultant voltage V and the angle Φ can be
derived as follows.
=
+ 
=
=
=
()
+ (
)
=
+ 
=
ℎ =
+ 
The impedance in an AC circuit is similar to a resistance in a DC circuit. The unit for
impedance is ohms(
Ω)
Phase angle:
∅=

(
)
∅=

(


)
∅=

(
)
∅=

(

)
Power Factor:
The power factor in an AC circuit is defined as the cosine of the angle between voltage and
current ie., cosΦ
Problem:
A series RL circuit has a resistor 36
Ω and impedance of circuit is 10 Ω, then find power factor
Impedance Triangle:
We can derive a triangle called the impedance triangle from the phasor diagram of an RL
series circuit as shown
The impedance triangle is right angled triangle with R and X
L
as two sides and impedance as
the hypotenuse. The angle between the base and hypotenuse is Φ. The impedance triangle
enables us to calculate the following things.
1. Impedance =
+ 
2. Power Factor =/
3. Phase angle ∅=

(
)
4. Whether current is leading or lagging.
Problem:
A 200 V, 50 Hz, inductive circuit takes a current of 10A, lagging 30 degree. Find (i) the
resistance (ii) reactance (iii) inductance of the coil.
Solution:
=
̅
=


=20
Ω
)=
(
∅
)
=20 ∗ 
(
30
)
=17.32
Ω
ii)
=sin
(
∅
)
=20 ∗ sin
(
30
)
=10
Ω
)=
2
=
10
2 ∗ 3.14 ∗ 50
=0.0318
Problem:
A 230v, 50 Hz, is applied a series connected resistor 30 ohms and inductor 0.5mH, than
find X
L
, current through the circuit, voltage across each component, and also draw
phasor diagram between current and voltage.
Explain the behavior of AC through RC Series Circuit:
Consider an AC circuit with a resistance R and a capacitance C connected in series as shown
in the figure. The alternating voltage v is given by
=
sin ()
The current flowing in the circuit is i. The voltage across the resistor is V
R
and that across the
capacitor is
= is in phase with I
=
lags current by 90 degrees
With the above information, the phasor diagram can be drawn as shown.
The current I is taken as the reference phasor. The voltage V
R
is in phase with I and the
voltage V
C
lags behind the current by 90⁰. The resultant voltage V can be drawn as shown in
the figure. From the phasor diagram we observe that the voltage lags behind the current by an
angle Φ or in other words the current leads the voltage by an angle Φ.
The waveform and equations for an RC series circuit can be drawn as below.
=
sin ()
=
sin (+ ∅)
From the phasor diagram, the expressions for the resultant voltage V and the angle Φ can be
derived as follows.
=
+ 
=
=
=
()
+ (
)
=
+ 
=
ℎ =
+ 
The impedance in an AC circuit is similar to a resistance in a DC circuit. The unit for
impedance is ohms(
Ω).
Phase angle:
∅=

(
)
∅=

(


)
∅=

(
)
∅=


=

(

)
Impedance triangle:
Phasor algebra for RC series circuit.
=+ 0=∠0
̅
=−  
=∠ − ∅
̅
=
=
∠∅
Problem:
A Capacitor of capacitance 79.5µF is connected in series with a non inductive resistance
of 30 across a 100V, 50Hz supply. Find (i) impedance (ii) current (iii) phase angle
Solution:
=

=
∗.∗∗.∗

=40
Ω
) =
+ 
=
√
30
+ 40
=50
Ω
) =/=100/50=2
) ℎ =

=



=53
Problem:
A non-inductive resistor of 10
Ω
is in series with a capacitor of 100µF across a 250V,
50Hz ac supply. Determine the current taken by the capacitor and power factor of the
circuit
Problem:
A circuit consists of R and C reactance is 30
Ω
connected in series. Determine the value of
R for which power factor of the circuit is 0.8. Draw the phasor Diagram.
Behavior AC with R-L-C Series circuit:
Consider an AC circuit with a resistance R, an inductance L and a capacitance C connected in
series as shown in the figure. The alternating voltage v is given by
=
sin ()
The current flowing in the circuit is i. The voltage across the resistor is V
R,
the voltage
across the inductor is V
L
and that across the capacitor is V
C
.
V
R
=IR is in phase with I
V
L
=IX
L
leads the current by 90 degrees
V
C
=IX
C
lags behind the current by 90 degrees
With the above information, the phasor diagram can be drawn as shown. The current I is taken
as the reference phasor. The voltage V
R
is in phase with I, the voltage V
L
leads the current by
90⁰ and the voltage V
C
lags behind the current by 90⁰. There are two cases that can occur
V
L
>V
C
and V
L
<V
C
depending on the values of X
L
and X
C
. And hence there are two possible
phasor diagrams. The phasor V
L
-V
C
or V
C
-V
L
is drawn and then the resultant voltage V is
drawn.
From the phasor diagram we observe that when V
L
>V
C
, the voltage leads the current by an
angle Φ or in other words the current lags behind the voltage by an angle Φ. When V
L
<V
C ,
the
voltage lags behind the current by an angle Φ or in other words the current leads the voltage by
an angle Φ.
From the phasor diagram, the expressions for the resultant voltage V and the angle Φ can be
derived
as follows.
=
+ (
− 
)
=
()
+ (
− 
)
=
()
+ (
− 
)
=
ℎ  =
+ (
− 
)
ℎ  ∅=

(
− 
)= 


− 

=

− 
From the expression for phase angle, we can derive the following three cases
Case (i): When X
L
>X
C
The phase angle Ф is positive and the circuit is inductive. The circuit behaves like a series RL
circuit.
Case (ii): When
<
The phase angle Ф is negative and the circuit is capacitive. The circuit behaves like a
series RC circuit.
Case (iii): When
=
The phase angle Ф = 0 and the circuit is purely resistive. The circuit behaves like a pure
resistive circuit.
The voltage and the current can be represented by the following equations. The angle Φ is
positive or negative depending on the circuit elements.
=
sin ()
=
sin (± ∅)
Phasor algebra for RLC series circuit.
=+ 0=∠0
̅
=+ 
(
− 
)
=∠∅
̅
=
=
∠ − ∅
Problem:
A 230 V, 50 Hz ac supply is applied to a coil of 0.06 H inductance and 2.5 resistance
connected in series with a 6.8 µF capacitor. Calculate (i) Impedance (ii) Current (iii)
Phase angle between current and voltage (iv) power factor
Solution:
=2=1 ∗ 3.14 ∗ 50 ∗ 0.06=18.84
Ω
=

=
∗.∗∗.∗

=468
Ω
) =
+ (
− 
)
=
2.5
+ (18.84 − 468)
=449.2
Ω
=
=


=0.512
∅=


=

.

=−89.7
 =cos
(
∅
)
=cos
(
−89.7
)
=0.0056 
Problem:
A resistance R, an inductance L=0.01 H and a capacitance C are connected in series.
When an alternating voltage v=400sin( 3000t-20º)is applied to the series combination,
the current flowing is 10 2 sin(3000t-65º). Find the values of R and C.
Solution:
∅=65
− 20
=45
=2==3000 ∗ 0.01=30

(
∅
)
=
(
45
)
=1

(
∅
)
=

=1
− 
=
Z=
=


√
=
√
+ 
=20
Ω
=
− =30 − 20=10
Ω
=

=>=

=
∗
=33.3
Problem:
A coil of pf 0.6 is in series with a 100µF capacitor. When connected to a 50Hz supply, the
potential difference across the coil is equal to the potential difference across the
capacitor. Find the resistance and inductance of the coil.
Power:
In an AC circuit, the various powers can be classified as
1. Real or Active power or Average power.
2. Reactive power
3. Apparent power
Real or active power in an AC circuit is the power that does useful work in the circuit.
Reactive power flows in an AC circuit but does not do any useful work. Apparent power is the
total power in an AC circuit.
Instantaneous Power:
The instantaneous power is product of instantaneous values of current and voltages and it can be
derived as follows
=
=
sin ( + 
)*
sin (+ 
)
From trigonometric expression:
cos
(
 − 
)
− cos
(
 + 
)
=2sin
(
)
sin ()
=
(
cos (
− 
) − cos
(
2+ 
+ 
))
=
cos (
− 
) −
cos
(
2
)
The instantaneous power consists of two terms. The first term is called as the constant power
term and the second term is called as the fluctuating power term.
Average Power:
From instantaneous power we can find average power over one cycle as following.
=

∫
(
cos (
− 
) −
cos
(
2+ 
+ 
)
)()

=

cos (
− 
) ∗ (2− 0)
−

∫
−
cos
(
2
)
()

=
cos (
− 
)=
√
∗
√
=

∗ 

cos (
− 
)
As seen above the average power is the product of the RMS voltage and the RMS current.
Problem:
Calculate the instantaneous power and average power absorbed by the passive linear
network. If =
(
+ 
)
 and=
(
+ 
)
.
Problem:
Calculate the average power absorbed by an impedance Z=30-j70 ohms when applied a
voltage
=
∠
is applied across it.
Problem:
Problem:
Problem:
Problem:
Real Power:
The power due to the active component of current is called as the active power or real power. It
is denoted by P.
=∗ 
(
∅
)
=

(
∅
)
Real power is the power that does useful power. It is the power that is consumed by the
resistance. The unit for real power in Watt(W).
Reactive Power:
The power due to the reactive component of current is called as the reactive power. It is denoted
by Q.
=∗ 
(
∅
)
=
X
sin
(
∅
)
Reactive power does not do any useful work. It is the circulating power in the L and C
components. The unit for reactive power is Volt Amperes Reactive (VAR).
Apparent Power:
The apparent power is the total power in the circuit. It is denoted by S.
=  =
=
+ 
The unit for apparent power is Volt Amperes (VA).
Power Triangle:
From the impedance triangle, another triangle called the power triangle can be derived as shown.
The power triangle is right angled triangle with P and Q as two sides and S as the hypotenuse.
The angle between the base and hypotenuse is Φ. The power triangle enables us to calculate the
following things.
Apparent Power =
+ 
 =cos
(
∅
)
=
=
 
 
The power Factor in an AC circuit can be calculated by any one of the following
Methods
=    
=


=
=
 
 
Problem:
A coil having a resistance of 7 and an inductance of 31.8mH is connected to 230V, 50Hz
supply. Calculate (i) the circuit current (ii) phase angle (iii) power factor (iv) power
consumed v) Reactive power vi) Apparent power.
Solution:
=2=2 ∗ 3.14 ∗ 50 ∗ 31.8 ∗ 10

=10
Ω
=
+ 
=
√
7
+ 10
=12.2
Ω
) =
=

.
=18.85
) ∅=

=tan

(

)=−53

)=cos(∅)=cos
(
−53
)
=0.537 
) = 
(
∅
)
=230 ∗ 18.85 ∗ 0.537=2.484
) =sin
(
∅
)
=230 ∗ 18.85 ∗ 0.795=3.462 
) =
+ 
=
√
2.48
+ 3.46
=4.25
Problem:
A current of (120-j50)A flows through a circuit when the applied voltage is (8+j12)V.
Determine (i) impedance (ii) power factor (iii) power consumed and reactive power.
Solution:
=8 +  12
̅
=120 − 50
) 
̅
=
̅
=
 

=0.02 + 0.11=0.11∠79.7
=0.11
Ω
∅=79.7
) =cos
(
∅
)
=cos
(
79.7
)
=0.179 
)=
∗
=
(
8 + 12
)(
120 + 50
)
=360 + 1840
=+ 
=360
=1860
Problem:
A parallel circuit comprises of a resistor of 20
Ω
in series with an inductive reactance 15
Ω
in one branch and a resistor of 30
Ω
in series with a capacitive reactance of 20
Ω
in the other
branch. Determine the current and power dissipated in each branch if the total current
drawn by the parallel circuit is  ∠ − ⁰
Solution:
=20 + 15
=30 − 20
=10∠ − 30
According to KCL
=

=
(
8.66 − 5
)
∗
(

)
(

)
(

)
=3.8 − 6.08=7.17∠ − 60
=− 
(
8.66 − 5
)
−
(
3.8 − 6.08
)
=4.86 + 1.08=4.98∠ − 12.5
=
=7.17
∗ 20=1028.2
=
=4.98
∗ 30=744
Problem:
A circuit having a resistance of 20 and inductance of 0.07H is connected in parallel with a
series combination of 50 resistance and 60µF capacitance. Calculate the total current, when
the parallel combination is connected across 230V, 50Hz supply.
Problem:
An impedance coil in parallel with a 100µF capacitor is connected across a 200V, 50Hz
supply. The coil takes a current of 4A and the power loss in the coil is 600W. Calculate (i)
the resistance of the coil (ii) the inductance of the coil (iii) the power factor of the entire
circuit.
Solution:

=
=

=50
Ω
=
=600
=

=

=37.5
Ω
=

− 
=√50
− 37.5
=33.07
Ω
=

=
.
∗.∗
=0.105
=

=
∗.∗∗∗

=31.83
Ω
=+ 
=37.5 + 33.07
=−
=−31.83
=

=
(
..
)(
.
)
(
..
)
(
.
)
=27 − 32.72=42.42∠ − 50.5
∅=−50.5
=cos
(
∅
)
=cos
(
−50.5
)
=0.6365
Problem:
A series RLC circuit is connected across a 50Hz supply. R=100
Ω,
L=159.16mH and
C=63.7µF. If the voltage across C is 
∠
− 
V. Find the supply voltage
Solution:
=2=2 ∗ 3.14 ∗ 50 ∗ 159 ∗ 16 ∗ 10

=50
Ω
=

=
∗.∗∗.∗

=50
Ω
=
(
−
)
=150
∠
− 90
=−150
=−


=−


=3
∠
0
=+ 
(
− 
)
=100 + 
(
50 − 50
)
=100
Ω
==3 ∗ 100=300
Problem:
An alternative voltage of 160+j120 is applied to a circuit and the current in the circuit
given by 6+j8 A, Find,
i) Value of elements in circuit. ii) the power factor of circuit iii) power consumed.
Problem:
A series circuit consists of non-inductive resistor of 10
Ω
, an inductor having a reactance
of 50
Ω
, and a capacitor having a reactance of 30
Ω
. It is connected to a 230v ac supply.
Calculate 1) the current 2) the voltage across each component. 3) Draw to scale a phasor
diagram showing the supply voltage and current and voltage across each component.
Problem:
The voltage applied to a circuit is =
(
 + 
)
, and current flowing the
circuit is =
(
 + 
)
. Determine the impedance, resistance, reactance, power
and power factor of the circuit.
Problem:
A sinusoidal source supplies 1000KVR reactive power to load =
∠
− 
.
Determine a) the power factor b) the apparent power delivered to the load, and c) the
rms voltage.
Two marks Questions
1. Define RMS, Average and Form Factor.
2. Find the average value of sine wave.
3. Find RMS value of sinusoidal wave.
4. Define Real, Reactive and Apperant power.
5. Define power factor and find the power factor when Z=30∠45
0.
6. Write the formulae to calculate Real, Reactive and Apparent power.
7. Calculate the Reactive power when 30V DC is applied to Z=30+j60.
8. Write the relation between Real, Reactive and Apparent power.
9. Find the impedance of series connected circuit contains
a. R=30 ohms and C=0.003F
b. R=45 ohms and L=0.04H
When supplied voltage is 32V DC.
10. Draw impedance triangle and explain.
11. Draw power triangle and explain.
12. From the given Impedance triangle and current, How to find the Real, Reactive and Apparent
power.
Unit-IV
Resonance is a condition in an RLC circuit in which the capacitive and inductive reactance’s are
equal in magnitude, thereby resulting in a purely resistive impedance.
Q1)Explain about Series Resonance and derive an expression for its bandwidth.
The most prominent feature of the frequency response of a circuit may be the sharp peak (or
resonant peak) exhibited in its amplitude characteristic. The concept of resonance applies in
several areas of science and engineering. Resonance occurs in any system that has a complex
conjugatepair of poles; it is the cause of oscillations of stored energy from one formto another. It
is the phenomenon that allows frequency discrimination incommunications networks. Resonance
occurs in any circuit that has atleast one inductor and one capacitor.
Resonant circuits (series or parallel) are useful for constructing filters, as their transfer functions
can be highly frequency selective. They are usedin many applications such as selecting the
desired stations in radio andTV receivers.
Consider the series RLC circuit shown in the frequencydomain. The input impedance is
Z = H(ω)
=

=
R + jωL
+

….(1)
Z
=
R + j(ωL
-

)
..(2)
Resonance results when the imaginary part of the transfer function iszero, or
Im(Z) =
ωL -

= 0 ..(3)
The value of ω that satisfies this condition is called the resonant frequencyω
0
. Thus, the
resonance condition is
ω
0
L =

..(4)
..(5)
Or Since
ω
0
= 2πf
0
, f
0
=

√

Hz ..(6)
Note that at resonance: The impedance is purely resistive, thus, Z = R. In otherwords, the LC
series combination acts like a short circuit, andthe entire voltage is across R. The voltage Vs and
the current I are in phase, so that the power factor is unity. The magnitude of the transfer
function H(ω) = Z(ω) isminimum. The inductor voltage and capacitor voltage can be much
morethan the source voltage.The frequency response of the circuit’s current magnitude
ω
0
=
√

rad/s
I = | I | =

(

)
(7)
is shown in Fig. the plot only shows the symmetry illustrated inthis graph when the frequency
axis is a logarithm.
The average power
dissipated by the RLC circuit isP(ω) =
I
2
RThe highest power dissipated occurs at resonance,
when I = Vm/R
, so That
P(ω
0
)
=
..
(8)
At certain frequencies ω = ω1, ω2, the dissipated power is half themaximum value; that is,
P ( ω
1
) = P
(ω
2
) =
(
)
⁄

=

..(9)
Hence, ω1 and ω2 are called the half-power frequencies. The half-power frequencies are
obtained by settingZ equal to
√
2R and writing
+ (  −
 
)
Solving for ω, we obtain
ω
1
=
−

+
(

)
+

..(10)
ω
2
=

+
(

)
+

We can relate the half-power frequencies with the resonant frequency. From Eqs. (5) and (10),
ω
0
=
√
12
It is seen that the resonant frequency is the geometric mean of the half power frequencies. Notice
that ω1 and ω2 are in general not symmetricalaround the resonant frequency ω0, because the
frequency response is notgenerally symmetrical. However, as will be explained shortly,
symmetryof the half-power frequencies around the resonant frequency is often areasonable
approximation.Although the height of the curve in Fig.,it is determined by R, the width of the
curve depends on other factors. The width of theresponse curve depends on the bandwidth B,
which is defined as thedifference between the two half-power frequencies,
B = ω
2
– ω
1
(11)
This definition of bandwidth is just one of several that are commonly used.Strictly speaking, B in
Eq. (14.35) is a half-power bandwidth, because itis the width of the frequency band between the
half-power frequencies. The “sharpness” of the resonance in a resonant circuit is measured
quantitatively by the quality factor Q. At resonance, the reactive energyin the circuit oscillates
between the inductor and the capacitor. The qualityfactor relates the maximum or peak energy
stored to the energy dissipatedin the circuit per cycle of oscillation:
Q = 2π
     
         
..(12)
It is also regarded as a measure of the energy storage property of a circuitin relation to its energy
dissipation property. In the series RLC circuit,the peak energy stored is
LI
2
, while the energy
dissipated in one periodis
(I
2
R)(1/f). Hence,
Q = 2π
(
)
=

..(13)
Q =
=

..(14)
Notice that the quality factor is dimensionless. The relationship betweenthe bandwidth B and the
quality factor Q is obtained by substituting Eq.10 into Eq. (11) and utilizing Eq. (14).
B =
=
Or B =
CR Thus, The qualityfactor of a resonant circuit is the ratio of its resonant frequency to
its bandwidth. As illustrated in Fig, the higher the value of Q, the moreselective the circuit is but
the smaller the bandwidth. The selectivity of an RLC circuit is the ability of the circuit to respond
to a certain frequencyand discriminate against all other frequencies. If the band of frequencies
to be selected or rejected is narrow, the quality factor of the resonantcircuit must be high. If the
band of frequencies is wide, the quality factormust be low.A resonant circuit is designed to
operate at or near its resonantfrequency. It is said to be a high-Q circuit when its quality factor is
equal to or greater than 10. For high-Q circuits (Q ≥10), the half power frequencies are, for all
practical purposes, symmetrical around theresonant frequency and can be approximated as
ω
1
≅
ω
0
-
; ω
1
≅
ω
0
+
High-Q circuits are used often in communications networks.We see that a resonant circuit is
characterized by five related parameters:the two half-power frequencies ω1 and ω2, the resonant
frequencyω0, the bandwidth B, and the quality factor Q.
Problem:
In the circuit in Fig. 14.24, R = 2 &, L = 1 mH, and C = 0.4 μF.(a) Find the resonant frequency
and the half-power frequencies. (b) Calculatethe quality factor and bandwidth. (c) Determine the
amplitude of the current at ω0, ω1, and ω2.
Solution:
(a) The resonant frequency is
ω
0 =
√

=
√


∗.∗

=
50 krad/s
The lower half-power frequency is
ω
1
=−

+
(

)
+

-1 +
√
1 + 2500 krad/s = 49 krad/s
Similarly, the upper half-power frequency is
ω
2
=
1 +
√
1 + 2500 krad/s = 51 krad/s
(b) The bandwidth is
B = ω2 −ω1 = 2 krad/s or
B
=
=


=
2krad/s
The quality factor is
Q =
=

=
25
Q2.)Explain about Parallel Resonance and derive an expression for bandwidth.
The parallel RLC circuit in Fig. 14.25 is the dual of the series RLCcircuit. So we will avoid
needless repetition. The admittance is
Y = H(ω) =
=
+
jωC
+

..(16)
Y =
+ j
(
ωC
-

) ..(17)
Resonance occurs when the imaginary part of Y is zero
,i.e
ωC
-

= 0
ω
0 =
√

rad/s
which is the same as Eq. (5) for the series resonant circuit. Thevoltage |V| is sketched in Fig. as
a function of frequency. Noticethat at resonance, the parallel LC combination acts like an open
circuit, sothat the entire currents flows through R. Also, the inductor and capacitorWe can see
this from the fact thatcurrent can be much more than the source current at resonance.We exploit
the duality between Figsby comparingEq. (17) with Eq. (2). By replacing R, L, and C in the
expressionsfor the series circuit with 1/R, 1/C, and 1/L respectively, we obtain for
the parallel circuit
ω
1
=
-

+
(

)
+

;
ω
2
=

+
(

)
+

.. (18)
B = ω
2
- ω
1
=

..(19)
Q =

=
ω
0
RC =

..(20)
Using Eqs. (18) and (19), we can express the half-power frequenciesin terms of the quality
factor. The result is
ω
1
= ω
0
1 + (

)
−


;
ω
2
= ω
0
1 + (

)
+


..
(21)
Again, for high-Q circuits (Q ≥10)
ω
1
≅
ω
0
-
; ω
2
≅
ω
0
+
Table below presents a summary of the characteristics of the series andparallel resonant circuits.
Besides the series and parallelRLC consideredhere, other resonant circuits exist..
Summary of Characteristics of resonant RLC circuits
Characteristic
Series Circuit Parallel Circuit
Resonant frequency , ω
0
1
√

1
√

Quality Factor,Q
or

ω
0
RC
or
0
Bandwidth B
0
Q
0
Q
Half Power frequencies ω
1
, ω
2
ω
0
1 + (

)
±


ω
0
1 + (

)
±


For Q
≥
10, ω
1
, ω
2
ω
0
±
2
ω
0
±
2
Problem:
In the parallel RLC circuit in Fig. 14.27, let R = 8 k&, L = 0.2 mH, andC = 8 μF. (a) Calculate
ω0, Q, and B. (b) Find ω1 and ω2. (c) Determinethe power dissipated at ω0, ω1, and ω2.
ω
0
=
√

=
√
. ∗ 

∗  ∗

=

=
25 krad/s
Q =

=
∗ 
√
 ∗ 
∗ . ∗

=
1600
B=

=
15.625 rad/s
Due to the high value of Q, we can regard this as a high-Q circuit.Hence
,
ω
1
= ω
0
-
=
25,000 – 7.812 = 24,992 rad/s
ω
2
= ω
0
+
=
25,000 + 7.812 = 25,008 rad/s
(c) At
ω= ω
0
,
Y = 1/R or Z= R = 8KΩ Then
I
0
=
=
∟

=
1.25 ∟-90 mA
Since the entire current flows through R at resonance, the average powerdissipated at ω = ω0 is
P =
| Io|
R =
(1.25 x 10

)
(8 x 10
) = 6.25mW
P =


=

∗∗ 
=
6.25 mW
At ω= ω1, ω2,
P =


=
3.125 mW
Q3) Explain Thevenin’s Theorem with a suitable example.
Network Theorems are useful in determining the unknown values of current, Resistance, Voltage
etc in Electric Networks.
Thevenin’s theorem states that any two output terminals ( A & B ) of an active linear network
containing independent sources (it includes voltage and current sources) can be replaced by a
simple voltage source of magnitude V
th
in series with a single resistor R
th
where R
th
is the
equivalent resistance of the network when looking from the output terminals A & B with all
sources (voltage and current) removed and replaced by their internal resistances and the
magnitude of V
th
is equal to the open circuit voltage across the A & B terminals.
Simple Steps to Analyze Electric Circuit through Thevenin’s Theorem
1. Open the load resistor.
2. Calculate / measure the Open Circuit Voltage. This is the Thevenin Voltage (V
TH
).
3. Open Current Sources and Short Voltage Sources.
4. Calculate /measure the Open Circuit Resistance. This is the Thevenin Resistance (R
TH
).
5. Now, Redraw the circuit with measured open circuit Voltage (V
TH
) in Step (2) as voltage
Source and measured open circuit resistance (R
TH
) in step (4) as a series resistance and
connect the load resistor which we had removed in Step (1). This is the Equivalent Thevenin
Circuit of that Linear Electric Network or Complex circuit which had to be simplified and
analyzed by Thevenin’s Theorem. You have done.
6. Now find the Total current flowing through Load resistor by using the Ohm’s Law I
T
= V
TH
/
(R
TH
+ R
L
).
Example:
Find V
TH
, R
TH
and the load current flowing through and load voltage across the load resistor in
fig (1) by using Thevenin’s Theorem.
Fig 1
Step 1.
Open the 5kΩ load resistor
Fig 2
Step 2.
Calculate / measure the Open Circuit Voltage. This is the Thevenin Voltage (V
TH
). We have
already removed the load resistor from figure 1, so the circuit became an open circuit as shown
in fig 2. Now we have to calculate the Thevenin’s Voltage. Since 3mA Current flows in both
12kΩ and 4kΩ resistors as this is a series circuit because current will not flow in the 8kΩ resistor
as it is open. So 12V (3mA x 4kΩ) will appear across the 4kΩ resistor. We also know that
current is not flowing through the 8kΩ resistor as it is open circuit, but the 8kΩ resistor is in
parallel with 4k resistor. So the same voltage (i.e. 12V) will appear across the 8kΩ resistor as
4kΩ resistor. Therefore 12V will appear across the AB terminals. So,
V
TH
= 12V
Fig 3
Step 3.
Open Current Sources and Short Voltage Sources. Fig (4)
Fig 4
Step 4.
Calculate /measure the Open Circuit Resistance. This is the Thevenin Resistance (R
TH
)We have
Reduced the 48V DC source to zero is equivalent to replace it with a short in step (3),as shownin
figure (3) We can see that 8kΩ resistor is in series with a parallel connection of 4kΩ resistor
and 12k Ω resistor. i.e.:
8kΩ + (4k Ω || 12kΩ) ….. (|| = in parallel with)
R
TH
= 8kΩ + [(4kΩ x 12kΩ) / (4kΩ + 12kΩ)]
R
TH
= 8kΩ + 3kΩ
R
TH
= 11kΩ
Fig 5
Step 5.
Connect the R
TH
in series with Voltage Source V
TH
and re-connect the load resistor. This is
shown in fig (6) i.e. Thevenin circuit with load resistor. This the Thevenin’s equivalent circuit
Step 6.
Now apply the last step i.e Ohm’s law . calculate the total load current & load voltage as shown
in fig 6.
I
L
= V
TH
/ (R
TH
+ R
L
)= 12V / (11kΩ + 5kΩ) → = 12/16kΩ
I
L
= 0.75mA
And V
L
= I
L
x R
L
= 0.75mA x 5kΩ
V
L
= 3.75V
Thevenin’s theorem is applied to ac circuits in the same way as they are to dc
circuits.Thevenin's theorem, as applied to ac circuits, provides a method for reducing any circuit
to an equivalent form that consists of an equivalent ac voltage source in series with an equivalent
impedance. The only additional effort arises from the need to manipulate complex numbers. The
frequency-domain version of a Thevenin equivalent circuit is depicted in Fig, where a linear
circuit is replaced by a voltage source in series with an impedance.
Problem:
Obtain the Thevenin equivalent at terminals a-b of the circuit in
Solution:
We find Z
Th
by setting the voltage source to zero. As shown in Fig. (a) below, the 8-Ωresistance
is now in parallel with the −j6 reactance, sothat their combination gives
Z
1
= -j6 // 8 =
∗

=
2.88 – j3.84 Ω
Similarly, the 4-Ωresistance is in parallel with the j12 reactance, andtheir combination gives
Z
2
= 4 // j12 =
∗

=
3.6 + j1.2 Ω
The Thevenin impedance is the series combination of Z
1
and Z
2
; that is,
Z
Th
= Z
1
+ Z
2
= 6.48 – j2.64 Ω
To find V
Th
, consider the circuit in Fig. (b). Currents I
1
andI
2
are obtained as
I
1
=
∟

A, I
2
=
∟

A
Applying KVL around loop bcdeab in Fig.(b) gives
V
Th
– 4I
2
+ (-j6) I
1
=0
Or V
Th
= 4I
2
+ j6I
1
=
∟

+
∟()

=
37.95∟3.43
0
+ 72∟201.87
0
= -28.936 –j24.55 = 37.95 ∟220.31
0
V
Q4) Explain Norton’s Theorem with suitable example.
This is another useful theorem to analyze electric circuits like Thevenin’s Theorem, which
reduces linear, active circuits and complex networks into a simple equivalent circuit. The main
difference between Thevenin’s theorem and Norton’s theorem is that, Thevenin’s theorem
provides an equivalent voltage source and an equivalent series resistance, while Norton’s
theorem provides an equivalent Current source and an equivalent parallel resistance.
Norton’s Theorem may be stated as
Any Linear Electric Network or complex circuit with
Current and Voltage sources can be replaced by an equivalent circuit containing of a single
independent Current Source I
N
and a Parallel Resistance R
N
.
Norton’s Equivalent Circuit
Simple Steps to Analyze Electric Circuit through Norton’s Theorem
1. Short the load resistor
2. Calculate / measure the Short Circuit Current. This is the Norton Current (I
N
)
3. Open Current Sources, Short Voltage Sources and Open Load Resistor.
4. Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (R
N
)
5. Now, Redraw the circuit with measured short circuit Current (I
N
) in Step (2) as current Source
and measured open circuit resistance (R
N
) in step (4) as a parallel resistance and connect the
load resistor which we had removed in Step (3). This is the Equivalent Norton Circuit of that
Linear Electric Network or Complex circuit which had to be simplified and analyzed. You
have done.
6. Now find the Load current flowing through and Load Voltage across Load Resistor by using
the Current divider rule. I
L
= I
N
/ (R
N
/ (R
N
+ R
L
))
Example:
Find R
N
, I
N
, the current flowing through and Load Voltage across the load resistor in fig (1) by
using Norton’s Theorem.
Step 1.
Short the 1.5Ω load resistor as shown in (Fig 2).
Step 2.
Calculate / measure the Short Circuit Current. This is the Norton Current (I
N
).
We have shorted the AB terminals to determine the Norton current, I
N.
The 6Ω and 3Ω are then
in parallel and this parallel combination of 6Ω and 3Ω are then in series with 2Ω.
So the Total Resistance of the circuit to the Source is:-
2Ω + (6Ω || 3Ω) ….. (|| = in parallel with).
R
T
= 2Ω + [(3Ω x 6Ω) / (3Ω + 6Ω)] → I
T
= 2Ω + 2Ω = 4Ω.
R
T
= 4Ω
I
T
= V / R
T
I
T
= 12V / 4Ω= 3A..
Now we have to find I
SC
= I
N
… Apply CDR… (Current Divider Rule)…
I
SC
= I
N
= 3A x [(6Ω / (3Ω + 6Ω)] = 2A.
I
SC
= I
N
= 2A.
Step 3.
Open Current Sources, Short Voltage Sources and Open Load Resistor.
Step 4.
Calculate /measure the Open Circuit Resistance. This is the Norton Resistance (R
N
)
We have Reduced the 12V DC source to zero is equivalent to replace it with a short in step (3),
as shown in figure (4) We can see that 3Ω resistor is in series with a parallel combination of 6Ω
resistor and 2Ω resistor. i.e.:
3Ω + (6Ω || 2Ω) ….. (|| = in parallel with)
R
N
= 3Ω + [(6Ω x 2Ω) / (6Ω + 2Ω)]
R
N
= 3Ω + 1.5Ω
R
N
= 4.5Ω
Step 5.
Connect the R
N
in Parallel with Current Source I
N
and re-connect the load resistor. This is shown
in fig (6) i.e. Norton Equivalent circuit with load resistor.
Step 6.
Now apply the last step i.e. calculate the load current through and Load voltage across load
resistor by Ohm’s Law as shown in fig 7.
Load Current through Load Resistor…
I
L
= I
N
x [R
N
/ (R
N
+ R
L
)]= 2A x (4.5Ω /4.5Ω +1.5kΩ) → = 1.5A
I
L
= 1. 5AAndLoad Voltage across Load Resistor…
V
L
= I
L
x R
L
= 1.5A x 1.5Ω = 2.25V
Norton’s theorem is also applied to ac circuits in the same way as they are to dc circuits. The
only additional effort arises from the need to manipulate complex numbers. The Norton
equivalent circuit is illustrated in Fig.below, where a linear circuit s replaced by a current source
in parallel with an impedance. Keep in mind that the two equivalent circuits are related as
V
Th
=
Z
N
I
N
,
Z
Th
=
Z
N
just as in source transformation. V
Th
is the open-circuit voltage while I
N
is the short-circuit
current.
Obtain current Io in Fig. below using Norton’s theorem.
Solution:
Our first objective is to find the Norton equivalent at terminals a-b. Z
N
is found in the same way
as Z
Th
. We set the sources to zero as shownin Fig. (a). As evident from the figure, the (8 − j2)
and (10 + j4)impedances are short-circuited, so that
Z
N
= 5Ω
To get I
N,
we short-circuit terminals a-b as in Fig. (b) andapply mesh analysis. Notice that
meshes 2 and 3 form a supermesh because of the current source linking them. For mesh 1,
-j40 + (18 +j2) I
1
- (8 – j2) I
2
- (10 + j4) I
3
=0 …(1)
For the supermesh,
(13- j2)I
2
+ (10 + j4) I
3
– (18 + j2) I
1
= 0 …(2)
At node a, due to the current source between meshes 2 and 3,
I
3
= I
2
+ 3 …(3)
Adding Eqs. (1) and (2) gives
−j40 + 5I
2
= 0 ⇒I
2
= j8
From Eq. (3),
I
3
= I
2
+ 3 = 3 + j8
The Norton current is
I
N
= I
3
= (3 + j8) A
Figure (c) shows the Norton equivalent circuit along with the impedanceat terminals a-b. By
current division
I
0
=

I
N
=


=
1.465∟38.48
0
A
Q5) Explain Superposition Theorem with a suitable example.
The superposition principle states that the voltage across (or current through) an element in a
linear circuit is the algebraic sum of the voltages across (or currentsthrough) that element due to
each independent source acting alone.
The principle of superposition helps us to analyze a linear circuit withmore than one independent
source by calculating the contribution of eachindependent source separately. However, to apply
the superposition principle,we must keep two things in mind:
1. We consider one independent source at a time while all otherindependent sources are turned
off. This implies that wereplace every voltage source by 0 V (or a short circuit), andevery current
source by 0 A (or an open circuit). This way weobtain a simpler and more manageable
circuit.Other terms such as killed, made inactive, deadened,or set equal to zero are often used to
conveythe same idea.
2. Dependent sources are left intact because they are controlledby circuit variables.With these in
mind, we apply the superposition principle in three steps:
Steps to Apply Super position Principle:
1. Turn off all independent sources except one source. Find theoutput (voltage or current) due to
that active source using nodal ormesh analysis.
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all thecontributions due to the independent
sources.Use the superposition theorem to find v in the circuit
Solution:
Since there are two sources, let
v = v1 + v2
wherev1 and v2 are the contributions due to the 6-V voltage source and
the 3-A current source, respectively. To obtain v1, we set the currentsource to zero, as shown in
Fig. Applying KVL to the loop in Fig.gives
12i1 − 6 = 0 _⇒i1 = 0.5 A
Thus,v1 = 4i1 = 2 V
We may also use voltage division to get v1 by writing
v
1
=

(6) = 2V.To get v2, we set the voltage source to zero, as in FigUsingcurrent division,i
3
i
3
=
 
(3) =
2A
Hence,v2 = 4i3 = 8 V
And we find
v = v1 + v2 = 2 + 8 = 10 V
Since ac circuits are linear, the superposition theorem applies to ac circuitsthe same way it
applies to dc circuits.
Use the superposition theorem to find Io in the circuit
Let I
0
=
+

Where
and

are due to the voltage and current sources respectively. To find
, consider the
circuit given below. If we let Z be the parallel combination of –j2 and 8 + j10, then
Z =
( )

=
0.25 – j2.25
And current
is
=


=

..
Fig (a) Fig (b)
= -2.353 + j2.353...(2)
To get

, consider the circuit in Fig (b) . For mesh 1,
(8 + j8) I
1
– j10 I
3
+ j2I
2
=0…(3)
For mesh 2, (4-j4) I
2
+ j2 I
3
= 0 …(4)
For mesh3, I
3
=5 ...(5)
From eqs (4) & (5),
(4 –j4)I
2
+ j2 I
1
+ j10 =0
Expressing I
1
in terms of I
2
gives
I
1
= (2 + j2) I
2
-5 …(6)
Substituting Eqs. (5) and (6) into Eq. (3), we get
(8 + j8)[(2 + j2)I
2
− 5] − j50 + j2I
2
= 0
or
I
2
=


=
2.647 –j 1.176
Current

is obtained as

= -I
2
= -2.647 + j 1.176 …(7)
From eqs (2) & (7),
I
0
=
+

= -5 + j 3.529 =6.12∟144.78
0
A
Q6) Explain Maximum Power Transfer Theorem and derive an expression for Pmax.
In many practical situations, a circuit is designed to provide power to aload. While for electric
utilities, minimizing power losses in the processof transmission and distribution is critical for
efficiency and economicreasons, there are other applications in areas such as communications
where it is desirable to maximize the power delivered to a load. We nowaddress the problem of
delivering the maximum power to a load whengiven a system with known internal losses. It
should be noted that thiswill result in significant internal losses greater than or equal to the power
delivered to the load. The Thevenin equivalent is useful in finding the maximum power a
linear circuit can deliver to a load. We assume that we can adjust the loadresistance RL. If the
entire circuit is replaced by its Thevenin equivalent except for the load, as shown in Fig.the
power delivered to the loadis
P =
R
L
=
(


)
R
L
..(1)
For a given circuit, V
Th
and R
Th
are fixed. By varying the load
resistanceRL, the power delivered to the load varies as sketched in Fig. 4.49. Wenotice from Fig.
4.49 that the power is small for small or large values ofRL but maximum for some value of RL
between 0 and∞. We now wantto show that this maximum power occurs when RL is equal to
RTh. Thisis known as the maximum power theorem.
Maximum power transfer theoremstates that maximum power is transferred to the load when
the load resistance equals the Thevenin resistance as seen from the load (R
L
= R
Th
).
To prove the maximum power transfer theorem, we differentiatep in Eq. (4.21) with respect to
RL and set the result equal to zero. Weobtain


=
Vth
()
( )
()
=
Vth
(Rth+Rl−2Rl)
(Rth+Rl)
3
This implies that0= (R
Th
+ R
L
− 2RL) = (R
Th
− R
L
) ..(2)
which yields
R
L
= R
Th
..(3)
showing that the maximum power transfer takes place when the load resistance R
L
equals the
Thevenin resistance R
Th
. We can readily confirm that he maximum power is given by showing
that

<0.The source and load are said to be matched when R
L
= R
Th
. The maximum power
transferred is obtained by substituting Eq.(3) into Eq. (1), for
P
max
=

 
..(4)
Equation (4) applies only when R
L
= R
Th
. When R
L
≠R
Th
, wecompute the power delivered to the
load using Eq. (1).
PROBLEM:
Find the value of RL for maximum power transfer in the circuit of Fig. Find the
maximum power
.
Solution:
We need to find the Thevenin resistance R
Th
and the Thevenin voltage V
Th
across the terminals a-
b. To get R
Th
, we use the circuit in Fig. (a)and obtain
R
Th
= 2 + 3 + 6 // 12 = 5 +
∗

=
12Ω
To get V
Th
, we consider the circuit in Fig. (b). Applying mesh analysis,
−12 + 18i
1
− 12i
2
= 0, i
2
= −2 A
Solving for i1, we get i
1
= −2/3. Applying KVL around the outer loopto get V
Th
across terminals
a-b, we obtain
−12 + 6i
1
+ 3i
2
+ 2(0) + V
Th
= 0 _⇒V
Th
= 22 V
For maximum power transfer, R
L
= R
Th
= 9Ω
and the maximum power is
P
max
=

 
=

∗
=
13.44W
Maximum Power Transfer Theorem for AC Circuits:
When a load is connected to a circuit, maximum power is transferred to the load when the load
impedance is the complex conjugate of the circuit's output impedance. The complex conjugate of
R - jX
C
is R + jX
L
, where the resistances and the reactances are equal in magnitude. The output
impedance is effectively Thevenin's equivalent impedance viewed from the output terminals.
When Z
L
is the complex conjugate of Z
out
, maximum power is transfered from the circuit to the
load. An equivalent circuit with its output impedance and load is shown below
Q7)Prove Reciprocity Theorem.
The ratio of excitation to response remains invariant in a reciprocal network with respect
to an interchange between the points of application of excitation and measurement of the
response.
Under Basic Electrical Engineering In many electrical networks it is found that if the positions
of voltage source and ammeter are interchanged, the reading of ammeter remains the same.
Suppose a voltage source is connected to a passive network and an ammeter is connected to other
part of the network to indicate the response. Now any one interchanges the positions of ammeter
and voltage source that means he or she connects the voltage source at the part of the network
where the ammeter was connected and connects ammeter to that part of the network where the
voltage source was connected. The response of the ammeter means current through the ammeter
would be the same in both the cases. This is where the property of reciprocity comes in the
circuit. The particular circuit that has this reciprocal property, is called reciprocal circuit. This
type of circuit perfectly obeys reciprocity theorem.
Verify reciprocity theorem for the T-circuit.
Let us find response I2 of voltage source E in the position shown in Figure
Z
eff
= (Z
2
// Z
3
) + Z
1 =
(
∗

)
+ Z
1
=
    

I
1
=

I
2
= I
1
.


=
.

Case 2
When positions of source, i.e. E, and response, i.e. I are interchanged asshown in next Figure
Z
eff
= (Z
1
// Z
3
) + Z
2
=
    

=

=
.


=
.

From the above, it can be seen that I
2
=
Hence theorem is verified. The ratio

or
is called the transfer impedance and is given by
Z
T
=
12+23+31
3
=
Z
1
+ Z
2
+


Problems:
1) Find Vx by first finding Vth and Rth.
(Ans: Vx = 2V)
2) For the circuit given below find V
AB
by first finding Thevenins equivalent circuit.
(Ans: Vab=17V)
3) Find the voltage across 100Ω resistor by finding Thevenins equivalent circuit.
(Ans: V
100
= 22.9 V)
4) Find the current in the 50Ω resistor by drawing Nortons equivalent circuit.
(Ans:I
N
= 10.7 A)
5) Find the current through 10 resistor using superposition theorem
(Ans: I = 7/15 A)
6) Verify Reciprocity theorem for the network shown below, with source and response
positions being ab and cd.
7) Calculate current in impedance Z
3
using Superposition Theorem.
(Ans: I
3
= 9.65∟-75
0
A)
8) Find current in Z3 using Thevenin’s theorem in above network.
9) Find current in Z3 using Norton’s theorem in above network.
10) Find the value of RL that results in maximum power being transferred to RL.
Unit V : Fundamentals of Electrical Machines :
Construction, Principle, Operation and Applications of (i) Single Phase Transformer (ii) Single Phase
Induction motor (iii) DC Motor.
Single Phase Transformer :
To overcome losses, the electricity from a generator is passed through a step up transformer, which
increases the voltage. Throughout the distribution system, the voltages are changed using step-down
transformers to voltages suitable to the applications at industry and homes.
Construction:
Where:
V
P
- is the Primary Voltage
V
S
- is the Secondary Voltage
N
P
- is the Number of Primary Windings
N
S
- is the Number of Secondary Windings
Φ (phi) - is the Flux Linkage
Elements of Transformer:
Two coils having mutual inductance:
Laminated steel core.the two coils are insulated from each other and the steel core. Some suitable
container(Tank) for the assembled core and windings. Medium(Transformer oil) for insulating the core
and its windings from its container.
Suitable bushings( porcelain, oil filled or capacitor type), for insulating and bring out the terminals of
windings from the tank.
Transformer core is constructed of transformer sheet steel laminations assembled to provide a
continuous magnetic path with a minimum airgap included. The steel used is of high silicon content,
sometimes heat treated to produce a high permeability and low hysteresis loss at the usual operating
flux densities. The eddy current loss is minimized by laminating the core, the laminations being insulated
from each other by a light coat of core-plate varnish or by an oxide layer on the surface. The thickness of
laminations vary from 0.35mm for a frequency 50hz to 0.5mm for frequency 25Hz. The joints if the core
laminations in the alternate layers are staggered in order to avoid the presence of narrow gaps right
through the cross section of the core. Such staggered joints are said to be imbricated.
The transformers can be classified into two types depending on the manner in which the coils are
wound on the core.
i) Core type: Winding surrounded by a considerable part of the core.
ii) Shell type: Core surrounded by a considerable portion of windings.
iii) Spirakore type: In both core and shell type transformers, the individual laminations are cut
in the form of long strips of Es,Ls and Is
Another way of classifying transformer is depending on the method of cooling employed.
a) Oil filled self cooled:
The assembled winding and cores of such transformers are mounted in the welded, oil
tight steel tank provided with steel cover. After putting the core at its proper place, the
tank is filled with purified high quality insulating oil. The oil serves to convey the heat
from the core and the winding to the case from where it is radiated out to the
surroundings. For small size, the tanks are usually smooth surfaced, but for large sizes,
the cases are frequently corrugated or fluted to get greater heat radiation area without
increasing the cubical capacity of the tank still large size are provided with radiators or
pipes.
b) Oil filled water cooled: the winding and the core are immersed in the oil, but there is
mounted near the surface of the oil a cooling coil through which cold water is kept
circulating. The heat is carried away by this water. The largest transformers such as
those used with high voltage transmission lines are constructed in the manner.
c) Air-blast type: for voltage below 25KV, transformers can be built for cooling by means of
an air blast. The transformer is not immersed in oil, but is housed in a thin sheet-metal
box open at both ends through which air is blown from the bottom to the top by means
of a fan or blower.
Conventional Representation of Transformer in Electrical Circuits:
Principle of operation:
Transformer is a static piece of electric device.
If transformer electric power I one circuit to electrical power of same frequency to another
circuit.
The transformation is done by the process of mutual induction, where the two electric circuits
are in mutual inductive influence of each other without any physical contact.
The first coil in which electric energy is fed from AC supply is called primary winding and the
other from which energy is drawn out is called secondary coil.
It secondary wind is a closed circuit a current flows in it.
The working of the transformer can be explained as two methods:
1) Step up transformer:
If the primary coil has 3 loops and secondary coil has 30, the voltage
is stepped up 10 times.
2) Step down transformer:
If primary coil has 30 loops and secondary coil has 3 loops, the voltage is stepped down 10
times.
Emf Equation of A Transformer:
N
1
= no. of primary turns,
N
2
=No. of secondary turns.
∅
= B
m
*A : maximum flux in core in webers.
f=frequency of AC i/p In Hz.
Average rate of change of flux=
∅

= 4∅

or volts.
Rate of change of flux per turn ( Induced emf in volts)=
 

= 4∅
.
  =


= 1.11
  


= 1.11 ∗ 4∅
= 4.44
∅

      ℎ   = 4.44∅
= 4.44∅

4.44
       
= 4.44
∅

=
= 4.44∅
= 
=
= (   )
>
=> > 1 =>  
<
=> < 1 =>  
Transformer - Losses And Efficiency
Losses In Transformer
In any electrical machine, 'loss' can be defined as the difference between input power and output
power. An electrical transformer is an static device, hence mechanical losses (like windage or friction
losses) are absent in it. A transformer only consists of electrical losses (iron losses and copper losses).
Transformer losses are similar to losses in a DC machine, except that transformers do not have
mechanical losses.
Losses in transformer are explained below -
(I) Core Losses Or Iron Losses
Eddy current loss and hysteresis loss depend upon the magnetic properties of the material used for the
construction of core. Hence these losses are also known as core losses or iron losses.
Hysteresis loss in transformer: Hysteresis loss is due to reversal of magnetization in the
transformer core. This loss depends upon the volume and grade of the iron, frequency of magnetic
reversals and value of flux density. It can be given by, Steinmetz formula:
W
h
= ηB
max
1.6
fV (watts)
where, η = Steinmetz hysteresis constant
V = volume of the core in m
3
Eddy current loss in transformer: In transformer, AC current is supplied to the primary winding
which sets up alternating magnetizing flux. When this flux links with secondary winding, it produces
induced emf in it. But some part of this flux also gets linked with other conducting parts like steel
core or iron body or the transformer, which will result in induced emf in those parts, causing small
circulating current in them. This current is called as eddy current. Due to these eddy currents, some
energy will be dissipated in the form of heat.
(Ii) Copper Loss In Transformer
Copper loss is due to ohmic resistance of the transformer windings. Copper loss for the primary winding
is I
1
2
R
1
and for secondary winding is I
2
2
R
2
. Where, I
1
and I
2
are current in primary and secondary winding
respectively, R
1
and R
2
are the resistances of primary and secondary winding respectively. It is clear that
Cu loss is proportional to square of the current, and current depends on the load. Hence copper loss in
transformer varies with the load.
Efficiency Of Transformer
Just like any other electrical machine, efficiency of a transformer can be defined as the output power
divided by the input power. That is efficiency = output / input .
Transformers are the most highly efficient electrical devices. Most of the transformers have full load
efficiency between 95% to 98.5% . As a transformer being highly efficient, output and input are having
nearly same value, and hence it is impractical to measure the efficiency of transformer by using output /
input. A better method to find efficiency of a transformer is using, efficiency = (input - losses) / input = 1
- (losses / input).
Condition For Maximum Efficiency
Let,
Copper loss = I12R1
Iron loss = Wi
Hence, efficiency of a transformer will be maximum when copper loss and iron losses are equal.
That is Copper loss = Iron loss.
VOLTAGE REGULATION OF A SINGLE PHASE TRANSFORMER:
The voltage regulation is the percentage of voltage difference between no load and full load voltages of
a transformer with respect to its full load voltage.
Explanation:
Say an electrical power transformer is open circuited, means load is not connected with secondary
terminals. In this situation, the secondary terminal voltage of the transformer will be its secondary
induced emf E
2
. Whenever full load is connected to the secondary terminals of the transformer, rated
current I
2
flows through the secondary circuit and voltage drop comes into picture. At this situation,
primary winding will also draw equivalent full load current from source. The voltage drop in the
secondary is I
2
Z
2
where Z
2
is the secondary impedance of transformer. Now if at this loading condition,
any one measures the voltage between secondary terminals, he or she will get voltage V
2
across load
terminals which is obviously less than no load secondary voltage E
2
and this is because of I
2
Z
2
voltage
drop in the transformer.
Expression of Voltage Regulation of Transformer
Expression of Voltage Regulation of Transformer,
represented in percentage, is
Voltage Regulation of Transformer for Lagging Power Factor
Now we will derive the expression of voltage regulation in detail.
Say lagging power factor of the load is cosθ
2
, that means angle
between secondary current and voltage is θ
2
Here, from the above diagram,
Angle between OC & OD may be very small, so it can be neglected and OD is considered nearly equal to
OC i.e.
Voltage regulation of transformer at lagging power factor,
Voltage Regulation of Transformer for Leading Power Factor
Let's derive the expression of voltage regulation with leading
current, say leading power factor of the load is cosθ
2
, that means
angle between secondary current and voltage is θ
2
.
Here, from the above diagram,
Angle between OC & OD may be very small, so it can be neglected and OD is considered nearly equal to
OC i.e.
Voltage regulation of transformer at leading power factor,
Why is Transformer Rated in KVA, not in KW ?
Copper losses ( I²R) depends on Current which passing through transformer winding while Iron Losses or
Core Losses or Insulation Losses depends on Voltage. So the Cu Losses depend on the rated current of
the load so the load type will determine the power factor, that is why the rating of Transformer in kVA,
and not in kW.
Solved Example:
A 40KVA single phase transformer has 400 turns on primary and 100 turns on secondary the primary is
connected to 200V, 50Hz supply, Determine
1) The secondary voltage on open circuit.
2) The current flowing through the two windings on full load.
3) The maximum value of flux.
Solution: Tr, rating=40KVA, N
1
=400,N
2
=100
Primary induced voltage V
1
=200V,
1) Secondary voltage on open circuit: V
2
=
=
∗ 
= 2000 ∗
100
400
= 500
2)Primary Current (I
1
): at full load,
=
∗
= 40 ∗


=20A
     
=
∗
= 40 ∗


=80A
3)Maximum value of flux
   = 4.44
∅
∅
=
4.44 ∗  ∗ 
=
200
4.44 ∗ 50 ∗ 200
= 0.022 
Single Phase Induction Motor:
Single phase induction motor perform a great varity of useful services in the home , the office , factory,
business establishments on the farm and many other places where electricity is available.
There are different type of Single Phase Induction Motor:
1) Split Phsae Motor:
a. Resistance start motors.
b. Capacitor start motors.
c. Permanent –split(single value) capacitor motor.
d. Two value capacitor motor.
2) Shaded pole induction motor.
3) Reluctance start induction motor.
4) Repulsion start induction motor.
Construction and Working Principle:
An induction motor is simply an electric transformer whose primary winding is stationary and secondary
winding is free to rotate.
This motor has
1) Its stator is provided with a single phase winding and
2) A centrifugal switch is used in same types of motors in order to cutout a winding , used only for
starting purpose. It has distributed stator winding and a squirrel cage rotor. When fed from a
single phase supply , its stator winding , produces a flux(field) which is only alternating i.e., one
which alternates along one space axis only. Now an alternating or pulsating flux acting on a
stationary squirrel cage rotor cannot produce rotation(only a revolving flux can ). That is why a
single phase motor is not self starting.
However, if the motor of such machine is given an initial start by a small external force, in either
direction , then immediately a torque arises and the motor accelerates to its final speed. This
behaviors is explained in two ways:
1) Double-field revolving theory.
2) Cross-field theory:
This theory makes use of the idea that an alternating uni - axis quantity can be represented
by two oppositely rotating vectors of half magnitude. Accordingly, an alternating sinusoidal
flux can be represented by two revolving fluxes, each equal to half the value of the
alternating flux and each rotating synchronously ( = 120 ⁄ ) in opposite direction.
However if the rotor is started somehow say, in the clockwise direction, the clockwise
torque starts increasing and the same time, the anticlockwise torque starts decreasing.
Hence there is a certain amount of net torque in the clockwise direction which accelerated
the motor to full speed. Depending on different starting methods, the different type of
induction motor are mentioned above.
Equivalent Circuit of Single Phase Induction Motor:
The single phase induction motor has been imagined to be
made up of
1) One stator winding and
2) Two imaginary rotor. The stator impedance is
Z
1
=R
1
+j X
1
. the impedance of each rotor is
(r
2
+jx
2
) where r
2
and x
2
represent half the actual rotor values in stator terms.(x
2
stands
for half the standstill reactance of the rotor, as referred to stator) Iron loss has been
neglected, the exciting branch is shown consisting of exciting reactance only. Each rotor
has been assigned half the magnetizing reactance(x
m
: half the actual reactance).
Impedance of the forward running motor
=
(

)
(

)
and it runs with a slip ‘s’
Impedance of backward running motor
=
(


)

(

)
And it runs with a slip of (2-s).
Under standstill condition, V
f
is 90 to 95% of applied voltage.
Forward torque is synchronous watts
=
Backward Torque is
=

Total Torque =
− 
DC Motor:
An electrical motor is a machine which converts electrical energy in to mechanical energy. Its action is
based on the principle that when a current carrying conductor is placed in a magnetic field, it
experiences a mechanical force whose direction is given by FLEMINGS LEFT hand rule and whose
magnitude is given by F=BIL Newton.
The same DC machine can be operated both like a generator as well as motor. So the construction of the
machine will be same for both DC motor and DC generator.
Construction of Dc Machine:
The different parts of the machine and their working are
explained as follows.
Yoke: The outer frame of the machine is called yoke and it
serves two purposes.
1) Mechanical support for the poles and a
protecting cover for the whole machine.
2) It carries magnetic flux produced by pole.
This is made of cast iron/cast steel or rolled steel.
Pole Cores and Pole Shoes: These are built of the thin laminations of annealed steel which are riveted
together under hydraulic pressure. The thickness of laminations vary from 1mm to 0.25mm. The
laminated poles are secured to the yoke by means of screws bolted through the yoke into the pole body.
The pole shoe serve two purposes.
1) The spread out the flux in the air gap and also being of large cross section, reduce the reluctance
of the magnetic path.
2) They support exciting (or) field coils.
Pole Coils: The field or pole coils are former wound and placed on the pole core. The current passing
through these coils electromagnetics the poles which produces the necessary flux that is cut by the
revolving armature.
Armature Core: The houses the armature coils and causes them to rotate and hence cut the magnetic
flux.
It is a cylindrical or drum shaped and built up of circular sheet steel discs or laminations approximately
0.5mm thickness.
Usually the laminations are performed for air ducts which permits axials flow of air through the
armature for cooling purpose. A complete circular lamination is made up of four or six or even eight
segmental laminations. The two keyways are notched in each segment and are dovetailed or wedge
shaped to make the laminations self locking in position. The purpose of laminations is to reduce the
eddy current losses.
Armature winding: These are usually former wound. These are first wound in the form of flat
rectangular coils and are them pulled into their proper shape in a coil puller. Various conductors of the
coils are insulated from each other. The conductors are placed in the armature slots which are lined with
tough insulating material, the slot insulation is folded over above the armature conductors placed in the
slot and is secured in place by special hard wooden or fiber wedge.
Commutator: The function of commutator is to facilitate collection of current( or supply) from armature
conductors. It rectifies AC current in armature conductors into DC current in external load circuit if
generator or rectifies DC current from supply to ac current in armature if motor.
Commutator segments are insulated from each other by mica. Number of segments are equal to
number of armature coils.
Brushes and Bearings: Their function is to collect current from commutator. These are made of carbon
or graphite.
Working Principle:
When the field magnets are excited and its armature conductors are supplied with current from the
supply mains, they experience a force tending to rotate the armature. Armature conductors under N-
pole are assumed to carry current downwards and those under S-pole carry current upwards. By
applying FLEMINGS LEFT hand rule, the direction of the force on each conductor can be found. It will be
seen that each conductor experiences a force F which tends to rotate the armature in anti-clockwise
direction. These forces collectively produce a driving torque which sets the armature rotating.
TORQUE EQUATION OF DC MOTOR
Under Electrical Motor The term torque as best explained by Dr. Huge d Young is the quantitative
measure of the tendency of a force to cause a rotational motion, or to bring about a change in rotational
motion. It is in fact the moment of a force that produces or changes a rotational motion.
The equation of torque is given by,
Where F is force in linear direction.
R is radius of the object being rotated,
and θ is the angle, the force F is making with R vector
The dc motor as we all know is a rotational machine, and torque of dc motor is a very important
parameter in this concern, and it’s of utmost importance to understand
the torque equation of dc motor for establishing its running
characteristics.
To establish the torque equation, let us first consider the basic circuit diagram of a dc motor, and its
voltage equation.
Referring to the diagram beside, we can see, that if E is the supply voltage, E
b
is the back emf produced
and I
a
, R
a
are the armature current and armature resistance respectively then the voltage equation is
given by,
But keeping in mind that our purpose is to derive the torque equation of dc motor we multiply both
sides of equation (2) by I
a
.
Now I
a
2
.R
a
is the power loss due to heating of the armature coil, and the true effective mechanical
power that is required to produce the desired torque of dc machine is given by,
The mechanical power P
m
is related to the electromagnetic torque T
g
as,
Where ω is speed in rad/sec.
Now equating equation (4) & (5) we get,
Now for simplifying the torque equation of dc motor we substitute.
Where, P is no of poles,
φ is flux per pole,
Z is no. of conductors,
A is no. of parallel paths,
and N is the speed of the D.C. motor.
  =
2
60
− − − − − − − − − − − (7)
Substituting equation (6) and (7) in equation (4), we get:
=

2
The torque we so obtain, is known as the electromagnetic torque of dc motor, and subtracting the
mechanical and rotational losses from it we get the mechanical torque.
Therefore,
=
mechanical losses.
This is the torque equation of dc motor. It can be further simplified as:
=
∅
ℎ 
=

2
Which is constant for a particular machine and therefore the torque of dc motor varies with only flux φ
and armature current I
a
.
The Torque equation of a dc motor can also be explained considering the
figure below.
Here we can see Area per pole
=

=
=

2
Current / conductor
=
/ 
Therefore, force per conductor =
= 
/
Now torque
=
. = 
. /
=

2
Hence the total torque developed of a dc machine is,
=

2
This torque equation of dc motor can be further simplified as:
=
∅
where
=


Which is constant for a particular machine and therefore the torque of dc motor varies with only flux φ
and armature current I
a
.