BASIC ENGINEERING – II
Part A ELECTRICAL ENGINEERING
Prepared by
Dr.R.Jegatheesan
Professor, EEE Dept.
AC & DC circuits
Circuit parameters, Ohms law, Kirchhoff’s laws. Average and RMS values,
concept of phasor representation. RLC serious circuits and series resonance.
RLC parallel circuits (includes simple problems in DC & AC circuits). Introduction
to three phase system – types of connection, relationship between line and phase
values. (qualitative treatment only).
Electrical machines
Definition of mmf, flux and reluctance, leakage flux, fringing, magnetic materials
and B-H relationship. Problems involving simple magnetic circuits. Faraday’s
laws, induced emfs and inductances, brief idea on Hysteresis and eddy currents.
Working principle, construction and applications of DC machines and AC
machines (1-phase transformers, 3-phase induction motors, single phase
induction motors – split phase, capacitor start and capacitor start & run motors).
Wiring & lighting
Types of wiring, wiring accessories, staircase & corridor wiring, Working and
characteristics of incandescent, fluorescent, SV & MV lamps. Basic principles of
earthing, simple layout of generation , transmission & distribution of power.
TEXT BOOKS
1. Muthusubramanian R, Salaivahanan S, Muraleedharan K A, Basic Electrical, Electronics
and Computer Engineering, Tata McGraw – Hill, 1999
2. Mehta V K, Principles of Electronics, S Chand & Co, 1980
ELECTRIC CIRCUITS
Electric circuits are broadly classified as Direct Current (D.C.) circuits and
Alternating Current (A.C.) circuits. The following are the various elements that
form electric circuits.
D.C. Circuits
Elements
Voltage source
A.C. Circuits
Representation
-
+
Elements
Voltage source
Current source
Current source
Resistor
Resistor
Inductor
Capacitor
Representation
-
~ +
DC CIRCUITS
The voltage across an element is denoted as E or V. The current through the
element is I.
Conductor is used to carry current. When a voltage is applied across a
conductor, current flows through the conductor. If the applied voltage is
increased, the current also increases. The voltage current relationship is shown
in Fig. 1.
I
V
Fig. 1 Voltage – current relationship
It is seen that I  V. Thus we can write
I=GV
where G is called the conductance of the conductor.
(1)
I=GV
(1)
Very often we are more interested on RESISTANCE, R of the conductor, than the
conductance of the conductor. Resistance is the opposing property of the
conductor and it is the reciprocal of the conductance. Thus
R=
1
1
or G =
G
R
Therefore
I=
(2)
V
R
(3)
The above relationship is known as OHM’s law. Thus Ohm law can be stated as
the current flows through a conductor is the ratio of the voltage across the
conductor and its resistance. Ohm’s law can also be written as
V=RI
R=
V
I
(4)
(5)
The resistance of a conductor is directly proportional to its length, inversely
proportional to its area of cross section. It also depends on the material of the
conductor. Thus
R=ρ

A
(6)
where ρ is called the specific resistance of the material by which the conductor is
made of. The unit of the resistance is Ohm and is represented as Ω. Resistance of
a conductor depends on the temperature also. The power consumed by the
resistor is given by
P=VI
(7)
When the voltage is in volt and the current is in ampere, power will be in watt.
Alternate expression for power consumed by the resistors are given below.
P = R I x I = I2 R
(8)
V V2
P=Vx
=
R
R
(9)
KIRCHHOFF’s LAWS
There are two Kirchhoff’s laws. The first one is called Kirchhoff’s current law,
KCL and the second one is Kirchhoff’s voltage law, KVL.
Kirchhoff’s current law deals with element currents meeting at a junction, which
is a meeting point of two are more elements.
Kirchhoff’s voltage law deals with element voltages in a closed loop also called
as closed circuit.
Kirchhoff’s current law
Kirchhoff’s current law states that the algebraic sum of element currents meeting
at a junction is zero.
Consider a junction P wherein four elements, carrying currents I1, I2, I3 and I4, are
meeting as shown in Fig. 2.
I3
I2
I1
Fig. 2 Currents meeting at a junction
P
I4
Note that currents I1 and I4 are flowing out from the junction while the currents I2
and I3 are flowing into the junction. According to KCL,
I1 – I2 - I3 + I4 = 0
(10)
The above equation can be rearranged as
I1 + I4 = I2 + I3
(11)
From equation (11), KCL can also stated as at a junction, the sum of element
currents that flows out is equal to the sum of element currents that flows in.
Kirchhoff’s voltage law
Kirchhoff’s voltage law states that the algebraic sum of element voltages around
a closed loop is zero.
Consider a closed loop in a circuit wherein four elements with voltages V 1, V2, V3
and V4, are present as shown in Fig. 3.
V1
+
-
-
V2
V4
Fig. 3 Voltages in a closed loop
+
+
+
V3
-
Assigning positive sign for voltage drop and negative sign for voltage rise, when
the loop is traced in clockwise direction, according to KVL
V1 - V2 - V3 + V4 = 0
(12)
The above equation can be rearranged as
V1 + V4 = V2 + V3
(13)
From equation (13), KVL can also stated as, in a closed loop, the sum of voltage
drops is equal to the sum of voltage rises in that loop.
Find the currents I 1 , I 2 , I 3 and I 4 in the circuit shown.
I4
10 A
12 A
2A
I2
I3
I1
5A
Find the voltages V 1 , V 2 and V 3 in the circuit shown.
+ 8V +
V1
-
V2 +
5V
+
+ 2V - +
V3
-
-
Resistors connected in series
Two resistors are said to be connected in series when there is only one common
point between them and no other element is connected in that common point.
Resistors connected in series carry same current. Consider three resisters R 1, R2
and R3 connected in series as shown in Fig. 4. With the supply voltage of E,
voltages across the three resistors are V1, V2 and V3.
R1
+
+
V1
-
I
R2
+
V2
R3
-
+
V3
E
-
-
Fig. 4 Resistors connected in series
As per Ohm’s law
V1 = R1 I
V2 = R2 I
V3 = R3 I
(14)
I
R1
+
V1
-
+
R2
V2
-
+
V3
E
+
Re q
R3
I
-
-
+
E
-
Fig. 4 Resistors connected in series
Applying KVL,
E = V1 + V 2 + V3
(15)
= (R1 + R2 + R3) I = Re q I
(16)
Thus for the circuit shown in Fig. 4,
E = Re q I
where E is the circuit voltage, I is the circuit current and Re
(17)
q
is the equivalent
resistance. Here
Re q = R1 + R2 + R3
(18)
This is true when two are more resistors are connected in series. When n numbers of
resistors are connected in series, the equivalent resistor is given by
Re q = R1 + R2 + …………. + Rn
(19)
Voltage division rule
Consider two resistors connected in series. Then
R1
+
+
V1
R2
I
-
+
E
V2
-
V1 = R1 I
V2 = R2 I
E = (R1 + R2) I and hence I = E / (R1 + R2)
Total voltage of E is dropped in two resistors. Voltage across the resistors are
given by
V1 =
R1
E
R1 R2
V2 =
R2
E
R1 R2
and
(20)
(21)
Resistors connected in parallel
Two resistors are said to be connected in parallel when both are connected
across same pair of nodes. Voltages across resistors connected in parallel will be
equal.
Consider two resistors R1 and R2 connected in parallel as shown in Fig. 5.
I1
R1
A
I2
R2
I
+
E
-
Fig. 5 Resistors connected in parallel
As per Ohm’s law,
I1 =
E
R1
(22)
E
I2 =
R2
R1
I1
Re q
A
I2
E
R1
I2 =
E
R2
R2
I
+
I1 =
E
I
E
Equivalent circuit
Applying KCL at node A
I = I1 + I 2 = E (
1
1

)
R1 R2
(23)
From the equivalent circuit shown
I=
E
Req
where E is the circuit voltage, I is the circuit current and Re
(24)
q
is the equivalent
resistance. Comparing eq. (23) and (24)
1
1
1


R eq R1 R2
(25)
1
1
1


R eq R1 R2
From the above
Thus R e q 
(25)
R  R2
1
 1
Req
R1 R 2
R1 R 2
R1  R 2
(26)
When n numbers of resistors are connected in parallel, generalizing eq. (25),
Re q can be obtained from
1
1
1
1


 .......... ...... 
R eq R1 R2
Rn
(27)
Current division rule
I1
R1
A
I2
I1 =
R2
I
I2 =
+
E
E
R1
E
R2
(22)
-
Fig. 5 Resistors connected in parallel
Referring to Fig. 5, it is noticed the total current gets divided as I 1 and I2. The
branch currents are obtained as follows.
From eq. (23)
E =
R1 R2
I
R1  R2
(29)
Substituting the above in eq. (22)
I1 =
R2
I
R1  R2
(30)
I2 =
R1
I
R1  R2
Example 1
Three resistors 10Ω, 20Ω and 30Ω are connected in series across 100 V supply.
Find the voltage across each resistor.
Solution
10Ω
+
+
V1
-
I
20Ω
+
V2
100V
Current I = 100 / (10 + 20 + 30) = 1.6667 A
Voltage across 10Ω = 10 x 1.6667 = 16.67 V
Voltage across 20Ω = 20 x 1.6667 = 33.33 V
Voltage across 30Ω = 30 x 1.6667 = 50 V
30Ω
-
+
V3
-
-
Example 2
Two resistors of 4Ω and 6Ω are connected in parallel. If the supply current is 30 A,
find the current in each resistor.
Solution
I1
I2
4Ω
6Ω
30 A
Using the current division rule
Current through 4Ω =
6
x 30  18 A
46
Current through 6Ω =
4
x 30  12 A
46
Example 3
Four resistors of 2 ohms, 3 ohms, 4 ohms and 5 ohms respectively are connected
in parallel. What voltage must be applied to the group in order that the total power
of 100 W is absorbed?
Solution
Let RT be the total equivalent resistor. Then
1
1 1 1 1 60  40  30  24 154
    

RT 2 3 4 5
120
120
Resistance RT =
120
 0.7792 Ω
154
Let E be the supply voltage. Then total current taken = E / 0.7792 A
Thus (
E
) 2 x 0.7792  100 and hence E2 = 100 x 0.7792 = 77.92
0.7792
Required voltage =
77.92  8.8272 V
Example 4
When a resistor is placed across a 230 V supply, the current is 12 A. What is the
value of the resistor that must be placed in parallel, to increase the load to 16 A?
Solution
R1
R1
12 A
12 A
+
230 V
-
16 A
4A
+
R2
230 V
-
To make the load current 16 A, current through the second resistor = 16 –12 = 4 A
Value of second resistor R2 = 230/4 = 57.5 Ω
Example 5
A 50 Ω resistor is in parallel with a 100 Ω resistor. The current in 50 Ω resistor is
7.2 A. What is the value of third resistor to be added in parallel to make the line
current as 12.1A?
Solution
7.2 A
50 Ω
7.2 A
100 Ω
+
E
50 Ω
3.6 A 100 Ω
1.3 A
Supply voltage E = 50 x 7.2 = 360 V
+
R
360 V
-
Current through 100 Ω = 360/100 = 3.6 A
When the line current is 12.1 A, current through third resistor = 12.1 – (7.2 + 3.6)
= 1.3 A
Value of third resistor = 360/1.3 = 276.9230 Ω
Example 6
A resistor of 3.6 ohms is connected in series with another of 4.56 ohms. What
resistance must be placed across 3.6 ohms, so that the total resistance of the
circuit shall be 6 ohms?
Solution
3.6 Ω
4.56 Ω
R1
RT = 6 Ω
3.6 ║ R1 = 6 – 4.56 = 1.44 Ω
Thus
3.6 x R1
 1.44;
3.6  R1
Therefore
3.6  R1
3.6

 2.5;
R1
1.44
Required resistance R1 = 3.6/1.5 = 2.4 Ω
3.6
 1.5
R1
Example 7
A resistance R is connected in series with a parallel circuit comprising two
resistors 12 Ω and 8 Ω respectively. Total power dissipated in the circuit is 70 W
when the applied voltage is 22 V. Calculate the value of the resistor R.
Solution
12 Ω
R
8Ω
+
22 V
Power dissipated = 70 W
-
Total current taken = 70 / 22 = 3.1818 A
Equivalent of 12 Ω ║ 8 Ω = 96/20 = 4.8 Ω
Voltage across parallel combination = 4.8 x 3.1818 = 15.2726 V
Voltage across resistor R = 22 – 15.2726 = 6.7274 V
Value of resistor R = 6.7274/3.1818 = 2.1143 Ω
Example 8
The resistors 12 Ω and 6 Ω are connected in parallel and this combination is
connected in series with a 6.25 Ω resistance and a battery which has an internal
resistance of 0.25 Ω. Determine the emf of the battery if the potential difference
across 6 Ω resistance is 6 V.
12 Ω
6.25 Ω
6Ω
Voltage across 6 Ω = 6 V
0.25 Ω
E
Solution
Current in 6 Ω = 6/6 = 1 A
Current in 12 Ω = 6/12 = 0.5 A
Therefore current in 0.25 Ω = 1.0 + 0.5 = 1.5 A
Using KVL E = (0.25 x 1.5) + (6.25 x 1.5) + 6 = 15.75 V
Therefore battery emf E = 15.75 V
Example 9
A circuit consist of three resistors 3 Ω, 4 Ω and 6 Ω in parallel and a fourth
resistor of 4 Ω in series. A battery of 12 V and an internal resistance of 0.6 Ω is
connected across the circuit. Find the total current in the circuit and the terminal
voltage across the battery.
3Ω
Solution
4Ω
4Ω
6Ω
0.6 Ω
12 V
4 Ω ║ 6 Ω = 24/10 = 2.4 Ω
1.4 Ω ║ 3 Ω = 7.2/5.4 = 1.3333 Ω
Total circuit resistance = 4 + 0.6 + 1.3333 = 5.9333 Ω
Circuit current = 12/5.9333 = 2.0225 A
Terminal voltage across the battery = 12 – (0.6 x 2.0225) = 10.7865 V
Example 10
An electrical network is arranged as shown. Find (i) the current in branch AF (ii)
the power absorbed in branch BE and (iii) potential difference across the branch
CD.
A
13 Ω
B
11 Ω
C
24 V
22 Ω
18 Ω
1Ω
F
E
5Ω
14 Ω
D
9Ω
Solution
Various stages of reduction are shown.
A
13 Ω
B
11 Ω
C
24 V
22 Ω
18 Ω
1Ω
F
A
14 Ω
14 Ω
D
E
13 Ω
B
1
11 Ω
C
24 V
22 Ω
1Ω
F
2
18 Ω
E
7Ω
13 Ω
11 Ω
24 V
22 Ω
18 Ω
1Ω
F
A
2
7Ω
E
13 Ω
B
18 Ω
C
24 V
22 Ω
1Ω
F
18 Ω
E
3
13 Ω
A
18 Ω
B
C
24 V
22 Ω
F
E
13 Ω
A
3
18 Ω
1Ω
B
A
22 Ω
B
24 V
24 V
22 Ω
22 Ω
9Ω
1Ω
F
4
F
E
A
24 V
6
11 Ω
1Ω
F
E
5
1Ω
E
Current in branch AF = 24/12 = 2 A from F to A
Using current division rule current in 13 Ω in Fig. 4= 1 A
Referring Fig. 3, current in branch BE = 0.5 A
Power absorbed in branch BE = 0.52 x 18 = 4.5 W
Voltage across BE = 0.5 x 18 = 9 V
Voltage across CE in Fig. 1 =
7
x 9 = 3.5 V
18
Referring Fig. given in the problem, using voltage division rule, voltage
across in branch CD =
5
x 3.5  1.25 V
14
Example 11
Using Kirchhoff’s laws, find the current in various resistors in the circuit shown.
3Ω
6Ω
25 V
45 V
4Ω
Solution
Let the loop current be I1 and I2. We can find element currents in terms of loop
currents.
6Ω
3Ω
C
D
B
25 V
45 V
4Ω
I1
I2
A
6Ω
3Ω
C
B
25 V
D
45 V
4Ω
I1
I2
A
Considering the loop ABCA, KVL yields
6 I1 + 4 (I1 – I2) – 25 = 0
For the loop CDAC, KVL yields
3 I2 – 45 + 4 (I2 – I1) = 0
Thus 10 I1 - 4 I2 = 25
- 4 I1 + 7 I2 = 45
On solving the above I1 = 6.574 A;
I2 = 10.1852 A
Current in 4Ω resistor = I1 – I2 = 6.574 – 10.1852 = - 3.6112 A
Thus the current in 4Ω resistor is 3.6112 A from A to C
Current in 6 Ω resistor = 6.574 A; Current in 3 Ω resistor = 10.1852 A
Example 12
Find the current in 5 Ω resistor in the circuit shown.
5Ω
6Ω
3Ω
B
C
A
25 V
4Ω
D
45 V
Solution
Let the loop current be I1, I2 and I3.
5Ω
Three loops equations are:
- 25 + 6 (I1 + I3) + 4 (I1 – I2) = 0
I3
6Ω
3Ω
C
- 45 + 4 (I2 – I1) + 3 (I2 + I3) = 0
A
5 I3 + 6 (I3 + I1) + 3 (I2 + I3) = 0
25 V
I1
 10  4 6 
 4 7

3


 6
3 14
I3 = -14 A
4Ω
45 V
I2
I 1 
25 
I  = 45 : On solving
 2
 
I 3 
 0 
Current in 5 Ω resistor = 14 A from A to C
Example 13
Using mesh analysis find the current i0 and the voltage v a b in the circuit shown.
Answers: 1.7777 A
53.331 V
Example 14
In the circuit shown, determine Vx and the power absorbed by 12 Ω resistor.
1Ω
+
Vx
1.2 Ω
4Ω
2Ω
8Ω
+
12 V
3Ω
-
Answers: 2 V
1.92 W
6Ω
12 Ω
MAGNETIC CIRCUITS
Introduction
A substance, which when suspended freely, points in the direction of north and
south is called a MAGNET. Magnet attracts iron fillings. It is also called as
permanent magnet. A current passing though a conductor (or a coil) also can
produce magnetic effect and it is called as Electromagnet.
A permanent magnet has one north pole and one south pole. The imaginary lines
which travel from north pole to south pole outside the magnet are called
magnetic lines of force. They are drawn by plotting successive directions pointed
out by a small compass needle in the magnetic field. Magnetic lines of forces are
shown in Fig. 1 and they pass through the magnet.
S
N
Fig. 1 Magnetic lines of forces
Flux, Magneto Motive Force and Reluctance
The magnetic lines of force in the magnetic field is called Flux. Its unit is
Weber(Wb). 1 Wb = 108 magnetic lines. Flux is denoted by φ .
Magnetic flux per unit cross sectional area is called Flux density and it is
expressed in Weber / metre2. Flux density is denoted by B.
Magneto Motive Force (mmf) is the source of producing flux in the magnetic
circuit. It can be explained through Electromagnet. When a current of I ampere is
passed through a coil of N turns, results in a mmf of N I . This N I ampere turns is
called the mmf and its unit is ampere turns (AT).
Reluctance is the property of magnetic circuit that opposes the setting of flux.
Reluctance, S =
mmf
flux
Its unit is ampere turns / weber.
The following table shows the similarities between magnetic and electric circuits.
Sl.
Magnetic circuit
Electric circuit
1
Magnetic flux, φ webers
Electric current, I ampere
2
Magneto motive force, AT
EMF, E volts
3
Reluctance, S AT / Wb
Resistance, R ohm
No.
4
φ=
mmf
reluctance
Current =
emf
resistance
Leakage flux and Fringing effect
Usually we assume that all the flux lines take path of the magnetic medium. But,
practically, some flux lines do not confine to the specified medium. It is because,
to prevent the leakage flux, there is no perfect magnetic insulator. Some flux lines
can pass through air also.
The flux which do not follow the desired path in a magnetic circuit is known as
leakage flux.
All the magnetic flux which complete the desired magnetic circuit are the useful
flux.
To account for the leakage flux, leakage coefficient is introduced. Leakage
coefficient, denoted by λ is defined as follows.
Leakage coefficient, λ =
total flux
useful flux  leakage flux φ  φ


useful flux
useful flux
φ
Leakage coefficient will be greater than ONE.
An air gap is often introduced in the magnetic circuit out of necessity. When
crossing an air gap, the magnetic lines of force have a tendency to bulge out.
This is because the magnetic lines of force repel each other when they are
passing through non-magnetic material. This phenomenon is known as fringing.
It is shown in Fig. 2
N
S
Area at
iron path
Area at
air-gap path
Fig. 2 Fringing effect
Fringing effect increases the effective area of cross section of the air-gap and as
a result the flux density in the air-gap is reduced.
Problems involving simple magnetic circuits
Before doing problems involving magnetic circuits it is necessary to know some
more terms associated with the magnetic circuit.
Magnetic field intensity, (also called as Magnetizing force) denoted as H, is the
mmf per unit length of magnetic flux path. Thus,
H=
NI

Flux density is proportional to magnetic field intensity. Thus B  H . The constant
of proportionality is called permeability, µ. Thus B = µ H or
µ=B/H
Permeability of vacuum or free space is denoted as µ 0. Its value is
4 π x 10-7. Permeability of any other medium is given by
µ = µ 0 µr
where µr is called the relative permeability of the medium.
An expression for Reluctance, S can be obtained as follows.
S
NI NI
NI


;
φ Ba μ 0 μ r H a
Since H 
NI

we get S 

a μ0 μr
Permeance, P is the reciprocal of Reluctance.
I
I
An iron core coil with a small air gap is shown in Fig. 3.
Coil has N turns.
Current through coil = I
Mean radius of magnetic path = Rm
RR
m
m
Cross section of core is circular with diameter d
Length of air gap = ℓ
g
Note that
Fig. 3 Iron core coil
1.Tolal reluctance of magnetic path = reluctance of iron path + reluctance of air gap path.
2. Total source mmf = mmf required to establish flux in iron path + mmf required to
establish flux in air gap.
3. mmf = reluctance x flux OR mmf = H x length of magnetic path
Example 1
A toroidal air core coil with 2000 turns has a mean radius of 25 cm. The diameter
of each turn is 6 cm. If the current in the coil is 10 A, find (a) MMF (b) flux and (c)
flux density. (Toroidal coil consists of copper wire wrapped around a cylindrical
core)
I
Solution
Given Air core coil N = 2000 turns;
I
Rm = 25 cm; d = 6 cm.
(a) MMF = N I = 2000 x 10 = 20000 AT
(b) Flux = MMF / Reluctance
Reluctance, S =

a μ0 μr
Rm
R
m
; Since it is air core = µr = 1
ℓ = 2 π x 0.25 = 1.5708 m; a = π r2 = π x 0.032 = 0.002827 m2
Reluctance, S =
Flux, φ =
1.5708
 4.4217 x 108 AT / Wb
7
0.002827x 4 π x 10
20000
= 4.5231 x 10-5 Wb
8
4.4217 x 10
φ 4.5231x 10 5

(c) Flux density, B =
= 0.016 Wb / m2
a
0.002827
I
Example 2
The flux produced in the air gap between two magnetic poles is 0.05 Wb. If the
cross sectional area of the air gap is 0.2 m2, find (a) flux density, (b) magnetic
field intensity, (c) reluctance and (d) permeance of the air gap. Find also the mmf
dropped in the air gap, given the length of air gap to be 1.2 cm.
Solution
Given Flux, φ = 0.05 Wb;
(a) Flux density, B =
a = 0.2 m2; ℓ g = 0.012 m
0.05
 0.25 Wb / m2
0.2
(b) Depending on the data H can be calculated either from H =
H=
NI
or

B
B
0.25
5


1.9894
x
10
AT / m
; Magnetic field intensity, H =
μ
μ 4 π x 10 7
(c) Reluctance, S =

0.012

 4.7746 x 10 4 AT / Wb
7
a μ 0 0.2 x 4 π x 10
(d) Permeance, P =
1
1

 2.0944 x 10-5 Wb / AT
4
S 4.7746 x 10
MMF = H x ℓ g = 1.9894 x 105 x 0.012 = 2.3873 x 103 AT
OR
MMF = Flux x Relectance = 0.05 x 4.7746 x 10 4 = 2.3873 x 103 AT
Example 3
A ring has mean diameter of 15 cm, a cross section of 1.7 cm 2 and has a radial
gap of 0.5 mm cut in it. It is uniformly wound with 1500 turns of insulated wire and
a current of 1 A produces a flux of 0.1 mWb across the gap. Calculate the relative
permeability of iron on the assumption that there is no magnetic leakage.
Solution
Given Dm = 15 cm; a = 1.7 cm2; ℓ g = 0.5 mm; N = 1500 turns; I = 1 A;
φ = 0.1 mWb
I
I
Dm = 15 cm; a = 1.7 cm2; ℓ g = 0.5 mm;
N = 1500 turns; I = 1 A; φ = 0.1 mWb
MMF produced = 1500 x 1 = 1500 AT
Rm
R
m
Total reluctance = MMF / Flux = 1500 / 0.0001 = 1500 x 104 AT / Wb
Total reluctance = Reluctance of iron path + Reluctance of air gap
Reluctance of air gap =

0.0005
6


2.3405
x
10
AT / Wb
4
7
a μ 0 1.7 x 10 x 4 π x 10
Reluctance of iron path = 15 x 106 – 2.3405 x 106 = 12.6595 x 106 AT / Wb
Length of iron path = π x 15 x 10-2 – 0.05 x 10-2 = 47.0739 x 10-2 m
47.0739 x 10 2
2203.5423x 106


Thus 12.6595 x 10 =
a μ 0 μ r 1.7 x 10 4 x 4 π x 10 7 x μr
μr
6

Thus µr = 2203.5423 / 12.6595 = 174.0623
Example 4
A series magnetic circuit has an iron path of length 50 cm and an air gap of 1mm.
The cross section of the iron is 6.66 cm2 and the exciting coil has 400 turns.
Determine the current required to produce a flux of 0.9 mWb in the circuit. The
following points are taken from the magnetization curve for the iron.
Flux density (Wb / m2):
1.2
Magnetizing force (AT / m):
500 1000 2000 4000
1.35
1.45
1.55
Solution
Given ℓi = 0.5 m; ℓg= 1 x 10-3 m; a = 6.66 x 10-4 m2; N = 400; Flux = 0.9 mWb

1 x 10-3

 1.1949 x 106 AT / Wb
Reluctance of air gap =
4
7
a μ 0 6.66 x 10 x 4 π x 10
Required air gap mmf = 0.9 x 10-3 x 1.1949 x 106 = 1075.4 AT
Flux density in the iron path = 0.9x 10-3 / (6.66 x 10-4) = 1.3514 Wb / m2
From the given data, for a flux density of 1.3514 Wb / m2
corresponding value of H = 1000 + (1000 x 0.0014 / 0.1) = 1014 AT / m
Required iron path mmf = 1014 x 0.5 = 507 AT
Total mmf required = 1075.4 + 507 = 1582.4 AT
Current required = 1582.4 / 400 = 3.956 A
Example 5
An iron rod of 1 cm radius is bent to a ring of mean diameter 30 cm and wound
with 250 turns of wire. Assume the relative permeability of iron as 800. An air gap
of 0.1 cm is cut across the bent ring. Calculate the current required to produce a
useful flux of 20000 lines if (a) leakage is neglected and (b) leakage factor is 1.1.
Solution
Given r = 1 cm; Dm= 0.3 m; N = 250; µr = 800;
I
I
ℓg = 0.001 m; Flux φ = 20000 / (108) = 0.2 mWb
Leakage is neglected Flux in iron path = flux in air gap
Area of cross section, a = π x 10-4 = 0.0003142 m2
Reluctance of air gap =
Rm
R
m

0.001

 2.5327 x 106 AT / Wb
7
a μ 0 0.0003142x 4 π x 10
Required air gap mmf = 0.0002 x 2.5327 x 106 = 506.54 AT
Length of iron path = (π x 0.3) – 0.001 = 0.9415 m
Reluctance of iron path =

0.9415

 2.9807 x 106 AT / Wb
7
a μ 0 μr 0.0003142x 4 π x 10 x 800
Required iron path mmf = 0.0002 x 2.9807 x 106 = 596.14 AT
Total mmf required = 506.54 + 596.14 = 1102.68 AT
Current required = 1102.68 / 250 = 4.4107 A
Leakage factor is 1.1
As in previous case, required air gap mmf = 0.0002 x 2.5327 x 10 6 = 506.54 AT
To maintain useful flux of 0.2 mWb in the air gap,
flux required in the iron path = 1.1 x 0.2 = 0.22 mWb
Required iron path mmf = 0.00022 x 2.9807 x 10 6 = 655.754 AT
Total mmf required = 506.54 + 655.754 = 1162.294 AT
Current required = 1162.294 / 250 = 4.6492 A
Find the emf of the battery in the circuit shown.
8A
0.6 Ω
0.1 Ω
E
0.4 Ω
0.6 Ω
Example 6
The magnetic circuit shown in Fig. 4 has the following dimensions: ℓ 1 = 10 cm,
ℓ2 = ℓ3 = 18 cm, cross sectional area of ℓ1 path = 6.25 x 10
area of ℓ2 and ℓ3 paths = 3 x 10
-4
-4
m2, cross sectional
m2, length of air gap = 1 mm. Taking the relative
permeability of the material as 800, find the current in the 600 turn exciting coil to
establish a flux of 100 x 10
ℓ2
-6
Wb in the air gap, neglecting leakage and fringing.
ℓ1
Fig. 4 - Example 6
ℓ3
Solution
Given ℓ1 = 10 cm; ℓ2 = ℓ3 = 18 cm; a1 = 6.25 x 10- 4 m2; a2 = a3 = 3 x 10=4 m2
ℓg = 1mm; µr = 800; N = 600; φ 1 = 100 x 10- 6 Wb
Reluctance of path 1:

1 x 10-3

 1.2732 x 106 AT / Wb
Reluctance of air gap =
4
7
a μ 0 6.25 x 10 x 4 π x 10
Length of iron path = 10 – 0.1 = 9.9 cm
Reluctance of iron path =

0.099

a μ 0 μr
6.25 x 10  4 x 4 π x 10 7 x 800
 0.1576 x 10 6 AT / Wb
Thus R1 = (1.2732 + 0.1576) x 106 = 1.4308 x 106 AT / Wb
MMF1 = 100 x 10- 6 x 1.4308 x 106 = 143.08 AT
Reluctance of path 2:
Reluctance of iron path =

0.18

a μ 0 μr
3 x 10  4 x 4 π x 10 7 x 800
 0.5968 x 10 6 AT / Wb
Flux in path 1 will divide equally; Thus flux = 50 x 10 -6 Wb
MMF2 = 50 x 10-6 x 0.5968 x 106 = 29.84 AT
Since path 2 and path 3 are in parallel, it is required to consider mmf for only one
of them. Thus,
Total MMF = 143.08 + 29.84 = 172.92 AT
Exciting current required = 172.92 / 600 = 0.2882 A
Electrical equivalent of the magnetic circuit considered is shown below.
-6
0.1576 x 106 AT/Wb
50 x 10 Wb
172.92 AT
0.5968x 10 AT/Wb
50 x 10 Wb
-6
6
0.5968x 10 AT/Wb
-6
100 x 10 Wb
1.2732 x 106 AT/Wb ((air gap)
6
Hysteresis loss
B
M
Hmax, Bmax
N
P
O
S
H
R
- Hmax, - Bmax
Q
Fig. 6 – Circuit for B-H curve
Fig. 7 – Hysteresis loop
Consider an iron bar which can be magnetized as shown in Fig. 6. Magnetizing
force, H can be varied by controlling the current through the coil. Corresponding
values of flux density B can be noted. First the B-H curve will follow OM shown if
Fig. 7. Now if H is decreased gradually, B will not decrease along MO. Instead it
will decrease along MN. Even when H is zero, B has a definite value ON. This
implies that even on removing the magnetizing force, H, the iron bar is not
getting demagnetized completely. The value of ON measures the retentivity of the
material.
To demagnetize the iron bar, the magnetizing force has to be applied in the
reverse direction. Flux density, B becomes zero at P. The value of H as measured
by OP is known as coercive force. If H is further increased, the curve will follow
the path PQ. By taking H back from – Hmax, a similar curve QRSM is obtained. It is
seen that B always lags behind H. This lagging character of B with respect to H is
called hysteresis and the complete loop is called hysteresis loop. Different
magnetic material will have different hysteresis loop. Fig. 8 shows the hysteresis
loop of cast steel and alloyed steel.
M
N
- Hmax P
O
S
Hmax
R
Q
Fig. 8 – Hysteresis loop of cast steel and alloyed steel
Area of hysteresis loop gives the hysteresis loss per unit volume of the material.
About 4% addition of Slican to steel give rise to reduction in hysteresis loop area
and hence hysteresis loss.
Eddy current loss
Whenever a conducting material cuts the magnetic flux (armature core in the
case of rotating machines) an emf is induced in the core. This emf sets up
current through the solid mass. Such current is known as eddy current. Flow of
eddy current results in eddy current loss.
The eddy current loss is proportional to square of the thickness of the material.
This loss can be minimized by using a laminated core, which offers high
resistance for the flow of eddy current.
Faraday’s Laws of Electromagnetic Induction
When a current flows in a conductor, magnetic field is produced. The reverse
phenomenon, whereby an Electro Motive Force (EMF) and hence current is
produced in an electric circuit by some action on magnetic field. This is called
electromagnetic induction. Consider the setup shown in Fig. 9.
A
A
B
G
K
Fig. 9 – Static induced emf
When the switch, K is closed from the open position, there will be induced
voltage and hence current in coil B as indicated by the galvanometer G. When the
key is opened from the closed position, the current flow will be in the opposite
direction. This illustrates the production of static induced emf.
Consider the setup shown in Fig. 10.
G
B
N
S
A
Fig. 10 – Dynamic induced emf
When the conductor AB is moved from the top position in the downward
direction, it cuts the magnetic field at right angle. An emf is induced in the
conductor resulting current flow as indicated by the galvanometer. When the
conductor is moved from the bottom position in the upward direction, there will
be current flow in the opposite direction. This illustrates the production of
dynamic induced emf.
The results of the above two experiments can be summed up into two laws,
known as Faraday’s Laws of Electromagnetic Induction.
First Law: Whenever the flux linking with a coil changes, a static emf is induced
in it and as such the emf lasts only for the time the change is taking place.
OR
When a moving conductor cuts the magnetic field, an emf induced in it which is
called as dynamic emf.
Second Law: The magnitude of the induced emf is equal to the rate of change of
flux linkage. Flux linkage = Flux x Number ot turns having units as Wb. turns.
Induced emf
An emf is induced in a coil or conductor whenever there is a change in flux
linkages. The change in flux linkages can occur in two ways.
(i)
The coil is stationary and the magnetic field is changing. Resulting
induced emf is known as static induced emf. Transformer works on this
principle.
(ii)
The conductor is moved in a stationary magnetic field in such a way
that there is change in flux linkage. Resulting induced emf is known as
dynamic induced emf. Generator works on this principle.
Static induced emf
In this case, the coil is held stationary and the magnetic field is varied. The
induced emf may be self induced or mutually induced.
A
A
B
G
K
Fig. 9 – Static induced emf
Two coils are wound over a magnetic specimen. Coil A is energized using a
battery. If switch K is initially closed, then a steady current of I ampere will flow
through the coil A. It produces a flux of φ Wb. Let us assume that the entire flux
links coils A and B. When the switch is suddenly opened, the current reduces to
zero and the flux linking both the coils becomes zero. As per Faraday’s law, emf
is induced in both the coils A and B. Such emfs are known as static induced
emfs. Static induced emf can be classified into two categories, namely self
induced emf and mutually induced emf.
Self induced emf
If a single coil carries a current, flux will be set up in it. If the current changes,
the flux will change. This change in flux will induce an emf in the coil. This kind of
emf is known as self induced emf. In other words, self induced emf is the emf
induced in a circuit when the magnetic flux linking it changes because of the
current changes in the same circuit.
The magnitude of this self induced emf e = N
dφ dψ

dt
dt
Mutually induced emf
Mutually induced emf is the emf induced in one circuit due to change of flux
linking it, the flux being produced by the current in another circuit.
Referring to Fig. 9, when a change in current though coil A occurs, we find the
flux linking coil B changes. Hence, an emf is induced in coil B and it is called as
mutually induced emf.
Dynamic induced emf
G
B
N
S
A
Fig. 10 – Dynamic induced emf
Consider the experimental setup shown in Fig. 10.The magnetic poles, produce a
stationary flux density of B Wb. / m2. Let the conductor length be ℓ meters. The
conductor is moved at right angle to the field. Let the distance moved in dt
second be dx meters.
Area swept by the conductor in dt sec. = ℓ dx m2
Magnetic flux cut by the conductor = B ℓ dx Wb.
Taking the conductor has one turn, corresponding
flux linkage, ψ = B ℓ dx Wb Turn
Rate of change of flux linkage = B ℓ
dx
dt
According to Faraday’s Law, this is the induced emf, e in the conductor.
Thus induced emf, e = B ℓ v volts
where v = linear velocity =
dx
dt
Let the conductor be moved with velocity v m / sec. in an inclined direction,
making an angle ө to the direction of field. Then
Induced emf, e = B ℓ v sin ө volts
This is the basic principle of working of a generator.
Force on current carrying conductor
I
F
B
N
S
A
Fig. 11 – Force on current carrying conductor
Consider the setup shown in Fig. 11. When a current of I ampere flows in the
conductor from A to B, it will experience a force, F given by
F = B ℓ I Newton
This relation is true if the conductor is at right angle to the magnetic field. In case
if the conductor is an inclined direction, making an angle ө to the direction of
field, then
F = B ℓ I sin ө Newton
This is the basic principle of working of a motor.
Self inductance, L
Self inductance of a coil, L is the rate of change of flux linkages with respect to
the current in it. Its unit is Henry. Thus
L=
dψ
dφ
= N
Henry
dI
dI
Equation for self inductance
Consider a magnetic circuit shown in Fig. 12.
I
Fig. 12 Self inductance
With usual notations
Magnetizing force, H =
NI
AT / m

Flux density, B = µ0 µr H = µ0 µr (
Magnetic flux, φ = µ0 µr (
NI
) Wb. / m2

I
NI
) a Wb.

N2 I
Flux linkage = N φ = µ0 µr (
) a Wb. Turns

μ 0 μr N2 a
dφ
φ
N2
Self inductance, L = N
=N =
=
dI
I

(  / a μ 0 μr )
N2
=
Reluctance
dψ
dφ
Thus self inductance L =
= N
=
dI
dI
N2
Reluctance
Expression for self induced emf in terms of self inductance
The magnitude of self induced emf, e = N
Thus self induced emf, e = N
=L
dφ
dt
dφ d I
x
dI dt
dI
dt
The self induced emf in a circuit is directly proportional to the rate of change of
current in the same circuit.
Mutual inductance
Mutual inductance between two circuits is defined as the flux linkages of one
circuit per unit current in the other circuit.
Coil 1
I1
Coil 2
Reluctance of the magnetic circuit = S
Flux in coil 1 φ1 =
N1 I 1
N I
 1 1
Reluctance
S
Assuming that all the flux φ1 links the entire coil 2,
flux linkage of circuit 2 due to current in circuit 1 ψ21
Mutual inductance M =
ψ 21 N1 N2

I1
S
N1 N2 I 1
=
S
Consider two air core coils having self inductances L 1 and L2 that are closer to
each other as shown in Fig. 12. When current passes through coil 1, flux φ
11
is
produced in coil 1. Only a part of this flux links with coil 1 and the remaining flux
links both the coils 1 and 2. Generally, the flux linking both the coils is useful and
it is called mutual flux and represented by φ . The other part of the flux is called
21
leakage flux represented by φ . When the coil 2 carries current, flux produced in
ℓ1
it is φ
and φ
22
12
and leakage flux is φ
ℓ2
and the mutual flux is φ . Fluxes φ , φ , φ
12
are shown in Fig. 13.
φ 12
I2
I1
φ 1 φ 2
φ 21
Fig. 13 Two coils in proximity
ℓ1
21
ℓ2
The operation of many useful devises which utilizes mutual inductance
phenomenon depends upon how close the coils are coupled to each other. A
fraction of total flux produced by a coil links both the coils and this coefficient
represented by k. The coefficient of coupling depends on the relative position of
coils 1 and 2. Thus, coefficient of coupling, k =
φ21
φ 11

φ 12
φ22
. It is to be noted that
coefficient of coupling is always ≤ 1. If both the coils are far apart, then k = 0. On
the other hand if both the coils are wound over the same core, then k = 1. Similar
to the definition of self inductance, mutual inductances can be written as
M12 = N1
d φ 12
and
dI2
M21 = N2
d φ 21
d I1
Using energy criteria, it can be proved that M12 = M21 = M
Then M2 = N1 N2
d φ 12 d φ 2 1
dI2
d I1
= N1 N2 k
= k2 N1
Thus M = k
L1 L 2
dφ22
dI2
d φ 11
d I1
N2
k
d φ 11
d I1
dφ22
dI2
= k2 L 1 L 2
Certain formulae
Static induced emf:
e=N
dφ
volts
dt
e=L
dI
volts
dt
Dynamic induced emf:
e = B ℓ v sin ө volts
Force on a current carrying conductor:
F = B ℓ I sin ө Newton
Self inductance:
N2
L=
Reluctance
dφ
L=N
Henry
dI
Mutual Inductance:
M=
N1 N2
S
M= k
L1 L 2
Henry
Example 7
A coil of resistance 150 Ω is placed in a magnetic flux of 0.1 m Wb. It has 500
turns and a galvanometer of 450 Ω resistance is connected in series with it. The
coil is moved from the given field to another field of 0.3 m Wb. In 0.1 sec. Find the
average induced emf and the average current through the coil.
Solution
Given Rc = 150 Ω; φ 1 = 0.1x 10-3 Wb.; N = 500 turns; Rg = 450 Ω; φ 2 = 0.3x 10-3 Wb.;
t = 0.1 sec.
(0.3 x 10 3  0.1x 10 3 )
dφ
Induced emf, e = N
= 500 x
= 500 x 2 x 10-3 = 1.0 Volt
0.1
dt
Current, I = induced emf / total resistance = 1.0 / (150 + 450) = 0.001667 A
Example 8
A conductor of length 100 cm moves at right angle to a uniform magnetic field of
flux density 1.5 Wb. / m2 with a velocity of 30 m / sec. Calculate the emf induced
in it.
Find also the value of induced emf when the conductor moves at an angle of 600
to the direction of the magnetic field.
Solution
Given ℓ = 1.0 m; ө = 900 ; B = 1.5 Wb. / m2; v = 30 m / sec.;
ө = 600
Induced emf, e = B ℓ v = 1.5 x 1.0 x 30 = 45 V
With ө = 600. Induced emf, e = B ℓ v sin ө = 45 x sin 60 0 = 38.9711 V
Example 9
A conductor of 10 cm long lies perpendicular to a magnetic field of strength 1000
AT / m., Find the force acting on it when it carries a current of 60 A.
Solution
Given ℓ = 0.1 m; ө = 900 ; H = 1000 AT / m; I = 60 A
Flux density, B = µ0 H = 4 π x 10-7 x 1000 = 0.001257 Wb. / m2
Force, F = B ℓ I = 0.001257 x 0.1 x 60 = 0.00754 Newton
Example 10
An air cored toroidal coil has 480 turns, a mean length of 30 cm and a crosssectional area of 5 cm2. Calculate (a) the inductance of the coil and (b) the
average induced emf, if a current of 4 A is reversed in 60 m sec.
Solution
Given N = 480 turns; ℓ = 0.3 m; a = 5 x 10-4 m2; dI = 8 A; dt = 60 x 10-3 sec.
Inductance, L = N2 / Reluctance
Reluctance, S = ℓ / ( a µ0) =
0.3
 0.4775 x 109 AT/Wb
4
7
5 x 10 x 4 π x 10
4802
Inductance, L =
 0.4825 x 10 3  0.4825 mH
9
0.4775 x 10
Induced emf, e = L
dI
8
 0.4825 x 10 3 x
 0.06433 V
dt
60 x103
Example 11
A current of 5 A when flowing through a coil of 1000 turns establishes a flux of
0.3 m Wb. Determine the self inductance if the coil.
Solution
Given I = 5 A; N = 1000 turns; φ = 0.3x 10-3 Wb.;
0.3 x 103
dφ
 1000 x
 0.06 H
Self inductance, L = N
5
dI
Example 12
A coil has a self inductance of 30 mH. Calculate the emf in the coil when the
current in the coil (a) increases at the rate of 300 A / sec. (b) raises from 0 to 10 A
in 0.06 sec.
Solution
Given L = 30 x 10-3 H;
(a) Induced emf, e = L
dI
 30 x 10 3 x 300  9 V
dt
(b) Induced emf, e = L
dI
10
 30 x 10 3 x
 5V
dt
0.06
Example 13
The number of turns in a coil is 250. When a current of 2 A flows in this coil, the
flux in the coil is 0.3 m Wb. When this current is reduced to zero in 2 m sec., the
voltage induced in another coil is 63.75 V. If the coefficient of coupling between
the two coils is 0.75, find the self inductances of the two coils, mutual inductance
and the number of turns in the second coil.
Solution
Given N1 = 250; I1 = 2 A; φ 1 = 0.3x 10-3 Wb.; dI1 = 2 A ; dt1 = 2 m sec; e2 = 63.75 V;
k = 0.75
N1 = 250; I1 = 2 A; φ 1 = 0.3x 10-3 Wb.; I1’ = 0 ; dt = 2 m sec; e2 = 63.75 V; k = 0.75
dφ 1
0.3 x 10 3
Self inductance, L1 = N1
 250 x
 0.0375H
d I1
2
Induced emf in coil 2, e2 = M
dI 1
2
 Mx
 63.75
dt
0.002
N1 = 250
Thus mutual inductance, M = 63.75 mH
Since M = k
2
k = 0.75
L1 L 2
2
0.06375 = 0.75 x 0.0375 x L2
Thus self inductance of coil 2, L2 = 0.1927 H
I1 = 2 A; I1’ = 0
φ1 = 0.3 m Wb
φ2 = 0.225 m Wb
’
φ1 = 0;
φ2’ = 0;
dt = 2 m sec.
Flux φ 2  k φ1 = 0.75 x 0.3x 10-3 Wb = 0.225 x 10-3 Wb; φ 2  k φ1  0
'
dφ 2
0.225 x 10 3
Also, e2 = N2 x
 N2 x
 63.75
dt
2 x 10 3
Thus N2 = 567
'
FUNDAMENTALS OF AC
Electrical appliances such as lights, fans, air conditioners, TV, refrigerators, mixy,
washing machines and industrial motors are more efficient when they operate
with AC supply. The required AC voltage is generated by AC generator also called
as alternator.
A waveform is a graph in which the instantaneous values of any quantity are
plotted against time. A periodic waveform is the one which repeats itself at
regular intervals. A waveform may be sinusoidal or non sinusoidal. Examples of a
few periodic waveforms are shown in Fig.1.
(a) Sinusoidal waveform
(b) Rectangular waveform
Fig. 1
(c) Sawtooth waveform
Alternating waveform is a waveform which reverses its direction at regular
intervals. Sinusoidal and rectangular waveforms shown above are alternating
waveforms. Let us see more details about sinusoidal waveform.
x(t)
xm
ωt
Fig. 2
Fig. 2 shows a sinusoidal waveform, which can be called as a sinusoid. It can
represent a voltage or current. Its equation can be written as
x(t) = xm sin (ωt + φ)
Thus a sinusoid is described in terms of
i)
its maximum value
ii)
its angular frequency, ω and
iii)
its phase angle φ
(1)
It is evident that sinusoid repeats in a cyclic manner. The number of cycles it
makes in one second is called the frequency (f). Thus the unit for frequency is
cycles per second which is also commonly known as hertz (Hz). Electric supply
has a frequency of 50 or 60 Hz. In communication circuit, the frequency will be in
the order of Mega Hz.
The time taken by the sinusoid to complete one cycle is called the period (T) of
the sinusoid. When the supply frequency is 50 Hz, the sinusoid makes 50 cycles
in one second. Thus the period is 1/50 = 0.02 second. The frequency and the
period are related as
T=
1
f
or f =
1
T
(2)
The angular frequency of sinusoid is represented by ω and its unit is radians per
second. In one cycle the angle covered is 2π radians. When the frequency is f
cycles per second, the angle covered in one second will be 2πf radians. Thus
ω=2πf
While drawing a sinusoid, instead of ωt, time t can be taken in the x-axis.
(3)
Example 1
Consider the voltage sinusoid
v(t) = 70 sin ( 60 t + 20 0 ) V
Find the amplitude, phase, angular frequency, frequency, period and the value of
voltage at time t = 0.25 s.
Solution
Amplitude v m = 70 V
Phase φ = 20 0
Angular frequency ω = 60 rad / s
Frequency f =
Period T =
ω
60
= 9.5511 Hz

2π
2π
1
1
=
= 0.1047 s
f
9.5511
Voltage value at t = 0.25 s is
v (0.25) = 70 sin ( 60 x 0.25 x
180
+ 20 0 ) = 24.59 V
π
The two sinusoids shown in Fig. 3 are x(t) = x m sin ωt and x(t) = xm sin(ωt + φ)
x(t)
x(t) = xm sin(ωt)
x(t) = xm sin(ωt + φ)
ωt
Fig. 3
The sinusoid x(t) = xm sin(ωt + φ) leads the sinusoid x(t) = x m sin ωt by an angle
of φ. The sinusoids can also be written as
x(t) = xm sin(θ + φ)
(4)
The average value of the periodic waveform can be obtained as:
Average value =
Area under one complete cycle
Period
Average value is also called as mean value.
(5)
The Root Mean Square (RMS) value of periodic waveform is:
RMS value =
Area under squared curve for one cycle
Period
(6)
Form Factor is defined as
Form Factor =
RMS Value
Average value
(7)
Peak Factor is defined as
Peak Factor =
Peak Value
R M S value
(8)
Consider a current waveform described by
i(t) = Im sin θ
(9)
Its positive half cycle and negative half cycle of such sinusoids are negative of
each other. Hence the area in one cycle is zero. For such sinusoidal wave form
the average value is the average value over half cycle.
Thus


0
0
Area of the curve =  Im sin θ dθ = Im (  cos θ)
Iav =
2 Im
= 0.6366 Im
π
= Im ( 1 1) = 2 Im
(10)
When we square the waveform i(t) = Im sin θ, the first and the second half of the
cycle will be same. Therefore while computing the R M S value of i(t) = I m sin θ it
is enough to consider only one half cycle.

Area of square curve =
I
2
m
sin 2 θ dθ
0
2
Im
=
2


0
2
Im
sin 2
(θ(1  cos 2 θ) dθ =
)
2
2

0
2
Im
π 2
=
[π – 0] =
Im
2
2
2
Im
Mean square value =
2
RMS value
=
Im
2
= 0.7071 Im
(11)
(12)
Form factor =
0.7071 Im
RMS Value
 1.11
=
Average value
0.6366 Im
(13)
Peak factor =
Im
Peak Value
 1.414
=
0.7071 Im
R M S value
(14)
We may be calculating average and RMS values of waveforms in which inclined
straight line variations are present. Consider the waveform shown in Fig. 4. Its
square curve is shown in Fig. 5. Area A1 of the square curve can be calculated as
follows.
v2
v
V2m
Vm
A1
t
x
x
Fig. 4
t
Fig. 5
Equation of the straight line is: v =
x
2
Vm2 2
vm
t3
Area A1 =  2 t dt = 2
x 3
0 x
x
=
0
2
m
2
vm
v
t ; Then v2 =
x
x
t2
1 2
Vm x
3
It can be verified that the above result is true for the waveform shown in Fig. 6
also.
v
Vm
x
Fig. 6
t
Example 2
Find the average and RMS
5A
values of the waveform
shown in Fig. 7
t
4
2
Fig. 7
Solution
Iav =
1
1 1
x area of the triangle  x x 5 x 2  2.5 A
2
2 2
i2
25
The square curve is shown in Fig. 8.
Area of square curve =
1
x 25 x 2  16.6663
3
2
Fig. 8
Mean square value =
RMS value =
16.6663
 8.3332
2
8.3332  2.8867 A
Example 3
Find the average and RMS value of the waveform shown in Fig. 9.
v
Vm
π
t
π/3 2π/3
Fig. 9
Solution
Area of positive half cycle =
Average value =
2π
1π
π
1π
Vm  Vm 
Vm =
Vm
23
3
23
3
2
Vm  0.6667 Vm
3
V m2
The square curve is shown in Fig. 10.
Area of square curve
A3
A1
π 2
π 2 π 2 5
=
Vm 
Vm 
Vm  π Vm2
9
3
9
9
Mean Square value =
A2
5 2
Vm ;
9
π
3
2π
3
Fig. 10
Thus RMS value = 0.7454 Vm
π
Example 4
Find the average and RMS values of the half wave rectified sine wave shown in
Fig. 11.
i
Im
0
π
Solution
2π
Fig. 11
As seen earlier, area of half sine wave = 2 Im
Total area = 2 Im + 0 = 2 Im
Average value Iav =
2 Im
= 0.3183 Im
2π
As seen earlier, area of square of half sine wave =
Total area of square curve =
Mean of square curve =
RMS value IRMS = 0.5 Im
π 2
Im
2
1 2
1 π 2
2
= 0.25 Im
Im = Im
4
2π 2
π 2
Im
2
θ
Example 5
Find the average and RMS values of the full wave rectified sine wave shown in
Fig. 12.
i
Im
0
Solution
π
2π
Fig. 12
As seen earlier, area of half sine wave = 2 Im
Total area = 2 Im + 2 Im = 4 Im
4 Im
2
=
Im = 0.6366 Im
2π π
Average value Iav =
As seen earlier, area of square of half sine wave =
2
Total area of square curve = π Im
Mean of square curve =
RMS value IRMS =
Im
2
1 2
1
2
= Im
π Im
2
2π
= 0.7071 Im
π 2
Im
2
θ
If the waveform is the sum of several waveforms, its AVERAGE value is the sum
of average values of its components and its RMS values can be obtained as
follows.
Let
W = W1 + W2 + W3 and their RMS values be W1 RMS, W2 RMS and W3 RMS respectively.
Then
WRMS =
W12RMS  W22RMS  W32RMS
Example 6
A conductor carries simultaneously a direct current of 10 A and a sinusoidal
alternating current with a peak value of 10 A. Find the RMS value of the conductor
current.
Solution
Conductor current i(t) = (10 + 10 sin ωt) A
Here W1 = 10 A and W2 = 10 sin ωt A
Therefore W1 RMS = 10 A; W2 RMS = 7.071 A
RMS value of conductor current =
102  7.0712  12.2474A
Cycle Test 1 28 – 02 – 2011
1. Briefly explain the fringing effect.
2. Define Magnetic field intensity.
3. State and explain Kirchhoff’s current law and Kirchhoff’s voltage law
with suitable examples.
4. Using Kirchhoff’s laws, find the voltages E1 and E2.
E2
8Ω
-
+
E1
+
3A
+
2V
-
+
1A
5V
-
5. Find the power delivered by the voltage source.
3kΩ
18 k Ω
6kΩ
20 k Ω
-
+
12 V
8kΩ
6. Find the ampere turns required to produce a flux of 0.4 mWb. in the
air gap of 0.5 mm in a circular magnetic circuit. The iron ring has a
cross section of 4 mm2 and 63 cm mean length. Assume the relative
permeability of iron as 800. Neglect flux leakage.

0.5 x 10-2
7


9.9472
x
10
AT / Wb
Reluctance of air gap =
6
7
a μ 0 4 x 10 x 4 π x 10
Reluctance of iron path =

a μ 0 μr
63 x 10-2

 1.5667 x 108 AT / Wb
6
7
4 x 10 x 4 π x 10 x 800
Total mmf required = 0.4 x 10-3 x 25.6142 x 107 = 1.0246 x 105 AT
SINGLE PHASE AC CIRCUITS
PHASORS
Consider a linear ac circuit having one or more sinusoidal inputs having same
frequency as shown in Fig. 1. The amplitudes and phase angles of the inputs may
be different while their frequency should be same.
Sinusoidal
Sinusoidal
inputs of
outputs of
Linear ac circuit
same
same
frequency
frequency
Fig. 1
The output what we may be interested may be voltage across an element or
current through an element. The output waveform will be sinusoidal with the same
frequency as the input signals. This could be easily verified experimentally.
The steady-state analysis of such circuits can be carried out easily using phasors.
A sinusoid is fully described when its maximum value, angular frequency and
phase are specified.
A question may arise whether we should always deal with such sinusoidal time
function to represent voltage and current in ac circuits. When all the inputs are
sinusoidal time function with the same angular frequency ω, the voltage or the
current in any part of the circuit will also be of sinusoidal time function with the
SAME ANGULAR FREQUENCY ω. Hence it is redundant to carry information of
ω, while representing voltages and currents in ac circuits. This idea gives birth to
the concept of PHASORS.
The phasor corresponding to sinusoid x(t) = x m cos (ωt + φ) is X =
xm
2
φ
(1)
In case x(t) is expressed as x(t) = x m sin (ωt + φ), it can be written as
x(t) = x m sin (ωt + φ) = x m cos (ωt + φ –
X =
xm
2
 
π
) and the corresponding phasor is
2
π
2
(2)
In a similar way we can state:
If x(t) = - x m cos (ωt + φ) its phasor is X =
If x(t) = - x m sin (ωt + φ) its phasor is X =
xm
2
xm
2
φ- π
(3)
π
2
(4)
 
Eqs. (1) to (4) are useful to find the phasor for a given sinusoid.
- sin
Fig. 1 is useful to locate
the quadrant in which
the phasor lies
- cos
cos
sin
Fig. 1 Quadrants for Phasor
A few sinusoids and the corresponding phasors are;
x (t) =
1
2 150 cos ( ωt + 15 0 )
X = 150 15 0
1
x (t) =
2 150 cos ( ωt - 75 0 )
X = 150 - 75 0
x (t) =
2 100 sin ωt
X = 100 - 90 0
x (t) =
2 100 sin (ωt + 30 0 )
X = 100 - 60 0
x (t) =
2 100 sin ( ωt - 150 0 )
2
3
4
5
2
3
4
X = 100 - 240 0
5
x (t) = -
2 80 cos ( ωt + 30 0 )
X = 80 210 0
x (t) = -
2 80 sin (ωt - 30 0 )
X = 80 60 0
6
7
6
7
X
X
7
5
X
. The above Phasors are
shown in Fig. 2
X
6
Fig. 2 Phasors of given sinusoids
X
X
3
X
4
2
1
The important motivation for the use of phasors is the ease with which two or
more sinusoids at the same frequency can be added or subtracted. In the
sinusoidal steady state, all the currents and voltages are of same frequency.
Hence phasors can be used to combine currents or voltages. KCL and KVL can
be easily interpreted in terms of phasor quantities.
A phasor is a transformed version of a sinusoidal voltage or current waveform
and it contains the amplitude and phase angle information of the sinusoid.
Phasors are complex numbers and can be depicted in a complex plane. The
relationship of phasors on a complex plane is called a phasor diagram.
Example 1
Using phasor concept, find the sum of 4 voltages given by :
v1 =
2 50 sin ωt
v2 =
2 40 sin (ωt + π / 3)
v3 =
2 20 sin (ωt – π / 6)
v4 =
2 30 sin (ωt + 3π / 4)
Solution
In phasors corresponding to the sinusoids are:
V 1 = 50 - 90°
=
0.0 - j 50.0
V 2 = 40 - 30°
=
34.6410 - j 20.0
V 3 = 20 - 120° =
V 4 = 30 45°
=
V1 + V 2 + V 3 + V 4
-10.0 - j 17.3205
21.2132 + j21.2132
= 45.8542 - j66.1073
= 80.4536  - 55.25 0
Corresponding sinusoid is obtained as
vT =
2 80.4536 cos (ω t - 55.25 0 ) = 113.7786 cos (ω t - 55.25 0 )
= 113.7786 sin (ω t + 34.75 0 )
OPERATOR j
Operator j is useful in dealing with COMPLEX NUMBERS.
j  1900
j2  1900 x 1900  11800   1
j3   j  1  900
j4  1
j3
3 +j 4
j4
3
3
COMPLEX NUMBERS
4+j3
- 3.3 + j 2.2
-3–j4
2.4 – j 4.4 are a few complex numbers.
They can be represented either in RECTANGULAR
FORM
or
POLAR
FORM. The numbers shown above are in rectangular form. There is one
phasor corresponding to each complex number as shown below.
4  j 3  5 36.870
j3
4
-3
-j4
- 3  j 4  5  - 126.90
- 3.3  j 2.2  4 146.30
j 2.2
- 3.3
2.4
- j 4.4
2.4  j 4.4  5  - 61.4 0
SINGLE ELEMENT IN STEADY STATE
Voltage-current relationship of resistor, inductor and capacitor can be obtained in
phasor form. Such phasor representations are useful in solving ac circuits.
RESISTOR
Let the voltage v(t) across the resistor terminals be
v (t) = V m cos ωt
(5)
The current through it is given by
i(t) 
V
v(t)
 m cos ω t
R
R
(6)
Expressing the equations (5) and (6) in phasor form we get
V 
Vm
I 
Vm
2
0 0
2R
0 0
(7)
(8)
The impedance of an element is defined as the ratio of the phasor voltage across
it to the phasor current through it. Thus
Z 
V
I
(9)
Z 
For a resistor
V
 R  00
I
(10)
Thus in the case of a resistor, voltage-current relationship is
V = RI
(11)
Representation of resistor in time frame and its phasor form are shown in Fig. 3.
i(t)
+
R
v(t)
I
-
R
+
V
-
Fig. 3 Representation of a resistor
It is to be noted that as seen by the Eqns. (7) and (8),both the voltage V and the
current I have the same phase angle of 0 0 . The phasor diagram showing the
voltage and current in a resistor is shown in Fig. 4.
I
V
Fig. 4 Phasor diagram - resistor
I
V
Fig. 4 Phasor diagram - resistor
In the phasor diagram shown in Fig. 4, importance must be given to the phase
angles of the voltage V and the current I. The lengths of the phasors depend on
their magnitude and the scale chosen. In no occasion length of a voltage phasor
and the length of a current phasor can be compared since they have different
units. The scale for current phasors will be like
1 cm = x Volts while the scale for the voltage phasors will be like
1 cm = y Ampere.
The impedance of the resistor is R 0°. In a general network where R is
embedded, the phasor corresponding to the voltage across R and the phasor
corresponding to the current through R are always in phase.
INDUCTOR
For the inductor, the voltage-current relationship is
v(t)  L
di(t)
dt
(12)
In steady state, let the current through it be
i (t) = Im cos ωt
Then v(t)  L
(13)
di(t)
= - ωL Im sin ωt
dt
(14)
Expressing the above two equations in phasor form we have
I 
Im
2
0 0
and V  ω L
(15)
Im
2
900
(16)
The impedance of the inductor is given by
Z 
V
= ωL 90 0 = j ωL = j X L
I
(17)
where X L = ωL
(18)
Thus, the terminal relationship of an inductor in phasor form is
V = j XL I
(19)
Representation of inductor in time frame and its phasor form are shown in Fig. 5.
i(t)
L
+
j XL
I
v(t)
+
-
V
-
Fig. 5 Representation of a inductor
It is to be noted that as seen by the Eqns. (15) and (16), the voltage V leads the
current I by a phase angle of 90 0 . The phasor diagram showing the voltage and
current in an inductor is shown in Fig. 6.
V
I
Fig. 6 Phasor diagram - inductor
V
I
Fig. 6 Phasor diagram - inductor
It is to be noted that the voltage V leads the current I by 90° or we can also state
that the current I lags the voltage V by 90°. The steady state impedance
corresponding to the inductance L is jX L where X L = ωL. The quantity X L is
known as the INDUCTIVE REACTANCE.
CAPACITOR
For the capacitor, the voltage-current relationship is
i(t)  C
dv(t)
dt
(20)
In steady state, let the voltage across it be
v (t) = Vm cos ωt
Then i(t)  C
dv(t)
= - ωC Vm sin ωt
dt
(21)
(22)
Expressing the above two equations in phasor form we have
V 
Vm
2
0 0
and I  ω C
(23)
Vm
2
900
(24)
The impedance of the capacitor is given by
Z 
=
V
I
1
j
- 90 0 = ωC
ωC
where XC =
= - j XC
(25)
1
ωC
(26)
Thus, the terminal relationship of a capacitor in phasor form is
V = - j XC I
(27)
Representation of capacitor in time frame and its phasor form are shown in Fig. 7.
i(t)
C
+
v(t)
I
-
+
Fig. 7 Representation of a capacitor
- jX C
V
-
It is to be noted that as seen by the Eqns. (23) and (24), the current I leads the
voltage V by a phase angle of 90 0 . The phasor diagram showing the voltage and
current in a capacitor is shown in Fig. 8.
I
V
Fig. 8 Phasor diagram of - capacitor
It is to be noted that the current I leads the voltage V by 90° or we can also state
that the voltage V lags the current I by 90°. The steady state impedance
corresponding to the capacitance C is - jXC where XC =
1
. The quantity XC is
ωC
known as the CAPACITIVE REACTANCE.
It is conventional to say how the current phasor is relative to voltage phasor.
Thus for the resistor, the current phasor is in phase with the voltage phasor. In
an inductor, the current phasor lags the voltage phasor by 90 0. In the case of a
capacitor, the current phasor leads the voltage phasor by 90 0 .
Example 2
The voltage of v =
2 80 cos (100 t - 55 0 ) V is applied across a resistor of 25Ω.
Find the steady state current through the resistor.
Solution
Here V = 80   55 0
and R = 25 Ω
V 80   55 0

 3.2   55 0 A
Thus, current I =
R
25
Current i(t) = 4.5255 cos (100 t - 55 0 ) A
Example 3
The voltage of v =
2 20 sin (50 t - 25 0 ) V is applied across an inductor of
0.1 H. Find the steady state current through the inductor.
Solution
Phasor voltage V = 20 - 25 0 - 90 0 = 20 - 115 0 V
Impedance Z = j ω L = j 50 x 0.1 = 5 90 0 Ω
V 20   1150
0
Current I =


4


205
A
0
Z
5 90
Converting this to the time domain
Current i(t) = 5.6569 cos (50 t - 205 0 ) A
= - 5.6569 cos (50 t - 25 0 ) A
Example 4
The voltage of v =
2 12 cos (100 t - 25 0 ) V is applied across a capacitor of
50 F. Find the steady state current through the capacitor.
Solution
Phasor voltage V = 12 - 25 0 V
Impedance Z = - j
1
1
 j
  j 200 Ω
6
ωC
100 x 50 x 10
12   25 0
V
0
0
Current I =

A

0.06

65
A

60

65
mA
Z 200   900
Converting this to the time domain
Current i(t) = 84.8528 cos (100 t + 65 0 ) mA
ANALYSIS OF RLC CIRCUITS
An ac circuit generally consists of resistors, inductors and capacitors connected
in series, parallel and series-parallel combinations. Often we need to simplify the
circuit by finding the equivalents. Further to this, we have to make use of KVL,
KCL, source transformation, voltage division and current division what we
discussed in previous chapter, by replacing resistors by impedances and dc
voltages and currents by voltage phasors and current phasors.
A coil used in ac circuit will have its own resistance in addition to the inductive
reactance due to its inductance. One such coil is shown in Fig. 9.
R
j XL
Fig. 9 A coil in an ac circuit
It is clear that the resistance R and inductive reactance j X L are connected in
series. The impedance of this coil is
Z = R + j XL
(28)
Now consider a case where a resistance R and a capacitance having a capacitive
reactance - j X C are connected in series as shown in Fig. 10.
R
- j XC
Fig. 10 A resistance and a capacitance in series
The impedance of the circuit is
Z = R - j XC
(29)
IMPEDANCE AND ADMITTANCE
The steady state impedance ( a complex quantity ) can be written in two forms,
namely Rectangular form and Polar form as
Rectangular form: Z = R + j X
Polar form:
Z =
Z 
If two impedances Z 1 and Z 2 are connected in series as shown in Fig. 11, then
the equivalent impedance is
Z eq = Z1 + Z 2
(30)
Z1
Z2
Z eq
Fig. 11 Two impedances connected in series
This could be generalized to n number of impedances connected in series as
Z e q = Z 1 + Z 2 + …………….+ Z n
(31)
If n number of impedances Z 1 , Z 2 , …., Z n are connected in parallel as shown in
Fig. 12, the equivalent impedance is obtained from
1
1
1
1


 .......... . 
Z eq
Z1
Z2
Zn
(32)
Z eq
Z1
Z2
……
Zn
Fig. 12 n impedances connected in parallel
If two impedances Z 1 and Z 2 are connected in parallel, then the equivalent
resistance is obtained from
1
1
1


Z eq
Z1
Z2

(33)
Z1  Z 2
Z1 Z 2
Therefore
Z eq =
Z1 Z2
Z1  Z 2
(34)
While dealing with the parallel circuit, it is also useful to define another quantity
called ‘admittance’. ADMITTANCE is defined as the reciprocal of the impedance
and it is denoted by Y. Thus
Y =
1
Z
(35)
When the admittance Y is written in rectangular form as
Y = G+jB
(36)
G is called as ‘conductance’ and B is called as ‘susceptance’. The unit of G, B
and Y are mho or siemens and is denoted by
.
When two impedances Z 1 and Z 2 are connected in parallel, referring to Eqn. (33)
the equivalent admittance Y e q is given by
Y eq = Y1 + Y 2
(37)
where Y 1 and Y 2 are the admittances corresponding to the impedances Z 1 and
Z 2 respectively.
When n number of admittances Y 1 ,Y 2 ,…….., Y n are connected in parallel, Eqn.
(3.44) can by generalized as
Y e q = Y 1 + Y 2 +…………..+ Y n
(38)
If there are n equal impedances Z are connected in series, then the equivalent
impedance is
Z eq = n Z
(39)
Similarly if there are n equal admittances Y are connected in parallel, then the
equivalent admittance is
Y eq = n Y
(40)
RL CIRCUIT
Having studied how to combine the series and parallel impedances we shall now
see how the RL, RC and RLC circuits can be analyzed.
Let us consider a simple circuit in which a resistor and an inductor are connected
in series as shown in Fig. 13.
j XL
R
I
+
VR
-
+
VL
-
E
Fig. 13 RL circuit
Taking the supply voltage as reference
E  E 0 0
(41)
Circuit impedance Z  R  j X L  Z θ
(42)
E 0 0
E
E
Circuit current I 


θ
Z
Z θ
Z
(43)
= I θ
where
I 
E
Z
Further VR  R I  R I   θ
VL  j X L I  X L I   θ  900
Using KVL, we get V R + V L = E
(44)
(45)
(46)
(47)
(48)
The phasor diagram for this RL circuit can be got by drawing the phasors E, I, V R
and V L as shown in Fig. 14.
VL
E = E 00 V
II== I  - θ
90 0
θ
I=
VR = R I   θ
IV=L = XL
E
VL
I   θ  900
VR
I=
I
Fig.14 Phasor diagram of RL circuit
Consider the triangle formed by the phasors V R , V L and E. Recognizing that
VR  R I ,
by
I
VL  X L I and E  Z I
if each side of the triangle is divided
Z will form a triangle as shown in Fig, 15. This triangle
then R, X L and
is known as the IMPEDANCE TRIANGLE.
Z
By knowing any two of
XL
θ
R
Fig. 15 Impedance diagram of RL circuit
4 quantities, other two can
be calculated
RC CIRCUIT
Let us now consider the circuit in which a resistor and a capacitor are connected
in series as shown in Fig. 16.
- j XC
R
I
+
VR
-
+
VC
-
E
Fig. 16 RC circuit
Taking the supply voltage as reference
E  E 0 0
(49)
Circuit impedance Z  R  j X C  Z   θ
(50)
E 0 0
E
E
Circuit current I 


θ
Z
Z θ
Z
(51)
= I θ
(52)
where
I 
E
Z
(53)
Further VR  R I  R I θ
(54)
VC   j X C I  X C I θ  900
(55)
Using KVL, we get V R + V C = E
(56)
The phasor diagram for this RC circuit can be got by drawing the phasors V R , V C ,
E and I as shown in Fig. 17.
VR
I
θ
90
0
R
VC
θ
E
Z
XC
Fig. 18 Impedance triangle of RC circuit
VC
Fig. 17 Phasor diagram of RC circuit
Consider the triangle formed by the phasors V R , V C and E. Recognizing that
VR  R I ,
by
I
VC  X C I and E  Z I
then R, X C and
if each side of the triangle is divided
Z will form a triangle as shown in Fig, 18. This triangle
is known as the IMPEDANCE TRIANGLE.
GE 0106 BASIC ENGINEERIN II
SURPRISE TEST 1 March 2010
1. When the current through a coil of 1000 turns is increased by 5 A, the flux in it
increases by 0.3 m Wb. Determine the self inductance of the coil.
2. The coefficient of coupling between coil A of 1200 turns and coil B of 1000 turns is
0.6. A current of 12 A in coil A establishes in it a flux of 0.12 m Wb. If the current in
coil A changes from 12 A to -12 A in 0.02 sec., what would be the emf induced in
coil B?
3. For the wave form shown, find the average value, RMS value, Form factor and
Peak factor.
v
12 V
0
2 sec.
5 sec.
t (sec.)
When the current through a coil of 1000 turns is increased by 5 A, the flux in it
increases by 0.3 m Wb. Determine the self inductance of the coil.
0.3 x 10 3
dφ
L=N
 1000 x
 0.06 H
dI
5
The coefficient of coupling between coil A of 1200 turns and coil B of 1000 turns is
0.6. A current of 12 A in coil A establishes in it a flux of 0.12 m Wb. If the current in
coil A changes from 12 A to -12 A in 0.02 sec., what would be the emf induced in
coil B?
NA = 1200;
NB = 1000; k = 0.6; When dI = 12 A, d φ A = 0.12 m Wb.
When dI = 24 A, d φ A = 0.24 m Wb.; Corresponding dt = 0.02 sec.
d φ B = 0.6 x 0.24 = 0.144 m Wb
eB = N B
dφB
0.144 x 103
 1000 x
 7.2 V
dt
0.02
For the wave form shown, find the average value, RMS value, Form factor and
Peak factor.
v
12 V
0
Average value =
2 sec.
5 sec.
t (sec.)
1 1
[
x 5 x 12 ] = 6 V
5 2
Area of square curve = (
1
1
144
x 2 x 144 )  ( x 3 x 144 ) 
x 5  240
3
3
3
Mean square =
240
 48 ;
5
Form Factor =
RMS value
6.9282

 1.1547
Average value
6
Peak Factor =
Peak value
12

 1.7321
RMS value
6.9282
RMS value =
48  6.9282 V
RLC CIRCUITS
Analysis of RLC circuits is the series, parallel and series-parallel combination of
RL and RC circuits. Equivalent of RLC circuit will be R, or RL or RC circuit as
illustrated in the examples to be discussed.
POWER AND POWER FACTOR
Let E 0 0 be the supply voltage in an AC circuit. The supply current may lag or
lead the supply voltage. Let the supply current be I - θ. The supply current can
be resolved into two components (i) A component Ip in phase with the voltage
and (ii) A component Iq at right angle to the voltage as shown in Fig. 19.
θ
Iq
E 00
Ip
I - θ0
Fig. 19 Power and Power factor
Current Ip is called the active or in-phase component while Iq is known as reactive
or quadrature component. As seen from Fig. 19
Ip = I cos θ and
(57)
Iq = I sin θ
(58)
It is to be noted that
I cos θ, I sin θ and I form three sides of a right angle triangle as in Fig. 20.
Ip
θ
I
Iq
Fig. 20 Components of current
Active Power (P)
Active power is the real power consumed by the circuit. This is due to the inphase component.
Active or real power P = E Ip
= E
I cos θ Watts
(59)
Reactive Power (Q)
The power associated with the reactive component of current Iq is known as
reactive power. Its unit is Volt Ampere Reactive (VAR).
Reactive power Q = E Iq
= E
I sin θ VAR
(60)
Apparent Power and Power Factor
The product of voltage and current, E
I is called as Apparent Power, denoted
by S . Its unit is Volt Ampere (VA).
Apparent power S = E
I
VA
(61)
Similar to Fig. 20, real power P, reactive power Q and apparent power S form
three sides of a right angle triangle as shown in Fig. 21.
P
θ
S
Q
P= E
I cos θ
Q= E
I sin θ
S= E
I
Fig. 21 Components of power
Power Factor (pf) is the ratio of real power to apparent power.
Thus power factor =
E I cos θ
E I
= cos θ
(62)
By the above definition, it is not possible to distinguish whether the load is
inductive or capacitive. If the load is inductive, the current is lagging the voltage
and the nature of the power factor is LAGGING. On the other hand if the load is
capacitive, the current is leading the voltage and hence the nature of the power
factor is LEADING.
Whenever power factor is furnished, it must be clearly stated whether it is lagging
or leading. For inductive load, the power factor is cos  lagging; for capacitive
load, the power factor is cos  leading; for resistive load since the voltage and
current are in-phase, power factor angle  is zero and the power factor is said to
be UNITY.
Power associated with R, L and C can be obtained as follows.
In the case of resistor, p.f. angle is zero and hence
P= E
I cos θ = E
Q E
I sin θ = 0
I  Z I I  I
2
R
(63)
(64)
In the case of pure inductor and pure capacitor, p.f. angle = 900 and hence
P= E
Q = E
I cos θ = 0
I sin θ = E
(65)
I  Z I I  I
2
X
(66)
Example 5
In a series circuit containing pure resistance and pure inductance, the current
and voltage are: i(t) = 5 sin (314t +
5π
2π
) and v(t) = 20 sin (314t +
). (i) What is
6
3
the impedance of the circuit? (ii) What are the values of resistance, inductance
and power factor? (iii) What is the power drawn by the circuit?
Solution
Current I =
Phasor current and phasor voltages are
5
2
Impedance Z =
5
120-90 =
2
300;
Voltage V =
20
2
150-90 =
20
2
600
V 20 60

 4 300 Ω  ( 3.4641  j 2) Ω
I
5 30
Resistance R = 3.4641 Ω
XL = 2 Ω;
314 L = 2; L =
2
H;
314
Inductance L = 6.3694 mH
Angle between voltage and current = 300
p.f. = cos 300 = 0.866 lagging
Power P = V I cos θ =
20
2
x
5
2
x 0.866 = 43.3 W
V
I
Example 6
An inductive coil takes 10 A and dissipates 1000 W when connected to a supply
of 250 V, 25 Hz. Calculate the impedance, resistance, reactance, inductance and
the power factor.
Solution
10 A
2
P = I R ; Resistance R =
Z 
250
 25 Ω ;
10
1000 W
θ
1000
= 10 Ω
100
250 V, 25 Hz
From impedance triangle X =
25 2  102  22.9128Ω
Thus impedance Z = (10 + j 22.9128) Ω = 25  66.420 Ω
Resistance R = 10 Ω
Inductance L =
Reactance (Inductive) XL = 22.9128 Ω
XL
22.9128

 0.1459 H
2πf
2 π x 25
From impedance triangle, power factor =
|Z|
R
10

 0.4 lagging
Z
25
R
X
Example 7
A resistance is connected in series with a coil. With a supply of 250 V, 50 Hz, the
circuit takes a current of 5 A. If the voltages across the resistance and the coil are
125 V and 200 V respectively, calculate (i) impedance, resistance and reactance
of the coil (ii) power absorbed by the coil and the total power. Draw the phasor
5A
diagram.
XC O
RC O
R
125 V
200 V
250 V 50 Hz
Resistance R =
ZC O =
125
 25 Ω
5
200
 40 Ω ;
5
Since
Z C O = 40 Ω,
Since
Z T = 50 Ω;
ZT =
Fig. 21 Example 7
250
 50 Ω
5
R C O  j XC O  40 ; R 2C O  X2C O  1600
(25  R C O )  j XC O  50
625 + 50 RC O + R 2C O  X2C O  2500 i.e. 50 RC O = 2500 – 625 -1600 = 275
Resistance of the coil RC O = 5.5 Ω
Also X2C O  1600 - 5.52  1569.75
Reactance of the coil XC = 39.62 Ω
Impedance of the coil ZC O = (5.5 + j 39.62) = 40 82.10 Ω
Power absorbed by the coil PC O = 52 x 5.5 = 137.5 W
Total power PT = (52 x 25) + 137.5 = 762.5 W
Total impedance ZT = (30.5 + j 39.62) = 50 52.410Ω
I R C O  5 x 5.5  27.5 V ;
I XC O  5 x 39.62  198.1 V
Phasor diagram is shown in Fig. 22.
250 V
52.410
200 V
125 V
198.1 V
27.5 V
I
Fig. 22 Phasor diagram-Example 7
Example 8
When a resistor and a seriesly connected inductor coil, are supplied with 240 V, a
current of 3 A flows lagging behind the supply voltage by 37 0. The voltage across
the coil is 171 V. Find the value of the resistor, resistance and reactance of the
inductor coil.
XCO
RCO
R
Solution
3  - 370 A
171 V
240 V
Fig. 23 Example 8
Supply voltage E = 240 00 V;
Circuit impedance Z =
E
= 80  370 Ω = (63.8908 + j 48.1452) Ω
I
Thus R + RCO = 63.8908 Ω
For the coil, Z CO 
RCO =
Supply current I = 3  - 370 A
and
XCO = 48.1452 Ω
171
 57 Ω ; From impedance triangle of the coil
3
572  48.14522  30.5129 Ω
Value of resistor R = 63.8908 – 30.5129 = 33.3779 Ω
Resistance of coil RCO = 30.5129 Ω;
Reactance of coil XCO = 48.1452 Ω
Example 9
When a voltage of 100 V at 50 Hz is applied to choking coil 1, the current taken is
8 A and the power is 120 W. When the same supply is applied to choking coil 2,
the current is 10 A and the power is 500 W. Find the current and power when the
supply is applied to two coils connected in series.
Solution
Resistance R1 =
120
 1.875 Ω
82
Impedance Z 1 
Resistance R2 =
100
 12.5 Ω ;
8
Therefore X1 =
12.5 2  1.8752  12.3586 Ω
500
5Ω
10 2
Impedance Z 2 
100
 10 Ω ;
10
Total resistance RT = 6.875 Ω;
Therefore X2 =
102  5 2  8.6603 Ω
Total reactance XT = 21.0189 Ω
Total impedance ZT = (6.875 + j 21.0189) = 22.1147 71.890 Ω
Total current IT 
100
 4.5219 A
22.1147
Power PT = 4.52192 x 6.875 = 140.5771 W
Example 10
A resistance of 100 ohm is connected in series with a 50 µF capacitor. When the
supply voltage is 200 V, 50 Hz, find the (i) impedance, current and power factor (ii)
the voltage across resistor and across capacitor. Draw the phasor diagram.
Solution
Resistor R = 100 Ω;
10 6
Reactance of the capacitor XC =
= 63.662 Ω
2π x 50 x 50
Impedance Z = (100 – j 63.662) = 118.5447 - 32.480
Taking the supply voltage as reference, E = 200 00 V
200 0 0
E
0


1.6871

32.48
A
Current I =
Z 118.5447  32.480
Power factor = cos 32.480 = 0.8436 leading
Voltage across resistor VR = 100 x 1.6871 32.480 = 168.71 32.480 V
Voltage across capacitor VC = - j 63.662 x 1.6871 32.480 = 107.4042 - 57.520 V
Phasor diagram is shown in Fig. 24.
VR = 168.71 32.48 V
0
I =1.6871 32.480 A
32.480
900
E = 200 00 V
VC = 107.4042 - 57.520 V
Fig. 24 Phasor diagram - Example 10
Example 11
In a circuit, the applied voltage of 150 V lags the current of 8 A by 40 0. (i) Find the
power factor (ii) Is the circuit inductive or capacitive? (iii) Find the active and
reactive power.
Solution
Power Factor = 0.766 leading
Circuit is capacitive.
Active Power P = 150 x 8 x 0.766 = 919.2 W
Reactive Power Q = 150 x 8 x 0.6428 = 771.36 VAR
Example 12
Find the circuit constants of a two elements series circuit which consumes 700 W
with 0.707 leading power factor. The applied voltage is V = 141.4 sin 314 t volts.
Solution
P  700 W; V 
I 
141.4
 99.9849 V ; cos θ = 0.707; Since Power P = V I cos θ
2
700
 9.9025 A and
99.9849 x 0.707
Z 
99.9849
 10.0969 Ω
9.9025
From the impedance triangle
Resistance R = Z cos θ  10.0969 x 0.707  7.1385 Ω
Reactance XC = Z sin θ  10.0969 x 0.707  7.1385 Ω
Capacitance C =
1
 446.132 μ F
314 x 7.1385
Example 13
A series R-C circuit consumes a power of 7000 W when connected to 200 V, 50 Hz
supply. The voltage across the resistor is 130 V. Calculate (i) the resistance,
impedance, capacitance, current and p.f. (ii) Write the equation for the voltage
and current.
- j XC
R
Solution
130 V
(
130 2
) R  7000
R
Power P = 7000 W
200 V 50 Hz
Fig. 25 Circuit – Example 13
1302
Resistance R =
 2.4143 Ω ;
7000
Current I 
130
 53.8458 A
2.4143
Since 200 x 53.8458 x cos θ = 7000, p.f. = 0.65 leading; θ = 49.460
From impedance triangle, reactance XC = R tan θ = 2.4143 x 1.1691 = 2.8226 Ω
R
Impedance Z = (2.4143 – j 2.8226) Ω = 3.7143  - 49.460Ω
1
Capacitance C =
 1127.72 μF
2 π x 50 x 2.8226
Current I = 53.8458  49.460 A;
θ
Power Factor = 0.65 leading
Taking supply voltage as reference
v(t) =
2 x 200 cos (2π x 50 t)  282.8427 cos 314.16 t
i(t) =
2 x 53.8458 cos (2π x 50 t  49.460 )  76.15 cos ( 314.16 t  49.460 ) A
XC
Example 14
A coil of resistance 10 Ω and inductance 0.1 H is connected in series with a 150
μF capacitor across 200 V, 50 Hz supply. Calculate (i) inductive reactance,
capacitive reactance, impedance, current and power factor and (ii) the voltage
across the coil and capacitor.
0.1 H
10 Ω
150 μF
Solution
Data are marked in Fig. 26
200 V 50 Hz
Fig. 26 Circuit – Example 14
Inductive reactance XL = 2 π x 50 x 0.1 = 31.4159 Ω
Capacitive reactance XC =
106
 21.2207 Ω
2 π x 50 x 150
Impedance Z = (10 + j 31.4159 – j 21.2207) = (10 + j 10.1952) Ω = 14.2808  45.550 Ω
Taking supply voltage as reference
Current I =
200
 14.0048   45.55 0 A
14.280845.55
Power factor = 0.7003 lagging
Impedance ZCoil = (10 + j 31.4159) = 32.9691  72.340 Ω
Voltage VCoil = ZCoil x I = 461.7257  26.790 V
Voltage VCap = ZCap x I = - j 21.2207 x 14.0048  - 45.550 = 297.1916  - 135.550 V
Example 15
In the circuit shown in Fig. 27, the current leads the voltage by 50 0. Find value of
R and the voltages across each circuit element. Draw the phasor diagram.
10 m H
R
Solution
5 μF
Circuit is capacitive
Power factor angle θ = 500
200 V 500 Hz
Fig. 27
Taking the supply voltage as reference V = 200  00 volts
XL = 31.4159 Ω; XC = 63.662 Ω
R
Impedance Z = R + j 31.4159 – j 63.662 = R – j 32.2461
500
32.2461 Ω
From the impedance triangle
tan θ =
32.2461
 1.19175 ; Thus
R
Resistance R = 27.0578 Ω
Impedance Z = (27.0578 – j 32.2461) = 42.0944  - 500 Ω
Current I =
V
200
=
 4.7512 500 A
Z
42.0944  50
Voltage across R VR = R x I = 128.557  500 V
Voltage across L VL = j 31.4159 x 4.7512  500 = 149.2632  1400 V
Voltage across C VC = - j 63.662 x 4.7512  500 = 302.4708  - 400 V
Phasor diagram is shown in Fig. 28.
I
VL
VR
V
V = 200 0 0
II == 4.7512 500 A
VI R== 128.557 500 V
IV=L = 149.2632 1400 V
0
IV=C = 302.4708  - 40 V
I=
VC
Fig. 28 Phasor diagram – Example 15
Example 16
A 230 V, 50 Hz voltage is applied to a coil of L = 5 H and R = 2 Ω in series with a
capacitance C. What value must C have in order that the p.d. across the coil shall
be 250 V?
j XL
2Ω
Solution
Refer Fig. 29
- j XC
250 V
XL = 2 π x 50 x 5 = 1570.7963 Ω
230 V 500 Hz
Fig. 29 Circuit – Example 16
Coil impedance ZCO = (2 + j 1570.7963) = 1570.7975  89.930 Ω
I 
250
 0.1592 A ;
1570.7975
Therefore
ZT 
22 + XT2 = 1444.72362;
i.e. 1570.8 – XC = ± 1444.7; Thus
230
 1444.7236 Ω ;
0.1592
Thus XT = ± 1444.7 Ω
XC = 126.1 Ω
If XC = 126.1 Ω circuit is INDUCTIVE;
If XC = 3015.5 Ω circuit is CAPACITIVE;
C=
C=
or
XC = 3015.5 Ω
1
 25.243 μ F
2π x 50 x 126.1
1
 1.056 μF
2π x 50 x 3015.5
Example 17
A resistance R, an inductance L = 0.5 H and a capacitance C are connected in
series. When a voltage v = 350 cos(3000 t – 200) volt is applied to this series
combination, the current flowing is 15 cos(3000 t – 600) amperes. Find R and C.
Solution
Supply voltage V = 247.4873  - 200 Volts
Current I = 10.6066  -600 A
Inductive reactance XL = 3000 x 0.5 = 1500 Ω
Impedance Z = R + j (XL – XC) =
V 247.4873  20

 23.3333400 Ω
I
10.6066  60
= (17.8743 + j 14.9984) Ω
Thus R = 17.8743 Ω
XL – XC = 14.9984; Thus XC = 1500 – 14.9984 = 1485.0016 Ω
Therefore C =
1
 0.2245 μ F
3000 x 1485.0016
Example 18
Calculate the admittance Y, the conductance G and the susceptance B of a circuit
consisting of 10 Ω in series with an inductor of 0.1 H when the frequency is 50 Hz.
Solution
Inductive reactance XL = 2 π x 50 x 0.1 = 31.4159 Ω
Circuit impedance Z = (10 + j 31.4159) Ω
Circuit admittance Y =
1
= (0.0092 – j 0.0289) = 0.03033  -72.34 mho
Z
Thus Conductance G = 0.0092 mho; Susceptance B = 0.0289 mho Inductive
Example 19
An impedance of (7 + j 5) Ω is connected in parallel with another circuit having
impedance of (10 – j 8) Ω. The supply voltage is 230 V, 50 Hz. Calculate (i) the
admittance, conductance and susceptance of the combined circuit and (ii) the
total current taken from the mains and the p.f.
Solution
Impedance Z1 = (7 + j 5) Ω; Admittance Y1 = (0.0946 – j 0.0676) mho
Impedance Z2 = (10 - j 8) Ω; Admittance Y2 = (0.061 – j 0.0488) mho
Total admittance YT = Y1 + Y2 = (0.1556 – j 0.0188) mho = 0.1567  - 6.890 mho
Conductance G = 0.1556 mho; Susceptance B = 0.0188 mho Inductive
Taking the supply voltage as reference, V = 230  00 volts
Total current taken I = Y V = 36.041  - 6.890 A
p.f. = 0.9928 lagging
Example 25
Consider RLC series circuit with R = 100 Ω, L = 1.0 H and C = 1.0 µF. It is
connected to 500 V variable frequency supply. For a range of ω = 800 to 1200 rad.
per sec. in steps of 100 rad. per sec., compute the values of X L, XC, Z and I and
plot them. Mark the region of leading and lagging power factor. For ω = 1000 rad.
per sec., find the values of voltages across the inductance and capacitance.
Solution
The calculated values are shown in the table.
ω rad. / sec. XL Ω XC Ω
800
900
1000
1100
1200
800
900
1000
1100
1200
1250
1111
1000
909
833
Z Ω
Z Ω
I A
100 – j 450
100 – j 211
100
100 + j191
100 + 367
461
233
100
216
380
1.08
2.15
5.0
2.31
1.32
R = 100 Ω
L = 1.0 H
C = 1.0 μF
Z Ω
Z Ω
I A
100 – j 450
100 – j 211
100
100 + j191
100 + 367
461
233
100
216
380
1.08
2.15
5.0
2.31
1.32
ω rad. / sec. XL Ω XC Ω
800
900
1000
1100
1200
800
900
1000
1100
1200
1250
1111
1000
909
833
1300
*
1200
X
XC
o
XC
XL
I
*
1100
X
XL
1000
*
ΙZΙ
6.0
500
5.0
400
4.0
ΙIΙ
200
2.0
100
1.0
*
X
900
600
3.0
*
leading
p.f.
800
C = 1.0 μF
ΙZΙ
300
X
900
o
L = 1.0 H
o
X
800
R = 100 Ω
lagging
p.f.
1000
1100
Fig. 35 Resonance charecteristics
ω
1200
When ω = 1000 rad. per sec., XL = 1000 Ω and XC = 1000 Ω. Taking the supply
voltage as reference, current I = 5 0 0 A .
Therefore
VL = j 1000 x 5 = j 5000 V
VC = - j 1000 x 5 = - j 5000 V
Even though the supply voltage is 500 V, voltages across the inductance and
capacitance are 5000 V.
It is to be noted that VR = 100 x 5 = 500 V and VR + VL + VC = 500 V The phasor
diagram is shown in Fig. 36.
VL
VR
I
VC
Fig. 36 Phasor diagram
SERIES RESONANCE
An R – L – C series circuit is said to be at resonance when the applied voltage
and the resulting current are in phase. Thus at resonance condition, the
equivalent impedance of the circuit will be a resistance only. Since the supply
voltage and the circuit current are in phase, the power factor of the resonant
circuit is unity.
Consider an R – L – C series circuit connected to a variable frequency supply as
shown in Fig. 37.
R
C
L
Variable frequency
AC supply
R
j XL
- j XC
Variable frequency
AC supply
Fig. 37 Resonance Circuit
Impedance of the circuit Z = R + j (XL – XC)
(67)
While the frequency is increased from a low value, X L increases and XC
decreases. At one particular frequency, called resonance frequence, XL and XC
are equal and the total impedance will be equal to R. This angular frequency is
designated as ω0. For other frequencies
Z  R 2  (X L  X C ) 2
(68)
Thus at resonance frequency, the impedance is minimum and is equal to R and
the current is at the maximum value given by
E
R
. When ω < ω0 , XC > XL and
hence p.f. is of leading nature. When ω > ω 0 , XL > XC and hence the p.f. is lagging
in nature. At resonance frequency since the circuit impedance is of resistance
only, the p.f is unity. Expression for angular resonance frequency ω 0 can be
obtained as follows. At resonance condition, XL = XC. Therefore
ω0 L =
1
ω0 C
;
Thus ω0 =
Resonance frequency f0 =
1
rad. / sec.
LC
1
2π
1
LC
Hz
(69)
(70)
Resonance condition can be achieved by varying the values of either L or C also.
QUALITY FACTOR
In Example 26, it was found that, at resonance condition, voltage across L or C is
much larger than the supply voltage. The property of developing high voltage
during resonance condition is defined by QUALITY FACTOR, also referred as Q
Factor.
VL
Q Factor, Q =
Thus Q =
Also Q =
E
XL I
R
I
XC I
R

I

VC
E
at resonance condition.
XL ω0 L

R
R

XC
1

R
ω0 C R
(71)
(72)
(73)
BANDWIDTH
At resonance condition, the circuit current is maximum and is given by I0 =
E
R
.
The entire power input is absorbed by R and this power is given by P 0 = I02 R. For
all other frequencies around ω0, circuit current is less than I0 and hence the
power absorbed by the circuit will be less than P0. The power absorbed will be
when the circuit current is
P0
2
1
I0
2
BANDWIDTH is defined as that range of frequencies around the resonance
frequency ω0, within which the power absorbed by the circuit is greater than or
equal to
P0
, where P0 is the power absorbed by the circuit at resonance condition
2
OR within which the circuit current is greater than or equal to
circuit current at resonance condition.
I0
, where I0 is the
2
I
I0
I0
2
ω1
ω2
ω
Fig. 38 Bandwidth
Referring to Fig. 38,
Bandwidth ωB W = ω2 - ω1
It can be shown that ωB W =
(74)
R
L
(75)
Also from eq. (72), it can be seen that
Q=
ω0 L
R
=
ω0
ωB W
(76)
Example 26
A series R-L-C circuit with R = 10 Ω, L = 10 mH and C = 1 μF is connected to 200 V
variable frequency supply. Calculate the resonance frequency. Also find the
circuit current and the voltages across the elements. Determine the Quality factor
and Bandwidth.
Solution
ω02
1
106
=

 108 ; Therefore ω0 = 104 rad. / sec.
L C 0.01 x 1
Resonant frequency f0 =
ω0
= 1591.55 Hz.; At resonance XL = XC = ω0 L = 100 Ω
2π
Circuit current at resonance I0 =
200
= 20 A
10
Voltage across resistor VR = 10 x 20 = 200 V
Voltage across L and C:
VL  VC  XL I 0  100 x 20  2000 V
Quality factor Q =
X L 100

 10
R
10
Bandwidth ωB W =
R
10

 1000 rad. / sec. = 159.155 Hz
L 0.01
Example 27
A series R-L-C circuit is connected to a 200 V, 50 Hz supply. When L is varied, the
maximum current obtained is 0.4 A. At that condition, the voltage across the
capacitor is 330 V. Find the circuit constants.
Solution
Resistor R = 200 / 0.4 = 500 Ω
VL  VC  XL I 0  330 ; Therefore XL = XC = 330 / 0.4 = 825 Ω
ω0 = 2 π x 50 = 314.1593
Inductance L =
XL
825

 2.626 H
ω 0 314.1593
Capacitance C =
1
1

F  3.8583 μF
ω 0 X C 314.1593 x 825
Example 20
When a 240 V, 50 Hz supply is applied to a resistor of 15 Ω which is in parallel
with an inductor, total current is 22.1 A. What value must the frequency has for
the total current to be 34 A?
Solution
Current in resistor and
current in inductor will
have a phase difference
of 900
240 V Ω, 50 Hz
15 Ω
XL
Fig. 30 – Example 20
16 A
IR 
240
 16 A ;
15
Therefore XL =
162 + I L
2
= 22.12; Thus I L = 15.245 A
240
 15.7429 Ω ;
15.245
Thus L =
Ι IL Ι
15.245
 0.05011 H
2 π x 50
22.1 A
16 A
With new frequency
162 + I L
2
= 342;
New frequency f =
Thus I L = 30 A and XL =
8
 25.4089Hz
2 π x 0.05011
240
8Ω
30
Ι IL Ι
34 A
Example 21
A coil of resistance 15 Ω and inductance 0.05 H is connected in parallel with noninductive resistor of 20 Ω. Find (i) the current in each branch and the total current
supplied and (ii) the phase angle of the combination when a voltage of 200 V at 50
Hz is applied. Draw the phasor diagram.
Solution
X
L
= 2 π x 50 x 0.05
= 15.708 Ω
15 Ω
200 V Ω, 50 Hz
20 Ω
XL
Taking supply voltage as
reference
Fig. 31 – Example 21
200
 6.3594 - j 6.6596  9.2083  - 46.320 A
Current in the coil I1 =
15  j15.708
Current in non-inductive resistor I2 = 10 A
Total current I T  16.3594 - j 6.6596  17.663  - 22.150 A
Current lags the voltage by 22.150
0
46.32
I2
22.150
V
Fig. 32 – Phasor diagram - Example 22
IT
I1
Example 22
A coil of inductance 6 mH and resistance 40Ω is connected across a supply of
100 V, 800 Hz. Also across the supply is a circuit consisting of 4 μF in series with
50 Ω resistor. Find (i) the total current taken from the supply and (ii) the phase
angle between the currents in the coil and the capacitance. Draw the phasor
diagram.
Solution
X
L
= 2 π x 800 x 0.006
= 30.1593 Ω
40 Ω
100 V
800 Hz
6 mH
50 Ω
4 μF
6
XC =
10
 49.7359 Ω
2π x 800 x 4
Fig. 33 – Example 22
Z 1  40  j 30.1593  50.0957 37.020 Ω
Z 2  50  j 49.7359  70.5242   44.850 Ω
Current I1 =
100
 1.5938 - j 1.2019  1.9962  - 37.020 A
40  j 30.1593
Current I2 =
100
 1.0053 - j 1.0  1.418 44.85 0 A
50  j 49.7359
Total current IT = (2.5991 – j 0.2019) A = 2.6069  - 4.440 A
Angle between currents I1 and I2 = 81.870
I2
44.850
V (Ref)
0
4.44
0
IT
37.02
I1
Fig. 34 – Phasor diagram - Example 22
Example 23
Two circuits, with impedances Z1 = (10 + 15) Ω and Z2 = (6 – j 8) Ω are connected
in parallel. If the total current supplied is 15 A, find the power consumed by each
impedance.
Solution
Taking the supply current as reference, IT = 15 0 0 A
Current I1 =
6  j8
x 15  (1.9672  j 8.3607 )  8.589   76.760 A
16  j 7
Current I2 =
10  j15
x 15  (13.0328  j 8.3607 )  15.484 32.680 A
16  j 7
Power P1 = |I1|2 R1 = 8.5892 x 10 = 737.7092 W
Power P2 = |I2|2 R2 = 15.4842 x 6 = 1438.5255 W
Example 24
Two coils are connected in parallel across a voltage of 200 V, 50 Hz. The coils
have resistances of 10 Ω and 5 Ω and inductances of 0.023 H and 0.035 H
respectively. Find (i) current in each coil and total current and (ii) p.f. of the
combination.
Solution
XL1 = 2 π x 50 x 0.023 = 7.2257 Ω; XL2 = 2 π x 50 x 0.036 = 10.9956 Ω
Z1 = (10 + j 7.2257) Ω = 12.3374 35.850 Ω
Z2 = (5 + j 10.9956) Ω = 12.079 65.550 Ω
200 V
Current I1 = V / Z1 = 16.2109   35.850 A
50 Hz
Current I2 = V / Z2 = 16.5577   65.850 A
Total current IT = 31.6743   50.860 A
p.f. of the combination = 0.6312 lagging
10 Ω
5Ω
XL1
XL2
THREE PHASE SYSTEM
In general, generation, transmission and utilization of electric power is more
economical in three phase system compared to single phase system.
The windings of three phase alternators are designated as AA’, BB’ and CC’. The
voltages generated in these windings are
e AA'  E m cos ωt
e BB'  E m cos( ω t  1200 )
(77)
e CC'  E m cos( ω t  2400 )
The phasor descriptions of three voltages are shown in Fig. 39. Here
E AA' is
taken as reference. Each voltage phasor is lagging the previous one by 120 0 .
E CC '
E A A'  E0 0
EB B'  E  1200
E AA'
E BB'
E C C'  E  2400
E AA'  EBB'  E CC'  E 
Fig. 39 Phasor representation of 3 phase voltages
Em
2
Generally E AA' is written as E A . Other phasors are represented likewise. Thus
E A  E0 0
E B  E  1200
(78)
E C  E  2400
The three generator windings are connected either in STAR (wye) or in DELTA.
STAR CONNECTED GENERATOR
Fig. 40 shows the winding connections of star connected generator. The
generator is connected to a 3 phase load.
A
IA
EA
O
N
C
L
E AB
IB
B
A
D
IC
Fig. 40 Star connected generator
E A , E B and E C are called the PHASE VOLTAGES. E AB , E BC and E CA are called
the LINE VOLTAGES or line to line voltages. The current flowing in each phase is
called PHASE CURRENT
CURRENT.
and the current flowing in each line called LINE
Let Il and Ip h be the magnitude of line current and phase current and E l and
E p h be the magnitude of line voltage and phase voltage. In case of star connected
system
Line current = Phase current
i.e. Il = Ip h
(79)
Taking E A as the reference, the voltage phasors are shown in Fig. 41.
EC
- EB
E AB
E A  E ph 0 0
E B  E ph   1200
EA
EB
E C  E ph   2400
Fig. 41 Voltage phasors - Star system
The relationship between line voltage and phase voltage can be obtained as
follows.
E AB  E A  EB  Eph  Eph (0.5  j0.866)
 Eph (1.5  j0.866)
 3 Eph 300
The above result can be seen from the Fig. 41.
Similar expression can be obtained for E BC and E CA also. Collectively, we have
E AB 
3 E ph 30 0
E BC 
3 E ph   90 0
E CA 
3 E ph 1500
Thus E l = E AB
(80)
 EBC  E CA

3 Eph
Therefore for star connected system
El 
3 E ph
(81)
I l  Iph
Power supplied by the
three phase alternator
 3 x phase power
 3 E ph Iph cos θ
(82)

(83)
3 E l Il cos θ
Above results are true for star connected load also, except that the power is
consumed by the load.
DELTA CONNECTED GENERATOR
Delta connected generator is shown in Fig. 42.
C'
A
IA
ICA
I AB
A'
C
B'
IBC
L
O
IB
A
D
B
IC
Fig. 42 Delta connected generator
I A , I B and I C are called LINE CURRENTS. I AB , I BC and I CA are called PHASE
CURRENTS. The voltage across each phase is called PHASE VOLTAGE
and
voltage across two lines is called LINE VOLTAGE or line to line to line voltage.
In case of delta connected system, line voltage is equal to phase voltage. i.e.
E l = E ph
(84)
Taking I BC as reference, current phasors are shown in Fig. 43.
I AB
I BC
IBC  Iph 0 0
ICA  Iph   1200
I AB  Iph   2400
I CA
IA
Fig. 43 Current phasors – Delta connected system
Considering the junction point formed by A and C '
I A = I C A - I A B = I p h ( - 0.5 - j 0.866 ) - I p h ( - 0.5 + j 0.866 )
= -j
3 I ph
The above result can be seen from Fig. 43. Similar expression can be obtained for
I B and I C . Collectively, we have
IA 
3 Iph   90 0
IB 
3 Iph   2100
IC 
3 Iph 30 0
(85)
Therefore Il = I A  IB  IC 
3 Iph
Thus for delta connected system
E l = E ph
Il =
(86)
3 I ph
Power supplied by the
three phase alternator
 3 x phase power
 3 E ph Iph cos θ
(87)

(88)
3 E l Il cos θ
Above results are true for delta connected load also except that the power is
consumed by the load.
Working principle, construction and applications of DC Generator
The dc generator is rotating electrical machine which converts mechanical
energy into electrical energy. The generator is usually driven by a steam turbine
or water turbine which is called as prime mover.
The dc generator operates on the principle based on the Faraday’s Law of
electromagnetic induction. The generator should have (i) magnetic field (ii)
conductors capable of carrying current (iii) movement of conductors in the
magnetic field.
Necessary magnetic field is produced by field coil. The set of conductors in
which the voltage is induced, is called the armature.
The voltage induced in the coil will be as shown in Fig. 15.
Fig. 15 EMF induced in an armature coil
Depending on how the Armature and Field windings are connected, we have
different types of dc generators. They are shown in Fig. 18.
YY
Y
L
A
Z
O
O
G
L
A
G
A
A
AA
AA
D
Fig. 18 (b) Shunt generator
Fig. 18 (a) Series generator
Y
D
ZZ
YY
Y
A
Z
O
G
ZZ
L
Z
A
AA
D
ZZ
YY
L
A
O
A
G
D
Fig. 18 (c) Short shunt compounded
generator
AA
Fig. 18 (d) Long shunt compounded
generator
Application of dc generators
Shunt generators are used in supplying nearly constant loads. They are used for
charging batteries and supplying the fields of synchronous machines.
Series generators are used as boosters for adding voltage to transmission lines
to compensate for the line drop.
Cumulative compound generators are used for drives which require constant dc
voltage supply.
Differential compound generators are used in arc welding.
Working principle, construction and applications of DC Motor
Whenever a current carrying conductor is kept in a stationary magnetic field, an
electromotive force is produced. This force is exerted on the conductor and
hence is moved away from the field. This is the principle used in dc motors.
Construction of dc motor is exactly similar to dc generator.
In a dc motor, both the armature and the field windings are connected to a dc
supply. Thus, we have current carrying armature conductors placed in a
stationary magnetic field. Due to electromagnetic torque exerted on the armature
conductors, the armature starts revolving. Thus, electrical energy is converted
into mechanical energy in the armature.
When the armature is in motion, we have revolving conductors in a stationary
magnetic field. As per Faraday’s Law of electromagnetic induction, an emf is
induced in the armature conductors. As per Lenz’s law, this induced emf opposes
the voltage applied to the armature. Hence it is called back emf. There will be
small voltage drop due to armature resistance. Thus, the applied voltage has to
overcome the back emf in addition to supplying the armature voltage drop. The
input power is used to produce necessary torque for the continuous rotation of
the armature.
Depending on how the Armature and Field windings are connected, we have
different types of dc motors. They are shown in Fig. 19.
Depending on how the Armature and Field windings are connected, we have
different types of dc motors. They are shown in Fig. 19.
+
+
YY
YY
YY
DC
supply
voltage
A
DC
supply
voltage
M
M
Z
M
AA
AA
M
Fig. 19 (a) Shunt motor
Fig. 19 (a) Series motor
YY
+
Y
+
DC
supply
voltage
A
-
Z
G
AA
ZZ
-
-
Y
A
DC
supply
voltage
YY
Z
A
ZZ
G
ZZ
AA
Fig. 19 (c) Short shunt compounded motor
Fig. 19 (d) Long shunt compounded motor
Application of dc motors
DC series motors are used in electric trains, cranes, hoists, conveyors etc. where
high starting torque is required.
Shunt motors are used where the speed has to remain constant under loaded
condition.
Compound motors are used for driving heavy tools for intermittent heavy loads
such as rolling mills, printing machines etc.
Working principle, construction and applications of 1- phase transformer
The transformer works on the principle of electromagnetic induction. The
induced emf in a transformer comes under the classification of statically induced
emf.
The transformer is a static apparatus used to transfer electrical energy from one
circuit to another. The two circuits are magnetically coupled. One of the circuits,
namely Primary, is energized by connecting it to an ac supply at specific voltage
magnitude, frequency and waveform. Then we have a mutually induced voltage
available across the second circuit, namely Secondary, at the same frequency
and waveform but with a desired voltage magnitude. These aspects are indicated
in Fig. 20.
EMF induced in primary side E1 = N1
dφ
dt
Since same flux is linking both the primary and secondary coils
EMF induced in primary side E2 = N2
Voltage ratio
dφ
dt
E 1 N1

E 2 N2
Since losses in the transformer are very less, Voltampere in both the sides are
equal. i.e.
E 1 I1 = E 2 I2
Then the current ratio
I 1 E 2 N2


I 2 E1 N1
Apart from primary and secondary windings, transformer has a good magnetic
core.
The transformer core is generally laminated and is made out of a good magnetic
material such as transformer steel or silicon steel. Such a material has high
relative permeability and low hysteresis loss. There are two types of transformer
cores. They are known as Core Type and Shell type. In core type transformer, L –
shaped stampings as shown in Fig. 21 are used. One core type transformer is
shown in Fig. 22.
Fig. 21 L – type stampings
Laminated core of a shell type transformer is shown in Fig. 23. In this E – type
and I type laminations are used. Fig. 24 shows a shell type transformer.
Fig. 23 Laminated core of shell type transformer
Application of transformers
The transformers are classified as Step-up transformers and Step-down
transformers. When the secondary voltage is more than the primary voltage,
transformer is called a step-up transformer. In step-down transformer, the
secondary voltage is less than the primary voltage.
Transformers are used in the following applications:
(i)
Power transformers located in Power Plants are used to step-up the
generated voltage to a high transmission voltage.
(ii)
Transformers are used in distribution circuit to step-down voltages to
the desired level.
(iii)
Almost all electronic circuits use transformers.
(iv)
Potential transformers are used to measure high voltages and current
transformers are used to measure high currents.
(v)
Furnace transformers and welding transformers are some special
applications of transformers.
Working principle, construction and applications of 3- phase
induction motor
When a three phase balanced voltage is applied to a three phase
balanced winding, a rotating magnetic field is produced. This field
has a constant magnitude and rotates in space with a constant speed.
If a stationary conductor is placed in this field, an emf will be induced
in it. By creating a closed path for the current to flow, an
electromagnetic torque can be exerted on the conductor. Thus the
conductor is put in rotation.
A three phase balanced voltage is applied across the three phase balanced stator
winding. A rotating magnetic field is produced. This magnetic field completes its
path through the stator, the air gap and the rotor. The rotor conductors, which
are stationary at the time of starting, are linked by time varying magnetic field.
Therefore emf is induced in the rotor conductors. Since the rotor circuit forms a
closed path, rotor current is circulated. Thus the current carrying conductors are
placed in a rotating magnetic field. Hence an electromotive force is exerted on
the rotor conductors and the rotor starts rotating.
According to Lenz’s law, the nature of the induced current is to oppose the cause
producing it. Here the cause is the relative motion between the rotor conductors
and the rotating magnetic field. Hence the rotor rotates in the same direction as
that of the rotating magnetic field.
In practice, the rotor speed never equals to the speed of the rotating magnetic
field. The difference in the two speeds is called the slip. The current drawn by the
stator gets adjusted according to the load on the motor.
Three phase induction motors are used in industry for very many purposes. They
are used in lathes, drilling machines, agricultural and industrial pumps,
compressors and industrial drives.
The important parts of a three phase induction motor are schematically
represented in Fig. 25. Broadly classified, they are stator and rotor which are
described below.
Stator is the stationary part of the motor. The stator core consist of high grade,
low loss electrical sheet-steel stampings assembled in the frame. Slots are
provided on the inner periphery of the stator to accommodate the stator
conductors. Required numbers of stator conductors are housed in the slots.
These conductors are arranged to form a balanced three phase winding. The
stator winding may be connected in star or delta.
Rotor is the rotating part of the induction motor. The air gap between the stator
and rotor is as minimum as possible. The rotor is also in the form of slotted
cylindrical structure. There are to types of rotors, namely Squirrel Cage rotor and
Slip-ring or Wound rotor.
Fig. 26 shows the construction of a squirrel cage rotor.
Fig. 26 Squirrel cage rotor of three phase induction
motor
In this type, each rotor slot accommodates a rod or bar made of good conducting
material. These rotor bars are short circuited at both ends by means of end rings
made of the same metal as that of rotor conductors. Thus the rotor circuit forms a
closed path for any current to flow through.
Fig. 27 shows the rotor of slip-ring induction motor. In this case conductors are
housed in rotor slots. These conductors are connected to form a star connected
balanced three phase winding. The rotor is wound to give same number of poles
as the stator. The three ends of the rotor winding are connected to the three sliprings. The brushes are riding over the slip-rings. Slip-rings are short circuited at
the time of starting. External resistances can be connected to control the speed
of the motor. Although the wound rotor motor costs more than a squirrel cage
motor, it has the features of controlling the torque and the speed.
Starting
resistance and
speed
controller
Fig. 27 Rotor of slip-ring induction motor
Working principle, construction and applications of single phase induction motor
Single phase induction motors are used in variety of applications at home,
factory, office and business establishments. Single phase induction motor is not
self starting. Additional arrangement has to be made to make it self-starting. This
could be achieved by using two windings, main winding and starting winding,
with large phase difference between the currents carried by them. This kind of
split-phase motor produces a revolving flux and hence makes the motor self
starting. Depending on the circuit element connected in series with the starting
winding, the split-phase motors are classified into
(i)
Resistance-start induction motor
(ii)
Capacitance-start induction motor
(iii)
Capacitance-start-and-run motor
Resistance-start induction motor
Im
Main winding
Starting winding
Single phase
a.c. supply
Is
S
Rotor
Fig. 28 Resistance start induction motor
Resistance start induction motor is shown in Fig. 28. The starting winding has a
high resistance connected in series with it. The current flowing through it is given
by Is. The main winding has low resistance and high reactance and it carries
current Im. Current in starting winding is Is. The torque developed by the motor is
proportional to sin α where α is the angle between Im and Is as shown in Fig. 29.
For obtaining high torque, angle α should be as high as possible. Here θ is the
power factor angle. The centrifugal switch S disconnects the starting winding
when the motor speed reaches 80% of full load speed.
V
α
Is
θ
Im
I
Fig. 29 Phasor diagram of Resistance start induction motor
Capacitor-start induction motor
In the capacitor-start induction motor, a capacitor is connected in series with the
starting winding as shown in Fig. 30.
Im
Main winding
Starting winding
Single phase
a.c. supply
Is
C
Rotor
S
Fig. 30 Capacitor start-induction motor
Is
α
θ
V
I
Im
Fig. 31 Phasor diagram of capacitor-start induction motor
The phasor diagram of capacitor-start induction motor is shown in Fig. 31.
The following are the advantages of capacitor-start induction motor:
(i)
(ii)
Increase in starting torque
Better starting power factor
Capacitor-start-and- run motor
Capacitor-start-and-run motor is similar to that of the capacitor-start motor
except that the capacitor in the starting winding circuit remains there through out
the operation of the motor. The advantages of this type of motor are:
(i)
Low noise in the motor while running
(ii)
Higher power factor
(iii)
Higher efficiency
(iv)
Improved over-load capacity
HOUSE WIRING
House wiring deals with the distribution system arranged within the domestic
premises. Wiring requirement varies with customer to customer. House wiring
generally done on either 230 V single phase or 400 V three phase supply. In the
latter case, total load is divided among the three phases. An earth wire is also run
connecting all the power plugs from where large quantity of electrical energy is
tapped by using electrical appliances like heater, electric iron, hot plate, air
conditioner etc.
Wiring materials and accessories
The following are the wiring material used for house wiring:
Switches
Lamp holders
Socket out-lets
Ceiling roses
Switch boards
Miniature circuit breaker
Wires
Conduits
Fuse unit
The accessories used for house wiring are:
Screw driver
Hammer
Cutting pliers
Drilling machine
Nose pliers
Test lamp
Wood saw
Wire stripper
Hack saw
Knife
Types of wiring
The type of wiring depends on environment, durability, safety, appearance and
cost.
Cleat Wiring: In this system, V I R (Vulcanised India Rubber) conductor are
supported in porcelain cleats. It is much cheaper; but will not provide good
appearance.
Wooden Casing Capping: This system is more commonly used. It consists of
rectangular wooden blocks, called casing. It has two grooves into which the
wires are laid. Two or three wires of same polarity may be run in one groove.
Wires of opposite polarity are not run in the same groove. The wooden casing at
the top is covered by means of capping and is screwed on it. Nowadays the
wooden casing and cappings are replaced by plastic to give good appearance
and long life.
Conduit Wiring: In this system of wiring, V I R conductors are run inside metallic
pipes called conduit. The conduits are buried into the walls. This system of wiring
provides mechanical protection and good appearance. Nowadays instead of
metal, PVC pipes are used.
Staircase Wiring
In staircase wiring a single lamp, placed at the middle of the staircase, is
controlled by switches at two places, one at the beginning of the staircase and
the other at the end of the staircase. For this purpose two-way switches are
required. The wiring circuit is shown below.
L
N
P
1
S1
2
1
2
S2
Position of switch S1 Position of switch S2 Condition of lamp
1
1
ON
1
2
OFF
2
1
OFF
2
2
ON
Corridor Wiring
The diagram for corridor wiring is shown below.
S3
S1
S2
2
S4
2
P
1
1
L1
L2
N
Moving from left to right:
Enters
Closes S1
L1 ON
Reaches S2
Put S2 to 2
L1 OFF and L2 ON
Reaches S3
Put S3 to 2
L2 OFF and L3 ON
Reaches S4
Opens S4
L3 OFF
Moving from right to left:
Enters
Closes S4
L3 ON
Reaches S3
Put S3 to 1
L2 ON and L3 OFF
Reaches S2
Put S2 to 1
L1 ON and L2 OFF
Leaves
Opens S1
L1 OFF
L3
Incandescent lamp
This works on the principle that any hot body radiates energy. An electric current
passes through a thin filament, heating it to a temperature that produces light.
The enclosing glass bulb contains either a vacuum or an inert gas to prevent
oxidation of the hot filament.
Incandescent bulbs are made in a wide range of sizes and voltages, from 1.5 volts to
about 300 volts. They require no external regulating equipment and have a low
manufacturing cost, and work well on either alternating current or direct current. As a
result the incandescent lamp is widely used in household and commercial lighting, for
portable lighting such as table lamps, car headlamps, and flashlights, and for decorative
and advertising lighting.
Fluorescent lamp
A fluorescent lamp or fluorescent tube is a gas-discharge lamp that uses
electricity to excite mercury vapor. The excited mercury atoms produce shortwave ultraviolet light that then causes a phosphor to fluoresce, producing visible
light. A fluorescent lamp converts electrical power into useful light more
efficiently than an incandescent lamp. Lower energy cost typically offsets the
higher initial cost of the lamp. The lamp is more costly because it requires a
ballast to regulate the flow of current through the lamp.
While larger fluorescent lamps have been mostly used in commercial or institutional
buildings, the compact fluorescent lamp is now available in the same popular sizes and
is used as an energy-saving alternative in homes.
Sodium vapor lamp
A Sodium vapor lamp is a gas discharge lamp which uses sodium in an excited
state to produce light. There are two varieties of such lamps: low pressure and
high pressure. Because sodium vapor lamps cause less light pollution than
mercury-vapor lamps, many cities that have large astronomical observatories
employ them.
Mercury vapor lamp
A mercury-vapor lamp is a gas discharge lamp that uses mercury in an excited
state to produce light. The arc discharge is generally confined to a small fused
quartz arc tube mounted within a larger borosilicate glass bulb. The outer bulb
may be clear or coated with a phosphor; in either case, the outer bulb provides
thermal insulation, protection from ultraviolet radiation, and a convenient
mounting for the fused quartz arc tube.
Mercury vapor lamps (and their relatives) are often used because they are
relatively efficient. Phosphor coated bulbs offer better color rendition than either
high- or low-pressure sodium vapor lamps. Mercury vapor lamps also offer a very
long lifetime, as well as intense lighting for several special purpose applications.
Earthing
Earthing provides safe discharge of electric current due to leakages and faults to
ground.
All metallic parts of electrical appliances shall be connected by earth wire made
of very good conductor and finally the earth wire is connected to ground.
Earthing can be done through G.I. pipe or G.I. plate buried in the ground and
surrounded by charcoal and common salt to provide good conductivity. To
ensure safety earth resistance should be checked now and then and it is kept at a
very low value.