As per Revised Syllabus of
Visvesvaraya Technological University
®
Basic Electrical
Engineering - at a Glance
(One day Revision Book)
Uday A. Bakshi
M.E. (Electrical)
Formerly Lecturer in Department of Electronics Engg.
Vishwakarma Institute of Technology
Pune
Mrs. Varsha U. Bakshi
B.E. (Electronics)
Assistant Director,
Noble Institute of Computer Training
Pune
®
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TT able of Contents
Module - 1
®
Chapter - 1
D.C. Circuits
(1 - 1) to (1 - 22)
Chapter - 2
Electromagnetism
(2 - 1) to (2 - 18)
Module - 2
Chapter - 3
D.C. Machines
(3 - 1) to (3 - 28)
Chapter - 4
Measuring Instruments
(4 - 1) to (4 - 8)
Module - 3
Chapter - 5
Single Phase A.C. Circuits
(5 - 1) to (5 - 40)
Chapter - 6
Domestic Wiring
(6 - 1) to (6 - 12)
Module - 4
Chapter - 7
Three Phase Circuits
(7 - 1) to (7 - 24)
Chapter - 8
Synchronous Generators (Alternators)
(8 - 1) to (8 - 12)
Module - 5
Chapter - 9
(9 - 1) to (9 - 16)
Transformers
Chapter - 10 Three Phase Induction Motor
(ii)
(10 - 1) to (10 - 14)
1
D.C. Circuits
Chapter at a Glance
1.
Relation between Charge and Current
I=
Where
I
=
Q
t
Amperes
Average current flowing
while
Q = Total charge transferred
t = Time required for transfer of charge.
2.
Electric Potential and Potential Difference
Electrical Potential =
3.
Resistance
4.186 joules = 1 calorie
So finally,
Where
4.
R =
and
1 joule = 0.24 calorie
rl
a
l = Length in metres,
a = Cross-sectional area in square metres
r = Resistivity in ohms-metres,
R = Resistance in ohms
Ohm's Law
· Ohm's Law is, I =
5.
Work done W
=
Charge
Q
V
amperes
R
or
V = IR volts or
V
= Constant = R ohms
I
Electrical Work
Electrical work W = V ´ Q
\
W = VIt
J But I =
J
Q
t
Where t = Time in seconds
(1 - 1)
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6.
1-2
D.C. Circuits
Electrical Energy
\
Electrical energy E = Power ´ Time = V I t
joules
1 Wh = 1 watt ´ 1 hour = 1 watt ´ 3600 sec = 3600 watt-sec i.e. J
and
7.
3
6
1 kWh = 1000 Wh = 1 ´ 10 ´ 3600 J = 3.6 ´ 10 J
Current Division in Parallel Circuit of Resistors
\
é R1 ù
I2 = ê
ú IT
ë R1 + R 2 û
\
é R2 ù
I1 = ê
ú IT
ë R1 + R 2 û
IT
I2
I1
V
+
_
–
R1
R2
Fig. 1.1
Important Theory Questions and Answers
Ø
+
Explain the ideal and practical voltage source.
VTU : Mar.-02, Marks 4
· Ideal voltage source is defined as the energy source which gives constant voltage across
its terminals irrespective of the current drawn through its terminals.
IL
+
–
+
Vs
–
Load
Vs
VL
_
(a) Symbol
VL
+
VL = V s
Vs
IL
0
(b) Circuit
(c) Characteristics
Fig. 1.2 Ideal voltage source
· But practically, every voltage source has small internal resistance shown in series with
voltage source and is represented by R se as shown in the Fig. 1.3.
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1-3
D.C. Circuits
· Because of the R se , voltage across terminals decreases with increase in the load current
and it is given by expression,
VL = – (R se ) I L + Vs = Vs – IL Rse
· For ideal voltage source,
possible.
Internal
resistance Rse
+
–
Vs
Rse = 0 and for practical voltage source
IL
RL
VL
+
VL
_
it is as small as
Ideal
Vs
Practical
When there is no
load, IL = 0 and
VL = Vs
IL
0
(a) Circuit
(b) Characteristics
Fig. 1.3 Practical voltage source
Ø
+
Explain the ideal and practical current source.
VTU : Aug.-01, 03, Marks 5
· Ideal current source is the source which gives constant current at its terminals
irrespective of the voltage appearing across its terminals.
IL
Is
Is
Load
VL
_
(a) Symbol
IL
+
Is
IL = Is
VL
0
(b) Circuit
(c) Characteristics
Fig. 1.4 Ideal current source
· But practically, every current source has high internal resistance, which is in parallel
with current source and it is represented by R sh . This is shown in the Fig. 1.5.
· Because of R sh , current through its terminals decreases slightly with increase in voltage
at its terminals.
· For ideal current source, R sh = ¥ while for practical current source it is as high as
possible.
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1-4
Internal
resistance
Is
IL
Ish
Rsh
Load
IL
+
VL
_
D.C. Circuits
Ideal
Is
Practical
VL
(b) Characteristic
0
(a) Circuit
IL + Ish = Is
Thus as Ish
increases, IL
decreases.
IL < I s
Fig. 1.5 Practical current source
Ø
+
State Ohm's law and its limitations.
VTU : Jan.-08, July-08, Marks 6
· The Ohm's law gives relationship between the potential difference (V), the current (I)
and the resistance (R) of a d.c. circuit.
· It states that, the current flowing through the electric circuit is directly proportional to
the potential difference across the circuit and inversely proportional to the resistance of
the circuit, provided the temperature remains constant.
V
I µ
Where I is the current flowing in amperes, the V is the
· Mathematically,
R
voltage applied and R is the resistance of the conductor, as
R
shown in the Fig. 1.6.
+
–
· The Ohm's law can be defined as, the ratio of potential
difference (V) between any two points of a conductor to the
current (I) flowing between them is constant, provided that
the temperature of the conductor remains constant.
I
V
Fig. 1.6 Ohm's law
· The limitations of the Ohm's law are,
1) It is not applicable to the nonlinear devices such as diodes, zener diodes, voltage
regulators etc.
2) It does not hold good for non-metallic conductors such as silicon carbide. The law for
such conductors is given by,
V = k Im
Ø
where k, m are constants.
+ VTU : July-07, 11, Marks 6;
Jan.-09, 11, 13, June-13, Marks 5
State and explain Kirchhoff's laws.
· There are two Kirchhoff's laws.
1. Kirchhoff's Current Law (KCL)
· The law can be stated as,
The total current flowing towards a junction point is equal to the total current
flowing away from that junction point.
· Another way to state the law is,
The algebraic sum of all the current meeting at a junction point is always zero.
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1-5
D.C. Circuits
· The word algebraic means considering the signs of various currents.
åI
at junction point = 0
Sign convention : Currents flowing towards a junction point are assumed to be positive
while currents flowing away from a junction point assumed to be negative.
· Consider a junction point in a complex network as shown in the
Fig. 1.7. The currents I1 and I 2 are positive as entering the
junction while I 3 and I 4 are negative as leaving the junction.
· Applying KCL,
I3
O
å I at junction O = 0
I1 + I 2 - I 3 - I 4 = 0
I2
I1
I4
I1 + I 2 = I 3 + I 4
i.e.
Fig. 1.7 Junction point
· The law is very helpful in network simplification.
2. Kirchhoff's Voltage Law (KVL)
"In any network, the algebraic sum of the voltage drops across the circuit elements of any
closed path (or loop or mesh) is equal to the algebraic sum of the e.m.f. s in the path"
Ø
In other words, "the algebraic sum of all the branch voltages, around any closed path or
closed loop is always zero."
Around a closed path
åV= 0
Important Solved Examples
Example 1.1 Find the equivalent resistance across the terminals PQ of the network shown in the
+
Fig. 1.8.
P
50 W
50 W
100 W
50 W
100 W
100 W
100 W
Q
Fig. 1.8
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VTU : June-83
Basic Electrical Engineering
Solution :
1-6
D.C. Circuits
Replacing the lowest parallel combination of 100 W we get,
P
P
50 W
50 W
50 W
100 W
50 W
Series
50 W
»
50 W
100 W
100 W
100 W
100 W
Q
Q
Parallel
100´100
= 50 W
100+100
Fig. 1.8 (a)
P
50 W
P
50 W
Parallel
100´100
= 50 W
100+100
50 W
100 W
Series
50 W
»
100 W
100 W
Q
Q
Fig. 1.8 (b)
P
50 W
RPQ = 50 + 50 = 100 W
RPQ
Series
50 W
Q
Fig. 1.8 (c)
Example 1.2 Two voltmeters A and B, having resistances of 5.2 kW and 15 kW respectively are
connected in series across 240 V supply. What is the reading on each voltmeter ?
+
Series
Solution : The arrangement
is shown in the Fig. 1.9.
\ R eq = RA + R B
= 5.2 + 15 = 20.2 kW
VTU : June-81
I
Voltmeter
B
Voltmeter
A
»
240 V
RB
5.2 kW
VA
15 kW
VB
240 V
Fig. 1.9
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I
RA
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Basic Electrical Engineering
1-7
\
I=
D.C. Circuits
V
240
= 0.01188 A
=
R eq
20.2 ´ 10 3
According to Ohm's law,
VA = I ´ RA = 0.01188 ´ 5.2 ´ 10 3 = 61.7821 V
and
VB = I ´ R B = 0.01188 ´ 15 ´ 10 3 = 178.2179 V
Thus reading on voltmeter A is 61.7821 V and that on B is 178.2179 V.
Example 1.3 Find
the
equivalent
resistance
3W
between the two points A and B shown in the
4W
2W
Fig. 1.10.
4W
Solution : Identify combinations of series
and parallel resistances.
The resistances 5 W and 6 W are in series, as
going to carry same current.
1W
A
5W
6W
B
7W
So equivalent resistance is 5 + 6 = 11 W
Fig. 1.10
While the resistances 3 W , 4 W, and 4 W are in
parallel, as voltage across them same but current divides.
\ Equivalent resistance is,
1
1 1 1 10
= + + =
R
3 4 4 12
\
R =
12
= 1.2 W
10
Replacing these combinations redraw the figure as shown in the Fig. 1.10 (a).
Now again 1.2 W and 2 W are in series so equivalent resistance is 2 + 1.2 = 3.2 W while
11 W and 7 W are in parallel.
Using formula
R1 R2
11 ´ 7
77
equivalent resistance is
=
= 4 . 277W .
R1 + R2
11 + 7
18
Replacing the respective combinations redraw the circuit as shown in the Fig. 1.10 (b).
2 W Series 1.2 W
Parallel
1W
A
11 W
B
A
1W
3.2 W
4.277 W
7W
Parallel
Fig. 1.10 (a)
Fig. 1.10 (b)
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B
Basic Electrical Engineering
1-8
D.C. Circuits
Now 3.2 and 4.277 are in parallel.
\
3 . 2 ´ 4 . 277
= 1.8304 W
3 .2 + 4 . 277
Replacing them by
\
RAB = 1+ 1.8304 = 2.8304 W
Example 1.4
Find the current in all the branches of the network shown in the Fig. 1.11.
+
VTU : Aug.-95
80 A
A
30 A
F
0
W
.02
0.0
60 A
2W
0.01 W
E
70 A
0.0
B
0.01 W
1W
0.0
3W
C
60 A
D
120 A
Fig. 1.11
Solution : Let current through the branch AB be
I amperes.
Hence applying KCL at various nodes, the
various branch currents can be obtained as
shown in the Fig. 1.11 (a).
80 A
I–80
30 A
A
– +
I
60 A
F
+
–
–
+ B
I–50
I–60
–
+
C
E +
–
Applying KVL to the loop ABCDEFA,
70 A
I–120
+–
D
I
120 A
Fig. 1.11 (a)
- I ´ 0.02 - (I - 60) ´ 0.01 - I ´ 0.03 - (I - 120) ´ 0.01 - (I - 50) ´ 0.01 - (I - 80) ´ 0.02 = 0
\ - I [0.02 + 0.01 + 0.3 + 0.01 + 0.01 + 0.02] + 0.6 + 1.2 + 0.5 + 1.6 = 0
\
\
- 0.1 I + 3.9 = 0
I = 39 A
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60 A
Basic Electrical Engineering
1-9
D.C. Circuits
Hence the various branch currents are,
Branch
Example 1.5
Current
Direction
AB
39 A
from A to B
BC
– 21 A
from C to B
CD
39 A
from C to D
DE
– 81 A
from E to D
EF
– 11 A
from F to E
FA
– 41 A
from A to F
Find the value of R and the current flowing through it in the network shown in
the Fig. 1.12, when the current in the branch OA is zero.
+
A
4W
W
1W
B
1.5
0A
VTU : Oct.-85; Mar.-94
R
O
1.5 W
10 V
C
Fig. 1.12
Solution : Step 1 : The circuit diagram is given.
A
Step 2 : Mark all the branch currents
I1-I2
4W
1.5
Step 3 : Mark all the voltage polarities
O
I1-I2
B
1
I2
I2
I1
0A
10 V
W
The step 2 and 3 are combined and shown in the
Fig. 1.12 (a).
R
I1-I2
I2
1.5 W I1
C
Fig. 1.12 (a)
Step 4 : Apply KVL to various loops.
Loop AOCA, -1.5 (I 1 - I 2 ) + I 2 R + 0 = 0
i.e.
-1.5 I 1 + I 2 (1.5 + R) = 0
…(1)
Loop AOBA,
0 + I 2 ´ 1 - 4 (I 1 - I 2 ) = 0
i.e.
- 4 I1 +5 I2 = 0
…(2)
Loop BOCB
-I 2 ´ 1 - I 2 R - 1.5 I 1 + 10 = 0
i.e.
- 1.5 I 1 - I 2 (1 + R) = –10
…(3)
From equation (2),
I1 =
5
I = 1.25 I 2
4 2
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…(4)
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1 - 10
D.C. Circuits
Substituting in equation (1) we get,
-1.5 (1.25 I 2 ) + I 2 (1.5 + R) = 0
- 1.875 I 2 + I 2 (1.5 + R) = 0
i.e.
\
-1.875 I 2 = - I 2 (1.5 + R)
R = 0.375 W
\
Substituting in equation (3) we get,
1.5 + R = 1.875
i.e.
-1.5 (1.25 I 2 ) - I 2 (1 + 0.375) = –10
\
- 3.25 I 2 = –10
i.e.
I 2 = + 3.0769 A
… Current through R
Example 1.6 Find the VCE and VAG for the circuit shown in Fig. 1.13.
A
B
E
F
6W
+
8W
–
+
5W
20 V
10 V
5W
–
+
40 V
–
7W
9W
D
C
H
G
Fig. 1.13
Solution : Assume the two currents as shown in the Fig. 1.13 (a)
6W
A I1
+
20 V
B
–
I1
+
+
–
–
–
D
–
I2
+
9W
+
–
+
I1 C
H I
2
G
7W
Applying KVL to the two loops,
and - 8 I 2 - 5 I 2 - 7 I 2 + 40 = 0
- 6 I 1 - 5 I 1 - 9 I 1 + 20 = 0
\
I 1 = 1 A and I 2 = 2 A
i) Trace the path C-E,
\
VCE = – 5 V = 5 V with C negative
ii) Trace the path A-G,
C
5W
+
10 V
(5I2)
– 10 V +
–
B
H
5W
Fig. 1.13 (b)
\
VAG = 30 V with A positive
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40 V
I3
–
Fig. 1.13 (a)
(5I1)
– 5 V+
F
+
+
5W
–
+ 10 V
–
5W
I2
8W
E
E
Basic Electrical Engineering
1 - 11
D.C. Circuits
Example 1.7 Find the current in the branch A - B in the d.c. circuit shown in the Fig. 1.14,
using Kirchhoff's laws.
16 A
A
1W
1W
5A
4A
1W
1W
1W
B
7A
Fig. 1.14
Solution : The various branch currents are shown in the Fig. 1.14 (a).
Applying KCL at
various nodes
16 A
A
16 –I1–I2
I2
D
5A
–
I2
+
+
1W
I1
1W
–
16 –I1–I2
+
1W
I2–5
C
–
+
4
+
–
1W
I2–5
I1
1W
–
(16 – I1– I2 – 4) = (12 – I1– I2)
B
7A
Fig. 1.14 (a)
Applying KVL to loop ADBA
–I2 – (I2 – 5) + I1 = 0
\
I1 – 2I2 = –5
Applying KVL to the loop ACBA,
– (16 – I1 – I2) – (12 – I1 – I2) + I1 = 0
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Basic Electrical Engineering
1 - 12
D.C. Circuits
\ – 16 + I1 + I2 – 12 + I1 + I2 + I1 = 0
3I1 + 2I2 = 28
\
… (2)
Add (1) and (2), 4I1 = 23
\
I1 = 5.75 A
... This is the current through branch AB.
Example 1.8 Find the value of 'R' so that 1 A would flow in it, for the network in the Fig. 1.15.
R
1W
6W
+
10 W
1A
12 V
2W
–
Fig. 1.15
Solution : The various branch currents are
shown in the Fig. 1.15 (a).
Loop ABGH,
i.e.
– I2 – 12 – 10 I1 = 0
10 I1 + I2 = – 12
I1
A
… (1)
+
+
+
–
– 6 (I1 – I2 – 1) – 2 (I1 – I2) + 12 + I2 = 0
– 8 I1 + 9 I2 = – 18
1
–
10
Loop BCEFGB,
i.e.
+
+
I2
–
H
… (2)
I1
R
I1 – I2 C 1
B
D
–
–
I1 – I2 –1 6
+
2W
–
12 V
G
F
Fig. 1.15 (a)
Solving, I1 = – 0.9183 A and I2 = – 2.8163 A
\ Current through 6 W = I1 – I2 – 1 = 0.898 A
\ Drop across 6 W = 6 ´ Current through 6 W = 6 ´ 0.898 = 5.388 V
Same is drop across R = R ´ 1 = 5.388 as
R = 5.388 W
\
Example 1.9
For the circuit shown below, find the current
through each of the three resistors.
+
VTU : Dec.-12, Marks 10
Solution : Step 1 : Show the branch currents.
Step 2 : Apply KVL to the two loops.
–20I 1 + 20 – 100(I 1 + I 2 ) = 0
20 W
100 W
20 V
10 W
60 V
Fig. 1.16
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1 - 13
i.e. 120I 1 + 100I 2 = 20
D.C. Circuits
… (1)
20 W
+100(I 1 + I 2 ) – 60 + 10I 2 = 0
i.e. 100I 1 + 110I 2 = 60
Step 3 :
+
I1
–
100 W
… (2)
–
I2
Solving, I 1 = – 1.1875 A, I 2 = 1.625 A
+
\Current through 10 W = I 2 = 1.625 A ®
I
+
–
I1 + I2
– +
60 V
10 W
\Current through 100 W = I 1 + I 2 = 0.4375 A ¬
20 V I
– + 1
II
I2
Fig. 1.16 (a)
\Current through 20 W = I 1 = – 1.1875 A i.e 1.1875 A ¬
Example 1.10 Find the current in the 8 W resistor in the following circuit using Kirchoff's laws.
+
10 W
5W
15 V
15 V
10 W
VTU : May-13, Marks 8
8W
25 V
10 W
5W
Fig. 1.17
Solution : Step 1 : Show the branch currents.
5W
I1
–
+
I1
+
15 V –
10 W
–
+
I2
+
15 V
–
I2
I
–
(I1 – I2)
–
+
II
25 V
+
10 W
8W
+
–
I1
–
+
5W
I2
–
+
10 W (I1 – I2)
I1
Fig. 1.17 (a)
Step 2 :
Apply KVL to loop I and II.
–5I 1 – 15 – 8I 2 – 5I 1 – 10I 1 + 15 = 0
i.e.
+20I 1 + 8I 2 = 0
–10(I 1 – I 2 ) + 25 – 10(I 1 – I 2 ) + 8I 2 + 15 = 0
Step 3 :
–20I 1 + 28I 2 = – 40
i.e.
Solving equation (1) and (2),
I 1 = 0.444 A,
I 2 = – 1.111 A
\ Current through 8 W = – 1.111 A
i.e.
1.111 A ­
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… (1)
… (2)
Basic Electrical Engineering
1 - 14
D.C. Circuits
Example 1.11 A 20 V battery with an internal resistance of 5 W is connected to a resistance of
‘x' W. If an additional 6 W resistance is connected across the battery, find the value of ‘x' so
+
that the power supplied externally by the battery remains the same.
Solution : Case i] Consider the resistance x alone as
shown in the Fig. 1.18 (a) .
20
Now
I =
5+x
20 V
VTU : Aug.-95
5W
I
I
Series
I
x
20 ö 2
P = I 2 ´ R = æç
…(1)
Fig. 1.18 (a)
÷ ´ ( x)
è5+ xø
This is power supplied by battery, to x W.
Case ii] Now 6 W resistance is connected in addition to x as shown in the Fig. 1.18 (b).
\
20 V
5W
I
20 V
5W
xW
I
6W
I
Parallel
(6 || x) W
Fig. 1.18 (b)
Fig. 1.18 (c)
Combining x and 6 W which are in parallel we get circuit as shown in the Fig. 1.18 (c).
Now
I =
20
=
5 + ( 6 || x)
20 ( 6 + x)
20
=
30 + 11x
6x
5+
6
+
x
(
)
Hence power supplied by battery is,
2
é 20 ( 6 + x) ù
P = ê
ú ´ ( 6 || x)
ë( 30 + 11x) û
as P = I 2 R
2
é 20 ( 6 + x) ù
6x
P = ê
\
ú ´ 6+x
(
)
ë( 30 + 11 x) û
Power supplied must remain same, so equating equations (1) and (2),
2
2
é 20 ( 6 + x) ù
6x
é 20 ù
ê(5 + x) ú ´ x = ê( 30 + 11 x) ú ´ ( 6 + x)
û
ë
ë
û
\
\
400 x
(5 + x)
2
=
400 ( 6 + x) 6 x
( 30 + 11x)
2
( 30 + 11 x) 2
i.e.
(
900 + 660 x + 121 x 2 = ( 36 + 6 x) 25 +10 x + x 2
= 6 x ( 6 + x)(5 + x) 2
)
900 + 660 x + 121 x 2 = 900 + 360 x + 36 x 2 + 150 x + 60 x 2 + 6 x 3
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…(2)
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1 - 15
6 x 3 - 25 x 2 - 150 x = 0
\
i.e.
D.C. Circuits
(
)
x 6 x 2 - 25 x - 150
= 0
The value of x cannot be zero.
+25 ±
\
x =
(25 2 ) - 4 ´ 6 ´ ( -150)
2´ 6
25 ± 65
= 7.5 W
x =
12
\
Example 1.12
...Neglecting negative value
A circuit consists of two parallel resistors having resistance of 20 W and 30 W
respectively connected in series with 15 W. If current through 15 W resistor is 3 A, Find :
i) Current in 20 W and 30 W resistors ii) The voltage across the whole circuit iii) The total
power and power consumed in all resistances.
Solution : The arrangement is shown in the Fig. 1.19.
I1
20 W
15 W
Total current I = 3 A
R eq =
\
I =
\
(20|| 30) + 15 =
V
R eq
i.e.
3=
20 ´ 30
+ 15 = 27 W
20 + 30
I2
I=3A
30 W
V
27
V
Fig. 1.19
V = 81 V
... Voltage across each circuit
I1 = I ´
30
3
= 3 ´ = 1.8 A
20 + 30
5
... Current through 20 W
I2 = I ´
20
2
= 3 ´ = 1.2 A
20 + 30
5
... Current through 30 W
P = V ´ I = 81 ´ 3 = 243 W
... Total power
P20W = I 21 ´ 20 = (1.8) 2 ´ 20 = 64.8 W
P30W = I 22 ´ 30 = (1.2) 2 ´ 30 = 43.2 W
P15W = I 2 ´ 15 = ( 3) 2 ´ 15 = 135 W
Cross check is P = P20 + P30 + P15
Example 1.13
A particular battery when loaded by a resistance of 50 W gives the terminal
voltage of 48.6 V. If the load resistance is increased to 100 W, the terminal voltage is
observed to be 49.2 V.
Determine, i) E.M.F. of battery ii) Internal resistance of battery
Also calculate the load resistance required to be connected to get the terminal voltage of
+
(49.5V)
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Basic Electrical Engineering
1 - 16
D.C. Circuits
Solution : Let E.M.F of the battery is E volts and internal resistance r W. The two cases
are shown in the Fig. 1.20 (a) and (b).
IL1
r
E
+
_
50 W
IL2
r
RL1
Vt1 = 48.6 V
+
_
E
100 W
(a)
Vt2 = 49.2 V
RL2
(b)
Fig. 1.20
From Fig. 1.20 (a) we can write,
Vt1 = E - I L1 r
…(1)
From Fig. 1.20 (b) we can write,
Vt2 = E - I L2 r
…(2)
Applying Ohm's law to the load resistance,
and
Vt1 = I L1 R L1
i.e.
I L1 =
Vt2 = I L2 R L2
i.e.
I L2 =
Vt1
48. 6
=
= 0.972 A
R L1
50
Vt 2
R L2
=
49. 2
= 0.492 A
100
Substituting in equations (1) and (2),
48.6 = E - 0 .972 r
…(3)
49.2 = E - 0. 492 r
Subtracting equations (3) from equations (4), 0.6 = 0.48 r
r = 1.25 W
\
E = 49.815 V
\
Now the required terminal voltage is
Vt3
=
\
Vt3
= E - I L3 r
\
49.5 = 49.815 - I L3 ´ 1.25
\
I L3
Now
\
\
Vt3
49.5
R L3
49.815 - 49.5
= 0.252 A
1.25
=
I L3 R L3
… Internal resistance
… Battery voltage
r
49.5 V.
=
…(4)
E
49.815 V
IL3
1.25 W
RL3
Vt3 = 49.5 V
Fig. 1.20 (c)
= 0.252 R L3
= 196.428 W
…Load resistance
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1 - 17
D.C. Circuits
Important Multiple Choice Questions with Answers
Resistance
Q.1
The resistance is _________ proportional to length and _________ proportional to area
of cross-section.
a) directly, directly
c) inversely, directly
Q.2
b) directly, inversely
d) none of these
[Ans. : b]
1 joule = _________ calories
a) 0.21
b) 0.24
c) 0.28
d) 0.22
[Ans. : b]
Q.3
A wire of resistance R is stretched to double its length. The new resistance of the wire is
VTU : June-10
_________.
R
R
a)
b) 2 R
c) 4 R
d)
[Ans. : c]
2
4
Q.4
The resistance of a conductor having length l, area of cross section a and resistivity r is
VTU : June-13
given as _____ .
ra
rl
l
a) R =
b) R =
c) R = rla
d) R =
l
a
ar
[Ans. : b]
Q.5
Resistance of a wire always increases if _____ .
+
+
a) temperature is reduced
+
VTU : June-13
b) temperature is increased
c) number of free electrons available become less
d) number of free electrons available become more
[Ans. : b]
Energy Sources
Q.1
A circuit without any energy source is called _________ .
a) passive
Q.2
b) active
b) zero
c) constant
d) none of these
[Ans. : b]
For ideal current source, internal resistance is _________ ohms.
a) infinite
Q.4
D) distributed
[Ans. : a]
For ideal voltage source, internal resistance is _________ ohms.
a) infinite
Q.3
c) linear
b) zero
c) constant
A practical voltage source is represented by _________ .
d) none of these
[Ans. : a]
+
VTU : Dec.-11
a) a resistance in parallel with an ideal voltage source
b) a resistance in series with an ideal current source
c) a resistance in series with an ideal voltage source
d) none of the above
[Ans. : c]
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Basic Electrical Engineering
1 - 18
D.C. Circuits
Ohm's Law
Q.1
Validity of Ohm's law requires that the_____.
a) voltage should remain constant
c) resistance must remain constant
b) current should remain constant
d) power must remain constant
+
VTU : Feb.-09
[Ans. : b]
Q.2
For application of Ohm's law, _________ of circuit must remain constant.
a) voltage
Q.3
b) inductance
d) inductance
[Ans. : c]
c) capacitance
d) diode
[Ans. : d]
+
The Ohm's law can not be applied to _____ .
a) resistance
Q.5
c) resistance
The Ohm's law cannot be applied to _________.
a) resistance
Q.4
b) current
b) inductance
c) capacitance
VTU : Jan.-13
d) diode
The condition for the validity under Ohm's law is that the _____ .
+
[Ans. : d]
VTU : June-13
a) temperature should remain constant
b) current should be proportional to voltage
c) resistance must be wire wound type
d) all of the above
Q.6
[Ans. : a]
+
A linear resistor is one which obeys _____ .
a) Ampere's law
b) Lenz's law
c) Ohms law
VTU : June-13
d) Kirchhoff's law
[Ans. : c]
Series Circuit
Q.1
A series circuit consists of 4.7 k W, 5.6 k W , 9 k W and 10 kW resistors. Which resistor has
the most voltage across it ?
a) 4.7 k W
b) 5.6 kW
c) 9 kW
d) 10 kW
+
VTU : Aug.-09
[Ans. : d]
Q.2
In a series circuit, _________ remains same.
a) current
b) voltage
c) resistance
d) none of these
[Ans. : a]
Q.3
In a series circuit, the equivalent resistance is ___ of all the individual resistances.
a) smallest
b) same as
c) largest
d) none of these
[Ans. : c]
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Basic Electrical Engineering
Q.4
1 - 19
The voltage drop across 8 W resistance is ___ .
1W
8W
a) 100 V
4W
2W
225 V
+
Q.5
D.C. Circuits
–
b) 80 V
c) 220 V
d) 120 V
[Ans. : d]
The voltage across the short circuit is _________ .
a) infinite
b) one
c) zero
d) none of these
[Ans. : c]
Parallel Circuit
Q.1
The load increases means load resistance _________.
a) increases
b) decreases
c) remains constant
d) none of these
[Ans. : b]
Q.2
In a parallel circuit, _________ remains same.
a) current
b) voltage
c) resistance
d) none of these
[Ans. : b]
Q.3
In a parallel circuit, the equivalent resistance is _________ of all the individual
resistances.
a) smallest
b) same as
c) largest
d) none of these
[Ans. : a]
Q.4
If 20 resistances, each of 1 W are connected in parallel then the equivalent resistance is
_________ .
a) 0.01 W
Q.5
b) 0.05 W
c) 20 W
d) 0.02 W
[Ans. : b]
The total current drawn by the circuit shown from the supply is _________ .
9W
150 V
2W
6W
3W
Fig. 1.21
a) 10 A
Q.6
b) 5 A
c) 1 A
d) 15 A
[Ans. : d]
If the 3 W resistance is removed from the circuit shown in the Fig. 1.21 the current drawn
by the circuit is _________ .
a) 14.285 A
b) 9.185 A
c) 2 A
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d) 1.185 A [Ans. : a]
Basic Electrical Engineering
Q.7
1 - 20
D.C. Circuits
The voltage across the parallel circuit shown in the Fig. 1.22 is _______ .
1W
2W
3W
10 A
4W
Fig. 1.22
a) 1.8 V
Q.8
b) 4.8 V
c) 8.4 V
d) 2.8 V
[Ans. : b]
Two resistors R1 and R2 give combined resistance of 4.5 W when in series and 1W when
VTU : Dec.-11
in parallel, the resistances are _________ .
+
a) 2 W and 2.5 W
b) 1 W and 3.5 W
c) 1.5 W and 3 W
d) 4 W and 0.5 W.
[Ans. : c]
Q.9
The total resistance of parallel circuit is ___________.
a) less than the smallest resistance
b) more than the smallest resistance
c) more than the highest resistance
d) none of these
+
VTU : June-12
[Ans. : a]
Kirchhoff's Laws
Q.1
The algebraic sum of all the currents at a junction point is always zero is the statement of
_________ law.
a) KVL
Q.2
b) Lenz's
c) Faraday's
d) KCL
[Ans. : d]
The Fig. 1.23 shows a part of a closed electrical circuit. The potential drop between A
VTU : June-10
and B is _________ .
+
A
6V
4A
2W
1W
B
Fig. 1.23
a) 18 V
Q.3
b) – 18 V
c) 4 V
Kirchhoff's voltage law applies to circuit with _________.
a) linear elements only
d) – 4 V
+
[Ans. : a]
VTU : Dec.-11
b) non-linear elements only
c) linear, non-linear, active and passive elements
d) linear, non-linear, active, passive, time varying as well as time invariant elements.
[Ans. : a]
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Basic Electrical Engineering
Q.4
1 - 21
D.C. Circuits
The polarity of voltage drop across a resistor is determined by _________.
+
a) the value of resistor
b) the value of current
c) direction of current in resistor
d) the polarity of source
VTU : Jan.-14
[Ans. : c]
Electrical Power
Q.1
If 100 V is applied across a 200 V, 100 W bulb, the power consumed will be,_____ .
a) 100 W
b) 50 W
c) 25 W
+
d) 12.5 W
VTU : Feb.-09
[Ans. : c]
Q.2
The power dissipation in each of three parallel branches is 1 W. The total power
VTU : Aug.-09
dissipation of the circuit is _________ .
+
a) 1 W
Q.3
Q.6
b) 1250 W
c) 312.5 W
b) three fourth
c) one fourth
+
VTU : Feb.-10
d) 3125
. W
[Ans. : c]
d) 0.707 times.
[Ans. : c]
Which of the following statements is true both for a series and a parallel circuit ?
+
a) Resistances are additive
b) Powers are additive
c) Currents are additive
d) Voltage drops are additive
VTU : Feb.-10
[Ans. : b]
Refer to the Fig. 1.24. Which of the following statement is true ?
Lamp 1 will be less brighter than Lamp 2.
Lamp 1 will be more brighter than Lamp 2.
Both the lamp will glow with equal brightness.
None of the above.
[Ans. : b]
Lamp 1
60 W, 240 V
a) kWh
b) Wh
Fig. 1.24
VTU : Jan-13
c) Watt -second
Lamp 2
100 W, 240 V
240 V
The practical unit of electrical energy is _____ .
+
Q.8
[Ans. : c]
+
a)
b)
c)
d)
Q.7
d) 9 W
The voltage applied across an electric iron is halved. The power consumption of the iron
VTU : Feb.-10
reduces to _________ .
a) one half
Q.5
c) 3 W
The resistance of a 200 W, 250 V lamp is ________ .
a) 625 W
Q.4
b) 4 W
d) Joule second
[Ans. : a]
Three resistors of 4 W, 6 W and 9 W are connected in parallel in a network. Maximum
VTU : Jan.-14
power will be consumed by ________ .
+
a) 4 W resistor
b) 6 W resistor
c) 9 W resistor
d) all resistor
[Ans. : a]
Electrical Energy
Q.1
+
The practical unit of electrical energy is _________ .
a) kW-hr
b) watt-hr
c) watt-second
TM
VTU : Feb.-10
d) joule-second
TECHNICAL PUBLICATIONS - An up thrust for knowledge
[Ans. : a]
Basic Electrical Engineering
Q.2
1 - 22
A 2 W resistor is connected in series with parallel combination of 10 W and 15 W resistors.
Then heat dissipated in kWsec for 1 hour in circuit, when current of 2 A flowing in 2 W
resistor is _________.
a) 115.2
Q.3
D.C. Circuits
b) 1.152
c) 11.52
d) 115200 [Ans. : a]
Energy consumed by a heater of rating 1000 W by operating it for a period of
VTU : Dec.-11
2 hours will be _________ .
+
a) 1 kWh
b) 2 kWh
c) 2.5 kWh
d) 4 kWh
[Ans. : b]
Current Division in Parallel Circuit of Resistors
Q.1
In a circuit shown, the current through 5 W resistance is _________ .
100 V
a) 15 A
Q.2
+
_
5W
20 W
b) 20 A
c) 25 A
d) 4 A
[Ans. : b]
The current drawn by the resistance of 8 W in the circuit shown is _________ .
+
100 V
8W
12 W
–
10 W
a) 1.555 A
Q.3
b) 2.555 A
c) 5.5555 A
d) 4.5555 A
+
The current in 5 ohm resistor is ___________.
[Ans. : c]
VTU : June - 12
5W
3A
10 W
a) 2 A
b) 3 A
c) 1 A
d) 1.5 A
[Ans. : a]
Concept of the Terminal Voltage of a Cell
Q.1
The practical voltage source has e.m.f. of E volts and internal resistance is r ohms. If it
supplies a load current of I amperes, the terminal voltage is _________.
a) E
Q.2
æ Eö
b) ç ÷
èrø
c) Ir
d) E - Ir
[Ans. : d]
A practical voltage source of 100 V is connected across 10 W resistance and the terminal
voltage across 10 W is found to be 80 V, then the internal resistance of the source is
_________ .
a) 2.5 W
b) 10 W
c) 5 W
d) 8 W
[Ans. : a]
qqq
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2
Electromagnetism
Chapter at a Glance
1.
Faraday's Laws of Electromagnetic Induction
e = -N
2.
df
dt
volts
Magnetic Flux and Magnetic Flux Density
1 weber = 10
B=
3.
or Tesla
Magnetomotive Force (M.M.F.)
ampere turns
Reluctance and Permeance
S=
Permeance =
6.
m2
Ampere turns
NI
=
AT / m
Length
l
m. m. f. = N I
5.
Wb
lines of force.
Magnetic Field Strength
H=
4.
f
a
8
l
l
=
A/Wb
ma m0 mr a
1
Reluctance
Permeability
m=
B
H
i.e.
B
= mH
(2 - 1)
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Basic Electrical Engineering
m0 =
2-2
B
in vacuum = 4p´ 10 -7 H/m
H
m = m 0m r
7.
H/m
Relation between Flux, M.M.F. and Reluctance
M. M. F.
F
=
Reluctance
S
Flux f =
8.
Electromagnetism
Nature of the Induced E.M.F.
e = B l v sin q volts
9.
Force on a Current Carrying Conductor in a Magnetic Field
F = B I l sin q newtons
10. Self Inductance
L=
\
\
Nf
I
e = -L
dI
dt
volts
L=
N× NI
N2
=
I ×S
S
L=
N2 m 0 m r a
N2 m a
=
l
l
henries
…QS =
henries
11. Mutually Induced E.M.F.
\
\
æN f ö
M=ç 2 2÷
è I1 ø
M=
N 2 f2
I1
M=
N2 K1 f1
I1
henries
and
TM
e2 = - M
d I1
volts
dt
TECHNICAL PUBLICATIONS - An up thrust for knowledge
l
ma
Basic Electrical Engineering
2-3
M=
K1 N1 N2
S
\
M=
N1 N2
S
\
M=
N1 N2
æ l ö
ç
÷
èma ø
Electromagnetism
… For K1 = 1
=
\
M=
K 2 N1 N2
S
\
M=
N1 N2
S
N1 N2 a m
N1 N2 a m 0 m r
=
l
l
…QS =
l
ma
… For K2 = 1
12. Coefficient of Coupling or Magnetic Coupling Coefficient
K=
M
L1 L 2
13. Energy Stored in a Magnetic Field
E=
1 2
LI
2
joules
Important Theory Questions and Answers
Ø
State and explain the Faraday's laws of electromagnetic induction.
+ July-03, 04, Marks 5; July-06, Marks 8; July-07, June-13, Marks 6
1. First Law : Whenever the number of magnetic lines of force (flux) linking with a
coil or circuit changes, an e.m.f. gets induced in that coil or circuit.
2. Second Law : The magnitude of the induced e.m.f. is directly proportional to the
rate of change of flux linkages (flux ´ turns of coil).
Flux linkages = Flux ´ Number of turns of coil
· Consider a coil having N turns. The initial flux linking with a coil is f1 .
\
Initial flux linkages = Nf1
· In time interval dt, the flux linking with the coil changes from f1 to f2 .
\
Final flux linkages = Nf2
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Basic Electrical Engineering
\
2-4
Rate of change of flux linkages =
Electromagnetism
Nf2 - Nf1
dt
· Now as per the first law, e.m.f. will get induced in the coil and as per second law the
magnitude of e.m.f. is proportional to the rate of change of flux linkages.
Nf2 - Nf1
Nf2 - Nf1
e µ
i.e.
e = K´
\
dt
dt
\
e = N
df
dt
(df = f2 - f1 )
· Thus such an induced e.m.f. is mathematically expressed alongwith its sign as,
e = -N
Ø
df
dt
volts
State and explain Lenz's law.
Statement : The direction of
an induced e.m.f. produced by
the electromagnetic induction
is such that it sets up a
current which always opposes
the cause that is responsible
for inducing the e.m.f.
· Consider a solenoid
shown in the Fig. 2.1.
+
VTU : July-03, 04, July-06; July-07, Marks 4
Repulsive force due to
induced e.m.f. and
current in coil
Coil
N
NN
S
S
Bar magnet
Direction of motion
as
G
Fig. 2.1 Lenz's law
· According to Lenz's law, the
direction of current due to induced e.m.f. is so as to oppose the cause producing it.
The cause is motion of bar magnet towards the coil.
· So e.m.f. will set up a current through coil in such a way that the end of solenoid
facing bar magnet will become N-pole.
· Hence two like poles will face each other experiencing force of repulsion which is
opposite to the motion of bar magnet as shown in the Fig. 2.1.
· If the same bar magnet is moved away from the coil, then induced e.m.f. will set up a
current in the direction which will cause, the end of solenoid facing bar magnet to
behave as S-pole.
· Because of this, two unlike poles will face each other and there will be force of
attraction which is opposite to the direction of movement of magnet.
Ø
+
What is reluctance ? State its units.
VTU : Dec.-03, 05, Marks 2
· In an electric circuit, current flow is opposed by the resistance of the material, similarly
there is opposition by the material to the flow of flux which is called reluctance.
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Basic Electrical Engineering
2-5
Electromagnetism
· It is defined as the resistance offered by the material to the flow of magnetic flux
through it. It is denoted by 'S'.
· It is measured in amperes per weber (A/Wb) or ampere-turns per weber (AT/Wb).
Ø
State Fleming's right hand rule. Mention its application.
+ VTU : Jan.-03, July-03,04; Feb.-05; July-04,11, Dec.-11, Marks 3
· The direction of dynamically induced e.m.f. is given by Fleming's right hand rule.
· If three fingers of a right hand, namely thumb, index finger and middle finger are
outstretched so that everyone of them is at right angles with the remaining two, and if
in this position index finger is made to point in the direction of lines of flux, thumb in
the direction of the relative motion of the conductor with respect to flux then the
outstretched middle finger gives the direction of the e.m.f. induced in the conductor.
· The rule is commonly used in the d.c. generators.
Ø
State Fleming's left hand rule. Mention its application.
+ VTU : Jan.-03, 11; July-03, 04,11; Feb.-05, Marks 3
· The rule states that, ‘Outstretch the three fingers of the left hand namely the first finger,
middle finger and thumb such that they are mutually perpendicular to each other. Now
point the first finger in the direction of magnetic field and the middle finger in the
direction of the current then the thumb gives the direction of the force experienced by
the conductor'.
· The rule is commonly used in the d.c. motors.
Ø
Define mutual inductance and state its unit. Derive the expression for mutual inductance.
+ VTU : Jan.-03; Feb.-05; July-04, Marks 8
· Let
N1 = Number of turns of coil A,
N2 = Number of turns of coil B
I1 = Current flowing through coil A
f1 = Flux produced due to current I1
f2 = Flux linking with coil B
· According to Faraday's law, the induced e.m.f. in coil B is,
d f2
e2 = - N2
dt
f
Now
f2 = 2 ´ I1
I1
\
\
Rate of change of f2 =
f2
´ Rate of change of current I 1
I1
d f2
f dI
= 2× 1
I 1 dt
dt
TM
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Basic Electrical Engineering
\
e 2 = - N2 ×
2-6
Electromagnetism
f2 dI1
æ N f ö dI 1
= -ç 2 2÷
×
I1 dt
è I 1 ø dt
æN f ö
· Here ç 2 2 ÷ is called coefficient of mutual inductance denoted by M.
è I1 ø
· Coefficient of mutual inductance is defined as the property by which e.m.f. gets
induced in the second coil because of change in current through first coil. It is also
defined as the flux linkages of the coil per ampere current in other coil.
Ø
Derive an expression for the energy stored in a magnetic field.
+ Jan.-03, 04, 06,13, 14, Marks 6; July-04, 06, 09, 12, 13; Feb.-05, Marks 8
· Let the induced e.m.f. in a coil be,
dI
e = -L
dt
· This opposes a supply voltage. So supply voltage ‘V' supplies energy to overcome this,
which ultimately gets stored in the magnetic field.
dI
dI
V = – e = - é-L ù = L
\
êë dt úû
dt
\
Power supplied = V ´ I
= L
dI
´ I
dt
\ Energy supplied in time dt is,
E = Power ´ Time = L
dI
´ I ´ dt
dt
= L dI ´ I joules.
\ Integrating above, the total energy stored is,
I
ù
é I2 ù
é I2
E = ò L dI I = L ò dI I = L ê ú = L ê - 0ú
2
2
û
ë û0
ë
0
0
I
\
E=
1 2
LI
2
I
joules
Important Solved Examples
Example 2.1
A coil is wound uniformly with 300 turns over a steel of relative permiability 900,
having a mean circumference of 40 mm and corss-sectional area of 50 mm2. If a current of 5
A is passed through the coil, find
i) m.m.f.
ii) reluctance of the ring and
iii) flux
Solution : Given : N = 300, mr = 900, l = 40 mm = 40 × 10
TM
+
–3
m,
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2-7
2
i)
–6
Electromagnetism
2
a = 50 mm = 50 × 10 m , I = 5 A
m.m.f. = NI = 300 × 5 = 1500 AT
ii)
S =
l
40 ´ 10 -3
= 70.7355 ´ 103 AT/Wb
=
7
6
m 0m r a
4p ´ 10 ´ 900 ´ 50 ´ 10
This is reluctance of the ring.
iii)
S =
m. m. f.
f
Q
f =
m. m. f
1500
=
= 21.2057 mWb
S
70.7355 ´ 10 3
… Flux
A ring shaped core is made up of two parts of same material. Part one is a
Example 2.2
magnetic path of length 25 cm and with cross sectional area 4 cm 2 , whereas part two is of
length 10 cm and cross sectional area of 6 cm 2 . The flux density in part two is 1.5 Tesla. If
the current through the coil, wound over core, is 0.5 Amp., calculate the number of turns of
coil. Assume m r is 1000 for material.
Solution : The
Fig. 2.2.
arrangement
B2
is
shown
+
in
the
f = 1.5 ´ 6 ´ 10 - 4 = 9 ´ 10 - 4 Wb
\
Key Point
The flux f is same through both the parts,
as series circuit.
2
a2 = 6 cm
l2 = 10 cm
N
B2 = 1.5 T
I
2
a1 = 4 cm
l1 = 25 cm
Fig. 2.2
S = S I + S II
=
mr = 1000
I
0.5 A
f
=
a2
VTU : Dec.-05
l1
l2
=
+
m 0m r a 1 m 0m r a 2
25 ´ 10 - 2
4p ´ 10 - 7 ´ 1000 ´ 4 ´ 10 - 4
+
10 ´ 10 - 2
4p ´ 10 - 7 ´ 1000 ´ 6 ´ 10 - 4
= 629988.3164 AT/Wb
f =
\
\
9 ´ 10 - 4 =
m. m. f
NI
=
S
S
N ´ 0.5
629988.3164
N = 1133.979 » 1134
... Number of turns
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2-8
Electromagnetism
Example 2.3 A coil of 300 turns wound on a core of non magnetic material has an inductance of
10 mH. Calculate i) flux produced by a current of 5 A. ii) the average value of the emf
induced when a current of 5 A is reversed in 8 millisecond.
Solution : N = 300, L = 10 mH,
i) L =
Nf
I
i.e. f =
+
VTU : Aug.-03, Marks 5
I = 5 A.
LI 10 ´ 10 -3 ´ 5
= 166.667 mWb
=
N
300
ii) Current is reversed i.e. becomes –5 A in dt = 8 ms.
[-5 - 5]
dI
= 12.5 V
= -10 ´ 10 -3
e = -L
\
dt
8 ´ 10 -3
Key Point dI is change in current i.e. [final current - initial current] which is [–5 – 5] = –10 A.
2
Example 2.4 Two 200 turns, air cored solenoids, 25 cm long have a cross-sectional area of 3 cm
each. The mutual inductance between them is 0.5 mH. Find the self inductance of the coils and
+
the coefficient of coupling.
Solution :
l
= 25 cm = 25 ´ 10 -2 m,
VTU : Jan.-90
a = 3 cm = 3 ´ 10 -4 m 2
2
M = 0.5 mH
and
N 1 = N 2 = 200
The self inductance of both the coils will be same.
N2
L1 = L2 =
S
where
S =
=
\
L =
Now
K =
l
m0 a
as m r = 1 for air core
25 ´ 10 -2
4 p ´ 10 -7
´
3 ´ 10 -4
= 6.6314 ´ 10 8
AT/Wb
( 200) 2
= 6.031 ´ 10 -5 H = 60.31 mH
6.6314 ´ 108
0 . 5 ´ 10 -6
M
M
M
=
=
=
= 0.00828
L
L1 L2
60. 31 ´ 10 -6
L2
Example 2.5 Two identical 1000 turn coils X and Y lie in parallel planes such that 60 % of the
flux produced by one coil links with the other. A current of 5 A in X produces a flux of
5 ´ 10 -6 Wb in itself. If the current in X changes from +6 A to –6 A in 0.01 sec, what will
be the magnitude of the e.m.f. induced in Y ? Calculate the self inductance of each coil.
+
Solution : N 1 = N 2 = 1000, I 1 = 5 A,
f1 = 5 ´ 10 -6 Wb
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Electromagnetism
N 2 f2
I1
Now
M =
But
f2 = 0.6 f1
\
M =
as 60 % flux link with other
0.6 N 2 f1
0.6 ´ 1000 ´ 5 ´ 10 -6
=
= 6 ´ 10 -4 H
I1
5
So induced e.m.f. in Y will be,
e2 = - M
d I1
é- 6 - 6 ù
= - 6 ´ 10 -4 ê
= 0.72 V.
dt
ë 0.01 úû
As coils are identical, both will have same value of self inductance as,
N 1 f1 1000 ´ 5 ´ 10 -6
=
= 0.001 H = 1 mH
L1 = L2 =
I1
5
Example 2.6 A magnetic core is in the form of a closed ring of mean length 20 cm and
cross-sectional area 1 cms 2 . Its relative permeability is 2400. A coil of 2000 turns is
uniformly wound around it. Find the flux density set up in the core if a current of 66 mA is
passed through the coil. Find the energy stored in the magnetic field set up.
Find the inductance of the coil, if an air gap of 1 mm is cut in the ring perpendicular to the
+
direction of the flux.
VTU : May-05
Solution : Given l = 20 cm, a = 1 cm 2 , m r = 2400, N = 2000, I = 66 mA
Case 1 :
S =
l
20 ´ 10 -2
= 663.1455´ 10 3 AT/Wb
=
m 0m r a
4p ´ 10 -7 ´ 2400 ´ 1 ´ 10 -4
m.m.f = NI = 2000 ´ 66 ´ 10 -3 = 132 AT
\
f =
NI
132
=
= 1.9905 ´ 10 -4 Wb
S
663.1455 ´ 10 3
\
B =
f 1.9905 ´ 10 -4
=
= 1.9905 Wb/m 2 i.e. T
4
a
1 ´ 10
L =
(2000) 2
Nf
N2
= 6.03185 H or L =
=
3
I
S
663.1455 ´ 10
E =
1 2
1
LI = ´ 6.03185 ´ (66 ´ 10 -3 ) 2 = 13.1373 mJ
2
2
\
… Flux density
… Energy stored
Case 2 : New air gap is cut of length lg = 1 mm in the ring.
\
li = Iron length = l - lg = 20 ´ 10 -2 - 1 ´ 10 -3 = 0.199 m
\
S = S i + Sg =
lg
li
+
m 0m r a m 0 a
… m r = 1 for air gap
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=
2 - 10
1 é li
1
ù
+l =
m 0 a êëm r g úû 4p´ 10 -7 ´ 1 ´ 10 -4
Electromagnetism
é 0.199 + 1 ´ 10 -3 ù
êë 2400
úû
= 8.6175 ´ 10 6 AT/Wb
\
L =
…Total reluctance
(2000) 2
N2
= 0.4641 H
=
S
8.6175 ´ 10 6
…New inductance.
Example 2.7 Two coils A and B, have self inductances of 120 mH and 300 mH respectively. A
current of 1 A through coil 'A' produces flux linkage of 100 mWb turns in coil 'B'. Calculate
i) mutual inductance between the coil.
ii) average e.m.f. induced in coil 'B' if current of 1 A in coil 'A' is reversed at a uniform rate
+
in 0.1 sec. Also find coefficient of coupling.
VTU : Dec.-04
Solution : LA = 120 mH, L B = 300 mH
I A = 1 A produces N B fB = 100 mWb
i)
M =
N B fB
100 ´ 10 -6
=
= 100 mH
IA
1
…Mutual inductance
dI A
dt
The current in coil A is reversed i.e. it is –1 A in 0.1 sec.
\
D I = (New value – Original value) = (–1 –1) = –2 A
ii)
e B = -M
and
D t = 0.1 sec
\
\
dI A
dt
=
DI
-2
=
= – 20 A/sec
Dt 0.1
e B = -100 ´ 10 6 ´ ( -20) = 2 mV
K =
M
=
LA L B
… Induced e.m.f. in B
100 ´ 10 -6
120 ´ 10 -6 ´ 300 ´ 10 -6
= 0.527
… Coefficient of coupling
Example 2.8 Two identical coils P and Q, each with 1500 turns, are placed in parallel planes
near to each other, so that 70% of the flux produced by current in coil P links with coil Q. If
a current of 4 A is passed through any one coil, it produces a flux of 0.04 mWb linking with
itself. Find the self inductances of the two coils, the mutual inductance and coefficient of
+
coupling between them.
Solution : N P = N Q = 1500, fQ = 0.7 fP
VTU : Dec.-03
... 70 % linking
Let I P = 4 A and fP = 0.04 mWb
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\
LP =
2 - 11
Electromagnetism
N P fP 1500 ´ 0.04 ´ 10 -3
= 15 mH
=
IP
4
Let I Q = 4 A then fQ = 0.04 mWb
N Q fQ 1500 ´ 0.04 ´ 10 -3
= 15 mH
\
LQ =
=
IQ
4
M =
And
K =
N Q fQ
IP
=
N Q 0.7 fP
M
=
LP LQ
IP
=
1500 ´ 0.7 ´ 0.04 ´ 10 -3
= 10.5 mH
4
10.5 ´ 10 -3
(
15 ´ 10 -3
)
2
= 0.7
Example 2.9 If a current of 5 A flowing in coil with 1000 turns wound on a ring of
ferromagnetic material produces a flux of 0.5 mWb in the ring. Calculate i) self inductance of
coil ii) e.m.f. induced in the coil when current is switched off and reaches zero value in
2 millisec. iii) mutual inductance between the coils, if a second coil with 750 turns is wound
+
uniformly over the first one.
VTU : May-03
f = 0.5 mWb, N = 1000, I = 5 A
Solution :
Nf
1000 ´ 0.5 ´ 10 -3
= 0.1 H
=
I
5
i)
L =
ii)
e = -L
dI
é 0 -5 ù
= - 0.1ê
ú = 250 V
dt
ë 2 ´ 10 -3 û
iii) Let
N 2 = 750 of other coil
As other coil is wound on first, all the flux produced by coil 1 links with the second coil.
as
\
f2 = K 1 f1 = f1
K1 = 1
\
M =
N [K f ] 750 ´ 0.5 ´ 10 -3
N 2 f2
= 2 1 1 =
= 0.075 H
I1
I1
5
Example 2.10 Two windings connected in series are wound on a ferromagnetic ring having
cross-sectional area of 750 mm 2 and a mean diameter of 175 mm. The two windings have
250 and 750 turns, while the relative permeability of material is 1500. Assuming no leakage of
flux, calculate the self inductances of each winding and the mutual inductance as well.
Calculate e.m.f. induced in coil 2 if current is coil 1 in increased uniformly from zero to 5 A
+
in 0.01 sec.
Solution :
l = Length of magnetic circuit = p´ d mean
l = p´ 175 ´ 10 - 3 = 0.5497 m
a = 750 mm 2 = 750 ´ 10 - 6 m 2 = 7.5 ´ 10 - 4 m 2
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Electromagnetism
N 1 = 250, N 2 = 750, m r = 1500
Self inductance,L =
Nf
NI
but f =
I
S
\
L =
N NI
N2
=
IS
S
We have,
S =
l
ma
\
S =
0 . 5497
l
=
7
m 0 mr a
4p´ 10
(1500) 7. 5 ´ 10 -4
(
)
(
)
= 388833.2 AT/Wb
\
L1 =
N 21
( 250) 2
= 0.1607 H
=
S
388833.2
L2 =
N 22
(750) 2
= 1.4466 H
=
S
388833.2
The mutual inductance between the two windings is given by,
M =
\
N1 N2
( 250) (750)
=
= 0.4822 H
S
388833.20
M = 0.4822 H
E.M.F. induced in coil 2 is,
e 2 = -M
dI 1
(5 - 0)
= -0 . 4822 ´
= –241.1 V
dt
0 . 01
Example 2.11 The winding of an electromagnet is wound with 96 turns and has a resistance of
50 ohms. The exciting voltage is 250 V and the flux linking the coil is 5 mWb. Find the
energy stored in the magnetic field. Then if the current is reversed in 0.1 sec, what emf is
+
induced in the coil ?
Solution : N = 96, R = 50 W, V = 250 V, f = 5 mWb.
V 250
I =
=
=5A
R
50
\
L =
Nf 96 ´ 5 ´ 10 -3
=
= 96 mH
I
5
1 2
1
LI = ´ 96 ´ 10 -3 ´ (5) 2 = 1.2 J
2
2
The current is reversed i.e. Ifinal = – 5 A, dt = 0.1 s
\
W =
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Basic Electrical Engineering
e = -L
=
2 - 13
Electromagnetism
-I
dI
ù
éI
= -96 ´ 10 -3 ê final initial ú
dt
dt
û
ë
-96 ´ 10 -3 [-5 - 5]
= 9.6 V
0.1
2
Example 2.12 An iron ring of 10 cm in diameter and 8 cm
2
in cross-section is wound with
300 turns of wire. For a flux density of 1.2 Wb/m and relative permeability of 500, find the
+
exciting current, the inductance and the energy stored.
Solution :
2
VTU : May-07
2
d = 10 cm, a = 8 cm , N = 300, B = 1.2 Wb/m , m r = 500
l = p´ d = p´ 10 cm = 0.3141 m
S =
l
0. 3141
= 624.882 ´ 10 3 AT/Wb
=
–7
–4
m 0m r a
4p´ 10 ´ 500 ´ 8 ´ 10
f = B ´ a = 1.2 ´ 8 ´ 10 – 4 = 9.6 ´ 10 – 4 Wb
\
f =
\
9.6 ´ 10 – 4 =
\
NI
S
300 ´ I
624.882 ´ 10 3
I = 2A
\
L =
( 300) 2
N2
= 0.14402 H
=
S
624.882 ´ 10 3
E =
1 2 1
LI = ´ 0.14402 ´ ( 2) 2 = 0.288 J
2
2
Example 2.13 An air cored solenoid 1 m in length and 10 cm in diameter has 5,000 turns.
Calculate : i) the self inductance and ii) the energy stored in the magnetic field when current
+
of 2 A flows in solenoid.
Solution :
\
l = 1 m, d = 10 cm, N = 5000, m = m 0 as air cored
p 2 100 p
a =
cm 2 = 7.854 ´ 10 – 3 m 2
d =
4
4
l
1
= 101. 3209 ´ 10 6 AT Wb
=
S =
–
7
–
3
m 0a
4p´ 10 ´ 7.854 ´ 10
(5000) 2
N2
= 0.2467 H
=
S
101.3209 ´ 10 6
i)
L =
ii)
I = 2A
1
1
E =
LI 2 = ´ 0.2467 ´ 2 2 = 0.4934 J
2
2
\
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Basic Electrical Engineering
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Electromagnetism
Important Multiple Choice Questions with Answers
Magnetic Field and Magnetic Lines of Force
Q.1
The region around the magnet in which magnetic influence can be experienced is called
_______ .
a) flux
b) line of force
c) strength
d) magnetic field
[Ans. : d]
Q.2
The direction of flux internal to the magnet is from_____.
a) N-pole to S-pole
b) S-pole to N-pole
c) circular
d) none of the above
[Ans. : d]
Magnetic Flux and Magnetic Flux Density
Q.1
1 weber = ____ lines of force.
a) 10
Q.2
8
b) 10
6
c) 10
3
-8
d)10
[Ans. : a]
The unit of flux density is _____ .
a) weber
b) AT
c) tesla
d) none of these
[Ans. : c]
Magnetic Field Strength
Q.1
Q.2
The magnetic field strength H is given by_____ .
Nl
Il
NI
b)
c)
a)
I
N
l
d) N I l
[Ans. : c]
The unit of magnetic field strength H is_____ .
a) weber
b) AT
c) tesla
d) AT/m
[Ans. : d]
d) Wb/AT
[Ans. : a]
Reluctance and Permeance
Q.1
The S.I. unit of reluctance is________ .
a) AT/Wb
Q.2
Q.3
c) AT/m
The reluctance is _____ to the relative permeability of the magnetic circuit.
a) inversely proportional
b) directly proportional
c) not dependent
d) none of the above
[Ans. : a]
The reluctance in a magnetic circuit is analogous to ____ in an electric circuit.
a) voltage
Q.4
b) AT
b) current
c) resistance
d) e.m.f.
c) susceptance
d) resistance
[Ans. : b]
[Ans. : c]
The _______ is reciprocal of the reluctance.
a) permeability
b) permeance
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Electromagnetism
Permeability
Q.1
The unit of permeability is _______ .
a) H/m
Q.2
b) Hm
c) weber
d) tesla
[Ans. : a]
b) easy to pass the flux, through the material
c) none of these
[Ans. : b]
For a magnetic material, the value of m r is _______ .
a) low
b) zero
c) high
d) one
[Ans. : c]
d) one
[Ans. : d]
Higher value of m r means _______ .
a) difficult to pass the flux, through the material
Q.3
Q.4
For free space, the value of m r is_______ .
a) low
b) zero
c) high
Magnetic Field due to Circular Conductor i.e. Solenoid
Q.1
The direction of magnetic field due to straight current carrying conductor is given by___ .
a) Fleming's left hand rule
c) Fleming's right hand rule
Q.2
b) N pole
c) positive of battery
d) none of these
[Ans. : b]
A coil wound around a core to produce a magnet is called ______ .
a) pole
Q.4
[Ans. : b]
According to right hand thumb rule applied to solenoid, the thumb points in the direction
of _____ .
a) S pole
Q.3
b) Right hand thumb rule
d) None of the above
b) transformer
c) solenoid
d) none of these
[Ans. : c]
An electromagnet with a circular core is called ______ .
a) compensating winding b) transformer
c) solenoid
d) toroid
[Ans. : d]
Magnetic Circuit
Q.1
The reluctance in a magnetic circuit is analogous to ______ in an electric circuit.
a) voltage
Q.2
c) resistance
d) e.m.f.
[Ans. : c]
d) e.m.f.
[Ans. : b]
d) e.m.f.
[Ans. : d]
The flux is analogous to _______ of the electric circuit.
a) voltage
Q.3
b) current
b) current
c) resistance
The m.m.f. is analogous to ______ of the electric circuit.
a) voltage
b) current
c) resistance
Faraday's Laws of Electromagnetic Induction
Q.1
The e.m.f. is obtained from the magnetic flux by________ .
a) Ampere
b) Faraday
c) Oersted
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d) Ohms
[Ans. : b]
Basic Electrical Engineering
Q.2
2 - 16
A coil of 2000 turns, produces a flux of 1 mWb.
e.m.f. induced is ________ V.
a) - 40
Q.3
b) 20
df
dt
b) N
c) 40
d) 60
df
di
c) – N
df
dt
d) L
[Ans. : c]
b) Coulomb's law
df
di
+
The law that finds application in electrolysis _____ .
a) Faraday's law
Q.5
The flux is reversed in 0.1 sec then
The e.m.f. induced in a coil of N turns is _______ .
a) N
Q.4
Electromagnetism
c) Ohm's law
[Ans. : c]
VTU : June-13
d) Lenz's law
[Ans. : a]
According to Faraday's law of electromagnetic induction an emf is induced in a conductor
VTU : June-13
whenever it _____ .
+
a) lies in a magnetic field
b) lies perpendicular to the magnetic field
c) cuts the magnetic flux
d) moves parallel to the direction of magnetic field.
[Ans. : c]
Lenz's law
Q.1
According to ________ the induced e.m.f. opposes the cause producing it.
a) Ohm's law
Q.2
b) Faraday's law
c) Kirchhoff's law
d) Lenz's law
[Ans. : d]
"In all cases of electromagnetic induction, an induced voltage will cause a current to flow
in a closed circuit in such a direction that the magnetic field which is caused by that
current will oppose the change that produces the current" is the original statement of
VTU : June-13
_____ .
+
Q.3
a) Lenz's law
b) Faraday's law of magnetic induction
c) Fleming's law of induction
d) Ampere's law
Which law is synonymous to the occurrence of diamagnetism ?
a) Ampere's law
b) Maxwell's law
c) Coulomb's law
+
[Ans. : a]
VTU : June-13
d) Lenz's law
[Ans. : d]
Nature of the Induced E.M.F.
Q.1
Q.2
When e.m.f. is induced due to physical movement of the coil then it is called ______ .
a) statically induced e.m.f.
b) forcefully induced e.m.f.
c) dynamically induced e.m.f.
d) magnetically induced e.m.f.
The dynamically induced e.m.f. can be found in _______ .
a) generator
Q.3
[Ans. : c]
b) transformer
c) bulb
The magnitude of statically induced e.m.f. depends on _____.
a) the coil resistance
b) the flux magnitude
c) the rate of change of flux
d) all of these
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d) none of these
[Ans. : a]
+
VTU : Jan.-14
[Ans. : c]
Basic Electrical Engineering
2 - 17
Electromagnetism
Self Inductance
Q.1
An e.m.f. of 7.2 volts is induced in a coil of 6 mH. Then the rate of change of current is :
VTU : Jan.-09
_______ .
+
a) 12 A/s
Q.2
b) 120 A/s
Q.4
b)
NI
f
c)
Nf
I
d)
VTU : Jan.-13
I
Nf
[Ans. : c]
The self inductance L is _______ number of turns.
a) directly proportional to square of
b) inversely proportional to square of
c) directly proportional to
d) none of the above
[Ans. : a]
A current of 20 A is reversed in 0.1 sec through an inductor of 1 H then e.m.f. induced is
VTU : Jan.-13
_____ volts.
+
a) 200
Q.5
+
The self inductance L is given by _______ .
a) N f I
Q.3
[Ans. : c]
d) 12000 A/s.
c) 1200 A/s
b) - 200
c) - 400
d) 400
+
Inductance opposes _______ in current in a circuit.
a) only increase
b) only decrease
c) change
[Ans. : d]
VTU : June-12
d) none of these
[Ans. : c]
Coefficient of Coupling or Magnetic
Coupling Coefficient
Q.1
The maximum value of coefficient of coupling is ________ .
a) 100 %
Q.2
b) more than 100 %
b) unity
c) very high
b) AT/Wb
c) unitless
The flux linkage between the coils is maximum when m = ?
a) 1 L1 L 2
Q.5
d) none of these
[Ans. : a]
d) 0.5
[Ans. : b]
d) H/m
[Ans. : c]
The unit of coefficient of coupling is ______ .
a) amperes
Q.4
c) 90 %
VTU : July-09, Jan.-11
If the entire flux produced by one coil links with the other then its coefficient of coupling
is ________ .
a) zero
Q.3
+
b)
c) L1 = L 2
L1 L 2
+
VTU : July-11
d) L1 L 2 .
[Ans. : b]
If coefficient of coupling between two coils is increased, mutual inductance between the
VTU : June-12
coils ___________.
+
a) is increased
b) is decreased
c) remains unchanged
d) none of these
[Ans. : a]
qqq
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Notes
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3
D.C. Machines
Chapter at a Glance
1.
E.M.F. Equation of D.C. Generator
E =
2.
fP N Z
60A
e.m.f. equation with A = P for Lap and A = 2 for Wave
Types of D.C. Generators
D.C. generators
Separately
excited
Self
excited
Shunt
Series
Compound
Cumulative
Differential
Fig. 3.1 Types of d.c. generators
3.
Shunt Generator
Ia = IL + Ish
IL
Ish
Ra
G
Rsh
Vt
F1
Load
F2
Ia
A1
G
E
IL
LOAD
A2
Fig. 3.2
Fig. 3.3 D.C. shunt generator
(3 - 1)
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–
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Ish =
3-2
D.C. Machines
Vt
R sh
E = Vt + IaRa + Vbrush
4.
Series Generator
Ia = Ise= IL where Ise = Current through series field winding.
E = Vt + IaRa + IaR
se
+ Vbrush
S1
Ia
S2
IL
A1
G
E
+
Vt
LOAD
–
A2
5.
Compound Generator
a) Long Shunt :
Fig. 3.4 Series generator
Ia = Ish + IL
Ish
IL
Vt
=
R sh
Ish
S2
Ise
F1
S1
A1
Ia
where
Rsh = Resistance of shunt field winding
F2
G
E
LOAD
+
Vt
–
IL
A2
Fig. 3.5 Long shunt compound generator
E = Vt + IaRa + Ia Rse + Vbrush where Rse = Resistance of series field winding
b) Short Shunt :
S2
Ia = IL + Ish
Ish =
Ish
E - Ia R a
R sh
F1
S1
Ia
A1
G
F2
Ise
A2
E
IL
LOAD
+
Vt
–
Ia
Fig. 3.6 Short shunt compound generator
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Now,
\
6.
3-3
Ise = IL hence E = Vt + Ia Ra + IL Rse + Vbrush
Ish =
Vt + I L R se
R sh
Efficiency of a D.C. Machine
% Mechanical efficiency (h m ) =
% Electrical efficiency (h e ) =
%h=
E g Ia
´ 100
Output of driving machine
Vt I L
´ 100
E g Ia
% Commercial efficiency (h c ) =
7.
D.C. Machines
Vt I L
´ 100
Output of the driving machine
Total output
´ 100
Total input
Back E.M.F. in a D.C. Motor
Eb =
fPNZ
60 A
volts
V = Eb + Ia Ra + Brush drop
Ia =
8.
V- E b
Ra
Power Equation of a D.C. Motor
VIa = Eb Ia + I 2a Ra
9.
Torque Equation of a D.C. Motor
P = T ´ w Watts
Ta =
PZ
PZ
1
= 0.159 f Ia .
fI ´
A
2p a
A
Nm
· So on no load, motor produces a torque Ta0 which satisfies the friction, windage and
iron losses of the motor.
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Power developed (Eb0 ´ Ia0)
where
and
=
D.C. Machines
Friction, windage and, iron losses.
Eb0 = Back e.m.f. on no load.
Ia0 = Armature current drawn on no load.
10. Types of D.C. Motors
D.C. motor
D.C. shunt motor
D.C. compound motor
D.C. series motor
Long shunt
Short shunt
Fig. 3.7 Types of d.c. motors
11. Torque and Speed Equations
T µ f Ia
T µ Ia
…For shunt motors
T µ Ia f µ I 2a
Eb µ f N
i.e.
…For series motors
N µ
Eb
f
Important Theory Questions and Answers
Ø
Draw the neat sketch representing the cut section of a d.c. machine. Explain the important
features of different parts involved there on.
+ VTU : Jan.-03, July-04, 06, 08, 11 Marks 5; June-10, Marks 8
· Fig. 3.8 shows a cross section of typical d.c. machine (See Fig. 3.8 on next page)
·
It consists of the following parts :
Yoke
a) Functions :
1. It serves the purpose of outermost cover of the d.c. machine.
2. It provides mechanical support to the poles.
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D.C. Machines
F2 F1
–
+
Yoke
Inter polar axis
Field winding
Flux produced
N
S
S
Commutator
Pole shoe
Polar or field axis
Pole core
Armature winding
Armature
core
N
Armature
tooth
Armature
slot
Shaft
Brush
BASE
Fig. 3.8 A cross-section of typical d.c. machine
3. It forms a part of the magnetic circuit. It provides a path of low reluctance for
magnetic flux.
b) Choice of material : It is prepared by using cast iron. For large machines rolled steel,
cast steel, silicon steel is used.
Poles
Each pole is divided into two parts namely, I) Pole core and II) Pole shoe
a) Functions of pole core and pole shoe :
1. Pole core basically carries a field winding which is necessary to produce the flux.
2. It directs the flux produced through air gap to armature core, to the next pole.
3. Pole shoe enlarges the area of armature core to come across the flux, which is
necessary to produce larger induced e.m.f.
b) Choice of material :
It is made up of magnetic material like cast iron or cast steel.
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D.C. Machines
· As it requires a definite shape and size, laminated construction is used. The laminations
of required size and shape are stamped together to get a pole which is then bolted to
the yoke.
Field Winding (F1 - F2)
· The field winding is wound on the pole core with a definite direction.
a) Functions : To carry current due to which pole core, on which the field winding is
placed behaves as an electromagnet, producing necessary flux.
As it helps in producing the magnetic field i.e. exciting the pole as an electromagnet it is
called field winding or exciting winding.
b) Choice of material : It is made up of aluminium or copper.
Armature
· The armature is further divided into two parts namely,
I) Armature core II) Armature winding
I) Armature core :
Armature core is cylindrical in shape mounted on the shaft.
a) Functions :
1. Armature core provides house for armature winding i.e. armature conductors.
2. To provide a path of low reluctance to the magnetic flux produced by the field
winding.
b) Choice of material :
It is made up of magnetic material like cast iron or cast steel.
II) Armature winding : Armature winding is nothing but the interconnection of the
armature conductors, placed in the slots provided on the armature core periphery.
a) Functions :
1. Generation of e.m.f. takes place in the armature winding in case of generators.
2. To carry the current supplied in case of d.c. motors.
3. To do the useful work in the external circuit.
b) Choice of material :
It has to be made up of conducting material, which is copper.
Commutator
· The basic nature of e.m.f. induced in the armature conductors is alternating. This needs
rectification in case of d.c. generator, which is possible by a device called commutator.
a) Functions :
1. To facilitate the collection of current from the armature conductors.
2. To convert internally developed alternating e.m.f. to unidirectional ( d.c.) e.m.f.
b) Choice of material : It is also made up of copper segments.
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Brushes and Brush Gear
· Brushes are stationary and resting on the surface of the commutator.
a) Function : To collect current from commutator and make it available to the stationary
external circuit.
b) Choice of material :
Brushes are normally made up of soft material like carbon.
Bearings
· Ball-bearings are usually used as they are more reliable. For heavy duty machines, roller
bearings are preferred.
Ø
What is back e.m.f. ? State its significance.
+ VTU : Feb.-05; Jan.-11; July-07, 08, 11, Marks 5
· In a d.c. motor, electrical input i.e. the supply voltage is the cause for the armature
current and the motoring action and hence this induced e.m.f. opposes the supply
voltage. This e.m.f. tries to set up a current through the armature which is in the
opposite direction to that, which supply voltage is forcing through the conductor.
· As this e.m.f. always opposes the supply voltage, it is called back e.m.f.
Significance of Back E.M.F.
· Due to the presence of back e.m.f. the d.c. motor becomes a regulating machine.
· Back e.m.f. is proportional to speed, Eb µ N.
· Back e.m.f. regulates the flow of armature current and it automatically alters the
armature current to meet the load requirement.
· At start the speed N of the motor is zero hence the back e.m.f. is also zero.
Ø
Derive the expression of armature torque develped in a d.c. motor.
+ VTU : Jan.-03, 05, 09, 13, 14; July-03, 07, Dec.-11, Marks 6
· Consider a wheel of radius R meters acted upon
circumferential force F newtons as shown in the Fig. 3.9.
· Angular speed is,
w =
by
a
R
F
2 pN
rad/sec
60
Fig. 3.9
· Workdone in one revolution is,
W = F ´ Distance travelled in one revolution = F ´ 2 p R Joules
P = Power developed =
\
Rotation
F´ 2 p R
Workdone
F ´ 2p R
æ 2 pNö
=
=
= ( F ´ R) ´ ç
÷
Time
Time for 1 rev
æ 60 ö
è 60 ø
çN÷
è ø
P = T ´ w Watts
· Let Ta be the gross torque developed by the armature of the motor.
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· The gross mechanical power developed in the armature is Eb Ia.
2 pN
Power in armature = Armature torque ´ w
i.e.
Eb I a = T a ´
60
fPNZ
Eb =
But Eb in a motor is given by,
60 A
2 pN
fPNZ
\
´ Ia = Ta ´
60 A
60
Ta =
\
Ø
PZ
PZ
1
= 0.159 f Ia .
f Ia ´
A
2p
A
Nm
Explain the various characteristics of d.c. shunt motor.
+ VTU : Jan.-04, 06, 07, 08, 10; Feb.-05; July-05, Marks 5
i) Torque - Armature current characteristics :
· For a constant values of Rsh and supply
voltage V, Ish is also constant and hence
flux is also constant.
\
T
Ta
Tsh
Ta µ Ia
· As load increases, armature current
increases, increasing the torque developed
linearly.
Ta0
· On no load Tsh = 0. The current required is
Ia0 on no load to produce Ta0 and hence
Tsh graph has an intercept of Ia0 on the
current axis.
0
Tf
Loss
torque
Ia0
Ia
Fig. 3.10 T Vs Ia for shunt motor
ii) Speed - Armature current characteristics :
· From the speed equation we get,
N µ V – I a Ra
N
as f is constant.
· So as load increases, the armature
current increases and hence drop IaRa
also increases.
N0
Constant speed line
· Hence for constant supply voltage,
V – Ia Ra decreases and hence speed
reduces.
· But as Ra is very small, for change in Ia
from no load to full load, drop IaRa is
very small and hence drop in speed is
Ia
Fig. 3.11 N Vs Ia for shunt motor
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also not significant from no load to full
load.
D.C. Machines
N
Constant speed line
iii) Speed - Torque characteristics :
· This curve shows that the speed almost
remains
constant
though
torque
changes from no load to full load
conditions. This is shown in the
Fig. 3.12.
T
Fig. 3.12 N Vs T for shunt motor
Ø
Explain the various characteristics of d.c. series motor.
+ VTU : Jan.-04, 07, 08, 10; Feb.-05; July-05, Marks 5
i) Torque - Armature current characteristics :
· For the series motor the series field winding is carrying the entire armature current
hence,
T
Ta µ f Ia µ I 2a
TµIa
· Thus torque in case of series motor is
proportional to the square of the
armature current. This relation is
parabolic in nature as shown in the
Fig. 3.13.
Ta
Tsh
2
TµIa
Tf
· As load increases, armature current
increases and torque produced increases
proportional to the square of the
armature current upto a certain limit.
0
Ia0
Point of
saturation
Ia
Fig. 3.13 T Vs Ia for series motor
· Saturation means though the current through the winding increases, the flux produced
remains constant. Hence after saturation the characteristics take the shape of straight
line as flux becomes constant, as shown.
ii) Speed - Armature current characteristics
From the speed equation we get,
Eb
V- I a R a -I a R se
N µ
µ
f
Ia
as f µ Ia in case of series motor
· The values of Ra and Rse are so small that the effect of change in Ia on speed overrides
the effect of change in V – Ia Ra – Ia Rse on the speed.
· Hence in the speed equation, Eb @ V and can be assumed constant.
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D.C. Machines
N
· So speed equation reduces to,
1
Ia
Nµ
· So speed - armature current characteristics
is rectangular hyperbola type as shown
in the Fig. 3.14.
iii) Speed - Torque characteristics
· In case of series motors,
T µ I 2a
and
N µ
0
1
Ia
Ia
Fig. 3.14 N Vs Ia for series motor
· Hence we can write,
Nµ
1
T
· Thus as torque increases when load
increases, the speed decreases.
N
· On no load, torque is very less and
hence speed increases to dangerously
high value.
· Thus the nature of the speed - torque
characteristics is similar to the nature of
the
speed - armature
current
characteristics.
T
· The speed - torque characteristics of a
series motor is shown in the Fig. 3.15.
Ø
Fig. 3.15 N Vs T for series motor
State the applications of d.c. shunt and series motors.
+ VTU : Jan.-04, 06, 07, 08, 10; Feb.-05; July-05, Marks 2
Types of motor
Characteristics
Applications
Shunt
Speed is fairly constant and
medium starting torque.
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2)
3)
4)
5)
6)
Blowers and fans
Centrifugal and reciprocating pumps
Lathe machines
Machine tools
Milling machines
Drilling machines
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Series
Ø
3 - 11
High starting torque. No
load condition is dangerous.
Variable speed.
D.C. Machines
1)
2)
3)
4)
5)
Cranes
Hoists, Elevators
Trolleys
Conveyors
Electric locomotives
Explain the necessity of starter for a d.c. motor. With a neat sketch explain the working of
three point starter for a d.c. motor.
+ VTU : Jan.-03, 04, 07; July-04, 05, 06, 09; Feb.-05, Marks 8
· At the starting instant the speed of the motor is zero, (N = 0). As speed is zero, there
cannot be any back e.m.f.
Eb at start = 0
\
· The voltage equation of a d.c. motor is, V = Eb + Ia Ra
at start,
V = IaRa
as Eb = 0
Ia =
\
V
Ra
… At start
· As armature resistance is very small, the armature current at start is very high.
· So at start, motor is showing a tendency to draw an armature current which is 15 to 20
times more than the full load current.
· Such high current drawn by the armature at start.
1. It may affect the performance of the other equipments connected to the same line.
2. Such excessively high armature current, blows out the fuses.
3. A large armature current flowing for a longer time may burn the insulation of the
armature winding.
· To restrict this high starting armature current, a variable resistance is connected in
series with the armature at start. This resistance is called starter or a starting resistance.
So starter is basically a current limiting device.
· The Fig. 3.16 shows three point starter. (See Fig. 3.16 on next page)
· The starter is basically a variable resistance, divided into number of sections. The
contact points of these sections are called studs and brought out separately shown as
OFF, 1, 2, … upto RUN.
· There are three main points of this starter :
1. ‘L’ ® Line terminal to be connected to positive of supply.
2. ‘A’ ®
To be connected to the armature winding.
3. ‘F’ ® To be connected to the field winding.
· The OLR and NVC are the two protecting devices of the starter.
· Initially the handle is in the OFF position. The d.c. supply to the motor is switched on.
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R2
R1
Starting
resistance
Soft iron piece
1
OFF
2
R3
3
D.C. Machines
R4
4
5
R5
No volt coil
Run
Starter handle
Spring
Over load
release
L
+
F
A
Lever Triangular
iron piece
F1
V
D.C.
F2
–
A1
M
A2
Fig. 3.16 Three point starter
· Then handle is slowly moved against the spring force to make a contact with stud
No. 1. At this point, field winding gets supply through the parallel path provided to
starting resistance, through NVC. While entire starting resistance comes in series with
the armature and armature current which is high at start, gets limited.
· As the handle is moved further, it goes on making contact with studs 2, 3, 4 etc.,
cutting out the starting resistance gradually from the armature circuit. Finally when the
starter handle is in ‘RUN’ position, the entire starting resistance gets removed from the
armature circuit and motor starts operating with normal speed.
Important Solved Examples
Example 3.1
A 4 pole, lap connected D.C. generator has 600 armature conductors and runs at
1200 r.p.m. This generator has a total flux of 24 Wb in it.
i) Calculate the e.m.f. induced in the above D.C. generator.
ii) Find the speed at wich it should be driven to produce the same e.m.f. when wave connected.
+
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D.C. Machines
Solution : P = 4, lap hence A = P = 4, Z = 600, N = 1200 r.p.m.
Here total flux fT = f´ P = 0.24 Wb is given.
Key Point
fT
0.24
=
= 0.06 Wb
P
4
\
f =
i)
Eg =
ii) For wave connection,
A = 2
\
Eg =
fPN¢ Z
60A
\
N¢ =
60 ´ 2 ´ 720
= 600 r.p.m.
0.06 ´ 4 ´ 600
Example 3.2
fPNZ 0.06 ´ 4 ´ 1200 ´ 600
=
= 720 V
60A
60 ´ 4
i.e. N¢ =
60A ´ Eg
fPZ
…New speed.
A 4 pole, lap wound, d.c. generators has 42 coils with 8 turns per coils. It is driven
at 1120 r.p.m. If useful flux per pole is 21 mWb, calculate the generated e.m.f. Find the speed
at which it is to be driven to generate the same e.m.f. as calculated above, with wave wound
armature.
Solution :
P=4
–3
f = 21 mWb = 21 ´ 10 Wb
N = 1120 r.p.m.
Coils = 42 and turns/coil = 8
Total turns = Coils ´ Turns/coil = 42 ´ 8 = 336
Z = 2 ´ total turns = 2 ´ 336 = 672
i) For lap wound,
A=P
\
fN Z
21 ´ 10 -3 ´ 1120 ´ 672
=
= 263.424 V
60
60
E =
ii) For wave wound, A = 2
and
E = 263.424 V
\
E =
fP N Z
120
i.e.
263.424 =
21 ´ 10 -3 ´ 4 ´ N ´ 672
120
N = 560 r.p.m.
Example 3.3
A shunt generator supplies a load of 10 kW at 200 V, through a pair of feeders of
total resistance 0.05 W. The armature resistance is 0.1 W. The shunt field resistance is 100W.
Find the terminal voltage and the generated e.m.f.
Solution :
The arrangement is shown in the Fig. 3.17.
P = VL ´ IL
i.e.
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IL =
\
IL
3 - 14
D.C. Machines
P
VL
10 ´ 10 3
=
= 50 A
200
Vt = VL + IL ´ Rfeeder
Feeder
0.05 W
IL
Ia
IL
+
Ish
G
Eg
= 200 + 50 ´ 0.05
VL
200 V
Rsh Vt
100 W
Load
–
= 202.5 V
This is voltage across field winding.
Vt
202.5
Ish =
=
\
R sh
100
Fig. 3.17
= 2.025 A
\
Ia = IL + Ish = 50 + 2.025 = 52.025 A
Eg = Vt + Ia Ra = 202.5 + 52.025 ´ 0.1 = 207.7025 V
\
Example 3.4 A dc shunt generator supplies a load of 7.5 kW of at 200 V. Calculate the induced
e.m.f. if armature resistance is 0.6 W and field resistance is 80 W .
Solution :
IL
\
Ra = 0.6 W, Rsh = 80 W
IL
P 7.5 ´ 10 3
= 37.5 A
=
=
Vt
200
Ish =
+
Vt
200
= 2.5 A
=
R sh
80
+
Ish
Ia
G
AU : May-09
Ra
Vt
Rsh
200 V
Ia = IL + Ish = 37.5 + 2.5 = 40 A
Load P =
7.5 kW
_
\ Eg = Vt + IaRa
Fig. 3.18
= 200 + 400 ´ 0.6 = 224 V
… Induced e.m.f.
Example 3.5 A short-shunt cumulative compound DC generator supplies 7.5 kW at 230 V. The
shunt field, series field and armature resistances are 100, 0.3 and 0.4 ohms respectively.
Calculate the induced e.m.f. and the load resistance.
Solution :
IL =
Ish =
P
7.5 ´ 10 3
= 32.608 A
=
Vt
230
Vt + I L R se
230 + 32.608 ´ 0.3
=
= 2.3978 A
R sh
100
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\
3 - 15
D.C. Machines
Ia = IL + Ish
IL = Ise
= 32.608 + 2.3978
+
Rse
= 35 A
\ Eg = Vt + IaRa + ILRse
= 230 + 35 ´ 0.4 + 32.608 ´ 0.3
= 253.7824 A
0.3 W
Ish
Ia
Rsh
G
100 W
Vt
Load
P = 7.5 kW
Ra
0.4 W
Vt = 230 V
_
Fig. 3.19
Example 3.6 A short-shunt cumulative compound D.C. generator supplies 48 kW at 240 V. The
shunt field, series field and armature resistances are 120, 0.015 and 0.03 ohms respectively.
+
Calculate the induced e.m.f. and the load resistance.
Solution :
IL
P
48 ´ 10 3
= 200 A
=
=
Vt
240
I sh
V + I L R se
= t
R sh
=
IL = Ise
240 + 200 ´ 0.015
= 2.025 A
120
120 W
+
Rse
Ish
Rsh
0.015 W
Ia
G
Ra
Vt
= 240 V
P = 48 kW
Load
0.03 W
–
\
I a = I L + I sh = 202.025 A
\
Eg = Vt + I a R a + I L R se = 240 + 202.025 × 0.03 + 200 × 0.015 = 249.06 V
RL =
AU : Dec.-11
Fig. 3.20
Vt
240
=
= 1.2 W
IL
200
Example 3.7 A DC motor connected to a 460 V supply has no armature resistance of 0.15 ohms.
Calculated (1) the value of back e.m.f. when the armature current is 120 A (2) the value of
armature current when the back e.m.f. is 447 V.
Solution :
1)
V = 460 V, Ra = 0.15 W
Ia = 120 A
\
Eb = V – IaRa = 460 – 120 ´ 0.15 = 442 V
2)
Eb = 447 V
\
Ia =
V - Eb
460 - 447
=
= 86.667 A
Ra
0.15
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D.C. Machines
Example 3.8 A 4 pole, d.c. motor has lap connected armature winding. The flux per pole is
30 mWb. The number of armature conductors is 250. When connected to 230 V d.c. supply it
draws an armature current of 40 A. Calculate the back e.m.f. and the speed with which motor
is running. Assume armature resistance is 0.6 W.
Solution :
P = 4, A = P = 4 as lap, V = 230 V, Z = 250
–3
f = 30 mWb = 30 ´ 10 Wb, Ia = 40 A
From voltage equation, V = Eb + Ia Ra
\
Eb = 206 V
And
Eb =
\
N = 1648 r.p.m.
fPNZ
60 A
i.e.
206 =
230 = Eb + 40 ´ 0.6
i.e.
30 ´ 10 -3 ´ 4 ´ N ´ 250
60 ´ 4
Example 3.9 A 240 V, 4 pole, shunt motor running at 1000 r.p.m. gives 15 H.P. with an
armature current of 50 A and a field current of 1.0 A. The armature winding is
wave-connected and has 540 conductors. Its resistance is 0.1 W and drop at each brush is 1 V.
Find a) Useful torque ; b) Total torque ; c) Useful flux per pole and d) Rotational losses.
+
Solution : V = 240 V,
P = 4,
N = 1000 r.p.m.,
VTU : Feb.-2000
Pout = 15 H.P.
Ia = 50 A, Ish = 1 A, Ra = 0.1 W, wave connected so A = 2 , Z = 540,
Vbrush = 1V/brush
a) Useful torque is shaft torque,
P
P
Tsh = out = out
w
æ 2pN ö
ç 60 ÷
è
ø
=
15 ´ 735 . 5 ´ 60
= 105.35 N-m
2p ´ 1000
b) Total torque is armature torque,
Ta =
Now
Eb = V – IaRa – Brush drop
=
\
E bI a
Power developed by armature
=
2
p
N
æ
æ 2pN ö
ö
ç 60 ÷
ç 60 ÷
è
è
ø
ø
Ta =
240 – (50 ´ 0.1) – (1´ 2) = 233 V
233 ´ 50
= 111.2493 N-m
2
æ p ´ 1000 ö
ç
÷
60
è
ø
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as 1 H.P. = 735.5 W
Basic Electrical Engineering
c)
Eb =
\
f =
d)
3 - 17
D.C. Machines
fP N Z
and A = 2 for wave connection
60 A
60 A E b
60 ´ 2 ´ 233
=
= 12.95 mWb
PNZ
4 ´ 1000 ´ 540
Rotational losses,
Lost torque =
- Tsh ) =
Rotational losses
æ 2pN ö
ç 60 ÷
è
ø
Rotational losses
æ 2pN ö
ç 60 ÷
è
ø
\
(Ta
\
Rotational losses =
(111 . 2493 - 105 . 35) ´ 2p ´ 1000
= 617.77 watts
60
A 200 V, 4 pole, lap wound, d.c. shunt motor has 800 conductors on its armature.
Example 3.10
The resistance of the armature winding is 0.5 W and that of shunt field winding is 200 W. The
motor takes a current of 21 A, the flux per pole is 30 mWb. Find the speed and the gross
+
torque developed in the motor
Solution : P = 4,
V = 200 V,
f = 30 mWb =
For shunt motor, Ish =
Ia
=
A = P = 4,
30 ´ 10 -3
Wb,
Z = 800,
=
\
190 =
\
N =
V
200
=1A
=
R sh
200
IL
Rsh
Ra
fP N Z
60 A
+
Ia
Ish
IL – I sh = 21 – 1 = 20 A
M
V=200 V
–
Fig. 3.21
30 ´ 10 - 3 ´ 4 ´ N ´ 800
60 ´ 4
475 r.p.m
Ta = 0.159 ´ fI a ´
PZ
A
= 0.159 ´ 30 ´ 10
Example 3.11
Rsh = 200 W
IL = 21 A
Eb = V – Ia Ra = 200 – 20 ´ 0.5 = 190 V
Now, E b
Ra = 0.5 W,
VTU : Dec.-96
–3
… From torque equation
´ 20 ´
4 ´ 800
4
= 76.38 N-m.
A 4 pole, 250 V series motor has wave connected armature with 1254 conductors.
The flux per pole is 22 mWb when the motor is taking 50 A. The armature and series field coil
resistance are respectively 0.3 W and 0.2 W. Calculate the speed and torque of the motor and
+
also the power developed in watts.
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VTU : Feb.-05, Marks 10
Basic Electrical Engineering
3 - 18
D.C. Machines
P = 4, Z = 1254, f = 22 mWb, Ia = 50 A, A = 2
Solution :
Eb = V – Ia(Ra + Rse)
= 250 – 50(0.3 + 0.2)
\
\
250 V
fPNZ
=
60A
225 =
Rse
0.2 W
= 225 V
Eb
Ia = Ise
+
0.3 W
22 ´ 10 -3 ´ 4 ´ N ´ 1254
60 ´ 2
M
Ra
–
Fig. 3.22
N = 244.6716 r.p.m.
…Speed
4 ´ 1254
PZ
Ta = 0.159 f Ia é ù = 0.159 ´ 22 ´ 10 -3 ´ 50 ´
êë A úû
2
= 438.65 Nm
Pm = Ta ´ w = Ta ´
…Torque
244.6716
2pN
= 438.65 ´ 2p t ´
60
60
= 11239.068 W = 11.239 kW
… Power developed
Pm = EbIa = 225 ´ 50 = 11250 W
OR
The difference is due to use of approximated figures in the expression of Ta.
A 120 V d.c. shunt motor has an armature resistance of 0.2 W and shunt field
Example 3.12
resistance of 60 W. It runs at 1800 r.p.m. when it takes full load current of 40 A. Find the
speed of the motor while it is operating at half the full load, with load terminal voltage remaing
+
same.
VTU : Jan.-03, Marks 6
Solution :
+
IL1
Ia1
40 A
V = 120 V
+
T1
Ish
M
Rsh
IL2
Ia2
T2 = T1 / 2
Ish
Full
load
V = 120 V
–
M
Rsh
Half
load
–
(a) N1 = 1800 r.p.m.
(b) N2 = ?
Fig. 3.23
Ish =
V
120
=2A
=
R sh
60
\
f µ Ish hence f is constant
\
T µ f I a µ Ia
…Constant
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Basic Electrical Engineering
3 - 19
\
T1
I
= a1
T2
I a2
\
Ia2 = 0.5 Ia1
Now
IL1 = Ish + Ia1
\
Ia2 = 0.5 ´ 38 = 19 A
i.e.
D.C. Machines
T1
I
= a1
0.5 T1 I a2
…(1)
i.e. Ia1 = 40 – 2 = 38 A
\
Eb1 = V – Ia1Ra = 120 – 38 ´ 0.2 = 112.4 V
\
Eb2 = V – Ia2Ra = 120 – 19 ´ 0.2 = 116.2 V
N µ
But
Eb
µ Eb
f
\
N1
E b1
=
N2
E b2
\
N2 =
Key Point
…f is constant
116.2
´ 1800 = 1860.854 r.p.m.
112.4
…Half load speed
Remember that load decides torque, which decides armature current and not directly
line current. Torque condition gives new armature current.
A 250 V d.c. shunt motor has an armature resistance of 0.5 W and shunt field
Example 3.13
resistance of 250 W. When driving a load at 600 r.p.m., the torque of which is constant, the
armature takes 20 A. If it is desired to raise the speed from 600 to 800 r.p.m., what resistance
must be inserted in the field circuit? Assume the magnetization curve to be a straight line.
+
VTU : Aug.-08, Marks 10
Solution :
Ia1 =
20 A
Ia2
+
Ish1
Ra =
M
0.5 W
Rsh
250 W
Ra
V = 250 V
0.5 W
+
Rx
M
Ish2
V = 250 V
Rsh
–
–
(a) N1 = 600 r.p.m.
(b) N2 = 800 r.p.m.
Fig. 3.24
I sh1
V
250
=1A
=
=
R sh
250
E b1 = V - I a 1 R a = 250 - 20 ´ 0 . 5 = 240 V
T µ fI a µ I sh I a
\
T1
I
I
= sh 1 ´ a 1 = 1
T2
I sh 2 I a 2
...Torque is constant
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Basic Electrical Engineering
1
20
´
I sh 2 I a 2
\
= 1
Nµ
3 - 20
i.e. I a 2 I sh 2 = 20
...(1)
Eb Eb
µ
f
I sh
E b 1 I sh 2
N1
=
´
E b 2 I sh 1
N2
\
D.C. Machines
600
240 I sh 2
=
´
Eb2
800
1
i.e.
\
Eb2
= 320
I sh 2
But
E b2 = V - I a 2 R a = 250 - 0 . 5 I a 2
\
æ 20 ö
E b2 = 250 - 0 . 5 ç
÷
è I sh 2 ø
...(2)
250 Using in equation (2),
...From equation (1)
10
I sh 2
I sh 2
= 320
250 I sh 2 - 10 = 320 I 2sh 2 i.e. 320 I 2sh 2 - 250 I sh 2 + 10 = 0
\
Solving,
I sh2 = 0.7389 A, 0.0422 A
Neglecting lower value,
I sh2 = 0.7389 A
But,
I sh2 =
V
R sh + R x
R x = 88.3407 W
\
i.e.
0.7389 =
250
250 + R x
...External resistance required
Important Multiple Choice Questions with Answers
Working Principle of a D.C. Machine as a Generator
Q.1
The generator works on the principle of ________ .
a) statically induced e.m.f.
c) dynamically induced e.m.f.
Q.2
Q.3
b) mutual induction
d) Kirchhoff's laws
[Ans. : c]
The direction of induced e.m.f. in a generator is given by ________ .
a) Fleming's right hand rule
b) right hand thumb rule
c) Fleming's left hand rule
d) Cork screw rule.
[Ans. : a]
In Fleming's right hand rule, the index finger indicates ________ .
a) E. M. F.
c) direction of relative motion
b) lines of flux
d) current
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[Ans. : b]
Basic Electrical Engineering
Q.4
3 - 21
D.C. Machines
If the angle between the plane of flux and plane of relative motion is 0º then the induced
e.m.f. is ________ .
a) zero
b) maximum
c) infinite
d) none of these
[Ans. : a]
Q.5
If B is the flux density, l is active length of conductor and v is velocity of conductor, then
induced e.m.f. is given by,
a) E = Bl 2 v
b) E = B2 l v
d) E = B l v2 [Ans. : c]
c) E = B l v
Constructional Details of a D.C. Machine
Q.1
+
A commutator is made up of _______.
a) iron lamination
b) copper segments
c) both iron lamination and copper segments d) none of the above.
Q.2
Q.3
Q.4
+
a) mica
d) carbon.
b) copper
+
a) a.c.
b) d.c.
c) pulsating
d) d.c. superimposed over a.c.
Q.9
[Ans. : a]
c) commutator
d) brushes. [Ans. : c]
________ provides mechanical support to the d.c. machine.
b) Armature
c) Yoke
d) Bearings [Ans. : d]
c) cast steel
d) cast iron [Ans. : d]
Yoke is made up of ________ .
b) aluminium
Air ducts are provided in armature core to ________ .
b) cool the machine
d) none of these
[Ans. : b]
The generation of e.m.f. takes place in ________ of a d.c. machine.
a) armature winding
b) field winding
c) pole core
d) interpoles
[Ans. : a]
The brushes are made up of ________ .
a) copper
Q.10
VTU : Jan.-10
+
b) equalizer rings
a) increase the core area
c) to accommodate the winding
Q.8
[Ans. : d]
The components of a d.c. generator which plays vital role in providing direct current is
VTU : Jan.-10
_____ .
a) copper
Q.7
c) cast iron
VTU : July-09
The nature of current flowing in the armature of a d.c. machine is _____ .
a) Poles
Q.6
[Ans. : b]
The material for commutator brushes is always _______ .
a) dummy coils
Q.5
VTU : Jan.-09, 11
b) iron
c) silver
d) carbon
[Ans. : d]
d) 16
[Ans. : c]
A. d.c. machine having ________ poles has four magnetic circuits.
a) 8
b) 2
c) 4
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Basic Electrical Engineering
Q.11
Q.12
Q.13
3 - 22
The armature of a d.c. machine is laminated because ________ .
c) to reduce copper loss
d) to reduce mechanical loss
a) it is easy to collect e.m.f.
b) they are rotating
c) to avoid wear and tear of commutator
d) to reduce friction losses
[Ans. : c]
The function of commutator is ______ .
b) production of flux
d) convert a.c. e.m.f. to d.c.
[Ans. : d]
Practically number of commutator segments is ______ .
b) less than armature coils
d) less than field coils
[Ans. : d]
The bearings used to support the rotor shaft are ______ .
b) roller bearings
c) magnetic bearings d) ball bearings
+
The yoke of a DC machine is made of ________ .
a) silicon steel
Q.17
[Ans. : b]
The brushes are made up of soft material because ________ .
a) bush bearings
Q.16
VTU : June-10
b) to reduce eddy current loss
a) more than field coils
c) equal to armature coil
Q.15
+
a) to reduce hysteresis loss
a) collection of voltage
c) production of torque
Q.14
D.C. Machines
b) soft iron
c) aluminum
[Ans. : d]
VTU : June-11
d) cast steel
+
Carbon brushes are used in a DC machine because_____ .
[Ans. : d]
VTU : June-11
a) carbon lubricates and polishes the commutator b) contact resistance is decreased
c) carbon is cheap
d) none of these
[Ans. : d]
Q.18
Q.19
a) to collect current from conductors
b) to change d.c. to a.c.
c) to conduct the current to brushes
d) to change a.c. to d.c.
The rotating part of d.c. machine is called ________.
a) armature
Q.20
c) frame
b) aluminium
c) cast steel
b) copper
d) cast iron
c) mica
b) field
[Ans. : d]
+
VTU : June-12
+
[Ans. : d]
+
+
c) armature
[Ans. : a]
VTU : Jan-13
VTU : June-13
d) steel
The rotating part of a dc machine is called the _____ .
a) rotor
VTU : Dec.-11
d) yoke
The field coils of a dc machine are made of _____ .
a) carbon
Q.22
b) field system
Yoke is made up of ___.
a) copper
Q.21
+
The function of a commutator in a d.c. generator is_____ .
[Ans. : b]
VTU : June-13
d) stator
[Ans. : c]
Types of Armature Winding
Q.1
For a 'P' pole lap wound armature of D.C. machine the number of parallel paths are equal
VTU : July.-09; Jan.-11
to _____ .
+
a) 2
b) 2P
c) P
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d) P/2.
[Ans. : c]
Basic Electrical Engineering
Q.2
3 - 23
The number of parallel paths equal to number of poles in ________ winding.
a) wave
Q.3
D.C. Machines
b) distributed
c) concentrated
d) lap
In wave type winding, the number of parallel paths of armature winding is ______ .
+
a) P
Q.4
b) 2
c) 4
b) 4
d) Zero
[Ans. : a]
b) low voltage, high current
d) high voltage, high current
[Ans. : b]
In generators, wave winding is preferred for ______ .
b) low voltage, high current
d) high voltage, high current
[Ans. : c]
For 'P' pole lap wound armature DC machine, number of parallel paths _____ .
+
a) 2
Q.8
[Ans. : b]
In generators, lap winding is preferred for ______ .
a) low voltage, low current
c) high voltage, low current
Q.7
P
2
d)
c) 1
a) low voltage, low current
c) high voltage, low current
Q.6
VTU : June-10
______ number of conductors constitute one turn.
a) 2
Q.5
[Ans. : d]
b) 2P
c) P
VTU : Jan.-13
d) P/2
[Ans. : c]
The number of parallel paths in the armature winding of a four pole, wave connected dc
machine having 28 coil-sides is ________ .
a) 28
b) 14
c) 4
d) 2
[Ans. : d]
E.M.F. Equation of D.C. Generator
Q.1
The e.m.f generated by a given d.c. generator depends upon ______.
+
a) flux only
b) speed only
VTU : Jan.-09,Jan.-11
c) flux and speed
d) terminal voltage.
[Ans. : c]
Q.2
In an e.m.f. equation of a d.c. generator, Z indicates ______ .
a) conductors
Q.3
b) brushes
b) 4
c) 1
d) Zero
[Ans. : c]
In a d.c. generator, if speed of prime mover is halved and flux per pole is doubled, the
induced e.m.f. will ______ .
a) remain constant
c) increase by 4 times
Q.5
d) commutator segments [Ans. : a]
The brush drop in d.c. machine is about ______ V/brush.
a) 2
Q.4
c) field turns
b) increase by 2 times
d) none of these
[Ans. : a]
E.M.F. of d.c. machine is inversely proportional to ________.
+
a) flux/pole
paths
d) number of parallel
[Ans. : d]
b) poles
c) conductors
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
VTU : June-12
Basic Electrical Engineering
Q.6
D.C. Machines
The emf generated by a d.c. generator depends on ________.
a) flux only
Q.7
3 - 24
b) speed only
c) flux and speed
+
VTU : Jan.-13, 14
d) terminal voltage
The emf generated in a dc generator depends upon _____ .
a) brush contact drop
b) commutation
c) number of parallel paths
d) terminal voltage
+
[Ans. : c]
VTU : June-13
[Ans. : c]
Self Excited Generator
Q.1
In a separately excited d.c. generator ______ is not necessary.
a) armature
Q.2
Q.3
b) residual magnetism
c) field flux
d) rotor
[Ans. : b]
Which is the cause of failure to excite self excited generator ?
a) Absence of residual flux
b) Wrong field connections
c) Driven in opposite direction
d) All of the above
[Ans. : d]
The dc generator having residual magnetism gives zero induced emf, the speed will be
_____ .
VTU : June-13
+
a) zero
b) very small
c) rated one
d) any
[Ans. : a]
Shunt Generator
Q.1
Residual magnetism is necessary in a d.c._______.
a) shunt generator
c) shunt motor
Q.2
+
VTU : Jan.-09
b) separately excited generator
d) series motor.
[Ans. : a]
In a shunt generator, which of the following relations is true ?
a) Ia = I L + Ish
b) I L = Ia + Ish
c) I L - Ia = Ish
d) I L + Ia + Ish = 0
[Ans. : a]
Q.3
The field winding of _____generator has thin wire of large number of turns.
a) series
b) shunt
c) compound
d) none of these
[Ans. : b]
Series Generator
Q.1
The field winding of ________ generator has thick wire of less number of turns.
a) series
b) shunt
c) compound
d) none of these
[Ans. : a]
Compound Generator
Q.1
When the fluxes produced by series and shunt field windings help each other, the
compound generator is called ______ .
a) differential compound
c) cumulatively compound
b) long shunt compound
d) short shunt compound
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
[Ans. : c]
Basic Electrical Engineering
3 - 25
D.C. Machines
Applications of Various Types of D.C. Generators
Q.1
______ generator is used for boosters on d.c. feeders.
a) D.C. series
c) Cumulatively compound
Q.2
b) D.C shunt
d) None of these
[Ans. : a]
______ generator is used for battery charging.
a) D.C. series
c) Cumulatively compound
b) D.C shunt
d) None of these
[Ans. : b]
Efficiency of a D.C. Machine
Q.1
The efficiency of a DC generator means its ________ .
a) electrical efficiency
b) overall efficiency
c) mechanical efficiency
d) none of the above.
[Ans. : d]
Principle of Operation of a D.C. Machine as a Motor
Q.1
The direction of force in a motoring action is determined by ______ .
a) Fleming's right hand rule
c) Fleming's left hand rule
b) end rule
d) right hand thumb rule
[Ans. : c]
Back E.M.F. in a D.C. Motor
Q.1
The back e.m.f. of a motor at the moment of starting is ______.
a) zero
Q.2
b) maximum
c) low
+
VTU : Jan.-09
d) optimum. [Ans. : a]
The relationship between the applied voltage and back e.m.f. in D.C. motors is _____ .
+
a) V = Eb + Ia Ra
b) V = Eb – Ia Ra
c) V = Eb
VTU : July-09
d) none of these.
[Ans. : a]
Q.3
The back e.m.f. in a motor is due to ______ .
a) generating action
c) reverse action
Q.4
[Ans. : a]
In a 240 V d.c. motor, the back e.m.f. is 220 V and Ra = 0.5 W then its armature current
is ______ .
a) 20 A
Q.5
b) motoring action
d) none of these
b) 10 A
c) 80 A
+
The d.c. motor equation is ______ .
a) V = Eb + IaRa
d) 40 A
b) V = Eb – IaRa
c) Eb = IaRa – V
[Ans. : d]
VTU : June-10
d) none of these
[Ans. : a]
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
Basic Electrical Engineering
Q.6
3 - 26
D.C. Machines
+
The current drawn by armature of a d.c. motor is ______ .
a) V / Ra
b) Eb / Ra
c) (V - Eb ) / Ra
VTU : Dec.-11
d) (Eb - V) / Ra .
[Ans. : c]
Q.7
a) zero
Q.8
Q.9
b) maximum
c) minimum
+
c) V
VTU : Jan.-13
d) 40 A
[Ans. : d]
+
The back e.m.f. in a dc motor is given as _______.
b) V – Ia Ra
VTU : June-12
d) optimum [Ans. : a]
In a 240 V d.c. motor, Eb = 220 V, Ra = 0.5 W, Ia is ______ .
a) 20 A
b) 10 A
c) 80 A
a) V + IaRa
Q.10
+
At the moment of starting a d.c. motor, its back emf is ________.
VTU : Jan.-14
d) IaRa
[Ans. : b]
+
The speed of the d.c. motor is _______.
VTU : Jan.-14
a) directly porportional to both its back emf and flux
b) inversely proportional to both its back emf and flux
c) directly porportional to flux but inversely proportional to its back emf
d) directly proportional to its back emf but inversely porportional to flux
[Ans. : d]
Torque Equation of a D.C. Motor
Q.1
An electrical equivalent of gross mechanical power developed in a d.c. motor is ______ .
a) VI L
Q.2
c) Eb Ia
d) Eb Ish
[Ans. : c]
A 4 pole d.c. motor has lap winding with 360 conductors. It takes armature current of 20A
and flux is 10 mWb. It develops gross torque of ______ .
a) 11.44 Nm
Q.3
b) Eb I L
b) 1.44 Nm
c) 8 Nm
d) 5.4 Nm
[Ans. : a]
As the speed of a d.c. motor increases, the armature current ______ .
a) increases
b) decreases
c) remains same
d) none of these
[Ans. : b]
Q.4
The armature torque is ______ that of shaft torque.
a) less than
b) same as
c) greater than
d) none of these
[Ans. : c]
Q.5
The no load power drawn by a d.c. motor is used to overcome ______ losses.
a) constant
Q.6
b) variable
c) copper
Torque in d.c. motor is proportional to ________.
a) only flux
b) only Ia
c) both flux and Ia
d) none of these
d) friction
+
[Ans. : a]
VTU : June-12
[Ans. : c]
TM
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Basic Electrical Engineering
3 - 27
D.C. Machines
Torque and Speed Equations
Q.1
The speed of a d.c. motor is ______ .
b) directly proportional to Ia
d) inversely proportional to Ra
a) directly proportional to Eb
c) inversely proportional to flux
Q.2
+
The torque of a shunt motor is proportional to_____ .
a) armature current
b) applied voltage
c) square of the armature current
d) none of these.
[Ans. : c]
VTU : Dec.-11
[Ans. : a]
Characteristics of D.C. Shunt Motor
Q.1
Q.2
Q.3
Which D.C. motor will be preferred for constant speed line shafting _____ .
+
VTU : July-09
a) cumulatively compound motor
b) differentially compound motor
c) shunt motor
d) series motor.
[Ans. : c]
The speed of a d.c. shunt motor ________ from no load to full load.
a) falls slightly
b) improves slightly
c) remains unchanged
d) falls rapidly.
+
VTU : Jan.-10
[Ans. : a]
______ motor has constant speed characteristics.
a) D.C. series
b) D.C. compound
c) D.C. shunt
d) None of these
[Ans. : c]
Q.4
______ motor has best speed regulation.
a) D.C. series
b) D.C. compound
c) D.C. shunt
d) None of these
[Ans. : c]
Q.5
When constant speed and medium starting torque is necessary, ______ motor is used.
a) d.c. series
Q.6
b) d.c. compound
c) d.c. shunt
d) none of these
A DC motor is still used in industrial applications because it is ____.
a) cheap
c) provides fine speed control
+
b) simple in construction
d) none of these
[Ans. : c]
VTU : June-11
[Ans. : c]
Characteristics of D.C. Series Motor
Q.1
______ motor cannot be started on no load.
a) D.C. series
b) D.C. compound
c) D.C. shunt
d) None of these
[Ans. : a]
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Q.2
Q.3
3 - 28
D.C. Machines
For the movement of a train d.c. ______ motors are used.
+
a) shunt
d) none of these
b) series
c) compound
+
The speed of a series motor at no-load is_____ .
a) zero
b) 1500 r.p.m.
VTU : June-10
c) 3000 r.p.m.
[Ans. : b]
VTU : Dec.-11
d) infinity.
[Ans. : d]
Characteristics of D.C. Compound Motor
Q.1
______ motor is used for rolling mills.
a) D.C. series
b) D.C. compound
c) D.C. shunt
d) None of these
[Ans. : b]
Q.2
As load current increases, the speed of ______ motor increases.
a) shunt
c) series
Q.3
b) cumulative compound
d) differential compound
[Ans. : d]
______ motor is not suitable for any practical application.
a) Shunt
c) Series
b) Cumulative compound
d) Differential compound
[Ans. : d]
Necessity of Starter
Q.1
The function of a starter in a d.c. motor is to ________ .
+
VTU : Jan-10,11
a) control its speed
b) increase its starting torque
c) limit the starting current to a safer value
d) reduce armature reaction effect. [Ans. : c]
qqq
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4
Measuring Instruments
Important Theory Questions and Answers
Ø
With the help of neat diagram explain the construction and principle of operation of
dynamometer type wattmeter.
+ VTU : Jan.-04, 07, 08, 09, 11; July-04, 05, 08, 11, Dec.-11, Marks 8
· The Fig. 4.1 shows the construction of the dynamometer type wattmeter.
Scale
Scale
Pointer
F1
F1 – F2 = Fixed coil
L
I1
F2
Supply
Fixed coil
or current coil
connections
M
C
Load
V
Moving coil
F1
or pressure coil
connections
Moving coil High resistance
R
R
F2
Fixed coil
(b)
(a)
Air friction
damping
Fig. 4.1 Construction of dynamometer type wattmeter
· It consists of a fixed coil. It is divided into two halves F1 and F2 positioned parallel to
each other. The distance between them can be adjusted to provide uniform magnetic
field required for the operation. These coils are air cored to avoid hysteresis losses.
These are clamped in place against the coil supports made up of ceramic.
· The moving coil is wound on a non-metallic former which is pivoted centrally between
the fixed coils. It is made highly resistive by connecting high resistance in series with it.
· A pointer is connected to the moving system madeup of aluminium.
· The fixed coil is called a current coil as it is connected in series with the load to carry
the current I1 which is main current.
(4 - 1)
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4-2
Measuring Instruments
· The moving coil is connected across the supply, carrying current I 2 proportional to the
voltage hence it is called pressure coil or voltage coil.
· The controlling torque is provided by the springs.
· The damping is provided by the air friction damping. The eddy current damping is not
used as it may distort the operating magnetic field.
Working :
While operation, the wattmeter is connected in the circuit as shown in the Fig. 4.2.
· When current passes through the fixed and
moving coils, both coils produce the
magnetic fields.
· The field produced by fixed coil is
proportional to the load current while the
field produced by the moving coil is
proportional to the voltage.
· As the deflecting torque is produced due
to the interaction of these two fields, the
deflection is proportional to the power
supplied to the load.
Wattmeter
Supply
M
L
C
V
I1
Load
I2
Fig. 4.2 Connections of wattmeter
· Thus the wattmeter indicates the power consumption of the load.
· It can be used for a.c. and d.c both.
D.C. working :
· For the air cored fixed coils the flux density B is proportional to the current through the
coils i.e. B µ I1 .
· While the current through pressure coil is proportinal to the voltage i.e. I 2 µ V.
· The deflecting torque is due to the interaction of the two fluxes hence proportinal
to B I 2 .
· Td µ BI 2 µ I1 V µ POWER as the d.c. power is the product of voltage and current.
A.C. working :
· In a.c. circuit the value of the instantaneous torque is proportional to the product of the
instantaneous voltage (v) and the current (i).
· Let f is the power factor angle of the load then the voltage and current are given by,
v = Vm sin wt and i = I m sin (wt - f) .
· Due to the inertia of the moving system, the deflection is proportinal to the average
value of the torque produced.
· Td µ average of (vi) µ average [ Vm sin wt ´ I m sin (wt - f) ]µ [V I cos f] µ POWER
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4-3
Measuring Instruments
· In a.c. operation, V and I are the r.m.s. values of the voltage and current respectively.
· Due to spring control, these instruments have uniform scale and q µ POWER.
Ø
With the help of neat diagram explain the construction and principle of operation of single
phase energy meter.
+
VTU : Jan.-03, Marks 10; July-03, 06, 07, 09; Feb.-05; Jan.-06, 10,
June-12, 13, Jan.-13, 14, Marks 8
· The Fig. 4.3 shows the induction type single phase energy meter.
To recording
mechanism
I2
Pressure
coil
M2
L1
A.C. Supply
Cu shading
bands
Shunt magnet
f2
Load
L2
Braking
magnet
Disc
Current
coil
(Less turn)
f1
M1
I1
Series magnet
Fig. 4.3 Induction type single phase energy meter
· It consists of two electromagnets whose core is made up of silicon steel laminations.
· The coil of one of the electromagnets, called current coil, is excited by load current
which produces flux. This is called a series magnet. The coil has few turns of heavy
guage wire.
· The coil of another electromagnet is connected across the supply and it carries current
proportional to supply voltage. This coil is called pressure coil. This is called shunt
magnet. This coil has large number of turns of fine wire.
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Measuring Instruments
· The flux produced by shunt magnet is brought in exact quadrature with supply voltage
with the help of copper shading bands placed over the central limb, whose position is
adjustable.
· The moving system consists of a light aluminium disc mounted on a light alloy shaft.
This disc is positioned in between series and shunt magnets. It is supported between
jewel bearings. The moving system runs on hardened steel pivot. A pinion engages the
shaft with the counting mechanism.
· The braking system consists of a permanent magnet placed near the aluminium disc for
braking mechanism. This magnet is used to control the speed of the disc.
· The registering mechanism records continuously a number which is proportional to the
revolutions made by the aluminium disc. By a suitable system, a train of reduction
gears, the pinion on the shaft drives a series of pointers.
Working :
· The current coil produces the alternating flux f1 which is proportional and in phase
with the current through the current coil.
· The pressure coil carries the current and produces the flux f2 which proportional to the
supply voltage V and lags behind it by 90° which is achieved by the copper shading
bands.
· Major portion of the flux f2 crosses the narrow gap between the central and the side
limbs of the shunt magnet and only small amount passes through the disc which is the
useful flux.
· Both the fluxes f1 and f2 induce e.m.f.s in the disc which produce the eddy currents in
the disc.
· The interaction between these fluxes and the eddy currents produce the necessary
driving torque and the disc starts rotating.
· The speed of disc is controlled by the C shaped magnet called braking magnet. When
the peripheral portion of the disc rotates in the air gap, eddy currents are induced in
the disc which oppose the cause producing them i.e. relative motion of disc with
respect to magnet. Hence braking torque Tb is generated. This is proportional to
speed N of the disc.
· By adjusting position of this magnet, desired speed of disc is obtained.
· Spindle is connected to recording mechanism through gears which record the energy
supplied.
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4-5
Measuring Instruments
Important Multiple Choice Questions with Answers
Classification of Measuring Instruments
Q.1
Q.2
Integrating meters are used for the measurement of ______ .
+
a) current
c) power
d) energy
c) integrating
d) none of these
b) voltage
VTU : Jan.-09
[Ans. : d]
A voltmeter is _______ instrument.
a) recording
b) indicating
[Ans. : b]
Q.3
A X-Y plotter is ________ instrument.
a) recording
b) indicating
c) integrating
d) none of these
[Ans. : a]
Q.4
An energymeter is _________ instrument.
a) recording
b) indicating
c) integrating
d) none of these
[Ans. : c]
Controlling System
Q.1
In the measuring instruments, under equilibrium condition, controlling torque (Tc) and
VTU : July-09
deflecting torque (Td) are _____ .
+
a) Tc = Td
b) Tc > Td
c) Tc < Td
d) none of these
[Ans. : a]
Q.2
Without _____ torque, pointer will swing beyond it's final position with indefinite deflection.
a) controlling
b) deflecting
c) damping
d) none of these
[Ans. : a]
Q.3
______ is used to obtain the controlling torque.
a) Air
Q.4
b) Springs
c) Fluid
d) Magnets [Ans. : b]
When pointer deflects and attains a final steady state position then ______ .
a) only damping torque acts
c) only deflecting torque acts
b) only controlling torque acts
d) both controlling and deflecting torques act
[Ans. : d]
Q.5
When pointer of an indicating instrument comes to rest in the final deflection position then
______ .
a) only controlling torque acts
c) both torques act
b) only deflecting torque acts
d) none of these
+
VTU : Aug.-11
[Ans. : c]
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4-6
Measuring Instruments
Damping System
Q.1
If the pointer moves very slowly to its final position without oscillation, the system is said
to be ______ .
a) critically damped
b) underdamped
c) overdamped
d) none of these
[Ans. : c]
Q.2
In practice slightly ______ systems are preferred.
a) critically damped
b) underdamped
c) overdamped
d) none of these
[Ans. : b]
Q.3
The damping force due to fluid is ______ that of air force.
a) greater than
b) less than
c) same as
d) none of these
[Ans. : a]
Q.4
In eddy current damping, the disc is made up of ______ .
a) iron
b) copper
c) aluminium
d) nickel alloy
[Ans. : c]
Dynamometer Type Wattmeter
Q.1
In a dynamometer wattmeter the fixed coil is ______ .
a) current coil
c) current or potential coil
Q.2
+
VTU : Jan.-09, 13
b) potential coil
d) none of the above
[Ans. : a]
An electrodynamometer type instrument can be employed for measurement of ______ .
VTU : Jan.-10
______ .
+
a) d.c. voltages
b) a.c. voltages
c) d.c. as well as a.c. voltages
d) d.c. voltages but for a.c. volatges, rectification is necessary
Q.3
[Ans. : c]
______ instrument is used as a calibration instrument.
a) Moving coil
b) Moving iron
c) Dynamometer
d) None of these
[Ans. : c]
Q.4
Q.5
The dynamometer type wattmeter is used to measure ______ .
VTU : June-10
b) only ac power
c) both d.c. and a.c. power
d) both active and reactive power [Ans. : c]
The pointer in the dynamometer type wattmeter is made of ______.
+
a) copper
Q.6
+
a) only d.c. power
b) aluminum
c) phosphor bronze
The dynamometer type wattmeter is used to measure ___ .
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VTU : June-10
d) platinum [Ans. : b]
+
VTU : Jan.-11
Basic Electrical Engineering
Q.7
Q.8
Q.9
4-7
Measuring Instruments
a) only D.C. power
b) only a.c. power
c) both a.c. and d.c. power
d) both active and reactive power [Ans. : c]
+
The moving coil in a dynamometer wattmeter is connected ______.
a) in series with the fixed coil
b) across the supply
c) in series with the load
d) across the load
VTU : Dec.-11
[Ans. : b]
The type of wattmeter commonly used for measurement of power in ac circuit is
VTU : June-12
___________.
+
a) rectifier type
b) dynamometer type
c) moving iron type
d) thermo-couple type
Dynamometer type instruments can be used for ______ .
a) A.C. only
b) D.C. only
c) both A.C. and D.C. power
d) none of these
+
[Ans. : b]
VTU : Jan.-14
[Ans. : c]
Single Phase Energy Meter
Q.1
Q.2
Q.3
Q.4
The average torque acting on the aluminium disc of an energymeter is proportional to the
VTU : Jan.-09
_______ consumed by the circuit.
+
a) current
b) voltage
c) power
d) none of the above
In the energy meter, constant speed of rotation of disc is provided by _____.
+
a) shunt magnet
b) series magnet
c) braking magnet
d) none of these
VTU : Jan.-10
a) braking torque is zero
b) braking torque is equal to operating torque
c) braking torque is maximum
d) operating torque is constant
[Ans. : b]
In energymeter, _______ effect is used.
b) induction
c) hall
d) magnetic [Ans. : b]
c) breaking
d) damping [Ans. : b]
In energymeters _______ torque is absent.
b) controlling
In energymeters, the number of revolutions in time t of disc is proportional to the ______ .
a) current
Q.7
[Ans. : c]
+
a) deflecting
Q.6
VTU : Jan.-09
In an energy meter, the moving system attains the steady speed when, ______ .
a) thermal
Q.5
[Ans. : c]
b) voltage
c) power
d) energy
[Ans. : d]
The graph of % error against load current is called ______ for an energymeter.
a) characteristic curve
c) calibration curve
b) transfer curve
d) none of these
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[Ans. : c]
Basic Electrical Engineering
Q.8
4-8
The rotation of the disc without any current through the current coil is called ______ .
a) creeping
Q.9
b) cogging
c) crawling
d) none of these
+
One unit of electrical energy is equivalent to ___________.
a) 1 kWh
Q.10
Measuring Instruments
b) 3600 W-sec
c) 100 Wh
[Ans. : a]
VTU : June-10
d) 10 kWh [Ans. : a]
In the energy meter, constant speed of rotation the disc is provided by _______ .
+
a) shunt magnet
b) series magnet
VTU : June-10, Jan.-13
c) braking magnet
d) none of these
[Ans. : c]
Q.11
In the energy meter, constant speed of rotation of the disc is provided by ___ .
+
a) shunt magnet
b) series magnet
c) braking magnet
VTU : Jan.-11
d) none of these
[Ans. : c]
Q.12
Q.13
Q.14
Q.15
Q.16
Under no load condition, the revolution of the disc due to kinetic energy of an energy
VTU : July-11
meter can be blocked by ______ .
+
a) brake magnet
b) electromagnet
c) creeping hole with brake magnet
c) copper shading band
The voltage coil of a single phase energy meter ______.
+
[Ans. : c]
VTU : Dec.-11
A) is highly resistive
B) is highly inductive
C) is highly capacitive
D) has a phase angle equal to load p.f. angle. [Ans. : a]
The meter constant of energy meter is given by ______.
+
A) rev./kW
D) rev./kVA [Ans. : c]
B) rev./watt
C) rev./kWh
VTU : Dec.-11.
In energy meter, constant speed of rotation of disc is provided by ________.
+
A) shunt magnet
B) series magnet
C) braking magnet
D) none of these
VTU : June-12
[Ans. : c]
The most commonly used induction type instrument is ______.
a) voltmeter
b) ammeter
c) watt-hour meter
d) wattmeter
[Ans. : c]
qqq
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5
Single Phase A.C. Circuits
Chapter at a Glance
1.
Generation of A.C. Voltage
q = wt
2.
Standard Definitions Related to Alternating Quantity
f =
1
T
w = 2 pf
Amplitude =
3.
radians
Hz
radians/sec.
e = Em sin w t volts
and
Effective Value or R.M.S. Value
I
r.m.s.
= 0.707 Im
\
V
r.m.s.
= 0.707 Vm
Average Value
Iav = 0.637 Im
6.
radians/sec.
Equation of an Alternating Quantity
\
5.
2p
T
Peak to Peak value
2
e = Em sin q volts
4.
w=
or
and
Vav = 0.637 Vm
Form Factor (Kf)
Form factor,
Kf =
Kf =
R. M. S. value
Average value
0.707 I m
= 1.11
0.637 I m
for sinusoidally varying quantity
(5 - 1)
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7.
Single Phase A.C. Circuits
Crest or Peak Factor (Kp)
Peak factor
8.
5-2
Kp =
Maximum value
R. M. S. value
Kp =
Im
= 1.414
0.707 I m
for sinusoidal waveform
Mathematical Representation of Phasor
If
e = Em sin (w t ± f) then polar form is,
E = E Ð ± f where E is r.m.s. value =
Em
2
.
Important Note : To obtain polar form from the instantaneous equation, express the given
equation in sine form instead of cosine form.
e = Em cos (w t ± f ) then express it as,
If,
e = Em sin (w t + 90º ± f )
\ Phase of alternating quantity = 90º ± f .
Key Point
If
To obtain phase, express the equation in sine form if given in cosine as,
e = E m cos (wt)
then
e = E m sin (wt + 90º) as sin(90º + q) = cos q
Thus the phase is 90° and not zero.
9.
Multiplication and Division of Phasors
and
Q = X 2 + jY2
P = X1 + jY1
Then
P´ Q =
[r1 Ðf1 ]´ [r2 Ðf2 ] = [r1 ´ r2 ]Ðf1 + f2
r Ðf
r
P
= 1 1 = 1 Ðf 1 - f 2
Q
r2 Ðf2 r2
10. A.C. through Pure Resistance
Im =
Vm
R
\
Pav = V
\
Pav = V × I
rms
and
×I
rms
I
f = 0º
V
V and I in phase
watts
2
watts = I R
watts
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5-3
Single Phase A.C. Circuits
11. A.C. through Pure Inductance
V
V
p
i = Im sin æç w t - ö÷ where Im = m = m and XL = w L = 2 p f L W
w L XL
2ø
è
\
V
· In purely inductive circuit, current lags voltage by 90°.
90º
I lags V by 90º
XL = w L = 2 p f L W
I
· Pure inductance never consumes power.
12. A.C. through Pure Capacitance
Vm
p
p
i =
sin æç w t + ö÷ = Im sin æç w t + ö÷
\
2ø
2ø
è
è
æ 1 ö
ç
÷
wC
è
ø
where
·
Im
V
= m
XC
1
1
XC =
=
W
wC 2 p f C
and
In purely capacitive circuit, current leads voltage by 90º.
· Pure capacitance never consumes power.
13. Impedance
Z = R + j0 = RÐ0º ohms.
Z=
VÐ0º
V
= Ð90º = X L Ð90º = 0 + j X L ohms
IÐ - 90º
I
Z=
VÐ0º
V
= Ð - 90º = X C Ð - 90º = 0 - j X C ohms
IÐ + 90º
I
14. A.C. through Series R-L Circuit
Z = R + j XL W
Z = |Z| Ð f W
where|Z| =
\
R 2 + X 2L ,
–1 é X ù
f = tan ê L ú
ë R û
P = V I cos f watts
P = V I cos f
where V and I are r.m.s. values
watts
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I leads V by 90º
I
90º
V
Basic Electrical Engineering
\
5-4
Single Phase A.C. Circuits
S = VI VA
Q = V I sin f
VAR
V I cos f
True Power
= cos f
=
VI
Apparent Power
Power factor =
R
Z
cos f =
Power factor = cos f where f is the angle between supply voltage and current.
· Nature of power factor always tells position of current with respect to voltage.
15. A.C. through Series R-C Circuit
Z=
(R) 2 + (X C ) 2
(magnitude) is the impedance of the circuit.
Z = R – j XC W
Z = R – j XC = | Z | Ð – f
where|Z|=
\
R 2 + X 2C , f = tan
–1
P = V I cos f watts
Apparent power,
é- X C ù
ê R ú
û
ë
where V and I are r.m.s. values
S=VI
VA
True or average power, P = V I cos f W
Q = V I sin f VAR
Reactive power,
For any single phase a.c. circuit, the average power is given by,
P = V I cos f
watts
where V, I are r.m.s. values
cos f = Power factor of circuit
cos f is lagging for inductive circuit and cos f is leading for capacitive circuit.
16. A.C. through Series R-L-C Circuit
Sr. No.
Circuit
Impedance (Z)
Polar
f
p.f. cos f
Remark
Rectangular
1.
Pure R
R Ð 0º W
R + j 0W
0º
1
Unity p.f.
2.
Pure L
XL Ð 90º W
0 + j XL W
90º
0
Zero lagging
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Basic Electrical Engineering
5-5
Single Phase A.C. Circuits
3.
Pure C
XC Ð – 90º W
0 – j XC W
– 90º
0
4.
Series RL
|Z| Ð + fº W
R + j XL W
0º Ð f Ð 90º
cos f
Lagging
5.
Series RC
|Z| Ð – fº W
R – j XC W
– 90ºÐ f Ð 0º
cos f
Leading
6.
XL > XC Lagging
R+jXW
|Z| Ð ± fº W
Series RLC
Zero leading
f
X = X L – XC
cos f
XL < XC Leading
XL = XC Unity
17. A.C. Parallel Circuit
G 2 + B2 ,
Y = G m j B, |Y| =
G = Conductance =
R
Z2
,
B
–1
f = tan
B
G
= Susceptance =
X
Z2
Important Theory Questions and Answers
Ø
With a neat sketch briefly explain how an alternating voltage is produced when a coil is
VTU : Mar.-01; July-03; Aug.-05, Marks 6
rotated in a magnetic field.
+
· The basic principle of an a.c. generation is the principle of electromagnetic induction.
The sine wave is generated according to Faraday’s law of electromagnetic induction.
· It consists of a permanent magnet having two poles. A single turn rectangular coil is
kept in the vicinity of the permanent magnet.
Permanent
magnet
N
Shaft
Axis of rotation
Direction of rotation
a
N
Flux lines
(v) city
lo
ve
d
Slip-rings
C1
Q
P-Q-brushes
P
b
C2
c
ity
loc
ve (v)
S
S
R
I
Fig. 5.1 Single turn alternator
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v = Direction of
instantaneous
velocity
Basic Electrical Engineering
5-6
Single Phase A.C. Circuits
· The coil is made up of two conductors namely a-b and c-d. Such two conductors are
connected at one end to form a coil.
· The coil is so placed that it can be rotated about its own axis.
· The remaining two ends C1 and C2 of the coil are connected to the rings mounted on
the shaft called slip rings. Slip rings are also rotating members of the alternator.
· The two brushes P and Q are resting on the slip rings. The brushes are stationary and
just making contact with the slip rings. The overall construction is shown in the Fig. 5.1.
Ø
Define R.M.S. value of an alternating quantity. Obtain the relation between r.m.s. value and
the maximum value of an alternating quantity.
+ VTU : July-03, 04, 06; Jan.-04, 07; Feb.-05; July-11, Jan.-13, 14, Marks 4
· The effective or r.m.s. value of an alternating current is given by that steady current (D.C.)
which, when flowing through a given circuit for a given time, produces the same amount of
heat as produced by the alternating current, which when flowing through the same circuit for
the same time.
· Consider sinusoidally varying alternating current and square of this current as shown in
the Fig. 5.2.
i,i
i
2
2
2
2
i = Im sin q
2
i = Imsinq
i
p
0
time
2p
dq
Base
p
Fig. 5.2 Waveform of current and square of the current
Step 1 :
The current
i = Im sin q
Area of square curve over half cycle =
2
p
òi
2
2
i = I 2m sin q
Step 2 : Square of current
dq and length of the base is p .
0
Step 3 : Average value of square of the current over half cycle
p
Area of curve over half cycle
=
Length of base over half cycle
òi
0
=
TM
2
p
dq
=
1
p
p
2
ò i dq
0
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=
1
p
p
ò Im sin
0
2
2
q dq
Basic Electrical Engineering
I 2m
p
=
p
ò
0
5-7
p
I 2m é sin 2 q ù
é1 - cos 2 q ù
dq
=
q
ú
ê
2 úû 0
2 pêë
2
û
ë
Single Phase A.C. Circuits
I 2m
I2
[ p] = m
2p
2
=
Step 4 : Root mean square value i.e. r.m.s. value can be calculated as,
\
I
r.m.s.
= 0.707 Im
· The above result is also applicable to sinusoidal alternating voltages.
\
Ø
V
r.m.s.
= 0.707 Vm
Define average value of an alternating quantity. Obtain the relation between average value
and the maximum value of an alternating quantity.
+ VTU : July-03, 04, 06; Jan.-04, 07; Feb.-05; July-11, Marks 4
· The average value of an
alternating quantity is defined
as that value which is obtained
by
averaging
all
the
instantaneous values over a
period of half cycle.
Current
i
p
0
Time
2p
q
· Consider sinusoidally varying
current, I = Im sin q
dq
· Consider
the
elementary
Fig. 5.3 Average value of an alternating current
interval of instant ‘dq ’ as
shown in the Fig. 5.3. The
average instantaneous value of current in this interval is say, ‘i’ as shown.
· The average value can be obtained by taking ratio of area under curve over half cycle
to length of the base for half cycle.
\
Iav =
Area under curve for half cycle
Length of base over half cycle
p
ò i dq
Iav =
=
\
0
p
=
1
p
p
òi dq=
0
1
p
p
ò Im sin q dq
=
0
Im
p
p
ò sin q =
0
Im
I
2 Im
[ – cos p + cos 0] = m [ 2 ] =
= 0.637 Im
p
p
p
Iav = 0.637 Im
and
Vav = 0.637 Vm
TM
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Im
[- cos q]p0
p
Basic Electrical Engineering
Ø
5-8
Single Phase A.C. Circuits
+ VTU : July-09, Marks 2
Define form factor.
· The form factor of an alternating quantity is defined as the ratio of r.m.s. value to the
average value,
Form factor,
Ø
Kf =
R. M. S. value
Average value
+ VTU : Mar.-01; Aug.-03, Marks 2
Define peak factor.
· The peak factor of an alternating quantity is defined as ratio of maximum value to the
r.m.s. value.
Peak factor
Ø
Kp =
Maximum value
R. M. S. value
Derive an expression for the instantaneous power in a pure resistor energised by sinusoidal
VTU : Aug.-02; Mar.-04 Marks 4
voltage.
+
v,i
v = Vm sin wt
i = Im sin wt
0
p
2p Time
0
I
V
Both in phase
(a)
(b)
Fig. 5.4 A.C. through purely resistive circuit
· The phasors are drawn in phase and there is no phase difference in between them.
Phasors represent the r.m.s. values of alternating quantities.
· The instantaneous power in a.c. circuits can be obtained by taking product of the
instantaneous values of current and voltage.
V I
2
P = v × i = Vm sin(w t)×Im sin wt = VmIm sin (wt) = m m (1 – cos 2 w t)
2
\
Ø
P =
Vm I m
V I
– m m cos (2 w t)
2
2
Derive the expression for the instantaneous power in a pure capacitor energised by sinusiodal
voltage. Draw the wave shapes of current, voltage and power.
+ VTU : Jan.-01, 03; May-10, Marks 8
· The expression for the instantaneous power can be obtained by taking the product of
instantaneous voltage and current.
p
P = v × i = Vm sin (w t) × Im sin æç w t + ö÷
2ø
è
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Basic Electrical Engineering
5-9
p
as sin æç wt + ö÷ = cos w t
2ø
è
= Vm Im sin (w t) cos (w t)
\
P=
Vm I m
sin (2 w t)
2
Single Phase A.C. Circuits
as 2 sin w t cos w t = sin 2 w t
· Thus, power curve is a sine wave of frequency double that of applied voltage.
· The Fig. 5.5 shows waveforms of current, voltage and power.
p,v,i
v
p
e
+v
i
e
+v
e
0
Pa v= 0
+v
Time
–v
e
–v
e
Fig. 5.5 Waveforms of voltage, current and power
· The areas of positive and negative loops are exactly the same and hence, average power
consumption is zero.
Ø
1.
Draw the power triangle and define active power, reactive power and apparent power. State
VTU : July-03, 06, Marks 6
their units.
+
Power Triangle :
· Power triangle can be obtained as shown in the Fig. 5.6.
· If we multiply voltage equation by current I, we get the
power equation.
V I = VR I + VL I i.e. V I = V cos fI + V sin fI
VI
VLI = VI sin f
f
2. Real or True or Active Power (P) :
· It is defined as the product of the applied voltage and
the active component of the current.
VRI = VI cos f
Fig. 5.6 Power triangle
· It is real component of the apparent power. It is measured in unit watts (W) or
kilowatts (kW).
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5 - 10
Single Phase A.C. Circuits
3. Apparent Power (S) :
· It is defined as the product of r.m.s. value of voltage (V) and current (I). It is denoted
by S.
\
S = VI VA
· It is measured in unit volt-amp (VA) or kilo volt-amp (kVA).
4. Reactive Power (Q) :
· It is defined as product of the applied voltage and the reactive component of the
current.
· It is also defined as imaginary component of the apparent power.
· It is represented by ‘Q’ and it is measured in unit volt-amp reactive (VAR) or
kilovolt-amp reactive (kVAR).
Q = V I sin f
VAR
Power Factor (cos f)
Ø
Define power factor and explain its significance in a.c. circuit.
+ VTU : Jan.-02, 04; July-04 Marks 5; July-09 Marks 2
· It is defined as factor by which the apparent power must be multiplied in order to
obtain the true power.
· It is the ratio of true power to apparent power.
Power factor =
V I cos f
True Power
=
= cos f
VI
Apparent Power
· It is also defined as the ratio of resistance to the impedance.
· Nature of power factor always tells position of current with respect to voltage.
Ø
For a.c. circuit consisting of R and C, draw the phasor diagram and show that the current
VTU : Mar.-99; Aug.-02, Jan.-13, Marks 4
leads the voltage.
+
· Consider a circuit consisting of pure resistance
R-ohms and connected in series with a pure capacitor
of C-farads as shown in the Fig. 5.7.
C
R
i
i
· Circuit draws a current I, then there are two voltage
drops,
a) Drop across pure resistance
VR = I × R
VR
VC
v = Vmsin wt
Fig. 5.7 Series R-C circuit
b) Drop across pure capacitance VC = I × XC
1
and I, VR, VC are the r.m.s. values
where
XC =
2 pf C
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5 - 11
Single Phase A.C. Circuits
· Following are the steps to draw the phasor diagram :
1) Take current as reference phasor.
2) In case of resistance, voltage and current are in phase. So, VR will be along
current phasor.
3) In case of pure capacitance, current leads voltage by 90º i.e. voltage lags
current by 90º so VC is shown downwards i.e. lagging current by 90º.
4) The supply voltage being vector sum of these two voltages VC and VR
obtained by completing parallelogram.
· The phasor diagram and voltage
triangle are shown in the
Fig. 5.8.
VR
O
f
90º
· From the phasor diagram the
current leads the voltage by
angle f which is decided by the
circuit components R and C.
A
I
O
f
VR = IR
VC
A
VC = I X C
V
B
B
V
(a) Phasor diagram
VC
(b) Voltage triangle
Fig. 5.8
A.C. through Series R-L-C Circuit
Ø
Derive an expression for impedance, phase angle and power for series R-L-C circuit energised
VTU : July-06; Jan.-09, June-13, Marks 8
by sinusoidal voltage.
+
· Consider a circuit consisting of resistance
R ohms pure inductance L henries and
capacitance C farads connected in series
with each other across a.c. supply. The
circuit is shown in the Fig. 5.9.
R
L
VR
VL
VC
90º
90º
I
C
VL
I
· The a.c. supply is given by,
v = Vm sin wt.
VR
I
VC
v = Vmsin wt
Fig. 5.9 R-L-C series circuit
a) Drop across resistance R is
VR = I R
b) Drop across inductance L is
VL = I XL
c) Drop across capacitance C is VC = I XC
VL
VL > VC
V
(VL – VC)
O
f
B
B
V
I
A VR
O
f
VR
(VL – VC)
A
I lags V
VC
Fig. 5.10 Phasor diagram and voltage triangle for XL > XC
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I
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5 - 12
Single Phase A.C. Circuits
VL
O
A
f
VR
(VC – VL)
VC
I
O
f
VR
A
I
(VC – VL)
V
B
B V
I leads V
VC > V L
Fig. 5.11 Phasor diagram and voltage triangle for XL < XC
VL
VC = VL
O
VR = V
I
VC
Fig. 5.12 Phasor diagram for XL = XC
For RLC series circuit impedance is given by,
Z=R+jX
where X = XL – XC = Total reactance of circuit
1)
If
X L > X C,
X is positive and circuit is inductive.
2)
If
XL < XC,
X is negative and circuit is capacitive.
3)
If
XL = XC,
X is zero and circuit is purely resistive.
R
éX - X C ù
2
2
tan f = ê L
ú , cos f = Z and Z = R + (X L - X C )
R
û
ë
· The average power consumed by the circuit is,
Pav = Average power consumed by R + Average power
consumed by L + Average power consumed by C
· But, pure L and C never consume any power.
\
2
Pav = Power taken by R = I R = I (I R) = I VR
· But,
VR = V cos f in both the cases
\
P = V I cos f
W
· Thus, for any condition, XL > XC or XL < XC, in general power can be expressed as,
P = V I cos f
watts
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5 - 13
Single Phase A.C. Circuits
Important Solved Examples
Example 5.1
A sinusoidal voltage of 50 Hz has a maximum value of 200 2 volts. At what time
measured from a positive maximum value will the instantaneous voltage be equal to 141.4
volts ?
Solution : f = 50 Hz,
Vm = 200 2 V,
v1 = 141.4 V
The equation of the voltage is,
v = Vm sin(2p ft) = 200 2 sin(2p´ 50 t) V
For v = v 1 14 1.4 = =200 2 sin(2p´ 50 ´ t 1 )
t 1 = 1.666 ´ 10 -3 sec
\
... Use radian mode for sin
T 1
But this time is measured from t = 0. At positive maximum, time is = = 5 ´ 10 -3 sec so
4 4f
3
sec is before
t = t 1 = 1.666 ´ 10
V
positive maximum.
From Fig. 5.13.
t m - t 1 = 5 ´ 10 -3 - 1.666 ´ 10 -3
141.4 V
As the waveform is symmetrical,
at the time of 3.314 ´ 10 -3 sec
measured after positive maximum
value, the instantaneous voltage
will be again 141.4 V.
Example 5.2
t
t2
t1
= 3.314 ´ 10 -3 sec
T sec
tm = ––
4
Fig. 5.13
A sinusoidal wave of frequency 50 Hz has its maximum value of 9.2 Amps. What
will be its value at (a) 0.002 sec after the wave pass through zero in positive direction. (b)
0.0045 sec after the wave passes through positive maximum. Show the values of current in a
neat sketch of the wave form.
Solution : The waveform is shown in the Fig. 5.14.
Now I m = 9.2 A and f = 50 Hz
i = I m sin 2 p ft = 9.2 sin 100 p t A
\
a) At t = 0.002 sec,
i = 9.2 sin (100 p´ 0.002) = 5.4076 A
... Use sin in radians
1
b) At t = 0.0045 sec after positive maximum as shown, time period T = = 0.02 sec so
f
T 0.02
3
positive maximum occurs at t = =
= 5 ´ 10 sec. After this 0.0045 sec means value
4
4
of i at t = 5 ´ 10 -3 + 0.0045 = 9.5 ´ 10 -3 sec from t = 0.
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5 - 14
(
i = 9.2 sin 100 p´ 9.5 ´ 10 -3
\
) = 1.4391 A
0.0045
sec
i
9.2 A
5.4076 A
1.4391 A
0
0.002 sec
Single Phase A.C. Circuits
0.01
sec
–3
5x10
sec
–9.2 A
0.02
sec
t
–3
9.5x10
sec
T = 1 = 1 = 0.02 sec
f 50
Fig. 5.14
Example 5.3
An alternating voltage has an effective value of 70.7106 V and frequency of
60 Hz. Find its average value, form factor, crest factor assuming it to be purely sinusoidal.
Solution :
Effective value means R.M.S. value.
Er.m.s. = 70.7106 V,
f = 60 Hz
Key Point : The frequency does not affect the r.m.s. or average values.
Em =
\
2 ´ E r.m.s. =
2 ´ 70.7106 = 100 V
Eav = 0.637 Em = 0.637 ´ 100 = 63.7 V
Kf =
r. m. s.
70.7106
=
= 1.11
Average
63.7
Kp =
Maximum
100
=
= 1.414
r. m. s.
70.7106
Example 5.4 Calculate
effective
values
of
the
the
average
and
saw
tooth
waveform shown in Fig. 5.15.
Voltage V
200 V
The voltage completes the cycle by
falling back to zero instantaneously after
regular interval of time.
0
2
4
Cycle
Solution : Let us calculate equation for
the instantaneous value of the voltage.
Fig. 5.15
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8
Time
Basic Electrical Engineering
5 - 15
Single Phase A.C. Circuits
The voltage increases linearly from 0 to 200 V in two seconds. So slope between 0 to
2 seconds is,
200 - 0
= 100
=
2
\ Equation for the instantaneous value is, v(t) = 100 t
The average value =
Area under curve
=
Base
2
ò0
2
(100 t) dt 1 é
t2 ù
= ê100 ú = 50 ´ 2 = 100 volts
2
2 ê
2ú
û0
ë
The r.m.s. value = Root of the mean of square
2
2
2
ò (100 t) dt
0
=
2
=
ét 3 ù
1
´ (100) 2 ´ ê ú
2
3
ëê úû 0
=
2
5000 ´
8
3
= 115.47 volts
Example 5.5 A resultant current wave is made-up of two components :
A direct current of 10 A and a sinusoidal alternating current of 50 Hz with a peak value of
10 A.
i)
Draw the resultant current wave and other waveforms (Mark all parameters).
ii) Write down an expression for the resultant current wave, taking t = 0 at a point
where the a.c. component is zero and rising in a positive direction.
iii) Calculate the average value of current of the resultant wave over one complete cycle.
iv) Determine the r.m.s. value of the resultant wave.
v) Calculate the peak factor and form factor.
Solution : i) The resultant is shown in the Fig. 5.16.
ii) For d.c., Idc = 10 A
Current
For a.c.
i = Im sin q = 10 sin q
So the resultant is,
iR = Idc + i = 10 + 10 sin q
This is the expression for the resultant
wave.
iii) Now iR = 10 + 10 sin q
The average value can be obtained as,
1
iR(average) =
2p
Resultant current
20 A
2p
ò iR
D.C.
10 A
0
dq
p
0
t
A.C. component
–10 A
Fig. 5.16
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=
=
1
2p
5 - 16
2p
ò [10+ 10 sin q] d q =
0
Single Phase A.C. Circuits
1
[10 q - 10 cos q]20 p
2p
1
[10 ( 2 p - 0) - 10 (cos 2 p - cos 0) ] = 10 A
2p
iv) The r.m.s. value is given by,
1
2p
iR (r.m.s.) =
1
2p
=
1
2p
=
0
1
2p
2p
ò (10 + 10 sin q)
2p
ò [100 + 200 sin q + 100 sin
2
q] d q
0
2p
ò
0
æ 1 - cos 2 q ö
[100 + 200 sin q + 100 ç
÷] d q
2
ø
è
2p
0
=
1
[ 300 ´ 0 p] =
2p
150 = 12.2474 A
12 . 2474
R. M. S.
= 1.2247
=
10
Average
Maximum
20
=
= 1.633
R. M. S.
12 . 2474
For the waveform shown find the
form factor.
Solution :
v(t)
V
Find r.m.s. value.
To find equation of line, two points are (0, 0)
and (1, V).
y2 -y1
V -0
–V
Slope =
=
=V
\
x2 - x1
1 -0
\
dq
0
1 é
100
100
100 ( 2 p - 0) - 200 (cos 2 p - cos 0) +
( 2 p - 0) (sin 4 p - sin 0) ù
2 p ëê
2
4
ûú
Peak factor =
Step 1 :
2
=
Form factor =
Example 5.6
2
ò iR d q =
1 é
æ q sin 2 q öù
100 q - 200 cos q + 100 ç ÷
ê
2pë
4 øúû
è2
=
v)
2p
1
2
3
v(t) = Vt,
… 0 £t < 1
Fig. 5.17
(y = mx)
For 1 £ t £ 2, two points are (1, – V) and (2, 0)
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\
Slope =
5 - 17
Single Phase A.C. Circuits
0 - ( -V )
=+V
2 -1
But as it does not pass through origin, y = mx + C
i.e.
v(t) = Vt + C
Putting (2, 0), 0 = 2V + C
\
\
i.e. C = – 2 V
… 1 £t £2
v(t) = Vt – 2V = V(t – 2),
Vr.m.s. =
Area of curve over a squared wave cycle
Length of base over a cycle
2
1
2
ò v ( t ) dt
=
2
2
2
ò v ( t ) dt + ò v ( t ) dt
0
0
=
2
1
2
Example 5.7 A nonsinusoidal voltage has a form factor of 1.25 and crest factor of 1.63. If its
average value is 50 V, calculate its i) r.m.s. value and ii) Maximum value.
Solution :
Kf = 1.25, Kp = 1.63, Vav = 50
R. M. S.
R. M. S.
i.e. 1.25 =
Kf =
Average
50
KP =
Max
R. M. S.
i.e.
1.63 =
Max
62.5
i.e.
i.e.
+ MU : May-03
R.M.S. = 62.5 V
Maximum = 101.875 V
Example 5.8 Find the r.m.s. and average value of the waveform shown in the Fig. 5.18.
i
10
8
6
4
2
0
1
2
3
4
5
6
7
8
9 10
t
Fig. 5.18
Solution :
Iav =
IR.M.S. =
0 + 2 + 4 + 6 + 8 + 10 + 8 + 6 + 4 + 2
=5A
10
0 2 + 2 2 + 4 2 + 6 2 + 8 2 + 10 2 + 8 2 + 6 2 + 4 2 + 2 2
= 5.8309 A
10
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Example 5.9 Find
5 - 18
Single Phase A.C. Circuits
the r.m.s. value of the resultant current in a wire which carries
simultaneously a direct current of 10 A and a sinusoidal alternating current with a peak
+ VTU : Aug.-95
value of 10 A.
Soltuion : When a wire carries combination of d.c. and other alternating signals then,
Ir.m.s. =
I 2dc + I 21 r.m.s. + I 22 r.m.s. +...
In this example, Idc = 10 A,
\
Ir.m.s. =
(10) 2
Ir.m.s. =
+ (7.071) 2
Im
2
=
10
= 7.071 A
2
= 12.2474 A
A voltage v = 141 sin {314 t + p / 3} is applied to
Example 5.10
iii) Capacitance of 100 mF.
Find in each case rms value of current and power dissipated.
Draw the phasor diagram in each case.
i) Resistor of 20 ohms
ii) Inductance of 0.1 henry
Solution : Comparing given voltage with v = Vm sin ( w t + q) we get,
V
Vm = 141 V and hence V = Vr.m.s. = m = 99.702 V
2
w = 314 and hence f =
w
p
= 50 Hz, q = = 60º
2p
3
Hence the polar form of applied voltage becomes,
V = 99.702 Ð 60º V
Case 1 :
R = 20 W
I
Irms = 4.9851 A
\
The phase of both V and I is same for pure resistive
circuit. Both are in phase.
P = VI = 99.702 ´ 4.9851 = 497.0244 W
The phasor diagram is shown in the Fig. 5.19 (a).
Case 2 :
L = 0.1 H
\
\
V
V 99. 702 Ð 60º
=
= 4.9851 Ð 60º A
=
20 Ð 0º
R
V and I
in phase
I
60º
0
Fig. 5.19 (a)
V
I lags V
by 90º
XL = w L = 314 ´ 0 . 1 = 31.4 W
I =
|V| 99.702
= 3.1752 A
=
XL
31.4
60º 90º
This is r.m.s. value of current. It has to lag the applied
voltage by 90º in case of pure inductor.
0
30º
I
Hence phasor diagram is shown in the Fig. 5.19 (b).
Fig. 5.19 (b)
The individual phase of I is – 30º.
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5 - 19
Single Phase A.C. Circuits
In polar form I can be represented as 3 . 1752 Ð - 30º A.
Pure inductor never consumes power so power dissipated is zero.
Case 3 :
C = 100 mF
\
XC
=
1
1
=
= 31.8471 W
wC 314 ´ 100 ´ 10 -6
\
I
=
|V|
99.702
=
= 3.1306 A
XC
31.8471
This is r.m.s. value of current.
V
150º
I
It has to lead the applied voltage by 90º in case of
pure capacitor.
I leads V
by 90º
60º
90º
0
Hence phasor diagram is shown in the Fig. 5.19 (c).
The individual phase of I is 150º. In polar form I
can be represented as 3.1306 Ð + 150º. A Pure
capacitor never consumes power and hence power
dissipated is zero.
Fig. 5.19 (c)
Example 5.11 A series RL circuit takes 400 W at a power factor of 0.8 from a 120 V,
+ VTU : Dec.-86
supply. Calculate the values of R and L.
Solution : Given,
P = 400 W,
50 Hz
f = 50 Hz, V = 120 V,
cos f = 0.8
P = V I cos f
Now,
\
400 = 120 × I × 0.8
\
I = 4.167 A
|V|
120
=
= 28.8 W
| I | 4.167
\
|Z| =
and
cos f = 0.8,
\
hence, f = 36.86º
Z = 28.8 Ð 36.86º W
… f positive as inductive
= 23.04 + j 17.276 W
Compare with R + jXL,
R = 23.04 W
\
XL = 2 p f L = 17.276
But,
\
L =
17.276
= 0.055 H
2 p´ 50
Example 5.12 An inductive coil draws a current of 2 A, when connected to a 230 V, 50 Hz
supply. The power taken by the coil is 100 watts. Calculate the resistance and inductance of
+ VTU : Jan.-03, Marks 6
the coil.
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Basic Electrical Engineering
5 - 20
Single Phase A.C. Circuits
Solution : P = 100 W, I = 2 A, V = 230 V, f = 50 Hz
Key Point : XL component does not consume any power. Only resistance r consumes
power.
\
P
\
100
|Z|
But
= I2 r
= (2)2 × r
=
115 =
\
2
(115)
r + j XL i.e. |Z| =
= 625 + (XL)
2
\
L =
XL
r
r 2 + (X L ) 2
( 25) 2 + ( X L ) 2
XL = 112.2497 W
i.e.
… Resistance
V 230
=
= 115 W
I
2
Z =
\
i.e. r = 25 W
I
i.e.
X 2L
but
XL = 2 p fL
= 12600
V
Fig. 5.20
XL
112 . 2497
=
= 0.3573 H
2pf
2p´ 50
…Inductance
Example 5.13 A series circuit with a resistance R = 10 W and inductance 20 mH has a current of
i = 2 sin 500 t. Obtain the total voltage across the series circuit and angle by which the current
+
lags the voltage.
Solution :
\ I m = 2 A,
JNTU : Jan.-10
i(t) = 2 sin 500 t A = I m sin w t A
I=
Im
2
= 1.4142 A (RMS)
w = 500 rad/sec
R
L
10 W
20 mH
I
\ X L = w L = 500 ´ 20 ´ 10 - 3 = 10 W
V
\ Z = R + j X L = 10 + j 10 W = 14.1421 Ð 45º W
Fig. 5.21
The phase of the current is 0º. i.e. I = 1.4142 Ð 0º A
V = I ´ Z = 1.4142 Ð 0º ´ 14.1421 Ð 45º = 20 Ð 45º V
\
\ Angle by which current lags voltage = 45º
Example 5.14 A 20 W resistance and 30 mH inductance are connected in series and the circuit is
fed from a 230 V, 50 Hz AC supply. Find
i) Reactance across the inductance, impedance, admittance, current
ii) Voltage across the resistance iii) Voltage across the inductance
iv) Real, reactive and active powers v) Power factor
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Basic Electrical Engineering
5 - 21
Single Phase A.C. Circuits
Solution : The circuit is shown in the Fig. 5.22.
X L = 2 p f L = 2 p ´ 50 ´ 30 ´ 10 -3 = 9.4248 W
i)
\
Z = R + j X L = 20 + j 9.4248 W
= 22.1094 Ð 25.2317º W
Y =
R
L
20 W
30 mH
V = 230 V
50 Hz
... Impedance
1
1
=
Z
22.1094 Ð 25.2317º
Fig. 5.22
= 0.04522 Ð - 25.2317º ª
\
I =
... Admittance
V
230 Ð 0º
=
= 10.4028 Ð - 25.2317º A
Z
22.1094 Ð 25.2317º
... Current
ii)
VR = I R = 10.4028 Ð -25.2317º ´ 20 Ð 0º = 208.056 Ð -25.2317º V
iii)
VL = I ´ (j X L ) = 10.4028 Ð -25.2317º ´ j 9.4248
= 10.4028 Ð -25.2317º ´ 9.4248 Ð90º = 98.0443 Ð 64.7683º V
P = V I cos f = 230 ´ 10.4028 ´ cos ( - 25.2317º ) = 2164.365 W
iv)
Q = V I sin f = 230 ´ 10.4028 ´ sin ( - 25.2317º ) = – 1019.9359 VAR
The negative sign indicates lagging nature of reactive volt-amperes.
S = V I = 230 ´ 10.4028 = 2392.644 VA
v) Power factor = cos f = cos ( - 25.2317º ) = 0.90459 lagging
Example 5.15 A capacitor having a capacitance of 10 mF is connected in series with a non
inductive resistance of 120 W across 100 V, 50 Hz. Calculate the power, current and the phase
difference between current and voltage.
+
JNTU : Jan.-10, June-11
Solution : The circuit is shown in the Fig. 5.23.
XC =
1
1
= 318.3098 W
=
2 p f C 2 p ´ 50 ´ 10 ´ 10 - 6
Z = R - j X C = 120 - j 318.3098 W = 340.178 Ð - 69.344º W
V
100 Ð 0º
\ I=
=
= 0.2939 Ð 69.344º A
Z
340.178 Ð - 69.344º
I
C
R
10 mF
120 W
Phase difference between V and I = f = 69.344º leading
\
P = V I cos f = 100 ´ 0.2939 ´ cos (69.344º ) = 10.3697 W
or
P = I 2 R = (0.2939) 2 ´ 120 = 10.369 W
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100 V
50 Hz
Fig. 5.23
Basic Electrical Engineering
5 - 22
Single Phase A.C. Circuits
Example 5.16 A metal filament lamp rated 750 W, 110 V is to be connected in series with a
capacitor across a 220 V, 50 Hz supply. Calculate
i) The capacitance required
ii) The power factor.
+
JNTU : Aug.-08, Jan.-10
Solution : The arrangement is shown in the Fig. 3.16.7.
Bulb
For bulb, P = 750 W, V = 110 V
P 750
= 6.8181 A
=
V 110
\
I=
\
R=
\
Z = R - j X C = |Z| Ð f
\
|Z| =
R2
+ X 2C
\
|I| =
V
=
|Z |
\
R 2 + X 2C =
,
220 V, 50 Hz
Fig. 5.24
|I| = 6.8181 A
220
2
R + X 2C
= 6.8181
220
= 32.267
6.8181
\ (16.1333) 2 + (X C ) 2 = 32.267
i.e.
260.2833 + X 2C = 1041.159
\
X 2C = 780.876
i.e.
X C = 27.9441
\
1
= 27.9441
2p f C
i.e.
C=
1
2 p´ 50 ´ 27.9441
\
C = 113.9092 mF
\
Z = 16.133 - j 27.9441 W = 32.2669 Ð - 60º W
\
R
I
V
110
= 16.1333 W
=
I
6.8181
C
cos f = cos ( - 60º ) = 0.5 leading
Example 5.17 In
a
R-C
series
circuit,
... power factor
voltage
across
the
combination
is
given
by
40 sin ( 2000 t + 45º )R = 10 W . The current leads the voltage by p / 3 radians. Find the value of
+
C. Also find the expression for current.
JNTU : Jan.-10, June-11
v(t) = 40 sin (2000 t + 45º) = Vm sin ( w t + f1 )
\ Vm = 40 V i.e. V =
w = 2000 rad/sec,
40
= 28.2842 V (RMS)
2
f1 = 45º
C
R
Solution : The circuit is shown in the Fig. 5.25.
I
10 W
v(t)
(a)
Fig. 5.25
\ V = 28.2842 Ð 45º V
... Polar form
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Basic Electrical Engineering
Current leads by f =
5 - 23
Single Phase A.C. Circuits
I
p
rad = 60º
3
The phase of the current f2 = 105º
\ cos f = cos 60º = 0.5 leading
But cos f =
\
\
\
\
R
=
Z
V
60º
f2
45º
... power factor
R
2
R + X 2C
10
0.5 =
10 2 + X 2C
i.e.
(b)
Fig. 5.25
X C = 17.3205 W
17.3205 =
100 + X 2C = ( 20) 2
1
2000 C
1
=
wC
i.e.
XC
i.e.
C = 28.8675 mF
Z = R - j X C = 10 - j 17.3205 W = 20 Ð - 60º W
|V| 28.2842
= 1.4142 A (RMS)
=
|Z|
20
\
|I| =
\
Im =
\
i(t) = I m sin ( w t + f2 ) = 2 sin (2000 t + 105º)A
2I =
2 ´ 1.4142 = 2 A with f2 = 105º
Example 5.18 A current of average value 18.019 A is flowing in a circuit to which a voltage of
peak value 141.42 V is applied. Determine i) Z1 = R ± jX
ii) Power
p
Given : V lags I by radians.
6
+ VTU : Aug.-05, Marks 6; Aug.-08, Marks 5
Solution : Iav = 18.019 A, Vm = 141.42 V
For sinusoidal quantity,
R. M. S.
= 1.11
Kf =
Average
\
\
I r.m.s.
= 1.11
I av
I r.m.s.
= 1.11
18 . 019
i.e.
Irms = 20 A
Vrms =
Vm
2
=
141 . 42
2
IÐf
= 100 V
f=
V lags I by p/6 radians i.e.
I leads V by p/6 radians
\
V
=
100 Ð 0º V,
I = 20 Ð
I leads V
p rad
6
Fig. 5.26
p
= 20 Ð 30º A
6
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V Ð 0º
Basic Electrical Engineering
i)
Z1
=
5 - 24
Single Phase A.C. Circuits
V 100 Ð 0º
=
= 5 Ð – 30º W
20 Ð 30º
I
= 4.3301 – j 2.5 W = R – j X W
P = VI cos f = 100 × 20 × cos (– 30º) = 1732.0508 W
ii)
Example 5.19 A circuit having a resitance of 12 W, in inductance of 0.15 H and a capacitance of
100 µF in series is connected across a 100 V, 50 Hz supply. Calculate the impedance, current,
the phase difference between the current and supply voltage.
+ VTU : Aug.-03, Marks 6
Solution :
XL = 2 p fL = 2 p × 50 × 0.15 = 47.1238 W
XC =
\
R
1
1
= 31.8309 W
=
2pfC 2p´ 50 ´ 100 ´ 10 -6
12 W
I
Z = R + j X L – j XC
C
0.15 H 100 mF
100 V, 50 Hz
= 12 + j 47.1238 – j 31.8309
\
L
Fig. 5.27
Z = 12 + j 15.2929 W = 19.4389 Ð 51.8795º W
… Impedance
V
100 Ð 0º
=
= 5.1443 Ð – 51.8795ºA
… Current
Z 19.4389Ð 51.8795º
As voltage is assumed reference, the phase difference between voltage and current is
51.8795º such that current lags voltage.
cos f = cos (51.8795º) = 0.6173 lagging
\
I =
Key Point As XL > XC, the circuit has lagging p.f.
Example 5.20
A circuit consists of a resistance of 10 W, an inductance of 16 mH and a
capacitance of 150 µF connected in series. A supply of 100 V at 50 Hz is given to the circuit.
Find the current, power factor and power consumed by the circuit. Draw the vector diagram.
+ VTU : Jan.-06, Marks 8
Solution :
XL = 2 p fL = 2 p × 50 × 16 × 10– 3= 5.0265 W
XC =
1
1
=
= 21.2206 W
2p fC 2p´ 50 ´ 150 ´ 10 -6
\
Z = R + j XL – j XC = 10 + j 5.0265 – j 21.2206
\
Z = 10 – j 16.1941 W = 19.0328 Ð – 58.304º W
\
V
100Ð0°
I =
=
=5.254 Ð+58.304º A
Z 19 . 0328Ð - 58 . 304°
cos f =
R
10
=
= 0.5254 leading
Z 19 . 0328
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10 W
I
R
16 mH 150 mF
L
C
100 V, 50 Hz
Fig. 5.28
… X C > XL
Basic Electrical Engineering
\
5 - 25
P = VI cos f
Single Phase A.C. Circuits
VL
(26.409 V)
= 100 × 5.254 × 0.5254
I Leads V
by 58.3º
= 276.045 W
VR (52.54 V)
O
|VR| = IR = 5.254 × 10
I
58.3º
= 52.54 V
|VL| = IXL = 5.254 × 5.0265
VC + VL
V = V R + V L + VC
= 100 V
= 26.409 V
|VC| = I XC = 5.254 × 21.2206
VC
(111.493 V)
= 111.493 V
V = VR + VL + VC
Fig. 5.28 (a)
Example 5.21 A choke coil and pure resistance are connected in series across 230 V, 50 Hz, a.c.
supply. If the voltage drop across coil is 190 V and across resistance is 80 V while current
drawn by the circuit is 5 A. Calculate, i) Internal resistance of coil ii) Inductance of coil
iii) Resistance R iv) Power factor of the circuit v) Power consumed by the circuit.
+ JNTU : May-04
Coil
Solution :
V = 230 V ,
V
I
ZT =
ZT =
=
2
i.e. 38 =
2
= (R + r) + (XL)
From (2)
(38)2 = r2 + (XL)2
\
Fig. 5.29
VL
190
=
= 38 W
I
5
(46)
… (1)
r 2 + (X L ) 2
2
… (2)
… (3)
VR = 80 V = I R
R =
… (4)
VR
80
= 16 W
=
I
5
2
2
2
\ From (3) 2116 = (R) + 2 R r + r + (XL)
Substituting (4) in (3)
VR = 80 V
230 V, 50 Hz
(R + r) 2 + (X L ) 2
From (1)
Now
VL = 190 V
230
= 46 W
5
ZL = r + j XL
R
I=5A
I=5A
Impedance of coil ZL =
XL
2
2116 = (16) + 2 ´ 16 ´ r + (38)
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2
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Basic Electrical Engineering
\
5 - 26
Single Phase A.C. Circuits
r = 13 W
r 2 + (X L ) 2
Now
38 =
Now
XL = 2 p f L
i.e.
2
2
i.e. (38) = (13) + (XL)
L =
2
XL = 35.707 W
i.e.
XL
35 . 707
=
= 0.1136 H
2 pf
2p´ 50
ZT = (R + r) + j (XL) = (16 + 13) + j (35.707) = 29 + j 35.707 W
\
cos f =
(R + r) 29
=
= 0.6304 lagging
ZT
46
Power consumed P = V I cos f = 230 ´ 5 ´ 0.6304 = 724.96 W
Example 5.22 A coil A having a resistance of 10 ohms and inductance of 0.2 henry is connected
in series with another coil B having a resistance of 30 ohms and inductance 0.1 H. The two
coils in series are fed from 200 V, 50 Hz supply. Determine the voltage across each coil,
power dissipated in each coil and the power factor of the combined series circuit . Draw the
+ JNTU : May-06
phasor (vector) diagram.
Solution :
R1
Z1 = R1 + jXL1,
L1
XL2 = 2p fL 2 = 2p´ 50 ´ 0.1 = 31.4159 W
IT
Z1
Z2
Coil 1
Coil 2
Z1 = 10 + j 62.8318 W
V = 200 V
50 Hz
= 63.6226 Ð 80 . 957 º W
Fig. 5.30
Z2 = 30 + j 31.4159 W
= 43.439 Ð 46. 32º W
ZT = Z1 + Z2 = 40 + j 94.2477 W = 102 . 3847 Ð67 º W
\
200Ð 0º
V
= 1.9534 Ж 67º A
=
102. 3847 Ð67 º
ZT
fT = 67º lagging
V1 = IT Z1 = 1.9534Ð - 67 º ´ 63.6226Ð80.957º = 124.2803Ð13.97ºV
V2 = IT Z2 = 1.9534Ð - 67º ´ 43.439Ð46.32º = 84.8537Ж20.68º V
P1 = I T
P2 = I T
L2
Z2 = R2 + j XL2
XL1 = 2pfL 1 = 2p´ 50 ´ 0.2 = 62.8318 W
IT =
R2
2
´ R 1 = (1 . 9534) ´ 10 = 38.1577 W
2
´ R 2 = (1 . 9534) ´ 30 = 114.4731 W
2
2
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Basic Electrical Engineering
Cross check :
5 - 27
Single Phase A.C. Circuits
P1 + P2 = PT = V IT cos fT
= 200 ´ 1.9534 ´ cos( -67 º )
V1
+13.97°
O
= 152.6 W
– 67°
Power factor = cos fT
– 20.680°
V = V1 + V 2
V2
= cos (– 67º)
The phasor
Fig. 5.30 (a).
Example 5.23
= 0.3907 lagging
diagram is shown
I
in
the
Fig. 5.30 (a)
Find the reading of an ammeter when the voltmeter across 3 W resistor in the
+ VTU : July-88
circuit shown in the Fig. 5.31 is 36 V.
V
IT
A
I1
3W
–j3 W
I2
5W
j2 W
V
Fig. 5.31
Solution : The voltage across 3 W resistance is 36 V.
\
\
|V3 W| = |I1| × R
36 = I1 × 3
I1 = 12 A magnitude (r.m.s. value)
\
The voltmeter always reads r.m.s. value.
Now, the impedance of branch 1 is,
Z1 = 3 – j 3 W = 4.2426 Ð – 45º W
Let I1 be reference having angle 0º i.e. I1 = 12 Ð 0º A
The two branches are in parallel so voltage across them is same, equal to the supply
voltage V.
V = I1 Z1 = 12 Ð 0º × 4.2426 Ð – 45º = 50.9116 Ð – 45º V
\
Hence, angle of voltage is – 45º. Same is the voltage across branch 2 as in parallel.
Z2 = 5 + j2 W = 5.3851 Ð + 21.8º W
\
I2 =
50.9116Ð - 45º
V
=
= 9.454 Ð – 66.8º A
5. 3851 Ð + 21.80º
Z2
In rectangular form,
I1 = 12 Ð 0º A = 12 + j 0 A
I2 = 9.454 Ð – 66.8º A = 3.724 – j 8.689 A
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Basic Electrical Engineering
\
5 - 28
Single Phase A.C. Circuits
IT = I1 + I2 = 12 + j 0 + 3.724 – j 8.689
= 15.724 – j 8.689 A = 17.965 Ð – 28.92º A
\
Ammeter reading = | IT | = 17.965 A
Example 5.24
… r.m.s. value
Two circuits with impedances of Z1 = 10 + j 15 W and Z2 = 6 – j8 W are
connected in parallel. If the supply current is 20 A, what is the power dissipated in each
+ VTU : Aug.-03, Marks 8; Feb.-06, Marks 10; July-06, Marks 6
branch ?
Solution : Using current division rule,
Z2
I1 = I T ×
and
\
Z1 + Z2
Z1 =
10 + j 15
Z2
I2
10 + j 15 = 18.0277 Ð 56.309º W
6–j8
IT
20 A
Z2 = 6 – j 8 = 10 Ð – 53.1301º W'
Z 1 + Z2 =
Z1
I1
Z1
I2 = I T ×
Z1 + Z2
V
10 + j 15 + 6 – j 8
Fig. 5.32
= 17.4642 Ð 23.6293º W
Assuming IT as reference,
10Ð - 53.1301º
I1 = 20 Ð 0º ×
= 11.452 Ð – 76.7594º A
\
17.4642Ð23.6293º
18.0277 Ð 56 309º
= 20.6453 Ð 32.7607º A
17.4642Ð23.62930
Now in Z1, only resistive part (10 W) consumes power,
P1 = I 21 R1 = (11.45)2 × 10 = 1311.483 W
\
I2 = 20 Ð 0º ×
and
In Z2, only resistive part (6 W) consumes power,
2
P2 = I 22 R2 = (20.6453) × 6 = 2557.3704 W
\
Example 5.25 Two impedances (150 + j 157) W and (100 – j 110) W are connected in parallel
across a 220 V, 50 Hz supply. Find the total current and power factor.
+ VTU : Feb.-05, Marks 5
Solution : Z1 = 150+ j157 W = 217.138 Ð 46.306º W
I1
Z2 = 100– j110 W = 148.660Ж 47.726º W
\
\
ZT = Z1 || Z2 =
ZT
=
Z1 ´ Z2
Z1 + Z2
Z1
150 + j 157 W
Z2
IT
[217.138 Ð 46.306º ] ´[148.660 Ð - 47.726º ]
150 + j 157 + 100 - j 110
I2
100 – j 110 W
220 V,50 Hz
Fig. 5.33
=
32279.735Ð - 1.42°
250 + j 47
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Basic Electrical Engineering
=
\
\
IT =
5 - 29
Single Phase A.C. Circuits
32279.735 Ð - 1 42º
= 126.8959 Ð – 12.067º W
254.3796 Ð 10.647 °
V
220 Ð 0º
=
= 1.7337 Ð + 12.067º A
ZT
126.8959Ð -12.067º
cos f T = cos (12.067º) = 0.978 leading
Example 5.26 A circuit having a resistance 20 W
20 W
0.07 H
50 W
60 mF
and inductance of 0.07 H is connected in parallel
with a series combination of 50 W resistance and
60 mF capacitance. Calculate the total current,
when the parallel combination is connected across
230 V, 50 Hz, supply as shown in Fig. 5.34.
230 V, 50 Hz
+ VTU : Feb.-07, Marks 8
Solution :
Fig. 5.34
Z 1 = 20 + jX L = 20 + j2 p f L = 20 + j 21.9911 W = 29.7255 Ð47.71º W
Z 2 = 50 - jX C = 20 - j
1
= 50 - j53.0516 W = 72.9Ð - 46.7 º W
2p f C
Now Z 1 || Z 2 ,
\
\
Z1 ´ Z2
29.72 Ð47.71 ´ 72. 9Ð – 46.7
=
Z1 + Z2
(20 + j 21. 99) + (50 – j 53. 05)
2166.588 Ð1. 01
2166.588 Ð 1.01
=
Z =
= 28.29 Ð24.93º W
76.58 Ð – 23. 92
70 – j 31.06
Z =
Now total current, I =
230 Ð 0º
V
=
= 8.13Ð - 24.93º A
Z 28. 29 Ð 24. 93
The total current I lags the supply voltage by 24.93º.
Exercise
Ø
Which of the following devices work at i) unity ii) lagging and iii) leading p.f. a. Fluorescent
lamp b. Electric iron c. Incandescent lamp d. Condenser bank e. Induction motor
+ VTU : July-05, Marks 4
a) Electric iron - lagging.
b) Flourescent lamp - It takes lagging current and p.f. of 0.5.
\ Capacitor is required to improve p.f. to 0.9.
c) In candescent lamp - lagging.
d) Condenser bank to improve p.f. - leading.
e) Induction motor - The p.f. is high at full load but decreases to a low value at small
loads.
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Basic Electrical Engineering
5 - 30
Single Phase A.C. Circuits
Multiple Choice Questions
Introduction to A.C. Fundamentals
Q.1
The main advantages of a.c. is ________ .
a)
b)
c)
d)
Q.2
a.c. motors are expensive.
the a.c. voltages can be raised or lowered.
a.c. transmission is very costly.
none of the above.
[Ans. : b]
The main advantage of purely sinusoidal waveform is ________ .
a) it is the only alternating waveform.
b) it is the only standard waveform.
c) any other waveform can be resolved into series of sinusoidal waveforms of
different frequencies.
d) it produces distorted waveforms when applied.
[Ans. : c]
Generation of A.C. Voltage
Q.1
The generation of a.c. is according to ________ .
a) Faraday's law
c) Ohm's law
Q.2
b) 90º
b) 90º
b) 90º
[Ans. : b]
c) 180º
d) 270º
[Ans. : d]
c) 180º
d) 270º
[Ans. : c]
A coil is rotating in the uniform field of a 10-pole generator. In one revolution of the coil,
VTU : June-12
the number of cycles generated by voltage is ____.
+
a) 10
Q.6
d) 270º
The e.m.f. in a coil is zero when q = ________ .
a) 45º
Q.5
c) 180º
The e.m.f. in a coil is at its negative maximum when q = _________ .
a) 45º
Q.4
[Ans. : a]
The e.m.f. in a coil is at its positive maximum when q = _________ .
a) 45º
Q.3
b) Thevenin's theorem
d) Kirchhoff's law .
b) 5
c) 2.5
d) 4
[Ans. : b]
A coil is rotating in the uniform magnetic field of a 8-pole generator. In one revolution of
VTU : Jan.-14
the coil, the number of cycles generated by voltage is ____.
+
a) one
b) two
c) four
d) eight
[Ans. : c]
Standard Definitions Related to Alternating Quantity
Q.1
The time period of a sinusoidal wave form with 200 Hz frequency is _________ .
+
a) 0.05 s
b) 0.005 s
c) 0.0005 s
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
VTU : Aug.- 09
d) 0.5 s
[Ans. : b]
Basic Electrical Engineering
Q.2
5 - 31
Single Phase A.C. Circuits
The time taken by an alternating quantity to complete one cycle is called
a) frequency
b) speed
c) waveform
d) time period
[Ans. : d]
Q.3
The unit of frequency is ________ .
a) rad/sec
Q.4
b) seconds
c) hertz
d) volts
[Ans. : c]
c) w = 2pf
d) w =
1
f
[Ans. : c]
The relation between f and w is ________.
a) f =
1
w
b) w =
2p
f
Equation of an Alternating Quantity
Q.1
A sinusoidal voltage varies from zero to maximum of 250 V. The voltage at the instant of
VTU : Feb.- 10
60º of the cycle will be __________ .
+
a) 150 V
Q.2
d) 108.25 V [Ans. : b]
+
b) 50 Hz
c) 25 Hz
VTU : Feb.- 10
d) 100 Hz. [Ans. : b]
The equation of an alternating current is given by, i = 14.1421 sin 100 p t then the time
taken by it to complete three cycles is ________ .
a) 0.02 sec
Q.4
c) 125 V
An a.c. voltage is given by V = 40 sin 314 t. The frequency is______ .
a) 75 Hz
Q.3
b) 216.5 V
b) 0.06 sec
c) 0.08 sec
d) 0.01 sec [Ans. : b]
An instantaneous value of an alternating current having r.m.s. value of 7.071 A at 120º is
________ .
a) 4.33 A
b) 6.123 A
c) 8.66 A
d) 3.061 A [Ans. : c]
Q.5
An instantaneous value of an alternating voltage having 50 Hz frequency and maximum
value of 100 V at 0.01 sec is ________ .
100
a) 100 V
b) 100 2 V
c)
V
d) 0 V
2
[Ans. : d]
Q.6
A sinusoidal voltage has a magnitude of 200 V at 150º then its maximum value
is ________.
a) 100 V
Q.7
d) 300 V
[Ans. : b]
b) 0.015 sec
c) 0.02 sec
d) 0.07 sec [Ans. : a]
An alternating voltage is given by V = 100 sin (314 t – 30º) volts. The frequency is
VTU : June-10
________ .
+
a) 25 Hz
Q.9
c) 200 V
An alternating current is given by i = 20 sin 100 p t. The time taken by the current to
achieve – 20 A second time, measuring from t = 0 is ________.
a) 0.035 sec
Q.8
b) 400 V
b) 50 Hz
c) 60 Hz
An A.C. voltage is given by 100 sin 314 t. The frequency is ___.
a) 50 Hz
b) 75 Hz
c) 25 Hz
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
d) 100 Hz
[Ans. : b]
d) 100 Hz
[Ans. : a]
+ VTU : Jan.-11
Basic Electrical Engineering
Q.10
5 - 32
An alternating emf is given by e = 200 sin314 t. The instantaneous value of emf at
VTU : July-11
t = 1/200 sec is ______ .
+
a) 150 V
Q.11
Single Phase A.C. Circuits
b) 175 V
c) 200 V
d) 225 V.
[Ans. : c]
An alternating current is given by i (t ) = I m sin 2w t. Then frequency of the alternating
current is, ___________ .
VTU : July-11
+
a) w p Hz
b) 2 w p Hz
c) w 2p Hz
d) None of these.
[Ans. : a]
Effective Value or R.M.S. Value
Q.1
A sinusoidal voltage is represented as 141.42 sin 314 w t. r.m.s. value of voltage and
VTU : Feb.-09
frequency are respectively ________ .
+
a) 141.42 V, 314 Hz
Q.2
c) 200 V, 100 Hz
c) 141.421
b) 200 2
d) 400
[Ans. : a]
The concept of effective value is based on ________.
a) photoelectric effect
c) friction
Q.4
d) 100 V,100 Hz.
[Ans. : b]
The equation of an alternating quantity is v = 282.84 sin 100 p t then its r.m.s. value is
________.
a) 200
Q.3
b) 100 V, 50 Hz
b) heating effect
d) none of the above
[Ans. : a]
The voltage of domestic a.c. supply is 230 V. This value represents ________.
a) peak value
b) average value
c) r.m.s. value
d) mean value
[Ans. : c]
Q.5
The voltmeter in a.c. circuit always measures ________ values.
a) average
b) maximum
c) r.m.s.
d) none of these
[Ans. : c]
Q.6
The r.m.s. value of an alternating current is
10
A then its peak to peak value is
2
________ .
a) 10 A
Q.7
b) 30 A
c) 5 A
Definition of root-mean square value is ________.
d) 20 A
+
[Ans. : d]
VTU : Dec.-11
a) Square root of area under the square curve over half cycle to length of base over half
cycle.
b) Average value by
2. c) Ratio of maximum value to average value
d) None of the above.
Q.8
[Ans. : a]
The equation of an alternating current is i = 42.42 sin 628t. The effective value will be
VTU : Dec.-11
________.
+
a) 27 A
b) 30 A
c) 2.7 A
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
d) 3 A
[Ans. : b]
Basic Electrical Engineering
5 - 33
Single Phase A.C. Circuits
Average Value
Q.1
The peak value of a time wave is 400 V, its average value is _______.
+
a) 254.6 V
Q.2
b) 282.6 V
VTU : Aug.-09 ;Jan 11
c) 400 V
d) 565.5 V [Ans. : a]
The average value of sinusoidally varying voltage is ______ than its r.m.s. value.
a) more
b) less
c) same as
d) none of the above
[Ans. : b]
Q.3
The peak value of a sine wave is 400 V, its average value is ___.
a) 254.8 V
Q.4
c) 400 V
d) 565 V
The average value of sine wave over a one complete cycle is ____.
a) zero
Q.5
b) 282.6 V
b) + 1
c) – 1
+ VTU : June-12
d) 1/2
[Ans. : a]
+ VTU : Jan-14
The average value of sin q over cycle is ____.
a) + 1
b) – 1
c) zero
[Ans. : a]
d) 1/2
[Ans. : c]
Form Factor (Kf)
Q.1
The form factor of purely sinusoidal waveform is ________ .
a) 1.11
Q.2
b) 1.21
c) 1.414
The form factor is the ratio of _____ .
d)
2
[Ans. : a]
+ VTU : June-13
a) average value to rms value
b) rms value to average value
c) peak value to average value
d) peak value to rms value
[Ans. : b]
Crest or Peak Factor (Kp)
Q.1
Which of the following wave has least value of peak factor ?
a) Sine wave
c) Triangular wave
Q.2
b) Square wave
d) Full wave rectified sine wave.
The peak factor of a sinusoidally varying voltage is ________ .
a) 1.414
b) 1.11
c) 0.866
+
[Ans. : b]
VTU : June-10
d) 0.707
[Ans. : a]
R.M.S. Value of Combined Waveform
Q.1
A wire carries 5 A d.c. and alternating current of 15 sin w t A then the effective value of
the resultant current is ________.
a) 5 A
b) 15 2 A
c) 20 A
TM
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d) 11.72 A [Ans. : d]
Basic Electrical Engineering
5 - 34
Single Phase A.C. Circuits
Phasor Representation of an Alternating Quantity
Q.1
The phasor rotates in ________ direction.
a) anticlockwise
b) clockwise
c) circular
d) none of the above
[Ans. : a]
Concept of Phase of an Alternating Quantity
Q.1
An alternating current is 14.142 sin (100 p t – 30º) A and an alternating voltage is
p
282.842 sin (100 p t + ) V then the phase difference between V and I is _______.
4
p
a) 75º
b) 30º
c) rad
d) 15º
[Ans. : a]
4
Q.2
If two sinusoidal quantities are in phase quadrature then the phase difference between
them is ________ .
a) 45º
Q.3
Q.4
b) zero
c) 180º
d) 90º
If i1 = A sin (w t) and i2 = B sin (w t + 30º) then ________ .
b) i1 and i2 are in phase
a) i1 leads i2 by 30º
d) None of the above
c) i2 leads i1 by 30º
[Ans. : d]
[Ans. : c]
The negative maximum of a cosine wave occurs at ________ .
a) 45º
b) 90º
c) 180º
d) 270º
[Ans. : c]
Addition and Subtraction of Alternating Quantities
Q.1
For addition and subtraction, a.c. quantity is expressed in________ system.
a) spherical
b) cylindrical
c) rectangular
d) polar
[Ans. : c]
Multiplication and Division of Phasors
Q.1
For multiplication and division, a.c. quantity is expressed in ________ system.
a) cylindrical
Q.2
b) spherical
c) rectangular
d) polar
[Ans. : d]
Given (8 + j 6) ´ ( -10 - j 7.5), then the result in polar form is ________ .
a) 12.5 -143.1
b) 125 -106.2º
c) 125 +106.2º
d) 12.5 143.1.
[Ans. : a]
Q.3
+ VTU : Jan.-14
If two phasors A = 60 Ð 40º, B = (6 + j0), then A/B = _______.
a) 360 Ð 40º
b) 60 Ð 40º
c) 10 Ð 40º
d) 10 Ð – 40º
[Ans. : c]
A.C. through Pure Resistance
Q.1
The power factor of pure resistive circuit is _______ .
a) zero
b) unity
c) lagging
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
+
VTU : Jan.-09,11
d) leading. [Ans. : b]
Basic Electrical Engineering
Q.2
5 - 35
In a purely resistive circuit, the average power Pav is ______ the peak power, Pmax.
+
a) double
Q.3
b) one-half of
c) one-fourth
d) equal to [Ans. : b]
b) voltage leads current
d) none of the above
[Ans. : a]
The average power consumption in a pure resistance is, _______ .
a) Vrms Im
Q.5
VTU : Jan.-10
For a pure resistive circuit, the voltage and current relation is, _______ .
a) in phase
c) voltage lags current
Q.4
Single Phase A.C. Circuits
b) Vrms I 2rms
c) I 2m R
d)
b) unity
[Ans. : d]
+ VTU : Jan-13
The power factor of a pure resistive circuit is _____ .
a) zero
Vm I m
2
c) lagging
d) leading
[Ans. : b]
A.C. through Pure Inductance
Q.1
Inductive reactance of a coil of inductance 0.5 H at 50 Hz is ______ .
+
a) 15.7 ohm
Q.2
b) 157 ohm
c) 50 ohm
VTU : Jan.-09;11
d) 25 ohm. [Ans. : b]
In a pure inductor, the current _______ voltage by 90º.
a) leads
b) lags
c) in phase
d) none of the above
[Ans. : b]
Q.3
The inductive reactance of an inductor L is given by, _______ .
a) w L
Q.4
L
f
c) fL
2pL
f
[Ans. : a]
d) infinite
[Ans. : c]
d)
The average power consumption in a pure inductor is, _______ .
a) maximum
Q.5
b)
b) minimum
c) zero
+
The average power consumption is a pure inductor is _____ .
a) maximum
b) minimum
c) zero
VTU : Jan.-13
d) infinite
[Ans. : c]
A.C. through Pure Capacitance
Q.1
Q.2
a) lags behind the voltage by 90º
b) leads the voltage by 90º
c) remains in phase with voltage
d) none of these
VTU : Jan.-10
[Ans. : b]
In a pure capacitor, the voltage _______ current by 90º
a) leads
Q.3
+
In a pure capacitive circuit, the current _______ .
b) in phase
c) lags
d) none of the above
[Ans. : c]
The capacitive reactance of a capacitor C is given by, _______ .
a)
1
fC
b)
1
2pw C
c)
TM
2pw
C
TECHNICAL PUBLICATIONS - An up thrust for knowledge
d)
1
2pfC
[Ans. : d]
Basic Electrical Engineering
Q.4
5 - 36
Single Phase A.C. Circuits
The average power consumption in a pure capacitor is, _______ .
a) zero
b) infinite
c) negative
d) none of the above
[Ans. : a]
Q.5
The relation between frequency and capacitive reactance is, _______ .
a) square
Q.6
c) inverse
d) linear
[Ans. : c]
The reactance of a capacitor at 50 Hz is 5 W. If the frequency is increased to 100 Hz, the
VTU : July-11
new reactance is, ______ .
+
a) 5 W
Q.7
b) direct
b) 2.5 W
c) 10 W
In a pure capacitive circuit, the current will _____ .
d) 25 W
[Ans. : b]
+ VTU : June-13
a) lag behind the voltage by 90°
b) lead the voltage by 90°
c) remain in phase with voltage
d) none of these
[Ans. : b]
A.C. through Series R-L Circuit
Q.1
In a certain RL circuit, VR = 2 V and VL = 3 V. The magnitude of total voltage is_____ .
+
a) 2 V
Q.2
b) 3 V
c) 5 V
VTU : July-09
d) 3.6 V
[Ans. : d]
The impedance of the series R-L circuit is, _______ .
a) R– jXL
b) R+ jXL
c) XL+jR
d) none of these
[Ans. : b]
Q.3
The power factor for the series R-L circuit is, _______ .
a) zero
Q.4
Q.5
a) decreases
b) remains constant
c) increases
d) none of these
[Ans. : c]
b) VI sin f
c) IZ
d) VI
[Ans. : d]
b) VI sin f
c) IZ
d) VI
[Ans. : a]
c) zero
d) leading
[Ans. : b]
The power factor of d.c. circuit is, _______ .
b) unity
A series R-L circuit of 6 + j 8 W carries a current of 5 A then its power consumption is,
_______ .
a) Zero
Q.9
[Ans. : c]
The true power in single phase a.c. circuit is given by, _______ .
a) lagging
Q.8
d) leading
The apparent power in single phase a.c. circuit is given by, _______ .
a) VI cos f
Q.7
c) lagging
The phase difference between V and I for the series R-L circuit _______ as XL increases.
a) VI cos f
Q.6
b) unity
b) 30 W
c) 150 W
d) 40 W
[Ans. : c]
The instantaneous voltage and current for a.c. circuit are, v = 100 sin 377 t V and i = 25
pö
æ
sin ç377t - ÷ A then the power consumption of the circuit is, _______ .
3
ø
è
a) 125 W
b) 625 W
c) 225 W
d) 526 W [Ans. : b]
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
Basic Electrical Engineering
Q.10
5 - 37
Single Phase A.C. Circuits
The reactive power in a single phase a.c. circuit is given by _______ .
+
a) EI cos f
b) EI sin f
c) EI
VTU : June-10
d) none of these
[Ans. : b]
Q.11
The maximum and minimum values of power factor can be ______.
a) +1 and –1
b) +1 and –5
c) +1 and 0
+ VTU : Dec.-11
d) +5 and –5
[Ans. : b]
Q.12
a) leading
Q.13
+ VTU : June-13
In an R - L series circuit the pf is _____ .
b) lagging
c) zero
d) unity
The power factor of an ac circuit is equal to _____ .
a) cosine of the angle
c) unity for a resistive circuit
[Ans. : c]
+ VTU : June-13
b) sine of the phase angle
d) unity for a reactive circuit
[Ans. : c]
A.C. through Series R-C Circuit
Q.1
Q.2
Q.3
The impedance of A.C. circuit is 50 Ж 60º ohm. Then the circuit is _______ .
+
a) resistive
b) capacitive
c) inductive
d) none of the above.
VTU : Jan.-09
[Ans. : b]
By adding more resistance to an RC circuit ________.
+ VTU : Dec.-11
a) the real power increases
b) the real power decreases
c) the power factor decreases
d) the phase difference increases. [Ans. : a]
When the frequency of the applied voltage in series RC circuit is increased the
VTU : July-09
capacitance reactance _________.
+
a) increase
b) decreases
c) becomes zero
d) remains same
[Ans. : b]
Q.4
The impedance of the series R-C circuit is, _______ .
a) R + jXC
Q.5
c) R + j 2 p fC
XC
j
[Ans. : d]
d) lagging
[Ans. : c]
d) R +
The power factor for the series R-C circuit is, _______ .
a) zero
Q.6
b) XC – j R
b) unity
c) leading
The power consumption of series R-C circuit _______ if XC increases.
a) remains same
b) decreases
c) increases
d) none of these
[Ans. : a]
Q.7
A series R-C circuit of 6 – j8 W carries a current of 10 A then its power consumption is,
_______ .
a) 60 W
b) 600 W
c) 100 W
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
d) 80 W
[Ans. : b]
Basic Electrical Engineering
Q.8
5 - 38
As the power factor angle increases, _______ .
a) the active power increases
c) the active power decreases
Q.9
d) 45º
[Ans. : d]
c) 50 Hz
d) 2 kHz
d) 60 Hz
[Ans. : a]
b) increased
c) kept same
d) removed [Ans. : b]
b) 0.65
c) 0.55
d) 0.75
[Ans. : d]
If the voltage across R is 25 V and across C is 100 V in a series R-C circuit then the
supply voltage is _______ .
a) 125 V
Q.14
c) 90º
If the total current drawn by the circuit is 20 A and a component in phase with the
voltage is 15 A then the power factor of the circuit is ________.
a) 0.8
Q.13
b) 0º
For improving power factor, the resistance in the circuit must be _______ .
a) decreased
Q.12
[Ans. : c]
The capacitive reactance of a 20 mF capacitor is 7.9577 W at a frequency of ______ .
a) 1 kHz
Q.11
b) the reactive power decreases
d) the apparent power increases
If the active and reactive power components of a circuit are equal then the power factor
angle is, _______ .
a) 30º
Q.10
Single Phase A.C. Circuits
b) 25 V
c) 100 V
d) 103.077 V
[Ans. : b]
A series R.C. circuit of 6 – j8 W carries a current of 10 A then its power consumption is
VTU : Jan.13
_____ .
+
a) 60 W
b) 600 W
c) 100 W
d) 80 W
[Ans. : b]
A.C. through Series R-L-C Circuit
Q.1
The series R-L-C circuit will have lagging power factor if XL is _______ XC.
a) less than
Q.2
c) same as
d) none of these
[Ans. : b]
The power consumption of R-L-C series circuit is, _______ .
a) VI
Q.3
b) greater than
b)
2
V
Z
c) I R
2
d) I V
[Ans. : c]
The power factor of the circuit shown in the Fig. 5.35 is _______ .
10 W
100 mH
25 mF
V, 50 Hz
Fig. 5.35
a) 0.112 lagging
b) unity
c) zero
d) 0.112 leading
[Ans. : d]
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
Basic Electrical Engineering
Q.4
Q.5
5 - 39
Single Phase A.C. Circuits
+
The p.f. is lagging when _______ .
a) voltage lags the current
b) current lags the voltage
c) voltage lags the power
d) current lags the power
The voltage of the applied source in the circuit of Fig. 5.36 is ____.
R
40 V
L
C
50 V
50 V
VTU : June-10
[Ans. : b]
+ VTU : June-12
Fig. 5.36
a) 50 V
Q.6
b) 100 V
c) 40 V
d) 140 V
[Ans. : c]
In the circuit shown in Fig. 5.37 the potential difference across the various elements are
Jan.-14
shown. What is the source voltage, V ?
+
50 V
50 V
50 V
V
Fig. 5.37
a) 50 V
b) 100 V
c) zero
d) 150 V
[Ans. : a]
A.C. Parallel Circuit
Q.1
The admittance is _______ the impedance.
a) equal to
b) square of
c) reciprocal of
d) none of these
[Ans. : c]
Q.2
The admittance of series R-L circuit is given by, _______ .
a) G + jB
Q.3
c) B + jG
d) G – jB
[Ans. : d]
c) watt
d) ampere
[Ans. : a]
The unit of admittance is _______ .
a) mho
Q.4
b) B – jG
b) ohm
The net susceptance of a parallel resonating circuit is _______ .
a) maximum
b) zero
c) unity
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
d) negative [Ans. : b]
Basic Electrical Engineering
Q.5
5 - 40
+ VTU : June-12
The power taken by the circuit shown is ____.
a) 480 W
b) 1920 W
c) 1200 W
240 V
Q.6
Single Phase A.C. Circuits
d) none of these
XL
= 30 W
R = 30 W
+ VTU : Jan.-13
The admittance is _______ impedance.
a) equal to
[Ans. : b]
b) square of
c) reciprocal of
d) square root of
[Ans. : c]
qqq
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
6
Domestic Wiring
Chapter at a Glance
Important Theory Questions and Answers
Ø
Mention and describe in brief various types of wiring systems.
+
VTU : Jan.-03, 06, 08, 10, 11;
July-03, 04, 06, 07, 08; Feb.-05, June-12, 13, Marks 5
· Various types of wiring used in practice are:
1. Cleat wiring
2. Casing capping
3. Surface wiring
4. Conduit wiring
5. Metal sheathed wiring
Cleat Wiring
· In this type wires are clamped between porcelain cleats.
· The cleats are made up of two halves.
· One half is grooved through which wire passes while the other fits over the first.
Screw
· The whole assembly is
then mounted on the wall
or wooden beam with the
help of screws.
· The lower half of the
porcelain cleat is known as
base having grooves for
conductors while the upper
half is known as cap which
is shown in the Fig. 6.1.
Upper half
of cleat
Wire
Lower half
of cleat
Wooden beam or wall
Fig. 6.1 Cleat wiring
Advantages :
1. This method is one of the cheapest method. 2. It is most suitable for temporary work.
3. It can be very quickly installed. 4. It can be recovered without any damage of material.
5. Inspection and changes can be made very easily.
6. Erection does not require skilled labour.
(6 - 1)
TM
TECHNICAL PUBLICATIONS - An up thrust for knowledge
Basic Electrical Engineering
6-2
Domestic Wiring
Disadvantages :
1. This method does not give attractive appearance.
2. Dust and dirt collects on the cleats.
3. Maintenance cost is very high.
Casing Capping
· In this method,
casing
is
a
rectangular
strip
made from teak
wood or now a
days made up of
P.V.C.
Capping
Screw
Wires
· It has two grooves
into
which
the
wires are laid.
Casing
Fig. 6.2 Casing capping
· Then casing is covered with a rectangular strip of wood or P.V.C. of the same width,
called capping. The capping is screwed into casing by means of screws fixed at every
15 cm.
· The casing is fixed to the walls and apart from it by 3.5 mm with the help of porcelain
discs or cleats.
Advantages :
1. Neat and clean appearance.
2. Its installation is easy compared to some other methods of wiring.
Disadvantages :
1. The requirement of skilled labour for the installation.
2. The method is costly.
Surface Wiring
· In this type, the wooden battens are
fixed on the surface of the wall, by
means of screws and rawl plugs.
Screw
Wooden batten
· The metal clips are provided with the
battens at regular intervals.
· The wire runs on the batten and is
clamped on the batten using the metal
clips.
Wire
Advantages :
1. A good appearance.
Clip
Fig. 6.3 Wooden batten wiring
2. Simple to erect.
3. Cheaper in cost.
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Domestic Wiring
Disadvantages :
1. Wires are always exposed to atmosphere and hence subjected to dust, dirt, fumes and
other dangerous atmospheric conditions.
2. Bending and breaking of batten and hence of the wires may occur after some time.
Conduit Wiring
· In this method, metallic tubes called as
conduits are used to run the wires.
Metal conduit
Wall
· The conduits are made up of mild steel
which is annealed so that it can be bent
without breaking. The standard length
of conduit is generally 4 m.
Saddle
Wires
Fig. 6.4 Surface conduit wiring
· The conduits are threaded at both ends with one coupler attached.
· The conduits are supplied with black enamel coating on its internal and external
surface.
· The conduits are to be erected completely before laying any cable in it. The rigid
conduits are always terminated at outlets into a box which may be round, square or
octagonal.
· Inspection boxes are used to facilitate the pulling of conductors while junction boxes are
used to house the junctions of the conductors.
Advantages :
1. It is durable.
2. It has a long life. 3. It requires very less maintenance.
Disadvantages :
1. The repairs are very difficult in case of concealed conduit wiring.
2. This method is most costly. 3. The erection requires highly skilled labour.
Metal Sheathed Wiring
· In this type of wiring, vulcanised india rubber (V.I.R.) conductors covered with lead
alloy sheath are used. It is similar to C.T.S. or T.R.S. wiring.
· The insulated conductor is covered with a metal sheath which protects the wiring
system from mechanical injury and atmospheric conditions.
· The wires with metal sheath on it are run on the wooden batten which is fixed on the
wall with the help of screws. The wire is clamped on the batten using metal clips.
Advantages :
1. It protects the conductors from mechanical injury.
2. It provides better appearance.
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Domestic Wiring
Disadvantages :
1. It is expensive as compared to C.T.S. or T.R.S. wiring.
2. It is unsuitable in corrosive environment.
Ø
With the help of circuit diagram, explain the two way and three way control of lamps.
+ VTU : Jan.-03, 06, 08, 13, July-03, 04, 06, 07, 08; Feb.-05, Marks 5
Two Way Control of Lamps
· This is also called as
staircase wiring as it is
commonly used for stair
cases and corridor lighting.
· It consist of two way
switches. A two way switch
operates always in one of
the two possible positions.
A
P
Single phase
A.C. Supply
B
1
1
2
2
2 way switches
N
· The circuit is shown in the
Fig. 6.5.
· Assume that lamp is on first
floor. Switch A is on first
floor and B is on second
floor. In the position shown
in the Fig. 6.5, the lamp is
OFF.
· When
person
changes
position of switch A from (1)
to (2) then lamp gets phase
through switches A and B
and it gets switched ON as
shown in Fig. 6.6.
· When person reaches on
second floor, the lamp is
required to be switched OFF
as shown in Fig. 6.6.
· So person will change switch
B from (2) to (1), due to
which
phase
connection
reaching to the lamp gets
opened and lamp will be
switched OFF as shown in
Fig. 6.7.
Lamp
Fig. 6.5 Control of one from two points
A
P
B
1
1
2
2
Connection
N
Lamp ON
Fig. 6.6 'ON' Position of lamp
A
P
B
1
1
2
2
No connection
N
Lamp OFF
Fig. 6.7 'OFF' position of lamp
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· Thus ON and OFF of one lamp can be controlled from two positions with the help of
two way switches.
Three Way Control of Lamps
· It consist of two way switches A and B and one intermediate switch C.
· The intermediate switch can have positions to connect points 1-4, 3-2, as shown or 1-2
and 3-4 shown dotted. The switch A is on first floor and switch B is on third floor say.
· In the position shown in Fig. 6.8, the lamp is ON.
Intermediate
switch C
A
1
2
3
4
P
2
Connection
B
1
Connection
1
2
N
Lamp (ON)
Fig. 6.8 Three way control of lamps
· When person from floor 2 changes switch C position to have connections 1-2, and 3-4
then there is open circuit in the connection and lamp gets switched OFF.
· Now if the person from third floor changes the position of switch B from 1 to 2, then
again Lamp gets supply through position 2 of A, 3-4 of C and 2 of B. The lamp gets
switched ON.
· Again if switch A position is changed, lamp gets switched OFF.
· Thus we have the control of lamp from three different positions which is called three
way control of lamps.
Ø
Explain the necessity of earthing.
+ VTU : Jan.-04, 07, 11; Feb. 05; July-05, 09; May-10, Marks 4
· When a person, standing on the earth touches the machine, current I gets an alternative
path through the body of the person to earth from the insulation resistance, finally to
the neutral of the supply.
I body =
V
R i + R body + R E
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Fuse
I
· When the insulation becomes
Line
Machine
weak or defective or if one of
the windings is touching to
the frame directly due to
Person
some fault then R i i..e.
V
insulation resistance becomes
almost zero resistance of
body and earth are not very
Earth
Neutral
high
and
hence
I body
increases to such a high
value that the person receives
Fig. 6.9 Machine is not earthed
a fatal shock. Such a current
is called a leakage current. Hence when the machine is not earthed, there is always a
danger of the shock, under certain fault conditions.
· In case of earthing, the frame of the machine is earthed as shown in the Fig. 6.10.
· When the person touches the
frame, and if there is a leakage due
to fault condition, due to earthing a
leakage current takes a low
resistance path i.e. path from frame
to earth, bypassing the person.
Fuse
Line
I
Machine
V
Person
· As earthing resistance is very very
Earth
low compared to the body of the Neutral
person, current
prefers low
Earthing of
Path of leakage current
resisstance path. Thus I body is
machine
negligibly small compared to earth
Fig. 6.10 Machine is earthed
current. So entire leakage current
passes through the earthing contact bypassing the body of the person. The value of
I body is not sufficient to cause any shock to the person.
Ø
Explain the plate earthing alongwith a neat diagram.
+ VTU : Jan.-04, 07, 14, Feb.-05; July-05, 09; May-10, Dec.-11, Marks 6
· The earth connection is provided with the help of copper plate or galvanized iron (G.I.)
plate. The copper plate size is 60 cm ´ 60 cm ´ 3.18 mm. The plate is embedded
3 meters ( 10 feet ) into the ground. The plate is kept with its face vertical.
· The plate is surrounded by the alternate layer of coke and salt for minimum thickness
of about 15 cm. The earth wire is drawn through G.I. pipe and is perfectly bolted to the
earth plate. The nuts and bolts must be of copper plate and must be of galvanized iron
for G.I. plate.
· The earth lead used must be G.I. wire or G.I. strip of sufficient cross-sectional area to
carry the fault current safely. The earth wire is drawn through G.I. pipe of 19 mm
diameter, at about 60 cm below the ground.
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· The G.I. pipe is fitted with a funnel on the top. In order to have an effective earthing,
salt water is poured periodically through the funnel.
· The earthing efficiency, increases with the increase of the plate area and depth of
embedding. If the resistivity of the soil is high, then it is necessary to embed the plate
vertically at a greater depth into the ground.
Cast iron cover
30 cm x 30 cm
60 cm
3m
G.I. pipe
12.7 mm diameter
Funnel covered
with wire mesh
19 mm
diameter
15 cm
Layer of coke and salt
90 cm
60 cm x 60 cm x 6.3 mm
G.I. plate
90 cm
Fig. 6.11 Plate earthing
Ø
Explain the pipe earthing alongwith a neat diagram.
+ VTU : July-03; Jan.-06, 09, 11, June-10, Marks 8
· In this method of earthing a G.I. pipe of 38 mm diameter and 2 meter ( 7 feet ) length
is embedded vertically into the ground. This pipe acts as an earth electrode. The depth
depends on the condition of the soil.
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Domestic Wiring
· The earth wires are fastened to the top section of the pipe above the ground level with
nut and bolts.
· The pit area around the pipe is filled with salt and coal mixture for improving the
condition of the soil and earthing efficiency. The schematic arrangement of pipe
earthing system is shown in the Fig. 6.12.
Cement concrete
2.75 m
Funnel with
wire mesh
60 cm
Lug
19mm
diameter
15 cm
12.7 mm
diameter
15 cm
12 mm diameter
Main G.I. pipe
2m
Alternate layers
of charcoal and salt
Fig. 6.12 Pipe earthing
· In summer season, soil becomes dry. In such case salt water is poured through the
funnel connected to the main G.I. pipe through 19 mm diameter pipe. This keeps the
soil wet.
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Domestic Wiring
Important Multiple Choice Questions with Answers
Types of Wiring Systems
Q.1
________is the cheapest wiring method.
a) cleat
Q.2
b) casing-capping
c) conduit
d) surface
[Ans. : b]
b) casing-capping
c) conduit
d) surface
[Ans. : b]
b) casing-capping
c) conduit
d) surface
[Ans. : d]
b) casing-capping
c) conduit
d) surface
[Ans. : c]
b) casing-capping
c) conduit
d) surface
[Ans. : c]
b) casing-capping
c) conduit
d) surface
[Ans. : c]
d) surface
[Ans. : b]
_________method is used for the damped places.
a) cleat
Q.10
[Ans. : a]
_________method is most costly and requires skilled labour.
a) cleat
Q.9
d) surface
_________method makes the wiring electric shock proof.
a) cleat
Q.8
c) conduit
_________method gives full mechanical protection to the wires.
a) cleat
Q.7
b) casing-capping
In this method, the wires are exposed to the atmosphere.
a) cleat
Q.6
[Ans. : a]
_________method is used for the voltages upto 250 V.
a) cleat
Q.5
d) surface
_________method is popularly used for the residential and domestic purposes.
a) cleat
Q.4
c) conduit
________method is used for the temporary wiring.
a) cleat
Q.3
b) casing-capping
b) metal sheathed
c) conduit
Most modern wiring system for domestic and commercial installation is _________.
+ VTU : Jan.-14
a) cleat wiring
b) Wooden-Batten wiring
c) wooden-casing wiring
d) conduit wiring
[Ans. : d]
Wiring Schemes
Q.1
Two way control of lamp is also called ________wiring.
a) staircase
b) godown
c) flexible
d) none of these
[Ans. : a]
Q.2
__________switch is important in 3 way control of lamp.
a) single pole
b) intermediate
c) multipole
d) none of these
[Ans. : b]
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Domestic Wiring
Introduction to Fuse
Q.1
Q.2
Q.3
a) current limiting device
b) protective device
c) voltage limiting device
d) none of these.
b) high resistivity and high melting point
c) high resistivity and low melting point
d) low resistivity and low melting point.
[Ans. : c]
Operation of _________depends on the selection of its proper rating.
Q.13
b) element
c) factor
d) none of these [Ans. : b]
b) rated current
c) fusing current
d) none of these [Ans. : c]
b) rated current
c) fusing current
d) none of these [Ans. : a]
b) 5
c) 8
d) 2
[Ans. : d]
b) melting point
c) permittivity
d) none of these [Ans. : b]
The fuse material must have high_________.
b) melting point
c) resistivity
d) none of these [Ans. : a]
The minimum fusing current of a fuse unit is 2.1 A and fusing factor is 1.1. Then it rated
VTU : Aug.-11
carrying current of fuse element is________ .
+
a) 2.2 A
Q.12
d) none of these [Ans. : a]
The fuse material must have low__________.
a) conductivity
Q.11
c) neutral
For the household fuse the fusing factor is generally___________.
a) conductivity
Q.10
b) parallel
The ratio of the minimum fusing current and the current rating of the fuse is
called__________.
a) 10
Q.9
d) none of these [Ans. : c]
The minimum value of the current at which the fuse melts is called_________.
a) fusing factor
Q.8
c) fuse
The part of the fuse which melts when high current flows through it is called the
fuse________.
a) fusing factor
Q.7
b) MCB
The fuse is always connected in _________of the circuit.
a) body
Q.6
+ VTU : Feb.-10
a) low resistivity and high melting point
a) series
Q.5
[Ans. : b]
The material used for fuse wire should be of _______ .
a) switch
Q.4
+ VTU : Jan.-09, Aug.-09, Feb.-11
A fuse is a _______ .
b) 2.13 A
c) 1.909 A
The primary function of a fuse is to________ .
a) protect the appliance
b) open the circuit
c) prevent excessive current
d) protect the line
Ratio minimum fusing current/current rating in fuse is _____ .
a) fusing factor
b) rated current
d) 0.5238 A
[Ans. : c]
+ VTU : Dec.-11
+
[Ans. : b]
VTU : Jan.-13
c) fusing current d) melting current
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[Ans. : a]
Basic Electrical Engineering
Q.14
6 - 11
Domestic Wiring
The fuse wire for smaller current rating (up to 10 A) are made of_____ .
+
a) lead-tin alloy
b) copper
c) lead
VTU : Jan.-14
d) aluminium
[Ans. : a]
Earthing
Q.1
A good earthing should provide ________ resistance in earthing path.
+
a) low
Q.2
Q.3
d) none of these. [Ans. : a]
+
VTU : Feb.-10
a) good conductor of electricity
b) mechanically strong
c) both (a) and (b)
d) mechanically strong but bad conductor of electricity.
[Ans. : a]
The resistance of the earthing wire is ___________.
b) moderate
c) very small
d) none of these [Ans. : c]
A good earthing should provide________ resistance in earthing path.
a) low
Q.5
c) medium
The earth wire should be _______ .
a) very high
Q.4
b) high
VTU : Aug.-09
b) lough
c) medium
+
VTU : Feb.-11
d) none of these [Ans. : a]
Earthing of electrical installation is carried out to protect ___________.
+
VTU : June-12
a) equipments from damage
b) personnel against electric shock
c) equipments from short circuit
d) all of these
[Ans. : d]
Methods of Earthing
Q.1
The current causing an electric shock is called ___________current.
a) skin
Q.2
c) insulating
d) none of these [Ans. : b]
For the current levels of about __________, person with dry skin gets a mild shock.
a) 2.5 A
Q.3
b) leakage
b) 1.5 mA
c) 100 mA
d) 2.5 mA
[Ans. : d]
Coke can be used as a sandwich between salt of an earthing system, to _______ .
a) by pass the current
b) avoid melting of the salt
c) improve conductivity
d) to hole moisture content.
[Ans. : c]
Electric Shock
Q.1
The effect of electric current on vital human organs depends upon _________.
a) magnitude of current
b) duration of current
c) frequency of current
d) all of these
+ VTU : June-12
[Ans. : d]
qqq
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Notes
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7
Three Phase Circuits
Chapter at a Glance
1.
Relations for Star Connected Load
VL =
3 Vph for star connection and I L = I ph
Thus line voltage is 3 times the phase voltage in star connection.
· Remember that Z ph relates I ph and V ph hence angle f is always between I ph and V ph
and not between the line values and Z ph =
Vph
I ph
. The line values do not decide the
impedance angle or power factor angle.
· Every line voltage leads the respective phase voltage by 30°.
P =
3 V L I L cos f
watts
For star connection, to draw phasor diagram, use
VRY = VR - VY ,
2.
VYB = VY - VB
and
VBR = VB - VR
Relations for Delta Connected Load
IL =
3 I ph
and V ph = V L
… for delta connection
Thus line current is 3 times the phase current in delta connection.
Thus for delta connection, to draw phasor diagram, use
IR = IRY - IBR ,
IY = IYB - IRY
and
IB = IBR - IYB
· Every line current lags the respective phase current by 30°.
(7 - 1)
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P=
7-2
3 V L I L cos f
Three Phase Circuits
watts
· The expression for power is same but values of line currents are different in
star and delta connected load which must be correctly determined to obtain
power.
3.
Wattmeter
W = Vpc ´ I c ´ cos
· If
I c = I ph and Vpc = Vph then
I c Ù Vpc = I ph Ù Vph = f and then
only wattmeter reads per phase
power which is Vph I ph cos f .
4.
(Ic Ù Vpc )
watts
Ic
M
L
Current coil
C
V
Pressure coil
Vpc
Fig. 7.1
Two Wattmeter Method
W = W1+W 2 = Three phase power =
\
3 VL I L cos f
W 1 = VL I L cos ( 30 - f)
W 2 = VL I L cos ( 30 + f)
· It can be observed that whether load is star of delta, the expressions for W 1 and W 2
remain same.
In case of leading power factor loads , readings of W 1 and W 2 are interchanged
compared to lagging power factor load.
For star or delta lagging p.f. load, W1 = VLILcos ( 30 - f) and W 2 = VLILcos ( 30 + f)
For star or delta leading p.f. load, W1 = VLILcos ( 30 + f) and W 2 = VLILcos ( 30 - f)
For star or delta unity p.f. load, cos f = 1 and f = 0°,
5.
W1 = W 2 = VLILcos 30°
Power Factor Calculation by Two Wattmeter Method
ìï
é 3(W1 - W 2 ) ù üï
p.f. cos f = cos í tan -1 ê
úý
ïî
ë (W1 + W 2 ) û ïþ
· The power factor cos f is always positive but its nature must be determined by
observing sign of tan f .
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6.
7-3
Three Phase Circuits
Effect of P.F. on Wattmeter Readings
Range of p.f.
Range of 'f '
W1 sign
W2 sign
Remark
cos f = 0
f = 90°
positive
negative
W1 = W2
0 Ð cos f Ð 0.5
90° Ð f Ð 60°
positive
negative
cos f = 0.5
f = 60°
positive
0.5 Ð cos f Ð 1
60° Ð f Ð 0°
positive
positive
cos f = 1
f = 0°
positive
positive
0
W1 = W2
Important Theory Questions and Answers
Ø
List the advantages of three phase system over single phase system.
+
VTU : July-03, 04; Jan.-04, 08, 11, 14, Marks 4
1) Three phase alternator occupies less space and has less cost too than single phase
having same rating.
2) For a transmission and distribution, three phase system needs less copper or less
conducting material than single phase system.
3) Three phase motors are self starting.
4) Three phase system give steady output.
5) Power factor of single phase motor is poor than three phase motors of same rating.
Ø
Obtain the relationship between line and phase values of current in a three phase balanced
Feb.-05; July-05; 07, 09; June-10, 12, 13, 14; Jan.-11,
star connected system.
+
Dec.-11, Marks 8
· Consider the balanced star connected load as shown in the Fig. 7.2.
IR
R
R
IR
VRY
Three
phase Y
supply
h
IY
VY
IB
VBR VYB
B
Zp
VR
Zp
IB
B
N
Zp
h
VB
Fig. 7.2 Star connected load
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h
IY
Y
Basic Electrical Engineering
7-4
Line voltages, VL = VRY = VYB = VBR
Phase voltages, Vph = VR = VY = VB
and
and
Three Phase Circuits
· From the Fig. 7.2 we can write,
VRY = VRN + VNY But VNY = - VYN (Generally
suffix N is not used for phase voltages)
· The phasor diagram to get VRY is shown in the
VB
-VY=Vp
O
\
Ø
VL =
30º
120º
· Perpendicular is drawn from point A on vector
OB representing VL .
VRY=VL
B
h
60º
Fig. 7.3. The VY is reversed to get - VY and then
it is added to VR to get VRY.
· From triangle OAB, cos 30º =
IL = IR = IY = IB
I ph = I R = I Y = I B
Line currents,
Phase currents,
C
A
VR=Vp
h
VY
Fig. 7.3
OC (VRY 2)
=
i.e.
OA
VR
3 (VL 2)
=
Vph
2
3 Vph for star connection and I L = I ph
Obtain the relationship between line and phase values of current in a three phase balanced
delta connected system.
+ VTU : Jan.-03; July-04, 05, 07, Marks 8; July-08, Marks 10;
Jan.-10, Dec.-11, Marks 6
· Consider the balanced delta connected load as shown in the Fig. 7.4.
IR
R
R
VRY
Three
phase Y
supply
IY
VBR
VYB VBR
B
VRY
Zp
Zp
h
IYB
IBR
IB
B
Zp
h
h
IRY
Y
VYB
Fig. 7.4 Delta connected load
· Line voltages : VL = VRY = VYB = VBR
· Phase voltages : Vph = VRY = VYB = VBR
Line currents : I L = I R = I Y = I B
Phase currents : I ph = I RY = I YB = I BR
· The phasor diagram to obtain line current I R by carrying out vector subtraction of
phase currents I RY and I YB is shown in the Fig. 7.5.
· IBR is reversed to get – IBR and then added to IRY to get IR.
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Three Phase Circuits
· Similarly OB bisects angle between - I BR and I RY
which is 60º.
\
Ð BOA = 30°
and
· From triangle OAB,
OC I R 2
=
· cos 30° =
I RY
OA
IBR
120º
I
OC = CB = L
2
O
I 2
3
= L
I ph
2
i.e.
IRY = Ip
A
30º
h
60º C
-IBR=Ip
IYB
h
B
IR=IL
Fig. 7.5
\
Ø
IL =
and V ph = V L
3 I ph
… for delta connection
Show that in a three phase, balanced circuit, two wattmeters are sufficient to measure the total
three phase power and power factor of the circuit .
+ VTU : Jan.-03, 04, 06, 08, 09, 13, July-03, Marks 8
W1
IR
R
Load 1
W2
IY
Y
R
N
B
IB
B
Load 3
Load 2
Y
Fig. 7.6
VB
· Consider star connected load and two
wattmeters connected as shown in the Fig. 7.6.
(
W 1 = I R ´ VRB ´ cos I R Ù VRB
W 2 = I Y ´ VYB ´ cos
· From Fig. 7.7,
and
)
and
(IY Ù VYB )
I R Ù VRB = 30 – f
I Y Ù VYB = 30 + f
\
W 1 = I R VRB cos ( 30 - f)
\
W 1 = VL I L cos ( 30 - f)
and
W2 =
VR
f
f
IY
30º
30º
-VB
VY
I Y VYB cos ( 30 + f)
IR
VYB
Fig. 7.7
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Basic Electrical Engineering
7-6
Three Phase Circuits
W 2 = VL I L cos ( 30 + f)
\
W 1 + W 2 = VL I L [cos ( 30 - f) + cos ( 30 + f)]
=
VL I L [cos 30 cos f + sin 30 sin f + cos 30 cos f - sin 30 sin f]
= 2 VL I L cos 30 cos f = 2VL I L
W1 + W2 =
\
Ø
3
cosf
2
3 VL I L cos f = Total 3 phase power
How power factor can be obtained from two wattmeter readings ?
+
VTU : Jan.-02, 05, 06, Marks 5; July-03, 07, 08, Marks 6
· For balanced, lagging p.f. load, W1 = VL I L cos ( 30 - f) and W 2 = VL I L cos ( 30 + f)
W1 + W 2 =
\
W1 – W 2
3VL I L cosf
= VL I L sin ( f)
W1 - W 2
=
W1 + W 2
\
\
Ø
f
VL I L sin f
3 VL I L cos f
=
tan f
i.e.
3
tan f =
3(W1 - W 2 )
(W1 + W 2 )
é 3(W1 - W 2 ) ù
= tan -1 ê
ú
ë (W1 + W 2 ) û
é 3(W1 - W 2 ) ù ïü
ïì
p.f. cos f = cos í tan -1 ê
úý
ïî
ë (W1 + W 2 ) û ïþ
Discuss the effect of variation of power factor on wattmeter readings.
+
VTU : July-02, 07, Marks 6; Jan.-10, Marks 6; June-10, Marks 5
· Consider different cases,
Case i) cos f = 0 f = 90°
1
\ W1 = VL I L cos ( 30 - 90) = + VL I L
2
i.e.
and
1
W 2 = VL I L cos ( 30 + 90) = - VL I L
2
W1 + W 2 = 0
W1 =
Case ii)
W2
but
W 2 = – W1
cos f = 0.5, f = 60°
\ W1 = VL I L cos ( 30 - 60) = VL I L cos30
and W 2 = VL I L cos ( 30 + 60) = 0
W1 + W2 = W1 = Total power.
\
· For all power factors between 0 to 0.5 W 2 shows negative and W1 shows positive, for
lagging p.f.
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7-7
Case iii) cos f = 1, f = 0°
\ W1 = VL I L cos ( 30 + 0) = VL I L cos 30
and
· Both W1 and W 2 are equal and positive.
Three Phase Circuits
W 2 = VL I L cos ( 30 - 0) = VL I L cos 30
· For all power factors between 0.5 to 1 both wattmeter gives +ve reading.
Important Solved Examples
Example 7.1
A star connected load consists of 6 W resistance in series with an 8 W inductive
reactance in each phase.A supply voltage of 440 V at 50 Hz is applied to the load. Find the
line current, power factor and power consumed by the load .
Zph = 6 + j 8 W = 10 Ð 53.13º W,
Solution :
Vph =
Iph =
VL
3
Vph
Z ph
VL = 440 V
440
= 254.034 V
3
=
=
254 . 034
= 25.4034 A
10
IL
R
R
Iph
Vph
VL = 440 V
3 Phase
50 Hz
+ VTU : Jan.-04, Marks 8
= VL/Ö 3
Y
Zph
N
Zph
Zph
B
Y
B
Fig. 7.8
\
IL = Iph = 25.4034 A
cos f =
\
P =
... Line current
R
6
=
= 0.6 lagging
Z 10
3VL I L cosf =
... Inductive
3 ´ 440 ´ 25.4034 ´ 0.6
= 11615.99 W = 11.616 KW
Example 7.2
... Power
Three inductive coils each having resistance of 16 ohm and reactance of 12 ohm are
connected in star across a 400 V, three-phase 50 Hz supply. Calculate :
i) Line voltage,
iv) Phase current,
ii) Phase voltage,
v) Power factor,
iii) Line current,
vi) Power absorbed.
Draw phasor diagram.
Solution : Rph = 16 W, XL = 12 W per ph, Star connection VL = 400 V
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7-8
Three Phase Circuits
Zph = Rph + j XL = 16 + j 12 W = 20 Ð + 36.86° W
\
Using rectangular to polar conversion on calculator.
VL = 400 V
i) Line voltage
ii) Phase voltage
Vph =
iii)
Iph =
For star connection,
VL
3
=
400
= 230.94 V
3
=
230.94
= 11.547 A
20
Vph
Z ph
IL = Iph
VB
VBR
i.e.
IB
…Star VL =
3 Vph
Line current = 11.547 A
–VY
VRY
36.86º
–VR
36.86º
IY
36.86º
VR
IR
–VB
VY
VYB
Fig. 7.9
iv)
Phase current = 11.547 A
cos f =
v) Power factor
R ph
Z ph
=
16
= 0.8 lagging
20
…Inductive hence lagging
vi) Power absorbed
P = 3 VL I L cos f = 3 ´400 ´ 11.547 ´ 0.8 = 6400 W
The phasor diagram can be shown as in the Fig. 7.9.
Example 7.3 A balanced three phase star connected load of 100 kW takes a leading current of
80 amp, when connected across 3 f , 1100 volt, 50 Hz supply. Find the value of resistance phase
and capacitance phase of load and p.f. of load. Find the total kVA and kVAR of the circuit.
Solution :
PT = 100 kW, I L = 80 A, VL = 1100 V, f = 50 Hz, Star
Vph
=
VL
3
=
1100
= 635.0852 V
3
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7-9
Three Phase Circuits
I ph = I L = 80 A
PT =
\
\
\
100 ´ 10 3 =
3 VL I L cos f
3 ´ 1100 ´ 80 ´ cos f
cos f = 0.656 leading i.e. f = 48.9984°
Vph = 635.0852 Ð 0° V,
I ph = 80 Ð + 48.9984° A
Vph
\
Z ph =
\
R ph = 5.2082 W, X Cph = 5.9911 W
\
C ph =
I ph
= 7.9385 Ð - 48.9984° W = 5.2082 - j 5.9911 W
1
2p´ f ´ X C
= 531.313 mF
ph
p.f. = 0.656 leading
VA =
3 VL I L =
3 ´ 1100 ´ 80
= 152420.4711 VA = 152.4204 kVA
VAR =
3 VL IL sin f
= 152.4204 ´ 103 ´ sin (48.9984º)
= 115030.397 VAR = 115.0303 kVAR
Example 7.4 Each of the star connected load consists of a non-reactive resistance of 100 W in
parallel with a capacitance of 31.8 µF. Calculate the line current, power absorbed, the total
kVA and power factor when connected to a 416 V, 3 phase, 50 Hz, supply.
+ MU : May-11
Solution : Star connected load, R = 100 W, C = 31.8 µF
1
XC =
= 100 W
\
2 p fC
\
Zph = R || – j XC =
=
R
100 x ( -j 100)
[100 - j 100]
10000 Ð - 90º
141.4213 Ð - 45º
C
Zph
Fig. 7.10
= 70.7107 Ð – 45º W
VL = 416 V, Vph =
VL
3
= 240.177 V (be the reference)
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\
Iph =
\
7 - 10
Vph
=
Z ph
Three Phase Circuits
240.177 Ð 0º
= 3.3966 Ð 45º A
70.7107 Ð - 45º
IL = Iph = 3.3966 A
P =
Total kVA =
… Line current
3 VL IL cos f =
3 ´ 416 ´ 3.3966 ´ cos (– 45º) = 1730.5543 W
3 VL IL = 2.4473 kVA
Power factor = cos (– 45º) = 0.7071 leading
Example 7.5
Three 100 W resistors are connected in i) Star and ii) Delta across a 415 V, 50 Hz,
3-phase supply. Calculate the line and phase currents and the power consumed in each case.
+ VTU : Feb.-05, Marks 8
Solution :
IL
IL
Ip
VL = 415 V
h
R
Vph =
N
R
VL
Ö3
Ip
VL = 415 V
h
R
R
Vph = VL
R = 100 W
R
R
R = 100 W
(a) Star
(b) Delta
Fig. 7.11
(a) Star :
\
Vph =
Iph =
VL
3
Vph
Z ph
415
= 239.6 V, Zph = Rph = 100 W
3
=
=
239.6
= 2.396 A
100
\
IL = IPh = 2.396 A
\
P =
=
3VL I L cos f
... cos f = 1 as load is resistive
3 ´ 415 ´ 2. 396 ´ 1 = 1722.2474 W
(b) Delta : Vph = VL = 415 V
Vph
415
= 4.15 A
=
Iph =
\
100
Z ph
\
IL =
3 I ph =
\
P =
3VL I L cos f =
3 ´ 4.15 = 7.188 A
3 ´ 415 ´ 7.188 ´ 1 = 5166.7422 W
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7 - 11
Three Phase Circuits
Example 7.6 Each phase of delta connected load consists of a 50 mH inductor in series with a
parallel combination of 50 W resistor and 50 µF capacitor. The load is connected to a three
phase 550 V, 800 rad/sec ac supply. Find : 1) Phase current 2) Line current 3) Power drawn
+ MU : Dec.-09
4) Power factor 5) Reactive power.
Solution : The circuit of each phase is shown in
the Fig. 7.12.
w = 800 rad/sec,
\
XL
VL = 550 V
50 W
50 mH
1
= 25 W
= w L = 40 W, XC =
wC
R || – j XC =
50 mF
Zph
50 ´ ( -j 25)
1250 Ð - 90º
=
(50 - j 25)
55.9017 Ð - 26.56º
Fig. 7.12
= 22.3606 Ð – 63.44º W = 10 – j 20 W
Zph = + j XL + (R || – j XC) = j 40 + 10 – j 20 W
\
= 10 + j 20 W = 22.3606 Ð + 63.44º W, f = 63.44º
Vph = VL = 550 Ð 0º V
Vph
550 Ð 0º
= 24.5968 Ð – 63.44º A
22. 3606 Ð 63.44º
i)
Iph =
ii)
IL =
3 Iph =
iii)
P =
3 VL IL cos f =
Z ph
=
… Assuming Vph as reference
3 ´ 24.5968 = 42.6029 A (Magnitude)
3 ´ 550 ´ 42.6029 ´ cos (63.44º) = 18.1468 kW
cos f = cos (63.44º) = 0.4471 lagging
iv)
v)
Q =
3 VL IL sin f = 36.3016 kVAR
Three coils each having resistance of 10 W and the inductance of 0.02 H are
Example 7.7
connected in star across 440 V, 50 Hz, three phase supply. Calculate
i) Phase voltage ii) Phase current iii) Line current and iv) Total power consumed.
VTU : Nov.-87, Marks 8
+
Solution :
The given supply voltage is always line voltage
\
VL = 440V, R ph = 10 W and
\
L ph = 0.02 W
X Lph = 2pf L ph = 2p´ 50 ´ 0 . 02 = 5.2831 W
\
Z ph = R ph + j X Lph = 10 + j 6 . 2831 W = 11 . 81 Ð + 32 . 14°W
\
Z ph
= 11.81 W
and
f = 32 . 14° lagging
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7 - 12
i) Phase voltage
Vph =
ii) Phase current
I ph =
iii) Line current
VL
3
=
Vph
Z ph
440
= 254.03 V
3
=
as star connected
3 VL I L cos f =
P =
... As star connected
254 . 03
= 21.51 A
11 . 81
I L = I ph = 21.51 A
iv) Power consumed
Three Phase Circuits
3 ´ 440 ´ 21.51 ´ cos ( 32.14°) = 13880.68 W
Example 7.8 Prove that a three phase balanced load draws three times as much power when
connected in delta, as it would draw when connected in star.
Solution : Let load is three phase balanced with per phase impedance of Zph W. Let VL
be the line voltage available which remains same whether load is connected in star or
delta. What changes is the phase voltage and hence phase and line current values
depending on star and delta connection of the load.
Case 1 : Star connection of load
VL
Vph1 =
= Phase voltage for star connection
\
3
\
and
Iph1 =
Vph1
Z ph
IL1 = Iph1 =
Vph1
… For star connection
Z ph
cos f depends on components of Zph and remains same for any connection of the load.
Vph1
Pstar =
cos f
… VL is constant
\
3 VL IL1 cos f = 3 ´ VL ´
Z ph
=
=
3 VL ´
(VL / 3)
cos f
Z ph
… as Vph1 = VL /
3
VL2
cos f watts
Z ph
Case 2 : Delta connection of load
Vph2 = VL = Phase voltage for delta connection
\
and
Iph2 =
IL2 =
Vph2
Z ph
=
VL
Z ph
3 Iph2 =
3 VL
Z ph
… For delta connection
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Three Phase Circuits
cos f remains same for both star and delta connection
Pdelta =
3 VL IL2 cos f
3 VL ´
=
VL2
= 3
Example 7.9
Z ph
… VL is constant
3 VL
cos f
Z ph
… As IL2 =
cos f = 3 Pstar
3 VL / Zph
… Proved
Three identical choke coils are connected as a delta load to a three phase supply. The
line current drawn from the supply is 15 A and total power consumed is 75 kW. The kVA
input to the load is 10 kVA. Find out
i) Line and phase voltage,
ii) Impedance/phase, iii) Reactance/phase,
iv) Resistance/phase,
v) Power factor
vi) Phase current,
vii) Inductance (if frequency is 50 Hz)/phase.
Solution : Coils are in delta connection
IL = 15 A, PT = 7.5 kW, kVA = 10 kVA
3 VL I L i.e. 10 ´ 10 3 =
Now
VA =
\
VL =
i)
VL = Vph = 385 V
ii)
|Zph| =
\
|Zph| =
\
7. 5 ´ 10 3 =
\
\
\
iii)
iv)
v)
10 ´ 10 3
15 3
3 VL ´ 15
= 385 V = Vph
Vph
but
Iph =
and
PT =
I ph
385
= 44.456 W
8. 66
IL
3
=
3 VL I L cos f
3 ´ 385 ´ 15 ´ cos f
cos f = 0.75
f = 41.42° lagging so + ve
Zph = 44.456 Ð + 41.42° = 33.33 + j 29.41 W
XLph = 29.41 W
Rph = 33.336 W
cos f (power factor) = 0.75 lagging
TM
15
= 8.66 A
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vi)
7 - 14
Three Phase Circuits
Iph = 8.66 A
Lph =
vii)
X Lph
=
2 pf
29 . 4109
= 93.617 mH
2 p´ 50
Example 7.10 Three pure elements are connected in star, draw x kVAR. What will be the value
of elements that will draw the same kVAR when connected in delta across the same supply .
+ MU : Dec.-95, 99
Solution : Let reactance of pure element be X1 per phase.
Case 1 : Star connection
Vph1
VL
, Iph1 =
= IL1 (magnitude)
Vph1 =
X1
3
\
kVAR1 =
3 VL IL1 sin f =
3 VL
IL1
VL
X1
X1
Vph1
X1
X1
… sin f = 1, f = 90º
=
3 VL
VL
3 X1
=
VL2
(a)
Fig. 7.13 (a)
…(7.1)
X1
IL2
Case 2 : Delta connection
VL = Vph2 , Iph2 =
IL2 =
=
kVAR2 =
=
=
Vph2
X2
X2
X2
3 Iph2
3 Vph2
X2
VL
X2
(magnitude)
(b)
Fig. 7.13 (b)
3 VL IL2 sin f
Vph2
3 VL ´ 3
X2
…Vph2 = VL
VL2
3 VL 3 VL
=3
X2
X2
…(7.2)
For same kVAR, equate equations (7.1) and (7.2),
\
VL2
3 VL2
=
X1
X2
i.e.
X2 = 3 X1
Example 7.11 Three resistances each having value of 40 W are connected in i) Star and ii) Delta
across 400 V, 3 phase supply. Calculate the power taken from the supply. If one of the three
resistances get open circuited, find the total power taken from the supply in each case.
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7 - 15
Three Phase Circuits
Solution :
40 W
Case i) Star
VL = 400 V, Rph = 40 W, cos f = 1
Vph =
\ Iph =
\
P =
VL
3
40 W
40 W
= 230.94 V
(a)
Vph
R ph
= 5.7735 A = IL
3 VL IL cos f =
Open
3 ´ 400 ´ 5.7735 ´ 1
Rph
= 4 kW
One of the resistance is open in star connection.
VL
400
=
=5A
IL =
\
R ph + R ph
80
\ Power consumed = 2 ´ I 2L Rph
Rph
IL
Series
VL
(b)
40 W
2
= 2 ´ (5) ´ 40 = 2 kW
Case ii)
VL = 400 V
40 W
VL
40 W
Delta
VL = Vph = 400 V, Iph =
Vph
R ph
= 10 A
(c)
\
IL =
3 Iph = 10
\
P =
3 VL IL cos f = 12 kW
3 A
Rph
…cos f = 1
When one of the resistances is open, VL
appears across each of the remaining Rph.
\
IL =
Rph
VL
VL
(d)
VL
= 10 A
R ph
Fig. 7.14
Power consumed = 2 ´ I 2L Rph
\
2
= 2 ´ 10 ´ 40 = 8 kW
Example 7.12
Two wattmeters are connected to measure the input of a 15 H.P., 50 Hz, 3-phase
induction motor at full-load. The full-load efficiency and p.f. are 0.9 and 0.8 lagging
respectively. Find the readings of the two wattmeters.
Solution :Pout = 15 H.P.,
Now
%h =
h = 0.9,
+ VTU : Feb.-2000, Marks 7
cos f = 0.8
Pout
´ 100
Pin
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7 - 16
15 ´ 735 . 5
Pin
\
0.9 =
\
Pin = 12258.33 W
But
Pin =
3 VL I L cos f
12258 . 33 =
3 VL I L ´ 0.8
\
\
VL I L = 8846.9
and
Three Phase Circuits
as 1 H.P. = 735.5 watts
f = cos –1 0.8 = 36.86°
\
W1 = VL I L cos( 30 – f) = 8846.69 cos ( 30 - 36.86) = 8783.1737 watts
and
W2 = VL I L cos( 30+ f) = 8846.69cos ( 30 + 36.86) = 3476.569 watts
Example 7.13 A 3-phase 10 kVA load has P.F. of 0.342. The power is measured by two
watt-meter method. Find the reading of each wattmeter when i) P.F. is leading ii) P.F. is
+
lagging.
MU : Dec.-10
Solution : VL IL = 10 ´ 103 VA, cos f = 0.342, f = 70º
i) For leading power factor
W1 = VLIL cos (30 + f) = 10 ´ 103 cos (30 + 70) = – 1736.4817 W
W2 = VLIL cos (30 – f) = 10 ´ 103 cos (30 – 70) = 7660.4444 W
ii) For lagging power factor
W1 = VLIL cos (30 – f) = 7660.4444 W
W2 = VLIL cos (30 + f) = – 1736.4817 W
Example 7.14
A 3 phase motor load has a p.f. of 0.397 lagging. Two wattmeters connected to
measure power show the input as 30 kW. Find reading of each wattmeter.
Solution :
+ MU : May-11
PT = 30 kW, cos f = 0.397 lagging, f = 66.609º
PT =
3 VL IL cos f
i.e.
VL IL = 43628.4838
\
W1 = VL IL cos (30 – f) = 43628.4838 cos (30 – 66.609º) = 35.0216 kW
\
W2 = VL IL cos (30 + f) = 43628.4838 cos (30 + 66.609º) = – 5.0216 kW
Cross check that PT = W1 + W2.
Example 7.15
Each of the two wattmeters connected to measure the input to a three phase
circuit, reads 20 kW. What does each instrument reads, when the loads p.f. is 0.866 lagging
with the total three phase power remaining unchanged in the altered condition ?
+ VTU : Jan.-03, Marks 6
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7 - 17
Solution : W1 = 20 kW and W2
ìï
cos f1 = cos í tan -1
îï
Three Phase Circuits
= 20 kW.
ì
é 3 ( W1 - W2 ) ù
3( 0) ù ü
-1 é
ê
ú = cos í tan ê
3 úý
ë 40 ´ 10 û þ
ë ( W1 + W2 ) û
î
= cos 0º = 1
PT = Total power = W1 + W2 = 40 kW
\
... Original p.f.
where W1¢ and W2¢ are new readings
W1¢ + W2¢ = 40 kW
cos f2 = 0.866
... New power factor
\
ìï
é 3 ( W1¢ - W2¢ ) ù üï
cos f2 = cos í tan -1 ê
úý
ë ( W1¢ + W2¢ ) û þï
îï
\
ìï
é 3 ( W1¢ - W2¢ ) ù üï
0.866 = cos í tan -1 ê
úý
3
ïî
ë 40 ´ 10
û ïþ
\
cos
–1
0.866 =
\
tan [30º] =
\
W1¢ - W2¢
Adding,
and
é 3 ( W1¢ - W2¢ ) ù
tan -1 ê
ú
3
ë 40 ´ 10
û
3 ( W1¢ - W2¢ )
40 ´ 10 3
= 13333.333 while
2 W1¢ = 53333.333
i.e.
... cos
–1
0.866 = 30º
... Taking tan of both sides
W1¢ + W2¢ = 40 ´ 10 3
W1¢ = 26666.667 W
W2¢ = 40 ´ 10 3 - W1¢ = 13333.334 W
Example 7.16 Two wattmeters connected to measure the input to balanced three-phase circuit
indicates 2500 and 500 W, respectively. Find the total power supplied, and the power factor of
the circuit
i) When both readings are positive and
ii) When the latter reading is obtained after reversing the connections to the current coil.
+ VTU : Dec.-09, Marks 10
Solution : W1 = 2500 W,
W2 = 500 W
Case 1 : Both positive
\
ìï
é 3 (W1 - W2 ) ù üï
cos f = cos í tan -1 ê
= 0.6546
(W1 + W2 ) ú ýï
ïî
ë
ûþ
PT = W1 + W2 = 3000 W
Case 2 : W2 is negative i.e. W2 = – 500 W
ì
é 3 (2500 - ( -500) ù ü
cos f = cos í tan -1 ê
\
ú ý = 0.3592
ë (2500 - 500) û þ
î
PT = W1 + W2 = 2500 – 500 = 2000 W
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7 - 18
Three Phase Circuits
Important Multiple Choice Questions with Answers
Introduction
Q.1
In a 6 phase supply system, the voltages are separated from each other by _____ .
a) 120º
b) 90º
c) 180º
d) 60º
[Ans. : d]
Advantages of Three Phase System
Q.1
The advantage of star-connected supply system is that _____ .
+
a) line-current is equal to phase current
b) line voltage is equal to
c) two voltage can be used
d) it is simple arrangement
VTU : Jan.-14
3 phase voltage
[Ans. : b]
Generation of Three Phase Voltage System
Q.1
a) 30º apart
Q.2
b) 60º apart
b) infinity
[Ans. : d]
c) line current
d) phase current.
[Ans. : a]
In a three phase balanced system, the voltages are displaced by an angle of_____ from
each other.
b)180º
c) 120º
d) 360º
[Ans. : c]
In a three phase balanced supply system, the sum of the instantaneous values of the
VTU : June-10
three voltages at any instant is ___________ .
+
a) maximum
Q.5
d) 120º apart.
+
a) 90º
Q.4
c) 90º apart
VTU : Jan.-09, 11, 13
The algebraic sum of instantaneous phase currents in a three phase balanced system
VTU : Jan.-09
is ______ .
a) zero
Q.3
+
In a three phase system, the e.m.f.'s are :
b) zero
c) minimum
d) none of these
[Ans. : b]
In a 3 f system, if the instantaneous value of phase R and Y are + 60 V and – 40V
VTU : July-11
respectively, then instantaneous voltage of phase B is _____ .
+
a) – 20 V
b) 40 V
c) 120 V
d) none of the above.
[Ans. : a]
Q.6
Q.7
The rated voltage of a 3 phase system is given as _____ .
+ VTU : Dec.11
a) r.m.s. phase voltage
b) peak phase voltage
c) r.m.s. line-to-line voltage
d) peak line-to-line voltage.
The phase sequence RBY denotes that ___________ .
[Ans. : c]
+ VTU : Dec.-11
a) e.m.f. of phase-B lags that of phase-R by 120º
b) e.m.f. of phase-B leads that of phase-R by 120º
c) both (a) and (b) are correct
d) none of these.
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[Ans. : a]
Basic Electrical Engineering
Q.8
7 - 19
Three Phase Circuits
+ VTU : Jan-14
The phase sequence R Y B denotes that ___________ .
a) the e.m.f. of R leads Y by 120º
b) the e.m.f. of Y lags R by 120º
c) the e.m.f. of Y leads by B 120º
d) all of these
[Ans. : d]
Important Definitions Related to Three Phase System
Q.1
The direction of rotation of the three phase machines depends on _____ .
a) line voltages
b) phase sequence
c) phase currents
d) phase voltages
[Ans. : b]
Balanced Load
Q.1
In a balanced three phase load, the power factor of the three phases are_____ .
+
a) different
b) same
c) zero
VTU : July-09
d) none of these.
[Ans. : b]
Q.2
In a 3 f balanced star connected load, the neutral current is equal to :
+
a) Zero
b) I phase
c) I Line
VTU : Jan.-10
d) Unpredictable.
[Ans. : a]
Relations for Star Connected Load
Q.1
For star connection, the line current is _____that of phase current.
a) greater than
b) less than
c) same as
d) none of these
[Ans. : c]
Q.2
For star connection, the phase voltage is _____ times the line voltage.
a)
Q.3
3
c)
2
d)
1
2
[Ans. : b]
b) VR – VY
c)
VR
VY
d) VR ´ VY
[Ans. : b]
In star connection, the line voltage leads phase voltage by _____ .
a) 120º
Q.5
1
3
The line voltage VRY is given by _____ in a star connected system.
a) VR + VY
Q.4
b)
b) 180º
c) 60º
d) 30º
[Ans. : d]
In a star connected system VR = 220 Ð 30º V and VY = 220 Ð – 90º V then the line
voltage VRY = _____V.
a) 381.05 + Ð 60º
b) 381.051 +Ð – 60º c) 381.051 Ð 90º
d) 381.051Ð 120º
[Ans. : a]
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Basic Electrical Engineering
Q.6
Q.8
+
Vph
3
b) VL =
c) VL = Vph
3Vph
The voltage VAB = 50Ð30º volts. Then, VBA is _______ volts.
a) 50Ð – 180º
b) 50 Ð – 150º
c) 50Ð – 30º
d) none of these
[Ans. : b]
+
VTU : June-10
d) 50Ð – 210º
[Ans. : b]
The angle between line voltage and phase voltage for a balance star connected
VTU : Aug.-11
circuit is ________ .
+
a) 30º
Q.9
Three Phase Circuits
In a three phase balanced star system, the relation between the line voltage VL and the
VTU : June-10
phase voltage Vph is ________ .
a) VL =
Q.7
7 - 20
b) 30º± f
c) 60º
d) 120º
[Ans. : a]
Three inductive coils each having an impedance of 17.7 W are connected in star. The
circuit is fed from a 3-phase, 400 V, 50 Hz supply. The current (line) drawn by the circuit
VTU : June-12
is equal to ____.
+
a) 22.6 A
Q.10
c) 13 A
d) none of these
[Ans. : c]
For a 3-phase star connected balanced circuit having inductive load, the angle between
VTU : June-12
the line currents and corresponding line voltages is equal to ____.
+
a) 30°
Q.11
b) 39.14 A
b) 30° – f
c) 30° + f
d) f
[Ans. : c]
In a 3 phase balanced star - connected load, neutral current is equal to ____.
+ VTU : June-13
a) zero
b) IP
c) IL
d) unpredictable
[Ans. : a]
Relations for Delta Connected Load
Q.1
In delta connected system, the relation between the line current IL and phase current Iph
is _______ .
VTU : July-09
+
a) IL = Iph
b) IL = Iph / 3
c) IL =
3 Iph
d) IL = 3 Iph .
[Ans. : c]
Q.2
The relationship between the line and phase voltage of a D-connected circuit is given
VTU : Jan.-10,11, June-13
by, _____________ .
+
a) VL = VP
Q.3
c) VL = VP 2
d) VL =
2
V .
p P [Ans. : a]
In delta connection, the relation between line current and phase current is_____.
a) IL =
Q.4
b) VL = 3 VP
1
I
3 ph
b) IL =
c) IL = Iph
3 Iph
d) none of these
[Ans. : b]
In delta connection, the relation between line voltage and phase voltage is _____.
a) VL = Vph
b) VL =
c) VL =
3 Vph
TM
1
V
3 ph
TECHNICAL PUBLICATIONS - An up thrust for knowledge
d) none of these
[Ans. : a]
Basic Electrical Engineering
Q.5
7 - 21
Three Phase Circuits
In delta connection shown in the Fig. 7.15 the line current IR is given by _____.
IR
R
IRY
IBR
Y
B
IYB
Fig. 7.15
b) IBR – I RY
a) I RY – I BR
Q.6
Q.7
c) I RY + I BR
d) none of these
The relation between line and phase quantities in a delta connection is ________ .
+ VTU : Aug.-11
a) El =
3 E ph , Il = I ph
c) E l =
3 Eph , I l =
b) E l = Eph , I l =
3 I ph
d) E l = Eph , I l = I ph .
3 I ph
[Ans. : b]
+ VTU : Jan.-13
In a 'D' connected system relation between IL and Iph is ____.
b) I L = I ph / 3
a) I L = I ph
c) I L = 3 × I ph d) I L = 3 I ph
[Ans. : c]
Power Triangle for Three Phase Load
Q.1
The power taken by a 3 phase load is given by the expression_____ .
+ VTU : July-09,Jan.-11
a) 3VLIL cos f
b)
3 VLIL cos f
c) 3VLIL sin f
d)
3 VLIL sin f
[Ans. : b]
Q.2
The power factor angle is the angle between_____.
a) line voltage and phase current.
c) phase voltage and line current.
b) line voltage and line current.
d) phase voltage and phase current.
[Ans. : d]
Q.3
The total apparent power in a three phase system is given by _____.
a) VL IL
b) Vph Iph
c)
3 VL IL
d)
3 Vph Iph
[Ans. : c]
Q.4
The total reactive volt-amperes in a three phase system are given by _____.
a)
3 VL IL
b)
3 VL IL cos f
c) VL IL
d)
3 VL IL sin f
[Ans. : d]
Q.5
The cos f can be expressed as _____.
a)
Q.6
Active power
Apparent power
b)
Active power
Reactive power
c)
Reactive power
Apparent power
d) None of these
[Ans. : a]
The device used for the power factor improvement is ________ .
a) induction motor
b) alternator
c) synchronous condenser d) fan [Ans. : c]
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Basic Electrical Engineering
Q.7
7 - 22
Which of the following apply to power in a purely reactive circuit ?
a) P = 0, Q = 0
c) P = 0 and Q is maximum
Q.8
Q.10
b)
3 P
d) P
b) P =
3 VL IL
c) P =
d) P =
3 Vph Iph cos f
3 VL IL cos f
The total active power in a 3 ph. system is _____ .
b)
3 V LI L
c) VLIL
3 VL IL cosf
The power in a 3 phase system is given by
d)
[Ans. : a]
+ VTU : June-12
a) P = VL IL cos f
[Ans. : c]
+ VTU : Jan-13
3 VL IL sin f
[Ans. : b]
3 VL, IL cos f, where f is the phase angle
+ VTU : June-13
between _____ .
Q.12
[Ans. : c]
+
P
c)
3
Active power drawn by a 3-phase balanced load is given by ____.
a)
Q.11
b) P is maximum and Q = 0
d) P and Q both are maximum
A 3 phase star connected load consumes P watts of power from a 400 V supply. If the
same balanced load is connected in delta across that same supply, then power
VTU : Dec.-11
consumption is ________ .
a) 3 P
Q.9
Three Phase Circuits
a) line voltage and line current
c) line voltage and phase current
b) phase voltage and phase current
d) phase voltage and line current [Ans. : b]
Three equal impedances are first connected in delta across a 3 - phase balanced supply.
If the same impedances are connected in star across the same supply_____ .
+ VTU : June-13
Q.13
a) phase currents will be one - third
b) line currents will be one - third
c) power consumed will be one - third
d) none of these
Three phase apparent power is equal to _____ .
a)
3 VL IL
b)
3 VL IL cos f
c)
3 VL IL sin f
[Ans. : c]
+ VTU : Jan-14
d) VL IL
[Ans. : c]
Wattmeter
Q.1
The wattmeter senses the angle between _____ .
a) the phase voltage and phase current. b) the line voltage and line current.
c) the current in current coil and voltage across pressure coil.
d) none of the above.
Q.2
For a wattmeter shown in the Fig. 7.16 its reading is _____ .
R
W
cos f
lagging
p.f.
load
Y
B
Fig. 7.16
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[Ans. : c]
Basic Electrical Engineering
7 - 23
a) VL IL sin f
c) VL IL cos f
Q.3
Three Phase Circuits
b) VL IL cos(30 + f )
d) VL IL cos (30 – f )
[Ans. : b]
For a wattmeter shown in the Fig. 7.17 its reading is _____ .
W
R
cos f
lagging
p.f.
load
Y
B
Fig. 7.17
a) VL IL cos f
b) VL IL sin f
c) Vph Iph cos f
d) Vph Iph sin f
[Ans. : c]
Two Wattmeter Method
Q.1
W1 and W2 are the readings of two wattmeters used to measure power of a 3 f balanced
VTU : Jan.-10
load. The active power drawn by the load is _______ .
+
a) W1 + W2
b) W1 - W2
c)
3 (W1 + W2 )
d)
3 (W1 - W2 ).
[Ans. : a]
Q.2
In a star connected circuit, the load impedance per phase is 10 + j0 W while the line
voltage is 440 V then the two wattmeter readings are _____ .
a) 9680 W, 2000 W
c) 9680 W, 3200 W
Q.3
b) 10000 W, 1800 W
d) 16766.25 W, 16766.25 W
[Ans. : d]
The sum of the two-wattmeters readings in a 3 phase balanced system is _____ .
+
a) Vph Iph cos f
b) 3VL IL cos f
c)
3 VL IL cosf
VTU : Dec.-11
d) none of these.
[Ans. : c]
Power Factor Calculation by Two Wattmeter Method
Q.1
In the measurement of three-phase power by two wattmeter method, if the two wattmeter
VTU : Jan.-09
readings are equal, then the p.f. of the circuit is ______ .
+
a) 0.8 lag
Q.2
b) 0.8 lead
c) zero
d) unity
[Ans. : d]
In a two wattmeter method, W1= 4000 W and W2 = 1000 W, then the power factor of the
circuit is _____ .
a) 0.72
b) 0.69
c) 0.55
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d) 0.51
[Ans. : b]
Basic Electrical Engineering
7 - 24
Three Phase Circuits
Effect of P.F. on Wattmeter Readings
Q.1
In the 2 wattmeter method of measuring 3 phase power, the two wattmeters indicate
equal and opposite readings when the load power factor angle is_____degrees lagging.
+
a) 60
Q.2
b) 0
b) 0.8 leading
c) zero
b) 0.8 to 1
c) 0.5 to 0.75
b) 0
c) 0.866
d) 1
[Ans. : d]
When power factor is 0.5, the wattmeter reading is such that _____ .
b) w1 is +ve,w2 is – ve
d) w1 = 2 w2
[Ans. : c]
c) w1 +ve, w2 = 0
When two wattmeters are connected in a 3-phase circuit to measure its total power
consumption, one of the wattmeter would read zero, when the load power factor is ____.
b) unity
c) 0.5 lagging
+ VTU : June-12
d) zero
[Ans. : c]
+ VTU : Jan-13
If the two wattmeters show equal reading, power factor is _____ .
a) zero
Q.8
d) 0 to 0.5 [Ans. : d]
+
a) 0.2 lagging
Q.7
[Ans. : d]
When the two wattmeters used to measure a three phase power, give equal readings,
VTU : June-10,Jan.-11
then the p.f. of the circuit is ______ .
a) w1 = w2
Q.6
d) unity
If one of the two wattmeter reading is negative then the range of power factor is_____.
a) 0.5
Q.5
[Ans. : d]
+
a) 0.5 to 1
Q.4
d) 90.
In the measurement of 3 f power by two wattmeters, if the two wattmeters readings are
VTU : Jan.-10
equal, the power factor of the circuit is _______ .
a) 0.8 lagging
Q.3
c) 30
VTU : July-09
b) 0.5
c) unity
d) 0.866
[Ans. : c]
In a three-phase power measurement by two wattmeter method, both wattmeters reads
VTU : Jan-14
the same value. The power factor of the load must be _____ .
+
a) unity
b) 0.707 lag
c) 0.707 lead
d) zero
[Ans. : a]
Reactive Volt-Amperes by Two Wattmeter Method
Q.1
The total reactive volt-amperes by two wattmeter readings is given by ________.
a) W1 – W2
b)
3 (W1 – W2)
c) W1 + W2
d)
3 (W1 + W2)
[Ans. : b]
qqq
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Synchronous Generators
(Alternators)
8
Chapter at a Glance
1.
Working Principle of Alternator
\
f =
Ns =
So
2.
Hz (cycles per sec).
120 f
P
where f = Required rated frequency
Armature Winding
\
Pole pitch = 180º electrical = slots per pole (no. of slots/P) = n
b = 1 slot angle =
\
3.
PN
120
180º
n
E.M.F. Equation of an Alternator
\
a = b ´ Number of slots by which coils are short pitched.
or
a = 180º – Actual coil span of the coils.
\
a
Kc = cos æç ö÷
è2ø
where a = Angle of short pitch
æ mb ö
sin ç
÷
è 2 ø
Kd =
b
m sin æç ö÷
è2ø
Eph = 4.44 Kc Kd f f Tph volts
Eline =
3 ´ E ph =
3 ´ 4.44 Kc Kd f f Tph volts ( For star connected )
(8 - 1)
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Basic Electrical Engineering
8-2
Synchronous Generators (Alternators)
Important Theory Questions and Answers
Ø
State the advantages of rotating field over rotating armature used in alternators.
+ VTU : Jan.-07, 08; July-05, 06, 09, Dec.-11, Marks 6
· For stationary armature large space can be provided to accommodate large number of
conductors and the insulation.
The stationary armature avoids the interaction of mechanical and electrical stresses.
· The problem of sparking at the slip rings can be avoided by keeping field rotating
which is low voltage circuit and high voltage armature as stationary.
· Rotating field makes the overall construction very simple.
· The ventilation arrangement for high voltage side can be improved if it is kept
stationary.
· It is easier to collect large currents at high voltages from a stationary part.
Ø
Discuss the different types of rotors used in the alternators. Mention their characteristic features
and applications.
+ VTU : Jan.-03, 04, 06, 09, 10, 13; July-03, 04, 08, 10, 11, June-13, Marks 8
1. Salient Pole Type Rotor :
· Poles are projected out from the surface of
the rotor.
· The poles are built up of thick steel
laminations. The poles are bolted to the
rotor as shown in the Fig. 8.1.
· The field winding is provided on the pole
shoe.
· These rotors have large diameters and
small axial lengths.
Field winding
N
Projected
pole
S
S
Bolt
Mechanical
support
Shaft
N
Fig. 8.1 Salient pole type rotor
· As mechanical strength of salient pole type
is less, this is preferred for low speed alternators ranging from 125 r.p.m. to 500 r.p.m..
· Water turbines and I.C. engines.
2. Smooth Cylindrical Type Rotor :
· The rotor consists of smooth solid steel cylinder,
having number of slots to accommodate the field coil.
The slots are covered at the top with the help of steel
or manganese wedges.
· The unslotted portions of the cylinder itself act as the
poles. The poles are not projecting out.
S
Slot
Field coil
N
D.C.
Pole
N
Shaft
S
· These rotors have small diameters and large axial
lengths. This is to keep peripheral speed within limits. Fig. 8.2 Smooth cylindrical rotor
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Basic Electrical Engineering
8-3
Synchronous Generators (Alternators)
· Mechanically very strong and thus preferred for high speed alternators ranging between
1500 to 3000 r.p.m..
· Steam turbines, electric motors.
Ø
Derive the e.m.f. equation of an alternator. Explain the significance of winding factors.
+ VTU : Feb.-05; Jan.-09, 11, 13; July-08, 10, June-12, Marks 6
· Let f = Flux per pole, in Wb
P = Number of poles
Ns = Synchronous speed in r.p.m.
Z = Total number of conductors
\
Zph =
f = Frequency of induced e.m.f. in Hz
Zph = Conductors per phase
Z
as number of phases = 3.
3
· For one revolution of a conductor, eavg per conductor =
fP
P Ns
=f
60
æ 60 ö
ç
÷
N
è sø
P Ns
P Ns
i.e.
= 2f
f=
· But
120
60
\ eavg per conductor = 2 f f volts
Flux cut in one revolution
Time taken for one revolution
\eavg per conductor =
\
… (8.1)
From equation (8.1)
e.m.f. per turn = 2 ´ (e.m.f. per conductor) = 2 ´ (2 f f) = 4 f f volts.
Kf =
R. M. S.
= 1.11
Average
… For sinusoidal
\ R.M.S. value of Eph = Kf ´ Average value = 1.11 ´ 4 f f Tph
Eph = 4.44 f f Tph volts
Ø
Explain voltage regulation of an alternator and state its significance.
+ VTU : Jan.-03, 06, 08, 10; Aug.-03, Marks 6
The voltage regulation of an alternator is defined as the change in its terminal voltage
when full load is removed, keeping field excitation and speed constant, divided by the
rated terminal voltage.
· So if , Vph = Rated terminal voltage
and E ph = No load induced e.m.f. then voltage
regulation is defined as,
% Reg =
E ph - Vph
Vph
´ 100
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Basic Electrical Engineering
8-4
Synchronous Generators (Alternators)
Terminal
voltage
· For lagging and unity p.f. conditions there is
always drop in the terminal voltage hence
regulation values are always positive.
· While for leading capacitive load conditions,
the terminal voltage increases as load current
increases. Hence regulation is negative in
such cases.
· The ideal value of the regulation is zero.
Less
the
regulation,
better
is
the
performance of an alternator.
Leading p.f.
Vph = Eph
Unity p.f.
Lagging p.f.
IL
0
Fig. 8.3 Load characteristics of an
alternator
Important Solved Examples
Example 8.1 Find the number of armature conductors in series per phase of three phase, 50 Hz ,
10 pole alternator having 90 slots. The winding is to be star connected to give a line voltage
of 11 kV, when the flux is 160 mWb. The winding factor is unity. Also find the voltage
regulation when the full load terminal voltage is 10.60 kV.
VTU : Feb.-05, Marks 8
P = 10, Slots = 90, Eline = 11 kV, f = 160 mWb, Kd = 1, VL = 10.6 kV.
Solution :
Eph =
E line
3
=
11 ´ 10 3
= 6350.853 V
3
Eph = 4.44 Kc Kd f f Tph
But
\
+
6350.853 = 4.44 ´ 1 ´ 1 ´ 160 ´ 10 -3 ´ 50 ´ Tph
\
Tph = 178.79 » 179
\
Zph = 2 ´ Tph = 2 ´ 179 = 358
\
%R =
=
Example 8.2
E ph - Vph
Vph
... Conductors in series per phase
´ 100
where Vph =
VL
3
=
10.6
= 6.1199 kV
3
6350.853 - 6119.9128
´ 100 = 3.773 %
6119.9128
A 3 phase Y connected alternator driven at 900 r.p.m. is required to generate a
line voltage of 460 V at 60 Hz on open circuit. The stator has two slots / pole / phase and
four conductors per slot. Calculate the number of poles and the useful flux per pole, if the
+ VTU : Aug.-05, Marks 8
winding factor is 0.966.
Solution :
Ns = 900 r.p.m., Eline = 460 V, f = 50 Hz
m = 2 slots / pole / ph,
Assuming full pitch coil, Kc = 1.
4 conductors/slot, Kd = 0.966
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Basic Electrical Engineering
Ns =
\
P =
8-5
Synchronous Generators (Alternators)
120 f
P
120 f 120 ´ 60
=
=8
Ns
900
... Poles
\ Slots / phase = m´ P = 2 ´ 8 = 16
\
Total slots = 16 ´ 3 = 48
\
Z = Total conductors = 48 ´ 4 = 192
\
Zph =
\
Tph =
Eph =
Z ph
= 32
2
E line
3
=
460
= 265.5811 V
3
Eph = 4.44 Kc Kd f f Tph
But
\
Z 192
=
= 64
3
3
265.5811 = 4.44 ´ 1 ´ 0.966 ´ f´ 60 ´ 32
\
f = 0.03225 Wb = 32.25 mWb
... flux
Example 8.3 A 3 f , 16 pole, Y-connected alternator has 144 slots on the armature periphery. Each
slot contains 10 conductors. It is driven at 375 r.p.m. The line value of e.m.f. available across
the terminals is observed to be 2.657 kV. Find the frequency of the induced e.m.f. and flux per
+
pole.
Solution :
P = 16,
E line = 2.657 kV,
VTU : Feb.-10, Marks 8
N s = 375 r.p.m., 10 conductors/slot.
Z = Total slots ´ conductors slot = 144 ´ 10 = 1440
\
Z ph =
\
Tph =
E ph =
Z 1440
=
= 480
3
3
Z ph
2
E line
3
=
480
= 240
2
=
2.657 ´ 10 3
= 1534.01966 V
3
n = Slots Pole =
144
= 9
16
m = Slots Pole Phase =
\
... 2 conductors, 1 turn
9
= 3,
3
b=
180°
= 20°
n
3 ´ 20° ö
mb
sin æç
÷
è 2 ø
2
=
=
= 0.95979
b
20° ö
æ
m sin
3 sin ç
÷
2
è 2 ø
sin
Kd
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Basic Electrical Engineering
8-6
Synchronous Generators (Alternators)
Kc = 1
Ns =
\
f =
... Full pitch coil
120 f
P
i.e.
f=
Ns ´ P
120
375 ´ 16
= 50 Hz
120
... Frequency
E ph = 4.44 Kc Kd f f Tph
\
1534.01966 = 4.44 ´ 1 ´ 0.95979 ´ f ´ 50 ´ 240
\
f = 0.03 Wb
... Flux per pole
Important Multiple Choice Questions with Answers
Concept of Slip Rings and Brush Assembly
Q.1
For connecting rotating parts to stationary ,_______ are used.
a) brushes
b) shaft
c) slip rings
d) commutator
[Ans. : c]
Advantages of Rotating Field over Rotating Armature
Q.1
The rated voltage of alternators used is power station is usually _____ .
+ VTU : Jan.-09
a) 440 kV
Q.2
b) 220 kV
c) 110 kV
The field winding of an alternator is excited by _____ .
a) d.c.
b) a.c.
d) 11 kV
[Ans. : d]
+ VTU : July-09, 11
c) both d.c. and a.c.
d) none of these.
[Ans. : a]
Q.3
Q.4
In alternators ______ is a rotating member.
a) field winding
b) armature winding
c) compensating winding
d) none of these.
The rotor of the synchronous generator has
a) 4 slip rings
b) 3 slip rings
c) 2 slip rings
[Ans. : a]
+ VTU : Dec.-11
d) no slip rings.
[Ans. : c]
Q.5
Q.6
In modern alternators, the rotating part is ________.
a) field
b) armature
c) field and armature
d) none of these
[Ans. : a]
The field winding of an alternator is excited by _____ .
a) d.c.
b) a.c.
+ VTU : June-12
c) a.c. and d.c.
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+ VTU : Jan.-13
[Ans. : a]
Basic Electrical Engineering
Q.7
8-7
Synchronous Generators (Alternators)
In synchronous generators _____ .
a) the field poles are stationary and the armature conductures rotate
b) the armature conductures are stationary and the field poles rotate
c) field and armature both are stationary
d) none of these
[Ans. : b]
Construction of Alternators
Q.1
Q.2
A salient pole field construction is used for alternator having _____ .
+ VTU : Jan.-09, 11, 13; June-10
a) low and medium speed
b) large speed
c) very large speed
d) none of these.
[Ans. : a]
The non-salient pole field construction is used for _____alternator.
+ VTU : July-09
a) low speed
d) none of these.
b) medium speed
c) high speed
[Ans. : c]
Q.3
Q.4
Q.5
Q.6
The salient pole type rotors have _____ .
a) smaller diameter
b) larger diameter
c) smaller axial length
d) both (b) and (c)
[Ans. : d]
The most suitable rotor for a turbo-alternator designed to operate at high speed is _____ .
+ VTU : Jan.-10
a) salient pole type rotor
b) smooth cylindrical type rotor
c) squirrel cage rotor
d) either of the above.
[Ans. : b]
The range of speed for salient pole synchronous machines is _______.
a) above 500 r.p.m.
b) above 1000 r.p.m.
c) 125 to 500 r.p.m.
d) None of these
[Ans. : c]
________ rotor construction is used for turbo alternators.
a) Nonsalient
Q.7
+ VTU : Jan.-10
b) Salient
c) Squirrel cage
d) Slip ring [Ans. : a]
________ construction has small diameter and large axial lengths.
a) Sailent pole
b) Non-sailent pole
c) Projected pole
d) None of these.
[Ans. : b]
Q.8
Salient pole type rotor has _____ diameter and ______ axial lengths.
a) large, large
b) small, small
c) large , small
d) small, large
[Ans. : c]
Q.9
Non-salient pole type rotor has _____ diameter and ______ axial lengths.
a) large, large
b) small, small
c) large , small
d) small, large
[Ans. : d]
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Q.10
8-8
Synchronous Generators (Alternators)
Air gap is _____ for salient pole rotors.
a) uniform
b) non-uniform
c) axial
d) none of these
[Ans. : b]
Q.11
Air gap is ______ for nonsalient pole rotors.
a) uniform
b) non-uniform
c) axial
d) none of these
[Ans. : a]
Q.12
Q.13
The stator of an alternator is identical to that of a _____ .
+ VTU : Aug.-11
a) d.c. generator
b) three phase induction motor
c) single phase induction motor
d) none of these.
[Ans. : b]
High speed alternators are driven by ______.
a) diesel engine
b) hydraulic turbines
c) steam turbines
d) none of these.
[Ans. : c]
Q.14
+ VTU : Dec.-11
The salient pole type rotors are _____ .
a) smaller in axial length
b) larger in axial length
c) smaller in diameter
d) larger in diameter and smaller in axial length.
[Ans. : d]
Q.15
Q.16
In a synchronous machine, the stator frame is made of _____ .
+ VTU : June-13
a) Stain steel
b) CRCoGS
c) Cast iron or welded steel plates
d) Laminated silicon steel
[Ans. : d]
The stator core of a synchronous machine is laminated so as to reduce
_____ .
+ VTU : June-13
a) Eddy current loss
b) hysteresis
c) both eddy current and hysteresis loss
d) the size and weight of the machine
Q.17
[Ans. : a]
+ VTU : June-13
The stator slot insulations in synchronous machine is made of _____ .
A) mica cloth
B) fibre glass
C) polister sheets D) any of these
[Ans. : b]
Q.18
The machine that supplies dc to the rotor is called the _____ .
a) rectifier
Q.19
c) convertor
d) invertor
[Ans. : b]
A 4-pole. 1200 rpm alternater will generate an emf at a frequency of _____ .
+ VTU : June-13
a) 60 Hz
Q.20
b) exciter
+ VTU : June-13
b) 50 Hz
c) 40 Hz
d) 25 Hz
[Ans. : c]
The current from an alternator is taken out to external load circuit through _____ .
+ VTU : June-13
a) commutator segments
b) slip-rings
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8-9
c) carbon brushes
Synchronous Generators (Alternators)
d) solid connection
[Ans. : d]
Working Principle of Alternator
Q.1
A 4 pole, 1200 r.p.m. alternator generates e.m.f. at a frequency of ____.
+
a) 25 Hz
Q.2
b) 40 Hz
c) 50 Hz
VTU : Jan.-09, 11, 13
d) 60 Hz.
The frequency of e.m.f. generated by an alternator depends upon the alternator speed,
N (r.p.m.) and number of poles on the alternator, P and is given by ______.
+
a)
Q.3
[Ans. : b]
PN
60
b)
60 N
P
c)
PN
120
d)
VTU : Jan.-10
120 N
.
P
[Ans. : c]
The number of cycles generated in a 6 pole alternator in one revolution is _____.
+
a) 3
b) 6
c) 50
VTU : July-09
d) none of these
[Ans. : a]
Q.4
For an alternator, 1º mechanical = ___________ electrical.
a) Pº
Q.5
b) 2 Pº
c) 4 Pº
d)
Pº
2
[Ans. : d]
The maximum value of the synchronous speed for an alternator is _________ .
a) 1500 r.p.m.
b) 2000 r.p.m.
c) 3000 r.p.m.
d) 4000 r.p.m.
[Ans. : c]
Q.6
The synchonous speed for an alternator is given by_________.
a)
120 f
P
b)
120 P
f
c) 120 f P
d) None of these.
[Ans. : a]
Q.7
The number of cycles of e.m.f. generated in a 4 pole alternator per revolution is _______.
+
a) 4
Q.8
Q.10
c) 50
d) 100
[Ans. : b]
An 8 pole alternator runs at 600 r.p.m. The frequency of the induced e.m.f. is _____ .
a) 40 Hz
Q.9
b) 2
VTU : June-10
b) 50 Hz
c) 60 Hz
The frequency of e.m.f. generated depends on _____ .
a) speed
b) number of poles
c) flux
d) both (A) and (B).
+
VTU : June-10
d) 70 Hz
[Ans. : a]
+ VTU : Dec.-11
[Ans. : d]
The frequency of voltage generated by an alternator having 8-poles and rotating at
VTU : June-12
250 r.p.m is ________.
2
a) 60 Hz
b) 50 Hz
c) 25 Hz d) 16 Hz
3
[Ans. : d]
+
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Basic Electrical Engineering
Q.11
8 - 10
Synchronous Generators (Alternators)
An alternator has a phase sequence of RYB for its phase voltage. In case the direction of
rotation of alternator is reversed, the phase sequence will become ________.
a) RBY
b) RYB
c) YRB
+ VTU : June-12
d) none of these
[Ans. : a]
Armature Winding
Q.1
Q.2
Usually modern alternators have short pitched windings so as to ______.
+
a) increase the machine rating
b) improve the voltage waveform
c) improve generated voltage
d) none of these
Q.4
[Ans. : b]
One pole pitch means _______ degrees electrical.
a) 90º
Q.3
VTU : Jan.-09
b) 120º
c) 180º
d) 360º
[Ans. : c]
The disadvantages of a short pitched coils in an alternator is that _________ .
+ VTU : Aug.-11
a) harmonics are introduced
b) waveform become non sinusoidal
c) voltage round the coil is reduced
d) none of the above.
[Ans. : c]
Alternators have short-pitched winding so as to ________.
a) increase machine rating
b) improve the voltage waveform
c) improve generated voltage
d) none of these
+ VTU : June-12
[Ans. : b]
E.M.F. Equation of an Alternator
Q.1
For full pitch coil, the pitch factor Kp is _____ .
a) 1
b) greater than 1
+
VTU : July-09; Jan.-11,13
c) less than 1
d) none of these.
[Ans. : a]
Q.2
Q.3
The ratio of the phasor sum of the e.m.f.s induced in all the coils distributed in a number
of slots under one pole to the arithmetic sum of the e.m.f.s induced is known as ______.
a) breadth or distribution factor
b) coil-span factor
c) pitch factor
d) winding factor.
VTU : Jan.-10
[Ans. : a]
If a is angle of short pitch then coil span factor is ________ .
a) sin a
Q.4
+
b) cos
a
2
c) cos 2a
d) cos
2
a
[Ans. : b]
The e.m.f. equation of an alternator is Eph = ______ .
a) 4.44 f f Kd
b) 4.44 f Tph Kc
c) 4.44 Kc Kd f f Tph d) None of these.
[Ans. : c]
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Q.5
8 - 11
If b is the slot angle, the distribution factor is __________ .
æ bö
sin çm ÷
è 2ø
a)
æbö
m sin ç ÷
è2ø
Q.6
Synchronous Generators (Alternators)
b)
sin(mb)
m sin (b)
æbö
m sinç ÷
è2ø
c)
æm bö
sin ç
÷
è 2 ø
d) None of these.
c) equal to
d) none of these.
[Ans. : a]
The distribution factor is always ______ unity.
a) more than
b) less than
[Ans. : b]
Q.7
If an alternator has 48 slots with 4 poles then the slot angle is ______ .
a) 12º
Q.8
b) 45º
c) 4º
d) 15º
[Ans. : d]
+ VTU : Dec.-11
The distribution factor is defined as the ratio of _____ .
a) arithmetic sum of coil e.m.f.'s to phasor sum of coil e.m.f.'s
b) phasor sum of e.m.f. per coil to the arithmetic sum of coil e.m.f.'s
c) phasor sum of coil e.m.f.'s to the arithmetic sum of coil e.m.f.'s
d) phasor sum of coil e.m.f.'s to the per phase voltage.
Q.9
Full pitch windings have coil span of _____ .
a) 180º
b) 90º
c) 270º
[Ans. : c]
+ VTU : Jun-13
d) 360º
[Ans. : a]
Voltage Regulation of an Alternator
Q.1
When an alternator is loaded, its terminal voltage _____ .
a) increases
b) decreases
c) does not change
+
VTU : June-10
d) none of these
[Ans. : d]
qqq
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Basic Electrical Engineering
8 - 12
Synchronous Generators (Alternators)
Notes
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9
Transformers
Chapter at a Glance
1.
E.M.F. Equation of a Transformer
E1 = 4.44 f f m N1 volts
E2 = 4.44 f f m N2 volts
E2
N2
=
=K
E1
N1
V2
I
= 1 =K
V1
I2
kVA rating of a transformer =
I1 full load =
V1 I1
V I
= 2 2
1000
1000
kVA rating ´ 1000
V1
… (1000 to convert kVA to VA)
kVA rating ´ 1000
I2 full load =
V2
2.
...1000 to express in kVA
Equivalent Resistance of Transformer
R1e = R1 + R¢2 = R1 +
R2
K2
2
R2e = R2 + R1¢ = R2 + K R1
High voltage side ® Low current side ® High resistance side
Low voltage side ® High current side ® Low resistance side
(9 - 1)
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3.
9-2
Magnetic Leakage in a Transformer
X1e = X1 + X¢2
where
X2
X¢2 =
where
X2e = X2 + X1¢
4.
Transformers
K2
2
X1¢ = K X1
…K=
N2
N1
Equivalent Impedance
Z1e = R1e + j X1e
Z2e = R2e + j X2e
2
Z2e = K Z1e
5.
and
Z1e =
… (9.11.6)
K2
Voltage Regulation of Transformer
% Voltage regulation =
6.
Z 2e
E 2 - V2
´ 100
V2
%R=
I 2 R 2e cos f ± I 2 X 2e sin f
´ 100
V2
%R=
I1 R1e cos f ± I 1 X1e sin f
´ 100
V1
Efficiency of a Transformer
%h =
n
(VA rating) ´ cos f
´ 100
(VA rating) ´ cos f + Pi + PCu
= Fraction by which load is less than full load =
New I2 = n (I2)F.L.
2
New PCu = n (PCu )F.L.
%h =
n (VA rating) cos f
n (VA rating) cos f+ Pi + n 2 (PCu ) F. L.
TM
´ 100
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Actual load
Full load
Basic Electrical Engineering
9-3
Transformers
Pi = I 22 R 2e = PCu
Copper losses = Iron losses
Pi = PCu
Pi
(PCu )F. L.
I2m = (I2)F.L.
% h max =
i.e.
V2 I 2m cos f
´ 100
V2 I 2m cos f+ 2Pi
as PCu = Pi
Important Theory Questions and Answers
Ø
Explain the principle of operation of single phase transformer.
+ VTU : Jan.-03, 08, 13; July-03, 04, 08, Dec.-11, June-12, 13, Marks 4
Secondary
winding
Primary
winding
Primary
voltage
I
t
A.C.
supply
T
N1
Load
N2
I
Laminated magnetic iron core
Flux(f)
Secondary
voltage
Voltage
level changes
but frequency i.e. time
period T
remains same
T
Fig. 9.1 Basic transformer
· The transformer works on the principle of mutual induction which states that when
two coils are inductively coupled and if current in one coil is changed uniformly
then an e.m.f. gets induced in the other coil.
· It consists of two inductive coils which are electrically separated but linked through a
common magnetic circuit. The two coils have high mutual inductance. The basic
transformer is shown in the Fig. 9.1.
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9-4
Transformers
· One of the two coils is connected to a source of alternating voltage. This coil in which
electrical energy is fed with the help of source is called primary winding (P).
· The other winding is connected to load. The electrical energy transformed to this
winding is connected to the load. This winding is called secondary winding (S).
· The primary winding has N1 number of turns while the secondary winding has N2
number of turns.
· When primary winding is excited by an alternating voltage, it circulates an alternating
current. This current produces an alternating flux ( f) which completes its path through
common magnetic core as shown dotted in the Fig. 9.1. Thus an alternating flux links
with the secondary winding.
· As the flux is alternating, mutually induced e.m.f. gets developed in the secondary
winding.
Ø
With neat sketch explain the constructional details of core and shell type transformers.
+ VTU : Mar.-01; July-03, 07, 08, Dec.-11, Marks 8
1. Core type transformer : It has a single magnetic circuit. The core is rectangular
having two limbs. The winding encircles the core.
Yoke
Core
Core
L.V. insulation
P
S
L.V. winding
H.V. insulation
H.V. winding
Limb
Flux
(a) Representation
(b) Construction
Fig. 9.2 Core type transformer
· The coils used are of cylindrical type, wound in helical layers with different layers
insulated from each other by paper or mica.
· Both the coils are placed on both the limbs. The low voltage coil is placed inside near
the core while high voltage coil surrounds the low voltage coil.
· Core is made up of large number of thin laminations.
2. Shell type transformer : It has a double magnetic circuit. The core has three limbs.
· Both the windings are placed on the central limb. The core encircles most part of the
windings.
· The coils used are generally multilayer disc type or sandwich coils.
· The core is laminated.
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9-5
Transformers
Core
H.V. winding
L.V. winding
P
Side limb
Core
S
Centre limb
Flux
(a) Representation
(b) Construction
Fig. 9.3 Shell type transformer
Ø
Derive the e.m.f. equation of a transformer.
+ VTU : Jan.-03, 08, 14; July-03, 04, 08, June-13, Marks 6
Flux f
fm
f = f m sin wt
p
0
– fm
1
4f
2p
wt
1
2f
1
T = ––
f
Fig. 9.4 Sinusoidal flux
· The various quantities which affect the magnitude of the induced e.m.f. are :
and
f = Flux
f m = Maximum value of flux
N1 = Number of primary winding turns N2 = Number of secondary winding turns
f = Frequency of the supply voltage
E1 = R.M.S. value of the primary induced e.m.f.
E2 = R.M.S. value of the secondary induced e.m.f.
· From Faraday's law of electromagnetic induction the average e.m.f. induced in each
turn is proportional to the average rate of change of flux.
df
Average e.m.f. per turn = Average rate of change of flux =
dt
df
Change in flux
Now,
=
dt
Time required for change in flux
· Consider the 1/4th cycle of the flux as shown in the Fig. 9.4. In 1/4th time period, the
change in flux is from 0 to f m.
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\
9-6
df fm - 0
=
= 4 f f m Wb/sec
dt
æ 1 ö
ç ÷
è4fø
Transformers
th
as dt for 1/4
time period is 1/4f seconds
\ Average e.m.f. per turn = 4 f f m volts
For sinusoidal quantity,
Form Factor =
R. M. S. value
= 1.11
Average value
\ R.M.S. value = 1.11 ´ Average value
\ R.M.S. value of induced e.m.f. per turn = 1.11 ´ 4 f f m = 4.44 f f m
· There are N1 number of primary turns hence the R.M.S. value of induced e.m.f. of
primary denoted as E1 is,
E1 = N1 ´ 4.44 f f m volts
· While as there are N2 number of secondary turns the R.M.S. value of induced e.m.f. of
secondary denoted E2 is,
E2 = N2 ´ 4.44 f f m volts
· The expressions of E1 and E2 are called e.m.f. equations of a transformer.
E1 = 4.44 f f m N1 volts
E2 = 4.44 f f m N2 volts
Ø
Explain the various losses in a transformer and how to minimize them ? On what factors they
depend ? Give the equations for these losses.
+ VTU : Feb.-05; Jan.-07, 08, 09, 10, 11; July-08, 11, Marks 6
· In a transformer, there exists two types of losses.
i) Core losses ii) Copper losses.
1. Core or Iron losses
· Due to alternating flux set up in the magnetic core of the transformer, it undergoes a
cycle of magnetisation and demagnetisation.
Hysteresis loss = Kh B1m. 67 f v
watts
· The induced e.m.f. in the core tries to set up eddy currents in the core and hence
responsible for the eddy current losses. The eddy current loss is given by,
Eddy current loss = Ke B2m f 2 t 2 watts/unit volume
· The core or iron losses are also called constant losses.
· The iron losses are minimized by using high grade core material like silicon steel
having very low hysteresis loop and by manufacturing the core in the form of
laminations.
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9-7
Transformers
2. Copper losses
· The copper losses are due to the power wasted in the form of I2R loss due to the
resistances of the primary and secondary windings. The copper loss depends on the
magnitude of the currents flowing through the windings.
· Total Cu loss = I12 R1 + I 22 R 2
= I12 (R1 + R¢2 ) = I 22 (R 2 + R¢1 ) = I12 R1e = I 22 R 2e .
· Copper losses are proportional to the square of the current and square of the kVA
rating as voltage is constant.
2
2
PCu µ I µ (kVA)
So,
· The copper losses are kept minimum by designing the windings with low resistance
values.
Important Solved Examples
Example 9.1
A single phase 2200 / 250 V, 50 Hz transformer has a net core area of 36 sq.cm
and a maximum flux density of 0.6 Wb/m2. Calculate the number of turns of primary and
+ JNTU : [H] : May-12
secondary windings.
2
Solution : E1 = 2200 V, E2 = 250 V, f = 50 Hz, a = 36 cm , Bm = 6 Wb/m
2
fm = B m ´ a = 6 ´ 36 ´ 10 –4 = 0.0216 Wb
E1 = 4.44 f m f N 1 i.e. N1 =
E1
E2
Example 9.2
=
N1
N2
i.e. N2 =
2200
= 458.79 » 459
4.44 ´ 50 ´ 0.0216
E2
250
´ 459 = 52.15 » 52
´ N1 =
E1
2200
The maximum flux density in the core of 250/3000 volts, 50 Hz single phase
transformer is 1.2 webers per square meter. If the e.m.f. per turn is 8 volts determine primary
+ JNTU : [H] : Aug.-06
and secondary turns and area of the core.
Bm = 1.2 T, E1 = 250 V, E2 = 3000 V, f = 50 Hz
E1
E
250
3000
e.m.f./turn =
= 2 = 8 i.e.
=8=
N1
N2
N1
N2
Solution :
\
N1 = 31
and
E1 = 4.44 f fm N1
\
N2 = 375
i.e.
E1
= 4.44 × 50 × fm
N1
æ E1 ö
ç
÷
8
è N1 ø
= 0.036 Wb
=
fm =
4.44 ´ 50 4.44 ´ 50
f
f
0.03603
= 0.03003 m2
i.e. a = m =
B = m
1.2
a
Bm
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Example 9.3
9-8
Transformers
The primary winding of a 50 Hz single phase transformer has 480 turns and is fed
from 6400 V supply. The secondary winding has 20 turns. Find the peak value of flux in the
+
core and the secondary voltage.
Solution :
f = 50 Hz, N1 = 480, E1 = 6400 V, N2 = 20
E1
N1
N
20
× 6400 = 266.667 V
=
i.e.
E 2 = 2 E1 =
E2
N2
N1
480
E1 = 4.44 f fm N1
Example 9.4
JNTU : [H] : Aug.-06
i.e.
fm =
6400
= 0.06 Wb
4.44 ´ 50 ´ 480
The number of turns on the primary and secondary windings of a single phase
transformer are 350 and 35 respectively. If the primary is connected to a 2.2 kV, 50 Hz supply,
+ JNTU : [H] : May-06
determine the secondary voltage.
Solution :
N1 = 350, N2 = 35, E1 = 2.2 kV, f = 50 Hz
E1
N1
=
E2
N2
\
E2 =
N2
35
× E1 =
´ 2.2 ´ 10 3 = 220 V
N1
350
Example 9.5 A 1 f transformer has 1000 turns on its primary and 400 turns on the secondary
side. An a.c. voltage of 1250 V, 50 Hz is applied to its primary side, with the secondary open
circuited. Calculate : i) The secondary e.m.f. ii) Maximum value of flux density, given that the
effective cross-sectional area of core is 60 cm 2 .
+ VTU : Feb.-10, Marks 4
Solution : N 1 = 1000, N 2 = 400, E 1 = 1250 V, f = 50 Hz
E1
N1
1250 1000
i)
=
i.e.
=
E2
N2
E2
400
\
E 2 = 500 V
ii) Area of core A = 60 cm 2
E 1 = 4.44 fm f N 1
\
fm = 5.6306 mWb
\
Bm =
Example 9.6
i.e. 1250 = 4.44 ´ fm ´ 50 ´ 1000
fm
5.6306 ´ 10 - 3
=
= 0.9384 Wb m 2
A
60 ´ 10 - 4
A 15 kVA, 2200/110 V transformer has R1 = 1.75 W, R2 = 0.0045 W. The leakage
reactances are X1 = 2.6 W and X2 = 0.0075 W. Calculate,
a) Equivalent resistance referred to primary
b) Equivalent resistance referred to secondary
c) Equivalent reactance referred to primary
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9-9
d)
Equivalent reactance referred to secondary
e)
Equivalent impedance referred to primary
f)
Equivalent impedance referred to secondary
Transformers
g) Total copper loss.
Solution : The given values are, R1 = 1.75 W, R2 = 0.0045 W, X1 = 2.6 W, X2 = 0.0075 W
110
1
=
= 0.05
K =
2200 20
R2
a)
R1e = R1 + R¢2 = R1 +
b)
R2e = R2 + R¢1 = R2 + K R1
K2
= 1 . 75 +
0 . 0045
( 0 . 05) 2
= 3.55 W
2
2
= 0.0045 + (0.05) ´ 1 . 75 = 0.00887 W
X2
c)
X1e = X1 + X¢2 = X1 +
d)
X2e = X2 + X¢1 = X2 + K2 X1
K2
= 2.6 +
0 . 0075
( 0 . 05) 2
= 5.6 W
= 0 . 0075 + ( 0 . 05) 2 ´ 2 .6 = 0.014 W
e)
\
f)
\
Z1e = R1e + j X1e = 3.55 + j 5.6 W
|Z1e| =
3 . 55 2 + 5 . 6 2 = 6.6304 W
Z2e = R2e + j X2e = 0.00887 + j 0.014 W
|Z2e| =
( 0 . 00887) 2 + ( 0 . 014) 2 = 0.01657 W
g) To find full load copper loss, calculate full load current.
(I1)F.L. =
kVA ´ 1000 25 ´ 1000
= 11.3636 A
=
2200
V1
2
\ Total copper loss = [(I1)F.L.] R1e = (11.3636) 2 ´ 3.55 = 458.4194 W
This can be cross checked as,
kVA ´ 1000 25 ´ 1000
(I2)F.L. =
= 227.272 A
=
110
V2
Total copper loss = I 21 R 1 + I 22 R 2
= (11.3636) 2 ´ 1.75 + (227.272) 2 ´ 0.0045
= 225.98 + 232.4365 = 458.419 W
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9 - 10
Transformers
Important Multiple Choice Questions with Answers
Introduction
Q.1
+ VTU : June-13
Transformer is used _____ .
a) to step up the voltage
b) to step down the voltage
c) on dc
d) to step up or step down the voltage
[Ans. : d]
Working Principle
Q.1
A transformer is a ______ device.
a) a.c.
b) d.c.
c) both a.c. and d.c.
d) none of these.
[Ans. : a]
Q.2
Q.3
Q.4
A transformer works on the principle of ______ .
a) Faraday's law
b) mutual induction
c) ferrari
d) superposition
[Ans. : b]
The primary and secondary windings of a transformer are ______ coupled to each other.
a) electrically
b) magnetically
c) electrically and magnetically
d) none of these.
[Ans. : b]
The frequency of secondary voltage is ______ that of the primary voltage.
a) greater than
b) less than
c) same as
d) none of these.
[Ans. : c]
Q.5
The ______ on both sides of a transformer remains same.
a) voltage
b) current
c) power
d) impedance.
[Ans. : c]
Q.6
The flux produced in the core is ______ .
a) directly proportional to the supply frequency
b) directly proportional to the supply voltage
c) inverse proportional to the square of the frequency
d) none of the above
Q.7
[Ans. : b]
The flux in the transformer core is ______ .
a) rotating
b) partly rotating
c) partly alternating
d) purely alternating.
[Ans. : d]
Q.8
If transformer is connected to d.c. supply ______ .
a) primary may burn out .
c) the primary impedance will increase.
b) primary voltage will increase.
d) none of these.
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[Ans. : a]
Basic Electrical Engineering
Q.9
9 - 11
Transformers
A transformer transfers electrical energy from primary to secondary usually with a change
in
a) frequency
b) power
c) voltage
d) time period.
[Ans. : c]
Q.10
Q.11
Q.12
Which of the following does not change in an ordinary transformer ?
a) Voltage
b) Current
c) Frequency
d) All of these
+ VTU : June-12
[Ans. : c]
In a transformer electrical power is transferred from primary to secondary
VTU : June-13
_____ .
+
a) through air
b) by magnetic nux
c) through insulating medium
d) none of these
[Ans. : b]
+ VTU : June-13
The two windings of a transformer are _____ .
a) conductively linked
b) inductively linked
c) not linked at all
d) electrically linked
[Ans. : b]
Parts of Transformer and Construction
Q.1
The core of the transformer is laminated to reduce ______ .
a) eddy current loss
c) copper loss
Q.2
Q.4
VTU : Jan.-09, 11
b) hysteresis loss
d) friction loss.
[Ans. : a]
The vertical portion on which coils are wound in a transformer is called ______ .
a) core
Q.3
+
b) yoke
c) joint
d) limb.
[Ans. : d]
Generally ______ is used for laminations of a transformer core.
a) high grade silicon steel
b) copper
c) iron
d) manganin.
[Ans. : a]
The core provides ______ path to the flux produced.
a) low resistance
b) low reluctance
c) low voltage
d) none of these
[Ans. : b]
Types of Single Phase Transformers
Q.1
Q.2
In a core type transformer ______ .
a) the core encircles the winding.
b) the winding encircles the core.
c) the limb encircles the yoke.
d) none of the above.
[Ans. : b]
______ construction has a double magnetic circuit.
a) core type
b) shell type
c) berry type
d) three phase.
[Ans. : b]
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Basic Electrical Engineering
9 - 12
Transformers
E.M.F. Equation of a Transformer
Q.1
Q.2
An ideal transformer does not change ______ .
a) voltage
b) current
c) power
d) none of the above.
b) 2 A
Q.6
b) 100
b) 4.44 f f m
c) 2 f f m
a) directly proportional
b) equal
c) inversely proportional
d) none of these
d) 0.1
[Ans. : b]
d) f f m
[Ans. : a]
[Ans. : c]
The transformer rating is expressed on VA because ______ .
on both sides it is constant.
losses are independent of load power factor.
the frequency is constant on the load side.
the flux in the core remains constant.
[Ans. : b]
For a 250/25 V transformer having 1 kVA rating, the full load primary current is ______ .
b) 4 A
c) 0.4 A
d) 0.04 A
[Ans. : b]
For a 10 kVA, 2000/200 V transformer, the half load secondary current is ______ .
b) 5 A
c) 25 A
d) 10 A
[Ans. : c]
For 400 V/100 V transformer, the secondary turns are 16 then the primary turns are
______ .
a) 4
Q.10
c) 0.01
The turns ratio is ______ to current ratio.
a) 50 A
Q.9
[Ans. : a]
The average e.m.f. per turn in a transformer is ______ .
a) 40 A
Q.8
d) 100 A
+
a)
b)
c)
d)
Q.7
c) 10 A
A transformer steps up the voltage by a factor of 100. The ratio of current in the primary
VTU : Jan.-10,11
to that in the secondary is ______ .
a) 4 f f m
Q.5
[Ans. : c]
+
a) 1
Q.4
VTU : Jan.-09, June-10
If an ammeter in the secondary of a 100/10 V transformer reads 10 A, the current in the
VTU : July-09
primary would be _____ .
a) 1 A
Q.3
+
b) 64
c) 16
d) 8
[Ans. : b]
The value of flux used in an e.m.f. equation of a transformer is ______ .
a) r.m.s.
b) average
c) maximum
d) instantaneous
[Ans. : c]
Q.11
For a 50 Hz transformer, the primary turns are 100 and maximum flux in the core is
0.08 Wb then the primary induced e.m.f. is ______ .
a) 1856 V
Q.12
b) 1276 V
c) 176 V
d) 1776 V
[Ans. : d]
In a step-up transformer ______ remain constant.
+
a) voltage
d) none of these
b) current
c) power
VTU : June-10
[Ans. : c]
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Q.13
Q.14
9 - 13
Transformers
Transformation ratio in a transformer is equal to ______ .
+
E
a) 1
E2
I
d) 2
I1
N
b) 1
N2
N
c) 2
N1
b) N2 N1
c) I 2 I1
[Ans. : c]
+ VTU : Dec.-11
The transformation ratio of a transformer is ________.
A) V1 V2
VTU : June-10
d) all of these.
[Ans. : b]
Q.15
A single phase transformer has 250 turns on primary and 1000 turns on the secondary
winding. If the primary winding is connected across a 230 V, 50 Hz, single phase supply,
VTU : June-12
the voltage induced in the secondary winding is ________.
+
a) 920 V
Q.16
b) 2 A
[Ans. : a]
c) 10 A
d) 100 A
[Ans. : a]
The frequency of secondary voltage is ________ that of primary voltage.
+ VTU : Jan.-13
b) less than
c) same as
d) double
[Ans. : c]
+ VTU : June-13
A transformer does not transform _____ .
a) power
Q.19
d) 690 V
+ VTU : Jan.-13
a) greater than
Q.18
c) 1840 V
If secondary current of 100/10 V transformer is 10 A, then primary current is
a) 1 A
Q.17
b) 230 V
b) voltage
c) current
d) impedance
[Ans. : a]
A single phase, 5 kVA, 200 V/100 V, trasformer has rated primary and secondary currents
VTU : Jan.-14
at rated voltage _____ .
+
a) 25 A and 50 A
b) 50 A and 25 A
c) 12.5 A and 62.5 A d) 62.5 A and 12.5 A
[Ans. : a]
Ideal Transformer on No Load
Q.1
The no load primary current I0 in transformer _____ .
a) is in phase with V1
b) leads V1 by 90º
c) lags behind V1 by 90º
+
VTU : July-09
d) lags V1 by an angle between 0º and 90º.
[Ans. : d]
Q.2
The _____ component of no load current is required to produce flux in the core.
a) wattful
b) power
c) core loss
d) magnetizing
[Ans. : d]
Transformer on Load (M.M.F. Balancing on Load)
Q.1
+
The flux in transformer core ______ .
a) increases with load
b) decreases with load
c) remains constant irrespective of load
d) none of the above.
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VTU : Jan.-09
[Ans. : c]
Basic Electrical Engineering
Q.2
9 - 14
Transformers
+ VTU : Dec.-11
The magnitude of mutual flux in a transformer is
a) low at low loads and high at high loads
b) high at low loads and low at high loads
c) same at all loads
d) varies at low loads and constant at high loads.
[Ans. : c]
Equivalent Resistance of Transformer
Q.1
For a transformer, the turns ratio is 10 : 1 then its primary resistance of 10 W will be
______ when referred to secondary.
a) 1 W
Q.2
b) 0.01 W
c) 0.1 W
d) 10 W
[Ans. : c]
A high voltage side is ______ .
a) low current, low impedance side,
c) low current, high impedance side,
b) high current, low impedance side,
d) none of the above
[Ans. : c]
Voltage Regulation of Transformer
Q.1
+
The regulation of a transformer is defined as ______ .
VTU : Jan.-10
a) rise in terminal voltage when loaded
b) fall in terminal voltage when loaded
c) change in secondary terminal voltage from no-load to full-load as a percentge of
secondary no load terminal voltage.
d) change in flux from no-load to full-load.
Q.2
For leading power factor loads, the regulation of transformer is ______ .
a) positive
Q.3
[Ans. : c]
b) negative
c) zero
d) unity.
[Ans. : b]
For better performance of transformer, the regulation must be ______ .
a) high
b) infinite
c) very low
d) none of these
[Ans. : c]
Q.4
The voltage regulation of a transformer is zero for ______ load.
a) lagging
Q.5
b) resistive
c) inductive
d) capacitive
[Ans. : d]
If the transformer regulation is positive, ______ load is connected to the transformer.
a) capacitive
b) inductive
c) resistive
d) none of these
[Ans. : b]
Losses in a Transformer
Q.1
Losses which do not occur in transformer are _____.
a) copper losses
b) magnetic losses
+
c) friction losses
VTU : July-09; Feb.-11
d) none of these.
[Ans. : c]
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Basic Electrical Engineering
Q.2
Q.3
9 - 15
+
The losses which vary with load in a power transformer are ______ .
a) friction and windage losses
b) copper losses
c) eddy current losses
d) hysterisis losses
[Ans. : b]
b) magnetic hum
d) electric energy.
[Ans. : a]
A transformer core is laminated to reduce the ______ .
a) hysteresis loss
c) leakage reactance
Q.5
VTU : Jan.-10
The core losses are dissipated in the form of ______ .
a) heat
c) light
Q.4
Transformers
b) eddy current loss
d) all of these.
[Ans. : b]
Which loss is not common between rotating machines and transformer ?
a) copper loss
b) eddy current loss
c) core loss
d) friction loss.
[Ans. : d]
Q.6
The transformer efficiency is maximum when ______ .
a) Pi > PCu
b) Pi < PCu
c) Pi = PCu
d) none of these.
[Ans. : c]
Q.7
Which loss is variable in a transformer ?
a) eddy current
Q.8
b) copper
Q.10
Q.11
Q.12
+
[Ans. : b]
b) 200 W
c) 400 W
VTU : June-10
d) 300 W
[Ans. : b]
When the supply frequency of a transformer is doubled then the hystersis losses ?
+ VTU : Aug.-11
a) remain same
b) doubled
c) reduced by 50%
d) hystersis loss equal to eddy current loss.
Transformer cores are laminated in order to
[Ans. : b]
+ VTU : Dec.-11
a) simplify its construction
b) minimize eddy current loss
c) reduce cost
d) reduce hysteresis loss.
[Ans. : b]
A transformer has full load copper loss of 800 W and core loss of 600 W. Total loss at
VTU : June-12
no load will be approximately ________.
+
a) 1400 Watts
b) 1100 Watts
c) 1000 Watts
d) 600 Watts
[Ans. : d]
+ VTU : Jan-13
The core of a transformer is laminated to reduce _____ .
a) eddy current
Q.13
d) friction.
A transformer has 200 W iron loss at full load. The iron loss at half full load is_____.
a) 100 W
Q.9
c) hysteresis
b) hysteresis current
c) copper loss
d) friction loss
[Ans. : a]
The core of a transformer is assembled with laminated sheets so as to _____ .
a) reduce hysteresis loss
b) reduce eddy current loss
b) both hysteresis and eddy current loss
d) copper loss
+ VTU : Jan.-14
[Ans. : b]
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Basic Electrical Engineering
Q.14
9 - 16
Transformers
If the full load core loss of a transformer is 100 W, its core loss at half load will be
VTU : Jan-14
_____ .
+
a) 200 W
b) 100 W
c) 50 W
d) 25 W
[Ans. : b]
Efficiency of a Transformer
Q.1
The copper loss of certain transformer at half-full load is measured as 200 W. Then the
VTU : Jan.-09, 11, 13
copper loss at full load will be ______ .
+
a) 100 W
Q.2
b) 200 W
If copper loss of a transformer at 1/4
would be _____.
a) 100 W
Q.3
c) 400 W
th
d) 800 W
[Ans. : d]
full load is 100 W, then its full load copper loss
+
b) 400 W
c) 800 W
VTU : July-09
d) 1600 W. [Ans. : d]
A transformer has maximum efficiency at full load when iron losses are 1600 W then its
half load copper losses are ______ .
a) 1600 W
b) 6400 W
c) 400 W
d) none of these
[Ans. : c]
Q.4
The full load copper losses of transformer are 500 W then the copper losses on full load
at 0.8 power factor lagging are ______ .
a) 1000 W
Q.5
Q.6
b) 250 W
a) high, high
b) high, low
c) low, high
d) low, low
[Ans. : c]
+
b) 800 W
c) 200 W
d) 1600 W [Ans. : c]
A transformer is working at its maximum efficiency with iron-loss of 500 W, then its
VTU : Dec.-11
copper-loss will be
+
b) 250 W
c) 300 W
d) 400 W.
[Ans. : a]
The efficiency of a transformer at full load 0.8 pf lag is 95 %. The efficiency at 0.8 pf lead
VTU : June-12
is ________.
+
a) 99 %
Q.9
[Ans. : d]
The full load copper loss for a transformer is 800 W, then the copper loss at half the full
VTU : Aug.-11
load is
a) 500 W
Q.8
d) 500 W
Regulation and efficiency of a transformer should be respectively _____ .
a) 400 W
Q.7
d) 125 W
b) 95.5 %
c) 95 %
d) 90 %
A transformer operates at maximum efficiency, when _____ .
a) core losses minimum
b) copper loss minimum
c) core loss = copper loss
d) none of these
[Ans. : c]
+ VTU : Jan.-14
[Ans. : c]
qqq
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10
Three Phase Induction Motor
Chapter at a Glance
1.
Slip of Induction Motor
s =
%s =
Ns - N
Ns
… (Absolute slip)
Ns - N
´ 100
Ns
… (Percentage slip)
N = Ns (1 – s)
\
2.
\
… (From the expression of slip)
s = 1 at start
Effect of Slip on the Rotor Frequency
fr =
sf
Important Theory Questions and Answers
Ø
What is rotating magnetic field ? What is the speed of rotating magnetic field ?
+ VTU : Jan.-03, 05, 07, 10; July-03, 05, 06, 08, Marks 4
· The Fig. 10.1 shows the phasor diagram with fR
as reference. The directions shown are the
assumed positive directions of the three fluxes.
The flux in opposite direction to the directions
shown in treated as negative.
fB
120º
fR
120º
120º
fY
Fig. 10.1
(10 - 1)
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Basic Electrical Engineering
10 - 2
Three Phase Induction Motor
· The equations of the three fluxes are,
fR = fm sin q, fY = fm sin (q – 120º), fB = fm sin (f – 240º)
· The total flux fT is the vector sum of fR , fY and fB for various values of q.
· If phasor diagram is drawn for various values of q, it can be seen that the magnitude of
fT is always 1.5 fm but it rotates in space. Such a magnetic field is called rotating
magnetic field.
This shows that when a three phase stationary windings are excited by balanced three phase
a.c. supply then the resulting field produced is rotating magnetic field. Though nothing is
physically rotating, the field produced is rotating in space having constant amplitude.
· For a standard frequency whatever speed of R.M.F. results is called synchronous speed,
in case of induction motors. It is denoted as Ns.
\
Ns =
where
120f
= Speed of R.M.F.
P
f = Supply frequency in Hz
P = Number of poles for which winding is wound.
· The direction of rotating magnetic field depends on the phase sequence of the three
phase supply.
Ø
Discuss the important features of squirrel cage and phase wound rotor constructions in an
+ VTU : Jan.-03, 05, Marks 6; Jan.-08, Marks 8
induction motor.
1. Squirrel Cage Rotor
Copper or
aluminium bars
End ring
(b) Symbolic representation
(a) Cage type structure of rotor
Fig. 10.2 Squirrel cage rotor
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Basic Electrical Engineering
10 - 3
Three Phase Induction Motor
2. Slip Ring Rotor or Phase Wound Rotor
R
Slip rings
Shaft
R
B
Y
Brush
B
Y
External star
connected rheostat
Star connected
rotor winding
Rotor
frame
Fig. 10.3 Slip rings or wound rotor
Sr. No.
Ø
Wound or slip ring rotor
Squirrel cage rotor
1.
Rotor consists of a three phase winding
similar to the stator winding.
Rotor consists of bars which are shorted at
the ends with the help of end rings.
2.
Construction is complicated.
Construction is very simple.
3.
Slip rings and brushes are present to add
external resistance.
Slip rings and brushes are absent.
4.
The construction is delicate and due to
The construction is robust and maintenance
brushes, frequent maintenance is necessary. free.
5.
The rotors are very costly.
Due to simple construction, the rotors are
cheap.
6.
Rotor copper losses are high hence
efficiency is less.
Rotor copper losses are less hence have
higher efficiency.
7.
Used for lifts, hoists, cranes, elevators,
compressors etc.
Used for lathes, drilling machines, fans,
blowers, water pumps, grinders, printing
machines etc.
Explain the working principle of three phase induction motor.
+ VTU : Jan.-07, 09, 10, 11, 13, 14; July-03, 04, 05, 06, 07, 08, 11, 12, Marks 6
· Induction motor works on the principle of electromagnetic induction.
· When a three phase supply is given to the three phase stator winding, a rotating
magnetic field of constant magnitude is produced. The speed of this rotating magnetic
field is synchronous speed, Ns r.p.m.
120 f
= Speed of rotating magnetic field.
Ns =
P
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Basic Electrical Engineering
10 - 4
Three Phase Induction Motor
· This rotating field produces an effect of rotating poles around a rotor.
· Let direction of rotation of this rotating magnetic field is clockwise as shown in the
Fig. 10.4 (a).
S
Direction of
R.M.F.
Direction of
R.M.F.
S
Stator
Stator
Stator
R.M.F.
R.M.F
Flux due to
induced
rotor current
R.M.F.
Rotor
Rotor
Rotor conductors
S
Rotor
Induced current in rotor conductor
(a)
(b)
(c)
Fig. 10.4
· Now at this instant rotor is stationary and stator flux R.M.F. is rotating.
· As rotor forms closed circuit, induced e.m.f. circulates current through rotor called rotor
current as shown in the Fig. 10.4 (b). Direction of this current is going into the paper
denoted by a cross as shown in the Fig. 10.4 (b).
· Any
current
carrying
conductor
produces its own flux. So rotor
produces its flux called rotor flux. The
direction of rotor flux is clockwise as
shown in the Fig. 10.4 (c).
R.M.F
Direction
S
Stator
Cancellation
of two fluxes
(low flux area)
Addition
of fluxes
(high flux area)
· Both the fluxes interact with each as
shown in the Fig. 10.4 (d).
Rotor
· On left of rotor conductor, two fluxes
are in same direction hence add up to
get high flux area.
Rotor
conductor
Mechanical
force
Fig. 10.4 (d) Interaction of fluxes
· On right side, two fluxes cancel each other to produce low flux area.
Ø
Derive the expression for the slip and frequency of rotor currents.
+ VTU : Jan.-09, July-07, 11,12, Marks 6
· In case of induction motor, the speed of rotating magnetic field is,
120 f
… (10.1)
Ns =
P
· The frequency of this induced e.m.f. at start is same as that of supply frequency.
· The rotor is wound for same number of poles as that of stator i.e. P.
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Basic Electrical Engineering
10 - 5
Three Phase Induction Motor
· If fr is the frequency of rotor induced e.m.f. and rotor currents, in running condition at
slip speed Ns – N then there exists a fixed relation between (Ns – N), fr and P similar
to equation (10.1).
120 f r
, Rotor poles = Stator pole = P
… (10.2)
(Ns – N) =
P
· Dividing equation (10.2) by equation (10.1) we get,
(120f r / P)
Ns - N
N -N
f
but s
=
= Slip s i.e. s = r
(120f / P)
Ns
Ns
f
\
Ø
fr =
sf
Explain why an induction motor needs starter ?
+ VTU : Jan.-04, 06; July-03, 04, 05, 06, 10, Dec.-11, Marks 4
· In a three phase induction motor, magnitude of induced e.m.f. in the rotor circuit
depends on the slip of the induction motor.
· At start the value of slip is at its maximum equal to unity.
· The magnitude of induced e.m.f. at start is maximum as slip speed i.e. relative speed
between rotor and the rotating magnetic field is maximum.
· As rotor conductors are short circuited in most of the motors, due to squirrel cage
construction, this e.m.f. circulates very high current through rotor at start.
· Hence as rotor current is high at start, consequently stator draws a very high current
of the order of 5 to 8 times full load current at start.
· Due to such high current at start there is possibility of damage of the motor winding.
· Similarly due to sudden in rush of current, other appliances connected to the same line
may be subjected to voltage spikes which may affect their working.
· To avoid such effects it is necessary to limit current drawn by the motor at start. Hence
starter is necessary for an induction motor.
· Starters not only limit the starting current but also provide protection to the induction
motor against over loading and low voltage conditions. The starters also provide single
phasing protection too.
Ø
With a neat circuit diagram explain a star-delta starter for a 3 phase induction motor.
+ VTU : July-03, 04, 05, 06, 13; Jan.-04, 06, 14, Marks 8
· It uses TPDT [Tripple Pole Double Throw Switch] which connects the stator winding in
star at start and then in delta while normal running.
· Initially when switch is in the START position, the stator winding gets connected in
star.
Vph =
VL
3
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Basic Electrical Engineering
10 - 6
Three Phase Induction Motor
3 Phase supply
R
Y
B
R
Y
B
1
Stator
winding
1
2
3
3
2
R
B
TPDT
switch
RUN
Delta
Y
START
Star
Squirrel
cage rotor
Fig. 10.5 Star-delta starter
· Due to this, starting current also gets reduced by factor 1 / 3.
· When motor attains 50 to 60 % of normal speed, switch is thrown in the RUN position.
· Hence winding gets connected in delta.
Vph = VL = Rated voltage
· Each phase of the winding gets rated voltage.
· The operation of switch can be automatic by using relays which ensures that motor will
not start with switch in RUN position.
Important Solved Examples
Example 10.1
A 3 phase induction motor has 6 poles and runs at 940 r.p.m. on full load. It is
supplied from an alternator having 4 poles and running at 1500 r.p.m. Calculate the full load
slip and the frequency of the rotor currents of the induction motor.
+
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VTU : Aug.-03, Marks 7
Basic Electrical Engineering
10 - 7
Three Phase Induction Motor
Solution : Alternator supplies induction motor,
Alternator
PA = 4, NO = 1500 r.p.m.
Induction motor
PM = 6, N = 960 r.p.m.
Fig. 10.6
120 f
1500 ´ 4
Ns =
= 50 Hz
i.e. f =
PA
120
For alternator,
For an induction motor, f = 50 Hz is stator frequency,
\
Ns =
\
%s =
\
fr
Example 10.2
120 f 120 ´ 50
=
= 1000 r.p.m.
PM
6
Ns - N
1000 - 960
´ 100 =
´ 100 = 4 %
Ns
1000
= s f = 0.04 ´ 50 = 2 Hz
If the electromotive force in the stator of an 8 pole induction motor has a frequency
of 50 Hz and that in the rotor 1.5 Hz, at what speed is the motor running and what is the
+
slip ?
Solution :
VTU : Aug.-09, Aug.-04,05, Marks 6
f = 50 Hz, P = 8, fr = 1.5 Hz
120 f
120 ´ 50
Ns =
= 750 r.p.m.
=
P
8
fr = s f
\
s =
\
i.e. 1.5 = s ´ 50
1.5
= 0.03
50
... Slip
i.e. 3 %
N = Ns (1 – s) = 750 (1 – 0.03) = 727.5 r.p.m.
Example 10.3
... Speed
A 4 pole, 3 f, 50 Hz induction motor runs at a speed of 1470 rpm. Find the
synchronous speed, the slip and frequency of the induced e.m.f. in the rotor under this
+
condition.
Solution :
P = 4, f = 50 Hz, N = 1470 r.p.m.
120 f 120 ´ 50
=
= 1500 r.p.m.
Ns =
P
4
s =
N s - N 1500 - 1470
=
= 0.02
Ns
1500
i.e. 2 %
fr = s f = 0.02 ´ 50 = 1 Hz
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VTU : Feb.-10, Marks 6
Basic Electrical Engineering
10 - 8
Three Phase Induction Motor
Important Multiple Choice Questions with Answers
Introduction
Q.1
+
An induction motor works with ______ .
a) d.c. only
b) a.c. only
c) both a.c. and d.c.
d) none of the above.
VTU : Feb.-09,11
[Ans. : b]
Rotating Magnetic Field
Q.1
The stator winding of three phase induction motor produces ______ magnetic field.
a) alternating
Q.2
Q.3
b) rotating
Q.5
b) 1200
c) 1800
d) 1500
[Ans. : d]
The direction of rotation of an induction motor depends on ______ .
a) phase sequence
b) supply frequency
c) supply voltage
d) none of these
[Ans. : a]
If any two phases of an induction motor are interchanged then ______ .
a) motor will burn
c) motor speed will reduce
Q.6
d) constant [Ans. : b]
The magnitude of rotating magnetic field is ______ times the maximum flux of any
individual phase.
1
a) 1.5
b) 3
c) 2.5
d)
3
[Ans. : a]
For a 4 pole, 50 Hz, three phase induction motor, the synchronous speed is
______
r.p.m.
a) 1000
Q.4
c) pulsating
b) motor will stop
d) direction of rotation will change [Ans. : d]
When a 3-f supply is given to the stator of 3-f induction motor, a ________ magnetic field
VTU : June-10
is produced.
+
a) stationary
b) alternating
c) rotating
d) none of these
[Ans. : c]
Q.7
Q.8
The speed at which the rotating magnetic field produced by stator currents rotates is
VTU : June-12
________.
+
a) synchronous speed
b) rotor speed
c) greater than synchronous speed
d) none of these
Synchronous speed of three phase induction motor is given by _____ .
+
a) Ns = 120 f P
Q.9
[Ans. : a]
b) 120 f / P
c) 120 P / f
d) f P/120
The relation between Ns, f and P of three-phase inductor is _____ .
P
a) Ns =
120 f
b) Ns =
120 P
f
c) f =
TM
PNs
120
d) f =
PNs
P
TECHNICAL PUBLICATIONS - An up thrust for knowledge
VTU : Jan.-13
[Ans. : b]
+
VTU : Jan.-14
[Ans. : c]
Basic Electrical Engineering
10 - 9
Three Phase Induction Motor
Construction
Q.1
The frame of induction motor is usually made of _____ .
a) silicon steel
Q.2
Q.3
c) aluminium
d) bronze
+
a) supply frequency
b) motor speed
c) supply voltage
d) both (a) and (b)
[Ans. : b]
VTU : Jan.-10
[Ans. : d]
______ rotor is permanently short circuited.
b) Wound
c) Squirrel cage
d) Cup type [Ans. : c]
In squirrel cage rotor, the slots are skewed ______ .
a) to reduce losses
c) to reduce magnetic hum
Q.5
VTU : July-09, Feb.-11
The number of poles in a 3 f induction motor is determined by the ______ .
a) Slip ring
Q.4
b) cast iron
+
b) to give support
d) to reduce friction
[Ans. : c]
The slip rings are usually made up of ______ .
a) copper
b) iron
c) carbon
d) phosphor-bronze
[Ans. : d]
Q.6
The rotor power factor is ______ in nature.
a) unity
Q.7
c) zero
b) slip ring
c) squirrel cage
d) universal [Ans. : b]
b) less than the stator poles.
d) zero
[Ans. : c]
For a delta connected slip ring rotor, the number of slip rings required are ______ .
a) 3
Q.10
[Ans. : d]
In an induction motor, the number of rotor poles is ______ .
a) greater than the stator poles
c) equal to the stator poles.
Q.9
d) lagging
In ______ motor, the speed can be controlled from rotor side.
a) a.c. series
Q.8
b) leading
b) 2
c) 1
d) 0
[Ans. : a]
______ induction motor is preferred when maintenance is the main consideration.
a) Slip ring
b) Wound rotor
c) Split phase
d) Squirrel cage
[Ans. : d]
Q.11
Q.12
The air gap between the stator and the rotor of a 3-f . I.M. ranges from _______.
+
a) 2 cm to 4 cm
b) 0.4 mm to 4 mm
c) 1 cm to 2 cm
d) 4 cm to 6 cm
VTU : June-10
[Ans. : b]
Phase wound induction motors are less extensively used than squirrel cage induction
VTU : Aug.-11
motors because _______.
+
a) slip rings are required on the rotor circuit
b) rotor windings are generally star connected
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Basic Electrical Engineering
10 - 10
Three Phase Induction Motor
c) they are costly and require greater maintenance
d) none of the above
Q.13
[Ans. : c]
+
In a three phase induction motor _______.
VTU : Jan.-14
a) rotor conductors are open circuited
b) rotor conductors are short circuited
c) stator winding is open
Q.14
d) none of these
[Ans. : b]
The number of slip rings in a three phase wound rotor induction motor is _______.
a) 3
b) 4
c) 9
d) 2
+
VTU : Jan-14
[Ans. : a]
Working Principle
Q.1
Synchronous speed of three phase induction motor is given by ____ .
a) Ns = 120 f P
b) Ns = 120 f / P
c) Ns = 120 P / f
+
VTU : July-09
d) Ns = f P/120
[Ans. : b]
Q.2
Q.3
Q.4
The rotor of a 3 f induction motor rotates in the same direction as that of stator rotating
VTU : Jan.-10
field. This can be explained by ______ .
+
a) Faraday's laws of electromagnetic induction
b) Lenz's law
c) Newton's law of motion
d) Flemings right hand rule
Speed of the induction motor is ______ that of Ns.
a) greater than
b) less than
c) same as
Q.7
Q.8
b) Lenz's law
c) Faraday's law
d) Ohm's law
[Ans. : b]
If the synchronous speed of the 50 Hz induction motor is 750 r.p.m., it has ______ stator
poles.
a) 8
Q.6
d) none of these
[Ans. : b]
The rotor of an induction motor rotates in the same direction as that of rotating magnetic
field, according to ______ .
a) Coulombs law
Q.5
[Ans. : b]
b) 4
c) 2
The slip of an induction motor at stand still is ______ .
d) 6
+
a) zero
b) one
c) infinity
d) none of the above.
[Ans. : a]
VTU : Feb.-09,11
[Ans. : b]
An induction motor under full load has a slip of about ______ .
+
a) 0.03
d) zero
b) 0.1
c) 0.3
VTU : Jan.-09
[Ans. : a]
A 4 pole, 440 V, 50 Hz induction motor is running at a slip 4 %. The speed of motor is
VTU : July-09
____.
+
a) 1260 r.p.m.
b) 1440 r.p.m.
c) 1500 r.p.m.
d) 1560 r.p.m.
[Ans. : b]
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Basic Electrical Engineering
Q.9
10 - 11
If Ns is the synchronous speed and 's' is the slip, then actual running speed of an
VTU : Jan.-10
induction motor will be ______ .
+
a) Ns
Q.10
b) s Ns
b) infinite
b) 0.02
b) N
Q.19
Q.21
[Ans. : a]
d) 1500
[Ans. : b]
c) Ns – N
d) slip is one
+
[Ans. : b]
VTU : June-10
d) N – Ns
[Ans. : c]
+
b) 4 %
c) 5 %
d) 3.33 %
[Ans. : b]
The difference between synchronous speed and actual speed is 100 rmp and the
synchronous speed is 1500 r.p.m., then the value of slip is _______.
b) 10 %
c) 6.66 %
d)15 %.
[Ans. : c]
d) 1.
[Ans. : d]
When the rotor of a 3 f induction motor is blocked, the slip is
b) 0.5
c) 0.1
The rotor of a 3 phase induction motor always runs at _______.
+
VTU : Dec.-11
a) synchronous speed
b) less than synchronous speed
c) more than synchronous speed
d) none of these.
Slip of an induction motor at standstill is _______.
a) zero
Q.20
d) 0.08
A supply of 50 Hz is given to a 3-f I.M. having 4 poles. If the I.M. runs at 1440 rpm the
VTU : June-10
slip is ________ .
a) zero
Q.18
c) slip is zero
In a 3-f induction motor, the slip speed is given by _______.
a) 2 %
Q.17
c) 1495
b) slip is negative
a) 3 %
Q.16
[Ans. : d]
The rotor speed is more than the synchronous speed in a three phase induction motor
when ______ .
a) Ns
Q.15
d) 1
c) 0.01
b) 1455
a) slip is positive
Q.14
c) 100
For a 4 pole, 50 Hz induction motor, the full load slip is 0.03 hence its full load speed is
______ r.p.m.
a) 1420
Q.13
d) (Ns - 1) s [Ans. : c]
For a 6 pole, 50 Hz induction motor, the full load speed is 950 r.p.m. hence full load slip
is ______ .
a) 0.05
Q.12
c) (1 - s) Ns
The value of slip is ______ at start.
a) zero
Q.11
Three Phase Induction Motor
b) unity
c) greater than unity
[Ans. : b]
+
VTU : Dec.-11
d) negative. [Ans. : a]
If the rotor terminals of a 3 phase slip ring induction motor are not short-circuited and the
VTU : Dec.-11
supply is given to the stator, the motor will _______.
+
a) not start
b) start running
c) run at high speed
d) run at low speed
When an induction motor is at standstill its slip is ________.
a) zero
b) 0.5
c) 1
d) infinity
[Ans. : c]
+
VTU : June-12
[Ans. : d]
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Basic Electrical Engineering
Q.22
+
b) sNs
b) 1
c) ¥
[Ans. : b]
+
VTU : Jan-13
d) – 1
[Ans. : b]
+
b) 1440 rpm
c)1500 rpm
D) 1560 rpm
[Ans. : b]
+
Speed of an induction motor is _____ that of Ns.
a) greater than
Q.26
d) (Ns – 1)s
A 4 pole, 440 V, 50 Hz induction motor is running at a slip of 4 % the speed of motor is
VTU : Jan-13
_____ .
a) 1260 rpm
Q.25
c) (1 – s) Ns
The slip of an induction motor at standstill is _____ .
a) 0
Q.24
Three Phase Induction Motor
If Ns is synchronous speed and 's' is the slip, then the actual running speed of an
VTU : June-12
induction motor will be ________.
a) Ns
Q.23
10 - 12
b) less than
c) same as
VTU : Jan-13
D) double
[Ans. : d]
If the starting winding of a single phase induction motor is left in the circuit _____ .
+
a) the motor will run faster
VTU : June-13
b) the motor will run slower
c) there will be undue sparking
d) the auxiliary winding will get over-heated due to continuous flow of current and may get
damaged
[Ans. : d]
Q.27
Q.28
Q.29
Which of the following types of motors are not single phase ac motors ?
+
a) Induction type motors
b) Commutator type motors
c) Synchronous type motors
d) Schrage motors
VTU : June-13
[Ans. : a]
Which of the following types of motors are not the induction motors ?
+
VTU : June-13
a) Repulsion motors
b) Split phase motors
c) Shaded pole motors
d) Repulsion start induction motors [Ans. : a]
+
When speed of induction motor is zero, its slip is ______ .
a) zero
b) 0.5
c) one
VTU : Jan-14
d) infinity
[Ans. : c]
Effect of Slip on the Rotor Frequency
Q.1
A 4 pole, 50 Hz induction motor runs at a speed of 1440 r.p.m. The frequency of the
VTU : Jan.-09, Feb.-11
rotor induced e.m.f. is ______ .
+
a) 3 Hz
Q.2
b) 2.5 Hz
c) 2 Hz
d) 1 Hz.
[Ans. : c]
The relation between rotor frequency (f') and stator frequency (f) is given by___.
+
a) f' = s f
b) f' = f / s
c) f' =
sf
VTU : July-09
d) f' = (1 – s) f.
[Ans. : a]
Q.3
A 3 f , 440 V, 50 Hz, induction motor has 4 % slip. The frequency of rotor e.m.f. is
VTU : Jan.-10
______ .
+
a) 200 Hz
b) 50 Hz
c) 2 Hz
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d) 0.2 Hz
[Ans. : c]
Basic Electrical Engineering
Q.4
10 - 13
A 440 V, 50 Hz induction motor has a slip of 5 % then the frequency of rotor currents is
______ .
a) 1.5 Hz
Q.5
Three Phase Induction Motor
b) 2.5 Hz
c) 0.5 Hz
The frequency of rotor current or e.m.f. is given by _______.
B) f2 = f1 s
A) f2 = sf1
C) f2 = (1- s) f1
d) 50 Hz
+
[Ans. : b]
VTU : Dec.-11
D) f2 = s f1. [Ans. : a]
Applications
Q.1
Q.2
External resistance is connected to the rotor of a 3 f phase wound induction motor in
VTU : July-11
order to ________.
+
a) reduce starting current
b) collector current
c) as a star connected load
d) none of these
[Ans. : a]
Initial starting current drawn by a 3-phase induction motor in terms of full load current on
application of rated voltage (approximately) is ________.
a) equal to full load current
b) 2 times
c) more than 10 times
d) 5 times (approx.)
+
VTU : June-12
[Ans. : d]
qqq
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Basic Electrical Engineering
10 - 14
Notes
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Three Phase Induction Motor