Test- Machine With Solution
Q.1 A Transformer has its maximum efficiency of 99% at 10KVA at 0.9 pf. The
Cu-losses at 5 kW and 0.8 pf are
(a) 91 W
(b) 30.2 W
(c) 35.6 W
(d) 23.5 W
Ans. (c)
Exp. At a load S L  xS0  10 kVA, S0  Rated & pf cos  = 0.9,
Maximum efficiency max  0.99

xS 0Cos
xS 0Cos  Pi  x 2PcF
Pi & PCF are Iron loss & full load Cu losses.
Pi  x 2PCF
for max


0.99 
10  0.9
10  0.9  Pi  Pi
Pi  0.091kW
x
S L 10

S0 S0
p i  x 2PCF
2

 10 
0.091     PCF
 S 

PCF  9.1  104S2 kW
At a load PL  5kW, Cos  0.8
SL 
x
PL
5

 6.25 kVA
Cos 0.8
S L 6.25

S0
S0
Cu losses = x 2PCF
2
 6.25 
4 2

  9.1  10 S 0
 S0 
 0.0356 kW
 35.6W
Common Data for Q.2, Q.3 & Q.4
A 10KVA, 1000V/100V Single phase transformer has
R1  2, X 1  5, R 2  0.03 and X 2  0.05
Q.2 The secondary terminal voltage (approximately) at 0.707 lag pf when delivering full load current with primary voltage held fixed at 1 kV.
(a) 80.3 V
(b) 90.4 V
(c) 96.2 V
(d) 98.5 V
Ans. (b)
Q.3 Determine the pf of rated load, supplied at 100V such that the terminal
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Test- Machine With Solution
voltage observed on reducing the load to zero is still 100V.
(a) 0.89 lead
(b) 0.707 lead
(c) 0.92 lag
Ans. (a)
N
1
Exp. Transformation rotio K  2 
N1 10
R2 
X 2 
(d) upf
R2
0.03

 3
2
2
K
1
10
 
X2
 5
K2
Total resistance referred to primary
R 01  R1  R2  2  3  5
X 01  X1  X 2  5  5  10
Rated
S0  10 kVA
V1  1 kV, V2  100V
S0  V1I 1  V2 I 2

I1  10A & I 2  100A
Base impedance referred to primary
V
1000
Z1B  1 
 100
I1
10
per unit impedances.
R
5
R pu  1 
 0.05
Z1B 100
X pu 
X 01
10

 0.1
Z1B 100
At rated current & pf Cos (lag), The voltage regulation,
VR  R puCos  X puS in 
At Cos  0.707 lag
   45  Sin   0.707
VR  0.05  0.707  0.1  0.707
 0.10605  10.605%
Secondary voltage referred to primary
V
V2  2  10V2
K
V  V2
V .R  1
V2
 0.10605 
1000  V2
V2
 V2  904.12V
 V2  90.412V
It no-load voltage is equal to full load voltage, i.e. Vnl  V fl  V .R  0
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Test- Machine With Solution
At leading pf c o s 
VR  R puCos  X puSin   0
 tan  
R pu
X pu

0.05
 0.5
0.1
   26.57  pf  Cos  0.895 lead
Q.4 Determine the maximum applied voltage on primary side if rated load is
delivered at secondary at lag pf
(a) 1000 V
(b) 988 V
(c) 1231 V
(d) 1112 V
Ans. (d)
Exp.
At lag pf & rated load
VR  R puCos   X puSin 
d
VR   0
d
VR max if
 tan  
X pu
R pu

0.1
2
0.05
   63.45
VR max  0.05  cos 63.45  0.1  sin 63.45
 0.112  11.2%
V2  100V rated  V2 
V .R . 
V2
 1000V
K
V1  V2
V2
 VRmax 
V1max  V2
V2
V1max  1000
1000
 1112V
 0.112 
 V1max
Q.5 consider the following statements regarding Distribution Transformer.
1. It operates at rated load in general.
2. Its magnetising flux m is smaller than the ordinary power transformer
of the same rating.
3. Its Iron losses are large as compared to the ordinary power transformer
of the same rating.
4. Its size is large as compared to the ordinary power trasformer of same
rating.
5. It gives maximum efficiency at rated load
The currect statements are
(a) 1, 2 and 3
(b) 3 and 4
(c) only 3
(d) 3, 4 and 5
Ans. (c)
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Test- Machine With Solution
Exp.
Distribution transformer operates opproximately 70% of its rated load
mostly
So
max at x  0.7
Pi  x 2PCF  Pi  0.49PCF
while ordinary power transformer operates mostly at its rated load
So
max at x  1
Pi  x 2PCF  Pi  PCF
So Iron loss in distribution transformer are comparatively smaller.
So peak flux density Bm in distrebution Transformer is designed to be small.
V
some for both.
f
If area of cross-section of core = A
m  Bm A
Magnetising flux m 
As Bm small so A is large in distribution transormer.
Common Data for Q.6 & Q.7
A 500 V/100 V, 10 kVA, T
wo winding transformer is to be used as an autotransformer to supply a 100V load from 600V source.
Q.6 Its kVA rating as an auto transformer.
(a) 12 kVA
(b) 8 kVA
(c) 60 kVA
(d) 48 kVA
Ans. (a)
Q.7 Its efficiency at rated load and 0.9 pf(lag) is 98% as a 2-winding transformer. Then efficiency as an auto transformer
(a) 96.8 %
(b) 99.6 %
(c) 99.2 %
(d) 98.3%
Ans. (d)
Exp.
10 kVA
voltages
500/100 V
currents
20 A/100 A
Auto transformer as 600V/100V
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Test- Machine With Solution
V1  600V , I1  20 A
V2  100V , I 2  120 A
Sauto  V1I1  V2I 2  12 kVA
At rated load x  1
Efficiency of Tw transformer
Cos

SCos  Pi  PCF
0.98 
10  0.9
10  0.9  Pi  PCF

Pi  PCF  0.184 kW
As current & voltages in each winding is same in both two-winding & auto
trans
former configuration the losses Pi  PCF same
auto 
Sauto Cos
Sauto Cos  Pi  PCF
12  0.9
12  0.9  0.184
 98.32%

Q.8
The value of load resistance for maximum power delivered to R L is
(a) 20 
Ans. (a)
(b) 5 
Exp. Transformation Ratio K 
(c) 12 
(d) 1.25 
N2
2
N1
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Test- Machine With Solution
load resistance refarred to primary
R
R
R1L  L2  L
4
K
For maximum power delivered
R1L  Z s  3  j 4  5
 RL  20
Q.9 A 1:2:4 three winding transformer is connected as shown in figure below
The reading of voltmeter
(a) 50 V
(b) 250 V
Ans. (d)
Exp.
(c) 150 V
(d) 350 V
OR
Using KVL
V0  50  100  200  0

V0  350V
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Test- Machine With Solution
Q.10 The efficiency of a 100 kVA transformer at upf is 98% at full as well as
half load. The maximum efficiency is obtained at a load
(a) 70.7 kVA
(b) 50 kVA
(c) 90.6 kVA
(d) 82.5 kVA
Ans. (a)
Exp.
x S Cos

x S Cos  Pi  x 2PCF
Rated So  100 kVA
  98%  0.98
pf Cos  1
At full load x  1
1  100  1
0.98 
1  100  1  Pi  12  PCF

Pi  PCF  2.04kW  1
At half full load x  0.5
0.5  100  1
0.98 
0.5  100  1  Pi  (0.5)2  PCF

Pi  0.25PCF  1.02  2
By solving (1) & (2)
Pi  0.68kW,
PCF  1.36kW
For max Pi  x 2PCF
x 
Pi
0.68

 0.707
PCF
1.36
S L  xS 0  0.707  100
 70.7 kVA
Q.11 A 4-pole Alternator has 72 slots carrying a 6-phase distributed winding.
Each coil has span from 1 to 10. The winding factor for 3rd harmonic is
1
1
cot150
cot150
(a)
(b)
(c)
(d)
0
3 cos15
6 sin150
3
6
Ans. (d)
Exp.
S  72, P  4, m  6
S
3
SPP q 
mp
P   4 
 


 10
S
72
18
3q 
3  3  10
Sin
2 
2

3
3  10
qSin
3Sin
2
2
Sin
K d3
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Test- Machine With Solution
1
3 2Sin15
Coil span from 1 to 16
i.e. coil span  16  1  15 slots

 15  150
Angle of chording   180  150  30
Pitch factor K P 3  Cos
3
3  30
 Cos
2
2
1
2
Winding factor

K w  Kd3 KP3 
1
6sin15
Common Data for Q.12 & Q.13
A 220V, 10 kW, DC shunt motor is operating at a speed of 1000 rpm when
taking 11A from supply. Its armature resistance is 2 and field resistance is
220.
Q.12 The torque developed by motor
(a) 20 Nm
(b) 23.2 Nm
(c) 19.1 Nm
(d) 18 Nm
Ans. (c)
Q.13 If armature terminals are reversed suddenly and a resistance of 40 is
connected in series with armature. The Braking Torque developed is
(a) 200 Nm
(b) 23.2 Nm
(c) 19.1 Nm
(d) 230.2 Nm
Ans. (c)
Exp.
Shunt Motor
supply V  220V , I  11A
V
220
Field current I f  R  220  1A
f
arm current I a  I  I f  10 A
Back emf E  V  I a Ra  220  10  2  200V
E  K n  N Where K n 
ZP
60 A
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Test- Machine With Solution
As I f  Constant    Constant
K n 
E
200

 0.2 V
rpm
N 1000
Toque developed T  K a  I a Where K a 
ZP
2 A
30
30
Kn Ia 
 0.2  10


 19.1Nm
If arm. terminals are interchanged
T 
Braking current
V E
220  200
Ib 

Ra  Rb
2  40
 10A
Braking torque developed
30
Tb  K a  I b 
Kn Ib

30
 0.2  10

 19.1Nm
Q.14 An Isolated Induction Generator without Capacitor Bank can supply the
power to
(a) 3   Induction motor at no load
(b) DC motor through 3   rectifier
(c) Overexcited synchronuous motor
(d) Under excited synchronuous motor
Ans. (c)
Exp.
Induction gen. can operate only at leading power factor. so load connected
should be of leading pf i.e. overexcited syn. motor.
Q.15 If the voltage Regulation of a DC generator is –95% approximately. It is
(a) DC Shunt Generator
(b) DC Series Generator
(c) Level Compound Generator (d) Differentially Compound Generator
Ans. (b)
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Test- Machine With Solution
Q.16 Match the lists
List-I
A. DC Shunt Generator
B. Level Compound Generator
C. Cumulative Compound Generator
D. Differential Compound Generator
A
B
C
D
(a)
3
4
1
2
(b)
2
3
4
1
(c)
3
1
4
2
(d)
4
3
2
1
Ans. (b)
1.
2.
3.
4.
List-II
–8%
12%
0%
4%
Q.17 A DC generator is supplying rated load at 230V. If the terminal voltage
observed on reducing the load to zero is still 230V then it is a
(a) Under Compound Generator
(b) Differential Compound Generator
(c) Series Generator with large number of field turns
(d) Level Compound Generator
Ans. (d)
Q.18 A DC generator is operating at constant speed and constant field excitation. If it is supplying maximum power at a voltage of 220 it’s generated emf is
(a) 110 V
(b) 220 V
(c) 440 V
(d) depends upon the armature resistance
Ans. (c)
Q.19 A DC series motor is driving a load which is propotional to the cube of
speed. The total resistance of motor is 2 and motor takes 5A and runs at
1000rpm, when operating from 200V supply. The resistance to be inserted in
series with armature to reduce the operating speed to 600rpm, is (Neglect
saturation).
(a) 10.23 
(b) 61.4 
(c) 5.2 
(d) 2.5 
Ans. (b)
Exp. DC series motor
Total resistance R  Ra  Rse  2
V  200V At I a  5 A, N  1000rpm
back emf E  V  I a R  200  5  2  190V
Flux by series field se  I a
 se  K se I a
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Test- Machine With Solution
E  K n se N  K n K se I a N
E
190
Constant K n K se  I N  5  1000
a
 0.038 V
A-rpm
Torque T  N 3
3

T
N
 3
T
N
Torque develope T  K ase I a  K a K se I a 2
or
T  Ia2

3
 I a 
 N
   
N 
 Ia 
2
2

At
3
 I a 
 600 
  I a  2.32A
  
5
 1000 
I a  2.32A
E   K n K se I a N 
Let
 0.038  2.32  600
 52.9V
Ra  Ra  Rext
E   V  I a Ra
52.9  200  2.33  Ra

Ra  63.4

Rext  61.4
Q.20 A 5 kW, 220 V, 4-pole, lap connected DC motor has 400 armature conductors. At full load, the flux per pole is 25 mWb and rotational losses are
200W. The armature resistance is 1  . The speed at full load is
(a) 1280 rpm
(b) 1350 rpm
(c) 1160 rpm
(d) 1400 rpm
Ans. (c)
Exp.
At full load, Net mechanical output =5 kW
Gross power o / p  5000  200  5200W
At full load current
E  V  I a Ra  220  I a  1
power output P0  EI a
5200   220  I a  I a
 I a 2  220I a  5200  0
I a  26.9A, 193.1A
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Test- Machine With Solution
So
I a  26.9A, (193.1A large so it not considerable
E  220  26.9  193.1V
 ZN  P 
E 
 
60  A 
Lap connected so number of paralled paths
A P 4
0.025  400  N  4 
193.1 
 
60
4

Speed N  1158.6rpm
Q.21 A DC shunt motor is driving constant power load at rated speed while
drawing rated armature current. If the terminal voltage is halved but field
current is kept constant, then (approximately).
(a) I a gets doubled & N is halved
(b) Both I a & N gets doubled
(c) I a gets doubled but N remains same
(d) I a gets halved & N gets doubled
Ans. (a)
Exp.
Power output P0  EI a constant
field current I f = const    constant
back emf E  V  I a Ra
Let
Ra  0  E  V
P  V I a  constant
V
V 
As
P  V I a  V I a
2

I a  2 I a
E
V
Speed N  K   K 
n
n
V
V
N
N 


K n  2K n  2
Common Data for Q.22 & Q.23
The stator of a 3   , 4-pole SRIM is connected to 50Hz supply. At the rotor
terminals a frequency of 30Hz is required.
Q.22 The possible speed (s) of rotor is (are)
(a) 600 rpm
(b) 2400 rpm
(c) 600 or 2400 rpm
(d) 600 & 900 rpm
Ans. (c)
Q.23 For a frequency of 30Hz across rotor terminals
(a) Two emfs of equal magnitudes and opposite phase sequence can be
obtained
(b) Two emfs of different magnitude and same sequence
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Test- Machine With Solution
(c) Only one possible emf
(d) The emf of any magnitude and sequence
Ans. (a)
Q.24 A 6-pole, 50Hz, 3   I.M. has a rotor resistance of 0.2 per phase and
maximum torque is obtained at a speed of 875rpm while full load speed is
960rpm. The resistance to be added in each rotor phase to obtain 80% of full
load troque at starting is
(a) 0.16 
(b) 0.05 
(c) 0.22 
(d) 0.192 
Ans. (d)
Exp.
120 f
Ns 
 1000rpm
P
speed for Tmax, N m  875rpm
So slip sm 
N s  Nm
 0.125
Ns
Full load speed N fl  960 rpm
s fl  0.04
if stator impedence is neglected
sm 

R2
 0.125
X2
0.2
 0.125
X2
 X 2  1.6
T 
3 E22R2 / s
 s  R2  2
2
 s   X 2
At full load & R20  0.2
3E 22
0.2 / 0.04
T fl 
 s  0.2  2
2

  1.6
0.04
 3E 2 
 0.18   2 
 s 
At starting s  1 & R2  R20  Rext
3E 22
R2
Tst 
 0.8T fl
 s R22  X 22
 3E22 
3E22
R2


 0.8  0.18 

s R22  1.62
 s 
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13
Test- Machine With Solution

R22
R2
 0.145
 2.56
 R22  6.9R2  2.56  0
R2  0.39, 6.5
Higher value 6.5 creates excess power loss
So it is not acceptable
So
R2  0.39
 Rext  0.19
Q.25 An alternator has a synchronuous reactance of 20% and neglegible resistance. The pf of rated load at which it gives zero voltage regulation is
(a) upf
(b) 0.1 lead
(c) 0.995 lead
(d) 0.99 lag
Ans. (c)
Exp.
Resistance Ra  0 , X s  20%  0.2 pu
Rated current = I a , Rated voltage = V
V
Base impedance Z B  I
a
Xs
pu reactance X s  Z
B
Xs Ia
I R
 0.2 , Ra ( pu )  a a
V
V
zero voltage regulation, i.e. E = V
At leading pf
X s( pu ) 
2
2
E 2  VCos  I a Ra   V sin   I a X s 
2
2

I a Ra  
Ia X s  
 V  Cos 
 Sin 

V  
V  

2
E 2  V 2  Cos  R pu


2
2
  Sin  X pu 

 1  Cos   Sin  X pu
2


2

 Cos2  Sin 2  2X puSin  X 2pu  1
X pu
0.2
 0.1
2
2
 Cos  0.995 leading
 Sin 

Q.26 An alternator has armature resistance of 2% and synchronuous reactance of 20%. The pf at which it gives voltage regulation of 10%.
(a) 0.956 lead
(b) 0.944 lag
(c) 0.823 lag
(d) 0.853 lead
Ans. (b)
Exp.
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14
Test- Machine With Solution
E V
 100  10
V
 E  1.1V
E  V , i.e. pf Cos lag
V .R. 
As
2
2
E 2  VCos  I a Ra   VSin  I a X s 
 E 2  V 2  Cos  R pu


2
  Sin  X spu 
2


1.12  Cos  0.022  Sin  0.22



1.21  Cos2  Sin 2  0.04Cos  0.0004 0.4Sin  0.04
0.4Sin  0.04Cos  0.1696
Cos  10Sin  4.24
1  RCos , 10  RSin

R  10.05 ,   84.3ο
R Cos Cos  Sin Sin   4.24

Cos      0.422
     65
   149.4 ,19.3
pf angle   90    19.3  Cos  0.944
Q.27 A lossless salient pole synchronuous motor has reluctance along q-axis
twice that of along d-axis and operating in such a manner as excitation voltage
is equal to the terminal voltage. The ratio of reluctance torque to the electromagnetic torque developed when it is delivering maximum load is
(a) 0.5
(b) 0.75
(c) 0.25
(d) 1
Ans. (a)
Exp.
The power o/p developed
EV
V  1
1 
P 
Sin  


 Sin 2
Xd
2  X q X d 
E  V , Rq  2Rd
N 2
N 2
X

,
X

d
Reactances q
Rq
Rd
 Xq 
P 
Xd
2
V2
V2
Sin  
Sin 2
Xd
2X d
P
Torque developed T  
s
T 
V2
V2
Sin  
Sin 2
s X d
2 s X d
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Test- Machine With Solution
electromagnetic torque Tem 
V
V2
Sin 
s X d
2
Reluctance torque Trel  2 X Sin 2
s d
Trel Sin 2
Ratue T  2Sin   Cos
em
dP
dT
 0 or
0
For maximum load
d
d
dP V 2

Cos  Cos 2   0
d
Xd
Cos2  Cos  0
 2Cos 2  Cos  1  0
1  1  4
Cos 
4
3 1 1


 As cos   ve 
4
2
Trel 1

Tem 2
Q.28 A 3   synchronuous motor, connected to   bus, is operating with
normal excitation with increase in load
1. Armature current increases
2. pf becomes lagging
3. pf becomes leading
4. load angle increases
5. Reactive power flows from Bus to motor
6. Reactive power flows from motor to Bus
from these correct statements are
(a) 1, 3, 4 & 6
(b) 2, 4, & 5
(c) 1, 3, & 6
(d) 1, 2, 4 & 5
Ans. (d)
Exp.
The V curres of syn. motor
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Test- Machine With Solution
At load P1 , I f 01 is nominal excitation (at upf) and current I a 01
If load is increased P2  P1
Current increases I a 02  I a 01
For P2, I f 02 is nominal. But I f in constant at I f 01 so for P2 , I f 01 gives under
excited syn. motor, i.e. at lag pf.
Syn. motor at lag pf consumes VAR from bus.
Q.29 A synchronuous generator fed from   bus is delivering half-full load. If
an increase in its field current causes a reduction in its armature current,
then generator is
(a) delivers active power to bus and absorbs reactive power from bus
(b) absorbs active power from bus and delivers reactive power to bus
(c) absorbs active and reactive power from bus
(d) delivers active and reactive power to bus
Ans. (a)
Exp.
Increase in field current causes a reduction in armature current so syn.
generator is operating in under excited condition, i.e. at leading pf .
Gen. delivering active power to the bus.
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Test- Machine With Solution
At lead pf , generator is supplying -ve VAR to bus hence it is consuming +ve
VAR from the bus.
Q.30 A 3   lossless cylindrical rotor synchronous motor is connected to
  bus, it is operating in under excited condition and at constant pf, if load
on the motor is increased, the excitation voltage variation is represented by
(a) Curve-1
Ans. (a)
Exp.
For syn. motor
(b) Curve-2
(c) Curve-3
(d) Curve-4
E  V  j I a Xs
At lag pf cos  i.e I a lags behind V by angle 
Power P  V I a cos
V & cos  Cosnstant so if load P increases I a increases and locus of phasor
E is given by straight line as shown.
E is decreasing, reaches a minimum, i.e E min and then increasing continuously.
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