Electrical Machines Test - 3 - PDF Flipbook
Electrical Machines Test - 3
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Electrical
Machines
Test-03Solutions
ELECTRICAL MACHINES
1. Match the following
List - I (Test)
A. No load and Blocked rotor
B. Sumpner's test
C. Swinburne’s test
List-II
1. Transformer
2. Induction machine
3. Synchronous machine
4. DC machine
Codes:
ABC
a) 2 1 4
b) 1 2 3
c) 3 2 1
d) 4 1 2
Answer: (a)
2. Keeping in view the requirement of parallel operation, which of
the 3-phase connections given below are possible?
a) delta-delta to delta-star
b) delta-delta to star-delta
c) star- star to delta-star
d) delta-star to star-delta
Answer: (d)
1
Solution:
∆ − & − ∆ transformers both belong to ±300 phasor group
hence they can be used for parallel operation.
3. Which one of the following statements is correct?
In a 3-phase induction motor, the torque developed is maximum
when the rotor circuit resistance per phase is equal to
a) rotor leakage reactance per phase at standstill
b) slip times the rotor leakage reactance per phase at standstill
c) stator resistance per phase
d) stator leakage reactance per phase
Answer: (b)
Solution:
At maximum torque, s = 2
2
4. A 6-pole, 3-phase alternator running at 1000 rpm supplies to an
8-pole 3-phase induction motor which has a rotor current of
frequency 2 Hz. The speed at which the motor operates is
a) 1000 rpm
b) 960 rpm
c) 750 rpm
d) 720 rpm
Answer: (d)
Solution:
Supply frequency f = 1000×6 = 50
120
as sf. i.e., slip frequency = 2 Hz
2
⇒ s = 0.04
∴ = 120×5 = 750 rpm
8
∴ = (1 − )
= 750(1 − 0.04)
= 720 rpm
5. A three-winding transformer is connected to an AC voltage
source as shown in the figure. The number of turns are as
follows: N1 = 100, N2 = 50, N3 = 50. If the magnetizing current
is neglected, and the currents in two windings are 2̅ = 2∠300
and 3̅ = 2∠1500 , then what is the value of the current 1̅ , in
amperes?
a) 1∠900
b) 1∠2700
c) 4∠900
d) 4∠2700
Answer: (a)
Solution:
1̅ = 2∠300 50 + 2∠1500 50
100 100
= 1∠300 + 1∠1500 = 1∠900
3
6. A squirrel-cage induction motor having a rated slip of 4% on
full load has a starting torque same as the full load torque.
Which one of the following statements is correct?
The starting current is
a) equal to the full load current
b) twice the full load current
c) four times the full load current
d) five times the full load current
Answer: (d)
Solution:
∝ 2 ⇒ = � � 2 = 1
⇒ = 5
7. In the measurement of Xd, Xq (in ohms), following data are
obtained by the slip test on a salient pole machine:
Idmax = 10 A Idmin = 6.5 A
Vdmax = 30 V Vdmin = 25 V
Which one of the following is correct?
a) Xd = 3, Xq = 3.86
b) Xd = 4.615, Xq = 2.5
c) Xd = 3, Xq = 2.5
d) Xd = 4.61, Xq = 3.86
Answer: (b)
4
Solution:
= 30 = 4.615
6.5
= 25 = 2.5
10
8. In a single phase induction motor driving a fan load, the reason
for having a high resistance rotor is to achieve
a) low starting torque
b) quick acceleration
c) high efficiency
d) reduced size
Answer: (b)
Solution:
Starting torque ∝ rotor resistance
9. A 1 kVA, 230V/100V single phase 50Hz transformer having
negligible winding resistance & leakage inductance is operating
under saturation, while 250V, 50 Hz sinusoidal supply is
connected to the high voltage winding. A resistive load is
connected to the low voltage winding which draws rated current.
Which one of the following quantities will not be sinusoidal?
a) Voltage induced across the low voltage winding.
b) Core flux
c) Load current
d) current drawn from the source
Answer: (d)
5
Solution:
In a single phase transformer the induced emf is always
sinusoidal under loaded condition, because of sinusoidal flux.
To produce sinusoidal flux during saturation condition of
transformer, the current drawn from supply should be a peaky
wave.
10. A small 3-phase induction motor has a short-circuit current 5
times of full-load current and full-load slip 5%. If starting
resistance starter is used to reduce the impressed voltage to 60%
of normal voltage, the starting torque obtained in terms of full
load torque would be
a) 30 %
b) 45 %
c) 55 %
d) 80 %
Answer: (b)
Solution:
= � �2 ×
= (0.6)2 × (5)2 × (0.05)
= 0.45
or, = 45%
6
11. Consider the following statements:
Assertion (A): Large synchronous machines are constructed
with armature winding on stator.
Reason (R): Stationary armature winding would have reduced
armature reactance.
Of these statements
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true but R is not the correct explanation of
A.
c) A is true but R is false.
d) A is false but R is true.
Answer: (c)
Solution:
Reason given is not correct. Armature reactance has nothing to
do with stationary or rotating. Large synchronous machines are
constructed with armature winding on stator so as the ease of
evacuation of higher power, ease of cooling etc.
12. A hydraulic turbine having rated speed of 250 rpm is
connected to a synchronous generator. In order to produce
power at 50 Hz, the numbers of poles required in the generator
are
a) 6
b) 12
c) 16
d) 24
7
Answer: (d)
Solution:
= 120 ⇒ = 120
= 120×50 = 24
250
13. Which of the following parameters in an induction motor
influences the magnetizing reactance to the maximum extent?
a) Axial length of the rotor stack
b) Axial length of the stator stack
c) Radial length of air gap
d) Number of slots on the stator
Answer: (c)
Solution:
Radial length of air gap influences the reluctance
(∝1/inductance) of the magnetic circuit.
14. What does the SCR (short circuit ratio) of a synchronous
machine yield?
a) 1
XS(unsaturated)p.u
b) 1 in Ohm
XS(unsaturated)
c) 1
XS(adjusted)p.u
d) 1 in Ohm
XS(adjusted)
Answer: (c)
8
Solution:
By definition SCR = (For Vt rated on O.C)If1
(For Ia rated on S.C)If2
= Xs 1 in p.u
adjusted
Using O.C.C. and S.C.C
If1 = Vt rated ; If2 = Ia rated
k1 k2
where, k1 and k2 are constants
Xs adjusted is for If and Vt
Xs unsaturated is for If = const. and
Xs adjusted < Xs unsaturated
15. When a 3-phase alternator is suddenly short-circuited at its
terminals, the initial value of the short-circuit current is limited
by which one of the following
a) Sub-transient reactance xd”
b) Transient reactance xd’
c) Synchronous reactance xs
d) Sum of xd”, xd’ and xs
Answer: (a)
Solution:
With both field circuit and damper winding present, for sudden
armature circuit changes, the reactance viewed from the
armature terminals is the d-axis sub-transient reactance " .
9
16. It is desired to measure parameters of 230V/115 V, 2 kVA,
single-phase transformer. The following watt-meters are
available in a laboratory:
W1: 250 V, 10 A, Low Power Factor
W2: 250 V, 5 A, Low Power Factor
W3: 150 V, 10 A, High Power Factor
W4: 150 V, 5A, High Power Factor
The watt-meters used in open circuit test and short circuit test of
the transformer will respectively be
a) W1 and W2
b) W2 and W4
c) W1 and W4
d) W2 and W3
Answer: (d)
Solution:
For open circuit test [on LV side]: The current is of 4 to 8% of
full load current ≅ 1.4 A
And is of Low power factor ≅ 0.2
.'. W2 is required
For short circuit test [on HV side]:
The current is full load current ≅ 8.7A
The power factor is high ≅ 0.5 to 0.6
.'. W3 is required
10
17. Match List-I (Machines) with List-II (Tests) and select the
correct answer:
List-I List-II
A. Transformer 1. Slip test
B. DC Motor 2. Blocked Rotor Test
C. Alternator 3. Sumpner’s Test
D. Induction Motor 4. Swinburne’s Test
Codes:
ABC D
a) 3 4 1 2
b) 4 3 2 1
c) 3 4 2 1
d) 4 3 1 2
Answer: (a)
Data for Question (Q.18):
A synchronous motor is connected to an infinite bus at 1.0 p. u
voltage and draws 0.6 pu at unity power factor. Its synchronous
reactance is 1.0 p.u and resistance is negligible.
18. Keeping the excitation voltage same, the load on the motor is
increased such that the motor current increases by 20%. The
operating power factor will become
a) 0.995 lagging
b) 0.995 leading
c) 0.791 lagging
d) 0.848 leading
11
Answer: (a)
Solution:
|IaXs|2 = E2 + V2 − 2EV cos δ
Ia = 1.2 × 0.6
= 0.72 p. u
(0.72 × 1)2 = 12 + (1.1662)2 − 2 × 1 × 1.1662 cos δ
cos δ = 0.789, δ = 37.850
Ia = V∠00−E∠−δ
jXs
= 1∠00−1.1662∠−37.850
1∠900
Ia = 0.712∠−6.310
∴ New power factor = cos 6.31
= 0.994 lag
19. An 8-pole single phase induction motor is running at 690 rpm.
What is its slip with respect to forward and backward fields,
respectively?
a) 0.08, 2.0
b) 0.08, 1.92
c) 1.92, 0.08
d) 2.0, 0.08
Answer: (b)
Solution:
Forward slip Sf = 750−690 = 60 = 8%
750 750
Backward slip Sb = 2 – Sf = 1.98 p.u.
12
20. In a split phase motor, the running winding should have
a) high resistance and low inductance
b) high resistance as well as high inductance
c) low resistance and high inductance
d) low resistance as well as low inductance
Answer: (c)
21. The series winding of a cumulatively compounded dc motor is
short-circuited while driving a load at rated torque. This results
in
a) Reduction in both the armature current and motor speed.
b) Increase in the armature current and reduction in the motor
speed.
c) Increase in both the armature current as well as the motor
speed.
d) Reduction in the armature current and increase in the motor
speed.
Answer: (c)
Solution:
The armature current, Ia = Te
ka(∅sh+∅se)
Rated torque = . If series winding is short-circuit
Flux, ∅es = 0
Now, armature current is Ia1 = Te > Ia. Hence, armature
ka∅sh
current increases
For ∅ = 0, motor will operate as a shunt motor
13
22. Assertion (A): A synchronous motor operating from constant
voltage and constant frequency source has a substantially
constant resultant air-gap flux.
Reason (R): If the d.c field current in this motor can set up the
required resultant air-gap flux, the lagging reactive volt-amperes
drawn from a.c source is zero.
a) Both A and R are true and R is the correct explanation of A
b) Both A and R are true but R is NOT the correct explanation
of A
c) A is true but R is false
d) A is false but R is true
Answer: (b)
Solution:
Refer V-curves of synchronous motor. The reactive current-
drawn depends upon the value of field current i.e. under
excitation or over excitation.
23. A 3-phase squirrel-cage induction motor is started by means of
a star/delta switch. What is the starting current of the motor?
a) 3 times the current with direct on line starting
b) 1/3 times the current with direct on line starting
c) 1/√3 times the current with direct on line starting
d) √3 times the current with direct on line starting
Answer: (b)
14
Solution:
For comparison starting current of motor generally taken on line
side not motor side. If it is considered on motor side option (c)
i.e. √3 times the current with DL will be correct
24. For a lossless two winding single-phase transformer on NO-
LOAD, the following four phasor diagrams A, B, C and D are
considered. Standard symbols are being used to represent the
voltage, current and flux
Which is the correct phasor-diagram?
15
a) A
b) B
c) C
d) D
Answer: (d)
Solution:
• Magnetizing current (Im) lags the applied voltage (V1).
• Voltage reduced in primary (E1) is equal and opposite to V1.
• E2 = V2 (emf induced in secondary).
25. Which one of the following statements is not correct in respect
of synchronous machines?
a) In salient pole machines, the direct-axis synchronous
reactance is greater than the quadrature axis synchronous
reactance.
b) The damper bars help the motor to self-start.
c) Short circuit ratio is the ratio of field current required to
produce the rated voltage on open circuit to the rated
armature current.
d) The V-curve of a synchronous motor represents the variation
in the armature current with field excitation at a given output
power.
Answer: (c)
16
Solution:
SCR (short circuit ratio) is defined as the ratio of field current
required to generate rated voltage on open circuit, to the field
current required to circulate rated armature current on three-
phase short circuit.
26. What is the rotor copper loss of a 3-phase 550 volt, 50 Hz, 6
poles induction motor developing 4.1 kW at the shaft with
mechanical loss of 750 W at 970 rpm?
a) 175 W
b) 150 W
c) 100 W
d) 250 W
Answer: (b)
Solution:
Rotor copper loss, = �1− � × Mechanical power developed
= 120× = 120×50 = 1000
6
= − = 1000−970 = 0.03
1000
∴ Rotor copper loss= 0.03 × (4100 + 750) = 150
1−0.03
27. Two transformers with identical voltage ratings are working in
parallel to supply a common load. The percentage impedance of
one transformer is higher compared to that of the other. The load
sharing between the two transformers will
a) be proportionate to their percentage impedances.
b) be independent of their percentage impedances.
17
c) be inversely proportional to their respective impedances.
d) depend on the resistance to leakage reactance ratio of each
transformer.
Answer: (c)
Solution:
kVA shared ∝ leakage 1
impedance
28. The rated slip of an induction motor at full-load is 5% while
the ratio of starting current to full load current is four. The ratio
of the starting torque to full load torque would be
a) 0.6
b) 0.8
c) 1.0
d) 1.1
Answer: (b)
Solution:
= 5%, = 4
. = � � 2 .
.
= (4)2 × 0.05
= 0.8
18
29. Assertion (A): AC armature windings are short chorded by
selecting value of coil span more than the pole pitch.
Reason (R): Short chording is done to eliminate harmonics in
the induced e.m.f.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and B are true but R is NOT the correct explanation
of A.
c) A is true but R is false.
d) A is false but R is true.
Answer: (d)
Solution:
Short chorded means coil span less than pole pitch hence
assertion itself is not correct.
30. Consider the following statements:
The armature torque in a dc motor is a function of
1. Field flux
2. Armature current
3. Speed
4. Damping
Which of these statements is/are correct?
a) 1, 2, 3 and 4
b) 3 and 4 only
c) 1 and 2 only
d) 4 only
Answer: (c)
19
Solution:
T = K∅Ia
f = flux
Ia = armature current
20
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