Electrical Machines Test - 5 - PDF Flipbook
Electrical Machines Test - 5
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GATE
EEE
Electrical
Machines
Test-05Solutions
ELECTRICAL MACHINES
1. A balanced three-phase voltage is applied to a star-connected
induction motor, the phase to neutral voltage being V. The stator
resistance, rotor resistance referred to the stator, stator leakage
reactance, rotor leakage reactance referred to the stator, and the
magnetizing reactance are denoted by rs, rr, xs, xr and Xm
respectively. The magnitude of the starting current of the motor
is given by:
a) V
�(rs+rr)2+(xs+xr)2
b) V
�rs2+(rs+Xm)2
c) V
�(rs+rr)2+(Xm+xr)2
d) V
�rs2+(Xm+xr)2
Answer: (a)
Solution:
The approximate equivalent circuit of the induction motor per
phase when referred to stator is
1
The magnitude of starting current,
=
�( + )2+( + )2
�∵ during starting slip (s) = 1, ∴ rr �1s − 1� = 0�
2. In dc machine, the armature winding are placed on the (rotor
because of the necessity for
a) electromechanical energy conversion
b) generation of voltage
c) commutation
d) development of torque
Answer: (c)
3. In a three-phase induction motor with increase in load from light
to heavy
a) Both stator and rotor power factor increase.
b) Both stator and rotor power factor decrease.
c) Stator power factor increases while rotor power factor
decrease.
d) Stator power factor decreases while rotor power factor
increases.
Answer: (a)
Solution:
As load is increased, slip becomes smaller, so drop is 2 2 in
rotor decreases which further decreases angle between 2 and
s 2, power factor of rotor increases. Similarly power factor of
stator is increased.
2
4. Maximum temperature limit for class F insulation is
a) 1300
b) 1200
c) 1050
d) 1550
Answer: (d)
5. The copper loss of a certain transformer at half-fully load is
measured as 400 watts. The copper loss at full load will be
a) 800 watts
b) 200 watts
c) 400watts
d) 1600watts
Answer: (d)
6. What does the vertical line between open circuit characteristic
and zero power factor characteristic of a synchronous generator
represent?
a) Voltage drop due to synchronous reactance at rated current.
b) Voltage drop due to leakage reactance at rated current.
c) Induced voltage in armature due to resultant flux.
d) Induced voltage in armature due to field flux alone.
Answer: (b)
3
Solution:
7. Two d.c series motors are connected in series to produce a
torque T. Now if the motors are connected in parallel, the torque
produced will be
a) T/4
b) T/2
c) 2 T
d) 4 T
Answer: (a)
Solution:
For DC motors:
Torque, T = kϕIa
ϕ → flux and Ia → current
for series motor ϕ ∝ Ia
∴ T = k∙Ia∙Ia
Or T ∝ 2
4
For series connection:
Draw input power = V.I
TA = TB = kϕ. I = T
Or T = kI2 ……. (i)
For parallel connection:
Each motor will have the half value of current now (if the drawn
power is constant).
∴ Tparallel ∝ ϕ ∙ I (for each motor)
2
Or Tparallel ∝ I ∙ I (as ϕ ∝ I)
2 2
Or, Tparallel = 2 ……….(ii)
4
From equation (i) and (ii),
Tparallel =
4
5
8. The torque-speed characteristics of motor (TM) and load (TL) for
two cases are shown in the figures (a) and (b). The load torque is
equal to motor torque at points P, Q, R and S
The stable operating points are
a) P and R
b) P and S
c) Q and R
d) Q and S
Answer: (b)
Solution:
6
Equilibrium point P:
In a motor, developed torque rotates the rotor in its direction.
Let +ve developed torque rotate the rotor in the clockwise
direction. Then +ve speed is speed in the clockwise direction.
When operating at P, let the speed in anti-clockwise direction
momentarily increase, reaching the point P'. Then developed
torque (which tends to rotate the rotor in the clockwise
direction) increases, while load torque remains unchanged. So
speed in anti-clockwise direction decreases, and the operating
point tends to return to P. P is stable.
Equilibrium point Q:
An increase of anti-clockwise speed while operating at Q takes
the point of operation to Q'. Developed torque (which tends to
oppose anti clockwise rotation) decreases, and speed in anti-
clockwise direction further increases. Q is unstable.
Equilibrium point R:
While operating at R, let speed in clockwise direction increase.
Both load torque (which resists motion, and developed torque
7
(which causes motion) increase AB is the new load torque and
AC is new developed torque. There is a net developed torque,
and speed in clockwise direction increases further. R is
unstable.
Equilibrium point S:
While operating at S, let speed in clockwise direction decrease.
Both load toque and developed torque decreases, but there is a
net developed torque and speed in clockwise direction tends to
increase. S is stable.
9. T is the load torque of a dc series motor having linear
magnetization and negligible armature resistance speed of the
motor is
a) inversely proportional to √T
b) directly proportional to √T
c) inversely proportional to T2
d) directly proportional to T2
Answer: (a)
10. In an induction motor, what will happen if air-gap is
increased?
a) Speed will reduce
b) Efficiency will improve
c) p.f. will be lowered
d) Break-down torque will reduce
Answer: (c)
8
Solution:
To keep the flux constant for increased air gap, more
magnetizing current would be required.
Flux, ϕ = mmf
reluctance
mmf ∝ current,
reluctance ∝ airgap
to keep flux (ϕ) constant, mmf i.e., current has to increase. This
in turn reduce no load power factor which consequentially leads
to reduce full load p.f.
11. For battery charging, which of the following DC generators is
used?
a) DC series generation
b) DC shunt generator
c) Short shunt compound generator
d) Long shunt compound generator
Answer: (b)
Solution:
An application of DC shunt generators is to charge the batteries.
12. An under-excited synchronous motor is operating at no load on
an infinite bus of voltage 1.0 p.u. and drawing current of 0.8 p.u.
Its synchronous reactance is 0.5 p.u. What is the p.u. value of its
excitation emf?
a) 1.4
b) 1.0
c) 0.6
9
d) 0.4
Answer: (c)
Solution:
Motor is at no load, so Ef (excitation emf) and VBUS are in phase
and torque angle δ = 0.
The drop in Xs is demagnetizing
� ⃗ = � ⃗ − ⃗ � ⃗ = ∠ − 900�
= 1 .
⃗ = (0.8)(0.5) = 0.4 .
� � ⃗ � = 1 − 0.4 = 0.6 .
13. A single phase transformer has specification as 250 kVA,
11000 V/415 V, 50 Hz. What are the approximate value of
primary and secondary currents?
a) Primary = 22.7 A, Secondary = 602 A
b) Primary = 602 A, Secondary = 22.7 A
c) Primary = 301A, Secondary = 22.7A
d) Primary = 11.4 A, Secondary = 301 A
Answer: (a)
10
14. In Figure, the characteristic that corresponds to the variation of
synchronous reactance of synchronous motor with field current
is
a) Curve I
b) Curve II
c) Curve III
d) Curve IV
Answer: (b)
Solution:
Zs ≃ Xs = EOC
ISC
If = constant
Upto knee point both OCC & SCC are linear.
∴ Zs is constant for unsaturated position. But above knee OCC is
non-linear & SCC is linear. So Zs decreases during saturated
condition.
11
15. In which type of motor Rheostat dynamic braking is very
effective
a) Cumulatively compounded dc motor.
b) dc series motor.
c) dc shunt motor.
d) Separately excited dc motor.
Answer: (d)
Solution:
Direction of the field separately excited can be easily changed
by changing terminals.
16. For a transformer to be tested at full load conditions but
consuming only losses from the mains, we do
a) load fest
b) open-circuit and short-circuit tests
c) back to back test
d) none of these
Answer: (c)
17. Vacuum cleaners are
a) Permanent magnet f.h.p. motors.
b) Single-phase capacitor motor with direct-online start.
c) Reluctance motor.
d) Universal motor.
Answer: (d)
12
Solution:
Universal motor are dc/ac series motor which provide high
starting torque, can run at high speed and are light weight and
compact.
18. Air gap at the pole tips of a dc machine is kept more than that
at the centre of the pole mainly to reduce
a) reactance voltage
b) effect of armature reaction
c) losses of armature core
d) noise of the machine
Answer: (b)
19. The voltage at the terminals of a dc series generator running at
rated rpm and no load will be
a) more than the rated voltage
b) full rated voltage
c) a very small voltage
d) zero
Answer: (c)
20. The field current of a synchronous motor is increased while its
load is constant. How will its power angle and power factor
change?
a) Power angle decreases and power factor improves
b) Power angle remains same throughout but Power factor
improves
13
c) Power angle increases while its power factor gradually
decreases
d) Power angle and power factor both increase
Answer: (b)
Solution:
Power factor changes with change in excitation whereas power
angle changes with change in active power or steam supply
input.
21. The iron loss in a 100 kVA transformer is 1 kW and full load
copper losses are 2 kW. The maximum efficiency occurs at a
load of
a) 100 kVA
b) 70.7 kVA
c) 141.4 kVA
d) 50 kVA
Answer: (b)
Solution:
Load for maximum efficiency,
�Sη�max = Sfull load�PcPuifl
= 100kVA�21 = 70.7 kVA
22. The purpose of providing dummy coils in a wave would dc
machine armature is to
a) enhance armature
b) amplify voltage
14
c) mechanical balance for the rotor
d) reduce eddy current loss
Answer: (c)
23. In an ideal transformer, the no-load primary current I0
a) lags behind V1 by slightly less than 900
b) is in phase with V1
c) lead V1 by 90°
d) lags V1 by an angle lying between 00 and 900
Answer: (a)
24. Consider the following statements:
A ceiling fan fails to start because
1. Main winding is open-circuited.
2. Auxiliary winding is open-circuited.
3. Capacitor is short-circuited.
4. Supply terminals are reversed.
5. Main and auxiliary winding connections are interchanged.
Which of these statements are correct?
a) 1, 2 and 5
b) 2, 3, 4 and 5
c) 1, 2 and 3
d) 1, 3 and 4
Answer: (c)
15
Solution:
When supply terminal are changed or main and auxiliary
windings connections are interchanged, motor will not be failed
to start. In other three cases motor will not start either due to
absence of main or auxiliary winding.
25. At a certain speed and flux, the voltage generated by a dc
generator is 230 volts. If the speed is increased by 20 % and the
flux is simultaneously reduced by 10 %, then the voltage will be
a) increased by 10 %
b) reduced by 20 %
c) increased by 8 %
d) decreased by 8 %
Answer: (c)
26. Which of the following information is obtained by Potier's
method using open circuit characteristic and zero power factor
characteristic?
a) Field current equivalent of armature reaction
b) Synchronous reactance
c) Leakage reactance
d) Leakage reactance and field current equivalent of armature
reaction.
Answer: (d)
16
Solution:
27. Changing the electrical connectors to reverse direction of
rotation of a motor running at full speed is failed
a) slugging
b) plugging
c) dynamic braking
d) brush shifting
Answer: (b)
28. Hysteresis losses are present in iron core coil when
a) the current in the coil is alternating
b) the current is unsymmetrical alternating only
c) the current in the coil is d.c only
d) the current in the coil is sinusoidal only
Answer: (a)
Solution:
When current is AC then magnetization and demagnetization
process happens which causes hysteresis losses in the iron core.
17
29. To calculate the full-load voltage regulation of a synchronous
generator at any power factor using mmf method, it is required
to have
a) Open-circuit and short-circuit characteristics.
b) Open-circuit and zero power factor characteristics.
c) Open-circuit characteristics, field current at rated short-circuit
current and armature resistance.
d) Short-circuit characteristics, zero power factor characteristics
and armature resistance.
Answer: (c)
Solution:
• Open circuit test, short circuit test to find field current
required to circulate rated armature current.
• Resistance of stator winding per phase.
30. When the speed becomes more than the synchronous speed
during hunting, the damper bars develop
a) synchronous motor torque
b) single-phase induction motor torque
c) induction motor torque
d) induction generator torque
Answer: (d)
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