Electric Drives Questions and Answers – DC Motors – Three Phase Induction Motors

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This set of Electric Drives Multiple Choice Questions & Answers (MCQs) focuses on “DC Motors – Three Phase Induction Motors”.

1. A three-phase slip ring induction motor is fed from the rotor side with the stator winding short-circuited. The frequency of the current flowing in the short-circuited stator is ____________
a) Slip frequency
b) Supply frequency
c) The frequency corresponding to rotor speed
d) Zero
View Answer

Answer: a
Explanation: The relative speed between rotor magnetic field and stator conductors is sip speed and hence the frequency of induced e.m.f is equal to slip frequency.
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2. An 8-pole, 3-phase, 50 Hz induction motor is operating at a speed of 720 rpm. The frequency of the rotor current of the motor in Hz is __________
a) 2
b) 4
c) 3
d) 1
View Answer

Answer: a
Explanation: Given a number of poles = 8. Supply frequency is 50 Hz. Rotor speed is 720 rpm. Ns = 120×f÷P=120×50÷8 = 750 rpm. S=Ns-Nr÷Ns = 750 – 720÷750 = .04. F2=sf=.04×50=2 Hz.

3. Calculate the phase angle of the sinusoidal waveform z(t)=78sin(456πt+2π÷78).
a) π÷39
b) 2π÷5
c) π÷74
d) 2π÷4
View Answer

Answer: a
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, α represents a phase difference.

4. Calculate the moment of inertia of the disc having a mass of 54 kg and diameter of 91 cm.
a) 5.512 kgm2
b) 5.589 kgm2
c) 5.487 kgm2
d) 5.018 kgm2
View Answer

Answer: b
Explanation: The moment of inertia of the disc can be calculated using the formula I=mr2×.5. The mass of the disc and diameter is given. I=(54)×.5×(.455)2=5.589 kgm2. It depends upon the orientation of the rotational axis.

5. Calculate the moment of inertia of the thin spherical shell having a mass of 73 kg and diameter of 36 cm.
a) 1.56 kgm2
b) 1.47 kgm2
c) 1.38 kgm2
d) 1.48 kgm2
View Answer

Answer: a
Explanation: The moment of inertia of the thin spherical shell can be calculated using the formula I=mr2×.66. The mass of the thin spherical shell and diameter is given. I=(73)×.66×(.18)2=1.56 kgm2. It depends upon the orientation of the rotational axis.
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6. A 50 Hz, 4poles, a single phase induction motor is rotating in the clockwise direction at a speed of 1425 rpm. The slip of motor in the direction of rotation & opposite direction of the motor will be respectively.
a) 0.05, 0.95
b) 0.04, 1.96
c) 0.05, 1.95
d) 0.05, 0.02
View Answer

Answer: c
Explanation: Synchronous speed, Ns=120×50÷4=1500 rpm. Given a number of poles = 4. Supply frequency is 50 Hz. Rotor speed is 1425 rpm. S=Ns-Nr÷Ns=1500-1425÷1500=.05. Sb=2-s=1.95.

7. The frame of an induction motor is made of _________
a) Aluminum
b) Silicon steel
c) Cast iron
d) Stainless steel
View Answer

Answer: c
Explanation: The frame of an induction motor is made of cast iron. The power factor of an induction motor depends upon the air gap between stator and rotor.

8. The slope of the V-I curve is 5°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .3254 Ω
b) .3608 Ω
c) .3543 Ω
d) .3443 Ω
View Answer

Answer: d
Explanation: The slope of the V-I curve is resistance. The slope given is 5° so R=tan(5°)=.3443 ω. The slope of the I-V curve is reciprocal of resistance.

9. In an induction motor, when the number of stator slots is equal to an integral number of rotor slots _________
a) There may be a discontinuity in torque slip characteristics
b) A high starting torque will be available
c) The maximum torque will be high
d) The machine may fail to start
View Answer

Answer: d
Explanation: When the number of stator slots is an integral multiple of a number of rotor slots the machine fails to start and this phenomenon is called cogging.
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10. A 3-phase induction motor runs at almost 1000 rpm at no load and 950 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?
a) 30 revolution per minute
b) 40 revolution per minute
c) 60 revolution per minute
d) 50 revolution per minute
View Answer

Answer: d
Explanation: Supply frequency=50 Hz. No-load speed of motor = 1000 rpm. The full load speed of motor=950 rpm. Since the no-load speed of the motor is almost 1000 rpm, hence synchronous speed near to 1000 rpm. Speed of rotor field=1000 rpm. Speed of rotor field with respect to rotor = 1000-950 = 50 rpm.

11. Calculate the active power in a 487 H inductor.
a) 2482 W
b) 1545 W
c) 4565 W
d) 0 W
View Answer

Answer: d
Explanation: The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in case of the inductor so the angle between V & I is 90°. P = VIcos90 = 0 W.

12. Calculate the active power in a 788 ω resistor with 178 A current flowing through it.
a) 24.96 MW
b) 24.44 MW
c) 24.12 MW
d) 26.18 MW
View Answer

Answer: a
Explanation: The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in case of the resistor so the angle between V & I is 90°. P=I2R=178×178×788=24.96 MW.

Sanfoundry Global Education & Learning Series – Electric Drives.

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To practice all areas of Electric Drives, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn