# Electric Drives Questions and Answers – Induction Motors – Basics Principles of Speed Control

This set of Electric Drives Multiple Choice Questions & Answers (MCQs) focuses on “Induction Motors – Basics Principles of Speed Control”.

1. In the chopper circuit, commutation times are _______
a) Current commutation is equal to voltage commutation
b) Current commutation is less as compared to that of voltage commutation
c) Current commutation is more as compared to that of voltage commutation
d) Both commutation techniques are not comparable

Explanation: In current commutation, commutation time=CVr÷Io. In voltage commutation, commutation time= CVs÷Io. Hence, the commutation time of the current commutation is less as compared to voltage commutation.

2. The generated e.m.f from 4-pole armature having 1 conductors driven at 1 rev/sec having flux per pole as 10 Wb, with wave winding is ___________
a) 30 V
b) 40 V
c) 70 V
d) 20 V

Explanation: The generated e.m.f can be calculated using the formula Eb = Φ×Z×N×P÷60×A, Φ represent flux per pole, Z represents the total number of conductors, P represents the number of poles, A represents the number of parallel paths, N represents speed in rpm. In wave winding number of parallel paths are 2. Eb = 10×4×1×60÷60×2 = 20 V.

3. The unit of voltage is Pascal.
a) True
b) False

Explanation: The voltage is equal to one volt when 1 A of current flows through 1 Ω resistor. It is mathematically represented as I×R. It is expressed in terms of a volt(V).

4. Calculate the moment of inertia of the sphere having a mass of 8.4 kg and radius of 61 cm.
a) 3.124 kgm2
b) 3.125 kgm2
c) 4.545 kgm2
d) 5.552 kgm2

Explanation: The moment of inertia of the egg can be calculated using the formula I=Σmiri2. The mass of egg and radius is given. I=(8.4)×(.61)2=3.125 kgm2. It depends upon the orientation of the rotational axis.

5. The most suitable control-motor application is __________
a) AC shunt motor
b) DC separately motor
c) AC one-phase induction motor
d) DC shunt motor

Explanation: DC separately motor has definite full-load speed, so they don’t ‘run away’ when the load is suddenly thrown off provided the field circuit remains closed. The speed for any load within the operating range of the motor can be readily obtained.
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6. In a DC series motor, the e.m.f developed is proportional to _______
a) N×Ia
b) N×Ia2
c) N×Ia3
d) N×Ia.5

Explanation: In a DC series motor, the e.m.f developed is equal to KmΦN. In a DC series, the motor field winding is connected in series with the armature so the flux in the field winding is proportional to current. Eb = KmΦN α Ia×N.

7. Calculate the value of the time period if the frequency of the signal is .07 sec.
a) 14.28 sec
b) 14.31 sec
c) 14.23 sec
d) 14.78 sec

Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T = 1÷F=1÷.07=14.28 sec.

8. The slope of the V-I curve is 13.89°. Calculate the value of resistance.
a) .247 Ω
b) .345 Ω
c) .231 Ω
d) .222 Ω

Explanation: The slope of the V-I curve is resistance. The slope given is 13.89° so R=tan(13.89°)=.247 Ω. It behaves like a normal resistor.

9. In a DC shunt motor, the e.m.f developed is proportional to ___________
a) Ia
b) Ia2
c) Ia3
d) Iao

Explanation: In a DC shunt motor, the e.m.f developed is equal to KmΦN. In a DC shunt, the motor field windings are connected separately and excited by a constant DC voltage. E = KmΦN α Ia°.

10. Calculate the power factor angle during the resonance condition.
a) 0°
b) 10°
c) 80°
d) 90°

Explanation: During the resonance condition, the reactive power generated by the capacitor is completely absorbed by the inductor. Only active power flows in the circuit. Net reactive power is equal to zero and Φ=0°.

11. Calculate the value of the duty cycle if the system is on for 5 sec and off for inf sec.
a) 0
b) .4
c) .2
d) .1

Explanation: Duty cycle is Ton÷Ttotal. It is the ratio of time for which the system is active and the time taken by the signal to complete one cycle. D = Ton÷Ttotal=5÷inf=0.

12. Calculate the value of the frequency of the AC supply in India.
a) 0 Hz
b) 50 Hz
c) 49 Hz
d) 60 Hz

Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. AC supply magnitude is variable. It changes with time so the frequency of AC supply is 50 Hz.

13. DC series motor cannot run under no load.
a) True
b) False

Explanation: DC series motor cannot be run under no load condition because at no-load speed of the motor is very which can damage the shaft of the motor. There should be some load that should be connected to it.

14. Calculate the value of the frequency if the time period of the signal is .2 sec.
a) 5 Hz
b) 4 Hz
c) 2 Hz
d) 3 Hz

Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. It is expressed in Hz. F = 1÷T = 1÷.2 = 5 Hz.

Sanfoundry Global Education & Learning Series – Electric Drives.

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