This set of Electric Drives Multiple Choice Questions & Answers (MCQs) focuses on “Speed Control of Series Motor”.

1. A particular current is made up of two components: a 10 A and a sine wave of peak value 14.14 A. The average value of current is _________

a) Zero

b) 24.14 A

c) 10 A

d) 14.14 A

View Answer

Explanation: Average value of DC current is 10 A. Average value of AC current is 0 A as it is alternating in nature. The average value of current is 10+0 = 10 A.

2. A 38-pole, 3-phase, 80 Hz induction motor is operating at a speed of 12 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 75.2

b) 76.1

c) 79.2

d) 79.6

View Answer

Explanation: Given a number of poles = 8. Supply frequency is 50 Hz. Rotor speed is 720 rpm. N

_{s}=120×f÷P=120×80÷38 = 252.63 rpm. S=N

_{s}-N

_{r}÷N

_{s}= 252.63-12÷252.63 = .952. F

_{2}= sf = .952×80 = 76.1 Hz.

3. Calculate the phase angle of the sinusoidal waveform i(t)=sin(.6πt+π÷.88).

a) 100π÷88

b) 100π÷8

c) 100π÷8

d) π÷88

View Answer

Explanation: Sinusoidal waveform is generally expressed in the form of V=V

_{m}sin(ωt+α) where V

_{m}represents peak value, ω represents angular frequency, α represents a phase difference.

4. Calculate the moment of inertia of the satellite having a mass of 79 kg and diameter of 83 cm.

a) 13.65 kgm^{2}

b) 13.60 kgm^{2}

c) 12.67 kgm^{2}

d) 13.82 kgm^{2}

View Answer

Explanation: The moment of inertia of the satellite can be calculated using the formula I=mr

^{2}. The mass of the satellite and diameter is given. I=(79)×(.415)

^{2}=13.60 kgm

^{2}. It depends upon the orientation of the rotational axis.

5. A particular voltage is made up of two components: a 5 A and a cosine wave of peak value 7.8 A. The average value of voltage is _________

a) Zero

b) 35.14 A

c) 78 A

d) 5 A

View Answer

Explanation: Average value of DC voltage is 5 A. Average value of AC current is 0 A as it is alternating in nature. The average value of current is 5+0 = 5 A.

6. Armature reaction is demagnetizing in nature due to a purely capacitive load in the synchronous generator.

a) True

b) False

View Answer

Explanation: Due to a purely capacitive load, armature current is in phase with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in phase with the field flux. It will try to increase the net magnetic field.

7. Armature reaction is magnetizing in nature due to a purely resistive load in the synchronous generator.

a) True

b) False

View Answer

Explanation: Due to a purely resistive load, armature current is in quadrature with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in quadrature with the field flux. It will try to increase the net magnetic field.

8. The slope of the V-I curve is 4.9°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .03 Ω

b) .08 Ω

c) .04 Ω

d) .07 Ω

View Answer

Explanation: The slope of the V-I curve is resistance. The slope given is 4.9° so R=tan(4.9°)=.08 Ω. The slope of the I-V curve is reciprocal of resistance.

9. Calculate the velocity of the satellite if the angular speed is 87 rad/s and radius is 7.4 m.

a) 643.8 m/s

b) 642.4 m/s

c) 641.9 m/s

d) 643.2 m/s

View Answer

Explanation: The velocity of the satellite can be calculated using the relation V=Ω×r. The velocity is the vector product of angular speed and radius. V=Ω×r = 87×7.4 = 643.8 m/s.

10. A 3-phase induction motor runs at almost 70 rpm at no load and 50 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) 20 revolution per minute

b) 30 revolution per minute

c) 40 revolution per minute

d) 50 revolution per minute

View Answer

Explanation: Supply frequency=50 Hz. No-load speed of motor = 70 rpm. The full load speed of motor=50 rpm. Since the no-load speed of the motor is almost 70 rpm, hence synchronous speed near to 70 rpm. Speed of rotor field=70 rpm. Speed of rotor field with respect to rotor=70-50= 20 rpm.

11. Calculate the active power in a 9.854 H inductor.

a) 4.98 W

b) 0 W

c) 8.59 W

d) 1 W

View Answer

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 reactive power in a 45 Ω resistor with 1.78 A current flowing through it.

a) 28.8 VAR

b) 23.4 VAR

c) 25.82 VAR

d) 0 VAR

View Answer

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 0°. Q=VIsin(0°)=0 VAR.

13. Calculate the value of the frequency if the inductive reactance is 72 Ω and the value of the inductor is 7 H.

a) 1.63 Hz

b) 1.54 Hz

c) 1.78 Hz

d) 1.32 Hz

View Answer

Explanation: The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation X

_{L}= 2×3.14×f×L. F = X

_{L}÷2×3.14×L = 72÷2×3.14×7 = 1.63 Hz.

14. Calculate the active power in a 2 Ω resistor with 8 A current flowing through it.

a) 125 W

b) 128 W

c) 123 W

d) 126 W

View Answer

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 0°. P=I

^{2}R=8×8×2=128 W.

**Sanfoundry Global Education & Learning Series – Electric Drives.**

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