Electrical Drive Starting Method MCQ [Free PDF] – Objective Question Answer for Electrical Drive Starting Method Quiz

21. 100 V, 2 A, 90 rpm separately excited dc motor with armature resistance (Ra) equal to 8 ohms. Calculate back emf developed in the motor when it operates on 3th/4 of the full load. (Assume rotational losses are neglected.

A. 100 V
B. 87 V
C. 88 V
D. 90 V

Answer: C

Back emf developed in the motor can be calculated using the relation

Eb = Vt-I×Ra.

In question, it is asking for 3th/4 load, but the data is given for full load so current becomes 3th/4 of the full load current

= 2÷1.33 = 1.5 A.

100 V is terminal voltage it is fixed so

Eb = 100-1.5×8 = 88 V. 

 

22. The slope of the V-I curve is 16.8°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

A. .324 Ω
B. .301 Ω
C. .343 Ω
D. .398 Ω

Answer: B

The slope of the V-I curve is resistance. The slope given is 16.8° so

R=tan(16.8°)=.301 Ω.

The slope of the I-V curve is reciprocal to resistance. 

 

23. Calculate the value of the torque when 1 N force is applied perpendicular to a 1 m length of chain fixed at the center.

A. 1 N-m
B. 3 N-m
C. 2 N-m
D. 4 N-m

Answer: A

Torque can be calculated using the relation

T = (length of chain) × (Force applied)
= r×F×sin90.

F is given as 1 N and r is 1 m

then torque is 1×1 = 6942 N-m. (the angle between F and r is 90 degrees). 

 

24. A 3-phase induction motor runs at almost 140 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. 80 revolution per minute
C. 90 revolution per minute
D. 70 revolution per minute

Answer: C

Supply frequency=50 Hz.
No-load speed of motor= 140 rpm.
The full load speed of the motor=50 rpm.

Since the no-load speed of the motor is almost 140 rpm, hence synchronous speed is near 140 rpm.

Speed of rotor field=140 rpm. Speed of rotor field with respect to the rotor

= 140-50 = 90 rpm. 

 

25. Calculate the active power in a .7889 H inductor.

A. .123 W
B. .155 W
C. 0 W
D. .487 W

Answer: C

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 the case of the inductor so the angle between V & I is 90°.

P = VIcos90° = 0 W. 

 

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

A. 2.4 W
B. 3.4 W
C. 2.2 W
D. 3.2 W

Answer: D

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 the case of the resistor so the angle between V & I is 90°.

P=I2R=2×2×.8=3.2 MW. 

 

27. The unit of angular frequency is Hz.

A. True
B. False

Answer: A

Angular frequency is defined as the rate of change of angular displacement with respect to time. Angular displacement is generally expressed in terms of a radian. The unit of angular frequency is Hz.

ω=2×3.14×f. 

 

28. Calculate the active power in a .56 F capacitor.

A. 37.8 W
B. 0 W
C. 15.4 W
D. 124.5 W

Answer: B

The capacitor 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 the case of the capacitor so the angle between V & I is 90°.

P = VIcos90° = 0 W.

Current leads to the voltage in the case of the capacitor. 

 

29. Calculate the value of the angular acceleration of the motor using the given data: J = .01 kg-m2, load torque= 790 N-m, motor torque= 169 N-m.

A. -62100 rad/s2
B. -62456 rad/s2
C. -34056 rad/s2
D. -44780 rad/s2

Answer: A

Using the dynamic equation of motor

J*(angular acceleration) = Motor torque – Load torque:

.01*(angular acceleration) = 169-790=-621,

angular acceleration=-62100 rad/s2.

The motor will decelerate and will fail to start. 

 

30. A 14-pole, 3-phase, 50 Hz induction motor is operating at a speed of 99 rpm. The frequency of the rotor current of the motor in Hz is __________

A. 39.5
B. 40
C. 38.45
D. 39.9

Answer: C

Given a number of poles = 14.

The supply frequency is 50 Hz.

The rotor speed is 699 rpm. Ns=120×f÷P

=120×50÷14 = 428.57 rpm.

S=Ns-Nr÷Ns

=428.57-99÷428.57=.769.

F2=sf=.769×50=38.45 Hz. 

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