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

61. Calculate the moment of inertia of the stick about its center having a mass of 1.1 kg and a length of 2.9 m.

A. .66 kgm2
B. .77 kgm2
C. .88 kgm2
D. .47 kgm2

Answer: B

The moment of inertia of the stick about its center can be calculated using the formula

I=ML2÷12.

The mass of the stick about its center and length are given.

I=(1.1)×.0833×(2.9)2=.77 kgm2.

It depends upon the orientation of the rotational axis. 

 

62. Calculate the useful power developed by a motor using the given data: Eb = 4V and I = 52 A. Assume frictional losses are 3 W and windage losses are 2 W.

A. 203 W
B. 247 W
C. 211 W
D. 202 W

Answer: A

Useful power developed by the motor can be calculated using the formula

P = Eb*I -(rotational losses) = 4*52 – (5) = 203 W.

If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor. 

 

63. Calculate the value of the frequency if the capacitive reactance is .1 Ω and the value of the capacitor is .02 F.

A. 71.25 Hz
B. 81.75 Hz
C. 79.61 Hz
D. 79.54 Hz

Answer: C

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

Xc = 1÷2×3.14×f×C. F

= 1÷Xc×2×3.14×C

= 1÷.1×2×3.14×.02 = 79.61 Hz. 

 

64. The slope of the V-I curve is 6.9°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

A. .38 Ω
B. .59 Ω
C. .34 Ω
D. .12 Ω

Answer: D

The slope of the V-I curve is resistance.

The slope given is 6.9° so R=tan(6.9°)=.12 Ω.

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

 

65. Calculate the active power in an 8764 H inductor.

A. 8645 W
B. 6485 W
C. 0 W
D. 4879 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.

Voltage leads the current in the case of the inductor. 

 

66. Calculate the active power in a 543 F capacitor.

A. 581 W
B. 897 W
C. 0 W
D. 892 W

Answer: C

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. 

 

67. Calculate the active power in a 32 H inductor.

A. 28 W
B. 189 W
C. 4 W
D. 0 W

Answer: D

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. 

 

68. Calculate the active power in an 8965 Ω resistor with.23 A current flowing through it.

A. 547.12 W
B. 474.24 W
C. 554.78 W
D. 123.88 W

Answer: B

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=I2R=.23×.23×8965

=474.24 W. 

 

69. Which one of the following methods would give a lower than the actual value of regulation of the alternator?

A. ZPF method
B. MMF method
C. EMF method
D. ASA method

Answer: B

MMF method is an optimistic method of voltage regulation as it gives lower than the actual value of voltage regulation. MMF method will give the values that are lesser than the actual value. 

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