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

Electrical Drive Starting Method MCQ

 

1. The ferrite cores are used for ____________ transformers.

A. Small transformers
B. Medium transformers
C. Large transformers
D. Medium and small transformers

Answer: A

Ferrite cores are used for cores of small transformers used in communication circuits at high frequencies and low energy levels. Because ferrites have high resistivity they will have lower eddy current losses. 

 

2. A 2-pole, 3-phase, 50 Hz induction motor is operating at a speed of 400 rpm. The frequency of the rotor current of the motor in Hz is __________

A. 43.33
B. 42.54
C. 43.11
D. 41.47

Answer: A

Given a number of poles = 2.

The supply frequency is 50 Hz.

The rotor speed is 400 rpm.

Ns = 120×f÷P=120×50÷2 = 3000 rpm.

S=Ns-Nr÷Ns = 3000-400÷3000 = .8666.

F2=sf=.8666×50=43.33 Hz. 

 

3. Calculate the phase angle of the sinusoidal waveform w(t)=.45sin(87πt+8π÷787).

English Conversation About Computer

A. 2π÷39
B. 8π÷787
C. 5π÷74
D. 42π÷4

Answer: B

The sinusoidal waveform is generally expressed in the form of

V=Vmsin(ωt+α)

where

Vm represents the peak value
ω represents the angular frequency
α represents a phase difference. 

 

4. Calculate the moment of inertia of the disc having a mass of 1 kg and radius of 1 m.

A. 1 kgm2
B. .5 kgm2
C. 2 kgm2
D. 3 kgm2

Answer: B

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=(1)×.5×(1)2=.5 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 3 kg and a diameter of 6 cm.

A. .0178 kgm2
B. .0147 kgm2
C. .0398 kgm2
D. .0144 kgm2

Answer: A

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=(3)×.66×(.03)2=.0178 kgm2.

It depends upon the orientation of the rotational axis. 

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6. The power factor of a synchronous motor __________

A. Improves with an increase in excitation and may even become leading at high excitations
B. Decreases with increase in excitation
C. Is independent of its excitation
D. Increase with loading for a given excitation

Answer: A

From the inverted V-curve, we can see when the power factor is leading power factor decreases when the excitation increases and it is over-excited conditions and when the power factor is lagging if the motor power factor is increasing if excitation increases. 

 

7. The frame of a synchronous motor is made of _________

A. Aluminum
B. Silicon steel
C. Cast iron
D. Stainless steel

Answer: C

The frame of a synchronous motor is made of cast iron. The power factor of a synchronous motor depends upon maximum power transfer capability. 

 

8. The slope of the V-I curve is 45°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

A. 2 Ω
B. 3 Ω
C. 4 Ω
D. 1 Ω

Answer: D

The slope of the V-I curve is resistance.

The slope given is 45° so R=tan(45°)=1 Ω.

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

 

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

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

Answer: C

EMF method is a pessimistic method of voltage regulation as it gives higher than the actual value of voltage regulation. EMF method will the values that are greater than the actual value. 

 

10. A 3-phase induction motor runs at almost 1100 rpm at no load and 640 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. 430 revolution per minute
B. 440 revolution per minute
C. 460 revolution per minute
D. 450 revolution per minute

Answer: C

Supply frequency=50 Hz.
No-load speed of motor = 1100 rpm.
The full load speed of the motor = 640 rpm.

Since the no-load speed of the motor is almost 1100 rpm, hence synchronous speed is near 1100 rpm.

Speed of rotor field=1100 rpm.

Speed of rotor field with respect to the rotor

= 1100-640 = 460 rpm. 

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