Dynamics of Electric Drives MCQ [Free PDF] – Objective Question Answer for Dynamics of Electric Drives MCQ Quiz

41. Calculate the time period of the waveform x(t)=24sin(24πt+π÷4).

A. .064 sec
B. .047 sec
C. .083 sec
D. .015 sec

Answer: C

The fundamental time period of the sine wave is 2π.

The time period of x(t) is 2π÷24π=.083 sec.

The time period is independent of phase shifting and time-shifting. 

 

42. The turn-off times of the devices in the increasing order is ___________

I. MOSFET
II. BJT
III. IGBT
IV. Thyristor

A. I, III, II, IV
B. I, II, III, IV
C. III, I, II, IV
D. III, II, IV, I

Answer: A

Increasing turn-off time implies decreasing speed and the majority of carrier devices do not have any minority charge carrier storage so they have less turn-off time hence MOSFET has the least turn-off time. So, the increasing order of turn-off time is, MOSFET < IGBT < BJT < Thyristor. 

 

43. Which of the following devices should be used as a switch for high power and high voltage application?

A. GTO
B. MOSFET
C. TRIAC
D. Thyristor

Answer: D

A thyristor is used for high-power applications but it has a limited frequency range and cannot be used at high frequencies. A thyristor is a unidirectional, bipolar and semi-controlled device. 

 

44. Calculate the useful power developed by a motor using the given data: Pin = 3000 W, Ia = 60 A, Ra = .4 Ω. Assume frictional losses are 200 W and windage losses are 400 W.

A. 970 W
B. 960 W
C. 980 W
D. 990 W

Answer: B

Useful power is basically the shaft power developed by the motor that can be calculated using the formula

Psh = Pdev-(rotational losses)

Pdev = Pin-Ia2Ra = 3000-602(.4)=1560 W.

The useful power developed by the motor is

Psh = Pdev-(rotational losses)=1560 –(200+400)=960 W. 

 

45. Calculate the phase angle of the sinusoidal waveform y(t)=55sin(4πt+π÷8).

A. π÷8
B. π÷5
C. π÷7
D. π÷4

Answer: A

The sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α)

where

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

 

46. The axis along which no emf is produced in the armature conductors is called as _________

A. Geometrical Neutral Axis (GNA.
B. Magnetic Neutral Axis (MNA)
C. Axis of rotation
D. Axis of revolution

Answer: B

The axis along which no emf is produced in the armature conductors is called Magnetic Neutral Axis (MNA).

The coil undergoing commutation must lie along the magnetic neutral axis so that no emf is induced and there is no sparking at the time of commutation. 

 

47. The generated e.m.f from 25-pole armature having 200 conductors driven at 10 rev/sec having flux per pole as 20 mWb, with two parallel paths is ___________

A. 400 V
B. 500 V
C. 200 V
D. 300 V

Answer: B

The generated can be calculated using the formula

Eb = Φ×Z×N×P÷60×A,

Where

Φ represents 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

Eb = .02×25×200×600÷60×2= 500 V. 

 

48. The unit of the flux is Weber.

A. True
B. False

Answer: A

Flux is the total amount of magnetic field lines passing through a given area. Φ is a dot product of magnetic flux density and area. The unit of the flux is Weber (WB). 

 

49. Which of the following motor can be referred to as a universal motor?

A. DC shunt motor
B. DC compound motor
C. Permanent magnet motor
D. DC series motor

Answer: D

DC series motor can operate on DC and AC. It is a universal motor. Universal motors are those motors that can operate on both DC and AC. DC shunt motor can only operate on DC because of pulsating torque in AC. 

 

50. The phase difference between voltage and current in the inductor.

A. 45°
B. 90°
C. 80°
D. 55°

Answer: B

In the case of an inductor, the voltage leads the current by 90°, or the current lags the voltage by 90o. The phase difference between voltage and current is 90°. 

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