Energy Relations During Motor Starting MCQ [Free PDF] – Objective Question Answer for Energy Relations During Motor Starting Quiz

11. Calculate the active power in a 157.1545 H inductor.

A. 4577 W
B. 4567 W
C. 4897 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. 

 

12. Calculate the active power in a 1.2 Ω resistor with 1.8 A current flowing through it.

A. 3.88 W
B. 3.44 W
C. 3.12 W
D. 2.18 W

Answer: A

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 90o.

P=I2R=1.8×1.8×1.2=3.88 W. 

 

13. Calculate the total heat dissipated in a resistor of 12 Ω when 9.2 A current flows through it.

A. 2.01 KW
B. 3.44 KW
C. 1.01 KW
D. 2.48 KW

Answer: C

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=9.2×9.2×12=1.01 kW. 

 

14. Calculate mark to space ratio if the system is on for 9 sec and the time period is 11 sec.

A. 4.6
B. 4.8
C. 4.5
D. 4.9

Answer: C

Mark to space is Ton÷Toff. It is the ratio of the time for which the system is active and the time for which is inactive.

M = Ton÷Toff=9÷(11-9) = 4.5. 

 

15. Calculate the angular frequency of the waveform y(t)=69sin(40πt+4π).

A. 40π Hz
B. 60π Hz
C. 70π Hz
D. 20π Hz

Answer: A

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

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

where

Vm represents peak value
Ω represents the angular frequency
α represents a phase difference
Ω can be directly calculated by comparing the equations.

Ω = 40π Hz. 

 

16. The generated e.m.f from 22-pole armature having 75 turns driven at 78 rpm having flux per pole as 400 mWb, with lap winding is ___________

A. 76 V
B. 77 V
C. 78 V
D. 79 V

Answer: C

The generated e.m.f 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.

One turn is equal to two conductors. In lap winding, the number of parallel paths equals the number of poles.

Eb = .4×22×75×2×78÷60×22 = 78 V. 

 

17. Calculate the phase angle of the sinusoidal waveform u(t)=154sin(9.85πt-π÷89).

A. -78π÷9
B. -12π÷5
C. -π÷89
D. -2π÷888

Answer: C

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. 

 

18. Calculate the moment of inertia of the stick about its end having a mass of 22 kg and a length of 22 cm.

A. .088 kgm2
B. .087 kgm2
C. .089 kgm2
D. .086 kgm2

Answer: B

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

I=ML2÷3.

The mass of the stick about its end and length is given.

I = (22)×.33×(.11)2 =.087 kgm2.

It depends upon the orientation of the rotational axis. 

 

19. 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. 

 

20. 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. 

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