# Electric drive Armature Voltage Control MCQ [Free PDF] – Objective Question Answer for Electric drive Armature Voltage Control Quiz

1. Calculate the frequency of the waveform x(t)=45sin(40πt+5π).

A. 24 Hz
B. 27 Hz
C. 23 Hz
D. 20 Hz

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

The frequency of x(t) is

40π÷2π=20 Hz.

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

2. The generated e.m.f from 16-pole armature having 57 turns driven at 78 rev/sec having flux per pole as 5 mWb, with lap winding is ___________

A. 44.16 V
B. 44.15 V
C. 44.46 V
D. 44.49 V

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 is equal to the number of poles.

Eb = .005×16×57×2×4680÷60×16=44.46 V.

3. Calculate the phase angle of the sinusoidal waveform x(t)=42sin(4700πt+2π÷3).

A. 2π÷9
B. 2π÷5
C. 2π÷7
D. 2π÷3

The sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, and α represents a phase difference.

4. Calculate the moment of inertia of the rod about its end having a mass of 39 kg and a length of 88 cm.

A. 9.91 kgm2
B. 9.92 kgm2
C. 9.96 kgm2
D. 9.97 kgm2

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

I=ML2÷3.

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

I=(39)×.33×(.88)2=9.96 kgm2.

It depends upon the orientation of the rotational axis.

5. Calculate the moment of inertia of the rod about its center having a mass of 11 kg and a length of 29 cm.

A. .091 kgm2
B. .072 kgm2
C. .076 kgm2
D. .077 kgm2

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

I=ML2÷12.

The mass of the rod about its center and length is given.

I=(11)×.0833×(.29)2=.077 kgm2.

It depends upon the orientation of the rotational axis.

6. Calculate the shaft power developed by a motor using the given data: Eb = 404V and I = 25 A. Assume frictional losses are 444 W and windage losses are 777 W.

A. 8879 W
B. 2177 W
C. 8911 W
D. 8897 W

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

P = Eb*I-(rotational losses)

= 404*25- (444+777) = 8879 W.

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

7. Calculate the value of the frequency if the capacitive reactance is 13 Ω and the value of the capacitor is 71 F.

A. .0001725 Hz
B. .0001825 Hz
C. .0001975 Hz
D. .0001679 Hz

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÷13×2×3.14×71 = .0001725 Hz.

8. The slope of the V-I curve is 27°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

A. .384 Ω
B. .509 Ω
C. .354 Ω
D. .343 Ω

The slope of the V-I curve is resistance. The slope given is 27° so

R=tan(27°)=.509 ω.

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

9. Calculate the active power in a 7481 H inductor.

A. 1562 W
B. 4651 W
C. 0 W
D. 4654 W

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.

10. Calculate the active power in a 457 F capacitor.

A. 715 W
B. 565 W
C. 545 W
D. 0 W

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 the voltage in the case of the capacitor.

11. Calculate the active power in a 181 H inductor.

A. 2448 W
B. 1789 W
C. 4879 W
D. 0 W

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 17 ω resistor with 18 A current flowing through it.

A. 5508 W
B. 5104 W
C. 5554 W
D. 5558 W

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 0°.

P=I2R

=18×18×17=5508 W.

13. Swinburne’s test can be conducted on ___________

A. Series motor
B. Shunt motor
C. Compound motor
D. Shunt and compound motor

The test is practically applicable for machines that have flux constant like the shunt and compound machines as this is a no-load test and the DC series motor should not be run at no-load because of high speed.

14. The generated e.m.f from 20-pole armature having 800 conductors driven at 30 rev/sec having flux per pole as 60 mWb, with 16 parallel paths is ___________

A. 1900 V
B. 1840 V
C. 1700 V
D. 1800 V

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

Eb = .06×20×1800×800÷60×16 = 1800 V.

15. The unit of active power is Watt.

A. True
B. False

The active power in the electrical circuits is useful to power. It determines the power factor of the system. It is expressed in terms of Watt.

P=VIcosΦ.

16. Calculate the mass of the ball having a moment of inertia of 4.5 kgm2 and a radius of 14 cm.

A. 229.59 kg
B. 228.56 kg
C. 228.54 kg
D. 227.52 kg

The moment of inertia of the ball can be calculated using the formula I=∑miri2.

The moment of inertia of the ball and radius is given.

M=(4.5)÷(.14)2 = 229.59 kg.

It depends upon the orientation of the rotational axis.

17. The field control method is suitable for constant torque drives.

A. True
B. False

The field control method is generally used for obtaining speeds greater than the base speed. It is also known as the flux weakening method. It is suitable for constant power drives.

18. What is the unit of the intensity?

A. Watt/m2
B. Watt/m
C. Watt/m4
D. Watt/m3

Intensity is defined as the amount of power incident in a particular area. It is mathematically expressed as

I = Power incident (Watt)÷Area(m2).

19. Calculate the value of the frequency of the signal that completes half of the cycle in 70 sec. Assume the signal is periodic.

A. 0.00714 Hz
B. 0.00456 Hz
C. 0.00845 Hz
D. 0.00145 Hz

The frequency is defined as the number of oscillations per second. It is reciprocal to the time period. It is expressed in Hz. The given signal completes half of the cycle in 70 seconds then it will complete a full cycle in 140 seconds. F = 1÷T=1÷140=.00714 Hz.

20. The slope of the V-I curve is 26°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

A. .487 Ω
B. .482 Ω
C. .483 Ω
D. .448 Ω

The slope of the V-I curve is resistance. The slope given is 26° so

R=tan(26°) = .487 Ω.

The slope of the V-I curve is resistance.

21. For large DC machines, the yoke is usually made of which material?

A. Cast steel
B. Cast iron
C. Iron
D. Cast steel or cast iron

Yoke in DC machines is made up of cast steel. Yoke provides structural support and mechanical strength to the machine. It helps in carrying the flux from the North pole to the South pole.

22. Calculate the terminal voltage of the Permanent Magnet DC motor having a resistance of 2 Ω and a full load current of 5 A with 20 V back e.m.f.

A. 30 V
B. 25 V
C. 20 V
D. 31 V

A permanent magnet DC motor is a special type of motor in which flux remains constant. The terminal voltage can be calculated using the relation

Vt = Eb+IaRa = 20+5×2 = 30 V.

23. Armature reaction is demagnetizing in nature due to purely lagging load.

A. True
B. False