# Electric Drives Questions and Answers – DC Motors – Direct Control of Armature-Terminal Voltage

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This set of Electric Drives Puzzles focuses on “DC Motors – Direct Control of Armature-Terminal Voltage”

1. Calculate the frequency of the waveform x(t)=45sin(40πt+5π).
a) 24 Hz
b) 27 Hz
c) 23 Hz
d) 20 Hz

Explanation: 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

Explanation: The generated e.m.f can be calculated using the formula Eb = Φ×Z×N×P÷60×A, Φ represent 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 a 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

Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, α represents a phase difference.
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4. Calculate the moment of inertia of the rod about its end having a mass of 39 kg and length of 88 cm.
a) 9.91 kgm2
b) 9.92 kgm2
c) 9.96 kgm2
d) 9.97 kgm2

Explanation: 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 length of 29 cm.
a) .091 kgm2
b) .072 kgm2
c) .076 kgm2
d) .077 kgm2

Explanation: 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

Explanation: 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

Explanation: 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

Explanation: 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 of resistance.

9. Calculate the active power in a 7481 H inductor.
a) 1562 W
b) 4651 W
c) 0 W
d) 4654 W

Explanation: 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 case of the inductor so the angle between V & I is 90°. P=VIcos90 = 0 W. Voltage leads the current in 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

Explanation: 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 case of the capacitor so the angle between V & I is 90°. P = VIcos90 = 0 W. Current leads the voltage in 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

Explanation: 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 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

Explanation: 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=18×18×17=5508 W.

Sanfoundry Global Education & Learning Series – Electric Drives.

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