# Electric Drives Questions and Answers – Methods to Reduce the Energy Loss During Starting

This set of Electric Drives Questions & Answers for Exams focuses on “Methods to Reduce the Energy Loss During Starting”.

1. Calculate the angular frequency of the waveform y(t)=69sin(40πt+4π).
a) 40π Hz
b) 60π Hz
c) 70π Hz
d) 20π Hz

Explanation: 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 angular frequency, α represents a phase difference. Ω can be directly calculated by comparing the equations. Ω = 40π Hz.

2. 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

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 = .4×22×75×2×78÷60×22 = 78 V.

3. 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

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.

4. Calculate the moment of inertia of the stick about its end having a mass of 22 kg and length of 22 cm.
a) .088 kgm2
b) .087 kgm2
c) .089 kgm2
d) .086 kgm2

Explanation: 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.

5. Calculate the moment of inertia of the stick about its center having a mass of 1.1 kg and length of 2.9 m.
a) .66 kgm2
b) .77 kgm2
c) .88 kgm2
d) .47 kgm2

Explanation: 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 is given. I=(1.1)×.0833×(2.9)2=.77 kgm2. It depends upon the orientation of the rotational axis.
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6. 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

Explanation: 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.

7. Calculate the value of the frequency if the capacitive reactance is .1 Ω and the value of the capacitor is .02 F.
a) 71.25 Hz
b) 81.75 Hz
c) 79.61 Hz
d) 79.54 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÷.1×2×3.14×.02 = 79.61 Hz.

8. The slope of the V-I curve is 6.9°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .38 Ω
b) .59 Ω
c) .34 Ω
d) .12 Ω

Explanation: The slope of the V-I curve is resistance. The slope given is 6.9° so R=tan(6.9°)=.12 Ω. The slope of the I-V curve is reciprocal of resistance.

9. Calculate the active power in an 8764 H inductor.
a) 8645 W
b) 6485 W
c) 0 W
d) 4879 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 543 F capacitor.
a) 581 W
b) 897 W
c) 0 W
d) 892 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 32 H inductor.
a) 28 W
b) 189 W
c) 4 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 an 8965 Ω resistor with .23 A current flowing through it.
a) 547.12 W
b) 474.24 W
c) 554.78 W
d) 123.88 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=.23×.23×8965=474.24 W.

13. Which one of the following methods would give a lower than the actual value of regulation of the alternator?
a) ZPF method
b) MMF method
c) EMF method
d) ASA method

Explanation: MMF method is an optimistic method of voltage regulation as it gives lower than the actual value of voltage regulation. MMF method will give the values that are lesser than the actual value.

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