# Electric Drives Questions and Answers – Speed Control of Shunt or Separately Excited Motors

This set of Electric Drives Question Paper focuses on “Speed Control of Shunt or Separately Excited Motors”.

1. The advantage of the double squirrel cage induction motor over a single cage rotor is that its _______
a) Efficiency is higher
b) Power factor is higher
c) Slip is larger
d) Starting current is lower

Explanation: The starting current flows through the outer cage which has higher resistance and hence starting current is lower. This is one the important advantage of the double squirrel cage induction motor over a single cage rotor.

2. A 16-pole, 3-phase, 60 Hz induction motor is operating at a speed of 150 rpm. The frequency of the rotor current of the motor in Hz is __________
a) 20
b) 40
c) 30
d) 10

Explanation: Given a number of poles = 16. Supply frequency is 60 Hz. Rotor speed is 150 rpm. Ns=120×f÷P=120×60÷16 = 450 rpm. S=Ns-Nr÷Ns=450-150÷450=.666. F2=sf=.666×60=40 Hz.

3. Calculate the amplitude of the sinusoidal waveform z(t)=715sin(165πt+2π÷468).
a) 710
b) 715
c) 716
d) 718

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. R.M.S value of the sinusoidal waveform V=48sin(6.5πt+89π÷46.8).
a) 33.94 V
b) 33.56 V
c) 33.12 V
d) 33.78 V

Explanation: R.M.S value of the sinusoidal waveform is Vm÷2½ = 48÷2½ = 33.94 V and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vm.

5. The short circuit test on a 3-φ induction motor is conducted at a rotor speed of _______
a) Zero
b) < Ns
c) > Ns
d) Ns

Explanation: Short-circuit test in an induction motor is also called a Blocked rotor test so it is conducted at zero speed. Net input power taken is equal to the variable losses.

6. If induction motor air gap power is 10 KW and mechanically developed power is 8 KW, then rotor ohmic loss will be _________ KW.
a) 1
b) 2
c) 3
d) 4

Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=10-8=2 KW.

7. A 3-phase induction machine draws active power P and reactive power Q from the grid. If it is operated as a generator, then P and Q will be _________
a) Positive and Negative
b) Negative and Positive
c) Positive and Positive
d) Negative and Negative

Explanation: Induction generator will be delivering power to the bus to generate flux it will consume reactive power from the bus. Since active power is delivered the active power drawn will be negative but reactive power is absorbed and hence active power absorbed is positive.

8. The slope of the V-I curve is 39.1°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .81 Ω
b) .36 Ω
c) .75 Ω
d) .84 Ω

Explanation: The slope of the V-I curve is resistance. The slope given is 39.1° so R=tan(39.1°)=.81 Ω. The slope of the I-V curve is reciprocal of resistance.

9. If induction motor air gap power is 48 KW and gross developed power is 28 KW, then rotor ohmic loss will be _________ KW.
a) 10
b) 20
c) 30
d) 40

Explanation: Rotor ohmic losses are due to the resistance of armature windings. Net input power to the rotor is equal to the sum of rotor ohmic losses and mechanically developed power. Rotor ohmic losses=Air gap power-Mechanical developed power=48-28=20 KW.

10. The power factor of a squirrel cage induction motor is ___________
a) Low at light load only
b) Low at heavy loads only
c) Low at the light and heavy loads both
d) Low at rate load only

Explanation: At light loads, the current drawn is largely a magnetizing current due to the air gap and hence the power factor is low.

11. Calculate the active power in a .154 H inductor.
a) 22 W
b) 14 W
c) 45 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 time period of the waveform z(t)=24sin(πt+π÷4)+4sin(2πt+π÷6).
a) 2 sec
b) 3 sec
c) 4 sec
d) 1 sec

Explanation: The fundamental time period of the sine wave is 2π. The time period of z(t) is L.C.M {2,1}=2 sec. The time period is independent of phase shifting and time shifting.

13. Calculate the total heat dissipated in a resistor of 50 Ω when 1.4 A current flows through it.
a) 98 W
b) 92 W
c) 91 W
d) 93 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 90°. P=I2R=1.4×1.4×50=98 W.

14. Calculate mark to space ratio if the system is on for 26.3 sec and off for 24.2 sec.
a) 1.086
b) 1.042
c) 1.214
d) 1.876

Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff=26.3÷24.2=1.086.

Sanfoundry Global Education & Learning Series – Electric Drives.

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