# Electric Drives Questions and Answers – Induction Motors Speed Control – Rotor Voltage Injection

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This set of Electric Drives written test Questions & Answers focuses on “Induction Motors Speed Control – Rotor Voltage Injection”.

1. A 32-pole, 3-phase, 70 Hz induction motor is operating at a speed of 112 rpm. The frequency of the rotor current of the motor in Hz is __________
a) 40.2
b) 46.1
c) 40.1
d) 40.6

Explanation: Given a number of poles = 32. Supply frequency is 70 Hz. Rotor speed is 112 rpm. Ns = 120×f÷P=120×70÷32 = 262.5 rpm. S=Ns-Nr÷Ns = 262.5-112÷262.5=.573. F2=sf=.573×70=40.1 Hz.

2. A 20-pole, 3-phase, 90 Hz induction motor is operating at a speed of _______ rpm. The frequency of the rotor current of the motor in Hz is 20.
a) 418.56
b) 420.12
c) 421.23
d) 422.45

Explanation: Given a number of poles = 20. Supply frequency is 90 Hz. Ns=120×f÷P=120×90÷20 = 540 rpm. S=20÷90 = .222 Hz. Nr=(1-s)Ns = 420.12 rpm. Rotor speed is 420.12 rpm.

3. Calculate the amplitude of the sinusoidal waveform z(t)=.27sin(.369πt+.142π÷4.25).
a) .287
b) .270
c) .216
d) .287

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. By comparing the waveform z(t) with the generalized sinusoidal expression we can see Vm=.27 and ω=.369π rad/s.

4. R.M.S value of the sinusoidal waveform q(t)=2.11cos(87.25t+78π÷478.23).
a) 1.49 V
b) 1.56 V
c) 1.12 V
d) 1.78 V

Explanation: R.M.S value of the sinusoidal waveform is Vm÷2½ and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vm. The r.m.s value is Vrms = 2.11÷2½ = 1.49 V.

5. The no-load circuit test on a 3-Φ induction motor is conducted at a rotor speed of _______
a) Zero
b) < Ns
c) > Ns
d) Ns

Explanation: No-load circuit test in an induction motor is also called an open circuit test so it is conducted at synchronous speed. Net input power taken is equal to the no-load rotational losses.

6. If induction motor air gap power is 67 KW and mechanically developed power is 26 KW, then rotor ohmic loss will be _________ KW.
a) 41
b) 42
c) 43
d) 44

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=67-26=41 KW.

7. Calculate the line voltage in star connection when phase voltage=45 V.
a) 77.9 V
b) 77.6 V
c) 77.2 V
d) 77.8 V

Explanation: The line voltage in case of star connection is 1.73 times of phase voltage. It leads the phase voltage by an angle of 30°. VL-L=1.73×45=77.9 V.

8. The slope of the V-I curve is 1.1°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .019 Ω
b) .036 Ω
c) .075 Ω
d) .084 Ω

Explanation: The slope of the V-I curve is resistance. The slope given is 1.° so R=tan(1.1°)=.019 Ω. The slope of the I-V curve is reciprocal of resistance.

9. If induction motor rotor power is 157.5 KW and gross developed power is 79.9 KW, then rotor ohmic loss will be _________ KW.
a) 77.5
b) 77.6
c) 76.9
d) 77.1

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=157.5-79.9=77.6 KW.

10. The power factor of a squirrel cage induction motor generally is ___________
a) .6-.8
b) .1-.2
c) .2-.4
d) .5-.7

Explanation: At light loads, the current drawn is largely a magnetizing current due to the air gap and hence the power factor is low. The power factor of a squirrel cage induction motor generally is .6-.8.

11. Calculate the active power in a .89 H inductor.
a) 1.535 W
b) 0 W
c) 2.484 W
d) 1.598 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 i(t)=sin(πt+6π÷4)+sin(πt+5π÷6).
a) 2 sec
b) 4 sec
c) 5 sec
d) 3 sec

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

13. Calculate the total heat dissipated in a rotor resistor of 14.23 Ω when .65 A current flows through it.
a) 6.45 W
b) 6.01 W
c) 6.78 W
d) 6.98 W

Explanation: The rotor 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=.65×.65×14.23=6.01 W.

14. Calculate mark to space ratio if the system is on for 4.3 sec and off for 78.2 sec.
a) .054
b) .047
c) .039
d) .018 