# Electric Drives Questions and Answers – Power Rating Determination of Electric Motors for Different Applications

This set of Electric Drives Interview Questions and Answers for Experienced people focuses on “Power Rating Determination of Electric Motors for Different Applications”.

1. Which one of the following motor is a 1-Φ AC motor?
a) Synchronous motor
b) Series motor
c) Shunt motor
d) Capacitor run

Explanation: Capacitor run motor is a 1-Φ AC motor. The capacitor is used to provide a 90° phase difference between the currents of main and auxiliary winding.

2. Current rating of the IM depends upon the cross-section area of the conductors.
a) True
b) False

Explanation: The current rating of the Induction motor depends upon the cross-section area of the conductors. Resistance ∝ 1÷Area, I ∝Area.

3. Voltage rating of the IM depends upon the insulation level.
a) True
b) False

Explanation: The voltage rating of the IM depends upon the insulation level. The insulation level is also decided on the basis of the over-voltages.

4. 450-seconds rated motor is preferable for __________
a) Light duty cranes
b) Medium duty cranes
c) Intermittent duty cranes
d) Heavy duty cranes

Explanation: Light duty cranes are used in power plants, compressor plants. 450-seconds rated motor is preferable for light-duty cranes. They can safely operate for 450-seconds without exceeding the specified temperature.

5. Calculate the 3-∅ power developed in the IM using the following data: VL=20 V, IL=5 A, cosΦ=.8.
a) 138.56 W
b) 128.41 W
c) 163.78 W
d) 177.59 W

Explanation: The power developed in the IM is √3VLILcos∅. The power is independent of time and frequency. P=√3VLILcos∅=√3×20×5×.8=138.56 W.
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6. The unit of the moment of inertia is Joule.
a) True
b) False

Explanation: The moment of inertia is taken as the sum of the product of the mass of each particle with the square of their distance from the axis of the rotation. The unit of the moment of inertia is kg×m2=kgm2.

7. Calculate mark to space ratio if the system is on for 20 sec and off for 3 sec.
a) 6.66
b) 4.48
c) 8.2
d) 5.6

Explanation: Mark to space ratio 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=20÷3=6.66.

8. Calculate the average power dissipated if the value of the duty cycle is .5 and Pmax=12 W.
a) 6 W
b) 5 W
c) 4 W
d) 3 W

Explanation: The average power dissipated can be calculated using the relation Pavg=D×Pmax. The value of the average power is .5×12=6 W.

9. Full form of PWM.
a) Pulse width modulation
b) Pulse run modulation
c) Power width modulation
d) Pulse wisdom mode

Explanation: PWM stands for pulse width modulation. This technique is used to alter the voltage level of the output of the inverter. We can control the voltage level using the PWM technique.

10. What is the correct relationship between the phase voltage and line voltage in Delta connection with the positive sequence?
a) VL=Vp
b) VL=√3Vp
c) VL=√2Vp
d) IL=√2Ip

Explanation: In the delta connection line voltage is the same as the phase voltage. The line current is √3 times of the phase current and leads by 30°.

11. What is the relationship between line and phase current in Star connection with the positive sequence?
a) VL=Vp
b) VL=√9Vp
c) VL=√2Vp
d) IL=Ip

Explanation: In the star connection line current is the same as the phase current. The line voltage is √3 times of the phase voltage and leads by 30°.

12. Choose the correct relationship in delta connection with the positive sequence.
a) IL=Ip
b) VL=√3Vp
c) VL=Vp
d) IL=√3Ip

Explanation: In the delta connection line voltage is the same as the phase voltage. The line current is √3 times of the phase current and leads by 30°.

13. Choose the correct relationship in Star connection with the positive sequence.
a) VL=Vp
b) VL=√3Vp
c) VL=√2Vp
d) IL=√3Ip

Explanation: In the star connection line current is the same as the phase current. The line voltage is √3 times of the phase voltage and leads by 30°.

14. Calculate the 3-∅ power developed in the IM connected in star using the following data: Vp=7 V, IL=7 A, cosΦ=.6.
a) 88.5 W
b) 88.4 W
c) 88.8 W
d) 88.2 W

Explanation: The power developed in the IM is √3VLILcos∅. The power is independent of time and frequency. In star connection VL=√3Vp=12.12 V. P=√3VLILcos∅=3×20×5×.8=88.2 W.

Sanfoundry Global Education & Learning Series – Electric Drives.

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