# Induction Motor MCQs Part 15

Induction Motor MCQs Part 15

## If supply voltage of a 3-phase induction motor is increased two times, the torque of motor is

If supply voltage of a 3-phase induction motor is increased two times, the torque of motor is

1. Increased two times
2. Decreased two times
3. Decreased four times
4. Increased four times

Correct answer: 4. increased four times

## The mechanical power developed by a 3-phase induction motor running at 2% slip having 1000 W input is:

The mechanical power developed by a 3-phase induction motor running at 2% slip having 1000 W input is:

1. 30 W
2. 950 W
3. 500 W
4. 970 W

Solution: Mechanical power developed by motor = (1 – s ) * Power input to rotor = (1-0.03)*1000 = 970 W

## A 3-phase, 12 pole, star-connected induction motor runs at 600 V, 50 Hz. The motor has rotor resistance of 0.04 Ω/phase and has a stand-still reactance of 0.5 Ω/phase. The speed of motor at maximum torque is

A 3-phase, 12 pole, star-connected induction motor runs at 600 V, 50 Hz. The motor has rotor resistance of 0.04 Ω/phase and has stand-still reactance of 0.5 Ω/phase. The speed of motor at maximum torque is:

1. 285 RPM
2. 460 RPM
3. 680 RPM
4. 720 RPM

Solution:

Synchronous speed (Ns) = 120f/P = 500 RPM

Motor speed corresponding to maximum torque

N = (1 – s) Ns = (1 – 0.08) * 500 = 460 RPM

## A 3-phase, 60 Hz, 8 pole induction motor running with a slip of 3% is taking 20 kW. Stator losses amount 0.5 kW. The total torque developed is

A 3-phase, 60 Hz, 8 pole induction motor running with a slip of 3% is taking 20 kW. Stator losses amount 0.5 kW. The total torque developed is:

1. 120 Nm
2. 109 Nm
3. 207 Nm
4. 352 Nm

Solution:

Stator output = Stator input – Stator losses = 20 – 0.5 = 19.5 kW

Stator output = Rotor input = 19.5 kW

———————-

Rotor Cu loss = s * Rotor input = 0.03 * 19.5 = 0.585 kW

Rotor output = 19.5 – 0.585 = 18.915 kW

———————-

Ns = 120 * f/P = 120 * 60/8 = 900 RPM

N = (1 – s) * Ns = (1 – 0.03) * 900 = 873 RPM

———————-

Total torque (T) = 9.55 * Rotor output/N

T = 9.55 * 18.915 * 1000 / 873 = 207 Nm

## A 100 HP 3 phase induction motor with a synchronous speed of 750 RPM

A 100 HP 3 phase induction motor with a synchronous speed of 750 RPM at 50 HZ has 96 stator slots with 4 conductors per slot and 120 rotor slots with 2 conductors per slot. If the full-load efficient is 0.92 and power factor is 0.85 lagging. The current per phase when stator is star-connected (rotor is wound 3-phase) to a 3-phase, 600 V supply will be:

1. 50 A
2. 92 A
3. 108 A
4. 162 A