Induction Motor MCQs Part 15

Contents

- 1 If supply voltage of a 3-phase induction motor is increased two times, the torque of motor is
- 2 The mechanical power developed by a 3-phase induction motor running at 2% slip having 1000 W input is:
- 3 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
- 4 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
- 5 A 100 HP 3 phase induction motor with a synchronous speed of 750 RPM

## 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

- Increased two times
- Decreased two times
- Decreased four times
- 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:

- 30 W
- 950 W
- 500 W
- 970 W

Correct answer: 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:

- 285 RPM
- 460 RPM
- 680 RPM
- 720 RPM

Correct answer: 2. 460 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:

- 120 Nm
- 109 Nm
- 207 Nm
- 352 Nm

Correct answer: 3. 207 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:

- 50 A
- 92 A
- 108 A
- 162 A

Correct answer: 2. 92 A

Solution:

Line current (IL) = P/(√3 * VL cos Φ η = 92 A

Phase current = Line current = 92 A