# Induction Motor MCQs Part 14

Induction Motor MCQs Part 14

## Which of following is correct regarding the rotor speed in a 3 -phase induction motor? It is

Which of following is correct regarding the rotor speed in a 3 -phase induction motor? It is

1. Smaller than synchronous speed
2. Greater than synchronous speed
3. Equal to synchronous speed
4. None of these

Correct answer: 1. Smaller than synchronous speed

## If the air gap between the rotor and stator of a 3-phase induction motor is increased, then

If the air gap between the rotor and stator of a 3-phase induction motor is increased, then

1. Leakage reactances are decreased
2. Leakage reactances are increased
4. None of the above

Correct answer: 2. Leakage reactances are increased

## A 3 phase, 8 pole, 60 Hz induction motor has equivalent resistor of 0.08 Ω/phase. If stalling speed is 600 RPM. How much resistance must be included per phase inorder to obtain maximum starting torque:

A 3 phase, 8 pole, 60 Hz induction motor has equivalent resistor of 0.08 Ω/phase. If stalling speed is 600 RPM. How much resistance must be included per phase inorder to obtain maximum starting torque:

1. 0.11 Ω
2. 0.16 Ω
3. 0.19 Ω
4. 0.26 Ω

Solution:

Synchronous speed Ns = 120f/P

Ns = 120 * 60/8 = 900 RPM

Slip (s) = Ns – N/Ns = (900 – 600)/900 = 0.33

Since torque is max (stalling) s = R2/X2

OR X2 = R2/s

X2 = 0.08/0.33 = 0.24 Ω

Let resistance to be included per phase is r ohms. In order to obtain maximum starting torque:

R2 + r = X2

0.08 + r = 0.24

r = 0.16 Ω

## A 20 HP, 6 pole, 50 Hz slip ring induction motor runs at 950 RPM at full-load with a rotor current of 20 A. The copper loss in short-circuiting gear is 250 W and mechanical losses amount to 1000 W. The resistance per phase of 3-phase rotor winding is

A 20 HP, 6 pole, 50 Hz slip ring induction motor runs at 950 RPM at full-load with a rotor current of 20 A. The copper loss in short-circuiting gear is 250 W and mechanical losses amount to 1000 W. The resistance per phase of 3-phase rotor winding is:

1. 0.225 Ω
2. 0.501 Ω
3. 0.822 Ω
4. 0.998 Ω

Solution:

Synchronous speed (Ns) = 120f/P = 120 * 50/6 = 1000 RPM

———————

Slip (s) = Ns – N/Ns = (1000 – 950/1000) = 0.05

———————

Output = 20 HP = 20 * 746 = 14920 W

———————

Total mechanical output (Pm) = 14920 + 250 + 1000 = 16170 W

———————

Rotor Cu loss = Pm * s/(1-s) = 16170 * 0.05/1-0.05 = 851 W

———————

Loss in rotor winding = 851 – Losses in short-circuiting gear

3 * I’2² R2 = 851 – 250 = 601

———————

R2 = 601 / 3 * (20)² = 0.501 Ω

## An induction motor has rotor resistance of 0.003 Ω/phase. If resistance is increased to 0.0006 Ω per phase, the maximum torque will be

An induction motor has rotor resistance of 0.003 Ω/phase. If resistance is increased to 0.0006 Ω per phase, the maximum torque will be:

1. Remain the same
2. Be reduced to half
3. Increase by 50%
4. Increase by 2 times

Correct answer: 1. Remain the same

Solution:

The torque of 3-phase induction motor for fixed supply voltage under normal running conditions is

T = KsR2/(R2² + s² X2²)

——————-

For maximum torque R2 = sX2

Therefore, above equation bcecomes

T∝1/2X2

Therefore, change of rotor resistance R2 doesn’t effect value of maximum torque.