Class_30/10/2026

EV 605/EE 605

Class_30/10/2026

A 400 V, 50 Hz motor operates at 30 Hz with constant V/f. The required voltage is:

1: A 400 V, 50 Hz motor operates at 30 Hz with constant V/f. The required voltage is:

400 V

320V

240 V

300 V

Why is voltage boost typically required at very low frequencies when implementing open-loop $V/f$ speed control for an induction motor?

2: Why is voltage boost typically required at very low frequencies when implementing open-loop $V/f$ speed control for an induction motor?

To counteract the high rotor slip frequency experienced during startup transients.

To prevent core saturation caused by excessive leakage flux linkage in the stator slots.

To increase the mechanical pull-out torque to match high starting load friction.

A. To compensate for the significant voltage drop across the stator resistance.(Your answer) Correct answer At low frequencies, the stator leakage reactance is very small, making the stator resistance drop comparable to the terminal voltage, which severely reduces air-gap flux.

With voltage reduction at constant frequency, the slip at maximum torque is approximately:

3: With voltage reduction at constant frequency, the slip at maximum torque is approximately:

Proportional to voltage

Inversely proportional to voltage

Proportional to voltage squared

Unchanged

A 400 V, 50 Hz motor is supplied at 400 V, 25 Hz. Neglecting saturation initially, its air-gap flux is

4: A 400 V, 50 Hz motor is supplied at 400 V, 25 Hz. Neglecting saturation initially, its air-gap flux is

0.5 times rated

4 times rated

Equal to rated

2 times rated

A motor has 4% slip at 50 Hz. Under constant V/f control, frequency reduces to 25 Hz while slip frequency remains unchanged. Its new slip is

5: A motor has 4% slip at 50 Hz. Under constant V/f control, frequency reduces to 25 Hz while slip frequency remains unchanged. Its new slip is

2%

8%

4%

16%

An induction motor is driven above its synchronous speed by a prime mover while connected to a fixed-frequency grid. In this regenerative braking region, the slip is:

6: An induction motor is driven above its synchronous speed by a prime mover while connected to a fixed-frequency grid. In this regenerative braking region, the slip is:

Negative ( s<0), and the motor acts as an induction generator

Between zero and one ( 0<s<1), maintaining normal motoring action.

Greater than unity ( s>1), causing high dynamic braking currents.

Zero ( s=0), resulting in zero torque and floating rotor operation.

In rotor resistance speed control of a wound-rotor induction motor, increasing the external rotor resistance

7: In rotor resistance speed control of a wound-rotor induction motor, increasing the external rotor resistance

Decreases the slip at which maximum torque occurs.

Increases both the maximum torque magnitude and the synchronous speed.

Shifts the peak torque point to a higher slip value without changing the maximum torque magnitude.

Improves the overall drive efficiency across all speed ranges.

During plugging (reverse-current braking) of an induction motor, the phase sequence of the stator supply is reversed while the motor is running. What is the value of slip during this transition?

8: During plugging (reverse-current braking) of an induction motor, the phase sequence of the stator supply is reversed while the motor is running. What is the value of slip during this transition?

Slip is exactly 1 ( s=1).

Slip is exactly zero ( s=0).

Slip lies between 1 and 2 ( 1

Slip is negative ( s<0)

A 4-pole, 50 Hz induction motor runs at 1440 rpm. It is reconnected for 6 poles. What supply frequency maintains the same speed if slip frequency remains unchanged?

9: A 4-pole, 50 Hz induction motor runs at 1440 rpm. It is reconnected for 6 poles. What supply frequency maintains the same speed if slip frequency remains unchanged?

76 Hz

72 Hz

74 Hz

75 Hz