Induction Motor Drives Questions and Answers

Answer: A

Correct answer: allows the reduction ratio to be chosen

An autotransformer starter can be tapped at 50, 65 or 80 per cent, so the reduction can be matched to the load rather than fixed. Star-delta gives one ratio only, and it requires all six winding ends brought out to the terminal box, which the autotransformer does not. At a tapping \( k \) the starting current and the torque both fall as \( k^2 \) at the motor, so the autotransformer gives more torque than star-delta for the same line current.

Answer: B

Phase-angle control of back-to-back thyristors raises the voltage gradually, limiting the current at the cost of torque. It cannot give more torque per ampere, which is where a full converter with frequency control differs fundamentally.

Answer: C

Correct answer: fan and pump loads

Reducing the stator voltage lowers the torque as \( V^2 \) and so lets the machine settle at a higher slip. That is workable only where the load torque itself falls steeply with speed, as it does for a fan or centrifugal pump, whose torque varies roughly as the square of the speed. A constant-torque load such as a hoist would simply stall, and a load whose torque rises as the speed falls is worse still.

Answer: D

Correct answer: large rotor loss at reduced speed

Speed is reduced by increasing the slip, and the rotor copper loss is \( sP_g \). Running at 50 per cent slip means half the air gap power is burnt in the rotor, inside the machine, where it is hard to remove. This restricts the method to small machines and to loads whose power demand falls quickly with speed, which limits the running time spent at high slip.

Answer: A

Since the synchronous speed is proportional to frequency, lowering the frequency lowers the whole characteristic without increasing the slip. That is what makes it an efficient method, unlike voltage or resistance control.

Answer: B

Flux is roughly the voltage divided by the frequency, so holding the voltage while lowering the frequency drives the iron into saturation and the magnetising current up. Reducing the voltage in proportion is what avoids it.

Answer: C

Correct answer: approximately constant flux in the machine

The stator emf is \( E \approx 4.44 f N \phi \), so holding \( V/f \) constant holds \( \phi \) constant. Constant flux means the machine retains its full torque capability at every frequency and neither saturates at low frequency nor is under-fluxed at high. The approximation is in neglecting the stator resistance drop, which is why a voltage boost is added at low frequency.

Answer: D

At a few hertz the resistive drop is a large fraction of the applied voltage, so the flux would collapse without an addition. The boost is either a fixed offset or is computed from the current, which is the more accurate arrangement.

Answer: A

The converter cannot exceed the supply voltage, so beyond that point the flux falls as the frequency rises and the available torque falls with it. The region is one of constant power, exactly as field weakening is in a dc drive.

Answer: B

Correct answer: from standstill up to base frequency

Below base frequency the drive raises the voltage with the frequency, keeping the flux and so the torque capability constant; the power rises in proportion to the speed. Above base frequency the voltage is already at its ceiling and cannot rise further, so the flux falls as \( 1/f \) and the machine enters the constant-power, field-weakening region with torque falling as \( 1/f \).