Induction Motor Drives Questions and Answers

Answer: C

Correct answer: the rotor resistance

The maximum torque is

in which the rotor resistance does not appear. Adding rotor resistance therefore moves the peak but does not raise or lower it. The applied voltage, the rotor reactance and the supply frequency all do affect it, the voltage as its square.

Answer: D

Correct answer: the square of the voltage

At a fixed slip the rotor current is proportional to the applied voltage, and the torque is proportional to the product of flux and rotor current, both of which scale with \( V \). Hence

A ten per cent drop in supply voltage costs about nineteen per cent of the torque, which is why an induction motor on a weak supply may fail to start a load it handles easily at rated voltage.

Answer: A

Correct answer: in an unstable region with most load characteristics

Between standstill and \( s_{max} \) the torque rises as the speed falls. If the load torque exceeds the motor torque at some point in that range, the machine slows, its torque falls further, and it stalls; there is no stable intersection with a constant-torque load. Stable running is on the other side of the peak, between \( s_{max} \) and zero slip, where the torque falls steeply as the speed rises and any speed disturbance is self-correcting.

Answer: B

Correct answer: generates and returns power to the supply

Driven above \( N_s \) the slip becomes negative, the rotor emf reverses, and the torque opposes the direction of rotation. The machine acts as an induction generator, delivering real power to the supply while still drawing its magnetising reactive power from it. This is the regenerative braking region, used on hoists and on wind turbines with cage generators.

Answer: C

Correct answer: one third of its direct-on-line value

In star each winding sees \( V_L/\sqrt{3} \), so the phase current falls by \( \sqrt{3} \); and the line current in star equals the phase current, whereas in delta it is \( \sqrt{3} \) times the phase current. The two factors combine to give

The torque, proportional to the square of the voltage, falls to one third as well. The one-ninth figure is a common trap: it is the ratio of the phase currents multiplied out incorrectly.

Answer: D

Torque follows the square of the phase voltage, and the phase voltage falls by \( \sqrt{3} \), so the torque falls by three. A load needing more than a third of full-load torque to break away cannot be started this way.

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.