Induction Motor Drives Questions and Answers

Answer: C

Access to the rotor circuit through brushes and slip rings is what the method requires, and a cage rotor offers none. That mechanical complication is why cage machines with converters have largely displaced the wound rotor machine.

Answer: D

Correct answer: the slip power is wasted in the resistance

The rotor circuit carries \( sP_g \), and in this scheme all of it is dissipated as heat in the external resistors. At 30 per cent slip that is nearly a third of the air gap power thrown away, so the efficiency is roughly \( 1 - s \). The method does reduce the speed below synchronous, and it leaves the maximum torque untouched; it is the wasted energy that condemns it, and the slip power recovery schemes exist precisely to reclaim it.

Answer: A

Correct answer: to the supply through a converter

Instead of burning the slip power in resistors, a rectifier and inverter take it from the slip rings and feed it back to the mains. The machine then runs at reduced speed with an efficiency close to that at full speed, because the slip power is recovered rather than lost. The scheme needs a wound rotor with slip rings; it cannot be applied to a cage machine.

Answer: B

Correct answer: the slip power over the intended speed range

The converter handles only what flows in the rotor circuit, which is \( sP_g \). A drive intended to work between full speed and 70 per cent of it needs a converter rated for about 30 per cent of the machine rating, which is the great economy of the scheme compared with a full-rated stator-side inverter. Narrowing the speed range shrinks the converter further; widening it enlarges it.

Answer: C

Because the diode rectifier in the rotor circuit passes power in one direction only, slip power can leave the rotor but not enter it. A Scherbius arrangement with a bidirectional converter covers both sides of synchronous speed.

Answer: D

Correct answer: can pass slip power in both directions

A Kramer drive uses a diode rectifier in the rotor, so slip power can only flow outward and the machine runs below synchronous speed. A Scherbius drive uses a fully controlled converter, or a cycloconverter, which can also inject power into the rotor at slip frequency and so drive the machine above synchronous speed. Both act on the rotor circuit and both need a converter; the difference is the direction of flow.

Answer: A

The stiff dc link and the switching pattern together define the voltage applied, and the machine draws whatever current results. It is the arrangement used in almost every modern drive, generally with pulse width modulation.

Answer: B

A large link inductor makes the source stiff in current, and the machine voltage is whatever the machine and the imposed current produce. The arrangement is robust and inherently regenerative but is now largely confined to very large drives.

Answer: C

Correct answer: synthesise a nearly sinusoidal current in the machine

The inverter switches the dc link voltage on and off at a high carrier frequency, varying the width of the pulses so that their average follows a sinusoid. The winding inductance smooths what is left, and the current in the machine is close to sinusoidal with the harmonics pushed up around the carrier, where they are easily filtered and contribute little torque ripple. Varying the modulation depth and the fundamental frequency together is also how a PWM drive realises constant V/f.

Answer: D

Higher frequency means smaller current excursions between switchings but more switchings to pay for. The choice is a compromise, influenced also by audible noise and by the stresses the fast edges place on the winding insulation.