Induction Motor Drives Questions and Answers

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 \( 1/f \).

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.