Induction Motor Drives Questions and Answers

Answer: C

Correct answer: interchanging any two supply leads

Swapping any two of the three leads reverses the phase sequence, and the direction of the rotating field follows the sequence. Reversing all three restores the original sequence and changes nothing - it is the same rotation relabelled. Opening the rotor circuit merely removes the torque, and raising the frequency speeds the machine up in the direction it was already going.

Answer: D

Correct answer: unity

\[ s = \frac{N_s - N}{N_s}, \] so at standstill \( N=0 \) and \( s=1 \). The slip runs from unity at rest down to a few per cent at full load and to zero only at synchronous speed, which a motor never reaches because it would then have no induced rotor current and no torque. Slip goes negative only when the machine is driven above synchronous speed and generates.

Answer: A

Correct answer: a few per cent

A normal cage motor runs at two to five per cent slip at full load, so a four-pole machine on a 50 Hz supply turns at about 1440 rpm against a synchronous 1500. The slip is small because the rotor resistance is low, and it must be non-zero because the rotor current, and hence the torque, is produced by the relative motion.

Answer: B

Reduced voltage means reduced flux, so the same current gives less torque and the machine slips further to carry its load, drawing yet more current. Sustained undervoltage is therefore a thermal hazard as well as a performance one.

Answer: C

Establishing the air gap flux takes reactive current whatever the load, and at no load that is nearly all the machine draws. It is typically a quarter to a third of rated current, which is why light loading gives such a poor power factor.

Answer: D

Continuous frequency control with the voltage tracking it gives efficient operation at any speed, and rated torque is available throughout the constant-flux region. That combination is what displaced the dc drive from most new work.

Answer: A

Correct answer: a little over 90 per cent overall

A modern motor of a few kilowatts is around 90 to 93 per cent efficient and the inverter around 96 to 98, so the product is a little over 90 per cent at rated load. The figure falls at light load, which is why the flux reduction discussed elsewhere in this topic is worth doing, and no drive is anywhere near 100 per cent since both the copper losses and the switching losses are unavoidable.

Answer: B

Correct answer: fan power falls with the cube of the speed

For a fan on a fixed system the flow varies with speed, the pressure with the square of the speed, and the power with the cube: \[ \frac{P_2}{P_1} = \left(\frac{N_2}{N_1}\right)^{3}. \] Running at 80 per cent speed therefore takes about half the power, and at 50 per cent about an eighth. That cube law is the whole economic case for variable-speed fan and pump drives.

Answer: C

Correct answer: dissipates the pressure it drops

A damper does not reduce the work the fan does; it throttles the flow by adding resistance, and the pressure dropped across it is converted straight to heat and turbulence. The fan still runs at full speed against a higher pressure and absorbs almost as much power as before. A drive reduces the speed instead, so the cube law applies and the energy is never generated at all.

Answer: D

Rated current at low frequency gives rated copper loss with almost no air flow, so the machine overheats unless it is cooled independently or a larger frame is used. It is one of the most common errors in variable-speed retrofits.