Induction Motor Drives Questions and Answers

Answer: A

Correct answer: \( 120f/P \)

The stator winding produces a field that rotates at the synchronous speed. A pair of poles is traversed in one cycle of the supply, so with \( P \) poles the field completes \( 2f/P \) revolutions per second, and in revolutions per minute

With \( f = 50 \) Hz a four-pole machine gives \( 120 \times 50 / 4 = 1500 \) rpm and a six-pole machine 1000 rpm. The other expressions either use 60 in place of 120, which is the per-pole rather than per-pole-pair figure, or invert the roles of \( f \) and \( P \) and so make the speed fall as the frequency rises.

Answer: B

Correct answer: \( (N_s - N)/N_s \)

Slip measures how far the rotor lags the rotating field, expressed as a fraction of the synchronous speed:

At standstill \( N = 0 \) and \( s = 1 \); at synchronous speed \( N = N_s \) and \( s = 0 \). Dividing the difference by \( N \) rather than \( N_s \) would make the slip infinite at standstill, and \( N/N_s \) is the per-unit speed, which is \( 1 - s \) rather than \( s \).

Answer: C

Correct answer: the slip times the supply frequency

The rotor conductors cut the rotating field at the relative speed \( N_s - N \), and the frequency of the emf induced in them is proportional to that relative speed:

At standstill \( s = 1 \) and the rotor frequency equals the supply frequency, which is why a wound rotor at rest behaves as a transformer secondary. At normal running slips of a few per cent the rotor frequency is only a hertz or two, and that low frequency is why the rotor reactance is small in normal operation.

Answer: D

Correct answer: the slip times the air gap power

Of the power \( P_g \) crossing the air gap, the fraction \( s \) is dissipated in the rotor resistance and the remaining \( (1-s) \) appears as mechanical power:

The loss is therefore fixed by the slip alone, not by the rotor resistance, and any scheme that runs the machine at a large slip wastes a correspondingly large share of the air gap power. This is the whole objection to rotor resistance and stator voltage control.

Answer: A

Since a fraction \( s \) of the air gap power is lost in the rotor, at best \( 1 - s \) reaches the shaft. A machine held at fifty per cent slip therefore cannot exceed fifty per cent efficiency however good the rest of it is.

Answer: B

Correct answer: transferred from stator to rotor across the gap

The air gap power is what remains of the input after the stator copper and core losses have been taken out, and it is the power carried across the gap by the field. It then divides between rotor copper loss and mechanical power in the ratio \( s : (1-s) \). It is not the terminal input, which is larger by the stator losses, and not the shaft output, which is smaller by the rotor loss and the friction and windage.

Answer: C

Torque is the air gap power referred to synchronous speed, which is why air gap power is sometimes called torque in synchronous watts. Using the mechanical power and the rotor speed gives the same torque, since both are reduced by the same factor.

Answer: D

Correct answer: unity

Starting means standstill, so \( N = 0 \) and

Slip zero is synchronous speed, where no torque is developed at all. The slip of maximum torque and the rated full-load slip are both particular operating points on the characteristic and are neither of them the starting condition.

Answer: A

Correct answer: the ratio of rotor resistance to rotor reactance

Maximum torque occurs where the rotor circuit transfers most power, which is where the effective rotor resistance \( R_2/s \) equals the standstill rotor reactance:

Because it depends on the ratio, adding external rotor resistance moves the peak to a higher slip without changing its height. A machine with \( R_2 = 0.1\,\Omega \) and \( X_2 = 0.5\,\Omega \) peaks at 20 per cent slip.

Answer: B

Correct answer: increases the starting torque without changing the maximum torque

The maximum torque depends on \( V \) and \( X_2 \) but not on \( R_2 \), so the height of the peak is fixed. What the added resistance does is raise \( s_{max} = R_2/X_2 \), sliding the peak towards standstill. If the resistance is chosen so that \( R_2 = X_2 \), the peak lands exactly at \( s = 1 \) and the machine starts with its maximum torque. This is the classic reason for using a wound rotor machine on a hard-starting load.