Three Phase Induction Motors Questions and Answers

Answer: A

The slip speed of an induction motor is the difference between the speed of the rotating magnetic field and the speed of the rotor.

\[ N_{slip} = N_s - N_r = sN_s \]

Because the rotor always runs a little slower than the synchronous speed, \(N_s - N_r\) is positive, so the order in option A is the one that gives a positive slip. Option B gives the negative of it, and the rotor speed or the synchronous speed alone is not a slip speed. The answer is option A.

Answer: B

In an induction motor the rotor input \(P_2\) divides into the rotor copper loss \(sP_2\) and the gross mechanical power \((1 - s)P_2\).

\[ \begin{aligned} P_m = (1 - s)P_2 = (1 - 0.04) \times 200 \\ &= 192\ \text{kW} \end{aligned} \]

The rotor copper loss is \(0.04 \times 200 = 8\) kW, and 200 − 8 = 192. The value 208 kW adds the loss, 196 kW subtracts 4 kW instead of 4 per cent, and 184 kW subtracts twice the loss. The answer is 192 kW, option B.

Answer: A

The synchronous speed of a 4 pole motor on a 50 Hz supply is

\[ \begin{aligned} N_s = \frac{120f}{P} = \frac{120 \times 50}{4} \\ &= 1500\ \text{rpm} \end{aligned} \]

\[ \begin{aligned} s = \frac{N_s - N}{N_s} \\ &= \frac{1500 - 1380}{1500} = \frac{120}{1500} \\ &= 0.08 = 8\% \end{aligned} \]

The values 6%, 4% and 2% would need speeds of 1410, 1440 and 1470 rpm. The answer is 8%, option A.

Answer: D

In an induction motor the air gap power \(P_g\) is divided into the rotor copper loss \(sP_g\) and the mechanical power developed \(P_m = (1 - s)P_g\). If the core and stator copper losses are very small compared with the gap power, then the input is nearly equal to the gap power, and if the developed power is much greater than the no load (friction and windage) losses, the output is about equal to the developed power.

\[ \begin{aligned} \eta \approx \frac{P_m}{P_g} = 1 - s = 1 - 0.15 \\ &= 0.85 = 85\% \end{aligned} \]

The option 15% is the slip itself, 100% would be a motor with no slip, and 95% would be a slip of 0.05. The answer is 85%, option D.

Answer: D

Rotor resistance control cannot be used for the speed control of a squirrel cage induction motor. The bars of a cage rotor are short-circuited inside the rotor, so no external resistance can be put in the rotor circuit. A wound rotor has slip rings through which resistance can be added.

Stator voltage control, stator frequency control and the control of the stator current can all be applied to a cage motor, since they act on the stator. So the fourth option is the answer.

Answer: C

The frequency of the rotor current is 60 Hz. The rotor frequency is the slip times the supply frequency, and when the rotor is stationary the slip is 1.

\[ f_r = s f = 1 \times 60 = 60\ \text{Hz} \]

The value 0 Hz would be for a rotor running at synchronous speed, where the slip is zero, 30 Hz would be for a slip of 0.5, and 120 Hz would be for a slip of 2. So the third option is correct.

Answer: A

The stator frame of an induction motor can be made of cast iron. The frame gives mechanical support and protection, and cast iron is cheap, strong in compression and easy to cast into a frame, while the laminated core inside carries the flux.

The stator winding and the rotor winding are made of copper or aluminium conductors, and brushes are made of carbon. So the first option is the answer.

Answer: D

The wrong statement is that the direction of rotation cannot be reversed. The direction of rotation of a 3-phase induction motor can be reversed by interchanging any two of the three supply lines, which reverses the direction of the rotating magnetic field.

The other statements are true: it is self-starting, it has a high efficiency, and it always runs below synchronous speed, since a rotor at synchronous speed would have no induced current and so no torque. So the fourth option is the answer.

Answer: B

Low power factor is the statement that is not correct for a squirrel cage induction motor, according to the key. A squirrel cage motor has a simple, rugged construction, a high efficiency and a low cost, and these are the usual advantages given for it.

The power factor of an induction motor is in fact low on light load, so the statement is not wholly wrong, which makes the question arguable. The answer follows the key, and the second option is the one given.

Answer: D

The slip is 6 %. The synchronous speed of a 4-pole motor on a 50 Hz supply is 1500 rpm, and the slip is the difference between this and the rotor speed, divided by the synchronous speed.

\[ N_s = \frac{120 \times 50}{4} = 1500\ \text{rpm} \]

\[ s = \frac{1500 - 1410}{1500} = 0.06 = 6\% \]

The values 3 %, 4 % and 5 % would give speeds of 1455, 1440 and 1425 rpm. So the fourth option is correct.