Discussion
A three-phase 50 Hz, 12 pole, 200 kW slip-ring induction motor drives a fan whose torque is proportional to the square of speed. At full load, the motor slip is 0.045. The rotor resistance measured between any two slip rings is \(61\ \mathrm{m\Omega}\). What is the speed of the motor?
Answer: A
The synchronous speed of the 12 pole motor on a 50 Hz supply is
\[ N_s = \frac{120 \times 50}{12} = 500\ \text{rpm} \]
At a slip of 0.045 the rotor speed is
\[ N = (1 - 0.045) \times 500 = 477.5\ \text{rpm} \]
\[ \begin{aligned} \omega = \frac{2\pi \times 477.5}{60} \\ &= 50\ \text{rad/s} \end{aligned} \]
The power rating and the rotor resistance are not needed to find the speed. The values of 100 and 200 rad/s are too high for a 12 pole motor, since even the synchronous speed is only 52.4 rad/s. The answer is 50 rad/s, option A.
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