Discussion
If the rotor input of an SCIM running with a slip of 4% is 200 kW, the gross power developed by its rotor is ______.
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.
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