Discussion
For a three-phase induction motor, if the gap power is very high as compared to the sum of the core loss and stator copper loss, and developed power is very high as compared to no load power, then what is the efficiency when it operates with a slip of 0.15?
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.
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