Discussion
The rotor copper loss of an induction motor is
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:
\[ P_{cu,r} = s P_g, \qquad P_{mech} = (1-s) P_g \]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.
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