Discussion
If that machine has rotational losses of 500 W, its shaft output is
Answer: C
Correct answer: 7.5 kW
The converted power is \( EI_a \), and the rotational losses come out of it:
\[ P_{mech} = 200 \times 40 = 8000 \text{ W} \]\[ P_{shaft} = 8000 - 500 = 7500 \text{ W} = 7.5 \text{ kW} \]The input to the armature was 8.8 kW, so the efficiency of the armature circuit is about 85 per cent, the losses being 800 W of copper and 500 W of rotational loss.
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