Discussion
A dc generator delivers 20 kW with total losses of 2 kW. Its efficiency is
Answer: A
Correct answer: 91 per cent
Efficiency is output over input, and the input is the output plus the losses:
\[ \eta = \frac{20}{20 + 2} = \frac{20}{22} \approx 0.909 = 91\% \]Dividing the output by itself plus losses is the generator convention; for a motor the input is measured and the output is input less losses. The 90 per cent answer comes from dividing the losses by the output rather than by the input.
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