Discussion
A 500 kW load at 0.75 power factor draws an apparent power of about
Answer: A
Correct answer: 667 kVA
\[ S = \frac{P}{\cos\phi} = \frac{500}{0.75} = 667\ \mathrm{kVA}. \] The apparent power is what the cables, switchgear and transformer must carry, so a poor power factor makes the whole installation work harder for the same useful output - and it is why the utility bills on kVA or penalises a low power factor.
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