Discussion
A load of 500 kW at a power factor of 0.8 imposes an apparent demand of
Answer: D
Correct answer: 625 kVA
Apparent power is the real power divided by the power factor: \[ \frac{500}{0.8} = 625\ \mathrm{kVA}. \] The network must be sized for the 625 kVA, not the 500 kW, because it is the current that heats the conductors - which is exactly why demand is billed in kVA rather than kW.
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