Discussion
Improving that load's power factor from 0.8 to 0.95 reduces the apparent demand to about
Answer: A
Correct answer: 526 kVA
\[ \frac{500}{0.95} = 526\ \mathrm{kVA}. \] Correcting the power factor from 0.8 to 0.95 cuts the apparent demand by about a hundred kVA for the same useful output, and at 350 per kVA that is some 35 000 a month - which is what pays for the capacitor bank, usually within a year or two.
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