Discussion
Correcting that 500 kW load from 0.75 to 0.95 power factor reduces its apparent power to about
Answer: B
Correct answer: 526 kVA
\[ \frac{500}{0.95} = 526\ \mathrm{kVA}. \] The correction releases about 141 kVA of capacity in the cables and transformer - over a fifth - without any change to the real power delivered. That released capacity is often worth more than the tariff saving, since it can defer an upgrade.
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