Discussion
The capacitor rating needed for that correction, with \( \tan\phi \) of 0.882 and 0.329, is about
Answer: C
The difference of the tangents is 0.553, and multiplying by the 500 kW active power gives 277 kilovar. Note that this exceeds the 141 kVA reduction in apparent power, since the two are different quantities added at right angles.
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