Discussion
A 1,000 kW synchronous motor at 0.9 power factor leading supplies reactive power of about
Answer: B
Correct answer: 484 kVAr
The apparent power is
\[ S = \frac{P}{\cos\phi} = \frac{1000}{0.9} = 1111 \text{ kVA} \]and with \( \sin\phi = \sqrt{1 - 0.81} = 0.436 \),
\[ Q = S\sin\phi = 1111 \times 0.436 \approx 484 \text{ kVAr} \]Because the power factor is leading, that 484 kVAr is supplied to the network, offsetting the lagging demand of induction motors elsewhere in the plant.
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