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GATE 2017 EE – Question 49

Power Systems · Voltage and Frequency control, Power factor correction · 2 marks · Multiple choice

A load is supplied by a 230 V, 50 Hz source. The active power $P$ and the reactive power $Q$ consumed by the load are such that 1 kW $\leq P \leq$ 2 kW and 1 kVAR $\leq Q \leq$ 2 kVAR. A capacitor connected across the load for power factor correction generates 1 kVAR reactive power. The worst case power factor after power factor correction is

  1. 0.447 lag
  2. 0.707 lag
  3. 0.894 lag
  4. 1

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Correct answer: (B) 0.707 lag

Explanation

After the capacitor supplies 1 kVAR, the net reactive power is $Q - 1$, which lies between 0 and 1 kVAR. The power factor is worst when $P$ is smallest and the net $Q$ is largest, that is $P = 1$ kW and $Q_{net} = 1$ kVAR. Then $\cos\phi = \frac{1}{\sqrt{1^2 + 1^2}} = 0.707$ lagging.