GATE 2016 EE (EE2) – Question 51

Power Systems · Voltage and Frequency control, Power factor correction · 2 marks · Numerical answer

An industry consumes 500 kVA at 0.8 lagging pf. A synchronous motor taking 100 kW is added to make the total pf unity. The motor’s power factor is ________.

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Correct answer: 0.31 to 0.33

Explanation

The original real power is 500×0.8=400 kW and reactive power is 500×0.6=300 kvar lagging. The added motor must absorb −300 kvar, i.e. supply 300 kvar leading, to cancel it.

Its apparent power is √(100²+300²)=316.228 kVA. Thus motor pf is 100/316.228=**0.316228 leading**, within 0.31–0.33.

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