Gate 2016 EE -Set 2 – Question 41

Given that the rating is 500kvA and $$\cos \phi =0.8$$

Active power, $$P_{1}=500\times 0.8=400kW$$

Reactive power, $$Q_{1}=500\times 0.6=300kvAR$$

To raise the power factor to unity, the motor should supply 300kvAR

The motor ratings are now, 100kW, 300kvAR

$$\phi =\tan ^{-1}\left ( \frac{Q}{P} \right )=\tan ^{-1}\left ( \frac{300}{100} \right )$$

$$\Rightarrow \phi =71.56^{\circ}$$

$$\cos \phi =\cos (71.56)=0.316$$