Question

A synchronous motor improves the power factor of a 300kW load from 0.6 lagging to 0.8 lagging and simultaneously the motor carries a load of 100 kW. The power factor at which the motor operates is ______

• A. 0.7

Answer 1

The correct option is A 0.7
Load, $P_1=300kW,motorload=100kW$
P.f. of load, $cos{\varphi }_1=0.6lagging$
P.f. of load, $cos{\varphi }_2=0.8lagging$
Combined load $P=P_1+P_2=300+100=400kW$
Leading kVAR taken by the motor
$=P_{1}tan{\varphi }_1-P_{2}tan{\varphi }_2$
$=300tan\left(co{s}^{-1}0.6\right)-400tan\left(co{s}^{-1}0.8\right)$
= 400 - 300 = 100 kVAR
Rating of the motor $=\sqrt{(100{)}^2+(100{)}^2}=141.42kVA$
$P.fofmotor,cos{\varphi }_m=\frac{100}{141.42}=0.707$ leading