A single phase motor is connected to 400V, 50Hz supply. The motor draws a current of 31.7A at a power factor 0.7 lag. The capacitance required in parallel with motor to raise the power factor of 0.9 lag (in micro farads) is __________

Correct Answer: 94.62
Active power drawn by the motor=VIcosФ = 400*31.7*0.7 = 8876 W Reactive power = VIsinФ=400*31.7*sin(45.57.29) = 9055.3 VAR New power factor=cosθ2 = 0.9 θ2=cos-1(0.9) Q2=8876*tan(25.84) = 4298.855 VAR Change in reactive power=9055.3-4298.855 = 4756.4 VAR Qc = V2/Xc = V2*2πfC C=4756.4/(4002*2π*50) = 94.62μF.