A single phase motor connected to 400 V, 50 Hz supply takes 30 A at a power factor of 0.7 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.9 lagging.

• A. 89.5 μF
• B. 69.5 μF
• C. 99.5 μF
• D. 79.5 μF

Answer: (A)

For single phase induction motor,
P = VIcosφ
P = 400*30*0.7
P = 8400 W
Q1 = P*tan(φ) = 8400*1.020
Q = 8568.82 VAR
Now power factor change to 0.9
So, φ' = 25.84 degree
Q = 8400*tan(25.84)
Q = 4067.95
Qnet = Q - Q'
Qnet = 8568.82 - 4067.95
Qnet = 4500.8
C = V²/(2ᴨf)
C = (400*400)/(2ᴨ*50)
C = 8.95*10⁻⁵
C = 89.5 μF