A single phase motor connected to 400 V, 50 Hz supply takes 25 A at a power factor of 0.75 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.95 lagging.

• A. 72.55 pF
• B. 62.55 pF
• C. 82.55 pF
• D. 92.55 pF

Answer: (C)

For single phase induction motor,
P = V*I*cosφ
= 400*25*0.75
= 7500 W
Q1 = P*tan(φ) = 7500*0.80
= 6015 VAR
Now power factor change to 0.95
So, φ' = 25.84 degree
Q = 7500*tan(18.19)
= 2464.424
Qnet = Q - Q'
= 6015 - 2464.424
= 4500.8
C = V²/(2*ᴨ*f*Q)
= (400*400)/(2*ᴨ*50*3575.76)
= 8.255*10^(-5)
= 82.55 μF