A Single phase 220V, 50Hz, motor takes a supply current of 50A at a P.F (Power factor) of 0.6. The motor power factor has to be improved to 0.9 by connecting a capacitor in parallel with it. Calculate the required capacity of Capacitor in both kVAR and Farads.

(1) Calculate total power：
Motor Input = Power = V x I x Cosθ
                                   = 220V x 50A x 0.6
                                   = 11kW

(2) Calculate kVAR needed：
Actual     P.F = Cosθ1 = 0.6
Required P.F = Cosθ2 = 0.9
θ1 = Cos^-1 (0.6) = 53.13 ; Tan (53.13) = 1.3333
θ2 = Cos^-1 (0.9) = 25.84 ; Tan (25.50) = 0.4843
Required Capacitor kVAR to improve P.F from 0.60 to 0.90
= P ( Tan(Cos^-1(0.6)) – Tan(Cos^-1(0.9)) )
= 11kW (1.3333– 0.4843)
= 9.339 kVAR

(3) Convert kVAR to Capacitance
C in μF = kVAR / (2 π f VV) x 1,000,000,000
Putting the Values in the above formula
= 9.339kVAR / (2 x π x 50 x 220 x 220) x 1,000,000,000
= 9.339 / 15,205,308.44 x 1,000,000,000
= 9.339 / 1,520,530,844,000,000,000
= 614.2μF

電容單位 Farad(F)
Values of capacitors are usually specified in
farads (F),
microfarads (μF) 10^-3
nanofarads (nF) 10^-6
picofarads (pF) 10^-9
femtofarads (fF) 10^-12
attofarads (aF) 10^-15

1 mF (millifarad, one thousandth (10^-3) of a farad) 1F = 1,000 mF
1 μF (microfarad, one millionth (10^-6) of a farad)    1F = 1,000,000 μF
1 nF (nanofarad, one billionth (10^-9) of a farad)      1F = 1,000,000,000 nF
1 pF (picofarad, one trillionth (10^-12) of a farad)     1F = 1,000,000,000,000 pF

亦可由其它單位灌計算出來：