This document provides a formula and example for calculating the capacitance needed for power factor correction of an induction motor. It explains that adding a capacitor in parallel with a single phase motor reduces the current supplied by the grid. This connection decreases power loss in the supply wires and voltage drop, increasing the voltage at the motor terminals. Power factor correction of an induction motor from 0.8 to unity can reduce wire losses by 36%.

1. Power Factor Correction Formula
Q = Reactive Power ( Vars, Joules per second)
V = System Voltage(Volts)
Xc = Capacitive Reactance (Ohms) = 1/(2π FC) Farads
F = System Frequency (Hertz)
C = Capacitance (Farads)
Q = V2
/ Xc Vars , Xc = V2
/ Q (Ohms)
1/(2π FC) = V2
/ Q ,
C = Q/(2πFV2
) Farads
Example:
Q = 250 Vars
V = 120 volts
F = 60 Hertz
C = 250/(2π60x1202
) = 46 x10-6
Farads or 46 micro Farads
The capacitor is connected in parallel with the single phase motor. The capacitor supplies Vars
to the motor and the motor supplies Vars to the capacitor. This connection reduces the current
supplied by the grid. Power loss in the supply wires is reduced and voltage drop decreases
increasing voltage at the motor terminals . Induction motor Power Factor Correction from 0.8 to
unity reduces wire losses by 36%.
Daniel Gray, P.Eng.
Port Elgin, Ontario
N0H 2C2
Cell: 519-377-8006
Home: 519-389-4558