Question

An inductive circuit draws a power 550 W from a 220V-50Hz source. The power factor of the circuit is 0.8. The current in the circuit lags behind the voltage. Show that a capacitor of about `(1)/(42 pi) xx10^(-2) F` will have to be commected in the circuit to bring its power factor to unity.

Answer 1

Circuit is inductive, i.e. there is no capacitor in the circuit,
Given Power factor:
`cos(phi)=(R )/(sqrt(R^(2)+X_(L)^(2))=0.8 implies X_(L)=3/4 R`
Now power : `P=(E_(v)^(2))/(Z^2)R`
`550=((220)^(2)R)/(R^(2)+((3)/(4)R)^(2)) implies R = 56.32 Omega`
Now we have to bring the power factor to unity. For this
`X_(C )=X_(L) implies (1)/(omega C)=3/4 R implies C=4/(3 omega R)`
`implies C=(4)/(3xx2 pi 50 xx56.32) =(1)/(4224 pi) F`.