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An inductive circuit draws a power 550 W...

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.

Text Solution

Verified by Experts

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`.
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