A 60 cycle AC, circuit has a resistance of `200 Omega` and inductor of 100 mH. What is the power factor? What capacitance placed in the circuit will make the power factor unity?

Explanation

`Z=sqrt(R^(2)+(omegaL)^(2))`
`=sqrt((200)^(2)+(2pi60 xx 100 xx 10^(-3))^(2)) = 203.52Omega`

Power factor, `cosphi = R/Z= (200)/(203.52)=0.983`
Now, for the power factor to be unit, `phi=0` that means
`X_(L) = X_(C)` or `omegaL=1(omegaC)`
`therefore C=1/((omega)^(2)L)=1/((2pi60)^(2)xx 100 xx 10^(-3))`
`therefore C=70.36muF`