A circuit draws 330 W from a 110 V , 60 Hz AC line. The power factor is 0.6 and the current lags the voltage. The capacitance of a series capacitor that will result in a power factor of unity is equal to

Options

A. `31 mu F `
B. ` 54 mu F `
C. `151 mu F `
D. `201 mu F `

Answer

B

Explanation

Resistance `(R) = (V^(2))/(P) = (110 xx 110)/( 330) = (110)/(3) Omega `
When ` cos phi = 0.6 rArr (R)/(sqrt(R^(2) + X_(L)^(2))) = 0.6 `
`rArr X_(L)^(2) = (R^(2))/((0.6)^(2)) - R^(2) rArr X_(L) = (4 R)/(3)`
When cos ` phi = 1 `
i.e. `X_(L) = X_(C)`
`rArr = (4 R)/(3) = (4)/(3) xx (110)/(3) = (440)/(9) Omega rArr (1)/(2 pi f C) = (440)/(9)`
`rArr C = (9)/(440 xx 2 xx 3.14 xx 60) = 54 mu F `