A capacitor having capacitance C is put in series with a 20 `Omega` resistance . The power factor is equal to 0.5 . Find the value of C if AC supply is 90 V - 200 Hz.

Explanation

R = 20 `Omega`
`phi = (R)/(Z) = 0.5`
` Z = (R)/(phi) = (20)/(0.5) = 40 Omega`
E = 90 V, f = 200 Hz
`Z^(2) = R^(2) + X_(C)^(2)`
`Z^(2) - R^(2) = X_(C)^(2)`
`(40)^(2) - (20)^(2) = X_(C)^(2)`
34.64 = `X_(C) `
34.64 = `(1)/(omega C) `
`2 pi f C = (1)/(34.64)`
C = `(1)/(2 xx 3.14 xx 200 xx 34.64)`
= `2 xx 10^(-5)` F