A `100 mu F` capacitor in series with a `40 Omega` resistance is connected to a `100 V_(1) 60 Hz` supply. Calculate (i) the reactance (ii) the impedance and (iii) maximum current in the circuit.

Explanation

Here, `C = 10 mu F = 10 xx 10^(-6) = 10^(-5)F`
(i) Capacitive reactance,
`X_(C) = (1)/(omega C) = (1)/(2pi v C) = (1)/(2 xx 3.14 xx 50 xx 10^(-5)) = 318.5 Omega`
(ii) Impedance of CR circuit.
`Z_(CR) = sqrt(R^(2) + X_(C)^(2)) = sqrt((50)^(2) + (318.5)^(2))`
`= 322.4 Omega`
(iii) Current, `I_(rms) = (E_(rms))/(Z_(CR)) = (220)/(322.4) = 0.68 A`