A `10 mu g` capacitor is in series with a `50 Omega` resistance and the combination is connected to a `220V`, `50Hz` line. Calculate (i) the capactive reactance, (ii) the impedance of the circuit and (iii) the current in the circuit.

Explanation

Here, `C = 10 mu F = 10 xx 10^(-6) = 10^(-5) F`
`R = 50` ohm, `E_("rms") = 220 V, v = 50 Hz`
(i) Capacitive reactance,
`X_(C ) = (1)/(omega C) = (1)/(2 pi 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`