A circuit containing a 80 mH inductor and a `60 mu F` capacitor in series is connected to a 230 V 50 Hz supply. The resistance of the circuit is negligible.
Obtain the current amplitude and rms values.

Explanation

(a) For `V=V_(0)sin omega t`
`I=(V_(0))/(|omegaL-(1)/(omegaC)|) sin (omega t +(pi)/(2)), " if "R=0`
where - sign appears if `omegaL gt 1//omegaC`, and + sign appears if `omegaL lt 1//omegaC.`
`I_(0)=11.6A, I_("rms")=8.24A`
(b) `V_("Lrms")=207V, V_("Crms")=437V`
(c) Whatever be the current I in L, actual voltage leads current by `pi//2`. Therefore, average power consumed by L is zero.
(d) For C, voltage lags by `pi//2`. Again, average power consumed by C is zero.
(e) Total average power absorbed is zero.