A circuit containing an `80mH` inductor and a `60muF` capacitor in series is connected to a `230V-50Hz` Supply. The resistance in the circuit is negligible .
(a) Obtain the current amplitude and rms currents.
(b) Obtain the rms values of voltage across inductor and capacitor.
(c ) What is the average power transferred to the inductor and to the capacitor?
(d) What is the total power absorbed by the circuit?

Explanation

Inductance, L=80mH=`80xx10^(-3)H`
Capacitan ce, `C=60muF=60xx10^(-6)F`
Supply voltage, V=230V
Frequency, V= 50Hz
Angular frequency , `oemga=2nv=100`n rad/s
peak voltage, `v_(0)=Vsqrt(2)=23-sqrt(2)V`.
(a) Maximum current is given as:
`I_(0)=(V_(0))/((omegaL-( 1)/(omegaC)))`
`=(230sqrt(3))/((100pixx80xx10^(-3)-(1)/(100pixx6010^(-6)))`
`=(230sqrt(2))/((8pi-(1000)/(6pi)))=-11.63A`
the negative sign appears because `omegaLlt(1)/(omegaC)`.
Amplitude of maximum current. `|I_(0)|=11.63A`
Hence, rms value of current. `I=(I_(0))/sqrt(2)=(-11.63)/sqrt(3)=-8.22A`
(b) Potential difference across the inductor,
`V_(L)=IxxomegaL`
`=8.22xx100nxx80xx10^(-8)`
`=206.61V`
Potential difference across the capacitor,
`V_(C )=Ixx(1)/(omegaC)`
`=8.22xx(1)/(100pixx60xx10^(-6))=436.3V`
(c) Average power consumed by the inductor is zero as actual voltage leads the current `(pi)/(2)`
(d) Average power consumed by the capacitor is zero as voltage lags current by `(pi)/(2)`.
(e) The total power absorbed (averaged over one cycle ) is zero.