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?

A circuit containing an `80mH` inductor and a `60muF` capacitor in ser

1.	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?
Answer» 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.

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