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

Here, `L = 50 mH = 80 xx 10^(-3) H, C = 60 mu F = 60 xx 10^(-6) F, R = 0, E_(v) = 230 V, E_(0) = sqrt2 xx E_(v) = sqrt2 xx 230 V`
`v = 50 Hz, omega = pi v = 100 pi rad//s`
(a) `I_(0) = ? I_(v) = ?`
`I_(0) = (E_(0))/((omega L - (1)/(omega C))) = (230 sqrt2)/((100 pi xx 60 xx 10^(-6) - (1)/(100 pi xx 60 xx 10^(-6)))) = (230 sqrt2)/((8 pi - (1000)/(6 pi))) = (230 sqrt2)/(-27.91) = 11.63` amp.
`I_(v) = (I_(0))/(sqrt2) = (-11.63)/(1.414) = - 8.23` amp
Negative sign appear as `omega L lt (1)/(omega C) :.` e.m.f. lags behind the current by `90^(@)`
(b) Across L, `V = I_(v) xx omega L = 8.23 xx 100 pi xx 80 xx 10^(-3) = 206.74` volt
Across C, `V = I_(v) xx (1)/(omega C) = 8.23 xx (1)/(100 pi xx 60 xx 10^(-6)) = 436.84` volt
As voltages across L and C `180^(@)` out of phase, therefore, they get subtracted.
That is why applied r.m.s. voltage = 436.84 - 206.74 = 230.1 volt
(c ) Average power transferred over a complete cycle by the source to inductor is always zero because of phase difference of `pi//2` between voltage and current through L.
(d) Average power transferred over a complete cycle by the source to the capacitor is also zero becasue of phase difference of `pi//2` between voltage and current through C.
(e) Total average power absorbed by the circuit is also, therefore zero