A series AC circuit contains an inductor `(20 mH)`, a capacitor `(100 muF)`, a resistor `(50 Omega)` and an AC source of `12 V, 50 Hz`. Find the energy dissipated in the circuit in `1000 s`.

Explanation

The time period of the source is
`T = 1/v = 20 ms`.
The given times 1000 s is much larger than the time period. Hence we can write the average power dissipated as
`P_(av) = V_(rms) i_(rms) cosvarphi`
where `cosvarphi = R/Z` is the power factor. Thus,
`P_(av) = V_(rms) V_(rms)/Z R/Z = (R V_(rms)^2/(z^2)`
`= ((50 Omega) (12 V)^2))/Z^2`
`= 7200/Z^2 OmegaV^2`. ... (i)
The capacitive reactance `1/(omegaC) = 1/(2pi xx 50 xx 100 xx 100^(-6)) Omega`
`= 100/(pi) Omega`.
The inductive reactance `= omegaL`
`= 2pi xx 50 xx 20 xx 10^(-3) Omega = 2pi Omega`.
The net reactance is ` X = 1/(omegaC) - omegaL`.
`= 100/(pi) Omega - 2pi Omega = 25.5 Omega`.
Thus `Z^2 = (50 Omega)^2 + (25. 5 Omega)^2 = 3150 Omega^2`.
From (i), average power `P_(av) = (7200 Omega-V^2)/(3150 Omega^2) = 2.286 W`.
The energy dissipated in `1000 s = P_(av) xx 1000 s`
`= 2.3 xx 10^3 J`