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

`P_(av) = V_(rms) I_(rms) cos phi`
In LCR
` cos phi = (V_R)/(V) = (IR)/(IZ)`
` rArr P_(av) = V_(rms) ( (V_(rms))/(Z)) R/Z = (V_(rms)^2 . R)/(Z^2)`
Here `V_(rms) = 12V`
` R = 50 Omega , X_L =omega L = 2pi v L`
` = 2pi (50 Hz) (20 xx 10^(-3) H)`
` X_C = (1)/(omega C) = (1)/(2pi v C) = (1)/(2pi (50HZ) (100 xx 10^(-6) F))`
`= ((100)/(pi))Omega `
` rArr Z = sqrt(R^2 + (X_L - X_C)^2) = 56Omega`
Energy used in 1000 s is `P_(Av) t= ((V_(rms)^2 R)/(Z^2)) t`
`= ((12^2 xx 50)/(56^2)) xx 1000 = 2.3 xx 103 J`