A capacitor of capacitance `100 muF` and a coil of resistance `50 Omega` and inductance `0.5 H` are connected in series with `a 110 V, 50 Hz AC` source. Find the rms value of the current.

Explanation

The resistance of the circuit is `R = 50 Omega`.
The reactance of the capacitor `= 1/(omegaC)`
`= 1/((2pi xx 50 s^(-1)) (100 xx 10^-6) F)) = 31.8 Omega`
The reactance of the inductor `= omegaL`
`= (2pi xx 50 s^(-1)) (0.5 henry = 157 Omega`
`= 31.8 Omega - 157 Omega = - 152.2 Omega`.
Hence, the impedance `Z = sqrt(R^2 + X^2)`
`= sqrt((50 Omega)^@ + (125.2 Omega)^2) = 134.6 Omega`.
The rms current `= E_(rms)/Z = (110 V)/(134.6 Omega) = 0.82 A`.