In a series `LCR` circuit with an AC source, `R = 300 Omega, C = 20 muF, L = 1.0 henry, epsilon_(rms) = 50 V` and `v = 50/(pi) Hz`. Find (a) the rms current in the circuit and (b) the rms potential differences across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.

Explanation

Given `R=300 Omega`, `C=20 muF`
`=(20 xx 10^-6)F`,
`L=1 Henry`,
`E=50 V`, `v=50//piHz`
(a) `i_0 =E_0/z`
and `Z=sqrt(R^2 + (X_e -X_L)^2)`
`=sqrt((300)^2 + (1//2pifC-2pifL)^2)`
`=sqrt((300)^2 + ((1)/(2pi xx 50//pi xx 20 xx 10^-6 -2pi xx 50/pi xx 1))^2`
`=sqrt((300)^2 + (10^4/20 -100)^2) = 500`
:. `i_0 = E_0/Z = 50/500 = 0.1A`
(b) Potential across the capacitor
`=i_0 xx X_C= 0.1 xx 500 = 50V`
Potential difference across the resistor
`=i_0 xx R = 0.1 xx 300 = 30 V`
Potential difference across the inductor
`=i_0 xx X_L = 0.1 xx 100`
`=10V`
R.M.S. potential `=50V`
Net sum of all the potential drops
`=50 V + 30 V +10V`
`=90V`
Sum of potential drops = R.m.s potential applied.