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

`V_(0) =50sqrt2V`
`omega =2pi f = 2pi xx (50)/(pi) =100 rad/sec`
`X_(L) = omega_(L) =100 xx 1 =100 Omega`
`X_(C) = (1)/(omegaC) =(1)/(100 xx 20 xx 10^(-6))500Omega`
`Z =sqrt(R^(2) + (X_(C) -X_(L)^(2)))`
`= sqrt((300)^(2) + (500 -100)^(2) =500Omega`
`i_(0) = (V_(0))/(Z) = (50sqrt2)/(500) = 0.1sqrt2A`
`i_(rms) =(i_(0))/sqrt2 =(0,1sqrt2)/(sqrt2 =0.1 )A`
`V_(R) = i_(rms) R =0.1 xx 300 =30 V`
`V_(L) =i_(rms) X_(L) =0.1 xx 100 =10V`
`V_(C) =i_(rms) X_(C) =0.1 xx 500 =50V`
`UC = (1)/(2) C V_(C)^(2) xx 20 xx 10^(-6) xx (50)^(2) =25 mJ`
`UL = (1)/(2) L i^(2)_(rms) = (1)/(2) xx 1xx (0.1)^(2) =5mJ`
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