In a series `LCR` circuit with an `ac` source of `50 V,R=300 Omega`,frequency `v=50/piHz`.The average electric field energy, stored in the capacitor and average magnetic energy stored in the coil are `25 mJ` and `5 mJ` respectively.The `RMS` current in the circuit is `0.10 A`.Then find

Options

A. capacitance `C` of capacitor is `20 mu F`
B. inductance `L` of inductor is `2 H`
C. peak voltage of source is `50 V`
D. the sum of rms voltage across the three elements is `35.4 V`

Answer

A,B,C,D

Explanation

Let `I_(r)` be the rms current through the circuit then
`I_(r ) = 2A, (I_(r ))/(omega C) = 20 V, I_(r ) omega C = 20 V` and `I_(r ) R = 10 V`
solving we get
`R = 50 Omega, C = (1)/(pi) xx 10^(-3) F` and `L = (1)/(10 pi) H`
`:. V_(s)` = source voltage =
`I_(r ) = sqrt(R^(2) + (omega L - (1)/(omega C))^(2))`
`= sqrt((I_(r ) R)^(2) + (I_(r) omega L - (I)/(omega C))^(2))`
`= sqrt(10^(2) + (20 - 20)^(2)) = 10` volts
After in inductor is shorted
`I_(r ) = (V_(s))/(sqrt(R^(2) + (1)/(omega^(2) C^(2)))) = (10)/(sqrt(25 + 10)) = (2)/(sqrt(5)) A`
`v_(1) = I_(r ) R = 2 sqrt(5)` volts, `v_(2) = (I_(r ))/(omega C) = 4 sqrt(5)` volts