An `L-C-R` series circuit with `100Omega` resistance is connected to an `AC` source of `200V` and angular frequency `300rad//s`. When only the capacitance is removed, the current lags behind the voltage by `60^@`. When only the inductance is removed the current leads the voltage by `60^@`. Calculate the current and the power dissipated in the `L-C-R` circuit

Options

A. `200 W`
B. `100 W`
C. `50 W`
D. `400 W`

Answer

D

Explanation

Av.Electric field energy =
`((1)/(2) CV_("rms")^(2)) = 25 xx 10^(-3) J`
`:. (1)/(2) C (I_("rms") X_(C ))^(2) = 25 xx 10^(-3) J`
`:. (1)/(2) C I_(0)^(2) (1)/(2 pi^(2) v^(2) c^(2)) = 25 xx 10^(-3) J`
`:. C = 20 mu F`
Average magnetic energy `((1)/(2) LI_("rms")^(2)) = L = 2 H`
`:. L = (2 xx 5 xx 10^(-3))/((.10)^(2)) L = 1` henry
`V_(R ) = I_("rms") R = V_(C ) = 50 V = V_(L) = I_("rms") L`
`= (0.10) .300 = (.10) xx (1)/(2 pi ((50)/(sqrt(t))).20 xx 10^(-6))`
`= (.10) xx 2 pi (50)/(sqrt(2)) (I)`
`V_(R ) = 30 V = V_(C ) = 50 V = V_(L) = 10 W`
rsm voltageof source `E_("rms") = (50)/(sqrt(2))`
`:. E_("rms") = 35.36 V`