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. 100 W
B. 200 W
C. 400 W
D. 800 W

Answer

C

Explanation

(i) When C is removed, L and R are in series and the phase difference between L and R is given by
`tan theta = (omega L)/R :. Omega L = R tan theta = 100 xx tan 60^(@)` .....(i)
(ii) When L is removed, C and R are in series.
`:.` The phase difference between C and R is given by
`tan theta' = 1/(RC omega) :. 1/(omegaC) = R tan theta' = 100 tan 60^(@)`
Thus `omega L = 1/(omega C)`
`:.` The impedance of the circuit is given by
`Z = sqrt(R^(2)(omega L -(1)/(omega L)^(2)))`
`=sqrt(R^(2)) = R = 100 Omega`
and Power = `(V_(rms)^(2))/R = (200 xx 200)/(100) = 400 W`.