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

Answer

D

Explanation

Here, `R = 100 Omega, E_(upsilon) = 200` volt, `omega = 300 rad//s`
When capacitor is removed,
`tan phi = (X_(L))/(R ) = tan 60^(@) = sqrt3`
`X_(L) = sqrt3 R`
`Z = (R^(2) + (X_(L) -X_(C ))^(2))`
`= sqrt(100^(2) + 0) = 100 Omega`
`I_(upsilon) = (E_(upsilon))/(Z) = (200)/(100) = 2 A`
`P = E_(upsilon) I_(upsilon) cos phi = E_(upsilon) I_(upsilon) ((R)/(Z))`
`= (200) (2) xx (100)/(100) = 400 W`