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

Explanation

Here, `R = 100 Omega, E_(v) = 200 V, omega = 300 rad//s`
When capacitance is removed,
`tan phi = (Z_(L))/(L) = tan 60^(@) = sqrt3 :. X_(L) = sqrt3 R`
When only the inductance is removed
`tan phi = (X_(C ))/(R ) = tan 60^(@) = sqrt3 :. X_(C ) = sqrt3 R`
As `X_(L) = X_(C )`, the LCR circuit is in resonance.
`:. Z = R = 100 Omega`
`I_(v) = (E_(v))/(Z) = (200)/(100) = 2 A`
At resonance, `P = E_(v) I_(v) cos 0^(@)`
`= 200 xx 2 xx 1 =400 W`