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An L-C-R series circuit with 100Omega re...

An `L-C-R` series circuit with `100Omega` resisance 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

Text Solution

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When capacitance is removed, then
`tanphi=X_L/R`
or `tan60^@=X_L/R`
`:. X_L=sqrt3R`
When inductance is removed, then
`tanphi=X_C/R`
or `tan60^@C=X_C/R`
`:. X_C=sqrt3R`
From Eq.(i) and (ii), we see that `X_C=X_L`
So, the `L-C-R` circuit is in resonance
Hence, `Z=R`
`:. i_("rms")=V_("rms")/Z=200/100=2A`
`(:P:)=V_("rms")i_("rms")cosphi`
At resonance current and voltage are in phase,
or `phi=0^@`
`:. (:P:)=(200)(2)(1)=400W`
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