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

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

Text Solution

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When capacitance is removed,
`tanphi=(X_(L)/R)` or `tan60^(@)=X_(L)/R`
`therefore X_(L) = sqrt(3)R`.............(i)
When inductance is removed,
`tanphi=(X_(C))/R` or `tan60^(@)=X_(C)/R`
`therefore X_(C) = sqrt(R)`...........(ii)
From Eqs. (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`
Power dissipated (P) = `V_(rms)I_(rms)cosphi`
At resosnance, current and voltage are in phase or `phi=0^(@)`
`therefore (P) = (200)(2)(1)=400W`
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