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

An `L-C-R` series circuit with `100Omega` resisance is connected to an

1.	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
Answer» 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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