• # question_answer An LCR circuit contains resistance of 100 ohm and a supply of 200 volt at 300 radian angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by $60{}^\circ$. If on the other hand, only inductor is taken out the current leads by $60{}^\circ$ with the applied voltage. The current flowing in the circuit is: A)  1 amp                        B)  1.5 ampC)  2 amp                        D)  2.5 amp

According to the question. $\tan {{60}^{o}}=\frac{\omega L}{R}$and $\tan {{60}^{o}}=\frac{1/\omega c}{R}$ $\therefore$$\omega L=\frac{1}{\omega c},$(case of resonance) Now, $z=\sqrt{{{R}^{2}}+\left( \omega L-\frac{1}{\omega c} \right)}=R$ $\therefore$${{I}_{rms}}=\frac{{{E}_{rms}}}{z}=\frac{{{E}_{rms}}}{R}=\frac{200v}{100\Omega }=2\,Amp.$ Hence, the correction option is [c].