• # question_answer A 200 V variable frequency AC source is connected to a series combination of L = 5 H, $C=80\mu F\text{ }$and $R=40\,\Omega .$ Calculate (i) angular frequency of source to get the maximum current in the circuit (ii) current amplitude at resonance (iii) power dissipation in the circuit

Given,   ${{E}_{rms}}=200\,V,$            $L=5\,H,$ $C=80\mu F=80\times {{10}^{-6}}F$ and $R=40\,\Omega$ (i) For the maximum current in the circuit, ${{X}_{L}}={{X}_{C}}$ $\Rightarrow$   $\omega L=\frac{1}{\omega C}$ $\therefore$ Resonant frequency, ${{\omega }_{r}}=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5\times 80\times {{10}^{-6}}}}=50\,rad/s$ (ii) Impedance,   $Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$             $Z=R=40\,\Omega$                 $[\because \,{{X}_{L}}={{X}_{C}}]$ Current, ${{I}_{rms}}=\frac{{{E}_{rms}}}{Z}=\frac{200}{40}=5\,A$ Current amplitude at resonance,             ${{I}_{0}}={{I}_{rms}}\sqrt{2}=5\times 1.414=7.07\,A$ (iii) Power dissipated in circuit, $P={{E}_{rms}}{{I}_{rms}}\cos 0{}^\circ =200\times 5\times 1=1000\,W$