• # question_answer In the circuit shown in the figure, the AC source gives a voltage $V=10\sin (1000\,t).$Neglecting source resistance, the voltmeter and ammeter reading will be A) $20\sqrt{\frac{25}{538}}\,V$and $\frac{10}{\sqrt{538}}A$ B)  $10\sqrt{\frac{250}{528}}\,V$and $\frac{10}{\sqrt{538}}A$ C)  $\sqrt{\frac{250}{538}}\,V$and $\frac{1}{\sqrt{538}}A$ D)  $\sqrt{\frac{125}{539}}\,V$and $\frac{1}{\sqrt{538}}\,A$

Impedance of the circuit is $Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$ $R=8\,\Omega$ (given) ${{X}_{L}}=\omega L=1000\times 10\times {{10}^{-3}}=10\,\Omega$ ${{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2000\times 20\times {{10}^{-6}}}=25\,\Omega$ $\Rightarrow$$Z=\sqrt{64+{{(15)}^{2}}}=\sqrt{64+225}=\sqrt{269}$ ${{I}_{\max }}=\frac{{{V}_{0}}}{Z}=\frac{10}{\sqrt{269}}A$ ${{I}_{rms}}=\frac{{{\operatorname{I}}_{max}}}{\sqrt{2}}=\frac{10}{\sqrt{538}}A$ ${{V}_{rms}}={{I}_{rms}}\times \sqrt{{{5}^{2}}+{{(15)}^{2}}}$ $=\frac{10}{\sqrt{538}}\times \sqrt{25+225}=10\sqrt{\frac{250}{538}}\,V$