Physics - EMI/AC Question with Solution | TestHub

$$\begin{gathered} Given,\mathrm{I}={\mathrm{I}}_0\cos \left(\mathrm{\omega t}\right), Peak or maximum current {\mathrm{I}}_0=1\mathrm{A} \\ angular frequency,\mathrm{\omega }=500 \mathrm{rad} {\mathrm{s}}^{-1} \\ capaci\tan ce \mathrm{C}=20\mathrm{\mu F} \\ resis\tan ce\hspace{0.17em}R=10\mathrm{\Omega } \\ battery voltage \mathrm{V}=50\mathrm{V} \end{gathered}$$If q represent the charge on capacitor’s upper plate:
$\mathrm{I}\left(\mathrm{t}\right)={\mathrm{I}}_0\cos (\mathrm{\omega }\mathrm{t})\mathrm{ }$
$\Rightarrow \mathrm{q}\left(\mathrm{t}\right)=\frac{{\mathrm{I}}_0}{\mathrm{\omega }}\sin \left(\mathrm{\omega }\mathrm{t}\right)$
$\therefore$Max charge$\left({\mathrm{q}}_{\mathrm{m}\mathrm{a}\mathrm{x}}\right)=\frac{1\mathrm{A}}{500\frac{\mathrm{R}\mathrm{a}\mathrm{d}}{\mathrm{s}}}=2\times {10}^{-3}\mathrm{C}$
Charge of upper plate at$\mathrm{t}=\frac{7}{6\mathrm{\omega }}$
$=\frac{{\mathrm{I}}_0}{\mathrm{\omega }}\sin \left(\frac{7\mathrm{\pi }}{\mathrm{\omega }}\right)\mathrm{ }=-\frac{{\mathrm{I}}_0}{2\mathrm{\omega }}$
When capacitor is fully charged then charge on upper plate$=50\mathrm{v}\times 20\mathrm{\mu }\mathrm{f}$
$=10\times {10}^{-3}\mathrm{C}$
$\Rightarrow \mathrm{Q}=1\times {10}^{-3}-\left(\frac{{\mathrm{I}}_0}{2\mathrm{\omega }}\right)=2\times {10}^{-3}\mathrm{C}$
Voltage across capacitor when A and D are connected$=\frac{1\times {10}^{-3}\mathrm{C}}{20\times {10}^{-6}\mathrm{\mu }\mathrm{f}}=50\mathrm{V}$
Total voltage across$\mathrm{R}=100\mathrm{V}$
Current is anticlockwise