An inductor and two capacitors are connected in the circuit as shown in figure. Initially capacitor A has no charge and capacitor B has CV charge. Assume that the circuit as no resistance at all. At t = 0, switch S is closed, then [given LC $=\frac{2}{{\mathrm{\pi }}^2\times {10}^4}{\mathrm{s}}^2\text{ and }\mathrm{CV}=100\mathrm{mC}]$

Let at any time, charge and current in the circuit wire are as shown.

$\mathrm{I}=\frac{{\mathrm{dq}}_1}{\mathrm{dt}}$

Applying Kirchhoff s law:

$\begin{array}{l} \frac{\mathrm{CV}-{\mathrm{q}}_1}{\mathrm{C}}-\frac{{\mathrm{q}}_1}{\mathrm{C}}-\mathrm{L}\frac{\mathrm{dI}}{\mathrm{dt}}=0\Rightarrow \frac{\mathrm{CV}-2{\mathrm{q}}_1}{\mathrm{LC}}=\mathrm{L}\frac{{\mathrm{d}}^2{\mathrm{q}}_1}{{\mathrm{dt}}^2}\\ \Rightarrow \frac{{\mathrm{d}}^2{\mathrm{q}}_1}{{\mathrm{dt}}^2}=-\frac{2}{\mathrm{LC}}\left({\mathrm{q}}_1-\frac{\mathrm{CV}}{2}\right)\\ \Rightarrow {\mathrm{q}}_1-\frac{\mathrm{CV}}{2}=\mathrm{Asin}(\mathrm{\omega t}+\mathrm{\delta }) ........\left(\mathrm{i}\right)\end{array}$

where $\mathrm{\omega }=\sqrt{\frac{2}{\mathrm{LC}}}=100\mathrm{\pi rad}/\mathrm{s}$

$\text{ At }\mathrm{t}=0,{\mathrm{q}}_1=0\Rightarrow 0-\frac{\mathrm{CV}}{2}=\mathrm{Asin}\mathrm{\delta } ......\left(\mathrm{ii}\right)$

Differentiating $\text{ (i), }\frac{{\mathrm{dq}}_1}{\mathrm{dt}}-0=\mathrm{A\omega cos}(\mathrm{\omega t}+\mathrm{\delta })$

$\Rightarrow \mathrm{I}=\mathrm{A\omega cos}(\mathrm{\omega t}+\mathrm{\delta }) .......\left(\mathrm{iii}\right)$

$\text{ At }\mathrm{t}=0,\mathrm{I}=0\Rightarrow 0=\mathrm{A\omega cos}\mathrm{\delta }\Rightarrow \mathrm{\delta }=\frac{\mathrm{\pi }}{2}$

From (ii), $\mathrm{A}=-\frac{\mathrm{CV}}{2}$

From(i), ${\mathrm{q}}_1=\frac{\mathrm{CV}}{2}-\frac{\mathrm{CV}}{2}\sin \left(\mathrm{\omega t}+\frac{\mathrm{\pi }}{2}\right)$

$\begin{array}{l}\Rightarrow {\mathrm{q}}_1=\frac{\mathrm{CV}}{2}(1-\cos \mathrm{\omega t}) .......\left(\mathrm{iv}\right)\\ \Rightarrow {\mathrm{q}}_1=50(1-\cos 100\mathrm{\pi t})\mathrm{mC}\end{array}$

$\text{ Now }{\mathrm{q}}_2=\mathrm{CV}-{\mathrm{q}}_1=\frac{\mathrm{CV}}{2}(1+\cos \mathrm{\omega t}) ......\left(\mathrm{v}\right)$

               $=50(1+\cos 100\mathrm{\pi t})$mC

From (iii), $\mathrm{I}=-\mathrm{A\omega sin}\mathrm{\omega t}$

Current in the circuit is maximum for $\mathrm{\omega t}=\frac{\mathrm{\pi }}{2}$.

And for $\mathrm{\omega t}=\frac{\mathrm{\pi }}{2}$, we see from (iv) and (v) that ${\mathrm{q}}_1={\mathrm{q}}_2=\frac{\mathrm{CV}}{2}$