In the circuit shown, switches $S_1$ and $S_3$ have been closed for 1 sec and $S_2$ remained open. Just after 1 second is over, switch $S_2$ is closed and $S_1,S_3$ are opened. After that instant:

 

Maximum charge on the capacitor is

At any instant

$\begin{array}{l}\frac{1}{2}L{I}^2+\frac{q^2}{2C}={\left(1-{\mathrm{e}}^{-1}\right)}^2\\ \frac{1}{2}{\left(\frac{\mathrm{dq}}{\mathrm{dt}}\right)}^2+\frac{{\mathrm{q}}^2}{2}={\left(1-{\mathrm{e}}^{-1}\right)}^2\\ \frac{\mathrm{dq}}{\mathrm{dt}}=\sqrt{2{\left(1-{\mathrm{e}}^{-1}\right)}^2-{\mathrm{q}}^2}\\ {\int }_{\left(1-{\mathrm{e}}^{-1}\right)}^{\mathrm{q}} \frac{dq}{\sqrt{{\left(\sqrt{2}\left(1-{\mathrm{e}}^{-1}\right)\right)}^2-{\mathrm{q}}^2}}={\int }_0^{\mathrm{t}} \mathrm{dt}\\ \mathrm{q}=\sqrt{2}\left(1-{\mathrm{e}}^{-1}\right)\sin \left(\mathrm{t}+\frac{\pi }{4}\right)\\ {\mathrm{q}}_{max}=\sqrt{2}\left(1-{\mathrm{e}}^{-1}\right)\\ \mathit{I}\mathit{I}\mathbf{ }\mathit{m}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\\ \frac{1}{2}L{I_{max}}^2=\frac{{q_{max}}^2}{2C}=\frac{1}{2}\left(1\right){\left(1-{\mathrm{e}}^{-1}\right)}^2+\frac{1}{2\left(1\right)}{\left(1-{\mathrm{e}}^{-1}\right)}^2={\left(1-{\mathrm{e}}^{-1}\right)}^2\\ C=1F, L=1H then \\ \frac{{q_{max}}^2}{2\left(1\right)}={\left(1-{\mathrm{e}}^{-1}\right)}^2\\ {q_{max}}^2=2{\left(1-{\mathrm{e}}^{-1}\right)}^2\\ {\mathrm{q}}_{max}=\sqrt{2}\left(1-{\mathrm{e}}^{-1}\right)\end{array}$