The capacitor shown in fig. is in steady state. The energy stored in the capacitor is
Medium
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Since both resistance is same so current in $\mathrm{R}$ would be $\mathrm{I}.$ So current at $\mathrm{x}$ would be $21$. The whole current will pass through $\mathrm{R}$ (in steady state) so potential difference across $\mathrm{xy}$ would be
$\mathrm{V}_{\mathrm{xy}}=(2 \mathrm{I})(\mathrm{R})$
so energy stored $=\frac{1}{2} \mathrm{C}\left(\mathrm{V}_{\mathrm{xy}}\right)^{2}$
$\boxed{{\text{U}} = \frac{1}{2}{\text{C}} \times 4{{\text{I}}^2}{{\text{R}}^2} = 2{\text{C}}{{\text{I}}^2}{{\text{R}}^2}}$
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