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A parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to potential difference $2V$ . The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
zero
$\frac{{25C{V^2}}}{6}$
$\frac{{3C{V^2}}}{2}$
$\frac{{9C{V^2}}}{2}$
Solution
$\mathrm{V}_{\text {connmen }}=\frac{\mathrm{C}_{2} \mathrm{V}_{2}-\mathrm{C}_{1} \mathrm{V}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
$\mathrm{V}_{\text {connmen }}=\frac{(2 \mathrm{C}) 2 \mathrm{V}-\mathrm{CV}}{2 \mathrm{C}+\mathrm{C}}=\mathrm{V}$
$\mathrm{U}_{\text {combination }}=\frac{1}{2}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}_{\mathrm{C}}^{2}$
$=\frac{1}{2}(2 \mathrm{C}+\mathrm{C}) \mathrm{V}^{2}=\frac{3}{2} \mathrm{CV}^{2}$
