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Two identical capacitors have same capacitance $C$. One of them is charged to the potential $\mathrm{V}$ and other to the potential $2 \mathrm{~V}$. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :
$\frac{1}{4} \mathrm{CV}^2$
$2 \mathrm{CV}^2$
$\frac{1}{2} \mathrm{CV}^2$
$\frac{3}{4} \mathrm{CV}^2$
Solution
$\mathrm{V}_{\mathrm{C}}-\frac{\mathrm{q}_{\text {net }}}{\mathrm{C}_{\text {net }}}=\frac{\mathrm{CV}+2 \mathrm{CV}}{2 \mathrm{C}}$
$\mathrm{V}_{\mathrm{C}}=\frac{3 \mathrm{~V}}{2}$
Loss of energy
$=\frac{1}{2} \mathrm{CV}^2+\frac{1}{2} \mathrm{C}(2 \mathrm{~V})^2-\frac{1}{2} 2 \mathrm{C}\left(\frac{3 \mathrm{~V}}{2}\right)^2$
$=\left(\frac{\mathrm{CV}^2}{4}\right)$
