Two identical capacitors have the same capacitance ' $C$ '. One of them is charged to potential ' $V_1$ '…

Two identical capacitors have the same capacitance ' $C$ '. One of them is charged to potential ' $V_1$ ' and the other to $V_2$. The negative ends of the capacitors are connected together. When positive ends are also connected, the decrease in energy of the combined system is
  1. $\frac{1}{4} \mathrm{C}\left(\mathrm{V}_1-\mathrm{V}_2\right)^2$
  2. $\frac{1}{2} \mathrm{C}\left(\mathrm{V}_1^2+\mathrm{V}_2^2\right)$
  3. $\frac{1}{2} \mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)$
  4. $\frac{1}{2} \mathrm{C}\left(\mathrm{V}_1+\mathrm{V}_2\right)^2$

Solution

$\mathrm{Q}_1=\mathrm{CV}_1, \mathrm{Q}_2=\mathrm{CV}_2, \mathrm{Q}=\mathrm{Q}_1+\mathrm{Q}_2=2 \mathrm{CV}$ After connection, $\begin{aligned} & \mathrm{CV}_1+\mathrm{CV}_2=2 \mathrm{CV} \\ & \therefore \mathrm{V}=\frac{\mathrm{V}_1+\mathrm{V}_2}{2} \end{aligned}$ Decrease in energy $=\frac{1}{2} \mathrm{CV}_1^2+\frac{1}{2} \mathrm{CV}_2^2-\frac{1}{2}(2 \mathrm{C})\left(\frac{\mathrm{V}_1+\mathrm{V}_2}{2}\right)^2$ $=\frac{1}{4} \mathrm{C}\left(\mathrm{V}_1-\mathrm{V}_2\right)^2$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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