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

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

Solution

$\mathrm{U}_{\mathrm{i}}=\frac{1}{2} \mathrm{CV}_1^2+\frac{1}{2} C V_2^2=\frac{1}{2} \mathrm{C}\left(\mathrm{~V}_1^2+\mathrm{V}_2^2\right)$
When the capacitors are joined, common potential, $\mathrm{V}=\frac{\mathrm{CV}_1+\mathrm{CV}_2}{2 \mathrm{C}}=\frac{\mathrm{V}_1+\mathrm{V}_2}{2}$ $\therefore \quad$ Final energy of the system, $\begin{aligned} \mathrm{U}_{\mathrm{f}} & =\frac{1}{2}(2 \mathrm{C}) \mathrm{V}^2=\frac{1}{2} 2 \mathrm{C}\left(\frac{\mathrm{~V}_1+\mathrm{V}_2}{2}\right)^2 \\ & =\frac{1}{4} \mathrm{C}\left(\mathrm{~V}_1+\mathrm{V}_2\right)^2 \end{aligned}$ $\begin{aligned} \therefore \quad \text { Decrease in energy } & =\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}} \\ & =\frac{1}{4} \mathrm{C}\left(\mathrm{~V}_1-\mathrm{V}_2\right)^2 \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 1)

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