Question

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 the positive ends are also connected, the decrease in energy of the combined system is

• A. $\frac{1}{4}C\left(V_1^2-V_2^2\right)$
• B. $\frac{1}{4}C\left(V_1^2+V_2^2\right)$
• C. $\frac{1}{4}C(V_1-V_2{)}^2$
• D. $\frac{1}{4}C{\left(V_1^2+V_2^2\right)}^2$

Answer 1

Answer: (C)

For Initial $E=\frac{1}{2}C{V}_1^2+\frac{1}{2}C{V}_2^2$
For Final $E=2\times \frac{1}{2}C{\left(\frac{V_1+V_2}{2}\right)}^2$
$\Delta E=\frac{1}{2}C{V}_1^2+\frac{1}{2}C{V}_2^2-2\times \frac{1}{2}C{\left(\frac{V_1+V_2}{2}\right)}^2$
$\Rightarrow \Delta E=\frac{1}{4}C{\left(V_1-V_2\right)}^2$