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

Answer 1

Answer: (C)

$\frac{1}{4}C(V_1-V_2{)}^2$

Refer to image

For Initial $E=\frac{1}{2}C{V}_1^2+\frac{1}{2}C{V}_2^2$

For Final $E=2\times \frac{1}{2}C{\left(\frac{V_1+V_2}{2}\right)}^2$

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

$\Rightarrow \Delta E=\frac{1}{4}C(V_1-V_2{)}^2$