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

Answer 1

Answer: (C)

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

Answer 2

For Initial $E =\frac{1}{2} CV _{1}^{2}+\frac{1}{2} CV _{2}^{2}$
For Final $E=2 \times \frac{1}{2} C\left(\frac{V_{1}+V_{2}}{2}\right)^{2}$
$\Delta E =\frac{1}{2} CV _{1}^{2}+\frac{1}{2} CV _{2}^{2}-2 \times \frac{1}{2} C \left(\frac{ V _{1}+ V _{2}}{2}\right)^{2}$
$\Rightarrow \Delta E =\frac{1}{4} C \left( V _{1}- V _{2}\right)^{2}$