Question

The energy of a charged capacitor is $U$. Another identical capacitor is connected parallel to the first capacitor, after disconnecting the battery. The total energy of the system of these capacitors will be

• A. $\frac{U}{4}$
• B. $\frac{U}{2}$
• C. $\frac{3U}{2}$
• D. $\frac{3U}{4}$

Answer 1

Answer: (B)

$\frac{U}{2}$

Common potential $=\frac{C_1{V}_0+C_2\times 0}{C_1+C_2}=\frac{C_2{V}_0}{C_1+V_2}$
$U_{before}=\frac{1}{2}{C}_1{V}_0^2$
$U_{after}=\frac{1}{2}{C}_1{\left[\frac{C_1{V}_0}{C_1+C_2}\right]}^2+\frac{1}{2}{C}_2{\left[\frac{C_1{V}_0}{C_1+C_2}\right]}^2$
$=\frac{1}{2}{\left[\frac{C_1{V}_0}{C_1+C_2}\right]}^2(C_1+C_2)$
$\Rightarrow \frac{U_{before}}{U_{after}}=\frac{C_1+C_2}{C_1}$
Here, $C_1=C_2=C$
$\therefore \frac{U_{before}}{U_{after}}=\frac{2C}{C}$
$\Rightarrow U_{after}=\frac{U}{2}$.