Question

A condenser of capacity $C_1$ is charged to a potential $V_0$. The electrostatic energy stored in it is $U_0$. It is connected to another uncharged condenser of capacity $C_2$ in parallel. The energy dissipated in the process is

• A. $\frac{C_1}{C_1+C_2}{U}_0$
• B. $\frac{C_2}{C_1+C_2}{U}_0$
• C. $\left(\frac{C_1-C_2}{C_1+C_2}\right)U_0$
• D. $\frac{C_1{C}_2}{2(C_1+C_2)}{U}_0$

Answer 1

The correct option is B $\frac{C_2}{C_1+C_2}{U}_0$

Energy stored

$U_0=\frac{1}{2}{C}_1{V}_0^2\dots \left(1\right)$

Where, $U_0=$ Initial energy stored

It is known that energy dissipated while connecting $C_1$ to another condenser of capacity $C_2$ in parallel is given as

$\frac{1}{2}\left(\frac{C_1{C}_2}{C_1+C_2}\right)[V_1-V_2{]}^2$

$\left[\right(V_1\text{and}{V}_2\text{are initial potential differences across two capacitors}\left)\right]$

Here,
$V_1=V_0{V}_2=0$ ( uncharged capacitor )

$\therefore$ Energy dissipated $=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}{V}_0^2$

$=\frac{C_2}{C_1+C_2}{U}_0$ $\left(\text{from}\right(1\left)\right)$
Hence, option (a) is correct.