Question

A capacitor $C$ is charged by a battery of $V$ volts . Then it is connected to an uncharged capacitor $2C$ as shown in figure. No match the following two columns.

Column-1 Column-2

$\text{(a)}$ After closing the switch energy stored in $C$.	$\text{(p)}\frac{1}{9}C{V}^2$
$\text{(b)}$ After closing the switch energy stored in $2C$.	$\text{(q)}\frac{1}{6}C{V}^2$
$\text{(c)}$ After closing the switch loss of energy during redistribution of charge.	$\text{(r)}\frac{1}{18}C{V}^2$
$\text{(s)}$ None of these

• A. $\text{(a)}\to \text{(p)};\text{(b)}\to \text{(r)};\text{(c)}\to \text{(s)}$
• B. $\text{(a)}\to \text{(r)};\text{(b)}\to \text{(p)};\text{(c)}\to \text{(s)}$
• C. $\text{(a)}\to \text{(s)};\text{(b)}\to \text{(q)};\text{(c)}\to \text{(r)}$
• D. $\text{(a)}\to \text{(r)};\text{(b)}\to \text{(q)};\text{(c)}\to \text{(r)}$

Answer 1

The correct option is B $\text{(a)}\to \text{(r)};\text{(b)}\to \text{(p)};\text{(c)}\to \text{(s)}$

After closing the switch, the common potential in parallel.

$V=\frac{\text{Total Charge}}{\text{Total Capacity}}=\frac{CV}{3C}=\frac{V}{3}$

Energy stored in capacitor of capacitance $C$ , $U_C=\frac{1}{2}C{\left(\frac{V}{3}\right)}^2=\frac{1}{18}C{V}^2$

Energy stored in capacitor of capacitance $2C$ , $U_{2C}=\frac{1}{2}\times 2C\times {\left(\frac{V}{3}\right)}^2=\frac{1}{9}C{V}^2$

Loss in energy $\Delta U=U_i-U_f$

$\therefore \Delta U=\frac{1}{2}C{V}^2-\frac{1}{2}\times 3C\times {\left(\frac{V}{3}\right)}^2=\frac{1}{3}C{V}^2$

Hence, option (b) is the correct answer.