Question

A $2\mu \text{F}$ capacitor is charged as shown in figure. The percentage of its stored energy dissipated after switch $\text{S}$ is turned to position $\text{2}$ is $(8\mu F$ is the capacitance of air-filled parallel plate capacitor$)$:

• A. $0\%$
• B. $20\%$
• C. $75\%$
• D. $89\%$

Answer 1

The correct option is D $89\%$

When the $\text{S}$ is connected to $\text{1}$:

Energy stored is, $U_i=\frac{1}{2}C{V}^2$

(Here $C=2\mu F)$

$\Rightarrow U_i=\frac{1}{2}\times 2\times {10}^{-6}\times V^2=V^2\mu \text{J}$

Initial charge, $Q_i=2V\mu C$


When switch $\text{S}$ is turned to position $\text{2}$:


Let, the common potential be $V^{\prime }$.

Capacitance of $8\mu \text{F}$ capacitor will be $C_2=K{C}_0=2\times 8=16\mu \text{F}$

For $2\mu \text{F}$ capacitor, capacitance $C_1=2\mu F$

Now, the net charge:
$\Rightarrow q_{net}=(C_1+C_2)V^{\prime }\mu C$

From the conservation of charge:
$q_{net}=Q_i=(2+16)V^{\prime }$
$\Rightarrow 2V=18V^{\prime }$
$\therefore V^{\prime }=\frac{V}{9}$

Total final energy stored in system,
$U_f=(\frac{1}{2}\times C_1\times V^{\prime 2})+(\frac{1}{2}\times C_2\times V^{\prime 2})$

$\Rightarrow U_f=[\frac{1}{2}\times 2\times \frac{V^2}{81}]+[\frac{1}{2}\times 16\times \frac{V^2}{81}]$

$\Rightarrow U_f=\frac{V^2}{81}+\frac{8V^2}{81}$

$\Rightarrow U_f=\frac{9V^2}{81}=\frac{V^2}{9}$

Energy dissipated,

$\Delta U=V^2-\frac{V^2}{9}=\frac{8V^2}{9}$

$\%$ of energy dissipated:

$\%\Delta U=\frac{\left(\frac{8V^2}{9}\right)}{V^2}\times 100=\frac{800}{9}\approx 89\%$

Hence, option (d) is the correct answer.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: In problems involing change in position of switch}\\ \text{focus on the charge in equivalent capacitance due to}\\ \text{charge in circuit.}\end{array}$