Question

A capacitor of $2\mu F$ is charged as shown in the diagram. When the switch $S$ is turned to position $2$, the percentage of its stored energy dissipated is

• A. $75\%$
• B. $80\%$
• C. $0\%$
• D. $20\%$

Answer 1

The correct option is B $80\%$

Initially the energy stored in the capacitor can be given by
$E_i=\frac{1}{2}C{V}^2=\frac{1}{2}\times 2\times (\frac{Q}{2}{)}^2=\frac{Q^2}{4}$

Now we connect to point 2 and the circuit becomes

Now the charge ${}^{\prime }{q}^{\prime }$ flows from one capacitor to other till both reach the same potential
We know that $V=\frac{Q}{C}$
So equating the potential
$V_1=V_2\Rightarrow \frac{Q-q}{2}=\frac{q}{8}\Rightarrow q=\frac{4Q}{5}$
Charge left on $2\mu F$ capacitor
$Q_1=Q-q=Q-\frac{4Q}{5}=\frac{Q}{5}$
Potential of $2\mu F$ capacitor
$V_1=\frac{Q}{C}=\frac{Q}{10}$
Charge on $8\mu F$ capacitor
$Q_2=\frac{4Q}{5}$
Potential of $8\mu F$ capacitor $=$ Potential of $2\mu F$ capacitor $=\frac{Q}{10}$
Final energy stored in both tha capacitors
$E_f=\frac{1}{2}{C}_1{V}^2+\frac{1}{2}{C}_2{V}^2$
$E_f=\frac{1}{2}(C_1+C_2)V^2$
$E_f=\frac{1}{2}(2+8)\frac{Q^2}{100}=\frac{Q^2}{20}$
Percentage of energy lost $=\frac{E_i-E_f}{E_i}\times 100\frac{\frac{Q^2}{4}-\frac{Q^2}{20}}{\frac{Q^2}{4}}\times 100=80\text{\%}$