Physics - Capacitor Question with Solution | TestHub

Initial energy stored in the capacitor $C_1$ is:

$U_i=\frac{1}{2}{C}_1{V}^2$

When the switch is turned to position 2, the charge is shared between $C_1$ and $C_2$. The common potential $V_c$ is:

$V_c=\frac{C_{1}V}{C_1+C_2}$

The final energy of the system is:

$U_f=\frac{1}{2}(C_1+C_2)V_c^2=\frac{1}{2}(C_1+C_2){\left(\frac{C_{1}V}{C_1+C_2}\right)}^2=\frac{1}{2}\frac{C_1^2{V}^2}{C_1+C_2}$

The energy dissipated is:

$\Delta U=U_i-U_f=\frac{1}{2}{C}_1{V}^2-\frac{1}{2}\frac{C_1^2{V}^2}{C_1+C_2}=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}{V}^2$

The percentage of stored energy dissipated is:

$\text{Percentage}=\frac{\Delta U}{U_i}\times 100=\frac{\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}{V}^2}{\frac{1}{2}{C}_1{V}^2}\times 100=\frac{C_2}{C_1+C_2}\times 100$

Given $C_1=2\mu \text{F}$ and $C_2=8\mu \text{F}$:

$\text{Percentage}=\frac{8}{2+8}\times 100=\frac{8}{10}\times 100=80\%$

Answer: $80\%$