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

Answer 1

Answer: (D)

Initial energy stored in capacitor $2\mu F:U_i=\frac{1}{2}2(V{)}^2=V^2$

Final voltage after switch $2$ is on: $V_f=\frac{C_1{V}_1}{C_1+C_2}=\frac{2V}{10}=0.2V$

Final energy in both the capacitors, $U_f=\frac{1}{2}(C_1+C_2)V_f^2=\frac{1}{2}\times 10\times {\left(\frac{2V}{10}\right)}^2=0.2V^2$

Therefore, energy dissipated $=\frac{V^2-0.2V^2}{V^2}\times 100=80$%