Question

A $2\mu F$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is (The $8\mu F$ capacitor is initially uncharged)

• A. 0%
• B. 20%
• C. 75%
• D. 80%

Answer 1

The correct option is D

80%

$2\mu FgetschargedtoapotentialV.Whenconnectedtothe8\mu Fcapactor,thecommonpotentialis{V}_{com}=\frac{c_1{v}_1+c_2{v}_2}{c_1+c_2}=\frac{2v+8\left(0\right)}{10}=\frac{v}{5}{U}_i=\frac{1}{2}\times 2\times {10}^{-6}\times =v^2=v^2\times {10}^{-6}{U}_f=\frac{1}{2}\times 10\times {10}^{-6}\times \frac{v^2}{25}=0.2v^2\times {10}^{-6}$\\
Thus 80% of the energy is dissipated