Question

a 2 micro farad capacitor is charged as shown .the percentage of its stored energy dissipated after switch s is turned to position 2 is?
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Answer 1

$$\begin{gathered} Whencapacitorof2\mu FisconnectedtoavoltageVthenthecharging \\ Q=CV=2\times {10}^{-6}V \\ Whenthischargedcapacitorisconnectedtoacapacitorof8\mu Finparallel \\ thenthechargedcapacitorlosesitschrageto8\mu Fconnectedinparallel. \\ Sonetcapaci\tan ceofthecombinationis; \\ {C}_0=2+8=10\mu F \\ TotalchargeQ=2\times {10}^{-6}V \\ Nowtheamountofchargelostto8\mu Fcapacitorisgivenby; \\ Q\text{'}=\frac{8\mu F}{10\mu F}\times 2\times {10}^{-6}V=1.6\times {10}^{-6}V \\ Thustheenergylostto8\mu Fisgivenby, \\ U\text{'}=\frac{1}{2}\times \frac{Q{\text{'}}^2}{C\text{'}}=\frac{1}{2}\times \frac{1.6\times {10}^{-6}V\times 1.6\times {10}^{-6}V}{8\times {10}^{-6}}=0.16\times {10}^{-6}{V}^2 \\ Totalenergystoredwas=U=\frac{1}{2}\times \frac{Q^2}{C}=\frac{1}{2}\times \frac{2\times {10}^{-6}V\times 2\times {10}^{-6}V}{2\times {10}^{-6}}=1\times {10}^{-6}{V}^2 \\ Therefore\%ofenergylost=\frac{U\text{'}}{U}\times 100=\frac{0.16\times {10}^{-6}{V}^2}{1\times {10}^{-6}{V}^2}\times 100=16\% \end{gathered}$$