Question

n identical capacitors are joined in parallel and are charged to potential V.now they are separated and joined in series.then total energy and potential energy will be??

Answer 1

$$\begin{gathered} \mathrm{In}\mathrm{series}\mathrm{combination},\mathrm{the}\mathrm{potential}\mathrm{differences}\mathrm{are}\mathrm{additive}. \\ \mathrm{Thus}, \\ \mathrm{V}\text{'}={\mathrm{V}}_1+{\mathrm{V}}_2+....{\mathrm{V}}_{\mathrm{n}}=\mathrm{nV} \\ \mathrm{Now}, \\ \mathrm{Initial}\mathrm{energy},\mathrm{U}=\frac{1}{2}\left({\mathrm{nC}}_0\right){\mathrm{V}}^2 \\ \mathrm{where},{\mathrm{C}}_0\mathrm{is}\mathrm{capacitance}\mathrm{of}\mathrm{each}\mathrm{capacitor}. \\ \mathrm{Final}\mathrm{energy},\mathrm{U}\text{'}=\frac{1}{2}\left(\frac{{\mathrm{C}}_0}{\mathrm{n}}\right)(\mathrm{nV}{)}^2 \\ =\frac{1}{2}{\mathrm{C}}_0({\mathrm{nV}}^2)=\mathrm{nU} \end{gathered}$$