Physics - Capacitor Question with Solution | TestHub

Initial energy of the combined system $${\mathrm{U}}_1=\frac{1}{2}{\mathrm{CV}}_1^2+\frac{1}{2}{\mathrm{CV}}_2^2=\frac{\mathrm{C}}{2}\left({\mathrm{V}}_1^2+{\mathrm{V}}_2^2\right)$$ On joining the two condensers in parallel, common potential, $$\mathrm{V}=\frac{{\mathrm{V}}_1+{\mathrm{V}}_2}{2}$$ $\therefore$ Final energy of the combined system: $${\mathrm{U}}_2\equiv \frac{1}{2}(\mathrm{C}+\mathrm{C}){\left(\frac{{\mathrm{V}}_1+{\mathrm{V}}_2}{2}\right)}^2\text{. }$$ $\therefore$ - Decrease in energy will be: $$\begin{array}{rl}\Delta \mathrm{U}& ={\mathrm{U}}_{\mathrm{L}}-{\mathrm{U}}_2\\ & =\frac{\mathrm{C}}{2}\left({\mathrm{V}}_1^2+{\mathrm{V}}_2^2\right)-\frac{1}{2}(\mathrm{C}+\mathrm{C}){\left(\frac{{\mathrm{V}}_1+{\mathrm{V}}_2}{2}\right)}^2\\ & =\frac{1}{4}\mathrm{C}{\left({\mathrm{V}}_1-{\mathrm{V}}_2\right)}^2\end{array}$$