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A capacitor $C$ is fully charged with voltage $V _{0}$ After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{ C }{2} .$ The energy loss in the process after the charge is distributed between the two capacitors is$.........$$CV _{0}^{2}$
$0.166$
$0.5$
$0.33$
$0.25$
Solution
$\frac{ CV _{0}- q }{ C }=\frac{ q }{ C / 2}=\frac{2 q }{ C }$
$V _{0}=\frac{3 q }{ C } \Rightarrow q =\frac{ CV _{0}}{3}$
$U _{ i }=\frac{1}{2} CV _{0}^{2}$
$U _{ f }=\frac{\left(\frac{2 CV _{0}}{3}\right)^{2}}{2 C }+\frac{\left(\frac{ CV _{0}}{3}\right)^{2}}{2\left(\frac{ C }{2}\right)}$
$=\frac{1}{2} CV _{0}^{2}\left[\frac{4}{9}+\frac{2}{9}\right]=\frac{1}{2} CV _{0}^{2}\left(\frac{2}{3}\right)$
Heat loss $=\frac{1}{2} CV _{0}^{2}-\left(\frac{2}{3}\right)\left(\frac{1}{2} CV _{0}^{2}\right)$
$=\frac{1}{6} CV _{0}^{2}$
