A capacitor of $2\,\, \mu F$ is charged as shown in the diagram. When the switch $S$ is turned to position $2,$ the percentage of its stored energy dissipated is ......$\%$
NEET 2016, Diffcult
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Initially, the energy stored in $2 \,\mu F$ capacitor is
$U_{i}=\frac{1}{2} C V^{2}=\frac{1}{2}\left(2 \times 10^{-6}\right) V^{2}=V^{2} \times 10^{-6}\, \mathrm{J}$
Initially, the charge stored in $2\, \mu \mathrm{F}$ capacitor is
$Q_{i}=C V=\left(2 \times 10^{-6}\right)\, V=2 V \times 10^{-6}$ coulomb. When switch $S$ is turned to position $2,$ the charge flows and both the capacitors share charges till a common potential $V_{c}$ is reached.
$V_{C}=\frac{\text { total charge }}{\text { total capacitance }}=\frac{2 V \times 10^{-6}}{(2+8) \times 10^{-6}}=\frac{V}{5} \text { volt }$
Finally, the energy stored in both the capacitors
$U_{f}=\frac{1}{2}\left[(2+8) \times 10^{-6}\right]\left(\frac{V}{5}\right)^{2}=\frac{V^{2}}{5} \times 10^{-6} \,\mathrm{J}$
$\%$ loss of energy, $\Delta U=\frac{U_{i}-U_{f}}{U_{i}} \times 100 \%$
$=\frac{\left(V^{2}-V^{2} / 5\right) \times 10^{-6}}{V^{2} \times 10^{-6}} \times 100 \%=80 \%$
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