A $2\,\mu F$ Capacitor is charged as shown in the figure. The percentage of its store energy dissipated after the switch $S$ is turned to position $2$, is.....$\%$
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$\%$ Heat $ \mathrm{loss}=\frac{\Delta \mathrm{U}}{\mathrm{U}_{\mathrm{i}}} \times 100$
$=\frac{\frac{1}{2} \frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}(\mathrm{V}-0)^{2}}{\frac{1}{2} \mathrm{C}_{1} \mathrm{V}^{2}} \times 100$
$=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}} \times 100=\frac{8}{10} \times 100=80 \%$
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