A 2,mu F capacitor is charged as shown in the | vedclass

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Question

$U_{i}=\frac{1}{2}(2) V^{2}=V^{2}$

After connecting $S$ to $2$

$V_{	ext {common }}=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}=\frac{2 V+8 	im

Vedclass · Accepted Answer

$80$

Vedclass · Answer

$U_{i}=\frac{1}{2}(2) V^{2}=V^{2}$

After connecting $S$ to $2$

$V_{	ext {common }}=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}=\frac{2 V+8 	imes 0}{2+8}=\frac{V}{5}$

$U_{f}=\frac{1}{2}(2+8)\left(\frac{V}{5}ight)^{2}=\frac{V^{2}}{5}$

Percent energy dissipated $=\frac{U_{i}-U_{f}}{U_{i}} 	imes 100$

$=\frac{V^{2}-\frac{V^{2}}{5}}{V^{2}} 	imes 100=80 \%$

A $2\,\mu F$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position $2$, is.....$\%$

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