A 2,μ F capacitor is charged as shown in figure. percentage of its stored energy dissipated after s

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1. Electric Charges and Fields

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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.....$\%$

A

$0$

B

$20$

C

$75$

D

$80$

Solution

$U_{i}=\frac{1}{2}(2) V^{2}=V^{2}$

After connecting $S$ to $2$

$V_{\text {common }}=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}=\frac{2 V+8 \times 0}{2+8}=\frac{V}{5}$

$U_{f}=\frac{1}{2}(2+8)\left(\frac{V}{5}\right)^{2}=\frac{V^{2}}{5}$

Percent energy dissipated $=\frac{U_{i}-U_{f}}{U_{i}} \times 100$

$=\frac{V^{2}-\frac{V^{2}}{5}}{V^{2}} \times 100=80 \%$

Standard 12

Physics

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(AIIMS-2017)