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

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2. Electric Potential and Capacitance

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A $2 \ \mu F$ capacitor is charged as shown in 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

$80 \%$

(IIT-2011)

Solution

Initial energy stored in capacitor $2 \mu F : \quad U _{ i }=\frac{1}{2} 2(V)^2= V ^2$

Final voltage after switch 2 is on: $V _{ f }=\frac{ C _1 V_1}{ C _1+ C _2}=\frac{2 V}{1 O }=0.2 V$

Final energy in both the capacitors,

$U _{ f }=\frac{1}{2}\left( C _1+ C _2\right) V _{ f }^2=\frac{1}{2} \times 10 \times\left(\frac{2 V}{10}\right)^2=0.2 V^2$

Therefore, energy dissipated $=\frac{ V ^2-0.2 V^2}{V^2} \times 100=80 \%$

Standard 12

Physics

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