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A 2muF capacitor is charged as shown in ...

A `2muF` capacitor is charged as shown in the figure. The percentage of its stored energy disispated after the switch S is turned to poistion 2 is

A

`0%`

B

`20%`

C

`75%`

D

`80%`

Text Solution

Verified by Experts

`U_(i)=(1)/(2)(2)V^(2)=V^(2)`
After connecting S to 2.
`V_("common")=(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))=(2V+8xx0)/(2+8)=(V)/(5)`
`U_(f)=(1)/(2)(2+8)((V)/(5))^(2)=(V^(2))/(5)`
Percent energy dissipated `=(U_(i)-U_(f))/(U_(i))xx100`
`=(V^(2)-(V^(2))/(5))/(V^(2))xx100=80%`
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