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

A capacitor of `2muF` is charged as shown in the figure. When the switch S is turned to position 2, the percentage of its stored energy dissipated is-

A

0

B

0.2

C

0.75

D

0.8

Text Solution

Verified by Experts

The correct Answer is:
D
`q_1=c_1v=2v=q` (say)
the charge will remain constant after switch is shifted from possition 1 to possition 2
`Ui=1/2q^(2)/c_1=q^(2)/(2xx2)=q^(2)/4`
`U _f=1/2 q^(2)/c _f=q^(2)/(2xx10)=q^(2)/20 therefore "Energy dissiplated"=U_i-U_f=q^(2)/5`
This energy dissiplated `(q^(2)/5)` is 80% of the initial stored energy `(-q^(2)/4)`
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