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A capacitor of 2 mu F is charged as show...


A capacitor of `2 mu F` is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is:

A

`0%`

B

`20%`

C

`75%`

D

`80%`

Text Solution

Verified by Experts

The correct Answer is:
D
Initial energy stored in capacitor `2muF`
`U_(s)= (1)/(2) 2(V)^(2)=V^(2)`
`=(1)/(2) xx 2 xx 10^(-6) xx V^(2)`
`=V^(2) xx 10^(-6)`
Final voltage after switch 2 is ON
`V_(1)= (C_(1)V_(1))/(C_(1) + C_(2))= (2C)/(10)= 0.2V`
Final energy in both capacitors
`U_(1)= (1)/(2) (C_(1) +C_(2)) V_(1)^(2)= (1)/(2)xx 10 xx 10^(-5) xxx ((2V)/(10))^(2)= 0.2V^(2)`
So energy dissipated `=(V^(2)- 0.2V^(2))/(V^(2))xx 100= 80%`
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