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Capacity of a capacitor is 48mu F. When ...

Capacity of a capacitor is `48mu F`. When it is charged from `0.1` C to `0.5` C , change in the energy stored is

A

2500 J

B

`2.5xx10^(-3)J`

C

`2.5xx10^(6) J`

D

`2.42xx10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaE=E_(2)-E_(1)=Q_(2)^(2)/(2C)-Q_(1)^(2)/(2C)`
`=1/(2C)[Q_(2)^(2)-Q_(2)^(2)]=((0.25-0.01)/(2xx48xx10^(-6)))`
= 2500 J
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