1.

Capacity of a capacitor is `48mu F`. When it is charged from `0.1` C to `0.5` C , change in the energy stored isA. 2500 JB. `2.5xx10^(-3)J`C. `2.5xx10^(6) J`D. `2.42xx10^(-2)`

Answer» Correct Answer - 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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