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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. `2.5 xx 10^(-3) J`C. `2.5 xx 10^(6) J`D. `2.42 xx 10^(-2) J` |
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Answer» Correct Answer - A Change in energy `DeltaU = (1)/(2) [(q_(1)^(2) - q_(2)^(2))/(C)]` `= (1)/(2) [ ((0.5)^(2) - (0.1)^(2))/(48 xx 10^(-6))]` `= (1)/(2)[ (0.25 - 0.01)/(48 xx 10^(-6))]` `= (1)/(2) [(0.24)/(48 xx 10^(-6))]` `= (1)/(2) [ (10^(4))/(2)]` = 2500 J |
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