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A body of capacity `4 muF` is charged to `80V` and another body of capacity `6muF` is charged to `30V`. When they are connected the energy lost by `4 muF` capacitor isA. `7.8 mJ`B. `4.6 mJ`C. `3.2 mJ`D. `2.5 mJ` |
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Answer» Correct Answer - A Initial energy of body of capacitance `4 muF` is `U_(i) = (1)/(2) xx (4 xx 10^(-6))(80)^(2) = 0.0128 J` Final potential on this body after connection is `V = (4 xx 80 + 6 xx 30)/(4 + 6) = 50 V1`. So final energy on it `U_(f) = (1)/(2) xx 4 xx 10^(-6) (50)^(2) = 0.005 J` Energy lost by this body `= U_(i) - U_(f) = 7.8 mJ` |
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