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A `20F` capacitor is charged to `5V` and isolated. It is then connected in parallel with an uncharged `30 F` capacitor. The decrease in the energy of the system will beA. `25 J`B. `200 J`C. `125 J`D. `150 J` |
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Answer» Correct Answer - D Charge stored on `20 F` capacitor `= 20 xx 5 = 100 C` When `20F` capacitor is connected to `30F` capacitor, `V = (q)/(C_(1) + C_(2)) = (100)/(20 + 30) = 2V` Energy of the system will be `E_(f) = (1)/(2) xx (30 + 20) xx (2)^(2)` `= 100 J` Energy before isolation `E_(i) = (1)/(2) xx 20 xx (5)^(2) = 250 J` Therefore, decrease in energy `= 250 - 100 = 150 J` |
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