InterviewSolution
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InterviewSolution
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A `2muF` capacitor is charged as shown in the figure. The percentage of its stored energy disispated after the switch S is turned to poistion 2 is A. `80%`B. `0%`C. `20%`D. `75%` |
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Answer» Correct Answer - A Initial energy stored in capacitor `2 muF` `U_(i) = (1)/(2)2(V)^(2) = V^(2)` Final voltage after switch `2` is `ON` `V_(f) = (C_(1)V_(1))/(C_(1)+C_(2)) = (2V)/(10) = 0.2 V` Final energy in both the capacitors `U_(f) = (1)/(2) (C_(1) + C_(2)) V_(f)^(2) = (1)/(2) 10 ((2V)/(10))^(2) = 0.2 V^(2)` So energy dissipated `= (V^(2) - 0.2 V^(2))/(V^(2)) xx 100 = 80%` |
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