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Two capacitors of `25 mu F` and `100 mu F` are connected in series to a source of `120 V`. Keeping their charges uncharged, they are separated and connected in parallel to eachother. Find out (i) pot. Diff. between the plates of each capacitor (ii) energy loss in the process. |
Answer» Correct Answer - `38.4 V, 0.052 J` Here, `C_(1) = 25 muF, C_(2) = 100 muF, C_(s) = 20 muF`, `q = C_(s) V_(s) = 2400 mu C` on each capacitor `C_(p) = 125 mu F, V_(p) = (2400+2400)/(125) = 38.4 V` Loss of energy `= (1)/(2) C_(s) V_(s)^(2) - (1)/(2) C_(p) V_(p)^(2)` |
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