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A capacitor of capacity `C` is connected with a battery of potential `V` in parallel. The distance between its plates is reduced to half at one, assuming that the charge remains the same. Then to charge the capacitance upto the potential `V` again, the energy given by the battery will beA. `CV^(2)//4`B. `CV^(2)//2`C. `2CV^(2)//4`D. `CV^(2)` |
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Answer» Correct Answer - D |
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