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A capacitor of capacitance of `2muF` is charged to a potential difference of `200V`, after disconnecting from the battery, it is connected in parallel with another uncharged capacitor. The final common potential is `20V` then the capacitance of second capacitor is :A. `2muF`B. `4muF`C. `18muF`D. `16muF` |
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Answer» Correct Answer - C |
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