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A condenser of capacitance `10 muF` has been charged to `100V`. It is now connected to another uncharged condenser in parallel. The common potential becomes `40 V`. The capacitance of another condenser isA. `15 muF`B. `5 muF`C. `10 muF`D. `16.6 muF` |
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Answer» Correct Answer - A By using `V = (C_(1)V_(1) + C_(2)V_(2))/(C_(1) + C_(2))` `rArr 40 = (10 xx 100 + C_(2) xx 0)/(10 + C_(2)) rArr C_(2) = 15 muF` |
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