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Two capacitances of capacity `C_(1)`and `C_(2)` are connected in series and potential difference `V` is applied across it. Then the potential difference across `C_(1)` will beA. `V(C_(2))/(C_(1))`B. `V(C_(1) + C_(2))/(C_(1))`C. `V (C_(2))/(C_(1) + C_(2))`D. `V(C_(1))/(C_(1) + C_(2))` |
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Answer» Correct Answer - C Charge flowing ` = (C_(1)C_(2))/(C_(1) + C_(2))V` Potential diff.across `C_(1) = (C_(1)C_(2)V)/(C_(1) + C_(2)) xx (1)/(C_(1)) = (C_(2)V)/(C_(1) + C_(2))` |
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