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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. `VC_(2)/C_(1)`B. `(V(C_(1)+C_(2)))/C_(1)`C. `(VC_(2))/((C_(1)+C_(2)))`D. `(VC_(1))/((C_(1)+C_(2)))` |
Answer» Correct Answer - C `V_(1)=Q/C_(1)=(C_(s)V)/C_(1)=((C_(1)C_(2))/(C_(1)+C_(2)))V/C_(1)=(VC_(2))/(C_(1)+C_(2))` |
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