A capacitor `C_(1)` is charged to a p.d.V. The charging battery is then removed and the capacitor is connected to an uncharged capacitor `C_(2)`. The final p.d. across the combination isA. `VC_(1)/(C_(1)+C_(2))`B. `VC_(2)/(C_(1)+C_(2))`C. `V(C_(1)C_(2))/(C_(1)+C_(2))`D. `V/(C_(1)+C_(2))`

A capacitor `C_(1)` is charged to a p.d.V. The charging battery is the

1.	A capacitor `C_(1)` is charged to a p.d.V. The charging battery is then removed and the capacitor is connected to an uncharged capacitor `C_(2)`. The final p.d. across the combination isA. `VC_(1)/(C_(1)+C_(2))`B. `VC_(2)/(C_(1)+C_(2))`C. `V(C_(1)C_(2))/(C_(1)+C_(2))`D. `V/(C_(1)+C_(2))`
Answer» Correct Answer - A The charge on condenser `C_(1)=Q_(1)=C_(1)V` If it is connected across uncharged condenser of capacity `C_(2)`, then p.d. across `C_(2)` is given by , `"potential"=Q_(1)/C_(2)=VC_(1)/(C_(1)+C_(2))`

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A capacitor `C_(1)` is charged to a p.d.V. The charging battery is then removed and the capacitor is connected to an uncharged capacitor `C_(2)`. The final p.d. across the combination isA. `VC_(1)/(C_(1)+C_(2))`B. `VC_(2)/(C_(1)+C_(2))`C. `V(C_(1)C_(2))/(C_(1)+C_(2))`D. `V/(C_(1)+C_(2))`

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