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

Question

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

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 is
A. `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))`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

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))`

5 views Answered 2023-02-05 15:29