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A capacity of capacity C(1) is charged u...

A capacity of capacity `C_(1)` is charged up to `V` volt and then connected to an uncharged capacitor of capacity `C_(2)`. Then final potential difference across each will be

A

`(C_2V)/(C_1+C_2)`

B

`(C_1V)/(C_1+C_2)`

C

`(1+C_2/C_1)`

D

`(1-C_2/C_1)V`

Text Solution

Verified by Experts

The correct Answer is:
B
`V_"common"=(C_1V_1+C_2V_2)/(C_1+C_2)=(because V_2=0)`
`rArr V_"common"=(C_1V)/(C_1+C_2)`
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