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A capacitor of capacitance C(0) is charg...

A capacitor of capacitance `C_(0)` is charged to a potential `V_(0)` and is connected with another capacitor of capacitance C as shown. After closing the switch S, the common potential across the two capacitors becomes V, The capacitance C is given by

A

`(C_0(V_0-V))/V_0`

B

`(C_0(V-V_0))/V_0`

C

`(C_0(V+V_0))/V_0`

D

`(C_0(V_0-V))/V`

Text Solution

Verified by Experts

The correct Answer is:
D
From charge conservation
`C_0V_0=C_0V+CVimpliesCV=C_0V_0-C_0V`
`impliesC=(C_0(V_0-V))/(V)`
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