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

A capacitor of capacity `C_(1)` is charged to the potential of `V_(0)`. On disconnecting with the battery, it is connected with a capacitor of capacity `C_(2)` as shown in the adjoining figure. The ratio of energies before and after the connection of switch S will be

A

`(C_1+C_2)//C_1`

B

`C_1//(C_1+C_2)`

C

`C_1C_2`

D

`C_1//C_2`

Text Solution

Verified by Experts

The correct Answer is:
A
`U_i=1/2C_1V_0^2" & "Q=CV_0`
`U_f=1/2(Q^2)/(C_(eq))=(C_1^2V_0^2)/(2(C_1+C_2))` [as Q = constant]
`U_i/U_f =(C_1V_0^2)/(C_1^2V_0^2)xx(C_1+C_2)=((C_1+C_2))/(C_1)`
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