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In the circuit shown, the capacitor C1 i...

In the circuit shown, the capacitor `C_1` is initially charged with charge `q_0`. The switch S is closed at time t = 0 . The charge on `C_2` after time t is

A

`(q_0C_1)/(C_1+C_2) (1-e^((-t(C_1+C_2))/(C_1C_2R)))`

B

`(q_0C_2)/(C_1+C_2) (1-e^((-t(C_1+C_2))/(C_1C_2R)))`

C

`(q_0C_1)/(C_1+C_2) (1-e^((-t)/(C_2R)))`

D

`(q_0C_2)/(C_1+C_2) (1-e^((-t)/(C_2R)))`

Text Solution

Verified by Experts

The correct Answer is:
B
b. `I = (dq)/(dt)` , apply Kirchhoff's law: `(q_0-q)/C_1 = IR+q/C_2`
`q_0/C_1-q[(C_1+C_2)/(C_1C_2)] = (dq)/(dt)R`
`int_(0)^(q) (dq)/(q_0/C_1-q((C_1+C_2)/(C_1C_2))) = int_(0)^(t) (dt)/R`
Solve to get `q = (q_0C_2)/(C_1+C_2)[1-e^(-t(C_1+C_2)/(C_1C_2R))]`.
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