Initially the switch `'S'` is open for a ling time. Now the switch `'S'` is closed at `t=0(R_(1)=2Omega, R_(2)=2Omega, C_(1)=1F` and `C_(2)=1F)` Then

Options

A. The current through the wire `AB` at `t=2` sec is `1//2e` ampere
B. The current through the wire at `t=2` sec is `1//e` ampere.
C. The charge on the capacitor `C_(2)`, which has initially unchargd at `t=2` sec is `(1- 1/(2e))` coulimbs
D. The charge on the capacitor `C_(1)` at `t=2` sec is `(1- 1/(2e))` coulombs

Answer

A, C, D

Explanation

When the switch is closed then at `t=0` the distribution of the charge is shown in the figure.

By applying Kirchooff's Law `I_(1)^(')=((1-q))/2`
`int_(1//2)^(q) (dq)/(1-q)=int_(0)^(1) (dt)/2`
`q=(1-1/2e^(-t//2))`
`I_(1)^(')=(dq)/(dt)=1/4 e^(-t//2)`