The switch S shown in figure is kept closed for a long time and is then opened at `t=0`.Find the current in the middle `10(Omega)`resistor at `t=1.0ms`.

Explanation

Charge stored on capacitor in long time
`q_(0) = CE = 25 xx 10^(-6)xx12 = 300 xx 10^(-6) = 3 xx 10^(-4)"coul"`
Now discharging `RC` circuit

Charge on capacitor at any time `t`
`q = q_(0) e^(-t//RC)`
`i = (-dq)/(dt) = (q_0)/(RC) e^(-t//RC)`
`= (3xx10^(-4))/(10xx25xx10^(-6)) e^(-(10^(-3))/(10xx25xx10^(-6)))`
`=1.2 e^(-4)=1.2/(e^4) A`.