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

In steady state condition,
Potential difference across the capacitor
`= 25 mu F `
` Potential difference across the 10 Omega
resistance
` = 12/5 xx 10 = 24V `
` Q = (Q_0) e^(t/RC) = V xx C (e ^-t/RC))`
` = 24 xx 25 xx (10^-6)e^(-4) `
` = 24 xx 25 xx (10^-6) xx 0.0183 `
` 10.9 xx (10^-6) C `
Charge by the capacitor after time t'
Current in the 10 Omega resistor
` = (10.9 xx (10^-6)C/1 xx (10^-3) sec) = 11 mA. `