Question
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 $1.0\Omega$ resistor at t = 1ms.

Answer

In steady state condition, no current passes through the $25\mu\text{F}$ capacitor, $\therefore$ Net resistance $=\frac{10\Omega}{2}=5\Omega$
Potential difference across the capacitor = 5 Potential difference across the $10\Omega$ resistor$=\frac{12}{5}\times10=24\text{V}$
$\text{q}=\text{Q}\Big(\text{e}^{\frac{-\text{t}}{\text{RC}}}\Big)=\text{V}\times\text{C}\Big(\text{e}^{\frac{-\text{t}}{\text{RC}}}\Big)$ $=24\times25\times10^{-6}\bigg[\text{e}^{\frac{-1\times10^{-3}}{10\times25\times10^{-4}}}\bigg]$
$=24\times25\times10^{-6}\text{e}^{-4}$
$=24\times25\times10^{-6}\times0.0183=10.9\times10^{-6}\text{C}$
Charge given by the capacitor after time t. Current in the $10\Omega$ resistor $=\frac{10.9\times10^{-6}\text{C}}{1\times10^{-3}\text{sec}}=11\text{mA}.$

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