Question

The switch S shown in figure (32−E35) is kept closed for a long time and is then opened at t = 0. Find the current in the middle 10 Ω resistor at t = 1 ms.

Answer 1

Initially, the switch S was closed; so, the capacitor was getting charged. Also, the two resistors are connected in parallel. The equivalent resistance of the circuit,
$R=\frac{1}{{\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{10}}}=5\mathrm{\Omega }$

Initially, the switch was closed and the capacitor was getting charged. So, the two resistances were in parallel connection. Hence, their effective resistance will be 5 Ω.
Potential difference across the 10 Ω resistance, $V=\frac{12}{5}\times 10=24\mathrm{V}$

When the switch is opened, the decay of charge through the capacitor,
$$\begin{gathered} Q=Q_0{\mathrm{e}}^{-\frac{t}{RC}}=V\times C{\mathrm{e}}^{-\frac{t}{RC}} \\ \frac{t}{RC}=\frac{1\times {10}^{-3}}{10\times 25\times {10}^{-6}}=\frac{1000}{250}=4 \\ \Rightarrow Q=24\times 25\times {10}^{-6}\times {\mathrm{e}}^{-4} \\ =24\times 25\times {10}^{-6}\times 0.0183 \\ =10.9\times {10}^{-6}\mathrm{C} \end{gathered}$$
Current in the 10 Ω resistor,
$i=\frac{Q}{t}=\frac{10.9\times {10}^{-6}\mathrm{C}}{1\times {10}^{-3}\sec }=11\mathrm{mA}$