Topic: Network Theory
In the circuit shown below, switch $S_{1}$ and $S_{2}$ are in open and close position respectively for long time. At $t=t_{0}$ switch $S_{1}$ is closed and switch $S_{2}$ is opened. What would be the current through $R_{1}$ immediately after the transition of switches?
- A. 0 mA
- B. 1 mA
- C. 0.5 mA
- D. 2 mA
Answer: B. 1 mA
Explanation: At $t=t_{0}^{-}$, steady state is reached and the capacitor acts as an open circuit. $V_c(t_{0}^{-}) = V_c(t_{0}^{+}) = -5\text{ V}$.
At $t=t_{0}^{+}$, $S_{1}$ is closed and $S_{2}$ is opened. Using nodal analysis at node A (between the 10 k$\Omega$ resistors and $R_1$):
$\frac{V_A-10}{10} + \frac{V_A}{10} + \frac{V_A+5}{5} = 0$
$\frac{2V_A}{5} = 0 \Rightarrow V_A = 0\text{ V}$
So, $V_{5\text{ k}\Omega} = I_{5\text{ k}\Omega}(5\text{ k}\Omega) = 5\text{ V}$
$I_{5\text{ k}\Omega} = 1\text{ mA}$.