Topic: Network Theory
In a given network (shown in Figure), a steady state is reached with switch k open. At t=0 which is closed. Determine the values of $I_{1}$, $I_{2}$ and $I_{3}$ at $t=0^{+}$
- A. 1 A, 1/3 A, 0 A
- B. 1/3 A, 1/3 A, 1/3 A
- C. 0 A, 0 A, 0 A
- D. 1 A, 1 A, 0 A
Answer: A. 1 A, 1/3 A, 0 A
Explanation: For $t<0$, steady state is reached. The inductor acts as a short circuit and capacitors act as open circuits. Total resistance $= 2\Omega + 4\Omega = 6\Omega$. $I_L(0^-) = 6\text{V} / 6\Omega = 1\text{A}$.
By voltage division logic given in the reference, the voltage across the 0.5 F capacitor is $V_{C1}(0^-) = 4/3\text{V}$.
At $t=0^+$, the inductor current cannot change instantaneously, so $I_1 = 1\text{A}$. Also, $I_3 = 0\text{A}$ (as KVL violates otherwise, according to the official solution logic). The current $I_2 = \frac{V_{C1}(0^+)}{4\Omega} = \frac{4/3}{4} = 1/3\text{A}$.