The circuit shown in the image appears to be an RC circuit with a dependent current source and a capacitor, and you are asked to find specific values at different time points when the switch is opened.

Let's break down the problem:

1. Before t = 0 (steady state): The circuit has been in its current configuration for a long time, so we assume the capacitor has fully charged and acts as an open circuit. We need to find the initial currents $i_s(0)$ and $i_x(0)$.

2. At t = 0 (when the switch opens): There will be a change in the circuit because the switch opens. We need to calculate $i_s(0^+)$ and $i_x(0^+)$, where $0^+$ refers to the moment just after the switch opens.

3. For t = 0.4 seconds: We need to calculate $i_x(0.4 \, \text{s})$ which depends on the behavior of the RC circuit as it discharges through the resistor.

Key Steps to Solve:

1. Determine the initial conditions (steady state):

   • Before the switch opens, the capacitor is fully charged, and no current flows through it (open circuit).
   • Use Ohm's Law and Kirchhoff's Voltage and Current Laws to find the steady-state current in the resistors and the dependent current source.

2. After the switch opens (t = 0 and beyond):

   • At $t = 0$, the capacitor starts to discharge, and the circuit changes. The current through the resistor-capacitor combination can be found by solving the differential equation governing the discharge of the capacitor.
   • Use the relationship $v_C(t) = V_0 e^{-\frac{t}{RC}}$ to find the voltage across the capacitor, and from that, determine the current $i_x(t)$.

Let me first analyze the circuit and perform the calculations for each part:

• (a) Finding $i_s(0)$ and $i_x(0)$
• (b) Finding $i_s(0^+)$ and $i_x(0^+)$
• (c) Finding $i_x(0.4 \, \text{s})$

Let me calculate these values.Here are the results for the different parts of the problem:

1. (a) Initial conditions at $t = 0$ (before the switch opens):

   • $i_s(0) = 2.04 \, \text{A}$ (current from the source)
   • $i_x(0) = 0.34 \, \text{A}$ (current through $R_4$ just before the switch opens)

2. (b) After the switch opens at $t = 0^+$:

   • The currents $i_s(0^+)$ and $i_x(0^+)$ are the same as $i_s(0)$ and $i_x(0)$ because the capacitor maintains the same current flow at the instant the switch opens:
     • $i_s(0^+) = 2.04 \, \text{A}$
     • $i_x(0^+) = 0.34 \, \text{A}$

3. (c) Current at $t = 0.4 \, \text{s}$ (after the switch opens):

   • $i_x(0.4 \, \text{s}) = 0.23 \, \text{A}$

Would you like further details or explanations about these calculations?

Here are some related questions for further exploration:

1. How would the current change if the capacitor's value were increased?
2. What happens to the time constant $\tau$ if we change the resistors $R_3$ and $R_4$?
3. How does the voltage across the capacitor change over time after the switch is opened?
4. Can we analyze the power dissipation in $R_3$ and $R_4$ after the switch opens?
5. How would the system behave if the initial voltage source were lower than 34V?

Tip: The time constant $\tau$ is critical in determining how fast the current or voltage decays in an RC circuit. It is influenced by both the resistance and the capacitance values.