To solve this problem, we need to analyze the circuit at the moment just before $t = 0$, denoted as $t = 0^-$. Here's the step-by-step solution:

1. Understanding the Circuit Before $t = 0$

• The switch has been closed for a long time, so the capacitor would have been fully charged, and the circuit reached a steady state.
• In the steady state for DC circuits, a capacitor acts as an open circuit. Therefore, no current flows through the capacitor branch.

2. Finding the Voltage Across the Capacitor Before $t = 0$

• The entire voltage $Y = 35 \text{ V}$ would appear across the capacitor.
• This is because in the steady state, the voltage across the capacitor remains constant.

3. Current in the Circuit at $t = 0^-$

• The circuit simplifies to a series combination of the 2 kΩ and 3 kΩ resistors because the capacitor is an open circuit at steady state.

Let's calculate the current $i(0^-)$ just before the switch opens:

• Total resistance in series: $R_{\text{total}} = 2 \text{ k}\Omega + 3 \text{ k}\Omega = 5 \text{ k}\Omega$

• Current $i(0^-)$ is given by Ohm's Law: $i(0^-) = \frac{Y}{R_{\text{total}}} = \frac{35 \text{ V}}{5 \text{ k}\Omega} = \frac{35 \text{ V}}{5000 \Omega} = 0.007 \text{ A} = 7 \text{ mA}$

Thus, $i(0^-) = 7 \text{ mA}$.

Would you like further details on the calculation or to discuss the circuit's behavior after $t = 0$? Here are some related questions:

1. What happens to the current $i(t)$ immediately after $t = 0$?
2. How does the capacitor discharge over time after the switch opens?
3. What is the time constant of the circuit after the switch opens?
4. How can we determine the voltage across the capacitor for $t > 0$?
5. How would the circuit change if the switch remained closed?

Tip: When a capacitor is involved in DC circuits, always consider the behavior in both steady-state (acting as an open circuit) and transient conditions (depending on the initial voltage across it).