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To solve for \( v_C(0^-) \), the voltage across the capacitor at the moment just before \( t = 0 \), we can use the properties of capacitors and inductors in steady-state conditions for \( t < 0 \).

### Key points:

1. **Steady-state behavior** of capacitors and inductors:
   - At \( t = 0^- \), the circuit has reached steady state.
   - A **capacitor** in steady state behaves like an open circuit because the voltage across it has stabilized (no current flows through it).
   - An **inductor** in steady state behaves like a short circuit because the current through it has stabilized (voltage drop across it is zero).

2. **Initial conditions at \( t = 0^- \)**:
   - For the capacitor, since it's an open circuit, the current \( i_C(0^-) \) is zero.
   - For the inductor, since it's a short circuit, we need to analyze the current through it at \( t = 0^- \), which is constant.

#### Steps:
1. **Determine the steady-state behavior of the circuit** for \( t = 0^- \):
   - The 3 A current source will drive current through the 10 Ω resistor and continue flowing through the circuit.
   - Since the capacitor behaves as an open circuit, the current cannot flow through it, and thus the current from the 4 A source will flow only through the inductor's branch and the 5 Ω resistor.

2. **Calculate the voltage across the capacitor**:
   - For \( t = 0^- \), the voltage \( v_C(0^-) \) can be calculated using Kirchhoff’s Voltage Law (KVL) around the left loop involving the 10 Ω resistor and the capacitor.

Since no current flows through the capacitor in steady state, the voltage across the capacitor will be the same as the voltage across the 10 Ω resistor, driven by the 3 A current source.

   \[
   v_C(0^-) = I \times R = 3 \, \text{A} \times 10 \, \Omega = 30 \, \text{V}
   \]

Thus, \( v_C(0^-) = 30 \, \text{V} \), and the correct answer is:

\[
\boxed{a. \, 30 \, \text{V}}
\]

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Would you like further details or have any questions? Here are five related questions to deepen your understanding:

1. How does the behavior of an inductor differ from a capacitor in a DC steady-state?
2. How would the circuit behavior change if there were no 4 A current source?
3. What happens to the voltage across the capacitor immediately after \( t = 0 \) (i.e., \( v_C(0^+) \))?
4. How can you calculate the energy stored in the capacitor just before \( t = 0 \)?
5. What would happen if the capacitor had an initial charge at \( t = 0^- \)?

**Tip:** In DC steady state, inductors and capacitors are often simplified to short circuits and open circuits, respectively, to make analysis easier.

For this circuit, what is v_C(0^-), the voltage across the capacitor at time t = 0^-?

This circuit problem focuses on calculating the steady-state voltage across a capacitor at time t = 0^-. Using Kirchhoff's Voltage Law and steady-state behaviors of inductors and capacitors, we solve for v_C(0^-). Key concepts include Ohm's Law and capacitor/inductor behavior in DC circuits.

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