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Where did the answers for il and vc come from

The circuit shown represents a typical first-order RL or RC circuit at \( t = 0^- \), just before a switch is closed or a condition changes. Let's break down where the values for the inductor current \( i_L(0^-) = 2A \) and capacitor voltage \( v_C(0^-) = 4.8V \) come from based on the circuit elements:

### Step 1: Understanding the steady-state at \( t = 0^- \)

- **Inductors in steady state**: Inductors act as short circuits in DC steady-state conditions. That means before any switching happens (at \( t = 0^- \)), the current through the inductor would be constant, and the voltage across it would be zero.
  
- **Capacitors in steady state**: Capacitors act as open circuits in DC steady-state conditions. That means before switching (at \( t = 0^- \)), the voltage across the capacitor would be constant, and no current would flow through it.

### Step 2: Finding \( i_L(0^-) \) and \( v_C(0^-) \)

1. **Inductor Current \( i_L(0^-) = 2 \, \text{A} \)**:
   - Since the inductor behaves like a short circuit in DC steady state, the 24V voltage source is driving current through the resistor network (a 2Ω resistor and a 10Ω resistor in series).
   - Applying **Ohm's Law** and **Kirchhoff's Voltage Law (KVL)**:
     \[
     I = \frac{V_{\text{source}}}{R_{\text{total}}} = \frac{24}{2 + 10} = \frac{24}{12} = 2 \, \text{A}.
     \]
   - Therefore, the current through the inductor, \( i_L(0^-) \), is 2A because, in steady state, all of the current supplied by the voltage source flows through the inductor.

2. **Capacitor Voltage \( v_C(0^-) = 4.8 \, \text{V} \)**:
   - The voltage across the capacitor can be found using **voltage division** since the capacitor is open in steady-state DC conditions. The voltage divider formula between the 2Ω and 10Ω resistors is:
     \[
     v_C(0^-) = V_{\text{source}} \times \frac{R_2}{R_1 + R_2} = 24 \times \frac{10}{2 + 10} = 24 \times \frac{10}{12} = 4.8 \, \text{V}.
     \]
   - So, the voltage across the capacitor at \( t = 0^- \) is 4.8V.

### Summary:
- \( i_L(0^-) = 2A \) comes from the current through the inductor, which is the same as the current supplied by the 24V source in steady-state DC conditions.
- \( v_C(0^-) = 4.8V \) is the voltage across the capacitor, calculated using the voltage divider formula between the 2Ω and 10Ω resistors.

Would you like more details or have any questions?

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Here are five relative questions that extend this topic:

1. How would the current \( i_L(t) \) change immediately after \( t = 0 \)?
2. How does the circuit behave for \( t > 0 \) if a switch is involved?
3. What happens to the voltage across the inductor immediately after \( t = 0^+ \)?
4. How would you find the time constant for this RL circuit?
5. How would the initial conditions change if the circuit had been in operation for a longer time?

**Tip**: In circuits with capacitors and inductors, always consider their behavior in steady-state DC conditions: capacitors become open circuits, and inductors become short circuits.

Where did the answers for il and vc come from?

Explore how to calculate the inductor current iL(0^-) and capacitor voltage vC(0^-) in a DC steady-state RL circuit. This problem applies key principles of circuit analysis, including Ohm's Law, voltage division, and the behavior of inductors and capacitors in steady state. Ideal for Grades 11-12 and college-level physics or electrical engineering students.

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