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say we have a circuit like this

When is the inductor treated as a wire? When is it treated as an inductor? Times: t= 0- (which in this case, i imagine is the same as saying t < 0), t= infinity, t = 0+

In this circuit analysis, we deal with the behavior of the inductor in three different time frames: \( t = 0^- \) (before the switch is closed), \( t = 0^+ \) (just after the switch is closed), and \( t \to \infty \) (after a long time has passed). Here's how the inductor behaves in each case:

### 1. **At \( t = 0^- \) (before the switch closes, or \( t < 0 \))**:
   - The circuit is unconnected because the switch is open.
   - The inductor has no current flowing through it because the circuit is incomplete.
   - The inductor can be ignored in this case since the switch has not been closed.

### 2. **At \( t = 0^+ \) (immediately after the switch is closed)**:
   - At the moment the switch is closed, \( t = 0^+ \), the current through the inductor cannot change instantaneously (this is one of the fundamental properties of inductors: they resist instantaneous changes in current). 
   - The current through the inductor will remain as it was just before \( t = 0 \), which is \( i(0^-) = 0 \), because no current was flowing through the circuit before the switch was closed.
   - In this initial moment after the switch closes, the inductor behaves like an open circuit, because its opposition to a sudden change in current is equivalent to having infinite impedance. So, we treat it as an open circuit right at \( t = 0^+ \).

### 3. **As \( t \to \infty \) (after a long time)**:
   - After a long period of time, the inductor reaches a steady state. In a DC circuit (which this is, as \( V_s \) is constant), the current through the inductor eventually becomes constant.
   - When the current is constant, the inductor behaves like a short circuit (a simple wire) because the voltage across the inductor is given by \( v_L = L \frac{di(t)}{dt} \), and \( \frac{di(t)}{dt} = 0 \) in steady state.
   - So, at \( t = \infty \), the inductor is treated as a wire, or a short circuit.

### Summary:
- **\( t = 0^- \):** No current flows, circuit is incomplete.
- **\( t = 0^+ \):** Inductor behaves as an open circuit (no current flows instantly).
- **\( t \to \infty \):** Inductor behaves as a short circuit (wire), and current is steady.

Do you want me to solve for \( i(t) \) explicitly in this RL circuit for all time \( t \)?

### Related Questions:
1. How is the current through the inductor calculated at any time \( t \) after the switch is closed?
2. How does the time constant \( \tau \) affect the rate at which the current changes in this circuit?
3. What would happen if the value of the inductor \( L \) were doubled?
4. How does the presence of two resistors in the circuit influence the steady-state current?
5. What is the mathematical relationship between the voltage across the inductor and the current through it at any time \( t \)?

### Tip:
Remember, inductors resist changes in current, so they act as open circuits at the instant of a switch and as wires once the current stabilizes over time.

Say we have a circuit like this: When is the inductor treated as a wire? When is it treated as an inductor? Times: t= 0- (which in this case, i imagine is the same as saying t < 0), t= infinity, t = 0+.

Explore how an inductor behaves at different moments (t = 0-, t = 0+, t → ∞) in an RL circuit with resistors R1 and R2. Learn how the current changes over time and when the inductor is treated as a wire or an open circuit.

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