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.