LR Circuits Right After Switching and Long After

An inductor's defining behavior is V = L(dI/dt): it generates a voltage proportional to how fast current through it is changing. It cannot tolerate a sudden jump in current, because that would require infinite dI/dt and infinite voltage. So the instant a switch closes, if the inductor had zero current before, it still has zero current after. It behaves exactly like a cut wire in that moment, forcing current to find another path. This is the intuition for the whole topic: freeze the inductor's current at its starting value and solve the rest of the circuit with ordinary resistor rules.

Long after switching, everything has settled into a steady state: currents are constant, so dI/dt = 0, so the inductor's voltage drop is zero. A component with zero voltage drop but nonzero current is just a wire. Replace every inductor with a plain wire, redraw the circuit, and solve it with Ohm's law and Kirchhoff's rules like any resistor network. The hard part, the exponential rise or decay of current over time, never has to be touched for these two snapshots.

This is the mirror image of how a capacitor behaves, because a capacitor obeys I = C(dV/dt) and cannot tolerate a sudden jump in voltage instead of current. Right after switching, a capacitor holds whatever voltage it had before, so if it starts uncharged it acts like a plain wire (zero voltage drop). Long after, it is fully charged and current into it stops, so it acts like a broken wire. Inductor and capacitor trade behaviors at the two extremes because one fights changes in current and the other fights changes in voltage.