How does an RL circuit respond to a step voltage input?

In an RL circuit with step input, current builds up exponentially: I(t) = (Vs/R)(1 - e^(-Rt/L)), where L/R is the time constant. The inductor initially opposes current change, acting like an open circuit at t=0, then gradually allows current to reach its final value (Vs/R). Inductor voltage starts at Vs and decays to zero. This behavior is crucial in switching power supplies, motor starting circuits, and relay driver design.