Instantaneous voltage division in an RL integrator: At the instant of a rising pulse edge, how is the input voltage divided between R and L?

Introduction / Context:
RL integrators respond to fast edges by momentarily placing most of the input across the inductor because inductors initially oppose changes in current. Understanding the instantaneous division right at the edge is key for predicting spike shapes and amplitudes.

Given Data / Assumptions:

• RL integrator configuration (series R and L).
• Observe behavior at t = 0+ for a rising step (pulse edge).
• Ideal components; no core saturation or parasitics.

Concept / Approach:
At the instant of a rising edge, the current through an inductor cannot change instantaneously. Therefore i_L(0+) ≈ i_L(0-) and initially the current is essentially the pre-edge value. With nearly zero change in current at t = 0+, the resistor drop v_R = i * R is near zero, so the inductor must support almost the entire applied voltage.

Step-by-Step Reasoning:
At t = 0+: di/dt is large, but i itself has not yet increasedv_R(0+) = i(0+) * R ≈ 0v_L(0+) = Vin - v_R(0+) ≈ VinTherefore, the input is across L at the instant of the rising edge

Verification / Alternative check:
Time constants: i(t) rises with τ = L/R. Only after a finite time will current build and a significant voltage appear across R; initially it is negligible.

Why Other Options Are Wrong:

• All across R or 63% splits: These describe later times, not the instant of the edge.

Common Pitfalls:
Confusing immediate (t = 0+) conditions with steady-state values or mixing RC intuition with RL behavior. Inductors initially behave like opens to current changes, forcing the applied voltage across themselves.

Final Answer:
all the input voltage is across the inductor