RL step response timing In a simple series RL circuit excited by a DC step, the inductor current rises toward its final value exponentially with time constant τ = L / R. After exactly two time constants have elapsed, the current reaches approximately what percentage of its final steady-state value?

Introduction / Context:Time constants are quick design tools for estimating how fast circuits respond. For RL and RC first-order systems, the exponential approach to steady state follows a consistent pattern that is widely memorized for 1τ, 2τ, 3τ, and so on. This question targets the numerical recall linked to two time constants.

Given Data / Assumptions:

• Series RL with total resistance R and inductance L.
• Step input of DC voltage applied at t = 0.
• Ideal components; no saturation or parasitics.

Concept / Approach:For a step from 0 to I_final, the current is i(t) = I_final * (1 − e^(−t/τ)), where τ = L / R. At t = 2τ, the fraction to final is 1 − e^(−2). Numerically, e^(−2) ≈ 0.1353, so the fraction is about 0.8647 or 86% of the final value.

Step-by-Step Solution:

Verification / Alternative check:Known milestones: at 1τ → 63%; at 2τ → 86%; at 3τ → 95%; at 4τ → 98%; at 5τ → 99%+. These standard approximations validate the calculation.

Why Other Options Are Wrong:

Common Pitfalls:Mixing RC and RL forms (they share the same percentages); misremembering the milestone values; confusing current rise with voltage across the inductor (which decays).

Final Answer:86%