RL step response: A 220 Ω resistor and a 60 mH inductor in series are connected to a 20 V DC source via a switch. What is the inductor current two time constants after closing the switch?

Introduction / Context:In a first-order RL circuit, current grows exponentially toward a final value after a DC step is applied. The time constant τ governs how quickly the current approaches steady state. This question practices using τ to find the current at a specified multiple of τ.

Given Data / Assumptions:

• R = 220 Ω, L = 60 mH, V = 20 V DC.
• Initial current i(0) = 0 A (switch just closed).
• Find i(t) at t = 2τ, where τ = L / R.

Concept / Approach:

For a rising RL current: i(t) = I_final * (1 − e^(−t/τ)), where I_final = V / R. Compute τ, then evaluate i at t = 2τ.

Step-by-Step Solution:

Verification / Alternative check:

At t = 3τ the current would be ~95% of final (≈ 86.4 mA); at 2τ we expect ≈ 86% of final, matching ≈ 78–79 mA.

Why Other Options Are Wrong:

91 mA is the final current, not the value at 2τ. 57 mA corresponds to an earlier time. 400 mA exceeds the DC limit V/R.

Common Pitfalls:

Using e^(−tR/L) but mixing R and L units; forgetting I_final; rounding too early.

Final Answer:

78 mA