RL time constant to near steady state: A 320 µH inductor is in series with a 3.3 kΩ resistor. Approximately how long does it take the current to build up to essentially its final value after a DC step (use about 5 time constants as 'full')?

Introduction / Context:In a first-order RL circuit driven by a DC step, the current approaches its final value exponentially with time constant τ = L / R. A common engineering rule is that about 5τ is effectively 'full' (over 99% of the final value). This question tests time-constant calculation and unit handling.

Given Data / Assumptions:

• L = 320 µH = 320 × 10^-6 H.
• R = 3.3 kΩ = 3300 Ω.
• Use 5τ ≈ 5 * (L / R) for near-steady-state timing.

Concept / Approach:Compute τ = L / R, then multiply by 5. Careful prefix conversion is crucial to avoid errors of 10^3 or 10^6 in the answer.

Step-by-Step Solution:

Verification / Alternative check:At 4τ, response is ≈ 98.2%; at 5τ, ≈ 99.3%, which is typically considered essentially full in practical design.

Why Other Options Are Wrong:

• 0.48 ms or 0.48 s or 48 s: These are orders of magnitude too large; they arise from mishandling µH/kΩ prefixes.

Common Pitfalls:

• Treating microhenry as millihenry or forgetting to convert kilohms to ohms.

Final Answer:0.48 µs