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?
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A78 mA
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B91 mA
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C57 mA
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D400 mA
Answer
Correct Answer: 78 mA
Explanation
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:
τ = L / R = 0.06 H / 220 Ω ≈ 0.0002727 s.I_final = V / R = 20 / 220 = 0.09091 A ≈ 90.9 mA.At t = 2τ: i(2τ) = I_final * (1 − e^(−2)) ≈ 90.9 mA * 0.8647 ≈ 78.6 mA.Rounded to the nearest listed value: 78 mA.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