RL step response sizing: A DC source Vs = 12 V is applied to a series RL circuit with R = 60 Ω and L = 24 mH. After exactly 2 time constants following the step, what is the current in the inductor (current rise value)?

Introduction / Context:The RL step response describes how current builds up when a DC voltage is applied. Designers use the time constant τ to predict rise time and choose component values that meet transient requirements in drivers, relays, and filters.

Given Data / Assumptions:

• Supply Vs = 12 V DC.
• Series elements: R = 60 Ω, L = 24 mH.
• We evaluate current at t = 2τ after the step is applied.

Concept / Approach:For a series RL driven by a DC step, the current is i(t) = I_final * (1 − e^(−t/τ)), where I_final = Vs / R and τ = L / R. Substituting the given values yields the numerical current at t = 2τ.

Step-by-Step Solution:Compute time constant: τ = L / R = 0.024 H / 60 Ω = 0.0004 s = 0.4 ms.Final current: I_final = Vs / R = 12 V / 60 Ω = 0.2 A = 200 mA.Evaluate exponent at t = 2τ: e^(−2) ≈ 0.135335.Current at t = 2τ: i(2τ) = 0.2 A * (1 − 0.135335) = 0.2 * 0.864665 ≈ 0.172933 A.Rounded to three significant figures: 173.0 mA.

Verification / Alternative check:Rule of thumb: After 1τ current ≈ 63.2% of final; after 2τ ≈ 86.5%; after 3τ ≈ 95.0%. With I_final = 200 mA, 0.865 * 200 mA ≈ 173 mA, confirming the computation.

Why Other Options Are Wrong:79.9 mA: near 40% of final, much earlier than 2τ.126.4 mA: around 63% of final, corresponds to 1τ.198.6 mA: almost full-scale; closer to 3–4τ or steady state, not 2τ.

Common Pitfalls:Mixing time constant definitions or forgetting to convert millihenries to henries. Always compute τ = L / R with consistent SI units.

Final Answer:173.0 mA