RL decay timing: For a series RL circuit with L = 20 H and R = 10 Ω, what is the approximate time required for a 0.4 A inductor current to decay essentially to zero after a switch opens?

Introduction / Context:Inductor current decays exponentially after the driving source is removed. Engineers often estimate a “practical zero” using the time constant concept to plan protection networks (snubbers), decide measurement windows, or ensure safe discharge intervals in power circuits.

Given Data / Assumptions:

• Series RL path with L = 20 H, R = 10 Ω.
• Initial current I0 = 0.4 A (any nonzero value decays with the same time constant).
• Looking for “complete” or “practical” decay (≈ 1% or less of I0).

Concept / Approach:The RL time constant is τ = L / R. Current decays as i(t) = I0 * e^(−t/τ). A common practical rule: after 5τ, i(t) ≈ 0.0067 * I0 (less than 1%), often treated as “essentially zero.”

Step-by-Step Solution:

Verification / Alternative check:Exact check: at t = 10 s, i(10) = 0.4 * e^(−10/2) = 0.4 * e^(−5) ≈ 0.4 * 0.0067 ≈ 2.7 mA, effectively zero for most purposes.

Why Other Options Are Wrong:

• 2 s (1τ): Leaves about 36.8% of I0; far from zero.
• 4 s (2τ): Leaves about 13.5% of I0.
• 8 s (4τ): Leaves about 1.8% of I0; closer, but the standard rule of thumb uses 5τ.

Common Pitfalls:Confusing τ with full decay time; τ is just the characteristic time, not the endpoint. Also, forgetting that the initial current magnitude does not change τ.

Final Answer:10 seconds