← CA-5 · CAPACITORS, INDUCTORS & FIRST-ORDER CIRCUITS

RL step response

YOU WILL BE ABLE TO

Define the RL time constant τ = L/R.
Write the current step response of an RL circuit.
Apply the universal first-order formula to inductor current.
Analyse switching events using current continuity.

Swap the capacitor for an inductor and the whole of Lesson 3 replays in the dual key. Close the switch at t = 0: the inductor's current cannot jump (Lesson 2), so it starts at zero and climbs toward its final value — which is set by the resistor alone, because a settled inductor is a short. Same exponential, same landmarks, one changed formula:

\tau = \frac{L}{R}

RL time constant

The dual step: at t = 0 the switch closes and the current climbs toward V_s/R with time constant τ = L/R.

Note the R moved to the denominator — and check it against intuition rather than memory: MORE resistance now means a SMALLER final current to reach, so the transient finishes sooner; more inductance means more opposition to changing current, so slower. The response from rest, and the universal form for any starting point:

i(t) = \frac{V_s}{R}\left(1 - e^{-t/\tau}\right)

i(t) = i(\infty) + \left[i(0) - i(\infty)\right]e^{-t/\tau}

RL step · universal form

WORKED EXAMPLE

RL charging, start to finish

In the figure: V_s = 10 V, R = 100 Ω, L = 100 mH, switch closes at t = 0 with i(0) = 0.
Final current: inductor → short, so i(∞) = 10/100 = 100 mA. Time constant: τ = 0.1/100 = 1 ms.
$i(t) = 100(1 - e^{-t/1\,\mathrm{ms}})$ mA. At t = 1 ms: 63.2 mA. At 5 ms: ≈ 99.3 mA — done.
The inductor's voltage tells the complementary story: $v_L = V_s\,e^{-t/\tau}$ starts at the FULL 10 V (all the supply lands on L at the first instant, since i = 0 drops nothing across R) and decays to zero as the inductor fades into a wire.

i(∞) = 100 mA, τ = 1 ms, i(1 ms) ≈ 63.2 mA

WORKED EXAMPLE

The opening switch, quantified

Lesson 2 warned about interrupting inductor current; the universal formula now makes it precise. Suppose the figure's circuit has settled at 100 mA, and the switch opens — but a protection path lets the current decay through a total loop resistance of 1 kΩ.
New time constant: τ = L/R = 0.1/1000 = 0.1 ms. Current continuity: i(0⁺) = 100 mA, i(∞) = 0.
$i(t) = 100\,e^{-t/0.1\,\mathrm{ms}}$ mA. The inductor voltage at the first instant: i·R = 0.1 × 1000 = 100 V across the decay path — from a 10 V circuit.
That factor of ten is the design trade in every flyback arrangement: a higher decay resistance kills the current faster but at proportionally higher voltage. ME-6 turns this arithmetic into the standard protection diode.

A 10 V circuit briefly generating 100 V — safely, because the path was planned

PRACTICE

Find the time constant of a 50 mH inductor in series with 500 Ω.

An RL circuit driven from rest heads toward a final current of 80 mA. What is the current after exactly one time constant, and after five?

An RL circuit has i(0) = 0, i(∞) = 40 mA, τ = 2 ms. How long until the current reaches 30 mA?

State the RL duals of these RC facts: (a) "τ = R_Th·C"; (b) "a capacitor charges toward the open-circuit (Thevenin) voltage."