Mini Physics
Part of Mini EducationSwitch subject

UY1: R-L Circuit

Derive RL current growth and decay using Kirchhoff's loop rule and the RL time constant.

  • University Physics Year 1
On this page
Why this matters + quick links

This page gives the UY1 working model/result for R-L Circuit. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

1) At a glance

  • In an RL circuit, current cannot change instantaneously because the inductor opposes changes in current (v_L = L di/dt).
  • Growth (source applied):
Ldi/dt + Ri = E
  • Decay (source removed):
Ldi/dt + Ri = 0
  • Time constant:
τ = L/R

Prerequisites: Self-Inductance & Inductors, Magnetic-Field Energy In Inductor
Next uses: Phasors & Alternating Currents, L-R-C Series Circuit

2) Setup

Sign convention used here:

  • Traverse loop in current direction.
  • Resistor drop is + Ri.
  • Inductor drop (self-induced emf opposing change) is + L di/dt.

Assume R and L are constant.

Common traps (initial conditions + signs)
  • Current through an ideal inductor is continuous: i(0⁺) = i(0⁻), even when a switch flips.
  • The inductor voltage sign depends on your passive sign convention. A safe check is: the inductor “pushes back” against changes in current.
  • Time constant is τ = L/R (not RC).

3) Core derivation/explanation

Growth (switch to source at t = 0)

Equation:

Ldi/dt + Ri = E, i(0) = 0

Solution:

i(t) = (E/R)(1-e^(-t/τ)), τ = L/R

So i rises from 0 to E/R.

Decay (source removed)

Equation:

Ldi/dt + Ri = 0, i(0) = I₀

Solution:

i(t) = I₀e^(-t/τ)

Energy in inductor:

U_L = (1/2)Li²

which dissipates through R during decay.

Checks (sanity)

  • Growth: i(0) = 0 and i(∞) = E/R.
  • Decay: i(t) must decrease exponentially toward 0 with the same τ = L/R.

4) Worked example(s)

Given R = 8.0 Ω, L = 0.40 H, E = 12 V:

  • τ = L/R = 0.050 s.
  • Final current I_f = E/R = 1.5 A.
  • At t = 0.10 s = 2τ:

i = 1.5(1-e⁻²) = 1.30 A

If source is then removed with i(0) = 1.30 A, after another 0.10 s:

i = 1.30e⁻² = 0.176 A

5) Practice set (with hints + answers)

  1. R = 5 Ω, L = 0.25 H, E = 10 V. Find τ and final current. Hint: τ = L/R, I_f = E/R. Answer: τ = 0.050 s, I_f = 2.0 A.

  2. For Q1, find i at t = τ during growth. Hint: i = I_f(1-e⁻¹). Answer: i = 1.26 A.

  3. A decaying RL current has I₀ = 3.0 A and τ = 0.20 s. Find i(0.40 s). Hint: two time constants. Answer: i = 3.0e⁻² = 0.406 A.

6) Summary + next steps

RL circuits are first-order transient systems. Once you identify the correct loop equation and initial condition, growth and decay are immediate from the same time constant τ = L/R.

Next: Phasors & Alternating Currents Previous: L-R-C Series Circuit Back To Electromagnetism (UY1)