LR Circuits: Solving the Current ODE
- Set up the LR circuit differential equation from Kirchhoff’s loop rule
- Solve the ODE to obtain i(t) = (ε/R)(1 − e^(−Rt/L)) and the decay solution
- Analyze inductor behavior at t = 0 and t → ∞ and contrast it with capacitors
The circuit that cannot change its mind instantly
Connect a battery ε, resistor R, and inductor L in series and close the switch. The inductor’s back-EMF −L·di/dt opposes the rising current, so i cannot jump — it must grow continuously from zero. The loop rule at every instant reads ε − iR − L·di/dt = 0: the same “rate proportional to remaining gap” structure as the charging RC circuit, so the solution will again be an exponential approach, this time of the current toward its final value ε/R.
Solve ε − iR − L·di/dt = 0 with i(0) = 0 to find the current in an LR circuit after the switch closes.
- 1.Rearrange: L·di/dt = ε − iR, so di/dt = (R/L)(ε/R − i).
- 2.Separate variables: di/(ε/R − i) = (R/L)·dt.
- 3.Integrate from the start (i = 0 at t = 0): −ln[(ε/R − i)/(ε/R)] = Rt/L.
- 4.Exponentiate and solve: ε/R − i = (ε/R)e^(−Rt/L), so i(t) = (ε/R)(1 − e^(−Rt/L)).
- 5.The time constant is τ = L/R, so i(t) = (ε/R)(1 − e^(−t/τ)); the voltage across the inductor is V_L = L·di/dt = ε·e^(−t/τ), starting at the full ε and dying to zero.
- 6.Check the limits: i(0) = 0 (the inductor blocks the initial jump) and i → ε/R as t → ∞ (steady current, inductor acts like a wire) ✓.
The inductor’s two limiting behaviors — opposite to the capacitor’s
For transient snapshots, an inductor is the capacitor’s mirror image. At t = 0, current cannot jump, so an inductor that started at zero current momentarily acts like an open circuit — no current in its branch. As t → ∞, the current is steady, di/dt = 0, the back-EMF vanishes, and the ideal inductor acts like a bare wire. Compare: the capacitor starts as a wire and ends as an open. Swap the two components and the transient story runs exactly backward.
Decay, and the danger of opening the switch
If the battery is switched out and the inductor drives its current i₀ through a resistor, the loop rule −iR − L·di/dt = 0 gives the pure decay i(t) = i₀e^(−t/τ), with the stored ½Li₀² dissipated in R. But try to interrupt an inductive current abruptly — opening a switch — and di/dt becomes enormous, so the back-EMF ε = −L·di/dt spikes to values far above the battery voltage. That spike arcs across switch contacts; it is why inductive circuits use flyback diodes, and why the ODE’s insistence on continuous current has very practical teeth.
A 12 V battery, a 6.0 Ω resistor, and a 30 mH inductor are in series. Find τ, the final current, the current one time constant after closing the switch, and the initial rate di/dt.
- 1.Time constant: τ = L/R = (30 × 10⁻³)/(6.0) = 5.0 × 10⁻³ s = 5.0 ms.
- 2.Final current (inductor a wire): i_∞ = ε/R = 12/6.0 = 2.0 A.
- 3.At t = τ: i = 2.0(1 − e⁻¹) = 2.0 × 0.632 ≈ 1.3 A.
- 4.Initial slope from the loop rule at t = 0 (where i = 0): ε = L·di/dt, so di/dt = ε/L = 12/(30 × 10⁻³) = 4.0 × 10² A/s.
- 5.Check: an initial rate of 400 A/s sustained for one τ = 5.0 ms would give 2.0 A — the final current. The tangent-line-at-zero always aims at the asymptote one time constant away.
The LR time constant is τ = L/R — resistance in the denominator — while the RC constant is τ = RC with resistance in the numerator. Increasing R makes an RC circuit slower but an LR circuit faster (the resistor eats the stored energy sooner). Writing τ = LR, or importing RC intuition about R, is the classic transient-circuit error.
Immediately after the switch closes in a series battery–resistor–inductor circuit, the current is:
Long after the switch closes, the voltage across the ideal inductor is:
Answer the 2 checkpoints as you read.
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