04 · Intermediate

RL transients

Apply the same first-order method to one-inductor linear networks. Carry inductor current continuously through switching, reduce the t>0 resistive/source network seen by L, derive L diL/dt+Rth iL=Vth and τ=L/Rth, solve source-free and driven current responses, derive inductor voltage from di/dt, explain why interrupting current can demand a large voltage, verify energy dissipation, and state when ideal one-inductor reasoning fails.

165–210 minIntermediate

In this lesson

Determine iL(0+) from the pre-switch state using inductor-current continuity.
Find iL(∞) from the post-switch DC network where the ideal inductor has zero terminal voltage.
Derive the Thévenin-reduced RL equation L diL/dt+Rth iL=Vth.
Derive τ=L/Rth and iL(t)=i∞+[i0−i∞]e^(−t/τ).
Derive inductor voltage from vL=L diL/dt with a consistent passive reference.
Solve source-free current decay, current build-up under a DC source and arbitrary nonzero initial conditions with one formula.
Explain why opening an inductor-current path can produce a large voltage and why practical circuits need a valid energy-return/dissipation path.
Compute Rth correctly when dependent sources are present using a test source if required.
Verify source-free decay by showing resistor dissipation equals the initial magnetic energy ½LI0².
State limits of the ideal model including winding resistance, saturation, core loss, hysteresis, parasitic capacitance and additional storage states.

Starting question

A current flows in an inductor when a switch changes position. The topology changes instantly, but the magnetic energy cannot disappear instantly. What current must exist at 0+, what current will the new circuit eventually support, and what resistance determines how quickly the transition occurs?

04.1Reduce the post-switch network seen by the inductor

Remove L temporarily and look into its two terminals for the t>0 network. For a linear resistive/source remainder, use the Thévenin pair Vth,Rth. Reconnect the inductor with passive references so current enters the positive voltage terminal.

Figure 4.0 — Thévenin equivalent seen by the inductor

The one question is: which current and voltage references correspond exactly to the differential equation?

Current iL enters the positive reference terminal of vL, so the passive convention gives vL=L·diL/dt. Traversing the loop with iL gives Vth−Rth iL−vL=0.

V_th−R_thi_L−v_L=0

v_L=L di_L/dt

Therefore

L di_L/dt+R_thi_L=V_th

For Rth>0, the stable time constant is

τ=L/R_th

04.2Initial and final currents

For finite non-impulsive voltage, inductor current is continuous:

i_L(0⁺)=i_L(0⁻)=I₀

At DC steady state the ideal inductor has vL=0. In the Thévenin form,

i_L(∞)=I_∞=V_th/R_th

The complete current is therefore

i_L(t)=I_∞+[I₀−I_∞]e^−tRth/L

Figure 4.1 — RL transients follow the same first-order logic as RC

The one question is: which quantity changes role when C is replaced by L?

The solution structure is unchanged; the continuous state is now inductor current and equivalent resistance appears in the denominator of the time constant.

04.3Inductor voltage is derived from the current slope

di_L/dt=−[I₀−I_∞]/τ · e^−t/τ

v_L(t)=L di_L/dt=R_th[I_∞−I₀]e^−t/τ

At 0+,

v_L(0⁺)=V_th−R_thI₀

which matches KVL in the post-switch circuit if the inductor is momentarily represented by the current source I0 needed to preserve its state.

04.4Source-free current decay

If an inductor initially carrying I0 discharges through a resistor R with no independent source,

Figure 4.2 — Source-free RL decay

The one question is: how should the references be chosen while the inductor returns its stored energy?

The initial current keeps its reference direction. With the passive voltage reference shown on L, vL becomes negative during decay: the inductor delivers power to the resistor.

V_th=0, R_th=R, τ=L/R

i_L(t)=I₀e^−tR/L

Under passive references,

v_L(t)=−R I₀e^−tR/L

so pL=vLiL is negative: the inductor returns its stored magnetic energy to the resistor.

w_L(t)=½L I₀²e^−2tR/L

04.5Current build-up under a DC source

For a DC source Vs driving a series R–L network from iL(0)=0,

Figure 4.3 — Current build-up in a series RL circuit

The one question is: where are iL, vR and vL in the driven loop?

The same current flows through R and L. At 0+, iL=0 so vR=0 and vL=Vs; at steady state vL→0 and vR→Vs.

I_∞=V_s/R, τ=L/R

i_L(t)=V_s/R [1−e^−tR/L]

v_L(t)=V_se^−tR/L

At 0+, all source voltage appears across the ideal inductor because current is still zero. At long time, inductor voltage tends to zero and the resistor carries Vs/R.

04.6Why interrupting inductor current can create a large voltage

The inductor law can be rearranged as

v_L=L Δi/Δt

for an average change over a short interval. For a fixed current change, demanding an extremely small Δt demands a large voltage magnitude.

An ideal open switch would try to force current to zero instantly while providing no current path. The ideal lumped model then requires an unbounded voltage impulse. Real circuits resolve this through parasitic capacitance, arcing, semiconductor breakdown, a clamp/snubber/flyback path, or other nonideal behavior.

04.7Worked nonzero-initial-current example

Pedagogical example: L=40 mH, Rth=10 Ω, Vth=20 V, and I0=0.5 A.

I_∞=20/10=2 A

τ=40 mH/10 Ω=4 ms

i_L(t)=2−1.5e^−t/4ms A

Voltage is

v_L(t)=10(2−0.5)e^−t/4ms=15e^−t/4ms V

Initial slope:

di/dt|₀₊=(2−0.5)/4ms=375 A/s

and v/L=15 V/40 mH=375 A/s, confirming the result independently.

04.8Energy check for passive decay

For source-free RL decay, resistor power is

p_R=R I₀²e^−2Rt/L

Integrating,

∫₀^∞p_Rdt=R I₀²·L/(2R)=½L I₀²

which equals the initial inductor energy. This independently verifies the transient and the sign interpretation.

04.9Model limits and special cases

The simple exponential assumes one independent magnetic storage state and a linear time-invariant network. Important departures include winding resistance, nonlinear inductance/saturation, hysteresis, frequency-dependent core loss, skin/proximity effects, parasitic capacitance, and mutual coupling to other coils.

If Rth=0 with a nonzero ideal constant Vth, the equation is L di/dt=Vth: current ramps rather than approaches a finite exponential equilibrium. Thus τ=L/Rth is meaningful only when a resistive restoring path produces a stable first-order pole.

If another independent C or L is present, the system may be second order or higher even if one component dominates.

04.10Mastery exercises

L=50 mH carrying 0.8 A discharges through 25 Ω. Find τ, iL(t), vL(0+) and initial energy.
For exercise 1, find time until current reaches 0.1 A.
L=30 mH, R=15 Ω, Vs=12 V, i(0)=0: write i(t) and vL(t).
A 20 mH inductor starts at −0.2 A and is connected to Vth=6 V through Rth=10 Ω. Find i(t) and initial slope.
Why is τ proportional to L but inversely proportional to R?
Why can opening an inductor path create a large voltage?
How should Rth be found when dependent sources exist?
Verify the source-free energy balance by integration.

Solutions

1. τ=0.05/25=2 ms; i=0.8e^(−t/2ms) A; vL(0+)=−RI0=−20 V; w0=½(0.05)(0.8²)=16 mJ.

2. 0.1=0.8e^(−t/2ms) ⇒ t=2ms ln8≈4.159 ms.

3. I∞=0.8 A, τ=2 ms; i=0.8[1−e^(−t/2ms)] A; vL=12e^(−t/2ms) V.

4. I∞=0.6 A, τ=2 ms; i=0.6−0.8e^(−t/2ms) A; initial slope=(0.8)/(2ms)=400 A/s.

5. Larger L requires more volt-seconds to change current; larger R creates a larger opposing/resistive voltage for a given current and therefore faster decay.

6. Current continuity demands a large di/dt only if a correspondingly large voltage is available; forcing Δt toward zero forces voltage magnitude upward.

7. Deactivate independent sources, retain dependent sources, apply a test source at the inductor port and use Rth=Vt/It.

8. ∫0∞R I0²e^(−2Rt/L)dt=½LI0².

Chapter 05 will unify RC and RL through poles, time constants, frequency response and design/measurement criteria, showing exactly how the time-domain exponential and the first-order filter of AC-02 are the same system viewed from two domains.