FIG.28 · CIRCUITS

RL CIRCUITS AND BACK-EMF

The coil's version of hesitation. Open a switch and pay in kilovolts.

§ 01

A coil resists its own current change

Replace the capacitor from FIG.27 with an inductor and you get the mirror circuit: a source, a resistor, and a coil in series. Close the switch and you might expect the current to snap to its Ohm's-law value V₀/R the way a bulb snaps to full brightness. It doesn't. The current eases up, climbing an exponential that slows as it approaches the steady state, and — same as the capacitor's voltage — it never quite arrives in finite time.

The reason is Lenz, not Ohm. A rising current threads a rising magnetic flux through its own coil. By Faraday's law, a changing flux induces an EMF, and by Lenz's rule that EMF has a sign: it opposes the change. A coil whose current is on the rise presents a voltage back against the source, as if fighting to keep things the way they were. Joseph Henry discovered this phenomenon on the same laboratory timetable as Faraday and named what an engineer now calls back-EMF. The SI unit of inductance — the henry — is his.

A capacitor cannot change its voltage discontinuously because V = Q/C and charge takes time to arrive. An inductor cannot change its current discontinuously because V = L · dI/dt and an infinite dI/dt would demand an infinite voltage. The two primitives are duals: the capacitor is an electrical spring, the inductor is an electrical flywheel, and both hold their state until something pays the toll to change it.

§ 02

The ODE

Take the loop. Battery V₀, resistor R, inductor L, all in series. Apply Kirchhoff's voltage law — the sum of voltage drops around a closed loop equals zero, or equivalently, the source EMF equals the sum of the drops across each element:

EQ.01
V0  =  IR  +  LdIdtV_0 \;=\; I\,R \;+\; L\,\dfrac{dI}{dt}

That is a first-order linear ODE in I(t). Rearrange to dI/dt = (V₀ − I·R)/L, separate variables, integrate with the initial condition I(0) = 0, and you land on the clean exponential:

EQ.02
I(t)  =  V0R(1et/τL),τL  =  LRI(t) \;=\; \dfrac{V_0}{R}\,\bigl(1 - e^{-t/\tau_L}\bigr), \qquad \tau_L \;=\; \dfrac{L}{R}

Compare to the RC charge-up, V_c(t) = V₀(1 − e^(−t/RC)), and the duality lines up term by term: V_c → I, V₀ → V₀/R, RC → L/R. The coil plays the role the capacitor played, but the variable being held in check has flipped from voltage to current. This is why τ_L has seconds as its units — check it: henrys/ohms = V·s/A ÷ V/A = s. The arithmetic of dimensional analysis always has the final word.

§ 03

Rise — the 63 % mark

The current climbs slowly at first because the inductor is fighting hardest when the current is rising fastest. As the current approaches its asymptote V₀/R, the back-EMF dies away, and the resistor eventually absorbs the whole source voltage at steady state. At that point the inductor is a wire — a zero-volt short — and all the drop sits on R.

FIG.28a — current rises on 1 − e^(−t/τ); back-EMF decays on e^(−t/τ); together they always sum to V₀
loading simulation

Watch the amber I(t) trace climb and the green back-EMF trace fall. They are two faces of the same exponential. At t = τ_L the current has reached 63.2 % of its final value — the universal marker, same as RC. At , 86.5 %. At , 99.3 %. That "5 τ settling time" rule of thumb every datasheet recites applies here too, because the mathematics is the same mathematics.

Slide the L slider. A bigger inductor means a bigger τ_L, which means a lazier ramp — the coil's flywheel weighs more, so it takes longer to spin up. Cut L in half and everything speeds up linearly. This is a knob engineers use every day when sizing inductors for switching supplies.

§ 04

Decay — the mirror

Cut the source off while current is flowing and leave a resistor in the path to let the current continue. The loop equation becomes 0 = I·R + L · dI/dt, the same ODE with no forcing term. The solution is the naked exponential:

EQ.02b
I(t)  =  I0et/τLI(t) \;=\; I_0 \, e^{-t/\tau_L}
FIG.28b — source removed; current freewheels through R and decays exponentially
loading simulation

At t = τ_L the current has dropped to 1/e ≈ 36.8 % of its initial value. At 5τ_L, below 1 %. Every joule of the ½LI² that was stored in the magnetic field drains out through the resistor as I²R heat — the same way a spinning flywheel with a brake loses its kinetic energy to the brake pad. The inductor, for its part, is now the source: it pushes current in the direction it was already flowing, because that is what "opposing the change" means when the change is a fall toward zero.

§ 05

Stored energy — ½ L I²

The question "where did the energy go while the current was ramping?" has a clean answer: into the magnetic field the coil was building. Integrate the source's delivered power P = V · I across the ramp and the usual factor of ½ drops out:

EQ.03
U  =  0ILidi  =  12LI2U \;=\; \int_0^I L\,i\,di \;=\; \tfrac{1}{2}\,L\,I^2

This is the same ½LI² derived in FIG.24 (Energy in Magnetic Fields), here re-seen from the circuit side. An ideal inductor is a perfect reservoir: nothing is dissipated in L itself, and every joule that went in came from the source, doing work against the back-EMF. FIG.24 also showed that this energy lives in the field's volume at density u = B²/(2μ₀). FIG.35 (Poynting's theorem, §07.4) will show that when the current changes, energy flows through the surrounding space to or from that reservoir — the radiation picture. For now hold onto the circuit-level fact: the inductor holds ½LI² and will hand it back the instant you give it a path.

§ 06

Flyback — the spark-plug reveal

Here is where the coil's refusal to change its current becomes spectacularly visible. Steady current I_steady = V₀/R is flowing through an RL circuit. You yank the switch open. What should happen to the current? In a lump-element world with an open switch, it should go to zero instantly. But the inductor refuses — V = L · dI/dt says that forcing dI/dt → ∞ demands V → ∞, and physics offers a compromise: whatever voltage the coil can build, it will, until the open gap ionises and the current gets to keep flowing through an arc.

FIG.28c — open the switch on an inductive circuit. dI/dt goes huge; V_L = L·dI/dt goes huge; a lilac arc forms across the gap
loading simulation

Click "open switch" and watch what happens. The animation models a 10 mH coil carrying 3 A with a 20 µs mechanical cut. That gives dI/dt ≈ −1.5 × 10⁵ A/s and V_L ≈ −1500 V — a kilovolt-scale spike across a coil that moments ago was dropping 12 V. The spike jumps the switch gap as an arc, the air ionises, and the current finishes its ramp-down through the spark instead of through the wire. That arc is not a malfunction. That arc is the physics refusing to let the magnetic energy disappear in zero time.

The engineering corollary is safety. Relay coils, motor windings, solenoids — anything with meaningful L — will kick out a huge spike every time you de-energise them. A flyback diode across the coil, reverse-biased in normal operation, clamps the spike the instant V_L reverses and lets the current decay through the diode at a safe voltage. Every MOSFET driver, every IGBT inverter, every CMOS microcontroller pin driving a solenoid has one. Skip it and the semiconductor dies the first time the coil kicks.

§ 07

Where it shows up

RL transients sit inside every piece of switched electronics you own. Switched-mode power supplies — the brick on your laptop cable — cycle an inductor through ramp-and-release twenty to a hundred thousand times a second, transferring the ½LI² reservoir to a capacitor on the output. Relays and solenoids live and die on their flyback; the diode across the coil is mandatory. Plasma generators and induction heating rely on the same spike in deliberately extreme form to ionise gas or to drive eddy currents in a workpiece. Ignition coils are flyback running on purpose, scaled up with a turns ratio. And the most quietly consequential: every ignition spark, every time a car engine fires, is one cylinder of one coil L · dI/dt clipped out of the air.