Inductance and RL Circuits

An inductor is simply a coil of wire, and its inductance measures how strongly it resists any change in the current flowing through it. The name comes from Faraday's induction: the coil induces a voltage in itself, which is why this is also called self-inductance. That makes inductors good at smoothing out current, storing energy for a moment, and producing big voltage kicks when a current gets cut off. The formula $\mathcal{E}_L = -L\,dI/dt$ looks technical, but it's just $L$ (how stubborn the coil is, in henries), $dI/dt$ (how fast you're trying to change the current) and the minus sign (it pushes back against the change). It's the circuit version of inertia; bear with it.

Electrical inertia. Close the switch on a big electromagnet and the current takes a noticeable moment to build; open it and a spark jumps across the switch. A coil resists changes in its own current. The reason is Faraday's law turned on itself: the coil's current makes a field, the field threads the coil's own turns, so changing the current changes the flux, which induces an EMF in the very coil carrying it.

Inductance. The field, and so the flux through each turn, is proportional to the current. The constant of proportionality for the total flux linkage is the inductance: $N\Phi_B = LI$. Then Faraday's law gives

By Lenz's law this back EMF opposes the change: it pushes against a rising current and pushes a falling one forward. In a circuit the potential drops by $L\,dI/dt$ across an inductor in the direction of the current, just as it drops by $IR$ across a resistor: nothing for a steady current, a lot for a fast change. One henry is one volt per (ampere per second).

A solenoid's inductance, derived. A long solenoid of length $\ell$, cross-section $A$ and $n$ turns per meter has field $B = \mu_0nI$ inside (Ampère's law). Each turn links $\Phi_B = \mu_0nIA$, and there are $N = n\ell$ turns:

The turns count twice: more turns make more field, and each extra turn links that field again. Like a capacitor's $C = \varepsilon_0A/d$, the answer is pure geometry. Fill the coil with iron and $\mu_0$ becomes $\mu_r\mu_0$, with $\mu_r$ in the hundreds or thousands, which is why real inductors and transformers have iron or ferrite cores.

Energy in the field. Driving current $I$ against the back EMF takes power $P = I\cdot L\frac{dI}{dt}$, so building the current from $0$ up to $I$ stores

the twin of a capacitor's $\frac12CV^2$. Where is it stored? For the solenoid, $\frac12(\mu_0n^2A\ell)I^2 = \frac{(\mu_0nI)^2}{2\mu_0}\cdot A\ell = \frac{B^2}{2\mu_0}\times\text{volume}$. The energy lives in the field itself, at $u_B = B^2/2\mu_0$ joules per cubic meter, the partner of the electric field's $\frac12\varepsilon_0E^2$.

The RL circuit. A battery $\mathcal{E}$, a resistor $R$ and an inductor $L$ in series, with the switch closed at $t = 0$. Going around the loop (Kirchhoff's voltage law):

Take a look at this equation: it's the same linear first-order equation as the RC circuit from the DC circuits unit, with $L/R$ where $RC$ used to be. Separate it, $\frac{dI}{\mathcal{E}/R - I} = \frac{R}{L}\,dt$, integrate, $-\ln\left(\frac{\mathcal{E}}{R} - I\right) = \frac{R}{L}t + C$, and use $I(0) = 0$:

At $t = 0$ the current is zero, so the whole battery voltage sits across the inductor and the current starts rising at $\mathcal{E}/L$. As it grows, the resistor takes more of the voltage and the inductor less, so the rise slows. After one $\tau$ the current is at 63% of $\mathcal{E}/R$, after $5\tau$ over 99%. Once it stops changing, the inductor is just a wire.

Switching off. Replace the battery with a wire and the inductor keeps the current going: $L\,dI/dt + IR = 0$, so $I = I_0e^{-t/\tau}$. The resistor burns $\int_0^\infty I_0^2e^{-2t/\tau}R\,dt = \frac12I_0^2R\tau = \frac12LI_0^2$, exactly the stored energy.

But simply break the circuit and the current has nowhere to go. It must drop to zero in microseconds, so $dI/dt$ is enormous and $L\,dI/dt$ can reach thousands of volts: that's the spark at the switch. Engineers tame it with a flyback diode across relay coils and motor windings, which gives the current a path to decay through. A car's ignition coil exploits it instead, interrupting a few amperes and stepping the kick up with a second winding to throw a 20,000 V spark from a 12 V battery.

Two rules for circuits. An inductor's current can't jump, since that would need infinite voltage, just as a capacitor's voltage can't. So just after a switch moves, every inductor carries the same current it did just before; long after, at DC steady state, every inductor is a short circuit.

A common mistake. It's tempting to think the voltage across an inductor can't be bigger than the battery's, since the battery is the only source in the circuit. It can: at switch-off, $L\,dI/dt$ can be far larger, and that's exactly the spark at the switch. Two other slips: writing $\mu_0N^2A\ell$ (with $N$ the total turns, the formula is $\mu_0N^2A/\ell$), and mixing up $L/R$ with $RC$.

Inductors sit wherever current has to be smoothed, stored or switched. The boost converter in a laptop charger or an LED driver pumps energy into an inductor and releases it at a higher voltage, thousands of times a second. A car's ignition coil interrupts a few amperes to throw a 20,000 V spark, while relay and motor drivers carry flyback diodes so the same kick doesn't destroy them. MRI magnets store megajoules in their fields and chokes on power and audio lines block high-frequency noise. The wire loops cut into the road at traffic lights detect a car because its steel body changes the loop's inductance.