Energy Stored in an Inductor

An inductor stores energy in the magnetic field it creates, and the half in $E = \tfrac{1}{2}LI^2$ has the same origin as the half in a capacitor's $\tfrac{1}{2}CV^2$ or a spring's $\tfrac{1}{2}kx^2$. Building the current is not free: the coil opposes any change in current with a back-EMF of $L\,di/dt$, so the source must push against it the whole way up. That opposition grows in step with the current, from nothing at the start to its full value at the end, so the average is half the maximum and the work done is half of what a naive product would give. Integrate $P = vi = Li\,(di/dt)$ from zero to $I$ and the half appears formally.

A 100 mH choke carrying 3 A holds $0.5 \times 0.1 \times 9 = 0.45\ \text{J}$. That sounds negligible until you ask how fast it can be released: interrupt that current in 10 µs and you are dissipating 45 kW while it lasts. Because the energy goes as the square of the current, a coil at twice the current holds four times as much — an important sizing fact, since it means an inductor's usable rating is set by the current it can carry without saturating far more than by its inductance.

The release is what makes inductors interesting. Open a switch on an energised coil and the current cannot stop instantly, because stopping it instantly would require infinite $di/dt$; instead the collapsing field drives the terminal voltage to whatever value is needed to keep charge moving — through the air across the opening contacts, if that is the only path left. A car's ignition coil is this effect engineered on purpose, turning 12 V into a 30 kV spark by breaking a primary current. A switch-mode power supply does the same thing gently and tens of thousands of times a second, filling an inductor from the input and emptying it into the output, which is how a buck converter changes DC voltage with efficiency a resistor could never approach.

The mistakes here are mostly about symmetry that is not really there. A capacitor holds its energy at rest and will still bite you next week; an inductor's store exists only while the current flows, and cutting the current does not park the energy, it forces it out somewhere in the next few microseconds. Design for where it goes — a flyback diode, a snubber, a clamp — or the switch contacts and the transistor will volunteer. Second, $L$ in an iron-cored or ferrite-cored coil is not constant with current: as the core approaches saturation the inductance falls, sometimes by half, so $\tfrac{1}{2}LI^2$ computed with the datasheet's small-signal inductance overstates the energy a saturating part actually holds, and the current rises much faster than expected. Third, on AC the current to use is the peak, not the RMS, because it is the instantaneous current that sets the instantaneous stored energy. And do not read the stored energy as a loss — it is returned to the circuit, unlike $I^2R$ heat, which is gone.