Inductor Energy: E = ½LI²
Energy is stored in the magnetic field around the inductor while current flows. Unlike a capacitor (which stores energy in an electric field), an inductor’s energy depends on current squared. Interrupting the current forces the energy out as a voltage spike.
I — Current through the inductor. Energy scales with I² — doubling the current quadruples the energy.
SW — Switch. Opening it interrupts the current and forces the stored energy out as a voltage spike (back-EMF).
V — Supply voltage driving current through the inductor.
Inductor Energy Calculator
A triangle solver for E = ½LI². Enter any two of inductance, current, or energy — the calculator solves for the third. Work forwards (“how much energy is stored in this inductor?”) or backwards (“what inductance stores 1 mJ at 2 A?”). An inductor stores energy in its magnetic field, proportional to current squared. That energy must go somewhere when the current is interrupted — if there is no designed path, it creates a destructive voltage spike.
The Three Formulas
L = 2E / I² — find inductance from energy and current
I = √(2E / L) — find current from energy and inductance
Leave one field empty. The calculator fills it. Energy scales with I² — doubling the current quadruples the stored energy, just as doubling the voltage quadruples capacitor energy.
SMPS Power Inductor (47 µH, 3 A)
Wh = 0.000212 / 3600 = 58.8 nWh
0.212 mJ — this energy cycles in and out of the magnetic field at every switching period (typically 100 kHz–1 MHz). At 500 kHz, the inductor absorbs and releases 0.212 mJ every 2 µs. The MOSFET, diode, and capacitor in the buck converter manage this energy transfer continuously. Green safety status — standard circuit protection is sufficient.
Relay Coil (100 mH, 50 mA)
0.125 mJ. Seems tiny, but when the transistor driving the relay switches off, this energy must discharge. The inductor tries to maintain current flow by spiking the voltage across the switch. Without a flyback diode, a 100 mH coil at 50 mA can produce a voltage spike of hundreds of volts — enough to destroy a MOSFET or BJT. A simple diode across the coil clamps the spike to ~0.7 V and dissipates the 0.125 mJ as heat. For dedicated inductive load protection design, see the Inductive Load Calculator.
Motor Winding (10 mH, 20 A)
Wh = 2.0 / 3600 = 0.556 mWh
2 joules. Red safety status — significant energy that produces dangerous arc voltages when a contactor opens. Industrial motor circuits use arc suppressors, snubber networks (RC across the contactor), or MOV (metal oxide varistor) clamps to absorb this energy safely.
Large Power Reactor (1 H, 100 A)
Wh = 5000 / 3600 = 1.39 Wh
5 kJ — equivalent to a small capacitor bank. Power line reactors, large electromagnets, and superconducting coils store serious energy. Interrupting 100 A through 1 H without protection generates extreme voltages. These systems use dedicated energy dump circuits, crowbar diodes, or resistor banks to safely dissipate the stored energy during shutdown.
Reverse Solve — Find Inductance
“I Need 1 mJ of Energy at 2 A”
500 µH at 2 A stores exactly 1 mJ. This reverse solve is useful for energy-harvesting circuits and magnetic energy storage sizing — start with the energy budget and find the inductance.
“I Need 10 J of Energy at 5 A”
800 mH at 5 A. A large iron-core inductor or transformer winding. Red safety status — 10 J at 5 A requires proper arc suppression.
Reverse Solve — Find Current
“A 47 µH Inductor Stores 0.5 mJ — What Current?”
4.61 A. This tells you the peak current the inductor reaches if it stores 0.5 mJ — useful for verifying that the inductor’s saturation current rating is not exceeded. For detailed current analysis in inductor circuits, see the Inductor Current Calculator.
Why Current Squared Matters
47 µH at 2 A → 94 µJ (4× energy)
47 µH at 5 A → 587.5 µJ (25× energy)
47 µH at 10 A → 2.35 mJ (100× energy)
Increasing current from 1 A to 10 A multiplies stored energy by 100×. This is why high-current inductors need much more robust flyback protection than low-current ones — the energy that must be managed grows quadratically.
Inductor vs Capacitor Energy — The Duality
Capacitor: E = ½CV² — energy stored in electric field, proportional to voltage squared
An inductor resists changes in current (produces voltage spikes when interrupted).
A capacitor resists changes in voltage (produces current surges when shorted).
Both are energy storage elements. The inductor stores magnetically at a rate determined by current; the capacitor stores electrically at a rate determined by voltage. In switching power supplies, energy shuttles between the two every switching cycle. For the capacitor side of this duality, see the Capacitor Energy Calculator.
Safety Assessment
Yellow — energy above 100 mJ OR current above 5 A. A flyback diode or snubber should be used to absorb the energy when switching off.
Red — energy above 1 J AND current above 1 A. Significant stored energy that produces dangerous voltage spikes and arcing. Use contactor arc suppressors, MOVs, or dedicated energy dump circuits.
The Back-EMF Problem
An inductor maintains current flow. When a switch opens, the inductor drives the voltage at the switch as high as needed to keep current flowing — potentially thousands of volts across a gap that was previously at supply voltage. This back-EMF (electromotive force) is the inductor releasing its stored energy. The energy is fixed (E = ½LI²); the voltage spike is limited only by whatever path exists for the current.
Protection strategies absorb the energy in a controlled way: a flyback diode clamps the voltage to ~0.7 V above supply and dissipates the energy as heat. An RC snubber limits the dV/dt and absorbs the energy in the resistor. A TVS (transient voltage suppressor) or MOV clamps to a defined voltage. The calculator tells you how much energy the protection circuit must handle. The underlying V = IR relationship behind these protection circuits is covered by Ohm’s Law.
Frequently Asked Questions
Last updated: March 2026