Inductor Stored Energy Calculator
Computes the energy stored in an inductor from inductance in henrys and current in amperes using half L I squared.
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Energy Stored in an Inductor
Whenever current runs through an inductor, energy collects in the magnetic field around its coil. You find the instantaneous energy with E = ½ L I². Here L is the inductance in henries (H), I is the current in amperes (A), and the answer comes out in joules (J). Take a 10 mH choke carrying 2 A: it holds 0.5 × 0.01 × 2² = 0.02 J, or 20 mJ.
A capacitor's energy tracks voltage squared. An inductor behaves differently, with its energy tied to the square of the current. Cut off an inductive current suddenly and you get a big voltage spike (v = L · di/dt), because the magnetic energy has nowhere to go but out. If you want to see where the formula comes from, integrate the instantaneous power p(t) = v(t)·i(t) over time. Boylestad's Introductory Circuit Analysis and Sedra & Smith's Microelectronic Circuits both work it through.
Applications
Storing energy this way is exactly how switch-mode power supplies work (buck, boost, flyback converters all rely on it). The same idea drives automotive ignition coils that step a 12 V battery up to tens of kilovolts, the capacitor-inductor pulse circuits inside defibrillators, and the superconducting magnets in MRI scanners, which hold megajoules. When that stored energy gets dumped abruptly, harmonic and surge problems can follow; the IEC 61000 family and IEEE 519 standards deal with those.
FAQ
Why does an inductor need a flyback diode? Open the switch feeding an inductor and the stored ½LI² has to go somewhere. A freewheeling diode gives the current a route to bleed off as heat, rather than arcing across the switch contacts.
Does core saturation affect the formula? It does. Once the core saturates, effective inductance falls off and the plain ½LI² ends up overestimating the stored energy. For non-linear cores, reach for the integral ∫ i·dλ instead.
How is this different from capacitor energy? A capacitor keeps ½ C V² in an electric field and fights changes in voltage. An inductor keeps ½ L I² in a magnetic field and fights changes in current.
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The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.