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Inductor Energy Calculator

Find the energy stored in an inductor.

How it is calculated

The energy held in the magnetic field of an inductor is E = ½ · L · I². L is the inductance in henries (H) and I is the current in amperes (A); the result is in joules (J), shown with six decimals so small values stay visible. Inductance must be greater than zero. Because the current is squared, the sign of the current does not matter, and doubling the current quadruples the energy. The formula assumes an ideal inductor whose inductance stays constant; a core that saturates at high current will store less than predicted.

Worked example

An inductor of 10 mH carries 2 A. Convert the inductance: 10 mH = 0.01 H. Square the current: 2² = 4. Then E = ½ × 0.01 × 4 = 0.02. The stored energy is 0.020000 J.

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Common mistakes

  • Typing millihenries or microhenries as if they were henries. 10 mH must be entered as 0.01, otherwise the answer is a thousand times too large.
  • Forgetting to square the current. Doubling I does not double the energy, it makes it four times larger.
  • Entering the current in milliamps without converting to amperes.
  • Expecting the stored energy to be available instantly. It is stored in the field, and an abrupt interruption of current produces a voltage spike instead of a quiet release.

Sources

About this calculator

This calculator finds the magnetic energy stored in an inductor using E = ½ · L · I², where L is inductance in henries and I is the current flowing through the coil in amperes. That energy is stored in the magnetic field generated by the current, analogous to how a capacitor stores energy in an electric field via E = ½CV².

This relationship governs how much energy a choke, transformer winding, or solenoid can hold and release, which is critical when sizing inductors for switch-mode power supplies, flyback converters, and buck/boost regulators — engineers need to know the peak energy an inductor must handle without saturating its core.

Electrical and electronics engineers designing power supply circuits, hobbyists building DC-DC converters, and students in electromagnetism courses working through inductor problems use this formula.

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Comments (1)

  • Editorial team

    In a flyback or boost design, the number here is the energy the inductor handles in each switching cycle, so multiplying it by the switching frequency gives a rough ceiling on the power it can pass. For example, 100 µH at 5 A is 0.001250 J, and at 100 kHz that bounds the throughput near 125 W before any losses. Use the peak current, not the average. Which converter topology are you sizing an inductor for?

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