Inductor Energy Calculator

How it works

Current through an inductor builds a magnetic field, and building that field takes energy. Because the inductor opposes any change in current with a voltage L × dI/dt, the work done to raise the current from zero to I adds up to E = ½ × L × I², where L is the inductance in henries, I the current in amps and E the energy in joules.

The energy depends on the square of the current, so doubling the current stores four times the energy, and the direction of the current makes no difference. Rearranged, the current needed to hold a given energy is I = √(2E ÷ L).

Results are shown with the SI prefix that keeps the number readable, such as mJ or mA.

Formula

E = ½ × L × I²       (joules)
I = √(2 × E ÷ L)     (amps, for a target energy)
1 mH = 10^-3 H;  1 µH = 10^-6 H

Example

A 10 mH inductor (0.01 H) carrying 2 A stores E = ½ × 0.01 × 2² = 0.02 J, shown as 20 mJ.

To store 1 J in the same 10 mH inductor takes I = √(2 × 1 ÷ 0.01) = √200 = 14.1421 A.

Assumptions and limitations

• The inductance is constant. Iron- and ferrite-cored inductors saturate above their rated current, after which the inductance falls and E = ½ × L × I² overstates the stored energy; the rated or saturation current is given in the manufacturer's data sheet.
• The winding's resistance is ignored. It does not change the stored energy, but it dissipates power continuously while current flows.
• The stored energy is released when the current is interrupted, which can produce a large voltage spike across the switch. The calculator does not assess that hazard.
• Results are estimates for learning and circuit sketching, not a substitute for the manufacturer's data sheet or a qualified engineer's design review.

Frequently asked questions