Stored energy in an inductor, solved three ways. Enter any two of energy, inductance and current — the calculator finds the third, converts units, and shows what the same energy looks like in a capacitor.
E = ½ · L · I²
Result
—
volts
Worked examples
Given
Calculation
Result
L = 10 mH, I = 2 A
E = ½ × 0.01 × 2²
0.02 J = 20 mJ
E = 50 mJ, I = 1 A
L = 2 × 0.05 / 1²
0.1 H = 100 mH
E = 20 mJ, L = 10 mH
I = √(2 × 0.02 / 0.01)
√4 = 2 A
L = 47 µH, I = 300 mA
E = ½ × 47×10⁻⁶ × 0.3²
2.115 µJ = 0.002115 mJ
L = 1 H, I = 5 A
E = ½ × 1 × 5²
12.5 J = 0.00347222 Wh
Scope boundary: what this tool does not do
This calculator assumes inductance L is already known and computes the stored energy. It does not compute inductance from coil geometry (turns, diameter, length) — for that, use our solenoid inductance calculator (Wheeler and Nagaoka formulas). It also does not compute AC inductive reactance XL = 2πfL. For the capacitor counterpart E = ½CV², see our capacitor energy calculator.
Methodology
The energy stored in the magnetic field of an ideal inductor is E = ½LI². It follows from integrating power over time: with v = L·di/dt, the instantaneous power is p = v·i = L·i·di/dt, and ∫₀ᴵ L·i di = ½LI². The calculator evaluates this relation in SI units internally (henries, amperes, joules) and converts inputs and outputs through the selected unit prefixes. The capacitor-equivalent readout inverts E = ½CV² to C = 2E/V² so you can compare magnetic and electric energy storage at a chosen voltage. All computation runs locally in your browser; no data leaves the page.
Frequently asked questions
What is the formula for the energy stored in an inductor?
E = ½LI², where E is energy in joules, L is inductance in henries, and I is current in amperes. Example: L = 10 mH (0.01 H) and I = 2 A gives E = ½ × 0.01 × 4 = 0.02 J, which is 20 mJ. This calculator solves the formula in all three directions.
How do I find inductance from stored energy and current?
Rearrange the formula to L = 2E/I². Example: E = 50 mJ (0.05 J) with I = 1 A gives L = 2 × 0.05 / 1 = 0.1 H, which is 100 mH. Note the current must be non-zero, otherwise inductance is undefined.
How do I find current from stored energy and inductance?
Use I = √(2E/L). Example: E = 20 mJ (0.02 J) with L = 10 mH (0.01 H) gives I = √(2 × 0.02 / 0.01) = √4 = 2 A. Because current is squared in the energy formula, doubling the current quadruples the stored energy.
How much energy does a small inductor store?
Take a typical 47 µH inductor carrying 300 mA: E = ½ × 47×10⁻⁶ × 0.09 = 2.115×10⁻⁶ J, which is 2.115 µJ or 0.002115 mJ. Small-signal inductors store microjoules; energy scales with the square of current, so even a modest current rise adds up fast.
How much energy does a large power inductor store?
A 1 H inductor at 5 A stores E = ½ × 1 × 25 = 12.5 J. In watt-hours that is 12.5 / 3600 = 0.00347222 Wh (about 3.47 mWh). For scale, 1 joule equals 1/3600 watt-hour, or 277.778 µWh.
How does inductor energy compare to capacitor energy?
Capacitors store energy as E = ½CV². The same 0.02 J stored in our 10 mH / 2 A example could be held by a 1.6 mF capacitor charged to 5 V, since C = 2E/V² = 2 × 0.02 / 25 = 0.0016 F. This page shows a live capacitor-equivalent readout for any voltage you enter.
Why is the current squared in the inductor energy formula?
Energy is the integral of power over time: the inductor voltage is v = L·di/dt and power is v·i = L·i·di/dt, whose integral from 0 to I is ½LI². Practically: at 10 mH, raising the current from 2 A to 4 A quadruples the stored energy from 0.02 J to 0.08 J.
Which units can I use?
Energy: joules (J), millijoules (mJ), microjoules (µJ). Inductance: henries (H), millihenries (mH), microhenries (µH). Current: amperes (A), milliamperes (mA). Conversions: 1 mJ = 1000 µJ, 1 mH = 1000 µH, 1 A = 1000 mA — so 0.5 H is 500 mH and 0.447214 A is 447.214 mA.
Can this calculator solve for L or I instead of E?
Yes — switch tabs to 'Find L' or 'Find I'. Two examples: E = 1 J with I = 2 A needs L = 2×1/4 = 0.5 H (500 mH); E = 1 mJ with L = 10 mH gives I = √(2×0.001/0.01) = 447.214 mA. The solver refuses impossible cases: I = 0 when solving for L, and L ≤ 0 when solving for I.
Does this tool compute inductance from coil geometry?
No — it assumes inductance is already known and computes the stored energy. To compute inductance from physical dimensions (turns, diameter, length) use our solenoid-inductance-calculator, which implements the Wheeler and Nagaoka formulas. For the capacitor counterpart E = ½CV², see our capacitor-energy calculator.
We use cookies and third-party services (including Google AdSense) to personalize content, analyze traffic, and serve relevant ads. By clicking "Accept", you consent to our use of cookies as described in our Privacy Policy.
Report an issue
2 taps, no typing needed — diagnostics are sent automatically.