Library / Circuits, electronics and noise / ELN-10
Circuits, electronics and noise
Magnetic energy stored in an inductor
Energy held in the magnetic field of a single linear inductor carrying current I.
- Ref
- ELN-10
- Rev
- v1
- Field
- Circuits, electronics and noise
- Result
- W [J]
A linear inductor carrying current I stores W = ½LI² in its magnetic field. Doubling the current quadruples the stored energy. The formula holds while the core stays linear, so a saturated core stores less energy than this. The stored energy is the energy a flyback switch must dissipate or return when the current is interrupted.
Energy held in the magnetic field of a single linear inductor carrying current I. It rises with the square of the current, and a saturated core stores less than this.
Calculator
- Energy if the current doubles0.36 J
- Flux linkage, L I0.06 Wb
- Current that would store 1 J, √(2/L)10 A
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A 0.02 H inductor carrying 3 A stores 0.09 J; at twice the current it would store 0.36 J.
- Whatever opens the circuit must absorb this energy: the stored 0.09 J is why a flyback diode or snubber is needed across a relay coil.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J) | Local step | Change in result |
|---|---|---|---|---|---|---|
| I Current | 0.001 to 100 A | 1e-8 | 100 | 100 | +10% | +21 % |
| L Inductance | 1e-6 to 10 H | 4.5e-6 | 45 | 45 | +10% | +10 % |
Sweep of Current
| Current (A) | Stored energy (J) |
|---|---|
| 0.001 | 1e-8 |
| 0.004217 | 1.778e-7 |
| 0.01778 | 3.162e-6 |
| 0.07499 | 5.623e-5 |
| 0.3162 | 0.001 |
| 1.334 | 0.01778 |
| 5.623 | 0.3162 |
| 23.71 | 5.623 |
| 100 | 100 |
Sweep one variable
Variables and how each one contributes
Stored energy
What the equation gives you. Everything else feeds it.
Inductance
The inductance, in H. A larger inductance stores more energy at the same current, in direct proportion. Raising it by 10 % (to 0.022) moves the result by +10 %, from 0.09 to 0.099 J.
Current
The current through the inductor, in A. Raises the energy with the square of the current. Raising it by 10 % (to 3.3) moves the result by +21 %, from 0.09 to 0.1089 J.
Worked example with the values above
- Square the current: I² = 3² = 9 A².
- Multiply by L: 0.02 × 9 = 0.18.
- W = ½ × 0.18 = 0.09 J.
Assumptions and limits
- Ideal linear capacitors and inductors with constant values.
- Leakage current, series resistance and saturation are neglected.
Common mistakes
- Forgetting to square the current.
- Using this when the core has saturated.
- Forgetting the factor of one half.
Statistics and uncertainty
The stored energy is quadratic in the current, so a 1 % error in the current gives about 2 % in the energy. Inductance falls near saturation, so the real energy is lower at high current.
Where it is used
- Timing, smoothing and switched-capacitor circuits.
- Energy storage in inductors and capacitors for converters and pulse circuits.
Maths you need
- Capacitance: C = Q/V
- Charge conservation: charge on isolated nodes does not change
- Energy in a capacitor or inductor: ½CV² and ½LI²