When engineers and hobbyists search for the power of inductor formula, they are usually looking for one of two entirely different equations. In transient DC circuits (like switching power supplies), we calculate instantaneous power to determine energy transfer rates. In steady-state AC circuits (like motor starters or power factor correction banks), we calculate reactive power to determine the magnetic field's energy exchange with the grid. Confusing the two is a primary cause of blown MOSFETs and undersized AC chokes.
This guide breaks down both formulas, defines every variable, tracks units through real-world worked examples, and highlights the specific assumptions that dictate when these equations hold true.
The Core Power of Inductor Formulas Defined
An inductor opposes changes in current by storing energy in a magnetic field. The mathematical expression for power depends on whether you are analyzing a dynamic transient event or a continuous AC cycle.
1. Instantaneous Power (Transient & DC Switching)
The instantaneous power p(t) absorbed or delivered by an inductor at any exact microsecond is the product of the voltage across it and the current through it. Since v(t) = L(di/dt), we substitute to get the fundamental transient formula:
p(t) = L × i(t) × (di/dt)
2. Reactive Power (Steady-State AC)
In an AC circuit, an ideal inductor stores energy during one quarter-cycle and returns it to the source during the next. The average real power consumed is zero. However, the rate of this energy exchange is measured as reactive power Q:
Q = Irms2 × XL = Irms2 × (2πfL)
Symbol Definition Table
| Symbol | Name | SI Unit | Typical Practical Range |
|---|---|---|---|
| p(t) | Instantaneous Power | Watts (W) | 0.1 W to 500 W (switching supplies) |
| Q | Reactive Power | Volt-Amps Reactive (VAR) | 10 VAR to 500 kVAR (grid/motors) |
| L | Inductance | Henrys (H) | 1 μH to 10 H |
| i(t) | Instantaneous Current | Amperes (A) | mA to 100+ A |
| Irms | RMS Current | Amperes (A) | 0.1 A to 1000+ A |
| di/dt | Current Slew Rate | Amperes per second (A/s) | 105 to 109 A/s |
| XL | Inductive Reactance | Ohms (Ω) | 0.1 Ω to 10 kΩ |
| f | Frequency | Hertz (Hz) | 50/60 Hz (mains) to MHz (RF) |
Rearranged Forms
Depending on your design constraints, you will frequently need to isolate variables. Here are the algebraic rearrangements for both domains:
- Solving for Inductance (Transient): L = p(t) / [i(t) × (di/dt)]
- Solving for Slew Rate: di/dt = p(t) / [L × i(t)]
- Solving for Inductance (AC Reactive): L = Q / [Irms2 × 2πf]
- Solving for RMS Current Limit: Irms = √[Q / (2πfL)]
- Solving for Frequency: f = Q / [Irms2 × 2πL]
Assumptions, Realistic Magnitudes, and Fatal Unit Mistakes
When the Formulas Apply (and Their Assumptions)
Both equations assume an ideal, linear inductor. This means the component has zero DC resistance (DCR), zero parasitic capacitance, and a magnetic core that does not saturate. In reality, every inductor has winding resistance that dissipates real power as heat (P = I2R), which the reactive formula completely ignores. Furthermore, the transient formula assumes the inductance L remains constant; if the core saturates, L drops precipitously, causing di/dt to spike and rendering the linear formula invalid.
Realistic Answer Magnitudes
If your calculation yields an answer outside these typical bounds, re-check your math:
- Signal/RF Inductors: Instantaneous power is typically in the microwatts (μW); reactive power is in the milliVAR (mVAR) range.
- DC-DC Converter Power Inductors: Instantaneous power during the switch-on phase ranges from 5 W to 200 W.
- AC Mains Chokes / PFC Banks: Reactive power is measured in kiloVAR (kVAR). A 50 kVAR capacitor bank requires an inductor (reactor) of similar magnitude to tune or limit inrush.
Fatal Unit Mistakes That Break the Math
⚠ The milliHenry & Microsecond Trap: The most common error in the transient formula is mixing prefixes. If L is in μH (10-6) and time is in μs (10-6), the di/dt term must be calculated in raw Amperes per second, not Amperes per microsecond. Failing to convert the time denominator to seconds inflates the power result by a factor of one million. Always convert all inputs to base SI units (Henrys, Amperes, Seconds) before multiplying.
⚠ Peak vs. RMS in AC: The reactive power formula Q = I2XL strictly requires Irms. If your oscilloscope reads a peak-to-peak AC current, you must divide by 2√2 (approx 2.828) to find Irms before squaring it. Squaring peak current will overstate your reactive power by a factor of 2.
Worked Example 1: Instantaneous Power in a Buck Converter
Scenario: You are designing a 12V-to-3.3V buck converter. During the MOSFET's 'on' time, the current through the main power inductor ramps linearly. You need to know the peak instantaneous power transferred to the inductor's magnetic field at the exact moment the switch turns off.
Given Parameters:
- Inductance (L) = 4.7 μH = 4.7 × 10-6 H
- Current ramps from 1.5 A to 5.5 A
- Ramp time (Δt) = 2.2 μs = 2.2 × 10-6 s
Step 1: Calculate the current slew rate (di/dt)
di/dt = Δi / Δt
di/dt = (5.5 A - 1.5 A) / (2.2 × 10-6 s)
di/dt = 4.0 A / 0.0000022 s = 1,818,181.8 A/s
Step 2: Identify the peak instantaneous current i(t)
The peak power occurs at the peak current, just before the switch turns off.
i(t) = 5.5 A
Step 3: Apply the instantaneous power formula
p(t) = L × i(t) × (di/dt)
p(t) = (4.7 × 10-6 H) × (5.5 A) × (1,818,181.8 A/s)
p(t) = 0.0000047 × 5.5 × 1818181.8
p(t) = 47.0 W
Verification & Unit Tracking:
[H] × [A] × [A/s] = [V·s/A] × [A] × [A/s] = [V] × [A] = [Watts]. The magnitude (47 W) is highly realistic for the transient energy storage phase of a 15A-class buck converter inductor.
Worked Example 2: Reactive Power in an AC Line Reactor
Scenario: You are installing a 3-phase AC motor drive and need to size a line reactor (inductor) to limit harmonic distortion. You need to calculate the reactive power the inductor will introduce into the system to ensure it doesn't excessively drop the voltage or ruin the power factor.
Given Parameters:
- Inductance (L) = 2.5 mH = 0.0025 H
- RMS Current (Irms) = 18 A
- Line Frequency (f) = 60 Hz
Step 1: Calculate Inductive Reactance (XL)
XL = 2πfL
XL = 2 × 3.14159 × 60 Hz × 0.0025 H
XL = 376.99 × 0.0025 = 0.942 Ω
Step 2: Apply the reactive power formula
Q = Irms2 × XL
Q = (18 A)2 × 0.942 Ω
Q = 324 A2 × 0.942 Ω
Q = 305.2 VAR
Verification & Unit Tracking:
[A]2 × [Ω] = [A]2 × [V/A] = [V·A] = [VAR]. A reactive power of ~305 VAR per phase (approx 915 VAR total for 3-phase) is standard for a 10 HP motor drive line reactor. For deeper reading on AC power triangles, refer to the All About Circuits guide on Reactive Power.
Frequently Asked Questions
Why does the power of inductor formula yield zero average real power in AC?
In a purely inductive AC circuit, the voltage waveform leads the current waveform by exactly 90 degrees. When you multiply the instantaneous voltage and current over a full 360-degree cycle, the positive power (energy stored in the magnetic field) perfectly cancels out the negative power (energy returned to the source). Therefore, the integral of p(t) over time is zero. The inductor consumes no net real power (Watts); it only exchanges reactive power (VAR). For a visual breakdown of this phase relationship, Electronics Tutorials provides excellent waveform diagrams.
How do I calculate inductor power loss if the formula assumes an ideal component?
The formulas above calculate energy storage and exchange, not heat dissipation. To find the real power lost as heat, you must account for two non-ideal factors: DCR (DC Resistance) and Core Losses. Calculate copper loss using Pcu = Irms2 × DCR. Core loss (hysteresis and eddy currents) is usually derived from the manufacturer's datasheet graphs (e.g., Steinmetz equation parameters) based on the specific core material (like Ferrite or powdered iron) and the operating frequency. Total real power dissipation is the sum of these two losses.
Does the power of inductor formula apply when the magnetic core saturates?
No. The formula p(t) = L × i(t) × (di/dt) assumes L is a constant value. When an inductor's core reaches magnetic saturation, the permeability of the core drops toward that of free air, causing the effective inductance L to collapse dramatically. If you attempt to use the nominal inductance value in the formula while the core is saturated, your calculated power and slew rate will be wildly incorrect. In saturation, di/dt spikes uncontrollably, which is why switching regulators use peak-current-mode control to turn off the MOSFET before saturation occurs.
