Formula for Reactance of Inductor: Solved Examples & Reference

The formula for reactance of inductor is X_L = 2πfL. It calculates the opposition an inductor presents to alternating current (AC) at a specific frequency, measured in ohms (Ω). Unlike DC resistance, which is fixed by the wire's physical dimensions, inductive reactance scales linearly with both the frequency of the AC signal and the inductance value of the component.

Below is the complete breakdown of the equation, a real-world reference matrix for common component values, and step-by-step solved problems with strict unit tracking to prevent calculation errors on the bench.

The Core Equation, Symbol Definitions, and Rearranged Forms

The fundamental relationship governing inductive reactance is expressed as:

X_L = 2 × π × f × L

Every variable in this equation must be converted to its base SI unit before multiplying. Plugging in prefixes like 'milli' or 'micro' directly without converting to scientific notation is the primary cause of calculation failures in circuit design.

Rearranged Forms for Design Work

When designing filters or audio crossovers, you rarely solve for X_L directly. Instead, you know the target reactance and frequency, and need to select the component. Use these algebraic rearrangements:

• To find required Inductance (L): L = XL / (2 × π × f)
• To find Cutoff Frequency (f): f = XL / (2 × π × L)

Real-World Inductive Reactance Reference Matrix

What does a realistic answer magnitude look like? It depends entirely on the application domain. RF chokes operating at low frequencies yield milliohm reactances, audio crossover inductors yield single-digit ohms, and mains filtering chokes yield tens to hundreds of ohms.

The table below maps standard off-the-shelf inductor values (like those found in the Bourns 2100 series or Würth WE-PD lines) across three common frequency domains. Use this as a sanity check for your own calculations.

Note: At 60 Hz, a 10 μH inductor presents only 3.8 milliohms of reactance. In a 120V AC mains circuit, this is effectively a dead short. This matrix proves why high inductance values (10 mH+) are mandatory for 50/60 Hz line filtering, while microhenry values are reserved for high-frequency switching converters.

Step-by-Step Worked Examples with Unit Tracking

Abstract formulas fail on the workbench without rigorous unit tracking. Below are two common scenarios solved with explicit intermediate steps.

Example 1: Sizing a Ripple Choke for a Rectified Mains Supply

Scenario: You are designing a linear power supply. The full-wave rectified 60 Hz AC mains produces a 120 Hz ripple frequency. You need to drop at least 15 Ω of reactance at this ripple frequency to adequately filter the DC output. What inductance value is required?

Step 1: Identify knowns and target.

• Target Reactance (X_L) = 15 Ω
• Frequency (f) = 120 Hz
• Unknown: Inductance (L) in Henrys

Step 2: Select the rearranged formula.

L = X_L / (2 × π × f)

Step 3: Substitute and solve.

• L = 15 / (2 × 3.14159 × 120)
• L = 15 / 753.98
• L = 0.01989 Henrys

Step 4: Convert to standard component units.

0.01989 H × 1000 = 19.89 mH. You would select a standard 22 mH power choke (e.g., a Bourns 2100-22K) to ensure the reactance exceeds the 15 Ω minimum threshold.

Example 2: Calculating Reactance of an RF Choke at Switching Frequencies

Scenario: You are debugging a buck converter switching at 400 kHz. The schematic calls for a 4.7 μH inductor. What is the actual inductive reactance at the switching node?

Step 1: Convert all units to base SI.

• L = 4.7 μH = 4.7 × 10^-6 H (or 0.0000047 H)
• f = 400 kHz = 400 × 10³ Hz (or 400,000 Hz)

Step 2: Apply the primary formula.

• X_L = 2 × π × f × L
• X_L = 2 × 3.14159 × 400,000 × 0.0000047
• X_L = 2,513,272 × 0.0000047
• X_L = 11.81 Ω

At 400 kHz, this tiny 4.7 μH surface-mount inductor presents nearly 12 ohms of reactance, which is sufficient to shape the current ramp in a typical 5A switching regulator.

Assumptions, Parasitics, and Fatal Unit Mistakes

The equation X_L = 2πfL is an idealization. According to standard AC circuit theory documented by resources like Electronics Tutorials, this formula only holds true under specific conditions. Ignoring these conditions leads to catastrophic circuit failures, particularly in high-frequency or high-power designs.

When the Formula Applies (and When It Breaks)

1. Sinusoidal Steady-State: The formula assumes a pure, continuous sine wave. For square waves (like those from a 555 timer or microcontroller PWM), you must calculate the reactance for the fundamental frequency and each odd harmonic individually, as square waves contain infinite odd-integer frequency components.
2. Below Self-Resonant Frequency (SRF): Real inductors possess parasitic parallel capacitance between their wire windings. As frequency increases, this capacitance creates a parallel LC tank circuit. At the SRF, the inductor's impedance peaks. Above the SRF, the component acts as a capacitor, and the formula X_L = 2πfL becomes entirely invalid. Always check the manufacturer datasheet (e.g., Würth Elektronik or Coilcraft) to ensure your operating frequency is at least 10x to 100x below the SRF.
3. Ignoring DC Resistance (DCR): Inductive reactance (X_L) is only the imaginary part of total impedance (Z). The total opposition to current is Z = √(DCR² + X_L²). At very low frequencies where X_L approaches zero, the physical copper resistance (DCR) dominates the circuit behavior.

The 'Prefix Trap': Unit Mistakes That Break Calculations

The most common error when using the formula for reactance of inductor is failing to convert prefixes to base scientific notation.

For deeper analysis on how core saturation and parasitic elements alter ideal inductor behavior in practical power supply designs, refer to application notes like the Texas Instruments Inductor Basics guide (SNVA038). Always verify your theoretical X_L calculations against the specific component's impedance vs. frequency graph in the datasheet before finalizing a PCB layout.