Inductive Reactance (X_L = 2πfL)

XL=2πfLX_L = 2\pi f L
sig figs

Worked example: 10 mH at 60 Hz → X_L = 3.76991 ohm. Press Try an example to run it live, then adjust anything.

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Capacitive Reactance (X_C = 1/2πfC): Both use ac frequency.Series RLC Impedance: Both use inductive reactance.Energy Stored in an Inductor: Both use inductance.LC Resonant Frequency: Both use inductance.Mutual Inductance of Coupled Coils: Both use inductance.RL Cutoff Frequency: Both use inductance.
4 more relatedRL Time Constant (τ = L/R): Both use inductance.Two Inductors in Parallel: Both use inductance.Two Inductors in Series: Both use inductance.Q Factor of a Series Resonant Circuit: Both use inductance.
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Inductive Reactance (X_L = 2πfL) explained

LXLf

An inductor opposes change in current, not current itself, so how hard it opposes depends on how fast you are asking the current to change. On a sine wave of frequency ff the current has to reverse 2f2f times a second, and the steepness of that reversal rises in proportion to the frequency, hence XL=2πfLX_L = 2\pi f L, climbing linearly. The 2π2\pi is there because a full cycle is 2π2\pi radians, and the group 2πf2\pi f is angular frequency ω\omega, which is why the relation is more compactly written XL=ωLX_L = \omega L. At DC, where nothing changes, the reactance is zero and the coil is just its own winding resistance.

A 10 mH coil offers 2π×60×0.01=3.8 Ω2\pi \times 60 \times 0.01 = 3.8\ \Omega at 60 Hz mains, 63 Ω at 1 kHz, and 628 Ω at 10 kHz. That strong frequency preference makes an inductor a natural low-pass element: put it in series with a load and the low frequencies get through while the high ones are held back. A loudspeaker crossover does exactly this, an inductor in series with the woofer passing bass and blocking treble, while a capacitor does the complementary job for the tweeter. The same principle is at work in a mains filter choke and in the smoothing inductor of a power supply.

Reactance is what an inductor contributes to the circuit's impedance, Z=R+jXLZ = R + jX_L, where the jj records that the voltage across an inductor leads its current by 90°. This quarter-cycle lag is the physical heart of the matter: current lags voltage in a coil because the coil resists getting started, and it is the reason an inductive motor load drags a plant's power factor down and needs correcting with capacitors. On the site's LC resonance page the inductive and capacitive reactances become equal and, because they are opposite in sign, cancel: the whole subject of tuning is contained in that cancellation.

Reactance is measured in ohms, and that is the single most misleading fact about it. It is not resistance and it does not behave like resistance in three important ways. First, it dissipates no power: the energy goes into the magnetic field on one quarter-cycle and comes back out on the next, so a pure reactance heats nothing, and a wattmeter across it reads zero. Second, it does not add arithmetically to resistance. A coil with 30 Ω of winding resistance and 40 Ω of reactance presents 302+402=50 Ω\sqrt{30^2 + 40^2} = 50\ \Omega, never 70, because the two are 90° apart and combine in quadrature. Third, it is frequency-specific, so a single number is meaningless without the frequency it was computed at, and on a supply carrying harmonics, the fifth harmonic sees five times the reactance the fundamental does, which is why harmonic currents can produce voltage distortion far out of proportion to their size. Finally, this page gives the reactance alone; to find the current from a supply voltage you need the full impedance, resistance included.

Inductive Reactance (X_L = 2πfL) formula

XL=2πfLX_L = 2\pi f L
Where
  • XLX_L= Inductive reactance (Ω)
  • ff= Frequency (Hz)
  • LL= Inductance (mH)
Revised 16 October 2026
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formula.expert. (2026, October 16). Inductive Reactance (X_L = 2πfL) Formula & Calculator. https://formula.expert/solvers/inductive-reactance