Example 1
f = 50 Hz, L = 0.1 H -> XL ≈ 31.4 Ω.
Inductive reactance calculator
Type f and L. The calculator computes XL = 2πfL. 50 Hz and 0.1 H is about 31.4 Ω. 60 Hz and the same L is about 37.7 Ω. f and L must be positive.
Capacitive reactance: Xc = 1/(2πfC). Resonance: f = 1/(2π√LC).
Result
Frequency f (Hz) and Inductance L (H). The result shows up here.
How it works
Inductive reactance is XL = 2π f L. At 50 Hz and 0.1 H you get about 31.4 Ω. At 60 Hz and the same L about 37.7 Ω. At 50 Hz and 0.01 H about 3.14 Ω. Higher f, larger XL. At 1000 Hz and 0.01 H about 62.8 Ω. At 1e6 Hz and 1e-5 H also about 62.8 Ω.
The form has two fields: f in hertz and L in henrys. Type 10 mH as 0.01, not as 10. The result is in ohms. A comma, a period, and 1e-2 are the same L.
f and L must be positive. Zero will not run. At f = 0 an ideal coil would have XL = 0 (DC), but the calculator needs f > 0. Both fields need a number before 31.4 Ω can appear.
Xc = 1/(2πfC) is on the neighbouring page and falls with f. Here XL rises with f. Resonance when XL = Xc is on f = 1/(2π√(LC)). Here the coil alone, no R.
Reactance XL at one f and one L. XL enters |Z| in AC. This calculator does not build a network. It takes one f and one L.
Type 50 and 0.1, click Calculate, and match about 31.4 Ω. Then jump to 60 Hz at the same L and see 37.7 Ω. The header symbol does not choke the coil.
How to use
- In the first field enter f in hertz, for example 50. US mains is 60.
- In the second field enter L in henrys, for example 0.1. 10 mH is 0.01, not 10.
- Click Calculate. The calculator computes 2πfL. 50 Hz and 0.1 H give about 31.4 Ω.
- f and L must be positive. Zero will not run. Both fields need a number.
- For Xc, open capacitive reactance. Resonance when XL = Xc is on f = 1/(2π√(LC)).
Formula
XL = 2πfL
f > 0, L > 0. Unit of XL: ohm (Ω).
XL, f, and L of the coil
Coil reactance rises with frequency: XL = 2πfL. At 50 Hz and 0.1 H you get about 31.4 Ω. At 60 Hz and the same 0.1 H about 37.7 Ω.
- XL
- Inductive reactance in ohms. 2π × 50 × 0.1 ≈ 31.4 Ω. Not Ohm resistance R.
- f
- Frequency in hertz. Mains 50 Hz. Higher f, larger XL. f > 0.
- L
- Inductance in henrys. 0.1 H. Type 10 mH as 0.01. L > 0.
- π
- The constant in 2πfL. At 50 Hz and 0.1 H the factor 2π·50 = 100π ≈ 314.
Real-life examples
Example 2
f = 60 Hz, L = 0.1 H -> XL ≈ 37.7 Ω.
Example 3
f = 50 Hz, L = 0.01 H -> XL ≈ 3.14 Ω.
Example 4
f = 50 Hz, L = 1 H -> XL ≈ 314 Ω.
Example 5
f = 1000 Hz, L = 0.01 H -> XL ≈ 62.8 Ω.
Example 6
f = 20 Hz, L = 0.1 H -> XL ≈ 12.6 Ω.
Example 7
f = 1e6 Hz, L = 1e-5 H -> XL ≈ 62.8 Ω.
Example 8
f = 50 Hz, L = 0.5 H -> XL ≈ 157 Ω.
Ways to use this calculator
- You compute about 31.4 Ω from 50 Hz and 0.1 H.
- You keep L and jump from 50 to 60 Hz: about 37.7 Ω.
Frequently asked questions
How much XL at 50 Hz and 0.1 H?
XL = 2π × 50 × 0.1 ≈ 31.4 Ω. At 60 Hz and 0.1 H about 37.7 Ω.
Which units do I type?
f in hertz, L in henrys. 10 mH is 0.01 H. Result in ohms.
What if L = 0?
Zero will not run. f and L must be positive. Both fields need a number.
How is XL different from Xc?
XL = 2πfL rises with f. Xc = 1/(2πfC) falls with f. Another card.
When is XL = Xc?
At LC resonance. Frequency is on the f = 1/(2π√(LC)) page.
How much at 1000 Hz and 0.01 H?
XL ≈ 62.8 Ω. At 1e6 Hz and 1e-5 H also about 62.8 Ω.
Does a comma in 0.1 H work?
Yes. 0,1 and 0.1 are the same L. 1e-2 too. A comma and a period mean the same value.
Is this Ohm resistance R?
No. XL is reactance. In AC it enters |Z|. It does not replace R on the Ohm page.
What about f = 0?
DC: an ideal coil has XL = 0. The calculator needs f > 0.
How much at 50 Hz and 0.01 H?
XL ≈ 3.14 Ω. Ten times smaller L, ten times smaller XL.
Knowledge sources
The formula is the school one. Units follow SI; NIST SP 330 and BIPM define the measures, not your result.
Page updated in 2026.