Reactance Calculator (Capacitive & Inductive)
Work out the reactance of a capacitor (Xc = 1/2πfC) or an inductor (XL = 2πfL), or solve for the frequency or the part value. Add a parasitic to see where a real part resonates.
How to Use
- Choose Capacitive for a capacitor or Inductive for an inductor.
- Choose what to find: the reactance, the frequency, or the capacitance or inductance.
- Enter the other two values and pick a unit beside each one (pF to F, nH to H, Hz to GHz, mΩ to MΩ).
- The answer appears in the highlighted field and the first readout; change its unit to convert it.
- Optionally add the part’s parasitic (a capacitor’s series inductance or an inductor’s winding capacitance) to see its resonant frequency, then check Show Work.
Worked Example
A capacitor. 100 nF at 1 kHz: Xc = 1 / (2π × 1,000 Hz × 0.0000001 F) = 1,592 Ω (1.592 kΩ). The impedance is −j1.592 kΩ: the current leads the voltage by 90°, and 1 V across it drives 628.3 µA.
An inductor, solved for the part. A crossover inductor that matches an 8 Ω driver at 2 kHz: L = XL / (2πf) = 8 / (2π × 2,000) = 636.6 µH. At 4 kHz the same inductor is 16 Ω, so it passes less and less of the treble.
The common mistake: forgetting the part is not ideal. A 100 nF ceramic with 1 nH of series inductance (ESL) looks like 0.1592 Ω at 10 MHz by the ideal formula. But it resonates at 1 / (2π√(1 nH × 100 nF)) = 15.92 MHz, and at 10 MHz its net reactance is only 0.0963 Ω: it behaves like a 165.2 nF capacitor. Above 15.92 MHz it acts as an inductor.
Show Work
Formulas
Background
Reactance comes from energy storage. A capacitor stores energy in its electric field and an inductor in its magnetic field (Faraday’s induction, 1831), and each hands the energy back every half cycle. Because they give it back, an ideal capacitor or inductor dissipates no average power, and its current is a quarter-cycle (90°) away from its voltage. The word reactance was proposed by the French engineer Édouard Hospitalier in 1893 and adopted by the American Institute of Electrical Engineers in 1894.
In 1893 Charles Proteus Steinmetz showed engineers how to treat AC circuits with complex numbers, writing impedance as Z = R + jX. Reactance is the imaginary part: −1/(ωC) for a capacitor and +ωL for an inductor, where ω = 2πf. With that, AC circuits could be solved with the same algebra as DC circuits, which is still how filters, matching networks and power-factor corrections are worked out.
Real parts are not ideal. KYOCERA AVX’s Circuit Designer’s Notebook models a capacitor as its capacitance in series with a small inductance: as the frequency approaches the series resonance the part appears larger than its marked value, and beyond resonance it is inductive. Würth Elektronik’s ANP074 and Coilcraft’s document 671 describe the inductor’s mirror image: winding capacitance in parallel, a self-resonant frequency, and the advice to keep it about ten times above the working frequency.
About This Calculator
This page replaces the separate capacitive and inductive reactance calculators. For either part it finds the reactance, the frequency or the capacitance or inductance from the other two, writes the answer into the field you are solving for (in the unit you pick), and shows the complex impedance, ω and susceptance. The log-log curve marks your operating point across six decades, and an optional parasitic adds the real part’s resonance to the curve.
Formulas follow KYOCERA AVX’s Circuit Designer’s Notebook (Xc, series resonance, effective capacitance), Würth Elektronik ANP074 (XL, Q and self-resonance of an inductor) and ROHM TechWeb’s reactance and resonance articles. Everything runs in your browser.
Related calculators: LC Resonance & Q Factor Calculator, RC Filter Calculator, and RL Filter Calculator.
Frequently Asked Questions
What is the formula for capacitive reactance?
Xc = 1 / (2π × f × C), with f in hertz and C in farads. A 100 nF capacitor at 1 kHz has Xc = 1 / (2π × 1,000 × 0.0000001) = 1,592 Ω. At 10 kHz it falls to 159.2 Ω: ten times the frequency, one tenth of the reactance.
What is the formula for inductive reactance?
XL = 2π × f × L. A 10 mH inductor at 1 kHz has XL = 2π × 1,000 × 0.01 = 62.83 Ω, and the reactance rises in step with frequency, so at 2 kHz it is 125.7 Ω. That is why an inductor passes DC and blocks high frequencies, the opposite of a capacitor.
How is reactance different from resistance and impedance?
Reactance opposes AC like resistance does, but it stores energy and gives it back each cycle instead of turning it into heat, and it shifts the current by 90°. Impedance combines the two: Z = R + jX, |Z| = √(R² + X²). A 1 kΩ resistor in series with 100 nF at 1 kHz is √(1,000² + 1,592²) = 1,880 Ω, not 2,592 Ω.
What size capacitor do I need for a given reactance?
C = 1 / (2π × f × Xc). For 1 Ω at 100 MHz that is 1.592 nF. Above its series resonance a real capacitor stops acting like one: a 100 nF part with 1 nH of series inductance resonates at 15.92 MHz, and at 10 MHz it already behaves like 165 nF (KYOCERA AVX’s effective-capacitance formula).
At what frequency are Xc and XL equal?
At resonance, f = 1 / (2π√(LC)). A 10 mH inductor and a 100 nF capacitor both have 316.2 Ω of reactance at 5,033 Hz. Below that the capacitor dominates; above it the inductor does.
How do I use the Reactance Calculator (Capacitive & Inductive)?
Just type your numbers. The answer shows up right away — there is no button to press. Change anything and it updates by itself.
Does it cost anything or need an account?
No. The tool is completely free, there is no account to create, and it keeps working offline after the page first loads.
Is anything I type uploaded?
No. The tool works entirely on your device, so the values you enter never leave your browser.
Common Use Cases
Bypass capacitor
For no more than 1 Ω of reactance at 100 MHz, C = 1 / (2π × 100 MHz × 1 Ω) = 1.592 nF.
Audio coupling capacitor
Into a 10 kΩ input, keeping Xc to a tenth of that (1 kΩ) at 20 Hz needs 7.958 µF; a 10 µF part gives 795.8 Ω.
Mains-frequency capacitor
100 µF is 26.53 Ω at 60 Hz and 31.83 Ω at 50 Hz.
Switching-regulator inductor
A 22 µH inductor at 500 kHz has 69.12 Ω of reactance.
Speaker crossover
An inductor whose reactance equals an 8 Ω driver at 2 kHz is 8 / (2π × 2,000) = 636.6 µH.
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