Free Capacitive Reactance Calculator
Find the capacitive reactance of a capacitor at a given frequency using Xc = 1/(2π·f·C). Enter the signal frequency in hertz and the capacitance in microfarads (µF), and the calculator returns the reactance in ohms, in kilohms, and the angular frequency ω = 2πf.
Enter the signal frequency in hertz and the capacitance in microfarads (µF) to find the capacitive reactance Xc = 1/(2π·f·C). Frequency and capacitance must both be greater than zero.
Xc = 1/(2π·f·C). Capacitance is converted from µF to farads (µF × 10⁻⁶). Reactance falls as frequency or capacitance rises — a capacitor blocks DC (Xc → ∞ as f → 0) and passes high frequencies.
Quick answer
Capacitive reactance is Xc = 1/(2π·f·C), where f is frequency in hertz and C is capacitance in farads; the result is in ohms (Ω). For a 10 µF capacitor at 60 Hz, Xc = 1/(2π × 60 × 10×10⁻⁶) ≈ 265.26 Ω. Reactance is inversely proportional to both frequency and capacitance, so it falls as either rises and rises toward infinity as frequency approaches zero (DC).
Formula & method
Xc = 1 / (2 · π · f · C)
- Xc — Capacitive reactance, in ohms (Ω)
- f — Signal frequency, in hertz (Hz)
- C — Capacitance, in farads (F); µF × 10⁻⁶
- π — Pi, approximately 3.14159
Capacitive reactance in ohms. C must be in farads (multiply µF by 10⁻⁶) and f in hertz. The 2πf term is the angular frequency ω, so equivalently Xc = 1/(ω·C).
ω = 2 · π · f so Xc = 1 / (ω · C)
Angular frequency ω is in radians per second. Reactance is the magnitude of the capacitor's opposition to AC; the current actually leads the voltage by 90°, so the impedance is Zc = −j·Xc.
Examples
- Input
- f = 60 Hz, C = 10 µF
- Result
- Xc ≈ 265.258 Ω
- Why
- Convert capacitance: 10 µF = 10 × 10⁻⁶ = 0.00001 F. Then Xc = 1 / (2π × 60 × 0.00001) = 1 / (2 × 3.14159 × 60 × 0.00001) = 1 / 0.00376991 ≈ 265.258 Ω. This is the default case shown when the calculator loads.
- Input
- f = 1000 Hz, C = 1 µF
- Result
- Xc ≈ 159.155 Ω
- Why
- 1 µF = 1 × 10⁻⁶ = 0.000001 F. Xc = 1 / (2π × 1000 × 0.000001) = 1 / 0.00628319 ≈ 159.155 Ω. Notice that raising the frequency lowers the reactance — the capacitor opposes high-frequency signals less.
- Input
- f = 60 Hz, C = 1 µF
- Result
- Xc ≈ 2652.58 Ω
- Why
- 1 µF = 0.000001 F. Xc = 1 / (2π × 60 × 0.000001) = 1 / 0.000376991 ≈ 2652.58 Ω. With ten times less capacitance than the first example, the reactance is ten times larger, confirming the inverse relationship between Xc and C.
- Input
- f = 120 Hz, C = 2.2 µF
- Result
- Xc ≈ 602.860 Ω
- Why
- 2.2 µF = 2.2 × 10⁻⁶ = 0.0000022 F. Xc = 1 / (2π × 120 × 0.0000022) = 1 / 0.00165876 ≈ 602.860 Ω. At the low end of the audio band this reactance is significant, which is why coupling-capacitor values affect bass response.
When to use this tool
- Designing or analyzing RC filters, coupling, and bypass networks where the capacitor's reactance sets the cutoff or corner frequency.
- Working out how much AC current a capacitor will pass at a given frequency, since AC current I = V / Xc for an ideal capacitor.
- Choosing capacitor values for crossovers, snubbers, or power-factor correction where reactance at the operating frequency matters.
- Checking homework or lab results in an AC circuits, electronics, or physics course that uses Xc = 1/(2π·f·C).
Common mistakes
- Forgetting to convert microfarads to farads. You must multiply µF by 10⁻⁶ before applying Xc = 1/(2π·f·C); using the raw µF number makes the reactance a million times too small.
- Dropping the 2π factor and dividing by f·C only. The angular frequency is 2πf, not f, so leaving out 2π makes your reactance about 6.28 times too large.
- Confusing reactance with resistance. Reactance opposes AC and depends on frequency, while resistance is constant; they combine as impedance Z = √(R² + Xc²), not by simple addition.
- Entering zero (or leaving a field blank) for frequency or capacitance. At f = 0 (DC) or C = 0 the formula divides by zero, so the reactance is undefined — physically a capacitor blocks DC entirely.
Frequently asked questions
What is the formula for capacitive reactance?
Capacitive reactance is Xc = 1/(2π·f·C), where f is the frequency in hertz, C is the capacitance in farads, and the result is in ohms. Because 2πf equals the angular frequency ω, you can also write it as Xc = 1/(ω·C).
What is the capacitive reactance of a 10 µF capacitor at 60 Hz?
About 265.26 ohms. Convert 10 µF to 0.00001 F, then Xc = 1/(2π × 60 × 0.00001) = 1/0.00376991 ≈ 265.26 Ω. This is the default example the calculator shows on load.
Why does capacitive reactance decrease as frequency increases?
Because frequency appears in the denominator of Xc = 1/(2π·f·C). A higher frequency means the voltage reverses faster, so the capacitor charges and discharges more readily and opposes the current less. That is why a capacitor blocks DC (very high Xc) but passes high-frequency AC (low Xc).
What is the difference between reactance and impedance?
Reactance (Xc) is only the frequency-dependent opposition of the capacitor, measured in ohms. Impedance (Z) is the total opposition of a circuit and combines resistance and reactance as a complex quantity, Z = R − jXc, with magnitude √(R² + Xc²). For an ideal capacitor alone, the impedance is purely reactive: Zc = −jXc.
How do I convert microfarads to farads for this calculation?
Multiply the value in microfarads by 10⁻⁶ (one millionth). For example, 10 µF = 10 × 10⁻⁶ = 0.00001 F and 0.1 µF = 1 × 10⁻⁷ F. This calculator performs the conversion automatically when you enter capacitance in µF.
Is capacitive reactance the same as the capacitor's resistance?
No. An ideal capacitor has no resistance and dissipates no power; reactance is its energy-storing opposition to AC and changes with frequency. Unlike resistance, reactance also introduces a 90° phase shift — the current through a capacitor leads the voltage across it by a quarter cycle.
Sources & references
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