Capacitive reactance: what a capacitor does to AC
A capacitor blocks DC but passes AC, and how easily depends on the frequency. Its reactance, XC = 1 / (2πfC), plays the part resistance plays in Ohm's law — the current is V / XC — but with two differences: it falls as the frequency rises, and it dissipates no power, the current running a quarter cycle ahead of the voltage. This calculator gives XC for any capacitor and frequency, draws it against frequency, and with a source voltage and a series resistor works out the current, the impedance, the phase and the RC corner frequency.
How to use the capacitive reactance calculator
- Type the capacitance as it is printed (100n,
4.7u, 10uF) and the frequency (50, 1k, 10M). XC appears at once and the graph shows how it changes two decades either side. - Optional: a source voltage (RMS) gives the current the capacitor passes, and its phase.
- Optional: a series resistor turns it into an RC circuit: the drawing adds the resistor, the tiles show the impedance, the phase angle and the frequency where XC = R (the filter's corner), and the graph marks it.
- Other frequencies tabulates the same capacitor at mains, audio and RF frequencies; Calculation writes the maths out; click the capacitor in the drawing for the reverse calculation (which C gives this reactance elsewhere).
The formula
XC = 1 / (2π f C)
A capacitor's current is how fast its charge changes: I = C × dV/dt. A sine wave of amplitude V at frequency f changes at a peak rate of 2πf × V, so the peak current is 2πfC × V, and the ratio of voltage to current — the thing that behaves like a resistance — is 1 / (2πfC). The 2πf is the angular frequency ω, so you will also see XC = 1 / (ωC). Units: farads × hertz is 1 / ohms, so the result is in ohms.
Your values XC = 1 / (2π × 1 kHz × 100 nF) = 1.592 kΩ.
Why it depends on frequency
Double the frequency and the voltage swings twice as often, so twice the charge moves per second for the same swing: twice the current, half the reactance. On a graph with logarithmic axes XC is a straight line falling one decade of ohms for every decade of frequency. That is why a capacitor that is effectively open at 50 Hz can be a near short at 10 MHz, and why a decoupling capacitor only works above some frequency: below it, XC is too high to matter.
| Frequency | XC | Behaves like |
|---|---|---|
| 50 Hz | 31.8 kΩ | an open circuit to mains hum |
| 1 kHz | 1.59 kΩ | a few kilohms: audio passes with loss |
| 100 kHz | 15.9 Ω | tens of ohms |
| 10 MHz | 159 mΩ | a fraction of an ohm: a short for RF noise |
Current and phase
I = V / XC, leading V by 90°
Ohm's law still works with XC in place of R, as long as both are RMS or both peak. But the current is largest when the voltage is changing fastest — as it crosses zero — not when it is largest, so the current wave runs a quarter cycle (90°) ahead of the voltage. Over a whole cycle the capacitor takes energy for half the time and gives it back for the other half: the product V × I is reactive power, measured in var, and no heat is produced.
With a resistor in series: impedance and the RC corner
|Z| = √(R² + XC²) φ = atan(XC / R)
Resistance and reactance do not simply add: the resistor's voltage is in phase with the current and the capacitor's is 90° behind it, so they combine like the two sides of a right triangle. The result, the impedance Z, sets the current (I = V / |Z|) and the phase angle sits between 0° (resistor dominates, high frequency) and 90° (capacitor dominates, low frequency). The frequency where XC = R, f = 1 / (2πRC), is the RC filter's corner: there the output is 70.7% (−3 dB) and the phase 45°. Take the output across C for a low-pass filter, across R for a high-pass; the RC time constant calculator looks at the same circuit in time instead of frequency.
Your values, step by step
- Reactance: 1 / (2π × 1 kHz × 100 nF) = 1.592 kΩ; at ten times the frequency it would be 159.2 Ω.
Worked example: 100 nF at 1 kHz
XC = 1 / (2π × 1000 × 100 × 10⁻⁹) = 1.592 kΩ. At 10 kHz the same capacitor is 159.2 Ω, at 100 Hz 15.92 kΩ. Put 1 V RMS across it at 1 kHz and 628.3 µA flows, a quarter cycle ahead of the voltage, with no heating. Add a 1.6 kΩ resistor in series and the corner lands at 994.7 Hz, almost exactly 1 kHz: the classic RC low-pass for audio work.
Questions
Is reactance negative?
In the complex notation engineers use, a capacitor's impedance is −jXC (and an inductor's +jXL), the minus sign carrying the 90° lead. Its size, which is what this calculator gives, is positive. When a capacitor and an inductor are in series the two cancel, which is what resonance is.
Does a capacitor really pass AC?
No charge crosses the gap. The plates charge and discharge in step with the voltage, so current flows in the wires on both sides as if it went through, and to the rest of the circuit that is the same thing.
What about DC?
f = 0 makes XC infinite: once charged, a capacitor passes no steady current. That is why one in series blocks DC while letting the signal through (a coupling capacitor), and one across the supply shorts out high-frequency noise while ignoring the DC (decoupling).
Why does my big electrolytic not work at high frequency?
Real capacitors have inductance and resistance in their leads and plates. Above a few hundred kilohertz an electrolytic's ESL takes over and its impedance rises again, so a small ceramic capacitor is put in parallel to cover the higher frequencies.