Capacitive Reactance (X_C = 1/2πfC)

No charge crosses the gap between a capacitor's plates, yet a capacitor plainly passes alternating current. What actually happens is that charge piles onto one plate and off the other, and if the supply reverses before much has accumulated, the opposing voltage never gets large and current keeps flowing freely. The faster the alternation, the less charge accumulates per half-cycle and the less the capacitor pushes back — so its opposition falls as frequency rises, $X_C = 1/(2\pi f C)$. This is the exact mirror of the inductor, which opposes more as frequency rises, and the two mirror each other in sign as well as in shape.

A 1 µF capacitor presents $1/(2\pi \times 60 \times 10^{-6}) = 2.65\ \text{k}\Omega$ at 60 Hz, 159 Ω at 1 kHz, and 0.16 Ω at 1 MHz — from a substantial obstacle to a near short circuit across that range. That is the whole basis of bypassing: a 100 nF capacitor from a supply rail to ground is 26 kΩ at mains frequency, so it draws nothing, but a fraction of an ohm at the megahertz where a digital chip's switching noise lives, where it acts as a local reservoir. In a loudspeaker crossover the same behaviour puts a capacitor in series with the tweeter, blocking bass and passing treble.

At DC the relation gives infinity, and that is the honest answer: once charged, a capacitor is an open circuit, which is why this page refuses $f = 0$ rather than returning a number. That property is as useful as the frequency dependence — a coupling capacitor passes an audio signal while blocking the DC bias on either side of it, and every AC-coupled amplifier stage depends on it. The complementary behaviours of $X_L$ and $X_C$ meet on the LC resonance page, where at one frequency they are equal and, being opposite in sign, cancel entirely.

The first trap is the one shared with inductive reactance: ohms are not resistance. A capacitor dissipates no power — the current leads the voltage by 90°, energy flows in and back out each quarter-cycle, and a wattmeter reads zero. It follows that reactance never adds arithmetically to resistance: 30 Ω of resistance in series with 40 Ω of capacitive reactance is 50 Ω of impedance. The second trap is specific to this page. Inductive and capacitive reactances are opposite in sign, $+jX_L$ against $-jX_C$ — so in a series circuit they subtract, and the net is $X = X_L - X_C$ before you take the quadrature sum with $R$. Add them and you will get an answer that is not merely wrong but wrong in the direction that hides resonance completely. Third, a real capacitor is not pure: its equivalent series resistance does dissipate, which is what heats an electrolytic under ripple current and eventually kills it, and above its self-resonant frequency a capacitor's own lead inductance takes over and it starts behaving inductively — a bypass capacitor used above that point is doing nothing you intended.