A resistor opposes current the same way at every frequency. A capacitor does not: it blocks DC completely, lets high-frequency AC through easily, and opposes low-frequency AC strongly. That frequency-dependent opposition is called capacitive reactance, and it is the reason capacitors are used to couple audio signals, filter power supplies, and even drop mains voltage without a transformer. This post derives the formula and works a complete example.

Figure 1: A 5 V AC source driving a 100 nF capacitor (a), and the reactance of that capacitor against frequency (b). Every ×10 of frequency divides the reactance by 10. (Schematic, not to scale.)

Capacitive reactance

The reactance of a capacitor is:

X_C = 1 / (2πfC)

where f is the frequency in hertz and C is the capacitance in farads. X_C is measured in ohms, just like resistance, and it obeys an AC form of Ohm's law: I = V / X_C. But there are two big differences from a resistor. First, at f = 0 (DC) the reactance is infinite — a capacitor is an open circuit to DC, which is why it can block DC while passing AC. Second, the current through a capacitor leads the voltage by 90°: the current is largest while the voltage is still rising, because current is what charges the plates.

Worked example — 100 nF driven by 5 V

Take a C = 100 nF capacitor driven by a 5 V RMS sine wave, and find the current at three frequencies.

Step 1 — 100 Hz: X_C = 1 / (2π × 100 × 100 × 10⁻⁹) = 15,915 Ω. Current I = 5 / 15915 = 0.314 mA. At mains-like frequencies the little capacitor is almost an open circuit.

Step 2 — 1 kHz: X_C = 1 / (2π × 1000 × 100 × 10⁻⁹) = 1,592 Ω. Current I = 5 / 1592 = 3.14 mA — ten times the current at ten times the frequency.

Step 3 — 10 kHz: X_C = 1 / (2π × 10000 × 100 × 10⁻⁹) = 159 Ω. Current I = 5 / 159 = 31.4 mA — a hundred times the 100 Hz current.

Step 4 — power: an ideal capacitor dissipates no average power at all, because the current and voltage are 90° apart (P = V × I × cos 90° = 0). Energy just shuttles in and out of the electric field. The reactive power is Q = V × I: at 1 kHz that is 5 × 3.14 mA = 15.7 mVAR, and at 10 kHz it is 157 mVAR — power that flows back and forth but is never consumed.

A practical use — the capacitive dropper

Because a capacitor drops voltage without dissipating power, it can replace a resistor in a low-current mains supply — the classic capacitive dropper in LED night lights. Take a 470 nF capacitor on 230 V, 50 Hz mains: X_C = 1 / (2π × 50 × 470 × 10⁻⁹) = 6,773 Ω, so the current is I = 230 / 6773 = 34 mA — enough to run a string of LEDs, with almost no heat. A resistor doing the same job would burn over 7 W. Warning: this is a mains circuit — it needs an X-rated capacitor, a discharge resistor and a fuse, and it is not a beginner project. The calculation shows why it works; the safety design is a separate subject.

Key points to remember

• Capacitive reactance X_C = 1 / (2πfC): it falls as frequency rises, and it is infinite at DC — a capacitor blocks DC and passes AC.
• AC Ohm's law: I = V / X_C. The current leads the voltage by 90°.
• Worked example: 100 nF at 5 V RMS gives 0.314 mA at 100 Hz, 3.14 mA at 1 kHz and 31.4 mA at 10 kHz — ×10 frequency, ×10 current.
• An ideal capacitor dissipates no average power; the reactive power Q = V × I just flows back and forth.
• Practical use: a 470 nF capacitor drops 230 V mains to 34 mA with almost no heat — the capacitive dropper. Mains work needs proper safety design.