Worked examples: mains-frequency X_C and a rail bypass cap

Two problems. First, the most common real-world calculation — what is X_C for a given cap on mains. Second, the frequency sweep a bypass cap actually sees on a 5 V rail.

Example 1 — 10 µF on 50 Hz mains

A 10 µF capacitor is placed across a 230 V RMS, 50 Hz supply. Find its capacitive reactance, the RMS current through it, and the reactance at 5 kHz for comparison.

1. Reactance at 50 Hz.
   $$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \cdot 50 \cdot 10 \times 10^{-6}} \approx 318\,\Omega$$
2. RMS current. $I_{rms} = V_{rms}/X_C = 230 / 318 \approx 0.72\,\text{A}$. Not negligible — this is why shunting a cap across mains is done with care (the cap must also tolerate the 325 V peak).
3. Reactance at 5 kHz (100× higher). Same cap, frequency up by a factor of 100:
   $$X_C = \frac{1}{2\pi \cdot 5000 \cdot 10 \times 10^{-6}} \approx 3.18\,\Omega$$
   Same 100× drop. A cap that looks like a 318 Ω resistance to the 50 Hz mains looks like a 3 Ω near-short to 5 kHz audio riding on the same line.

Drag the frequency marker left and right to read $X_C$ off the line at any frequency. Move the capacitance slider to see how different caps just shift the line up (smaller C) or down (bigger C) without changing its −1 slope.

Example 2 — bypass cap on a 5 V logic rail

A microcontroller draws short bursts of current at ~10 MHz during every clock edge. A 100 nF ceramic bypass cap sits between the rail and ground to supply those bursts locally without dragging down the rail. How much reactance does this cap present to the 10 MHz burst, and to the 50 Hz ripple that might also reach the rail from an upstream rectifier?

1. Reactance at 10 MHz.
   $$X_C = \frac{1}{2\pi \cdot 10 \times 10^6 \cdot 100 \times 10^{-9}} \approx 0.16\,\Omega$$
   A fraction of an ohm — near-short at that frequency. Burst-current pulses can flow straight from cap to MCU without rippling the rail upstream.
2. Reactance at 50 Hz.
   $$X_C = \frac{1}{2\pi \cdot 50 \cdot 100 \times 10^{-9}} \approx 31.8\,\text{k}\Omega$$
   Effectively open at mains frequency. The bypass cap doesn't burden the 50 Hz rail ripple because at that frequency it's barely a load.
3. Why this topology works. The cap's job is to shunt high-frequency content to ground — exactly where its reactance is smallest. That's why fast-switching chips need many small bypass caps (each tuned for its own frequency band) rather than one big one.