Electronics · Math shelf · Optional — for the curious · build 2026.09.23-1644

τ = RC

The idea

τ = R × C. A resistor and a capacitor together make a clock. Their product — τ, the Greek letter tau — is the circuit's time constant, in seconds.

Ohms × farads = seconds?!

It sounds like nonsense until you build the units from their definitions (the What's a Coulomb? sheet is the toolbox here):

The units, canceling on the page. Volts strike volts, then coulombs strike coulombs — and Ω × F has nowhere left to go but seconds.

The volts cancel, and coulombs divided by coulombs-per-second leaves plain seconds. The units aren't being cute — they're telling you these two parts genuinely measure out time together.

What τ means on the bench

Charge a capacitor through a resistor and it does not fill at a steady rate. It sprints at first, then eases off: the fuller it gets, the less voltage is left across the resistor, so the less current flows in. The closer it gets to the target, the slower it approaches — and τ is the yardstick of that approach:

After…	The cap has reached
1 τ	63% of the supply voltage
2 τ	86%
3 τ	95%
4 τ	98%
5 τ	99.3% — engineers call this "fully charged"

The charging curve: a sprint, then a long glide. Each τ covers 63% of whatever distance remains — which is why it never quite arrives, and why 5τ counts as arrived.

Discharging is the mirror image: pull the supply away, give the charge a path through a resistor, and the voltage falls to 37% of where it started after each τ. Same clock, run downhill — this is the fade you'll build in the capacitor lesson.

Worked examples

10 kΩ and 100 µF. τ = 10,000 × 0.0001 = 1 second. A one-second glide from a two-part circuit — no chip, no code.
1 kΩ and 100 µF. τ = 0.1 s — ten times less resistance, ten times faster. The fade becomes a quick wink.
The pot strikes again. A 10 kΩ pot as the resistor sweeps τ from ≈0 to 1 s. In lesson 5 the knob controlled brightness; give it a capacitor and the same knob controls time.

Where this is headed

The 555 timer lessons run entirely on this equation: the chip just watches an RC charge climb past a threshold, resets it, and repeats. Pick R and C, and you've picked the blink rate — τ = RC is the metronome inside every blinker and beeper you'll build.

Capacitors remember

A capacitor keeps its charge after the supply is off — that's its whole job. At our 5 V it's harmless, but build the habit now: mains-powered equipment contains capacitors that stay dangerous long after the plug is pulled, and the people who repair it treat every big capacitor as loaded until proven empty.

The math, for the curious — always optional

Why 63%? The gap to the target shrinks by the same fraction every τ — the signature of exponential decay. The fraction that remains after one τ is 1 ÷ e ≈ 0.37, where e ≈ 2.718 is the same constant that runs compound interest and population growth. So the cap covers 1 − 0.37 = 63% per τ. Two τ leave 0.37 × 0.37 ≈ 14% of the gap; that's the 86% row in the table.

The energy in the capacitor — and where the ½ comes from

A charged capacitor holds energy, and the formula has a famous fraction in it:

E = ½ × C × V²

That ½ isn't decoration — it's the ghost of a genuine piece of calculus, and you can see it with a triangle. Remember the definitions: capacitance means the voltage climbs in proportion to the charge delivered (V = Q ÷ C — a straight line), and a volt is a joule per coulomb, so each bit of charge costs whatever the voltage is when it arrives. The first charge arrives when the capacitor is nearly empty — almost free. The last arrives against the full voltage V — full price. The average price over the whole fill is exactly half of V:

The triangle that replaces calculus. Price-per-coulomb climbs in a straight line from 0 to V, so the total cost is the area under it: ½ × Q × V. Substitute Q = C × V and out falls E = ½CV².

Worked: the 1000 µF can charged to 5 V holds ½ × 0.001 × 25 = 12.5 mJ. (Compare a battery, which delivers every coulomb at full voltage — no ½. The capacitor's fraction exists because it charges itself uphill.)

For the very curious — the missing half

Here's a theorem with a sting: charge a capacitor from a fixed supply through any resistor, and the supply spends Q × V of energy — but the capacitor stores only ½ Q V. Exactly half is lost as heat in the resistor, no matter what the resistance is. Big R wastes it slowly, small R wastes it fast; the total is always half. (And yes — the smooth version of the triangle argument is integration: E = ∫V dq. When calculus arrives in your math life, you have already met it.)

Practice

Answers at the bottom.

τ for 1 kΩ with 1,000 µF?
A 10 kΩ / 100 µF pair starts charging from 0 V toward 5 V. About what voltage after 1 τ? How long until "fully charged"?
You want τ ≈ 2 s using a 100 µF capacitor. What resistance — and how do you build it from a kit with nothing bigger than 10 kΩ? (The Series & Parallel sheet has the trick.)
Same 100 µF cap, but you want the fade faster: τ ≈ 0.05 s. What R?
Your supply's other setting is 3.3 V. How much energy does the 1000 µF can hold at 3.3 V — and why is it less than half of the 5 V figure?

Answers

1. 1,000 × 0.001 = 1 s. 2. 63% of 5 V ≈ 3.2 V after 1 τ (= 1 s); fully charged at 5 τ ≈ 5 s. 3. R = 2 ÷ 0.0001 = 20 kΩ — two 10 kΩ in series. 4. R = 0.05 ÷ 0.0001 = 500 Ω — a 470 Ω from the kit lands at τ ≈ 0.047 s, close enough (and the next sheet, Tolerance & Error, is about why "close enough" is an engineering answer). 5. ½ × 0.001 × 3.3² ≈ 5.4 mJ — well under half of 12.5 mJ, because energy rides on V²: the voltage fell to 0.66 of 5 V, so the energy fell to 0.66² ≈ 0.44 of it.