Electronics
RC time constant explained
The RC time constant, τ (tau), is the resistance times the capacitance: τ = R × C, in seconds. Charge a capacitor through a resistor and it reaches about 63% of the supply voltage after one τ, and counts as full after about 5τ.
Why it charges fast, then slowly
Round the loop, the resistor's voltage and the capacitor's add up to the supply. At switch-on the capacitor is empty, so the resistor takes the whole supply and the current starts at its biggest, V ÷ R. As the capacitor's voltage rises, less is left across the resistor, so the current shrinks and the filling slows down.
A capacitor's voltage can't jump: charge has to flow in first. At the first instant it still reads 0 V.
Live experiment
Ride the charging curve
A 9 V battery charges a capacitor through a resistor. Pick R and C, press Switch on, and watch the voltage climb, with τ marked where it reaches 63% (5.7 V) and 5τ where it's full. Bigger R or C, slower, but every curve ends at 9 V.
“RC charging” is lesson 2 of 5 in Capacitor Caves, Electronics World 3. Free with an account. Lessons unlock in order, so you’ll do the 16 before it first.
τ = R × C
Each time constant closes 63% of the gap that's left:
| After | Charged to |
|---|---|
| 1τ | 63% |
| 2τ | 86% |
| 3τ | 95% |
| 5τ | 99.3% |
A bigger R lets charge in more slowly; a bigger C needs more charge for each volt. Either makes τ longer, but the capacitor still ends at the full supply. The exact curve is VC = V × (1 − e−t/τ).
A worked example
100 kΩ charges 47 µF from 9 V.
After 4.7 s the capacitor is at 0.63 × 9 V ≈ 5.7 V. It's full after about 5τ ≈ 24 s.
Discharging, and where it's used
Discharging runs the same clock the other way: through a resistor, the voltage keeps 37% of what it had each τ, so about 5% is left after 3τ. Charging and discharging follow one rule: each τ closes 63% of the gap between where the capacitor is and where it's heading.
RC delays are everywhere. A power-on delay holds a chip in reset for a moment after you switch on, a camera flash takes seconds to recharge, and an RC low-pass filter smooths out anything faster than its time constant.
Free lesson
Now build it yourself
Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.
“RC charging” is lesson 2 of 5 in Capacitor Caves, Electronics World 3. Free with an account. Lessons unlock in order, so you’ll do the 16 before it first.
Or start from the very first lesson, no account neededWork it out
- RC time constant calculatorτ, charge time, cutoff frequency and the voltage at any moment.
- Capacitor energy calculatorJoules and charge in a capacitor, and how much your circuit can use.
- RC filter cutoff frequency calculatorThe cutoff from R and C, the part for a cutoff you want, and the gain at any frequency.
- RC time constant practice problemsτ = R × C, the 63% rule, charging and discharging, with the working.
Practice questions
Three quick questions from the lessons. Pick an answer to check it.
Show the answer
Answer: 63%. After one τ it has closed 63% of the gap to the supply. 37% is the 100 − 63 that's still missing, and 99% takes about 5τ.
- Why not “37%”? 37% is what's still missing after one τ, the 100 − 63. It's also what's left after one τ of discharging.
- Why not “50%”? Half full comes a little sooner, at about 0.69τ. By one full τ the capacitor is past half, at 63%.
- Why not “99%”? That takes about five time constants. Each τ closes 63% of the gap that's left, so it creeps up and is only nearly full after 5τ.
Show the answer
Answer: 37%. In each τ it closes 63% of the gap to 0 V, so 37% is left.
- Why not “0%”? A discharging capacitor never quite reaches 0 V. Each τ removes 63% of what's left, so after 5τ less than 1% remains.
- Why not “63%”? 63% is how much it has lost after one τ. What's left is 100 − 63 = 37%.
Show the answer
Answer: 14%. 37% is left after 1τ, and 37% of that after 2τ, 0.37 × 0.37 ≈ 14%. 5% is after 3τ.
- Why not “5%”? 5% is what's left after 3τ: 0.37 × 0.37 × 0.37 ≈ 0.05. After only 2τ, more is left.
- Why not “37%”? 37% is left after just one τ. The second τ takes 63% of that too, leaving 37% of 37%, about 14%.
“RC charging” is lesson 2 of 5 in Capacitor Caves, Electronics World 3. Free with an account. Lessons unlock in order, so you’ll do the 16 before it first.
Questions people ask
What is the RC time constant?
It's the resistance times the capacitance, τ = R × C, in seconds. It's the time a capacitor charging through a resistor takes to reach about 63% of the supply voltage, or a discharging one to fall to about 37% of where it started.
Why is one time constant 63%?
The charging voltage follows V × (1 − e^(−t/τ)). At t = τ that's 1 − e⁻¹ ≈ 0.632 of the supply. Each further τ closes 63% of the gap that's left: 86% after 2τ, 95% after 3τ.
How long does a capacitor take to fully charge?
In theory it never quite gets there. In practice it counts as full after about five time constants, when it's at 99.3% of the supply. So 100 kΩ and 47 µF (τ = 4.7 s) is full after roughly 24 s.
Do ohms times farads really give seconds?
Yes. An ohm is a volt per amp and a farad is a coulomb per volt, so Ω × F = coulombs per amp, and a coulomb is an amp for one second.
Related topics
- How does a capacitor work?Two plates and a gap: how a capacitor stores charge and energy, and why it blocks steady DC.
- Low-pass filter explainedA resistor and a capacitor that pass the bass and fade the treble, and where the cutoff sits.
- How does an inductor work?A coil that fights changes in its current, the high-voltage spike it makes, and how to tame it.