Electronics
Capacitor discharge explained
A charged capacitor with a resistor across it discharges fast at first, then slower and slower. In each time constant, τ = R × C, its voltage falls to 37% of what it had, so it's at 5% after 3τ and under 1% after 5τ. The curve is V = V0 × e−t/τ.
Why it slows down
The capacitor's own voltage pushes the current through the resistor. At the start the voltage is highest, so the current is biggest, V0 ÷ R, and the charge drains quickly. As the voltage falls, the push falls, the current shrinks and the draining slows. It's like a hot drink cooling: fast while it's hot, slowly as it nears room temperature.
So a capacitor loses the same fraction of what's left in equal times, not the same number of volts. If it halves in 2 s, it halves again in the next 2 s.
The 37% rule
| After | Left |
|---|---|
| 1τ | 37% |
| 2τ | 14% |
| 3τ | 5% |
| 5τ | under 1% |
Example: 2200 µF charged to 24 V, emptying through 10 kΩ.
After 3τ, 66 s, it's at 0.05 × 24 = 1.2 V.
Live experiment
Run an LED on a capacitor
Hold the button and a 9 V battery fills a capacitor while it lights Pip, a red LED. Let go and Pip runs on the capacitor alone, fading fast and then slower. Try a bigger capacitor: it keeps him glowing longer.
“RC discharging” is lesson 3 of 5 in Capacitor Caves, Electronics World 3. Free with an account. Lessons unlock in order, so you’ll do the 17 before it first.
One rule for charging and discharging
Both follow one rule: in each τ, the voltage closes 63% of the gap between where it is and where it's heading. Charging heads for the supply, so after 1τ it has done 63%. Discharging heads for 0 V, so after 1τ, 37% is left. Halving takes about 0.69τ.
The stored energy, ½ × C × V², falls twice as fast as the voltage, because it goes with V².
Bleeder resistors
Power supplies put a bleeder resistor straight across their big capacitors, so they empty after switch-off. Choose it with the 5%-after-3τ row: to fall to 5% within 15 s, 3 × R × C must be under 15 s. But a smaller bleeder wastes more while the supply runs, P = V² ÷ R, so pick the largest one that's quick enough.
An LED doesn't empty a capacitor all the way. It stops conducting below about 2 V, so the glow fades out with charge still on the plates. That's the little light that lingers on a charger after you unplug it.
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 discharging” is lesson 3 of 5 in Capacitor Caves, Electronics World 3. Free with an account. Lessons unlock in order, so you’ll do the 17 before it first.
Or start from the very first lesson, no account neededWork it out
Practice questions
Three quick questions from the lessons. Pick an answer to check it.
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 only takes away 63% of what's left.
- Why not “63%”? 63% is the share it loses in one τ. What's left is 100 − 63 = 37%.
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 the share still missing after one τ, or what's left when discharging, not how far a charging capacitor has got.
- Why not “50%”? The curve isn't a straight line, so one τ isn't the halfway point. It's at 63%, and half comes at about 0.69τ.
- Why not “99%”? 99% takes about five time constants, not one.
Show the answer
Answer: quadruples. Energy goes with V². Twice the voltage gives 2² = 4 times the energy. That's why camera flashes charge their capacitors to hundreds of volts.
- Why not “halves”? More voltage means more stored energy, never less.
- Why not “stays the same”? Stored energy depends on the voltage, E = ½ × C × V², so it can't stay the same.
- Why not “doubles”? Energy goes with V², not V, so twice the voltage gives four times the energy.
“RC discharging” is lesson 3 of 5 in Capacitor Caves, Electronics World 3. Free with an account. Lessons unlock in order, so you’ll do the 17 before it first.
Questions people ask
What is the formula for capacitor discharge?
V = V₀ × e^(−t/τ), where V₀ is the starting voltage and τ = R × C. The current follows the same curve, starting at V₀ ÷ R. After one τ the voltage is 37% of where it started.
How long does a capacitor take to discharge?
In theory, for ever: it loses the same fraction in each τ. In practice it's under 1% after 5τ. So 2200 µF through 10 kΩ, τ = 22 s, is near enough empty after about 110 s.
How long until a capacitor's voltage halves?
About 0.69τ, because e^(−0.69) ≈ 0.5. It halves again in the next 0.69τ, and so on: from 12 V it's 6 V, then 3 V, then 1.5 V, at equal intervals.
What is a bleeder resistor?
A resistor straight across a power supply's big capacitor, so it empties after switch-off. A smaller one empties faster but wastes more power while the supply is on, P = V² ÷ R: 10 kΩ across 24 V wastes about 0.058 W.
Related topics
- RC time constant explainedτ = R × C: how long a capacitor takes to charge through a resistor, and the 63% rule.
- How does a capacitor work?Two plates and a gap: how a capacitor stores charge and energy, and why it blocks steady DC.
- Energy stored in a capacitorE = ½ × C × V²: how many joules a capacitor holds, why voltage counts double, and how much you can use.
- RL circuit time constant explainedHow fast a coil's current builds: τ = L ÷ R, and why more resistance speeds it up.