Capacitor Time Constant (RC): Charge and Discharge Explained

The time constant of a resistor-capacitor circuit is τ = R x C. With a 10 kΩ resistor and a 100 µF capacitor, τ = 10,000 x 0.0001 = 1 second. After one time constant the capacitor charges to 63.2% of the supply voltage. After five it reaches about 99.3%, so this circuit is effectively full in 5 seconds. Discharging to half voltage takes RC x ln 2 = 0.69 seconds.

The capacitor charging time calculator and the capacitor discharge time calculator do these sums for any values. Here is where the numbers come from.

The RC time constant formula

τ = R x C

R is resistance in ohms and C is capacitance in farads. The answer is in seconds. The Greek letter τ (tau) is the usual symbol, and the RC time constant article on Wikipedia covers the derivation.

Unit prefixes trip people up more than the formula. 10 kΩ is 10,000 ohms. 100 µF is 0.0001 farads. Multiply them and you get 1 second.

Charging: the 63% rule

When a capacitor charges through a resistor from a fixed supply voltage V, its voltage follows this curve:

V(t) = V x (1 - e^(-t/τ))

At t = τ, the bracket becomes 1 - e^-1, which is 0.632. So the capacitor is at 63.2% of the supply after one time constant, no matter what R and C are. Each further time constant closes 63.2% of the gap that is left.

Time constant table

Time	Charge (% of supply)	Discharge (% of start voltage left)
1τ	63.2%	36.8%
2τ	86.5%	13.5%
3τ	95.0%	5.0%
4τ	98.2%	1.8%
5τ	99.3%	0.7%

For the 10 kΩ and 100 µF example, each row is one more second.

The 5τ rule

The charging curve never quite reaches 100% in theory. In practice, after 5 time constants the capacitor is at 99.3%, close enough that engineers call it fully charged. For the example circuit, that is 5 seconds.

If you need a different target, solve the charging formula for time:

t = -τ x ln(1 - Vtarget / V)

Discharging a capacitor

A charged capacitor emptying through a resistor follows the mirror curve:

V(t) = V0 x e^(-t/τ)

After one time constant, 36.8% of the starting voltage remains. After five, 0.7% remains.

To find the time to reach any voltage, solve for t:

t = τ x ln(V0 / Vtarget)

Half voltage is a common target. Then V0 / Vtarget = 2, and t = τ x ln 2. With τ = 1 second, that is 0.69 seconds. This is the capacitor’s half-life, and it is the same at every step: the voltage halves again every 0.69 seconds.

Charging with a constant current

Not every circuit uses a resistor. If a current source charges the capacitor at a fixed current I, the voltage rises in a straight line instead of a curve. The time to change the voltage by ΔV is:

t = C x ΔV / I

Double the current and the time halves. Double the capacitance and the time doubles.

Energy stored in a capacitor

The energy a capacitor holds is:

E = ½ x C x V²

E is in joules when C is in farads and V in volts. Because voltage is squared, doubling the voltage stores four times the energy. That is also why reaching 63.2% of the voltage after one time constant means only about 40% of the final energy is stored at that point (0.632 squared).

Where people go wrong

• Mixing units. Kilohms with microfarads give milliseconds, not seconds. Convert everything to ohms and farads first, or keep track of the prefixes.
• Forgetting other resistance. The supply, wires and the capacitor itself add resistance, which lengthens τ a little.
• Using the charging resistor for discharge when the discharge path is different.

The FAQ answers what the capacitor time constant is, how long a capacitor takes to charge and how long it takes to discharge.