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Electronics · Circuits

RC Time Constant Calculator

Find τ = R × C, the 63.2% charge point, and the ~5τ time a capacitor needs to charge or discharge.

The series resistor.
The capacitor being charged.
Try a circuit
Time constant (τ = R × C)
1000ms

Reaches 63.2% of the source in one τ; treated as fully charged (~99.3%) after 5τ ≈ 5 s.

Charging curve
Capacitor charging toward the source voltage over five time constantsV source0t = 0t = 5τ

Fraction of the source voltage as the capacitor charges. It reaches 63.2% at one τ and about 99.3% by 5τ, where it is considered fully charged.

The RC time constant is τ = R × C. With a 10 kΩ resistor and a 100 µF capacitor, τ = 10 000 × 0.0001 = 1 second. After one time constant a charging capacitor reaches 63.2% of the supply voltage, and it is treated as fully charged after about 5τ — here, 5 seconds.

What the RC time constant is

The time constant, written with the Greek letter τ (tau), sets how quickly a capacitor charges or discharges through a resistor. It is simply the resistance multiplied by the capacitance: a bigger resistor limits current, and a bigger capacitor stores more charge, so either one makes the circuit respond more slowly. One τ is the time to reach 63.2% of the way to the final voltage — not the time to finish.

τ = R × C

R in ohms and C in farads give τ in seconds — 1 Ω × 1 F = 1 s

V(t) = Vsource(1 − e^(−t ÷ τ))

Charging voltage over time; at t = τ the exponential term leaves 1 − e⁻¹ ≈ 0.632

Worked example

Take a 10 kΩ resistor in series with a 100 µF capacitor across a supply.

  1. 1
    Convert the resistance to ohms. 10 kΩ = 10 000 Ω.
  2. 2
    Convert the capacitance to farads. 100 µF = 100 × 10⁻⁶ = 0.0001 F.
  3. 3
    Multiply R by C. τ = 10 000 × 0.0001 = 1 s — the time constant.
  4. 4
    Find the 63.2% point. After one τ (1 s) the capacitor charges to 63.2% of the supply voltage.
  5. 5
    Estimate the full-charge time at 5τ. 5 × 1 s = 5 s to reach about 99.3%, which counts as fully charged.

Charge and discharge at each time constant

Percentages are of the source voltage. Charging follows 1 − e⁻ⁿ; discharging follows e⁻ⁿ. Both use the same τ.

Elapsed timeCharging (% of source)Discharging (% of source)
63.2%36.8%
86.5%13.5%
95.0%5.0%
98.2%1.8%
99.3%0.7%

Reading the results

Charging vs. discharging. Charging climbs along 1 − e^(−t/τ) toward the supply, while discharging decays along e^(−t/τ) toward zero. The shapes mirror each other and share the same τ, so a capacitor takes as long to empty as it does to fill.

Why ~5τ counts as “done.” The exponential never mathematically reaches 100%, but by 5τ it is within 0.7% of the target. Engineers therefore treat about five time constants as fully charged or fully discharged for practical purposes.

Filters and timing. The same τ sets the corner frequency of an RC filter, f = 1 ÷ (2π × τ), and the delay in timing or debounce circuits. Increase R or C to slow the response and lower the cutoff; decrease them to speed it up.

What does 63.2% mean in the RC time constant?
One time constant is defined as the time for a charging capacitor to reach 1 − e⁻¹ ≈ 63.2% of the supply voltage. It comes straight from the formula V(t) = V(1 − e^(−t/τ)): setting t = τ makes the exponent −1, leaving 63.2%. On discharge, τ is the time to fall to e⁻¹ ≈ 36.8%.
Why is a capacitor considered fully charged after 5 time constants?
The charge curve approaches the supply voltage but never exactly reaches it. After 5τ it is at about 99.3%, close enough that the remaining 0.7% is negligible for most circuits, so 5τ is the standard rule of thumb for full charge or discharge.
Does the time constant depend on the supply voltage?
No. τ = R × C depends only on the resistance and capacitance, not on voltage. A higher supply charges the capacitor to a higher final voltage, but it still reaches 63.2% of that value in exactly one τ.
How do I make the circuit charge faster or slower?
Change R or C. Reducing either value shortens τ and speeds up charging and discharging; increasing either lengthens τ and slows them down. Because τ is their product, halving R has the same effect as halving C.
Is the discharge time the same as the charge time?
Yes, both use the same τ = R × C. Charging rises as 1 − e^(−t/τ) and discharging falls as e^(−t/τ), so a capacitor takes the same amount of time to reach 63.2% charged as it does to drop to 36.8% of its starting voltage.
How does τ relate to an RC filter’s cutoff frequency?
The cutoff (corner) frequency of a first-order RC filter is f = 1 ÷ (2π × τ). A longer time constant means a lower cutoff frequency, so the same R and C that set a slow charge time also set where the filter starts attenuating.