RC Time Constant Calculator
Work out the time constant τ of a resistor–capacitor (RC) circuit from the resistance and capacitance, plus the time it takes the capacitor to charge to 63.2%, 86.5%, 95% and (effectively fully) 99.3% of the supply voltage.
Enter Values
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How to use this calculator
- Enter the series resistance R in kilohms (kΩ).
- Enter the capacitance C in microfarads (µF).
- Read the time constant τ (shown in milliseconds and seconds) and the charge-level times at 1τ, 2τ, 3τ and 5τ.
How it works
The time constant is τ = R·C with R in ohms and C in farads. Here R is converted from kΩ (×1000) and C from µF (×1e-6). During charging the capacitor voltage rises as Vc = Vs·(1 − e^(−t/τ)): it reaches 63.2% of the supply after one time constant, ~86.5% after two, ~95% after three and ~99.3% after five — which is why a capacitor is normally treated as fully charged after about 5τ. The same τ sets the discharge rate, where the voltage decays as Vs·e^(−t/τ).
Worked example
Worked example. For R = 10 kΩ and C = 100 µF: τ = 10,000 Ω × 0.0001 F = 1 s (1,000 ms). The capacitor reaches 63.2% after 1 s, and is effectively fully charged after 5τ = 5 s.
Common mistakes
- Mixing up units — R must be treated in ohms and C in farads before multiplying; the tool handles the kΩ and µF conversions for you, so enter the raw kΩ and µF values.
- Assuming the capacitor charges instantly to the supply — it approaches it exponentially and only reaches ~99.3% after five time constants.
- Confusing nanofarads (nF) or picofarads (pF) with microfarads (µF); convert to µF first (1 nF = 0.001 µF).
Frequently asked questions
Why is 63.2% the number for one time constant?
After t = τ the exponential term e^(−t/τ) = e^(−1) ≈ 0.368, so the capacitor has charged to 1 − 0.368 = 0.632, i.e. 63.2% of the supply voltage.
How long until the capacitor is 'fully' charged?
It never reaches 100% exactly, but after 5τ it is at ≈99.3% of the supply, which is treated as fully charged for practical purposes.
Does the same time constant apply to discharging?
Yes. When the capacitor discharges through the same resistance the voltage falls as Vs·e^(−t/τ), reaching 36.8% after 1τ and about 0.7% after 5τ.
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