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How to use
In a first-order RC circuit, resistance R times capacitance C sets the time constant τ=RC. After one τ, a capacitor charging from zero reaches about 63.2% of the supply, while a discharging capacitor retains about 36.8%. The tool also reports cutoff frequency fc=1/(2πτ) and 5τ settling time.
- Enter R and choose Ω, kΩ or MΩ.
- Enter C with its capacitance unit, then time and voltage.
- Calculate τ, fc, and charging/discharging percentages and voltages at that time.
How it is calculated
Example
Example: R=10 kΩ and C=100 µF give τ=1 s and fc≈0.1592 Hz. At t=1 s with 5 V, charging voltage is ≈3.1606 V and discharging voltage ≈1.8394 V.
Important notes
The equations assume an ideal step source, lumped R and C, and no extra source resistance, ESR, leakage or loading. Real circuits vary with tolerances and conditions; do not rely on this calculator alone for safety-critical design.
Worked examples and interpreting results
Resistance 10 kΩ and capacitance 100 µF give time constant one second. One second after ideal charging begins, voltage reaches about 63.212% of its final value.
| Case | Calculation | Result |
|---|---|---|
| Time constant | 10000 × 100 × 10⁻⁶ | 1 s |
| Charge percentage after one second | (1 − e⁻¹) × 100 | 63.212056 % |
How to check the result
Select kΩ and µF and time 1. Charging assumes an initially uncharged capacitor, constant source and ideal components; discharge follows exponential decay from an initial voltage.
Frequently asked questions
Why 63.2% after one τ?
Because 1−e⁻¹≈0.6321 in the exponential charging equation.
Why is 5τ used?
After 5τ, charging reaches about 99.33% and discharge retains about 0.67%, a common practical settling reference.
Does fc apply to every RC circuit?
1/(2πRC) is the ideal one-pole RC −3 dB corner; actual behavior also depends on topology and loading.