How to use this calculator
- Enter or choose resistance.
- Enter or choose resistance unit.
- Enter or choose capacitance.
- Enter or choose capacitance unit.
- Enter or choose capacitor is.
- The time constant τ updates as you type. Use Copy result or Share to keep it.
Method
τ = R × C; charging: V(t) = V × (1 − e^(−t/τ)); discharging: V(t) = V × e^(−t/τ); f_c = 1 ÷ (2πRC)
- R
- resistance in ohms
- C
- capacitance in farads
- τ
- time constant in seconds
- V
- supply voltage (charging) or starting voltage (discharging)
- t
- time since the switch
Worked example
- τ = 10,000 Ω × 0.0001 F = 1 s.
- After t = 1 s = 1τ: V = 5 × (1 − e⁻¹) = 5 × 0.63212 = 3.1606 V.
- f_c = 1 ÷ (2π × 1) ≈ 0.1592 Hz.
A 10 kΩ resistor with a 100 µF capacitor has τ = 1 s; after one second the capacitor has charged to 3.16 V of 5 V.
What the result means
After 1τ the capacitor has covered 63.2% of the change, after 3τ 95%, and after 5τ 99.3%, which is the usual “fully charged” rule. The same τ sets the −3 dB corner of a first-order RC filter.
Assumptions and limits
- Ideal resistor and capacitor driven by a step from a stiff voltage source.
- Discharging starts from the voltage entered and decays towards zero.
- Capacitor leakage, ESR, source resistance and tolerance (often ±10–20% for electrolytics) change the real timing.
Questions
Why 63%?
After one time constant the remaining gap is e⁻¹ ≈ 36.8% of the original, so 63.2% of the change has happened.
How long until the capacitor is fully charged?
Mathematically never, but after 5τ it is within 0.7% of the final voltage, which is treated as complete in practice.