How to use it
- Choose charging (from 0 V towards a supply) or discharging (from a starting voltage towards 0 V).
- Enter the voltage, the resistance in kΩ and the capacitance in µF.
- Enter the moment you are interested in, in seconds from the start.
- Optionally change the target level to see when the capacitor gets there (90 % by default).
Formulas
With τ = R × C (kΩ × µF = milliseconds, so 10 kΩ × 100 µF = 1000 ms = 1 s):
- Charging: V = Vs × (1 − e^(−t/τ)), I = (Vs ÷ R) × e^(−t/τ)
- Discharging: V = V0 × e^(−t/τ), I = (V0 ÷ R) × e^(−t/τ)
- Charge Q = C × V, energy E = ½ × C × V²
- Time to a level p: t = −τ × ln(1 − p) when charging, t = −τ × ln(p) when discharging
| Time | Charging, % of Vs | Discharging, % of V0 |
|---|---|---|
| 1τ | 63.21 | 36.79 |
| 2τ | 86.47 | 13.53 |
| 3τ | 95.02 | 4.98 |
| 4τ | 98.17 | 1.83 |
| 5τ | 99.33 | 0.67 |
Worked example
A 100 µF capacitor charges from 5 V through 10 kΩ. τ = 10 kΩ × 100 µF = 1 s. After 1 second:
- V = 5 × (1 − e^−1) = 3.1606 V, that is 63.21 % of the supply;
- I = 0.5 mA × e^−1 = 0.1839 mA;
- Q = 100 µF × 3.1606 V = 316.06 µC, E = ½ × 100 µF × 3.1606² = 0.4995 mJ;
- it reaches 90 % after −1 × ln 0.1 = 2.3026 s.
Discharged from 5 V instead, the same capacitor holds 1.8394 V after 1 second.
Tips
- Units matter: 1 kΩ × 1 µF = 1 ms, 1 MΩ × 1 µF = 1 s.
- The formulas assume an ideal capacitor. Leakage and the source’s internal resistance slow real circuits a little.
- Energy grows with the square of the voltage: at 63 % of the voltage only 40 % of the final energy is stored.