1RC Timing: 10kΩ + 1µF at 5V
Inputs
Result
τ = 10000 × 0.000001 = 0.01s = 10ms. 5τ = 50ms. E = 0.5 × 0.000001 × 25 = 12.5µJ.
Time Constant (τ)
10.000 ms
Full Charge
50.000 ms
e.g. 0.000001 = 1µF
Time Constant (τ)
10.000 ms
Full Charge (5τ)
50.000 ms
Energy
12.50 µJ
Charge (Q)
5.00 µC
63% Charge
10.000 ms
Inputs
Result
τ = 10000 × 0.000001 = 0.01s = 10ms. 5τ = 50ms. E = 0.5 × 0.000001 × 25 = 12.5µJ.
Inputs
Result
Parallel: 10 + 22 + 47 = 79µF.
The time constant τ = R × C. After one τ, a capacitor charges to 63.2% of supply voltage. After 5τ, it reaches 99.3% (considered fully charged). For 10kΩ + 1µF: τ = 10ms.
| Time Constants | % Charged | % Remaining | Voltage (5V supply) |
|---|---|---|---|
| 1τ | 63.2% | 36.8% | 3.16V |
| 2τ | 86.5% | 13.5% | 4.32V |
| 3τ | 95.0% | 5.0% | 4.75V |
| 5τ | 99.3% | 0.7% | 4.97V |
Parallel: Ctotal = C1 + C2 + C3 (add up). Series: 1/Ctotal = 1/C1 + 1/C2 + 1/C3 (reciprocal sum). Parallel increases capacitance; series decreases it but increases voltage rating.
E = ½ × C × V². A 1000µF capacitor at 5V stores 12.5mJ. Energy increases with the square of voltage—doubling voltage quadruples energy.
Q = C × V. Charge in coulombs equals capacitance in farads times voltage. A 100µF cap at 10V holds 1mC (1 millichoulomb).
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Last Updated: Mar 20, 2026
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