RC Circuit Time Constant Calculator
Try an example
Time constant
1 s
τ = R × C = 10 kΩ × 100 µF. After one time constant a charging capacitor reaches 63.2% of the supply voltage; after 5τ (5 s) it is within 1% of full and is usually treated as fully charged.
Cutoff frequency f = 1 / (2πRC) = 159.2 mHz, the −3 dB point of an RC low-pass or high-pass filter built from these parts.
| Time | Elapsed | Charging: voltage reached | Discharging: voltage left |
|---|---|---|---|
| 1τ | 1 s | 63.2% | 36.8% |
| 2τ | 2 s | 86.5% | 13.5% |
| 3τ | 3 s | 95% | 4.98% |
| 4τ | 4 s | 98.2% | 1.83% |
| 5τ | 5 s | 99.3% | 0.674% |
The capacitor is
At the chosen time
1.967 V on the capacitor
After 500 ms (0.5τ): 39.3% of 5 V, with 303.3 µA flowing through the resistor as the capacitor charges.
Reaching 90% of the supply takes 2.303 s (2.3τ).
Horizontal axis: time in multiples of τ (1 s each). Vertical axis: percent of the supply voltage.
Calculate the time constant of a resistor-capacitor circuit, its cutoff frequency, how far a capacitor has charged or discharged after any time, and how long it takes to reach a target voltage, with values in any unit from picofarads to farads.
Formula
Worked Example
10,000 Ω × 0.0001 F = 1 s. Half a time constant in, the capacitor holds 5 × (1 − e^−0.5) = 1.967 V while 303 µA still flows through the resistor. Reaching 90% takes τ × ln 10 = 2.3 time constants.
Frequently Asked Questions
What is the RC time constant?
The product of resistance and capacitance, τ = R × C, in seconds when R is in ohms and C in farads. It is the time a charging capacitor takes to reach 63.2% of the supply voltage, or a discharging one to fall to 36.8% of where it started.
How long does a capacitor take to charge fully?
In theory it never quite gets there, because each time constant closes 63.2% of the remaining gap. After 5τ it is at 99.3%, which is usually treated as fully charged; after 7τ it is at 99.9%.
How is the time constant related to cutoff frequency?
For a first-order RC filter, f = 1 ÷ (2πRC). At that frequency the output is 70.7% of the input amplitude, the −3 dB point. Taking the output across the capacitor gives a low-pass filter; across the resistor, a high-pass filter.