About the RC Time Constant Calculator
This RC time constant calculator works out τ = R × C for a resistor–capacitor circuit, then shows the capacitor voltage and current at any moment while it charges from a supply or discharges through the resistor. It plots the full exponential curve over five time constants, tells you how long it takes to reach a chosen percentage of the supply, and gives the −3 dB cutoff frequency of the same components used as a low-pass or high-pass filter.
It is handy for designing power-on reset and delay circuits, switch debouncing, 555 timer and LED fade circuits, snubbers, and simple audio or sensor filters. Resistance is entered in kilohms and capacitance in microfarads, so the time constant comes out directly in milliseconds.
The math assumes an ideal step input, an ideal capacitor with no leakage, and a charging source with negligible internal resistance (add it to R if it is not). Real capacitors often have ±10–20% tolerance, so treat timing results as nominal.
With the default inputs, the time constant τ is 1,000 ms. Change any value above to recalculate instantly.
How to use the rc time constant calculator
- 1Choose whether the capacitor is charging or discharging.
- 2Enter the resistance in kilohms and the capacitance in microfarads.
- 3Enter the supply (or starting) voltage and the time you are interested in.
- 4Set a target percentage to see how long the circuit takes to get there.
- 5Read τ, the voltage and current at time t, and the filter cutoff frequency.
Formula and method
When a capacitor charges through a resistor, the current is largest at the start and falls as the capacitor voltage approaches the supply, producing an exponential curve. The time constant τ = R·C is the time to reach about 63.2% of the way to the final voltage; after 3τ it is 95% there and after 5τ over 99.3%, which is usually treated as fully charged. Discharge follows the mirror-image curve, falling to 36.8% after one τ.
Solving the exponential for time gives how long it takes to reach a chosen fraction p of the supply. Current is the voltage across the resistor divided by R. The same R and C used as a first-order filter have a −3 dB cutoff at fc = 1 ÷ (2πRC). With R in kΩ and C in µF, R·C comes out in milliseconds.
- τ
- Time constant
- R
- Resistance (kΩ)
- C
- Capacitance (µF)
- V
- Supply voltage or initial capacitor voltage
- Vc(t)
- Capacitor voltage at time t
- p
- Target voltage as a fraction of V
- fc
- Cutoff frequency (Hz)
Worked examples
10 kΩ and 100 µF charging from 5 V
τ = 10 kΩ × 100 µF = 1,000 ms (1 s). After one time constant the capacitor reaches 5 × (1 − e⁻¹) ≈ 3.16 V, 63.2% of the supply, and the current has fallen to 0.18 mA. Reaching 90% takes −1,000 × ln(0.1) ≈ 2.30 s. As a filter the cutoff is 1 ÷ (2π × 1 s) ≈ 0.159 Hz.
Switch debounce: 47 kΩ, 0.1 µF discharging from 3.3 V
τ = 47 × 0.1 = 4.7 ms. After 10 ms the voltage has decayed to 3.3 × e^(−10/4.7) ≈ 0.39 V. Falling to 30% of 3.3 V (the logic-low threshold of many inputs) takes −4.7 × ln(0.3) ≈ 5.66 ms.
Audio low-pass filter: 1 kΩ and 0.1 µF
τ = 1 kΩ × 0.1 µF = 0.1 ms, so fc = 1 ÷ (2π × 0.0001 s) ≈ 1,592 Hz. Frequencies well above that are attenuated at 20 dB per decade.
Frequently asked questions
What is the RC time constant?+
It is τ = R × C, the time a capacitor takes to charge to about 63.2% of the supply voltage through a resistor, or to discharge to about 36.8% of its starting voltage. With R in ohms and C in farads, τ is in seconds.
How long does it take a capacitor to fully charge?+
In theory it never reaches 100%, but after 5 time constants it is above 99.3%, which is treated as fully charged. A 10 kΩ resistor and 100 µF capacitor (τ = 1 s) are effectively full after about 5 seconds.
How do I calculate the cutoff frequency of an RC filter?+
Use fc = 1 ÷ (2πRC). At that frequency the output is 3 dB down, about 70.7% of the input voltage. For 10 kΩ and 10 nF the cutoff is about 1.59 kHz.
What percentage is reached after each time constant?+
Charging reaches about 63.2% after 1τ, 86.5% after 2τ, 95.0% after 3τ, 98.2% after 4τ and 99.3% after 5τ. Discharging leaves 36.8%, 13.5%, 5.0%, 1.8% and 0.7% respectively.
Why does the charging current decrease over time?+
The current is set by the voltage across the resistor, which is the supply minus the capacitor voltage. As the capacitor charges that difference shrinks, so the current falls exponentially toward zero.