RC Time Constant Calculator
Calculate the RC time constant, charging and discharging voltage over time, and the filter cutoff frequency of a resistor–capacitor circuit.
How to use this calculator
- 1Enter the resistance and capacitance with their units — mixing kΩ and µF by hand is where τ usually goes wrong.
- 2Set the supply voltage, and choose whether the capacitor is charging or discharging.
- 3Enter a moment in time to see the voltage, current and charge at that instant.
How the calculation works
τ = R × C charging: V(t) = V₀(1 − e^(−t/τ)) discharging: V(t) = V₀e^(−t/τ) f_c = 1 ÷ (2πRC) E = ½CV²- τ
- Time constant, in seconds — the time to reach 63.2% of the way to the final value
- R
- Series resistance, in ohms
- C
- Capacitance, in farads
- V₀
- Supply voltage when charging, or starting voltage when discharging
- f_c
- Cutoff frequency, where output power falls to half the input — the −3 dB point
Ohms multiplied by farads gives seconds exactly. That is not a coincidence or an approximation: a farad is a coulomb per volt and an ohm is a volt per amp, and an amp is a coulomb per second, so the units cancel to seconds.
The 63.2% figure is 1 − 1/e. It is not a rounded convention but the exact value of the exponential after one time constant, which is why it appears in every datasheet.
Charging and discharging share the same τ. A capacitor takes as long to fill as to empty through the same resistance, since the same exponential governs both.
The cutoff frequency is where the reactance of the capacitor equals the resistance. Setting 1/(2πfC) = R and solving for f gives f = 1/(2πRC) directly.
Worked example
10 kΩ with 100 µF, charging from 5 V
- 1.Convert to SI: 10 kΩ is 10,000 Ω, and 100 µF is 0.0001 F.
- 2.τ = 10,000 × 0.0001 = 1 second exactly.
- 3.After 1 second, t/τ = 1, so the capacitor has reached 1 − e⁻¹ = 0.6321, or 63.21%.
- 4.Voltage: 5 × 0.6321 = 3.1606 V.
- 5.Practically full at 5τ = 5 seconds, by which point it is at 99.33%.
- 6.As a low-pass filter the cutoff is 1 ÷ (2π × 1) = 0.1592 Hz — this pair passes almost nothing but DC.
Result: τ = 1 s, 3.1606 V after 1 second
A 1 kΩ / 100 nF filter, discharging
- 1.τ = 1,000 Ω × 100 × 10⁻⁹ F = 0.0001 s, or 100 µs.
- 2.After 100 µs, t/τ = 1 exactly, so the capacitor has fallen to e⁻¹ = 36.79% of its starting voltage.
- 3.Voltage: 12 × 0.3679 = 4.4146 V.
- 4.Note the symmetry: charging reaches 63.21% in one τ, discharging falls to 36.79% — the two add to exactly 100%.
- 5.Cutoff frequency: 1 ÷ (2π × 0.0001) = 1,591.5 Hz, a practical audio-range filter.
Result: τ = 100 µs, 4.4146 V remaining
Why 63.2%, and why it never quite finishes
A capacitor charging through a resistor does not fill at a steady rate. The current that charges it is driven by the voltage difference between the supply and whatever the capacitor already holds — so as the capacitor fills, the driving voltage shrinks, the current falls, and charging slows. That feedback produces an exponential approach, not a straight line.
One time constant is the moment the capacitor reaches 1 − 1/e of the way to its final value, which is 63.212%. The figure looks arbitrary but is exact: it is what the exponential gives at t = τ. The same logic makes each subsequent τ close 63.2% of the *remaining* gap, which is why the milestones are 63.2%, 86.5%, 95.0%, 98.2% and 99.3%.
Strictly the capacitor never reaches the supply voltage — the approach is asymptotic. Engineering settles this by convention: 5τ, at 99.33%, is treated as fully charged, because the remaining 0.67% is smaller than component tolerances anyway.
The same circuit is a timer and a filter
An RC pair looks like two different components depending on whether you are thinking in time or in frequency, but it is one circuit and one behaviour. In the time domain it charges with time constant τ. In the frequency domain it passes slow changes and attenuates fast ones, with the transition at f = 1/(2πRC).
These are the same statement. A signal changing faster than the capacitor can follow gets smoothed; a signal slower than τ passes through essentially untouched. The cutoff frequency is simply where the capacitor's reactance equals the resistance — set 1/(2πfC) = R and the formula falls out.
Which filter you get depends only on where you take the output. Across the capacitor gives a low-pass filter, since the capacitor holds slow signals and shorts fast ones. Across the resistor gives a high-pass filter. Nothing else about the circuit changes.
Where the energy goes when you charge a capacitor
A capacitor charged to voltage V stores ½CV² joules. The half is not a fudge: voltage rises from zero to V as charge accumulates, so the average voltage during charging is V/2, and the energy is charge times that average.
The consequence surprises people. Charging a capacitor through a resistor from a fixed supply always wastes exactly as much energy in the resistor as it stores in the capacitor — 50% efficiency, regardless of the resistance. A smaller resistor charges faster but dissipates the same total heat; it just does it sooner.
This is why switching regulators exist. Getting past 50% requires ramping the supply voltage or using an inductor to move charge, not simply choosing a better resistor.
What this assumes, and where it stops
Assumptions
- An ideal capacitor and an ideal resistor: no leakage current, no equivalent series resistance, no dielectric absorption.
- A step change in supply voltage at t = 0, with the capacitor starting fully discharged when charging or fully charged when discharging.
- The source has negligible internal resistance compared with the series resistor.
- Component values are exact, ignoring tolerance.
Limitations
- A single resistor and single capacitor in series. Multi-stage filters, RLC circuits and active filters behave differently and are not covered.
- Real capacitors drift: electrolytics commonly carry ±20% tolerance and lose capacitance with age and heat, so a computed τ is a nominal figure, not a guarantee.
- Assumes DC step response. Continuous AC analysis needs impedance and phase, which are not reported here beyond the cutoff frequency.
- The cutoff frequency assumes the filter is unloaded. Connecting a low-impedance load across the output shifts it.
Common questions
What is the RC time constant?
The product of resistance and capacitance, τ = RC, measured in seconds. It is the time a capacitor takes to charge to 63.2% of the supply voltage, or to discharge to 36.8% of its starting voltage. Ohms times farads gives seconds exactly, with no conversion factor.
How long does a capacitor take to fully charge?
Strictly it never does — the approach is exponential and asymptotic. In practice 5 time constants is treated as fully charged, at which point it has reached 99.33% of the supply. That convention is used because the remaining fraction is smaller than normal component tolerances.
Why is it 63.2% after one time constant?
Because 1 − 1/e = 0.63212. The charging curve is V₀(1 − e^(−t/τ)), so at t = τ the exponent is exactly −1. It is a precise mathematical value rather than a rounded rule of thumb, which is why the same figure appears in every datasheet.
How do I find the cutoff frequency of an RC filter?
Use f = 1 ÷ (2πRC), which is the frequency where output power drops to half the input — the −3 dB point. Taking the output across the capacitor makes it a low-pass filter; across the resistor makes it a high-pass filter. The cutoff is the same either way.
Sources
Formula and content last reviewed on .
Results are estimates for information only, not professional advice.
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