RC Time Constant Calculator
Calculate the RC time constant τ = R × C in seconds from resistance and capacitance, plus the capacitor charging and discharging voltage at any time.
RC Time Constant Calculator
Calculator
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Formula: τ = R × C | V(t) = V₀(1 - e^(-t/τ)) charging | V(t) = V₀ × e^(-t/τ) discharging
Worked example — τ = 10 ms; effectively fully charged after 50 ms
Formula
τ = R × C | V(t) = V₀(1 - e^(-t/τ)) charging | V(t) = V₀ × e^(-t/τ) discharging
The RC time constant τ is the time to charge to 63.2% (or discharge to 36.8%). After 5τ, the circuit is >99% complete.
Worked Examples
Example 1: 10 kΩ resistor with 1 µF capacitor
Problem:R = 10 kΩ, C = 1 µF
Solution:τ = R × C = 10,000 × 0.000001 = 0.01 s = 10 ms Charged to 63.2% at 10 ms, 86.5% at 20 ms, 95.0% at 30 ms, 99.3% at 50 ms (5τ)
Result:τ = 10 ms; effectively fully charged after 50 ms
Example 2: Voltage on the capacitor at a given time
Problem:R = 47 kΩ, C = 100 nF, supply V₀ = 5 V, t = 2 ms after switch-on
Solution:τ = 47,000 × 100×10⁻⁹ = 4.7 ms V(t) = V₀(1 − e^(−t/τ)) = 5 × (1 − e^(−2/4.7)) = 5 × (1 − 0.653) = 1.73 V
Result:1.73 V after 2 ms (about 35% of the supply)
Example 3: Choosing R for a required delay
Problem:A timer must reach 2/3 of V₀ (the 555 threshold) in 1 s using C = 10 µF
Solution:1 − e^(−t/τ) = 0.667 → t/τ = ln 3 = 1.0986 → τ = 1 / 1.0986 = 0.910 s R = τ / C = 0.910 / 10×10⁻⁶ = 91,000 Ω
Result:R ≈ 91 kΩ (use the E24 value 91 kΩ)
Frequently Asked Questions
Why is 5τ important?
After 5 time constants, the capacitor is 99.3% charged/discharged. Engineers use this as the practical "fully complete" threshold.
What percentage of charge is reached at each time constant?
At 1τ the capacitor reaches 63.2% of full charge, at 2τ it reaches 86.5%, at 3τ it reaches 95.0%, at 4τ it reaches 98.2%, and at 5τ it reaches 99.3%. The exponential curve means the rate of change slows dramatically after the first few time constants. For discharge, the voltage drops to 36.8% at 1τ, 13.5% at 2τ, 5.0% at 3τ, and so on — the complement of the charging percentages.
How are RC circuits used as filters?
RC circuits act as frequency-selective filters because of the capacitor's frequency-dependent impedance. A low-pass RC filter (resistor in series, capacitor to ground) passes low frequencies and attenuates high frequencies above the cutoff frequency fc = 1/(2πRC). A high-pass RC filter (capacitor in series, resistor to ground) does the opposite. These simple filters are used in audio equalizers, noise suppression, power supply smoothing, signal coupling, and anti-aliasing before analog-to-digital conversion.
What is the difference between RC and RL time constants?
Both RC and RL circuits exhibit exponential charging behavior, but the energy storage element differs. RC circuits store energy in an electric field (capacitor), while RL circuits store energy in a magnetic field (inductor). The RL time constant is τ = L/R (inductance divided by resistance), whereas RC is τ = R×C. In practice, RC circuits are more common in signal processing and timing applications, while RL circuits are used in motor drives, switching power supplies, and applications where inductors are unavoidable.
How do I calculate the delay until the capacitor reaches a specific threshold voltage?
Invert the charging equation. For a rise from zero toward V0, the time to reach a threshold Vth is t = -tau x ln(1 - Vth / V0). Reaching half the supply takes 0.693 tau, and reaching two thirds takes about 1.099 tau. When charging between two thresholds, subtract the two times: a 555 timer charges from one third to two thirds of the supply, and the difference works out to ln(2) x tau, or 0.693 tau, which is where the familiar 0.693 R C interval comes from.
Why does multiplying ohms by farads give an answer in seconds?
The units genuinely cancel. An ohm is a volt per ampere, and a farad is a coulomb per volt. Multiplying gives coulombs per ampere, and since an ampere is one coulomb per second, the product is seconds. That is why a 10 kilohm resistor with a 1 microfarad capacitor gives 10,000 x 0.000001 = 0.01 second. It also explains the practical shorthand that megohms times microfarads gives seconds directly, and kilohms times microfarads gives milliseconds.
How is the cutoff frequency related to the time constant?
They are two views of the same network. The corner frequency is fc = 1 / (2 pi R C), which is simply 1 / (2 pi tau). At that frequency the capacitive reactance equals the resistance, the output is 3 dB below the passband, and the phase shift is 45 degrees. A tau of 10 ms therefore corresponds to about 15.9 Hz. A long time constant means slow transient response and a low corner frequency, so speeding up the step response necessarily widens the bandwidth and admits more noise.
How does the driving source impedance change the effective time constant?
The capacitor charges through everything in series with it, not just the nominal resistor. The output resistance of a logic gate, op amp, or signal generator adds to R, so the true time constant is (R + Rsource) x C. With a 1 kilohm resistor fed from a 600 ohm generator the transient is 60 percent slower than calculated. The same applies on the load side: an oscilloscope probe adds capacitance in parallel with C, typically around 10 pF or more for a passive probe, which slows high-impedance nodes.
Why do electrolytic capacitors make RC timing unreliable?
Aluminium electrolytics are often specified with tolerances as wide as minus 20 to plus 80 percent, they lose capacitance as the electrolyte dries with age and heat, and their leakage current can partially discharge the node you are trying to time. Class II ceramics such as X7R additionally lose a large fraction of their capacitance under DC bias. For timing that must hold its value, use film or C0G/NP0 ceramic capacitors, which offer tight tolerance, low leakage, and stable behaviour with voltage and temperature.
Which combination of R and C should I choose when many give the same time constant?
The product fixes tau, but the split sets the practical behaviour. A large resistor with a small capacitor draws little current and suits low-power timing, but the node becomes high impedance, so amplifier bias current, board leakage, and capacitive pickup all introduce error. A small resistor with a large capacitor is quieter and less sensitive to leakage, but loads the driving source and forces a physically larger capacitor. In signal filtering, R is usually chosen first from the source and load impedances, then C follows from the desired corner frequency.
References
Background & Theory
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Reviewed for accuracy by Daniel Agrici, Founder & Lead Developer · Editorial policy
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