Capacitor Charge Time Calculator — RC Time Constant and Time to a Target Voltage
Find how long a capacitor takes to charge or discharge to a target voltage through a resistor. Solves for time, the inverse of a voltage-at-a-time curve.
Capacitor Charge Time Calculator
Background.
A capacitor charging through a resistor does not reach its target and stop — it approaches the supply voltage along an exponential curve, quickly at first and then ever more slowly, and in principle never quite arrives. This page answers the practical version of that: given a resistor, a capacitor, a supply and a voltage you care about, how long does it actually take? It works in both directions, charging up from empty and discharging down from a starting voltage, and it reports the answer both in seconds and in time constants.
The time constant is the whole story compressed into one number: τ = RC, with ohms times farads giving seconds directly. After one time constant a charging capacitor has covered 63.21% of the gap to the supply; after another, 63.21% of what remains; and so on. That fraction never changes, which is why the curve has the same shape whether τ is a microsecond or an hour. Five time constants is the conventional 'fully charged' point, and it is worth knowing that the convention is approximate: 1 − e⁻⁵ works out to 99.3262%, so 0.6738% is still missing at 5τ.
The default here is a worked example from OpenStax University Physics — a relaxation oscillator in which a 50 mF capacitor charges from a 100 V source until a neon lamp fires at 80 V. Note the resistance: 101 Ω, not 100 Ω, because the textbook folds the source's 1 Ω internal resistance into the external resistor before computing anything. That detail is the most common reason a measured charge time disagrees with a calculated one, and it is why this page's resistance field asks for the total series resistance rather than just the resistor you soldered in. The answer is τ = 5.05 s and t = 8.13 s.
Charging and discharging are not symmetrical, and the difference catches people out. Charging to 80% of the supply takes 1.609 time constants, because you are covering 80% of the whole gap. Discharging to 80% of the starting voltage takes only 0.223 time constants, because you are losing just 20%. Same circuit, same 80%, times differing by more than seven to one. Discharging to half — the RC equivalent of a half-life — always takes τ ln 2 = 0.693 time constants, whatever the values are.
A note on what this page is for, so you land in the right place. It solves for **time**: you supply the voltage you care about and it returns how long. If you have the opposite question — what voltage will the capacitor be at after a given interval — or you need the frequency-domain behaviour of the same network, its cutoff frequency, gain and phase shift, use the RC circuit calculator instead. Same two components, opposite direction, and neither page can substitute for the other.
One genuine limitation, stated here rather than buried. This models exactly one ideal resistor and one ideal capacitor driven by an ideal DC source, starting fully discharged or fully charged. It does not model a load in parallel with the capacitor, which changes both the final voltage and the effective time constant; it does not model leakage, equivalent series resistance or dielectric absorption; and it assumes the source can supply the initial current, which for a large capacitor and a small resistor can be very large indeed. If you are calculating a discharge time for safety purposes, add margin, verify with a meter, and remember that dielectric absorption can let a capacitor recover a surprising voltage after it has been shorted and released.
What is capacitor charge time calculator?
An RC circuit is a resistor and a capacitor in series with a DC source, and it is the simplest circuit that has a time dimension at all. When the source is connected, current flows to charge the capacitor, but as the capacitor's voltage rises the voltage across the resistor falls, so the current falls too — and the current is what charges the capacitor. That self-limiting feedback produces an exponential approach rather than a straight line. Solving the loop equation gives q(t) = Q(1 − e^(−t/τ)), where τ = RC has units of seconds and Q is the final charge. Dividing by C converts it to voltage, which is the form used here. The reverse process, with the source removed and the capacitor discharging through the resistor, gives v(t) = V₀e^(−t/τ) — the same time constant, pure decay. Both are first-order exponentials, and the time constant is the only parameter either of them has. That makes RC networks the standard way to create a delay, debounce a switch, set an oscillator's frequency, or shape a filter's response, and it makes τ the number an engineer reaches for first. OpenStax's own examples of the pattern are intermittent windscreen wipers, pacemakers and strobe lights, all of which are relaxation oscillators built around a capacitor charging to a threshold.
How to use this calculator.
- Choose charging or discharging. They are different curves, not two ways of saying the same thing.
- Enter the total series resistance, including the source's internal resistance and any wiring — not just the resistor's marked value.
- Enter the capacitance and pick its unit. Ohms times farads gives seconds, so 10 kΩ with 100 nF is one millisecond.
- Enter the supply voltage when charging, or the starting voltage when discharging.
- Enter the target voltage. It must be strictly between zero and the supply, because the curve is asymptotic at both ends.
- Read the time to target, then check the time-constant count in the verdict: under 1τ you are on the steep part of the curve, past 5τ you are in the flat tail where leakage matters more than the calculation.
- For a discharge you are relying on for safety, add margin and verify with a meter — the ideal curve does not include leakage or dielectric absorption.
The formula.
Start with the time constant, because everything else is a multiple of it. τ = RC, and the units work out without any conversion factor: an ohm times a farad is a second. Ohms and farads are both awkward sizes in practice, so the pairings worth memorising are kilohms with microfarads giving milliseconds, and megohms with picofarads also giving milliseconds. For the default, 101 Ω × 50 × 10⁻³ F = 5.05 s.
Now the curve. Charging, the capacitor voltage is v(t) = V_s(1 − e^(−t/τ)). At t = 0 the exponential is 1 and the voltage is zero; as t grows the exponential decays and the voltage climbs towards V_s without ever reaching it. To find the time for a given voltage, rearrange: divide by V_s, subtract from one, take the natural logarithm and multiply by −τ. That gives t = −τ ln(1 − V_t/V_s), which is exact — no iteration is involved.
Working the default: V_t/V_s = 80/100 = 0.8, so 1 − 0.8 = 0.2, and ln(0.2) = −1.6094379124. Multiplying by −5.05 gives 8.1276614578 seconds. OpenStax computes the same example and prints 8.13 s. The ratio t/τ = 1.609 is the number of time constants, and it depends only on the voltage ratio — change the resistor and the capacitor together to keep τ the same and this number does not move at all.
Discharging is the same algebra with a simpler curve: v(t) = V₀e^(−t/τ), so t = −τ ln(V_t/V₀). Falling from 100 V to 80 V is ln(0.8) = −0.2231435513, giving 1.1268749341 s — about a seventh of the charging time to the same 80% figure, because charging to 80% means covering most of the gap while discharging to 80% means losing very little of it. The special case worth remembering is the half-way point: falling to 50% always takes τ ln 2 = 0.693τ, exactly analogous to a radioactive half-life.
The two reference voltages the page reports are pure numbers, independent of the circuit. After one time constant a charging capacitor sits at 1 − e⁻¹ = 63.2120558829% of the supply, and a discharging one at e⁻¹ = 36.7879441171% of where it started. OpenStax states the first as 0.632 and the corresponding current fall as 0.368. Five time constants gives 1 − e⁻⁵ = 99.3262% — the page prints that figure rather than the usual rounded 99%, because the difference matters when you are deciding whether a capacitor is safe to touch.
One guard is worth explaining because it is a refusal rather than a result. A charging capacitor reaches exactly the supply voltage only after infinite time, so a target equal to the supply makes the logarithm's argument zero and the time infinite. The page rejects that input with a message rather than printing an infinity. The same applies to a discharge target of exactly zero.
Rounding happens once, at the end, to ten decimal places, and all intermediate arithmetic runs at forty significant digits on a private numeric context so that nothing elsewhere in the site can change this page's precision.
A worked example.
The relaxation oscillator from OpenStax University Physics: a 50 mF capacitor charges from a 100 V source through a 100 Ω resistor plus 1 Ω of source internal resistance, and a neon lamp fires when the capacitor reaches 80 V. The total series resistance is therefore 101 Ω, not 100 Ω, and the time constant is τ = 101 × 0.05 = 5.05 seconds. To reach 80 V — 80 percent of the supply — the time is t = −5.05 × ln(1 − 0.8) = −5.05 × ln(0.2) = −5.05 × (−1.6094379124) = 8.1276614578 seconds, which is 8127.6614577922 milliseconds. The textbook rounds this to 8.13 s, and that is what this calculator's figure rounds to. Expressed in time constants the answer is 1.6094 τ, which sits in the ordinary working range between one and five. For reference, after exactly one time constant the capacitor would be at 63.2120558829 V — 63.21 percent of the supply, a figure that depends on nothing but the exponential itself — and full settling at five time constants would take 25.25 seconds. The lamp fires well before that, at 8.13 s, which is why a relaxation oscillator flashes at a useful rate instead of waiting for the capacitor to top out.
Frequently asked questions.
How long does it take a capacitor to charge?
What is the RC time constant and why is it in seconds?
Why does my measured charge time not match the calculation?
How long does a capacitor take to discharge to a safe voltage?
Why is discharging to 80% so much faster than charging to 80%?
Is a capacitor really fully charged after 5 time constants?
Does the supply voltage change how long charging takes?
References& sources.
- [1]OpenStax (Rice University), University Physics Volume 2, §10.5 'RC Circuits'. Primary source for every equation on this page: Equation 10.8 'q(t) = Cε(1 − e^(−t/RC)) = Q(1 − e^(−t/τ))', Equation 10.9 'τ = RC' with the note that 'the units of RC are seconds, units of time', Equation 10.11 'q(t) = Q e^(−t/τ)' for discharging, and the reference point '1 − e⁻¹ = 1 − 0.368 = 0.632'. Also the source of this page's default inputs: Example 10.8 computes 'τ = (R + r)C = (101 Ω)(50 × 10⁻³ F) = 5.05 s' and 't = −5.05 s · ln(1 − 80 V/100 V) = 8.13 s'. Retrieved 2026-07-29.
- [2]OpenStax (Rice University), University Physics Volume 2, §8.1 'Capacitors and Capacitance'. Source of C = Q/V and the farad as 'one coulomb per one volt' — the relation that converts the published charge-versus-time equations into the voltage-versus-time form this page solves. Retrieved 2026-07-29.
- [3]Engineering LibreTexts / James M. Fiore, Introduction to Circuit Analysis, capacitor and inductor chapters. Independent second authority, consulted to check the behaviour rather than to derive it: C = Q/V, the element laws, and the constraints on instantaneous change that make the exponential approach unavoidable. Cross-checked on a second scale — 10 kΩ with 100 nF gives τ = 1 ms and a half-way time of τ ln 2 = 0.693 ms, which this calculator returns. Retrieved 2026-07-29.
- [4]MIT OpenCourseWare, 6.002 Circuits and Electronics, Spring 2007 (Prof. Anant Agarwal). Course materials on first-order RC transients and the time constant, used as a third independent treatment. Landing page verified to load and to list the course number, term and instructor. Retrieved 2026-07-29.
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