Audited ·Last updated 31 Jul 2026·4 citations·Tier 2·0 uses

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

Charging or discharging
Include the source's internal resistance and any wiring. The textbook example this page defaults to adds a 1 Ω internal resistance to a 100 Ω resistor to get 101 Ω — leaving it out changes the answer by 1%.
Resistance unit
Ohms multiplied by farads gives seconds, so the unit pairing matters: 10 kΩ with 100 nF is 1 ms, and so is 10 MΩ with 100 pF.
Capacitance unit
When charging, this is the voltage the capacitor is heading towards. When discharging, it is the voltage it starts at.
V
Must be strictly between zero and the supply. A charging capacitor approaches the supply voltage asymptotically and never actually reaches it, so asking for 100% of the supply has no finite answer.
V
Time to reach the target
8.1277
t = −τ ln(1 − V_t/V_s) when charging, or −τ ln(V_t/V₀) when discharging. A closed-form answer, not an iteration.
Time in milliseconds
8,127.6615 ms
Time constant τ = RC
5.05 s
Time constant in milliseconds
5,050 ms
Settling time (5τ)
25.25 s
Voltage after one time constant
63.2121 V
Target as a share of the supply
80.00
Reading the result
Charging through 101 Ω into 50000 µF gives a time constant of τ = RC = 5.05 s. Reaching 80 V from a 100 V supply — 80% of the way — takes 8.127661 s, which is 1.6094 time constants. That sits in the usual working range of one to five time constants — past the steepest part of the curve but before it flattens out. After exactly one time constant the capacitor sits at 63.2121 V — 63.2120558829% of the supply — and full settling at 5τ takes 25.25 s. One ideal resistor, one ideal capacitor and an ideal DC source, starting from fully discharged. Add any source or wiring resistance into the resistance figure — the published example this page defaults to does exactly that, folding a 1 Ω internal resistance into a 100 Ω resistor to get 101 Ω. Leakage, ESR and anything else drawing current from the capacitor are not modelled.

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.

  1. Choose charging or discharging. They are different curves, not two ways of saying the same thing.
  2. Enter the total series resistance, including the source's internal resistance and any wiring — not just the resistor's marked value.
  3. Enter the capacitance and pick its unit. Ohms times farads gives seconds, so 10 kΩ with 100 nF is one millisecond.
  4. Enter the supply voltage when charging, or the starting voltage when discharging.
  5. Enter the target voltage. It must be strictly between zero and the supply, because the curve is asymptotic at both ends.
  6. 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.
  7. 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.

τ = R × C charging: t = −τ ln(1 − V_t ⁄ V_s) discharging: t = −τ ln(V_t ⁄ V₀)

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.

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.

modecharging
capacitance50
target Voltage80
resistance Unitohm
supply Voltage100
capacitance UnitmF
resistance101

Frequently asked questions.

How long does it take a capacitor to charge?
There is no single answer, because a capacitor approaches the supply voltage asymptotically and never quite reaches it — you have to say what voltage you care about. The time to any particular voltage is t = −τ ln(1 − V_t/V_s), where τ = RC. Useful landmarks: one time constant gets you to 63.21% of the supply, two to 86.47%, three to 95.02%, five to 99.33%. In practice engineers call the capacitor charged at five time constants and accept the missing 0.67%. If you need a specific voltage — a threshold, a trigger, a logic level — enter it above and the page gives the exact time rather than a landmark.
What is the RC time constant and why is it in seconds?
τ = RC, and the units genuinely work out: an ohm is a volt per amp, a farad is a coulomb per volt, and a coulomb is an amp-second, so multiplying them leaves seconds. OpenStax states it plainly — 'the units of RC are seconds, units of time'. The pairings worth memorising are kilohms × microfarads = milliseconds, and megohms × picofarads = milliseconds. The time constant is the only parameter the curve has: double it and everything takes twice as long, but the shape of the curve — the percentages at each landmark — never changes.
Why does my measured charge time not match the calculation?
The usual reason is resistance you did not count. The source's internal resistance, the wiring and any current-sensing resistor all add to R, and the textbook example this page defaults to shows exactly that: a 1 Ω internal resistance added to a 100 Ω resistor changes the answer from 8.05 s to 8.13 s. After that the usual suspects are capacitor tolerance, which for electrolytics can be −20%/+80% and moves the answer by the same proportion; anything else drawing current from the capacitor, which changes both the final voltage and the effective time constant; and, for Class 2 ceramics, DC-bias derating, which can remove most of the marked capacitance at working voltage.
How long does a capacitor take to discharge to a safe voltage?
Enter the discharge mode with your bleeder resistance and the voltage you consider safe, and the page gives the time. But treat the answer as a lower bound and add margin, for three reasons. Leakage and dielectric absorption are not modelled: a capacitor that has been discharged and disconnected can recover a measurable voltage on its own over minutes, which is a genuine hazard in high-voltage work. Real bleeder resistors can fail open without any visible sign. And this model assumes the resistor is actually connected for the whole time. Always verify with a meter before touching terminals rather than relying on a computed time.
Why is discharging to 80% so much faster than charging to 80%?
Because the two mean different things. Charging to 80% of the supply means covering 80% of the whole gap, which takes −ln(0.2) = 1.609 time constants. Discharging to 80% of the starting voltage means losing only 20%, which takes −ln(0.8) = 0.223 time constants. Same circuit, same percentage, and the times differ by more than seven to one. The genuinely symmetrical comparison is the half-way point: charging to 50% and discharging to 50% both take τ ln 2 = 0.693 time constants.
Is a capacitor really fully charged after 5 time constants?
No, and the page prints the exact figure rather than the convention. After five time constants a charging capacitor has reached 1 − e⁻⁵ = 99.3262% of the supply, leaving 0.6738% to go. After ten it is at 99.9955%. It never reaches 100% in the ideal model, and in a real circuit it never gets there at all because leakage sets a floor slightly below the supply. Five time constants is the settled point by convention because 0.67% is usually below anything you care about — but if you are working to a tolerance tighter than that, use the exact target instead of the rule of thumb.
Does the supply voltage change how long charging takes?
Only through the ratio. The time depends on V_t/V_s, not on either voltage separately, so charging to 8 V from a 10 V supply takes exactly as long as charging to 80 V from a 100 V supply. What the supply voltage does change is the current: the initial surge is V_s/R, which for a large capacitor and a small resistor can be enormous and can exceed what the source or the capacitor can survive. The time constant is blind to that, so check the peak current separately when the resistance is small.

References& sources.

  1. [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. [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. [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. [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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