Capacitor Charge Time Calculator
Calculates how long a capacitor takes to charge fully through a resistor, using the RC time constant. A capacitor is considered effectively fully charged after five time constants (5 × R × C).
Last updated: May 2026
Compare with similar
About this calculator
When a capacitor charges through a resistor, it does not fill instantly — it follows an exponential curve governed by the RC time constant, written as the Greek letter tau (τ) and equal to resistance × capacitance. After one time constant the capacitor reaches about 63.2% of the supply voltage; after two, 86.5%; after three, 95%; after four, 98.2%; and after five time constants it reaches about 99.3%, which engineers treat as "fully charged" for practical purposes. That is why full charge time ≈ 5 × R × C. This calculator takes resistance in ohms and capacitance in microfarads (the most common practical unit), converts the capacitance to farads by dividing by one million, and multiplies by five. For a 1,000 Ω resistor and a 1,000 µF capacitor, one time constant is 1 second (1000 × 0.001), so full charge takes about 5 seconds. The same 5RC rule describes discharge: after five time constants a charging capacitor is nearly full and a discharging one is nearly empty. RC timing underpins countless circuits — timer and oscillator circuits, debounce filters for switches, power-supply soft-start, camera flashes, and the low-pass and high-pass filters that shape audio and signals. Keep units consistent: ohms with farads gives seconds, so the microfarad-to-farad conversion is essential. Note this model assumes a simple series resistor-capacitor charged from a constant voltage; real circuits with multiple components, leakage, or non-ideal sources will deviate, and very large electrolytic capacitors have tolerances of ±20% or more that affect the actual time.
How to use
Example 1 — Timing circuit. A 1,000 Ω resistor charges a 1,000 µF capacitor. Enter 1000 and 1000. Result: 5 seconds. Verify: 1000 µF = 0.001 F; τ = 1000 × 0.001 = 1 s; 5 × 1 = 5 s. ✓ Example 2 — Fast filter. A 470 Ω resistor charges a 10 µF capacitor. Enter 470 and 10. Result: about 0.0235 seconds (23.5 ms). Verify: 10 µF = 0.00001 F; τ = 470 × 0.00001 = 0.0047 s; 5 × 0.0047 = 0.0235 s. ✓ Such a short time constant is typical for signal-filtering rather than visible timing.
Frequently asked questions
Why is a capacitor considered fully charged after 5 time constants?
Capacitor charging is exponential, so technically it never reaches 100% — it just gets ever closer. After one time constant (τ = R × C) it reaches 63.2%, and after five it reaches about 99.3%, which is close enough that the remaining 0.7% is negligible for virtually all practical purposes. Engineers therefore adopt the "5RC" convention as the effective full-charge time. The same logic applies to discharge: after five time constants the voltage has fallen to about 0.7% of its starting value. If your application needs a tighter tolerance, you can use more time constants, but the gains shrink rapidly because each additional τ only closes part of the remaining gap.
What units do I need to enter?
Enter resistance in ohms and capacitance in microfarads (µF), the unit most capacitors are labelled in. The calculator converts microfarads to farads internally by dividing by one million, because the time-constant formula requires farads to produce an answer in seconds. If your capacitor is rated in nanofarads (nF) or picofarads (pF), convert first: 1,000 nF = 1 µF, and 1,000,000 pF = 1 µF. Likewise convert kilo-ohms to ohms (1 kΩ = 1,000 Ω) and megohms to ohms (1 MΩ = 1,000,000 Ω). Getting the units right is the single most important step — a microfarad/farad slip changes the answer by a factor of a million.
Does this work for capacitor discharge too?
Yes. The time constant τ = R × C governs both charging and discharging, and the 5RC rule applies equally: after five time constants a discharging capacitor has dropped to about 0.7% of its initial voltage, effectively empty. The exponential shape is mirrored — charging rises toward the supply voltage while discharging falls toward zero, but both reach the 99.3%/0.7% mark at five τ. So the number this calculator gives you is also a good estimate of how long a capacitor takes to discharge through the same resistance. This matters for safety: high-voltage capacitors can hold a dangerous charge long after power is removed, and the discharge time tells you how long bleeder resistors need.
What mistakes do people make with RC timing?
The most frequent is a unit error — entering capacitance in farads when it is actually microfarads, or leaving resistance in kilo-ohms — which throws the result off by orders of magnitude. Another is ignoring component tolerance: electrolytic capacitors are often ±20%, so the real charge time can differ noticeably from the calculated value. People also forget that the formula assumes a single series resistor and an ideal constant-voltage source; additional parallel resistance, the source’s internal resistance, or capacitor leakage all change the effective time constant. Finally, some assume the capacitor reaches full supply voltage at exactly 5τ — it reaches 99.3%, which is close but not literally 100%.
When does the simple 5RC model not apply?
The model assumes a basic series resistor-capacitor circuit charged from a constant DC voltage with ideal components. It breaks down when the source has significant internal resistance, when multiple resistors and capacitors interact (you must first reduce them to an equivalent R and C), or when the supply voltage is not constant. It also does not describe AC behaviour, where you need capacitive reactance and impedance instead. Non-ideal effects — dielectric absorption, leakage current, and large tolerances in electrolytics and supercapacitors — make the real charge time deviate. For precision timing, designers measure the actual circuit or use dedicated timer ICs rather than relying on nominal component values.