Why is one time constant equal to 63.2% charge?

After a time equal to τ, the exponential term e^(−1) ≈ 0.368 remains, so the capacitor has charged to 1 − 0.368 = 0.632, or 63.2% of the supply voltage. This is a fixed property of the exponential curve regardless of the component values.

How long until the capacitor is fully charged?

Strictly the capacitor never reaches 100%, but after 5τ it is at about 99.3%, which is treated as fully charged for practical purposes. That is why the calculator reports 5τ as the full-charge time.

Does the same time constant apply to discharging?

Yes. During discharge the voltage follows V(t) = V₀·e^(−t/τ) with the same τ = R·C, falling to 36.8% of its starting value after one time constant.

What if I change the resistor or capacitor?

The time constant scales linearly with both, so doubling either R or C doubles τ and the charge time. Increase resistance or capacitance to slow the circuit down, or reduce them to speed it up.

RC Time Constant Calculator

Resistance

Capacitance

Capacitance Unit

Source Voltage

Elapsed Time

Time Constant (τ)0.100000 s

Charge to 63.2% (1τ)0.100000 s

Full Charge (~5τ)0.500000 s

Charge at Elapsed Time63.21%

Voltage at Elapsed Time3.1606 V

When a resistor charges a capacitor, the voltage rises along an exponential curve governed by the RC time constant. This tool computes that time constant (τ = R·C), shows how long it takes to reach the characteristic 63.2% and roughly 99% charge levels, and reports the exact charge percentage and capacitor voltage at any elapsed time you specify. It is the go-to calculation for timing circuits, debounce filters, and power-up delays.

Formula

τ = R · C ; V(t) = V₀ · (1 − e^(−t/τ))

τ: Time constant (seconds)
R: Series resistance (ohms)
C: Capacitance (farads)
V₀: Source voltage (volts)
t: Elapsed charging time (seconds)

How it works

Enter the series resistance in ohms and the capacitance with its unit (F through pF); the calculator multiplies them to find τ.
Enter the source voltage and an elapsed time in seconds.
It applies V(t) = V₀·(1 − e^(−t/τ)) to report the charge fraction and capacitor voltage at that instant, and notes that about 5τ are needed for a practically full charge.

Worked example

A 1 kΩ resistor charges a 100 µF capacitor from a 5 V source; what is the state after 0.1 s?

τ = 1000 × 0.0001 = 0.1 s.
Elapsed time 0.1 s equals exactly one time constant.
Charge fraction = 1 − e^(−1) = 0.6321, so 63.21%.
Capacitor voltage = 5 × 0.6321 ≈ 3.1606 V.

τ = 0.1 s, the capacitor is 63.21% charged, sitting at about 3.16 V after 0.1 s.

Frequently asked questions

Why is one time constant equal to 63.2% charge?: After a time equal to τ, the exponential term e^(−1) ≈ 0.368 remains, so the capacitor has charged to 1 − 0.368 = 0.632, or 63.2% of the supply voltage. This is a fixed property of the exponential curve regardless of the component values.
How long until the capacitor is fully charged?: Strictly the capacitor never reaches 100%, but after 5τ it is at about 99.3%, which is treated as fully charged for practical purposes. That is why the calculator reports 5τ as the full-charge time.
Does the same time constant apply to discharging?: Yes. During discharge the voltage follows V(t) = V₀·e^(−t/τ) with the same τ = R·C, falling to 36.8% of its starting value after one time constant.
What if I change the resistor or capacitor?: The time constant scales linearly with both, so doubling either R or C doubles τ and the charge time. Increase resistance or capacitance to slow the circuit down, or reduce them to speed it up.

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