Calcylator
RC Circuits

The RC time constant:
how fast a capacitor charges or discharges

Know what τ means in time, how far a capacitor has charged after each one, and how to turn it into a timer or a filter.

Calcylator Editorial Team

Updated · 5 min read

What the time constant measures

When you connect a resistor and a capacitor in series to a supply, the capacitor does not fill instantly. Charge flows through the resistor, so a bigger resistance or a bigger capacitor makes the process slower. The product of the two is a time, and it gives the natural pace of the circuit.

Time constant =τ = R × C
R:
resistance in ohms (Ω)
C:
capacitance in farads (F)
τ:
time constant in seconds
Ohms times farads equals seconds.
  • R

    10 kΩ = 10,000 Ω

  • C

    100 µF = 0.0001 F

Time constant

1 second

10,000 × 0.0001 = 1.0 s.

The charging curve and the 63 percent rule

Voltage across a charging capacitor follows an exponential curve towards the supply voltage Vs.

Charging voltage =V(t) = Vs × (1 − e^(−t ÷ τ))
Vs:
supply voltage
t:
time since the switch closed, in seconds
e:
Euler's number, about 2.718
TimeCharge reachedVoltage on a 5 V supply
1 τ63.2%3.16 V
2 τ86.5%4.32 V
3 τ95.0%4.75 V
4 τ98.2%4.91 V
5 τ99.3%4.97 V

The curve is steep at first and then flattens, because the current through the resistor shrinks as the capacitor voltage approaches the supply. Theoretically it never quite arrives, which is why engineers quote five time constants as full for most purposes.

Discharging and time to reach any level

A capacitor discharging through the resistor decays by the same pattern in reverse: V(t) = V₀ × e^(−t ÷ τ). After one τ it holds 36.8 percent of its starting voltage, which is the complement of 63.2 percent.

Time to reach a fraction x of the supply =t = −τ × ln(1 − x)
x:
target fraction, for example 0.9 for 90%
  • τ

    1 s

  • Target

    90% of the supply

Time needed

≈ 2.3 s

−1 × ln(1 − 0.9) = −ln(0.1) = 2.303 s, or 2.3 τ.

This form is useful for designing delays: choose the threshold of the next stage, work out the fraction, and solve for t.

Choosing R and C for a delay

Design starts with the delay you want and the voltage level at which the next stage reacts. A microcontroller input that switches at about 70 percent of its supply sees the capacitor reach that point after −ln(1 − 0.7) = 1.2 time constants.

  • Wanted delay

    0.5 s to reach 90%

  • Capacitor chosen

    10 µF

Resistor

≈ 22 kΩ

τ = 0.5 ÷ 2.303 = 0.217 s; R = 0.217 ÷ 0.00001 = 21.7 kΩ. The nearby standard 22 kΩ gives τ = 0.22 s and a delay of 0.507 s.

Choose C first, because capacitor values are coarser and more expensive to vary, then trim R. Keep R large enough not to load the source, and C small enough that leakage does not dominate.

Using τ for delays and filters

A reset delay is a direct application. With 4.7 kΩ and 10 µF, τ = 4,700 × 0.00001 = 0.047 s, or 47 ms, so a signal reaches about 63 percent of final value in about 47 ms and settles in roughly 235 ms.

The same RC pair also forms a simple low-pass filter. The cutoff frequency, where output is about 70.7 percent of input, is f_c = 1 ÷ (2π × R × C). For the 10 kΩ and 100 µF pair that is 1 ÷ (2π × 1) ≈ 0.16 Hz, so only very slow changes pass.

  • Larger τ means smoother, slower response and lower cutoff frequency.
  • Smaller τ means faster response but more noise passing through.

Real-world limits and check points

  • Capacitor tolerance can be ±20 percent for electrolytic types, so real timing can shift by that much.
  • Leakage current and temperature change effective values, which matters for long delays.
  • The source and any load add their own resistance, changing the effective R.
  • Voltage across a capacitor must stay under its rating, and electrolytics are polarised.

Compute τ with a calculator, then use the 63 percent and five-τ rules to estimate times without further maths.

Current at switch-on and where the energy goes

At the moment of closing the switch the capacitor is empty and the whole supply sits across the resistor. With 5 V and 10 kΩ the starting current is 5 ÷ 10,000 = 0.5 mA, and it falls away exponentially with the same time constant.

The finished capacitor holds ½ × C × V² = ½ × 100 µF × 25 = 1.25 mJ. The supply delivered twice that, 2.5 mJ, and the other 1.25 mJ was lost as heat in the resistor, whatever the resistance. A larger resistor only makes the process slower, not more efficient.

This is also why a very small resistance produces a large inrush current, so supplies and switches need to be rated for it.

Everyday circuits that rely on τ

  • Switch debounce: a 10 kΩ resistor with a 100 nF capacitor gives τ = 1 ms, so the contact bounce of a few milliseconds is smoothed before it reaches the input.
  • Power-on reset: a slow RC rise holds a chip in reset until the supply has settled, using the threshold method above to set the delay.
  • Bleeder resistor: a 100 kΩ resistor across a 1,000 µF capacitor gives τ = 100 s, so about 500 s, or just over eight minutes, to discharge safely after power-off.
  • Sensor smoothing: a low-pass RC on a noisy analogue line trades response speed for steadier readings.

Reading τ from a scope trace is straightforward. Capture the charge curve, mark the final level, and find the time at which the signal reaches 63.2 percent of the total change. That time is τ, and dividing by R gives C, a useful way to check the value of an unmarked capacitor.

In each case the time constant sets the speed of the response, and the choice between larger or smaller τ is a trade between noise rejection and responsiveness.

Remember finally that the same mathematics describes many other systems. A cooling cup of tea, a thermometer settling to room temperature and a battery voltage recovering after load all follow a curve of this shape, with a time constant of their own. Recognising the pattern makes the 63 percent and five-constant rules useful well beyond electronics, and it is a quick way to estimate when a slowly settling reading can be trusted.

Common questions

What is the RC time constant formula?

The time constant is τ = R × C, with resistance in ohms and capacitance in farads giving seconds. A 10 kΩ resistor with a 100 µF capacitor gives 1 second. It sets how fast the circuit charges.

Why is the time constant 63 percent?

After one τ, a charging capacitor reaches 1 − e⁻¹ ≈ 63.2 percent of the supply voltage. This comes from the exponential charging curve. A discharging one falls to 36.8 percent in one τ.

How long does it take to fully charge a capacitor?

In theory never, but about five time constants brings it within 1 percent of the supply, at 99.3 percent. With τ = 1 s, that is roughly 5 seconds. Four time constants gives 98.2 percent.

How do I calculate time to reach a certain voltage?

Use t = −τ × ln(1 − x), where x is the fraction of supply voltage. For 90 percent, t = 2.303 × τ. With τ = 1 s, that is about 2.3 seconds.

What is the cutoff frequency of an RC filter?

It is f = 1 ÷ (2π × R × C). For 10 kΩ and 100 µF that is about 0.16 Hz. At this frequency the output is about 70.7 percent of the input amplitude, which is −3 dB.

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