RC Time Constant – Charging, Discharging and the Cutoff Frequency
Put a resistor in series with a capacitor and you have built the most useful circuit in electronics. Its behaviour is governed by a single number, the time constant τ = R · C, which with R in ohms and C in farads comes out directly in seconds. A 10 kΩ resistor with a 100 µF capacitor gives τ = 1 s; a 4.7 kΩ resistor with 10 µF gives 47 ms. This RC time constant calculator evaluates that product and then solves the full first-order response around it — voltage, current, elapsed time, charge, energy and the filter corner — rearranging the exponential for whichever quantity you leave unknown.
Where 63.2 % comes from
The capacitor voltage follows V(t) = V_f + (V_i − V_f)·e^(−t/τ), which for the usual case of an empty capacitor charging towards a supply collapses to V(t) = V₀·(1 − e^(−t/τ)), and for a discharge to V(t) = V₀·e^(−t/τ). Substituting t = τ makes the exponent exactly −1, so the exponential becomes 1/e ≈ 0.368. That is the whole story behind the famous figure: one time constant leaves 36.8 % of the journey outstanding, so 63.2 % of it is done. The number belongs to the exponential function, not to any particular resistor, which is why every RC network on earth shares it.
The same reasoning gives the rest of the settling table, each row computed from 1 − e^(−n) rather than looked up: 86.47 % at 2τ, 95.02 % at 3τ, 98.17 % at 4τ and 99.33 % at 5τ. Because the curve is asymptotic the capacitor never actually arrives, so 5τ is a convention, not a fact — it simply leaves an error of 0.67 %, smaller than the tolerance of the parts themselves.
Current runs the other way
While the voltage rises, the current falls: I(t) = (V_f − V_i)/R · e^(−t/τ). The instant the switch closes the capacitor is still empty, so the full supply appears across the resistor and the inrush current is simply I₀ = V₀/R — the largest current the circuit will ever see. As the capacitor fills, less voltage is left for the resistor and the current decays on exactly the same exponential as the voltage rises. Sizing the series resistor is therefore as much about surviving t = 0 as about hitting a timing target.
A 4.7 kΩ resistor charging a 10 µF capacitor from 9 V gives τ = 47 ms. After 30 ms the exponent is 30/47 = 0.6383, so V = 9 × (1 − e^(−0.6383)) = 4.246 V and I = (9/4700) × e^(−0.6383) = 1.011 mA. The capacitor holds 42.46 µC and 90.15 µJ at that moment, having covered 47.18 % of its journey.
Reading the same circuit as a filter
Swap to the frequency domain and the identical R and C become a first-order filter with its corner at f_c = 1/(2πRC) = 1/(2πτ). Take the output across the capacitor and you have a low-pass; take it across the resistor and you have a high-pass. A 1 kΩ resistor with 100 nF corners at 1.592 kHz, where the output is 1/√2 of the input — the −3.01 dB that names the point — with 45° of phase shift. Beyond the corner a single pole rolls off at 20 dB per decade, but the asymptote is only accurate well away from the corner, so the calculator uses the exact magnitude 1/√(1 + (f/f_c)²) instead. At 5 kHz, about 3.142 times the corner, that gives −10.362 dB and −72.343°.
Charge, energy and the missing half
A charged capacitor holds Q = C·V and E = ½·C·V². A 1000 µF capacitor at 24 V therefore carries 24 mC and 288 mJ, and through a 10 Ω series resistor it draws a 2.4 A inrush. There is a well-known surprise hiding here: the supply had to push Q coulombs through V volts, delivering CV² joules, yet the capacitor only keeps half of it. The resistor dissipates the other half — always exactly half, whatever the resistance. A small resistor burns it quickly and a large one slowly, but no series resistor beats 50 % efficiency.
τ = RC rises with both factors, so the worst cases are the all-low and all-high corners. A 10 kΩ ±5 % resistor with a 100 µF ±20 % electrolytic gives a nominal 1 s that can legitimately land anywhere between 0.76 s and 1.26 s. Never trust a bare RC network for precision timing — use a crystal, or design for the band.
Designing to a target
Most real work runs the formula backwards. Given a required delay, R = τ/C and C = τ/R size the missing part: a 2.5 s time constant with a 47 µF capacitor calls for 53.19 kΩ, which is not a value anyone stocks. The nearest E24 part is 51 kΩ, delivering 2.397 s — a −4.12 %error, comfortably inside the capacitor’s own tolerance. To hit a specific voltage rather than a specific τ, invert the exponential: t = −τ·ln(1 − V_target/V₀) for charging and t = −τ·ln(V_target/V₀) for discharging, so reaching 90 % of a 5 V rail through 220 kΩ and 1 µF takes 0.507 s. Useful shortcuts fall out of the same logarithm: the half-life is τ·ln2 ≈ 0.693τ and the 10–90 % rise time that oscilloscopes measure is τ·ln9 ≈ 2.197τ.