RC and RL time constant and the time to charge or discharge to any percent; series RLC resonant frequency, impedance, phase, Q factor and bandwidth.
τ = 1 s: after one τ the capacitor voltage has risen to 63.2% of 5 V, and it takes 2.303 s to reach 90%. After 5τ it is within 1% of the final value.
An RC circuit responds with time constant τ = RC and an RL circuit with τ = L/R: after one τ the capacitor voltage or inductor current has covered 63.2% of the way to its final value, and reaching a fraction p takes t = −τ ln(1 − p). For a series RLC circuit the calculator finds the resonant frequency f₀ = 1/(2π√(LC)), the impedance and phase at your signal frequency from the reactances 2πfL and 1/(2πfC), and Q = (1/R)√(L/C).
The default, 10 kΩ with 100 µF on a 5 V supply, gives τ = 1 s and 2.30 s to reach 90%. RC time constants set timer delays, switch debounce periods and filter cut-off frequencies, 1/(2πRC). With 10 Ω, 10 mH and 1 µF in series, f₀ = 1,591.55 Hz and Q = 10.
Components are ideal and the input is a step: no capacitor leakage and no inductor winding resistance beyond R. After 5τ the response is within 0.67% of its final value (e⁻⁵), the usual rule for "fully charged".
Checked against: Python 3.8 math: τ = 10⁴ × 10⁻⁴ = 1 s; t = τ ln 10; E = ½CV² = ½·10⁻⁴·25
Checked against: Python 3.8 math: 5(1 − e⁻¹) = 3.1606028
Checked against: Python 3.8 math: ln 2 = 0.6931472 (τ = 1 s)
Checked against: Python 3.8 math: τ = L/R = 10⁻⁴ s, I = 5/100, i(50 µs) = 0.05(1 − e^−0.5) = 0.0196735
τ = R × C, in seconds when R is in ohms and C in farads. It is the time a charging capacitor takes to reach 63.2% (1 − e⁻¹) of the supply voltage, or a discharging one to fall to 36.8%. 10 kΩ with 100 µF gives τ = 1 s; 1 kΩ with 1 µF gives 1 ms.
About five time constants. After 5τ the voltage is 99.33% of the supply (1 − e⁻⁵), which engineers treat as fully charged; in theory it never quite gets there. Reaching 90% takes τ ln 10 ≈ 2.303τ and 99% takes τ ln 100 ≈ 4.605τ, so a 10 kΩ, 100 µF circuit reaches 99% in 4.6 s.
f₀ = 1/(2π√(LC)), with L in henries and C in farads; in a series circuit R does not shift it. 10 mH with 1 µF resonates at 1,591.55 Hz, and 100 µH with 100 pF at 1.59 MHz. At f₀ the inductive and capacitive reactances are equal and cancel, so the impedance falls to R alone and the current peaks.
Q = (1/R)√(L/C), the ratio of either reactance at resonance to the resistance. It sets the −3 dB bandwidth, Δf = f₀/Q: 10 Ω, 10 mH and 1 µF give Q = 10 and a 159 Hz band around 1,592 Hz. Above Q = 0.5 the circuit rings (underdamped); exactly 0.5 is critically damped, reached here at R = 2√(L/C) = 200 Ω.
τ = L/R. It is the time for the current to rise to 63.2% of its final value V/R after the switch closes, or to fall to 36.8% once the source is removed and the coil discharges through R. 10 mH with 100 Ω gives τ = 100 µs, so on 5 V the current reaches 99% of 50 mA in about 460 µs.
Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 7 worked examples whose answers come from independent sources; for example, “10 kΩ and 100 µF charging to 90%” is checked against Python 3.8 math: τ = 10⁴ × 10⁻⁴ = 1 s; t = τ ln 10; E = ½CV² = ½·10⁻⁴·25.
OpenStax University Physics Volume 2, §10.5 RC circuits; §14.4 RL circuits; §15.3 RLC series circuits with AC; HyperPhysics — Series RLC resonance and Q.
7 worked examples with independently sourced answers ship with this calculator. They run in the test suite; you can run them here too.
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