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.
Mis à jour Exemples vérifiés : 7
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".
Source de vérification : Python 3.8 math: τ = 10⁴ × 10⁻⁴ = 1 s; t = τ ln 10; E = ½CV² = ½·10⁻⁴·25
Source de vérification : Python 3.8 math: 5(1 − e⁻¹) = 3.1606028
Source de vérification : Python 3.8 math: ln 2 = 0.6931472 (τ = 1 s)
Source de vérification : 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.
La précision dépend de vos données et des hypothèses de la méthode. Le calcul décimal utilise 50 chiffres significatifs, mais les estimations, méthodes numériques et données sources peuvent être moins précises ; l’arrondi affiché ne supprime pas ces limites. Exemples résolus vérifiés à partir de sources indépendantes : 7. Par exemple, « 10 kΩ and 100 µF charging to 90% » est vérifié à l’aide de 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.
Ce calculateur comprend 7 exemples résolus dont les réponses proviennent de sources indépendantes. Ils font partie de la suite de tests et peuvent aussi être exécutés ici.
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