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RC time constant and RLC circuit calculator

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.

更新日 検証済みの例：7

Accepts 4.7k
V
%
Percent of the final value (charging) or of the starting value (discharging)
その他の設定
Auto picks s, ms, µs or ns from the size of τ
試す
Time constant τ
s
Time constant τ: 1 s
有効桁数：6；最も近い値へ、等距離ならゼロから遠い値へ
Time to reach the level
2.30259s
Final value
5V
Energy stored when fully charged
0.00125J

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.

Capacitor voltage over five time constants

01234012345Time (s)Capacitor voltage (V)1τ: 63.2%90% at 2.303 s
計算方法 S

  1. Time constant

    τ=RC=(10,000)(0.0001)=1 s\tau = RC = (10{,}000)(0.0001) = 1\ \mathrm{s}

  2. Rise towards the final value

    x(t)=X∞(1−e−t/τ),X∞=5 Vx(t) = X_\infty\left(1 - e^{-t/\tau}\right),\quad X_\infty = 5\ \mathrm{V}

  3. Time to reach 90%

    t=−τln⁡(1−0.9)=2.30259 st = -\tau\ln(1 - 0.9) = 2.30259\ \mathrm{s}

  4. Stored energy when fully charged

    E=12CV2=0.00125 JE = \tfrac12 CV^2 = 0.00125\ \mathrm{J}

RC time constant and RLC circuit calculatorについて

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".

計算例

10 kΩ and 100 µF charging to 90%

Circuit
RC
Resistance R
10
Resistance unit
kΩ
Capacitance C
100
Capacitance unit
µF
Process
Charging / rising
Source voltage
5 V
Time to reach this level
90%
Time constant τ
1 s
Time to reach the level
2.30259 s
Energy stored when fully charged
0.00125 J

照合元：Python 3.8 math: τ = 10⁴ × 10⁻⁴ = 1 s; t = τ ln 10; E = ½CV² = ½·10⁻⁴·25

Capacitor voltage after one τ

Circuit
RC
Resistance R
10
Resistance unit
kΩ
Capacitance C
100
Capacitance unit
µF
Process
Charging / rising
Source voltage
5 V
Time to reach this level
90%
Value at time
1 s
Value at the chosen time
3.1606 V

照合元：Python 3.8 math: 5(1 − e⁻¹) = 3.1606028

Discharging to half: t = τ ln 2

Circuit
RC
Resistance R
10
Resistance unit
kΩ
Capacitance C
100
Capacitance unit
µF
Process
Discharging / decaying
Source voltage
5 V
Time to reach this level
50%
Time to reach the level
0.693147 s

照合元：Python 3.8 math: ln 2 = 0.6931472 (τ = 1 s)

RL: 100 Ω and 10 mH

Circuit
RL
Resistance R
100
Resistance unit
Ω
Inductance L
10
Inductance unit
mH
Process
Charging / rising
Source voltage
5 V
Time to reach this level
90%
Show times in
s
Value at time
50 µs
Time constant τ
0.0001 s
Final value
0.05 A
Value at the chosen time
0.019673 A

照合元：Python 3.8 math: τ = L/R = 10⁻⁴ s, I = 5/100, i(50 µs) = 0.05(1 − e^−0.5) = 0.0196735

よくある質問

What is the RC time constant?

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.

How long does it take a capacitor to fully charge?

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.

How do you calculate the resonant frequency of an RLC circuit?

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.

What is the Q factor of a series RLC circuit?

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.

What is the time constant of an RL circuit?

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.

「RC time constant and RLC circuit calculator」の精度はどのくらいですか？

精度は入力値と計算方法の前提に依存します。十進演算には有効数字50桁を使いますが、推定、数値計算手法、元データの精度はそれより低い場合があります。表示の丸め処理でこれらの制約がなくなるわけではありません。 独立した出典の解答と照合した計算例：7。 例えば、「10 kΩ and 100 µF charging to 90%」はPython 3.8 math: τ = 10⁴ × 10⁻⁴ = 1 s; t = τ ln 10; E = ½CV² = ½·10⁻⁴·25と照合しています。

この計算方法の出典は何ですか？

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.

この計算機について

τRC=RC, τRL=LR,v(t)=V(1−e−t/τ);f0=12πLC, ∣Z∣=R2+(XL−XC)2, Q=1RLC\tau_{RC} = RC,\ \tau_{RL} = \frac{L}{R},\quad v(t) = V(1 - e^{-t/\tau});\qquad f_0 = \frac{1}{2\pi\sqrt{LC}},\ |Z| = \sqrt{R^2 + (X_L - X_C)^2},\ Q = \frac1R\sqrt{\frac{L}{C}}

出典

  1. OpenStax University Physics Volume 2, §10.5 RC circuits; §14.4 RL circuits; §15.3 RLC series circuits with AC
  2. HyperPhysics — Series RLC resonance and Q

出典と照合済み

この計算機には、独立した出典の解答を使った計算例が 7 件あります。テストに組み込まれており、ここでも実行できます。

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