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.
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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".
Multiply resistance by capacitance: τ = R × C, in seconds when R is in ohms and C in farads. For 10 kΩ and 100 µF, τ = 10,000 × 0.0001 = 1 s. A capacitor charging through that resistor reaches 63.2% of the supply after one τ and counts as full after five time constants (5 s).
Charging from zero, the capacitor voltage follows V(t) = Vs(1 − e^(−t/τ)). Discharging from a starting voltage V₀, it follows V(t) = V₀ e^(−t/τ). Each time constant closes 63.2% of the gap that remains:
The ideal curve never reaches 100%, which is why 5τ is the working definition of fully charged or fully discharged. OpenStax derives both curves.
Solve the curve for time, with the level p written as a fraction:
Charging to 90% takes τ ln 10 = 2.303τ, exactly as long as discharging to 10% left. Reaching 99% takes τ ln 100 = 4.605τ, and the halfway point comes at τ ln 2 = 0.693τ.
On a 5 V supply, the 10 kΩ, 100 µF circuit reaches 4.5 V (90%) after 2.303 s and 4.95 V (99%) after 4.605 s. One second in, it sits at 5 × (1 − e^(−1)) = 3.16 V.
In an RL circuit the current, not a voltage, rises along the same curve toward I = V/R, with τ = L/R.
For 100 Ω and 10 mH on 5 V, τ = 0.01 ÷ 100 = 100 µs. The current heads for 50 mA, is 19.67 mA after 50 µs and reaches 99% after about 460 µs.
Resistance acts in opposite directions in the two circuits. A larger R slows an RC circuit, because τ = RC grows, but speeds up an RL circuit, because τ = L/R shrinks. The coil's winding resistance is part of R (OpenStax on RL circuits).
The first row is the 10 kΩ, 100 µF circuit charging to 90%. The second discharges the same pair to 10% remaining. The third is 10 Ω with 10 mH, rising to 90% of its final current.
| Scenario | Time constant | Time to target |
|---|---|---|
| Default | 1 s | 2.30259 s |
| RC decay | 1 s | 2.30259 s |
| RL rising | 1 ms | 2.30259 ms |
All three take 2.30259τ, because charging to 90% and decaying to 10% both cover nine tenths of the change.
A charged capacitor holds E = ½CV², and an inductor carrying its final current holds E = ½LI² with I = V/R. At 5 V, 100 µF stores ½ × 0.0001 × 25 = 1.25 mJ. The 10 mH coil at 50 mA stores ½ × 0.01 × 0.05² = 12.5 µJ.
A series RLC circuit driven by a sine wave resonates at f₀ = 1/(2π√(LC)). There the inductive reactance XL = 2πfL equals the capacitive reactance XC = 1/(2πfC), the two cancel, and the impedance falls to R alone.
At any other frequency, the impedance is |Z| = √(R² + (XL − XC)²) and the phase is φ = arctan((XL − XC)/R). Worked through for 10 Ω, 10 mH and 1 µF:
Below resonance the circuit is capacitive: the phase is negative and current leads voltage. At 2 kHz the same parts are inductive, with |Z| = 47.16 Ω and φ = +77.76°, so voltage leads current. At f₀ the current peaks at 5 ÷ 10 = 0.5 A.
The quality factor is Q = (1/R)√(L/C), and the −3 dB bandwidth is f₀/Q. In the example, Q = (1/10) × √(0.01/0.000001) = 10, so the peak is 159.15 Hz wide. R leaves f₀ where it is; it sets how sharp the peak is.
The same parts decide how the circuit responds to a step. Compare α = R/(2L) with ω₀ = 1/√(LC):
HyperPhysics and OpenStax treat series resonance in more depth.
The same τ sets the −3 dB cut-off of a first-order RC low-pass or high-pass filter: fc = 1/(2πRC). 10 kΩ with 100 nF gives τ = 1 ms and a cut-off of 159.15 Hz.
Reduce several resistors or capacitors to one equivalent with the series and parallel calculator before working out τ. The Ohm's law calculator gives the V/R current and power.
Checked against: τ = 10⁴ × 10⁻⁴ = 1 s; t = τ ln 10; E = ½CV² = ½·10⁻⁴·25
Checked against: 5(1 − e⁻¹) = 3.1606028
Checked against: ln 2 = 0.6931472 (τ = 1 s)
Checked against: Independent calculation
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.
This calculator includes 7 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.