Audited 31 Jul 2026·Last updated 28 Oct 2026·5 citations·Tier 1·0 uses

RC Circuit Calculator — Time Constant, Cutoff Frequency and Filter Response

Find an RC network's time constant, corner frequency, capacitor voltage after a step, 10–90 % rise time, and low- or high-pass gain and phase at any frequency.

RC Circuit Calculator

The series resistance of the network. If a source has significant output impedance, add it here — a 600 Ω signal generator driving a 1 kΩ resistor is a 1.6 kΩ network, not a 1 kΩ one.
Resistance unit
Nominal value from the part marking. Class-2 ceramics (X7R, Y5V) lose a large fraction of this under DC bias and with age, so the real corner frequency of a built circuit routinely sits well above the calculated one.
Capacitance unit
The voltage the step goes to when charging, or the voltage the capacitor starts at when discharging. Zero and negative values are accepted; the percentage output does not depend on this field.
V
Measured from the instant the step arrives. Zero is legal and returns the starting condition.
Time unit
Transient
The frequency at which the gain and phase are reported. It has no effect on the time constant or the transient — those are properties of the network, not of the signal.
Hz
Filter arrangement
Time constant τ
1
τ = R × C in milliseconds: the time a charging capacitor takes to reach 63.2 % of the step, or a discharging one to fall to 36.8 % of its starting voltage.
Cutoff (corner) frequency
159.1549 Hz
Capacitor voltage at t
3.1606 V
Percent of the step
63.21
10 %–90 % rise time
2.1972 ms
Settling time (5τ)
5 ms
Gain at the signal frequency
-16.0722 dB
Gain as a ratio
0.1572
Phase shift
-80.9569°
Reading of the result
At 1000 Hz this low-pass (output taken across the capacitor) network has a corner frequency of 159.1549 Hz and passes 0.157177 of the input (-16.0722 dB), with the output lagging the input by 80.9569°. That is past the corner, in the −3 dB to −20 dB transition region where a first-order network rolls off at 20 dB per decade. Component tolerance dominates this figure in practice: a ±5 % resistor with a ±10 % ceramic moves the corner frequency by roughly ±15 %, which is far larger than any digit shown here.

Background.

How long does a capacitor take to charge through a resistor? The answer turns on a single product. Multiply the resistance in ohms by the capacitance in farads and you get the time constant τ, in seconds — an ohm times a farad is exactly a second. A 10 kΩ resistor feeding a 100 nF capacitor gives 1 ms; swap the capacitor for 1 µF and it becomes 10 ms. After one τ the capacitor has covered 63.2 % of the way to the supply, after five it is within 0.7 % of it, and in theory it never quite arrives.

The same pair of parts also sorts signals by frequency. A capacitor's opposition to alternating current, its reactance, falls as frequency rises, so a resistor and capacitor in series form a voltage divider whose ratio depends on frequency. Take the output across the capacitor and slow signals pass while fast ones are shunted away: a low-pass filter. Take it across the resistor and the reverse happens: a high-pass filter. Either way the dividing line sits at 1/(2πRC), which for the 1 ms pair is 159.15 Hz. Time constant and corner frequency are one fact written in two languages, which is why the capacitor that slows a logic edge is also the one that strips high-frequency noise from it.

One set of entries gives both views: τ and the corner, the capacitor voltage at any moment after a step, the 10–90 % rise time, the five-τ settling time, and the gain and phase at a frequency you choose.

All of it assumes ideal parts — a source with no internal resistance, no load on the output, and a capacitor exactly at its marked value. Capacitors are where real boards stray most, since a common ceramic can lose much of its printed capacitance once a DC voltage sits across it; the tolerance notes below show how far a built circuit may land from the figures.

What is rc circuit calculator?

An RC circuit is any arrangement in which current through a resistor charges or discharges a capacitor. Because it stores energy in only one place, the capacitor's electric field, it is a first-order system: its response to a sudden change is a single exponential, and its response to a sine wave has a single corner. The time constant τ = RC is the time needed to close 63.2 % of whatever gap remains between the capacitor's voltage and its final value. That proportion holds at every point on the curve, so the second τ closes 63.2 % of the remaining 36.8 %, bringing the total to 86.5 %, and so on. The corner frequency f_c = 1/(2πRC) is where the output voltage has fallen to 1/√2, about 0.7071, of the input, so the output power is half the input power. You meet RC networks as switch debouncers, power-on reset delays, supply decoupling, anti-aliasing filters ahead of an analogue-to-digital converter, coupling capacitors between amplifier stages, the timing element of a 555 timer, and snubbers across relay contacts and switching transistors.

How to use this calculator.

  1. Enter the resistance and pick Ω, kΩ or MΩ. If the source driving the network has its own output resistance, add it to the resistor: a 600 Ω signal generator feeding a 1 kΩ resistor behaves as a 1.6 kΩ network.
  2. Enter the capacitance and its unit as marked on the part. Small ceramics are often coded in picofarads with three digits — 104 means 10 followed by four zeros, 100,000 pF, which is 100 nF or 0.1 µF — and European schematics write the same part as 100n.
  3. For a timing question, set the step voltage, the time since the step with its unit, and whether the capacitor is charging from 0 V or discharging from the supply. A negative or zero step is accepted, and the percentage output ignores the voltage entirely.
  4. For a filter question, enter the signal frequency in hertz and choose low-pass (output across the capacitor) or high-pass (output across the resistor). The frequency has no effect on τ or on any of the transient results.
  5. Compare the 10–90 % rise time, not τ, against a scope reading or a datasheet edge figure; it is 2.1972 times longer. Use the five-τ figure when the capacitor must have effectively finished moving, such as the wait before reading a sensor after power-up.

The formula.

τ = R·C f_c = 1 ⁄ (2πRC) V(t) = V₀(1 − e^(−t/τ)) |H| = 1 ⁄ √(1 + x²)

Charging. Around the loop, the step voltage V₀ splits between the resistor and the capacitor: V₀ = iR + v, where v is the capacitor voltage. The capacitor's current is i = C·dv/dt, so V₀ = RC·dv/dt + v. That first-order differential equation, starting from 0 V, has the solution v(t) = V₀(1 − e^(−t/τ)) with τ = RC. Discharging from V₀ through the same resistor gives v(t) = V₀·e^(−t/τ). These are equations 10.8, 10.9 and 10.11 in §10.5 of OpenStax University Physics Volume 2. The percentage output is simply (1 − e^(−t/τ)) or e^(−t/τ) multiplied by 100, which is why it does not depend on the supply.

Rise time. Setting v/V₀ to 0.1 and to 0.9 and subtracting the two times gives t₉₀ − t₁₀ = τ·ln(0.9/0.1) = τ·ln 9 ≈ 2.1972τ. The same algebra answers any threshold question: the time to reach a fraction p of the step is τ·ln(1/(1 − p)).

Frequency response. A capacitor's reactance is X_C = 1/(2πfC). With the output across the capacitor, the divider ratio has magnitude 1/√(1 + (f/f_c)²), where f_c = 1/(2πRC) is the frequency at which X_C equals R. With the output across the resistor, f and f_c swap places. Writing x for f/f_c (low-pass) or f_c/f (high-pass), both cases become |H| = 1/√(1 + x²), and the gain in decibels is −10·log₁₀(1 + x²), Fiore's equation 10.14. The low-pass output lags the input by arctan(f/f_c); the high-pass output leads it by arctan(f_c/f). At f = f_c, x = 1, the gain is 1/√2 (−3.0103 dB) and the phase is 45° in either direction.

Two consequences fall out of the algebra. The squared low-pass and high-pass gains always add to exactly 1, and their phases are always 90° apart, so the two outputs split the input's power between them. And because f_c = 1/(2πτ), rise time and bandwidth are locked together: t_r = ln 9/(2πf_c) ≈ 0.35/f_c, the rule used to read an oscilloscope's bandwidth as an edge speed.

The model leaves out the capacitor's equivalent series resistance and inductance, which take over at high frequency; dielectric absorption, which lets a discharged capacitor creep back up a little; leakage; the source's output resistance unless you add it to R; and any load on the output node, which forms a second divider.

A worked example.

Example

Take the defaults: a 10 kΩ resistor, a 100 nF capacitor, a 5 V step, and the question of where the capacitor stands 1 ms later while charging. In base units R = 10,000 Ω and C = 0.0000001 F, so τ = 10,000 × 0.0000001 = 0.001 s, or 1 ms. The corner frequency is 1 ÷ (2π × 0.001) = 159.1549 Hz. The 1 ms elapsed time is exactly one time constant, so the exponential term is e⁻¹ = 0.3679 and the capacitor has covered 1 − 0.3679 = 0.6321 of the step: 5 × 0.6321 = 3.1606 V, or 63.2121 % of the supply. The rise from 10 % to 90 % takes 1 ms × ln 9 = 2.1972 ms, and the five-τ settling time is 5 ms, by which point the capacitor is at 99.3262 % of 5 V. On the filter side, a 1,000 Hz signal against a 159.1549 Hz corner gives f/f_c = 6.2832, which is exactly 2π because f/f_c = 2πfRC and f × R × C here is 1,000 × 0.001 = 1. Then 1 + 6.2832² = 40.4784, its square root is 6.3623, and the low-pass gain is 1 ÷ 6.3623 = 0.1572, or −16.0722 dB. The output lags by arctan(6.2832) = 80.9569°. In practical terms a 1 V sine at 1 kHz leaves this low-pass network at about 0.157 V, a little under a sixth of its amplitude and nearly a quarter-cycle late. That places it between the −3 dB corner and the −20 dB mark a single RC section reaches one decade higher, at 1,591.5 Hz.

capacitance100
elapsed Time Unitms
transient Modecharging
signal Frequency1,000
resistance Unitkilohm
filter Typelow-pass
supply Voltage5
capacitance UnitnF
resistance10
elapsed Time1

Frequently asked questions.

How long does a capacitor take to fully charge through a resistor?
Mathematically it never finishes: each τ closes 63.2 % of whatever gap is left, so the gap shrinks without reaching zero. Engineers call it done at five time constants, 99.3262 % of the step, which is the settling figure shown. The stages along the way are 63.2 % at 1τ, 86.5 % at 2τ, 95.0 % at 3τ and 98.2 % at 4τ. Precision work counts further: 7τ reaches 99.909 %, and settling to within half a step of a 12-bit converter needs ln 8192 ≈ 9.0τ.
How do I work out the time to reach a particular capacitor voltage?
Rearrange the charging curve: t = τ × ln(V₀ ÷ (V₀ − V)), where V₀ is the step and V the target. Reaching half the step takes τ × ln 2 = 0.693τ. Reaching two-thirds takes τ × ln 3 = 1.0986τ, which is where the familiar 1.1RC pulse length of a 555 timer in monostable mode comes from, since its comparator trips at two-thirds of the supply. For a discharging capacitor falling to V, use t = τ × ln(V₀ ÷ V).
Why is the 10–90 % rise time longer than the time constant?
Because it measures a different stretch of the curve. τ runs from the moment of the step to the 63.2 % point; the rise time skips the fast first 10 % but continues through the slow climb from 63 % to 90 %, and that climb costs more time than the skipped start saves. The result is τ × ln 9 = 2.1972τ. Scope specifications and logic datasheets quote edges as 10–90 % times, so using τ in their place understates an edge by a factor of about 2.2.
Where does the rule 'rise time ≈ 0.35 ÷ bandwidth' come from?
From this network. For a single RC pole the 10–90 % rise time equals ln 9 ÷ (2π × f_c), and ln 9 ÷ 2π is 0.3497, so rise time ≈ 0.35 ÷ f_c. A 159.15 Hz corner rises in 2.1972 ms; a single-pole 100 MHz front end rises in about 3.5 ns. The constant is exact only for one pole; instruments with sharper roll-offs above their bandwidth use a somewhat larger figure, so check the maker's number for a specific scope.
Why is the corner called the −3 dB point, and how can I find it on the bench?
At the corner the output voltage is 1/√2 = 0.7071 of the input, so output power, which goes with voltage squared, is exactly half. Half power is 10 × log₁₀(0.5) = −3.0103 dB, which everyone rounds to −3 dB. On a two-channel scope you can find it by sweeping a signal generator until the output amplitude drops to 0.7071 of the input, or until the two traces sit 45° apart — the phase at the corner is 45° for either arrangement.
How steep is the roll-off of a single RC filter?
Gentle: 20 dB per decade, about 6 dB per octave, once you are well past the corner. At twice the corner frequency a low-pass passes 0.4472 of the input (−6.99 dB), at ten times it passes 0.0995 (−20.04 dB), and at a hundred times about −40 dB. Stacking two sections doubles the slope, but the pair's −3 dB point falls to 0.6436 f_c even with buffered identical sections, and lower without buffers, so sharp filters are usually built as active stages.
Why doesn't my measured cutoff frequency match the calculated one?
Usually the capacitor. Ordinary tolerances already matter: a ±5 % resistor with a ±10 % capacitor puts a nominal 159 Hz corner anywhere from about 138 Hz to 186 Hz. Class-2 ceramics such as X7R, X5R and Y5V also lose capacitance with DC voltage across them and slowly with age, which pushes the real corner upward. Other suspects are the source's output resistance adding to R and a load across the output. Where the corner must hold, use C0G (NP0) ceramic or film capacitors and 1 % resistors.
What happens to the timing when a load is connected across the capacitor?
The load becomes part of the network. A resistive load R_L across the capacitor puts it in parallel with R as far as the capacitor is concerned, so τ shrinks to (R ∥ R_L) × C and the corner rises, while the final voltage drops to V₀ × R_L ÷ (R + R_L). A 10 kΩ / 100 nF network loaded by another 10 kΩ settles at half the step with τ = 0.5 ms. Even a load ten times R cuts τ to 0.9091 of its unloaded value, so buffer the output when timing matters.
How do I choose R and C for a particular cutoff frequency?
Pick the capacitor first, because standard capacitor values are sparser than resistor values, then solve R = 1 ÷ (2π f_c C). For a 1 kHz corner with 10 nF, R = 15,915 Ω, so a 15.8 kΩ 1 % part gives 1,007.3 Hz and a 16 kΩ part gives 994.7 Hz. For a coupling capacitor feeding an amplifier, the amplifier's input resistance plays the part of R: 1 µF into 10 kΩ makes a high-pass with its corner at 15.92 Hz.
Does a higher supply voltage make the capacitor charge faster?
Not in the ideal model. τ depends only on R and C, and the percentage of the step reached at a given moment is the same for a 1.8 V, 5 V or −12 V step; only the voltage itself scales. A −12 V step reaches −7.5854 V after one time constant, the same 63.2 % of the way, and a 0 V step stays at 0 V. Real class-2 ceramics are the exception, because their capacitance, and therefore τ, changes with the voltage across them.

References& sources.

  1. [1]OpenStax (Rice University), University Physics Volume 2, §10.5 'RC Circuits'. Source of the transient equations used here: Eq. 10.8 q(t) = Cε(1 − e^(−t/RC)) = Q(1 − e^(−t/τ)) with V_C(t) = ε(1 − e^(−t/τ)); Eq. 10.9 τ = RC; Eq. 10.11 q(t) = Qe^(−t/τ) for discharge; and the printed statement that at t = τ the charge has reached 1 − e^(−1) = 1 − 0.368 = 0.632 of maximum. Open access; retrieved 2026-07-29.
  2. [2]James M. Fiore, AC Electrical Circuit Analysis: A Practical Approach, §10.3 'Bode Plots' (Engineering LibreTexts). Source of the frequency-domain equations: Eq. 10.15 f_c = 1/(2πRC) for both arrangements; Eq. 10.14 lag-network gain A'_v = −10·log₁₀(1 + f²/f_c²); Eq. 10.11 lead-network gain 1/√(1 + f_c²/f²); the printed −45° lag and +45° lead at the critical frequency; and the statement that break frequencies are 'also known as the half-power points, or corner frequencies'. Open access; retrieved 2026-07-29.
  3. [3]OpenStax, University Physics Volume 2, §15.3 'RLC Series Circuits with AC'. Source of the capacitive reactance the divider rests on, Eq. 15.3 X_C = 1/(ωC), together with Eq. 15.9 φ = tan⁻¹((X_L − X_C)/R) and Eq. 15.11 Z = √(R² + (X_L − X_C)²) for the phase and magnitude algebra. Open access; retrieved 2026-07-29.
  4. [4]James M. Fiore, AC Electrical Circuit Analysis: A Practical Approach, §8.2 'Series Resonance' (Engineering LibreTexts). Cited as an independent restatement of the half-power definition this page's −3.0103 dB band edge rests on: Eq. 8.2 f₀ = 1/(2π√LC) and Eq. 8.3/8.4 BW = f₂ − f₁ with Q = f₀/BW, where f₁ and f₂ are defined as the half-power frequencies. Open access; retrieved 2026-07-29.
  5. [5]IEC 60050, International Electrotechnical Vocabulary (IEV Online / Electropedia), entry 131-12-46 'reactance', X = Im(Z) — the normative source for the vocabulary used on this page. ACCESS NOTE: electropedia.org returns HTTP 403 to automated retrieval, so the entry number and term were confirmed from the IEC's indexed entry title but the definition text was not opened. No figure on this page is taken from it.

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τ = R·C f_c = 1 ⁄ (2πRC) V(t) = V₀(1 − e^(−t/τ)) |H| = 1 ⁄ √(1 + x²)
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