# What is the RC filter cutoff?

RC filter calculator: the cutoff frequency f_c = 1 ÷ (2πRC) of a low-pass or high-pass RC filter, or the R or C for a cutoff, with the gain in dB and the phase at a signal frequency.

- Page: https://www.acalculator.org/physics/rc-filter-calculator
- JSON spec: https://www.acalculator.org/physics/rc-filter-calculator.json
- Version: 668326a77ffa

## Default answer

Example with the default inputs (Find Cutoff frequency, Filter Low-pass, Resistance (R) 1 kΩ, Capacitance (C) 100, Capacitance unit nF, Signal frequency (f) 10,000 Hz): f_c = 1.59155 kHz; the time constant RC is 0.0001 s.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| find | Find | Which value to work out from f_c = 1 ÷ (2πRC). |
| type | Filter | Low-pass takes the output across the capacitor; high-pass takes it across the resistor. |
| R | Resistance (R) | The resistor in series with the capacitor. |
| C | Capacitance (C) | The capacitor, in the unit chosen below. |
| Cu | Capacitance unit | The unit of the capacitance you type. |
| fc | Cutoff frequency (f_c) | The frequency where the output falls to 1 ÷ √2 of the input (−3 dB). |
| f | Signal frequency (f) | Optional: a frequency to see the gain and phase of the filter at. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Answer | The value worked out, to 6 significant figures, in the largest unit that keeps it at 1 or more. |
| cutoffHz | Cutoff frequency (Hz) | f_c = 1 ÷ (2πRC), in hertz. |
| omega | Cutoff angular frequency (rad/s) | ω_c = 2πf_c = 1 ÷ (RC), in radians per second. |
| tau | Time constant RC (s) | τ = R × C, in seconds. |
| resistance | Resistance (Ω) | R in ohms. |
| capacitance | Capacitance (F) | C in farads. |
| gain | Gain at f (Vout ÷ Vin) | The output amplitude over the input amplitude at the signal frequency. |
| db | Gain at f (dB) | 20 × log₁₀ of the gain: −3.01 dB at the cutoff. |
| phase | Phase at f (°) | How far the output leads (+) or lags (−) the input, in degrees. |
| reactance | Capacitive reactance at f (Ω) | X_C = 1 ÷ (2πfC), in ohms. |

## Method

f_c = 1 ÷ (2πRC); R = 1 ÷ (2πf_cC); C = 1 ÷ (2πf_cR); with x = f ÷ f_c: low-pass gain 1 ÷ √(1 + x²), phase −arctan x; high-pass gain x ÷ √(1 + x²), phase +arctan(1 ÷ x); dB = 20 log₁₀ gain.

## Assumptions

- An ideal first-order RC filter: an ideal resistor and capacitor in series, driven by a sine wave from a source with no resistance of its own, and an output that draws no current.
- The cutoff (corner, −3 dB, half-power) frequency is where the capacitive reactance equals the resistance.

## Worked examples

1. find = cutoff, type = low, R = 1,000, C = 100, Cu = nF gives cutoffHz = 1,591.549431, omega = 10,000, tau = 0.0001, result = f_c = 1.59155 kHz. Source: OpenStax, University Physics Volume 2, §15.3 RLC Series Circuits with AC (Equations 15.9 to 15.11: φ, I₀ = V₀ ÷ Z, Z = √(R² + (X_L − X_C)²); V_R = I₀R, V_C = I₀X_C), https://openstax.org/books/university-physics-volume-2/pages/15-3-rlc-series-circuits-with-ac (retrieved 2026-10-19).
2. find = resistance, type = high, C = 10, Cu = uF, fc = 60, f = 60 gives R = 265.258238, result = R = 265.258 Ω, gain = 0.707107, db = -3.0103, phase = 45, reactance = 265.258238. Source: OpenStax, University Physics Volume 2, §15.2 Simple AC Circuits (Equation 15.3, X_C = 1 ÷ ωC; ω = 2πf; Example 15.1: a 10 µF capacitor at 60 Hz has X_C = 265 Ω), https://openstax.org/books/university-physics-volume-2/pages/15-2-simple-ac-circuits (retrieved 2026-10-19): X_C = 265 Ω for 10 µF at 60 Hz, so R = X_C there; OpenStax, University Physics Volume 2, §15.3 RLC Series Circuits with AC (Equations 15.9 to 15.11: φ, I₀ = V₀ ÷ Z, Z = √(R² + (X_L − X_C)²); V_R = I₀R, V_C = I₀X_C), https://openstax.org/books/university-physics-volume-2/pages/15-3-rlc-series-circuits-with-ac (retrieved 2026-10-19).
3. find = capacitance, type = low, R = 10,000, fc = 1,000 gives C = 0, result = C = 15.9155 nF. Source: OpenStax, University Physics Volume 2, §15.3 RLC Series Circuits with AC (Equations 15.9 to 15.11: φ, I₀ = V₀ ÷ Z, Z = √(R² + (X_L − X_C)²); V_R = I₀R, V_C = I₀X_C), https://openstax.org/books/university-physics-volume-2/pages/15-3-rlc-series-circuits-with-ac (retrieved 2026-10-19).
4. find = cutoff, type = low, R = 1,000, C = 100, Cu = nF, f = 10,000 gives gain = 0.157177, db = -16.072235, phase = -80.956939. Source: OpenStax, University Physics Volume 2, §15.3 RLC Series Circuits with AC (Equations 15.9 to 15.11: φ, I₀ = V₀ ÷ Z, Z = √(R² + (X_L − X_C)²); V_R = I₀R, V_C = I₀X_C), https://openstax.org/books/university-physics-volume-2/pages/15-3-rlc-series-circuits-with-ac (retrieved 2026-10-19).

## FAQ

### How do you calculate the cutoff frequency of an RC filter?

f_c = 1 ÷ (2πRC), with R in ohms and C in farads. A 1 kΩ resistor with a 100 nF capacitor gives f_c = 1 ÷ (2π × 1,000 × 0.0000001) = 1,591.55 Hz, about 1.59 kHz.

### What happens at the cutoff frequency?

The capacitive reactance equals the resistance, so the output is 1 ÷ √2 = 70.7% of the input: −3.01 dB, or half the power. The output leads the input by 45° in a high-pass filter and lags it by 45° in a low-pass filter.

### What is the difference between a low-pass and a high-pass RC filter?

The same resistor and capacitor in series, with the output taken from a different part. Across the capacitor, low frequencies pass and high ones are cut (low-pass). Across the resistor, high frequencies pass and low ones are cut (high-pass). Both have the same cutoff, 1 ÷ (2πRC).

### How do you choose R and C for a cutoff frequency?

Pick one and solve for the other: R = 1 ÷ (2πf_cC) or C = 1 ÷ (2πf_cR). For a 60 Hz cutoff with a 10 µF capacitor, R = 1 ÷ (2π × 60 × 0.00001) = 265.258 Ω, the reactance of 10 µF at 60 Hz.

### What is the RC time constant?

τ = R × C, in seconds. A capacitor charging through R reaches 63.2% of its final charge after one time constant. The cutoff is f_c = 1 ÷ (2πτ), so a 0.1 ms time constant gives 1,591.55 Hz.

### How fast does an RC filter roll off?

Well past the cutoff, a first-order RC filter loses about 20 dB for each factor of 10 in frequency (6 dB per octave). A low-pass filter at 10 times its cutoff passes 1 ÷ √101 = 9.95% of the input, −20.04 dB.

## Sources

- OpenStax, University Physics Volume 2, §15.2 Simple AC Circuits: Equation 15.3, capacitive reactance X_C = 1 ÷ (ωC), ω = 2πf; Example 15.1 (a 10 µF capacitor at 60 Hz: X_C = 265 Ω), retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-2/pages/15-2-simple-ac-circuits
- OpenStax, University Physics Volume 2, §15.3 RLC Series Circuits with AC: Equations 15.9 (phase angle), 15.10 (I₀ = V₀ ÷ Z) and 15.11 (Z = √(R² + (X_L − X_C)²)); V_R = I₀R and V_C = I₀X_C, retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-2/pages/15-3-rlc-series-circuits-with-ac
- OpenStax, University Physics Volume 2, §10.5 RC Circuits: Equation 10.9, the time constant τ = RC; 63.2% of full charge after one τ, retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-2/pages/10-5-rc-circuits
- OpenStax, University Physics Volume 2, §9.5 Electrical Energy and Power: Equation 9.13, P = V² ÷ R, retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-2/pages/9-5-electrical-energy-and-power
- OpenStax, University Physics Volume 1, §17.3 Sound Intensity: Equation 17.12, a level in decibels is 10 log₁₀ of a power ratio; twice the power is 3.01 dB more, retrieved 2026-10-19. https://openstax.org/books/university-physics-volume-1/pages/17-3-sound-intensity
- HyperPhysics (Georgia State University), RC High Pass Filter: breakpoint angular frequency ω = 1 ÷ (RC); a voltage ratio in decibels is 20 log₁₀(Vout ÷ Vin), retrieved 2026-10-19. https://hyperphysics.gsu.edu/hbase/Electric/filcap.html