How this instrument works
An RC low-pass filter has two honest descriptions built from the same resistor and capacitor: how fast it responds in time, τ = RC, and where it starts cutting in frequency, f_c = 1 ⁄ (2πRC). This instrument computes both from a single pair specifically so the two can be checked against each other — enter Resistance and Capacitance once, and let the same circuit answer for itself twice, from two different domains.
The link between them is exact: f_c = 1 ⁄ (2πτ), for any first-order RC pair, no exceptions. That 2π is not decoration — it converts an angular frequency, ω_c = 1/RC in radians per second, into an ordinary frequency in hertz, since ω = 2πf. Multiply this page's two outputs together and the product is always 1 ⁄ (2π) ≈ 0.159155, whatever R and C you enter, which is the self-check the quiz framing is built around.
The trap most people fall into is assuming a bigger resistor or capacitor raises the cutoff frequency, when it is the opposite: f_c is inversely proportional to both, so doubling either one halves the cutoff and doubles the time constant. Both formulas also describe a single first-order pole — one resistor, one capacitor, no load resistance pulling on the output and no stray wiring inductance — conditions a real bench measurement rarely matches exactly, which is why a built filter's measured corner drifts slightly from the ideal number.
- Enter Resistance — the unit menu switches between Ω and kΩ.
- Enter Capacitance — choose nF or µF from its unit menu.
- Read Cutoff frequency, computed as f_c = 1 ⁄ (2πRC), in Hz or kHz.
- Read RC time constant, τ = RC, in milliseconds.
- Multiply the two results together: the product should land on 1 ⁄ (2π) ≈ 0.159155 every time — if it doesn't, recheck the R or C you entered.
Worked example — 10 kΩ and 100 nF, read cutoff-first
Take Resistance = 10 kΩ (10,000 Ω) and Capacitance = 100 nF (1×10⁻⁷ F) — the same pair the main RC filter calculator uses, but read here cutoff-first. Cutoff frequency = 1 ⁄ (2π × 10,000 × 1×10⁻⁷) = 159.154943092 Hz, the frequency at which this single-pole filter's output has fallen to about 70.7% of its input amplitude.
The identical pair gives RC time constant = 10,000 × 1×10⁻⁷ = 0.001 s, or 1 millisecond — the time for a step response to climb to about 63.2% of its final value. Multiply the two results: 159.154943092 Hz × 0.001 s = 0.159154943092, which is 1 ⁄ (2π) to nine decimal places — proof both numbers describe one physical circuit, not two unrelated calculations.
Questions
Why does this calculator return two results instead of one?
Because a single resistor-capacitor pair has both a time-domain identity, τ = RC, and a frequency-domain identity, f_c = 1 ⁄ (2πRC), and the two are locked together by f_c = 1 ⁄ (2πτ). Showing both turns the page into a self-check: multiply the pair together and the product should always land on 1 ⁄ (2π) ≈ 0.159155, regardless of the R and C you chose.
What happens to the cutoff frequency if I double the resistance?
It halves, not doubles. Cutoff frequency is inversely proportional to resistance, f_c ∝ 1 ⁄ R, so raising a 10 kΩ resistor to 20 kΩ against the same 100 nF capacitor drops the cutoff from 159.15 Hz to 79.58 Hz. People expect a bigger component to mean a bigger number, but here it lowers the frequency at which the filter starts attenuating.
Does the RC time constant mean the capacitor is fully charged after that long?
No — τ = RC is only the time to reach about 63.2% of the final voltage in a step response, following V(t) = V_final(1 − e^(−t/τ)). A capacitor is usually treated as practically full after about 5τ, roughly 99.3%, but the exponential model never reaches exactly 100% in finite time.
Where does the 2π in the cutoff-frequency formula actually come from?
From converting between angular frequency and ordinary frequency. The RC circuit's output falls to -3 dB once angular frequency ω reaches 1 ⁄ RC; since ω = 2πf, solving for f gives f_c = 1 ⁄ (2πRC). It is unit bookkeeping, not a separate physical effect — expressed in radians per second instead of hertz, the cutoff is simply ω_c = 1 ⁄ RC, with no 2π at all.
Is the relationship f_c = 1 ⁄ (2πτ) specific to a low-pass filter?
No — it holds for the RC high-pass configuration too, since both share the identical first-order pole set by the same R and C. What changes between low-pass and high-pass is which node's voltage is treated as the output, across the capacitor or across the resistor, not the cutoff frequency or the time constant.