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

RLC Circuit Calculator — Resonant Frequency, Q Factor, Bandwidth and Impedance

Resonant frequency, Q, bandwidth, exact half-power edges, damping ratio, and impedance and phase at any frequency for a series or ideal parallel RLC circuit.

RLC Circuit Calculator

Circuit topology
Total series resistance for the series topology, or total loading across the tank for the parallel topology. This is the only element that sets Q once L and C are fixed.
Resistance unit
Nominal inductance. Real inductors have a self-resonant frequency of their own, above which they behave capacitively — the arithmetic here is only valid well below it.
Inductance unit
Nominal capacitance. Class-2 ceramics lose a large share of their marked value under DC bias, which moves the resonant frequency up.
Capacitance unit
The frequency at which impedance, phase and the two reactances are reported. It does not affect the resonant frequency, Q, bandwidth or damping — those are properties of the components alone.
Hz
Resonant frequency f₀
159.1549
f₀ in hertz, the frequency at which the inductive and capacitive reactances are equal and cancel.
Impedance at the analysis frequency
47.1587 Ω
Impedance phase
77.7575°
Quality factor Q
10
−3 dB bandwidth
15.9155 Hz
Lower half-power frequency
151.396 Hz
Upper half-power frequency
167.3115 Hz
Damping ratio ζ
0.05
Damped ringing frequency
158.9559 Hz
Inductive reactance X_L
125.6637 Ω
Capacitive reactance X_C
79.5775 Ω
Reading of the result
ζ = 0.05 is below 1, so the circuit is underdamped — it rings at 158.9559 Hz before settling, and its Q of 10 gives a −3 dB bandwidth of 15.9155 Hz around 159.1549 Hz. Model: R, L and C all in series, driven by an ideal voltage source. Ideal components only — no ESR, no core loss, and no inductor self-resonance, above which this arithmetic describes a different component from the one on your bench.

Background.

Connect an inductor to a capacitor and you have built something that wants to oscillate. Energy held in the capacitor's electric field drains into the inductor's magnetic field and back again, at a rate fixed by the two component values alone. Add resistance and some of that energy turns into heat on every swing. Those three ingredients are behind the tuned stage of a radio receiver, the LC filter on a switching regulator's output, the tank of an oscillator, and the ringing on a scope trace when a fast edge meets stray inductance and capacitance.

Give this calculator R, L and C with their units, say whether they are wired in series or as a parallel tank, and it reports the natural frequency f₀, the quality factor Q, the −3 dB bandwidth and its two edges, the damping ratio, and the frequency a disturbance actually rings at. A second group of outputs describes the circuit at a frequency you choose: the impedance magnitude, its phase angle, and the inductive and capacitive reactances. A one-line reading of the damping sums up whether the circuit rings or settles.

The topology choice changes more than people expect, because resistance plays opposite roles in the two arrangements. In series, current flows through R, so a smaller resistance means less loss and a sharper peak. In a parallel tank, R is a path across the tank that bleeds energy away, so a larger resistance means a sharper peak. With the default 100 mH and 10 µF, a 10 Ω resistor gives Q = 10 in series but only 0.1 in parallel; the parallel tank needs 1 kΩ to match.

Everything assumes ideal components. Real coils have winding resistance, core loss and a self-resonant frequency of their own, and class-2 ceramic capacitors lose a large share of their marked value under DC bias. Those effects shift a physical resonance by percentages, so read the long decimal tails as arithmetic, not as a prediction of what a bench measurement will show.

What is rlc circuit calculator?

An RLC circuit combines resistance, inductance and capacitance, and because it has two places to store energy it is a second-order system: it can oscillate. Resonance is the frequency at which the inductor's reactance, which rises with frequency, exactly equals the capacitor's, which falls, so the two cancel and the network behaves like the bare resistance. The quality factor Q measures how sharp that resonance is. It has two equivalent readings: the resonant frequency divided by the bandwidth between the half-power points, and 2π times the energy stored divided by the energy lost per cycle. The damping ratio ζ = 1/(2Q) describes the same property in time: below 1 a disturbance rings while decaying, at exactly 1 the circuit is critically damped and settles as fast as possible without overshoot, and above 1 it is overdamped and returns slowly without oscillating.

How to use this calculator.

  1. Pick the topology. Series means one current flows through all three parts; parallel means all three share one voltage, with R connected across the inductor and capacitor.
  2. For a parallel tank, make sure R is the load or loss across the tank. If your resistance is the coil's own winding resistance, sitting in series with the inductor inside a parallel circuit, that arrangement is not modelled and its resonance is lower.
  3. Enter R, L and C with their unit menus — Ω, kΩ or MΩ; nH, µH, mH or H; pF, nF, µF, mF or F. Each value must be greater than zero.
  4. Set the analysis frequency to the frequency of the signal you care about: the impedance, phase and two reactances are reported there. It has no effect on f₀, Q, bandwidth or damping, which belong to the components alone.
  5. Use Q and the bandwidth to judge selectivity, and the damping ratio to judge transient behaviour. The sign of the phase tells you whether the circuit looks inductive or capacitive at your chosen frequency.

The formula.

f₀ = 1 ⁄ (2π√(LC)) Q_s = (1/R)√(L/C) Q_p = R√(C/L) ζ = 1 ⁄ (2Q) BW = f₀ ⁄ Q

Resonance comes from setting the two reactances equal, 2πfL = 1/(2πfC), which gives f₀ = 1/(2π√(LC)) with L in henries and C in farads. Resistance does not appear, so R moves the sharpness of the peak but never its position. A useful intermediate is the characteristic impedance Z₀ = √(L/C), the reactance of either part at f₀. The quality factors are then simple ratios: Q = Z₀/R in series and Q = R/Z₀ for the ideal parallel tank, so for the same three parts the two Q values multiply to exactly 1. The damping ratio is ζ = 1/(2Q) and the bandwidth is BW = f₀/Q. The half-power edges are computed exactly, f₁,₂ = f₀(√(1 + 1/(4Q²)) ∓ 1/(2Q)), rather than as the textbook shortcut f₀ ± BW/2 that Fiore's text gives for higher-Q circuits. The exact pair always differs by f₀/Q and always multiplies to f₀², because resonance is the geometric mean of the two edges; the shortcut drifts at moderate Q and goes negative below Q = 0.5. A step or pulse makes an underdamped circuit ring at f_d = f₀√(1 − ζ²), and at or above ζ = 1 the damped frequency is reported as zero because nothing oscillates. Impedance follows OpenStax: X_L = 2πfL and X_C = 1/(2πfC); in series Z = R + j(X_L − X_C), with |Z| = √(R² + (X_L − X_C)²) and phase arctan((X_L − X_C)/R). For the parallel tank the same algebra runs on admittance, Y = 1/R + j(2πfC − 1/(2πfL)), and |Z| = 1/|Y|. Phase is voltage relative to current: positive means inductive, negative capacitive. Left out of the model are winding resistance, core and dielectric loss, capacitor ESR and ESL, skin effect and inductor self-resonance.

A worked example.

Example

Take the default series circuit: 10 Ω, 100 mH and 10 µF, examined at 200 Hz. In base units L = 0.1 H and C = 10⁻⁵ F, so LC = 10⁻⁶ and √(LC) = 10⁻³. The resonant frequency is 1 ÷ (2π × 10⁻³) = 159.1549 Hz. The characteristic impedance is √(0.1 ÷ 10⁻⁵) = √10,000 = 100 Ω, so Q = 100 ÷ 10 = 10 and ζ = 1 ÷ 20 = 0.05. The bandwidth is 159.1549 ÷ 10 = 15.9155 Hz, between edges at 151.396 Hz and 167.3115 Hz; their product, 25,330.2959, equals 159.1549². A disturbance rings at 159.1549 × √(1 − 0.05²) = 158.9559 Hz. At 200 Hz, ω = 2π × 200 = 1,256.6371 rad/s. The inductor's reactance is 1,256.6371 × 0.1 = 125.6637 Ω and the capacitor's is 1 ÷ (1,256.6371 × 10⁻⁵) = 79.5775 Ω, leaving a net 46.0862 Ω of inductive reactance. The impedance is √(10² + 46.0862²) = 47.1587 Ω at a phase of arctan(46.0862 ÷ 10) = 77.7575°. 200 Hz is about 26% above resonance and well past the upper edge at 167.3 Hz, so the circuit is outside its passband there: it presents 4.7 times its resonant impedance, and the voltage across it leads the current by 77.8°, most of the way to a quarter-cycle.

capacitance10
inductance100
inductance UnitmH
topologyseries
analysis Frequency200
resistance Unitohm
capacitance UnituF
resistance10

Frequently asked questions.

Why doesn't changing the resistance move the resonant frequency?
Because resonance is defined by the reactances cancelling, and neither reactance depends on R. Resistance changes how quickly energy is lost, which widens or narrows the peak and lowers the frequency a disturbance rings at, f₀√(1 − ζ²). The undamped f₀ itself stays at 1/(2π√(LC)) whatever resistor you fit.
Should I choose series or parallel?
Follow how the parts are connected to the source. A series circuit has its lowest impedance at resonance, equal to R, and draws the most current there, which suits notch-to-ground traps and series-tuned filters. An ideal parallel tank has its highest impedance at resonance, also equal to R, and blocks there, as in an oscillator tank or a tuned amplifier load.
My parallel tank's only resistance is the coil's winding resistance. How far off is this?
That circuit puts R in series with L, and Fiore gives its resonance as f₀√(1 − CR²/L). The correction depends on the coil's own Q, Z₀/R: the square root equals √(1 − 1/Q²). At a coil Q of 10 the true resonance is 0.5% lower than the ideal figure; at a Q of 3 it is 5.7% lower.
What resistance makes the circuit critically damped?
Critical damping is ζ = 1, which is Q = 0.5. In series that needs R = 2√(L/C); for the default 100 mH and 10 µF, that is 200 Ω. In the parallel tank it needs R = ½√(L/C), or 50 Ω. Smaller series resistance, or larger parallel resistance, makes the circuit ring; the opposite makes it sluggish.
Does a Q of 0.707 mean a step response with no overshoot?
No. Q = 0.707 (ζ = 0.707) gives the flattest possible frequency response, the Butterworth case, but a step still overshoots by about 4.3%. Only ζ of 1 or more, Q of 0.5 or less, settles with no overshoot at all. With the default L and C, a 141.42 Ω series resistor lands at ζ = 0.7071.
Why aren't the half-power frequencies symmetric around f₀?
On a linear frequency scale the response is lopsided: resonance sits at the geometric mean of the edges, √(f₁f₂), not halfway between them. For the default circuit the exact lower edge is 151.396 Hz, while the shortcut f₀ − BW/2 gives 151.1972 Hz. The gap is small at Q = 10 but grows as Q falls, and below 0.5 the shortcut turns negative.
How do I tell which side of resonance my signal is on?
Compare the two reactances, or read the phase. In series, a positive phase means X_L is larger and you are above f₀; a negative phase means you are below it. A parallel tank does the reverse: it looks inductive below resonance and capacitive above, so at 200 Hz the default parts with 1 kΩ across them show −77.7575°.
I entered f₀ as the analysis frequency. Why isn't the phase exactly zero?
The resonant frequency is irrational, so any value you type is rounded, and the circuit sits a hair off resonance. Typing the default circuit's f₀ to ten decimal places still gives an impedance of 10 Ω and a phase of about 3.358 × 10⁻¹¹ degrees, which is zero for every practical purpose.
Why can't I set the resistance to zero for a perfect resonator?
A lossless circuit would have infinite Q, zero bandwidth and edges that coincide, none of which can be shown as numbers or built in hardware. Every real coil, capacitor and trace has some loss, so enter a small resistance that represents it; values as low as a millionth of the chosen unit are accepted.
How long does an underdamped circuit keep ringing?
Each cycle the oscillation's amplitude shrinks by roughly a factor of e^(−π/Q), so it falls to about 37% of its starting value after about Q/π cycles. The default circuit, with Q = 10, takes about 3.2 cycles to get there while ringing at 158.9559 Hz; a radio tank with a Q in the hundreds rings for many tens of cycles.

References& sources.

  1. [1]James M. Fiore, AC Electrical Circuit Analysis: A Practical Approach, §8.2 'Series Resonance' (Engineering LibreTexts). Source of f₀ = 1/(2π√LC) (Eq. 8.2), the series quality factor as Q = X₀/R_T (Eq. 8.12) and Q = (1/R_T)√(L/C) (Eq. 8.13), and the bandwidth relations BW = f₂ − f₁ with Q = f₀/BW (Eq. 8.3/8.4). Also the source of the half-power approximations f₁ ≈ f₀ − BW/2 and f₂ ≈ f₀ + BW/2 (Eq. 8.10/8.11), printed as valid for higher-Q circuits — this page computes the exact form instead and says so. Open access; retrieved 2026-07-29.
  2. [2]James M. Fiore, AC Electrical Circuit Analysis: A Practical Approach, §8.3 'Parallel Resonance' (Engineering LibreTexts). Source of the parallel quality factor Q = R_T/X_L (Eq. 8.19) and Q = R_T√(C/L) (Eq. 8.20), and of Eq. 8.21 f₀ = (1/2π√LC)√(1 − CR²/L) for the parallel topology in which R is in series with the coil — the arrangement this page explicitly does not model. Open access; retrieved 2026-07-29.
  3. [3]OpenStax (Rice University), University Physics Volume 2, §15.3 'RLC Series Circuits with AC'. Source of X_C = 1/(ωC) (Eq. 15.3), X_L = ωL (Eq. 15.8), the phase φ = tan⁻¹((X_L − X_C)/R) (Eq. 15.9) and the magnitude Z = √(R² + (X_L − X_C)²) (Eq. 15.11). Open access; retrieved 2026-07-29.
  4. [4]OpenStax, University Physics Volume 2, §14.6 'RLC Series Circuits'. Source of the damped angular frequency ω′ = √(1/LC − (R/2L)²) (Eq. 14.46) and of the printed damping conditions — 1/LC > R²/4L² underdamped, 1/LC = R²/4L² critically damped, 1/LC < R²/4L² overdamped — which this page expresses through ζ = 1/(2Q) with the boundary at exactly ζ = 1. Open access; retrieved 2026-07-29.
  5. [5]OpenStax, University Physics Volume 2, §15.4 'Power in an AC Circuit'. Source of the average power P_ave = ½I₀V₀cos φ (Eq. 15.12) and of the identification of cos φ = R/Z as the power factor — the definition that makes the −3 dB edges 'half-power' frequencies and the reason the impedance phase output matters. Open access; retrieved 2026-07-29.

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f₀ = 1 ⁄ (2π√(LC)) Q_s = (1/R)√(L/C) Q_p = R√(C/L) ζ = 1 ⁄ (2Q) BW = f₀ ⁄ Q
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