RLC Resonant Frequency & Q Calculator

Understanding the RLC resonance calculator — concept and principle

This calculator finds the resonant frequency of an inductor–capacitor pair, the quality factor Q and bandwidth of a series RLC circuit, and the characteristic impedance √(L/C). It is used to tune radio and filter circuits, design notch and band-pass filters, analyse oscillators and wireless-charging coils, and predict where unwanted ringing will appear.

Like a swing, the circuit has one natural rhythm: 1,591.55 Hz. A small R keeps Q high (10).

How it is calculated

Resonant frequency: f₀ = 1 / (2π√(LC)), with L in henries (mH ÷ 1000) and C in farads (µF × 10⁻⁶).
Quality factor (series): Q = (1/R) · √(L/C).
Bandwidth: BW = R / (2πL) = f₀ / Q, the width between the two −3 dB (half-power) frequencies.
Characteristic impedance: Z₀ = √(L/C), the reactance of L (or C) at resonance.

The principle

An inductor's reactance 2πfL rises with frequency, and a capacitor's reactance 1/(2πfC) falls. At one frequency they are equal and, being opposite in sign, cancel. Setting 2πfL = 1/(2πfC) and solving gives f₀ = 1/(2π√(LC)). At that point a series circuit looks like a pure resistance R, so the current is at its maximum and in phase with the voltage. Energy swings back and forth between the magnetic field of L and the electric field of C, and R removes a little each cycle. Q compares the stored energy with the energy lost per cycle: a high Q (small R relative to √(L/C)) gives a sharp, narrow peak.

Worked example

Inputs: R = 10 Ω, L = 10 mH = 0.01 H, C = 1 µF = 10⁻⁶ F.
LC = 10⁻⁸, √(LC) = 10⁻⁴ s, so f₀ = 1 ÷ (2π × 10⁻⁴) = 1,591.55 Hz (ω₀ = 10,000 rad/s).
Z₀ = √(0.01 ÷ 10⁻⁶) = √10,000 = 100 Ω.
Q = 100 ÷ 10 = 10.
Bandwidth = 10 ÷ (2π × 0.01) = 159.155 Hz, which equals 1,591.55 ÷ 10.

Each point is the calculator re-run with only “L (mH)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

L (mH)	Resonant frequency (Hz)
2.5	3,183.098862
5	2,250.79079
7.5	1,837.762985
10	1,591.549431
12.5	1,423.525087
15	1,299.494669
20	1,125.395395

How to read the result

Q below about 0.5 means the circuit is overdamped and does not ring; Q of a few to tens is typical for practical LC filters; crystals and cavity resonators reach thousands or more. In a series circuit at resonance, the voltage across L and across C each equals Q times the source voltage, so a Q of 10 on a 10 V source puts about 100 V across the capacitor.

Common mistakes

Using the series Q formula for a parallel tank. For a parallel RLC with R across the tank, Q = R · √(C/L), the inverse relationship.
Forgetting coil resistance: the winding's own resistance adds to R and lowers Q.
Mixing µF, nF and pF; an error of 1000 in C shifts f₀ by a factor of about 31.6.

Keep in mind: These are ideal formulas. Real inductors have resistance and self-capacitance, capacitors have ESR, and parasitic wiring shifts f₀ at high frequency. High-Q series circuits can develop dangerous voltages across L and C.

Frequently asked questions

Is the resonant frequency the same in series and parallel circuits?

For ideal components, yes, f₀ = 1/(2π√(LC)) in both. With significant coil resistance, the parallel resonance shifts slightly.

What is the difference between f₀ and ω₀?

ω₀ is in radians per second and equals 2πf₀. Here ω₀ = 1/√(LC) = 10,000 rad/s.

How do I lower the resonant frequency?

Increase L or C. Because f₀ depends on the square root, quadrupling LC halves the frequency.

Sources: C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, McGraw-Hill, chapter on frequency response and resonance · The ARRL Handbook for Radio Communications, American Radio Relay League

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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