Library / Electrical / ELE-27
Electrical
Resonant frequency
Frequency at which inductive and capacitive reactance are equal.
- Ref
- ELE-27
- Rev
- v1
- Field
- Electrical
- Result
- f0 [Hz]
L is the inductance, entered in millihenries, and C the capacitance, in microfarads. At f₀ the inductive reactance 2πf₀L equals the capacitive reactance 1/(2πf₀C), so the two cancel in a series circuit. ω₀ = 2πf₀ = 1/√(LC) is the angular form.
Frequency at which inductive and capacitive reactance are equal. A tuned circuit responds most strongly there.
Calculator
- Angular frequency ω₀1,000 rad/s
- Characteristic impedance √(L/C)10 Ω
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The defaults tune to 159 Hz, where each reactance is 10 Ω. The same maths gives the natural frequency of a mass on a spring.
- To raise f₀ by a factor of 10 the product LC must fall by a factor of 100, which is why radio tuning uses very small values.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (Hz) | Local step | Change in result |
|---|---|---|---|---|---|---|
| C Capacitance | 0.001 to 100,000 µF | 5.033 | 50,330 | 50,320 | +10% | −4.65 % |
| L Inductance | 0.001 to 100,000 mH | 1.592 | 15,920 | 15,910 | +10% | −4.65 % |
Sweep of Capacitance
| Capacitance (µF) | Resonant frequency (Hz) |
|---|---|
| 0.001 | 50,330 |
| 0.01 | 15,920 |
| 0.1 | 5,033 |
| 1 | 1,592 |
| 10 | 503.3 |
| 100 | 159.2 |
| 1,000 | 50.33 |
| 10,000 | 15.92 |
| 100,000 | 5.033 |
Sweep one variable
Variables and how each one contributes
Resonant frequency
What the equation gives you. Everything else feeds it.
Inductance
The inductance, in mH. Inverse square root: four times the inductance halves the resonant frequency. A 10 % rise in L lowers f₀ by about 5 %.
Capacitance
The capacitance, in µF. Inverse square root, the same as L: a 10 % rise in C lowers f₀ by about 5 %.
Worked example with the values above
- Convert units: L = 10 mH = 0.01 H, C = 100 µF = 1×10−4 F.
- Product and root: √(LC) = √1×10−6 = 0.001.
- f₀ = 1 / (2π × 0.001) = 159.2 Hz.
Assumptions and limits
- Ideal inductor and capacitor with negligible resistance and stray capacitance.
- Lumped components, with the wavelength much longer than the circuit.
Common mistakes
- Entering L and C in H and F when the boxes expect mH and µF (or the reverse).
- Forgetting the 2π and quoting ω₀ as if it were a frequency in Hz.
Statistics and uncertainty
f₀ ∝ (LC)−½, so each relative tolerance is halved. With 5 % capacitors and 5 % inductors, f₀ is uncertain by about 3.5 %. For a narrow filter that may exceed the bandwidth, so one component is usually made adjustable.
Where it is used
- Tuning a radio receiver or a filter to a chosen frequency.
- Estimating the ringing frequency of an LC circuit.
Maths you need
- Reactance: ωL and 1/ωC
- Units and prefixes: milli, micro