How to Calculate the Resonant Frequency

Finds the resonant frequency as f = 1 ÷ (2π√(LC)). At this frequency the inductive and capacitive reactances cancel and the impedance falls to its minimum.

Explanation

A circuit holding both a coil and a capacitor has one special frequency at which $X_L$ and $X_C$ cancel exactly. The impedance collapses to just $R$ , its minimum, and the current peaks. This is resonance.

f_0 = \dfrac{1}{2\pi \sqrt{LC}}

Where it comes from

Simply set $X_L = X_C$ .

2\pi f L = \dfrac{1}{2\pi f C} \quad \Longrightarrow \quad f^2 = \dfrac{1}{4\pi^2 LC} \quad \Longrightarrow \quad f_0 = \dfrac{1}{2\pi\sqrt{LC}}

Example

With $L = 10$ mH and $C = 1$ μF,

f_0 = \dfrac{1}{2\pi \sqrt{0.01 \times 10^{-6}}} = \dfrac{1}{2\pi \times 10^{-4}} = 1591.5\ \text{Hz}

At that frequency $X_L = X_C = 100$ Ω, and they duly cancel.

This is how a radio tunes

Every station in range arrives at the aerial at once. The tuned circuit is what plucks out one of them.

Turning the variable capacitor changes $C$ , which slides the resonant frequency, until it matches the station you want. Only the current at that frequency swells; everything else stays small. Turning the tuning knob was always just turning that capacitor.

Watch out

Because the current peaks at resonance, it can become dangerously large in an unprotected circuit. The voltage across the coil or capacitor can even exceed the supply voltage. Any real design has to check for this.