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LNGR SPKRHIFI. TWO WAYS. HANDCRAFTED.
Knowledge / From components to crossovers
LNGR · Learning workshop

RLC & resonance

An inductor and capacitor can exchange energy. Resistance helps determine how pronounced resonance becomes.

Try the experiment ↓
L ↔ Cf₀ = 1/(2π√LC)

Functional illustration · equations and topology in the text

Start without prior knowledgeUnderstand the model limitsDE / EN
01

The mental picture

A swing alternates between kinetic and potential energy. An LC system exchanges energy between magnetic and electric fields. Losses damp the exchange, but suitable external excitation can still create a large response.

02

The principle

In a series RLC circuit, inductor and capacitor reactances have opposite signs. At resonance they cancel in the total impedance, leaving only R to limit current in the ideal model. This does not mean the voltages across L and C are zero.

03

Try it yourself

Interactive experiment

RLC & resonance

Reduce R at fixed L and C. Resonance frequency stays the same, but the current peak becomes taller and narrower.

Model limits: Ideal series RLC circuit driven by a sine. Frequency axis spans f0/10 to 10f0; all losses are represented by R.

04

A worked example

10 mH and 10 µF resonate at about 503 Hz. With 10 Ω and 5 V RMS, resonant current is 0.5 A. Q is about 3.16; voltages across the reactive components can exceed source voltage.

A closer look · equations and assumptions
Formula notes
f_0=\frac{1}{2\pi\sqrt{LC}}
Z=R+j\left(\omega L-\frac{1}{\omega C}\right)
Q_{\mathrm{series}}=\frac{\omega_0L}{R}
I=\frac{U}{|Z|}

For the series circuit, half-power bandwidth is Δf = R/(2πL), giving Q = f0/Δf. Other topologies have different Q relationships. A parallel resonator can have high input impedance at resonance. “Resonance” alone does not tell you whether current peaks or dips.

A common pitfall. Do not confuse series and parallel arrangements. Low resistive loss can allow large resonant currents and voltages.

05

Workshop takeaway

Resonance is a tool, not a quality badge. Frequency, damping and where the output is taken determine its effect.

Sources & transparency

Original illustrations and educational calculations. No measurements of a specific product. Edited 11 October 2026.

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

13 · Idealised driver: Re = 6 Ω, Le = 0.6 mH. Find Zobel R and C for the Re-plus-Le model.

Show worked answer
  1. Zobel R = Re = 6 Ω.
  2. Zobel C = Le/Re².
  3. C = 16.67 µF. This model has no mechanical resonance.
Calculation
C_Z=\frac{L_e}{R_e^2}=16.67\,\mu F

15 · A series RLC branch has L = 1 mH and C = 10 µF. Find resonance frequency.

Show worked answer
  1. L · C = 10⁻⁸.
  2. Square root = 0.0001.
  3. Resonance = 1591.55 Hz.
Calculation
f_0=\frac{1}{2\pi\sqrt{LC}}=1591.55\,Hz

16 · How do you move this resonance to half frequency while retaining L?

Show worked answer
  1. Resonance is inversely proportional to square root of C.
  2. Quadruple C for half frequency.
  3. New C = 40 µF.
Calculation
C_{new}=4C=40\,\mu F

17 · A measured peak at 2 kHz is 4 dB above target. What voltage ratio is needed for ideal local correction?

Show worked answer
  1. Correction is -4 dB.
  2. Ratio = 10^(-4/20).
  3. Ratio = 0.6310; bandwidth and phase must also fit.
Calculation
k=10^{-4/20}=0.6310

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20 exercises · PDF ↓Teacher solutions · PDF ↓