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Physics · Simulator 017

RLC Resonance

Drive a resistor, inductor and capacitor in series with an AC source. Sweep the frequency to find resonance, where the current peaks and stays in phase with the voltage, and see how the resistance sets the sharpness of the peak.

How the model works

In a series RLC circuit driven by V₀ sin(ωt), the steady-state current is I₀ sin(ωt − φ), with

X_L = ωL, X_C = 1/(ωC), Z = √(R² + (X_L − X_C)²)
I₀ = V₀ / Z, tan φ = (X_L − X_C) / R

At resonance ω₀ = 1/√(LC), the reactances cancel. Then Z = R, the current is largest, and φ = 0. Below resonance the capacitor dominates and the current leads the voltage (φ < 0); above it the inductor dominates and the current lags (φ > 0).

The quality factor Q = (1/R)√(L/C) measures sharpness: the half-power bandwidth is Δf = f₀ / Q. Radio tuners use high-Q circuits to pick out one station.

Ideal components and steady state only (start-up transients are not shown). Real inductors have their own resistance.

Guided experiments

  1. Find resonance. L = 10 mH, C = 1 µF. Prediction: f₀ = 1/(2π√(LC)) ≈ 1592 Hz, where current and voltage line up.
  2. Lower the resistance. R from 50 Ω to 10 Ω. Prediction: the peak current rises fivefold and Q goes from 2 to 10, so the curve becomes much narrower.
  3. Off resonance. Drive at 800 Hz. Prediction: capacitive (X_C > X_L). The current peaks before the voltage, a negative phase of about −71° with R = 50 Ω.

Challenge

An AM radio uses L = 240 µH. What capacitance tunes it to 1000 kHz? If R = 12 Ω, what is the bandwidth?

Reveal the explanation

C = 1/((2πf)² L) = 1/((6.283×10⁶)² × 240×10⁻⁶) ≈ 106 pF. Then Q = (1/12)√(240×10⁻⁶ / 106×10⁻¹²) ≈ 125, so Δf ≈ 1000 kHz / 125 = 8 kHz, close to the 9–10 kHz channel spacing used for AM.

References

  • OpenStax, University Physics Volume 2, §15.5 “Resonance in an AC Circuit”.