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Resonance in a series RLC circuit

Curricula: Sek II, APC (extension)

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Aim

Measure the resonance curve of a series RLC circuit.

The bench

Source: sine, amplitude 5 V (RMS 3.54 V), 159.15 Hz — the resonant frequency of the circuit. Ammeter in AC mode, a 10 Ω resistor, a 0.1 H inductor (winding 1 Ω) and a 10 µF capacitor in series. Voltmeters in AC mode — across the inductor and across the capacitor.

Procedure

  1. Record the current and the voltages across the inductor and the capacitor.
  2. Vary the frequency from 50 to 500 Hz and record the current.
  3. Plot a graph of \(I(f)\).
  4. Compare the voltages across \(L\) and \(C\) with the source voltage at resonance.
Expected result

At resonance the current is about 299 mA, and the voltage across the inductor and across the capacitor is about 29.9 V each, 8–9 times the RMS voltage of the source (3.54 V); the voltages across \(L\) and \(C\) are in antiphase and cancel each other. Far from resonance the current is small: at 100 Hz, about 36 mA. The resonant frequency \(f_0 = 1 / (2\pi\sqrt{LC})\) ≈ 159 Hz.

Questions

  1. How will the resonance curve change if the resistor is replaced with a 100 Ω one?
  2. Why can the voltage across the capacitor be greater than the source voltage?
Answers
  1. The resonant frequency stays at about 159 Hz, because it depends only on \(L\) and \(C\). The peak current falls to about 3.54 V / 102 Ω ≈ 35 mA (the winding, the source and the ammeter add about 2 Ω), and the curve becomes low and wide: at 100 Hz the current is still about 25 mA, 70% of the peak.
  2. At resonance the reactances of the inductor and the capacitor are equal, about 100 Ω each, and their voltages are in antiphase and cancel, so the source drives the current only through about 12 Ω of resistance. The current of 0.299 A through the capacitor's 100 Ω gives 0.299 A · 100 Ω ≈ 29.9 V across it, 8.5 times the source's 3.54 V. The energy swings back and forth between the inductor and the capacitor, and the source only makes up the losses in the resistance.