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Damped oscillations in an LC circuit

Curricula: APC, Sek II

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Aim

Observe free damped oscillations in an LC circuit and measure their period.

The bench

A 6 V source charges a 10 µF capacitor through a 10 kΩ resistor. The switch (open) connects a 1 H inductor with a winding resistance of 10 Ω to the capacitor. Oscilloscope across the capacitor: CH1, 2 V/div, 20 ms/div.

Procedure

  1. Set the speed chip in the top panel to ×0.1.
  2. Close the switch and watch the waveform.
  3. Freeze the display with the RUN/STOP button, measure the time of five periods and divide it by 5.
  4. Open the switch so that the capacitor charges again, and repeat with a 40 µF capacitor.
Expected result

After the switch is closed, the voltage across the capacitor oscillates sinusoidally and decays. The period is about 20 ms — one division, so five periods are five divisions; the frequency is about 50 Hz: \(T = 2\pi\sqrt{LC}\). The damping is set by the winding resistance and the charging resistor. With 40 µF the period doubles — to about 40 ms.

Questions

  1. What energy conversions take place during one period?
  2. Why do the oscillations decay?
Answers
  1. Twice in each period the energy passes from the electric field of the capacitor, \(C \cdot U^2 / 2\), to the magnetic field of the inductor, \(L \cdot I^2 / 2\), and back. When the capacitor voltage is at its peak the current is zero, and when the voltage passes through zero the current is at its peak; at the start the capacitor holds about 10 µF · (6 V)² / 2 = 0.18 mJ.
  2. Each period part of the energy turns into heat in the resistances of the circuit: the 10 Ω winding, through which the whole oscillating current flows, and the 10 kΩ charging resistor, which stays connected across the capacitor through the source. The energy left for the oscillations falls, and the amplitude falls with it.