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Phase shift

Curricula: Sek II

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Aim

See the phase shift between current and voltage in a capacitor.

The bench

Source: sine, amplitude 5 V, 100 Hz. A 10 µF capacitor and a 10 Ω shunt in series; the voltage across the shunt is proportional to the current. Oscilloscope: CH1 — the source (2 V/div), CH2 — the shunt (0.1 V/div), DUAL mode, 2 ms/div.

Procedure

  1. Compare the positions of the maxima of the two sine waves.
  2. Measure the time shift and convert it to degrees: \(360^\circ \cdot \Delta t / T\).
  3. In the menu on the oscilloscope screen, select X-Y mode.
Expected result

The period is 10 ms — 5 divisions. The current (CH2) leads the voltage by almost a quarter of a period — about 2.4 ms, almost 90°: besides the capacitance, the circuit contains the shunt and the source's internal resistance. The amplitude across the shunt is about 0.31 V. In X-Y mode you get an ellipse with its axes almost along the grid.

Questions

  1. Why does the current in a capacitor lead the voltage?
  2. What will the X-Y figure look like if the capacitor is replaced by a resistor?
  3. How will the phase shift change if an inductor is used instead of the capacitor?
Answers
  1. The current is the rate at which charge flows onto the plates, and the voltage is proportional to the charge, \(U = Q / C\). The current is therefore greatest when the voltage changes fastest, as it passes through zero, and zero when the voltage is at its peak and momentarily stops changing, so the current peaks a quarter of a period earlier: 2.5 ms at 100 Hz. On the bench it is 2.4 ms because \(X_C\) ≈ 159 Ω is in series with the 10 Ω shunt and the 0.5 Ω internal resistance of the source.
  2. In a resistor current and voltage are in phase, so the ellipse closes into a straight line through the centre, rising to the right. Its tilt depends on the resistance and on the channel scales.
  3. The current would lag the voltage: the CH2 maxima would come after the CH1 maxima by a little less than a quarter of a period, almost 90°. In an inductor the self-induced EMF opposes the change of current, so the current peaks only after the voltage has passed its own peak; the winding resistance, the shunt and the internal resistance keep the angle below 90°.