AC voltage on the oscilloscope

Curricula: GCSE (ac and dc), A-level, Sek I

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Aim

Measure the period, frequency, amplitude and RMS value of an alternating voltage.

The bench

AC voltage source: sine wave, amplitude 5 V, 50 Hz. A voltmeter in AC mode and an oscilloscope (CH1, 2 V/div, 5 ms/div) are connected to its output.

Procedure

1. Use the screen grid to find the period and the peak-to-peak value of the signal.
2. Compare with the automatic measurements on the oscilloscope's menu strip (Vpp, Vrms, frequency, period).
3. Switch the voltmeter to DC mode, then back to AC.
4. Set 100 Hz on the source, then select the "Square" waveform.

Expected result

The period is 4 divisions (20 ms), the frequency is 50 Hz; the peak-to-peak value is 5 divisions (10 V). The voltmeter in AC mode shows about 3.54 V — the amplitude divided by \(\sqrt{2}\); in DC mode — about 0 (the mean of a sine wave). At 100 Hz a period takes 2 divisions. For a square wave with an amplitude of 5 V the RMS value is 5 V.

Questions

1. Why does the voltmeter show 3.54 V and not 5 V?
2. How many periods of 50 Hz fit on the screen at 5 ms/div?

Answers

1. In AC mode the voltmeter shows the RMS value: the DC voltage that would heat a resistor equally. For a sine wave it is \(U_{rms} = U_0 / \sqrt{2}\) = 5 V / 1.414 ≈ 3.54 V; 5 V is the amplitude, which the voltage reaches only at the peaks.
2. The screen is 10 divisions wide, so at 5 ms/div it shows 10 × 5 = 50 ms. A period of 50 Hz lasts \(T = 1 / f\) = 20 ms, so 50 / 20 = 2.5 periods fit on the screen.