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RMS value

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

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Aim

Compare the effect of alternating and direct current and understand the meaning of the RMS value.

The bench

At the top — a 7.07 V DC voltage source and a 12 V / 5 W lamp. At the bottom — a sinusoidal voltage source with an amplitude of 10 V, 50 Hz, and an identical lamp, with a voltmeter in AC mode above it. On the right — an oscilloscope connected to the AC voltage source (CH1, 5 V/div, 5 ms/div).

Procedure

  1. Compare the brightness of the lamps.
  2. Record the voltmeter reading and the oscilloscope's automatic measurements (Vpp, Vrms).
  3. Switch the voltmeter to DC mode.
  4. On the AC source, select the square wave and compare the brightness again.
Expected result

The lamps burn equally bright: a sine wave with an amplitude of 10 V heats the filament just as 7.07 V DC does. The AC voltmeter reads about 6.9 V: part of the voltage is dropped across the source's 0.5 Ω internal resistance. The peak-to-peak value on the oscilloscope is about 19.5 V. In DC mode the voltmeter reads about 0. For a square wave with an amplitude of 10 V the RMS value is 10 V, and the lower lamp burns brighter.

Questions

  1. Why is the RMS value of a sine wave \(\sqrt{2}\) times smaller than its amplitude?
  2. What DC voltage is needed to make the lamps equally bright after switching to the square wave?
Answers
  1. The heating power in a resistance is \(u^2 / R\). For a sine wave \(u^2 = U_m^2 \cdot \sin^2(\omega t)\), and \(\sin^2\) averages to one half over a period, so the DC voltage that heats the same satisfies \(U^2 = U_m^2 / 2\), that is \(U = U_m / \sqrt{2}\): 10 V / √2 ≈ 7.07 V on this bench.
  2. 10 V. A square wave with an amplitude of 10 V is at +10 V or −10 V at every moment, and the filament heats equally for either polarity, so the RMS value of the square wave equals its amplitude.