All About Circuits
All About Circuits

AC Electric Circuits

Peak, Average, and RMS Measurements

10 questions By Tony R. Kuphaldt

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  • Question 4 of 10

    Determine the RMS amplitude of this sinusoidal waveform, as displayed by an oscilloscope with a vertical sensitivity of 0.2 volts per division:

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  • Question 5 of 10

    Determine the RMS amplitude of this square-wave signal, as displayed by an oscilloscope with a vertical sensitivity of 0.5 volts per division:

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  • Question 6 of 10

    Suppose two voltmeters are connected to source of “mains” AC power in a residence, one meter is analog (D’Arsonval PMMC meter movement) while the other is true-RMS digital. They both register 117 volts while connected to this AC source.

    Suddenly, a large electrical load is turned on somewhere in the system. This load both reduces the mains voltage and slightly distorts the shape of the waveform. The overall effect of this is average AC voltage has decreased by 4.5% from where it was, while RMS AC voltage has decreased by 6% from where it was. How much voltage does each voltmeter register now?
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Comments (3)
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    Fabrizio_EL April 03, 2020

    Hello, sorry but i doubt the asnwer to question #5, or at least i would ask for a better explanation.
    For what i know, RMS amplitude of such a waveform is 1 x sqrt(0,5), while 0,5 seems to be average amplitude.

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    • RK37 April 06, 2020
      The answer is correct. The signal—call it x(t)—is a square wave that transitions between 0.5 V and -0.5 V. To find the RMS value, we square x(t), then integrate it over a given period of time T, then divide by T, then take the square root. Let's say that x(t) has a period of 2 seconds. After squaring x(t), we have a signal that is a constant voltage of 0.25 V, because 0.5^2 = 0.25 and (-0.5)^2 = 0.25. If we integrate for T = 2 seconds, the result is 0.25 * 2 = 0.5. Dividing by T, we have 0.5/2 = 0.25, and taking the square root of 0.25 yields 0.5. This works for any number that you choose for T. The important concept is the following: If you square a signal like the one shown in the diagram, you end up with a constant value. The average value of a constant value is the same constant value, and thus when you take the square root, you "undo" the square operation and return to the original amplitude.
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      • F
        Fabrizio_EL April 14, 2020
        Many thanks, actually my (huge) mistake was to consider that waveform from 0 to 1, since i didn't notice the "0" line through it. So i used the formula Vrms = sqrt( d% * Vhigh^2 + (1-d%) * Vlow^2 ), with d = 0,5, Vhigh = 1, Vlow = 0. Using instead this formula with the correct values Vhigh = 0,5 and Vlow = -0,5 leads to your same result. And now that you explained with simple integral calculus i can see the equivalence (mathwise) between the two methods. Thank you again for being so clear.
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