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 1 of 10

    How is it possible to assign a fixed value of voltage or current (such as “120 volts”) to an AC electrical quantity that is constantly changing, crossing 0 volts, and reversing polarity?
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  • Question 2 of 10

    Suppose a DC power source with a voltage of 50 volts is connected to a 10 Ω load. How much power will this load dissipate?

    Now suppose the same 10 Ω load is connected to a sinusoidal AC power source with a peak voltage of 50 volts. Will the load dissipate the same amount of power, more power, or less power? Explain your answer.
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  • Question 3 of 10

    Suppose that a variable-voltage AC source is adjusted until it dissipates the exact same amount of power in a standard load resistance as a DC voltage source with an output of 120 volts:

    In this condition of equal power dissipation, how much voltage is the AC power supply outputting? Be as specific as you can in your answer.
    Reveal answer

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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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