All About Circuits
All About Circuits

AC Electric Circuits

Peak, Average, and RMS Measurements

10 questions By Tony R. Kuphaldt

PDF Version
Page 4 of 4 0 of 10 answers revealed (0%)

  • Question 10 of 10

    In calculating the thickness of insulators for high-voltage AC power lines, which type of measurement is the best to use for voltage: peak, average, or RMS? Explain why.
    Reveal answer

Related Tools:

Published under the terms and conditions of the Creative Commons Attribution License
Comments (3)
Log in to comment
  • F
    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.

    Like. Reply
    • 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.
      Like. Reply
      • 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.
        Like. Reply

You May Also Like