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A.C. Voltage waveforms and harmonics?

  1. Sep 18, 2011 #1
    Hi, I have the following question:

    An A.C. voltage, V comprises of a fundamental voltage of 100V rms at a frequency of 120Hz, a 3rd harmonic which is 20% of the fundamental, a 5th harmonic which is 10% of the fundamental and at a phase angle of 1.2 radians lagging.

    (1) Write down an expression for the voltage waveform

    (2) The voltage at 20ms (milliseconds)

    (3) Given an ideal V = 100V rms, what is the percentage error at 20ms

    I have an idea for the first and second parts but I get a huge percentage error in (3) which makes me doubt whether the first 2 parts were correct, can anyone help please?

    (1) Since V = Vrms x sqrt2 = 141.4V at 120Hz

    3rd harmonic = 20% of 141.4 = 28.3V at 360Hz

    5th harmonic = 10% of 141.4 = 14.1V at 600Hz

    v = [141.1sin(240πt)] + [28.3sin(720πt] + [14.1sin(1200πt+1.2)] π=pi


    (2) V at 20ms = 36.8 + 20.1 + 13.7 = 70.6V


    (3) Error is [(100-70.6)/100]X100 = 29.4% !? That seems like a massive error, where has my equation gone wrong?

    Help anyone!?
     
  2. jcsd
  3. Sep 18, 2011 #2

    uart

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    That 5th harmonic is 1.2 radians leading, not lagging.
     
  4. Sep 18, 2011 #3

    uart

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    BTW. You calculated the error incorrectly, however there is no reason why the percentage error could not be huge when calculated at a single point.

    For example, what would the percentage error by at t=0?
     
  5. Sep 19, 2011 #4
    Well spotted, it should be -1.2 but re-calculating V at 20ms it's still 70.5V. How have I calculated the error incorrectly?
     
  6. Sep 19, 2011 #5

    uart

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    Using your equations I get (2) V at 20ms = 82.9 + 26.9 - 13.1.
     
    Last edited: Sep 19, 2011
  7. Sep 20, 2011 #6
    Sorry my calculator wasn't in radians! Would that then make the error 3.3%?
     
  8. Jan 24, 2012 #7
    Hi
    Im stuck on the same question. Ive been using 141.4sin(2pi x f(120) x 0.02), which gives me 9.41v
    28.3Sin(2pi x 360 x 0.02) = 21.45v
    14.1sin(2pi x 600 x .02 - 1.2) = 18.08v

    Then added these up to give combined voltage at 20ms of 48.93v
    This is obviously a diffenrent answer to answers above, could someone please point out if I am doing something wrong.
    Many thanks
     
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