AC Power Worksheet - Q39

Discussion in 'Homework Help' started by Spoon, May 16, 2010.

  1. Spoon

    Thread Starter Member

    May 18, 2009
    12
    0
    Hi,

    I'm currently doing the AC power worksheet on this website, found at url
    http://www.allaboutcircuits.com/worksheets/acpower.html

    When calculating the capacitor value I keep getting exactly double the value.

    I calculated the initial power factor to be 0.707 (lagging), which agrees with the solution.

    The reactance of the inductor is Xl = 1.0824∏.
    Xc = 1 / (2∏fC)
    therefore C = 1 / (1.0824∏*2∏*60)
    C = 7 . 8 * 10^-4
    C = 780 uF which is exactly double the required value of 390uF

    Thanks,
    Steve
     
  2. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
    312
    You must be missing the final C=(C*.5) in the equation ;)

    What 'Question #' in the sheet are you referring to?
     
  3. Spoon

    Thread Starter Member

    May 18, 2009
    12
    0
    as stated in the title, Question #39 ... thanks
     
  4. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
    312
    Jeez, I cant read today..or yesterday.. I apologize. Gimme a minute.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If you were placing C in series with the load then 780uF would be correct. In the parallel placement you only have to compensate for the reactive part of the load current - 390uF is correct in that case..
     
    Spoon likes this.
  6. Spoon

    Thread Starter Member

    May 18, 2009
    12
    0
    Thanks, "only have to compensate for the reactive part of the load current". Was doing this wrong, silly mistake. I agree with the answer now
     
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