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Calculating Average Power

  1. Mar 18, 2013 #1
    1. Given that v(t) = -6 + 12 cos(2π200t + π/4) - 9 cos(2π400t + π/6) volts. The signal is applied to a 14 kΩ resistor. Find the average power dissipated on the resistor in mW. Round your answer off to two decimal places.



    2. Pavg = (sqrt(Vdc^2 + vp^2/2))/R



    3. I used the above equation just P=Vrms^2/R to calculate the power, this is wrong and I'm not sure what to do.
     
    Last edited by a moderator: Mar 18, 2013
  2. jcsd
  3. Mar 18, 2013 #2

    berkeman

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    Staff: Mentor

    What's vp? How did you combine the two cos() terms to get one AC term?
     
  4. Mar 18, 2013 #3
    I did not combine them i just added them together. vp is meant to be the the voltage peak. It would be 12 and -9.
     
  5. Mar 18, 2013 #4

    berkeman

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    That's not correct. You need to combine the AC components to get a resultant AC waveform. What if the phases of the two AC waveforms are exactly 180 degrees apart? You will not get a very big AC waveform then.

    The frequencies of the AC components are the same, so you can use Phasors to solve this question...
     
  6. Mar 18, 2013 #5
    Is that the only way to do this question, I do not believe we have done phasors in class, also how are the frequencies the same?
     
  7. Mar 18, 2013 #6

    berkeman

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    Oh, oops, you are right. They are off by a factor of two.
     
  8. Mar 18, 2013 #7
    I just got the right answer I used power spectrum and did Vrms^2 and added all the individual signal powers up. On the sinusoidal waveforms I divided by 2 Vrms^2/2 I am not sure why, it said that in my notes. Why did they divide by 2?

    Thanks for the help
     
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