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Average power on a resistor

  1. Jan 26, 2016 #1
    1. The problem statement, all variables and given/known data
    $$R=5 \Omega , C =50 \mu F, L=5 mH$$ are connected in series on voltage source $$u(t)=150 \sin (1000t)+100 \sin(2000t)$$. Find the average power on resistor R.

    2. Relevant equations $$
    P=\frac{U^2}{R} $$,
    $$ U_{ef}= \sqrt{ \frac{1}{T} \int_{0}^{T} u^2(t) dt } $$


    3. The attempt at a solution

    I got that rms of u(t) is 127.48 , and now i am not sure which omega should I use to calculate Xc and XL.
    The period of function u(t) is $$2 \pi/1000$$, so that implies I should use $$\omega=1000$$ but I can't get the correct solution. (I don't get correct solution when I use 2000 either). I am obviously doing something , can you help?
     
    Last edited by a moderator: Jan 26, 2016
  2. jcsd
  3. Jan 26, 2016 #2

    cnh1995

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    Homework Helper

    I believe superposition theorem will be helpful here.
     
  4. Jan 26, 2016 #3

    gneill

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    Staff: Mentor

    What result value are you getting? Show your calculation. Your rms value for the voltage source looks good (but be sure to include units when you show a result!).


    P.S.: Use a capital omega for ohms: \Omega yields ##\Omega## in LaTeX.
     
  5. Jan 26, 2016 #4
    For omega=1000, I get Xc= -j20 ohms and XL= j5 ohms . So the impedance is Z=5-j15. Now I need to find the current. Do I just use rms value of voltage and divide it by |Z|?
    I did that, got I=8.06 A, and then power is 325 W, but that's wrong.
     
  6. Jan 26, 2016 #5

    gneill

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    Staff: Mentor

    Heed @cnh1995 's clue. You can use superposition here to work out the individual contributions of the two source voltage components.

    It's a simple matter to write the rms values for the two source voltages directly from their definitions. Then find the individual rms currents due to each (using ##I = u/|Z|## as you've written, where here u is the individual rms voltage of a source), and then power dissipated by the resistor due to each. Then sum the results.
     
  7. Jan 26, 2016 #6
    Thank you very much, I get the correct result.
     
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