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Average power on a resistor

  • Thread starter crom1
  • Start date
57
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1. Homework Statement
$$R=5 \Omega , C =50 \mu F, L=5 mH$$ are connected in series on voltage source $$u(t)=150 \sin (1000t)+100 \sin(2000t)$$. Find the average power on resistor R.

2. Homework 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:

cnh1995

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I believe superposition theorem will be helpful here.
 

gneill

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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.
 
57
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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.
 

gneill

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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.
 
57
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Thank you very much, I get the correct result.
 

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