How Is Average Power Calculated in a Series RLC Circuit?

AI Thread Summary
The discussion focuses on calculating the average power in a series RLC circuit with given resistance, capacitance, and inductance values, using a voltage source with two frequency components. The user initially calculates the RMS voltage but struggles with determining the correct angular frequency to use for calculating reactance values. They receive guidance to apply the superposition theorem to analyze the contributions from each frequency component separately. By calculating the individual RMS currents and the power dissipated by the resistor for each component, the user ultimately arrives at the correct average power result. This approach highlights the importance of using superposition in circuits with multiple frequency sources.
crom1
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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.

Homework Equations

[/B]$$
P=\frac{U^2}{R} $$,
$$ U_{ef}= \sqrt{ \frac{1}{T} \int_{0}^{T} u^2(t) dt } $$

The Attempt at a Solution


[/B]
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?
 
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I believe superposition theorem will be helpful here.
 
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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.
 
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.
 
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.
 
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Thank you very much, I get the correct result.
 
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