How Is Average Power Calculated in a Series RLC Circuit?

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Homework Help Overview

The problem involves calculating the average power in a series RLC circuit with given resistance, capacitance, and inductance values, connected to a voltage source composed of two sinusoidal components. Participants are exploring the implications of the circuit's parameters on power calculation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the rms voltage and the appropriate angular frequency for calculating reactance. There is mention of the superposition theorem as a potential method to analyze the contributions of the voltage components. Questions arise regarding the correct calculation of impedance and current, as well as the proper application of formulas for power.

Discussion Status

The discussion is active, with participants sharing calculations and questioning their results. Some guidance has been offered regarding the use of superposition and the need to calculate individual contributions to power from each voltage source. There is no explicit consensus, but progress is being made towards understanding the problem.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance they can receive. There is an emphasis on ensuring correct units and interpretations of results.

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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