Average power on a resistor

In summary, a voltage source with values $$R=5 \Omega , C =50 \mu F, L=5 mH$$ connected in series has an average power of 325 W on resistor R when supplied with a voltage of $$u(t)=150 \sin (1000t)+100 \sin(2000t)$$ using superposition theorem to calculate the individual contributions of each source voltage component. The rms values of the two source voltages were used to find the individual rms currents and power dissipated by the resistor, which were then summed to get the final result.
  • #1
crom1
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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.

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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  • #2
I believe superposition theorem will be helpful here.
 
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  • #3
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.
 
  • #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.
 
  • #5
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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  • #6
Thank you very much, I get the correct result.
 

What is average power on a resistor?

The average power on a resistor is the amount of energy that is dissipated or consumed by a resistor over a given period of time. It is measured in watts and is a key factor in determining the performance and durability of a resistor.

How is average power on a resistor calculated?

The average power on a resistor can be calculated by multiplying the voltage across the resistor by the current flowing through it. This is known as the "power formula" and is represented by the equation P = VI, where P is power in watts, V is voltage in volts, and I is current in amperes.

What factors affect the average power on a resistor?

The average power on a resistor is affected by the voltage and current passing through it, as well as the resistance of the resistor itself. Other factors that can influence the average power include the ambient temperature, the type of material the resistor is made of, and the frequency of the electrical signal passing through the resistor.

Why is the average power on a resistor important?

The average power on a resistor is important because it helps determine the maximum power that a resistor can handle without overheating or being damaged. It also affects the efficiency of a circuit and can impact the overall performance of electronic devices.

How can the average power on a resistor be managed?

The average power on a resistor can be managed by selecting a resistor with the appropriate power rating for the intended application. This can ensure that the resistor can handle the amount of power passing through it without overheating. Additionally, using multiple resistors in parallel can also distribute the power and reduce the average power on each individual resistor.

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