How to Calculate Average Power Dissipation on a Resistor?

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Discussion Overview

The discussion revolves around calculating the average power dissipation on a resistor given a specific voltage signal. Participants explore the application of formulas related to power calculation, particularly in the context of alternating current (AC) signals and their components.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a voltage function and seeks to find the average power dissipated on a 14 kΩ resistor, using the formula Pavg = (sqrt(Vdc^2 + vp^2/2))/R.
  • Another participant questions the meaning of vp and how to combine the two cosine terms into one AC term.
  • Clarifications are made regarding vp being the peak voltage, with values of 12 and -9 volts mentioned.
  • Concerns are raised about the necessity of combining AC components to obtain a resultant waveform, especially considering phase differences.
  • One participant suggests using Phasors to solve the problem, but another expresses uncertainty about having covered Phasors in class.
  • A later reply corrects a misunderstanding about the frequencies of the AC components, noting they are off by a factor of two.
  • One participant reports successfully obtaining the correct answer using power spectrum analysis and questions the division by 2 in the calculation of Vrms.

Areas of Agreement / Disagreement

Participants express differing views on the approach to combining AC components and the use of Phasors. There is no consensus on the best method to calculate the average power dissipation, and some misunderstandings about the frequency of the components are clarified but not fully resolved.

Contextual Notes

Participants reference specific formulas and methods without fully resolving the mathematical steps involved in combining AC components or the rationale behind certain calculations, such as the division by 2 in the power calculation.

eatsleep
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1. Given that v(t) = -6 + 12 cos(2π200t + π/4) - 9 cos(2π400t + π/6) volts. The signal is applied to a 14 kΩ resistor. Find the average power dissipated on the resistor in mW. Round your answer off to two decimal places.



2. Pavg = (sqrt(Vdc^2 + vp^2/2))/R



3. I used the above equation just P=Vrms^2/R to calculate the power, this is wrong and I'm not sure what to do.
 
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eatsleep said:
1. Given that v(t) = -6 + 12 cos(2π200t + π/4) - 9 cos(2π400t + π/6) volts. The signal is applied to a 14 kΩ resistor. Find the average power dissipated on the resistor in mW. Round your answer off to two decimal places.



2. Pavg = (sqrt(Vdc^2 + vp^2/2))/R



3. I used the above equation just P=Vrms^2/R to calculate the power, this is wrong and I'm not sure what to do.

What's vp? How did you combine the two cos() terms to get one AC term?
 
berkeman said:
What's vp? How did you combine the two cos() terms to get one AC term?

I did not combine them i just added them together. vp is meant to be the the voltage peak. It would be 12 and -9.
 
eatsleep said:
I did not combine them i just added them together. vp is meant to be the the voltage peak. It would be 12 and -9.

That's not correct. You need to combine the AC components to get a resultant AC waveform. What if the phases of the two AC waveforms are exactly 180 degrees apart? You will not get a very big AC waveform then.

The frequencies of the AC components are the same, so you can use Phasors to solve this question...
 
berkeman said:
That's not correct. You need to combine the AC components to get a resultant AC waveform. What if the phases of the two AC waveforms are exactly 180 degrees apart? You will not get a very big AC waveform then.

The frequencies of the AC components are the same, so you can use Phasors to solve this question...

Is that the only way to do this question, I do not believe we have done phasors in class, also how are the frequencies the same?
 
eatsleep said:
Is that the only way to do this question, I do not believe we have done phasors in class, also how are the frequencies the same?

Oh, oops, you are right. They are off by a factor of two.
 
berkeman said:
Oh, oops, you are right. They are off by a factor of two.

I just got the right answer I used power spectrum and did Vrms^2 and added all the individual signal powers up. On the sinusoidal waveforms I divided by 2 Vrms^2/2 I am not sure why, it said that in my notes. Why did they divide by 2?

Thanks for the help
 

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