Power Dissipation in a Resistor: Calculate Average

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Discussion Overview

The discussion revolves around calculating the average power dissipated in a resistor given a specific current and voltage function. The scope includes mathematical reasoning and conceptual clarification related to power calculations in electrical circuits.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant presents the current and voltage equations and attempts to calculate power using the formula p=ui, expressing confusion about the result.
  • Another participant suggests using the root-mean-square (rms) values for voltage and current to find the average power, indicating that the power factor for a resistor is 1.
  • A request for clarification on the term "rms" and the notation for angular frequency (ω) is made, indicating a lack of understanding of these concepts.
  • Further clarification is provided about rms and an alternative method for finding the average value of the power function by integrating over a specified interval.
  • One participant expresses uncertainty about the concept of root-mean-square and requests a step-by-step solution to the problem.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the understanding of rms or the method to solve the problem, as there are multiple requests for clarification and differing levels of familiarity with the concepts involved.

Contextual Notes

There are unresolved assumptions regarding the participants' familiarity with electrical engineering concepts such as rms and integration techniques. The discussion reflects varying levels of understanding and the need for foundational explanations.

Who May Find This Useful

This discussion may be useful for students learning about electrical power calculations, particularly those unfamiliar with root-mean-square values and integration in the context of sinusoidal functions.

dortec
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A resistor draws a current i=8sinwt at a voltage of v=200sinwt. Calculate the average power dissipated in the resistor.

What i did is p=ui = 1600sin^2 (wt) and i got stuck:P i don't think it's the right equation.. the answer should be 800W .. and that's nothing like it.
Could someone help me ?
 
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Hi dortec! :smile:

(have an omega: ω and try using the X2 tag just above the Reply box :wink:)

Hint: for a resistor, the power factor, cosφ, is 1, and so Paverage = VrmsIrms :wink:
 
Can you rephrase what you just said please. i actually didnt understand. wat's rms? and what do u mean by have an omega.. can u re-explain please:S
 
Hi dortec! :smile:

rms means "root-mean-square" … do you know what that is?

(if you don't, you can get the same result by finding the average value of sin2ωt, which you get by integrating it from 0 to 2π)

(and where you typed w, I assumed you would have preferred ω. :wink:)
 
nope i don't know what root-mean-square:S and my bad for the omega thing:P.. can u please show me how to solve it?
 
ok, never mind if you haven't done root-mean-square …

just find the average value of your original P(t) = 1600sin2ωt, which you get by integrating it from 0 to 2π/ω, and dividing by … ? :smile:

(alternatively, just write 1600sin2ωt in terms of cos2ωt and/or sin2ωt, and then the average is obvious :wink:)
 

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