Rms

Discussion in 'Homework Help' started by Zulkifal, Feb 7, 2016.

Is my answer alright

  1. yes

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  2. no

    1 vote(s)
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Multiple votes are allowed.
  1. Zulkifal

    Thread Starter New Member

    Feb 5, 2016
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    Need help please
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    So you just want people to vote on whether your answer is correct? Okay.
     
  3. Zulkifal

    Thread Starter New Member

    Feb 5, 2016
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    yes and if its wrong help me figure it out please...
     
  4. Zulkifal

    Thread Starter New Member

    Feb 5, 2016
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    my bad its 1/20 so the answer would be 9.7467
     
    Last edited: Feb 7, 2016
  5. WBahn

    Moderator

    Mar 31, 2012
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    You're at least closer.

    How much total energy would be dumped into a 1 Ω resistor over the course of the waveform?

    What is the average power being dumped into at 1 Ω resistor?

    What DC voltage would dump that same average power into that same 1 Ω resistor?
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    We can't help you figure out what you did wrong unless you show us what you did.
     
  7. Zulkifal

    Thread Starter New Member

    Feb 5, 2016
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    Sir Correct me if i'm wrong
    power being dumped is p=v^2/r p=(9.7467)^2/1kiloohm
    9.7467 Dc voltage would dump the same power
    I figured out the rms value to find dc equivalent voltage as average of ac is zero so i take the Rms by formula v=sqrt1/Tintegral(v)^2dt
     
  8. MrAl

    Well-Known Member

    Jun 17, 2014
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    Isnt the period 20, not 10 ?
     
    Zulkifal likes this.
  9. WBahn

    Moderator

    Mar 31, 2012
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    This would be true only if Vrms = 9.7467 V, and it doesn't. You can't just assume that your answer is correct when the goal is to determine that your answer is correct.

    Don't cram everything together, the equation "v=sqrt1/Tintegral(v)^2dt" is almost unreadable.

    Assuming you mean v = sqrt( (1/T) integral((v^2)dt) ), then please SHOW how you used this to get 9.7467 V. We have absolutely no way to point out where you are going wrong as long as you refuse to show your work. That should not be a difficult concept to understand.
     
  10. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I agree with WBahn that you haven't shown your work very well, but I can see in the image attached to post #1 that your remaining problem is a simple arithmetic mistake.

    You are evaluating:
    RMS1.png

    when you should be evaluating:
    RMS2.png
     
  11. RBR1317

    Active Member

    Nov 13, 2010
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    So I wondered if it would work for a 1Ω resistor, would it also work for a 33Ω resistor? And indeed it does, but it would still be simpler to calculate the RMS voltage as ½√350.

    RMS-33-resistor.png
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    Agreed. Suggesting that he use a specific resistor value was just intended to make the concepts extremely concrete. But the concept is that it works for ANY resistor value (including a non-specified symbol resistance R) and you can easily show that there is no need to even use that. It all falls out from the definition of "effective" power (not even "rms" power -- the "rms" is a direct consequence of the definition of "effective" power).
     
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