what is the power dissipated in 2 ohm resistor?

Discussion in 'Homework Help' started by JITHINSMMB, Jan 21, 2013.

  1. JITHINSMMB

    Thread Starter New Member

    May 17, 2010
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    1
    what is the power dissipated in 2 ohm resistor when input current is

    i(t)=2+((2√2)cos(2t+45°))+2sin2t

    please help?
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    This is the Homework Help, not Homeword Done For You forum. So you need to show whatever work you have done in your best attempt to date to solve the problem. If nothing else, describe what you think the approach needs to involve, even if you don't see how to bring it all together just yet.
     
  3. JITHINSMMB

    Thread Starter New Member

    May 17, 2010
    9
    1
    calculated rms value of current and multiplied with 2 ohm.

    p=(√6^2)*2= 12w
    but answer is wrong!!!
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Okay, so let's take a step back.

    How did you come up with that as the RMS current?
     
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  5. JITHINSMMB

    Thread Starter New Member

    May 17, 2010
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    first simplified above equation to 2+2cos2t
    by adding rms value of each term, I got √6
     
  6. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    You summed real power vs imaginary power directly.

    That method is incorrect, which is why Power Factor correction is important.
     
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  7. JITHINSMMB

    Thread Starter New Member

    May 17, 2010
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    1
    thanks for your reply,
    then how can I proceed?
     
  8. WBahn

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    Mar 31, 2012
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    Is the correct answer 20W?
     
  9. JITHINSMMB

    Thread Starter New Member

    May 17, 2010
    9
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    correct answer is 3W , :confused::confused::confused:
     
  10. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Do you know what imaginary/reactive power is?

    Is that topic something your course has covered?
     
  11. JITHINSMMB

    Thread Starter New Member

    May 17, 2010
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    yes , I know it.
    till now I did not get expected help, so please close the thread.
     
    Last edited: Jan 21, 2013
  12. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Could you do us, and others with a question similar to yours, a favor and post your work to get the correct answer?
     
  13. WBahn

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    Mar 31, 2012
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    Yeah, I came up with that doing it in my head and almost as soon as I posted that I realized that I had screwed it up because one of the cross terms doesn't go away. I get 12W, the same as you. I got this by performing the square on the three terms directly.

    I've confirmed that you combined the two trig functions properly and I, too, get that the current is

    i(t) = 2A+2A*cos(2t)

    I even plotted it in the time domain in Excel and found the average power numerically and got 6W.

    So I think that your answer is correct. I don't know how they are getting 3W.

    Actually, I think I do. If this were the voltage across the resistor, and not the current through it, then you would have 3W.
     
  14. WBahn

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    Mar 31, 2012
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    I'm not following how you came to the conclusion that he did this.
     
  15. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Math error on my part. :rolleyes:
     
  16. WBahn

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    Mar 31, 2012
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    It seems to be going around. :rolleyes:

    The OP's the only one that seems to have their math act together on this one!
     
  17. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Letting Mathematica do the hard work, I get:

    [​IMG]]
     
  18. WBahn

    Moderator

    Mar 31, 2012
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    Yep. I think the author just screwed up and divided the square of his function by 2 instead of multiplying it by 2. The reason the mistake didn't get caught is because the author can't be bothered to track their units, so they forfeited the ability to spot the problem.
     
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