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Rms current in microwave

  1. Mar 7, 2009 #1
    1. The problem statement, all variables and given/known data
    peak power consumed by microwave is 1180 W when operated at 120 Vrms. What rms current does that microwave draw?


    2. Relevant equations
    I'm not getting the correct answer, I think I am getting I and V confused with Irms and Vrms and Imax.

    3. The attempt at a solution
    I=P/V
    Peak=I^2R
    Vrms=.707V
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 7, 2009 #2

    lanedance

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    so waht answer do you get & how?
     
  4. Mar 7, 2009 #3
    I found a value for R with P=V^2/R and tried plugging that into P=I^2R, then I took that I multiplied by .707
     
  5. Mar 8, 2009 #4

    lanedance

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    best to work all in rms or all in peak, never combine the 2

    note Vrms = Vp/sqrt(2) where Vp is peak V
    and Irms = Ip/sqrt(2)

    If P = VI, what is the relation between peak & avg power

    If we say

    Pavg = Vrms.Irms
    Pp = Vp.Ip
     
    Last edited: Mar 8, 2009
  6. Mar 8, 2009 #5
    thats the part I don't understand, how to combine the two equations to get the relationship between the two, are they equal because power is conserved?
     
  7. Mar 8, 2009 #6

    lanedance

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    P=VI is always true & its fine to use either (rms or peak), so you know you are working in peak or rms exclusively

    if you combine the definitions below they give
    Pp = 2.Pavg
    this is probably where the confusion comes in

    so convert to avg power then use P = VI
     
  8. Mar 8, 2009 #7
    the answer is just not coming out:
    I plugged in Ppeak=2(Vrms*Irms)
    which is 1180=2(120)*2Irms
    1180/240=2Irms
    Irms=2.458

    am I not getting something? or is it a number thing?
     
  9. Mar 8, 2009 #8

    lanedance

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    ? you have added an extra factor of two here, any reason?

     
  10. Mar 8, 2009 #9
    I assumed I had to multiply 2 times each factor 2(VI). I got it now... thank you so much for your help!
     
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