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Simple Power Problem HELP!

  1. Feb 6, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    How much power is dissipated by R1? (R1=8 Ω, R2=17 Ω, V=6 V.)

    Sorry the picture is so big!!!

    2. Relevant equations



    3. The attempt at a solution

    P = E^2/R = 8^2/6 = 10.67W what am I doing wrong?
     
  2. jcsd
  3. Feb 6, 2008 #2
    Do I have to add the resistors??
     
  4. Feb 6, 2008 #3
    No P=IV so the change in power is the dissipation of power so delta P=IV-IV
    so use v=IR and get P=I^2R -I^2R for the points after and at (before) the resistor, also remember always final-initial
     
  5. Feb 7, 2008 #4
    So I would do...
    I = E/Req = 6/25
    P = 0.24^2 * 8 - 0.24^2 * 17 = -0.5184
    Negative? Why would it be negative?
     
  6. Feb 7, 2008 #5
    Can someone explain this....

    No P=IV so the change in power is the dissipation of power so delta P=IV-IV
    so use v=IR and get P=I^2R -I^2R for the points after and at (before) the resistor, also remember always final-initial

    I don't understand what Vi and Vf are?
     
  7. Feb 7, 2008 #6
    so i need to multiply the voltage drop across the resistors by the current. but how do i calculate the voltage drop?
     
  8. Feb 7, 2008 #7
    solved.....
     
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