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Power dissipation in resistors

  1. Jan 14, 2012 #1
    Calculate the power dissipated in R1, R2 and R



    [/b] Fig1.jpg



    Previously i worked out the Current I1, I2 and I3 so i planned to use this to calculate the power dissipated;

    For current i got;

    I1= 0.8409 A
    I2= -0.4545 A
    I3= 0.39 A

    I used I^2/R to get the following answers

    R1 = 0.8409^2 / 1 = 0.7071 W
    R2 = -0.4545^2 / 2 = -0.4131 W
    R3 = 0.39^2 / 3 = 0.4563 W
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 14, 2012 #2
    Think about these I*V=watts V=I*R
     
  4. Jan 14, 2012 #3

    gneill

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    Staff: Mentor

    A negative real number squared gives a positive result!
     
  5. Jan 14, 2012 #4
    Of course............

    R1 = 0.8409^2 / 1 = 0.7071 W
    R2 = -0.4545^2 / 2 = 0.2066 W
    R3 = 0.39^2 / 3 = 0.4563 W

    Am i on the right lines with this?
     
  6. Jan 14, 2012 #5
    You were right first time I should have checked but your notation fooled me / usually means divide * is multiply.
     
  7. Jan 14, 2012 #6

    gneill

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    First, the power is p = V2/R or p = I2R. Since you are using currents and resistances the second equation applies here. So you should be MULTIPLYING the square of the currents by the resistance values, not dividing as you've indicated. Recheck the results with this in mind.
     
  8. Jan 14, 2012 #7
    so

    p=(I^2)R

    thus

    (0.8409^2)*1 = 0.70711 W
    (-0.4545^2)*2 = 0.41314 W
    (0.39^2)*3 = 0.4563 W

    Hopefully thats it!
     
  9. Jan 14, 2012 #8

    gneill

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    Yes, that looks good for the given resistances and currents.
     
  10. Jan 14, 2012 #9
    Thanks for your help
     
  11. Jan 16, 2012 #10
    charger, can you show me how you worked out those currents?
     
  12. Jan 16, 2012 #11

    NascentOxygen

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    Staff: Mentor

    Discussion of this circuit in this thread.
     
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