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Power dissipation in a circuit (Kirchoff's Laws)

  1. Mar 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Three resistors, R1 = 27 Ω, R2 = 42 Ω, and R3 = 53 Ω, are connected in a multiloop circuit, as shown in the figure. Determine the amount of power dissipated in the three resistors.



    2. Relevant equations

    Junction Rule
    Loop Rule


    3. The attempt at a solution

    I am getting mired in the math.

    i2 + i3 = i1

    V2 + i2R2 - i1R1 = 0

    V1 + i1R1 - i3R3 = 0

    I try using these equations to reduce it down to one variable, but I can't get there.
     

    Attached Files:

  2. jcsd
  3. Mar 4, 2013 #2

    SteamKing

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    Substitute the values for the voltages and resistors. You have three equations and three unknowns. You should be able to solve this system (try using Cramer's rule after making the substitutions)
     
  4. Mar 4, 2013 #3
    Substituting the values helped me reduce it, but it produced the wrong answer. This leads me to believe my original equations are wrong. Is there anything that looks bad about them?
     
  5. Mar 4, 2013 #4
    i believe the 1st equation will be i1+i2+i3=0... try using that
     
  6. Mar 4, 2013 #5
    That isn't working. I'm not saying it isn't right, but if it is, then something else is messed up.
     
  7. Mar 4, 2013 #6
    the third equation is completely wrong!! how did u arrive at that first tell me
     
  8. Mar 4, 2013 #7
    I'm going around the circuit. Other than that, I don't have a good rational. How do you determine it?
     
  9. Mar 4, 2013 #8
    i don't think that's gonna work... there's a cell and emf in the middle... instead of that, try considering the bottom loop....
     
  10. Mar 4, 2013 #9
    So it would be:

    V1 + V2 +i2R2 - i3R3 = 0

    Is that right?
     
  11. Mar 4, 2013 #10
    yeah.... doo they work out?
     
  12. Mar 4, 2013 #11
    I get something sort of close to the final answer, but it isn't right.

    For power, I use:

    P = i^2(R)
     
  13. Mar 4, 2013 #12
    Hold up. I found an arithmetic error in my process. I think this will work.
     
  14. Mar 4, 2013 #13
    Yep, it worked. Thanks for your help.
     
  15. Mar 4, 2013 #14
    good job
     
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