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Kirchhoff Problem

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data
    I have been given the above information and asked to find the current in the edge ac. s is the sink, t is the source.

    2. Relevant equations
    Kirchoff's First Law, current in = current out.
    Kirchoff's Second Law, the sum of resistance.current for all vertices in a cycle = 0.

    3. The attempt at a solution
    Firstly I have assumed that all currents connected to the sink go into the sink, and all connected to the source go away from the source. I have also assumed the current flows from c to b, and c to a. I have made these assumptions to make sure there cannot be an infinite cycle in the graph and I simply cannot see a way to solve this if the graph is not directed.

    So from kirchoff 1 i have:
    Icb - 10 - Ibs = 0
    Itc - Ica - Icb = 0
    Ita + Ica + 10 - Ias = 0

    And from kirchoff 2:
    cycle tca : 6Itc + Ica - 3Ita = 0
    cycle cab : Ica - 20 - 3Icb = 0
    cycle bas : 20 + Ias - 2Ibs = 0

    However I cannot seem to solve these simultaneously to get any closer to my result,and I'm not sure my initial assumptions about the direction of flow of current are valid, any hints/tips would be appreciated!
     
    Last edited by a moderator: Nov 14, 2009
  2. jcsd
  3. Nov 14, 2009 #2
    I was able to solve them simultaneously using the matrix solver on my HP50 calculator:

    Ibs = -21.6667
    Ias = -63.3333
    Icb = -11.6667
    Ica = -15.0000
    Ita = -58.3333
    Itc = -26.6667

    The magnitude of the sink and source currents is 85.
     
  4. Nov 14, 2009 #3
    So are my assumptions about the direction of current correct?
    and why are these values all negative?
    I mean I'm not even sure im using the correct method! :S
     
  5. Nov 14, 2009 #4
    @ tiny-tim : OK, thanks a lot! I'll work on that!!
     
    Last edited by a moderator: Nov 14, 2009
  6. Nov 14, 2009 #5

    tiny-tim

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    Welcome to PF!

    Hi Dench! Welcome to PF! :smile:

    {have an omega: Ω :wink:)
    Yes, the graph does have to be directed, but there's no need to make the correct assumptions … if you guess wrong, the current will come out negative instead of positive … you won't lose any marks for it! :wink:

    (and obviously, if you use I1 I2 I3 etc, you have to draw arrows, but it you use Icb etc, meaning from c to b, then you don't.)
    You have 6 unknowns and 6 equations, so if you just slug your way through, everything should come out fine. :smile:

    (The Electrician, please don't give out full answers!)
     
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