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Homework Help: Finding current using Kirchhoff's circuit rules

  1. Oct 29, 2012 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    Kirchhoff's loop rule and (maybe) Kirchhoff's current law

    3. The attempt at a solution

    So I attempted to apply Kirchhoff's loop rule to the circuit, and ended up with a system of linear equations:

    24 - 6I[itex]_{1}[/itex] - 3I[itex]_{3}[/itex] = 0 [itex]\rightarrow[/itex] 2I[itex]_{1}[/itex] - I[itex]_{3}[/itex] = 8
    36 - 6I[itex]_{1}[/itex] - 6I[itex]_{2}[/itex] = 0 [itex]\rightarrow[/itex] I[itex]_{1}[/itex] + I[itex]_{2}[/itex] = 6
    12 - 3I[itex]_{3}[/itex] - 6I[itex]_{2}[/itex] = 0 [itex]\rightarrow[/itex] 2I[itex]_{2}[/itex] + I[itex]_{3}[/itex] = 4

    but every time I try to solve it, I end up with everything dropping out. I'm not sure what I'm missing here, but I've been banging my head against this one for about 4 hours now, so any help would be appreciated!
    Last edited: Oct 29, 2012
  2. jcsd
  3. Oct 29, 2012 #2


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    Homework Helper

    Check the signs.
    You also have a relation between the currents. (Nodal Law) Use that and two loop equations.

    Last edited: Oct 29, 2012
  4. Oct 29, 2012 #3
    Thanks, that makes more sense now.

    I might have eventually realized that my system was cancelling out because it was using mutually derived equations, but I probably never would have realized that I was treating I[itex]_{3}[/itex] as if it were flowing in two directions at once.

    Using two of the equations from the original system in the nodal rule equation I[itex]_{1}[/itex] = I[itex]_{2}[/itex]+ I[itex]_{3}[/itex] made it all fall together though. I'm still not sure how I'd figure out what direction I[itex]_{3}[/itex] was flowing in if it weren't given in the problem, but at least I'm FINALLY done with this problem.

    tl;dr you are the best, thanks
  5. Oct 29, 2012 #4


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    Homework Helper

    You need not find out the direction of the currents in advance. If you have not chosen the proper direction you would get negative value for the current.

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