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Kirchhoff's Rules

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Sorry for edit, accidentally posted.

    Circuit is show in my attachment. I have 3 loops with three unknown currents.

    8 Ohms of resistance down the left side.
    6 Ohms of resistance and a 4V battery down the center.
    4 Ohms of resistance and a 12V battery down the right side.

    I honestly have no idea if I'm even approaching this correctly.
    2. Relevant equations

    Kirchhoff's Rules

    3. The attempt at a solution

    See attachment:

    Loop A (Big Loop)
    -8I1 +12V +4I1 +4I3 = 0
    -4I1 +4I3 = -12V

    Loop B (Left)
    -8I1 +4V -6I1 +6I2 = 0
    -14I1 + 6I2 = -4V

    Loop C (Right)
    -6I3 - 6I2 - 4V + 12V - 4I3 = 0
    -10I3 - 6I2 = -8V

    combine in a matrix
    I1 I2 I3 V
    a -4 0 -4 -12
    b -14 6 0 -4
    c 0 -6 -10 8

    rref()

    I1 = 1.083
    I2 = 1.861
    I3 = -1.917

    Im not entirely sure, but I think these are voltages for the loops, so from there I would set
    V= IR for each of the loops to find the current, but this produces the wrong answer. This is the easiest circuit in my textbook, and I have been searching the web for problems which is only making matters worse I think.

    Any help would be greatly appreciated.
     

    Attached Files:

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    Last edited: Nov 18, 2009
  2. jcsd
  3. Nov 18, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    You don't have three independent loops. Only two are independent, equation for the third loop is a linear combination of two others. Third equation that you need to solve is about currents in the upper node.

    But second opinion won't hurt, could be I am missing something. It happens when I stick my head outside of Chemistry forum :grumpy:

    --
    methods
     
  4. Nov 18, 2009 #3

    Delphi51

    User Avatar
    Homework Helper

    I don't think it is right. For instance, the -8I1 +12V +4I1 +4I3 = 0
    should be -8I1 +12V + 4I3 = 0 because I1 does not flow through the 4 ohm resistor.

    Recommend you just use two currents. Say, I1 as you have it down the left side and I2 in place of your I3 up on the right side. Then the current down the center is I2 - I1. This gives you a system of two equations to solve so quite a bit simpler. Just do the left and right loops to get them.
     
  5. Nov 18, 2009 #4
    I'm still not sure that I know what I'm doing, but sticking to two loops helped; I have the correct answers now. This is such a headache, you guy have no idea how long Ive been stuck on that simple little problem, onto the next one.

    Thanks for the help.
     

    Attached Files:

  6. Nov 19, 2009 #5

    Delphi51

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    Homework Helper

    Super! Glad to see you using that matrix method for solving linear systems.
    I had a lot of trouble with that problem, too. I find it hard to get the sign of each term right.
     
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