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Help with kirchoff's laws

  1. Dec 30, 2007 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data

    [​IMG]


    Find the values of [itex]I_C,I_B,I_L[/itex]

    2. Relevant equations
    [itex]\sum V=\sum IR[/itex]
    [itex]\sum I=0[/itex]


    3. The attempt at a solution

    By Kirchoff's first law:
    [tex]I_C-I_B-I_L=0[/tex]

    by Kirchoff's 2nd law:
    [tex]0.1I_C+1.2I_L=14[/tex]

    [tex]0.011I_B+1.2I_L=12[/tex]

    When I used these to solve for the unknowns I got
    [itex]I_C=21.25A \ \ I_B=11.35A \ \ I_L=9.9A[/itex]

    Now the solution says to use the first two equations and this one:
    [itex]0.1I_C +0.011I_B=2[/itex]

    and that gives a different answer to what I had calculated. Did I take the incorrect loops?
     
  2. jcsd
  3. Dec 30, 2007 #2
    Your third equation should be -0.011*ib+1.2*il=12. Do you know why?
     
  4. Dec 30, 2007 #3

    rock.freak667

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    The only way I see that would work is if I reversed the direction of the current [itex]I_B[/itex] which I think should not really matter because when solved if the direction is wrong the answer will be negative.
     
  5. Dec 30, 2007 #4

    dynamicsolo

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    What (dau)^4 is saying is that the current I_B is passing through the 0.011 ohm resistor in the same direction that you are tracing currents in that loop. So the change in potential across that resistor will be a "drop" of 0.011·I_B in that equation. (The way you have potential changes traced in your other loop is consistent.)

    You'll then find that

    [tex]0.1I_C+1.2I_L=14[/tex]
    and
    [tex]-0.011I_B+1.2I_L=12[/tex]

    leads to

    [itex]0.1I_C +0.011I_B=2[/itex]
     
    Last edited: Dec 30, 2007
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