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Two Loops, Two Batteries: Find the Currents

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Question: Find the current in the circuit shown.
    http://i1307.photobucket.com/albums/s599/JustOn4Fun/PhysicsFigure_zps39d4c66d.png

    2. Relevant equations
    Not sure if equations are necessary since it's a question testing the use of the following rules:
    Junction Rule
    The Loop Rule

    3. The attempt at a solution

    At Junction A: I1-I2-I3=0
    At Junction B:-I1+I2+I3=0
    Loop 1:15V-I3R-I1R=0
    Loop 2:-9V-I2R+I3R=0

    I1=I2+I3

    15V-100I3-100I1=0
    15/100=I3+I1=.15

    -9V-100I2+I3=0
    I3-I2=9/100=.09

    (I3+I1)-(I3-I2)=.06
    I1+I2=.06

    I think I'm having a brain fart or something because it feels like I'm just messing with the problem but not seeing something that's right under my nose. Thank you for the help!
     
  2. jcsd
  3. Feb 10, 2013 #2

    ehild

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    Cancel one of the currents from the equations. For example, substitute I2+I3 for I1 into the first loop equation.

    ehild
     
  4. Feb 10, 2013 #3
    I tried what you said and got the following:
    15V-I3R-(I2+I3)R=0
    and when the I3R cancels out, I'm left with 15V-I2R=0 giving me I2=.15

    The answers are: I1=0.070A, I2=-0.010A, and I3=0.080A

    What am I doing wrong?
     
  5. Feb 10, 2013 #4

    ehild

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    I3 does not cancel out.

    ehild
     
  6. Feb 10, 2013 #5
    Ah, yeah bad mistake. I got...
    15V-2I3R-I2R=0
    15V=2I3R-I2R
    Unfortunately, I'm not seeing it haha
     
  7. Feb 10, 2013 #6

    ehild

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    Now you have two equation with two unknowns.
    15V=2I3R-I2R
    -9V-I2R+I3R=0

    Isolate I2 from one of them and substitute into the other one.

    ehild
     
  8. Feb 10, 2013 #7

    ehild

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    A sign error again.
    Now you have two equation with two unknowns.
    15V-2I3R-I2R=0
    -9V-I2R+I3R=0

    Isolate I2 from one of them and substitute into the other one.

    ehild
     
  9. Feb 10, 2013 #8
    Ah, I got it now. Thank you very much for being patient with me.
     
  10. Feb 10, 2013 #9

    ehild

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    You are welcome:smile:

    ehild
     
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