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Kirchoff's loop rule

  1. Sep 11, 2013 #1

    Sastronaut

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

    Compute the current through R2 of the circuit in the figure.

    2. Relevant equations
    kirchoff's loop rule
    V=IR


    3. The attempt at a solution
    Junction: I1+I2=I3

    Alpha loop: V1-I3R3-I5R5=0
    I1=(V1-I5R5)/R3=?

    Beta loop: V2-I8R8-I5R5=0

    *without one of the currents provided I am not exactly sure where to proceed from here...any help would be greatly appreciated! Anything that is a different color than black in the figure is what I added to the circuit diagram! Thank You PF!
     

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    Sastronaut, Sep 11, 2013
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  3. Sep 11, 2013 #2

    Pythagorean

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    Gold Member

    You have two of the currents (in terms of other variables) so you can find the third.
     
    Pythagorean, Sep 11, 2013
  4. Sep 11, 2013 #3

    Sastronaut

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    I have been trying to figure out the algebra but I am always left with two variable (grouping of I1, I2, or I3). I am not sure how I can get it down to just one variable for current...
     
    Sastronaut, Sep 11, 2013
  5. Sep 11, 2013 #4

    Pythagorean

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    start here:

    I3 = I1 + I2

    and you know what I1 and I2 are, yeah? So....

    I3 = ?
     
    Pythagorean, Sep 11, 2013
  6. Sep 11, 2013 #5

    Sastronaut

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    I3=(-5v+(5ΩI8))/-8Ω

    ?????
     
    Sastronaut, Sep 11, 2013
  7. Sep 11, 2013 #6

    Pythagorean

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    I'm not sure what you did there... I1 and I2 will be in terms of I3 and your knowns.

    Actually, I'm not quite sure what you did for your alpha/beta loops either. There is no I5 or R5 or I8/R8. Don't try to plug in your values yet, just use the names of the constants.
     
    Pythagorean, Sep 11, 2013
  8. Sep 11, 2013 #7

    Sastronaut

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    I think I figured it out I have a value of 1.076A through R2. really appreciate the help :)
     
    Sastronaut, Sep 11, 2013
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