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Kirchhoff's Rules help

  1. Dec 22, 2004 #1
    okay I got a question asking me to determine the currents in I1, I2, and I3 in a circuit assuming that the resistance of each battery is 1 ohm (I drew a seperate 1ohm resistor after every battery to represent this) and the terminal voltage of the 6V battery

    now I have drawn a diagram of the circuit and labeled every part of it, I am assuming I have to use Kirchhoff's Rules to solve this problem

    now here is where I get stuck, do I make 3 loops since there are 3 batteries or will 2 loops be fine
    (using the diagram loop 1 will probably be 'abefa' and loop 2 will be 'abcdefa', do I need loop 3 'ebcde')

    also at which point should I apply Kirchhoff's junction rule?

    Attached Files:

  2. jcsd
  3. Dec 22, 2004 #2


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    You have 3 unknowns, so you need three equations. One of these is the junction rule for the three currents (at either of the two junctions...you'll see that they both give you the same equation). The other two will be two loop equations.

    If you write the third loop equation, you'll find that it is actually just a combination of the two previous ones, and so, gives you nothing new.
  4. Dec 22, 2004 #3
    okay still seems I'm screwing up somewhere...

    junction is at point b on my graph so.... I1 = I2+I3

    loop 1 is 'bcdefab' => +12-9(I1)-15(I3)+6-19(I3) = 0

    loop 2 is 'befab' => +11(I2)-12-15(I3)+6-19(I3) = 0

    I1 = 18-34(I3) / 9 ==> I1 = 2-3.8(I3)

    I2 = -6-34(I3) / -11 ==> I2 = 0.55+3.1(I3)

    I3 = I1 - I2

    I3 = (2-3.8(I3))-(0.55+3.1(I3)) ==> I3 = 1.45 - 6.9(I3)

    I'm guessing I'm screwing up somewhere in the veginning with either the loops of the junction cause that's th only place I can think of where I migth be going wrong unless I'm not noticing something
  5. Dec 22, 2004 #4


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    What makes you think you're wrong ? Everything looks okay to me, so far.

    You've got : I3 = 1.45 - 6.9(I3)

    Or, 7.9(I3) = 1.45 ==>I3 = (1.45/7.9) A
  6. Dec 22, 2004 #5
    that gives me

    I3 = 0.18A

    the answer key tells me I am wrong as it says I3 should be 0.055A

    so yeah basically I'm stuck and not quite sure where I'm going wrong
  7. Dec 22, 2004 #6
    When I solved it, i got i3 = 0.185 A. I believe the answer key is wrong.
  8. Dec 22, 2004 #7
    not that I don't trust you but is there someone else that could confirm this as well, I would like to have a few confirmations that they key is actually wrong before I decide to go against it, thnx ;)
  9. Dec 23, 2004 #8


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    I get 0.18A for I3 also. Make sure that you've got the question right... You might have gotten the battery orientation wrong or something.
  10. Dec 23, 2004 #9
    okay tripple checked everything, I got nothing wrong so it seems that the answer key is bogus, thnx for the help
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