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Kichoff Simultaneous Equations

  1. Jan 18, 2009 #1
    1. The problem statement, all variables and given/known data

    I what to know how I form an equation for the middle loop, as there is no EMF in that part. Or, how do I go about forming the equations? Also, am I right to say that I should have 3 equations?

    2. Relevant equations

    http://www.theground.co.uk/CCF18012009_00000.jpg

    Thanks!

    Jase
     
  2. jcsd
  3. Jan 18, 2009 #2

    Doc Al

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    Staff: Mentor

    In any loop, the sum of the voltage drops (IR) and the EMFs must be zero.

    One way is to assign each branch its own current. You'll need as many equations as branches.
     
  4. Jan 18, 2009 #3
    Thanks!

    So the middle section, we'd have 0V = (Ix20) + (Ix10) + (Ix8) + (Ix20)?
     
  5. Jan 18, 2009 #4

    Doc Al

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    Staff: Mentor

    Yes, but each segment of that loop will have a different current. And take care with signs when you add up the voltage drops.
     
  6. Jan 22, 2009 #5
    Thanks for the reply.

    I'm trying it again. I've updated the image.

    I have managed to have do this with 2 loops, but the the 3 loops. Using Kichoff's second law, I'll have 3 equations I understand.

    What I am confused about is where I need to lable the current changes to enable me to use Kichoff's first law.

    Thanks!
     
  7. Jan 22, 2009 #6

    Doc Al

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    I would label each segment of the circuit with its own current. I see six separate segements, thus I'd have six currents. Then I'd use 3 loops and 3 nodes to produce my equations. (It's not as hard as it sounds.)
     
  8. Jan 22, 2009 #7
    Another thing I found helpful when I took this class is, once you follow Doc Al's advice, to set up all the equations in a matrix. It helps make the algebra easier! :smile:
     
  9. Jan 22, 2009 #8
    I would assume that the 6 currents would be that of where the resitors are. where would the 6th be?

    For the first loop on the left I have:
    200V = 10.I1 (thus I1 = 10A).

    2nd loop I have:
    0V = 8.I2 + 10.I1 + 20.I3 + 20.I5

    3rd:
    -125V = 30.I4 + 20.I5

    For the nodes, and K's 1st Law:

    I3 + I4 = I5

    And I don't understand how to get the other 2 when I have 2 currents coming out of the node: I5 = I4 + I2 ???

    I've got to understand this, doing a Product Design degree. Thanks.
     
  10. Jan 22, 2009 #9

    Doc Al

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    Staff: Mentor

    The 6th current will be through E1.

    Good.

    Good.

    Good.

    Good.

    That works.

    Get one more node equation (which will involve the 6th current).
     
  11. Jan 22, 2009 #10
    Thanks!

    So I6 + I2 = I1.

    I should be able to do the algebra from these.

    Could you tell me why you would have I6 there, because of the EMF? As I thought that that current through the EMF would be the same as I1 - or is it because the bottom node is split?

    Thanks again!
     
  12. Jan 22, 2009 #11

    Doc Al

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    Staff: Mentor

    That works.
    Because the current splits. (From the above equation, the only way that I6 = I1 would be if I2 = 0.)

    Whenever you have a node, assume that each segment intersecting that node has its own current.
     
  13. Jan 22, 2009 #12
    I'm understanding it now. Many Thanks again! :)
     
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