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Kirchoff's Circuit Law

  1. Jan 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Determine the values of the other currents in Figure P1.37, given that ia = 2 A, ic =−3 A,
    ig = 6A,and ih= 1A.

    ealHzUE.png


    2. Relevant equations
    ∑currents entering a node = ∑currents exiting the node


    3. The attempt at a solution

    I used Kirchoff's Circuit Law on each of the outer nodes:

    if = ig + ih = 6 + 1 = 7A

    ia + id = if
    2 + id = 7
    id = 5A

    ic + ih= ie
    -3 + 1 = ie
    ie = -2A

    ib = ia + ic = 2 - 3 = -1A

    I think I may have went wrong somewhere, because when I apply KCL to one of the inner nodes I get this:

    ie + ig = id (?)
    -2 + 6 ≠ 5
     
  2. jcsd
  3. Jan 23, 2014 #2

    gneill

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    There's only one "inner node" where B, D, G, and E meet; despite there being two connection dots there, they are all one single conducting path with no components in between.

    So, your node equation needs to account for ##i_b##, too.
     
  4. Jan 23, 2014 #3
    Aha, in that case it will be
    -2 + 6 = -1 + 5
    4 = 4

    How do I know that the node includes ##i_b##, is that current not affected by the current ##i_d##?

    Is the point connecting B and D not a node?
     
  5. Jan 23, 2014 #4

    Simon Bridge

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    The two inner nodes are actually one node.
    You missed a current: two currents go into the node and two currents go out of the node.

    [edit]gah: too slow!
     
  6. Jan 23, 2014 #5

    gneill

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    Staff: Mentor

    So long as there is a continuous conducting path it is all one node. A node is not a dot on the diagram, it's the entire continuous conducting path.

    To make it more clear, you can shift component E over a bit as follows:

    attachment.php?attachmentid=65959&stc=1&d=1390527029.gif

    The wiring in green is continuously connected, so it's all one node.
     

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  7. Jan 23, 2014 #6
    Ah, yes that confused me a bit. It makes sense now thanks!
     
  8. Jan 23, 2014 #7

    Simon Bridge

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    Great - the other way to look at it is to add an extra current arrow ##\small{i_0}## in the gap between the "two" center nodes. You can tell which direction it has to point in by looking at the other currents - but it actually doesn't matter:

    Current in = current out:
    ... so on the left node you get ##i_0 = i_b+i_d##
    ... and for the right node you get ##i_0=i_g+i_e##

    ... eliminate the ##\small{i_0}## term gives you ##i_b+i_d = i_g+i_e##

    and it all comes out in the wash ;)

    It's just easier to realize that the nodes for Kirkoffs laws actually include the entire wire between components. Thus two drawn node-symbols, on the diagram, that have no components between them, are one and the same.
     
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