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Kirchoff's law for current

  1. Oct 10, 2015 #1
    1. The problem statement, all variables and given/known data
    See the picture. I have to find ## i ## and ## u_{ab} ##.

    2. Relevant equations Well, as the thread name says, Kirchoff law ## \sum_{k=1}^{n} i_k=0 ## , that is algebraic sum of currents in one point is 0.


    3. The attempt at a solution
    Well, first I tried to set up few equations from points A,B,C,D and it didn't seemed to lead me nowhere. And I am confused about ## R=0 ## part, does that mean that current flows only where ## R=0 ## or it doesn't ?
    Does the part where ## R=3 \Omega ## and ## R= 4 \Omega ## matters while calculating ## i ## or only when we are trying to find ## u_{ab} ##?
     

    Attached Files:

  2. jcsd
  3. Oct 10, 2015 #2

    gneill

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    Staff: Mentor

    The first thing to do is to consider the circuit topology rather than its geography. That is, consider the wiring to be perfect wires and replace that R = 0 with a wire, too. Identify the nodes. How many essential nodes are there?

    Yes, the 3 Ω and 4 Ω resistors will matter -- the current flowing through them creates the potential drop that is your ##u_{ab}##.
     
  4. Oct 10, 2015 #3
    Well, I guess 4 of them-A,B,C,D . And I wrote 4 equations but I have 5 unknowns - ##i , i_1 , i_2 ##and currents that flow from D to A and from C to B. (also current from D to C)
    How should I find those two currents ?

    EDIT: Is it something like this: If ## i_x ## is current from A to D, and ## i_y ## from B to C:

    From node A

    ## 10=i+i_x ##

    And from node B: ## i+4=i_y ## and from there ##i_x+i_y= 14 ## and since ## u_{ad}= u_{bc} \Rightarrow 4 i_y = 3 i_x ##

    solving that we get ## i_x = 6 ## and ## i_y = 8 ## so ## i=4 A##
     
    Last edited: Oct 10, 2015
  5. Oct 10, 2015 #4
    If the above is correct, then I have one more question. For example when we have parallel resistors, we would calculate the current flowing through them with ## I=I_1+I_2 ## and ## I_1 R_1 = I_2 R_2 ## , and conclusion is that where the resistance is bigger, the current is smaller. Then why does any current flow from A to D (and from D to A) ?
    Why doesn't the ##10 A## just go from A to B, and why doesn't ## i_1## go from D to C when there is resistance from D to A , and in the problem they say R=0 ?
     
  6. Oct 10, 2015 #5

    gneill

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    Staff: Mentor

    Topologically there are only two nodes in the circuit. Points A and B are locations on one node (call it node AB). Since R = 0, DC is also a single node. Remember, a node is the entirety of any path that is comprised of perfect conductor (wires).

    There's a total of 14 A flowing into node AB (that's a given). By Kirchoff, that current must also leave the node somehow. The sum must be zero. The only paths available are via the two resistors. Now, since both resistors are connected at both of their ends (they are in parallel) thanks to R being zero, you can apply the current division principle to work out how much of that 14 A goes through each. You've stated the equations underlying the principle in your post above (#4).

    Fig1.png

    You can also conclude that the total of 14 A must leave the circuit via ##i_1## and ##i_2##, but without additional information you can't tell how the 14 A will be divided between them.

    Going back to the "geographic" view of the circuit, what can you say about the potential difference between A and B (that is, what is ##u_{ab}##?
     
  7. Oct 10, 2015 #6
    Thank you for your answer, it's much clearer now. Finding voltage (potential difference ) is easy when you know the currents.## u_{ab} ## is clearly 0, but I was looking for ## u_{ac} ## (my mistake , I wrote in #1 ##u_{ab}##) which is also very easy to find, its ## i_4 \cdot 4 \Omega## or ## i_3 \cdot 3 \Omega = 24 V##.
     
  8. Oct 10, 2015 #7

    gneill

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    Staff: Mentor

    Yup. Well done.
     
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