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Nodal Analysis - KVL Loop question.

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data
    I've obtained the node equations that's in the matrix for rows 1, 2, and 4. I got 3 as well but I'm not sure if my analysis was correct.

    http://img252.imageshack.us/img252/3151/nodeq.jpg [Broken]


    3. The attempt at a solution

    http://img684.imageshack.us/img684/9803/dsc0010van.jpg [Broken]

    So basically this is from the bottom-left of the circuit.

    v1 = -2 + Vx, therefore

    Vx = v1 + 2 (I wrote -2 in the drawing by mistake)

    So doing KVL: -Vx + 2 + 5io + v2 = 0

    => (-v1 -2) + 2 + v3 + v2 = 0

    So, -v1 + v2 + v3 = 0 [which is row 3 of the matrix]

    Note that io = v3/5, so 5io is just v3.

    Is this correct reasoning?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 12, 2012 #2

    gneill

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    Staff: Mentor

    Hmm. Looks to me like you've got a supernode situation since some nodes are directly tied together with voltage sources. Can you identify the boundaries of the supernode? Which nodes are left as independent nodes?
     
  4. Apr 12, 2012 #3
    I've done all of that, the supernode encompasses everything from v1 to v3. So I did KCL for that supernode and then KCL at node 4.

    I just need to know if I've done the KVL for the loop in the bottom-left hand correctly. My answers all work out to the solutions given, but I just had a funny feeling about the KVL as I was struggling to get it initially because of that 2V source and the direction of it.
     
  5. Apr 12, 2012 #4

    gneill

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    Staff: Mentor

    Okay. Well the KVL you wrote looks fine.

    Since you know that 5i0 = v3, would it not have been simpler to just write KVL from node v1 to v2 as:

    v2 = v1 - v3

    and rearrange?
     
  6. Apr 12, 2012 #5
    Hmm, you're right! I guess only practice helps recognise things like that easily. Thanks once again.
     
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