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Simple Nodal Analysis

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    2. Relevant equations

    KCL
    KVL
    V = IR

    3. The attempt at a solution

    I used 4 nodes at the top, one reference node at the bottom.

    I have it labeled V1, V2, V3, V4, starting from the far left -> right.

    My nodal equations are:

    At v1:

    + 7V

    At v2:

    0 = (v2-v1) / 2 + v2 / 4 + (v2 - v3) / 4

    At v3:

    (v3-v2) / 4 + v3 / 8 + (v3 - v4) / 8 = 0

    At v4:


    0 = (v4-v3)/ 8 - 1mA

    so we know V0 is v2-v3 = v0.

    I'm not sure if this is correct so far.
     
  2. jcsd
  3. Feb 13, 2007 #2

    mjsd

    User Avatar
    Homework Helper

    your image is too small to be legible
     
  4. Feb 13, 2007 #3
    So far it's correct.

    Try clicking on the (thumbprint) image to view a larger one.
     
  5. Feb 13, 2007 #4
    For v4, is it 8V in my case? I'm stuck with the simult. equations, and I'm just curious. I have a 1mA current going through the 8k resistor, so i'm believing its 8V? Can anyone verify this?
     
  6. Feb 13, 2007 #5
    I have simplified it to this:

    V1 = 7V (like before)

    for node at V2...

    4V_2 - 2V_1 - V_3 = 0

    for node at V3...

    4V_3 - 2V_2 - V_4 = 0

    for node at V4...

    V_3 - V_4 = 8

    So, all of the eq put together...

    4V_2 - 2V_1 - V_3 = 0
    4V_3 - 2V_2 - V_4 = 0
    V_3 - V_4 = 8
    V_0 = V_2 - V_3


    I'm not sure what to do after here.
     
  7. Feb 13, 2007 #6
    No, v4 is not 8V, but either one of (v4-v3) or (v3-v4) is and it is not difficult to figure out which one is.

    The nodal equation for node 4 is incorrect. Please check that against your first post. You have above 3 equations with 3 unknowns (v2, v3, v4). It becomes thus a mathematical problem now to resolve the 3 unknowns. The final step, of course, would be to put these values into finding v0 = v2-v3.
     
  8. Feb 13, 2007 #7
    Oh, woops, it should be v_4 - v_3 = 8, as in my first post.

    Its this math part that kills me, I'll be back in a few:)

    Thanks so far, doodle.
     
  9. Feb 14, 2007 #8
    Nevermind, it's correct. I just checked.

    Thanks for all those that helped :)
     
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