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Nodal analysis: simple confusion

  1. Oct 15, 2012 #1
    My question about this excercise is the 80+20 part in the denominator. Why and when do I do that? I tried doing it but got the wrong answer
     

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  2. jcsd
  3. Oct 15, 2012 #2

    gneill

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    There is a single series-connected path from node 1 through the two resistors and the voltage source to the reference node. So the total resistance along that path is the sum of the two resistances. That makes the current through that branch (V1 - 24V)/(80Ω + 20Ω).

    There's no need to introduce extra nodes along a series branch; just add up the voltage sources and resistance values.
     
  4. Oct 15, 2012 #3
    Ok!!

    How about that +125 in node 3. I had it as V3-125. Why is it positive
     

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  5. Oct 15, 2012 #4

    gneill

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    It's positive because of the polarity of the 125V source along the path from node 3 to the reference node. The potential at its "-" terminal is effectively 125V below the reference potential. Since the potential of node 3 is measured with respect to the reference node, this makes the total potential across the resistor V3 + 125V.

    When in doubt, do a "KVL walk" along the path and add up the various potential changes. Starting at the reference node:

    -125V + i3*R = v3

    So the current is (v3 + 125V)/R
     
  6. Oct 15, 2012 #5
    When i multiply to reduce the fractions, which one should i aim for? Any of the deno?
     
  7. Oct 15, 2012 #6

    gneill

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    In what situation? Sounds like an exercise in fractions and algebra.
     
  8. Oct 15, 2012 #7
    Can i use any denominator to simplify the eacuation
     
  9. Oct 15, 2012 #8

    gneill

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    Surely it depends upon the equation and what the goal of the simplification is. You'll have to be more specific.
     
  10. Oct 15, 2012 #9
    I have a doubt when asked to find the current at each resistor. Sometimes the book uses the same ecuaction established in the nodal analysis. For example, (V1-128)/5 + V1/60 etc. is my ecuation after the analysis. Well, some times they use it like it was established I=(V1-128)/5 but sometimes they change it like so, I=(128-V1)/5... .??
     
  11. Oct 15, 2012 #10

    gneill

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    A sign change usually means that it was altered in order to match the assumed direction of the current being found.

    When you write the equations for nodal analysis it's usually the case that you just assume, for example, that all currents are flowing out of each node (or into each node). This is done in order to make writing the equations a straightforward procedure where you don't have to think about "actual" current directions beforehand.

    However, it is also often the case that the circuit drawing will have an indication of what direction should be assumed for a given current when you report the value. The match up can be done by adjusting the sign of of the appropriate nodal analysis term to correspond to the requested direction assumption.
     
  12. Oct 15, 2012 #11
    Awesome! Thanks
     
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