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Nodal analysis

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img822/5921/homeworkprobsg22.jpg [Broken]

    Determine the values of the node voltages Va and Vb


    2. Relevant equations

    Nodal analysis, V = IR

    maybe KVL and/or KCL


    3. The attempt at a solution

    I got the wrong answer / not sure what the reference node is used for (I know it's the one connecting to ground though, does it have to do with the current?)


    Vb - Va = 12V

    Vb = Va + 12V


    KCL at a:

    1.5A + I = Va/6Ω

    I = -1.5A + Va/6Ω


    KCL at B: 3.5A + I + Vb/3Ω = 0


    Plugging Vb in to KCL at b:

    3.5A + I + (Va + 12)/3Ω = 0


    Now plugging in KCL at a, and then solving for Va:

    3.5A -1.5A + Va/6Ω + (Va + 12V)/3Ω = 0

    2A + (Va + 2Va + 24V)/6Ω = 0

    2A + (3Va + 24V)/6Ω = 0

    (3Va + 24V)/6Ω = -2A

    3Va + 24V = -12V

    3Va = -30V

    thus

    Va = -10V which is supposedly wrong and would make Vb = 2V also wrong


    Where did I go wrong?


    Thank you.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 8, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    (-12 - 24) is not -30
     
    Last edited by a moderator: May 6, 2017
  4. Feb 8, 2013 #3
    Besides the numerical error gneill pointed out, I would say a few things.

    First, you didn't identify the current "I" on your schematic. When you're asking for help, make everything about your problem clear. If those who would help you have to exert extra effort to decipher things you should have labeled, they may just decide not to bother. Apparently the current "I" is the current through the 12 volt source.

    Now, when you're doing nodal analysis you need to adhere to certain conventions. In particular you must decide what sign to give currents at the nodes. I prefer to use the convention that currents leaving a node are taken to be positive. It's not necessary to use this convention; one may equally well choose to take currents entering the node to be positive, but you must be consistent.

    You were not consistent, and therefore incurred some compensating errors.

    At this point:

    you have given a positive sign to the current "I".

    And here:

    you have also given a positive sign to "I". The first occurrence was at node a and the second at node b. The current "I" must have different signs at node a and node b.

    The compensating error is that in this equation:

    1.5A + I = Va/6Ω

    if we rearrange it, we have:

    1.5A + I - Va/6Ω = 0

    The current in the 6Ω resistor going to ground is taken to be negative, but in this equation:

    3.5A + I + Vb/3Ω = 0

    the current in the 3Ω resistor going to ground is taken to be positive.

    As I say, you lucked out and you had compensating errors. You may not be so lucky the next time.
     
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