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Nodal analysis with voltage source

  1. Nov 17, 2015 #1
    1. The problem statement, all variables and given/known data

    Use nodal analysis to calculate Va, Vb, and the currents IE1 and IE2 without using source conversions

    I have tried this many, many times, and I must be missing something


    2. Relevant equations

    The equations I came up with:

    VA-10/ 1.1 + VA/1.2 +VA-VB/1.4 = 0

    VB+2/1.5 + VB-VA/1.4 +.5 = 0





    3. The attempt at a solution

    This results in

    VA (1/1.1+1/1.2 + 1/1.4) - VB (1/1.4) = 10/1.1
    -VA ( 1/ 1.4) + VB (1/1.5+1/1.4) = -1.833

    when put in a matrix, I am getting the wrong answers, because VA is supposed to be -4.81 and VB is supposed to be -1.80

    Thanks for your help another exam one.jpg another exam one.jpg

     
    Last edited by a moderator: Nov 17, 2015
  2. jcsd
  3. Nov 17, 2015 #2

    gneill

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    Staff: Mentor

    Hi Stevenfred, Welcome to Physics Forums!

    Your diagram doesn't specify where you've chosen to put VA and VB. Can you clarify that? I suspect that you've got some issues dealing with the voltage sources in the branches, but we should start with a firm definition of the nodes.
     
  4. Nov 17, 2015 #3
    Sorry that slipped my mind, and thank you for the welcome. Node A is the left side directly under R2 and Node B is on the right under R4.
     
  5. Nov 17, 2015 #4

    NascentOxygen

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    Staff: Mentor

    The voltage on one side of R1 is Va. The voltage on the other side of R1 is -10V. The voltage difference gives you the voltage across R1, and the voltage difference is Va - (-10).

    Now, try again to find the voltage difference across R3.
     
  6. Nov 17, 2015 #5

    gneill

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    Staff: Mentor

    Okay, no problem. So your circuit looks like this:
    Fig1.png
    For the first equation, note that the 10 V source effectively increases the potential to the reference node. That is, if you assume a current ##I## flowing from VA towards the reference node, a "KVL walk" of the branch would be: ##VA - I~R1 +10 = 0##, making ##I = (VA + 10)/1.1## . You've also left out the 2 V source in the branch between VA and VB. Can you redo the first equation now?

    For the second equation, I don't understand your first term. I see only the 1.5k resistor in that branch so the 2V value is a mystery, and the second term leaves out the 2 V source in the VB to VA branch. The .5 term for the current is fine though.

    You should get in the habit of using parentheses to group your terms so that the order of operations is clear. VB - VA/1.4 is something quite different from (VA - VB)/1.4 :smile:
     
  7. Nov 18, 2015 #6
    Thank you all for your help, I figured it out .:smile:
     
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