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Use the Node Voltage method to solve

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Gl4flF7.jpg
    From the above circuit find VCB and VDG

    Know Data:
    VE = 0 (ground)

    2. Relevant equations
    KCL and Ohm's Law are used in Node Voltage Method.

    3. The attempt at a solution
    I decided I would determine the node voltages then work out VCB and VDG afterwards.
    1. KCL and Ohm's law at Node C
    ## 15 + V_F + V_D = 3 V_C## -(1)​

    2. Super Node at EG
    2.1 KCL and Ohm's law for all currents entering
    ## 2(V_F - V_G) = 2 V_E - V_D - V_B##
    ## 2(V_F - V_G) + V_D + V_B = 2 V_E## -(2)​
    2.2 Potentials at Nodes E and G
    ## V_G = -5V## -(3)​

    3. KCL and Ohm's law at Node D
    ## V_C = 2 V_D## -(4)
    This is as far as I got I am unsure what to do next so I can get 5 equations so they can all be solved simultaneously.
     
  2. jcsd
  3. Nov 5, 2014 #2

    ehild

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    The problem highly simplifies if you replace the series and parallel resistances with their resultants.
     
  4. Nov 5, 2014 #3
    So if I replaced the two between F and G, and the one between C and F right?
     
  5. Nov 5, 2014 #4

    ehild

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    Yes, and the two resistors between C and E, and also the two resistors in series with the 15 V battery.

    In the simplifed circuit, you can find the voltage VC with the node voltage method . Knowing Vc, you can calculate the currents, and knowing the currents, you can determine all node voltages.
     
  6. Nov 5, 2014 #5
    Simplified Circuit.
    upload_2014-11-6_1-31-21.png
    This is what I have done. Not sure if correct?
    upload_2014-11-6_1-28-52.png
    How would I find VB? Is VA really = 15V.
    NOTE: VD = 2VC so it can be eliminated.
     
  7. Nov 5, 2014 #6
    Actually I think I have it just wait.
     
  8. Nov 5, 2014 #7
    Simplified Circuit.
    upload_2014-11-6_1-31-21-png.75137.png
    This is what I have done.
    upload_2014-11-6_1-51-21.png
    I think I have made a mistake somewhere.
    NOTE: VB is meant to be negative at end where = 76.67V
     
  9. Nov 5, 2014 #8

    ehild

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    Your solution is very complicated...

    The circuit is equivalent with the one in the picture, except B and D nodes disappearing. You can solve it for Vc. nodvoltmet.JPG
     
  10. Nov 5, 2014 #9
    Thanks I have now managed to solve this!
     
  11. Nov 5, 2014 #10

    ehild

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    No, VA is not 15 V, as we count the potential with respect to E. VA-VB=15 V
     
  12. Nov 5, 2014 #11
    I figured Va = Vb + 15 and it worked!
     
  13. Nov 5, 2014 #12

    ehild

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    Splendid! What did you get for VC?
    Have you tried to solve the problem also with the very simple equivalent circuit?
     
  14. Nov 5, 2014 #13
    I got VC = 2.5 V
     
  15. Nov 5, 2014 #14

    ehild

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    Using the equivalent circuit, the node equation for the currents at C is: ##\frac{15-Vc}{2}=\frac{Vc}{2}+\frac{Vc-(-5)}{1.5}## You need only this equation to solve for Vc, which is the same you got.
     
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