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Circuit Analysis- Node Voltage Problem

  1. Aug 24, 2014 #1
    In the attached picture, I have the problem and the equations I've come up with to solve for the requested variables. I have found seven equations for the seven unknowns, however, after plugging them into a calculator my answer is wrong. Which equation(s) is/are wrong, and why?
    Relevant equations
    Kirchoff's Current Law- sum of currents entering node = sum of currents leaving it
    Ohm's Law- Voltage = Current * Resistance
     

    Attached Files:

  2. jcsd
  3. Aug 24, 2014 #2

    NascentOxygen

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    Hi CyrilAmethyst. Welcome to the famous Physics Forums. :smile:

    I can't spot any mistakes there. Have you substituted your results to find whether any inconsistency shows up? There is always the possibility that the published answer is wrong. :mad:
     
    Last edited: Aug 24, 2014
  4. Aug 24, 2014 #3

    rude man

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    Homework Helper
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    I also think your equations are correct. However, you are making life hard for yourself. You have really only 2 unlnown nodes, so two equations suffice. (The 12V node is obvious and the node at the bottom of the 1 ohm resistor is obviously 5(12 - v1). So just sum currents at v1 and v2.
     
  5. Aug 24, 2014 #4
    NascentOxygen: Thanks for the welcome! I've found help on here from other people posting their problems so I figured I'd see if I could find some help myself. There isn't a published answer, just two of my fellow classmates are coming up with something different. Having substituted the values in, I'm not seeing any blatant inconsistencies, so I think I'm going to have to turn in what I have (with the solved equations, rather) and move on.

    rude man: But wouldn't you need the value of Vo to solve that? And to solve for Vo you'd need at least one more equation, and every other possible equation I can think of involves the different currents, so you'd need another... and that's sort of how it escalated to seven equations. While I'll likely have already turned in the assignment by the time you respond, it'd be great if you could show me where I could have had an easier time for the future!

    Thank you both so much for your help, regardless!

    EDIT: Also, is there a 'thanking' system on these forums, to give you guys credit for answering my question?
     
    Last edited: Aug 24, 2014
  6. Aug 25, 2014 #5

    NascentOxygen

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    What's your answer for V2?

    B
    e
    l
    o
    w


     
    Last edited: Aug 25, 2014
  7. Aug 25, 2014 #6

    rude man

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    Yes, but you know Vo = 12 - v1.

    Your two equations, for future reference, would be

    (12 - v1)/R2 = (v1 - v2)/R8 + v1/R4 + 3

    (v1 - v2)/R8 + 3 = [v2 + 5(12 - v1)]/R1

    I have used R1 = 1 ohm, R2 = 2 ohms, etc. for clarification. As always I favor labeling all components and saving numbers for the very end only. That allows units checking term-by-term.

    Note that I have substituted -5(12 - v1) for the voltage at the bottom of R1. You could have let that voltage = v3, then the second equation would have been (v1 - v2)/R8 + 3 = (v2 - v3)/R1 and a third equation would go v3 = -5(12 - v1).
    [/quote]

    EDIT: Also, is there a 'thanking' system on these forums, to give you guys credit for answering my question?[/QUOTE]

    There should be a 'thanks' button somewhere; however, it was just announced that the PF sites are being revamped ("PF 4.0"), with a new 'like' button in lieu of the old 'thanks' one. Stay tuned!

    Meanwhile - yer' welcome!
     
  8. Aug 25, 2014 #7
    In the end I got V1=~7V and V2=~24V. I can't check the exact numbers because the paper's already gone in, but several of my colleagues settled on V1=~-10 and V2=~-100. However, one of them seemed to use the same equations as I did at first glance, so perhaps I just made a computational error.

    So basically, rude man, you used simply two Kirchoff's Current Law equations, just substituted all of the currents with their equivalents in terms of V1 and V2? I do appreciate all of the help!
     
  9. Aug 25, 2014 #8
    I get this:

    attachment.php?attachmentid=72440&stc=1&d=1408997391.png
     

    Attached Files:

    • Ans.png
      Ans.png
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  10. Aug 25, 2014 #9

    rude man

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    No, I never use KCL. I think those fake currents are dumb! I did what you partially did - sum currents to zero at v1 and v2, that's all! That's all I ever do, and I've been doing it for 40 years! It minimizes the number of equations (providing you lump all series components together in one impedance).
     
  11. Aug 25, 2014 #10

    rude man

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    I got v1 = -10.9V, v2 = -100.4V using my two equations. So Vo = 12 - (-10.9) = + 22.9V.
     
    Last edited: Aug 25, 2014
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