Superposition and Nodal Analysis

Discussion in 'Homework Help' started by jegues, Sep 14, 2010.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    See attached figure for problem statement.

    I wasn't sure how to use superposition to obtain Vo in this circuit, so I skipped that part of the question and figured I'd give it a shot using Nodal Analysis. (See other figure attached for my attempt)

    I assigned a ground node at the bottom, and I labeled all my other nodes. From here I assigned currents in all branches(I defined 5 currents from io to i4) and applied KCL to nodes A and B.

    From there I'm using Ohm's Law to replace the currents (io to i4) with voltages and resistances.

    How does everything look so far? I'm kinda confused how I'm going to find Vo from this considering its not in my equation.

    Vo should be the voltage across the branch with V3 and 2R connected in series, but how do I define it in solely terms of input voltages (V1, V2, V3)?

    Should I use nodal analysis to development my equations and solve for R in terms of V1, V2 and V3?

    If so then I could write KVL on the right-most loop of the my circuit (the open circuit loop) and with this define Vo in terms of V1, V2, V3 right?

    Any ideas on what route I should take? Are there any problems with my KCL equations or Ohm's Law equations?

    Any tips/suggestions/advice would be greatly appreciated!
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    In superposition you simply replace V2 and V3 with a short.
    And then solve for Vo1. Next you short V1 and V3 and put V2, and solve for Vo2.
    And finally you put V3 and short V1, V2 and solve for Vo3.
    And Vo = Vo1+Vo2+Vo3

    As for nodal, expression for I1, I3 and I4 are incorrect, for example
    I1 = (VA - V1)/2R
    And Vo = V3+ I4 *2R
     
  3. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    Okay first I will continue with my attempt using nodal analysis then after I will proceed with my superposition attempt.

    Let me try writing my equations for I1, I3 and I4 again.

    I1 = (VA - V1)/2R

    I3 = (VB - V2)/2R

    I4 = (VB-V3)/3R

    Are these correct now? Can I continue and plug these into my two KCL equations defined previously?
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    If you written I4=I2+I3 so you assume that V3 is greater then V2.
    So I4 = (V3-Vb)/3R
     
  5. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    Okay I've fixed this error as well and continued with the voltage analysis solving for VB. (See figure for my work, 2 pages)

    Does anyone see any errors?

    NOTE: there is an error on the first page, it says I4 = (VB - V3)/3R. I forgot to change it there, but I changed it in all my calculations for my KCL equations and Isolating VA and solving VB.

    If this correct, then I know I4 because I4 = (VB-V2)/3R, and if I know I4 then I can solve for Vo in solely terms of input voltages correct?
     
  6. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    If this holds true, finishing off... (See figure attached for my last page of work)

    Does this look okay?
     
  7. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    Here is my attempt for using superposition to solve the same circuit. (See figure attached, 3 pages)

    The Vo I obtained using superposition is not the same as the Vo I obtained using nodal analysis, so I'm assuming I must have made a mistake(if not several) in either my node analysis, or superposition, or both.

    I'd love it for someone to take a peak at my work so I can get some sort of feedback whether I'm following the correct procedure or not.

    Any tips/suggestions/comments/corrections you have me would be greatly appreciated.

    Thanks again!
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Your solution for VB using nodal equation is correct.
    And if it's comes to Vo I made a small error
    If current I4 is positive (flows from V3, when V3>VB) then Vo = V3 - I4*2R.
    So Vo = (V1 + 2*V2 + 4*V3)/8 = 0.125*V1 + 0.25*V2 + 0.5*V3

    As for superposition you make a error only in Vo1 .
    For my it's strange that you use nodal equation to solve Vo1 if we have this simple circuit with only one voltage source.
    Which must supply all current to the circuit.

    But your nodal equation are well written.

    -VA/(2*R) = (VA - V1)/(2*R) + (VA - VB)/R (1)

    (VA - VB)/R = VB/(2*R) + VB/(3*R) (2)

    So if we solve this properly

    VB = (3*V1)/16


    and

    Vo1 = (3*V1)/16 * 2R/3R = V1/8 = 0.125*V1
     
    Last edited: Sep 15, 2010
  9. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    Thank you for the response Jony130, with this I was able to finish both problems successfully.

    I shorted out the voltage sources and took a look at my circuit and I couldn't see how I could simplify things and obtain a resistor with equivalent resistance, so I decided to go with the backup plan which was nodal analysis.

    How would you solve Vo1 with the other two voltages sources shorted out without using nodal analysis? I couldn't see how to go about doing it (not comfortably atleast).
     
  10. Jony130

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    Feb 17, 2009
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    OK, no problem. I redrawn the schematic to much more simpler form.
    And now we clearly see that this circuit is quite simple.
    And I hope you see it too ?
     
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    Last edited: Sep 15, 2010
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