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Nodal analysis with supernode

  1. May 9, 2014 #1

    Maylis

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    ImageUploadedByPhysics Forums1399685955.565254.jpg

    Hello, I am having some difficulty solving this problem and I have doubts regarding my expression with the supernode. Basically, I said that current is negative if the current in my first KCL equation in the first term

    V1 - vs / ZL1

    Here is my attempt

    ImageUploadedByPhysics Forums1399686096.520212.jpg


    And here is a photo to part B that I will start working on now

    ImageUploadedByPhysics Forums1399686146.239933.jpg
     
  2. jcsd
  3. May 9, 2014 #2

    Maylis

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    I caught an error in my third KCL equation with not including v2/R2, but I'm still working on the power delivery part
     
  4. May 9, 2014 #3

    Maylis

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    Here is my work for part b and c...I'm just not finding anything that will get me the current going into the dependent voltage source

    ImageUploadedByPhysics Forums1399688591.474179.jpg
     
  5. May 10, 2014 #4

    gneill

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    Were you given component values and source voltage specifications for this problem? Otherwise the algebra will soon become horrendous, solving for the individual node voltages symbolically.

    If you were given values, use them while constructing your node equations. Then you'll only be carrying around the variables for the node voltages as you simplify the complex constants.
     
  6. May 10, 2014 #5

    Maylis

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    Far chance if that happening. The problem is intended to be horrendous. My professor is evil

    Although part b gives values to d numerical calculations.
     
  7. May 10, 2014 #6

    The Electrician

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    Looking at the third term in your third equation, you have (V1-Vs)/ZL1, which is apparently intended to be the current in the dependent source. Is that your intention? That expression is the current in L1; how do you get the current in the dependent source from that?

    But, besides that, the source in the lower right is a voltage source, not a current source. You can't write a usual node equation at node 3. In a another thread you said "I remember the professor saying to never do KCL at the output of an op-amp". The reason you can't do that is that the output pin of an (ideal) opamp is a voltage source. That's why you can't write a node equation at node 3 of this problem; you'll need a constraint equation there.

    And, finally, where is the equation for node 4?
     
  8. May 10, 2014 #7

    Maylis

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    Ok here is my next attempt.

    ImageUploadedByPhysics Forums1399773016.474095.jpg

    ImageUploadedByPhysics Forums1399773047.736749.jpg
     
  9. May 11, 2014 #8

    The Electrician

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    I recommend numbering equations to match the number of the corresponding node. I would number your equation 3 as number 4.

    Your equations 4 and 5 should be used to derive a single equation, V3 = 10((V1-V2)/R1), or casting in the form the problem requires:

    (10/R1)*V1 - (10/R1)*V1 - V3 = 0

    I would call this equation 3.
     
  10. May 12, 2014 #9

    Maylis

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    How does this look?

    ImageUploadedByPhysics Forums1399925398.253017.jpg

    ImageUploadedByPhysics Forums1399925411.527537.jpg
     
  11. May 12, 2014 #10

    The Electrician

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    Looks pretty good.
     
  12. May 12, 2014 #11

    Maylis

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    ImageUploadedByPhysics Forums1399931148.972598.jpg

    Ok so this is part b and c. How will I find if it is absorbing or delivering power using my equations from part (a)?
     
  13. May 12, 2014 #12

    Maylis

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    Nodal with supernode

    I tried doing a mesh to find all the loop currents but I think solving a system of linear equations with complex numbers is way too complicated

    ImageUploadedByPhysics Forums1399933615.959925.jpg

    I had read in this thread

    https://www.physicsforums.com/showthread.php?t=195567

    that you can split the system into its real part and imaginary part, but I am a bit confused on that. Well, do I keep everything on the right side of the equation the same? because zero is not imaginary, so how could I have imaginary parts on the right??
     
  14. May 12, 2014 #13

    Maylis

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  15. May 12, 2014 #14

    The Electrician

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    What sort of tools do you presently use to solve systems of equations with pure real coefficients?

    Do you have access to any mathematical software such as Matlab, Mathcad, etc.?

    Do you have a calculator that can do complex arithmetic? What calculator do you have?
     
  16. May 12, 2014 #15

    Maylis

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    Normally I use a system solver on my calculator the ti 36x pro, but it cannot do complex systems of equations. All these questions that I have been doing are from previous exams so there is no access to Matlab
     
  17. May 13, 2014 #16

    The Electrician

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    This method doesn't work.
     
  18. May 13, 2014 #17

    The Electrician

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    You say these problems are from previous exams; are you just doing them for practice?

    When you say there is no access to Matlab, do mean that during an exam you wouldn't be allowed to use Matlab? Because if you're just doing this for practice you could use any software you wanted, couldn't you? Maybe you're saying that if this were a real exam, you wouldn't be allowed to use Matlab so you want to learn how to do the problems without Matlab.

    Can you use any calculator you want on an exam? Because solving the problem in this thread really needs to use some method that can do complex arithmetic, including complex system solving. I see TI86 calculators for sale on eBay for $10, and they can do complex system solving. Even though it's discontinued, it will still do what you need. Maybe you should get one of those older TI calculators that can do complex arithmetic.
     
  19. May 13, 2014 #18

    Maylis

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    Oh well the exam is today so not much I can do about it anyway
     
  20. Feb 16, 2015 #19
    First of all, verify that if you can go with Nodal or supernode circuit analysis...
    if you cant find a way, then this post is specially for you. Supernode analysis
    if still face the problem in analyzing the circuit, then i can send you the whole process (step by step) thanks.
     
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