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Find voltage between two nodes

  1. Mar 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Find voltage between node 1 and node 2


    2. Relevant equations
    What is the algorithm?


    3. The attempt at a solution
    R1=4 Ohms
    R2=5 Ohms
    R3=8 Ohms
    R4=3 Ohms
    R5=6 Ohms
    R6=7 Ohms
    Eэ=20 V
    E3=17 V
    E5=15 V
     

    Attached Files:

  2. jcsd
  3. Mar 24, 2011 #2
    use balanced wheat stone bridge thing
     
  4. Mar 24, 2011 #3
    Do I need second Kirhgoff rules' equations for every loop?
     
  5. Mar 24, 2011 #4
  6. Mar 24, 2011 #5

    tiny-tim

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    hi builder_user! :smile:
    (Kirchhoff!)

    'fraid so! :wink:
     
  7. Mar 24, 2011 #6
    What's the next step after Kirchgoff's rules?What do I need to find?I know current in every branch.
     
  8. Mar 24, 2011 #7

    gneill

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    Staff: Mentor

    You want the voltage between nodes 1 and 2. Use the current you found for the first loop (leftmost) using Kirchhoff to determine the voltage across R1. Then do the obvious voltage sum.
     
  9. Mar 24, 2011 #8
    i1=3.7A

    U=Eэ-i1*R1=5.2V?

    And the final result -- 20(Eэ)+5.2(U)=25.2V?
     
    Last edited: Mar 24, 2011
  10. Mar 24, 2011 #9

    gneill

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    That's not the value I get for i1. Better check your derivation.

    EDIT: My error. Redoing my sums I see that i1 is indeed about 3.7A.
     
    Last edited: Mar 24, 2011
  11. Mar 24, 2011 #10
    Ok.But what's the next?
     
  12. Mar 24, 2011 #11

    gneill

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    The next what? :confused: Wasn't it the voltage between nodes 1 and 2 that you were looking for?
     
  13. Mar 24, 2011 #12
    but i's only I1.
    Or R1*I1 - is the result?
     
  14. Mar 24, 2011 #13

    gneill

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    Very sorry. I see that your figure of 3.7A for i1 and about 5.2V for the voltage between nodes 1 and 2 look okay.
     
  15. Mar 24, 2011 #14
    A-a-a-a-a!OMG!I'm already started to do all my work from the beginig...It's about 15 pages A4!
     
  16. Mar 24, 2011 #15

    gneill

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    Again, sorry about that.

    15 pages of A4? That seems like a lot for just three loops. What method are you using to solve the equations?
     
  17. Mar 24, 2011 #16
    It's a big work like coursework.I have 8 exercises need to do with this scheme(different trasformations like triangle-star and different methods of finding currents like method of equivalent generator(I don't know how does it called in English.translate from russian),node voltage method and etc.).And another 7 schemes.One of them is my previous topic's scheme.

    I got this result(3.7) by different methods.So if it's wrong - all my work is wrong
     
    Last edited: Mar 24, 2011
  18. Mar 24, 2011 #17

    gneill

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    I see, so this is just one exercise for the given schematic.

    It seems to me that a KVL loop approach is the most straight forward for this particular question, since you really only need to solve for the current in the first loop. Have you learned the method to directly write (by inspection) the matrix form of the equations? That allows you solve for the currents using a matrix method (such as Cramer's Rule), which takes only a few lines.
     
  19. Mar 24, 2011 #18
    if it's cramer's method...
    I know this method but I only use it in programming.I solve all equtations in MathCAd.And then I show result without numbers and then the final result.The result without numbers is very big sometimes.
     
    Last edited: Mar 24, 2011
  20. Mar 24, 2011 #19

    gneill

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    Okay, even better.

    Attached is an example, using MathCad, of solving the loops by writing the loop equations in matrix form (which can be done by inspection, and is (almost) foolproof). It's very quick.
     

    Attached Files:

  21. Mar 24, 2011 #20
    Intersting.I didn't know about it in Mathcad
    all results are the same as the results in my method of countour currents

    thank you for your help.
     
    Last edited: Mar 24, 2011
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