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Find potential between A and B

  1. Dec 4, 2008 #1
    1. The problem statement, all variables and given/known data
    Find potential between A and B


    2. Relevant equations



    3. The attempt at a solution
    The R1 is confusing me. If it hadnt been there simple voltage divider would have taken care of things quite easily...I cant make logic of how to solve it now...How do I deal with R1 ?
     

    Attached Files:

  2. jcsd
  3. Dec 4, 2008 #2
    What methods of circuit analysis have you studied so far?

    Do you know how to do loop, mesh or node analysis?

    Do you know about Thevenin equivalents?
     
  4. Dec 4, 2008 #3
    Everything except the Thevenin Equivalents. The circuit I have attached is the result of applying superposition principle. Actually there was another Voltage source (V') connected between R3 and the common joint of R3,R4 and R7, with positive towards the right. I had to find the voltage across AB (One Port). SO by using the superposition principle and suppressing the V' this is the circuit I got. I applied loop analysis one by one i.e. once on this circuit and then by suppressing the V source but I don't get the right answer. And to make matters worse when I apply the loop analysis without suppressing any of the sources I get a completely different answer.

    The values given are :-
    V = 18V
    V' = 13V
    R1=R2=R3=3ohm
    R4=R5=6ohm
    R6=20ohm
    R7=10ohm

    My answers :-
    1. Suppressing the V' then current through R7 = 0.09A
    2. Suppressing the V then current through R7 = 0.571A
    3. Solving without superposition then current through R7 = 0.125A
     
  5. Dec 4, 2008 #4
    I would be better able to help you if you will post the original circuit, including the V' source.
     
  6. Dec 4, 2008 #5
    here is the original circuit
     

    Attached Files:

  7. Dec 4, 2008 #6
    I won't be able to see the jpg until it gets approved, which could take hours.

    If you'll upload it to:http://www.freeimagehosting.net/

    and post the URL here, I will be able to get it sooner.
     
  8. Dec 4, 2008 #7
  9. Dec 5, 2008 #8
    Ok, I got it. This problem is a natural for the loop method. You don't need to suppress any voltage sources. I labeled the loop currents as I1 for the bottom left loop, I2 for the top loop, and I3, I4 for the next two loops to the right of the I1 loop.

    Can you write the loop equations for those loops and show your work?

    Just to inspire you, I got a current of 1.5 amps in R7, and that makes 15 volts at the A-B port. I have been known to make mistakes though. :-(
     
  10. Dec 5, 2008 #9
    For loop 1:-

    [tex]3(i_1 + i_4) + 6(i_1 - i_2) = 18[/tex]

    For loop 2:-

    [tex]6(i_2 - i_1) + 3(i_2 + i_4) + 20(i_2 + i_3) = 13[/tex]

    For loop 3:-

    [tex]6(i_3 - i_4) + 20(i_3 + i_2) +10i_3 = 0[/tex]

    For loop 4:-

    [tex]3i_4 +3(i_4 + i_1) + 3(i_4 + i_2) + 6(i_4 - i_3) = 13[/tex]
     
  11. Dec 5, 2008 #10
  12. Dec 5, 2008 #11
    You have it exactly right. Your equations give the same currents I got, even though I selected my loops slightly differently.

    What currents do you get when you solve the system?
     
    Last edited: Dec 5, 2008
  13. Dec 5, 2008 #12
    I get i3 =0.125
     
  14. Dec 5, 2008 #13
    How are you solving it? I get:

    I1 = 3.9333
    I2 = 2.4
    I3 = -1.5
    I4 = -1.0
     
  15. Dec 5, 2008 #14
    Simlifying the equations then making i1 as the subject from last equation and substituting it in two of above and then solving simultaneously to get i2 , i3 and i4....must be some algebraic mistake
     
  16. Dec 5, 2008 #15
    I used a matrix solver after setting up the matrix and excitation vector like this; just collect the coefficients for each current:

    Code (Text):
    [  9  -6   0  3 ]  [ I1 ]     [ 18 ]
    [ -6  29  20  3 ]* [ I2 ]  =  [ 13 ]
    [  0  20  36 -6 ]  [ I3 ]     [  0 ]
    [  3  3   -6 15 ]  [ I4 ]     [ 13 ]
    then use Matlab, or a calculator with matrix capabilities to solve it.

    If you have the matrix set up properly, this method guarantees no algebra mistakes! :-)

    Given the magnitude of the voltage sources and the resistor values, I think .125 amps is definitely too small.
     
  17. Dec 5, 2008 #16
    which loops did you make?
     
  18. Dec 5, 2008 #17
    I used yours.
     
  19. Dec 5, 2008 #18
    Look at your 2nd equation, for example. You can collect terms and rewrite it as:

    -6*I1 + 29*I2 + 20*I3 + 3*I4 = 13

    and similarly for the others. Thats how you get the entries in the matrix.
     
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