Mutual Inductance - Mesh analysis

Discussion in 'Homework Help' started by ydgmms, Dec 4, 2007.

  1. ydgmms

    Thread Starter Member

    Oct 6, 2007
    10
    0
    I'm desperately trying to figure out this problem with ZERO luck.

    I'm thinking its either my mesh equations or my calculating Vo after I obtain I2.

    Mesh1 I get: 1*J1 + j6*J1 + j2*J2 = 10 angle 0
    Mesh2 I get: j2*J1 + j4*J2 + 4*J2 - j3*J2 = 0

    In Matrix(scilab input format, cuz I dont know how to do a matrix here)
    a=[ 1+6j , 2j ; 2j , 4+j ]
    b=[10;0]

    So then, to calculate J1 and J2, I do

    c=a/det(a)
    c =

    0.2416534 - 0.0206677i 0.0794913 + 0.0063593i
    0.0794913 + 0.0063593i 0.0524642 - 0.1558029i

    d=c*b
    d =

    2.4165342 - 0.2066773i
    0.7949126 + 0.0635930i

    Now I'm stuck here, getting my J2 into Vo. I try V=IR, so
    d(2,1)*-3j =

    ans =

    0.1907790 - 2.3847377i

    Which is not Vo according to the book...

    Anyone? :(
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Your problem is that the calculation c=a/det(a) does not give the inverse of a. Doesn't Scilab have a function like c = Inv(a) or c = Inverse(a)?
     
  3. ydgmms

    Thread Starter Member

    Oct 6, 2007
    10
    0
    The example problem I was trying to follow, had the coeffecient matrix over its determinate. ?
     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    If you will scan or photograph the page(s) from the example you're trying to follow and post them, I'll explain what they're doing.

    In the meantime, if you will substitute c = inverse(a), (or whatever the correct syntax in Scilab is) instead of c = a/det(a), you will get the correct answer.
     
  5. ydgmms

    Thread Starter Member

    Oct 6, 2007
    10
    0
    You are absolutely correct sir. I tried it with the way I SHOULD have the entire time. and within a minute I had the correct answer. Pissed me off to no end, i spent 2 hours last nite and i dunno how long monday and sunday nites.

    i did c=a\b, then d=c(2,1)*-3j and got the proper Vo


    Code ( (Unknown Language)):
    1.  
    2. -->a
    3.  a  =
    4.  
    5.     1. + 6.i    2.i
    6.     2.i         4. + i
    7.  
    8. -->b
    9.  b  =
    10.  
    11.     10.
    12.     0.
    13.  
    14. -->c=a\b
    15.  c  =
    16.  
    17.     0.5246423 - 1.5580286i
    18.   - 0.7949126 - 0.0635930i
    19.  
    20. -->c(2,1)*-3*%i
    21.  ans  =
    22.  
    23.   - 0.1907790 + 2.3847377i
    24.  
    25. -->
    26.  
    And when I convert ans into a phasor, I get the proper Vo.

    Thanks.. I dunno what the example problem and I were doing. :D
     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    See the attachment for an explanation.
     
  7. ydgmms

    Thread Starter Member

    Oct 6, 2007
    10
    0

    Huh..interesting. I've never tried to figure out the inverse of a matrix by hand. Thats what threw me off. Never been taught it, always just been told 'use a calculator or MATLAB(scilab for us poor college students)'. So I thought they were doing something different when they had the matrix over its determinate.

    I had noticed the sign change of the -j5 values. But thought it may be a misprint, as during the course we've seen a few in the book.


    Oh well. Thanks for the explanation. I think I'll stick to scilab and how I normally solve coeffecient matrices.
     
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