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Final tomorrow loop/mesh analysis using supermeshes

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data
    1 for picture38... apply loop analysis to find the current through the 2 ohm resistor
    2 for picture39... apply loop analysis to find the voltage across the 10 ohm resistor


    2. Relevant equations
    v=ir
    kvl

    3. The attempt at a solution
    i'm using a, b, c etc as current labels as well as loop names
    1)loop a
    4a+4(a-b)+4(a-c)=0
    condenses to 3a-2b=0

    supermesh c&b
    -10+8(c-a)+4(b-a)+2b=0

    using 2=c-b

    condenses to 7b-6a=-3

    i get b=1/3... which is so not right...

    so wat am i doing wrong??

    2)
    loop a
    4a+75+a+10(a-b)=0

    2=c-b

    supermesh c&b

    -25+2c+10(b-a)+3c=0

    wat am i doing wrong again?

    i guess supermesh is my problem... or maybe the algebra... but i dont get it... at all.. got an exam tomorrow... so please help quick
    thankyou
     

    Attached Files:

  2. jcsd
  3. Feb 5, 2008 #2
    Solution to first problem..

    Hi mate, I've drawn up a dodgy solution in paint for you, it should help you :)
     

    Attached Files:

  4. Feb 5, 2008 #3
    Number 2)

    I got for loop a:
    4a + a + 10a - 10b +75 = 0
    15a - 10b = -75 -----(1)

    then for supermesh:
    -25 + 2c + 10b - 10a + 3c = 0
    -10a + 10b + 5c = 25 ----(2)

    then at node N (the node to which the 2A source arrow points)

    2 + b - c = 0 therefore c = b+2 ----(3)

    sub (3) into (2) then you have two easy eqs to solve

    Using KVL at the node is basically your control eq.. it lets you reduce the ueqs down so you can solve them.
     
    Last edited: Feb 5, 2008
  5. Feb 6, 2008 #4
    thankyou andrew... that rhymes :P
    i'm gonna save this for future reference.. the solutions that is..
     
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