Branch Current Method

Discussion in 'Homework Help' started by Bill Allman, Apr 20, 2009.

  1. Bill Allman

    Thread Starter New Member

    Apr 20, 2009
    1
    0
    Doing a quick refresher on Branch current method. I'm fine until I reach the final part of the lesson and he figures the current values from these final equations.

    -1I1 + 1I2 - 1I3 = 0 KCL
    4I1 + 2I2 + 0I3 = 28 KVL
    0I1 - 2I2 - 1I3 = -7 KVL

    I1 = 5A
    I2 = 4A
    I3 = -1A

    Could someone show the algebraic equation that shows how we get 5 amps, 4 amps, and -1 amp from these equations above.
     
  2. guitarguy12387

    Active Member

    Apr 10, 2008
    359
    12
    You have to simultaneously solve each equation. I usually do it by the substitution method. Solving one equation in terms of the other variables and then back substitute. Or if you have a TI-89, it can do it quick and easy.
     
  3. Jack Bourne

    Active Member

    Apr 30, 2008
    39
    0
    I have seen some people solve it by using Matricies. I personally do not know how to do it this way so I would either use my calculator as it has a solve function on it or just use the substitution method.
    So Basically, you've got these equations 4I1 + 2I2 = 28, - 2I2 - 1I3 = -7, -1I1 + 1I2 - 1I3 = 0 . These are easy to use as you can have I2 = 14-2I1 -> into second -2(14-I1) = I3-7 so -I3= 21+2I2 -> into third and it gives you the answers.
     
  4. jasperthecat

    Member

    Mar 26, 2009
    20
    0
    Another way of solving the set of equations is to use the linear programming approach This is relatively straight forward for 3 or 4 unknowns . You aim to get 2 co-efficients as zero on one line and one coefficient as zero on another.

    -1 +1 -1 = 0

    +4 +2 0 = 28

    0 -2 -1 = -7

    Multiply equation 1 by 4


    -4 +4 -4 = 0

    +4 +2 0 = 28

    0 -2 -1 = -7
    Add equation 1 to eqn 2


    0 +6 -4 = 28

    +4 +2 0 = 28

    0 -2 -1 = -7

    Multiply eqn3 by 3

    0 +6 -4 = 28

    +4 +2 0 = 28

    0 -6 -3 = -21


    Add eqn 1 & 3

    0 0 -7 = 7

    +4 +2 0 = 28

    0 -6 -3 = -21

    From eq 1 gives I3 = -1

    Substitute in eqn 3

    I2 = 4


    Substitute in 2
    4 I1 = 20

    I1 = 5

    J
     
  5. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Matrix algebra is another technique for solving such a system of linear equations.

    hgmjr
     
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