Branch Current Method

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Bill Allman

Joined Apr 20, 2009
1
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.
 
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.
 

Jack Bourne

Joined Apr 30, 2008
39
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.
 

jasperthecat

Joined Mar 26, 2009
20
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
 
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