newb question on KCL algebra

Thread Starter

naitek

Joined May 31, 2008
15
Hello; I am trying to find how to go about solving KCL, KVL not the equations, but the math.
The first equation results from KCL applied at node 1, the second equation results from KCL applied at node 3 and the third equation results from KCL applied at node 4. Collecting terms this becomes:


i understand getting the equations, i just dont know how to go about using Gaussian elimination or some other way mathematically to ge tmy answer. thanks

-
 

WBahn

Joined Mar 31, 2012
26,073
Since you mentioned Gaussian elimination, go Google Gaussian elimination and read up on it. Then come back and make you best attempt to apply it to the system of equations above and we can help you find and fix any mistakes you make.
 

Fibonacci

Joined May 23, 2014
25
Hello; I am trying to find how to go about solving KCL, KVL not the equations, but the math.
The first equation results from KCL applied at node 1, the second equation results from KCL applied at node 3 and the third equation results from KCL applied at node 4. Collecting terms this becomes:


i understand getting the equations, i just dont know how to go about using Gaussian elimination or some other way mathematically to ge tmy answer. thanks

-
Simplifying, you can express in matricial form the equations:
(1/9)V1-(1/30)V3 = -55/9 ... I
-(1/30)V1+(7/75)V3-(3/50)V4 = 107/15 ... II
-(3/50)V3+(19/150)V4 = -19/5 ... III

[1/9, -1/30, 0
-(1/30),(7/75),-(3/50)
0,-(3/50),(19/150)], this is known like coefficient matrix, "A".
Also we have the variable vector x= [V1;V3;V4] and the results vector
b=[-55/9;107/15;-19/5].
You can find the answers for the vector x solving the following matrix equation: x =A^-1*b, where A^-1 is the inverse matrix. It is easy, only you need to have a computer that performs the operations, for example, the TI-89 calculator, MATLAB, etc.
 
Top