- #1

Dafe

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## Homework Statement

With conductances [tex]c_{1}=1, c_{2}=c_{3}=2[/tex], multiply matrices to find

[tex] A^TCAx = f [/tex].

For [tex] f = (1,0,-1) [/tex] find a solution to [tex] A^TCAx = f [/tex].

Write the potentials [tex] x [/tex] and currents [tex] y = -CAx [/tex] on the triangle graph, when the current source [tex] f [/tex] goes into node 1 and out from node 3.

## Homework Equations

## The Attempt at a Solution

[tex]

A^TCA =

\left[ \begin{array}{ccc} 3 & -1 & -2 \\ -1 & 3 & -2 \\ -2 & -2 & 4 \\ \end{array} \right]

[/tex]

After elimination I get:

[tex]

A^TCA =

\left[ \begin{array}{cccc} 3 & -1 & -2 & 1 \\ 0 & 8 & -8 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right]

[/tex]

Which to me means that the loop is broken, and that [tex] x_{3} [/tex] is a free variable.

The book chooses [tex] x_{3}=7/8 [/tex]. Is there a good reason for that value?

Maybe I am missing some important point here?

Thanks.