Engineering Solving Circuits: Finding V_1-V_3

[VinnyCee]

Homework Statement

Determine voltages V_1 through V_3 in the circuit below.

http://img264.imageshack.us/img264/9311/chapter3problem161oz.jpg

Homework Equations

KCL, V = iR

The Attempt at a Solution

So I added some variables to represent currents and a super-node. The variables are in red and the super-node in light-blue.

http://img264.imageshack.us/img264/1875/chapter3problem16part21ki.jpg

V_A\,=\,V_2

V_3\,=\,13\,V

I_1\,=\,\frac{V_1\,-\,V_3}{\frac{1}{2}\Omega}\,=\,2\,V_1\,-\,2\,V_3

I_2\,=\,\frac{V_1\,-\,0}{1\Omega}\,=\,V_1

I_3\,=\,\frac{V_2\,-\,0}{\frac{1}{4}\Omega}\,=\,4\,V_2

I_4\,=\,\frac{V_2\,-\,V_3}{\frac{1}{8}\Omega}\,=\,8\,V_2\,-\,8\,V_3

Now I use KCL at the super-node:

I_1\,+\,I_2\,+\,I_3\,+I_4\,=\,2\,A

(2\,V_1\,-\,2\,V_3)\,+\,(V_1)\,+\,(4\,V_2)\,+\,(8\,V_2\,-\,8\,V_3)\,=\,2\,A

3\,V_1\,+\,12\,V_2\,-\,10\,V_3\,=\,2\,A

3\,V_1\,+\,12\,V_2\,-\,10(13\,V)\,=\,2\,A

3\,V_1\,+\,12\,V_2\,=\,132

And get the voltage equation from inside the super-node:

V_1\,-\,V_2\,=\,2\,V_A

V_1\,-\,V_2\,-\,2\,V_2\,= \,0

V_1\,-\,3\,V_2\,=\,0

Now put into a matrix and rref to get V_1 and V_2:

\left[\begin{array}{ccc}3&12&132\\1&-3&0\end{array}\right]\,\,\longrightarrow\,\,\left[\begin{array}{ccc}1&0&\frac{132}{7}\\0&1&\frac{44}{7}\end{array}\right]

So I get these for V_1 through V_3:

V_1\,=\,\frac{132}{7}\,V\,\approx\,18.86\,V

V_2\,=\,\frac{44}{7}\,V\,\approx\,6.286\,V

V_3\,=\,13\,V

Does this look right?

 

[doodle]

Yup, looks right to me.