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.

 

[piercas]

I would say that your solution looks correct. You have used the appropriate equations and methods to solve for the unknown voltages in the circuit. Your answer also makes sense intuitively, as V_1 and V_2 are both positive and smaller than V_3, which is the voltage source. Good job on solving the circuit!