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## CIRCUIT ANALYSIS: 4 resistor, Current src, Voltage src, V.C.V.S. - Find node Voltages

1. The problem statement, all variables and given/known data

Determine voltages $V_1$ through $V_3$ in the circuit below.

2. Relevant equations

KCL, V = iR

3. 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.

$$V_A\,=\,V_2$$

$$V_3\,=\,13\,V$$

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

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

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

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

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?
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