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 P: 492 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?