# KCL for this Supernode

I have an exercise (and merely an exercise) from my engineering Circuits class. Either way, the example they wrote up for a supernode makes perfect sense, but the solution to the exercise does not.

I thought that since the answer is expressed as a matrix equation perhaps I simply needed to simplify in some way, but when I used wolfram-alpha to do this simplification, the numerical solutions were wildly incorrect, meaning there must be a problem with my actual systems of equations. The circuit goes as thus: simulate this circuit – Schematic created using CircuitLab

The book offers this solution as a matrix equation: $$\left( \begin{array}{cc} \frac{3}{5} & -\frac{1}{2} & \frac{11}{20} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{3}{2} & \frac{3}{2} & 1\\ 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \end{array} \right) \left( \begin{array}{cc} v_1 \\ v_2\\ v_3\\ v_4 \end{array} \right) = \left( \begin{array}{cc} 0 \\ 0\\ 10\\ 40 \end{array} \right)$$ Which I can't begin to reconcile with my systems of equations, which go: $$\frac{1}{10}v_1 + \frac{1}{20}v_2 +\frac{1}{2}(v_3-v_2)+\frac{1}{2}(v_1-v_4)=0$$ $$-2(v_2-v_3)+\frac{1}{4}v_4+\frac{1}{2}(v_4-v_1)+\frac{1}{2}(v_2-v_3)=0$$ And of course $$v_1-v_3=10$$ $$v_2-v_4=40$$ According to WolframAlpha, my Systems of Equations have cleaner (but incorrect) answers, meaning I won't actually bother to try to rewrite them as the book has them.

For starters, I don't understand how the numerator on these fractions ends up as anything but one....

If anyone can be kind enough to point me in the right direction, thanks a bunch!

• In your first equation the $\frac{v_2}{20}$ term should be $\frac{v_3}{20}$ and then your equation will agree with the first row of the matrix. – Alfred Centauri Oct 25 '13 at 22:29
• In your second equation, the $\frac{v_4}{4}$ term should be $\frac{v_4}{2}$ and then your equation will agree with the second row of the matrix. – Alfred Centauri Oct 25 '13 at 22:31
• Also, I think the current source should be labelled $2\cdot I_{R1}$ where the reference direction for $I_{R1}$ is from node 2 to node 3. – Alfred Centauri Oct 25 '13 at 22:32

$$\frac{1}{10}v_1 + \frac{1}{20}v_3 +\frac{1}{2}(v_3-v_2)+\frac{1}{2}(v_1-v_4)=0$$
Group the terms involving, for example, $v_1$
$$v_1(\frac{1}{10} + \frac{1}{2}) = v_1 \frac{3}{5}$$