# 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

## 1 Answer

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

I've left several comments pointing out careless errors in your equations but this deserves an answer even if the answer is elementary algebra.

Take, for example your first (corrected) equation:

$$\frac{1}{10}v_1 + \frac{1}{20}v_3 +\frac{1}{2}(v_3-v_2)+\frac{1}{2}(v_1-v_4)=0$$

Group the terms involving, for example, $v_1$

$$v_1(\frac{1}{10} + \frac{1}{2}) = v_1 \frac{3}{5}$$