Resistor Coefficient Voltage Matrix Doesn't have a Solution

Question

Follow the link to the circuit that I need to solve for all the resistor voltage drops.

I have to make 5 linearly independent equations in order to solve for all 5 unknown voltages.

Using the loops as drawn in the image following the link I get the equations:

1. iR1*4000 + iR2*2000 = 8
2. -iR2*2000 + iR3*1000 = 4
3. iR4*6000 + iR5*1000 = -4
4. iR1*4000 + iR3*1000 = 12
5. -iR2*2000 + iR3*1000 + iR4*6000 - iR5*1000 = 0

And the Matrix:

4000  2000     0     0     0     8

   0 -2000  1000     0     0     4

   0     0     0  6000  1000    -4

4000     0  1000     0     0    12

   0 -2000  1000  6000 -1000     0

When I row reduce this matrix, I get a row of all zeros which indicates that there isn't one unique solution to this matrix. I cant figure out what I am doing wrong as this is impossible. I could probably use different loops but I still need to understand why this configuration does not work. I have the understanding that all of my equations are linearly independent.

Note: There are better ways to solve it but the Question asks to solve it with exactly a 5 x 6 Matrix.

Answer 1

In the following schematic, \$I_1\$, \$I_2\$, and \$I_3\$ are the interior loops, which I consider to be walked "clockwise" starting always at their lower left corner. Those are set up with KVL. There is also \$V_\text{A}\$ and \$V_\text{B}\$. Those are set up with KCL. This would make for 5 equations and 5 unknowns:

^{simulate this circuit – Schematic created using CircuitLab}

So, it's my belief that you were being asked to solve for all three loop currents using KVL and both unknown node voltages using KCL. The equations would be:

$$\begin{align*} 0\:\text{V}+8\:\text{V}-4\:\text{k}\Omega\,\cdot\,I_1-2\:\text{k}\Omega\,\cdot\left(I_1-I_2\right)&= 0\:\text{V}\\\\ 0\:\text{V}-2\:\text{k}\Omega\,\cdot\left(I_2-I_1\right)-1\:\text{k}\Omega\,\cdot\,I_2+4\:\text{V}&=0\:\text{V}\\\\ 0\:\text{V}-4\:\text{V}-6\:\text{k}\Omega\,\cdot\,I_3-1\:\text{k}\Omega\,\cdot\,I_3&=0\:\text{V}\\\\ \frac{V_\text{A}}{4\:\text{k}\Omega}+\frac{V_\text{A}}{2\:\text{k}\Omega}+\frac{V_\text{A}}{1\:\text{k}\Omega}&=\frac{8\:\text{V}}{4\:\text{k}\Omega}+\frac{0\:\text{V}}{2\:\text{k}\Omega}+\frac{-4\:\text{V}}{1\:\text{k}\Omega}\\\\ \frac{V_\text{B}}{6\:\text{k}\Omega}+\frac{V_\text{B}}{1\:\text{k}\Omega}&=\frac{-4\:\text{V}}{6\:\text{k}\Omega}+\frac{0\:\text{V}}{1\:\text{k}\Omega}\\\\ \end{align*}$$

I'll let you sort out the matrix from the above statements. It will have a solution.

The results will be \$I_1= 2\frac27\:\text{mA}\$, \$I_2= 2\frac67\:\text{mA}\$, \$I_3= -\frac47\:\text{mA}\$, \$V_\text{A}=-1\frac17\:\text{V}\$, and \$V_\text{B}=-\frac47\:\text{V}\$.

From that you can, of course, completely work out all necessary voltage drops across the resistors, too.