Circuit Problem

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.

  • 1
    $\begingroup$ You can do with three unknown currents only (f.i. those in $R_1,R_3,R_4$). The other currents are found by difference. $\endgroup$ – Yves Daoust May 20 at 22:01
  • $\begingroup$ I am ok with your equations 1,2,3, and that is all loops you need. Beyond that, donÄt forget Kirchhoff's law, i.e., $I_1-I_2-I_3=0$ and $I_4=I_5$ $\endgroup$ – Hagen von Eitzen May 20 at 22:05


The matrix of the system is


In fact, as there is no resistor in the edge $V_2$, the loop of $R_4$ is stand-alone (as shown by the matrix structure). And the determinant is obviously positive.

By superposition:

Set $V_2=0$ (short-circuit), and $V_1$ sees an equivalent resistance


Then set $V_1=0$, and $V_2$ sees a resistance

$$\frac1{\dfrac1{R_1}+\dfrac1{R_2}}+R_3$$ on one side and $$R_4+R_5$$ on the other. With that you can find all currents.


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