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I initially asked this question on math.stackexchange.com, but nobody answer so I thought I could try to ask here, this at least to some degree related to electrical engineering.

Please help me to correct my understanding.

$$C^{-1}y+Ax=b \\ A^Ty =f$$

This pair is equilibrium equations, \$C^{-1}+Ax=b\$ represents Ohm's law, derived from:

$$e=b-Ax$$

Vector \$x\$ represent potential on each node in a graph, on each node present abstract force which repel abstract flow. Flow goes to lower potential (which repel less). Act of multiplication \$Ax\$ produces potential difference. I just add all columns vectors adjusted by corresponding potential on each node, this is certainly should give me potential differences on each edge. But formula telling me:

$$Ax=b-e$$

This is my problem. What \$e\$ stands for? I know \$b\$ - potential differences, I can find potential at each node, I know \$x\$, I can find differences. What does \$e\$ mean? In another words if I know potential at each node, then \$C^{-1}y=Ax\$?

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  • \$\begingroup\$ You have to tell more to get some useful help. Where did you get those equations from? Normally you should find there also a definition of the symbols used. BTW Just by looking at the dimensions and assuming that \$A\$ is not dimensionless if \$x\$ represents potential \$b\$ cannot mean potential differences. \$\endgroup\$ – Curd May 25 '17 at 14:16
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Just a guess: it could be that you are dealing with some vector/matrix notation of nodal analysis or mesh analysis.

Nodal analysis:

  • \$A\$ would represent conductivities (usually denoted by \$G\$),
  • \$x\$ would be node potentials (usually denoted by \$v\$) and
  • \$b\$ and \$e\$ would mean branch currents and (independent) current sources or vice versa (usually denoted by \$i\$).

Mesh analysis:

  • \$A\$ would represent resistances (usually denoted by \$R\$),
  • \$x\$ would be mesh currents (usually denoted by \$i\$) and
  • \$b\$ and \$e\$ would mean branch voltages and (independent) voltage sources or vice versa (usually denoted by \$v\$).

In both cases \$Ax = b - e\$ would in fact represent Ohm's law.

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  • \$\begingroup\$ Hello Curd thank you for response! I get these equations from book of prof.Gilbert Strand introduction to Applied Mathematics, i think this is about nodal analysis, could you please clarify, as you telling in comment "if x represents potential b cannot mean potential differences", why? when we multiply this would be potential on node a - potential on node b, what else potential difference could mean? \$\endgroup\$ – Anatoly Strashkevich May 25 '17 at 14:46
  • \$\begingroup\$ as I wrot above: just from looking at dimensions (units): Elements of \$A\$ probably have unit Ohm (resistance) or Siemens (conductivity); i.e. they are not just pure numbers. Therefore if you multiply a quantity \$x\$ by \$A\$ the result can not have the same dimension as \$x\$. I.e. \$x\$ and \$Ax\$ can not have both dimension potential (unit Volt) which is what you suggested. \$\endgroup\$ – Curd May 25 '17 at 16:22

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