Understanding equilibrium equations

Question

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\$?

Answer 1

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.