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We have a simple resistor network \$g\$ with \$5\$ nodes and \$6\$ edges (resistors) as given below:

enter image description here

We assume all resistors have the same resistance (say \$1\$ Ohm, so homogeneously distributed resistances). Now suppose we want to know which edges would carry a non-zero current if we applied a \$1\$ Volt potential difference across the nodes \$1\$ and \$4.\$ Once we apply this constant potential difference, there will be current flowing either from node \$1\$ towards \$4\$ or the other way around. Assuming so far I haven't made a mistake in my description (from circuit theory point of view), if we adopt a "positive" direction for the current (I guess it doesn't matter which way is chosen), how do we determine the current direction along each edge? In other words, having now induced a current flow through the resistor network, my originally undirected graph will become a directed one, but I don't know how the edge directionality ought to be defined consistently.

One hypothetical situation of current flow once a voltage is applied between nodes \$1\$ to \$4\$ could be: (intuitively I don't expect there to be current flowing between 4 -> 5 because probably the potential difference between those nodes is 0 given the difference was applied across 1 and 4.)

enter image description here

Once I can determine the direction of flow across each edge then I can write down the incidence matrix \$A\$ where then for an edge that goes from node \$x\$ to node \$y\$, the row corresponding to that edge has \$−1\$ in column \$x\$ and \$1\$ in column \$y\$ with all other entries in that row being \$0.\$ Having the matrix \$A,\$ to answer my original question (namely which edges are carrying non-zero current), for the current vector \$\mathbf{i}\$, I invoke the Kirchhoff's law, namely that:

$$A^T \mathbf{i} = \mathbf{0},$$

where the current vector has a dimension equal to number of edges in \$g,\$ then finding an eigen-basis for the nullspace of \$A^T,\$ I would know which entries of \$\mathbf{i}\$ are non-zero. I admit I am not entirely sure about this approach, as in whether it's really the way to find an answer to my original question.

Summary of questions:

  1. How do I determine the direction of edges, in other words direction of current flow along each resistor, as described in the first paragraph?

  2. Given the original question of: "which edges will carry a non-zero current once a potential difference is applied across nodes \$1\$ and \$4,\$," is my approach as described in the last paragraph at all sound?

I've specifically chosen a very small and relatively simple network in this question for the purpose of illustration.

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  • \$\begingroup\$ Sorry I wasn't aware MathJax isn't available on electronics SE. If readability is poor I can remove the $ signs. \$\endgroup\$ – user929304 Mar 11 at 16:05
  • \$\begingroup\$ It is available, you just need to know how to use it. See my example edit and finish it yourself \$\endgroup\$ – Eugene Sh. Mar 11 at 16:06
  • \$\begingroup\$ @EugeneSh. Done. Brilliant that it's available, many thanks! I simply assumed it worked similar to physics SE. \$\endgroup\$ – user929304 Mar 11 at 16:12
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    \$\begingroup\$ When you put a voltage source between 1 & 4. Is the resistor 1 to 4 still there? \$\endgroup\$ – StainlessSteelRat Mar 11 at 16:39
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    \$\begingroup\$ Just clarifying the starting conditions. \$\endgroup\$ – StainlessSteelRat Mar 11 at 20:58
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if we adopt a "positive" direction for the current (I guess it doesn't matter which way is chosen), how do we determine the current direction along each edge? In other words, having now induced a current flow through the resistor network, my originally undirected graph will become a directed one, but I don't know how the edge directionality ought to be defined consistently.

You can't necessarily know which direction current is flowing without completing the analysis of the circuit (calculating the value of every current, magnitude as well as direction). This is because every current path in a circuit influences the others, except in specific short or open-circuit cases.

(In complex circuits we assume that — or rather, engineer them so that — some influences are minimal and approximate the short or open cases; for example, we assume that the power supply "rails" have a constant voltage (and design the power supply and distribution so that this assumption is almost correct) regardless of the current taken from them, so we can analyze each powered sub-circuit independently, and insofar as they aren't independent we call these interactions “noise” rather than trying to analyze/simulate the entire system except in unusual circumstances.)


Therefore, don't insist on getting the correct direction of current flow as an early step in analyzing a circuit. Instead, pick a direction arbitrarily (or guess if you're doing it by hand), then complete your analysis. If the answer turns out to be negative, then you can reverse the edge of your directed graph if you care.

Note also that if you analyze a circuit under different conditions (simple example: a battery or capacitor may be either charging or discharging) then the current flow may be reversed, but if you want to see how the circuit behaves it's more useful to have a consistent edge direction that sometimes results in a negative current (or voltage or power) than to have only positive values.

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  • \$\begingroup\$ Thanks a lot, this has clarified a few things for me. If I may ask, could you please elaborate a bit further what is meant by completing the analysis once an arbitrary direction is chosen? Maybe if it is easier with a simple example similar to the one i had used for the question. For instance, in the shown example, I see at least 3 paths for current to flow from 1 to 4 (directly, 1-2-3-4, 1-3-4, and so forth), so I already don't know which one should be chosen. My guess is that according to charge conservation, all 3 are chosen, and similarly at all subsequent nodes, so e.g. the current ... \$\endgroup\$ – user929304 Mar 11 at 21:02
  • \$\begingroup\$ ... flowing from 1 to 3 will be split into a current going from 3-2 and 3 to 4, or is that not the case? I'm trying to learn how systematically at each node the splitting direction goes, given we have chosen an arbitrary positive direction. Thanks in advance for any additional clarifications. \$\endgroup\$ – user929304 Mar 11 at 21:04
  • \$\begingroup\$ @user929304 "…what is meant by completing the analysis…" — you need to set up a system of equations corresponding to the circuit and solve it. There are various techniques to do this, and I'm not the person to write up an introduction to the subject. As one of the earlier comments said — "Try looking up Nodal Analysis and Mesh Analysis." My answer is only saying that you can't practically find the direction of the currents without also finding their magnitude, whatever technique you use or invent. This is because every current path is affected by every other. \$\endgroup\$ – Kevin Reid Mar 11 at 22:02

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