Electronic – Direction of current flow in resistor networks

graphhomeworkkirchhoffs-lawsNetworkresistors

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.

Best Answer

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.