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enter image description hereI am trying to calculate the equivalent resistance from a random resistance network formed by carbon nanotubes (CNTs.)

I have a MATLAB program where the CNTs are randomly distributed inside a volume. The code basically calculates the distance between the CNTs and if the distance is less than the tunneling distance (user defined) it achieves percolation (meaning that two opposite faces are connected through a path.) Then with matrices, I calculated the voltage drop on every CNTs (resistor) so now I have a vector of electrical potentials which apparently with it I can calculate the electrical current inside the volume and then the equivalent resistance.

However, I have been trying to understand how can I do this.

enter image description here

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  • \$\begingroup\$ So your partially processed data can be viewed as a network of resistors and that you have already solved for the voltages of all the nodes in between the resistors? If this much is already available, you have the current through the resistors also. If you select any two nodes, you have a potential difference between them also available with you. At how many points are the external potential applied? \$\endgroup\$
    – AJN
    Commented May 5, 2021 at 9:33
  • \$\begingroup\$ If you have a graph with edges and vertices then you can select any pair of vertices you want and work out the current. May I assume you have a formula that computes the current given any pair of vertices (separated by one edge) and their voltages? \$\endgroup\$
    – jonk
    Commented May 5, 2021 at 9:38
  • \$\begingroup\$ So I have the voltage drop in each resistor, but nothing else! How can I calculate the current? ant the external potential is applied in both faces. \$\endgroup\$
    – Sebas Tian
    Commented May 5, 2021 at 12:33
  • \$\begingroup\$ The problem is every time I generate the Network, the graphs will change so I am after an equation or a mathematical operation with the voltage drop matrices from which I can calculate the equivalent resistance \$\endgroup\$
    – Sebas Tian
    Commented May 5, 2021 at 12:34
  • \$\begingroup\$ You say that "potential is applied at the faces". Since you have solved for the voltages at every node, you know the current going into each resistor that is touching a face ((Vface-Vothernode)/R). Just add up the currents going into the material and divide the potential difference by the sum of the currents. It should give you the equivalent resistance; right ? \$\endgroup\$
    – AJN
    Commented May 5, 2021 at 12:42

1 Answer 1

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resistor network

A generalised picture of the resistor network (As I have understood it) is shown above. V1-V2 is the applied potential difference between the two faces (seems to be 20 V in your picture). V3, V4, V5, V6 etc are the node voltages you have already solved for.

To find the equivalent resistance, you need to divide the applied potential difference by the current flowing into the material. For that you need to sum up the currents of all the resistors which are directly connected to a selected face (say, the face with potential V1). R_eqv = V1 / (i1+i2+i3+i4).

To find those currents, divide the potential difference across each resistor (directly connected to the selected face) by its resistance. e.g. (V1-V3)/R1.

I don't think you need to worry about all the internal node voltages or any other internal details to calculate resistance since resistance is something which is seen by the measuring equipment which can only see the faces and the net current flowing into the material.

I don't know if the above assumption would hold good if anything inside the material acts as a voltage source (With quantum and nano things, you can never tell :| ). If there is some such phenomenon, you can still find equivalent resistance by varying the applied voltage and dividing the difference in applied voltage by difference in current. e.g. find current for 20 V applied voltage. Then repeat for 21 V applied voltage. The equivalent resistance is then 1 V / (difference in current).

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  • \$\begingroup\$ Thank you very much. You've been extremly helpufl. \$\endgroup\$
    – Sebas Tian
    Commented May 13, 2021 at 8:47

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