I am having trouble understanding the concept of equivalent resistances. In the finite grid network pasted here, i know that if i want the equivalent resistance between the nodes at which the current source is attached (node 3 and 9), i do node analysis, calculating the node voltages. then at point node 9, where the current should leave the circuit, i am able to find the equivalent resistance by dividing the voltage difference between node 3 and 9, by the current injected by the current source. enter image description here

However, suppose in this same network, i wanted the equivalent resistance between 3 and any other node say node 5? now the net current sum at node 5 will be 0 as a node gathers no current, so i am unsure of the denominator.

As a hack, i could disconnect the current sources output port from node 9 and place it node 5 (image 2) and perform the procedure. However, then i am not sure about this commentor here saying that the equivalent resistance is independent of where we attach the current source. in that sense we should have been able to calculate the resistance with the current source attached at 3 and 9 as earlier as well.

enter image description here


  • I guess what i am asking is, that how to calculate equivalent reistance between A and C when I have applied the voltage/current source between A and B. or does this question not make sense, and to measure equivalent resistance btw A and C you need to apply the source btw A and C.
  • If it can be done, how would you calculate the current at C when the source is between AB. ( the current at the node C sums to 0)
  • \$\begingroup\$ Have you tried delta to star conversion? \$\endgroup\$
    – Kaswechiha
    Commented Jan 24, 2021 at 5:12
  • \$\begingroup\$ @Kaswechiha I dont know much about this. but from the [electrical4u.com/… here it doesnt seem that 3 branches form a loop for my case, and a lot of the nodes are 4 connection ports, not 3. \$\endgroup\$ Commented Jan 24, 2021 at 5:57
  • \$\begingroup\$ I bet symmetry is your friend ;) \$\endgroup\$
    – carloc
    Commented Jan 24, 2021 at 10:17
  • \$\begingroup\$ @carloc i think the resistance values were distracting, i have removed those. also please read the question carefully, i am asking if i can use the first circuit to calculate the equivalent resistance between 3 and 5, or if i need to connect the source btw 3 and 9 for this question to make sense \$\endgroup\$ Commented Jan 25, 2021 at 0:50

1 Answer 1


What the commenter in your linked problem likely meant, is that it doesn't matter where you put the current source, the resistance across that current source will be the same everywhere due to symmetry reasons because all resistors have the same value.

So I think he actually isn't really trying to say what you're trying to do.

In general, what you are trying to do can be considered impossible except in very specific cases. These cases can be deduced as follows:

Consider you can write a reduced form of the nodal analysis: i.e. combining resistors together until you end up with just the following two nodes, while using the common \$V_3\$ node as the ground node:

$$Y\cdot \left(\begin{matrix} V_{5a} \\ V_{9a} \end{matrix}\right) = \left(\begin{matrix} 1 \\ 0 \end{matrix}\right)$$

Remember that \$R_{a}\$ is actually just \$V_{9a}/1A\$. We don't have access to \$V_{5a}\$ in what you want to do. We are actually interested in the solution of

$$Y\cdot \left(\begin{matrix} V_{5b} \\ V_{9b} \end{matrix}\right) = \left(\begin{matrix} 0 \\ -1 \end{matrix}\right)$$

More specifically, we are actually only interested in \$V_{5b}\$, as \$V_{5b}/1A\$ will be the resistance.

Now the easiest would be to just solve the second set of equations (which is what you get if you just analyze it with the current source between 3 and 5). However, what you appear to want to do is start with the solution of the first set of equations, \$(V_{5a}, V_{9a})\$, and then immediately jump to the solution of the second set of equations.

It is possible by realizing that in both situations, the Y-matrix is identical, while the right-hand side is just a permutation, or

$$\left(\begin{matrix} 0 \\ -1 \end{matrix}\right) = \left(\begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix}\right)\cdot\left(\begin{matrix} 1 \\ 0 \end{matrix}\right)$$

This allows us to link both of these sets of equations together, such that you find that:

$$\left(\begin{matrix} V_{5b} \\ V_{9b} \end{matrix}\right) = \left(Y^{-1}\cdot\left(\begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix}\right)\cdot Y\right)\cdot\left(\begin{matrix} V_{5a} \\ V_{9a} \end{matrix}\right) = F'\cdot\left(\begin{matrix} V_{5a} \\ V_{9a} \end{matrix}\right)$$

(\$F\$ contains unit-less values)

Now, saying that we could determine \$V_{5b}\$ only from knowing \$V_{9a}\$ implies that \$F'_{11} = 0\$, because else we would need to know \$V_{5a}\$ as well, since \$V_{5b} = F'_{11} V_{5a} + F'_{12} V_{9a}\$. It also means that you have a way to calculate \$F'_{12}\$, which is maybe not as straightforward as you might like. We then have that:

$$F'_{11} = 0 \Rightarrow Y_{22}\cdot Y_{21} + Y_{11}\cdot Y_{12} = 0$$

$$F'_{12} = \frac{Y_{12}^2 + Y_{22}^2}{Y_{11}Y_{22}-Y_{12}Y_{21}}$$

I think at this point that you can see that this case is not obviously satisfied. In your specific case, it will even never happen without a lot of open-circuits. In your example, the Y-matrix can be reduced to

$$Y = \left(\begin{matrix} g_{3-5}+g_{5-9} & -g_{3-9} \\ -g_{3-9} & g_{3-9}+g_{5-9} \end{matrix}\right)$$

If all \$g\$ are positive, then \$F'_{11} \neq 0\$. The only exception would be a situation where \$g_{5-9}=0\$, or in other words, that there is no conductive path between the both of them. In that case, \$F'_{12} = \frac{g_{3-9}}{g_{3-5}}\$ and this is trivial because it basically requires you to know the resistance \$g_{3-9}\$...

  • \$\begingroup\$ Am I correct in uderstanding that Y is the L matrix mentioned here :(equation 5 and 6). if so, wouldnt it always be invertible. and wouldnt a non invertible matrix imply we cant get the solution of V for a circuit, implying either no or multiple solutions, which we know is not true for an electrical circuit. \$\endgroup\$ Commented Jan 26, 2021 at 22:27
  • \$\begingroup\$ I think I was half asleep when I was writing this answer, because there are a few other points that also don't make sense (along with your remark). I'll edit it. \$\endgroup\$
    – Sven B
    Commented Jan 28, 2021 at 8:40
  • \$\begingroup\$ I took a closer look and changed my answer. \$\endgroup\$
    – Sven B
    Commented Jan 28, 2021 at 10:29

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