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Given 6 points that are connected with each other with a resistor of resistance $R$, find the resistance between any two points. (Answer: $R/3$)

Illustration of the problem

(All the conducting wires have the same resistance $R$.)

I know that such a wording immediately implies that these 6 points are absolutely identical, which makes it possible to apply symmetry arguments that will help in reducing the network to a simpler one.

That is, after choosing any two points in the network, the remaining four points will still be identical, so we can swap any of them and the network will remain the same. Thus, we can remove the resistors which are connected between these other four points, since the points are identical.

However, we can also swap the two chosen points, and the system will still remain the same. So why can't we also remove the resistor between the two chosen points?

I'm told of the following analogy: The system of these 6 points is like a system of 6 absolutely similar balls painted, say, in white. By choosing two points, we paint them in black, thus the system loses some symmetry level, but its certain elements are still symmetrical under certain rearrangements.

Specifically, any two of the white balls can be swapped without changing the system in any way, so all the white balls are identical and we can ignore any resistors between them. But swapping the two black balls still won't change the system, so why can't we follow the same logic and ignore the resistor between them as well?

I'll generalize the question a little bit: why don't we care about other symmetries in the system?

(I'm looking forward a simple explanation, without involving advanced math, as I'm just a self-taught beginner and I'm only familiar with calculus. So I try to avoid matrices and anything advanced which students learn in advanced courses of electronics. I just want to get the idea and the concept itself.)

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Can you please include a diagram. Helps clarify the question and avoid misunderstanding... –  mehfoos Jul 23 '13 at 12:01
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WalterWhite, it's not only difficult to tell what's being asked here, the preface paragraph is unclear (particularly without a diagram or three) and, on first read, illogical. I've voted to close. –  Alfred Centauri Jul 23 '13 at 12:36
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What's the purpose of voting to close the question? Just inform me to clarify the question a little bit and I'll try to make it more clear. No need to close things immediately. By the way, the problem itself is taken from an olympiad. –  stuck_with_problem Jul 23 '13 at 12:47
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@WalterWhite, if it were a minor clarity issue, I would do just that. But, it's not a minor clarity issue; the first paragraph is a mess. You've asked for simple explanation but you've written the equivalent of "spaghetti code" for a question. –  Alfred Centauri Jul 23 '13 at 13:08
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@AlfredCentauri - the first paragraph is a mess? Really? I gave a question as it is written originally. Then I presented the solution to the question using symmetry in detail, so that people will be able to understand what confuses me in that solution. Then I even generalized my question a bit and asked why we do not take all the symmetries into account. However, apparantely you and the moderator didn't understand the question properly (for some magic reason) and decided to close. Smart decision, what can I say. I hope that you're happy now. Thanks for all the people who helped anyway. –  stuck_with_problem Jul 23 '13 at 17:28

4 Answers 4

up vote 5 down vote accepted

We can't remove the resistor between the two points we've chosen because they're not at the same voltage.

OK, let's unpack that a little. Imagine that you actually have a resistor network (any resistor network) built and want to measure its resistance with an ohmmeter. To do that, you need to choose two of the points in the network and connect the leads of the ohmmeter to them. The ohmmeter will then pass a small DC current through the network, measure the voltage difference $\Delta V$ between its leads and the current $I$ flowing through the network between them, and calculate the resistance $R$ of the network using Ohm's law: $$R = \frac{\Delta V}I$$

Now, since we're feeding a fixed DC current into the network, and since we only have passive resistive components in it, the network will very rapidly (essentially instantaneously) settle into a steady state where each node is at a constant voltage and each resistor has a constant current flowing through it.

Specifically, the node to which we've connected the negative lead of the ohmmeter is pulled down to some fixed voltage $V^-$, while the node to which we've connected the positive lead is pulled up to some voltage $V^+ > V^-$. Every other node $i$ of the network will be at some intermediate voltage $V_i$ between $V^-$ and $V^+$. Using your colored balls analogy, it's as if we've chosen two balls, colored one of them white and one black, and colored the rest of the balls with different shades of gray according to their equilibrium voltage, as determined by Ohm's law and Kirchhoff's first law.

Indeed, we can mechanically solve for the equilibrium current $I$ through the system simply by writing down the expressions for the current flow through each resistor $ij$ given by Ohm's law: $$I_{ij} = \frac{V_i - V_j}{R_{ij}}$$ and the conservation of current at each node $i$ given by Kirchhoff's first law: $$\sum_j I_{ij} = I^0_i = 0 \text{ for all }i \notin \{a,b\},$$ together with the known voltages $V_a = V^+$ and $V_b = V^-$ and excess currents $I^0_a = -I^0_b = I$ for the endpoint nodes $a$ and $b$, and solving the resulting system of linear equations for $I$.

However, if we want to simplify the system before solving it, we can apply two useful observations:

  • First, if two nodes have the same voltage, no current can flow between them: $V_i = V_j$ $\implies$ $I_{ij} = 0$. (Check this using Ohm's law above!) Thus, we can completely ignore any resistors between such nodes. In fact, we can even effectively collapse such nodes together into a single node (as if they were connected by a wire with zero resistance), as long as we remember to account for the fact that we may end up with several resistors in parallel between two nodes.

  • Second, if we have two nodes $i$ and $j$ such that $R_{ik} = R_{jk}$ for all nodes $k$ (where we take $R_{ik} = \infty$ if $i$ and $k$ are not connected) and $I^0_i = I^0_j$, then we can swap the labels of those two nodes without changing any of the parameters of the system. Thus, by symmetry, the solution must have $V_i = V_j$, since otherwise swapping the labels would change the solution without changing the parameters (which is a contradiction if the system is well defined and thus uniquely solvable).

In your example network, every node is connected to every other node by identical resistors, and so $R_{ik} = R_{jk}$ for all nodes $i$, $j$, $k$. For all but the two chosen endpoint nodes, we also have $I^0_i = I^0_j = 0$, and so all the other nodes except the endpoints can be exchanged without changing the system. Thus, we may ignore any resistors between them and even collapse them all into a single node.

However, the reason we cannot exchange the two chosen endpoints is that we broke the symmetry when we connected the measurement leads to them: those points have current flowing into them from outside the network, which will pull them to different voltages. In particular, the voltage difference will cause a non-zero current to flow across any resistor connecting those two nodes, and so such resistors cannot be ignored when calculating the total current flow across the system.

(If there was no externally supplied current, all the nodes in the network would indeed be symmetrical, and we could correctly deduce that no current would flow between any of them. But that scenario is completely useless for calculating the resistance, since we'd just end up with the indeterminate form $R = \Delta V/I = 0/0$.)

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+1 I think the first sentence in the penultimate paragraph may be precisely what WalterWhite is looking for. –  Alfred Centauri Jul 23 '13 at 14:34
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Thank you for the detailed explanation. I now understand that the symmetry is broken the moment you choose the points. It may sound very obvious, but it wasn't obvious for me. Thanks again. –  stuck_with_problem Jul 23 '13 at 17:30

Given $N \ge 2$ nodes fully interconnected with $\frac{1}{2}N(N-1)$ resistors of resistance $R$ (i.e. each two nodes connected to each other with a resistance $R$), the resistance between any two nodes is $\frac{2}{N}R$.

This follows directly from allocating a voltage "$+1$" and "$-1$" to the two nodes under consideration, which results in the $N-2$ other nodes (all directly connected to both nodes under consideration) acquiring a zero voltage. As no current will flow through resistances connecting nodes at the same potential, all resistances not directly connected to the two nodes under consideration can be removed.

In other words: the only conductance paths contributing are a single "one hop" of resistance $R$ and $N-2$ "double hops" of resistance $2R$. The total parallel conductance of all these path combined is $\frac{1}{R} + \frac{N-2}{2R} \ = \frac{N}{2R}$. The reciprocal of this quantity is the effective resistance between the two nodes.

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As best as I can tell, the question is this: "so why can't we follow the same logic and remove the resistor between the two chosen points?". –  Alfred Centauri Jul 23 '13 at 13:32
    
I think you're right. Have added a phrase to explain this point. –  Johannes Jul 23 '13 at 13:37

There is a nice argument for these type of questions. The proof follows from the linearity of the system. Here's how it works:

  1. Assume I enter the current $I$ from one node, and extract $\frac{I}{5}$ from all other nodes. Because of the symmetry, I know that exactly $\frac{I}{5}$ will flow through the adjacent wires(edges).
  2. I am now looking at another node, assume I enter $\frac{I}{5}$ current from all other nodes and extract $I$ from this one. Again using symmetry, I know that exactly $\frac{I}{5}$ will flow through wires connected to that node.
  3. Now consider the superposition of these two cases. I am injecting $\frac{6I}{5}$ from one node and extracting the same from the other vertex. No current will exit or enter from other nodes. Also we know the current passing through the wire connecting these two nodes is exactly $\frac{2I}{5}$(it follows from linearity).

Ergo using the definition of equivalent resistance and the potential difference between these two nodes, we will have:

$$I_{\text{total}}R_{\text{eqv}}=\frac{2I}{5}R \Rightarrow R_{\text{eqv}}=\frac{R}{3}$$

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As best as I can tell, the question is this: "so why can't we follow the same logic and remove the resistor between the two chosen points?". –  Alfred Centauri Jul 23 '13 at 13:32
    
@AlfredCentauri The question has changed a lot since I posted my answer. "so why can't we follow the same logic and remove the resistor between the two chosen points?" is obvious, because they don't have the same potential! –  Ali Jul 23 '13 at 20:08

so why can't we follow the same logic and remove the resistor between the two chosen points?

OK, I'll directly address your question in part. Specifically, why can't we remove the resistor between the two chosen points?

The answer is an elementary application of Ohm's Law. Pick any two nodes. There is a resistor that directly connects those two nodes.

(1) Place a voltage source across the two nodes. There is now a voltage across the resistor between the two nodes and thus, by Ohm's Law, a current through that resistor.

(2) This current adds to the total current through the voltage source.

(3) If you remove the resistor, you change the current through the voltage source and thus, you change the resistance seen by the voltage source.

Conclusion: you cannot remove the resistor between the chosen nodes without changing the equivalent resistance between those nodes.

Clearly, there is a flaw in the logic that led you to the conclusion that the resistor can be removed. But honestly, the way the logic is written, it's not clear what the logic is or how it led you to this conclusion.

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