I was solving problems from Irodov when I came across the method called point of symmetry method. I cant find this method anywhere. So what is this method? Here is the link to the problem solution http://irodovphysicssolutionrks.blogspot.in/2013/12/problem3150_16.html
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asked Jun 8, 2015 at 15:44
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2$\begingroup$ For symmetric networks one can simplify by removing branches that carry no current, which is probably what he does here. Please note, though, that all of these symmetric problems are just brain-teasers with very little practical value. You should instead focus on understanding how to solve general circuits without any symmetries by using Kirchhoff's laws and linear algebra (assuming that this is for a EE degree?). $\endgroup$– CuriousOneJun 8, 2015 at 15:51
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1$\begingroup$ In general cases, when using known transformations wont help you, you can add a test generator of arbitrary emf between points you're calculating resistance. You can do that in this example but its not necessary since its easy to spot nodes on the same potential ( hence no current flows between the edge connecting them. ) $\endgroup$– user16688Jun 8, 2015 at 16:04
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3 Answers
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The idea of the linked page is this: you can often find the equivalent resistance of certain highly symmetrical assemblies of resistors by adding "phantom" wires between points with equal potential. This has the advantage of giving quick, near-automatic answers without the use of Kirchhoff's laws, but it is very limited and works in very few cases.
This page explains it in (much!) better detail than the page in OP's link.
answered Apr 23, 2016 at 3:47
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2$\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ Apr 23, 2016 at 4:21
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1$\begingroup$ @JonCuster I think that you are being a little too harsh about this comment which does answer the question. The answer given by ritvik1512 assumes that the current through each of the resistors connected to node $A$ is the same. What Soham is saying is that the other ends of those resistors are at the same potential and so they can considered to be in parallel ie connected together with "phantom" wires. Repeating the process from node $B$ gives you a series arrangement between nodes $A$ and $B$ of 3 parallel resistors then 6 parallel resistors then 3 parallel resistors. $\endgroup$– FarcherApr 23, 2016 at 6:20
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$\begingroup$ @JonCuster I've expanded my answer. I think the question specifically asked for an explanation of the nonstandard method given in the link, so I believe this is an appropriate answer. $\endgroup$ Apr 23, 2016 at 6:37
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I'm not too sure, what you mean by "Point of Symmetry" method, and the link doesn't make much sense as well.
Anyways, I'll list the approach, I usually use, see if it is of any help to you.
Suppose the cube is:
Kirchhoff's current law, which states that the sum of the currents entering and exiting a node is zero, is essential in the analysis.
The first step is to recognize that at a node where equal resistances exist, current entering the node will be distributed equally between the number of output branches - in this case, three. For convenience sake, I assigned an input current of $3$ amperes at the corner labeled "A," so that 1 amp will flow through each output branch. Note that $1 A$ flows through each branch.
On the far side of each of those branches is another node with two output branches. Again, due to symmetry, the input current will divide evenly so that $\frac{1}{2}A$ flow into each branch. Looking at the cube's output node labeled "B," it is apparent that the same situation exists as with "A."
Now that you know the current through each branch, and you know that each branch has a single $1 \Omega$ resistor in it, Ohms law allows you to calculate the voltage across each resistor.
The next step is to sum the voltage from input node "A" to output node "B." Any path you take travels along three edges, and you can find the total potential difference.
Finally apply the Ohm's Law to get the resultant resistance.
Finally, I agree with the comments above. You must learn how to deal with any type of circuits in general.
Hope it helps!
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Your friends for this problem are the laws of Ohm and Kirchhoff. Both laws are linear and you can build a system of linear equations that you solve. In the end I get 5/6 times the resistor along the sides of the cube. Other answers are also possible because of the boundary conditions of the system. An article on my blog https://pa1ejo.wordpress.com/2018/01/27/resistor-cube/ describes how to code it in MATLAB, the model values are compared to measurements of a soldered resistor cube.
Ejo Schrama
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