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Equivalent resistance between A and B

Hi, I've been doing some current electricity problems by using Kirchhoff's laws. I've tried applying KVL(Kirchhoff Voltage law) to this circuit, but to no avail. There happens to be too many variables to work on with my approach. I would appreciate if I could get some help regarding solving this problem and similar ones.

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  • $\begingroup$ This is a symmetry question, often the central resistor carries no current due to symmetry, i.e. both side are at the same potential. $\endgroup$ – PhysicsDave Jan 6 at 20:43
  • $\begingroup$ But here the central R does have current. The way to solve is to see 2 paths. Path 1 is A to D to F to B, note that D to F equates to 0.5R so this paths 2.5R. Path 2 is A to C to E to B, in this path we get 3R (note central R has zero I). Combine the 2 paths in parallel. $\endgroup$ – PhysicsDave Jan 6 at 20:52
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The way to solve is to see 2 paths. Path 1 is $A$ to $D$ to $F$ to $B$, note that $D$ to $F$ equates to $ 0.5R$ so this path is $ 2.5R$. Path 2 is $A$ to $ C$ to $ E$ to $B$, in this path we get $3R$ (note central $R $ has zero $I$). Combine the 2 paths in parallel.

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    $\begingroup$ Well, it gives the right answer, but how does it work? Why does D to F equate to 0.5 $R?$ Why do you say that the "central $R$ has zero $I"?$ None of the resistors has zero current. Would you please spell out your method more fully? $\endgroup$ – Philip Wood Jan 6 at 23:13
  • $\begingroup$ The method used was to look at it as 2 circuits in parallel even though there are common resistors in both paths. So if we unhook the wires for path 2 and look at path 1 we get path 1 resistance and similarly for path2 we unhook path 1. When considering path 2 only, R in the middle is zero current. Because we can isolate 2 paths within the circuit they merely become like 2 resistors in parallel. $\endgroup$ – PhysicsDave Jan 7 at 2:12
  • $\begingroup$ Thank you. Some further clarification if you please. (a) You say that path 2 is A to C to E to B. On this path, what is the route from C to E: via F or via D or some sort of composite? (b) Why is it that, on route 1, "D to F equates to 0.5 R" ? $\endgroup$ – Philip Wood Jan 7 at 18:17
  • $\begingroup$ For C to E include the 5 central resistors, there are 2 parallel branches, 2R, 2R and for the R due to symmetry, the single R has no voltage, so you are left with 2 2Rs in parallel which is R. For D to F look at the 5 resistors as 3 parallel branches of 2R,2R and R which becomes 0.5R. The trick here is the symmetry, it says I in AC and BE is equal also AD and BF are equal currents so therefore the circuit can be broken into 2 separate parts, i.e 2 separate current branches even though there are common components. $\endgroup$ – PhysicsDave Jan 8 at 0:32
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Here are two hints. (1) Apply Kirchhoff I at the junctions on the diagram itself, by labelling the currents $x,$ $y,$ $(x-y)$ (or whatever is right), and so on. In other words, don't waste time writing formal Kirchhoff I equations, and don't call the currents $"I_1",$ $"I_2"$ etc.

(2) Look for symmetries in the circuit. In this case the right hand end is an upside down version of the left hand end, and that enables you to label everything in terms of three unknown currents. Find their values using Kirchhoff II.

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