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A electric circuit (in the picture) is given where all the resistance are of 1 ohm. I have to find its equivalent resistance. (9,10, 11th points are conncections)

My attempt: I think electron flow will follow 2 paths: 1 2 9 3 10 4 11 5 6 and 1 8 7 6. So the resistances of these 2 paths are in series combination. So, equiavalent resistance of path 1 and 2 respictively are 2 and 3 ohm. As these 2 paths are in parallel the equivalent resistance would be 6/5 ohm.

Am I right? I am assuming that 3 short-circuits are present (2 9 3, 3 10 4, 4 11 5) in those 3 subcycles of the whole circuit.

I think my attempt is wrong. Any hint?

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closed as off-topic by AccidentalFourierTransform, Jon Custer, Chris, Emilio Pisanty, stafusa Feb 6 '18 at 0:31

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  • $\begingroup$ Just to clarify, is 9 10 and 11 are connections? or the wires crossing over, without touching other? $\endgroup$ – Thanushan Jun 7 '17 at 11:24
  • $\begingroup$ Conncetion they are. @Thanushan $\endgroup$ – Mockingbird Jun 7 '17 at 11:31
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    $\begingroup$ That is a short circuit over $R_{23}$, $R_{34}$, and $R_{45}$ meaning they can go out of the problem. $\endgroup$ – mikuszefski Jun 7 '17 at 11:36
  • $\begingroup$ ....that would be unusual for that type of problem, though. Therefore, the question by @Thanushan $\endgroup$ – mikuszefski Jun 7 '17 at 12:03
  • $\begingroup$ I think that if there were no connections, there would be no need to mark it with node numbers. So probably the crossed wires are connected. $\endgroup$ – NonStandardModel Jun 7 '17 at 13:02
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Current will take all possible paths. There are more than 2 possible paths, and they are not connected in series or parallel.

First try simplifying the circuit. The resistors on the top row are all shorted out, so they can be removed without affecting the circuit. The diagonal resistors are connected at the top and bottom to the vertical resistors, so these are in parallel; there are 2 resistors in parallel at the LH and RH branches, and 3 in the middle two branches.

You can then apply Kirchhoff's Rules to the simplified circuit, making use of symmetry. Alternatively, a combination of 3 resistors connected to the same node can be replaced by 3 resistors arranged in a triangle, using the $Y-\Delta$ Transformation. All of the resistors are then in series or parallel.

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I would use the loop current method to come up with a system of equations. They use Cramer's method to solve for the current in the large loop. Once you have the current in the large loop, call it $I_1$, the equivalent resistance is $E/I_1$.

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