# Effective resistance of a weird looking electric circuit [closed]

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?

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

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• Just to clarify, is 9 10 and 11 are connections? or the wires crossing over, without touching other? – Thanushan Jun 7 '17 at 11:24
• Conncetion they are. @Thanushan – Mockingbird Jun 7 '17 at 11:31
• That is a short circuit over $R_{23}$, $R_{34}$, and $R_{45}$ meaning they can go out of the problem. – mikuszefski Jun 7 '17 at 11:36
• ....that would be unusual for that type of problem, though. Therefore, the question by @Thanushan – mikuszefski Jun 7 '17 at 12:03
• 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. – NonStandardModel Jun 7 '17 at 13:02

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.
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 $E/I_1$.