# Finding equivalent resistance?

The image shown right below is a schematic diagram of the circuit. There we need to find the equivalent resistance through X and Y.

I reduce the inner most resistors , because I figure out that there is a Wheatstone bridge out there, the resistance that I removed are shown below. How can I proceed further , I have no idea , I am stuck on this .How could I proceed further and find the equivalent resistance through X and Y ?

• It's like the author of the book, hates students. – efox29 May 13 '15 at 6:15
• apply a delta to star conversion to the outer resistors and see what you get. – Sada93 May 13 '15 at 6:16
• Delta to star means? – On the way to success May 13 '15 at 6:19
• @efox29 that's not too much hate. Had he really hated students, each resistor would have had different values. – K. Rmth May 13 '15 at 6:37
• @K.Rmth that would just be sadistic – Sada93 May 13 '15 at 6:38

It's quite easy if you exploit the symmetry: Because of the radial symmetry the 5 marked nodes have the same potential. Therefore there will be no current going throught the 5 marked resistors, i.e. you can treat them as being absent.

So the final resistance will be (R + R / 2) / 5

• Why the potential is same when the symmetry comes to act. – On the way to success May 13 '15 at 6:53
• Damn, Didn't see that. Its certainly easier than simplifying the circuit. – Sada93 May 13 '15 at 6:59
• Very witty answer and also a reminder to not jump to the maths without trying to simplify first. – K. Rmth May 13 '15 at 7:15
• @On the way to success: symmetry means that conditions at node "top left" are equal to conditions at node "top right" (or node "bottom left", "bottom, center" or "bottom right"). They are not distinguishable. Therefore they all must have the same potential. – Curd May 13 '15 at 8:23
• @On the way to success: R / 2 is the resistance of two resistors in parallel. R + (...) is the resistance of a network consisting of a resistor in series with a subnetwork. (...) / 5 is the resistance of a network consisting of 5 equal subnetworks in parallel. That should be enough hints. – Curd May 13 '15 at 10:31

As this is some kind of homework, you will not get a full answer here.

However:

First of all, if there is a current flowing from X to Y, it first definitively has to pass the inner resistors. So, why do you think you can remove them?

And there is a hint: You should already have noticed that the circuit is symmetric. Use this! For example, the current through each of the inner five resistors is exactly 1/5 of the total current. What current will flow through the middle resistors? What current through the outer?

Finally, the solution is a very simple, and you can calculate it without writing down anything.

• I removed because I could see a Wheatstone bridge out there – On the way to success May 13 '15 at 6:51