Effective resistance in a pentagram of resistors

[bassist11]

Homework Statement

So I have a pentagram of resistors, one in each branch of resistance "r". Each joint is also connected to the two joints across from it but none of the wires inside the pentagram are connected. I need to find the resistance between any two adjacent joints.

Homework Equations

The Attempt at a Solution

Basically, I know I have to assign currents to each branch first and assume I know these currents because in the end they will divide out anyways. I know Kirchhoff's laws and how to add resistors in parallel and series. I'm just stuck on how the currents divide. Our teacher has given us an answer of 2/5 resistance and if you divide the current going in at one point into four due to symmetry, you won't get that.

Help?

 

[flatmaster]

I'm not sure what the circuit is you're describing, but it sounds like you might be in one of the configurations that is neither parallel nor series. In this case, you need Kirchhoff.

 

[eachus]

It is a "trick" problem. Either you see the solution or you don't. Letter the points of the pentagram ABCDE in order around the perimeter. Now, if you want to calculate the current from A to B, assuming they are the only connections to an external current source, start by ignoring the triangle CDE. You have one path of resistance r and three paths of resistance 2r. Now notice that the symmetry of the circuit means that you can't tell vertices C, D, and E apart. Each is the mid-point of one of the 2r paths, and each is part of triangle CDE. By renaming the verticies you can see that current won't flow in triangle CDE. (Or manybe you can't see it, a lot of students have trouble doing the mental gymnastics.)

So A and B are directly connected by a resistor of 1 r, and indirectly by three circuits of resitance 2 r. The easiest way to proceed now is to look at the currents. If r is 1 ohm, and the voltage from A to B is one volt, you will have a current of 1 Ampere through the direct path, and three 1/2 Ampere currents through the other vertices. Total current 5/2 Ampere, effective resistance 2/5 r.

 

[bassist11]

That helps a ton^ When I was working it earlier, I kind of came to the same conclusion about no current flowing through that upper triangle. I think I just went wrong though when I divided the current into four when it comes in.

Would I be right in saying that the effective resistance is the same between any two points?

Our physics teacher has just been piling these "novelty" resistor problems on us.. We've done the cube, two infinite chains, a square, a diamond, you name it..