Dodecagon Resistance

If you apply $1$ V to one node and ground another node, then by symmetry, all the remaining nodes will be at $0.5$ V. So the total current going out through the grounded node will be $0.5$ times the number of nodes that are at $0.5$ V added to $1$ times the node to which $1$ V was applied. So $I = (10*0.5 + 1*1)/300 = 1/50$ . So, the effective resistance will be: $R_e = \boxed{50} \mbox{ohms}$

• Pick any two nodes, and mark them as A and B.
• Apply $1V$ to node A and ground node B.
• Remove from the network all the resistors which are not connected to A or B. You should get the following circuit:
• Resistors you've just removed were previously connected between any two of the red points. It can easily be calculated that the potential of each of the red nodes is $0.5V$ , meaning that the voltage between any two of the red points is $0V$ .
• Try putting all the resistors you've removed back in their place. You will find that each time you connect a resistor back in the circuit, it will end up being connected between equipotential points, which means that no current will flow through it. It turns out that all the previously removed resistors have no impact on the equivalent resistance between A and B, and so it follows that the resistance of the dodecagon network is actually equivalent to the one of the network in the picture I posted here. So, the total resistance is:

$\dfrac{1}{R_{tot}} = 10 \times \dfrac{1}{2R} + \dfrac{1}{R} = \dfrac{12}{2R} = \dfrac{6}{R} \rightarrow R_{tot} = \dfrac{R}{6} = \boxed{50 \Omega}$

Using exactly the same procedure, this can be generalized for networks represented as regular polygons with $n$ sides, where the total resistance between any two selected vertices is:

$\boxed{R_{tot} = \dfrac{2R}{n}}$