Digging into the nanoworld

Since the circuit is symmetrical about line AB, we fold it about this line and form a circuit shown to the right.

Initially, let the resistance of each wire be $2R$ .

When the circuit is folded, the resistance of orange wires is $R$ and that of the blue wire is $2R$ .

Finding out the equivalent resistance of new circuit is easy and it comes out to be $\dfrac{40R}{7}$ .

Hence, the ratio is $\dfrac{20}{7} = \fbox{2.8571}$ .

As only the ratio is asked, we can safely assume the resistance of each arm to be 1 unit. The circuit is a symmetric one; so we can divide it into two halves connected in series. The red dotted line shows an equipotential region from where it is divided.

IMAGE

The resistance between A and B in the first figure gets halved as it is divided and so are the similar resistances within which the red line passes.

Now, in the equivalent circuit, we again look for equipotential surfaces (the red lines in the figure below). We see that resistances connected between two equipotential surfaces are same and are in parallel. So, the simplified circuit becomes as follows:

IMAGE

Net resistance in the above figure= $\frac{10}{7}$

So, the total resistance in the original circuit becomes $2 \times \frac{10}{7} =2.83$ units

Since we had assumed the resistance of each arm to be 1 unit, the required ratio becomes $2.83$

Let's apply a voltage V to the circuit and measure the total current that pass through it. By the definition of resistance, $R_{eq} = \frac{V}{i} (I)$ . For convention, let's denote by $r$ the resistance of an edge.

By the symmetry of the circuit we must have the following disposition of currents, where $a$ and $k$ are currents yet to be determined: image1 . However, by Kirchoff's Law of Currents $\frac{i}{2} = a + k (II)$

Using Kirchoff's Law of Tensions (KLT) on the hexagon in this figure image2 we have the following: $ar + ar + ar = kr + 2kr + kr \rightarrow 3a = 4k \rightarrow k = \frac{3a}{4}$

Substituting into $(II)$ , $\frac{i}{2} = a + \frac{3a}{4} \rightarrow \frac{i}{2} = \frac{7a}{4} \rightarrow a = \frac{2i}{7} (III)$

Now, let's apply KLT to the path in green shown in this figure image3 . We have: $\frac{ir}{2} + \frac{ir}{2} + ar + ar + ar + \frac{ir}{2} + \frac{ir}{2} = V \rightarrow 2ir + 3ar = V$ . Using $(III) : 2ir + \frac{6ir}{7} = V \rightarrow \frac{14ir}{7} + \frac{6i}{7} = V \rightarrow \frac{20ir}{7} = V$

Finally, using our first equation $(I) : R_{eq} = \frac{V}{i} = \frac{\frac{20ir}{7}}{i} = \frac{20r}{7}$

The problem asked $\frac{R_{eq}}{r} = \frac{\frac{20r}{7}}{r} = \frac{20}{7}$