Don't resist it! (2)

Label the cube circuit as shown in the left figure. By symmetry, the three vertices adjacent to $A$ , $C'$ and $C''$ have the same voltage with respect to $A$ and $B$ . Therefore, $C'$ and $C''$ can be considered as a point $C$ . Similarly, the three vertices adjacent to $B$ , $D'$ and $D''$ can be considered as a point $D$ . Then the equivalent circuit of the cube circuit is as the right figure. And the resultant resistance between $A$ and $B$ is given by:

$\begin{aligned} R_{AB} & = \left(r_{AC} + r_{CD} + r_{DB}\right)|| r_{AB} \\ & = \big(1||1 + 1||(1||1+1+1||1)||1 + 1||1\big) || 1 \\ & = \left(\frac 12 + 1||\left(\frac 12+1+\frac 12\right)||1 + \frac 12\right) || 1 \\ & = \left(\frac 12 + 1||2||1 + \frac 12\right) || 1 \\ & = \left(\frac 12 + \frac 25 + \frac 12\right) || 1 \\ & = \frac 75 || 1 = \frac {\frac 75\times 1}{\frac75 + 1} \approx \boxed{0.583} \end{aligned}$

Using the symmetry of the cube with respect to side $\overline {AB}$ , we can deduce that there are five different intensities (colors) we have to relate.

We now apply Kirchhoff's laws of nodes and closed loops to build enough equations to find them:

$\begin{aligned} \text{Node A:}&\quad\quad I_T&=&I_1+2I_2 \\ \text{Node N:}&\quad\quad I_2&=&I_3+I_4 \\ \text{Node P:}&\quad\quad I_5&=&2I_4 \\ \text{Loop NPQM:}&\quad\quad I_3&=&2I_4+I_5 \\ \text{Loop ANMB:}&\quad\quad I_1&=&2I_2+I_3 \\ \end{aligned}$

We get: $I_5=2I_4;\quad I_3=4I_4;\quad I_2=5I_4;\quad I_1=14I_4;\quad I_T=24I_4$

$\large R=\dfrac{V_{AB}}{I_T}=\dfrac{1\Omega \cdot I_1}{I_T}=\dfrac{14I_4}{24I_4}=\boxed{\dfrac{7}{12}\Omega}$