Current Electricity Part 4

Each of the six segments (three outer edges and the three inner spokes from the central junction) is a resistor of equal length, so each segment has resistance $R_1 = \frac{R}{6}$.The figure is a tetrahedron: apex $D$ (top vertex) and base vertices $A$, $B$, $C$ (the central junction), with every pair of vertices joined by one segment of resistance $R_1$.By left-right symmetry, the nodes $C$ and $D$ are equidistant from $A$ and $B$, so they are at the same potential.Hence no current flows through the segment $CD$, and nodes $C$ and $D$ can be merged.Therefore, between $A$ and $B$ there are three parallel branches:(i) the direct edge $AB$ of resistance $R_1$,(ii) the path $A \to C \to B$ of resistance $2R_1$,(iii) the path $A \to D \to B$ of resistance $2R_1$.So $R_{AB} = R_1 \parallel 2R_1 \parallel 2R_1$.$\frac{1}{R_{AB}} = \frac{1}{R_1} + \frac{1}{2R_1} + \frac{1}{2R_1} = \frac{2}{R_1}$$\Rightarrow R_{AB} = \frac{R_1}{2} = \frac{1}{2} \times \frac{R}{6} = \frac{R}{12}$Comparing with $R_{AB} = \frac{R}{n}$ gives $n = 12$.Hence the correct option is 12.