Current Electricity Part 4

Concept:Use symmetry in the regular hexagon to identify points at equal potential, then simplify the network using series and parallel combinations.Explanation:Let the corners be A, B, C, D, E, F in order, with the centre O.Current enters at A and leaves at D (opposite corners).Due to symmetry about axis AD:Points B and F have the same potential.Points C and E have the same potential.The centre O lies on the axis AD, so no current flows through OB, OF, OC, OE (these wires are open-circuit for the main path).Now consider the path through the hexagon rim:From A to B: resistance $r$.From B to C: there are two parallel paths:Direct wire B–C: $r$.Via centre: $r$ (B–O) + $r$ (O–C) = $2r$.So equivalent resistance $R_{BC} = \frac{r \times 2r}{r+2r} = \frac{2r}{3}$.Then A–B–C–D: $r + \frac{2r}{3} + r = \frac{8r}{3}$.Similarly, the other rim path A–F–E–D gives the same $\frac{8r}{3}$.Direct path A–O–D: $r + r = 2r$.These three paths (ABCD, AFED, AOD) are in parallel between A and D.So total resistance $R$ satisfies:$$\frac{1}{R} = \frac{1}{\frac{8r}{3}} + \frac{1}{2r} + \frac{1}{\frac{8r}{3}} = \frac{3}{8r} + \frac{1}{2r} + \frac{3}{8r} = \frac{3+4+3}{8r} = \frac{10}{8r}.$$Thus $R = \frac{8r}{10} = \frac{4r}{5}$.Answer:$\frac{4}{5}r$, which corresponds to option B.