A regular hexagon is formed by six wires each of resistance $r\Omega$ and the corners are joined to the centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be
Answer: (B) $\dfrac45r$
Label the corners $A, B, C, D, E, F$ in order, with the current entering at $A$ and leaving at the opposite corner $D$; $O$ is the centre. Take $V_A = 1$, $V_D = 0$.
By symmetry $V_O = \dfrac12$, $V_B = V_F = x$ and $V_C = V_E = 1 - x$.
KCL at $B$ (neighbours $A$, $C$, $O$):
$$(1 - x) + (1 - x - x) + \left(\tfrac12 - x\right) = 0\Rightarrow 4x = 2.5\Rightarrow x = 0.625$$
Current leaving $A$ (each wire $r$):
$$I = \frac{(1 - 0.625) + (1 - 0.625) + (1 - 0.5)}{r} = \frac{1.25}{r}$$
$$R_{AD} = \frac1I = \frac{r}{1.25} = \frac45r$$
Solution by Sreeraj P, M.Sc Physics