Let us assume we have
connected a battery of emf V
across points 1 and 6. if potential
of point 1 is V, then the potential
of point 6 is 0.
From symmetry, it is clear that
the current entering at point1 will equally divide the resistance. Hence, the potential differences
across resistances, conncted
across (1,8) , (1,2), and (1,4), will be equal .
Similarly the current coming out at point 6 will be equal
to current ennering at point 1. By symmetry, we can say the current in the resistances connected across points (7,6),
(3,6), and (6,5) will be equal. Hence, potential difference
across these resistance should be equal. Let us assume the
potentials at points 3,5, and 7 be x then the potentials at points 2,4, and 8 should be (V - x). At point 7,
`([x- (V - x)])/(R ) + (x -0)/(R ) +([x- (V - x)])/(R ) = 0`
or ` x (2)/(5)V (i)`
At point 1, `I =(3[V - (V -x)])/(R ) = (3x)/(R ) (ii)`
From (i) and (ii)
` (V)/(I) = (5R)/(6)`
Hence, effective resistance will be 5R/6.
