Correct Answer - A
(a) The circuit is symmetrical about the axis POQ.
The circuit above the axis POQ represents balanced
wheatstone bride. Hence the central resistance 2R is
ineffective. Similarly in the lower part (below the axis
POQ) the central resistance 2R is ineffective.
Therefore the equivalent circuit is drawn.
`:. 1/(R_(PQ)) = 1/(4R) + 1/(4R) + 1/(2r) = (r+r+2r)/(4Rr)`
`R_(PQ) = (2Rr)/(R+r)`.