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(B) The given circuit is a hexagonal network with a central node. We can solve this using the symmetry method.By observing the symmetry about the axis

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$0.5\,R$

Vedclass · Answer

(B) The given circuit is a hexagonal network with a central node. We can solve this using the symmetry method.By observing the symmetry about the axis $AB$,the potential at the nodes above and below the axis are equal.This allows us to simplify the circuit by folding it along the axis $AB$.After folding,the resistors in the upper half and lower half combine in parallel.The upper branch effectively becomes two resistors of $R$ in series,giving $2R$. The lower branch also becomes $2R$.The central horizontal resistor remains $R$.Now,we have three parallel branches between $A$ and $B$: one with resistance $2R$,one with $R$,and one with $2R$.The equivalent resistance $R_{AB}$ is given by:$\frac{1}{R_{AB}} = \frac{1}{2R} + \frac{1}{R} + \frac{1}{2R}$$\frac{1}{R_{AB}} = \frac{1 + 2 + 1}{2R} = \frac{4}{2R} = \frac{2}{R}$Therefore,$R_{AB} = \frac{R}{2} = 0.5\,R$.

Find out the equivalent resistance between $A$ and $B$.

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