Five equal resistances each of value $R$ are connected in a form shown alongside. The equivalent resistance of the network
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$\frac{1}{{{R_{BD}}}} = \frac{1}{{2R}} + \frac{1}{R} + \frac{1}{{2R}} \Rightarrow {R_{BD}} = \frac{R}{2}$
Between $A$ and $C$ circuit becomes equivalent to balanced Wheatstone bridge so ${R_{AC}} = R$.
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