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(A) The given circuit can be simplified by identifying the series and parallel combinations.Let the central node be $O$. The resistors connected to $O

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$r$

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(A) The given circuit can be simplified by identifying the series and parallel combinations.Let the central node be $O$. The resistors connected to $O$ are in parallel with the outer loop.However,looking at the symmetry,the circuit can be redrawn as a Wheatstone bridge or by using symmetry arguments.By applying a potential difference between $A$ and $B$,we can see that the resistors form a bridge circuit.Specifically,the equivalent resistance $R_{AB}$ is calculated as follows:$R_{AB} = \frac{(r+r)(r+r)}{(r+r)+(r+r)} = \frac{2r \cdot 2r}{2r+2r} = \frac{4r^2}{4r} = r$.

Find the equivalent resistance between points $A$ and $B$ in the network of resistors,each of resistance $r \Omega$.

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