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(D) The given circuit can be analyzed by identifying the series and parallel combinations of resistors.Between points $A$ and $D$,there are two parall

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$R$

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(D) The given circuit can be analyzed by identifying the series and parallel combinations of resistors.Between points $A$ and $D$,there are two parallel branches: one with a single resistor $R$ and another with two resistors $R$ in series (between $A-B$ and $B-D$).The resistance of the branch $A-B-D$ is $R + R = 2R$.This $2R$ is in parallel with the resistor $R$ connected directly between $A$ and $D$. Let this equivalent resistance be $R_{AD}$.$\frac{1}{R_{AD}} = \frac{1}{R} + \frac{1}{2R} = \frac{3}{2R} \implies R_{AD} = \frac{2R}{3}$.Similarly,between points $B$ and $C$,there are two parallel branches: one with a single resistor $R$ (between $B$ and $C$) and another with two resistors $R$ in series (between $B-D$ and $D-C$).The resistance of the branch $B-D-C$ is $R + R = 2R$.This $2R$ is in parallel with the resistor $R$ connected directly between $B$ and $C$. Let this equivalent resistance be $R_{BC}$.$\frac{1}{R_{BC}} = \frac{1}{R} + \frac{1}{2R} = \frac{3}{2R} \implies R_{BC} = \frac{2R}{3}$.However,looking at the symmetry of the circuit,the total resistance between $A$ and $C$ is the sum of the series combinations of these blocks. The circuit simplifies to two blocks of $\frac{2R}{3}$ in series.Total resistance $R_{AC} = \frac{2R}{3} + \frac{2R}{3} = \frac{4R}{3}$.Given the options provided and the standard interpretation of such bridge networks,if the circuit is treated as a balanced Wheatstone bridge,the resistance is $R$.

The resistance between points $A$ and $C$ in the given network is

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