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(D) The circuit is symmetric. Let the potential at $A$ be $V_A$ and at $B$ be $V_B$. By applying symmetry or nodal analysis,we can simplify the circui

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$\frac{8}{7}\,\Omega$

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(D) The circuit is symmetric. Let the potential at $A$ be $V_A$ and at $B$ be $V_B$. By applying symmetry or nodal analysis,we can simplify the circuit.Looking at the circuit,we can identify two identical triangular sections connected in series/parallel combinations.By simplifying the network,the equivalent resistance $R_{eq}$ is calculated as follows:$1$. The two resistors in the middle branch form a parallel combination with the upper branch,which simplifies to $\frac{2}{3}\,\Omega$.$2$. The total resistance between $A$ and $B$ is the parallel combination of the upper path (having resistance $1 + \frac{2}{3} + 1 = \frac{8}{3}\,\Omega$) and the lower path (having resistance $1 + 1 = 2\,\Omega$).$3$. $R_{eq} = \frac{(\frac{8}{3}) 	imes 2}{(\frac{8}{3}) + 2} = \frac{\frac{16}{3}}{\frac{14}{3}} = \frac{16}{14} = \frac{8}{7}\,\Omega$.

In the network shown in the following figure,each resistance is $1\,\Omega$. The effective resistance between $A$ and $B$ is

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