In the given circuit,the equivalent resistanc | vedclass

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(C) The circuit exhibits symmetry about the axis passing through $A$ and $B$. Due to this symmetry,the points $L$,$M$,and $N$ (as labeled in the solut

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$\frac{R}{3}$

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(C) The circuit exhibits symmetry about the axis passing through $A$ and $B$. Due to this symmetry,the points $L$,$M$,and $N$ (as labeled in the solution image) are at the same potential.Since these points are at the same potential,no current flows through the resistors connected between $L-M$ and $M-N$. Thus,these resistors can be removed from the circuit.After removing these resistors,the circuit simplifies into three parallel branches between $A$ and $B$:$1$. The top branch consists of two resistors of resistance $R$ in series,giving a total of $2R$.$2$. The middle branch consists of two resistors of resistance $R$ in series,giving a total of $2R$.$3$. The bottom branch consists of two resistors of resistance $R$ in series,giving a total of $2R$.Now,we have three resistors of $2R$ connected in parallel.The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{2R} + \frac{1}{2R} = \frac{3}{2R}$Therefore,$R_{eq} = \frac{2R}{3}$.

In the given circuit,the equivalent resistance between the points $A$ and $B$ is (in $ohm$):

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