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(D) Let the resistance of each of the $6$ segments be $r$. Since the total resistance of the wire is $R$,we have $6r = R$,which implies $r = R / 6$.Lo

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$12$

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(D) Let the resistance of each of the $6$ segments be $r$. Since the total resistance of the wire is $R$,we have $6r = R$,which implies $r = R / 6$.Looking at the circuit,we can identify that the structure is equivalent to a Wheatstone bridge configuration. The two resistors connected to the top node and the two resistors connected to the central node form a balanced bridge between points $A$ and $B$.Specifically,the path from $A$ to $B$ consists of two parallel branches:$1$. $A$ branch with two resistors in series: $r + r = 2r$.$2$. Another branch with two resistors in series: $r + r = 2r$.$3$. $A$ direct resistor between $A$ and $B$ with resistance $r$.Thus,the equivalent resistance $R_{AB}$ is given by:$\frac{1}{R_{AB}} = \frac{1}{2r} + \frac{1}{2r} + \frac{1}{r} = \frac{1}{r} + \frac{1}{r} = \frac{2}{r}$.Substituting $r = R / 6$:$R_{AB} = \frac{r}{2} = \frac{R / 6}{2} = \frac{R}{12}$.Comparing this with $R / n$,we get $n = 12$.

$A$ wire of resistance $R$ is bent into a triangular pyramid as shown in the figure,with each segment having the same length. The resistance between points $A$ and $B$ is $R / n$. The value of $n$ is

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