A wire of resistance $R$ is bent into a triangular pyramid as shown in figure with each segment having same…

A wire of resistance $R$ is bent into a triangular pyramid as shown in figure with each segment having same length. The resistance between points $A$ and $B$ is $R / n$. The value of $n$ is :
  1. 16
  2. 14
  3. 10
  4. 12

Solution

As $r=\frac{R}{6}$

(As balanced wheat stone bridge is formed) Now, Equivalent resistance between A and B can be written as
$\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{2 \mathrm{r}}+\frac{1}{2 \mathrm{r}}+\frac{1}{\mathrm{r}}=\frac{2}{\mathrm{r}}$
$\mathrm{R}_{\mathrm{AB}}=\frac{\mathrm{R}}{12}$

Asked in: JEE Main 2025 (07 Apr Shift 1)

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