JEE Physics 2023 Question with Solution

Given: A regular $n$-sided polygon is made of uniform wire of total resistance $R$.

Find: The equivalent resistance between two adjacent corners.

Let the resistance of each side be

$$r = \frac{R}{n}$$

because the total wire is divided equally among $n$ sides.

Between adjacent corners $A$ and $B$, there are two parallel paths:

1. Direct side $AB$ with resistance $r$
2. The remaining $(n-1)$ sides with total resistance $(n-1)r$

So, the equivalent resistance is

$$\frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{(n-1)r}$$

Therefore,

$$R_{eq} = \frac{r\,(n-1)r}{r + (n-1)r}$$ $$R_{eq} = \frac{(n-1)r^2}{nr}$$ $$R_{eq} = \frac{(n-1)r}{n}$$

Now substitute

$$r = \frac{R}{n}$$

Hence,

$$R_{eq} = \frac{(n-1)}{n}\cdot \frac{R}{n} = \frac{(n-1)R}{n^2}$$

Therefore, the equivalent resistance is $\frac{(n-1)R}{n^2}$. The extracted solution working gives this value, although the solution incorrectly labels it as Option 4 / D. With the given options, this value corresponds to Option A.

Given: Total resistance of the polygonal wire is $R$ and all $n$ sides are equal.

Find: Equivalent resistance across two adjacent vertices.

Because the wire is uniform, each side has the same resistance:

$$r = \frac{R}{n}$$

If the terminals are connected across adjacent corners, current can go from one corner to the other in two ways:

• through one side directly, resistance $r$
• through the rest of the polygon, resistance $(n-1)r$

These two paths join the same pair of nodes, so they are in parallel. Using the parallel combination formula,

$$R_{eq} = \frac{r\cdot (n-1)r}{r + (n-1)r}$$

The denominator becomes

$$r + (n-1)r = nr$$

So,

$$R_{eq} = \frac{(n-1)r^2}{nr} = \frac{(n-1)r}{n}$$

Substituting $r = \frac{R}{n}$,

$$R_{eq} = \frac{(n-1)}{n}\cdot \frac{R}{n} = \frac{(n-1)R}{n^2}$$

Therefore, the correct option by value is A.