Question

Equivalent resistance between the adjacent corners of a regular \(n\)-sided polygon of uniform wire of resistance \(R\) would be :

• A. \(\frac{(n-1) R}{(2 n-1)}\)
• B. \(\frac{(n-1) R}{n}\)
• C. \(\frac{n^2 R}{n-1}\)
• D. \(\frac{(n-1) R}{n^2}\)

Hint:

When calculating equivalent resistance in networks, break down the network into series and parallel combinations. Remember the formulas for equivalent resistance in series and parallel.

Answer 1

The Correct Option is D

Step 1: Analyze the Resistance of Each Side

Let \(r\) be the resistance of each side of the \(n\)-sided polygon. Since the total resistance of the wire is \(R\), and there are \(n\) sides, the resistance of each side is:

\[ r = \frac{R}{n} \]

Step 2: Consider Adjacent Corners A and B

When we consider the equivalent resistance between adjacent corners A and B, the polygon can be viewed as two resistors in parallel:

• One resistor with resistance \(r\) (between A and B).
• The other resistor with resistance \((n-1)r\) (the remaining part of the polygon).

Step 3: Calculate the Equivalent Resistance

The equivalent resistance (\(R_{eq(AB)}\)) between A and B is given by the formula for parallel resistors:

\[ \frac{1}{R_{eq(AB)}} = \frac{1}{r} + \frac{1}{(n-1)r} \] \[ R_{eq(AB)} = \frac{r \times (n-1)r}{r + (n-1)r} = \frac{(n-1)r^2}{nr} = \frac{(n-1)r}{n} \]

Step 4: Substitute the Value of \(r\)

Substitute \(r = \frac{R}{n}\) back into the equation:

\[ R_{eq(AB)} = \frac{(n-1)\left(\frac{R}{n}\right)}{n} = \frac{(n-1)R}{n^2} \]

Conclusion:

The equivalent resistance between the adjacent corners of the polygon is:

\[ \frac{(n-1)R}{n^2} \quad \text{(Option 4)}. \]

Answer 2

Suppose resistance of each arm is r, then r=R/n 
Req(AB)​=R1​+R2​R1​R2​​ 
r+(n−1)rr(n−1)r​ 
=nrr(n−1)r​ 
=nn−1​r 
=n2(n−1)R​