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

Question

Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance R would be : (A) ((n-1) R/(2 n-1)) (B) ((n-1) R/n) (C) (n2 R/n-1) (D) ((n-1) R/n2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/976172c5a24cb806ba7e140f4cc31709-.png" /> Suppose resistance of each arm is r, then r=R / n Re q(A B)=(R1 R2/R1+R2) (r(n-1) r/r+(n-1) r) =(r(n-1) r/n r) =(n-1/n) r =(( n -1) R/n2)

Q. Equivalent resistance between the adjacent corners of a regular $n$ -sided polygon of uniform wire of resistance $R$ would be :

$\frac{( n - 1 ) R}{( 2 n - 1 )}$

$\frac{( n - 1 ) R}{n}$

$\frac{n ^{2} R}{n - 1}$

$\frac{( n - 1 ) R}{n ^{2}}$

Suppose resistance of each arm is $r$ , then $r = R / n$
$R_{e q (A B)} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$
$\frac{r ( n - 1 ) r}{r + ( n - 1 ) r}$
$= \frac{r ( n - 1 ) r}{n r}$
$= \frac{n - 1}{n} r$
$= \frac{( n - 1 ) R}{n ^{2}}$

Q. Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance R would be :

(2n−1)(n−1)R​

n(n−1)R​

n−1n2R​

n2(n−1)R​