Question

$\text{A}$ and $\text{B}$ are two points on a uniform ring of resistance $R$ made of material of resistivity $\rho$ as shown in the figure. Find the equivalent resistance between $\text{A}$ and $\text{B}$ interms of $R$.

• A. $\frac{3R}{4}$
• B. $\frac{3R}{16}$
• C. $\frac{R}{4}$
• D. $\frac{R}{16}$

Answer 1

The correct option is B $\frac{3R}{16}$

Let, $l$ be the circumference of ring, $A$ be the area of cross-section and $R$ be the resistance of the uniform ring.

We know that, Resistance depends upon the length of the material. Since the area of cross-section is uniform, we can say that $R\propto l$

Equivalent diagram of the given figure is as given below,


From the diagram, it is evident that, $R_1$ and $R_2$ are parallel to each other and the equivalent resistance $R_{eq}$ is given by $$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$$ Since, $R\propto l$ we can write that, $R_1\propto l_1$ and $R_2\propto l_2$.

Thus, $R_{eq}=\frac{k{l}_1{l}_2}{l_1+l_2}\text{where},k=\frac{\rho }{A}$

From the figure , $l_1+l_2=l$ also $l_1=\frac{l}{4}$ and $l_2=\frac{3l}{4}$

Substituting the data in $R_{eq}$ we get,

$R_{eq}=\frac{k\times \frac{l}{4}\times \frac{3l}{4}}{l}$

$\Rightarrow R_{eq}=\frac{\rho }{A}\times \frac{3l}{16}$

$\Rightarrow R_{eq}=\frac{3R}{16}$

Hence, option (b) is the correct answer.

Alternate method:

we know , $R_{eq}=\frac{R\theta (2\pi -\theta )}{4{\pi }^2}$

Since, $\text{A}$ and $\text{B}$ are perpendicular, $\theta ={90}^0\text{or}\frac{\pi }{2}.$

Thus, $R_{eq}=\frac{R\times \frac{\pi }{2}\times (2\pi -\frac{\pi }{2})}{4{\pi }^2}$

$\Rightarrow R_{eq}=\frac{3R}{16}$.