Question

A finite square grid with each link having resistance $r$ is fitted in a resistanceless conducting circular wire. The equivalent resistance between $\text{A}$ and $\text{B}$ will be:
$($Given, $r=80/7\Omega )$

• A. $6\Omega$
• B. $8\Omega$
• C. $12\Omega$
• D. $15\Omega$

Answer 1

The correct option is A $6\Omega$

The circular wire is a resistance less conducting wire, hence all points lying on it will have the same potential.

Also, the square grid have symmetric arrangements of component segments (links) with respect to point $\text{A}$ and $\text{B}$, thus all points at symmetric/equivalent position w.r.t $\text{A}$ and $\text{B}$ will have same electric potential.


We can observe from above diagram that points $(9,10),(11,12),(13,14)$ & $(15,16)$ are at equivalent or identical position w.r.t $\text{A}$.

Hence, $(9,10,11,12,13,14,15,16)$ will have same potential.

Similarly, points $(5,6,7,8)$ are at identical position w.r.t $\text{B}$. Hence, they will be at same potential.

Also points $(1,2,3,4)$ will be at same potential, as they are at symmetric position w.r.t $\text{B}$.

Again points $(17,18,19,20)$ being at symmetrical location w.r.t $\text{B}$ will have same potential.

Now redrawing the circuit with each segment having resistance $r$.


Here $a\to (9,10,11,12,13,14,15,16)$
$b\to (5,6,7,8)$
$c\to (17,18,19,20)$
$d\to (1,2,3,4)$

We can just count the number of respective resistance connected between these points and redrawing the circuit with effective resistance.


The equivalent resistance $R_{eq}$ between $\text{A}$ and $\text{B}$ will be

$R_{eq}=\frac{r}{8}+\frac{\left(3r/8\right)\left(r/4\right)}{\left(3r/8\right)+\left(r/4\right)}+\frac{r}{4}$

$\Rightarrow R_{eq}=\frac{r}{8}+\frac{3r}{20}+\frac{r}{4}$

$\Rightarrow R_{eq}=\frac{21r}{40}=\frac{21}{40}\times \frac{80}{7}(\because r=80/7\Omega )$

$\therefore R_{eq}=6\Omega$

Hence, option $\left(a\right)$ is correct.

Why this question ? It intends to challenge the identification of points located at symmetric or equivalent position in a complex geometric pattern and offers a lot of learning !!! Tip: Point $\text{B}$ in the grid is a central point, and due to conducting circular wire circumscribing grid, the corner points become equivalent to point $\text{A}$. Thus whole grid can be simplified by observing equipotential points.