Question

Twelve resistors each of resistance $r$ are connected together so that each lies along the edge of the cube as shown in the figure. The equivalent resistance between points $1$ and $4$ is

• A. $\frac{5r}{12}$
• B. $\frac{7r}{12}$
• C. $\frac{11r}{12}$
• D. $\frac{13r}{12}$

Answer 1

The correct option is B $\frac{7r}{12}$



From the diagram, we can see that points $(2,5)$ are symmetrically located with respect to points $1$ and $4$. So they are at same potentials.

Similarly, points $(3,8)$ are also symmetrically located with respect to points $1$ and $4$. So they are again at same potentials.

On this basis redrawing the circuit,


Simplifying above circuit by using concept of series and parallel combination,


On successive reduction in circuit, we get,


Since, both are in parallel combination, so equivalent resistance will be

$R_{eq}=\frac{\left(7r/5\right)\times r}{\left(7r/5\right)+r}=\frac{7}{12}r$

The equivalent resistance between points $1$ and $4$ is $7r/12.$

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

$$\begin{array}{c}\text{Why this question?}\\ \text{Key Idea-This question gives you a}\\ \text{hint that how symmetry can simplify}\\ \text{complex resistors network.}\end{array}$$