In the network of resistors each of value Rshown the figure, the equivalent resistance between the junctions Aand Eis

In the network of resistors each of value shown the figure, the equivalent resistance between the junctions <img alt="\mathrm{A}" src="

In the network of resistors each of value $\mathrm{R}$ shown the figure, the equivalent resistance between the junctions $\mathrm{A}$ and $\mathrm{E}$ is

Option: 1
$5 \mathrm{R} / 7$

Option: 2
$5 \mathrm{R} / 12$

Option: 3
$7 \mathrm{R} / 5$

Option: 4
$7 \mathrm{R} / 12$

Answers (1)

Symmetry of the circuit shows that B and D are at the same potential and $\mathrm{F}$ and $\mathrm{H}$ are at another potential. So the circuit can be redrawn as shown in figure.

The equivalent resistance in the middle line between $\mathrm{B}$ and $\mathrm{H}$ is.
$\mathrm{\frac{R}{2}+R+\frac{R}{2}=2 R}$ . The total equivalent resistance between $\mathrm{B}$ and $\mathrm{H}$ is $\mathrm{R^{\prime}}$ such that.
$\mathrm{\frac{1}{\mathrm{R}^{\prime}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{2 \mathrm{R}} \Rightarrow \mathrm{R}^{\prime}=\frac{2}{5} \mathrm{R}}$

The equivalent resistance between $\mathrm{A}$ and $\mathrm{E}$ along path $\mathrm{A B E}$ is
$\mathrm{\frac{\mathrm{R}}{2}+\frac{2}{5} \mathrm{R}+\frac{\mathrm{R}}{2}=\frac{7}{5} \mathrm{R}}$

Total equivalent resistance between $\mathrm{A}$ and $\mathrm{E}$ is the resistance $\mathrm{R}$ and (7/5)R in parallel that is
$\mathrm{\mathrm{Req}=\frac{\mathrm{R} \times \frac{7}{5} \mathrm{R}}{\mathrm{R}+\frac{7}{5} \mathrm{R}}=\frac{7 \mathrm{R}}{12}}$