Question

A wire of resistance $\text{R}$ is bent to form a square $\text{ABCD}$ as shown in the figure. The effective resistance between $\text{E}$ and $\text{C}$ is
($\text{E}$ is mid-point of arm $\text{CD}$)

• A. $\frac{3}{4}\text{R}$
• B. $\frac{1}{16}\text{R}$
• C. $\frac{7}{64}\text{R}$
• D. $\text{R}$

Answer 1

The correct option is C $\frac{7}{64}\text{R}$

Circuit can be reduced as

Here ${\text{R}}_{\text{DA}}={\text{R}}_{\text{AB}}={\text{R}}_{\text{BC}}=\text{R}/4$ and ${\text{R}}_{\text{DE}}={\text{R}}_{\text{EC}}=R/8$
Now ${\text{R}}_{\text{ED}},{\text{R}}_{\text{DA}},{\text{R}}_{\text{AB}},{\text{R}}_{\text{BC}}$ are in series.

$\therefore {\text{R}}_s=\frac{R}{8}+\frac{R}{4}+\frac{R}{4}+\frac{R}{4}=\frac{R+2R+2R+2R}{8}=\frac{7R}{8}$

$\therefore {\text{R}}_{eq}=\frac{\left(\frac{7R}{8}\right)\left(\frac{R}{8}\right)}{R}=\frac{7R}{64}$