Question

In the diagram, resistance between any two junctions is R. Equivalent resistance across terminals A and B is

• A. $\frac{11R}{7}$
• B. $\frac{18R}{11}$
• C. $\frac{7R}{11}$
• D. $\frac{11R}{18}$

Answer 1

The correct option is D $\frac{11R}{18}$


$R_{CD}=\frac{R\times 2R}{R+2R}=\frac{2R}{3}$

Similarly, $R_{DB}=\frac{R\times 2R}{R+2R}=\frac{2R}{3}$

Now the equivalent circuit becomes

$R_{CB}=\frac{\frac{4R}{3}\times R}{\frac{4}{3}R+R}=\frac{4R}{7}$

Here, R and $\frac{4R}{7}$ are in series, so

$\frac{4R}{7}+R=\frac{4R+7R}{7}=\frac{11R}{7}$ A

$R_{AB}=\frac{\frac{11}{7}R\times R}{\frac{11}{7}R+R}=\frac{11R}{18}$