Question

A set of identical resistors $R$ is wired in combinations shown above.
What is the equivalent resistance of this combination?

• A. $\frac{32}{13}R$
• B. $\frac{13}{32}R$
• C. $\frac{10}{7}R$
• D. $\frac{7}{10}R$
• E. $2R$

Answer 1

The correct option is A $\frac{32}{13}R$

In figure (a) : $R_1=\frac{R\times R}{R+R}=\frac{R}{2}$

$\therefore$ $R^{\prime }=R_1+R=\frac{3R}{2}$

Also $R_2=R+R=2R$

Similarly $R_3=R+R=2R$

In figure (b) : $R_4=\frac{R^{\prime }\times R}{R^{\prime }+R}=\frac{\frac{3R}{2}\times R}{\frac{3R}{2}+R}=\frac{3R}{5}$

In figure (c) : $R_5=\frac{R_4\times R_2}{R_4+R_2}=\frac{\frac{3R}{5}\times 2R}{\frac{3R}{5}+2R}=\frac{6R}{13}$

Thus equivalent resistance of the circuit $R_{eq}=R_3+R_5=\frac{6R}{13}+2R=\frac{32R}{13}$