Question

Five equal resistances each of value $R$ are connected to form a network as shown in figure. The equivalent resistance of the network between the points $A$ and $B$ is

• A. $\frac{1}{2}R$
• B. $2R$
• C. $\frac{5}{8}R$
• D. $\frac{8}{5}R$

Answer 1

Answer: (C)

The given circuit is redrawn as shown in the figure. The resistances connected between $BC$ and $CD$ are in series. Therefore its equivalent resistance is $2R.$ The resistance $2R$ and the resistance connected between $BD$ are in parallel. Let its equivalent resistance be $R_1$. $\therefore \frac{1}{R_1}=\frac{1}{2R}+\frac{1}{R}$
or $R_1=\frac{2}{3}R$
The resistance $R_1$ and resistance connected between $AD$ are in series. Let its equivalent resistance be $R_2$.
$\therefore R_2=R+\frac{2}{3}R=\frac{5}{3}R$
The resistance $R_2$ and resistance connected between $AB$ are in parallel.
Hence the equivalent resistance between $AB$ is $R_{eq}$
$\therefore \frac{1}{R_{eq}}=\frac{1}{\frac{5}{3}R}+\frac{1}{R}=\frac{3}{5R}+\frac{1}{R}$
or $R_{eq}=\frac{5R}{8}$