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. $2 \,R$
• 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 $B C$ and $C D$ are in series. Therefore its equivalent resistance is $2 R .$ The resistance $2 R$ and the resistance connected between $B D$ are in parallel. Let its equivalent resistance be $R_{1}$. $\therefore \frac{1}{R_{1}}=\frac{1}{2 R}+\frac{1}{R}$
or $R_{1}=\frac{2}{3} R$
The resistance $R_{1}$ and resistance connected between $A D$ 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 $A B$ are in parallel.
Hence the equivalent resistance between $A B$ is $R_{eq}$
$\therefore \frac{1}{R_{e q}}=\frac{1}{\frac{5}{3} R}+\frac{1}{R}=\frac{3}{5 R}+\frac{1}{R}$
or $R_{e q}=\frac{5 R}{8}$