Question

Six resistors of $3 \Omega$ each are connected along the sides of a hexagon and three resistors of $6 \Omega$ each are connected along $A C, A D$ and $A E$ as shown in the figure. The equivalent resistance between $A$ and $B$ is equal to

• A. $2\, \Omega$
• B. $6\, \Omega$
• C. $3\, \Omega$
• D. $9\, \Omega$

Answer 1

Answer: (A)

Resistances $R_{A F}$ and $R_{F E}$ are in series combination. Therefore their equivalent resistance $R^{\prime}=R_{A F}+R_{F E}=3+3=6\, \Omega$.
Now the resistance $R_{A E}$ and equivalent resistance $R^{\prime}$ are in parallel combination.
Therefore relation for their equivalent resistance
$\frac{1}{R^{\prime \prime}}=\frac{1}{R^{\prime}}+\frac{1}{R_{A E}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$
$ \Rightarrow R^{\prime \prime}=3 \Omega .$
We can calculate in the same manner for $R_{E D}, R_{A C}, R_{D C}$ etc.
and finally the circuit reduces as shown in the figure. Therefore, the equivalent resistance between $A$ and $B$
$=\frac{(3+3) \times 3}{(3+3)+3}=\frac{18}{9}=2\, \Omega$.