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