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
AIPMT 1994, Medium
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Resistances $R _{ AF }$ and $R _{ FE }$ are in series combination. Therefore their equivalent resistance $R = R _{ AF }+ R _{ FE }=3+3=6 \Omega$. Now the resistance $R _{ AE }$ and equivalent resistance $R$ ' are in parallel combination. Therefore relation for their equivalent resistance.
$\frac{1}{R^{*}}=\frac{1}{R^{\prime}}+\frac{1}{R_{A E}}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3} \Rightarrow R^{\prime \prime}=3 \Omega$.
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 $B$
$=\frac{(3+3) \times 3}{(3+3)+3}=\frac{18}{9}=2 \Omega$
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