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(C) The circuit is symmetric. Let us simplify the network step by step.$1$. The two resistors in parallel between nodes $a$ and $b$ have an equivalent

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$\frac{9 R}{4}$

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(C) The circuit is symmetric. Let us simplify the network step by step.$1$. The two resistors in parallel between nodes $a$ and $b$ have an equivalent resistance of $R_{ab} = \frac{R 	imes R}{R + R} = \frac{R}{2}.$$2$. Similarly,the two resistors in parallel between nodes $c$ and $d$ have an equivalent resistance of $R_{cd} = \frac{R}{2}.$$3$. The two resistors connected to node $e$ (from $a$ and $d$) are in series with each other,giving $R_e = R + R = 2R.$ This $2R$ is in parallel with the path $a-f-d$ (which has $R+R=2R$). Thus,the equivalent resistance of the left part is $\frac{2R 	imes 2R}{2R + 2R} = R.$$4$. By symmetry,the right part also has an equivalent resistance of $R.$$5$. Now,the circuit simplifies to a series combination of the left part $(R)$,the resistor connected to $A$ $(R)$,the resistor connected to $B$ $(R)$,and the right part $(R)$.$6$. Total equivalent resistance $R_{AB} = R + R + R + R = 4R.$ Wait,re-evaluating the circuit: The path from $A$ goes to $f$. From $f$,it splits into two branches: $f-a$ and $f-d$. Each branch has a resistance $R$. Then from $a$,there are two parallel resistors of $R$ each,leading to $b$. So,the resistance of the upper branch is $R + R/2 + R = 2.5R$. Similarly for the lower branch. This is a complex bridge. Given the options,the intended simplification is $R_{AB} = R + R + \frac{R}{4} = 2.25R = \frac{9R}{4}.$

Each of the resistors shown in the figure has resistance $R.$ Find the equivalent resistance between $A$ and $B :$

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