Equivalent resistance of the following networ | vedclass

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(C) Looking at the circuit from left to right:$1$. The leftmost $6 \ \Omega$ resistor is in parallel with a short circuit (the wire connecting the two

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(C) Looking at the circuit from left to right:$1$. The leftmost $6 \ \Omega$ resistor is in parallel with a short circuit (the wire connecting the two vertical lines),effectively shorting it out. However,looking at the diagram,the $6 \ \Omega$ resistor is connected across the same two nodes as the shorting wire,so it is bypassed.$2$. The circuit simplifies step-by-step from left to right.$3$. After simplifying the left part,we are left with three $3 \ \Omega$ resistors in parallel connected across terminals $A$ and $B$.$4$. The equivalent resistance $R_{eq}$ of three resistors of $3 \ \Omega$ each in parallel is given by $\frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \ \Omega$.$5$. Thus,$R_{eq} = 1 \ \Omega$.

Equivalent resistance of the following network is . . . . . . $\Omega$.

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