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(A) In the given circuit,the battery of $2 \ V$ is connected to a triangular network of three resistors,each of $3 \ \Omega$.Looking at the circuit,th

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(A) In the given circuit,the battery of $2 \ V$ is connected to a triangular network of three resistors,each of $3 \ \Omega$.Looking at the circuit,the current from the battery splits at the top node. One path goes through one $3 \ \Omega$ resistor,and the other path goes through two $3 \ \Omega$ resistors connected in series.Thus,the resistance of the first branch is $R_1 = 3 \ \Omega$.The resistance of the second branch is $R_2 = 3 \ \Omega + 3 \ \Omega = 6 \ \Omega$.These two branches are in parallel with each other across the $2 \ V$ battery.The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \ \Omega^{-1}$.Therefore,$R_{eq} = 2 \ \Omega$.The total current $I$ flowing from the battery is:$I = \frac{V}{R_{eq}} = \frac{2 \ V}{2 \ \Omega} = 1 \ A$.

The current flowing through the circuit shown in the figure is ........... $A$.

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