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(D) In the given circuit,the battery is connected to two branches in parallel. One branch contains a single $3\,\Omega$ resistor,and the other branch

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(D) In the given circuit,the battery is connected to two branches in parallel. One branch contains a single $3\,\Omega$ resistor,and the other branch contains two $3\,\Omega$ resistors in series.$1$. The resistance of the first branch is $R_1 = 3\,\Omega$.$2$. The resistance of the second branch is $R_2 = 3\,\Omega + 3\,\Omega = 6\,\Omega$.$3$. These two branches are in parallel,so 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}$$\Rightarrow R_{eq} = 2\,\Omega$.$4$. Using Ohm's law,the total current $I$ supplied by the $2\,V$ battery is:$I = \frac{V}{R_{eq}} = \frac{2\,V}{2\,\Omega} = 1\,A$.

The current in the following circuit is ........... $A$.

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