A battery of 6V is connected to the circuit as shown below. The...

The given circuit in the question is a case of a balanced Wheatstone bridge. Therefore, the circuit can be simplified as given below.

The equivalent resistance for the given circuit will be,

${R}_{eq} = \dfrac{\left(3 + 3\right) \times \left(6 + 6\right)}{\left(3 + 3\right) + \left(6 + 6\right)} + 2 = 6Ω$

Therefore, the current supplied by the battery is

$i = \dfrac{V}{{R}_{eq}} = \dfrac{6}{6} = 1A$