The total current supplied to the circuit as shown in the figure by the $$5$$ V battery is ______ A.

Resistance of top resistors ($$R_{1}, R_{2}$$) = 5 Ω each

Resistance of bottom resistors ($$R_{3}, R_{4}$$) = 2.5 Ω and 5 Ω

$$\frac{1}{R_{top}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$$

$$R_{top} = \frac{5}{2} = 2.5\text{ Ω}$$

$$\frac{1}{R_{bot}} = \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{2.5} + \frac{1}{5} = \frac{2}{5} + \frac{1}{5} = \frac{3}{5}$$

$$R_{bot} = \frac{5}{3} \approx 1.67\text{ Ω}$$

$$R_{total} = R_{top} + R_{bot} = 2.5 + 1.67 = 4.17\text{ Ω}$$

$$I = \frac{5\text{ V}}{4.17\text{ Ω}} \approx 1.2\text{ A}$$