Find the current
flowing out of the
battery in the
circuit.

The top ones being parallel gives 1/(1/30 + 1/40) = 120/7 ohms. The rightmost path in series gives 30 ohms, and then the parallel pair on the right is 1/(1/50 + 1/30) = 75/4 ohms. I = V/R => 9 / (75/4+120/7) [tex]\approx[/tex] .25A
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