Circuit Resistance

Initial Resistors in Series Simplification: $R_a = R_2 + R_3 = 9+9 = 18Ω$

Resistors in Parallel Simplification: $R_b = \Big( \frac{1}{R_a} + \frac{1}{R_4}\Big)^{-1} = \Big( \frac{1}{18} + \frac{1}{9} \Big)^{-1} = \Big( \frac{1}{6} \Big)^{-1} = 6Ω$

Final Resistor in Series Simplification: $R_{circuit} = R_1 + R_b + R_5 = 9+6+9 = 24Ω$

Current Calculation: $I = \frac{V}{R_{circuit}} = \frac{30V}{24Ω} = \boxed{1.25 A}$

Let each resistor be $R=9 \Omega$ . The equivalent resistance connected to battery $V$ is given by:

$\begin{aligned} R_{eq} & = R_1 + (R_2+R_3) || R_4 + R_5 \\ & = R + (R+R) || R + R \\ & = 2R + \frac {2R^2}{3R} \\ & = \frac 83 R = 24 \ \Omega \\ \implies I & = \frac V{R_{eq}} = \frac {30}{24} = \frac 54 = \boxed{1.25} \text{ A} \end{aligned}$