Resistors!

$\text{First, R8 and R9 are in parallel combination, on solving that you'll get 1 equivalent resistance for both R8 and R9. Let it be }r_1 \\ \frac{1}{R_p} = \frac{1}{R8} + \frac{1}{R9} \\ \implies \frac{1}{r_1} = \frac{1}{1} + \frac{1}{2} = \frac{3}{2} \\ \implies r_1 = \frac{2}{3} \Omega\\ \text{Now R4, }r_1\text{, R5, R7, R6 are in series combination, now find equivalent resistance for those, and let it be } r_2\\ R_s = R4 + r_1 + R5 + R6 + R7 = 1 + \frac{3}{2} + 2 + 1 + 1\\ \implies r_2 = \frac{17}{3} \Omega\\ \text{Now, }r_2\text{ and R2 are in parallel connection, Let their equivalent resistance be }r_3 \\ \frac{1}{R_p} = \frac{1}{r_2} + \frac{1}{R2} = \frac{3}{17} + \frac{1}{2}\\ \implies r_3 = \frac{34}{23} \Omega\\ \text{Now, R1, R3 and }r_3\text{ are in series combination. So the total resistance would be.} \\ R_s = R1+ R3 + r_3 = 2 + 2 + \frac{34}{23} = \frac{126}{23}\Omega\\ \text{Therefore, Total resistance is } \frac{126}{23} \Omega\\ \text{Now, using Ohm's law} \\ V = IR \\ \implies I = \frac{V}{R} \\ \implies I = \frac{10}{\frac{126}{23}} = 1.8254\textrm{ A} \\$

If you know any other method of finding this, feel free to post it