Calculate the equivalent resistance of the bottom branch.

The 5Ω and 10Ω resistors are in series in the bottom branch.

$$R_{bottom} = 5\,\Omega + 10\,\Omega = 15\,\Omega$$

The voltage across the bottom branch is 12V. Using Ohm's Law ($V = IR$):

$$I_1 = \frac{V}{R_{bottom}} = \frac{12\,\text{V}}{15\,\Omega} = 0.8\,\text{A}$$

The voltage across the top branch is 12V, and the resistance is 10Ω. Using Ohm's Law:

$$I_2 = \frac{V}{R_{top}} = \frac{12\,\text{V}}{10\,\Omega} = 1.2\,\text{A}$$

$$I_{total} = I_1 + I_2 = 0.8\,\text{A} + 1.2\,\text{A} = 2.0\,\text{A}$$

The sum of the current $I_1$ and $I_2$ is $\boxed{\text{2.0 A}}$.