Given R1 = 6 and R2 = 12 .

The formula for two resistors in parallel is:

$$R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}$$

Given $R_1 = 6 \text{ } \Omega$ and $R_2 = 12 \text{ } \Omega$.

$$R_{eq} = \frac{6 \text{ } \Omega \times 12 \text{ } \Omega}{6 \text{ } \Omega + 12 \text{ } \Omega}$$ $$R_{eq} = \frac{72 \text{ } \Omega^2}{18 \text{ } \Omega}$$ $$R_{eq} = 4 \text{ } \Omega$$

Ohm's Law states $V = I \times R_{eq}$, so $I = \frac{V}{R_{eq}}$.
Given the voltage $V = 12 \text{ V}$ and the equivalent resistance $R_{eq} = 4 \text{ } \Omega$.

$$I = \frac{12 \text{ V}}{4 \text{ } \Omega}$$ $$I = 3 \text{ A}$$

The total electric current is $\text{3 A}$.