Calculate the equivalent resistance (R_p) of the two parallel resistors.

The $4 \text{ } \Omega$ and $3 \text{ } \Omega$ resistors are in parallel.

$$R_p = \frac{R_1 \times R_2}{R_1 + R_2}$$ $$R_p = \frac{4 \text{ } \Omega \times 3 \text{ } \Omega}{4 \text{ } \Omega + 3 \text{ } \Omega}$$ $$R_p = \frac{12 \text{ } \Omega^2}{7 \text{ } \Omega}$$ $$R_p \approx 1.714 \text{ } \Omega$$

The parallel combination ($R_p$) is in series with the $2 \text{ } \Omega$ resistor.

$$R_{total} = R_p + R_{series}$$ $$R_{total} = 1.714 \text{ } \Omega + 2 \text{ } \Omega$$ $$R_{total} = 3.714 \text{ } \Omega$$

The voltage source is $60 \text{ V}$.

$$I = \frac{V}{R_{total}}$$ $$I = \frac{60 \text{ V}}{3.714 \text{ } \Omega}$$ $$I \approx 16.15 \text{ A}$$

The total electric current is $\text{16.15 A}$.