Solve for me this physics

The formula for two resistors in parallel is $\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3}$.

$$\frac{1}{R_p} = \frac{1}{4 \, \Omega} + \frac{1}{6 \, \Omega}$$

To add these fractions, find a common denominator, which is 12.

$$\frac{1}{R_p} = \frac{3}{12 \, \Omega} + \frac{2}{12 \, \Omega}$$ $$\frac{1}{R_p} = \frac{3+2}{12 \, \Omega}$$ $$\frac{1}{R_p} = \frac{5}{12 \, \Omega}$$ $$R_p = \frac{12}{5} \, \Omega$$ $$R_p = 2.4 \, \Omega$$

The equivalent resistance $R_p$ is in series with $R_1$. The formula for resistors in series is $R_{total} = R_1 + R_p$.

$$R_{total} = 2 \, \Omega + 2.4 \, \Omega$$ $$R_{total} = 4.4 \, \Omega$$

The total resistance of the circuit is $\boxed{\text{4.4 } \Omega}$.

Ohm's Law states $V = I R$, so $I_{total} = \frac{V}{R_{total}}$.

$$I_{total} = \frac{12 \, \text{V}}{4.4 \, \Omega}$$ $$I_{total} = \frac{120}{44} \, \text{A}$$ $$I_{total} = \frac{30}{11} \, \text{A}$$ $$I_{total} \approx 2.73 \, \text{A}$$

The total current flowing from the battery is $\boxed{\text{2.73 A}}$.

The total current $I_{total}$ flows through $R_1$ and then through the parallel combination $R_p$. So, the voltage across $R_p$ is $V_p = I_{total} \times R_p$.

$$V_p = \frac{30}{11} \, \text{A} \times 2.4 \, \Omega$$ $$V_p = \frac{30}{11} \times \frac{12}{5} \, \text{V}$$ $$V_p = \frac{6 \times 12}{11} \, \text{V}$$ $$V_p = \frac{72}{11} \, \text{V}$$ $$V_p \approx 6.55 \, \text{V}$$

Using Ohm's Law for $R_2$: $I_2 = \frac{V_p}{R_2}$.

$$I_2 = \frac{\frac{72}{11} \, \text{V}}{4 \, \Omega}$$ $$I_2 = \frac{72}{11 \times 4} \, \text{A}$$ $$I_2 = \frac{18}{11} \, \text{A}$$ $$I_2 \approx 1.64 \, \text{A}$$

The current flowing through $R_2$ is $\boxed{\text{1.64 A}}$.

Since $R_2$ and $R_3$ are in parallel, the potential difference across $R_3$ ($V_3$) is the same as the potential difference across $R_2$ and the parallel combination ($V_p$).

$$V_3 = V_p$$ $$V_3 = \frac{72}{11} \, \text{V}$$ $$V_3 \approx 6.55 \, \text{V}$$

The potential difference across $R_3$ is $\boxed{\text{6.55 V}}$.