Calculate the potential difference of the 5 ohm and the 10 ohm resistors

For two resistors in parallel, the equivalent resistance $R_{p}$ is given by the formula:

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

Substituting $R_1 = 5 \Omega$ and $R_2 = 10 \Omega$:

$$R_{p} = \frac{5 \Omega \times 10 \Omega}{5 \Omega + 10 \Omega}$$ $$R_{p} = \frac{50 \Omega^2}{15 \Omega}$$ $$R_{p} = \frac{10}{3} \Omega \approx 3.33 \Omega$$

The parallel combination of the $5 \Omega$ and $10 \Omega$ resistors is in series with the $3 \Omega$ and $2 \Omega$ resistors. The ammeter indicates that the total current flowing from the source is $6 \text{ A}$. In a series circuit, the current is the same through all components.
Therefore, the current flowing through the parallel combination is $6 \text{ A}$.

$$I_{p} = 6 \text{ A}$$

Using Ohm's Law, $V = I \times R$, where $V$ is the potential difference, $I$ is the current, and $R$ is the equivalent resistance of the parallel combination.

$$V_{p} = I_{p} \times R_{p}$$ $$V_{p} = 6 \text{ A} \times \frac{10}{3} \Omega$$ $$V_{p} = 2 \times 10 \text{ V}$$ $$V_{p} = 20 \text{ V}$$

Since the $5 \Omega$ and $10 \Omega$ resistors are in parallel, the potential difference across each of them is the same as the potential difference across their equivalent resistance.

The potential difference across the $5 \Omega$ resistor is $\boxed{\text{20 V}}$.
The potential difference across the $10 \Omega$ resistor is $\boxed{\text{20 V}}$.