Solve in English

The 6 $\Omega$ and 4 $\Omega$ resistors are in parallel.

$$\frac{1}{R_p} = \frac{1}{6 \Omega} + \frac{1}{4 \Omega}$$ $$\frac{1}{R_p} = \frac{2}{12 \Omega} + \frac{3}{12 \Omega} = \frac{5}{12 \Omega}$$ $$R_p = \frac{12}{5} \Omega = 2.4 \Omega$$

The parallel combination $R_p$ is in series with the 2 $\Omega$ resistor.

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

The total voltage is $V_{total} = 12 \text{ V}$.

$$I_{total} = \frac{V_{total}}{R_{total}} = \frac{12 \text{ V}}{4.4 \Omega} \approx 2.727 \text{ A}$$

The reading of the ammeter is approximately $\boxed{\text{2.73 A}}$.

9. (b) Calculate the effective resistance of the circuit.

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

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

9. (c) Calculate the potential difference across the 6 $\Omega$ resistor.

The potential difference across the 6 $\Omega$ resistor is the same as the potential difference across the parallel combination, $V_p$.

$$V_p = I_{total} \times R_p$$

Using the more precise value for $I_{total} = \frac{12}{4.4} = \frac{30}{11} \text{ A}$:

$$V_p = \frac{30}{11} \text{ A} \times 2.4 \Omega = \frac{30}{11} \times \frac{12}{5} \text{ V} = \frac{6 \times 12}{11} \text{ V} = \frac{72}{11} \text{ V} \approx 6.545 \text{ V}$$

The potential difference across the 6 $\Omega$ resistor is approximately $\boxed{\text{6.55 V}}$.

9. (d) Calculate the current flowing through the 4 $\Omega$ resistor.

The potential difference across the 4 $\Omega$ resistor is $V_p = \frac{72}{11} \text{ V}$.

$$I_{4\Omega} = \frac{V_p}{R_{4\Omega}} = \frac{\frac{72}{11} \text{ V}}{4 \Omega} = \frac{72}{11 \times 4} \text{ A} = \frac{18}{11} \text{ A} \approx 1.636 \text{ A}$$

The current flowing through the 4 $\Omega$ resistor is approximately $\boxed{\text{1.64 A}}$.