Using Norton's theorem, calculate the current flowing through the 11.2 Ω resistor in FIGURE 2. Also calculate the voltage drop across this resistor.

To find $I_N$, we remove the load resistor ($R_{load} = 11.2 \, \Omega$) and place a short circuit across its terminals. Let the terminals be B and ground. This means the voltage at terminal B is $0 \, \text{V}$.
The circuit now consists of the $20 \, \text{V}$ source, $R_1 = 20 \, \Omega$, $R_2 = 12 \, \Omega$, and $R_3 = 8 \, \Omega$.
Let Node A be the junction between $R_1$, $R_2$, and $R_3$.
When terminal B is shorted to ground, $R_2$ is connected between Node A and ground, and $R_3$ is also connected between Node A and ground. Thus, $R_2$ and $R_3$ are in parallel.
The equivalent resistance of $R_2 \parallel R_3$ is:

$$R_{23} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{12 \, \Omega \times 8 \, \Omega}{12 \, \Omega + 8 \, \Omega} = \frac{96 \, \Omega^2}{20 \, \Omega} = 4.8 \, \Omega$$

The total current ($I_{total}$) flowing from the source is through $R_1$ and then through the parallel combination of $R_2$ and $R_3$:

$$I_{total} = \frac{V}{R_1 + R_{23}} = \frac{20 \, \text{V}}{20 \, \Omega + 4.8 \, \Omega} = \frac{20 \, \text{V}}{24.8 \, \Omega} = \frac{200}{248} \, \text{A} = \frac{25}{31} \, \text{A}$$

This current $I_{total}$ reaches Node A and splits between $R_2$ and $R_3$. The Norton current $I_N$ is the current flowing through the short circuit, which is the current that would flow through $R_2$ in this configuration (since the short is across the terminals where $R_{load}$ was, and $R_2$ is connected to that node). Using the current divider rule:

$$I_N = I_{total} \times \frac{R_3}{R_2 + R_3} = \frac{25}{31} \, \text{A} \times \frac{8 \, \Omega}{12 \, \Omega + 8 \, \Omega} = \frac{25}{31} \, \text{A} \times \frac{8}{20} = \frac{25}{31} \, \text{A} \times \frac{2}{5} = \frac{10}{31} \, \text{A}$$ $$I_N \approx 0.3226 \, \text{A}$$

To find $R_N$, we turn off the independent voltage source (replace it with a short circuit) and look into the terminals where the load resistor was connected (terminals B and ground).
With the voltage source shorted, $R_1$ is connected from Node A to ground. $R_3$ is also connected from Node A to ground. So, $R_1$ and $R_3$ are in parallel.

$$R_{13} = \frac{R_1 \times R_3}{R_1 + R_3} = \frac{20 \, \Omega \times 8 \, \Omega}{20 \, \Omega + 8 \, \Omega} = \frac{160 \, \Omega^2}{28 \, \Omega} = \frac{40}{7} \, \Omega$$

Now, looking from terminals B to ground, $R_2$ is in series with the parallel combination of $R_1$ and $R_3$.

$$R_N = R_2 + R_{13} = 12 \, \Omega + \frac{40}{7} \, \Omega = \frac{12 \times 7 + 40}{7} \, \Omega = \frac{84 + 40}{7} \, \Omega = \frac{124}{7} \, \Omega$$ $$R_N \approx 17.7143 \, \Omega$$

Now we connect the load resistor $R_{load} = 11.2 \, \Omega$ in parallel with the Norton equivalent circuit ($I_N$ in parallel with $R_N$). We use the current divider rule:

$$I_{load} = I_N \times \frac{R_N}{R_N + R_{load}}$$ $$I_{load} = \frac{10}{31} \, \text{A} \times \frac{\frac{124}{7} \, \Omega}{\frac{124}{7} \, \Omega + 11.2 \, \Omega}$$

Convert $11.2 \, \Omega$ to a fraction: $11.2 = \frac{112}{10} = \frac{56}{5}$.

$$I_{load} = \frac{10}{31} \, \text{A} \times \frac{\frac{124}{7}}{\frac{124}{7} + \frac{56}{5}} = \frac{10}{31} \, \text{A} \times \frac{\frac{124}{7}}{\frac{124 \times 5 + 56 \times 7}{35}}$$ $$I_{load} = \frac{10}{31} \, \text{A} \times \frac{\frac{124}{7}}{\frac{620 + 392}{35}} = \frac{10}{31} \, \text{A} \times \frac{\frac{124}{7}}{\frac{1012}{35}}$$ $$I_{load} = \frac{10}{31} \, \text{A} \times \frac{124}{7} \times \frac{35}{1012} = \frac{10}{31} \, \text{A} \times \frac{124 \times 5}{1012}$$ $$I_{load} = \frac{6200}{31372} \, \text{A} = \frac{1550}{7843} \, \text{A}$$ $$I_{load} \approx \boxed{\text{0.1976 A}}$$

Using Ohm's Law:

$$V_{load} = I_{load} \times R_{load}$$ $$V_{load} = \frac{1550}{7843} \, \text{A} \times 11.2 \, \Omega = \frac{1550}{7843} \, \text{A} \times \frac{56}{5} \, \Omega$$ $$V_{load} = \frac{310 \times 56}{7843} \, \text{V} = \frac{17360}{7843} \, \text{V}$$ $$V_{load} \approx \boxed{\text{2.213 V}}$$