Apply Kirchhoffs law to FIGURE 1 below, and determine the value of the current in each branch of the circuit. Using Norton's theorem, calculate the current flowing through the 11.2 Q resistor in FIGURE 2. Also calculate the voltage drop across this resistor.

We sum the voltage drops around the loop in the clockwise direction.
The current $I_1$ flows through the $1 \, \Omega$ resistor.
The current in the common branch BE is $(I_1 - I_2)$ flowing from B to E.
The voltage sources are $E_1 = 15 \, \text{V}$ (positive at A, negative at F) and $E_2 = 12 \, \text{V}$ (positive at B, negative at E).

Starting from point A and moving clockwise:

• Voltage drop across $1 \, \Omega$ resistor (AF): $I_1 \times 1 \, \Omega$.
• Voltage source $E_1$: Moving from A to F, we go from the positive terminal to the negative terminal of $E_1$. This is a voltage drop, so $-15 \, \text{V}$.
• Voltage drop across $0.2 \, \Omega$ resistor (BE): We are moving from E to B. The current $(I_1 - I_2)$ flows from B to E. So, moving from E to B, we are going against $(I_1 - I_2)$. This results in a voltage rise, so $-(I_1 - I_2) \times 0.2 \, \Omega$.
• Voltage source $E_2$: Moving from E to B, we go from the negative terminal to the positive terminal of $E_2$. This is a voltage rise, so $+12 \, \text{V}$.

Summing these voltages to zero:

$$I_1(1) - 15 - (I_1 - I_2)(0.2) + 12 = 0$$ $$I_1 - 15 - 0.2I_1 + 0.2I_2 + 12 = 0$$ $$0.8I_1 + 0.2I_2 - 3 = 0$$ $$0.8I_1 + 0.2I_2 = 3$$

Multiplying by 10 to clear decimals:

$$8I_1 + 2I_2 = 30$$

Dividing by 2:

$$4I_1 + I_2 = 15 \quad (*\text{Equation 1})$$

We sum the voltage drops around the loop in the clockwise direction.
The current $I_2$ flows through the $0.2 \, \Omega$ resistor in branch CD.
The current in the common branch BE is $(I_1 - I_2)$ flowing from B to E.
The voltage sources are $E_2 = 12 \, \text{V}$ (positive at B, negative at E) and $E_3 = 9 \, \text{V}$ (positive at C, negative at D).

Starting from point B and moving clockwise:

• Voltage drop across $0.2 \, \Omega$ resistor (BE): We are moving from B to E. The current $(I_1 - I_2)$ flows from B to E. This is a voltage drop, so $(I_1 - I_2) \times 0.2 \, \Omega$.
• Voltage source $E_2$: Moving from B to E, we go from the positive terminal to the negative terminal of $E_2$. This is a voltage drop, so $-12 \, \text{V}$.
• Voltage drop across $0.2 \, \Omega$ resistor (CD): $I_2 \times 0.2 \, \Omega$.
• Voltage source $E_3$: Moving from C to D, we go from the positive terminal to the negative terminal of $E_3$. This is a voltage drop, so $-9 \, \text{V}$.

Summing these voltages to zero:

$$(I_1 - I_2)(0.2) - 12 + I_2(0.2) - 9 = 0$$ $$0.2I_1 - 0.2I_2 - 12 + 0.2I_2 - 9 = 0$$ $$0.2I_1 - 21 = 0$$ $$0.2I_1 = 21$$ $$I_1 = \frac{21}{0.2} = 105 \, \text{A} \quad (*\text{Equation 2})$$

We already found $I_1$ from Equation 2. Now substitute $I_1 = 105 \, \text{A}$ into Equation 1:

$$4(105) + I_2 = 15$$ $$420 + I_2 = 15$$ $$I_2 = 15 - 420$$ $$I_2 = -405 \, \text{A}$$

The currents in the branches are:

• Current in branch AF (and FE, ED): This is the mesh current $I_1$. $$I_{AF} = I_1 = \boxed{\text{105 A}}$$
• Current in branch BE: This is the difference between the mesh currents, $(I_1 - I_2)$, flowing from B to E. $$I_{BE} = I_1 - I_2 = 105 \, \text{A} - (-405 \, \text{A}) = 105 \, \text{A} + 405 \, \text{A} = \boxed{\text{510 A}}$$
• Current in branch CD (and BC): This is the mesh current $I_2$. $$I_{CD} = I_2 = \boxed{\text{-405 A}}$$ The negative sign indicates that the actual current flows in the opposite direction to the assumed clockwise direction for $I_2$, i.e., from D to C.