You're on a roll — To determine the current through the $3 \Omega$ resistor, we will use Kirchhoff's Voltage Law (KVL) with mesh analysis. Step 1: Define mesh currents. We define three clockwise mesh currents: $I_1$: Current in the top-left mesh. $I_2$: Current in the bottom-left mesh. $I_3$: Current in the right mesh. Step 2: Apply KVL to each mesh. For Mesh 1 (top-left loop): Starting from the bottom-left corner of Mesh 1 and moving clockwise: $$ -6 \text{ V} + (I_1)(2 \Omega) + (I_1 - I_2)(3 \Omega) = 0 $$ $$ 2I_1 + 3I_1 - 3I_2 = 6 $$ $$ 5I_1 - 3I_2 = 6 \quad (\text{Equation 1}) $$ For Mesh 2 (bottom-left loop): Starting from the bottom-left corner of Mesh 2 and moving clockwise: $$ -6 \text{ V} + (I_2 - I_1)(3 \Omega) + (I_2 - I_3)(4 \Omega) = 0 $$ $$ 3I_2 - 3I_1 + 4I_2 - 4I_3 = 6 $$ $$ -3I_1 + 7I_2 - 4I_3 = 6 \quad (\text{Equation 2}) $$ For Mesh 3 (right loop): Starting from the top-right corner of Mesh 3 and moving clockwise: $$ (I_3 - I_1)(4 \Omega) + (I_3 - I_2)(4 \Omega) = 0 $$ $$ 4I_3 - 4I_1 + 4I_3 - 4I_2 = 0 $$ $$ -4I_1 - 4I_2 + 8I_3 = 0 $$ Dividing by 4: $$ -I_1 - I_2 + 2I_3 = 0 \quad (\text{Equation 3}) $$ Step 3: Solve the system of linear equations. From Equation 3, we can express $2I_3$ in terms of $I_1$ and $I_2$: $$ 2I_3 = I_1 + I_2 $$ So, $4I_3 = 2(I_1 + I_2)$. Substitute this into Equation 2: $$ -3I_1 + 7I_2 - 2(I_1 + I_2) = 6 $$ $$ -3I_1 + 7I_2 - 2I_1 - 2I_2 = 6 $$ $$ -5I_1 + 5I_2 = 6 \quad (\text{Equation 4}) $$ Now we have a system of two equations with two variables ($I_1$ and $I_2$): 1. $5I_1 - 3I_2 = 6$ 4. $-5I_1 + 5I_2 = 6$ Add Equation 1 and Equation 4: $$ (5I_1 - 3I_2) + (-5I_1 + 5I_2) = 6 + 6 $$ $$ 2I_2 = 12 $$ $$ I_2 = \frac{12}{2} = 6 \text{ A} $$ Substitute the value of $I_2$ into Equation 1: $$ 5I_1 - 3(6 \text{ A}) = 6 $$ $$ 5I_1 - 18 = 6 $$ $$ 5I_1 = 6 + 18 $$ $$ 5I_1 = 24 $$ $$ I_1 = \frac{24}{5} = 4.8 \text{ A} $$ Step 4: Determine the current through the $3 \Omega$ resistor. The $3 \Omega$ resistor is shared by Mesh 1 and Mesh 2. The current through it is the difference between $I_1$ and $I_2$. If we consider the current flowing downwards (in the direction of $I_1$), it is $I_1 - I_2$. $$ I_{3\Omega} = I_1 - I_2 $$ $$ I_{3\Omega} = 4.8 \text{ A} - 6 \text{ A} $$ $$ I_{3\Omega} = -1.2 \text{ A} $$ The negative sign indicates that the actual current flows upwards through the $3 \Omega$ resistor, opposite to the assumed direction of $I_1$ through that branch. The "size" of the current refers to its magnitude. The size of the current through the $3 \Omega$ resistor is $\boxed{\text{1.2 A}}$. Send me the next one 📸