Let's solve for \( I_x \) in two steps using mesh analysis. **Given:** - \( I_a = 2 \) A - \( I_b = 5 \) A - \( V_a = 7 \) V - \( R_1 = 1.6 \) Ω - \( R_2 = 1.7 \) Ω - \( R_3 = 0.7 \) Ω **Step 1: Express \( I_x \) in terms of mesh currents** From the diagram, the current \( I_x \) through \( R_2 \) is the difference between mesh currents \( I_a \) and \( I_b \): \[ I_x = I_a - I_b \] **Explanation:** This is because \( I_a \) flows downward through \( R_2 \), and \( I_b \) flows upward (opposite direction) through \( R_2 \). --- **Step 2: Substitute the values** Substitute the given values for \( I_a \) and \( I_b \): \[ I_x = I_a - I_b = 2 - 5 = -3 \] **Explanation:** A negative value means the net current through \( R_2 \) is in the direction of \( I_b \) (upward), not \( I_a \) (downward). --- ### **Final Answer** \[ \boxed{-3.00} \] (Rounded to two decimal places as required)

Let's solve the circuit using mesh analysis with two loops (meshes). We'll assume the left mesh as our reference and follow these steps: --- ### **Given Data:** - \( I_a = 2\,A \) (current in left mesh) - \( I_b = 5\,A \) (current in right mesh) - \( V_a = 7\,V \) - \( R_1 = 1.6\,\Omega \) - \( R_2 = 1.7\,\Omega \) - \( R_3 = 0.7\,\Omega \) --- ### **Step 1: Define mesh currents** - **Left mesh current:** \( I_a \) - **Right mesh current:** \( I_b \) --- ### **Step 2: Write KVL equations for each mesh** --- #### **Left mesh:** The left mesh contains \( V_a \), \( R_1 \), and \( R_3 \). The mesh current \( I_a \) flows through \( R_1 \) and \( R_3 \). - Voltage across \( R_1 \): \( R_1 \times I_a \) - Voltage across \( R_3 \): since \( R_3 \) is between the two meshes, the current through \( R_3 \) is \( I_a - I_b \). Applying KVL: \[ V_a - R_1 I_a - R_3 (I_a - I_b) = 0 \] Substitute values: \[ 7 - 1.6 I_a - 0.7 (I_a - I_b) = 0 \] --- #### **Right mesh:** The right mesh contains \( R_2 \), \( R_3 \), and a voltage source \( V_a \). The current \( I_b \) flows through \( R_2 \), and \( R_3 \) is shared between the two meshes. Applying KVL: \[ - R_2 I_b - R_3 (I_b - I_a) + V_b = 0 \] But since no \( V_b \) is given, and the circuit indicates the voltage source \( V_a \) is connected to the left mesh only, the right mesh equation simplifies to: \[ - R_2 I_b - R_3 (I_b - I_a) = 0 \] Substituting the values: \[ - 1.7 I_b - 0.7 (I_b - I_a) = 0 \] --- ### **Step 3: Write the equations explicitly** **Equation 1 (Left mesh):** \[ 7 - 1.6 I_a - 0.7 I_a + 0.7 I_b = 0 \] Simplify: \[ 7 - (1.6 + 0.7) I_a + 0.7 I_b = 0 \] \[ 7 - 2.3 I_a + 0.7 I_b = 0 \] \[ 2.3 I_a - 0.7 I_b = 7 \] --- **Equation 2 (Right mesh):** \[ - 1.7 I_b - 0.7 I_b + 0.7 I_a = 0 \] Simplify: \[ - (1.7 + 0.7) I_b + 0.7 I_a = 0 \] \[ - 2.4 I_b + 0.7 I_a = 0 \] --- ### **Step 4: Solve the system of equations** Rewrite: 1. \( 2.3 I_a - 0.7 I_b = 7 \) 2. \( 0.7 I_a - 2.4 I_b = 0 \) --- ### **Step 5: Solve for \( I_a \) and \( I_b \)** From equation 2: \[ 0.7 I_a = 2.4 I_b \Rightarrow I_a = \frac{2.4}{0.7} I_b \approx 3.43 I_b \] Substitute into equation 1: \[ 2.3 (3.43 I_b) - 0.7 I_b = 7 \] Calculate: \[ 7.89 I_b - 0.7 I_b = 7 \] \[ (7.89 - 0.7) I_b = 7 \] \[ 7.19 I_b = 7 \] Solve for \( I_b \): \[ I_b = \frac{7}{7.19} \approx 0.97\,A \] Now, find \( I_a \): \[ I_a = 3.43 \times 0.97 \approx 3.33\,A \] --- ### **Final Answer:** \[ \boxed{I_x = I_a - I_b = 3.33 - 0.97 = 2.36} \] **Rounded to two decimal places:** **2.36** --- **Note:** The value of \( I_x \) is the current difference between the two mesh currents, which matches the physical current through the shared resistor \( R_2 \).