Let's solve for \( I_2 \) and \( \Delta V_{ab} = V_a - V_b \) using **Kirchhoff's laws**. --- ### **Step 1: Assign values** Given: - \( V_1 = 15\,V \) - \( V_2 = 9\,V \) - \( R_1 = R_5 = 50\,\Omega \) - \( R_2 = R_6 = 150\,\Omega \) - \( R_3 = 81\,\Omega \) - \( R_4 = 70\,\Omega \) --- ### **Step 2: Label Currents** From the diagram: - \( I_1 \) goes down through \( R_1 \), \( R_2 \) - \( I_2 \) to the right through \( R_3 \), \( R_4 \) - \( I_3 \) goes down through \( R_5 \), \( R_6 \) At the top junction: \( I_1 = I_2 + I_3 \) --- ### **Step 3: Write Loop Equations** #### **Loop 1 (left loop: V1, R1, R2, R3):** \[ - V_1 + I_1 R_1 + I_1 R_2 + (I_1 - I_2) R_3 = 0 \] \[ -15 + 50I_1 + 150I_1 + 81(I_1 - I_2) = 0 \] \[ -15 + 200I_1 + 81I_1 - 81I_2 = 0 \] \[ -15 + 281I_1 - 81I_2 = 0 \] \[ 281I_1 - 81I_2 = 15 \quad \text{(1)} \] --- #### **Loop 2 (right loop: V2, R4, R5, R6, R3):** \[ - V_2 + I_3 R_5 + I_3 R_6 + (I_3 - I_2) R_4 + (I_3 - I_1) R_3 = 0 \] \[ -9 + 50I_3 + 150I_3 + 70(I_3 - I_2) + 81(I_3 - I_1) = 0 \] \[ -9 + 200I_3 + 70I_3 - 70I_2 + 81I_3 - 81I_1 = 0 \] \[ -9 + (200+70+81)I_3 - 70I_2 - 81I_1 = 0 \] \[ -9 + 351I_3 - 70I_2 - 81I_1 = 0 \] \[ 351I_3 - 70I_2 - 81I_1 = 9 \quad \text{(2)} \] --- #### **Junction Rule:** \[ I_1 = I_2 + I_3 \implies I_3 = I_1 - I_2 \quad \text{(3)} \] --- ### **Step 4: Substitute \(I_3\) into Eqn (2):** \[ 351I_3 - 70I_2 - 81I_1 = 9 \] \[ 351(I_1 - I_2) - 70I_2 - 81I_1 = 9 \] \[ 351I_1 - 351I_2 - 70I_2 - 81I_1 = 9 \] \[ (351I_1 - 81I_1) - (351I_2 + 70I_2) = 9 \] \[ 270I_1 - 421I_2 = 9 \quad \text{(4)} \] --- ### **Step 5: Solve the Two Equations** Eqn (1): \[ 281I_1 - 81I_2 = 15 \] Eqn (4): \[ 270I_1 - 421I_2 = 9 \] Let's solve these two equations: #### **Multiply (1) by 421:** \[ 421 \cdot (281I_1 - 81I_2) = 421 \cdot 15 \\ 118301I_1 - 34101I_2 = 6315 \] #### **Multiply (4) by 81:** \[ 81 \cdot (270I_1 - 421I_2) = 81 \cdot 9 \\ 21870I_1 - 34001I_2 = 729 \] #### **Subtract (second from first):** \[ (118301I_1 - 34101I_2) - (21870I_1 - 34001I_2) = 6315 - 729 \\ (118301 - 21870)I_1 + (-34101 + 34001)I_2 = 5586 \\ 96431I_1 - 100I_2 = 5586 \] But let's solve for \( I_2 \): From Eqn (1): \[ 281I_1 - 81I_2 = 15 \\ 81I_2 = 281I_1 - 15 \\ I_2 = \frac{281I_1 - 15}{81} \] Plug into Eqn (4): \[ 270I_1 - 421I_2 = 9 \\ 270I_1 - 421 \left( \frac{281I_1 - 15}{81} \right) = 9 \\ 270I_1 - \frac{421 \cdot 281I_1 - 421 \cdot 15}{81} = 9 \\ 270I_1 - \frac{118301I_1 - 6315}{81} = 9 \\ \] Multiply both sides by 81: \[ 81 \cdot 270I_1 - (118301I_1 - 6315) = 9 \cdot 81 \\ 21870I_1 - 118301I_1 + 6315 = 729 \\ -96431I_1 = 729 - 6315 \\ -96431I_1 = -5586 \\ I_1 = \frac{-5586}{-96431} = 0.0579\,\text{A} \] Now, plug \( I_1 \) back to get \( I_2 \): \[ I_2 = \frac{281I_1 - 15}{81} = \frac{281 \times 0.0579 - 15}{81} = \frac{16.2679 - 15}{81} = \frac{1.2679}{81} = 0.0157\,\text{A} \] --- ### **Step 6: Calculate \( \Delta V_{ab} \) (from a to b)** Path: from a to b is across \( R_4 \) (current \( I_2 \) to the right) and across \( R_6 \) (current \( I_3 \) downward). First, calculate \( I_3 \): \[ I_3 = I_1 - I_2 = 0.0579 - 0.0157 = 0.0422\,\text{A} \] From \( a \) to \( b \): \[ \Delta V_{ab} = V_a - V_b = -I_2 R_4 - I_3 R_6 \] \[ = -(0.0157)(70) - (0.0422)(150) = -1.099 - 6.33 = -7.43\,\text{V} \] --- ## **Final Answers** **1. \( I_2 = \boxed{0.0157\,\text{A}} \) (or \( 15.7\,\text{mA} \) to the right)** **2. \( \Delta V_{ab} = V_a - V_b = \boxed{-7.43\,\text{V}} \)** If you need more significant digits or further clarification, let me know!