Let's solve the circuit step by step using **Kirchhoff’s Voltage Law (KVL)**. --- ## **Step 1: Assign Current Direction** Assume the current \( I \) flows clockwise (from \( V_1 \) up through \( R_2 \), down through \( R_3 \), left through \( R_4 \), and up through \( R_1 \)). --- ## **Step 2: Write KVL Equation** Starting at the bottom left (after \( R_1 \)), moving clockwise: 1. **Through \( V_1 \) (up):** \( +24\,V \) 2. **Through \( R_2 \) (right):** \( -I \cdot 150 \) 3. **Through \( R_3 \) (down):** \( -I \cdot 330 \) 4. **Through \( V_2 \) (down):** \( -12\,V \) 5. **Through \( R_4 \) (left):** \( -I \cdot 180 \) 6. **Through \( R_1 \) (up):** \( -I \cdot 270 \) Sum (KVL, total voltage drop = ): \[ (+24) - (I \cdot 150) - (I \cdot 330) - 12 - (I \cdot 180) - (I \cdot 270) = \] --- ## **Step 3: Combine Terms** Combine all terms involving \( I \): \[ 24 - 12 - I(150 + 330 + 180 + 270) = \] \[ 12 - I(930) = \] --- ## **Step 4: Solve for \( I \)** \[ 12 = 930I \] \[ I = \frac{12}{930} \] \[ I = .012903\, \text{A} \] --- ## **Step 5: Write Final Answers** **(a) Current through \( R_3 \):** \[ I_{R_3} = .0129\, \text{A} \] **(b) Current through \( R_4 \):** \[ I_{R_4} = .0129\, \text{A} \] > **Both are the same, since it’s a single-series loop.** --- ## **Final Answers (to three decimal places):** \[ \boxed{ \begin{aligned} &\text{(a) Current through } R_3 = .0129\, \text{A} \\ &\text{(b) Current through } R_4 = .0129\, \text{A} \end{aligned} } \]