Let's solve the problem **step by step** using **Kirchhoff’s Voltage Law (KVL)**. --- ## **Step 1: Label the currents** Let the current through \( R_1 \) (top branch) be \( I_1 \) (to the right), and through \( R_2 \) (bottom branch) be \( I_2 \) (to the right). --- ## **Step 2: Apply Kirchhoff’s Voltage Law** **Loop 1 (top loop: \( E_1 \) → \( R_1 \) → right side → back to \( E_1 \)):** \[ E_1 - I_1 R_1 = \] **Loop 2 (bottom loop: \( E_3 \) → \( R_2 \) → left side → back to \( E_3 \)):** \[ E_3 + I_2 R_2 = \] But actually, these loops are not independent; both resistors share a node. Let's consider the entire loop (start at the bottom left, go through \( E_1 \), \( R_1 \), down, \( R_2 \), and \( E_3 \)): \[ E_1 - I_1 R_1 - I_2 R_2 - E_3 = \] But more efficiently, let's use the **series circuit rule** (since both batteries are in series and both resistors are in series): - The total emf in the loop: \( E_1 - E_3 = 9\,V - 6\,V = 3\,V \) - Total resistance: \( R_1 + R_2 = 22\,\Omega + 18\,\Omega = 40\,\Omega \) --- ## **Step 3: Find the Current** \[ I = \frac{\text{Total emf}}{\text{Total resistance}} = \frac{3\,V}{40\,\Omega} = .075\,A \] --- ## **Step 4: Determine Direction** - The current flows from the higher emf battery (\( E_1 \)) towards the lower emf (\( E_3 \)). - In \( R_1 \), current is to the right. - In \( R_2 \), current is also to the right (since it's a single loop). --- ## **Step 5: Final Answers** ### **Current through \( R_1 \):** - **Magnitude:** \( .075\,A \) - **Direction:** Right ### **Current through \( R_2 \):** - **Magnitude:** \( .075\,A \) - **Direction:** Right --- ## **Summary Table** | Resistor | Magnitude (A) | Direction | |----------|:-------------:|:---------:| | R1 | .075 | Right | | R2 | .075 | Right | --- **Final Answer:** - **Current through R1: Magnitude = .075 A, Direction = towards Right** - **Current through R2: Magnitude = .075 A, Direction = towards Right**