Let's solve the problem step by step: --- ## **Step 1: Find the Equivalent Resistance** The circuit shows a **series** connection of three resistors: 4 Ω, 10 Ω, and 12 Ω. The total (equivalent) resistance \( R_{eq} \) in series is: \[ R_{eq} = R_1 + R_2 + R_3 = 4\,\Omega + 10\,\Omega + 12\,\Omega = 26\,\Omega \] **Explanation:** When resistors are in series, their resistances simply add together. --- ## **Step 2: Find the Total Current in the Circuit** The total voltage from the two batteries is: \[ V_{total} = V_1 + V_2 = 10.\,V + 10.\,V = 20.\,V \] Using Ohm's Law (\( I = \frac{V}{R} \)): \[ I = \frac{V_{total}}{R_{eq}} = \frac{20.\,V}{26\,\Omega} \approx .769\,A \] **Explanation:** In a series circuit, the same current flows through all resistors. The total voltage divides across the resistors according to their resistances. --- ## **Final Answers** - **Current through each resistor:** \( \boxed{.769\,A} \) (same for all, since they are in series)