## Step 1: Find the Equivalent Resistance Let's analyze the circuit: **Resistor Arrangement:** - The 8. V battery is connected to a combination of resistors. - The top branch has a **4 Ω** resistor. - The middle branch has a **5 Ω** and a **10 Ω** resistor **in series**. - The bottom branch has an **8 Ω** and a **4 Ω** resistor **in series**. - All three branches are **in parallel** with each other. ### Calculate Series Combinations - **Middle branch:** \( 5\ \Omega + 10\ \Omega = 15\ \Omega \) - **Bottom branch:** \( 8\ \Omega + 4\ \Omega = 12\ \Omega \) ### Calculate Parallel Combination Let the resistances in the three branches be \( R_1 = 4\ \Omega \), \( R_2 = 15\ \Omega \), and \( R_3 = 12\ \Omega \): \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \] \[ \frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{15} + \frac{1}{12} \] \[ \frac{1}{R_{eq}} = \frac{15}{60} + \frac{4}{60} + \frac{5}{60} = \frac{24}{60} \] \[ R_{eq} = \frac{60}{24} = 2.5\ \Omega \] **Explanation:** We first combined the resistors in each branch in series, then calculated the equivalent resistance for the parallel branches using the parallel formula. --- ## Step 2: Find the Total Current Now, use Ohm's law to find the total current through the battery: \[ I = \frac{V}{R_{eq}} = \frac{8.\ \text{V}}{2.5\ \Omega} = 3.2\ \text{A} \] **Explanation:** The total current is found by dividing the battery voltage by the equivalent resistance of the circuit. --- ## Final Answers - **Equivalent Resistance:** \( 2.5\ \Omega \) - **Total Current through the Battery:** \( 3.2\ \text{A} \)