Let's break this down **in two steps** with explanations for each, and then provide the final answers. --- ### **Step 1: Find the Equivalent Resistance of the Resistor Network** **Explanation:** - The circuit has a combination of series and parallel resistors. - Let's label the resistors as in the diagram: - \( R_1 = 1.00 \, \Omega \) - \( R_2 = 2.00 \, \Omega \) - \( R_3 = 1.00 \, \Omega \) - \( R_4 = 2.00 \, \Omega \) - \( R_5 = 1.00 \, \Omega \) #### **Analyzing the Network:** - \( R_2 \) and \( R_3 \) are **in series**, so their combination is: \( R_{23} = R_2 + R_3 = 2.00 + 1.00 = 3.00 \, \Omega \) - \( R_4 \) and \( R_5 \) are **in series**, so their combination is: \( R_{45} = R_4 + R_5 = 2.00 + 1.00 = 3.00 \, \Omega \) - Now, \( R_{23} \) and \( R_{45} \) are **in parallel**: \[ \frac{1}{R_{eq, 2345}} = \frac{1}{R_{23}} + \frac{1}{R_{45}} = \frac{1}{3.00} + \frac{1}{3.00} = \frac{2}{3.00} \] \[ R_{eq, 2345} = \frac{3.00}{2} = 1.50 \, \Omega \] - Finally, \( R_1 \) is **in series** with this combination: \[ R_{total} = R_1 + R_{eq, 2345} = 1.00 + 1.50 = 2.50 \, \Omega \] **Final Equivalent Resistance:** \[ \boxed{2.50 \, \Omega} \] --- ### **Step 2: Find the Current Through the Battery and Each Resistor** **Explanation:** - The total voltage is \( V = 14.0 \, V \) - The total resistance from step 1 is \( R_{total} = 2.50 \, \Omega \) - **Current from the battery:** \[ I_{total} = \frac{V}{R_{total}} = \frac{14.0}{2.50} = 5.60 \, \text{A} \] #### **Current through Each Resistor:** - \( R_1 \) (in series with the battery): current is \( 5.60 \, \text{A} \) - Voltage drop across \( R_1 \): \( V_1 = I_{total} \times R_1 = 5.60 \times 1.00 = 5.60 \, V \) - Voltage remaining for the parallel section: \( V_{parallel} = 14.0 - 5.60 = 8.40 \, V \) **For both parallel branches:** - Each branch has a total resistance of \( 3.00 \, \Omega \) - Current through each branch: \( I_{branch} = \frac{V_{parallel}}{3.00} = \frac{8.40}{3.00} = 2.80 \, \text{A} \) **Resistors in each branch:** - \( R_2 \) and \( R_3 \) (series): \( I_2 = I_3 = 2.80 \, \text{A} \) - \( R_4 \) and \( R_5 \) (series): \( I_4 = I_5 = 2.80 \, \text{A} \) --- ## **Final Answers** **(a) Current through the battery:** \[ \boxed{5.60 \, \text{A}} \] **Current through each resistor:** - \( R_1: 5.60 \, \text{A} \) - \( R_2: 2.80 \, \text{A} \) - \( R_3: 2.80 \, \text{A} \) - \( R_4: 2.80 \, \text{A} \) - \( R_5: 2.80 \, \text{A} \) **(b) Equivalent resistance of the resistor network:** \[ \boxed{2.50 \, \Omega} \]