## Step 1: Find the Equivalent Resistance First, analyze the circuit: - \( R_1 = 4\ \Omega \) (in series with the battery) - \( R_2 = 6\ \Omega \) (in series after \( R_1 \)) - After \( R_2 \), the circuit splits into two parallel branches: - Branch 1: \( R_3 = 2\ \Omega \) (upper branch) - Branch 2: Two resistors in series: \( R_4 = 6\ \Omega \) and \( R_5 = 8\ \Omega \) (lower branch) Calculate the resistance of the lower branch: \[ R_{\text{lower}} = R_4 + R_5 = 6\ \Omega + 8\ \Omega = 14\ \Omega \] Now, combine \( R_3 \) and the lower branch in parallel: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_3} + \frac{1}{R_{\text{lower}}} = \frac{1}{2} + \frac{1}{14} = \frac{7}{14} + \frac{1}{14} = \frac{8}{14} = \frac{4}{7} \] \[ R_{\text{parallel}} = \frac{7}{4} = 1.75\ \Omega \] Add all series resistances: \[ R_{\text{eq}} = R_1 + R_2 + R_{\text{parallel}} = 4\ \Omega + 6\ \Omega + 1.75\ \Omega = 11.75\ \Omega \] **Explanation:** We simplified the circuit by combining series and parallel resistors step-by-step to find the total equivalent resistance. --- ## Step 2: Find the Current in \( R_5 \) (\( 8\ \Omega \)) Total current from the battery: \[ I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{11.\ \text{V}}{11.75\ \Omega} \approx .936\ \text{A} \] Current splits between the two parallel branches. The voltage across both branches is the same as the voltage drop after \( R_1 \) and \( R_2 \): First, calculate voltage across the parallel branches: \[ V_{\text{parallel}} = I_{\text{total}} \times (R_1 + R_2) = .936\ \text{A} \times (4 + 6)\ \Omega = .936 \times 10 = 9.36\ \text{V} \] \[ V_{\text{parallel}} = 11.\ \text{V} - 9.36\ \text{V} = 1.64\ \text{V} \] But this is not correct. Let's use the correct method: Voltage drop across \( R_1 \) and \( R_2 \): \[ V_{R_1} = I_{\text{total}} \times R_1 = .936 \times 4 = 3.744\ \text{V} \] \[ V_{R_2} = I_{\text{total}} \times R_2 = .936 \times 6 = 5.616\ \text{V} \] \[ V_{\text{parallel}} = 11.\ \text{V} - (3.744 + 5.616) = 11. - 9.36 = 1.64\ \text{V} \] The voltage across \( R_5 \) (since it's in series with \( R_4 \)): \[ I_{R_5} = \frac{V_{\text{parallel}}}{R_4 + R_5} = \frac{1.64}{14} \approx .117\ \text{A} \] --- ## **Final Answers** **1. The equivalent resistance of the circuit is:** \[ \boxed{11.75\ \Omega} \] **2. The current in \( R_5 = 8\ \Omega \) is:** \[ \boxed{.117\ \text{A}} \]