Let’s solve the circuit step by step: ### **Given:** - \( V_A = V_B = 14.0\, \text{V} \) - \( R_1 = R_2 = R_3 = R_4 = R_5 = 75.0\, \Omega \) - \( I_1 = 0.210\, \text{A} \) flows through \( R_1 \) (to the left). --- ## **Step 1: Find \( I_2 \) (Current through \( R_2 \))** **Let's label the loop currents as follows:** - Let \( I_2 \) be the current through \( R_2 \) (downward as positive). - \( I_1 \) is already given (left through \( R_1 \)). ### **Loop Equations** #### **Loop 1: VA → R1 → R2 → VA** The loop equation: \[ V_A - I_1 R_1 - I_2 R_2 = 0 \] \[ 14.0 - (0.210)(75.0) - I_2 (75.0) = 0 \] \[ 14.0 - 15.75 - 75.0 I_2 = 0 \] \[ -1.75 - 75.0 I_2 = 0 \] \[ 75.0 I_2 = -1.75 \] \[ I_2 = -\frac{1.75}{75.0} = -0.0233\, \text{A} \] #### **So:** \[ \boxed{I_2 = -0.0233\, \text{A}} \] (The negative sign means the actual direction is **upward** through \( R_2 \), opposite to the assumed direction.) --- ## **Step 2: Find \( V_4 \) (Voltage across \( R_4 \))** \( R_4 \) is in the branch parallel to \( R_2 \). The voltage across \( R_4 \) is the same as the voltage across the combination of \( R_3 \) and \( R_5 \). First, find the voltage at the node between \( R_1 \) and \( R_2 \): ### **Voltage drop across \( R_1 \):** \[ V_{R_1} = I_1 \cdot R_1 = 0.210 \times 75.0 = 15.75\, \text{V} \] ### **Voltage at the node after \( R_1 \):** \[ V_{\text{node}} = V_A - V_{R_1} = 14.0 - 15.75 = -1.75\, \text{V} \] ### **Voltage across \( R_4 \):** The voltage across \( R_4 \) is the difference between the node voltage (after \( R_1 \)) and the voltage at the bottom (which is ground, or 0V): \[ V_4 = V_B - V_{\text{node}} \] \[ V_4 = 14.0 - (-1.75) = 15.75\, \text{V} \] #### **So:** \[ \boxed{V_4 = 15.75\, \text{V}} \] --- ## **Step 3: Find \( P_5 \) (Power dissipated by \( R_5 \))** \( R_5 \) is in series with \( R_4 \) and \( R_3 \), all of which are in parallel with the \( R_2 \) branch. ### **Current through \( R_5 \):** The voltage across the \( R_4 + R_5 \) branch is \( V_4 = 15.75\, \text{V} \), and \( R_5 = 75.0\, \Omega \). First, the total resistance in this path is \( R_3 + R_4 + R_5 = 75.0 + 75.0 + 75.0 = 225.0\, \Omega \). The current through this branch: \[ I_{\text{branch}} = \frac{V_4}{R_3 + R_4 + R_5} = \frac{15.75}{225.0} = 0.0700\, \text{A} \] The current through \( R_5 \) is the same as in this branch. ### **Power dissipated by \( R_5 \):** \[ P_5 = I^2 R_5 = (0.0700)^2 \cdot 75.0 = 0.0049 \cdot 75.0 = 0.3675\, \text{W} \] #### **So:** \[ \boxed{P_5 = 0.368\, \text{W}} \] --- ## **Final Answers** ### 1. \( I_2 = \boxed{-0.0233\, \text{A}} \) ### 2. \( V_4 = \boxed{15.75\, \text{V}} \) ### 3. \( P_5 = \boxed{0.368\, \text{W}} \) Let me know if you'd like a circuit diagram or further clarification!