Let's solve the problem step by step using **Kirchhoff’s Junction Loop Rules**. --- ## Step1: **Label Currents** Let: - \( I \): Main current from battery through \( r \). - \( I_1 \): Current through \( R_1 \). - \( I_2 \): Current through \( R_2 \). - \( I_3 \): Current through \( R_3 \). - \( I_4 \): Current through \( R_4 \) (ammeter reading). - \( I_5 \): Current through \( R_5 \). **From the diagram:** - At junction A: \( I = I_1 + I_2 \) - At junction D: \( I_2 = I_4 + I_5 \) - At junction B: \( I_1 = I_3 + I_4 \) - At junction E: \( I_5 + I_3 = I \) --- ## Step 2: **Write Loop Equations** ### **Loop 1**: Battery → A → R1 → R4 → D → battery \[ \epsilon - I r - I_1 R_1 - I_4 R_4 - I_2 R_2 = \] But since currents split, let's define mesh currents. ### **Mesh Analysis** Let: - \( I_1 \): through loop A-B-D-A (R1, R4) - \( I_2 \): through loop A-E-D-A (R2, R5) Let’s assign: - Mesh 1 (left mesh): Battery (\(\epsilon\)), \(r\), \(R_2\), \(R_5\) - Mesh 2 (upper right mesh): \(R_1\), \(R_4\) - Mesh 3 (lower right mesh): \(R_3\), \(R_5\), \(R_4\) ### **Assign mesh currents:** Let: - Mesh 1: \( I_a \) (counterclockwise, battery, \(r\), \(R_2\), \(R_5\)) - Mesh 2: \( I_b \) (clockwise, \(R_1\), \(R_4\), sharing \(R_4\) with mesh 3) - Mesh 3: \( I_c \) (clockwise, \(R_3\), \(R_5\), sharing \(R_5\) with mesh 1 and \(R_4\) with mesh 2) #### **Write mesh equations:** ### **Mesh 1 (A → D → E → A):** \[ \epsilon - I_a r - I_a R_2 - (I_a - I_c)R_5 = \] \[ 36. - 1.5I_a - 20I_a - 25(I_a - I_c) = \] \[ 36. - 1.5I_a - 20I_a - 25I_a + 25I_c = \] \[ 36. - (1.5+20+25)I_a + 25I_c = \] \[ 36. - 46.5I_a + 25I_c = \] \[ 46.5I_a - 25I_c = 36. \tag{1} \] --- ### **Mesh 2 (A → B → D → A):** \[ I_b \text{ goes through } R_1 \text{ and } R_4 \text{ (shared with Mesh 3)}. \] \[ I_b R_1 + (I_b - I_c)R_4 = \] \[ 10I_b + 30(I_b - I_c) = \] \[ 10I_b + 30I_b - 30I_c = \] \[ 40I_b - 30I_c = \] \[ 4I_b - 3I_c = \implies I_b = \frac{3}{4}I_c \tag{2} \] --- ### **Mesh 3 (B → C → E → D → B):** \[ I_c \text{ through } R_3, R_5, R_4 \] \[ I_c R_3 + I_c R_5 + (I_c - I_b)R_4 = \] \[ 15I_c + 25I_c + 30(I_c - I_b) = \] \[ 15I_c + 25I_c + 30I_c - 30I_b = \] \[ 70I_c - 30I_b = \] \[ 7I_c - 3I_b = \implies I_c = \frac{3}{7}I_b \tag{3} \] --- ## Step 3: **Solve the System** From (2): \( I_b = \frac{3}{4}I_c \) From (3): \( I_c = \frac{3}{7}I_b \implies I_b = \frac{7}{3}I_c \) But \( I_b = \frac{3}{4}I_c \) and \( I_b = \frac{7}{3}I_c \) can't be simultaneously true unless zero. Let's double-check mesh equations. #### Mesh 2 (A → B → D → A): The mesh current through \(R_1\) is \( I_b \), through \(R_4\) is \( I_b - I_c \), so the sum is: \[ I_b R_1 + (I_b - I_c) R_4 = \] \[ 10I_b + 30(I_b - I_c) = \] \[ 10I_b + 30I_b - 30I_c = \] \[ 40I_b - 30I_c = \] \[ 4I_b - 3I_c = \implies I_b = \frac{3}{4}I_c \] #### Mesh 3 (B → C → E → D → B): Through \(R_3\): \(I_c\), through \(R_5\): \(I_c - I_a\), through \(R_4\): \(I_c - I_b\): \[ I_c R_3 + (I_c - I_a) R_5 + (I_c - I_b) R_4 = \] \[ 15I_c + 25(I_c - I_a) + 30(I_c - I_b) = \] \[ 15I_c + 25I_c - 25I_a + 30I_c - 30I_b = \] \[ (15+25+30)I_c - 25I_a - 30I_b = \] \[ 70I_c - 25I_a - 30I_b = \tag{3} \] Now, let's use equations: 1. \( 46.5I_a - 25I_c = 36. \) 2. \( I_b = \frac{3}{4}I_c \) 3. \( 70I_c - 25I_a - 30I_b = \) Plug \(I_b\) from (2) into (3): \[ 70I_c - 25I_a - 30 \left( \frac{3}{4}I_c \right) = \] \[ 70I_c - 25I_a - 22.5I_c = \] \[ (70 - 22.5)I_c = 25I_a \] \[ 47.5I_c = 25I_a \] \[ I_a = \frac{47.5}{25}I_c = 1.9I_c \] Now substitute \(I_a = 1.9I_c\) into equation (1): \[ 46.5I_a - 25I_c = 36. \] \[ 46.5(1.9I_c) - 25I_c = 36. \] \[ 88.35I_c - 25I_c = 36. \] \[ 63.35I_c = 36. \] \[ I_c = \frac{36.}{63.35} = .568 \text{ A} \] Now: \[ I_a = 1.9I_c = 1.9 \times .568 = 1.08 \text{ A} \] \[ I_b = \frac{3}{4}I_c = .426 \text{ A} \] --- ## Step 4: **Answers** ### (a) **Ammeter reading** (current through \( R_4 \)): Current through \( R_4 \) is \( I_b - I_c = .426 - .568 = -.142 \) A **Negative sign** means direction is from **D→B**. **Ammeter reading:** \[ \boxed{.142 \text{ A, direction D}\rightarrow\text{B}} \] --- ### (b) **Terminal voltage of the battery** (\( V_{AD} \)): \[ V_{\text{terminal}} = \epsilon - I_a r = 36. - (1.08)(1.5) = 36. - 1.62 = 34.4 \text{ V} \] \[ \boxed{34.4 \text{ V}} \] --- ### (c) **Total power delivered by the battery to the external circuit**: Power delivered to external circuit (excluding internal resistance): \[ P = I_a \, V_{\text{terminal}} = 1.08 \times 34.4 = 37.2 \text{ W} \] \[ \boxed{37.2 \text{ W}} \] --- ## **Final Answers (to three significant figures):** 1. **Ammeter reading:**                                                                                                                             \[ \boxed{.142~\text{A},~\text{direction D}\rightarrow\text{B}} \] 2. **Terminal voltage:** \[ \boxed{34.4~\text{V}} \] 3. **Total power delivered to external circuit:** \[ \boxed{37.2~\text{W}} \]