Let's break down the problem step by step: ## **Given Data** - \(\math{E}_1 = 3.97\,V\) - \(\mathcal{E}_2 = 1.06\,V\) - \(R_1 = 6.87\,\Omega\) - \(R_2 = 2.16\,\Omega\) - \(R_3 = 3.45\,\Omega\) ## **Step 1: Assign Currents** Let: - \(I_1\) = current in the left loop (through \(R_1\), from \(\mathcal{E}_1\)) - \(I_2\) = current in the right loop (through \(R_2\), from \(\mathcal{E}_2\)) Assume the direction of currents as shown in the image (standard clockwise). Since \(R_3\) is shared, the current through \(R_3\) is \(I_1 - I_2\). ## **Step 2: Write Loop Equations (Kirchhoff's Law)** **Left loop:** \[ \mathcal{E}_1 - I_1 R_1 - (I_1 - I_2)R_3 = \] **Right loop:** \[ \mathcal{E}_2 - I_2 R_2 - (I_2 - I_1)R_3 = \] Or equivalently: \[ \mathcal{E}_2 - I_2 R_2 - (I_2 - I_1)R_3 = \] ## **Step 3: Plug in Values** **Equation 1:** \[ 3.97 - 6.87I_1 - 3.45(I_1 - I_2) = \] \[ 3.97 - 6.87I_1 - 3.45I_1 + 3.45I_2 = \] \[ 3.97 - (6.87 + 3.45)I_1 + 3.45I_2 = \] \[ 3.97 - 10.32I_1 + 3.45I_2 = \] \[ 10.32I_1 - 3.45I_2 = 3.97 \quad (1) \] **Equation 2:** \[ 1.06 - 2.16I_2 - 3.45(I_2 - I_1) = \] \[ 1.06 - 2.16I_2 - 3.45I_2 + 3.45I_1 = \] \[ 1.06 + 3.45I_1 - (2.16 + 3.45)I_2 = \] \[ 1.06 + 3.45I_1 - 5.61I_2 = \] \[ 3.45I_1 - 5.61I_2 = -1.06 \quad (2) \] ## **Step 4: Solve the Equations** Multiply (2) by \(\frac{10.32}{3.45}\) to eliminate \(I_1\): \[ 3.45I_1 - 5.61I_2 = -1.06 \] \[ 10.32I_1 - 16.78I_2 = -3.17 \] Now subtract equation (2) from equation (1): \[ (10.32I_1 - 3.45I_2) - (10.32I_1 - 16.78I_2) = 3.97 - (-3.17) \] \[ (-3.45I_2) + 16.78I_2 = 3.97 + 3.17 \] \[ 13.33I_2 = 7.14 \] \[ I_2 = \frac{7.14}{13.33} \approx .536\,A \] Substitute \(I_2\) back into equation (1): \[ 10.32I_1 - 3.45(.536) = 3.97 \] \[ 10.32I_1 - 1.848 = 3.97 \] \[ 10.32I_1 = 3.97 + 1.848 = 5.818 \] \[ I_1 = \frac{5.818}{10.32} \approx .563\,A \] ## **Step 5: Find Current through \(R_3\)** \[ I_3 = I_1 - I_2 = .563 - .536 = .027\,A \] ## **Step 6: Power Dissipated in Each Resistor** \[ P = I^2 R \] ### (a) \(R_1\) \[ P_1 = I_1^2 R_1 = (.563)^2 \times 6.87 \approx 2.18\,W \] ### (b) \(R_2\) \[ P_2 = I_2^2 R_2 = (.536)^2 \times 2.16 \approx .624\,W \] ### (c) \(R_3\) \[ P_3 = (I_1 - I_2)^2 R_3 = (.027)^2 \times 3.45 \approx .0025\,W \] ## **Step 7: Power of Each Battery** **Battery 1:** \[ P_{\text{batt 1}} = \mathcal{E}_1 \cdot I_1 = 3.97 \times .563 \approx 2.24\,W \] **Battery 2:** \[ P_{\text{batt 2}} = \mathcal{E}_2 \cdot I_2 = 1.06 \times .536 \approx .57\,W \] --- ## **Final Answers** | Part | Value | Units | |------|-------|-------| | (a) | 2.18 | W | | (b) | .624 | W | | (c) | .0025| W | | (d) | 2.24 | W | | (e) | .57 | W | --- **If you need the answers in more significant digits, recalculate with more precision. Let me know if you have any questions or need further clarification!**