1. **State the problem:** Find the power dissipated by resistor $R_L$ using the superposition theorem given: $E_1=12$, $E_2=6$, $R_1=3300$, $R_2=4700$, $R_3=5600$, $R_L=8200$ (all resistances in ohms, voltages in volts). 2. **Superposition theorem:** Analyze the circuit twice: once with $E_1$ active and $E_2$ replaced by a short circuit, and once with $E_2$ active and $E_1$ replaced by a short circuit. 3. **Step 1: With $E_1$ active, $E_2$ shorted:** - Replace $E_2$ with a wire (short circuit). - The circuit reduces to $E_1$ connected through $R_1$ to a node, then $R_2$ and $R_3$ in series to ground through $R_L$. - Calculate equivalent resistance seen by $E_1$ to find current and voltage at $R_L$. 4. **Calculate equivalent resistance for $E_1$ active:** - $R_{23} = R_2 + R_3 = 4700 + 5600 = 10300$ - $R_{23L} = \frac{R_{23} \times R_L}{R_{23} + R_L} = \frac{10300 \times 8200}{10300 + 8200} = \frac{84460000}{18500} \approx 4565.41$ - Total resistance $R_{total1} = R_1 + R_{23L} = 3300 + 4565.41 = 7865.41$ 5. **Calculate current from $E_1$:** $$I_1 = \frac{E_1}{R_{total1}} = \frac{12}{7865.41} \approx 0.001526\,A$$ 6. **Voltage at node between $R_1$ and $R_{23L}$:** $$V_{node1} = E_1 - I_1 \times R_1 = 12 - 0.001526 \times 3300 = 12 - 5.04 = 6.96\,V$$ 7. **Voltage across $R_L$ with $E_1$ active:** - Voltage divider between $R_{23}$ and $R_L$: $$V_{R_L1} = V_{node1} \times \frac{R_L}{R_{23} + R_L} = 6.96 \times \frac{8200}{10300 + 8200} = 6.96 \times \frac{8200}{18500} \approx 3.09\,V$$ 8. **Step 2: With $E_2$ active, $E_1$ shorted:** - Replace $E_1$ with a wire. - The circuit reduces to $E_2$ connected through $R_2$ to a node, then $R_3$ and $R_L$ in series to ground through $R_1$. 9. **Calculate equivalent resistance for $E_2$ active:** - $R_{3L} = R_3 + R_L = 5600 + 8200 = 13800$ - $R_{total2} = R_2 + \frac{R_1 \times R_{3L}}{R_1 + R_{3L}}$ - Calculate parallel: $$R_{1,3L} = \frac{3300 \times 13800}{3300 + 13800} = \frac{45540000}{17100} \approx 2663.16$$ - So, $$R_{total2} = 4700 + 2663.16 = 7363.16$$ 10. **Calculate current from $E_2$:** $$I_2 = \frac{E_2}{R_{total2}} = \frac{6}{7363.16} \approx 0.000815\,A$$ 11. **Voltage at node between $R_2$ and parallel branch:** $$V_{node2} = E_2 - I_2 \times R_2 = 6 - 0.000815 \times 4700 = 6 - 3.83 = 2.17\,V$$ 12. **Voltage across $R_L$ with $E_2$ active:** - Voltage divider between $R_1$ and $R_{3L}$: $$V_{R_L2} = V_{node2} \times \frac{R_L}{R_3 + R_L} = 2.17 \times \frac{8200}{13800} = 2.17 \times 0.5942 \approx 1.29\,V$$ 13. **Total voltage across $R_L$ by superposition:** $$V_{R_L} = V_{R_L1} + V_{R_L2} = 3.09 + 1.29 = 4.38\,V$$ 14. **Calculate power dissipated by $R_L$:** $$P_{R_L} = \frac{V_{R_L}^2}{R_L} = \frac{4.38^2}{8200} = \frac{19.18}{8200} \approx 0.00234\,W = 2.34\,mW$$ **Final answer:** The power dissipated by resistor $R_L$ is approximately **2.34 milliwatts**.