1. **Problem statement:** Calculate the total power dissipated in the circuit with resistors $R_1=10\ \Omega$, $R_2=3\ \Omega$, $R_3=8\ \Omega$ and voltage sources $V_1=5\ V$, $V_2=8\ V$. Given options are 1.1W, 3.0W, 7.1W, and 21W. 2. **Understanding the circuit:** The note says $R_T = R_1 + R_2 = 9 + 3 = 11\ \Omega$, but $R_1$ is given as 10, so we assume $R_1=9\ \Omega$ for this calculation. The total resistance in series is $R_T = 9 + 3 = 12\ \Omega$ if we consider original values, but we follow the note $R_T=11\ \Omega$. 3. **Calculate total voltage:** Since $V_1=5V$ and $V_2=8V$, total voltage $V_T = V_1 + V_2 = 5 + 8 = 13\ V$. 4. **Calculate total current:** Using Ohm's law, $I = \frac{V_T}{R_T} = \frac{13}{11} \approx 1.18\ A$. 5. **Calculate total power dissipated:** Power $P = I^2 R_T = (1.18)^2 \times 11 \approx 1.39 \times 11 = 15.29\ W$. 6. **Check options:** None match 15.29W, so re-examine assumptions. 7. **Alternative approach:** If $R_T = R_1 + R_2 + R_3 = 10 + 3 + 8 = 21\ \Omega$, then $I = \frac{13}{21} \approx 0.619\ A$. 8. **Power with total resistance:** $P = I^2 R_T = (0.619)^2 \times 21 \approx 0.383 \times 21 = 8.04\ W$. 9. **Closest option:** 7.1W (option c) is nearest to 8.04W. **Final answer:** The nearest total power dissipated is **7.1W** (option c).