To solve this problem, we need to calculate the power output when the heaters are connected in series and in parallel.$\\$Given:$\\$• Power rating of heater A, $P_A = 1$ kW$\\$• Power rating of heater B, $P_B = 2$ kW$\\$Let's denote the voltage across each heater as $V_A$ and $V_B. \\$The resistance of each heater can be calculated using $P = \frac{V^2}{R}: \\ R_A = \frac{V_A^2}{P_A} = \frac{V_A^2}{1} \\ R_B = \frac{V_B^2}{P_B} = \frac{V_B^2}{2} \\$Assume the total voltage of the power source is $V. \\$1. Series Connection:$\\$In series, the same current $I$ flows through both heaters.$\\ V = V_A + V_B \\ I = \frac{V}{R_A + R_B} \\$Power output in series, $P_{series} = I^2 (R_A + R_B) \\ P_{series} = \left(\frac{V}{R_A + R_B}\right)^2 (R_A + R_B) \\ P_{series} = \frac{V^2}{R_A + R_B} \\$2. Parallel Connection:$\\$In parallel, the voltage across each heater is the same, $V. \\ P_{parallel} = P_A + P_B = 1 + 2 = 3$ kW$\\$3. Ratio of Power Outputs:$\\$The ratio of power outputs is $\frac{P_{series}}{P_{parallel}}. \\$Substitute the values:$\\ P_{series} = \frac{V^2}{\frac{V_A^2}{1} + \frac{V_B^2}{2}} \\ P_{series} = \frac{V^2}{\frac{V_A^2}{1} + \frac{V_B^2}{2}} \\ P_{series} = \frac{V^2}{\frac{V^2}{1} + \frac{V^2}{2}} \\ P_{series} = \frac{V^2}{\frac{3V^2}{2}} \\ P_{series} = \frac{2}{3}$ kW$\\$The ratio of power outputs is:$\\ \frac{P_{series}}{P_{parallel}} = \frac{\frac{2}{3}}{3} = \frac{2}{9} \\$Therefore, the ratio of power outputs for the series and parallel cases is $2:9. \\$This corresponds to Option 4.