To solve this problem, we need to analyze the circuit in two scenarios:$\\$(i) All bulbs are glowing.$\\$(ii) Two bulbs from section A and one from section B are glowing.$\\$Let's denote the resistance of each bulb as $R. \\$Scenario (i): All bulbs are glowing.$\\$• There are two parallel branches, each with three bulbs in series.$\\$• The equivalent resistance of each branch is $3R. \\$• The total equivalent resistance $R_{eq1}$ is given by:$\\ R_{eq1} = \frac{3R \cdot 3R}{3R + 3R} = \frac{9R^2}{6R} = \frac{3R}{2}. \\$• The power consumed $P_1$ is given by:$\\ P_1 = \frac{E^2}{R_{eq1}} = \frac{E^2}{\frac{3R}{2}} = \frac{2E^2}{3R}. \\$Scenario (ii): Two bulbs from section A and one from section B are glowing.$\\$• In section A, two bulbs are in series, so their equivalent resistance is $2R. \\$• In section B, one bulb is glowing, so its resistance is $R. \\$• These two branches are in parallel, so the total equivalent resistance $R_{eq2}$ is:$\\ R_{eq2} = \frac{2R \cdot R}{2R + R} = \frac{2R^2}{3R} = \frac{2R}{3}. \\$• The power consumed $P_2$ is given by:$\\ P_2 = \frac{E^2}{R_{eq2}} = \frac{E^2}{\frac{2R}{3}} = \frac{3E^2}{2R}. \\$Now, we find the ratio of power consumption:$\\ \frac{P_1}{P_2} = \frac{\frac{2E^2}{3R}}{\frac{3E^2}{2R}} = \frac{2}{3} \cdot \frac{2R}{3E^2} \cdot \frac{3E^2}{2R} = \frac{4}{9}. \\$Therefore, the ratio of power consumption is $4:9. \\$This corresponds to Option 1.