Six similar bulbs are connected as shown in the figure with a DC source of emf E, and zero internal resistance. The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section A and one from section B are glowing, will be:

Let resistance of each bulb be R. Initially, 3 in parallel in series with 3 in parallel, equivalent resistance \( R_{eq1} = \frac{R}{3} + \frac{R}{3} = \frac{2R}{3} \). Power \( P_1 = \frac{E^2}{2R/3} = \frac{3E^2}{2R} \). In second case, 2 in parallel in section A and 1 in section B, \( R_{eq2} = \frac{R}{2} + R = \frac{3R}{2} \). Power \( P_2 = \frac{E^2}{3R/2} = \frac{2E^2}{3R} \). Ratio \( P_1 : P_2 = 9 : 4 \).