A $$6.0$$ volt battery is connected to two light bulbs as shown in figure. Light bulb 1 has resistance $$3$$ ohm while light bulb 2 has resistance $$6$$ ohm. Battery has negligible internal resistance. Which bulb will glow brighter?

The brightness of a bulb is determined by its electric power consumption $$P = \frac{V^2}{R}$$ when connected in a parallel configuration where the potential difference across each bulb is identical.

Using the parallel circuit condition: $$V_1 = V_2 = V = 6.0\text{ V}$$

Evaluating power consumption for each bulb:

$$P_1 = \frac{V^2}{R_1} = \frac{6^2}{3} = 12\text{ W}$$

$$P_2 = \frac{V^2}{R_2} = \frac{6^2}{6} = 6\text{ W}$$

$$P_1 > P_2 \implies \text{Bulb 1 glows brighter}$$