\n– The brightness of a bulb (assumed to be resistive) is proportional to the power dissipated, P. Power can be calculated as P = V\u00b2\/R or P = I\u00b2R, where V is the voltage across the bulb, I is the current through the bulb, and R is its resistance. For a given bulb (fixed R), maximum brightness corresponds to maximum voltage or maximum current.
\n– Assuming each battery provides a voltage V:
\n – Diagram (a) (one battery, one bulb): Voltage across bulb = V.
\n – Diagram (b) (two batteries in series, one bulb): The voltages of batteries in series add up. Total voltage = 2V. This voltage is across the bulb.
\n – Diagram (c) (two batteries in parallel, one bulb): Batteries in parallel maintain the same voltage as a single battery (assuming identical ideal batteries). Total voltage = V. This voltage is across the bulb. (Parallel connection increases total current capacity, but not voltage).
\n – Diagram (d) (one battery, two bulbs in series): The voltage V from the battery is divided between the two bulbs. Assuming identical bulbs, voltage across each bulb = V\/2.
\n– Comparing the voltage across the bulb in each case: (a) V, (b) 2V, (c) V, (d) V\/2.
\n– Since the power dissipated (and thus brightness) is proportional to V\u00b2, the bulb will glow brightest when the voltage across it is highest, which is in case (b) with two batteries in series (2V).
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