Power In Series And Parallel Bulbs

Each bulb's resistance follows from its rating at the common voltage through (R = V^2/P), so the lower-wattage bulb has the larger resistance: the 40 W bulb's resistance is (V^2/40), which is 2.5 times that of the 100 W bulb's (V^2/100). In a series connection both bulbs carry the same current (I), so the power each actually dissipates is (P = I^2 R), which is proportional to resistance. The higher-resistance 40 W bulb therefore dissipates more power and glows brighter. The second option is wrong because the wattage rating applies only at rated voltage, not in this series arrangement. The third option is wrong because equal current with unequal resistance gives unequal power. The fourth option misstates the cause, since the series current is identical for both bulbs. This is a classic JEE conceptual result combining (P = V^2/R) for ratings and (P = I^2R) for the actual circuit. A plausibility check confirms it: in series the shared current makes power track resistance, and the dimmer-rated bulb, being more resistive, paradoxically outshines its higher-rated partner. The conclusion neatly reverses in a parallel connection, where both bulbs experience the full supply voltage and (P = V^2/R) once again applies, so the 100 W bulb then glows brighter; recognising which formula is appropriate to the connection is the central insight this problem is designed to test.