Answer :
Final answer:
In a series circuit with a 220 V supply, the 100 W bulb will be brighter than the 60 W bulb due to its higher resistance, leading to greater power dissipation (P=I^2R). The 60 W bulb's lower resistance results in lower power dissipation, thus being dimmer.
Explanation:
When comparing light bulbs of different power ratings connected in series, it's crucial to understand that series circuits have the same current flowing through all components. Therefore, the brightness of a bulb is dependent on its power dissipation, which is the product of the square of the current and the resistance of the bulb (P=I^2R). Given the same current through each bulb in a series circuit, the bulb with the highest resistance will have the highest power dissipation and thus be brighter.
The 60 W bulb will have a lower resistance than the 100 W bulb because power is proportional to the reciprocal of the resistance when the voltage is constant (P = V^2/R). Consequently, when connected in series to a 220 V supply, the 60 W bulb will have less resistance, lower power dissipation, and therefore, it will be dimmer than the 100 W.
Illumination in this scenario also relates to how voltage, current, and resistance interact according to Ohm's law (V=IR) and the power equation (P=IV). Therefore, when working with series circuits, it is crucial to consider the intrinsic properties of the components to understand the overall behavior of the system.