Final answer:
In a series circuit with a 60-W and a 100-W bulb connected to a 120-V outlet, the 60-W bulb will be brighter because it has a higher resistance, leading to a greater voltage drop across it and hence more power dissipation.
Step-by-step explanation:
When two light bulbs with different power ratings, a 60-W bulb and a 100-W bulb, are connected in series to a 120-V outlet, the bulb with the higher resistance will draw less current. Power (P) is related to voltage (V) and current (I) by the equation P = VI. In a series circuit, the current flowing through all components is the same. However, for a given power rating, the resistance (R) can be calculated using P = V^2/R, which implies R = V^2/P. Therefore, the 60-W bulb has a higher resistance than the 100-W bulb because it is designed to draw less current to operate at its rated power when connected to the same voltage.
The brightness of the bulb in a series circuit will depend on the power dissipation, which is a product of the current flowing through the bulb and the voltage drop across it. In this case, both bulbs get the same current, but due to the different resistances, the voltage drop across them will be different. The bulb with more resistance (60-W) will have a greater voltage drop across it and will thus dissipate more power, making it brighter. Consequently, when connected in series to a 120-V outlet, the 60-W light bulb will be brighter than the 100-W bulb because it draws more current due to its lower resistance and will have a greater voltage drop across it, leading to more power dissipation.