Final answer:
In a series circuit, when a switch is opened, it interrupts the circuit and the light bulb turns off. For holiday lights wired in series, the burnout of one bulb either turns off all others or slightly increases the voltage for the remaining bulbs, depending on the design of the circuit.
Step-by-step explanation:
When a switch in a series circuit is opened, the electrical path is interrupted and current cannot flow to the light bulb. Consequently, the light bulb that the electricity is flowing to goes off (Option B). In a series circuit, all components share the same current path. Therefore, if one component such as a switch opens the circuit, the flow of electricity stops, and any light bulbs connected in series will turn off.
If we are talking about holiday lights wired in series, when one bulb that operates like an open switch when it burns out, all other lights will go out as well. For a series string operating on 120 V with 40 identical bulbs, under normal conditions, each bulb would have an operating voltage of 120 V divided by 40, which results in 3 V per bulb. However, if one bulb short circuits (in newer versions), current continues to flow, and the remaining bulbs would receive a slightly higher voltage than normal because the total resistance of the circuit decreases, which in this case would be 120 V divided by 39 bulbs, around 3.08 V per bulb.
For the scenario where two different rated household light bulbs are connected in series, the bulb with the higher resistance (typically the one with lower wattage rating, in this case, the 60 W bulb) will be brighter because it will dissipate more power as heat and light than the one with lower resistance.