Final answer:
In a series circuit, when a lightbulb is missing or burned out, all other bulbs will also go out due to an open circuit in older strings or will continue to light at a slightly higher voltage in newer strings with shunts. Incandescent lightbulbs dim due to increased resistance just before their filaments break.
Step-by-step explanation:
When a lightbulb is missing or broken in a series circuit, it behaves similarly to an open switch, breaking the circuit and preventing current from flowing. As a result, all other lightbulbs in the circuit will not light up. This is because in a series circuit, the electrical path is singular and all components share the same current flow. If one bulb in an old string of lights that operates on 120 volts with 40 identical bulbs burns out, the entire string goes dark as the circuit is open. Each bulb would normally operate at 3 volts (120 volts divided by 40 bulbs).
Newer versions of holiday lights are designed with shunts that create a short circuit when a bulb burns out, which allows the current to continue to flow to the remaining bulbs. This means that if one bulb burnt out, the remaining bulbs would still light up. For such a string still operating on 120 volts with now 39 bulbs, each would operate at approximately 3.08 volts (120 volts divided by 39 bulbs).
Finally, incandescent lightbulbs often grow dim just before their filaments break because of increased resistance. As bulbs age, the filament gets thinner and more fragile. This decrease in filament thickness results in a higher resistance, lower current, and thus less light output before the filament eventually breaks and the bulb goes out.