Final answer:
In a parallel circuit, if a lightbulb goes missing or breaks, other bulbs continue to function, but in a series circuit, if one bulb breaks, it can cause the entire string to go dark. In a string of 40 bulbs at 120 V, each bulb normally gets 3V. With shunts in modern strings, a bulb can burn out without affecting the others, but this fails if too many bulbs burn out.
Step-by-step explanation:
When it comes to holiday lights, parallel circuits behave differently from series circuits. If a lightbulb goes missing or breaks in a parallel circuit, the rest of the bulbs will continue to function normally as each bulb has its own path to the voltage source. However, in a series circuit, if one bulb breaks, it can disrupt the entire circuit.
In a series circuit, bulbs that behave like an open switch when they burn out will cause all other bulbs in the series to go dark because the electrical connection is broken. For a string of 40 identical bulbs operating on 120 V, the normal operating voltage of each would be 120V/40 = 3V per bulb. If a bulb is designed to short circuit when it burns out, current will continue to flow through the other bulbs, and with one less bulb, the operating voltage for the remaining 39 bulbs would be 120V/39, which results in a slightly higher voltage per bulb than before.
Additionally, some holiday lights use a shunt that allows current to flow around a burnt-out bulb, but if too many bulbs burn out, even this mechanism can fail. Also, an individual lamp's brightness in a series circuit can depend on the power rating of the bulbs connected, with the higher power rated bulb typically glowing dimmer due to the voltage drop across the other bulb in the series.