Final answer:
If one bulb in a series circuit burns out and opens the circuit, the other bulbs will go out. For a 120 V circuit with 40 bulbs, each bulb normally operates at 3 V. If a burnt-out bulb short circuits, the operating voltage of each of the remaining 39 bulbs slightly increases to 3.08 V.
Step-by-step explanation:
When bulbs are connected in a series circuit, if one bulb burns out, it disrupts the entire circuit. In the case of holiday lights wired in series, if a bulb that breaks the electrical connection when it burns out (acts like an open switch), all the other bulbs will also go out. Consequently, for a series circuit with 40 identical bulbs operating at 120 V, the normal operating voltage of each bulb would be the total voltage divided by the number of bulbs, which is 3 V per bulb (120 V / 40 bulbs).
For newer versions of holiday lights where bulbs short circuit when they burn out (acting like a closed switch), the total operating voltage would be divided by the remaining bulbs, which means the operating voltage of each of the remaining 39 bulbs would increase slightly. It would be 3.08 V per bulb (120 V / 39 bulbs), assuming all other conditions remain the same and ignoring any change in resistance due to the short circuiting of the burned-out bulb.
In conclusion, if one bulb in a series circuit burns out and acts like an open switch, the other bulbs will go out, too. If a bulb instead short circuits when it burns out, the remaining bulbs will see a slight increase in voltage.