In old series-wired holiday lights, if one bulb burns out, all go out, each having a normal voltage of 3 V. In newer versions, if a bulb shorts out, the rest stay lit, each with slightly increased voltage of 3.08 V. In series, the 60 W bulb will likely be brighter due to a higher resistance.
Step-by-step explanation:
The question involves understanding series circuits and how they operate when bulbs burn out. In old strings of holiday lights wired in series, if one bulb burns out, it acts like an open switch, causing all other bulbs to go out because the circuit is interrupted. With a 120 V supply and 40 identical bulbs, the normal operating voltage of each bulb is 120 V divided by 40, which is 3 V per bulb.
In contrast, in the newer versions of holiday lights, if a bulb burns out, it creates a short circuit, acting like a closed switch. This allows the current to continue flowing through the other bulbs. If a string operates on 120 V and, after one bulb burns out, has 39 remaining identical bulbs, the new operating voltage of each would be 120 V divided by 39, resulting in approximately 3.08 V per bulb.
When two household lightbulbs rated 60 W and 100 W are connected in series, the brightness of the bulbs depends on the current passing through them. In a series circuit, the current is the same through all components. However, the bulb with the higher resistance, which in this case is likely to be the 60 W bulb, will have a higher voltage drop across it and will be brighter.