Final answer:
If batteries are connected in series, the brightness of identical bulbs will generally increase as the total voltage across the circuit increases. In a series circuit like holiday lights, if a bulb using an old version that breaks the circuit burns out, all lights go out. In newer versions, the rest of the bulbs remain lit because they short circuit, bypassing the burnt-out bulb.
Step-by-step explanation:
When batteries are connected in series, the total voltage of the system increases because the voltages of individual batteries are added together. If we have identical bulbs and the batteries are connected in series, the brightness of the bulbs would generally increase, given that the bulbs are designed to operate safely at the increased voltage without burning out. This is due to the fact that a greater voltage across the bulb leads to an increased electric current, assuming the resistance of the bulb remains constant, and thus more power is dissipated by the bulb, making it brighter.
For example, if you have a flashlight with two AAA batteries connected in series and you add another AAA battery, the voltage supplied to the bulb increases, and so does its brightness. In the case of holiday lights wired in series, if an old version of the bulb (one that acts like an open switch when it burns out) is used and one bulb burns out, all of the bulbs go out since the circuit is broken. However, newer versions of bulbs (those which act like a closed switch) will continue to allow the other bulbs to light up even if one burns out, though the voltage across each of the remaining bulbs would slightly increase due to the lower total resistance in the circuit.