Final answer:
The brightness of identical light bulbs connected in a circuit depends on whether they are wired in series or parallel. Opening switch S interrupts current flow, affecting the brightness of the bulbs based on their connection relative to the switch. In older series-connected holiday lights, all bulbs go out if one burns out, while in newer versions, the lights remain on courtesy of a shunt.
Step-by-step explanation:
Brightness of Bulbs in Electric Circuits
The question posed involves understanding the functioning of bulbs in an electric circuit, specifically focusing on their brightness, the effect of opening a switch, and the current change through the bulbs. Brightness in bulbs is typically related to the power consumed by the bulb, which is a product of voltage across and current through the bulb. Since the bulbs are identical and connected in a particular arrangement (not detailed here), we can assume that they have the same brightness if connected in parallel or have varied brightness if in series depending on their position in the circuit.
Opening the Switch S and Current Changes
When switch S is opened, the current flow is interrupted. In a series circuit, this would cause all bulbs to go out as the circuit becomes open. In a parallel circuit, only the section of the circuit involving the opened switch would be affected. The current in bulbs not in line with the opened switch would remain unchanged, while bulbs in series with the opened switch would cease to have current flowing through them and would therefore no longer emit light.
In terms of holiday lights, older strings connected in series would have all the lights go out if one bulb burns out, which is akin to an open switch. For such strings at 120 V with 40 bulbs, each bulb normally operates at 3 V (120 V divided by 40 bulbs). Newer versions with a short circuit design would remain lit except for the malfunctioning bulb, and the operating voltage for the remaining bulbs would be slightly higher, that is, 120 V divided by 39 bulbs.