Final answer:
If the circuit described has bulbs in parallel, opening the switch at B would result in only bulb B turning off while bulbs A and C remain lit. The brightness of the remaining bulbs would not change and the current in bulb B would drop to zero.
Step-by-step explanation:
When the switch at B is opened in a circuit with three identical light bulbs A, B, and C, the behavior of the bulbs will depend on the configuration of the circuit.
If the bulbs are arranged in a parallel circuit, opening the switch at B will effectively remove bulb B from the circuit while bulbs A and C will continue to be lit, as they are still connected to the power source through separate paths. Therefore, the correct answer is that only bulb B will turn off.
However, if the bulbs are in a series circuit, opening the switch will interrupt the flow of current through the entire circuit, and all the bulbs will turn off. Since the specific wiring configuration is not provided, this information is based on common circuit arrangements and how they operate.
If this were the case of holiday lights wired in series where opening one bulb is like opening a switch, then if any bulb (such as bulb B) were to burn out or be removed, all bulbs would turn off because the electrical connection would be broken.
However, this scenario seems to describe bulbs in parallel, given that all three bulbs are initially lit despite a switch being open.
As for the brightness of bulb A compared to B and C when switch S is closed, all bulbs would typically have similar brightness in a parallel circuit. If switch S is opened, the brightness of the remaining lit bulbs (A and C) should not change significantly, as each bulb receives full voltage from the source.
In this imagined scenario, the current in bulb B would drop to zero when switch S is opened, while the currents in bulbs A and C would remain the same as they operate on separate branches of the parallel circuit.