Final answer:
The brightness of identical bulbs in a circuit depends on their configuration and Ohm's law. Before opening switch S, bulbs A and B are equally bright and C maintains equal brightness. After opening S, A goes out and B remains lit with unchanged brightness, while C stays the same.
Step-by-step explanation:
Understanding Bulb Brightness in an Electric Circuit
When assessing the brightness of bulbs in an electric circuit, it's important to consider the circuit configuration and Ohm's law. Ohm's law states that the current (I) through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor; the formula is I = V/R. In a circuit with identical light bulbs, bulb A, which is in parallel with bulb B and switch S, would initially have the same brightness as bulb B since they share the same voltage and have identical resistances. Bulb C, being in series with the parallel combination of A and B, would also have the same brightness in this scenario.
After opening switch S, bulb A is removed from the circuit and only bulb B would be lit, whereas bulb C would remain unchanged in terms of brightness. Regarding current change, bulb A would have no current flowing through it once switch S is opened, whereas the current in bulb B would remain the same as it was before the switch was opened, as it's in series with bulb C and the power source.
It is essential to understand that brightness in incandescent light bulbs depends on the power dissipated in the bulb, typically represented by the formula P = I²R, where P is the power, I is the current, and R is the resistance. Therefore, changes in current due to circuit alterations will affect the brightness of the bulbs. Assuming the two cells in the flashlight schematic are the only sources of voltage and that they are in series with the bulb, the total electromotive force (emf) provided to the circuit will be the sum of the individual emfs of the two cells, and the current flowing through the circuit (and therefore the bulb) would be calculated by adding those emfs and dividing by the total resistance (sum of internal resistances and bulb resistance).