Final answer:
Closing the switch in a circuit with two identical lightbulbs, assuming they are connected in parallel and no other significant changes happen to the circuit, is likely to maintain the intensity of both bulbs at the same level because the voltage across each bulb and the current through each one remains constant. C. intensity of both bulbs increases
Step-by-step explanation:
When the switch in the circuit is closed, and assuming the two lightbulbs are identical and connected in parallel, the intensity of both bulbs will likely increase. This is because closing the switch completes the circuit, allowing current to flow. The lightbulb's intensity is a function of the power consumed by the bulb, which is directly related to the current flowing through it. If the bulbs were previously in a series configuration with a switch that bypasses one of the bulbs when closed, closing the switch would decrease the resistance of the circuit, hence increasing the current and the brightness of both bulbs.
For identical bulbs connected in parallel, the voltage across each bulb is equal to the source voltage. Therefore, when the switch is closed, the voltage across each bulb does not change, and the current through each bulb does not change, assuming there is no significant internal resistance in the source or added resistance in the circuit. The unchanged current means the power consumption and resulting brightness of each bulb also stay the same.
However, it's essential to note that if the question refers to a series circuit, or if closing the switch significantly changes the circuit configuration, the effects on the intensity of the light bulbs could vary. The scenarios described above are based on the assumption that closing the switch does not alter the circuit configuration aside from completing the circuit. Furthermore, the actual effect could also depend on the details of the circuit not provided in the question, such as the presence of additional components or the nature of the power source.