Final answer:
The current of a lightbulb with a power of 60 W and a resistance of 300 Ω, using the formula I = √(P / R), is approximately 0.4472 A. The given options in the question do not match this value, but the closest option would be 0.2 A, which is not the precise value.
Step-by-step explanation:
To determine the current of a lightbulb with a power rating of 60 W and a resistance of 300 Ω, we can use Ohm’s Law and the power formula combined. The power formula is P = I^2 * R, where P is power in watts, I is current in amps, and R is resistance in ohms.
Rearranging this formula to solve for I, we get I = √(P / R). Substituting in the given values, we get I = √(60 W / 300 Ω). Here, I = √(0.2) which yields a current of approximately 0.4472 A. However, this value doesn’t match any of the multiple-choice options given.
Perhaps there has been a mistake in the options provided or in the interpretation of the question. Normally, when such discrepancies occur, it's essential to revisit the calculations and assumptions, and if they are correct, discuss with the teacher. However, based on the given choices, none of them correspond to the correct calculation. The closest rounded value to the proper calculation would be 0.2 A, which is option a, although it doesn't accurately represent the calculated current.