Final answer:
Approximately 3.75×1019 electrons pass through the light bulb each second when a 6.00 A current is running through a 12-gauge copper wire.
Step-by-step explanation:
The question is asking how many electrons pass through a light bulb each second when a current of 6.00 A flows through a 12-gauge copper wire of 2.05 mm diameter, given that copper has 8.5×1028 free electrons per cubic meter. To find the answer, we apply the relationship between current (I), charge (q), and time (t), which is I = q/t. Here, q is the charge passing through the light bulb in coulombs, and since electrons each have a charge of approximately -1.602×10-19 coulombs, we calculate q by multiplying the number of electrons by this charge value.
First, we find the total charge passing through the light bulb each second by multiplying the current by the time (which is 1 second in this case): q = I×t = 6.00 A×1 s = 6.00 C. Now we can find the number of electrons by dividing the total charge by the charge of an individual electron: Number of electrons = q/e = 6.00 C / 1.602×10-19 C/electron ≈ 3.75×1019 electrons.
Thus, approximately 3.75×1019 electrons pass through the light bulb each second when a 6.00 A current flows through the wire.