A flashlight bulb operating at a voltage of 4.5 V has a resistance of 8.0 W.How many electrons pass through the bulb filament per second (e = 1.6 x 10^-19 C)? A)3.7 x 10¹⁶ B)1.8 x 10²¹ C)9.4 x 10¹⁷ D)3.5 x 10¹⁸

Question

Quizplus · Accepted Answer

To calculate the number of electrons passing through the bulb filament per second, we need to use the formula:

I = V/R

where I is the current flowing through the bulb filament.

Substituting the given values, we get:

I = 4.5 V / 8.0 Ω

I = 0.5625 A

Now, using the formula:

n = I / q

where n is the number of electrons passing through the bulb filament per second, and q is the charge on an electron (1.6 x 10^-19 C).

Substituting the given values, we get:

n = 0.5625 A / 1.6 x 10^-19 C

n = 3.5 x 10^18 electrons per second

Therefore, the correct answer is D) 3.5 x 10^18.

A Flashlight Bulb Operating at a Voltage of 4