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 ' 10 - 19 C) ?

1.11*10¹⁹ electrons

Explanation:

The formula for electric energy is given as,

E = QV ... Equation 1

Where E = Electric Energy, Q = charge, V = Voltage.

Make Q the subject of the equation

Q = E/V ... Equation 2

Given: E = 8 W, V = 4.5 V.

Substitute into equation 2

Q = 8/4.5

Q = 1.778 C

using,

n = Q/e ... Equation 3

Where n = number of electron, e = Charge of one electron.

Given: e = 1.6*10⁻¹⁹ C

Substitute into equation 3

n = 1.778/1.6*10⁻¹⁹

n = 1.11*10¹⁹ electrons