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

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