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)?

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Answer:

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

The number of electrons that should be pass through the bulb fi-la-ment per second is 3.5*10^18.

ohm law:

Since the voltage = 4.5 V

And, the resistance is = 8.0 W

So here we apply the ohm law

V = iR

i = V/R

= 4.5/8

= 0.56

Now t = 1 seconds

So, i = q/t

q/t = 0.56

q = 0.56 C

Now the electrons should be

= 0.56/1.6*10^-19

= 3.5*10^18

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