Question

The charge that passes through the filament of a certain light bulb in 5.00 s is 3.0 C. a. What is the current in the light bulb? b. How many electrons pass through the filament of the light bulb in a time interval of 1.0 min?

Answer 1

Answer:

(a) I = 0.6 A

(b) 22.5 * 10^19 electrons

Explanation:

Parameters given:

(a) Current is given as the time Tate of charge flow. Mathematically,

I = Q/t

Charge, Q = 3 C

Time, t = 5 secs

I = 3/5 = 0.6A

(b) The current flowing through the filament is 0.6A. In 1 min (60 secs), the charge will be:

Q = I*t

Q = 0.6 * 60 = 36 C

An electron has a charge, Qe, of 1.6022 * 10^(-19) C. Hence, the number of electrons will be:

N = Q/Qe

N = 36/(1.6022 * 10^(-19))

N = 22.5 * 10^19 electrons

Answer 2

The current in the bulb is 0.60 ampere (A). The number of electrons (n) to pass through the filament of the light bulb is $\mathbf{ 2.25 \times 10^{20} \ electrons}$

The current passing through a light bulb can be determined by using the  formula:

$\mathbf{I = \dfrac{\Delta Q}{\Delta t}}$

Given that:

• Charge ΔQ =  3.00 C
• time Δt = 5.00 s

$\mathbf{I = \dfrac{3.0 C}{5.00 \ s}}$

I = 0.60 A

Recall that:

Q = number of electron (n) × charge on the electron (e)

If we replace the value of Q into $\mathbf{I = \dfrac{\Delta Q}{\Delta t}}$; we get:

$\mathbf{I = \dfrac{n\times e}{\Delta t}}$

Making, the number of electrons (n) the subject of the formula, we have:

$\mathbf{n = \dfrac{I \times \Delta t}{e}}$

where;

• e = 1.60 × 10⁻¹⁹ C

$\mathbf{n = \dfrac{0.6 \ A \times 60 \ sec}{1.60 \times 10^{-19} \ C}}$

$\mathbf{n = 2.25 \times 10^{20} \ electrons}$

Therefore, we can conclude that the number of electrons (n) to pass through the filament of the light bulb is $\mathbf{ 2.25 \times 10^{20} \ electrons}$

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