I = 3 A Example 17.1 Turn on the Light Goal Apply the concept of current. Proble

Question

I = 3 A
Example 17.1 Turn on the Light Goal Apply the concept of current.

Problem The amount of charge that passes through a filament of a certain lightbulb in 1.79 s is 1.67 C.

(a) Find the current in the bulb.

(b) Find the number of electrons that pass through the filament in 5.28 s. Strategy Substitute into Equation 17.1 for part (a), then multiply the answer by the time given in part (b) to get the total charge that passes in that time. The total charge equals the number N of electrons going through the circuit times the charge per electron.
Solution (a) Compute the current in the lightbulb. Substitute the charge and time into Equation 17.1. = 1 A
(b) Find the number of electrons passing through the filament in 5.28 s. The total number of electrons times the charge per electron equals the total charge, I?t. (1) Nq = I ?t Substitute and solve for N. N(1.60 10-19 C/electron) = I (5.28 s)
N = 2 electrons

Remarks In developing the solution, it was important to use units to ensure the correctness of equations such as Equation (1). Notice the enormous number of electrons passing through a given point in a typical circuit.

Explanation / Answer

this problem can be solved very easily if you are able to keep track of your units

you are given:
(6.93 * 10^ 21) / 2.5 (electrons/minute) = 2.772 * 10^21 (electrons/minute)

you simply need to convert to amperes, or (C/second)
the factors you need are given by:
1 (minute) = 60 (seconds)
1 (electron) = 1.60 * 10^-19 (C)

[2.772 *10^21 (electrons/minute)]
* [ 1/60 (minute/second)]
* [1.60 * 10^-19 (C/electron)]
= 7.392 (C / second) = 7.392 A notice how the units canceled out

you might need to drop the last digit if you are interested in significant figures

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