Question

The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons transported from the positive electrode to the negative electrode from the time the battery is first used until it is essentially dead?

3.7 x 10^18
5.3 x 10^23
4.4 x 10^22
1.6 x 10^19
1.2 x 10^23

Answer 1

`N = 1.2 * 10^(23)`

Step-by-step explanation:

As we know that the current through the battery is given as

`i = (0.88 A) e^(-t/6)`

here from above equation we know that current will become zero when time elapsed is very large

so here we can say that charge will flow through the battery from t = 0 to t = infinite

now we have

`q = \int i dt`

`q = \int (0.88 A) e^(-t/6) dt`

`q = 0.88* -6(3600)* (e^(-t/6))`

`q = -1.90 * 10^(4) (0 - 1)`

`q = 1.90 * 10^(4) C`

As we know that

`q = Ne`

`N = (q)/(e)`

`N = (1.90 * 10^4)/(1.6 * 10^(-19))`

`N = 1.2 * 10^(23)`