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

Answer:

$N = 1.2 \times 10^{23}$

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

$q = \int i dt$

$q = \int (0.88 A) e^{-t/6} dt$

$q = 0.88\times -6(3600)\times (e^{-t/6})$

$q = -1.90 \times 10^{4} (0 - 1)$

$q = 1.90 \times 10^{4} C$

As we know that

$q = Ne$

$N = \frac{q}{e}$

$N = \frac{1.90 \times 10^4}{1.6 \times 10^{-19}}$

$N = 1.2 \times 10^{23}$