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?
a. 3.7 x 10^18
b. 5.3 x 10^23
c. 4.4 x 10^22
d. 1.6 x 10^19
e. 1.2 x 10^23

Answer 1

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

$I(t)=0.88e^{-t/6\times3600s}$

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

$q=\int\limits {I} \, dt$

Here the limits of integration is from 0 to infinite. So,

$q=\int\limits {0.88e^{-t/6\times3600s}}\, dt$

$q=0.88\times(-6\times3600)(0-1)$

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

$N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}$

N = 1.2 x 10²³