The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is $I=(0.75\mathrm{A})e^{-4[6\mathrm{h})}$. What is the total number of electrons transported from the positive electrode to the negative electrode by the charge escalator from the time the battery is first used until it is completely dead?

The current as a function of time is $I=(0.75\mathrm{A})e^{-4[6\mathrm{h})}$. We have to find the total number of electrons transported from the positive electrode to the negative electrode by the charge escalator from the time the battery is first used until it is completely dead

The dependence of the current on time is $I\left(t\right)=C{e}^{-\frac{t}{\tau }}$
where, $C=0.75\mathrm{A}$and$\tau =6\mathrm{h}=21600\mathrm{s}$.
The total charge by integrating over the time and take the latter running from zero to infinity.

The integral will be,
$Q\left(t\right)=\int C{e}^{-\frac{t}{\tau }}$
Taking the values,$C=0.75\mathrm{A}$And $\tau =21600\mathrm{s}$
$Q_{tot}=0.75A\times 21600s\times {\left[e^{-\frac{t}{T}}\right]}_0^{\mathrm{\infty }}=16200\mathrm{C}$
So, the number of electrons is,
$Q=ne\Rightarrow n=\frac{Q}{e}$
Finally, $n=\frac{16200C}{1.6\times {10}^{-19}}=1.01\times {10}^{23}.$

Therefore, total number of electrons transported from the positive electrode to the negative electrode by the charge escalator from the time the battery is first used until it is completely dead is$1.01\times {10}^{23}$