Complete 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?
Answer:
The total number of electrons transported is [tex]n=2.54*10^{14}\ electrons[/tex]
Explanation:
Charge is generally expressed as
[tex]q = \int\limits^{\infty}_0 {I} \, dt[/tex]
Where I is the current and it is given as
[tex]I = (0.88A) e^{{ -t *6 hr} } = (0.88A) e^{-21600t} }[/tex]
Substituting into equation above d
[tex]q = \int\limits^{\infty}_0 {0.88 e^{-21600t}} \, dt[/tex]
[tex]q = 0.88[-\frac{e^{-21600t}}{21600} ]\left \ {{\infty} \atop {0}} \right.[/tex]
[tex]q = - \frac{0.88}{21600} [-1][/tex]
[tex]q = 4.074*10^{-5}C[/tex]
This charge can also be expressed as
[tex]q = n *e[/tex]
Where is the number of electron
Making n the subject
[tex]n = \frac{q}{e}[/tex]
[tex]= \frac{4.074*10^{-5}}{1.60*10^{-19}}[/tex]
[tex]=2.54*10^{14}\ electrons[/tex]
