Question

The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is: I = ( 0.600 A)e^(-t/6hr).

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 (expressed as a number of electrons)?

Answer 1

Concepts and reason

The concepts required to solve this question are charge and current.

First, determine the total number of charge flowing in the circuit in the entire time.

Finally, find the total number of electrons by dividing the total charge by charge on one electron.

Fundamentals

The current flowing through a circuit is equal to the rate at which the charge is flowing through the circuit.

The relation between the current (I) and the charge (Q) flowing in a circuit is as follows:

$I = \frac{{dQ}}{{dt}}$

The number of electrons (n) contained in a given amount of charge (Q) can be determined as follows:

$n = \frac{Q}{e}$

Here, e is the charge of one electron, Q is the charge, and n is the number of electrons.

The relation between the current (I) and the charge (Q) flowing in a circuit is as follows:

$I = \frac{{dQ}}{{dt}}$

Rearrange the above expression for Q.

$dQ = Idt$

Integrate the above expression from 0 to $\infty$.

$\begin{array}{c}\\\int_0^\infty {dQ} = \int_0^\infty {Idt} \\\\Q = \int_0^\infty {Idt} \\\end{array}$

Substitute $\left( {0.600{\rm{ A}}} \right){e^{ - \left( {\frac{t}{{6{\rm{ h}}}}} \right)}}$ in the above expression.

$\begin{array}{c}\\Q = \int_0^\infty {\left( {0.600{\rm{ A}}} \right){e^{ - \left( {\frac{t}{{6{\rm{ h}}}}} \right)\left( {\frac{{{\rm{1 h}}}}{{3600{\rm{ s}}}}} \right)}}dt} \\\\ = - \left( 6 \right)\left( {3600{\rm{ s}}} \right)\left( {0.600{\rm{ A}}} \right)\left[ {{e^{ - \left( {\frac{t}{{6{\rm{ h}}}}} \right)\left( {\frac{{{\rm{1 h}}}}{{3600{\rm{ s}}}}} \right)}}} \right]_0^\infty \\\\ = \left( { - 12960{\rm{ C}}} \right)\left( {{e^{ - \infty }} - {e^0}} \right)\\\\ = 1.296 \times {10^4}{\rm{ C}}\\\end{array}$

The number of electrons (n) contained in a given amount of charge (Q) can be determined as follows:

$Q = ne$

Substitute $1.296 \times {10^4}{\rm{ C}}$ for Q and $1.6 \times {10^{ - 19}}{\rm{ C}}$ for e in the above expression.

$\begin{array}{c}\\Q = ne\\\\n = \frac{Q}{e}\\\\n = \frac{{1.296 \times {{10}^4}{\rm{ C}}}}{{1.6 \times {{10}^{ - 19}}{\rm{ C}}}}\\\\ = 8.1 \times {10^{22}}{\rm{ electrons}}\\\end{array}$

Ans:

The number of electrons transported are $8.1 \times {10^{22}}{\rm{ electrons}}$.