91Ó°ÊÓ

The total amount of charge that entered a wire at time is $t$.

And is given by the expression$Q=\left(20\mathrm{C}\right)\left(1-e^{-4(20\mathrm{s})}\right)$, where $t$ is in seconds.

The charge as a function of the time is given by
$Q\left(t\right)=C\left(1−e^{−\frac{t}{\mathrm{τ}}}\right)$
The current will be just the time derivative of this expression, which is
$I=\frac{dQ}{dt}$

$=\frac{C}{−\mathrm{τ}}\frac{d}{dt}\left(1−e^{−\frac{t}{\mathrm{τ}}}\right)$

$=\frac{C}{\mathrm{τ}}{e}^{−\frac{t}{\mathrm{τ}}}$

Hence, the current is calculated as,

The maximal value of the current is achieved when the portion underneath the brackets is unity. It will occur because of $t→∞$.

The maximal value of the current is reached when the part under the brackets is unity, which means that the maximal current is $I_{\max }=10\mathrm{A}$.

Expect that this will happen when $t→∞$.

Therefore, the maximum value of the current is$10A$

Graph the interval $0≤t≤10s$,$I$versus $t$, will show the ensuing linear relation.

Conspiring this process with a graph will show the ensuing linear relation:

Current can be calculated by: $I=(10\mathrm{A})e^{\frac{-t}{2.0\mathrm{s}}}$

The graphical representation is,

Graph for the interval $0≤t≤10\mathrm{s}$is