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{\bf Question}

An AC current $I$, in a circuit with inductance $L$ and resistance
$R$ is given by

$$L\ds\frac{dI}{dt}+RI=E,$$

where $L,\ R$ and $E$ are constant. Find $I$, given that $I=0$
when $t=0$.
\medskip

{\bf Answer}

$L\ds\frac{dI}{dt}+RI+E$

Could do by variables separable method or

$\ds\frac{dI}{dt}+\ds\frac{R}{L}I=\ds\frac{E}{L}$ cf
$\ds\frac{dI}{dt}+PI=Q$

Use $P=\ds\frac{R}{L},\ Q=\ds\frac{E}{L}$.

Thus integrating factor

$R=\exp\left(\ds\int P \,dt\right)=\exp\left(\ds\int\ds\frac{R}{L}
\,dt\right)=e^{\frac{R}{L}t}$

$\Rightarrow
e^{\frac{R}{L}t}\ds\frac{dI}{dt}+\ds\frac{R}{L}e^{\frac{R}{L}t}I
=\ds\frac{E}{L}e^{\frac{R}{L}t}$

$\Rightarrow
\ds\frac{d}{dt}\left(e^{\frac{R}{L}t}I\right)=\ds\frac{E}{L}e^{\frac{R}{L}t}$

$\Rightarrow e^{\frac{R}{L}t}I=\ds\frac{E}{L}e^{\frac{R}{L}t}$

$\Rightarrow
e^{\frac{R}{L}t}I=\ds\frac{E}{L}\ds\frac{L}{R}e^{\frac{R}{L}t}+c$

$\Rightarrow I=\ds\frac{E}{R}+ce^{\frac{R}{L}t}$

where $c$ is constant

Now if $I=0$ when $t=0$

$0=\ds\frac{E}{R}+ce^0 \Rightarrow c=-\ds\frac{E}{R}$

Thus \un{$I=\ds\frac{E}{R}\left(1-e^{\frac{R}{L}t}\right)$}
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