RL 11-22-2020

We are asked to solve the differential equation:

$\dot{I} + I = \mathrm{e}^{-t} \ ; \ I(0) = 0$ $\implies \mathrm{e}^{t}\dot{I} + \mathrm{e}^{t}I = 1 \implies \frac{d}{dt}\left(\mathrm{e}^{t} I\right) = 1 \implies I = (t+c)\mathrm{e}^{-t}$ $\because I(0) = 0 \implies c = 0 \implies I = t\mathrm{e}^{-t}$

To compute the maximum value of current:

$\dot{I} = (1-t)\mathrm{e}^{-t}=0$ $\implies t = 1$

The second derivative check shows that:

$\ddot{I} (1) = -\frac{1}{\mathrm{e}} <0$

Therefore, $t=1$ corresponds to a maxima. Therefore, the maximum current occurs at $t=1$ and its value is:

$\boxed{I_{\mathrm{max}} = I(1) = \frac{1}{\mathrm{e}}}$

Let's take a Laplace Transform approach on this RL circuit. Per KVL, if the circuit be represented by:

$Li'(t) + Ri(t) = V_{S}(t)$ ,

or $i'(t) + i(t) = e^{-t}, i(0)=0$

then the Laplace Transform yields:

$[sI(s) - i(0)] + I(s) = \frac{1}{s+1}$ ;

or $(s+1)I(s) = \frac{1}{s+1};$

or $I(s) = \frac{1}{(s+1)^2};$

or $i(t) = te^{-t}$ . Taking the first derivative of $i(t)$ equal to zero gives:

$i'(t) = (1-t)e^{-t} = 0 \Rightarrow t = 1$

and the second derivative at $t=1$ gives:

$i''(t) = (t-2)e^{-t} \Rightarrow i''(1) = -e^{-1} < 0$ (hence, a maximum over $t \ge 0$ ).

Thus, $\boxed{i_{MAX} = i(1) = \frac{1}{e}}$ amps.