Now let's consider your lack of using units.

The final current in the inductor is not 5. It is 5 A. Because you didn't include the units, you messed up and said that the current (though you thought it was the initial current) was 5.0 mA. That's an error of three orders of magnitude!

Nearly all physical quantities have units and those units are an integral part of the quantity. It is meaningless to say that someone is 72 tall or that they weigh 100. Those give you zero information about the quantities described. You can't even make comparisons with them. Is someone that is 72 taller or shorter than someone that is 32? You have zero idea.

Beyond that, if you religiously track your units throughout your work, then you will find that you will catch most (not all) of your mistakes shortly after you make them instead of hours later (or not at all). In the academic realm this means you lose points unnecessarily. In the real world it means things get damaged and people get injured and killed unnecessarily.

Sadly, the people that write text books almost universally come from a purely academic background with little to no industry experience. As a result they have the academic mindset and are very sloppy with units (even when they've gone to lengths to say how important they are -- but they really only mean that it's important on the final answer). This attitude then infects the vast majority of students learning from those texts.

Consider the equation you started with (and that I'm guessing was given to you as is)

i = 5.0(1 - e^(-5000t))

This tells us nothing about the magnitude of the current -- is it 5 nA, 5 mA, 5 A, 5 kA? Any of those might be reasonable depending on the specifics.

This tells us nothing about the time scale of the decay -- is t in microseconds, milliseconds, seconds, hours? Again, any of those might be reasonable depending on the specifics.

So we are left with having to guess -- and engineering is NOT about guessing (at least not about this kind of stuff).

I'm guessing (see -- having to guess, which is major-time not good) that this is 5 A and that the time is in seconds.

Since the argument to any transcendental function has to be dimensionless, that means that the 5000 has units of inverse-seconds. Thus this equation should be

\(

i(t) \; = \; 5.0 \, A \, \(1 \; - \; e^{-5000 \, s^{-1} \; t} \)

\)

In addition to being correct, by having the units we gain the option to write things in a manner that is easier for us to comprehend, such as:

\(

i(t) \; = \; 5.0 \, A \, \(1 \; - \; e^{\frac{-t}{200 \, \mu s}} \)

\)

Now we can see immediately that we are dealing with a system that has a 200 μs time constant. If someone tells us that t = 1.2 ms we don't have to guess whether, perhaps, the t in the equation was supposed to be in milliseconds or not. Give me a time in picosecond or minutes and I can unambiguously tell you what the value of the exponent is in that equation.

I can also do the math without having to go to base units, which tend to be either so big or so small that humans are more likely to make mistakes. For instance, if you want the voltage across the inductor, then

\(

v(t) \; = \; L \frac{di(t)}{dt}

v(t) \; = \; 2.0 \, mH \frac{d}{dt}\(5.0 \, A \, \(1 \; - \; e^{\frac{-t}{200 \, \mu s}} \) \)

v(t) \; = \; \( 5.0 \cdot 2.0 \) \, mH \cdot A \frac{d}{dt}\(1 \; - \; e^{\frac{-t}{200 \, \mu s}} \)

v(t) \; = \; 10.0 \, mH \cdot A \( - \(\frac{-1}{200 \, \mu s} \) \) e^{\frac{-t}{200 \, \mu s}}

v(t) \; = \; \frac{10.0}{200} \, \frac{mH \cdot A}{\mu s} \, e^{\frac{-t}{200 \, \mu s}}

v(t) \; = \; \frac{10.0}{200} \, \frac{1000 \, m}{1} \, \frac{mH \cdot A}{\mu s} \, e^{\frac{-t}{200 \, \mu s}}

v(t) \; = \; 50.0 \, \frac{m \cdot mH \cdot A}{\mu s} \, e^{\frac{-t}{200 \, \mu s}}

v(t) \; = \; 50.0 \, \frac{H \cdot A}{s} \, e^{\frac{-t}{200 \, \mu s}}

\)

Note that m·m cancels μ since (1 milli)(1 milli) = (1 micro).

From the constitutive equation for inductance itself we see that

\(

1 \, H \; = \; 1 \, \frac{V \cdot s}{A}

\)

Hence

\(

v(t) \; = \; 50.0 \, \frac{H \cdot A}{s} \, e^{\frac{-t}{200 \, \mu s}}

v(t) \; = \; 50.0 \, \frac{H \cdot A}{s} \, \frac{V \cdot s}{H \cdot A} \, e^{\frac{-t}{200 \, \mu s}}

v(t) \; = \; 50.0 \, V \, e^{\frac{-t}{200 \, \mu s}}

\)

And if the units hadn't worked out to a voltage at the end, we would have known that we had made a mistake somewhere along the way and gone and tracked it down. This is MUCH better than just taking on a V onto the end of the final answer because that is what we want the units to be.

Note that I did the above in much more detail than is normally needed just to show how it all works out.