It says in the question that the current is 0 prior to the switch opening. Is that what you mean

It isn't what I meant, but it is perfectly valid to use that explicit condition (I forgot it was there). If that statement hadn't been there, the answer would have been along the lines that, provided the switch had been closed "long enough", that all of the current would go through the wire since no current could go through the resistor (the wire forces a condition of zero volts across both the resistor and the inductor). Although, having just said that, it turns out that the statement giving that explicit condition does actually need to be there because there would be nothing preventing some or all of the current flowing through the (ideal) inductor even with the switch closed.

Okay, so we have the current in the inductor just before the switch opens. But that does not, by itself, tell us what the current will be in the inductor just after the switch opens. So what will that current be and why?

 

B + (A - B)e^-t/tau

Yep. And I think you now see how that general form comes about. Good job.

The way I think of it is that we have a term that gives us the final value, B, and then a term that turns on at t=0 and that term needs a coefficient that turns on the initial value (A) and turns off the final value (-B).

 

Yep.

Now, what if we wanted the initial value to be A (not A+B) but the final value to remain B?

Ok i think something just clicked. In the question it says relating to the current source. So instead of I(∞), replace that with I1 = 4A, so I(t) = 4 + (I(0)=0 - 4)e^-t/tau where tau = R/L = 4/20 = 1/5
I(t) = 4 - 4e^(-L/R(t)) = 4 - 4e^(-5t)?

 

Ok i think something just clicked. In the question it says relating to the current source. So instead of I(∞), replace that with I1 = 4A, so I(t) = 4 + (I(0)=0 - 4)e^-t/tau where tau = R/L = 4/20 = 1/5
I(t) = 4 - 4e^(-L/R(t)) = 4 - 4e^(-5t)?

Close, but I'm going to yell at you (gently) about two things.

First, you haven't given any basis for replacing I(∞) with I1=4A. I could make a tiny modification to that circuit, such as putting a second resistor in series with the inductor, and the final inductor current would only be a fraction of the current source output. So you need to justify why 4A is the final current in the inductor. Hint: Whatever the final current level is in the inductor, is it changing? If not, then what is the voltage across the inductor? Given this, what is the voltage across the resistor? Given this, what is the current in the resistor? Given this, what must the current in the inductor be? A similar chain of reasoning would let you find the final inductor current even if I put that second resistor in there.

The second thing I'm going to yell at you about is units. Anyone that has seen more than a few of my posts already knows that I am a Units Nazi. People can and do die because engineers can't be bothered to properly track their units. Tracking units is perhaps the single most powerful error detection tool available to the engineer and to not avail yourself of it at all times is, in my opinion, tantamount to criminal negligence (and, at times, courts and juries have agreed in wrongful death and similar cases).

We all make mistakes when working through the math and, most of the time, those mistakes mess up the units. If we are tracking the units, we can catch this right away and quickly fix it. If, instead as is so commonly the case, people ignore units entirely and then just tack onto the final result the units they want the answer to have, we completely abdicate our moral, ethical, and legal responsibility to exercise due diligence when performing our jobs. Sadly, few engineers are taught the power of tracking units and even fewer are ever expected to actually track their units. I believe that is because most instructors and professors and textbook authors have an exclusively academic background and have little to no real-world engineering experience; for them, a missed mistake costs some points on an assignment or perhaps some embarrassment in a publication. In the real world, it means space probes crash into planets, airliners run out of fuel, and numerous other incidents some of which are fatal.

So how does this apply to what you have done here? Well, let's see.

\(
I(t) = I_f \; + \; \( I_i \; - \; I_f\)e^{-\frac{t}{\tau}}
\)

Now, you've said that

\(
\tau \; = \; \frac{R}{L}
\)

resulting in

\(
I(t) = I_f \; + \; \( I_i \; - \; I_f\)e^{-\frac{L}{R}t}
\)

Exponents, in general, have to be dimensionless. More particularly, the arguments to ANY transcendental function, which includes trigonometric and logarithmic/exponential functions, must be dimensionless. This means that

\(
-\frac{L}{R}t
\)

Has to be dimensionless. Is it? Let's see.

\(
-\frac{L}{R}t \; = \; -\frac{20H}{4\Omega}t
\)

What are the units on henries and ohms. We can get these from the constitutive equations for each:

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

L has units of inductance, (H), v has units of voltage, (V), di has units of current, (A), and dt has units of time, (s).

\(
1V \; = \; 1 H \frac { \( 1A \) } { \( 1s \) }
\)

So L has units of

\(
1 H \; = \; 1\frac{Vs}{A}
\)

Similarly, from Ohm's Law (V=IR), resistance has units of

\(
1\Omega \; = \; 1\frac{V}{A}
\)

Plugging these back into our exponent we have

\(
-\frac{L}{R}t \; = \; -\frac{20 \frac{Vs}{A} } {4 \frac{V}{A} } t \; = \; (5s)t
\)

Since t has units of time, this has units of time-squared and is therefore not dimensionless. Thus, we KNOW that this answer is wrong and there is no point going any further until we sort it out.

With the help of the units, we can almost immediately spot the problem. You know that tau is a time constant and has to have units of time. What are the units on your time constant, namely R/L?

You are getting a time constant of 0.2s when the actual time constant is 5s, a factor of 25 different. That can easily be enough to cause a design to fail, possibly catastrophically and dangerously.

Even if you had calculated the time constant correctly, it is very common and easy to multiply the variable 't' by the time constant instead of dividing by it. I still do it on an all-too-regular basis. But if the units are there, you will catch this immediately. It also removes all ambiguity as to whether the values of 't' should be in seconds or milliseconds or minutes or hours or whatever. They have to be in whatever units ultimately make the exponent dimensionless.

Always, always, ALWAYS track your units. Sadly, that often means that you have to spend a bit of time up front filling in the units where the instructor or author couldn't be bothered to do their job properly. But it will mean far fewer mistakes get through and, from time to time, you may well catch errors made by them. I've found errors in textbooks this way as well as in data sheets and in the work of customers and colleagues. Errors that would have gone uncaught for who knows how long if I hadn't taken the time to go back and properly add in the units into someone else's work.

For completeness, in addition to ALWAYS tracking your units -- ALWAYS, not some of the time, but ALWAYS -- the other most powerful error detection tool is simply asking if the answer makes sense. That's why I wanted you to look at the circuit and determine what the initial and final currents have to be.