Electronic – Value of resistor and the current in branch with the capacitor

Slim Intro to Differentials

Take a falling rock and imagine that it starts in your hand with zero speed and you release it over the edge of a high cliff. Before it hits the bottom, how far will the rock fall given any particular amount of time after it was released?

In algebra, you readily learn how to solve distance (\$x\$), time (\$t\$), and speed (\$v\$) problems with a simple relationship:

$$x=v\cdot t$$

But the above equation doesn't really help in the falling-rock case above. That's because the speed is continually changing as it falls. The above equation is great if you already know the average speed and the time over which that average was computed. But then, you already must know the distance, by then, too. So the whole thing presumes an answer you don't have. This is a failing of algebra. It works great when you already have averages. But it works more poorly when everything is always-changing.

You could use the above equation over shorter times, though. So, for example, you could start by saying that \$v=0\$ for the first second and compute \$x_{t=0}=0\$. Then, knowing the idea that \$v=g\cdot t\$ (the speed increases with time, using the constant \$g\$ as a factor to get a value), you could estimate \$x_{t=1}=g\cdot 1\:\text{s}=g\$ for the 2nd second. So by the first two seconds you estimate \$x_\text{total}=x_{t=0}+x_{t=1}\$. You could continue on this way, adding up all the pieces to see where that gets you.

But you already know that by the first second you already got a wrong starting answer. And, in fact, you will always have an estimated difference ever smaller (behind) where the falling rock is actually at. There are lots of other ways to go with this process. You could assume the speed at the end of the period rather than at the beginning of it, or you could work out the average midpoint speed and use that. You could also use a \$\frac12\$ second, instead, to reduce the errors. Or a \$\frac14\$ second. But for all the added work you do, the upshot is that algebra is about averages. And these presume you already know the end of the story.

So what do you do?

The differential viewpoint is to create a new type of variable that algebra doesn't have. This variable can only hold infinitely small values, which below I'll call infinitesimals. An infinitesimal is as \$\frac1{\infty}\$, \$\frac2{\infty}\$, \$\frac{4.5}{\infty}\$, or \$\frac{-\sqrt{2}}{\infty}\$. They are smaller than any possible finite number. All of these values are very, very close to zero but they are not zero. The new kind of variable gets a special "mark" that is used so that anyone else reading your writing will understand the meaning. This mark, today, is to place the letter d in front of the usual variable name.

Before I return to the above falling-rock situation, imagine that you divide up some finite distance, \$x\$, into an infinite number of equally tiny values (infinitesimal fragments of \$x\$) that can be re-added back up to \$x\$. Then each of these tiny values might be called \$\text{d}x\$. All we've really done is take some length and divide it up into extremely small conceptual pieces. Adding them back must result in the original length, so it follows that \$x=\int\text{d}x\$. (I've added a new symbol here, \$\int\$, to indicate a special kind of sum -- an infinite sum -- which you need if you want to get back a finite number from a bunch of infinitely small values.) This just says, "If I break up some length \$x\$ into an infinite number of tiny segments of length \$\text{d}x\$, then when I add them back together I must get back my original length, \$x\$," by definition.

There's something subtle I missed above, but is worth adding now. We should really think of \$\int \text{d}x\$ as adding up a lot of tiny bits back into some distance traveled by adding those bits up. Not as an exact \$x\$. By this, I mean, that all those bits of distance add up to some traveled distance. But it leaves out the place where it all started, \$x_0\$, let's say. So actually, we also need to know \$x_0\$ before we can work out the final value of \$x\$. So in calculus we note that fact by instead writing \$x=\int\text{d}x+x_0\$. This is just another way of saying that if we start at \$x_0\$ and then advance from there through the infinite sum of infinitely (but not zero) many tiny distance motions away from \$x_0\$, then the ending place will be the sum of the starting point plus all the tiny distances added up.

We can now re-phase the falling-rock question this way:

$$ \text{d} x=v_{_t}\cdot \text{d} t $$

It's still actually the same equation. But it now reads a little differently. It says that the infinitely small distance traveled by the rock's fall is equal to the speed the rock has, at that moment in time, multiplied by the infinitely small bit of time we've allowed for it to travel.

Note that if you shrink down the time the rock falls to an infinitely small moment, then the velocity of the rock, \$ v_{_t} = g \cdot t \$, is an exact value and no longer an average. We can always get a precision value for \$v_{_t}\$ because of the way gravity works. But we can only get a precision value for a very, very tiny moment of time. The new equation above expresses this, perfectly.

Of course, it's still useless. We can't use it for anything practical. To do that, we need to sum up all those bits. So, to get that we just sum both sides like this:

$$\begin{align*} \int \text{d}x&=\int \left(v_{_t}\cdot \text{d}t\right)\\ \therefore x&=\int \left(v_{_t}\cdot \text{d}t\right)\\ &=\int \left(g\cdot t\cdot \text{d}t\right)\\ &=g\cdot \int \left(t\cdot \text{d}t\right) \end{align*}$$

I was able to pull \$g\$ outside the infinite sum (integral), because it is a constant. And you already know that \$k\cdot a_1+k\cdot a_2 + ... + k \cdot a_n= k \cdot (a_1+a_2+...+a_n)\$. So that's why I was able to do that here.

Also, I've applied a multiplication to the variable representing just an infinitesimal of time. Usually, in textbooks, this multiplication is left out by some unexplained magic. This is because Dedekind and Weierstrass, the authors of what is taught as calculus today, refused to accept the idea of infinitesimal variables. They felt that, despite how well they worked for physicists for the prior century in solving some of the more difficult problems, such ideas were simply "non-mathematical." (Not well-grounded.)

It would wait for Abraham Robinson's non-standard analysis to bring these ideas back into the field of mathematics in the early 1960's and ground them in already-accepted ideas. But by then, the damage had been done. So, while it is true that today infinitesimal variables are well-grounded mathematics. They aren't broadly taught.

Now, we've one problem to resolve. How do we figure out the mess, \$\int \left(t\cdot \text{d} t\right)\$? It's really not that hard. Pause for a moment and think. This is a product, which "looks like" an area calculation, right? Let's say a "rectangle" that is \$t\$ high and \$\text{d}t\$ wide? It could be, couldn't it? Maybe?

Let's assume we start at \$0\$ and proceed towards a final end-point at \$t\$ and see where that takes us:

On the left, I've started from \$t=0\$ and have started to place each successive rectangle adjacently, with the first one placed left-most, the next one just beside it to the right, and so on. I've shown only the first four of these rectangles in Figure 1.A. There are an infinite number of them, though. On the right side, in FIgure 1.B, you can see the finished result. You have to imagine it, but I've completed the entire "shape" on the left side, with the infinite number of very thin rectangles placed side by side until you finally reach the end, at \$t\$. As you can see, the total area of all of these tiny rectangles added up makes up a triangle. And you can see that the area of this triangle must be \$\frac12 t^2\$. The result is that we can say:

$$\begin{align*} x&=g\cdot \int_0^t \left(t\cdot \text{d}t\right)=\frac12\:g\cdot t^2 \end{align*}$$

This is a good moment to say that the above is a definite integral, which means that I knew the range over which the sum was performed (\$0\$ to \$t\$, here.) There is also an indefinite integral, which is similar but doesn't include a known range. The indefinite integral focuses on the overall shape (in this case, the triangular shape which could also be extended backwards, too, before \$0\$ if you continued the line slope with a ruler) of the result, but leaves out the details of exactly where to start and end for a later time. So the indefinite integral here would be:

$$\begin{align*} x&=g\cdot \int \left(t\cdot \text{d}t\right)=g\cdot \left(\frac12 t^2 + C_0\right)=\frac12\:g\cdot t^2+C_1 \end{align*}$$

\$C_0\$ represents an unknown constant to be determined later (where also \$C_1=g\cdot C_0\$, above.) For example, if the rock had started with some initial speed, then \$C_1\$ would be set to that starting speed (and can be positive if in the direction gravity will accelerate the object or else negative if in a direction opposite the direction of gravity's acceleration.)

But in our example case from the beginning where the initial speed was zero, we can just say \$C_1=0\$ and find the common answer to the problem as:

$$\begin{align*} x&=\frac12\:g\cdot t^2 \end{align*}$$

Obviously, situations can be more complex than this. It's just an introduction to the idea of infinitesimals and the creation of a new algebraic variable which can only hold infinitely small, but non-zero, values.

The only case where the infinitesimal variable can be thought of as exactly zero is when the range over which the integral to be performed is also zero -- meaning that the starting and ending point are both the same place. Otherwise, the infinitesimal variable should be thought of as an infinitely small, uniform step-wise piece, while going from the starting point of a range to the ending point of that range; a value that is infinitely small, while not quite zero.

Once you grasp this concept, you face the problem of being able to sum up lots of small bits. Much of calculus is about how to perform those sums. Or, put another way, how to imagine stacking up tiny areas side by side as indicated above and looking more closely at the area within the resulting geometric shape that describes that final sum. As you can well imagine, some of those shapes won't be as simple as a triangle! And that's where your imagination begins.

I want to return to this earlier equation:

$$ \text{d} x=v_{_t}\cdot \text{d} t $$

We should be able to divide both sides by \$\text{d} t\$, yes? It's just a variable, after all, even if it is a weird variable for algebra. So let's do that:

$$ \frac{\text{d} x}{\text{d} t}=v_{_t} $$

Or, put another way, \$v_{_t}=\frac{\text{d} x}{\text{d} t}=\frac{\text{d}}{\text{d} t} x\$. Now here's what's kind of interesting. I've divided one infinitesimal variable by a different infinitesimal value and the result is a finite value, the speed! We can do that? Yes, we can. If it helps, just think of this like dividing \$\frac1{\infty}\$ by \$\frac4{\infty}\$ to get \$\frac14\$. The \$\infty\$ parts cancel out and you can get a finite value from that. A better way to look at the meaning here is this: "How do the tiny changes made in one of these variables compare with the tiny changes made in another one of them?" It's just a ratio. And it is quite possible for one change to be 4 times another's change. That's not a conundrum. It actually makes sense.

So feel free to do this. These variables are, in fact, variables just like those you are used to in algebra. You can also cancel them, similarly, since the ratio of an infinitesimal variable, taken relative to itself, must be 1. So, they do cancel out.

One final thing before I try and answer your question. There is another idea, probably far too early to introduce to you, but worth pointing towards anyway. Let's go back to the original algebraic equation and the subsequent differential equation for the falling-rock situation:

$$\begin{align*} x&=v\cdot t\\ \text{d}x &= v_{_t}\cdot \text{d}t \end{align*}$$

The first equation can only be approximate. The second above is exact (so long as time itself is independent of the speed of the object.) And I think you may be beginning to see a difference in perspective arriving, just in the very way I speak about things here. Note that I added the caveat of "so long as time itself is independent of the speed of the object," as a note. Once you acquire this viewpoint, these kinds of caveats will be ever in your mind, as you read and think about equations and expressions and problems you are considering. It is the beginning of a new way to see the world, itself. And it is life-changing.

By the way, isn't there a hint of something "simple" there? Like some kind of easy transformation of one into the other? There is, in fact. And it will be worth learning about, at some point. This process has a broad term for it, called implicit differentiation. I'll use that process, below, in the answer area.

I would have written more, or less, both better I think, but I didn't have the time for either.

Towards an Answer to Your Question

I happen to know (because I've spent time to learn it) that implicit differentiation can be used to turn \$y=e^x\$ into \$\text{d} y = e^x\:\text{d}x\$. You could go look that up in a book, in fact. Though it will only be in a small subset of textbooks that you will see it exactly that way. Usually, it is written as \$\frac{\text{d}}{\text{d}x}\, e^x=e^x\$. But you can just multiply both sides by \$\text{d}x\$ (and substitute \$y\$ for \$e^x\$ on the left side) in order to get my result at the beginning of this paragraph.

I often use the capital D as an "algebraic operator" that can consume algebra and turn it into new algebra. So the way this works is something like this:

$$\begin{align*} D\left[y\right.&=\left. e^{_{x}}\right]\\\\ D\left[y\right]&=D\left[ e^{_{x}}\right]\\\\ \text{d}\,y&=e^{_{x}}\cdot D\left[ x\right]\\\\ \text{d}\,y&=e^{_{x}}\cdot \text{d} \, x \end{align*}$$

This explains the production I suggested at the start of the first paragraph in this section. It's an algorithmic process for chewing up algebra equations and turning them into differential equations. It is a powerful machine and it just works. It will take time for you to follow why, as it requires some additional calculus: the product and chain rules, for example.

So now I have to apologize, but I'm going to speed ahead a bit towards an answer for you, using this D-operator algorithm:

$$\begin{align*} D\left[q\right.&=\left.3\left(1-e^{_{-t}}\right)\right]\\\\ D\left[q\right] &= D\left[3\left(1-e^{_{-t}}\right)\right]\\\\ \text{d} q&= 3\cdot D\left[1-e^{_{-t}}\right]\\\\ &= 3\cdot \left(D\left[1\right]-D\left[e^{_{-t}}\right]\right)\\\\ &= 3\cdot \left(0-e^{_{-t}}\cdot D\left[-t\right]\right)\\\\ &= 3\cdot \left(0-e^{_{-t}}\cdot -D\left[t\right]\right)\\\\ &= 3\cdot \left(-e^{_{-t}}\cdot -\text{d} t\right)\\\\ &\therefore\\\\ \text{d} q &= 3\cdot e^{_{-t}}\cdot \text{d}t\\\\ &\text{dividing both sides by d}t,\\\\ \frac{\text{d} q}{\text{d}t} &= 3\cdot e^{_{-t}} \end{align*}$$

But that's current! The rate of change of charge with respect to time is, in fact, just current. Since \$I=\frac{\text{d} q}{\text{d}t}\$, it follows that \$I_\text{C}=3\cdot e^{_{-t}}\$ in your case. And at \$t=0\$, this means the current is \$3\:\text{A}\$.

Which means that (A) in your question is true!

And from there, the rest can be examined.