Charging of capacitor in RC circuit

I answered this SE EE question on Dec 24, 2020 and edited just now.

My answer

I answered this question two years ago. Now I added some additional pictures and explanations to my answer hoping to get the rating it deserves. I have used more water analogies to illustrate the circuit evolution and the problem solved by the active version. The circuit diagrams are conceptual - they are "geometrically drawn" and visualized by voltage bars and current loops.

Perfect C integrator

If we supply a capacitor by an "ideal" current source (pump), there is no problem - the voltage (water level) linearly increases.

Imperfect RC integrator at ideal load conditions

But we want to drive the integrator, as usual, by a voltage source. So, we connect a resistor R to convert the input voltage into current. And indeed, at the first moment everything is fine - the output voltage begins to change linearly.

Imperfect RC integrator at non-ideal load conditions

But after a while a problem appears - the voltage VC across the capacitor begins not linearly changing because it is subtracted from the input voltage and the current decreases - I = (VIN - VC)/R.

So this was my answer "why not linear"... but I will continue to show how we can made it linear. In fact, this is the goal in most cases of practice; exponential relation, with few exceptions, is undesirable. 

Active RC integrator

Obviously, we have to compensate somehow the "undesired" voltage drop. For this purpose, we can connect a variable voltage source in series to the capacitor and with the same polarity as the input voltage source (travelling the loop)... and adjust its voltage equal to the voltage drop across the capacitor. As a result, the voltage drop will be removed and the current will be as in the beginning - I = (VIN - VC +VC)/R = VIN/R.

The hydraulic analogy on the right is represented by a little unusual communicating vessels and corresponds to the op-amp circuit below. Here are some explanations about the text inside the figure:

The little man on the left is a "helper" and the capacitor on the right is a "thief":) So the "thief" steals voltage... but the "helper" restores it and adds it to the input voltage. The left vessel is a constant pressure source. The right vessel is a "bottomless vessel" - when its water level tries to increase, the little man oh the right lowers the vessel thus keeping up a "hydraulic virtual ground".

Op-amp inverting integrator

If we feel bored doing this tedious job, we assign it to an op-amp. This is the idea of the "op-amp inverting integrator".

I created this Corel Draw picture in the 90's (the element designations do not correspond to the generally accepted ones). Here are some explanations for the inscriptions inside the figure. The circled op-amp (including the bipolar power supply) is a "helping voltage source"; UOUT (VOUT) is a "copy" and UC (VC) is the "original" voltage. So, the op-amp "copies" the capacitor voltage and adds it in series to the input voltage EIN (VIN).

Conclusion

This is a great idea that we can see everywhere. We can figuratively call it "removing a disturbance by an anti-disturbance".

My comments

1. @LvW, An engineer that is an inventor even will not try to explain why the relation is not linear; it will simply make it linear with a clever trick:) This is the great idea of the negative feedback - it does not care what the disturbance is; it simply compensates for it with an equivalent anti-disturbance. See my answer...
2. @Elliot Alderson, For the purposes of this excellent intuitive explanation, a capacitor is a voltage source... like a battery.... but rechargeable battery. What is confusing is your formal and sterile thinking. It would be good to gain some practical experience to start feeling what is happening in circuits ...
3. @Hearth, Thanks for the interpretation. There is no problem; we all know each other very well and understand what we are talking about. A little humor is never superfluous. I just saw the downvoter's reaction and made the connection with the great principle. BTW it can be implemented without negative feedback, e.g., by a "negative capacitor". Really, this is not the exact answer but it is closely related to it; this is the answer to the next question that logically follows, "How do we make the capacitor charge linearly?" In your place, I would first admire the idea and then criticize...
4. @Elliot Alderson, My answer is closely related to this question; so it is useful. If you take the trouble to follow the link above, you will see a 5-step scenario; the third step is dedicated to this question. Showing how something nonlinear can become linear is an indirect (and more original) way to explain what causes this nonlinearity over time...
5. @tlfong01, I think it has become clear that this cannot be done through intuition but through mathematics. Unfortunately, intuition only tells us that the current is progressively decreasing but it can't tell us exactly how ... But nevertheless, I keep thinking about this phenomenon of "communicating vessels". But now the question arose how to make a -1 ohm resistor. Interesting topics have no end...
6. @tlfong01, Nice resource... I have a rule: qualitative things should be explained by qualitative means and quantitative things should be calculated by quantitative tools. I have noticed that processes in nature where the difference between two pressure-like quantities determines through a flow-like quantity the second pressure-like quantity, lead to an exponent.
7. @tlfong01, This is all true but for "normal" people :) You know that there are some slightly obsessed people who want to know these things :) I think this is some kind of mental illness :) Have a nice day!
8. @tlfong01, Aren't you impressed by my geometric interpretation? It derives directly from the hydraulic analogy of "communicating vessels". If necessary, should I translate the captions inside the illustration? Have you come across many such illustrations on the web? If so, tell me to contact the author and express my admiration. I have browsed about 100 million pages and finally managed to find something similar. He "opened my eyes" what a transimpedance amplifier is.
9. I have explained in detail the idea of such an interpretation in the Wikibooks page...
10. The circled op-amp in my picure is a "helping voltage source" (this is the text); Uout (VOUT) is a "copy" and Uc (VC) is an "original"...
11. So, the op-amp "copies" the capacitor voltage and adds it in series to the input voltage (EIN)...
12. I just know what is important and what not in circuits. Here the important is to make the exponent linear and to explain why it is an exponent is not so important...
13. OK, I will be glad to see your progress:)
14. I browsed through your RP resources. It strikes me that you handle very specific information. I have concentrated on the basic ideas behind circuits and I am not interested in details. Therefore, we have differences of opinion. My advice is not to delve too deeply into my explanations but to do your specific job. It is difficult for you to go down to my level because you solve specific problems.
15. On the other hand, for me these many details are quite burdensome, although often interesting. My conclusion is that it is best for everyone to stay at their level and only if necessary to go to the other. Cheers!