Why does an ideal capacitor give rise to a rectangular cyclic voltammogram (CV)? This question is sort of a sequel to my previous question about cyclic voltammetry (CV).  One of the responses made reference to the fact that an ideal capacitor gives rise to a rectangular cyclic voltammogram.  Can you please help me understand why this is the case?  In other words, why does an ideal capacitor reach a constant current $I$ as soon as a voltage $V$ is applied?

I indeed see nearly ideal CVs in many literature articles (CVs which are rather rectangular with rounded corners).  In other figures, though, I see relative deviation from "rectangles with rounded corners," in that I see abrupt peaks, spikes, or valleys.

For example, below I have plotted two figures from Khomenko, Electrochimica Acta 2005, 50, 2499-2506.  Just very roughly and "hand wavy," what might be the qualitative reason for the "rectangle with rounded corners" behavior of Figure 8 (left) and the "abrupt peaks" behavior of Figure 4 (right)?  Could it be that the sample in Figure 8 (left) is relatively unreactive toward apllied potential, whereas the sample in Figure 4 (right) undergoes redox (Faradaic) reactions -- indicating the presence of so-called pseudocapacitance -- when an external potential is applied?



Please know that I am not looking for an answer specific to the article to which I linked.  I am only asking this question in the context of basic, qualitative aspects of cyclic voltammetry.
The kind of analysis truly required is outside the scope of chemistry and would be best handled in the careful hands of an electrical engineer. I'll give a brief attempt that exploits an easy-to-use circuit simulator.

We have to first think of an equivalent circuit that we can use as a model to determine its behavior. I suggest the following:



where my variable is R1's value for its resistance, which I shall set to 0, 10 and 100 ohm. If we were to watch this happen in time then we would see the following:




Quickly converting these to Current vs Voltage and running two other simulations at different resistances and we get:



These results are due to setting up differential equations and solving them appropriately.

You can play with the circuit I made for yourself here.