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It is known that the voltage drop across a capacitor is a continuous function of time. This means that, for each instant t0, we may write:

V(t0-) = V(t0+)

This relationship is very used in the time domain analysis of RC circuits for instance, and it is due to the fact that if V was not continuous, it would mean infinite current across the capacitor, which is not physically possible.

But which is the mathematical proof of that? I remember I saw it on a book but I do not find it any more. I remember that it was a consequence of the relationship i = C dV/dt....

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  • $\begingroup$ I have edited the question. Please note that the device is called a "capacitor". "Capacitance" is a property of the capacitor. $\endgroup$ – Aaron Stevens Oct 15 '19 at 22:37
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To visualize this it might be helpful to re-write the equation in terms of voltage as a function of time, or

$$v_{C}(t)=\frac{1}{C}\int i(t)dt+v_{C}(0)$$

Or, for the case where $i(t)=constant=I_{C}$,

$$\Delta v_{C}= \frac{1}{C}I_{C}\Delta t+v_{0}$$

Meaning in this case if the voltage varies linearly as a function of time at constant current, then as $\Delta t\to 0$ $\Delta v\to 0$ unless $I_{C}\to ∞$. Which is another way of saying $V(t0-) = V(t0+)$ unless $I_{C}\to ∞$

Hope this helps.

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Well, you wrote your answer yourself !

Since

$i=C dV/dt$

it means that for the current $i$ to be finite, $V$ may not have "jumps". A function that has jumps does not have a derivative at that point, in a rigorous mathematical point a view.

If one does "hand-waving" about that, one might (erroneously) speak of the "(some kind of pseudo)-derivative" of a function with a jump, but then this object would be infinite.

There are rigorous ways to speak of things like that that mathematicians call "distributions" but that would lead us too far.

Just see that a finite current implies a finite derivative of $V$ and thus you need

V(t0-) = V(t0+)

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