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In an resistor-capacitor circuit, the voltage stored is measured by taking the two values in Farads and Ohms - and the curve looks like this:

RC circuit voltage

What would this curve be called, logarithmic? exponential (inverse?) I have not a clue in how to explain the nature of the growth to somebody without showing them the actual picture.

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up vote 0 down vote accepted

It's in general impossible to say what a curve is just from looking at a picture of it, but your curve could be a logistic curve, with equation

$$y(t) = \frac{1 - e^{-t}}{1+e^{-t}}$$

and which looks like this.

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Thank you so much, although the average person may not know what one is - it fits perfectly for what I am trying to explain. To calculate voltage from time in my graph, you use a similar formula ($Vs * (1 - e^{-t/τ}$) - excellent! – Johnny Nero. Dec 1 '11 at 9:49

Since it is a resistor-capacitor circuit, this means that the differential equation describing your system must be of the following type:

$$R\frac{dQ}{dt}+\frac{Q}{C}=U \; ,$$

in which $R$ is the resistance, $C$ the capacity, $Q$ the charge stored in the capacitor at some time, $U$ the voltage applied on the system by some external source like a DC battery.

Solving this system gives

$$Q(t)=CU+(Q_0-CU)e^{-\frac{t}{RC}} \; ,$$

in which $Q_0$ is the charge stored at time $0$. If you need the voltage stored, just divide by $C$ to get:

$$V(t)=U+(V_0-U)e^{-\frac{t}{RC}} \; ,$$

with $V_0=Q_0/C$. The figure you show is giving the current flowing through the system, but that is related to the voltage by $V=IR$. Also the $V$ in the picture is what I called $U$.

As Oltarus already mentioned, this is an exponential curve, and the phenomenon it describes is often termed "exponential relaxation".

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This should be the accepted answer. – J. M. Dec 1 '11 at 11:21
Thank you Raskolnikov. I may be missing something simple, however if I apply the formula.. C=0.000330, R=4700, U=5, Q0 = 0?, it will equal ~0.00000000069510576900. – Johnny Nero. Dec 2 '11 at 6:24
First of all, you can't come out a number as a result since the formula is time dependent. So, unless you did specify a time, you did something wrong. Second, you also don't mention any units, so the numbers don't mean much. – Raskolnikov Dec 2 '11 at 9:09

That is clearly an exponential curve:

$$i(t)=c-e^{a \cdot (b-t)}$$

for the graph to pass by $(0;0)$, you must have:

$$a \cdot b = \ln(c)$$

The precise values of $a$ and $b$ are very difficult to give, because I have not enough informations about the curve.

Example with $a=0.6$ and $b=5$ (and $c=e^3$)

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Thank you, even though this specific math diverges away from electronics, it and Wolfram make me want to mess with it more. – Johnny Nero. Dec 1 '11 at 10:02
No problem, my pleasure! – SteeveDroz Dec 1 '11 at 10:21

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