What does the slope of an IV plot for a RC circuit represent?

So this is the circuit I am analyzing, a simple charging and discharging of a 100 µF capacitor:

Just out of curiosity, I ran the simulation, hit the switch to charge the capacitor and took the liberty of entering the data points for the voltage across the capacitor and the current through it (by recording the values in the respective multimeters over time) into my graphing calculator, and this is what was generated:

It is clear the IV relationship is linear, which is what I expected since I've been told by numerous sources that both resistors and capacitors are linear devices, but I cannot seem to figure out what the slope (which by linear regression displayed above is about -840 x 10^-6 (unknown units)) represents. It seems like the units are volts per amp, which by Ohm's Law is equivalent to resistance, ohms, but that would be nowhere close to the total resistance of the circuit, which is 2 kΩ. ChatGPT says the slope represents the time constant, but for this circuit that would be 2000 Ω x 100 µF = 200 ms which is also nowhere close to the apparent slope. Even taking the reciprocal of -840 x 10^-6, you get -1190, which doesn't seem to line up with any of the variables involved.

• The slope should be the negative reciprocal of the series resistance Rs since it's derived from Ic = (12 - Vc)/Rs. Rs in your case is 2000 ohms, but the number of significant figures suggests that you're using "real" models for the components; is there a way to simulate with ideal components?
– vir
Apr 15 at 4:17
• It looks like the problem was not that they weren't ideal components but that the time step "TMAX" under the "Interactive" tab on Multisim was too small. When I set TMAX to a higher value, 0.01, I got the slope to be -500.0 x 10^-6, the negative reciprocal of 2000. Horray! Apr 15 at 21:58

The unknown units you mentioned in your question are Siemens. also known as $$\\mho\$$. If you divide it into 1 you will get Ohms.

Of course when you relate the voltage across the capacitor to the current in the resistor (or capacitor, since they are in series with each other), you will get a line (constant slope.) Since $$\I=\frac{V-V_c}{R}\$$ and therefore $$\\text{d}I=-\frac{\text{d}\,V_c}{R}\$$, it follows that $$\\frac{\text{d}\,V_c}{\text{d}I}=-R\$$. Of course, if you consider current for the y-axis, then you are working out $$\\frac{\text{d}\,I}{\text{d}V_c}=-\frac1{R}\:\mho\$$.

Yeah, the slope is negative.

In this case, during the charging cycle you should get $$\-500\:\mu\mho\$$ because the two resistors add up to $$\2\:\text{k}\Omega\$$. Granted. You came up with a different value, which I can't exactly explain. But I don't know the exact details, either, and the number you got doesn't worry me that much given what little I know.

Also note that time isn't involved in the way you arranged the chart. You have removed time from any consideration by contrasting the loop current with the voltage across the capacitor. If, on the other hand, you were to try and look at things treating time, linearly, then your results would show a curve.

For example, having a look at what the circuit load looks like from the voltage supply perspective gets something like this:

It starts at $$\2\:\text{k}\Omega\$$, as expected given the two resistors in series that you have. But the apparent value climbs over time because the current declines over time. You can see how it appears to track the capacitor voltage curve, but inversely so.

But above, time is on the x-axis. So these are curves and not straight lines, anymore.

Anyway, great question to ask. +1. Keep it up. You are thinking about things and wondering about what you see. This is a quality I value/respect. So I think you will do well. Best wishes!

• Thanks for the thorough answer. I figured out how to get the slope to be -500.0 micro-Siemens on the dot! See my comment to the question for more details. I am wondering about the graph you showed me though. What do you mean by "apparent value" - are you saying the effective resistance is changing over time? Apr 15 at 22:16
• @polimorphism As the capacitor charges up, the voltage across the resistor diminishes, lowering the current through it. From the perspective of the voltage source, this "looks like" a higher resistance. Eventually, when the capacitor is fully charged (in infinite time, technically) there will be no current at all. And in that case, the voltage supply "sees" infinite resistance. So a way of summing all that up is that the effective resistance changes over time while charging. Apr 16 at 0:39
• One more question: what would the area under this curve represent? Since it would have units of amps x volts being watts, my guess is that it would have something to do with power, but talking about something like "total power" as a cumulative sum doesn't jibe well with my mind for some reason. Cumulative energy makes sense, and total power at an instant makes sense, but cumulative power doesn't really. Apr 17 at 22:11
• @polimorphism Which curve? The time-based ones I show in my answer would be area units in Webers and Henries. So it must be the linear curve you are asking about. In that case the area units are watts. The confusion may come from the fact that watts includes time (Joules/s) and yet there's no time involved in the graph. But it is the total power dissipated throughout the charging process, however long that actually takes or over whatever range you select. So it's okay. Turns out that for some range, this power is independent of the time required. Apr 17 at 22:32
• @polimorphism It's not a useful area to consider. If the capacitor is small, it takes very little time to reach 99%. If it is very large, it takes a great deal longer. But the curve doesn't show these differences. The number of Joules absorbed by the resistor will be quite different, one to the other. Yet this curve will not show any difference between the two situations, because time has been eliminated and power involves time. Another way to see the lack of value is to realize that it takes infinite time to fully charge the cap. On average, that means 0 W. Always. Apr 18 at 6:39

It's a straight line because the current comes from the the voltage drop across R1, and that plus the voltage on the capacitor add to make the supply voltage.

So the fundamental truth represented here is kirchoff's loop law.