# Understanding the ordinate at $R=0$ in a capacitor experiment [closed]

I've done an experiment to study a capacitor's discharge time and calculate an unknown capacitor's capacitance.

With the help of an oscilloscope, I measured the time it took for the capacitor's voltage to reach half the maximum voltage. I repeated this procedure for different resistors and I got this graph:

I've spent hours trying to explain the coordinate at $$R=0$$ ($$b=-0.8369$$), but I can't come up with a plausible explanation!

• This is possibly an artifact of the best fit line, as it doesn't exactly give the relation. Maybe more data points will push the value closer to 0.Is it possible for you to take more measurements? Commented May 26, 2020 at 16:18
• Unfortunately, no. That was all I could take with the time I had... Commented May 26, 2020 at 16:21
• The most curious thing is the uncertainty given by the linear adjustment for this value is pretty small: 0,03 to be more specific. Commented May 26, 2020 at 16:27
• At first, I thought it could be explained by the resistance in the wires and equipment, but in that case the value would be positive, not negative! Commented May 26, 2020 at 16:29

Curve fitting for data is valid only when both the quantities being considered have any physical meaning. Also, curve fitting must be taken with a pinch of salt. By fitting and extending a curve, you are only obtaining a likely value, which often, is not the right value. A diode's I-V characteristics is such an example where extending the linear region beyond a certain point would not really reflect on what actually happens when you conduct the experiment. However, in your case, it is a case of bad fitting. When I say "bad fitting", extending it to $$R=0$$ does not yield right answers. However, it is possibly good enough for $$R=3000$$.

As one of the comments to the OP's question said, if you take more samples, the expected tension at $$R=0$$ will probably approach zero as well.

A curve fit is a curve that approximately represents the data points you have. The emphasis is on approximately and not accurately. The accuracy increases with the number of data points you have already, although that might be an oversimplification. So in your case, the number of data points is simply not sufficient. As an example, for the $$R=7000$$ case, your line is clearly missing the point. It is close enough, but surely not exact. Same thing at $$R=0$$. Here, however, the approximation led to physically infeasible values.

• It clearly isn't a good fitting for $R=0$, but why? That's the part that's eluding me... As far as I know, it should be a linear fit for all values. Commented May 26, 2020 at 17:17

Assuming you mean "voltage" when you say "tension", a capacitor discharges with

$$V(t) = V_0 \cdot e^{-t/(RC)}$$

You have determined the time $$t_{50}$$ to the half way discharge point, i.e. $$e^{-t_{50}/(RC)} = 0.5$$. We can solve for $$C$$

$$C = \frac{t_{50}}{R \cdot ln(2)}$$

You can use this formula to create a capacitance from each individual measurement that should be a constant (i.e. a straight horizontal line), which can do curve fit, or simply take the mean. If it's NOT a straight horizontal line (roughly), than something is wrong with your experiment.

Looking at $$R=0$$ is not helpful, since the discharge time will also be close to zero so the get $$0/0$$ which is undefined and any small type of noise or parasitic effect will throw off your data.

• If I calculate the capacitance for each individual measurement, I'll obtain a huge error for all the values, but if I extrapolate for much higher values of resistance, it is practically constant. There's some kind of "negative" resistance causing the whole graph to "be moved downwards". Like I said before, I'd understand if the value was positive (because I'm not taking into account the resistance of the wires), but a negative value is really weird... EDIT: And, yes, I meant voltage (electric tension) Commented May 26, 2020 at 17:21
• @PedroNogueira, while some languages use a word cognate with the English tension for potential difference, that usage in English is long obsolete. You might sometimes hear the term "high tension wire", but otherwise we use "voltage" or "potential difference". Commented May 26, 2020 at 17:37
• What do you mean by "huge" error ? How do you know what the error is? Eyeballing your data, you get discharge times that are "too fast" as the resistance decreases by quite a bit, It's likely that yo have a significant problem in your measurement setup, but it's hard diagnose without having the details. Could be something non-linear: you draw a lot more current when the resistance gets smaller. Easy to check: keep the resistor constant (at a small value) and vary the voltage. If the discharge time varies with voltage, you have a linearity problem Commented May 26, 2020 at 17:38
• Thanks for the explanation The Photon. Changed! About the "huge" error, Hilmar I'm able to assess it, because I measured the real capacitance using a multimeter. The capacitance measured using the multimeter is 519nF, the one obtained from the slope is 578nF and the ones obtained for the individual values start with 281nF (for the smallest resistance) and end with 456nF (for the biggest one). Anyway, I'll try your advice the next time I have the chance. Commented May 26, 2020 at 17:54