2
$\begingroup$

I am a high school physics teacher and came across some issues today that I am unable to explain.

We made a circuit with an $11$ V power source connected to a $100~k\Omega$ resistor and $100~\mu F$ capacitor. We used a Vernier voltage probe to observe the potential across the capacitor increase. The potential increased to about $6$ V and stayed relatively constant. I'm really unsure of why it would not go up to $11$ V.

We also observed the capacitor discharging by removing the power source from the circuit. It seemed to discharge down to about 0.2V and stay around there. I short circuited the capacitor to completely drain it and the potential went down to zero but then slowly built back up to 0.2V. We discovered that the voltage probe itself seems to be charging the capacitor and that just doesn't make sense to me.

Big thanks in advance to anybody that can help me make sense of either or both of these issues.

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ I assume it's a parallel circuit, the voltage is DC, and it's an electrolytic capacitor, but you should state that info explicitly in the question. Also, did you take any current readings? What's the current limit of your power supply? $\endgroup$
    – PM 2Ring
    Feb 20, 2019 at 8:21
  • $\begingroup$ Yes, the info @PM2Ring requested would be very helpful. Also the model of the voltmeter. $\endgroup$
    – noah
    Feb 20, 2019 at 9:31
  • $\begingroup$ What is the internal resistance of your Vernier voltage probe? You could explain your result, if it is of the order of 100k (voltage divider). $\endgroup$
    – flaudemus
    Feb 20, 2019 at 9:58
  • $\begingroup$ It's a series circuit while charging and when discharging the only two elements are the resistor and capacitor. Of course it's DC if the goal is to charge the capacitor. I really don't know if it's an electrolytic capacitor - how can I tell? That makes sense about the internal resistance of the probe - I'll look into that. $\endgroup$
    – steve george
    Feb 21, 2019 at 2:58
  • 1
    $\begingroup$ Being electrolytic should be obvious as they are polarized -- meaning you must obey the negative and positive terminals (or wires) on an electrolytic capacitor. Thus, it is normal to have the negative terminal very clearly identified, usually with minus signs on the body of the capacitor on the side with the negative terminal. $\endgroup$
    – K7PEH
    Feb 21, 2019 at 18:30

1 Answer 1

Reset to default
1
$\begingroup$

For one, at $11\,\text{V}$ you are using the probe outside its specifications (up to $10\,\text{V}$). The $6\,\text{V}$ issue might have to do with the black lead of the sensor being directly connected to ground through the interface connector (or whatever potential your measurement device like a tablet or laptop is at). This could possibly also cause the charging of the capacitor when the power is disconnected. From the information given this is really mostly speculation. Best would be to test again with a voltage differential probe that doesn't have this property, if you have one handy. Or just a regular multimeter will do.

With an internal resistance of $1\,\text{M}\Omega$, the probe acting as a voltage divider should not affect the results to such a degree. However, you can easily eliminate this possibility by using a different resistor and measuring again.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.