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Questions regarding capacitor and errors

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apep

New Member
Hello, I am new here, and got two question regarding capacitor and errors...

Question 1:
I recently did an experiment where I charged and discharged a capacitor to estimate its capacitance (by getting voltage vs. time and the time constant).
Now, ideally, the charging and discharging rates would be same and produce same shaped exponential growht/decay curves on voltage vs. time graph... however, the charging occurred more rapidly and has a time difference of about 100 s (in comparison to discharge, which was slow)...
What could be the source of the error?... why did the charging and discharging rates differ with charging being so much faster?

Question 2:
In a similar experiment, I changed the resistor that was on the circuit and decreased the resistance of the circuit (500 kΩ to 10 kΩ) to have two different estimates of the time constant and hence, estimate two of same capacitance... Now, the first one came out to be close to the actual capacitance value of about 33µF, but the second estimated capacitance (from decreased resistance) was about 600 µF... (6 V battery was used, by the way).
Again, what could be the source of error? As the estimation was based on the graph (first question), could it be from the human errors from recording the data or is there other electronic sources of error?

Thank you in advance.
 

Sceadwian

Banned
Your methods aren't defined, so it's impossible to draw a conclusion, but a difference of 19 times the actual capacitance means something is obviously wrong. You need to fully flesh out all of your data and your circuit along with the test methods or there's no place to start from.

The scientific method is a pain in the ass, but it works if you use it.
 
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Grossel

Well-Known Member
I agreed to Sceadwian. Those numbers seems to be weird.

Can you post your test circuit here?
What kind of voltmeter did you use?
Have you tried with different (brand) voltmeter?
 

apep

New Member
I have attached the image of the data and the circuit...
The measurements were taken using a stop watch and a multimeter (as voltmeter)...

In both cases the charging and discharging trends do not match... (Q1)

In the case of R1 ~ 10 kΩ, the time constant is about 6 s from my calculation (using 2/3 of stable value voltage) and based on the total resistance, the capacitance was calculated to be about 600 µF. The first trend actual got similar capacitance to the expected value (got about 34µF)... (Q2)
 

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Grossel

Well-Known Member
Hi.

You'r results is just as expected. When discharging, the cap discharges through R1 and R2 in series - while charging through only R1 = faster.
 

MrAl

Well-Known Member
Most Helpful Member
Hi,

Yes, if you want to do it this way make R2 much smaller than R1 and you'll see approximately the same charge and discharge times.
 

RCinFLA

Well-Known Member
Assuming the Vc is voltage across cap, you should be charging cap to 6 vdc supply. In higher R1 of 1 meg ohm you get 1.65v. For lower R1 of 10k you are getting 3.1 vdc.

Your capacitor is leaky. For first case it is showing about 4.35 uA's of leakage current. For R1=10k case it is showing 290 uA's of leakage.

It is likely an electrolytic cap you are using. It will have some leakage current but second case is shows very high leakage. Make sure you have cap polarity correct as electrolytic caps are polarized. 1 meg ohm charge and discharge resistor is quite high value and you have to have a very low leakage electrolytic cap for it not to effect results. If you have polarity correct then the cap is bad as 290 uA of leakage for 3v of bias is very high, even for an electrolytic cap.

Also, keep in mind any DVM you put across cap will have an input resistance of likely 10 meg ohms so with a 1 meg charging resistor you will have a 10 meg ohm in parallel with cap that will effect charge profile a bit.
 
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