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Charge of two capacitors in parallel, should be easy.

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Wollowstone

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It should be easy. It is about a 12V DC source and two triplets resistance-capacitor-resistance in parallel. The theory and spice predict that in the long run ( many times more than 10*R*C), the two capacitors will charge at 12V, but in practice, only one does! C = 0.33uF for each triplet, each r=51.1kΩ for the first triplet, while R = 5.62M for the second triplet, the difference of voltage between the capacitor is greater than 1 V. For smaller C, r and R, the difference is smaller, but it is still there ( 0.4 V for C=182nF, r=10kΩ, R = 226kΩ). The voltage at each capacitor falls back at 0V when the source get disconnected. Can eve add a third parallel path made of a single resistance, and while this changes each voltage, it does not really change the difference of observed voltages.

Someone has a simple explanation?
 

Reloadron

Well-Known Member
Most Helpful Member
Your crystal ball should have warned you in advance. A really good crystal ball will rent a replacement for the weak.
What about the strong? :)

Ron
 

Wollowstone

New Member
Here is a picture


MontageTest2CondensateursEnChargeDécharge.png
The capacitors are not necessary electrolytic ones ( I used ceramic and film too). In the reported case, with numerical results, I used a 12V instead of 9V here shown.
 

audioguru

Well-Known Member
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How are you measuring the voltage across the capacitor? A multimeter input has a resistance of maybe 10M ohms that loads down the voltage of the capacitor because then it is part of a voltage divider.
 

Wollowstone

New Member
I measure the voltages with an oscilloscope with the probes not at points "a" and "b", but the other side of the capacitor C, between the cap and the resistance. The probes are grounded at the common ground. They add to RL. No?
 

ronsimpson

Well-Known Member
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Many meter have a "resistance" of 11meg ohms, when in DC measure mode.
So you have 5.62meg x 2 = 11.24meg of resistors and probably 11meg inside the meter. So your meter should read 1/2 as much as you think. Reading might be 6 volts.
upload_2017-10-10_16-12-25.png
In the case of 51.1k resistors the error is small. (about 1% off)

I don't know what meter you are using. I think your meter is causing the effect you see.

Your scope probe has the same problem. (maybe worse)
 

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Wollowstone

New Member
In the case of 51.1k resistors the error is small. (about 1% off)

I don't know what meter you are using. I think your meter is causing the effect you see.
The probes are supposed to be at 20MΩ, but I see that you meant. That is why I tried with smaller resistances, but I forgot about the 1% error within which the observed data can sit with the explanation that you supplied. Thanks for the explanation.
 

Ratchit

Well-Known Member
It should be easy. It is about a 12V DC source and two triplets resistance-capacitor-resistance in parallel. The theory and spice predict that in the long run ( many times more than 10*R*C), the two capacitors will charge at 12V, but in practice, only one does!
I don't believe it. As long as the battery voltage is sustained, the two parallel DC circuits will act independently of each other. You don't need SPICE to tell you that. Check to see that the battery voltage is constant.

C = 0.33uF for each triplet, each r=51.1kΩ for the first triplet, while R = 5.62M for the second triplet, the difference of voltage between the capacitor is greater than 1 V. For smaller C, r and R, the difference is smaller, but it is still there ( 0.4 V for C=182nF, r=10kΩ, R = 226kΩ). The voltage at each capacitor falls back at 0V when the source get disconnected. Can eve add a third parallel path made of a single resistance, and while this changes each voltage, it does not really change the difference of observed voltages.

Someone has a simple explanation?
Try energizing each capacitor without the other and notice that there is no difference whether they are connected together or not. By the way, the capacitor with the two large resistors has a time constant of 3.7 seconds. That means about 20 seconds to fully energize.
When you refer to voltage between caps, do you really mean voltage across the plates? Finally, notice that caps energize. They do not charge. A cap energized at 100 volts contains the same charge as it did at zero volts. Inquire if you have any questions about this.

Ratch
 

Wollowstone

New Member
notice that there is no difference whether they are connected together
I tried with many different capacitors and resistance with a consistant behavior (that is, the cap in the triplet with the large resistance was observed to a lightly less voltage than the other cap). I also tried with relatively small caps+resistors with the same behavior (but with less voltage difference).

It seems that the error in the readings is consistently larger with the larger resistances, which could explain the observed difference though, which was something that I hadn't considered before !

Thanks for you contribution, it is always nice to see that someone could be interested in your problem. :)
 

crutschow

Well-Known Member
Most Helpful Member
Finally, notice that caps energize. They do not charge.
Does that mean you don't charge a battery either, you "energize" it? :rolleyes:
 

Ratchit

Well-Known Member
Does that mean you don't charge a battery either, you "energize" it? :rolleyes:
That is correct. One pumps as many charge carriers (electrons) out of a battery as one puts into it. That gives the battery a net charge change of zero. What one really does is "charge" the battery with energy, so one might as well say the battery is being energized. In fact, doesn't a particular brand of battery made by rabbits call itself an "energizer"?

Ratch
 

Ratchit

Well-Known Member
I tried with many different capacitors and resistance with a consistant behavior (that is, the cap in the triplet with the large resistance was observed to a lightly less voltage than the other cap). I also tried with relatively small caps+resistors with the same behavior (but with less voltage difference).

It seems that the error in the readings is consistently larger with the larger resistances, which could explain the observed difference though, which was something that I hadn't considered before !

Thanks for you contribution, it is always nice to see that someone could be interested in your problem. :)
As others have pointed out, those large resistors compete with the impedance of your voltmeter for voltage drop. If you have a high voltage probe for your meter, use that to measure the voltage. Of course, the sensitivity of your voltage meter will be less, but maybe you can still measure 9-12 volts. Otherwise, put a 100 Meg or so resistor in series with your probe, effectively making it a high impedance voltmeter. You will have to calculate the voltage from the reading and resistor value you use, but that should be within your capabilities. Otherwise, it is difficult to diagnose the problem second hand without being on site to see the setup and results. Keep trying to get the answer.

Ratch
 

crutschow

Well-Known Member
Most Helpful Member
What one really does is "charge" the battery with energy, so one might as well say the battery is being energized.
Well, good look with trying to convince the world to say that capacitors and batteries are being energized instead of charged. :banghead:
At this point I think you have total of one person who is doing that (counting yourself of course).
 

Ratchit

Well-Known Member
Well, good look with trying to convince the world to say that capacitors and batteries are being energized instead of charged. :banghead:
At this point I think you have total of one person who is doing that (counting yourself of course).
If my reasoning is correct. I don't mind being a single point of shining light.

Ratch
 

tomizett

Active Member
To be fair to Ratchit, the signature at the bottom of each post does give fair warning!
 

Reloadron

Well-Known Member
Most Helpful Member
Does one of those bunnies figure into this?

Ron
 

throbscottle

Well-Known Member
That is correct. One pumps as many charge carriers (electrons) out of a battery as one puts into it. That gives the battery a net charge change of zero. What one really does is "charge" the battery with energy, so one might as well say the battery is being energized. In fact, doesn't a particular brand of battery made by rabbits call itself an "energizer"?

Ratch
That's it, I'm not going to connect a battery to anything any more. I'm going to pump some charge carriers from it!
 
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