Two very different ways of charging and discharging a capacitor.

A "very different procedure". Guess you try to understand something and dont want to be disturbed, right? So check this.http://everycircuit.com/circuit/6113042325307392 BTW, dont make things more complicate as they are.

@hurz ... You are showing here that you have no idea about an RC network.

and the original circuit from 2cent http://everycircuit.com/circuit/6013597189406720

You are a novice in this area Fraudz, ... No point in starting in with your diversionary tactics already. Go and learn about RC networks and then you will see that the procedure between the left and right similar outcomes here is very different.... which I said but you deny.

Im sure you will find a way to overcomplicate it and make a difference. So far, as i proofed its identical. It doesn't matter if RC or CR. Lets see what you mess around to make it a "very different procedure" as you said.

If you change the ground (or reference, as it should probably be called) to one side or another of the square wave source it makes it a high-pass or a low-pass resistor network seemingly paradoxically, as high-pass and low-pass are opposites, but really it’s like hooking it up forwards vs backwards, or measuring across the resistor vs measuring across the capacitor. Really the same circuit, just a different reference, and a very different result. But there are still the same voltages and currents across every component.

By changing the where the reference (ground symbol) goes you change what the lines above the wires are doing as you change what they are measuring. But that is to be expected, as they are now measuring something much different. If you measured everything with volt meters instead of the scope (which uses the ground to connect its second terminal to) and then changed the reference, nothing on the scope would change because you are measuring across components now, not between a wire and the reference point.

@jason9 ... nice to see you thinking it through. Keep on and you might get to see the very big difference in how the cap is charged up.

The capacitor is charged from the resistor through the capacitor to ground vs through the capacitor to the resistor to ground, correct? Is this the “very big difference”? Either this “very big difference” doesn’t seem so big to me (my definition of big may not be exactly the same as yours) or I’m missing something.

Now its alread a "very big different procedure" Please dont overcomplicate things. It doesn't help anybody, just sounds like "predominantly a range of factors" fraudster language

@jason9 ... You are on the right lines, but please stipulate left and right circuit, and left and right plate of a cap so that we are on the same page, and when you do that then first ask yourself 'What is the voltage at each side of the cap in question in the first instant of charging. For instance: What is the voltage on the right plate of the cap of the left circuit when it first starts to charge? (using earth as our fixed zero reference point throughout).

Of course if you use earth as the reference point there will be a very big difference if it is in a different point of the circuit. But that does not mean the circuit behaves any different. It just means you are measuring across a new set of two points (one plate of the cap and one wire versus that same plate of the cap and another wire).

And I also have no clue what “stipulate” means.

Shouldn't you call "earth" a ground or a 0V reference? B/c to me saying earth is like a wire going into literal earth.

@sshsslfun ... the words: earth, ground, zero reference, can all be used interchangeably in a board circuit like this. Americans tend to use 'ground' whilst England uses 'earth'. We don't need to separate those words in to different meanings in this EC situation here.

@jason9 ... Keep zero earth as the fixed reference in our discussion until we decide to change that, otherwise we will end up talking about different things. Let me help you here: highlight both the left cap plate wire and the right cap plate wire (stick to the left circuit until we have finished with it). Slide your finger upwards on your scope so that all the traces are combined now. Then look at the trace of that right cap plate wire and see its voltage and shape. What do you make of that right plate voltage at the very start of the charging of every pulse?

It looks like a square wave would after being passed through an RC high-pass. In other words, a spike followed by a curve to zero, then repeat but inverted, then repeat again but inverted again, etc.

this is the voltage across the generator and the voltage across the resistor, correct. Lets see what comes next

Look just at the very start of that spike of that right plate... it shoots up vertically in the first instant that the 10v pulse is applied. So how does that right plate get to be zooming high at the same instant as the left plate. I will answer that, otherwise it will take us a long time: The current at a capacitor is proportional to the "change in input volts divided by the change in time happening". In other words it is proportional to the 'speed' at which the input volts is changing. That left plate input pulse changes from zero (off) to 10v (on) in the first instant as seen on the leading edge of the square pulse.... so the current likewise shoots up high as just described due to that speed. That high current can quickly now pass from one plate of the cap to the other and so act as though the plates are a closed switch, ie as though the two plates are joined as just one bit of metal.. Now: if the left plate is at 10v, then so is that joined right plate! Thus, the right plate is seen to likewise shoot up instantly as you saw. (The current does likewise if you wish to see it by highlighting the resistor). So: Both plates are high volts at the same time .... and that means there is no voltage between them (no p.d.). Look now at the very start of the curve there at the start of the pulse ... It reads as zero on the scope meter (volts across the plates), and that is not surprising, since I have just said that there is no p.d. The next part then becomes even more interesting... but I must go now until tomorrow. You have enough to think about there for now, along with 'Why does this charge curve now progress in the way that it does if both plates are already high'?

lame duck "i have to go" i guess jason does understand this without any problem, even your wording "both plates are already high" is not fair and should navigate jason into a failure. You try to give him the illusion the cap is fully charged, but its NOT its completely empty and thats why the curve does look how it looks like. BTW, why do you think jason has much enough to think about. I have the feeling you are going home, cuz you have no idea how to continue in your explanation about "very big different procedure" in charge. lame duck

@hurz, I was able to see that “both plates are already high” did not mean it was fully charged, as they are the same voltage meaning zero volts across the capacitor meaning uncharged. @2ctiby, I know how capacitors work, so it is no mystery to me how the voltage on one plate goes way high when the voltage on the other plate does. You didn’t have to explain to me. I understand that capacitors will maintain the same voltage difference across their plates (including no voltage difference) unless there is sufficient current maintained for with sufficient time (enough total coulombs) to change the voltage significantly.

right jason, can you imagine what comes tomorrow as a "very big" surprise, i cant.

@jason9 ... In the previous section above, I explained how both plates in the left circuit cap start at the same instant with a similar high voltage, and that is a key point to note, and so as I mentioned, have zero p.d. between them at that starting point. That was a result of a high current being introduced due to the speed of voltage input from zero to10v. of the square wave. Now: when that 10v first input is reached (top of the leading vertical edge), the 10v now holds constant as is seen by the horizontal top of the 10v pulse input. In other words, there is suddenly now no 'speed' of voltage change any more. The input voltage holds at a constant 10v, the 'change' of input V now holds at zero, so now the current must also be zero, as mentioned above (being proportional to the change in Vin divided by change in time). The present current then dwindles away towards zero amount through the resistor, and is proportional to the similar dwindling of the voltage seen on the right plate there on the trace. As this right plate voltage dwindles down towards zero (at a rate which is determined by Resistor and Capacitor values, ie: RC x 5 approx.), there is thus an equivalent ongoing rise of p.d. between the plates. eg: if the right plate goes down to 7v from a 10v start, then that means a difference of 3v between the plates. If the right plate then dwindles down to 4v, there is now a difference of 6v between the plates, since the left plate remains held at 10v etc. So: The dwindling down towards zero of the right plate is accompanied by an increase of difference between the two plates. (increase of p.d. between the plates). This then is the curve seen there of the volt meter across the plates... an increasing voltage due to the decreasing right plate voltage. (The meter is showing the difference of volts between the plates, not the actual voltage of a plate).

@jason9 ... That is the main conclusion here for the charging up of the cap in the left circuit: Two key points: 1: It charges because of the right plate dwindling down to zero after 2: having been started with a sharp high volts input at both plates ... and that can be seen by the trace of the right plate. That described procedure for the rise in p.d. across the plates is very different to the procedure involved for the rise in p.d. of the circuit on the right, even though the charging results are the same. We can go through the right charging circuit procedure if you wish, and see how it is very different, or you can do that yourself and ask as you go if you wish. You will see on the right circuit that there is 1: Not a dwindling of a plate voltage for the charge up,  and 2: No such initiation of a similar high voltage start on both plates. Hence two very different procedures to attain similar results.

All of my descriptions above have been entirely about the charge up of the cap. Don't confuse any of that with the discharge of a cap which has not yet been involved there.

I give you credit for publishing this circuit and sharing your observations about the differences between the two circuits. I’m sorry that your circuit was hijacked by EC’s most notorious and obnoxious troll hence the reluctance of so many users to publish informative circuits. But rest assured that your work and the work of so many other members of the EC community are greatly appreciated.

Thanks Lenz, I aimed to stick to the main points rather than get side tracked.

Troll Lenny, it seems you did not read what @2ctiby wrote so far. He completely missed that point to show us what the "very big difference" is between the two circuits. Next time before you try to bully against any member on EC, read first then ask! BTW its not fair to call @jason9 or @sshsslfun trolls, cuz they are not. You are a troll. Again, read first then ask. @2ctiby, we are waiting for the "very big difference". Troll Lenny, try to be silent as long you have no idea what the topic is all about. @2ctiby, why dont you write your explanation into the circuit descriptions, cuz anybody is then able to make a copy of what you have written? Would make it easier to point out where you are right or wrong.

Oh Warner, my long time trollish master, 2ctiby has made his position crystal clear on several occasions but your personal vendetta against 2ctiby has blinded your objectivity as evidenced by your ad homein attacks.

@2ctiby, before you start now and continue the bullying of Lenny, please tell him, you just talked about the left circuit. So far you did not mentioned the right one! Isn't that a way of being respectless to not read what you have said and just use your effort to bully against other EC members? I would be very dissapointed about Lenny. He just ignors you to have the oportunity of bullying. What a bad character.

Lol...😂

Lenny fact is, you havn't read what @2ctiby wrote. In other words, you give a shit on what he said!

The left circuit charges because of 1: its right cap plate there dwindling down to zero after 2: having been started with a sharp high volts input at both plates ... and that can be seen by the trace of its right plate. That described procedure above, for the rise in p.d. across the plates of the left circuit is very different to the procedure involved for that rise in p.d. of the circuit on the right, even though the charging results are the same. You will see on the right circuit that there is 1: Not a dwindling of a plate voltage for the charge up,  and 2: No such initiation of a similar high voltage start on both plates. Hence two very different procedures to attain similar results.... I have already said that above here, but this is a reminder for hurz who seems not to have read that.

To remind you: "We can go through the right charging circuit procedure if you wish, and see how it is very different, or you can do that yourself and ask as you go if you wish."

If you have any questions about how the circuit on the right works hurz, then do ask.... ( "...ask as you go if you wish."). If instead you know how the circuit on the right is charging, then you will already realise that it is not using the same procedure as that of the left circuit which I described. If instead you disagree with any technical point that I have made, then I would like to hear it.

You seem to look at everything relative to the reference with few exceptions. I look at stuff relative to the reference too, but I also look at the voltages relative to neighboring components and across components as well. You see big differences from changes in the location of the reference, and I don’t. I believe it to simply be because of the way we view circuits. We each have our own way of interpreting those wavy lines. I understand what you are saying, and I see how it is a big difference for you, but both scenarios are identical to me. I’m not saying either one of us is right or wrong, just that we see things different.

Wrong, it does use the exact same procedure, the capacitor does not know there are 10V. The capacitor just has two pins one to the generator the other connected to the resistor. any "high" voltage he cant see. There is no third pin to ground. This is what you measure, seen from outside but not the capacitor. Ground is just there for you to put nodes on the scope, its not there for the capacitor to see any differences. forget ground and your outside observation. If you would be a capacitor left hand left plate and right hand right plate you cant see in which of both circuits you are built in! You can just see current and see a delta voltage across left to right hand/plate. You are not able to see 10V to ground. You wont even know where ground is. Cuz this is just an outside definition you made, but the capacitor does not know about that! If, he even might see -1million volt to the sky. Your observation as to declare this as a "difference in procedure" is just wrong. There is no difference.

@jason9 ...The procedure on the right is such that the cap voltage increases from zero earth of the battery/pulse. The procedure on the left is such that the cap voltage increases not from that zero reference, but instead it increases due to a dwindling present initial high voltage that was implanted at its right plate as described. You will need to go away and think that through, I can not help you further than that without using a calculus description.

@hurz ... utter crap from start to finish here from you, with no technical points of discussion.

@2cent , i have not expected you understand anything. If you want to make the world more complex for yourself as it is needed, do so. Mess around with your life. But dont try to teach this bullshit to the world, as long you do not see whats relevant and what is not.

put the reference point in your left circuit on top of the generator and now where is your "high 10V" now? Gone... gone as your absurd theories http://everycircuit.com/circuit/4966920370257920

@Fraudz ... Look at the authentic formula for finding Vout for a capacitor RC and you will see that we have to start with Vin ... If you are going to change the goalposts so that the standard Vin is no longer the voltage applied across the network, then you are just creating nonsense. You are really struggling here to divert attention from my production in order to hide from your lack of knowledge in this area. Only a few will be hoodwinked by your futile ploy. Many more are likely to loose yet more respect for you when you lie and scheme rather than face your inabilities.

Are you angry about yourself and your messy life? "i have to go" i think that was the moment you realised, its nonsense what you are talking. Here one more time for you: Keep it simple!

@2ctiby “no technical points of discussion” referring to @hurz saying the capacitor does now know or care about ground but only knows about voltage between its two plates and the current going through it is complete nonsense to me! How could you say that? @hurz was literally JUST talking about technical stuff! Do you have some filter on you filtering out what you think isn’t technical because it comes from @hurz? Him talking about what matters/doesn’t matter to the cap and thus whether or not it changes its charging procedure was very relevant and technical! How can you not see that? I thought you were better than that.

@jason9 ... The technical points which I was referring to were those which I made myself, ie: I was inviting him to say which of those points he disagreed with.  Both you and hurz are welcome to state any point that I made here about the cap and how it relates to the described plate traces seen on the scope there. The 'crap' comment referred to the fact that no trace analysis point of mine was discussed.

The cap on the left circuit is charged by the action of its right plate gradually becoming a lower voltage, having been forced high in the first instance. The cap on the right circuit is charged by a different action where one plate remains at zero and the other plate simply rises to give the charge there. Hence two very different actions/methods of producing similar outcomes regarding p.d. across the cap, and current through. ... Hurz obviously did not realize that these two important different actions are happening, and was quick to make wrong conclusions along with derogatory remarks, and then to divert attention from his mistake by setting off along other lines which detract from my presentation about these differences.

Listen: you still talking about voltages while you watching a single point! Thats just wrong and makes me LOL cuz you even miss the very basics in electronic but want to teach jason something. Ridiculous. YOU CANT MEASURE A VOLTAGE BY REFERENCING A SINGLE POINT ONLY! This "high 10V" are across the generator they are not across the capacitor. The capacitor is to say it correct and not wrong as you do, in its very first moment in its impedance or resistance 0 Ohm and a short, and a short is for voltages 0 V, at zero ohm you cant have a voltage drop. So the capacitor in both cases does start with zero voltage and zero ohm, and it will charge up in both cases upto 10V (after 5tau or 5*RC its 9.93V but if we wait long enough it will come very very close to 10V) ACROSS both plates and both capacitors in both circuits. There is no "difference in procedure" between both capacitors. The capacitor does not know in which circuit 'he' is built in. You can at any point in time exchange both capacitors and the circuits will continues without any "differences" !!! nothing i can agree to see "very big differences in procedure" not as you explain. Epic fail. Check this before we continue to listen any bullshit you are talking about electronic http://everycircuit.com/circuit/5719698684248064

The cap on the left circuit is charged by the action of its right plate gradually becoming a lower voltage, having been forced high in the first instance. The cap on the right circuit is charged by a different action where one plate remains at zero and the other plate simply rises to give the charge there. Hence two very different actions/methods of producing similar outcomes regarding p.d. across the cap, and current through. ... Hurz obviously did not realize that these two important different actions are happening, and was quick to make wrong conclusions along with derogatory remarks, and then to divert attention from his mistake by setting off along other lines which detract from my presentation about these differences.

you can exchange both capacitors while charging at any point in time and wont see any difference!

You won't see any difference, because they are both charging up from zero p.d. across each cap and in the same time interval, but they operate in very different ways, as described above, and can be seen on the scope traces there, eg the right plate of the left circuit shows the initial spike high which then exponentially falls, thus showing the rise in p.d. across the cap, unlike the non-earth held plate on the right circuit which simply charges up from zero, since both plates there on the right start at zero due to the initial high current which makes the right cap appear as a closed switch briefly and so puts the full voltage entirely across the resistor at the start there, with zero on the cap side of that right circuit resistor.

There is no initial spike at the right plate in the left circuit! What you mean is the voltage across the resistor. The voltage across the capacitor is in both circuits without any spikes. There are three different voltages with three components. The sum of all voltages is always 0V. If you know two of them you can calculate the thrid. If you have the voltage across the resistor you know its made as sum of voltage generator and voltage capacitor. When will you learn that ground is just a reference of 0V and not an absolut physical effect. Its just the black/negative wire of our scopemeter which we define as zero volt! If we move this point its suddenly not 0Volt cuz we define another point as zero. So as the voltage across the capacitors in both circuits is identical the charging procedure is also identical. http://everycircuit.com/circuit/4726900417888256 check in this circuit where you can find your spiky voltage -> across the resistors! Before or after the voltage source who cares. The charging of the capacitors in both circuits is identical

The charging is agreed identical, I have kept saying that, but the left cap does that by a reduction from an initial 'spike' voltage of the right plate which occurs due to the initial high current making the cap in the left circuit act like a closed switch and so putting the entire voltage initially across the resistor with the high voltage side being at the left of the resistor, ie at the right plate of the cap in the left circuit. That high plate voltage which is now similar to the left plate voltage then dwindles exponentially down to zero, and so shows a p.d. rise across the cap. That exponential right plate of the left circuit voltage decrease is similar to the current decrease which occurs after the initial similar current 'spike' there which was initiated due to the high 'speed' alteration from zero to 10v of the pulse source, ie i= dV/dt (ignoring the C and R values for now, just to demonstrate the point). It is really a differentiation thing happening, but I have been avoiding the maths. I can't keep repeating this explanation. You just need to re-read it through until you follow it. Highlight the resistor to see the initial current spike along with the pulse input 10v there which caused it....( join the scope traces with an upward finger push on android/iOS) ... Notice the familiar overall pattern of differentiation by the current as an extra feature... each of those spikes is a similar show of what I have just said about dV/dt, but don't get side tracked by the negative values here, I will explain those at a later date if this charging is absorbed fully.

There is no initial spike voltage at the right plate! Cant be, cuz this is just one point! What do you mean? Please be precise

I know that you are keen to point out that things would be different if the reference points or input voltages etc were to be re-positioned, but you need to shelve that for now and just stick to my seen circuits that I have presented here. In other words: stick to the earth zero reference which both myself and EC are using here as being the (black) meter lead unless I say different (eg when I talk about the p.d. across the cap). So: Now that you have seen the initial current spike on the leading vertical edge of each square wave as simply shown there on the EC scope, get rid of all highlights and renew just two highlight to be seen together (not separated on the scope). These two highlights need to be the 10v input wire on the far left between the pulse source and the left cap plate, thus showing the square input pulse, ...  and the right cap plate wire in that same left circuit which goes from that cap to the resistor. You then see the trace of the right cap voltage compared with the shown square pulse, both simply referenced to the same earth of that circuit. See there that the initial vertical bit of that right plate wire trace looks exactly like the current trace shape that you saw earlier, ie: an initial spike which (almost) shoots up to 10v vertically with the leading edge of the 10v pulse input, and then it falls towards zero exponentially, just like the current. That right plate voltage drop which you are seeing there is the reason for the exponentially similar rise which occurs across the cap plates. (it's an integration then seen here across the cap in this case. The left plate stays put at 10v as per pulse, and 10 -Vr = Vp where Vr is the right plate voltage at each point in time as it drops, and Vp is the resulting voltage at that time across the plates).

Note that it does not matter what value you use for the input pulse voltage Vin ... it does not have to be 10v. The above holds true whatever value is used there, and the exponential time taken for both the right plate drop and the integrated voltage across the plates will always be approx 5RC from start to completion. The same applies to the more commonly seen circuit on the right, where the 'across cap' integrated voltage charge is measured at the wire between the resistor and the cap in reference to earth there, where no drop of plate voltage is involved in the charge up, and there the Vacross cap is just the rise above zero at any instant of time... which works out to be the same in both left and right circuits, and can be determined by standard formula.... and in that formula when used in either circuit here, Vin is the 10v square wave pulse with earth reference, and Vout is the voltage across the cap, where in the circuit on the right that is cap to resistor wire... using earth reference. The left circuit Vout is the cap to resistor wire ... using the 10v pulse input as its reference.... Again showing important differences between the two circuits, but with similar outcomes. In both cases the capacitor is 'seeing' the voltage p.d. across its terminals.

No, the funny thing is you do not know how to measure voltages and use the wrong voltages to buildup your absured theories about the "very diffrent" charge procedure. First you talking about the Voltage at the right plate (which is the voltage across the resistor or across generator+capacitor) then next you talking about thensame point on the right circuit between RC but now its the voltage across the capacitor or across generator+resistor. To buildup your theorie you used the voltage across the resistor on the left circuit and on the right circuit you have used the voltage across the cap! Left you take the HPF, differentiating voltage (the spiky one) and on the right circuit you take the LPF integrating voltage. You use two different voltages to explain you absurd theorie about the different way of charge. Isn't that funny? Poblem is, you measure wrong voltages to build wrong theories.

I can't help you any further unless you ask just one specific query at a time which you think is wrong... eg: your first query seems to be about the voltage on the right plate of the cap in the left circuit. Let's look at that again: The voltage on that right plate wire (using earth as the other terminal of the meter like I said), can be seen on the scope as shooting up with the square pulse left leading edge. Now, before the square pulse was switched on, there was no charge at all on the cap plates, so the right cap plate was no potential (left circuit the whole discussion here). Then when the 10v pulse is switched on, the left cap plate is instantly connected to that 10v, so surely this far you agree that the left plate is now 10v as shown by the vertical left pulse leading edge. ... Hence, that increase of 0 to 10 volts happened with great speed. Current through the circuit is directly related to that speed, so we therefore see a similar spike shootup of current also when the 10v pulse first occurs at the leading square edge. In that first instant then, the cap now appears as though it has suddenly become a closed switch, conducting infinitely high current which is limited to a maximum here due to the resistor (V= iR). If you re-draw a new similar left circuit with a closed switch instead of a capacitor there, you will see that the full 10v is now at the left terminal of the resistor, ie 10v at the right side of the switch. In other words, the right side of our capacitor is now also at 10v, just like the left side, due to the current which 'closed the switch'. Look now at the trace on the scope for that left circuit cap right side plate wire. Do you agree that it shows a sudden high increase along with the current and the leading pulse edge. (It doesn't quite reach full 10v but ignore that lack here). ... So: when the pulse is first switched on, there is a sudden shoot up of the right plate to become (almost) equal voltage to the left plate, the left of which is held at a fixed 10v being connected to the pulse. Is there any of that which you disagree with so far?

Nobody can help you, stop writing so much and try to understand. You do repeat again and again the same mistake. Your absured theorie "very different way of procedure in charge the capacitors" is based of TWO DIFFERENT VOLTAGES ! Cant you see that? Because the voltages are different you argue its a different procedure? This is just bullshit, wakeup and see the mess you have created, do you think this is helpful for anybody. Again, you use to different voltages to argue the charge procedure is different? Left the voltage across the resistor and on the right circuit the voltage across the capacitor?????????? TWO DIFFERENT VOLTAGES? Please use the same voltage in both circuuts and compare, anything else is bullshit. You compare apples to pears.

Look carefully: Your mention of me saying "Left voltage across resistor etc" would be better seen as me saying " voltage of the cap-to-resistor to earth" in both left and right circuits ... thus similar , not apples to pears, but we need to stick just with the left circuit until all is clear there first. We will not be able to continue with this if you can not answer my previous guiding questions one at a time, so again: Is there any of that which you disagree with so far? You really need to consider deeply that scope exponential drop as seen for the right cap wire in the left circuit.

We are talking about your headline of this thread, is that clear?

we can not continue if you do not know how voltages are measured correctly, this will be a mess as always.

"Left voltage across resistor etc" would be better seen as me saying " voltage of the cap-to-resistor to earth"in both left and right circuits ...

You have to compare the voltage across the capacitor in BOTH circuits. And not across the resistor in the left and across the cap in the right

again you continue to compare to different voltages

holy shit

how many times do i have to repeat? In the left circuit you neasure the voltage across the resistor while in the right you measure the cap. Vr left Vc right

Left Vr

right Vc

dont you get that? Base on two different voltages you build your absurd theorie.

right?

"Two very different ways of charging"

because Vr looks totaly different then Vc you build your absurd theorie the charging is very different.

is this scientific? NO

this is messy behaviour. I cant help you, you are lost

My offer to help you ends here. You are obviously trying to divert attention from the fact that you missed the original point of mine about the spike and reduction of voltage in the right cap of the left circuit as seen on the scope trace, which you will not even acknowledge exists.

you dont get it. You contine to talk about two different voltages. Any help is waste of time. You made a big mistake and dont want to see that. Based on this mistake you developed an absurd theorie "very big difference on charging" but you compare two circuit but two different voltages out of this circuits. That is just dumb. Measure in both circuits the voltage across the cap and you can continue. Else better stop and look for a anither hobby. Or do it privat nobody must see this bullsh

how do I delete other people's circuits

If someone remarks: "What an excellent man you are!" and this pleases you more than his saying, "What a bad man you are!" know that you are still a bad man.

@faceblast - you cannot because the ability would extend you you as well. You can, however, block their comments and circuits from appearing at all of your computer alone:--------------------------------- http://everycircuit.com/circuit/6697380226007040