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modified 6 years ago

The paradox about capacitors

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02:54:41

Capacitors with the same capacitance and the same serial resistor charge and discharge equally fast, no matter how high its charging voltage is. Try it yourself!

published 7 years ago

7 years ago

In german readable form Q=CU

7 years ago

@hurz ... Your formula of Q=CU here has no relevance for the charge/discharge timing of a capacitor which LeButch correctly described. Feel free to adjust your formula to make it relevant.

7 years ago

CU sounds like Q.

7 years ago

So Q=Q

7 years ago

@kiani ... There is a difference between a formula being correct, and a formula being relevant. The hurz formula there is correct but irrelevant regarding the timing calculation of a capacitor, unlike your formula above (C * V = i * t) which is just plain incorrect.

7 years ago

Do you not agree that,,,,,,,,,,,,,,,,,,,,,, i = C (dV/dt) !! If this is correct then,, practically CV=it is too.

7 years ago

Messy is back, do you still see 4 different voltages in a circuit of 3 components: Resistor, Capacitor and Battery? Shut the fuck up

7 years ago

@kiani, last time he told me a triangle has 4 sides. 4 voltages in a circuit of three components. But this was only valid in one special circuit, in a second fully symetric circuits there where just 3 voltages. But the circuit with 4 voltages was his special circuit. What an idiot

7 years ago

Q=C*U is a valid formula and can help to understand this circuit here, but only if you do not see 4 fantasy voltages. Hallucinations do not help you to see the real world.

7 years ago

I also think it is both of course correct AND relevant. Since we are not trying to work out timing, but rather the above example and Q=CU shows indeoendace of element time.

7 years ago

It is clear in a blured way that @ 2ctiby thinks maths. And not electronics engineering.. As an example. A shop keeper looks at the items in his shop and he notes down and calculate his profit in a blink of an eye. While a chef looks at the same items and he can taste the meal he is going to prepare...

7 years ago

i = C dV/dt is correct, but that is often very different from i= CV/t (you used the latter). eg: look at say a normal integrator or differentiator RC circuit where we see regular current spikes. Those spikes are caused by the rapid sudden change of input voltage switching on/off, eg as input pulse rise/fall edges. ie: it is that "change" of voltage in unit time which is relevant, regardless of the actual value of voltage. ie: The "speed" of V, not the value of the V itself is the main feature, (and Kirchhoff remains intact because the numerous voltages/currents do not act concurrently in time), and that is the point of the statement by LeButch: 'The charging time does not depend on V' ... (it depends on the rate of change of V, and an appropriate formula shows it).

7 years ago

Of course what you say is one way of looking at it,, however.......http://everycircuit.com/circuit/6631457158856704

7 years ago

@kiani ... Your link here is diverting away from the points which I made for the LeButch presentation. In your link, it is not an RC circuit as per our discussion. The link has a constant current source which presents a whole different aspect regarding integration, so let's just stick to what I said: Your presented formula was wrong and that of hurz was not relevant to the charge timing point which LeButch was describing.

7 years ago

Here an electrinics problem solvrd with my equation.....plz. Chk........http://everycircuit.com/circuit/6628158556864512

7 years ago

@kiani ... The formula: i= CdV/dt can be integrated for easier usage and combined with the resistance in an RC circuit. The outcome of that is a useful formula which can be used to give us the V, i, or t at any instant during the charge/discharge of the cap. You may come across different manipulations of this same formula, depending on whether you are looking to find current or voltage. If it is just the full charge/discharge time that we are after, then we can simply use the standard: t = 5RC The discharge formula for any point of i is: i = io* (e^-[t/RC]) where io is the max i of the RC network. Feel free to use that formula to find your required structure. (you may perhaps prefer to choose a C first then work from there, rather than finding a C). You can then compare your new result with your link result and so consider any difference. You really need your own new thread for this now.

7 years ago

kiani's Q=CU is correct and useful. What we can see is Q = 0.01mAs = 10uF * 1V and for the second circuit Q = 0.1mAs = 10uF * 10V so what we see mathematical the second circuit with 10V is just the first circuit but multiplied with 10 on both sides! So proofed, the time to fully charge the capacitor is identical in both cases and the voltage is irrelevant. Thats the easy way. Lets do it with a more complex looking formula. We can use u(t)=U*[1-e^(-t/tau)]for the voltage across the capacitor where U is the maximum voltage. We can already see the voltage is independed cuz the right part of the formula is a % value from 0 to 1 and there is no impact to "t".

7 years ago

Yes this is useful and a step ahead, cause it dhowes we don't need to calculate something we don't need, cause the system runs independent.. I suppose that is why RC is called Time Constant.......,, cause time is constant...... All roads end up in Rome!

7 years ago

The easy way to find the full charge/discharge time regardless of volts input is 5RC like I said, but to find any given time, Volts, current point in the charge/discharge is going to require a formula on the lines of my description, not just Q=CV, and not the wrong formula mentioned originally by kiani.

7 years ago

Ok, let me say something bland,, So we all agree on different points... My formulea is rule of thumb formulea, ehich might not be correct fully but is not wrong either ( if one knowd what they are doing).

7 years ago

@Hurz. Would it be right to say any storage element, must have a time independancy somewhere, cause that id how they do their storying.. Like memory which tskrs time to make and time taken out, means storage.... So conclusion. Is. The black holes contain the memory of universe. Snd anything and everything that exsisted in and eith time... Now the wuestion arises, that can memory exist again ( in and with time),, in other words become time!? My answer is,, it has no choice but to exist again.

7 years ago

if you understand its voltage independed to the point 5RC any idiot should see its also valid for any other percent of voltage independent of max voltage or you dont understand it at all. But that wasn't the point LeButch made, that was just an addon i made. @kiani, my Q=CU was translation from your CV= it cuz C is capacitance V is voltage and in german U. Q is Ampere seconds and that was what you wrote with i * t , right Ampere seconds. Q=C*U=i*t @kiani you made a thumb of rule calculation, but your formula is absolut identical to my one, i dont see a translation error, only Mr. @2cent does know it all better

7 years ago

This is poetry.

7 years ago

With higher voltages, the inrush current is higher, so the cap charges faster. Same thing for the discharge. Thats why they charge equally fast.

7 years ago

Q=CV as mentioned by hurz. That formula needs adapting for t variations when time is involved, (which is what the original post is considering). Kiani showed a t adaptation in his formula, but it was a wrong adaptation which did not consider the "change" in volts as opposed to the actual value of the volts. That is where hurz and kiani differed...hurz was shown unresolved, kiani was shown resolved wrongly for describing the charging progression. The relevance of dV as opposed to just V is important to the understanding of how and why the spikes occur which affect the charge/discharge of a cap. Both were lacking in describing that all important aspect.

7 years ago

@LeButch ... yes to some extent. That extra high current inrush occurs because of the sudden change of volts (ie dV is high, and i is proportional to dV) at switching on/off. However, because the time taken for the instant switching is virtually zero (ie: dt tends to 0), then even 1v change of instant input will cause a virtually infinite current, ... thus a 100v change makes no difference. The reason being that in either case, the max current will be achieved as limited by the series resistor in that circuit. That is why we talk of the cap current being like a closed switch at the outset, with max current.

7 years ago

Once the switch is closed, the input voltage stays put at its constant high (d.c. square wave)... ie: the dV has now suddenly changed to zero. Then since dV is zero, the current must also be zero (i is proportional to dV as before), so the spike action is halted and the residual current decreases exponentially towards zero ... as described by the formulae which I am referring to, hence the shape of the current trace on the scope, until a new action occurs when a switch off suddenly occurs at the falling edge of the input voltage. That shows differently according to whether it is a snip or a return path for the discharge. The converse of the current decrease is seen as the rise in voltage charge across the cap.

7 years ago

@2ctiby, thank you for sharing...👍

7 years ago

Lenny tops the chaos, at least one understands sentences like, "Then since dV is zero, the current must also be zero, so the spike action is halted and the residual current decreases exponentially towards zero" can you explain Lenny? Its ZERO and then it decreaaes down to ZERO?

7 years ago

Lennzul. Is busy re_writing defenition of voltage as Energy,,, he is on to unified formulrea for the whole thing...

7 years ago

@lenzrulz. Thanks ... Perhaps I need to clarify that detail for hurz: Whilst the switch is in the full closed position, there is then regarding the input, no 'change in voltage' (ie: 'dV' of the input is zero whilst the input voltage holds at 1v unchanging... top schematic section). Before that, whilst the switch was in the process of making its brief closure, the rate of change in input volts was high, ie: it went from 0v to 1v in a twinkling of an eye time, and so caused a creation of a high current spike. Now that dV has become zero however, that 'creation of current' suddenly becomes zero also. The now ensuing drop of (as opposed to creation of) current from its spike peak value, occurs exponentially in time as it drops towards zero current value, all as seen by the scope current trace if the top resistor is highlighted in the original schematic shown above here.

7 years ago

Kiani used to make respectful comments until he became Warner’s (“hurz”) bootlicker, I guess you are the company you keep...🤔

7 years ago

@2cent long drivel answer as always, instead of saying sorry that was nonsense, if something is already zero it cant go down to zero again. Such explanation makes it impossible to take you serious. BTW, have you noticed what you saying is completley out of topic? We are talking about this circuit here from @LeButch, it was just about chargeup a capacitor in a low pass configuration, NOT dischrage. Think about and be less messy and shorter and more precise in answer and some might follow you in feature and understand what you want to say. Bootlicker Lenny did not understand anything. This fraudster stinks 🤧

7 years ago

Here we see Hurz making a complete fool of himself. He can not follow my detail (which would be reasonable if he were to admit it, or just keep quiet), but instead he angrily calls it nonsense and thinks that the described lowering of current is the cap voltage discharge. That lowering of current is just where the voltage is rising across the cap as the current drops ... ie: it concerns the voltage charging only, as can also be seen on the schematic. I have made no mention at all about cap discharge. Hurz is completely lost regarding an RC circuit like this passive integrator. He is embarrassing himself. He just does not understand i= CdV/dt ... I am happy to answer any questions from anyone regarding any aspect of my detail.

7 years ago

BTW, how many voltages do you see here per circuit, 3 or 4. Already at that point nobody can follow you. Nobody can follow sentences like "its zero volt and now it decreases to zero". Cant agree at least on this ITS bullshit? No you cant.

7 years ago

I never said "its zero to zero" anywhere... That is all in your wild imagination. I described the dV in an appropriate way which you are incapable of following. You don't seem to know the difference between V and dV, and you even thought that discharge was involved.

7 years ago

Try asking a straight specific technical question by quoting any sentence of my description and I will answer them one by one.

7 years ago

My question, and its my only one i have to RC circuits as you understand them: How can something decrease to zero if its already zero? Thats all. BTW, was it wikipedia which get it wrong or was it you?

7 years ago

Warner (“hurz”) excels at throwing red herrings at his opponents to mask his embarrassing statements. Nowhere did 2ctiby say “it’s zero to zero”, unfortunately Warner lacks elementary reasoning skills as evidenced by his snide remarks. He should pay attention to what is actually written rather than what he imagines is written.

7 years ago

Nothing anywhere throughout my description mentions "zero to zero". Instead, I said that the current decreases from its maximum current value exponentially down to zero current. That starting max current is simply V=iR ie: i=V/R so in the top schematic of 1v and 100k, the starting current for the ensuing mentioned drop is 100uA in theory (ignore a few uA difference here). Look at the top current trace and confirm that curve drop to zero from approx 100uA

7 years ago

That current drop corresponds to the similar but opposite rise in volts ... ie: the charge up voltage of the cap. Notice that the starting point of that max current is at the top of a linear spike which is on the left of the seen curve there. That top of the spike where the curve starts is important. It is the point where the switch is classed as fully closed, as opposed to the switch being in the process of closing.

7 years ago

Whilst the switch was in its process of closing, the change in volts with respect to the extremely small time taken was very high. The change in V was 1v .... called: dV Let's say the change in time was 1ms ... called dt Then dV/dt = 1/0.001 ie: 1000 volts per second. Mathematically, the current = that rate of change. ie: i = dV/dt so here the current would be 1000A. But it can not go more than the the limiting resistor allows, so the max current is our 100uA

7 years ago

"ie: the dV has now suddenly changed to zero. Then since dV is zero, the current must also be zero (i is proportional to dV as before), so the spike action is halted and the residual current decreases exponentially towards zero"

7 years ago

At the end of that first 1ms the switch is now suddenly classed as fully closed, rather than being in the process of closing. The input voltage of 1v now remains constant, not changing (unlike when it was in the process of heading up to that 1v during the switch closure activity). So now dV is zero ... ie: there is no change in input volts, regardless of time passing. The top of the square wave horizontal stays at a constant unchanging 1v (very different to that rapid vertical increase from the leading edge 0v to 1v). Notice that V is held at 1v, but dV is now 0 ... the change in voltage has gone from very high as per vertical edge, to zero suddenly now for the horizontal unchanging 1v input. That point where the dV changes to zero, ie: at the top of the rising edge of the input voltage, is where the current suddenly ceases and becomes 'zero created current' as opposed to the 'rapidly created current' of the spike seen at the leading edge of the square wave 1v vertical input.

7 years ago

to explain to you both, this is nonsense, i is not proportional to dv, its di is prop to dv. and i is not zero at that point its i = 1V/10kOhm and will then decrease down to zero exponentially, even it will not reach zero, anyway. di is first high positive and then it turns over to be negative, the point where it crosses zero is by the way for its understanding irrelevant. Please separate i and di carfully before you write this bullshit above. Any answer for you should be very short, nobody is willing to read your nonsense, even Lenny does not, cuz he is lost anyway, Lenny do us a coffee nobody expects more from you.

7 years ago

Although the current has now ceased and is not producing any further current (ie further i = 0 now), we still have to deal with the high (max) current which was there at the top of the leading edge spike. If this circuit had been a simple resistor without a cap, then that current would disappear instantly, but here with a cap in an RC circuit it can not go that instantly. Instead it must drop to zero gradually, ie exponentially as per the RC values. So... we now see that dwindling current coming down to zero from its max starting point at the top of the left spike. (That place where i was zero, meaning that there was no more creation of current, not forgetting the current already there which we are now dwindling down). Whilst this current is dwindling down, as seen by the exponential current curve, the voltage is being created (integrated) across the cap in a similar but opposite way. Hence that voltage across the cap shows an integrated voltage curve of the current. The formula for that integration is the one we are familiar with where the current i has been replaced by V/R and re-arranged so that the output V is presented in terms of input V rather than by i (although a similar formula can be used for i just as easily).

7 years ago

Charge up formula: Vt = Vbatt(1-e^(-t/RC)) where Vt is the output volts across the cap.

7 years ago

Note: hurz is wrong when he says that it is di rather than i that we are dealing with. We are dealing with i = CdV/dt not di in any of the above description of mine.

7 years ago

i just read the formula and this is correct. Your "zero to zero" is not and dont mix i and di in feature. You answer in four comments for such a simple circuit, is it so promplematic for you to understand, do you think Lenny can follow you? Lenny where is my coffee, not even this you can do.

7 years ago

I have nothing further to say other than i = CdV/dt viewers can check it out. It is not di there.

7 years ago

you deal with ONLY a capacitor you always forget the hole circuit which includes a resistor and a battery.

7 years ago

And your comments are long and longer and loooooong to hide your little knowledge? Keep facts short with arguments if you wish somebody does follow you

7 years ago

and remember only @2ctiby is right Wiki is wrong and hurz is always wrong

7 years ago

what about 1V/100k is equal 100uA? @2ctiby is always right, no shit you mean 10uA. Oh no you messed up 100k its just 10k. Zero down to zero, WOW you are the biggest messy i have ever seen

7 years ago

Yes, LeButch top schematic resistor is 10k seen which gives the 100uA which I mentioned, not 100k resistor ... a little typo compared to a mass of your ignorance in all things RC and calculus which you know nothing about and pretend to know all.

7 years ago

in your crude syntax i is first high then its zero and then its 100uA to go down to zero. Very funny. so far i had the hope you have enough knowledge to bring math together with simple electronics, but this chain, zero, high, zero, 100uA is a epic fail which i call bullshit in mega long packed comments not worse to read. Keep shot and switch on your brain before offending users who read this nonsense you cant explain.

7 years ago

Hurz wouldn't know where to start if we mention integration.... insults and diversion would ensue from the start. What a dick head. ... and where has anyone ever come across di in the mentioned description... he won't be showing us any time soon.

7 years ago

you see, instead of short technical arguments you have just rude and nasty ones to be a great foolmouth.

7 years ago

in the moment you realise you cant explain your own bullshit your bloodpressure seems to be very high, you reapeat yourself, this is like hospitalism.

7 years ago

healthy/reality check: you see still 4 voltages in an RC circuit like the first one?

7 years ago

@Fraudz ...Let's see you integrate something simple so that we can continue laughing at your pretend knowledge.... how about an RC integrator circuit... Then again, don't bother. We all saw how you thought that I had mentioned cap discharge when it was never introduced. Complete Fraudster ... watch this space ... di coming soon.

7 years ago

Its always the same, you leave a chaos in your explanations full of errors and if im the only one who is reading these long lasting mess and find unlogic bullshit you cant explain. I started to explain what you might mean end used feerly some math "di" and i in senese of i(t) sure. Is this to much math for you to follow, just because you do not see that in you wiki folumlar? If you dont mean di then its even worse what you mean with "zero current followed down to zero current" high dv with an intermezzo down to zero and then upto 100uA to then slowy dwidling down to zero again. @2ctiby this is a linear circuit and if this is not helpful for you to see you talking nonsense, sorry what else can help you, MATH CANT if you do not understand this. You really lousy math knowledge if its not a cooking recipe you are lost.

7 years ago

What a load of horse pucky from Fraudz who thought that discharge was involved because of the curve shape. We need say no more.

7 years ago

You think anybody can understand your messy presentation? You cant answer question and mix i(t) with di and i and all this "zero down to zero" which you will never have said but what else can i do to copy and paste it here, and you ignore any explanatiins, funny. So why all this find its just chaos, more chaos they any chaos circuit kiani presented. Messy

7 years ago

Warner ("hurz") is tenaciously fixated on propagaing his imaginary "it's zero to zero" assertion despite multiple comments clarifying thus issue. His pigheadedness gains him the notoriety of being EC's most obnoxious troll that has ever slithered around a community. And he wonders why no one gravitates to him socially?...🤔

7 years ago

Fraudz continues to lie, decieve, divert and abuse day after day, year after year. EC is doomed to being a very sad site until he is booted out permanently.

7 years ago

Lenny can only bully and jump on threads he by far dont understand. @2cent prepares presentaions which nobody reads or understand messes around and if making mistkes dont point that out he will be impertinent, cause we all are wrong, wiki is wrong im wrong, just @2cent is correct and nice.

7 years ago

I understand more than your pay grade and your bully assertion is hilarious...😂

7 years ago

go and make us some coffee if you can, dont tell me you know how to make coffee, or i will skipe this request, cuz you are a fraudster the only thing you can is lying.

7 years ago

Your Achilles heel’ is your lack of tenable evidence that I’m lying hence your rhetoric, the only question remaining is which type of coffee would you prefer?

7 years ago

you are always looking for others Achilles heel, is that what fraudsters do, right? Where is the weak point of a system so you can intrude the easiest and safest way, right. Tell us more.

7 years ago

You would never understand...😂

7 years ago

right, your presence here on EC in relation with electronics is absolut incomprehensible

7 years ago

i = CdV/dt for an RC circuit ... but Fraudz says that it should be di and not just i ... Now then Fraudz, explain where di fits in to anything in an RC circuit.

7 years ago

Make use of di freely in any circuit. You will be surpised its present in all circuits, Mr. Messy

7 years ago

Nonsense hurz. The current i needs to be integrated when dealing with RC circuits for the charging voltage formula which I was describing. Your di is a differential, and is thus back to front and unusable for that RC topic ... Stop plucking calculus bits out of the air from the internet ... you are only going to look stupid by doing that.

7 years ago

Warner has redefined the word “stubborn” given his incredulous attitude concerning 2ctiby’s evidential argument.

7 years ago

di is not relevant for RC circuits, you get in new trouble Mr mess

7 years ago

@hurz ... detail the use of your suggested di for RC networks, then we can discuss what you present amicably.

7 years ago

Could "i" have a go at this.. With permission..... i=c(dv/dt),,,, now. Just look ....t=c(dv/di)

7 years ago

We can not separate the t here. Start again, and present that correct i=c(dv/dt) in terms of dv= Why can't we separate the t here ? ... The reason is because dt means "the change in time" and we can not separate "the change in" phrase from the actual time yet. (but we can eventually, albeit not by simply shifting it over to a different parameter such as the i ).

7 years ago

i or I is nrver instant. So in real world there is only ever di.. / d x

7 years ago

When dx small enough, then i bacomes I.

6 years ago

@kiani ... No. Your understanding of calculus is very limited, as we saw when you thought that d could be used as an algebraic term. You would be better off avoiding your pre-thought ideas about calculus, and learn it properly before using it wrongly.

6 years ago

In an RC, V..circuit there are 5 variables.....R,,,C,,,V,,,i,,,,t,,,, now one could derive a formulea in terms of any variable..... Right!? One could derive that. t=CdV/di,, this formulea works, "i" hsve checked it on a cct. On EC snd made measurements....anf it confirms.....now lets see your calculus derive an expression for i in terms of other varuanles. Plz.

6 years ago

http://everycircuit.com/circuit/5136243550322688

6 years ago

kiani, he is not willing/able to fillow any external ideas and thoughts. Everything in a discussiin must come from him and you have only to agree or directly to disagree and get then one more a repeat copy of what hensaid before. He is not able to take your position, which means his knowledge is limited to a cooking recite he knows, but dont go a half step left or right, hes lost. You must follow him word by word and agree everything with "yes sir." thats how "fun learning" works with @2cent fraudster.

6 years ago

Yes he seems to be a slave to rigidity of mathematics... he is doing it the wrong way,, instead of relating problems yo maths. He is relating maths. To problems. The mathematicians call "fiddle factor" a "constant". And then they are happy.

6 years ago

i dont even think hes a mathematicians. For a while he was very keen to know what my degree is all about, i told him exactly while he tried to down grade it in any sentences. But i think he was trying to bring it down to something whats similar to his "grade". Many fail hes doing, its little knowledge in math pushing somewherr into everycircuit for any price, but the corner connections ignoring and interfaces to practical examples are completly missing. Its like a Chef knows how to make a super jammy sauce and is going to MacD to teach them how to make it, but forget to which burger it taste good. If he adapts his sauce it will taste jammy to any burger, but he did not even tried cuz he is lost to bring it together. Burger and its sauce!

6 years ago

Warner my friend, you cannot reason your way out of a wet paper bag...😂 You have zero reasoning skills plus you are no (calculus) match with 2ctiby and that my friend is a bitter pill you can’t swallow.

6 years ago

Well lenzul your calculus seems to be as good as your electronics.

6 years ago

Well, not everyone has so good electronics like you do, Kiani...

6 years ago

Lenny, nobody can take you serious, for what?

6 years ago

@buchLet you havrn't tasted my cooking.!

6 years ago

Warner, nor does anybody take you seriously my friend...😎

6 years ago

Heh, Kiani didnt get it...

6 years ago

No you didn't get it, bucho.

6 years ago

Lebucho is bored.

6 years ago

hmm, kiani to me it looks a bit different, why?

6 years ago

Sometimes i go soft on horrible kids, they must have had a tough life..just eating and sleeping, and have their clothes washed by mumy and dady.

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