Capacitors, basic questions - Page 1
Forum section: electronics / beginners

author: Richardcavell
author_rank: Contributor
timestamp: November 22, 2024, 09:03:12 am
content: Hi.I’m learning about capacitors. I’m watching a simulation on falstad that is available from the circuits menu->Basic->CapacitorsWhen the circuit switch is to the left,1. Is it correct to say that the capacitor is being charged up?2. Conceptually, should I be thinking about the voltage as rising or should I be thinking about the charge rising?3. Is it the case that the voltage across the capacitor increases as it gets charged? And that means that the voltage across the resistor gets less?When the circuit switches to the right,4. Would a capacitor always discharge in the opposite direction to how it was charged?5. if the capacity is electrolytic, is the bottom terminal negative?6. How reproducible is the amount of charge that a capacitor can store? If I cycle it multiple times, is there variation?

author: golden_labels
author_rank: Super Contributor
timestamp: November 22, 2024, 09:23:38 am
content: Quote from: Richardcavell on November 22, 2024, 09:03:12 amYesQuote from: Richardcavell on November 22, 2024, 09:03:12 amEither. Both statements describe the same phenomenon equally well. Though some linguists would argue in strict sense charge is being collected, not rising.Quote from: Richardcavell on November 22, 2024, 09:03:12 amYes.Quote from: Richardcavell on November 22, 2024, 09:03:12 amYes.Quote from: Richardcavell on November 22, 2024, 09:03:12 amOn a schematic, which terminal is which depends entirely on how you draw it. The convention is to draw lower voltages towards bottom, regardless of the component. But that’s not always possible or convenient.If you ask about electrolytic capacitors, their internal structure makes them suitable to be charged in one direction only. In that sense they are polarized devices and have clearly marked cathode (positive) and anode (negative). Note there do exist non-polarized (bipolar) electrolytics capacitors too. But they are for niche applications.Quote from: Richardcavell on November 22, 2024, 09:03:12 amIgnoring aging, without external influences the capacity of a single capacitor is very reproducible. However, mind that it varies heavily across speciments and may be influenced to great extent even by the voltage you apply to the capacitor, not to mention other effects.

author: PGPG
author_rank: Frequent Contributor
timestamp: November 22, 2024, 12:33:23 pm
content: Quote from: Richardcavell on November 22, 2024, 09:03:12 amImagine that inside capacitor it has some parasitic serial resistance (for ceramic capacitors these may be single mΩ, for electrolytic it could be hundreds of mΩ). Energy dissipated (during charging and discharging) in this resistor is converted into heat and heats the capacitor. This is what you loose during charging discharging. If you cycle too often you can overheat capacitor. Because of these electrolytic capacitors destined to work in DC/DC converters (like 200kHz) are not the same as destined to work in filtering 100Hz from bridge rectified 50Hz supply.I saw a regular (non-electrolytic) capacitor swollen from heat dissipation (it was working in a 2kW radio transmitter).

author: MrAl
author_rank: Super Contributor
timestamp: November 22, 2024, 03:36:26 pm
content: Quote from: Richardcavell on November 22, 2024, 09:03:12 am#1 and #2 It depends on the context. That is, what you need to analyze at the time. If you need to analyze the circuit with respect to the charge, then charge is a concern, but if you need to know what the voltage is doing, then voltage is the main concern. In circuit analysis, the voltage and current are usually the choice of variables because they can show what is happening with the capacitor along with other voltage measurements. For example, if you were using Nodal Analysis you would be concerned mostly with the voltages and currents in the circuit.#3 Yes, we say the capacitor gets "charged". We think of that as if something is entering the capacitor that does not leave unless we discharge the capacitor. That something we call "charge". Capacitors are a little more complicated than that, but you often do not need to think of anything more than that above. If you want to know more, you'd have to study capacitors themselves and how they work in detail.#4 If the current flow was into one of the capacitor terminals when it was becoming charged, then to discharge would mean drawing the current out of that same terminal. It's almost like filling a beaker with water, then to empty it you have to have the water flow out not in.#5 An electrolytic capacitor can only be charged with a current flowing into the positive terminal following conventional current flow, which means charge goes into the positive terminal to charge it. To discharge the charge (current) has to flow out of the positive terminal. Note this is using conventional current flow which means current flows from the positive of a battery around the circuit and then into the negative terminal of the battery. Also know that if we ground the positive terminal of the capacitor we can apply a negative voltage to the minus terminal of the capacitor to get the same charging effect. So in that case it's almost like we are using the negative terminal of the capacitor to charge it, but the polarity is the same as before as the positive terminal is still more positive than the negative terminal.#6 This depends on the context too, as to what you are trying to find out. Are you:A. Trying to figure out if the capacitor loses energy all by itself, orB. Trying to figure out if you can use a capacitor over and over again without it degrading.For A, capacitor are not 100 percent efficient. Therefore, if you pump energy into it, you cannot get that same amount of energy out of it because there will be a loss. They sometimes call that dielectric absorption. For B, you can use a capacitor over and over again, which means charge and discharge it over and over again, but they can wear out, especially electrolytics. Electrolytics are a problem in power supplies because they leak electrolyte and that causes the capacitance to become much lower and also causes the ESR to increase. The ESR is the equivalent series resistance of the capacitor, and that's like there is always a small value resistor in series with the capacitor. That resistor eats up energy. It can be on the order of 0.001 Ohms up to around 10 Ohms but maybe more.

author: Harry_22
author_rank: Frequent Contributor
timestamp: November 24, 2024, 08:05:15 am
content: Hi!If you begin to operate with such a concept as the charge then it would be most correct to start studying the capacitor from the electrostatic position.In radio engineering the capacitor usually begin to be studied with an RC circuit.I made a special for you in Microcap.PSFreeware Microcap is here. Just unzip it into the folder and run any MC12.https://drive.google.com/file/d/1MvPEWrPnJActRC2H1dEmYp3nr6EGssbp/view?usp=drive_link

author: bostonman
author_rank: Super Contributor
timestamp: November 25, 2024, 02:26:52 am
content: Since we are on the topic of basic questions regarding capacitors, I've often wondered about something (even though I should know the answer at this stage of my experience/education).When voltage is first applied to a capacitor, it's a "short' which causes power supplies to dip at power on.Is the "short" actually shorted to ground or is it because the capacitor is "empty" like an empty barrel and the inrush current is quickly filling it?

author: SteveThackery
author_rank: Super Contributor
timestamp: November 25, 2024, 11:08:40 am
content: Quote from: bostonman on November 25, 2024, 02:26:52 amThat question shows you might have the wrong mental model, because a capacitor is not analogous to a barrel. When water flows into a barrel it fills it up, but water doesn't flow out of the barrel at the same time. When charging a capacitor, current flows into one terminal and exactly the same amount of current flows out of the other terminal.I'm going to post a diagram later on showing my own mental model, which I think is more accurate.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 25, 2024, 11:58:56 am
content: Sorry for the really shit drawing! (Thumbnail below - click to expand.)I think this model is more closely analogous to a capacitor. Imagine this whole system is filled with water. There is a chamber with a diaphragm dividing it into two halves. The diaphragm will stretch, and it is held in the middle position by two springs, as shown.When a small amount of water is pushed into the left hand pipe, the diaphragm begins to move to the right, which has three effects:1/ A equal amount of water is pushed out of the right hand pipe.2/ The springs acting on the diaphragm act to "push back" against the incoming flow of water. That is, they create aback pressure(a voltage) proportional to the volume of water added to the left hand compartment.3/ When the back pressure from the springs equals the pressure of the incoming water, the flow will stop.4/ Increasing the pressure of the incoming water will cause more water to flow into the left hand compartment until the back pressure from the springs again equals the pressure of the incoming water. At that point the flow will again stop.5/ When water flows into the left hand compartment, the same amount flows out of the other side.6/ If the water pressure is removed from the left hand compartment, the springs will push the diaphragm back to the centre position, causing water to squirt out of the left hand pipe, and the same volume of water to be sucked into the right hand compartment.Thus a capacitor acts like a "springy barrier" to the flow of electric charge. The back pressure from the springs is proportional to the amount of charge pushed into the capacitor.There is one shortcoming with this model: the shape of the chamber suggests that, given enough pressure, the chamber can actually fill right up and store no more water (charge) regardless of the pressure applied. This is not how capacitors really work. With a capacitor, you can push more and more charge into it until the diaphragm splits open (the dielectric breaks down).Nevertheless, the important concept is that a capacitor acts like a springy barrier to the flow of charge, and charging a capacitor involves current flowing into one terminal and an equal amount flowing out of the other terminal.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 25, 2024, 11:59:11 am
content: Quote from: SteveThackery on November 25, 2024, 11:08:40 amI hope this comment doesn't sidetrack this discussion and confuse the OP. I have some of the same kinds of questions as they do.So the exact amount of current flows into and out of a capacitor. Just like a wire.However, unlike a wire, the current doesn't flow in and outinstantaneously, isn't that correct? Because the capacitor does store charge; the current that gets released is delayed from what flows in, right? In fact, isn't that one of the fundamental properties of a capacitor?Since a capacitor does store energy, it kind of is like a barrel, but perhaps a barrel with an outlet and with some baffles inside that impede the flow of water. Or something like that ...QuoteThat might be very helpful.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 25, 2024, 12:01:47 pm
content: Quote from: Analog Kid on November 25, 2024, 11:59:11 amActually no, that's not right. There is no delay. Also, the energy is stored in the sprung diaphragm (the dielectric). Hopefully my post will help.

author: Analog Kid
author_rank: Super Contributor
timestamp: November 25, 2024, 12:24:15 pm
content: Hmm; now I'm a little confused. And I'm certainly not disputing you on this, but since capacitors have this behavior:how could current flow out of a charging capacitor with no delay? Wouldn't that be behaving like a wire (or a non-capacitive, non-inductive) resistance?, where current flows instantaneously? isn't that what the time constant of charge/discharge all about?

author: golden_labels
author_rank: Super Contributor
timestamp: November 25, 2024, 03:45:21 pm
content: Quote from: bostonman on November 25, 2024, 02:26:52 amThere is no need for quotation marks around word “short”, because it really is a short. Of course physical capacitors have parasitic resistances and reactances, and so do wires and the power source. But as long as we talk about an idealized capacitance, it is a short without quoting. This leads to multiple paradoxes, if we apply simplified models while forgetting infinities are on the stage.(1)IMO the most neat way to see how it’s a short would be to look at the situation in the frequency domain. But I’m aware this isn’t going to be the right answer here.So, sticking to the time domain, imagine a syringe, filled with air, with its outlet blocked. In the begining air is not compressed. That is: the syringe stores no energy in air pressure. We can push the piston without effort, experiencing basically no resistance. This is our short. But as we push the pison, air gets compressed and energy is stored in the syringe. If we push with the same force, the resistance makes us move the piston less and less, reducing the amount of energy that flows into the syringe. If we let go, the syringe discharges, pushing the piston back. That initial lack of effort is our short.A capacitor has two ends, because it’s more akin to a rubber membrance blocking a pipe. At first the membrane shows no resistance (short, as if there was no membrane at all), but almost instantly it starts to push against incoming fluid.Quote from: Analog Kid on November 25, 2024, 11:59:11 amSee the pipe with membrane comparison. Fluid enters and leaves the pipe on both ends equally, at the same instant.Quote from: Analog Kid on November 25, 2024, 11:59:11 amYou mixed energy and charge, treating them as the same thing. It’s a barrell that collectsenergy. But what flows is thecharge. The barrel analogy doesn’t work for the charge. The charge is indeed collected, but not in a barrel-like manner.(1)For example connecting two capacitors, one of them charged, with a 0 Ω wire. Or modeling behavior under AC signal by evaluating behavior at a single point of the sinusoid.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 25, 2024, 03:52:11 pm
content: I can tell that you haven't properly read and absorbed my post. Go back and read it again until it's really in your head. At that point you will understand why your question just above this post doesn't make sense.(BTW, technically you shouldn't talk about current "flowing", but I'm going to keep using it because we are only talking about mental models here, not physics.)First of all, if you had fully understood my post you would realise that current cannot flow into one side of the capacitor without an identical amount flowing out of the other side, simultaneously - with no delay. Stop thinking of a capacitor as a water barrel - it's simply the wrong analogy. Go back to my diagram and you will see that watermustflow out of the right hand pipe when water flows into the left hand pipe.DO NOT PROCEED UNTIL THAT IS BURNED INTO YOURSKULL. Sorry, that sounded rude. Into your brain. It's important because you won't understand capacitors until you've understood that.Now to the question.....Imagine a restriction in the pipe somewhere to the left of the capacitor in my diagram (see the pic below). When water flows through a restriction, it creates a pressure drop across it. The pressure drop is proportional to the current flowing through it.1/ So, start with the capacitor discharged and a restriction in the pipe on the left. This is the equivalent of a resistor in series with the capacitor. The whole thing is at rest: no water flowing and the capacitor is in the discharged state, so no pressure drop across the restriction or the capacitor.2/ Let's imagine we apply 10psi of water pressure to the end of the left hand pipe. So that is 10psi at the left hand end of the restriction. Take a look at the capacitor: the diaphragm is in the central position, so the springs are not creating any back pressure. So there is 0 psi at the entrance to the capacitor, and that means there is 0 psi at the right hand end of the restriction.3/ Therefore there is 10psi pressure drop across the restriction. As a result water will flow quickly through it and into the left hand chamber of the capacitor, pushing the diaphragm to the right which in turn pushes the same amount of water out of the right hand pipe.4/ As the diaphragm moves to the right the springs begin to apply an opposing force to it, trying to push it to the left, which increases the pressure in the left hand chamber, and thus at the right hand end of the restriction. Let's say after one second of flow the back pressure is now 2psi.5/ That means there is only 8psi pressure drop across the restriction, so the current flowing through it reduces (remember, the current in the restriction is proportional to the pressure drop across it). So, the current begins to reduce from its starting point. Water continues to flow, but more slowly, so the diaphragm continues to move to the right, compressing the springs further and thus increasing the back pressure.6/ As the back pressure increases, the pressure drop across the restriction reduces, so the flow of water through the restriction also reduces.7/ Hopefully you can see by now that when charging a capacitor (moving the diaphragm to the right against the spring pressure) through a restriction, the water flow starts high (limited only by the restriction) and then reduces more and more as the back pressure from the capacitor increases.SummaryIt is important to understand thatthose curves you showed apply to a capacitor and resistor in series. The current starts high and falls away as the back pressure from the capacitor increases. The pressure drop across the capacitor starts at zero (0 psi) and rises quickly at first because the current is at its highest. As time passes the pressure across the restriction starts to fall (due to the back pressure from the capacitor), so the pressure across the capacitor rises more gradually. When the current is almost zero the pressure across the capacitor remains almost constant.That is why you see those curves: the current line starts high, then falls away. The voltage across the capacitor starts at zero and rises rapidly due to the high current. As the current reduces the voltage rises more gradually until it almost levels out.And to repeat one more time: those curves have nothing to do with these "delays" you keep mentioning. Capacitors on their own have no delays. There is only current flowing and pressure rising. Whenever the current is restricted, the voltage across the capacitor rises in a curve. How quickly it rises depends on the voltage applied to the circuit, the resistance of the resistor, and the capacitance of the capacitor.So only when a resistor gets involved can we say that there is a delay before the capacitor reaches a certain voltage.

author: Harry_22
author_rank: Frequent Contributor
timestamp: November 25, 2024, 07:04:33 pm
content: One boundary condition needs to be imposed. Just as current cannot exist outside a conductor, water cannot flow out of a pipe open side.This boundary condition is violated only in the case of high voltage or, by analogy, high pressure, when breakdown/breakthrough occurs.Your capacitor model is very plausible!If the membrane reaches the stop when the pressure/voltage increases, then a breakdown of the membrane/dielectric occurs. In this case we have to use a capacitor for a higher pressure/voltage.

author: bostonman
author_rank: Super Contributor
timestamp: November 26, 2024, 03:33:24 am
content: The time you took to explain this assures me that my question wasn't that silly as I initially believed.QuoteJust to clarify, my analogy about an "empty barrel" was on the idea of an empty hot water tank (say a fifty-gallon). When you first open the valve, water gushes in and possibly causes a drop in water pressure at the water main. As the tank fills, the water flow decreases, and, when the tank is near full, the water flow drops drastically.The image I had was whether A: the "empty barrel" (capacitor) was causing the water flow (input current) to create a "short"; or B: the hot water tank had a giant hole on the bottom and water was dumping out causing a "short".Your explanations helped.So technically I should be able to put a current sense resistor (milli-ohms) on both the positive and negative side of the capacitor and see exactly the same current at turn on?I'm uncertain if I've overlooked caring about why a capacitor is a short at turn on, whether it was never taught, or I just didn't take time to understand, but a few years ago this question occurred to me and I was always hesitant to ask anyone.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 26, 2024, 03:51:55 am
content: Quote from: bostonman on November 26, 2024, 03:33:24 amNot just technically but actually. Try it! You can use a simulation or real components - either will do.Quote from: bostonman on November 26, 2024, 03:33:24 amIt's probably more accurate to say itbehaveslike a short circuit at turn on. That is, for the first infinitely small moment in time it does not create any resistance to current flow - any back pressure. But as soon as any measurable amount of current has passed into the capacitor (and the same amount has passed out of the other terminal) the capacitor begins to charge up and, in doing so, creates a back pressure (an opposing EMF) which opposes the incoming flow of current. And, of course, you now know the rest.

author: SteveThackery
author_rank: Super Contributor
timestamp: November 26, 2024, 04:13:40 am
content: I think I would just like to close my contribution to this thread (others will continue, I'm sure) by reminding you that what I have described is just a model. Like all models, it is not reality. The well known saying "All models are wrong, but some are useful" applies to mine, I hope.Mine has its limitations. As I've said before, when the diaphragm in the chamber is pressed hard against the right hand wall, no more water will flow into it regardless of the pressure (voltage) applied. This is not the case with real capacitors. With real ones you can push more and more charge into the capacitor until the diaphragm (sorry, the dielectric) breaks down and the whole shebang turns into a low value resistor.One aspect where my model works well is in showing that the energy is stored in the sprung diaphragm, not in the water per se. In a real capacitor, the energy is stored in the dielectric, not the electrons crammed onto the plates.Real capacitors have lots of secondary characteristics which my model doesn't address and which we haven't talked about. All capacitors alter their capacitance to a greater or lesser extent when their temperature changes. Some capacitors, remarkably, change their capacitance when the voltage applied to them changes. Some capacitors behave like little microphones, and some will - just occasionally - behave like minute loudspeakers, making faintly audible squeaks and buzzes. Oh, and of course the big one: some capacitors - electrolytics and tantalums - will only work in one direction.So you can see that the world of capacitors is a big one with loads more to learn.

author: Harry_22
author_rank: Frequent Contributor
timestamp: November 26, 2024, 07:03:36 am
content: All this is wonderful, of course, but we missed one point during the brainstorm; OP didn’t appear here anymore.Does he understand everything?

author: bostonman
author_rank: Super Contributor
timestamp: November 26, 2024, 03:03:39 pm
content: QuoteMaybe it was a homework or test problem that the answer is no longer needed.

author: PGPG
author_rank: Frequent Contributor
timestamp: November 26, 2024, 05:33:24 pm
content: Quote from: bostonman on November 26, 2024, 03:33:24 amIn electronic world - yes, yes, yes and one more yes, but looking at it from electrostatic point of view...Imagine at the beginning the discharged capacitor connected with one end to your GND level that you assume as reference. Whole capacitor has a potential 0V related to your reference.Now you charge the capacitor from let us assume 10V source that has its '-' connected to GND and '+' we connect to capacitor.Capacitor after being charged has average potential of 5V (one electrode is 0 and second is 10V). So where from is this potential. The only possibility is that not exactly the same amount of charge went into one its electrode that went out of the second one

author: TimFox
author_rank: Super Contributor
timestamp: November 26, 2024, 05:40:47 pm
content: The first thing I learned about capacitor circuits in Electronics 101 (in a physics department) is that you can't change the voltage across a capacitor instantaneously.(Similarly, you can't change the current through an inductor instantaneously.)Of course, these simple statements are usually applied to ideal capacitors, such as used in coupling the output from a first amplifier to the input of a second amplifier.Eventually, with current flow charging the capacitor, the voltage across it does change.

author: bostonman
author_rank: Super Contributor
timestamp: November 27, 2024, 12:13:49 am
content: I'm not being sarcastic or asking this to question the answers provided by everyone, but if a capacitor is a DC block, then how is the same DC current still measured on both sides?

author: TimFox
author_rank: Super Contributor
timestamp: November 27, 2024, 12:20:44 am
content: Thought experiment: connect a DC power supply, switch, (optional) resistor, DC ammeter (preferably needle, not digital), capacitor under test, and second DC ammeter in series.Start by shorting out the capacitor (with switch open), remove the short, and close the switch.Observe the meters: since they are at the two terminals of the supply, they should move together (quick rise followed by fall).

author: SteveThackery
author_rank: Super Contributor
timestamp: November 27, 2024, 12:36:11 am
content: Quote from: PGPG on November 26, 2024, 05:33:24 pmThis is the most ridiculous post in the whole thread. On what planet does a capacitor with a 10V potential difference across its terminals have anything to do with averages? It's absurd! A ten ohm resistor with 1 amp of current in it has an "average potential" (whatever that means) of 5V?!! So where can I find this 5V? Where do I connect my multimeter?Should I buy a 24V battery for my car because its "average potential" is the 12V that my car needs?Your final sentence is a non sequitur - it is not a logical conclusion from the first (false) argument. Delete the whole post - you are just embarrassing yourself.

author: PGPG
author_rank: Frequent Contributor
timestamp: November 27, 2024, 08:57:37 am
content: Quote from: SteveThackery on November 27, 2024, 12:36:11 amWhat is the minimum voltage were average starts to makes sense?What is your body average voltage when you check socket phase pin with neon mains voltage tester (a small screwdriver).Neon tester won't light up until the current flows so where is this current then going?RS485 IC datasheets specify that average voltage between A and B lines differs no more than 0.1V (or 0.2V) between two states and it is important parameter as these difference is a source of common mode potential at output that can be a reason to violate EMC limits.On what I have written you can look from electrostatic perspective or assuming parasitic capacitors (to stay in electronic and elements with lumped constants (sorry, I'm not sure how to say it in English).So the capacitor from my example have also some (very, very little, but higher than 0) capacitance between it (as a whole) and GND reference. And these capacitance is also charged from +10V source but as this capacitance second end is simply the whole GND than this current goes through connection between 10V supply and top capacitor pin but no goes through this our capacitor second pin (but goes through supply second pin).Quote from: SteveThackery on November 27, 2024, 12:36:11 amFortunately it is only your opinionAnd at the beginning I have written that staying in electronic world you can ignore this what I then write.

author: PGPG
author_rank: Frequent Contributor
timestamp: November 27, 2024, 10:39:28 am
content: Quote from: bostonman on November 27, 2024, 12:13:49 amAfter some time I understood that "DC block" means "DC blocking" and not "DC box" - what I was considering what author could have in mind.What you understand as DC current:1. Unidirectional current with changing value, or2. Unidirectional current with constant value?If 1.As long as we are in normal electronics current always flows in a closed circuit and you can measure it at any point of that circuit. Both sides of capacitor are simply two points of the same circuit so in both of them you can measure the current and get exactly the same result. If there will be more elements in the same circuit (resistors, induction coils, other capacitors, voltage sources) you will be able to measure current at any of the connections and you will get the same result.When you connect voltage source through some resistance (in daily life we are not dealing with superconductors) to capacitor than DC current will flow forever. At the beginning it will be higher and then lower and lower shortly reaching immeasurable values.If 2.In real life you will never get constant DC current continuously flowing through capacitor because it means it is charged and charged and at some moment its voltage reaches the limit its dielectric can survive. A breakdown occurs and the capacitor ceases to be a capacitor.But if in simulation you connect DC constant current source to capacitor then probably (I practically don't use simulation so I'm not sure if connection will not be questioned) you will get continuously flowing constant DC current through capacitor. It is possible only because ideal capacitor can be charged to infinity.Concept that capacitor blocks DC current should be understood that constant DC current can't flow through capacitor infinitely. It can flow through capacitor only for limited time until capacitor is charged to its limit voltage.The practical use in electronic is that you can have two points in circuit with different DC voltages and connecting them with capacitor you allow AC signal to be moved from one point to the other while DC voltages can stay different (DC current not flows through this capacitor = it blocks DC current while allows AC to go through it).

author: golden_labels
author_rank: Super Contributor
timestamp: November 27, 2024, 11:46:21 am
content: Quote from: bostonman on November 27, 2024, 12:13:49 amBecause you don’t apply DC. Word “apply” already implies a change, doesn’t it? That is AC. If we imagine a capacitor sees true non-zero DC, and note that is impossible in reality and pretty useless in theory,(1)there is indeed no current observed.If there is any current, it must be equal on both ends(2)of a component. This is a rule in the lumped element model. Don’t forget that.(1)While speaking of its behavior in an electronic circuit as an ideal capacitor. In other models it may be more useful: for example dielectric breakdown or biasing affecting real capacitance.(2)Strictly speaking all currents must sum to 0, which should also answer the question of components with more than 2 ends.