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It is known that:

  • AC current can flow through capacitors
  • A wire has some inherent capacitance
  • A capacitor is the same as an open circuit with plates at either end, and the size of the plates corresponds to the capacitance

So if you have a circuit like this:

enter image description here

then it is equivalent to a circuit like this:

enter image description here

where C1 is miniscule, and equivalent to the capacity of the wires. If current can flow through circuit 2, why can't it flow through circuit 1 using the same logic? Does this imply that an open switch doesn't prevent the flow of electricity?

Edit

Andy Aka said AC current does flow through an open switch. So can it be said that:

a) it is impossible to turn a lightbulb off

and

b) it is impossible to turn a lightbulb on

My explanation for statement a) is that, if an open switch is identical to a closed one, in that current can flow through it, then it should be impossible for an open switch to stop electricity from flowing through a lightbulb.

enter image description here

For statement b), given that the following two circuits are equivalent, that current should flow over the gap in circuit 3 instead of through the lightbulb, since that open switch has no resistance to the AC current. Therefore the lightbulb would never be powered.

enter image description here

enter image description here

Since these two statements are obviously false, the only explanation I can think of is that either Andy Aka's answer is false, or that an open circuit still has a finite resistance. (ignoring sparking)

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    \$\begingroup\$ Finite reactance, not resistance. Capacitors are reactive. \$\endgroup\$
    – Hearth
    Commented Jun 23, 2023 at 13:37
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    \$\begingroup\$ Hint: a) the more capacitance the capacitor has, the more AC current could flow through it; b) open switch has incredibly low (but still non-zero) capacitance. \$\endgroup\$ Commented Jun 24, 2023 at 0:43
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    \$\begingroup\$ For RF applications, this is an actual problem. You add a "switch", and all you get is 40 dB attenuation. \$\endgroup\$ Commented Jun 24, 2023 at 2:59

6 Answers 6

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This is your question: -

If AC current can flow through a capacitor, why can't it flow through an open circuit?

However, you have made a contradiction in saying this: -

A wire has some inherent capacitance

If both wires have capacitance then an AC current will flow because it's no longer a true open-circuit.

Does this imply that an open switch doesn't prevent the flow of electricity?

Correct. An open circuit (real) switch will have open-contact capacitance and, although it might only be 0.1 pF, an AC current will flow if there's an AC voltage across the switch contacts.

So, how much current might flow? If we take a UK household supply of 230 volts, 50 Hz and, we assume the open-circuit capacitance is 0.1 pF then, the current that flows is this: -

$$I_{RMS} = 230 \times 2\pi\times 50\times 0.1\times 10^{-12} = 7.23\text{ nA}$$

As one would expect it's not a lot of current because, a 0.1 pF capacitor creates an impedance of around 32 GΩ. If it were a bigger capacitor, the current would be proportionately bigger.

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  • \$\begingroup\$ Can you elaborate? At the end you say it doesn't but then you say an AC current will flow if there is a voltage difference. Also why wouldn't AC current flow if both wires have capacitance? Thanks. \$\endgroup\$ Commented Jun 23, 2023 at 8:47
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    \$\begingroup\$ @stickynotememo I misread it. I meant to say yes it does. I shall fix. \$\endgroup\$
    – Andy aka
    Commented Jun 23, 2023 at 8:49
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    \$\begingroup\$ @ctrl-alt-delor I'm sorry, but that is incorrect. \$\endgroup\$
    – Andy aka
    Commented Jun 24, 2023 at 16:07
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    \$\begingroup\$ @ctrl-alt-delor: Ideal wires don't have capacitance, so in the OP's circuit diagram with an open circuit, no current will flow. But it's an inaccurate diagram / model for a physical circuit made of real lengths of metal, especially when we're interested in the unavoidable parasitic capacitances. We use the same word (wires) for those real-world objects, so we need to specify "ideal" vs. "physical" or "real" wires" which do have capacitance (to ground, and to any other conductors they're not fully shielded from.) \$\endgroup\$ Commented Jun 24, 2023 at 19:07
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    \$\begingroup\$ Sorry I meant much (or in most cases significant) capacitance. \$\endgroup\$ Commented Jul 15, 2023 at 18:52
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Your statement implies that AC can either flow or not, it's like 1 or 0 with no options in between.

If AC flows through some small capacitance, it does not make it fully short circuit. You can consider it as a "current leak" which may be so little so it will not be enough to turn on a bulb or even be considered at all.

The same with resistance of an open switch in an air. It has a very high resistance, but it's not infinite. On another hand it's high enough to not consider it in most situations.

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  • \$\begingroup\$ So you say that there is a finite amount of current pass across a capacitor? I was taught that AC can flow and DC can't - no matter the size of the capacitor. If there is indeed a relation between the size of the capacitor and the "resistance" that it has to AC flow, what is it? How can it be calculated? As for your last sentence about open switches having a resistance, I think we can ignore sparking for the purposes of this question, and so the open switch would have infinite resistance until it starts sparking, not "high enough to not consider"). Please correct me if I'm wrong. \$\endgroup\$ Commented Jun 24, 2023 at 0:54
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    \$\begingroup\$ @stickynotememo The capacitor has a similar effect to a resistor, except that it doesn't dissipate power and its "resistance" (properly, reactance) depends on the AC frequency. \$\endgroup\$
    – Hearth
    Commented Jun 24, 2023 at 2:36
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    \$\begingroup\$ @stickynotememo spark is a short circuit. I'm not talking about that. I'm talking about a very high resistance in open air which is negligible in a matter if an opened switch can turn on a bulb. The current can flow, but when the current is extremely low, then you may still consider "it does not flow at all" in some situations. Not every amount of current can turn on a bulb or create a short circuit. The current has a value, not just "can flow" and "can't flow". \$\endgroup\$
    – nochkin
    Commented Jun 24, 2023 at 3:35
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    \$\begingroup\$ @stickynotememo "So you say that there is a finite amount of current pass across a capacitor? I was taught that AC can flow and DC can't - no matter the size of the capacitor." Uh, why do you think there are capacitors of different size if size doesn't matter? In a handwavy kind of way, AC resistance is 1/(2 pi f C) for a given frequency f and a given capacity C. Handwavy because current runs 90° of phase before voltage as opposed to a real resistor. \$\endgroup\$
    – user107063
    Commented Jun 24, 2023 at 15:09
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    \$\begingroup\$ @stickynotememo yes. Lots of stuff that we treat as infinite out of convenience is actually "only" Gohm or so. Most of the time it doesn't matter. Then you try to do something that's sensitive to very small currents and it does matter. \$\endgroup\$
    – hobbs
    Commented Jun 24, 2023 at 19:27
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The idea that current can't flow at all in a circuit with an "open" (such as a capacitor or a switch) is an oversimplification given to introductory students. A more accurate statement is that constant, sustained current can't flow in a circuit with an open.

(The following description is classical. I'm sure an electrical engineer or quantum physicist would cringe at this explanation. Still, I think it's close enough to answer your question.)

In this edited picture from your question, the red dot represents the voltage at the output of the generator. Since this is AC, the voltage at the red dot will be varying continuously. In classical terms, that point will have a constantly changing surplus or deficiency of electrons.

enter image description here]

Now, in an idealized circuit, the voltage at the blue dot would have the same voltage because "it's the same point electrically."

But no change actually happens instantaneously. In a real circuit, it will take some time for the electrons to migrate along the wire. The generator will be "pushing electrons" onto the area of the red dot, making it more negative than the charge at the blue dot. That difference in charge causes negatively charged electrons to repel each other, causing some amount of current to flow until enough electrons have reached the blue dot, so that the charge distribution is even along the length of wire between the red and blue dot. Once charge is distributed evenly along the length of wire, flow of current will cease.

But that is the trick. Because this is AC, charge will never be evenly distributed along the length of the wire. The charge at the blue dot will forever by lagging behind the amount of charge at the red dot because it takes an extremely small, but finite time for the charge to equalize between those two points. This is the idea behind the "time constant" you may have been introduced to in your studies of capacitance.

And that's the idea of the capacitor: The large plates that are rolled up inside the capacitor are meant to act as a reservoir of electrons. It takes some time to "charge" plates with excess electrons. And after the plates are fully charged, it takes some time for the extra electrons to drain off of them when the voltage reverses. That's why capacitors are said to oppopse a change in voltage: When voltage reverses—for a very short time—the electrons coming off one of the capacitor plates can make up for electrons the generator is no longer providing.

You would see the same thing (almost) in a DC circuit: Once a voltage source is applied to one end of the capacitor, it will take some time (roughly 5 "time constants") for enough electrons to move from the voltage source to the capacitor plate for change distribution to equalize. At that point, current will stop flowing because the capacitor is acting like an open.

So don't think that capacitors allow current flow in the same way that complete, continuous circuits allow current flow. Current never flows "though" a capacitor in the same way it flows through a closed circuit. A capacitor truly is an "open." But on extremely small time scales, some current will flow nonetheless because electrons repel each other, so they will flow along the wire, either towards or away from the capacitor, in order to equally distribute charge. And when AC is used (provided the frequency is fast compared to the capacitance), the capacitor will always be in the process of charging or discharging, so it looks like current flows through the capacitor.

With regard to the circuit diagrams in the original post: Will the light bulb light? To be pedantic, the only way to know would be to know the capacitance of the capacitor, the capacitance of the switch, the resistance of the bulb, and the frequency of the AC. Then run through the impedece calculations. To be practical (if limiting the discussion to residential wiring) we know from experience that opening a switch does in fact turn off a light. So we know that the capacitance of the switch is so vanishingly small that the displacement current isn't sufficient to power the light. And that makes sense. The plates of capacitors are designed to amass electric charge. The poles of switches are designed to minimize capacitance.

A word on jargon: It's very common for non-engineer technicians like me to be taught that current actually flows through a capacitor. (They are often taught by other non-engineers who are just passing on what they were taught.) I think that is more oversimplification since, for all intents and purposes, that how it acts for a technician who doesn't have to actually design circuits. Even engineers will talk about the capacitor "passing" high frequencies, but I think that is either speaking metaphorically or just engineer jargon. So you will find textbooks and other authorities speaking of AC current flow "through" a capacitor. The jargon may be so ingrained that they just don't think about it any more, so they may not be aware of the confusion that their wording can cause.

Pulling out my old Navy Electricity and Electronics Training Series Module 2 on alternating current, it has this to say (printed page number 4-8):

Since the plates of the capacitor are changing polarity at the same rate as the ac voltage, the capacitor seems to pass an alternating current. Actually, the electrons do not pass through the dielectric, but their rushing back and forth from plate to plate causes a current flow in the circuit. It is convenient, however, to say that the alternating current flows "through" the capacitor. You know this is not true, but the expression avoids a lot of trouble when speaking of current flow in a circuit containing a capacitor. (p. 124 of 250 of the online PDF)

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  • \$\begingroup\$ +1. So am I correct in summarising your answer in that a capacitor connected to a AC circuit would at least "appear" that it's allowing current to flow through it, and it would at least function as a closed switch, but if the size of the capacitor is too small (as in the case of parasitic capacitance), the capacitor will fill up and stop current flowing at some point, before the AC wave reaches its peak. This would explain my proposition b), since the capacitance of the open switch would fill up and force rest of electricity to flow through the lightbulb instead. Is this understanding correct? \$\endgroup\$ Commented Jun 24, 2023 at 1:06
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    \$\begingroup\$ @stickynotememo. I think you're correct, but your wording makes me wonder. For example: "Aka said AC current does flow through an open switch." No. Andy said current flows. He didn't say it flows through the switch. My post was trying to explain the difference. Or "...that open switch has no resistance to the AC current...bulb would never be powered" Eh: A charging capacitor has no resistance, but it uses energy to build an electric field between the plates, so it acts like a load. That is the reactance or impedance you may have come across. Bulb may light. Depends on the balance. \$\endgroup\$ Commented Jun 24, 2023 at 3:28
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    \$\begingroup\$ @stickynotememo. See edit tacked on to end of my post on jargon. \$\endgroup\$ Commented Jun 24, 2023 at 3:29
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When you look to extremes, you have to be very careful about how you are modeling your circuit.

An open circuit is an open circuit. By definition, no current flows through it. Full stop. Next question.

But what does an open circuit really look like when you make one? You quickly find that the physical circuit you build and call an "open circuit" looks an awful lot like a capacitor with a plate on each side (the wire) and a dielectric between them (air/vacuum/sulfur hexafloride... take your pick).

What does this mean? Well, the first thing it means is that we can never construct an ideal open circuit. It will always have some capacitance to it. It may be an astonishingly small capacitance, but it will be there. So do not try to blindly apply idealized circuits to physical circuits, nor blindly assume your intuition about physical circuits always applies to idealized circuits. One always has to ask "is this model a good model of what is going on?" And, as you can see from your question, this is one of the corner cases where the models get wishy-washy.

The second set of idealized questions is what is AC and DC in the first place, and how do they interact with a capacitor? An AC signal is one which we model as \$sin(\omega t + \theta)\$. That is it is a sinusoid with some frequency (we use \$\omega\$, the angular frequency to save on writing. \$\omega = 2\pi f\$ where f is the frequency in cycles. We just get tired of writing the \$2\pi\$). And, skipping a ton of mathematics, we can jump to the concept of impedence, which is exactly like the concept of resistance except it is allowed to be imaginary to account for all of the funny phasing behaviors that can happen with AC signals. The imaginaryness of it is important, but we can handwave here and only speak to the magnitudes of these things. The impedance of a capacitor is \$\frac 1 {j\omega C}\$, where \$j\$ is our imaginary number (we don't use \$i\$ because it's already used for current in all sorts of things). So right away we can see that if the capacitance of the capacitor, \$C\$, goes up, its impedance goes down and that has an effect remarkably similar to resistance going down. Larger and larger capacitors look more and more like a short circuit. Likewise, as \$\omega\$ does up (increasing frequency), the impedance goes down. At higher frequencies, capacitors look more and more like shorts. This can be a problem for circuits, as it introduces many "leakages" which have to be accounted for as one gets into higher and higher frequencies. But your problem is interested in the opposite direction. As C goes down (acting more and more like an open circuit), impedance goes up. As it approaches 0 capacitance, impedance approaches infinity. Even AC signals "don't like" to go through small capacitors. And as frequency goes down (towards DC), impedance also goes up. But what about in the extremes? The points that are giving you pause. An "ideal open circuit" has not just a small capacitance, but 0 capacitance. This puts a 0 in the denominator, making the impedance equations I am using invalid - I can't say there's a current due to the impedance of that capacitor because my equation no longer applies. But, using the language I used earlier, we can speak of the impedance "approaching" infinity (something that is defined in a very formal way using calculus). The same logic applies for DC. DC can also be thought of as being on the same spectrum as AC if you think about it as a sinusoidal signal with \$\omega = 0\$. And doing so puts a zero in the denominator, making the equations invalid. But what does it really mean to be DC? When one gets very formal about it, the only signals that can truly be idealized DC are signals which are infinite in duration. A true 5V DC signal is one which was 5V at the creation of the universe, and 5V long after the heat death of the universe. In this extreme sense, you can appreciate things like how it doesn't matter how large a capacitor is, no DC current flows through it. You really have to think about the long game.

In practice, we work with DC where we flip a switch or connect a battery, and "DC" flows through the circuit. This is not idealized DC. This is a practical definition of DC. And, if you've learned Fourier transforms, you can learn how to take the "step functions" that occurs when you flip a switch from off to on, and later from on to off, and spread the energy out across all of the frequencies. And, one quickly finds that most of the signal is in a frequency band so astonishingly much lower than the response time of any element in the circuit that we handwave away the frequency all together, and model it as-if it were all \$\omega = 0\$. But when you're asking abut the curious edge between a capacitor and an open circuit, its important to remember that these signals do have a frequency to them. It's astonishingly small, but it's there. It's even calculable. So with "practical DC" you will find there is indeed some leakage current across any practical open circuit. The current is just practically 0.

What you explore is not an idle curiosity. It is something we have to worry about when designing practical DC devices. Often when we start up a circuit, there is an "inrush current" which is quite a lot larger than the steady state "DC" current we design for, because of these AC components. Circuits must be designed to withstand this inrush current, or they degrade and stop functioning.

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If you have an AC voltmeter and are reasonably adept at working around mains voltage, try this: with all lights and other devices in a room off, measure the voltage across the wires going to a light fixture. You will probably measure 1-2V AC, even with everything off. The miniscule current flowing through the parasitic capacitance sets up an impedance divider and results in the small voltage.

This voltage is too small to drive incandescent light bulbs or most other devices, but some cheap LED lights can actually give off some (dim) light in this state, leading to the curious phenomenon that you really can't turn these lights off!

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  • \$\begingroup\$ Does a measure voltage across the terminals really show current flow? You could set up a voltmeter across a 9v battery and measure 9v without there actually being any current flow. \$\endgroup\$ Commented Jun 24, 2023 at 1:00
  • \$\begingroup\$ @stickynotememo: When you measure with a voltmeter, there is current flow. It's just very, very small. \$\endgroup\$ Commented Jun 25, 2023 at 4:44
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    \$\begingroup\$ The measurement of a nonzero AC voltage on a circuit that is switched off demonstrates that the switch has a non-infinite AC impedance that is capable of passing current, which is essentially the original question. \$\endgroup\$
    – d3jones
    Commented Jun 25, 2023 at 11:39
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Now the confusion here occurs if we are looking at it literally as an electron flow, if you would explore more deeply, the electron flow even for a DC current is not continous.

There is no point of looking at the AC flow from the perspective of electron flows, the best thing to do if so look at AC current as merely an induction caused by changing magnetic field. This would very easily explain the flow of AC current through a capacitor rather than considering merely an electron flow.

Now the same question can be asked for even transformers, since they also are not strictly closed circuits as per definition, yet we observe the AC current flow through them. Hence the best way to understand AC current is to understand it as merely an induction of the changing magnetic field. In fact even the sinosoidal wave viz used to represent AC current has been taken due to the changing magnitude and changing direction of magnetic fields in generators, this would be a better way to understand AC current rather than understanding it on the basis of electron flow through a conductor. I hope this helps you in your understanding.

Now let us look at the capacitor, now in an open circuit generally like a switch, the magnetic field build up is not enough for it to induce AC current on to the other lead like a switch. So the capacitor is designed in such a way that the magnetic field can be built up on one lead, and this built up magnetic field changes it can induce AC current onto the other plate of the capacitor. Thus giving us the charge and discharge cycle and allowing AC current to pass through the capacitor.

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  • \$\begingroup\$ Could you explain what you mean by "even for a DC current is not continuous"? \$\endgroup\$
    – Hearth
    Commented Dec 2, 2023 at 3:26
  • \$\begingroup\$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. \$\endgroup\$
    – Community Bot
    Commented Dec 2, 2023 at 4:34

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