Capacitor makes the circuit open?

[samieee]

We know charge is accumulated on the conductor plates of capacitor. Here is a circuit (image) with voltage source, resistor and capacitor. Now due to the capacitor the circuit is actually open so flow of charge aka current is zero. Then how can charge be accumulated on the plates of capacitor?

 

[DragonPetter]

We know charge is accumulated on the conductor plates of capacitor. Here is a circuit (image) with voltage source, resistor and capacitor. Now due to the capacitor the circuit is actually open so flow of charge aka current is zero. Then how can charge be accumulated on the plates of capacitor?

It is only open at steady state, meaning after the circuit is in this configuration for a long time so that only the DC component of your voltage source is present.

Before the circuit is in the state of your schematic, there is no charge accumulated on the plates and so there is no voltage across the capacitor, this is known as an initial condition. Once the voltage is applied, charge flows through the resistor and begins accumulating on the plate. Initially, since there is no charge on the plate and so no voltage, all of the voltage is dropped across the resistor. As the plates begin filling up with charge, their voltage increases and so less voltage is across the resistor and so less current is flowing. Once enough charge has been accumulated on the plates so that its voltage is equal to the voltage of the source, the voltage across the resistor will be 0, and so no more current will flow. This is when it is considered an open, and in stead state -- the charge is already accumulated.

So, you should know that the capacitor is only an open to DC voltage/current, and not to AC.

 

[samieee]

Thanks for your reply.

Once the voltage is applied, charge flows through the resistor and begins accumulating on the plate.

Though voltage is applied the circuit is in open condition so the current flowing through resistor should be zero isn't it?

 

[DragonPetter]

Thanks for your reply.

Though voltage is applied the circuit is in open condition so the current flowing through resistor should be zero isn't it?

The circuit is only in the open condition once enough charge has accumulated on a capacitor so that its voltage is equal to the DC voltage applied.

Remember the voltage on a capacitor is V = Q/C, so as more charge is added, its voltage increases.

To your question specifically, if the capacitor is already in the charged condition, then the current flow through the resistor will be 0.

 

[sophiecentaur]

Thanks for your reply.

Though voltage is applied the circuit is in open condition so the current flowing through resistor should be zero isn't it?

Eventually, yes, with DC applied.

 

[samieee]

The circuit is only in the open condition once enough charge has accumulated on a capacitor so that its voltage is equal to the DC voltage applied.

But before that? Isnt it open in all conditions due to the dielectric between the plates?

 

[DragonPetter]

But before that? Isnt it open in all conditions due to the dielectric between the plates?

No. If there was initially no charge on the capacitor, and then you put it in a circuit with current, the current coming into it would get "captured" by the capacitor, and the charge of the current will start to accumulate on the capacitor. This actually means that the capacitor is acting more like a short circuit rather than an open circuit in the very beginning. Once the capacitor has captured enough charge, its voltage increases til it cannot capture any more charge, and this happens over a long time. When it is finally filled with charge that it can't take anymore, it acts like an open circuit.

The dielectric simply determines how much charge the capacitor can hold at a given voltage. It determines what the capacitance value of the capacitor is, and so you can see from what I said before V = Q/C, if you rearrange it to how much charge it can hold per a given voltage C = Q/V. Its basically telling you how much energy the capacitor can store inside its dielectric.

 

[sophiecentaur]

But before that? Isnt it open in all conditions due to the dielectric between the plates?

I think you are using the word 'open' wrongly which is leading you to a false conclusion. Any 'open' circuit - even the gap between switch contacts or bare ends of two wires - has a finite Capacitance, which can allow signal leakage so the 'open-ness' is all relative to the frequency involved and the particular situation involved.
An Open circuit is a practical description of a situation in which any current which may be flowing is negligible in the context of the other, more significant, currents when the open circuit has been closed.

It is very easy to try to categorise things too rigidly and to become 'obsessed' with getting the names right. This is the beauty of using Maths to describe things in Science because it involves using actual Numbers and Quantities, rather than Absolute terms to 'classify' things.

 

[DragonPetter]

This might be a useful analogy. Imagine you are pumping a really small tire with a hand pump that is connected with a hose.

Charge is air, and the flow of air is current.

The diameter and length of the hose is the resistor, it determines how much air you can pass through it at a given pressure on the air.

The tire is the capacitor. When it is flat, it has no air in it, and when it is full, it has the most air you can fill it with. The elasticity of the tire rubber is like the dielectric, how much it can stretch determines how much air it can hold at a given pressure. If you fill it til the tire can't hold anymore and pops, its like breaking through the dielectric.

The pump applies pressure that is like the voltage. If you start pushing down the pump handle, it will create a pressure that pushes air through the hose and into the tire.

As you begin pushing down the handle, the tire is empty (capacitor is not charged) and so it takes all of the air into it with no resistance. In this sense, it is acting like a short circuit, because there is no pressure in it so it looks like the hose isn't even blocked by anything and freely flows air to the outside pressure. The only resistance you see is from the hose, since it can only pass a certain amount of air at a given pressure.

As the handle moves down, the air keeps filling up the tire, and so the pressure in the tire builds up. This means it is becoming harder to fill the tire with more air, and so the air flow slows down, and more pressure is in the tire and less in the hose.

Once the tire fills up completely, you won't be able to push down the handle anymore because you aren't strong enough. You don't have enough pressure to push down the handle, you can sit on the handle and it won't move. The air still won't move since the pressure in the tire is just as strong. In this way, it looks like an open circuit, because no more air will flow.

It is the same way in a capacitor. Your voltage source begins pushing current through, and the capacitor is empty, so it looks like a short circuit. Once you have filled the capacitor with charge, its voltage is equal to the voltage you use to push the charge, and so you can no longer push anymore through, and it looks like an open.

 

[sophiecentaur]

Umm.
That is a very long winded analogy and I wonder just how useful it is for explaining things simply in terms of the actual electrical Capacitance across a gap in a wire vs the capacity of a high value Capacitor.
Sometimes less is more. But then, you really needed to included the case then the hose is blocked off before the tyre, in order to show the quantitative difference between a Capacitor and an open circuit.

 

[DragonPetter]

Umm.
That is a very long winded analogy and I wonder just how useful it is for explaining things simply in terms of the actual electrical Capacitance across a gap in a wire vs the capacity of a high value Capacitor.
Sometimes less is more. But then, you really needed to included the case then the hose is blocked off before the tyre, in order to show the quantitative difference between a Capacitor and an open circuit.

The point is that a capacitor does act exactly as an open circuit or a short circuit in specific conditions, and not in all conditions (t = infinity/~5 time constants and t= 0).

And I think the analogy serves to give the concept of "capacitance" pretty well since the OP seems to be confused with the relationship of charge accumulation and capacitance, and how this leads to the short/open conditions. It shows that you only reach an open after charge has been pumped, not in all conditions, and it also shows that in fact it is acting like a short when you initially begin pumping.

 

[sophiecentaur]

You actually made my point for me because that first sentence of yours expresses the whole idea perfectly in just over one line of writing.

PS I am very cranky about analogies because they so often let you down in the end! I guess I should get over it!

 

[DragonPetter]

You actually made my point for me because that first sentence of yours expresses the whole idea perfectly in just over one line of writing.

PS I am very cranky about analogies because they so often let you down in the end! I guess I should get over it!

Yes, but explaining it directly did not seem to help a lot and I'm still not sure that it would until a better idea of what is going on is there. That's why I decided to make an analogy.

 

[DragonPetter]

PS I am very cranky about analogies because they so often let you down in the end! I guess I should get over it!

I know analogies are bad, but they help make pictures/animations in your head that give you the intuitive feelings of things.

Anyway, if an analogy is mathematically equivalent to what it is being an analogy of, is it still an analogy or at least a bad analogy? Differential equations for mechanical and electrical systems can have identical relationships.

 

[sophiecentaur]

I take your point but every analogy needs it own Caveat if you want to prevent problems. I so often read responses that have grabbed the wrong bit of the analogy and ignored the important message. Having said that, I just lurve the mass on spring analogy for almost every oscillator description.

 

[sophiecentaur]

I know analogies are bad, but they help make pictures/animations in your head that give you the intuitive feelings of things.

Anyway, if an analogy is mathematically equivalent to what it is being an analogy of, is it still an analogy or at least a bad analogy? Differential equations for mechanical and electrical systems can have identical relationships.

AH yes - once you introduce the maths model and the correspondence can be spelled out, you get my vote.

 

[Averagesupernova]

PS I am very cranky about analogies because they so often let you down in the end! I guess I should get over it!

I think that it is generally understood that ALL analogies only go so far. An analogy is a good place to start from. The fault lies within the person who expects it to solve all of the problems, not within the analogy. Analogies have a tendency to put a persons mind in a certain mode that better prepares them the real thing rather than jumping right into the real thing. I have seen the question asked about how computers know how to do stuff. My reply to that is: How do mousetraps know how to catch mice? It is such a brief reply that it can hardly be considered an analogy but it activates a certain mode of thinking which is important.

 

[psparky]

I liked the analogy with the tire and air pump...good stuff.

Don't mind Sophie...he always gets grumpy with those...haha!

Let's face it...an interesting analogy AND a mathematical model work best for learning.

If you have only one of the two...you are typically missing something!

My firehose and water analogy leading into P=IV was a perfect example of this!

 

[psparky]

Because I'm pickin on Sophie today...let me sum up you capacitor question:

i(t)=C*dv/d(t)

A perfect mathematical model.

Any questions?

Alright...I'll throw in a small analogy as well...what the heck.

That statements says...the change in voltage in respect to time multiplied by the capacitance...equals the current.

So when you first put your voltage source across your uncharged zero volt capacitor...the change in voltage is large compared to time...so you get a fairly large current...as it charges...the voltage across the cap starts to climb and the change is much less...so the current gets to be less. Then when the cap reaches its full charge...there is no change in voltage...there fore there is no current!

Any questions?

 

[psparky]

Might as well explain inductors as well...the work exactly the opposite. They resist current at start up and short in steady state DC.

v(t)=L*di/dt

Before they get energized...they have zero current thru them. When you turn the switch on...the change in current is fairly large...then a fairly large voltage is across them...to have a voltage...I must have a resistance. As time passes...the change in current is much less...so the voltage across is much less...at steady state there is no change in current...therefore no voltage...and the inductor acts like a short.

 

[samieee]

According to boylestad's "Introductory Circuit Analysis", chap-10(capacitors) section 10.3:

Before continuing it is important to note that the entire flow of charge is through the battery and resistor- not through the region between the plates. In every sense of definition there is an open circuit between the plates of capacitor

http://books.google.com.bd/books?id=cH0KitWACEcC&pg=PA274&dq=boylestad+eleventh+edition&hl=en&sa=X&ei=wBx2T9uLOcrKrAfWiqWLDQ&redir_esc=y#v=onepage&q&f=truepage 398

Those are the words that made me confused. I thought if a circuit is open then how can charge be accumulated or flowed? Or 'flow of charge and accumulation of charge is different thing'-is the point I am not getting?

 

[psparky]

According to boylestad's "Introductory Circuit Analysis", chap-10(capacitors) section 10.3:http://books.google.com.bd/books?id=cH0KitWACEcC&pg=PA274&dq=boylestad+eleventh+edition&hl=en&sa=X&ei=wBx2T9uLOcrKrAfWiqWLDQ&redir_esc=y#v=onepage&q&f=truepage 398

Those are the words that made me confused. I thought if a circuit is open then how can charge be accumulated or flowed? Or 'flow of charge and accumulation of charge is different thing'-is the point I am not getting?

i(t)=C*dv/dt

It's impossible to sum it up any better than this :)

 

[DragonPetter]

According to boylestad's "Introductory Circuit Analysis", chap-10(capacitors) section 10.3:http://books.google.com.bd/books?id=cH0KitWACEcC&pg=PA274&dq=boylestad+eleventh+edition&hl=en&sa=X&ei=wBx2T9uLOcrKrAfWiqWLDQ&redir_esc=y#v=onepage&q&f=truepage 398

Those are the words that made me confused. I thought if a circuit is open then how can charge be accumulated or flowed? Or 'flow of charge and accumulation of charge is different thing'-is the point I am not getting?

I think his point is that charge does not physically travel through the capacitor. This does not mean that charge on both ends of the capacitor cannot change, which a change in charge over time is current. Remember that the separation of charge over a distance defines an electric field, and so there is a voltage and current across the capacitor when the charge transfers between the positive and negative plates of a capacitor, even if charge does not pass through the capacitor. This is only in AC, transient conditions.

 

[sheshank]

Hello friend
I found myself with the same question some time around three years back...I managed to find the answer. I think I can explain what I understood...
First let's understand the concept of Capacitors. Capacitors are nothing on their plates. Everything lies in the di-electric layer between them. The Chemical composition of few di-electrics in such a way that they transfer the force (EMF) but refuse to transfer electrons (Current). So, how does the force gets transmitted to the other end? This process is common in almost all the conduction phenomena. The charge of opposite polarity develops on either sides of the material. Now the electrons on the negative terminal of the batter respond to move through the material. It depends on the type of solid how much current flows. In case of di-electrics, it doesn't flow. The force gets transmitted, but electrons aren't allowed to pass. As a result the faces of the di-electric facing the positive and negative terminal of the batter stay the way as they are - after the force is transmitted. That is such that they have charge deposited on its surfaces. We use this property of the dielectric. We interface the surface of the capacitor with conductive plates.
As the plates are conductive they also place themselves with the same charge as on the surface of the dielectric. So, after a DC voltage is applied the charge is developed in the process as just described. Now, if the voltage is so heavy, then dielectric break down occurs and the current flows. This means that the chemical properties of the dielectric in between the capacitor plates has been chemically remodified. This is what we term as 'burning of capacitors'. Usually its these capacitors and transformers that burn in any circuit causing a very bad smell.

Supposing that the voltage applied is in the range of the capacitor, then what happens? The charge develops and this we call as the charged capacitor. The amount of charge i.e.. capacity of a capacitor is mainly based on the surface area the conducting plates are in contact with and the type of dielectric material between the plates. Once the dielectric is charged it has some energy stored (inside the dielectric because of the charge accumulation on the surface) its ready to be dissipated. Now, when something like a closed circuit with a combination of resistors is attached to the capacitor, the capacitor immediately discharges exponentially (You can compute this in any way using integral calculus.).

As it decreases exponentially, there is some time lag between the charge and slow discharge.(Even though it immediately charges but slowly discharges). This is used in RC and ringing circuits to produce frequencies of various scales. This is the base for the present Nanotechnology and Wireless communications.

I think I solved your question. If you have a query about how the AC travels through the capacitor post another question and I'll try to explain in the best way I can.

 

[vk6kro]

There is a simple experiment which usually settles questions like this one.

You get a large electrolytic capacitor and a flashlight bulb and put them in series across a power supply.
Observe polarity for the electro.

When you connect the circuit, the bulb will light up briefly and then gradually dim and then go dark.

Remove the power source and connect the two leads that went to the power source, together.

The flashlight bulb will again light up briefly while the capacitor discharges, and then it will go dark.

I avoid plumbing analogies because the discussion then becomes about plumbing.
These analogies don't work anyway, because household plumbing mostly involves water leaving the system, not going in circles like electrical circuits.

 

[sophiecentaur]

Hello friend
I found myself with the same question some time around three years back...I managed to find the answer. I think I can explain what I understood...
First let's understand the concept of Capacitors. Capacitors are nothing on their plates. Everything lies in the di-electric layer between them. The Chemical composition of few di-electrics in such a way that they transfer the force (EMF) but refuse to transfer electrons (Current). So, how does the force gets transmitted to the other end? This process is common in almost all the conduction phenomena. The charge of opposite polarity develops on either sides of the material. Now the electrons on the negative terminal of the batter respond to move through the material. It depends on the type of solid how much current flows. In case of di-electrics, it doesn't flow. The force gets transmitted, but electrons aren't allowed to pass. As a result the faces of the di-electric facing the positive and negative terminal of the batter stay the way as they are - after the force is transmitted. That is such that they have charge deposited on its surfaces. We use this property of the dielectric. We interface the surface of the capacitor with conductive plates.
As the plates are conductive they also place themselves with the same charge as on the surface of the dielectric. So, after a DC voltage is applied the charge is developed in the process as just described. Now, if the voltage is so heavy, then dielectric break down occurs and the current flows. This means that the chemical properties of the dielectric in between the capacitor plates has been chemically remodified. This is what we term as 'burning of capacitors'. Usually its these capacitors and transformers that burn in any circuit causing a very bad smell.

Supposing that the voltage applied is in the range of the capacitor, then what happens? The charge develops and this we call as the charged capacitor. The amount of charge i.e.. capacity of a capacitor is mainly based on the surface area the conducting plates are in contact with and the type of dielectric material between the plates. Once the dielectric is charged it has some energy stored (inside the dielectric because of the charge accumulation on the surface) its ready to be dissipated. Now, when something like a closed circuit with a combination of resistors is attached to the capacitor, the capacitor immediately discharges exponentially (You can compute this in any way using integral calculus.).

As it decreases exponentially, there is some time lag between the charge and slow discharge.(Even though it immediately charges but slowly discharges). This is used in RC and ringing circuits to produce frequencies of various scales. This is the base for the present Nanotechnology and Wireless communications.

I think I solved your question. If you have a query about how the AC travels through the capacitor post another question and I'll try to explain in the best way I can.

Unfortunately, what you write about dielectrics is not relevant to the basic function of a capacitor because two plates IN A VACUUM still behave as a Capacitor.

Also, as you introduced the idea of exponential decay, why carry on with an extra verbal description when the Maths has done it all for you?

 

[sophiecentaur]

I avoid plumbing analogies because the discussion then becomes about plumbing.
These analogies don't work anyway, because household plumbing mostly involves water leaving the system, not going in circles like electrical circuits.

Hear Hear.

 

[Averagesupernova]

I avoid plumbing analogies because the discussion then becomes about plumbing.
These analogies don't work anyway, because household plumbing mostly involves water leaving the system, not going in circles like electrical circuits.

But there gets to be a point where we just assume that certain things are happening within a circuit that 99.9% of us will never prove or disprove, or even need to understand. It is faith. A designer who specializes in voltage controlled oscillators for instance doesn't worry about the battery chemistry in the battery of the portable receivers that the VCOs will go into. He may be concerned about the things about that battery that affect the VCO but these things are spec'd and someone besides the VCO designer deals with it. The same thing can be said about the water outside of the building. In the big picture, the water does go round and round in our environment, but obviously not with the same tight controls within our electrical circuits. But for plumbig analogies, we could just was well assume that there is a magic pump outside of the house that collects all spilled water, filters it and pumps it back around. In this analogy, the magic pump is the battery that the VCO designer could not care any less about. It just doesn't concern him beyond the specs that are required. To the VCO designer the battery is a magic pump.
-
In analogies, the question is always asked that says: "What if this or that happens? This doesn't follow what happens in the electrical circuit!" And it may well be true, but at that point, the analogy ends. There will likely be a day when how we think of electricity, semiconductors, or matter in general will probably no longer work. We will then have determined that how we thought of it up until then was just an analogy. To an electrician who installs wiring in new houses the wires he installs and hooks up just as well be carring invisible magic green marbles. It will work fine for him. But when that electrician becomes a semiconductor designer, well, the marbles will fail him.
-
I don't see why people rip on analogies. It is like someone selling me a set of combination wrenches and I suddenly expect them to be able to do machine work on engine internals. The failure isn't with the wrenches, it is with my unreasonable expectations of a simple set of tools.

 

[psparky]

But there gets to be a point where we just assume that certain things are happening within a circuit that 99.9% of us will never prove or disprove, or even need to understand. It is faith. A designer who specializes in voltage controlled oscillators for instance doesn't worry about the battery chemistry in the battery of the portable receivers that the VCOs will go into. He may be concerned about the things about that battery that affect the VCO but these things are spec'd and someone besides the VCO designer deals with it. The same thing can be said about the water outside of the building. In the big picture, the water does go round and round in our environment, but obviously not with the same tight controls within our electrical circuits. But for plumbig analogies, we could just was well assume that there is a magic pump outside of the house that collects all spilled water, filters it and pumps it back around. In this analogy, the magic pump is the battery that the VCO designer could not care any less about. It just doesn't concern him beyond the specs that are required. To the VCO designer the battery is a magic pump.
-
In analogies, the question is always asked that says: "What if this or that happens? This doesn't follow what happens in the electrical circuit!" And it may well be true, but at that point, the analogy ends. There will likely be a day when how we think of electricity, semiconductors, or matter in general will probably no longer work. We will then have determined that how we thought of it up until then was just an analogy. To an electrician who installs wiring in new houses the wires he installs and hooks up just as well be carring invisible magic green marbles. It will work fine for him. But when that electrician becomes a semiconductor designer, well, the marbles will fail him.
-
I don't see why people rip on analogies. It is like someone selling me a set of combination wrenches and I suddenly expect them to be able to do machine work on engine internals. The failure isn't with the wrenches, it is with my unreasonable expectations of a simple set of tools.

Word up Homey Claus.

Where's the "Like" button on this thing?

 

[sophiecentaur]

But there gets to be a point where we just assume that certain things are happening within a circuit that 99.9% of us will never prove or disprove, or even need to understand. It is faith. A designer who specializes in voltage controlled oscillators for instance doesn't worry about the battery chemistry in the battery of the portable receivers that the VCOs will go into. He may be concerned about the things about that battery that affect the VCO but these things are spec'd and someone besides the VCO designer deals with it. The same thing can be said about the water outside of the building. In the big picture, the water does go round and round in our environment, but obviously not with the same tight controls within our electrical circuits. But for plumbig analogies, we could just was well assume that there is a magic pump outside of the house that collects all spilled water, filters it and pumps it back around. In this analogy, the magic pump is the battery that the VCO designer could not care any less about. It just doesn't concern him beyond the specs that are required. To the VCO designer the battery is a magic pump.
-
In analogies, the question is always asked that says: "What if this or that happens? This doesn't follow what happens in the electrical circuit!" And it may well be true, but at that point, the analogy ends. There will likely be a day when how we think of electricity, semiconductors, or matter in general will probably no longer work. We will then have determined that how we thought of it up until then was just an analogy. To an electrician who installs wiring in new houses the wires he installs and hooks up just as well be carring invisible magic green marbles. It will work fine for him. But when that electrician becomes a semiconductor designer, well, the marbles will fail him.
-
I don't see why people rip on analogies. It is like someone selling me a set of combination wrenches and I suddenly expect them to be able to do machine work on engine internals. The failure isn't with the wrenches, it is with my unreasonable expectations of a simple set of tools.

Analogies are just fine - in retrospect. When you actually 'know' the real thing, then the analogy makes perfect sense. But that's because you can see where the analogy does and doesn't fit.
The water-in-pipes analogy is fine as long as no one imagines the Kinetic Energy is involved in 'Electricity'. From what I read on these pages, the MAJORITY of people trying to get a better understanding of Electricity have KE at the heart of their picture. They can't help wanting the electrons to be like drops of water hurtling through a pipe and doing their work of heating things up and moving things. For them, it has been two steps forward and three steps back. And they don't realize it.
Likewise, with Photons, everyone starts off believing that the word "particle" implies all sorts of things about size, shape and even frequency. Talk to anyone who 'really' knows their stuff about photons as 'particles' and you can see that they have had to shed the 'particle analogy' way back in their learning process.
Analogies are best kept to within one's own mind or in one-to-one private conversations where the 'teacher' can maintain just where and when the analogy is valid and can help steer the 'learner' from massive pitfalls.
And in reply to the wrench 'analogy'. There are many people who keep using a wrench as a hammer because they don't know any better. They are not even aware that the chair they have built with it will soon fall to pieces. How were they being "unreasonable" to use a wrench if the 'expert' gave them the wrench in the first place?