Capacitor Theory Thoughts - Comments Requested

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crossman

Senior Member
Location
Southeast Texas
Okay, I would really appreciate some thoughts on this... I have been spending hours thinking about this, and want to get some insight on how the rest of you would look at this.

First, we have a capacitor on a constant DC source:

capdcsource.jpg


A graph of the voltages, currents, and power would look like this:

capdcgraph.jpg


1) Electrons flow from the negative side of the source into one plate and it becomes negatively charged. At the same time, electrons are "pulled" off the other plate into the positive side of the source and this plate becomes positively charged.

2) The negative plate and the positive plate have an increasing potential difference across them during charging.

3) The energy in the current flow during charging is converted into an electrostatic field.

4) The potential difference between the plates of the capacitor is in opposition to the source voltage. It is of opposite polarity.

5) As the difference of potential across the plates rises, it opposes the source voltage to a greater and greater extent, and the current will decrease.

6) After the capacitor accumulates enough electrons on one plate and enough "holes" on the other plate, the voltage on the capacitor will be equal to and opposite of the source voltage and current will stop flowing because the two voltages cancel.

7) We could speak of the "opposition to current flow" that is in this circuit... and we could measure it in Ohms if we take the source voltage and divide it by the current flow.

8) If we do this at each instant in time, the Ohm value will be increasing as the current flow is decreasing. The Ohm value will start near zero and rise to near infinity as the capacitor charges.

9) The "variable nature" of the opposition to current flow of a capacitor could be mimicked with a variable resistor and a graph exactly similar to the graph above could be produced by varying the resistance in a precise manner.

10) The source voltage and the current are ?in phase? in the variable-resistive circuit.

11) The DC source had to expend a real amount of energy to push and pull the electrons through the opposition contained in the resistor.

12) If we assumed the variable resistor in (10) to be a load in our building and the voltage source to be a DC service from the utility, and we let the variable resistor mimic a charging capacitor, the energy expended in the source to make the current flow would be converted to heat in the resistor. This heat would increase the overall energy content of our building.

13) Anytime that voltage and current are in phase, the power is given in Watts.

14) Concerning the variable Ohms scenario, the "graph-capable" wattmeter in the diagram would give us a watt reading at each instant in time. This watt reading would be constantly changing until the current stopped flowing at some very large value of resistance. The Watt graph would start at a high level and fall to zero just as the current starts high and falls to zero. This would be represented by the green power graph in the image above.

15) A watt-hour meter would give a reading equal to the amount of energy used in the time period.

Back to the capacitor:

16) During the charging process, the voltage and current are "in phase? just like they are in the variable resistor.

17) The power that is measured at any given instant is equal to the source voltage multiplied by the current.

18) The DC source had to expend a real amount of energy to push and pull the electrons which charged the capacitor. This energy is converted into the electrostatic field in the capacitor.

19) If we assumed the capacitor to be a load in our building and the voltage source to be a DC service from the utility, and we let the capacitor charge, the energy expended in the source to make the current flow would be converted into an electrostatic field in the capacitor. This electrostatic field would increase the overall energy content of our building.

20) The "graph-capable" wattmeter in the diagram would give us a watt reading at each instant in time. This watt reading would be constantly changing until the current stopped flowing at the point where the capacitor voltage was equal and opposite to the source voltage. It would start at a high level and fall to zero just as the current starts high and falls to zero. This would be represented by the green power graph in the image above.

21) A watt-hour meter would give a reading equal to the amount of energy used in the time period.

Is there anything above that is incorrect?

[ January 28, 2005, 04:16 PM: Message edited by: crossman ]
 

Ed MacLaren

Senior Member
Re: Capacitor Theory Thoughts - Comments Requested

Is there anything above that is incorrect?
I would say that everything is correct in principle, subject to the following observation.

If there is negligible resistance in the circuit the charging will take place so quickly that a meter will not have time to respond. Power will also be "negligible".

Ed

[ January 28, 2005, 04:31 PM: Message edited by: Ed MacLaren ]
 

crossman

Senior Member
Location
Southeast Texas
Re: Capacitor Theory Thoughts - Comments Requested

You are right, I didn't think of that. Under theoretical perfect conditions, the charging would be very fast.

Okay, would any of my statements change if we let a little resistance exist in the wires?

quote:
--------------------------------------------------------------------------------
Originally posted by Ed MacLaren:
Power will also be "negligible".
--------------------------------------------------------------------------------

I see it differently on the power part. Power wouldn't be neglible because with absolutely no resistance, the current flow would be near infinite for the short duration that it is flowing. And a near infinite current can produce a lot of power in a short time.

Whew, I was sweating it for a moment, thought the whole thing was falling apart before my eyes.

[ January 28, 2005, 04:58 PM: Message edited by: crossman ]
 

dereckbc

Moderator
Staff member
Location
Plano, TX
Re: Capacitor Theory Thoughts - Comments Requested

Crossman, it would be easier to examine if you assinged some values to the capacitor, some sort of resistance to the line, and a voltage to the source. For example, assumming there is no resistance in the line, the instant you closed the circuit the voltage accross the capacitor is zero, and current is @ max for 0 seconds. No time constants to work with. Add some values of voltage, resistance, and capacitance, and now you can figure some time constants and currents. Try something like 1-ohm, 1-micro-farrad, and 1-volt to KISS it. It takes 5TC to charge a capacitor. 1TC = RC

[ January 28, 2005, 05:12 PM: Message edited by: dereckbc ]
 

crossman

Senior Member
Location
Southeast Texas
Re: Capacitor Theory Thoughts - Comments Requested

Oh man! It took me awhile to put all that together! Let me see if I have time to do it this weekend.

Still, assuming a little bit of resistnace in the wires, do you see anything wrong with anything I said?
 

crossman

Senior Member
Location
Southeast Texas
Re: Capacitor Theory Thoughts - Comments Requested

BTW... I am going somewhere with all this, I just want to see if yall are in agreement with the statements first.
 

rattus

Senior Member
Re: Capacitor Theory Thoughts - Comments Requested

Crossman,

This is the classic problem of charging a capacitor through a resistance with a step function. That is you close the switch at time zero and predict the charging current and capacitor voltage.

I used to throw this at applicants for engineering positions. Most could muddle through it, some could not. Guess who did not get hired.

The charging current is given by,

i(t) = (V/R)e^^(-t/RC)

Energy is converted to heat in the resistor, Ditto when discharging the cap. Energy of the fully charged cap is given by,

W = CV^2/2

The same amount of energy is lost in the resistor which makes the process 50% efficient.

Please note that this applies to a step function, not an AC source.

As R decreases to very small values, the power increases but the energy conversion is the same.

If R = 0, then the power is infinite for zero time and the energy is radiated. This has never been done the lab though.

Crossman, you should note that this is done with time domain analysis.

[ January 28, 2005, 06:54 PM: Message edited by: rattus ]
 

crossman

Senior Member
Location
Southeast Texas
Re: Capacitor Theory Thoughts - Comments Requested

Originally posted by rattus:
I used to throw this at applicants for engineering positions. Most could muddle through it, some could not. Guess who did not get hired .
I wouldn't get hired! :)

As I think about it, it makes sense that the resistor would have half the power and the capacitor would have half. Since it is a series circuit, the decreasing current is the same in each component. And as the voltage of the capacitor is increasing, the voltage on the resistor is decreasing... err...

Oops! Darn it, at first it seemed intuitive, but now it doesn't.

I'll have to do some research on this so I can understand it. Thank you Rattus for showing me this things. I learn something just about every time I look at the forum!

Now, do you see any problems with my original post?
 

physis

Senior Member
Re: Capacitor Theory Thoughts - Comments Requested

it makes sense that the resistor would have half the power and the capacitor would have half. Since it is a series circuit, the decreasing current is the same in each component.
But a capacitor isn't even a conductor. Resistors dissipate heat because of friction.
 

physis

Senior Member
Re: Capacitor Theory Thoughts - Comments Requested

3) The energy in the current flow during charging is converted into an electrostatic field.

4) The potential difference between the plates of the capacitor is in opposition to the source voltage. It is of opposite polarity.

5) As the difference of potential across the plates rises, it opposes the source voltage to a greater and greater extent, and the current will decrease.

6) After the capacitor accumulates enough electrons on one plate and enough "holes" on the other plate, the voltage on the capacitor will be equal to and opposite of the source voltage and current will stop flowing because the two voltages cancel.

7) We could speak of the "opposition to current flow" that is in this circuit... and we could measure it in Ohms if we take the source voltage and divide it by the current flow.
[3] I want to address this but I'm not thinking clearly about it at the moment.

[4] The polarity is the same as the source. The electric fields of the electrons on the plate will repel other electrons but that is not the same as being of opposite polarity.

[5] The energy of the source will push electrons as close together as it can on the plate. When they're as tightly packed as the source can get them the voltage on the capacitor will match the voltage of the source. The capacitor doesn't begin opposing the source as it becomes more charged. When it's charged it is essencially removed from the circuit, no more current flows on it's leads. For DC.

[6] Again, the polarity is the same as the source. There wont be any canceling.

[7] Dereck already mentioned that you need an R value in the circuit. I think the only thing that would impede the charging of an ideal capacitor with 0 Ohms in the circuit is the limit of the speed of light.

Big list. I'll save some for later. :)
 

Ed MacLaren

Senior Member
Re: Capacitor Theory Thoughts - Comments Requested

4) The potential difference between the plates of the capacitor is in opposition to the source voltage. It is of opposite polarity.
I guess you would have to say that the supply and the capacitor are connected "series-opposing".
If they were "series-aiding" there would be a short circuit with large current flow.

Ed

Cap4.gif
 

rattus

Senior Member
Re: Capacitor Theory Thoughts - Comments Requested

6) Strictly speaking, "holes" exist only in semiconductors, Instead we have fewer electrons in the conductive energy band. That is we have an excess on one plate and a deficit on the other.

7) It is not that simple. You must use calculus to find the current.

8) The resistance, R, does not change. You cannot describe the current in this way.

9) Current could be mimicked, but your resistor would have to reach infinity. The changing current is the result of changing voltage across the resistor. Not a good analogy.

10) Phase has no meaning in this example.

11) Yes.

12) Keep it Simple Sam. Current through a resistor always converts electrical energy to heat which will add to the AC load.

13) "Phase" is inappropriate here. However, the product, v(t)i(t), indicates the energy flow into the circuit. This energy will NOT return to the source. I think using the Watt as the unit of power here is OK. "Apparent" power is only used with AC.

14) You could do that as a lab experiment to demonstrate that the math is correct, but it is much easier to take the v(t)i(t) product.

15) The AC watt-hour meter would not work here, although there must be instruments which would measure the power. Again, this is easily calculated.

Back to the capacitor:

16) Again, "phase" is meaningless here. The voltage is a "step function" which is a function of time.

17) Yes, v(t)i(t) = p(t)

18) Half the energy makes it to the cap; the other half is lost in R.

19) Keep it Simple Sam. A cap would not be much of a load. Anytime you charge a cap, it stores energy. A can of gasoline stores energy. Batteries store energy. A tanker truck filled with gasoline stores a lot of energy.

20) Yes

21) Yes, but not the conventional watt-hour meter.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Re: Capacitor Theory Thoughts - Comments Requested

Originally posted by crossman: Is there anything above that is incorrect?
Plenty, I'm afraid. :(

1) . . . . At the same time, electrons are "pulled" off the other plate into the positive side of the source and this plate becomes positively charged.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">You can put it that way. But I prefer to describe it as electrons being ?pushed away from? the second plate by the sudden appearance of electrons on the first plate. I like to explain this in terms of the ?opposite charges attract, like charges repel? rule.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
2) The negative plate and the positive plate have an increasing potential difference across them during charging.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">True. Your original diagram showed no resistors in the circuit. But there is resistance in the wires. Voltage cannot change instantaneously in a capacitor, but current can. At the moment you turn this thing on, you get a huge current. All of the voltage in the source is dropped in the wires. As time goes on, voltage in the capacitor increases, current decreases, and the voltage dropped along the wires decreases.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
3) The energy in the current flow during charging is converted into an electrostatic field.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Not strictly true, but close. A ?field? is not ?energy.? More precisely, some of the energy supplied by the source is dissipated in the form of heating in the wires. This ends when the current drops to zero. The rest of the source?s energy is converted into electrostatic potential energy that is stored by virtue of the existence of the electric field.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
10) The source voltage and the current are ?in phase? in the variable-resistive circuit.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Let me be as gentle as I can, while remaining crystal clear: NO, NO, Sorry, but NO! The concept of ?phase? is absolutely restricted to periodic functions. If the voltage does not go up and down in a pattern that repeats itself forever, then the word ?phase? can not be used to describe it. In the remainder of your posted discussion, any statement that uses the word ?phase? is meaningless. You have not yet told us where you are planning to lead this discussion, but let me advise you that if your desired path has need of the word ?phase,? then I recommend you abandon the path right here and now.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
11) The DC source had to expend a real amount of energy to push and pull the electrons through the opposition contained in the resistor.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">True. And the rate at which that ?real energy? was provided by the source is called the ?real power? supplied by the source.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
12) If we assumed the variable resistor in (10) to be a load in our building and the voltage source to be a DC service from the utility, and we let the variable resistor mimic a charging capacitor, the energy expended in the source to make the current flow would be converted to heat in the resistor. This heat would increase the overall energy content of our building.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">True, but you are no longer modeling a capacitor. There would be no heat generated within a capacitor. There would be a limited amount of heat generated in the wires, as I said earlier.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
13) Anytime that voltage and current are in phase, the power is given in Watts.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">True, but irrelevant. The word ?phase? does not belong in this discussion.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">Back to the capacitor:

16) During the charging process, the voltage and current are "in phase? just like they are in the variable resistor.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Not true, even if you were to disregard my statement about the use of the word ?phase? in a DC system. Your plot did not show the voltage in the capacitor as a function of time. If it did, it would have shown that the voltage in the capacitor is rising at the same time the current through the capacitor is falling. There is no way to even think about that as being ?in phase.?</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
17) The power that is measured at any given instant is equal to the source voltage multiplied by the current.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">True, but irrelevant.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
18) The DC source had to expend a real amount of energy to push and pull the electrons which charged the capacitor. This energy is converted into the electrostatic field in the capacitor.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Not true. See my earlier statements. Also, let me caution you not to use the word ?real? in two different contexts within the same discussion. The common English context of ?real? is ?actual, not fictional.? The electrical engineering context of ?real? is ?associated with power dissipated by resistive loads.?</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
19). . . . This electrostatic field would increase the overall energy content of our building.
</font>
  • <font size="2" face="Verdana, Helvetica, sans-serif">Only in the same sense as using one battery to recharge another battery. You have the same amount of energy that you started with (less any losses in the wires during charging), it is just stored in a different component.</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Re: Capacitor Theory Thoughts - Comments Requested

Originally posted by rattus:
6) Strictly speaking, "holes" exist only in semiconductors, Instead we have fewer electrons in the conductive energy band. That is we have an excess on one plate and a deficit on the other.

7) It is not that simple. You must use calculus to find the current.

8) The resistance, R, does not change. You cannot describe the current in this way.
I disagree with your comments on these three points.

For #6, current flowing in any system whatsoever can be described in terms of electrons going one direction or ?holes? going the other direction. The version using ?holes? is called ?conventional current flow,? and is still taught in schools.

For #7, crossman?s statement, as written, is true. Ohms at any instant in time can be calculated by dividing the volts at that instant by the amps at that instant. You are right in stating that calculus is needed, in order to obtain an expression of current as a function of time or an expression of ohms as a function of time.

For #8, the overall impedance of the circuit is changing throughout the time period shown on the graph. The resistance ?R? of the wires is a constant. The reactance ?X? of the capacitor is not constant. The voltage in the capacitor, as a function of time, is equal to the source voltage minus an exponentially decreasing function. The current through the capacitor is equal to the value of the capacitance times the rate at which that voltage is changing. If you divide these two, you will get the impedance of the capacitor, as a function of time. It is also an exponential function, but with a positive coefficient of time in the exponent. That is, it increases without bound, as time goes on.
 

rattus

Senior Member
Re: Capacitor Theory Thoughts - Comments Requested

From Charlie B.

"For #6, current flowing in any system whatsoever can be described in terms of electrons going one direction or ?holes? going the other direction. The version using ?holes? is called ?conventional current flow,? and is still taught in schools."

*Charlie B: Yes, conventional current assumes positive charges, but in reality electrons are flowing in the opposite direction. A "hole" is the absence of an electron in the valence band of a semiconductor such as silicon, germanium, or gallium arsenide. This from an aged semiconductor engineer. This is a simply a misapplication of the term "hole".

"For #7, crossman?s statement, as written, is true. Ohms at any instant in time can be calculated by dividing the volts at that instant by the amps at that instant. You are right in stating that calculus is needed, in order to obtain an expression of current as a function of time or an expression of ohms as a function of time."

*Charlie B, Ohm's law presumes that R is constant. You can express VI characteristic as a function of time, but it is improper to describe this as a changing resistance. The solution to this problem involves the product of V/R and an exponential function of time. e.g.,

i(t) = V/R(exp^^-t/RC)

where V, R, and C are constant.

To describe this as a changing resistance is misleading. It is like describing a motor by an impedance. In the steady state, its load can be represented as a single impedance, but to describe this impedance as a function of time is misleading.

"For #8, the overall impedance of the circuit is changing throughout the time period shown on the graph. The resistance ?R? of the wires is a constant. The reactance ?X? of the capacitor is not constant. The voltage in the capacitor, as a function of time, is equal to the source voltage minus an exponentially decreasing function. The current through the capacitor is equal to the value of the capacitance times the rate at which that voltage is changing. If you divide these two, you will get the impedance of the capacitor, as a function of time. It is also an exponential function, but with a positive coefficient of time in the exponent. That is, it increases without bound, as time goes on."

*Charlie B., Impedance and reactance have no meaning in time domain analysis which this clearly is. You need a frequency to compute reactance and there is none in this problem.

The only things changing are i(t), v(t) and t itself.

I thought we had settled this already with my reference to Dr. Heizer.
 

crossman

Senior Member
Location
Southeast Texas
Re: Capacitor Theory Thoughts - Comments Requested

Guys, first let me thank you for all the wonderful thoughts and for your time on this. I didn?t know it was going to explode in my face!! I went back and made some corrections and added some thoughts and these are all in bold on the regurgitated list way down below. I would not blame you if you didn?t read it because is very time consuming.

What I was trying to show is that a capacitor on an AC circuit in the first quadrant is very similar to a capacitor in a DC circuit. I don?t know if I managed that.

So if you want to skip down to the list, I certainly welcome you to, but I am going to cut to the root of my problem right now?..

First, the thinking that follows originated in the ?Is power constant in three-phase thread?? I saw some things that I intuitively thought were wrong, using RMS values in instantaneous equations, some sweeping assumptions based on RMS realities that were then applied to instantaneous calculations, and I wanted to know what was really happening.

My answer can be found by discussing the power in a capacitor. In an AC circuit, over time, the power coming into a capacitor is equal to the power going out of a capacitor and we say that the capacitor has no wattage. I agree wholeheartedly with this.

Take a gas generator (AC) with a certain amount of internal resistance, the gen feeding a cap through conductors with some small resistance (I don?t want to get bogged down talking about the resistance). The generator has to burn a certain amount of gasoline to charge the cap in the first quadrant. This energy is converted to another form in the cap and the energy is stored. In the second quadrant, the gen voltage is decreasing and the cap voltage is higher than and opposite of the gen voltage polarity, so the cap discharges back through the gen winding. The energy that was stored in the cap is converted to heat in the gen winding (this may not be entirely correct? there may be some sort of inductive type of reaction in the gen windings also. I don?t know.) Suffice it to say the energy isn?t used in the cap.

The result is, the gen burned gas, but after one half of a cycle, there is no energy in the cap any more, so the cap essentially was a bust as far as doing any work for me. This is where we get the notion of apparent power, VARs, PF. We burned the gas, apparently we did some work, but after half a cycle, poof it is gone. We still have to pay for the gas, but the effort was wasted as far as getting any useful work done.

Same thing happens in the third as the first, and fourth as the second quadrant, only directions of voltage and current are reversed. As long as we continue in time, the cap is using up no wattage over the long haul. Takes it in, dumps it back out.

But now let?s go back to the first quadrant. I am saying that the power going into that capacitor can be measured in watts, or watt hours over the time of the first quadrant. It takes energy to rearrange the molecules in the dielectric that help produce the electrostatic charge. These are particles that are being physically moved. Doesn?t it take energy to do this? Real energy?

Say I jerk the cap out of the circuit at the end of the first quadrant? the cap is charged to the AC peak voltage. And then I go hook up a light bulb to the cap and now the light uses up the watts. So all of a sudden the gasoline wasn?t wasted after all? As long as the energy was stored, the gas was wasted? Remember, the assumption is that no power is used in a capacitor. The power in a capacitor is zero? Even in the first quadrant?

Wouldn?t a watt hour meter hooked up to the circuit in this scenario actually register a reading for charging the cap? I?m talking about where we jerked the cap out of the circuit at the end of the first quadrant?

Let?s go to a resistor in a theoretically perfect world where all energy transfers are 100% efficient. Our generator is supplying current to the resistor. The resistor is putting off heat. I am using the heat to make steam. And I am using the steam to turn an AC generator and I am using that generator to charge a capacitor in the first quadrant. So the resistor is using watts in that first quadrant?

In this case, and speaking of only what happened in the first quadrant of both cases, the resistor is using watts, yet the end result is exactly the same as in the other first quadrant cap charging scenario. In both cases, I have converted the gasoline into a charged cap. But in one case we had watts, but the other we didn?t? Only because there was an intermediate heat transformation? What is so special about heat? Isn?t heat just another form of storing energy and transferring energy?

If I put each of these first quadrant scenarios in an impervious box, with the gasoline generator on the outside, the supply conductors going into the box, and you could only measure things on the outside of the box, is there any way you could tell which box had the cap and which one had the resistor/gen/cap setup? Again, this is on a theoretical basis with perfect and instantaneous energy transfer. And by impervious, I am saying no energy can come out, and the only energy that can go in is through the conductors.

I know it is far-fetched, but I am talking theory here. You may say that the definition of a Watt in an RC circuit is ?the energy used in the resistive portion of the circuit.? But that is only true for a special case. That is only true when we go for at least 1/2 of a cycle. Otherwise, the energy taken in is more than the energy going back to the source. The energy cancels only if we have 1/2 of a cycle or integer multiples thereof.

There is more than one definition of a watt. A watt isn?t even necessarily an electrical unit. The Watt that we use with our phasors/vectors in sine wave calculations is a special case, that, like I said, only applies to half cycles and integer multiples of 1/2 cycles. This watt that ?is only produced when a resistor converts electrical energy to heat? doesn?t apply and cannot be used in computations that are less than 1/2 of a cycle. But that doesn?t mean that a watt meter won?t register in the first quadrant of a cap circuit. I think it will. I suppose if you had a theoretical bi-directional wattmeter that could read instantaneous values, it would be reading positive in the first quadrant and negative in the second quadrant.

That is certainly what happens when you multiply the instantaneous voltage times the instantaneous current and plot power for a bunch of different instants in the first half cycle. Positive power in the first quadrant, negative power in the second. There is the same amount of positive power as there is negative power? and it cancels out to zero for each half-cycle. What units are on this power? It certainly isn?t volt-amps. The volt-amps of a capacitor are not zero. It certainly isn?t VARs. The VARs of a cap are not zero. That only leaves watts! The instantaneous current x the instantaneous source voltage in a cap circuit gives you the watts for that instant. Otherwise, tell me what the unit is?. The unit for instantaneous current x instantaneous voltage in a cap circuit is??

How about a theoretical 100% efficient 100% power factor motor that is running a 100% efficient air compressor and storing the energy as compressed air in a tank? So that motor used watts? But all we did was store the energy in another form. Kind of like storing the energy in a capacitor in the first quarter cycle. What is the difference?

Oh man, I am brain-dead now. I am sure I have blown this whole mess out of proportion and I apologize for that. If all this is inconsequential and meaningless then tell me so and I will drop it. Thank you for your patience.


From the first post in this thread? changes based on you guy?s comments:

Note: Assume some resistance in the conductors

1) changes for Charlie B: Upon connection of the capacitor to the DC source, electrons begin flowing in a clockwise direction in the circuit. Electrons flow out of the negative side of the source. Electrons flow into the positive side of the source. Electrons from the negative side of the source flow into the upper plate in the diagram and it becomes negatively charged. The electrons accumulating on this negative plate repel an equal number of the electrons on the other plate toward the positive side of the source. At the same time, the positive side of the source is attracting these electrons. The bottom plate in the diagram will become positively charged.

2) The negative plate and the positive plate have an increasing potential difference across them during charging.

3) changes for Charlie B: A certain portion of the energy in the current flow during charging will be converted to heat in the conductors. A percentage of the remainder of the energy will result in a change of the arrangement of the molecules in the dielectric of the capacitor. Before charging, the dielectric molecules are positioned at random. During, the dielectric molecules become increasingly aligned? the positive portion of the molecules will be aligned toward the negative plate, and the negative portion of the molecules will align with the positive plate. It takes energy to move these molecules in the dielectric

4) The potential difference between the plates of the capacitor is in opposition to the source voltage. It is of opposite polarity.

Note for physis: It is my thinking that this is a series circuit and the source voltage is trying to push the current in a clockwise direction while the voltage on the capacitor is trying to push the electrons counter-clockwise through the circuit. These voltages therefore can be called ?series-opposing? sources and they are of opposite polarity in their relation to the series circuit. This same line of reasoning applies to (5) and (6) below.

5) As the difference of potential across the plates rises, it opposes the source voltage to a greater and greater extent, and the current will decrease.

6) for Rattus: After the capacitor accumulates enough electrons on one plate and enough ?absence of electrons? (replaced the term ?holes? although I feel it is a legitimate term to use in this case) on the other plate, the voltage on the capacitor will be equal to and opposite of the source voltage and current will stop flowing because the two voltages cancel.

7) Changed for Rattus to specify that I am talking about only individual instants: We could speak of the "opposition to current flow" that is in this circuit... and we could express this opposition at any instant if we take the source voltage at that instant and divide it by the current flow at that instant. The unit of this opposition would be volts/amps and this is also called ?ohms.? Notice that I am not calling this reactance. This is simply a way to express the opposition based in units of volts and amps at given instants in time

8) For rattus: If we take the voltage and current values as defined from the graph in the first post of this thread and do the calculation in (7) at several different instants in time, the opposition will be relatively low if the current is high. The opposition will be relatively high if the current is low. The volts/amps (which is another way to express ohms) value will start near zero and rise to near infinity as the capacitor charges. I am not calling this resistance. I am calling it opposition to current flow.

9) For Rattus: Considering the voltage and current graph, the "variable nature" of the opposition to current flow of a capacitor could be mimicked with a variable resistor. A graph exactly similar to the graph above could be produced by varying the resistance in a precise manner.

Question for all: We are considering two circuits here. One with a capacitor, one with a variable resistor mimicking the graph of the source voltage and the current of the capacitor. If I enclosed the capacitor in a perfectly impervious box and the resistor in another perfectly impervious box, are there any measurements you could make to determine which box contains which? The source is on the outside of the box in each case and the conductors pass through. By impervious box, I mean that no energy can escape from the box, the only energy that can enter the box is through the conductors from the DC source, and you cannot make any measurements inside the box.

10) For Charlie b and rattus: While current is flowing in the variable resistor, the direction of the current flow in the circuit coincides with the polarity of the DC source voltage. In other words, the source voltage is such as to cause electrons to flow in the clockwise direction and the current flow is in this same direction.

11) changed for Charlie b, I took out ?real? because I don?t want any confusion: The DC source had to expend energy to push and pull the electrons through the opposition contained in the resistor and in the conductors.

12) changed for Charlie b and rattus? to indicate that the variable resistor could produce the exact current graph of the capacitor, but the resistor would produce heat but the capacitor would not. If we assumed the variable resistor in (10) to be a load in our building and the voltage source to be a DC service from the utility, and we let the variable resistor mimic the voltage and current graph of a charging capacitor, the energy expended in the source to make the current flow would be converted to heat in the resistor. This heat would increase the overall energy content of our building.

13) changed for Charlie b and rattus: The polarity of the source voltage in this circuit is trying to push the current in a clockwise direction in the circuit. The current IS flowing in the clockwise direction. At any instant that the current is flowing in the direction that the source voltage is pushing, we can multiply voltage times current and express the power in watts.

14) Concerning the variable Ohms scenario, the "graph-capable" wattmeter in the diagram would give us a watt reading at each instant in time. This watt reading would be constantly changing until the current stopped flowing at some very large value of resistance. The Watt graph would start at a high level and fall to zero just as the current starts high and falls to zero. This would be represented by the green power graph in the image above.

15) A DC capablewatt-hour meter would give a reading equal to the amount of energy used in the time period.

Back to the capacitor:

16) changed for rattus and Charlie b: During the charging process, the source voltage is trying to push current in a clockwise direction. The current IS flowing in the clockwise direction which is the direction of the polarity of the source.

17) The power that is measured at any given instant is equal to the source voltage multiplied by the current. Charlie b and rattus, this relates back to number (13). What do you think for the capacitor? What units would be used to express the power at each instant?

18) changed for Charlie b and rattus: The DC source had to expend energy to push and pull the electrons which charged the capacitor. A 1/2 of the energy in the current flow during charging will be converted to heat in the conductors. The remainder of the energy will result in a change of the arrangement of the molecules in the dielectric of the capacitor.

19) If we assumed the capacitor to be a load in our building and the voltage source to be a DC service from the utility
that is outside the building, and we let the capacitor charge, a portion of the energy expended in the source to make the current flow would be used to rearrange the alignment of the molecules in the dielectric of the capacitor. This rearrangement of the molecules would increase the overall energy content of our building.


Quote from Charlie B: Only in the same sense as using one battery to recharge another battery. You have the same amount of energy that you started with (less any losses in the wires during charging), it is just stored in a different component. Just wanted to reiterate that the source is outside of our building, the cap is in our building, when the cap is charged, our building contains more energy than it did before the cap charged.

20) The "graph-capable" DC wattmeter in the diagram would give us a watt reading at each instant in time. This watt reading would be constantly changing until the current stopped flowing at the point where the capacitor voltage was equal and opposite to the source voltage. It would start at a high level and fall to zero just as the current starts high and falls to zero. This would be represented by the green power graph in the image above.

21) A DC watt-hour meter would give a reading equal to the amount of energy used in the time period.
 

rattus

Senior Member
Re: Capacitor Theory Thoughts - Comments Requested

Crossman, let me pick on one thing at a time.

"There is more than one definition of a watt. A watt isn?t even necessarily an electrical unit. The Watt that we use with our phasors/vectors in sine wave calculations is a special case, that, like I said, only applies to half cycles and integer multiples of 1/2 cycles. This watt that ?is only produced when a resistor converts electrical energy to heat? doesn?t apply and cannot be used in computations that are less than 1/2 of a cycle. But that doesn?t mean that a watt meter won?t register in the first quadrant of a cap circuit. I think it will. I suppose if you had a theoretical bi-directional wattmeter that could read instantaneous values, it would be reading positive in the first quadrant and negative in the second quadrant."

The Watt is the unit of power used in electrical work. Horsepower is the unit of power used in mechanical work. 746 Watts is equivalent to 1 hp.

A watt is defined as an energy flow of 1 Joule/Sec. One definition, no more.

Power does not flow; energy flows.

A Watt is a Watt is a Watt whether AC, DC, transient, or whatever. No special cases.

You are being confused by the concepts of apparent and real power which are used by the power industry. Forget VAs and VARs for the moment.

It is acceptable to describe energy flow rates in and out of a circuit as positive and negative power in Watts.

[ January 30, 2005, 11:08 AM: Message edited by: rattus ]
 

crossman

Senior Member
Location
Southeast Texas
Re: Capacitor Theory Thoughts - Comments Requested

Rattus, thanks for the reply. I misinterpreted some of what was being said in the other thread. This got my mind to examining every little thing I could about it. After your reply, I think I have a handle on it now.

appreciated!
 
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