How many times have we all heard the phrase, "This capacitor is charged."? Does that statement mean a capacitor can store a charge? Does it mean that it contains more coulombs of negative electrons than it did before? Believe it or not, the answer to both previous questons is NO. Why? Because for every coulomb of charge accumulated on one plate of the capacitor, the same amount of charge is removed from the opposite plate. This results in a NET charge of ZERO, no matter what voltage is applied to the capacitor. The capacitor can be imbalanced with respect to charge, and this causes a voltage to appear across it. This voltage in turn creates an electrostatic field which stores energy. So, whenever a capacitor is imbalanced chargewise, and has a voltage across it. It is storing energy, not charge. Therefore we should refer to a capacitor in this condition as "energized", not "charged". The amount of energy stored by the capacitor is E=?CV?. Ratch
Posted on 2003-07-02 21:30:37 by Ratch
Energized:
Electrical energy (Voltage) has been applied to the device. (And of course felt across the device across both leads)

Charged:
Electrical current has been made to flow into the device. (And of course an equal amount has left the device via the lead opposite the one it flowed into)

Capacitance:
The capacity to store electrical charge in an electrostatic field.

Charge:
Coulombs (of charge) = Current times Time.
C = I * t (C in this case stands for Charge, not capacitance)

Saying that a capacitor does not hold charge is like saying that the rubber band pulled back across one's finger does not store the abitlity to propel itself across the room when the appropriate end is released. If I point it at your eye and release the correct end, will it not discharge it's stored energy into your eye? Energy is not energized, nor is it deenergized. Energy is charged and discharged.

Capacitors charge and discharge in response to changes of electrical potential across it's leads in what appears to be an attempt to neutralize the changes in Voltage. The exact charge of a Capacitor (in Coulombs) is equal to 1/2 the difference of the count of electrons of each plate divided by the count of electrons in one Coulomb of charge. The net charge of the capacitor is expressed in terms of it's capacity to pump electrons into a curcuit, and pumps only move particles. And the obvious is true, for every electron it pushes into a circuit it must also accept an electron back into it's other plate. When the count of electrons on either plate is the same, there remains no more capacity to pump or discharge.

It is the inductor that does not become charged, but rather becomes energized. An inductor will develop a voltage (potential energy) across it's leads in response to changes in current through it in what appears to be an attempt to neutalize changes in current. This is accompished by building and collapsing an elecromagnetic field around the inductor. This phenomenon is equivalent to inertia.

Capacitors resist change in voltage by applying current.
A capacitor cannot apply current unless it is first charged by a Voltage.
Voltage is pressure, and pressure is required to push electrons into the capacitor, and to evacuate electrons from the opposite plate.
A capacitor is said to be fully charged when the current flow into it has tapered off to zero, at which time it's electrostatic field has it's maximum potential to apply current.

Inductors resist change in current by applying voltage.
An inductor cannot apply voltage unless it is first energized by a Current.
Current is electron flow, and electron flow is required to build an electromegnetic field around an inductor.
An inductor is said to be fully energized when the voltage across it has tapered off to zero, at which time it's electromagetic field has it's maximum potential to apply voltage.
Posted on 2005-01-28 19:16:07 by gluespill
I prefer to see a cap as an integrator.
Posted on 2005-01-29 07:38:02 by lifewire
I suppose that any very short section of the charging curve of a capacitor could be viewed as an integral. Replace the feedback resistor with a capacitor on an inverting differential amplifier circuit and you'd see an output voltage across the cap that is nearly an exact integral from beginning to end of it's charge period.

But then again, the cap could be used to create a differentiator by replacing the input resistor in the same circuit with a capacitor instead of using it in the feedback loop. The curve of the output voltage of the circuit would be very similar to that of a differential. In a perfect universe, the cap would charge instantaneously, and the output of the circuit would be a straigth line up (or down depending on the polarity applied) to the maximum voltage, and then straight back down to zero.

The cap can't perform integration by itself, it must be made part of a circuit to function as one, so the circuit would actually be the integrator, not just the cap. What components does your circuit have that gives you an integral charge curve? Just a DC supply and a resistor?
Posted on 2005-01-29 10:33:25 by gluespill
gluespill,

Energized:
Electrical energy (Voltage) has been applied to the device. (And of course felt across the device across both leads

Electrical energy is NOT voltage. Voltage is applied to a capacitor and energy is stored within as an electrostatic field.

Charged:
Electrical current has been made to flow into the device. (And of course an equal amount has left the device via the lead opposite the one it flowed into)

You mean electrical charge, not current flows into the device, don't you? Anyway the same could be said of a resistor, but you would not call that charging a resistor, would you?

The capacity to store electrical charge in an electrostatic field.

The capacity to store electrical ENERGY in an electrostatic field.

Charge:
Coulombs (of charge) = Current times Time.
C = I * t (C in this case stands for Charge, not capacitance)

Coulombs of displacement equals the integral (accumulation) of I(t)*dt. Q=I*t is only valid if the current is constant.

Saying that a capacitor does not hold charge is like saying that the rubber band pulled back across one's finger does not store the abitlity to propel itself across the room when the appropriate end is released. If I point it at your eye and release the correct end, will it not discharge it's stored energy into your eye? Energy is not energized, nor is it deenergized. Energy is charged and discharged.

From the American Heritage Dictionary:
en?er?gize v. en?er?gized, en?er?giz?ing, en?er?giz?es. --tr. 1. To give energy to; activate or invigorate

As I explained very clearly above, a charge flow into a capacitor causes its energy to change and its net charge to be unchanged. How can you say it is "charged" then?

The exact net charge of a Capacitor (in Coulombs) is equal to 1/2 the difference of the count of electrons of each plate divided by the count of electrons in one Coulomb of charge.

The net charge on a capacitor is the same before and after a voltage is applied. That is a fact. The stored energy changes, and the voltage difference between the plates changes.

Capacitors resist change in voltage by applying current.

No they don't. Caps resist current by building up a back voltage.

A capacitor cannot apply current unless it is first charged by a Voltage.

Strictly speaking, current is never applied. Voltage is. Can't have current without voltage, except may in a superconductor.

... it's electromagetic field has it's maximum potential to apply voltage.

The EM is determined by the current, not the voltage.

The cap can't perform integration by itself, it must be made part of a circuit to function as one, so the circuit would actually be the integrator, not just the cap. What components does your circuit have that gives you an integral charge curve? Just a DC supply and a resistor?

A cap directly integrates (accumulates) current over time resulting in charge. Ratch
Posted on 2005-01-31 12:00:13 by Ratch
Please do not confuse the activity and behaviour of an inductor with that of a capacitor, and vice versa.
They both act in very predictable and opposite fashions in both AC & DC circuits.

A capacitor will oppose change in voltage across itself by adding current to or absorbing current from the circuit.
The current flow into or out of a capacitor remains at zero until the voltage across its leads either increases or decreases.
Once the voltage across a capacitor has attempted to change, the current into and out of the capacitor will increase momentarily to a non-zero value.

An inductor will oppose change in current through itself by generating a voltage across itself.
The voltage across an inductor will remains at zero until the current through it either increases or decreases.
Once the current through the inductor has has attempted to change, the voltage across the inductor increases momentarily to a non-zero value.

Capacitors resist changes in voltage across their leads. That is why we sometimes put them in parallel with the output of DC rectifier circuits in order to filter some of the reslutling ripple. When the voltage at the output of the rectifier circuit increases momentarily due to ripple, the capacitor charges slightly, which keeps the extra little bit of current from reaching the load. Then when the voltage at the output of the rectifier circuit decreases on the back edge of the ripple, the capacitor reacts by discharging a little current into the load. The load sees a more constant voltage as a result and therefore develops a more constant voltage drop than if the capacitor were omitted.

If capacitors opposed changes in current, then they would have the ability to keep current constant in a circuit. That priveledge, however, belongs to the inductor. The inductor is just a piece of looped wire. Being wire, it can have constant current flow through it and can react to any changes in the current through it. A capacitor can only receive current flow in one direction for for a finite length of time.

Since the capacitor is composed of two plate surfaces isulated from one another, the capacitor does not have the ability to conduct current through itself. The electrons that come in on one lead of a capacitor can only leave the capacitor on that same lead. They do not typically jump across the plates and exit on the other lead. The capacitor will allow electrons to be conducted into one lead as long as there is sufficient voltage to do so. As more and more electrons are added to one of the plates, it becomes more and more difficult to add them. As the capacitor becomes fuller, more voltage is required to add more electrons, just as when a metal tank is filled with air by applying air pressure to its opening, it becomes more and more difficult to add more air as the tank grows fuller.

A capicitor cannot truly conduct current, and it has no knowledge of how much current is flowing elsewhere in a circuit. It only knows how tightly packed its electrons are, and whether or not enough pressure (voltage) is being applied to add more electrons and/or to keep the ones it currently has packed inside contained. If the pressure (voltage) outside has gone down and there is a discharge path, the capacitor will release electrons back into the circuit until its electrons are only as tightly packed as the outside pressure (voltage) is capable of maintaining.

Inductors resist changes in current going through it. That is why we sometimes put them in series with the output of DC rectifier circuits in order to filter some of the reslutling ripple. When the current being supplied to the load increases momentarily due to ripple, the inductor builds it's electromagetic field slightly larger than before which produces a small voltage across the inductor opposite in polarity to that of the rectifier circuit output. This opposes the output voltage just enough to keep the overall voltage applied to the load from increasing. Then when the current being supplied to the load decreases on the back edge of the ripple, the inductor reacts by collapsing some of its electromagnetic field developing a small voltage with the same polarity as the rectifier output. This aids or complements the output voltage just enough to keep the overall voltage applied to the load from decreasing.

An easy way to observe the behaviour of these two components is to examine them in DC circuits. Start the observation with no power applied to the circuit, and then record what happens starting at the exact moment the power is applied and ending when all change in the circuit has stopped.

Start with an inductor and a resistor in series with a DC supply and a switch. With the switch open, take readings across all devices in the circuit. (It's obviously not really a circuit until the switch is thrown, but it is a logical circit nonetheless.)

You should measure zero volts across both the inductor and across the resistor.
You should read the full DC supply voltage across both the DC supply and also across the switch.
So, with no current flowing through the inductor, it has no voltage across it. Makes sense.

Next, put an analog voltmeter (one with a needle, not digital) or an oscilloscope across the inductor.
While watching the voltmeter/scope very closely, close the switch to energize the circuit.

At the very moment that you close the switch, you should read nearly the full DC supply voltage across the inductor. The polarity of the inductor's voltage should be opposite to that of the supply. In other words, the lead of the inductor closest to the positive termianal of the DC supply should have measured positive, and the other lead negative. This will occur very quickly, and might not be noticeable if too small a value of inductor is used. You will probably only see the needle or scope trace jump towards the value of the DC supply voltage, after which it will return to zero volts.

The inductor is just a piece of wire wrapped in loops, so once it has finished it's 'inducting' in this purely DC circuit, it reverts to acting exactly like the wire that it is. For all practical purposes, wires have no voltage drop across them, and neither does the inductor in this circuit after the circuit has stabilized.

The interesting thing, though, is that at the moment you closed the switch, the inductor was acting the same as an open switch, and a perfect circuit with a perfect measuring device would have measured the full DC source voltage across the inductor at the moment the switch was closed. At the very moment after closing the switch, a very small amount of current will start to flow through the inductor which results in a very small electromagnetic field being generated around the inductor. This current increases over time, and as it does, so also does the electromagnetic field around the inductor increase in intensity.

At some point, (likely in only a few milliseconds depending on the values of components and DC supply chosen) the current through the inductor stabilizes to the value that would be flowing through the resistor if the inductor were not even in the circuit. At this point the only difference between this circuit and one without the inductor is the presence of the electromagnetic field around the inductor.

So again, with no current through the inductor, there is no voltage drop across it.
At the moment the switch is closed, the inductor has the full source voltage across it for a moment with polarity opposite that of the supply.
Current flow increases up to the maximum limited only by the series resistor.
When the current flow has stabilized at maximum, again there is no current flow through the inductor,
at which point there will also be no voltage drop measured across the inductor.

You could perform the same procedure, but with the voltage measuring device connected across the resistor. In this case you should measure zero volts across the resistor at the moment the switch is thrown. In a short amount of time you should see the voltage drop across the resitor climb to the same value as the DC source. The resistor initally has no voltage drop across it when the switch is closed because the inductor is acting like an open switch preventing any current from flowing anywhere in the circuit. But as the inductor builds its magnetic field and current flow increases, the resistor develops an increasing voltage drop until it reaches maximum.

Please skip this paragraph if you find MIT like blather to be offensive.
<begin MIT like blather>At any moment in time, the sum of all vector voltages in the loop will be equal to zero.]
At any moment in time, the current will be the same value and phase at any point in the circuit.]
Between the moment that the switch is closed and the moment that the current stabilizes, the current though any single component, including the DC source, will be out of phase with the voltage across it.<end MIT like blather>

The next observation I would suggest would be to change the current in the circuit and observe what happens. This can be accomplished by connecting a resistor across the leads of the switch. While the switch is closed, the additional resistor has no effect on current. However, at the moment you open the switch, the resistor becomes a series component in the circuit. Adding resistance in a series circuit will result in a decrease in current. More resistance means less current assuming the supply voltage remains constant.

Before opening the switch in the energized and stabilized circuit, connect your voltage measurement device across the inductor. Then, once again watching the voltmeter/scope very closely, open the switch. You should observe a brief voltage drop across the inductor, but the polarity should be opposite to the polarity you observed when you first energized the circuit. This time the inductor lead closest to the DC supply's positive terminal will now be negative for a moment, while the other is positive. The current will once again stabilize and the current through the circuit will only be limited by the two resistors in the circuit. Once again the inductor is acting like the wire that it is in the stabilized circuit.

So recapping:
Initially, before energizing the circuit, with no current through the inductor, there is no voltage drop across it.
At the moment the switch is closed, the inductor has the full source voltage dropped across it for a moment with polarity opposite that of the supply.
Current flow increases up to the maximum limited only by the single series resistor.
When the current flow has stabilized at maximum, again there is no current flow through the inductor, at which point there will also be no voltage drop measured across the inductor.
We then increased the resistance in the circuit wich caused a decrease in current.
When the current decreased, there was a brief voltage drop across the inductor, but this time with the same polarity of the DC supply.
Once the current stabilized to the new lower value, the voltage across the inductor was once again zero volts.

In conclusion:
The only time the inductor is acting like an inductor and not a piece of wire is when the current through it attempts to change.

When the current through the inductor attempts to increase, the inductor develops a voltage across itself opposite in polarity to that of the DC supply. This briefly reduces the overall applied voltage (like putting one of the batteries of a flashlight in backwards) and briefly prevents some of the change.

When the current through the inductor attempts to decrease, the inductor develops a voltage drop across itself with the same polarity as the DC supply. This briefly increases the overall applied voltage (like batteries working together in series) in the circuit which briefly prevents some of the change.

The inductor does this by building its field to oppose current increases, and by collapsing its field to oppose current decreases.

An inductor opposes changes in the amount of current flow. It does this by generating a voltage across itself to either aid the supply when current has attempted to decrease, or to oppose the supply when the current has attempted do increase.

Posted on 2005-01-31 20:47:43 by gluespill
gluespill,
I have not mentioned inductors in my previous posts within this thread. You are preaching to the choir when you describe the electrical behavior of inductors and capacitors. I have had many university courses associated with this subject. My conflict is not about what happens. It is about what it is called. I contend that supplying energy to a capacitor should be call energizing, not charging. Charging sounds too much like one is bestowing a charge into the capacitor, which is not the case. As far as charge is concerned, energizing only imbalances the charge on the plates of the cap, but it does not change the net charge. If you can address just those points I made, it would simplify the discussion. If you have something else to eludicate, may I suggest you start a new thread. Ratch
Posted on 2005-02-01 06:44:50 by Ratch
No Ratch, I will not address any of your specific points. I do not have the same level of skill that you posess for picking fly poop out of pepper. I'll leave that to you.
Posted on 2005-02-01 11:42:48 by gluespill
Well my only problem with glues lengthy description of an inductor is he keeps using the term "resistance" when in actuallity the term for opposition of current flow in an inductor is "reactance".
Does anyone remember the phrase "ELI the ICE man" ? Remember that means voltage leads the current through an inductor and vice-versa for a capacitor. So the opposition to current flow is a net result of the phase shift that occures in an inductor. This comes in the form of a stored electro-magnetic field that will impead the current flow inside the wire by being at some phase opposite to that current flow.
Posted on 2006-10-22 21:28:10 by mrgone
mrgone,

Wow, long time since this thread has been active.

OK, first let's establish some definitions.  Current is the movement of charge per unit of time.  In other words, current is charge flow.  Although most folks say current flow, what they really mean is charge flow.  As I pointed out in other threads in this forum, current flow means charge flow flow, which is redundant and ridiculous.

Resistance and reactance act to reduce current by two different methods.  Resistance requires energy to move the charge carriers so current can exist.  Since voltage is energy per unit of charge, this causes a reduction of voltage and converts the electrical energy into heat.  Inductive reactance is different.  That is caused my a moving magnetic field cutting across the coil windings causing a back-voltage.  This too reduces the current, because less voltage is available to move the charge carriers. But it does not expend energy.  One thing to keep in mind is that the back-voltage is caused by the changing magnetic field.  Therefore establishing a high DC current in a high inductance coil will not impede the charge flow after the current is established, even though a strong magnetic field is present.  It is only when the current changes that a voltage will be induced which impedes or assists the charge flow.  The energy will be taken from the charge carriers to establish the field and given back when the field collapes, for a net energy loss of zero if resistance is neglected.

So the opposition to current flow is a net result of the phase shift that occures in an inductor

Not quite correct.  The opposition to current change is because of the back/forward voltage, caused by the changing magnetic field, works for/against the applied voltage.  We talk about phase when sinusoidal voltage/currents are present, but if the applied voltage is a ramp or sawtooth, what is the phase of that? Ratch

Posted on 2006-10-22 23:12:47 by Ratch
You know when you think about it, theoretically that's true that reactance is only a result of a changing voltage (AC). If there were only opposition to the flow of electrons when power is first applied in a DC circuit and afterwards it was purely resistive determined by the resistance of the wire itself than why would the inductance of a coil still have a bearing on the amount of power consumed?
Let me clarify that by using an example and then give my explanation of the event. Let's suppose we have a common solenoid. Simply a coil with some sort of metalic or semi-metalic core. For simplicity sake we will suppose a nail wraped with several turns of wire. In a sence we have a coil or an inductor...agreed? Now if I move a piece of metal near the solenoid it will attract it and hold it to the solenoid/inductor. Now if in the case of a solenoid I have say a spring opposing this attraction (possibly a relay would be a clearer example), this situation will consume power. The kinetic energy required to energize the relay will be felt by the power source. Bear in mind the current flow is in only one direction. So why would the atuator of a relay cause extra power to be consumed if the only opposition to current flow is resistive? That would not make sense.
Here's my explanation. Even in the situation of DC, the electro-magnetic field opposes the current flow and is at some phase angle determined by the inductance of the coil. When a relay is energized, the actuator increases the inductance therefore creating a larger reactance or impeadance to the flow of current. Bear in mind, the actuator is not physically attached to the wire in the coil, so there is no way for the resistance to increase as this is performed through the elecro-magnetic field. So if the electro-magnetic field is changed by increasing the inductance, then we must have a larger opposing field which would cause larger power consumption.
Posted on 2006-10-23 08:30:15 by mrgone
mrgone,

Why are we discussing inductance in this thread, which it is named Charging Capacitors?

You know when you think about it, theoretically that's true that reactance is only a result of a changing voltage (AC). If there were only opposition to the flow of electrons when power voltage is first applied in a DC circuit and afterwards it was purely resistive determined by the resistance of the wire itself than why would the inductance of a coil still have a bearing on the amount of power energy consumed?

Pure inductance, no matter how large, does not consume any energy on a net total basis.  It consumes energy when the field is being established, and gives it all back when the field collapes, for a net total of zero.

Let me clarify that by using an example and then give my explanation of the event. Let's suppose we have a common solenoid. Simply a coil with some sort of metalic or semi-metalic core. For simplicity sake we will suppose a nail wraped with several turns of wire. In a sence we have a coil or an inductor...agreed?

Yes.

Now if I move a piece of metal near the solenoid it will attract it and hold it to the solenoid/inductor. Now if in the case of a solenoid I have say a spring opposing this attraction (possibly a relay would be a clearer example), this situation will consume power energy. The kinetic energy required to energize the relay will be felt by the power source. Bear in mind the current flow is in only one direction. So why would the atuator of a relay cause extra power energy to be consumed if the only opposition to current flow is resistive? That would not make sense.

I think you are referring to a spring tensioned solenoid plunger.  No energy will be lost in the circuit on a net total basis, except for wire resistance and mechanical friction.  It will take more energy to activate the spring tensioned plunger, and some of the energy will stored as the potential energy of a stretched spring.  The rest of the energy will go into establishing the magnetic field.  When the solenoid is released.  All the energy will return to the circuit.

Here's my explanation. Even in the situation of DC, the electro-magnetic field opposes the current charge flow and is at some phase angle determined by the inductance of the coil.

In sinusoidal (AC), the voltage and current can be out ot phase with each other.  But the voltage and current are both sinusoidally shaped.  For a non-sinusoidal excitation, the voltage and current do not have the same shape, so "phase" has no meaning.

When a relay is energized, the actuator increases the inductance therefore creating a larger reactance or impeadance to the flow of current charge

Yes, it does.  A coil with a metal core will have more inductance than one with an air core.

Bear in mind, the actuator is not physically attached to the wire in the coil, so there is no way for the resistance to increase as this is performed through the elecro-magnetic field. So if the electro-magnetic field is changed by increasing the inductance, then we must have a larger opposing field which would cause larger power energy consumption.

As I said before, the amount of inductance and thereby reactance will not increase energy consumption, because there is no energy consumption due to reactance in the first place

I am not quite sure about what you are trying to explain.  By the way, power is the rate of energy conversion per unit of time.  A lot of folks say "apply power" when they really mean apply electrical energy.  Of course, the electrical energy source has to have the capacity to supply energy at a sufficient rate.  That capacity can be referred to as power.  Ratch

Posted on 2006-10-23 11:40:32 by Ratch
Quote
Why are we discussing inductance in this thread, which when it is named Charging Capacitors?
Posted on 2006-10-23 19:17:43 by gluespill
Yeah that is pretty funny  :D Maybe we can start a new thread and carry on our lively debate...lol.

In sinusoidal (AC), the voltage and current can be out ot phase with each other.  But the voltage and current are both sinusoidally shaped.  For a non-sinusoidal excitation, the voltage and current do not have the same shape, so "phase" has no meaning.

However I would like to take issue with you Ratch about this one statement. First I would like to say that it sounds as if we 3 here(glue also) are trained from the same text books. I mean, what I hear you guys saying is what I was taught. What I'm trying to do is think outside the box alittle bit. Do you agree that magnetism is still to this day one of the most widely misunderstood phenomena?
With that being said, let me address your statement. What is the phase angle of DC? Well, there is no angle because it is zero. Never the less there is a magnetic field that opposes the current flow(flow of electrons & that is another debate...lol). I dare to state that the direction of current moving through the wire is different than that of the opposing field or there would be no load other than the resistance of the wire but we know this is not the case. Can you not see that two forces of energy opposing each other, though not changing in phase can still be out of phase with each other? I find this to be an interesting speculation and perhaps, who knows....we might discover something here. We'll split the prophets 20% for Ratch & 20% for Glue and 60% for me.. :lol:
Posted on 2006-10-24 11:54:54 by mrgone
mrgone,

Do you agree that magnetism is still to this day one of the most widely misunderstood phenomena?

At the macro level, I think that electrostatics and magnetics are one of the most understood fields of physics.  There is a huge knowledge base of cause and effect.  Especially after Maxwell established a firm mathematical foundation for the unification of these two phenomena.  At the quantum level, things are more unknown, as are many other quantum things.

With that being said, let me address your statement. What is the phase angle of DC? Well, there is no angle because it is zero.

I tried to point out in my previous post, a phase angle only has definition when the comparison is done between two waves that have the same shape, but not necessarily the same amplitude.  A recurrent sinusoidal wave is the only one where the voltage and current is sinusoidal in its steady state period, assuming a linear circuit with energy storage elements (capacitors and inductors).  Exciting such a circuit with a DC voltage will cause a current wave shaped differently than a constant DC value.  Different shapes mean no phase definition is possible.

Never the less there is a magnetic field that opposes the current flow(flow of electrons & that is another debate...lol). I dare to state that the direction of current moving through the wire is different than that of the opposing field or there would be no load other than the resistance of the wire but we know this is not the case. Can you not see that two forces of energy opposing each other, though not changing in phase can still be out of phase with each other?

With respect to inductance, there is a magnetic field that cuts across the coil wire and generates a voltage that opposes a change of current, regardless whether the current is trying to increase or decrease.  You could say that this is back-voltage is "out of phase" with the applied voltage.  But that is a different meaning of the word "phase".  The phase of AC analysis refers to the angle of lag or lead between steady state sinusoidal voltage and current, not the voltage and counter-voltage occurring when the current is changing in an inductor.

You can divide up the profits any way you like,  Division into zero of any non-zero amount is still zero.  :lol: Ratch

P.S.  I will be going on vacation for around 3 weeks starting tomorrow.  If I don't answer right away, you will know why.
Posted on 2006-10-24 16:09:35 by Ratch
Boy you are hard core text books aren't you? Believe me, I've read them. Yeah you need a vacation. It might help you think outside the box alittle...lol.
Posted on 2006-10-25 08:12:39 by mrgone
mrgone - your post about the relay stimulated my thinking and had me trying to work out the whole electromagnetic/electrostatic thing in my head again.  in my studies i have found much discussion about 'phase' outside the realm of classical sinusoidal ac current flow.  my understanding is that every force has both direction and magnitude, both of which can change from moment to moment. with voltage, i perceive the direction as the polarity of the voltage, and magnitude as the amount of voltage applied or felt.  with current, i see the direction as the direction current is flowing and magnitude as the measure of how much current is flowing in that direction.  anything that has direction and magnitude can be expressed as a vector.  any two vectors can be compared, and their difference is what is commonly referred to as phase.

ratch will point out that 'current flow' is redundant, but that's just he language convention we've adopted.  why do people drive on a parkway and park on a driveway?  language convention.  it is a phrase that most of us embrace and use frequently.  i think we get it that current doesn't flow, it's just what we call it.  semantics.  a forum is about communicating, and if we can get past the quoting and correcting, we might learn something from one another, those of us that don't already know everything, that is.

ratch will point out that current is not a force, but if it helps me to relate my thoughts to you, so what?

Fourier discovered that all wave forms regardless of shape color or creed can be expressed as the sum of of multiple underlying sinusoidal wave forms.  even the square wave is composed of multiple sine waves.

i'm pretty sure that electrical energy is dissipated in the forms of heat, light, or mechanical motion, not just heat.  anytime the momentum of a mass is changed, i.e. a relay arm is actuated, work is being performed.  the electrical energy required to actuate a relay could theoretically be captured and stored for later use after the relay de-energizes, but i have never seen a relay circuit capable of doing so.  sure, some of the energy is stored in the spring, but when the relay arm is released from the magnetic field holding it in place, the energy stored in the spring is dissipated mechancially and transferred back into the physical device, not back into the circuit.

according to the laws of conservation, energy is never lost or created, it is simply transferred from one place in one form or other into another place and or form.  when heat, light, or mechanical energy is dissipated, something is always there to recieve the energy.  when we accelerate our car in a straight line, we take energy from the fuel and transfer it into the road.  since the planet is so much larger than the car, the planet doesn't seem to move and we are left with the illusion that only the car has moved.  but for every action, there is an equal and opposite reaction, so the planet does move ever so infinitesimally in the opposite direction of the car.  when we brake, we give the energy back to the planet by pushing it in the opposite direction that we did when accelerating.

so, back to the relay.  at the moment when you energize the relay with a dc source, i see alot of changes taking place.  the coil of the relay, being that it is an inductor, opposes the attempted change in current by instantly offering the opposite amount of voltage back to the source.  the thing that is allowing it to push back is the energy it is receiving from the circuit which it is using to build its field of magnetic lines of force.  it can only build a certain intensity of this field depending on its value of inductance, and the amount of votage being applied to it.  as the field grows, its ability to oppose the attempted change in current decreases until a balance is found between the field and the voltage applied.  once this balance is found it no longer opposes the the change in current from zero up to the applied voltage, but its magnetic field remains constant as long as the applied voltage does not change.]

magnetic lines of force travel through some material more easily than others.  iron conducts magnetic lines of force much better than air does, so as the coil's magnetic lines of force grow enough past the air to find the iron relay arm, they begin to concentrate in the arm rather than in the air around the arm.  kind of like where the majority of current will flow through the path of least resistance, but magnetic lines of force instead of current.  once the arm has concentrated enough lines of force to overcome the physical resistance of its spring, the lines contract and pull the arm with it.  if there were enough lines of force, the arm could even overcome the physical resistance of the contacts and crush through them, or cause the arm to bend.  in the event that enough lines of force were generated to cause this physical transformation, i wonder how that energy could be given back to the circuit?

i think if you spend enough time living in text-book world, you forget about real world applications.  Your relay is definitely a good example of real world.  the relay circuits that i've seen deactivate the relay by simply breaking the circuit.  sometimes a diode is put in parallel with the coil to shunt the current resulting in the immediate collapse of its magnetic field.  when work is performed in the real world, energy is expended from a souce that is not typically recoverable (at least not immediately).  when the relay arm is pulled down and it collides with the contact underneath it, physical energy is transferred into the stationary contact, and into whatever that contact happens to by physically attached to.  the same thing happens when the arm is released, once the arm snaps back and makes contact with whatever is restricting the amount it can travel back, another impact occurs where physical energy is again transferred to a physical device.  lots of energy getting used up there.  if you put a 'larger' or 'stiffer' spring in the same relay, i think it would take more energy to pull the arm down.  i'm pretty sure that energy is being expended to hold the arm down against the spring, also.  but it does make one think.

now take a tank circuit with a cap and inductor in parallel with each other and apply a sinusoidal voltage to it.  if the circuit is tuned correctly, the cap will discharge into the inductor for half the sinusoidal cycle, and then the indutor will discharge in the opposit direction back into the capacitor for the second half of the cycle.  on paper, where there is no resistance or capacitor plate leakage, and the efficiency of the inductor is 100%, i.e. it has no eddy currents, etc, you can theoretically remove the source and the two components would continue to discharge back and forth into each other for eternity at the same tuned frequency.

yep, i'd have to agree that pure the implementation of pure inductance does not consume any energy, and that it gives back everything that it recieves.  i think this is an important understanding.  however, a relay is a far cry from a purely inductive device where all energy is returned to the circuit.  i would say that a great deal of the energy used actuate the relay is not given back to the circuit.  i would tend to classify a relay/solenoid as a simple engine capable of doing work, and work requires the expenditure of energy.

the phase angle between the voltage (at a given point in circuit with respect to another point) and the current (flowing past that given point) can always be determined regardless of what shape waveforms are employed.

your thoughts stimulated my thinking, and i appreciate having the chance to express how i see things

current flow
current flow
current flow
current flow

Posted on 2006-10-25 22:20:09 by gluespill
Hey glue, sorry I have not responded in a while. I figured with Ratch being on vacation there would be no one to argue with...lol. I believe you could tell him the sky is blue and he would dissagree...lol. I guess that's what makes him Ratch.
Well that was a very interesting analysis. I guess my point was, that if you have a direct current flowing through a coil, we assume the electron flow to be of a phase that is zero. But there is an electro-magnetic field also being produced and initially the impeadance or opposition to the current flow by the electro-magnetic field is great but there is a ballancing that occurs. So possibly the only opposition at that point is resistive but if we measure the power loss I believe we will find that it is actually greater than IR. I believe there is still opposition to the current by the field. Anytime you have an electromagnetic feild near a piece of wire, there will be current flow. So is this current flow now exactly in phase (meaning same zero direction) as the current or electron flow?
If this is the case that there is still some small opposition, then might the coils be wound in a more efficient manner? I guy name Guanella did alot of interresting phase shift techniques by using all kinds of bizzar windings on ferrite and iron.
I don't know, I'm still of a mind that magnetism is still a very mysterious phenomena and is to this day largely misunderstood.
Let's take for instance a simple magnet. This magnet is producing energy like a battery... right? I mean it will attract objects made of iron like a nail and it takes physical energy to remove it. Yet, to this day, I know of no way to convert this energy into electricity without some sort of mechanical motion. Doesn't this strike you as a curriosity? Some questions that come to mind are.... Does the magnetic intensity ware down? I believe the answer is yes. Then why does it break down? Is there a method to recharge it? What happens to the atomic structure of the magnet as it breaks down in intensity. Would it break down faster if energy from the magnet was being used without motion, such as a nail remaining against the magnet. And for the "pizza resistance" (lol) Why or is it possible to convert this magetic energy field into a usable energy form such as electricity without motion? I'm sure Ratch will have all the answers when he gets back from vacation....lol.

Thinking out side the box,mrgone
Posted on 2006-11-02 10:03:48 by mrgone
mrgone,

Boy you are hard core text books aren't you? Believe me, I've read them. Yeah you need a vacation. It might help you think outside the box alittle...lol.

And why not?  They are readily available and contain just about all the basic technical knowledge from the past and the present.  Presentations from cutting edge science papers will sooner or later appear in textbooks if they have merit.  You should not disparage textbooks, you should embrace them.

What does thinking outside the box mean in this context?  If it means looking at things in a different way or doing somethiing out of the ordinary, then fine, providing that action is congruent with proven scientific laws and knowledge.  Otherwise it is science fiction.  Ratch
Posted on 2006-11-08 21:03:20 by Ratch
mrgone,

I believe you could tell him the sky is blue and he would dissagree...lol.

Not unless I could prove it.  You have never observed me make a statement without backup or reasoned discourse.

I guess my point was, that if you have a direct current flowing through a coil, we assume the electron flow to be of a phase that is zero.

Still don't believe that phase has no meaning for DC?  Try Googling for DC Phase and see how many entries you get pertaining to electrical technology.  Then try AC Phase and notice over 30 million hits.

Anytime you have an electromagnetic feild near a piece of wire, there will be current flow.

Only if the magnetic field is moving at right angles to the wire.  A stationary wire and steady field don't produce any voltage.

So is this current flow now exactly in phase (meaning same zero direction) as the current or electron flow?

Depends on the immittance of the primary and secondary circuits.  Sounds like you are referring to transformers.  Lots of study material on those items is available in textbooks and internet.

If this is the case that there is still some small opposition, then might the coils be wound in a more efficient manner? I guy name Guanella did alot of interresting phase shift techniques by using all kinds of bizzar windings on ferrite and iron

A good power transformer is very efficient in that it does not allow very much magnetic flux to escape, and keeps eddy currents down to a minimum.

I don't know, I'm still of a mind that magnetism is still a very mysterious phenomena and is to this day largely misunderstood.

Unless you are talking about how magnetism interacts with the sunspots, or how a interstellar hydrogen ramscoop could be designed, applied magnetics is a very mature technology.

Let's take for instance a simple magnet. This magnet is producing energy like a battery... right?

Wrong, unless you move it at right angles to a conductor such as a wire.  By the way, you are referring to a permanent magnet, right?

I mean it will attract objects made of iron like a nail and it takes physical energy to remove it.

So, where is the wonderment?  Gravity attracts all mass, and it takes physical energy to return objects back to their original position.

Yet, to this day, I know of no way to convert this energy into electricity without some sort of mechanical motion. Doesn't this strike you as a curriosity?

No, there is no energy to convert, unless you supply it with some mechanical motion.

Does the magnetic intensity ware down? I believe the answer is yes. Then why does it break down? Is there a method to recharge it?

Yes, magnets lose their attraction with use and age.  The ferrous molecules of a permanent magnetic lose their collective polar alignment with age and vibration.  They can be restored by inserting them in a strong magnetic field to realign their molecules.

What happens to the atomic structure of the magnet as it breaks down in intensity.

All the polar molecules don't line up in the same direction anymore.

Would it break down faster if energy from the magnet was being used without motion, such as a nail remaining against the magnet

No energy is involved if no motion exists.

And for the "pizza resistance" (lol) Why or is it possible to convert this magetic energy field into a usable energy form such as electricity without motion?

Do you mean piece de resistance?  You can't.  No more than you can convert gravity into usable energy without motion.

I'm sure Ratch will have all the answers when he gets back from vacation....lol.

You betcha.  Ratch

Posted on 2006-11-08 21:50:38 by Ratch