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Author Topic: MH's ideal coil and voltage question  (Read 487975 times)

poynt99

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Re: MH's ideal coil and voltage question
« Reply #1200 on: June 23, 2016, 01:37:47 AM »
This is interesting.

I do see where MH is coming from, although I disagree with calling the voltage drop across a resistor "cemf".

Kirchhoff's KVL always holds, so the problem is perplexing, especially in light of the descriptions on self induction, whereby an opposing "self-induced emf" results from the changing flux in the coil. "Opposing" in what manner?

One argument is that if the EMF and CEMF were equal, no circuit current would flow. I think we may be arguing with an incorrect assumption in mind. We can envision the voltage source and the induced emf as two generators back to back, but is this accurate? A voltage source and a resistor in parallel also have voltages that are back to back, yet current flows. Isn't an inductor very much like a frequency-dependent resistor? If so, then maybe it makes sense for it to have a voltage drop across it.

The self-induced emf is present, but it is perhaps not what we might expect. When we use the term "emf" we expect it to mean there is or will be a resulting measurable voltage. When the voltage source is connected and the magnetic field begins building, the coil immediately starts inducing a cemf across its terminals. The coil has now become a generator; but there is a "twist". The load seen by the coil's terminals is the voltage source, and it would appear as a short circuit to the coil.

I made a statement a few posts back in that an induced current always has an associated induced voltage. I retract that and restate it this way: An induced current usually has an associated induced voltage. The fact is that a coil can have a current induced in it, even if the coil is shorted. This I believe is the scenario with the self-induced emf and current with our simple voltage source and inductor. So I am saying that the cemf induced in the coil goes to 0V, while the self-induced opposing current goes to some value, determined by the inductance and rate of flux change.

In terms of KVL, it holds fine, and the "drop" across the inductor is determined by the voltage source, just as in the case for the resistor, but it is NOT the cemf, if indeed emf is equated with voltage.

Is this explanation crazy? Perhaps, but I've not seen any other that makes sense to me.

tinman

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Re: MH's ideal coil and voltage question
« Reply #1201 on: June 23, 2016, 01:43:08 AM »
PW:

When current flows through a resistor, is there an electric field inside that resistor?

The answer of course is there is an electric field inside the resistor.  When we do a line integral on the electric field from one end to the other end of the resistor we get an EMF.

There is no rule that says that the EMF being generated by the resistor cannot be called a CEMF.

MileHigh

There is no EMF being generated by the resistor.
A resistor is a current flow regulator,and determines how much current will flow through it,in relation to the voltage placed across it.

Reading your last 5 or 6 posts,with springs,shopping trollies,and the likes,i can see you are totally lost when it comes to equal action/reaction,when there is motion-or,potential or kinetic energy storage involved.
There will be no flow of current when the CEMF is equal to that of the applied EMF.
A 2kg lifting force will not raise a 2kg mass. The mass will become weightless as far as the ground it was sitting on is concerned,but there will be no motion. This is an equal and opposite force.
In order for that 2kg mass to rise/move/accelerate,there must be a difference between the two forces,where the applied force(the lifting force)is greater than the weight of the mass. Over time,this higher applied force is stored in the mass,and can become either kinetic or potential energy.

The same applied to water flow through a pipe. If the water is being pumped up hill by a pump,once that water in the pipe reaches a height where the head pressure equals that of which the pump can produce,there will be no flow of water through the pipe.

So like i(and many others have tried to tell you),once the CEMF (back pressure) equals that of the applied EMF(forward pressure),there will be no flow of current.

When a force meets an equal and opposite force--there will be no motion-period.


Brad

hoptoad

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Re: MH's ideal coil and voltage question
« Reply #1202 on: June 23, 2016, 07:19:15 AM »
snip...
The coil has now become a generator; but there is a "twist". The load seen by the coil's terminals is the voltage source, and it would appear as a short circuit to the coil.
snip...
The voltage source would only be seen as a short circuit if it had zero impedance  i.e. an 'ideal' voltage/current source.
Since the title of this thread incorporates an 'ideal coil', then having an ideal voltage/current source is fine for hypothesis.

But in the practical real electronics world, nothing can actually be ideal outside of hypothesis, including the supply. It will have impedance of some value, the value of which is dependent on the nature of the source.
Cheers

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #1203 on: June 23, 2016, 08:29:17 AM »
Reading your last 5 or 6 posts,with springs,shopping trollies,and the likes,i can see you are totally lost when it comes to equal action/reaction,when there is motion-or,potential or kinetic energy storage involved.
There will be no flow of current when the CEMF is equal to that of the applied EMF.
A 2kg lifting force will not raise a 2kg mass. The mass will become weightless as far as the ground it was sitting on is concerned,but there will be no motion. This is an equal and opposite force.
In order for that 2kg mass to rise/move/accelerate,there must be a difference between the two forces,where the applied force(the lifting force)is greater than the weight of the mass. Over time,this higher applied force is stored in the mass,and can become either kinetic or potential energy.

The same applied to water flow through a pipe. If the water is being pumped up hill by a pump,once that water in the pipe reaches a height where the head pressure equals that of which the pump can produce,there will be no flow of water through the pipe.

So like i(and many others have tried to tell you),once the CEMF (back pressure) equals that of the applied EMF(forward pressure),there will be no flow of current.

When a force meets an equal and opposite force--there will be no motion-period.

Brad

I am not "lost" in any way, shape, or form.  I will revisit the "lost" theme with respect to you in a subsequent posting.

Quote
There will be no flow of current when the CEMF is equal to that of the applied EMF.

There certainly will be current flow for an inductor, it's happens in real life.  But there will be no current flow for a capacitor.  You have to think and analyze these types of situations on a case by case basis.

Quote
A 2kg lifting force will not raise a 2kg mass. The mass will become weightless as far as the ground it was sitting on is concerned,but there will be no motion. This is an equal and opposite force.

I am not sure what that has to do with the discussion but a 2 kg lifting force on a 1 kg mass will still result in an equal and opposite "force" of 2 kg.

Quote
In order for that 2kg mass to rise/move/accelerate,there must be a difference between the two forces,where the applied force(the lifting force)is greater than the weight of the mass. Over time,this higher applied force is stored in the mass,and can become either kinetic or potential energy.

Only a moderate amount of descrambling was required.  But once you factor acceleration into the system then the forces are equal and opposite.  We are talking basic Newton's Laws here.

Quote
So like i(and many others have tried to tell you),once the CEMF (back pressure) equals that of the applied EMF(forward pressure),there will be no flow of current.

When a force meets an equal and opposite force--there will be no motion-period.

As has been explained to you, for the case of an inductor, current flows and the CEMF is equal and opposite to the EMF.  This is the basic application of KVL or in the physical world Newton's Third Law of Motion.

When a force meets an equal and opposite force, there will be no motion, or, there will be acceleration.

MileHigh

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #1204 on: June 23, 2016, 08:39:02 AM »
Brad:

Are you a lost soul or are you a man with the conviction to stand up for what you say and back it up with a logical argument?  Put meaning to your statements and show conviction or just be a space cadet?  Which one is it going to be?

Let's look at two cases.

1)  The unanswered request for you to put substance behind your claim with an actual example.

The ball is now in your court.  You say that the CEMF must be lower than the EMF for current to flow?  I have never seen any concrete examples of that from you.  Now is the time.  Show us some examples where the CEMF is lower than the EMF with all the specifics and all of the numbers crunched to explain how much current flows.

Forget about the motor example, and keep it simple and use a coil.  Give some examples providing all of the specifics and the EMF and CEMF values, the current flow, the whole nine yards.

You say the correct model is that the CEMF is less than the EMF?  Go ahead and give some examples with all of the details laid out so we can see if your model works or not.

2)  Who really has lost their marbles?

You have a one-volt voltage source connected across a one-Henry inductor, which gives you one amp of current per second flowing through the inductor.

So where is the EMF and the CEMF?  The one-volt voltage source is the EMF.  The CEMF is the one amp of current per second flowing through the inductor causing a one-volt voltage drop across the inductor.

The one-volt voltage drop across the inductor is the CEMF.   Look at that, the EMF and the CEMF are equal and opposite, and current flows through the inductor.

That's the real deal and that's the way it is modeled.

There is no difference between the EMF and the CEMF and current flows through the inductor.

So Brad, you allege that I have "lost my marbles" with respect to the example above.  Then you break down that example and show us exactly where and why it is wrong.  If you can't do that then you are the one that has lost your marbles.

MileHigh

tinman

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Re: MH's ideal coil and voltage question
« Reply #1205 on: June 23, 2016, 10:54:50 AM »
Brad:

Are you a lost soul or are you a man with the conviction to stand up for what you say and back it up with a logical argument?  Put meaning to your statements and show conviction or just be a space cadet?  Which one is it going to be?

Let's look at two cases.

1)  The unanswered request for you to put substance behind your claim with an actual example.

The ball is now in your court.  You say that the CEMF must be lower than the EMF for current to flow?  I have never seen any concrete examples of that from you.  Now is the time.  Show us some examples where the CEMF is lower than the EMF with all the specifics and all of the numbers crunched to explain how much current flows.

Forget about the motor example, and keep it simple and use a coil.  Give some examples providing all of the specifics and the EMF and CEMF values, the current flow, the whole nine yards.

You say the correct model is that the CEMF is less than the EMF?  Go ahead and give some examples with all of the details laid out so we can see if your model works or not.

2)  Who really has lost their marbles?

You have a one-volt voltage source connected across a one-Henry inductor, which gives you one amp of current per second flowing through the inductor.

So where is the EMF and the CEMF?  The one-volt voltage source is the EMF.  The CEMF is the one amp of current per second flowing through the inductor causing a one-volt voltage drop across the inductor.

The one-volt voltage drop across the inductor is the CEMF.   Look at that, the EMF and the CEMF are equal and opposite, and current flows through the inductor.

That's the real deal and that's the way it is modeled.

There is no difference between the EMF and the CEMF and current flows through the inductor.

So Brad, you allege that I have "lost my marbles" with respect to the example above.  Then you break down that example and show us exactly where and why it is wrong.  If you can't do that then you are the one that has lost your marbles.

MileHigh

This one volt voltage drop across the inductor is horse radish,and also makes things more confusing than they need to be.
The inductor dose not cause the voltage drop,as the voltage applied across the inductor by the source is exactly what will be across the inductor. You might as well say there is a 1 volt voltage drop across a 1 volt battery--how stupid dose that sound ::)

MH
You are doing nothing but adding some sort of idiotic confusion to everything everyone has learned.
I dont give a rats ass what you say,if the CEMF is equal to the applied EMF,then no current will flow through that inductor.
And this cods wallop about a resistor creating an equal and opposite CEMF to that of the applied EMF,is nothing short of insane.

Quote
You say that the CEMF must be lower than the EMF for current to flow?


Yep,as that is what stops the current through an inductor shooting straight up to a maximum value determined by the inductors winding resistance,when a voltage is placed across.
Same with water flowing through pipes-as i stated before. Once the head pressure pushing back against the pump equals the pumps maximum pumping pressure,there will be no water flow.
The pressure is your voltage,and the water flow rate is your current.


Say what you will MH,but i am no longer interested in what you have to say.


Brad

hoptoad

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Re: MH's ideal coil and voltage question
« Reply #1206 on: June 23, 2016, 11:38:22 AM »
snip...
There is no difference between the EMF and the CEMF and current flows through the inductor.
snip...
MileHigh
When an inductor is first connected to a Voltage Source at time=0 and counting, the EMF and CEMF are EQUAL and their is NO current flow.

At end of TC 1 (TC = time constant of the inductor) the CEMF will be .37 of The EMF and supply current is flowing.
At end of TC 2 the CEMF will be .14 of The EMF and supply current is flowing.
At end of TC 3 the CEMF will be .5 of the EMF and supply current is flowing.
At end of TC 4 the CEMF will be .2 of the EMF and supply current is flowing.
At end of TC 5 the CEMF will be .0 of the EMF = non existent and supply current is flowing.

Any moment after that, the current through the inductor will be steady, the EMF will be steady and there will be NO CEMF.

MH, for someone who insists that others use the correct nomenclature when referring to circuitry, you seem to be pretty liberal with how you apply it yourself. Voltage drop across a component is not CEMF. EMF is not CEMF. Constant EMF is sustainable, constant CEMF is not.

Here's a link showing the relationship of CEMF in a series LR circuit, with the resistance external to the inductor and the presumption of an ideal inductor. Step through the exercise and answer the questions.

https://www.wisc-online.com/learn/career-clusters/stem/ace5903/an-inductor-opposing-a-current-change

Cheers

tinman

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Re: MH's ideal coil and voltage question
« Reply #1207 on: June 23, 2016, 12:06:09 PM »
 author=poynt99 link=topic=16589.msg486943#msg486943 date=1466638667]


In terms of KVL, it holds fine, and the "drop" across the inductor is determined by the voltage source, just as in the case for the resistor, but it is NOT the cemf, if indeed emf is equated with voltage.

Is this explanation crazy? Perhaps, but I've not seen any other that makes sense to me.


Quote
Isn't an inductor very much like a frequency-dependent resistor? If so, then maybe it makes sense for it to have a voltage drop across it.

How can it be Poynt. A resistor has no inductance,produces no magnetic field,and the value of resistance in a resistor dose not change over time.

Quote
Kirchhoff's KVL always holds, so the problem is perplexing, especially in light of the descriptions on self induction, whereby an opposing "self-induced emf" results from the changing flux in the coil. "Opposing" in what manner?

Opposing as in-producing a current that flows in opposition to that of which the applied EMF is trying to create.
As we discussed earlier,it is what stops the current going straight to a maximum value,once a voltage is placed across that inductor--what else would limit the rate of current increase in that inductor when a voltage is placed across it,if it is not the self induced EMF?.

Quote
One argument is that if the EMF and CEMF were equal, no circuit current would flow. I think we may be arguing with an incorrect assumption in mind. We can envision the voltage source and the induced emf as two generators back to back, but is this accurate?

No,that would not be accurate as far as i am concerned.
The applied EMF would be seen as a prime mover,and the self induced EMF(CEMF) would be the generator attached to the prime mover. The load drawn from the generator could be seen as the CEMF,as there is always losses,the CEMF (power dissipated in load),will always be less than the power being supplied to the generator by the prime mover-->(or we may see something different at the other place  ;) )

Quote
A voltage source and a resistor in parallel also have voltages that are back to back, yet current flows.

A resistor is not a voltage source,and has no voltage across it until such time one is applied to it from a voltage source. At that point in time,the resistor now becomes a resistive heater-nothing more.

Quote
The self-induced emf is present, but it is perhaps not what we might expect. When we use the term "emf" we expect it to mean there is or will be a resulting measurable voltage. When the voltage source is connected and the magnetic field begins building, the coil immediately starts inducing a cemf across its terminals. The coil has now become a generator; but there is a "twist". The load seen by the coil's terminals is the voltage source, and it would appear as a short circuit to the coil.

The voltage source will not be seen as a load,as the self induced EMF is less than the applied EMF,due to transformer losses,and there for,a potential difference still exists,and the higher potential is the voltage source. So the inductor is still seen as the load,because it has the lowest value of the two potentials
If the self induced EMF was equal to the applied EMF,then there is no potential difference between the two,and current will not flow unless there is a potential difference.

Quote
I made a statement a few posts back in that an induced current always has an associated induced voltage. I retract that and restate it this way: An induced current usually has an associated induced voltage. The fact is that a coil can have a current induced in it, even if the coil is shorted. This I believe is the scenario with the self-induced emf and current with our simple voltage source and inductor.

Yes,this was told to us on this thread by verpies many pages back,and some others here decided to tell him to stop confusing the subject at hand,by introducing things not associated with the question.
But now we have seen a full circle,and we are right back to the point verpies was making :D

Quote
So I am saying that the cemf induced in the coil goes to 0V, while the self-induced opposing current goes to some value, determined by the inductance and rate of flux change.

Is that the case?-im not sure.
If the CEMF value is 0,then how can a current be produced without voltage,when that coil(as discussed above)is not an ideal coil with an ideal short across it's terminals. Only an ideal coil with an ideal wire across it's terminals will allow for a current to flow through it,without a voltage across it.
Has anyone ever wound a coil with a center tap,and measured the voltage across half of the inductors winding's,while being subjected to a rising and falling current--will it be equal to 1/2 of the applied EMF?.


Brad

tinman

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Re: MH's ideal coil and voltage question
« Reply #1208 on: June 23, 2016, 12:47:04 PM »
 author=hoptoad link=topic=16589.msg486958#msg486958 date=1466674702]


Quote
At end of TC 1 (TC = time constant of the inductor) the CEMF will be .37 of The EMF and supply current is flowing.
At end of TC 2 the CEMF will be .14 of The EMF and supply current is flowing.
At end of TC 3 the CEMF will be .5 of the EMF and supply current is flowing.
At end of TC 4 the CEMF will be .2 of the EMF and supply current is flowing.
At end of TC 5 the CEMF will be .0 of the EMF = non existent and supply current is flowing.

It is good to see there is some one else that is not falling for MHs gobble doc.

But now 2 questions for you Hoptoad.
1-why dose the magnetic field that is inducing the CEMF,slowly decrease in change over time in an inductor from T=0-->moment of the applied voltage across the coil.
2-why is the CEMF equal to the applied EMF at the moment the voltage is placed across the coil?.

Quote
MH, for someone who insists that others use the correct nomenclature when referring to circuitry, you seem to be pretty liberal with how you apply it yourself.

As i have said many times,MH changes things when it suit's his needs.
Do as i say--not as i do. ;)

Quote
Voltage drop across a component is not CEMF. EMF is not CEMF. Constant EMF is sustainable, constant CEMF is not.

Indeed. ;)

Quote
Here's a link showing the relationship of CEMF in a series LR circuit, with the resistance external to the inductor and the presumption of an ideal inductor. Step through the exercise and answer the questions.

I doubt that will happen. MH is on a !!MH!! is right saga,and nothing gets in the way.


Brad

Johan_1955

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Re: MH's ideal coil and voltage question
« Reply #1209 on: June 23, 2016, 12:59:36 PM »
But now 2 questions for you Hoptoad.
1-why dose the magnetic field that is inducing the CEMF,slowly decrease in change over time in an inductor from T=0-->moment of the applied voltage across the coil.
2-why is the CEMF equal to the applied EMF at the moment the voltage is placed across the coil?.

Brad, you know already why, because you'''re did demonstrate it before, its the change!

And Change we need all: Europe complete HOPEFULLY for a

citfta

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Re: MH's ideal coil and voltage question
« Reply #1210 on: June 23, 2016, 01:00:14 PM »
MileHigh,

You are apparently using the voltage drop across the coil as YOUR definition for CEMF.   That is not what the rest of the electronics world uses as that definition.  For the rest of us CEMF is the generated voltage that opposes the applied voltage.  The CEMF is generated by the increasing magnetic field of the coil as the current rises.  If your claim that the CEMF equals the EMF were true then no current would flow and that means the could NOT BE any CEMF.  Sorry, but your argument makes no sense at all.  I haven't read all the posts in this thread but it appears you are the only one that believes CEMF can equal EMF.  I really don't think all the rest of us are wrong and you are the only one correct about this.

Respectfully,
Carroll

hoptoad

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Re: MH's ideal coil and voltage question
« Reply #1211 on: June 23, 2016, 02:00:57 PM »
snip...
But now 2 questions for you Hoptoad.
1-why dose the magnetic field that is inducing the CEMF,slowly decrease in change over time in an inductor from T=0-->moment of the applied voltage across the coil.
2-why is the CEMF equal to the applied EMF at the moment the voltage is placed across the coil?.
snip...
The magnetic field of the inductor is still increasing through each time constant until it is 100% maximum value derived from steady current at the end of TC5. Only the RATE of increase in the source current/magnetic field is getting less with each TC.

The CEMF does not arise from the source current/magnetic field of the inductor per se, it arises from Changes in the source current/magnetic field of the inductor. The level of cemf is most dependent on the Rate of change of the source current/magnetic field not necessarily the strength of the magnetic field or amount of source current. At actual time =0, in a real world inductor, nothing happens really. But at 0+ picoseconds to microseconds the RATE of change is highest causing the maximum cemf to arise.

But cemf is not self sustaining because it is an emergent phenomena with a value based on rate of change of current/magnetic field, and there can be no more change if emf = cemf, so the cemf begins to drop, and as it does, more source current flows. The diminishing cemf still opposes the Rate Of Change of the source, but not the actual flow of current per se. In opposing the rate of change it diminishes the cause of its own existence. So the cemf diminishes in the familiar TC curve we see in all inductors.

Cheers

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #1212 on: June 23, 2016, 02:43:33 PM »
Well, I see we are now all unwitting players in some kind of Monty Python comedy sketch.  We are a bunch of crazy people in a tizzy over Kirchhoff's voltage law.  It's like we are all in electronics kindergarten, and it's the very first day.  We are all running around with runny noses and bumping into walls.

We have our knickers in a super twist.  It's the Mother of all Wedgies, and some of us have our brains in our cojones.

The question is, who has the cojones to state the truth?

This one volt voltage drop across the inductor is horse radish,and also makes things more confusing than they need to be.
The inductor dose not cause the voltage drop,as the voltage applied across the inductor by the source is exactly what will be across the inductor. You might as well say there is a 1 volt voltage drop across a 1 volt battery--how stupid dose that sound ::)

MH
You are doing nothing but adding some sort of idiotic confusion to everything everyone has learned.
I dont give a rats ass what you say,if the CEMF is equal to the applied EMF,then no current will flow through that inductor.
And this cods wallop about a resistor creating an equal and opposite CEMF to that of the applied EMF,is nothing short of insane.
 
Yep,as that is what stops the current through an inductor shooting straight up to a maximum value determined by the inductors winding resistance,when a voltage is placed across.
Same with water flowing through pipes-as i stated before. Once the head pressure pushing back against the pump equals the pumps maximum pumping pressure,there will be no water flow.
The pressure is your voltage,and the water flow rate is your current.

Say what you will MH,but i am no longer interested in what you have to say.

Brad

Quote
This one volt voltage drop across the inductor is horse radish,and also makes things more confusing than they need to be.
The inductor dose not cause the voltage drop,as the voltage applied across the inductor by the source is exactly what will be across the inductor. You might as well say there is a 1 volt voltage drop across a 1 volt battery--how stupid dose that sound

Gustav Robert Kirchhoff is rolling in his grave and wants to start walking again.  This is one of those mind-blowing moments with you.  Lock yourself in a room with half a dozen books on basic electronics don't come out until you have read through them and understood them.

Sorry, but you sound incredibly stupid.  You have been playing with electronics all this time, for years, and you can't understand what a bloody voltage drop is?

Quote
I dont give a rats ass what you say,if the CEMF is equal to the applied EMF,then no current will flow through that inductor.

But you have been repeatedly asked to give an example with a coil where the CEMF is less than the EMF resulting in current flow and you refuse to do it because you can't do it.

Quote
Same with water flowing through pipes-as i stated before. Once the head pressure pushing back against the pump equals the pumps maximum pumping pressure,there will be no water flow.
The pressure is your voltage,and the water flow rate is your current.

No, because that example would be like connecting a capacitor across an EMF source.  The head pressure in the pipe would be like the voltage across the capacitor, and no current flows.  The equivalent and proper example would be a long pipe.  Water flows through the long pipe and at the junction of the pump and the long pipe, the maximum pumping pressure of the pump and the head pressure in the pipe are the same.  Then as you travel along the long pipe the pressure in the pipe decreases.

So Brad,

1) The unanswered request for you to put substance behind your claim with an actual example remains unanswered because you simply have no clue what to do and you can't show an example.

2) You could not break down my example and show us exactly where and why it is wrong because in fact it is correct.  You didn't even try.

So no progress for you, a double fail on your part.

MileHigh

minnie

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Re: MH's ideal coil and voltage question
« Reply #1213 on: June 23, 2016, 03:03:45 PM »



   Well with our ideal set up of inductor the current and time lead us
 to a situation of infinity.
                John.

hoptoad

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Re: MH's ideal coil and voltage question
« Reply #1214 on: June 23, 2016, 03:13:18 PM »


   Well with our ideal set up of inductor the current and time lead us
 to a situation of infinity.
                John.
Indeed. Anybody got an ideal inductor?