Author Topic: MH's ideal coil and voltage question (Read 487933 times)

MileHigh · « **Reply #1170 on:** June 22, 2016, 07:43:04 AM »

Okay, now it's brain twister time. I am going to use an analogy to model this EMF/CEMF business and if somebody gets it, great. If not, better luck next time.

Look at the attached graphic, we are going to use the Force-Voltage analogy for the first setup. Look at how the variables are modeled.

The setup is as follows: Imagine a long trough 50 meters long and 1/2 meter deep and 1/2 meter wide. You can fill the trough with water, or oil, and do experiments with moving boats or carts or masses and stuff like that, as they move back and forth in the trough. They have setups like that for testing model ships. See the attached pic.

Here is the voltage source-resistor model:

You have a wheeled cart that can move through the trough and the trough is filled with viscous oil. You push the cart though the viscous oil with a constant force of 100 Newtons.

The "EMF" is the 100 Newtons of force applied to make the cart move.
The "CEMF" is the 100 Newtons of resistance that the cart experiences from the viscous oil.

The "CEMF" is equal and opposite to the "EMF."

That's all there is to it, and the velocity of the cart corresponds to the current flow.

A constant rate of power is expended by the "voltage source" by applying a constant force to the cart of 100 Newtons times the velocity of the cart and this power is dissipated in the viscous oil and heats it up.

------------------

Look at the attached graphic, we are going to use the Force-Current analogy for the second setup. Look at how the variables are modeled. We are switching to the Force-Current analogy because inductance is modeled like a spring.

Now we have to change the setup. We empty the viscous oil out of the trough. We fill the trough with a long spring, and the there is no friction between the bottom of the spring and the trough.

Here is the voltage source-inductor model:

You push a wheeled cart down the trough at a constant velocity of one meter per second, and the front of the cart is connected to the long spring. The other end of the spring is fixed to the back wall of the trough. As the cart moves the spring slowly gets compressed.

The "EMF" is the the measurement of the constant velocity of the cart moving forward.
The "CEMF" is the the measurement of the constant velocity of the spring being compressed backwards.

The "CEMF" is equal and opposite to the "EMF."

The "Current" in this case is very interesting. It corresponds to the force that the cart impresses on the compressible spring. The further the cart moves down the trough the more force has to be impressed on the spring and the higher the mechanical power required to keep moving forward at a constant velocity.

The power expended by the "voltage source" all goes into the compressed spring. The spring stores (integrates) all of that expended power and stores it as potential energy.

I know that this stuff usually gets blank stares, but it is what it is.

MileHigh

hoptoad · « **Reply #1171 on:** June 22, 2016, 10:33:04 AM »

Quote from: MileHigh on June 22, 2016, 07:43:04 AM

snip..
The power expended by the "voltage source" all goes into the compressed spring.
I know that this stuff usually gets blank stares, but it is what it is.
MileHigh

The power expended by the "voltage source" (cart) all goes into the compressed spring AND the springs retaining (trough) wall.
Without the trough wall the spring would simply move with the cart rather than be compressed by it. So any moving/pushing force from the cart is expended by both the spring AND trough wall.

In an 'ideal' scenario with an ideal cart without friction, and no internal spring losses, etc, not only would the cart and spring have to be ideal, but the wall would also have to be ideal and capable of total power reflection and no power absorption/dissipation.

Cheers.

tinman · « **Reply #1172 on:** June 22, 2016, 11:18:20 AM »

Quote from: MileHigh on June 22, 2016, 06:58:42 AM

Loner:

MileHigh

Quote

When it comes to electronics, you are not as smart as you think. Beyond that, don't you dare allege that I am a troll. I am making logical points and I am sincerely trying to debate with people, even if the debate can get heated. It takes two to tango. I am not buying your "wise man coming to share some pearls of wisdom" vibe, at all. If you have some issues with me, then say them straight to my face and we will debate them. Fair enough?

Gee MH,you attack anyone that dose not agree with you.
There is also the fact that i did not see Loner mention your name once,but you think he is referring to you--guilty conscience maybe?.

Quote

Sure, let's get real and use a little common sense and start off with a resistor.

You have a one-volt voltage source connected across a one-ohm resistor, which gives you one amp of DC current.

So where is the EMF and the CEMF? The one-volt voltage source is the EMF. The CEMF is the one amp flowing through the one-ohm resistor causing a one-volt voltage drop across the resistor.

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

Now let's repeat the whole process for an inductor:

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.

So, you are wrong. There is no difference between the EMF and the CEMF and current flows for both the resistor and the inductor. Simple enough.

If you have a technical comment to the above discussion I am all ears. And again, if you want to say something to me then just say it.

Yep-you have finally lost your marbles.

Brad

MileHigh · « **Reply #1173 on:** June 22, 2016, 12:31:50 PM »

Quote from: tinman on June 22, 2016, 11:18:20 AM

Gee MH,you attack anyone that dose not agree with you.
There is also the fact that i did not see Loner mention your name once,but you think he is referring to you--guilty conscience maybe?.

Yep-you have finally lost your marbles.

Brad

Brad:

You have the concepts of "attack" and "defend" mixed up and backwards in your head. Stop playing your "on stage" "dullest tool in the toolbox" games. Your silly comment is just another stunt that backfired.

You think that I have lost my marbles? 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.

MileHigh

MileHigh · « **Reply #1174 on:** June 22, 2016, 12:40:49 PM »

Quote from: hoptoad on June 22, 2016, 10:33:04 AM

The power expended by the "voltage source" (cart) all goes into the compressed spring AND the springs retaining (trough) wall.
Without the trough wall the spring would simply move with the cart rather than be compressed by it. So any moving/pushing force from the cart is expended by both the spring AND trough wall.

Cheers.

The retaining wall at the end of the trough is stationary and does not move. So think about the ramifications of that.

tinman · « **Reply #1175 on:** June 22, 2016, 12:56:23 PM »

Quote from: MileHigh on June 22, 2016, 07:43:04 AM

Okay, now it's brain twister time. I am going to use an analogy to model this EMF/CEMF business and if somebody gets it, great. If not, better luck next time.

Look at the attached graphic, we are going to use the Force-Voltage analogy for the first setup. Look at how the variables are modeled.

The setup is as follows: Imagine a long trough 50 meters long and 1/2 meter deep and 1/2 meter wide. You can fill the trough with water, or oil, and do experiments with moving boats or carts or masses and stuff like that, as they move back and forth in the trough. They have setups like that for testing model ships. See the attached pic.

Here is the voltage source-resistor model:

You have a wheeled cart that can move through the trough and the trough is filled with viscous oil. You push the cart though the viscous oil with a constant force of 100 Newtons.

The "EMF" is the 100 Newtons of force applied to make the cart move.
The "CEMF" is the 100 Newtons of resistance that the cart experiences from the viscous oil.

The "CEMF" is equal and opposite to the "EMF."

That's all there is to it, and the velocity of the cart corresponds to the current flow.

A constant rate of power is expended by the "voltage source" by applying a constant force to the cart of 100 Newtons times the velocity of the cart and this power is dissipated in the viscous oil and heats it up.

------------------

Look at the attached graphic, we are going to use the Force-Current analogy for the second setup. Look at how the variables are modeled. We are switching to the Force-Current analogy because inductance is modeled like a spring.

Now we have to change the setup. We empty the viscous oil out of the trough. We fill the trough with a long spring, and the there is no friction between the bottom of the spring and the trough.

Here is the voltage source-inductor model:

You push a wheeled cart down the trough at a constant velocity of one meter per second, and the front of the trough is connected to the long spring. As the cart moves the spring slowly gets compressed.

The "EMF" is the the measurement of the constant velocity of the cart moving forward.
The "CEMF" is the the measurement of the constant velocity of the spring being compressed backwards.

The "CEMF" is equal and opposite to the "EMF."

The "Current" in this case is very interesting. It corresponds to the force that the cart impresses on the compressible spring. The further the cart moves down the trough the more force has to be impressed on the spring and the higher the mechanical power required to keep moving forward at a constant velocity.

The power expended by the "voltage source" all goes into the compressed spring.

I know that this stuff usually gets blank stares, but it is what it is.

MileHigh

MH
I think you are a little lost when it comes to understanding what CEMF is in an inductor-or anything for that matter.

Quote verpies post 230: The current would not increase if CEMF = EMF

You do know that CEMF is a function of reactance, not of resistance,and so i do not know what all that garble about CEMF across a resistor was all about

To say that there is a voltage drop of 1 volt across a 1 ohm resistor that is supplied a voltage of 1 volt from a voltage source,is just horse radish talk. There is no drop in voltage at all across the resistor,as the potential difference across the resistor is the very same as that supplied from the source--no drop in voltage. If we had two 1 ohm resistors in series,and we applied 1 volt across that resistor series,then we would see a voltage drop across each resistor of 1/2 a volt. Using the word !voltage drop! across a single resistor with a voltage of 1 volt placed across it from a voltage source,is just bollocks talk.
And then to try and say that resembles an equal and opposite CEMF is just more bollocks.
There is no self induction in a bloody resistor that is non inductive.

To quote Poynt,post 1231: The voltage across the coil terminals does not change, it is determined by the voltage source. But the induced cemf is in series opposing with the voltage source Vin, and its value is determined by the frequency of Vin and the inductance L.
So, from this perspective the induced cemf is usually not equal to the applied emf (Vin). It is usually lower.
Poynt post 1233 Quote: I don't think he will assume that. I believe he knows that even though the induced emf (cemf) is opposite in polarity to that of the applied voltage, it will almost always be less, and therefore there will still be a net applied emf and resulting current.The EMF is applied by the user, while the cemf is self-induced; VL = L x di/dt. So according to the equation, the induced cemf can be any value and will vary depending on the frequency and inductance. The applied EMF never changes its value.As I mentioned above, the induced cemf can be any value, and is dependent on the input frequency and the inductance. I don't see any rules being broken here so help me out. Where am I going wrong?

And from Loner,post 1258:

Quote

IF the Applied voltage WERE EXACTLY equal to the "CEMF" there would be NO CURRENT. Second, IF there were NO Current, there would be NO CEMF!!!

So the first thing you did there,was attack him,because you believed he was talking about you. How do you know this to be the case?

I dont know what planet you are on MH,but the CEMF value is not the same as the EMF that created it.
Quote;Inductance is the name given to the property of a component that opposes the change of current flowing through it and even a straight piece of wire will have some inductance. Inductors do this by generating a self-induced emf within itself as a result of their changing magnetic field
So,if the CEMF was the same value as the EMF,then the current produced by the CEMF that flows in the opposite direction to that of the current induced by the EMF would be the same value,and there for,there would be no flow of current.

Quote: In an electrical circuit, when the emf is induced in the same circuit in which the current is changing this effect is called Self-induction, ( L ) but it is sometimes commonly called back-emf as its polarity is in the opposite direction to the applied voltage. Lenz’s Law tells us that an induced emf generates a current in a direction which opposes the change in flux which caused the emf in the first place, the principal of action and reaction.

So MH,Self-induced emf is the e.m.f induced in the coil due to the change of flux produced by linking it with its own turns. Only when total flux linkage/flux cutting is achieved,will the CEMF be equal to the EMF that created it--like in an ideal coil maybe

Brad

tinman · « **Reply #1176 on:** June 22, 2016, 01:11:36 PM »

Quote from: MileHigh on June 22, 2016, 12:31:50 PM

Brad:

You have the concepts of "attack" and "defend" mixed up and backwards in your head. Stop playing your "on stage" "dullest tool in the toolbox" games. Your silly comment is just another stunt that backfired.

You think that I have lost my marbles? 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.

MileHigh

MH
First we can look at this backwards,with using me example of a DC motor(or any electric motor for that matter).Lets say it's a 12 volt DC PM motor. When you place 12 volts across that motor,the current will be at it's highest,as the magnetic fields from the stator at this point in time,are cutting the rotor windings at it's lowest rate of change over time,and so the produced CEMF is also at it's lowest value. As the motor speeds up,the current draw begins to drop,becaust the rate of change of the magnetic fields has increased,thus increasing the CEMF value. At maximum RPM,the magnetic fields rate of change to that of the rotor windings is at it's greatest,and then so to is the CEMF value.
The reverse happens with the inductor,where the rate of change of the induced magnetic field is greatest the moment a voltage is applied across the inductor. This results in the highest value of CEMF produced,and so the lowest value of current flow. Over time,the rate of change of the magnetic field begins to decrease,and so the CEMF value also begins to decrease. This results in an increase of current flow through the inductor.

MH
It is the CEMF value to that of the EMFs value, that determines the current flow rise over time in an inductor. The CEMF is the current limiter of the inductor,be it in a coil,or an electric motor.

Brad

hoptoad · « **Reply #1177 on:** June 22, 2016, 01:16:45 PM »

Quote from: MileHigh on June 22, 2016, 06:58:42 AM

snip...
So where is the EMF and the CEMF? The one-volt voltage source is the EMF. The CEMF is the one amp flowing through the one-ohm resistor causing a one-volt voltage drop across the resistor.
The one-volt voltage drop across the resistor is the CEMF. Look at that, the EMF and the CEMF are equal and opposite, and current flows through the resistor.
snip...

In the unideal world, the one-volt voltage source is the EMF. = Correct
The CEMF is the one amp flowing through the one-ohm resistor causing a one-volt voltage drop across the resistor.= Incorrect
There is no CEMF, there is simply an applied EMF with an 'equal and opposite' reaction of heat dissipated by the resistor.

Quote

Now let's repeat the whole process for an inductor:
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.

This is where the difference between an inductor and resistor has to take account of changing current flow. At the very moment of connection, the inductance as well as the resistance will oppose the flow due to CEMF that arises from inductance, but as the coil reaches its maximum current and minimum change in current, the cemf will disappear leaving only the resistance to oppose the current and limiting it to 1 amp.
The resistor will exhibit an instantaneous 1 amp at connection for the first and ongoing seconds, but the inductor must necessarily inhibit the instantaneous flow on connection, to less than 1 amp (average) for the first second until the change in current ceases and becomes a steady flow of 1 amp per second thereafter. To maintain a steady state of 1 amp under steady current/voltage, the coil itself must still have 1 ohm resistance.

Quote

So, you are wrong. There is no difference between the EMF and the CEMF and current flows for both the resistor and the inductor. Simple enough
If you have a technical comment to the above discussion I am all ears. And again, if you want to say something to me then just say it.
MileHigh

If there were no difference in the reaction characteristics of resistors and inductors to current/voltage, in particular, changing current/voltage, we could just use resistors for everything that we currently use a coil for.
But as you already know, inductors are prized for their characteristic reaction to changing currents/voltages not so much for steady (DC) current/voltages. Though they are used to often to provide a steady (limited) current due their propensity for opposing changes in current.
I'd love to see a resistor that could do the job of a coil/inductor in most applications. Especially a resistor to replace a generator coil.

Cheers

tinman · « **Reply #1178 on:** June 22, 2016, 01:20:17 PM »

Quote from: MileHigh on June 22, 2016, 12:31:50 PM

Brad:

You have the concepts of "attack" and "defend" mixed up and backwards in your head. Stop playing your "on stage" "dullest tool in the toolbox" games. Your silly comment is just another stunt that backfired.

When it comes to electronics, you are not as smart as you think. Beyond that, don't you dare allege that I am a troll. I am making logical points and I am sincerely trying to debate with people, even if the debate can get heated. It takes two to tango. I am not buying your "wise man coming to share some pearls of wisdom" vibe, at all. If you have some issues with me, then say them straight to my face and we will debate them. Fair enough?

Now,i re read the post again MH,and not once did i see your name mentioned.
But regardless of that,you went on the attack,and proceeded to tell Loner how he is not as smart as he thinks he is.
Then i see you say to him to tell you straight to your face about any issues he has with you--remembering that your name was never mentioned once in his post.
You also say he alleged that you were a troll,and yet(once again) ,your name was never mentioned.
So with no mention of your name at all,how can you say you were just defending your self?-->defending your self against what allegations ?.

I think Loner is a very smart man,and he also agrees with myself,Poynt and verpies-Added hoptoad--and im sure a whole lot of others,when it comes to !what! is CEMF.

Brad

MileHigh · « **Reply #1179 on:** June 22, 2016, 01:45:20 PM »

I am going to repeat my question to you Brad:

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.

hoptoad · « **Reply #1180 on:** June 22, 2016, 01:51:25 PM »

Quote from: MileHigh on June 22, 2016, 12:40:49 PM

The retaining wall at the end of the trough is stationary and does not move. So think about the ramifications of that.

The wall doesn't have to move to absorb energy and be a part of the energy dissipation equation. But in your analogy, without including the wall there will be no compression of the spring, merely movement of the spring in the same direction as the cart.

When you lob a tennis ball against a wall, some energy is lost in sound through the air, and some energy of the ball is lost through heat by the distortion and rebounding of the balls shape and some will be lost to the wall itself in the form of pressure, heat and internal sound.

Any pressure of one object on another will create heat. The greater the pressure, the greater the heat. As the cart pushes the spring, the spring pushes the wall. Each component places pressure on the other and generates heat, the amount of which, however miniscule or large, being dependent on the pressure exerted.

Of course I'm explaining in terms of the unideal world, you know, reality, not ideal. Thats why I said you need an ideal wall to go with your ideal cart and ideal spring, for an ideal scenario, in which one can endlessly postulate about the 'ideal XYZ'.
However reality provides a way of supplanting postulation with knowledge through empirical evidence gained by observational research.

MileHigh · « **Reply #1181 on:** June 22, 2016, 02:09:55 PM »

Hoptoad:

Here is the short definition for CEMF:

<<<
https://en.wikipedia.org/wiki/Counter-electromotive_force#cite_ref-Graf_1-0

The counter-electromotive force (abbreviated counter EMF, or CEMF),[1] also known as the back electromotive force, is the voltage, or electromotive force, that pushes against the current which induces it.
>>>

If you reread my example for the resistor and the inductor keeping the definition for CEMF in mind it should make sense.

For the trough, the best way of looking at it is to keep things simple. You have a concrete trough that is hollowed out of the earth like a long and narrow swimming pool. All four walls of the trough are fixed and immovable. You put a long spring into the trough, and then a moveable cart moves forward with a constant velocity and pushes against the spring.

With that simple model all of the power expended to move the cart gets stored in the spring, and only in the spring.

MileHigh

picowatt · « **Reply #1182 on:** June 22, 2016, 02:11:39 PM »

Quote from: MileHigh on June 22, 2016, 06:58:42 AM

There is no difference between the EMF and the CEMF and current flows for both the resistor and the inductor. Simple enough.

MH,

CEMF is typically used with reference to inductance.

As such, I believe most of us understand CEMF to be more so along the following:

A current flowing thru a conductor creates a magnetic field. A portion of that magnetic field induces a rate of change dependent voltage into that same conductor. The polarity of that induced voltage opposes the initial current flow thru the conductor. It is that induced voltage, in opposition to the initial current flow, that is referred to as CEMF.

Can you provide a reference citing an example of your usage of CEMF with regard to resistors?

PW

tinman · « **Reply #1183 on:** June 22, 2016, 02:25:48 PM »

Quote from: MileHigh on June 22, 2016, 01:45:20 PM

I am going to repeat my question to you Brad:

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.

MH

We have already done this with your ideal coil and ideal voltage.
We know the coil has a resistance value of 0 ohms--or no resistance.
What do you think it is that stops the current going straight to an infinite value,as soon as the voltage is placed across it?.
Why dose it take 3 seconds to reach a current of only 2.4 amp's,and not shoot up to an infinite amount of current flowing through that inductor.
What is the !reactance! in inductive reactance?-->what is reacting to what?
What is !self inductance! ?

Quote

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.

I have the feeling that this is just your way of making me waste my time on something that is already very clear--except to you it seems.
So i have voted not to fall for this !waste of time! ploy by you,and instead,i give you a video that should make it very clear to you. As you will see,the circuit to show the effective CEMF is quite simple,and im sure i could show what he explains in the video.

I hope you listen to this video very carefully,and then you will understand that if the CEMF was equal to the EMF that produced it,there would be no current flow,as there would be no voltage drop/or potential difference across the inductor.

https://www.youtube.com/watch?v=RGTRXlarKww

But even then,i have a feeling we are going to see you !once again! try and change physics to suit your needs.

Brad

hoptoad · « **Reply #1184 on:** June 22, 2016, 02:58:19 PM »

Quote from: MileHigh on June 22, 2016, 02:09:55 PM

Hoptoad:
Here is the short definition for CEMF:
<<<
https://en.wikipedia.org/wiki/Counter-electromotive_force#cite_ref-Graf_1-0

The counter-electromotive force (abbreviated counter EMF, or CEMF),[1] also known as the back electromotive force, is the voltage, or electromotive force, that pushes against the current which induces it.
>>>
If you reread my example for the resistor and the inductor keeping the definition for CEMF in mind it should make sense.

For the trough, the best way of looking at it is to keep things simple. You have a concrete trough that is hollowed out of the earth like a long and narrow swimming pool. All four walls of the trough are fixed and immovable. You put a long spring into the trough, and then a moveable cart moves forward with a constant velocity and pushes against the spring.

With that simple model all of the power expended to move the cart gets stored in the spring, and only in the spring.

MileHigh

I know what CEMF is and I know that even in the real world, technically every single circuit that has even a miniscule current will have some inductance, because a moving charge creates a magnetic field, and therefore exhibits a cemf opposing further changes to its own motion. I also know that in most real world circuits, inductance is disregarded when talking about resistors and semiconductors because the inductance of a resistor/semiconductor is so miniscule as to be irrelevent to its purposeful characteristics. When we want inductance we deliberately set about making it with a coil, not e.g. a carbon resistor.

As for your 'all four walls are fixed and immovable' in your cart analogy, if any single one of the walls is used as a backstop for the spring, then that wall becomes part of the energy dissipation system.( In reality, since all four walls are all part of the same entire mass of the trough, then the whole trough is part of the energy dissipation.)
You can not compress any spring from one side only. Try it without an anchor point like one of the walls, your spring would just move with the force applied to one side.

Go ahead, try to compress a spring by applying a force to one side without a backstop becoming an essential part of the system of components needed for the energy transfer required to compress it.

You don't need a concrete trough. Just a spring, your hand and .... whatever else is NEEDED to compress the spring.
Now I agree that most of the energy, even in a real world system, will get stored in the spring, but not all. The spring itself, unless ideal, will dissipate some of the compression energy it receives through heat due to lattice stress and displacement.
Only in an idealized world can perfect power transmission and storage take place. Sadly we don't live in that world, we live in the real world, with circuit losses incurred and components that will never be 'ideal'.

News:

User Menu

Author Topic: MH's ideal coil and voltage question (Read 487933 times)

MileHigh

Re: MH's ideal coil and voltage question

hoptoad

Re: MH's ideal coil and voltage question

tinman

Re: MH's ideal coil and voltage question

MileHigh

Re: MH's ideal coil and voltage question

MileHigh

Re: MH's ideal coil and voltage question

tinman

Re: MH's ideal coil and voltage question

tinman

Re: MH's ideal coil and voltage question

hoptoad

Re: MH's ideal coil and voltage question

tinman

Re: MH's ideal coil and voltage question

MileHigh

Re: MH's ideal coil and voltage question

hoptoad

Re: MH's ideal coil and voltage question

MileHigh

Re: MH's ideal coil and voltage question

picowatt

Re: MH's ideal coil and voltage question

tinman

Re: MH's ideal coil and voltage question

hoptoad

Re: MH's ideal coil and voltage question