MH's ideal coil and voltage question

[MileHigh]

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.

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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]

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]

Loner:





MileHigh

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?.

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]

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]

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.