MH's ideal coil and voltage question

[poynt99]

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]

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]

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]

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.

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.

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.

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.

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]

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