MH's ideal coil and voltage question

[tinman]

I am also going to add post 15 from AC

In an ideal voltage source the source Emf would be fixed and an ideal inductor would have virtually no losses. It seems to me no current could flow because the moment a charge tried to moved due to the ideal voltage source Emf the ideal inductor would produce an equal and opposite Cemf to oppose it. Ideally if the source Emf is always instantaneously opposed by the inductors Cemf then nothing can move, a stalemate.

There is also a number of others that believe the same.

Just to make it clear,my first answer was incorrect

Quote: you cannot place an ideal voltage across an ideal inductor.
The reason being,at T=0,when the ideal voltage is placed across the ideal inductor,the current would rise instantly to a value of infinity. The reason this cannot happen,is because an ideal inductor dose not dissipate any power in the form of heat,due to the fact that it has no resistance or hysteresis loss,as it is an ideal inductor. If an ideal voltage was placed across an ideal inductor(in theory),it would result in an explosion the likes the universe has not seen since the creation of it-the big bang all over again.

It is clear that this is not the outcome,and i retract that answer,but it will remain as a reference as to where i started,and where i am now.

I am happy with my later answer,and until !!proven!! otherwise,remains my answer.

At T=0,no current flows,as the CEMF induced current keeps the EMF induced current in check.

If you happen to find the mechanism that allows current to flow,even though the CEMF is equal to that of the applied EMF Poynt,then we would have learned something together-along with everyone else on this thread.

Brad

[partzman]

Perhaps we could stand back for a moment and consider the heretical idea of what we have if we remove Cemf from our single inductor analysis! 

The formula for Emf is EMF = L*dI/dt and the formula for Cemf is Cemf = -L*dI/dt. Logically as has been stated, if Cemf = Emf then no current will flow in our inductor.  But our inductor does exhibit inductance so therefore the Cemf must less than the Emf by a varying amount depending on the magnitude of the inductance. Does anyone have a derivation for this relationship?

OTOH, we can calculate the inductance of an air core single layer solenoid (neglecting fringing) with 𝑳 = 𝝻o × 𝒏^2 × 𝝿 × 𝙍^2 × 𝒍 where 𝒏 = 𝑵/ 𝒍. I used unicode characters to eliminate the confusion between small L for length of coil and I for current. We have no Cemf used here and the inductance is dependent on the coil's physical properties and the permeability of space.

Without the CEMF,  we could say however that we have a fixed counter-magneto motive force or Cmmf due to the physical position of each wind to the others.  Mmf = N*I and Cmmf = -N*I.  The H field around the wire (synonymous with the flux field) is
H = N*I/𝒍 which tends to cancel or buck between adjacent windings and aids on the outside of the windings.  As we move our winds closer together, the flux cancellation is greater yielding a higher inductance and as we move the windings farther apart, we have less cancellation resulting in a lower inductance. No Cemf required.

pm

[poynt99]

PM,

Let's not confuse the applied emf with the cemf. The induced cemf is L x di/dt, and it is negative due to Lenz's law. The applied emf is Vin, and is set by the input voltage, not the inductor.

The inductor voltage is often referred to as an induced emf, and that is fine as long as "induced" prefaces "emf". It is easier to just say cemf. The emf in our case is of course referring to the applied emf, or voltage source.

[poynt99]

With a non-ideal inductor, the voltage across the inductor is at a maximum upon t=0. The resistor may as well not be in the circuit at this instant (it could be replaced with a short). This is when the applied emf=cemf, and at this very instant, current begins to flow. Seems to me this is evidence that current can and does flow when the cemf=emf.

The only difference with the ideal inductor of course is that the full cemf is always across the inductor, and so it makes sense that the linear increase of current continues as long as the emf is applied, just as it does for the brief time in the non-ideal case just after t=0.

[MileHigh]

Quote from: MileHigh on June 23, 2016, 02:43:33 PM

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?

MileHigh

That graphic you made is cringe-worthy and nobody wants to touch it.  All that you are doing is showing is how foolish or ignorant or stupid you can be.  Like I told you, buy yourself a few books on basic electronics and lock yourself in a room for a month and read them and understand them.  You definitely deserve the trash talk in this case.