Author Topic: MH's ideal coil and voltage question (Read 490442 times)

tinman · « **Reply #1350 on:** June 27, 2016, 03:19:06 PM »

I am also going to add post 15 from AC

Quote

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 · « **Reply #1351 on:** June 27, 2016, 04:05:04 PM »

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 · « **Reply #1352 on:** June 27, 2016, 04:41:26 PM »

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 · « **Reply #1353 on:** June 27, 2016, 04:46:32 PM »

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 · « **Reply #1354 on:** June 27, 2016, 06:37:19 PM »

Quote from: tinman on June 27, 2016, 11:26:30 AM

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.

picowatt · « **Reply #1355 on:** June 27, 2016, 06:58:25 PM »

Just a few thoughts:

1. By stating that we are using an inductor of 5Hy, we have, by definition, stated that the inductor will produce a CEMF of 4 volts when the RATE OF CHANGE of the current flowing thru it is .8 amps per second. Any inductor that does not produce 4 volts of CEMF when the RATE OF CHANGE of the current flowing thru it is .8 amps per second is not, by definition, a 5Hy inductor.

2. When the 5Hy inductor is connected across the 4 volt Vsource, current flows until it reaches a rate of change of .8 amps per second at which time a CEMF of 4 volts is generated. When the inductor's CEMF equals 4 volts, the rate of change of the current flow drops towards zero amps per second. As the current flow's rate of change becomes less than .8 amps per second, the CEMF produced is also less than 4 volts. When the CEMF produced is less than 4 volts, current flow will again increase until it once again reaches a rate of change of .8 amps per second, causing the CEMF to again be 4 volts, which again reduces the rate of change of the current flow toward zero. This action continues for as long as the 4 volts is applied across the inductor.

3. Although the action above is described in a step wise fashion, those familiar with the use of negative feedback in analog circuits will easily visualize the above action as being a smooth, continuous, and self-regulating action that maintains a continuous .8 amps per second of rate of change thru the inductor.

4. Unlike a "normal" inductor, the magnitude of the current flowing thru an ideal inductor having zero DC resistance has no effect upon, that is, produces no deviation from, the as defined .8 amps per second rate of change necessary to generate 4 volts of CEMF (as evidenced by a linear increase in current).

5. With regard to a "normal" inductor that does have resistance, as the magnitude of the current flow changes, the IR related voltage drop produced by that current also changes. That voltage drop does affect, that is, causes the rate of change to deviate from, the as defined .8 amps per second rate of change necessary to generate a CEMF of 4 volts (as evidenced by a deviation from a linear increase in current).

6. Being more so an energy storage and retrieval mechanism, the reactance of an inductor is not a "dissipative" mechanism. Only whereby the inductor deviates from ideal (having resistance, etc.) is any energy stored or retrieved lost to dissipative mechanisms.

PW

minnie · « **Reply #1356 on:** June 27, 2016, 09:46:16 PM »

tinman,
how about measuring it? My guess is that you'd have to be quicker than C.
John.

tinman · « **Reply #1357 on:** June 28, 2016, 01:04:07 AM »

Quote from: MileHigh on June 27, 2016, 06:37:19 PM

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.

I dont care how you wish to word it MH,as i am not bound by incorrect definitions such as it seems the EE world is,nor am i too worried about what you think. When a value drops,to me it means it is now lees than the supply value. If you have a drop in HP from that that is supplied,then you have less than that of what the motor is putting out. We get a drop in HP from the engine to the wheels,due to friction in the drive line-so less out than in as far as HP go's.

The resistor causes no drop in voltage to that being supplied-if the supply can provide the required voltage. So,regardless if that resistor is across the battery or not,there is no drop in voltage-the voltage remains at 2 volts.

Brad

tinman · « **Reply #1358 on:** June 28, 2016, 01:19:14 AM »

Quote from: picowatt on June 27, 2016, 06:58:25 PM

Just a few thoughts:

1. By stating that we are using an inductor of 5Hy, we have, by definition, stated that the inductor will produce a CEMF of 4 volts when the RATE OF CHANGE of the current flowing thru it is .8 amps per second. Any inductor that does not produce 4 volts of CEMF when the RATE OF CHANGE of the current flowing thru it is .8 amps per second is not, by definition, a 5Hy inductor.

2. When the 5Hy inductor is connected across the 4 volt Vsource, current flows until it reaches a rate of change of .8 amps per second at which time a CEMF of 4 volts is generated. When the inductor's CEMF equals 4 volts, the rate of change of the current flow drops towards zero amps per second. As the current flow's rate of change becomes less than .8 amps per second, the CEMF produced is also less than 4 volts. When the CEMF produced is less than 4 volts, current flow will again increase until it once again reaches a rate of change of .8 amps per second, causing the CEMF to again be 4 volts, which again reduces the rate of change of the current flow toward zero. This action continues for as long as the 4 volts is applied across the inductor.

3. Although the action above is described in a step wise fashion, those familiar with the use of negative feedback in analog circuits will easily visualize the above action as being a smooth, continuous, and self-regulating action that maintains a continuous .8 amps per second of rate of change thru the inductor.

4. Unlike a "normal" inductor, the magnitude of the current flowing thru an ideal inductor having zero DC resistance has no effect upon, that is, produces no deviation from, the as defined .8 amps per second rate of change necessary to generate 4 volts of CEMF (as evidenced by a linear increase in current).

5. With regard to a "normal" inductor that does have resistance, as the magnitude of the current flow changes, the IR related voltage drop produced by that current also changes. That voltage drop does affect, that is, causes the rate of change to deviate from, the as defined .8 amps per second rate of change necessary to generate a CEMF of 4 volts (as evidenced by a deviation from a linear increase in current).

6. Being more so an energy storage and retrieval mechanism, the reactance of an inductor is not a "dissipative" mechanism. Only whereby the inductor deviates from ideal (having resistance, etc.) is any energy stored or retrieved lost to dissipative mechanisms.

PW

PW

I guess that it could be looked at that way,but dose not resolve the fact that current still flows when there is no potential difference.

Also,why is this not true for a DC PM motor-such as the ones used in my video.
If the BackEMF is equal to the applied EMF,then the motor will not draw any current-or current will not flow through that motor. This i have-and can do again more accurately show on my bench.

BackEMF and CEMF are one in the same.
Inductive kickback is BackEMF. We are now saying that the inductive kickback from our ideal coil,will now give back the same amount of energy it took to create it,when the CEMF value is equal to the EMF value across that coil,as we have no I/R losses.

As far as i can work out,if the feedback is going to be the same as the applied,then the net power flowing is 0--we just end up with a tank circuit that keeps self oscillation.

Brad

tinman · « **Reply #1359 on:** June 28, 2016, 01:37:09 AM »

Anyway,here is a quick and dirty test i did on the BackEMF value the magnets play a part in,in regards to a DC PM motor.

It was going to be a much more accurate test,but lack of the needed equipment means we only got a !round about! value-you will see why.

https://www.youtube.com/watch?v=-Jf_daUOdy0

Brad

picowatt · « **Reply #1360 on:** June 28, 2016, 02:05:57 AM »

Quote from: tinman on June 28, 2016, 01:19:14 AM

PW

I guess that it could be looked at that way,but dose not resolve the fact that current still flows when there is no potential difference.

Simply stated, if there is no potential difference, no current flows. If no current flows, there is no CEMF... If there is no CEMF, there is a potential difference. If there is a potential difference, current flows. Ad infinitum...

Think about it a bit. An inductor's CEMF is a self regulating feedback mechanism dependent on the RATE OF CHANGE of the current flowing thru the inductor.

As for the rest, I would never state that an inductor's CEMF, a motor's BEMF, and "inductive kickback" (or flyback) are all identical.

PW

3Kelvin · « **Reply #1361 on:** June 28, 2016, 05:30:06 AM »

Hello together,

i think that there is no electric field in the ideal Coil (super conductor coil).
If there where a(n) electric field inside, the current have to go to infinity.
In other words, over a super conductive wire is no voltage drop measurable.
A super conductive coil has N turns to build up the 5 Henry.

P lost= i²*R
I= U/R
If R = 0 also U=0

But this is only my personal assumption.

Love + Peace
3K

tinman · « **Reply #1362 on:** June 28, 2016, 01:30:18 PM »

Quote from: MileHigh on June 25, 2016, 06:42:34 AM

PW:

Pushing 100 thread pages to get Brad and others to understand how an inductor works and you get this from Brad:

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

[Part 2 from Brad]

The CEMF is created by the changing magnetic field,which is due to the increasing current over time,that was induced when the voltage was placed across the inductor
This CEMF creates a current flow that is in the opposite direction to that of which the applied voltage induced.

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

However, it appears to be pretty clear that Brad believes that there is a "tangible reverse current" due to the CEMF somewhere in the mix when you connect a battery or EMF source to a coil.

Hence my frustration, and myself and Brad had a little mutual trash talk. I am very tempted to just walk away.

MileHigh

Lenz's law MH
Quote: If an induced current flows, its direction is always such that it will oppose the change which produced it.
Quote: Lenz's law states that when an emf is generated by a change in magnetic flux according to Faraday's Law, the polarity of the induced emf is such, that it produces an current that's magnetic field opposes the change which produces it.

This is something you should know by now MH.

Brad

tinman · « **Reply #1363 on:** June 28, 2016, 02:32:20 PM »

Quote from: MileHigh on June 27, 2016, 06:37:19 PM

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.

Please feel free to take a DMM,and measure the voltage drop you speak of in a parallel circuit such as depicted in my diagram,and then post your results here,along with a diagram as to how you measured a voltage drop across the resistor.
Also please remember,that a voltage is a value of potential difference between two points,and a voltage is not a voltage drop.

Quote Fundamentals of electricity

Quote

Wires carrying current always have inherent resistance, or impedance, to current flow. Voltage drop is defined as the amount of voltage loss that occurs through all or part of a circuit due to impedance.

We could even add your CEMF producing resistor to this,and say that if there was an EMF drop of 2 volts across the resistor,then your equal and opposite CEMF would return that 2 volt's,and so the total loss across the resistor is 0--nothing. And so by your own analogy(the CEMF producing resistor),an equal and opposite reaction happens,and the voltage drop is counter acted,and so,no voltage drop took place.
Aint that a hoot MH--you killed your own claim.

Brad

poynt99 · « **Reply #1364 on:** June 28, 2016, 02:42:10 PM »

Quote from: tinman on June 28, 2016, 01:19:14 AM

I guess that it could be looked at that way,but dose not resolve the fact that current still flows when there is no potential difference.

There is no "potential difference" with a resistor yet current flows.

Quote

BackEMF and CEMF are one in the same.
Inductive kickback is BackEMF. We are now saying that the inductive kickback from our ideal coil,will now give back the same amount of energy it took to create it,when the CEMF value is equal to the EMF value across that coil,as we have no I/R losses.

Brad, IK is not the same as cemf or bemf.

Quote

As far as i can work out,if the feedback is going to be the same as the applied,then the net power flowing is 0--we just end up with a tank circuit that keeps self oscillation.

With standard feedback in amplifiers, even with 100% feedback the signal passes through at unity gain. Something to think about perhaps.

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