MH's ideal coil and voltage question

[tinman]

Going back to my simplified example...

Given:
L=5H
V=+4V

What happens:
At t<0, XL=0
At t=0, XL=infinity, and the current rises from 0A linearly at 0.8A/s.

At t=1, XL=5 Ohms
At t=2, XL=2.5 Ohms
At t=4, XL=1.25 Ohms
i.e. with every doubling of "t", "XL" goes to 1/2 of previous value. XL never reaches 0 Ohms.

At t>0 when I_L begins rising, an induced cemf of -4V is produced, based on Vcemf=L x di/dt. The cemf remains steady at -4V as long as di/dt stays at 0.8A/s. This of course occurs simultaneous with the application of 4V and the rise of I_L.

The question is, if emf=cemf, would it not makes sense that this result is created through an equalizing process? The amps/s rate ultimately being determined by the applied voltage and the inductance values.

The equalization I refer to comes about via the simultaneous process of an applied emf that wants to drive a current, vs. a reactionary process that wants to lower that current. To me this is very much like a negative feedback mechanism commonly used in linear amplifiers. Your amplifier may have an inherent gain of 1000, but through the application of negative feedback, the gain is reduced to some desired level, such as 100. In the case of our self-inductance, the negative feedback mechanism is the self-induced current and B field, which happens to oppose the B field resulting from the applied voltage. It is not quite an exact analogy, but conceptually similar. It all happens in real time, simultaneously, and only the end result is observable.

Perhaps it comes down to an applied emf, vs. an induced E field. We know the applied emf is 4V, but can we break the induced cemf (equivalent to the E field) down any further? Well, for a multi-turn inductor we can divide the cemf by the number of turns to obtain the induced cemf per turn. To me this would represent the actual value of the E field circulating around the inductor. So if we have a total cemf of 4V, and a 1000-turn coil, the actual induced E field would be 4/1000 = 4mV. This does not sound like much, but with R=0, 4mV could drive a significant current in a single loop, and each loop would carry the same current. I'm sure the induced current can be derived from the E field (or B field) and rate of change.

Some things to perhaps think about anyway. Sorry there are no definitive answers here. Still thinking about this.

and the current rises from 0A linearly at 0.8A/s.

So we have a coil that is ideal--free from resistive losses.
We have a current that is going to follow a straight linear rise at 800mA/s
Our current dose not follow an exponential curve,as it would with a coil that has a resistance value.

This being true,it must also be true that the magnetic fields change it time would remain a constant value to that of the linear current rise,and that value would be the value it was at T=0--the instant the voltage was placed across the coil.
If the magnetic field is increasing/changing in time at a constant value,then the CEMF must also be doing the same. If the CEMF value is the same as the applied EMF,and the coil has no resistance,and cannot dissipate energy,then the reverse current flow must also be the same as the induced current flow.

The only reason we get an exponential current curve,is because real world inductors have a resistance value,and so some of the energy is dissipated as heat,and so the difference between the EMF and CEMF value. But an ideal coil has no resistance,and there for cannot dissipate energy,and so there is no loss associated with the CEMF value as there is in real world coils.

I really hope you have a closer look at this,because as i stated long ago,things are not the same when the coil/inductor is void of resistance.


Brad

[poynt99]

Perhaps consider the following:

a) emf = cemf = the E field times the number of turns.
b) If an inductor has 1000 turns, then the effective cemf to emf ratio is 1/1000.
c) per turn, the cemf is therefore much less than the applied emf.

[tinman]

Perhaps consider the following:

a) emf = cemf = the E field times the number of turns.
b) If an inductor has 1000 turns, then the effective cemf to emf ratio is 1/1000.
c) per turn, the cemf is therefore much less than the applied emf.

And how can that be,when the EMF also see the same amount of turns.
Also,if this were true,dose this mean an ideal 1 turn coil will have an EMF to CEMF ratio of 1:1,and no current can flow.

I do not think that is right.
If the ratio was 1/1000,then the current would rise much faster.

There is also the fact that,the more turns you have,with the same current flow value,the greater the magnetic field. This results in a greater magnetic field,but the rate of change over time is still a linear constant-along with the induced current.

Are we skipping equal and opposite reactions,even though there is no dissipated power?


Brad

[picowatt]

Mmm-interesting.

A question for you all.

Regarding the BackEMF in a DC PM motor,such as the ones i am using in the previous video.
Is the bulk of the BackEMF a result of the PMs,or a result of the rising and falling current in the rotor winding's ?

I will await a day or two,before i post my video showing the result's.

MH-->what do you think?


Brad

I'll venture a guess.

I see 6 separate test conditions.

1.  Locked rotor, no PM, no pole pieces.
     Armature acts only as an inductor with CEMF, no BEMF, max I determined by R_DC

2.  Locked rotor, no PM but with pole pieces installed.
     Similar to above but with larger inductance, CEMF only, no BEMF, max I determined by R_DC

3.  Locked rotor, with PM and with pole pieces installed.
     As above but with larger inductance or shift in BH curve (saturation), CEMF only, no BEMF, max I determined by R_DC

4.  Spinning rotor, no PM, no pole pieces (no magnetic or conductive material in proximity to rotor, spun via external means)
     Rotor acts as an inductor with modulated inductance and R_DC, some noise, CEMF only, no BEMF, max I determined by R_DC

5   Spinning rotor, no PM but with pole pieces installed, rotor spins as attraction motor
     Rotor still has inductance but those effects are swamped by the BEMF now present.  Max I determined by BEMF.

6.  Spinning rotor, with PM and pole pieces installed, rotor spins as normal PM motor.
     As above, max I determined by BEMF, more torque (or RPM) available for a given BEMF (current draw)


PW     

[picowatt]

So we have a coil that is ideal--free from resistive losses.
We have a current that is going to follow a straight linear rise at 800mA/s
Our current dose not follow an exponential curve,as it would with a coil that has a resistance value.

This being true,it must also be true that the magnetic fields change it time would remain a constant value to that of the linear current rise,and that value would be the value it was at T=0--the instant the voltage was placed across the coil.
If the magnetic field is increasing/changing in time at a constant value,then the CEMF must also be doing the same. If the CEMF value is the same as the applied EMF,and the coil has no resistance,and cannot dissipate energy,then the reverse current flow must also be the same as the induced current flow.

The only reason we get an exponential current curve,is because real world inductors have a resistance value,and so some of the energy is dissipated as heat,and so the difference between the EMF and CEMF value. But an ideal coil has no resistance,and there for cannot dissipate energy,and so there is no loss associated with the CEMF value as there is in real world coils.

I really hope you have a closer look at this,because as i stated long ago,things are not the same when the coil/inductor is void of resistance.


Brad

Tinman,

Are you stating that you believe that the only amount of current that can flow thru an inductor with zero DC resistance is the amount of current that flows at T=0?

PW