MH's ideal coil and voltage question

[partzman]

No, I do not see a problem.  I see an inherent negative feedback mechanism and the reason di is limited to and maintained at precisely .8A/s.

Consider what happens upon application of the 4 volt EMF.  Current begins to flow thru the inductor and...

1.  As di reaches a rate of change equal to .8A/s, the CEMF becomes equal to 4 volts.

2.  As per "B)" in the drawing, as the CEMF becomes equal to the EMF, di also reduces towards zero.

3.  But if di becomes less than .8A/s, the CEMF also becomes less than 4 volts which allows di to increase. 

4.  Return to number 1 above (loop forever)

Although described above in a rather step-wise fashion, it is, in reality, a smooth and continuous process and a classic example of a negative feedback loop.

Thanks for taking the time to post the drawings.  By all means, enjoy the 4th!

PW

Added:

Also note that I believe it more appropriate for L2 to be replaced with a simple conductor (wire).  Having an inductor inside a "model" of an inductor is redundant and could lead to confusion.

I would replace L2 with a wire and then place a dotted line box surrounding both the CEMF Vsource and the wire with di indicated.  The entire dotted line box would then be labeled as L2.

(I would also label the value for CEMF1 as X_v or X_v1 because the only time the CEMF=4volts is when di=.8A/s)

I don't have the time to comment on the above except to say that you still have a math fudge with your definition of Cemf=L*di/dt in order for your model to work.

So, I've attached a new model that may satisfy all requirements.  No time to go into detail but I think it is self explanatory.

pm

[picowatt]

But then, perhaps not...

Partzman,

There was no "math fudge" needed, everything worked perfectly just as it was.

As for your new model that apparently requires 3 separate types of EMF, you've completely lost me...

Your notation "D", in particular, is most confusing.

PW

[tinman]

An ideal inductor sitting at rest, is essentially an ideal piece of wire. So when an ideal emf is placed across it, the current wants to instantly jump to infinity, right at t=0.

Due to the self-induced cemf however (the negative feedback), current is limited to rise at 0.8A/s (in our case with 4V and 5H).

There are two reasons the induced cemf = the applied emf:

1) The obvious reason is that the ideal voltage source is holding the inductor terminal voltage at 4V.

2) The not so obvious reason is due to the self-induction feedback process trying to be explained here.

To explain further, try thinking about the process right from the very beginning. As I said above, the moment Vin is applied, the current will try to instantly go to infinity. We know however that simultaneously, the inductor is self-inducing an opposing emf which is going to cause the current to increase at a set rate determined by Vin and L. This is the feedback mechanism that makes an inductor what it is. The inductor doesn't limit current, it limits the change in current. Any change in current is reflected in a corresponding opposing emf, which opposes the applied emf. In the case when R=0, the emf and cemf are always equal. The cemf settles or equalizes to the same value as the applied emf.

And so in a real coil,the CEMF dose not equal the EMF,due to the resistance value of the windings-as i said some time back.
If CEMF did equal the EMF in a real coil,then why not the same feed back as an ideal coil free from winding resistance?.
We know that in a real coil,the current is not limited,and will rise to the steady state value,and so the CEMF cannot be equal to the EMF.


Brad

[poynt99]

And so in a real coil,the CEMF dose not equal the EMF,due to the resistance value of the windings-as i said some time back.

Yes, with a real coil at t=0 the cemf=emf, and yet, current flows. I've mentioned this a number of times now.

If CEMF did equal the EMF in a real coil,then why not the same feed back as an ideal coil free from winding resistance?.

There is feedback as well with real coils. You can see the current is rising at 0.8A/s for the first 200ms when R=1 Ohm. After this point the voltage begins to trade off with the resistor until finally ALL the voltage is across the R. It started out with ALL the voltage across the L. Do you wish me to repost the sim result showing the 1 Ohm trace tracking the 1p Ohm trace for the first 200ms?

We know that in a real coil,the current is not limited,and will rise to the steady state value,and so the CEMF cannot be equal to the EMF.

I'm not sure how you came to that conclusion, but it is wrong. What do you mean the current is not limited? Of course it is; in fact it is limited in two different ways: 1) a maximum of 0.8A/s current rate of rise, 2) a maximum current of Vin/R.

[tinman]

Yes, with a real coil at t=0 the cemf=emf, and yet, current flows. I've mentioned this a number of times now.
There is feedback as well with real coils. You can see the current is rising at 0.8A/s for the first 200ms when R=1 Ohm. After this point the voltage begins to trade off with the resistor until finally ALL the voltage is across the R. It started out with ALL the voltage across the L. Do you wish me to repost the sim result showing the 1 Ohm trace tracking the 1p Ohm trace for the first 200ms?
I'm not sure how you came to that conclusion, but it is wrong. What do you mean the current is not limited? Of course it is; in fact it is limited in two different ways: 1) a maximum of 0.8A/s current rate of rise, 2) a maximum current of Vin/R.

As i said, the CEMF cannot be equal to the EMF like it is with the ideal coil.
What i ment by current flow not being limited,was to that like the ideal coil,where the current is !apparently! limited to 800mA /second. As the winding resistance increases,our current trace follows a greater curve.

If you look close enough at the current,you will see that it first rises right along side the voltage,when the V is placed across the coil. I (and Mags) suspect this to be because of the parasitic capacitance of a real coil.
This being the case,then at T=0,the CEMF is not equal to the EMF, because if it was,then there would be no current flow at T=0.

At vox T=0, a voltage is placed across the inductor.
At the very same time,a current rises to a certain value,where this value will depend on the value of winding capacitance-and so dose the time period for this brief current flow.
The current will fall to 0, once the  capacitance has been charged.
Shortly after that,the current will rise ,and track it's  exponential curve path.

That is  what i have found so far in my tests.


Brad