MH's ideal coil and voltage question

[minnie]

Scratchbot, hang on in here.
   What's needed is some good theoretical work to be done. I've been looking since
about 2005 and I haven't seen a proven watt, plenty of crap,though!
  Check out old Leif Holmlid of Gothenburg, that's the sort of area something
could come from.
            John.

[tinman]

 author=MileHigh link=topic=16589.msg486819#msg486819 date=1466523762]

Since we know that an action produces an equal and opposite reaction, then the coil has to be reacting equally and opposite to the battery.  Same thing for a resistor.
So even though the battery is imposing its voltage on the coil, you need to be able to shift your perspective and go "inside" the coil and realize that the coil is pushing back with the same CEMF.

No it is not.
You are lost when you think that the total EMF is going to be converted only to a CEMF.
Every action dose not always equal !one only! opposite and equal reaction. The sum of reactions will equal the action.
E.G-3 pool balls on a pool table. Two placed side by side,and one at the other end of the table. The single ball is hit with the queue,and that single ball strikes the two at the other end of the pool table right in the center of the two balls. Each of the two balls takes of in different directions. We now have two reactions that summed together equal the action--the single ball that struck them-minus some loss due to sound.

You can take a ridiculously simple example from the physical world:  If you are balancing a water jug on your head, then your head is pushing up with the same force as the water jug is pushing down.  It is as simple as that.

It's not that simple at all.

Or you just have enough common sense to realize the applied EMF and the CEMF are identical without even having to think about it.

No they are not,as if they were,no current would flow,as no voltage would exist across the coil.
!OH!-unless that coil was !!ideal!! ,where the flux linkage/flux cutting was actually ideal

Kirchhoff's voltage law (KVL) The sum of all the voltages around the loop is equal to zero. This law is also called Kirchhoff's second law, Kirchhoff's loop (or mesh) rule, and Kirchhoff's second rule.

The applied EMF and the CEMF from the coil must add up to zero.  Therefore, the CEMF must be equal and opposite to the EMF.

Yea--good one MH--only we do not have a loop,we have a coil attached to a voltage supply.


Brad

[tinman]

 author=poynt99 link=topic=16589.msg486830#msg486830 date=1466530196]

I don't think he will assume that. I believe he knows that even though the induced emf (cemf) is opposite in polarity to that of the applied voltage, it will almost always be less, and therefore there will still be a net applied emf and resulting current.

Of course i will not assume that.This is the very reason an electric motor will draw a lot of current at start up,and the current will reduce as the motor gains speed.
MH is just lost again.

The EMF is applied by the user, while the cemf is self-induced; VL = L x di/dt. So according to the equation, the induced cemf can be any value and will vary depending on the frequency and inductance. The applied EMF never changes its value.
I believe that when there is an induced current, there is always an associated induced emf.
Yes, but this is a series circuit as well as a parallel circuit, and the induced cemf would be as I depicted.
As I mentioned above, the induced cemf can be any value, and is dependent on the input frequency and the inductance. I don't see any rules being broken here so help me out. Where am I going wrong?

You are not going wrong Poynt,and i think MH is having a bad day.
As you would have seen in this thread,he laughed at some of us when we said that in an ideal coil,the CEMF would be equal to the applied EMF. But now-now he is saying the same thing he laughed at us for saying.
If the coil was indeed ideal,then that would be the case-->the CEMF would equal the applied EMF.
We know this is the case,and we can prove it with a thought experiment using a simple DC motor.

We apply 12 volts across a 12 volt DC motor.
The motor will draw maximum current on startup. As the rotor windings pick up speed,they cut through the stator magnetic fields quicker,and so see a magnetic field that has a greater change in time. This results in a larger value of CEMF,and so the total voltage seen across the coil is reduced,and so the current is reduced. Once running at it's top rated RPM,with no load attached to the motor,the least amount of current will be flowing through the motor,as the CEMF is now at it's greatest. Our motor is also acting as a generator<--something to remember in time to come,but say no more at this time,on this thread.
Now,if there were no resistive or frictional losses(the motor is ideal),how much current would need to flow through that motor to keep it spinning?


Brad.

[tinman]

 author=MileHigh link=topic=16589.msg486824#msg486824 date=1466527005]

I am puzzled why you would say the CEMF is not equal to the applied EMF.

Says the man that laughed at myself and Mags when we stated that in an ideal coil,the CEMF would be equal and opposite to that of the applied EMF.

As you say over there-You have drunk too much cool aid MH

then the CEMF is always equal and opposite to the applied EMF.

There you go Mag's--aint that a hoot

Brad

[MileHigh]

Yea--good one MH--only we do not have a loop,we have a coil attached to a voltage supply.

Brad

After six years, and all of the discussions that have taken place, that's just another jaw-dropping ridiculous outrageous goof that you have made that would make any poor electronics teacher want to smash his head against the wall and smash his Stradivarius violin into a thousand splinters.

I just don't know what to say.  People are aghast?  The sky is falling?  We are at a private resort, Brad's Bizarro Retreat?  Hit the resent button and hope for no BSOD?

It's just beyond belief sometimes.  Somebody else can pick up the pieces with you.