MH's ideal coil and voltage question

[verpies]

Besides, take a 9 volt and 12 volts battery and connect them in parallel and what do you get.

Very high current. Equivalent to a 3V shorted battery.

However the same does not apply to a 9V battery inserted into an ideal inductor, because the inductor is not a voltage source - it is a current source. 
The internal impedance of an ideal current source is infinite, while the internal impedance of an ideal voltage source is zero.

That's why a current source connected to a voltage source does not result in any infinities.

[tinman]

MH did know the answer to the question, and could easily have plotted out the current trace as I have encouraged you to do, but he wanted you and others to work it through yourselves.

MH's question is thought-provoking and well thought out, although the wording could have been geared more towards the experimenters here rather than the techies. At any rate, I can assure you that MH knew all along exactly how the circuit currents behave, given the input he specified. After all, he is the one that formulated the question.

Brad, you really are being silly with what you said there. I know you and MH don't seem to be getting along well, but he deserves more benefit of doubt than you are affording him here on this electronics question.

You mean like the benefit of doubt that he gave me regarding the ICE having resonant systems?. Guessed you missed all that. This is the very same situation. He told me i had no idea what i was talking about,when Internal combustion engines are my forte--my area of expertise.
The difference is,i backed up my knowledge with provided fact's,and this is something no one here can do with an actual test,as we are talking about ideals we do not have. What we are doing is placing a theory based around !best guesses!.

So i have given MH no more than he has given me,and in fact,i have never used the foul language he has toward me.
Not once did i see you,or any other EE guy here tell MH to calm it down when the roll was reversed,but i see you are quick to jump on me when i do the same that has been done to me.

I have seen this very same thing with other members that disagree with MHs analogy.
It's an !agree with me! or your wrong attitude MH has--plane and simple.
As i said in the other thread,i will now treat him as he treats me.

As i said,there is a pattern that is followed on this forum,and that is the EE guys stick together--bar one,that being (as i have always said) verpies. I would also put vortex1 in there with verpies,but he dosnt frequent this forum much-sadly.
As verpies said in reply to this question,Quote : The equivalent circuit model for an ideal inductor is not an inductor with a wire shorting across its ends.
verpies-Just because most of the world does it wrong does not mean that we have to.


Brad

[tinman]

Sounds to me like you are agreeing with me.

The source is trying to drive the opposite polarity energy into the inductor. When the source is at -3V, it offers up a place for the inductor to dump some of its stored energy. Given more time at -3V, the inductor would fully lose its "positive" energy, and begin to energize in the "negative" direction.

Since it isn't long enough for that to occur, the energy in the inductor simply decreases to a lower "positive" level, and there is only one place for that lost energy to go. You can even see it in the sim.

Well this is how i see it Poynt.

We have a loop of water pipe that represents our ideal inductor loop. In that loop of water pipe we have a pump. Our pump is our voltage source,and the water in that pipe is our current. We start the pump,and this puts pressure(our voltage) on the water(our current) the water starts to flow around our loop. We have a bypass valve in the pump,so as when we switch the pump off(0 volts),the water can still flow in the direction it was(our current is now flowing with the pump off). While the water is flowing,we start the pump up so as it spins in the opposite direction. The pump(our now reversed voltage) wants to now push the water in the opposite direction to that of which it is already flowing. The energy in the moving water is not added to,or stored in the pump--it is working against the pump,and the pump will draw more current to stop the flow of water that is flowing in the wrong direction, before it can start to move the water(current) in the right direction to that of what the pump wants to move it in.

That is how i see it.

Brad

[verpies]

@Tinman

That's a mechanical/hydraulic analogy.  This type of mechanical thinking is a long lost skill among physicists and will lead you to the correct results but the analogy must be precise.  Unfortunately yours is not precise enough.

We have a loop of water pipe that represents our ideal inductor loop. In that loop of water pipe we have a pump.

Let's make it a positive displacement pump- like a Lobe Pump.
I chose the Lobe Pump for this analogy because in such pump, the direction and speed of the impeller has 1:1 correspondence to the direction and speed of the water current.

Our pump is our voltage source and the water in that pipe is our current.

Corrections:
The water in the pipe symbolizes electric charge.  The motion of this charge symbolizes electric current.
BTW: The mass/inertia of the water symbolizes inductance.

The force/torque applied to the pump's impeller symbolizes the voltage. 
It is important not to conflate the pump with the voltage, because the pump itself is not the force - it is only a mechanism to transfer the force to the water/charge.

We start the pump,and this puts pressure(our voltage) on the water(our current) the water starts to flow around our loop.

Generaly, I agree.
I would write: "...this puts force/pressure on the water causing its acceleration and motion (current)"

We have a bypass valve in the pump,so as when we switch the pump off (0 volts), the water can still flow in the direction it was (our current is now flowing with the pump off).

Yes, the water can flow even when the force (torque) applied to the impeller is zero, but a bypass valve is not necessary for this flow to continue, as e.g. the impeller of a lobe pump will continue to rotate under the current of water already flowing through this pump.

This "bypass valve" is an extraneous component that should have tipped you off, that the analogy is not precise enough.

While the water is flowing,we start the pump up so as it spins in the opposite direction. The pump(our now reversed voltage) wants to now push the water in the opposite direction to that of which it is already flowing.

Here is where you reap the bad fruit of your analogy ;(
Precisely the cause of the conceptual error is in the phrase: "we start the pump up so as it spins in the opposite direction".
The correct phrase should have been: "we apply a torque to the pump's impeller in the opposite direction".

Note, that the application of opposite force/torque to the impeller, does not immediately result in the reversal of the impeller's direction (nor water's direction).  I remind you of the 1:1 correspondence between the direction of the impeller and the direction of the water, (their speed also).

I am sure that now you have the tools to complete the rest of the analogy by yourself.

[tinman]

@Tinman

That's a mechanical/hydraulic analogy.  This type of mechanical thinking is a long lost skill among physicists and will lead you to the correct results but the analogy must be precise.  Unfortunately yours is not precise enough.
Let's make it a positive displacement pump- like a Lobe Pump.
I chose the Lobe Pump for this analogy because in such pump, the direction and speed of the impeller has 1:1 correspondence to the direction and speed of the water current.
Corrections:
The water in the pipe symbolizes electric charge.  The motion of this charge symbolizes electric current.
BTW: The mass/inertia of the water symbolizes inductance.

The force/torque applied to the pump's impeller symbolizes the voltage. 
It is important not to conflate the pump with the voltage, because the pump itself is not the force - it is only a mechanism to transfer the force to the water/charge.
Generaly, I agree.
I would write: "...this puts force/pressure on the water causing its acceleration and motion (current)"


This "bypass valve" is an extraneous component that should have tipped you off, that the analogy is not precise enough.


Note that the application of opposite force/torque to the impeller does not immediately result in the reversal of the impeller's direction (nor water's direction).  I remind you of the 1:1 correspondence between the direction of the impeller and the direction of the water, (their speed also).

I am sure now you have the tools to complete the rest of the analogy by yourself.

Here is where you reap the bad fruit of your analogy
Precisely the cause of the error is in the phrase "we start the pump up so as it spins in the opposite direction".
The correct phrase should have been: "we apply a torque to the pump's impeller in the opposite direction".

I see no difference in what i said,and what you have stated above.
The motor that drives the positive displacement pump you have used,will still encounter the force of the water moving in the opposite direction to that of the applied torque to the pump.
You have simply separated the pump from the motor. But any force placed upon the pump will be transferred to the motor.

Yes, the water can flow even when the force (torque) applied to the impeller is zero, but a bypass valve is not necessary for this flow to continue, as e.g. the impeller of a lobe pump will continue to rotate under the current of water already flowing through this pump.

As the motion of the pump/motor combo i was using to represent the voltage(our force),then i included the bypass valve to represent no motion of the motor/pump--the equivalent of the 0 volt level in the question. Having the impeller still moving,would be seen as a resistance against the flow of water,and in our ideal loop,we have no resistance to the flow of the current,and so the bypass valve was included for that reason also.
I believe that this is a close representation to what we have,than having the impeller being rotated by the water,resulting in a resistance to that flow of water.

Brad