MH's ideal coil and voltage question

[verpies]

As I understand it in the analogy the torque of the pump is the voltage.
Am I correct?

Yes.
It is interesting to consider, which way the energy flows when the pump rotates against that torque.

[MileHigh]

If the question in this thread is ever answered properly or not, I strongly advise anybody interested to go back and read it again from the beginning.  You can decide for yourselves.

As an example, I somehow doubt we are going to get any more challenges about the concept of a time-variable ideal voltage source.  Nor will we get any admissions that this was all just silliness and a useless and nonsensical waste of time and energy from the main players.

I will also repeat the main goals for this thread:

1.  Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands the concepts and understands what he is doing.
2.  Brad admits that he was wrong when he stated that my response to the harder question was wrong.

I would really hope that Brad successfully achieves those two goals.

[verpies]

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.

The difference is huge because now the torque is not synonymous with the rotational direction of the impeller and the pump can rotate with or against the torque.

Thus, we can have a typical 4-quadrant operation:
1) The pump moving CW with a CW torque.
2) The pump moving CW against a CCW torque.
3) The pump moving CCW with a CCW torque.
4) The pump moving CCW against a CW torque.

In two of these cases, the energy is transferred to the pump, and in two other cases, the energy is transferred from the pump.
Can you assign the direction of energy flow in these four cases?

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.

In an ideal pump, as well as in an ideal pipe, there is no resistance to the water's motion and lately we've been considering an analogy of an ideal L circuit here, so all the analogical hydraulic components must be ideal, too.
Thus, the water can move the impeller without any friction ...and the impeller can move the water without any friction, too.

[MileHigh]

Yes.
It is interesting to consider, which way the energy flows when the pump rotates against that torque.

I am just going to offer up simplified analogies and perhaps they will help.

The water pump in this case never impedes the water flow, but what it does do is create an increase in pressure in that water itself.   So you don't necessarily have to think about a tangible water pump, just what it does.  So if the pressure is 10 psi on one side of the pump, then the pressure is 14 psi on the other side of the pump.  Or perhaps the pump drives the pressure the other way:  the pressure is 10 psi on one side of the pump, and 7 psi on the other side of the pump.

That's the type of water pump we are talking about in this example, a constant-pressure pump.  You don't care about the water flow rate or even the direction of the flow, the only thing that counts is that the pump establishes a difference in water pressure from one side to the other side.  You can even block the water flow, and the pump still does the same thing.

However there is another type of water pump you can also imagine, a constant-flow pump.  It's easy to imagine perhaps a large diesel engine with a transmission that lowers the shaft output speed that drives some small pistons that pump water at a certain flow rate.

In this pump, you don't care at all what the water pressure is on either side of the pump.  The only thing  you care about is that the pump pumps say six gallons of water per minute.  The pressure at the output side of the pump could be 100 psi or 1000 psi, it doesn't matter and the pump will pump at a constant water flow rate.

[wattsup]

If the question in this thread is ever answered properly or not, I strongly advise anybody interested to go back and read it again from the beginning.  You can decide for yourselves.

As an example, I somehow doubt we are going to get any more challenges about the concept of a time-variable ideal voltage source.  Nor will we get any admissions that this was all just silliness and a useless and nonsensical waste of time and energy from the main players.

I will also repeat the main goals for this thread:

1.  Brad gets up the learning curve and understands the original question and then answers it correctly all by himself and clearly demonstrates that he understands the concepts and understands what he is doing.
2.  Brad admits that he was wrong when he stated that my response to the harder question was wrong.

I would really hope that Brad successfully achieves those two goals.

Where did the bold words in your post come from? Seems this is the first time this is used in front of the words "ideal voltage source" since this was not in your initial question of ideal voltage on an ideal inductor. If so then why is that graph showing a straight line at 4 volts for 3 seconds? Why would you equate a time variable with ideal voltage, when ideal voltages do not change?

Hmmmmmmmmm (with my wet index finger pointing upwards) did the wind change directions? Or do I sense a hurricane is about to form.

wattsup