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Overunity Machines Forum



MH's ideal coil and voltage question

Started by tinman, May 08, 2016, 04:42:41 AM

Previous topic - Next topic

0 Members and 21 Guests are viewing this topic.

Can a voltage exist across an ideal inductor that has a steady DC current flowing through it

yes it can
5 (25%)
no it cannot
11 (55%)
I have no idea
4 (20%)

Total Members Voted: 20

Magluvin

Quote from: partzman on May 14, 2016, 03:59:43 PM
Mags,

I am not the one you asked for an answer on this but I will go ahead and offer my opinion.  It appears from your comment highlighted above, you assume that if an inductance is 100% efficient, it will not allow any current to flow. At 100% efficiency, the emf is totally cancelled by the cemf thus resulting in zero current flow or infinite inductance.  It must have even the smallest amount of resistance to perform like a real inductor.

When we apply the formula delta I = E*t/L or rearrange the formula to solve for inductance of emf, we assume ideal conditions such as an ideal inductance.  The answer we get from using this formula on an inductance with resistance is very close to what we would measure on the bench. The less dc resistance the coil has, the closer the answer is. From this we can deduce that an ideal 5h inductance is just that, 5h with no impediment from any resistance.

partzman

Yes you do understand my point. Thanks.

As for the rest of it, how can we deduce that the CEMF that counters the input isnt equal to the input?  Maybe there is something a miss here. Why isnt the cemf ideal also when it comes to the ideal inductor? Where is the calculation of some loss that keeps the cemf lower than the input? 

Again, the sim says what the sim says. Look at the graph again and expand that time out for hours or even days. There will always be a curve in the real world no matter the level of resistance. Who is to say that with absolute zero resistance that the mechanism that is at work in the inductor that impedes the input still has some sort of loss in order for it to be that the cemf is not equaled to the input applied, all of the time? How does that formula account for that? This is the question. ;) Maybe that possibility is ignored some how?  Like how would we know for sure without actual hands on testing of such a device? Yet we talk about it as if it is just standard thinking without going any deeper.  Just think a little deeper. How deep can we see beyond just some formula dropped on us and think thats all we need to know?

Mags

MileHigh

You are dealing with the same problems in understanding, and the same "rules" that are seemingly made up on the fly using incorrect logic, and the same lack of understanding in how an inductor actually works.

When you connect a real or ideal voltage source to a real or an ideal inductor, or to a resistor, then the EMF imposed on any of those three devices results in a CEMF in the device that is always equal and opposite to the EMF.

The idea that the CEMF must be a bit lower than the EMF for current to flow is 100% wrong, it's a "rule" that has been made up because it "sounds right."

The resistor responds to the EMF with a resultant current flow and that power is burnt off as heat.

The real or ideal inductor responds to the EMF with a resultant current flow and that power is stored in the magnetic field and if there is a resistance then a small amount of that power is burned off as heat.

Why does the inductor do what it does?   Why does the shopping cart move when you push on it?  Read about an inductor and understand its dynamic response to changing voltage conditions.  This is a device whose response is determined by differential and integral equations.  Learn intuitively how an inductor responds and learn and understand the equations that back up the intuitive understanding of how the inductor responds.  A shopping cart's response is intuitive, you have to get to the point where the inductor's response is intuitive.

I will repeat that this notion that "if the CEMF is the same as the EMF then current will not flow" is a fallacy.  It does not apply to a resistor, and it does not apply to an inductor.

What about a capacitor?  If the CEMF from the capacitor is the same as the EMF applied to it, does any current flow?  The answer is no, there is no current flow.  The capacitor is simply not the same as the resistor or the inductor.  That's just the way it is and if you want to understand electronics then you have to understand these things.

What about "ideal CEMF?"

You guys define "ideal CEMF" first, and then perhaps we can discuss it.  What does that actually mean?  Or is that just a meaningless term that has been made up on the fly?  Lay your cards on the table about this "ideal CEMF" business.

partzman

Quote from: Magluvin on May 14, 2016, 04:38:23 PM

Who is to say that with absolute zero resistance that the mechanism that is at work in the inductor that impedes the input still has some sort of loss in order for it to be that the cemf is not equaled to the input applied, all of the time? How does that formula account for that? This is the question. ;)

Mags

That's a fair enough question. Let's see if we can logically work out an answer. Obviously the formula(s) don't seem to supply a suitable answer because they assume ideal components. So, maybe we can look at one factor that determines the inductance L of a real air coil for example.  This would be the geometry of the coil such as turns, spacing, length of winding and diameter of windings.  Apart from any outside influence, this geometry would determine the effect of the self induced emf or cemf produced by the time rate change of current or dI/dt on the applied emf which is basically the inductance of the coil.  If we now reduce the resistance of the coil windings to zero, we still maintain the same geometry.  So are we to say that the laws of induction due to this particular geometry are going to change because of zero resistance? Or will this example coil become an ideal inductor with pure inductance as determined by it's geometry?

Regarding simulation of an ideal inductor, our current limited versions only offer us the ability to compare a reasonable resistance to a very small resistance.  I was able to get LtSpice to simulate the 5H inductor with a dcr of 1e-320 without blowing up.  The point of this is, the current slope is straighter or has less droop over time with every decrease in dcr. So, could we assume that a trend is established that says if we continue to decrease the dcr to zero, we will experience an infinite straight current line with a perfect dI/dt representing the ideal inductance?

partzman

Edit

poynt99

question everything, double check the facts, THEN decide your path...

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Magluvin

Working on my bike. The quick release lever shaft broke and the nut to nut fasteners on one end of the axle came loose and tightened the axle solid. Couldnt even push it back as the threads of the axle were digging into the axle holder slots. Was on my way back from the beach to get a little sun and see the beautiful women in bikinis everywhere. Then tragedy. About 3 miles from my shop or home. Some guys in a truck offered me a ride.  ;D   So now I am in fix the bike mode.

The bike is electric. About 80lb. So walking it back was a hard option without the wheel working in any way that it should.

I looked over what both of you posted. Ill go over it again after.  Thanks for the explanations. That is all Ive been asking for. Lets see if it all fits.

Back later.

Mags