MH's ideal coil and voltage question

[tinman]

 author=MileHigh link=topic=16589.msg487063#msg487063 date=1466779069]

It's like you have regressed and you are back to some kind of whackadoo "Great Pumpkin" fantasy.

Lol--really
Lets have a look at your next statement.

So like if the 5th time constant current is 100 amps and the 1st time constant current is 25 amps then the reverse current produced by the CEMF is 75 amps.   So does that mean when you first apply the voltage across the coil the current is -75 amps?  It's "Attack from Planet Bizarro and the Pumpkin Patch Creatures."

I think you had better do a refresher course in time constants,and the % of current at the end of  the first time constant will be, of that of the end peak current flow at the fifth time constant-or the steady state current flow
Here is a blunt hint. The current will rise to 63.2% of it's steady state value after the first time constant. So in reference to your 100 amps steady state current flow ,after the first time constant,the current flow will be 63.2  amps.So the reverse current value at the end of the first time constant is 36.8 amps-not 75 amps as you say. God only knows where or how you came up with 25amps after the first time constant. I suspect it was from the same place you found that resistor that generates it's own CEMF

I can't see anything.  I thought the current was -75 amps at the start.

If you keep coming up with those oddball numbers,derived from you wackadoo mathematics,then yes--you will never see nothing.

I am totally confused.  I tried every setting on my secret decoder ring and it's not able to unscramble what you are saying.
It's like you have completely regressed and this entire thread never happened.

Indeed--you totally are.
It's like you have gone from some one with some intelligence,to some one that has defiantly drank to much coolaid. First you bring us the resistor that generates it's own CEMF,and now a new time constant calculation for inductors

If the coil is a real coil then you have to account for the resistance.  Then it becomes a slightly more difficult problem.  Using standard mathematical techniques you solve for the circuit and you get the standard exponential equation that we all know that with a tiny bit of algebraic rearranging of some variables gives you a nice convenient time constant to work with.

It would seem that you are unable to work with those standard mathematical techniques,and derive the correct % of current flow for each time constant.

That's all there is too it, a coil integrates on voltage to give you current just like a shopping cart integrates on force to give you velocity.  It's just Mother Nature in action.
All of the stuff in your head about "battling currents" is a model that simply does not work.  It's crazy talk.  It's like something that you found in a pumpkin patch.

It is not pumpkin patch talk that dose not work,it is actual fact.

When a voltage is placed across a coil, the current will change quickly from zero. This  change creates an expanding magnetic field around the coil, and when this happens,it induces a voltage back into the coil. This self induced voltage is known  as back EMF,or CEMF,and creates a current flow in the OPPOSITE direction to the current flow that induced it in the first place. The result of this is that it slows the rate of change of the current that was induced from the voltage being placed across the coil(the impedance). If this initial rate of change were to continue in a linear fashion, the current would reach its maximum or steady state value in 5 time constants,where the time constant is T=L/R seconds.

MH
Every time you try and make me look bad,you only end up making your self look stupid.


Brad

[MileHigh]

Even a hovercraft will still experience some resistance, however miniscule, from the air it passes through. It will also experience viscosity however miniscule. You can approach zero resistance and zero viscosity but never reach it except in an 'ideal' vacuum. However, in a vacuum, the hovercraft's mode of operation will be negated. Their is no 'ideal' in anything. As I said ,yes, we use ideal equations because they work for real world designing purposes, but the products of our designs will always have a real error margin built in to them however well they approximate the ideal design.

I've agreed with almost everything you've posted on real circuits except for your erroneous assertion that a voltage drop is the same as cemf. I disagree that we can know exactly how an ideal 'anything' will act, because their is no such thing as an ideal in an unideal world.
We can imagine, postulate and use provable math, ignore infinitesimally small values, reach high values of probability and a consensus of agreement - yet we can never 'know' it. We can only know what is real. The whole question of 'ideal' is no different to 'God'. We use all the evidence available to us to argue in favour or against, but will never 'know'. Not being able to know 'ideal' or 'perfection' is not mysticism, its reality.
Cheers

Yes, but there is no point in talking about the fact that there will be a very small resistance.  I already qualify that in my comments.  The first paragraph is not necessary.

To make things simpler, I already agreed that it makes sense to not call a voltage drop across a resistor CEMF.  I stated that at least TWICE.  So why are you and Brad stating that?   Why?

We do know how an ideal inductor will work.

[tinman]

When you first connect a conductor to a source nothing happens instantly,, I read somewhere at one time that this condition is called the relaxation time period and can be calculated.

If you had a long enough wire you could turn it on and then back off before this time period has been reached,, I read a small article on some people who used a large spool of coaxial cable to run this kind of test,, but I would think that the capacitance of that cable would also have an influence on such tests.

What information is propagating and how is it propagating,, has been one of those little niggles I think about from time to time.

Webby
We can switch an inductor on and off fast enough so as no current flows.
The propagation speed would have to be the speed of light-would it not?.
The reason you can see this with a long piece of wire,is because it has resistance and inductance,and so,it has impedance.
Even with a long straight piece of wire,you must form a loop to apply a voltage,and so now you have a single turn coil.


Brad

[tinman]

Yes, but there is no point in talking about the fact that there will be a very small resistance.  I already qualify that in my comments.  The first paragraph is not necessary.

To make things simpler, I already agreed that it makes sense to not call a voltage drop across a resistor CEMF.  I stated that at least TWICE.  So why are you and Brad stating that?   Why?

We do know how an ideal inductor will work.

Perhaps you could tell us all,as to why the current dose not shoot straight up to it's steady state value,when a voltage is place across it
What impedes on that current flow?


Brad

[MileHigh]

Brad:

I think you had better do a refresher course in time constants,and the % of current at the end of  the first time constant will be, of that of the end peak current flow at the fifth time constant-or the steady state current flow
Here is a blunt hint. The current will rise to 63.2% of it's steady state value after the first time constant.

That's a real doozie Brad.  Let's see, I probably first learned about a time constant in 1977.  Do you really think I don't know what the approximate values are?  I was tired and just threw up some arbitrary numbers for illustrative purposes.

So here you are making a complete fool of yourself, a spectacle of yourself, by "pretending" that I don't know the approximate value of the first time constant.  That's you cynically being an asshole for all to see.

You are such a little imp, I think you need a good spanking.

I suspect it was from the same place you found that resistor that generates it's own CEMF

Bad little imp boy!  You read multiple times how I agreed that it is more appropriate to avoid that term for a resistor.  Little imp boy needs another spanking, count down from 200.

First you bring us the resistor that generates it's own CEMF,and now a new time constant calculation for inductors

Now count down from 300.

It would seem that you are unable to work with those standard mathematical techniques,and derive the correct % of current flow for each time constant.

Now count down from 400.  That's what you call beet red for being a bad boy.

When a voltage is placed across a coil, the current will change quickly from zero. This  change creates an expanding magnetic field around the coil, and when this happens,it induces a voltage back into the coil. This self induced voltage is known  as back EMF,or CEMF,and creates a current flow in the OPPOSITE direction to the current flow that induced it in the first place.

No it doesn't create a current flow in the OPPOSITE direction.  That is a nonsensical idiotic fantasy that does not happen and you will never find a single stitch of evidence for that.  It's a bad-old-days belligerent Brad fantasy.

The EMF source is say a battery, and the CEMF source from the coil looks like another battery.  That's two EMF sources facing each other at the same potential and so NO CURRENT FLOWS towards the battery driving the coil.  You have about as much electronics common sense as lumpy gravy.  Your whackadoo fantasy story is completely and utterly ridiculous and after all the work that was done in this thread you should be embarrassed for regressing once again.  Like I have told you, GET SOME ELECTRONICS BOOKS AND LOCK YOURSELF IN A ROOM FOR A MONTH AND READ THEM AND UNDERSTAND THEM.

very time you try and make me look bad,you only end up making your self look stupid.

Anybody can read this thread from the beginning through the end with a doorman's clicker, the type used to count the number of people in the club.  Every time you have "a moment" the person can register a click.  Seriously, you wouldn't want to know what the count was when they got to this very posting.

MileHigh