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MH's ideal coil and voltage question

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

Previous topic - Next topic

0 Members and 1 Guest 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

picowatt

Quote from: MileHigh on July 06, 2016, 10:03:50 PM
PW:

Were you a buddy of the late George Martin?  lol

Here is a news clip where you see the guts in some detail:

https://www.youtube.com/watch?v=k4_BUIM7gY0

Perhaps Poynt will make the next Sticky Fingers or Abbey Road in it!

MileHigh

Cool video, nice closeup of the isoloop...

I can assure you from experience, the razor blades were only used for cutting tape...

PW

MileHigh


synchro1

Imagine two opposite electrodes of a coil each end connected to one of the two poles of a battery positive and negative; Now, if we disconnect the negative side rapidly, the magnetic field collapses, the voltage polarity reverses and the electro motive force flows into the positive side of the battery like two batteries in series one with a higher voltage.

Disconnecting the positive side of the coil would cause a collapse of the magnetic field and a flow of BEMF into the negative side of the power battery. One direction has to be the opposite of the original current flow!

verpies

Quote from: tinman on July 06, 2016, 07:20:53 PM
So with a real coil,we have a small time delay between applied EMF ,and produced CEMF. This is enough time to create an offset between the EMF induced current,and the CEMF induced current.
No instantaneous reactions exist in a quantized world

picowatt

Quote from: tinman on July 06, 2016, 07:20:53 PM

The confusion for me is,PW keeps assuming that the EMF will produce a current of 800mA/second,an so some how the CEMF induced current is less than this 800mA.

One last time:

Before the _fixed_ 4volt EMF is applied to the 5H inductor, there is no CEMF being generated so CEMF=0

At T=0, the 4volt EMF is applied to the inductor and current rapidly rises:


1. As di reaches .8A/s,  CEMF = EMF

2  If CEMF = EMF,  di < .8A/s

3. If di < .8A/s,  CEMF < EMF

4. If CEMF < EMF,  di increases (return to #1, loop forever)


Read thru the loop (1-4) ten times or so, perhaps it will come in to focus...  This is an example of a negative feedback loop.

(A negative feedback loop is a common concept in electronics and there are probably a hundred, if not many more, negative feedback loops operating in the electronic circuits you use daily.  Although many people think of gain setting when they think of negative feedback, the most common and numerous circuits throughout your house that use negative feedback are most likely current sources.)

As far as how much current flows at T=0, it does not really matter.  If one electron flows at T=0, then 1 picosecond later, a current flow of something just shy of 5 million electrons would be equivalent to .8A/s and that would cause the CEMF to be equal to 4 volts. (5 million electrons flowing being equal to .8 picoamperes or thereabouts, I'm tired so...)

It is the rate of change, not the absolute amount of current, that determines the CEMF  Start out with as many electrons in as narrow a slice of time as desired, but when from time slice to time slice the current rises (changes) at the equivalent rate of .8A/s, the generated CEMF will be 4 volts.

Remember, we are discussing inductance, ideal inductance, so please don't add the effects of a resistor and capacitor to the circuit being discussed.  That comes later.

PW