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Author Topic: MH's ideal coil and voltage question  (Read 477935 times)

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #375 on: May 14, 2016, 05:39:59 PM »
Well they wont be two seperate coils in parallel MH,because one winding will be CW and the other CCW.
So they will be in series,not parallel

Brad

If it really was a toroid, then there will be two junction points where the voltage source makes contact with the toroid as shown in the diagram.  Current will flow say from top to bottom in each half coil.

There will be some flux cancellation at the two junction points.  But excluding that, for all practical intents and purposes each half of the toroid will be a separate and independent coil.  Because they are separate and independent, the winding direction is meaningless.

The two half-toroid coils will look like two separate coils in parallel across the voltage source.

MileHigh

.....................................................

This is an addendum.....

Okay so Verpies is talking about an ideal toroidal inductor where there is total flux coupling from coil turn to coil turn no matter what the physical configuration.  It's almost like there is a "magic ideal core" to the toroid directing the flux.

In this case if you connect up the voltage source as per my diagram, the CW and the CCW turns due to the current splitting up in different directions will cause total flux cancellation and it will look like a direct short.

Note as per the toroidal inductor diagram, if the connections are not 180 degrees apart, you get a kind of "dial up an ideal inductor value" configuration.

I can get this but Verpies I think it had to be described better.  I hope I am right, and please correct me if I am wrong.

tinman

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Re: MH's ideal coil and voltage question
« Reply #376 on: May 14, 2016, 05:44:25 PM »
 author=poynt99 link=topic=16589.msg484272#msg484272 date=1463239770]




Quote
If it has 0V across it, then yes it is equivalent to a short circuit. However, this is a voltage source that starts out with +4V across it, therefore it must be drawn as an ideal voltage source or function generator.

And who states that this is the case?
Regardless of whether there is a voltage or not,the fact that there is no internal series resistance means that the current can still flow through that voltage source unimpeded.
When we drop down to 0 volts (as stated in MHs question),the current will continue to flow,and at a steady state. As verpies stated,you cannot measure a voltage across a shorted(looped) ideal inductor.

Quote
It is a voltage source (set to voltage X) with a series resistor equaling 0 Ohms.
There is no paradox. You are misinterpreting an ideal voltage source as a zero Ohm wire. That is not the case unless as I said, the voltage is set to 0V.

And MHs question requires a o voltage at one point in the test. At that point,the current still flow's,as there is no resistance that impedes that current flow through the now looped(shorted )circuit. Now,as virpies stated,you cannot place a voltage across a shorted ideal inductor,and you have just stated that at 0 volts,then yes,the ideal voltage source is like a 0 ohm wire. So now how do you once again increase the voltage after 0 volt's,as you are now trying to place a voltage across an ideal shorted(looped) inductor?

See the paradox now?.

Brad

Quote
You would have a good quality capacitor. What would happen? Nothing if it wasn't connected to anything. If connected to some load it would happily discharge.

Now we place that ideal inductor across that ideal capacitor--what happens?
Can you measure a voltage across any part of that circuit?.

picowatt

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Re: MH's ideal coil and voltage question
« Reply #377 on: May 14, 2016, 05:44:42 PM »
Figure 3 represents MHs circuit.


Brad

Figure 2 represents MH's question, and is an accurate representation of an ideal voltage source and inductor.

An ideal inductor is "theoretically" made from ideal wire having no resistance or capacitance.  It is not "shorted" by a wire having no resistance.

A straight length of ideal wire still has inductance and behaves accordingly.  Coiled up, that same length of ideal wire has increased inductance.  At no time is there any resistance in the ideal wire and at no time is there a requirement to short the inductor with another ideal wire to model it. 

PW

verpies

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Re: MH's ideal coil and voltage question
« Reply #378 on: May 14, 2016, 05:45:18 PM »
An ideal voltage source is not properly illustrated as a short circuit.
So what is the resistance of a 0V voltage source?
....or if you connect an ideal capacitor, charged to 11V, to an ideal 1V voltage source, then what current will flow through the capacitor at t0?

verpies

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Re: MH's ideal coil and voltage question
« Reply #379 on: May 14, 2016, 05:48:57 PM »
What would happen if we reduced the internal series resistance of a charged cap(say 1000uF with 10 volts across it) to a value of 0(added),and then used that as our ideal voltage source,and placed the ideal inductor across it ?.
An ideal LC circuit will be formed and sinusoidal current will appear.
An infinite current will not flow because the ideal inductor will still possess non-zero reactance and impedance despite having zero resistance.

tinman

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Re: MH's ideal coil and voltage question
« Reply #380 on: May 14, 2016, 05:53:58 PM »
If it really was a toroid, then there will be two junction points where the voltage source makes contact with the toroid as shown in the diagram.  Current will flow say from top to bottom in each half coil.

There will be some flux cancellation at the two junction points.  But excluding that, for all practical intents and purposes each half of the toroid will be a separate and independent coil.  Because they are separate and independent, the winding direction is meaningless.

The two half-toroid coils will look like two separate coils in parallel across the voltage source.

MileHigh

And here we have another MH paradox added,so as he is not found to be wrong again.

Correct answer.
The two coils will be in series-not parallel.
The top of one coil (coil A) will produce say a north field,and the top of the other coil(coil B) will produce a south field. The opposite end of each coil will of course produce the opposite field. As it is a toroid,we can follow a circular path. We will have a field configuration that represents a series connection-North to south to north to south. If it were parallel,then we would have north/north at one end,and south /south at the other.

You only have to replace the two coils with batteries,where north may represent positive,and south represent negative,to see that it is a series arrangement ,and not a parallel arrangement.



Brad

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #381 on: May 14, 2016, 05:58:16 PM »
And here we have another MH paradox added,so as he is not found to be wrong again.

Correct answer.
The two coils will be in series-not parallel.
The top of one coil (coil A) will produce say a north field,and the top of the other coil(coil B) will produce a south field. The opposite end of each coil will of course produce the opposite field. As it is a toroid,we can follow a circular path. We will have a field configuration that represents a series connection-North to south to north to south. If it were parallel,then we would have north/north at one end,and south /south at the other.

You only have to replace the two coils with batteries,where north may represent positive,and south represent negative,to see that it is a series arrangement ,and not a parallel arrangement.



Brad

I did an addendum to my posting #382 to account for Verpies' explanation of his configuration.

poynt99

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Re: MH's ideal coil and voltage question
« Reply #382 on: May 14, 2016, 06:00:32 PM »
And who states that this is the case?
That an ideal 0V source is equivalent to an ideal conductor? By deduction, that is what it is. It is common sense. It is impractical of course (except in simulators where it can be used to measure current), but it is a precise equivalent. Doesn't an ideal non-inductive conductor always have 0V across it? Well, so does an ideal voltage source set to 0V.

Quote
Regardless of whether there is a voltage or not,the fact that there is no internal series resistance means that the current can still flow through that voltage source unimpeded.
When we drop down to 0 volts (as stated in MHs question),the current will continue to flow,and at a steady state.
Yep, I'm not arguing against that.

Quote
As verpies stated,you cannot measure a voltage across a shorted(looped) ideal inductor.


And MHs question requires a o voltage at one point in the test. At that point,the current still flow's,as there is no resistance that impedes that current flow through the now looped(shorted )circuit. Now,as virpies stated,you cannot place a voltage across a shorted ideal inductor,and you have just stated that at 0 volts,then yes,the ideal voltage source is like a 0 ohm wire. So now how do you once again increase the voltage after 0 volt's,as you are now trying to place a voltage across an ideal shorted(looped) inductor?
I'm not sure what verpies is saying, but the fact is that when the voltage drops to 0V (even after it was at some non-zero level), you will measure 0V across the inductor.

Quote
See the paradox now?.
No.

Quote
Now we place that ideal inductor across that ideal capacitor--what happens?
Can you measure a voltage across any part of that circuit?.
Of course you can.

poynt99

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Re: MH's ideal coil and voltage question
« Reply #383 on: May 14, 2016, 06:04:15 PM »
So what is the resistance of a 0V voltage source?
It is whatever the internal resistance of that source is.

Quote
....or if you connect an ideal capacitor, charged to 11V, to an ideal 1V voltage source, then what current will flow through the capacitor at t0?
Infinite current of course.

partzman

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Re: MH's ideal coil and voltage question
« Reply #384 on: May 14, 2016, 06:14:47 PM »
author=poynt99 link=topic=16589.msg484272#msg484272 date=1463239770]

And MHs question requires a o voltage at one point in the test. At that point,the current still flow's,as there is no resistance that impedes that current flow through the now looped(shorted )circuit. Now,as virpies stated,you cannot place a voltage across a shorted ideal inductor,and you have just stated that at 0 volts,then yes,the ideal voltage source is like a 0 ohm wire. So now how do you once again increase the voltage after 0 volt's,as you are now trying to place a voltage across an ideal shorted(looped) inductor?

See the paradox now?.

Brad


If a voltage source has resistance, this resistance is equivalent to being in series with an ideal voltage generating means. The voltage generating means is free to vary while still maintaining the same series resistance. The same exists in an ideal voltage source except the series resistance is zero.  So, you are correct that if we short the ideal voltage source, infinite current will flow. You are also correct that if the voltage source is zero when connected to an ideal inductor, any previous current flowing will continue without change.

However, we are allowed to vary the voltage generating means with a zero series resistance connected to an ideal coil which also has a zero series resistance and due to the nature of the self inductance of the coil, we will experience a change in current depending on the circuit values.

partzman

Edit-Changed to "connected to"

 

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #385 on: May 14, 2016, 06:15:32 PM »
<<<  Now,as virpies stated,you cannot place a voltage across a shorted ideal inductor,  >>>

Again, I believe I understand what Verpies meant because he responded to my questions about this.

A "shorted ideal inductor" just becomes a dead short circuit by his definition.  Half of the ideal turns cancel out the flux of the other half of the ideal turns and you end up with a "ideal zero ohm resistor with no inductance."

By his definition a "shorted ideal inductor" simply disappears from the circuit and looks like an ideal wire with zero inductance.

verpies

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Re: MH's ideal coil and voltage question
« Reply #386 on: May 14, 2016, 06:18:16 PM »
It is not "shorted" by a wire having no resistance.
An open inductor makes no more sense than a shorted capacitor and these components are complete conjugates of each other.

Open conductor cannot:
- store energy,
- maintain the level of flux that penetrates it,
- obey the Lenz law,
- generate magnetic flux density gradient that can do mechanical work. 

In fact an open inductor loses all of the properties of inductors except an empty voltage potential under varying flux conditions, that cannot even be measured without completing the inductor's circuit.

A straight length of ideal wire still has inductance and behaves accordingly. 
Yes, but what does it matter if k=1 ?

At no time is there any resistance in the ideal wire and at no time is there a requirement to short the inductor with another ideal wire to model it. 
The requirement to close the inductor's circuit is there every time you want an inductor to act as an inductor and do more than nothing.
That's why some good simulators will not even allow you to have an open inductor in a simulation and out of those they don't have problems with open capacitors.

I remind you that inductors and capacitors are complete conjugates (col: opposites) of each other.

tinman

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Re: MH's ideal coil and voltage question
« Reply #387 on: May 14, 2016, 06:28:31 PM »
That an ideal 0V source is equivalent to an ideal conductor? By deduction, that is what it is. It is common sense. It is impractical of course (except in simulators where it can be used to measure current), but it is a precise equivalent. Doesn't an ideal non-inductive conductor always have 0V across it? Well, so does an ideal voltage source set to 0V.


No.
Of course you can.

Quote
Regardless of whether there is a voltage or not,the fact that there is no internal series resistance means that the current can still flow through that voltage source unimpeded.
When we drop down to 0 volts (as stated in MHs question),the current will continue to flow,and at a steady state.

Yep, I'm not arguing against that.

Quote
I'm not sure what verpies is saying, but the fact is that when the voltage drops to 0V (even after it was at some non-zero level), you will measure 0V across the inductor.

But we agree that the inductor is now a shorted loop,and current is flowing through this circuit loop that includes the ideal voltage source(not to be confused with the ideal voltage,which now has a value of 0 volts)

To quote verpies
Quote
It is impossible to connect such voltage source across a shorted ideal inductor, because in such case  this voltage source would see a load, which does not have any resistance nor reactance

So now i ask--
At 0 volts we both agree that we are in a shorted(looped) condition,and a steady current is flowing through this shorted ideal inductor loop,and due to the 0 resistance value,it will continue to flow infinitely. How is this short removed just by turning up the voltage on the ideal voltage source?
If it is not removed,then how are you placing a voltage across this !now shorted! ideal inductor?

The paradox being-quote verpies-->It is impossible to connect such voltage source across a shorted ideal inductor.

Is there some sort of two kinds of !shorted!?.
A short is a short-is it not?.


Brad

tinman

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Re: MH's ideal coil and voltage question
« Reply #388 on: May 14, 2016, 06:30:48 PM »
I wonder if MH would have said that the current continued to flow through his circuit during the 0 volt time of his question?

I think not,as it would have been something he would never have seen,due to not understanding the meaning of Ideal.
But we will never know,as he never would answer his own question ::)


Brad

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #389 on: May 14, 2016, 06:38:30 PM »
I wonder if MH would have said that the current continued to flow through his circuit during the 0 volt time of his question?

I think not,as it would have been something he would never have seen,due to not understanding the meaning of Ideal.
But we will never know,as he never would answer his own question ::)

Brad

You think wrong. I intentionally stated that the voltage becomes zero volts to see if you guys would get that.

You're just being ridiculous when you say I do not understand the meaning of "ideal."

You guys apparently are slowly but surely getting up the learning curve.

Do you now accept the fact that an ideal voltage source can output a voltage waveform that changes in time?