MH's ideal coil and voltage question

[tinman]

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]

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]

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.

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.

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.

See the paradox now?.

No.

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]

So what is the resistance of a 0V voltage source?

It is whatever the internal resistance of that source is.

....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]

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"