Joule Thief 101

[Johan_1955]

Not that it matters but, back to the shopping cart for a moment...

The frictionless cart will still resist any movement because it has mass, especially if it is full as stated.  Yes, it will move more easily due to being frictionless but, F=MA still applies.

I realize that MH was trying to make a point about electronics so my post makes no difference, I just did not want folks to forget about inertia in the cart example.

Bill

Bill, Imagine the friction with the mother in Law from Sencellus in the trolly;-))


Mass meat's ..................!?

[MileHigh]

Yes MH,, these things are tough aren't they.

Are we only looking at your arms and shopping cart?? not if you have to speed up because it is accelerating away from you,, if it only moved and you only had to move your arms the same distance to keep applying the force,, then we are staying with just looking at your arms and the shopping cart.

My response to your questions in post #2590:

<<< This is a real-life example.  The person moves along with the shopping cart as it accelerates. >>>

I know, it's too much to expect that you would remember a response to your own question that was made earlier today.

And you are still worried about arms and legs?  Are the wheels made of rubber or steel?  Are there Brussels sprouts in the shopping cart?  It might be critical information.

Abandon hope.

MileHigh

[MileHigh]

For reference, this is the harder version of the question I answered that Brad made reference to in post 2607:

<<<
Here is the harder version of the question and the answer:

You have an ideal voltage source and an ideal coil of 5 Henrys.  At time t=0 seconds the coil connects to the ideal voltage source.  The voltage source waveform is 20*t^2.  So as the time t increases, the voltage increases proportional to the square of the time.

The question is what happens starting at t = 0

The answer:

The current through the ideal coil starts from zero at time t = 0 and then increases with this formula:  i = 1.33*t^3.

Time..........Voltage.........Current
0...............0.................0
1...............20...............1.33
5...............500.............166.67
10.............2000............1333.33
20.............8000............10666.67
50.............50000..........166666.7

Brad, you need to try to get up the learning curve such that you get to the point where you come back and acknowledge the answer given above is correct.
>>>

And poor Brad thinks I am talking about "DC current" when I am talking about a rising voltage waveform proportional to t-squared and the resultant rising current waveform that is proportional to t-cubed.  The mind boggles.

Quote Brad from post #2607:

<<< Have you lost your marbles MH ?
This whole thing you have been peddling is about how you can place a voltage across an ideal coil,and a DC current will flow through that coil.See below >>>

What's even more of a joke is when I started the process of answering the easier question these were Brad's responses:

<<<
You are the epic failure others claim you to be.
You are a total disaster
Your a fraud.
You epic failure.
You are now the laughing stock of this forum.
>>>

Brad:  Everything you read in your own quoted text above in reality applies to you yourself.  You have been bluffing your way through this whole thing.  It's a farce and a fiasco.

MileHigh

[Magluvin]

For reference, this is the harder version of the question I answered that Brad made reference to in post 2607:

<<<
Here is the harder version of the question and the answer:

You have an ideal voltage source and an ideal coil of 5 Henrys.  At time t=0 seconds the coil connects to the ideal voltage source.  The voltage source waveform is 20*t^2.  So as the time t increases, the voltage increases proportional to the square of the time.

The question is what happens starting at t = 0

The answer:

The current through the ideal coil starts from zero at time t = 0 and then increases with this formula:  i = 1.33*t^3.

Time..........Voltage.........Current
0...............0.................0
1...............20...............1.33
5...............500.............166.67
10.............2000............1333.33
20.............8000............10666.67
50.............50000..........166666.7

Brad, you need to try to get up the learning curve such that you get to the point where you come back and acknowledge the answer given above is correct.
>>>

And poor Brad thinks I am talking about "DC current" when I am talking about a rising voltage waveform proportional to t-squared and the resultant rising current waveform that is proportional to t-cubed.  The mind boggles.

Quote Brad from post #2607:

<<< Have you lost your marbles MH ?
This whole thing you have been peddling is about how you can place a voltage across an ideal coil,and a DC current will flow through that coil.See below >>>

What's even more of a joke is when I started the process of answering the easier question these were Brad's responses:

<<<
You are the epic failure others claim you to be.
You are a total disaster
Your a fraud.
You epic failure.
You are now the laughing stock of this forum.
>>>

Brad:  Everything you read in your own quoted text above in reality applies to you yourself.  You have been bluffing your way through this whole thing.  It's a farce and a fiasco.

MileHigh

"And poor Brad thinks I am talking about "DC current" when I am talking about a rising voltage waveform proportional to t-squared and the resultant rising current waveform that is proportional to t-cubed.  The mind boggles."


What is an ideal voltage source? Is it not a source that can hold its voltage no matter the load???  So if the ideal inductor is put across the ideal voltage source there should be no rising voltage waveform, as you put it, because the ideal voltage source will maintain its voltage no matter the load. No?? And if the ideal voltage source, as you put it, is directly across the ideal inductors leads.  The only thing that could possibly rise is current, that is if current flows at all considering the argument of cemf being ideal also, that Brad, AC and I have posed issue with.

Mags

[MileHigh]

"And poor Brad thinks I am talking about "DC current" when I am talking about a rising voltage waveform proportional to t-squared and the resultant rising current waveform that is proportional to t-cubed.  The mind boggles."


What is an ideal voltage source? Is it not a source that can hold its voltage no matter the load???  So if the ideal inductor is put across the ideal voltage source there should be no rising voltage waveform, as you put it, because the ideal voltage source will maintain its voltage no matter the load. No?? And if the ideal voltage source, as you put it, is directly across the ideal inductors leads.  The only thing that could possibly rise is current, that is if current flows at all considering the argument of cemf being ideal also, that Brad, AC and I have posed issue with.

Mags

The issue of an ideal voltage source varying in time has already been discussed several times on this thread.  Just look at the original question, the ideal voltage source varies with time.  I don't know how ideas like "an ideal voltage source cannot change in time" take hold but apparently they do.  Somebody says it and nobody thinks to question it.

<<< The only thing that could possibly rise is current, that is if current flows at all considering the argument of cemf being ideal also >>>

We are beating an old horse to death at this time.  It's a second strange idea, it's baffling and it and the "fixed ideal voltage source" idea would not last more than eight minutes apiece on a real electronics forum before they were sliced to pieces.

Look at this clip that discusses ideal inductors and brush up on your inductor concepts:

https://www.youtube.com/watch?v=B8CPGiK59f8