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Does it matter if the resistor is on top or below a voltage source...

by Vishera
Tags: matter, resistor, source, voltage
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jim hardy
#19
Apr8-14, 12:17 AM
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mBG was warmer....
meBigGuy
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Apr8-14, 01:51 AM
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Quote Quote by jim hardy View Post
that's my lotto series - 42, 21, 14, 7, 3, 2
That's an interesting sequence, but I'm terrible at decoding sequences. I would expect to see a 6.
mjhilger
#21
Apr8-14, 09:15 AM
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Spoiler

First identify the impedance of the Black Box. Then match that impedance with a load and measure the temp of the box at steady state. Then change the load resistor to a value say half what it was before. If it is a Thevenin source, then the source impedance will dissipate more power than the load power and you should see the box temp rise. A Norton impedance sharing the same voltage as the load will only dissipate a lower power than the load and the temp should go down a little.

The mBG was warmer was a nice clue!
jim hardy
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Apr8-14, 09:57 AM
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wow, mj , - that'd work.

Could you do it in a single step?

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

mBG re missing 6:
when i lived in Florida i'd buy one ticket a week in their lottery which was a six number draw.
So i figured that would be editmake up for my missing factor,
mjhilger
#23
Apr8-14, 10:06 AM
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I suppose if you short the output, the Norton box will either stay same temp or cool as no power would be dissipated by its impedance. While the Thevenin would push all power through its impedance and certainly warm. Either way I think you would have to know the starting temp and an ending temp, so two steps?

I'm making an assumption that our power sources are ideal and stay same temp regardless the load.
jim hardy
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Apr8-14, 10:31 AM
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mj you've got it .....

what would your method show with box unloaded?

Spoiler
assume you have at hand a good calorimiter

a single measurement will show whether the internal resistance is being warmed by a Norton current source in parallel,
or not being warmed by a Thevenin voltage source in series which cannot push any current through the infinite external load


good thinking ! I didn't get that far first time...

old jim
mjhilger
#25
Apr8-14, 10:44 AM
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You are right of course. I just thought without a starting point, you don't really know the change.

I like puzzles!
meBigGuy
#26
Apr8-14, 09:26 PM
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My favorite question regards connecting two 1F capacitors together.

One capacitor is charged to 2V, the other discharged. When you connect them together, one would assume the resulting voltage would be 1 volt. But, if that were true, the system energy (1/2 C*V^2 each cap), would have to reduce from 2 joules (on one capacitor) to 1 joule (1/2 on each capacitor). How can that be?
NascentOxygen
#27
Apr15-14, 05:29 PM
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Quote Quote by jim hardy View Post
mj you've got it .....

what would your method show with box unloaded?
Okay....so in summary, whenever you needed to use the Norton battery you'd find it had already gone flat!

So the company manufacturing Norton batteries would go out of business, customers would all be moving to buy Thévenin cells for their excellent longevity?
jim hardy
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Apr16-14, 12:39 AM
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Quote Quote by meBigGuy View Post
My favorite question regards connecting two 1F capacitors together.

One capacitor is charged to 2V, the other discharged. When you connect them together, one would assume the resulting voltage would be 1 volt. But, if that were true, the system energy (1/2 C*V^2 each cap), would have to reduce from 2 joules (on one capacitor) to 1 joule (1/2 on each capacitor). How can that be?
Hmm ... if i visualize a parallel plate capacitor
C = [itex]\frac{εA}{D}[/itex]
and paralleling two of them is just like joining the two sets of plates at an edge , doubling the area...
Clearly charge is conserved but not energy.
must be like a physical inelastic collision where momentum is conserved but not energy ?


Thought experiment:
What would happen if you grabbed one capacitor's set of plates and slowly stretched them to twice their initial area? Using an insulated capacitor-stretcher, of course.
jim hardy
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Apr16-14, 12:51 AM
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Quote Quote by NascentOxygen View Post
Okay....so in summary, whenever you needed to use the Norton battery you'd find it had already gone flat!

So the company manufacturing Norton batteries would go out of business, customers would all be moving to buy Thévenin cells for their excellent longevity?


good one !
I'll think of it every time i see an "Energizer Bunny" commercial.


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