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Does it matter if the resistor is on top or below a voltage source... 
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#20
Apr814, 01:51 AM

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#21
Apr814, 09:15 AM

P: 93

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! 


#22
Apr814, 09:57 AM

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PF Gold
P: 3,686

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 


#23
Apr814, 10:06 AM

P: 93

Spoiler
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. 


#24
Apr814, 10:31 AM

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PF Gold
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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 


#25
Apr814, 10:44 AM

P: 93

Spoiler
You are right of course. I just thought without a starting point, you don't really know the change.
I like puzzles! 


#26
Apr814, 09:26 PM

P: 1,084

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? 


#27
Apr1514, 05:29 PM

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Thanks
P: 5,363

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


#28
Apr1614, 12:39 AM

Sci Advisor
PF Gold
P: 3,686

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 capacitorstretcher, of course. 


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