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

1. Apr 6, 2014

### Vishera

Does it matter if the resistor is on top or below a voltage source after doing a source transformation?

According to this picture from wikipiedia, the resistor is on top of the voltage source:

But even if the resistor (Z) was on the other side of the voltage source, it would still be equivalent to the current source parallel to Z, right? Because the current-voltage characteristic is still the same?

2. Apr 6, 2014

### mjhilger

Totally the same. Most designers use the lower line as the lowest voltage -V or gnd; as such the resistor is shown on the high side the the voltage source, but the results between A & B are just the same as you determined.

3. Apr 6, 2014

### phinds

Your choice of description ("on top or below") seems to me to be not only non-standard but inaccurate. you are talking about a series circuit and its equivalent parallel circuit. Rather than saying the resistor in the first diagram is "on top", it is better described as "in series with".

4. Apr 6, 2014

### jim hardy

it is so difficult to make ourselves slow down and speak unambiguously...
A question well stated is half answered.

"on top (above?) or below voltage source" refers to before the transformation. After the transformation it is alongside a current source.

do you mean: "if i started with the resistor drawn below instead of above the voltage source, that is between V- and B instead of between V+ and A, would the result of my transformation be the same, as shown on the right of my drawing?

Last edited: Apr 6, 2014
5. Apr 6, 2014

### davenn

incorrect

in the left diag, the resistor is in series with the load across A and B

in the right diag, the resistor is in parallel with the load across A and B

Dave

6. Apr 6, 2014

### phinds

The two circuits are legitimate equivalents so I'm not sure that distinction matters. The load sees identical circuits. The second is the Norton equivalent of the first (or, if you prefer, the first is the Thevenin equivalent of the second)

7. Apr 6, 2014

### mjhilger

I took the OP at the position that he understood the series Thevenin and parallel Norton equal. I understood the question to be "Does the series component ordering of the resistor and voltage source matter in the Thevenin circuit?" The answer to that question is NO. I took him to understand that the parallel Norton circuit was as stated and he just wanted clarification on the ordering. Hopefully if the OP was asking a different question, he will restate it more clearly.

8. Apr 7, 2014

### Vishera

Here is what I originally meant in my question.

I know from Wikipedia that these two circuits are equivalent:
1.
2.

However, I am unsure if these two circuits are equivalent:
1.
2.

9. Apr 7, 2014

### phinds

yes, as far as the outside world (the load) is concerned, they are equivalent, but that does assume you take your ground point as being to the right of the resistor in the series circuit.

This is more clear if you simply draw the resistor vertically above or below the voltage source.

10. Apr 7, 2014

### jim hardy

Nice clarification, Vishera !

Last edited: Apr 7, 2014
11. Apr 7, 2014

### davenn

I really must be having a brain fade !

that really doesn't make sense to me

Dave

12. Apr 7, 2014

### phinds

Dave, I don't really know how to respond to that. Do you understand what Thevenin and Norton equivalent circuits are?

13. Apr 7, 2014

### mjhilger

Dave, intro to Electrical Engineering introduces Thevenin & Norton circuits. Thevenin uses ideal voltage source with 0 (zero) impedance, Norton uses ideal current source with infinite impedance. So when you determine the impedance for a Thevenin you short any voltage source and calculate the impedance. For the Norton you open circuit any current source and calculate the impedance. If you follow this on the diagrams above you will see that between A & B the impedance is the same. The Norton current is the Thevenin voltage divided by the impedance; or the Thevenin voltage is the Norton current * impedance. This is from early introduction to electrical engineering analysis for circuits.

14. Apr 7, 2014

### davenn

Yeah I do

was just having a "senior moment" Am on the same page now

mjhilger .... sorry for the disagreement

cheers
Dave

15. Apr 7, 2014

### jim hardy

the old technician's riddle -

You are given a sealed black box and told only "it contains an equivalent source".
The box has two terminals with some voltage between them.
Without either opening or X-raying the box, but anything else goes
How do you figure out whether it contains a Thevenin or a Norton ?

16. Apr 7, 2014

### meBigGuy

I know the answer to this. 42! Am I getting warm?

Last edited: Apr 7, 2014
17. Apr 7, 2014

### jim hardy

:thumbs:

42, you say ?

that's my lotto series - 42, 21, 14, 7, 3, 2

18. Apr 7, 2014

### phinds

HA ! I never heard that one before, but I like it.

spoiler altert
Put a very low-ohm, low power resistor across them. A Thevenin equivalent will just cause the resistor to have a very small voltage across it whereas the Norton equivalent will make it explode (or at least burn out very quickly).

19. Apr 8, 2014

### jim hardy

mBG was warmer....

20. Apr 8, 2014

### meBigGuy

That's an interesting sequence, but I'm terrible at decoding sequences. I would expect to see a 6.

21. Apr 8, 2014

### mjhilger

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. Apr 8, 2014

### jim hardy

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 [STRIKE]be[/STRIKE] editmake up for my missing factor,

Last edited: Apr 8, 2014
23. Apr 8, 2014

### mjhilger

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. Apr 8, 2014

### jim hardy

mj you've got it .....

what would your method show with box unloaded?

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. Apr 8, 2014

### mjhilger

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

I like puzzles!

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