Need Help Understanding Thevenin's Theorem Example

[guerillary]

I just started learning electronics and am completely stumped by a circuit analysis example using Thevenin's Theorem on page 77 of "Practical Electronics for Inventors", Third Edition.

In the attached book excerpt, the circuit diagram looks to me as if resistors R1 and R2 are in series, however the description of proper analysis (highlighted) says the resistors are in parallel.

Am hoping someone here can please help me understand why the resistors are in parallel as opposed to in series.

THANKS!

 

[erikl]

Quite simple: check the picture text (Replace source with short, ...) and/or read the text in front of your highlighted text thoroughly: ... replacing the dc source (V_BAT) with a short ...

 

[guerillary]

Thanks for the quick reply yet I'm such a beginner that your answer is still hard for me to understand. Even after replacing the source with a short, resistors R1 & R2 still look like they are in series with each other to me.

Might there be an even more basic / diluted / novice way to explain the answer to me?

Sorry for any inconvenience, I appreciate your help.

 

[Audioguru]

If the battery is shorted then one end of each resistor connects to each other and the short connects their other ends together. Then they are in parallel.
A battery usually has a VERY low resistance like a short anyway so the resistors are in parallel for AC signals.

 

[guerillary]

Thanks Audioguru but I'm still at a loss to understand this. If the very definition of resistors in series is that they are connected to each other at one end, how are said resistors suddenly in parallel if their connection to each other is unchanged, yet there is a voltage short on their each of their ends that is not connected to one another?

From what I understand of resistors in parallel, both ends of the resistors are connected to the + & - voltage rails, so no ends of those resistors are directly connected together in series.

I know I'm the one who is misunderstanding here but am just struggling to understand why - why resistors directly connected together at one end are suddenly / considered to be in parallel to each other when there's a voltage short on their other ends.

 

[ads-ee]

It's because the A-B terminals where you are finding the Thevenin equivalent is measured is across a single resistor, therefore the Thevenin equivalent of the resistors look like they are in parallel. i.e. shorting the supply makes the A-B look at both resistors from both ends of both resistors (together in parallel). You would have to take the A-B across one end of each resistor with the other end of the resistors connected together to make it series.

Maybe another source for calculating this might help.

Here are a couple of my ASCII art schematics

[FONT=Courier New]
Series
 -----vvvv----vvvv-----o A
 |     R1      R2
---
 -
 |
 ----------------------o B

Parallel (notice how I moved the supply to the bottom :-))
 ----------------------o A
 |           |
 >           >
 > R1        > R2
 >           >
 |           |
 -----|'|'-------------o B

Now short the supply as the text in you book says to determine the equivalent resistance...
[/FONT]

 

[guerillary]

Thanks so much ads-se, the pdf is especially helpful to me. Although I'm still not 100% there, I realize the big issue is my lack of understanding source transformations and their effect on whether resistors are in series or parallel to a resulting voltage / current source. The book I'm reading "Practical Electronics for Inventors" (3rd Ed.) barely covers this topic, which apparently seems important to a solid grasp of Thevenin's (and Norton's) theorems. Now I should be en route to making sense of this all. Much appreciated.

 

[ark5230]

Redrawn from above suggestions

 

[guerillary]

Yes, thank you that re-drawing helps a lot!