Working out Thevenin Equivalent

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eddysd
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Hi eddysd. Can you show your work so far? How did you arrive at Rth = 300Ω ?
 
I used 1/R = 1/R2 + 1/R3 and then added that to R1.
 
and why are you taking r2 as parallel to r3 ?

why are you NOT taking r1 as parallel to r3 ?

Seems to me like you need to go back and study what parallel and series MEAN.

EDIT: actually, I may be wrong about that. Perhaps what you actually need is to go back and study what a Thevenin equivalent IS. What becomes an open circuit? What becomes a closed circuit? What are the steps you are supposed to follow to GET the equivalent circuit?
 
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I think the idea is that the load is on the right of the Vout and not actually shown in that image judging by the lecturers previous notes, I apologise for not mentioning this in the original post. The lecturer told me that you follow the current round, and when there is a choice of direction, that part of the circuit is parallel, would this not make R3 parallel with R2?

EDIT: Posted this before noticing your edit!
 
eddysd said:
I used 1/R = 1/R2 + 1/R3 and then added that to R1.

Ah. Remember that you want to find the equivalent resistance when all sources are suppressed and you are "looking into" the network from its output end:

attachment.php?attachmentid=46130&stc=1&d=1334178675.gif
 

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gneill said:
Ah. Remember that you want to find the equivalent resistance when all sources are suppressed and you are "looking into" the network from its output end:

attachment.php?attachmentid=46130&stc=1&d=1334178675.gif

gneill, just as a general note, I ASKED him what becomes open circuit and what becomes closed circuit, so that he would THINK about it.

You TOLD him what the answer is so he doesn't HAVE to think about it.

In general it's more help to someone in the long run if you get them to think, not feed them answers.
 
OK, so would I add R1 and R2 then take the parallel with R3? Even if I have worked out Rth, since I don't know the current how would I work out Vth?
 
eddysd said:
I think the idea is that the load is on the right of the Vout and not actually shown in that image judging by the lecturers previous notes, I apologise for not mentioning this in the original post. The lecturer told me that you follow the current round, and when there is a choice of direction, that part of the circuit is parallel, would this not make R3 parallel with R2?

Another way to judge whether two components are in parallel or in series (or neither) is as follows:

Two components are in parallel when they share exactly two nodes.

Two components are in series when they are the only components that share a single node.
 
gneill said:
Another way to judge whether two components are in parallel or in series (or neither) is as follows:

Two components are in parallel when they share exactly two nodes.

Two components are in series when they are the only components that share a single node.

Thanks, that is a much better way of looking at it than my lecturer's!
 
So now I have worked out the Thevenin equivalent resistance to be 270Ω, but I still don't know how to work out the equivalent voltage. This isn't actually for coursework or homework, but is revision for an exam, so I would be grateful if anyone could give me a step by step method to work it out.
 
No, this is what I was trying to do, but can't think of a way without knowing the current