Find the Thevenin equivalent circuit

In summary: On my calc (and I am a bit rusty on this) the voltage across R3 should be the voltage between terminals a and b, multiplied by (R1 + R2)/R1.2. You could also use ohms law and just multiply the voltage across R1 and R2 by the resistances.The resistance is indeed ##R_1||(R_2+R_3)##. Do you understand where you made the mistake in the previous post about the resistance?
  • #36
Yes. Re-read post #19.
 
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  • #37
The Electrician said:
##R_{123} ≠ R_{th}##

##R_{123}## is the sum of all the resistors.
Right, I got it now. ##V_{th} = 16.77v##
 
  • #38
Two years ago you said in a post: https://www.physicsforums.com/threads/independent-study-for-linear-algebra.880866/

"Hello, I just completed a first course in linear algebra and really enjoyed my studies. So much so that I want to pursue it more in the fall as an independent study, i am a EE major in college and was curious what directions might be useful for applications in that field."

Here's how you could solve the current problem quickly using linear algebra. Using a calculator that can do linear algebra makes it very convenient.

Thev.png
 

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  • #39
The Electrician said:
Here's how you could solve the current problem quickly using linear algebra. Using a calculator that can do linear algebra makes it very convenient.

View attachment 228374
How did you think to do this? the matrix algebra aside, you could have still done these calculations by hand. Where does this method come from?
 
  • #40
icesalmon said:
How did you think to do this? the matrix algebra aside, you could have still done these calculations by hand. Where does this method come from?

The first thing I said in the demonstration is "Perform a nodal analysis." It's just a nodal analysis with the two nodal equations shown as the rows of a 2x2 matrix.
 
  • #41
The Electrician said:
The first thing I said in the demonstration is "Perform a nodal analysis." It's just a nodal analysis with the two nodal equations shown as the rows of a 2x2 matrix.

Oh that portion of your post was blocked out:
thumbnail.jpg

I don't remember using nodal analysis this way, interesting. Thanks!
 

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  • #42
icesalmon said:
I believe I have ##I_L = 18mA\frac{3.9kΩ}{2.06kΩ}## so ##I_L = 34mA##
I, then, have ##V_{th} = 34mA*(2.7kΩ) = 91.8V##
The answer: ##V_{th} = 16.77V##

If ##R_{123} = R_{th} = 2.06kΩ## and ##I = 18mA## than I'm doing some arithmetic wrong here

The other have explained it well. Now try to get 16.77 using the second method. Draw the circuit the same way but instead of the blue part elements draw a line so it's short-circuited that way. You lose the ##R_1## element since it's parallel to the wire and then find the current that goes through the left part again. Then you multiply it with ##R_{th}## and you get the same result.
 
  • #43
diredragon said:
Draw the circuit the same way but instead of the blue part elements draw a line so it's short-circuited that way. You lose the ##R_1## element since it's parallel to the wire and then find the current that goes through the left part again. Then you multiply it with ##R_{th}## and you get the same result.
That I did, with this method I got ##I_L = 8.16A## so ##V_{th} = I_LR_{th}## = ##16.7V##. Thanks
 
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