Thevenin equivalent with 2.7 kΩ and parallel 4.7 kΩ, 3.9 kΩ resistors

[The Electrician]

There is probably a follow-on problem which wants the student to use the derived Thevenin equivalent to then determine the current through the (load?) resistor R when the 47k and 180V are then hooked back up.

 

[icesalmon]

The correct equation for the current divider is $I_L=I\frac{R_3}{R_{123}}$. Then $V_{th}=I_LR_1$.

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

 

[icesalmon]

The problem wants the Thevenin equivalent at the a,b terminals with the 47 kΩ and 180 volt source not present, as in post #12.

This is correct, $V_{th}$ is the voltage across that blue terminal, in this case $R = R_L$ and the 180V source need to be removed

 

[icesalmon]

It's more likely that the authors drew it this way so students learn that they need to remove any load before finding the Norton or Thevenin equivalent.

The outlined procedure in the text for thevenin's theorem explicitly states this must be done but on later problems the circuits are drawn more abstractly so the reader is almost forced to redraw the circuit in order to progress

 

[The Electrician]

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

$R_{123} ≠ R_{th}$

$R_{123}$ is the sum of all the resistors.

 

[The Electrician]

Yes. Re-read post #19.

 

[icesalmon]

$R_{123} ≠ R_{th}$

$R_{123}$ is the sum of all the resistors.

Right, I got it now. $V_{th} = 16.77v$

 

[The Electrician]

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.

 

[icesalmon]

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?

 

[The Electrician]

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.

 

[icesalmon]

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:

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

 

[diredragon]

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.

 

[icesalmon]

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