- #36
The Electrician
Gold Member
- 1,391
- 207
Yes. Re-read post #19.
Right, I got it now. ##V_{th} = 16.77v##The Electrician said:##R_{123} ≠ R_{th}##
##R_{123}## is the sum of all the resistors.
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 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
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 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.
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
That I did, with this method I got ##I_L = 8.16A## so ##V_{th} = I_LR_{th}## = ##16.7V##. Thanksdiredragon 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.