- #1
The_Lobster
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There's a problem in my book where it asks me to find the Thevenin and Norton equivalent circuit of the attached circuit, as viewed from AB. I believe I know the answer, but just had to check to be sure this is not a trick question?
Isn't the Thevenin eq. circuit as seen from AB exactly the one drawn, without R1 and R2? And the Norton one just a source transformation from the one we see?
Thanks in advance!
Isn't the Thevenin eq. circuit as seen from AB exactly the one drawn, without R1 and R2? And the Norton one just a source transformation from the one we see?
Thanks in advance!