The Norton equivalent

EE 314 · Lecture 08, Norton's theorem · back to the slide

This is the circuit from the lecture, running. Change it and the answer changes with it:

Symbols again. Put in $I_N = 1.2$ mA and $R_N = 5$ k, add the same 1 k load, and it reads 1 V — the same load, the same answer, as the Thevenin form with 6 V and 5 k.