Norton's Theorem

Problem-Solving Strategy for applying the Norton's Theorem

STEP 1. It is essentially the same as that for Thévenin’s theorem, with the exception that we are dealing with the short-circuit current instead of the open-circuit voltage.

PROBLEM:

Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed. 

If there are any errors in solution, please let me know in the comments

SOLUTION:

In the figure above, we can get

$I_{sc}=8mA$

Then we can determine $R_{Th}$ by replacing the voltage source with short circuit and the current sources with open circuits. Its value is

$R_{Th}=\frac{1}{\frac{1}{3k}+\frac{1}{6k}+\frac{1}{2k}}=1k\Omega$

Therefore, $I_o=8mA\times\frac{1k}{1k+4k}=1.6mA$

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