Norton's Equivalent Circuit

[johnsy1312]

Finding the Norton Equivalent circuit for the circuit attached


I attempted this solution but i am unsure if my Norton's current is correct:

$R_N = \frac{1}{\frac{1}{6}+\frac{1}{12}+\frac{1}{12}}=3ohms$


$I'_N = \frac{12*2}{12+4}=1.5A, I''_N = 2A$

$I_N = 2A - 1.5A = 0.5A$

 

[gneill]

Finding the Norton Equivalent circuit for the circuit attached


I attempted this solution but i am unsure if my Norton's current is correct:

$R_N = \frac{1}{\frac{1}{6}+\frac{1}{12}+\frac{1}{12}}=3ohms$


$I'_N = \frac{12*2}{12+4}=1.5A, I''_N = 2A$

$I_N = 2A - 1.5A = 0.5A$

Can you explain the logic behind your Norton current calculations? In particular, what motivates your calculation of $I'_N$ ?