Norton's Equivalent Circuit

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  • Thread starter johnsy1312
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Finding the Norton Equivalent circuit for the circuit attached


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

[itex]R_N = \frac{1}{\frac{1}{6}+\frac{1}{12}+\frac{1}{12}}=3ohms[/itex]


[itex]I'_N = \frac{12*2}{12+4}=1.5A, I''_N = 2A[/itex]

[itex]I_N = 2A - 1.5A = 0.5A[/itex]
 

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  • #2
gneill
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Finding the Norton Equivalent circuit for the circuit attached


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

[itex]R_N = \frac{1}{\frac{1}{6}+\frac{1}{12}+\frac{1}{12}}=3ohms[/itex]


[itex]I'_N = \frac{12*2}{12+4}=1.5A, I''_N = 2A[/itex]

[itex]I_N = 2A - 1.5A = 0.5A[/itex]

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

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