Is Your Calculation of Norton's Current Correct in the Circuit Analysis?

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SUMMARY

The discussion focuses on calculating Norton's current in a circuit analysis context. The user calculates the Norton equivalent resistance, R_N, as 3 ohms using the formula R_N = 1/(1/6 + 1/12 + 1/12). The user also computes two current values, I'_N as 1.5A and I''_N as 2A, leading to a final Norton current, I_N, of 0.5A. The inquiry seeks clarification on the logic behind the calculation of I'_N.

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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:

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
 

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johnsy1312 said:
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## ?
 

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