Thevenin and Norton Equivalents

I was trying to solve a simple electronics Thevenin/Norton equivalent problem, and I understand the concept as I solved many others correctly. However, I seem not to understand how this one can be solved, since my 3 approaches return different results.

I tried to explain the question and my reasoning as clearly as possible, as you can see in the image below. I am hoping you can point out what is wrong and right about those reasoning. Your theory is sound, just a few minor errors.

For your 1st approach, to compute the norton current, you short the two terminals of the circuit you wish transform (A & B for your circuit). The short circuit current is the same current through the output 5 V voltage source. When the output is shorted, the intermediate node you labeled as Vy in later analysis must be 5 V. Apply KCL at you will find $$I_N = -4 \textrm{ mA}$$.

For the 2nd approach, your KCL at node $V_y$ is correct just an aritmetic error on the solution. $$V_y = 3\textrm{ V}$$ and the output thevin voltage is then $$V_{th} = -2 \textrm{ V}$$

For the third approach the two 1k resistors in parallel form an equivalent resistance of 500 ohms not 2 kOhms, which yields the same thevin voltage as $$V_{th} = -2 \textrm{ V}$$

The complete Norton/Thevin solution is then, simulate this circuit – Schematic created using CircuitLab

• Hello and thank you for your answer. I understood all you said but still can't figure out what is wrong in reasoning like this: 'in approach 1, current from source is divided between the 2 resistors and the wire (R=0), so it fully 'prefers' the wire and I_AB = 1 mA. What is wrong in that? – J.O'Connor May 27 '17 at 20:36
• Hi that is correct if you were to put a short across a current source with parallel resistance, all the current would flow through the short. The only piece you missed is that the 5V source causes -5 mA to flow through it while the current source causes +1 mA. The net sum is -4 mA. – sstobbe May 27 '17 at 20:43