Why is my circuit nodal analysis not giving the correct I_sc value?

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SUMMARY

The discussion focuses on the challenges faced in calculating the short-circuit current (I_sc) in a circuit using nodal analysis. The user attempted to solve for I_sc by applying super node analysis at nodes 1 and 4, but arrived at an incorrect value of 8mA. The correct approach involves recognizing that node 2 is shorted to ground, leading to the conclusion that V2 equals 0. The user utilized Wolfram Alpha to verify their results, indicating the need for accurate application of Kirchhoff's Current Law (KCL) in the analysis.

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Homework Statement


Capture.png


The Attempt at a Solution


Essentially I am having trouble with solving for I_sc when I remove the load and replace it with a wire.

I used a super node at 1 and 4 and also in my diagram I have defined the currents with directional arrows. I have been on this question for a long time and cannot get the correct answer which is I_sc = 8mA. I know it is wrong because I plugged it into wolfram alpha and ended up with the results as seen below.

From KCL:
photo_18.jpg


Where V1=a, V2=b, V3=c, I_sc=x and solving it gets:
idtsO.png
 
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V2 = 0, because node 2 is shorted to ground.
 

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