# Will I1 current be same as the current through the controlled source here?

I was solving this question related to two port parameters. I am confused as to what will be $$\I_1\$$ when $$\V_1=0\$$ i.e. the input port is short circuit-ed, since the controlled current source, $$\gv_x\$$ will be parallel to the short. From one perspective, it seems like it will form a loop in which only $$\gv_x\$$ will flow, hence making $$\I_1 = gv_x\$$ and $$\Y_{12}=g\$$. And from another it seems like magnitude of $$\I_1\$$ will be difference of current through the $$\1\Omega\$$ resistor and the current drawn by the current source i.e. $$\|I_1|=|I_{1\Omega}-gv_x|\$$ as per KCL at the node connecting the controlled source, the 1 ohm resistor and the short. Another perspective I thought of was, since the two input terminals will get shorted, all the current flowing through 1 ohm resistor will flow through the short and the controlled source won't matter, since current prefers path of least resistance and an ideal current source has infinite resistance, so no current would be drawn by the current source.

I find circuit analysis quite confusing, I am not even confident as to whether all these scenarios I described above make complete sense.Any tips on how to get better at it?

When in doubt always stick to the basic principles you know.

You, correctly, mentioned KCL, that's the point!

So, why in the world should not apply to the black contour below?

$$I_1=g\,v_\mathrm{x}\pm I_\mathrm{R1\Omega}$$