Electronic – Kirchhoff’s laws with current source in a circuit

Whichever gives an easy set of equations. If you are doing resistor networks, count if there are more loops or more nodes. KVL if there are more loops, KCL if there are more nodes.

In more advanced circuits, like transistors, there is normally a very specific mode that lends itself to your problem space.

I think you may have an issue with one of the signs. Going around the loop, I get:

12 - 4 i - 2 Vo + 4 - 6 i = 0

Looks like you got the sign on the 6i term flipped.

So, if we take Vo = -6 i, you can calculate:

12 - 4 i + 12 i + 4 - 6 i = 0
16 + 2 i = 0
i = -8
Vo = 48

If you're stuck on understanding how to write this equation, the way to remember it is 'what goes up must come down'. Sum up the voltage across each element as you go around the loop. Since you go into the - side of the 12 volt source, you gain 12 volts. Then you pass through a resistor, and you lose 4 i volts. Then you go the wrong way through the dependent source and you lose 2 Vo volts. Then you go through the 4v source and gain 4 volts. Then you go through the final resistor and lose 6 i volts. You're back where you started, so all of that adds up to zero. Then you figure out what Vo is - if the current is flowing in the direction of the arrow, it will create a voltage drop in opposition to how Vo is marked, so Vo = -6 i. After that, finding i is just algebra.