Why will i calculate the wrong current when i transform part of circuit to Thevenin circuit?

Here is the circuit in this question

The original circuit is in the left hand side,as we can see,there are two norton circuits in it,so i transform them to the thevenin equivalent circuit,just like the circuit in the right hand side.

Now i can use the KVL to calculate the $$\i\$$ value

KVL:$$\-12=4i+4i+24+4i+12,\$$ so $$\-48=12i\$$,i can know $$\i=-4\$$

However,the answer for the $$\i\$$ is $$\2\$$, i want to ask why?can anyone tell me the reason that why will calculate the wrong answer?

The solution:

$$\i_1=\frac{-12-(-24)-(-12)}{4+4+4}=2\$$

• Use Superposition theorem, to find the current due to each source and add all. If current due to only 12V source(6A,3A are open) I = -1A, current due to 6A current source alone (12V is shorted,3A is open) I = +4A, current due to 3A current source alone (12V is shorted,6A is open) I = -1A, Now add all current algebrically = +4A-1A-1A= 2A. Apr 28, 2020 at 16:33