I have to calculate the current I and the voltage between a and b using the superposition theorem.
I can solve this directly by calculating the voltage between points, but I have to do it with that theorem. Problem is that I can't figure out where does the current go in each case (shorting the $6V$ source or the $20V$ source), and everything I've tried gives me the wrong results.
If I short the $20V$ source, I'm pretty sure the current goes up from the $6V$ source, then right past the $5 \ \Omega$ resistor, then clockwise through the short circuit, then passes from $b$ to $a$ through the $2 \ \Omega$ resistor and back to the $6 V$ source. So I calculate the current as $ I'=\frac{6 V}{2 \Omega}=3A $ and $ V_{ab} = 6 V $. Easy but not sure if correct.
Now when I short the $6V$ source instead, I think the current goes down from the source, then to the right, then counterclockwise right past the $5 \ \Omega$ resistor, down to the node $a$, where it splits in two: one part goes through the $2 \ \Omega$ resistor and then reaches the source through the negative side, while the other part goes through the $3 \ \Omega$ resistor and then reaches the source through the positive side. Is this correct? I tried to apply Kirchhoff's laws but I got the wrong result (I need $I''$ to be equal to $12 A$ so that $ I = 12A - 3A = 9A $, which is the correct result).
It's been a long time since I last saw this topic, I may be too rusty and overlooking something trivial.
Edit: I drew the paths of the current I described above: when the $20V$ source is shorted on the left, and when the $6V$ source is shorted on the right. If I understand correctly, the current will go through the short circuit only whenever there is another branch.