In this exercise is asked to calculate "I" using the overlap theorem.
Something is wrong with my resolution, because the final answer is wrong.
Please help my find out what it is, i posted a photo of the resolution. TY.

Mod: rotated image and cleaned up.E

 

In your second diagram, the calaculation shows 3A flowing in 6Ω. Is that right?

 

In your second diagram, the calaculation shows 3A flowing in 6Ω. Is that right?

Yes sir.

 

So no part of that 3A flows through the 2Ω and 8Ω resistors?

 

Overlap theorem ?

Just for my information, long time since I have done this,

Is this the one where you one by one replace current sources by an open circuit, and voltage sources by a short,

then sum the current / volts in the circuits ?

 

So no part of that 3A flows through the 2Ω and 8Ω resistors?

Some part i think. But what are you trying to say?

 

Overlap theorem ?

Just for my information, long time since I have done this,

Is this the one where you one by one replace current sources by an open circuit, and voltage sources by a short,

then sum the current / volts in the circuits ?

Yes it is.

 

Some part i think. But what are you trying to say?

What I am trying to say is that in the second calculation you calculaye 18V as if all of the 3A is flowing in the 6Ω resistor but as the current is shared with the other branch then that 18V is incorrect.