Please ignore the poor diagram as I just used paint to draw the rough circuit.

So here's the current source:

which I transformed into voltage source:

Did I do it right?

 

the ampmeter is not placed across the load, the first drawing is wrong. in the second drawing, the voltmeter is in series with the load, that is wrong. measure3 amps in series with the load, and volts across the load.

 

This looks like homework-related, so probably should have been posted in the Homework Help forum.

The quality of the schematic is more than adequate -- but that all people would post something as good (or at all).

Also, thank you for posting your attempt at a solution. Very rare and very appreciated.

Finally, yes, your transformation is correct. The resistor could have gone in either the top or the bottom (or the left) branch. Also, you could have swapped the polarity of the voltage supply and then specified the voltage as -12V. All of them are equal. By convention, a circuit such as this would typically be drawn with a voltage source that has the negative terminal at the bottom and (and so would have a value of -12V) and the resistor would be in the top branch. But that is purely by common convention because that's what people are used to seeing. It is technically correct as you have drawn it.

 

the ampmeter is not placed across the load, the first drawing is wrong. in the second drawing, the voltmeter is in series with the load, that is wrong. measure3 amps in series with the load, and volts across the load.

There's no meters in either circuit. The first is a current source in parallel with it's output resistance and the second is an equivalent voltage source in series with its output resistance.

 

This looks like homework-related, so probably should have been posted in the Homework Help forum.

The quality of the schematic is more than adequate -- but that all people would post something as good (or at all).

Also, thank you for posting your attempt at a solution. Very rare and very appreciated.

Finally, yes, your transformation is correct. The resistor could have gone in either the top or the bottom (or the left) branch. Also, you could have swapped the polarity of the voltage supply and then specified the voltage as -12V. All of them are equal. By convention, a circuit such as this would typically be drawn with a voltage source that has the negative terminal at the bottom and (and so would have a value of -12V) and the resistor would be in the top branch. But that is purely by common convention because that's what people are used to seeing. It is technically correct as you have drawn it.

Thank you for the reply!
Can i know that if i rearrange the circuit to this:

Would it still be right?
My intuition says that it wouldn't be but this guy at 21:30 says that it is.

 

Yes, this transformation is also correct.

 

Yes, this transformation is also correct.

But isn't in that case current can't flow to the resistor because + side is on the opposite side of the resistor

 

But isn't in that case current can't flow to the resistor because + side is on the opposite side of the resistor

I don't get what you are trying to say?
If you short the circuit terminals the current will flow for sure. And remember that the internal circuit behavior of a circuit does not matter here. The "outside world" behavior is important here.
And this two simple circuits behave the same way from the "outside world" point of view.