the questions are in the picture

please help .

the main question is :

how did we know that i = 5A and not i = -5A ?

is the case in the middle -5A ?

 

Your diagram does not explicitly define what the current i is in relation to the circuit. Does it mean the "conventional" current flowing in a clockwise direction around the loop? The indicated polarity of voltage Vr on resistor (?) element A would suggest this is the case.

Again, it is not explicitly clear what Vx means on your circuit. Does it mean that voltage which would be measured with a voltmeter connected across the source with the positive meter probe connected to the uppermost terminal of the source and the negative probe connected to the lowermost terminal of the source?

If the answer to both the aforementioned questions is true then i would be +5A in the [leftmost] circuit. These are the magnitude & polarity values i which will satisfy Kirchoff's voltage law for the circuit.

 

Does it mean the "conventional" current flowing in a clockwise direction around the loop?

the conventional .

Does it mean that ....

yes .

now if we flipped only the source at the left side , would that make i = -5A ?

thanks a lot for your help .

 

Yes- If you flipped only the source then the current i would be -5A. The voltage across the element A would now have the opposite polarity to that shown on the original figure. I'm unsure whether the indicated Vr of 9V was given in the original question or you were given A is a 1.8 ohm resistor?