#### PG1995

##### Full Member level 5

In the following linked pictures I have marked the sections which I have found more troublesome with numbers (1, 2, 3, 4, 5). The aim is to find Thevenin equivalent. Please help me. It would be really kind and nice of you. Thank you

1: http://img843.imageshack.us/img843/4292/norton1.jpg

2: http://img847.imageshack.us/img847/2450/norton2h.jpg

At "1" in link #1, it says

**(Vo - 0)/2**. It is for the branch in which Ix is flowing. Ix is flowing upward - from the ground toward Vo which simply means that Vo is at lower potential than the ground. Therefore, it should be (in my view):

**(0 - Vo)/2**. Do you see the problem? I understand that if we do that the nodal equation will also change.

At "2" it says

**Vo = -4 V**. What does this mean? In what situation does the current flows from ground to some other point (at -ve potential) in real world?

At "3" having

**Rth = -4 ohm**is completely incomprehensible for me. Please help me with it. I know the author gives some explanation. But one ohm resistance is equivalent one volt used per coulomb. So having a -ve sign with resistance doesn't make sense because it's directionless quantity.

At "4", can we take the direction of current CCW instead of CW? (To see what is author doing here in this circuit please have a look on link #2).

At "5", why has the author used

**-4i**when he has used +ve sign for

**+9i**?