Discussion in 'Homework Help' started by PG1995, Jul 2, 2011.

1. ### PG1995 Thread Starter Distinguished Member

Apr 15, 2011
813
6
Hi

This is confusing me very much. I would try to ask it now and try my best to convey my confusion. Please have a look on the following link: http://img708.imageshack.us/img708/3581/img0002ri.jpg

The scan also has my comments there which are an attempt to explain what is troubling me.

The Circuit #2 could be found here in full context here: http://img843.imageshack.us/img843/4292/norton1.jpg (Original link)

As you see the author in the "Original link" says that the circuit on left of the terminals a-b is supplying power but I don't get it. From the viewpoint of current source Io the terminal "a" should be +ve because current source is pointing toward it; that means its +ve terminal is connected with "a". But from the viewpoint of the circuit on the left of terminal a-b the terminal "a" is at -4V potential which I don't get. Do you see my confusion?

Thank you very much for all the help.

2. ### blah2222 Distinguished Member

May 3, 2010
582
38
Well, from the example if you use a 1A current source for Io, then this would produce a -4 V voltage at Vo ('a'), with respect to ground ('b'). The current source is still passing current from 0 V to -4 V so the potential around the current source is +4 V.

Hope that helped.

3. ### PG1995 Thread Starter Distinguished Member

Apr 15, 2011
813
6
Thank you, blah.

I'm sorry but I don't get it. I think the current source is passing current from -4V to 0V. The current source has its arrow toward 'a' and 'a' has potential of -4V. Please clear the confusion for me. Thanks.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,878
1,368
As I wrote on another forum the polarity of a voltage across current source doesn't matter at all. Current will be flow in the direction indicated by the arrow.
This may sound strange, but this is how ideal current source behave.

Last edited: Jul 2, 2011