If you look at this website: http://www.ibiblio.org/kuphaldt/electricCircuits/DC/DC_10.html


Search for the word "downstream", you'll see a basic DC circuit with 3 unknown currents. There is a node in the middle that I called Vx. Using the equations:

-I1 + I2 -I3 = 0
I1 = (Vx-28)/4
I2 = -Vx/2
I3 = (Vx-7)

We get:
-(Vx-28)/4 -Vx/2 - (Vx-7) = 0
-Vx + 28 - 2Vx -4Vx + 28 = 0
-7Vx = -56
Vx = 8

I1 = (Vx-28)/4 = -5A
I2 = -Vx/2 = -4A
I3 = (Vx-7) = -1A

BUT if you scroll down on this site and look at their solution, they found I1 and I2 to be positive. Who is correct, me or this website? Thanks!

 

It just means the original assumptions about current direction were wrong. Try working the problem assuming that all currents are positive and entering the node. The currents with negative signs are actually leaving the node instead of entering the node.

 

It has been the bane of electronics engineers, hobbyists and the like for years!

Dave

 

You may wanna check your calculations as
From your listing

-I1 + I2 -I3 = 0

and answers are

I1 = (Vx-28)/4 = -5A
I2 = -Vx/2 = -4A
I3 = (Vx-7) = -1A

say substitute I1 and I3 in which gives

5 + I2 + 1 = 0

I2 = -6A.

 

Sorry..

I1 = (Vx-28)/4 = -5A
I2 = -Vx/2 = -4A
I3 = (Vx-7) = 1A


So I have the exact opposite answer than they got.. how does that work?

 

Here is the schamatic comparing my results to the website's

 

Just think of Vx being smaller then sources ( thats why currents are in those directions) Now your equations will turn to
I1 = (28-Vx)/4
I2 = -Vx/2
I3 = (7-Vx)

That should solve the problem.

 

It shouldn't matter what direction I initially assume the currents to be flowing in. anyway, someone on Yahoo Answers helped me with this.