# Is this website correct? (direction of DC current in basic example)

#### Amenably_Amendable

Joined Jan 21, 2008
13
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! #### Papabravo

Joined Feb 24, 2006
13,923
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.

#### Dave

Joined Nov 17, 2003
6,970

#### kotophey

Joined Jan 20, 2008
3
You may wanna check your calculations as

-I1 + I2 -I3 = 0

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.

#### Amenably_Amendable

Joined Jan 21, 2008
13
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?

#### Amenably_Amendable

Joined Jan 21, 2008
13
Here is the schamatic comparing my results to the website's

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#### kotophey

Joined Jan 20, 2008
3
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.