Norton's Theorem: Solving Current Across Terminals a-b

[DEU.Osterhagen]

Homework Statement

Obtain the norton equivalent of the circuit on terminals a-b. use the result to find the current i.
see attached image.

Homework Equations

1.Shorting the terminal when finding norton
2. Open current source and short voltage source when finding norton resistance

The Attempt at a Solution

By 2. We obtain
Rn = 6+4=10.

When Terminals a and b are short circuited we find current across it using nodal analysis
2+((12-v)/6)=v/4
v=9.6V

So that
In= (12-9.6)/6 =0.4A (current source facing upward)

This is the part where i am confused.
So connecting back the original components across terminal a-b (5 ohm and 4A in parallel) to the norton equivalent.
By current division We obtain


i = (10/(10+5))(4A + 0.4A) = 2.93A

The real answer is 2.4A , it's kinda odd because 0.4A norton current source is facing downward with this answer.


Edit. Pls move this to introductory physics

 

[ehild]

The short-circuit current flows downward if the source current points upward. You assumed the short-circuit current flowing upward, that means the Norton-current pointing downward, that is negative.

ehild

 

[NascentOxygen]

Hi, and welcome to the forums.

So that
In= (12-9.6)/6 =0.4A (current source facing upward)

Norton's Theorem does not involve I_in! His formula requires I_out.

 

[DEU.Osterhagen]

I can see it now.

I've read the book and realized that whenever the short circuit current face upward, then the norton current which represent the rest of the circuit will face downward and vice versa.

In this case, i made the short circuit current point upward.
Then the norton current source should point downward.

 

[NascentOxygen]

Well, okay. But aren't you going to be tripped up if ever the source is drawn horizontally?