I calculated the Norton equivalent circuit of a given circuit and I found the result is the same as my book provide, except for the sign.
Here is the circuit:
simulate this circuit – Schematic created using CircuitLab
So the circuit I've to analyse to calculate the Norton current is this:
Doing a KCL at the node 1:
\$i_N=-\frac{Av_d}{R}\$
Now, my book says that it's \$i_N=+\frac{Av_d}{R}\$, in fact it draws the current \$i_N\$ in the opposite direction.
My question is: why?
Best Answer
This is just a matter of references*.
When you start to tackle the problem you short circuit your port and draw a reference for the Norton current. It's just like saying "this is what positive Norton current is to me".
After solving the circuit you get a result that is negative? No problem: it just means that the current is actually flowing in the opposite way, i.e. out of node 1. Your answer and your book's are just the same: the book says that output current is positive, and gets a positive current that is then going out, while you say that input current is positive, and get a negative current that is then going out.
Note that an answer giving only the current would be incomplete: if I told you that the current flowing in the controlled source is 1A you should ask me: is it going from plus to minus or vice versa?
You can think of an analogy in kinematics: the first thing you do before solving a problem is choosing a reference system*. If you don't choose it and tell your teacher that the ball speed is 10m/s he should yell at you.
*I believe that's the correct English word, please comment if it's not.