In the given circuit, while finding Norton or Short circuit current, we short terminal A and B. So I have assumed that is(t) = short circuit current.
Is it correct to assume so?
I did it because as terminal A and B are short circuit, all the current is(t) has zero resistance path to ground. Is my analysis correct?
Electrical – Use of Norton theorem to find short circuit current
circuit analysisnorton
Related Solutions
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.
This is easy to solve:
You keep combining sources until you are left with a single source. In this case, working with Thevenin sources (as apposed to Norton) seems easier.
V1,R1 is a Thevenin source you can see by inspection. So is V2,R2. Combine them.
Now add R3 to the resistance of that source. That is the Thevenin equivalent looking left into the right side of R3.
Add R4 to this and force I4 thru the result.
Best Answer
We're using superposition to establish the short circuit curent. When ignoring \$i_1\$ we can calculate the impedance between terminals A and B:
$$ Z_{AB} = j \omega L_2 + \frac{1}{j \omega C + 1/R} $$
leading to a current of $$ i_{u1} = \frac{u_1}{Z_{AB}} = \frac{u_1}{j \omega L_2 + \frac{1}{j \omega C + 1/R}}$$
This is the first partial current. Now, if we consider \$i_1\$ and replace \$u_1\$ with a short circuit, \$i_1\$ is connected directly between terminals A and B. Thus:
$$ i_{short} = i_1 + i_{u1} $$
So you are half correct. The short circuit current consists of a part \$i_1\$ and a part \$i_{u1}\$ contributed by the voltage source.