Calculating Short Circuit Current when Nortonizing a Circuit

Question

My objective is to find v (voltage drop across 8 ohm) in the below circuit. I was able to thevenize the circuit successfully (V thevenin = 12 V and R thevenin = -2 ohm) and find v. But I find it difficult to apply Norton's Theorem in the below circuit.

^{simulate this circuit – Schematic created using CircuitLab}

The Norton equivalent resistance is easy to find (-2 ohm). To find the short circuit current, I shorted the 8 ohm resistor. This will also short the 4A current source and will hence become redundant (No interaction with left side of circuit). This reduces to the following circuit:

^{simulate this circuit}

From the above circuit I wrote the following:

\$i_{N}=-(10-i)=i-10\$

\$\Longrightarrow i=10+i_{N}\$

Applying KVL to the mesh:

\$-2i+4i=0\$

\$i=0\$

\$\Longrightarrow i_{N}=-10 A\$

This is not the same as:

\$\dfrac{V_{th}}{R_{th}}=-\dfrac{6}{7}A\$

What have I done wrong? Have I overlooked something?

Edit:

I will show here how I calculated R thevenin. I first detached the 8 ohm resistor from the circuit and nullified all independent sources. Since the circuit has dependent sources, I attach a 1 V voltage source across the terminals AB.

^{simulate this circuit}

\$R_{Th}\$ then becomes \$\dfrac{1}{i_{0}}\$. The idea is to find \$i_{0}\$ which is the same as \$i\$.

You can read more here: https://www.allaboutcircuits.com/technical-articles/thevenin-theorem-dependent-source-circuits/

Answer 1

Guidances only:

1) Shorting the 8 Ohm resistor doesn't make the 4 amperes source redundant. 4 A is a part of the short circuit current. The rest comes from the controlled source.

2) Norton's theorem isn't a method how the equivalent should be calculated, Norton's theorem states that the equivalent current source with parallel resistor exists if the DC circuit is linear. It's possible that your teacher has declared "If I one day happen to order you to apply Norton's theorem to solve a certain voltage or current, then you should calculate the open circuit voltage and the short circuit current!". I do not believe such declaration really exists, but that's a belief.

3) when the circuit has controlled sources, you have a danger to use for ex. same i in different situations where the circuit is different due the needed opens and shorts to derive the equivalents. Your i is totally different when the circuit is the original, 8 Ohm is taken away or when 8 Ohm is replaced with a wire.

I recommend you to stop with equivalent circuits to keep every controlled current and voltage in unique state. You get 2 equations with 2 unknowns i and v easily. Simply write i=10A+4i and the other equation for the currents at the right end of your 4i source. With Norton's or Thevenin's equivalent sources - when applied right - you often generate only new unknowns to a circuit which has controlled sources and you will not advance at all.