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:
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.
\$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/
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