why isn't Norton transform applicable on ideal sources?
Consider two equivalent circuits:







simulate this circuit &ndash; Schematic created using CircuitLab

These are related by the equations:

$$ R_{th} = R_{no} \\
V_{th} = I_{no} R_{no} \\
I_{no} = V_{th} / R_{th} $$

Let's say \$V_{th} = 1\:\mathrm V\$. Then:

$$ I_{no} = 1/R_{th} $$

When \$R_{th} = 0\$, what do you get?



You can, if you want, think about the one-sided limit as \$R_{th} \searrow 0\:\Omega\$. As you approach that limit, \$I_{no}\$ approaches infinity and \$R_{no}\$ approaches zero. That gives you some insight into what happens as a voltage source becomes arbitrarily close to ideal, but you still can't divide by zero when it becomes exactly ideal.

What you are doing, by choosing to ignore the undefinedness of dividing by zero, is called a formal calculation, and while that may sound like a great idea, but it's not:


  In mathematical logic, a formal calculation is a calculation which is systematic, but without a rigorous justification.


You can also read "without a rigorous justification" as "without any engineering utility". What you end up with is a mathematically impossible circuit. Just to get you thinking, here are some more mathematically impossible circuits:







simulate this circuit

If you wonder why these are impossible, the answer is simply "because mathematics says so". It seems odd, because attempting to build these circuits isn't impossible, but remember that schematics are simplifications of real behavior and only the properties of the circuit that are drawn are modelled.